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Q: How to implement CAPTCHA in a form using JQuery I am using JQuery.
I have got the JQuery Dialog Login form, now I want to implement the captcha on the form if users inputs is invalid.
I am using JQuery to check all the form validation and finally my main form is submitted.
Please suggest the some good example or the page links.
A: Captcha in pure javascript/jQuery doesn't exist since JS is disabled for robot. It'd be useless.
Anyway, you can implement it with PHP or any dynamic language.
A quick search would have give you a lot of results
For example :
http://plugins.jquery.com/project/simpleCaptcha
http://www.webdesignbeach.com/beachbar/ajax-fancy-captcha-jquery-plugin (w/o jQuery)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,884
|
\section{Introduction}
\label{1}
Double beta decay is the most rare nuclear process measured so far which presents a great interest, especially for testing the lepton number conservation (LNC) and understanding the neutrino properties. Within the Standard Model (SM) it conserves the lepton number and can occur through several decay modes, with emission of two neutrinos/anti-neutrinos ($2\nu$) in the final states. However, theories beyond SM predict that this process may also occur without conservation of the lepton number, and hence, without emission of neutrinos/anti-neutrinos, through the so called neutrinoless ($0\nu$) decay modes. According with the number and type of leptons we may have the following DBD modes: i) two neutrino double-electron decay ($2\nu\beta^-\beta^-$); ii) neutrinoless double-electron decay ($0\nu\beta^-\beta^-$); iii) two neutrino double-positron decay ($2\nu\beta^+\beta^+$); iv) neutrinoless double-positron decay ($0\nu\beta^+\beta^+$); v) two neutrino electron capture positron emitting decay ($2\nu EC\beta^+$); vi) neutrinoless electron capture positron emitting decay ($0\nu EC\beta^+$); vii) two neutrino double electron capture decay ($2\nu ECEC$) and viii) neutrinoless double electron capture decay ($0\nu ECEC$).
Complete information about the achievements in the study of DBD can be found in several recent excellent reviews \cite{AEE08}-
\cite{VES12}, which also contain a comprehensive list of references in domain. The lifetimes for the DBD modes can be factorized, in good a approximation, as follows:
\begin{eqnarray}
\left( T^{2\nu}_{1/2} \right)^{-1}=G^{2\nu}(E_0, Z)\mid M^{2\nu}\mid^2~~;~~
\left( T^{0\nu}_{1/2} \right)^{-1} \nonumber\\
=G^{0\nu}(E_0, Z)\mid M^{0\nu}\mid^2 \left(< \eta_{l}> \right)^2 \ ,
\end{eqnarray}
\noindent
where $G^{(2\nu, 0\nu)}$ are the PSF and $M^{(2\nu, 0\nu)}$ the nuclear matrix elements (NME) for the $2\nu$, $0\nu$ decay modes respectively, and $<\eta_l>$ is a beyond SM parameter which contains information about the neutrino properties in the (most common) hypothesis that the $0\nu$ $\beta\beta$ decay (DBD) occurs through the neutrino exchange mechanism. As seen, PSF and NME are key quantities for estimating/predicting DBD lifetimes and/or for deriving neutrino properties, hence it is important to calculate them with high accuracy. Moreover, the PSF values largely fix the order of magnitude of the DBD lifetimes. So far, much effort has been and is still devoted to the precise calculation of the NME \cite{ROD07}-\cite{RAH10}, while PSF have been less discussed because it was considered that they were computed with enough precision \cite{PR59}-\cite{SC98}. Recently, the PSF were recalculated with a more rigorous method, by using exact electron Dirac wave functions (w.f.) and taking into account the finite nuclear size and electron screening effects \cite{KI12}-\cite{SM13}. The authors found differences between their results and those calculated previously with approximate electron w.f., especially for heavier nuclei and for $\beta^+\beta^+$ decay modes. This justifies a more careful independent re-evaluation of the PSF involved in DBD, using more accurate methods.
The purpose of this work is to give a complete, up-date list of the PSF values for the i)-vii) DBD modes mentioned above, in all nuclei of interest and possible transitions to final states. The decay mode $0\nu ECEC$ can not occur to the order of approximation that is presently considered in the literature. We developed high precision routines to compute the relativistic (Dirac) electron w.f. taking into account the nuclear finite size and screening effects. In addition to the previous calculations, we use a Coulomb potential derived from a realistic proton density distribution in nucleus, improve the precision of the numerical routines used to solve the Dirac equations and integrate the PSF expressions, and use $Q$-values reported recently \cite{audi12}. These ingredients proved to be important, leading in many cases to significant corrections to the present available PSF values, which are discussed as well. Accurate PSF values (besides the NME) are necessary ingredients both for theorists to improve the DBD lifetime predictions and constraint the neutrino parameters, and for experimentalists to plan their set-ups.
\section{Theoretical framework}
To compute PSF for DBD decay modes we need first to obtain the w.f. of the electron(s)/positron(s) emitted or electron(s) captured in the decay, which are distorted by the Coulomb potential of the nucleus. Older calculations have used a non-relativistic approach where the distortion of the w.f. by the Coulomb field was considered through Fermi (Coulomb) factors obtained (for the emitted particles) by taking the square of the ratio of the Schr\"odinger scattering solution for a point charge $Z$ to a plane wave, evaluated at the origin \cite{PR59}-\cite{HS84}. In a better approximation, the Fermi factors are calculated using a relativistic treatment of the electron/positron w.f, but with approximate Dirac functions (the Fermi factor is defined as the square of the ratio of the values of the Dirac w.f. of the electron at the nuclear surface) and without the inclusion of screening effects \cite{DOI83,SC98}. Recently, Kotila and Iachello (KI) recalculated the PSF using exact Dirac electron/positron w.f. and including the screening effect \cite{KI12}-\cite{KI13}. In this work we adopt this more rigorous relativistic treatment of KI but with the inclusion of additional ingredients, as it is described in the following.
\subsection{The radial wave functions}
For free states we use relativistic scattering electron/positron w.f. solutions of the Dirac equation in a central (Coulomb) potential:
\begin{equation}
\Psi^+_{\epsilon\kappa\mu}(r) = \left(\begin{array}{l}g_{\kappa}(\epsilon,r)\chi^\mu_\kappa\\if_\kappa(\epsilon,r)\chi^\mu_{-\kappa}\end{array}\right)~~
\end{equation}
for $~\beta^- ~\rm{decay}$ and
\begin{equation}
\Psi^-_{\epsilon\kappa\mu} =\left(\begin{array}{l}if_{\kappa}(\epsilon,r)\chi^{-\mu}_{-\kappa}\\-g_\kappa(\epsilon,r)\chi^{-\mu}_\kappa\end{array}\right)~~i\end{equation}
for $\beta^+~ \rm{decay}$
\noindent
where $\kappa=(l-j)(2j+1)$ is the relativistic quantum number and $\chi^\mu_\kappa$ are spherical spinors. The quantities
$g_{\kappa}(\epsilon,r)$ and $f_{\kappa}(\epsilon,r)$ are the large and small components of the radial w.f. which satisfy the radial equations:
\begin{eqnarray}
\label{dirac}
{dg_{\kappa}(\epsilon,r)\over dr}=
-{\kappa\over r}g_{\kappa}(\epsilon,r)
+{\epsilon-V+m_ec^2\over c\hbar}f_{\kappa}(\epsilon,r)\\
{df_{\kappa}(\epsilon,r)\over dr}=
-{\epsilon-V-m_ec^2\over c\hbar}g_{\kappa}(\epsilon,r)
+{\kappa\over r}f_{\kappa}(\epsilon,r) \nonumber
\end{eqnarray}
where $V$ can be negative/positive (for $\beta^-/\beta^+$).
These functions are normalized so that they have
the following asymptotic behavior:
\begin{eqnarray}
\label{asi}
\left(\begin{array}{l}g_{k}(\epsilon,r)\\f_{k}(\epsilon,r)\end{array}\right)\sim
{\hbar e^{-i\delta_k}\over pr}
\left(\begin{array}{l}\sqrt{{\epsilon+m_ec^2\over 2\epsilon}}
\sin(kr-l{\pi\over 2}-\eta\ln (2kr)+\delta_k)\\
\sqrt{{\epsilon-m_ec^2\over 2\epsilon}}
\cos(kr-l{\pi\over 2}-\eta\ln (2kr)+\delta_k)\end{array}\right)
\end{eqnarray}
Here
$c$ is the speed of the light, $m_e/\epsilon$ are the electron mass/energy,
$k={p/\hbar}$ is the electron wave number, $\eta=Ze^2/\hbar$ (with $Z=\pm Z$ for $\beta^{\mp}$ decays), $v$ is the Sommerfeld parameter, $\delta_\kappa$ is the phase shift and $V$ is the Coulomb interaction energy between the electron and the daughter nucleus.
For the continuum spectrum, the radial function is normalized to the
asymptotic form of the Coulomb function. The phase shifts are obtained by
matching the inner numerical solution to the analytic function.
The bound states w.f. for the electron
\begin{equation}
\Psi^b_{\epsilon_n\kappa\mu}(r) = \left(\begin{array}{l}g_{n,\kappa}(r)\chi^\mu_\kappa\\if_{n,\kappa}(r)\chi^\mu_{-\kappa}\end{array}\right)
\end{equation}
are solutions of the Dirac equation (\ref{dirac}) and correspond
to the eigenvalues $\epsilon_{n}$ ($n$ is the radial
quantum number). The quantum number $\kappa$ is related to
the total angular momentum $j_\kappa=\mid\kappa\mid-1/2$. These wave functions are normalized such that
\begin{equation}
\int_0^\infty [g^2_{n,\kappa}(r)+f^2_{n,\kappa}(r)]dr=1.
\end{equation}
An asymptotic solution is obtained by means of the WKB approximation
and by considering that the potential $V$ is
negligible small:
\begin{equation}
{f_{n,\kappa}\over g_{n,\kappa}}={c\hbar\over \epsilon+m_ec^2}\left({g'_{n,\kappa}\over g_{n,\kappa}}+{\kappa\over r}\right)
\end{equation}
where
\begin{equation}
{g'_{n,\kappa}\over g_{n,\kappa}}=-{1\over 2}\mu'\mu^{-1}-\mu
\end{equation}
with
\begin{equation}
\mu=\left[{\epsilon+m_ec^2\over \hbar^2 c^2}(V-\epsilon+m_ec^2)+{\kappa^2\over r^2}\right]^{1/2}.
\end{equation}
In our calculations we use $n$=0 and $n$=1 number of nodes, for
the orbitals $1s_{1/2}$ and $2s_{1/2}$, $\kappa$ being -1.
The eigenvalues of the discrete spectrum are obtained by matching
two numerical solutions of the Dirac equation: the inverse solution
that starts from the asymptotic conditions and the direct one that starts at $r$=0.
\subsection{The Coulomb potential}
The nuclear size corrections are usually taken into account by considering an unscreened potential $V$ obtained for a uniform charge
distribution in a sphere of radius $R_A$ \cite{DOI83}, \cite{KI12}:
\begin{equation}
V(Z,r)=\left\{\begin{array}{ll}-{Z{\alpha\hbar c}\over r}, & r \ge R_A,\\
-Z(\alpha\hbar c)\left({3-(r/R_A)^2\over 2R_A}\right), & r<R_A, \end{array}
\right.
\label{cd}
\end{equation}
where generalized atomic units are used. The values of
$\hbar$, of the electron
charge $e$ and its mass $m_e$ are considered as unity. The energy
unit is $E_0=me^4/h^4=27.2114$ eV, the Bohr radius is
$a_0=\hbar^2/m_ee^2$=0.529177 $\AA$ and the speed of the light
in vacuum is $c$=137.036 (the inverse of the fine structure constant).
In this work we take into account the influence of the nuclear structure by using a potential $V(r)$ derived from a realistic proton density distribution
in the nucleus. This is done by solving the Schr\"odinger equation for a Woods-Saxon potential. In this case:
\begin{equation}
V(Z,r)=\alpha\hbar c\int{\rho_e(\vec{r'})\over \mid \vec{r}-\vec{r'}\mid}
d\vec{r'}
\label{vpot}
\end{equation}
where the charge density is
\begin{equation}
\rho_e({\vec{r}})=\sum_i (2j_i+1)v_i^2 \mid\Psi_i(\vec{r})\mid^2
\end{equation}
$\Psi_i$ is the proton (Woods-Saxon) w. f. of the spherical single particle state
$i$ and $v_i$ is its occupation amplitude. The factor $(2j_i+1)$ reflects the
spin degeneracy.
The difference between the behavior of the constant charge density $\rho_e$
and the realistic charge density is displayed in Fig. \ref{dens} for
the daughter nucleus $^{150}$Sm.
We computed the Coulomb potential with formula (\ref{vpot}).
In this case, the differences given
by the charge densities are translated in a shift of 0.5 MeV energy
in the potential at $r=0$. This difference in energy vanishes when
$r$ increases, but is able to affect the values of the w.f.
The screening effect is taken into account by multiplying the expression of $V(r)$ with a function $\phi(r)$, which is the solution of the Thomas Fermi equation:
$d^2\phi/dx^2 = \phi^{3/2}/\sqrt x$, with $x=r/b$, $b\approx 0.8853a_0Z^{-1/3}$ and $a_0$ = Bohr radius. It is
calculated within the Majorana method \cite{sal}. The screening effect is taken into account in the same manner as in Ref. \cite{KI12}.
The modality in which the screening function modifies
the Coulomb potential depends on the specific mechanism and its
boundary conditions.
\begin{figure}
\includegraphics[width=0.5\textwidth]{figure1.eps}
\caption{
Profile of the realistic proton density $\rho_e$ for $^{150}$Sm
(thick line)
compared with that given with the constant density approximation (dot-dashed line).
\label{dens}}
\end{figure}
\begin{figure}
\includegraphics[width=0.5\textwidth]{figure2.eps}
\caption{The absolute values of the electron phase space parameter
$f_i(\epsilon,r)$ and $g_i(\epsilon,r)$
($i$=-1,1) are plotted with a full line for the daughter nucleus
$^{150}$Sm at an energy $\epsilon$=2 MeV
as function of the distance $r$. The deviations that arise when only the constant charge
distribution is taken into account, $\delta g_i(\epsilon,r)$
and $\delta f_i(\epsilon,r)$, are displayed as well
with a dashed line and their scales are on the right.
A thin line locates the nuclear surface $R=r_0\times A^{1/3}$.}
\label{furi}
\end{figure}
In the case of the $\beta^-\beta^-$ process, the potential used to obtain the electron w.f. is:
\begin{equation}
rV_{\beta^-\beta^-}(Z,r)=(rV(Z,r)+2)\times\phi(r)-2
\end{equation}
to take into account the fact that DBD releases a final
positive ion with charge +2. Here, the product $\alpha\hbar c$=1, for
atomic units. In this case, the charge
number $Z=Z_0+2$ corresponds to the daughter nucleus,
$Z_0$ being the charge number of the parent nucleus.
In the case of the $\beta^+\beta^+$ process,
the potential used to obtain the electron wave functions
is:
\begin{equation}
rV_{\beta^+\beta^+}(Z,r)=(rV(Z,r)-2)\times\phi(r)+2
\end{equation}
where the final configuration is characterized by an ion with charge -2.
In this case, the daughter nucleus has the charge number $Z=Z_0-2$.
In both approaches, at $r$=0 the potential is unscreened because
$\phi(0)$=1. Asymptotically $\phi(r)$ tends to 0 and we are left with the
charge number of the final system.
In the case of the $EC$ process, the potential used to obtain the electron w.f. reads:
\begin{equation}
rV_{EC}(Z,r)=(rV(Z,r)+1)\times\phi(r)-1
\end{equation}
\noindent
and the charge number $Z=Z_0$ corresponds to the parent nucleus. In the case of the
$\beta^+$ process, the potential used to obtain the positron wave functions reads:
\begin{equation}
rV_{\beta^+}(Z,r)=(rV(Z,r)-1)\times\phi(r)+1
\end{equation}
to take into account that in the final
configuration we have an ion with charge -1.
In this case the daughter nucleus has the charge number $Z=Z_0-1$.
In the case of the ECEC process, the potential used to obtain the electron wave functions is:
\begin{equation}
V_{ECEC}(Z,r)=V(Z,r)\times\phi(r)
\end{equation}
and $Z=Z_0$, the final system being neutral.
The solutions of the Dirac equation for free states at $\epsilon$=2 MeV in the case
of $^{150}$Sm are plotted in Fig. \ref{furi}. With a dashed line
we plotted the deviations $\delta g_i(\epsilon,r)=g_i(\epsilon,r)-\tilde{g}_i(\epsilon,r)$ and
$\delta f_i(\epsilon,r)=f_i(\epsilon,r)-\tilde{f}_i(\epsilon,r)$ ($i$=-1,1) that are due to the
constant density approximation. The quantities $g_i$ and $f_i$ are
obtained for a realistic charge distribution while $\tilde{g}_i$ and
$\tilde{f}_i$ are calculated by considering a constant charge density
in the nucleus. The deviations exhibit an oscillatory behavior leading
to uncertainties on the nuclear surface. The nuclear surface is marked with a vertical line
in the figure.
\subsection{Calculation of the phase space factors}
\subsubsection{Double Electron and Double Positron decay modes}
To compute the PSF, we have to obtain the electron phase factors $f_{jk}^{(0)}$
\begin{equation}
f_{11}^{(0)}=\mid f^{-1-1}\mid^2+\mid f_{11}\mid^2+\mid f^{-1}_{~~~1}\mid^2
+\mid f^{~-1}_1\mid^2
\end{equation}
with
\begin{eqnarray}
f^{-1-1}=g_{-1}(\epsilon_1)g_{-1}(\epsilon_2)~;~f_{11}=f_1(\epsilon_1)f_1(\epsilon_2),\\
f^{-1}_{~~~1}=g_{-1}(\epsilon_1)f_1(\epsilon_2)~;~ f_{1}^{~-1}=f_1(\epsilon_1)g_1(\epsilon_2)
\end{eqnarray}
\noindent
from the solutions of the Dirac equation by considering s-wave states and neglecting the neutrino mass.
The values of the $f$ and $g$ functions are approximated with the solutions on the nuclear surface (the method I from \cite{KI12}).
\begin{eqnarray}
g_{-1}(\epsilon)=g_{-1}(\epsilon,R_A)~;~ f_1(\epsilon)=f_1(\epsilon,R_A)
\end{eqnarray}
\noindent
where $R_A=1.2A^{1/3}$ fm.
For the $2\nu\beta\beta$ decay mode and for transitions to ground states, the PSF expression reads:
\begin{eqnarray}
G^{\beta\beta}_{2\nu}(0^+\rightarrow0^+)={2\tilde{A}^2\over 3\ln 2 g_A^4(m_ec^2)^2}
\int_{m_ec^2}^{Q^{\beta\beta}+m_ec^2} d\epsilon_1 \nonumber\\
\times \int_{m_ec^2}^{Q^{\beta\beta}+2m_ec^2-\epsilon_1} d\epsilon_2
\int_{0}^{Q^{\beta\beta}+2mc_e^2-\epsilon_1-\epsilon_2}
d\omega_1 \nonumber\\
\times f_{11}^{(0)}w_{2\nu}(\langle K_N\rangle^2+\langle L_N\rangle^2+
\langle K_N\rangle \langle L_N\rangle)
\label{fomg}
\end{eqnarray}
\noindent
where
$Q^{\beta\beta}=M(A,Z_0)-M(A,Z_0-2)-4m_ec^2$ is the kinetic energy released in the process.
$\langle K_N\rangle$, $\langle L_N\rangle$ are expressions that depend on the electron/positron
$(\epsilon_{1,2})$ and neutrino ($\omega_{1,2}$) energies, and on the g.s. energy of
the parent nucleus and on the excited states energy of the intermediate nucleus \cite{HS84}-\cite{KI12}.
\begin{equation}
\langle K_N\rangle=
{1\over \epsilon_1+\omega_1+\langle E_N\rangle-E_I}+
{1\over \epsilon_2+\omega_2+\langle E_N\rangle-E_I}
\label{ie1}
\end{equation}
\begin{equation}
\label{ie2}
\langle L_N\rangle=
{1\over \epsilon_1+\omega_2+\langle E_N\rangle-E_I}+
{1\over \epsilon_2+\omega_1+\langle E_N\rangle-E_I}
\end{equation}
Here, the difference in energy in the denominator can be
obtained from the approximation
$\tilde{A}^2=[W_0/2+\langle E_N\rangle -E_I]^2$, where
$\tilde{A}=1.12A^{1/2}$ (in MeV) gives the energy of the giant Gamow-Teller resonance in the intermediate nucleus. The quantity $W_0$ is
related to the $Q$ value of the process and
\begin{equation}
w_{2\nu}={g_A^4(G\cos\theta_C)^4\over 64\pi^7\hbar}
\omega_1^2\omega_2^2(p_1c)(p_2c)\epsilon_1\epsilon_2.
\end{equation}
\noindent
where $\omega_1$ and $\omega_2=Q^{\beta\beta}-\epsilon_1-\epsilon_2-\omega_1+2m_ec^2$ are
the neutrino energies.
The PSF are finally renormalized to the electron rest energy and are reported in $[yr^{-1}]$.
The PSF for the $2\nu\beta\beta$ decay mode and for transitions to excited $0^+_1$ states is calculated with a formula similar to (\ref{fomg}),
but replacing $Q^{\beta\beta}$ by $Q(0^+_1) = Q^{\beta\beta} - E_x(0^+_1)$, which is the kinetic energy released in this transition. $E_x(0^+_1)$ is the energy of the excited $0^+_1$ state of the daughter nucleus $x$.
For the $2\nu\beta\beta$ decay mode and for transitions to excited $2^+_1$ states, the PSF formula reads \cite{SC98}-\cite{SM13}:
\begin{eqnarray}
G^{\beta\beta}_{2\nu}(0^+\rightarrow2^+_1)={2\tilde{A}^6\over \ln 2 g_A^4(m_ec^2)^2}
\int_{m_ec^2}^{Q^{\beta\beta}{(2^+_1)}+m_ec^2} d\epsilon_1
\nonumber\\
\times \int_{m_ec^2}^{Q^{\beta\beta}{(2^+_1)}+2m_ec^2-\epsilon_1} d\epsilon_2\nonumber \int_{0}^{Q^{\beta\beta}{(2^+_1)}+2mc_e^2-\epsilon_1-\epsilon_2} d\omega_1 \nonumber \\
\times
f_{11}^{(0)}w_{2\nu}(\langle K_N\rangle-\langle L_N\rangle)^2
\end{eqnarray}
where $Q(2^+_1) = Q - E_x(2^+_1)$.
For the $0\nu\beta\beta$ decay and for transitions to g.s. the PSF reads:
\begin{equation}
\label{int1}
G^{\beta\beta}_{0\nu}(0^+\rightarrow0^+) = {2\over{4g^4_AR_A^2 \ln2}}\int_{m_ec^2}^{Q^{\beta\beta}+m_ec^2}
f_{11}^{(0)}w_{0\nu}d\epsilon_1
\end{equation}
where $R_A=1.2A^{1/3}$ is the nuclear radius and
\begin{equation}
w_{0\nu }={g_A^4(G\cos\theta_C)^4\over 16\pi^5}
(m_ec^2)^2 (\hbar c^2) (p_1c) (p_2c) \epsilon_1\epsilon_2
\end{equation}
In our calculations, the Fermi constant is $G=1.16637\times10^{-5}$ GeV$^{-2}$
and $\cos\theta_C$=0.9737.
In Eq.(\ref{fomg}) it is convenient to redefine the PSF by a renormalization that eliminates the constant $g_A$ and correlate (by dividing by $4R_A^2$) the dimension of $G_{0\nu}$ with the NME which are dimensionless. Thus, the PSF are also reported in $[yr^{-1}]$. A similar expression is employed in the PSF calculation for the transitions to excited $0^+_1$ states, but replacing $Q^{\beta\beta}$ by $Q^{\beta\beta}(0^+_1)$.
The formula used for the calculation of the PSF for $2\nu\beta^+\beta^+$ is similar to that used for $2\nu\beta^-\beta^-$ decay, but $\epsilon_{1,2}$ are now the positron energies. Also, we use the same approximations as described above,
to evaluate the radial positron w. f. ($g$ and $f$) at the nuclear surface and replace the excitation energy $E_N$ in the intermediate odd-odd nucleus by a suitable average energy.
\subsubsection{The EC$\beta^+$ case}
For the $EC\beta^+$ decays the energy released in the process is $Q^{EC\beta}=M(A,Z_0)-M(A,Z_0-2)-2mc_e^2$.
If the numerical solutions of the Dirac equation are obtained
in Bohr units $a_0$, the probability that an electron is found on the surface
of a nucleus of radius $R_A$ can be defined as:
\begin{equation}
B^2_{n,\kappa}={1\over 4\pi (m_ec^2)^3}\left({\hbar c\over a_0}\right)^2
\left({a_0\over R_A}\right)^2
[g_{n,\kappa}^2(R_A)+f_{n,\kappa}^2(R_A)]
\end{equation}
The PSF expression for $2\nu\beta\beta$ decay modes is
\begin{eqnarray}
G^{EC\beta^+}_{2\nu}=
{2A^2\over 3\ln2}{(G\cos\theta)^4\over 16\pi^5\hbar}
(m_ec^2)\sum_{i=0,1}B^2_{i,-1} \nonumber \\
\times \int_{m_ec^2}^{Q^{EC\beta}+\epsilon_{i,-1}+m_ec^2}
\int_0^{Q^{EC\beta}+\epsilon_{i,-1}-\epsilon_p+m_ec^2}
\nonumber \\
\times [g_{-1}^2(\epsilon_p)+f_{1}^2(\epsilon_p)]\nonumber \\
\times
(\langle K_N\rangle^2+\langle L_N\rangle^2+
\langle K_N\rangle\langle L_N\rangle)\omega_1^2\omega_2^2
p_pc\epsilon_pd\omega_1d\epsilon_p
\end{eqnarray}
where $\epsilon_{n,\kappa}$ are the binding energies
of the electron while $p_p$ and $\epsilon_p$ are
the linear momentum and the energy of the positron.
Here, the expressions for
$\langle K_N\rangle$ and $\langle L_N\rangle$ are similar to those from Eqs. (\ref{ie1})-(\ref{ie2}), but where
$\epsilon_1$ is replaced by $\epsilon^c_{i,-1}=m_ec^2-\epsilon_{i,-1}$, the energy of the captured electron
and $\epsilon_2$ is replaced by $\epsilon_p$, the energy of the emitted positron.
For the $0\nu\beta\beta$ decay process the PSF expression is:
\begin{eqnarray}
G^{EC\beta^+}_{0\nu}={1\over 4 R_A^2}{2\over \ln2}
{(G\cos\theta)^4\over 4\pi^3}(\hbar c^2)(m_ec^2)^5
\nonumber \\ \times
\sum_{i=0,1}B^2_{i,-1}
[g_{-1}^2(\epsilon_{p,i})+f_{1}^2(\epsilon_{p,i})]p_pc\epsilon_p
\end{eqnarray}
where $\epsilon_{p,i}$ denotes the maximal value of the positron associated to the state $i$.
\subsubsection{The $2\nu ECEC$ case}
The PSF expression is defined as:
\begin{eqnarray}
G_{2\nu}^{ECEC}={2\tilde A^2\over 3\ln 2}
{(G\cos\theta)^4\over 16\pi^3\hbar}
(mc_e^2)^4
\nonumber \\ \times
\sum_{i,j=0,1}B^2_{i,-1}B^2_{j,-1}
\int_0^{Q^{ECEC}+\epsilon_{i,-1}+\epsilon_{j,-1}}
\nonumber \\ \times
(\langle K_N\rangle^2+\langle L_N\rangle^2+
\langle K_N\rangle\langle L_N\rangle)\omega_1^2\omega_2^2d\omega_1
\end{eqnarray}
\noindent
where $Q^{ECEC}=M(A,Z_0)-M(A,Z_0-2)$ is the energy released in the process.
The expressions for
$\langle K_N\rangle$ and $\langle L_N\rangle$ are similar to those from Eqs. (\ref{ie1})-(\ref{ie2}), but where
$\epsilon_{(1,2)}$ are replaced by $\epsilon^c_{(i,j)-1}=m_ec^2-\epsilon_{(i,j)-1}$, the energies of the captured electrons.
\section{Numerical details}
\label{3}
The numerical solutions of the Dirac equation were obtained by using the
power series method of Ref. \cite{buh}. We built up a code that use
a numerical algorithm
similar to that described in Ref. \cite{sal1}, the normalization to unity for free states being
done as indicated in Ref. \cite{sal2}.
In the numerical procedure, the potential energy
as function of the distance $r$ is approximated
with a spline cubic function that interpolates
values defined by an increment $x$.
The radial w. f. is expanded as an infinite power series
that depends on the increment and the coefficients
of the spline function. Therefore, the values
of the w. f. are calculated step by step in the
mesh points. The accuracy of the solutions depends
on the increment and the number of terms in the series
expansion. We used an increment interval of 10$^{-4}$ fm
and at least 100 terms in the series expansion. These values
exceed the convergence criteria of Ref. \cite{sal1}.
At very large distances, the behavior of the w. f.
must resemble to that of the Coulomb function. This last
condition provides a way to renormalize the amplitude to unity
and to determine the phase shift.
For discrete states, the asymptotic behavior of the w.f. gives a boundary for the inverse solutions. The eigenvalue
is obtained when the direct
solutions and the inverse ones match together.
In order to find the bound states of the electron, a procedure
which differ from that given in Ref. \cite{sal1} is used. We start to
compute numerical solution of the Dirac equation for a total
energy close to $m_ec^2$, and we span an interval of 0.3 MeV under
this energy in steps of 0.0002 MeV. In this range of energies, all the possible
bound solutions are found. In the Ref. \cite{sal1}, the interval
of the allowed solutions is fixed by initial conditions,
but its lower limit is
an approximate one and sometimes the equations cannot be solved
numerically.
For the PSF computation,
all integrals are performed accurately with Gauss-Legendre quadrature in 32 points.
We calculated up to 49 values of the radial functions in the Q value
energy interval, values that are interpolated with spline functions.
In our calculations we used up-dated values of $Q$ reported recently in Ref. \cite{audi12}.
\section{Results and discussion}
\label{4}
Our results are tabulated in tables \ref{tab1} - \ref{tab6}.
For comparison with other previous results we refer mainly to those reported by KI \cite{KI12}-\cite{KI13} because both methods (KI and ours) calculate the PSF more precisely than other previous ones, i.e. using exact electron w.f. obtained by solving numerically the Dirac equation with inclusion of the finite nuclear size and electron screening effects. In addition, in our calculations we use a Coulomb potential derived from a realistic proton density distribution in the daughter/parent nucleus, improve the numerical precision of the routines that we developed both to solve the Dirac equation and to integrate the PSF values, and use updated $Q$-values \cite{audi12}. These additional ingredients lead in many cases to differences as compared with KI's results, that we qualitatively discuss in the following. For the $\beta^-\beta^-$ decay mode our results are in good agreement with KI's results \cite{KI12}, the differences between their values and ours are within 3\% for $G_{0\nu}$ and for the transitions to final g.s., and within 10\% for transitions to excited $0^+_1$ states, with two exceptions, $^{110}Te$ and $^{130}Te$ where the differences are within 18\% and 38\%, respectively. For $G_{2\nu}$ the differences are within 7\%, except the nuclei $^{128}Te$ (20\%) and $^{238}U$ (our result is 7 times larger than the KI's one). We mention that $\beta^-\beta^-$ decay modes are experimentally the most interesting DBD because the $2\nu$ decay mode is already measured for eleven nuclei for transitions to g.s., and for two nuclei for transitions to excited $0^+_1$. Also, the $0\nu$ channel is intensively searched for checking the LNC and extracting information about neutrino properties. Although not yet measured, the $\beta^-\beta^-$ decays to excited $2^{+}_1$ states are interesting as well, to probe alternative mechanisms for $0\nu$ decay to the most common one, i.e. by exchange of a virtual light neutrino between two nucleons inside the nucleus. For these transitions KI have not reported PSF values. Comparing our PSF values with older results \cite{DOI83}, \cite{SC98}, we found differences in the range (17-86)\%, for both $G_{0\nu}$ and $G_{2\nu}$, unless $^{76}Ge$, $^{116}Cd$ and $^{128}Te$ where our results are several times larger. For the $\beta^+\beta^+$ decay modes our PSF values are in agreement with KI's results within (3-12)\% for $G_{0\nu}$ and within (6-23)\% for $G_{2\nu}$. For $EC\beta^+$ decay mode the differences between KI's results and ours are within 35\% both for $G_{0\nu}$ and $G_{2\nu}$ for the most part of the isotopes, unless a few cases where PSF values amount to about 70\%. There are, however, some heavier nuclei where our results differ by factors of 5-10 from other previous results. Finally, for the $ECEC$ decay mode the differences between our $G_{2\nu}$ values and the KI ones, are within (30-55)\%, unless the heavier nuclei where the differences are significantly larger. It is worth to mention that we got significant differences as compared with KI's results especially for the decays where the $Q$-values are small. In these cases the numerical precision in the PSF integration is important and must be treated carefully. The use of a realistic Coulomb potential can also lead to differences between KI and our results, especially for double-positron emitting and/or positron emitting electron(s) captured decays. The use of recently reported $Q$ values \cite{audi12} may bring other differences between our results and previous ones. For example, we first computed the PSF with the same $Q$-values used by KI and then with the $Q$-values taken from \cite{audi12} and got differences of (3-4)\% between the two sets of results. We mention that, in our opinion, differences of 10\% or more in the PSF values are significant for precise DBD calculations/estimations, hence the evaluation of the PSF for many DBD decay modes and/or transitions is still a challenge and it is necessary to investigate it thoroughly.
\section{Conclusions}
\label{5}
We recalculated with a more precise method the PSF involved in the $\beta^-\beta^-$, $\beta^+\beta^+$,
$EC\beta^+$ and $ECEC$ DBD modes. In addition to the previous (most accurate) method \cite{KI12}-\cite{KI13}, we use a Coulomb potential derived from a realistic proton density distribution in nucleus, improve the precision of our numerical routines used to solve the Dirac equations and to integrate the PSF expressions, and use $Q$ values taken from a recent mass evaluation \cite{audi12}. These ingredients lead in many cases to significant corrections to the present available PSF values, which are discussed as well. Accurate values of the PSF (besides the NME) are necessary ingredients in the DBD study, both for theorists to improve the lifetimes predictions and constraint the neutrino parameters, and for experimentalists, to plan their set-ups.
\\
{\it \bf Acknowledgments}. This work was supported by a grant of the Romanian Ministry of National Education, CNCS UEFISCDI, project PCE-2011-3-0318, Contract no. 58/28.10/2011.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction} \label{sec:intro}
Cellular foams \cite{gibson_mechanics_1982},
semiflexible fiber and polymer networks \cite{broedersz_modeling_2014}, and many recently-developed mechanical metamaterials \cite{bertoldi_flexible_2017, goodrich_principle_2015, rocks_designing_2017} all belong, in idealized form, to a general class of disordered lattices. Such lattices can range in size from microscopic scaffolds for biological tissue growth \cite{janmey2008} to modern architectural structures \cite{knippers2016}. In each case, one can further idealize the material or structure as a mathematical network of connections between slender beams that intersect at various points within the material. From an engineering perspective, such materials are promising because of their light weights and their tunable, designable properties: a Poisson ratio from the auxetic \cite{hanifpour_mechanics_2018, reid_auxetic_2018,goodrich_principle_2015} to the incompressible limits \cite{goodrich_principle_2015}, a targeted local response to a remote perturbation \cite{rocks_designing_2017}, or the ability to change shape \cite{bertoldi_flexible_2017}. A disadvantage of these materials is that those that are constructed from stiff materials can degrade progressively through successive abrupt failures of the beams during loading \cite{driscoll_role_2016,hanifpour_mechanics_2018,zhang_fracturing_2018}. To design optimized structures and safely use them for structural applications, it is necessary to assess the most likely locations of fracture. Such predictive understanding would further enable the design of a material to fail in a prescribed way.
Fracture experiments have been conducted previously on printed, disordered auxetic materials \cite{hanifpour_mechanics_2018} and laser-cut, disordered honeycomb two-dimensional (2D) lattices \cite{driscoll_role_2016}. In these studies, very different fracture behaviors (ductile versus brittle) have been obtained by changing the loading direction \cite{hanifpour_mechanics_2018} or tuning the rigidity \cite{driscoll_role_2016}. In the latter study, a clear change arose in the spatial organization of fractures: they either can be dispersed throughout a system or be localized in the form of a narrow crack. Therefore, although some tunable parameters for controlling failure behavior have been identified, what determines these particular failure locations remains an open question. According to Griffith theory \cite{griffith_vi._1921}, damage in brittle materials focuses at the tip of a crack. However, factors such as material disorder \cite{roux_rupture_1988,alava_role_2008, shekhawat_damage_2013,curtin_brittle_1990,kahng_electrical_1988}, material rigidity \cite{driscoll_role_2016}, and the connectivity (specifically, mean degree) of networks \cite{driscoll_role_2016,zhang_fiber_2017} can affect the spatial organization of damage. As one tunes each of these factors, one can make failures spread throughout a system (diffuse damage), rather than forming a narrow crack (localized damage).
Zhang \textit{et al.} \cite{zhang_fracturing_2018} showed recently that failures can also be delocalized in topological Maxwell lattices (in which freely-rotating joints that are linked by rigid struts are on the verge of mechanical instability) \cite{mao2018}.
They performed numerical experiments on the tensile fracture of deformed square and kagome lattices, demonstrating that stress and fracture concentrate on self-stress domain walls, even in the presence of damage that is introduced elsewhere in the system. In another recent paper, Tordesillas \textit{et al.} \cite{tordesillas_interdependent_2018} studied damage locations in discrete-element simulations of concrete samples under uniaxial tension. From a network-flow analysis of the contact-network topology and contact capacities of a specimen, the authors determined the location of the principal interacting macrocracks. In their samples, they observed that secondary macrocracks develop in the pre-failure regime after damage occurs elsewhere, but before the formation of a dominant macrocrack, which sets the ultimate failure pattern of a sample.
In the present paper, we investigate where damage occurs in disordered lattices that consist of identical-width beams, with a network topology specified by the contacts measured from a real, quasi-2D granular packing (see Fig.~\ref{fig:method}).
We identify a common property, a large value of {\it geodesic edge betweenness centrality} \cite{girvan_community_2002}, that is shared by the failure locations of progressive damage events of our tested samples.
Even without modeling the physical interactions between nodes, this property provides a diagnostic for identifying likely failure locations. Such an indicator would permit assessing these locations in
a structure, without studying its detailed energetic or stress states.
Ultimately, the choice of a granular-inspired geometry for the disordered lattice will provide a route toward generalizing these studies across inherently different systems, which are linked by their network topology.
\begin{figure}
\center
\includegraphics[width=\linewidth]{Figure1.pdf}
\caption{
(a) Force chains (cyan) recorded in a two-dimensional assembly of frictional photoelastic disks (red), which we image via a circular polariscope \cite{daniels_photoelastic_2017}. Brighter particles carry stronger forces.
(b) Contact network (yellow), which we extract using an open-source photoelastic solver \cite{jekollmer_pegs:_2018}, overlaid on the reconstructed ``pseudo-image'' \cite{daniels_photoelastic_2017}.
(c) Network representation in which each particle center is a node (orange dots) and each load-bearing contact is an edge (blue lines) \cite{papadopoulos_network_2018}.
(d) Corresponding physical sample that we laser-cut from an acrylic sheet, with the edges represented by beams that intersect at crosslinks (which correspond to the nodes in the network). }
\label{fig:method}
\end{figure}
For each network, we laser-cut an acrylic sheet using a contact network that matches the one observed in a packing, and we then test its behavior under compressive or tensile loading.
Because the set of contacts in a packing forms a network that is embedded in a plane, a lattice does as well. Such a lattice network consists of edges (representing the beams of the lattice) that intersect at nodes, which occur at the crosslinks of the lattice. Conceptually similar structures occur for streets and intersections in the study of road networks \cite{crucitti_centrality_2006,lee2014},
connections between internet routers, plant veins \cite{katifori_quantifying_2012}, fungi \cite{lee2017}, and many other spatial systems \cite{barthelemy_spatial_2011, barthelemy_morphogenesis_2018}.
Network analysis provides useful approaches --- including measures, algorithms, and theory --- for characterizing complex spatial systems at multiple scales, ranging from local features to mesoscale and macroscale ones, and examining how they evolve \cite{newman2018,papadopoulos_network_2018}. As discussed by Smart \textit{et al.} \cite{smart_granular_2008}, it is appealing to investigate what insights network analysis and associated topics (e.g., graph theory and algebraic topology) can yield on novel physical systems, especially in comparison to traditional approaches. For example, this perspective was adopted by Tordesillas \textit{et al.} \cite{tordesillas_interdependent_2018} to study quasi-brittle failure using network flow. Such approaches have also been useful for the study of mesoscale structures, such as dense communities of nodes, in granular systems \cite{bassett_influence_2012}. Therefore, network analysis appears to be a promising route to identify common analytical tools that are capable of relating failure behaviors across a variety of disordered systems.
One important approach in network analysis is the calculation of ``centrality'' measures to ascertain the most important nodes, edges, and other subgraphs in a network \cite{newman2018,faust1994}. One particularly popular type of centrality, known as betweenness centrality, measures whether one or more parts of a network lie on many short paths; it has been employed to characterize the importances of nodes \cite{freeman_set_1977}, edges \cite{girvan_community_2002}, and other subgraphs. The most common type of betweenness centrality uses geodesic (i.e., strictly shortest) paths.
Recently, in a study of granular materials, Kollmer \textit{et al.} \cite{kollmer2019betweenness} showed that there is a positive correlation between the geodesic betweenness centrality value of a node and the pressure on the corresponding particle. Previously, Smart \textit{et al.} \cite{smart_effects_2007} reported that edges with large geodesic betweenness centrality exert a strong influence on heat transport in granular media. Inspired by these investigations, we selected from among the variety of network measures \cite{newman2018,barthelemy_morphogenesis_2018} and focus on calculating geodesic edge betweenness centrality (GEBC). See \eqref{ebc-def} in Materials and Methods for its definition.
As was reported in Berthier \textit{et al.} \cite{berthier_rigidity_2018}, one can control the compressive and tensile failure behaviors of a disordered lattice by tuning the mean degree of its associated network. This control parameter provides a way to create systems with a variety of failure behaviors, ranging from ductile-like to brittle-like failure. In the present paper, we show for samples across the spectrum from brittle-like to ductile-like failure (see ``Mechanical testing protocol'' in Materials and Methods) that individual beam failures occur predominantly on edges with GEBC values that are above the mean of the network. From this result, we conclude that GEBC is a useful diagnostic for forecasting possible failure locations in our contact networks. We demonstrate the ability of an GEBC-based test, which consists of comparing the geodesic edge betweenness centrality value of an edge to a threshold value, to discriminate between beams that fail and those that remain intact.
This finding, together with the work of \cite{smart_effects_2007}, suggests that
betweenness centrality is a useful measure for capturing essential physical properties in disordered systems. Our study also confirms that tools from network analysis give a promising paradigm for the study of fracture.
The effectiveness of GEBC, which depends on network topology rather than on specifying mechanical interactions, is unexpected. This motivates a deeper analysis to determine which behaviors do not depend primarily on the detailed physical properties of a system, but instead depend on its geometry (and associated network topology).
Our use of unweighted networks focuses our investigation on network topology, and we compare results for both a measure (specifically, GEBC) that ignores the physics
and a well-known scalar electrical analogy of elasticity known as a random-fuse network (RFN) model \cite{de_arcangelis_random_1985,kahng_electrical_1988}.
The RFN model identifies the most stressed beams as the edges with the largest currents, as determined by solving Kirchhoff's laws, for a given voltage drop across the boundaries.
We show that the RFN model, even with its incorporation of physical considerations, does not markedly improve performance over GEBC.
This indicates that one can capture essential features of the lattice failure behavior by geometric (rather than physical) considerations.
\section{Results}\label{results}
\subsection{Spatial heterogeneity and changes with applied strain of geodesic edge betweenness centrality}
We examine the ability of geodesic edge betweenness centrality $\tilde{E}$ (see ~\eqref{ebc-def} in Materials and Methods) to forecast the specific locations at which our samples fail. For each initial (and subsequently altered) network, we find that geodesic edge betweenness takes a broad range of values across the network. In Fig.~\ref{fig:EBC}a, we show the probability density function of the initial geodesic edge betweenness $\tilde{E}_0$ for each initial network at each value of the mean degree $\overline{z}_0$. To facilitate notation, we use the subscript $0$ to designate our initial networks and the quantities that we measure and compute with them.
In all cases, the distribution of values is approximately exponential, and it is largely independent of $\overline{z}_0$. Because each failure event (with associated edge removals) results in a new set of shortest paths, we obtain a new distribution of geodesic edge betweennesses for each altered network. Just as stress redistributes after damage \cite{alava_statistical_2006,zapperi_plasticity_1997,hidalgo_fracture_2002,berthier_damage_2017}, geodesic edge betweenness (due to its nonlocal nature) also redistributes in a system. In Fig.~\ref{fig:EBC}b, we show a characteristic example of redistribution after a failure event. The redistributions are system-wide: some edges are ``reloaded'', becoming more important with respect to the others (i.e., $\tilde{E}_{s+1}>\tilde{E}_{s}$, when going from strain step $s$ to strain step $s+1$),
others are ``unloaded'' (i.e., $\tilde{E}_{s+1}<\tilde{E}_{s}$), and some edges (in lavender) have the same (or almost the same) value. By contrast, removal of unimportant edges (i.e., those with small values of geodesic edge betweenness) results in small (in amplitude) changes.
\begin{figure}
\center
\includegraphics[width=\linewidth]{Figure2.pdf}
\caption{Characterization of geodesic edge betweenness (re)distribution:
(a) Probability density function (PDF) of the initial geodesic edge betweenness centrality $\tilde{E}_0$ for the different initial networks. We show the PDFs for several values of mean degree.
(b) Changes in geodesic edge betweenness centralities $\tilde{E}$ after a failure event that occurs at the red ellipse at a compressive strain of $\varepsilon \approx 1.98\%$ on a network with mean degree $\overline{z}_0 = 2.40$. The values of the lavender edges change by less than $10^{-2}$.
}
\label{fig:EBC}
\end{figure}
Damage occurs progressively through a sequence of tensile or compressive loading. In Fig.~\ref{fig:Maps}, we show examples of damage progression for three values (one per row) of $\overline{z}_0$. Within each row, a sample progresses from its initial, intact network $G_0$ (an unweighted and undirected graph) through an altered network at which approximately 50\% of its beams have failed, and then to the network immediately before the final failure. In the image immediate after the last one that we show in each row, there is no longer a set of beams that connects the top and bottom boundaries of the sample. We color each edge in a network according to the value of $\tilde{E}_s$ at that strain step.
\begin{figure*}
\centering
\includegraphics[width=\linewidth]{Figure3.pdf}
\caption{ Example images of the spatial distribution of normalized geodesic edge betweenness centralities $\tilde{E}$ (given by the color bar), which we plot at a particular applied strain $\varepsilon$ (see ``Mechanical testing protocol'' in Materials and Methods for a discussion of strain steps) for samples that are subject to compression.
The rows show samples with
(top) $\overline{z}_0 = 2.40$;
(middle) $\overline{z}_0 = 3.00$;
(bottom) $\overline{z}_0 = 3.60$;
Within each row, we show the progression (of strain steps) in $\varepsilon$ from
(left) initial, intact networks $G_0$ with adjacency matrix ${\bf A}_0$; to
(center) the step at which $50\%$ of breakages have occurred
(with $\varepsilon \approx 3.39 \%$, $\varepsilon \approx 1.68 \%$, and $\varepsilon \approx 1.90\%$ from top to bottom); and finally to
(right) the strain step immediately before a system-spanning failure
(with $\varepsilon \approx 9.56 \%$, $\varepsilon \approx 3.66 \%$, and $\varepsilon \approx 1.95\%$ from top to bottom).
}
\label{fig:Maps}
\end{figure*}
Geodesic edge betweenness is spatially heterogeneous across a network, and we observe that large values (bright colors) can occur throughout a network. These locations shift both in space and in time, due to the disordered structure (which arises from geometry) of the lattice.
By contrast, for a regular lattice, the importance of edges decreases with their distance from the geometric center of a system \cite{barthelemy_morphogenesis_2018}. The introduction of disorder --- such as by rewiring, addition, or removal of edges --- results in more complicated distributions and can lead to geographically central edges with smaller importance than elements that are farther from the geometric center \cite{lion_central_2017}.
Importantly, although the topologies of the networks underlying our lattices
are inherited from uniaxially compressed granular packings, we not observed a preferred orientation for edges with GEBC values that are above the mean.
Granular packings encode their preparation history in the form of anisotropic stresses \cite{majmudar_contact_2005,bililign2019}, but this anisotropy is not readily identifiable from the contact network (which is unweighted).
The GEBC values at a given strain step illustrate the broad distribution of values, as we observed in the exponential probability density function of $\tilde{E}_0$ (see Fig.~\ref{fig:EBC}).
Even in these small systems, some edges have values up to $20$ times the mean of the system; these are ones that are particularly important for connecting different parts of a network. Many other edges occur only infrequently as shortest-path connectors. The variations in spatial distribution along the rows of Fig.~\ref{fig:Maps} highlight the importance of the removed edges, as we emphasized in Fig.~\ref{fig:EBC}b.
Importantly, although $\tilde{E}$ tends to decrease with distance from the geometric center, this need not be true for specific samples. For the near-final networks (in the right-most column in Fig.~\ref{fig:Maps}) at $\overline{z}_0=2.40$ and $\overline{z}_0=3.00$, the maximum of $\tilde{E}$ is located near the left boundary of the sample, rather than near the middle. In both cases, the largest values of $\tilde{E}$ occur on edges that connect the top and bottom parts of the network, and these are also the next beams that will break (and lead to the final cascade of failures).
\subsection{Geodesic edge betweenness centralities of failed edges}
\begin{figure*}
\center
\includegraphics[width=\linewidth]{Figure4.pdf}
\caption{
(a) Cumulative distribution function of geodesic edge betweenness centrality of failed edges of all experiments. We show the probability density function in the inset.
(b) Fraction of edges in the network for which $\tilde{E}>\tilde{E}_{\mathrm{th}}$ (blue dots, left axis) and fraction of failed beams for which $\tilde{E}\leq\tilde{E}_{\mathrm{th}}$ (orange diamonds, right axis). (c) Magnification of the crossover point between the CCDF and the failure rate (and the envelope of results of individual samples).
}
\label{fig:CDFs}
\end{figure*}
Such observations suggest that there is a correlation between large values of $\tilde{E}$ and future failure locations. To assess the generality of this finding, for each breaking beam, we calculate the GEBC $\tilde{E}_f$ during step $s-1$ immediately before its failure at step $s$.
For all of our samples and for all non-large failure events (which we take to mean that no more than three beams are involved), we enumerate the immediately-preceding values of $\tilde{E}$ for the failed edges. In Fig.~\ref{fig:CDFs}a, we show the cumulative distribution function (CDF) of this set of values, together with the corresponding probability density function (PDF) in the inset. We fit the PDF with an exponential with mean $\tilde{E}_f^\ast \approx 10.3$ (with $R^2 \approx 0.96$).
There is a corresponding gradual increase for $\tilde{E}_f \gtrapprox 10$ of the CDF, suggesting that few failing edges have a value that is significantly larger than the mean. We observe such large values of $\tilde{E}$ only when the samples are near full failure; at this point, only a few paths are available to connect the top and bottom boundaries of the network. One can see this situation in the right column of Fig.~\ref{fig:Maps}. Focusing on $\tilde{E}_f = 1$, we see that about $76\%$ of the breakages occur on edges with values of $\tilde{E}_f$ that are above the mean. Because only a small subset of the network's edges have $\tilde{E}>1$ (see the distribution in Fig.~\ref{fig:EBC}a), even the value of $\tilde{E}$ alone is a valuable diagnostic for forecasting failure locations.
We can refine this diagnostic by directly considering the population of edges that exceed a threshold value $\tilde{E}_{\mathrm{th}}$. We illustrate this population by plotting the complementary cumulative distribution function (CCDF) on the left vertical axis in Fig.~\ref{fig:CDFs}b. Because the proportion of edges that satisfy $\tilde{E}>\tilde{E}_{\mathrm{th}}$ evolves after each edge removal and differs across initial networks, we choose each point of the curve to be the maximum value that we encounter among all networks. The success rate of this diagnostic is the fraction of failed beams that satisfy $\tilde{E} > \tilde{E}_{\mathrm{th}}$, and the failure rate is the fraction for which $\tilde{E} \leq \tilde{E}_{\mathrm{th}}$. We show the latter in orange diamonds in the right vertical axis of Fig.~\ref{fig:CDFs}b for all non-large failure events among all tested samples, regardless of the tensile or compressive nature of the applied loading. In Fig.~\ref{fig:CDFs}c, we focus on the point at which the CCDF and the failure-rate curves cross; this intersection occurs at $\tilde{E}_{\mathrm{th}} \approx 1.1$, corresponding to a value on the CCDF curve (i.e., the fraction of edges for which $\tilde{E} \gtrapprox 1.1$) of about $0.34$ and a failure rate of about $0.26$. This intersection point indicates that considering all edges with above-mean geodesic edge betweenness values provides a reasonable population of edges to consider, but one can choose other values in a trade-off between forecast failure rate and the fraction of examined edges.
The above general results exhibit sample-to-sample variation. To highlight this, we include an envelope of the failure rate in Fig.~\ref{fig:CDFs}c. To obtain this envelope, we determine a failure rate curve for each of the $14$ samples (see Materials and Methods). We obtain each curve by examining the failure events that occur on each initial intact network. For each threshold value, we track the best (lower point) and worst (upper point) failure-rate value among the $14$ curves. The envelope is the set of points between these lower and upper bounds for each threshold value. Although the scatter is non-negligible, for a threshold of $\tilde{E}_{\mathrm{th}} = 1$, we still obtain success rates above 65\% for all samples.
\subsection{Test sensitivity and specificity}
Performing sensitivity and specificity analysis \cite{simundic_measures_2009} allows a more detailed determination of the suitability of using $\tilde{E} > \tilde{E}_{\mathrm{th}}$ to identify beams that are likely to fail. We define the outcome of this test as a true positive (TP), false positive (FP), true negative (TN), or false negative (FN) according to the state of the beam (failing or remaining intact). See our summary in Table~\ref{tab:TestDef}.
{\it Sensitivity} is defined as the probability of obtaining a positive test result for the population of failed beams (i.e., the proportion of true positives), so $\mathrm{sensitivity} = \mathrm{TP}/(\mathrm{TP} + \mathrm{FN})$.
Similarly, {\it specificity} is the probability of obtaining a negative test result for the population of intact beams (i.e., proportion of true negatives), so $\mathrm{specificity} = \mathrm{TN}/(\mathrm{TN} + \mathrm{FP})$.
These two measures quantify the success of our test for correctly identifying beams that will fail or remain intact.
\begin{figure}
\center
\includegraphics[width=\linewidth]{Figure5.pdf}
\caption{Evaluation of our test's accuracy. (a) Sensitivity and specificity versus the threshold $\tilde{E}_{\mathrm{th}}$. (b) Receiver operating characteristic (ROC) curve summarizing the $(\mathrm{sensitivity}, 1-\mathrm{specificity})$ pairs that we obtain for different values of $E_c$. The dashed diagonal line indicates the behavior of a test that cannot discriminate between failing and intact beams.
}
\label{fig:ROC}
\end{figure}
We calculate sensitivity and specificity in considering all non-large failure events of all experiments as a function of the threshold $\tilde{E}_{\mathrm{th}}$, and we show the results in Fig.~\ref{fig:ROC}a. As expected, sensitivity and specificity show opposite trends: as one lowers the threshold, the true-positive fraction (sensitivity) increases, but so does the false-positive fraction, so that the specificity (i.e., the true-negative fraction) decreases. As one increases the threshold, the opposite occurs: we obtain a lower true-positive fraction (sensitivity decreases), and the false-positive fraction decreases (specificity increases). There is a crossover between sensitivity and specificity at $\tilde{E}_{\mathrm{th}} \approx 1.1$, which is close to the value $1$ that we used above.
Computing a receiver operating characteristic (ROC) curve \cite{zweig_receiver-operating_1993} provides additional insight into the choice of $\tilde{E}_{\mathrm{th}}$. As we show in Fig.~\ref{fig:ROC}b, we measure sensitivity and specificity as a function of $\tilde{E}_{\mathrm{th}}$. A test with perfect forecasting of failing versus intact beams would go through the upper-left corner (in which sensitivity and specificity are both $1$), and a test without any predictive power would follow the dashed diagonal line. (Anything below this line gives a result that is worse than random guessing and indicates a test direction that is opposite to what should have been chosen.)
To obtain a global estimate of the accuracy of the test that goes beyond visual examination, we compute the area under the curve of the ROC curve. This ranges from $0.5$ (no discrimination) to $1$ (perfect accuracy). The value for the curve in Fig.~\ref{fig:ROC}b is $0.79$, indicating a good capability of our test to discriminate between beams that will remain intact versus those that will fail.
\begin{table}
\centering
\caption{Definition of the outcome of a test.}
\begin{tabular}{c| c c}
Test: Is $\tilde{E}>\tilde{E}_{\mathrm{th}}$? & Beam fails & Beam does not fail\\
\hline
Positive & True Positive (TP) & False Positive (FP) \\
Negative & False Negative (FN) & True Negative (TN)\\
\hline
\end{tabular}
\label{tab:TestDef}
\end{table}
\section{Discussion} \label{discussion}
\subsection{Other network diagnostics and approaches for forecasting failures}
Motivated by the results in \cite{smart_effects_2007} and the geometric origin of our samples, we have focused on using geodesic edge betweenness centrality for forecasting failures in them. However, other network measures are also worth considering as possible diagnostics for forecasting failure locations. In particular, it is desirable to take advantage of the fact that the various flavors of betweenness are correlated with other quantities in certain types of networks. In some networks, for example, geodesic node betweenness can scale approximately with node degree \cite{barthlemy_betweenness_2004}. To give another example, Scellato \textit{et al.} \cite{scellato_backbone_2006} studied the relation between GEBC and a quantity known as {\itshape information centrality} in networks based on the road systems of several cities.
For the edge $e_{ij}$ of a network, information centrality is
\begin{equation}
J_{ij} = \frac{F[G]-F[G']}{F[G]}\,,
\end{equation}
where
\begin{equation}
F[G] = \dfrac{1}{N(N-1)}\sum_{i,j=1,\ldots,N;\,i\neq j}\dfrac{d_{ij}^{\mathrm{Eucl}}}{d_{ij}}
\end{equation}
is the
{\itshape efficiency} of an $N$-node graph $G$, the graph $G'$ results from removing edge $e_{ij}$ from $G$, the quantity $d_{ij}$ is the distance between nodes $i$ and $j$ (e.g., from the shortest number of steps between $i$ and $j$ in an unweighted graph), and $d_{ij}^{\mathrm{Eucl}}$ is the Euclidean distance between those nodes.
We investigate the relation between geodesic edge betweenness centrality and information centrality using a modified expression for efficiency, where we set $d_{ij}^{\mathrm{Eucl}}=1$ for all pairs $(i,j)$ with $i \neq j$. To calculate information centrality, we use a code from \cite{noauthor_brain_nodate}. Information centrality gives an indication of the perturbation of transmission across a network when one removes an edge. In other words, we ask the following question: how harmful is a beam failure for connections across a sample? When considering all edges of all of our networks (both initial and altered), we obtain a correlation (with a Spearman rank-correlation coefficient of $0.55 \pm 0.014$) of information centrality with geodesic edge betweenness centrality. For failed edges, the Spearman correlation coefficient ($0.77 \pm 0.01$) is even larger. Motivated by these calculations, we checked and confirmed that considering values above the mean for information centrality yields similar results as using geodesic edge betweenness centrality as a test for potential failures. Therefore, information centrality is an alternative to geodesic edge betweenness centrality to probe systems for likely failure locations. Measures based on shortest paths are not always highly correlated with each other (and the extent of such correlation also depends on network type) \cite{lee2014}, so different measures related to betweenness can give complimentary insights.
\subsection{Comparison with other diagnostics}
To disentangle the roles of physical and geometric effects, we evaluate the importance of introducing physical considerations by repeating our analysis using an RFN-based test. (See ``Random-fuse network (RFN) model'' in Materials and Methods.)
In place of {$\tilde{E}$}, we determine the current $\tilde{I}$ that flows through a network.
As with GEBC, we determine the fraction of edges in a network with a
current above a threshold current $\tilde{I}_{\mathrm{th}}$ and the fraction of failing beams for which $\tilde{I}_f\leq \tilde{I}_{\mathrm{th}}$ (where $\tilde{I}_f$ denotes the current of an edge during step $s-1$ immediately before its failure at step $s$).
For threshold values in the range $[0,2.5]$, we find that this test performs somewhat better than the test using GEBC (see Fig.~S1a).
However, in the range $[0,1.4]$, the fraction of edges with a current above the threshold is slightly larger than the fraction of edges with a GEBC above the same threshold. Consequently, the trade-off between the forecast failure rate and the fraction of possible edges differs between the two tests for the same threshold value. This trade-off diminishes the advantage.
We also find that the current and GEBC values of the failed edges are positively correlated, with a Pearson correlation coefficient of $R \approx 0.81$ (see Fig.~S1b).
This suggests that failing beams, most of which have a larger stress than the mean value in the network (i.e., $\tilde{I}_f>1$ for these edges) lie on many shortest paths, explaining the small improvement in forecast capability from using the RFN-based test.
This finding is similar to the results of Kollmer {\it et al.} \cite{kollmer2019betweenness}, who observed in 2D packings of frictional particles that ones with large {\it node} betweenness centralities are statistically likely to be highly stressed. Consequently, it is appropriate to examine betweenness centralities (of both nodes and edges) to capture important mechanical properties of physical systems.
Interestingly, as we show in Fig.~S2, the two tests do not systematically misdiagnose failure locations (i.e., yield false-negative outputs) for the same edges. Using the RFN-based test, we observe a correlation between beam angle and current flow (see Fig.~S3a) and find a bias towards misdiagnosed edges that are roughly perpendicular to the loading direction (see Fig.~S3b). It appears that failures occur at edges at all distances from the boundaries (see Fig.~S4a). However, for some threshold values, most GEBC-misdiagnosed edges tend to occur near boundaries (see Fig.~S4b), where large values of GEBC are less frequent.
The dependency of geodesic edge betweenness centrality with distance from a sample's geometric center, even altered by the presence of disorder, is an important feature of networks that are embedded in a plane. To test whether we can circumvent this limitation of the test, we calculate geodesic edge betweenness centralities on a collection of modified networks. For a given network $G$ and for each edge $e_{ij}$, we generate a duplicated network such that the edge $e_{ij}$ is at the geometric center. To construct such a graph, we first duplicate the network with mirror symmetries with respect to each boundary. We then calculate the geodesic edge betweenness centrality of $e_{ij}$ by considering only a section of this duplicated network that is approximately centered on $e_{ij}$. We repeat this procedure for each edge of the network $G$ to obtain centrality values for the network, centered at that edge. Using this approach yields a cumulative distribution for $\tilde{E}_f$ and a test ($\tilde{E}_f>\tilde{E}_{\mathrm{th}}$) success rate similar to the original networks.
Indeed, while boundary edges can have large $\tilde{E}$ values,
the distributions of $\tilde{E}$ are more homogeneous than in the original networks, such that the test can misdiagnose edges in other locations.
Consequently, the use of a duplicated network does not improve the forecasting ability of our approach.
Developing methods to appropriately consider the role of boundaries remains a central question for planar graphs --- not only for granular materials, but also for other applications, such as determining high-traffic edges in road networks \cite{lee2014} and nutrient-transportation networks \cite{lee2017} --- and more generally in spatially-embedded networks.
In our comparison of these GEBC-based and RFN-based tests, we observe a correlation between the physical and geometric properties of failing beams. Interestingly, a test that includes a minimal set of physical ingredients (i.e., the RFN-based test) performs only somewhat better than a test (the GEBC-based test) that is based on geometric considerations. Because these two tests have rather different limitations --- with less successful forecasting of near-perpendicular edges versus near-boundary edges --- it is useful to employ both as complementary approaches. Finally, it is worth noting that although the RFN-based test is faster computationally than the GEBC-based one, neither approach requires significant computational resources for the system sizes that we consider in this study.
\section{Conclusions} \label{conclusions}
The idea, proposed in papers such as \cite{smart_granular_2008} and reviewed in \cite{papadopoulos_network_2018} in the context of granular and particulate systems, to use network analysis to achieve insights on novel physical systems seems very promising for studies of fracture.
In this paper, we explored the application of centrality measures (based on shortest paths) to forecast failure locations in physical samples. Many other tools from network analysis, such as those based on exploration of mesoscale structures, also promise to yield fascinating insights into investigations of physical networks. In particular, examining how they can contribute to forecasting not only where, but also when, failures occur in disordered networks is a central point for future studies.
In the present investigation, we found that calculations based on shortest paths can be very helpful for forecasting failure locations in disordered lattices. Specifically, calculating the geodesic betweennesses of the edges in a network permits one to assess which edges are more prone to failure than others. Considering only edges with values above the mean geodesic edge betweenness of a network allows one to discard a large fraction of edges as unlikely failure locations. This feature of the test makes it very valuable, particularly as it avoids a detailed analysis of energetics.
Combined with \cite{smart_effects_2007,kollmer2019betweenness}, our work provides evidence that betweenness centrality successfully identifies physical properties in both granular packings and lattices that are derived from them. Similarly, analyses inspired by rigidity percolation in granular materials have identified that our disordered lattices undergo a ductile--brittle failure transition as a function of connectivity $\overline{z}_0$, as determined by counting degrees of freedom and constraints \cite{berthier_rigidity_2018}.
However, we have focused on testing the sensitivity and specificity of our approach in the context of disordered lattices that we generated from force networks in quasi-2D granular packings, and this does not ensure its success for other network structures. Indeed, it has been established that the origin of a disordered structure --- whether from numerical spring networks, frictionless jammed sphere packings, or diluted networks --- can strongly affect elastic responses and the rigidity transition \cite{ellenbroek_non-affine_2009,ellenbroek_rigidity_2015}. Therefore, we expect that both the granular origins of our samples (and hence their geometry) may affect the particular failure behavior that we have observed in this study. Collectively, these investigations (and ours) point towards a need to understand the importance of network topology and geometry themselves, regardless of their manifestation as a granular packing or a lattice. This is an important step towards distinguishing between geometric effects and system-specific physical effects. Therefore, an important future direction is to examine networks obtained by other means, such as an over-constrained granular packing, a biological system such as leaf-venation patterns, or by randomly pruning a crystalline lattice.
Our work also opens the door for structure design and the purposeful setting of desired failure locations. One can build particular network topologies into designed materials that permit the constraining of failures to regions of a sample or, by contrast, promote desirable patterns of damage spreading to ensure the robustness of structures.
\section{Methods} \label{sec:methods}
\subsection{Experimental samples} \label{sec:samples}
We conduct experiments on a set of disordered structures that we derived from experimentally-determined force networks in granular materials, as done in Berthier \textit{et al.} \cite{berthier_rigidity_2018}. The methodology to create these experimental samples is inspired by the work of \cite{driscoll_role_2016,goodrich_principle_2015,reid_auxetic_2018, ellenbroek_non-affine_2009,ellenbroek_rigidity_2015}, who performed a similar process numerically. We begin from observed force-chain structures in a quasi-2D photoelastic granular material. The granular packings consist of $N=824$ bidisperse circular discs (of two distinct radii $r_1$ and $r_2$, with $r_1/r_2 = 1.4$) in approximately equal numbers, as shown in Fig.~\ref{fig:method}a. We uniaxially load each packing under a series of finite displacements of one wall, generating multiple realizations of both packings and force networks.
Using an open-source photoelastic solver \cite{daniels_photoelastic_2017,jekollmer_pegs:_2018}, we identify all load-bearing contacts in a system, yielding a network of physical connections between particles that we use to generate a disordered lattice (see Fig.~\ref{fig:method}b). We construct a network by assigning each particle center as a node of a graph $G$ and then placing an edge between two nodes wherever we observe a load-bearing contact. The network is associated with an $N \times N$ binary (i.e., unweighted) adjacency matrix ${\bf A}$, with elements
\begin{equation}\label{matrix}
A_{ij}=\left\{
\begin{array}{ll}
1\,, \, \text{if particles}\,\, i\,\,\text{and}\,\, j \,\, \text{are load-bearing}\,,\\
0\,,\, \text{otherwise}\,.\\
\end{array}
\right.
\end{equation}
We showed an example network in Fig.~\ref{fig:method}c. It is undirected because each contact is bidirectional, and its associated adjacency matrix is therefore symmetric about the diagonal ($A_{ij} = A_{ji}$).
We laser-cut the physical samples from acrylic plastic sheets (with an elastic modulus of about $3$ GPa) of thickness $h = 3.17$~mm. Each edge becomes a beam of width $1.5$~mm; beams intersect at crosslinks that correspond to the centers of particles (i.e., the nodes), and the particles' radii sets the length of the beams. We adopt the term ``crosslink'' from the study of fiber networks \cite{zhang_fiber_2017,broedersz_modeling_2014}, which
consist of filaments (bonds) that are bound via crosslinkers that either allow energy-free rotations or associate angular variations to a finite cost of energy (as ``welded'' crosslinks). We showed an example sample in Fig.~\ref{fig:method}d; note that samples that we construct multiple times based on the same mathematical networks by cutting from different sheets of material are not perfectly identical due to small details of processing during cutting.
A simple characteristic of a network is its {\itshape mean degree} $\overline{z}$ (also known as the ``connectivity'' or ``coordination number''), which is equal to the mean number of edges per node. That is,
\begin{equation}
\overline{z} = \frac{1}{2N}\sum_{i,j}^N A_{ij}\,.
\end{equation}
It is known that the bulk properties of amorphous solids \cite{wyart2005, berthier_rigidity_2018} are influenced strongly by $\overline{z}$.
We use the subscript $0$ denote initial networks (i.e., networks before any subsequent modifications from lattice beam failures).
We study $6$ different initial networks, with mean degrees $\overline{z}_0 = \{2.40, 2.55, 2.60, 3.00, 3.35, 3.60\} \pm 0.02$, which we draw from two different initial granular configurations.
We do a total of $14$ experiments, which we test each network at least once in compression and once in tension; for the networks with $\overline{z}_0 = 2.60$ and $\overline{z}_0 = 3.00$, we do an additional tensile test on
a second set of fully intact samples.
To obtain a sample that is as close as possible to the isostatic value $\overline{z}_{\mathrm{iso}} = 3.00$ of an infinite friction packing \cite{henkes_rigid_2016}, we prune a network that initially has a value of $\overline{z}_0 = 3.60$ by progressively removing its contacts with the smallest force values.
\subsection{Mechanical testing protocol} \label{sec:exper}
We perform compression and tension tests using an Instron 5940 Single Column system with a $2$~kN load cell. We use a displacement rate of $1.0$~mm/min for tension experiments and $1.5$~mm/min for compression experiments. In compression, we confine a sample between two parallel acrylic plates to constrain out-of-plane buckling. We record each experiment using a Nikon D850 digital camera at a frame rate of 24 or 60 fps. During the course of each experiment, beams break throughout a sample as damage progresses. Using the time series of measured compressive and tensile forces, we identify each {\it failure event}, which corresponds to a set of one or more breakages that occur simultaneously. Our frame rates are insufficient to distinguish multiple, successive breakage events that occur within a single failure event, but they are sufficient to easily separate the failure events from each other. In all cases, we are able to determine the locations of individual beam failures by examining the images collected immediately following a recorded drop in force. The failure events occur sequentially, deteriorating the structure until complete failure of a sample. This corresponds to having a crack going through the sample from one lateral side to the other, such that there is no set of beams that connects the top and bottom boundaries.
As damage progresses, the adjacency matrix ${\bf A}$ (and the associated network $G$; see \eqref{matrix}) that encodes the network structure changes following each failure event. When the beam that connects nodes $i$ and $j$ fails, we set $A_{ij} = A_{ji} = 0$ to record this event. We thus do a series of computations on networks that are based on measurements at a particular strain step $s$, which is associated to an applied strain value $\varepsilon$. We distinguish between the initial network $G_0$ (with adjacency matrix ${\bf A}_0$), which is associated with the fully intact sample, and altered networks $G_s$ (with associated adjacency matrices ${\bf A}_s$).
Note that, as characterized in \cite{berthier_rigidity_2018}, both tensile and compressive loading of samples with $\overline{z}_0 < \overline{z}_{\mathrm{iso}}$ will fail from breakages that are well-separated in time and are spatially spread in a sample (i.e., ductile-like failure). By contrast, it was shown in \cite{berthier_rigidity_2018} that for $\overline{z}_0 > \overline{z}_{\mathrm{iso}}$, a few temporally separated breakages take place before the samples break
abruptly and all of the failed beams are localized, forming a narrow crack (i.e., brittle-like failure).
Therefore, for the samples with $\overline{z}_0 = 3.35$ and $\overline{z}_0 = 3.60$, the deterioration of a sample's structure occurs via both small (one to three breakages at a time) and large (more than three simultaneous breakages) failure events. In our analysis, we remove corresponding edges from the networks as failures take place, and we then perform fresh calculations of centrality.
Our analysis of failure locations excludes the large events, because we are specifically interested in the progression of failures. Our results are qualitatively similar for samples tested in tension versus in compression, so we do not distinguish between these two loading conditions in our analysis.
\subsection{Geodesic edge betweenness centrality (GEBC)} \label{sec:EBC}
Because failures in our samples consist mostly of breaking beams (rather than the thicker crosslinks), we focus on an edge-based counterpart of geodesic node betweenness centrality \cite{girvan_community_2002}. This measure gives insight into the importance of edges in a network in terms of how often they are on shortest paths between origin and destination nodes. Considering an edge $e_{ij}$ that links nodes $i$ and $j$ in a graph $G$, we calculate a symmetric {\itshape geodesic edge betweenness centrality} matrix based on the fraction of shortest paths that traverse an edge when considering all origin--destination pairs of nodes in a network (including nodes $i$ and $j$) \cite{barthelemy_morphogenesis_2018}:
\begin{equation}\label{ebc-def}
E_{ij} = \sum_{s\neq t} \frac{\sigma_{st}(e_{ij})}{\sigma_{st}}\,,
\end{equation}
where $\sigma_{st}$ is the number of shortest paths from node $s$ to $t$, and $\sigma_{st}(e_{ij})$ is the number of those paths that include the edge $e_{ij}$. We compute $E_{ij}$ using open-source code from \cite{noauthor_brain_nodate} (which uses an algorithm that is a slight modification of the one in \cite{brandes_faster_2001}). This measure can be computationally costly for large networks. The computation time is ${\cal O}(Nm)$ for sparse networks, where $N$ and $m$ denote the numbers of nodes and edges, respectively, of a network \cite{girvan_community_2002}. All of our graphs $G$ are undirected and unweighted, but one can also study notions of edge betweenness centralities for directed and weighted graphs.
It is common to normalize $E_{ij}$ by $\tfrac{1}{2}N(N-1)-1$ (i.e., by the number of edges, other than the one under consideration) \cite{chen_preliminaries_2015} or by $(N-1)(N-2)/2$ (i.e., the number of node pairs) \cite{barthelemy_morphogenesis_2018} to ensure that geodesic edge betweenness values lie between $0$ and $1$. However, because we will compare the relative importance of edges to others in a given network and as successive edge removals occur, we use a different normalization. In our calculations, for a given network at strain step $s$ and characterized by its adjacency matrix ${\bf A}_s$ (where $s=0$ for the initial network), we define the normalized geodesic edge betweenness matrix $\tilde{{\bf E}}_{s} = {\bf E}_{s}/\overline{E}_{s}$, where $\overline{E}_{s}$ is the mean over all edges of the network $G_s$. To study the importance of the failing beams, for each strain step, we compute the matrix $\tilde{\bf E}_{s}$ and extract the values $\tilde{E}_f$ of the edges that fail in the next failure event.
\subsection{Random-fuse network (RFN) model} \label{sec:RFN}
We create a RFN \cite{de_arcangelis_random_1985,kahng_electrical_1988} in which each fuse matches an edge of the network for one of our samples.
All fuses have identical conductance, because the beams have identical thickness.
We load the top and bottom boundaries by applying a fixed voltage to the top nodes and connecting the bottom nodes to ground (zero voltage). We determine edge voltages and their associated currents by solving Kirchhoff's laws. Analogous to our examination of GEBC, we normalize the current by the mean over all edge currents in a network. We define a current matrix at strain step $s$ by $\bold{\tilde{I}}_s = \bold{I}_s/\overline{I}_s$, where $s=0$ denotes the initial network and associated quantities. We then proceed as with GEBC: for each strain step $s$, we calculate the matrix $\tilde{\bf I}_s$ and extract the normalized current (i.e., stress) $\tilde{I}_f$ of the edges that fail in the next failure event.\\
\textbf{Acknowledgements:} We gratefully acknowledge Jonathan Kollmer for sharing the granular force network data that was collected in \cite{berthier_rigidity_2018}. This research was supported by the James S. McDonnell Foundation. We thank two anonymous referees for their many helpful comments, and we are also particularly thankful to Referee 2 for the suggestion to use the RFN-based test to perform a deeper analysis.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}\label{sec:intro}
In this paper we wish to study the completely isometric automorphisms of a free product of noncommutative disc algebras amalgamated over the identity. In particular it will be established that such automorphisms have a rigid structure. This will tell us that one can think of such a free product as the appropriate non-commutative analogue of an algebra of holomorphic functions.
The noncommutative disc algebra, $\mathfrak A_n$, introduced by Popescu \cite{Pop0}, is the universal operator algebra generated by a row contraction, $(\mathfrak s_1,\cdots, \mathfrak s_n)$ such that $\sum_{i=1}^n \mathfrak s_i\mathfrak s_i^* \leq I$, and the identity. $\mathfrak A_1 = A(\mb D)$, the disc algebra, and for $n\geq 2$ this should be thought of as a multivariable noncommutative generalization of the disc algebra. This algebra is also completely isometrically isomorphic to the norm closed algebra generated by the left creation operators and the identity on the full Fock space.
Popescu generalized this from one row contraction to several which he called a sequence of contractive operators, namely $(\mathfrak s_{i,1}, \cdots, \mathfrak s_{i,n_i}), 1\leq i \leq m$ with each tuple a row contraction \cite{Pop1, Pop2}. In this paper, such an object will be called a {\bf set of row contractions} since the author believes this to be clearer. Popescu goes on to prove that the universal operator algebra generated by a set of row contractions and the identity is the free product of $\mathfrak A_{n_1}, \cdots, \mathfrak A_{n_m}$ amalgamated over the identity. This algebra is denoted $*_{i=1}^m \mathfrak A_{n_i}$ or $\mathfrak A_{n}^{*m}$ if they all have the same size.
He also shows that these free products are completely isometrically isomorphic to some universal operator algebras studied by Blecher and Paulsen \cite{BlechPaul, Blech} which they denote $OA(\Lambda, \mathcal{R})$.
It will be established that every completely isometric automorphism of such a free product of $\mathfrak A_n$ is induced by a biholomorphism of the character space. Thus, such an automorphism will be given by a completely isometric automorphism of each noncommutative disc algebra and a permutation of these algebras. This is due to Rudin, Ligocka and Tsyganov \cite{Rudin, Lig, Tsyganov}, that a biholomorphism of a product space is just a product of biholomorphisms in each space and a permutation of the spaces. This will all be shown in Section 2.
The remainder of the paper concerns multivariable dynamics and operator algebras.
Specifically, a multivariable dynamical system $(X,\sigma)$ is a locally compact Hausdorff space $X$ along with $n$ proper continuous maps $\sigma = (\sigma_1,\cdots, \sigma_n)$ of $X$ into itself. To such a system one can construct two non-selfadjoint operator algebras. The first being the {\bf semicrossed product algebra} $C_0(X) \times_\sigma \mb F_n^+$ which is the universal operator algebra generated by $C_0(X)$ and $\mathfrak s_iC_0(X), 1\leq i\leq n$, where the operators $\mathfrak s_1,\cdots, \mathfrak s_n$ are contractions and satisfy the following covariance relations
\[
f \mathfrak s_i = \mathfrak s_i (f\circ\sigma_i) \ \ {\rm for } \ \ f\in C_0(X) \ \ {\rm and} \ \ 1\leq i\leq n.
\]
Similarly define the {\bf tensor algebra} to be the same as above but with the added condition that the generators are row contractive $\|[\mathfrak s_1 \cdots \mathfrak s_n]\| \leq 1$. These algebras should be thought of as encoding an action of the free semigroup algebra $\mb F_n^+$ on $C_0(X)$. If $n=1$ these algebras are the same and it has been proven that two such algebras are algebraically isomorphic if and only if their associated dynamical systems are conjugate, there exists a homeomorphism between the spaces intertwining the maps. This study was first initiated by Arveson \cite{Arv1} in 1967, later taken up in \cite{ArvJos, Peters, HadHoo} with the introduction of the semicrossed product given by Peters and finally proven in full generality by Davidson and Katsoulis \cite{DavKat1} in 2008.
For the multivariable case, Davidson and Katsoulis in \cite{DavKat2} introduced a new notion of conjugacy. We say that two dynamical systems $(X,\sigma)$ and $(Y,\tau)$ are {\bf piecewise conjugate} if there is a homeomorphism $\gamma: X \rightarrow Y$ and an open cover $\{V_\alpha : \alpha\in S_n\}$ such that $\gamma \circ\tau_i \circ\gamma^{-1}|_{V_\alpha} = \sigma_{\alpha(i)}|_{V_\alpha}$. They went on to show that if the tensor or semicrossed product algebras of two systems are isomorphic then the dynamical systems are piecewise conjugate.
A converse to this was then established by them for the tensor algebra case in many contexts, $X$ is totally disconnected or $n\leq 3$ for instance (the $n=4$ case was established later in \cite{Ramsey}), most of which gave that two algebras being algebraically isomorphic was enough to produce a completely isometric isomorphism between them. For the semicrossed product case it was shown that the converse fails as in the following example.
\begin{example}{\cite[Example 3.24]{DavKat2}}
Let $X = \{1,2\}$ and consider the following self maps on $X$:
\[
\sigma_1 = \operatorname{id}, \ \ \sigma_2(1) = 2, \sigma_2(2) = 1, \ \ \tau_1(x) = 1, \ \ \tau_2(x) = 2.
\]
Hence, $(X,\sigma)$ and $(X,\tau)$ are piecewise conjugate. With some work they show that $C^*_e(C(X) \times_\sigma \mathbb F^2_+) = M_2(C^*(\mathbb F^3))$, while $C^*_e(C(X) \times_\tau \mathbb F^2_+) = \mathcal O_2$, the Cuntz algebra. These C$^*$-algebras are not isomorphic so the semicrossed products cannot be completely isometrically isomorphic.
\end{example}
However, this just implies that piecewise conjugacy is not strong enough to be an invariant for the semicrossed product algebras. If one broadens the theory to a C$^*$-dynamical system, which is comprised of a C$^*$-algebra and $n$ $*$-endomorphisms, then Kakariadis and Katsoulis have shown that the C$^*$-dynamical system forms a complete invariant for the semicrossed product algebra up to what they call outer conjugacy when the C$^*$-algebra has trivial center \cite{KakKats}. Additionally, much work has gone into describing the C$^*$-envelope of the tensor algebra and semicrossed product algebra of a dynamical system \cite{Duncan0, DavFulKak2}. The interested reader should look at the recent survey article by Davidson, Fuller and Kakariadis \cite{DavFulKak} which presents these universal operator algebras in the more general context of semicrossed products of operator algebras by semigroup actions. The study of these semigroup actions can be found in \cite{DavFulKak2} and \cite{DuncanPeters}.
To deal with the obstruction posed by the above example, Section 3 introduces a form of conjugacy, called partition conjugacy, that lies between piecewise conjugacy and conjugacy. Theorem \ref{Thm:conjugacy} shows that if two systems are partition conjugate then their semicrossed product algebras are completely isometrically isomorphic. Conversely, Section 4 shows that a c.i. isomorphism of semicrossed product algebras induces a c.i. isomorphism of free products of noncommutative disc algebras. The rigidity proven in Section 2 is then passed along which allows us to show that semicrossed product algebras are c.i. isomorphic if and only if their related dynamical systems are partition conjugate.
Finally, Section 5 introduces a representation theory that is a combination of the nest representation theory from \cite{DavKat2} and Duncan's edge-colored directed graph representation theory from \cite{Duncan2}. This allows us to distinguish the pre-images of the maps at a given point, which in turn proves that the tensor and semicrossed product algebras are c.i.i if and only if the ranges of the maps of the dynamical system are pairwise disjoint.
\section{Automorphisms of free products}
Consider the free product of noncommutative disc algebras $*_{i=1}^m \mathfrak A_{n_i}$ with generators $(\mathfrak s_{i,1}, \cdots, \mathfrak s_{i,n_i}), 1\leq i \leq m$, a set of row contractions.
Popescu \cite[Theorem 2.5]{Pop1}, by way of a dilation result, shows that the generators of the free product can be taken to be row isometries of Cuntz type, in other words each $\mathfrak s_{i,j}$ is an isometry and
\[
\mathfrak s_{i,1}\mathfrak s_{i,1}^* + \cdots + \mathfrak s_{i,n_i}\mathfrak s_{i,n_i}^* = I, \ \ 1\leq i\leq m,
\]
in some concrete representation, $*_{i=1}^m \mathfrak A_{n_i} \subseteq B(\mathcal{H})$.
This is quite natural as the C$^*$-envelope of $\mathfrak A_d$ is the Cuntz algebra $\mathcal{O}_d$, with $\mathcal{O}_1 = C(\mb T)$.
Let $\mathcal{C}$ be the commutator ideal of $\mathfrak A_n$, the closed ideal generated by $\mathfrak s_i\mathfrak s_j - \mathfrak s_j\mathfrak s_i, 1\leq i,j\leq n$. Then $\mathcal{A}_n = \mathfrak A_n/\mathcal{C}$ is the universal operator algebra generated by a commuting row contraction and the identity.
A reproducing kernel Hilbert space (RKHS) $\mathcal{H}$ is a Hilbert space of functions on a specific topological space $X$ such that at every $\lambda\in X$ point evaluation at $\lambda$ is a continuous linear functional. This implies that there is a kernel function $k_\lambda \in \mathcal{H}$ such that $\langle f, k_\lambda\rangle = f(\lambda)$, so it reproduces the value of $f$ at $\lambda$. The RKHS is uniquely determined by its kernel functions and has a natural identification with certain holomorphic functions on $X$. Finally, the multiplier algebra, $\operatorname{Mult}(\mathcal{H})$, is the operator algebra of all functions on $X$ given by
\[
\operatorname{Mult}(\mathcal{H}) = \{M_f : M_f(h) = fh \in \mathcal{H}, \forall h\in \mathcal{H}\}.
\]
The Drury-Arveson space, $\mathcal{H}^2_n$, is the RKHS on the open unit ball in $\mb C^n$, namely $\mb B_n$, with kernel functions
\[
k_\lambda(z) \ = \ \frac{1}{1 - \langle z,\lambda\rangle}, \ \ \lambda \in \mb B_n.
\]
Because $1\in \mathcal{H}^2_n$ then for all $M_f\in \mathcal{M}_n =: \operatorname{Mult}(\mathcal{H}^2_n)$ we have that $M_f(1) = f\cdot 1 = f\in \mathcal{H}^2_n$. Hence, the multipliers are naturally identified as holomorphic functions. One should note though, for $n\geq 2, \mathcal{M}_n$ is contained in but not equal to $H^\infty(\mb B_n)$ since the operator norm on $\mathcal{M}_n$ is not equal to the supremum norm. Arveson \cite{Arv2} developed this theory and showed by way of a dilation theory for commuting row contractions that $\mathcal{A}_n = \mathfrak A_n/\mathcal{C}$ is completely isometrically isomorphic to the norm closed algebra $\overline{\operatorname{Alg}}\{1, M_{z_1}, \cdots, M_{z_n}\}$. Lastly, this implies that $\mathcal{A}_n$ is naturally contained in $C(\overline{\mb B}_n) \cap H^\infty(\mb B_n)$ but not equal to when $n\geq 2$.
The development of the Drury-Arveson space was spurred on by the obstructions one faces in the polydisc $\mb D^n$. Specifically, while Ando \cite{Ando} showed that two commuting contractions dilate to two commuting unitaries, giving that the bidisc algebra $A(\mb D^2) = C(\overline{\mb D}^2)\cap H(\mb D^2)$ is the universal operator algebra generated by two commuting contractions, Parrott and later Kajser and Varopoulos \cite{Parrott, Varopoulos} showed that there are three commuting contractions that do not dilate to three commuting isometries. Hence, the universal operator algebra generated by $n$ commuting contractions when $n\geq 3$ is a mysterious algebra that is not the polydisc algebra $A(\mb D^n)$. However, Ito \cite[Proposition I.6.2]{NagyFoias} proved that a set of commuting isometries dilates to a commuting set of unitaries and so the polydisc algebra is the universal operator algebra generated by $n$ commuting isometries.
For the context of a set of row contractions we would like to develop an operator algebra of holomorphic functions on $\times_{i=1}^m \mb B_{n_i}$, which one could call the {\bf polyball}.
\begin{definition}
For $n_1,\cdots, n_m \in \mb N$ let $\mathcal{H}^2_{n_1,\cdots, n_m}$ be the RKHS on $\times_{i=1}^m \mb B_{n_i}$ given by the kernel functions
\[
k_\lambda(z) = \prod_{i=1}^m \frac{1}{1 - \langle z_i, \lambda_i\rangle_{\mb C^{n_i}}}, \ \ \lambda \in \times_{i=1}^m \mb B_{n_i}.
\]
Thus, $\mathcal{H}^2_{n_1,\cdots,n_m} = \mathcal{H}^2_{n_1}\otimes \cdots\otimes \mathcal{H}^2_{n_m}$ and is then naturally identified as a space of holomorphic functions on the polyball.
As well, if $M_{z_{i,j}}, 1\leq j\leq n_i, 1\leq i\leq m$ are the coordinate multipliers in $\operatorname{Mult}(\mathcal{H}^2_{n_1,\cdots, n_m})$ then define
\[
\mathcal{A}_{n_1,\cdots, n_m} \ = \ \overline{\operatorname{Alg}}\{1, M_{z_{1,1}}, \cdots, M_{z_{m, n_m}}\}.
\]
\end{definition}
As before, since $1\in \mathcal{H}^2_{n_1,\cdots, n_m}$ then $\mathcal{A}_{n_1,\cdots, n_m}$ can also be naturally identified as an operator algebra of holomorphic functions. Furthermore, the generators of $\mathcal{A}_{n_1,\cdots, n_m}$ form a set of row contractions and so by the universal property there exists a completely contractive homomorphism
\[
\pi : *_{i=1}^m \mathfrak A_{n_i} \rightarrow \mathcal{A}_{n_1,\cdots, n_m}
\]
mapping $\mathfrak s_{i,j}$ to $M_{z_{i,j}}$.
Except in special cases, $\ker \pi$ will not be equal to the commutator ideal $\mathcal{C}$ of the free product because $*_{i=1}^m \mathfrak A_{n_i}/\mathcal{C}$ is the universal operator algebra generated by a commuting set of commuting row contractions.
Going back to the noncommutative context, recall that the character space of a Banach algebra $\mathcal{B}$, denoted $M(\mathcal{B})$, is the space of all multiplicative linear functionals endowed with the weak$^*$ topology. Popescu \cite{Pop1} proved that the characters of $*_{i=1}^m \mathfrak A_{n_i}$ are exactly the point evaluations $\rho_\lambda$ where $\lambda \in \times_{i=1}^m \overline{\mb B}_{n_i}$ and $\rho_\lambda(\mathfrak s_{i,j}) = \lambda_{i,j}$. If $\varphi \in \operatorname{Aut}(*_{i=1}^m \mathfrak A_{n_i})$, a bounded automorphism, then it induces a homeomorphism of the character space by
\[
\rho_\lambda \in M(*_{i=1}^m \mathfrak A_{n_i}) \ \ \mapsto \ \ \varphi^*(\rho_\lambda) := \rho_\lambda \circ \varphi.
\]
Since the boundary of the closed polyball will be mapped to itself, $\varphi^*$ is also a homeomorphism of the open polyball.
Thus, for every $\lambda\in \times_{i=1}^m \mb B_{n_i}$ there exists $\mu\in \times_{i=1}^m \mb B_{n_i}$ such that $\varphi^*(\rho_\lambda) = \rho_\mu$ and so we abuse notation and define $\varphi^*$ to be a map from $\times_{i=1}^m \mb B_{n_i}$ into itself defined by
\[
\varphi^*(\lambda) = \mu \ \ \Leftrightarrow \ \ \varphi^*(\rho_\lambda) = \rho_\mu,
\]
which is a homeomorphism on the polyball. Notice that $M(\mathcal{A}_{n_1,\cdots, n_m})$ is also homeomorphic to $\times_{i=1}^m \overline{\mb B}_{n_i}$.
\begin{proposition}\label{Prop:characterextension}
Let $\pi : *_{i=1}^m \mathfrak A_{n_i} \rightarrow \mathcal{A}_{n_1,\cdots, n_m}$ be the canonical quotient map, then
\[
\ker \pi = \bigcap_{\lambda\in \times_{i=1}^m \mb B_{n_i}} \ker \rho_\lambda.
\]
\end{proposition}
\begin{proof}
Let $\lambda\in \times_{i=1}^m \mb B_{n_i}$ and $\rho_\lambda^{\mathcal{A}} \in M(\mathcal{A}_{n_1,\cdots,n_m})$. Then $\rho_\lambda^{\mathcal{A}} \circ\pi$ is a multiplicative linear functional, i.e. a character, of $*_{i=1}^m \mathfrak A_{n_i}$. Hence, there exists $\mu \in \times_{i=1}^m \mb B_{n_i}$ such that $\rho_\lambda^\mathcal{A}\circ \pi = \rho_\mu$.
However,
\[
\lambda_{i,j} = \rho_\lambda^\mathcal{A}(M_{z_{i,j}}) = \rho_\lambda^\mathcal{A}\circ\pi(\mathfrak s_{i,j}) = \rho_\mu(\mathfrak s_{i,j}) = \mu_{i,j}.
\]
Thus, $\lambda = \mu$ and so $\ker \pi \subset \ker \rho_\lambda$ for all $\rho_\lambda \in M(*_{i=1}^m \mathfrak A_{n_i})$.
Conversely, if $A\in *_{i=1}^m \mathfrak A_{n_i}$ such that $\rho_\mu(A) = 0$ for every $\mu$ in the polyball then $\pi(A) \in \ker \rho_\lambda^\mathcal{A}$ for every $\lambda$ in the polyball. But $\pi(A) = M_f$ where $f$ is a continuous function on $\times_{i=1}^m \overline{\mb B}_{n_i}$ with
\[
0 =\rho_\lambda^\mathcal{A}(\pi(A)) = \rho_\lambda^\mathcal{A}(M_f) = f(\lambda).
\]
Therefore, $f = 0$ and $\pi(A) = 0$.
\end{proof}
\begin{corollary}\label{Cor:holomorphicautos}
If $\varphi \in \operatorname{Aut}(*_{i=1}^m \mathfrak A_{n_i})$ then it induces $\tilde\varphi \in \operatorname{Aut}(\mathcal{A}_{n_1,\cdots,n_m})$, such that $\tilde\varphi \circ\pi = \pi \circ\varphi$.
\end{corollary}
\begin{proof}
By the previous proposition, since $\varphi^*$ is a bijection of $\times_{i=1}^m \mb B_{n_i}$, we have
\[
\varphi^{-1}(\ker \pi) \ = \ \varphi^{-1}\big(\bigcap_{\lambda\in \times_{i=1}^m \mb B_{n_i}} \ker \rho_\lambda\big)
\ = \ \bigcap_{\lambda \in \times_{i=1}^m \mb B_{n_i}} \ker \varphi^*(\rho_\lambda)
\]
\[
= \ \bigcap_{\lambda \in \times_{i=1}^m \mb B_{n_i}} \ker \rho_\lambda \ = \ \ker \pi.
\]
Therefore, $\varphi$ induces the required automorphism of the quotient by $\pi$.
\end{proof}
From this theory we can now prove that $\varphi^*$ is not just a homeomorphism but also a biholomorphism by a standard argument using the character space.
\begin{theorem}
If $\varphi \in \operatorname{Aut}(*_{i=1}^m \mathfrak A_{n_i})$ then $\varphi^*$ is a biholomorphism and so is in $\operatorname{Aut}(\times_{i=1}^m \mb B_{n_i})$.
\end{theorem}
\begin{proof}
By Corollary \ref{Cor:holomorphicautos} there exists $\tilde\varphi \in \operatorname{Aut}(\mathcal{A}_{n_1,\cdots, n_m})$ induced by $\varphi$. Define the following map $H:\times_{i=1}^m \mb B_{n_i} \rightarrow \times_{i=1}^m \mb C^{n_i}$ by
\[
H(\lambda) = (\tilde\varphi\circ\pi(\mathfrak s_{1,1})(\lambda), \cdots, \tilde\varphi\circ\pi(\mathfrak s_{m,n_n})(\lambda)), \ \ \lambda\in \times_{i=1}^m \mb B_{n_i}.
\]
This is a holomorphic map on $\times_{i=1}^m \mb B_{n_i}$ into itself because each $\tilde\varphi\circ\pi(\mathfrak s_{i,j})$ is a holomorphic function.
Moreover, by the proof of Proposition \ref{Prop:characterextension}
\[
H(\lambda) = (\rho_\lambda^\mathcal{A}(\tilde\varphi\circ\pi(\mathfrak s_{1,1})), \cdots, \rho_\lambda^\mathcal{A}(\tilde\varphi\circ\pi(\mathfrak s_{m,n_m})))
\]
\[
= (\rho_\lambda^\mathcal{A}\circ\pi\circ\varphi(\mathfrak s_{1,1}), \cdots, \rho_\lambda^\mathcal{A}\circ\pi\circ\varphi(\mathfrak s_{m,n_m})
\]
\[
= (\varphi^*(\rho_\lambda(\mathfrak s_{1,1})), \cdots, \varphi^*(\rho_\lambda(\mathfrak s_{m,n_m}))) = \varphi^*(\lambda) \in \times_{i=1}^m \mb B_{n_i},
\]
because $\varphi^*(\rho_\lambda)$ is completely contractive, it's a character.
Do this again for $\varphi^{-1}$ to get that $\varphi^*$ is a biholomorphism.
\end{proof}
It remains to be shown that the map $\varphi \mapsto \varphi^*$ is a bijection. To this end we first look at the structure of the automorphism group of the polyball.
It is well known that $\operatorname{Aut}(\mb D^n) = \operatorname{Aut}(\mb D)^n \rtimes S_n$, where the symmetric group has the natural permutation action, see \cite{Rudin}. A generalization of this due to Ligocka and Tsyganov \cite{Lig, Tsyganov} says that if there is biholomorphism of a product space onto another product space, with each space having a $C^2$ boundary, then there are the same number of spaces and by rearranging the order of the second product space the biholomorphism is just a product of individual biholomorphisms.
Specifically, the automorphisms of $\times_{i=1}^m \mb B_{n_i}$ are given by automorphisms in each of the spaces along with permutations among the homeomorphic spaces. Specifically,
\[
\operatorname{Aut}(\times_{i=1}^n (\mb B_i)^{k_i}) = \times_{i=1}^n (\operatorname{Aut}(\mb B_i)^{k_i} \rtimes S_{k_i}),
\]
where the automorphisms of the ball are fractional linear transformations \cite{Rudin2}.
In \cite{DavPitts, Pop3}, Davidson-Pitts and Popescu separately proved that the isometric automorphisms of $\mathfrak A_n$ are completely isometric and in bijective correspondence with the automorphisms of the ball, $\mb B_n$, in a natural way. Indeed, both papers show that the following construction of Voiculescu \cite{Voic} is the appropriate way to create an automorphism of $\mathfrak A_n$ from an automorphism of the ball.
Let $\phi \in \operatorname{Aut}(\mb B_n)$. Then there exists $X = \left[\begin{array}{cc} x_0 & \eta_1^* \\ \eta_2 & X_1 \end{array}\right] \in U(1,n)$ such that
\[
\phi(\lambda) \ \ = \ \ \frac{X_1\lambda + \eta_2}{x_0 + \langle \lambda, \eta_1\rangle} \ \ for \ \ \lambda \in \mb B_n.
\]
Voiculescu then defines an automorphism of the Cuntz-Toeplitz algebra $\mathcal{E}_n$, with generators the left creation operators, $L_1,\cdots, L_n$, on the full Fock space, by
\[
\varphi(L_\zeta) = (\overline{x}_0 I - L_{\overline{\eta}_2})^{-1}(L_{\overline{X}_1\zeta} - \langle \zeta, \overline{\eta}_1\rangle I ),
\]
where $L_\zeta = \sum_{i=1}^n \zeta_i L_i$. One can think of the above map as a non-commutative fractional linear map. This automorphism restricts to the non-commutative disc algebra, $\mathfrak A_n$. Lastly, this is the natural way to do this because $\varphi^* = \phi$, \cite[Lemma 4.10]{DavPitts}. In what follows, note that $\operatorname{Aut}_{ci}(\mathcal{B})$ denotes the completely isometric automorphisms of $\mathcal{B}$.
\begin{lemma}\label{Lemma:surjective}
Let $\phi \in \operatorname{Aut}(\times_{i=1}^m \mb B_{n_i})$. There exists $\varphi \in \operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i})$ such that $\varphi^* = \phi$.
\end{lemma}
\begin{proof}
As was described before any automorphism of the polyball is the composition of a permutation and automorphisms in each component.
First, consider a permutation $\alpha \in S_{n_1\cdots n_m}$ such that $n_{\alpha(i)} = n_i$ for all $1\leq i\leq m$. Define the map $\varphi_\alpha$ on the generators of the free product by
\[
\varphi_\alpha(\mathfrak s_{i,j}) = \mathfrak s_{\alpha(i),j}, \ \ 1\leq j\leq n_i, 1\leq i\leq m.
\]
By the universal property of the free product this extends to a completely contractive endomorphism of $*_{i=1}^m \mathfrak A_{n_i}$. As well, $\varphi_{\alpha^{-1}}$ is the inverse of $\varphi_\alpha$ on the generators and so $\varphi_{\alpha^{-1}} = \varphi_\alpha^{-1}$. Thus, $\varphi_\alpha \in \operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i})$ and $\varphi_{\alpha}^* = \alpha^{-1}$.
Second, for a fixed $1\leq k\leq m$ consider $\tilde\varphi \in \operatorname{Aut}_{ci}(\mathfrak A_{n_k})$. Then define $\varphi$ on the generators by
\[
\varphi(\mathfrak s_{i,j}) = \mathfrak s_{i,j}, 1\leq j\leq n_i, i\neq k, \ \ {\rm and} \ \ \varphi|_{\mathfrak A_{n_k}} = \tilde\varphi.
\]
Again by the universal property $\varphi$ extends to a completely contractive homomorphism of $*_{i=1}^m \mathfrak A_{n_i}$. In the same way construct a c.c. homomorphism from $\tilde\varphi^{-1}$ to see that this $\varphi\in \operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i})$ with $\varphi^*|_{|\mb B_{n_i}} = \operatorname{id}, i\neq k$ and $\varphi^*|_{\mb B_{n_k}} = \tilde\varphi^*$. Therefore by the fact that $\operatorname{Aut}_{ci}(\mathfrak A_{n_k}) \simeq \operatorname{Aut}(\mb B_{n_k})$ the conclusion follows.
\end{proof}
Now we turn our attention to proving the injectivity of the map $\varphi \mapsto \varphi^*$. First we need a few results. Consider the map $\gamma_{\bf z}$ on $*_{i=1}^m \mathfrak A_{n_i}$, where ${\bf z} = (z_1,\cdots, z_m), z_i = (z_{i,1},\cdots, z_{i,n_i}) \in \overline{\mb D}^{n_i}$ and
\[
\gamma_{\bf z}(\mathfrak s_{i,j}) = z_{i,j}\mathfrak s_{i,j}, \ \ 1\leq i\leq m, 1\leq j\leq n_i.
\]
Call such a map a {\bf gauge endomorphism}. When all the $z_{i,j}$ are in $\mb T$ these are the well known {\bf gauge automorphisms}. Let $\gamma_z$ denote the gauge automorphism where every $z_{i,j} = z \in \mb T$. The following is a Fourier series theory for these algebras that is only slightly modified from the work of Davidson and Katsoulis in \cite{DavKat2}.
The map $\Phi_k(a) = \int_{\mb T} \gamma_z(a) \overline{z}^k dz$ is a completely contractive projection of $*_{i=1}^m \mathfrak A_{n_i}$ onto the span of all words in the $\mathfrak s_{i,j}, 1\leq i\leq m, 1\leq j\leq n_i$, of length $k$. From this we can define a Fourier-like series for our algebra. Define the $k$th Cesaro mean by
\[
\sum_k(a) = \sum_{i=0}^k (1 - \frac{i}{k}) \Phi_i(a) = \int_{\mb T} \gamma_z(a)\sigma_k(z) dz,
\]
where $\sigma_k$ is the Fejer kernel. Hence, a slight modification of the Fejer Theorem in Fourier analysis gives that
\[
a = \lim_{k\rightarrow \infty} \sum_k(a) = a, \ \ \forall a\in *_{i=1}^m \mathcal{A}_{n_i}.
\]
Therefore, for $a\in *_{i=1}^m \mathfrak A_{n_i}$ we have that $\Phi_k(a) = 0, \forall k\geq 0$ if and only if $a = 0$.
\begin{lemma}\label{Lemma:injective}
The gauge endomorphisms of $*_{i=1}^m \mathfrak A_{n_i}$ are completely contractive and injective.
\end{lemma}
\begin{proof}
By universality $\gamma_{\bf z}$ is a completely contractive homomorphism. For every $k\geq 0$ we have that $\gamma_{\bf z}$ leaves the subspace spanned by words of length $k$ invariant and moreover, is injective on this subspace. Hence, $\gamma_{\bf z}\circ\Phi_k = \Phi_k\circ\gamma_{\bf z}$.
Now, if $a\in *_{i=1}^m \mathfrak A_{n_i}$ and $a\neq 0$ then there exists $k\geq 0$ such that $\Phi_k(a) \neq 0$. This implies that $\Phi_k(\gamma_{\bf z}(a)) = \gamma_{\bf z}(\Phi_k(a)) \neq 0$ and so $\gamma_{\bf z}(a) \neq 0$, either. Therefore, $\gamma_{\bf z}$ is injective.
\end{proof}
We are now in a position to prove the analogous result to that in \cite{DavPitts, Pop3}.
\begin{theorem}\label{Thm:Autos}
$\operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i}) \simeq \operatorname{Aut}(\times_{i=1}^m \mb B_{n_i})$.
\end{theorem}
\begin{proof}
Let $\varphi \in \operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i})$ such that $\varphi^* = \operatorname{id}$. This implies that $\varphi(s_{i,j}) = \mathfrak s_{i,j} + C_{i,j}$ where $C_{i,j} \in \ker \pi$ with $\pi$ the canonical quotient map of $*_{i=1}^m \mathfrak A_{n_i}$ onto $\mathcal{A}_{n_1,\cdots, n_m}$. If we can show that $C_{i,j} = 0$ for $1\leq i\leq m, 1\leq j\leq n_i$ then the result is established.
First, consider the completely contractive projection onto the commutative algebra corresponding to the 1st component, $\pi_1 : *_{i=1}^m \mathfrak A_{n_i} \rightarrow \mathcal{A}_{n_1}$, which is obtained by the universal property. Clearly, $\ker \pi \subset \ker \pi_1$.
Now consider the unital map $\nu : *_{i=1}^m \mathfrak A_{n_i} \rightarrow M_2(*_{i=1}^m \mathfrak A_{n_i})$ given by
\[
\nu(\mathfrak s_{1,j}) = \left[\begin{array}{cc} \mathfrak s_{1,j} & 0 \\ 0 & \mathfrak s_{1,j}\end{array}\right], \ \ 1\leq j\leq n_1 \ \ {\rm and}
\]
\[
\nu(\mathfrak s_{i,j}) = \left[\begin{array}{cc} 0 & \frac{1}{\sqrt{2}} \mathfrak s_{i,j} \\ 0 & \frac{1}{\sqrt{2}} \mathfrak s_{i,j} \end{array}\right],\ \ 2\leq i \leq n, 1\leq j\leq n_i.
\]
The row contractive generators are sent to row contractions so by the universal property $\nu$ is a completely contractive homomorphism.
For example
\[
\nu(\mathfrak s_{2,1}\mathfrak s_{1,1}\mathfrak s_{2,1} + \mathfrak s_{1,1}^2) = \left[\begin{array}{ll} \mathfrak s_{1,1}^2 & \frac{1}{2}\mathfrak s_{2,1}\mathfrak s_{1,1}\mathfrak s_{2,1} \\ 0 & \frac{1}{2}\mathfrak s_{2,1}\mathfrak s_{1,1}\mathfrak s_{2,1} + \mathfrak s_{1,1}^2\end{array}\right].
\]
Note that the (1,1) entry of $\nu(a)$ is $\pi_1(a)$.
Let $z_1 = (1, \cdots, 1) \in \overline{\mb D}^{n_1}$, $z_i = (\frac{1}{\sqrt 2}, \cdots, \frac{1}{\sqrt 2}) \in \overline{\mb D}^{n_i}$, $2\leq i\leq m$ and ${\bf z} = (z_1,\cdots, z_m)$. By the above theory, If $C\in \ker \pi$ then $\nu(C) = \left[\begin{array}{cc} 0 & \gamma_{\bf z}(C) \\ 0 & \gamma_{\bf z}(C) \end{array}\right]$. Thus,
\[
\nu(\mathfrak s_{1,j} + C_{1,j}) = \left[\begin{array}{cc} \mathfrak s_{1,j} & \gamma_{\bf z}(C_{1,j}) \\ 0 & \mathfrak s_{1,j} + \gamma_{\bf z}(C_{1,j}) \end{array}\right], 1\leq j \leq n_i.
\]
But, $[\nu(\mathfrak s_{1,1} + C_{1,1}), \cdots, \nu(\mathfrak s_{1,n_1} + C_{1,n_1})]$ is a row contraction which implies that \newline $[\mathfrak s_{1,1}, \gamma_{\bf z}(C_{1,1}), \cdots, \mathfrak s_{1,n_1}, \gamma_{\bf z}(C_{1,n_1})]$ is also a row contraction. However, we had assumed that $\mathfrak s_{1,1}\mathfrak s_{1,1}^* + \cdots + \mathfrak s_{1,n_1}\mathfrak s_{1,n_1}^* = I$ and so $\gamma_{\bf z}(C_{1,j}) = 0, 1\leq j\leq n_1$. Finally, by Lemma \ref{Lemma:injective} $\gamma_{\bf z}$ is injective and so $C_{1,1} = \cdots C_{1,n_1} = 0$.
Repeating the above argument for $i \in \{2,\cdots, m\}$ yields that $\varphi = \operatorname{id}$.
\end{proof}
\begin{corollary}
If $\varphi : *_{i=1}^{m_1} \mathfrak A_{n_i} \rightarrow *_{j=1}^{m_2} \mathfrak A_{k_j}$ is a completely isometric isomorphism, then $m_1 = m_2$ and there exists $\alpha \in S_{m_1}$ such that $\varphi = \times_{i=1}^{m_1} \varphi_i$ where $\varphi_i$ is a completely isometric isomorphism from $\mathfrak A_{n_i}$ onto $\mathfrak A_{k_{\alpha(i)}}, 1\leq i\leq m_1$.
\end{corollary}
\begin{proof}
As before, this isomorphism induces a homeomorphism of the character spaces, $\times_{i=1}^{m_1} \overline{\mb B}_{n_i}$ and $\times_{j=1}^{m_2} \overline{\mb B}_{k_j}$, that is a biholomorphism on their interiors. By \cite{Lig, Tsyganov}, this implies that $m_1 = m_2$ and there exists a permutation $\alpha \in S_{m_1}$ such that $n_{i} = k_{\alpha(i)}, 1\leq i\leq m_1$. As in Lemma \ref{Lemma:surjective} we can define a completely isometric isomorphism $\varphi_\alpha$ that encodes this permutation.
Therefore, $\varphi_\alpha^{-1}\circ\varphi\in \operatorname{Aut}_{ci}(*_{i=1}^m \mathfrak A_{n_i})$ and the conclusion follows.
\end{proof}
\section{Partition conjugacy}
Consider the following form of conjugacy for multivariable dynamical systems.
\begin{definition}
Two dynamical systems, $(X,\sigma)$ and $(Y,\tau)$ are said to be {\bf partition conjugate} if there exists a homeomorphism $\gamma: X\rightarrow Y$ and clopen sets $V_{i,j} \subset X, 1\leq i, j \leq n$ such that
\begin{enumerate}
\item $\cup_{i=1}^n V_{i,j} = X$ and $V_{i,j} \cap V_{i', j} = \emptyset $ when $ i\neq i'$.
\item $\cup_{j=1}^n V_{i,j} = X$ and $V_{i,j} \cap V_{i, j'} = \emptyset $ when $ j\neq j'$.
\item $\sigma_i|_{V_{i,j}} = \gamma^{-1} \circ \tau_j \circ \gamma|_{V_{i,j}}$, $1\leq i,j\leq n$.
\item $\sigma_i^{-1}(\sigma_i(V_{i,j})) = V_{i,j} = \gamma^{-1}(\tau_j^{-1}(\tau_j\circ\gamma(V_{i,j}))).$
\end{enumerate}
\end{definition}
One should note that when $X$ is connected this is just conjugacy. In the general setting this is stronger than piecewise conjugacy and weaker than conjugacy. An equivalent condition to (4) is that $\sigma_i(V_{i,j}) \cap \sigma_i(V_{i,j'}) = \emptyset$ for $j\neq j'$ since if $x$ is in this intersection then $\sigma_i^{-1}(x) \subset V_{i,j} \cap V_{i,j'}$, a contradiction.
\begin{theorem}\label{Thm:conjugacy}
If $(X,\sigma)$ and $(Y,\tau)$ are partition conjugate then $C_0(X) \times_\sigma \mb F_n^+$ and $C_0(Y)\times \mb F_n^+$ are completely isometrically isomorphic.
\end{theorem}
\begin{proof}
Let $\gamma, \{V_{i,j}\}_{1\leq i,j\leq n}$ be the given partition conjugacy of the two systems.
Suppose $\mathfrak s_1, \cdots, \mathfrak s_n$ and $\mathfrak t_1, \cdots, \mathfrak t_n$ are the generators of $C_0(X) \times_\sigma \mb F_n^+$ and $C_0(Y) \times_\tau \mb F_n^+$ respectively. Define a covariant representation of $(X,\sigma)$ by
\[
\varphi(f) = f \circ \gamma^{-1}
\]
\[
\varphi(\mathfrak s_i) = \sum_{j=1}^n \mathfrak t_{j} \chi_{\gamma(V_{i,j})}.
\]
From the fact that
\[
\chi_{\tau_j\circ\gamma(V_{i,j})} \mathfrak t_j = \mathfrak t_j \chi_{\tau_j\circ\gamma(V_{i,j})} \circ \tau_j = \mathfrak t_j \chi_{\tau_j^{-1}\circ\tau_j\circ\gamma(V_{i,j})} = \mathfrak t_j \chi_{\gamma(V_{i,j})}
\]
we can get that for $j\neq j'$
\[
(\mathfrak t_j\chi_{\gamma(V_{i,j})})^* \mathfrak t_{j'} \chi_{\gamma(V_{i,j'})} \ \ = \ \ \chi_{\gamma(V_{i,j})}\mathfrak t_{j}^*\mathfrak t_{j'} \chi_{\gamma(V_{i,j'})}
\]
\[
= \ \ \mathfrak t_j^*\chi_{\tau_j\circ\gamma(V_{i,j})} \chi_{\tau_{j'}\circ\gamma(V_{i,j'})} \mathfrak t_{j'} \ \ = \ \
\mathfrak t_j^*\chi_{\gamma^{-1}\circ\sigma_i(V_{i,j})} \chi_{\gamma^{-1}\circ\sigma_i(V_{i,j'})}\mathfrak t_{j'} \ \ = \ \ 0.
\]
Thus, $\varphi(\mathfrak s_i)$ is an isometry,
\[
\varphi(\mathfrak s_i)^*\varphi(\mathfrak s_i) = {\left(\sum_{j=1}^n \mathfrak t_{j} \chi_{\gamma(V_{i,j})}\right)}^*\sum_{j=1}^n \mathfrak t_{j} \chi_{\gamma(V_{i,j})}
= \sum_{j=1}^n \chi_{\gamma(V_{i,j})}\mathfrak t_j^*\mathfrak t_j \chi_{\gamma(V_{i,j})} = \sum_{j=1}^n \chi_{\gamma(V_{i,j})} = I.
\]
Additionally, $\varphi$ satisfies the covariance relations.
\[
\varphi(f)\varphi(\mathfrak s_i) = (f\circ\gamma^{-1}) \sum_{j=1}^n \mathfrak t_j \chi_{\gamma(V_{i,j})}
\]
\[
= \sum_{j=1}^n \mathfrak t_j \chi_{\gamma(V_{i,j})} (f\circ\gamma^{-1} \circ\tau_j)
\]
\[
= \sum_{j=1}^n \mathfrak t_j \chi_{\gamma(V_{i,j})} (f\circ\sigma_i\circ\gamma^{-1}) = \varphi(\mathfrak s_i)\varphi(f\circ\sigma_i).
\]
Therefore, by the universal property $\varphi$ extends to a completely contractive homomorphism of $C_0(X) \times_\sigma \mb F_n^+$ into $C_0(Y) \times_\tau \mb F_n^+$.
Similarly, we can define another map $\theta : C_0(Y) \times_\tau \mb F_n^+ \rightarrow C_0(X) \times_\sigma \mb F_n^+$ by
\[
\theta(f) = f\circ\gamma
\]
\[
\theta(\mathfrak t_j) = \sum_{i=1}^n \mathfrak s_i \chi_{V_{i,j}}.
\]
In exactly the same way $\theta(\mathfrak t_j)$ is an isometry which satisfies the covariance relations. Hence, $\theta$ is a completely contractive homomorphism. Therefore, $\theta = \varphi^{-1}$ because it is the inverse on the generators and so $\varphi$ is a completely isometric isomorphism.
\end{proof}
For the $n=2$ discrete case we can specify exactly what this partition conjugacy looks like by giving a partition of the vertices in $X$ where partition conjugacy implies conjugacy on each equivalence class.
Let $(X,\sigma)$ be a countable discrete dynamical system and $x\in X$. Define the equivalence class of $x$ by putting $z\in [x]$ if $\sigma_i(x) = \sigma_j(z)$, $i,j \in \{1,2\}$ and extending transitively. That is, $[x]$ is the smallest set containing $x$ that satisfies
\[
[x] = \cup_{1\leq i, j \leq 2} \sigma_i^{-1}(\sigma_j([x])).
\]
Note that the following proposition will not be true if there are three or more maps in the dynamical system.
\begin{proposition}
Let $(X,\{\sigma_1,\sigma_2\})$ and $(Y,\{\tau_1,\tau_2\})$ be partition conjugate countable discrete dynamical systems.
If $X = [x]$, a single equivalence class, then the systems are conjugate.
\end{proposition}
\begin{proof}
Without loss of generality assume that $X=Y$.
Assume that the systems are not conjugate. Thus, there exists $x_1, x_2\in X$ such that
\[
\sigma_i(x_1) = \tau_i(x_1), i=1,2,
\]
\[
\sigma_1(x_2) = \tau_2(x_2), \sigma_2(x_2) = \tau_1(x_2)
\]
\[
\sigma_1(x_i) \neq \sigma_2(x_i), i = 1,2.
\]
However, $x_1,x_2 \in [x]$ which implies that there exist $i_1,\cdots, i_{2m} \in \{1,2\}$ and points $z_0,\cdots, z_m \in [x] = X$ such that
\[
z_0 = x_1, z_m = x_2
\]
\[
\sigma_{i_{2j-1}}(z_{j-1}) = \sigma_{i_{2j}}(z_{j}), 1\leq j\leq m.
\]
We may also assume that the $z_i$ are all distinct and that the $\sigma_{i_{2j}}(z_j)$ are distinct as well so that the connection between $x_1$ and $x_2$ has no redundancy. This implies that $i_{2j} \neq i_{2j+1}, 1\leq j\leq m-1$.
For ease of notation, let $\sigma_{i_1}(z_0) = z$ and $i_0, i_{2m+1} \in \{1,2\}$ such that $i_0\neq i_1$ and $i_{2m} \neq i_{2m+1}$. Now,
\[
z_0 \in \sigma_{i_1}^{-1}(\{z\}), \ \ z_0 \notin \sigma_{i_0}^{-1}(\{z\}), \ \ z_1\in \sigma_{i_2}^{-1}(\{z\}), \ \ {\rm and} \ \ z_1 \notin \sigma_{i_3}^{-1}(\{z\})
\]
There are two cases. First, when $i_1 = i_2$ we have $\{z_0,z_1\} \in \sigma_{i_1}^{-1}(\{z\})$ and not in $\sigma_{i_0}^{-1}(\{z\})$ which implies that $\{z_0,z_1\} \in \tau_{i_1}^{-1}(\{z\})$ by the pre-image condition since $\tau_{i_1}(x_1) = z$. Thus, $\tau_{i_1}(z_1) = \sigma_{i_1}(z_1)$ and $\tau_{i_0}(z_1) = \sigma_{i_0}(z_1)$.
The second case is when $i_1 \neq i_2$ and so is equal to $i_3$. This implies that $z_1 \notin \sigma_{i_1}^{-1}(\{z\})$. Hence, by the pre-image condition $z_1 \notin \tau_{i_1}^{-1}(\{z\})$ since $z_0 \notin \sigma_{i_2}^{-1}(\{z\})$. Hence, $\tau_{i_1}(z_1) = \tau_{i_3}(z_1) = \sigma_{i_1}(z_1)$ and $\tau_{i_0}(z_1) = \sigma_{i_0}(z_1)$. Either case the conclusion is the same.
For the rest of the proof repeat this argument to get that $\tau_{i_1}(z_j) = \sigma_{i_1}(z_j)$ and $\tau_{i_0}(z_j) = \sigma_{i_0}(z_j)$ for $2\leq j\leq m$. This will contradict the assumption that $\sigma_1(x_2) = \tau_2(x_2), \sigma_2(x_2) = \tau_1(x_2)$ and $\sigma_1(x_2) \neq \sigma_2(x_2)$.
\end{proof}
\section{The characterization theorem}
Suppose $\Theta$ is a completely isometric isomorphism of semicrossed product algebras. This will be shown to induce a c.i. isomorphism of some canonical quotient algebras which in turn induces a c.i. isomorphism of free products of noncommutative disc algebras, thus reducing to the theory of Section 2. It will be shown that this rigidity passes back through to the semicrossed product algebras.
\begin{remark}\label{iandci}
Suppose that $\Theta : C_0(X) \times_\sigma \mb F_n^+ \rightarrow \mathcal{B}$ is a contractive homomorphism. But then $\Theta(C_0(X))$ and $\Theta(\mathfrak s_1),\cdots, \Theta(\mathfrak s_n)$ form a covariant representation of $(X,\sigma)$ since the generators remain contractions. Therefore, by the universal property $\Theta$ is actually completely contractive. In this and the next sections everything will be stated as being c.c or completely isometric but one should remember that it does not weaken the conclusions to say contractive or isometric.
\end{remark}
\begin{definition}
Given a multivariable dynamical system $(X,\sigma)$ and and a closed subset $X'\subseteq X$, the {\bf sub-dynamical system} $(X',\sigma)$ is the locally compact Hausdorf space $X'$ along with $n$ partially defined proper continuous functions $\sigma_1,\cdots, \sigma_n$ where $\sigma_i$ is defined at $x\in X'$ if and only if $\sigma_i(x) \in X'$.
\end{definition}
One can develop this in general but we are interested only in the case of a finite number of points, so assume $|X'| < \infty$.
For such a closed subset $X'$ of $X$ consider the closed ideal $J_{(X',\sigma)}$ of $C_0(X) \times_\sigma \mb F_n^+$ generated by $f\in C_0(X)$ such that $f(x) = 0, \forall x\in X'$.
Define
\[
\pi_{(X',\sigma)} : C_0(X) \times_\sigma \mb F_n^+ \rightarrow (C_0(X) \times_\sigma \mb F_n^+)/J_{(X',\sigma)}
\]
to be the canonical quotient map. This quotient algebra is generated by $C(X')$ and $\pi_{(X',\sigma)}(\mathfrak s_i), 1\leq i\leq n$ so it can be concretely represented as a subalgebra of $M_{|X'|}(B(\mathcal{H})) = B(\mathcal{H}^{(|X'|)})$ where we will label $\mathcal{H}^{(|X'|)} = \oplus_{x\in X'} \mathcal{H}_x$. Define $P_x$ to be the projection onto $\mathcal{H}_x$ for every $x\in X'$ and
\[
\theta_{x,y} : B(\mathcal{H}^{(|X'|)}) \rightarrow B(\mathcal{H}_x, \mathcal{H}_y) = B(\mathcal{H})
\]
to be the compression map onto the $(y,x)$ entry of each matrix. Of course, such a $\theta_{x,y}$ is linear and contractive. As well, this satisfies
\[
\theta_{x,y}(P_y A P_x) = \theta_{x,y}(A), \ \forall A\in B(\mathcal{H}^{(|X'|)}).
\]
Such a map becomes useful since $\theta_{x,y}(\pi_{(X', \sigma)}(\mathfrak s_i)) \neq 0$ if and only if $\sigma_i(x) = y$, for any $x,y\in X'$.
First, we want to show that this quotient has some form of canonicity. To achieve this we turn to operator algebras of finite directed graphs. Let $G = (V, E, r, s)$ be a finite directed graph with $V$ the vertex set, $E$ the edge set, and $r$ and $s$ the range and source maps. The {\bf Cuntz-Krieger algebra} of $G$, $\mathcal{O}(G)$, is the universal C$^*$-algebra generated by Cuntz-Krieger families $\{S,P\}$ on $\mathcal{H}$ where this family $\{S,P\}$ consists of a set $\{P_v :v\in V\}$ of mutually orthogonal projections on $\mathcal{H}$ and a set $\{S_e : e\in E\}$ of partial isometries on $\mathcal{H}$ such that
\\ \indent $\bullet$ ${S_e}^* S_e = P_{s(e)}$ for all $e\in E$,
\\ \indent $\bullet$ ${S_e}^* S_f = 0$ if $e\neq f$, and
\\ \indent $\bullet$ $P_v = \sum_{\{e\in E: r(e) = v\}} S_e{S_e}^*$ whenever $v$ is not a source.
The norm closed subalgebra of $\mathcal{O}(G)$ generated by the $\{S_e\}$ and the $\{P_v\}$ is called the {\bf tensor algebra} and denoted $\mathcal{T}_+(G)$. The tensor algebra was first introduced by Muhly and Solel in \cite{MuhlySolel} under the name {\em quiver algebra} in the context of C$^*$-correspondences and Toeplitz algebras and was shown to be equivalent to the above definition in \cite{FowlerRaeburn}. A good review of all things graph theory can be found in \cite{Raeburn}.
For $G = (V, E, r,s)$ with $E = E_1\dot\cup \cdots \dot\cup E_n$ Duncan \cite{Duncan1} defines an edge-colored Cuntz-Krieger family $\{S,P\}$ on $\mathcal{H}$ where $\{\{S_e : e\in E_i\}, P\}$ is a Cuntz-Krieger family for $(V, E_i, r,s)$ for $1\leq i\leq n$.
To every finite sub-dynamical system, $(X', \sigma)$, one can associate a graph. In particular, let $V= X'$ and whenever $x, y\in X'$ with $\sigma_i(x) = y$ define $e\in E_i$ with $s(e) = x$ and $r(e) = y$. Taking $E = \dot\cup_{i=1}^n E_i$ we see that $G = (V,E)$ is a finite directed graph with a finite number of edges. For each $1\leq i\leq n$ let $G_i = (V,E_i)$.
\begin{lemma}
For a sub-dynamical system $(X', \sigma)$ of $(X,\sigma)$ with $|X'| < \infty$, the quotient $(C_0(X) \times_\sigma \mb F_n^+)/J_{(X',\sigma)}$ is completely isometrically isomorphic to the free product $*_{i=1}^n \mathcal{T}_+(G_i)$ amalgamated over $\operatorname{span}\{P_v : v\in V\} \simeq C(X')$.
\end{lemma}
\begin{proof}
Recall that we are considering this quotient as a subalgebra of $M_{|X'|}(B(\mathcal{H}))$ and that $P = \{P_x : x\in V = X'\}$ is a set of mutually orthogonal projections. For each $e\in E_i$, set $S_e = P_{r(e)}\pi_{(X',\sigma)}(\mathfrak s_i)P_{s(e)}$ noting that $\theta_{s(e),r(e)}(S_e)$ is a contraction. Now, for $y\in X'$ and $1\leq i\leq n$ let $\{e_1,\cdots, e_k\} = r^{-1}(\{y\}) \cap E_i$ we have that $P_y\pi_{(X',\sigma)}(S_i)(\sum_{j=1}^k P_{s(e_j)})$ is a contraction and thus
\[
[\theta_{s(e_1), y}(\pi_{(X',\sigma)}(\mathfrak s_i)), \ \cdots\ , \ \theta_{s(e_k), y}(\pi_{(X',\sigma)}(\mathfrak s_i))]
\]
is a row contraction. It should be noted that for $\pi_{(X',\sigma)}(\mathfrak s_i)$ there is at most one entry in each column since the $\sigma_i$ are functions.
Considering the $S_e$ as operators on $B(\mathcal{H})$ we get a set of row contractions indexed over $r(E) \subseteq X'$. As was mentioned before, Popescu \cite{Pop1} showed that this dilates to a sequence of row isometries $\{V_e : e\in E\}$ on $B(\mathcal{K})$ such that $\sum_{e \in E_i, r(e) = x} V_eV_e^* = I$ for each $x\in r(E_i), 1\leq i\leq n$.
Construct operators $R_i \in M_{|X'|}(B(\mathcal{K}))$ such that the $P_y R_i P_x = V_e$ when $s(e) = x$ and $r(e) = y$ for $e\in E_i$. The operator algebra generated by $C(X')$ and the $R_i$ is generated by an edge-colored Cuntz-Krieger family, a fact that will be used later.
Now, for $1\leq i\leq n$ and $y\in X'$
\[
P_yR_i = R_i\big(\sum_{\sigma_i(x) = y} P_x\big).
\]
Thus for $f\in C_0(X)$ we have that
\[
\pi_{(X',\sigma)}(f) R_i = R_i \pi_{(X',\sigma)}(f\circ\sigma_i).
\]
Thus, $\pi_{(X',\sigma)}$ and $R_1,\cdots, R_n$ form a covariant representation of $(X,\sigma)$. Hence, by the universal property there exists $\theta : C_0(X) \times_\sigma \mb F_n^+$ onto $\overline\operatorname{Alg}\{C(X'), R_1,\cdots, R_n\}$ such that $\theta(\mathfrak s_i) = R_i, 1\leq i\leq n$. Furthermore, for every $f\in C_0(X)$ such that $f|_{X'} = 0$, $\theta(f) = \pi_{(X',\sigma)}(f) = 0$ and so $\theta(J_{(X',\sigma)}) = 0$.
In other words, $(C_0(X) \times_\sigma \mb F_n^+)/J_{(X',\sigma)}$ is completely isometrically isomorphic to $\overline\operatorname{Alg}\{C(X'), R_1,\cdots, R_n\}$ and the conclusion follows.
\end{proof}
\begin{definition}
Let $\mathcal{B} \subseteq M_k(B(\mathcal{H}))$ be an operator algebra. The {\bf entry algebra} of $\mathcal{B}$ is the operator algebra $E(\mathcal{B}) \subset B(H)$ generated by $B_{i,j}, 1\leq i,j \leq k$, where $B = [B_{i,j}]_{i,j=1}^k$ for every $B\in\mathcal{B}$.
\end{definition}
Applying this to our context we recover a familiar object.
\begin{corollary}\label{Cor:freeproduct}
If $X' = \{x_1,\cdots, x_k\} \subset X$ then $E((C_0(X) \times_\sigma \mb F_n^+)/J_{(X',\sigma)})$ is equal to $*_{i=1}^m \mathfrak A_{n_i}$.
\end{corollary}
\begin{proof}
From the proof of the previous proposition we see that
\[
\{\theta_{x,y}(\pi_{(X',\sigma)}(\mathfrak s_i)) : x\in X', \sigma_i(x) = y\}, \ \ y\in X', \ 1\leq i\leq n
\]
is a set of row isometries. By the proof of the previous proposition these row isometries are free from each other.
\end{proof}
In all of the following theory we take $X=Y$ and assume that any completely isometric isomorphism, $\Theta$, is the identity on $C_0(X)$. This we can do without loss of generality because $\Theta$ maps $C_0(X)$ to $C_0(Y)$ $*$-isomorphically and $(Y,\tau)$ is conjugate to $(\gamma(Y), \gamma\circ\tau\circ\gamma^{-1})$ for any homeomorphism $\gamma$ implying that the semicrossed products of those two systems are completely isometrically isomorphic.
\begin{lemma}\label{Lemma:entryalgebra}
If $\Theta : C_0(X) \times_\sigma \mb F_n^+ \rightarrow C_0(X) \times_\tau \mb F_n^+$ is a completely isometric isomorphism and $X' \subseteq X$ with $|X'| < \infty$ then $\Theta$ induces a completely isometric isomorphism $\varphi_{X'}$ of the entry algebras $E((C_0(X)\times_\sigma \mb F_n^+)/J_{(X', \sigma)})$ and $E((C_0(X) \times_\tau \mb F_n^+)/J_{(X', \tau)})$.
\end{lemma}
\begin{proof}
Such a completely isometric isomorphism $\Theta$, because it fixes $C_0(X)$, takes $J_{(X,\sigma)}$ isomorphically onto $J_{(X',\tau)}$. Hence, it induces a completely isometric isomorphism, $\Theta_{X'}$, between the quotients such that $\pi_{(X',\tau)}\circ \Theta = \Theta_{X'} \circ \pi_{(X',\sigma)}$. This implies that $\Theta_{X'}|_{C(X')} = \operatorname{id}$ and so
\[
\Theta_{X'}(P_y\pi_{(X',\sigma)}(\mathfrak s_i)P_x) = P_y \pi_{(X', \tau)}(\Theta(\mathfrak s_i))P_x \in P_y M_{|X'|}(E((C_0(X)\times_\tau \mb F_n^+)/J_{(X',\tau)}))P_x
\]
provides the required map on the generators of the entry algebra. In other words, define $\varphi_{X'}$ on the generators of the entry algebra by
\[
\varphi_{X'}(\theta_{x,y}(\pi_{(X',\sigma)}(\mathfrak s_i))) \ \ = \ \ \theta_{x,y}(\pi_{(X',\tau)}(\Theta(\mathfrak s_i))).
\]
$\Theta(\mathfrak s_i)$ is a contraction so $\big(\theta_{y,x}(\pi_{(X',\tau)}(\Theta(\mathfrak s_i)))\big)_{x\in X'}$ is a row contraction.
Recall that Davidson and Katsoulis showed that completely isometrically isomorphic algebras implied piecewise conjugate dynamical systems \cite{DavKat2} which shows that $\varphi_{X'}$ is non-zero on the generators.
This map is well defined since the structure of the entry algebras was shown to be a free product in the previous corollary. By the universal property of the free product of noncommutative disc algebras $\varphi_{X'}$ is a completely contractive homomorphism. In the same way, $\Theta_{X'}^{-1}$ induces a completely contractive homomorphism from $E((C_0(X)\times_\tau \mb F_n^+)/J_{(X',\tau)})$ to $E((C_0(X)\times_\sigma \mb F_n^+)/J_{(X',\sigma)})$ that is the inverse on the generators. Therefore, $\varphi_{X'}$ is a completely isometric isomorphism.
\end{proof}
The existence of this entry algebra isomorphism is entirely dependent on the form of these algebras. For instance, if $U_1$ and $U_2$ are the generators of $A(\mb D) * A(\mb D)$ then the algebra generated by $U_1\oplus U_1$ and the algebra generated by $U_1\oplus U_2$ are both completely isometrically isomorphic to $A(\mb D)$ but their entry algebras are $A(\mb D)$ and $A(\mb D)*A(\mb D)$ respectively, which we know to be not isometrically isomorphic. Remarkably this is even false for C$^*$-algebras, Plastiras \cite{Plastiras} produces two non-isomorphic C$^*$-algebras that are isomorphic after tensoring with $M_2$.
\begin{proposition}\label{Prop:permutation}
For a completely isometric isomorphism $\Theta : C_0(X) \times_\sigma \mb F_n^+ \rightarrow C_0(X) \times_\tau \mb F_n^+$ and $x\in X$ there exists a unique permutation $\alpha_{\Theta, x} \in S_n$ such that for every finite $X' \subset X$ with $x\in X'$ we have
\[
\pi_{(X', \tau)}(\Theta(\mathfrak s_i))P_x \in \overline{\operatorname{Alg}}\{C(X'), \pi_{(X',\tau)}(\mathfrak t_{\alpha_{\Theta,x}(i)})\}P_x, \ \ 1\leq i\leq n.
\]
\end{proposition}
\begin{proof}
Fix $x\in X$. First consider $X' = \{x, \sigma_1(x), \cdots, \sigma_n(x)\} = \{x, \tau_1(x), \cdots, \tau_n(x)\}$, because $(X,\sigma)$ and $(X,\tau)$ are piecewise conjugate.
Lemma \ref{Lemma:entryalgebra} shows that $\Theta$ induces a completely isometric isomorphism $\varphi_{X'} : E((C_0(X)\times_\sigma \mb F_n^+)/J_{(X',\sigma)}) \rightarrow E((C_0(X)\times_\tau \mb F_n^+)/J_{(X',\tau)})$. Corollary \ref{Cor:freeproduct} gives that $\varphi_{X'}$ is a completely isometric isomorphism of free products of noncommutative disc algebras. By the corollary to Theorem \ref{Thm:Autos} this implies that both entry algebras are made up of the same noncommutative disc algebras and after rearrangement $\varphi_{X'}$ is just a product of noncommutative disc algebra completely isometric automorphisms.
Therefore, there exists a permutation $\alpha\in S_{n\times|X'|}$ such that for each $1\leq i, j\leq n$, the row contraction $\big( \theta_{y,\sigma_j(x)}(\pi_{(X',\sigma)}(\mathfrak s_i))\big)_{y\in X'}$ is mapped into the noncommutative disc algebra generated by the row contraction $\big( \theta_{y,\tau_{j'}(x)}(\pi_{(X',\tau)}(\mathfrak t_{i'}))\big)_{y\in X'}$ where $\alpha(i,\sigma_i(x)) = (i', \tau_{j'}(x))$.
Define the required permutation $\alpha_{\Theta,x} \in S_n$ by $\alpha(i, \sigma_i(x)) = (\alpha_{\Theta,x}(i), \tau_j(x)), 1\leq i\leq n$. Thus, if $E_{y,z}$ is the matrix unit in $B(\mathcal{H}^{|X'|})$ taking $\mathcal{H}_z$ onto $\mathcal{H}_y$, then
\[
\pi_{(X',\tau)}(\Theta(\mathfrak s_i))P_x = \Theta_{X'}(\pi_{(X',\sigma}(\mathfrak s_i))P_x
\]
\[
= \Theta_{X'}(P_{\sigma_i(x)}\pi_{(X',\sigma)}(\mathfrak s_i)P_x)
= P_{\sigma_i(x)}\pi_{(X',\tau)}(\Theta(\mathfrak s_i))P_x
\]
\[= \theta_{x, \sigma_i(x)}(\pi_{(X',\tau)}(\Theta(\mathfrak s_i)) E_{\sigma_i(x),x}
= \varphi_{X'}(\theta_{x, \sigma_i(x)}(\pi_{(X',\sigma)}(\mathfrak s_i)))E_{\sigma_i(x),x}
\]
\[\in \overline\operatorname{Alg}\{C(X'), \theta_{x, \tau_j(x)}(\pi_{(X',\tau)}(\mathfrak t_{\alpha_{\Theta,x}(i)}))E_{\tau_j(x),x}, 1\leq j\leq n\}P_x
\]
\[
= \overline\operatorname{Alg}\{C(X'), \pi_{(X',\tau)}(\mathfrak t_{\alpha_{\Theta,x}(i)})P_x.
\]
Finally, we need to establish that this permutation is unique. Let $X'' \subset X$ be any finite set such that $x\in X''$ and let $\tilde X = X'' \cup X'$.
Again, because $\Theta$ is the identity on $C_0(X)$
\[
\pi_{(\tilde X, \tau)}(\Theta(\mathfrak s_i))P_x = \theta_{x,\sigma_i(x)}(\pi_{(\tilde X, \tau)}(\Theta(\mathfrak s_i)))E_{\sigma_i(x),x}
\]
\[
= \varphi_{\tilde X}(\theta_{x,\sigma_i(x)}(\pi_{(\tilde X,\sigma)}(\mathfrak s_i)))E_{\sigma_i(x),x}
\]
\[
\in \overline\operatorname{Alg}\{ C(\tilde X), \theta_{x,y}(\pi_{(\tilde X,\tau)}(\mathfrak t_{j}))E_{y,x}, \forall y\in \tilde X\}P_x
\]
\[
= \overline\operatorname{Alg}\{ C(\tilde X), \pi_{(\tilde X, \tau)}(\mathfrak t_j)\}P_x
\]
for some $1\leq j\leq n$. However, $P_{X'}(\pi_{(\tilde X,\sigma)})P_{X'}$ is identical to $\pi_{(X', \sigma)}$ (which is true as well when $\sigma$ is replaced by $\tau$) and so $j$ must be equal to $\alpha_{\Theta,x}(i)$ by the first argument. Finally, compression to $\mathcal{H}^{(|X''|)} \subset \mathcal{H}^{(|\tilde X|)}$ gives the desired conclusion.
\end{proof}
Out of the proof of the previous proposition we get what will become the pre-image condition:
\begin{corollary}\label{Cor:pre-image}
For $x,y\in X$ if $\sigma_i(x) = \sigma_i(y)$ then $\alpha_{\Theta,x}(i) = \alpha_{\Theta, y}(i)$ and \\ if
$\tau_j(x) = \tau_j(y)$ then $\alpha^{-1}_{\Theta,x}(j) = \alpha^{-1}_{\Theta, y}(j)$.
\end{corollary}
\begin{proof}
If $z= \sigma_i(x) = \sigma_i(y)$ then for $X' = \{x,y,z\} \subset X$ we have that
\[
(\phi_{x,z}(\pi_{(X',\sigma)}(\mathfrak s_i)), \ \phi_{y,z}(\pi_{(X',\sigma)}(\mathfrak s_i))
\]
is a row contraction of elements in $E((C_0(X) \times_\sigma \mb F_n^+)/J_{(X',\sigma)})$. As such $\varphi_{X'}$ will map this into the same noncommutative disc algebra generated by the row contraction
\[
(\phi_{x,z}(\pi_{(X',\tau)}(\mathfrak t_j)), \phi_{y,z}(\pi_{(X',\tau)}(\mathfrak t_j)), \phi_{z,z}(\pi_{(X',\tau)}(\mathfrak t_j)))
\]
for some $1\leq j\leq n$. The previous proposition dictates that $\alpha_{\Theta,x}(i) = j = \alpha_{\Theta, y}(i)$.
The last conclusion follows from the fact that $\alpha_{\Theta^{-1},x} = \alpha_{\Theta,x}^{-1}$.
\end{proof}
We now will show that these permutations $\alpha_{\Theta,x}$ change continuously with respect to $x$.
\begin{proposition}\label{Prop:continuous}
Let $\alpha_\Theta : X \rightarrow S_n$ be defined as $\alpha_\Theta(x) = \alpha_{\Theta,x}$.
Then $\alpha_\Theta$ is a continuous function where $S_n$ has the discrete topology.
\end{proposition}
\begin{proof}
Let $\{x_k\} \in X$ be a sequence converging to $x\in X$ such that $\alpha_\Theta(x_k) = \alpha \in S_n, \forall k\geq 1$. Assume that $\alpha_{\Theta, x} \neq \alpha$ and so there exists $1\leq i\leq n$ such that $\alpha_{\Theta, x}(i) \neq \alpha(i)$.
In the same way as in the previous section we can define the gauge automorphisms of $C_0(X) \times_\tau \mb F_n^+$ to be $\gamma_z, z = (z_1,\cdots, z_n) \in \mb T^n$ by letting $\gamma_z(\mathfrak s_i) = z_iS_i$ and extending to a completely isometric isomorphism by the universal property.
Again, define for $z = (z,\cdots, z) \in \mb T^n$
\[
\Phi_k(a) \ \ = \ \ \int_{\mb T} \gamma_z(a) \overline{z}^k dz,
\]
the completely contractive projection onto words of length $k$ of the generators $S_1,\cdots, S_n$. Finally, the sum of these projections converges to $a$ by way of the Cesaro means, giving us that $\Phi_k(a) = 0, \forall k\geq 1$ if and only if $a = 0$.
Now, look at $\Theta(\mathfrak s_i)$ where we have $\alpha_{\Theta, x}(i) \neq \alpha(i)$. $\Theta(\mathfrak s_i) \notin C_0(X)$ since $\Theta^{-1}|_{C_0(X)} = \operatorname{id}$ there exists $k\geq 1$ such that
\[
\Phi_k(\Theta(\mathfrak s_i)) = \sum_{w\in \mb F_n^+, |w| \leq k} \mathfrak s_w f_w \neq 0.
\]
Claim: $f_w(x) \neq 0$ if and only if $w = \alpha_{\Theta, x}(i)^k$.
\vskip 6 pt
\noindent Verification: Let $X' = \{x\} \cup \{ \sigma_w(x) : w\in \mb F_n^+, |w| \leq k\} \subseteq X$.
Hence,
\[
\pi_{(X',\sigma)}(\mathfrak s_wf_w)P_x = \pi_{(X',\sigma)}(\mathfrak s_w)f_w(x)P_x = 0 \ \ \ \textrm{if and only if} \ \ \ f_w(x) = 0
\]
because $\pi_{(X',\sigma)}(\mathfrak s_w)P_x \neq 0$.
Now, Proposition \ref{Prop:permutation} says that $\pi_{(X', \tau)}(\Theta(\mathfrak s_i))P_x\neq 0$ is in the algebra generated by $T_{\alpha_{\Theta,x}(i)}$ times $P_x$.
Next, we can define the projections $\Phi_k$ for the quotient as well such that $\Phi_k \circ\pi_{(X',\tau)} = \pi_{(X',\tau)} \circ\Phi_k$.
Moreover, if $A\in \overline\operatorname{Alg}\{C(X'), \pi_{(X',\tau)}(T_j)\}$ then so is $\Phi_k(A)$.
Thus,
\[
\pi_{(X',\tau)}(\Phi_k(\Theta(\mathfrak s_i)))P_x = \Phi_k(\Theta(\pi_{(X',\sigma)}(\mathfrak s_i)))P_x \in \overline\operatorname{Alg}\{C(X'), \pi_{(X',\tau)}(\mathfrak t_{\alpha_{\Theta,x}(i)})\}P_x.
\]
The claim follows.
Lastly, the $f_w$'s are continuous functions so there is a $m\geq 1$ such that $f_w(x_m) \neq 0$ for $w = \alpha_{\Theta, x}(i)^k$ as well. The previous claim is still true when we replace $x$ with $x_m$. However, this implies that $\alpha = \alpha_{\Theta, x_m}(i) = \alpha_{\Theta, x}(i)$, a contradiction. Therefore, $\alpha_\Theta$ is a continuous function.
\end{proof}
Finally, the second main result of the paper, that the semicrossed product algebra completely characterizes multivariable dynamical systems up to partition conjugacy.
\begin{theorem}
The dynamical systems $(X,\sigma)$ and $(Y,\tau)$ are partition conjugate if and only if $C_0(X) \times_\sigma \mb F_n^+$ and $C_0(Y) \times_\tau \mb F_n^+$ are completely isometrically isomorphic.
\end{theorem}
\begin{proof}
Theorem \ref{Thm:conjugacy} proves one direction. So assume that $\Theta : C_0(X) \times_\sigma \mb F_n^+ \rightarrow C_0(Y) \times_\tau \mb F_n^+$ is a completely isometric isomorphism. Again, without loss of generality we can assume that $X = Y$ and that $\Theta|_{C_0(X)} = id|_{C_0(X)}$.
Define
\[
V_{i,j} \ = \ \{x\in X: \alpha_{\Theta, x}(i) = j\}, \ 1\leq i,j\leq n,
\]
where the $\alpha_{\Theta,x}$ is the canonical permutation obtained from $\Theta$ and $x$ found in Proposition \ref{Prop:permutation}. Since this runs through all of $X$ we have that
\[
\dot\cup_{j=1}^n V_{i,j} = X, 1\leq i\leq n, \ \ {\rm and} \ \ \dot\cup_{i=1}^n V_{i,j} = X, 1\leq j\leq n.
\]
Furthermore, $\alpha_\Theta^{-1}(\{\gamma\})$ is clopen for every $\gamma \in S_n$ by Proposition \ref{Prop:continuous} and so
\[
V_{i,j} = \cup_{\gamma(i) = j, \gamma\in S_n} \alpha_\Theta^{-1}(\{\gamma\})
\]
is clopen as well. In the proof of the same proposition it was shown that for every $x\in V_{i,j}$ we have that $\sigma_i(x) = \tau_j(x)$.
Lastly, if $x\in \sigma_i^{-1}(\sigma_i(V_{i,j})$ then there exists $y\in V_{i,j}$ such that $\sigma_i(x) = \sigma_i(y)$. Corollary \ref{Cor:pre-image} then says that $\alpha_{\Theta,x}(i) = \alpha_{\Theta, y}(i) = j$ and so $x\in V_{i,j}$. Repeat the same argument to obtain $\tau_j^{-1}(\tau_j(V_{i,j})) = V_{i,j}$ as well. Therefore, $\{V_{i,j}\}$ is the desired partition and the two dynamical systems are partition conjugate.
\end{proof}
\section{The tensor algebra}
We conclude this paper by showing when the tensor algebra and the semicrossed product algebra of a dynamical system $(X,\sigma)$ are completely isometrically isomorphic, namely if and only if the $\sigma_i$ have pairwise disjoint ranges. This is achieved by the following representation theory.
Introduced independently in \cite{KatsKribs} and \cite{Solel}, nest representations form an important class in the representation theory of non-selfadjoint operator algebras. To be precise a nest representation is a representation whose lattice of invariant subspaces forms a nest, that is, is linearly ordered. For instance, irreducible representations are nest representations
In \cite{DavKat1} and \cite{DavKat2}, 2-dimensional nest representations with lattice of invariant subspaces equal to $\{\{0\}, \mb Ce_1, \mb C^2\}$ were used to great effect. We generalize this idea in the following manner:
\begin{definition}
Consider the following subalgebra of the upper triangular matrices, $\mathcal{T}_m$
\[
\mathcal{N}_m \ \ = \ \ \left\{ \left[\begin{array}{cccc} x_{1,1} & x_{1,2}& \cdots & x_{1,m} \\ & x_{2,2} \\ &&\ddots \\ &&&x_{m,m} \end{array}\right] : x_{i,i}, x_{1,i} \in \mathbb C, 1\leq i\leq m \right\}.
\]
If $\mathcal{A}$ is an operator algebra, let $\operatorname{rep}_{\mathcal{N}_m}(\mathcal{A})$ denote the collection of completely contractive homomorphisms $\rho : \mathcal{A} \rightarrow \mathcal{T}$ such that compression to the $(i, j)$ entry, call it $\rho_{i,j}$, is a norm 1 linear functional, for all $1\leq i = j\leq m$ and $i=1, 2\leq j\leq m$.
\end{definition}
The avid reader will note that this is no longer a nest representation for $m\geq 3$.
Notice that compression to any of the diagonal entries is a non-trivial homomorphism into $\mb C$, that is, a character in $\mathcal{M}(\mathcal{A})$. Recall that the only characters of $C_0(X)$ are point evaluations. Thus , for $\rho \in \mathcal{M}(C_0(X) \times_\sigma \mb F_n^+)$, there exists $x\in X$ such that $\rho$ restricted to $C_0(X)$ is point evaluation at $x$. In \cite{DavKat2} it was shown that the set of characters corresponding to point evaluation at $x\in X$ is homeomorphic to $\overline{\mb D}^k$ where $k$ is the number of maps that fix the point $x$.
Hence, for $\rho \in \operatorname{rep}_{\mathcal{N}_m}(C_0(X) \times_\sigma \mathbb F^n_+)$ there exist $x_1,\cdots, x_m \in X$ such that $\rho_{i,i}(f) = f(x_i), \forall f\in C_0(X)$. Note that $\rho(C_0(X))$ is diagonal because $\rho$ is completely contractive and so a $*$-homomorphism on $C_0(X)$.
For $1\leq i\leq m$ consider the map $\theta_i(A) = \left[\begin{array}{cc} \rho_{1,1}(A) & \rho_{1,i+1}(A) \\ 0 & \rho_{i+1,i+1}(A) \end{array}\right]$. $\theta$ is a nest representation and by \cite[Lemma 3.15]{DavKat2} there exists $1\leq j\leq n$ such that $\sigma_j(x_i) = x_1$.
In light of this, for ${\bf x} = (x_1,\cdots, x_m)$ denote $\operatorname{rep}_{x, {\bf x}}(C_0(X) \times_\sigma \mb F_n^+)$ to be the subset of $\operatorname{rep}_{\mathcal{N}_{m+1}}$ such that $\rho_{1,1}$ is a character corresponding to $x$ and for $1\leq i\leq m$, $\rho_{i+1,i+1}$ is a character corresponding to $x_i$. In other words, if $\operatorname{rep}_{x, \bf x}(C_0(X) \times_\sigma \mb F_n^+)$ is non-empty then $\{x_1,\cdots, x_m\} \subset \cup_{i=1}^n \sigma^{-1}_i(\{x\})$.
The following example was inspired by \cite{Duncan2} where they define a similar 3x3 representation to identify when two edges of an edge-colored directed graph belong to different colors.
\begin{example}\label{Example:Colours}
Let $(X,\sigma)$ be a dynamical system with $x, x_1,\cdots, x_m\in X$ for $m\leq n$ and suppose that $\sigma_i(x_i) = x$ for $1\leq i\leq m$. Define a representation $\rho$ by
\[
\rho(f) = \left[\begin{array}{cccc} f(x) & 0 & \cdots & 0 \\ & f(x_1) \\ & & \ddots \\ &&& f(x_m) \end{array}\right] \ \ {\rm and}
\ \ \rho(\mathfrak s_k) = \left[\begin{array}{cccc} 0 & \delta_{1k} & \cdots & \delta_{mk} \\ &0 \\ &&\ddots \\ &&&0 \end{array}\right]
\]
where $\delta_{ik}$ is the Kronecker delta function. Thus, for $A = \sum_{w\in \mb F_n^+} \mathfrak s_w f_w$ where all but finitely many of the $f_w \in C_0(X)$ are 0
\[
\rho(A) = \left[\begin{array}{cccc} f_0(x) & f_1(x_1) & \cdots & f_m(x_m) \\ & f_0(x_1) \\ &&\ddots \\ &&&f_0(x_m) \end{array}\right]
\]
is a completely contractive homomorphism onto $\mathcal{N}_{m+1}$ which clearly satisfies the $\rho_{i,j}$ norm 1 condition. Therefore, $\rho$ extends to a c.c. homomorphism on all of $C_0(X) \times_\sigma \mb F_n^+$ and so, for ${\bf x} = (x_1,\cdots, x_m)$, we have $\rho \in \operatorname{rep}_{x, \bf x}(C_0(X) \times_\sigma \mathbb F^n_+)$.
\end{example}
Although not necessary to prove the main theorem of this section we have the following converse to the previous example.
\begin{proposition}\label{Prop:Colours}
Let $\mathcal{A} = C_0(X) \times_\sigma \mb F_n^+$, $x,x_1,\cdots, x_m \in X$ and ${\bf x} = (x_1,\cdots, x_m)$. If $\operatorname{rep}_{x,\bf x}(\mathcal{A})$ is non-empty then there exist distinct $1\leq i_1, \cdots, i_m \leq n$ such that $\sigma_{i_1}(x_1) = \cdots = \sigma_{i_m}(x_m) = x$.
\end{proposition}
\begin{proof}
Let $\rho \in \operatorname{rep}_{x,\bf x}(\mathcal{A})$. For each $1\leq i\leq m$ consider the compression to the 1st and $i+1$st rows and columns
\[
\theta_i = \left[\begin{array}{cc} \rho_{1,1} & \rho_{1,i+1} \\ 0 & \rho_{i+1,i+1}\end{array}\right].
\]
Let $\mathfrak s_{i_1}, \cdots, \mathfrak s_{i_k}$ be those generators not in the kernel of $\rho_{1,i+1}$ and let $t = \max_{j=1}^k \|\theta_i(\mathfrak s_{i_j})\|$.
If $t < 1$ then consider the covariant representation of $(X,\sigma)$ given by $\theta_i(C_0(X))$ and
\[
T_j = \left[\begin{array}{cc} \rho_{1,1}(\mathfrak s_j) & (2-t)\rho_{1,i+1}(\mathfrak s_j) \\ 0 & \rho_{i+1,i+1}(\mathfrak s_j)\end{array}\right].
\]
Note that by the triangle inequality this still gives $\|T_j\| \leq 1$. Hence, by the universal property there exists a completely contractive homomorphism $\theta'_i$ of $\mathcal{A}$ onto $\mathcal{T}_2$, the upper triangular matrices taking $\mathfrak s_j$ to $T_j$. Moreover, we have that
\[
\theta'_i = \left[\begin{array}{cc} \rho_{1,1} & (2-t)\rho_{1,i+1} \\ 0 & \rho_{i+1,i+1}\end{array}\right].
\]
Now, contractive implies that $|(2-t)\rho_{1,i+1}| \leq 1$ but then $|\rho_{1,i+1}| \leq \frac{1}{2-t} < 1$, a contradiction since it is a norm 1 linear functional.
Hence, there must be at least one $1\leq j\leq k$ such that $\|\theta_i(\mathfrak s_{i_j})\| = 1$ and because $\rho_{1,i}(\mathfrak s_{i_j}) \neq 0$ this implies that $\rho_{1,l}(\mathfrak s_{i_j}) = 0$ for all $l \neq 1, i_j$.
Therefore, $\sigma_{i_j}(x_i) = x$ and $\theta_k(\mathfrak s_{i_j})$ is diagonal for all $k\neq i$.
\end{proof}
\begin{theorem}
Let $(X,\sigma)$ be a dynamical system, then $\mathcal{A}(X,\sigma)$ and $C_0(X) \times_\sigma \mathbb F^n_+$ are completely isometrically isomorphic if and only if the ranges of the $\sigma_i$ are pairwise disjoint. \end{theorem}
\begin{proof}
Suppose that there are points $x, x_1, x_2\in X$ such that $\sigma_1(x_1) = \sigma_2(x_2) = x$. If $x_1 = x_2 = x$ then there is a point fixed by more than one map and Davidson and Katsoulis have shown in \cite[Corollary 3.11]{DavKat2} that $\mathcal{A}(X,\sigma)$ and $C_0(X) \times_\sigma \mb F_n^+$ are not even algebraically isomorphic due to their differing character spaces.
Assume then that $x_1 \neq x$. If ${\bf x} = (x_1, x_2)$ Example \ref{Example:Colours} says that $\operatorname{rep}_{x, \bf x}(C_0(X) \times_\sigma \mb F_n^+)$ is non-empty. Again we turn to \cite{DavKat2} to see that a completely isometric isomorphism $\gamma$ between these algebras induces a bijection $\gamma_c$ between the character spaces that corresponds to the homeomorphism $\gamma_s$ that arises from $\gamma$ being a $*$-automorphism of $C_0(X)$. Without loss of generality we may take $\gamma_s$ to be the identity since conjugacy among dynamical systems implies completely isometrically isomorphic among the tensor algebras. Hence, there must exist a $\rho \in \operatorname{rep}_{x,\bf x}(\mathcal{A}(X,\sigma))$ when the algebras are c.i.i.
In light of this, consider, as in the proof of Proposition \ref{Prop:Colours}, the nest representation $\theta_1$ given by the compression to the upper left 2x2 corner of $\rho$. Let $i_1,\cdots, i_k$ be the indices such that $\rho_{1,2}(\mathfrak s_{i_j}) \neq 0$, implying that $\sigma_{i_j}(x_1) = x$ and, since $x_1\neq x$, that $\rho_{2,2}(\mathfrak s_{i_j}) = 0$ (since it is a character and $\sigma_{i_j}$ does not fix $x_1$). Let $t = \|[\theta_1(\mathfrak s_{i_1}) \cdots \theta_1(\mathfrak s_{i_k})]\|$. Because the generators of $\mathcal{A}(X,\sigma)$ are row contractive and $\theta_1$ is c.c. then $t\leq 1$. If $t< 1$ then consider the slight adjustment to $\theta_1$ given by
\[
\theta'_1 = \theta_1 + \left[\begin{array}{cc} 0 & (\frac{t}{\sqrt k}) \rho_{1,2} \\ 0 & 0\end{array}\right].
\]
This map, $\theta'_1$, is a homomorphism with $\|[\theta'_1(\mathfrak s_1) \cdots \theta'_1(\mathfrak s_n)]\| \leq 1$ which implies by the universal property of the tensor algebra that $\theta'_1$ is completely contractive. But then $\|(1 + \frac{t}{\sqrt k}) \rho_{1,2}\| \leq 1$ which implies $\|\rho_{1,2}\| < 1$, a contradiction.
Assume then, that
\[
\|[ \rho_{1,1}(\mathfrak s_{i_1}) \ \rho_{1,2}(\mathfrak s_{i_1}) \cdots \rho_{1,1}(\mathfrak s_{i_k}) \ \rho_{1,2}(\mathfrak s_{i_k}) ] \|
= \|[\theta_1(\mathfrak s_{i_1}) \cdots \theta_1(\mathfrak s_{i_k})]\| = 1.
\]
However, $[\rho(\mathfrak s_1) \cdots \rho(\mathfrak s_n)]$ is a row contraction as well thus $\rho_{1,3}(\mathfrak s_i) = 0$ for $1\leq i\leq n$, contradicting the fact that $\rho_{1,3}$ is a norm 1 functional. Therefore, when the ranges of the $\sigma_i$ overlap then the tensor and semicrossed product algebras cannot be completely isometrically isomorphic.
Conversely, suppose that the ranges of the $\sigma_i$ are pairwise disjoint. Because $\sigma_i$ is proper and continuous then $\sigma_i(X)$ is closed in $X$. Hence, there exists $f_1,\cdots, f_n \in C_0(X)$ such that $f_i f_j = 0$ for $i\neq j$ and $f_i(x) = 1, \forall x\in \sigma_i(X)$.
If $\mathfrak s_1,\cdots, \mathfrak s_n$ are the generators of $C_0(X) \times_\sigma \mb F_n^+$ then for $i\neq j$
\[
\mathfrak s^*_j \mathfrak s_i \ = \ (f_j\circ \sigma_j) \mathfrak s_j^* \mathfrak s_i (f_i\circ\sigma_i)
\ = \ \mathfrak s_j^* f_j f_i \mathfrak s_i \ = \ 0.
\]
Thus, $[\mathfrak s_1 \cdots \mathfrak s_n]$ is a row contraction which implies that there exists a completely contractive homomorphism from $\mathcal{A}(X,\sigma)$ onto $C_0(X) \times_\sigma \mb F_n^+$ taking generators to generators. This map is the inverse of the canonical quotient from the semicrossed product onto the Tensor algebra and therefore the two algebras are completely isometrically isomorphic.
\end{proof}
The following example is due to Duncan \cite{Duncan2}, it illustrates that there are topologically isomorphic tensor and semicrossed product algebras that are not completely isometrically isomorphic.
\begin{example}
Consider the four point set $X = \{1,2,3,4\}$ and the pair of maps
\[
\sigma_1(1) = 2, \sigma_1(2) = \sigma_1(3) = \sigma_1(4) = 3 \ \ {\rm and} \ \ \sigma_2(1) = 2, \sigma_2(2) = \sigma_2(3) = \sigma_2(4) = 4.
\]
Thus, $\mathcal{A}(X,\sigma)$ and $C(X) \times_\sigma \mb F_2^+$ are isomorphic as Banach algebras simply by sending the generators to the generators. Indeed, by the previous theorem we already know that for $X' = \{2,3,4\}$, $\mathcal{A}(X', \sigma) = C(X') \times_\sigma \mb F_2^+$. Now if $U_1, U_2$ are the generators of $A(\mb D) * A(\mb D)$ and $L_1,L_2$ are the generators of $\mathcal{A}_2$, then
\[
\left[\begin{array}{cc} I & \\ &0 \end{array}\right] \mapsto \left[\begin{array}{cc} I & \\ & 0 \end{array}\right], \ \ \left[\begin{array}{cc} 0 & \\ & I \end{array}\right] \mapsto \left[\begin{array}{cc} 0 & \\ & I \end{array}\right]
\]
\[
{\rm and} \ \ \left[\begin{array}{cc} 0& U_i \\ & 0 \end{array}\right] \mapsto \left[\begin{array}{cc} 0& L_i \\ & 0 \end{array}\right], i=1,2,
\]
is an isomorphism between the algebras they generate.
It should be noted that the two maps have overlapping range so the two algebras are not (completely) isometrically isomorphic.
\end{example}
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series={CBMS Regional Conference Series in Mathematics},
volume={103},
publisher={Published for the Conference Board of the Mathematical
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Providence, RI},
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pages={vi+113},
isbn={0-8218-3660-9},
}
\bib{Ramsey}{thesis}{
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title={Algebraic characterization of multivariable dynamics},
organization={University of Waterloo},
type={Master's Thesis},
date={2009},
}
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author={Rudin, W.},
title={Function theory in polydiscs},
publisher={W. A. Benjamin, Inc., New York-Amsterdam},
date={1969},
pages={vii+188},
}
\bib{Rudin2}{book}{
author={Rudin, W.},
title={Function theory in the unit ball of ${\bf C}\sp{n}$},
series={Grundlehren der Mathematischen Wissenschaften [Fundamental
Principles of Mathematical Science]},
volume={241},
publisher={Springer-Verlag, New York-Berlin},
date={1980},
pages={xiii+436},
isbn={0-387-90514-6},
}
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author={Solel, B.},
title={You can see the arrows in a quiver operator algebra},
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volume={77},
date={2004},
number={1},
pages={111--122},
issn={1446-7887},
}
\bib{NagyFoias}{book}{
author={Sz.-Nagy, B.},
author={Foias, C.},
author={Bercovici, H.},
author={K{\'e}rchy, L.},
title={Harmonic analysis of operators on Hilbert space},
series={Universitext},
edition={2},
edition={Revised and enlarged edition},
publisher={Springer, New York},
date={2010},
pages={xiv+474},
isbn={978-1-4419-6093-1},
}
\bib{Tsyganov}{article}{
author={Tsyganov, Sh.},
title={Biholomorphic mappings of the direct product of domains},
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volume={41},
date={1987},
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pages={824--828, 890},
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},
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},
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}
\end{biblist}
\end{bibdiv}
\end{document}
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Q: Java 8 stream within stream parallelism I have a fairly complicated process that requires several levels of nested for loops.
Actions are only performed for one specific set of conditions. In other words:
for(){
if(){
for(){
if(){
//Something happens RIGHT HERE
}
//And maybe here
}
}
}
No else statements, just one solid code path iterating through a bunch of different types of objects.
My question is, if I were to replace this logic with streams (considering how many operations the CPU has to perform to complete this cycle, I think paralleling the process is the way to go) and I have a stream within a stream within a stream (BWOOOMMMPPPPPPP INCEPTION NOISE) and I parallelstream() the top level, will the streams beneath that top level still run in sequence, but just within their own respective threads?
toplevelItems.parallelstream().forEachOrdered{
//Does this stuff run in series within as many threads as there are toplevelItems
otherObjects.stream().forEach{
//or Naw?
stillOtherObjects.stream().forEach{
A: Anything below the top parallelStream() is going to be done in a serial fashion. I assume you are working with something like a List<List<>>, so you will be creating a new stream below the top level which has no connection to the top one.
|
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{"url":"https:\/\/matheducators.stackexchange.com\/questions\/20971\/how-to-explain-continuity-and-differentiability-to-highschool-students","text":"# How to explain continuity and differentiability to highschool students? [closed]\n\nI would say continuity is the idea that points numerically close to the input point of a function agree with the value of the function at that point. Now, suppose I introduce the idea of a derivative, it takes two close by points of the function and computes the slope.\n\n$$\\lim_{x \\to a} f(x) = f(a)$$\n\nHowever, if a function is continuous i.e: really close by points numerically agree, how can there be a variation in the function to define the derivative?\n\nExcuse me if the question is poorly phrased.\n\n\u2022 I find that an important feature in demonstrating what something is is to also demonstrate what it is not. Particularly, continuity has a definition but it conceptually straightforward to demonstrate discontinuity and then conceptually define continuity as not discontinuous. Jun 6 '21 at 18:12\n\u2022 How can a limit exist if a function is not constant? Jun 7 '21 at 18:11\n\u2022 This seems more like a math question than a question about teaching math.\n\u2013\u00a0user507\nJun 7 '21 at 23:14\n\n# Continuity\n\n## Method 1\n\nOne of the most intuitive ways that I like to think about what it means for a function to be \"continuous\" is by the \"drawing test\": can you draw the function's graph on a piece of paper without ever removing your pen\/pencil from the paper? A continuous function is a function whose graph is an unbroken curve.\n\n## Method 2\n\nAnother way that might be helpful in understanding what it means for a function to be continuous is to actually look at the types of discontinuities for a function. You can read Wikipedia's article on the types of discontinuities for more information. In particular, there are three types of discontinuities:\n\n1. A removable discontinuity:\n1. A jump discontinuity:\n1. An infinite (or essential) discontinuity:\n\nAs you can see, in neither of these scenarios can we draw the function's graph without removing our pencil from the page. Hence, the function is not continuous.\n\nHere's a nice visual summary of each discontinuity:\n\n# Differentiability\n\nDifferentiability is a more technical and lengthier topic than continuity, as differentiability brings about discussions regarding smoothness and multi-variability. I'll leave it up to you to teach\/introduce differentiability to your students as per the specified curriculum and course outline. The topic is rich, as even in a typical Calculus 3 course, differentiability is studied in depth.\n\nThat being said, sticking to functions with a single variable, and proceeding in a fashion as we did above, we know intuitively now that a continuous function is one that has no \"breaks\" in its graph. Similarly, a differentiable function from $$\\mathbb{R}^2$$ to $$\\mathbb{R}$$ is a function that not only has no breaks in its graph, but also has a well-defined plane that is tangent to the graph at each point. Hence, we expect a differentiable function to not have any sharp folds, corners, or \"peaks\" in the graph; the graph must be smooth.\n\n\u2022 While not disagreeing with your general approach, I would be wary of introducing the concept of an essential discontinuity at high-school, and would only do so for well-prepared students and with carefully chosen examples. Cases like f(x) = 1\/x (which is implied by your picture) are best avoided, since f(x) = 1\/x is in fact continuous everywhere it is defined. The canonical example of a function with an essential discontinuity is the piecewise defined function g(x) = sin(1\/x) when x \u2260 0 and g(x) = x when x = 0. Jun 9 '21 at 7:55\n\u2022 @StephanKubicki Furthermore, a true infinite discontinuity (an infinite discontinuity at a point in the domain) can often be regarded as a jump discontinuity, but I'm not sure that essential discontinuities are actually infinite discontinuities. Jun 12 '21 at 16:37\n\nFor continuity, I would generally follow one or both methods suggested by CuriosityCalls' answer (modulo my reservations about essential discontinuities, which I expressed in my comment to that answer).\n\nFor differentiability, in one sense this is straightforward for students who have already encountered the definition of the derivative as $$\\lim_{h\\rightarrow 0} \\frac{f(x+h)-f(x)}{h}$$. In this case, $$f(x)$$ is differentiable just in case that limit exists.\n\nOf course in practice, at the high-school level, matters are not so simple, since we often want to avoid having to compute that limit to determine whether the derivative exists. At this level, it is typically sufficient for students to know that derivatives have the intermediate value property, and hence a derivative can not have a jump discontinuity. This is enough to justify simple cases like $$f(x)= |x|$$ not having a derivative at $$x=0$$, and the intuitive idea that a function is not differentiable at a point if its graph has a \"sharp corner\".\n\nHowever, to avoid incorrect generalisation, I would also consider showing students the function $$g(x) = \\left\\{ \\begin{array}{ll} x\\sin\\left(\\frac{1}{x}\\right) & x \\neq 0 \\\\ 0 & x = 0 \\end{array} \\right.$$\n\nand observe that while the derivative does not have a jump discontinuity at 0, the derivative does not exist at 0. This is just to emphasise that what matters for whether a function is differentiable at $$x=a$$ is whether the limit of the difference quotient exists; showing that a (hypothetical) derivative is not continuous does NOT prove that the derivative does not exist. The test described above is just a handy shortcut method that can be used when the functions are suitably nice.","date":"2022-01-23 08:32:55","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\": 0, \"katex\": 0, \"math-container\": 9, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7782894372940063, \"perplexity\": 284.51694055214494}, \"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-2022-05\/segments\/1642320304217.55\/warc\/CC-MAIN-20220123081226-20220123111226-00427.warc.gz\"}"}
| null | null |
Introverted Boss (hangul: 내성적인 보스; RR: Naeseongjeogin Boseu), es una serie de televisión surcoreana transmitida del 16 de enero del 2017 hasta el 14 de marzo del 2017 por medio de la cadena tvN.
La serie fue creada por "Studio Dragon" y Choi Jin-hee, contó con la participación invitada de Han Chae-ah, Kim Jun-su, Heo Young-ji, Kim Byung-man, Choi Dae-chul, Lee Byung-joon, Kang Nam-gil, entre otros...
Historia
La serie se centra en Eun Hwan-ki, quien a pesar de ser el CEO de "Brain" una de las más importantes empresas de relaciones públicas, es una persona extremadamente tímida, por lo que debido a su personalidad, incluso sus empleados no lo conocen bien.
Por otro lado Chae Ro-woon, es una joven energética y agradable, que comienza a trabajar en la compañía de Hwan-ki para descubrir la verdadera razón por la cual su hermana mayor se quitó la vida y vengarse de los responsables. Al inicio Ro-woon quiere destruir la empresa, sin embargo conforme va a conociendo a los empleados y a Hwan-ki se da cuenta de que estaba equivocada.
Poco a poco Ro-woon comienza a enamorarse del tímido Hwa-ki, a quien intentará ayudar a salir de su zona de confort, relacionándose con los demás y experimentando nuevas cosas.
Reparto
Personajes principales
Personajes recurrentes
Otros personajes
Apariciones especiales
Episodios
La serie estuvo conformada por 16 episodios, los cuales fueron emitidos cada lunes y martes a las 23:00 horas (zona horaria de Corea (KST)).
Premios y nominaciones
Producción
Creada por "Studio Dragon" y Choi Jin-hee, la serie fue dirigida por Song Hyun-wook contó con la participación del escritor Joo Hwa-mi.
La producción estuvo a cargo de So Jae-hyeon, junto con los productores ejecutivos Lee Jin-ho y Lee Jae-gil.
También conocida como "My Shy Boss" y "Sensitive Boss".
La música fue compuesta por Uhm Ki-yub, mientras que la cinematografía fue realizada por Han Dong-hyun y la edición estuvo en manos de Kang Yoon-hee.
Contó con el apoyo de las compañías de producción "KBS Media" y "Introverted Boss SPC", fue distribuida por tvN.
Emisión en otros países
Referencias
Enlaces externos
(inglés)
Introverted Boss Official Website (coreano)
Allkpop - TV Serie - Introverted Boss (inglés)
Soompi English > Introverted Boss (inglés)
Introverted Boss (episodes review)
Soompi | Noticias en Español | Serie | Introverted Boss (español)
Series de televisión de Corea del Sur
Series de televisión en coreano
Series de televisión
Series de televisión iniciadas en 2017
Series de televisión de Corea del Sur iniciadas en 2017
Series de televisión finalizadas en 2017
Series de televisión de Studio Dragon
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{"url":"https:\/\/planetmath.org\/integercontraharmonicmeans","text":"# integer contraharmonic means\n\nLet $u$ and $v$ be positive integers.\u2009 There exist nontrivial cases where their contraharmonic mean\n\n $\\displaystyle c\\;:=\\;\\frac{u^{2}\\!+\\!v^{2}}{u\\!+\\!v}$ (1)\n\nis an integer, too.\u2009 For example, the values\u2009 $u=3,\\;v=15$\u2009 have the contraharmonic mean\u2009 $c=13$.\u2009 The only \u201ctrivial cases\u201d are those with\u2009 $u=v$,\u2009 when\u2009 $c=u=v$.\n\n $u$ $v$ $c$ $2$ $3$ $3$ $4$ $4$ $5$ $5$ $6$ $6$ $6$ $6$ $7$ $7$ $8$ $8$ $8$ $9$ $9$ $...$ $6$ $6$ $15$ $12$ $28$ $20$ $45$ $12$ $18$ $30$ $66$ $42$ $91$ $24$ $56$ $120$ $18$ $45$ $...$ $5$ $5$ $13$ $10$ $25$ $17$ $41$ $10$ $15$ $26$ $61$ $37$ $85$ $20$ $50$ $113$ $15$ $39$ $...$\n\nThe nontrivial integer contraharmonic means form Sloane\u2019s sequence http:\/\/oeis.org\/search?q=A146984&language=english&go=SearchA146984.\n\nFor any value of $u>2$, there are at least two greater values of $v$ such that\u2009 $c\\in\\mathbb{Z}$.\n\nProof.\u2009 One has the identities\n\n $\\displaystyle\\frac{u^{2}\\!+\\!((u\\!-\\!1)u)^{2}}{u+(u\\!-\\!1)u}\\;=\\;u^{2}\\!-\\!2u% \\!+\\!2,$ (2)\n $\\displaystyle\\frac{u^{2}\\!+\\!((2u\\!-\\!1)u)^{2}}{u+(2u\\!-\\!1)u}\\;=\\;2u^{2}\\!-\\!% 2u\\!+\\!1,$ (3)\n\nthe right hand sides of which are positive integers and different for\u2009 $u\\neq 1$.\u2009 The value\u2009 $u=2$\u2009 is an exception, since it has only\u2009 $v=6$\u2009 with which its contraharmonic mean is an integer.\n\nIn (2) and (3), the value of $v$ is a multiple of $u$, but it needs not be always so in to $c$ be an integer, e.g. we have\u2009 $u=12,\\;v=20,\\;c=17$.\n\nProposition 2.\u2009 For all\u2009 $u>1$, a necessary condition for $c\\in\\mathbb{Z}$\u2009 is that\n\n $\\gcd(u,\\,v)>1.$\n\nProof.\u2009 Suppose that we have positive integers $u,\\,v$ such that\u2009 $\\gcd(u,\\,v)=1$.\u2009 Then as well,\u2009 $\\gcd(u\\!+\\!v,\\,uv)=1$,\u2009 since otherwise both $u\\!+\\!v$ and $uv$ would be divisible by a prime $p$, and thus also one of the factors (http:\/\/planetmath.org\/Product) $u$ and $v$ of $uv$ would be divisible by $p$; then however\u2009 $p\\mid u\\!+\\!v$ would imply that\u2009 $p\\mid u$\u2009 and\u2009 $p\\mid v$, whence we would have\u2009 $\\gcd(u,\\,v)\\geqq p$.\u2009 Consequently, we must have\u2009 $\\gcd(u\\!+\\!v,\\,uv)=1$.\n\nWe make the additional supposition that $\\displaystyle\\frac{u^{2}\\!+\\!v^{2}}{u\\!+\\!v}$ is an integer, i.e. that\n\n $u^{2}\\!+\\!v^{2}=(u\\!+\\!v)^{2}\\!-\\!2uv$\n\nis divisible by $u\\!+\\!v$.\u2009 Therefore also $2uv$ is divisible by this sum.\u2009 But because\u2009 $\\gcd(u\\!+\\!v,\\,uv)=1$, the factor 2 must be divisible by $u\\!+\\!v$, which is at least 2.\u2009 Thus\u2009 $u=v=1$.\n\nThe conclusion is, that only the \u201cmost trivial case\u201d\u2009 $u=v=1$\u2009 allows that\u2009 $\\gcd(u,\\,v)=1$.\u2009 This settles the proof.\n\nProposition 3.\u2009 If $u$ is an odd prime number, then (2) and (3) are the only possibilities enabling integer contraharmonic means.\n\nProof.\u2009 Let $u$ be a positive odd prime.\u2009 The values\u2009 $v=(u\\!-\\!1)u$\u2009 and\u2009 $v=(2u\\!-\\!1)u$\u2009 do always.\u2009 As for other possible values of $v$, according to the Proposition 2, they must be multiples of the prime number $u$:\n\n $v=nu,\\quad n\\in\\mathbb{Z}$\n\nNow\n\n $\\mathbb{Z}\\ni\\frac{u^{2}\\!+\\!v^{2}}{u\\!+\\!v}\\;=\\;\\frac{(n^{2}\\!+\\!1)u}{n\\!+\\!1},$\n\nand since $u$ is prime, either\u2009 $u\\mid n\\!+\\!1$\u2009 or\u2009 $n\\!+\\!1\\mid n^{2}\\!+\\!1$.\n\nIn the former case\u2009 $n+1=ku$,\u2009 one obtains\n\n $c=\\frac{(n^{2}\\!+\\!1)u}{n\\!+\\!1}\\;=\\;\\frac{(k^{2}u^{2}\\!-\\!2ku\\!+\\!2)u}{ku}\\;=% \\;ku^{2}\\!-\\!2\\!+\\!\\frac{2}{k},$\n\nwhich is an integer only for\u2009 $k=1$\u2009 and\u2009 $k=2$, corresponding (2) and (3).\n\nIn the latter case, there must be a prime number $p$ dividing both $n\\!+\\!1$ and $n^{2}\\!+\\!1$, whence\u2009 $p\\nmid n$.\u2009 The equation\n\n $n^{2}\\!+\\!1\\;=\\;(n\\!+\\!1)^{2}\\!-\\!2n$\n\nthen implies that\u2009 $p\\mid 2n$.\u2009 So we must have\u2009 $p\\mid 2$,\u2009 i.e. necessarily\u2009 $p=2$.\u2009 Moreover, if we had\u2009 $4\\mid n\\!+\\!1$\u2009 and\u2009 $4\\mid n^{2}\\!+\\!1$,\u2009 then we could write\u2009 $n\\!+\\!1=4m$,\u2009 and thus\n\n $n^{2}\\!+\\!1\\;=\\;(4m\\!-\\!1)^{2}\\!+\\!1\\;=\\;16m^{2}\\!-\\!8m\\!+\\!2\\not\\equiv 0\\pmod% {4},$\n\nwhich is impossible.\u2009 We infer, that now\u2009 $\\gcd(n\\!+\\!1,\\,n^{2}\\!+\\!1)=2$,\u2009 and in any case\n\n $\\gcd(n\\!+\\!1,\\,n^{2}\\!+\\!1)\\;\\leqq\\;2.$\n\nNevertheless, since\u2009 $n\\!+\\!1\\geqq 3$\u2009 and\u2009 $n\\!+\\!1\\mid n^{2}\\!+\\!1$,\u2009 we should have\u2009 $\\gcd(n\\!+\\!1,\\,n^{2}\\!+\\!1)\\geqq 3$.\u2009 The contradiction means that the latter case is not possible, and the Proposition 3 has been proved.\n\nProposition 4.\u2009 If\u2009 $(u_{1},\\,v,\\,c)$\u2009 is a nontrivial solution of (1) with\u2009 $u_{1},\u2009 then there is always another nontrivial solution\u2009 $(u_{2},\\,v,\\,c)$\u2009 with\u2009 $u_{2}.\u2009 On the contrary, if\u2009 $(u,\\,v_{1},\\,c)$\u2009 is a nontrivial solution of (1) with\u2009 $u,\u2009 there exists no different solution\u2009 $(u,\\,v_{2},\\,c)$.\n\nFor example, there are the solutions\u2009 $(2,\\,6,\\,5)$\u2009 and\u2009 $(3,\\,6,\\,5)$;\u2009 $(5,\\,20,\\,17)$\u2009 and\u2009 $(12,\\,20,\\,17)$.\n\nProof.\u2009 The Diophantine equation (1) may be written\n\n $\\displaystyle u^{2}\\!-\\!cu\\!+\\!(v^{2}\\!-\\!cv)\\;=\\;0,$ (4)\n\nwhence\n\n $\\displaystyle u\\;=\\;\\frac{c\\!\\pm\\!\\sqrt{c^{2}\\!+\\!4cv\\!-\\!4v^{2}}}{2},$ (5)\n\nand the discriminant of (4) must be nonnogative because of the existence of the real root (http:\/\/planetmath.org\/Equation) $u_{1}$.\u2009 But if it were zero, i.e. if the equation \u2009$c^{2}\\!+\\!4cv\\!-\\!4v^{2}=0$\u2009 were true, this would imply for $v$ the irrational value $\\frac{1}{2}(1\\!+\\!\\sqrt{2})c$.\u2009 Thus the discriminant must be positive, and then also the smaller root $u$ of (4) gotten with \u201c$-$\u201d in front of the square root is positive, since we can rewrite it\n\n $\\frac{c\\!-\\!\\sqrt{c^{2}\\!+\\!4cv\\!-\\!4v^{2}}}{2}\\;=\\;\\frac{c^{2}\\!-\\!(c^{2}\\!+% \\!4cv\\!-\\!4v^{2})}{2(c+\\sqrt{c^{2}\\!+\\!4cv\\!-\\!4v^{2}})}\\;=\\;\\frac{2(v\\!-\\!c)v% }{c\\!+\\!\\sqrt{c^{2}\\!+\\!4cv\\!-\\!4v^{2}}}$\n\nand the numerator is positive because\u2009 $v>c$.\u2009 Thus, when the discriminant of the equation (4) is positive, the equation has always two distinct positive roots $u$.\u2009 When one of the roots ($u_{1}$) is an integer, the other is an integer, too, because in the numerator of (5) the sum and the difference of two integers are simultaneously even.\u2009 It follows the existence of $u_{2}$, distinct from $u_{1}$.\n\nIf one solves (1) for $v$, the smaller root\n\n $\\frac{c\\!-\\!\\sqrt{c^{2}\\!+\\!4cu\\!-\\!4u^{2}}}{2}\\;=\\;\\frac{2(u\\!-\\!c)u}{c\\!+\\!% \\sqrt{c^{2}\\!+\\!4cu\\!-\\!4u^{2}}}$\n\nis negative.\u2009 Thus there cannot be any\u2009 $(u,\\,v_{2},\\,c)$.\n\nProposition 5.\u2009 When the contraharmonic mean of two different positive integers $u$ and $v$ is an integer, their sum is never squarefree.\n\nProof.\u2009 By Proposition 2 we have\n\n $\\gcd(u,\\,v)\\;=:\\;d\\;>\\;1.$\n\nDenote\n\n $u\\;=\\;u^{\\prime}d,\\quad v\\;=\\;v^{\\prime}d,$\n\nwhen\u2009 $\\gcd(u^{\\prime},\\,v^{\\prime})\\,=\\,1$.\u2009 Then\n\n $c\\;=\\;\\frac{(u^{\\prime\\,2}\\!+\\!v^{\\prime\\,2})d}{u^{\\prime}\\!+\\!v^{\\prime}},$\n\nwhence\n\n $\\displaystyle(u^{\\prime}\\!+\\!v^{\\prime})c\\;=\\;(u^{\\prime\\,2}\\!+\\!v^{\\prime\\,2}% )d\\;\\equiv\\;[(u^{\\prime}\\!+\\!v^{\\prime})^{2}\\!-\\!2u^{\\prime}v^{\\prime}]d.$ (6)\n\nIf $p$ is any odd prime factor of $u^{\\prime}\\!+\\!v^{\\prime}$, the last equation implies that\n\n $p\\nmid u^{\\prime},\\quad p\\nmid v^{\\prime},\\quad p\\nmid[\\;\\;],$\n\nand consequently\u2009 $p\\mid d$.\u2009 Thus we see that\n\n $p^{2}\\mid(u^{\\prime}\\!+\\!v^{\\prime})d\\;=\\;u\\!+\\!v.$\n\nThis means that the sum $u\\!+\\!v$ is not squarefree.\u2009 The same result is easily got also in the case that $u$ and $v$ both are even.\n\nNote 1.\u2009 Cf.\u2009 $u\\!+\\!v=c\\!+\\!b$\u2009 in $2^{\\circ}$ of the proof of this theorem (http:\/\/planetmath.org\/ContraharmonicMeansAndPythagoreanHypotenuses) and the Note 4 of http:\/\/planetmath.org\/node\/138this entry.\n\nProposition 6.\u2009 For each integer \u2009$u>0$\u2009 there are only a finite number of solutions\u2009 $(u,\\,v,\\,c)$\u2009 of the Diophantine equation (1).\u2009 The number does not exceed $u\\!-\\!1$.\n\nProof.\u2009 The expression of the contraharmonic mean in (1) may be edited as follows:\n\n $c\\;=\\;\\frac{(u\\!+\\!v)^{2}-2uv}{u\\!+\\!v}\\;=\\;u\\!+\\!v-\\frac{2u(u\\!+\\!v\\!-\\!u)}{u% \\!+\\!v}\\;=\\;v\\!-\\!u+\\frac{2u^{2}}{u\\!+\\!v}$\n\nIn to $c$ be an integer, the quotient\n\n $w\\;:=\\;\\frac{2u^{2}}{u\\!+\\!v}$\n\nmust be integer; rewriting this last equation as\n\n $\\displaystyle v\\;=\\;\\frac{2u^{2}}{w}\\!-\\!u$ (7)\n\nwe infer that $w$ has to be a http:\/\/planetmath.org\/node\/923divisor of $2u^{2}$ (apparently\u2009 $1\\leqq w\u2009 for getting values of $v$ greater than $u$).\u2009 The amount of such divisors is quite restricted, not more than $u\\!-\\!1$, and consequently there is only a finite number of suitable values of $v$.\n\nNote 2.\u2009 The equation (7) explains the result of Proposition 1 ($w=1$,\u2009 $w=2$).\u2009 As well, if $u$ is an odd prime number, then the only factors of $2u^{2}$ less than $u$ are 1 and 2, and for these the equation (7) gives the values\u2009 $v:=(2u\\!-\\!1)u$\u2009 and\u2009 $v:=(u\\!-\\!1)u$\u2009 which explains Proposition 3.\n\n## References\n\n\u2022 1 J. Pahikkala: \u201cOn contraharmonic mean and Pythagorean triples\u201d.\u2009 \u2013 Elemente der Mathematik 65:2 (2010).\n Title integer contraharmonic means Canonical name IntegerContraharmonicMeans Date of creation 2013-12-04 10:25:44 Last modified on 2013-12-04 10:25:44 Owner pahio (2872) Last modified by pahio (2872) Numerical id 45 Author pahio (2872) Entry type Topic Classification msc 11Z05 Classification msc 11D45 Classification msc 11D09 Classification msc 11A05 Synonym integer contraharmonic means of integers Related topic ComparisonOfPythagoreanMeans Related topic DivisibilityInRings Related topic Gcd Defines contraharmonic integer","date":"2018-11-18 19:21:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 204, \"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\": 0.9178080558776855, \"perplexity\": 1784.6833376639984}, \"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-2018-47\/segments\/1542039744561.78\/warc\/CC-MAIN-20181118180446-20181118202446-00539.warc.gz\"}"}
| null | null |
Kinangop Constituency is an electoral constituency in Kenya. It is one of five constituencies in Nyandarua County. The constituency has five wards, all of which elect Members of County Assembly (MCA's) for the Nyandarua County. The constituency was established for the 1988 elections.
See the constituency website for more information:- www.kinangopconstituency.com
It is not to be confused with Kinango Constituency located in Kwale County.
Main ethnic group: Kikukyu
Main economic activity: Agriculture (Dairy and crop farming)
Main crops: Irish potatoes, Cabbages, Carrots, Fruits, Kales, Wheat, Barley
Main dairy companies: KCC, Tuzo, Brookside, Kinangop dairies and other small dairy companies
Members of Parliament
Wards
References
External links
Constituencies in Nyandarua County
Constituencies in Central Province (Kenya)
1988 establishments in Kenya
Constituencies established in 1988
|
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"redpajama_set_name": "RedPajamaWikipedia"
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| 23
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Vote for the PrepXtra girls basketball player of the week for Feb. 19-24.
Webb's Haylee Luttrell voted PrepXtra girls basketball player of the week.
The Webb Senior eared 47 percent of the vote for her 16 points in the 40-19 win over CPA in a Division II-A state quarterfinal.
Iva Parton, of Pigeon Forge, came in second place with 41 percent of the vote for her 21 points in the 46-42 consolation victory over Northview Academy and 14 points in a region quarterfinal win over Fulton.
Alcoa's Madison McClurg finished in third place with nine percent of the vote for her 13 points in the Tornadoes dominating 53-18 win over Fulton for the district title.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,036
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\section{Hypergeometric Functions and Two Lemmas}
We first discuss hypergeometric functions, since we will use them
in some of our proofs.
The hypergeometric function ${}_2 F_1(a,b;c;z)$ is defined
for $|z| < 1$ and for $c$ not a non-positive integer by
\[
{}_2 F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n n!} z^n
\]
(see \cite{NIST:DLMF}, eq.~15.2.1), where
$(a)_n = a(a+1) \cdots (a+n-1) = \Gamma(a+n)/\Gamma(a)$.
Note that $(a)_0 = 1$.
(Note that if $c$ is a non-positive integer then not all terms in the
sum are defined, which is why such values of $c$ are excluded.)
A hypergeometric function may be analytically
continued to a single valued analytic function on $\mathbb{C}$ minus
the part of the real axis from $1$ to $\infty$. This analytic continuation
is called the principal branch of the hypergeometric function.
For $\Rp c > \Rp b > 0$, we have that
\begin{equation}\label{eq_euler_hypergeo_int}
{}_2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 x^{b-1} (1-x)^{c-b-1} (1-zx)^{-a} dx
\end{equation}
(see \cite{NIST:DLMF}, eq.~15.6.1).
Also, for $|\arg(1-z)|<\pi$ we have
\begin{equation}\label{eq_euler_hypergeo_transf}
{}_2F_1(a,b;c;z) = (1-z)^{c-a-b} {}_2F_1(c-a, c-b;c;z)
\end{equation}
(see \cite{NIST:DLMF}, eq.~15.8.1).
Kummer's quadratic transformation states that for $|z|<1$ we have
\begin{equation}\label{eq_gauss_hypergeo_transf}
{}_2F_1(a,b;2b;4z/(1+z)^2) = (1+z)^{2a} {}_2F_1(a,a+\frac{1}{2}-b;
b + \frac{1}{2}; z^2)
\end{equation}
(see \cite{NIST:DLMF}, eq.~15.8.21).
If $\Rp c > \Rp (a+b)$ then the power series defining
$_{2}F_{1}$ converges absolutely on the circle $|z|=1$
(see \cite{NIST:DLMF}, 15.2.(i)), and thus uniformly on $|z| \le 1$.
In addition, for $\Rp c > \Rp (a+b)$ the value of the hypergeometric
function at $1$ (that is, the sum of the hypergeometric series for
$z=1$) is given by
\begin{equation}\label{eq_hypergeo_value_1}
{}_2F_1(a,b;c;1) =
\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}
\end{equation}
(see \cite{NIST:DLMF}, eq.~15.4.20).
The following lemma is well known, although we do not know if anyone has
found the sharp constant before (see e.g.~\cite{Zhu_Ap}, Theorem 1.7).
\begin{lemma} \label{lemma:1minusrep}
Let $p>1$ and $0<r<1$. Then
\[
\begin{split}
\frac{1}{2\pi} \int_0^{2\pi} \frac{1}{|1-re^{i\theta}|^p} d\theta &=
(1-r^2)^{1-p} {}_2F_1\left(1-\frac{p}{2}, 1-\frac{p}{2};1;r^2\right)
\\ &\le
\frac{\Gamma(p-1)}{\Gamma(p/2)^2} (1-r^2)^{1-p}.
\end{split}
\]
Furthermore, the bound is sharp, in the sense that the integral in question,
divided by $(1-r^2)^{1-p}$, is always less than or equal to
$\tfrac{\Gamma(p-1)}{\Gamma(p/2)^2}$, but the quotient approaches
$\tfrac{\Gamma(p-1)}{\Gamma(p/2)^2}$ as $r \rightarrow 1$.
\end{lemma}
In the case $p=2$, the equality says that
\[
\frac{1}{2\pi} \int_0^{2\pi} \frac{1}{|1-re^{i\theta}|^2} d\theta =
(1-r^2)^{-1}.
\]
\begin{proof}
The integral in question is equal to
\[
\frac{1}{\pi} \int_0^\pi (1-2r\cos\theta + r^2)^{-p/2}\, d\theta.
\]
Making the substitution $x = (\cos \theta + 1)/2$, we see that the
integral is equal to
\[
\begin{split}
&\ \frac{1}{\pi} (1+r)^{-p} \int_0^1 \left(1 -
\frac{4r}{(1+r)^2}x\right)^{-p/2} x^{-1/2} (1-x)^{-1/2} dx
\\ &=
(1+r)^{-p} {}_2F_1(p/2, 1/2;1; 4r/(1+r)^2)
\end{split}
\]
by equation \eqref{eq_euler_hypergeo_int}. Now using equation
\eqref{eq_gauss_hypergeo_transf}, we see this is equal to
\[
{}_2F_1(p/2, p/2;1;r^2).
\]
Equation \eqref{eq_euler_hypergeo_transf} shows this is equal to
\[
(1-r^2)^{1-p} {}_2F_1(1-(p/2), 1-(p/2); 1; r^2).
\]
The bound now follows from equation \eqref{eq_hypergeo_value_1},
since the series representation shows that
$_2F_1(1-(p/2), 1-(p/2); 1; r^2)$ increases from $r=0$ to $r=1$.
The remark about $p=2$ is true because
$_2F_1(1, 1;1;x) =(1-x)^{-1}.$
\end{proof}
The following lemma is likely well known, at least without
the sharp constant,
although we do not know of a specific place where it appears in the
literature.
\begin{lemma}\label{yam1}
Suppose that $s < 1$ and $m+s > 1$ and that $k > -1$.
Let $0 \le x < 1$. Then
\[
\int_0^1 \frac{(1-y)^{-s}}{(1-xy)^m} y^k \, dy \le C_1(s,m,k) (1-x)^{1-s-m}
\]
where $C_1(s,m,k) < \infty$ is defined by
\[
C_1(s,m,k) = \frac{\Gamma(k+1)\Gamma(1-s)}{\Gamma(2+k-s)}
\max_{0 \le x \le 1} {}_2F_1( 2+k-s-m, 1-s; 2+k-s; x).
\]
Furthermore, the bound is sharp
in the sense that the integral in question,
divided by $(1-x)^{1-s-m}$, is always less than or equal to
$C_1(s,m,k)$, and furthermore $C_1(s,m,k)$ is the smallest
constant with this property.
\end{lemma}
\begin{proof}
By \eqref{eq_euler_hypergeo_int}, we have
\[
\int_0^1 \frac{(1-y)^{-s}}{(1-xy)^m} y^k \, dy =
\frac{\Gamma(k+1)\Gamma(1-s)}{\Gamma(2+k-s)} {}_2F_1(m,k+1;2+k-s;x)
\]
Now \eqref{eq_euler_hypergeo_transf} gives that this is equal to
\[
\frac{\Gamma(k+1)\Gamma(1-s)}{\Gamma(2+k-s)}
(1-x)^{1-s-m} {}_2F_1(2+k-s-m,1-s;2+k-s;x).
\]
Now since $s+m-1 > 0$, the function
${}_2F_1(2+k-s-m,1-s;2+k-s;z)$
converges uniformly on
$|z| \le 1$, and so
${}_2F_1(2+k-s-m,1-s;2+k-s;x)$ is bounded for $|x| \le 1$.
Thus,
the above displayed expression is less than or equal to
\( C_1(s,m,k) (1-x)^{1-s-m}.\)
\end{proof}
Note that if $2+k > s+m$ and $2+k > s$, then the hypergeometric function
${}_2F_1(2+k-s-m, 1-s; 2+k-s; x)$ is increasing on $[0,1)$, and
so the maximum in the bound occurs at $x=1$. By
\eqref{eq_hypergeo_value_1},
$C_1(s,m,k)$ becomes
\[
\frac{\Gamma(2+k-s) \Gamma(s+m-1) \Gamma(k+1) \Gamma(1-s)}
{\Gamma(1+k) \Gamma(m) \Gamma(2+k-s)}
=
\frac{\Gamma(s+m-1) \Gamma(1-s)}
{\Gamma(m)}.
\]
\section{Bounds on Integral Means of Bergman Projections}
As discussed above, we make the following definition.
\begin{definition} Let $f \in L^p(\mathbb{D})$ for $0 < p \le \infty$. Define
\[
M_p(r,f) = \left\{ \frac{1}{2\pi} \int_0^{2\pi} |f(re^{i\theta})|^p d\theta
\right\}^{1/p}
\]
for $0 < p < \infty$ and
\[ M_p(r,f) = \esssup_{0 \le \theta < 2\pi} |f(re^{i\theta})| \]
for $p = \infty$.
\end{definition}
Note that for $f \in L^p(\mathbb{D})$,
the integral means $M_p(r,f)$ are defined for almost
every $r$ such that $0 < r < 1$, and in fact the
function $M_p(\cdot, f)$ is in
$L^p(r \, dr)$ on $[0,1)$. For
$0 < p < \infty$ this follows immediately from Fubini's theorem
(see \cite{Rudin_Big}, Theorem 7.12),
and for $p = \infty$ it may be proved either directly with the aid
of Fubini's theorem, or by noting that
$M_\infty(r,f) = \lim_{p \rightarrow \infty} M_p(r,f)$.
We say that a function
$f \in L^p(\partial \mathbb{D})$ is
in the Sobolev space
$W^{k,p}(\partial \mathbb{D})$ if
for every $1 \leq n \leq k$
and every function $g \in C^\infty(\partial \mathbb{D})$,
there is a function $h_n$ such that
$\int_0^{2\pi} f(e^{i\theta}) \tfrac{d^n}{d\theta^n} g(e^{i\theta}) \, d\theta
= (-1)^{n}
\int_0^{2\pi} h_n(e^{i\theta}) g(e^{i\theta}) \, d\theta,$
and furthermore $f$ and each $h_n$ are in $L^p(\partial \mathbb{D}).$
Then $h_n$ is unique (see \cite[Chapter 5]{Evans}), so
we denote $h_n$ by $\tfrac{d^n}{d\theta^n} f$.
It is well known that if $f \in W^{k,p}$
for $1 < p < \infty$ then $f \in C^{k-1}$(see for example
\cite[Section 5.6]{Evans}). In fact, since the dimension here is $1$,
this assertion is not difficult to show directly, and also follows
for $p=1$.
We next define an auxiliary operator which we will use to help
bound the Bergman projection.
\begin{definition}
Let $f \in L^1(\partial \mathbb{D}).$ Define
\[
\mathcal{P}_r^{(n)}(f)(\theta) = \frac{(n+1)!}{2\pi} \int_0^{2\pi}
\frac{f(e^{i\phi})r^n e^{-in\phi} }{(1-re^{i(\theta-\phi)})^{2+n}} d\phi.
\]
\end{definition}
We now have the following theorem, which gives a bound on the
$L^p$ norm of $\mathcal{P}_r^{(n)}(f)$.
\begin{theorem}\label{thm:Prn_bound}
Let $1 \le p \le \infty$, and let $k$ be an integer such that $0 \le k \le n$.
Assume that $f$ is in the Sobolev space $W^{k,p}(\partial \mathbb{D})$.
Then
\[
\|\mathcal{P}_r^{(n)}(f)\|_p \le
\frac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}
r^{n-k} (1-r^2)^{k-n-1} \left\| \frac{d^k}{d\theta^k}
\left[ f(e^{i\theta}) e^{-in\theta}\right]\right\|_p,
\]
where $\|\cdot\|_p$ denotes the $L^p(\partial \mathbb{D})$ norm.
\end{theorem}
\begin{proof}
First assume that $p < \infty$.
Performing integration by parts $k$ times gives
\[
\begin{split}
\mathcal{P}_r^{(n)}(f)(\theta) &= \frac{(n+1)!}{2\pi} \int_0^{2\pi}
\frac{f(e^{i\phi}) r^n e^{-in\phi}}{(1-re^{i(\theta-\phi)})^{2+n}} d\phi
\\ &=
r^{n-k} e^{-ik\theta} \frac{(n-k + 1)!}{2\pi i^k} \int_0^{2\pi}
\frac{\frac{d^k}{d\theta^k}[f(e^{i\phi})
e^{-in\phi}]}{(1-re^{i(\theta-\phi)})^{2+n-k}} d\phi.
\end{split}
\]
This is legitimate since $f$ is in $W^{k,p}$, and thus all its derivatives
except possibly the $k^{th}$ are continuous.
We have also used the fact that both
$f(e^{i\phi})$ and $(1-re^{i(\theta - \phi)})^{-1}$ are periodic
in $\phi$ with period $2\pi$.
The above displayed equation, Lemma \ref{lemma:1minusrep} and
H\"{o}lder's inequality immediately gives the case $p = \infty$.
If $p < \infty$,
let $m=n+2-k$, and let
$g(e^{i\theta}) = {\frac{d^k}{d\phi^k}[f(e^{i\phi}) e^{-in\phi}]}$.
Note that
\[
(r^{n-k} (n-k+1)!)^{-p} \|\mathcal{P}_r^{(n)}(f)\|_p^p \le
\int_0^{2\pi} \left| \int_0^{2\pi}
\frac{|g(e^{i\phi})|}{|1-re^{i(\theta - \phi)}|^m} \,
\frac{d\phi}{2\pi}
\right|^p \frac{d\theta }{2\pi}.
\]
But the right hand side of the above inequality equals
\[
\int_0^{2\pi} \left| \int_0^{2\pi}
\frac{|g(e^{i\phi})|}{|1-re^{i(\theta - \phi)}|^{m/p}}
\frac{1}{|1-re^{i(\theta - \phi)}|^{m/q}}
\, \frac{d\phi}{2\pi}
\right|^p \frac{d\theta }{2\pi},
\]
where $q$ is the conjugate exponent to $p$.
By H\"{o}lder's inequality, this is less than or equal to
\[
\int_0^{2\pi} \int_0^{2\pi}
\frac{|g(e^{i\phi})|^p}{|1-re^{i(\theta - \phi)}|^m} \, \frac{d\phi}{2\pi}
\left( \int_0^{2\pi}\frac{1}{|1-re^{i(\theta - \phi)}|^{m}}
\, \frac{d\phi}{2\pi}
\right)^{p/q} \frac{d\theta}{2\pi}.
\]
And by Lemma \ref{lemma:1minusrep}, this is at most
\[
\left(\frac{\Gamma(m-1)}{\Gamma(m/2)^2}\right)^{p-1}
\int_0^{2\pi} \int_0^{2\pi}
\frac{|g(e^{i\phi})|^p}{|1-re^{i(\theta - \phi)}|^m} \,
\frac{d\phi}{2\pi}
(1-r^2)^{(1-m)(p-1)}
\frac{d\theta}{2\pi},
\]
where we have used the fact that $p/q = p-1$.
Now Tonelli's theorem shows that this equals
\[
\begin{split}
& \phantom{={}} \left(\frac{\Gamma(m-1)}{\Gamma(m/2)^2}\right)^{p-1}
(1-r^2)^{(1-m)(p-1)}
\int_0^{2\pi} |g(e^{i\phi})|^p
\int_0^{2\pi} \frac{1}{|1-re^{i(\theta - \phi)}|^m}
\frac{d\theta}{2\pi} \frac{d\phi}{2\pi}
\\
&\le
\left(\frac{\Gamma(m-1)}{\Gamma(m/2)^2}\right)^{p-1}
\frac{\Gamma(m-1)}{\Gamma(m/2)^2}
(1-r^2)^{(1-m)(p-1)}
(1-r^2)^{1-m} \int_0^{2\pi} |g(e^{i\phi})|^p \frac{d\phi}{2\pi}
\\
&=
\left(\frac{\Gamma(m-1)}{\Gamma(m/2)^2}\right)^p
(1-r^2)^{(1-m)p}
\|g\|_p^p,
\end{split}
\]
where we have again applied Lemma \ref{lemma:1minusrep}.
This proves the result for $p < \infty$.
(Note that in the case $p = 1$, the above proof still works and really
only involves Lemma \ref{lemma:1minusrep} and Tonelli's theorem, but
not H\"{o}lder's inequality.)
\end{proof}
For $f \in L^1(\mathbb{D})$, recall that the Bergman projection of $f$ is
defined by
\[
\mathcal{P} f(z) = \frac{1}{\pi}
\int_{\mathbb{D}} \frac{f(w)}{(1-\conj{w}z)^{2}} dA(w),
\]
and thus
\begin{equation}\label{eq:bp_deriv}
\frac{d^n}{dz^n} (\mathcal{P} f)(z) = \frac{(n+1)!}{\pi}
\int_{\mathbb{D}} \frac{f(w)\conj{w}^n}{(1-\conj{w}z)^{2+n}} dA(w).
\end{equation}
Therefore, if $z=re^{i\theta}$, we have that
\[
\begin{split}
\frac{d^n}{dz^n}(\mathcal{P} f)(z) &= \frac{(n+1)!}{\pi}\int_0^1 \rho \int_0^{2\pi}
\frac{f(\rho e^{i\phi})\rho^n e^{-in\phi}}{(1-r \rho e^{i(\theta -
\phi)})^{2+n}}\, d\phi \, d\rho
\\
&= \frac{(n+1)!}{\pi}
\int_0^1 \rho \int_0^{2\pi}
\frac{f_\rho(e^{i\phi}) \rho^n e^{-in\phi}}
{(1-r \rho e^{i(\theta -
\phi)})^{2+n}}\, d\phi \, d\rho,
\\
&=
2
\int_0^1 \rho r^{-n} \mathcal{P}_{r\rho}^{(n)}f_\rho (e^{i\theta})\, d\rho,
\end{split}
\]
where $f_\rho(e^{i\theta}) = f(\rho e^{i\theta})$.
\begin{theorem}\label{thm:BPn_bound}
Let $1 \le p \le \infty$, and
let $k$ and $n$ be integers such that $0 \le k \le n$.
Suppose that $f \in L^1(\mathbb{D})$, and that the restriction of $f$ to
almost every circle of radius less than $1$
centered at the origin is in $W^{k,p}$.
Then the following inequality holds:
\[
\begin{split}
M_p \left(r, \frac{d^n}{dz^n}(\mathcal{P} f(z))\right ) &\le
2 \frac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}
\quad \times \\
&\ r^{-k} \int_0^1 \rho^{n+1-k}
M_p\left( \frac{d^k}{d\theta^k} (e^{-in\theta} f),
\rho \right) (1-r^2\rho^2)^{k-n-1}
d\rho.
\end{split}
\]
\end{theorem}
We make the following remark about this theorem: since
\begin{equation}\label{eq:simple_deriv_inequality}
\left| \frac{d^k}{d\theta^k} (e^{-in\theta} f) \right|
\le
\sum_{j=0}^k \binom{k}{j} n^{k-j} \left| \frac{d^j}{d\theta^j} f \right|,
\end{equation}
it is not hard to use the above theorem to bound $M_p(r, (\mathcal{P} f)^{(n)})$
strictly in terms of the integral means of the first $k$ derivatives of
$f$ in the $\theta$ direction.
\begin{proof}
Again, first assume that $p < \infty$. We have that
\[
\begin{split}
M_p \left(r, \frac{d^n}{dz^n}(\mathcal{P} f(z))\right )
= \left( \frac{1}{2\pi} \int_0^{2\pi} \left|
2 \int_0^1 \rho r^{-n} \mathcal{P}_{r\rho}^{(n)}f_\rho (e^{i\theta}) d\rho
\right|^p d\theta
\right)^{1/p}.
\end{split}
\]
By Minkowski's inequality, this is less than or equal to
\[
2 \int_0^1 \left( \frac{1}{2\pi} \int_0^{2\pi}
\rho^p r^{-pn} |\mathcal{P}_{r\rho}^{(n)}f_\rho (e^{i\theta})|^p
d\theta \right)^{1/p} d\rho.
\]
By Theorem \ref{thm:Prn_bound}, this is less than or equal to
\[
2 \frac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}
r^{-k} \int_0^1 \rho^{n+1-k} \left\|
\frac{d^k}{d\theta^k} (e^{-in\theta} f_\rho) \right\|_p
(1-r^2\rho^2)^{k-n-1}
d\rho
\]
which equals
\[
\begin{split}
2 \frac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}
r^{-k} \int_0^1 \rho^{n+1-k}
M_p\left( \frac{d^k}{d\theta^k} (e^{-in\theta} f), \rho
\right) (1-r^2\rho^2)^{k-n-1}
d\rho.
\\
\end{split}
\]
The proof is slightly easier in the case $p = \infty$, as we
do not need Minkowski's inequality.
Alternately, to see that the theorem still holds for $p = \infty$,
we can take the
limit in the bound as $p = \infty$, using the monotone convergence
theorem and the fact that $M_p(r,f)$ increases with $p$.
\end{proof}
We now discuss Lipschitz and Lebesgue-Lipschitz classes, since they are
relevant to some corollaries which we are about to prove.
A function $f$ is said to be Lipschitz of order $\alpha$
for $0 < \alpha \le 1$ if there is
some constant $A$ such that $|f(x) - f(y)| \le A|x-y|^\alpha$ for all
$x$ and $y$ in its domain.
The class of all such functions is denoted by $\Lambda_\alpha$.
For a function $f$ defined on the unit circle, we define its
integral modulus of continuity of order $p$ for $p < \infty$ by
\[
\omega_p(t,f) = \sup_{0 < h \le t}
\left[ \frac{1}{2\pi} \int_0^{2\pi}
|f(x+h) - f(x)|^p \right]^{1/p}.
\]
If $\omega_p(t,f) = O(t^\alpha)$ for some $\alpha$ such that
$0 < \alpha \le 1$, we say that $f$ belongs to the
Lebesgue-Lipschitz class $\Lambda_{\alpha,p}$.
For $p = \infty$, we define $\Lambda_{\alpha,\infty} = \Lambda_\alpha$.
We will need
Theorem 5.4 in
\cite{D_Hp}, which states that an analytic function is in
$H^p$ for $1 \le p < \infty$
and has boundary values in $\Lambda_{\alpha}^p$ if and only if
the integral means of its derivative satisfy
$M_p(r,f') = O((1-r)^{-1+\alpha})$.
We will also use Theorem 5.1 from the same reference, which says that
an analytic function is in
$H^\infty$ and has boundary values in
$\Lambda_{\alpha,\infty}$ if and only if
the integral means of its derivative
satisfy $M_\infty(r,f') = O((1-r)^{-1+\alpha})$.
(As stated, the theorem has the function being continuous in
$\overline{\mathbb{D}}$ in place of its being in $H^\infty$, but any
analytic function in $H^\infty$ with Lipschitz (or even continuous)
boundary values is continuous in $\overline{\mathbb{D}}$.)
\begin{theorem}\label{thm:bp-lipschitz}
Let $1 \le p \le \infty$ and $0 < \alpha < 1$ and $n \geq 1$.
Suppose that $f$ is measurable in $\mathbb{D}$ and that the restriction of $f$ to
almost every circle of radius less than $1$ centered at the origin is
in $W^{n,p}$.
Suppose that
$M_p \left( \frac{d^{n}}{d\theta^{n}} f, r \right) = O( (1-r)^{-1+\alpha})$
Then $\mathcal{P}(f)^{(n-1)} \in H^p$, and in fact the boundary values of
$\mathcal{P}(f)^{(n-1)}$ are
in the Lebesgue-Lipschitz space $\Lambda_{\alpha,p}$ .
\end{theorem}
\begin{proof}
Note that the assumptions imply that
$M_p(r, \frac{d^n}{d\theta^n}(e^{-i\theta}f)) = O((1-r)^{-1+\alpha})$,
and that $f \in L^1(\mathbb{D})$.
By the above theorem,
\[
\begin{split}
M_p(r, \mathcal{P}(f)^{(n)}) &\le C \int_0^1 (1-\rho)^{-1+\alpha} (1-r^2 \rho^2)^{-1}
\rho \, d \rho \\
&\le C \int_0^1 (1-\rho)^{-1+\alpha} (1-r \rho)^{-1} \rho \, d \rho
\end{split}
\]
for $r$ near
enough to $1$, where $C$ is a constant.
By Lemma \ref{yam1}, the above expression is less than or equal to
$C (1-r)^{\alpha}$ for $r$ near enough to $1$, where
$C$ is another constant.
But this implies that
$\mathcal{P}(f)^{(n-1)} \in H^p$ and has boundary values in $\Lambda_{\alpha}^p$.
\end{proof}
We now state a corollary related to our original motivation for
studying this problem. If we are given an $f \in A^p$ with
unit norm,
where $1 < p < \infty$ and $p$ has conjugate exponent $q$, then
there is a function $k \in A^q$ (unique up to a positive scalar multiple)
such that $f$ solves the extremal problem of maximizing
$\Rp \int_{\mathbb{D}} g \overline{k} \, d\sigma$ among all functions $g$
of unit $A^p$ norm.
The broad question that first motivated our study was:
if we know that $f$ has certain regularity, can
we say anything about regularity properties for $k$?
The next corollary is an example of this.
\begin{corollary}
Let $2 \le p < \infty$,
let $p-1 \le s \le \infty$,
let $q$ be the conjugate exponent to $p$, and let $0 < \alpha < 1$.
Let $f$ be analytic and suppose that
$f \in H^s$ with boundary values in
$\Lambda_{\alpha}^s$, and that
$\|f\|_{A^p}=1$.
Let $k$ a function in $A^q$ such that $f$ solves the
extremal problem of finding a function $g$ of unit $A^p$ norm
maximizing $\Rp \int_{\mathbb{D}} g \overline{k} \, d\sigma$.
Then $k \in H^{s/(p-1)}$ and the boundary values of $k$ are
in $\Lambda_{\alpha, s/(p-1)}$.
\end{corollary}
\begin{proof}
By the above mentioned Theorem 5.4 from \cite{D_Hp}, we have that
$M_s(r,f') = O((1-r)^{-1+\alpha})$.
Now, if we write $f = u + iv$, we have
\[
\frac{\partial}{\partial \theta} (|f|^{p-2} f) =
(p-2)|f|^{p-4}[u u_\theta + v v_\theta]f + |f|^{p-2} f_\theta.
\]
The absolute value of the above expression is bounded by
\(
(p-1)|f|^{p-2} |f'|,
\)
where we have used the
fact that $f_{\theta} = iz f'$ and the
Cauchy-Schwarz inequality applied to
$\langle u_\theta, v_\theta \rangle$ and $\langle u, v \rangle.$
Thus we have
\[
M_{s/(p-1)}\left(r,\frac{\partial}{\partial \theta} (|f|^{p-2} f)\right)
\le (p - 1) \|f\|_{H^s}^{p-2} M_s(r,f')
\le C (1-r)^{-1 + \alpha},
\]
where in the first inequality we have used
H\"{o}lder's inequality, and in the second we have used the hypothesis
about the growth of the integral means of $f'$.
Also, it is clear that $M_{s/(p-1)}(r, |f|^{p-2} f)$ is bounded.
By Theorem \ref{thm:bp-lipschitz}, this implies that
$\mathcal{P}(|f|^{p-2} f) \in H^{s/(p-1)}$ and that
$\mathcal{P}(|f|^{p-2} f)$ has boundary values
in $\Lambda^{s/(p-1)}_{\alpha}$.
But since $\mathcal{P}(|f|^{p-2} f)$ is a constant multiple of $k$, the
corollary holds.
\end{proof}
The next corollary is an analogous result for higher regularity.
\begin{corollary} Let $n \geq 0$ be an integer, and suppose that
$n + 1 \leq p < \infty$ and that $p-1 \leq s \leq \infty$.
Also suppose that $f^{(n-1)} \in \Lambda^s_\alpha$, where
$0 < \alpha < 1$, and that
$\|f\|_{A^p}=1$.
Let $k$ a function in $A^q$ such that $f$ solves the
extremal problem of finding a function $g$ of unit $A^p$ norm
maximizing $\Rp \int_{\mathbb{D}} g \overline{k} \, d\sigma$.
Then $k^{(n-1)} \in H^{s/(p-1)}$ and the boundary values of $k^{(n-1)}$ are
in $\Lambda_{\alpha, s/(p-1)}$.
\end{corollary}
\begin{proof}
We have that $\partial^n/\partial \theta^n (f^{p/2} \overline{f}^{(p/2)-1})$
equals
\[
\sum_{k=0}^n \binom{n}{k} \partial_\theta^k (f^{p/2})
\partial_\theta^{n-k} (\overline{f})^{p/2}.
\]
Now each term of $\partial_\theta^k (f^{p/2})$ is of the form
$C f^{\alpha} g_1 g_2 \cdots g_m$ where $C$ is a constant,
each $g_m$ is some $\theta$
derivative of $f$ of order at most $n$, and $\alpha + m = p/2$. Also, only
one $g_j$ can be a $k^{\text{th}}$ derivative.
Each term of $\partial_\theta^k (\overline{f}^{p/2-1})$ is of the form
$C \overline{f}^{\alpha} g_1 g_2 \cdots g_m$ where each $g_j$ is some
derivative of $\overline{f}$ of order at most $n-k$, and $\alpha + m = p/2$.
Also, only
one $g_j$ can be a $(n-k)^{\text{th}}$ derivative.
Thus, each term of
$\partial^n/\partial \theta^n (f^{p/2} \overline{f}^{(p/2)-1})$
is of the form
$C f^{\alpha} \overline{f}^\beta g_1 g_2 \cdots g_m$, where
$\alpha + \beta + m = p - 1$ and each $g_j$ is a $\theta$ derivative of
either $f$ or $\overline{f}$ of order at most $n$.
Note that because $p \geq n + 1$, we have that
$\alpha + \beta$ is nonnegative.
Then $f^\alpha \conj{f}^{\beta} \in L^{s/(\alpha + \beta)}$ and each $g_j$
is in $H^s$, unless $g_j$ is an $n^{\text{th}}$ derivative. Thus, all terms
(except for the exceptional ones with an $n^{\text{th}}$ derivative)
are in $H^{s/(p-1)}$, by H\"{o}lder's inequality.
Also, since $M_p(r,\partial^n_\theta f) = O((1-r)^{-1+\alpha})$,
H\"{o}lder's inequality again shows that
the terms with an $n^{\text{th}}$ derivative
have integral means of order $s/(p-1)$ that are
$O((1-r)^{-1+\alpha})$. The result now follows as in the above corollary.
\end{proof}
These corollaries are similar to
Theorem 4.3 in \cite{tjf2}, which is proved by very different methods.
That theorem is only proved
for $p$ an even integer. It requires us assume that
$f \in H^s$ for $p-1 < s < \infty$,
and yields that $k \in H^{s/(p-1)}$.
Whether Theorem 4.3 from
\cite{tjf2} holds when $p$ is not an even integer is still an
open question.
\section{Bounds on Sobolev norms of Bergman Projections}
We now illustrate how our previous results can be used to
bound certain weighted $L^p$ norms of derivatives of
Bergman projections by other weighted $L^p$ norms of
$\theta$ derivatives of the original function.
We will need the following lemma.
\begin{lemma}\label{yam2}
Suppose $1<p<\infty$ and that $j,k > -1$ and $m > 0$
and $u < 1$, and that
$u > 1 - mp$.
Set $w = u + (m-1)p$.
For a measurable function $f$ define
\[
g(x) = \int_0^1 \frac{|f(y)|}{(1-xy)^m} y^k \, dy,
\]
where we allow $g(x)$ to take on $\infty$ as a value.
Then
\[\|g\|_{L^p(x^j(1-x)^{-u}\, dx)} \le
C_2 \|f\|_{L^p(x^k(1-x)^{-w}\, dx)},\]
where $L^p$ spaces in the bound
are on the interval $[0,1]$, and where
\[
\begin{split}
&C_2 = C_2(p,m,k,j,u) = \\
&\inf_{\substack{
0 < b < m\\ \text{$a$ satisfies all of
\eqref{eq:4aconditions}}}}
C_1(aq, (m-b)q, k)^{1/q} C_1(ap+(m-b)p+u-(p/q), bp, j)^{1/p}.
\end{split}
\]
\end{lemma}
\begin{proof}
Let $q$ be the conjugate exponent to $p$.
Choose $b$ so that $0 < b < m$.
First note that the above conditions imply that
\begin{subequations}
\begin{align}
1-m-\frac{u}{p} & < \frac{1}{q}\\
1-\frac{u}{p}-m & < 1 - \frac{u}{p} - (m - b)\\
\frac{1}{q}-(m-b) & < \frac{1}{q} \\
\frac{1}{q} - (m-b) &< 1 - \frac{u}{p} - (m-b).
\end{align}
\end{subequations}
Thus we can find a number $a$ satisfying
\begin{subequations}\label{eq:4aconditions}
\begin{align}
1-m-\frac{u}{p} < & a \label{eq:4aconditions-agtnob}\\
\frac{1}{q}-(m-b) < & a \label{eq:4aconditions-agtb}\\
& a < \frac{1}{q} \label{eq:4aconditions-altnob}\\
& a < 1 - \frac{u}{p} - m + b.\label{eq:4aconditions-altb}
\end{align}
\end{subequations}
We may assume without loss of generality that $f \ge 0$, since if
the inequality holds for $|f|$ it holds for $f$. Now
\[
\begin{split}
&\phantom{={}}\int_0^1 \frac{f(y)}{(1-xy)^m}y^k\, dy
\\
&=
\int_0^1 \frac{f(y)(1-y)^{a}}{(1-xy)^{b}}
\frac{(1-y)^{-a}}{(1-xy)^{m-b}} y^k dy
\\ &\le
\left[ \int_0^1 \frac{ |f(y)|^p (1-y)^{ap}}{(1-xy)^{bp}} y^k dy
\right]^{1/p} \times
\left[ \int_0^1 \frac{(1-y)^{-aq}}{(1-xy)^{(m-b)q}} y^k dy\right]^{1/q},
\end{split}
\]
by H\"{o}lder's inequality.
But by Lemma \ref{yam1}, the above expression is less than or equal to
\[
C_{1,1}^{1/q}
(1-x)^{(1/q)-a-(m-b)}
\left[ \int_0^1 \frac{ |f(y)|^p (1-y)^{ap}}{(1-xy)^{bp}} y^k dy
\right]^{1/p},
\]
where $C_{1,1} = C_1(aq, (m-b)q, k)$.
This is valid because $aq + (m-b)q > 1$ and $aq < 1$, which follow from
inequalities \eqref{eq:4aconditions-agtb} and \eqref{eq:4aconditions-altnob}.
So then
\[
\begin{split}
&\phantom{={}} \|g\|^p_{L^p(x^j (1-x)^{-u}\, dx)}
\\& =
\int_0^1 \left| \int_0^1 \frac{f(y)}{(1-xy)^m}y^k\, dy\right|^p
(1-x)^{-u} x^j \, dx \\&\le
C_{1,1}^{p/q}
\int_0^1 (1-x)^{(p/q)-ap-(m-b)p}
\int_0^1 \frac{|f(y)|^p (1-y)^{ap}}{(1-xy)^{bp}} y^k \, dy \,
(1-x)^{-u} x^j \, dx\\
&=
C_{1,1}^{p/q}
\int_0^1 |f(y)|^p (1-y)^{ap} \int_0^1
\frac{(1-x)^{(p/q)-ap-(m-b)p-u}}{(1-xy)^{bp}} \,x^j \, dx \, y^k \,dy,
\end{split}
\]
by Tonelli's theorem for nonnegative functions.
Applying the previous lemma again
we see that this is less than or
equal to
\[
\begin{split}
&\phantom{={}}
C_{1,1}^{p/q} C_{1,2}
\int_0^1 |f(y)|^p (1-y)^{ap} (1-y)^{1+(p/q)-ap-(m-b)p-u-bp} y^k \, dy \\
&= C_{1,1}^{p/q} C_{1,2}
\int_0^1 |f(y)|^p (1-y)^{-w} y^k \, dy \\
&= C_{1,1}^{p/q} C_{1,2} \|f\|^p_{L^p(x^k(1-x)^{-w}\, dx)}
\end{split}
\]
where $C_{1,2} = C_1(ap+(m-b)p+u-(p/q), bp, j)$
This works because $u + (m-b)p + ap - \tfrac{p}{q} < 1$ and
$u + mp + ap - \tfrac{p}{q} > 1$,
which follow from
inequalities \eqref{eq:4aconditions-altb} and \eqref{eq:4aconditions-agtnob},
respectively.
\end{proof}
Note that the proof works even if $g$ is equal to $\infty$ for
some $x$, since
H{\"o}lder's inequality holds even if the left or right sides are infinite,
and Tonelli's theorem holds even if some of the integrals involved
are infinite.
One important case is when $j=k=m=1$ and $u=0$.
In this case we can
choose $a=1/(pq)$ and $b=1/p$, and then we see that
$
C_2(p,1,1,1,0) \le C_1(1/p, 1, 1)^{1/q} C_1(1/q, 1, 1)^{1/p}.$
(We have tried to find a choice of $a$ and $b$ yielding a better
bound on $C_2$, but were not able).
But by the remarks after Lemma \ref{yam1}, this is equal to
\[
\left[ \frac{\Gamma(1/p) \Gamma(1/q)}{\Gamma(1)} \right]^{1/q}
\left[ \frac{\Gamma(1/q) \Gamma(1/p)}{\Gamma(1)} \right]^{1/p}
= \Gamma\left(\frac{1}{p}\right) \Gamma\left(1 - \frac{1}{p} \right)
= \frac{\pi}{\sin(\pi / p)}
\]
by the reflection formula for the $\Gamma$ function.
It is interesting to note that the bounds in the following theorem do
not depend on $p$.
\begin{theorem}\label{yam3}
Let $1 \le p \le \infty$ and $1<s<\infty.$
Suppose that $0 \le k \le n$, where $n$ and $k$ are integers. Also
suppose that $j -k > -1$ and $1-(n+1-k)s < u < 1$ and set $w = u + (n-k)s$.
Also suppose that
the restriction of $f$ to
almost every circle of radius less than $1$ centered at the origin
is in $W^{k,p}$, and
that $f$ is in $L^1(\mathbb{D})$.
Then
\[
\begin{split}
&\phantom{={}}
\left\{ \int_0^1 [M_p(r, (\mathcal{P} f)^{(n)})]^s \, (1-r)^{-u} r^j \,
dr \right\}^{1/s}
\\ &\le
C_3(s,n-k,j-k,u)
\left\{
\int_0^{1} \left[M_p \left(\frac{d^k}{d\theta^k}(e^{-in\theta}f),r
\right)\right]^s
(1-r)^{-w} r^{n-k+1} \, dr
\right\}^{1/s}
\end{split}
\]
where
\[
C_3(s,n-k,j-k,u) = \frac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}
C_2(s,n-k+1,n-k+1,j-k,u).
\]
\end{theorem}
\begin{proof}
Define
\[
g(r) = \int_0^1 \rho^{n+1-k} M_p\left( \frac{d^k}{d\theta^k}
(e^{-in\theta}f),\rho \right)
(1-\rho r)^{k-n-1} d\rho.
\]
Then by Theorem \ref{thm:BPn_bound} and the fact that
$(1-\rho^2 r^2)^{k-n-1} \le (1 - \rho r)^{k-n-1}$ we have
\[
M_p(r, (\mathcal{P} f)^{(n)}) \le
C r^{-k} g(r),
\]
where $C = 2 \tfrac{\Gamma(n+1-k)\Gamma(n+2-k)}{\Gamma((n+2-k)/2)^2}.$
But by Lemma \ref{yam2},
\[
\begin{split}
&\phantom{={}} \left( \int_0^1 |g(r)|^s \, (1-r)^{-u} \, r^{j-k} dr \right)^{1/s}
\\ &\le C_2(s,n-k+1,n-k+1,j-k,u) \quad \times \\
&\qquad \qquad \left[
\int_0^{1} \left[
M_p \left(\frac{d^k}{d\theta^k}(e^{-in\theta}f),r \right)
\right]^s (1-r)^{-w} r^{n-k+1} \, dr
\right]^{1/s}.
\end{split}
\]
\end{proof}
By using Equation \eqref{eq:simple_deriv_inequality}, it is not
hard to modify the bound in the theorem so that it only involves the
integral means of the first $k$ derivatives of $f$ in the
$\theta$ direction.
Note that if we take $1 < s < \infty$ and $n=k$ and $j=1+n$ and
$u=0$, then by the remarks after Lemma \ref{yam2} we see that
$C_3(s,0,1,0) \le 2\pi/\sin(\pi s)$.
If we also take $p=s$ and note that $r \, dr \, d\theta = dA$,
we have the following corollary.
\begin{corollary}
For $1 < p < \infty$ and $n \ge 0$, if $f \in L^1(\mathbb{D})$ and
$f$ is in $W^{n,p}$ when restricted to almost every circle of radius less than
$1$ centered at the origin, we have
\[
\| \mathcal{P}(f)^{(n)} \|_{L^p(r^n dA)} \le
2\frac{\pi}{\sin(\pi/p)} \left\| \frac{d^n}{d\theta^n}
(e^{-in\theta}f) \right\|_{L^p(dA)}.
\]
\end{corollary}
Here is another corollary, which follows from taking
$p=s$, replacing $n$ with $n + k$ where $n, k \ge 0$,
and letting
$j=1 + k$,
and $u=0$.
\begin{corollary}
For $1 < p < \infty$ and integers $n,k \ge 0$, if
$f \in L^1(\mathbb{D})$ and
$f$ is in $W^{k,p}$ when restricted to almost every circle of radius less than
$1$ centered at the origin we have
\[
\| \mathcal{P}(f)^{(n + k)} \|_{L^p(r^k dA)} \le
C_3(p,n,1,0) \left\| \frac{d^k}{d\theta^k}(e^{-i(n+k)\theta}f)
\right\|_{L^p(r^n (1-r)^{-np} \, dA)}.
\]
\end{corollary}
If $k = 0$, the right hand side above simplifies to
\[
C_3(p,n,1,0) \| f \|_{L^p(r^n (1-r)^{-np} \, dA)}.
\]
Now, if we take $b = 1/p$ and $a= -n + 1/(pq)$ in the definition of
$C_2$, we see that
\[
\begin{split}
&\phantom{={}} C_3(p,n,1,0) \\
& \le
2 \frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(1 + (n/2))} \ \times \\
&\phantom{={}} C_1(aq, (n+1-b)q, n+1)^{1/q}
C_1(ap + (n+1-b)p - (p-1), bp, 1)^{1/p} \\ &=
2 \frac{\Gamma(n+1) \Gamma(n+2) \Gamma(1/p)^{1/q} \Gamma(nq+(1/q))^{1/q}
\Gamma(1/q)^{1/p} \Gamma(1/p)^{1/p}}
{\Gamma(1 + (n/2))\Gamma(nq+1)^{1/q} \Gamma(1)^{1/p}}
\\ &=
2 \frac{ \Gamma(n+1) \Gamma(n+2)
\Gamma(1/p)\Gamma(1/q)^{1/p} \Gamma(nq + (1/q))^{1/q}}
{\Gamma(1 + (n/2)) \Gamma(nq+1)^{1/q}}.
\end{split}
\]
\section{A Counterexample}
We now give an example of a function $f$ such that $M_2(r,f)$ is
bounded but $\mathcal{P} f$ is not in $H^2$.
In fact, the function in our example can be chosen so that
$f \in C^{\infty}(\mathbb{D})$ and so that
$M_\infty (r, f) \rightarrow 0$ as $r \rightarrow 1^{-}$.
This function is called a ``counterexample'' since it is a counterexample
to the conjecture that $\mathcal{P}(f) \in H^2$ if $M_2(r,f)$ is bounded.
Note that for the special case where $f$ is of the form
$|g|^{p-2}g$ where $g$ is an analytic function, then
$\mathcal{P}(f) \in H^2$ if $M_2(r,f)$ is bounded (see the introduction).
We first derive some general formulas for the Bergman projection of a
function.
Suppose that $f \in L^2(\mathbb{D})$.
Note that for almost every $r$ in $[0,1]$, $f$ restricted to the circle
of radius $r$ has a Fourier series
since it is in $L^2([0,2\pi))$ for almost every $r$.
Thus we can write
\[
f(re^{i\theta}) = \sum_{n=-\infty}^\infty a_n(r) e^{in\theta},
\]
where for a.e.~$r$ convergence holds in $L^2(0, 2\pi)$.
Here
\[
a_n(r) = \frac{1}{2\pi} \int_0^{2\pi} f(re^{i\theta}) e^{-in\theta}\, d\theta.
\]
Note that the functions $a_n(r)$ are measurable by Fubini's theorem.
Also, by Fubini's theorem
\[
\int_{\mathbb{D}} |f(z)|^2 \, \frac{dA(z)}{2\pi} =
\frac{1}{\pi}
\int_0^1 \int_{0}^{2\pi} |f(r e^{i\theta})|^2 \, d\theta \, r\, dr.
\]
But since, for almost every fixed $r$, the Fourier series in $\theta$
of $f(re^{i\theta})$ converges in $L^2(0, 2\pi)$, we have
\begin{equation}\label{eq:disc_fourier_norm}
\int_{\mathbb{D}} |f(z)|^2 \, \frac{dA(z)}{2\pi} =
\frac{1}{\pi}
\int_0^1 \sum_{n=0}^\infty |a_n(r)|^2 \, r \, dr.
\end{equation}
We now prove the following lemma relating what we have said to
calculating the Bergman projection of $f$.
\begin{lemma}
Suppose that $f \in L^2(\mathbb{D})$. Then we can write
\[
f(re^{i\theta}) = \sum_{n=-\infty}^\infty a_n(r) e^{in\theta}
\]
for a.e.~$r$,
where for a.e.~$r$, convergence holds in $L^2(0, 2\pi)$.
Also, the Bergman projection of $f$ is given by
\[
(\mathcal{P} f)(z) = \frac{1}{\pi} \sum_{n=0}^\infty
\left[ \int_0^1 (n+1)a_n(r) r^{n+1} dr \right]
z^n.
\]
\end{lemma}
\begin{proof}
Let $z=r e^{i\theta}$ and $w=\rho e^{i\phi}$. Note that
\[
\begin{split}
(\mathcal{P} f)(z) &= \int_{\mathbb{D}} \sum_{n=0}^\infty (n+1) z^n \conj{w}^n f(w)
\, d\sigma(w)
\\ &=
\frac{1}{\pi}
\int_0^1 \int_0^{2\pi} \sum_{n=0}^\infty \sum_{m=-\infty}^\infty
(n+1) a_m(\rho) e^{im\phi} r^{n} \rho^{n+1} e^{in\theta}e^{-in\phi}
\, d\phi \, d\rho
\\
\end{split}
\]
by Fubini's theorem. For fixed $z$ and $\rho$, the sum
$ \displaystyle \sum_{n=0}^\infty (n+1) z^n \conj{w}^n$
converges uniformly on $[0,2\pi]$, and thus for
fixed $z$ and
almost every fixed $\rho$, the sum
$\displaystyle
\sum_{n=0}^\infty (n+1) z^n \conj{w}^n f(w)$
converges in $L^2([0,2\pi])$.
Also, for almost every fixed $\rho$
the sum
$ \displaystyle \sum_{m=-\infty}^\infty a_m(\rho) e^{im\phi}$
converges in $L^2(0,2\pi)$.
Thus, we can
move the integral over $\phi$ inside the two summations to see that
\[
(\mathcal{P} f)(z)
= \frac{1}{\pi}
\int_0^1 \sum_{n=0}^\infty (n+1) a_n(\rho) \rho^{n+1} r^n e^{in\theta} \, d\rho.
\]
Now, we wish to apply the dominated convergence theorem to move the sum
outside the integral. To see that we can do this, note that
for each $\rho$ and each $N \ge 0$ we
have by the Cauchy-Schwarz inequality that
\[
\left| \sum_{n=0}^N (n+1) a_n(\rho) \rho^{n} r^n e^{in\theta} \right|
\le
\left(\sum_{n=0}^\infty |a_n(\rho)|^2 \right)^{1/2}
\left(\sum_{n=0}^\infty (n+1)^2 r^{2n} \rho^{2n} \right)^{1/2}.
\]
But the second sum can be bounded by $(1+r^2)/(1-r^2)^3$
independently of $\rho$
and the first sum is
integrable in with respect to the measure $\rho \, d \rho$ by
equation \eqref{eq:disc_fourier_norm}.
Thus we may apply the dominated convergence theorem to see
that
\[
(\mathcal{P} f)(z)
= \frac{1}{\pi}
\sum_{n=0}^\infty
\left[ \int_0^1 (n+1) a_n(\rho) \rho^{n+1} \, d\rho \right]
z^n.
\]
\end{proof}
Of course, this theorem shows that the formula for the Bergman
projection is valid if $f \in L^p$ for $p > 2$,
since $L^p$ is then a subset of $L^2$. The formula also holds for
any $f \in L^p$ for $p > 1$. This can be shown by using
the fact that the Fourier series of an $L^p$ function converges to
that function in $L^p$ for $1 < p < \infty$, by using
H\"{o}lder's inequality instead of the Cauchy-Schwarz inequality, and
by using Fubini's theorem and the Hausdorff-Young inequality to show that
$\sum_{n=0}^\infty |a_n(\rho)|^q$ is integrable with respect to
$\rho \, d \rho$. However, we will really only need the
formula to hold for bounded functions, since the
functions to which we need to apply the theorem will all be bounded.
To construct a function $f$ such that $M_\infty(r,f) \rightarrow 0$ as
$r \rightarrow 1^{-}$ and $\mathcal{P} f \not\in H^2$,
we will use the following lemma. Note that the constant $1/4$ in
the lemma is
not sharp and could be replaced any number strictly between $0$ and $1$.
\begin{lemma}
There is an increasing sequence
$0=b_0, b_1, b_2, \ldots \rightarrow 1$ and an increasing sequence
of non-negative integers $m_1, m_2, \ldots$ such that
\[
\int_{b_{n-1}}^{b_n} (m_n + 1) r^{m_n + 1} \, dr \ge \frac{1}{4}.
\]
\end{lemma}
\begin{proof}
We prove this by induction. Let $b_0=0, b_1=1/\sqrt{2}$, and $m_1 = 0$.
Then we have
\[
\int_{b_0}^{b_1} (m_1 + 1)r^{m_1+1} \, dr =
\int_0^{1/\sqrt{2}} r \, dr = \frac{1}{4}.
\]
Now, suppose we have found an increasing sequence of constants
$b_0, \ldots, b_n < 1$ and an increasing sequence of non-negative integers
$m_1, \ldots m_n$ satisfying the above condition.
Note that for each $k \ge 0$,
\[
\int_{b_n}^1 (k+1)r^{k+1} \, dr = \frac{k+1}{k+2}(1-b_n^{k+1}).
\]
Now, as $k \rightarrow \infty$, this approaches $1$, so there is some $k$
such that
\[
\int_{b_n}^1 (k+1)r^{k+1} \, dr = \frac{k+1}{k+2}(1-b_n^{k+1}) \ge \frac{1}{2}.
\]
We choose $m_{n+1}$ to be the smallest such $k$.
Now, the above inequality
implies that there is some constant $b$ such that
$b_n < b < 1$ and
\[
\int_{b_n}^{b} (k+1)r^{k+1} \, dr = \frac{1}{4}.
\]
We then choose $b_{n+1} = b$.
\end{proof}
We now have the following theorem in which we construct bounded functions
whose Bergman projections are not in $H^2$.
\begin{theorem}
Let the $b_n$ be defined as in the previous lemma.
Let $\{c_n\}_{n=1}^\infty$ be a bounded sequence such that
$\sum_{n} |c_n|^2 = \infty$. Define
\[
a_j(r) = c_j \chi_{[b_{j-1}, b_j)}(r)
\]
and
\[
f(re^{i\theta}) = \sum_{j=1}^\infty a_j(r) e^{im_j \theta}.
\]
Then $f$ is bounded but $\mathcal{P} f$ is not in $H^2$.
\end{theorem}
\begin{proof}
For each $r$ there is exactly one $j$ such that
$r \in [b_{j-1}, b_j)$, which implies that
$M_\infty(r,f) = c_j$, where $j$ is the number such that
$r \in [b_{j-1}, b_j)$. Note that this implies that
$f$ is bounded. Thus we have that
\[
(\mathcal{P} f)(z) =
\frac{1}{\pi}
\sum_{n=1}^\infty c_n \left[ \int_{b_{n-1}}^{b_n} (m_n + 1) r^{m_n + 1} \, dr
\right] z^{m_n}.
\]
But this means that the $m_n{}^{\text{th}}$ term in the Taylor
series of $\mathcal{P} f$ is at least $c_n/4$ in absolute value, so that the
Taylor coefficients of $\mathcal{P} f$ are not square summable.
\end{proof}
In the theorem, if we choose the sequence $\{c_n\}$ so that
it approaches $0$ (but is not square summable), then the function
$f$ defined in the statement of the theorem will approach $0$ uniformly
as $z$ approaches the boundary of the disc, but its Bergman projection
will not be in $H^2$. Thus, we have the following corollary.
\begin{corollary}
There is a bounded function $f(z)$ in $\mathbb{D}$ that approaches $0$ uniformly
as $|z| \rightarrow 1$, such that $\mathcal{P} f \not \in H^2$.
\end{corollary}
We note in passing that if we define
$a_j(r) = c_j \phi_j(r)$, where $\phi_j$ is a $C^\infty$
bump function with support in
$(b_{j-1}, b_j)$ that is equal to $1$ on a sufficiently large part of
$(b_{j-1}, b_j)$
then we can even construct $f$ so that it is in $C^\infty$ and
approaches $0$ uniformly as $|z| \rightarrow 1$.
\nocite{Olver:2010:NHMF}
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Clomipramine, Anafranil, is a tricyclic antidepressant used to treat depression and Obsessive-Compulsive Disorder, OCD. Usage, dosage, side effects.
Suicidality in Children and Adolescents — Antidepressants increased the risk of suicidal thinking and behavior (suicidality) in short-term studies in children and adolescents with major depressive disorder (MDD) and other psychiatric disorders. Anyone considering the use of Anafranil or any other antidepressant in a child or adolescent must balance this risk with the clinical need. Patients who are started on therapy should be observed closely for clinical worsening, suicidality, or unusual changes in behavior. Families and caregivers should be advised of the need for close observation and communication with the prescriber. Anafranil is not approved for use in pediatric patients. Pooled analyses of short-term (4 to 16 weeks) placebo-controlled trials of 9 antidepressant drugs (SSRIs and others) in children and adolescents with major depressive disorder (MDD), obsessive compulsive disorder (OCD), or other psychiatric disorders (a total of 24 trials involving over 4400 patients) have revealed a greater risk of adverse events representing suicidal thinking or behavior (suicidality) during the first few months of treatment in those receiving antidepressants. The average risk of such events in patients receiving antidepressants was 4%, twice the placebo risk of 2%. No suicides occurred in these trials.
Download the complete FDA warning concerning antidepressant use in children and adults.
Clomipramine hydrochloride (Anafranil) is a tricyclic antidepressant used to treat depression and obsessive-compulsive disorder (OCD). It may also be used to treat other conditions as determined by your doctor.
Clomipramine is indicated for the treatment of obsessions and compulsions in patients with Obsessive-Compulsive Disorder (OCD). The obsessions or compulsions must cause marked distress, be time-consuming, or significantly interfere with social or occupational functioning, in order to meet the DSM-III-R (circa 1989) diagnosis of OCD.
The effectiveness of clomipramine for long-term use (i.e. for more than 10 weeks) has not been systematically evaluated in placebo-controlled trials. The physician who elects to use clomipramine for extended periods should periodically re-evaluate the long term usefulness of the drug for the individual patient.
Clomipramine is contraindicated in patients with a history of hypersensitivity to clomipramine or other tricyclic antidepressants.
Clomipramine HCl should not be given in combination, or within 14 days before or after treatment, with a monoamine oxidase (MAO) inhibitor. Hyperpyretic crisis, seizures, coma, and death have been reported in patients receiving such combinations.
Clomipramine is contraindicated in patients with existing liver or kidney damage and should not be administered to patients with a history of blood dyscrasias.
Clomipramine is contraindicated in patients with glaucoma, as the condition may be aggravated due to the atropine-like effects of the drug.
Seizure was identified as the most significant risk of clomipramine use.
Caution should be used in administering clomipramine to patients with a history of seizures or other predisposing factors. e.g., brain damage of varying etiology, alcoholism, and concomitant use with other drugs that lower the seizure threshold.
Rare reports of fatalities in association with seizures have been reported by foreign post-marketing surveillance, but not in U.S. clinical trials. In some of these cases, clomipramine had been administered with other epileptogenic agents: in others, the patients involved had possibly predisposing medical conditions. Thus a causal association between clomipramine treatment and these fatalities has not been established.
Physicians should discuss with patients the risk of taking clomipramine HCl while engaging in activities in which sudden loss of consciousness could result in serious injury to the patient or others, e.g., the operation of complex machinery, driving, swimming, climbing.
Suicide: Since depression is a commonly associated feature of OCD, the risk of suicide must be considered. Prescriptions for clomipramine HCl should be written for the smallest quantity of capsules consistent with good patient management, in order to reduce the risk of overdose.
Cardiovascular: Tricyclic antidepressants, particularly in high doses, have been reported to produce sinus tachycardia, changes in conduction time and arrhythmias. A few instances of unexpected death have been reported in patients with cardiovascular disorders. Myocardial infarction and stroke have also been reported with drugs of this class. Therefore, clomipramine should be administered with extreme caution to patients with a history of cardiovascular disease, especially those who have a history of conduction disorders, those with circulatory lability and elderly patients. It also has a hypotensive action which may be detrimental in these circumstances. In such cases, treatment should be initiated at low doses with progressive increases only if required and tolerated, and the patients should be under close surveillance at all dosage levels. Monitoring of cardiac function and the ECG is indicated in such patients.
Psychosis, Confusion, And Other Neuropsychiatric Phenomena: Patients treated with clomipramine have been reported to show a variety of neuropsychiatric signs and symptoms including delusions, hallucinations, psychotic episodes, confusion, and paranoia. Because of the uncontrolled nature of many of the studies, its is impossible to provide a precise estimate of the extent of risk imposed by treatment with clomipramine. As with tricyclic antidepressants to which it is closely related, clomipramine may precipitate an acute psychotic episode in patients with unrecognized schizophrenia.
Mania/Hypomania: During premarketing testing of clomipramine in patients with affective disorder, hypomania or mania was precipitated in several patients. Activation of mania or hypomania has also been reported in a small proportion of patients with affective disorder treated with marketed tricyclic antidepressants, which are closely related to clomipramine.
Central Nervous System: More than 30 cases of hyperthermia have been recorded by nondomestic post-marketing surveillance systems. Most cases occurred when clomipramine was used in combination with other drugs. When clomipramine and a neuroleptic were used concomitantly, the cases were sometimes considered to be examples of a neuroleptic malignant syndrome.
Sexual Dysfunction: The rate of sexual dysfunction in male patients with OCD who were treated with clomipramine in the premarketing experience was markedly increased compared with placebo controls (i.e., 42% experienced ejaculatory failure and 20% experienced impotence, compared with 2.0% and 2.6% respectively, in the placebo group). Approximately 85% of males with sexual dysfunction chose to continue treatment.
Weight Changes: In controlled studies of OCD, weight gain was reported in 18% of patients receiving clomipramine, compared with 1% of patients receiving placebo. In these studies, 28% of patients receiving clomipramine had a weight gain of at least 7% of their initial body weight, compared with 4% of patients receiving placebo. Several patients had weight gains in excess of 25% of their initial body weight. Conversely, 5% of patients receiving placebo had weight losses of at least 7% of their initial body weight.
Surgery: Prior to elective surgery with general anesthetics, therapy with clomipramine HCl should be discontinued for as long as is clinically feasible, and the anesthetist should be advised.
The safety and effectiveness in children below the age of 10 have not been established. Therefore, specific recommendations cannot be made for the use of clomipramine in children under the age of 10.
No unusual age-related adverse events have been identified in this elderly population, but the data is insufficient to rule out possible age-related differences, particularly in elderly patients who have concomitant systemic illnesses or who are receiving other drugs concomitantly.
Patients should be warned that, while taking clomipramine, their responses to alcoholic beverages, other CNS depressants (e.g. barbiturates, benzodiazepines or general anesthetics) or anticholinergic agents (e.g. atropine, biperiden, levodopa) may be exaggerated. When tricyclic antidepressants are given in combinations with anticholinergics or neuroleptics with an anticholinergic action, hyperexcitation states or delirium may occur, as well as attacks of glaucoma. Tricyclic antidepressants should not be employed in combination with anti-arrhythmic agents of the quinidine type.
Clomipramine should not be used with MAO inhibitors.
Since clomipramine may diminish or abolish the antihypertensive effects of guanethidine, clonidine, reserpine, methyldopa, patients requiring concomitant treatment for hypertension should be given antihypertensives of a different type (e.g. diuretics, beta-blockers).
Clomipramine should be discontinued prior to elective surgery, for as long as clinically feasible, since little is known about the interaction between clomipramine and general anesthetics.
If administered concomitantly with estrogens, the dose of clomipramine should be reduced since steroid hormones inhibit the metabolism of clomipramine.
Because clomipramine is highly bound to serum proteins, the administration of clomipramine to patients taking other drugs that are highly bound to protein (i.e. warfarin, digoxin) may cause an increase in plasma concentrations of these drugs, potentially resulting in adverse effects. Conversely, adverse reactions may result from the displacement of protein bound clomipramine by other highly bound drugs.
The most commonly observed adverse events associated with the use of clomipramine HCl and not seen at an equivalent incidence among placebo-treated patients were gastrointestinal complaints, including dry mouth, constipation, nausea, dyspepsia, and anorexia; nervous system complaints, including somnolence, tremor, dizziness, nervousness, and myoclonus; genitourinary complaints, including changed libido, ejaculatory failure, impotence, and micturition disorder; and other miscellaneous complaints, including fatigue, sweating, increased appetite, weight gain, and visual changes.
The following list of adverse reactions have also been observed with clomipramine. These are listed in order of decreasing frequency.
Neurological: Extrapyramidal effects such as ataxia, also headache, delirium, speech disorders, muscle weakness, muscle hypertonia, tinnitus, paresthesias of the extremities, convulsions, EEG changes, hyperpyrexia. Peripheral neuropathy has been reported with other tricyclic antidepressants.
Cardiovascular: Hypotension, particularly orthostatic hypotension with associated vertigo, sinus tachycardia, palpitations. A quinidine-like effect and other reversible ECG changes in patients with normal cardiac status (such as flattening or inversion of T-waves, depressed S-T segments). Arrhythmias, hypertension, conduction disorders (e.g. widening of QRS complex, PQ changes, bundle-branch block), syncope.
Behavioral: Drowsiness, fatigue, restlessness, confusion accompanied by disorientation (particularly in geriatric patients and patients suffering from Parkinson's disease), anxiety states, agitation, sleep disturbances, insomnia, nightmares, aggravated depression, hypomania or manic episodes, disturbed concentration, visual hallucinations, impaired memory, aggressiveness, yawning, depersonalization, activation of latent psychosis, delusions.
Hematologic: Leukopenia, agranulocytosis, thrombocytopenia, eosinophilia and purpura. One case of pancytopenia has been reported.
Gastrointestinal: Vomiting, abdominal pain, diarrhea, taste perversion, elevated transaminases, obstructive jaundice, hepatitis with or without jaundice.
Endocrine: Weight loss, breast enlargement and galactorrhea in the female, inappropriate antidiuretic hormone (ADH) secretion syndrome, gynecomastia in the male, changes in blood sugar levels, increase in prolactin levels, menstrual irregularity.
Allergic: Allergic skin reactions (skin rash, urticaria), photosensitization, pruritus, edema, drug fever.
Since children may be more sensitive than adults to acute overdosage with tricyclic antidepressants, and since fatalities in children have been reported, effort should be made to avoid potential overdose particularly in this age group.
Signs and symptoms vary in severity depending upon factors such as the amount of drug absorbed, the age of the patient, and the time elapsed since drug ingestion. Blood and urine levels of clomipramine may not reflect the severity of poisoning: they have chiefly a qualitative rather than quantitative value, and they are unreliable indicators in the clinical management of the patient. The first signs and symptoms of poisoning with tricyclic antidepressants are generally severe anticholinergic reactions. CNS abnormalities may include drowsiness, stupor, coma, ataxia, restlessness, agitation, delirium, severe perspiration, hyperactive reflexes, muscle rigidity, athetoid and choreiform movement, and convulsions. Cardiac abnormalities may include arrhythmia, tachycardia, ECG evidence of impaired conduction, and signs of congestive heart failure, and in very rare cases, cardiac arrest. Respiratory depression, cyanosis, hypotension, shock, vomiting, hyperpyrexia, mydriasis, oliguria or anuria, and diaphoresis may also be present.
Patients in whom overdosage is suspected should be admitted to hospital without delay. No specific antidote is available and treatment is essentially symptomatic and supportive. Gastric lavage or aspiration should be performed promptly and is recommended up to 12 hours or even more after the overdose, since the anticholinergic effect of the drug may delay gastric emptying. Administration of activated charcoal may help to reduce absorption of the drug. As clomipramine is largely protein bound, forced diuresis, peritoneal dialysis and hemodialysis are unlikely to be of value.
In the alert patient, the stomach should be emptied promptly by lavage. In the obtunded patient, the airway should be secured with a cuffed endotracheal tube before beginning lavage (do not induce emesis). Instillation of activated charcoal slurry may help reduce absorption of CMI.
External stimulation should be minimized to reduce the tendency for convulsions. If anticonvulsants are necessary, diazepam and phenytoin may be useful. Adequate respiratory exchange should be maintained, including intubation and artificial respiration, if necessary. Respiratory stimulants should not be used.
In severe hypotension or shock, the patient should be placed in an appropriate position and given a plasma expander, and, if necessary, a vasopressor agent by intravenous drip. The use of corticosteroids in shock is controversial and may be contraindicated in case of overdosage with tricyclic antidepressants. Digitalis may increase conduction abnormalities and further irritate an already sensitized myocardium. If congestive heart failure necessitates rapid digitalization, particular care must be exercised. Hyperpyrexia should be controlled by whatever external means are available, including ice packs and cooling sponge baths, if necessary. Hemodialysis, peritoneal dialysis, exchange transfusions, and forced diuresis have generally been reported as ineffective because of the rapid fixation of clomipramine in tissues.
The slow intravenous administration of physostigmine salicylate has been used as a last resort to reverse severe CNS anticholinergic manifestations of overdosage with tricyclic antidepressants; however, it should not be used routinely, since it may induce seizures and cholinergic crises.
After you start taking this medicine, several weeks may pass before you feel the full benefit.
Do not miss any doses. If you miss a dose of this medicine, take it as soon as possible. If it is almost time for your next dose, skip the missed dose and go back to your regular dosing schedule. Do not take 2 doses at once. If you take 1 dose daily at bedtime, do not take the missed dose the next morning.
Additional Information:: If your symptoms do not improve after taking this medicine for 4 weeks, inform your doctor. Do not share this medicine with others for whom it was not prescribed. Do not use this medicine for other health conditions. Keep this medicine out of the reach of children.
Dosage should be individualized according to the requirements of each patient. Treatment should be initiated at the lowest recommended dose and increased gradually, noting carefully the clinical response and any evidence of intolerance. During the initial dose titration phase, the total daily dose of clomipramine should be divided and served with meals to reduce gastrointestinal side-effects.
Steady-state plasma levels may not be achieved until 2 to 3 weeks after a dosage adjustment. It may thus be advisable to wait 2 to 3 weeks after the initial dose titration phase, before attempting further dosage adjustments. It should be kept in mind that a lag in therapeutic response usually occurs at the onset of therapy, lasting from several days to a few weeks. Increasing the dosage does not normally shorten this latent period and may increase the incidence of side effects.
Adults-Initial Dosage: Clomipramine therapy should be initiated at daily doses of 25 mg. Dosage may be increased by 25 mg increments, as tolerated, at 3 to 4 day intervals up to a total daily dose of 150 mg by the end of 2 weeks. Thereafter, the dose may be gradually increased over a period of several weeks to 200 mg. Doses in excess of 200 mg daily are not recommended for outpatients. Occasionally, in more severely depressed hospitalized patients, dosages up to 300 mg daily may be required.
Children & Adolescents-Initial Dosage: As with adults, the starting dose is 25 mg daily and should be gradually increased (also given in divided doses with meals to reduce gastrointestinal side effects) during the first 2 weeks, as tolerated, up to a daily maximum of 3 mg/kg or 100 mg, whichever is smaller. Thereafter, the dosage may be increased gradually over the next several weeks up to a daily maximum of 3 mg/kg or 200 mg, whichever is smaller. As with adults, after titration, the total daily dose may be given once daily at bedtime to minimize daytime sedation.
Elderly and Debilitated Patients: In general, lower dosages are recommended for these patients. Initially, 20 to 30 mg daily in divided doses is suggested, with very gradual increments, depending on tolerance and response. Blood pressure and cardiac rhythm should be checked frequently, particularly in patients who have unstable cardiovascular function.
Maintenance/Continuation Treatment (Adults, Children, and Adolescents): While there are no systematic studies that answer the question of how long to continue clomipramine, OCD is a chronic condition and it is reasonable to consider continuation for a responding patient. Although the efficacy of clomipramine after 10 weeks has not been documented in controlled trials, patients have been continued in therapy under double-blind conditions for up to 1 year without loss of benefit. However, dosage adjustments should be made to maintain the patient on the lowest effective dosage, and patients should be periodically reassessed to determine the need for treatment. During maintenance, the total daily dose may be given once daily at bedtime.
Tablets:: Anafranil is available as capsules of 10, 25, 50, and 75 mg for oral administration.
The information in this monograph is not intended to cover all possible uses, directions, precautions, drug interactions or adverse effects. This information is generalized and is not intended as specific medical advice. If you have questions about the medicines you are taking or would like more information, check with your doctor, pharmacist, or nurse. Last updated 1/05.
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Petr Fifka (* 27. ledna 1967) je český politik, farmaceut a manažer, od října 2021 poslanec Poslanecké sněmovny PČR, v letech 2014 až 2022 zastupitel hlavního města Prahy, v letech 2002 až 2006 a opět v letech 2010 až 2014 zastupitel městské části Praha 4, člen ODS.
Život
Vystudoval Farmaceutickou fakultu Univerzity Karlovy v Hradci Králové (získal titul PharmDr.). Poté studoval management na Open University. Od roku 1992 pracuje na odborných a manažerských pozicích v mezinárodním farmaceutickém průmyslu.
Petr Fifka žije v Praze, konkrétně v části Praha 4. Je ženatý a má tři děti.
Politické působení
V roce 2000 se stal členem ODS. V komunálních volbách v roce 2002 byl za ODS zvolen zastupitelem městské části Praha 4. Ve volbách v roce 2006 nekandidoval. Znovu byl zvolen ve volbách v roce 2010. Ve volbách v roce 2014 se již o mandát zastupitele městské části neucházel.
V komunálních volbách v roce 2014 byl za ODS zvolen zastupitelem hlavního města Prahy. Ve volbách v roce 2018 mandát zastupitele hlavního města obhájil. Ve volbách v roce 2022 již nekandidoval.
Ve volbách do Poslanecké sněmovny PČR v roce 2013 kandidoval z 30. místa za ODS v Praze, ale neuspěl. Ve volbách do Poslanecké sněmovny PČR v roce 2021 kandidoval z pozice člena ODS na 10. místě kandidátky koalice SPOLU (tj. ODS, KDU-ČSL a TOP 09) v Praze. Získal 4 616 preferenčních hlasů a stal se poslancem.
Reference
Čeští farmaceuti
Čeští manažeři
Poslanci Parlamentu České republiky (2021–2025)
Členové Zastupitelstva hlavního města Prahy
Čeští místní politici
Členové ODS
Absolventi Farmaceutické fakulty Univerzity Karlovy
Narození v roce 1967
Narození 27. ledna
Žijící lidé
Muži
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,319
|
Q: Realtime progress from Windows service I'm working on an application which I think would be better suited to being a Windows service. The only issue I have come up against in planning to convert it is that it has a window which displays activity. Now I know it's bad practice (and deprecated in Win7) to show a GUI from a service so my question is, what is the best way to show progress?
My first idea was a polling method using RPC and MIDL but an operation could start and finish in a second or two it would be very inaccurate to using polling.
What are my other options for this?
Thanks,
J
EDIT: My question is more about the communication method, I plan to split it into the service and a task tray icon but I want to display a window with progress bars to show the progress of tasks running in the service but polling would be too slow unless it was sub-second which seems a waste, is there a way to push progress to the task tray app?
A: Create a service that does the work, and an optional GUI application that, if running, sits in the system tray and allows you to open a window watching the progress. You can communicate the progress over e.g. shared memory or named pipes - but remember that the service is running while noone is logged in, so it should in no way depend on being able to display progress.
Push mechanisms:
*
*Shared memory, a mutex and an event (CreateEvent). Use the PulseEvent call whenever the service updates state, have the systray app wait on the event (with e.g. MsgWaitForMultipleObjects)
*Named pipes (CreateNamedPipe) - These are also waitable for the systray app
*TCP/UDP - avoid this, you'll just run into overeager firewalls that don't understand the concept of localhost
A: Create an icon in the system tray. When your service is doing its thing, change the icon to something animated, and change the tooltip of the icon (when the user hovers the mouse) to display progress complete in the balloon text.
EDIT: On second thought, I'd probably not animate the icon. Just give it a different look. Make it red, or something.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,831
|
Q: Postgres how to know what are all the indexes going to be updated on update indexed json field Users table design is as follows,
id - integer
workspace_id - integer
data - json
created_at - timestamp
updated_at - timestamp
Stored data in json field as follows,
{"age": 21, "city": "London", "name": "ABC", "test_filed1": "text"}
I have created index like follows
CREATE INDEX concurrently index_user_name ON users using gist(workspace_id, (data ->> 'name') gist_trgm_ops)
Likewise I have few indexes for optimise the query. I want to know if I update any one field in json which is not indexed, will it update index in postgres as field is part of same json column? For instance age field is not indexed, will it update index_user_name index when I update age alone? Any way to run update query alone with any option to know what are indexes going to be updated?
A: The index will get maintained. There is no optimization to suppress maintenance when a value derived from the column doesn't change even though the column value itself changes. There was an attempt to add such an optimization a few years ago, but it did not succeed.
So it will not go through HOT update, and not only will that index get maintained, but so will all the other indexes on the table.
I don't know of a good way of discovering this fact without just knowing it, or digging into the internals (like with pg_waldump, or with pageinspect)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,417
|
DHCD Divisions
DHCD provides a variety of services to individuals and organizations interested in developing housing or businesses in District communities.
Rental Conversion and Sale Division - The Rental Conversion and Sale Division (CASD) administers the Rental Housing Conversion and Sale Act of 1980, as amended (D.C. Law 3-86) (the Conversion Act) and the Condominium Act of 1976 (D.C. Law 9-82) (the Condominium Act).
The Conversion Act regulates, among other things: tenant opportunity to purchase rights, tenant first rights of refusal, offer of sale notices, notices of transfer and the conversion of property to cooperatives or condominiums.
The Condominium Act regulates condominium formation and registration of condominium units before a developer may offer units to interested buyers. CASD also administers the Structural Defect Warranty Claim Program.
The Rental Accommodations Division - The Rental Accommodations Division (RAD) administers the Rental Housing Act of 1985 (D.C. Law 6-10), commonly known as "rent control." The Act's purposes are to: (1) protect low- and moderate-income tenants from erosion of their income from increased housing costs; (2) provide incentives for the construction of new rental units and rehabilitate vacant units; (3) continue to improve administrative mechanisms for resolving disputes between landlords and tenants; and (4) prevent the erosion of moderately priced rental housing while providing landlords with a reasonable rate of return on their investments. RAD regulates residential rental housing licensing and registration, rent adjustment procedures, rent adjustment petitions and adjudication, eviction control, informal mediation, compliance and enforcement, technical assistance to stakeholders, and community outreach.
RAD can be reached on 202-442-9505. Residents may also visit the DHCD Housing Resource Center located on the ground floor at 1800 Martin Luther King, Jr. Avenue, SE, Washington, DC 200020.
The DHCD Housing Resource Center is open Monday through Friday from 8:30 am – 3:30 pm. In addition, residents can access DCHousingSearch.org from the DHCD Housing Resource Center. DCHousingSearch.org is a FREE listing service that provides easy access to information about housing opportunities within the District of Columbia.
Development Finance Division - DHCD's Development Finance Division (DFD) provides funding for the development of rental, homeownership and community facility developments that serve District of Columbia neighborhoods. As both the creation and preservation of affordable housing units are important to DHCD, DFD plays a prominent role in helping the agency achieve its annual multifamily housing production goals.
Property Acquisition and Disposition Division - The Property Acquisition and Disposition Division (PADD) stabilizes neighborhoods by decreasing the number of vacant and abandoned residential properties in the District, and transforming vacant and/or abandoned residential properties into homeownership opportunities or District of Columbia residents at all income levels.
PADD has three main functions:
Encourage property owners to rehabilitate and/or occupy their vacant and abandoned residential property;
Acquire vacant, abandoned and deteriorated properties through negotiated friendly sale, eminent domain, donation or tax sale foreclosure when owners are unwilling or unable to maintain their properties; and
Dispose of properties in the PADD inventory by selling the properties to individuals or developers to be rehabilitated into high quality affordable and market-rate single-family and/ or multifamily for-sale housing in District neighborhoods.
PADD disposes of properties by three methods:
Solicitation of Offers;
Lottery; and
PADD, through the Homestead and Home Again Programs, has successfully acquired and disposed of hundreds of vacant, abandoned and deteriorated properties in the District and created affordable housing opportunities for District residents.
Residential and Community Services Division - The Residential and Community Services Division (RCSD) provides funding for programs focused on housing needs and neighborhood revitalization. RCSD works through Community Based Organizations (CBO) to provide comprehensive housing counseling services, small business technical assistance and façade improvement opportunities. RCSD administers the District's Home Purchase Assistance Program, Employer Assisted Housing Program and Negotiated Employee Affordable Home Purchase Program, which provide financial assistance for low and moderate-income households and District Government employees for the purpose of first-time home purchase. The Division also provides rehabilitation resources in the form of grants and loans that address health, safety and building code violations, to income eligible owner-occupant and rental units, in order to preserve homeownership.
Portfolio and Asset Management Division - The Portfolio and Asset Management Division (PMD) manages the allocation of Low Income Housing Tax Credits (LIHTC), which provides 9 percent tax credits to developers of new or rehabilitated rental housing for the production of housing affordable to low- and moderate-income persons at or below 60 percent of Median Family Income. It also provides portfolio management oversight to outstanding loans in the division. Established in FY 2008, the division monitors the status of existing loans to ensure compliance with loan covenants and collections of loans that are due and conducts the reviews of the risks and relationships of potential borrowers to protect the Department's assets.
Office of Program Monitoring - The Office of Program Monitoring (OPM) conducts oversight and reviews of DHCD projects and funding recipients. Its core functions include the following types of oversight: (1) contract compliance – completing various federally required compliance reviews as part of the underwriting and project development process; (2) quality assurance – monitoring the compliance of DHCD funded sub-recipients with federal HOME Investments Partnership Program (HOME) and Community Development Block Grant Program (CDBG) funding requirements; and (3) compliance monitoring – ensuring projects developed by DHCD through the Housing Production Trust Fund (HPTF), CDBG, HOME and LIHTC programs remain in compliance with federal and local program requirements throughout the duration of the projects period of affordability.
The compliance staff monitors compliance with these programs through a review of annual reports and periodic on-site visits to properties. These on-site visits consist of a file review and physical inspection of a percentage of the assisted units. As the monitoring entity for the IRS on the LIHTC Program and HUD on the HOME, CDBG and ESG Programs, DHCD reports directly to them on issues of non-compliance. OPM also monitors Environmental Review, Davis Bacon, Relocation, Fair Housing and Section 3 requirements as they relate to these programs.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,137
|
//=============================================================================
// System : Sandcastle Help File Builder Utilities
// File : SyntaxFilterInfo.cs
// Author : Eric Woodruff (Eric@EWoodruff.us)
// Updated : 11/10/2009
// Note : Copyright 2009, Eric Woodruff, All rights reserved
// Compiler: Microsoft Visual C#
//
// This file contains the class used to provide information about the loaded
// language syntax filter build components.
//
// This code is published under the Microsoft Public License (Ms-PL). A copy
// of the license should be distributed with the code. It can also be found
// at the project website: http://SHFB.CodePlex.com. This notice, the
// author's name, and all copyright notices must remain intact in all
// applications, documentation, and source files.
//
// Version Date Who Comments
// ============================================================================
// 1.8.0.3 11/10/2009 EFW Created the code
//=============================================================================
using System;
using System.Collections.ObjectModel;
using System.Globalization;
using System.Xml.XPath;
namespace SandcastleBuilder.Utils.BuildComponent
{
/// <summary>
/// This class contains information about the loaded language syntax filter
/// build components.
/// </summary>
public class SyntaxFilterInfo
{
#region Private data members
//=====================================================================
private string id, generatorNode, languageNode;
private int sortOrder;
private Collection<string> alternateNames;
#endregion
#region Properties
//=====================================================================
/// <summary>
/// This read-only property returns the ID of the syntax filter
/// </summary>
public string Id
{
get { return id; }
}
/// <summary>
/// This read-only property returns the sort order of the syntax filter
/// </summary>
public int SortOrder
{
get { return sortOrder; }
}
/// <summary>
/// This read-only property returns an optional list of alternate names
/// that will map to this syntax filter.
/// </summary>
public Collection<string> AlternateNames
{
get { return alternateNames; }
}
/// <summary>
/// This read-only property returns the XML to use for the generator
/// XML node in the build component configuration file.
/// </summary>
public string GeneratorXml
{
get { return generatorNode; }
}
/// <summary>
/// This read-only property returns the XML to use for the language
/// XML node in the build component configuration file.
/// </summary>
public string LanguageXml
{
get { return languageNode; }
}
#endregion
/// <summary>
/// Constructor
/// </summary>
/// <param name="component">The XPath navigator containing the syntax
/// filter component's configuration information</param>
internal SyntaxFilterInfo(XPathNavigator component)
{
XPathNavigator item;
string attrValue;
id = component.GetAttribute("id", String.Empty);
if(String.IsNullOrEmpty(id))
throw new InvalidOperationException("Syntax filter configuration contains no 'id' attribute");
attrValue = component.GetAttribute("sortOrder", String.Empty);
// Sort order is optional
if(String.IsNullOrEmpty(attrValue))
sortOrder = Int32.MaxValue;
else
sortOrder = Convert.ToInt32(attrValue, CultureInfo.InvariantCulture);
// Alternate names are optional
alternateNames = new Collection<string>();
attrValue = component.GetAttribute("alternateNames", String.Empty);
if(!String.IsNullOrEmpty(attrValue))
foreach(string n in attrValue.Split(new char[] { ',' }, StringSplitOptions.RemoveEmptyEntries))
{
attrValue = n.Trim();
if(attrValue.Length != 0)
alternateNames.Add(attrValue.ToLowerInvariant());
}
item = component.SelectSingleNode("generator");
if(item == null)
throw new InvalidOperationException("Syntax filter configuration contains no 'generator' node");
generatorNode = item.OuterXml;
item = component.SelectSingleNode("language");
if(item == null)
throw new InvalidOperationException("Syntax filter configuration contains no 'generator' node");
languageNode = item.OuterXml;
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,193
|
{"url":"http:\/\/mathhelpforum.com\/algebra\/92522-question-has-been-bugging-me-days.html","text":"# Math Help - This question has been bugging me for days!\n\n1. ## This question has been bugging me for days!\n\nMy question, as follows:\n\nIf f 64=25 sq KM how many times would 63 equal that #?\n\n2. do you mean function of 64 is 25 sqkm and how many sqkm would be function of 63?\n\n3. Originally Posted by MathPerson345\nMy question, as follows:\n\nIf f 64=25 sq KM how many times would 63 equal that #?\ndo you mean function of 64 is 25 sqkm and how many sqkm would be function of 63?\nIt appears as if MathPerson345 means this:\n\n$64 \\times f = 25 Sq.Km.$\n\nso\n\n$f = \\frac {25}{64}$\n\nand MathPerson345 wants\n\n$63 \\times g = 25$\n\nthe revised value is:\n\n63 x 0.386825396 = 25\n\n4. Originally Posted by MathPerson345\nMy question, as follows:\n\nIf f 64=25 sq KM how many times would 63 equal that #?","date":"2014-08-31 01:06:08","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\": 3, \"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\": 0.827254056930542, \"perplexity\": 3908.1158469761804}, \"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-2014-35\/segments\/1408500835844.19\/warc\/CC-MAIN-20140820021355-00107-ip-10-180-136-8.ec2.internal.warc.gz\"}"}
| null | null |
When something says equip next to it, it means it is already equipped.
If you are losing a fight, head back out into the world and grind to level up and get cash for better weapons.
Try and attack from the side or the back or you will get hit.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,045
|
Die Droga krajowa 72 (kurz DK72, pol. für ,Nationalstraße 72' bzw. ,Landesstraße 72') ist eine Landesstraße in Polen. Sie führt von Konin in östlicher Richtung über Turek, Poddębice, Łódź und Brzeziny bis Rawa Mazowiecka und verläuft weitestgehend parallel zur Autobahn A2. Die Gesamtlänge beträgt 189 km.
Geschichte
Nach der Neuordnung des Straßennetzes 1985 wurden einige Abschnitte des heutigen Straßenverlaufes in das Netz der neuen Landesstraßen aufgenommen. Der Abschnitt von Konin bis Uniejów wurde als Teil der Landesstraße 469 geführt. Das Teilstück von Uniejów bis Balin war Teil der damaligen Landesstraße 473. Der Abschnitt von Balin bis Łódź wurde der Landesstraße 709, der Abschnitt zwischen Łódź und Rawa Mazowiecka der Landesstraße 72 zugeordnet. Mit der Reform der Nummerierung vom 9. Mai 2001 wurden die Abschnitte zwischen Konin und Łódź Teile der Landesstraße 72, die in Richtung Westen bis Konin verlängert wurde.
Wichtige Ortschaften entlang der Strecke
Konin
Tuliszków
Turek
Uniejów
Poddębice
Aleksandrów Łódzki
Łódź
Brzeziny
Jeżów
Głuchów
Rawa Mazowiecka
Siehe auch
Liste der Landesstraßen in Polen
Weblinks
Website der GDDKiA (polnisch)
Website des Programmes Drogi Zaufania (polnisch)
Einzelnachweise
72
72
72
Droga krajowa 072
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,375
|
\section{Totally ramified case}
Let $K$ be a henselian discrete
valuation field.
Let $L$ be a totally ramified
Galois extension of $K$
and let $G={\rm Gal}(L/K)$
be the Galois group.
For a rational number $r>1$,
the upper ramification group
$G^r$ defined in \cite[Definition 3.4]{AS}
equals the subgroup
defined in \cite[Chapitre IV, Section 3]{CL}
denoted $G^{r-1}_{\rm cl}$,
by \cite[Proposition 3.7 (3)]{AS}.
Assume that $L$ is wildly ramified
and
let $r>1$ be the largest rational
number such that
the subgroup $G^r$
of the wild inertia subgroup
$P\subset G$
is non-trivial.
Let $E$ be the residue field
and $e=e_{L/K}$
be the ramification index.
We give a description
of the canonical injection
\begin{equation}
G^{r\vee}={\rm Hom}_{{\mathbf F}_p}
(G^r,{\mathbf F}_p)
\to
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1}
,E)
\label{eqchclE}
\end{equation}
for the ${\mathbf F}_p$-vector space
$G^r$,
in the case where $L$ is totally ramified
over $K$.
The injection (\ref{eqchclE})
is a special case of (\ref{eq1}).
We begin with a description
of extensions of vector spaces
over a field of characteristic $p>0$
by ${\mathbf F}_p$-vector
spaces.
\begin{lm}\label{lmapol}
Let $F$ be a field of characteristic $p>0$.
{\rm 1.}
Let $G\subset F$
be a finite subgroup of the additive group.
Then, the polynomial
\begin{equation}
a_1=\dfrac{\prod_{\sigma\in G}(X-\sigma)}
{\prod_{\sigma\in G,\,
\sigma\neq 0}(-\sigma)}
\label{eqa1}
\end{equation}
$\in F[X]$ is a unique additive
separable polynomial
such that the coefficient
of degree $1$ is $1$ and that
the sequence
\begin{equation}
\begin{CD}
0@>>> G
@>>>
{\mathbf G}_a
@>{a_1}>>
{\mathbf G}_a
@>>>0
\end{CD}
\label{eqapol}
\end{equation}
is exact.
{\rm 2. (\cite[Proposition 2.1.6
(2)$\Rightarrow$(3)]{red})}
Let $E$ be an $F$-vector space of finite
dimension and
let $0\to G\to H\to E\to 0$
be an extension of $E$ by
an ${\mathbf F}_p$-vector space $G$
of finite dimension, as smooth group
schemes over $F$.
Define a morphism
\begin{equation}
[H]\colon
G^\vee=
{\rm Hom}_{{\mathbf F}_p}
(G,{\mathbf F}_p)
\to
{\rm Ext}(E,{\mathbf F}_p)
=E^\vee
={\rm Hom}_F(E,F)
\label{eqH}
\end{equation}
by sending a character
$\chi\colon G\to {\mathbf F}_p$
to the linear form $f\colon E\to F$
such that there exists
a commutative diagram
$$\begin{CD}
0@>>> G@>>> H@>>> E@>>>0\\
@.@V{\chi}VV@VVV@VVfV@.\\
0@>>>{\mathbf F}_p
@>>> {\mathbf G}_a
@>{x^p-x}>> {\mathbf G}_a
@>>> 0.
\end{CD}$$
If $H$ is connected, then
the morphism
$[H]\colon
G^\vee
\to E^\vee$ is an injection.
\end{lm}
\proof{
1.
By \cite[Lemma 2.1.5]{red},
$a=\prod_{\sigma\in G}(X-\sigma)
\in F[X]$ is an additive
separable polynomial
such that (\ref{eqapol})
with $a_1$ replaced by $a$
is exact.
Since the coefficient in $a$
of degree $1$ is
$\prod_{\sigma\in G,\,
\sigma\neq 0}(-\sigma)$,
the assertion follows.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\medskip
Let $K$ be a henselian discrete
valuation field and
$L$ be a totally ramified
Galois extension
of degree $e$
of Galois group $G$.
Let $\alpha\in L$
be a uniformizer
and let $E=F$ denote the residue field.
The minimal polynomial
$f\in {\cal O}_K[X]$
is an Eisenstein polynomial
and the constant term $\pi=f(0)$
is a uniformizer of $K$.
We define a closed immersion
$T={\rm Spec}\, {\cal O}_L
\to Q={\rm Spec}\, {\cal O}_K[X]$
by sending $X$ to $\alpha$.
For a rational number $r>1$
such that $er\in{\mathbf Z}$,
define a dilatation
$$Q^{[r]}_T=
{\rm Spec}\, {\cal O}_L[X]
\Bigl[\dfrac f{\alpha^{er}}\Bigr]
\to Q_T=
{\rm Spec}\, {\cal O}_L[X].$$
The generator
$f$ of the kernel
$I={\rm Ker}({\cal O}_K[X]
\to {\cal O}_L)$
defines a basis over ${\cal O}_L$
of the conormal module
$N_{T/Q}=I/I^2$
and $\alpha^{er}$
defines a basis
of the $E$-vector space
${\mathfrak m}_L^{er}/
{\mathfrak m}_L^{er+1}$.
As subspaces of
$N_{E/Q}=J/J^2$
for $J={\rm Ker}({\cal O}_K[X]
\to E)=
(X, f)
=(X,\pi)$,
we have an equality
\begin{equation}
N_{T/Q}\otimes_{{\cal O}_L}E
={\mathfrak m}_K/
{\mathfrak m}_K^2
\label{eqNTQ}
\end{equation}
since $f$ is an Eisenstein polynomial.
The basis $f$ of
$N_{T/Q}\otimes_{{\cal O}_L}E$
corresponds to
the uniformizer
$\pi=f(0)
\in {\mathfrak m}_K/
{\mathfrak m}_K^2$.
By sending
$S$ to $f/\alpha^{er}$,
we define an isomorphism
${\cal O}_L[X,S]/
(f-\alpha^{er}S)
\to {\cal O}_L[X]
[f/\alpha^{er}]$.
Since $f$ is an Eisenstein polynomial,
the reduced closed fiber
$Q^{[r]}_E=
{\rm Spec}\, ({\cal O}_L[X]
[f/\alpha^{er}]
\otimes_{{\cal O}_L}E)_{\rm red}$
is identified with
${\rm Spec}\, E[S]$.
By this identification
and (\ref{eqNTQ}),
we define an isomorphism
\begin{align}
Q^{[r]}_E\to
{\rm Hom}_E({\mathfrak m}_L^{er}
/{\mathfrak m}_L^{er+1},
N_{T/Q}\otimes
_{{\cal O}_L}E)^\vee
\to&\,
{\rm Hom}_E({\mathfrak m}_L^{er}
/{\mathfrak m}_L^{er+1},
{\mathfrak m}_K/
{\mathfrak m}_K^2)^\vee
\nonumber
\\
&
=
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}
\label{eqQr}
\end{align}
of smooth group schemes of
dimension 1 over $E$.
Let $Q^{(r)}_T\to
Q^{[r]}_T$ be the normalization
and define a section
$T\to Q^{(r)}_T$
to be the unique lifting of
the section
$T\to Q_T$
defined by sending $X$ to $\alpha$.
Let $Q^{(r)}_E$
denote the reduced part
of the closed fiber
$Q^{(r)}_T\times_T{\rm Spec}\, E$
and let
$Q^{(r)\circ}_E\subset Q^{(r)}_E$
denote the connected
component containing
the image of the closed point
of $T$ by the section
$T\to Q^{(r)}_T$.
\begin{pr}\label{prcl}
Let $K$ be a henselian discrete
valuation field with residue field
$F$ of characteristic $p>0$.
Let $L$ be a totally ramified
Galois extension
of degree $n=e$
with residue field $E=F$
and let $G={\rm Gal}(L/K)$
be the Galois group.
Let $\alpha\in L$
be a uniformizer
and let $f\in {\cal O}_K[X]$
be the minimal polynomial.
Decompose
$f=\prod_{i=1}^n(X-\alpha_i)$
so that $\alpha_n=\alpha$
and ${\rm ord}_L(\alpha_i-\alpha_n)$
is increasing in $i$.
{\rm 1.}
Let $r>1$ be the largest rational number
such that $G^r\neq 1$.
Then, we have
\begin{equation}
er=
{\rm ord}_Lf'(\alpha)
+{\rm ord}_L(\alpha_{n-1}-\alpha_n).
\label{eqr}
\end{equation}
Define an injection
$\beta\colon G^r\to {\mathbf G}_a$
by $\beta(\sigma)
\equiv\dfrac{\sigma(\alpha)-\alpha}
{\alpha_{n-1}-\alpha_n}
\bmod {\mathfrak m}_L$
and
an additive polynomial
$b_1\in E[X]$
by $b_1=
\prod_{\sigma\in G^r}
(X-\beta(\sigma))
/\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))$.
Define an isomorphism
${\mathbf G}_a\to
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}$ by
$f'(\alpha)
(\alpha_{n-1}-\alpha_n)/f(0)
\in {\mathfrak m}_L^{e(r-1)}$
and identify
$Q^{[r]}_E$ with
${\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}$
by the isomorphism
{\rm (\ref{eqQr})}.
Then, there exists an isomorphism
\begin{equation}
\begin{CD}
0@>>>
G^r@>>>
Q^{(r)\circ}_E
@>>>
Q^{[r]}_E
@>>>
0
\\
@.@|@AAA@AA{f'(\alpha)
(\alpha_{n-1}-\alpha_n)/f(0)}A@.\\
0@>>>
G^r@>\beta>>
{\mathbf G}_a@>{b_1}>>
{\mathbf G}_a@>>>0
\end{CD}
\label{eqGrcl}
\end{equation}
of exact sequences.
{\rm 2.}
Let $i>0$ be
the largest integer such that
$G_{i,{\rm cl}}={\rm Ker}(G\to
{\rm Aut}({\cal O}_L/{\mathfrak m}_L^{i+1}))
\neq 1$.
Then, we have
\begin{equation}
i=
{\rm ord}_L(\alpha_{n-1}-\alpha_n)-1.
\label{eqi}
\end{equation}
Let $K\subset M\subset L$
be the intermediate extension
corresponding to $G_{i,{\rm cl}}\subset G$
and let $U^i_L=1+{\mathfrak m}^i_L
\subset L^\times$
and $U^i_M=1+{\mathfrak m}^i_M
\subset M^\times$ be the
multiplicative subgroups.
Let
$N^i\colon U^i_L/U^{i+1}_L
\to
U^i_M/U^{i+1}_M$
denote the morphism induced by the norm
$N_{L/M}\colon L^\times \to M^\times$
and
$T^i\colon U^i_L/U^{i+1}_L
={\mathfrak m}^i_L/{\mathfrak m}^{i+1}_L
\to
U^i_M/U^{i+1}_M
={\mathfrak m}^i_M/{\mathfrak m}^{i+1}_M$
be the isomorphism
induced by the trace
${\rm Tr}_{L/M}\colon L\to M$.
Define an isomorphism
${\mathbf G}_a\to
U^i_L/U^{i+1}_L$
by sending $1$
to the class of
$\alpha_{n-1}/\alpha_n\in U^i_L$.
Then, the diagram
\begin{equation}
\begin{CD}
0@>>>
G_{i,{\rm cl}}@>{\sigma
\mapsto \sigma(\alpha)/\alpha}>>
U^i_L/U^{i+1}_L
@>{(T^i)^{-1}\circ N^i}>>
U^i_L/U^{i+1}_L
@>>>
0
\\
@.@|@AA{\alpha_{n-1}/\alpha_n}A
@AA{\alpha_{n-1}/\alpha_n}A@.\\
0@>>>
G_{i,{\rm cl}}@>\beta>>
{\mathbf G}_a@>{b_1}>>
{\mathbf G}_a@>>>0
\end{CD}
\label{eqGicl}
\end{equation}
is an isomorphism of exact sequences.
\end{pr}
\proof{
1.
We have (\ref{eqr}) by
\cite[Lemma 3.3.1.5]{red}.
We have a commutative diagram
(\ref{eqGrcl}) with $b_1$ and
$f'(\alpha)
(\alpha_{n-1}-\alpha_n)$
replaced by
$b=\prod_{\sigma\in G^r}
(X-\beta(\sigma))$
and $c=
\prod_{i=1}^m
(\alpha_n-\alpha_i)\cdot
(\alpha_{n-1}-\alpha_n)^{n-m}$
for $m=\#G-\#G^r$
by \cite[Lemma 3.3.1.1]{red},
since the canonical isomorphism
$N_{T/Q}\to N_{E/S}\otimes E$
maps $f$ to $f(0)$.
Since $b=
\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))\cdot b_1$
and
$c=\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))\cdot f'(\alpha)
(\alpha_{n-1}-\alpha_n)$,
we obtain (\ref{eqGrcl}).
2.
Since ${\cal O}_L=
{\cal O}_K[\alpha]$
and ${\rm ord}_L(\alpha_i-\alpha_n)$
is increasing,
the equality (\ref{eqi}) follows from
the definition of $G_{i,{\rm cl}}$.
By \cite[Chapitre V, Proposition 8, Section 6]{CL},
the morphism
$(T^i)^{-1}\circ N^i\colon
U^i_L/U^{i+1}_L\to
U^i_L/U^{i+1}_L$
is defined by a separable additive
polynomial such that
the coefficient of degree 1 is 1
and the upper line of (\ref{eqGicl})
is exact.
Since $\sigma(\alpha)/\alpha
=1+(\sigma(\alpha)-\alpha_n)/
(\alpha_{n-1}-\alpha_n)\cdot(
\alpha_{n-1}/\alpha_n-1)$,
the left square is commutative.
Since the left square is commutative,
the right square is also commutative
by the uniqueness of $b_1$.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\begin{cor}\label{corcl}
{\rm 1.}
We have
\begin{equation}
er=
{\rm ord}_Lf'(\alpha)+(i+1).
\label{eqri}
\end{equation}
{\rm 2.}
There exists an isomorphism
\begin{equation}
\begin{CD}
0@>>>
G^r@>>>
Q^{(r)\circ}_E
@>>>
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}
@>>>
0
\\
@.@|@AAA@AA{f'(\alpha)\cdot
\alpha_n/f(0)}A@.\\
0@>>>
G_{i,{\rm cl}}@>>>
U^i_L/U^{i+1}_L
@>{(T^i)^{-1}\circ N^i}>>
U^i_L/U^{i+1}_L
@>>>
0
\end{CD}
\label{eqGricl}
\end{equation}
of exact sequences.
\end{cor}
\proof{
1.
The equality (\ref{eqri})
follows from (\ref{eqr})
and (\ref{eqi}).
2.
Combining (\ref{eqGrcl})
and (\ref{eqGicl}),
we obtain (\ref{eqGricl}).
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\medskip
By Lemma \ref{lmapol}.2,
the extension
in the upper line of (\ref{eqGricl})
defines a canonical injection
\begin{equation}
G^{r\vee}={\rm Hom}_{{\mathbf F}_p}
(G^r,{\mathbf F}_p)
\to
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1}
,
E).
\label{eqchcl}
\end{equation}
Assume that the residue
field of $F$ is perfect
and let $L$ be a Galois extension
of $K$.
Let $K^{\rm ur}\subset L$
denote the maximum
unramified extension
corresponding
to the inertia subgroup $I\subset G$.
For a rational number $r>1$,
we apply the construction
of (\ref{eqchcl})
to the totally ramified
extension
$M\subset L$ of $K^{\rm ur}$
corresponding to
$G^{r+}=\bigcup_{s>r}
G^s\subset I\subset G$
and to $H^r={\rm Gr}^rG=
G^r/G^{r+}\subset H
={\rm Gal}(M/K^{\rm ur})
=I/G^{r+}$.
Let $e'=e_{M/K}$ be the ramification
index and $E'\subset E$ be
the residue field of $M$.
We obtain an injection
\begin{align}
({\rm Gr}^rG)^\vee=
{\rm Hom}_{{\mathbf F}_p}
({\rm Gr}^rG,{\mathbf F}_p)
\to&\,
{\rm Hom}_{E'}(
{\mathfrak m}_M^{e'(r-1)}/
{\mathfrak m}_M^{e'(r-1)+1},E')
\label{eqgrch}
\\
&
\subset
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},E).
\nonumber
\end{align}
For abelian extensions,
we have the Hasse-Arf theorem.
\begin{thm}[{\rm \cite[Chapitre V,
Section 7, Th\'eor\`eme 1]{CL}}]\label{thmHA}
Let $K$ be a henselian discrete valuation
field with perfect residue field
and let $L$ be a finite abelian
extension of $K$.
Let $n\geqq 1$ be an integer
and $r$ be a rational number
satisfying $n<r\leqq n+1$.
Then, we have
$G^r=G^{n+1}$.
\end{thm}
\section{Tangent spaces
and a characterization
of ramification groups}
\begin{df}[{\rm \cite[Definition 1.1.8]{red}}]
\label{dfcot}
Let $K$ be a discrete valuation
field,
$S={\rm Spec}\, {\cal O}_K$
and $F$ be the residue field.
{\rm 1.}
For an extension $E$ of $F$,
let $L_{E/S}$ denote the cotangent complex
and
we call the spectrum
\begin{equation}
\Theta_{K,E}=
{\rm Spec}\, S(H_1(L_{E/S}))
\label{eqThA}
\end{equation}
of the symmetric algebra
over $E$
the tangent space of $S$ at $E$.
{\rm 2.}
If ${\cal O}_K\to {\cal O}_{K'}$
is a faithfully flat morphism of
discrete valuation rings,
we say that $K'$ is
an extension of
discrete valuation fields of $K$.
We say that
an extension $K'$
of
discrete valuation fields of $K$
is tangentially dominant if,
for a morphism $\bar F\to \bar F'$
of algebraic closures of the residue fields,
the morphism
$$
S(H_1(L_{\bar F/S}))
\to
S(H_1(L_{\bar F'/S'}))$$
is an injection.
\end{df}
The morphism
\begin{equation}
{\mathfrak m}_K/
{\mathfrak m}_K^2\otimes_F\bar F
=
H_1(L_{F/S})\otimes_F\bar F
\to
H_1(L_{\bar F/S})
\label{eqLFm}
\end{equation}
defined by the functoriality
of cotangent complexes
is an injection
by \cite[Proposition 1.1.3.1]{red}.
The injection (\ref{eqLFm})
is an isomorphism if
$F$ is perfect.
The distinguished triangle
$L_{S/{\mathbf Z}}
\otimes^L_{{\cal O}_S}
\bar F
\to L_{\bar F/{\mathbf Z}}
\to L_{\bar F/S}\to $
defines a canonical surjection
\begin{equation}
H_1(L_{\bar F/S})
\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}
\bar F
\label{eqLHO}
\end{equation}
\cite[Proposition 1.1.7.3]{red}
such that the composition with
(\ref{eqLFm}) is induced by
$d\colon
{\mathfrak m}_K/
{\mathfrak m}_K^2\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}F$.
If $K$ is of characteristic $p>0$,
(\ref{eqLHO})
is an isomorphism
by
\cite[Proposition 1.1.7.3]{red}.
If $K'$
is a tangentially dominant
extension of $K$,
the morphism
$H_1(L_{\bar F/S})
\to
H_1(L_{\bar F'/S'})$
is an injection.
\begin{pr}[{\rm \cite[Proposition 1.1.10]{red}}]
\label{prcot}
Let $K\to K'$
be an extension of discrete valuation
fields.
We consider the following conditions:
{\rm (1)}
The ramification index
$e_{K'/K}$ is $1$
and $F'={\cal O}_{K'}/
{\mathfrak m}_{K'}$
is a separable extension of
$F={\cal O}_K/
{\mathfrak m}_K$.
{\rm (2)}
The extension $K'$
is tangentially dominant over $K$.
{\rm (3)}
The ramification index
$e_{K'/K}$ is $1$.
\noindent
Then, we have
the implications
{\rm (1)}$\Rightarrow${\rm (2)}$\Rightarrow${\rm (3)}.
\end{pr}
\begin{thm}\label{thmfil}
Let $r>1$ be a rational number.
For finite Galois extensions $L$
of henselian discrete valuation fields $K$,
there exists a unique way
to define a normal subgroup
$G^r$ of the Galois group
$G={\rm Gal}(L/K)$
satisfying the following conditions:
{\rm (1)}
If the residue field of $K$
is perfect, then
$G^r=G^{r-1}_{\rm cl}$.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$.
Then the natural injection
$G'={\rm Gal}(L'/K')
\to G$ for $L'=LK'$
induces an isomorphism
$G^{\prime r}\to G^r$.
\end{thm}
For a separable closure
$\bar K$ of $K$,
extend the normalized discrete
valuation ${\rm ord}_K$
to $\bar K$.
For a rational number $r$,
set
${\mathfrak m}_{\bar K}^r
=\{x\in \bar K\mid {\rm ord}_K
x\geqq r\}
\supset
{\mathfrak m}_{\bar K}^{r+}
=\{x\in \bar K\mid {\rm ord}_K
x> r\}$.
The quotient
${\mathfrak m}_{\bar K}^r
/{\mathfrak m}_{\bar K}^{r+}$
is a vector space of dimension 1
over the residue field $\bar F$.
For $r>1$,
define $G^{r+}=\bigcup_{s>r}G^s$
and ${\rm Gr}^rG=G^r/G^{r+}$.
\begin{thm}\label{thmchar}
Let $r>1$ be a rational number.
For finite Galois extensions $L$
of henselian discrete valuation fields $K$,
for morphisms
$L\to \bar K$
to separable closures over $K$
and for the residue field $\bar F$
of $\bar K$,
there exists a unique way
to define an injection
\begin{equation}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
\to
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S})).
\label{eqgrL}
\end{equation}
satisfying the following conditions:
{\rm (1)}
Assume that the residue field of $K$
is perfect.
Let $E$ be the residue field of $L$,
$e=e_{L/K}$ be
the ramification index
and identify
$
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},
E)
$
with a subgroup of
$
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))$
by the injection
${\mathfrak m}_K/
{\mathfrak m}_K^2
\to
H_1(L_{\bar F/S})$
{\rm (\ref{eqLFm})}.
Then, the diagram
\begin{equation}
\begin{CD}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))\\
@|@AAA\\
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>{\rm (\ref{eqchcl})}>>
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},
E)
\end{CD}
\label{eqgrcl}
\end{equation}
is commutative.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$,
let $\bar K\to \bar K'$
be a morphism of separable
closures extending
$L\to L'=LK'$
and let $\bar F\to \bar F'$
be the morphism of residue fields.
Then, for the natural injection
$G'={\rm Gal}(L'/K')
\to G$,
the diagram
\begin{equation}
\begin{CD}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))\\
@VVV@VVV\\
{\rm Hom}({\rm Gr}^rG',{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F'}(
{\mathfrak m}_{\bar K'}^r/
{\mathfrak m}_{\bar K'}^{r+},
H_1(L_{\bar F'/S'}))
\end{CD}
\label{eqLL}
\end{equation}
is commutative.
\end{thm}
The uniqueness is a consequence
of the following
existence of a tangentially dominant
extension with perfect residue field.
\begin{pr}[{\rm \cite[Proposition 1.1.12]{red}}]\label{prperf}
Let $K$ be a discrete valuation field.
Then, there exists a
tangentially dominant extension
$K'$ of $K$ such that
the residue field $F'$ is perfect.
\end{pr}
\proof[Proof of Theorem {\rm \ref{thmfil}}]{
We show the uniqueness.
By Proposition \ref{prperf},
there exists a tangentially dominant
extension $K'$
of $K$ with perfect residue field.
Let $G'={\rm Gal}(L'/K')
\to G$ be the natural injection
for $L'=LK'$.
Then, by the conditions (1) and (2),
the subgroup $G^r\subset G$
is the image of $G^{\prime r-1}_{\rm cl}
\subset G'$.
To show the existence,
it suffices to prove that the subgroup
$G^r\subset G$ defined in \cite{AS}
satisfies the conditions (1) and (2).
The equality $G^r=G^{r-1}_{\rm cl}$
is proved in
\cite[Proposition 3.7 (3)]{AS}.
The condition (2) is satisfied by
\cite[Proposition 4.2.4 (1)]{red}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\proof[Proof of Theorem {\rm \ref{thmchar}}]{
We show the uniqueness.
If the residue field is perfect,
the morphism (\ref{eqgrL})
is uniquely determined by the commutative
diagram (\ref{eqgrcl})
since its right vertical arrow is an injection
induced by the injection (\ref{eqLFm}).
In general,
by Proposition \ref{prperf},
there exists a tangentially dominant
extension $K'$
of $K$ with perfect residue field.
Then, the morphism (\ref{eqgrL})
is uniquely determined by the commutative
diagram (\ref{eqLL})
since its right vertical arrow
is an injection.
To show the existence,
it suffices to prove that
the morphism
\cite[(4.20)]{red} satisfies
the conditions (1) and (2).
Assume that
the residue field is perfect.
To show the commutative diagram
(\ref{eqgrcl}),
we may assume that
$G^{r+}=1$
and ${\rm Gr}^rG=G^r$
by the construction of the morphisms.
Then, since
the construction of
{\rm (\ref{eqchcl})}
is a special case of
\cite[(4.20)]{red},
the condition (1) is satisfied.
The condition (2) follows from
\cite[(4.19)]{red}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\section{Abelian extensions}
\begin{thm}\label{thmfila}
Let $r>1$ be a rational number.
{\rm 1.}
For finite abelian extensions $L$
of henselian discrete valuation fields $K$,
there exists a unique way
to define a normal subgroup
$G^r$ of the Galois group
$G={\rm Gal}(L/K)$
satisfying the following conditions:
{\rm (1)}
If the residue field of $K$
is perfect, then
$G^r=G^{r-1}_{\rm cl}$.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$.
Then the natural injection
$G'={\rm Gal}(L'/K')
\to G$ for $L'=LK'$
induces an isomorphism
$G^{\prime r}\to G^r$.
{\rm 2.}
Let $L$ be a finite abelian extension
of a henselian discrete valuation field $K$
and let $n\geqq 1$ be the integer
satisfying $n<r\leqq n+1$.
Then, we have $G^r=G^{n+1}$.
\end{thm}
\begin{thm}\label{thmchara}
Let $n>1$ be an integer.
For finite abelian extensions $L$
of henselian discrete valuation fields $K$,
for morphisms
$L\to \bar K$
to separable closures over $K$
and for the residue fields $\bar F$ of
$\bar K$,
there exists a unique way
to define an injection
\begin{equation}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S})).
\label{eqgrLa}
\end{equation}
satisfying the following conditions:
{\rm (1)}
Assume that the residue field of $K$
is perfect
and let the notation be as
in Theorem {\rm \ref{thmchar} (1)}.
Then the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S}))\\
@|@AAA\\
{\rm Hom}(G^{n-1}_{\rm cl}
/G^n_{\rm cl},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^{n},
E)
\end{CD}
\label{eqgrcla}
\end{equation}
is commutative.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$
and let the notation be as
in Theorem {\rm \ref{thmchar} (2)}.
Then, the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S}))
\\
@VVV@VVV
\\
{\rm Hom}(G^{\prime n}/G^{\prime n+1},
{\mathbf F}_p)
@>>>
{\rm Hom}_{F'}(
{\mathfrak m}_{K'}^n/
{\mathfrak m}_{K'}^{n+1},
H_1(L_{\bar F'/S'}))
\end{CD}
\label{eqLLa}
\end{equation}
is commutative.
\end{thm}
\proof[Proof of Theorem {\rm \ref{thmfila}}]{
1. is proved in the same way as
Theorem \ref{thmfil}.
2. By 1, this
follows from the Hasse-Arf theorem
Theorem \ref{thmHA}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\proof[Proof of Theorem {\rm \ref{thmchara}}]{
This is a special case of
Theorem \ref{thmchar}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\medskip
Assume that $K$ is
a henselian discrete valuation field
of equal characteristic $p>0$ and
let $L$ be a finite abelian extension.
Then, by the Hasse-Arf theorem
Theorem \ref{thmfila}.2
and by the isomorphism
$H_1(L_{\bar F/S})\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F$
(\ref{eqLHO}),
for an integer $n>0$,
the injection
{\rm (\ref{eqgrLa})}
defines an injection
\begin{equation}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F).
\label{eqgrM}
\end{equation}
A decreasing filtration $(G^n_{\rm Ma})$
indexed by integers $n>0$
is defined in \cite[Definition 3.1.1]{M}
as a non-logarithmic modification
of a filtration $(G^n_{\rm Ka})$
defined in \cite[Definition (2.1)]{K}.
Further, for an integer $n>0$,
a canonical morphism
\begin{equation}
{\rm rsw}'\colon
{\rm Hom}(G^n_{\rm Ma}/G_{\rm Ma}^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)
\label{eqgrMY}
\end{equation}
is defined in
{\rm \cite[Definition 3.2.5]{M}}
except the case
$p=2$, $n=2$
and in
{\rm \cite[Definition 1.18]{Y}}
in the exceptional case
$p=2$, $n=2$,
as a modification of
the refined Swan conductor
defined in \cite[Corollary (5.2)]{K}.
As an application,
we give a new proof of
the equalities of the two filtrations
and the two morphisms,
different from that in
\cite{aml} and \cite{Y}.
\begin{cor}\label{corab}
Let $L$ be an abelian extension
of a henselian discrete valuation field $K$
of equal characteristic $p>0$
and let $n>1$ be an integer.
{\rm 1.({\cite[Th\'eor\`eme 9.10 (i)]{aml}
for $n>1$,
\cite[Theorem 3.1]{Y}}
for $n$ general)}
We have an equality
$G^n=G^n_{\rm Ma}$
of subgroups of $G$.
{\rm 2.(\cite[Th\'eor\`eme 9.10
(ii)]{aml} for $n>1$,
\cite[Corollary 2.13]{Y} for $n$ general)}
The injection
{\rm (\ref{eqgrM})}
is the same as ${\rm rsw}'$
{\rm (\ref{eqgrMY})}.
\end{cor}
The following proof
is by the reduction
to the logarithmic variant
\cite[Th\'eor\`eme 9.11]{aml}
in the classical case
where the residue field
is perfect.
\proof{
It suffices to show that the filtration
$(G^n_{\rm Ma})$
and the morphism
${\rm rsw}'$
satisfy the conditions in
Theorems \ref{thmfila} and \ref{thmchara}.
We show that the conditions (1)
are satisfied.
Assume that the residue field
$F$ is perfect.
Then, we have
$G^n=G^{n-1}_{\rm cl}$
and
$G^n_{\rm Ma}
=G^{n-1}_{\rm Ka}$.
Since $G^{n-1}_{\rm Ka}
=G^{n-1}_{\rm cl}$ in this case
by \cite[Th\'eor\`eme 9.11 (i)]{aml},
the condition (1)
in Theorem \ref{thmfila} is satisfied.
We identify
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K} F$
with
${\mathfrak m}_K/
{\mathfrak m}_K^2$
by
$d\colon {\mathfrak m}_K/
{\mathfrak m}_K^2
\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K} F$
and
${\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},$
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)$
with
${\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^n,\bar F)$
by the induced isomorphism.
Then, the morphism
(\ref{eqgrMY})
is identified with
the morphism
\begin{equation}
{\rm rsw}\colon
{\rm Hom}(G^{n-1}_{\rm cl}
/G^n_{\rm cl},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^n,F)
\label{eqgrK}
\end{equation}
defined in \cite[Corollary (5.2)]{K}.
Since the morphism
(\ref{eqgrK}) equals
(\ref{eqgrM}) by \cite[Th\'eor\`eme 9.11 (ii)]{aml},
the condition (1)
in Theorem \ref{thmchara} is satisfied.
We show that the conditions (2)
are satisfied.
For an extension
$K'$ of henselian discrete valuation field
of ramification index $1$,
the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n_{\rm Ma}/
G^{n+1}_{\rm Ma},{\mathbf F}_p)
@>{{\rm rsw}'}>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)\\
@VVV@VVV\\
{\rm Hom}(G^{\prime n}_{\rm Ma}
/G^{\prime n+1}_{\rm Ma},
{\mathbf F}_p)
@>{{\rm rsw}'}>>
{\rm Hom}_{F'}(
{\mathfrak m}_{K'}^n/
{\mathfrak m}_{K'}^{n+1},
\Omega^1_{{\cal O}_{K'}}
\otimes_{{\cal O}_{K'}}\bar F')
\end{CD}
\label{eqgrMM}
\end{equation}
is commutative.
Hence the condition (2) in Theorem \ref{thmchara}
is satisfied.
If $K'$ is tangentially dominant
over $K$,
then the morphism
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F
\to
\Omega^1_{{\cal O}_{K'}}
\otimes_{{\cal O}_{K'}}\bar F'$
is an injection.
Hence by the commutative
diagram (\ref{eqgrMM}),
the morphism
$G^{\prime n}/G^{\prime n+1}
\to G^n/G^{n+1}$
is a surjection.
By the descending induction
on $n$,
the condition (2) in Theorem \ref{thmfila}
is satisfied.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
}
\section{Totally ramified case}
Let $K$ be a henselian discrete
valuation field.
Let $L$ be a totally ramified
Galois extension of $K$
and let $G={\rm Gal}(L/K)$
be the Galois group.
For a rational number $r>1$,
the upper ramification group
$G^r$ defined in \cite[Definition 3.4]{AS}
equals the subgroup
defined in \cite[Chapitre IV, Section 3]{CL}
denoted $G^{r-1}_{\rm cl}$,
by \cite[Proposition 3.7 (3)]{AS}.
Assume that $L$ is wildly ramified
and
let $r>1$ be the largest rational
number such that
the subgroup $G^r$
of the wild inertia subgroup
$P\subset G$
is non-trivial.
Let $E$ be the residue field
and $e=e_{L/K}$
be the ramification index.
We give a description
of the canonical injection
\begin{equation}
G^{r\vee}={\rm Hom}_{{\mathbf F}_p}
(G^r,{\mathbf F}_p)
\to
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1}
,E)
\label{eqchclE}
\end{equation}
for the ${\mathbf F}_p$-vector space
$G^r$,
in the case where $L$ is totally ramified
over $K$.
The injection (\ref{eqchclE})
is a special case of (\ref{eq1}).
We begin with a description
of extensions of vector spaces
over a field of characteristic $p>0$
by ${\mathbf F}_p$-vector
spaces.
\begin{lemma}\label{lmapol}
Let $F$ be a field of characteristic $p>0$.
{\rm 1.}
Let $G\subset F$
be a finite subgroup of the additive group.
Then, the polynomial
\begin{equation}
a_1=\dfrac{\prod_{\sigma\in G}(X-\sigma)}
{\prod_{\sigma\in G,\,
\sigma\neq 0}(-\sigma)}
\label{eqa1}
\end{equation}
$\in F[X]$ is a unique additive
separable polynomial
such that the coefficient
of degree $1$ is $1$ and that
the sequence
\begin{equation}
\begin{CD}
0@>>> G
@>>>
{\mathbf G}_a
@>{a_1}>>
{\mathbf G}_a
@>>>0
\end{CD}
\label{eqapol}
\end{equation}
is exact.
{\rm 2. (\cite[Proposition 2.1.6
(2)$\Rightarrow$(3)]{red})}
Let $E$ be an $F$-vector space of finite
dimension and
let $0\to G\to H\to E\to 0$
be an extension of $E$ by
an ${\mathbf F}_p$-vector space $G$
of finite dimension, as smooth group
schemes over $F$.
Define a morphism
\begin{equation}
[H]\colon
G^\vee=
{\rm Hom}_{{\mathbf F}_p}
(G,{\mathbf F}_p)
\to
{\rm Ext}(E,{\mathbf F}_p)
=E^\vee
={\rm Hom}_F(E,F)
\label{eqH}
\end{equation}
by sending a character
$\chi\colon G\to {\mathbf F}_p$
to the linear form $f\colon E\to F$
such that there exists
a commutative diagram
$$\begin{CD}
0@>>> G@>>> H@>>> E@>>>0\\
@.@V{\chi}VV@VVV@VVfV@.\\
0@>>>{\mathbf F}_p
@>>> {\mathbf G}_a
@>{x^p-x}>> {\mathbf G}_a
@>>> 0.
\end{CD}$$
If $H$ is connected, then
the morphism
$[H]\colon
G^\vee
\to E^\vee$ is an injection.
\end{lemma}
\begin{proof}
1.
By \cite[Lemma 2.1.5]{red},
$a=\prod_{\sigma\in G}(X-\sigma)
\in F[X]$ is an additive
separable polynomial
such that (\ref{eqapol})
with $a_1$ replaced by $a$
is exact.
Since the coefficient in $a$
of degree $1$ is
$\prod_{\sigma\in G,\,
\sigma\neq 0}(-\sigma)$,
the assertion follows.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\medskip
Let $K$ be a henselian discrete
valuation field and
$L$ be a totally ramified
Galois extension
of degree $e$
of Galois group $G$.
Let $\alpha\in L$
be a uniformizer
and let $E=F$ denote the residue field.
The minimal polynomial
$f\in {\cal O}_K[X]$
is an Eisenstein polynomial
and the constant term $\pi=f(0)$
is a uniformizer of $K$.
We define a closed immersion
$T={\rm Spec}\, {\cal O}_L
\to Q={\rm Spec}\, {\cal O}_K[X]$
by sending $X$ to $\alpha$.
For a rational number $r>1$
such that $er\in{\mathbf Z}$,
define a dilatation
$$Q^{[r]}_T=
{\rm Spec}\, {\cal O}_L[X]
\Bigl[\dfrac f{\alpha^{er}}\Bigr]
\to Q_T=
{\rm Spec}\, {\cal O}_L[X].$$
The generator
$f$ of the kernel
$I={\rm Ker}({\cal O}_K[X]
\to {\cal O}_L)$
defines a basis over ${\cal O}_L$
of the conormal module
$N_{T/Q}=I/I^2$
and $\alpha^{er}$
defines a basis
of the $E$-vector space
${\mathfrak m}_L^{er}/
{\mathfrak m}_L^{er+1}$.
As subspaces of
$N_{E/Q}=J/J^2$
for $J={\rm Ker}({\cal O}_K[X]
\to E)=
(X, f)
=(X,\pi)$,
we have an equality
\begin{equation}
N_{T/Q}\otimes_{{\cal O}_L}E
={\mathfrak m}_K/
{\mathfrak m}_K^2
\label{eqNTQ}
\end{equation}
since $f$ is an Eisenstein polynomial.
The basis $f$ of
$N_{T/Q}\otimes_{{\cal O}_L}E$
corresponds to
the uniformizer
$\pi=f(0)
\in {\mathfrak m}_K/
{\mathfrak m}_K^2$.
By sending
$S$ to $f/\alpha^{er}$,
we define an isomorphism
${\cal O}_L[X,S]/
(f-\alpha^{er}S)
\to {\cal O}_L[X]
[f/\alpha^{er}]$.
Since $f$ is an Eisenstein polynomial,
the reduced closed fiber
$Q^{[r]}_E=
{\rm Spec}\, ({\cal O}_L[X]
[f/\alpha^{er}]
\otimes_{{\cal O}_L}E)_{\rm red}$
is identified with
${\rm Spec}\, E[S]$.
By this identification
and (\ref{eqNTQ}),
we define an isomorphism
\begin{align}
Q^{[r]}_E\to
{\rm Hom}_E({\mathfrak m}_L^{er}
/{\mathfrak m}_L^{er+1},
N_{T/Q}\otimes
_{{\cal O}_L}E)^\vee
\to&\,
{\rm Hom}_E({\mathfrak m}_L^{er}
/{\mathfrak m}_L^{er+1},
{\mathfrak m}_K/
{\mathfrak m}_K^2)^\vee
\nonumber
\\
&
=
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}
\label{eqQr}
\end{align}
of smooth group schemes of
dimension 1 over $E$.
Let $Q^{(r)}_T\to
Q^{[r]}_T$ be the normalization
and define a section
$T\to Q^{(r)}_T$
to be the unique lifting of
the section
$T\to Q_T$
defined by sending $X$ to $\alpha$.
Let $Q^{(r)}_E$
denote the reduced part
of the closed fiber
$Q^{(r)}_T\times_T{\rm Spec}\, E$
and let
$Q^{(r)\circ}_E\subset Q^{(r)}_E$
denote the connected
component containing
the image of the closed point
of $T$ by the section
$T\to Q^{(r)}_T$.
\begin{proposition}\label{prcl}
Let $K$ be a henselian discrete
valuation field with residue field
$F$ of characteristic $p>0$.
Let $L$ be a totally ramified
Galois extension
of degree $n=e$
with residue field $E=F$
and let $G={\rm Gal}(L/K)$
be the Galois group.
Let $\alpha\in L$
be a uniformizer
and let $f\in {\cal O}_K[X]$
be the minimal polynomial.
Decompose
$f=\prod_{i=1}^n(X-\alpha_i)$
so that $\alpha_n=\alpha$
and ${\rm ord}_L(\alpha_i-\alpha_n)$
is increasing in $i$.
{\rm 1.}
Let $r>1$ be the largest rational number
such that $G^r\neq 1$.
Then, we have
\begin{equation}
er=
{\rm ord}_Lf'(\alpha)
+{\rm ord}_L(\alpha_{n-1}-\alpha_n).
\label{eqr}
\end{equation}
Define an injection
$\beta\colon G^r\to {\mathbf G}_a$
by $\beta(\sigma)
\equiv\dfrac{\sigma(\alpha)-\alpha}
{\alpha_{n-1}-\alpha_n}
\bmod {\mathfrak m}_L$
and
an additive polynomial
$b_1\in E[X]$
by $b_1=
\prod_{\sigma\in G^r}
(X-\beta(\sigma))
/\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))$.
Define an isomorphism
${\mathbf G}_a\to
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}$ by
$f'(\alpha)
(\alpha_{n-1}-\alpha_n)/f(0)
\in {\mathfrak m}_L^{e(r-1)}$
and identify
$Q^{[r]}_E$ with
${\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}$
by the isomorphism
{\rm (\ref{eqQr})}.
Then, there exists an isomorphism
\begin{equation}
\begin{CD}
0@>>>
G^r@>>>
Q^{(r)\circ}_E
@>>>
Q^{[r]}_E
@>>>
0
\\
@.@|@AAA@AA{f'(\alpha)
(\alpha_{n-1}-\alpha_n)/f(0)}A@.\\
0@>>>
G^r@>\beta>>
{\mathbf G}_a@>{b_1}>>
{\mathbf G}_a@>>>0
\end{CD}
\label{eqGrcl}
\end{equation}
of exact sequences.
{\rm 2.}
Let $i>0$ be
the largest integer such that
$G_{i,{\rm cl}}={\rm Ker}(G\to
{\rm Aut}({\cal O}_L/{\mathfrak m}_L^{i+1}))
\neq 1$.
Then, we have
\begin{equation}
i=
{\rm ord}_L(\alpha_{n-1}-\alpha_n)-1.
\label{eqi}
\end{equation}
Let $K\subset M\subset L$
be the intermediate extension
corresponding to $G_{i,{\rm cl}}\subset G$
and let $U^i_L=1+{\mathfrak m}^i_L
\subset L^\times$
and $U^i_M=1+{\mathfrak m}^i_M
\subset M^\times$ be the
multiplicative subgroups.
Let
$N^i\colon U^i_L/U^{i+1}_L
\to
U^i_M/U^{i+1}_M$
denote the morphism induced by the norm
$N_{L/M}\colon L^\times \to M^\times$
and
$T^i\colon U^i_L/U^{i+1}_L
={\mathfrak m}^i_L/{\mathfrak m}^{i+1}_L
\to
U^i_M/U^{i+1}_M
={\mathfrak m}^i_M/{\mathfrak m}^{i+1}_M$
be the isomorphism
induced by the trace
${\rm Tr}_{L/M}\colon L\to M$.
Define an isomorphism
${\mathbf G}_a\to
U^i_L/U^{i+1}_L$
by sending $1$
to the class of
$\alpha_{n-1}/\alpha_n\in U^i_L$.
Then, the diagram
\begin{equation}
\begin{CD}
0@>>>
G_{i,{\rm cl}}@>{\sigma
\mapsto \sigma(\alpha)/\alpha}>>
U^i_L/U^{i+1}_L
@>{(T^i)^{-1}\circ N^i}>>
U^i_L/U^{i+1}_L
@>>>
0
\\
@.@|@AA{\alpha_{n-1}/\alpha_n}A
@AA{\alpha_{n-1}/\alpha_n}A@.\\
0@>>>
G_{i,{\rm cl}}@>\beta>>
{\mathbf G}_a@>{b_1}>>
{\mathbf G}_a@>>>0
\end{CD}
\label{eqGicl}
\end{equation}
is an isomorphism of exact sequences.
\end{proposition}
\begin{proof}
1.
We have (\ref{eqr}) by
\cite[Lemma 3.3.1.5]{red}.
We have a commutative diagram
(\ref{eqGrcl}) with $b_1$ and
$f'(\alpha)
(\alpha_{n-1}-\alpha_n)$
replaced by
$b=\prod_{\sigma\in G^r}
(X-\beta(\sigma))$
and $c=
\prod_{i=1}^m
(\alpha_n-\alpha_i)\cdot
(\alpha_{n-1}-\alpha_n)^{n-m}$
for $m=\#G-\#G^r$
by \cite[Lemma 3.3.1.1]{red},
since the canonical isomorphism
$N_{T/Q}\to N_{E/S}\otimes E$
maps $f$ to $f(0)$.
Since $b=
\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))\cdot b_1$
and
$c=\prod_{\sigma\in G^r,\sigma\neq 1}
(-\beta(\sigma))\cdot f'(\alpha)
(\alpha_{n-1}-\alpha_n)$,
we obtain (\ref{eqGrcl}).
2.
Since ${\cal O}_L=
{\cal O}_K[\alpha]$
and ${\rm ord}_L(\alpha_i-\alpha_n)$
is increasing,
the equality (\ref{eqi}) follows from
the definition of $G_{i,{\rm cl}}$.
By \cite[Chapitre V, Proposition 8, Section 6]{CL},
the morphism
$(T^i)^{-1}\circ N^i\colon
U^i_L/U^{i+1}_L\to
U^i_L/U^{i+1}_L$
is defined by a separable additive
polynomial such that
the coefficient of degree 1 is 1
and the upper line of (\ref{eqGicl})
is exact.
Since $\sigma(\alpha)/\alpha
=1+(\sigma(\alpha)-\alpha_n)/
(\alpha_{n-1}-\alpha_n)\cdot(
\alpha_{n-1}/\alpha_n-1)$,
the left square is commutative.
Since the left square is commutative,
the right square is also commutative
by the uniqueness of $b_1$.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\begin{corollary}\label{corcl}
{\rm 1.}
We have
\begin{equation}
er=
{\rm ord}_Lf'(\alpha)+(i+1).
\label{eqri}
\end{equation}
{\rm 2.}
There exists an isomorphism
\begin{equation}
\begin{CD}
0@>>>
G^r@>>>
Q^{(r)\circ}_E
@>>>
{\mathfrak m}_L^{e(r-1)}
/{\mathfrak m}_L^{e(r-1)+1}
@>>>
0
\\
@.@|@AAA@AA{f'(\alpha)\cdot
\alpha_n/f(0)}A@.\\
0@>>>
G_{i,{\rm cl}}@>>>
U^i_L/U^{i+1}_L
@>{(T^i)^{-1}\circ N^i}>>
U^i_L/U^{i+1}_L
@>>>
0
\end{CD}
\label{eqGricl}
\end{equation}
of exact sequences.
\end{corollary}
\begin{proof}
1.
The equality (\ref{eqri})
follows from (\ref{eqr})
and (\ref{eqi}).
2.
Combining (\ref{eqGrcl})
and (\ref{eqGicl}),
we obtain (\ref{eqGricl}).
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\medskip
By Lemma \ref{lmapol}.2,
the extension
in the upper line of (\ref{eqGricl})
defines a canonical injection
\begin{equation}
G^{r\vee}={\rm Hom}_{{\mathbf F}_p}
(G^r,{\mathbf F}_p)
\to
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1}
,
E).
\label{eqchcl}
\end{equation}
Assume that the residue
field of $F$ is perfect
and let $L$ be a Galois extension
of $K$.
Let $K^{\rm ur}\subset L$
denote the maximum
unramified extension
corresponding
to the inertia subgroup $I\subset G$.
For a rational number $r>1$,
we apply the construction
of (\ref{eqchcl})
to the totally ramified
extension
$M\subset L$ of $K^{\rm ur}$
corresponding to
$G^{r+}=\bigcup_{s>r}
G^s\subset I\subset G$
and to $H^r={\rm Gr}^rG=
G^r/G^{r+}\subset H
={\rm Gal}(M/K^{\rm ur})
=I/G^{r+}$.
Let $e'=e_{M/K}$ be the ramification
index and $E'\subset E$ be
the residue field of $M$.
We obtain an injection
\begin{align}
({\rm Gr}^rG)^\vee=
{\rm Hom}_{{\mathbf F}_p}
({\rm Gr}^rG,{\mathbf F}_p)
\to&\,
{\rm Hom}_{E'}(
{\mathfrak m}_M^{e'(r-1)}/
{\mathfrak m}_M^{e'(r-1)+1},E')
\label{eqgrch}
\\
&
\subset
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},E).
\nonumber
\end{align}
For abelian extensions,
we have the Hasse-Arf theorem.
\begin{theorem}{\rm ({\cite[Chapitre V,
Section 7, Th\'eor\`eme 1]{CL}})}
\label{thmHA}
Let $K$ be a henselian discrete valuation
field with perfect residue field
and let $L$ be a finite abelian
extension of $K$.
Let $n\geqq 1$ be an integer
and $r$ be a rational number
satisfying $n<r\leqq n+1$.
Then, we have
$G^r=G^{n+1}$.
\end{theorem}
\section{Tangent spaces
and a characterization
of ramification groups}
\begin{definition}{\rm({\cite[Definition 1.1.8]{red}})}
\label{dfcot}
Let $K$ be a discrete valuation
field,
$S={\rm Spec}\, {\cal O}_K$
and $F$ be the residue field.
{\rm 1.}
For an extension $E$ of $F$,
let $L_{E/S}$ denote the cotangent complex
and
we call the spectrum
\begin{equation}
\Theta_{K,E}=
{\rm Spec}\, S(H_1(L_{E/S}))
\label{eqThA}
\end{equation}
of the symmetric algebra
over $E$
the tangent space of $S$ at $E$.
{\rm 2.}
If ${\cal O}_K\to {\cal O}_{K'}$
is a faithfully flat morphism of
discrete valuation rings,
we say that $K'$ is
an extension of
discrete valuation fields of $K$.
We say that
an extension $K'$
of
discrete valuation fields of $K$
is tangentially dominant if,
for a morphism $\bar F\to \bar F'$
of algebraic closures of the residue fields,
the morphism
$$
S(H_1(L_{\bar F/S}))
\to
S(H_1(L_{\bar F'/S'}))$$
is an injection.
\end{definition}
The morphism
\begin{equation}
{\mathfrak m}_K/
{\mathfrak m}_K^2\otimes_F\bar F
=
H_1(L_{F/S})\otimes_F\bar F
\to
H_1(L_{\bar F/S})
\label{eqLFm}
\end{equation}
defined by the functoriality
of cotangent complexes
is an injection
by \cite[Proposition 1.1.3.1]{red}.
The injection (\ref{eqLFm})
is an isomorphism if
$F$ is perfect.
The distinguished triangle
$L_{S/{\mathbf Z}}
\otimes^L_{{\cal O}_S}
\bar F
\to L_{\bar F/{\mathbf Z}}
\to L_{\bar F/S}\to $
defines a canonical surjection
\begin{equation}
H_1(L_{\bar F/S})
\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}
\bar F
\label{eqLHO}
\end{equation}
\cite[Proposition 1.1.7.3]{red}
such that the composition with
(\ref{eqLFm}) is induced by
$d\colon
{\mathfrak m}_K/
{\mathfrak m}_K^2\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}F$.
If $K$ is of characteristic $p>0$,
(\ref{eqLHO})
is an isomorphism
by
\cite[Proposition 1.1.7.3]{red}.
If $K'$
is a tangentially dominant
extension of $K$,
the morphism
$H_1(L_{\bar F/S})
\to
H_1(L_{\bar F'/S'})$
is an injection.
\begin{proposition}{\rm ({\cite[Proposition 1.1.10]{red}})}
\label{prcot}
Let $K\to K'$
be an extension of discrete valuation
fields.
We consider the following conditions:
{\rm (1)}
The ramification index
$e_{K'/K}$ is $1$
and $F'={\cal O}_{K'}/
{\mathfrak m}_{K'}$
is a separable extension of
$F={\cal O}_K/
{\mathfrak m}_K$.
{\rm (2)}
The extension $K'$
is tangentially dominant over $K$.
{\rm (3)}
The ramification index
$e_{K'/K}$ is $1$.
\noindent
Then, we have
the implications
{\rm (1)}$\Rightarrow${\rm (2)}$\Rightarrow${\rm (3)}.
\end{proposition}
\begin{theorem}\label{thmfil}
Let $r>1$ be a rational number.
For finite Galois extensions $L$
of henselian discrete valuation fields $K$,
there exists a unique way
to define a normal subgroup
$G^r$ of the Galois group
$G={\rm Gal}(L/K)$
satisfying the following conditions:
{\rm (1)}
If the residue field of $K$
is perfect, then
$G^r=G^{r-1}_{\rm cl}$.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$.
Then the natural injection
$G'={\rm Gal}(L'/K')
\to G$ for $L'=LK'$
induces an isomorphism
$G^{\prime r}\to G^r$.
\end{theorem}
For a separable closure
$\bar K$ of $K$,
extend the normalized discrete
valuation ${\rm ord}_K$
to $\bar K$.
For a rational number $r$,
set
${\mathfrak m}_{\bar K}^r
=\{x\in \bar K\mid {\rm ord}_K
x\geqq r\}
\supset
{\mathfrak m}_{\bar K}^{r+}
=\{x\in \bar K\mid {\rm ord}_K
x> r\}$.
The quotient
${\mathfrak m}_{\bar K}^r
/{\mathfrak m}_{\bar K}^{r+}$
is a vector space of dimension 1
over the residue field $\bar F$.
For $r>1$,
define $G^{r+}=\bigcup_{s>r}G^s$
and ${\rm Gr}^rG=G^r/G^{r+}$.
\begin{theorem}\label{thmchar}
Let $r>1$ be a rational number.
For finite Galois extensions $L$
of henselian discrete valuation fields $K$,
for morphisms
$L\to \bar K$
to separable closures over $K$
and for the residue field $\bar F$
of $\bar K$,
there exists a unique way
to define an injection
\begin{equation}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
\to
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S})).
\label{eqgrL}
\end{equation}
satisfying the following conditions:
{\rm (1)}
Assume that the residue field of $K$
is perfect.
Let $E$ be the residue field of $L$,
$e=e_{L/K}$ be
the ramification index
and identify
$
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},
E)
$
with a subgroup of
$
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))$
by the injection
${\mathfrak m}_K/
{\mathfrak m}_K^2
\to
H_1(L_{\bar F/S})$
{\rm (\ref{eqLFm})}.
Then, the diagram
\begin{equation}
\begin{CD}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))\\
@|@AAA\\
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>{\rm (\ref{eqchcl})}>>
{\rm Hom}_E(
{\mathfrak m}_L^{e(r-1)}/
{\mathfrak m}_L^{e(r-1)+1},
E)
\end{CD}
\label{eqgrcl}
\end{equation}
is commutative.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$,
let $\bar K\to \bar K'$
be a morphism of separable
closures extending
$L\to L'=LK'$
and let $\bar F\to \bar F'$
be the morphism of residue fields.
Then, for the natural injection
$G'={\rm Gal}(L'/K')
\to G$,
the diagram
\begin{equation}
\begin{CD}
{\rm Hom}({\rm Gr}^rG,{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F}(
{\mathfrak m}_{\bar K}^r/
{\mathfrak m}_{\bar K}^{r+},
H_1(L_{\bar F/S}))\\
@VVV@VVV\\
{\rm Hom}({\rm Gr}^rG',{\mathbf F}_p)
@>>>
{\rm Hom}_{\bar F'}(
{\mathfrak m}_{\bar K'}^r/
{\mathfrak m}_{\bar K'}^{r+},
H_1(L_{\bar F'/S'}))
\end{CD}
\label{eqLL}
\end{equation}
is commutative.
\end{theorem}
The uniqueness is a consequence
of the following
existence of a tangentially dominant
extension with perfect residue field.
\begin{proposition}{\rm (\cite[Proposition 1.1.12]{red})}\label{prperf}
Let $K$ be a discrete valuation field.
Then, there exists a
tangentially dominant extension
$K'$ of $K$ such that
the residue field $F'$ is perfect.
\end{proposition}
\begin{proof}\hspace{-2mm}{\bf of Theorem \ref{thmfil}}
We show the uniqueness.
By Proposition \ref{prperf},
there exists a tangentially dominant
extension $K'$
of $K$ with perfect residue field.
Let $G'={\rm Gal}(L'/K')
\to G$ be the natural injection
for $L'=LK'$.
Then, by the conditions (1) and (2),
the subgroup $G^r\subset G$
is the image of $G^{\prime r-1}_{\rm cl}
\subset G'$.
To show the existence,
it suffices to prove that the subgroup
$G^r\subset G$ defined in \cite{AS}
satisfies the conditions (1) and (2).
The equality $G^r=G^{r-1}_{\rm cl}$
is proved in
\cite[Proposition 3.7 (3)]{AS}.
The condition (2) is satisfied by
\cite[Proposition 4.2.4 (1)]{red}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\begin{proof} \hspace{-2mm}{\bf of Theorem \ref{thmchar}}
We show the uniqueness.
If the residue field is perfect,
the morphism (\ref{eqgrL})
is uniquely determined by the commutative
diagram (\ref{eqgrcl})
since its right vertical arrow is an injection
induced by the injection (\ref{eqLFm}).
In general,
by Proposition \ref{prperf},
there exists a tangentially dominant
extension $K'$
of $K$ with perfect residue field.
Then, the morphism (\ref{eqgrL})
is uniquely determined by the commutative
diagram (\ref{eqLL})
since its right vertical arrow
is an injection.
To show the existence,
it suffices to prove that
the morphism
\cite[(4.20)]{red} satisfies
the conditions (1) and (2).
Assume that
the residue field is perfect.
To show the commutative diagram
(\ref{eqgrcl}),
we may assume that
$G^{r+}=1$
and ${\rm Gr}^rG=G^r$
by the construction of the morphisms.
Then, since
the construction of
{\rm (\ref{eqchcl})}
is a special case of
\cite[(4.20)]{red},
the condition (1) is satisfied.
The condition (2) follows from
\cite[(4.19)]{red}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\section{Abelian extensions}
\begin{theorem}\label{thmfila}
Let $r>1$ be a rational number.
{\rm 1.}
For finite abelian extensions $L$
of henselian discrete valuation fields $K$,
there exists a unique way
to define a normal subgroup
$G^r$ of the Galois group
$G={\rm Gal}(L/K)$
satisfying the following conditions:
{\rm (1)}
If the residue field of $K$
is perfect, then
$G^r=G^{r-1}_{\rm cl}$.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$.
Then the natural injection
$G'={\rm Gal}(L'/K')
\to G$ for $L'=LK'$
induces an isomorphism
$G^{\prime r}\to G^r$.
{\rm 2.}
Let $L$ be a finite abelian extension
of a henselian discrete valuation field $K$
and let $n\geqq 1$ be the integer
satisfying $n<r\leqq n+1$.
Then, we have $G^r=G^{n+1}$.
\end{theorem}
\begin{theorem}\label{thmchara}
Let $n>1$ be an integer.
For finite abelian extensions $L$
of henselian discrete valuation fields $K$,
for morphisms
$L\to \bar K$
to separable closures over $K$
and for the residue fields $\bar F$ of
$\bar K$,
there exists a unique way
to define an injection
\begin{equation}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S})).
\label{eqgrLa}
\end{equation}
satisfying the following conditions:
{\rm (1)}
Assume that the residue field of $K$
is perfect
and let the notation be as
in Theorem {\rm \ref{thmchar} (1)}.
Then the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S}))\\
@|@AAA\\
{\rm Hom}(G^{n-1}_{\rm cl}
/G^n_{\rm cl},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^{n},
E)
\end{CD}
\label{eqgrcla}
\end{equation}
is commutative.
{\rm (2)}
Let $K'$ be a tangentially dominant
extension of $K$
and let the notation be as
in Theorem {\rm \ref{thmchar} (2)}.
Then, the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
@>>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
H_1(L_{\bar F/S}))
\\
@VVV@VVV
\\
{\rm Hom}(G^{\prime n}/G^{\prime n+1},
{\mathbf F}_p)
@>>>
{\rm Hom}_{F'}(
{\mathfrak m}_{K'}^n/
{\mathfrak m}_{K'}^{n+1},
H_1(L_{\bar F'/S'}))
\end{CD}
\label{eqLLa}
\end{equation}
is commutative.
\end{theorem}
\begin{proof} \hspace{-2mm}{\bf of Theorem \ref{thmfila}}
1. is proved in the same way as
Theorem \ref{thmfil}.
2. By 1, this
follows from the Hasse-Arf theorem
Theorem \ref{thmHA}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\begin{proof} \hspace{-2mm}{\bf of Theorem \ref{thmchara}}
This is a special case of
Theorem \ref{thmchar}.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
\medskip
Assume that $K$ is
a henselian discrete valuation field
of equal characteristic $p>0$ and
let $L$ be a finite abelian extension.
Then, by the Hasse-Arf theorem
Theorem \ref{thmfila}.2
and by the isomorphism
$H_1(L_{\bar F/S})\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F$
(\ref{eqLHO}),
for an integer $n>0$,
the injection
{\rm (\ref{eqgrLa})}
defines an injection
\begin{equation}
{\rm Hom}(G^n/G^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F).
\label{eqgrM}
\end{equation}
A decreasing filtration $(G^n_{\rm Ma})$
indexed by integers $n>0$
is defined in \cite[Definition 3.1.1]{M}
as a non-logarithmic modification
of a filtration $(G^n_{\rm Ka})$
defined in \cite[Definition (2.1)]{K}.
Further, for an integer $n>0$,
a canonical morphism
\begin{equation}
{\rm rsw}'\colon
{\rm Hom}(G^n_{\rm Ma}/G_{\rm Ma}^{n+1},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)
\label{eqgrMY}
\end{equation}
is defined in
{\rm \cite[Definition 3.2.5]{M}}
except the case
$p=2$, $n=2$
and in
{\rm \cite[Definition 1.18]{Y}}
in the exceptional case
$p=2$, $n=2$,
as a modification of
the refined Swan conductor
defined in \cite[Corollary (5.2)]{K}.
As an application,
we give a new proof of
the equalities of the two filtrations
and the two morphisms,
different from that in
\cite{aml} and \cite{Y}.
\begin{corollary}\label{corab}
Let $L$ be an abelian extension
of a henselian discrete valuation field $K$
of equal characteristic $p>0$
and let $n>1$ be an integer.
{\rm 1.({\cite[Th\'eor\`eme 9.10 (i)]{aml}
for $n>1$,
\cite[Theorem 3.1]{Y}}
for $n$ general)}
We have an equality
$G^n=G^n_{\rm Ma}$
of subgroups of $G$.
{\rm 2.(\cite[Th\'eor\`eme 9.10
(ii)]{aml} for $n>1$,
\cite[Corollary 2.13]{Y} for $n$ general)}
The injection
{\rm (\ref{eqgrM})}
is the same as ${\rm rsw}'$
{\rm (\ref{eqgrMY})}.
\end{corollary}
The following proof
is by the reduction
to the logarithmic variant
\cite[Th\'eor\`eme 9.11]{aml}
in the classical case
where the residue field
is perfect.
\begin{proof}
It suffices to show that the filtration
$(G^n_{\rm Ma})$
and the morphism
${\rm rsw}'$
satisfy the conditions in
Theorems \ref{thmfila} and \ref{thmchara}.
We show that the conditions (1)
are satisfied.
Assume that the residue field
$F$ is perfect.
Then, we have
$G^n=G^{n-1}_{\rm cl}$
and
$G^n_{\rm Ma}
=G^{n-1}_{\rm Ka}$.
Since $G^{n-1}_{\rm Ka}
=G^{n-1}_{\rm cl}$ in this case
by \cite[Th\'eor\`eme 9.11 (i)]{aml},
the condition (1)
in Theorem \ref{thmfila} is satisfied.
We identify
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K} F$
with
${\mathfrak m}_K/
{\mathfrak m}_K^2$
by
$d\colon {\mathfrak m}_K/
{\mathfrak m}_K^2
\to
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K} F$
and
${\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},$
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)$
with
${\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^n,\bar F)$
by the induced isomorphism.
Then, the morphism
(\ref{eqgrMY})
is identified with
the morphism
\begin{equation}
{\rm rsw}\colon
{\rm Hom}(G^{n-1}_{\rm cl}
/G^n_{\rm cl},{\mathbf F}_p)
\to
{\rm Hom}_F(
{\mathfrak m}_K^{n-1}/
{\mathfrak m}_K^n,F)
\label{eqgrK}
\end{equation}
defined in \cite[Corollary (5.2)]{K}.
Since the morphism
(\ref{eqgrK}) equals
(\ref{eqgrM}) by \cite[Th\'eor\`eme 9.11 (ii)]{aml},
the condition (1)
in Theorem \ref{thmchara} is satisfied.
We show that the conditions (2)
are satisfied.
For an extension
$K'$ of henselian discrete valuation field
of ramification index $1$,
the diagram
\begin{equation}
\begin{CD}
{\rm Hom}(G^n_{\rm Ma}/
G^{n+1}_{\rm Ma},{\mathbf F}_p)
@>{{\rm rsw}'}>>
{\rm Hom}_F(
{\mathfrak m}_K^n/
{\mathfrak m}_K^{n+1},
\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F)\\
@VVV@VVV\\
{\rm Hom}(G^{\prime n}_{\rm Ma}
/G^{\prime n+1}_{\rm Ma},
{\mathbf F}_p)
@>{{\rm rsw}'}>>
{\rm Hom}_{F'}(
{\mathfrak m}_{K'}^n/
{\mathfrak m}_{K'}^{n+1},
\Omega^1_{{\cal O}_{K'}}
\otimes_{{\cal O}_{K'}}\bar F')
\end{CD}
\label{eqgrMM}
\end{equation}
is commutative.
Hence the condition (2) in Theorem \ref{thmchara}
is satisfied.
If $K'$ is tangentially dominant
over $K$,
then the morphism
$\Omega^1_{{\cal O}_K}
\otimes_{{\cal O}_K}\bar F
\to
\Omega^1_{{\cal O}_{K'}}
\otimes_{{\cal O}_{K'}}\bar F'$
is an injection.
Hence by the commutative
diagram (\ref{eqgrMM}),
the morphism
$G^{\prime n}/G^{\prime n+1}
\to G^n/G^{n+1}$
is a surjection.
By the descending induction
on $n$,
the condition (2) in Theorem \ref{thmfila}
is satisfied.
\nopagebreak\par\hspace*{\fill}$\square$\par\vskip2mm
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,190
|
Technology Filtered by - RF & Microwave
New Electronics strives to bring you all the latest technology news from the RF & Microwave sector. Advances in electronics are often fast-paced and innovative, so we know that as a design engineer you want to be kept up-to-date with current developments.
Below is a comprehensive list of all the latest electronics technology news from New Electronics.
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Communications Hardware
Interface/Interconnect
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,077
|
{"url":"http:\/\/w3.pppl.gov\/~hammett\/work\/1995\/pppl_gf_annual.html","text":"# Gyrofluid Contributions to the 1994 PPPL Annual Report:\n\nComparison of Gyrofluid Turbulence Simulations with Experiments\n\n****************************************************************\n\nSignificant advances have recently been made in comprehensive gyrofluid simulations of tokamak turbulence, leading to encouraging comparisons with experimental measurements. These simulations now include a number of effects important for realistic comparisons with experiments, including toroidal geometry and the associated toroidal curvature destabilization mechanisms\\cite(Beer94, Waltz92), turbulence-generated fine-scale poloidal flows\\cite{Dorland93,Beer94}, kinetic effects such as Landau-damping and gyro-averaging\\cite{Dorland93}, impurities and beams\\cite{Dorland93}, all in a high-resolution field-line coordinate system\\cite{Beer94b}.\n\nResearchers at the Institute for Fusion Studies at the University of Texas,\\cite{Dorland94, Kotschenreuther94} in collaboration with PPPL, have developed a model for the thermal conductivity $chi$ based on these nonlinear gyrofluid simulations\\cite{Beer94}, aided by a linear gyrokinetic code for more accurate determination of critical temperature gradients and quasilinear estimates of $\\chi_e\/\\chi_i$. Careful comparisons have been carred out with a wide range of TFTR experiments. This IFS-PPPL transport model contains several scaling trends that might be expected from various previous theories, but it is unique in being directly based on first-principles, detailed, toroidal simulations. The simulations indicate that a particular instability (the Ion Temperature Gradient instability, or ITG mode) is the dominant instability for most TFTR L-mode plasmas. However, even the ITG mode is found to be only marginally unstable in the inner half of many plasmas, leading naturally to a chi(r) that increases with minor radius over most of the plasma. The predicted chi(r) eventually gets too small very near the plasma edge, where some other transport mechanism presumably dominates, so our comparisons with experiments have focussed on the core region (r\/a < 0.85) using the temperature at r\/a=0.85 as a boundary condition. The present IFS-PPPL transport model focuses on regimes where the ITG instability is dominant, and uses a quasilinear estimate of $chi_e$. The nonlinear gyrofluid simulations have recently been extended to include a sophisticated model of trapped electrons\\cite{Beer94,Hammett94}, so we can now begin to study regimes dominated by the trapped electron mode, and can study particle transport in addition to heat transport.\n\nThis first-principles transport model (with no experimentally adjustable parameters) has been compared with the core region (r\/a < 0.85) of more than 50 TFTR L-mode discharges, typically predicting the ion and electron temperature profiles T_i(r) and T_e(r) within the experimental error bars throughout the confinement zone. An example of this is found in the power scan in (Fig. 1). The dramatic increase of the central ion temperature observed in supershots (from 5 keV to 30 keV) is also reproduced (Fig. 2), though in order to get the detailed temperature profile shapes correct it appears necessary to upgrade the transport model to include the collisionless trapped electron mode. A number of effects are responsible for the improved supershot performance but the key mechanism appears to be via high Ti\/Te (the ratio of ion to electron temperature), which raises the threshold for the ITG instability and lowers the conductive part of the ion heat flux so that the (experimentally measured) convection part dominates near the core. The low edge recycling of a supershot gives a high edge ion temperature, which eventually leads to enhanced confinement all the way in to the core.\n\nThis level of quantitative agreement with experiments is very encouraging, but there are a number of transport issues which remain under study, such as the edge region, favorable isotope scaling in supershots, Bohm scaling in L-mode, density transport, various perturbative and time-dependent phenomena, effects of elongation and triangularity, high beta, etc. Possible explanations for some of these effects are being pursued, one of them being the issue of Bohm vs. gyro-Bohm scaling. The marginal stability effects in the IFS-PPPL transport model (and using the measurements at $r\/a=0.85$ as a boundary condition) can partially mask the raw gyro-Bohm scaling of the model so it is closer to the Bohm scaling results on TFTR than a purely gyro-Bohm scaling. Results from flux-tube vs. full-torus simulations suggest the existence of a transition from Bohm to gyro-Bohm scaling\\cite{Hammett94}, but a detailed theory of this transition has not yet been worked out. D-IIID has recently found experimental evidence of such transitions in various regimes\\cite{Luce94}.\n\n*********************************** Figure Captions:\n\nFig. 1 IFS-PPPL transport model compared with a TFTR power scan. The predicted T_i(r) profiles agree reasonably well with the measured profiles as the power is varied a factor of 4. [From Ref. Dorland94.] Here and in Fig. 2, the model is used to predict the temperature in the main plasma region (r\/a < 0.85) using r\/a=0.85 as a boundary condition.\n\nFig. 2 IFS-PPPL transport model compared with a TFTR L-mode\/Supershot pair. The simulations capture much of the enormous variation in the ion temperature between Supershots and L-modes. Nonlinear simulations that include trapped electron dynamics will probably fit the data better. [From Ref. Dorland94.]\n\n**************************************************\n\nCitations:\n\nBeer94: Michael Alan Beer, \"Gyrofluid Models of Turbulent Transport in Tokamaks\", Ph.D. Dissertation, Princeton University (1994).\n\nWaltz92: R.E. Waltz, R.R. Dominguez, G.W. Hammett, \"Gyro-Landau fluid models for toroidal geometry\", Phys. Fluids B 4, 3138 (1992).\n\nBeer94b: M.A. Beer, S.C. Cowley, G.W. Hammett, \"Field-aligned coordinates for Nonlinear Simulations of Tokamak Turbulence\", PPPL-3040 (1994), submitted for publication.\n\nDorland93: W. Dorland, \"Gyrofluid Models of Plasma Turbulence\", Ph.D. Dissertation, Princeton University (1993).\n\nDorland94: W. Dorland, M. Kotschenreuther, M.A. Beer, G.W. Hammett, R.E. Waltz, R.R. Dominguez, P.M. Valanju, W.H. Miner, Jr., J.Q. Dong, W. Horton, F.L. Waelbroeck, T.Tajima, M.J. LeBrun. \"Comparisons of Nonlinear Toroidal Turbulence Simulations with Experiment\". 15th IAEA Conference, (Seville, Spain, September 1994). (IEAE-CN-60\/D-P-I-6).\n\nKotschenreuther94: M.~Kotschenreuther, W.~Dorland, M.~A.~Beer and G.~W.~Hammett, \"Quantitative Predictions of Tokamak Energy Confinement from First Principles Kinetic Simulations\", Invited Talk, APS-DPP meeting (Nov. 1994), submitted to Physics of Plasmas (1995).\n\nHammett94: G.W. Hammett, M.A. Beer, J.C. Cummings, W. Dorland, W.W. Lee, H.E. Mynick, S.E. Parker, R.A. Santoro, M. Artun, H.P. Furth, T.S. Hahm, G. Rewoldt, W.M. Tang, \"Advances in Simulating Tokamak Turbulent Transport\", 15th Int. Conf. on Plasma Physics and Controlled Nuclear Fusion Research, Seville, Spain, 1994 (IAEA-CN-60\/D-2-II-1).\n\nLuce94: T. Luce, Petty, et.al., ??, 15th IAEA (1994).","date":"2018-01-23 13:39:30","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.548010528087616, \"perplexity\": 5714.859480921863}, \"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\/1516084891976.74\/warc\/CC-MAIN-20180123131643-20180123151643-00641.warc.gz\"}"}
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\section{Introduction}\lb{intro}
The Einstein's General Theory of Relativity (see, e.g., \citet{2015Univ....1...38I} and references therein) is currently the best description of the gravitational interaction at our disposal. It has successfully passed all the experimental and observational checks with which it has been put to the test so far \citep{2014LRR....17....4W} at different scales ranging from the Earth's surrounding \citep{2011PhRvL.106v1101E} and our Solar System \citep{Nordvedt2001} to extragalactic realms \citep{2018Sci...360.1342C}, including also compact stellar corpses \citep{2018IAUS..337..128K,2018Natur.559...73A} and the main sequence stars orbiting the supermassive black hole in our Galaxy \citep{2018A&A...615L..15G}, not to mention the recent direct discovery of the gravitational waves with Earth-based laser interferometers \citep{2016Univ....2...22C}. None the less, the still unexplained issues of the dark matter in galaxies and clusters of galaxies along with the observed accelerated expansion of the Universe may pose challenges to it \citep{2016Univ....2...23D,2016Univ....2...11V}.
Given its nature of fundamental pillar of our knowledge of the natural world, it is of the utmost importance to always submit under empirical scrutiny new parts of the theoretical structure of general relativity even where violations are, perhaps, least expected, as in the weak-field and slow-motion regime. \citet{Ginz59} wrote: \virg{[\ldots] the history of physics has seen no end of cases in which the certain has turned out to be false. A theory so fundamental to modern science must be rigorously verified if it is to be applied with complete confidence to the further development of cosmology and other areas of physics.}. To this aim, in this paper we will show that it may be possible, at least in principle, to bring an aspect of the post-Newtonian approximation \citep{2014grav.book.....P} which has never been tested so far into the detectability domain within the Solar System.
In the first post-Newtonian approximation, the metric tensor $g_{\sigma\nu},~\sigma,\nu = 0,1,2,3$ describing the gravitational field generated by a given mass-energy distribution made of a system of $N$ gravitationally interacting rotating bodies of arbitrary shape and composition is parameterized in terms of the so-called gravitoelectric potential $\phi$, which is a generalization of the Newtonian potential $U$, denoted sometimes by $w$, and the gravitomagnetic vector potential $\mathbf{w}$ \citep{1989NCimB.103...63B,1991PhRvD..43.3273D,2003AJ....126.2687S}. The latter one, on which we will focus, is generated by matter current densities proportional to the off-diagonal components $T^{0j},~j=1,2,3$ of the energy-momentum tensor $T^{\sigma\nu},~\sigma,\nu = 0,1,2,3$ of the source.
Outside any body of the system, the gravitoelectric potentials admit multipole expansions in terms of certain mass and spin multipole moments \citep{1989AIHPA..50..377B}; see, e.g., \citet{1998CQGra..15.1971B} and references therein for their role and importance in several branches of general relativity like gravitational waves.
Let us consider a single isolated extended rotating body at rest as source of the gravitational field.
The gravitomagnetic acceleration experienced by a test particle orbiting it is \citep{1994PhRvD..49..618D,2015CeMDA.123....1M}
\begin{equation}
{\bds A}_\textrm{gm} = \rp{\bds v}{c^2}\bds\times{\bds B}_\textrm{gm}.\lb{accel}
\end{equation}
In the empty space outside the spinning body, its gravitomagnetic field ${\bds B}_\textrm{gm}$ can be conveniently expressed in terms of a gravitomagnetic potential function $\phi_\textrm{gm}$ as
\citep[Eq.~(30)]{2014CQGra..31x5012P}
\begin{equation}
{\bds B}_\textrm{gm} = -4\bds\nabla\bds\times\mathbf{w}= -{\bds\nabla}\phi_\textrm{gm}.
\end{equation}
By assuming a uniformly rotating homogeneous oblate spheroid at rest, $\phi_\textrm{gm}$ can be expanded in terms of its spin-multipole moments as
\citep[Eqs.~(31)-(32)]{2014CQGra..31x5012P}
\begin{equation}
\phi_\textrm{gm} = -\rp{30 G S}{r^2}\sum_{i=0}^{\infty}\rp{\ton{-1}^i}{\ton{2i + 3}\ton{2i + 5}}\ton{\rp{R_\textrm{e}\varepsilon}{r}}^{2i}P_{2i + 1}\ton{\xi}=-\rp{GS}{r^2}\qua{2\xi -\rp{6}{7}\ton{\rp{R_\textrm{e}\varepsilon}{r}}^2 P_3\ton{\xi} + \ldots }.\lb{phigm}
\end{equation}
According to \citet[Eq.~(27)]{2014CQGra..31x5012P}, the relation connecting the body's ellipticity $\varepsilon$ with the Newtonian even zonal harmonics
is
\begin{equation}
J_{2i} = \rp{3\ton{-1}^i}{\ton{2i + 1}\ton{2i + 3}}\varepsilon^{2i},
\end{equation}
so that, for $i=1$, one has
\begin{equation}
J_2 = -\rp{1}{5}\varepsilon^2.\lb{stronza}
\end{equation}
The term with $i = 0$ in \rfr{phigm} corresponds to the spin-dipole moment, and yields the usual Lense-Thirring orbital precessions \citep{1918PhyZ...19..156L,2012GReGr..44..719I} which have been the subject of intense experimental scrutiny so far; see, e.g, \citet{2013CEJPh..11..531R} and references therein.
Here, we will explicitly calculate the direct orbital effects arising from \rfr{accel} evaluated for the spin-octupole moment in \rfr{phigm}, corresponding to $i=1$. Then, we will show that, in principle, they could be detectable with a dedicated spacecraft-based mission to Jupiter.
To the present author's best knowledge, should his proposal be eventually successful, it would be the first time that a general relativistic higher spin multipole moment would be measured.
The paper is organized as follows. In Section~\ref{precess}, we analytically work out the rates of change, averaged over one orbital period, of the Keplerian orbital elements of a test particle affected by the post-Newtonian acceleration imparted by the gravitomagnetic spin-octupole moment of its primary, assumed uniformly rotating, homogeneous and spheroidal in shape. We treat it perturbatively by using the standard decomposition in radial, transverse and normal components and the Gauss equations for the variation of the osculating Keplerian orbital elements. We restrict a priori neither to any peculiar spatial orientation of $\bds{\hat{S}}$ nor to particular orbital configurations of the orbiter. We subsequently confirm the resulting analytical results by numerically integrating the equations of motion. The perspectives for measuring such effects around Jupiter, which is the fastest spinning and most oblate major body of the Solar System, are treated in Section~\ref{expe}. While Juno (Section~\ref{Juno}), currently orbiting the gaseous giant along a 53-day, highly eccentric orbit, is unsuitable because its expected post-Newtonian effects are too small, a hypothetical new Jovian probe, provisionally dubbed, with a touch of irony, IORIO (In-Orbit Relativity Iuppiter\footnote{\textit{Iupp\u{\i}t\u{e}r} is one of the forms of the Latin noun of the god Jupiter.} Observatory, or IOvis\footnote{\textit{I\u{o}vis} means \virg{of Jupiter} in Latin. } Relativity In-orbit Observatory), moving along a much faster, moderately eccentric orbit could, in principle, be successfully used (Section~\ref{IORIO}). Numerical simulations of the Earth-spacecraft range-rate measurements, which are the actual observable quantities in such an astronomical scenario, preliminarily confirm such expectations. We investigate also other general relativistic features of motion impacting the probe's range-rate. In Section~\ref{multipoles}, we assess the consequences of the mismodeling in the Newtonian potential coefficients of Jupiter, while Section~\ref{poleposition} is devoted to the impact of the uncertainty in the Jovian spin axis orientation.
For each of such sources of systematic errors, we quantitatively evaluate the level of improvement with respect to their current accuracies required to bring their range-rate signatures at least to the same level of the various post-Newtonian signatures of interest. Furthermore, we look also at the temporal patterns of the competing Newtonian signals with respect to the relativistic ones.
We summarize our findings and offer our conclusions in Section~\ref{conclu}.
For the benefit of the reader, Appendix~\ref{appena} contains a list of symbols and definitions of the quantities used throughout the text, while tables and figures are grouped in Appendix~\ref{appenb}. Finally, it is appropriate to note that the present work should be regarded just as a concept study preliminarily investigating to a certain level of detail a scenario which may be potentially able to measure the investigated effect, not as a formal mission proposal.
\section{The long-term orbital precessions}\lb{precess}
For the sake of simplicity, let us, first, assume a coordinate system whose fundamental $\grf{x,~y}$ plane coincides with the body's equator, so that its symmetry axis $\bds{\hat{S}}$ is aligned with the reference $z$ axis.
The long-term rates of change of the osculating Keplerian orbital elements of the test particle, obtained by averaging over one orbital revolution the right-hand-sides of the standard Gauss equations \citep{2011rcms.book.....K,2014grav.book.....P} evaluated onto the Keplerian ellipse as reference trajectory, turn out to be
\begin{align}
\dot a_\textrm{gm} & = 0\lb{adot}, \\ \nonumber \\
\dot e_\textrm{gm} & = \rp{225 e G S R^2 J_2\cos I\sin^2 I\sin 2\omega}{28 c^2 a^5\ton{1-e^2}^{5/2}}\lb{edot}, \\ \nonumber \\
\dot I_\textrm{gm} & = -\rp{225 e^2 G S R^2 J_2\cos^2 I\sin I\sin 2\omega}{28 c^2 a^5\ton{1-e^2}^{7/2}}\lb{Idot}, \\ \nonumber \\
\dot\Omega_\textrm{gm} & = \rp{45 G S R^2 J_2}{112 c^2 a^5\ton{1-e^2}^{7/2}}\qua{-2 \ton{2 + 3 e^2}\ton{3 + 5 \cos 2I} +
5 e^2 \ton{1 + 3 \cos 2I} \cos 2\omega}\lb{Odot}, \\ \nonumber \\
%
\dot\omega_\textrm{gm} & = -\rp{225 G S R^2 J_2\cos I}{112 c^2 a^5\ton{1-e^2}^{7/2}}\grf{ -2 e^2 - 2 \ton{8 + 7 e^2} \cos 2 I
+ \qua{-2 - 3 e^2 + \ton{2 + 7 e^2} \cos 2I} \cos 2\omega }.\lb{odot}
\end{align}
For the sake of simplicity, here and in the following we omit the brackets $\ang{\ldots}$ denoting the average over one orbital period.
Let us, now, remove the limitation on the orientation of the primary's spin axis allowing it to be arbitrarily directed in space. The resulting long-term rates of change of the Keplerian orbital elements are
\begin{align}
\dot a_\textrm{gm} &=0,\lb{dota} \\ \nonumber \\
\dot e_\textrm{gm} &= \rp{225 e G J_2 R^2 S}{14 a^5 c^2 \ton{1 - e^2}^{5/2}} \mathcal{E}\ton{I,~\Omega,~\omega;~\bds{\hat{S}}},\lb{dote} \\ \nonumber \\
\dot I_\textrm{gm} &= \rp{45 G J_2 R^2 S}{224 a^5 c^2 \ton{1 - e^2}^{7/2}}\mathcal{I}\ton{I,~\Omega,~\omega;~\bds{\hat{S}}},\lb{dotI} \\ \nonumber \\
\dot \Omega_\textrm{gm} &= \rp{45 G J_2 R^2 S}{56 a^5 c^2 \ton{1 - e^2}^{7/2}} \mathcal{N}\ton{I,~\Omega,~\omega;~\bds{\hat{S}}},\lb{dotO} \\ \nonumber \\
\dot\omega_\textrm{gm} &= \rp{45 G J_2 R^2 S}{56 c^2 a^5\ton{1 - e^2}^{7/2}}\mathcal{P}\ton{I,~\Omega,~\omega;~\bds{\hat{S}}}.\lb{doto}
\end{align}
with
\begin{align}
\mathcal{E} & = \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{k}}}\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{P}}}\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{Q}}},\lb{Ecoef}\\ \nonumber\\
\mathcal{I} \nonumber &= 5 e^2 \cos 3I \sin 2 \omega \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega} \qua{1 -7 {\hat{S}}_z^2 + \ton{-{\hat{S}}_x^2 + {\hat{S}}_y^2} \cos 2 \Omega - 2 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} -\\ \nonumber \\
\nonumber & -5 \cos 2I \ton{4 + 6 e^2 +e^2 \cos 2 \omega}\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}\qua{-1 + 3 {\hat{S}}_z^2 + \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \cos 2 \Omega + 2 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} - \\ \nonumber \\
\nonumber &- 5 e^2 {\hat{S}}_z \sin 3I \sin 2 \omega \qua{-3 + 5 {\hat{S}}_z^2 + 3 \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \cos 2 \Omega + 6 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} + \\ \nonumber \\
\nonumber & + 5 e^2 {\hat{S}}_z \sin I \sin 2 \omega \qua{3 -5 {\hat{S}}_z^2 + 5 \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2}\cos 2 \Omega + 10 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} - \\ \nonumber \\
\nonumber &- 5 e^2 \cos I \sin 2 \omega \ton{-{\hat{S}}_y \cos\Omega + {\hat{S}}_x \sin\Omega}\qua{3 - 5 {\hat{S}}_z^2 + 5 \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \cos 2 \Omega + 10 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} - \\ \nonumber \\
\nonumber &- 10 {\hat{S}}_z \ton{4 + 6 e^2 + e^2 \cos 2 \omega} \sin 2I \qua{-2 {\hat{S}}_x {\hat{S}}_y \cos 2 \Omega + \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \sin 2 \Omega} + \\ \nonumber \\
\nonumber & + \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \grf{5 e^2 \cos 2 \omega \qua{-1 - 5 {\hat{S}}_z^2 + 5 \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \cos 2 \Omega + 10 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega} + \right.\\ \nonumber \\
&\left. + 2 \ton{2 + 3 e^2} \qua{-1 -5 {\hat{S}}_z^2 + 5 \ton{{\hat{S}}_x^2 - {\hat{S}}_y^2} \cos 2 \Omega + 10 {\hat{S}}_x {\hat{S}}_y \sin 2 \Omega}},\lb{Icoef} \\ \nonumber \\
\mathcal{N} \nonumber & = -5 \cos^2 I \cot I\ton{-4 - 6 e^2 + 3 e^2 \cos 2 \omega} \ton{{\hat{S}}_y \cos\Omega - {\hat{S}}_x \sin\Omega}^3 + \\ \nonumber \\
\nonumber & + 5 \cos I \cot I \ton{{\hat{S}}_y \cos\Omega - {\hat{S}}_x \sin\Omega}^2\qua{3 {\hat{S}}_z \ton{4 + 6 e^2 - 3 e^2 \cos 2 \omega} \sin I + 4 e^2 \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}\sin 2 \omega } + \\ \nonumber \\
\nonumber & + 10 e^2 \csc I \sin 2\omega \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \qua{2 {\hat{S}}_z^2 \sin^2 I + \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}^2} + \\ \nonumber \\
\nonumber & + 5 {\hat{S}}_z \cos^2\omega \qua{\ton{4 + 5 e^2} {\hat{S}}_x^2 \cos^2\Omega + \ton{4 + 3 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 + 5 e^2} {\hat{S}}_y \ton{{\hat{S}}_x \sin 2\Omega + {\hat{S}}_y \sin^2\Omega}} + \\ \nonumber \\
\nonumber & + 5 {\hat{S}}_z \sin^2\omega \qua{\ton{4 + 7 e^2} {\hat{S}}_x^2 \cos^2\Omega + \ton{4 + 9 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 + 7 e^2} {\hat{S}}_y \ton{{\hat{S}}_x \sin 2\Omega + {\hat{S}}_y \sin^2\Omega}} - \\ \nonumber \\
\nonumber & - 2 \csc I \grf{8 {\hat{S}}_z \sin I - \cos I \ton{-8 - 12 e^2 + 5 e^2 \cos 2 \omega} \ton{{\hat{S}}_y \cos\Omega - {\hat{S}}_x \sin\Omega} + \right.\\ \nonumber \\
\nonumber & \left. + e^2 \qua{{\hat{S}}_z \sin I\ton{12 - 5 \cos 2 \omega} + 5\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \sin 2 \omega }} + \\ \nonumber \\
\nonumber & + 5 \cot I \ton{{\hat{S}}_y \cos\Omega - {\hat{S}}_x \sin\Omega} \grf{8 e^2 {\hat{S}}_z \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \sin I \sin 2 \omega + \right.\\ \nonumber \\
\nonumber &\left. + \cos^2\omega \qua{\ton{4 + 5 e^2} {\hat{S}}_x^2 \cos^2\Omega + 3 \ton{4 + 3 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 + 5 e^2} {\hat{S}}_y \ton{{\hat{S}}_x \sin 2\Omega + {\hat{S}}_y \sin^2\Omega}} + \right. \\ \nonumber \\
&\left. + \sin^2\omega\qua{\ton{4 + 7 e^2} {\hat{S}}_x^2 \cos^2\Omega + 3 \ton{4 + 9 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 + 7 e^2} {\hat{S}}_y \ton{{\hat{S}}_x \sin 2\Omega + {\hat{S}}_y \sin^2\Omega}}},\lb{Ocoef}\\ \nonumber \\
\mathcal{P} \nonumber & =
-40 e^2 {\hat{S}}_z \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}\cos^2 I \sin 2\omega \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega} - 20 \ton{1 + 2 e^2} \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{m}}} \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{k}}}\sin 2\omega -\\ \nonumber \\
\nonumber &- 5 \cos^2 I \cot I \ton{{\hat{S}}_y \cos\Omega - {\hat{S}}_x \sin\Omega}^2 \qua{3 {\hat{S}}_z \ton{4 + 6 e^2 - 3 e^2 \cos 2\omega} \sin I + 4 e^2 \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}\sin 2\omega } - \\ \nonumber \\
\nonumber &- 10 e^2 \cot I \sin 2\omega \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \ton{2 {\hat{S}}_z^2 \sin^2 I + \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}}^2} -5 \cos I \cot I\cos^2 \omega \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega}\times\\ \nonumber \\
\nonumber &\times \qua{\ton{4 +5 e^2} {\hat{S}}_x^2 \cos^2 \Omega +3 \ton{4 + 3 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 +5 e^2} {\hat{S}}_y \sin\Omega \ton{2 {\hat{S}}_x \cos\Omega +{\hat{S}}_y \sin\Omega}} -\\ \nonumber \\
\nonumber &- 5 {\hat{S}}_z \cos I \sin^2 \omega \qua{\ton{4 +7 e^2} {\hat{S}}_x^2 \cos^2 \Omega + \right.\\ \nonumber\\
\nonumber &\left. + \ton{4 + 9 e^2} {\hat{S}}_z^2 \sin^2 I + \ton{4 +7 e^2} {\hat{S}}_y\ton{{\hat{S}}_x \sin 2\Omega +{\hat{S}}_y \sin^2\Omega}} -\\ \nonumber \\
\nonumber &- 5 \cos I \cot I \sin^2 \omega \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega} \qua{\ton{4 +7 e^2} {\hat{S}}_x^2 \cos^2 \Omega +3 \ton{4 + 9 e^2} {\hat{S}}_z^2 \sin^2 I +\right.\\ \nonumber\\
\nonumber &\left. + \ton{4 +7 e^2} {\hat{S}}_y\ton{{\hat{S}}_x \sin 2\Omega +{\hat{S}}_y \sin^2\Omega}} -\\ \nonumber\\
\nonumber &- 10 \ton{\bds{\hat{S}}\bds\cdot\bds{\hat{k}}}\cos^2 \omega \grf{\qua{\ton{7 + 6 e^2} {\hat{S}}_x^2 + \ton{5 + 2 e^2} {\hat{S}}_y^2 \cos^2 I} \cos^2 \Omega + \ton{5 + 2 e^2} {\hat{S}}_z^2 \sin^2 I +\right.\\ \nonumber \\
\nonumber &\left. + 2 {\hat{S}}_y \cos\Omega \grf{\ton{5 + 2 e^2} {\hat{S}}_z \cos I \sin I +{\hat{S}}_x \qua{7 + 6 e^2 - \ton{5 + 2 e^2} \cos^2 I} \sin\Omega} +\right.\\ \nonumber \\
\nonumber &+\left. \sin\Omega \qua{-\ton{5 + 2 e^2} {\hat{S}}_x {\hat{S}}_z \sin 2I + \ton{\ton{7 + 6 e^2} {\hat{S}}_y^2 + \ton{5 + 2 e^2} {\hat{S}}_x^2 \cos^2 I} \sin\Omega}} -\\ \nonumber \\
\nonumber &- 10\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{k}}} \sin^2 \omega \grf{\qua{\ton{5 + 2 e^2} {\hat{S}}_x^2 + \ton{7 + 6 e^2} {\hat{S}}_y^2 \cos^2 I} \cos^2 \Omega + \ton{7 + 6 e^2} {\hat{S}}_z^2 \sin^2 I +\right.\\ \nonumber \\
\nonumber &+\left. 2 {\hat{S}}_y \cos\Omega \grf{\ton{7 + 6 e^2} {\hat{S}}_z \cos I \sin I +{\hat{S}}_x \qua{5 + 2 e^2 - \ton{7 + 6 e^2} \cos^2 I} \sin\Omega} +\right.\\ \nonumber \\
\nonumber &+\left. \sin\Omega \qua{-\ton{7 + 6 e^2} {\hat{S}}_x {\hat{S}}_z \sin 2I + \ton{\ton{5 + 2 e^2} {\hat{S}}_y^2 + \ton{7 + 6 e^2} {\hat{S}}_x^2 \cos^2 I} \sin\Omega}} + \\ \nonumber \\
\nonumber & + 2 \csc I \grf{-\cos^2 I \ton{-8 - 12 e^2 +5 e^2 \cos 2\omega} \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega} + \right.\\ \nonumber \\
\nonumber & +\left. 4 \ton{3 + 2 e^2} \sin^2 I \ton{-{\hat{S}}_y \cos\Omega + {\hat{S}}_x \sin\Omega} + \right.\\ \nonumber \\
\nonumber & +\left. 5 \cos I \qua{{\hat{S}}_z \ton{4 + 4 e^2 - e^2 \cos 2\omega} \sin I +e^2\ton{\bds{\hat{S}}\bds\cdot\bds{\hat{l}}} \sin 2\omega }} -\\ \nonumber \\
\nonumber &- 5 \cot I \grf{-\cos^3 I \ton{-4 - 6 e^2 +3 e^2 \cos 2\omega} \ton{{\hat{S}}_y \cos\Omega -{\hat{S}}_x \sin\Omega}^3 + \right.\\ \nonumber \\
\nonumber &\left. + {\hat{S}}_z \sin I \cos^2 \omega \qua{\ton{4 + 5 e^2} {\hat{S}}_x^2 \cos^2 \Omega + \ton{4 +3 e^2} {\hat{S}}_z^2 \sin^2 I + \right.\right.\\ \nonumber \\
&\left.\left. + \ton{4 +5 e^2} {\hat{S}}_y \sin\Omega \ton{2 {\hat{S}}_x \cos\Omega +{\hat{S}}_y \sin\Omega}}}.\lb{ocoef}
\end{align}
It can be noted that \rfrs{dota}{doto}, along with \rfrs{Ecoef}{ocoef}, reduce to \rfrs{adot}{odot} for ${\hat{S}}_x={\hat{S}}_y=0,~{\hat{S}}_z=1$.
Our analytical results are fully confirmed by a numerical integration of the equations of motions, as shown by Figure~\ref{fig0}. Indeed, the analytically computed annual shifts, calculated with \rfrs{dota}{ocoef} for an arbitrary orbital configuration referred to the Earth's mean equator at the epoch J2000.0 of a fictitious test particle orbiting a hypothetical primary with the same physical characteristic of Jupiter, agree with the numerically produced time series of the Keplerian orbital elements obtained by integrating the equations of motion including the acceleration of \rfr{accel} evaluated with \rfr{phigm} for $i=1$.
Finally, we recall that orbital effects proportional to $GSJ_2c^{-2}$ \citep{2015IJMPD..2450067I} arise also from the interplay between the well known Newtonian quadrupolar acceleration due to $J_2$ and the post-Newtonian Lense-Thirring acceleration proportional to $GS c^{-2}$. Their order of magnitude is the same of the direct rates of change treated in the present Section. None the less, such indirect, mixed effects are likely unmeasurable in actual data reductions since they cannot be expressed in terms of a dedicated, solve-for scaling parameter which could be explicitly estimated. It is so because, contrary to the direct effects derived from \rfr{accel}, they do not come from a distinct acceleration which can be suitably parameterized.
\section{Perspectives of measuring the post-Newtonian gravitomagnetic orbital precessions due to the spin-octupole moment of Jupiter}\lb{expe}
\subsection{Juno}\lb{Juno}
The spacecraft Juno is currently orbiting Jupiter, whose relevant physical parameters are reported in Table~\ref{tavolaJup}, along a highly elliptical trajectory characterized by the orbital parameters listed in Table~\ref{tavolaJuno}.
The huge oblateness of the gaseous giant and the large eccentricity of the probe may suggest, at first sight, to look at such a system as a unique opportunity, in principle, to put to the test for the first time the gravitomagnetic effects due to the spin-octupole moment of an extended body. Unfortunately, the resulting orbital precessions of Juno turn out to be too small, as shown by Table~\ref{tavolaJuno} and Figure~\ref{fig1} displaying the simulated Earth-spacecraft range-rate signatures at the perijove passages PJ03, PJ06. In fact, the directly observable quantity of Juno used to mapping the Jovian gravity field is the two-way Ka-band Doppler shift. The frequent maneuvers required to keep the alignment of the transmitting antenna with the Earth tend to destroy the dynamical coherence of the orbit, not allowing to obtain steady time series of the spacecraft's orbital elements. Thus, the analytical calculation based on them should be regarded just as useful and easily understandable tools to perform a-priori sensitivity analyses. The same considerations hold, in principle, also for any other spacecraft orbiting Jupiter and communicating with the Earth. The signatures in Figure~\ref{fig1} were obtained as follows. For each perijove passes, we numerically integrated the equations of motion of the Earth, Jupiter and Juno in Cartesian rectangular coordinates referred to the \textcolor{black}{International Celestial Reference Frame (ICRF)} with and without the disturbing post-Newtonian acceleration under investigation.
More specifically, in our simplified model the Earth is subjected to the Newtonian acceleration due to the Sun, while Jupiter feels only the Newtonian acceleration of the Sun; the equations of motions of both the planets were integrated in a Solar System barycentric coordinate system. The equations of motion of Juno were integrated in a Jovicentric coordinate system; they include the Newtonian accelerations of Jupiter and the Sun and the post-Newtonian acceleration of \rfr{accel}.
For each perijove passes, both the runs shared the same set of initial conditions which were retrieved from the WEB interface HORIZONS maintained by JPL, NASA, for given initial epochs which, in the present case, are December 11, 2016, h: 13:00 (PJ03) and May 19, 2017, h: 02:00 (PJ06), respectively. After each run, a numerical time series of the Earth-probe range-rate $\dot\rho(t)$ was produced by projecting the Juno's velocity vector onto the Earth-Jupiter unit vector; $\dot\rho_\textrm{pert}(t)$ includes also the effect of the perturbing gravitomagnetic acceleration, while $\dot\rho_\textrm{N}(t)$ is the purely classical one due to only the Newtonian monopoles of the Sun and Jupiter. In order to single out the effect of the post-Newtonian acceleration of interest, the differences of both the time series were computed obtaining the curves for $\Delta\dot\rho(t)=\dot\rho_\textrm{pert}(t) - \dot\rho_\textrm{N}(t)$ displayed in Figure~\ref{fig3}. Our method\footnote{In actual data reductions, the appropriate time and spatial coordinates transformations between the Solar System Barycentric coordinate system and the suitably constructed planetocentric coordinate systems for Jupiter and the Earth \citep{1989NCimB.103...63B} are fully modeled and implemented, among other things, according to the most recent IAU resolutions \citep{2011rcms.book.....K}.}, which will be used also in Section~\ref{IORIO} for other Newtonian and post-Newtonian accelerations, was successfully tested by reproducing the Newtonian range-rate signatures due to the odd zonals $J_3,~J_5,~J_7,~J_9$ at PJ03, PJ06 displayed in \citet[Extended Data Fig.~3]{2018Natur.555..220I}.
\subsection{A dedicated, new spacecraft}\lb{IORIO}
However, Jupiter can still be considered as a viable scenario to try to measure its \textcolor{black}{post-Newtonian} gravitomagnetic spin-octupole effects. Indeed, by keeping a hypothetical new spacecraft at about the same distance from it along a much faster orbit, it is possible to select suitable values for $I,~\Omega,~\omega$ allowing for quite large precessions. Tables~\ref{tavola1}~to~\ref{tavola2}, which refer to a Jovian equatorial coordinate system, deal with two different orbital configurations yielding nominal precessions for the node and the pericentre as large as $\simeq 10^2-10^3~\textrm{mas~yr}^{-1}$, which are remarkably large values. More specifically, for a mildly eccentric orbit with $r\simeq 1.015~\textcolor{black}{R}$ with $I=\omega=90~\textrm{deg}$, the gravitomagnetic node precession would be as large as $\dot\Omega_\textrm{gm}=400~\textrm{mas~yr}^{-1}$, while for $I=360~\textrm{deg},~\omega=270~\textrm{deg}$ and the same orbit radius as before one has even $\dot\Omega_\textrm{gm}=-1,600~\textrm{mas~yr}^{-1},~\dot\omega_\textrm{gm} = 4,000~\textrm{mas~yr}^{-1}$. Such an insight is confirmed by some numerical simulations of the Earth-probe range-rate signature. Indeed, by adopting the \textcolor{black}{ICRF} and a Juno-like spatial orientation for the previously considered almost circular, fast jovicentric orbit of the proposed spacecraft, Figure~\ref{fig2} shows that the size of its relativistic signature would reach the $\simeq 0.03~\textrm{mm~s}^{-1}$ level after just 1 d. It should be recalled that the Doppler measurement accuracy of Juno is $\simeq 0.003~\textrm{mm~s}^{-1}$ after $1,000$ s. Figure~\ref{fig3} preliminarily investigates the sensitivity to the individual orbital elements. It turns out that, while the gravitomagnetic range-rate signature is rather insensitive to the eccentricity, at least for small values of it, the pericentre and the true anomaly, the size of the orbit and the orientation of its orbital plane in space have a major impact. Indeed, if, on the one hand, a \textcolor{black}{sufficiently} low orbit is mandatory to increase the signal of interest, on the other hand, certain values of the inclination and the node may push it up to the $\simeq 0.3~\textrm{mm~s}^{-1}$ level for $a = 1.015~R,~e = 0.0049$.
However, caution is in order since dedicated studies will be required to further investigate our idea in terms of its actual feasibility from a practical and engineering point of view. We mention the threat posed, in principle, to the electronics of Jovian probes by the Io plasma torus\footnote{The intense volcanic activity of Io, which is the dominant source of plasma at Jupiter, pours material into Io's atmosphere which is lost to the Jovian magnetosphere near Io. Such a material is then ionized and trapped by the magnetic field forming a torus of plasma around Jupiter. The torus consists of different regions extending from $\simeq 4~R$ to $\simeq 10~R$ \citep{2017AGUFMSM33C2669H}.} and the potentially quite large $\Delta v$ required to implement a successful orbit insertion. Another crucial issue is represented by the impact of other competing dynamical effects, which would act as source of systematic errors potentially biasing the recovery of the relativistic effect of interest. In this regard, we remark that the proposed scenario would benefit of the notable improvement of our knowledge about both the Jupiter's spin pole position and the Newtonian part of its gravity field arising from the analysis of the full data record of Juno, which is scheduled to deorbit into the planet on\footnote{See https://www.jpl.nasa.gov/missions/juno/ on the Internet.} July 2021. Suffice it to say that, until now, just 2 (PJ03 and PJ06) out of a total of expected 25 perijove passages dedicated to gravity field determination have been fully analyzed \citep{2018Natur.555..220I}, while the results from PJ08,~PJ10,~PJ11 should be publicly released soon \citep{2018EGUGA..20.9150D}. Table~\ref{tavolaJup} displays, among other things, the best estimates and the associated realistic uncertainties for the even and odd zonal coefficients $J_{\ell},~\ell=2,3,4,\ldots,12$, and the tesseral and sectorial multipoles $C_{2,1},~S_{2,1},~C_{2,2},~S_{2,2}$. The RA and Dec. of $\bds{\hat{S}}$ are currently known to an accuracy of $\simeq 100~\textrm{mas}$, as shown in Table~\ref{tavolaJup}, while their rates of change are accurate to $\simeq 50~\textrm{mas~yr}^{-1}$ \citep{2018EGUGA..20.9150D}. As far as the first even zonal harmonic of the Jovian gravity field, from Fig.~2 of the poster presented by \citet{2018EGUGA..20.9150D} it seems that its most recent accuracy is $\simeq 4\times 10^{-9}$, corresponding to a relative accuracy of $\simeq 3\times 10^{-7}$. Moreover, it is not unrealistic to assume that the measurement accuracy $\upsigma_{\dot\rho}$ may be better than that of Juno, whose measurements are mostly taken only at its perijove passages, because of the comparatively much larger number $\mathrm{N}$ of data points due to the higher orbital frequency and lower eccentricity. Indeed, $\upsigma_{\dot\rho}$ scales as $1/\sqrt{\mathrm{N}}$.
In the following, we want to quantitatively assess such issues in connection with the full potential of the proposed mission concept as a tool to measure even more general relativistic features of motion ranging from the standard Schwarzschild-like one proportional to $GM c^{-2}$, to the so far never tested gravitoelectric effect proportional to $GMJ_2 c^{-2}$ \citep{1988CeMec..42...81S,Sof89,1991ercm.book.....B}, including also the gravitomagnetic Lense-Thirring frame-dragging \citep{1918PhyZ...19..156L} proportional to $GS c^{-2}$.
\subsubsection{The impact of the mismodeling in the Jovian gravity field's multipoles}\lb{multipoles}
Figures~\ref{figJ2}~to~\ref{figS22}, obtained with the same computational method previously outlined in Section~\ref{Juno}, depict the numerically simulated Newtonian \textcolor{black}{(blue dashed curves)} and post-Newtonian \textcolor{black}{(red continuous curves)} range-rate time series for a given orbital configuration of the probe which, as it will be shown below, should make the detection of the relativistic signals more favorable.
In order to better visualize the temporal patterns of the various effects, the classical signatures were produced by using fictitious values $\mathcal{C}^\ast$ of the Newtonian gravity field coefficients able to make their magnitudes roughly equal to those of the post-Newtonian time series of interest. If such figures $\mathcal{C}^\ast$ for the Jovian multipoles are smaller than their present-day uncertainties listed in Table~\ref{tavolaJup}, they can be interpreted as a measure of how much they should still be improved with respect to their current levels of accuracy in order to make the size of the Newtonian signatures at least equal to the relativistic ones. If, instead, $\mathcal{C}^\ast$ are larger than their present mismodeling, they can be viewed as a measure of the relative accuracy with which a given relativistic signal would be impacted right now. See Table~\ref{tabres} for a complete list of such improvement factors for all the Newtonian multipoles considered here in connection with the various relativistic effects. It turns out that the largest improvements-of the order of $\simeq \textcolor{black}{10}-500$, with a peak of $1,000$ for $J_{10}$-would be required to bring the Newtonian signals to the level of the post-Newtonian gravitomagnetic effect proportional to $GSJ_2 c^{-2}$. \textcolor{black}{A} much smaller improvement would be required to make the size of the classical multipole signatures comparable with the post-Newtonian gravitoelectric and gravitomagnetic effects proportional to $GMJ_2c^{-2},~GSc^{-2}$. As far as the Schwarzschild-type signature is concerned, the current level of accuracy in almost all the Jovian multipoles, with the exception of \textcolor{black}{$J_{10},J_{11},~J_{12},~S_{2,1},~S_{2,2}$}, would yield a bias at the $\simeq 1-10$ per cent level. A very important feature of all the curves displayed in Figures~\ref{figJ2}~to~\ref{figS22} is that the relativistic ones exhibit neatly different temporal patterns with respect to the Newtonian ones, making, thus, easier to detect them. It would not be so for different orbital geometries of the probe.
\subsubsection{The impact of the uncertainty in the Jupiter's pole position}\lb{poleposition}
The position of the Jovian spin axis, determined by its right ascension $\alpha$ and declination $\delta$ with respect to the ICRF \citep{2018EGUGA..20.9150D}, enters the Newtonian accelerations induced by the gravity field multipoles in a nonlinear way. It can be easily realized, e.g., by inspecting the analytical expressions of the long-term precessions of the Keplerian orbital elements due to some even and odd zonal harmonics calculated by \citet{2011PhRvD..84l4001I,2013JApA...34..341R,2014Ap&SS.352..493R} for an arbitrary orientation of $\bds{\hat{S}}$. Thus, the uncertainties $\upsigma_\alpha,~\upsigma_\delta$ have an impact on the general relativistic effects of interest through the Newtonian multipolar signatures. The latest determinations of $\alpha,~\delta$ along with the associated realistic uncertainties, of the order of $\upsigma_\alpha,~\upsigma_\delta\simeq 0.1~\textrm{arcsec}$ \citep{2018EGUGA..20.9150D}, are listed in Table~\ref{tavolaJup}.
Figure~\ref{figRADEC} depicts the numerically simulated mismodeled range-rate signals due to the first four even zonals of Jupiter induced by the present-day errors $\upsigma_\alpha,~\upsigma_\delta$.
They were obtained as described in the previous Section by using the nominal values of the even zonals and taking the differences between the time series computed with $\delta_\textrm{max} = \delta + \upsigma_\delta,~\delta_\textrm{min} = \delta - \upsigma_\delta$ (\textcolor{black}{green dashed} curves) and $\alpha_\textrm{max} = \alpha + \upsigma_\alpha,~\alpha_\textrm{min} = \alpha - \upsigma_\alpha$ (\textcolor{black}{orange continuous} curves), respectively.
It turns out that the largest residual signals are due to the uncertainty in the declination. The largest one occurs for $J_2$, with an amplitude which can reach $\Delta\dot\rho_{\upsigma_{\delta}}^{J_2}\lesssim 60~\textrm{mm~s}^{-1}$. The signatures of the odd zonals are completely negligible. It can be shown that an improvement of $\upsigma_\delta$ by a factor of 100 with respect to the current value of Table~\ref{tavolaJup} would bring the size of the Newtonian $J_2$-induced range-rate time series to the same level of the post-Newtonian one proportional to $GSJ_2c^{-2}$. Such an improvement seems to be quite feasible in view of the fact that it already occurred from the analysis of PJ03, PJ06 \citep[Tab.~1]{2018Natur.555..220I} to that of PJ08, PJ10, PJ11 \citep{2018EGUGA..20.9150D}. In any case, as already noticed in the previous Section, the temporal pattern of the classical $J_2$ signal is different from the relativistic ones.
\section{Summary and overview}\lb{conclu}
We analytically worked out the long-term rates of change of the Keplerian orbital elements of a test particle orbiting an extended spheroidal rotating body induced by its general relativistic gravitomagnetic spin-octupole moment to the first post-Newtonian order. We neither assumed a preferred orientation for the body's symmetry axis nor adopted a particular orbital configuration for the test particle. Thus, our results have a general validity, being applicable, in principle, to whatsoever astronomical and astrophysical scenario of interest. We successfully checked them numerically by integrating the equations of motion.
We applied them to Jupiter, which is the fastest spinning and most oblate major body of the Solar System, and some existing or hypothetical spacecraft orbiting it. While for Juno the gravitomagnetic precessions, of the order of $\simeq 0.2~\textrm{mas~yr}^{-1}$, are too small to be detectable, for a putative new probe orbiting the gaseous giant in, say, $0.12~\textrm{d}$ along a moderately eccentric orbit with $r\simeq 1.015~\textcolor{black}{R}$, the spin-quadrupole effects may be as large as $400-4,000~\textrm{mas~yr}^{-1}$ depending on the orbital geometry, within the measurability threshold with the current tracking technologies. We confirmed such expectations by numerically calculating in the ICRF the signature induced by the general relativistic spin-octupole moment of Jupiter on the Earth-satellite range-rate measurements which, in a real data analysis, would represent the actual observable quantity. Indeed, by conservatively assuming a range-rate experimental precision of $\simeq 0.003~\textrm{mm~s}^{-1}$ over 1,000 s, as for Juno, it turns out that the post-Newtonian effect of interest could overcome such a level after just 1 full orbital revolution reaching, say, $0.03-0.3~\textrm{mm~s}^{-1}$ after 1 d depending mainly on the orientation of the orbital plane in space. Furthermore, in order to explore the full potential of the proposed mission concept, we looked also at the post-Newtonian gravitoelectric effects proportional to $GMJ_2c^{-2}$, which have never been put to the test so far, and at the standard Lense-Thirring and Schwarzschild signatures, proportional to $GSc^{-2},~GMc^{-2}$, respectively.
The experimental uncertainties in the values of both the Newtonian coefficients of the multipolar expansion of the Jovian gravity field and in the orientation of the spin axis of Jupiter would induce mismodeled range-rate signatures in the Doppler measurements of the spacecraft acting as sources of competing systematic biases for the post-Newtonian signals of interest. At present, just 5 of the planned 25 perijove passes dedicated to mapping the planet's gravity field of the ongoing Juno mission, scheduled to end in July 2021, have been analyzed so far. Thus, if and when the proposed mission will be finally implemented, it will benefit of the analysis of the entire Juno data record yielding a much more accurate determination of the Jovian gravity field coefficients and pole position than now.
For a given orbital configuration of the spacecraft, we numerically simulated its mismodeled Newtonian range-rate signatures due to the gravity field coefficients and the spin axis position of Jupiter currently determined by Juno, and the predicted post-Newtonian signals. We determined the level of improvement of the Jovian multipoles and pole position with respect to their present-day accuracies still required to bring the competing classical effects to the level of the various relativistic ones. It turned out that the most demanding requirements pertain the measurability of the $GSJ_2c^{-2}$ signature, implying improvements by a factor of $\simeq \textcolor{black}{10}-500$ for most of the Jovian gravity coefficients considered, with a peak of $1,000$ for $J_{10}$.
The other relatively small post-Newtonian effects, proportional to $GMJ_2c^{-2},~GSc^{-2}$, require less demanding improvements by a factor of just $\simeq 5-50$ or less. The Schwarzschild signature would be measurable right now at a $\simeq 1-10\%$ level, apart from the impact of
\textcolor{black}{$J_{10},~ J_{11},~ J_{12},~S_{2,1},~S_{2,2}$}.
As far as the Jupiter's spin axis is concerned, an improvement by a factor of 100 would be required for its declination $\delta$ to make the size of the $J_2$-induced signature to the same level of the post-Newtonian $GSJ_2c^{-2}$ one. The uncertainty in the declination $\alpha$ is less important. The range-rate signals due to the odd zonals are affected by the errors in the pole position at a negligible level. A fundamental outcome of our analysis consists of the fact that the temporal patterns of the relativistic signatures turned out to be quite different from the classical ones, making, thus, easier, in principle, to separate the post-Newtonian from the Newtonian effects.
Finally, we remark once more that the present work is not a formal mission proposal; instead, it should be regarded just as a sort of expanded mission concept which need further, dedicated studies concerning, e.g., the practical feasibility of the suggested scenario taking into account several important technological and engineering issues.
\section*{Acknowledgements}
I am grateful to D. Durante for useful information.
\begin{appendices}
\section{Notations and definitions}\lb{appena}
Here, some basic notations and definitions used throughout the text are presented \citep{1991ercm.book.....B,2003ASSL..293.....B,2011rcms.book.....K,2014grav.book.....P}.
\begin{description}
\item[] $G:$ Newtonian constant of gravitation
\item[] $c:$ speed of light in vacuum
\item[] $g_{\sigma\nu}:$ spacetime metric tensor
\item[] $\phi,~w:$ gravitoelectric potential
\item[] $U:$ Newtonian gravitational potential
\item[] $\mathbf{w}:$ gravitomagnetic potential
\item[] $T^{\sigma\nu}:$ energy-momentum tensor of the source
\item[] $M:$ mass of the primary
\item[] $\mu\doteq GM:$ gravitational parameter of the primary
\item[] $S:$ magnitude of the angular momentum of the primary
\item[] ${\bds{\hat{S}}} = \grf{{\hat{S}}_x,~{\hat{S}}_y,~{\hat{S}}_z}:$ spin axis of the primary in some coordinate system
\item[] $\alpha:$ right ascension (RA) of the primary's spin axis with respect to the Earth's mean equator at epoch J2000.0
\item[] $\delta:$ declination (DEC) of the primary's spin axis with respect to the Earth's mean equator at epoch J2000.0
\item[] ${\hat{S}}_x = \cos\delta\cos\alpha:$ $x$ component of the primary's spin axis with respect to the Earth's mean equator at epoch J2000.0
\item[] ${\hat{S}}_y = \cos\delta\sin\alpha:$ $y$ component of the primary's spin axis with respect to the Earth's mean equator at epoch J2000.0
\item[] ${\hat{S}}_z = \sin\delta:$ $z$ component of the primary's spin axis with respect to the Earth's mean equator at epoch J2000.0
\item[] $R_\textrm{e}:$ equatorial radius of the primary
\item[] $R_\textrm{p}:$ polar radius radius of the primary
\item[] $\varepsilon\doteq\sqrt{1 - \ton{\rp{R_\textrm{p}}{R_\textrm{e}}}^2}:$ ellipticity of the oblate primary
\item[] $J_\ell,~\ell=2,~3,~4,\ldots:$ Newtonian zonal multipole mass moments of the primary's gravity field
\item[] $C_{2,1},~S_{2,1},~C_{2,2},~S_{2,2}:$ tesseral $\textcolor{black}{(m=1)}$ and sectorial $\textcolor{black}{(m=2)}$ multipole mass moments of degree $\ell = 2$ of the primary's gravity field
\item[] ${\bds B}_\textrm{gm}:$ post-Newtonian gravitomagnetic field in the empty space surrounding the rotating primary
\item[] $\phi_\textrm{gm}:$ gravitomagnetic potential function in the empty space surrounding the rotating primary
\item[] ${\bds A}_\textrm{gm}:$ post-Newtonian gravitomagnetic acceleration experienced by the test particle
\item[] ${\bds r}:$ instantaneous position vector of the test particle with respect to the primary
\item[] $r_\textrm{min}:$ pericentre distance of the test particle with respect to the primary
\item[] $r_\textrm{min}:$ apocentre distance of the test particle with respect to the primary
\item[] $r:$ instantaneous distance of the test particle from the primary
\item[] ${\bds{\hat{r}}}\doteq {\bds r}/r:$ versor of the position vector of the test particle
\item[] $\xi\doteq \bds{\hat{S}}\bds\cdot\bds{\hat{r}}:$ cosine of the angle between the primary's spin axis and the position vector of the test particle
\item[] $P_{2i + 1}\ton{\xi}:$ Legendre polynomial of degree $2i + 1$
\item[] $\bds v:$ velocity vector of the test particle
\item[] $f:$ true anomaly of the test particle's orbit
\item[] $a:$ semimajor axis of the test particle's orbit
\item[] $n_{\rm b} \doteq \sqrt{\mu/a^3}:$ Keplerian mean motion of the test particle's orbit
\item[] $P_{\rm b}\doteq 2\uppi/n_{\rm b}:$ orbital period of the test particle's orbit
\item[] $e:$ eccentricity of the test particle's orbit
\item[] $I:$ inclination of the orbital plane of the test particle's orbit to the reference $\grf{x,~y}$ plane of some coordinate system
\item[] $\Omega:$ longitude of the ascending node of the test particle's orbit referred to the reference $\grf{x,~y}$ plane of some coordinate system
\item[] $\omega:$ argument of pericentre of the test particle's orbit referred to the reference $\grf{x,~y}$ plane of some coordinate system
\item[] $\bds{\hat{l}}\doteq\grf{\cos\Omega,~\sin\Omega,~0}:$ unit vector directed along the line of the nodes toward the ascending node
\item[] $\bds{\hat{m}}\doteq\grf{-\cos I\sin\Omega,~\cos I\cos\Omega,~\sin I}:$ unit vector directed transversely to the line of the nodes in the orbital plane
\item[] $\bds{\hat{k}}\doteq\grf{\sin I\sin\Omega,~-\sin I\cos\Omega,~ \cos I}:$ unit vector perpendicular to the orbital plane directed along the orbital angular momentum
\item[] $\bds{\hat{P}}\doteq \bds{\hat{l}}\cos\omega + \bds{\hat{m}}\sin\omega:$ unit vector in the orbital plane directed along the line of apsides towards the pericentre
\item[] $\bds{\hat{Q}}\doteq -\bds{\hat{l}}\sin\omega + \bds{\hat{m}}\cos\omega:$ unit vector in the orbital plane directed transversely to the line of apsides
\end{description}
\section{Tables and figures}\lb{appenb}
\begin{table*}
\caption{Relevant physical parameters of Jupiter. Most of the reported values come from \citet{2003AJ....126.2687S,2010ITN....36....1P,2018Natur.555..220I,2018EGUGA..20.9150D} and references therein. In particular, the values and the uncertainties of $\alpha,~\delta$ determining the Jovian pole position at the epoch J2017.0 come from \citet{2018EGUGA..20.9150D}, while the multipoles of the gravity potential are retrieved from \citet[Tab.~1]{2018Natur.555..220I}.
}\lb{tavolaJup}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Parameter & Units & Numerical value \\
\hline
$\mu$ & $\textrm{m}^3~\textrm{s}^{-2}$ & $1.26713\times 10^{17}$\\
$S$ & $\textrm{kg~m}^2$~$\textrm{s}^{-1}$ & $6.9\times 10^{38}$\\
$\alpha$ & $\textrm{deg}$ & $268.057132\pm 0.000036$\\
$\delta$ & $\textrm{deg}$ & $64.497159 \pm 0.000045$\\
$R$ & $\textrm{km}$ & $71,492$\\
$J_2$ & $\ton{\times 10^{-6}}$ & $14,696.572\pm 0.014$\\
$J_3$ & $\ton{\times 10^{-6}}$ & $-0.042 \pm 0.010$\\
$J_4$ & $\ton{\times 10^{-6}}$ & $-586.609\pm 0.004$\\
$J_5$ & $\ton{\times 10^{-6}}$ & $-0.069 \pm 0.008$\\
$J_6$ & $\ton{\times 10^{-6}}$ & $34.198 \pm 0.009$\\
$J_7$ & $\ton{\times 10^{-6}}$ & $0.124 \pm 0.017$\\
$J_8$ & $\ton{\times 10^{-6}}$ & $-2.426 \pm 0.025$\\
$J_9$ & $\ton{\times 10^{-6}}$ & $-0.106 \pm 0.044$\\
$J_{10}$ & $\ton{\times 10^{-6}}$ & $0.172 \pm 0.069$\\
$J_{11}$ & $\ton{\times 10^{-6}}$ & $0.033 \pm 0.112$\\
$J_{12}$ & $\ton{\times 10^{-6}}$ & $0.047 \pm 0.178$\\
$C_{2,1}$ & $\ton{\times 10^{-6}}$ & $-0.013\pm 0.015$ \\
$S_{2,1}$ & $\ton{\times 10^{-6}}$ & $-0.003 \pm 0.026$\\
$C_{2,2}$ & $\ton{\times 10^{-6}}$ & $0.000\pm 0.008$ \\
$S_{2,2}$ & $\ton{\times 10^{-6}}$ & $0.000\pm 0.011$ \\
\hline
\end{tabular}
\end{center}
\end{table*}
\begin{table*}
\caption{Relevant orbital parameters of the spacecraft Juno currently orbiting Jupiter.
Here, $R$ is meant as the equatorial radius $R_\textrm{e}$ of Jupiter. The source for the orbital elements of Juno, referred to the Jovian equator, is the freely consultable database JPL HORIZONS on the Internet at https://ssd.jpl.nasa.gov/?horizons from which they were retrieved by choosing the time of writing this paper as input epoch. The values of the post-Newtonian gravitomagnetic precessions of Juno due to the spin-octupole moment of Jupiter, calculated by means of \rfrs{edot}{odot}, are listed as well.
}\lb{tavolaJuno}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Parameter & Units & Numerical value \\
\hline
$a$ & $R$ & $56.7633$\\
$e$ & $-$ & $0.9818125961521484$\\
$r_\textrm{min}$ & $R$ & $1.03238$\\
$r_\textrm{max}$ & $R$ & $112.494$\\
$I$ & $\textrm{deg}$ & $98.98696267273439$ \\
$\Omega$ & $\textrm{deg}$ & $270.7926907554042$ \\
$\omega$ & $\textrm{deg}$ & $163.1267988695804$ \\
$P_{\rm b}$ & $\textrm{d}$ & $52.8133$\\
\hline
$\dot e_\textrm{gm}$ & $\textrm{yr}^{-1}$ & $5\times 10^{-12}$ \\
$\dot I_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.004$\\
$\dot \Omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.172$\\
$\dot \omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.190$\\
\hline
\end{tabular}
\end{center}
\end{table*}
\begin{table*}
\caption{Relevant orbital parameters for a hypothetical spacecraft, referred to the Jovian equator, and its post-Newtonian gravitomagnetic precessions due to the spin-octupole moment of Jupiter, calculated by means of \rfrs{edot}{odot}. Cfr. with the other orbital configuration proposed in Table~\ref{tavola2}.
}\lb{tavola1}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Parameter & Units & Numerical value \\
\hline
$a$ & $R$ & $1.015$\\
$e$ & $-$ & $0.0049$\\
$r_\textrm{min}$ & $R$ & $1.01$\\
$r_\textrm{max}$ & $R$ & $1.02$\\
$P_{\rm b}$ & $\textrm{d}$ & $0.12$\\
$I$ & $\textrm{deg}$ & $90$ \\
$\omega$ & $\textrm{deg}$ & $90$ \\
\hline
$\dot e_\textrm{gm}$ & $\textrm{yr}^{-1}$ & $0.0$ \\
$\dot I_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.0$\\
$\dot \Omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $400$\\
$\dot \omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.0$\\
\hline
\end{tabular}
\end{center}
\end{table*}
\begin{table*}
\caption{Relevant orbital parameters for a hypothetical spacecraft, referred to the Jovian equator, and its post-Newtonian gravitomagnetic precessions due to the spin-octupole moment of Jupiter, calculated by means of \rfrs{edot}{odot}. Cfr. with the other orbital configuration proposed in Table~\ref{tavola1}.
}\lb{tavola2}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Parameter & Units & Numerical value \\
\hline
$a$ & $R$ & $1.015$\\
$e$ & $-$ & $0.0049$\\
$r_\textrm{min}$ & $R$ & $1.01$\\
$r_\textrm{max}$ & $R$ & $1.02$\\
$P_{\rm b}$ & $\textrm{d}$ & $0.12$\\
$I$ & $\textrm{deg}$ & $360$ \\
$\omega$ & $\textrm{deg}$ & $270$ \\
\hline
$\dot e_\textrm{gm}$ & $\textrm{yr}^{-1}$ & $0.0$ \\
$\dot I_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $0.0$\\
$\dot \Omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $-1,600$\\
$\dot \omega_\textrm{gm}$ & $~\textrm{mas~yr}^{-1}$ & $4,000.2$\\
\hline
\end{tabular}
\end{center}
\end{table*}
\begin{table*}
\caption{IORIO scenario: improvement factors $\textcolor{black}{\kappa}$ (if $>1$) required to each of the Jovian multipole coefficients with respect to their current accuracy levels (see \citet[Tab.~1]{2018Natur.555..220I} and Table~\ref{tavolaJup}) to make the size of the corresponding Newtonian range-rate signatures equal to the magnitude of the general relativistic ones; see Figures~\ref{figJ2}~to~\ref{figS22}. If, in a given row, $\textcolor{black}{\kappa}<1$, the current level of accuracy in the multipole of that row would allow right now to measure the corresponding relativistic effects with the relative accuracies as good as $\textcolor{black}{\kappa}$ themselves. For example, in the second row corresponding to $J_3$, there are two figures smaller than 1; it means that the present-day accuracy in $J_3$ would yield a mismodeled Newtonian signal impacting, say, the Schwarzschild-like one at $1.67\%$. Instead, the accuracy of $J_3$ should be improved by a factor of $12.5$ with respect to its current level in order to induce a mismodeled Newtonian signature having, at least, the same magnitude of the relativistic effect proportional to $GSJ_2c^{-2}$. From Table~\ref{tavolaJup}, it should be noted that the values of $J_{11},~J_{12},~C_{2,1},~S_{2,1},~C_{2,2},~S_{2,2}$ are statistically compatible with zero.
}\lb{tabres}
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Multipole & $GSJ_2c^{-2}$ & $GMJ_2 c^{-2}$ & $GSc^{-2}$ & $GMc^{-2}$ \\
\hline
$J_2$ & $70$ & $5$ & $2.5$ & $0.11$ \\
$J_3$ & $12.5$ & $1.1$ & $0.5$ & $0.0167$ \\
$J_4$ & $33$ & $3.3$ & $1.4$ & $0.033$ \\
$J_5$ & $10$ & $0.8$ & $0.58$ & $0.01$ \\
$J_6$ & $100$ & $5$ & $2.5$ & $0.12$ \\
$J_7$ & $50$ & $4.5$ & $3.3$ & $0.067$ \\
$J_8$ & $33$ & $4$ & $2.2$ & $0.05$ \\
$J_9$ & $100$ & $10$ & $3.3$ & $0.15$ \\
$J_{10}$ & $1,000$ & $33$ & $33$ & $0.98$ \\
$J_{11}$ & $500$ & $28.6$ & $20$ & $0.7$ \\
$J_{12}$ & $500$ & $50$ & $28.6$ & $1.1$ \\
$C_{2,1}$ & $50$ & $4$ & $2.8$ & $0.1$ \\
$S_{2,1}$ & $500$ & $20$ & $12.5$ & $\textcolor{black}{1}$ \\
$C_{2,2}$ & $333$ & $28.6$ & $18.2$ & $0.4$ \\
$S_{2,2}$ & $500$ & $33$ & $20$ & $1$ \\
\hline
\end{tabular}
\end{center}
\end{table*}
\clearpage
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.2 cm\epsfbox{Soct_dedt.pdf}&\epsfysize= 5.2 cm\epsfbox{Soct_dIdt.pdf}\\
\epsfysize= 5.2 cm\epsfbox{Soct_dNdt.pdf}&\epsfysize= 5.2 cm\epsfbox{Soct_dpdt.pdf}\\
\end{tabular}
}
}
\caption{Numerically computed time series of the post-Newtonian shifts experienced by the eccentricity $e$, inclination $I$, node $\Omega$ and pericentre $\omega$ of a fictitious test particle induced by the gravitomagnetic spin-octupole moment of a putative central body characterized by the same physical properties of Jupiter (see Table~\ref{tavolaJup}). They were obtained by numerically integrating the equations of motion of the orbiter in Cartesian rectangular coordinates referred to the Earth's mean equator at the epoch J2000.0 with and without the acceleration of \rfr{accel} calculated with \rfr{phigm} for $i=1$. Both runs shared the same set of arbitrary initial conditions $a_0 = 1.5~R,~e_0 = 0.3,~I_0 = 45~\textrm{deg},~\Omega_0 = 30~\textrm{deg},~\omega_0 = 50~\textrm{deg},~f_0 = 45~\textrm{deg}$. For each Keplerian orbital element, its time series calculated from the purely Newtonian run was subtracted from that obtained from the post-Newtonian integration in order to obtain the signatures displayed here. The resulting rates of change, in \textcolor{black}{yr$^{-1}$ and} $~\textrm{mas~yr}^{-1}$, turn out to agree with the analytically computed ones in \rfrs{dote}{doto} with \rfrs{Ecoef}{ocoef} which are $\dot e_\textrm{gm} = 2.835\times 10^{-8}~\textrm{yr}^{-1},~\dot I_\textrm{gm} = 56.05~\textrm{mas~yr}^{-1},~\dot\Omega_\textrm{gm} = -142.89~\textrm{mas~yr}^{-1},~\dot\omega_\textrm{gm} = 362.74 ~\textrm{mas~yr}^{-1}$.}\label{fig0}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{c}
\epsfysize= 7.5 cm\epsfbox{JunoPJ03.pdf}\\
\epsfysize= 7.5 cm\epsfbox{JunoPJ06.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of Juno due to the post-Newtonian gravitomagnetic spin-octupole moment of Jupiter at the first two perijove passages PJ03 (December 11, 2016) and PJ06 (May 19, 2017) dedicated to gravity science. They were obtained by numerically integrating the equations of motion of the Earth, Jupiter and Juno in Cartesian rectangular coordinates referred to the ICRF with and without the general relativistic acceleration of \rfr{accel}, calculated for \rfr{phigm} with $i=1$, starting from the same set of initial conditions retrieved from the Web interface HORIZONS maintained by JPL. Then, for each perijove passage, the range-rate time series computed from the purely Newtonian run was subtracted from that obtained from the post-Newtonian integration in order to yield the curves displayed here. Cfr. with the two-way Ka-band range-rate residuals of Juno for the same perijove passes displayed in the Extended Data Figure~1 of \citet{2018Natur.555..220I} whose ranges of variation amount to $\simeq 0.050~\textrm{mm~s}^{-1}$, with a root-mean-square value of $\simeq 0.015~\textrm{mm~s}^{-1}$.}\label{fig1}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{c}
\epsfysize= 10.0 cm\epsfbox{Soct_rrate.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signature $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter characterized by the ICRF-related orbital configuration $a = 1.015~\textcolor{black}{R},~e = 0.0049,~I = 90.63~\textrm{deg},~\Omega = 268.85~\textrm{deg},~\omega = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ induced by the post-Newtonian gravitomagnetic spin-octupole moment of Jupiter after 1 d. It was obtained as described in the caption of Figure~\ref{fig1}.}\label{fig2}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_a.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_e.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_I.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_N.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_o.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_f.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the post-Newtonian gravitomagnetic spin-octupole moment of Jupiter after 1 d. They were obtained as described in the caption of Figure~\ref{fig1} by allowing the orbital elements of the spacecraft to vary within certain ranges of values with respect to the reference orbital configuration used to produce Figure~\ref{fig2}.}\label{fig3}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J2_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J2_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J2_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J2_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian first even zonal harmonic $J_2$ of Jupiter after 1 d. In each panel, a fictitious value $J_2^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, for which the actual value of $J_2$ is, instead, used, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_2^\ast = 2.0\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_2^\ast = 2.8\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_2^\ast = 5.6\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_2^\ast = 1.26\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian first even zonal is $\upsigma_{J_2} = 1.4\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ2}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J3_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J3_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J3_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J3_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian first odd zonal harmonic $J_3$ of Jupiter after 1 d. In each panel, a fictitious value $J_3^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_3^\ast = 8.0\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_3^\ast = 9.0\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_3^\ast = 2.0\times 10^{-8}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_3^\ast = 6.0\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian first odd zonal is $\upsigma_{J_3} = 1.0\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ3}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J4_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J4_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J4_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J4_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian second even zonal harmonic $J_4$ of Jupiter after 1 d. In each panel, a fictitious value $J_4^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_4^\ast = 1.2\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_4^\ast = 1.2\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_4^\ast = 2.8\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_4^\ast = 1.2\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian second even zonal is $\upsigma_{J_4} = 4\times 10^{-9}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ4}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J5_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J5_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J5_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J5_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian second odd zonal harmonic $J_5$ of Jupiter after 1 d. In each panel, a fictitious value $J_5^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_5^\ast = 8.0\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_5^\ast = 9.6\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_5^\ast = 1.36\times 10^{-8}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_5^\ast = 6.4\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian second odd zonal is $\upsigma_{J_5} = 8\times 10^{-9}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ5}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J6_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J6_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J6_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J6_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian third even zonal harmonic $J_6$ of Jupiter after 1 d. In each panel, a fictitious value $J_6^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_6^\ast = 9.0\times 10^{-11}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_6^\ast = 1.8\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_6^\ast = 3.6\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_6^\ast = 7.2\times 10^{-8}\right)$. The present-day actual uncertainty in the Jovian third even zonal is $\upsigma_{J_6} = 9\times 10^{-9}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ6}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J7_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J7_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J7_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J7_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian third odd zonal harmonic $J_7$ of Jupiter after 1 d. In each panel, a fictitious value $J_7^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_7^\ast = 3.4\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_7^\ast = 3.74\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_7^\ast = 5.1\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_7^\ast = 2.55\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian third odd zonal is $\upsigma_{J_7} = 1.7\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ7}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J8_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J8_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J8_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J8_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian fourth even zonal harmonic $J_8$ of Jupiter after 1 d. In each panel, a fictitious value $J_8^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_8^\ast = 7.5\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_8^\ast = 6.25\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_8^\ast = 1.125\times 10^{-8}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_8^\ast = 5.0\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian fourth even zonal is $\upsigma_{J_8} = 2.5\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ8}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J9_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J9_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J9_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J9_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian fourth odd zonal harmonic $J_9$ of Jupiter after 1 d. In each panel, a fictitious value $J_9^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_9^\ast = 4.4\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_9^\ast = 4.4\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_9^\ast = 1.32\times 10^{-8}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_9^\ast = 3.0\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian fourth odd zonal is $\upsigma_{J_9} = 4.4\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ9}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J10_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J10_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J10_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J10_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian fifth even zonal harmonic $J_{10}$ of Jupiter after 1 d. In each panel, a fictitious value $J_{10}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_{10}^\ast = 6.9\times 10^{-11}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_{10}^\ast = 2.07\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GSc^{-2};~J_{10}^\ast = 2.07\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_{10}^\ast = 7\times 10^{-8}\right)$. The present-day actual uncertainty in the Jovian fifth even zonal is $\upsigma_{J_{10}} = 6.9\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ10}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J11_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J11_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J11_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J11_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian fifth odd zonal harmonic $J_{11}$ of Jupiter after 1 d. In each panel, a fictitious value $J_{11}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_{11}^\ast = 2.24\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_{11}^\ast = 3.92\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_{11}^\ast = 5.6\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_{11}^\ast = 1.568\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian fifth odd zonal is $\upsigma_{J_{11}} = 1.12\times 10^{-7}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ11}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_J12_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J12_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_J12_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_J12_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian fifth odd zonal harmonic $J_{12}$ of Jupiter after 1 d. In each panel, a fictitious value $J_{12}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~J_{12}^\ast = 3.56\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~J_{12}^\ast = 3.56\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~J_{12}^\ast = 6.23\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~J_{12}^\ast = 1.602\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian fifth odd zonal is $\upsigma_{J_{12}} = 1.78\times 10^{-7}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figJ12}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_C21_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_C21_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_C21_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_C21_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian tesseral coefficient $C_{2,1}$ of Jupiter after 1 d. In each panel, a fictitious value $C_{2,1}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~C_{2,1}^\ast = 3.0\times 10^{-10}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~C_{2,1}^\ast = 3.75\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~C_{2,1}^\ast = 5.25\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~C_{2,1}^\ast = 1.5\times 10^{-7}\right)$. The present-day actual uncertainty in the Jovian tessreral coefficient is $\upsigma_{C_{2,1}} = 1.5\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figC21}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_S21_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_S21_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_S21_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_S21_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian tesseral coefficient $S_{2,1}$ of Jupiter after 1 d. In each panel, a fictitious value $S_{2,1}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~S_{2,1}^\ast = 5.2\times 10^{-11}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~S_{2,1}^\ast = 1.3\times 10^{-9}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~S_{2,1}^\ast = 2.08\times 10^{-9}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~S_{2,1}^\ast = 2.6\times 10^{-8}\right)$. The present-day actual uncertainty in the Jovian tessreral coefficient is $\upsigma_{S_{2,1}} = 2.6\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figS21}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_C22_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_C22_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_C22_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_C22_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian sectorial coefficient $C_{2,2}$ of Jupiter after 1 d. In each panel, a fictitious value $C_{2,2}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~C_{2,2}^\ast = 2.4\times 10^{-11}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~C_{2,2}^\ast = 2.8\times 10^{-10}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~C_{2,2}^\ast = 4.4\times 10^{-10}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~C_{2,2}^\ast = 2.0\times 10^{-8}\right)$. The present-day actual uncertainty in the Jovian sectorial coefficient is $\upsigma_{C_{2,2}} = 8.0\times 10^{-9}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figC22}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.4 cm\epsfbox{IORIO_S22_SJ2.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_S22_MJ2.pdf}\\
\epsfysize= 5.4 cm\epsfbox{IORIO_S22_S.pdf}&\epsfysize= 5.4 cm\epsfbox{IORIO_S22_M.pdf}\\
\end{tabular}
}
}
\caption{Simulated range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the nominal post-Newtonian accelerations considered in the text and by the Newtonian sectorial coefficient $S_{2,2}$ of Jupiter after 1 d. In each panel, a fictitious value $S_{2,2}^\ast$ is used in the Newtonian signature just for illustrative and comparative purposes. Indeed, it is suitably tuned from time to time in order to bring the associated classical signature to the level of the nominal post-Newtonian effect of interest, so to inspect the mutual (de)correlations of their temporal patterns more easily. Upper-left corner: post-Newtonian gravitomagnetic spin-octupole moment $\left(GSJ_2c^{-2};~S_{2,2}^\ast = 2.2\times 10^{-11}\right)$. Upper-right corner: post-Newtonian gravitoelectric moment $\left(GMJ_2 c^{-2};~S_{2,2}^\ast = 3.3\times 10^{-10}\right)$. Lower-left corner: Lense-Thirring effect $\left(GS c^{-2};~S_{2,2}^\ast = 5.5\times 10^{-10}\right)$. Lower-right corner: Schwarzschild $\left(GM c^{-2};~S_{2,2}^\ast = 1.1\times 10^{-8}\right)$. The present-day actual uncertainty in the Jovian sectorial coefficient is $\upsigma_{S_{2,2}} = 1.1\times 10^{-8}$ \citep[Tab.~1]{2018Natur.555..220I}. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figS22}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\centerline{
\vbox{
\begin{tabular}{cc}
\epsfysize= 5.3 cm\epsfbox{IORIO_J2_RA_DEC.pdf}&\epsfysize= 5.3 cm\epsfbox{IORIO_J4_RA_DEC.pdf}\\
\epsfysize= 5.3 cm\epsfbox{IORIO_J6_RA_DEC.pdf}&\epsfysize= 5.3 cm\epsfbox{IORIO_J8_RA_DEC.pdf}\\
\end{tabular}
}
}
\caption{Numerically simulated impact of the present-day errors $\upsigma_\alpha = 0.13~\textrm{arcsec},~\upsigma_\delta = 0.16~\textrm{arcsec}$ \citep{2018EGUGA..20.9150D} in the position of the spin axis of Jupiter on the range-rate signatures $\Delta\dot\rho$, in $\textrm{mm~s}^{-1}$, of a hypothetical Jovian orbiter induced by the Newtonian accelerations due to the first four even zonals $J_2,~J_4,~J_6,~J_8$ after 1 d. It turns out that the uncertainties in the Jupiter's spin axis affect the odd zonals signatures in a completely negligible way. The adopted orbital configuration for the probe is $a_0 = 1.015~R,~e_0 = 0.0049,~I_0 = 50~\textrm{deg},~\Omega_0 = 140~\textrm{deg},~\omega_0 = 149.43~\textrm{deg},~f_0 = 228.32~\textrm{deg}$ }\label{figRADEC}
\end{center}
\end{figure*}
\end{appendices}
\clearpage
\section*{Erratum:The post-Newtonian gravitomagnetic spin-octupole moment of an oblate rotating body and its effects on an orbiting test particle; are they measurable in the Solar system?}\lb{errazzo}
\addcontentsline{toc}{section}{\nameref{errazzo}}
In the published version \citep{2019MNRAS.484.4811I} of this paper,
due to the unfortunate and misleading definition\footnote{It comes from equation 27 of \citet{2014CQGra..31x5012P} for $n=1$. It reproduces incorrectly equation (56) of \citet{2003AJ....125.1580K} which, in fact, contains $(-1)^{n+1}$ instead of $(-1)^n$ entering equation 27 of \citet{2014CQGra..31x5012P}. However, it is the post-Newtonian spin-octupole acceleration relying upon $\phi_\textrm{gm}$ of equation (32) of \citet{2014CQGra..31x5012P} that matters; it is independent of all such unnecessary definitions.} $J_2 = -\varepsilon^2/5$ of \rfr{stronza}, the dimensionless quadrupole-type parameter $J_2$ entering the analytically computed post-Newtonian spin-octupole orbital precessions of \rfrs{adot}{doto} differs from the even zonal harmonic $J_2$ usually determined in standard spacecraft-based geodetic and geophysical data reductions which is, indeed, positive. Moreover, also their magnitudes are different, as can be straightforwardly noted in the case of Jupiter. Indeed, as per the IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties available on the Internet at https://www.iau.org/administration/resolutions/general$\_$assemblies/, the nominal polar and equatorial radii of Jupiter amount to $66,854~\textrm{km}$ and $71,492~\textrm{km}$, respectively, yielding $-\varepsilon^2/5 = -0.0251$. Its size is larger than the Juno-based positive value $J_2=0.0147$, listed in Table~\ref{tavolaJup}, by a factor of $1.7084$. Actually, the Juno-based, positive value $J_2=0.0147$ of Table~\ref{tavolaJup} was erroneously used in calculating the gravitomagnetic precessions $\dot e_\textrm{gm},~\dot I_\textrm{gm},~\dot\Omega_\textrm{gm},~\dot\omega_\textrm{gm}$ quoted in Tables~\ref{tavolaJuno}~to~\ref{tavola2} and mentioned throughout the paper, and in producing\footnote{In case of Figure~\ref{fig0}, it is not relevant since its purpose was just confirming the analytical calculation of \rfrs{adot}{doto} with a numerical integration of the equations of motion in the case of a fictitious astronomical scenario: the numerical values actually adopted for the primary's physical properties are unimportant. } Figs~\ref{fig0}~to~\ref{fig3} and the post-Newtonian spin-octupole curves in the upper-left panels of
Figs~\ref{figJ2}~to~\ref{figS22} instead of $-\varepsilon^2/5$. As a consequence, there is a minus sign mistake in $\dot e_\textrm{gm},~\dot I_\textrm{gm},~\dot\Omega_\textrm{gm},~\dot\omega_\textrm{gm}$ of Tables~\ref{tavolaJuno}~to~\ref{tavola2}, and in the post-Newtonian gravitomagnetic spin-octupole signatures displayed in Figs~\ref{fig0}~to~\ref{fig3} and in the upper-left panels of Figs~\ref{figJ2}~to~\ref{figS22}. Moreover, the magnitudes of $\dot e_\textrm{gm},~\dot I_\textrm{gm},~\dot\Omega_\textrm{gm},~\dot\omega_\textrm{gm}$ in Tables~\ref{tavolaJuno}~to~\ref{tavola2}, and the amplitudes in Figs~\ref{fig0}~to~\ref{fig3} and of both the Newtonian and post-Newtonian curves in the upper-left panels of Figs~\ref{figJ2}~to~\ref{figS22} are smaller than the correct ones by a factor of $1.7084$. As such, all the curves of the post-Newtonian spin-octupole effect in Figs~\ref{fig0}~to~\ref{fig3} and in the upper-left panels of Figs~\ref{figJ2}~to~\ref{figS22} should be flipped, and their sizes, along with those of the Newtonian signatures in the upper-left panels of
Figs~\ref{figJ2}~to~\ref{figS22}, rescaled by a factor of $1.7084$. Luckily, it strengthens our conclusions since it increases the signal-to-noise ratio of the post-Newtonian spin-octupole effect. The mutual (de)correlations of the post-Newtonian spin-octupole signatures with the classical ones in the upper-left panels of Figs~\ref{figJ2}~to~\ref{figS22} change accordingly. In the captions of Figs~\ref{figJ2}~to~\ref{figS22}, the values of $J_\ell^\ast,~\ell=2,3,\ldots 12$ and $C^\ast_{2,1},~C^\ast_{2,2},~S^\ast_{2,1},~S^\ast_{2,2}$ associated with the post-Newtonian spin-octupole effects should be rescaled by a factor of $1.7084$. All the figures in the second column from the left of Table~\ref{tabres} should be reduced by a factor of $1.7084$, which is a fortunate circumstance since it implies smaller improvements in our knowledge of the Jovian Newtonian multipoles to detect the post-Newtonian spin-octupole signatures. Finally, it would likely be more correct to replace $GSJ_2c^{-2}$ with, say, $GS\varepsilon^2 c^{-2}$ throughout the paper to avoid further misunderstandings; in particular, it would be better to replace $J_2$ in the analytical precessions of \rfrs{adot}{doto} with $-\varepsilon^2/5$. The conclusions pertaining the other post-Newtonian effects remain unchanged.
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Extra-curricular activities are highly valued at High March and the girls enjoy a wide range of clubs which are held at both lunchtime and after school. From Dance to Choir, Orchestra, Judo, Netball, Maths, Board Games, Computers and Drama there is something on offer for all tastes and interests. For this term's programme of activities, please click on the link below.
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction}
\label{sec:1}
Black holes belong to the most fascinating objects in astrophysics and are well suited to explore the regime of strong gravity. We consider here black holes which are described by the six parameter family of Kerr-Newman-Taub-NUT-de Sitter spacetimes including mass, rotation, electric and magnetic charge, gravitomagnetic mass (or NUT charge), and the cosmological constant. Maybe the best way to explore the gravitational field of such objects is through the observation of the motion of small massive particles and light, which is described by the geodesic equation. The complete set of orbits can best be analyzed using analytical methods. Already in 1931, Hagihara \cite{Hagihara} used Weierstrass elliptic functions to analytically solve the geodesic equation in Schwarzschild spacetime. Later, Darwin \cite{Darwin1959,Darwin1961} solved the geodesic equations in Schwarzschild spacetime in terms of Jacobian elliptic functions. These methods and their generalization to hyperelliptic functions can be used to solve the geodesic equation in the six parameter spacetime under consideration. Although this requires only mathematics from the 19th century, surprisingly the geodesic equations in Schwarzschild-de Sitter spacetime were analytically solved only in 2008 \cite{HackmannLaemmerzahl2008,HackmannLaemmerzahl08b}. Here we will review these solution methods and provide a short literature guide.
\section{The Kerr-Newman-Taub-NUT-de Sitter space-time}\label{sec:spactime}
We consider here spherically or axially symmetric spacetimes with up to six parameters, which are part of the family of type D Pleba\'nski-Demia\'nski spacetimes. The metric is given by \cite{GriffithsPodolsky06,MankoRuiz2005}
\begin{multline} \label{PD_metric}
ds^2 = \frac{\Delta_r}{\rho^2} \left(dt - (a \sin^2\theta + 2 n \cos\theta) d\varphi\right)^2 - \frac{\rho^2}{\Delta_{r}} dr^2 \\
- \frac{\Delta_{\theta}}{\rho^2} \sin^2\theta (a dt - (r^2 + a^2 + n^2) d\varphi)^2 - \frac{\rho^2}{\Delta_{\theta}} d\theta^2
\end{multline}
with
\begin{align}
\rho^2 & = r^2 + \left(n - a \cos\theta \right)^2 \,, \nonumber \\
\Delta_{\theta} & = 1 + \tfrac{1}{3} a^2 \Lambda \cos^2\theta - \tfrac{4}{3} \Lambda a n \cos\theta\,, \\
\Delta_{r} & =
r^2 - 2 M r + a^2 - n^2 + Q^2_{\rm e} + Q_{\rm m}^2 - \tfrac{1}{3} \Lambda \left(r^4 + (6 n^2 + a^2) r^2 + 3 (a^2 - n^2) n^2\right)\,,\nonumber
\end{align}
where $M$ is the mass, $a=J/M$ the specific angular momentum, $\Lambda$ the cosmological constant, $n$ is the gravitomagnetic mass, $Q_{\rm e}$ is the electric, and $Q_{\rm m}$ the magnetic charge of a gravitating source.
\section{Analytical solution methods}\label{sec:AS}
The motion of test particles in the spacetime metric \eqref{PD_metric} is given by the geodesic equation
\begin{align}
\frac{d^2x^\mu}{ds^2} + \Gamma^\mu{}_{\nu\rho} \frac{dx^\nu}{ds} \frac{dx^\rho}{ds} = 0
\end{align}
where $(\Gamma^\mu{}_{\nu\rho})$ are the Christoffel symbols and $s$ is an affine parameter. As the metric \eqref{PD_metric} is axially symmetric there exists the two constants of motion
\begin{align}
E & = u_\mu \xi^\mu_{(t)}\,, \quad L_z = - u_\mu \xi^\mu_{(\varphi)}\,. \label{genEL}
\end{align}
connected to the Killing vectors $\xi_{(t)}$ and $\xi_{(\varphi)}$, which can be interpreted as the energy and the specific angular momentum in direction of the symmetry axes. Here $u$ denotes the four-velocity. If we have even spherical symmetry, i.e.~for $a=n=0$, these two constants of motion together with the restriction to the equatorial plane, which is without loss of generality, and the normalization $g_{\mu\nu}dx^\mu dx^\nu = \epsilon$, where $\epsilon=0$ for light and $\epsilon=1$ for massive particles, is sufficient to separate the geodesic equation. This yields then the differential equation
\begin{align}
\left(\frac{dr}{ds}\right)^2 & = E^2-\frac{\Delta_r}{r^2} \left(\epsilon + \frac{L^2}{r^2}\right)\,,
\end{align}
and the energy and angular momentum take the form
\begin{align}
E=g_{tt}\frac{dt}{ds}\,, \quad L=L_z=r^2\frac{d\varphi}{ds}\,.
\end{align}
If the spacetime is axially symmetric, we cannot in general restrict the motion of test particles to the equatorial plane. Therefore, we need an additional constant of motion. In 1968 Carter \cite{Carter68} surprisingly found such a constant of motion, which can be derived as a separation constant and is not connected to an obvious symmetry of the spacetime. With this constant the geodesic equation can again be separated and we get equations of motions in the form
\begin{align}
\rho^4 \left( \frac{dr}{ds}\right)^2 & = ((r^2+a^2+n^2)E-aL_z)^2 - \Delta_r(\epsilon r^2+K)\,, \label{axial:drds}\\
\rho^4 \left(\frac{d\theta}{ds}\right)^2 & = \Delta_\theta (K-\epsilon(n-a\cos\theta)^2) - \frac{(E(a\sin^2\theta+2n\cos\theta)-L_z)^2}{\sin^2\theta}\,, \label{axial:dthetads}
\end{align}
where $K$ is the Carter constant. If $n=0$ then $K=(aE-L_z)^2$ corresponds to motion restricted to the equatorial plane. From the expression of energy and angular momentum \eqref{genEL} we get the additional equations
\begin{align}
\rho^2 \frac{d\varphi}{ds} & = \frac{a}{\Delta_r} ((r^2+a^2+n^2)E-aL_z) - \frac{E(2n\cos\theta+a\sin^2\theta)-L_z}{\Delta_\theta \sin^2\theta}\,,\\
\rho^2 \frac{dt}{ds} & = \frac{r^2+a^2+n^2}{\Delta_r} ((r^2+a^2+n^2)E-aL_z)\nonumber\\
& \quad - \frac{a\sin^2\theta+2n\cos\theta}{\Delta_\theta\sin^2\theta} (E(2n\cos\theta+a\sin^2\theta)-L_z) \,.
\end{align}
Note that the equations \eqref{axial:drds} and \eqref{axial:dthetads} are still coupled by the factor $\rho^2$. This issue was solved by Mino in 2003 \cite{Mino03} by introducing a new affine parameter $\lambda$ defined by $\frac{ds}{d\lambda}=\rho^2$
If we now consider in the spherically symmetric case the differential equations for $r(\varphi)$,
\begin{align}
\left(\frac{dr}{d\varphi}\right)^2 & = \frac{r^4}{L^2} \left( E^2-\frac{\Delta_r}{r^2} \left(\epsilon + \frac{L^2}{r^2}\right) \right)
\end{align}
we see that on the right hand side we have a polynomial of degree three or four, if the cosmological constant vanishes, and of degree five or six in general. The same holds for the differential equations \eqref{axial:drds} and \eqref{axial:dthetads}. This kind of differential equations can be solved in terms of elliptic functions if the polynomial is of degree three or four, and in terms of hyperelliptic functions if it is of degree five or six.
\subsection{Solutions in terms of elliptic functions}
For differential equations of the general type
\begin{align}
\left(\frac{dx}{dy}\right)^2 & = P_{3,4}(x)\,, \quad x(y_0)=x_0\label{genellipticdiff}
\end{align}
where $P_{3,4}$ is a polynomial of degree three or four, there are basically two (equivalent) solution methods based on the Jacobian elliptic function ${\rm sn}\,$ and on the Weierstrass elliptic function $\wp$. The first can be defined as the inverse of an elliptic integral,
\begin{align}
z = \int_0^w \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}} \quad \Rightarrow \quad {\rm sn}\,(z;k)=w
\end{align}
where $0\leq k\leq 1$ is the modulus, $w\in [0,1]$, and $z \in \mathbb{R}$. The Weierstrass elliptic function is given as a series
\begin{align}
\wp(z;2\omega_1,2\omega_2) & = \frac{1}{z^2} + \sum_{\omega_{nm} \in \Omega} \left( \frac{1}{(z-\omega_{nm})^2} - \frac{1}{\omega_{nm}^2} \right)\,,
\end{align}
where $2\omega_1$,$2\omega_2$ are the periods of $\wp$ ($\frac{\omega_1}{\omega_2} \notin \mathbb{R}$) and $\Omega=\{ \omega_{nm} \in \mathbb{C}| \omega_{nm}=2n\omega_1+2m\omega_2\}$. It solves the initial value problem (see e.g.~\cite{Markushevich77, Hurwitz})
\begin{align}
\left(\frac{dx}{dy}\right)^2 = 4x^3-g_2x-g_3\,, \quad x(0)=\infty \label{Weierstrassform}
\end{align}
where $g_2 = 60 \sum_{\omega_{nm}\in \Omega} \omega_{nm}^{-4}$, $g_3 = 140 \sum_{\omega_{nm}\in \Omega} \omega_{nm}^{-6}$. Note that both ${\rm sn}\,$ and $\wp$ can be written in terms of the Riemann theta function
\begin{align}
\theta[\tau v+w](z;\tau) & = \sum_{m \in \mathbb{Z}^g} \exp( \pi i (m+v)^t (\tau (m+v) + 2z + 2w) )\,, \label{deftau}
\end{align}
where $z\in\mathbb{C}^g$, $\tau$ is a $g\times g$ symmetric matrix with positive definite imaginary part, $\tau v+w \in \mathbb{C}^g$ is the characteristic, and $g$ is the genus, here $g=1$.
The general differential equation \eqref{genellipticdiff} can be solved in terms of Jacobian elliptic functions by applying a substitution which converts the problem to the form
\begin{align}
\left(\frac{d\tilde{x}}{dy}\right)^2 & = (1-\tilde{x}^2)(1-k^2\tilde{x}^2)\,.
\end{align}
This substitution depends on the degree of $P_{3,4}$ and its number of complex zeros as well as on the type of orbit you want to obtain. For a list see e.g.~\cite{Abramowitz}.
To find a solution of \eqref{genellipticdiff} in terms of the Weierstrass elliptic function you have to convert the problem to the standard form \eqref{Weierstrassform}, which can be obtained by first converting a polynomial of degree four to degree three by $x=\xi^{-1}+x_P$ if $P_{3,4}=\sum_i a_ix^i$ and $P_{3,4}(x_P)=0$ and subsequently, or if $P_{3,4}$ was of degree three in the first place, substituting $\xi=\frac{1}{b_3}(4z+\frac{b_2}{3})$ if $P=\sum_i b_ix^i$ is the polynomial of degree three.
Note that you always have to take care of the initial condition, too.
\subsection{Solutions in terms of hyperelliptic functions}
To generalize the solution methods outlined in the previous section we first need to consider the Jacobi inversion problem
\begin{align}
y_i = \sum_{j=1}^g \int_{\infty}^{x_j} \frac{t^{i-1}dt}{\sqrt{P(t)}}\,, \quad i=1,\ldots,g \label{Jacobiinversion}
\end{align}
where $P(t)=4t^{2g+1}+\sum_{n=0}^{2g} a_nt^n$ is a polynomial of degree $2g+1$ and $g$ is the genus. Note that for $g=1$ and $a_2=0$ we recover \eqref{Weierstrassform}. The $g$ solutions $x_j(y_1,\ldots,y_g)$ of \eqref{Jacobiinversion} can be given in terms of generalized Weierstrass functions. These are defined by the theta function via the Kleinian sigma function
\begin{align}
\sigma(z;\omega_1,\omega_2) & = C e^{iz^t\kappa z} \theta[K_\infty](z;\omega_1^{-1}\omega_2)\,,\\
\wp_{ij}(z;\omega_1,\omega_2) & = - \frac{\partial}{\partial z_i} \frac{\partial}{\partial z_j} \log \sigma(z;\omega_1,\omega_2)\,,
\end{align}
where $z\in \mathbb{C}^g$, $\omega_i$ are $g \times g$ matrices such that $\omega_1^{-1}\omega_2$ is symmetric with positive definite imaginary part, $\kappa = \eta(2\omega_1)^{-1}$ with the periods of the second kind $2\eta$, $K_\infty$ is the vector of Riemann constants, and $C$ is a constant which does not matter here (for further details see e.g.~\cite{BuchstaberEnolskiiLeykin97}; note that $K_\infty=\tau(\frac{1}{2},\frac{1}{2})^t + (0,\frac{1}{2})^t$ if $g=2$). The solutions of \eqref{Jacobiinversion} are then given by the solutions of
\begin{align}
x^g + \sum_{i=1}^g \wp_{gi}(y_1,\ldots,y_g) x^{i-1} = 0\,.
\end{align}
Let us consider now the case $g=2$ and a general differential equation of the form
\begin{align}
\left(\frac{dx}{dy}\right)^2 & = P_{5,6}(x)\,, \quad x(y_0)=x_0\label{genhyperdiff}
\end{align}
where $P_{5,6}$ is a polynomial of degree five or six. If it is of degree six it can be reformulated as $\tilde{x} \frac{d\tilde{x}}{dy}=\sqrt{P_5(\tilde{x})}$ by a substitution $x=\tilde{x}^{-1}+x_P$, where $x_P$ is a zero of $P_{5,6}$ and $P_5$ is a polynomial of degree five. Such a differential equation, or, if $P_{5,6}$ was of degree five in the first place, can then be cast in the form
\begin{align}
t^{i-1}\frac{dt}{dy}=\sqrt{P(t)}\,, \quad i=1,2 \label{normhyperdiff}
\end{align}
by an appropriate normalization. We may then find the solution to this equation as the limiting case $x_2\to\infty$ of the Jacobi inversion problem \eqref{Jacobiinversion} in the following way: first observe that
\begin{align}
t & := x_1 = \lim_{x_2\to\infty} \frac{x_1x_2}{x_1+x_2} = \lim_{x_2\to\infty} \frac{\wp_{12}(y_1,y_2)}{\wp_{22}(y_1,y_2)}\nonumber\\
& = \lim_{x_2\to\infty} \frac{\sigma\sigma_{12}-\sigma_1\sigma_2}{\sigma_2^2-\sigma\sigma_{22}} (y_1,y_2)\,,\label{solx}
\end{align}
where $\sigma_i(z)$ is the derivative of $\sigma$ with respect to $z_i$. From \eqref{Jacobiinversion} with $g=2$ and $x_2\to\infty$ we may identify either $y_1$ or $y_2$ with our physical coordinate $y$ in the differential equation \eqref{normhyperdiff}, say $y_i$. Fortunately, we get automatically rid of the other $y_j$, $j\neq i$, by the same limiting process $x_2\to\infty$. This is because the set of zeros of the theta function $z\mapsto \theta[K_\infty]((2\omega)^{-1} z)$, which is a one dimensional submanifold of $\mathbb{C}^2$, is given by all vectors $z=(z_1,z_2)$ which can be written as $z_i=\int_\infty^x \frac{t^{i-1}dt}{\sqrt{P(t)}}$ with the same $x$ (see e.g.~\cite{Mumford83}). This is exactly true for the vector $(y_1,y_2)$, which means that we may write $y_j=f(y_i)$ for some function $f$. As the zeros of the theta function are also zeros of $\sigma$ we can simplify \eqref{solx} to
\begin{align}
t & = - \frac{\sigma_1}{\sigma_2} (y_1,f(y_1)) \quad \text{ or } \quad t = - \frac{\sigma_1}{\sigma_2} (f(y_2),y_2)\,.
\end{align}
Note that according to the given initial condition we actually have that $y_i$ is the physical coordinate minus a constant.
\section{Analytical solutions in the literature}
In this section we will collect applications of the methods outlined in section \ref{sec:AS} to geodesic motion in the Kerr-Newman-Taub-NUT-de Sitter spacetime as given in section \ref{sec:spactime}. For older literature we refer to Sharp \cite{Sharp1979} who collected most of the papers on geodesic motion in Kerr-Newman spacetime and subclasses, which were available at that time. Partly this is still quite complete but we also try to update his collection (with respect to analytical solutions). Note that we only consider analytical solutions to general timelike and lightlike geodesics (with an electric or magnetic charge, as applicable). In particular, we do not list the vast literature on equatorial motion in Kerr spacetime. Of course, we do not claim that our list is complete.
{\bf{Schwarzschild:}} Regarding analytical solution methods the list of Sharp already contained the complete set of solutions. Most notably, this includes the works by Hagihara \cite{Hagihara}, who derived the analytical solutions in terms of Weierstrass elliptic functions, and Darwin \cite{Darwin1959,Darwin1961}, who used Jacobian elliptic functions and integrals.
{\bf{Reissner-Nordstr\"om:}} Surprisingly, the analytical solutions to the geodesic equation in Reissner-Nordstr\"om spacetime seem to be considered first only in 1983 by Gackstatter \cite{Gackstatter1983} although it can be handled completely analogously to the Schwarzschild case. He studied bound timelike geodesics and light in terms of Jacobian elliptic integrals and functions. Recently, Slez\'{a}kov\'{a} \cite{Slezakova2006} gave a comprehensive analysis of arbitrary timelike, lightlike, and even spacelike geodesics. Grunau and Kagramanova \cite{Grunau2011} solved the equations of motion of electrically and magnetically charged particles in Reissner-Nordstr\"om spacetime in terms of Weierstrass elliptic functions.
{\bf{Taub-NUT:}} Timelike geodesics were studied by Kagramanova et al \cite{Kagramanova2010} in terms of Weierstrass elliptic functions.
{\bf{Kerr:}} Most of the older literature on Kerr spacetime is concerned with the much simpler particular case of equatorial geodesics. We refer to Sharp \cite{Sharp1979} here for these early works. Note that in terms of the proper time (or the corresponding affine parameter for light) the equations of motion are still coupled. Therefore, most of the analytical solutions before the introduction of the Mino time \cite{Mino03} implicitly included integrals over the latitude or the radius, see e.g. Kraniotis \cite{Kraniotis2005} or Slez\'{a}kov\'{a} \cite{Slezakova2006} for a review. As notable exception, \v{C}ade\v{z} et al \cite{Cadezetal1998} introduced already in 1998 a similar parameter (called $P$, see their equation (34)) as they considered the motion of light.
After the introduction of the Mino time, in 2009 Fujita and Hikida \cite{FujitaHikida09} used this new affine parameter to derive the analytical solution for bound timelike geodesics in terms of Jacobian elliptic functions. General timelike geodesics and lightlike motion were treated shortly after that by Hackmann \cite{HackmannDiss} in 2010.
Note that Kraniotis \cite{Kraniotis2011} also derived analytical solutions for lightlike geodesics in terms of hypergeometric functions.
{\bf{Kerr-Newman:}} Charged particle motion was considered by Hackmann and Xu \cite{Hackmannetal2013} in terms of Weierstrass elliptic functions.
{\bf{Schwarzschild-de Sitter}}, also called Kottler space-time: Note that on the level of the differential equation, lightlike geodesics in Schwarzschild-de Sitter are identical with the lightlike equations of motion for Schwarzschild, as the cosmological constant can be absorbed in the definition of just a single parameter. Analytical solution are given e.g. in Gibbons et al \cite{Gibbonsetal2008}. General timelike geodesics in Kottler spacetime can be treated in terms of hyperelliptic functions as elaborated by Hackmann and L\"ammerzahl \cite{HackmannLaemmerzahl2008,HackmannLaemmerzahl08b}.
{\bf{Reissner-Nordstr\"om-de Sitter:}} The equations of motion for general timelike geodesics were solved in \cite{Hackmannetal2008}. The motion of photons was very recently analytically calculated by Villanueva et al \cite{Villanuevaetal2013} for a negative cosmological constants using Weierstrass elliptic functions.
{\bf{Taub-NUT-de Sitter:}} Timelike motion was analyzed in \cite{Hackmannetal2009}.
{\bf{Kerr-de Sitter:}} The equations of motions for timelike geodesics were analytically solved by Hackmann et al \cite{Hackmannetal2009,Hackmannetal2010} in terms of hyperelliptic functions. Note that Kraniotis \cite{Kraniotis2011} also derived analytical solutions for lightlike geodesics in terms of hypergeometric functions.
{\bf{Kerr-Newman-Taub-NUT-de Sitter:}} The general solution for timelike geodesic was shortly outlined in \cite{Hackmannetal2009}.
\begin{acknowledgement}
We thank the German Research Foundation DFG for financial support within the Research Training Group 1620 ``Models of Gravity''.
\end{acknowledgement}
\input{ProcKSM.bbl}
\end{document}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,661
|
Marx 200: Message from Cuba: Revolution teaches
Cuban revolutionaries greet our party as we celebrate 200 years since Marx's birth together.
Cuba Embassy
Julio Pugot, Counsellor of the Cuban embassy in Britain speaks at a meeting to mark the 200th anniversary of Marx's birth held in April 2018 in Saklatvala Hall, Southall.
Comrade Pugot brings greetings from Cuba, where Comrade Miguel Mario Díaz-Canel Bermúdez has recently been elected president – the first leader of the country who did not directly take part in the revolution. Indeed, Comrade Díaz-Canel is the first who was born after the revolution took place, in 1960 in socialist Cuba – making him a true child of the revolution.
Comrade Raúl Castro continues to head the Communist Party of Cuba and the Cuban army, and this is a sign of strength and continuity in the revolution, the ambassador tells us.
Cuba continues to extend aid to workers in need throughout the world, despite the ongoing blockade that has been further entrenched by the regime of US president Donald Trump – following in the footsteps of the 11 presidents who preceded him.
Comrade Ranjeet, chairing the meeting for the CPGB-ML, thanks Comrade Pugot for speaking at the meeting and introduces the main speaker, Comrade Dan, noting that even the BBC and financial capital agreed, in the wake of the 2008 financial crisis, that Karl Marx had not only been the "greatest thinker of the millennium", but that he had much to say about the nature of capitalist economy and crisis.
However, in doing so, they sought to strip Marx of his revolutionary essence, pretending that, while he had analysed the workings and inherent flaws of capitalism, he had been unable to offer any solutions.
And yet, quite the reverse is true. For Marx's real conclusions were that workers must take control of the productive forces and institute the rule of the working class – the dictatorship of the proletariat; the rule of the majority – in place of the dictatorship of the financial oligarchy, and build a planned economy, capable of using the great productive power that presently exists to satisfy the needs of working masses and to build a sustainable and humane economic order.
Cuba blockade
Marx 200
Alicia Alonso: architect of Cuba's socialist ballet
Cuba 60: Hasta la victoria siempre!
China celebrates Marx's 200th birthday
Workers must continue to stand in solidarity with revolutionary Cuba
Marx 200: A brief overview of an extraordinary life's work
Why do Marxists support the Syrian government?
Economics 101: Capitalism
Tribute to Marx on his 200th birthday
Cuba and the October Revolution
Cuba's view of Venezuela: Cuban ambassador to Britain speaks
A true daughter of the masses, whose talent and art was put at the service of the people and the building of a new society.
20 Dec 2019, Lalkar
Progressive humanity continues to cheer the incredible achievements of the tiny island nation, made in the teeth of the imperialists' determined and unceasing aggression.
14 Feb 2019, Proletarian
CPGB-ML participates in major international conference with other communist parties from around the world.
19 Oct 2018, Proletarian
Comrade Fidel may have gone, but the Cuban revolution lives on, and deserves the support of progressive humanity.
15 Nov 2018,
Why does Marx's name still strike fear and loathing into the hearts and minds of the bourgeoisie so many years after his death? What is the legacy that refuses to die?
19 Nov 2018, Lalkar
A brief overview of what Marxism Leninism has to tell us about the struggles for national liberation currently being waged all over the world.
11 Jul 2018,
An essential introduction to the workings of the capitalist system. Only by recognising the cause of our problems will we be able to find real solutions.
In founding scientific socialism, Marx gave workers the ability to liberate themselves from capitalist slavery and revealed their historical mission.
6 May 2018, Proletarian TV
Cuba's ambassador Teresita Vincente Sotolongo gives a stirring tribute to the October Revolution and outlines its impact on the oppressed world.
5 Oct 2018, Proletarian TV
The media war against Venezuela is part of the US's drive to control the entire continent, and especially Venezuela's important oil and coltan reserves.
10 Apr 2019, Proletarian TV
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,481
|
using System;
class Square
{
public int x, y;
public ConsoleColor color = ConsoleColor.Green;
public Char character = '#';
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,190
|
from __future__ import absolute_import, division, print_function, unicode_literals
import unittest
from stone.cli_helpers import parse_route_attr_filter
class MockRoute():
"""Used to test filtering on a route's attrs."""
def __init__(self, attrs):
self.attrs = attrs
class TestCLI(unittest.TestCase):
def test_parse_route_attr_filter(self):
_, errs = parse_route_attr_filter('*=3')
self.assertNotEqual(len(errs), 0)
_, errs = parse_route_attr_filter('test')
self.assertEqual(len(errs), 1)
self.assertEqual(errs[0], 'Unexpected end of expression.')
_, errs = parse_route_attr_filter('hide=true)')
self.assertNotEqual(len(errs), 0)
_, errs = parse_route_attr_filter('(hide=true')
self.assertNotEqual(len(errs), 0)
_, errs = parse_route_attr_filter('hide=true and and size=1')
self.assertNotEqual(len(errs), 0)
# Test bool
expr, errs = parse_route_attr_filter('hide=true')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'hide': True})))
self.assertFalse(expr.eval(MockRoute({'hide': 'true'})))
# Test int
expr, errs = parse_route_attr_filter('level=1')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'level': 1})))
self.assertFalse(expr.eval(MockRoute({'level': 2})))
self.assertFalse(expr.eval(MockRoute({'level': '1'})))
self.assertFalse(expr.eval(MockRoute({})))
# Test float
expr, errs = parse_route_attr_filter('f=1.25')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'f': 1.25})))
self.assertFalse(expr.eval(MockRoute({'f': 3})))
self.assertFalse(expr.eval(MockRoute({'f': '1.25'})))
self.assertFalse(expr.eval(MockRoute({})))
# Test string
expr, errs = parse_route_attr_filter('status="alpha"')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'status': 'alpha'})))
self.assertFalse(expr.eval(MockRoute({'status': 'beta'})))
self.assertFalse(expr.eval(MockRoute({'status': 0})))
self.assertFalse(expr.eval(MockRoute({})))
# Test null
expr, errs = parse_route_attr_filter('status=null')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'status': None})))
self.assertFalse(expr.eval(MockRoute({'status': 'beta'})))
self.assertFalse(expr.eval(MockRoute({'status': 0})))
self.assertTrue(expr.eval(MockRoute({})))
# Test conjunction: or
expr, errs = parse_route_attr_filter('a=1 or b=1')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1})))
self.assertTrue(expr.eval(MockRoute({'b': 1})))
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 1})))
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 10})))
self.assertFalse(expr.eval(MockRoute({'a': '0', 'b': 0})))
self.assertFalse(expr.eval(MockRoute({'a': 0})))
self.assertFalse(expr.eval(MockRoute({})))
# Test conjunction: and
expr, errs = parse_route_attr_filter('a=1 and b=1')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 1})))
self.assertFalse(expr.eval(MockRoute({'a': 1})))
self.assertFalse(expr.eval(MockRoute({'b': 1})))
self.assertFalse(expr.eval(MockRoute({'a': 1, 'b': 10})))
self.assertFalse(expr.eval(MockRoute({'a': '0', 'b': 0})))
self.assertFalse(expr.eval(MockRoute({'a': 0})))
self.assertFalse(expr.eval(MockRoute({})))
# Test multiple conjunctions
expr, errs = parse_route_attr_filter('a=1 or a=2 or a=3')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1})))
self.assertTrue(expr.eval(MockRoute({'a': 2})))
self.assertTrue(expr.eval(MockRoute({'a': 3})))
self.assertFalse(expr.eval(MockRoute({'a': 4})))
# Test "and" has higher precendence than "or"
expr, errs = parse_route_attr_filter('a=1 or a=2 and b=3 and c=4')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1})))
self.assertFalse(expr.eval(MockRoute({'a': 2})))
self.assertTrue(expr.eval(MockRoute({'a': 2, 'b': 3, 'c': 4})))
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 3, 'c': 4})))
self.assertFalse(expr.eval(MockRoute({'a': 0, 'b': 3, 'c': 4})))
expr, errs = parse_route_attr_filter('a=2 and b=3 and c=4 or a=1')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1})))
self.assertFalse(expr.eval(MockRoute({'a': 2})))
self.assertTrue(expr.eval(MockRoute({'a': 2, 'b': 3, 'c': 4})))
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 3, 'c': 4})))
self.assertFalse(expr.eval(MockRoute({'a': 0, 'b': 3, 'c': 4})))
# Test parentheses for overriding precedence
expr, errs = parse_route_attr_filter('(a=1 or a=2) and b=3 and c=4')
self.assertEqual(len(errs), 0)
self.assertTrue(expr.eval(MockRoute({'a': 1, 'b': 3, 'c': 4})))
self.assertTrue(expr.eval(MockRoute({'a': 2, 'b': 3, 'c': 4})))
self.assertFalse(expr.eval(MockRoute({'a': 1})))
self.assertFalse(expr.eval(MockRoute({'a': 1, 'b': 3})))
if __name__ == '__main__':
unittest.main()
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Platì (Griechisch: Πλατύ) ist eine Gemeinde der Metropolitanstadt Reggio Calabria in der Region Kalabrien in Italien.
Lage und Daten
In Platì wohnen Einwohner (Stand ). Das Dorf liegt am Fuße des Gebirges Aspromonte.
Die Nachbargemeinden sind Ardore, Benestare, Careri, Ciminà, Oppido Mamertina, Santa Cristina d'Aspromonte und Varapodio. Zur Gemeinde gehört der Ort Cirella.
Einwohnerentwicklung
Mafia
Das Dorf gilt als Hochburg der 'Ndrangheta. Hier soll es unterirdische Gänge und Kammern mit getarnten oder versteckten Türen geben, welche den Mitgliedern der Verbrecherorganisation ein Verschwinden bei Gefahr ermöglichen. Trotzdem versucht der italienische Staat immer wieder, Herr der Lage zu werden. Beispielsweise stürmten am 13. November 2003 insgesamt über 1000 Carabinieri nachts das Dorf, wobei 131 Verdächtige festgenommen wurden.
Weblinks
Wo fast jeder mit einem Kriminellen verwandt ist Neue Zürcher Zeitung vom 12. September 2015
Einzelnachweise
Ort in Kalabrien
|
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Q: Python - Write to csv from a particular column number I have a csv file Temp.csv that I am copying over to another csv file Final.csv using the following piece of code.
dirname = os.path.dirname(os.path.abspath(__file__))
csvfilename = os.path.join(dirname, 'Final.csv')
tempfile = os.path.join(dirname, 'Temp.csv')
with open(csvfilename, 'wb') as output_file:
writer = csv.writer(output_file, delimiter=',')
writer.writerow(["Title","This"])
writer.writerow([])
with open(tempfile, 'r') as data_file:
for line in data_file:
line = line.replace('\n', '')
row = line.split(",")
writer.writerow(row)
The last 5 lines of the code are going to write to the Final.csv file from column A. I want them to write to final.csv from column D. Also I have multiple columns in the Temp.csv file. What is the best solution for this? Using python 2.7.10.
A: EDIT Answering comment. You just need to prepend 3 empty strings to your row
dirname = os.path.dirname(os.path.abspath(__file__))
csvfilename = os.path.join(dirname, 'Final.csv')
tempfile = os.path.join(dirname, 'Temp.csv')
with open(csvfilename, 'wb') as output_file:
writer = csv.writer(output_file, delimiter=',')
writer.writerow(["Title","This"])
writer.writerow([])
with open(tempfile, 'r') as data_file:
for line in data_file:
line = line.replace('\n', '')
row = line.split(",")
# there are better ways to do this but this is the most straight forward
new_row = ['', '', '']
new_row.extend(row)
writer.writerow(new_row)
A: You can try :
with open(tempfile, 'r') as data_file:
for line in data_file:
line = line.replace('\n', '')
row = line.split(",")
csvfilename.writerow([row[0]+","+row[3]])
|
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"redpajama_set_name": "RedPajamaStackExchange"
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\section{Star Formation Efficiencies \& Star Cluster Formation}
Both burst strengths b defined by the relative increase of the stellar mass
in the course of starbursts, ${\rm b:=\frac{\Delta S_{burst}}{S_{total}}}$, and
Star Formation Efficiencies (SFEs) defined as the
total stellar mass formed out of an available mass of gas, ${\rm SFE :=}$
${\rm \frac{\Delta S_{burst}}{G}}$, are difficult to determine. Reasonable estimates are only possible in young post-starbursts. As
long as a burst is active only lower limits can be given. Once a burst is over
or if a burst lasts longer than the most massive stars' lifetimes, the amount
of stars already died needs to be accounted for. For
the stellar and gaseous masses before the burst can only be estimated on the basis of Hubble types, HI observations, etc.
The strongest bursts are reported in mergers of
massive gas-rich galaxies with total burst durations of the order of a few
100 Myr.
Bursts in massive interacting galaxies are much stronger and last much longer
than those in isolated dwarf galaxies. Blue Compact Dwarf
galaxies (BCDs), e.g., feature bursts with durations of the order of a
few Myr, ${\rm b \ll 0.1}$, ${\rm SFE \leq 0.01}$, and a trend of decreasing
burst strengths for increasing total galaxy masses (including HI)
(Kr\"uger {\it et al.\ } 1995). Massive interacting galaxies feature bursts
stronger and more efficient by one to two
orders of magnitude, similar to the
progenitors of E+A galaxies in clusters, ULIRGS, SCUBA galaxies, and
optically identified starburst galaxies in the early universe.
The post-burst spiral -- spiral merger remnant NGC 7252 with
two long, gas-rich tidal tails pointing at an age of $\mathrel{\spose{\lower 3pt\hbox{$\mathchar"218$} 1$ Gyr after the
onset of the strong interaction and its blue and radially constant colors and very strong Balmer absorption line spectrum must have experienced a very strong and global starburst
increasing its stellar mass by as much as $\sim 40$ \%
between 600 and 1000 Myr ago. Conservative estimates still lead to a very high
SFE $\geq 30$\% (Fritze -- v. Alvensleben \& Gerhard 1994). A large number of Star Clusters (SCs)
formed throughout the main body,
many of them apparently so strongly bound that they managed to survive for
500 $-$ 900 Myr the violent relaxation phase that restructured the remnant into a de Vaucouleurs profile (cf. Fritze -- v. Alvensleben \& Burkert 1995 and
Schweizer 2002 for a recent review).
Most of these star clusters are young Globular Clusters (GCs)
based on their ages, luminosities, and radii. How many clusters were
already destroyed since the onset of the burst?
An analogous system at a younger age is NGC 4038/39 where the two spiral
disks just started overlapping. Its burst around the two nuclei,
along the tidal structures, and -- strongest -- in the optically obscured
disk -- disk overlap region is in its initial stage. Thousands of bright
Young Star Clusters (YSCs) are
seen with luminosities ranging from those of individual red supergiant stars to
${\rm M_V \geq -15}$. How many of these will survive for $\gg 1$ Gyr
and become Globular Clusters (GCs)?
Hydrodynamic modelling shows that the formation of long-lived strongly bound GCs
requires SFEs $\gg 10$ \%, originally thought
to only occur in the early universe. In normal SF in spirals,
irregulars, and starbursting isolated dwarfs like BCDs, GC formation should
not be possible. In the high pressure ISM with its strong shocks in spiral
-- spiral mergers, however, GC formation is observed in reality
and in high-resolution hydrodynamical models (Yuexing {\it et al.\ } 2004). Young
and intermediate age GCs hence are tracers of high SFE periods in their
parent galaxies.
A number of very fundamental questions are still open at present:
Does the amount of SF that goes into massive, compact, long-lived SCs scale with
burst strength and/or (local/global) SFE? Or is there a threshold in SFE, below
which only field stars and weakly bound, less massive SCs or OB-associations can
be formed that dissolve on timescales of $10^8$ yr, and above which GCs can be
formed or even become the dominant component? Does the same star and SC formation
mechanism work in vastly different environments and scale over a huge dynamical
range or are there two different modes of SF like ``normal'' and ``violent''?
SCs are seen to form in many environments from normal Irrs and
spirals through dwarf starbursts, spiral mergers, and ULIRGs, constrained to nuclear regions (e.g. in ULIRGs), over their main body (e.g. NGC 4038/39), and all along tremendous tidal tails (e.g.
Tadpole cf. de Grijs {\it et al.\ } 2003). The spatial extent of a starburst probably depends on the orbit and relative
orientations of the interacting galaxies, on whether or not they had massive
bulges and/or DM halos. Are all
these YSC systems similar or systematically different in terms of
masses, mass functions, sizes, compactness or degree of binding and, hence,
survival times.
SCs are {\bf S}imple {\bf S}tellar {\bf P}opulations ({\bf SSP}s) with all stars
having the same age and chemical composition. Evolutionary synthesis models
like {\bf GALEV} describe the evolution of
SCs over a Hubble time, from the youngest stages of 4 Myr all through the oldest
GC ages $\geq 14$ Gyr for 5 different
metallicities ${\rm -1.7 \leq [Fe/H] \leq}$ $+0.4$. The TP-AGB phase is very important for age-dating of SCs between 100 Myr and a few Gyr on the basis of their ${\rm
V-I}$ colors
(cf. Schulz {\it et al.\ } 2002). Gaseous emission in terms of
lines and continuum for the respective metallicities makes important
contributions to broad band fluxes and colors at young ages (Anders \& Fritze -- v. Alvensleben 2003). Lick absorption indices significantly help disentangle ages and
metallicities of older SCs (Lilly \& Fritze -- v. Alvensleben {\sl in prep.}). {\bf GALEV} models yield the
detailed spectral evolution of SCs from 90 \AA \ through 160 $\mu$m, luminosities,
M/L-ratios, and colors in many filter systems (Johnson, HST, Washington,
Stroemgren, \dots) and can be retrieved from
{\sl http://www.uni-sw.gwdg.de/$^{\sim}$galev/}\ .
\section{Analysing Star Cluster Systems}
The time evolution of luminosities, colors, and M/L-ratios significantly depends
on metallicity in a way that is different in different wavelengths regions. For young
SC systems, like in NGC 4038/39, extinction is an important issue. Older
starbursts, like in NGC 7252, are significantly less extincted.
An ESO -- ASTROVIRTEL project provides us with HST and VLT multi-$\lambda$ photometry for SC
systems from young to old that have 4 and more passbands observed. A dedicated Spectral Energy
Distribution (SED) analysis tool called AnalySED compares observed SC SEDs with an extensive grid
of $117 \, 000$ SSP model SEDs for 5 different metallicities,
1170 ages from 4 Myr through 14
Gyr, and 20 extinction values ${\rm 0 \leq E_{B-V} \leq 1}$.
We use Calzetti {\it et al.\ } 's (2000) starburst extinction law since internal
extinction is only an issue in ongoing starbursts.
A probability ${\rm p(n) \sim exp(-\chi^2)}$ is assigned to each model SED
by a
maximum likelihood estimator ${\rm \chi^2=\sum_{\lambda}(m_{\lambda}^{obs}-
m_{\lambda}^{model})^2/\sigma_{obs}^2}$. The best fit model is the one with the
highest probability. Probabilities are normalised to ${\rm \sum_n p(n)=1}$.
Summing models with decreasing probabilities until ${\rm \sum_n p(n)=0.68}$
provides $\pm 1 \sigma$ uncertainties for ages, metallicities, exticntion values, and masses of individual SCs (Anders {\it et al.\ } 2004a).
Testing AnalySED with artificial SCs, we found
that there are good and bad passband combinations,
slightly depending on the ages and metallicities of the clusters, and we
identified a combination of 4 passbands U, B, V or I, and H or K with
observational photometric accuracy $\leq 0.2$ mag as optimal for YSC
systems. We agree with the independent investigation by Cardiel
{\it et al.\ } (2003), that at typical photometric accuracies broad band photometry
with useful passband combinations is as powerful in disentangling ages and
metallicities as is spectroscopy with typical S/N.
The AnalySED tool is currently extended to also include Lick indices for
analyses of intermediate-age and old GC systems (Lilly {\it et al.\ }, {\sl in prep.},
cf. Lilly's poster on the accompanying CD-ROM).
In the dwarf starburst galaxy NGC 1569 we identify 169 YSCs on the ASTROVIRTEL
images, the bulk of them with ages $\leq 25$ Myr, low extinction and metallicities. Their
masses are typically in the range from $10^3$ to ${\rm 10^4 M_{\odot}}$,
only the two Super SCs have
masses in the mass range of GCs (cf. Fig.1 and Anders {\it et al.\ } 2004b).
\begin{figure}[ht]
\begin{center}
\includegraphics[width=6.cm]{Fritze_Fig1.ps}
\end{center}
\caption{Distribution of YSC masses in NGC 1569 (histogram) as compared to the
Milky Way GC mass function, normalised to the same number of clusters (Gaussian
curve).}
\end{figure}
We conclude that the starburst in the dwarf galaxy NGC 1569 did not form
many GCs, whereas the starburst in the spiral -- spiral merger NGC 7252 did.
Does SC formation produce a continuum in masses and binding energies or are
there different modes of SC formation that respectively produce open and
globular clusters? With increasing burst strength an increasing number
of SCs is formed. Does the statistical effect of having a higher chance to have
a more massive cluster within a larger sample explain the difference
between cluster masses in NGC 1569 and NGC 7252? Probably not. Mass Functions (MFs) are power laws for open
cluster systems and Gaussians for old
GC systems. Initial MFs derived from models for survival
and destruction of GCs in galactic potentials are controversial.
Vesperini (2001)
favors an initially Gaussian shape for GC MFs that is essentially conserved
by the competing destructions of low-mass GCs through tidal disruption and of high-mass GC by
dynamical friction. Gnedin \& Ostriker (1997) favor an initially power-law GC MF
that is secularly transformed into a Gaussian by higher destruction rates for
lower mass GCs. The MF of the young SC system in NGC 4038/39, that very probably
comprises a mixture of OB-associations, open, and GCs, as derived
from HST photometry, is also controversial. Whereas Zhang \& Fall (1999)
derive a power-law MF using reddening-free Q$_1$Q$_2$ indices, we find a
Gaussian MF (Fritze -- v. Alvensleben 1998, 1999). Both approaches have
their drawbacks. Zhang \& Fall excluded a
significant number of clusters from the ambiguous age range in the Q$_1~-~$Q$_2$
- plot, we assumed a uniform average reddening in the WFPC1 UVI data. If we exclude the same SCs as
Zhang \& Fall, we also find a power-law. The dust distribution in NGC
4038/39 is clearly patchy and a reanalysis of ASTROVIRTEL UBVIK data
from HST WFPC2 and VLT ISAAC underway (Anders {\it et al.\ } {\sl in prep.}).
The shape of the MF need not correspond to the shape
of the Luminosity Function (LF) for a young SC system,
as age spread effects can distort the shape of the LF with respect to that of the
underlying MF -- to the point of transforming a Gaussian MF into a
power-law LF up to the observational completeness limit.
The key to survival or destruction is the strength of a SC's
internal gravitational binding, as measured for Galactic GCs by their concentration parameters c. By definition, ${\rm c:= log
\frac{r_t}{r_c}}$ involves the tidal and core radii. Very young clusters need
not be tidally truncated yet and tidal radii could not be measured for the bulk of the
YSCs on top of the bright galaxy background in NGC 4038/39 anyway.
We therefore define the compactness of a young SC by the ratio between its mass
and half-light radius (cf. Anders {\it et al.\ } {\sl in prep.}), a robust quantity that can reliably be measured and is
predicted by dynamical SC evolution models not to significantly change over
a Hubble time. To this aim, we first have to
improve upon the determination of SC radii by using appropriate aperture
corrections. Improved SC radii, in turn, lead to improved SC photometry, and,
hence, to improved photometric masses
(cf. Poster by P. Anders on the accompanying CD-ROM).
\section{Conclusions and Perspective}
From the ages, masses, and radii of their SCs we know that major gas-rich mergers can form
significant secondary populations of GCs in their strong and global starbursts.
SFEs in mergers are higher by 1$-$2 orders of magnitude
than in normal SFing galaxies and (non-interacting) dwarf galaxy starbursts. Comparing good precision
photometry in at least 4 reasonably chosen passbands (e.g. UBVK) to GALEV
evolutionary synthesis models for SCs with given age, metallicity, extinction,
and mass by means of a dedicated SED analysis tool (AnalySED) allows to
reliably determine individual SC ages, metallicities, extinction values, and
masses, including their respective 1$\sigma$ uncertainties. The first dwarf
galaxy starburst analysed in detail this way shows only very few clusters
with masses in the range of GCs among its $\sim 170$ YSCs. Clearly, both more
major merger and dwarf galaxy starbursts need to be analysed in detail.
Pixel-by-pixel analyses (de Grijs {\it et al.\ } 2003) or integrated field spectroscopy can provide burst strengths and SFEs. From a comparison with
HST multi-$\lambda$
imaging of their YSC systems the relative ratios of SF going into
field stars, short-lived open clusters, and long-lived GCs, respectively, can be determined. A key
question is whether these quantities as
well as the intrinsic properties of the YSCs, like masses and half-mass radii,
depend on environment or not, in a smooth way or with some threshold. A
comparison of starbursts in dwarf and massive, interacting and non-interacting
starbursts should tell if SF and SC formation are universal processes or
depend on environment. GC age and metallicity distributions will allow to trace back a galaxy's violent (star) formation
history and constrain galaxy formation scenarios (Fritze --
v. Alvensleben 2004). This requires B through
NIR photometry and medium resolution spectra to measure Lick indices.
With only one observed color we cannot disentangle the age -- metallicity degeneracy of
intermediate-age and old stellar populations
and see if more than one GC population is hidden in
the red peak of many elliptical galaxies' bimodal color distributions.
\begin{acknowledgments} I gratefully acknowledge travel support, in part from the DFG (FR 916/10-2) and in part from the organisers.\end{acknowledgments}
\begin{chapthebibliography}{20}
\bibitem{} Anders, P., Fritze -- v. Alvensleben, U., 2003, A\&A 401, 1063
\bibitem{} Anders, P., Bissantz, N., Fritze -- v. Alvensleben, U., de Grijs, R., 2004a, MN 347, 196
\bibitem{} Anders, P., de Grijs, R., Fritze -- v. Alvensleben, U., Bissantz, N., 2004b, MN 347, 17
\bibitem{} Calzetti, D., Armus, L., Bohlin, R. C., Kinney, A. L., Koorneef, J., Storchi -- Berrgmann, T., 2000, ApJ 533,682
\bibitem{} Cardiel, N., Gorgas, J., S\'anchez -- Bl\'azquez, P., Cenarro, A. J., Pedraz, S., Bruzual, G., Klement, J., 2003, A\&A 409, 511
\bibitem{} Fritze -- v. Alvensleben, U., 1998, A\&A 336, 83
\bibitem{} Fritze -- v. Alvensleben, U., 1999, A\&A 342, L25
\bibitem{} Fritze -- v. Alvensleben, U., 2004, A\&A 414, 515
\bibitem{} Fritze -- v. Alvensleben, U., Burkert, A., 1995, A\&A 300, 58
\bibitem{} Fritze -- v. Alvensleben, U., Gerhard, O. E., 1994, A\&A 285, 775
\bibitem{} Gnedin, O. Y., Ostriker, J. P., 1997, ApJ 474, 223
\bibitem{} de Grijs, R., Lee, J., Mora Herrera, C., Fritze -- v. Alvensleben, U., Anders, P., 2003, New Astron. 8, 155
\bibitem{} Kr\"uger, H., Fritze -- v. Alvensleben, U., Loose, H.-H., 1995,
A\&A 303, 41
\bibitem{} Schulz, J., Fritze -- v. Alvensleben, U., M\"oller, C. S., Fricke, K. J., 2002, A\&A 392, 1
\bibitem{} Schweizer, F., 2002, IAU Symp. 207, 630
\bibitem{} Vesperini, E., 2001, MN 322, 247
\bibitem{} Yuexing, L., MacLow, M.-M., Klessen, R. S., 2004, ApJ 614, L29
\bibitem{} Zhang, Q., Fall, S. M., 1999, ApJ 527, L81
\end{chapthebibliography}
\end{document}
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{"url":"https:\/\/wormtooth.com\/20141130-tensor-product-of-two-linear-maps\/","text":"Let $A \\in End(V)$ and $B \\in End(W)$ be two linear maps. We can define naturally the tensor product $A \\otimes B$ of $A$ and $B$, from $V \\otimes W$ to $V \\otimes W$, sending $v \\otimes w$ to $Av \\otimes Bw$. In this post, I am going to realize $A \\otimes B$ as a matrix and relate the determinant and trace of $A \\otimes B$ to the ones of $A$ and $B$.\n\nLet $V$ and $W$ be two vector spaces over a field $K$ with $\\dim V=n$ and $\\dim W=m$. Let $e_1, \\dots, e_n$ be a basis of $V$ and let $f_1, \\dots, f_m$ a basis of $W$. Under the basis, a linear map $A: V \\to V$ can be realized as an $n \\times n$ matrix $(a_{ij})_{1 \\le i,j \\le n}$, where $a_{ij} \\in K$. Similarly, a linear map $B: W \\to W$ can be realized as a $m \\times m$ matrix $(b_{kl})_{1 \\le k, l \\le m}$. Now, $V \\otimes W$, as a vector space, has basis $\\{e_i \\otimes f_j: 1 \\le i \\le n, 1 \\le j \\le m\\}$. And, \\begin{align*} A \\otimes B (e_i \\otimes f_j) &= A(e_i) \\otimes B(f_j) \\\\ &= \\sum_{k}a_{ki}e_k \\otimes \\sum_{l}b_{lj}f_l \\\\ &= \\sum_{k,l}a_{ki}b_{lj} e_k \\otimes f_l. \\end{align*}\n\nFrom above calculation, we know that the $(k,l)$-th row, $(i,j)$-th column coefficient of the matrix of $A \\otimes B$ is $a_{ki}b_{lj}$. To write down the matrix explicitly, we need to give an order to $\\{(i,j):1 \\le i \\le n, 1 \\le j \\le m\\}$. For instance, we can sort the indexes in an ascending order by using the first coordinate as primary key and the second coordinate as the second key. Let\u2019s look at the simplest example. Let $A=\\left(\\begin{array}{cc} a & b \\\\ c & d \\end{array}\\right)$ and $B=\\left(\\begin{array}{cc} x & y \\\\ z & w \\end{array}\\right)$. Then $A \\otimes B = \\left(\\begin{array}{cccc} ax & ay & bx & by\\\\ az & aw & bz & bw\\\\ cx & cy & dx & dy\\\\ cz & cw & dz & dw \\end{array}\\right).$\n\nProposition. Let $\\alpha$ be the eigenvalue of $A$ with eigenvector $v$ and let $\\beta$ be the eigenvalue of $B$ with eigenvector $w$. Then $A \\otimes B$ has eigenvalue $\\alpha\\beta$ with eigenvector $v \\otimes w$.\n\nProof. This is an easy calculation: \\begin{align*} A \\otimes B (v \\otimes w) &= Av \\otimes Bw \\\\ &= \\alpha v \\otimes \\beta w \\\\ &= \\alpha\\beta \\, v \\otimes w. \\end{align*} This completes the proof. $\\square$\n\nSince we can express determinants and traces in terms of eigenvalues, the above proposition relates the determinants and traces of $A \\otimes B$, $A$ and $B$ together.\n\nCorollary. $(i) \\; \\det(A \\otimes B) = (\\det A)^m (\\det B)^n$.\n$(ii) \\; \\text{tr}(A \\otimes B) = \\text{tr}\\, A \\cdot \\text{tr}\\, B$.\n\nProof. Without loss of generality, we can assume is $K$ is algebraic closed, since we can replace $K$ with its algebraic closure without changing the determinants and traces.\n\nLet $\\alpha_1, \\dots, \\alpha_n$ be all eigenvalues(with multiplicities) of $A$ and let $\\beta_1, \\dots, \\beta_m$ be all eigenvalues(with multiplicities) of $B$. Then by the Proposition, $\\alpha_i\\beta_j$ are all the eigenvalues of $A \\otimes B$, for $1 \\le i \\le n$ and $1 \\le j \\le m$. Hence, \\begin{align*} \\det(A \\otimes B) &= \\prod_{i,j} \\alpha_i\\beta_j \\\\ &= \\left(\\prod_i \\alpha_i\\right)^m \\left(\\prod_j \\beta_j\\right)^n \\\\ &= (\\det A)^m(\\det B)^n, \\end{align*} And \\begin{align*} \\text{tr}(A \\otimes B) &= \\sum_{ij} \\alpha_i\\beta_j \\\\ &= \\left(\\sum_i \\alpha_i\\right)\\left(\\sum_j \\beta_j\\right) \\\\ &= \\text{tr} \\, A \\cdot \\text{tr} \\, B. \\end{align*} $\\square$","date":"2023-01-28 23:44:18","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\": 2, \"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\": 1.0000100135803223, \"perplexity\": 179.51939984161857}, \"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-2023-06\/segments\/1674764499695.59\/warc\/CC-MAIN-20230128220716-20230129010716-00477.warc.gz\"}"}
| null | null |
\section{Introduction}
The journal \textit{Monthly Notices of the Royal Astronomical Society} (MNRAS) encourages authors to prepare their papers using \LaTeX.
The style file \verb'mnras.cls' can be used to approximate the final appearance of the journal, and provides numerous features to simplify the preparation of papers.
This document, \verb'mnras_guide.tex', provides guidance on using that style file and the features it enables.
This is not a general guide on how to use \LaTeX, of which many excellent examples already exist.
We particularly recommend \textit{Wikibooks \LaTeX}\footnote{\url{https://en.wikibooks.org/wiki/LaTeX}}, a collaborative online textbook which is of use to both beginners and experts.
Alternatively there are several other online resources, and most academic libraries also hold suitable beginner's guides.
For guidance on the contents of papers, journal style, and how to submit a paper, see the MNRAS Instructions to Authors\footnote{\label{foot:itas}\url{http://www.oxfordjournals.org/our_journals/mnras/for_authors/}}.
Only technical issues with the \LaTeX\ class are considered here.
\section{Obtaining and installing the MNRAS package}
Some \LaTeX\ distributions come with the MNRAS package by default.
If yours does not, you can either install it using your distribution's package manager, or download it from the Comprehensive \TeX\ Archive Network\footnote{\url{http://www.ctan.org/tex-archive/macros/latex/contrib/mnras}} (CTAN).
The files can either be installed permanently by placing them in the appropriate directory (consult the documentation for your \LaTeX\ distribution), or used temporarily by placing them in the working directory for your paper.
To use the MNRAS package, simply specify \verb'mnras' as the document class at the start of a \verb'.tex' file:
\begin{verbatim}
\documentclass{mnras}
\end{verbatim}
Then compile \LaTeX\ (and if necessary \bibtex) in the usual way.
\section{Preparing and submitting a paper}
We recommend that you start with a copy of the \texttt{mnras\_template.tex} file.
Rename the file, update the information on the title page, and then work on the text of your paper.
Guidelines for content, style etc. are given in the instructions to authors on the journal's website$^{\ref{foot:itas}}$.
Note that this document does not follow all the aspects of MNRAS journal style (e.g. it has a table of contents).
If a paper is accepted, it is professionally typeset and copyedited by the publishers.
It is therefore likely that minor changes to presentation will occur.
For this reason, we ask authors to ignore minor details such as slightly long lines, extra blank spaces, or misplaced figures, because these details will be dealt with during the production process.
Papers must be submitted electronically via the online submission system; paper submissions are not permitted.
For full guidance on how to submit a paper, see the instructions to authors.
\section{Class options}
\label{sec:options}
There are several options which can be added to the document class line like this:
\begin{verbatim}
\documentclass[option1,option2]{mnras}
\end{verbatim}
The available options are:
\begin{itemize}
\item \verb'letters' -- used for papers in the journal's Letters section.
\item \verb'onecolumn' -- single column, instead of the default two columns. This should be used {\it only} if necessary for the display of numerous very long equations.
\item \verb'doublespacing' -- text has double line spacing. Please don't submit papers in this format.
\item \verb'referee' -- \textit{(deprecated)} single column, double spaced, larger text, bigger margins. Please don't submit papers in this format.
\item \verb'galley' -- \textit{(deprecated)} no running headers, no attempt to align the bottom of columns.
\item \verb'landscape' -- \textit{(deprecated)} sets the whole document on landscape paper.
\item \verb"usenatbib" -- \textit{(all papers should use this)} this uses Patrick Daly's \verb"natbib.sty" package for citations.
\item \verb"usegraphicx" -- \textit{(most papers will need this)} includes the \verb'graphicx' package, for inclusion of figures and images.
\item \verb'useAMS' -- adds support for upright Greek characters \verb'\upi', \verb'\umu' and \verb'\upartial' ($\upi$, $\umu$ and $\upartial$). Only these three are included, if you require other symbols you will need to include the \verb'amsmath' or \verb'amsymb' packages (see section~\ref{sec:packages}).
\item \verb"usedcolumn" -- includes the package \verb"dcolumn", which includes two new types of column alignment for use in tables.
\end{itemize}
Some of these options are deprecated and retained for backwards compatibility only.
Others are used in almost all papers, but again are retained as options to ensure that papers written decades ago will continue to compile without problems.
If you want to include any other packages, see section~\ref{sec:packages}.
\section{Title page}
If you are using \texttt{mnras\_template.tex} the necessary code for generating the title page, headers and footers is already present.
Simply edit the title, author list, institutions, abstract and keywords as described below.
\subsection{Title}
There are two forms of the title: the full version used on the first page, and a short version which is used in the header of other odd-numbered pages (the `running head').
Enter them with \verb'\title[]{}' like this:
\begin{verbatim}
\title[Running head]{Full title of the paper}
\end{verbatim}
The full title can be multiple lines (use \verb'\\' to start a new line) and may be as long as necessary, although we encourage authors to use concise titles. The running head must be $\le~45$ characters on a single line.
See appendix~\ref{sec:advanced} for more complicated examples.
\subsection{Authors and institutions}
Like the title, there are two forms of author list: the full version which appears on the title page, and a short form which appears in the header of the even-numbered pages. Enter them using the \verb'\author[]{}' command.
If the author list is more than one line long, start a new line using \verb'\newauthor'. Use \verb'\\' to start the institution list. Affiliations for each author should be indicated with a superscript number, and correspond to the list of institutions below the author list.
For example, if I were to write a paper with two coauthors at another institution, one of whom also works at a third location:
\begin{verbatim}
\author[K. T. Smith et al.]{
Keith T. Smith,$^{1}$
A. N. Other,$^{2}$
and Third Author$^{2,3}$
\\
$^{1}$Affiliation 1\\
$^{2}$Affiliation 2\\
$^{3}$Affiliation 3}
\end{verbatim}
Affiliations should be in the format `Department, Institution, Street Address, City and Postal Code, Country'.
Email addresses can be inserted with the \verb'\thanks{}' command which adds a title page footnote.
If you want to list more than one email, put them all in the same \verb'\thanks' and use \verb'\footnotemark[]' to refer to the same footnote multiple times.
Present addresses (if different to those where the work was performed) can also be added with a \verb'\thanks' command.
\subsection{Abstract and keywords}
The abstract is entered in an \verb'abstract' environment:
\begin{verbatim}
\begin{abstract}
The abstract of the paper.
\end{abstract}
\end{verbatim}
\noindent Note that there is a word limit on the length of abstracts.
For the current word limit, see the journal instructions to authors$^{\ref{foot:itas}}$.
Immediately following the abstract, a set of keywords is entered in a \verb'keywords' environment:
\begin{verbatim}
\begin{keywords}
keyword 1 -- keyword 2 -- keyword 3
\end{keywords}
\end{verbatim}
\noindent There is a list of permitted keywords, which is agreed between all the major astronomy journals and revised every few years.
Do \emph{not} make up new keywords!
For the current list of allowed keywords, see the journal's instructions to authors$^{\ref{foot:itas}}$.
\section{Sections and lists}
Sections and lists are generally the same as in the standard \LaTeX\ classes.
\subsection{Sections}
\label{sec:sections}
Sections are entered in the usual way, using \verb'\section{}' and its variants. It is possible to nest up to four section levels:
\begin{verbatim}
\section{Main section}
\subsection{Subsection}
\subsubsection{Subsubsection}
\paragraph{Lowest level section}
\end{verbatim}
\noindent The other \LaTeX\ sectioning commands \verb'\part', \verb'\chapter' and \verb'\subparagraph{}' are deprecated and should not be used.
Some sections are not numbered as part of journal style (e.g. the Acknowledgements).
To insert an unnumbered section use the `starred' version of the command: \verb'\section*{}'.
See appendix~\ref{sec:advanced} for more complicated examples.
\subsection{Lists}
Two forms of lists can be used in MNRAS -- numbered and unnumbered.
For a numbered list, use the \verb'enumerate' environment:
\begin{verbatim}
\begin{enumerate}
\item First item
\item Second item
\item etc.
\end{enumerate}
\end{verbatim}
\noindent which produces
\begin{enumerate}
\item First item
\item Second item
\item etc.
\end{enumerate}
Note that the list uses lowercase Roman numerals, rather than the \LaTeX\ default Arabic numerals.
For an unnumbered list, use the \verb'description' environment without the optional argument:
\begin{verbatim}
\begin{description}
\item First item
\item Second item
\item etc.
\end{description}
\end{verbatim}
\noindent which produces
\begin{description}
\item First item
\item Second item
\item etc.
\end{description}
Bulleted lists using the \verb'itemize' environment should not be used in MNRAS; it is retained for backwards compatibility only.
\section{Mathematics and symbols}
The MNRAS class mostly adopts standard \LaTeX\ handling of mathematics, which is briefly summarised here.
See also section~\ref{sec:packages} for packages that support more advanced mathematics.
Mathematics can be inserted into the running text using the syntax \verb'$1+1=2$', which produces $1+1=2$.
Use this only for short expressions or when referring to mathematical quantities; equations should be entered as described below.
\subsection{Equations}
Equations should be entered using the \verb'equation' environment, which automatically numbers them:
\begin{verbatim}
\begin{equation}
a^2=b^2+c^2
\end{equation}
\end{verbatim}
\noindent which produces
\begin{equation}
a^2=b^2+c^2
\end{equation}
By default, the equations are numbered sequentially throughout the whole paper. If a paper has a large number of equations, it may be better to number them by section (2.1, 2.2 etc.). To do this, add the command \verb'\numberwithin{equation}{section}' to the preamble.
It is also possible to produce un-numbered equations by using the \LaTeX\ built-in \verb'\['\textellipsis\verb'\]' and \verb'$$'\textellipsis\verb'$$' commands; however MNRAS requires that all equations are numbered, so these commands should be avoided.
\subsection{Special symbols}
\begin{table}
\caption{Additional commands for special symbols commonly used in astronomy. These can be used anywhere.}
\label{tab:anysymbols}
\begin{tabular}{lll}
\hline
Command & Output & Meaning\\
\hline
\verb'\sun' & \sun & Sun, solar\\[2pt]
\verb'\earth' & \earth & Earth, terrestrial\\[2pt]
\verb'\micron' & \micron & microns\\[2pt]
\verb'\degr' & \degr & degrees\\[2pt]
\verb'\arcmin' & \arcmin & arcminutes\\[2pt]
\verb'\arcsec' & \arcsec & arcseconds\\[2pt]
\verb'\fdg' & \fdg & fraction of a degree\\[2pt]
\verb'\farcm' & \farcm & fraction of an arcminute\\[2pt]
\verb'\farcs' & \farcs & fraction of an arcsecond\\[2pt]
\verb'\fd' & \fd & fraction of a day\\[2pt]
\verb'\fh' & \fh & fraction of an hour\\[2pt]
\verb'\fm' & \fm & fraction of a minute\\[2pt]
\verb'\fs' & \fs & fraction of a second\\[2pt]
\verb'\fp' & \fp & fraction of a period\\[2pt]
\verb'\diameter' & \diameter & diameter\\[2pt]
\verb'\sq' & \sq & square, Q.E.D.\\[2pt]
\hline
\end{tabular}
\end{table}
\begin{table}
\caption{Additional commands for mathematical symbols. These can only be used in maths mode.}
\label{tab:mathssymbols}
\begin{tabular}{lll}
\hline
Command & Output & Meaning\\
\hline
\verb'\upi' & $\upi$ & upright pi\\[2pt]
\verb'\umu' & $\umu$ & upright mu\\[2pt]
\verb'\upartial' & $\upartial$ & upright partial derivative\\[2pt]
\verb'\lid' & $\lid$ & less than or equal to\\[2pt]
\verb'\gid' & $\gid$ & greater than or equal to\\[2pt]
\verb'\la' & $\la$ & less than of order\\[2pt]
\verb'\ga' & $\ga$ & greater than of order\\[2pt]
\verb'\loa' & $\loa$ & less than approximately\\[2pt]
\verb'\goa' & $\goa$ & greater than approximately\\[2pt]
\verb'\cor' & $\cor$ & corresponds to\\[2pt]
\verb'\sol' & $\sol$ & similar to or less than\\[2pt]
\verb'\sog' & $\sog$ & similar to or greater than\\[2pt]
\verb'\lse' & $\lse$ & less than or homotopic to \\[2pt]
\verb'\gse' & $\gse$ & greater than or homotopic to\\[2pt]
\verb'\getsto' & $\getsto$ & from over to\\[2pt]
\verb'\grole' & $\grole$ & greater over less\\[2pt]
\verb'\leogr' & $\leogr$ & less over greater\\
\hline
\end{tabular}
\end{table}
Some additional symbols of common use in astronomy have been added in the MNRAS class. These are shown in tables~\ref{tab:anysymbols}--\ref{tab:mathssymbols}. The command names are -- as far as possible -- the same as those used in other major astronomy journals.
Many other mathematical symbols are also available, either built into \LaTeX\ or via additional packages. If you want to insert a specific symbol but don't know the \LaTeX\ command, we recommend using the Detexify website\footnote{\url{http://detexify.kirelabs.org}}.
Sometimes font or coding limitations mean a symbol may not get smaller when used in sub- or superscripts, and will therefore be displayed at the wrong size. There is no need to worry about this as it will be corrected by the typesetter during production.
To produce bold symbols in mathematics, use \verb'\bmath' for simple variables, and the \verb'bm' package for more complex symbols (see section~\ref{sec:packages}). Vectors are set in bold italic, using \verb'\mathbfit{}'.
For matrices, use \verb'\mathbfss{}' to produce a bold sans-serif font e.g. \mathbfss{H}; this works even outside maths mode, but not all symbols are available (e.g. Greek). For $\nabla$ (del, used in gradients, divergence etc.) use \verb'$\nabla$'.
\subsection{Ions}
A new \verb'\ion{}{}' command has been added to the class file, for the correct typesetting of ionisation states.
For example, to typeset singly ionised calcium use \verb'\ion{Ca}{ii}', which produces \ion{Ca}{ii}.
\section{Figures and tables}
\label{sec:fig_table}
Figures and tables (collectively called `floats') are mostly the same as built into \LaTeX.
\subsection{Basic examples}
\begin{figure}
\includegraphics[width=\columnwidth]{example}
\caption{An example figure.}
\label{fig:example}
\end{figure}
Figures are inserted in the usual way using a \verb'figure' environment and \verb'\includegraphics'. The example Figure~\ref{fig:example} was generated using the code:
\begin{verbatim}
\begin{figure}
\includegraphics[width=\columnwidth]{example}
\caption{An example figure.}
\label{fig:example}
\end{figure}
\end{verbatim}
\begin{table}
\caption{An example table.}
\label{tab:example}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
Sun & 1.00 & 1.00\\
$\alpha$~Cen~A & 1.10 & 1.52\\
$\epsilon$~Eri & 0.82 & 0.34\\
\hline
\end{tabular}
\end{table}
The example Table~\ref{tab:example} was generated using the code:
\begin{verbatim}
\begin{table}
\caption{An example table.}
\label{tab:example}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
Sun & 1.00 & 1.00\\
$\alpha$~Cen~A & 1.10 & 1.52\\
$\epsilon$~Eri & 0.82 & 0.34\\
\hline
\end{tabular}
\end{table}
\end{verbatim}
\subsection{Captions and placement}
Captions go \emph{above} tables but \emph{below} figures, as in the examples above.
The \LaTeX\ float placement commands \verb'[htbp]' are intentionally disabled.
Layout of figures and tables will be adjusted by the publisher during the production process, so authors should not concern themselves with placement to avoid disappointment and wasted effort.
Simply place the \LaTeX\ code close to where the figure or table is first mentioned in the text and leave exact placement to the publishers.
By default a figure or table will occupy one column of the page.
To produce a wider version which covers both columns, use the \verb'figure*' or \verb'table*' environment.
If a figure or table is too long to fit on a single page it can be split it into several parts.
Create an additional figure or table which uses \verb'\contcaption{}' instead of \verb'\caption{}'.
This will automatically correct the numbering and add `\emph{continued}' at the start of the caption.
\begin{table}
\contcaption{A table continued from the previous one.}
\label{tab:continued}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
$\tau$~Cet & 0.78 & 0.52\\
$\delta$~Pav & 0.99 & 1.22\\
$\sigma$~Dra & 0.87 & 0.43\\
\hline
\end{tabular}
\end{table}
Table~\ref{tab:continued} was generated using the code:
\begin{verbatim}
\begin{table}
\contcaption{A table continued from the previous one.}
\label{tab:continued}
\begin{tabular}{lcc}
\hline
Star & Mass & Luminosity\\
& $M_{\sun}$ & $L_{\sun}$\\
\hline
$\tau$~Cet & 0.78 & 0.52\\
$\delta$~Pav & 0.99 & 1.22\\
$\sigma$~Dra & 0.87 & 0.43\\
\hline
\end{tabular}
\end{table}
\end{verbatim}
To produce a landscape figure or table, use the \verb'pdflscape' package and the \verb'landscape' environment.
The landscape Table~\ref{tab:landscape} was produced using the code:
\begin{verbatim}
\begin{landscape}
\begin{table}
\caption{An example landscape table.}
\label{tab:landscape}
\begin{tabular}{cccccccccc}
\hline
Header & Header & ...\\
Unit & Unit & ...\\
\hline
Data & Data & ...\\
Data & Data & ...\\
...\\
\hline
\end{tabular}
\end{table}
\end{landscape}
\end{verbatim}
Unfortunately this method will force a page break before the table appears.
More complicated solutions are possible, but authors shouldn't worry about this.
\begin{landscape}
\begin{table}
\caption{An example landscape table.}
\label{tab:landscape}
\begin{tabular}{cccccccccc}
\hline
Header & Header & Header & Header & Header & Header & Header & Header & Header & Header\\
Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit & Unit \\
\hline
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
Data & Data & Data & Data & Data & Data & Data & Data & Data & Data\\
\hline
\end{tabular}
\end{table}
\end{landscape}
\section{References and citations}
\subsection{Cross-referencing}
The usual \LaTeX\ commands \verb'\label{}' and \verb'\ref{}' can be used for cross-referencing within the same paper.
We recommend that you use these whenever relevant, rather than writing out the section or figure numbers explicitly.
This ensures that cross-references are updated whenever the numbering changes (e.g. during revision) and provides clickable links (if available in your compiler).
It is best to give each section, figure and table a logical label.
For example, Table~\ref{tab:mathssymbols} has the label \verb'tab:mathssymbols', whilst section~\ref{sec:packages} has the label \verb'sec:packages'.
Add the label \emph{after} the section or caption command, as in the examples in sections~\ref{sec:sections} and \ref{sec:fig_table}.
Enter the cross-reference with a non-breaking space between the type of object and the number, like this: \verb'see Figure~\ref{fig:example}'.
The \verb'\autoref{}' command can be used to automatically fill out the type of object, saving on typing.
It also causes the link to cover the whole phrase rather than just the number, but for that reason is only suitable for single cross-references rather than ranges.
For example, \verb'\autoref{tab:journal_abbr}' produces \autoref{tab:journal_abbr}.
\subsection{Citations}
\label{sec:cite}
MNRAS uses the Harvard -- author (year) -- citation style, e.g. \citet{author2013}.
This is implemented in \LaTeX\ via the \verb'natbib' package, which in turn is included via the \verb'usenatbib' package option (see section~\ref{sec:options}), which should be used in all papers.
Each entry in the reference list has a `key' (see section~\ref{sec:ref_list}) which is used to generate citations.
There are two basic \verb'natbib' commands:
\begin{description}
\item \verb'\citet{key}' produces an in-text citation: \citet{author2013}
\item \verb'\citep{key}' produces a bracketed (parenthetical) citation: \citep{author2013}
\end{description}
Citations will include clickable links to the relevant entry in the reference list, if supported by your \LaTeX\ compiler.
\defcitealias{smith2014}{Paper~I}
\begin{table*}
\caption{Common citation commands, provided by the \texttt{natbib} package.}
\label{tab:natbib}
\begin{tabular}{lll}
\hline
Command & Ouput & Note\\
\hline
\verb'\citet{key}' & \citet{smith2014} & \\
\verb'\citep{key}' & \citep{smith2014} & \\
\verb'\citep{key,key2}' & \citep{smith2014,jones2015} & Multiple papers\\
\verb'\citet[table 4]{key}' & \citet[table 4]{smith2014} & \\
\verb'\citep[see][figure 7]{key}' & \citep[see][figure 7]{smith2014} & \\
\verb'\citealt{key}' & \citealt{smith2014} & For use with manual brackets\\
\verb'\citeauthor{key}' & \citeauthor{smith2014} & If already cited in close proximity\\
\verb'\defcitealias{key}{Paper~I}' & & Define an alias (doesn't work in floats)\\
\verb'\citetalias{key}' & \citetalias{smith2014} & \\
\verb'\citepalias{key}' & \citepalias{smith2014} & \\
\hline
\end{tabular}
\end{table*}
There are a number of other \verb'natbib' commands which can be used for more complicated citations.
The most commonly used ones are listed in Table~\ref{tab:natbib}.
For full guidance on their use, consult the \verb'natbib' documentation\footnote{\url{http://www.ctan.org/pkg/natbib}}.
If a reference has several authors, \verb'natbib' will automatically use `et al.' if there are more than two authors. However, if a paper has exactly three authors, MNRAS style is to list all three on the first citation and use `et al.' thereafter. If you are using \bibtex\ (see section~\ref{sec:ref_list}) then this is handled automatically. If not, the \verb'\citet*{}' and \verb'\citep*{}' commands can be used at the first citation to include all of the authors.
\subsection{The list of references}
\label{sec:ref_list}
It is possible to enter references manually using the usual \LaTeX\ commands, but we strongly encourage authors to use \bibtex\ instead.
\bibtex\ ensures that the reference list is updated automatically as references are added or removed from the paper, puts them in the correct format, saves on typing, and the same reference file can be used for many different papers -- saving time hunting down reference details.
An MNRAS \bibtex\ style file, \verb'mnras.bst', is distributed as part of this package.
The rest of this section will assume you are using \bibtex.
References are entered into a separate \verb'.bib' file in standard \bibtex\ formatting.
This can be done manually, or there are several software packages which make editing the \verb'.bib' file much easier.
We particularly recommend \textsc{JabRef}\footnote{\url{http://jabref.sourceforge.net/}}, which works on all major operating systems.
\bibtex\ entries can be obtained from the NASA Astrophysics Data System\footnote{\label{foot:ads}\url{http://adsabs.harvard.edu}} (ADS) by clicking on `Bibtex entry for this abstract' on any entry.
Simply copy this into your \verb'.bib' file or into the `BibTeX source' tab in \textsc{JabRef}.
Each entry in the \verb'.bib' file must specify a unique `key' to identify the paper, the format of which is up to the author.
Simply cite it in the usual way, as described in section~\ref{sec:cite}, using the specified key.
Compile the paper as usual, but add an extra step to run the \texttt{bibtex} command.
Consult the documentation for your compiler or latex distribution.
Correct formatting of the reference list will be handled by \bibtex\ in almost all cases, provided that the correct information was entered into the \verb'.bib' file.
Note that ADS entries are not always correct, particularly for older papers and conference proceedings, so may need to be edited.
If in doubt, or if you are producing the reference list manually, see the MNRAS instructions to authors$^{\ref{foot:itas}}$ for the current guidelines on how to format the list of references.
\section{Appendices and online material}
To start an appendix, simply place the \verb'
\section{Introduction}
\label{sec:intro}
The study of emission lines provides a powerful way to investigate a large variety of physical processes ocurring in galaxies. This is particularly true for H$\alpha$, which is often one of the most prominent lines in the optical interval. Indeed, star-formation may produce a powerful H$\alpha$ line emission from recombination in nebular regions ionized by the radiation field of young, massive stars \citep[e.g,][]{osterbrock2005}, and the flux of this line is often used as an indicator of the star formation rate \citep{kennicutt1998_SFR,starlightI,Calzetti2013,madau2014}. But other sources than star-formation can also ionize the gas and lead to H$\alpha$ emission, like nuclear activity driven by AGNs, where the ionizing photons are produced in an accretion disk around a massive black hole \citep[][]{veron-cetty-veron2000}, or shocks in the gas \citep[][]{dopita1995,alatalo2016}.
In some cases the origin of the emission is not clear, as is the case of the LINERS. This terminology was introduced by \cite{heckman1980} for objects with emission lines with widths similar to that found in Seyfert galaxies but showing lower excitation when compared to those galaxies. Later, \cite{trinchieri1991} and \cite{binette1994} proposed that, instead, the LINER emission could originate through ionization by evolved stars, like post-AGB stars or white dwarfs. \cite{cidfernandes2010} have argued that many galaxies with weak LINER-like emission lines could be {\it retired} galaxies, where the star formation ceased and which are now being ionized by old stars. The introduction of integral field studies helped to clarify this issue by showing that the LINER emission in some galaxies is actually extended and can not be produced by nuclear activity alone \citep[e.g. ][]{singh2013,ricci2014,belfiore2016}.
Many studies on the morphology of the H$\alpha$ emission were carried out aiming to derive structural or kinematical properties of the galaxies or of the gas emission. \cite{karachentsev2010} investigated the H$\alpha$ emission for the Local Volume, finding evidence of stellar formation in early-type objects both in knots and in the central regions, interpreting it as indication of accretion of intergalactic gas. \cite{nelson2012} compared spatially resolved maps in H$\alpha$ with maps of the continuum in the R band for a sample of $z\sim1$ star forming galaxies, showing that although both the maps are quite similar, H$\alpha$ emission is more extended and more clumpy than the continuous emission. \cite{gavazzi2018}, analyzed a sample of 147 early-type galaxies (ETG) from the ATLAS$^{3D}$ survey \citep{Cappellari2011} and found that 37\% of them exhibited some emission in H$\alpha$, with about 14\% of them exhibiting relatively strong emission, produced mainly by low mass S0 with stellar and gaseous fast rotating discs.
Since H$\alpha$ emission (and its luminosity) is an important proxy for both nuclear activity and formation of stars, spatially resolved maps of H$\alpha$ may provide important informations about ionization mechanisms, star formation rates, mass assembly, and other galaxy properties. The advent of integral field spectroscopy surveys, like ATLAS$^{3D}$ \citep{Cappellari2011}, SAMI Galaxy Survey \citep{croom2012}, CALIFA \citep{sanchez2012}, and MaNGA \citep{bundy2015}, enabled the investigation of the spatially resolved properties of galaxies and their relation to processes influencing the formation and evolution of galaxies, like quenching. With spatially resolved data it is possible to study the radial dependence of star formation histories \citep{perez2013,garciabenito2017,rosa2017,goddard2017, amorim2017} or evaluate star formation rates inside galaxies, both locally and globally \citep{canodiaz2016,rosa2016,kokusho2017,medling2018}.
The morphology of the H$\alpha$ has often being addressed via the distribution within a galaxy of the star formation rate or other star formation indicators. As part of a series of studies on the star formation morphology in galaxies of the Virgo Cluster, \cite{Koopmann2001} analyzed R-band and H$\alpha$ images, radial profiles, integrated fluxes, and concentration indices to establish links between the H$\alpha$ morphologies of Virgo Cluster spiral galaxies and the types of interaction(s) that may have affected them. These authors proposed a set of classes (combinations of normal, enhanced, anemic, and truncated) to describe the global galaxy star formation morphology. Analyzing a small sample of galaxies in the MaNGA prototype run (P-MaNGA), \cite{li2015} classified the radial profiles of their three diagnostic parameters (Dn4000), EW(H$\delta$), and EW(H$\alpha$)) as either 'centrally quiescent' or 'centrally star forming'. Recently, \cite{spindler2018} investigated the spatial distribution of star formation in a large sample of galaxies of the MaNGA survey \citep{bundy2015}, revealing the existence of two groups of galaxies, which they named 'Centrally Suppressed' and 'Unsuppressed' regarding the radial distribution of their specific star formation rates. Analogously, \cite{medling2018} studied the resolved star formation of a sample of galaxies in the SAMI survey, also finding a split between galaxy populations with centrally concentrated star formation and those with flatter star formation profiles.
In this paper we consider the classification of H$\alpha$ emission radial profiles for a sample of 86 almost face-on CALIFA survey galaxies. We focus on H$\alpha$ to avoid, in the classification stage, discriminating between emission due to star formation, AGN or old stars. We propose an objective procedure where we first divide galaxies in two classes - those where the peak of the H$\alpha$ emission is in the galaxy center and those where this peak is outside the center- and, after, we divide the galaxies in the first group in two by using the light concentration in the $r$-band to take in to account in this classification the bimodal nature of the galaxy population. Indeed, in the local universe, a bimodality is present in many galaxy properties, such as colors and stellar formation rates \citep{Strateva2001, omill2006, baldry2006, zibetti2017, Nelson2018}, by which they can be separated into two large groups, early-types and late-types. Early-types comprise essentially elliptical and lenticular galaxies with old stellar populations, red colors, high masses and almost none star formation \citep{Trager2000, conselice2006, vanDokkum2007}. In contrast, late-type galaxies are mostly spirals (with and without bars) and irregulars galaxies, characterized by younger populations, blue colors, lower stellar masses and active star formation, when compared to early-type galaxies \citep{Bell2003,abilio2006}. We also investigate whether our profile classification is correlated with other galaxy parameters or properties, in particular those inferred from population synthesis. The analysis of stellar populations through galaxy spectral synthesis \citep[e.g., ][]{tinsley76,tinsley80,leitherer99,bruzualecharlot2003,starlightI} allows to determine properties such as mean stellar ages, metallicity, stellar mass and star formation history that are useful to characterize the galaxies. Additionally, morphological types, brightness profiles \citep{deVaucouleurs1948,Sersic1963} and also concentration parameters have been shown to relate to many physical features of galaxies or of their galaxy populations \citep{okamura1984,graham2001,gini2003,conselice2014}. \cite{Strateva2001}.
This paper begins with a description of our sample in Section \ref{sec:Data}. There we describe some relevant details of the CALIFA Survey, the Pipe3D pipeline and how we deal with the dust extinction. In Section \ref{sec:analysis} we explain the structural and physical parameters used here, as well as the procedure to determine and classify the radial H$\alpha$ profiles. Section \ref{sec:results} present a comparative study between the profile classes and other physical parameters. In Section \ref{sec:discussion} we discuss these results, making clear the relations between profile classes and galaxy properties. Section \ref{sec:summary} summarizes the work presented here.
\section{Data}
\label{sec:Data}
The galaxies analyzed in this work were selected from the CALIFA Survey, with datacubes treated by the Pipe3D pipeline, both briefly described below. Additionally, we present our procedure to correct the H$\alpha$ emission for dust attenuation, as well as the criteria adopted to select the final sample.
\subsection{The CALIFA sample}
\label{sec:califa}
The Calar Alto Legacy Integral Field Area \citep*[CALIFA;][]{califa_surveyI} survey, in its second release \citep{califadr2}, observed 200 galaxies of all Hubble types in the nearby universe. The galaxy sample of this survey was selected from the SDSS-DR7 photometric catalogue \citep{sdssDR7}, with constraints on isophotal diameter ($45" < D_{25} < 80"$) and on the redshift range ($0.005 < z < 0.03$).
The objects were observed with the integral-field spectrograph
PMAS/PPak mounted on the 3.5 m telescope at the Calar Alto observatory and with two different spectral setups for each galaxy: a V500 grating with low-resolution ($\lambda/\Delta\lambda=850$ in 5000Å) in the wavelength range 3745Å-7500Å, and a V1200 grating with medium-resolution ($\lambda/\Delta\lambda=1650$ in 4500Å) in the wavelength range 3400Å-4750Å. A more complete description of the CALIFA survey can be found in \cite{walcher2014}.
\begin{figure*}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/ur_dist.png}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/age_dist.png}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/massgal_dist.png}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/zh_dist.png}
\caption[]{Some properties of the galaxy sample discussed in this paper, comprised of face-on galaxies from CALIFA/DR2.}
\label{distrib_propriedades}
\end{figure*}
\subsection{Pipe3D}
Pipe3D \citep{Pipe3d,I_Pipe3d} is an analysis pipeline developed to extract information from IFS datacubes of surveys like CALIFA, MaNGA and SAMI \citep{califa_surveyI, bundy2015, croom2012}.
Pipe3D depends on a spectral synthesis tool \citep[FIT3D][]{I_Pipe3d} and comprises the GSD156 library of simple stellar populations \citep{cid2013}, with 156 templates covering 39 stellar ages (from 1Myr to 14.1Gyr), and 4 metallicities ($Z/Z_{\odot{}}=0.2, 0.4, 1, \text{and} 1.5$). All the steps used to obtain the dataproducts are well explained in \cite{I_Pipe3d,Pipe3d}.
In this work we have started by using the results of Pipe3D for 200 galaxies from CALIFA/DR2 \citep{califadr2}, with the V500 setup. From the many dataproducts obtained with Pipe3D, we used in our analysis the H$\beta$, [OIII]$\lambda$5007, H$\alpha$ and [NII]$\lambda$6583 emission lines maps, all of them in units of $10^{-16}$erg s$^{-1}$ cm$^{-2}$.
\subsection{H$\alpha$ maps: source and dust attenuation}
\label{sec:maps}
Most of our analysis is based on H$\alpha$ maps. To take in to account the effect of extiction due to dust within the galaxy on this and other line intensities, we corrected the dust attenuation by using a screen model \citep[e.g.][]{calzetti1994}. We adopted the extinction law of \cite{cardelli1989}, with $R_V=3.1$. These assumptions lead to a relation between the extinction coefficient in the V-band and the H$\alpha$/H$\beta$ intensity ratio as:
\begin{equation}
A_V = 6.10 \times \log \Bigg[\frac{H\alpha}{H\beta}\Bigg] - 2.78.
\end{equation}
In order to correct the dust attenuation pixel by pixel, we have used the mean value of $A_V$, calculated in $r > 0.20 a$ (where $a$ is the semi-major axis of the galaxy image; see Sect. \ref{sec:analysis}), to avoid contamination of this estimate by non-stellar nuclear activity.
\subsection{Our sample}
From the 200 galaxies observed in CALIFA/DR2, we selected nearly face-on objects, i.e., those with ellipticity $\epsilon < 0.35$ in the V band (see next section). Our final sample contains 86 galaxies of different morphological types, which we divided in three morphological classes as shown in Table \ref{tabela_morf}. This sample covers a range of physical properties typical of those found in bright galaxies in the local universe: it has colors in the interval $1.4 < u-r < 3.4$, mean stellar ages in $\sim 10^{7.8} < t_* ~({\rm yr}) <10^{9.8}$, stellar masses in $\sim 10^{9.5} < M_*/M_{\odot} < 10^{11.8}$ and metallicities in $-0.4 < \log(Z/Z_\odot) < -0.03$. Figure \ref{distrib_propriedades} shows the distribution of these properties for the 86 galaxies in the sample. Table \ref{tab01}, in the Appendix, summarizes relevant properties for each galaxy.
\section{Analysis}
\label{sec:analysis}
In this section we describe some morphological estimators for the $V$-band emission and our definitions of the concentration of the $H\alpha$ emission. We also show how we measured the $H\alpha$ emission profiles and present a procedure for their classification.
\subsection{Morphological and Structural Parameters}
\begin{table}
\vspace{0.3cm}
\centering
\begin{tabular}{|c| c| c|}
\hline
Morph. Class & Hubble type & \# \\
\hline \hline
E\underline{ }S0 & E0-E7, S0 & 30 \\
S\underline{ }early & Sa, Sab, Sb & 23 \\
S\underline{ }late & Sbc, Sc, Scd, Sd & 33 \\ [1ex]
\hline
\end{tabular}
\caption[]{Morphological classes of our sample, with the respective Hubble Types and number of objects.}
\label{tabela_morf}
\end{table}
\begin{figure*}
\begin{center}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/concentracoes_cr.png}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/concentracoes_Cefe_ha.png}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/concentracoes_Cc_ha.png}
\caption[]{Distribution of the concentrations $c_r$, $c_e(H\alpha)$ and $c_c(H\alpha)$ for the galaxies in the sample.}
\label{figure: distribuicoes_concentracao}
\end{center}
\end{figure*}
We have determined quantitative morphological parameters for the images delivered by Pipe3D for the galaxies in our sample through their image moments $\mu_{pq}$. The raw and central image moments of order $(p+q)$ are given by \citep{flusser09}
\begin{equation}
M_{pq}= \sum_{i=1}^{n_x} \sum_{j=1}^{n_y} x_i^p y_j^{q} F(x_i,y_j)
\end{equation}
and
\begin{equation}
\mu_{pq}= \sum_{i=1}^{n_x} \sum_{j=1}^{n_y} (x_i - \bar{x})^p (y_j - \bar{y})^{q} F(x_i,y_j),
\end{equation}
where $(x_i,y_i)$ are the coordinates of the $i$-th pixel of the image, ($\bar{x},\bar{y}$) is the centroid of the emission, given by
\[\bar{x} = \frac{M_{10}}{M_{00}} ~~~~~~~~~~
\bar{y} = \frac{M_{01}}{M_{00}},\]
and $F(x_i,y_i)$ is the flux in that pixel.
Image moments are useful for obtaining the major and minor semi-axes \textit{a} and \textit{b}, the ellipticity $\epsilon$, the inclination angle $\theta$ and the mean radius $\bar{R}$ of the ellipse that fits the image of a galaxy. We apply the moments of the image in the V band map reconstructed by Pipe3D of each galaxy, so that the flux $F(x_i,y_i)$ and all the derived (ellipse) parameters are in $V$ band. The moments were calculated using all pixels with positive fluxes.
Since we have galaxies of different sizes, it is convenient to normalize the scales (and in particular the major semi-axis) of each galaxy by its effective radius, $R_e(V)$, which is defined as the radius which comprises half of the $V$-band galaxy's flux.
To help in the discussion on the structural properties of the H$\alpha$ emission for each galaxy, we have derived two concentration parameters. The first is the effective concentration, $c_e(H\alpha)$, which measures the ratio of the radius containing 50\% of the H$\alpha$ emission ($R_e(H\alpha)$) in the H$\alpha$ maps, to the major semi-axis obtained in the $V$ band, $a$, through the image moments, and, second, the central concentration, $c_c(H\alpha)$, which measures the ratio between the H$\alpha$ flux within $0.2a$ and the total H$\alpha$ flux:
\begin{equation}
c_{e}(H\alpha) = \frac{R_{e}(H\alpha)}{a}
\end{equation}
\begin{equation}
c_c(H\alpha) = \frac{I_{H\alpha}(0.2a)}{I_{H\alpha}(a)}
\end{equation}
The concentraction $c_r$ of the luminosity in the r band, as defined by \cite{Strateva2001}, was also calculated for the galaxies in the sample, and the distributions of these three concentrations are shown in Figure \ref{figure: distribuicoes_concentracao}. The concentration $c_r$ was used by \cite{Strateva2001} to investigate the bimodality of the galaxy population (which is also present in the figure), and we show below that it is also useful to analyze H$\alpha$ profiles.
\subsection{Radial Profiles of H$\alpha$}
\label{sec:profiles}
In this section we discuss, initially, our procedure to obtain radial H$\alpha$ profiles and, after, the criteria we adopted to classify them in to three classes.
Given the H$\alpha$ image of a galaxy, we first define its centroid and other geometric parameters as the same obtained by using the moments of the image in the $V$ band. Then, for each pixel, we compute its radius as
\[ a_i =\sqrt{\frac{[\Delta{x} \sin \theta+\Delta{y} \cos \theta]^2+[\Delta{x}\cos \theta+\Delta{y} \sin \theta]^2}{(1-\epsilon)^2}}, \]
where $\Delta{x}=(x_i-\bar x)$, $\Delta{y}=(y_i-\bar y)$, $\theta$ is the inclination angle and $\epsilon$ is the ellipticity.
We sorted these radii and binned them in 50 radial bins with approximately the same number of pixels, and computed for each of these bins the mean value of the H$\alpha$ emission. We then built the H$\alpha$ radial profile of each galaxy by plotting the mean flux in each bin as a function of the bin radius.
The resulting profiles are shown in Figures \ref{CE_profiles}, \ref{CL_profiles} and \ref{E_profiles} in the appendix \ref{profiles_ha_appendix}. From visual inspection, we first point out that they can be divided in two broad classes with respect to the maximum H$\alpha$ emission: those with maximum emission at the galaxy center, which we designate as $C$ (for central) profiles, and those where the peak of the emission is outside the galaxy center, which we designate as $EX$ (for extended) profiles. This is somewhat analogous to the two classes {\it centrally quiescent} and {\it centrally star-forming} of \cite{li2015} or the {\it unsuppressed} and {\it centrally suppressed} classes discussed by \cite{spindler2018}. Figure \ref{distribuicao_rmax} shows the distribution of the radius of maximum emission $R_{ME}$ for our sample. There are 63 and 23 galaxies of our sample in classes $C$ and $EX$, respectively.
\begin{figure}
\includegraphics[width=1.0 \columnwidth,angle=0]{fig/histograma_Rmax_emission22.png}
\caption[]{The distribution of the radius of maximum H$\alpha$ emission, $R_{ME}$, for our sample. For almost 3/4 of our sample the maximum emission is in the galaxy center. }
\label{distribuicao_rmax}
\end{figure}
A visual examination of $C$ profiles shows that they comprise a variety of behaviors, from those where most of the H$\alpha$ emission indeed comes from the center, to those where a significant fraction of this emission comes from more external regions of the galaxy. Figure \ref{concentration_by_type} shows the distribution of the concentrations $c_{e}(H\alpha)$, $c_{c}(H\alpha)$, and $c_r$ for the $C$ and $EX$ profile classes. From the examination of this and other figures, we decided to divide the $C$ profiles in two sub-classes, with respect to the $c_r$ concentration. Indeed, \cite{Strateva2001} has shown that this quantity is useful to discriminate between the red and blue galaxy populations, with $c_r = 2.63$ providing a divisory line. We then define the $CE$ class as that containing early-type objects with $C$ profiles, i.e., those with $c_r > 2.63$, whereas the $CL$ class comprises late-type galaxies ($c_r < 2.63$) with $C$ profiles. From several tests, we verified that sub-classification of $C$ profiles with $c_r$ provides a more sensible classification than using the other concentrations, $c_{e}(H\alpha)$and $c_{c}(H\alpha)$. Our sample has 36 and 27 galaxies in classes $CE$ and $CL$, respectively.
\begin{figure*}
\begin{center}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/histograma_Conc_r_tipos_c2}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/histograma_Conc_efetiva_tipos_c2}
\includegraphics[width=0.49 \columnwidth,angle=0]{fig/histograma_Conc_central_tipos_c2}
\caption[]{Distribution of the concentrations $c_r$, $c_e(H\alpha)$ and $c_c(H\alpha)$ for the galaxies in the sample.}
\label{concentration_by_type}
\end{center}
\end{figure*}
\begin{figure*}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfil_Ha_NGC5614_correcao_Avm}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfis_all_CE_norm}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfil_Ha_NGC0477_correcao_Avm}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfis_all_CL_norm}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfil_Ha_NGC6154_correcao_Avm}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/perfis_all_EX_norm}
\caption[]{Examples of the three H$\alpha$ profile types. From top to bottom: $CE$, $CL$ and $EX$ profiles. The left panels show a typical profile (that with the class median $c_r$), whereas the right panels shows all profiles of a given type together, normalized by their maximum H$\alpha$ emission.}
\label{perfil_ha_classificacao}
\end{figure*}
Figure \ref{perfil_ha_classificacao} shows the result of this profile morphological analysis. The figure shows H$\alpha$ profiles separated by these three classes, as well as the profile with median $c_r$ of each class. In general, $CE$ profiles present most of the emission comming from the galaxy center, without relevant emission outside the central region; $CL$ profiles tend to be more extended, indicating H$\alpha$ emission spread over larger radii. For type $EX$ profiles, by definition most of the emission comes from regions outside the galaxy center, in one or more bumps.
\begin{table}
\vspace{0.3cm}
\centering
\begin{tabular}{|c| c| c| c|}
\hline
& $CE$ & $CL$ & $EX$ \\
\hline \hline
E\underline{ }S0 & 28 & 1 & 1 \\
S\underline{ }early & 4 & 8 & 11 \\
S\underline{ }late & 4 & 18 & 11 \\ [1ex]
\hline
\end{tabular}
\caption[Tabela de contingência]{Contingency table for the H$\alpha$ profiles and morphological classes, where we show the number of galaxies in each profile and morphological class.}
\label{Contigencytable_perfil_morfologia}
\end{table}
\section{Results}
\label{sec:results}
In this Section we present some consequences of our profile classification, arguing that it is indeed meaningful, by examining relations between the profile types and other galaxy parameters: morphological classes, concentration measurements and stellar population parameters. Finally, we use diagnostic diagrams to determine the origin of the H$\alpha$ emission.
\subsection{Relations with morphological types and concentrations}
We start by investigating whether the profile types are related to the galaxy morphological classes. A simple $\chi^2$ test of the contingency table of morphological classes and profile types, Table \ref{Contigencytable_perfil_morfologia}, leads to $\chi^2 \simeq 53$, with a $p-value=8.5 \times 10^{-11}$, indicating that the profiles and the morphological classes are not independent. Figure \ref{perfil_ha_mosaic} shows the proportion of each morphological class for each profile type. Considering $C$ profiles, our two classes discriminate well between the E\underline{ }S0 (mostly in the $CE$ class) and S groups (mostly in the $CL$ class). $EX$ profiles are mostly comprised of spiral galaxies. On the other side, the S\_early and S\_late groups are both numerous in $CL$ and $EX$ profile types.
By analyzing the $CL$ profiles in Figure \ref{CL_profiles} individually, it may be noticed that $\sim75\%$ of the objects have at least a second H$\alpha$ emission peak along the galaxy's extension, probably coming from star formation in the disk of these objects. Since $CL$ profiles are dominated by spiral galaxies, both S\underline{ }early and S\underline{ }late, it is reasonable to associate these secondary H$\alpha$ emission peaks to star-forming arms or rings in the body of the galaxy.
In contrast, when we examine E\underline{ }S0 objects classified as CL and EX (NGC1349 and NGC5784, respectively), we notice that they also have emission peaks outside the center. As \cite{gomes2016a} argues, these early-type galaxies present evidence of recent stellar activity, showing that not all early-type galaxies are actually quenched.
\begin{figure}
\begin{center}
\includegraphics[width=0.98 \columnwidth,angle=0]{fig/quase_mosaico3_c2.png}
\caption[]{Fraction of morphological classes in each of the H$\alpha$ profiles' type.}
\label{perfil_ha_mosaic}
\end{center}
\end{figure}
The relation between concentrations and the three types of H$\alpha$ profiles is shown in Figure \ref{fig: concentracoes_violin} using violin diagrams (which depict the range of values and the probability of variables). The mean value of the concentration $c_r$ is larger for type $CE$ than for $CL$, by construction, and $CL$ and $EX$ types have similar $c_r$ values, corresponding approximately to the blue peak seen in the bimodal $c_r$ distribution in Figure \ref{figure: distribuicoes_concentracao}. The mean value of the central H$\alpha$ concentration decreases from $CE$ to $CL$ and to $EX$ profiles. The effective concentration, which is in fact measuring how extended is the line emission in the galaxy, increases from $CE$ to $CL$ to $EX$. Table \ref{tabela_propriedades_Estelares_perfil} presents the mean values of the concentration measurements for each profile type.
\begin{figure*}
\begin{center}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/Cr_boxplot_c2.png}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/Ccha_boxplot_c2.png}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/Cha_boxplot_c2.png}
\caption[]{Distribution of concentrations $c_r$, $c_c(H\alpha)$ and $c_e(H\alpha)$ with their mean value shown as a white dot inside the violin plot.}
\label{fig: concentracoes_violin}
\end{center}
\end{figure*}
\subsection{Stellar Population Properties}
To investigate the relations of the H$\alpha$ profiles with the properties of stellar populations, we have also used the mean values of some stellar population properties- age, metallicity and stellar mass- obtained from Pipe3d, since these quantities are good tracers of other properties of galaxies \citep[][for some examples]{gallazzi2005,gallazzi2008,peletier2013,rosa2014b,zibetti2017}. The mean values for these quantities for each type of profile are shown in Table \ref{tabela_propriedades_Estelares_perfil}, whereas their violin diagrams, by H$\alpha$ profile types, are shown in Figure \ref{perfis_stellar_properties}.
\begin{table*}
\vspace{0.3cm}
\centering
\begin{tabular}{|c| c| c| c| c| c| c|}
\hline
Profile & ${c_r}$ & ${c_c}(H\alpha)$ & ${c_e}(H\alpha)$ & log($t^*$/Gyr) & log($Z/Z_{\odot}$) & log($M_* / M_{\odot}$)\\
\hline \hline
CE & 3.07$\pm 0.03$ & 0.33$\pm0.02$ & 0.66$\pm 0.03$ & 9.49$\pm0.05$ & -0.23$\pm 0.01$ & 11.15$\pm0.07$ \\
CL & 2.22$\pm 0.04$ & 0.22$\pm0.03$ & 0.64$\pm 0.02$ & 8.90$\pm0.07$ & -0.29$\pm 0.01$ & 10.61$\pm0.10$ \\
EX & 2.29$\pm 0.07$ & 0.14$\pm0.03$ & 0.70$\pm 0.02$ & 9.12$\pm0.07$ & -0.31$\pm 0.01$ & 10.80$\pm0.09$ \\ [1ex]
\hline
\end{tabular}
\caption[]{Mean value and standard deviation for the concentrations and stellar population parameters for the three types of H$\alpha$ profiles.}
\label{tabela_propriedades_Estelares_perfil}
\end{table*}
Considering only $C$ types, the mean ages, metallicities and masses are larger for galaxies with $CE$ than with $CL$ profiles; comparing $CL$ and $EX$ types, we notice that the latter seems to represent a population less evolved than the former, in the sense that their mean ages and metallicities are smaller for $CL$ than for $EX$ classes. Stellar masses in our sample are larger for $CE$ than for $CL$ types, and are larger for $EX$ than $CL$ types.
\begin{figure*}
\begin{center}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/idade_boxplot_c2.png}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/metalicidade_boxplot_c2.png}
\includegraphics[width=0.65 \columnwidth,angle=0]{fig/Mass_boxplot_c2.png}
\caption[]{The violin plot for the stellar age, metallicity and stellar mass and for each type of H$\alpha$ profile.}
\label{perfis_stellar_properties}
\end{center}
\end{figure*}
\subsection{The source of H$\alpha$ emission}
\label{emission_lines_source}
The next step in our analysis is to investigate the nature of the H$\alpha$ emission associated with each type of profile, if star formation, some kind of nuclear activity or both. For this we have made use of the \cite{bpt1981} BPT diagram. Figure \ref{bpt_all_galaxies} presents BPT diagrams for each profile type. At the left side of this figure we show the mean BPT diagram for each galaxy of a given profile type, whereas in the right panels we show this diagram separately for the internal ($r < 0.2 a$, in red) and external ($r > 0.2a$, in blue) regions of galaxies, again separated by profile type. This figure also shows the empirical line proposed by \cite{kauffmann2003} (where galaxies below it are classified as pure star-forming objects), the model line proposed by \cite{kewley2001} (where galaxies above it are considered AGNs), and the line proposed by \cite{schawinski2007} to discriminate between LINERs (below the line) and Seyfert galaxies (above the line). Any object between Kauffmann's line and Kewley's line are considered as an object in transition, with emission characteristics of both stars and AGNs.
The BPT diagram for the galaxies as a whole shows interesting trends with the H$\alpha$ profile. $CE$ objects present emission coming from both the star-forming and transition regions, with just one galaxy in the LINER region of the diagram. The fraction of galaxies in the transition region decreases from type $CE$ to $EX$ to $CL$ and, at the same time, the proportion of objects in the star-forming region increases. Galaxies with H$\alpha$ profiles of type $EX$ in this panel are all close to the line proposed by \cite{kauffmann2003}, showing less scatter than the other profiles.
Panels at the right side of Figure \ref{bpt_all_galaxies} tell a more interesting story. Considering the central region of the galaxies ($r < 0.2 a$, in red), type $CE$ profiles have most of their galaxies, $\sim 47\%$, in the transition region, $\sim 36 \%$ in the star-formation region and the other $17\%$ coming from the LINER region of the BPT diagram. Profiles of type $CL$ have $63\%$, $22\%$ and $15\%$ of central region emission coming from star-forming, transition and LINER regions, respectively. Galaxies with $CL$ profiles in the star-forming region are distributed forming a tail towads low values of $[NII]/H\alpha$, suggesting that these objects have lower metallicity than the others in $CE$ and $EX$ classes. $EX$ galaxies present a larger homogeneity in the distribution of central emission, with $30\%$, $39\%$ and $30\%$ of their members in star-forming, transition and LINER regions, respectively.
\begin{figure*}
\begin{center}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_CE_profile.png}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_in_out_CE_profile.png}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_CL_profile.png}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_in_out_CL_profile.png}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_E_profile.png}
\includegraphics[width=0.85 \columnwidth,angle=0]{fig/bpt_in_out_E_profile.png}
\caption[]{BPT diagram for galaxies in the sample: left panels show the diagram for galaxy mean emission line ratios of each type of H$\alpha$ profile. Right panels present the BPT diagram for the different profile types separated by inner (in blue) and outer (in pink) regions, respectively.}
\label{bpt_all_galaxies}
\end{center}
\end{figure*}
\section{Discussion}
\label{sec:discussion}
The H$\alpha$ emission is by far one of the most important tools to investigate galaxy properties. In this paper, through an analysis of a sample of CALIFA galaxies, we present a classification of the H$\alpha$ radial profile in three classes: $CE$, $CL$ and $EX$. The central profiles $CE$ and $CL$, have the peak of the line emission in the galaxy centre, whereas in $EX$ profiles, the maximum H$\alpha$ emission is outside the center. We found useful to distinguish two classes of central profiles ($CE$ and $CL$), dominated by early- or late-type populations, accordingly with the value of the $c_r$ concentration \citep{Strateva2001}. The profiles' types exhibit good relation with some structural and physical parameters such as concentrations and stellar population properties.
The first aspect to point out is that the large majority of our sample, $\sim 75$\%, has central ($C$) profiles. Dividing this group by the $c_r$ value in early, $CE$, and late $CL$, subtypes, we noticed that $CE$ profiles are dominated by early-type galaxies, E and S0. Spiral galaxies in the $CL$ class are mostly blue objects (in terms of the $c_r$ criterion), while the $EX$ class contains the same fraction of blue and red late-type galaxies. In other words, most of the spirals with central profiles are blue galaxies, whereas those with an extended profile have a mix of red and blue objects. E\underline{ }S0 galaxies are less than 4\% of the $CL$ and $EX$ profile classes; early-type galaxies in these classes are probably objects still forming stars, not quenched yet \citep[e.g.][]{gomes2016a}.
Concerning the H$\alpha$ concentration parameters, we have shown that the central concentration $c_c$ is indeed useful to describe the emission profiles: for central profiles, $CE$ objects tend to have more H$\alpha$ emission in the central region than the $CL$s and $EX$s. The effective concentration does not show strong trends with the profile types, although it tends to be larger for $EX$ profiles than for the others.
Stellar population properties also show trends with the profile types. Stellar populations are older for $CE$ profiles. Comparing mean stellar ages for $CL$ to $EX$ profiles, we can verify that the latter presents older populations than the former. Mean metallicities decrease from $CE$ to $CL$ and to $EX$. These results indicate that $CE$ profiles are more likely to be associated to early-type objects than $CE$ and $EX$ profiles.
The BPT diagram also presents significant differences between the three classes. Considering the mean value of the line ratios (left panels of Fig. \ref{bpt_all_galaxies}), the most striking difference is the large fraction of transition objects for $CE$ profiles compared to the others, which have most of their galaxies in the star-forming part of the diagrams. When we consider the internal and external BPT diagrams, several interesting features appear. For all profiles the number of galaxies with LINER central emission increases. The same is true for objects in the transition region, at least for late-types. For these galaxies, the large majority of the emission in the external regions seems to be due to star formation.
Indeed, the second aspect we consider important to highlight here is the difference in the properties of late type galaxies in the $CL$ and $EX$ classes. $EX$ objects have a significant fraction with central emission in the transition region, whereas for $CL$ objects, the fraction where the emission can be ascribed to star-formation is larger. Also, $CL$ galaxies seem to have a larger range of physical properties (e.g. nebular abundances) than late types in the $EX$ class. This seems to indicate that $EX$ objects are more evolved than $CL$ galaxies.
Our results support the findings of \cite{li2015} and \citet{spindler2018}. The latter authors analyzed SFR profiles from the SDSS-IV MaNGA survey, showing that they can be divided in 'centrally suppressed' and 'unsuppressed' star formation, in close association with our C and EX profiles. Our conclusions are in general agreement with theirs; they found that both centrally suppressed and unsuppressed galaxies have a bimodal distribution, whereas our EX galaxies are mostly blue. This is probably a consequence of our small sample size.
The classification procedure adopted here is objective and can be extended to other IFU-like observational data. We are interested in using this approach to investigate galaxy samples in the local universe observed by the surveys S-PLUS \citep[][in prep.]{mendesdeoliveira2018}, J-PLUS \citep{cenarro2018}, and J-PAS \citep{benitez2014}. Despite being photometric surveys, they are imaging large areas of the sky with a large number of narrow and broad-band filters (12 for S-PLUS and J-PLUS, and 59 for J-PAS). They contain narrow filters centred on H$\alpha$ and it has been shown that the H$\alpha$ emission can indeed be reliably determined for nearby objects with the observational setup of these surveys \citep{vilella-rojo2015,logrono-garcia2018}.
\section{Summary}
\label{sec:summary}
We have determined the radial profile of the H$\alpha$ emission for a sample of 86 face-on galaxies from the CALIFA survey. After visual examination, we classified these profiles in three classes, $CE$, $CL$ and $EX$, taking in to account the position of the maximum H$\alpha$ emission (in the galaxy centre or not) and, for galaxies where the maximum of H$\alpha$ emission comes from the galaxy centre, if the dominant population is red or blue, by using the light concentration in the $r$-band, $c_r$. Our results show that, for $\sim 75$\% of the sample, the peak of the emission is in the galaxy centre.
Galaxy properties, like morphology, mean stellar ages and metallicities, and the nature of the line emission, correlate well with these three classes. For example, most of the objects in the $CE$ class are early-type galaxies. Late-type galaxies are spread between the $CL$ and $EX$ classes, with $CL$ objects presenting features of a less evolved population when compared to the $EX$ class.
The classification presented here helps to highlight the diversity of H$\alpha$ emission in the galaxies of the local universe and its close links with properties usually adopted to describe the galaxy population.
\section*{Acknowledgements}
PMN thanks the financial support from Brazilian funding agency CNPq (grant 142436/2014-3). She also thanks Rosa González Delgado, Enrique Perez and Ruben García-Benito for useful comments. PMN also thanks the Instituto de Astrof\'isica de Andaluc\'ia (IAA/CSIC) and the Instituto de Astronomia, Geofísica e Ciências Atmosféricas (IAG/USP) for their warm scientific environment that made this work possible. LSJ aknowledges support of CNPq (304819/2017-4) and FAPESP (2017/237666-0) to his work.
\bibliographystyle{mnras}
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Q: Is replacing existing DLL without re-registering it possible? I need to update an existing DLL file that is already in use for our client-software that is installed on 400+ computers. This DLL file exists today and is already registered with regsvr32.exe when the software was installed.
The problem is that I cannot easily run the regsvr32.exe command without changing the entire update process, which is a mayor hassle as it is ancient and no documentation exists on it.
So I wondered if I actually have to re-register the dll file, or if replacing the file is enough. I tried searching for more information on this, but so far came up blank.
A: It depends on how you created the DLL.
If you broke the binary compatibility you must re-register the DLL using regsvr32. If you mantained the binary compatibility copying the new DLL should be enough.
VB6 had an option called "Binary Compatibility" which helped to deal with these issues (http://msdn.microsoft.com/en-us/library/aa733715(v=vs.60).aspx). I don't know if VC++ or VS.NET provide something similar.
Greetings,
edu
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\section{Introduction}
Hopf algebroids~\cite{bohmHbk,BrzMilitaru,lu}
are generalizations of Hopf algebras,
which are roughly in the same relation to groupoids as Hopf algebras
are to groups.
They are {\bf bialgebroids} possesing a version of an antipode,
where an (associative) bialgebroid is
the appropriate generalization of a bialgebra.
Hopf algebroids comprise several structure maps
defined on a pair of associative unital algebras,
the {\bf total algebra} $H$ (generalization of a function algebra on
the space of morphisms of a groupoid),
and the {\bf base algebra} $A$ (generalization of a function algebra
on the space of objects (equivalently, units) of a groupoid).
The main structure on the total algebra of a bialgebroid is an $A$-bimodule
structure on $H$ and the coproduct $\Delta:H\to H\otimes_A H$.
The commutative Hopf algebroids (where both $H$ and $A$ are
commutative) are easy to define by a categorical dualization of
the groupoid concept.
They are used as a classical tool in
stable homotopy theory \cite{hovey,Ravenel}.
The noncommutative Hopf algebroids over
a commutative base ($H$ noncommutative, $A$ commutative) are also rather
straightforward to introduce; this theory has been studied from late 1980-s,
under the influence of the quantum group theory \cite{maltsiniotis}.
The most obvious examples are the convolution algebras of finite groupoids.
Bialgebroids and Hopf algebroids
over a noncommutative base are much more complicated to
define; several nonequivalent versions were developed since around 1992 by
Schauenburg~\cite{schauenburg}, Lu~\cite{lu},
B\"ohm~\cite{bohmnew,bohmHbk,bohmInternal,BohmSzlach},
Xu~\cite{xu} and others. Many examples of Hopf algebroids over
noncommutative bases have been studied in the contexts of
inclusions of von Neumann algebra factors~\cite{bohmHbk},
dynamical Yang-Baxter equation, weak Hopf algebras,
deformation quantization~\cite{xu},
noncommutative torsors, noncommutative differential calculus and
cyclic homology~\cite{kow,kowKraehmer} etc.
In 2012, Meljanac devised a new approach to some examples of
(topological) Hopf algebroids over a noncommutative base
restricting the codomain of the coproduct
map in a useful, but somewhat {\it ad hoc} way. To construct that
codomain, he chooses a subalgebra $\mathcal{B}$ in the tensor
square $H\otimes H$ of the total algebra $H$,
such that the intersection of $\mathcal{B}$ with the kernel $I_A$ of
the projection $\pi: H\otimes H\to H\otimes_A H$
to the tensor square over the noncommutative base algebra $A$
is a two-sided ideal $I_A\cap\mathcal{B}\subset\mathcal{B}$
(with an appropriate behaviour under the antipode map).
The appearance of the two-sided ideal is a novel
and somewhat unexpected feature reminding of the classical case where
the base algebra $A$ is commutative and $I_A$ is two-sided itself.
The approach is developed in collaborative works~\cite{tajron}
with most details in~\cite{tajronkov}.
These papers neglect two mathematical issues.
Firstly, there is no care taken about the implicit use of completions:
the values of the coproduct involve infinite sums,
hence the codomain should be a completed tensor pro\-duct.
Secondly, at the algebraic level, they do not state a complete
axiomatic framework for their version of Hopf algebroid, nor state
its precise relation to other definitions.
Instead they construct an interesting class of examples
and give a partial list of essential properties.
In our article~\cite{halgoid} with Meljanac, we treat
a somewhat wider class of examples
in a mathematically rigorous way,
using G.~B\"ohm's definition of a symmetric Hopf algebroid,
partly adapted to a formally completed tensor product.
For a different adaptation which is an internal version
to a category of filtered-cofiltered vector spaces (with a more
sensible completion) see~\cite{StojicPhD}
and for the connections to formal geometry see~\cite{SkodaStojicformal}.
These works took care of the first issue (completions), but went away from
the original two-sided ideal approach.
To address the second issue, we here propose a new set of axioms which
expresses the essence of the two-sided ideal approach and discuss
it in the context. The subalgebra $\mathcal{B}\subset H\otimes_{\bm k} H$
in the new axioms is named the {\bf balancing subalgebra} and the new
version of the Hopf algebroid is named the
Hopf algebroid with the balancing subalgebra $\mathcal{B}$.
This new version of Hopf algebroids over a noncommutative base,
namely with a balancing subalgebra,
is in Theorem~\ref{th:MLu} compared to the Hopf algebroids of Lu,
because her version of the axioms for the antipode map
involves a choice of a section which is
close in spirit to the choice of balancing subalgebra
in our axiomatics and in the original approach of Meljanac.
Our main result is Theorem~\ref{th:scalarM}
(based on rather nontrivial Lemma~\ref{lem:key})
stating that every scalar extension Hopf algebroid
can be cast into a Hopf algebroid
with a (suitable choice of) balancing subalgebra $\mathcal{B}$.
\section{Bialgebroids over noncommutative base}
Bialgebroids and Hopf algebroids over a noncommutative base algebra
are far less obvious to formulate~\cite{bohmHbk,BrzWis,lu,schauenburg,xu}.
To give it a try, recall that
a ${\bm k}$-bialgebra is compatibly a monoid (algebra)
and a comonoid (coalgebra) in the same
monoidal category, namely that of ${\bm k}$-modules,
where ${\bm k}$ is the commutative ground ring.
Thus, for a noncommutative base ring $A$,
it is natural to try replacing the monoidal category of ${\bm k}$-modules
with the monoidal category of $A$-bimodules.
The latter is neither symmetric nor braided monoidal
in general, so the usual compatibility condition
between the comonoids and monoids makes no sense.
Instead, it appears that
the monoid and the comonoid part of the left $A$-bialgebroid
structure live in different monoidal categories~\cite{bohmHbk}.
The monoid structure $(H,\mu,\eta)$ on $H$ is in
the monoidal category of $A \otimes A^\mathrm{op}$-bimodules; equivalently,
$(H,m)$ is an associative ${\bm k}$-algebra
and the unit map $\eta$ is a morphism of ${\bm k}$-algebras
$\eta: A\otimes A^\mathrm{op}\to H$
(we say that $(H,\mu,\eta)$ is an $A\otimes A^\mathrm{op}$-ring);
map $\mu :H\otimes_{\bm k} H\to H$ satisfies $(h.a)k = h(a.k)$,
$\forall a\in A$, $\forall h,k\in H$,
hence it induces the map also denoted $\mu:H\otimes_A H\to H$.
The unit $\eta$ is usually described in terms of its
left leg $\alpha := \eta(-\otimes 1_{A^\mathrm{op}}):A\to H$ and
right leg $\beta := \eta(1_A\otimes -):A^\mathrm{op}\to H$,
also called the source and target maps respectively; then,
their images commute because $\alpha(a)\beta(b) =
\eta(a\otimes b)=\beta(b)\alpha(a)$.
An $A\otimes A^\mathrm{op}$-ring $(H,\mu,\eta)$ is below
described as $(H,\mu,\alpha,\beta)$.
The comonoid structure of $(H,\Delta,\epsilon)$
is instead in the monoidal category
of $A$-bimodules (we say that $H$ is an $A$-coring).
\begin{definition}
An $A\otimes A^\mathrm{op}$-ring $(H,\mu,\alpha,\beta)$ and
an $A$-coring $(H,\Delta,\epsilon)$
on an $A$-bimodule $H$
form a {\bf left associative $A$-bialgebroid}
$(H,\mu,\alpha,\beta,\Delta,\epsilon)$
if they satisfy the following compatibility conditions:
\begin{itemize}
\item[(C1)] the underlying $A$-bimodule
structure of the $A$-coring structure is determined by the source map and
target map (part of the $A\otimes A^\mathrm{op}$-ring structure):
$r.a.r' = \alpha(r)\beta(r')a$.
\item[(C2)] formula $\sum_\lambda h_\lambda\otimes f_\lambda \mapsto
\epsilon(\sum_\lambda h_\lambda\alpha(f_\lambda))$ defines an action
$H\otimes A\to A$ which extends the left regular action $A\otimes A\to A$
along the inclusion
$A\otimes A\stackrel{\alpha\otimes A}\longrightarrow H\otimes A$.
\item[(C3)] the linear map $h\otimes (g\otimes k)\mapsto \Delta(h)(g\otimes k)$,
$H\otimes (H\otimes H)\to H\otimes H$, induces
a well defined action $H\otimes (H\otimes_A H)\to H\otimes_A H$.
\end{itemize}
\end{definition}
The condition (C1) implies that the kernel
$I_A = \operatorname{Ker}\pi$
of the projection map
$$\pi: H\otimes_{\bm k} H\to H\otimes_A H$$
of $H$-bimodules is a {\bf right ideal}
in the algebra $H\otimes_{\bm k} H$, generated by the set of elements
of the form $\beta(a)\otimes 1 - 1\otimes\alpha(a)$:
\begin{equation}\label{eq:IA}
I_A = \left\{\,\beta(a)\otimes_{\bm k} 1 - 1\otimes_{\bm k}\alpha(a)\,|\,\,a\in A\,\right\}\cdot
(H\otimes_{\bm k} H)
\end{equation}
The third condition (C3) is here stated in the form of Lu.
Let
$$
H\times_A H = \Big\{\sum b_i\otimes b'_i \in H\otimes_A H \ \big|\,
\sum_i b_i\otimes b'_i \alpha(a) = \sum_i b_i \beta(a)\otimes
b'_i, \,\,\forall a\in A \Big\},
$$
which is an $H$-subbimodule of $H\otimes_A H$, called the Takeuchi
product. Then $H\times_A H$ is, unlike $H\otimes_A H$,
an associative algebra
with respect to the componentwise product. This is not true for
$H\otimes_A H$: the componentwise rule is not well defined because
it may depend on the chosen representatives in $H\otimes_{\bm k} H$;
this is because $I = I_A$ is only a {\it right} ideal in general.
Ping Xu~\cite{xu} has shown that (C3) is equivalent to the joint assertion
of the following two requirements:
\begin{itemize}
\item[(C3a)]
$\mathrm{Im}\,\Delta\subset H\times_A H$,
\item[(C3b)] $\Delta$ as a map
from $H$ to $H\times_A H$ is a homomorphism of algebras.
\end{itemize}
Of course, (C3b) makes sense only because of (C3a). A short calculation shows
that $\pi^{-1}(H\times_A H)$ is also a subalgebra of $H\otimes H$.
Observe a commutative diagram of $A$-bimodules:
\begin{equation}\label{eq:diagspacestimes}
\xymatrix{
&&\pi^{-1}(H\times_A H)\ar[r]\ar[d]^{\pi|_{\pi^{-1}(H\times_A H)}}& H\otimes H\ar[d]^\pi\\
H\ar[rr]^{\Delta\,\,\, }&&
H\times_A H\ar[r]& H\otimes_A H
}
\end{equation}
All arrows except those into $H\otimes_A H$ are also homomorphisms of algebras.
The equation
$\sum_i b_i\otimes b'_i \alpha(a) = \sum_i b_i \beta(a)\otimes b'_i$
for elements in $H\otimes_A H$
is demanded in the quotient, hence it holds only up to elements in $I_A$;
if we take the same equation strictly in $H\otimes H$ to cut
some subalgebra (actually a left ideal) $H\tilde\times H\subset H\otimes H$,
then the projection
$\pi|_{H\tilde\times H}$ maps this subalgebra within $H\times_A H$,
but is not necessarily onto.
In the categorical language,
$H\times_A H$ is an end (kind of a categorical limit)
of a coend (kind of a colimit), not the other way around.
However, Meljanac in his examples
takes some other subalgebra $\mathcal{B}\subset H\otimes H$
(not a universal construction)
first and then passes to the quotient by
$\pi|_{\mathcal{B}}$ (hence a colimit), with a result which is still
an algebra (different from $H\times_A H$).
To achieve this, he needs that
\begin{itemize}
\item[(C3MI)] $I_A\cap\mathcal{B}$
is a two-sided ideal in $\mathcal{B}$.
\end{itemize}
In addition, he (implicitly) requires
\begin{itemize}
\item[(C3Ma)]
$\mathrm{Im}\,\Delta\subset \mathcal{B}/(I_A\cap\mathcal{B})$,
\item[(C3Mb)] $\Delta$ as a map
from $H$ to $\mathcal{B}/(I_A\cap\mathcal{B})$ is a homomorphism of algebras.
\end{itemize}
\begin{definition} A {\bf left $A$-bialgebroid with a
balancing subalgebra $\mathcal{B}$} comprises an
$A\otimes A^\mathrm{op}$-ring $(H,m,\alpha,\beta)$ and
an $A$-coring $(H,\Delta,\epsilon)$
on the same $A$-bimodule $H$ which satisfy the compatibility conditions (C1)
and (C2), and a (not necessarily unital) subalgebra
$\mathcal{B}\subset H\otimes H$ satisfying (C3MI), (C3Ma) and (C3Mb)
and called the {\bf balancing subalgebra}.
\end{definition}
{\it A left $A$-bialgebroid with a balancing subalgebra $\mathcal{B}$
is not necessarily
a left associative $A$-bialgebroid in the standard sense},
because (C3) does not always hold.
However, if $\mathcal{B}$ is the preimage $\pi^{-1}(H\times_A H)$
of the Takeuchi product under the natural projection
$\pi$ then (C3) follows.
Conversely, given a left associative $A$-bialgebroid
$H = (H,m,\alpha,\beta,\Delta,\epsilon)$, we call
$\pi^{-1}(H\times_A H)\subset H\times_{\bm k} H$
the {\bf trivial balancing subalgebra} of $H$
and it makes $H$ into a left $A$-bialgebroid with the
trivial balancing subalgebra.
Observe a commutative diagram of $A$-bimodules where all arrows except
those into $H\otimes_A H$ are homomorphisms of algebras:
\begin{equation}\label{eq:diagspaces1}
\xymatrix{
&&\mathcal{B}\ar[r]\ar[d]^{\pi|_{\mathcal{B}}}& H\otimes H\ar[d]^\pi\\
H\ar[rr]^{\Delta\,\,\, }&& \mathcal{B}/(I_A\cap\mathcal{B})\ar[r]&
H\otimes_A H
}
\end{equation}
\begin{proposition}\label{prop:equivifBexists}
Let $(H,\mu,\alpha,\beta,\Delta,\epsilon)$ be the data defining
an $A\otimes A^\mathrm{op}$-ring and $A$-coring satisfying (C1), (C2) and (C3a).
Suppose there exist a subalgebra
$\mathcal{B}\subset H\otimes H$ such that (C3MI) and (C3Ma) hold.
Then these data define a $A$-bialgebroid with the balancing
subalgebra $\mathcal{B}$ iff
they (without $\mathcal{B}$) define an associative $A$-bialgebroid;
in other words, (C3b) holds iff (CM3b) holds.
\end{proposition}
\begin{proof}
This is a rather simple observation: (C3a) and (C3Ma) together
imply that $\mathrm{Im}\,\Delta\subset\mathcal{B}/(I_A\cap\mathcal{B})\cap H\times_A H$ which
has the structure of a subalgebra of
$\mathcal{B}/(I_A\cap\mathcal{B})$ and also of $H\times_A H$;
the algebra structures on $\mathcal{B}/(I_A\cap\mathcal{B})$ and $H\times_A H$
are both defined factorwise, hence equal on the intersection.
Thus (C3b) and (C3Mb) hold iff $\Delta:H\to\mathcal{B}/(I_A\cap\mathcal{B})\cap
H\times_A H$ is an algebra map
respectively with the same multiplication on the codomain,
hence they are equivalent.
\end{proof}
\section{Hopf algebroids: antipode}
\begin{definition} A {\bf Hopf $A$-algebroid}
in the sense of J-L.~Lu \cite{lu} (or a Lu-Hopf algebroid)
is a left associative $A$-bialgebroid
$(H,\mu,\alpha,\beta,\Delta,\epsilon)$ with an antipode map $\tau:H\to H$,
which is a linear antiautomorphism satisfying
\begin{eqnarray}
\tau\beta = \alpha\label{eq:taubetaalpha} \\
\mu(\mathrm{id}\otimes_{\bm k}\tau)\gamma\Delta = \alpha\epsilon\label{eq:gam} \\
\mu(\tau\otimes_A\mathrm{id})\Delta = \beta\epsilon\tau\label{eq:nogam}
\end{eqnarray}
for some linear section $\gamma: H\otimes_A H\to H\otimes H$
of the projection $\pi: H\otimes H\to H\otimes_A H$.
\end{definition}
The reason for introducing $\gamma$ in (\ref{eq:gam}) is the fact that
$\mu(\mathrm{id}\otimes_A\tau)\Delta$ is not a well
defined map because $\mu(\mathrm{id}\otimes_{\bm k}\tau)(I_A)\neq 0$ in general.
Indeed, $I_A$ is a linear span of the set of all elements of the form
$\beta(a)h\otimes k-h\otimes\alpha(a)k$, where $a\in A$ and $h,k\in H$,
and
$\mu(\mathrm{id}\otimes\tau)(\beta(a)h\otimes k-h\otimes\alpha(a)k) =
\beta(a)h\tau(k) - h\tau(k)\tau(\alpha(a))$
which can be nonzero in general. No such problems occur
with~(\ref{eq:nogam}) because
$$\mu(\tau\otimes\mathrm{id})(\beta(a)h\otimes k-h\otimes\alpha(a)k) =
\tau(h)\tau(\beta(a))k - \tau(h)\alpha(a)k
\stackrel{(\ref{eq:taubetaalpha})}=0.$$
\begin{definition}
A {\bf Hopf $A$-algebroid with a balancing subalgebra} $\mathcal{B}$
is a left $A$-bi\-al\-ge\-broid
$(H,\mu,\alpha,\beta,\Delta,\epsilon)$
with a balancing subalgebra $\mathcal{B}$
together with an algebra antihomomorphism $\tau:H\to H$,
called the {\bf antipode}, such that
\begin{eqnarray}\label{eq:tauvanish}
\mu(\mathrm{id}\otimes_{\bm k}\tau)(I_A\cap\mathcal{B})=0\label{eq:tauideal} \\
\tau\beta = \alpha\label{eq:mtaubetaalpha} \\
\mu(\mathrm{id}\otimes_A\tau)\Delta = \alpha\epsilon\label{eq:mr} \\
\mu(\tau\otimes_A\mathrm{id})\Delta = \beta\epsilon\tau\label{eq:ml}
\end{eqnarray}
\end{definition}
Two equations are the same as before: (\ref{eq:mtaubetaalpha})
is identical to (\ref{eq:taubetaalpha}) and (\ref{eq:ml})
to (\ref{eq:nogam}).
Equation (\ref{eq:mr}) makes sense because of (\ref{eq:tauideal}).
Notice that there is no need for a choice of a section $\gamma$. Choice
of the subalgebra $\mathcal{B}$ is intuitively
a wider choice which accomplishes the same.
\begin{remark}
The map $\mu(\mathrm{id}\otimes_{\bm k}\tau): h\otimes h'\to h\tau(h')$
is linear but neither a homomorphism nor an antihomomorphism of algebras;
hence it is not suffi\-cient to check~(\ref{eq:tauideal}) on the algebra
generators (or even worse, on ideal generators) of $I_A\cap\mathcal{B}$.
This will be the central difficulty in Section~\ref{sec:scalar}.
\end{remark}
\begin{theorem} \label{th:MLu}
If a Hopf algebroid with a balancing subalgebra satisfies (C3{a})
then it admits a structure of a Lu-Hopf algebroid.
\end{theorem}
\begin{proof} Choose a vector space splitting of $H\otimes_A H$
into $\mathrm{Im}\,\Delta$ and the linear complement; for $\gamma$
take any linear section
of the projection $\pi:H\otimes H\to H\otimes_A H$
such that values $\gamma(p)$ over points $p\in\mathrm{Im}\,\Delta$
are in $\mathcal{B}$ (this can be done by (C3Ma))
and on the linear complement prescribe any linear choice for $\gamma$,
for instance $0$.
Condition (C3b) holds by (CM3b)
and Proposition~\ref{prop:equivifBexists}.
Then $\mu(\mathrm{id}\otimes_{\bm k}\tau)\gamma\Delta(h)=\mu(\mathrm{id}\otimes_A\tau)\Delta(h)$
as the right hand side is defined by choosing any representative of $\Delta(h)$
in $H\otimes_{\bm k} H$ and evaluating $\mu(\mathrm{id}\otimes_{\bm k}\tau)$.
Thus~(\ref{eq:gam}) holds, and the other conditions on the antipode
are identities.
\end{proof}
We do not expect the converse of Theorem~\ref{th:MLu} in general,
but we show below that every scalar extension is in both classes.
Clearly, every Hopf algebroid over a commutative base
is also in both classes.
\section{Scalar extension Hopf algebroids}
\label{sec:scalar}
Let $T$ be a Hopf ${\bm k}$-algebra with comultiplication
$\Delta_T:T\to T\otimes_{\bm k} T$ and antipode $S$.
Let $A$ be a braided-commutative left-right Yetter-Drinfeld
$T$-module algebra. That means that $A$ is a unital associative algebra
with a left $T$-action $\blacktriangleright:T\otimes A\to A$
which is Hopf ($t\blacktriangleright (ab) =
(t_{(1)}\blacktriangleright a)(t_{(2)}\blacktriangleright b)$)
and a right $T$-coaction $a\mapsto a_{[0]}\otimes a_{[1]}$
which is morphism of algebras $A\to A\otimes T^{\mathrm{op}}$
~(see~\cite{BrzMilitaru}),
satisfying the left-right Yetter-Drinfeld condition
$$(t_{(1)} \blacktriangleright a_{[0]}) \otimes (t_{(2)} a_{[1]})
= (t_{(2)} \blacktriangleright a)_{[0]} \otimes (t_{(2)} \blacktriangleright a)_{[1]} t_{(1)},
\,\,\,\,\,\,\,\forall t \in T, \forall a\in A.$$
and the braided commutativity
$x_{[0]}(x_{[1]}\blacktriangleright a) = a x$ for all $a,x\in A$.
\begin{lemma}\label{lem:bcalt}
Braided commutativity codition is equivalent to the condition
$((S d_{[1]})\blacktriangleright a) d_{[0]} = d a$ for all $d,a\in A$.
\end{lemma}
\begin{proof} This is rather standard. We show for convenience one
direction: assu\-ming braided commutativity,
$$\begin{array}{lcl}
d a &=& d_{[0]} ((d_{[1]} S d_{[2]})\blacktriangleright a) \\
&=& d_{[0][0]}(d_{[0][1]}\blacktriangleright ((S d_{[1]})\blacktriangleright a))\\
&=& ((S d_{[1]})\blacktriangleright a) d_{[0]}
\end{array}$$
The proof of another direction is similar.
\end{proof}
The smash product $H = A\sharp T$ can be equipped with the
structure of the total algebra of a Hopf $A$-algebroid (which is below
called scalar extension Hopf algebroid),
either in sense of Lu \cite{lu}, but also in the
sense of the symmetric Hopf algebroid of B\"ohm.
The main formulas are
$$
\alpha(a) = a\sharp 1,\qquad \beta(a) = a_{[0]}\sharp a_{[1]},\qquad
\Delta(a\sharp t) = (a\sharp t_{(1)}) \otimes_A (1\sharp t_{(2)}).
$$
We often identify $A\sharp 1 = \mathrm{Im}\,\alpha$ with $A$
and $1\sharp T$ with $T$.
By the Definition~\ref{eq:IA}
the right ideal $I_A\subset H\otimes_A H$ is generated by the set
of all elements of the form
$$
I(a) = \beta(a)\otimes 1 - 1\otimes\alpha(a) =
a_{[0]}\sharp a_{[1]}\, \otimes 1 - 1 \otimes a,\,\,\,\,\,\,\,\,
a\in A.
$$
There is also another set of generators $R(a)$ of $I_A$, more convenient for
our analysis below.
\begin{proposition}\label{prop:RagenIA}
In the case of scalar extension $H = A\sharp T$,
$I_A$ is generated as a right ideal of $H\otimes_{\bm k} H$
by the set of all elements of the form
\begin{equation}\label{eq:Ra}
R(a) = a \,\otimes 1 - S a_{[1]}\,\otimes a_{[0]},
\,\,\,\,\,\,\,\,\,a\in A.
\end{equation}
\end{proposition}
\begin{proof}
In the notation introduced before the proposition,
$$
I(a) = (a_{[0]}\sharp 1 - S a_{[0][1]}\otimes a_{[0][0]})(a_{[1]}\otimes 1)
= R(a_{[0]}) (a_{[1]}\otimes 1).
$$
Notice that $a_{[0]}\in A$. On the other hand,
$$
R(a) = (a_{[0][0]} \sharp a_{[0][1]}\otimes 1 - 1\otimes a_{[0]}) (S a_{[1]}\otimes 1)
= I(a_{[0]})(S a_{[1]}\otimes 1).
$$
Therefore, the right ideal generated by the sets of all $I(a)$ and the
right ideal generated by all $R(a)$ coincide.
\end{proof}
We notice that the elements
$R_\mu$ in~\cite{tajronkov} are $R_\mu = R(\hat{x}_\mu)$ where $\hat{x}_\mu$,
$\mu = 1,\ldots,\mathrm{dim}\,\mathfrak{g}$
are the generators of the universal enveloping algebra $U(\mathfrak{g})$
corresponding to the linear basis of the Lie algebra
$\mathfrak{g}$ in their work. See~\ref{ssec:comp}.
\vskip .1in
\noindent {\bf Lu's section. }
For any scalar extension bialgebroid, J-H.~Lu \cite{lu} introduces
a section $\gamma : H\otimes_A H\to H\otimes_{\bm k} H$
as the unique ${\bm k}$-linear extension of the formula
\begin{equation}\label{eq:gammasmash}
\gamma : h \otimes_A (a\sharp t) \mapsto
\beta(a) h \otimes (1\sharp t),
\end{equation}
where $h\in H$, $a\in A$ and $t\in T$.
This formula defines a section of the projection
$H\otimes_{\bm k} H\to H\otimes_A H$
because $a\sharp t = \alpha(a)(1\sharp t)$; hence
$h\otimes_A (a\sharp t) = \beta(a) h\otimes_A (1\sharp t)$.
Section $\gamma$ is well defined by the above formula,
as on the elements of the form
$\beta(b)h\otimes (c\sharp \tilde{t}) - h\otimes \alpha(b)(c\sharp\tilde{t})$
the formula gives $\beta(b)\beta(c)h\otimes(1\sharp \tilde{t}) -
\beta(b c)h\otimes(1\sharp\tilde{t}) = 0$.
In particular, the formula~(\ref{eq:gammasmash}) gives
\begin{equation}\label{eq:gammaDelta}
(\gamma\circ\Delta)(a\sharp t) = (a\sharp t_{(1)})\otimes (1\sharp t_{(2)}).
\end{equation}
\subsection{Subalgebra $W\subset H\otimes H$ where $H=A\sharp T$ is a scalar extension Hopf algebroid over $A$}
\label{ssec:W}
\begin{notation} Let $T$ be a Hopf algebra and $A$ a braided commutative
algebra in the category of left-right Yetter-Drinfeld $T$-modules.
For a scalar extension $A\sharp T$ let
$W\subset (A\sharp T)\otimes (A\sharp T)$
be the smallest unital subalgebra such that all elements of
the form $a\otimes 1$ and all elements of the form $S a_{[1]}\otimes a_{[0]}$
(where $a\in A\cong A\sharp 1\subset A\sharp T$) are in $W$. Let $W^+$
be the two sided ideal in $W$ generated by all elements of the
form $R(a) = a\otimes 1 - S a_{[1]}\otimes a_{[0]}$
where $a\in A$ (compare~(\ref{eq:Ra})).
Let $W_0^+\subset W$ be the linear subspace of $W$ spanned
by all elements of the form $(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(x'\otimes 1)$ where $x,x'\in A\cong A\sharp 1$.
\end{notation}
We formulate Lemma 1 and Lemma 2 which together imply $W_0^+ = W^+$.\vskip .1in
\begin{lemma}\label{lem:R1}
For $x,z\in A$ we have $(x\otimes 1)(z\otimes 1 - S z_{[1]}\otimes z_{[0]})\in W_0^+$.
\end{lemma}
\begin{proof} Multiplying out, and using
$x S(t) = S(t_{(1)}) t_{(2)} x S (t_{(3)})
= S (t_{(1)}) (t_{(2)}\blacktriangleright x)$ for $x\in A$, $t\in T$, we obtain
$$\begin{array}{lcl}
x z\otimes 1 - x S (z_{[1]})\otimes z_{[0]} &=&
x z \otimes 1 - S (z_{[1]}) (z_{[2]}\blacktriangleright x)\otimes z_{[0]}\\
&=& \mathrm{by\ braided\ commutativity} \\
&=& z_{[0]} (z_{[1]}\blacktriangleright x) \otimes 1 -
S (z_{[1]}) (z_{[2]}\blacktriangleright x)\otimes z_{[0]}\\
&=& (z_{[0]}\otimes 1 - S (z_{[0][1]}))\otimes z_{[0][0]})
(z_{[2]}\blacktriangleright x\otimes 1)
\end{array}$$
and the right hand side is clearly in $W_0^+$ as claimed.
\end{proof}
\begin{lemma} \label{lem:R2}
$R(x)R(z) = (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1 - S z_{[1]}\otimes z_{[0]})\in W_0^+$.
\end{lemma}
\begin{proof} Since $x\mapsto x_{[1]}\otimes x_{[0]}$
is a morphism of algebras $A\to T^{\mathrm{op}}\otimes_{\bm k} A$
and $T\otimes_{\bm k} A\hookrightarrow A\sharp T\otimes_{\bm k} A\sharp T
= H\otimes_{\bm k} H$
inclusion of algebras, we conclude that
$x\mapsto S x_{[1]}\otimes x_{[0]}$ is a morphism of algebras
$A\to H\otimes_{\bm k} H$
(with respect to the componentwise multiplication in $H\otimes_{\bm k} H$).
Therefore,
$$
(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1
- S z_{[1]}\otimes z_{[0]}) =
$$
$$
=
(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1)
+ x S z_{[1]}\otimes z_{[0]} -
x z\otimes 1 + x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]}
$$
$$
= (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) +
(-x\otimes 1) (z\otimes 1
- S z_{[1]}\otimes z_{[0]}) + (x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]}).
$$
The first and the third summands on the right hand side
are manifestly in $W_0^+$ while for the second summand
we apply Lemma~\ref{lem:R1}.
\end{proof}
\begin{corollary} \begin{itemize} \item[(i)] $(S x_{[1]}\otimes x_{[0]})(z\otimes 1-S z_{[1]}\otimes z_{[0]})\in W_0^+$,
\item[(ii)] $W^+_0$ is a two-sided ideal in $W$,
\item[(iii)] $W^+ = W_0^+$.
\end{itemize}
\end{corollary}
\begin{proof} (i) follows by subtracting the expression
in the statement of Lemma 2 from the expression in the statement of Lemma 1.
(ii) $W^+_0$ is a right ideal: it is obvious that
we can multiply by $z\otimes 1$ from the right;
this together with the assertion of Lemma~\ref{lem:R2}
implies that we can also multiply
by $S z_{[1]}\otimes z_{[0]}$ from the right.
(ii) $W^+_0$ is a left ideal: using Lemma~\ref{lem:R1},
$(x\otimes 1) R(z) (x'\otimes 1) \in W_0^+ (x'\otimes 1)$
what is in $W_0^+$ as it is a right ideal.
Combining with Lemma~\ref{lem:R2} we also get
$(S x_{[1]}\otimes x_{[0]})R(z)(x\otimes 1) \in W_0^+$.
For (iii) notice first that trivially $W_0^+\subset W^+$.
For the converse inclusion, $W^+\subset W_0^+$, it is sufficient
to observe that $R(a)\in W_0^+$, apply
(ii) and the definition of $W^+$.
\end{proof}
Let now $\tau$ be the antipode of the scalar extension Hopf
algebroid $A\sharp T$ over $A$. We know (cf.~\cite{BrzWis,lu}) that
\begin{equation}\label{eq:tause}
\tau(a\sharp t) = S(t) S^2(a_{[1]}) \cdot a_{[0]}.
\end{equation}
\begin{theorem}
$\mu (id\otimes_{\bm k}\tau) W^+ = \{0\}$.
\end{theorem}
\begin{proof}
By Corollary (iii) $W^+ = W_0^+$, which
is the span of the elements of the form
$$(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1),\,\,\,\,\mathrm{where\ }\,
x,z\in A.$$
We can easily compute $\mu (id\otimes\tau)$ on such an element as
$$
x z - S (x_{[2]}) z S^2(x_{[1]}) x_{[0]}
= x z - ((S x_{[1]})\blacktriangleright z) x_{[0]}
= 0.
$$
by the braided commutativity.
\end{proof}
\vskip .2in
\subsection{Subalgebra $\mathcal{B}$ and
two-sided ideal $\mathcal{B}^+\subset\mathcal{B}$}
\label{ssec:B}
In this section,
we want to show that every scalar extension Lu-Hopf algebroid
$H = A\sharp T$ is also a Hopf algebroid with a
(carefully chosen) balancing subalgebra $\mathcal{B}$.
Using the inclusion $T\otimes_{\bm k} T
\hookrightarrow A\sharp T\otimes_{\bm k} A\sharp T$,
we identify the image of the coproduct
$\Delta_T:T\to T\otimes_{\bm k} T$ of the Hopf algebra $T$
with a subalgebra of $H\otimes_{\bm k} H$ which will be denoted
by $\Delta_T(T)$.
\begin{definition}\label{def:B}
Let the subalgebra $\mathcal{B}\subset A \sharp T \otimes_{\bm k} A \sharp T$
be the subalgebra generated by $W$ and $\Delta_T(T)$
or, equivalently, by the set
$$\left\{ X \otimes 1 , \ SX_{[1]} \otimes X_{[0]} \, \mid \, X \in A \right\} \cup \Delta_T(T).$$
The elements of this set will be called the {\bf distinguished generators of}
$\mathcal{B}$.
Let $\mathcal{B}^+$ be the two-sided ideal in $\mathcal{B}$
generated by the subset
$$\left\{ X \otimes 1 - \ SX_{[1]} \otimes X_{[0]} \, \mid \, X \in A \right\}\subset\mathcal{B},$$
whose elements are called the {\bf distinguished generators of} $\mathcal{B}^+$.
\end{definition}
\begin{theorem}\label{th:scalarbialgM}
\begin{itemize}
\item[(i)] $\mathcal{B}^+ = I_A\cap\mathcal{B}$.
\item[(ii)] (CRMI) holds: $I_A\cap\mathcal{B}$ is a two-sided ideal
in $\mathcal{B}$.
\item[(iii)] (CRMa) holds:
$\mathrm{Im}\,\Delta\subset\mathcal{B}/(I_A\cap\mathcal{B})$
\item[(iv)] The scalar extension is a bialgebroid with the balancing subalgebra $\mathcal{B}$.
\item[(v)] The algebra $\mathcal{B} / \mathcal{B}^+$
is isomorphic to $\Delta_L(A \sharp T) \subset A \sharp T \otimes_A A \sharp T$.
\end{itemize}
\end{theorem}
\begin{proof}
(i) follows immediately by the Definition~\ref{def:B}
for $\mathcal{B}^+$ and Proposition~\ref{prop:RagenIA}.
(ii) and (iii) is immediate check knowing the generators and (i).
For (iv) use (ii), (iii), Proposition~\ref{prop:equivifBexists}
and the fact that the (C3b) is known for scalar extensions~\cite{BrzMilitaru,lu}.
\end{proof}
\begin{remark}\label{rem:summ}
For the simplicity of exposition, in the following multilinear
calculations we omit the summation signs and the corresponding labels
when generically denoting simple products of distinguished generators of $\mathcal{B}$: we use Sweedler-like symbols. The generic symbol $K_i\otimes L_i$ for a distinguished generator
in the following lemma and theorem is a symbol for a sum of the form
$$\sum_{\alpha_i} K_i^{\alpha_i}\otimes L_i^{\alpha_i},$$ whose components $ K_i^{\alpha_i}\otimes L_i^{\alpha_i}$ are equal to components of one of these sums: $X \otimes 1$, $ SX_{[1]}\otimes X_{[0]}$, $f_{(0)} \otimes f_{(1)}$, where $X \in A$ and $f \in T$. When we say $K_i$ is in $A \sharp 1$, it means that each $K_i^{\alpha_i} \in A \sharp 1$, because by this notation $K_i$ denotes each $K_i^{\alpha_i}$ at the same time. Similarly, the generic symbol $K \otimes L$ for a product of distinguished generators $K_i\otimes L_i$, $$\prod_{i =1}^n \sum_{\alpha_i} ( K_i^{\alpha_i}\otimes L_i^{\alpha_i}) =
\sum_{\alpha_1, \ldots, \alpha_n} \prod_{i =1}^n \left( K_i^{\alpha_i}\otimes L_i^{\alpha_i}\right) =
\sum_{\alpha_1, \ldots, \alpha_n} \prod_{i =1}^n K_i^{\alpha_i} \otimes \prod_{i =1}^n L_i^{\alpha_i},$$ is a symbol for a sum $$
\sum_{\alpha_1, \ldots, \alpha_n} \prod_{i =1}^n K_i^{\alpha_i} \otimes \prod_{i =1}^n L_i^{\alpha_i}$$ and $K$ and $L$ are generic symbols denoting each product of components $\prod_{i=1}^n K_i^{\alpha_i}$ and $\prod_{i=1}^n L_i^{\alpha_i}$ at the same time.
\end{remark}
\begin{lemma}\label{lem:key} Let $U$ be a product of distinguished generators of $\mathcal{B}$. Then $$\mu(\mathrm{id} \otimes \tau)(U) \in A \sharp 1.$$
\end{lemma}
\begin{proof} Denote the distinguished generators that $U$ is a product of
by $K_1 \otimes L_1, \ldots, K_n \otimes L_n$.
The antipode $\tau$ is an algebra antihomomorphism, hence
\begin{equation}\label{eq:Eexpanded}
\mu(\mathrm{id} \otimes \tau)(U) = K_1 K_2 \cdots K_n \tau(L_n) \cdots \tau(L_1).
\end{equation}
We claim that $K_i \cdot (Z \sharp 1) \cdot \tau(L_i) \in A \sharp 1$ for $Z \sharp 1 \in A \sharp 1$. We inspect this claim case by case as follows.
\begin{itemize}
\item[(a)] If $K_i \otimes L_i \in \left\{ X \otimes 1, \ SX_{[1]} \otimes X_{[0]} \,|\,X\in A\right\}$, then by~(\ref{eq:tause})
\begin{equation}\label{eq:kZtL}
K_i \cdot (Z \sharp 1) \cdot \tau(L_i) = K_i \cdot Z\sharp 1 \cdot S^2(L_{i{[1]}}) \cdot L_{i[0]}.
\end{equation}
The product $\cdot$ is in the smash product $A\sharp T$; $A$ is identified
there with $A\sharp 1$ and $T$ with $1\sharp T$.
There are now two subcases, (a1) and (a2).
\begin{itemize}
\item[(a1)]
For $K_i \otimes L_i = X \otimes 1$,
(\ref{eq:kZtL}) equals $XZ = XZ\sharp 1$ which is in $ A \sharp 1$.
\item[(a2)]
For $K_i \otimes L_i = SX_{[1]} \otimes X_{[0]}$,
(\ref{eq:kZtL}) equals
$SX_{[2]} \cdot Z \cdot S^2(X_{[1]}) \cdot X_{[0]} = ((SX)_{[1]} \blacktriangleright Z) \cdot (SX)_{[2]} S((SX)_{[3]}) \cdot X_{[0]} = ((SX)_{[1]} \blacktriangleright Z) \cdot X_{[0]} = XZ$
which is in $A \sharp 1$.
\end{itemize}
\item[(b)] If $K_i \otimes L_i = f_{(1)} \otimes f_{(2)}$, then $ K_i \cdot Z \sharp 1 \cdot \tau(L_i) = f_{(1)} \cdot Z\sharp 1 \cdot S( f_{(2)}) = (f \blacktriangleright Z) \sharp 1 $ which is again in $A \sharp 1$.
\end{itemize}
Therefore the claim $K_i \cdot Z \sharp 1 \cdot \tau(L_i) \in A \sharp 1$ follows.
In particular, for $Z=1$, $K_n \tau(L_n) \in A\sharp 1$ so the claim provides the step and the base for the induction on $p$, where $0\leq p\leq n-1$,
proving that
$$K_{n-p}\cdots K_n \tau(L_n)\cdots \tau(L_{n-p})\in A\sharp 1.$$ By~(\ref{eq:Eexpanded}) it follows that $\mu(\mathrm{id} \otimes \tau)(U) \in A \sharp 1$; in other
words $\mu(\mathrm{id} \otimes \tau)(U)$ is of the form $D \sharp 1$ where $D\in A$.
\end{proof}
\begin{theorem}\label{th:scalarM}
Let $\tau:A\sharp T\to A\sharp T$ given by the formula~(\ref{eq:tause})
be the antipode of the scalar extension as a Lu-Hopf algebroid.
Then
\begin{itemize}
\item[(i)] $\mu( \mathrm{id} \otimes \tau) \mathcal{B}^+ = \left\{ 0 \right\} $.
\item[(ii)] $\tau$
makes the corresponding $A$-bialgebroid with a balancing subalgebra
from Theorem~\ref{th:scalarM}
into a Hopf $A$-algebroid with a balancing subalgebra.
\end{itemize}
\end{theorem}
\begin{proof} (i)
A general element of $\mathcal{B}^+$ is a linear combination of the
elements of the form
$$ \prod_j M_{j} \otimes N_{j} \cdot \left( X \otimes 1 - SX_{[1]} \otimes X_{[0]} \right) \cdot \prod_k K_{k} \otimes L_{k},$$
where $M_{j} \otimes N_{j}$, $K_{k} \otimes L_{k}$ are generic symbols for some distinguished
generators of $\mathcal{B}$, and the midd\-le
factor is some distinguished generator in $\mathcal{B}^+$.
Notice that
$M_{j}$, $N_{j}$, $K_{k}$, $L_{k} \in A \sharp 1 \cup 1 \sharp T$, $X \in A$.
Here $M_{j}$ is a generic symbol denoting each $M_{j}^{\alpha}$ at the same time, see Remark~\ref{rem:summ}.
By the linearity of $\mu( \mathrm{id} \otimes \tau)$ it is sufficient to prove
the assertion for one element of the form above.
We decompose it in two pieces, multiply component by component and
denote so obtained products by generic symbols
$M$, $N$, $K$ and $L \in A \sharp T$.
$$(M \cdot X\sharp 1 \cdot K )\otimes (N \cdot L) - (M \cdot 1 \sharp SX_{[1]} \cdot K ) \otimes (N \cdot X_{[0]} \sharp 1 \cdot L )$$
We apply map $\mu( \mathrm{id} \otimes \tau)$ to this expression.
Thus we need to show that
$$M \cdot X\sharp 1 \cdot K \cdot \tau(L) \cdot \tau(N) = M \cdot 1 \sharp SX_{[1]} \cdot K \cdot \tau( L) \cdot \tau( X_{[0]} \sharp 1 ) \cdot \tau(N).$$
This directly follows by a simpler formula:
$$ X\sharp 1 \cdot E = 1 \sharp SX_{[1]} \cdot E \cdot \tau( X_{[0]} \sharp 1 ) ,$$
where $E$ denoted $K \cdot \tau(L) \in A \sharp T$.
By Lemma~\ref{lem:key}, $K\cdot\tau(L)\in A\sharp 1$.
Therefore to finish the proof of the assertion (i) it is
sufficient to show that for all $X$, $D \in A$ we have
$$ X\sharp 1 \cdot D \sharp 1 = 1 \sharp SX_{[1]} \cdot D \sharp 1 \cdot \tau( X_{[0]} \sharp 1 ),$$
where $\tau(X_{[0]} \sharp 1) = \alpha_R(X_{[0]}) = S^2(X_{[1]}) \cdot X_{[0]}$.
This amounts to show
$$XD \sharp 1 = \left((SX)_{[1]} \blacktriangleright D \right) \sharp \, (SX)_{[2]} \, S((SX)_{[3]}) \cdot X_{[0]} \sharp 1 ,$$
that is,
$$XD \sharp 1 = \left((SX)_{[1]} \blacktriangleright D\right) X_{[0]} \sharp 1,$$
which is equivalent to the braided commutativity for $A \sharp T$.
Therefore, (i) is proven.
For the part (ii), according to Theorem~\ref{th:scalarbialgM}, part (iv)
it remains only to check the axioms for the antipode.
The antipode requirements~(\ref{eq:mtaubetaalpha}) and~(\ref{eq:ml})
have the same meaning
as in the case of Lu-Hopf definition hence they are true.
Now, thanks to (ii) the right-hand side of the equation~(\ref{eq:ml}),
that is, $\mu( \mathrm{id} \otimes \tau) \Delta = \mathrm{id}$,
does not depend on the representatives of $\Delta(a t)
= (a\sharp t_{(1)})\otimes (1\sharp t_{(2)})$ in $H\otimes_{\bm k} H$
where $a\in A$ and $t\in T$. So we should compute that
$$
(a \sharp t_{(1)}) (S^2((a t_{(2)})_{[1]})\cdot (a t_{(2)})_{[0]}) = a\sharp t,
$$
which is the same computation as the one for Lu's choice of $\gamma$,
see~(\ref{eq:gammaDelta}).
Our result is stronger only in the sense that the additional freedom
in $\mathcal{B}_+$ is allowed here and that
$\mathcal{B}$ is a balancing subalgebra in the sense of bialgebroids.
\end{proof}
\subsection{Comparison with the examples of Meljanac}
\label{ssec:comp}
Meljanac has devised his method~\cite{tajronkov,tajron}
to study topological Hopf algebroids related to
a Lie algebra $\mathfrak{g}_\kappa$ with
the universal enveloping algebra $U(\mathfrak{g}_\kappa)$
in physics literature called the $\kappa$-Minkowski space.
The same Hopf algebroid from the point of view of Lu-Hopf algebroid
(and also in physics language) has been studied in~\cite{kappaPLB} and (in
just slightly more mathematical treatment) in~\cite{JLZSMWa}
Works~\cite{tajronkov,tajron} made it clear that their
construction applies to any finite dimensional Lie algebra
$\mathfrak{g}$ in characteristic zero. We comment below on
how our construction of $\mathcal{B}$
relates to theirs in this special case.
As stated in the introduction, we neglect here the issues
related to the adaptation of the notion of Hopf algebroid to
the completed tensor products~\cite{halgoid}.
We use the notation from~\cite{halgoid}.
Generators of the Lie algebra $\mathfrak{g}$
are denoted $\hat{x}_1,\ldots,\hat{x}_n$
with commutators $[\hat{x}_\mu,\hat{x}_\nu] = C_{\mu\nu}^\lambda\hat{x}_\lambda$
and the generators of
the symmetric algebra of the dual $S(\mathfrak{g}^*)$
by $\partial^1,\ldots,\partial^n$. The completed dual
$T = \hat{S}(\mathfrak{g}^*)$ is a topological Hopf algebra, namely the
coproduct $\Delta_T:\hat{S}(\mathfrak{g}^*)\to\hat{S}(\mathfrak{g}^*)\hat\otimes\hat{S}(\mathfrak{g}^*)$
may be identified (using the symmetrization map $S(\mathfrak{g})\cong U(\mathfrak{g})$ which
is an isomorphism of coalgebras~\cite{halgoid} and its dual isomorphism
of algebras $\hat{S}(\mathfrak{g}^*)\cong U(\mathfrak{g})^*$)
the transpose (dual) map to the multiplication $U(\mathfrak{g})\to U(\mathfrak{g})\otimes
U(\mathfrak{g})$. Now $A = U(\mathfrak{g})$ becomes a braided-commutative
Yetter-Drinfeld module algebra over $T$; this is related to
a right Hopf action of $U(\mathfrak{g})$ on $T$ which helps building
the smash product $H = T\sharp A$. Regarding that the coproduct of
$U(\mathfrak{g})$ is cocommutative, one can take instead a left Hopf action
(composing the right action with the antipode) of $U(\mathfrak{g})$ on $T$
and the smash product $A\sharp T$, what is the situation in this
article and in~\cite{halgoid}.
If $R(a)$ is defined by~(\ref{eq:Ra}), then one observes that
$R(\hat{x}_\mu)$ equals $R_\mu$ of~\cite{tajronkov,tajron}).
It is known that $[R_\mu,R_\nu] = C_{\mu\nu}^\lambda R_\lambda$
(Eq.~(32) in~\cite{tajronkov}). Moreover,
there are no other relations among $R_\mu$, hence the subalgebra
in $H\otimes_{\bm k} H$ generated by $R_\mu$ is isomorphic to the universal
enveloping algebra $U(\mathfrak{g})$, but with generators $R_\mu$
in place of $\hat{x}_\mu$; thus denote it by $U(R)$.
The relation~\cite{tajronkov}~(33) $[\hat{x}_\mu\otimes 1,
R_\nu] = C_{\mu\nu}^\lambda R_\lambda$ shows that the products of elements
in $U(R)$ with elements in
$A\sharp 1 \otimes_{\bm k} {\bm k}\subset
A\sharp T\otimes_{\bm k} A\sharp T$ span a subalgebra in $H\otimes_{\bm k} H$.
This is precisely
our subalgebra $W$ in this case. However,
their relations~(3.1),(3.2) in~\cite{tajronkov},
which they used to prove that it is a subalgebra of $H\otimes_{\bm k} H$,
do not have a simple analogue for general scalar extensions.
It is also not clear what is the exact
structure of the subalgebra generated by $R(a)$-s in general.
On the other hand, our Hopf algebraic definition~(\ref{eq:Ra}) of
$R(a)$ and the corresponding definition of $W$
in the subsection~\ref{ssec:W} along with the lemmas
therein guarantee that such general $W$ is still a subalgebra in
$H\otimes_{\bm k} H$.
The issues are substantially more complicated when we pass from
$W$ to $\mathcal{B}$.
In the enveloping algebra case, the subalgebra $\mathcal{B}$
(denoted $\hat{\mathcal{B}}$ in~\cite{tajronkov})
is defined in~\cite{tajronkov} rather easily as the subalgebra of
all elements of the form $\sum_i w_i \Delta_{\hat{S}(\mathfrak{g})}(t_i)$
where $w_i\in W$, $t_i\in T = \hat{S}(\mathfrak{g}^*)$ are arbitrary
(the sums may be infinite, in an appropriate completion).
Equations~(3.1)-(3.4) in~\cite{tajronkov} for general Lie algebra case,
give $[\Delta \partial^\mu,R_\nu] = 0$
and $[\hat{x}_\mu\otimes 1,\Delta(\partial_\nu)]\in\Delta_T(T)$ and
imply immediately that these elements form a subalgebra
in $H\otimes_{\bm k} H$, and that
$\mathcal{B}$ has a very simple structure of
all sums of products of the form:
an element in $U(R)$ times
an element in $A\sharp 1\otimes_{\bm k} {\bm k}
\subset H\otimes_{\bm k} H$
times an element of the form $\Delta_T(t_i)$ with $t_i \in T$.
Similar structure as sum of products in three subalgebras in
this fixed order
for general scalar extension Hopf algebroids: in that case
$P_i$ does not commute with elements in $W$ and multiple
products (e.g.\ of the form $w t w' t' w''$) of elements in $W$ and
elements in $\Delta_T(T)$ may
appear, as analysed in the subsection~\ref{ssec:B}.
Regarding that $\mu(id\otimes_{\bm k}\tau)$ is not an antihomomorphism,
the multiple products bring the main difficulty in
our proof that the antipode $\tau$ is well defined
(see Theorem~\ref{th:scalarM} (i)).
Analogous comparisons may be made for the ideal
$\mathcal{B}_+ = I_A\cap\mathcal{B}$
which is in~\cite{tajronkov} not defined as the intersection,
but rather constructed
exactly as $\mathcal{B}$ but with the enveloping algebra $U(R)$
replaced by its ideal $U_+(R)\subset U(R)$ of elements which
are not degree $0$ in the standard filtration of the
universal enveloping algebra.
The commutation relations (3.1)-(3.4) in~\cite{tajronkov} imply
that such $\mathcal{B}_+$ is indeed a two-sided ideal in $\mathcal{B}$.
Our approach also differs from~\cite{tajronkov} in insisting
that the coproduct is still defined as taking values in $H\otimes_A H$;
the two-sided ideal trick is used only to make sense of the requirement
and to check that the induced
map into $\mathcal{B}/(I_A\cap\mathcal{B})$ is a morphism of algebras.
Moreover they view $\mathcal{B}$ as an abstract algebra
constructed from its pieces $U(R)$, $A\sharp 1\otimes_{\bm k}\genfd$
and $\Delta_T(T)\cong T$.
In our approach, the coherently associative
tensor product of bimodules $\otimes_A$ is used to formulate
the coassociativity of the coproduct as in Lu's, Xu's and B\"ohm's
definitions. In~\cite{tajronkov} the coproduct is taking values in
$\mathcal{B}/(I_A\cap\mathcal{B})$ by the definition
and, in the spirit of their viewpoint, the higher analogues of
$\mathcal{B}^{(k)}\subset
H\otimes_{\bm k} H\otimes_{\bm k} \cdots\otimes_{\bm k} H$ of
$\mathcal{B}\subset H\otimes_{\bm k} H$
and the higher analogues $\mathcal{B}^{(k)}_+$ of $I_A\cap\mathcal{B}$
are constructed in order to deal with the (co)associativity issues.
These are interesting structures but in our view more cumbersome than
the familiar usage of the bimodule tensor product $\otimes_A$.
\nodo{
\section{Weak Hopf algebras}\label{s:wha}
It is well-known that from the data of any weak Hopf algebra
one can define a corresponding Lu-Hopf algebroid.
Upon looking at our axiomatics,
G.~B\"ohm has observed and sketched to us how to
construct a Hopf algebroid with balancing subalgebra
from a weak Hopf algebra.
We present her results in this section, starting with
a short review of weak Hopf algebras.
\subsection{Weak bialgebras, standard definitions}
\def\mathbb H{\mathbb H}
A weak ${\bm k}$-bialgebra $\mathbb H$ (see~\cite{bohmnillszlachwHa})
is a tuple $(\mathbb H,\mu,\eta,\Delta,\epsilon)$ where
$(A,\mu,\eta)$ is an associative unital ${\bm k}$-algebra,
$(\mathbb H,\Delta,\epsilon)$
is a coassociative counital ${\bm k}$-coalgebra
and the following compatibilities hold:
\begin{itemize}
\item[(i)] $\Delta$ is multiplicative, $\Delta(a b) =
\Delta(a)\Delta(b)$ for all $a,b\in \mathbb H$.
\item[(ii)] Weak multiplicativity of the counit
\begin{equation}\label{eq:wmult}
\epsilon(x y z) = \epsilon(x y_{(1)})\epsilon(y_{(2)} z),
\end{equation}
\begin{equation}\label{eq:wmultop}
\epsilon(x y z) = \epsilon(x y_{(2)})\epsilon(y_{(1)} z).
\end{equation}
\item[(iii)] Weak comultiplicativity of the unit:
\begin{equation}\label{eq:Delta2}\begin{array}{l}
\Delta^{(2)} (1) = (\Delta(1) \otimes 1)(1\otimes \Delta(1))
\\
\Delta^{(2)} (1) = (1 \otimes\Delta(1))(\Delta(1) \otimes 1)
\end{array}\end{equation}
where we denoted $\Delta^{(2)} := (id\otimes\Delta)\Delta =
(\Delta\otimes id)\Delta$.
\end{itemize}
For every weak bialgebra there are ${\bm k}$-linear maps
$\Pi^L,\Pi^R:A\to A$
with properties $\Pi^R\Pi^R = \Pi^R$ and $\Pi^L\Pi^L = \Pi^L$
and defined by
$$
\Pi^L(x) := \epsilon(1_{(1)} x) 1_{(2)},\,\,\,\,
\Pi^R(x) := 1_{(1)}\epsilon(x 1_{(2)}).
$$
These expressions are met below in two of the axioms for the antipode
of a weak Hopf algebra.
Now
$$\begin{array}{lcl}
\epsilon(x z) = \epsilon(x 1 z)
\stackrel{(\ref{eq:wmultop})}= \epsilon(x 1_{(2)})\epsilon(1_{(1)}z)
&=& \epsilon(\epsilon(x 1_{(2)})1_{(1)}z)
= \epsilon(\Pi^R(x)z),
\\
&=& \epsilon(x \epsilon(1_{(1)}z)1_{(2)})
= \epsilon(x\Pi^L(z)).
\end{array}$$
The images of the idempotents $\Pi^R$ and $\Pi^L$,
$$
\mathbb H^R := \Pi^R(\mathbb H),\,\,\,\,\,\,\,\,\mathbb H^L = \Pi^L(\mathbb H),
$$
are dual as ${\bm k}$-linear spaces via
the canonical nondegenerate pairing $\mathbb H^L\otimes\mathbb H^R\to k$
given by $(x,y) \mapsto \epsilon(y x)$.
The identities $\Pi^L(x\Pi^L(y)) = \Pi^L(x y)$
and $\Pi^R(\Pi^R(x)y) = \Pi^R(x y)$ hold.
Dually also $\Delta(\mathbb H^L)\subset \mathbb H\otimes\mathbb H^L$,
$\Delta(\mathbb H^R)\subset \mathbb H^R\otimes\mathbb H$,
and $\Delta(1)\in \mathbb H^R\otimes \mathbb H^L$.
\subsection{B\"ohm's recipes}
It is known that a weak Hopf algebra $\mathbb H$
can be regarded as a Lu-Hopf algebroid over $A:=\Pi^L(\mathbb H)$
where the source map $\alpha$ being the inclusion $\Pi^L(\mathbb H)\subset\mathbb H$
and where the target map is given by
\begin{equation}\label{eq:betawha}
\beta(a) = 1_{(1)} \epsilon(1_{(2)} a)\,\,\,\,\,\mbox{for}\,\,\,\,a\in A,
\end{equation}
and the comultiplication $\Delta'$ of the bialgebroid
is the comultiplication of the weak Hopf algebra
$\Delta:\mathbb H\to\mathbb H\otimes_{\bm k}\mathbb H$ followed by the
projection $\mathbb H\otimes_{\bm k}\mathbb H\to\mathbb H\otimes_{\Pi^L(\mathbb H)}\mathbb H$.
\begin{lemma}\label{lem:difunit}
$1\otimes 1-\Delta(1)\in I_A$.
\end{lemma}
\begin{proof}
By definition, $I_A$ is generated by all expressions of the form
$I(h) := \beta(\Pi^L(h))\otimes 1 - 1\otimes\alpha(\Pi^L(h))$ where $h\in\mathbb H$.
Now $\Pi^L(h) = \epsilon(1_{(1)} h) 1_{(2)}$
hence by~(\ref{eq:betawha})
\begin{equation}\label{eq:Ihwha}\begin{array}{lcl}
I(h) &=& \epsilon(1_{(1)} h) 1'_{(1)} \epsilon(1'_{(2)} 1_{(2)})\otimes 1
- 1 \otimes \epsilon(1_{(1)} h) 1_{(2)}\\
&\stackrel{(\ref{eq:wmultop})}{=}&\epsilon(1_{(2)}h)1_{(1)} \otimes 1
- 1 \otimes \epsilon(1_{(1)} h) 1_{(2)},
\end{array}
\end{equation}
where $1'_{(1)}\otimes 1'_{(1)}$ denotes another copy of $\Delta(1)$.
In other terms,
$$I(h) = \Pi^R(h)\otimes 1 - 1\otimes\Pi^L(h).$$
It is sufficient to prove that
$1\otimes 1 -\Delta(1) = I(1_{(2)})(1_{(1)}\otimes 1)$ because the right hand side manifestly belongs to $I_A$.
To this aim, set $h = 1'_{(2)}$ in~(\ref{eq:Ihwha})
and multiply the first tensor factor by $1'_{(1)}$ from the right and we get
$$\begin{array}{lcl}
I(1_{(2)}')(1_{(1)}'\otimes 1) &=& \epsilon(1_{(2)}1'_{(2)}) 1_{(1)}1_{(1)}'
\otimes 1 - 1_{(1)}' \otimes \epsilon(1_{(1)}1'_{(2)})1_{(2)}
\\
&=& \epsilon((1 1')_{(2)}) (1 1')_{(1)} \otimes 1
- 1_{(1)}\otimes \epsilon(1_{(2)}) 1_{(3)}
\\
&=& 1\otimes 1 - \Delta(1),
\end{array}$$
where in the middle line the axioms~(\ref{eq:Delta2}) on $\Delta^{(2)}$ were used for the second summand.
\end{proof}
\begin{lemma}\label{lem:Igen}
The right ideal $I_A$ is generated by $1\otimes 1 - \Delta(1)$.
\end{lemma}
\begin{proof}
Not sure that this holds (could we do without this ?).
We need to show that for every $h\in\mathbb H$ there exist $\tilde{h}\in\mathbb H\otimes_{\bm k}\mathbb H$,
so that $I(h) = [1\otimes 1 - \Delta(1)]\tilde{h}$, that is
$$
[1'_{(1)}\epsilon(1'_{(2)}1_{(2)})\otimes 1 - 1\otimes 1_{(2)}] \epsilon(1_{(1)}h)
= [1'_{(1)}\epsilon(1'_{(2)}1_{(2)})\otimes 1 - 1\otimes 1_{(2)}]
(1_{(1)}\otimes 1)\tilde{h}
$$
what looks hard to satisfy in general.
\end{proof}
It is clear that the subset
$$ \mathcal{B}:= \Delta(1)(\mathbb H\otimes\mathbb H)\Delta(1)\subset \mathbb H\otimes\mathbb H,$$
is a subalgebra with the unit $\Delta(1)$
and that $\operatorname{Im}\,\Delta\subset\mathcal{B}$.
Then the intersection $\mathcal{B}\cap I_A$ should be the trivial,
zero ideal of $\mathcal{B}$ ?
(it were obvious had we proven the Lemma~\ref{lem:Igen}) Hence
$\mathcal{B}\cong \mathcal{B}/(\mathcal{B}\cap I_A)$
and $\Delta:\mathbb H\longrightarrow\mathbb H\otimes_{\Pi^L(\mathbb H)}\mathbb H$
factorizes, indeed,
through an algebra homomorphism
$\mathbb H\longrightarrow\mathcal{B}/(\mathcal{B}\cap I_A)$.
\subsection{Antipode}
A weak ${\bm k}$-bialgebra $\mathbb H$ is a {\bf weak Hopf algebra}
if there is a ${\bm k}$-linear map $S:\mathbb H\to\mathbb H$
(which is then called an antipode)
such that for all $x\in\mathbb H$
$$
x_{(1)} S(x_{(2)}) = \epsilon(1_{(1)} x)1_{(2)},
$$
$$
S(x_{(1)})x_{(2)} = 1_{(1)} \epsilon(x 1_{(2)}),
$$
$$
S(x_{(1)})x_{(2)} S(x_{(3)}) = S(x)
$$
As in the case of the corresponding Lu-Hopf algebroid we
set the antipode of the Hopf algebroid with a balancing subalgebra
to be $\tau = S$.
Since $I_A \cap\mathcal{B} = \{0\}$
any linear map vanishes on it;
hence so does in particular the map of~(\ref{eq:tauvanish}).
The other antipode axioms (8)-(10) seem to hold (proof?).
To study: How about the minimal Lu-Hopf algebroid $A\otimes A^\mathrm{op}$?
Can we choose then $\mathcal{B}=A\otimes 1\otimes 1\otimes A^\mathrm{op}$?
}
\begin{footnotesize}
|
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"redpajama_set_name": "RedPajamaArXiv"
}
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Winning Percentage Based On Turnover Margin Since 2000: CFB.
LSU has won 8 games or more every season since 2000, yet have continually underachieved.
Variety is the key when it comes to Baylor University's sporting success. All 19 of its athletic programs have enjoyed at least one post-season showing since 2007, and its trophy room is packed with 65 Big 12 Conference championship trophies. These trophies represent the success of their basketball, baseball, tennis, football, soccer, softball and golf players, as well as the equestrian team.
Top 10 Winningest College Football Programs of All Time.
We all need to come together. Play Sporcle's virtual live trivia to have fun, connect with people, and get your trivia on.Join a live hosted trivia game for your favorite pub trivia experience done virtually. If a virtual private party is more your thing, go here for details.Whittingham has been on Utah's staff since 1994 and been at the helm of the program since 2005. During his time as head coach of the Utes, he has a 120-61 record, giving him a .663 winning percentage.Georgia Southern, on the other hand, has been appearing on the gridiron since 1924, moving up to the FBS Sun Belt in 2014 after winning six FCS National Championships (from 1985 to 2000).
LSU and Clemson are set to square off for a shot at winning the 2019-20 national championship. Here are the schools with the most national championships in college football since the 1936 season.Ranking The 16 Heisman Trophy Winners Since 2000.. and he proved them all wrong putting together a great career and being the most dynamic player in college football when he could get on the field. 10 Jameis Winston. When he received the Heisman in 2000, Weinke became the oldest player to win the award at 28 years old. 5 Matt Leinart - 2004.
In fact, last year was the first time in five seasons that a team other than Alabama or Clemson won the National Championship, giving teams hope that there's about to be a changing of the guard. With that in mind, we've decided to share our prediction of the 25 best college football teams ahead of the kickoff of the 2020 season.
Tips and Tricks from our Blog. Do you have a blog? Join our linker program. Watch our How-To Videos to Become a Stathead; Subscribe to the Play Index and get access to more data than you can imagine.
What they do best:Boise State has the highest winning percentage in the FBS since 2000, and it is in the top five in bowl wins, and they've built on their reputation as a Group of 5 buster. The.
The Best College Football Programs Of The 1990s Are.
Alabama had four Super Bowl-winning quarterbacks on its resume by the mid-1970s but hasn't produced a Super Bowl-winning quarterback since. Bart Star led the Packers to wins in the first two Super.
Ties were broken by the team with the best overall record since 2010.. for that first national championship since 2000.. made waves through a 25-game win streak in the College Football.
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A postseason ban prevented Urban Meyer's first Ohio State team from being more than a college football footnote while going 12-0. Quarterback Braxton Miller threw for more than 2,000 yards and ran.
The past 20 years have been amazing ones for the East Grand Rapids athletic program. EGR has won 56 state team championships in 12 different sports since 2000, and with those titles have come.
Answer Depends on what you looking for, since the FCS and FBS play a different amount of games per year you should be asking what college football team has the highest winning percentage? Answer.
LOOK! Ohio State football is No. 1 in all-time AP Poll.
Winning six national championships and appearing in 16 California State Championships and having the longest undefeated streak in high school football history winning 151 strait games from 1992-2005. They also produced player such as Maurice Jones-Drew and D. J Williams. Numbers speak for themselves. Truly the Best of the Best.
College First Season Seasons Wins Losses Ties Win% 1 Kentucky: 1903 116 2,293 706 1 .765 2 Kansas: 1899 121 2,274 859 0 .726 3 North Carolina: 1911 109 2,261 799 0 .739 4 Duke: 1906 114 2,176 887 0 .710 5 Temple: 1895 125 1,926 1,079 0 .641 6 Syracuse: 1901 119 1,904 908 0 .677 7 UCLA: 1920 100 1,887 852 0 .689 8 Notre Dame: 1898 122 1,880.
From 1901-1905, Yost coached his winning college football teams to 56 straight games without losing, while outscoring the competition 2,821 to 42. Also under Yost as the coach, Michigan won 10 conference championships and four national championships. Twenty players earned All-American honors while playing on Yost's winning Michigan football teams.
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Q: SwiftyJSON- Iterate through objects with ID which are not an array I am making a request to an API that sometimes returns an array of simple JSON objects, which I am parsing with a simple "for i in count" loop, where I use SwiftyJSON to append json[i]["city"]. For example:
[{"city":"Lakefront","code":"NEW","country":"United States","municipality":"New Orleans","isChild":false,"hasChild":false},
{"city":"Auckland - Auckland International Airport","code":"AKL","country":"New Zealand","municipality":"Auckland","isChild":false,"hasChild":false},
{"city":"Blenheim","code":"BHE","country":"New Zealand","municipality":"Blenheim","isChild":false,"hasChild":false}]
However, in certain cases, the API will return an object with multiple pairs of keys and object values, which I am having trouble parsing with SwiftyJSON. For example:
{"2":{"city":"New York","code":"NYC","country":"United States","municipality":"New York","isChild":false,"hasChild":true},
"32":{"city":"John F. Kennedy - NY","code":"JFK","country":"United States","municipality":"New York","isChild":true,"hasChild":false},
"414":{"city":"LaGuardia - NY","code":"LGA","country":"United States","municipality":"New York","isChild":true,"hasChild":false}}
In this second case, is there a way to loop through the first object with SwiftyJSON, and get the object value noting that I will not know the ID (2, 32, 414) in advance?
Thanks!
A: In your first case you have an array of dictionaries. In the second case you have a dictionary of dictionaries. I haven't used SwiftyJSON in long enough that I don't remember how it works, but that should be enough to get you going.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,055
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Release the Fog in Mad Rocket: Fog of War
We all have that one game that we use to measure all others. It doesn't matter if it wasn't the first of its genre or type, we don't value the history of gaming than but our own experience and emotion that this game created. When it comes to strategy games, I measure all to StarCraft. I know, it's a high standard but really it's also something that helps you define the efforts developers put into their games.
When it comes to Mad Rocket: Fog of War, I think that Ratatat Studio did fairly well in their endeavors to create a decent base-building strategy game. However, while StarCraft had the big storyline and, well, more elaborate gameplay, Mad Rocket: Fog of War is a simpler take on the genre that defined the gaming industry especially in the 1990s. Unlike Blizzard's hit title, Mad Rocket: Fog of War is available only for iOS and Android devices, though a PC version would be really nice to show up in the future.
The story takes place in the scientific base which is under threat of being attacked by the enemies at any moment. You arrive there as a new commander responsible to strengthen the defenses and build structures that will protect it. The base's scientists work on the Eternium research which produced Dark Fog, a resource that will help you shield your position from the enemy. Of course, the attacks are not one-sided and you will have a chance to retaliate as well so that makes the gameplay even more dynamic.
The game offers different units to choose from and use their special skills for defense and attack. The best practice is like with any strategy, to build the best possible team and then go into attack mode on the opponent's base. But while that may sound easy, it's far from it. After all, this is a strategy game and it depends on your strategizing and planning to defeat the enemy and expand your base.
The fog here is your doom as much as your pal since your enemies will use it as well to hide. This is especially important in multiplayer mode when you fight against players from all over the world, and not the game's AI. The best way to deal with the fog is to send the best scouts to reveal what waits for you out there, and then plan your move so that it results in the minimum loss.
As you level up, new weapons will unlock which are pretty cool and useful. However, the same applies to your enemies' weapons, so always be ready for the worst case scenario. Your success will be recorded in the Season Rankings so the better you are, the higher you'll get on the leaderboard.
Music in the game is really interesting and reminded me of '70s sci-fi TV shows like Star Trek or Lost in Space which really helped to highlight the scientific vibe about the game. Art is something that may look decent and not so different from others in the better titles for mobile devices, but it fits with the game dynamics really well.
What makes it great, is that game has a PEGI 3 rating and you can play it with your kids in multiplayer mode. It beats Monopoly, that's for sure.
This entry was posted in Review and tagged Mad Rocket: Fog of War.
Say Hello To Castle Burn
Red Dead Redemption 2 has the second most successful launch ever in terms of revenue
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{
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Warhawks QB joins two other Sun Belt players on 85-player watch list.
ULM junior quarterback Caleb Evans, who accounted for 3,447 yards total offense and 30 touchdowns last season, has been named to the 2018 Maxwell Award Watch List, according to an announcement made Monday by the Maxwell Football Club.
The Maxwell Award has been presented to college football's player of the year since 1937.
Evans joins two other Sun Belt players – Arkansas State quarterback Justice Hansen and Appalachian State running back Jalin Moore – on the 85-player watch list.
An honorable mention All-Sun Belt selection in 2017, Evans completed 61 percent (211 of 344) of his throws for 2,868 yards, 17 touchdowns and six interceptions. He ranked among the conference leaders in 300-yard passing games (third with 5), total offense (third at 287.2 yards per game), passing efficiency (fifth with a 144.2 rating) and passing yards (fifth at 239.0 ypg.). Evans also ranked second on the team in rushing with 137 carries for 579 yards (4.2 avg.) and a school single-season record 13 rushing TDs for a QB. His 13 rushing TDs led the Sun Belt and ranked No. 26 in the NCAA Football Bowl Subdivision.
The 6-foot-2, 208-pound Evans accounted for a 515 yards total offense and six TDs against Arkansas State in the home finale. His 515 yards total offense were the 10th-best single-game total in the NCAA FBS in 2017 and the third-highest single-game total in school history. His six TDs also tied the ULM single-game record (Kolton Browning was responsible for 6 TDs each vs. Troy and Western Kentucky in 2012). Evans set career highs for passing yards (454) and TD passes (4) against A-State. His 454 passing yards ranked as the eighth-best single-game total in ULM history.
Evans was named Sun Belt Offensive Player of the Week after accounting for 403 yards total offense and five TDs in leading ULM to a 52-45 victory over Appalachian State. He completed 24-of-32 throws for 356 yards and three scores against App State, and his 50-yard TD strike to RJ Turner with 23 seconds left proved to be the game winner.
The Mansfield, Texas, native hit 24-of-37 throws for 405 yards and three scores at Idaho.
Evans accounted for 456 total yards and four TDs in ULM's 45-27 win at Texas State. He completed 24-of-37 throws for 433 yards and three scores against the Bobcats.
Evans was selected College Football Performance Awards National Performer of Week 4 and Sun Belt Offensive Player of the Week after accounting for 472 total yards and six TDs to lead ULM to a 56-50 double-overtime win at UL Lafayette. He connected on 28-of-34 passes for 343 yards and a TD against UL Lafayette. Evans set career highs for carries (16), rushing yards (129), rushing TDs (tied school single-game record with 5) and TDs responsible for (6).
The award is named in honor of Robert W. "Tiny" Maxwell, who played guard at the University of Chicago before transferring to Swarthmore College where he became a Walter Camp All-American in 1905. His contributions to the game were extensive as a college and professional player, assistant coach, official and sportswriter.
Semifinalists for the Maxwell Award will be announced on Oct. 29. Three finalists for the award will be identified on Nov. 19, with the winner announced at the Home Depot College Football Awards Show on Dec. 6 in Atlanta.
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Q: How to get daily totals for a queryset new Python/Django programmer here. I have been playing with d3.js and I am having some trouble creating a multi-series line chart. I have narrowed down the problem, and I think it's because the data I am sending to d3 has multiple values per day.
My question, should you choose to answer it, is: How can I calculate the total for each day for each transaction type (credit or debit)?
Models.py
class Transaction(models.Model):
class Transaction(models.Model):
user = models.ForeignKey(Users, null=True)
date = models.DateField(null=True, blank=True)
description = models.TextField(max_length=75, null=True, blank=True)
original_description = models.TextField(max_length=75, null=True, blank=True)
amount = models.DecimalField(max_digits=12, decimal_places=2, null=True, blank=True)
transaction_type = models.TextField(max_length=75, null=True, blank=True)
category = models.TextField(max_length=75, null=True, blank=True)
account = models.ForeignKey(Account, null=True)
account_name = models.TextField(max_length=75, null=True, blank=True)
dashboard.py
import homepage.models as hmod
@view_function
def process_request(request):
params = {}
if request.user.is_authenticated():
user_session = request.user
userid = user_session.id
else:
return HttpResponseRedirect('/homepage/login/')
trans = hmod.Transaction.objects.all().filter(user_id=userid).values_list('date', 'transaction_type', 'amount', 'category', 'account_name').order_by('date')
x = 0
y = 0
for tran in trans:
if tran[1] == "credit":
x += 1
else:
y += 1
print(x)
print(y)
trans_json = json.dumps(list(trans), cls=DjangoJSONEncoder)
params['trans_json'] = trans_json
return templater.render_to_response(request, 'dashboard.html', params)
*x and y were just being used to find the total number of transactions of each type while I was struggling to figure it out.
html/script -- only an excerpt. I can include the whole script if that would help
//bring json data from dashboard.py
var data = ${trans_json}
...
// Group by transaction_type
var credit = data.filter(function(d) {
return d[1] == "credit";;
});
var debit = data.filter(function(d) {
return d[1] == "debit";
});
....
svg.selectAll('.line')
.data([debit, credit])
.enter()
.append('path')
.attr('class', 'line')
.style('stroke', function(d) {
return colors(Math.random() * 50);
})
.attr('clip-path', 'url(#clip)')
.attr('d', function(d) {
return line(d);
})
With my current setup, the data is returned as a long block of all transaction (grouped by "debit" or "credit) in the format of d="2014-09-12,debit,-9.85,Fast Food,CHECKING..."
But, when I return line(d), I get the ugly error of d="MNaN,160.57661666666664CNaN,160.57661666666664,NaN..."
I have created a line graph following http://bl.ocks.org/markmarkoh/8700606 in the past, but the data I used then only had one entry/day. That's why I'm thinking my error might be that a user tends to have multiple transactions per day making it hard to plot a line.
I am open to any help that you all can provide. If you can help me get the daily total for my queryset and pass it to the script, awesome. Or if there is a better way to accomplish what I wish, I'm happy to scrap and start from another angle.
Thanks!
Note If There are small syntax errors (i.e. ${trans_json}) that is probably because I'm using Mako as well in a stack of sorts called django-mako-plus https://github.com/doconix/django-mako-plus
That being said, however, I could have real syntax errors ^^ as well. I tried tagging django-mako-plus, but it seems that's not a tag yet.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,833
|
\section{Introduction}
Optical computing uses the speed and parallelism that light provides in order to process information efficiently \cite{ambs2010optical}. For half a century, considerable efforts have been carried out to develop an all-optical computer, but such a dream has never been realized due to the fierce competition of electronics. Still, optics has found successful applications in long-range communication and specialized computations. The success of Machine Learning has sparked a revival of interest in photonic neuromorphic computing \cite{denz2013optical}. Machine Learning algorithms have recently been applied to a large variety of tasks, ranging from image recognition to Natural Language Processing, and the neural networks used in Machine Learning are well-suited for optical computing. In particular, an optical phenomenon, called light scattering, has successfully been applied to Echo-State Networks \cite{dong2018scaling}, a class of networks part of the more general framework of Reservoir Computing.
In complex media, light does not propagate in a straight line, as refractive index inhomogeneities alter the direction of propagation. For example, turbulent airflows in the atmosphere and turbid biological tissues distort images and prevent the acquisition of a well-resolved image. This effect has been intensively studied in astronomy and biological microscopy, to image through or inside these complex media. Adaptive optics techniques can compensate the effect of scattering using wavefront shaping \cite{tyson2010principles, maurer2011spatial}. For even stronger distortion, when light propagates through a one-hundred-micron-thick layer of white paint for instance and gets scattered by small particles at random positions, direct imaging is no longer possible as the image has been scrambled by multiple scattering events. This process results in a complex interference figure, called speckle pattern. In this regime, it is possible to describe the scattering process by a linear multiplication with a dense matrix, called the Transmission Matrix (TM), which is characteristic of the medium and has been shown to be random \cite{popoff2010measuring}.
In other words, after the complex medium, the information of the incident electric field has been mixed but not lost. Instead of trying to reverse the detrimental effect of scattering to image through complex media, we want to explore the use of this mixing of information as a resource for computation.
Controlling the incident field by using Spatial Light Modulators (SLMs),
we can efficiently perform high-dimensional matrix multiplication and obtain an optical computing strategy that is potentially competitive with state-of-the-art electronic devices for various Machine Learning applications.
Random matrix multiplications are present in Echo-State Networks \cite{jaeger2001echo, jaeger2004harnessing} and Liquid-State Machines \cite{maass2002real}. They are Recurrent Neural Networks with fixed generic interconnection weights between neurons. Training these networks is considerably simpler compared to other Recurrent Neural Networks with tunable interconnection weights. Only the final linear layer is trained to predict the correct output by solving a simple linear regression problem. This class of networks represents a promising numerical tool for understanding and predicting temporal datasets, and they have found successful applications such as speech recognition \cite{triefenbach2010phoneme}, robot motor control \cite{antonelo2008event}, or financial forecasting \cite{lukovsevivcius2012reservoir}.
Reservoir Computing (RC) unifies and generalizes the previous approach, it transforms the previous recurrent network with fixed weights into a generic reservoir \cite{verstraeten2007experimental}. The reservoir is not necessarily described by a neural network, any system with rich and stable dynamics can be leveraged for Reservoir Computing. This opens up the possibility to use unconventional substrates \cite{tanaka2019recent}, where a physical reservoir receives an external excitation from an input and predictions are read out from an observation of the reservoir state. In the last decade, there has been a strong interest to find an efficient physical implementation, from dedicated electronic boards like Field-Programmable Gate Arrays \cite{antonik2015fpga} and memristive devices \cite{donahue2015design}, to photonic circuits \cite{van2017advances, bueno2018reinforcement} and carbon nanotubes \cite{dale2016evolving}.
\begin{figure*}[!t]
\centering
\includegraphics[width=\linewidth]{Light_scattering5.png}
\caption{Multiple light scattering for optical random projections.
(a) When light propagates in a complex medium, it undergoes many scattering events, modifying its direction of propagation.
(b) This scattering process results in a complex interference figure, called a speckle pattern, after the complex medium.
(c) Experimental setup for optical random projections. The light from a laser (Ti:Sapphire laser at 800 nm, MaiTai, Spectra Physics) is modulated by a Spatial Light Modulator (Meadowlark 512$\times$512 LCoS Reflective SLM), sent in a random scattering medium (a diffuser 0.5 mm thick), and collected on a camera (Allied Vision, Manta G-046). The speckle pattern collected on the camera is a random projection of the incident SLM image.}
\label{light_scattering}
\end{figure*}
In this study, we use optical scattering to implement Reservoir Computing algorithms for chaotic time series prediction. Here, only the iterative computation of the reservoir state is accelerated optically; these successive reservoir iterations represent the most computationally demanding operation, while we still perform on a conventional computer the other steps related to the final linear layer. This optical computing strategy can outperform electronic implementations by two orders of magnitude in speed and scale to very large dimensions that we cannot reach in electronics due to memory limitations. In this paper, we improve over the previous optical realization of binary Echo-State Networks presented in \cite{dong2018scaling} thanks to the introduction of binary encodings to increase binary RC performance, as well as introducing a new non-binary RC algorithm that uses spatial phase shaping of light instead of the previously reported on-or-off intensity modulation.
The choice of the Spatial Light Modulator (SLM) technology to convert digital information into an analog electric field is critical. We investigate here two approaches, either the use of a Digital Micromirror Device (DMD) or an SLM based on Liquid Crystal on Silicon technology (LCoS). The former, with DMDs, can only display binary images and this constraint needs to be addressed efficiently to preserve the performance of the associated Reservoir Computing algorithm. On the other hand, LCoS-SLMs can imprint 8-bit phase images on the optical electric field, which facilitates this digital to analog conversion, but they are relatively slow compared to Digital Micromirror Devices (few tens of Hertz versus up to 20kHz).
We tested here these optical networks on chaotic system prediction, with the Mackey-Glass equations. As the reservoir is a complex dynamical system itself, Reservoir Computing is intrinsically linked with chaos and it can be trained as a non-linear predictor of chaotic time series. For instance \cite{pathak2018model} reports state-of-the-art performance on the chaotic Kuramoto-Sivashinsky equation, while other applications in chaos synchronisation and cryptography have been demonstrated \cite{antonik2018using}, revealing the intricate link between Reservoir Computing and chaos.
The main body of this article is organized as follows. We introduce the concepts of multiple light scattering in Section 2 and Reservoir Computing in Section 3. We detail the optical implementation in Section 4. Section 5 and 6 present the experimental results with LCoS-SLM and DMD respectively, introducing new binary encoding strategies in the last section.
\section{Light scattering}
\subsection{General background}
When light encounters refractive index inhomogeneities, it gets scattered and its direction of propagation is modified. Fog and white paint are typical volumetric scattering samples, with water droplets and titanium dioxyde pigments being the respective scattering particles. When light propagates through thick scattering samples, the number of scattering events is tremendously high. It is impossible to precisely describe all the light propagation in the medium, and at the exit of the scattering medium, one typically observes a speckle figure. This image results from a complex interference process where all the different scattering paths are recombined. Thanks to the large number of scattering events at random positions, this speckle image is seemingly random and its statistical properties are well characterized \cite{goodman2007speckle}. It represents a signature of the particular disordered medium and, for a given incident field, will be different from one scattering sample to another.
Originally developed for astronomy and biological imaging, wavefront shaping techniques can control light and even perform imaging, in this multiple scattering regime. For example, it is possible to focus light after a scattering medium, by modulating many incident modes to generate a constructive interference at a specific point \cite{vellekoop2007focusing}. This has given rise to a wealth of applications in imaging and beyond \cite{rotter2017light}.
\subsection{The Transmission Matrix of complex scattering media}
One particular object of interest for our current study is the Transmission Matrix (TM) \cite{popoff2010measuring}. Light propagation, even in the multiple scattering regime, remains a linear process: the output over a set of detectors can be described as the product between the incident electric field on a set of input modes, and the TM. This matrix thus characterizes light propagation between input and output modes. In practice, due to the finite size of our optical devices and not measuring back-reflected light, we only measure a sub-part of a theoretical complete Scattering Matrix that would fully describe light propagation in the disordered medium, both in transmission and reflection \cite{rotter2017light}.
We place a Spatial Light Modulator (SLM) and a camera at the two sides of a multiple scattering medium, to collect transmitted light through the scattering sample. An SLM is a device that is commonly used for wavefront shaping, it imprints a given image as a phase or amplitude modulation on an incident light field. The SLM displays an input image on the coherent light coming from a laser and this modulated electric field will propagate through the complex material to form a speckle figure on the camera.
This speckle, although seemingly random, depends on the input image on the SLM. This link is made explicit in the Transmission Matrix, that describes the linear relationship between the input electric field defined by the SLM image and the output speckle pattern. The camera records the intensity of the output electric field, its image is thus given by:
\begin{equation}
b = |H a|^2
\end{equation}
where $a$ is the electric field on the plane of the SLM and $H$ is the TM describing the propagation of light from the SLM to the camera, through the scattering medium.
The TM can be measured experimentally and it has been proven to be i.i.d random in the multiple scattering regime \cite{popoff2010measuring}. The matrix dimensions are determined by the number of controlled input and output modes, which are fixed by the resolution of the associated optical devices. Nowadays, common Spatial Light Modulators (SLMs) and cameras typically contain a few million pixels, thus a Transmission Matrix can reach the gigantic size larger than $10^6 \times 10^6$. We cannot possibly hope to measure such a large matrix, as it would take a prohibitive time, and it would be impossible to store it in the memory of a computer. However, we can leverage the very large dimensionality of the TM without measuring it: by displaying a vector on the SLM as an input before the disordered medium, we effectively multiply this vector by the very large TM and record the modulus of the result on the camera. Therefore we want to use the statistical properties of multiple scattering \cite{goodman2007speckle} for optical computing.
\subsection{Spatial Light Modulators for optical computing}
To have an efficient optical computing strategy, light modulation needs to be performed efficiently. There are several technologies available, as they have been developed for the display industry. Three important criteria are the speed of the modulation, the total number of pixels on the SLM, and the encoding depth of the SLM image. We document here Reservoir Computing implementations using two kinds of Spatial Light Modulators: Digital Micromirror Devices (DMDs) and SLMs based on the Liquid-Crystal on Silicon technology (LCoS-SLMs).
A Digital Micromirror Device is made of an array of micromirrors that can be set in two positions. In one state, light will be sent to the scattering medium, in the other, light will be deflected towards a beam blocker and won't contribute to the output speckle. Hence, using a DMD, we can display binary 0/1 images. Developed by Texas Instruments since the end of the 80s, it is now a mature technology as it is present in most videoprojectors nowadays. The speed of the DMD is their main advantage for optical computing, as they can reach tens of kHz. Mirrors can be switched on and off very quickly to produce greyscale for consumer displays, as the eye integrates in time the intensity of every pixel of an image. This method cannot be used for speckle patterns because it comes from interferences that depend on the electric field and not the intensity. Averaging the intensity of different random speckle patterns in time decreases the contrast of the final speckle pattern.
In contrast, phase-only LC-SLMs use twisted nematic liquid crystals to modulate the phase of an incident electric field. The orientation of the liquid crystals in a small region can be modulated by an imposed electric field, and this induced change will modulate the index of reflection, thus the phase of the reflected light accordingly. Organized in arrays of pixels, these Spatial Light Modulators are able to imprint a digital phase image coming from the computer. Their speed is limited by the response time of the liquid crystal system, and the fastest version available today reaches about 500 Hz (without taking into account data transfer).
\subsection{Apart\'e on random projections}
Propagation through a multiply light scattering medium results in applying a random matrix multiplication on the original input vector, an operation known as a random projection. As a matter of fact, random projections have been studied extensively in computer science as a convenient way to transform the dimension of a set of vectors while preserving its structure. The Johnson-Lindenstrauss lemma \cite{johnson1984extensions} states that pairwise distances are preserved as long as the dimension after the random projection is logarithmic in the number of elements in the original set, regardless of the original dimension of the vectors. Hence, random projections have been used in data compression and to reduce computational complexity in Randomized Linear Algebra \cite{mahoney2011randomized, woodruff2014sketching}. Several machine learning algorithms also use random projections, such as Random Features and Reservoir Computing, and they are particularly suited for optical computing thanks to their robustness against analog noise \cite{saade2016random, dong2018scaling}. In a nutshell, a random projection is a convenient generic operation that preserves distances between vectors. Thus, even if the speckle appears to be random without structure, it still contains information about the vector displayed on the SLM.
\section{Reservoir Computing}
\begin{figure*}[!t]
\centering
\includegraphics[width=\linewidth]{Reservoir_Computing2.png}
\caption{Reservoir Computing principle. (a) A Recurrent Neural Network with fixed weights is used as a reservoir to encode information from a time-dependent input. It can be replaced by another dynamical system. The output weights are trained with a linear regression. (b) Reservoir Computing flowchart. The successive reservoir states are first computed, then used for training or validation.}
\label{fig_sim}
\end{figure*}
\subsection{Background}
Machine learning is a powerful tool to learn patterns and make inferences in complex problems based on a large number of examples. It relies on models with tunable parameters, for example Neural Networks with interconnection weights between neurons tuned to perform a particular task. In supervised learning, these weights are trained using a large number of examples, which consist in pairs of input data with the desired outputs. Thanks to increasing computational power and a large amount of available data to process, they have achieved state-of-the-art performance on very diverse tasks, such as image recognition \cite{krizhevsky2012imagenet}, Natural Language Processing, or recommender systems \cite{bell2007lessons} to name a few. Today machine learning is a blooming field, and there is a number of other machine learning approaches beyond neural networks such as kernel methods and decision trees. Computational efficiency is a major research direction, as there is a strong need to scale down the heavy machine learning computations in smaller devices.
Recurrent Neural Networks are notoriously hard to train \cite{pascanu2013difficulty}. Recurrent connections are a challenge for error back-propagation, which is the method of choice to train neural networks with feed-forward connections such as Convolutional Neural Networks \cite{lecun1998gradient}. Back-propagation through time is possible \cite{werbos1990backpropagation}, but this method faces the problem of local minima, as well as exploding and vanishing gradients \cite{pascanu2013difficulty}. As a possible solution to bypass this training issue, Echo-State Networks (ESNs) \cite{jaeger2001echo, jaeger2004harnessing} are Recurrent Neural Networks with randomly fixed internal weights. Only the output weights are trained for a particular task, reducing the training to a simple linear regression. The number of tunable parameters is thus smaller, but this does not necessarily mean that this Neural Network model is less expressive than fully-tunable Recurrent Neural Networks. It is very easy to increase the number of neurons and it has been proven that large networks can universally approximate any continuous function \cite{grigoryeva2018echo}.
\subsection{Recursive equation}
To operate an Echo-State Network, a time-dependent input is first fed to the network with fixed weights. After the computation of all the ESN states, the output weights are either learned with a training dataset containing the desired outputs, or used to obtain a predicted output on another dataset for validation.
Let $\{i(t), t = 0, ..., T\} \in \left(\mathbb{R}^d\right)^T$ be an input time series of dimension $d$ and of length $T$. The ESN will be initialized in a random state $x(0) \in \mathbb{R}^n$ and its state at time $t$ will be denoted $x(t)$; $n$ is the dimension of the network, i.e. the number of neurons or reservoir nodes.
Let $W_{\rm{res}}$ be the internal weight matrix and $W_{\rm{in}}$ the weight matrix between the input and the network. Both weight matrices are random and fixed for Echo-State Networks. The nonlinear activation function of every neuron will be denoted $f$.
The successive ESN states are computed using the following recursive equation:
\begin{equation}
x(t+1) = f(W_{\rm{in}} i(t) + W_{\rm{res}} x(t))
\label{esneq}
\end{equation}
In other words, an Echo-State Network is a large set of neurons randomly interconnected, that evolves dynamically driven by an external input. This leads to the more general framework of Reservoir Computing, where the neural network can be replaced by any non-linear dynamical system. An input time series is fed to the reservoir, and the reservoir can be any generic dynamical system. At any time, the current state of the reservoir depends on the previous values of input data, it encodes this information in its state.
Thanks to this general framework, we introduce three objects that will enable us to tune the dynamics of the reservoir for optical implementations: a leak rate $a$, a random bias vector $b$, and an encoding function $g$ into the ESN equation.
\begin{equation}
x(t+1) = (1-a) x(t) + a f\big(W_{\rm{in}} g(i(t)) + W_{\rm{res}} g(x(t)) + b\big)
\label{rceq}
\end{equation}
The leak rate $a$ is an important parameter that controls the speed of the dynamics of the reservoir without changing its long-term stability, the bias parameter $b$ controls the diversity of neurons inside the reservoir states. Finally, the encoding function will be important in the following as images on the Spatial Light Modulator need to be either binary or phase-only. Compared to the implementation of \cite{dong2018scaling} that did not use such a function but enforced contraints directly in the activation function, the network states are real which increases the performance considerably.
\subsection{Final linear layer}
The output $o(t) \in \mathbb{R}^k$ is computed using a linear combination with weights $W_{\rm{out}} \in \mathbb{R}^{k \times n}$, which can be written as:
\begin{equation}
o(t) = W_{\rm{out}} x(t)
\end{equation}
The optimal set of output weights is obtained by solving a linear regression problem, which minimizes the following error metric:
\begin{equation}
E = \frac{1}{k} \sum_t \| \tilde{o}(t) - W_{\rm{out}} x(t) \|^2
\label{error}
\end{equation}
where $\tilde{o}(t)$ is the target output at time $t$.
Linear regression is a well-studied problem and many libraries already provide an efficient solver. We use the Ridge solver of the scikit-learn library in Python as the ridge regularization is important to avoid overfitting when the number of parameters, which is proportional to the number of neurons, is larger than the number of examples. It correspond to the addition of a term $\alpha \|W_{\rm{out}}\|^2$ in Equation (\ref{error}).
\subsection{Physical implementations of Reservoir Computing}
The flexibility of RC, that can use any generic high-dimensional reservoir for computation, makes it very promising for physical implementations \cite{tanaka2019recent, van2017advances}. Many have been proposed originating from very different research areas, such as optical nano-circuits \cite{vandoorne2014experimental} or carbon nanotubes \cite{dale2016evolving}, one early RC implementation even observed the ripples at the surface of a bucket of water for pattern recognition \cite{fernando2003pattern}. The RC framework is robust against noise and changes in network topology (dense, sparse, and local connections can be used). In principle, various physical systems receive an external time-dependent excitation and follow non-linear dynamics, sending the sequential input into a high-dimensional feature space. To make a prediction, the output is obtained by a linear combination of the observed state of the reservoir.
Following the recent trend of specialized electronic processors for efficient machine learning \cite{sze2017hardware}, neuromorphic electronic circuits for RC have been developed, based on analog circuits, FPGAs \cite{antonik2015fpga} or memristive devices \cite{donahue2015design}.
Optical node arrays have also been proposed, where semiconductor optical amplifiers are used to perform the non-linear activation function \cite{vandoorne2014experimental}. Increasing the number of neurons means increasing the number and the density of optical nodes, which is a challenging manufacturing task to address.
On the other hand, some optical reservoir computers use a single physical node \cite{appeltant2011information, larger2012photonic, paquot2012optoelectronic}. They use temporal multiplexing and are based on a delay-fiber system to generate interconnections between units. They can be operated very fast at the GHz frequency and there is a very active community pursuing this line of research. In this case, a large number of neurons decreases the effective frequency of this time-multiplexed scheme as more nodes need to be sent one by one in the delay system.
Our strategy to increase the dimension of the reservoir is to use free-space optics, combined with high-dimensional cameras and SLMs. The manufacturing challenge of producing cameras and SLMs with a large number of pixels has already been solved thanks to their use in the display industry. By exploiting this commodity, it is possible to implement large-scale Optical Reservoir Computing, interconnection between units being either provided by a Diffractive Optical Element (DOE) \cite{bueno2018reinforcement, antonik2018performance} or a scattering medium \cite{dong2018scaling}. In the first case, connections are typically local and they can be engineered when designing the DOE. In the second case that we study here, the weight matrix is dense and random, which is closer to the original ESN definition. With these devices, we can reach very large sizes but the operating frequency is limited by the SLM, which is typically in the tens or hundreds of Hertz, or more often by the camera. An implementation using an LCoS-SLM in phase modulation to generate the reservoir couplings has also been proposed \cite{pauwels2018towards}.
\subsection{Analysis of the reservoir dynamics}
The reservoir in Reservoir Computing is a tunable dynamical system, that can be put in a stable or chaotic regime depending on a few parameters controlling the dynamics. Empirically, best performance in Reservoir Computing happens when the reservoir is "at the edge of chaos" \cite{schrauwen2009computational}, which corresponds to a stable regime close to a chaotic one that exhibits rich but stable dynamics. In practice, the dynamics is controlled empirically by a few parameters and some hyper-parameter search is often used to find a particular dynamical system suited for the task at hand. Note that this procedure is common in Machine Learning as we often need to search for the best model to use for a particular problem by tuning a set of hyper-parameters.
To describe the complex dynamics of the Reservoir, several properties and quantities of interest have been proposed. The first and most important one, called the Echo-State Property \cite{jaeger2001echo}, describes the stability of the dynamical system. It states that a reservoir needs to forget its initial state after a finite time to be used for RC. In other words, for a given input time series, the observed state of a reservoir after a warm-up phase shall not depend on the method used to initialize the network, which is not related with the task to solve. The reservoir needs to operate in a stable dynamical regime, in contrast to a chaotic regime that would be very sensitive to initial conditions. As a consequence, a chaotic dynamical system cannot be used for Reservoir Computing. Additionally, to ensure that the reservoir state does not depend on the initial conditions, we typically throw away the first states following the initialization, both for training and prediction.
Two other properties to understand the reservoir dynamics are the separation and approximation properties \cite{maass2002real}. The separation property means that two different input time series need to lead to different final reservoir states. A complex reservoir with rich dynamics is able to encode more information and differentiate a larger number of inputs. The approximation property states that a single input time series perturbed with noise should consistently be mapped to the reservoir state. These properties are useful to characterize in a simple way the high-dimensional dynamics of a reservoir. For instance, it is possible to derive quantitative measures of these properties on experimental Reservoir Computers, and even display the observables introduced in this subsection in a 3D space to visualize the performance of a particular RC implementation \cite{dale2018substrate}.
In the end, there are four requirements for a dynamical system to be effectively used in RC \cite{tanaka2019recent}: high-dimensionality of the reservoir, non-linearity in the dynamics, the Echo-State Property, and a balance between the separation and approximation properties which depends on the task to solve.
\begin{figure*}[!t]
\centering
\includegraphics[width=.8\linewidth]{Optical_implementation3.png}
\caption{Principle of our optical implementation. (a) Scheme of an image sent to the Spatial Light Modulator (DMD or SLM). The current input and reservoir state are encoded and displayed on the Spatial Light Modulator. (b) The optical setup is used to perform the random projection. All the other steps are performed on a computer. (c) Speckle subsampling on the camera image. The next reservoir state is determined by camera values on a grid with spacing larger than the speckle grain size to minimize correlations.}
\end{figure*}
\section{Optical Implementation}
\subsection{General principle}
The experimental setup is depicted in Figure \ref{light_scattering}. The coherent light from a laser is enlarged using a telescope to fill the active area of the SLM, in order to maximize the number of pixels that can modulate the electric field. We use the SLM to display an image sent by the computer, formed from the current reservoir state and input data. The SLM modulates the coherent electric field and reflects it towards a scattering medium, that is thick enough to ensure a complete mixing of the incident modes, so that the Transmission Matrix is dense and not structured.
The resulting speckle pattern is captured by a camera, determining the next reservoir state, that will be encoded and displayed back on the SLM with the next input. This feedback loop between the camera and the SLM corresponds to one Reservoir Computing iteration and it will be repeated as many times as there are reservoir states to compute.
The optical implementation with a DMD is developed by LightOn and available for researchers as a cloud service \cite{lighton}, with a very similar implementation as in \cite{saade2016random}. We built a similar optical experiment with an LCoS-SLM instead of a DMD, to investigate the use of phase modulation in Optical Reservoir Computing. The wavelength of the laser we used is at 800 nm (Ti:Sapphire laser, MaiTai, Spectra Physics, operated in continuous-wave mode), we expand the beam to fill the SLM, a Meadowlark 512$\times$512 LCoS Reflective SLM. The modulated electric field is then focused by a 20$\times$ objective with 0.4 numerical aperture on a 0.5 mm thick scattering material. The scattered field is collected by another similar objective and the resulting speckle is captured by a CCD camera (Allied Vision, Manta G-046). The distance from the medium to the camera is adjusted in order to obtain speckles grain size larger than the pixel size.
\subsection{Encoding information on the Spatial Light Modulator}
At every RC iteration, the input and the reservoir state are concatenated and displayed on the SLM, but due the physical constraints of the SLM, the displayed image needs to be either binary amplitude or phase-only. An encoding operation is introduced at this step, it corresponds to the function $g$ in (\ref{rceq}). In the case of phase encoding, $g$ is simply defined by $g(x) = e^{i \pi x}$. $x$ is between 0 and 1 after normalization and encoded in 8 bits, which correspond to 255 grey levels that are sufficient to avoid any loss of performance due to discretization. On the other hand, the binarization constraint introduced by the DMD is more critical and how to encode a real-valued number will be discussed in Section VI.
The relative area of the input and the reservoir state is also an important parameter to control. By performing a hyper-parameter search on a noiseless simulation model for the same time-series prediction task, we have seen that the RC implementation gets optimal performance when the input area is around ten times larger than the reservoir area. However, in this case the contribution of the reservoir is very small: one-tenth of the signal after one iteration and approximately one-hundredth after two. Such a small perturbation would be quickly lost due to experimental noise which would be detrimental for the memory capacity of the reservoir, so we chose to fix the reservoir area to be equal to the input area.
\subsection{Reading the camera image}
\begin{figure*}[!t]
\centering
\includegraphics[width=.8\linewidth]{Fig_SLM2.png}
\caption{Optical implementation results of Reservoir Computing with multiple light scattering and an LCoS-SLM. (a) Encoded phase profile on the SLM. The reservoir state (left) and current time series value (right) are displayed in the central region, while the remaining area is left constant to provide a bias term. Each component of the vector to display is encoded in phase between 0 and pi. (b) Intensity profile captured by the camera during one iteration of Reservoir Computing. (c) Example of prediction on the chaotic Mackey-Glass dataset. The time series is fed to the reservoir until time $t_0=0$, and from the reservoir state at time $t_0$, we predict the future evolution of the time series. The time axis is normalized with the Lyapunov exponent. (d) NMSE for different reservoir sizes.}
\label{slm_fig}
\end{figure*}
After the propagation in the scattering medium, the camera returns a speckle pattern to the computer. This image represents a random projection of the data displayed on the SLM. This random image presents small speckle grains that are a few pixels wide, their size is determined by the diffraction of the finite numerical aperture of the optical system. We choose a sampling grid larger than the speckle grain size in order to remove this local range correlation and make sure that the interconnection matrix is fully random. This property can be been tested by computing the distribution of singular values of the Transmission Matrix, that should follow the prediction of Random Matrix Theory and is modified when local correlations are present in the matrix \cite{popoff2010measuring}.
The intensity values of the speckle pattern follows an exponential distribution, as they are the absolute value square of a complex Gaussian random variable \cite{goodman2007speckle}. Its mean depends on the laser power, the thickness of the scattering medium, and the exposure time of the camera. We empirically observed better performance when using the square root of the intensity, i.e. the modulus of the electric field, instead of the raw intensity measured by the camera, probably because this operation regularizes the reservoir state distribution for the subsequent linear regression.
\subsection{Chaotic time series prediction}
Mackey and Glass proposed in 1977 their famous equation to model physiological feedback systems:
\begin{equation}
\frac{du(t)}{dt} = \beta \frac{u(t-\tau)}{1+u(t-\tau)^n} - \gamma u(t)
\end{equation}
The first term corresponds to a delayed response of the system, which tends to 0 as $u$ tends to either 0 or infinity to keep the model realistic, while the second term can be interpreted as a classical decay with rate $\gamma$. This time-delay differential equation, in appearance simple, displays chaotic behavior for certain ranges of parameters, for example $\beta = 0.2$, $\gamma = 0.1$, $\tau = 17$, $n = 10$ as in \cite{jaeger2004harnessing}. The maximal Lyapunov exponent in this case is $\Lambda_{\rm{max}}=0.006$. In general, the Lyapunov exponent gives a measure for the total predictability of a system, it characterizes quantitatively the rate of separation of infinitesimally close trajectories in dynamical system, namely, the minimum amount of the time for which trajectories are diverging by a factor of $e$. This chaotic time series is one of the standard models used to test Reservoir Computing algorithms \cite{jaeger2001echo}.
\section{Mackey-Glass time series prediction: Reservoir Computing with a phase SLM}
\subsection{Experimental parameters}
Two different Mackey-Glass series with $T = 2000$ time-steps are randomly generated, for training and testing the RC algorithm with the SLM. The prediction task consists in two phases. In the first phase, the algorithm constructs the reservoir from the given test data $u(t_0), u(t_0 - 1), \ldots$, then it predicts the next $\tau$ values, $u(t_0+1), \ldots, u(t_0+\tau)$, using already constructed reservoir. The larger the prediction time $\tau$, the harder the associated prediction task becomes. Besides, the larger the reservoir size, the better the prediction performance. Therefore, the tasks with large prediction times require large reservoir sizes. The RC algorithm with SLM is tested for three typical reservoir sizes, $n = 256, 1024, 4096$, and for multiple prediction time periods $\tau = 1,2, \ldots, 1000$.
The SLM screen is split into three regions of equal areas, corresponding to the input, the reservoir state, and the bias. The bias region stays constant during the whole RC iterations and it provides a reference speckle on the camera that increases the diversity inside the reservoir states. Figure \ref{slm_fig}a presents an example of the encoded phase profile on the SLM and the captured intensity speckle pattern on the camera during one iteration of RC. The input information at every iteration is a single value, since the Mackey-Glass time series is one-dimensional. Therefore, the same input value is encoded onto multiple SLM pixels to respect the equal importance ratios between input, reservoir and bias regions. The same macro-pixel strategy is applied for the reservoir information when the size of the reservoir is smaller than the number of available pixels on the SLM.
The SLM splitting ratios with other important parameters that affect the performance of the RC algorithm, included the ridge regression parameters and the leak rate, are obtained by grid search using the scikit-learn library \cite{scikit-learn}. The leak rate is set to 0.3, we forget the first 100 initial reservoir states in regression, and the training regularization parameter is set to 10000.
This high value of the regularization parameter helps to compensate experimental noise, especially short-term fluctuations coming from detector noise, SLM flickering or mechanical vibrations. There are also long-term deviations coming from the decorrelation of the scattering medium to take into account that also affect the system performance.
\subsection{Experimental results}
Figure \ref{slm_fig}c shows one of the successful prediction results. The dashed line is the test time series fed to the reservoir until $t_0 = 0$. Afterwords, the algorithm has been switched into prediction mode (solid line). The temporal axis is normalized by the maximal Lyapunov exponent $\Lambda_{\rm{max}}$.
As a benchmark of the prediction performance we use the $\rm{NMSE}$ (Normalized Mean Square Error) for $N$ values of $t_0$ defined by:
\begin{equation}
\text{NMSE} = \frac{E}{N \sigma^2},
\label{NMSE}
\end{equation}
where $E$ is the squared error on the prediction task presented in Equation (\ref{error}), normalized by $\sigma^2$ the variance of the Mackey-Glass time series. Consequently, the smaller the $\rm{NMSE}$ value, the better the prediction performance is. In order to obtain smoother $\rm{NMSE}$ curves, we calculated its average over ten independent experiments with nine hundred test time series for each. The performance curves we obtain with the SLM optical implementation for different reservoir sizes $n_{\rm{res}} = 256,\ 1024,\ 4096$ are collected in Figure \ref{slm_fig}d. As one can see, the larger the reservoir size, the better the algorithm performance. The small oscillations of the $\rm{NMSE}$ from short prediction times originate from the oscillations of the Mackey-Glass time series, which are faster than the Lyapunov time. For longer prediction times, the task becomes harder which explains why the performance reaches a kind of plateau. The task becomes exponentially harder since the Mackey-Glass time series is chaotic and the Lyapunov exponent defines the typical time scale for chaotic divergence. It should converge to 1 for much longer time series as the reservoir only outputs the mean value of Mackey-Glass.
The speed of the algorithm is determined by the working frequency of the SLM and camera, as well as the data transfer speed between the computer and the optical devices. In our particular case, the setup is performing around 70 iterations per second. The main advantage of our optical implementation is its scalability. Namely, we can easily reach large reservoir sizes as the speed of the implementation almost does not depend on the size of the reservoir \cite{dong2018scaling}.
\section{Quantized Reservoir Computing: binary encoding for DMD implementation}
\subsection{Quantization in Reservoir Computing}
\begin{figure}[!t]
\centering
\includegraphics[width=0.5\linewidth]{Encoding6.png}
\caption{Distance matrices to compare encoding strategies. (a) Distance matrix $\|g(x) - g(y)\|$ for $x$ and $y$ between 0 and 1, with no encoding here $g(x) = x$ (for control). (b) Distance matrix for basket encoding (\ref{basket_enc}) with $n_{\rm{bin}} = 10$. (c) Distance matrix for threshold encoding (\ref{threshold_enc}) with $n_{\rm{bin}}=10$. (d) Distance matrix for base-2 encoding with $n_{\rm{bin}}=4$.}
\label{encoding}
\end{figure}
DMDs are based on the actuation of micromirrors to modulate the incident wavefront, they can operate at much faster rates than LCoS-SLMs but only a binary modulation is possible. In Reservoir Computing, this introduces a necessary binarization step in the dynamics of the Reservoir, which induces a discretization and loss of information. This quantization task has also been studied for other applications in Machine Learning \cite{courbariaux2016binarized}, as electronic devices are generally more efficient to perform summations than multiplications. They show that neuronal activations or weights can be binarized without notably decreasing the performance of the model. Thus, the architecture choice and high number of neurons are able to compensate the loss introduced by quantization.
It has been shown that binary Echo-State Networks can successfully learn to predict the chaotic Mackey-Glass time series \cite{dong2018scaling}. In this first implementation, the activation function was a simple threshold function on the intensity, an operation which loses a lot of information. However, in Reservoir Computing, the binarization step still has a significant impact on the reservoir dynamics and performances are generally worse than models without binarization. One explanation of this degradation is that binarization considerably modifies the stability of the reservoir dynamics, as every quantization threshold is a discontinuity and very similar inputs can lead to well-separated states. Hence the usual theoretical tools based on Lipschitz continuity to prove the Echo-State Property do not hold for a quantized activation function \cite{jaeger2001echo}. The dynamics of binary models, and in particular the transition between stability and chaos, has been studied in \cite{schrauwen2009computational}.
\subsection{Binarization strategy}
\begin{figure*}[!t]
\centering
\includegraphics[width=.8\linewidth]{Fig_DMD3.png}
\caption{Optical implementation results of Reservoir Computing with multiple light scattering and a DMD. (a) Binary amplitude image displayed on the DMD image during a RC iteration. The encoded reservoir state and input are displayed on approximately half of the DMD area, while the rest provides a constant random bias. (b) Camera image defining the reservoir state at the next iteration. Each component of the real-valued vector to display on the DMD is encoded in 10 binary values using basket encoding. (c) Example of prediction on the chaotic Mackey-Glass dataset. The time series is fed to the reservoir until time $t_0=0$, and from the reservoir state at time $t_0$, we predict the future evolution of the time series. The time axis is normalized with the Lyapunov exponent. (d) NMSE for different realizations of Optical Reservoir Computing: DMD implementation with basket encoding, DMD implementation with binary ESN \cite{dong2018scaling}, and LCoS-SLM with phase encoding. Reservoir sizes are 512, 5120, and 1024 respectively.}
\label{dmd_fig}
\end{figure*}
To minimize the loss of information due to binarization, we use here the camera image to define the reservoir state. This change improves the RC performance considerably. Here the activation function $f$ is simply a modulus square operator, without a threshold operation. The binarization is performed in another step, it corresponds to the introduction of the encoding function $g$ in \ref{rceq} that operates componentwise, i.e. on each pixel in parallel. It does not necessarily modify the dynamics of the system, but we keep the reservoir state as rich as possible by storing the camera image instead of a quantized version.
Thanks to the introduction of this encoding function, we can also change the dimensionality of the encoding and encode a reservoir activation on several DMD pixels. With this strategy, the function $g$ encodes the given value of one camera pixel on $n_{\rm{bin}}$ binary DMD pixels. Higher values of $n_{\rm{bin}}$ increase the regularity of the encoding function, and make the RC dynamics smoother.
Such an encoding also requires more DMD pixels to display the reservoir state. Hence, an efficient binarization scheme needs to balance between two constraints: a limited dimension expansion, to allow very large reservoir to be displayed on the DMD, and sufficient regularity and precision of the encoding.
There are several possibilities to define the encoding function $g$. As a separable function, i.e. operating componentwise, it is uniquely defined by the image of a single real number. After normalization of the number to encode, we assume that this number lies between 0 and 1. Intuitively, an efficient binarization strategy should send two close values on similar binary encodings and distant values on dissimilar encodings. Hence, the characteristics of an encoding are well-represented by its distance matrix, defined by the set of $\|g(x) - g(y)\|$ for all pairs $(x,y) \in [0;1]^2$. These distances should be small close to the diagonal, where $x$ and $y$ are similar, and increase further away from the diagonal. Kernel methods also study how distances and scalar products are transformed by non-linear embeddings, thus there is a close link between these binary embeddings and kernel methods.
Inspired by the Random Binning Features developed by Rahimi and Recht \cite{rahimi2008random}, we propose here an encoding function $g$, called basket encoding, where each component $g_i(x)$ for $i = 1, \ldots, n_{\rm{bin}}$ is defined by:
\begin{equation}
g_i(x) = \left\{ \begin{array}{lr}
1, & \text{if } x \in \left[c_i-s, c_i+s\right] \\
0, & \text{else}
\end{array}\right.
\label{basket_enc}
\end{equation}
where the centers and size of the bins are defined by $c_i = \frac{2i-1}{2n_{\rm{bin}}}$ and $s = \frac{2 \lfloor n_{\rm{bin}} / 2 \rfloor - 1}{4n_{\rm{bin}}}$ respectively. These values are chosen to obtain a larger number of different binary encodings while keeping a regular distance matrix.
Another binarization strategy, previously used in \cite{dong2018scaling} and referred to as threshold encoding, is to encode the input by using a set of uniformly-spaced thresholds:
\begin{equation}
g'_i(x) = \left\{ \begin{array}{lr}
1, & \text{if } x > t_i \\
0, & \text{else}
\end{array}\right.
\label{threshold_enc}
\end{equation}
where the thresholds are defined by $t_i = \frac{i}{n_{\rm{bin}}}$ for $i = 1, \ldots, n_{\rm{bin}}$. This scheme is also regular and can be used for RC, but the number of different binary encodings is smaller in this case (10 possible binary encodings for threshold encoding compared to 15 for basket encoding). As a result, for a fixed binary encoding dimension $n_{\rm{bin}}$, it will contain less information about the original real-value.
Fig. \ref{binary_sim} presents the distance matrices for these three different binarization strategies. We observe that both basket encoding and threshold encoding are well-behaved, as distances close to the diagonal, which correspond to close original real values, are small while the distance increases smoothly as the original values get further apart. Additionally, the basket encoding provides more precision for a given bit depth, here $n_{\rm{bin}} = 10$.
We also present the distance matrix using the representation in base 2. This binary encoding is the one used in the memory of a computer, it is the most compact one as $n_{\rm{bin}}$ represent $2^{n_{\rm{bin}}}$ different numbers. However, the computer implicitly differentiates bits according to their position, from the most-significant to the least-significant bit; in RC all bits should have the same importance. For example, 7 and 8 (or $\frac{7}{16}$ or $\frac{8}{16}$ after normalization) are very close but their binary encodings in base 2, 0111 and 1000, are very different.
\subsection{Cloud implementation}
We use an optical device developed by LightOn, available on the cloud. It is based on the same principle as the LCoS SLM presented previously, but uses a DMD for binary wavefront modulation. This device performs the optical random projections using multiple light scattering presented in Section 2 and is open for researchers to use.
In every optical RC iteration, we need to use the camera image to compute the next DMD image to display. However, the LightOn device is optimized to send DMD images by batches to maximize the speed of data transfer. We therefore compute batches of reservoirs in parallel driven by different time series, between 500 and 3'000. In the end, as we set the time series length to 800, we typically obtain tens of thousand of examples for training and testing.
Figure \ref{dmd_fig}ab presents the images sent to the DMD and the new state of the reservoir captured on the camera. On the DMD, we encode the current input value and reservoir state in a $1140 \times 912$ binary image. We display the input using basket encoding and 1000 pixels on an area representing approximately one fourth of the total DMD surface. The reservoir state of dimension $n_{\rm{res}}$ is encoded in ten times more binary DMD pixels using basket encoding. The reservoir also occupies one fourth of the DMD, which leaves one half of the DMD as a bias.
The leak rate is set to 0.2, we forget the 50 initial reservoir states, and the regularization parameter is set to 0.1 for training.
\subsection{Prediction results}
We present in Figure \ref{dmd_fig}d the $\rm{NMSE}$ of binary Reservoir Computing, binary Echo-State Network and the phase Reservoir Computing obtained with an LCoS-SLM. Reservoir sizes are set at 512 for binary Reservoir Computing, and 5120 for binary ESN to obtain a fair comparison as the basket encoding expands the binary dimension 10 times. We observe that thanks to the basket encoding, this new binarized version of Reservoir Computing is performing better than the previous binary ESN implementation. Additionally, we see that this experimental binary Reservoir Computing is performing better than the other experimental implementation of Reservoir Computing based on phase modulation. The gap in performance between DMD and LCoS-SLM implementations is probably due to a difference in stability and SNR of the optical devices, which is higher for the one developed by LightOn. On the other hand, binary Echo-State Networks do not perform as well due to the binarization operation.
\begin{figure}[!t]
\centering
\includegraphics[width=.5\linewidth]{Results_binary_sim3.png}
\caption{NMSE of Mackey-Glass prediction with RC using different encoding strategies: (a) tanh activation, (b) basket encoding defined in (\ref{basket_enc}), (c) threshold encoding defined in (\ref{threshold_enc}), and (d) binary ESN of \cite{dong2018scaling}. Each curve is an average of 5 realizations (numerical simulation), reservoir sizes is 512 for all cases with binary encoding dimension equal to 10 for (b) and (c).}
\label{binary_sim}
\end{figure}
Figure \ref{binary_sim} presents simulation results to compare different encoding strategies. We see that the basket encoding performs better than the other two binary strategies, threshold encoding and the binary ESN without encoding. This proves that it is important to have a regular and precise encoding for binary Reservoir Computing.
The speed of the DMD depends on the batch size, as the LightOn device is more efficient when one sends a large number of images to be displayed in a burst. Hence, we achieve 640 Hz with a batch size of 3'000 and 285 Hz with a batch size of 500. We are up to 2 times faster than the previously-reported speed in \cite{dong2018scaling} thanks to hardware optimization. An even faster implementation should be possible by sending directly the camera image to be displayed on the DMD with dedicated electronics, thus avoiding the computer in the RC loop to reduce latency.
\section{Conclusion}
This study presents how multiple light scattering can be harnessed for Reservoir Computing. Thanks to the complex interference that results from the scattering of light inside a complex medium, we can generate the successive states of a reservoir responding to a given input time series. Optical Reservoir Computing algorithms have been demonstrated for chaotic time series prediction, based on two different SLM technologies. On the one hand, we use an LCoS SLM to perform Reservoir Computing based on phase modulation of the electric field. On the other hand, the DMD implementation represents a promising solution for efficient Reservoir Computing thanks to its high working frequency, but it can only display a binary image. Encoding and binarization strategies have been proposed to make binary Reservoir Computing perform on par with real-valued networks.
This optical computing strategy can also be applied to other machine learning tasks where random projections prove useful, be it for their distance conservation properties or to emulate any fully-connected neural network with randomly fixed weights. Optical random projections have been demonstrated for instance in image recognition \cite{saade2016random} and change-point detection \cite{keriven2018newma}.
\section*{Acknowledgements}
We would like to thank Antoine Boniface for lending his experimental setup to perform the SLM experiment and a careful review of the manuscript, Claudio Moretti for his help in designing figures, LightOn for providing OPU access and Charles Brossollet for technical support with the OPU. This material is based upon work supported by the Defense Advanced Research Projects Agency (DARPA) under Agreement No. HR00111890042. Sylvain Gigan and Jonathan Dong also acknowledge partial support from H2020 European Research Council (ERC) (Grant 724473).
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Mark Roberts (born 29 October 1975 in Irvine) is a Scottish football player and coach.
Roberts' playing career was mainly based in Scottish football, with spells at Kilmarnock, Falkirk, Airdrieonians, Partick Thistle, St Mirren, Airdrie United and Ayr United. He also had a spell at Shelbourne in the Republic of Ireland. In 2012, Roberts was appointed player/manager of Ayr United, a position he held until his departure from the club in 2014. He then had a stint as head coach of Queen's Park.
Playing career
Roberts spent his youth career with Bellfield Boys Club in Kilmarnock during the 1991–92 season. He signed for Kilmarnock F.C. at the start of the 1992–93 season and stayed for 8 seasons. Roberts made 101 league appearances and scored 15 league goals. During 1999, Roberts was loaned out to Raith Rovers by Kilmarnock, where he made 3 league appearances and scored one league goal.
Roberts signed for Falkirk at the start of the 2000–01 season and stayed for one season, where he made 25 league appearances and scored 9 league goals. He then signed for Airdrieonians at the start of the 2001–02 season and made 36 league appearances and scored 12 league goals. Airdrieonians went out of business at the end of the season. He then signed for Dublin club Shelbourne. He played in the League of Ireland Premier Division in the summer of 2002.
Roberts signed for Paisley club St Mirren in the January 2003 transfer window for the remainder of the 2002–03 season. He made 13 league appearances and scored 3 league goals for the Paisley club. Roberts signed for Airdrie United at the start of the 2003–04 season and stayed for two seasons, where he made 62 league appearances and scored 3 league goals.
Roberts signed for Partick Thistle at the start of the 2005–06 season. Roberts scored a brace in his first game against Dumbarton. Roberts scored 22 goals during his first season at Firhill, as Partick Thistle gained promotion to the Scottish First Division via the play-offs. Roberts was one of 5 nominees for the Scottish First Division Player of the Year Award for the 2006–07 season. Roberts made 104 league appearances for Partick Thistle in his 4 seasons at the club and scored 30 league goals, as well as 17 goals in other competitions. His 47 goals for Thistle ensured that the Jags fans would hold him in high regard long after he left the club, particularly as many saw him as the player that led Thistle to promotion to the First Division.
Roberts signed for Ayr United in January 2009. On 18 September 2010, Roberts made history when he scored three penalty kicks for Ayr in a league fixture with East Fife. Roberts was sent off after a confrontation whilst celebrating his third. On 18 January 2011, Roberts scored the goal that sent SPL side Hibernian out of the Scottish Cup.
On 27 February 2015, Roberts signed for Scottish League Two club Clyde, having already played three times for the club as a trialist.
After leaving a coaching role with Clyde in August 2016, Roberts resumed his playing career at Junior side Hurlford United. He left the club in August 2018 to become assistant manager at Queen's Park.
On 5 February 2020, Roberts rejoined Hurlford United before departing upon the expiry of his contract in the summer of 2020.
Roberts signed for Troon ahead of their debut in senior football in September 2020.
Coaching career
Ayr United
On 15 May 2012, it was announced that Ayr would not renew the contract of Brian Reid, and that Roberts would replace him as manager for the 2012–13 season. Despite, winning the League 1 manager of the month award in August of the same year, Roberts was sacked by Ayr United in December 2014, with the club sitting in ninth place out of 10 clubs in Scottish League One.
Queen's Park
Roberts joined Queen's Park in August 2018 as an assistant to Gus MacPherson. After MacPherson left the club to join St. Mirren in September 2018 he took caretaker charge and was then appointed head coach on 12 October. Roberts left Queen's Park by mutual consent in December 2019.
Honours
Kilmarnock
Scottish Football League First Division promotion 1992–93
Airdrieonians
Scottish Challenge Cup: 2001–02
Shelbourne
League of Ireland Premier Division champions 2001–02
Partick Thistle
Scottish Football League First Division play-off winners 2005–06
Ayr United
Scottish Football League First Division play-off winners 2010–11
Statistics
Manager
Spell at Queen's Park was initially caretaker until permanent appointment on 12 October 2018.
References
External links
(Partick Thistle & Ayr United)
(Kilmarnock to Airdrie United)
1975 births
Airdrieonians F.C. (1878) players
Airdrieonians F.C. players
Association football forwards
Ayr United F.C. managers
Ayr United F.C. players
Falkirk F.C. players
Kilmarnock F.C. players
League of Ireland players
Living people
Partick Thistle F.C. players
Raith Rovers F.C. players
St Mirren F.C. players
Clyde F.C. players
Hurlford United F.C. players
Scottish Football League managers
Scottish Football League players
Scottish football managers
Scottish footballers
Scottish Professional Football League managers
Scottish Premier League players
Shelbourne F.C. players
Scottish Professional Football League players
Scottish Junior Football Association players
Queen's Park F.C. non-playing staff
Queen's Park F.C. managers
Expatriate association footballers in the Republic of Ireland
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"redpajama_set_name": "RedPajamaWikipedia"
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The inspection of Lupin Bioresearch Centre by the United States Food and Drug Administration (USFDA) has been successfully completed without any observations.
New Delhi: Drug firm Lupin Wednesday said inspection by the US health regulator of its bioresearch centre in Pune has been completed without any observations.
"This is the fifth successive inspection without any 483 observations, indicating superior quality compliance maintained at this global bioresearch facility," the company said in a statement.
The inspection of Lupin Bioresearch Centre by the United States Food and Drug Administration (USFDA) has been successfully completed without any observations, it added.
The centre conducts the in-vivo and in-vitro bioequivalence studies in Pune. This inspection also included review of studies done at its new site at Lupin Research Park, Pune, the statement said.
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction and Models}
We consider a system ${\bf T}$ of $n$ sporadic real-time tasks. A
sporadic task $\tau_i$ in ${\bf T}$ releases an infinite number of
jobs that arrive with the minimum inter-arrival time constraint. A
sporadic real-time task $\tau_i$ is characterized by its
\emph{worst-case execution time} $C_i$, its \emph{minimum
inter-arrival time} (also called period) $T_i$ and its \emph{relative deadline} $D_i$.
In addition, each job of task $\tau_i$ has also a specified worst-case
self-suspension time $S_i$. When a job of task $\tau_i$ arrives at
time $t$, the job should finish no later than its \emph{absolute
deadline} $t+D_i$, and the next job of task $\tau_i$ can only be
released no earlier than $t+T_i$. If the relative deadline $D_i$ of
task $\tau_i$ in the task set is always equal to (no more than,
respectively) the period $T_i$, such a task set is called a
\emph{implicit-deadline} (\emph{constrained-deadline}, respectively)
task set (system). If $D_i > T_i$ for a certain task $\tau_i$ in ${\bf
T}$, then the task system is an \emph{arbitrary-deadline} task
system. The response time of a
job is defined as its finishing time minus its release (arrival)
time. The worst-case response time $WCRT_i$ is
the upper bound on the response times of all the jobs of task $\tau_i$.
The \emph{response time analysis} of a task $\tau_i$ under a scheduling
algorithm is to provide a safe upper bound on $WCRT_i$.
There are two typical models for self-suspending sporadic task
systems: 1) the dynamic self-suspension task model, and 2) the
segmented self-suspension task model. In the \emph{dynamic}
self-suspension task model, e.g.,~\cite{ECRTS-AudsleyB04,RTAS-AudsleyB04,RTCSA-KimCPKH95,MingLiRTCSA1994,LiuChen:rtss2014,Huang_2015,ChenNelissenHuang-ecrts16}, in addition to the worst-case execution time
$C_i$ of sporadic task $\tau_i$, we have also the worst-case
self-suspension time $S_i$ of task $\tau_i$. In the \emph{segmented} self-suspension
task model, e.g., \cite{RTCSA-BletsasA05,Kim2016,WC16-suspend-DATE,RTSS-ChenL14,Huang:multiseg,PH:rtss98}, the execution behaviour of a job of task $\tau_i$ is
specified by interleaved computation segments and self-suspension
intervals. The dynamic
self-suspension model provides a simple specification by ignoring the
juncture of I/O access, computation offloading, or
synchronization. However, if the suspending behaviour can be
characterized by using a segmented pattern, the segmented
self-suspension task model can be more appropriate.
This report considers a \emph{segmented} self-suspension task
model. If a task $\tau_i$ can suspend itself, a job of task $\tau_i$
is further characterized by the computation segments and suspension
intervals as an array
$(C_{i}^1,S_{i}^1,C_{i}^2,S_{i}^2,...,S_{i}^{m_i-1},C_{i}^{m_i})$,
composed of $m_i$ computation segments separated by $m_i-1$ suspension
intervals.
In this report, we will \emph{only} consider the following special
case in fixed-priority (FP) preemptive scheduling:
\begin{itemize}
\item Task $\tau_n$ is the lowest-priority task and is a segmented
self-suspension task. We will further assume that $D_n \leq T_n$.
\item There are $n-1$ higher-priority tasks, $\tau_1,
\tau_2, \ldots, \tau_{n-1}$. These $n-1$ tasks are indexed from the
highest priority $\tau_1$ to the lowest priority $\tau_{n-1}$. The
task set $\setof{\tau_1, \tau_2, \ldots, \tau_{n-1}}$ can be an
arbitrary-, constrained-, or implicit-deadline task
set.
\end{itemize}
Since we only have one self-suspending task in this report, we use $m$
to denote $m_n$ for notational simplicity, where $m \geq 2$. Moreover, the arrival time
of a computation segment is defined as the moment when the computation
segment is ready to be executed, after all its previous computation
segments and self-suspension intervals are done. In this report,
the response time $R_j$ of a computation segment $C_n^j$ is defined as
the finishing time of the computation segment minus the arrival time
of the computation segment.
We consider uniprocessor fixed-priority preemptive scheduling. We say
that a release pattern of the tasks in ${\bf T}$ is \emph{valid} if
the jobs of the tasks in ${\bf T}$ do not violate any of the temporal
characteristics regarding to the minimum inter-arrival time, worst-case execution
time, and worst-case self-suspension time. We say that a schedule is
\emph{feasible} if all the deadlines are met for a valid release
pattern of the tasks in ${\bf T}$. Moreover, a task system (set) is
\emph{schedulable} by a scheduling algorithm if the resulting schedule for any valid release pattern of ${\bf T}$
is always feasible. A \emph{schedulability test} of a scheduling
algorithm for a given task system is to validate whether the task
system is schedulable by the scheduling algorithm. A \emph{sufficient}
schedulability test provides only sufficient conditions for validating
the schedulability of a task system. A \emph{necessary}
schedulability test provides only necessary conditions to allow the
schedulability of a task system. An \emph{exact} schedulability test
provides necessary and sufficient conditions for validating the
schedulability.
By the assumption of fixed-priority preemptive scheduling, a
schedulability test of task $\tau_i$ can be done by removing all the
lower-priority tasks, $\tau_{i+1}, \ldots, \tau_n$. Since task
$\tau_n$ is the lowest-priority task, the schedulability test of the
$n-1$ higher-priority tasks under the given FP preemptive scheduling
can be done by using the well-known response time analysis,
\cite{DBLP:conf/rtss/Lehoczky90,DBLP:conf/rtss/LehoczkySD89}.
Therefore, the remaining problem is to validate whether the segmented
self-suspension task $\tau_n$ is schedulable by FP preemptive
scheduling (as the lowest-priority task). For this problem, the only
results in the literature were provided by Lakshmanan and Rajkumar
\cite{LR:rtas10} and Nelissen et al. \cite{ecrts15nelissen}.
Lakshmanan and Rajkumar proposed a pseudo-polynomial-time worst-case
response time analysis, by revising the well-known critical instant theorem originally defined in \cite{Liu_1973}. This has been recently disproved by Nelissen et
al. \cite{ecrts15nelissen}. The schedulability test by Nelissen et
al. \cite{ecrts15nelissen} requires exponential-time complexity even
for such a case when the task system has \emph{only one
self-suspending task}. Furthermore, Nelissen et
al. \cite{ecrts15nelissen} also assumed that all the tasks are with
constrained deadlines, i.e., $D_i \leq T_i$ for every task $\tau_i \in
{\bf T}$. The other solutions \cite{Huang:multiseg,PH:rtss98}
require pseudo-polynomial time complexity but are only sufficient
schedulability tests.
Regarding to computational complexity, it was shown by Ridouard et
al. \cite{DBLP:conf/rtss/RidouardRC04} that the scheduler design
problem for the segmented self-suspension task model is ${\cal
NP}$-hard in the strong sense.\footnote{Ridouard et
al. \cite{DBLP:conf/rtss/RidouardRC04} termed this problem as the
feasibility problem for the decision version to verify the existence
of a feasible schedule.} The proof in
\cite{DBLP:conf/rtss/RidouardRC04} only needs each segmented
self-suspending task to have one self-suspension interval with two
computation segments. It was reported by Chen et
al. \cite{suspension-review-jj} that the schedulability test problem
in several cases is also strongly $co{\cal NP}$-hard under
dynamic-priority scheduling, in which the priority of a job may change
over time. Such observations made in \cite{suspension-review-jj} are
based on the special cases to reduce from the schedulability test
problem of the ordinary constrained-deadline sporadic task systems
(without self-suspension), which has been recently proved to be
$co{\cal NP}$-hard in the strong sense by Ekberg and Wang
\cite{DBLP:conf/ecrts/Ekberg015} under earliest-deadline-first (EDF)
scheduling. For fixed-priority (FP) preemptive scheduling, in which a
task is assigned a fixed priority level, the computational complexity
of the schedulability test problem for the segmented self-suspension
task model is open.
{\bf Contribution:} This report provides the following results of the
schedulability test problem and the worst-case response time analysis
for self-suspending sporadic task systems:
\begin{itemize}
\item In Section~\ref{sec:computational-complexity-fp}, we prove that
the schedulability analysis for fixed-priority (FP)
preemptive scheduling even with only one segmented self-suspending
task as the lowest-priority task is $co{\cal NP}$-hard in the strong
sense when there are more than one self-suspension interval (or
equivalently more than two computation segments). The computational
complexity analysis is valid for both implicit-deadline and
constrained-deadline cases, when the priority assignment is given.
Our proof also shows that validating whether there exists a feasible
priority assignment is also $co{\cal NP}$-hard in the strong sense
for constrained-deadline segmented self-suspending task systems.
\item This report shows that the upper bound on the
worst-case response time derived from the MILP developed by Nelissen
et al. \cite{ecrts15nelissen} can be very far from
the actual worst-case response time in Section~\ref{sec:gap}.
\end{itemize}
\section{Proof for the Necessary Condition of Worst-Case Response Time}
\label{sec:proof-necessary}
Let $\sigma$ be a fixed-priority preemptive schedule for a valid release pattern
$RP$ of the task system ${\bf T}$. We consider two cases:
\begin{itemize}
\item Case 1 when all the jobs of task $\tau_n$ in schedule $\sigma$
have their response times no more than $T_n$: We pick an arbitrary
job $J$ of task $\tau_n$ from $\sigma$. For this case, removing all
the other jobs of task $\tau_n$ (except $J$) from $\sigma$ does not change the
schedule of the remaining jobs in $\sigma$.
\item Case 2 when there exists a job of task $\tau_n$ in schedule
$\sigma$, in which the response time of the job is larger than
$T_n$. Let $J$ be the first job in the schedule $\sigma$ in which
the response time of $J$ is strictly larger than $T_n$. For this
case, removing all the other jobs of task $\tau_n$ (except $J$) from $\sigma$
does not change the schedule of the remaining jobs in $\sigma$.
\end{itemize}
In both cases, for $j=1,2,\ldots,m$, suppose that the arrival time and
finishing time of the $j$-th computation segment of job $J$ is $g_j$
and $f_j$, respectively. In both cases, by definition, $g_1 \leq f_1
\leq g_2 \leq f_2 \leq \cdots \leq g_m \leq f_m$, and $f_m - g_1 \leq
WCRT_n$.
The following lemmas provide the necessary conditions for the
worst-case release patterns for both cases. Specifically, Condition 1
in Lemma~\ref{lemma:necessary-critical-wcrt} was also provided in
Lemma 2 by Nelissen et al. in \cite{ecrts15nelissen}. For the
completeness of this report and the correctness of the MILP, we also
include the proof of Condition 1 in
Lemma~\ref{lemma:necessary-critical-wcrt} here.
\begin{lemma}
\label{lemma:necessary-critical-wcrt}
If $WCRT_n \leq T_n$, then the worst-case response time of task
$\tau_n$ happens (as necessary conditions) when
\begin{enumerate}
\item[Condition 1:] all the higher-priority tasks $\tau_1, \tau_2,
\ldots, \tau_{n-1}$ only release their jobs in time intervals $[g_j,
f_j)$ for $j=1,2,\ldots,m$, and
\item[Condition 2:] $g_{j+1} - f_j$ is always $S_n^j$, $\forall j=1,2,\ldots,m-1$, and
\item[Condition 3:] all the jobs are executed with their worst-case execution times.
\end{enumerate}
\end{lemma}
\begin{proof}
We prove this lemma by showing that job $J$ defined at the opening
of this section can only increase its response time by following
these three conditions. By the assumption $WCRT_n \leq T_n$, we
consider Case 1.
We start with Condition 1. Suppose that the
schedule $\sigma$ has to execute certain higher-priority jobs in
time interval $[t_1, g_1)$ and the processor idles right before
$t_1$. That is, the processor does not idle between $t_1$ and
$g_1$. In this case, we can change the arrival time of the
computation segment $C_n^1$ of job $J$ from $g_1$ to $t_1$. This
change in the release pattern $RP$ does not change the resulting preemptive
schedule $\sigma$, but the response time of $J$ becomes $f_m - t_1 \geq
f_m - g_1$ since $t_1 \leq g_1$.
For $j=2,3,\ldots,m$, suppose that the schedule $\sigma$ has to
execute certain higher-priority jobs to keep the processor busy in time interval $[t_j, g_j)$
and the processor either idles or executes job $J$ right before
$t_1$. By definition, for $j=2,3,\ldots,m$, we know that $t_j \geq
f_{j-1}$; otherwise the $(j-1)$-th computation segment of job
$J$ cannot be finished at time $f_{j-1}$. Similarly, we can change
the arrival time of the computation segment $C_n^j$ of job $J$ from
$g_j$ to $t_j$. This change in the release pattern $RP$ does not change
the resulting preemptive schedule $\sigma$ nor the response time of $J$.
Let $g_j$ for $j=1,2,\ldots,m$ be the revised arrival time of the
computation segment $C_n^j$ of job $J$ changed above. After we
change the release pattern of job $J$, we know that the suspension
time between the two computation segments $C_n^j$ and $C_n^{j+1}$ of
job $J$ is exactly $g_{j+1} - f_j \leq S_n^j$ for
$j=1,2,\ldots,m-1$. Therefore, the revised release pattern remains valid.
Now, we can safely remove the higher jobs released before $g_1$,
after or at $f_m$, and in time intervals $[f_1, g_2)$, $[f_2, g_3),
\ldots, [f_{m-1}, g_m)$. Since these jobs do not (directly or
indirectly) interfere in the execution of job $J$ at all, removing them does not
have any impact on the execution of job $J$. Again, let $RP$ be the revised release pattern, and let $\sigma$ be its corresponding FP preemptive schedule. Now,
Condition 1 holds.
We now revise the release pattern of the higher-priority jobs and
$J$ for Condition 2. We start from $j=2$. If $g_j - f_{j-1} <
S_n^{j-1}$, then we greedily perform the following steps:
\begin{itemize}
\item First, any higher-priority jobs released after or at $g_j$ are delayed exactly by $S_n^{j-1} - (g_j - f_{j-1})$ time units.
\item Second, the $(j-1)$-th suspension interval of job $J$ is
increased to suspend for exactly $S_n^{j-1}$ time units, and
$g_\ell$ is set to $g_\ell + S_n^{j-1} - (g_j - f_{j-1})$ for
$\ell=j,j+1,\ldots,m$.
\end{itemize}
With the above two steps, the schedule $\sigma$ remains almost
unchanged by just adding $S_n^{j-1} - (g_j - f_{j-1})$ amount of
idle time. We repeat the above procedure for $j=2,3,\ldots,m$.
Again, let $\sigma$ be the above revised schedule with the revised
release pattern. Now, after the adjustment, Condition 2 holds, and
the response time of job $J$ in this schedule is larger than or
equal to the original one.
Condition 3 is rather trivial. If a job in schedule $\sigma$ has a
shorter execution time, we can increase its execution time to its
worst-case execution time. We can then adjust the release pattern
with a similar procedure like the operations for Condition 2 to
increase the response time of job $J$.
With the above discussions, we reach the conclusion of this lemma.
\end{proof}
\begin{lemma}
\label{lemma:necessary-critical-wcrt-larger}
If $WCRT_n > T_n$, the response time of task
$\tau_n$ in a release pattern that satisfies Conditions 1, 2, and 3
in Lemma~\ref{lemma:necessary-critical-wcrt} is larger than $T_n$.
\end{lemma}
\begin{proof}
By the assumption $WCRT_n > T_n$, we consider that there exists a
schedule $\sigma$ in which Case 2 (at the opening of the section)
holds. The rest of the proof is identical to the proof of
Lemma~\ref{lemma:necessary-critical-wcrt}.
\end{proof}
We now demonstrate a few properties based on
Lemma~\ref{lemma:necessary-critical-wcrt}. By
Lemma~\ref{lemma:necessary-critical-wcrt}, for obtaining the (exact)
worst-case response time of task $\tau_n$, we simply have to examine
all the release patterns that satisfy the three conditions in
Lemma~\ref{lemma:necessary-critical-wcrt}. Specifically, Condition 1
of Lemma~\ref{lemma:necessary-critical-wcrt} implies that we can set an
\emph{offset} variable $O_{i,j}$ (with $O_{i,j} \geq 0$) to define the release time of the
first job of task $\tau_i$, arrived no earlier than $g_j$. That is,
for a given $O_{i,j}$, the first job released by task $\tau_i$ no
earlier than $g_j$ is released at time $O_{i,j} + g_j$.
With the above definitions, we can have the following properties.
These properties can be used to reduce the search space for the
worst-case response time of task $\tau_n$. The first property was also
provided in Corollary 1 in the paper by Nelissen et
al. \cite{ecrts15nelissen}.
\begin{property}
\label{property:p1}
For a higher-priority task $\tau_i$, there must be at least one $O_{i,j}$ equal to $0$.
\end{property}
\begin{proof}
This is quite trivial. By Lemma~\ref{lemma:necessary-critical-wcrt},
either task $\tau_i$ does not release any job to interfere in any
computation segment of job $J$ or $\tau_i$ releases at least one job
to interfere in certain computation segments of job $J$. For the
former case, we can set $O_{i,1}$ to $0$, which does not decrease
the resulting worst-case response time. For the latter case, if all
$O_{i,j} > 0$ for $j=1,2,\ldots,m$, let $j^*$ be the earliest
computation segment of job $J$ where task $\tau_i$ releases some
jobs to interfere in. We can greedily set $O_{i,j^*}$ to $0$, which
does not reduce the resulting worst-case response time.
\end{proof}
\begin{property}
\label{property:p2}
If the period of task $\tau_i$ is small enough, then the following
property holds:
\begin{itemize}
\item {\bf Case when $j=1$}: If $T_i \leq S_n^1$, then $O_{i,1}$ is $0$.
\item {\bf Case when $j=m$}: If $T_i \leq S_n^{m-1}$, then $O_{i,m}$ is $0$.
\item {\bf Case when $2 \leq j \leq m-1$}: If $T_i \leq S_n^{j-1}$ and $T_i
\leq S_n^{j}$ then $O_{i,j}$ is $0$.
\end{itemize}
That is, task $\tau_i$ releases one job together with the $j$-th
computation segment of the job $J$. Moreover, task $\tau_i$ also
releases the subsequent jobs strictly periodically with period $T_i$
until the $j$-th computation segment of job $J$ finishes.
\end{property}
\begin{proof}
The first case is clear due to Condition 2 in
Lemma~\ref{lemma:necessary-critical-wcrt}, i.e., the suspension time
of the first suspension interval is exactly $S_n^1$, since the
release pattern of task $\tau_i$ to interfere in the first
computation segment of job $J$ is independent from the other
computation segments. The second case is similar.
For any $j$ with $2 \leq j \leq m-1$, the condition $T_i \leq
S_n^{j-1}$ implies that the release pattern of task $\tau_i$ to
interfere in the $(j-1)$-th computation segment of job $J$ is
independent from the release pattern to interfere in the $j$-th
computation segment. Similarly, the condition $T_i \leq S_n^{j}$
implies that the release pattern of task $\tau_i$ to interfere in
the $j$-th computation segment of job $J$ is independent from the
release pattern of task $\tau_i$ to interfere in the $(j+1)$-th computation
segment. Therefore, when $T_i \leq S_n^{j-1}$ and $T_i \leq
S_n^{j}$, the release pattern of task $\tau_i$ to interfere in the
$j$-th computation segment of job $J$ is independent from the other
computation segments.
Moreover, when the release pattern of task $\tau_i$ to interfere in
the $j$-th computation segment of job $J$ is independent from the
other computation segments, the worst-case release pattern of task
$\tau_i$ to interfere in the $j$-th computation segment of job $J$
is to release 1) one job together with the $j$-th computation
segment of the job $J$, and 2) the subsequent jobs strictly
periodically with period $T_i$ until the $j$-th computation segment
of job $J$ finishes.
\end{proof}
\begin{property}
\label{property:p3}
If $T_i \geq T_n$ and $WCRT_n \leq T_n$, then a higher-priority task $\tau_i$ only
releases one job together with one of the $m$ computation segments
of the job (under analysis) of task $\tau_n$.
\end{property}
\begin{proof}
This comes from Condition 1 in
Lemma~\ref{lemma:necessary-critical-wcrt} and Property
\ref{property:p1}.
\end{proof}
\begin{property}
\label{property:p4}
If $T_i - C_i$ is small enough, then the following property of task
$\tau_i$ holds:
\begin{itemize}
\item {\bf Case when $j=1$}: If $T_i - C_i \leq S_n^1$, then $O_{i,1}$ is $0$.
\item {\bf Case when $j=m$}: If $T_i - C_i \leq S_n^{m-1}$, then $O_{i,m}$ is $0$.
\item {\bf Case when $2 \leq j \leq m-1$}: If $T_i -C_i \leq
S_n^{j-1}$ and $T_i - C_i
\leq S_n^{j}$ then $O_{i,j}$ is $0$.
\end{itemize}
That is, task $\tau_i$ releases one job together with the $j$-th
computation segment of the job $J$. Moreover, task $\tau_i$ also
releases the subsequent jobs strictly periodically with period $T_i$
until the $j$-th computation segment of job $J$ finishes.
\end{property}
\begin{proof}
This is a simple extension of Property~\ref{property:p2}.
\end{proof}
\section{Computational Complexity}
\label{sec:computational-complexity-fp}
In this section, we will prove that the schedulability test problem
for FP preemptive scheduling even with only one
\emph{segmented} self-suspending task as the lowest-priority task in
the task system is $co{\cal NP}$-hard in the strong
sense. Specifically, we will also show that our reduction implies
that finding whether there exists a feasible priority assignment under
FP scheduling for constrained-deadline task systems is
also $co{\cal NP}$-hard in the strong sense. We will
first consider constrained-deadline task systems and then revise the
reduction to consider implicit-deadline task systems.
Our reduction is from the 3-PARTITION problem
\cite{b:Gary79}:\footnote{For notational consistency and brevity in
our reduction, we index the $3M$ integer numbers from $2$.}
\begin{definition}[3-PARTITION Problem] We are given a positive
integer $V$, a positive integer $M$, and a set of $3M$ integer
numbers $\setof{v_2, v_3, \ldots, v_{3M+1}}$ with the condition
$\sum_{i=2}^{3M+1} v_i = MV$, in which $1 \leq V/4 < v_i < V/2$ and $M
\geq 3$. Therefore, $V \geq 3$. \\{\bf
Objective:} The problem is to partition the given $3M$ integer
numbers into $M$ disjoint sets ${\bf V}_1, {\bf V}_2, \ldots, {\bf
V}_M$ such that the sum of the numbers in each set ${\bf V}_i$ for
$i=1,2,\ldots,M$ is $V$, i.e., $\sum_{v_j \in {\bf V}_i} v_j = V$. \hfill\myendproof
\end{definition}
The decision version of the 3-PARTITION problem to verify whether such
a partition into $M$ disjoint sets exists or not is known ${\cal
NP}$-complete in the strong sense \cite{b:Gary79} when $M \geq 3$.
\subsection{Constrained-Deadline Task Systems}
\label{sec:conp-hard-constrained}
\begin{definition}[Reduction to a constrained-deadline system]
\label{def:reduced-taskset-constrained}
For a given input instance of the 3-PARTITION problem, we construct
$n=3MV+2$ sporadic tasks as follows:
\begin{compactitem}
\item For task $\tau_1$, we set $C_1 = V, S_1 = 0, D_1=V, T_1=3V$.
\item For task $\tau_i$ with $i=2,3,\ldots, 3M+1$, we set $C_i = v_i,
S_i = 0, T_i=21MV$ and $D_i=3MV/2$ if $M$ is an even number or
$D_i=3MV/2+ V/2$ if $M$ is an odd number.
\item For task $\tau_{3M+2}$, we create a segmented self-suspending
task with $M$ computation segments separated by $M-1$
self-suspension intervals\footnote{The first version of the proof
uses $6V$ for $S_{3M+2}^j$ for $j=1,2,\ldots,M-1$. By applying
Property~\ref{property:p3}, we can also set $S_{3M+2}^j$ to $2V$
and $D_{3M+2} = M (4V + 1) -V + 2V(M-1)= 6MV+M-3V$.}, i.e., $m=m_{3M+2}=M$, in which $C_{3M+2}^j= V+1$ for
$j=1,2,\ldots,M$, $S_{3M+2}^j = 6V$ for $j=1,2,\ldots,M-1$,
$D_{3M+2} = M (4V + 1) -V + 6V(M-1)= 10MV+M-7V$, and $T_{3M+2}=21MV$.
\end{compactitem}
Due to the stringent relative deadline of task $\tau_1$, it
must be assigned as the highest-priority task. Moreover, the $3M$
tasks, i.e., $\tau_2, \tau_3, \ldots, \tau_{3M+1}$, created by using
the integer numbers from the 3-PARTITION problem instance are assigned
lower priorities than task $\tau_1$ and higher priorities than task
$\tau_{3M+2}$. Since the integer numbers in the 3-PARTITION problem
instance are given in an arbitrary order, without loss of generality,
we index the tasks in $\tau_2, \tau_3, \ldots, \tau_{3M+1}$ by the
given priority assignment, i.e., a lower-indexed task has higher
priority. (In fact, we can also assign all these $3M$ tasks with the
same priority level.) \hfill\myendproof
\end{definition}
For the rest of the proof, the task set created in
Definition~\ref{def:reduced-taskset-constrained} is referred to as
${\bf T}^{red}$.
\begin{lemma}
\label{lemma:feasibility-sporadic-tasks}
Tasks $\tau_1, \tau_2, \ldots, \tau_{3M+1}$ in ${\bf T}^{red}$ can
meet their deadlines under the specified FP
scheduling.
\end{lemma}
\begin{proof}
In FP scheduling, the segmented
self-suspending task $\tau_{3M+2}$ in ${\bf T}^{red}$ has no impact
on the schedule of the higher-priority tasks. Therefore, we can use
the standard schedulability test for FP
scheduling to verify their schedulability. The schedulability of
task $\tau_1$ is obvious since $C_1 \leq D_1$. For
$i=2,3,\ldots,3M+1$, task $\tau_i$ is schedulable under
FP scheduling since $C_i + \sum_{j=1}^{i-1}
\ceiling{\frac{D_i}{T_j}}C_j =
\ceiling{\frac{D_i}{T_1}}C_1+\sum_{j=2}^{i} C_j \leq
\ceiling{\frac{D_i}{3V}}V+MV = D_i$, where the last equality is due
to $\ceiling{\frac{D_i}{3V}} = \ceiling{\frac{3MV/2}{3V}} = M/2$
when $M$ is an even number and $\ceiling{\frac{D_i}{3V}} =
\ceiling{\frac{3MV/2+V/2}{3V}} = (M+1)/2$ when $M$ is an add number.
\end{proof}
The worst-case response time of task $\tau_{3M+2}$
happens by using one of the release patterns with the conditions in
Lemma~\ref{lemma:wcrt-pattern-constrained}:
\begin{lemma}
\label{lemma:wcrt-pattern-constrained}
The worst-case response time of task $\tau_{3M+2}$ in ${\bf
T}^{red}$ under FP scheduling happens under
the following necessary conditions:
\begin{enumerate}
\item Task $\tau_{3M+2}$ releases a job at time $0$. This job
requests the worst-case execution time per computation segment and
suspends in each self-suspension interval exactly equal to its
worst case.
\item Task $\tau_1$ always releases one job together with each
computation segment of the job (released at time $0$) of task
$\tau_{3M+2}$, and releases the subsequent jobs strictly
periodically with period $3V$ until a computation segment of task
$\tau_{3M+2}$ finishes. Task $\tau_1$ never releases any job when
task $\tau_{3M+2}$ suspends itself.
\item For $i=2,3,\ldots,3M+1$, task $\tau_i$ only releases one job
together with one of the $M$ computation segments of the job
(released at time $0$) of task $\tau_{3M+2}$.
\end{enumerate}
All the jobs and all the computation segments are executed with their
worst-case execution time specifications.
\end{lemma}
\begin{proof}
For task $\tau_1$ in ${\bf T}^{red}$, since $T_1 < S_n^j$ for any $j=1,2,\ldots,M-1$, the
release pattern of task $\tau_1$ is independent from the computation
segments. This is formally proved in Property~\ref{property:p2} in Section~\ref{sec:proof-necessary}.
Moreover, since $T_i > D_n$
for $i=2,3,\ldots,3M+1$, such a higher-priority task $\tau_i$ in ${\bf T}^{red}$ only
releases one job together with one of the $M$ computation segments of
the job (under analysis) of task $\tau_n$. This is formally proved in
Property~\ref{property:p3} in Section~\ref{sec:proof-necessary}.
By putting all the above conditions together, we reach the conclusion for
Lemma~\ref{lemma:wcrt-pattern-constrained}.
\end{proof}
For the $j$-th computation segment of task $\tau_{3M+2}$, suppose that
${\bf T}_j \subseteq \setof{\tau_2, \tau_3, \ldots, \tau_{3M+1}}$ is
the set of the tasks released together with $C_{3M+2}^j$ (under the
third condition in Lemma~\ref{lemma:wcrt-pattern-constrained}). For
notational brevity, let $w_j$ be $\sum_{\tau_i \in {\bf T}_j} C_i$. By
definition, $w_j$ is a non-negative integer. Together with the second
condition in Lemma~\ref{lemma:wcrt-pattern-constrained}, we can use
the standard time demand analysis to analyze the worst-case response
time $R_j$ of the $j$-th computation segment of task $\tau_{3M+2}$
(after it is released) under the higher-priority interference due to
$\setof{\tau_1} \cup {\bf T}_j$.
The response time $R_j$ of a computation segment $C_{3M+2}^j$ is defined as
the finishing time of the computation segment minus the arrival time
of the computation segment.
For a given task set ${\bf T}_j$ (i.e., a given non-negative integer $w_j$),
$R_j$ is the minimum $t$ with $t > 0$ such that
\[
C_{3M+2}^j + (\sum_{\tau_i \in {\bf T}_j} C_i) +\ceiling{\frac{t}{T_1}}C_1= V + 1 + w_j + \ceiling{\frac{t}{3V}}V = t.
\]
Since $R_j$ only depends on
the non-negative integer $w_j$, we use $R(w_j)$ to represent $R_j$ for
a given ${\bf T}_j$.
We know that $V
+ 1 + w_j + \ceiling{\frac{t}{3V}}V = t$ happens with $\ell\cdot 3V < t
\leq (\ell+1)\cdot 3V$ for a certain non-negative integer $\ell$. That is,
$V + 1 + w_j + \ceiling{\frac{t}{3V}}V > t$ when $t$ is $\ell\cdot 3V$
and $V + 1 + w_j + \ceiling{\frac{t}{3V}}V \leq t$ when $t$ is
$(\ell+1)3V$. We know that $\ell$ is
$\ceiling{\frac{V+1+w_j}{2V}}-1$. Moreover,
\begin{align*}
R(w_j) &= \ell\cdot 3V + V+(V+1+w_j - \ell\cdot 2V) \\
&= 2V+1+w_j + \ell\cdot V \\
&= V+1+w_j + \ceiling{\frac{V+1+w_j}{2V}} V.
\end{align*}
This leads to three cases that are of interest:
\begin{equation}
\label{eq:response-time-np-hard}
R(w_j) =
\begin{cases}
2V+1+w_j & \mbox{ if }w_j \leq V -1\\
4V+1 & \mbox{ if } w_j = V \\
V+1+w_j + \ceiling{\frac{V+1+w_j}{2V}} V& \mbox{ if } w_j > V
\end{cases}
\end{equation}
For example, if $w_j = 3V-1$, then $R(w_j)$ is $6V$; if $w_j$ is $3V$,
then $R(w_j)$ is $7V+1$.
With the above discussions, we can now conclude that the unique
condition when task $\tau_{3M+2}$ misses its deadline in the following
lemma.
\begin{lemma}
\label{lemma:schedulability-equivalent}
Suppose that ${\bf T}_j \subseteq \setof{\tau_2, \tau_3, \ldots,
\tau_{3M+1}}$ and ${\bf T}_i \cap {\bf T}_j = \emptyset$ when $i
\neq j$. Let $w_j = \sum_{\tau_i \in {\bf T}_j} C_i$. If a task
partition ${\bf T}_1, {\bf T}_2, \ldots, {\bf T}_M$ exists such that
$\sum_{j=1}^{M} R(w_j) > M(4V+1)-V$ with $R(w_j)$ defined in
Eq.~\eqref{eq:response-time-np-hard}, then task $\tau_{3M+2}$ misses
its deadline in the worst case; otherwise, task $\tau_{3M+2}$ always
meets its deadline.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma:wcrt-pattern-constrained}, task $\tau_{3M+2}$ in
${\bf T}^{red}$ is not schedulable under the fixed-priority
preemptive scheduling if and only if there exists a task partition ${\bf T}_1,
{\bf T}_2, \ldots, {\bf T}_M$ such that $\sum_{j=1}^{M} R(w_j) +
\sum_{j=1}^{M-1} S_{3M+2}^j = 6(M-1)V + \sum_{j=1}^{M} R(w_j) >
D_{3M+2} = M (4V + 1) + 6V(M-1) -V$. This concludes the proof.
\end{proof}
Instead of investigating the combinations of the task partitions, we
analyze the corresponding total worst-case response time
$\sum_{j=1}^{M} R(w_j)$ for the $M$ computation segments of task
$\tau_{3M+2}$ (by excluding the self-suspension time) by considering
different \emph{non-negative integer assignments} $w_1, w_2, \ldots,
w_{M}$ with $\sum_{i=1}^{M} w_i = MV$ and $w_i \geq 0$ in the following
lemmas.
\begin{lemma}
\label{lemma:wcrt-all-the-same}
If $w_1 = w_2 = \cdots = w_{M} = V$, then
\[
\sum_{j=1}^{M} R(w_j) = M(4V+1),
\]
where $R(w_j)$ is defined in Eq.~\eqref{eq:response-time-np-hard}.
\end{lemma}
\begin{proof}
This comes directly by Eq.~\eqref{eq:response-time-np-hard}.
\end{proof}
\begin{lemma}
\label{lemma:wcrt-one-different}
For any non-negative integer assignment for $w_1, w_2, \ldots, w_M$
with $\sum_{i=1}^{M} w_i = MV$, if there exists a certain index $j$
with $w_j \neq V$, then
\[
\sum_{j=1}^{M} R(w_j) \leq M(4V+1)-V,
\]
where $R(w_j)$ is defined in Eq.~\eqref{eq:response-time-np-hard}.
\end{lemma}
\begin{proof}
Let ${\bf X}$ be the set of indexes such that $0 \leq w_j < V$ for
any $j \in {\bf X}$. Similarly, let ${\bf Y}$ be the set of indexes
such that $V < w_j$ for any $j \in {\bf Y}$. If $j \notin {\bf
X}\cup{\bf Y}$, then $w_j$ is $V$.
If there exists $j$ in ${\bf Y}$ with $w_j > 2V$, since
$\sum_{i=1}^{M} w_i = MV$, there must exist an index $i$ in ${\bf
X}$ with $w_i < V$. We can increase $w_i$ to $w_i'=V$, which increases
the worst-case response time $R(w_i)$ by $2V-w_i$ (i.e., from
$2V+1+w_i$ to $4V+1$). Simultaneously, we
reduce $w_j$ to $w_j' = w_j - (V-w_i) > V$. Therefore,
$w_i+w_j=w_i'+w_j'$. Moreover, the reduction of $w_j$ to $w_j'$ also reduces the
worst-case response time $R(w_j)$ by case 1) $V-w_i$ if
$\ceiling{\frac{V+1+w_j'}{2V}}$ is equal to $\ceiling{\frac{V+1+w_j}{2V}}$, and by case 2) $V-w_i+V$ if $\ceiling{\frac{V+1+w_j'}{2V}}$ is not equal to $\ceiling{\frac{V+1+w_j}{2V}}$. In both
cases, we can easily see that the worst-case response time is not
decreased in the new integer assignment.
Moreover, the index $j$
remains in ${\bf Y}$ and the index $i$ is removed from set ${\bf
X}$. We repeat the above step until all the indexes $j$ in ${\bf
Y}$ are with $w_j \leq 2V$.
It is clear that ${\bf X}$ and ${\bf Y}$ are both non-empty after the above step.
For the rest of the proof, let ${\bf X}$ and ${\bf Y}$ be defined after finishing the above step.
Therefore, the condition
$w_j \leq 2V$ holds for any $j \in {\bf Y}$.
Due to the pigeon-hole
principle, when ${\bf Y}$ is not an empty set, ${\bf X}$ is also not
an empty set. Moreover, for an element $i$ in ${\bf X}$, there must
be a subset ${\bf Y}'\subseteq {\bf Y}$ and an index $\ell \in {\bf
Y}'$ such that
\[ \sum_{j \in {\bf Y}'}
(w_j-V) \geq V-w_i > \sum_{j \in {\bf Y}'\setminus\setof{\ell}} (w_j-V).
\]
That is, we want to adjust $w_i$ to $V$ (i.e., $w_i$ is increased by
$V-w_i$), and the set ${\bf Y}'\setminus\setof{\ell}$ is not
enough to match the integer adjustment $V-w_i$ and the set ${\bf
Y}'$ is enough to match the integer adjustment $V-w_i$. We now
increase $w_i$ to $V$, which increases the worst-case response time
$R(w_i)$ by $2V-w_i$. Simultaneously, we reduce $w_j$ to $V$ for
every $j \in {\bf Y}' \setminus \setof{\ell}$ and reduce $w_\ell$ to
$w_\ell'=w_\ell - (V-w_i-\sum_{j \in {\bf Y}'\setminus\setof{\ell}}
(w_j-V))$. Since $V < w_j \leq 2V$ for any $j \in {\bf Y}'$ before
the adjustment, the adjustment reduces the worst-case response time
$R(w_j)$ by $w_j - V$ if $j\neq \ell$ and reduces $R(w_\ell)$ by
$w_\ell - w'_\ell$. Therefore, the adjustment reduces $\sum_{j \in
{\bf Y}'} R(w_j)$ by exactly $V-w_i$. Therefore, the adjustment in
this step to change $w_i$ in ${\bf X}$ and $w_j$ in ${\bf Y}'$ increases the
overall worst-case response time by exactly $V$ time units.
By adjusting with the above procedure repeatedly, we will reach the
integer assignment $w_1 = w_2 = \cdots = w_{M} = V$ with bounded
increase of the worst-case response time. As a result, we can
conclude that $\sum_{j=1}^{M} R(w_j) \leq M(4V+1)-|{\bf X}|V$. By the
assumption $\sum_{i=1}^{M} w_i = MV$ and the existence of $w_j \neq
V$ for some $j$, we know that $|{\bf X}|$ must be at least
$1$. Therefore, we reach the conclusion.
\end{proof}
We use an example to illustrate how the procedure in Lemma
\ref{lemma:wcrt-one-different} operates. Suppose that $w_1=0, w_2 =
3.5V, w_3=0.4V, w_4=0.6V, w_5=1.5V, w_6=0$ when $M=6$ and $V$ is an
integer multiple of $10$. We will start from ${\bf
X}=\setof{1,3,4,6}$ and ${\bf Y} = \setof{2,5}$. As shown in
Table~\ref{tab:example-lemma-merge}, the operation makes
$\sum_{j=1}^{6}R(w_j)$ increase. Note that the conclusion
$\sum_{j=1}^{M} R(w_j) \leq M(4V+1)-|{\bf X}|V$ in
Lemma~\ref{lemma:wcrt-one-different} was for $|{\bf X}|=\setof{3,4}$
in this example when $w_j < 2V$ for any $j \in {\bf Y}$.
\begin{table}[t]
\centering\scalebox{0.7}{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c||c|c||c|c||c|}
\hline
$w_1$ & $R(w_1)$
& $w_2$ & $R(w_2)$
& $w_3$ & $R(w_3)$
& $w_4$ & $R(w_4)$
& $w_5$ & $R(w_5)$
& $w_6$ & $R(w_6)$
& ${\bf X}$ & ${\bf Y}$ & $\sum_{j=1}^{6}R(w_j)$\\
\hline
0 & $2V+1$ & $3.5V$ & $7.5V+1$ & $0.4V$ & $2.4V+1$ & $0.6V$ & $2.6V+1$
& $1.5V$ & $4.5V+1$ & $0$ & $2V+1$ & $\setof{1,3,4,6}$ &
$\setof{2,5}$ & $21V+6$\\
\hline
$V$ & $4V+1$ & $2.5V$ & $5.5V+1$ & $-$ & $-$ & $-$ & $-$
& $-$ & $-$ & $-$ & $-$ & $\setof{3,4,6}$ & $\setof{2,5}$
& $21V+6$\\
\hline
$-$ & $-$ & $1.5V$ & $4.5V+1$ & $-$ & $-$ & $-$ & $-$
& $-$ & $-$ & $V$ & $4V+1$ & $\setof{3,4}$ & $\setof{2,5}$
& $22V+6$\\
\hline
$-$ & $-$ & $V$ & $4V+1$ & $V$ & $4V+1$& $-$ & $-$
& $1.4V$ & $4.4V+1$& $-$ & $-$ & $\setof{4}$ & $\setof{5}$
& $23V+6$\\
\hline
$-$ & $-$ & $-$ & $-$ & $-$ & $-$ & $V$ & $4V+1$
& $1V$ & $4V+1$& $-$ & $-$ & $\emptyset$ & $\emptyset$
& $24V+6$\\
\hline
\end{tabular}}
\caption{An example of Lemma~\ref{lemma:wcrt-one-different}}
\label{tab:example-lemma-merge}
\end{table}
We can now conclude the $co{\cal NP}$-hardness.
\begin{theorem}
\label{thm:conp-hard-constrained}
The schedulability analysis for FP scheduling
even with only one segmented self-suspending task as the
lowest-priority task in the sporadic task system is $co{\cal NP}$-hard
in the strong sense, when the number of self-suspending intervals in
the self-suspending task is larger than or equal to $2$ and $D_i \leq
T_i$ for every task $\tau_i$.
\end{theorem}
\begin{proof}
The reduction in Definition~\ref{def:reduced-taskset-constrained}
requires polynomial time. Moreover, by
Lemmas~\ref{lemma:wcrt-pattern-constrained},~\ref{lemma:schedulability-equivalent},~\ref{lemma:wcrt-all-the-same},~and~\ref{lemma:wcrt-one-different},
a feasible solution of the 3-PARTITION problem for the input
instance exists if and only if task $\tau_{3M+2}$ is not schedulable
by the FP scheduling when $M \geq 3$. Therefore, this concludes the proof.
\end{proof}
\begin{corollary}
\label{cor:conp-hard-constrained-scheduler-design}
Validating whether there exists a feasible priority assignment is
$co{\cal NP}$-hard in the strong sense for constrained-deadline
segmented self-suspending task systems.
\end{corollary}
\begin{proof}
This comes directly from Theorem~\ref{thm:conp-hard-constrained} and
the only possible priority level for task $\tau_{3M+2}$ to be
feasible in ${\bf T}^{red}$.
\end{proof}
\subsection{Implicit-Deadline Task Systems}
The $co{\cal NP}$-hardness in the strong sense for testing the
schedulability of task $\tau_n$ under FP scheduling can be easily
proved with the same input as in ${\bf T}^{red}$ by changing the
periods of the tasks as follows:
\begin{compactitem}
\item For task $\tau_1$, we set $D_1=3V, T_1=3V$.
\item For task $\tau_i$ with $i=2,..., 3M+1$, we set $T_i=D_i=10MV+M-7V$.
\item For task $\tau_{3M+2}$, we set $T_{3M+2} = D_{3M+2} =
10MV+M-7V$.
\end{compactitem}
Assume that $\tau_{3M+2}$ is the lowest-priority task. It is not
difficult to see that all the conditions in
Lemma~\ref{lemma:wcrt-pattern-constrained} still hold for testing whether task $\tau_{3M+2}$ can meet its deadline or not (but not for the worst-case response time if task $\tau_{3M+2}$ misses the deadline). Therefore, the
schedulability analysis for FP scheduling even with only one segmented
self-suspending task as the lowest-priority task in the sporadic task
system is $co{\cal NP}$-hard in the strong sense, when the number of
self-suspending intervals in the self-suspending task is larger than or
equal to $2$ and $D_i = T_i$ for every task $\tau_i$.
However, the above argument does not hold if we assign task
$\tau_{3M+2}$ to the highest-priority level. Therefore, the above
proof does not support a similar conclusion for implicit-deadlien task
systems to that for constrained-deadline task systems in
Corollary~\ref{cor:conp-hard-constrained-scheduler-design}.
\section{MILP Approaches}
\label{sec:proof-milp}
Even though the properties in
Lemma~\ref{lemma:necessary-critical-wcrt} provide the necessary
conditions for the worst-case response time, finding the worst-case release
pattern is in fact a hard problem as shown in the analysis in
Section~\ref{sec:computational-complexity-fp}. However, if we can
tolerate exponential time complexity, is there a strategy that can
find the worst-case pattern based on
Lemma~\ref{lemma:necessary-critical-wcrt} safely without performing
exhaustive searches? One possibility is to model the problem as an MILP,
which has been already presented by Nelissen et al. \cite{ecrts15nelissen}.
The worst-case response time analysis by Nelissen et
al. \cite{ecrts15nelissen} is based on the following mixed-integer
linear programming (MILP):
\begin{subequations}
\label{eq:MILP}\footnotesize{
\begin{align}
\mbox{\bf maximize: } \;\;\;\;S_n + \sum_{j=1}^{m} R_j \label{eq:objective}\\
\mbox{\bf subject to:} \qquad\qquad\qquad&\nonumber\\
R_j = C_n^j + \sum_{i=1}^{n-1} N_{i,j} \times C_i, \qquad&
\forall j=1,\ldots,m \label{eq:R_j}\\
O_{i,j} \geq 0, \qquad& \forall i=1,\ldots,n-1, \forall
j=1,\ldots,m \label{eq:O-positive}\\
O_{i,j+1} \geq O_{i,j} + N_{i,j}\times T_i - (R_j + S_n^j), \qquad \label{eq:O-distance}&
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m-1\\
0 \leq N_{i,j} \leq \ceiling{\frac{R_j - O_{i,j}}{T_i}}, \qquad&
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m \label{eq:N-bounds}\\
N_{i,j} \mbox{ is an integer }, \qquad&
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m \label{eq:N-integer}\\
R_j \leq UB_{ss,j} \qquad &\forall j=1,2,\ldots,m, \label{eq:total-segment-upperbound} \\
S_n + \sum_{j=1}^{m} R_j \leq UB_{ss} \label{eq:total-upperbound}\\
\mbox{ Eq. \eqref{eq:Rj>rel-original} holds}.
\end{align}}
\end{subequations}
In the above MILP, the objective function $S_n + \sum_{j=1}^{m} R_j$
is the worst-case response time of task $\tau_n$, where $R_j$ is a variable
(as a real number) that represents the response time of
the $j$-th computation segment $C_n^j$ of task $\tau_n$. The variable
$O_{i,j}$ defines the \emph{offset} of the first job of a
higher-priority task $\tau_i$ released no earlier than the arrival
time of the $j$-th computation segment $C_n^j$ of task $\tau_n$. That
is, if the arrival time of $C_n^j$ is $t_j$, then the first job of
task $\tau_i$ released at or after $t_j$ is at time $t_j +
O_{i,j}$. The integer variable $N_{i,j}$ defines the maximum number of
jobs of a higher-priority task $\tau_i$ that are released to
\emph{successfully interfere} in the computation segment $C_n^j$ of task
$\tau_n$.
The three additional constraints, expressed by Eq. (9), Eq. (11), and
Eq. (16), in the MILP in \cite{ecrts15nelissen} are expressed here by
Eq.~\eqref{eq:total-segment-upperbound}, Eq.~\eqref{eq:total-upperbound}, and
Eq.~\eqref{eq:Rj>rel-original}, respectively. Here, $UB_{ss}$ is defined as the
upper bound on the worst-case response time of task $\tau_n$, and
$UB_{ss,j}$ is defined as the upper bound on the worst-case response
time of the $j$-th computation segment of task $\tau_n$. Later in
this section, we will show that the condition in Eq.~\eqref{eq:R_j}
may over-estimate the worst-case response time. Therefore, the
additional constraint (expressed by Eq. (16), in the MILP in
\cite{ecrts15nelissen}) is used to reduce the pessimism as follows: {\small
\begin{equation}
\label{eq:Rj>rel-original}
\forall i=1,2,\ldots,n-1, j=1,2,\ldots,m, \qquad R_j > rel_{i,j} + \sum_{\ell=1}^{n-1}\max\left\{0,
\floor{\frac{O_{\ell,j}+N_{\ell,j}T_\ell - rel_{i,j}}{T_\ell}}C_\ell\right\},
\end{equation}}where $rel_{i,j} = O_{i,j} + (N_{i,j}-1) T_i$. This means that
the (total) execution time of all the higher-priority jobs (by tasks $\tau_1, \tau_2, \ldots, \tau_{n-1}$) released
after $rel_{i,j}$ should be less than $R_j-rel_{i,j}$.
Here, we first explain why the MILP by utilizing only the constraints
from Eq.~\eqref{eq:R_j} to Eq.~\eqref{eq:N-integer} is already a safe
(but \emph{not tight/exact}) result based on
Lemma~\ref{lemma:necessary-critical-wcrt}. Therefore, this also leads to
the motivation to examine the pessimism by different combinations of
the additional constraints Eq.~\eqref{eq:total-segment-upperbound},
Eq.~\eqref{eq:total-upperbound}, and Eq.~\eqref{eq:Rj>rel-original} in
Section~\ref{sec:gap}.
\subsection{MILP by Using Lemma~\ref{lemma:necessary-critical-wcrt}}
We only consider the release patterns of the tasks in ${\bf T}$, where
all the three conditions in Lemma~\ref{lemma:necessary-critical-wcrt}
hold. Let $r_{i,j}$ be the arrival time of the first job of task
$\tau_i$ arrived after or at time $g_j$ in a concrete release pattern,
in which all the three conditions in
Lemma~\ref{lemma:necessary-critical-wcrt} hold. If task $\tau_i$ does
not release any job after or at time $g_j$, we set $r_{i,j}$ to
$\infty$.\footnote{With the discussions below, we will later set
$r_{i,j}$ to $f_j + T_j$ for such a case (but not release any job of
task $\tau_i$ at time $f_j+T_j$).} By the minimum inter-arrival time
constraint of task $\tau_i$, we know that task $\tau_i$ cannot release
any job in time interval $(r_{i,j+1}-T_i, r_{i,j+1})$. That is, in
this release pattern, there are \emph{at most}
$\floor{\frac{r_{i,j+1}-T_i - r_{i,j}}{T_i}}+1 \leq \frac{r_{i,j+1} -
r_{i,j}}{T_i}$ jobs from task $\tau_i$ that can interfere in the
$j$-th computation segment of job $J$.
Let $N_{i,j}$ be the number of jobs of a higher-priority task $\tau_i$
released in time interval $[g_j, f_j)$ in this release pattern. By
definition, $N_{i,j}$ is a non-negative integer. The maximum number
of jobs that task $\tau_i$ can release in time interval $[r_{i,j},
f_j)$ in this release pattern can be expressed by the following inequality:
\begin{equation}
0 \leq N_{i,j} \leq \max\left\{0, \ceiling{\frac{f_j-r_{i,j}}{T_i}}\right\}, \qquad \forall i=1,\ldots,n-1, j=1,\ldots,m. \label{eq:N-ij-1-pre}
\end{equation}
The reason to put $\max\left\{0,
\ceiling{\frac{f_j-r_{i,j}}{T_i}}\right\}$ instead of only
$\ceiling{\frac{f_j-r_{i,j}}{T_i}}$ in the right-hand side of
Eq.~\eqref{eq:N-ij-1-pre} is to avoid the case that
$\ceiling{\frac{f_j-r_{i,j}}{T_i}} < 0$, which is possible if $r_{i,j}
> f_j + T_i$.
There is one simple trick regarding to the setting of $r_{i,j}$. If
$r_{i,j} > f_j + T_i$, for this release pattern, we know that 1) task
$\tau_i$ does not release any job to interfere in the $j$-th
computation segment of job $J$ and 2) the number of jobs of task
$\tau_i$ that are released to interfere in the $(j-1)$-th computation segment
of job $J$ is purely dominated by $\max\left\{0,
\ceiling{\frac{f_{j-1}-r_{i,j-1}}{T_i}}\right\}$. Therefore, if
$r_{i,j} > f_j + T_i$, we can safely set $r_{i,j}$ to $f_j+T_i$
(but we do not change the release pattern to release a job of task
$\tau_i$ at time $f_j + T_i$ for such a case). With this, we can then rephrase
Eq.~\eqref{eq:N-ij-1-pre} into
\begin{equation}
0 \leq N_{i,j} \leq \ceiling{\frac{f_j-r_{i,j}}{T_i}}, \qquad \forall i=1,\ldots,n-1, j=1,\ldots,m. \label{eq:N-ij-1}
\end{equation}
By earlier discussions, we also have
\begin{equation}
N_{i,j} \leq \frac{r_{i,j+1} - r_{i,j}}{T_i}, \qquad \forall i=1,\ldots,n-1, j=1,\ldots,m-1. \label{eq:N-ij-2}
\end{equation}
By Condition 1 and Condition 3 in Lemma~\ref{lemma:necessary-critical-wcrt}, we also know that
\begin{equation}
\label{eq:f_j}
f_j \leq g_j + C_n^j + \sum_{i=1}^{n-1} N_{i,j} \times C_i \qquad \forall j=1,2,\ldots, m.
\end{equation}
Without loss of generality, we can set $g_1$ to $0$. By Condition 2 in
Lemma~\ref{lemma:necessary-critical-wcrt}, we have
\begin{equation}
\label{eq:g_j}
g_1 = 0 \mbox{ and } g_j = f_{j-1} + S_n^{j-1} \qquad \forall j=2,3,\ldots, m.
\end{equation}
Now we can conclude the following theorem:
\begin{theorem}
\label{thm:MILP-v1}
Suppose that $g_j, f_j, r_{i,j}$ are variables of real numbers and
$N_{i,j}$ are variables for non-negative integer numbers for
$i=1,2,\ldots,n-1$ and for $j=1,2,\ldots,m$. The optimal solution of
the following MILP is a safe upper bound on the worst-case response time of task
$\tau_n$ if $WCRT_n \leq T_n$.
\begin{subequations}
\label{eq:MILP-v1}{\small
\begin{align}
& \mbox{\bf maximize: } \;\;\;\;f_m \label{eq:objective-v1}\\
& \mbox{\bf subject to:} \nonumber\\
&\qquad\qquad\qquad r_{i,j} \geq g_j, \forall i=1,\ldots,n-1, \forall
j=1,\ldots,m \label{eq:r-versus-g}\\
& \qquad\qquad\qquad N_{i,j} \mbox{ is an integer },
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m \label{eq:N-integer-v1}\\
& \qquad\qquad\qquad \mbox{ and Conditions in Eqs.~\eqref{eq:N-ij-1},~\eqref{eq:N-ij-2},~\eqref{eq:f_j},~\eqref{eq:g_j} hold}.\nonumber
\end{align}}
\end{subequations}
\end{theorem}
\begin{proof}
This comes from the above discussions and
Lemma~\ref{lemma:necessary-critical-wcrt}. The release pattern that
has the maximum $f_m$ (provided that $g_1$ is set to $0$) by using
FP preemptive scheduling under all the constraints due to the three
conditions in Lemma~\ref{lemma:necessary-critical-wcrt} leads to the
worst-case response time if $WCRT_n \leq T_n$.
\end{proof}
However, the MILP in Eq.~\eqref{eq:MILP-v1} is not an exact response
time analysis (or schedulability test) due to the following reason:
the condition in Eq.~\eqref{eq:f_j} is only a safe upper bound on $f_j$, but does not provide the exact
$f_j$ under the release pattern. Suppose that $n$ is $2$. We have
$T_1=4$ and $C_1=2$. Consider $g_1=0$ and $r_{i,1}=0$, $C_n^1 = 2$,
and $S_n^1=8$. In this case, it implies that the suspension interval
$S_n^1$ has no impact when we analyze the worst-case finishing time of
the first computation segment.\footnote{This is also proved in
Property \ref{property:p2}.} It is clear that the exact
(worst-case) finishing time of $C_n^1$ is $4$ under this release
pattern. However, there is another feasible solution that satisfies
Eq.~\eqref{eq:f_j} by setting $N_{1,1}$ to $2$, $f_1$ to $6$, and
$r_{1,2}$ to $14$. Therefore, in fact, $f_1$ can have the following
cases:\footnote{For this case, it becomes infeasible when $N_{1,1}$ is
larger than $2$.}
\begin{itemize}
\item $f_1$ is $2$ when $N_{1,1}$ is $0$,
\item $f_1$ is $4$ when $N_{1,1}$ is $1$, and
\item $f_1$ is $6$ when $N_{1,1}$ is $2$.
\end{itemize}
However, due to the objective function for \emph{maximization}, the
optimal MILP solution is to set $f_1$ to $6$ instead of $4$ under this
MILP.\footnote{This also explains why the statement in the earlier version
of this report (\url{https://arxiv.org/abs/1605.00124v1}) was erroneous since it skipped the above discussion and directly concluded that the MILP returns the exact worst-case response time.}
\subsection{Connection to the MILP by Nelissen et al. in ECRTS 2015}
The MILP in Eq.~\eqref{eq:MILP-v1} looks different from the MILP in
Eq.~\eqref{eq:MILP}, but they are in fact equivalent. Suppose that
$R_j = f_j - g_j, \forall j=1,2,\ldots,m$ and $O_{i,j} = r_{i,j} -
g_j, \forall i=1,2,\ldots,n-1, \forall j=1,2,\ldots,m$. We can
rephrase the MILP in Eq.~\eqref{eq:MILP-v1} into the MILP in
Eq.~\eqref{eq:MILP} as follows:
\begin{itemize}
\item Clearly, the objective function in Eq.~\eqref{eq:objective-v1}
is identical to that in Eq.~\eqref{eq:objective}.
\item The condition in Eq.~\eqref{eq:f_j} leads to Eq.~\eqref{eq:R_j}.
\item The condition in Eq.~\eqref{eq:r-versus-g} is identical to Eq.~\eqref{eq:O-positive}.
\item The condition in Eq.~\eqref{eq:g_j} and Eq.~\eqref{eq:f_j} can be used to rephrase Eq.~\eqref{eq:N-ij-2} into
{\small
\begin{equation*}
N_{i,j} \leq \frac{r_{i,j+1} - r_{i,j}}{T_i} = \frac{g_j + R_j + S_n^j + O_{i,j+1} - (g_j + O_{i,j})}{T_i} = \frac{R_j + S_n^j + O_{i,j+1} - O_{i,j}}{T_i},
\end{equation*}} which is identical to the condition in Eq.~\eqref{eq:O-distance}.
\item Moreover, the condition in Eq.~\eqref{eq:N-ij-1} is identical to Eq.~\eqref{eq:N-bounds}.
\end{itemize}
Therefore, we have the following corollaries.
\begin{corollary}
The optimal solution of the MILP in Eq.~\eqref{eq:MILP} (even by excluding Eqs.~\eqref{eq:total-segment-upperbound}, \eqref{eq:total-upperbound},~or~\eqref{eq:Rj>rel-original}) is a safe
upper bound of the
worst-case response time of task $\tau_n$ if $WCRT_n \leq T_n$.
\end{corollary}
\begin{corollary}
If the optimal solution of the MILP in Eq.~\eqref{eq:MILP} (even by excluding Eqs.~\eqref{eq:total-segment-upperbound}, \eqref{eq:total-upperbound},~or~\eqref{eq:Rj>rel-original}), or
equivalently the MILP in Eq.~\eqref{eq:MILP-v1} is no more than
$T_n$, then $WCRT_n \leq T_n$.
\end{corollary}
\section{Response Time Analysis: How Far is the Gap?}
\label{sec:gap}
Since the MILP approach listed in Section~\ref{sec:proof-milp} does
not provide the exact worst-case response
time of task $\tau_n$, it is also meaningful to examine whether the
upper bound on the worst-case response time by using the MILP
approach in Section~\ref{sec:proof-milp} is always very close to (or
not too far from) the exact worst-case response time. Unfortunately,
we will demonstrate a task set, in which the derived worst-case
response time from the MILP in Eq.~\eqref{eq:MILP} is at least $\frac{4m+4}{9}$ times
the exact worst-case response time, where $m \geq 2$ is the number of
computation segments of task $\tau_n$.
We consider the following task set ${\bf T}^{MILP}$ with $n=m+4$ tasks,
where $q$ is a positive integer, $m$ is a positive integer with $m
\geq 2$, and $0 < \epsilon < 1/q$:
\begin{itemize}
\item For task $\tau_1$, we set $C_1 = 1, S_1 = 0, D_1=T_1=2$.
\item For task $\tau_2$, we set $C_2 = q, S_2 = 0, D_2=T_2=4q$.
\item For task $\tau_3$, we set $C_3 = 2q-1+\epsilon, S_3 = 0, D_3=T_3=8q$.
\item For task $\tau_i$ with $i=4,5,\ldots, m+3$, we set $C_i = 1-\epsilon,
S_i = 0, D_i=8qm, T_i= 16qm^2+(m-1)(2q-1)$.
\item For task $\tau_{m+4}$, we create a segmented self-suspending
task with $m$ computation segments separated by $m-1$
self-suspension intervals, i.e., $m_{n}=m$, in which $C_{m+4}^j=
1-\epsilon$ for $j=1,2,\ldots,m$, $S_{m+4}^j = 2q-1$ for
$j=1,2,\ldots,m-1$. The values of $D_{m+4}$ and $T_{m+4}$ are left
open, and our goal here is to find the minimum feasible $D_{m+4}$
that can be set when $T_{m+4}$ is large enough.
\end{itemize}
The following property is very useful when we need to calculate the
worst-case response time:
\begin{property}
\label{property:upper-bound-exact-v2}
For a given positive integer $x$, the minimum $t | t > 0$ such that
$x(1-\epsilon)+\ceiling{\frac{t}{2}} + \ceiling{\frac{t}{4q}}q
+\ceiling{\frac{t}{8q}}(2q-1+\epsilon) =t$ happens when $t$ is
$x\cdot 8q$.
\end{property}
\begin{proof}
This can be proved by simple arithmetics.
\end{proof}
By using Property~\ref{property:upper-bound-exact-v2}, it is not
difficult to obtain the exact worst-case response time by using
Lemma~\ref{lemma:necessary-critical-wcrt}.
\begin{lemma}
\label{lemma:task-MILP}
Tasks $\tau_1, \tau_2, \ldots, \tau_{m+3}$ in ${\bf T}^{MILP}$ can
meet their deadlines. The worst-case response time of task $\tau_n$
in ${\bf T}^{MILP}$ is $16qm+(m-1)(2q-1)$.
\end{lemma}
\begin{proof}
The schedulability of tasks $\tau_1, \tau_2, \tau_3$ comes by using
the standard time demand analysis, and the schedulability of tasks
$\tau_4, \tau_5, \ldots, \tau_{m+3}$ follows from
Property~\ref{property:upper-bound-exact-v2}. The constructed task
set ${\bf T}^{MILP}$ has the following properties based on
Lemma~\ref{lemma:necessary-critical-wcrt}: a) We should always
release the three highest priority tasks together with a computation
segment of task $\tau_n$ and release their subsequent jobs
periodically and as early as possible by respecting their minimum
inter-arrival times until this computation segment of task $\tau_n$
finishes. b) If the response time of task $\tau_n$ is no more than
$16qm^2+(m-1)(2q-1)$, then each task $\tau_i$ for $i=4,5,\ldots,m+3$
only releases one job to interfere in a computation segment of task
$\tau_n$.
Suppose that there are $\ell_j$ tasks among $\tau_4, \tau_5, \ldots,
\tau_{m+3}$ which interfere in the $j$-th computation segment of task
$\tau_n$. We know that $\ell_j$ is an integer and $\ell_j \geq 0$
for $j=1,2,\ldots,m$ and $\sum_{j=1}^{m} \ell_j = m$. Moreover, the
response time of $j$-th computation segment of task $\tau_n$ is
$(\ell_j+1)\cdot 8q$ by
Property~\ref{property:upper-bound-exact-v2}. Therefore, we know
that the worst-case response time of task $\tau_n$ is
\[
\left(\sum_{j=1}^{m} (\ell_j+1)\cdot 8q\right) + (m-1)(2q-1) = 16qm + (m-1)(2q-1).
\]
Since $16qm+(m-1)(2q-1) < T_i$ for $i=4,5,\ldots,n-1$, we know that
the above value is an upper bound by all the possible release
patterns that satisfy Lemma~\ref{lemma:necessary-critical-wcrt}. And,
there is a concrete release/execution pattern which leads the
response time of task $\tau_n$ exactly to this upper
bound. Therefore, this is the exact worst-case response time of task
$\tau_n$ in ${\bf T}^{MILP}$.
\end{proof}
\subsection{Excluding the Boundary Constraints by Eq.~\eqref{eq:total-segment-upperbound} and Eq. \eqref{eq:total-upperbound}}
We first investigate whether the MILP without the boundary constraints presented by Eq.~\eqref{eq:total-segment-upperbound} and Eq. \eqref{eq:total-upperbound}. We explore this specific condition under a special case, by further ignoring
the interference of the tasks $\tau_4, \tau_5, \ldots,
\tau_{m+3}$. Then, the worst-case response time of the $j$-th
computation segment of task $\tau_n$ (after the segment is released)
can be obtained by the following MILP.
\begin{subequations}
\label{eq:MILP-v3}{\small
\begin{align}
& \mbox{\bf maximize: } R_j \label{eq:objective-v3}\\
& \mbox{\bf subject to:} \nonumber\\
& R_j = 1- \epsilon + N_{1,j} \cdot 1 + N_{2,j} \cdot q + N_{3,j}
\cdot (2q-1+\epsilon), \qquad\label{eq:R-v3}\\
& O_{1,j} \geq 0, O_{2,j} \geq 0, O_{3,j} \geq 0 \qquad \label{eq:O-v3}\\
&0 \leq N_{i,j} \leq \ceiling{\frac{R_j - O_{i,j}}{T_i}},\qquad
\forall i=1,\ldots,3 \label{eq:N-v3}\\
& R_j > rel_{i,j} + \sum_{\ell=1}^{3}\max\left\{0,
\floor{\frac{O_{\ell,j}+N_{\ell,j}T_\ell - rel_{i,j}}{T_\ell}}C_\ell\right\}, \qquad
\forall i=1,\ldots,3\label{eq:R>rel-v3}\\
& rel_{i,j} = O_{i,j} + (N_{i,j}-1) T_i, \qquad
\forall i=1,\ldots,3\label{eq:rel-v3}\\
& N_{i,j} \mbox{ is an integer },\qquad
\forall i=1,\ldots,3\label{eq:N-integer-v3}
\end{align}}
\end{subequations}
\begin{lemma}
\label{lemma:upper-bound-pessmistic-v3}
By the assumption that $q$ is a positive integer $q \geq 1$ and $0 < \epsilon < 1/q$, the setting of
$R_j=8q^2+6q+1+q\epsilon$, $O_{1,j} = 0$, $O_{2,j} = \epsilon/4$, $O_{3,j} =
\epsilon/2$, $N_{1,j} = 4q^2+3q+1$, $N_{2,j} = 2q+2$, and $N_{3,j}
= q+1$ is a feasible solution of the MILP in Eq.~\eqref{eq:MILP-v3}.
\end{lemma}
\begin{proof}
The first condition in Eq.~\eqref{eq:R-v3} holds since
{\small\begin{align*}
& 1-\epsilon+4q^2+3q+1+(2q+2)\cdot q + (q+1)\cdot(2q-1+\epsilon)\\
= &1-\xcancel{\epsilon}+4q^2+3q+\xcancel{1}+2q^2+2 q + 2q^2 -q + q\epsilon +2q-\xcancel{1} + \xcancel{\epsilon}\\
= & 8q^2+6q+1+q\epsilon.
\end{align*}}
The conditions in Eqs.~\eqref{eq:O-v3},~\eqref{eq:N-v3},
and~\eqref{eq:N-integer-v3} clearly hold. In this case, the condition
in Eq.~\eqref{eq:rel-v3} sets $rel_{1,j} = O_{1,j} + (N_{1,j}-1)\times
2 = 8q^2+6q$, $rel_{2,j} = O_{2,j} + (N_{2,j}-1)\times 4q =
8q^2+4q+\epsilon/4$, and $ rel_{3,j} = O_{3,j} + (N_{3,j}-1)\times 8q
= 8q^2 + \epsilon/2$. Now, we verify whether the condition in
Eq.~\eqref{eq:R>rel-v3} holds:
\begin{itemize}
\item When $i=1$, we have
\[
8q^2+6q + 1 < R_j.
\]
\item When $i=2$, we have
\begin{align*}
& 8q^2+4q + \epsilon/4 + \max\left\{0, \floor{\frac{8q^2+6q+2 -
(8q^2+4q + \epsilon/4)}{2}}\right\}\\
&+ q + \max\left\{0, \floor{\frac{8q^2+8q+\epsilon/2 -
(8q^2+4q + \epsilon/4)}{8q}}(2q-1+\epsilon)\right\}\\
=\;\;\; & 8q^2+4q + \epsilon/4 + q+q+0 =8q^2+6q+\epsilon/4< R_j
\end{align*}
\item When $i=3$, we have
\begin{align*}
& 8q^2+ \epsilon/2 + \max\left\{0, \floor{\frac{8q^2+6q+2 -
(8q^2 + \epsilon/2)}{2}}\right\}\\
&+ \max\left\{0, \floor{\frac{8q^2+8q+\epsilon/4 -
(8q^2 + \epsilon/2)}{4q}}q\right\} + 2q+1-\epsilon\\
= & 8q^2+ \epsilon/2 + 3q+q+2q+1-\epsilon = 8q^2+6q+1-\epsilon/2< R_j
\end{align*}
\end{itemize}
Therefore, we reach the conclusion.
\end{proof}
Now, we can examine the MILP in Eq.~\eqref{eq:MILP}, when
excluding
Eq.~\eqref{eq:total-segment-upperbound} and
Eq. \eqref{eq:total-upperbound}:
\begin{subequations}
\label{eq:MILP-v4}{\small
\begin{align}
& \mbox{\bf maximize: } \;\;\;\;S_n + \sum_{j=1}^{m} R_j \label{eq:objective-v4}\\
& \mbox{\bf subject to:} \qquad\qquad\qquad&\nonumber\\
& R_j = C_n^j + \sum_{i=1}^{n-1} N_{i,j} \times C_i, \qquad
\forall j=1,\ldots,m \label{eq:R_j-v4}\\
& O_{i,j} \geq 0, \qquad \forall i=1,\ldots,n-1, \forall
j=1,\ldots,m \label{eq:O-positive-v4}\\
& O_{i,j+1} \geq O_{i,j} + N_{i,j}\times T_i - (R_j + S_n^j), \qquad \label{eq:O-distance-v4}
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m-1\\
& 0 \leq N_{i,j} \leq \ceiling{\frac{R_j - O_{i,j}}{T_i}}, \qquad
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m \label{eq:N-bounds-v4}\\
& R_j > rel_{i,j} + \sum_{\ell=1}^{n-1}\max\left\{0,
\floor{\frac{O_{\ell,j}+N_{\ell,j}T_\ell - rel_{i,j}}{T_\ell}}C_\ell\right\} \;\;
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m\label{eq:R>rel-v4}\\
& rel_{i,j} = O_{i,j} + (N_{i,j}-1) T_i \qquad
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m\label{eq:rel-v4}\\
& N_{i,j} \mbox{ is an integer }, \qquad
\forall i=1,\ldots,n-1, \forall j=1,\ldots,m \label{eq:N-integer-v4}
\end{align}}
\end{subequations}
\begin{lemma}
\label{property:upper-bound-pessmistic-v4}
Suppose that $q$ is a positive integer $q \geq 1$ and $0 < \epsilon <
1/q$. For any $j=1,2,\ldots,m$, the setting of
$R_j=8q^2+6q+1+q\epsilon$, $O_{1,j} = 0$, $O_{2,j} = \epsilon/4$,
$O_{3,j} = \epsilon/2$, $N_{1,j} = 4q^2+3q+1$, $N_{2,j} = 2q+2$, and
$N_{3,j} = q+1$ is a feasible solution of the MILP in
Eq.~\eqref{eq:MILP-v4} for ${\bf T}^{MILP}$ when $N_{i,j}=0, O_{i,j} = 0$ for all
$i=4,5,\ldots,n-1$. Therefore, the optimal solution of
Eq.~\eqref{eq:MILP-v4} is at least $S_n + m( 8q^2+6q+1+q\epsilon) =
(m-1)(2q-1) + m( 8q^2+6q+1+q\epsilon)$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma:upper-bound-pessmistic-v3}, we only need
to further verify whether the condition in
Eq.~\eqref{eq:O-distance-v4} holds when $i=1,2,3$. By the definition
$S_n^j = (2q-1)$, we know that $O_{i,j} + N_{i,j} \times T_i -
(R_j+S_n^j) = O_{i,j} + N_{i,j} \times T_i - (8q^2+8q+q\epsilon) <
0$ for $i=1,2,3$. Therefore, the condition in
Eq.~\eqref{eq:O-distance-v4} is by definition satisfied.
\end{proof}
Now, we can reach the conclusion that MILP in Eq.~\eqref{eq:MILP-v4}
can be very far from the exact worst-case response time by the
following theorem.
\begin{theorem}
\label{theorem-unbounded-q}
The result of the MILP in Eq.~\eqref{eq:MILP-v4} for task $\tau_n$
in task set ${\bf T}^{MILP}$ divided by the exact worst-case
response time of task $\tau_n$ is at least $\frac{
m(8q^2+6q+1+q\epsilon) + (m-1)(2q-1)}{16qm+(m-1)(2q-1)}$. The
ratio can become unbounded by the number of computation segments or
the number of tasks when $q$ is sufficiently large.
\end{theorem}
\begin{proof}
This follows directly from Lemmas~\ref{lemma:task-MILP} and
\ref{property:upper-bound-pessmistic-v4}.
\end{proof}
\begin{corollary}
\label{corollary-unbounded-q}
The result of the MILP in Eq.~\eqref{eq:MILP} by excluding the
boundary constraints presented by Eq.~\eqref{eq:total-segment-upperbound} and
Eq. \eqref{eq:total-upperbound} for task $\tau_n$ in task set ${\bf
T}^{MILP}$ divided by the exact worst-case response time of task
$\tau_n$ is at least $\frac{ m(8q^2+6q+1+q\epsilon) +
(m-1)(2q-1)}{16qm+(m-1)(2q-1)}$.
\end{corollary}
\begin{proof}
This follows directly from Theorem~\ref{theorem-unbounded-q}.
\end{proof}
\subsection{Improvements by the Boundary Conditions
Eq.~\eqref{eq:total-segment-upperbound} and Eq.~\eqref{eq:total-upperbound}}
We now discuss the complete MILP in Eq.~\eqref{eq:MILP}. Calculating
$UB_{ss,j}$ for task $\tau_n$ in ${\bf T}^{MILP}$ is rather
straightforward. This can be done by releasing all the jobs together
with $C_n^j$. That is, $UB_{ss,j}$ is the minimum $t | t > 0$ such
that $(m+1)(1-\epsilon)+\ceiling{\frac{t}{2}} +
\ceiling{\frac{t}{4q}}q +\ceiling{\frac{t}{8q}}(2q-1+\epsilon)=t$. By
Property~\ref{property:upper-bound-exact-v2}, we know that $UB_{ss,j}$
is $(m+1)\cdot 8q$.
Calculating $UB_{ss}$ is tricky. However, for task $\tau_n$ in ${\bf
T}^{MILP}$ we can easily conclude that $UB_{ss} \leq m(m+1)\cdot 8q
+ S_n = 8qm(m+1) + (m-1)(2q-1)$. If we can get a very tight upper
bound of $UB_{ss}$, then, there is no need of the MILP. Here is how
$UB_{ss}$ was proposed to be calculated by Nelissen et
al. \cite{ecrts15nelissen}:
\begin{quote}
Nelissen et al. \cite{ecrts15nelissen}:\footnote{The text is
reorganized to use the proper references and notation in this
paper.} \emph{Constraints \eqref{eq:total-segment-upperbound} and
\eqref{eq:total-upperbound} reduce the research space of the problem
by stating that the overall response time of $\tau_n$ and the
response time of each of its execution regions, respectively, cannot
be larger than known upper-bounds computed with simple methods such
as the joint and split methods presented in \cite{bletsas:thesis}.}
\end{quote}
\begin{lemma}
\label{lemma:upper-bound-milp-full}
When $q$ is set to $m$ and $m \geq 2$, $UB_{ss}$ derived from the
joint and split methods presented in \cite{bletsas:thesis} is at
least $8m^2(m+1) + (m-1)(2m-1)$ for ${\bf T}^{MILP}$.
\end{lemma}
\begin{proof}
The joint and split methods presented in \cite[Pages
131-141]{bletsas:thesis} are based on the following concept:
\begin{itemize}
\item A self-suspension interval of task $\tau_n$ can be converted
to computation demand. (joint)
\item A self-suspension interval of task $\tau_n$ can be treated as
self-suspension, by considering their suffered worst-case
interference independently. (split)
\end{itemize}
The following proof is only sketched since this can be easily proved
by a simple observation. If we consider a self-suspension interval
$S_n^j$ as computation (i.e., the joint approach), then, the
additional workload $(2q-1)$ (due to suspension as computation)
increases the worst-case response time by $(2q-1)8q = (2m-1)8m =
16m^2-8m$. If we simply treat these two consecutive computation
segments $C_n^{j-1}$ and $C_n^{j}$ by considering their suffered
worst-case interference independently (i.e., the split approach),
this treatment only increases the worst-case response time by at
most $8m^2+2m-1$. Therefore, the \emph{joint} approach is always
worse than the \emph{split} approach, when $m \geq 2$. This can be
formally proved by starting from $j=1$ to convert any joint
treatment to a split treatment in a stepwise manner.
Hence, $UB_{ss} = m(m+1)\cdot 8q + S_n = 8m^2(m+1) + (m-1)(2m-1)$ by
splitting all the computation segments.
\end{proof}
With the above
discussions, we can reach the following lemma:
\begin{lemma}
\label{lemma:lower-bound-milp-full}
When $q$ is set to $m$ and $m \geq 2$, the objective function of the
MILP in Eq.~\eqref{eq:MILP} for task $\tau_n$ in ${\bf T}^{MILP}$ is
at least $(m-1)(2m-1) + m( 8m^2+6m+1+m\epsilon)$.
\end{lemma}
\begin{proof}
Since $(m+1)\cdot 8q > 8m^2+6m+1+m\epsilon$ when $q$ is set to $m$,
we know that Eq.~\eqref{eq:total-segment-upperbound} is satisfied by
adopting the solution in
Lemma~\ref{property:upper-bound-pessmistic-v4}. Similarly, since
$8qm(m+1) + (m-1)(2q-1) > (m-1)(2m-1) + m( 8m^2+6m+1+m\epsilon)$
when $q$ is set to $m$, we also know that
Eq.~\eqref{eq:total-upperbound} is satisfied by adopting the
solution in
Lemma~\ref{property:upper-bound-pessmistic-v4}. Therefore, the
solution in Lemma~\ref{property:upper-bound-pessmistic-v4} is a
feasible solution of the MILP in Eq.~\eqref{eq:MILP}, in which we reach the conclusion of
this lemma.
\end{proof}
\begin{theorem}
\label{theorem-bounded-M}
The result of the MILP in Eq.~\eqref{eq:MILP} (i.e., the MILP in
\cite{ecrts15nelissen}) for task $\tau_n$ in ${\bf T}^{MILP}$ divided
by the exact worst-case response time of task $\tau_n$ is at least\\
$\frac{ m(8m^2+6m+1+m\epsilon) + (m-1)(2m-1)}{16m^2+(m-1)(2m-1)} \geq
\frac{4m+4}{9}$, when $m \geq 2$.
\end{theorem}
\begin{proof}
This follows directly from Lemmas~\ref{lemma:task-MILP} and
\ref{lemma:lower-bound-milp-full}, and
\begin{align*}
& \frac{ m(8m^2+6m+1+m\epsilon) + (m-1)(2m-1)}{16m^2+(m-1)(2m-1)}
= \frac{8m^3+8m^2-2m+1+ m^2\epsilon}{18m^2-3m+1} \\
= \;\;&\frac{4m+4}{9} + \frac{\frac{4m^2}{3} -\frac{10m}{9} + \frac{5}{9} +
m^2\epsilon}{18m^2-3m+1} > \frac{4m+4}{9}.
\end{align*}
\end{proof}
\section{Conclusions and Discussions}
This report shows that the schedulability analysis for fixed-priority
preemptive scheduling even with only one segmented self-suspending
task as the lowest-priority task is $co{\cal NP}$-hard in the strong
sense. Moreover, we also show that the upper bound on the worst-case
response time by using a mixed-integer linear programming (MILP)
formulation by Nelissen et al. \cite{ecrts15nelissen} can be at least
$\Omega(m)$ times the exact worst-case response time, where $m$ is the
number of computation segments of task $\tau_n$.
Therefore, how to analyze the worst-case response time tightly remains
as an open problem for self-suspending sporadic task systems even with
one self-suspending sporadic task as the lowest-priority task under
fixed-priority preemptive scheduling.
\vspace{0.2in}
{\bf Acknowledgements.} The author would like to thank
Dr. Geoffrey Nelissen from CISTER, ISEP, Polytechnic Institute of
Porto and Prof. Dr. Cong Liu from UT Dallas for their feedbacks on an
earlier version of this report, which help the author improve the presentation
flow and the clarity. This report is supported by DFG, as part of the
Collaborative Research Center SFB876 (http://sfb876.tu-dortmund.de/).
{\small
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,415
|
Q: How to combine routing and templating? So, I've created a webshop system. It all worked perfectly, except for that projects can always be improved. So someone told me that 'Templating' and 'Routing' would be nice improvements. I've written classes for both of them, and they work fine! Now, I do wish to know how to combine them? How can I combine these classes so that data from the routing (to determine the content) needs to be placed inside the template. How would I do this?
Templating class:
class Template
{
private $assignedValues = array();
private $tpl;
/*
** @description Creates one single instance of itself and checks whether the template file exists.
** @param $path [string] This is the path to the template
*/
public function __construct($_path = '')
{
if(!empty($_path)){
if(file_exists($_path)){
$this->tpl = file_get_contents($_path);
}
else{
echo '<b>Template Error:</b> File Inclusion Error.';
}
}
}
/*
** @description Assign a value to a part in the template.
** @param $_searchString [string] This is the part in the template that needs to be replaced
** @param $_replaceString [string] This is the code/text that will replace the part in the template
*/
public function assign($_searchString, $_replaceString)
{
if(!empty($_searchString)){
$this->assignedValues[strtoupper($_searchString)] = $_replaceString;
}
}
/*
** @description Shows the final result of the page.
*/
public function show()
{
if(count($this->assignedValues > 0)){
foreach ($this->assignedValues as $key => $value) {
$this->tpl = str_replace('{'.$key.'}', $value, $this->tpl);
}
}
echo $this->tpl;
}
/*
** @description Quickly load a part of the page
** @param $quickLoad [string] This is the name of the file that will be loaded and assigned
** @param $_searchString [string] This is the part in the template that needs to be replaced
*/
public function quickLoad($_searchString, $part)
{
if(file_exists(INCLUDES.DS.$part.'.php')){
$this->assign($_searchString,include(INCLUDES.DS.$part.'.php'));
}
else{
return "That file does not exist!";
}
}
}
And the routing class:
class Route
{
protected $controller = 'App';
protected $method = 'Call';
protected $params = [];
/*
** @description Loads the classes and methods which are referred to.
*/
public function __construct()
{
$url = $this->parseUrl();
if($this->checkUrl())
{
unset($url[0]);
if(isset($url[1]))
{
if (file_exists('core/classes/' . $url[1] . '.class.php'))
{
$this->controller = $url[1];
unset($url[1]);
}
}
require_once('core/classes/' . $this->controller . '.class.php');
$this->controller = new $this->controller;
if (isset($url[2]))
{
if (method_exists($this->controller, $url[2]))
{
$this->method = $url[2];
unset($url[2]);
}
}
$this->params = $url ? array_values($url) : [];
$this->arrayUrl($this->params);
call_user_func_array([$this->controller, $this->method], $this->params);
}
}
/*
** @description Check whether the URL part contains a string
*/
public function checkUrl($index = '0',$value = 'Route'){
$url = $this->parseUrl();
if($url[$index] == $value){
return true;
}
return false;
}
/*
** @description Splits the url into pieces.
*/
protected function parseUrl()
{
if(isset($_GET['url']))
{
return $url = explode('/', filter_var(rtrim(urldecode($_GET['url']), '/'), FILTER_SANITIZE_URL));
}
}
/*
** @description Sets arrays in routes.
*/
protected function arrayUrl($params = array())
{
foreach($params as $index => $param)
{
if (preg_match('/>/',$param))
{
$newParam = explode('>', $param);
unset($this->params[$index]);
$this->params['fields'][$newParam[0]] = $newParam[1];
}
else{
unset($this->params[$index]);
$this->params[] = $param;
}
}
print_r($this->params);
}
}
I can access my routes by URL's like this:
http://localhost:8080/Webshop/Route/User/logout
With: Class & Method.
This is a simple example that I already use, because no data needs to be showed using this method. It only logs out the user that is logged in. After that, you get redirected to the home page. But how could I use routing for other pages? For example, update some user data without having to create an update file?
EDIT:
This is an example of a page I use now (index.php):
<?php
/*
** @description Includes config.php once so that we can use classes, defines etcetera.
*/
require_once('core/preferences/config.php');
/*
** @description Instanciate new route object.
*/
$route = new Route();
/*
** @description Check if a route isset. When not, continue, else: run route
*/
if(!$route->checkUrl())
{
/*
** @description Instanciate new template object.
*/
$template = new Template(TEMPLATES_PATH .'/camerashop24.tpl.html');
/*
** @description Assign values.
*/
$template->assign('title', 'Home');
$template->assign('root', '');
$template->quickLoad('loginout', 'loginout');
$template->quickLoad('title_block', 'title_block');
$template->quickLoad('cart','cart');
$template->quickLoad('menu', 'menu');
$template->assign('page', 'Home');
$breadcrumbs = new Breadcrumbs($_SERVER["REQUEST_URI"],'');
$template->assign('breadcrumbs', $breadcrumbs->data());
$content = "";
foreach(explode(",",Config::get('settings/front_page_cat')) as $item) {
$content .= "<div id='blue-box' class='blue-box'><h2 style='color: white;'>" . strtoupper($item) . "</h2></div>";
$category = new Category();
$category->find($item);
if($category->exists($item)){
foreach (explode(",",$category->data()->products) as $item) {
$product = new Product($item);
$product->find($item);
$content .= '<a href="Product/' . $product->data()->type . '">' . $product->showProduct($product->data()->type,$product->data()->name,$product->data()->price) . '</a>';
}
}
}
$template->assign('text', $content);
$template->quickLoad('footer','footer');
/*
** @description Showing content.
*/
$template->show();
}
But, what I want as an answer, how can I show the data from the routing (select users profile for example) in this template without having to create a page for it like this one.
A: I think you're gonna have to modify your Routing class. Ideally you want the routing class to give YOU the ability to tell it what controller the route should use. Th controller you select would act as the middle man for processing. Each controller's method would represent a route, something like this:
/**
* URL = www.test.com/index.php/home
* URL = www.test.com/index.php/about
*
* ROUTE Class:
* - should define routes. Example:
*
* $route["/home"] = MainController/home
* $route["/about"] = MainController/about
* $route["/*"] = MainController/*
*
*/
class MainController
{
public function home()
{
$template = new Template(TEMPLATES_PATH . 'home.tpl.html');
$template->show();
}
public function about()
{
$template = new Template(TEMPLATES_PATH . 'about.tpl.html');
$template->show();
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,304
|
Q: AWS Redshift: How often are STL tables purged Does anyone know how often are system tables in Redshift purged? Is Redshift doing it internally?
I can't find anything in the official Redshift guide.
Thanks.
A: Found it after all:
To manage disk space, the STL log tables only retain approximately two to five days of log history, depending on log usage and available disk space. If you want to retain the log data, you will need to periodically copy it to other tables or unload it to Amazon S3.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,955
|
\section{Introduction}
In 1963, N.\ Cabibbo proposed a model for weak hadronic currents based on $SU(3)$ symmetry \cite{cab63}. The model relied on the validity of the $V-A$ theory, the conserved vector current (CVC) hypothesis and, in order to preserve the universality of the weak interactions, introduced $\theta_c$---the Cabibbo angle---which must be determined experimentally. With the advent of the standard model of quarks and leptons and their interactions, in present-day terminology, the hadronic weak current can be expressed directly in terms of quark fields and $\theta_c$ is the rotation angle between the first two generations and the only parameter relevant to hadronic physics involving the light quark sector.
Even from its conception, the Cabibbo model has been a key approach to describing hyperon semileptonic decays (HSD). It should be kept in mind that the model has never been intended to be exact since $SU(3)$ is a broken symmetry. For half a century, the departure of the exact symmetry limit has been scrutinized using several methods in order to find discrepancies between theory and experiment; the leading weak form factors in HSD are the usual probes (see Ref.~\cite{fmg} and references therein for a brief description about some methods used so far).
In particular, according to the Ademollo-Gatto theorem \cite{ag}, the leading vector form factors $f_1$ are protected against $SU(3)$ symmetry-breaking (SB) corrections to lowest order in $m_s-\hat{m}$, where $\hat{m}$ denotes the mean mass of the up and down quarks.
The purpose of the present paper is not to provide evidence of SB in $f_1$ {\it per se} but rather, using general properties shared by the electromagnetic and weak currents as members of an $SU(3)$ octet, to propose two sum rules involving $f_1$ which are valid in the symmetry limit. Departures from this limit are then evaluated using the $1/N_c$ expansion of QCD so the modified sum rules are provided to second order in SB. The analysis allows us to identify the flavor representations responsible for SB in the sum rules.
This paper is organized as follows. In Sec.~\ref{sec:srsu3} the two sum rules involving leading vector form factors are derived by taking the matrix element of the vector current between the baryons within a $V=1$ multiplet. The resulting expressions are valid in the exact $SU(3)$ symmetry limit. In the presence of SB, two modified expressions are then provided. In Sec.~\ref{sec:add} a brief discussion about the procedure used by M.\ Ademollo and R.\ Gatto is provided, followed by a survey on the $1/N_c$ expansion of QCD in Sec.~\ref{sec:sur}. In Sec.~\ref{sec:bff} the $1/N_c$ expansion for the baryon vector current including first- and second-order SB is constructed. Restrictions imposed in the nonrenormalization of the baryon electric charge fix several operator coefficients which also participate in $f_1$. Consequently, a few of them survive and are the ones which produce SB to second order. The sum rules are tested with some analytical results obtained in the framework of (heavy) baryon chiral perturbation theory and the results are encouraging. Some closing remarks are listed in Sec.~\ref{sec:sum}.
\section{\label{sec:srsu3}Sum rules for baryon vector form factors}
Motivated by the success of the Gell-Mann--Okubo formula for baryon octet masses, T.N.\ Pham observed that the matrix elements of the $V=1$ $V$-spin multiplet can be related to each other in the exact flavor $SU(3)$ symmetry limit \cite{pham}.
Starting from the two $I$-spin relations,
\begin{subequations}
\begin{equation}
\langle I=1,I_3=0 | \overline{u} \Gamma^n d | I=1,I_3=1 \rangle = -\langle I=1,I_3=-1 | \overline{u} \Gamma^n d | I=1,I_3=0 \rangle,
\end{equation}
and
\begin{equation}
\langle I=0,I_3=0 | \overline{u} \Gamma^n d | I=1,I_3=1 \rangle = \langle I=1,I_3=-1 | \overline{u} \Gamma^n d | I=0,I_3=0 \rangle,
\end{equation}
\end{subequations}
the rotated $V$-spin versions of the above relations read
\begin{subequations}
\label{eq:vspin}
\begin{equation}
\langle V=1,V_3=0| \overline{u} \Gamma^n s|V=1,V_3=1\rangle = -\langle V=1,V_3=-1| \overline{u} \Gamma^n s|V=1,V_3=0\rangle,
\end{equation}
and
\begin{equation}
\langle V=0,V_3=0| \overline{u} \Gamma^n s|V=1,V_3=1\rangle = \langle V=1,V_3=-1| \overline{u} \Gamma^n s|V=0,V_3=0\rangle,
\end{equation}
\end{subequations}
respectively, where the bilinear forms $\overline{q}_1 \Gamma^n q_2$ are given in terms of quark fields $q_i$ and the matrices $\Gamma_{\alpha\beta}^n$ are the ones which appear in applications of the Dirac theory, namely, $\Gamma^S=\openone$, $\Gamma_\mu^V=\gamma_\mu$, $\Gamma_\mu^A = \gamma_5\gamma_\mu$ and so on \cite{bd}.
The decomposition of the $SU(3)$ octet into eigenfunctions of $V^2$ read
\begin{eqnarray}
& & |\Xi^-\rangle = |V=1,V_3=1 \rangle, \\
& & \left| \frac12 \Sigma^0 + \frac{\sqrt{3}}{2} \Lambda \right\rangle = |V=1,V_3=0 \rangle, \\
& & |p\rangle = |V=1,V_3=-1 \rangle, \\
& & \left| \frac{\sqrt{3}}{2} \Sigma^0 - \frac12 \Lambda \right \rangle = |V=0,V_3=0 \rangle.
\end{eqnarray}
Relations (\ref{eq:vspin}) apply to the matrix elements of any $SU(3)$ octet $\Delta S=1$ operator. For instance, the GMO relation is straightforwardly obtained from Eq.~(\ref{eq:vspin}a) by relating $\langle B_2|\partial_\mu(\overline{u} \gamma_\mu s)|B_1\rangle$ to $[f_1^{SU(3)}]_{B_1B_2}(M_{B_2}-M_{B_1})$, where $f_1^{SU(3)}$ stands for the $SU(3)$ symmetric value of the leading vector form factor $f_1$ at zero recoil and $M_{B_i}$ is the mass of baryon $B_i$ \cite{pham}. Similarly, for $\Gamma^n=\Gamma_\mu^A$ some interesting relations for the axial-vector to vector form factor ratios $g_1/f_1$ can also be found \cite{pham}.
The analysis can be extended to the vector current by using $\Gamma^n=\Gamma_\mu^V$. As a result, two simple although nontrivial expressions are obtained, namely,
\begin{subequations}
\begin{equation}
\left \langle \frac12 \Sigma^0 + \frac{\sqrt{3}}{2} \Lambda \Big| \overline{u} \gamma_\mu s \Big| \Xi^- \right\rangle = -\left \langle p \Big| \overline{u} \gamma_\mu s \Big| \frac12 \Sigma^0 - \frac{\sqrt{3}}{2} \Lambda \right\rangle,
\end{equation}
and
\begin{equation}
\left \langle \frac{\sqrt{3}}{2} \Sigma^0 - \frac12 \Lambda \Big| \overline{u} \gamma_\mu s \Big| \Xi^- \right \rangle = \left \langle p \Big| \overline{u} \gamma_\mu s \Big| \frac{\sqrt{3}}{2} \Sigma^0 + \frac12 \Lambda \right\rangle.
\end{equation}
\end{subequations}
Thus, in the limit of exact $SU(3)$ symmetry and neglecting isospin breaking, there are two relations among vector form factors, namely,
\begin{subequations}
\label{eq:sumr1}
\begin{equation}
\left[ f_1^{SU(3)} \right]_{\Xi^-\Sigma^0} + \sqrt{3} \left[ f_1^{SU(3)} \right]_{\Xi^-\Lambda} + \frac{1}{\sqrt{2}} \left[ f_1^{SU(3)} \right]_{\Sigma^-n} + \sqrt{3} \left[ f_1^{SU(3)} \right]_{\Lambda p} = 0,
\end{equation}
and
\begin{equation}
\sqrt{3} \left[ f_1^{SU(3)} \right]_{\Xi^-\Sigma^0} - \left[ f_1^{SU(3)} \right]_{\Xi^-\Lambda} - \sqrt \frac32 \left[ f_1^{SU(3)} \right]_{\Sigma^-n} + \left[ f_1^{SU(3)} \right]_{\Lambda p} = 0.
\end{equation}
\end{subequations}
Violations to relations (\ref{eq:sumr1}) are expected to occur due to flavor $SU(3)$ symmetry-breaking effects.
After some rearrangements, relations (\ref{eq:sumr1}) can be expressed, in the presence of SB, as
\begin{subequations}
\label{eq:srules}
\begin{equation}
\frac14 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Sigma^0} + \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Lambda} - \frac14 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Sigma^-n} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Lambda p} = \delta_1^{\mathrm{SB}},
\end{equation}
and
\begin{equation}
\frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Sigma^0} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Lambda} + \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Sigma^-n} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Lambda p} = \delta_2^{\mathrm{SB}},
\end{equation}
\end{subequations}
where $\delta_k^{\mathrm{SB}}$ arise from SB and are formally of order $\mathcal{O}(\epsilon^2)$; hereafter, $\epsilon$ will be referred to as a measure of SB. The origin of these $\delta_k^{\mathrm{SB}}$ corrections will be explored in the context of the $1/N_c$ expansion of QCD.
\section{\label{sec:add}A brief review about the Ademollo-Gatto result}
By assuming that the vector currents and the electromagnetic current are members of the same unitary octet and that the breaking of the unitary symmetry behaves as the eighth component of an octet, M.~Ademollo and R.~Gatto set up an important theorem on the nonrenormalization for the strangeness-violating vector currents \cite{ag}.
Following Ademollo and Gatto, the $a$th component of the vector current $\mathcal{J}^a$ to first order in SB can be written as\footnote{The authors of Ref.~\cite{ag} inadvertently omitted to subtract singlet and octet pieces off in some terms. Conclusions remain unchanged, though.}
\begin{eqnarray}
\mathcal{J}^a + \epsilon \delta \mathcal{J}^a & = & a_0 \mathrm{Tr} (\overline{B}B\lambda^a) + b_0 \mathrm{Tr} (\overline{B} \lambda^a B) + \epsilon a \left[ \mathrm{Tr} (\overline{B}B \{\lambda^a,\lambda^8\}) - \frac18 \delta^{a8} \mathrm{Tr} (\overline{B}B \{\lambda^c,\lambda^c\}) \right] \nonumber \\
& & \mbox{} + \epsilon b \left[ \mathrm{Tr} (\overline{B} \{\lambda^a,\lambda^8\} B) - \frac18 \delta^{a8} \mathrm{Tr} (\overline{B} \{\lambda^c,\lambda^c\} B) \right] \nonumber \\
& & \mbox{} + \epsilon c [\mathrm{Tr} (\overline{B} \lambda^a B \lambda^8) - \mathrm{Tr} (\overline{B} \lambda^8 B \lambda^a)] + \epsilon g \mathrm{Tr} (\overline{B}B) \mathrm{Tr}(\lambda^a \lambda^8) \nonumber \\
& & \mbox{} + \epsilon h \left[ \mathrm{Tr} (\overline{B} \lambda^a B \lambda^8) + \mathrm{Tr} (\overline{B} \lambda^8 B \lambda^a) - \frac14 \delta^{a8} \mathrm{Tr} (\overline{B} \lambda^c B \lambda^c) - \frac65 d^{a8c}d^{cgh} \mathrm{Tr} (\overline{B} \lambda^g B \lambda^h) \right], \label{eq:vcag}
\end{eqnarray}
where $B$ represents the baryon matrix, $\lambda^a$ denote the Gell-Mann matrices and $a_0$, $b_0$, \ldots, $h$ are coupling constants. The parameter $\epsilon$ is introduced to keep track of the number of times the perturbation enters; at the end of the calculation $\epsilon$ can be set to one without any loss of generality.
The electromagnetic current is defined in the usual way as
\begin{equation}
\mathcal{J}_{\mathrm{em}} = \mathcal{J}^Q \equiv \mathcal{J}^3 + \frac{1}{\sqrt{3}} \mathcal{J}^8,
\end{equation}
so the baryon charges, including first-order SB, are readily obtained from Eq.~(\ref{eq:vcag}). They read
\begin{equation}
Q_n + \epsilon \delta Q_n = -\frac23 a_0 - \frac23 b_0 + \frac{4}{3 \sqrt{3}} \epsilon a - \frac{8}{3 \sqrt{3}} \epsilon b + \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g + \frac{\sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_p + \epsilon \delta Q_p = -\frac23 a_0 + \frac43 b_0 + \frac{4}{3 \sqrt{3}} \epsilon a + \frac{4}{3 \sqrt{3}} \epsilon b - \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g - \frac{7\sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_{\Sigma^+} + \epsilon \delta Q_{\Sigma^+} = - \frac23 a_0 + \frac43 b_0 - \frac{8}{3 \sqrt{3}} \epsilon a + \frac{4}{3 \sqrt{3}} \epsilon b + \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g + \frac{\sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_{\Sigma^0} + \epsilon \delta Q_{\Sigma^0} = \frac13 a_0 + \frac13 b_0 - \frac{2}{3 \sqrt{3}} \epsilon a - \frac{2}{3 \sqrt{3}} \epsilon b + \frac{2}{\sqrt{3}} \epsilon g + \frac{\sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_{\Sigma^-} + \epsilon \delta Q_{\Sigma^-} = \frac43 a_0 - \frac23 b_0 + \frac{4}{3 \sqrt{3}} \epsilon a - \frac{8}{3 \sqrt{3}} \epsilon b - \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g + \frac{\sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_{\Xi^-} + \epsilon \delta Q_{\Xi^-} = \frac43 a_0 - \frac23 b_0 + \frac{4}{3 \sqrt{3}} \epsilon a + \frac{4}{3 \sqrt{3}} \epsilon b + \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g - \frac{7 \sqrt{3}}{10} \epsilon h,
\end{equation}
\begin{equation}
Q_{\Xi^0} + \epsilon \delta Q_{\Xi^0} = -\frac23 a_0 - \frac23 b_0 - \frac{8}{3 \sqrt{3}} \epsilon a + \frac{4}{3 \sqrt{3}} \epsilon b - \frac{2}{\sqrt{3}} \epsilon c + \frac{2}{\sqrt{3}} \epsilon g + \frac{\sqrt{3}}{10} \epsilon h,
\end{equation}
and
\begin{equation}
Q_\Lambda + \epsilon \delta Q_\Lambda = -\frac13 a_0 - \frac13 b_0 + \frac{2}{\sqrt{3}} \epsilon a + \frac{2}{\sqrt{3}} \epsilon b + \frac{2}{\sqrt{3}} \epsilon g + \frac{9\sqrt{3}}{10} \epsilon h,
\end{equation}
along with the isospin relation
\begin{equation}
\frac12 (Q_{\Sigma^+} + Q_{\Sigma^-}) = Q_{\Sigma^0}. \label{eq:isos}
\end{equation}
Solving the system of linear equations yields
\begin{equation}
a_0 = -\frac12, \quad b_0 = \frac12, \quad a = b = c = g = h = 0,
\end{equation}
which explicitly shows that first-order SB corrections to the electric current vanish. For the $|\Delta S|=1$ weak vector currents, the flavor index is $a = 4\pm i5$; according to the original assumption, first-order SB corrections also vanish for the vector current. This is in essence the celebrated result discovered by Ademollo and Gatto \cite{ag}.
\section{\label{sec:sur}A survey on the $1/N_c$ expansion of QCD}
In the large-$N_c$ limit, the baryon sector has a contracted $SU(2N_f)$ spin-flavor symmetry, where $N_f$ is the number of light quark flavors \cite{dm315,gs}. Corrections to the large-$N_c$ limit can be given in terms of
$1/N_c$-suppressed operators with well-defined spin-flavor transformation properties \cite{gs}; this yields the so-called $1/N_c$ expansion of QCD. For $N_f=3$, the lowest lying baryon states fall into a representation of the
spin-flavor group $SU(6)$. For $N_c=3$, the $\mathbf{56}$ dimensional representation is involved.
The $1/N_c$ expansion of any QCD operator transforming according to a given $SU(2)\times SU(N_f)$ representation can be written in terms of $n$-body operators $\mathcal{O}_n$ as \cite{djm95}
\begin{equation}
\mathcal{O}_\mathrm{QCD} = \sum_n c_{(n)} \frac{1}{N_c^{n-1}} \mathcal{O}_n,
\end{equation}
where the operator coefficients $c_{(n)}$ have power series expansions in $1/N_c$ beginning at order unity and the $\mathcal{O}_n$ are polynomials in the spin-flavor generators $J^k$, $T^c$, and $G^{kc}$, which can be written as 1-body quark operators acting on the $N_c$-quark baryon states, namely,
\begin{subequations}
\label{eq:su6gen}
\begin{eqnarray}
J^k & = & \sum_\alpha^{N_c} q_\alpha^\dagger \left(\frac{\sigma^k}{2}\otimes \openone \right) q_\alpha, \\
T^c & = & \sum_\alpha^{N_c} q_\alpha^\dagger \left(\openone \otimes \frac{\lambda^c}{2} \right) q_\alpha, \\
G^{kc} & = & \sum_\alpha^{N_c} q_\alpha^\dagger \left(\frac{\sigma^k}{2}\otimes \frac{\lambda^c}{2} \right) q_\alpha,
\end{eqnarray}
\end{subequations}
where $q_\alpha^\dagger$ and $q_\alpha$ are operators that create and annihilate states in the fundamental representation of $SU(6)$ and $\sigma^k$ and $\lambda^c$ are the Pauli spin and Gell-Mann flavor matrices, respectively. Because the baryon matrix elements of the spin-flavor generators (\ref{eq:su6gen}) can be taken as the values in the nonrelativistic quark model, this convention is usually referred to as the quark representation \cite{djm95}.
\section{\label{sec:bff}The baryon vector form factor in the $1/N_c$ expansion}
Now, let $V^{0c}$ denote the flavor octet baryon charge \cite{jen96},
\begin{equation}
V^{0c} = \left\langle B_2\left|\left(\overline{q} \gamma^0 \frac{\lambda^c}{2} q\right)_{\mathrm{QCD}}\right|B_1 \right\rangle,
\end{equation}
where the subscript QCD indicates that the quark fields are QCD quark fields rather than the quark creation and annihilation operators of the quark representation. $V^{0c}$ is spin-0 and a flavor octet, so it transforms as $(0,\mathbf{8})$ under $SU(2)\times SU(3)$; its matrix elements between $SU(6)$ symmetric states give the values of the leading vector form factor $f_1$.
On general grounds, flavor SB in QCD is due to the strange quark mass $m_s$ and transforms as a flavor octet \cite{djm95}. To linear order in SB, the correction is obtained from the tensor product $(0,\mathbf{8})\times (0,\mathbf{8})$ so that the $SU͑(2͒)\times SU(3)$ representations involved are $(0,\mathbf{1})$, $(0,\mathbf{8})$, $(0,\mathbf{10+\overline{10}})$ and $(0,\mathbf{27})$ \cite{jl,rfm98}.
Let $V^c + \epsilon\delta V^c$ be the $(0,\mathbf{8})$ operator containing the most general first-order SB. Its $1/N_c$ expansion reads
\begin{eqnarray}
V^c + \epsilon \delta V^c & = & c_{(1)}^\mathbf{8} T^c + c_{(2)}^\mathbf{8} \frac{1}{N_c} \{J^r,G^{rc}\} + \epsilon N_c a_{(0)}^\mathbf{1} \delta^{c8} + \epsilon a_{(1)}^\mathbf{8} d^{c8e} T^e + \epsilon a_{(2)}^\mathbf{8} \frac{1}{N_c} d^{c8e} \{J^r,G^{re}\} \nonumber \\
& & \mbox{} + \epsilon a_{(3)}^\mathbf{10+\overline{10}} \frac{1}{N_c^2} \left( \{T^c,\{J^r,G^{r8}\}\} - \{T^8,\{J^r,G^{rc}\}\} \right) \nonumber \\
& & \mbox{} + \epsilon a_{(2)}^\mathbf{27} \frac{1}{N_c} \left[ \{T^c,T^8\} - \frac{N_f-2}{2N_f(N_f^2-1)} N_c(N_c+2N_f) \delta^{c8} - \frac{2}{N_f^2-1} \delta^{c8} J^2 - \frac{N_f-4}{N_f^2-4}(N_c + N_f) d^{c8e} T^e \right. \nonumber \\
& & \mbox{\hglue1.9truecm} \left. - \frac{2N_f}{N_f^2-4} d^{c8e} \{J^r,G^{re}\} \right] \nonumber \\
& & \mbox{} + \epsilon a_{(3)}^\mathbf{27} \frac{1}{N_c^2} \left[ \{T^c,\{J^r,G^{r8}\}\} + \{T^8,\{J^r,G^{rc}\}\} - \frac{4}{N_f(N_f+1)}(N_c+N_f) \delta^{c8} J^2 \right. \nonumber \\
& & \mbox{\hglue1.9truecm} \left. - \frac{2}{N_f+2}(N_c + N_f) d^{c8e} \{J^r,G^{re}\} - \frac{2}{N_f+2} d^{c8e} \{J^2,T^e \} \right]. \label{eq:qfirst}
\end{eqnarray}
A few remarks are in order here. First, notice that the series has been truncated at the physical value $N_c=3$ so up to three-body operators should be retained. Second, the flavor singlet and octet components of the $\mathbf{27}$ operators have been explicitly subtracted off, so that only the truly flavor-$\mathbf{27}$ components remain. Third, the operator coefficients $c_{(n)}^\mathbf{8}$ come along with $n$-body operators given in the exact $SU(3)$ limit whereas the operator coefficients $a_{(n)}^\mathbf{rep}$ come along with $n$-body operators given in the representation $\mathbf{rep}$ which explicitly breaks flavor symmetry. And last but not least, higher-order operators are generated from the already existing ones by anticommuting with $J^2$. There is no need to include them because their contributions to the expansion can be accounted for by redefining the operator coefficients. Therefore, the series stands the way it is.
The matrix elements of the operator $V^Q+\epsilon \delta V^Q$ between $SU(6)$ symmetric states give the actual values of baryon charges $Q_B$ including first-order flavor SB. At the physical values $N_f=N_c=3$, the baryon charges read
\begin{equation}
Q_n + \epsilon \delta Q_n = -\frac13 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1}-\frac{1}{\sqrt{3}} \epsilon a_{(1)}^\mathbf{8} - \frac{1}{2 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} + \frac{1}{3\sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_p + \epsilon \delta Q_p = c_{(1)}^\mathbf{8} + \frac12 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} + \frac{1}{3 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} - \frac{1}{3 \sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} + \frac{7}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} + \frac{7}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_{\Sigma^+} + \epsilon \delta Q_{\Sigma^+} = c_{(1)}^\mathbf{8}+\frac12 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} + \frac{1}{\sqrt{3}} \epsilon a_{(1)}^\mathbf{8} + \frac{1}{6 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} + \frac{1}{3\sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_{\Sigma^-} + \epsilon \delta Q_{\Sigma^-} = -c_{(1)}^\mathbf{8} - \frac16 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} - \frac{1}{\sqrt{3}} \epsilon a_{(1)}^\mathbf{8} - \frac{1}{2 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} - \frac{1}{3\sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_{\Sigma^0} + \epsilon \delta Q_{\Sigma^0} = \frac16 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} - \frac{1}{6 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} - \frac{1}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{30\sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_{\Xi^-} + \epsilon \delta Q_{\Xi^-} = -c_{(1)}^\mathbf{8} - \frac16 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} + \frac{1}{3 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} + \frac{1}{3 \sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} + \frac{7}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} + \frac{7}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_{\Xi^0} + \epsilon \delta Q_{\Xi^0} = -\frac13 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} + \frac{1}{\sqrt{3}} \epsilon a_{(1)}^\mathbf{8} + \frac{1}{6 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} - \frac{1}{3\sqrt{3}} \epsilon a_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{20 \sqrt{3}} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{30 \sqrt{3}} \epsilon a_{(3)}^\mathbf{27},
\end{equation}
\begin{equation}
Q_\Lambda + \epsilon \delta Q_\Lambda = -\frac16 c_{(2)}^\mathbf{8} + \sqrt{3} \epsilon a_{(0)}^\mathbf{1} + \frac{1}{6 \sqrt{3}} \epsilon a_{(2)}^\mathbf{8} - \frac{3}{20} \sqrt{3} \epsilon a_{(2)}^\mathbf{27} - \frac{1}{10} \sqrt{3} \epsilon a_{(3)}^\mathbf{27}.
\end{equation}
Although the isospin relation (\ref{eq:isos}) reduces by one the number of independent equations, it is straightforward to notice that the $a_{(n)}^\mathbf{27}$ operator coefficients are not independent. Indeed, a new coefficient $x_{(2)}^\mathbf{27}$ can be defined as
\begin{equation}
x_{(2)}^\mathbf{27} = a_{(2)}^\mathbf{27} + \frac23 a_{(3)}^\mathbf{27},
\end{equation}
so that the number of operator coefficients is also reduced by one.
Solving the system of linear equations yields
\begin{equation}
c_{(1)}^\mathbf{8} = 1, \qquad c_{(2)}^\mathbf{8} = 0, \qquad a_{(0)}^\mathbf{1} = 0, \qquad a_{(1)}^\mathbf{8} = 0, \qquad a_{(2)}^\mathbf{8} = 0, \qquad a_{(3)}^\mathbf{10+\overline{10}} = 0, \qquad x_{(2)}^\mathbf{27} = 0,
\end{equation}
which nicely reproduces the Ademollo-Gatto result. The coupling constants introduced in Eq.~(\ref{eq:vcag}) are related to the coefficients of the $1/N_c$ expansion (\ref{eq:qfirst}) at $N_c=3$ by
\begin{subequations}
\label{eq:coeffrel}
\begin{eqnarray}
& & a_0 = -\frac12 c_{(1)}^\mathbf{8} + \frac{1}{12} c_{(2)}^\mathbf{8}, \\
& & b_0 = \frac12 c_{(1)}^\mathbf{8} + \frac{5}{12} c_{(2)}^\mathbf{8}, \\
& & a = -\frac14 a_{(1)}^\mathbf{8} + \frac{1}{24} a_{(2)}^\mathbf{8}, \\
& & b = \frac14 a_{(1)}^\mathbf{8} + \frac{5}{24} a_{(2)}^\mathbf{8}, \\
& & c = \frac16 a_{(3)}^\mathbf{10+\overline{10}}, \\
& & g = \frac32 a_{(0)}^\mathbf{1}, \\
& & h = - \frac16 x_{(2)}^\mathbf{27},
\end{eqnarray}
\end{subequations}
and they correspond to well-defined flavor representations.
Next, SB at second order can be incorporated into $V^c$. The $1/N_c$ expansion of this contribution reads
\begin{eqnarray}
\epsilon^2 \delta V^c & = & \epsilon^2 b_{(0)}^{\mathbf{1}}N_c d^{c88} \openone + \epsilon^2 b_{(1)}^{\mathbf{8}} \delta^{c8} T^8 + \epsilon^2 e_{(1)}^{\mathbf{8}} f^{c8e} f^{8eg} T^g + \epsilon^2 g_{(1)}^{\mathbf{8}} d^{c8e} d^{8eg} T^g \nonumber \\
& & \mbox{} + \epsilon^2 h_{(1)}^{\mathbf{8}} (i f^{ceg} d^{8e8} T^g-i d^{ce8} f^{8eg} T^g-i f^{c8e} d^{eg8} T^g) + \epsilon^2 b_{(2)}^{\mathbf{8}} \frac{1}{N_c} \delta^{c8} \{J^r,G^{r8}\} + \epsilon^2 e_{(2)}^{\mathbf{8}} \frac{1}{N_c} f^{c8e} f^{8eg} \{J^r,G^{rg}\} \nonumber \\
& & \mbox{} + \epsilon^2 g_{(2)}^{\mathbf{8}} \frac{1}{N_c} d^{c8e} d^{8eg} \{J^r,G^{rg}\} + \epsilon^2 h_{(2)}^{\mathbf{8}} \frac{1}{N_c} (i f^{ceg} d^{8e8} - i d^{ce8} f^{8eg} - i f^{c8e} d^{eg8}) \{J^r,G^{rg}\} \nonumber \\
& & \mbox{} + \epsilon^2 b_{(2)}^{\mathbf{10+\overline{10}}}\frac{1}{N_c^2} d^{c8e} \left(\{T^e,\{J^r,G^{r8}\}\} - \{T^8,\{J^r ,G^{re}\}\} \right) \nonumber \\
& & \mbox{} + \epsilon^2 b_{(2)}^{\mathbf{27}} \frac{1}{N_c} \left[ d^{c8e}\{T^e,T^8\} - \frac{N_f-4}{N_f^2-4}(N_c+N_f) d^{c8e} d^{8eg} T^g - \frac{2N_f}{N_f^2-4} d^{c8e} d^{8eg} \{J^r ,G^{rg}\} \right] \nonumber \\
& & \mbox{} + \epsilon^2 b_{(3)}^{\mathbf{27}} \frac{1}{N_c^2} \left[ d^{c8e} \left( \{T^e,\{J^r,G^{r8}\}\} + \{T^8,\{J^r ,G^{re}\}\} \right) - \frac{2}{N_f+2}(N_c+N_f) d^{c8e} d^{8eg} \{J^r ,G^{rg}\} \right. \nonumber \\
& & \mbox{\hglue2.0truecm} \left. - \frac{2}{N_f+2} d^{c8e} d^{8eg} \{J^2,T^g\} \right] \nonumber \\
& & \mbox{} + \epsilon^2 b_{(3)}^{\mathbf{64}} \frac{1}{N_c^2} \left[ \{T^c,\{T^8,T^8\}\} -\frac{N_f-2}{N_f(N_f^2-1)}N_c(N_c+2N_f) \delta^{88} T^c - \frac{1}{2} f^{c8e}f^{8eg} T^g \right. \nonumber \\
& & \mbox{\hglue2.0truecm} - \frac{N_f-4}{2(N_f^2-4)}(N_c+N_f) d^{c8e}\{T^e,T^8\} - \frac{N_f-4}{2(N_f^2-4)}(N_c+N_f) d^{88e} \{T^c,T^e\} - \frac{2}{N_f^2-1} \delta^{88} \{J^2,T^c\} \nonumber \\
& & \mbox{\hglue2.0truecm} \left. - \frac{N_f}{N_f^2-4} d^{c8e}\{T^8,\{J^r,G^{re}\}\} - \frac{N_f}{N_f^2-4} d^{88e}\{T^c,\{J^r,G^{re}\}\} \right]. \label{eq:qsecond}
\end{eqnarray}
The matrix elements of the operator $\epsilon^2 \delta V^Q$ between $SU(6)$ symmetric states gives second-order SB effects to the baryon octet electric charges. For neutron, these corrections read
\begin{equation}
\epsilon^2 \delta Q_n = -\epsilon^2 b_{(0)}^\mathbf{1} + \frac12 \epsilon^2 b_{(1)}^\mathbf{8} + \frac{1}{12} \epsilon^2 b_{(2)}^\mathbf{8} + \frac13 \epsilon^2 g_{(1)}^\mathbf{8} + \frac16 \epsilon^2 g_{(2)}^\mathbf{8} - \frac19 \epsilon^2 b_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{15} \epsilon^2 b_{(2)}^\mathbf{27} - \frac{2}{45} \epsilon^2 b_{(3)}^\mathbf{27} + \frac{13}{120} \epsilon^2 b_{(3)}^\mathbf{64},
\end{equation}
and similar expressions are found for the rest of the baryon charges. Again, the operator coefficients associated with the $\mathbf{27}$ representation are not independent, so a new coefficient $y_{(2)}^\mathbf{27}$ can be defined as
\begin{equation}
y_{(2)}^\mathbf{27} = b_{(2)}^\mathbf{27} + \frac23 b_{(3)}^\mathbf{27}.
\end{equation}
Only 8 out of 12 operator coefficients of expansion (\ref{eq:qsecond}) are involved in the seven expressions for the baryon charges. By using the important property that the electric charge remains unrenormalized to all orders in perturbation theory, the system can be solved in terms of one coefficient, namely,
\begin{subequations}
\label{eq:rest}
\begin{eqnarray}
& & b_{(0)}^\mathbf{1} = b_{(2)}^\mathbf{8} = g_{(2)}^\mathbf{8} = y_{(2)}^\mathbf{27} = 0, \\
& & g_{(1)}^\mathbf{8} = -\frac{5}{24} b_{(3)}^\mathbf{10+\overline{10}}, \\
& & b_{(1)}^\mathbf{8} = -\frac{13}{72} b_{(3)}^\mathbf{10+\overline{10}}, \\
& & b_{(3)}^\mathbf{64} = \frac52 b_{(3)}^\mathbf{10+\overline{10}}.
\end{eqnarray}
\end{subequations}
Thus, under the working assumptions, the baryon vector current is given to second-order in flavor SB in terms of, in principle, five nontrivial operator coefficients.
The matrix elements of $V^{4\pm i5}+\epsilon^2 \delta V^{4\pm i5}$ between $SU(6)$ baryon states yields the actual expressions for the leading vector form factors. For the observed processes, one has
\begin{equation}
\left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Lambda p} = 1 + \frac34 e_{(1)}^\mathbf{8} + \frac{1}{12} g_{(1)}^\mathbf{8} - \frac12 h_{(1)}^\mathbf{8} + \frac38 e_{(2)}^\mathbf{8} + \frac{1}{24} g_{(2)}^\mathbf{8} - \frac14 h_{(2)}^\mathbf{8} + \frac{1}{18} b_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{10} y_{(2)}^\mathbf{27}, \label{eq:r1}
\end{equation}
\begin{equation}
\left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Sigma^-n} = 1 + \frac34 e_{(1)}^\mathbf{8} + \frac{1}{12} g_{(1)}^\mathbf{8} - \frac12 h_{(1)}^\mathbf{8} - \frac18 e_{(2)}^\mathbf{8} - \frac{1}{72} g_{(2)}^\mathbf{8} + \frac{1}{12} h_{(2)}^\mathbf{8} - \frac{1}{18} b_{(3)}^\mathbf{10+\overline{10}} - \frac{1}{30} y_{(2)}^\mathbf{27} + \frac{1}{30} b_{(3)}^\mathbf{64}, \label{eq:r2}
\end{equation}
\begin{equation}
\left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Lambda} = 1 + \frac34 e_{(1)}^\mathbf{8} + \frac{1}{12} g_{(1)}^\mathbf{8} - \frac12 h_{(1)}^\mathbf{8} + \frac18 e_{(2)}^\mathbf{8} + \frac{1}{72} g_{(2)}^\mathbf{8} - \frac{1}{12} h_{(2)}^\mathbf{8} + \frac{1}{18} b_{(3)}^\mathbf{10+\overline{10}} + \frac{1}{10} y_{(2)}^\mathbf{27}, \label{eq:r3}
\end{equation}
\begin{equation}
\left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Sigma^0} = 1 + \frac34 e_{(1)}^\mathbf{8} + \frac{1}{12} g_{(1)}^\mathbf{8} - \frac12 h_{(1)}^\mathbf{8} + \frac58 e_{(2)}^\mathbf{8} + \frac{5}{72} g_{(2)}^\mathbf{8} - \frac{5}{12} h_{(2)}^\mathbf{8} - \frac{1}{18} b_{(3)}^\mathbf{10+\overline{10}} + \frac{1}{30} y_{(2)}^\mathbf{27} + \frac{1}{30} b_{(3)}^\mathbf{64}. \label{eq:r4}
\end{equation}
Expressions (\ref{eq:r1})-(\ref{eq:r4}) are the most general ones which account for second-order SB effects in the hyperon semileptonic decay form factors. Substituting them into sum rules (\ref{eq:srules}) yields
\begin{equation}
\delta_1^{\mathrm{SB}} = \frac16 y_{(2)}^\mathbf{27}, \qquad \qquad \delta_2^{\mathrm{SB}} = -\frac16 b_{(3)}^\mathbf{10+\overline{10}} + \frac{1}{20} b_{(3)}^\mathbf{64}.
\end{equation}
Under restrictions (\ref{eq:rest}), the final form of sum rules (\ref{eq:srules}) become
\begin{subequations}
\label{eq:srulesfinal}
\begin{equation}
\frac14 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Sigma^0} + \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Lambda} - \frac14 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Sigma^-n} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Lambda p} = 0,
\end{equation}
and
\begin{equation}
\frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Sigma^0} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Xi^-\Lambda} + \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Sigma^-n} - \frac34 \left[ \frac{f_1}{f_1^{SU(3)}} \right]_{\Lambda p} = -\frac{1}{24} b_{(3)}^\mathbf{10+\overline{10}}.
\end{equation}
\end{subequations}
These findings are remarkable. Sum rule (\ref{eq:srulesfinal}a) is valid in the presence of second-order SB whereas (\ref{eq:srulesfinal}b) gets corrections mainly from the $\mathbf{10}+\overline{\mathbf{10}}$ flavor representation.
Different methods have been used to evaluate SB effects to vector form factors. One of them is baryon chiral perturbation theory (BChPT) in the works by Krause \cite{krause} and Anderson and Luty \cite{and}. The former presented the calculation in relativistic BChPT to order $\mathcal{O}(p^2)$ in the chiral expansion whereas the latter used heavy baryon chiral perturbation theory (HBChPT) up to (partially complete) order $\mathcal{O}(p^3)$. Both calculations included as baryonic degrees of freedom only the spin-1/2 octet. Later on, Villadoro \cite{villa} used HBChPT with both octet and decuplet baryon degrees of freedom and included (partially) up order $\mathcal{O}(p^3)$ corrections corresponding to subleading in $1/M_B$ terms. In the context of covariant BChPT, two schemes are representative. The first one used infrared regularization in the work by Lacour, Kubis, and Meissner \cite{meiss} and the second one used the extended-on-mass-shell (EOMS) renormalization scheme in the work by Geng, Camalich, and Vicente-Vacas \cite{geng}. Both works performed calculations to order $\mathcal{O}(p^3)$; while the former included only octet baryons as active degrees of freedom, the latter did include both octet and decuplet baryons. A more recent calculation within large-$N_c$ baryon chiral perturbation theory to order $\mathcal{O}(p^2)$ has been presented in Ref.~\cite{fmg}. In this approach loop graphs with octet and decuplet intermediate states are systematically incorporated into the analysis because both spin-1/2 and spin-3/2 baryons together form an irreducible representation of spin-flavor symmetry.
Sum rule (\ref{eq:srulesfinal}a) is analytically fulfilled by the $f_1/f_1^{SU(3)}$ ratios given in Eqs.~(65)--(68) of Ref.~\cite{fmg}; it is also satisfied when the decuplet fields are not explicitly retained
in the effective theory but integrated out. In consequence, this sum rule is fulfilled by all the expressions for the $f_1/f_1^{SU(3)}$ ratios obtained within (heavy) baryon chiral perturbation theory to order $\mathcal{O}(p^2)$ of Refs.~\cite{krause,villa,meiss,geng}. The only exception is found in Ref.~\cite{and} where there is a wrong sign in one of the loop diagrams. Terms of subleading order in $1/M_B$ computed in Refs.~\cite{villa,meiss,geng} also satisfy sum rule (\ref{eq:srulesfinal}a).
As for sum rule (\ref{eq:srulesfinal}b), its right-hand side yields
\begin{equation}
\frac32 D^2 \left[H(m_K,m_\eta)-H(m_\pi,m_K)\right] - \frac38 \mathcal{C}^2 \left[K(m_K,m_\eta,\Delta)-K(m_\pi,m_K,\Delta)\right], \label{eq:sr2}
\end{equation}
where $D$ and $\mathcal{C}$ are the usual $SU(3)$ invariant couplings, the functions $H(m_1,m_2)$ and $K(m_1,m_2,\Delta)$ come from loop integrals and $\Delta$ is the decuplet-octet baryon mass difference \cite{fmg}. Numerically, the quantities between square parentheses in Eq.~(\ref{eq:sr2}) are $0.056$ and $0.029$, respectively. As a side remark, the large-$N_c$ cancellations between loop diagrams in Eq.~(\ref{eq:sr2}) are guaranteed to occur as a consequence of the contracted spin-flavor symmetry which is present in the $N_c\to \infty$ limit.
\section{\label{sec:sum}Closing remarks}
To close this paper, it should be stressed that two sum rules for leading vector form factors in hyperon semileptonic decays, (\ref{eq:srulesfinal}), have been obtained by exploiting only the general properties shared by the weak currents and the electromagnetic current under flavor $SU(3)$ symmetry. The $1/N_c$ expansion of QCD has been used to evaluate the group theoretic structure of the corrections. The sum rules have been tested against symmetry-breaking patterns obtained within (heavy) baryon chiral perturbation theory to order $\mathcal{O}(p^2)$ are they are well fulfilled. A full test beyond that chiral order currently is not possible because the existing calculations present partially complete or rather contradictory results among them. In the near future, lattice QCD could also be used to explore these sum rules; both the $1/N_c$ and SB hierarchies should be manifest.
\section{acknowledgments}
The authors thank J.L.\ Goity for helpful communications. This work has been partially supported by Consejo Nacional de Ciencia y Tecnolog{\'\i}a (M\'exico) and Fondo de Apoyo a la Investigaci\'on (Universidad Aut\'onoma de San Luis Potos{\'\i}).
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{
"redpajama_set_name": "RedPajamaArXiv"
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This Crock Pot Beer Chicken Recipe was named ZipList's Most Cooked Recipe of 2012!
People really seem to be divided into two camps when it comes to cooking with alcohol. There are those that avoid it like the plague and there are those that use it in every recipe that calls for it. I like to use alcohol in some recipes when it calls for it. It always seems to enhance the flavor of the dish.
Beer's essential role in this fabulous chicken comes down to the fact that alcohol can bind with both fat and water molecules. (Look at me getting all scientific!) That allows it to carry the flavor of the spices into the chicken and make the chicken incredibly flavorful.
Add in a little alcohol, in this case dark beer, and you enhance the ability of the salt in the spices to penetrate the chicken breast. The result is more juicy and flavorful chicken.
There's frequently a concern that too much alcohol will be left in the dish and make everyone drunk. But that's really not the case in many recipes. There simply isn't enough alcohol in most recipes to have a measured effect on you. Not to mention the cooking method plays a major role in how much alcohol is left.
Since this chicken is cooked in a crockpot for several hours, about 95% of the alcohol will be cooked out of this chicken before you serve it.
I like to add garlic powder, oregano and black pepper to my chicken. It seems to be the combination that adds the most flavor but is still mild enough to be paired with nearly everything else. But you can definitely play around with the spices. Use onion powder and paprika, or sprinkle in some Italian seasoning. Get creative!
Using chicken breast in this Crockpot Beer Chicken Recipe ensures that it stays low in points, so you can pair it with just about anything – a baked potato, roasted potato skins, wild rice, roasted veggies or a tossed salad. It's quick and easy to put together and is perfect for a busy night or that Super Bowl Sunday.
A wonderful and easy Chicken Crock Pot Recipe that tastes great and has just 3 Points. This Slow Cooker Beer Chicken also makes a great Weight Watchers Super Bowl Recipe idea.
I made this once, and when I got home from work the crockpot was still plugged in but it was on the back porch. I knew immediately when I opened the door why my roomie had placed it outside after he got home before me. The smell was horrendous, no words can describe it but I'll try: Take a wild North Korean dog, skin it, leave the tail on, add some kimchi and cat urine, then heat it for 2 days.
Try Clean Print as as add on to Google.
I made this and it turned out dry, I know the Beer cooks off but we both could still taste it. I really wanted to like it too.
Skinless breast is always dry and tasteless. Use chicken thighs with bone and skin. If you don't eat the skin, you'll be left with a moist flavorful meal that's fit to eat. It won't add significantly more fat or calories.
I go to use your "green" or "environmentally friendly" print recipe button…. 18 pages later!!!!!!!! Like what the?!?! Where is the print recipe button for just the recipe and not the other 17 pages now wasted, using your green printer friendly icon?! What a joke!! And not to mention the paper and printer ink cost!!!
I just made this and it turned out wonderful… NOT dry at all. I used 2.5 lbs of frozen chicken breast and a 16 ounce can of Leinenkugel's Summer Shandy. I used basil instead of oregano as I didn't have any oregano on hand. I also added parsley flakes and thyme. I didn't measure my seasoning but am sure I used more than what was called for. The chicken was very flavorful and moist. I served it with rice and veggies for a great healthy meal!
I tried this with a few improvisations, I added a can of condensed cream of mushroom soup, a jar of pearl onions, and I used basil instead of oregano (didn't have any oregano at the time). I used some larger chicken breasts so I only put in 6 of them, I ised a bottle of Corona Extra. It came out amazing!!!!! I absolutely loved the taste, the chicken is moist and flavorful! I am definitely going to be adding some cornstarch to thicken up the juice into a gravy.
Crock beer chicken- hideous. Cannot believe I wasted 2# chicken on this recipe. Useless. Boiled chicken is all it is.
I agree…..I wondered about using so much oregano…1 Tbsp? It was so overbearing, Ruined my Sunday dinner.
No idea why this has so many negative reviews. Mine was moist and flavorful, not dry and bland like many of you commented. I used about 3 lbs. (8 boneless, skinless breasts), and a 12 oz. bottle of Shiner Bock beer. I kept the seasonings the same, but probably used more than directed, as I did not measure. Cooked on high for 4 hours, shredded the chicken, and served it over brown rice with a side of roasted broccoli. Instead of making a gravy as some suggested, I just spooned the juice over the chicken and rice on our plates to keep it low calorie. There was PLENTY of juice. No idea why anyone's would come out dry.
I made this today and it's almost done but the chicken breasts are white. Where does the golden brown happen? And why does everyone talk about shredding the chicken. That's not in the receipe or the picture?
Cooked on high for 5 hours today, and I poked some holes in the chicken while it was cooking. Still came out pretty dry and bland. I will cut the chicken breasts up next time. I wonder, would adding corn starch thicken up the juice? I loved the flavor, but it'd be much better as a sauce instead of a juice. But yeah, it's horribly dry.
VERY DISAPPOINTING! I saw this pinned on Pinterest and looked forward to doing it. It was typical mushy, bland crock-pot chicken. I feel like I wasted good Guinness on this. If I were to re-vamp the recipe then I might oven braze with a LOT more aromatics or even use Guinness in a marinade and then grill.
I gave this a shot, with some changes after reading the comments – I used a combo of bone-in legs and thighs, browned them nicely along with some olive oil, 10 cloves of fresh garlic, shallots, before putting in the crock pot and changed the spices to 1/2 tsp oregano, 1/2 tsp thyme, 1/4 tsp red chili flakes, and a little salt and lemon pepper. Cooked on low for 6 hours. It was fantastic. I used Lagunitas IPA, will not be using this again as it does make the dish more bitter, next time will use Newcastle or a brown ale, I think.
I thought this was great! Read my full review on my blog! Thanks!
Wow I can't believe the bad feed-back on this recipe! I have made this several times and my family loves it! It's a perfect recipe for shredded chicken. I use it as my taco meat. I also will add BBQ sauce to the leftovers and it's great for BBQ chicken pizza or on salads, over rice! I love this recipe!
I wish I would have read the reviews before making this. It looks and smells wonderful. Extremely bland and mushy even after I put on additional spices. Will not make again.
Made this tonight and after reading the reviews was leary. Went on low for lil over 2 hours and then put on grill w barbecue sauce. YUMMY!!!
This is amazing. Ridiculously juicy.
I made this but when the chicken was fully cooked I transferred it (just the chicken none of the liquid) to a baking pan and brushed each piece with BBQ sauce then put it under the broiler for 5 to 10 minutes to cook the sauce and it was amazing. There were no leftovers.
This recipe is awful. The picture with this article is very deceiving. I made this tonight following the recipe exactly and had boiled soft chicken sitting in a bath of cooked beer. It SMELLED good, but was EXTREMELY bland. I'm gonna have to get crafty to use up all the leftovers I now have. Total waste and I'm bummed.
I agree, inedible! I cook a lot, and make some amazing recipes, but this was dry, bland and tasted like chicken boiled in beer.
I think the real problem is the chicken itself. Modern day poultry just doesn't taste like the chicken i ate as a kid way back in the 20th century. The only people that can get flavor into chicken these days are the guys at KFC!
Absolutely awful. I made this exactly as directed and couldn't have been more disappointed.
My husband said this is he best chicken he has ever had!
I used Sam Adams Light and cooked on high 4 hours.
Sides were Greek Salad and beer bread.
Going to make this tonight. I read all comments and can't wait to see if mine turns out! Just looking for something simple and i make beer can chicken all the time!! Wish me luck!
Please share recipes as I'm the head cook, and I'm just getting started.
This was meh. I do think it would be better if you brown the chicken beforehand. Will not make again.
I'm not a fan of beer at all, but my aunt makes tasty beer BBQ in the summer so I thought I'd give it a try. My aunt says cheap beer is the best to cook with because it has the most flavor. I couldn't bring myself to be seen buying bargain basement beer so I ended up stopping at a local gas station and getting a large bottle of Heineken.
The recipe doesn't say how much beer to use, so I bought the 22oz bottle. I added fresh minced garlic, potatoes and mushrooms. I also used 2 chicken breasts and two drumsticks instead of all those chicken breasts since I was only cooking for two plus leftovers. I think the chicken came out moist because of this modification. Towards the 6.5 hr mark I added egg noodles and turned it into a meal. I shredded the chicken and the noodles soaked up all the remaining liquid. The result was amazing. I actually went back for seconds, which I never do. The picture definitely doesn't reflect the end result. All in all, I highly recommend this recipe.
Very disappointing! No clue how to pictured version is browned–you can't really brown meat in the crockpot. I used 4 large breasts, followed recipe and it was cooked, fall apart chicken laying in juice, pretty boring. Since it did fall apart I think I'll shred, mix with BBQ sauce, & eat on rolls.
I only used three large frozen chicken breasts (equaled about 1.75lbs) and the chicken came out so juicy and brown (used guinness). Cooked it for 4.5hrs. If you don't like oregano, I would half the amount. I have made chicken breasts in the crock pot that come out with this weird texture of grainy mush, but found out if you put the chicken in frozen, you get the right result. I would definitely make this again!!
I just made this and it was so stinkin' delicious. I did, however, half this recipe since it's just me and my hubby. I DID NOT half the beer though. I think this is why my chicken wasn't dry. 4 chicken breasts and one bottle of beer. So if you use 8 chicken breasts I would use two bottles or cans of beer. The beer I used was Pipeline Porter by Kona Brewing Co. A dark, rich beer with coffee undertones. OMG so good. This was delicious and I will definitely be making it again!
I have made this two times. Once using skinless bone-in chicken thighs seasoned with salt, pepper and garlic salt which I marinated in the refrigerator for an hour and combined in the crock pot with a teaspoon of dried oregano, a can of chicken broth and the juice of a lemon as well as the beer. The second time I used skinless, bone-in chicken thighs seasoned with salt, pepper and garlic salt which I marinated in the refrigerator for an hour and combined in the crock pot with a can of chicken broth, one large chopped onion and five sprigs of fresh thyme and the beer. Both times, this dish came out perfectly. I like to serve this with rice cooked in seasoned chicken broth. It is pure comfort food.
Oh dear, it looks so good and crunchy. I like dry chicken but after reading the reviews not sure I want to make this. Is there a way to get the chicken crunchy looking like in the photo?
Possibly putting it under the broiler for a couple of minutes after it's done?
I've made this with Guinness, Shipyard Pumpkin, and Budweiser. Delicious each time. I usually do 4-5 chicken breasts instead of 8.
All I got was boiled chicken drowning in beer. Followed exact directions. Now I have to go BBQ it to get it brown. What a waste of good chicken.
This was one of the worst meals I've ever made. I followed the recipe, and it tasted awful. Cross this one off the list – I'll never make it again. A total waste of time and money.
Wish I'd read this before I wasted the time and a chicken's life. Perhaps the worst meal I've ever made!
Ok, so I read the comments and I'm sorry for those who had dry chicken, but mine was fabulous! I used 5 large, frozen, Tyson breast and one bottle of Yuenling (sp) Lagar with garlic powder (lots) & oregano. I did what one person suggested and started it on high for one hour and then on low. I ran errands for about 4 hours and arrived home at dinner time. Chicken cooked 5 hours total. I bought hot side dishes while I was out. I'll definitely make again. Thanks!
This was terrible. Way too much oregano flavor. Chicken was dry. Won't do this again.
I was thinking of a chicken crock pot meal (cause I'm tired of potroast, love it, just want something new) and seen this idea about chicken in beer in the crockpot. I love the idea, but with all the comments, I sure don't want to go wrong.
All I have is Budweiser, 2-3 bottles of. It's my stepdad's, but I don't mind drinking it either. I prefer Blue Moon, or Amber Bock, neither are on hand. How would BW turn out? I thought of using potatoes, baby carrots, peas, corn, cream of celery (or C.o.Chicken), diced tomatoes, and maybe lemon juice (or something sweat, sliced lemons, oranges, cherries or pineapples)… but I never made a recipe like this before, so I don't know how this would turn out. I'm not much of a Chef, so I don't know how I'd put it all together, concerning proportions and cook time, and I have a 6 or 8 quart cooker… I think. Not labeled, Idk for sure, but it's big.
Be feeding a family of 5-6, with 3 kids, big appetites. Any suggested comments would be great. Opinions, for or against the idea, will be duelly noted. Will let you guys know the final meal and result.
I made this using bone in chicken thighs, but I took the skin off. I used Guinness, 4 cloves of fresh garlic, 1/4 teaspoon of chilli flakes and 2 teaspoons of mixed French herbs. After the chicken was done I took it off the bone and added water and chicken Bisto to make gravy and then served it over mashed potatoes for the non-dieters and sautéed cabbage for myself (the dieter). We all loved it and the flavour was fabulous!
1. Use skin-on, bone-in chicken thighs instead of skinless, boneless chicken breasts. You can remove the skin, bones, and fat after the chicken is cooked.
2. Cut up vegetables (as if you were making pot roast) and place them in the crock pot with the beer first. They will take longer to cook than the chicken (especially if you use potatoes).
3. Season and brown the chicken in a skillet before placing in it a crock pot.
4. Start checking the chicken for doneness after 2 hours. Add the herbs when the chicken is almost done.
5. Use the leftover broth to make a gravy for the chicken and vegetables.
Unfortunately I had a REALLY BAD result with this recipe. I followed it to the T and cooked it EXACTLY as instructed and the result was soooooo dry that my family didn't eat more than a bite (I am not a meat eater, but my husband and son are). I bought 2 pounds of organic chicken breasts and that set me back $18. Major bummer.
We have done this on the grill. It is awesome. Looking forward to trying it in the crock pot.
Tried this recipe and it blew the lid off the crock pot… Literally, the pressure from the beer blew the lid off and across the countertop. Chicken looked like it had been drowned, and was all wrinkly. It looked nothing like the photo above. Will never cook this again!!!
This turned out great! But in my fashion, I made a few changes that may have made a difference. I only used 5 chicken breasts instead of 8. I used Cayenne Pepper instead of Black Pepper, and all the other spices I did to taste–so way more than 1 tsp the recipe called for. Having done Crock Pot cooking in the past, you need to over do on certain spices in order for them to not get lost. Or do what someone else suggested, and add spices an hour before the end of cooking time–but that kind of defeats the point of putting it in the pot and forgetting about it. I happened to have a leftover can of Guinness (go figure), so I used it–but only about 3/4 since I reduced the number of breasts. I set the Crock Pot on low for about 1.5 hours, then switched it to high for the last three; so all-in-all, I only did 4.5 hours. The breasts were tender and succulent, and VERY flavorful. I also used the gentleman's recommendation of creating a gravy from the broth, which just added a nice touch. Served it with Quinoa and mixed vegetables. Happy eating!
I made this last night and it was YUMMY! I used 4 mostly-frozen chicken breasts (huge ones), 2 bottles of Coors Light, a heaping tsp of salt (and some extra sprinkled over the top of each breast), a full tsp of black pepper, a heaping tbsp of garlic powder (cuz we love garlic), a heaping tbsp of dried oregano leaves, one small diced onion. I put all of the ingredients in before the chicken and then ladled the mixture over top of the breasts once they were in the pot. I cooked everything for an hour on High to make sure the breasts thawed, then I turned it down to Low. I cooked it all for 8 hours total (1 on high, 7 on low). There was SO much liquid that at the 6 hour mark I took some out and kept it to add back in if things got too dry. I checked it every half hour then but had no problems. At the 7 hour mark, I turned the breasts over and ladled the juice on top of them. I checked every hour or so to make sure they weren't getting dry. Used a knife and fork to open one up and slice out a tiny piece for tasting. Once done, served with rice. The meat was SOOO juicy and flavourful!! Highly recommended.
Wish I had read all these comments first. . Chicken was overlooked and bland. Would not make it again.. I even added carrots then peas at the end.
delicious, thanks for the recipe! I also made a gravy out of my leftover 'broth'. How would I go about figuring out those points?
I made this tonight for the first time. It was outstanding! I used Guinness and only cooked it for 6 1/2 hours on low (any longer and I think it would have been dry). That Guinness gravy idea is genius! Will try that next time! Also, hubby thought it might be good with a little lemon juice added in for a citrus kick. A+ from me!
To get the browning as in the picture above, you could take the chicken out while making the gravy or sides and place under the broiler for a couple minutes (remember to baste before or you will dry it out), just to brown the outside….
I used Guinness (as suggested) NOT a good choice for us. Won't be making this again. The chicken was fall apart tender, but the taste was way too bitter.
get redbridge beer. i'm gluten free as well and redbridge is a "dark" beer and would probably work well!
I used Blue Moon Pale Ale, the seasonings from the recipe along with Mrs. Dash salt free table seasoning and some red pepper flakes and served it with sweet potatoe hash browns, so delicious. Thank you for the recipe.
I cooked my chicken on low for 3 hours in my crock pot and it came out slightly dry. I recommend cooking on a low heat and checking the meat for readiness after 2 hrs. Certainly not a recipe I would leave unattended for 6 hrs while at work.
This was pretty bland and the chicken was dry.. I cooked it at low and it was done within 3-4 hours– I then turned it down to warm because it would have been way over cooked. I even put a little extra seasoning and it was still bland :/ if I make it again, I will use thighs instead of breasts.
This was horrible! My chicken turned into jerkey and wasn't even a little bit edible. I may try it again with a shorter cooking time but I kind of doubt it. I followed the recipe to the letter and even used a very trustworthy crock pot so I'm not sure what happened but I did not have good results and ended up throwing the entire thing away.
Recipe sounds very good, but someone mentioned they can't see how the chicken can brown while seeping in beer. I'm wondering the same thing. Does the chicken, the way you gave the instructions, come out moist?
Great question! The browned bits are just the spices I coated the chicken with, which blackened when cooking. Other than that, yes, it is just plain, skinless boneless chicken breast. The long cook in the beer helped to keep in tender and moist.
No, that can't possibly be the spices that are the "browned bits" as it only uses garlic powder and some dried oregano, plus salt/pepper. Mine had no color at all, was dry as a bone and the sauce after thickening was bland also. I followed the recipe to the letter, used a good beer and still it didn't look like this! I subscribe to your blog and resent being told that this picture is from that recipe!! I may have to unsubscribe! Sorry for being so nasty, but I think you need to look at this picture and recipe again!
K=I made this last night!! Really good! Thank you so much for the recipe. I followed the recipe except I used thigh meat instead. Came out so moist and actually, it looked like the picture!! I think next time I make it, I am goign to try garlic salt too!
Awesome chicken. I used Guinness as as suggested. It was a hit with my guests.
I used a whole chicken. Easy and tasty. Makes wonderful left overs with the balance shredded and used or tacos with the corn relish.
Does the chicken have a strong beer taste?
Mine didn't, and I used Guinness, which is a strong beer. All the alcohol cooks off, so the remaining taste is beautifully mild.
Tried this last night. I followed the recipe, with these exceptions: used Bud instead of Guiness (Bud was what I had on hand); added 1/2 C chopped onions and a chopped carrot; used chicken leg quarters instead of breasts. The texture of the chicken was good, and the aroma of this dish cooking all day was great! However, it tasted bland, and if I were going to make it again, I would follow the reader suggestion to add the spices towards the end of the cooking not the beginning. The crockpot cooking method also results in a lot of broth, and it would be good if the recipe had some way of converting that to a sauce and then incorporating the sauce into the final dish.
Came out bland? You used Bud, what did you expect…? Not trying to be rude, but I mean, really, it's bizarre to think any American light lager is going to contribute much, if any, flavor.
Has anyone tried this while a whole chicken?
I did, very tasty, delish!! except for the amount of grease from the skin, had to remove a lot of the fat next day, lots of gravy Made curry with leftovers.
Made with thigh meat and recipe was to die for. Also added mushrooms and parsnips made very good flavor dusted the top with powdered chicken broth for an added kick! Five stars!!!….
Made this for dinner tonight… I added some minced garlic (no garlic powder to be found when I went to get it). Plain old yuengling light for the beer, and some other spices. Cooked on low from 9-5, and since I knew it'd be in a while, I added a bit of chicken broth I had in the fridge. Thanks so much for the recipe, because wow is this tasty! Almost no effort on the front end, and a wonderful dinner when we all got home from work and afterschool activities! Yum & thanks!
Never add herbs and/or spices to a crock pot until about an hour or so before the end of the cooking time.
If you put them in before that the meal (any crock pot meal) will be bland.
Beer left over from the 4th of July? I don't even have beer left over from Columbus Day (yesterday).
Although I am a little late for this post, I thought…hmmm, I don't even have beer left over on the 5th of July! Thanks @hongryhawg:disqus for the laugh this morning!
Ok, so I wouldn't recommend this for a "cook while you're at work" dish. I also wouldn't recommend it iwth breast meat. Though my pot shut off at the prescribed time, I came home to dry chicken with a very strange texture. The flavor was ok, but, for chicken that basically braised (sat in liquid) for 8 hours, it was almost inedibly dry. I"m going to try it again with bone-in thighs and less cook time and see if that helps.
My chicken was terribly dry as well… Had to choke it down. Very disappointed since it smelled so good cooking! Tossing this recipe!
It's because 6 hours is way too long to cook any chicken in the crock pot.
I've made so many crock pot recipes, and they all usually take 4 hours. So, your chicken was dry because it was overcooked.
This is in the crockpot right now! I added a few spices and used Shiner Bock, our fave household beer. :) Can't wait to see how it turns out. Love fix it and forget it meals on junior high football game nights.
I made this this this morning and used coors light (it is all we had, haha) hope it is just as good as you all are saying. !!
I made this in the crock pot and used frozen chicken breasts. I added another half a can of beer because it was on low for 10 hours and I was worried it would dry up. It was too much beer. I recommend only one can if you are using frozen. It was very good and I will definitely make it again, just with less beer!
Was this in a 6 quart slow-cooker? If so, how much room was leftover for veggies then? I want to try & make this with maybe baby red potatoes or diced sweet potatoes (maybe even Yukon golds if small sized), baby carrots, celery, corn, & peas. I'm thinking maybe add the veggies in about 90 mins. prior to serving? Think that would be long enough? They'll be canned peas & corn & the rest of the veggies would be fresh, not frozen.
Yes, this was in a 6 quart slow cooker. There was plenty of room for veggies, but depending on how many you add, you may need to use some extra beer.
Choose a firm robust potato like whole or halved small red potatoes; they should hold up better with the longer cooking times.
By a bit, but not too much. Just check the chicken, and as soon as it cooked through and gets to the point of falling apart, then it's ready.
Thanks! I actually ended up getting three boneless/skinless breasts, but they were huge- total of 2.61 lbs! They were done after about 3 hrs 15 min on high, put them on low while I made sides. The gravy idea someone mentioned earlier was very good! I put about 1 1/2 Tbs of cornstarch, brought to a boil, then simmered for a few minutes, and poured it over the chicken and also the mashed taters. Delicious, thanks!
Warning: possible dumb question ahead…..I want to make this tomorrow, but the chicken would most likely be in the crock for about 12 hours before dinner. Do you recommend then to throw it in frozen or will thawed be fine on low for 12 hours? As it might be in for that length of time, would you also recommend more beer to keep it moist or would 1 bottle still be sufficient?
GREAT question! I haven't tried leaving it in that long myself, but I would def use some extra beer…maybe another half bottle. And I think using the frozen chicken would help a bit too. Let us know how it goes!!
I liked the chicken, but the oregano was overpowering… next time I'm gonna experiment with the spices… I'm thinking rosemary instead of oregano. The bacon mashed potato side dish was a hit and sprinkling some cheddar was delish, but the corn salad is really more like salsa and we are gonna eat it with chips instead.
Agreed. The oregano is way too powerful. Don't think a tablespoon is needed. Maybe 1/2 a tsp. Thinking more garlic, pepper, a little kick of lemon pepper or even some chicken grill seasonings.
What kinds of veggies could i put in here with out them getting mushy also could I cook potatoes right in there with the chicken like you do with a pot roast or will it make them taste funny??
Any root veg should hold up well. Potatoes should be fine too, unless they cook too long.
Do you start with thawed or frozen chicken breasts? I need to make for the weekend. Thanks!
I used thawed, but haven't tried frozen, so I don't know if that would work or not.
One thing you have to keep in mind when using frozen meet in the crock pot is you will have more liquid as the meat thaws, so may want to add less liquid initially.
Does the chicken need to be thawed before putting in the crockpot?
I also used Newcastle….double delicious. I added two Texas style sausages to the recipe for additional flavor and served the chicken with sliced sausage…..enjoyed by all.
My chicken was done in 3 hours. Some slight differences in what I did…1st chicken breast only had 3 were frozen, so started 1st hour on high then went to low setting. Is the fact that I had less chicken the reason for a faster cooking time…new to crock pot cooking. I have a 6 qt Revel crock pot if it matters.
I made this for the family today and it was so yummy! I used Guiness (Guiness makes everything better) and the same seasonings in the recipe. When the chicken was done, I removed most of the "broth" and put it in a small sauce pan and thickened it up with a cornstarch slurry. Then I poured the "gravy" back over the chicken and kept it warm until we were ready to eat. Served it over mashed potatoes. Definitely will make this dish again. Thanks for sharing it!
Mmmmmmm!!! LOVE LOVE LOVE the Guiness gravy idea!! Fantastic. Thanks so much for sharing. Me + Guinness = BFF.
Any idea of Points for the gravy?
If you add the corn starch that is just an extra 60 calories for 2 tablespoons, I figured out calories, not points because I don't do weight watchers, I do 5:2, and it came out to 250 calories per serving using 10 skinless, but bone in, chicken thighs and a can of Guinness.
Just discovered your site. Awesome recipes! Thank you so much. And thanks for including gluten free stuff. I am going to make this one with gluten free beer.
How does the chicken get so nice and brown, as in the picture? Also, could you use boneless skinless thighs, and how much would that increase the point value? Thanks!
The color of the chicken will depend on the kind of beer you use. I used Guinness, which is very dark. It was sooooo yummy! If you use thighs, the Points + value will probably increase by a Point or two.
Very disappointed with the recipe. My chicken looked had no color whatsoever…….looked nothing like the photo. How can it possibly brown while floating in beer? I also agree with another poster who said the amount of oregano was way to much. I ended up removing as much of it as I could from the chicken pieces and made chicken salad.
You don't have to worry about giving it to kids, the alcohol with boil off. I cook with beer in my crockpot all the time.
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Q: GmailApp getMessages() needs an access token I'm struggling to get getMessages() to work.
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Be on guard against "urgent" requests and unsolicited "deals" on the Internet.
FDIC reports that criminals masquerading as legitimate businesses or government agencies are tricking consumers into divulging valuable personal information over the computer, phone or fax in order to drain bank accounts. Here are the latest tips from the Federal Deposit Insurance Corp (FDIC) for protecting against new schemes using electronic devices.
THINK TWICE before responding to "urgent" text messages.
A new scam involves a text message sent to cell phones and smartphones warning bank customers that their debit card had been blocked for security reasons. The message urges users to call a hotline to unblock their card, but instead they reach an automated response system asking for their card number, personal identification number (PIN) and other information.
"Unfortunately, this is enough information for thieves to create counterfeit cards and commit fraud," says Michael Benardo, Chief of the FDIC's Cyber-Fraud and Financial Crimes Section.
Smartphone users are now being targeted by scammers because they almost always have their phone handy and tend to respond to calls and e-mails quickly, so that they may not realize a message is fake until it is too late. Not only that, but fake web sites are also harder to spot on a small screen.
BE ON GUARD against unexpected pop-up windows on web sites, including your bank's.
If after you're logged onto your bank's web site - or any web site for that matter - you get an unexpected pop-up window asking for your name, account numbers and other personal information, that is likely a sign that a hacker has infected your PC with spyware and is trolling for enough information to commit identity theft and gain access to your bank account.
It's normal for Austin Bank to ask you for your log in ID and password when you first log in and to ask you to answer a 'challenge question' if you want to reset your password or start using a new computer. But Austin Bank will not ask you - through a pop-up window - to type your name and information such as your date of birth, mother's maiden name, bank account and cell phone numbers. The bank only needs that type of detailed personal information when the account is initially opened.
BE SUSPICIOUS of unsolicited offers to download games, programs and other "apps."
Those "deals" could contain malicious software directing you to fake web sites or install spyware used to steal information that can lead to theft. "You should consider using anti-virus software specifically designed for smartphones and other mobile devices," advises David M. Nelson, an FDIC fraud specialist.
Be aware that cyber criminals always look for ways to use new technology such as smartphones to try to commit fraud.
Stop and think before giving personal information in response to an unsolicited request, especially one marked as urgent, no matter who the source supposedly is.
Only install programs you know are from legitimate web sites, such as your Internet service provider, financial institution, wireless phone company or trusted app vendors.
For additional tips on avoiding Internet fraud, visit www.onguardonline.gov.
This information provided by Financial Education Corporation.
Hey there! Thanks for sharing such a nice write up. When it comes to computer issues, it is always best to get it fixed as soon as possible. These computer problems such as virus, crashes and even slow computers could easily mean a bigger and more serious issue. Nonetheless, thanks for sharing and appreciate it.
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{
"redpajama_set_name": "RedPajamaC4"
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\section*{Supplementary information}
In the supplemental material we look at the mechanisms involved in entanglement and simulations involving them, additional more general spin qubit simulations, fidelity simulations which relate to the first virtual system proposed and simulations of a dynamical decoupling protocol that could be incorporated as part of the virtual system to improve the resulting coherence times and fidelities.
\smallbreak
\section*{Acknowledgments}
We thank S. Mukhin (from MISIS), Y. Hardy and P. Madonsela for discussion. SB thanks CSIR-NLC, the URC Wits and National Research Foundation (SA) for the BRICS multilateral program for funding. SB also acknowledges financial support from the Ministry of Education and Science of the Russian Federation in the framework of the Increased Competitiveness Program of NUST MISiS (grant No. K $3-2018-043$).
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\section{Introduction}\label{sc-intr}
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Radiative transport equations \cite{agoshkov2012boundary,Davison-1957,Pomraning-1973,Lewis-Miller-1984,Mihalis-Mihalis-1999,graziani2006computational,Dautray-Lions-2000} describe the flows of particles, such as photons, neutrons, and electrons, as they pass through and interact with a background medium. These equations are used
in various applications, including astrophysics and nuclear reactor analysis.
In this paper, we consider the scaled, steady-state, linear transport equation
\begin{subequations}\label{eq-main}
\begin{alignat}{3}
\Omega \cdot \nabla \Psi(\Omega,x) + \left(\frac{\sig{s}(x)}{\varepsilon}+\varepsilon
\sig{a}(x)\right) \Psi(\Omega,x) &=
\frac{\sig{s}(x)}{\varepsilon}\overline{\Psi}(x)
+ \varepsilon q(x), & \quad
&(\Omega,x)\in S \times D,\\
\Psi(\Omega,x) &= \alpha(\Omega,x), & &(\Omega,x)\in \Gamma^{-}.
\end{alignat}
\end{subequations}
Here $D\subset \mathbb{R}^d$ $(d=1,2,3)$ is an open, bounded, and Lipschitz domain; $S$ is the projection of the unit sphere in $\mathbb{R}^3$ into $\mathbb{R}^d$ (the interval $[-1,1]$ for $d = 1$ and unit disk for $d = 2$); and $\Gamma^{-} = \{ (x,\Omega) \in S\times\partial
D\mid\Omega\cdot n(x)<0\}$, where $n(x)$ is the outward unit normal vector at any point $x \in \partial D$ where the boundary is $C^1$.
The \textit{angular flux}
$\Psi$ is the flux of particles at the location $x$ moving with unit speed in the direction $\Omega$, and the \textit{scalar flux} $\overline{\Psi} = \frac{1}{|S|}\int_{S}\Psi d\Omega$ is the average of $\Psi$ over $S$.%
\footnote{Often the quantity $\Phi=4\pi\overline{\Psi}$ is referred to as the scalar flux. The difference is simply a normalization factor from integration of the sphere. Here, we borrow the convention used in \cite{Lewis-Miller-1984}.}
The functions $\sig{s}$ and $\sig{a}$ are (known) non-dimensionalized scattering and absorption
cross-sections, respectively, and $q$ is a (known) non-dimensionalized source.
The function
$\alpha(\Omega,x)$ is the (known) incoming flux at $x\in \partial D$ moving in the direction $\Omega$.
The constant $\varepsilon>0$ is a scaling parameter which characterizes the relative strength of scattering.
Designing effective numerical methods for \eqref{eq-main} is a serious
challenge, and the intent of this paper is to address two of the main issues.
Firstly, for a three-dimensional problem, the unknown intensity $\Psi$ is a function
of three spatial and two angular variables; the discretization of this five-dimensional phase space
usually requires significant computational resources. Secondly,
when the parameter $\varepsilon$ is small, $\Psi$ is nearly independent of $\Omega$ and can be approximated by the solution of a diffusion equation in the variable $x$ only \cite{Habetler-Matkowsky-1975,bardos1984diffusion,bensoussan1979boundary}. That is, away from the boundary, $\Psi(\Omega,x) = \Psi^{(0)} (x) + O(\varepsilon)$ as $\varepsilon \to 0$, where $\Psi^{(0)} $ satisfies
\begin{equation}\label{eq-difflim}
-\nabla\cdot\left(\frac{1}{3\sig{s}}\nabla \Psi^{(0)} (x)\right) + \sig{a}
\Psi^{(0)}(x) = q(x), \quad x \in D,
\end{equation}
along with appropriate boundary conditions.
A numerical method for \eqref{eq-main} should preserve this asymptotic limit without having to resolve the length scales associated with $\varepsilon$ \cite{jin1999efficient}. In other words, in the limit $\varepsilon \to 0$, a discretization of the transport equation \eqref{eq-main} should become a consistent and stable discretization of the diffusion equation \eqref{eq-difflim}. Otherwise
a highly refined mesh is needed to approximate the solution accurately \cite{larsen1987asymptotic}.%
\footnote{{This issue is also known as ``locking" in the elliptic literature \cite{babuvska1992locking}.}}
Classical approaches for discretizing \eqref{eq-main} often involve separate
treatment of the angular and spatial variables, and a variety of options are available.
Among them, the $S_N$-DG method \cite{HHE2010, larsen1989asymptotic,adams2001discontinuous} has received significant attention due to it's robustness, computational efficiency, and convenient implementation. The $S_N$ method (see\cite{larsen2010advances} for a substantial review and additional references) is a collocation method in which the angular variable $\Omega$ is discretized into a finite number of directions and a quadrature rule is used to evaluate $\overline{\Psi}$.
The $S_N$ discretization preserves non-negativity of $\Psi$ and can incorporate the boundary conditions from \eqref{eq-main} in a straightforward way. It also preserves the characteristic structure of the advection operator in \eqref{eq-main}, which allows for the use of fast sweeping techniques for inverting the discrete form of the operator on the left-hand side of \eqref{eq-main}.
Discontinuous Galerkin (DG) methods are a class of finite element methods that construct numerical solutions using piecewise
polynomial spaces. The DG approach was introduced in \cite{reed1973triangular} for the express purpose of solving equations like \eqref{eq-main}, followed shortly thereafter by a rigorous analysis in \cite{lesaint1974finite}. Since then, DG methods have been applied to nonlinear
hyperbolic conservation laws and convection-dominated problems
\cite{cockburn2001runge}, elliptic problems \cite{arnold2002unified},
and equations with higher-order derivatives \cite{yan2002local,xu2010local}.
When used with upwind fluxes, DG methods preserve the characteristic structure of \eqref{eq-main} that enables sweeps. Moreover, if the approximation space can support globally continuous linear polynomials, then DG methods with upwind fluxes will yield accurate numerical solutions for $\Psi$ without the need to resolve $\varepsilon$ with the spatial mesh \cite{larsen1989asymptotic, adams2001discontinuous, guermond2010asymptotic}. However, this condition on the approximation space means that at least $P^1$ elements must be used for a triangular mesh and $Q^1$ elements for a rectangular mesh.%
\footnote{{This condition can be circumvented for non-upwind methods. In \cite{ragusa2012robust}, the authors made the piecewise constant DG method asymptotic preserving with parameters adjusting numerical fluxes under different regimes. Similar techniques were introduced in finite volume contexts \cite{jin1996numerical} as well and were recently used in \cite{guermond2019positive} to develop a positive, asymptotic preserving method.} }
In order to reduce memory costs in the upwind $S_N$-DG method, while still preserving the asymptotic diffusion limit and maintaining the characteristic structure needed for sweeps, we propose in this paper to couple the finite element spaces between different collocation angles in the discrete ordinate approximation. Since the solution becomes isotropic
in the diffusion limit ($\varepsilon \to 0$), we hypothesize that only a $P^1$ (for triangles) or $Q^1$ (for rectangles) approximation of the angular average is necessary. Thus, instead of using a tensor product
finite element space for the $S_N$-DG system, we seek the solution in a proper subspace,
in which all the elements have isotropic slopes. This choice of finite
element space yields a significant reduction in memory per spatial cell, as illustrated in \cref{tab-sndg-costper}.
\begin{table}[!h]
\centering
\begin{tabular}{c|c|c}
\hline
Unknowns per cell & Triangles ($P^1$) & Rectangles ($Q^1$) \\
\hline
Standard $S_N$-DG & $(d+1)n_\Omega$ & $2^{d}n_\Omega$ \\
\hline
low-memory $S_N$-DG &
${n_\Omega + d}$ &
${(n_\Omega -1)+ 2^{d}}$ \\
\hline
Memory cost ratio as $n_\Omega \gg 1$ & $d+1$ & $2^{d}$\\
\hline \end{tabular}
\caption{Memory costs of standard $S_N$-DG and the low-memory variation, both for triangles and rectangles, for spatial dimension $d$. The first two rows give the number of unknowns per angle per spatial cell for each approach. The last row is the asymptotic ratio of the memory costs by two methods when $n_\Omega$ becomes large. }
\label{tab-sndg-costper}
\end{table}
In the diffusion limit, the low-memory approach typically displays second-order accuracy. However, because the finite element representation of each ordinate is coupled to all the other ordinates, the overall accuracy of the low-memory approach for fixed $\varepsilon$ is only first-order.
To address this drawback, we propose a modification of the low-memory scheme
that uses local reconstruction to improve accuracy. As long as the reconstruction uses upwind information, the resulting transport operator can still be inverted with sweeps. While rigorous theoretic properties of this modified scheme are still under
investigation, we observe numerically that it recovers second-order accuracy for arbitrary fixed
$\varepsilon$ and captures the asymptotic diffusion limit. However, the method does generate some small numerical artifacts at the discontiuity of the cross section, which we point out in the numerical results of \Cref{sc-num}.
The rest of the paper is organized as follows. In \Cref{sc-background}, we
introduce the background and revisit the $S_N$-DG method. Low-memory
methods, including the original first-order approach and the second-order reconstructed scheme, are
detailed in \Cref{sc-lmdg}. Numerical tests are provided in \Cref{sc-num} to illustrate the behavior of both approaches. Finally, conclusions and future work are discussed in \Cref{sc-conclude}.
\section{The $S_N$-DG method}\label{sc-background}
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In this section, we review the $S_N$-DG scheme and discuss its asymptotic properties and implementation. Throughout the paper, we consider the case $\inf_{x\in D} \sig{s}(x) = \delta_s > 0$ and $\inf_{x\in D}\sig{a}(x)=\delta_{\rm{a}} >0$, unless otherwise stated. In general, the well-posedness of \eqref{eq-main} also holds for $\sig{a} \geq 0$
\cite{wu2015geometric}. In some places, we will also assume that the cross-section is piecewise constant, either to simplify the exposition or to make connections between first- and second-order forms of the diffusion limit. In the numerics, we often consider nonzero boundary conditions. However, in proofs we often assume that $\alpha = 0$. When $\alpha$ is nonzero but isotropic, many of the results still hold. However, when $\alpha$ is anisotropic, the diffusion equation requires a boundary layer correction in order to be uniformly accurate \cite{Habetler-Matkowsky-1975}. At the discrete level, this situation requires more sophisticated analysis \cite{larsen1989asymptotic,adams2001discontinuous,guermond2010asymptotic} than is presented here.
\subsection{Formulation}
Consider a quadrature rule with points $\{\Omega_j\}_{ j =1}^{n_\Omega}$ and positive weights $\{w_j\}_{ j =1}^{n_\Omega}$ such that
\begin{equation}
\frac{1}{|S|}\int_S f(\Omega) d\Omega \approx \sum_{j=1}^{n_\Omega} w_j f(\Omega_j), \quad
\forall f \in C(S).
\end{equation}
We assume the quadrature is exact for polynomials in $\Omega$ up to degree two%
\footnote{Level symmetric quadratures of moderate size will satisfy these properties. See, e.g., \cite{Lewis-Miller-1984} and references therein.}%
; that is,
\begin{eqnarray}
\label{eq-polyquad}
(i)~\sum_{j=1}^{n_\Omega} w_j = 1,\quad
(ii)~\sum_{j=1}^{n_\Omega} w_j \Omega_j = 0, \quad \text{and} \quad
(iii)~\sum_{j=1}^{n_\Omega} w_j \Omega_j \otimes \Omega_j = \frac{1}{3} \operatorname{Id}.
\end{eqnarray}
The $S_N$ method approximates the angular flux $\Psi$ at the quadrature points $\{\Omega_j\}_{ j =
1}^{n_\Omega}$ by a vector-valued
function $\psi(x) = (\psi_1(x),\psi_2(x),\dots,\psi_{n_\Omega}(x))$ whose components satisfy a coupled system with $n_\Omega$ equations
\begin{equation}
\label{eq:sn}
\Omega_j \cdot \nabla \psi_j(x) + \left(\frac{\sig{s}}{\varepsilon}+\varepsilon \sig{a}\right)\psi_j(x)
= \frac{\sig{s}}{\varepsilon} \overline{\psi}(x) + \varepsilon q(x),
\qquad
\overline{\psi}(x) = \sum_{j=1}^{n_\Omega} w_j \psi(\Omega_j, x).
\end{equation}
To formulate the upwind DG discretization of the $S_N$ system \eqref{eq:sn}, let $\mathcal{T}_h = \{K\}$ be a quasi-uniform partition of the domain $D$. We assume $D = \cup_{K\in \mathcal{T}_h} \mathrm{cl}(K)$ to avoid unnecessary technicalities. Let $\mathcal{F}_h = \cup_{K\in \mathcal{T}_h}\partial K$ be the collection of cell interfaces and let $\mathcal{F}_h^\partial$ be the collection of boundary faces. Given a cell $K$, we denote by $\nu_K$ the outward normal on $\partial K$ and for any $x \in \partial K$, let $v^{\rm{int}}(x) = \lim_{\delta \to 0^+}v(x - \delta \nu_K)$ and $v^{\rm{ext}}(x) = \lim_{\delta \to 0^+} v(x + \delta \nu_K)$. Given a face $F$, we denote by $\nu_F$ a prescribed normal (chosen by convention) and, for any $x \in F$, let $v^{\pm} = \lim_ {\delta \to 0^+} v(x \pm \delta \nu_F)$. For convenience, we assume trace values are identically zero when evaluated outside of $D$.
The standard $S_N$-DG method uses the tensor-product finite element space
\begin{equation}\label{eq-cV}
\mathcal{V}_h= \prod_{j=1}^{n_\Omega} V_h,\qquad V_h = \{v_j: v_j|_K \in Z_1(K)\},
\end{equation}
where for triangular or tetrahedral meshes, $Z_1(K)$ is the space $P^1(K)$ of linear polynomials
on $K$ and for Cartesian meshes $Z_1(K)$ is the space $Q^1(K)$ of multilinear polynomials
on $K$. The space $\mathcal{V}_h$
can be equipped with an inner product $(\cdot,\cdot)$ and associated norm $\|\cdot\|$ given by
\begin{equation}
(u,v) = \sum_{K\in \mathcal{T}_h} \sum_{j=1}^{n_\Omega} w_j \int_K u_j v_j dx
\qquad\text{and}\qquad \|v\| = \sqrt{(v,v)}.
\end{equation}
The semi-norm induced by jumps at the cell interfaces
is given by
\begin{equation}\label{eq-jump}
\llbracket v\rrbracket = \left({\sum_{F\in \mathcal{F}_h}\sum_{j=1}^{n_\Omega} w_j \int_F
|\Omega_j\cdot \nu_F| (v_j^{-}-v_j^{+})^2 dx}\right)^{1/2}.
\end{equation}
To construct the $S_N$-DG method, define the local operators
\begin{subequations}
\begin{align}
\label{eq-Ljk}
L_{j,K}(u,v) =& -\int_K u_{j} \Omega_j \cdot \nabla v_j dx + \int_{\partial K}
\widehat{u}_{j} \Omega_j\cdot \nu_K v^{\rm{int}}_j dx \\
&+ \int_K
\left(\frac{\sig{s}}{\varepsilon}+\varepsilon \sig{a}\right) u_jv_j dx,\nonumber
\\
\label{eq-Sjk}
S_{j,K}(u,v) =& \int_K\frac{\sig{s}}{\varepsilon}\overline{u} v_j dx, \quad \text{with}~\overline{u} = \sum_{j=1}^{n_\Omega} w_j u_j,
\\
Q_{j,K,\alpha}(v) =& \int_K \varepsilon q v_j dx - \int_{\partial K \cap
\mathcal{F}_h^{\partial}} \alpha \Omega_j\cdot \nu_K v_j^{\mathrm{int}} dx,
\end{align}
\end{subequations}
where $\widehat{u}_j (x)= \lim_{\delta \to 0^-} u(x + \delta \Omega_j) $ is the upwind trace at $x \in \partial K$, and is defined as zero when the limit is taken outside of $D$. Then set
\begin{equation}\label{eq-Bdef}
B(u,v) = L(u,v) - S(u,v),
\end{equation}
where
\begin{equation}
\label{eq-LandS}
L(u,v) = \sum_{K\in \mathcal{T}_h}\sum_{j=1}^{n_\Omega} w_jL_{j,K}(u,v)
\quad \text{and} \quad
S(u,v) = \sum_{K\in \mathcal{T}_h}\sum_{j=1}^{n_\Omega} w_j S_{j,K}(u,v),
\end{equation}
and let
\begin{equation}\label{eq-lalpha}
Q_{\alpha}(v) = \sum_{K\in \mathcal{T}_h}\sum_{j=1}^{n_\Omega}
w_j Q_{j,K,\alpha}(v).
\end{equation}
The $S_N$-DG method is then: \textit{find $\psi_{h}=(\psi_{h,1},\dots,\psi_{h,n_\Omega}) \in \mathcal{V}_h $ such that}
\begin{equation}\label{eq-sndg-scheme}
B(\psi_h,v) = Q_{\alpha}(v), \qquad \forall v \in \mathcal{V}_h.
\end{equation}
\subsubsection{Implementation}\label{sc-basis}
Recall that $n_\Omega$ is the number of discrete ordinates in the $S_N$ discretization. Let $n_x = |\mathcal{T}_h|$ be the number of mesh cells in $\mathcal{T}_h$ and let $n_P$
be the dimension of $Z_1(K)$. Then the dimension of $\mathcal{V}_h$ is $n_\Omega\cdot n_x \cdot n_P$.
Let $\{b^{p,r}:p = 1,\ldots,n_x,r=0,\ldots,n_P-1\}$ be a set of basis functions for $V_h$, with $b^{p,r}$ locally supported on $K_p \in \mathcal{T}_h$. Then the set $\mathbb{B} = \{\xi^{l,p,r}:l =1,\dots,n_\Omega,p=1,\dots,n_x,r=0,\dots,n_P-1\}$, where $\xi_j^{l,p,r}(x) = \delta_{l j} b^{p,r}(x)$ ($j = 1,\ldots n_\Omega$) and $\delta$ is the Kronecker delta,
gives a complete set of basis functions
for $\mathcal{V}_h$.
With this choice of basis functions, the variational formulation in \eqref{eq-sndg-scheme}, written as
\begin{equation}\label{eq-vari-LSQ}
L(\psi_h,v) = S(\psi_h,v) + Q_{\alpha}(v),\qquad \forall v\in \mathcal{V}_h,
\end{equation}
can be assembled into a linear system (detailed in \Cref{ap-sndg-matrix})
\begin{equation}\label{eq-sndg-mat}
\mathbf{L} \mathbf{\Psi} = \mathbf{M} \mathbf{P} \mathbf{\Psi} + \mathbf{Q}.
\end{equation}
In the above equation, $\mathbf{L}$ is an $(n_\Omega\cdot n_x\cdot n_P)\times (n_\Omega\cdot n_x\cdot
n_P)$ block diagonal matrix, where the $j$-th block ($j = 1,\ldots n_\Omega$) corresponds to the discretization of the operator $\psi_j \to \Omega_j \cdot \nabla \psi_j + \left(\frac{\sig{s}}{\varepsilon}+\varepsilon \sig{a}\right)\psi_j$; $\mathbf{M}$ is an injective $(n_\Omega\cdot n_x\cdot n_P)\times (n_x\cdot n_P)$ matrix,
$\mathbf{P}$ is an $(n_x\cdot n_P) \times (n_\Omega\cdot n_x\cdot n_P)$ matrix; $\mathbf{Q}$ is an $(n_\Omega\cdot n_x\cdot n_P)$ vector assembled from the source $q$ and the inflow boundary $\alpha$; and $\mathbf{\Psi} = (\psi^{l,p,r})$ is an $(n_\Omega\cdot n_x\cdot n_P)$ vector such that
$\psi_h = \sum_{l,p,r}\psi^{l,p,r}\xi^{l,p,r}$.
If upwind values are used to evaluate the numerical trace $\widehat{u}_j$, each block of $\mathbf{L}$ can be inverted efficiently with a sweep algorithm.
The system in \eqref{eq-sndg-mat} can be solved numerically with a Krylov method by first solving the reduce system
\begin{equation}\label{eq-sndg-phi}
\mathbf{\Phi} - \mathbf{P} \mathbf{L}^{-1}\mathbf{M} \mathbf{\Phi} = \mathbf{P} \mathbf{L}^{-1} \mathbf{Q}
\end{equation}
for the ${n_x\cdot n_P}$ vector $\mathbf{\Phi}:=\mathbf{P} \mathbf{\Psi}$. This equation is derived by applying $\mathbf{L}^{-1}$ and then $\mathbf{P}$ to \eqref{eq-sndg-mat}. In a second step $\mathbf{\Psi}$ is recovered from the relation
\begin{equation}\label{eq-sndg-final-sweep}
\mathbf{\Psi} = \mathbf{L}^{-1}\mathbf{M} \mathbf{\Phi} + \mathbf{L}^{-1} \mathbf{Q}.
\end{equation}
The following theorem is proven in \Cref{ap-sndg-mat-psi}.
\begin{THM} \label{thm-sndg-mat-psi}
The matrix
$\mathbf{I}_{n_x\cdot n_P}-\mathbf{P}\mathbf{L}^{-1}\mathbf{M}$ is invertible.
\end{THM}
\begin{REM}[Sherman--Morrison formula] According to the
Sherman-Morrison formula (see for example \cite[Section 2.1.3]{golub2012matrix}): given invertible matrices $\mathbf{B} = \mathbf{A} + \mathbf{U} \mathbf{V}$ and
$\mathbf{I} + \mathbf{V} \mathbf{A}^{-1}\mathbf{U}$,
\begin{equation}\label{eq-sndg-mat-sm}
\mathbf{B}^{-1} = \mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{U} (\mathbf{I}+\mathbf{V}\mathbf{A}^{-1}\mathbf{U})^{-1}\mathbf{V} \mathbf{A}^{-1}.
\end{equation}
The direct application of \eqref{eq-sndg-mat-sm}
with $\mathbf{A} = \mathbf{L}$, $\mathbf{U} = -\mathbf{M}$ and $\mathbf{V} = \mathbf{P}$, yields the formula in \eqref{eq-sndg-final-sweep} with $\mathbf{\Phi}$ given by \eqref{eq-sndg-phi}.
\end{REM}
\subsubsection{Asymptotic scheme}
As $\varepsilon \to 0$, the $S_N$-DG scheme gives a consistent
approximation to the asymptotic diffusion problem. For simplicity, we focus here on the zero inflow boundary condition $\alpha
= 0$. The analysis of more general boundary conditions can be found in
\cite{adams2001discontinuous, guermond2010asymptotic, guermond2014discontinuous, larsen1989asymptotic}.
We use an overline to represent isotropic subspaces. For example,
\begin{equation}
\overline{\mathcal{V}}_h = \{v = (v_1,\ldots,v_{n_\Omega}) \in \mathcal{V}_h: v_i = \overline{v}, \forall i\}.
\end{equation}
We further define $\mathcal{C}_{h,\mathrm{zero}}$ to be the space of continuous functions in $\overline{\mathcal{V}}_h$ that vanish on $\partial D$. $\overline{\mathcal{V}}_h^d = \{(\vphi_1,\dots,\vphi_d): \vphi_i \in \overline{\mathcal{V}}_h\}$ is used to represent the tensor product space of $\overline{\mathcal{V}}_h$ with an induced norm still denoted as $\|\cdot\|$. In particular, since $\overline{\mathcal{V}}_h$ and $V_h$ are isomorphic,
we often identify $\overline{\mathcal{V}}_h$ with $V_h$.
To facilitate the discussion, we also define
\begin{equation}
\label{eq-Jh}
J_h = \frac{1}{\varepsilon}\sum_{j=1}^{n_\Omega} w_j\Omega_j \psi_{h,j} = \sum_{j=1}^{n_\Omega} w_j\Omega_j \frac{\psi_{h,j}-\overline{\psi}_h}{\varepsilon},
\end{equation}
which is a vector field in $\mathbb{R}^d$.
The following result is proved in \cite{guermond2010asymptotic}\footnote{The result in \cite{guermond2010asymptotic} is actually stated for more generally. In particular it allows $\alpha$ to be nonzero and possibly anisotropic.}; see also \cite{adams2001discontinuous} and \Cref{thm-lmdg-asympscheme} in this paper.
\begin{THM}[Asymptotic scheme]\label{thm-sndg-asympscheme}
Suppose $\alpha = 0$.
Then as $\varepsilon \to 0$, $(\psi_h)_{\varepsilon>0}$ and $(J_h)_{\varepsilon>0}$ converge to ${\psi}_h^{(0)} = \overline{\psi}_h^{(0)} \in
\mathcal{C}_{h,\mathrm{zero}}$ and $J_h^{(0)}\in \overline{\mathcal{V}}_{h}^d$, respectively, that are the unique solution to the mixed problem:
\begin{subequations}\label{eq-sndg-mix}
\begin{gather}
\sum_{K\in \mathcal{T}_h} \int_K \left(-J_h^{(0)}\cdot \nabla \vphi +
\sig{a}{\psi}_h^{(0)} \vphi\right) dx= \int_D q\vphi
dx,\label{eq-sndg-lim1}\\
\sum_{K\in \mathcal{T}_h}\int_K\left(\frac{1}{3}\nabla {\psi}_h^{(0)} +
\sig{s}
J_h^{(0)}\right)\cdot \zeta dx =
0\label{eq-sndg-lim2},
\end{gather}
\end{subequations}
$\forall \vphi \in \mathcal{C}_{h,\mathrm{zero}}$ and $\forall \zeta \in \overline{\mathcal{V}}_h^{d}$.
\end{THM}
\section{Low-memory strategies}\label{sc-lmdg}
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In this section, we generalize the statement of \Cref{thm-sndg-asympscheme} slightly to allow for proper subspaces of $\mathcal{V}_h$ in the finite element formulation. Based on the analysis, a first-order low-memory scheme is constructed. We then apply the reconstruction technique to lift the accuracy of the method to second-order.
\subsection{Asymptotic schemes with subspaces of $\mathcal{V}_h$}\label{subsec-asymptotic-subspaces} The results of \cref{thm-sndg-asympscheme} suggest that, rather than $\psi_{h}$, it is the approximation of the integrated quantities $\overline{\psi}_h$ and $J_h$ that play an important role in the diffusion limit. In particular, the continuity requirement on $\overline{\psi}_h^{(0)}$ plays a crucial role. Indeed, as is well known \cite{adams2001discontinuous}, if the space $V_h$ is constructed from piecewise constants, then \eqref{eq-sndg-mix} implies that ${\psi}_h^{(0)}$ is a global constant and $J_h^{(0) }= 0$. This solution is clearly inconsistent with the diffusion limit. However, it is possible to construct a DG method: \textit{find $\psi_{h}=(\psi_{h,1},\dots,\psi_{h,n_\Omega}) \in \mathcal{W}_h $ such that}
\begin{equation}\label{eq-sndglm-scheme}
B(\psi_h,v) = Q_{\alpha}(v), \qquad \forall v \in \mathcal{W}_h
\end{equation}
based on a proper subspace $\mathcal{W}_h \subset \mathcal{V}_h$ that maintains the diffusion limit, but requires fewer unknowns for a given mesh $\mathcal{T}_h$.
\begin{THM}\label{thm-gene-uni-solvency}
For each $\varepsilon>0$ and linear subspace $\mathcal{W}_h \subset \mathcal{V}_h$, \eqref{eq-gene-scheme} has a unique solution. In particular, if $\alpha = 0$, the solution
satisfies the energy estimate
\begin{equation}\label{eq-stab}
\frac{1}{\varepsilon}\|\sig{s}^{\frac{1}{2}}(\psi_h - \overline{\psi}_h)\|^2
+ \frac{\varepsilon}{2}\|\sig{a}^\frac{1}{2}\psi_h\|^2
+\frac{1}{2}\llbracket \psi_h\rrbracket^2
\leq \frac{\varepsilon}{2\delta_{\rm{a}}}\|q\|^2.
\end{equation}
\end{THM}
The proof is based on coercivity of $B(\cdot,\cdot)$ and we refer to \cite{HHE2010} and \cite{guermond2010asymptotic} for details. Here, $\alpha = 0$ is assumed for simplicity. Energy estimates with general inflow boundary condition can be found in \cite[Lemma 4.2]{guermond2010asymptotic}. In \cite{HHE2010}, the case $\varepsilon = 1$ is studied and error estimates are derived using the coercivity with respect to a modified norm.
We next characterize sufficient conditions for $\mathcal{W}_h $. Define the spaces
\begin{equation}
\overline{\Omega \mathcal{W}_h}:= \{\sum_{j=1}^{n_{\Omega}} w_j \Omega_j v_j: v\in \mathcal{W}_h\} \subset \overline{\mathcal{V}}_h^d
\quad \text{and} \quad
\Omega\cdot \overline{\Omega \mathcal{W}_h} := \{\Omega\cdot \zeta:\zeta \in \overline{\Omega \mathcal{W}_h}\} \subset \mathcal{V}_h,
\end{equation}
where $\Omega\cdot \zeta:=(\Omega_1 \cdot \zeta, \ldots, \Omega_{n_\Omega} \cdot \zeta)$.
According to \eqref {eq-Jh}, $J_h\in \overline{\Omega \mathcal{W}_h}$. \Cref{thm-sndg-asympscheme} can now be generalized to the space $\mathcal{W}_h $.
\begin{THM}\label{thm-lmdg-asympscheme}
Suppose $\alpha = 0$. Suppose $\mathcal{W}_h \subset \mathcal{V}_h$ is a linear space such that
$\Omega\cdot \overline{\Omega \mathcal{W}_h} \subset \mathcal{W}_h$.
Then as $\varepsilon \to 0$, $({\psi_h})_{\varepsilon>0}$ and $({J}_h)_{\varepsilon>0}$ converge to ${\psi}_h^{(0)} = \overline{\psi}_h^{(0)}\in
\mathcal{C}_{h,\mathrm{zero}}\cap\mathcal{W}_h$ and ${J}_h^{(0)}\in\overline{\Omega \mathcal{W}_h}$, respectively, that are the unique solution to the mixed problem
\eqref{eq-sndg-mix},
$\forall \vphi \in \mathcal{C}_{h,\mathrm{zero}}\cap \mathcal{W}_h$ and $\forall \zeta \in \overline{\Omega \mathcal{W}_h}$.
\end{THM}
\begin{proof}
Because the proof follows the arguments in \cite[Section 4]{guermond2010asymptotic} closely,
we provide only
a brief outline, emphasizing where the condition on the space $\mathcal{W}_h$ plays a role.
1. The stability estimate in \eqref{eq-stab} provides the following three bounds:
\begin{equation}\label{eq-bound}
(i)~\|\psi_h \|^2\leq \frac{1}{\delta_{\rm{a}}^2}\|q\|^2
,\quad
(ii)~\|\psi_h -\overline{\psi}_h \|^2 \leq \frac{\varepsilon^2}{\delta_{\rm{a}}
\delta_{\rm{s}}}\|q\|^2
,\quad \text{and} \quad
(iii)~ \llbracket\psi_h\rrbracket^2 \leq
\frac{\varepsilon}{\delta_{\rm{a}}}\|q\| ^2.
\end{equation}
Bounds (i) and (ii) imply that $\psi_h $ converges (via a subsequence) to a function $ {\psi}_h^{(0)} \in \overline{\mathcal{V}}_h$. Bound (iii)
implies that $\psi_h^{(0)} \in \mathcal{C}_{h,\mathrm{zero}} \cap \mathcal{W}_h = \mathcal{C}_{h,\mathrm{zero}} \cap \overline{\mathcal{W}}_h$.
2. Since, from the definition in \eqref{eq-Jh},
\begin{equation}
\|J_h \| \leq \sum_{j=1}^{n_\Omega} w_j \frac{\|\psi_h-\overline{\psi}_h\|}{\varepsilon},
\end{equation}
where $\|J_h\|$ is the tensor product norm of $J_h$ in $\overline{\mathcal{V}}^d_h$,
the bound (ii) implies further that $(J_h )_{\varepsilon>0} \subset \overline{\Omega \mathcal{W}_h}$ is uniformly bounded and hence converges subsequentially to a limit $J_h^{(0)}\in \overline{\Omega \mathcal{W}_h}$.
3. The equation in \eqref{eq-sndg-lim1} is derived by testing \eqref{eq-sndglm-scheme} with $v =\vphi \in \mathcal{C}_{h,\mathrm{zero}}\cap \mathcal{W}_h$ and
using the fact that $\vphi$ is independent of $\Omega$ and continuous in $x$.
4. It is the derivation of \eqref{eq-sndg-lim2} which uses the condition $\Omega\cdot \overline{\Omega \mathcal{W}_h} \subset \mathcal{W}_h$. Specifically, if $v = \Omega\cdot \zeta$ with $\zeta \in \overline{\Omega\mathcal{W}_{h}}$, then this condition implies that $v \in \mathcal{W}_h$. Therefore, we can test \eqref{eq-sndglm-scheme} with this choice of $v$ to find that
\begin{equation}
\begin{aligned}
L(\psi_h,\Omega \cdot \zeta) - S(\psi_h,\Omega \cdot \zeta)
=&- \sum_{j=1}^{n_\Omega} w_j \sum_{K \in \mathcal{T}_h}
\int_K \psi_{h,j} \Omega_j \cdot \nabla (\Omega_j \cdot \zeta) dx \\
&+ \sum_{j=1}^{n_\Omega} w_j \sum_{K \in \mathcal{T}_h}
\int_{\partial K} \widehat\psi_{h,j} (\Omega_j \cdot \nu_K) (\Omega_j \cdot \zeta^{\rm{int}}) dx \\
&+ \sum_{j=1}^{n_\Omega} w_j \sum_{K \in \mathcal{T}_h}
\int_K \left(
\left( \frac{\sig{s}}{\varepsilon} + \varepsilon \sig{a} \right) \psi_{h,j}
- \frac{\sig{s}}{\varepsilon} \overline{\psi}_h \right) (\Omega_j \cdot \zeta) dx \\
=:&\ I + II + III.
\end{aligned}
\end{equation}
We combine $I$ and $II$, using the fact that $ \overline{\psi}^{(0)}_{h} \in \mathcal{C}_{h,\mathrm{zero}}$ and invoking \eqref{eq-polyquad}. This gives
\begin{equation}
\label{eq-fick1}
\begin{aligned}
\lim_{\varepsilon\to 0} (I + II)
&=\sum_{j=1}^{n_\Omega} w_j (\Omega_j \otimes \Omega_j) :
\sum_{K \in \mathcal{T}_h} \left(- \int_K \overline{\psi}^{(0)}_{h} \nabla \zeta dx + \int_{\partial K} \overline{\psi}^{(0)}_{h} \nu_K \otimes \zeta^{\rm{int}} dx \right) \\
& = \frac{1}{3} \operatorname{Id} : \sum_{K \in \mathcal{T}_h} \int_K \nabla \overline{\psi}^{(0)}_{h} \otimes \zeta dx
= \sum_{K \in \mathcal{T}_h} \int_K \frac{1}{3} \nabla \overline{\psi}^{(0)}_{h} \cdot \zeta dx.
\end{aligned}
\end{equation}
Since $\sum_{j=1}^{n_\Omega} w_j \overline{\psi}_{h} \Omega_j =0 $,
\begin{align}
\label{eq-fick2}
\lim_{\varepsilon\to 0} III =
\lim_{\varepsilon\to 0} \sum_{K \in \mathcal{T}_h}
\int_K
\left( \frac{\sig{s}}{\varepsilon} + \varepsilon \sig{a} \right) \sum_{j=1}^{n_\Omega}(w_j \psi_{h,j} \Omega_j )\cdot \zeta dx =
\int_K \sig{s} J^{(0)}_h \cdot \zeta dx .
\end{align}
Finally, the right-hand side of \eqref{eq-sndglm-scheme} is (for $\alpha = 0$)
\begin{equation}
\label{eq-L0-test}
Q_0(v) = \sum_{j=1}^{n_\Omega} w_j \sum_{K \in \mathcal{T}_h} \int_K \Omega_j \cdot \zeta q dx= 0.
\end{equation}
Combining \eqref{eq-fick1}, \eqref{eq-fick2}, and \eqref{eq-L0-test} recovers \eqref{eq-sndg-lim2}.
5. Uniqueness of the subsequential limits $\psi ^{(0)}_h$ and $J^{(0)}_h$ follows from the uni-solvency of \eqref{eq-sndg-mix}. Indeed if $(\widetilde \psi_h,\widetilde J_h)$ is the difference between any two solutions of \eqref{eq-sndg-mix}, then
\begin{equation}
3\sig{s} \|\widetilde J_h\|^2 + \sig{a}\|\widetilde \psi_h \|^2 = 0.
\end{equation}
Since $\sig{s}$ and $\sig{a}$ are assumed positive, it follows that $\widetilde \psi_h$ and $\widetilde J_h$ are identically zero.
\end{proof}
We then discuss the choice of $\mathcal{W}_h$ and the corresponding space pair, $\mathcal{S}_h := \mathcal{C}_{h,\mathrm{zero}}\cap \mathcal{W}_h$ and $\mathcal{J}_h := \overline{\Omega\mathcal{W}_h}$, in the diffusion limit. Let $Z_0(K)$ be the space spanned by constants on $K$. Then we define the piecewise constant space
$
\mathcal{V}_{h,0} = \{v\in \mathcal{V}_h:v_j|_K \in Z_0(K), \forall K \in \mathcal{T}_h\}
$ and its orthogonal complement
$\mathcal{V}_{h,1} = \{v \in \mathcal{V}_h: \int_K v_j dx = 0, \forall K \in \mathcal{T}_h\}.
$
The isotropic subspace of $\mathcal{V}_{h,r}$ is denoted by $\overline{\mathcal{V}}_{h,r}$ and the subsequent product space is denoted by $\overline{\mathcal{V}}_{h,r}^d$, $r = 0,1$.
1. When $\mathcal{W}_h = \mathcal{V}_{h,0}$ or $\mathcal{W}_h =\{v\in \mathcal{V}_h:v_j|_K \in P_1(K),\forall K \in \mathcal{T}_h\}$, we have $\mathcal{S}_h = \{0\}$, which implies $\psi_h^{(0)} = 0$ and $J_h^{(0)} = 0$.
2. When $\mathcal{W}_h = \mathcal{V}_{h,0} + \overline{\mathcal{V}}_{h,1} + \Omega \cdot \overline{\mathcal{V}}_{h,1}^d$, it can be shown that $\mathcal{S}_h = \mathcal{C}_{h,\mathrm{zero}}$, $\mathcal{J}_h =\overline{\mathcal{V}}_h^d$\footnote{Since $\overline{\mathcal{V}}_h^d\supset \mathcal{J}_h = \overline{\Omega \mathcal{W}_h} \supset \overline{\Omega \left(\Omega\cdot \overline{\mathcal{V}}_h^d\right)} = \overline{\mathcal{V}}_h^d$, which forces $\mathcal{J}_h = \overline{\mathcal{V}}_h^d$. Here we have used (iii) in \eqref{eq-polyquad} for the last equality.} and $\Omega\cdot \mathcal{J}_h \subset \mathcal{W}_h$. The asymptotic scheme is the same as that of the original $S_N$-DG method.
If $\sig{s}$ and $\sig{a}$ are both piecewise constant, then the asymptotic scheme has the primal form: \textit{find $\psi_h^{(0)} \in \mathcal{C}_{h,\mathrm{zero}}$, such that
\begin{equation}\label{eq-lmdg-cg}
\sum_{K\in \mathcal{T}_h} \int_K \left(\frac{1}{3 \sig{s}}\nabla {\psi}_h^{(0)}
\cdot \nabla \vphi+\sig{a}{\psi}_h^{(0)}\vphi\right) dx= \int_D q\vphi
dx,
\end{equation}
$\forall \vphi \in \mathcal{C}_{h,\mathrm{zero}}$.} This is the classical continuous Galerkin approximation, which is stable and second-order accurate.
3. When $\mathcal{W}_h =\mathcal{V}_{h,0} + \overline{\mathcal{V}}_{h,1}$, then $\mathcal{S}_h = \mathcal{C}_{h,\mathrm{zero}}$, $\mathcal{J}_h =\overline{\mathcal{V}}_{h,0}^d$ and $\Omega\cdot \mathcal{J}_h \subset \mathcal{W}_h$. With $P^1$ elements and triangular meshes, the asymptotic scheme is essentially the $P_N$ scheme suggested by Egger and Schlottbom in \cite{egger2012mixed} with $N = 1$. If $Q^1$ elements and Cartesian meshes are used, the scheme yields the same variational form as that in \cite{egger2012mixed}, while the space pair no longer satisfies the condition $\nabla \mathcal{S}_h \subset \mathcal{J}_h$.
From another point of view, suppose $\sig{s}$ and $\sig{a}$ are piecewise constant, the primal form is: \emph{find $\psi_h^{(0)} \in \mathcal{C}_{h,\mathrm{zero}}$, such that
\begin{equation}\label{eq-difflimPF}
\sum_{K\in \mathcal{T}_h} \int_K\left( \frac{1}{3\sig{s}} \Pi_0(\nabla \psi_h^{(0)}) \cdot \Pi_0(\nabla \vphi) + \sig{a}\psi_h^{(0)}\vphi\right)dx =
\sum_{K\in \mathcal{T}_h} \int_K q \vphi dx,
\end{equation}
$\forall \vphi \in \mathcal{C}_{h,\mathrm{zero}}$.}
For $P^1$ elements on triangular meshes, \eqref{eq-difflimPF} is identical to \eqref{eq-lmdg-cg}.
For $Q_1$ elements on Cartesian meshes, one can show that \eqref{eq-difflimPF} is unisolvent. Furthermore, $\|\Pi_0(\nabla \psi_h^{(0)})\|^2 + \|\psi_h^{(0)}\|^2 \leq \max(\frac{3\sig{s}}{2\sig{a}},\sig{a}^{-2}) \|q\|^2$, if $\sig{a}\geq \delta_{\mathrm{a}} >0$. While the accuracy is hard to analyze under the finite element framework. Assume a uniform square mesh with cell length $h$. Let $\sig{s}$ and $\sig{a}$ be globally constant. Then \eqref{eq-difflimPF} can be rewritten as a finite difference scheme under the Lagrange basis functions.
\begin{align}
-\frac{\psi_{i-1,j-1}+\psi_{i-1,j+1}-4\psi_{i,j}+\psi_{i+1,j-1}+\psi_{i+1,j+1}}{3\sig{s}\cdot 2h^2}
+ \sig{a}A[\psi_{i,j}] = & A[q_{i,j}],\\
A[\psi_j] := \frac{1}{36} \left(\psi_{i-1,j-1}+\psi_{i-1,j+1}+\psi_{i+1,j-1}+\psi_{i+1,j+1}\right) &\\
+ \frac{1}{9}\left(\psi_{i-1,j}+\psi_{i,j-1}+\psi_{i,j+1}+\psi_{i+1,j}\right)\ + \frac{4}{9}\psi_{i,j}.& \nonumber
\end{align}
The truncation error of the method is $\mathcal{O}(h^2)$.
\
At the first glance,
$\mathcal{W}_h = \mathcal{V}_{h,0} + \overline{\mathcal{V}}_{h,1} + \Omega \cdot \overline{\mathcal{V}}_{h,1}$ seems to be the natural choice for constructing the low-memory scheme that preserves the correct diffusion limit. However, coupling between angles requires special treatment for reducing the system dimension. The extra moments $\Omega \cdot \overline{\mathcal{V}}_{h,1}^{d}$ will make the resulting system even larger than that of the original $S_N$-DG method. Although it may be worth to include extra moments for problems with anisotropic scattering, for which a large system has to be solved anyway, we avoid this option for solving \eqref{eq-main}. We therefore explore the other choice $\mathcal{W}_h = \mathcal{V}_{h,0} + \overline{\mathcal{V}}_{h,1}$ in the rest of the paper.
\subsection{Low-memory scheme}
Based on the analysis and discussion of \Cref {subsec-asymptotic-subspaces}, we propose a scheme that uses the finite element space
\begin{equation}
\mathcal{V}_h^{\mathrm{lm}} = \mathcal{V}_{h,0}+\overline{\mathcal{V}}_{h,1}.
\end{equation}
The low-memory $S_N$-DG scheme
is written as follows: \textit{find $\psi_h \in \mathcal{W}_h$, such that
\begin{equation}\label{eq-gene-scheme}
B(\psi_h,v) = Q_{\alpha}(v), \qquad \forall v \in \mathcal{V}_h^\mathrm{lm},
\end{equation}}
where $B$ and $Q_\alpha$ are defined in \eqref{eq-Bdef} and \eqref{eq-lalpha}, respectively.
We now show that this scheme can be implemented using sweeps; i.e., a strategy analagous to the \eqref{eq-sndg-phi} and \eqref{eq-sndg-final-sweep}, which relies heavily on the fast inversion of the operator $\mathbf{L}$. For simplicity, we only consider the case $\sig{s}$ being piecewise constant. The implementation is based on the block matrix formulation \eqref{eq-sndg-mat} of the $S_N$-DG method:
\begin{equation}
\mathbf{L} = \left[\begin{matrix} \mathbf{L}_{00}&\mathbf{L}_{01}\\ \mathbf{L}_{10}&
\mathbf{L}_{11}\end{matrix}\right], \quad
\mathbf{S} = \left[\begin{matrix}\mathbf{M}_0\mathbf{P}_0&\\&\mathbf{M}_1\mathbf{P}_1\end{matrix}\right],\quad \text{and} \quad
\mathbf{Q} = \left[\begin{matrix}\mathbf{Q}_0\\\mathbf{Q}_1\end{matrix}\right].
\end{equation}
Here $\mathbf{L}_{rr'}$ are matrix blocks associated to $L(u,v)$ with $u\in \mathcal{V}_{h,r'}, v\in \mathcal{V}_{h,r}$. The sizes of $\mathbf{L}_{00}$, $\mathbf{L}_{01}$, $\mathbf{L}_{10}$ and $\mathbf{L}_{11}$ are $(n_\Omega\cdot n_x)\times (n_\Omega\cdot n_x)$, $(n_\Omega\cdot n_x)\times (n_\Omega\cdot n_x\cdot (n_P-1))$, $(n_\Omega\cdot n_x\cdot (n_P-1))\times (n_\Omega\cdot n_x)$ and $(n_\Omega\cdot n_x\cdot (n_P-1))\times (n_\Omega\cdot n_x\cdot (n_P-1))$, respectively.
The block $\mathbf{S}_{rr'} = \mathbf{M}_{r} \mathbf{P}_{r'}$ is associated to $S(u,v)$ with $u\in \mathcal{V}_{h,r'}, v\in \mathcal{V}_{h,r}$; it has the same size as $\mathbf{L}_{rr'}$. The matrices $\mathbf{M}_0$ and $\mathbf{P}_0$ have dimensions $(n_\Omega\cdot n_x)\times n_x$ and $n_x \times(n_\Omega\cdot n_x)$, respectively; the matrices $\mathbf{M}_1$ and $\mathbf{P}_1$ have dimensions $(n_\Omega\cdot n_x\cdot(n_P-1))\times (n_x\cdot(n_P-1))$ and $(n_x\cdot (n_P-1)) \times(n_\Omega\cdot n_x\cdot (n_P-1))$, respectively. The vector block $\mathbf{Q}_r$ is associated to $Q_{\alpha}(v)$ for $v\in \mathcal{V}_{h,r}$, with $\mathbf{Q}_0$ an $n_\Omega\cdot n_x$ vector and $\mathbf{Q}_1$ an $n_\Omega\cdot n_x\cdot (n_P-1)$ vector.
Recall from \Cref{sc-basis} that for each $p$, $\{b^{p,r}\}_{r=0}^{n_P-1}$ forms a basis for $Z_1(K_p)$, and $\xi^{l,p,r}_j = \delta_{lj}b^{p,r}$. We further assume $\{b^{p,r}\}_{r=0}^{n_P-1}$ is an orthogonal set and $\{b^{p,0}\}_{p=1}^{n_x}$ is a set of constant functions on $K_p$. Then $\mathbb{B}_0 = \{\xi^{l,p,0}:l = 1,\dots,n_\Omega, p = 1,\dots,n_x\}$ and $\mathbb{B}_1 = \{\xi^{l,p,r}:l = 1,\dots,n_\Omega, p = 1,\dots,n_x,r = 1,\dots,n_P-1\}$ are sets of basis functions for $\mathcal{V}_{h,0}$ and $\mathcal{V}_{h,1}$, respectively.
Let $\mathbb{B}_1^\mathrm{lm} = \{\eta^{p,r}: \eta^{p,r}_j = b^{p,r},j =1,\dots,n_\Omega, p = 1,\dots,n_x, r =1,\dots,n_P-1\}$. Then $\mathbb{B}_1^\mathrm{lm}$
is a set of basis for $\overline{\mathcal{V}}_{h,1}$. Hence $\mathcal{V}_h^\mathrm{lm} = \mathrm{span} \{\mathbb{B}_0,\mathbb{B}_1^\mathrm{lm}\}$.
The dimension of $\mathcal{V}_h^\mathrm{lm}$ is then $n_\Omega\cdot n_x + n_x\cdot (n_P-1)$.
Because $\eta^{p,r}_j = b^{p,r} = \sum_{l=1}^{n_\Omega}\delta_{lj}b^{p,r}= \sum_{l=1}^{n_\Omega} \xi_j^{l,p,r}$, there exists a mapping from $\mathbb{B}_1$ to $\mathbb{B}_1^\mathrm{lm}$
\begin{equation}
\eta^{p,r} = \sum_{l=1}^{n_\Omega}\xi^{l,p,r} = \sum_{l',p',r'=1}^{n_\Omega}\Sigma^{(p,r),(l',p',r')}\xi^{l',p',r'},
\end{equation}
where $\mathbf{\Sigma} = (\Sigma^{(p,r),(l',p',r')})$ is an $(n_x\cdot (n_P-1))\times
(n_\Omega\cdot n_x \cdot (n_P-1))$ matrix with components
$\Sigma^{(p,r),(l',p',r')} = \delta_{pp'}\delta_{rr'}$. The matrix $\mathbf{\Sigma}$ corresponds to a
summation operator that maps an angular flux to a scalar flux, while $\mathbf{\Sigma}^T$ copies
the scalar flux to each angular direction.
Let the solution of the low-memory method be represented by
$\mathbf{\Psi} = \left[\mathbf{\Psi}_0, \mathbf{\Sigma}^T\mathbf{\Phi}_1\right]^T$.
Using the fact $\mathbf{P}_1\mathbf{\Sigma}^T = \mathbf{I}_{n_x\cdot (n_P-1)}$, one can show $\mathbf{\Psi}$ satisfies the equations
\begin{subequations}\label{eq-lmdg-block}
\begin{gather}
\mathbf{L}_{00}\mathbf{\Psi}_0+\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{\Phi}_1 = \mathbf{M}_0\mathbf{P}_0\mathbf{\Psi}_0 +
\mathbf{Q}_0,\label{eq-lmdg-block0}\noeqref{eq-lmdg-block0}\\
\mathbf{\Sigma}\mathbf{L}_{10}\mathbf{\Psi}_0+\mathbf{\Sigma}\mathbf{L}_{11}\mathbf{\Sigma}^T\mathbf{\Phi}_1 =
\mathbf{\Sigma}\mathbf{M}_1\mathbf{\Phi}_1 + \mathbf{\Sigma}\mathbf{Q}_1.\label{eq-lmdg-block1}\noeqref{eq-lmdg-block1}
\end{gather}
\end{subequations}
As that in the original $S_N$-DG method, the system dimension of \eqref{eq-lmdg-block} can be reduced with the following procedure.
\
1. Solve for $\mathbf{\Phi}_1$ in terms of $\mathbf{\Psi}_0$ through \eqref{eq-lmdg-block1}:
\begin{equation}\label{eq-lmdg-psi1}
\mathbf{\Phi}_1 = \mathbf{B}_{11}^{-1}\mathbf{\Sigma} \left(-\mathbf{L}_{10} \mathbf{\Psi}_0 +
\mathbf{Q}_1\right), \qquad \mathbf{B}_{11} = \mathbf{\Sigma} \mathbf{L}_{11}\mathbf{\Sigma}^T -
\mathbf{\Sigma}\mathbf{M}_1.
\end{equation}
2. Substitute $\mathbf{\Phi}_1$ from \eqref{eq-lmdg-psi1} into \eqref{eq-lmdg-block0} to
obtain a closed equation for $\mathbf{\Psi}_0$:
\begin{equation}
\begin{aligned}\label{eq-lmdg-psi0}
\mathbf{\Psi}_0
- \mathbf{L}_{00}^{-1} \mathbf{M}_0 (\mathbf{P}_0 \mathbf{\Psi}_0)
&-\mathbf{L}_{00}^{-1} \mathbf{L}_{01}\mathbf{\Sigma}^T(\mathbf{B}_{11}^{-1} \mathbf{\Sigma}\mathbf{L}_{10} \mathbf{\Psi}_0)
= \mathbf{L}_{00}^{-1} (\mathbf{Q}_0-
\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\mathbf{Q}_1).
\end{aligned}
\end{equation}
3. Apply $\mathbf{P}_0$ and $\mathbf{\Sigma}\mathbf{L}_{10}$ to \eqref{eq-lmdg-psi0} to
obtain a closed system for $\mathbf{X}_0 = \mathbf{P}_0\mathbf{\Psi}_0$ and {$\mathbf{X}_1 = \mathbf{B}_{11}^{-1}\mathbf{\Sigma}\mathbf{L}_{10}\mathbf{\Psi}_0$}:
\begin{equation}\label{eq-lmdg-reducemat}
\begin{aligned}
\mathbf{K}
\left[\begin{matrix}\mathbf{X}_0\\\mathbf{X}_{1} \end{matrix}\right]=\left[\begin{matrix}
\mathbf{P}_0\\
\mathbf{\Sigma}\mathbf{L}_{10}
\end{matrix}\right]\mathbf{L}_{00}^{-1} (\mathbf{Q}_0-
\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\mathbf{Q}_1),
\end{aligned}
\end{equation}
where
\begin{equation}
\mathbf{K} = \left[\begin{matrix}
\mathbf{I}_{n_x}- \mathbf{P}_0\mathbf{L}_{00}^{-1} \mathbf{M}_0 & -\mathbf{P}_0\mathbf{L}_{00}^{-1}\mathbf{L}_{01}\mathbf{\Sigma}^T\\
- \mathbf{\Sigma}\mathbf{L}_{10}\mathbf{L}_{00}^{-1} \mathbf{M}_0 &\mathbf{B}_{11} -\mathbf{\Sigma}\mathbf{L}_{10}\mathbf{L}_{00}^{-1}\mathbf{L}_{01}\mathbf{\Sigma}^T \\
\end{matrix}\right].
\end{equation}
4. Solve for $\mathbf{X}_0$ and $\mathbf{X}_1$ in \eqref{eq-lmdg-reducemat}.
Then use \eqref{eq-lmdg-psi0} and \eqref{eq-lmdg-psi1} to obtain $\mathbf{\Psi}$:
\begin{subequations}
\label{eq-lm-step4}
\begin{align}
\label{eq-lm-step4-phi}
\mathbf{\Psi}_0 &= \mathbf{L}_{00}^{-1} \mathbf{M}_0 \mathbf{X}_0 + \mathbf{L}_{00}^{-1}\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{X}_1 + \mathbf{L}_{00}^{-1} (\mathbf{Q}_0-
\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\mathbf{Q}_1),\\
\label{eq-lm-step4-psi}
\mathbf{\Phi}_1 &= \mathbf{B}_{11}^{-1} \mathbf{\Sigma}\left(-\mathbf{L}_{10} \mathbf{\Psi}_0 + \mathbf{Q}_1\right).
\end{align}
\end{subequations}
\
Only Step 4 above is needed to implement the algorithm.
If one solves for $\mathbf{\Psi}_0$ directly from \eqref{eq-lmdg-psi0},
then an $(n_\Omega\cdot n_x)\times(n_\Omega\cdot n_x)$ matrix should be
inverted. While with \eqref{eq-lmdg-reducemat},
the matrix dimensions are reduced to $(n_x\cdot n_P)\times(n_x\cdot n_P)$.
Typically $n_P$ is much smaller than $n_\Omega$.
We state the following theorems on the invertibility of $\mathbf{B}_{11}$ and $\mathbf{K}$, whose proof can be found in \Cref{ap-B11} and \Cref{ap-K}, respectively.
\begin{THM}\label{lem-B11}
$\mathbf{B}_{11}$ is invertible.
Furthermore, if the quadrature rule is central symmetric, then
$\mathbf{B}_{11}$ is symmetric positive definite. Here, central symmetry means $\Omega_j$ and $-\Omega_j$ are
both selected in the quadrature rule and their weights are equal $w_j = w_{-j}$.
\end{THM}
\begin{THM}\label{thm-K}
$\mathbf{K}$ is invertible.
\end{THM}
\begin{REM}
Typically, the linear system in such context is solved using the Krylov
method, in which one needs to evaluate the multiplication
of a vector with $\mathbf{K}$ in each iteration. We can use
the following formula to avoid repeated evaluation
in the left multiplication of $\mathbf{K}$.
\begin{equation}\label{eq-nonrec-block}
\mathbf{K} \left[\begin{matrix}\mathbf{X}_0\\\mathbf{X}_{1} \end{matrix}\right]=
\left[\begin{matrix}
\mathbf{I}_{n_x} & \\
&\mathbf{B}_{11}\\
\end{matrix}\right]
\left[\begin{matrix}
\mathbf{X}_0 \\ \mathbf{X}_1
\end{matrix}\right]
-
\left[\begin{matrix}
\mathbf{P}_0 \\
\mathbf{\Sigma}\mathbf{L}_{10}\\
\end{matrix}\right]\mathbf{L}_{00}^{-1}
\left(\mathbf{M}_0\mathbf{X}_0+\mathbf{L}_{01}\mathbf{\Sigma}^T\mathbf{X}_1\right).
\end{equation}
\end{REM}
\begin{REM}
As demonstrated in \cite{heningburg2019hybrid}, the inversion of the block $\mathbf{L}_{00}$ in \eqref{eq-lmdg-reducemat}, rather than the full matrix $\mathbf{L}$ in \eqref{eq-sndg-phi}, results in a significant savings in terms of floating point operations (and hence time-to-solution). This savings will be partially offset by the need to invert the matrix $\mathbf{B} _{11}$ in \eqref{eq-lmdg-psi1}. However, since the overall effect on time-to-solution depends heavily on the details of implementation, we do not investigate this aspect of the low-memory method in the numerical results, but instead leave such an investigation to future work.
\end{REM}
\subsection{Reconstructed low-memory scheme}\label{sc-rlmdg}
Because the low-memory scheme couples the angular components of $\mathcal{V}_{h,1}$, it is only first-order for fixed $\varepsilon > 0$. To recover second-order accuracy (formally), we introduce a spatial reconstruction procedure to approximate the anisotropic parts of $\mathcal{V}_{h,1}$.
\subsubsection{Numerical scheme}\label{sc-rec}
We denote by
$\Pi_i$ the orthogonal projection from $\mathcal{V}_{h}$ to $\mathcal{V}_{h,i}$, $i = 0,1$.
The only information from the low-memory space $\mathcal{V}_h^{\mathrm{lm}}$ retains from $v \in \mathcal{V}_{h,1}$ is $\overline{\Pi_1(v)}$; the information contained in
$\Pi_1 (v) - \overline{\Pi_1 (v)}$ is missing. We therefore introduce an operator $R_\alpha^*v =
R_\alpha \Pi_0(v) - \overline{R_\alpha \Pi_0(v)}$, where $R_\alpha\Pi_0$ is an operator that returns the reconstructed
slopes using piecewise constants and the boundary condition $\alpha$, to rebuild the difference.
Then the reconstructed scheme is written as: \textit{find $\psi_h \in \mathcal{V}_h^\mathrm{lm}$ such that
\begin{equation}\label{eq-rlmdg-vari}
B(\psi_h + R_\alpha^*\psi_h, v) = Q_{\alpha}(v),\qquad
\forall v\in \mathcal{V}_h^\mathrm{lm}.
\end{equation}}
The reconstruction $\psi_h + R_\alpha^*\psi_h$ then gives a more accurate approximation to $\Psi$.
Equivalently, by assembling all boundary terms into the right hand side, the
reconstructed scheme can also be formulated as a Petrov--Galerkin method with
trial function space
\begin{equation}\mathcal{V}_h^{\mathrm{rlm}} = \{v + R_0^*v: v\in \mathcal{V}_{h}^{\mathrm{lm}}\}.\end{equation}
Since $R_0^* 0 = 0$, $\mathcal{V}_h^{\mathrm{rlm}}$ is in fact a linear space. With this formulation, the reconstructed method solves the following problem: \textit{find $\psi_{h,R_0} \in
\mathcal{V}_{h}^{\mathrm{rlm}}$, such that
\begin{equation}\label{eq-rlmdg-scheme}
B(\psi_{h,R_0},v) = \widetilde{Q}_{\alpha}(v), \qquad \forall v \in \mathcal{V}_h^\mathrm{lm}.
\end{equation}}
The use of different trial and test functions spaces make the analysis of this scheme less transparent.
Currently, we have no theoretical guarantee of unisolvency or the numerical diffusion limit. We observe, however, that the method recovers second-order convergence for several different test problem across a wide range of $\varepsilon$.
\
In this paper, we apply the reconstruction suggested in \cite{heningburg2019hybrid} to recover slopes for simplicity, although in general other upwind approaches can also be used\footnote{For example, one can apply upwind reconstruction with wider stencils to improve the accuracy with an increased computational costs.
Furthermore, the reconstruction can also be different at different spatial cells along different collocation angles, which may lead to an adaptive version of the reconstructed method. We postpone the discussion on numerical efficiency with different reconstruction methods to future work.}.
For illustration,
we consider a uniform Cartesian mesh
on $[0,1]\times [0,1]\times [0,1]$. The grid points are labeled from
$\frac{1}{2}$
to $n+\frac{1}{2}$ respectively. We denote by $u^0_{i,j,k}$ the cell average of
$u$ on the cell $K_{i,j,k}$ that
centers at $(x_i, y_j, z_k)$. Along each direction
$\Omega = (\Omega_x,\Omega_y,\Omega_z)$,
\begin{equation}(R_\alpha\Pi_0 (u))|_{K_{i,j,k}} = (\delta_x^{s_x}u^0_{i,j,k}) (x - x_i) +
(\delta_y^{s_y}u^0_{i,j,k}) (y - y_j) + (\delta_z^{s_z}u^0_{i,j,k}) (z -
z_k),\end{equation}
where $s_x = - \text{sign} (\Omega_x)$,
\begin{align}
\delta^-_x u^0_{i,j,k} &=
\begin{cases}
\dfrac{u^0_{i,j,k} - u^{0}_{i-1,j,k}}{h},
& 2\leq i\leq n, \\
\dfrac{u^{0}_{1,j,k}-\alpha(\Omega,(0,y_j,z_k))}{h/2},
& i = 1,
\end{cases} \\
\delta^+_x u^0_{i,j,k} &=
\begin{cases}
\dfrac{u^{0}_{i+1,j,k} - u^0_{i,j,k}}{h}
& 1\leq i\leq n-1, \\
\dfrac{\alpha(\Omega,(1,y_j,z_k))-u^{0}_{n,j,k}}{h/2},
& i = n.
\end{cases}
\end{align}
$\delta_y^\pm$ and $\delta_z^\pm$ are defined similarly. For numerical results in the next section, we only reconstruct the $P^1$ slopes to recover the second-order accuracy;
$Q^1$ type reconstruction gives similar results in terms of the
convergence rate.
\subsubsection{Implementation}
Let $\mathbb{B}_0^\mathrm{rlm}= \{\xi^{l,p,0}+R_0^*\xi^{l,p,0}:l=1,\dots,n_\Omega, p =1,\dots,n_x\}$
and $\mathcal{V}_h^{\mathrm{rlm}} = \mathrm{span} \{\mathbb{B}_0^\mathrm{rlm},\mathbb{B}_1^\mathrm{lm}\}$.
As in the first-order method, the total degrees of freedom is $n_\Omega\cdot n_x+ n_x\cdot(n_P-1)$. The
boundary terms are assembled into a vector $\mathbf{r}_\alpha$. Here we use
$\mathbf{\Psi} = \left[\mathbf{\Psi}_0, \mathbf{\Psi}_1 \right]^T
to represent the solution of the reconstructed method, where $\mathbf{\Psi}_1 = \mathbf{\Sigma}^T\mathbf{\Phi}_1+(\mathbf{I}_{n_\Omega\cdot
n_x\cdot(n_P-1)}-\mathbf{\Sigma}^T\mathbf{P}_1)(\mathbf{R}
\mathbf{\Psi}_0 +\mathbf{r}_\alpha)$. Note $\mathbf{P}_1\mathbf{\Sigma}^T = \mathbf{I}_{n_x\cdot (n_P-1)}$, which implies $\mathbf{P}_1\mathbf{\Psi}_1 = \mathbf{\Phi}_1$. The block matrix form can then be written as follows.
\begin{subequations}\label{eq-rlmdg-block}
\begin{eqnarray}
\mathbf{L}_{00}\mathbf{\Psi}_0+\mathbf{L}_{01}\mathbf{\Psi}_1&=& \mathbf{M}_0\mathbf{P}_0\mathbf{\Psi}_0 +
\mathbf{Q}_0,\\
\mathbf{\Sigma}\mathbf{L}_{10}\mathbf{\Psi}_0+\mathbf{\Sigma}\mathbf{L}_{11}
\mathbf{\Psi}_1 &=&
\mathbf{\Sigma}\mathbf{M}_1\mathbf{\Phi}_1 + \mathbf{\Sigma}\mathbf{Q}_1.
\end{eqnarray}
\end{subequations}
With
\begin{eqnarray}
\widetilde{\mathbf{L}}_{00} &=& \mathbf{L}_{00} + \mathbf{L}_{01} \mathbf{R},\\
\widetilde{\mathbf{L}}_{10} &=& \mathbf{L}_{10} + \mathbf{L}_{11}(\mathbf{I}_{n_\Omega\cdot n_x\cdot(n_P-1)}-\mathbf{\Sigma}^T\mathbf{P}_1)\mathbf{R},\\
\widetilde{\mathbf{Q}}_0 &=& \mathbf{Q}_0 - \mathbf{L}_{01}(\mathbf{I}_{n_\Omega\cdot
n_x\cdot(n_P-1)}-\mathbf{\Sigma}^T\mathbf{P}_1)\mathbf{r}_\alpha,\\
\widetilde{\mathbf{Q}}_1 &=& \mathbf{Q}_1 - \mathbf{L}_{11}(\mathbf{I}_{n_\Omega\cdot
n_x\cdot(n_P-1)}-\mathbf{\Sigma}^T\mathbf{P}_1)\mathbf{r}_\alpha,
\end{eqnarray}
one can rewrite \eqref{eq-rlmdg-block} as
\begin{subequations}\label{eq-rlmdg-block-tilde}
\begin{eqnarray}
\widetilde{\mathbf{L}}_{00}\mathbf{\Psi}_0+\mathbf{L}_{01}\mathbf{\Sigma}^T\left(\mathbf{\Phi}_1 -\mathbf{P}_1\mathbf{R}
\mathbf{\Psi}_0 \right)&=& \mathbf{M}_0\mathbf{P}_0\mathbf{\Psi}_0 +
\widetilde{\mathbf{Q}}_0,\label{eq-rlmdg-block0}\\
\mathbf{\Sigma}\widetilde{\mathbf{L}}_{10}\mathbf{\Psi}_0+\mathbf{\Sigma}\mathbf{L}_{11}\mathbf{\Sigma}^T\mathbf{\Phi}_1
&=&
\mathbf{\Sigma}\mathbf{M}_1\mathbf{\Phi}_1 + \mathbf{\Sigma}\widetilde{\mathbf{Q}}_1.\label{eq-rlmdg-block1}
\noeqref{eq-rlmdg-block0,eq-rlmdg-block1}
\end{eqnarray}
\end{subequations}
We follow the procedure as before to reduce the system dimension.
\
1. Solve for $\mathbf{\Phi}_1$ in terms of $\mathbf{\Psi}_0$ through \eqref{eq-rlmdg-block1}:
\begin{equation}\label{eq-rlmdg-psi1}
\mathbf{\Phi}_1 = \mathbf{B}_{11}^{-1} \mathbf{\Sigma}\left(-\widetilde{\mathbf{L}}_{10} \mathbf{\Psi}_0 +
\widetilde{\mathbf{Q}}_1\right), \qquad \mathbf{B}_{11} = \mathbf{\Sigma} {\mathbf{L}}_{11}\mathbf{\Sigma}^T -
\mathbf{\Sigma}\mathbf{M}_1.
\end{equation}
2. Substitute $\mathbf{\Phi}_1$ from \eqref{eq-rlmdg-psi1} into \eqref{eq-rlmdg-block0} to
obtain a closed equation for $\mathbf{\Psi}_0$:
\begin{equation}\label{eq-rlmdg-psi0}
\begin{aligned}
\mathbf{\Psi}_0
- \widetilde{\mathbf{L}}_{00}^{-1} \mathbf{M}_0 (\mathbf{P}_0 \mathbf{\Psi}_0)
&-\widetilde{\mathbf{L}}_{00}^{-1} {\mathbf{L}}_{01}\mathbf{\Sigma}^T(\mathbf{B}_{11}^{-1}
\mathbf{\Sigma}\widetilde{\mathbf{L}}_{10} +\mathbf{P}_1
\mathbf{R})\mathbf{\Psi}_0\\
&= \widetilde{\mathbf{L}}_{00}^{-1} (\widetilde{\mathbf{Q}}_0-
{\mathbf{L}}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\widetilde{\mathbf{Q}}_1).
\end{aligned}
\end{equation}
3. Applying $\mathbf{P}_0$ and $\mathbf{\Sigma}\widetilde{\mathbf{L}}_{10}$ to \eqref{eq-rlmdg-psi0}, to obtain a closed system for $\mathbf{X}_0 = \mathbf{P}_0\mathbf{\Psi}_0$ and $\mathbf{X}_1 = (\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\widetilde{\mathbf{L}}_{10}+\mathbf{P}_1\mathbf{R})\mathbf{\Psi}_0$:
\begin{equation}\label{eq-rlmdg-reducemat}
\begin{aligned}
\widetilde{\mathbf{K}}
\left[\begin{matrix}\mathbf{X}_0\\\mathbf{X}_{1} \end{matrix}\right]
&=\left[\begin{matrix}
\mathbf{P}_0\\
\mathbf{\Sigma}\mathbf{L}_{10}
\end{matrix}\right]\widetilde{\mathbf{L}}_{00}^{-1} (\widetilde{\mathbf{Q}}_0-
{\mathbf{L}}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\widetilde{\mathbf{Q}}_1),
\end{aligned}
\end{equation}
where
\begin{equation}
\widetilde{\mathbf{K}} = \left[\begin{matrix}
\mathbf{I}_{n_x}- \mathbf{P}_0\widetilde{\mathbf{L}}_{00}^{-1} \mathbf{M}_0 &
-\mathbf{P}_0\widetilde{\mathbf{L}}_{00}^{-1}{\mathbf{L}}_{01}\mathbf{\Sigma}^T\\
- \mathbf{\Sigma}\widetilde{\mathbf{L}}_{10}\widetilde{\mathbf{L}}_{00}^{-1} \mathbf{M}_0 &\mathbf{B}_{11}
-\mathbf{\Sigma}\widetilde{\mathbf{L}}_{10}\widetilde{\mathbf{L}}_{00}^{-1}{\mathbf{L}}_{01}\mathbf{\Sigma}^T \\
\end{matrix}\right].
\end{equation}
4. Solve for $\mathbf{X}_0$ and $\mathbf{X}_1$ in \eqref{eq-rlmdg-reducemat}.
Use \eqref{eq-rlmdg-psi0} and \eqref{eq-rlmdg-psi1} to recover $\mathbf{\Psi}$.
\begin{eqnarray}
\mathbf{\Psi}_0 &=& \widetilde{\mathbf{L}}_{00}^{-1} \mathbf{L}_{01}\mathbf{X}_0 + \widetilde{\mathbf{L}}_{00}^{-1}
\mathbf{M}_0 \mathbf{X}_1 + \mathbf{L}_{00}^{-1} (\widetilde{\mathbf{Q}}_0-
{\mathbf{L}}_{01}\mathbf{\Sigma}^T\mathbf{B}_{11}^{-1}\mathbf{\Sigma}\widetilde{\mathbf{Q}}_1),\\
\mathbf{\Phi}_1 &=& \mathbf{B}_{11}^{-1}\mathbf{\Sigma} \left(-\widetilde{\mathbf{L}}_{10} \mathbf{\Psi}_0 +
\widetilde{\mathbf{Q}}_1\right).
\end{eqnarray}
\
As the first-order method, only Step 4 is used in the implementation. Since only upwind information is used, $\widetilde{\mathbf{L}}_{00}$ is invertible and can be inverted with sweeps along each
angular direction. Note $\mathbf{B}_{11}$ is invertible, as has been pointed out in \Cref{ap-B11}. One can follow the argument in \Cref{ap-K} to show $\widetilde{\mathbf{K}}$ is invertible if the scheme \eqref{eq-rlmdg-vari} is unisolvent.
\section{Numerical tests}\label{sc-num}
\setcounter{equation}{0}
\setcounter{figure}{0}
\setcounter{table}{0}
In this section, we present numerical tests to examine performance of the methods.
\subsection{One dimensional tests (slab geometry)}
In slab geometries, the radiative transport equation takes the form (see, e.g., \cite[Page 28]{Lewis-Miller-1984}).
\begin{align}\label{eq-1d}
\mu \partial_x\psi (\mu,x) + \left(\frac{\sig{s}}{\varepsilon} +\varepsilon
\sig{a}\right) \psi(\mu,x) &=
\frac{\sig{s}}{2\varepsilon}\int_{-1}^1\psi(\mu',x)d\mu' + \varepsilon
q(x),\\
\psi(\mu,x_a) = \psi_l(\mu), \text{ if }\mu\geq0, &\quad \text{and} \quad \psi(\mu,x_b) = \psi_r(\mu), \text{ if }\mu<0,
\end{align}
where $x\in [x_a,x_b]$ and $\mu \in [-1,1]$. We will compare the $S_N$-$P^0$-DG scheme, $S_N$-$P^1$-DG scheme, low-memory
scheme (LMDG)
and the reconstructed scheme (RLMDG). Numerical error is evaluated in $L^1$ norm.
\begin{example}\label{examp-1dfab}
We first examine convergence rates of the methods using fabricated solutions. Let $\varepsilon = 1$, $\sig{s} = 1$,
$\sig{a} = 1$ and $D = [0,1]$.
Assuming the exact solution $\psi$, we compute the source term
$q$ and the inflow boundary conditions $\psi_l$ and $\psi_r$ accordingly.
With this approach, it may happen that $q$ depends on $\mu$. We use the $32$ points Gauss quadrature on $[-1,1]$ for $S_N$ discretization.
We consider the case $\psi = \cos x$ and $\psi = \cos(x+\mu)$. The
results are documented in \Cref{tab-1d-aniso}.
When $\psi$ is isotropic, the low-memory scheme
exhibits second-order convergence.
For the anisotropic case, the LMDG scheme
degenerates to first-order accuracy, while the RLMDG scheme
remains second-order accurate.
\begin{table}[h!]
\footnotesize
\centering
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline
&&\multicolumn{2}{c|}{$P^0$-DG}&\multicolumn{2}{c|}{$P^1$-DG}&\multicolumn{2}{c|}{LMDG
}&\multicolumn{2}{c}{RLMDG}\\
\hline
$\psi$&$h$&error&order&error&order&error&order&error&order\\
\hline
\multirow{2}{*}{$\cos x$}
&$1/20$ &3.79e-3& - & 2.48e-5& -& 2.24e-5& -& 9.14e-5&
- \\
&$1/40$ &1.91e-3& 0.99 & 6.27e-6& 1.98& 5.62e-6& 2.00& 2.31e-5&
1.99\\
&$1/80$ &9.55e-4& 1.00 & 1.58e-6& 1.99& 1.41e-6& 2.00& 5.80e-6&
1.99\\
&$1/160$&4.78e-4& 1.00 & 3.96e-7& 2.00& 3.52e-7& 2.00& 1.45e-6&
2.00\\
\hline
\multirow{2}{*}{$\cos (x+\mu)$}
&$1/20$& 4.70e-3& -& 2.11e-5& -& 3.20e-3& -& 7.74e-5&
- \\
&$1/40$& 2.36e-3& 0.99& 5.36e-6&1.98& 1.60e-3& 1.00& 1.95e-5&
1.99\\
&$1/80$& 1.18e-3& 1.00& 1.35e-6&1.99& 8.01e-4& 1.00& 4.89e-6&
1.99\\
&$1/160$& 5.91e-4& 1.00& 3.39e-7&1.99& 4.01e-4& 1.00& 1.23e-6&
2.00\\
\hline
\end{tabular}
\caption{Accuracy test for \Cref{examp-1dfab}.}\label{tab-1d-aniso}
\end{table}
\begin{figure}
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-nx-iso-eps-converted-to.pdf}
\caption{Number of mesh cells.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-mc-iso-eps-converted-to.pdf}
\caption{Degrees of freedom.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-sd-iso-eps-converted-to.pdf}
\caption{System dimension.}
\end{subfigure}
\\
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-nx-aniso-eps-converted-to.pdf}
\caption{Number of mesh cells.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-mc-aniso-eps-converted-to.pdf}
\caption{Degrees of freedom.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{1derr-sd-aniso-eps-converted-to.pdf}
\caption{System dimension.}
\end{subfigure}
\caption{Numerical efficiency in \Cref{examp-2daccu}. The first row is for isotropic test $\psi = \cos(x)$ and the second row is for anisotropic test $\psi = \cos(x+\mu)$. }\label{fig-efficiency}
\end{figure}
To better understand numerical efficiency, we analyze results in \Cref{tab-1d-aniso} by plotting $L^1$ error versus number of mesh cells, total degrees of freedom of the solution (memory costs), and number of equations in the reduced linear system (either \eqref{eq-sndg-phi}, \eqref{eq-lmdg-reducemat}, or \eqref{eq-rlmdg-reducemat}).
For the LMDG method, when the solution is isotropic, the method uses similar number of mesh cells as the $P^1$-DG method to reach the same accuracy. As a result, a reduced linear system of similar size is solved, but the degrees of freedom is smaller. For the anisotropic case, the LMDG method is first-order accurate. Compared with the $P^0$-DG method, it is able to reach similar accuracy on a coarser mesh. The reduced linear system is larger, but the number of degrees of freedom is indeed smaller.
For the RLMDG method, it seems to be less accurate compared with $P^1$-DG method, and a finer mesh has to be used to achieve the same accuracy. As a result, the solution degrees of freedom is similar to that of the $P^1$-DG method but the reduced system is even larger. However, we point out a more accurate reconstruction may solve this problem. For example, instead of using two cells, one can recover slopes in the interior region with a three-cell upwind reconstruction (which we call RLMDG$^*$). This new method is still second-order accurate, but its error is comparable to the $P^1$-DG method and significantly smaller than the current RLMDG method. Efficiency results for RLMDG$^*$ are depicted by green lines in \Cref{fig-efficiency} (they overlap with red lines in (d) and (f)). These results show that RLMDG$^*$ yields reduced systems of similar size to those of the $P^1$-DG method, but it uses less overall memory.
\end{example}
\begin{example}\label{examp-1ddifflim}
In the second numerical test, we examine the convergence rate and asymptotic preserving property of the methods. Let $\sig{s} = \sig{a} = 1$ in \eqref{eq-1d}. The computational domain
is set as $D =[0,{\pi}]$. We take $\psi_l =
\psi_r
= 0$ and $q = \frac{4}{3}\sin(x)$. The $32$-point Gauss quadrature
is used for $S_N$ discretization.
Numerical error at $\varepsilon = 10^{-5}$ and $\varepsilon = 1$ is listed in \Cref{tab-1d-eps10-5}, respectively.
The reference solutions are set as the numerical solutions
with $P^1$-DG scheme on a mesh with $1280$ cells.
One can see from \Cref{tab-1d-eps10-5}, the LMDG
scheme exhibits second-order convergence rate at $\varepsilon = 10^{-5}$, when the solution is
almost isotropic, while it converges at a first-order rate when $\varepsilon = 1$.
The RLMDG method is second-order in both cases.
Solution profiles of different schemes on a sparse
uniform mesh, with $h = \pi/8$, are shown in \Cref{fig-1dguermond-1}. When $\varepsilon = 10^{-5}$, both LMDG and RLMDG methods preserve the correct diffusion limit, unlike the $P^0$-DG method. When $\varepsilon = 1$, all schemes give valid approximations.
\begin{table}[h!]
\footnotesize
\centering
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline
&&\multicolumn{2}{c|}{$P^0$-DG}&\multicolumn{2}{c|}{$P^1$-DG}&\multicolumn{2}{c|}{LMDG}
&\multicolumn{2}{c}{RLMDG}\\
\hline
$\varepsilon$&$h$&error&order&error&order&error&order&error&order\\
\hline
\multirow{2}{*}{$10^{-5}$}
&$1/20$& 2.00e-0& - &1.89e-3&-& 1.89e-3&-&7.52e-3&-\\
&$1/40$& 2.00e-0&0.00&4.70e-4&2.01&4.71e-4&2.00&1.88e-3&2.00\\
&$1/80$& 2.00e-0&0.00&1.17e-4&2.01&1.16e-4&2.03&4.70e-4&2.00\\
&$1/160$& 1.99e-0&0.00&2.91e-5&2.00&3.06e-5&1.92&1.17e-4&2.00\\
\hline
\multirow{2}{*}{$1$}
&$1/20$& 1.06e-1&-& 2.91e-3&- &3.08e-2&-& 9.55e-3& -\\
&$1/40$& 5.38e-2&0.98& 7.72e-4&1.92&1.59e-2&0.95&2.60e-3&1.88\\
&$1/80$& 2.71e-2&0.99& 1.99e-4&1.95&8.09e-3&0.98&6.90e-4&1.91\\
&$1/160$& 1.35e-2&1.00& 5.03e-5&1.99&4.08e-3&0.99&1.80e-4&1.94\\
\hline
\end{tabular}
\caption{Accuracy test for \Cref{examp-1ddifflim}.}\label{tab-1d-eps10-5}
\end{table}
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{1deps1e-5-eps-converted-to.pdf}
\caption{$\varepsilon = 10^{-5}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{1deps1-eps-converted-to.pdf}
\caption{$\varepsilon = 1$.}
\end{subfigure}
\caption{Profiles of numerical scalar fluxes in \Cref{examp-1ddifflim}.}
\end{figure}
\end{example}
\begin{example}\label{examp-guermond}
We then consider a test
from \cite{ragusa2012robust} with discontinuous cross-sections.
The problem is defined on $[0,1]$ and is purely scattering, i.e.,
$\sig{a}\equiv 0$. The cross-section is $\sig{s} = \sig{s,1}=100$
on the left part of the domain $[0,0.5]$, and is $\sig{s} = \sig{s,2} =
100, 1000 \text{ or } 10000$ on the right part $[0.5,1]$.
The source term is constant $q = 0.01$. In the numerical test,
we set the mesh size to be $h =0.1$ and $h = 0.02$, and solutions are
depicted in \Cref{fig-1dguermond-1} and \Cref{fig-1dguermond-2}, respectively.
As one can see, unlike the $P^0$-DG scheme,
both LMDG and RLMDG schemes provide correct
solution profiles. Since the problem is diffusive, the LMDG scheme
gives accurate approximations that are almost indistinguishable with the
$P^1$-DG solutions. The reconstructed scheme has difficulty resolving the
kink at $x = 0.5$, likely because the reconstruction is no longer accurate at this point. This artifact can indeed be alleviated as the mesh is refined comparing \Cref{fig-1dguermond-1} and \Cref{fig-1dguermond-2}.
\begin{figure}[h!]
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d1-100-eps-converted-to.pdf}
\caption{$\sig{s,2} = 100$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d1-1000-eps-converted-to.pdf}
\caption{$\sig{s,2} = 1000$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d1-10000-eps-converted-to.pdf}
\caption{$\sig{s,2} = 10000$.}
\end{subfigure}
\caption{Profiles of numerical scalar fluxes in
\Cref{examp-guermond}, $h = 0.1$.}\label{fig-1dguermond-1}
\end{figure}
\begin{figure}[h!]
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d02-100-eps-converted-to.pdf}
\caption{$\sig{s,2} = 100$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d02-1000-eps-converted-to.pdf}
\caption{$\sig{s,2} = 1000$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{disccross3-0d02-10000-eps-converted-to.pdf}
\caption{$\sig{s,2} = 10000$.}
\end{subfigure}
\caption{Profiles of numerical scalar fluxes in
\Cref{examp-guermond}, $h = 0.02$.}\label{fig-1dguermond-2}
\end{figure}
\end{example}
\begin{example}\label{examp-disccross1}
In this numerical test, we solve a test problem from
\cite{larsen1989asymptotic} with discontinuous
cross-sections. We take $q = 0$ with the left inflow $\psi_l = 1$
at $x_a = 0$ and $\psi_r = 0$ at $x_b = 11$. Let
$\frac{\sig{s}}{\varepsilon} = \left\{\begin{matrix} 0,&0<x<1 \\100,&
1<x<11\end{matrix}\right.$ and
$\varepsilon{\sig{a}} = \left\{\begin{matrix} 2,&0<x<1 \\0,&
1<x<11\end{matrix}\right.$.
The $16$-point Gauss quadrature rule is used for angular discretization.
The spatial mesh is set as
$h = \left\{\begin{matrix}0.1,&0<x<1\\1,&1<x<11\\\end{matrix}\right.$
Profiles of the scalar flux obtained with various schemes are depicted in
\Cref{fig-disccross1-1}. The reference solutions are obtained with the $P^1$-DG
scheme on a refined mesh. The solution of the LMDG scheme is satisfactory. As before, the RLMDG scheme gives an
accurate approximation to the scalar flux, except for kinks near the
discontinuity. However, this numerical artifact can also be alleviated by suppressing the reconstruction across the discontinuity; see \Cref{fig-disccross1-2}.
\begin{figure}[h!]
\centering
\begin{subfigure}[h]{0.45\textwidth}
\includegraphics[width=\textwidth]{disccross1-eps-converted-to.pdf}
\caption{Numerical scalar fluxes.}\label{fig-disccross1-1}
\end{subfigure}
\begin{subfigure}[h]{0.45\textwidth}
\includegraphics[width=\textwidth]{disccross1_suppress-eps-converted-to.pdf}
\caption{With suppressed reconstruction.}\label{fig-disccross1-2}
\end{subfigure}
\caption{Profiles of numerical scalar fluxes in
\Cref{examp-disccross1}.
\end{figure}
\end{example}
\begin{example}\label{examp-disccross2}
This test is also from \cite{larsen1989asymptotic}, with $D = [0,20]$ and $\psi_l = \psi_r =0$. The cross-sections are
$\frac{\sig{s}}{\varepsilon} = \left\{\begin{matrix} 90,&0<x<10 \\100,&
10<x<20\end{matrix}\right.$
and
$\varepsilon{\sig{a}} = \left\{\begin{matrix} 10,&0<x<10 \\0,&
10<x<20\end{matrix}\right.$.
We solve the problem using the ${16}$-point Gauss quadrature rule and
the spatial mesh is uniform with $h=1$.
For this numerical test, the system has smaller changes among different directions.
Both the LMDG and RLMDG schemes give accurate approximations.
Solution profiles are give in \Cref{fig-disccross2}.
\begin{figure}[h!]
\centering
\includegraphics[width=0.45\textwidth]{disccross2-eps-converted-to.pdf}
\caption{Profiles of numerical scalar fluxes in
\Cref{examp-disccross2}.}\label{fig-disccross2}
\end{figure}
\end{example}
\subsection{Two dimensional tests}
We consider two dimensional problems on Cartesian meshes in this section.
\begin{example}\label{examp-2daccu}
We set $\varepsilon = 1$ and $\sig{s} = \sig{a} =1$ and test the accuracy with exact solutions $\psi = \sin(x+y)$ and
$\psi = (\Omega_x-3\Omega_y)^2\sin(2x+y)$.
As can be seen from \Cref{tab-2d-aniso},
for $\psi = \sin(x+y)$, both LMDG and RLMDG
schemes are second-order accurate. While for the anisotropic problem with $\psi = (\Omega_x-3\Omega_y)^2\sin(2x+y)$,
the RLMDG scheme is still second-order accurate and the LMDG
scheme is first-order accurate.
\begin{table}[h!]
\centering
\footnotesize
\hskip-1.0cm
\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}
\hline
\multicolumn{11}{c}{$\psi = \sin(x+y)$}\\
\hline
&\multicolumn{2}{c|}{$P^0$-DG}&\multicolumn{2}{c|}{$P^1$-DG}&\multicolumn{2}{c|}{$Q^1$-DG}&\multicolumn{2}{c|}{LMDG}&\multicolumn{2}{c}{RLMDG}\\
\hline
$h/\sqrt{2}$&error&order&error&order&error&order&error&order&error&order\\
\hline
$1/20$& 2.04e-2&- &1.45e-4&- &1.40e-4&- &1.24e-4&- &4.59e-4&-\\
$1/40$& 1.10e-2&0.89&3.42e-5&2.08&3.53e-5&1.98&3.12e-5&1.99&1.18e-4&1.96\\
$1/80$& 5.77e-3&0.94&8.28e-6&2.04&8.88e-6&1.99&7.82e-6&2.00&2.98e-5&1.98\\
$1/160$&2.96e-3&0.96&2.04e-6&2.02&2.26e-6&2.00&1.96e-6&2.00&7.51e-6&1.99\\
\hline
\multicolumn{11}{c}{$\psi = (\Omega_x - 2\Omega_y)^2 \sin(2x + y)$}\\
\hline
&\multicolumn{2}{c|}{$P^0$-DG}&\multicolumn{2}{c|}{$P^1$-DG}&\multicolumn{2}{c|}{$Q^1$-DG}
&\multicolumn{2}{c|}{LMDG}&\multicolumn{2}{c}{RLMDG}\\
\hline
$h/\sqrt{2}$&error&order&error&order&error&order&error&order&error&order\\
\hline
$1/20$& 7.84e-2 & & 1.64e-3& -& 1.39e-3& -& 5.04e-2& -& 4.81e-3& -\\
$1/40$& 4.18e-2 &0.91 & 4.12e-4& 2.00& 3.53e-4& 1.98& 2.57e-2& 0.97& 1.21e-3&1.99\\
$1/80$& 2.12e-2 &0.98 & 1.01e-4& 2.03& 8.87e-5& 1.99& 1.30e-2& 0.99& 3.05e-4&1.99\\
$1/160$& 1.07e-2 &0.97 & 2.52e-5& 2.00& 2.22e-5& 2.00& 6.51e-3& 0.99& 7.63e-5&2.00\\
\hline
\end{tabular}
\caption{2D accuracy test with fabricated solutions.}\label{tab-2d-aniso}
\end{table}
\end{example}
\begin{example}\label{examp-2ddifflim}
To examine the asymptotic preserving property, we consider the problem defined on $[-1,1]\times[-1,1]$ with
zero inflow boundary conditions. Let $\sig{s} =
\sig{a} = 1$. We assume $q =
(\frac{\pi^2}{6}+1)\cos(\frac{\pi}{2}x)\cos(\frac{\pi}{2}y)$. The
asymptotic solution is $\psi^{(0)}=
\cos(\frac{\pi}{2}x)\cos(\frac{\pi}{2}y)$. We
test with $\varepsilon= 1,2^{-6},2^{-10},2^{-14}$; the numerical results are given in \Cref{fig-2d}. For the $P^0$-DG and
$P^1$-DG schemes, solutions become zero near the
diffusion limit, while for the $Q^1$-DG scheme, LMDG scheme and RLMDG scheme, the correct asymptotic profile is maintained.
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P0-2e-0-eps-converted-to.pdf}
\caption{$P^0$, $\varepsilon = 1$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P0-2e-6-eps-converted-to.pdf}
\caption{$P^0$, $\varepsilon = 2^{-6}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P0-2e-10-eps-converted-to.pdf}
\caption{$P^0$, $\varepsilon = 2^{-10}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P0-2e-14-eps-converted-to.pdf}
\caption{$P^0$, $\varepsilon = 2^{-14}$.}
\end{subfigure}\\
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P1-2e-0-eps-converted-to.pdf}
\caption{$P^1$, $\varepsilon = 1$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P1-2e-6-eps-converted-to.pdf}
\caption{$P^1$, $\varepsilon = 2^{-6}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P1-2e-10-eps-converted-to.pdf}
\caption{$P^1$, $\varepsilon = 2^{-10}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{P1-2e-14-eps-converted-to.pdf}
\caption{$P^1$, $\varepsilon = 2^{-14}$.}
\end{subfigure}\\
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{Q1-2e-0-eps-converted-to.pdf}
\caption{$Q^1$, $\varepsilon = 1$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{Q1-2e-6-eps-converted-to.pdf}
\caption{$Q^1$, $\varepsilon = 2^{-6}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{Q1-2e-10-eps-converted-to.pdf}
\caption{$Q^1$, $\varepsilon = 2^{-10}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{Q1-2e-14-eps-converted-to.pdf}
\caption{$Q^1$, $\varepsilon = 2^{-14}$.}
\end{subfigure}\\
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{lmdg-2e-0-eps-converted-to.pdf}
\caption{LM, $\varepsilon = 1$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{lmdg-2e-6-eps-converted-to.pdf}
\caption{LM, $\varepsilon = 2^{-6}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{lmdg-2e-10-eps-converted-to.pdf}
\caption{LM, $\varepsilon = 2^{-10}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{lmdg-2e-14-eps-converted-to.pdf}
\caption{LM, $\varepsilon = 2^{-14}$.}
\end{subfigure}\\
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{rlmdg-2e-0-eps-converted-to.pdf}
\caption{RLM, $\varepsilon = 1$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{rlmdg-2e-6-eps-converted-to.pdf}
\caption{RLM, $\varepsilon = 2^{-6}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{rlmdg-2e-10-eps-converted-to.pdf}
\caption{RLM, $\varepsilon = 2^{-10}$.}
\end{subfigure}
~
\begin{subfigure}[b]{0.23\textwidth}
\includegraphics[width=\textwidth]{rlmdg-2e-14-eps-converted-to.pdf}
\caption{RLM, $\varepsilon = 2^{-14}$.}
\end{subfigure}
\caption{Profiles of numerical scalar fluxes in
\Cref{examp-2ddifflim}.}\label{fig-2d}
\end{figure}
\end{example}
\section{Conclusions and future work}\label{sc-conclude}
\setcounter{equation}{0}
\setcounter{figure}{0}
\setcounter{table}{0}
In this paper, we study a class of low-memory $S_N$-DG methods for the radiative transport equation. In our first method, we use the variational form of the original $S_N$-DG scheme with a smaller finite element space, in which functions have isotropic slopes. This method preserves the asymptotic diffusion limit and can still be solved with sweeps. It is first-order accurate and exhibits second-order convergence rate near the diffusion limit. The second method is a correction of the first method with reconstructed slopes, which also preserves the diffusion limit and is second-order accurate in general settings (numerically). A summary of different methods and their properties can be found in \Cref{tab-compare}.
Future work will focus on the efficiency boost of the low-memory methods. Possible directions include: (i) further reducing degrees of freedom by enriching piecewise constant space only with continuous linear elements; (ii) developing preconditioners for linear systems; (iii) comparing numerical efficiency of the methods with different reconstruction approaches, including adaptivity.
\begin{table}[h!]
\footnotesize
\centering
\begin{tabular}{c|c|c|c|c|c|c}
\hline
\multicolumn{2}{c|}{}&$P^0$-DG&$P^1$-DG&$Q^1$-DG&LMDG&RLMDG\\
\hline
\multicolumn{2}{c|}{Unisolvency when $\sig{a}\geq \delta_{\mathrm{a}} >0$}
&\multicolumn{4}{c|}{Yes}&\multirow{3}{*}{\thead{Unknown.\\ Numeri-\\cally:\\ Yes}}\\
\cline{1-6}
\multirow{2}{*}{\thead{Preserves \\interior \\diffusion
limit}}&1D&\multirow{2}{*}{No}&\multicolumn{3}{c|}{Yes}&\\\cline{2-2}\cline{4-6}
&2D& &\thead{Triangular: Yes \\Rectangular: No}&\multicolumn{2}{c|}{Yes}&\\
\hline
\multirow{2}{*}{\thead{Order of \\accuracy
}}&isotropic&\multirow{2}{*}{1}&\multicolumn{2}{c|}{\multirow{2}{*}{2}}&2&\multirow{2}{*}{2}\\\cline{2-2}\cline{6-6}
&anisotropic&&\multicolumn{2}{c|}{} &1&\\
\hline
\multirow{3}{*}{\thead{System \\dimension
}}&1D&\multirow{3}{*}{$n_x$}&\multicolumn{4}{c}{$2n_x$}\\\cline{2-2}\cline{4-7}
&2D&
&{$3n_x$}&\multicolumn{3}{c}{$4n_x$}\\\cline{2-2}\cline{4-7}
&3D& &{$4n_x$}&\multicolumn{3}{c}{$8n_x$}\\\cline{2-2}\cline{4-7}
\hline
\multirow{3}{*}{\thead{Solution \\dimension
}}&1D&\multirow{3}{*}{$n_\Omega \cdot
n_x$}&\multicolumn{2}{c|}{$2n_\Omega\cdot
n_x$}&\multicolumn{2}{c}{$n_\Omega\cdot n_x+n_x$}\\\cline{2-2}\cline{4-7}
&2D&
&{$3n_\Omega\cdot n_x$}&$4n_\Omega\cdot
n_x$&\multicolumn{2}{c}{$n_\Omega\cdot n_x+3n_x$}\\\cline{2-2}\cline{4-7}
&3D&&{$4n_\Omega\cdot n_x$}&$8n_\Omega\cdot
n_x$&\multicolumn{2}{c}{$n_\Omega\cdot n_x+7n_x$}\\
\hline
\end{tabular}
\caption{Comparison of different methods.}\label{tab-compare}
\end{table}
\vspace{-0.5cm}
\section*{Acknowledgment}
ZS would like to thank Oak Ridge National Laboratory for hosting his NSF
internship and to thank the staff, post-docs, interns and other visitors at ORNL for their
warm hospitality.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,009
|
using UnityEngine;
using System.Collections;
using BoothGame;
public class Spearman : Infantry, Ally {
private bool braced;
private bool turning;
private float turnDegrees;
private float startRotation;
public Spearman() : base() {
}
protected override Transform aggroCast() {
Debug.DrawRay(transform.position + new Vector3(0,1.7f,0), transform.forward * aggroRange);
// If we see an enemy within our aggro range, engage it in combat!
RaycastHit hit = new RaycastHit();
if (Physics.Raycast (transform.position + new Vector3(0,1.7f,0), transform.forward, out hit, aggroRange)) {
Component[] enemies = hit.transform.GetComponents(typeof(Enemy));
if (enemies.Length > 0 && ((Enemy)enemies[0]).getIsAlive()) return hit.transform;
}
return null;
}
public void turn(int degrees) {
turning = true;
startRotation = transform.rotation.eulerAngles.y;
turnDegrees = degrees;
}
protected override void charge() {
if (turning == true) {
transform.Rotate(new Vector3(0, turnDegrees/130.0f, 0));
if (Mathf.Abs(transform.rotation.eulerAngles.y - (startRotation + turnDegrees)) < 5) {
turning = false;
}
}
}
protected override void die() {
}
protected override void onFall() {
int i = random.Next(1,3);
mySounds.PlayOneShot(Resources.Load<AudioClip>("Human/falling"+i));
rigidbody.freezeRotation = false;
}
protected override void onCrash() {
}
protected override void specialBehavior() {
}
protected override void onGuardEngage() {
if (!braced) {
animation.Play("Brace");
braced = true;
}
Debug.DrawRay(transform.position, transform.up * 5, Color.yellow);
}
public void onMace() {
animation.CrossFade("Flail");
int i = random.Next(1,3);
mySounds.PlayOneShot(Resources.Load<AudioClip>("Human/falling"+i));
rigidbody.freezeRotation = false;
}
protected override void onMyStab() {
animation.CrossFade("Strike");
int i = random.Next();
if (i % 3 == 0) {
i = random.Next();
audio.PlayOneShot(Resources.Load<AudioClip>("Human/stab"+(i%3)));
}
}
public void brace() {
animation.Play("Brace");
braced = true;
}
void OnCollisionEnter(Collision c) {
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,673
|
Botanophila rupicapra is een vliegensoort uit de familie van de bloemvliegen (Anthomyiidae). De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1887 door Mik.
Bloemvliegen
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 2,149
|
Q: Routing the message with List Generic payloads using payload-type-router in spring-integration I'm trying to use the payload-type-router in xml based config of a legacy app and the application neither route to the specific channel nor throw any error at that instance. can we route the message of type List<?> using payload-type-router using XML configuration. I have n't found much examples in the doc as it only mentions routing using Integer, String classes but not with List.
<int:payload-type-router input-channel="run-router-channel">
<int:mapping type="com.foo.req.BlRunnerRequest" channel="runner-channel"/>
<int:mapping type="java.util.List" channel="run-channel"/> <!-- can spring integration support this approach-->
</int:payload-type-router>
Thanks in advance. What could be the alternatives for this kind of routing in XML ?
A: Yes. We can. I have just modified the unit test in the project to this:
<payload-type-router id="router" input-channel="routingChannel">
<mapping type="java.lang.String" channel="channel1" />
<mapping type="java.lang.Integer" channel="channel2" />
<mapping type="java.lang.Number[]" channel="channel3" />
<mapping type="java.lang.Long[]" channel="channel4" />
<mapping type="java.util.List" channel="channel5" />
</payload-type-router>
The I do this in the test method:
testService.foo(new GenericMessage<>(new ArrayList()));
PollableChannel channel5 = (PollableChannel) context.getBean("channel5");
assertThat(channel5.receive(100).getPayload()).isInstanceOf(List.class);
Probably something else is going on in your case and that request payload is really not a java.util.List.
Also router has this logic when it cannot deliver the message to some channel:
if (!sent) {
getDefaultOutputChannel();
if (this.defaultOutputChannel != null) {
this.messagingTemplate.send(this.defaultOutputChannel, message);
}
else {
throw new MessageDeliveryException(message, "No channel resolved by router '" + this
+ "' and no 'defaultOutputChannel' defined.");
}
}
Since I don't see a defaultOutputChannel configuration from your code snippet, that only means a MessageDeliveryException is thrown. If you don't see it in your case, then you swallow it somehow.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,807
|
This semester I was given the opportunity to contribute to the BlackBerry WebWorks Community API with two Cordova/PhoneGap plugins under the umbrella of UCOSP. Working with fellow UCOSP'r Jasmin Auger on both of these plugins, I learned the importance of reading a massive codebase and of developing code incrementally.
Learning the importance of reading a massive codebase, and developing code incrementally, went hand in hand with project development during the semester. The eight lines of code of the MessageBox plugin appeared only after reading hundreds of lines to code. From this I understood the "philosophy" of PhoneGap/Cordova and we discovered an already coded solution (in particular, the Notification object of PhoneGap) which was used to arrive at our small treasure. The BarcodeScanner took the bulk of the semester, and that is where I learned that developing code incrementally could save hours of time. I admit to committing a rookie mistake: considering how relatively quickly the MessageBox plugin was developed, I estimated that porting the BarcodeScanner plugin would take a short amount of time. So I coded everything at once, and thought that everything would work just fine. Wrong! I got inexplicable errors. So I went back to the drawing board and decided to build everything piece by piece, finally paying heed to what I was taught during introductory CS courses: "Develop your code incrementally. Constantly test." Not only did the plugin come along quickly, I felt a degree of attachment to the code I was writing: I was understanding it, and everything flowed logically.
There is nothing like reinforcing the lessons learned in CS courses with real world experience. It is even better when you can do it while contributing to the democratization of mobile development. The UCOSP experience has provided me with both.
Besides the BarcodeScanner and MessageBox plugins, another student ported the ExtractZipFile. The new term has already started for the four Canadian students, and they have started on porting some of our WebWorks 1.0 Community Extensions to the Cordova/WebWorks 2.0 plugin format. They are working on DeviceInfo, Thumbnail, GSECrypto, and Vibration.
Three students from American schools will be joining the team in February, and our kickoff Sprint will be held at Facebook HQ again in Palo Alto, California on February 7-9. If you're interested in following along with the code being written, we'll be tweeting with the hashtag #bbucosp at times, and all the code will be going into the public BlackBerry repositories on GitHub. We'd also appreciate your input and contributions to our open source projects. Contact me to get started: @timothywindsor on twitter or on GitHub.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 335
|
\section{Introduction}
Given a graph $G$, let $V(G)$, $E(G)$, and $\Delta(G)$ denote its vertex set, edge set, and maximum degree, respectively.
We use $C_n$ and $P_n$ for a cycle and a path, respectively, on $n$ vertices.
A graph $G$ is {\it $F$-free} for some graph $F$ if $G$ does not contain a subgraph isomorphic to $F$.
The {\it distance} between two edges $e_1$ and $e_2$ of $G$ is the distance between the two vertices corresponding to $e_1$ and $e_2$ in the line graph of $G$.
In these terms, a {\it matching} is a collection of edges with pairwise distance at least 2, and an {\it induced matching} is a collection of edges with pairwise distance at least 3.
Given a graph $G$, the {\it strong chromatic index} of $G$, denoted $\chi'_S(G)$, is the minimum $k$ such that $E(G)$ can be partitioned into $k$ induced matchings.
Since every edge has at most $2\Delta(G)^2-2\Delta(G)$ edges within distance 2, a greedy algorithm guarantees the following trivial bound: $\chi'_S(G)\leq 2\Delta(G)^2-2\Delta(G)+1$.
In 1985, Erd\H os and Ne\v set\v ril (see~\cite{bang2006six,faudree1990strong,halasz2012irregularities}) made the following renowned conjecture:
\begin{conjecture}[See \cite{bang2006six,faudree1990strong,halasz2012irregularities}]\label{conj:ErNe}
For a graph $G$,
\[\chi'_S(G)\leq
\begin{cases}
1.25 \Delta(G)^2 &\mbox{if $\Delta(G)$ is even }\\
1.25 \Delta(G)^2 -0.5 \Delta(G) +0.25 & \mbox{if $\Delta(G)$ is odd.}
\end{cases}
\]
\end{conjecture}
If Conjecture~\ref{conj:ErNe} is true, then it is sharp, as illustrated by blowing up each vertex of $C_5$ into an independent set of appropriate size.
Despite the steady interest of numerous researchers, the conjecture still seems to be far from reach.
We highlight two approaches with notable progress regarding Conjecture~\ref{conj:ErNe}.
One line of research focuses on reducing the coefficient of the leading term for graphs with sufficiently large maximum degree.
Molloy and Reed~\cite{molloy1997bound} proved an upper bound of $1.9993\Delta(G)^2$, which was reduced to $1.9653\Delta(G)^2$ by Bruhn and Joos~\cite{bruhn2018stronger}.
Recently, a significant improvement was made by Bonamy, Perrett, and Postle~\cite{bonamy2018colouring}, who showed $1.835\Delta(G)^2$.
Another line of research tackles the conjecture for small maximum degrees.
The only non-trivial case that is confirmed is $\Delta(G)=3$, which was resolved by Andersen~\cite{andersen1992strong} and independently by Hor\'ak, Qing, and Trotter~\cite{horak1993induced}.
The conjecture is open even for $\Delta(G)=4$.
Improving a result by Cranston~\cite{cranston2006strong}, Huang, Santana, and Yu~\cite{huang2018strong} recently proved that 21 colors suffice, whereas the conjectured bound is 20.
As a variation of Conjecture~\ref{conj:ErNe}, researchers also considered classes of graphs with forbidden subgraphs.
Unfortunately, the situation is not much better even for bipartite graphs.
The following conjecture by Faudree et al.~\cite{faudree1990strong} is still open:
\begin{conjecture}[\cite{faudree1990strong}]\label{conj:bipartite}
For a bipartite graph $G$, $\chi'_S(G)\leq \Delta(G)^2$.
\end{conjecture}
If the above conjecture is true, then it is tight as demonstrated by the complete bipartite graphs with appropriate part sizes.
As supporting evidence, the authors of \cite{faudree1990strong} proved that Conjecture \ref{conj:bipartite} is true for graphs where all cycle lengths are divisible by 4.
Note that all cycle lengths in bipartite graphs are divisible by 2.
Steger and Yu~\cite{STEGER1993291} verified Conjecture~\ref{conj:bipartite} for $\Delta(G)=3$, which is the only known non-trivial maximum degree case.
Mahdian~\cite{mahdian2000strong} strengthened Conjecture~\ref{conj:bipartite} by asserting the same conclusion holds when only a $5$-cycle is forbidden, opposed to forbidding all odd cycles.
\begin{conjecture}[\cite{mahdian2000strong}]\label{conj:c5}
For a $C_5$-free graph $G$, $\chi'_S(G)\leq \Delta(G)^2$.
\end{conjecture}
Generalizing a result of Mahdian~\cite{mahdian2000strong}, who investigated $C_4$-free graphs,
Vu~\cite{vu2002general} proved that the growth rate of the upper bound can actually be reduced by a logarithmic factor for $F$-free graphs, where $F$ is an arbitrary bipartite graph.
Namely, for a bipartite graph $F$, there exists a constant $C_F$ such that if $G$ is an $F$-free graph with sufficiently large $\Delta(G)$, then $\chi'_S(G)\leq C_F\frac{\Delta(G)^2}{\log\Delta(G)}$.
Moreover, this result is tight up to a multiplicative constant factor.
\bigskip
A natural lower bound on a coloring parameter is the corresponding clique number.
Given a graph $G$, a {\it strong clique} of $G$ is a set $S$ of edges such that every pair of edges in $S$ has distance at most $2$ in $G$.
The {\it strong clique number} of $G$, denoted $\SC(G)$, is the size of a maximum strong clique of $G$.
As a weakening of Conjecture~\ref{conj:ErNe}, Faudree et al.~\cite{faudree1990strong} made the following conjecture:
\begin{conjecture}[\cite{faudree1990strong}]\label{conj:SC}
For a graph $G$,
$$\SC(G)\leq
\begin{cases}
1.25 \Delta(G)^2 &\mbox{if $\Delta(G)$ is even }\\
1.25 \Delta(G)^2 -0.5 \Delta(G) +0.25 & \mbox{if $\Delta(G)$ is odd.}
\end{cases}$$
\end{conjecture}
If Conjecture~\ref{conj:SC} is true, then it is sharp by the same graph demonstrating the tightness of Conjecture~\ref{conj:ErNe}.
In contrast to the discouraging status quo for solving Conjecture~\ref{conj:ErNe}, there has been significant progress on Conjecture~\ref{conj:SC}.
The authors of~\cite{faudree1990strong} proved the existence of $\varepsilon>0$ such that $\SC(G)\leq (2-\varepsilon)\Delta(G)^2$ for sufficiently large $\Delta(G)$.
After successive improvements by Bruhn and Joos~\cite{bruhn2018stronger} and {\'S}leszy{\'n}ska-Nowak~\cite{sleszynska2016clique}, Faron and Postle~\cite{faron2019clique} recently proved that $\SC(G)\leq \frac{4}{3}\Delta(G)^2$.
We point out that Chung et al.~\cite{CHUNG1990129} proved that a graph where every pair of edges has distance at most 2 has at most $1.25\Delta(G)^2$ edges; this is different from the strong clique number since a strong clique does not necessarily contain all edges of the host graph.
We now redirect our attention to the strong clique number of graphs with forbidden subgraphs.
As supporting evidence for Conjecture~\ref{conj:bipartite}, Faudree et al.~\cite{faudree1990strong} proved that a bipartite graph $G$ has strong clique number at most $\Delta(G)^2$.
Note that this is tight as equality holds for the same graph demonstrating the tightness of Conjecture~\ref{conj:bipartite}.
It was recently revealed that forbidding all odd cycles is not necessary, as Cames van Batenburg, Kang, and Pirot~\cite{van2020strong} showed that a $C_5$-free graph $G$ has strong clique number at most $\Delta(G)^2$.
This enhancement verifies the strong clique version of Conjecture~\ref{conj:c5}, as well as Conjecture~\ref{conj:SC} with a much better upper bound.
Note that equality holds for complete bipartite graphs with appropriate part sizes.
Cames van Batenburg, Kang, and Pirot~\cite{van2020strong} also considered the class of graphs with other forbidden odd cycles.
They proved that a $C_3$-free graph $G$ has strong clique number at most $1.25\Delta(G)^2$, which is tight for blowups of $C_5$.
Note that this proves Conjecture~\ref{conj:SC} for $C_3$-free graphs.
They also proved that for $k\geq 3$, if $G$ is a $C_{2k+1}$-free graph where $\Delta(G)\geq 3k^2+10k$, then $\SC(G)\leq \Delta(G)^2$.
This implies the strong clique version of Conjecture~\ref{conj:bipartite}, as well as Conjecture~\ref{conj:SC} with a much better upper bound for graphs with a forbidden odd cycle and large maximum degree.
The authors of \cite{van2020strong} speculated that the situation is much different for the class of graphs with a forbidden even cycle.
In contrast to the quadratic upper bounds of all aforementioned conjectures, they put forth the below conjecture asserting a linear upper bound:
\begin{conjecture}[\cite{van2020strong}]\label{conj:general}
For a $C_{2k}$-free graph $G$ with $\Delta(G)\ge 1$, $\SC(G) \leq (2k-1)\Delta(G)-{2k-1\choose 2}$.
\end{conjecture}
If Conjecture~\ref{conj:general} is true, then it is sharp as exhibited by the following graph $H$: attach $\Delta(H)-(2k-2)$ pendent edges to each vertex of a complete graph on $2k-1$ vertices.
When $k=2$, however, Conjecture~\ref{conj:general} is false, since a $5$-cycle is $C_4$-free and the entire graph is a strong clique with five edges.
For other values of $k$, Conjecture~\ref{conj:general} seems plausible.
As evidence,
the authors of~\cite{van2020strong} proved the following:
\begin{theorem}[\cite{van2020strong}]
Let $G$ be a graph with $\Delta(G)\ge 1$.
\label{thm:van1}
\begin{itemize}
\item[\rm(i)] If $G$ is $C_4$-free and $\Delta(G)\geq 4$, then $\SC(G)\leq 3\Delta(G)-3$.
\item[\rm(ii)] For $k\geq 3$,
if $G$ is $C_{2k}$-free, then $\SC(G)\leq 10k^2\Delta(G)-10k^2$.
\item[\rm(iii)] For $k\ge 2$, if $G$ is $\{C_{2k}, C_{2k+1}, C_{2k+2}\}$-free, then $\SC(G)\leq (2k-1)\Delta(G)-(2k-3)$.
\end{itemize}
\end{theorem}
Note that (i) in the above theorem resolves Conjecture~\ref{conj:general} in the affirmative when $k=2$ and $\Delta(G)$ is not so small.
The authors of~\cite{van2020strong} also put forth the following conjecture for bipartite graphs with a forbidden even cycle.
\begin{conjecture}[\cite{van2020strong}]\label{conj:bipart}
For a $C_{2k}$-free bipartite graph $G$ with $\Delta(G)\ge 1$, $\SC(G) \leq k\Delta(G)-(k-1)$.
\end{conjecture}
If Conjecture~\ref{conj:bipart} is true, then it is sharp for the following graph: attach $p$ pendent edges to one vertex of degree $k-1$ in a complete bipartite graph $K_{k-1,p+k-1}$. As evidence for Conjecture~\ref{conj:bipart}, the following theorem was shown:
\begin{theorem}[\cite{van2020strong}]\label{thm:manycycles}
If $G$ is a $\{C_3, C_5, C_{2k}, C_{2k+2}\}$-free graph, then $\SC(G)\leq \max\{k\Delta(G), 2k(k-1)\}$.
\end{theorem}
Our first contribution is that we verify Conjecture~\ref{conj:bipart} in a much stronger form.
Theorem~\ref{thm:conj:biparite} resolves Conjecture~\ref{conj:bipart} in the affirmative.
\begin{theorem}\label{thm:conj:biparite}
For $k\ge 2$, if $G$ is a $C_{2k}$-free bipartite graph and $\Delta(G)\ge 1$, then $\SC(G) \leq k\Delta(G)-(k-1)$.
\end{theorem}
We strengthen Theorem~\ref{thm:conj:biparite} and obtain Theorem~\ref{thm:main:coro}, which improves aforementioned results by Cames van Batenburg, Kang, and Pirot.
Namely, we prove that the same conclusion can be reached by forbidding only $C_5$ and $\{C_5, C_3\}$ when $k\geq 4$ and $k=3$, respectively, opposed to forbidding all odd cycles.
\begin{theorem}\label{thm:main:coro}
For $k\ge 2$,
if $G$ is a $\{C_5,C_{2k}\}$-free graph and $\Delta(G)\ge 1$,
then the following holds:
\begin{itemize}
\item[\rm(i)] For $k \ge 4$, $\SC(G)\leq k\Delta(G)-(k-1)$.
\item[\rm(ii)] For $k\in\{2,3\}$, if $G$ is also $C_3$-free, then
$\SC(G)\leq k\Delta(G)-(k-1)$.
\end{itemize}
\end{theorem}
When $k\in \{2,3\}$, forbidding $C_3$ in (ii) is necessary as demonstrated by the following graph $H$:
attach $p$ pendent edges
to each vertex of a complete graph on $k+1$ vertices.
This graph is $\{C_5,C_{2k}\}$-free, yet $\SC(H)=(k+1)\Delta(H)-k(k+1)/2 >k\Delta(H)-(k-1)$ when $\Delta(H)$ is sufficiently large.
Our second contribution is that we almost prove Conjecture~\ref{conj:general}.
We are able to provide an upper bound that is off by only the constant term.
Note that Theorem~\ref{thm:general} is a strengthening of Theorem~\ref{thm:van1} (ii) and (iii).
\begin{theorem}\label{thm:general}
For $k\geq 3$, if $G$ is a $C_{2k}$-free graph and $\Delta(G)\ge 1$, then $\SC(G)\le (2k-1)\Delta(G)+(2k-1)^2$.
\end{theorem}
The paper is organized as follows.
We first provide some definitions and prove some lemmas in Section~\ref{sec:prelmi}.
In Section~\ref{sec:bipartite}, we prove Theorem~\ref{thm:conj:biparite}, which is used to show Theorem~\ref{thm:main:coro}.
The proof of Theorem~\ref{thm:general} is provided in Section~\ref{sec:general}.
\section{Preliminaries}\label{sec:prelmi}
We provide some definitions and useful observations in this section.
Given a graph $G$, let $S$ (resp. $W$) be a subset of the edges (resp. vertices) of $G$.
We use $G[S]$ (resp. $G[W]$) to denote the subgraph of $G$ induced by the edges in $S$ (resp. the vertices in $W$).
Let $G-S$ (resp. $G-W$) denote the graph obtained from $G$ by deleting the edges in $S$ (resp. vertices in $W$).
If $S = \{uv\}$ (resp. $W= \{v\}$), then denote $G-S$ by $G-uv$ (resp. $G-W$ by $G-v$).
Given a graph $G$ and $A, B\subseteq V(G)$,
let $E_G(A,B)$ denote the set of all edges in $G$
joining a vertex in $A$ and a vertex in $B$.
When we denote a cycle or a path, we drop commas for simplicity. For instance, a (directed) path $x_1,x_2,\ldots,x_n$ of length $n-1$ and a (directed) cycle $x_1,x_2,\ldots,x_n,x_1$ is denoted by $x_1x_2\ldots x_n$ and $x_1x_2\ldots x_nx_1$, respectively.
A {\it vertex cover} of a graph $G$ is a set $S\subseteq V(G)$ such that every edge of $G$ has an endpoint in $S$. The {\it vertex cover number} of $G$, denoted $\tau(G)$,
is the size of a minimum vertex cover of $G$.
The {\it matching number} of $G$, denoted $\alpha'(G)$, is the size of a maximum matching in $G$.
For a set of edges $M$, let $V(M)$ denote the set of all endpoints of edges in $M$.
The following is arguably the most famous theorem relating the matching number and the vertex cover number of bipartite graphs.
\begin{thm-konig}[\cite{konig1931grafok, egervary1931matrixok}]
\label{thm:konig}
If $G$ is a bipartite graph, then $\tau(G)=\alpha'(G)$.
Moreover, for a maximum matching $M$ of $G$, there is a minimum vertex cover that is a subset of $V(M)$.
\end{thm-konig}
We now prove two lemmas that will be often used in the proofs of our theorems.
\begin{lemma}\label{lem:matching:path}
Let $G$ be a bipartite graph, and $H$ be the subgraph of $G$ induced by a strong clique of $G$.
If $H$ has a matching $M$ of size $m$,
then $G[V(M)]$ contains a $P_{2m}$
containing all edges in $M$. Moreover, if $m\ge 4$, then $G[V(M)]$ contains
a $C_{2m-2}$
using at least $m-2$ edges in $M$.
\end{lemma}
\begin{proof}
Let $G$ be a bipartite graph with bipartition $(X,Y)$,
and let $M=\{x_1y_1,\ldots,x_{m}y_{m}\}$ be a matching of $H$ where $\{x_1, \ldots, x_m\}\subseteq X$ and $\{y_1, \ldots, y_m\}\subseteq Y$.
Construct an auxiliary directed graph $D$ where each vertex $w_i$ of $D$ represents the edge $x_iy_i$ in $M$
and $(w_i, w_j)$ is an arc of $D$ if $x_iy_j$ is an edge of $G$.
Note that $D$ is a semi-complete digraph since for distinct $i$ and $j$,
either $x_iy_j$ or $x_jy_i$ exists in $G$.
Since $D$ contains a tournament, which always has a Hamiltonian path,
we may assume that $w_1 w_2\ldots w_m$ is a directed path in $D$ by relabelling indices if necessary.
Thus $G[V(M)]$ has a path $y_1x_1 y_2 x_2\ldots y_m x_m$ of length $2m-1$ containing all edges in $M$.
Suppose that $m\ge 4$.
If $D$ is strongly connected, then $D$ contains a directed cycle of every length.
Thus, $D$ has a directed cycle of length $m-1$, which corresponds to a
$C_{2m-2}$
in $G[V(M)]$ using $m-1$ edges in $M$.
If $D$ is not strongly connected, then by the acyclic ordering of strongly connected components, there is a directed $(u,v)$-path of length $m-1$ for some $u,v$ where $(u,v)$ is an arc of $D$.
This corresponds to a $C_{2m-2}$
in $G[V(M)]$ using $m-2$ edges in $M$.
\end{proof}
For a strong clique $S$ of a graph $G$,
the graph $G$ is \textit{$S$-minimal} if
$S$ is not a strong clique of every proper subgraph of $G$. In other words, removing any vertex or edge of $G$ would violate that $S$ is a strong clique.
\begin{lemma}
\label{lem:minimal}
Let $S$ be a strong clique of a graph $G$.
If $G$ is $S$-minimal, then the following holds:
\begin{enumerate}[\rm(i)]
\item Every vertex of $G$ is incident with some edge in $S$.
\item The diameter of $G$ is at most $3$.
\item For every edge $uv \in E(G)\setminus S$, there are two edges $uu', vv'\in S$ such that $uv$ is the only edge joining $uu'$ and $vv'$.
\item If $S$ is a maximum strong clique of $G$, then for every edge $uv \in E(G)\setminus S$,
there is an edge $xy \in S$ whose distance from $uv$ in $G$ is at least $3$.
\end{enumerate}
\end{lemma}
\begin{proof}
(i) Suppose that $G$ has a vertex $v$ that is not incident with an edge in $S$.
Since the distance between two edges in $S$ is the same in both $G$ and $G-v$, $S$ is also a strong clique of $G-v$.
This is a contradiction to the assumption that $G$ is $S$-minimal.
(ii) Suppose that $G$ has two vertices $u$ and $v$ where the distance between $u$ and $v$ is at least $4$.
Then, for each $uu', vv' \in S$, the distance between $uu'$ and $vv'$ is at least $3$.
This is a contradiction to the assumption that $S$ is a strong clique of $G$.
(iii) Let $uv \in E(G) \setminus S$.
Suppose that $G$ has an edge other than $uv$ that joins $uu'$ and $vv'$ for every two edges $uu', vv' \in S$.
Then, for each pair of edges in $S$, the distance between them is the same in both $G-uv$ and $G$.
Thus $S$ is also a strong clique of $G-uv$.
This is a contradiction to the assumption that $G$ is $S$-minimal.
(iv) Suppose that $S$ is a maximum strong clique of $G$, and $uv \in E(G) \setminus S$.
If every edge in $S$ has distance at most $2$ from $uv$, then $S \cup \{uv\}$ is also a strong clique of $G$.
This is a contradiction to the assumption that $S$ is a maximum strong clique of $G$.
\end{proof}
We end this section with a result from~\cite{faudree1990strong}.
\begin{theorem}[\cite{faudree1990strong}]\label{thm:helppp}
For a bipartite graph $G$, $\SC(G) \leq \Delta(G)^2$.
\end{theorem}
\section{Proofs of Theorems~\ref{thm:conj:biparite} and \ref{thm:main:coro}}\label{sec:bipartite}
In this section, we first prove Theorem~\ref{thm:conj:biparite}, then show Theorem~\ref{thm:main:coro} by using Theorem~\ref{thm:conj:biparite}.
\begin{proof}[Proof of Theorem~\ref{thm:conj:biparite}]
Let $k\ge 2$, and $G$ be a $C_{2k}$-free bipartite graph with bipartition $(X,Y)$ and $\Delta(G)\ge 1$.
Let $H$ be the subgraph of $G$ induced by a maximum strong clique of $G$.
Recall that our goal is to show $|E(H)|\le k \Delta(G) -(k-1)$.
We first consider the case when $k=2$.
Let $M$ be a maximum matching of $H$ where
$M=\{x_1y_1,\ldots,x_my_m\}$ and $x_i \in X$ and $y_i \in Y$ for every $i \in \{1,2,\ldots,m\}$.
The theorem is trivial when $m=1$, so assume $m\ge 2$.
Without loss of generality assume $x_1y_2 \in E(G)$.
If either $\deg_H(y_1) \ge 2$ or $\deg_H(x_2) \ge 2$, then $G$ contains a $C_4$.
Thus, if $m=2$, then
$|E(H)|\le 2\Delta(G)-1$.
Now, we assume $m\ge 3$.
If $x_1$ has three neighbors $y_1,y_2,y_j$ in $V(M)$ for some $3\le j \le m$,
then since there must be an edge between $\{x_2,y_2\}$ and $\{x_j,y_j\}$,
either $x_1y_{2}x_{2}y_{j}x_1$ or $x_1y_{2}x_{j}y_{j}x_1$ is a $C_4$ in $G$, which is a contradiction.
Hence, $x_1$ has at most two neighbors $y_1$ and $y_2$ in $V(M)$.
Similarly, every vertex in $V(M)$ has at most two neighbors in $V(M)$.
This implies that $m=3$, and $G[V(M)]$ is a $6$-cycle $x_1y_2x_2y_3x_3y_1x_1$.
If $E(H)\setminus M$ is non-empty, then without loss of generality, let $x_1z \in E(H)\setminus M$.
Since $x_1z, x_3y_3 \in E(H)$, it follows that $x_3z \in E(G)$ and $x_1zx_3y_1x_1$ is a $C_4$, which is a contradiction.
Therefore, $E(H)=M$, so $|E(H)|=3 \le 2\Delta(G)-1$.
Now, we suppose that $k\ge 3$.
We may assume $G$ is $E(H)$-minimal by removing unnecessary vertices and edges of $G$.
If $\Delta(G)\leq k-1$, then $|E(H)|\le \Delta(G)^2 \le k \Delta(G)-(k-1)$ where the first inequality holds by Theorem~\ref{thm:helppp}, so we may assume that $\Delta(G) \ge k$.
By Lemma~\ref{lem:matching:path},
we may assume that $H$ does not contain a matching of size $k+1$, so
by the K\H onig--Egerv\'ary Theorem, $\tau(H)\le k$.
If either $\Delta(H) < \Delta(G)$ or $\tau(H) <k$, then $|E(H)|\le \tau(H)\Delta(H)< k \Delta(G)-(k-1)$.
So, we may assume that $\Delta(H)=\Delta(G)$ and $\tau(H)=k$.
Let $Z$ be a minimum vertex cover of $H$.
If each vertex in $Z$ has degree less than $\Delta(H)$ in $H$,
then $|E(H)|\le |Z|(\Delta(H)-1) < k \Delta(G)-(k-1)$. Hence, we may assume that there exists $z \in Z$ such that $\deg_H(z)=\Delta(H)=\Delta(G)$.
Without loss of generality, assume that $z \in X$.
Suppose that $zy \notin E(G)$ for some $y \in Y$.
Since $G$ is $E(H)$-minimal, $y$ is incident with an edge $xy$ of $H$ by Lemma~\ref{lem:minimal} (i).
Now, $x$ is adjacent to $z'$ for every $z' \in N_G(z)$ since $zz', xy \in E(H)$.
See the first figure of Figure~\ref{fig:thm1.3}.
\begin{figure}[h!]
\centering
\includegraphics[height=4cm,page=2]{fig-strong-clique.pdf}
\caption{Illustrations for Theorem~\ref{thm:conj:biparite}. Thick edges are guaranteed to be in $H$.}
\label{fig:thm1.3}
\end{figure}
Therefore, $\deg_G(x)\ge \deg_G(z)+1 =\Delta(G)+1$, which is a contradiction.
Hence, $Y=N_G(z)=N_H(z)$.
Suppose that $Z\cap Y\neq \emptyset$.
Let $Z_X=Z\cap X$ and $Z_Y=Z\cap Y$ so that $|Z|=|Z_X|+|Z_Y|=k$.
Since $Z$ is a vertex cover of $H$, each edge of $H$ is in either
$E_H(X, Z_Y)$ or $E_H(Z_X,Y\setminus Z_Y)$.
See the second figure of Figure~\ref{fig:thm1.3}.
Since each vertex in $Z_Y$ has degree at most $\Delta(G)$ in $H$, we know $|E_H(X, Z_Y)|\le |Z_Y|\Delta(G)$.
Also, since $|Y|=\Delta(G)$, we know $|E_H(Z_X,Y\setminus Z_Y)| \le |Z_X| (\Delta(G)-|Z_Y|)$.
Therefore,
\[
|E(H)|\le |Z_Y|\Delta(G)+|Z_X| (\Delta(G)-|Z_Y|) = (|Z_X|+|Z_Y|) \Delta(G)-|Z_X||Z_Y| \le k\Delta(G)-(k-1).
\]
The last inequality holds since both $Z_X$ and $Z_Y$ are not empty.
Now, suppose that $Z\subseteq X$.
If $x\in X\setminus Z$, then since $G$ is $E(H)$-minimal,
$x$ is incident with an edge $xy'$ of $H$ by Lemma~\ref{lem:minimal} (i).
This is a contradiction since $xy'$ is not covered by $Z$.
Therefore, $X\setminus Z=\emptyset$, so $X=Z$ and thus $|X|=k$.
Let $X'=\{x\in X\mid N_G(x)=Y\}$, and let $|X'|=\ell$.
Since $z\in X'$ and $G$ is $C_{2k}$-free, we know $1 \le \ell \le k-1$.
By the K\H onig--Egerv\'ary Theorem, since $\tau(H)=k$, there is a matching of size $k$ in $H$.
Thus, $H-X'$ has a matching of size $k-\ell$, and this matching is a strong clique of $G$.
Thus, it follows from Lemma~\ref{lem:matching:path} that there is a path $P$ of length $2(k-\ell)-1$ in $G-X'$ using all vertices in $X\setminus X'$.
Let $x$ and $y$ be the ends of $P$ in $X\setminus X'$ and $Y$, respectively.
Let $Y'=Y\setminus V(P)$.
Note that $|Y'|=\Delta(G)-(k-\ell) \ge \ell$.
If $x$ has a neighbor $y'$ in $Y'$,
then we can extend $P$ to a $C_{2k}$ by using all vertices in $X'$ and $\ell$ vertices in $Y'$ including $y'$, which is a contradiction.
See the third figure of Figure~\ref{fig:thm1.3}.
Hence, $x$ has no neighbor in $Y'$. Then,
\begin{align*}
|E(H)|&=|E_H(X',Y)|+|E_H(\{x\},Y)|+|E_H(X\setminus (X'\cup \{x\}),Y)|\\
&\le \ell\Delta(G) + (k-\ell)+(k-\ell-1)(\Delta(G)-1)\\
&\le (k-1)\Delta(G)+1\\
&\le k\Delta(G)-(k-1),
\end{align*}
where the last inequality holds since $\Delta(G)\ge k$.
This completes the proof.
\end{proof}
In order to prove Theorem~\ref{thm:main:coro}, we first show the following two lemmas.
\begin{lemma} \label{lem:H-minimal:clique}
Let $G$ be a $C_5$-free graph, and $H$ be the subgraph of $G$ induced by a maximum strong clique of $G$.
If $H$ is $C_3$-free, and $G$ is $E(H)$-minimal, then $G$ is bipartite.
\end{lemma}
\begin{proof}
We will show that $G$ does not contain an odd cycle.
We first show that $G$ has no $C_3$.
Suppose to the contrary that $G$ has a $C_3$. Let $xyzx$ be a $C_3$ of $G$ incident with the maximum number of edges of $H$.
Since $H$ has no $C_3$, we may assume that $xy \not \in E(H)$.
Since $G$ is $E(H)$-minimal,
there are edges $xx'$ and $yy'$ in $H$ whose distance is at least $3$ in $G-xy$ by Lemma~\ref{lem:minimal} (iii).
Moreover, since $xy\not\in E(H)$ and $E(H)$ is a maximum strong clique of $G$, there is an edge $uv\in E(H)$
whose distance from $xy$ is
at least $3$ in $G$ by Lemma~\ref{lem:minimal} (iv).
Thus, $x,y,z,x',y',u,v$ are all distinct.
From the pairwise distances between $xx',yy'$, and $uv$,
we may assume that $ux',vy'\in E(G)$ since $G$ is $C_5$-free.
Also, $zu, zv \not\in E(G)$ since if $zu \in E(G)$ and $zv \in E(G)$, then $zuvy'yz$ and $zvux'xz$, respectively, is a $C_5$ of $G$.
This further implies that $uv$ has distance at least $3$ to each of $xz$ and $yz$ in $G$.
So, $xz, yz \notin E(H)$. See the first figure of Figure~\ref{fig:H-minimal:clique}.
\begin{figure}[h!]
\centering
\includegraphics[height=2.5cm,page=4]{fig-strong-clique.pdf}
\caption{Illustrations for Lemma~\ref{lem:H-minimal:clique}}
\label{fig:H-minimal:clique}
\end{figure}
Also, $zx', zy'\not\in E(G)$ since if $zx'\in E(G)$ and $zy'\in E(G)$, then $zx'x$ and $zy'y$, respectively, is a $C_3$ of $G$ containing an edge of $H$,
which is a contradiction to the choice of $xyzx$.
See the second figure of Figure~\ref{fig:H-minimal:clique}.
Since $G$ is $E(H)$-minimal, there is an edge $zz'\in E(H)$ where $z'\not\in \{x,y,z,u,v,x',y'\}$ by Lemma~\ref{lem:minimal} (i).
See the third figure of Figure~\ref{fig:H-minimal:clique}.
Since the distance between $zz'$ and $uv$ must be at most $2$,
either $z'u\in E(G)$ or $z'v\in E(G)$.
In either case, $z'ux'xz$ or $z'vy'yz$ is a $C_5$ of $G$, which is a contradiction.
Therefore, $G$ is $C_3$-free.
Now, we prove that $G$ is bipartite.
Suppose to the contrary that $G$ is not bipartite, so let $C:x_1x_2\ldots x_{2m+1}x_1$ be a smallest odd cycle in $G$
with the maximum number of edges in $H$.
By the minimality of $|C|$, $C$ has no chords in $G$.
Moreover, $m=3$ since the diameter of $G$ is at most $3$ by Lemma~\ref{lem:minimal} (ii).
Since $C$ has no chords and $G$ is $C_5$-free, for each $i\in\{1,2,\ldots,7\}$, the distance between $x_i$ and $x_{i+3}$ is exactly $3$ in $G$ where addition in the indices is modulo 7.
Since $C$ has no chords in $G$, there are two consecutive edges, say $x_1x_2, x_2x_3$, of $C$ not in $H$.
Since $G$ is $E(H)$-minimal, there are edges $x_1y_1, x_4y_4$ of $H$ by Lemma~\ref{lem:minimal} (i).
Since
the distance between $x_1$ and $x_4$ is $3$,
it follows that $y_1y_4 \in E(G)$.
We also have $y_1,y_4 \notin\{x_1,x_4,x_5,x_6,x_7\}$ since
the distances between $x_1$ and $x_5$ and between $x_4$ and $x_7$ are exactly $3$.
Note that the cycle $x_1y_1y_4x_4x_5x_6x_7x_1$ has more edges of $H$ than $C$, because $x_1y_1, x_4y_4 \in E(H)$ but $x_1x_2, x_2x_3 \notin E(H)$, which is a contradiction to the choice of $C$.
Therefore, $G$ is bipartite.
\end{proof}
\begin{lemma} \label{lem:H:has:triangle}
Let $G$ be a $C_5$-free graph with $\Delta(G)\ge 1$, and $H$ be the subgraph of $G$ induced by a maximum strong clique of $G$.
If $H$ contains a $C_3$, then $\SC(G)\le 4\Delta(G)-3$.
\end{lemma}
\begin{proof}
Let $C:xyzx$ be a $C_3$ of $H$.
Suppose to the contrary that $H-\{x,y,z\}$ has two edges $uv$ and $u'v'$, where $u,v,u',v'$ are all distinct.
Since $E(H)$ is a strong clique of $G$ and $G$ is $C_5$-free,
we may assume that $ux,uy\in E(G)$.
Similarly,
we may assume that $u'$ is adjacent to two vertices of $C$.
Since $G$ is $C_5$-free, $u'$ is adjacent to both $x$ and $y$.
By the distance between $uv$ and $u'v'$, an edge of $G$ connects $\{u,v\}$ and $\{u',v'\}$.
In each case, however, we can find $C_5$, which is a contradiction.
Thus $H-\{x,y,z\}$ is a star, and let $v$ be its center vertex.
Then $\{v,x,y,z\}$ is a vertex cover of $H$, and so
$|E(H)|\le 4\Delta(H)-3\le 4\Delta(G)-3$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main:coro}]
Let $G$ be a $\{C_5,C_{2k}\}$-free graph with $\Delta(G)\ge 1$.
Let $H$ be the subgraph of $G$ induced by a maximum strong clique of $G$.
We may assume that $G$ is $E(H)$-minimal by removing unnecessary vertices and edges of $G$.
If $H$ does not contain a $C_3$, then by Lemma~\ref{lem:H-minimal:clique}, $G$ is bipartite, and so by Theorem~\ref{thm:conj:biparite}, it holds that $|E(H)|\le k\Delta(G)-(k-1)$.
If $H$ contains a $C_3$, then it is case (i) and so $k\ge 4$, and so $|E(H)|\le 4\Delta(G)-3\le k\Delta(G)-(k-1)$ by Lemma~\ref{lem:H:has:triangle}.
\end{proof}
\section{Proof of Theorem~\ref{thm:general}}\label{sec:general}
In this section, we prove Theorem~\ref{thm:general}.
Let $M$ be a matching of a graph $G$. For a vertex $x\in V(M)$,
an {\it $(x, M)$-path} is a path $P$ in $G[V(M)]$ starting with $x$ such that the last edge of $P$ is not in $M$ and for every distinct $u,v \in V(P)$,
if $uv$ is an edge in $M$, then $uv \in E(P)$.
For a vertex $x \in V(M)$, let $x'$ denote the neighbor of $x$ such that $xx'\in M$.
Note that $(x')'=x$.
A matching $M=\{x_1x_1',\ldots,x_mx_m'\}$ is {\it $x_1$-special} if
$x_i x_j \in E(G)$ if and only if $i=1\neq j$, $x_i' x_j' \in E(G)$ if and only if $2\le i<j \le m$, and $x_ix_j' \notin E(G)$ for $1\le i\neq j \le m$.
In other words, an $x_1$-special matching of size $m$ can be obtained from a complete graph on $m$ vertices by subdividing all edges incident with a vertex $x_1$ and adding a pendent edge to $x_1$. See Figure~\ref{fig:sp_matching} for an illustration.
We say $M$ is {\it special} if it is $x$-special for some vertex $x\in V(M)$.
\begin{figure}[h!]
\centering
\includegraphics[height=4cm,page=8]{fig-strong-clique.pdf}
\caption{An illustration for an $x_1$-special matching $M$. Note that the vertices not adjacent in the figure are not adjacent in $G$.} \label{fig:sp_matching}
\end{figure}
We will use the following observation frequently.
\begin{observation}\label{obs:special}
For a graph $G$, let $M$ be a matching of $G$ that is also a strong clique of $G$.
Suppose that $|M|=m\ge 3$.
If $x\in V(M)$, then the following holds:
\begin{enumerate}[(a)]
\item There is no $(x, M)$-path of length $2$ if and only if $M$ is $x$-special.
\item If $M$ is $x$-special, then for every $\ell\in \{1,\ldots,m\}\setminus\{2\}$, there is an $(x, M)$-path of length $\ell$.
\end{enumerate}
\end{observation}
\begin{lemma}\label{lem:path1}
For a graph $G$, let $M$ be a matching that is also a strong clique of $G$.
Suppose that $|M|=m\geq 2$ and $M$ is not $x$-special for some $x \in V(M)$.
If $x$ has a neighbor in $V(M)\setminus \{x,x'\}$,
then there is an $(x, M)$-path of length $\ell$ for every $\ell\in\{1,\ldots, m-1\}$.
\end{lemma}
\begin{proof}
We use induction on $\ell$ to prove Lemma~\ref{lem:path1}. Let $y$ be a neighbor of $x$ in $V(M)\setminus\{x, x'\}$.
If $\ell=1$, then $xy$ is an $(x, M)$-path of length $1$.
If $\ell=2$, then $m\ge 3$, and by Observation~\ref{obs:special}~(a),
there is an $(x, M)$-path of length $2$.
Assume that $\ell \ge 3$, so $m\ge 4$.
Let $M'=M\setminus\{xx'\}$. Note that $|M'|\ge 3$.
If there is an $(y, M')$-path of length $\ell-1$, then by prepending $xy$ to the path, we obtain an $(x, M)$-path of length $\ell$.
So, let us assume that no such path exists.
By the induction hypothesis, either
$M'$ is $y$-special or $y$ has no neighbors in $V(M')\setminus\{y, y'\}$.
By Observation~\ref{obs:special},
there are only two possible cases:
either $M'$ is $y$-special and $\ell-1=2$, or $M'$ is $y'$-special and $\ell-2=2$ (so $m\geq 5$).
For the first case, since $m\ge 4$ and $M'$ is $y$-special,
$y$ has two neighbors $z, w\in V(M')\setminus \{y,y'\}$ where $z'w' \in E(G)$.
See the first figure of Figure~\ref{fig:path}.
Since $xx'$ and $zz'$ are part of a strong clique of $G$, there must be an edge between $\{x,x'\}$ and $\{z,z'\}$.
In each case, there is an $(x, M)$-path of length $3=\ell$, which is $xzz'w'$, $xz'zy$, $xx'zy$, or $xx'z'w'$.
\begin{figure}[h!]
\centering
\includegraphics[height=2.5cm,page=9]{fig-strong-clique.pdf}
\caption{Illustrations for Lemma~\ref{lem:path1}} \label{fig:path}
\end{figure}
For the second case,
since $m\ge 5$ and $M'$ is $y'$-special,
$y'$ has three neighbors $z', w', r'\in V(M')\setminus \{y,y'\}$ where $z$, $w$, and $r$ are pairwise adjacent to each other.
See the second figure of Figure~\ref{fig:path}.
Since $xx'$ and $zz'$ are part of a strong clique of $G$, there must be an edge between $\{x,x'\}$ and $\{z,z'\}$.
In each case, there is an $(x, M)$-path of length $\ell=4$, which is $xzww'y'$, $xz'zwr$, $xx'zwr$, or $xx'z'zw$.
Thus, there is an $(x,M)$-path of length $\ell$ for every $\ell\in\{1,\ldots, m-1\}$, and this completes the proof.
\end{proof}
\begin{lemma}\label{lem:matching}
For a graph $G$, let $M$ be a matching of $G$ that is also a strong clique of $G$.
If $|M|=2m\geq6$, then $G$ contains a $C_{2m}$.
\end{lemma}
\begin{proof}
For every $x\in V(M)$,
let $W(x)$ be a maximum subset of $V(M)\cap N_G(x)$ such that $W(x)$ contains at most one endpoint of every edge in $M\setminus\{xx'\}$.
Also define $W'(x)=\{w'\mid w\in W(x)\}$.
\begin{claim}\label{claim:indep}
For every $x\in V(M)$, if $W(x)$ is not an independent set of $G$, then $G$ contains a $C_{2m}$.
\end{claim}
\begin{proof}
Let $x\in V(M)$ such that $W(x)$ is not an independent set of $G$.
First, suppose that $G[W(x)]$ contains a $P_3$, that is, $yz, zw \in E(G)$ for some $y,z,w \in W(x)$.
Let $M' = M\setminus\{xx',yy',zz'\}$, so $|M'|=2m-3\ge 3$.
Assume that there is a $(w, M')$-path $P$ of length $2m-4$ ending at $r\in V(M')\setminus\{w, w'\}$.
See the first figure of Figure~\ref{fig:matching1}.
Since $xx'$ and $rr'$ are part of a strong clique of $G$, there must be an edge between $\{x, x'\}$ and $\{r, r'\}$.
In each case, by adding $xyzw$, $x'xzw$, $r'xzw$, or $r'x'xw$ to $P$, we obtain a $C_{2m}$.
\begin{figure}[h!]
\centering
\includegraphics[height=3cm,page=10]{fig-strong-clique.pdf}
\caption{Illustrations for Claim~\ref{claim:indep}} \label{fig:matching1}
\end{figure}
Hence, let us assume that there is no such path.
By Observation~\ref{obs:special} and Lemma~\ref{lem:path1}, there are only two possible cases:
either $M'$ is $w$-special and $2m-4=2$,
or $M'$ is $w'$-special and $2m-5=2$.
Note that the second case is impossible.
For the first case, let $M'=\{ww', ss',rr'\}$ where $wr,ws,s'r' \in E(G)$.
See the second figure of Figure~\ref{fig:matching1}.
Since $xx'$ and $rr'$ are part of a strong clique of $G$, there must be an edge between $\{x,x'\}$ and $\{r,r'\}$.
In each case, there is a $C_6$, which is $xrr's'swx$, $xr's'swzx$, $xx'rwzyx$, or $xx'r's'swx$.
Therefore, $G$ contains a $C_{2m}$.
\smallskip
Now, suppose that $G[W(x)]$ has no $P_3$.
Since $W(x)$ is not an independent set of $G$,
$yz \in E(G)$ for some $y,z \in W(x)$. Let $M'=M\setminus\{xx',yy'\}$, so $|M'|=2m-2 \ge 4$.
Assume that there is a $(z,M')$-path $P$ of length $2m-4$ ending at $r \in V(M')\setminus \{z,z'\}$.
See the third figure of Figure~\ref{fig:matching1}.
Since $xx'$ and $rr'$ are part of a strong clique of $G$, there must be an edge between $\{x, x'\}$ and $\{r, r'\}$.
If $rx\notin E(G)$, then we obtain a $C_{2m}$ by adding $x'xyz$, $r'xyz$, or $r'x'xz$ to $P$, depending on the case.
So we may assume that $rx \in E(G)$.
By symmetry, we may assume that $ry\in E(G)$, which is a contradiction to our assumption since $ryz$ is a $P_3$ in $G[W(x)]$.
Hence, let us assume that there is no such path. By Observation~\ref{obs:special} and Lemma~\ref{lem:path1}, there are two possible cases: either
$M'$ is $z$-special and $2m-4=2$, or $M'$ is $z'$-special and $2m-5=2$.
Note that the second case is impossible.
For the first case,
$M'$ is a special matching of size $4$, so $G[V(M')]$ contains a $C_6$.
See Figure~\ref{fig:sp_matching}.
Therefore, $G$ contains a $C_{2m}$.
\end{proof}
\begin{claim}\label{claim:indep2}
For every $x\in V(M)$, if $G[W'(x)]$ contains a $P_3$, then
$G$ contains a $C_{2m}$.
\end{claim}
\begin{proof}
Suppose that for some $x\in V(M)$, $G[W'(x)]$ contains a $P_3$, that is, $y'z', z'w' \in E(G)$ for some $y',z',w' \in W'(x)$.
See the first figure of Figure~\ref{fig:matching2}.
If $m=3$, then $xyy'z'w'wx$ is a $C_{2m}$, so assume that $m\geq 4$.
Note that $\{y,z,w\}$ is an independent set of $G$ by Claim~\ref{claim:indep}.
We may further assume that $y'w' \notin E(G)$ since otherwise $W(y')$ is not an independent set of $G$, which is a contradiction to Claim~\ref{claim:indep}.
So, either $yw' \in E(G)$ or $wy' \in E(G)$.
Without loss of generality, assume that $yw' \in E(G)$.
\begin{figure}[h!]
\centering
\includegraphics[height=3cm,page=11]{fig-strong-clique.pdf}
\caption{Illustrations for Claim~\ref{claim:indep2}} \label{fig:matching2}
\end{figure}
Let $M'=M\setminus\{xx',yy',zz'\}$, so $|M'|=2m-3 \ge 5$.
Assume that there exists a $(w',M')$-path $P$ of length $2m-5$ ending at $r\in V(M')\setminus\{w, w'\}$.
See the second figure of Figure~\ref{fig:matching2}.
Since $xx'$ and $rr'$ are part of a strong clique of $G$, there must be an edge between $\{x,x'\}$ and $\{r,r'\}$.
In each case, by adding $xyy'z'w'$, $x'xzz'w'$, $r'xzz'w'$, or $r'x'xyw'$ to $P$, we obtain a $C_{2m}$.
Hence, let us assume that there is no such path.
By Observation~\ref{obs:special} and Lemma~\ref{lem:path1}, since $2m-5 \ge 3$, it must be that $M'$ is $w$-special and $2m-6=2$.
Then, since
$M'$ is a special matching of size $5$, $W(s)$ is not an independent set of $G$ for some vertex $s \in V(M')$.
See Figure~\ref{fig:sp_matching}.
Thus, $G$ contains a $C_{8}$ by Claim~\ref{claim:indep}.
Therefore, $G$ contains a $C_{2m}$.
\end{proof}
Let $x \in V(M)$ such that $|W({x})|$ is maximum.
Note that $|W'(x)|=|W(x)|\ge m \ge 3$.
From Claims~\ref{claim:indep}~and~\ref{claim:indep2}, we assume that
$W(x)$ is an independent set of $G$ and $W'(x)$ has no $P_3$.
\begin{claim}\label{claim:indep3}
If $W'(x)$ is not an independent set of $G$, then $G$ contains a $C_{2m}$.
\end{claim}
\begin{proof}
Suppose that $W'(x)$ is not an independent set of $G$, that is, $y'z' \in E(G)$ for some $y',z' \in W'(x)$.
Let $w' \in W'(x)\setminus \{y',z'\}$.
See the first figure of Figure~\ref{fig:matching3}.
Recall that $\{y,z,w\}$ is an independent set of $G$, and $w'y', w'z' \notin E(G)$ since $W'(x)$ has no $P_3$.
If $w$ is adjacent to both $y'$ and $z'$, then $W(z')$ is not an independent set of $G$, which is a contradiction to Claim~\ref{claim:indep}.
Hence, we may assume that either $wy' \notin E(G)$ or $wz'\notin E(G)$.
Without loss of generality, assume that $wy' \notin E(G)$, which implies that $yw' \in E(G)$.
Let $M'=M\setminus\{yy',zz',ww'\}$, so $|M'|=2m-3\geq 5$.
Suppose that $wz' \in E(G)$.
If $m=3$, then $xyw'wz'z$ is a $C_{2m}$, so assume that $m\ge 4$.
See the first figure of Figure~\ref{fig:matching3}.
Assume that there is an $(x, M')$-path $P$ of length $2m-5$ ending at $r\in V(M')\setminus\{x, x'\}$.
See the second figure of Figure~\ref{fig:matching3}.
Since $rr'$ and $zz'$ are part of a strong clique of $G$, there must be an edge between $\{r,r'\}$ and $\{z,z'\}$.
In each case, by adding $zz'y'yx$, $z'ww'yx$, $r'zz'wx$, or $r'z'y'yx$ to $P$, we obtain a $C_{2m}$.
\begin{figure}[h!]
\centering
\includegraphics[height=3cm,page=12]{fig-strong-clique.pdf}
\caption{Illustrations for Claim~\ref{claim:indep3}} \label{fig:matching3}
\end{figure}
Hence, let us assume that there is no such path.
By Observation~\ref{obs:special} and Lemma~\ref{lem:path1}, since $2m-5 \ge3$,
it must be that $M'$ is $x'$-special and $2m-6=2$.
Then, since $M'$ is a special matching of size $5$, $W(s)$ is not an independent set of $G$ for some vertex $s \in V(M')$.
See Figure~\ref{fig:sp_matching}.
Thus, $G$ contains a $C_{8}$ by Claim~\ref{claim:indep}.
Therefore, $G$ contains a $C_{2m}$.
Now, suppose that $wz'\notin E(G)$ so that $w'z\in E(G)$.
Assume that there is an $(x, M')$-path $P$ of length $2m-5$ ending at $r\in V(M')\setminus\{x, x'\}$.
See the third figure of Figure~\ref{fig:matching3}.
Since $rr'$ and $zz'$ are part of a strong clique of $G$, there must be an edge between $\{r,r'\}$ and $\{z,z'\}$.
In each case, by adding $zz'y'yx$, $z'zw'wx$, $r'zw'wx$, or $r'z'y'yx$ to $P$, we obtain a $C_{2m}$.
Hence, let us assume that there is no such path.
By Observation~\ref{obs:special} and Lemma~\ref{lem:path1}, there are three possible cases:
(i) $x$ has no neighbors in $V(M') \setminus \{x,x'\}$ and $2m-5=1$, (ii) $M'$ is $x$-special and $2m-5=2$, and (iii) $M'$ is $x'$-special and $2m-6=2$.
Note that the second case is impossible.
For the first case,
let $\{a',b'\}=W(x')\setminus \{y,y',z,z',w,w'\}$.
See the fourth figure of Figure~\ref{fig:matching3}.
Since $aa'$ and $yy'$ are part of a strong clique of $G$, there must be an edge between $\{a,a'\}$ and $\{y,y'\}$.
If $ya,yb \in E(G)$, then $yaa'x'b'by$ is a $C_6$.
Otherwise, we may assume without loss of generality that $ya \notin E(G)$, so one of the following is a $C_6$: $xx'a'yw'wx$, $xx'a'y'z'zx$, $xx'a'ay'yx$.
For the third case, since $G[W'(x')]$ contains a $P_3$, $G$ contains a $C_8$ by Claim~\ref{claim:indep2}.
Therefore, $G$ contains a $C_{2m}$.
\end{proof}
For simplicity, let $W(x)=\{u_1,\ldots,u_n\}$.
Note that $n\ge m \ge 3$.
By Claim~\ref{claim:indep3}, we may assume that $W'(x)$ is an independent set of $G$.
Hence, $G[W(x)\cup W'(x)]$ is bipartite.
By Lemma~\ref{lem:matching:path},
we may assume, by relabeling indices if necessary,
that $u_1u_1'u_2u_2'\ldots u_{m-1}u_{m-1}'u_m$ is a path in $G$.
Then, by adding two edges $u_mx$ and $xu_1$ to the path, we obtain a $C_{2m}$.
This proves Lemma~\ref{lem:matching}.
\end{proof}
Now, we prove Theorem~\ref{thm:general}.
\begin{proof}[Proof of Theorem~\ref{thm:general}]
Let $G$ be a $C_{2k}$-free graph with $\Delta(G)\geq 1$, and $H$ be the subgraph of $G$ induced by a maximum strong clique of $G$.
Let $M$ be a maximum matching of $H$ and let $|M|=m$.
If $m\geq 2k$, then $G$ contains a $C_{2k}$ by Lemma~\ref{lem:matching}.
Hence, $m\le 2k-1$.
If $\Delta(H)=1$, then $|E(H)|= |M|\le (2k-1)\Delta(G)$, so in what follows suppose that $\Delta(H)\ge 2$.
Define the following sets:
$Z=V(G)\setminus V(M)$,
$X=\{ v\in V(M)\mid N_H(v)\cap Z \neq \emptyset\}$, and $Y=\{ v\in V(M) \mid|N_H(v)\cap Z|\ge 2 \}$.
Also, let $D=\max\{ \deg_H(v)\mid v\in V(M)\setminus Y\}$, which implies $D\le \min\{ 2m,\Delta(H)\}$ since every vertex in $V(M)\setminus Y$ has at most one neighbor of $H$ in $Z$.
Now,
\[\sum_{v\in V(H)}\deg_H(v)
\le |Y|\Delta(H)+(2m-|Y|)D + |E_H(V(M),Z)|.\]
By the maximality of $M$, if $xx'\in M$, then $x$ and $x'$ cannot have distinct neighbors of $H$ in $Z$.
Thus every edge in $M$ has at most one endpoint in $Y$, so $|Y|\le m$.
Therefore, $|E_H(V(M),Z)|\le (\Delta(H)-1)|Y|+(2m-2|Y|)$.
Thus
$2|E(H)|\le 2|Y|\Delta(H) +(2mD-|Y|D+2m-3|Y|)$, and so
\begin{eqnarray}\label{eq:last}
&& |E(H)|\le |Y|(\Delta(H)-D/2-1) +m(D+1). \end{eqnarray}
Note that $\Delta(H)-D/2-1\ge 0$ since $\Delta(H)\ge \frac{\Delta(H)+2}{2}\ge 1+\frac{D}{2}$.
From \eqref{eq:last} and the fact that $|Y|\le m$, $D\le 2m$, and $m\leq 2k-1$, we conclude the following:
\begin{eqnarray*}
|E(H)|
&\leq& m(\Delta(H)-D/2-1)+m(D+1) \\
&=& m\Delta(H)+mD/2\\
&\leq & (2k-1)\Delta(H) + (2k-1)^2\\
&\leq & (2k-1)\Delta(G) + (2k-1)^2.
\end{eqnarray*}
\end{proof}
\begin{remark}
We can actually show $\SC(G)\le \frac{(2k-1)(4\Delta(G)+1)}{3}$; this is a better bound than the upper bound obtained from Theorem~\ref{thm:general} if $\Delta(G)\le 12k-8$.
We provide an outline of the proof, and use the notation in the proof of Theorem~\ref{thm:general}.
We may assume that $\Delta(H)\ge 2$.
By the maximality of $M$, there is no edge of $H$ between two vertices in $\{ y'\mid y\in Y\}$.
Thus, $V(M)\setminus\{ y'\mid y\in Y\}$ is a vertex cover of $H$.
If $|Y|\ge \frac{2m}{3}$, then $|E(H)|\le (2m-|Y|)\Delta(H)\le \frac{4(2k-1)}{3}\Delta(H)$.
If $|Y|<\frac{2m}{3}$, then, from \eqref{eq:last}, $|E(H)| \le \frac{2m}{3}(\Delta(H)-1-\frac{D}{2})+m(D+1)$,
since $\Delta(H)-1-\frac{D}{2}\ge 0$.
Hence, $|E(H)|\le \frac{2m}{3}\Delta(H) +\frac{2mD}{3}+\frac{m}{3} \le \frac{4(2k-1)\Delta(H)+(2k-1)}{3}$.
\end{remark}
\section*{Acknowledgements}
\small{Ilkyoo Choi was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education (No. NRF-2018R1D1A1B07043049), and also by the Hankuk University of Foreign Studies Research Fund.
Ringi Kim was supported by the National Research Foundation of Korea grant funded by the Korea government (No. NRF-2018R1C1B6003786), and also by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education (No. NRF-2019R1A6A1A10073887).
Boram Park was supported by the National Research Foundation of Korea grant funded by the Korea government (No. NRF-2018R1C1B6003577).}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,619
|
Q: Search based on DateTime I'm relatively new to ruby on rails. I do have a custom search form where I want to search with date and time. My database table does have a column with datetime as a datatype. I knew that we can extract date from Date(column name). I would like to know how to extract only time from datetime field. Please help me with this. Thanks in advance
A: Use date_format
mysql> select id, created_at from users limit 1;
+----+---------------------+
| id | created_at |
+----+---------------------+
| 8 | 2009-08-19 09:24:24 |
+----+---------------------+
1 row in set (0.00 sec)
mysql> select id, DATE_FORMAT(created_at, "%h/%i/%s") from users limit 1;
+----+-------------------------------------+
| id | DATE_FORMAT(created_at, "%h/%i/%s") |
+----+-------------------------------------+
| 8 | 09/24/24 |
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,412
|
Eumenes (auch Emenaios oder Eumenis; † 141) war von 131 bis zu seinem Tod Bischof von Alexandria. Dem Bericht des Eusebius von Caesarea zufolge konnte er sein Amt erst nach einer Vakanz von "einem Jahr und einigen Monaten" als Nachfolger des Justus antreten. Seine Amtszeit fällt in die Regierung der Kaiser Hadrian und Antoninus Pius. Marcus II. wurde sein Nachfolger.
Einzelnachweise
Bischof (2. Jahrhundert)
Person (Alexandria)
Geboren im 1. Jahrhundert
Gestorben 141
Mann
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 647
|
Legend Body Contouring improves cellulite appearance by selectively heating the fat cells in both the deep and superficial layers of the skin, causing them to secrete liquid fat and shrink in size. The heating also increases the blood circulation in the area to boost your metabolism, helping the lymphatic system dispose of the liquefied fat naturally.
There's typically no downtime and you can go about your normal routine after a session. Treatments are usually done weekly—with a complete program usually requiring six to eight sessions—for optimum results.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,382
|
TOKYO (10 p.m.)
Our Lives | TELLING LIVES
'Kurly in Kansai' host Ayana Wyse: 'I'm involved in the community, not the society'
by Rebecca Saunders
People who come to Japan can sometimes feel like they've been presented with a society akin to a Rubik's Cube. There are different colors and moving pieces, it takes time to figure out how it all makes sense. For Ayana Wyse, solving the puzzle hasn't just been a personal struggle. Instead, she has taken it on for her community.
Community organizer: Faced with limited options with regard to authentic cultural spaces, Ayana Wyse founded Black Creatives Japan as a way to find others like her looking to express themselves. | ALEX COOPER WEBSTER
From the suburbs of New York, the 33-year-old Wyse made the move to Japan nine years ago to teach in the Kansai region. She eventually moved to Osaka, a city she quickly fell in love with, and began carving out a life for herself there: She's a photographer, event organizer, DJ, part-time teacher and, most notably, a podcaster.
Though it was love, eventually Wyse looked for some outlet to express her opinion on living in Japan. It was around 2015 that she began thinking the best way to do that would be via an audio format.
"Sometimes you just don't want everyone looking at you," she tells The Japan Times via video chat, "you just want people to hear you."
In 2017, Wyse and her friend Alyse, launched the "Kurly in Kansai" podcast to fill what they saw as a void in the podcasting scene.
"I felt there weren't — or aren't — a lot of podcasts that have two black women hosting and talking about their experiences abroad," Wyse says.
The pair cover topics such as discrimination, culture and working at a Japanese company, occasionally bringing on guests that specialize in the topic they're talking about. The discussion is both casual and no-holds-barred with both hosts chiming in on current events and taking the odd detour into Japan-related anecdotes and geeky side-references.
Podcasts are uploaded monthly (sometimes twice a month) and generally run about an hour long. They can be found through Apple and Spotify, and Wyse also posts them to her YouTube channel, Yana_Yz.
"Our (target) audience is mainly black women, but it's for anyone who wants to hear a different perspective about life in Japan," she says. "We want everyone to enjoy it, but we may sometimes remind people that our experiences can be very different compared to those of other foreigners."
The alliterative title is a catchy way to illustrate what the two friends intended their podcast to focus on. "We — black women, me and Alyse — live in Kansai and we're talking about our experience," Wyse says. "And we're curly-headed," she adds with a laugh.
From Episode 11 onward, "Kurly in Kansai" got its own R&B-tinged theme tune, courtesy of singer and producer Devin Morrison. And Tokyoites, don't be discouraged by any digs you hear. "Kurly in Kansai" is about the love the hosts have for their part of Japan, with Wyse adding that she and Alyse share a "mutual disdain for the Tokyo hype."
Nights in Osaka
Hot topics: Ayana Wyse cohosts the 'Kurly in Kansai' podcast where her and her partner Alyse cover everything from cultural appropriation to Japan's recent tax hike. | ALEX COOPER WEBSTER
Wyse's love of music drew her into the Osaka reggae scene when she first arrived. Japanese music fans tend to dedicate themselves 100 percent to their chosen genres and Wyse was able to access the scene even though it was mainly made up of Japanese fans. However, she encountered some issues being a black woman amid a group of Japanese people who enjoyed reggae and its culture so devoutly.
"To me … I felt like I was like a token to them. That I was a cool accessory," Wyse says of the scene. "They didn't really outwardly show that to me, but every time I went to events, everybody who didn't know me was really surprised to see a dreadlocked, black woman at an event where everybody else was Japanese."
Cultural appropriation and being the one black person in a group of Japanese friends are the kinds of topics that wind up becoming key talking points on "Kurly in Kansai." The Dec. 1, 2017, podcast, "Cultural Appropriation in Japan," explores how Japanese culture is appropriated by the West, using examples such as white actress Scarlett Johansson's casting as the Japanese character Motoko Kusanagi in Hollywood's 2017 take on the classic Japanese manga "Ghost in the Shell," before moving on to discussing instances of the reverse.
In the West, fashion and music are now being scrutinized for mining minority cultures for their clout. But the practice isn't viewed as very problematic in Japan, which has a centuries-old tradition of importing ideas from overseas and making them their own. With an increasing population of non-Japanese in the country, however, the Japanese are starting to pay more attention as to where the culture they're experiencing is drawn from.
"I think (the Japanese) do appropriate, though mostly they try to appreciate the best way they can," Wyse says. "I feel like they take that little bit, then they just overdo it and it looks really bad. They try to understand it, but (that understanding) only scratches the surface."
Frustrated by the way reggae and black culture in general was being appreciated in Japan, Wyse decided to do something about it by starting a club night — Wonderground, a pan-African music event — with support from people she met via a group she founded called Black Creatives Japan (BCJ). Running since 2015, BCJ bills itself as a support collective that also operates as a networking community and platform for collaboration. Setting up BCJ and putting on music events, Wyse says, is a step toward being more than just another voice in the crowd. Her hope is that people like her will also be able to see their culture and experiences represented correctly.
"I don't necessarily want to be in the forefront all the time," Wyse says, "but I feel like no one's doing this, so I might as well step up. I just want to have something that I care about exist and, hopefully, (for) people who are also interested to join in."
Tweaking the rules
Through her experiences, Wyse has managed to solve at least one part of the puzzle that is Japanese society. Part of that solution is the realization that she cannot change the way things work here. Rather than being burdened by the everyday annoyances and pressures the majority of people living in the country feel, she has responded with a simple mantra.
"I say out loud, 'I'm not Japanese, so this doesn't affect me,'" she explains. "There are lots of things I ignore because I'm not Japanese. I don't necessarily try to follow all the rules."
Some of those societal rules and norms she's referring to, however, can be too much to ignore — how women are treated by men, for instance, and how difficult it can be to stand up for what you believe is right.
"I do try to defend myself and defend other fellow women. It's different for men," she says. "It's different for nonblack men and different for nonblack women, and so on. I definitely see things differently. But I try to step into other people's shoes so that I can have a broader view of what other foreigners experience in Japan."
The groups and events that Wyse has set up during her time here have enabled her to discover spaces that suit her, creating places where she — and others — don't have to alter themselves in order to fit some perceived standard.
However, feeling as though she doesn't have to conform to certain codes of Japanese society doesn't necessarily mean that she isn't part of it. Having been involved in organizing events, as well as volunteer work, Wyse isn't opting out of society — she's providing her own interpretation of it.
"I feel like I'm involved in the community, that's more like it — I'm involved in the community, not the society," she explains. "(Non-Japanese people are) meshing, but we're not fully involved or part of it like in some other societies."
Things change slowly in Japan. The puzzle may seem near impossible to solve, but each support group and podcast, like the twists and turns of a Rubik's Cube, gets us closer to a cohesive goal.
For more information on Black Creatives Japan, visit blackcreativesjapan.com. Also, check out "Kurly in Kansai" on Twitter: @KurlyInKansai
expats, living in Japan
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,213
|
Blondie in the Dough is a 1947 American comedy film directed by Abby Berlin and starring Penny Singleton, Arthur Lake, Larry Simms, and Marjorie Ann Mutchie. It is 21st of the 28 Blondie films.
Plot
Dagwood and his boss are frantically searching for an eccentric cookie tycoon in order to sign him to a contract. Unknown to them, Blondie has already met him and the man is enjoying himself making cookies in Blondie's kitchen while Dagwood is at work.
Cast
Penny Singleton as Blondie
Arthur Lake as Dagwood
Larry Simms as Baby Dumpling
Marjorie Ann Mutchie as Cookie
Daisy as Daisy the Dog
Jerome Cowan as George Radcliffe
Hugh Herbert as Llewellyn Simmons
Clarence Kolb as J.T. Thorpe
Danny Mummert as Alvin Fuddle
William Forrest as Bob Dixon
Eddie Acuff as Mr. Beasley
Norman Phillips Jr. as Ollie Shaw
Kernan Cripps as Harry Baxter
Fred F. Sears as Quinn
Alyn Lockwood as Mary
References
External links
1947 films
Columbia Pictures films
American black-and-white films
1947 comedy films
Blondie (film series) films
American comedy films
Films directed by Abby Berlin
1940s American films
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,945
|
{"url":"https:\/\/questioncove.com\/updates\/524c8507e4b0e0b65599280f","text":"Mathematics\nOpenStudy (megannicole51):\n\nA rod with density 3+sin(x) lies on the x-axis between x=0 and x=pi. Find the mass and center of mass of the rod.\n\nOpenStudy (anonymous):\n\nso here you integrate... from 0 to pi. $m = \\int\\limits_{0}^{\\pi}(3+\\sin x)dx$ and the center of mass of the rod is $CM = \\frac{ 1 }{ m }\\int\\limits_{0}^{\\pi}x(3+\\sin x)dx$\n\nOpenStudy (anonymous):\n\nyou there?\n\nOpenStudy (megannicole51):\n\nyeah im trying to figure it out sorry\n\nOpenStudy (anonymous):\n\nno it's okay. take your time\n\nOpenStudy (megannicole51):\n\nyay i got the right answers! :D thank you!\n\nOpenStudy (anonymous):\n\nyou're welcome!\n\nOpenStudy (megannicole51):\n\nhow do you find a mass of a triangle? is it the same or different?\n\nOpenStudy (anonymous):\n\nwhat's the density function?\n\nOpenStudy (megannicole51):\n\nhey sorry i fell asleep! so the function is 1+x grams\/cm^2 and this is what the pictures looks like that im given....|dw:1380761318589:dw|","date":"2021-08-05 19:03:31","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5726711750030518, \"perplexity\": 3502.4305657189097}, \"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-2021-31\/segments\/1627046156141.29\/warc\/CC-MAIN-20210805161906-20210805191906-00456.warc.gz\"}"}
| null | null |
Q: Почему нельзя увеличивать массив на ходу? void average()
{ int ctr=1;
int averagenum[ctr];
for (int i = 0; i<ctr;i++)
{
printf ("%d-ый элемент:", i);
scanf("%d", &averagenum[i]);
if(averagenum[i]!=0)
ctr++;
}
}
Почему так нельзя? Или можно, но я что-то не понял?
A: Вы выделяете память для массива из одного элемента (текущее значение ctr).
Но цикл в общем случае не останавливается после первой итерации и пытается писать в averagenum за пределами выделенной памяти. Результат печален...
То, что вы потом меняете значение ctr - массив не меняется.
Примерно как в ресторане вы бросаете кошелек официанту и говорите "гуляю на все", официант смотрит внутрь, берет деньги и тащит вам на ваш целый рубль три корочки хлеба :). После этого вы докладываете в кошелек пару тысяч - но заказ-то уже сделан - с чего официанту интересоваться, что у вас там происходит?...
A: Почему-то часто вижу такую ошибку.
Когда вы пишете что-то вроде:
int ctr=1;
int averagenum[ctr];
averagenum получает размер 1. Если потом поменять ctr, размер массива не изменится.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,689
|
using System;
using MonoTouch.UIKit;
using System.Drawing;
using DSoft.Datatypes.Calendar;
using DSoft.Datatypes.Calendar.Language;
using DSoft.Datatypes.Calendar.Enums;
using DSoft.UI.Calendar.Data;
using DSoft.Datatypes.Calendar.Data.Collections;
using DSoft.UI.Calendar.Themes;
using DSoft.UI.Calendar.Helpers;
using MonoTouch.Foundation;
using DSoft.Datatypes.Calendar.Interfaces;
using DSoft.Datatypes.Calendar.Data;
using MonoTouch.CoreGraphics;
namespace DSoft.UI.Calendar.Views
{
/// <summary>
/// Calendar cell view
/// </summary>
public class DSCalendarCell : DSTouchView
{
#region Fields
private DSCalendarCellInfo mCellInfo;
private UILabel mRightLabel;
private UIView mTodayHeader;
private DSCalendarEventCollection mEvents;
private DSCalendarEventsView mEventView;
private DSMoreEventsView mMoreEventsView;
private bool mSelected;
#endregion
#region Properties
/// <summary>
/// Gets a value indicating whether this instance is weekend.
/// </summary>
/// <value><c>true</c> if this instance is weekend; otherwise, <c>false</c>.</value>
private bool IsWeekend
{
get
{
switch(mCellInfo.Date.DayOfWeek)
{
case DayOfWeek.Sunday:
case DayOfWeek.Saturday:
return true;
default:
return false;
}
}
}
/// <summary>
/// Gets or sets the events for the cell
/// </summary>
/// <value>The events.</value>
internal DSCalendarEventCollection Events
{
get
{
return mEvents;
}
set
{
if (mEvents != value)
{
mEvents = value;
if (mEvents.Count != 0)
{
this.SetNeedsDisplay();
}
}
}
}
/// <summary>
/// Gets or sets the cell info.
/// </summary>
/// <value>The cell info.</value>
internal DSCalendarCellInfo CellInfo
{
get
{
return mCellInfo;
}
set
{
mCellInfo = value;
this.SetNeedsDisplay();
}
}
/// <summary>
/// Gets the more items.
/// </summary>
/// <value>The more items.</value>
internal int MoreItems
{
get
{
var count = 0;
if (Events != null && Events.Count > DSCalendarTheme.CurrentTheme.MaxEvents)
{
count = Events.Count;
}
return count;
}
}
/// <summary>
/// Gets or sets a value indicating whether this instance is selected.
/// </summary>
/// <value><c>true</c> if this instance is selected; otherwise, <c>false</c>.</value>
internal bool IsSelected
{
get
{
return mSelected;
}
set
{
if (mSelected != value)
{
mSelected = value;
this.SetNeedsDisplay ();
}
}
}
#endregion
#region Constuctors
/// <summary>
/// Initializes a new instance of the <see cref="DSoft.UI.Calendar.Views.DSCalendarCell"/> class.
/// </summary>
/// <param name="Frame">Frame.</param>
internal DSCalendarCell (RectangleF Frame) : base(Frame)
{
Initialize ();
}
#endregion
#region Overrides
/// <summary>
/// Draw the specified rect.
/// </summary>
/// <param name="rect">Rect.</param>
public override void Draw (RectangleF rect)
{
base.Draw (rect);
DrawCell(rect);
}
#endregion
#region Functions
private void Initialize()
{
this.Opaque = false;
this.BackgroundColor = UIColor.Clear;
this.AutosizesSubviews = true;
this.AutoresizingMask = UIViewAutoresizing.FlexibleWidth;
mRightLabel = new UILabel(RectangleF.Empty);
mRightLabel.BackgroundColor = UIColor.Clear;
mRightLabel.TextColor = UIColor.Gray;
mRightLabel.TextAlignment = UITextAlignment.Right;
mRightLabel.Font = DSCalendarTheme.CurrentTheme.CellTextFont;
this.AddSubview(mRightLabel);
mEventView = new DSCalendarEventsView(RectangleF.Empty);
mEventView.BackgroundColor = UIColor.Clear;
this.Add(mEventView);
if (DSCalendarTheme.CurrentTheme.MoreViewPosition == DSMoreEventsViewPositon.TopRight)
{
mMoreEventsView = DSCalendarTheme.CurrentTheme.MoreEventsView();
if (mMoreEventsView != null)
{
this.AddSubview(mMoreEventsView);
}
}
this.SingleTap += () =>
{
SingleTouchEvent();
};
this.DoubleTap += () =>
{
AddNewEvent();
};
}
/// <summary>
/// Draws the cell
/// </summary>
/// <param name="rect">Rect.</param>
private void DrawCell (RectangleF rect)
{
if (mCellInfo.IsFocus) {
if (mTodayHeader == null)
mTodayHeader = DSCalendarTheme.CurrentTheme.TodayCellHeaderView;
if (mTodayHeader != null) {
//set the frame
mTodayHeader.Frame = new RectangleF (0, 0, this.Frame.Width, 20).Integral();
if (mTodayHeader.Superview == null)
this.InsertSubviewBelow (mTodayHeader, mRightLabel);
}
}
var context = UIGraphics.GetCurrentContext ();
UIColor fillColor = UIColor.White;
var selectedColor = DSCalendarTheme.CurrentTheme.CellSelectedBackground;
if (!IsSelected || selectedColor == null) {
if (mCellInfo.IsFocus) {
fillColor = DSCalendarTheme.CurrentTheme.TodayCellBackground;
} else if (IsWeekend) {
fillColor = (mCellInfo.IsCurrent) ? DSCalendarTheme.CurrentTheme.WeekendCellBackground : DSCalendarTheme.CurrentTheme.InActiveWeekendCellBackground;
} else if (!mCellInfo.IsCurrent) {
fillColor = DSCalendarTheme.CurrentTheme.InActiveCellBackgrond;
} else {
fillColor = DSCalendarTheme.CurrentTheme.CellBackground;
}
} else {
fillColor = selectedColor;
}
fillColor.SetFill ();
context.FillRect (rect);
context.SetStrokeColor (DSCalendarTheme.CurrentTheme.CellBorderColor.CGColor);
context.SetLineWidth (DSCalendarTheme.CurrentTheme.GridBorderWidth);
context.StrokeRect (rect);
mRightLabel.Text = String.Empty;
//mRightLabel.BackgroundColor = UIColor.Red;
//
if (mCellInfo.IsFirstRow && DSCalendarTheme.CurrentTheme.DayStyle == CalendarDayStyle.FirstRow) {
//get the day string for this row
var aDay = EnumHelper.ConvertToWeekDay (mCellInfo.Date.DayOfWeek);
var aString = DSCalendarLanguage.CurrentLanguage.ShortStringForDay ((int)aDay) + " ";
mRightLabel.Text += aString;
}
mRightLabel.Text += mCellInfo.Date.Day.ToString ();
if (mCellInfo.ShowMonth) {
mRightLabel.Text += " " + DSCalendarEnglish.CurrentLanguage.ShortStringForMonth (mCellInfo.Date.Month - 1);
}
if (!IsSelected || selectedColor == null)
{
mRightLabel.TextColor = (mCellInfo.IsFocus) ? DSCalendarTheme.CurrentTheme.TodayCellTextColor
: (mCellInfo.IsCurrent) ? DSCalendarTheme.CurrentTheme.CellTextColor : DSCalendarTheme.CurrentTheme.InActiveCellTextColor;
}
else
{
mRightLabel.TextColor = DSCalendarTheme.CurrentTheme.CellSelectedTextColor;
}
float aPos = 0;
float yPos = 0;
float offSet = 4;
var fSize = this.StringSize(mRightLabel.Text,mRightLabel.Font,new SizeF(this.Frame.Width,20));
var fHeight = fSize.Height;
//is weekend
switch (DSCalendarTheme.CurrentTheme.DayLocation)
{
case CalendarDayLocation.TopRight:
{
aPos = offSet;
yPos = offSet;
mRightLabel.TextAlignment = UITextAlignment.Right;
}
break;
case CalendarDayLocation.BottomRight:
{
aPos = offSet;
yPos = this.Frame.Height - (fHeight + offSet);
mRightLabel.TextAlignment = UITextAlignment.Right;
}
break;
case CalendarDayLocation.TopLeft:
{
aPos = offSet;//mLeftLabel.Frame.Left + mLeftLabel.Frame.Width;
yPos = offSet;
mRightLabel.TextAlignment = UITextAlignment.Left;
}
break;
case CalendarDayLocation.BottomLeft:
{
aPos = offSet;//mLeftLabel.Frame.Left + mLeftLabel.Frame.Width;
yPos = this.Frame.Height - (fHeight + offSet);
mRightLabel.TextAlignment = UITextAlignment.Left;
}
break;
}
mRightLabel.Frame = new RectangleF(aPos,yPos,fSize.Width,fHeight).Integral();
RectangleF eventFrame;// = null;
if (yPos == offSet)
{
float top = ((mRightLabel.Frame.Top + mRightLabel.Frame.Height) + offSet) + 4;
eventFrame = new RectangleF(0,top,this.Frame.Width,this.Frame.Height - (top + 2));
}
else
{
eventFrame = new RectangleF(0,offSet,this.Frame.Width,this.Frame.Height - (mRightLabel.Frame.Height + offSet));
}
mEventView.CellDate = mCellInfo.Date;
mEventView.Events = mEvents;
mEventView.Frame = eventFrame.Integral();
if (DSCalendarTheme.CurrentTheme.MoreViewPosition == DSMoreEventsViewPositon.TopRight && MoreItems != 0)
{
if (mMoreEventsView != null)
{
var width = (this.Frame.Width - mRightLabel.Frame.Left)-4;
mMoreEventsView.IsToday = mCellInfo.IsFocus;
mMoreEventsView.RemainingItems = MoreItems;
mMoreEventsView.Frame = new RectangleF(mRightLabel.Frame.Left,0,width, DSCalendarTheme.CurrentTheme.MoreEventsViewHeight).Integral();
}
}
}
#region Internal Functions
internal void SingleTouchEvent()
{
this.IsSelected = true;
if (this.Superview != null && this.Superview is DSCalendarGridView)
{
((DSCalendarGridView)this.Superview).SelectedChanged(this);
}
}
internal void AddNewEvent ()
{
if (this.Superview != null && this.Superview is DSCalendarGridView)
{
var calendarGrid = (DSCalendarGridView)this.Superview;
calendarGrid.DataSource.AddEventForDate(this.mCellInfo.Date,calendarGrid.Superview as DSCalendarView, this);
}
}
/// <summary>
/// Called when an event is double tapped.
/// </summary>
/// <param name="AnEvent">An event.</param>
/// <param name="sender">Sender.</param>
internal void DoubleTappedEvent(DSCalendarEvent AnEvent, object sender)
{
if (this.Superview != null && this.Superview is DSCalendarGridView)
{
((DSCalendarGridView)this.Superview).DoubleTappedEvent(AnEvent, this);
}
}
#endregion
#endregion
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,892
|
Dillwynella texana is an extinct species of sea snail, a marine gastropod mollusk in the family Skeneidae.
Description
The shell contains four, spiral, smooth and shining whorls. The body whorl is nearly smooth but showing a slight tendency to bear furrows or lines radiating from the suture. The umbilicus is small. The aperture is round.
Distribution
This marine species was found as a fossil in the Lower Claiborne Eocene, Texas.
References
texana
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,284
|
Stefan Wolpe (født 25. august 1902, død 4. april 1972) var en tysk/amerikansk komponist født i Berlin.
Flygtede ved nazisternes magtovertagelse i 1933 til Wien, kom via Palæstina til USA i 1938, hvor han levede resten af sit liv.
Stort set alle hans værker er skrevet for mindre kammerbesætninger, og bærer præg af indflydelse fra Webern, de er serielt komponeret, rytmisk stærkt komplicerede og har en raffineret klangskønhed. Han har bl.a. skrevet 1 symfoni.
Udvalgte værker
Symfoni nr. 1 (1952) - for orkester
Passacaglia (19?) - for stort orkester
"2 Studier" (19?) - for stort orkester
Referencer
Komponister fra Tyskland
Personer fra Berlin
Tyskere i 1900-tallet
Amerikanere i 1900-tallet
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 107
|
// OpenPHACTS RDF Validator,
// A tool for validating and storing RDF.
//
// Copyright 2012-2013 Christian Y. A. Brenninkmeijer
// Copyright 2012-2013 University of Manchester
// Copyright 2012-2013 OpenPhacts
//
// Licensed 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.
//
package uk.ac.manchester.cs.datadesc.validator.server;
import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;
import java.io.InputStream;
import java.util.List;
import javax.servlet.http.HttpServletRequest;
import javax.ws.rs.QueryParam;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.Response;
import uk.ac.manchester.cs.datadesc.validator.rdftools.VoidValidatorException;
import uk.ac.manchester.cs.datadesc.validator.ws.WSRdfInterface;
import uk.ac.manchester.cs.datadesc.validator.ws.WSValidatorInterface;
import uk.ac.manchester.cs.datadesc.validator.ws.WsValidationConstants;
/**
*
* @author Christian
*/
public interface ValidatorWSInterface extends WSRdfInterface, WSValidatorInterface{
public Response validateHome(@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
public String getRdfDump() throws VoidValidatorException;
public Response getStatementList(@QueryParam(WsValidationConstants.SUBJECT) String subjectString,
@QueryParam(WsValidationConstants.PREDICATE) String predicateString,
@QueryParam(WsValidationConstants.OBJECT) String objectString,
@QueryParam(WsValidationConstants.CONTEXT) List<String> contextStrings,
@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
public Response getByResource(@QueryParam(WsValidationConstants.RESOURCE) String resourceString,
@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
public Response loadURI(@QueryParam(WsValidationConstants.URI) String address,
@QueryParam(WsValidationConstants.RDF_FORMAT) String formatName,
@Context HttpServletRequest httpServletRequest)throws VoidValidatorException;
public Response runSparqlQuery(@QueryParam(WsValidationConstants.QUERY)String query,
@QueryParam(WsValidationConstants.FORMAT)String formatName,
@Context HttpServletRequest httpServletRequest)throws VoidValidatorException;
public Response validate(@QueryParam(WsValidationConstants.TEXT) String text,
@QueryParam(WsValidationConstants.URI) String uri,
@QueryParam(WsValidationConstants.RDF_FORMAT) String rdfFormat,
@QueryParam(WsValidationConstants.SPECIFICATION) String specification,
@QueryParam(WsValidationConstants.INCLUDE_WARNINGS) Boolean includeWarning,
@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
public Response validateTurtleHtml(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail, //ToDo work out while this paramters is always null
@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
public String validateTurtle(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail) throws VoidValidatorException;
public Response validateTurtleHtml(@Context HttpServletRequest httpServletRequest) throws VoidValidatorException;
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,937
|
namespace google {
namespace cloud {
namespace dialogflow_cx_internal {
GOOGLE_CLOUD_CPP_INLINE_NAMESPACE_BEGIN
class EnvironmentsStub {
public:
virtual ~EnvironmentsStub() = 0;
virtual StatusOr<google::cloud::dialogflow::cx::v3::ListEnvironmentsResponse>
ListEnvironments(
grpc::ClientContext& context,
google::cloud::dialogflow::cx::v3::ListEnvironmentsRequest const&
request) = 0;
virtual StatusOr<google::cloud::dialogflow::cx::v3::Environment>
GetEnvironment(grpc::ClientContext& context,
google::cloud::dialogflow::cx::v3::GetEnvironmentRequest const&
request) = 0;
virtual future<StatusOr<google::longrunning::Operation>>
AsyncCreateEnvironment(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::CreateEnvironmentRequest const&
request) = 0;
virtual future<StatusOr<google::longrunning::Operation>>
AsyncUpdateEnvironment(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::UpdateEnvironmentRequest const&
request) = 0;
virtual Status DeleteEnvironment(
grpc::ClientContext& context,
google::cloud::dialogflow::cx::v3::DeleteEnvironmentRequest const&
request) = 0;
virtual StatusOr<
google::cloud::dialogflow::cx::v3::LookupEnvironmentHistoryResponse>
LookupEnvironmentHistory(
grpc::ClientContext& context,
google::cloud::dialogflow::cx::v3::LookupEnvironmentHistoryRequest const&
request) = 0;
virtual future<StatusOr<google::longrunning::Operation>>
AsyncRunContinuousTest(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::RunContinuousTestRequest const&
request) = 0;
virtual StatusOr<
google::cloud::dialogflow::cx::v3::ListContinuousTestResultsResponse>
ListContinuousTestResults(
grpc::ClientContext& context,
google::cloud::dialogflow::cx::v3::ListContinuousTestResultsRequest const&
request) = 0;
virtual future<StatusOr<google::longrunning::Operation>> AsyncDeployFlow(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::DeployFlowRequest const& request) = 0;
virtual future<StatusOr<google::longrunning::Operation>> AsyncGetOperation(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::longrunning::GetOperationRequest const& request) = 0;
virtual future<Status> AsyncCancelOperation(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::longrunning::CancelOperationRequest const& request) = 0;
};
class DefaultEnvironmentsStub : public EnvironmentsStub {
public:
DefaultEnvironmentsStub(
std::unique_ptr<
google::cloud::dialogflow::cx::v3::Environments::StubInterface>
grpc_stub,
std::unique_ptr<google::longrunning::Operations::StubInterface>
operations)
: grpc_stub_(std::move(grpc_stub)), operations_(std::move(operations)) {}
StatusOr<google::cloud::dialogflow::cx::v3::ListEnvironmentsResponse>
ListEnvironments(
grpc::ClientContext& client_context,
google::cloud::dialogflow::cx::v3::ListEnvironmentsRequest const& request)
override;
StatusOr<google::cloud::dialogflow::cx::v3::Environment> GetEnvironment(
grpc::ClientContext& client_context,
google::cloud::dialogflow::cx::v3::GetEnvironmentRequest const& request)
override;
future<StatusOr<google::longrunning::Operation>> AsyncCreateEnvironment(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::CreateEnvironmentRequest const&
request) override;
future<StatusOr<google::longrunning::Operation>> AsyncUpdateEnvironment(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::UpdateEnvironmentRequest const&
request) override;
Status DeleteEnvironment(
grpc::ClientContext& client_context,
google::cloud::dialogflow::cx::v3::DeleteEnvironmentRequest const&
request) override;
StatusOr<google::cloud::dialogflow::cx::v3::LookupEnvironmentHistoryResponse>
LookupEnvironmentHistory(
grpc::ClientContext& client_context,
google::cloud::dialogflow::cx::v3::LookupEnvironmentHistoryRequest const&
request) override;
future<StatusOr<google::longrunning::Operation>> AsyncRunContinuousTest(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::RunContinuousTestRequest const&
request) override;
StatusOr<google::cloud::dialogflow::cx::v3::ListContinuousTestResultsResponse>
ListContinuousTestResults(
grpc::ClientContext& client_context,
google::cloud::dialogflow::cx::v3::ListContinuousTestResultsRequest const&
request) override;
future<StatusOr<google::longrunning::Operation>> AsyncDeployFlow(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::cloud::dialogflow::cx::v3::DeployFlowRequest const& request)
override;
future<StatusOr<google::longrunning::Operation>> AsyncGetOperation(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::longrunning::GetOperationRequest const& request) override;
future<Status> AsyncCancelOperation(
google::cloud::CompletionQueue& cq,
std::unique_ptr<grpc::ClientContext> context,
google::longrunning::CancelOperationRequest const& request) override;
private:
std::unique_ptr<
google::cloud::dialogflow::cx::v3::Environments::StubInterface>
grpc_stub_;
std::unique_ptr<google::longrunning::Operations::StubInterface> operations_;
};
GOOGLE_CLOUD_CPP_INLINE_NAMESPACE_END
} // namespace dialogflow_cx_internal
} // namespace cloud
} // namespace google
#endif // GOOGLE_CLOUD_CPP_GOOGLE_CLOUD_DIALOGFLOW_CX_INTERNAL_ENVIRONMENTS_STUB_H
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,279
|
{"url":"https:\/\/cstheory.stackexchange.com\/questions\/47982\/generating-a-pseudo-random-rubiks-cube-in-on2-epsilon-time","text":"# Generating a pseudo random Rubik's cube in $O(n^{2+\\epsilon})$ time\n\nRecently I've begun considering how one could generate and solve an $$n \\times n\\times n$$ Rubik's cube for $$n$$ well over 10,000. To solve such a cube is feasible; easily implementable parallelizable algorithms to solve such large cubes can be created with the methods listed here:\n\nSuch can be executed in $$O(n^2)$$ time, considering that they rely on short sequences of moves that only displace a couple of cubes, and do not have to compute individual rotations. But scrambling is another story. Given that a randomly executed move takes time $$\\Omega(n)$$ and the diameter of the graph of the group of Rubik's cubes is $$\\Theta(n^2\/\\log n)$$, a naive scramble would take $$\\Omega(n^3\/ \\log n)$$, which is not at all feasible. Given that, is there an algorithm for scrambling large Rubik's cubes which:\n\n1. Takes $$O(n^{2+\\epsilon})$$ time\n2. Produces every possible reachable configuration with positive probability, and only produces such configurations\n3. Is not extremely skewed\n\u2022 Can you specify what a \"randomly executed move\" is, and why it takes $\\Omega(n)$ time? Dec 8 '20 at 9:30\n\nThis takes $$O(N^2)$$ time.","date":"2021-10-20 09:55:06","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\": 8, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.650833785533905, \"perplexity\": 793.7707659416825}, \"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-2021-43\/segments\/1634323585305.53\/warc\/CC-MAIN-20211020090145-20211020120145-00510.warc.gz\"}"}
| null | null |
If you specialize in consumable items—like food, medicine, and cosmetics—you know that plastic tube packaging plays a big part in it. As your trade becomes a full-grown business, you see fruits of your labor, like huge sales, kicking in. However, with money flowing in, it's time for you to consider the next step—outsourcing business processes including product packaging.
A growing business means increasing cost, and hiring co-packers is a game changer.
Profit by getting affordable turnkey solutions.
Studies show that the tube packaging industry is expected to exceed $9.94 billion by 2022. This means that the demand for plastic tube packaging is high. In fact, some companies like Consolidated Strategy Group are developing eco-friendly solutions for their packaging in response to calls for environmental sustainability.
This is the perfect opportunity for you to capitalize on outsourced packaging. With the ever-growing standards of packaging companies at competitive prices, you'll find players that can provide packaging to aid your brand.
Concentrate on product development and leave packaging to the experts.
Some of you might think that macro-managing the packaging process would diminish your product a bit. However, many experienced companies will study your product thoroughly, so you can be sure that quality control is still excellent. In fact, by giving co-packers control of the packaging process, you will now be able to concentrate on making the product itself better through product development.
For example, when you leave a turnkey provider for your pharmaceutical packaging, you can start developing the concentration of your drug. Even better, you will be able to develop your medicine and lessen the side effects.
Partner with a packaging company that allows you to customize.
The biggest fear of any company partnering with turnkey solution providers is losing their say in packaging. They tend to think that the company already has a template for their packaging. On the contrary, veteran packaging companies can readily customize their service according to the packaging you already have, or what you plan to have.
The great thing about plastic tubes is that they can be molded into various shapes, sizes, and colors. As your partner packaging company, we will take advantage of these qualities to provide you with your desired packaging. With us, not only do you get the packaging you want, you save more than if you package your product in-house!
With plastic tube packaging on the rise, now's the time for you to consider partnering with a competitive co-packer.
As one of the top cosmetic packaging companies in California for 40 years, Consolidated Strategy Group will help in your branding. You can concentrate on other business aspects, while we oversee the packaging solutions for you. Call us at 714-866-8301 for the best contract packages your company can ever have.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,452
|
Welcome to Connecticut Ballet's website!
Looking to find out about auditions?
Want to sign onto our mailing list?
Want to book a school program in your community?
Check out our 2018-2019 promo video. Click here.
• SUPPORT our production & educational programs by donating a used vehicle and receving a receipt for your tax-deductible contribution.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,788
|
# Gently by the Shore
Alan Hunter
_In Memoriam_
H.E. Hunter
I. Hunter
# Contents
Title Page
Dedication
CHAPTER ONE
CHAPTER TWO
CHAPTER THREE
CHAPTER FOUR
CHAPTER FIVE
CHAPTER SIX
CHAPTER SEVEN
CHAPTER EIGHT
CHAPTER NINE
CHAPTER TEN
CHAPTER ELEVEN
CHAPTER TWELVE
CHAPTER THIRTEEN
CHAPTER FOURTEEN
CHAPTER FIFTEEN
About the Author
By the Same Author
Copyright
CHAPTER ONE
EVEN THE SEA which lapped the August beaches of Starmouth looked grey at that hour of the morning. There was something mournful about it – it seemed to be grieving for the thronged crowds of noonday. Northwards it was embraced by the sprawl of the Albion Pier, much destroyed, much reconstructed, south-wards by the elegant iron-work of the Wellesley with its Winter Gardens, while facing it, across the wide promenade, lay the hectic holiday face of the town, a Victorian foundation in evil, dark-red brick with overlays of modern Marine Baroque. And the prevailing note was sadness. The dawn refused to ratify what man called gay. At this solemn hour, when PC Lubbock was observing his regulation speed between pier and pier, the Seaside Of The Midlands looked like a sleeping drunk stretched by the disapproving main.
He stopped, did PC Lubbock: he checked his well-regulated 2mph and conducted a survey of the morning scene. All was quiet, and most was still. On the pale-looking beach below him two or three figures were moving, slow, intent, each with a stick with which he occasionally stirred the marble-cold sand. Beyond them some sandpipers worked along the tide-line, further out some terns, and further still the gulls. Scavengers all were they, men and birds. PC Lubbock marked them with a permitting eye. He had seen them upon their lawful occasions for many a long year now and he gave them a favouring nod as he turned to pursue his jaunt.
But before he could get under way a change took place in the peaceful scene. A movement occurred, quite other than those he had come to expect from the deliberate trade of beach-combing. Towards him came leaping and capering, more animal than human, a strange, chattering figure, a figure that flailed its arms, a figure from whose splaying heels the sand shot up in clouds. PC Lubbock hesitated in his stride. There was something mindless and rather horrible about this bounding creature. Although he recognized it as Nits, a local halfwit, he couldn't help falling back a pace as it vaulted over the balustrade and dropped crouching at his feet.
'Well... and what d'you want, m'lad?' he demanded sternly, fixing his gaze on the halfwit's protruding green eyes.
They stared at him silently, seeming to strain towards him: the rest of the face sank backwards towards a toothy gape.
'What is it?' reiterated the Police Constable, raising his voice a degree.
Nits sucked in his lips as though preparing them for articulation. 'He... don't wake up,' he blurted in his slurring pipe.
'Eh? Who doesn't wake up?' asked the constable.
'The man... he don't wake up.'
Nits made an orang-outang-like gesture towards the direction from which he had come. 'All wet!' he whimpered, 'no clothes on... don't wake up.'
There was a pause while the trained mind arranged this information.
'You say it's _a man?_ ' PC Lubbock demanded suspiciously.
'A man – a man – a man!' Nits nodded his head with astonishing rapidity.
'You mean one like me?'
The head bobbed on as though worked by a piston.
With a stately cock of his leg, PC Lubbock stormed the balustrade and descended to the beach below. Through sand and through shingle went his boots, through shells and through seaweed, till he stood at last where the low slack water played old Harry with his spit and polish. And there he saw him, the man who didn't wake up, the man without clothes, the man who was all wet.
He had stood about five feet ten. He had weighed about 185. His hair had been pale brown, his eyes blue, his eyebrows slanting, his heavy features decidedly un-English. And he had acquired, probably rather late in life, a feature of the keenest police interest: a collection of four stab-wounds in the thorax.
PC Lubbock remarked a high percentage of these details. He glanced sharply at Nits, and sharply at the sea. Then, drawing his whistle with a flourish of professional adroitness, he blew a wailful blast to wake the morning air.
* * *
'There seems,' said Chief Inspector Gently, Central Office, CID, sagely, 'to be some as-yet-undiscovered connection between coastal resorts and homicide, Dutt. Have you noticed it?'
Detective Sergeant Dutt nodded dutifully, but without really listening to his senior. It had been hot in the train coming up. It was still hot in the train. Their third-class compartment was a little oven, and its atmosphere wasn't improved by the haze contributed by Gently's pipe.
'You've only to go back to the 'twenties,' continued Gently, with a damaging puff. 'There were the Crumbles murders – Field and Gray in '20, and Mahon in '24. Both classics, Dutt. Especially Mahon.'
'I was bashing me first beat in '24,' said Dutt reminiscently.
'Then there was Smith and the Brides in the Baths – Blackpool and Herne Bay were two of his spots – and coming the other way there's the Brighton Trunk Murders and Sidney Fox at Margate, and that other Starmouth business – slaughter in all shapes and sizes, and all of it going on by the sea. There's a link there somewhere, Dutt, you mark my words. The sea has a bad influence on potential homicides, whether it's recognized or not.'
'Dare say you're right, sir,' replied Dutt, staring out of the window.
'When I retire I shall write a monograph on it,' added Gently. 'There may be some implications which would help a good defence.'
He sank back into his seat and puffed away in silence. The train clattered on, wearying, somnolent. They were nearing the end of the run, four sun-beating hours of it, and both of them felt jaded and grimy. Outside stretched the marshes of East Northshire, very wide, very flat, their distance broken by nothing except the brick towers of windmills and the white handkerchief sails of yachts. Inside there was Gently's pipe and the sooty smell of third-class cushions...
'Well, it won't be so bad, sir,' said Dutt, trying to cheer himself up, 'it can't be worse than Southend or Margate.'
Gently smiled at a distant cow. 'It isn't,' he said, 'there's parts of it one grows to like.'
'You know the place, sir – you've been there before?'
'When I was ten,' admitted Gently, 'and that's further back than I like to remember.'
He thought about it, nevertheless. He could see himself now as he was then, a thoughtful child with sunburn and freckles, and those damned knickerbockers. A solitary child he had been, a bad mixer. It may have been the knickerbockers that made him antisocial.
'There isn't much difference between criminals and policemen,' he said, surprising Detective Sergeant Dutt.
They pulled in at Starmouth Ranelagh, a gloomy terminus where the smell of fish blended into a neat olfactory cocktail with the smell of soot, steam and engine oil.
'It hasn't changed,' mused Gently, 'that's just the smell it used to have.'
He reached down a battered leather suitcase and deposited himself and it upon the platform. Sergeant Dutt followed, carrying a similar case, while in his other hand he clasped the 'murder bag' with which a careful Central Office had equipped the expedition. Outside in the station yard the afternoon sun burned down stunningly. There was a taxi rank, and co-passengers clad in summer dresses and open-necked shirts were streaming towards it. Sergeant Dutt looked longingly, but Gently shook his head.
'It isn't far,' he said, 'they've got their headquarters just off the quay.'
'Don't know what they've gone and packed in the bag,' said Sergeant Dutt reprovingly, 'it's like a ton weight.'
'Probably a ball and chain for when we make the pinch,' replied his senior unfeelingly.
They left the station and plodded over a lift-bridge which carried the main road into the town. Below them a cloudy muddy-banked stream flowed pacifically, bearing on its bosom tugs with barges and smaller traffic. Further down two torpedo boats were moored at the quays, opposite them a lightship undergoing a refit, and one or two stream-drifters. Above the bridge was the yacht station, its staithe packed three-deep with visiting holiday-craft.
'It's a ruddy port!' exclaimed Sergeant Dutt, dropping his bags gratefully as Gently paused to admire the scene.
'Of course it's a port,' said Gently, 'where do you think your breakfast bloater hails from, Dutt?'
'Yus, but I thought it was like Margate – not like flipping Pompey!'
Gently grinned. 'There's a Margate side to it too,' he said. 'Look, Dutt – a ship-chandler's. Have you ever seen a ship being chandled?'
'Can't say as how I have, sir, come to think of it.'
'You should,' said Gently, 'your education is lacking. It's the duty of every intelligent citizen to see a ship being chandled, at least once...'
They proceeded across the bridge and down into the sun-baked street leading along the quays. Ahead of them now was the Town Hall, a handsome red-brick building in a style that was purely Dutch. In fact, the whole thing might have been Dutch, thought Gently, there was a strong Continental atmosphere. Coming in, now, through all those marshes with their cattle and windmills and sails... And then again it was full of overtones which kept him in a strange frame of mind. He couldn't settle himself to the idea of being out on a case. It was having been here so long ago that upset him, perhaps, the having known the place as a child his mind was baulking and refusing to come to grips with what he was doing. It showed itself in his facetiousness, in the way he twitted Dutt.
But it was no good: he was here on business only. Nostalgic memories didn't mix successfully with homicide, and he just had to shake himself into an alert and receptive state of mind.
'There's a cafe over there,' he said to Dutt, 'let's drop in for a cup of tea before we check in.'
'I was just going to mention it, sir,' panted the sweating Dutt, 'only you seemed to be in such a hurry!'
Gently clicked his tongue. 'I'm not in any hurry,' he said. 'There's nobody as patient as corpses, Dutt, especially when they've come out in a rash of stab-wounds...'
Superintendent Symms of the Starmouth Borough Police paced his office with military stride, a tall, spare man with close grey hair and a little clipped moustache. Inspector Copping, his man of parts, was being strong and silent in a corner.
'And that's it, gentlemen,' said the super, in tones as clipped as his moustache, 'we know nothing – we can find out nothing. We've got a corpse, and absolutely nothing else. There were no clothes and hence no laundry marks. You've had the prints and they're not on record. We've checked the Missing Persons' list for months without getting a lead and we've shown a slide at all the cinemas in town with no better result. In fact, gentlemen, it's a sticky sort of business, and I feel I ought to apologize for calling you in at all. But you understand how I'm placed. There are people above me who pretend to believe in miracles.'
Gently nodded gravely. 'It's our principal business to carry the can.'
'And you were specifically asked for, Gently – after that Norchester case of yours Central Office means only one person around here.'
'It was one of my luckier cases,' agreed Gently modestly.
'So you see, it was out of my hands.' The super paused, both in stride and speech. He was genuinely grieved at having to pass on such a stinker.
'It's a job for the file,' put in Inspector Copping from his corner, 'there's just no angle to it. He might have been jettisoned from a ship, or dumped there, or dumped somewhere else and washed up there. He might even have been shoved out of an aircraft and finished up there. There's no end to the ways he might've come – I've put in hours thinking up new ones.'
Gently nodded a mandarin nod and stuffed a clumsy hand into his pocket. They had some peppermint creams in that cafe, and he had bought a whole pound.
'The body was even _discovered_ by a halfwit... so far as we can make out he chivvied it around trying to wake it up.'
Gently made sympathetic noises over a peppermint cream.
'And then this blasted Lubbock got the seconds on him and tried three methods of artificial respiration.'
'He's been reprimanded,' said the super grimly, 'there'll be no more of that sort of thing from Lubbock.'
'And all the beachcombers for miles jamming around... it was like Bertram Mills'.'
There was a silence, during which the only sound was a sugary chewing from Gently.
'So you see that calling you in is simply a face-saver,' went on the super, recommencing to stalk. 'The lads higher up know there's no chance, but the thing got too much publicity. They daren't just sit tight and let it fade away.'
Gently shuffled a foot. 'Well, as long as you aren't expecting too much...'
'We aren't.' Inspector Copping laughed with a little conscious bitterness.
Gently laid a peppermint cream on the super's desk and appeared to study it, as though seeking inspiration. 'This halfwit who found him...' he began vaguely.
'They call him Nits,' supplied Copping. 'He's cracked all right – ought to be in a home. Real name's Gibson. Lives with his mother in one of the Grids.'
'And you checked up on him?'
'Naturally.'
'He wouldn't have been carrying a knife of any sort?'
Inspector Copping hesitated a moment and then plucked something from his pocket and threw it down on the desk in front of Gently. It was a cheap one-bladed penknife, and its one blade was broken. Gently poked it with a stubby finger.
'Of course there's no connection...?'
'None,' rapped Inspector Copping.
Gently picked it up. 'I'd like to keep it for the moment, all the same...' He opened and closed the little blade with a naïve curiosity. 'Did you find out anything else about him?' he asked. 'Has he got any friends – does anybody employ him?'
Inspector Copping grunted. 'He isn't employable. He hangs around the beach and people give him money, that's all. He spends it in the cinemas and amusement arcades. Everybody knows him, but nobody wants anything to do with him.'
'Has he ever given any trouble?'
'A visitor made a fuss about him once and we pinched him for begging. It took three men to bring him in. He's stronger than he looks.'
Gently revolved the peppermint cream with care. 'About the deceased,' he said, 'when did he die?'
'The report says between eleven and twelve p.m. on Tuesday.'
'When did you find him?'
'Lubbock saw him at five-ten a.m. on Wednesday.'
'So he'd only taken five hours to get where he was... it isn't very long. What was the state of the tide?'
'Low slack water. If he came in on the tide he must have grounded at about four.'
'That cuts it down another hour...' Gently stared at his white sugar tablet with elevated brows. 'The local currents... the ones just off-shore... what's their direction?'
Copping glanced at his superior.
'There's nothing just off-shore,' supplied the super, 'it's a perfectly safe beach at all states of the tide. There's a north-south current further out, about half a mile. It accounts for a few damn fools every season.'
'Do you know the speed of it?'
'Not precisely. Maybe six or seven knots.'
'So you give him an hour to get into the current and another hour to come back ashore he might have been put in eighteen miles north.'
'No.' The super shook his head. 'If he was put in from the shore it couldn't be more than five or six. The shore starts in westward just north of the town, and six miles up the coast is Summerness, beyond which it recedes very sharply. At Summerness the current would be two miles off-shore.'
'Two miles...' mused Gently. 'He wouldn't drift out that far in the time. It'd have to be lower down. What's up there in that direction?'
The super shrugged. 'It's a wide sand beach all the way, backed with marram hills and freshwater marshes. There are three villages and a lot of bungalows. A little way out of town there's the racecourse.'
'Has there been racing lately?'
'No. It's not due till next Tuesday.'
'I suppose you didn't do any checking up there?'
'What's the use?' interrupted Inspector Copping. 'It's a hundred to one against him having been put in there, and even if he was, what would we be looking for?'
'Someone might have seen something,' suggested Gently mildly, 'there's never any harm in asking questions.'
Inspector Copping's heavyish features flushed. 'The case has had publicity,' he said, 'we've asked for information both in the cinemas and the press. If anyone knew anything we should have heard by now – we've looked into everything that's come our way.'
'Please don't get the impression that we've been asleep,' said the super snappily, 'we may not be homicide experts, but at least we carry out our police duties with strict care and attention. You have made the suggestion that the body of the deceased was put into the sea somewhere between here and Summerness, but the suggestion rests merely on the fact that there is a north-south current. And the current may have brought it from some point at sea, and then again it may never have been in the current at all. It could even have drifted up from a southerly direction inside the current.'
Gently hunched his shoulders chastenedly and made a chessmove with the peppermint cream. 'It could even have been dropped off the pier,' he murmured.
'My guess is it came off a ship,' said Copping. 'There's no doubt about the fellow being a foreigner. Anyone could see that at a glance. The ethnologist who saw him reckoned he was a Slav of some sort, Central European. He could have gone overboard in the Wash somewhere and hooked on to that current.'
'And that would mean trying to pinpoint a ship of some or any nationality which was in the Wash about midnight on Tuesday,' said the super, 'and just suppose we found it, what good would it do us?'
'It'd be outside our jurisdiction,' said Copping brightly.
'Unless it was a British ship,' hazarded Gently.
'In which case we would have heard something before now,' said the super with a note of finality. 'No, Gently. I appreciate your attitude. It's your business to see that no stone is unturned and I can see that you propose to carry it out. But I think you'll have to agree in the long run that everything that can be done has been done. Where there's no identity, no apparent motive and no hopeful line of inquiry, then to proceed with a case is simply a formality. You must do it – that's your business: but I'm afraid that in this instance it will be a very thankless task.'
'And yet this man was murdered,' said Gently slowly. 'Somewhere there's someone who will kill more readily another time if we don't put a finger on him...'
'I know, I know!' snapped the super, 'but idealism is no use if there's no prospect of implementing it.'
Gently sighed and heaved himself out of the rather bleak chair which was maintained for visitors. 'There's nothing else you want to tell me?' he inquired.
'I've told you everything that we know.' The super paused, frowning. Then he looked at Gently a little more kindly. 'Don't think we're against you... I assure you it isn't that. If you can do anything with this affair I shall be the first to congratulate you, and Copping here will be the second.'
'Hear, hear,' responded Copping, though perhaps more from duty than conviction.
'I've arranged lodgings for you and the Sergeant in Nelson Street. There's a private office here you can use for interrogations. If you need a car you have only to ask for it, and any other assistance we can give.' The super stalked round his desk and held out his hand. 'The best of luck, Gently,' he said warmly, 'I only wish it had been someone with no reputation to lose.'
Gently shook the extended hand woodenly. 'I'd like to see the body,' he said.
'I've a full set of photographs and a copy of the pathologist's report for you,' replied the super. 'Copping will give them to you along with his own report.'
'I still want to see the body,' said Gently.
The super shrugged. 'Very well, then. Copping will take you round.'
They filed out in strict order of rank, Gently, Copping and Dutt, the latter having been a silent and respectful auditor of the conference in the office.
'We'll take a car,' said Copping, 'it isn't far to the mortuary, but you can put your bags in and I'll drop you at your lodgings.' He dodged into his office and came out with a file. 'These are the reports and the photographs – for what they're worth.'
Gently took them with a solemn nod.
The mortuary was a neat modern building of pastel-tone brick and had double doors of a reddish wood with lavish chromium-plated fitments. But it smelled exactly like all other mortuaries. Copping explained their errand to the sad-faced attendant. They were ushered into the dim and odoriferous interior.
'He's had company,' observed the attendant, indicating a second draped form, 'they pulled her out of the river up by the yacht-station.'
'You'd better watch they don't get into mischief,' said Copping callously.
The attendant laughed a ghoulish laugh and twitched the sheet from corpse number one.
' _Voilà,_ ' said Copping, 'the cause of all the trouble.'
Gently stepped forward and conducted a stolid examination of the wax-like body. It had no humanity now. There was nothing about it to suggest the warmth of life, the kindling of a soul. And the attentions of the pathologist had done little to help matters, though he had tidied up afterwards with needle and gut.
Sergeant Dutt made a hissing sound. 'No doubt about him being a foreigner, sir,' he said, 'there's a bit of the old Eyetye about him, if you ask me.'
'Age?' demanded Gently through his teeth.
'Early forties is their guess,' returned Copping.
'Much force?'
'One stab busted a rib. There's three in the lung and one in the heart. Penetration about four inches. Double-edged blade about three-quarters of an inch wide. And his wrists had been tied.'
'Poor beggar!' exclaimed the warm-hearted Dutt, 'they never give him a chance.'
'And those?' jerked Gently, indicating a group of brownish marks just above the pathologist's neat stitches.
'Burns,' said Copping, 'that's what the report says.'
Sergeant Dutt caught his breath. 'I've seen burns like that before, sir... during the war when I was in France...'
'I know,' said Gently, 'I've seen them too.'
He turned away from the slab and stood looking at the narrow window with its bar and pebble-glass pane.
'They didn't just want his life, they wanted something else too. I wonder what it was... I wonder why it was so important?'
Copping laughed harshly. 'When you know that you'll have solved the case,' he said. 'Let's get out of here. The smell gets on my stomach. You've done me out of my tea, bringing me to this place just before I knock off.'
CHAPTER TWO
BODY ON THE BEACH: YARD CALLED IN, ran the headline of the evening paper, Chief Inspector Gently To Take Charge, New Move In Riddle Of The Sands. It continued: 'There were fresh developments today in the murder mystery which has come to be known as "The Body On The Beach Murder". The Starmouth Borough Police acknowledged the gravity with which they view the case by calling upon the services of Scotland Yard. Chief Inspector Gently, well known in Northshire for his handling of the Sawmill Murders, has been assigned the task and this afternoon he arrived in Starmouth to take over the investigation. Superintendent Symms told our reporter in an interview today that sensational developments in the near future are not expected and that the arrival of Chief Inspector Gently was purely a routine step.'
There was also a photograph of Gently which the _Norchester Evening News_ had kept in cold storage from his last visit, but fortunately it wasn't recognizable...
All along the Front they were talking about it, from the bowling greens in the north to the funfair in the south. It was really making the week for them, holidaymakers and residents alike. Publicity it was, Publicity with a capital P – it dragged in excursionists to be plucked and made the holidaymakers feel that their stay would be truly memorable. For how often does a first-class murder turn up on one's doorstep during a holiday? A classic murder with stabbing, mystery, the Yard, and all that? They even had the spot marked X for them, thought Gently, as he turned away from the crowd which still milled excitedly on the beach: the Starmouth Borough Police, nothing if not thorough, had set up a ponderous post to mark the site of the discovery. Nobody took it seriously, that was the trouble... the police had already written it off as unsolvable, and everybody else looked on it as a bigger and better side-show. Even Gently himself was being infected by the feeling. He had been practically tipped off that he didn't have to exert himself.
And yet it was still there, up in the mortuary. That shrunken husk of what had once been a man. A foreigner, they all said, as though it were something subhuman – a foreigner whom they couldn't really care about, though he had been tortured, killed in cold blood and thrown into the sea, to be washed up, troublesome and unwanted, on their holiday shore... just a foreigner: one didn't bother too much about him.
But suppose one did bother, thought Gently, where did one begin on such an impossible business? He had taken the only step that suggested itself. He had got Dutt to phone headquarters to have the prints transmitted to Paris. Where did one go from there – what was one to try that the so-efficient Starmouth BP hadn't tried already? He sighed, and sat down heavily in a deckchair which still remained on the evening sand. He was still baulking, and he knew it. He still couldn't get his shoulder under the thing. There was something about just being in Starmouth, quite apart from anything else, that sapped his power of concentration. Those tight-fitting knickerbockers, for instance... And where were the donkeys...?
Behind him the lights blazed and jewelled as far as the eye could see, outlining buildings, flashing on signs, revolving on the sails of the windmill which reared further down. The two piers presented a strong contrast. The virile Albion seemed to burn and throb with illumination, to assert itself by sheer candlepower; the Wellesley contented itself with graceful and glittering outlines, making it appear, with its Winter Gardens, like an iced-cake shored-up above the sea. And there was the great evening medley of the Front, the undertone of the traffic, the beat of ten thousand feet, the shrieks and cries of ragamuffin children, the tinkle and soughing roar of mechanical music and the intermittent spang and crash of a shooting saloon not far away.
And then, of course, there was the sea, the sea that knew the secret, the heavy-looking evening sea that hissed and chuckled near that solitary post.
Gently took out his pipe and lit it. There had to be something, he told himself obstinately. After all, that man must necessarily have been murdered not very far away and murder under the best conditions is apt to leave traces. He blew out the match in a gust of smoke and held it poised in the air beside him. Except if it were done at sea, of course... but one mustn't begin by assuming that.
Suddenly, the match disappeared from his fingers. It went so quickly and so silently that for a moment Gently simply sat still in surprise. Then he jerked his head round to see by what agency the match had taken flight. But there was nothing to be seen. There was nobody within yards of the back of his chair. The nearest people to him were two uniformed Americans with their inamoratas and they were patently occupied with quite other things. Puzzled, he returned to his meditations. He puffed at his pipe, his empty fingers taking up the same position as before. And then, just as suddenly, with the lightest of twitches, the match reappeared in its former situation.
This time Gently got up. He got up with an alacrity unexpected in a bulky man of fifty summers. But his haste was quite needless, because the worker of these miracles was merely crouching behind the chair and it made no attempt at flight when Gently pulled away the chair and exposed it.
'And who may you be?' demanded Gently, realizing then whom it could be no other.
'I'm Nits – I'm Nits!' piped the halfwit, staring up painfully with his bulbous eyes. 'I know who you are, they told me who you are! I know – I know!'
Gently released the chair slowly and reseated himself, this time with his back to the sea. 'So you do, do you?' he said, 'and who did they tell you I am?'
'You're a policeman!' chattered Nits, 'you're a policeman, though you haven't got a hat. I know! They told me! You want to know about my man who wouldn't wake up.'
Gently nodded profoundly, keeping his eyes fixed on the halfwit's. 'And what else did they tell you about me?' he queried.
'They said I mustn't talk to you – ha, ha! – they said you might take me away and lock me up. But' – Nits assumed an expression of exaggerated cunning – 'I know you won't do that.'
'And how do you know I won't do that, Nits?'
'Because I haven't asked for any money. That's why they locked me up!'
Gently puffed at his pipe, still keeping the staring green eyes engaged. This was it, the solitary link – an idiot who ought to be in a home. Not even a rational creature, however stupid. Just an idiot, someone who couldn't testify anyway. As he sat there, smoking and brooding and watching the ragged Caliban crouched in the sand, he seemed to hear a mockery in the tinkled outburst of music and a laughter in the shuffling of feet on the promenade. What was the use of it? And who cared two hoots, really?
'So you found the man who wouldn't wake up...' he murmured.
Nits nodded in energetic glee.
'Just there, where they've put the post.'
Nits's head bobbed ceaselessly.
'And you tried to wake him up... then you went and told a policeman... and the policeman tried waking him up too.'
The head never wavered.
'You've no idea how he got there?'
The head changed direction agreeably.
'You didn't see anybody around before you found him?'
'My part,' said Nits, his features twisting into an absurd mask of aggression, 'nobody come on my part of the beach.'
It was just what was in Copping's report. The efficient Inspector had covered the ground admirably. Nits had told what he knew, and he didn't know anything: it just so happened that the corpse had been washed up on 'his' part of the beach.
'You were told not to talk to me,' said Gently wearily, 'who was it told you that?'
Nits grinned and chattered but made no intelligible reply.
'And why _did_ you talk to me, after being warned not to?'
The halfwit frowned ferociously and turned his trouser-pockets inside-out. 'The other man – he took it away!'
'Took _what_ away?'
'My knife – he took it!'
Gently smiled and felt for the little broken-bladed penknife. Nits gabbled with joy and snatched for it with the speed of a striking snake. But Gently had already experienced a sample of the halfwit's snatching and he held the knife carefully out of range.
'Who was it told you not to talk to me?' he demanded.
Nits chattered and tried another sudden grab.
'You get it when you tell me, not until.'
Nits made all sorts of fierce faces, but Gently merely made as if to return the knife to his own pocket.
'Jeff!' piped the halfwit suddenly, 'it was Jeff and Bonce – they told me.'
'And who are they?'
'I don't know – I don't know!'
'You know their names – you must know something else about them.'
'I just see them, that's all.'
'See them _where_?' persisted Gently, 'see them here – on the beach?'
But the halfwit relapsed into a mewing and gabbling, and refused to make himself any further intelligible. Gently sighed and tossed him the knife. It was plucked out of the air as though by the lash of a whip and Nits capered off, clutching it to his bosom, his two trouser-pockets still turned inside-out.
'Whoa – wait a minute!' called Gently, rising to his feet.
He produced a florin and held it out between thumb and finger. The halfwit paused in his flight, hesitated, and then came sidling back, spaniel-like, his chin tucked in until there seemed nothing of his face below the two bulging eyes. He didn't snatch at it, as Gently expected: he reached up and took the coin quietly from Gently's hand. Then he crept closer still, crouching, cringing almost, and stared up with his faceless eyes.
'The man who wouldn't wake up!' he piped, but in a sort of whisper.
Gently nodded silently.
'Different... different!'
'Different from... what?' murmured Gently.
'From when he was awake.'
_'From when he was awake!'_
Nits went into one of his fits of nodding.
'Hold it!' exclaimed Gently, feeling his universe beginning to rock, 'did you know him, Nits – did you know him when he was awake?'
'I knew him – I knew him!'
'But when did you know him – and where?'
Nits screwed his face up into an expression of rage and shook his head. Then he pointed to the tip of his almost non-existent chin.
'Hair!' he chattered, 'hair – when he was awake!'
The next moment he was capering over the beach again, leaving Gently with his eyebrows hoisted in almost comical surprise.
Twilight had become dusk and the lights which had sparkled like fugitive jewellery were now glowing and full. The blazing Front had a strange glamour about it, as though it belonged to a different world, and the holidaymakers too seemed to partake of the strangeness. Perhaps it was simply the multiplicity of lights destroying the shadows, perhaps only the sense of anonymity and freedom... they felt changed and in some way abnormal.
Gently picked his way through the promenade crowds and paused at the edge of the carriage-way. He felt changed also, though his changedness was due to something quite different. He'd got a lead, that was it. He'd found something to hang on to in this slippery orphan of a case.
Almost jauntily he crossed the carriage-way and directed his steps to a phone-box on the other side.
'Chief Inspector Gently... is Inspector Copping in, by any chance?'
The switchboard girl thought he might be if Gently would kindly hang on. Gently grunted and wedged himself into a supportable position in the corner of the box. Outside he could see the front of the amusement arcade from which blared much of the canned music which disturbed that part of the promenade – a striking blaze of light in the shape of three feathers, with a lurid red arrow snapping backwards and forwards as though working up to burst in through the door. And there was some jutting neonry which said LICENSED BAR... a ritzy sort of touch for an amusement arcade, thought Gently.
'Inspector Copping,' said the switchboard girl.
Gently jammed the door yet tighter-shut on the racket without. 'Gently here... I want something done,' he said. 'Look, Copping, can you get on to the pathologist who did the post-mortem? I want him to have another check.'
'Can't see what that's going to buy,' came Copping's voice plaintively, 'he didn't die of asthma.'
'I'm not interested in the way he died. I want a thorough examination of the skin of the face for spirit gum.'
'Spirit gum!'
'Or any other mucilage that may be present,' added Gently generously.
There was a pause at the other end, and then Copping came back: 'But what's he supposed to be now – a member of a touring company?'
Gently smiled at the leaping red arrow. 'Your guess is as good as mine...'
'And where did you dig up the idea, anyway?'
'Oh... it was a present for a good Central Office man. And by the way, Copping, you wouldn't know anything of two characters called Jeff and Bonce, would you?'
'Not to my knowledge. Are they hooked up to this business?'
'Could be,' admitted Gently, 'it's an even chance...'
There was a noise like a snort at the other end. 'But how do you _do_ it? I've been three days on this case!'
'Just luck, you know... you need it in homicide.'
'It looks like all the breaks were being saved up till you came. Are there any other small ways I can help?'
Gently brooded a moment. 'There's an amusement arcade down here... it's called "The Feathers" and it sports a licensed bar. What do you know about that?'
'Is that in the case too, or are you just being curious?'
'I've been tailing Nits... when he's finished collecting he makes for "The Feathers" like a homing pigeon.'
'Well, it's got a clean record. The proprietor is a man called Hooker – Louey Hooker. He lives in a flat at the back of the building, and he runs a bookie's business too. The office is under the flat and fronts on Botolph Street, which runs parallel with the Front.'
'A bookie's business.'
'That's right. They're still legal in this year of grace.'
Gently nodded at the undiscourageable arrow. 'Well... send me Dutt along, will you? And drag that pathologist away from whatever he's doing and put him to work.'
'You mean tonight?' inquired Copping in surprise.
'We're working, aren't we?' retorted Gently heartlessly.
He hung up and levered open the door of the phone-box. The year's hit-tune, mildly interruptive till then, leaped to meet him with a vengeful roar. Gently frowned and felt in his pocket for a peppermint cream. Mr Edison, he felt, hadn't been an unmixed blessing to mankind.
The interior of the amusement arcade was as aggressive as the exterior had promised. It was lit with a farrago of fluorescent tubes and popping bulbs, and the walls were panelled in a gooey pink plastic relieved by insets of 'teinte de boiled cabbage'. And there was a vigorous use of chromium plate in all departments. The décor man had obviously had a flair for it. Left and right of a central aisle the machines were deployed – all the latest attractions, space-flights, atom-bombing and the rest, with a few tried favourites still making a stand against the march of science. There was the crane that picked up a prize and dropped it down a shoot, Gently noticed... at least, it picked up a prize when Nits was operating it. The halfwit had apparently got the low-down.
Stationed behind a punchball machine, Gently watched the crouched, ragged figure insert coin after coin. Each time the descending grab would seize on one of the more substantial pieces of trash in the glass case. Sometimes it failed to grasp securely and nothing would rattle down the shoot except a few gaudy-coloured sweets, but always the grab dropped plumb on a prize in the first instance.
Gently lit his pipe and continued to watch. All round him machines were ringing and clattering. Any two of that crowd could be the two in question... at any moment they might spot Nits, or Nits them. And what then? he asked himself. Suppose he was lucky and stumbled on them? What they had said to Nits might have been no more than a joke, the sort of silly thing to be said to a halfwit. Of course it was odd that they had known him, Gently, on sight... but then, the picture in the evening paper might have jogged their memories. There had been bigger and better pictures of him in the same paper the year before.
No, he told himself, it wasn't much better than a hunch, after all...
The music changed to something plaintive and caressing, and as though it were a signal Nits crammed his collection of ballpoints and flash jewellery into his pockets and darted to the door. Gently moved forward also, but the halfwit came to a standstill short of the entrance, so he slid back again into the cover of the punch-ball machine. Was Nits expecting someone? It rather seemed like it. He stood by the door, apparently trembling, and strained his protruding eyes in the direction of the Wellesley Pier. Several people came in, but these were ignored. Nits didn't even glance at them as they pushed past. Then he gave a little whimper and a skip, like a dog sighting its master, and a moment later the object of his vigil appeared.
She was a blonde, a tall, big-bodied blonde. She didn't have to broadcast her vocation, either to Gently or the world. She wore a sleeveless green silk blouse, high-heels and a black hobble skirt, and walked with a flaunt that looked vaguely expensive.
'Geddart,' she said to Nits in the husky voice of sin, 'keep away from me, you dirty liddle so-and-so – how many more times must I tell you?'
'I've been a good boy!' piped Nits, frisking and cringing beside her as she hipped down the arcade.
'I don'd care – jusd keep away from me.'
'I got something – I got something! Look for you!' Nits pulled out his hoard of swag and tried to thrust it into the blonde's hands, but she snatched them away and the stuff tinkled on to the floor.
'I don't wand it!' she yelped, 'keep your dirdy muck to yourself! Don'd ever come near me again!' And she bustled away through the grinning crowd, leaving Nits to scrabble amongst the feet for his scattered treasure.
'That's my gal, Frenchy!' shouted someone, 'don't you have him if you don't fancy him!'
The blonde turned back and said something so filthy that even Gently was taken aback, then she swaggered through the swing-doors of the bar.
'Whoo- _whoo_!' was the cry, 'Frenchy's got the answer, don't you forget it!'
On the floor Nits chattered and sobbed with rage. 'I'll kill you – I'll kill you!' he babbled, 'I've been a good boy – I have – I have!'
Gently stooped and rescued a plastic ballpoint from under the heel of a bystander. 'Here,' he said, 'one you missed.'
Nits seized it and stuffed it into his pocket after the others. 'I'll kill you!' he whispered in an ecstasy of passion.
'Did she know him?' asked Gently, 'did she know the man who wouldn't wake up?'
Nits's green eyes burned at him like two malignant lamps and Gently, moving swiftly, moved only just in time. As it was the leaping halfwit sent his trilby flying. Then, recovering himself, Nits dived for the door and his turn of speed was something that Gently could only have sighed for in his palmiest days...
CHAPTER THREE
THE BAR WAS rather a contrast to the rest of the establishment. It had got missed out when the wielder of plastic and chromium-plate had gone his merry way. It was quite a large place and its dim, parchmented lighting made it seem larger still. It was also irregular in shape. There were corners of it that tucked away, and other corners which had been given an inglenook treatment. Opposite the swing-doors ran the bar counter, its supporting shelves well fledged with opulent looking bottles, and to the left of the counter was a door marked 'Private'. Further left again was a small exit door, leading probably into a side-street.
Gently eased himself through the swing-doors and stood still for a moment, adjusting his vision to the drop in candle-power. It seemed a fairly well-patronized place. Most of the tables and nookeries were occupied, and there were several customers perched on high stools at the counter. Also it seemed quiet in there, but that may have been due only to comparison with the racket going on outside the swing-doors.
He strolled across to the counter, where the blonde was taking charge of a noggin of straight gin.
'Chalk id up, Artie,' she crooned, 'and no chiselling, mind.'
'Who shall I chalk it up to?' asked the ferrety bartender with a wink.
'Don'd be cheeky, Artie – Louey don'd like it!'
She slunk away from the counter, and her eye fell on Gently for the first time. She recognized him, he knew – there was just that much of alert interrogation in her glance – and for a moment he thought she would say something. Then she shrugged a scantily-clad shoulder, gave her head a little toss, and swung away across the room to one of the nookeries.
Gently seated himself on a high stool and ordered an orange-squash.
'Who is she, Artie?' he asked the ferrety bartender.
Artie gave the squash-bottle a practised twist. 'Don't ask me – ask her,' he retorted sullenly.
'But I am asking you. What's her name?'
'It's Frenchy – and I'm not her boyfriend.'
'Her other name, Artie.'
'I'm telling you I don't know!'
'She mentioned a Louey...'
'What's that got to do with me?'
'She spoke as though you knew him...'
'Well, I don't. He must be someone new.'
Gently drank a mouthful of orange-squash and appeared to be losing himself in contemplation of the fruit-scum collected at the mouth of his glass.
'That'll be a bob,' said Artie, ' _if_ you don't mind.'
Gently drank some more and was still interested in the fruit-scum. 'You know, it's amazing,' he said casually, 'the number of people round here who know me without me knowing them... you seem to be the fifth, Artie, by my computation.'
The ferrety one stiffened. 'Don't know what you mean by that...'
'Never mind, never mind,' said Gently soothingly, 'we'll go into it some other time, shall we?' He slid off his stool and picked up the part-drunk glass of orange-squash.
'Hey!' clamoured Artie, 'that's still got to be paid for...!'
'Chalk it up,' returned Gently, 'and no chiselling, mind. Louey don'd like it...'
He ambled over to a small table by the wall and pulled up a seat with better padding than the high stool. There were other eyes on him besides Artie's; several customers at the counter had heard the conversation, and now turned to watch the bulky figure cramming itself into its chair. Not only at the counter either... out of the corner of his eye Gently could see Frenchy in her nookery, and two other figures near her. They were all giving him their attention...
''Ere!' whispered a sporty-looking individual to Artie, 'is that geezer a busy?'
'Yard,' clipped Artie from the corner of his mouth.
The sporty-looking type favoured Gently with a bloodshot leer. 'Nice bleedin' company we get here these days...'
Gently quaffed on imperturbably. He might have been entirely alone in the bar, so oblivious did he seem. He took out his pipe and emptied it with care into the ashtray; then he took out his tobacco and stuffed the bowl with equal care.
''E's set in for the night,' said the sporty-looking individual, 'blimey, you'll have to look sharp with them shutters at closing-time...'
'Why don't you offer him a light?' quipped his neighbour.
'What, _me_ – and him a busy? Give us another nip, Artie... there's a smell round here I don't like...'
Gently, however, lit his own pipe, and having lit it he entertained his audience with a scintillating display of smoke-rings. He could blow them single, double and treble, with combinations and variations. He had infinite patience, too. If one of his airy designs went wrong he had all the time in the world to try it out again...
The private door beside the bar opened and a man in seedy evening-dress appeared. He was a heavily built type of about forty with dark hair, a parrot-shaped face, and little pale eyes set very close together, and he smoked a cigarette in a gold-plated holder about as long as his arm. Gently surveyed him with mild interest through a pyramid of smoke. Faces of that shape must at all times be rarities, he thought.
'Oi – Peachey!' yipped the sporty-looking individual, and made a cautionary face while he thumbed over his shoulder in Gently's direction. Artie also hastened to breathe a word in the newcomer's ear. The man's two pale eyes reached Gently, paused and strayed uneasily away again. Gently's own slipped round to Frenchy. She was sitting up straight and shaking her peroxide head.
'Louey wants a fresh bottle,' said the newcomer hoarsely, 'gimme a white-label.'
Artie produced one from under the counter and handed it to him. He dived clumsily back through the door. Artie returned to his business of serving drinks without a further glance at Gently; there was an expression of satisfied malice on his face...
'You loog lonely for a big man,' said a voice at Gently's elbow, and he turned his head to see that Frenchy had slunk over to his table. She was smiling, at least with her mouth. Higher up it didn't show so much – by the time one got to her rather pretty warm-brown eyes it had gone completely. But she was smiling with her mouth.
Gently smiled too, somewhere between the South Lightship and Scurby Sands.
'I'm not lonely,' he said, 'there's too many people around who know me.'
Frenchy laughed, a throaty little gurgle. 'Thad's because the big man is famous... he geds his picture in the paper.'
'You think that makes people notice? Such a bad picture?'
'But of course... nobody talks about anything else except whad the police are doing.'
She pulled up another chair and sat down, not opposite Gently but to the side, where the table didn't hide anything. She slid forward and crossed her legs. They weren't terribly attractive, he noticed. The skin was a trifle coarse and the contours inclined to be knobbly – they were designed for strength rather than quality. But she managed them well, they were crossed with great competence. And the hobble skirt contrived to lose itself somewhere above the knee.
'Id musd be exciding,' she crooned, 'hunding down a murderer...'
Gently breathed an unambitious little smoke-ring.
'And difficuld too... especially one like this.'
Gently breathed two more, one exactly inside the other.
'I mean,' she continued, 'where does one begin to loog if one doesn'd know his name...?'
'What's _your_ name?' inquired Gently suddenly; 'all they call you round here is Frenchy.'
The brown eyes opened wide and the smile tailed off: but it was back again in a moment, and wider than ever.
'Surely you don'd suspecd me, Inspecdor...'
'I'm just asking your name.'
'Bud why should you wand to know thad...?'
'I'm curious, like all policemen.'
Frenchy seemed to consider the matter between half-closed lids. Gently stared at the table and smoked a few more puffs.
'If you wand to ask questions...' she began.
Gently favoured her with a glance.
'There are bedder places than this to ask them...'
She leaned forward over the table and balanced her chin in the palm of her hand. In effect the green silk blouse became an open peep-show.
'Afder all, it's your dudy,' she melted, 'and you know when girls dalk the besd...'
Gently sighed and felt in his pocket for a match. 'You're not local,' he said, 'you've had West End training... who brought you down here?'
For a moment he thought her scarlet nails were going to leap at his face. They angled for a strike, and the brown eyes burned with the merciless ferocity of a cat's. Then the fingers relaxed and the eyes narrowed.
'You filthy b— cop!' she hissed, all accent spent, 'I wouldn't let you touch me if you were the last bloody screw on God's earth, and that's the stinking truth!'
Gently shrugged and struck himself a fresh light. 'Where do you live?' he asked.
'Bloody well find out!'
'Tut, tut, my dear... it would save unnecessary police-work if you told me.'
'Well I'm not going to...!'
Gently held up a restraining hand. 'It doesn't really matter... now about our friend with the beard.'
She stopped in mid-flow, though whether on account of his casual remark or not Gently wasn't able to decide.
'Where did you meet him – here or in London?'
'Who?' she demanded sullenly.
'The deceased – the man who was stabbed.'
'Me!' she burst out, 'what have I got to do with it?'
'I don't know,' murmured Gently, 'I thought perhaps that was what you came across to tell me...'
Frenchy riposted with a stream of adjectives that fairly blistered the woodwork.
'Still, you might like to tell me about your movements on Tuesday night...' added Gently thoughtfully.
There was a pause, pregnant but not silent – silence was a strictly comparative term when only a pair of swing-doors separated them from the uproar without – and Gently occupied it usefully by prodding around in his pipe, which wasn't on its best behaviour. Over at the counter, he noticed, they were straining their ears to catch a word of what was taking place. And in Frenchy's nook two figures in the shadows leaned intently in his direction...
'You can't drag me into this, and you bloody well know it!' seethed Frenchy, with the aid of two other words. 'I never knew him – I didn't do nothing – I don't know nothing!'
Gently tapped his refractory pipe in the ashtray and drew on it tentatively.
'It's true!' she spat, 'why do you pick on me – who's been lying about me?'
'Who _might_ lie about you?' inquired Gently absently.
'How should I know? – anyone! A girl's got enemies. And I've got a right to know, haven't I? If someone's been making accusations—!'
'Nobody has accused anybody... yet.'
'Then what's it all about?'
Gently shrugged and forked about in his pipe again. 'If you're so far in the clear you shouldn't be afraid to tell me what you were doing on Tuesday night...'
'It's got nothing to do with it – I can't tell you plainer than that, can I?'
'Of course, if it's something you'd rather not officially acknowledge...'
Again the scarlet nails flexed and a flicker went over the brown eyes. But once more Frenchy controlled herself.
'I haven't got to tell you, flattie... you've got nothing on me!'
Gently nodded and turned out the fragments from his pipe. 'Between nine, say... and midnight...'
'All right, you bleeding copper!' Frenchy jumped to her feet and raised her voice to a scream. 'So he wants to know... he wants to know what I was doing on Tuesday night when someone was doing-in the bloke they found on the beach... I'm a naughty girl, and of course he picks on me!'
'That's right!' bawled the sporty-looking individual, sliding off his stool, 'you tell him, Frenchy, you tell him where to get off!'
'He doesn't know anything... he's just picking on me... maybe he's after something else too, the dirty so-and-so!'
'He wouldn't be the first, either!'
'And now he's looking for a chance to run me in... that's what it is...'
'Shame!' welled up from all over the bar.
'He comes from tarn just to pinch our Frenchy!' yapped the sporty-looking individual.
'They're a dirty lot... there isn't a man I'd call one amongst them... they're sent down here to find a murderer and all they can do is make trouble for girls like me.'
'It's all they're good for, chasing-up women!'
Gently looked up mildly from the refilling of his pipe. 'We don't seem to be getting very far with what you were doing on Tuesday night...' he murmured.
Frenchy rocked on her heels, fuming at him. 'I'll _tell_ you!' she screamed. 'I'll tell everybody, and they can bear me out. I was right here, that's where I was. I didn't shift an inch from this bar, and God help me!'
'It's the truth!' barked the sporty-looking individual, coming up, 'we saw her here, didn't we, boys?'
There was a unanimous chorus of assent.
'And after half past ten?' proceeded Gently.
'I was outside playing with the machines.'
'And after that?'
'Christ, can't a girl have any private life these days?'
'What was his name?' asked Gently amid laughter and jeering.
'Jeff!' shouted Frenchy, 'come and shake hands with a chief inspector.'
Gently glanced sharply at Frenchy's nook, where one of the two shadowy figures was getting reluctantly to his feet. He was a tall, well-made youth of sixteen or seventeen, not unhandsome of feature but with a weak, wide, thin-lipped mouth. He wore a Teddy boy ensemble of all one colour – plum red. It began with his bow tie and collar, descended through a straight-cut narrow-sleeved jacket and reached the ground via drain-pipe trousers and spats – a red of the ripest and fruitiest.
Gently eyed this vision curiously. It hovered uncertainly at some little distance.
'It was him?' inquired Gently, a shade of incredulity in his tone.
'Of course it was bloody well him... they all have to make a start, don't they?'
Gently beckoned to Jeff. 'Don't be frightened,' he said, 'it wasn't indictable...'
The Teddy boy came forward, flushing.
'Can you confirm what this woman says about Tuesday night?'
'I suppose so.'
'Would you care to describe it... I mean, the relevant parts?'
'There isn't anything to describe!' scowled Frenchy, 'he met me in the bar, that's all.'
Gently glanced at Jeff interrogatively.
'That's right... in the bar,' he said.
'And then?'
'And then we went to her... place.'
'Where is that?'
'It's a flat in Dulford Street.'
'And you spent the night?'
'I... actually... you see...'
'Of course he didn't!' Frenchy broke in, 'did you think I wanted his old man on my barrow? I turned him out at half past twelve... he'd done enough by then, anyway!'
There was a roar of laughter.
'And who is his old man?' inquired Gently smoothly.
'He's Wylie of Wylie-Marine.'
'You mean that big factory on the quays near the station...?'
'That's right, copper,' Frenchy sneered, 'you're good, aren't you?'
Gently drew a few slow puffs from his newly-filled pipe. Most of the occupants of the bar seemed to have drawn closer to a centre of such absorbing interest. But the second figure in Frenchy's nook wasn't joining in the general enthusiasm. On the contrary, he had shrunk back almost out of sight.
'And Bonce?' inquired Gently, inclining his head towards the nook.
'Bonce?' queried the Teddy boy. He had stuck his hands in his jacket pockets and seemed to be screwing himself up to an air of toughness.
'If you're Jeff, I take it that your shy friend is Bonce. What was he doing while you were getting off with Frenchy here?'
'Bonce!' shouted Frenchy, 'stop hiding yourself... the big noise is on to you too.'
All eyes turned towards the nook, where there was an uneasy stirring. Then there ventured forth a second version of the plum ensemble, shorter, clumsier and even more youthful looking than its predecessor. Bonce was no beauty. He had carroty hair, round cheeks, a snub nose and an inherent awkwardness. But he was sartorially correct. His outfit matched Jeff's down to the tie of the shoes.
'And what's _your_ name when you're at home?' queried Gently.
Bonce licked his lips and stared agonizedly. 'B-Baines, sir,' he brought out, 'Robert B-Baines.' He spoke with a Starmouth accent.
'And where do you live?'
'S-seventeen Kittle Witches Grid, sir.'
'Well, Baines, you've heard the account of Tuesday night your friend has given... I take it that you can endorse it?'
'Oh yes, sir!'
'You came here with him, in fact?'
'Yes, sir!'
'And you were with him until he departed with this woman here?'
'Yes, sir!'
'All the time?'
'Yes, sir!'
'Even when he was ingratiating himself with her?'
Bonce stared at him round-eyed.
'When he was getting off, I mean?'
'Oh yes, sir!... I mean... no, sir...'
'Well... which is it to be?'
'I... I...!' stuttered Bonce, completely floored.
'And when they had gone,' pursued Gently affably, 'what did you do then... when you were left on your own?'
'Don't you tell him!' screamed Frenchy before Bonce could flounder into a reply, 'it's all a have – you don't need to tell him nothing.'
'No, we haven't done anything,' blurted Jeff, trying to swagger, 'you keep quiet, Bonce.'
'He just comes in here trying to stir something up, trying to get people to say something he can pinch them for... that's how they work, the bleedin' Yard! I—!'
'CLOSING TIME!!!' roared a stentorian voice, a voice which drowned Frenchy, drowned the jazz and rattled empty glasses on some of the tables.
Every head spun round as though jerked by a string. It was as though a bomb had exploded over by the counter.
'FINISH YOUR DRINKS!!!' continued the voice, 'IT'S HALF PAST TEN!!!'
Gently peered round Frenchy's shapely form, which was hiding the owner of the voice from his view.
'DRINK UP, LADIES AND GENTS. YOU WOULDN'T WANT ME TO LOSE MY LICENCE!!!'
He was an enormous man, not so much in height, though he topped six feet, but enormous in sheer, Herculean bulk. His head was bald and seemed to rise to a point. His features were coarse and heavy, but powerful. There was a fleck in the pupil of one of his grey eyes and he had, clearly visible because of the sag of his lip, a gold tooth of proportions to match the rest of his person.
'BREAK IT UP NOW, LADIES AND GENTS. YOU CAN STILL AMUSE YOURSELVES WITH THE MACHINES!!!'
About fifty, thought Gently, and still in good fighting trim.
The owner of the voice moved ponderously across to Gently's table. He glowered at Frenchy and nodded towards the door.
'Get out!' he rumbled, 'you know I don't encourage your sort.'
Frenchy glared back defiantly for a moment, but she waggled off all the same; her parting shot was at Gently, not the gold-toothed one. It was unprintable.
'GET OUT!!!' detonated the big man, and Frenchy got.
His next target was Bonce.
' _How_ old do you say you are?'
'Eight-eighteen!' burbled Bonce.
'When was that – next Easter? Don't let me find you in this bar again.'
'B-But Louey, you never said anything before!'
'GET OUT!!!'
Bonce faded like a cock-crowed ghost.
Louey sighed draughtily. He picked up Gently's empty orange-squash glass and gave it his sad attention. Gently looked also. The hands that held the glass were like two hairy grappling-irons. On one of his crooked fingers Louey wore an out-size solitaire, on another a plain gold ring engraved with a bisected circle.
''Night, Louey,' leered the sporty-looking individual, passing by on his way to the door, 'watch your company – it ain't so healthy as it might be!'
Louey rumbled ominously and set down the glass again. 'Can't help it,' he said, turning apologetically to Gently, 'this time of the year you're bound to get some riff-raff... the best you can do is to keep kicking it out.'
Gently nodded sympathetically. He found Louey's gold tooth fascinating.
'There's girls like Frenchy... we know some of them, but there's fresh ones come up every summer. If they don't solicit you can't make too much of a fuss.'
Louey permitted himself a searching glance at Gently.
'And those kids... I suppose it's asking for trouble to have an arcade next to a bar.'
Gently rose to his feet and felt in his pocket for a coin.
'Here,' he said, 'I haven't paid for my drink.'
'Oh, never mind that!' Louey laughed comfortably, easily, as though he felt Gently to be an equal. 'Only too pleased to see you in here, Inspector... sorry if anything happened that shouldn't have done...'
'You needn't worry about that – it was nothing to do with you.' Gently paused and looked into Louey's deep-set eyes. They wore a deferential smile, but because of the fleck breaking into one of them the smile had a strangely hard quality, almost a sinisterness.
'There's only one thing bothers me,' mused Gently, picking up his shilling and re-pocketing it.
'And what is that, Inspector?'
'The way everyone around here knows me on sight... you, Mr Hooker, amongst the others.'
There was a rowdiness now along the promenade. There were drunks and near-drunks, quarrelsome and loutish roisterers. Alcohol had been added to the heady mixture of humanity about its annual purgation... the beer had begun to sing, and the whisky to argue. And they were largely youngsters, Gently noticed, it was the teenagers who did the shouting and singing. Banded together in threes and fours they swaggered about the Front, stupid with Dutch courage: lords of a pint, princelings of Red Biddy. Did nobody spank their children these days?
A burly figure shouldered across the carriage-way and joined him on the pavement.
'Have any luck, Dutt?' inquired Gently with interest.
'Yes, sir, I did, as a matter of fact.'
'Well, go on... don't spoil a good story.'
'I stood where you told me, sir, and kept an eye on the bookie's joint at the back. There wasn't no lights on there, but about quarter of an hour after you went in again the door opens and out hops a bloke in a dark suit.'
'Oh, he did, did he? I suppose he wasn't a freakish-looking cove with a parroty face?'
'No, sir, not this one. I got a good look at him under a street-lamp. He was about middling-size, dark hair, sort of slanty-eyed, and he'd got a long, straight conk. And there was a scar of some sort on his right cheek – knife or razor, I should say, sir.'
'Hmm,' mused Gently, 'interesting. And did you tail him?'
'Yes, sir – at least, I stuck to him all along the prom going south. But then he goes into the funfair and there was such a ruddy crowd there I didn't stand a chance. So after a bit I gives it up.'
'Ah well... we do our best,' sighed Gently.
'Do you think there's a hook-up there, sir – have we got something definite?'
Gently shook his head sombrely. 'I don't know, Dutt, and that's the truth. There's some racket goes on there, I'm pretty sure, but whether it connects with ours is beyond me for the moment. Anyway, I threw a scare into them... I'll tip off Copping to keep an eye lifting.'
'The bloke I was tailing looked a right sort,' said Dutt sagely.
'There's a lot of right sorts in there, Dutt,' agreed his senior, 'they'd keep the average policeman happy for weeks.'
CHAPTER FOUR
GENTLY WAS DREAMING what seemed to be a circular dream. It began at the stab wounds in the man who wouldn't wake up, took in all the principal characters at 'The Feathers' and wound up again with that stabbed torso. And it continued like that for round after round. Or was it all going on simultaneously? His dream-self found time to wonder this. There seemed to be two of him in the dream: he was both actor and producer. First (if there was a first), came the chest of the corpse, caught in a sort of golden glow, and he noticed with surprise that, although the stab-wounds were present, the pathologist's carvings were not. Next, his dream-camera lifted to take in Jeff, or rather the top part of Jeff: the rest of him dissolved into the haze which surrounded the corpse. He was shrugging his shoulders and saying something. Gently didn't know what it was he was saying, but he was acutely aware of the implication. Jeff _wasn't_ responsible. He might have done it, of course, that was beside the point. But he wasn't responsible. You couldn't possibly blame him.
As though to make it more emphatic the camera shifted to Bonce, who was blubbing and stuttering his innocence in the background. They couldn't help it. Gently fully agreed. They had done it at the behest of some irrevocable Fate, which was curious but in no way blameable. It was just how things were... And then Bonce shrank and his blubbing mouth disappeared. He had become Nits, and Nits had become nothing but two protruding green eyes, painfully straining. Gently knew what _he_ was saying. The halfwit's words piped clearly in his brain. 'I've been _a good_ boy,' they echoed, 'I've been _a good_ boy,' and Gently tried to ruffle his hair good-naturedly, but the head sank away under his hand...
Then it was Frenchy's rather knobbly knees trying hard to make themselves look attractive: the camera wouldn't lift to her face, it just kept focussed on those unfortunate knees. We aren't bad, they seemed to be pleading (and Gently heard a twang of Frenchy's croon, though there weren't any words): you'll see a lot worse than us on the beach. Of course, you've got to make allowances, but it's the same with everyone... honestly now, we aren't bad at all... you must admit it. And Gently admitted it. What was the use of struggling? He'd been round before and knew the rules of the game...
So the camera faded across to the parrot-faced man and Artie. They'd got a lot of empty bottles, squash-bottles, and Gently only had to see the bottles to know that he was the one who had emptied them. Not that they were being nasty about it, those two. On the contrary, they seemed to be almost sympathetic, in a sad sort of way. Gently had blotted his copy-book. He'd drunk through all those bottles of squash without paying for them. They knew he couldn't help it, but all the same... a man of his reputation... Gently felt in his pocket for some money. They shook their heads. It wasn't just paying for it that counted. It was the fact that he'd _done_ it at all...
And now Louey's gold tooth filled all the screen, a huge, glowing tooth with (and it seemed so natural that Gently realized he was expecting it) a glittering solitaire diamond set in the top and a bisected circle engraved underneath it. It doesn't matter, the tooth was saying, the inspector can do what he likes, he's always welcome. It's not the same with the inspector, he's an old friend of mine. Yes, he can do what he likes... he can do what he likes... it's not the same with the inspector... And then the glowing tooth became the glowing chest of the corpse again, and the dream was off anew. Or was it, after all? Wasn't it really simultaneous, flashing on and off like the arrow outside the arcade...?
Either way, the dreaming Gently perceived at last a change coming o'er the spirit of his dream. There was a word that kept getting interjected into the mechanism, and for some reason or none he didn't want to hear that word, he kept struggling not to hear it. But he did hear it. It persisted. It paid no attention either to himself or his characters, who were showing similar disapproval.
'Raouls! Otraouls!'
It was making Frenchy's knees jiffle and the empty bottles fall off the counter.
'Raouls! Raouls!'
Gently held Frenchy's knees still with one hand and tried to pick up bottles with the other, but he didn't seem to be getting anywhere.
'Raouls! Otraouls! Raouls!'
He made a final effort to shore-up his collapsing world, to ward off that frightful trump of doom. It was no use. Frenchy kicked the bottles from under his arm. There was a crash of glass which he knew to be the descent of every bottle in the bar and he was dragged back out of the dark or red-lit tunnel in the nick of time...
'Raouls! Otraouls!'
Gently snorted and rubbed his eyes. There really _was_ a sound like that. It was coming through his bedroom window, and getting louder every minute. He jumped out of bed and went to have a look. And then he remembered... over how many years? It was the boy with the hot rolls, that wandering voice of the morning... his very accent had been handed down intact.
Gently hammered on the communicating door. 'Dutt! Aren't you up?'
'Yessir. Been hup half an hour.'
'Half an hour!' Gently glanced at the watch propped up on his dressing-table. 'You're late, Dutt. You should have been up before.'
'Yessir.'
'We aren't on holiday, Dutt, when we're out in the country.'
'No, sir.'
'Discipline,' said Gently, shoving his feet into his bedroom slippers, 'that's the key to success, Dutt. Discipline and luck, but mostly discipline. Is Mrs Davis providing hot rolls for breakfast?'
'Well, sir, I really don't know...'
'Then find out, Dutt, find out, and if she isn't go down and buy half a dozen off that expert out there.'
Twenty minutes later a shining morning Gently put in his appearance at the breakfast table. The papers he had ordered lay fragrant on his plate and he turned them over as he stowed butter into his first roll. The case was still making front-page in the local. They had found a bigger and better photograph, one of which Gently was just a little proud. And they were up-to-date on his visit to the mortuary, and especially up-to-date on his calling out of the pathologist.
PATHOLOGIST RECALLED IN BODY-ON-THE-BEACH CASE, ran the local. GENTLY MOVES – PATHOLOGIST RECALLED – SENSATIONAL MIDNIGHT DEVELOPMENT, ran a London paper.
Gently shoved them across to Dutt. 'Nice press,' he said laconically.
'We'll have 'em round our necks today,' grumbled the sergeant.
Gently clipped the top off a boiled egg and took another bite from his roll. 'They make it seem so exciting,' he mumbled, 'as though we were shifting heaven and earth. I wonder what people would think if they knew how simple it all was?'
They were still finishing breakfast when Inspector Copping was ushered in. He bore an envelope in his hand and an almost reverential expression on his face.
'You were right!' he exclaimed, 'my God – and how! There wasn't only traces of gum on the face, there was crêpe hair too, and quite a bit of it considering. The super's blown up the pathy for not finding it the first time and the pathy's as sniffy as hell.'
'Wasn't his fault,' grunted Gently stickily, 'his job is finding out how they died...'
He wiped his hands on his serviette and thumbed open Copping's envelope. It contained the pathologist's report. He glanced over it.
'Must have been a full beard,' he mused, 'I'm glad he found some of the hair... it might have been a different colour.'
'You were even right about it not being spirit gum. He's going to do a thorough analysis when he's had some shut-eye.'
Gently shrugged. 'Don't wake him up specially. Have you got any artists down at headquarters?'
'Artists?' Copping stared.
'Somebody who can put a beard on some photographs.'
'Oh – _that!_ Our camera bloke can do it for you.'
'Then I'll want some copies of the Missing Persons' list and anybody you can spare to help Dutt go the rounds.'
'I'll have them laid on. But' – Copping looked doubtfully at the marmalade Gently was lavishing on his toast – 'what makes you so positive he came from the town?'
'I'm not,' grunted Gently, poising the piece of toast,
'it just seems to fit the picture, that's all.'
'What picture?' queried Copping.
'Mine,' retorted Gently, and he bit largely and well into the marmalady toast.
* * *
The super seemed a little off-hand that morning. He didn't seem as pleased as he ought to be with the progress being made. He congratulated Gently briefly on his discovery of the beard and asked some terse questions about what he proposed to do. Gently told him.
'You can have a couple of men,' said the super.
'There's something else... I mentioned it to Copping.'
'If it means more men, Gently, I'm afraid I can't spare them just now.'
'No hurry,' murmured Gently, 'I daresay it will keep. But it might be worth keeping an eye on the amusement arcade called "The Feathers".'
The super frowned. 'Well?' he snapped.
'I don't know quite what... vice, perhaps, for a start.'
'In that case it will have to wait. Vice is too common during the season in towns like this.'
'Could be something else... I thought it was worthwhile mentioning it.'
'I'll make a note of it, Gently. Is there anything else you want?'
'Not just at the moment.'
'Then I won't take up any more of your time.'
Outside the super's office Gently shook his head. 'Of course,' he said to Copping, 'I don't expect gratitude...'
'Oh, don't let the Old Man worry you,' returned Copping. 'He's got something else on his plate now, as well as homicide.'
'It must be fascinating, whatever it is.'
'It's forgery – a faked hundred-dollar bill. The super's panicking in case he has to run to the Central Office again. He's trying like mad to trace it to some American Forces personnel.'
Gently clicked his tongue. 'Why should American Forces personnel forge hundred-dollar bills to work off in Starmouth?'
'Search me – but if the super can get back to one of them he's in the clear.'
'Of course, I appreciate his point.'
Copping led the way to the photographer's shop, where Sergeant Dutt was watching the technician apply the final beard to half a dozen postcard prints. He had made two sets, profile and full-face, and the difference between the face bearded and the face unbearded was certainly striking.
Copping whistled when he saw them. 'No wonder we drew a blank the first time round... why do you think he dolled himself up that way?'
Gently shrugged. 'The usual reason – he didn't want somebody to recognize him.'
'But that's fantastic when you come to think of it. Nobody does that sort of thing outside spy thrillers.'
'Could be a spy thriller we're working on,' suggested Gently, dead-panned.
'Could be,' agreed Copping seriously.
It was a Saturday, a day of coming and going. As Gently plodded down Duke Street, which led from the dock side of the town to the Front, he was obliged to thread his way through a stream of parties and individuals lugging bags and suitcases, all of them in a hurry, all of them going one way. He surveyed them lugubriously. They were all good potential witnesses – any one of them might hold the clue he wanted, the unsuspected information. And now they were departing in their hundreds and thousands. They were splitting up and scattering to the four quarters of the Midlands.
On the Front it was the same. The beach had a patchy and unsettled look. Up and down the promenade chased laden cars, taxis and coaches, while the touts stood about in disconsolate groups, their function in abeyance. Everything had stopped. For a few hours the Pleasure Machine stood still. There were those who stayed on, but nobody paid them much attention: they were only there on sufferance, it seemed, until a new lot arrived and the machine began to turn again.
Gently crossed over by the Albion Pier and leaned on the balustrade overlooking the beach. In his breast pocket he could feel the stiff pasteboard of the two doctored photographs, and in the distance he could see the post set up by the Borough Police. If Nits knew him when he was alive, thought Gently, it was at least an even chance he met him here, on the Front... and if he met him on the Front it was ten to one he met him on this stretch, between the two piers. Because that was where 'his' part was, and beachcombers were jealous of their territories.
What next... where was the best prospect after that?
Did he drink, this false-bearded fugitive? Did he play bowls, or tennis, or eat a sandwich at one of the tea-shacks that prospered along the golden mile? Or buy himself a straw hat or sunglasses? Or an ice-cream?
Sunglasses, mused Gently, rummaging in his pocket for a peppermint cream – he'd want some sunglasses if he were playing hard-to-find. At least, he would if he hadn't bought them earlier, about the same time as he was buying crepe hair and adhesives. But it was no use making difficulties. There was a beach-gear stall only a dozen yards away. Gently swallowed the peppermint cream and presented himself at the counter.
'Police,' he said tonelessly, 'can you remember having seen this man during the last week or ten days?'
By lunchtime he'd got the usual mixed bag of possibles and improbables. There were people who thought they had, and those who weren't quite sure: there were numbers who were determined to recognize nothing shown them by a policeman. One gentleman, indeed, was completely positive. The deceased had been to his stall two days running – he'd bought some sun-tan lotion and a pair of frog-man flippers. 'When was that?' asked Gently eagerly. 'Yesterday and the day before,' responded the helpful one...
It was a dispiriting business. He'd been through it before many a time, and with similar results. But here and today it seemed particularly dejecting, as though the whole prospects of the case were tied up with his good or ill success that morning...
They weren't, of course. He was only probing a little of the surface. Elsewhere Dutt and his colleagues were at work on the lines of strongest probability. He glanced at his wristwatch and made for a phone-box. By now they ought to have made some progress.
He dialled, and got the switchboard girl.
'Chief Inspector Gently. Give me the desk.'
She gave him the desk and the duty sergeant answered slickly.
'Gently here... has Sergeant Dutt reported back yet?'
There was a buzz and a faraway question and answer.
'No, sir,' returned the duty sergeant, 'Bryce and Williams have come in – they're in the canteen having their lunch. I don't think they had much luck, sir. Shall I get them to speak to you?'
'No... don't bother them.' Gently made a rapid survey of the terrain without. 'When Dutt comes in get him to phone me at the Beachside Cafe... you got that?'
'The Beachside Cafe... what is the number, sir?'
'Find out,' retorted Gently peevishly, 'I'm a policeman, not the local directory.'
He hung up frowning and shouldered his way out of the box. So Bryce and Williams had drawn a blank also. Like himself. Like Dutt, probably. And there couldn't be so many chances left on that list...
He directed his steps to the Beachside Cafe. It was one of the smaller of the cafes on that part of the Front, a green-painted wooden structure with a sort of veranda that faced the sea. Gently sat himself at one of the veranda tables and ordered a _table d'hôte_ lunch. Three out of the four of them had drawn a blank... three out of four. Was it going to fold up on him, that little streak of luck – his 'dramatic midnight move', as the paper called it? But he'd been right... the man _had_ been wearing a false beard. And Nits had known about it, so the man must have been in Starmouth...
'Your soup, sir,' said the waiter at his elbow. Gently grunted and made room for the plate.
'Excuse me, sir, but aren't you Chief Inspector Gently?' faltered the waiter, hovering at a respectful distance.
Gently eyed him without enthusiasm. 'I might be,' he said.
'I recognized you from your picture in the paper, sir.'
'You're good at it,' said Gently, 'my mother wouldn't have done.'
'Naturally we're interested, sir, it all happening so close...'
Gently sighed and gave the waiter the benefit of a prolonged stare. 'You wouldn't like to be helpful, I suppose?' he asked.
'Of course, sir...' The waiter sounded as though he were conscious of being about to buy something.
'Really helpful?'
'If there's anything I can do...'
Gently produced his two doctored prints and shoved them under the waiter's nose. 'What did he have for lunch last time he was here, or don't you remember?'
The waiter gulped like a guilty schoolboy. 'Dover sole and chips, sir, and fruit salad to follow.'
'He had _what_ —!'
'Dover sole and chips, sir. I remember because it was on the Tuesday, which is the only day we have it.'
* * *
There was a razor-edged pause while Gently clutched at his chair to prevent it revolving quite so fast. The waiter flinched and edged back a pace.
'Now let's be calm about this,' said Gently sternly, 'it was Dover sole _and_ chips – not just Dover sole?'
'No, sir... it was always chips. He was very fond of them.'
'You mean he'd been here _before_?'
'Of course, sir. He came here regular.'
' _Regular!_ How long does it take someone to become a regular?'
The waiter looked worried. 'I think it was Thursday last week... might have been Wednesday. Anyway, he came every day after that, including Sunday... he sat at this table, sir. I thought perhaps you knew him.'
Gently laughed with a certain amount of hollowness. 'I do,' he said, 'in a manner of speaking. But I've still a lot to learn. What's your name?'
'Withers, sir.'
'Well, take that other chair, Withers.'
'Y-yes, sir.'
'Don't be nervous – I'll square you with your boss. And you can fetch in the roast beef when I'm ready for it – even Central Office men have to eat.'
'Yes, sir. Of course, sir!'
Withers pulled out the chair and lowered himself dubiously on to the edge of it. He had the unhappy air of someone who had bitten off more than he could chew. Gently crumbled some roll into his Brown Windsor and tested a mouthful. It seemed up to a fairish standard in provincial Brown Windsors.
'So he came here first on Thursday, Withers. Or it might have been Wednesday.'
'That's right, sir.'
'You haven't any preference.'
'N-no, sir... I just don't remember.'
Gently nodded intelligently and tried another spoonful of soup. 'Did he have any name that chanced to leak out?'
'He said to call him Max, sir.'
'Max, eh?' Gently rolled the word round his tongue. Now he'd even got a name for the fellow! 'Max anything or just Max?' he asked hopefully.
'Just Max, sir.'
Gently sighed. 'I felt it had to be. He had an accent, though, this Max?'
'Oh yes, sir.'
'What sort of an accent... did you recognize it?'
The waiter stirred tormentedly. 'Foreign, I'd say, sir.'
'Was it French, for instance?'
'Yes, sir, it might have been.'
'Or German?'
'No, I don't think so, sir.'
'Russian, maybe?'
'I wouldn't like to say it wasn't, sir.'
'You couldn't imitate something he said?'
The waiter shook his head and sent a haunted look towards the rear of the cafe. Gently shook his head also and reapplied himself to his soup. But why should he complain, he asked himself, why look such a regal gift-horse in the mouth? Ten minutes ago he had begun to despair and now he actually knew the dead man's name...!
'Describe it,' he said, 'describe Max coming in here and having lunch.'
'H-how do you mean, sir?' faltered the waiter.
'Tell me, man! Tell it as though he were just coming in at the door.'
The waiter twisted his hands together agonizedly and cleared his throat. 'H-he'd come in...' he began, 'he'd stand for a moment looking about... as though he expected to see somebody he knew...'
'Did he ever see that somebody?'
'No, I d-don't think so, sir.'
'What was he wearing?'
'He'd got a light grey suit, sir. On Sunday he wore a darker one, but the other days it was the light grey. And he had a blue bow tie.'
'Go on.'
'He carried an attaché case, sir, he had it with him every day except the last day... then there was his beard, that struck me as being funny... and the way he spoke...'
'What did he say?'
'When he first came in he asked me my name, sir. Then he sort of laughed and told me to call him Max.'
'Was there any reason for that?'
'It was because I called him "sir," sir. He said they didn't call people "sir" where he came from, and then he laughed again and patted me on the arm.'
'He was a friendly type, was he?'
'Oh yes, sir, quite a gent.'
'So he patted you on the arm. What happened then?'
'He ordered the chicken, sir, and sent me out for a bottle of wine... we aren't on the licence here, sir.'
'And what day were you serving chicken last week?'
'Wednesday, sir.'
'Ah!' said Gently with satisfaction. He laid down his spoon. 'We'll pause for a moment on that happy note... just pop along and see what the roast beef is doing.'
'Certainly, sir!'
'And fetch me a lager, Withers. The occasion seems to justify it.'
The waiter slipped from the chair and resumed his function with obvious relief. Gently smiled distantly at a paddling child. Another time Withers wouldn't be quite so forward in accosting chief inspectors who got their pictures in the papers...
And the name was Max. Max, in a light grey suit with a blue bow tie. Max, who came from somewhere where they didn't 'sir' people. Max, who was friendly. Max, who was quite a gent. Max, who had sat at that same table from Wednesday till Tuesday, eating his chicken, his Dover sole and chips, and drinking the wine Withers brought him from over the road... and Max, who had finished up as Exhibit A on the mortuary slab exactly a week after his first appearance. He was getting into focus, that one. Gently was beginning to see him, to fit him in. And over all there was his foreign-ness, pervasive and misty, his Franco-German-Russo-what-have-you foreign-ness...
Withers returned with Gently's roast beef and the lager. He seemed to have been gone a good deal longer than was strictly necessary, even allowing for the trip across the road. Gently raised his eyebrows to the unhappy man.
'Talked it over with the boss, Withers?' he inquired affably.
'I-I beg your pardon, sir!' stammered Withers, spilling some lager.
'Never mind, Withers... and don't be well-bred about the vegetables.'
The waiter served, and Gently picked up his knife and fork. It was odd, but he hadn't been feeling hungry when he came into the cafe...
'Sit down,' he mumbled to Withers, 'you'll give me indigestion, jiffling about like that.'
'I b-beg your pardon, sir, but really I ought to be getting on with my work... there isn't n-nothing I haven't told you, honest...'
Gently beamed at him over a mouthful of lager. 'Nonsense, Withers, we've only just begun...'
'It's making extra work for the others, sir,' persisted Withers, encouraged by the beam.
'Sit down!' retorted Gently with a slight touch of Bogartesque.
Withers sat down at great speed.
'... Now,' continued Gently, after a certain amount of plate-work, 'we got to him ordering the chicken and sending out for some wine. What sort of wine did he send for?'
'Just red wine, sir. I got him a brand they specialize in over the road.'
'I don't doubt it for a moment. Did he express his satisfaction?'
'N-no sir, not really.'
'Did he order the same wine the next day?'
'He asked if they hadn't got another brand... I couldn't understand the name he gave it.'
'What did it sound like?'
'It just sounded foreign, sir...'
'Like what sort of foreign?'
'I d-don't know... just gibberish.'
'Did you ask if they'd got it?'
'No, sir. I couldn't say the name.'
'So what did he have?'
'I got him Burgundy, sir, when he wanted a red, and Sauternes when he wanted a white.'
'And that was satisfactory?'
'He seemed a bit surprised at the price, sir.'
'He was a foreigner, Withers.'
'Yes, sir, I dare say that had something to do with it.'
Gently brooded a moment over a roast potato. Then he halved it meticulously and transported one half, suitably garnished with gravy, to a meditative mouth. 'What did he have for sweet, Withers?' he asked through the potato.
'Ice-cream, sir.'
'Not much to be deduced from that... was his coffee black?'
'Yes, sir.'
'Did he smoke...? Cigarettes...?'
'He bought a box, sir.'
'A box, Withers?'
'Twenty-five Sobranie, sir.'
Gently raised an eyebrow. 'And what particular variety?'
'Just Balkan Sobranie, sir. He bought a box every day after that...'
'He seems to have been a well-heeled foreigner, Withers.'
'Yes, sir. He never tipped less than half a crown.'
Gently finished his roast beef and motioned to have his plate removed. Withers took it adroitly and produced a cold sweet from a side-table. It was a trifle, a robustly constructed affair involving sliced pineapple, and Gently inserted a spoon in it with unabated gusto.
'Of course, he asked a few questions,' volunteered Withers, beginning to feel that Gently wasn't so bad after all. 'He wanted to know if we got many foreigners in Starmouth.'
'Mmph?' grunted Gently, 'what did you tell him?'
'I told him we scarcely saw one – not a right foreigner... just midlanders and such-like.'
'Yanks,' mumbled Gently.
'Well there... we don't count them.'
'Was he happy about the situation?'
'It didn't seem to worry him, sir. He said we might have him around for a bit... and later on, of course, he picked up with a woman...'
Gently made a choking noise over a segment of pineapple. 'What was that, Withers...!'
'He picked up with a fern, sir. Brought her in to lunch here on the Tuesday.'
Gently got rid of the pineapple with a struggle. 'So he did... did he! Just like that! Why the flaming hell didn't you say so sooner?'
'You never asked me, sir!' exclaimed Withers, surprised and apprehensive, 'it wasn't nobody really, sir... just one of the girls you get around here during the season...'
'Just one of the girls!' Gently gazed at the wilting waiter. Then he took himself firmly in hand and counted ten before firing the next question. 'You know her name? It wouldn't be Yvette, by any chance?'
'No, sir! I don't know her name! I've never had nothing to do with women of that class...'
'She's the little dark one with long slinky hair.'
'But this one's a blonde, sir – quite well set-up. And her hair is short.'
'Nice legs – smooth, rounded knees.'
'I d-didn't notice, sir...'
'Don't lie at this stage, Withers!'
'I thought they were bony, sir – I did, honest I did!'
'She speaks with an educated accent.'
'Not this one, sir – she's terribly common!'
'You'd recognize her again?'
'Of course, sir. Anywhere!'
A telephone began pealing at the counter inside the cafe and Gently relaxed his hypnotic attention from the freshly-shattered Withers. 'Go and take it,' he purred, 'it's probably for me.'
Withers departed like greased lightning. He was back inside seven seconds.
'A S-sergeant Dutt, sir, asking for you...'
Gently made the phone in even better time than Withers.
'Gently...!' he rapped, 'what's new with you, Dutt?'
'We've placed him, sir!' echoed Dutt's voice excitedly, 'he was missing from a lodging in Blantyre Road – disappeared on Tuesday evening and nothing heard since. The woman who let the room identified him straight away. His name was Max something – she didn't know what.'
A faraway look came into Gently's eyes. It was directed at the ceiling, but in reality it plumbed sidereal space and lodged betwixt two spiral nebulae.
'Get a car, Dutt,' he said, 'come straight down here and pick me up...'
'Yessir!' rattled Dutt, 'I'll be with you in ten minutes.'
'Ten minutes,' mused Gently, 'that'll just give me time to drink my coffee... won't it, Withers?'
CHAPTER FIVE
BLANTYRE ROAD WAS a shabby-genteel thoroughfare which began at the top of Duke Street and meandered vaguely in a diagonal direction until it joined the Front a good way south, where hotels had already begun to thin out. It was at its best at the top end. Just there it skirted a small park or garden, and the houses which faced it, Edwardian Rococo, had a wistful air of having known better times and more civilized people.
Outside one of these a crowd had collected. It spread along the pavement in both directions and was a model of quietness and patient expectancy. On the steps behind them the careful Copping had stationed a uniform-man, but his authority was somewhat vitiated by the presence of three gentlemen with cameras supported by four gentlemen without cameras – a contingent possessed of far more glamour than a mere police constable.
'Blimey!' exclaimed Dutt, as he, Gently and Copping came dramatically on the scene in the back of a police Wolseley, 'there wasn't a soul about here half an hour ago.'
'That woman must have blabbed,' snapped Copping, 'I sent Jennings down to try and stop it... blast her tattling tongue!'
'Of course, she's got a perfect right to...' murmured Gently.
The Wolseley made a three-point landing opposite the door and the police constable marched down to give them his official greeting.
'Sorry about this here, sir,' he apologized to Copping, 'that was all done before I arrived...'
'Never mind – never mind!' barked Copping, 'just keep those wolves there out of the house, that's all.'
He strode up the steps, an impressive figure. Gently followed with Dutt at a more sedate pace. The flashbulbs popped and the crowd rippled.
'How about a statement!' demanded a reporter, pushing up, notebook at the ready.
'Nothing about a statement!' boomed Copping, 'if you want a statement, come to headquarters for it.'
'A statement from you, then,' said the reporter, turning to Gently.
Gently shrugged and shook his head. 'Did you get one from Mrs Watts?' he inquired.
'We were actually getting one when the constable interfered...!'
'Then you probably know more than I do just at the moment...'
He pushed past and up the steps.
The interior of the house was as pleasingly period as the outside. Inside the front door was a long, narrow, but lofty hall, a good deal of it occupied by a disproportionately wide staircase. At the far end another door led into the back garden, a door equipped with panes of red and blue glass. There was a certain amount of upheaval apparent, quite incidental to the main theme – it was a lodging-house Saturday, one set of guests departed, the other not yet arrived. At the foot of the stairs lay a bundle of dirty sheets, in the dining-room, its door ajar, a heap of tablecloths and napkins... _Entr'acte,_ thought Gently. The phrase epitomized Starmouth on a Saturday.
Copping had marched ahead into Mrs Watt's private parlour, from whence could be heard issuing the landlady's strident and aggressive tones.
'I don't know why you're making all this fuss _now_ , I'm sure... I told the man who called round here on _Wednesday_... well, is it my fault if you didn't know about the beard?'
'There must be some mistake, mam,' came the discomfited voice of Copping, 'I'm sure O'Reilly...'
'Mistake, Inspector! I should just say there was a mistake. My daughter Deanna and my husband Ted both backed me up about it... "Beard or no beard," I says, "the man on that photograph is our number seven"... and that was on _Wednesday,_ Inspector, yet you come worrying me today of all days, a Saturday, and Race Week – it's too bad, it is really! If it's not making me all behind with my work, it's what my people are going to think with all that lot gawping outside...'
Dutt gave Gently a knowing wink. 'Aye, aye! I was waiting for him to run into that lot.'
'Somebody's boobed, Dutt.'
'Yessir... and it isn't you and me.'
Gently pushed in at the parlour door. It was a small but expensive room. The gilt-edge of Mrs Watts's season expended itself on radiograms, television sets, slow-burning stoves, carpets and furniture notable for its areas of glossy veneer. The available floor-space was a trifle restricted by these evidences of wealth. It occurred, where it occurred in small islands of gold mohair. On the largest of these, which adjoined the multi-tile hearth, Mrs Watts was conducting her attack, while a red-faced Copping had got himself wedged into a triangle between a radiogram and a television set.
'What do you send them round for?' continued the stalwart matron, snaking a glance at the new intruder _en_ _passant._ 'What's the idea of wasting our time asking questions when you aren't going to believe us anyway? Is that how you run the police in Starmouth? Is that why they keep putting the rates up?'
'I assure you, mam, if you'll let me explain...'
'Oh, I don't doubt, you'll be a wonderful one for explaining. And I dare say your explaining will get the work done by the time my people start coming in. If you ask _me_ , Mr Inspector, we need someone in Starmouth who can teach you your job... that body on the beach was a show-up for you, wasn't it just...?'
'Ahem!' coughed Gently, appropriating some mohair behind the door.
Mrs Watts shook her platinum locks and presented a square chin at him. 'And who's this?' she demanded of Copping, 'how many more have you brought down here to waste my time?'
'This is Chief Inspector Gently, mam!' explained the squirming Copping, 'he's in charge of the case... he wants to ask you a few questions.'
There was a pause while Mrs Watts digested this information. Then her expression underwent a change, passing from steely aggressiveness to steely affability. 'Well!' she said more placably, 'well! And aren't you the gentleman they've sent from Scotland Yard to clear up this body-on-the-beach business?'
Gently nodded gravely.
'The same Chief Inspector Gently that did that case at Norchester?'
'The same, Madam.'
'Well!' repeated Mrs Watts, 'of course, if I'd known that...' She favoured Gently with a smile in which steeliness was still the principal ingredient. 'Do please sit down, Inspector... I shall be pleased to be of any assistance. Deanna!' – her voice rose to a shout – 'Deanna, leave what you're doing and make a pot of tea, do you hear?'
There was a faint acknowledgement from without and Mrs Watts, satisfied, ushered Gently to the room's most dramatic and veneer-lavish chair. He contrived to avoid it, however, and it was Copping who became the victim...
'Now,' pursued Mrs Watts, 'I'd like you to know, Inspector—'
'Just a minute,' interrupted Gently, 'has the room been interfered with?'
'The _room_ , Inspector...?'
'Number seven – the room from which this man disappeared?'
Mrs Watts looked doubtful. 'I don't know what you mean, interfered with. I've changed the sheets and pillow-cases, and Ida (that's the maid) has polished and hoovered, but that's all... there's nothing been moved about.'
Gently sighed softly to himself. 'Well... we'll look in there later, if you don't mind. Now about the man himself...'
'I recognized him directly, Inspector. There was never any doubt.'
'You recognized him without the painted-in beard?'
'As soon as I clapped eyes on the photograph... "Yes," I says to the man, "that's our number seven. Only he's got a beard," I says, "a lot of it – all over his face."'
'And that was on Wednesday, the day after your lodger was missing?'
'That's right – Wednesday evening. Naturally, I didn't pay too much attention to him spending the night out... you can't be too particular about that sort of thing, Inspector. But when it got near tea-time and still no sign of him...'
'You rang the police and were shown the photographs. You acted very properly, Mrs Watts.'
'But the man didn't _believe_ me, Inspector – I could see he didn't!'
Copping made a rumbling noise. 'It was O'Reilly,' he brought out, 'he was going on transfer to Liverpool the next day... he didn't want to believe it...'
Gently nodded comfortably to one and the other. 'Everyone is human... even the police. And of course you recognized the touched-up photograph, Mrs Watts?'
'Naturally I did – and so did Deanna – and so did my husband Ted, who was in after his lunch.'
'You'd be prepared to swear to the identity in court?'
'I'd take my Bible oath on it, Inspector... and so will _they_.'
Gently nodded again and felt absently in the pocket where he stowed the peppermint creams. 'When was it he arrived?' he asked, struggling with the bag.
'It was last Wednesday week – in the morning, just after breakfast.'
'Go on. Describe what happened.'
'Well, I answered the door, Inspector, and there he stood. "I see you've got a room vacant," he says – only he had a queer way of slurring it, as though he were trying to be funny – "do you think I might see it?" he says. I mean, the cheek of it, Inspector! People are usually glad enough to _get_ rooms in the middle of the week at this time of the year, without being awkward about it. And him a foreigner too, and smelling as though he'd just walked off a fishing boat...!'
Gently paused in the act of transporting a peppermint cream to his mouth. 'A fishing boat?' he queried.
'Yes – that's just the way he smelt. Mind you, I don't want to accuse him of having been a dirty man. It was something that wore off later and the first thing he did was have a bath. But there's no doubt he had a fishy smell on that particular morning... well, I nearly slammed the door in his face!'
Mrs Watts pulled herself up in a way which reminded Gently of a baulking mule.
'How was he dressed... can you remember?' he asked.
'He'd got his light grey suit on – he nearly always wore that... a bit American, it was, with one of those fancy backs to the jacket.'
'Tie?'
'That was a bow.'
'Hat?'
'He never wore one that I can remember.'
'Did he have some luggage with him...?'
'He'd got a couple of cases, one bigger than the other... the big one is still in his room.'
'How about the other – what happened to that?'
'I suppose he took it with him, Inspector. He always did when he went out... he seemed to think there was something very precious about it.'
'Did you see him leave with it the last time you saw him?'
'No... I didn't see him after I'd given him his tea. Deanna saw him go out, perhaps she noticed. Deanna!' – Mrs Watts's voice rose piercingly again – 'come in here – the inspector wants to ask you a question!'
'Coming, Ma!' replied a sugary voice just without the door, and a moment later Deanna made her entrance bearing a chrome-and-plastic tea-tray.
'Put it down here, Deanna – I'll pour it out.' Mrs Watts was obviously proud of her daughter and wanted her to be admired. 'This is Chief Inspector Gently down here about the body on the beach... don't be afraid of him, my dear, there's no need to be shy.'
Deanna wasn't shy. She beamed at Gently with a mechanical smile which had haunting overtones of Mrs Watts in it, then seated herself next to him. She had a cat-like grace too studied to be pleasing. She was twenty-one or -two.
'My daughter's on the stage, Inspector,' chattered Mrs Watts, sploshing tea into straight-sided cups with lustred rims, 'she was in the pantomime last season... just in the chorus, you know.'
'I understudied the principal boy,' beamed Deanna.
'They're going to give her something bigger this year... of course, she's home with me during the summer.'
Gently accepted one of the straight-sided cups and stirred it with a spoon that had a knob of black plastic to its spindly shank. 'Getting back to your lodger...' he murmured.
'Of course, Inspector.' Mrs Watts handed a cup of tea to Dutt behind the television. 'Deanna dear, you saw him go out on Tuesday... the inspector wants to know if he had his case with him.'
'I don't really remember, Ma... I didn't know it was going to be important.'
'But it _is_ important, dear... you _must_ try to think.'
'I _am_ trying, Ma, but it isn't any good.'
'What time was it when he went out?' asked Gently.
Deanna curled round in her seat to him. 'I just can't remember, Inspector... isn't it awful of me?'
'What were you doing when you saw him?'
'Oh... I was going up to my room to get ready for the Tuesday dance at the Wellesley.'
'How long would that have taken you?'
'About an hour... aren't I terrible!'
'And then your boyfriend called for you?'
'Well yes, he did, Inspector!'
'And what time was that?'
'It was a quarter past eight... he was late.'
'Thank you, Miss Deanna.'
In his veneered throne Copping stirred restlessly. 'How about the visitor's book – what did he put in there?' he asked.
Mrs Watts's chin took on an ominous tilt. 'He didn't put anything in there. They don't, most of them, until they're going.'
'They should,' said Copping stoutly, 'they should make an entry as soon as they arrive.'
'Well, they _don't_ , Mr Nosey, and that's all there is to it. And if you're going to make trouble out of it you'll have to make trouble for everybody in Starmouth who lets rooms...'
Gently made a pacifying gesture. 'But surely he gave a name, Mrs Watts? Naturally, you would ask for that...'
'Of course I did, Inspector. And he gave it to me without any hanky-panky – only it was such a peculiar one that I couldn't even say it after him. So he just laughed in that rather nice way he had and told me to call him Max... and that's what we all called him.'
'Didn't you inquire his nationality?'
'He said he was an American but if he was, he hadn't been one for long, not with that accent.'
Gently sipped some tea and looked round for somewhere to put his cup. 'How long was he going to stay?' he asked.
'Just on to the end of this week – I hadn't any room for him after that. I'm usually full up right through, of course, but it just so happened through an illness...'
'Quite so, Mrs Watts. And did he pay up till the end of the week?'
'He did – it's one of the rules of this establishment.'
'There seemed to be no shortage of money with him?'
'Not him, Inspector. He'd got a whole wad of notes in his wallet – fivers, most of them.'
'Did he ask any questions before he took the room?'
'Well, the usual ones... how much it would be, if we'd got a separate bathroom and the like.'
'Did he ask about the other guests, for instance?'
'Yes, he did, now you come to mention it. He asked if they were all English and if they had all arrived the Saturday before.'
'And did that suggest anything to you?'
'He seemed a bit anxious about it... I thought he might be expecting to run into somebody he knew.'
'Somebody pleasant or somebody unpleasant?'
'Unpleasant, I suppose... if he really is the one you picked up on the beach.'
'Did he suggest that from the way he spoke?'
'Well no, Inspector, he didn't actually...'
Gently prized up a peppermint cream from the dwindling stock in his pocket. It induced that faraway look in his eye which Mrs Watts mistook for profound cerebration, but which in reality was connected with his solvency in terms of that important commodity... though Starmouth was pretty good peppermint cream country at most hours of the day and night.
'Was he a good mixer?' he asked absently.
'Oh, he got on with everyone, though I wouldn't say he made friends. But he got on with them. They all liked our Max.'
'Was he regular in his habits?' Gently yielded up his cup for a second fill from the hotel-plate teapot.
'I dare say he was... as people go when they're on holiday.'
'Tidy... a good lodger?'
'Oh yes... most of the time.' A frown hovered over the steely eyes as she handed Gently the freshly-filled cup. 'He left his room in a bit of a mess when he went out that last time, but probably he was in a hurry... you haven't always time to clear up after you.'
'A mess...!' Gently hesitated in the act of plying his plastic-knobbed spoon. 'What sort of a mess?'
'Well, if you ask _me_ , Inspector, he'd lost something and was trying to find it quickly, that's what it looked like. The wardrobe was open, the drawers pulled out of the dressing-table – right out, some of them – and if he hadn't up-ended his suitcase on to the floor then he'd given a good imitation of it. And the bed, too, I should say he'd had that apart, not to mention turning up a corner of the carpet. It was a proper mess, you can take it from me!'
Gently drew a long breath. 'But of course,' he said expressionlessly, 'of course you cleared it all up again, Mrs Watts?'
'I did, Inspector,' the regal matron assured him, 'I can't stand untidiness in my house, no matter from whom.'
'Ahh!' sighed Gently, 'I needn't have asked that one, need I...?'
The room faced back with a solitary and not-very-large sash window overlooking a small backyard. It was a typical lodging-house 'single', about eight by ten, not much more than a cupboard in which had to be packed the bed, wardrobe, dressing-table, chair and the tiny fitted wash-basin which tried to substantiate the terms Mrs Watts charged for such accommodation. The walls were papered in an irritable grained brown friezed with orange and green, the floor had a strip of carpet which echoed these colours. The bed and other furniture were of flimsy stained wood, late thirties in vintage, and the light-shade was a contraption of orange-sprayed glass with a golden tassel for the flies to perch on. In essence it bore a generic resemblance to the parlour downstairs, thought Gently. There was the same over-crowding and full-bodied vulgarity. It was only the cash index that varied so considerably.
Beside the bed stood an expensive looking suitcase, a rather jazzy affair styled in some sort of plastic with towelling stripes. Copping bent down to pick it up, but Gently laid a sudden hand on his arm. 'Watch it... I want this place printed,' he said.
'Printed?' Copping stared in surprise. 'There can't be much left to print after all this clearing-up...'
Gently shrugged. 'If there is, I want it.'
'But what does it matter – we've got three witnesses at least to identify him?'
'It isn't only him that interests us...'
He moved to the window, leaving Copping still staring.
The window was part open at the top. Immediately below it were the red pantiles roofing the outside offices, at the end of which could be seen part of a corrugated steel water-butt. The yard itself was no more than twenty yards long by ten wide. It was separated from its neighbours and the alley on which it backed by grimy brick walls. In the far corner a sad laburnum trembled, in the centre rotted a part-buried Anderson shelter, while close at hand there roosted three dustbins, one of them with its lid at a rakish angle...
Gently produced a not-perfectly-clean handkerchief and closed the window. 'Look,' he said to Copping, pointing to the catch.
Copping looked intelligently. 'It's broken,' he said.
Gently nodded and waited.
'Done from the outside – forced up with a chisel or something...'
Gently nodded again.
'Hell's bells – the room's been _burgled!_ ' exclaimed Copping, suddenly catching on. 'It wasn't the boyo who left it upside-down – it was somebody else – somebody looking for something he left behind here!'
'Which is why I'm printing the place...' murmured Gently helpfully.
'It's plain as a pikestaff – I can see the whole thing! He sneaked in up the alley – got in through that broken gate down there – climbed on to the roof by the water-butt and the down-pipe – forced up the catch!'
'Hold it,' interrupted Gently. 'Dutt, step up here a moment.'
Dutt, who had been lingering respectfully in the passage, came quickly to the window. Gently spoke to him without turning his head.
'Over there – where the coping's knocked off the wall... don't make it too obvious you're looking.'
'I can see him, sir,' muttered Dutt, 'if he'd just turn his loaf a fraction...'
'But who is it!' interrupted Copping, shoving in, 'is it someone you know—?'
'Back!' rapped Gently, 'keep away till Dutt has had a good look... there, you've scared him... he's off like a hare!'
Dutt raised himself from the stooping position he had taken up. 'It was him, sir,' he asserted positively, 'I saw the scar as he turned to run... you can't mistake a face like that.'
'I saw it too, Dutt, right down his cheek.'
'He must have copped a fair packet somewhere...'
'Also he has a strange interest in what goes on...'
'But who is he?' yapped Copping again, 'what's it all about, this I-spy stuff?'
Gently smiled at some spot that was miles behind Copping's head. 'It's just a little thing between Dutt and me,' he said, 'don't let it bother you... it's all over now. Suppose we do what you wanted and take a look in the suitcase?'
They retired from the window and a disgruntled Copping demonstrated how to open a suitcase before it had been printed. It was a charmingly well-filled suitcase. It contained an abundance of shirts and socks and underwear, besides some hairbrushes and toilet accessories which the tidy Mrs Watts had garnered from wash-bowl and dressing-table. And the contents were determined to be helpful. There were makers' labels attached to some of the clothes, names and patent numbers stamped on other items... even the suitcase itself had a guarantee label tied to the lining with blue silk. Gently had never seen such a helpful lot of evidence...
'It's American,' declared Copping brightly, 'look at this one – "Senfgurken Inc., NY" – and that razor – the toothbrush, even. It's all Yank stuff, right through.'
'And all brand new,' mused Gently.
'He must have bought it for the trip and he can't have been over here long. Or maybe he's a service-man on leave and fixed himself up at his P.X. Anyway, we know where to start looking. If his embassy doesn't know about him, the US Army will.'
'I wonder...' Gently breathed.
'Eh?' stared Copping.
'Of course, he said he was an American...'
Copping's stare became indignant. 'Who else but a Yank could get hold of this stuff? And who would want to fake up some American luggage, here in Starmouth? What's the point?'
Gently shrugged and dug up the last of his peppermint creams. 'That's what I'd like to know,' he said.
'He's a serviceman got in some bad company, you take my word. It's happened before in Starmouth... he's a deserter, that's my bet.'
Gently shook his head. 'It doesn't fit in. There's nothing American about Max except his clothes, and even they seem too good to be true. No... everything about him is wrong. He just won't add up into a good American.'
'He might add up into a bad one,' quipped Copping, but Gently didn't seem to be listening.
'The suit – his dark suit! What happened to that?'
'His dark suit?' echoed Copping.
'The one he wore on Sunday. Look in the wardrobe, Dutt. It may still be hanging there.'
Obediently Dutt pulled out his handkerchief and unlatched the wardrobe door. Sure enough a dark suit hung there, a shouldery close-waisted number in discreet midnight blue. Dutt turned back a lapel to show the tailor's label. It was of one Klingelschwitz, operating in Baltimore.
'Still American,' commented Copping, a shade triumphant.
'Go through the pockets,' ordered Gently dully.
Dutt went through them. There wasn't even any fluff. But as he was re-folding the trousers something small and bright fell from one of the turn-ups, a little disc of metal. Copping swooped on it and held it up.
'His lucky charm. He ought to have had it with him on Tuesday.'
'A circle with a line through it!' exclaimed Dutt, 'there's something familiar about that, sir – I've seen it before somewhere.'
'So have I.' A gleam came into Gently's eye. 'I saw it last night on the ring of a Mr Louis Hooker. I wonder if Louey has ever been to America...?'
CHAPTER SIX
THE SUPER WAS out when they arrived back at headquarters – rather to Gently's disappointment, because he would like to have bounced some of his findings on that sceptical man's desk. But the super was out: he had received a hot tip about his forgery scare, said the desk sergeant, and had departed with Bryce and two uniform men at a high rate of knots.
'He's got a warped sense of value,' pouted Gently to Copping. 'In some places it's homicide that gets top rating...'
'You're forgetting he handed that baby on to you,' grinned Copping, 'he's got an alibi now.'
'I still think a little bit of audience reaction is called for.'
They went into the canteen, where Copping did the honours. It was rather a dull place. The walls were distempered in a dingy neutral tint, the inadequate windows both at one end, the paint worn on lino-top tables and the bentwood chairs looking as though they had been rescued from a jumble sale.
'They've talked about refitting it for years,' Copping apologized, 'but somehow the finance committee never quite gets round to it... the food's all right, though. We made a stink about that a couple of months back.'
Gently examined a plate of sausages and beans apathetically. 'You have to make a stink at intervals if you want to keep them up to scratch...'
'Yes, but you should have seen what it was like before then!'
Gently shrugged and embarked on his sausages.
'We get in touch with the US authorities now?' inquired Copping, after a silence broken only by the incidental noises made by ingesting policemen.
'Nmp.' Gently pursued an errant bean round the rim of his plate.
'The military's got good records... they could tell us straight away.'
'Never mind. Some other time.'
'They'd know in town...'
'I know, but never mind.' Gently swallowed the tail-end of a sausage and grounded his knife and fork. 'Your print king,' he said, running his tongue round his lips, 'what's his name?'
'Dack's your man. Sergeant.'
'He's reliable... really?'
'You trained him, so he'd better be.'
Gently nodded and added a mouthful of strong tea to the sausages. 'Get him on the job. I can't spare Dutt just now. See that he does everything that might give something... _inside_ drawers as well as out... and then in the yard, at the _back_ of that down-pipe... he'll probably have to dismantle it. Don't wait for me. If you get results, rush some copies to town and check your own files.'
A smile spread over Copping's heavy features. 'What about Mrs W's new lodger?'
'Nothing about Mrs W's new lodger... he can sleep under the pier for all I care. When you've finished in there, seal it up and leave a uniform man in charge.'
'I don't pity the poor swine...! Where can I get you?'
'Oh... I'll look in later, or maybe ring.'
'You've got something else?'
'Could be,' returned Gently evasively, 'and then again, it couldn't.'
He drank some more tea while Copping indulged in speculative ratiocinations. 'It'd be easy to give the US military a ring... just to be sure.'
'No,' said Gently, kindly but firmly, 'we'll leave them to concentrate on Western Defence or whatever else it is they do in these parts...'
The Front had become its old gay self again by evening. Everybody hadn't arrived yet – there were still momentary appearances of towering coaches hailing from Coventry, Leicester, Wolves and Brum, dusty from long journeying, their passengers lolling and weary – but enough had already arrived, enough had checked in at their lodgings, deployed their belongings, washed, changed, tea'd, and now sallied forth, cash in hand – they really spent with a will on the Saturday night. Remote from it all, the sea looked cold. Nobody wanted the sea on that day of the week. It was there, it was the alleged attraction, but that was all... and in the setting sun it looked cold and hard.
More interesting was the local Evening and the two Londons. They proclaimed the wisdom of having chosen this week for the holiday instead of last week. Last week, of course, the body had been found and the Yard called in, but it was pretty obvious from the way things were going that it would be this week when the mystery was solved, the arrest made... BODY IDENTIFIED BY LANDLADY ran the local – Lodger Said to Have Worn False Beard: Missing Suitcase – and there was a photograph showing Gently's back and Copping posed at the top of the steps. The Londons didn't get it early enough to feature. They had to be content with a stop-press and no pics. But they did their best. They whooped it up joyfully. IT WAS ROGER THE LODGER – AND HIS WHISKERS WERE PHONEY, one was captioned, BODY ON THE BEACH – WHY SHAVE IT? asked the other. Yes... things were moving. It was obviously the right week to be in Starmouth, quite apart from the races.
'Can't help feeling we've been mucked about, sir,' observed Dutt, as the two of them turned the corner at the end of Duke Street, 'all these new people... thahsands of them... and we know for a start they haven't got nothink to do with it.'
Gently belched... those damned sausages! 'It's the ones who've gone that worry me,' he muttered.
'And then again, there's him we're going to pinch... could be any one of them, sir. This bloke coming along here, now, the one with the tasselled hat... I wouldn't put it past him.'
Gently clicked his tongue. 'You can't go on that sort of thing, Dutt.'
'I know, sir, but you can't help thinking about it. This isn't like the usual job – as a rule there's one or two to have a go at. But this time there's not a soul, not a blinking sausage' – Gently winced at this unkind reference – 'not a solitary bloke anywheres who you can lay your hand to your heart about. I mean, even that bloke with the scar, sir. What have we got on him, apart from him acting suspicious? I dare say he's up to something he wouldn't like us to know about, but honest now, what connection is that with the deceased? We've often put up pigeons like him on a job.'
Gently sighed, but the sigh was interrupted by a belch. 'This is why we get on so well together, Dutt,' he said bitterly, 'your cockney common sense is the best foil in the world for my forensic intuition...'
'Well, there you are, sir. I don't want to look on the black side...'
'Of course not, Dutt.'
'But you've got to admit it's still a bit speculative, sir.'
'Highly speculative, Dutt... which is why we're keeping firmly on the tail of any pigeons we put up.'
'Yessir. Of course, sir.'
'Including your man with a scar.'
'I wasn't presuming to criticize, sir...'
'No, Dutt, please don't... at least, not after I've been eating dogs in that damned canteen up there...!'
'I'm sorry, sir... they was perishing awful dogs.'
They came to a side street running along blankly under the shadow of a Babylonian cinema, a brick vault of Edwardian foundation and contemporary frontage.
'This is me, sir,' said Dutt, halting, 'I can work my way round and come out on the far side of Botolph Street.'
'There's cover there... you don't have to lean on a lamp-post?'
'There's a builder's yard with a gate I can get behind.'
'We don't want our pigeon frightened... if he's there. I'll give you twenty minutes to get set.'
'That'll be about it, sir.'
'And if he gives any trouble put cuffs on him. My forensic intuition suggests you'll be justified...'
Dutt turned off down the side street and Gently, with a dyspeptic grimace, crossed the carriageway and joined the noisy crowd jostling along the promenade. Everything was in full swing again, the lights, the canned music, the windmill sails, the crashing and spanging of the shooting saloon... a sort of fey madness, it seemed, a rash of inferno at the verge of the brooding ocean. He turned his back on it and leaned looking out at the cold water.
Dutt was right, of course. There was precious little connection. You could say Frenchy for certain, and that was all... and what did Frenchy add up to, even if you could prove it? A friendly foreigner dressed like a Yank and generous with his pound notes... he was natural meat for Frenchy. And of course she would lie. Of course she would dig up an alibi. Quite apart from anything else it was bad business for your last boyfriend to wind up a corpse on the beach.
And after Frenchy it was all surmise. There was nobody else who tied in at all, or not in a way that looked impressive when you wrote a report. He had wandered into town, this enigmatical foreigner, he had taken lodgings, he had found a cafe to his taste and a prostitute to his taste; and then he had been, in a short space of time, kidnapped, tortured, murdered and introduced into the sea, his room ransacked and plundered of something of value. There was a ruthlessness about that... it bore the stamp of organization. But there was no other handle. The organization persisted in a strict anonymity.
So he was left with his intuition, thought Gently, his intuition that made pictures and tried to fill them in, to make them focus, to eliminate their distressing areas of blankness. One didn't know, one simply felt. With the facts firmly grasped in the right hand one groped in the dark with the left... and if you were a good detective, you were lucky. Mere intellect was simply not enough.
He swallowed and grimaced again. If ever he ate another sausage...!
There was an air of restraint in the bar of 'The Feathers', as though everybody had been put on their best behaviour. It wasn't too full, either, considering it was Saturday night. The sporty type sat drinking whisky on a high stool, and one or two other less-than-salubrious characters whom Gently remembered from the previous night were scattered about the nearby tables. But there wasn't any Jeff and Bonce, and there wasn't any Frenchy... in fact, Gently noticed, there weren't any women in the bar at all, not of any kind.
He went across to the counter and settled himself on a stool, one from the sporty type.
Artie and the latter exchanged a leer, but there was no comment made.
'The usual?' inquired Artie, with a slight sneer in his voice.
Gently quizzed his ferrety features. 'You wouldn't have any milk, by any chance?'
' _Milk!_ ' Artie almost snorted the word. 'There's a milk-bar just down the road!'
'I'm serious... I want some milk.'
Artie eyed him balefully for a moment, then shrugged his shoulders and snatched a glass from under the counter. 'Boss's orders,' he sneered, 'got to treat policemen like gentlemen.' He ducked under the counter and disappeared through the adjacent door.
The sporty type tipped up the remains of his whisky. 'If you're looking for your girly, you won't find her here, guv,' he observed spiritously. 'Louey's had a purge – no women, no kids, and nothing out of line from no one... getting quite pally towards the coppers is Big Louey.'
Gently lifted his eyebrows. 'It's not a bad thing to be in most lines of business... what's yours?'
'What's mine?' The sporty type affected jocularity. 'Ho-ho! I'll keep on drinking what I'm drinking, and thank you very much!'
'I mean your business,' said Gently evenly.
'Oh, me business... I was going to say it was the first time a copper ever asked me... well, there you are! I'm what you might call a Turf Consultant.'
'You mean a tipster?'
'Now guv, when we're trying to add dignity to the profession...'
'And you make a living at it?'
'A bit of that and a bit of working with Louey. You don't run a bookie's business on your own.'
'Well, you seem to do all right at it.'
The sporty type squirmed a little, but was relieved of the necessity of making a reply by the return of Artie with the glass of milk. He slammed it down perilously in front of Gently.
'It's on the house... with Louey's compliments.'
Gently nodded and drank it slowly. He really needed that milk. Its soothing coolness flooded into his digestive chaos like a summons to order, nature's answer to a canteen sausage. He drained the last drop and regarded the filmy glass with a dreamy eye. There were just a few things in life...
'Louey got company?' he asked Artie.
'Nobody who's worried by policemen.'
'Tut, tut, Artie! I'm sure Louey wouldn't approve of that attitude... I was just wondering if he could spare me a few minutes.'
'Why ask?' retorted Artie, 'just walk right in like every other cop.'
Gently shook his head. 'You've got the wrong impression, Artie... you must have been rude to a policeman when you were a little boy.' He slid off the stool and went over to the door. Then he paused, hand on the knob. 'I suppose you didn't have sausages for tea, Artie?'
* * *
Louey's office was a comfortable room which exhibited a good deal of taste and some quiet expense. The walls were papered in two colours, maroon and grey, the floor was completely carpeted in grey to match and the pebble-grained glass windows, being on grey walls, had maroon curtains relieved by hand-blocked designs in dark blue. The furniture was in keeping. It was of discreet contemporary design showing Scandinavian influence. On the walls hung two coloured prints of race-horses after Toulouse-Lautrec, and under one of the windows stood a jardinière of cream wrought-iron containing a pleasant assortment of indoor plants. There was a short passage separating the office from the bar: it had the effect of reducing the canned crooners in the arcade to a distant, refined murmur.
Louey sat sprawled in a chair by his desk when Gently entered. He was nursing a cat on his knees, a black-and-white tom with a blue ribbon round its neck and a purr like an unoccupied buzzsaw. On another chair was seated the parrot-faced man, still garbed in his dubious evening-dress and still armed with his yard of gold-plated cigarette-holder. Louey greeted Gently with a smile from which his gold tooth shone.
'Pleased to see you, Inspector. I was wondering if you would honour us tonight.'
'Indeed? Then I won't be interrupting any business.'
Louey laughed his comfortable laugh and chivvied the tom with a huge hand. 'No business tonight... it's been a bad day for the punters. Not a favourite came home at Wolverhampton. A bad day, eh, Peachey?'
The parrot-faced man mumbled a nervous affirmative. He seemed equally apprehensive of both Gently and Louey. His small pale eyes wandered from one to the other, and he sat in his chair as though it were a penance to him.
'Peachey's my clerk,' explained Louey, seeming to linger on the words, 'he's a good boy... very useful... aren't you, Peachey? _Very_ useful! But sit down, Inspector, make yourself at home... as a matter of fact, we've just been talking about you.'
'Really?'
Louey smiled auriferously. 'The evening papers... probably exaggerated... still, we feel you deserve congratulations. The inspector has got a long way in twenty-four hours, hasn't he, Peachey – eh?'
Gently selected a chair upholstered in blue candy-stripe and swung it round, back to front. Then he seated himself heavily. Louey continued to smile.
'Will you have a drink...? Some more milk, if you prefer it?'
'No, thank you. I'll just smoke.'
Louey swept up a silver box from the desk and inclined his gigantic frame towards Gently.
'Try one of these... Russian. It's a taste I've acquired.'
'Thanks, but I smoke a pipe.'
'You watch your health, Inspector.'
Such a polite and obliging Louey, thought Gently, as he stuffed his pipe-bowl. Who would have expected such polish from the Goliath who had bawled out the bar last night? There seemed to be two of him... one for out there and one for in here, a Jekyll and Hyde Louey. He glanced around the room. Certainly it wasn't furnished by a moron...
'You like my office?' Louey leaned forward again with a lighter.
'It's not the usual sort of bookmaker's office.'
The gold tooth appeared. 'Perhaps I'm not the usual sort of bookmaker... eh? But most of my business is done in the outer office. I keep this one for myself and my friends.'
His eyes met Gently's, frank, steady, even the sinister effect of the fleck in the pupil seeming softened and modified. We are equals, they were trying to say, you are a man like myself: I recognize you. When we talk together there is no need for subterfuge...
'So you don't know that prostitute, Frenchy?' demanded Gently roughly – so roughly, in fact, that Peachey dropped his cigarette brandisher. But the grey eyes remained fixed unwaveringly upon his own.
'I'm afraid not, Inspector... apart from warning her to leave the bar once or twice.'
'Does _he_ know her?' Gently motioned towards Peachey with his head. Louey turned slowly towards his trembling clerk.
'Go on... tell the inspector.'
'I've s-spoken to her once or twice...!' Peachey had a whining, high-pitched voice, oddly reminiscent of Nits.
'Nothing else but that?'
'N-no... honest I haven't! Just in the bar... a joke...'
'You've never seen her with this fellow?' Gently whipped out one of the doctored photographs and shoved it under Peachey's nose. The unhappy clerk shot back a foot in his chair.
'Tell him,' rumbled Louey, 'don't waste the inspector's time.'
'No... n-never... I never seen him at all!'
'Then you know who he is?' snapped Gently.
'I tell you I never seen him!'
'Yet you recognize the photograph?'
'I never... I tell you!'
Louey broke in with his comfortable laugh and reached out a great hand to tilt the photograph in his direction.
'I think he can guess, Inspector... it isn't difficult, with all this talk of beards in the evening papers.'
'I'm asking Peachey!' Gently snatched the photograph out of Louey's fingers. 'You recognized him – didn't you? You didn't have to stop to work it out!'
'It's like Louey says!' burst out Peachey in desperation, 'I read about it in the papers... just like he says!'
Gently eased back in his candy-striped seat and laid the photograph on the corner of the desk. Louey studied it with interest, leaning his massive bald head a little to one side.
'They've touched it up neatly... the beard looks quite convincing.'
Gently felt for his matches but said nothing.
'No doubt he's a foreigner,' mused Louey, 'what part of the world would you say he came from... Inspector?'
Gently shrugged and struck a match.
'Of course, he could be a first-generation American... eh?'
Gently puffed a negative stream of smoke.
'Perhaps not. I've a feeling I'm wrong.'
Gently reached out to drop his match in an ashtray.
'Maybe Central European is nearer... or further east. Behind the Curtain, even?' Louey's eyes drifted slowly back to Gently, strong, assured.
'The Balkans?' suggested Gently quietly.
The grey eyes smiled approval. 'That would be my guess, too. Or perhaps we could be more definite... after all, the cast of feature is very distinctive. Shall we say Bulgarian?'
Gently nodded his mandarin nod.
'And – I think – a cultivated man... possibly Sofia?'
'As you say... possibly.'
Still smiling, Louey fondled the purring tom which continued to loll on his knees. It stretched itself and yawned contentedly. Then it flexed its claws with an exaggerated expression of unconcern, whisked its tail and tucked its head under one of its paws.
'Rain,' said Louey, 'it'll make the going soft... eh, Peachey?'
Peachey was sitting with his mouth open and giving an imitation of someone expecting an atomic bomb to explode.
'Then there's the other one...' murmured Gently, absently blowing a smoke-ring. 'You were saying, Inspector?'
'The man with the scar, doesn't he strike you as belonging to the same racial group?'
There was a pause broken only by the muted skirl of electronic jazz. Louey's fingers paused halfway along the tom's back. Even Gently's smoke-ring seemed to pause and hover, exactly between the three of them.
'Do I... know him, Inspector?' queried Louey in a finely-blended tone of frustrated helpfulness.
'You should do. He was here last night.'
'Last night? You mean here in the bar?'
'I mean here in the office – this one or the outer one.'
There was a further pause while Louey shook his head perplexedly. 'I don't know... it's rather puzzling. I'm afraid I'm not acquainted with a man with a scar – it's a conspicuous scar, I suppose, something that stands out?'
'Very conspicuous.'
'And he was here in the office?'
'He left at nine thirty-one.'
'Someone saw him leave?'
'Exactly.'
Louey looked hopelessly blank. 'If I knew his name, Inspector...'
'I intended to ask you for it.'
Louey sighed regretfully and reached out for the silver cigarette-box. 'He couldn't have been in here... I was here myself the whole evening. And as for the other office—' he hesitated in the act of selecting a primrose-coloured cigarette – 'Peachey!'
Peachey jerked as though yanked by a wire.
' _You_ were in the other office at half past nine... Peachey!'
'B-but boss—!'
'Now no excuses – you were working there till ten – you didn't leave the place except to fetch me something from the bar. He was getting out accounts, Inspector... we do a good deal of postal work.'
'But _boss_!' interrupted the anguished Peachey.
Louey pinned him with an unanswerable eye. 'Who was it, Peachey – who was the man with the scar? The inspector isn't asking these questions out of idle curiosity, you know...'
Poor Peachey gaped and gasped like a hooked cod.
'But wait a minute!' boomed Louey, 'half past nine – that must have been about the time I sent you for my whisky. Inspector' – his eye dropped Peachey as a terrier drops a rat – 'you were in the bar yourself just then, I believe. Did you notice Peachey come out, by any chance?'
Gently nodded reluctantly.
'Of course! Perhaps you can tell us at what time?'
'About half past nine... more or less.'
'Half past nine! Then it seems that Peachey _wasn't_ in the office when this man of yours was alleged to have left. Is that what you wanted to tell me, Peachey – is it?'
Peachey gulped apoplectically. 'That's right, boss! I wasn't there to s-see nobody!'
'And nobody looked in before that... none of our regulars about their accounts?'
'No, boss – no one at all!'
Louey extended a gigantic hand towards Gently. 'Sorry, Inspector... it doesn't look as though we can help much... does it?'
'No,' admitted Gently expressionlessly, 'it doesn't, does it?'
'Of course, this man may have looked in while the office was unoccupied.'
Gently shook his head. 'Let's not bother about that one, shall we?'
The grey eyes smiled approval again and Peachey sagged down into his chair, breathing heavily. Louey lit his cigarette, slowly, thoughtfully.
'You know, I've given this business a certain amount of thought, Inspector... one can't be indifferent, with the Press making so much of it... and there are certain points which seem to stand out.'
Gently hoisted an inquiring eyebrow, but said nothing.
'I admit in advance that I'm the merest amateur... naturally! But it's just possible that being outside it, away from the... tactical problems?... I'm in a more favourable position to study the strategy.'
'Go on,' grunted Gently.
Louey inhaled deeply and raised his head to blow smoke above Gently's face.
'There's this man... what is he doing here? A complete stranger – nobody knows him – the police don't know him (at least, I presume they don't?) – turning up one day at a popular English seaside resort – and disguised. What would bring him here? His motive is past guessing at. Why should anybody kill him when he got here? The motive is just as obscure.'
'Robbery,' suggested Gently, puffing some Navy Cut into a haze of Russian.
'Robbery?' The gold tooth showed lazily for a moment. 'You're forgetting, Inspector, he was reported to have been killed in cold blood. His hands were tied. Does that seem like robbery?'
'It seems like more than one person being involved.'
'Exactly... and that's my point! It wasn't the crime of an individual. All the facts are against it. The more you juggle with them, the more emphatic they become. It was an organized killing, an act carried out by a group of some description... who knows?'
The grey eyes slid up and fastened on Gently's, holding him, commanding him.
'A political killing, Inspector. The execution of a traitor... that's my reading of the situation. Your man was a fugitive. He chose Starmouth for his haven. But the organization he had betrayed found him out and exacted justice... doesn't that seem to fit what we know?'
Gently blew an exquisite ring.
'I think it does... better than any other interpretation. I hope I'm wrong – for your sake, Inspector. I believe these political killings are planned with a care which makes detection onerous and arrests unlikely. But the odds seem to lie that way... at least to my amateur way of thinking.'
The smile strayed back into the magnetic eyes and Louey part snuffed, part sucked a tremendous inhalation of smoke.
'I'd like you to know I appreciate your difficulties,' he concluded, spilling smoke as he talked. 'My admiration for your abilities won't be lessened, Inspector... what can be done by the police in these cases I am sure you will do.'
Gently nodded towards a peak in Darien. Then he reached for the photograph, pulled out his pen and drew on the matt surface a clumsy circle divided by a line. Without looking he handed it to Louey. The big man took it and stared at it.
'Is this something I should know about?' he inquired softly.
Gently lofted a careless shoulder. 'You were wearing it on your ring last night.'
' _My_ ring?' Louey extended his hand to display his solitaire.
'The one you were wearing last night.'
Louey hesitated a split second and then laughed. 'No, Inspector, you are mistaken... this is the only ring I wear. Tell him, Peachey, tell him... I wear this diamond to impress the clients... eh?'
The miserable Peachey contrived to nod.
'They like to do business with a man of substance... it's paid for itself over and over again.'
Gently turned towards him. There was a glint of excitement in the masterful, smiling eyes.
'So you see, you were mistaken, Inspector... you do see that, don't you?'
'Yes,' murmured Gently, 'I see it very plainly indeed.'
He didn't have far to go outside before he was joined by Dutt. The sergeant's cockney visage had a glum expression which told Gently all there was to know...
'No pigeon, Dutt... the dovecote was empty.'
'That's right, sir. Not a flipping feather.'
'I got the impression it might be. Everyone was so pleased to see me. A pity, Dutt. I get more and more interested in that laddie.'
'We could put out a portrait parley, sir. He shouldn't be difficult to pick up.'
'I wonder, Dutt. My feeling is that he's a bit of a traveller... it's the docks and airports that'll need an eye kept on them. On the other hand...'
'Yessir?'
'If he's the bird I think him, it's a matter of some curiosity why he's hung around here so long already.'
'You mean you know who he is, sir?'
'I wouldn't put my hand to my heart, Dutt. I'm of a suspicious character, like all good policemen. And then again... it doesn't do to overestimate. There's one thing, though: I want a sound sure ruling on the origin of that circle with a line through it.'
'You mean that little charm, sir?' queried Dutt, brightening.
'I do indeed, Dutt – that little charm.'
'Well, sir, I can tell you that right off the cuff... it came to me as I was standing there watching, sir. I knew I'd seen it before, like I said when we found it.'
'Go on, man... stop beating about the bush!'
'It's the sign of the TSK Party, sir – I come across it when I was attached to the Special.'
Gently halted under the blaze of one of the multicoloured standards that afforested the Front. 'And what,' he inquired, 'do we know about TSK Parties, Dutt?'
'Not a darn sight, sir,' replied Dutt, 'not if you put it like that. It's a sort of Bolshie outfit – they reckoned it picked up where the old Bolshie boys left off. They didn't even know wevver Joe was backing it or not – sort of freelance it was, if you get me. That Navy sabotage business was TSK. We had some US Federal men attached to us – they've had a lot of trouble with them in the States.'
'The States!' echoed Gently, 'It's always the States. Have you noticed, Dutt, how the American eagle keeps worrying us as we go about our quiet Central Office occasions?'
CHAPTER SEVEN
THE FLAP WAS still on at headquarters, in fact it had stepped up considerably during Gently's absence. There were lights on where they were usually off at such an hour, cars parked that ought to have been garaged and policemen due off duty, still buzzing around like (as Dutt rather coarsely described it) 'blue-arsed flies'. Gently, going down the corridor, was nearly bowled over by an impetuous Copping clutching a file.
'We've picked up the boyo who passed that note!' exclaimed the Borough Police maestro, sorting himself out. 'He's a skipper from up north – he's lousy with them – and what a yarn he's spun! They must think we're cracked, trying to pull gags like that. But the super'll give him a going over he won't forget in a hurry!'
Gently sniffed a little peevishly. 'Don't think I'm frivolous... I'm still trying to keep my mind on the crime before the last. Did your man get some prints?'
'Oh, the prints! He got a couple of sets that didn't tally with anything we've got.'
'A couple?'
'That's right... one lot on the suitcase and one on the window-frame. They turned up in other places, too, but those were the best impressions.'
'He compared them with Mrs W's and the rest, of course?'
'We know a little bit about the job...!'
'And you've sent them to town?'
'Right away, as per instructions.'
Gently fished out his wallet and extracted from it the doctored photograph. 'I want this printed now... is your man still around? He'll find mine on it amongst some others, but he needn't bother about them...'
Dutt was despatched with the photograph and Gently accompanied Copping to the super's office. That austere abode, always impressive, was now fairly crackling with forensic atmosphere. The super sat behind his desk as stiff as a ramrod. At a discreet distance a sergeant was ensconced at a table, taking down some details. At the same table sat a constable with a shorthand book and three pencils. On the door was a second constable, uneasily at ease. The focus of all this talent, a fresh-complexioned middle-aged man, had been arranged on a chair in the geometrical centre of the office: he sat there with a nervous awkwardness, like a member of an audience suddenly hoicked up on to the stage.
The super nodded to Gently as the latter entered and motioned him to take a seat. 'You'll excuse me, Inspector... I'm rather busy. I'd like a conference with you later, if you don't mind waiting.'
Gently inclined his head and sat down at the less congested end of the office. Copping delivered the file and appended himself to the end of the super's desk.
'Dalhoosie Road,' spelled out the sergeant. 'McKinky & Mucklebrowse Ltd, Potleekie Street, Frazerburgh. I think that's the lot, sir. It checks with the ship's papers.'
The super stiffened himself a few more degrees. 'Now, McParsons... I want you to listen very carefully to what I have to say. I'm charging you with being in possession and uttering a counterfeit United States banknote, and also with being in possession of four similar notes. Do you wish to say anything in answer to this charge? You are not obliged to say anything unless you wish to do so, but whatever you say will be taken down in writing and may be used in evidence.'
McParsons screwed up his weather-beaten face. 'But I tellt yer the whorl _lot_ , sir – I gi'ed ye all the evidence to prove I'm an honest man... what more do yer want noo?'
'It isn't evidence,' snapped the super, 'we didn't take it down and we're prepared to forget it. Think carefully, McParsons. You're in trouble, quite a lot of trouble, and the tale you told me down at the docks won't impress a jury – I can assure you of that. My advice to you is to forget it. The truth will help you a lot more, especially if it enables us to arrest the counterfeiters.'
'But losh, man, it _was_ the truth! I canna make up tales out of my heid.'
'Stop!' interrupted the super sharply. 'All you say now is evidence.'
'Then Gordamighty, let it be so – I'll noo complain o' ye puitin fause words into my mouth. It's jist the way I tellt it, nae more and nae less, so yer may as well scratch it doon on yer paper – it's all the evidence Andy McParsons can gi' ye.'
The super drilled at the same Andy McParsons for ten acetylene-edged seconds before replying... quite a feat, thought Gently, who was a connoisseur of superlatives. Then he snapped off a 'Right!' which seemed to suggest every bit of ten years and opened the file Copping had brought. The pages rustled accusingly.
'Starmouth Branch of the City & Provincial Bank... US banknote of one hundred dollar denomination, etc, etc... paid in by Joseph William Hackett, licensee of the "Ocean Sun"... see preceding report. Hackett on being questioned deposed that he changed the note for a seaman, a stranger to him... sparely built man, about five feet ten, aged about fifty, dressed in navy-blue suit and cap, fresh complexion etc... Scots accent. Detective Sergeant Haynes questioned Andrew Carnegie McParsons, Skipper of the steam-drifter _Harvest_ _Sea_ , at the yard of Wylie-Marine, where the said steam-drifter was undergoing a refit... denied all knowledge, etc—'
Gently coughed loudly and the super broke off to throw him a sharp stare. 'You had something to say, Inspector?'
'The name of the yard,' murmured Gently apologetically, 'could you repeat it, please?'
'Wylie-Marine, Inspector.'
'Thank you. I thought it sounded familiar.'
The super snorted and returned to his recitation.
'Afternoon of the fifteenth Hackett reported having seen aforementioned seaman in the neighbourhood of the yard of Wylie-Marine... proceeded to the same yard... Hackett picked out McParsons... McParsons admitted changing the note and was taken into custody... four similar notes of one hundred dollar denomination found in McParsons's possession.'
The super paused again and smoothed out the nicely typed report sheet.
' _Now_ ,' he said bitingly, 'we come to your story, McParsons.'
'But ye've had it a'ready,' replied the disconsolate skipper, 'hoo often maun I tell it to yer?'
'What you told us before you were charged will not be used as evidence. If you want to make a statement, now is the time.'
'Och, aye... ye're all for doing it by the buik, I ken that. Well, jist pit doon I had the notes fra ain Amurrican body... I see fine yer dinna believe a word of it.'
The super signalled to the shorthand constable. 'Begin at the beginning, McParsons. If this story of yours is to go on record we want the whole of it.'
McParsons sighed feelingly to himself. 'Aweel... ye'll have your way, there's noo doot. It was on the Tuesday then, the Tuesday last but one... we'd been in Hull a week, y'ken, wi' the boiler puffin' oot steam fra every crook an cranny... the engineer had puit in his report lang since, but auld Mucklebrowse is awfu' canny aboot runnin' up bills for repair... then awa' comes a wire to the agent tellin' us to puit out for Wylie's, me ainself to stay wi' the ship and the crew to take train back to Frazer. Sae we jist tuik aboard ain or twa necessaries and hung waitin' there for the evenin' tide. Noo the crew bodies was all ashore takin' their wee drap for the trip and Andy McParsons had jist come awa' fra the agent's, when along happens this Amurrican I tellt ye of... "Captain," says he (and morst respectful, the de'il take him!), "is that your ain ship lyin' there with steam up?" "It is," says I, "sae long as the rivets stick in the boiler." "Then ye're aboot goin' to sea," says he. "Aye," says I, "jist as soon as the laddies get back, which'll noo be a great while." "And you'll be goin' a long trip?" says he, gi'en a luik ower his shoulder. "Jist drappin' down the coast," says I, "we'll be sittin' tight in Starmouth before breakfast-time."
'Noo ye maun believe this, Supereentendent, or ye maun not – it's a' ain to the truth – but I hadna been gabbin' five minutes with this smooth-spaken cheil when he was jawin' me into stowin' him awa' in the _Harvest Sea._ "But wit's the trouble?" says I, "is it the police ye have stuck on yer sternsides?" "Naethin' of that, I swear," says he, "it's a private matter, an like to be the dearth of me if I canna get clear of this dock wi'out walkin' back off it. I'll pay ye," says he, "it's noo a question of money – but for the luve of the A'mighty let's gae doon into the cabin," and the puir loon luikit sae anxious I hadna the heart tae refuse.
'Weel, the short and the lang o't was we struck a bargain – twa hundred dollars and nae questions asked. I couldna take less, says I, since the crew maun be squared on tap, and in ony case it was a wee bit inconvenient tae get it in dollar notes, and sich big ains at that. "Och, but the crew mauna ken!" says he, "ain body's ower muckle – I canna bide more." "Then I doot the deal's off," says I, "for de'il a bit can ye be stowed awa' in sich a corckle-shell as this wi'out the crew being privy, not," says I, "unless we pop yer into a herring-bunker, where ye'll be wantin' a stomach lined wi' galvanized sheet to say the least o't." "Let it be sae," says he, "I've sleepit in places as bad or worrse." "Mon," says I, "if ye've nae been jowed around in a herring-bunker on the North Sea ye havena lived up till noo, sae dinna gae boastin'. Take yer ease in the cabin, where yell nose a' the fish ye'll be wantin' if there's a wee swell ootbye."
'But listen he wouldna, sae it was agreed he should ship in a bunker – though had I kent then whit I ken noo it'd been into the dorck wi' him, and nae mair argument – and he paid up his twa hundred dollars... not mentioning some wee discount business on three ither notes (I'd ta'n a bodle o' cash fra the agent and it rubbed against the grain tae say nay, ye understand). "Keep an eye lifted for strangers," says he, as I clappit him doon under the hatch, "dinna let a soul aboard ither than the crew bodies." "Dinna fash yersel,"says I, "I ken fine how to earn twa hundred dollars."
'Weel, there ye have it, Supereentendent. We drappit down here owernight and fetchit up at Wylie's before the toon was astir. I paid aff the crew bodies and saw them awa' to the station, then I lifted the hatch and huiked out the cargo. He wasna in the best o' shape, ye ken – it gi'es me a deal o' consolation thinkin' o't – but I gar him ha' a wash, whilk he did, and a swig at the borttle, whilk he didna, and betwixt doin' the ain and not doin' t'ither he was sune on his legs agin and marchin' off doon the quay. And that's the spae, evidence or testament of Andy Carnegie McParsons, the truth of whilk is kenned by him on the ain part and his creator in pairpetuity, whatever doots may occur in the more limited minds of his accusers.'
Saying which he folded his arms independently and returned the super some measure of that worthy's police-issue stare.
'And you expect to have this colourful account believed?' fired the latter corrosively.
'Och, noo! It's naethin' but the naked truth,' returned the Scot ironically, 'I dinna expec' the police to believe sich simple things.'
'I see nothing simple about it, McParsons. It has all the marks of being deliberately contrived. First this hypothetical American meets you just as you're about to put to sea – and when you're alone. Then, for reasons the most vague, he elects to spend a night on the North Sea immured in a herring-bunker rather than show himself to the crew. And finally he takes his leave when, once more, there are no witnesses. It's pretty thin, McParsons. It'll be cut to ribbons in court. If I were you I'd stop trying to shield whoever it is behind this racket and try to be helpful – we shall get them in the long run, you can depend on that.'
'Then for Gord's sake get them, Supereentendent, and dinna waste any mair time! Ye're noo the ain half sae anxious aboot it as I am sittin' here.'
'So you're sticking to your story?'
'Aye – onless ye can puit me up tae some lees whilk will suit yer better.'
The super glanced down at the file with something which might have been a low sigh. 'Very well,' he said dangerously, 'if you insist on having it that way... describe the man!'
'The Amurrican body?'
'Precisely, McParsons.'
'Weel, I doot I'm noo a policeman to be forever noticin' the crinks and crankles o' folk...'
The super snorted. 'Don't strain your imagination.'
'I willna, Supereentendent... it's me memory I'm jowin' the noo.'
'For instance... was he clean-shaven?' mumbled Gently, apparently studying his stubby fingernails. The Scot turned quickly towards him.
'Noo yer mention it, he wasna – he had a beard fra the temples doon.'
'He would have, wouldn't he?' demanded the super derisively.
'And his suit... Scots tweed?' suggested Gently.
'Na, man, it was ain o' they Yanky-doodle jobs, a' tap and noo bottom.'
'Dark?'
'Na... aboot the colour o' pipe-ash.'
'He was a youngish man?'
'Ower forty, ain or twa.'
'And he spoke with an educated accent?'
'Noo this cheil – he was Amurrican by adoption, ye ken... he spoke a fair smatterin' o' Sassenach, but he hadna it fra his mither.'
Gently felt once more in his breast-pocket for one of his doctored prints.
'Had he a beard like this one?'
McParsons rose excitedly to his feet. 'But yon's the man – the verra spittin' image! Sae ye kent him – ye kent him a' the while – it's jist a try-on, a' this chargin' and fulin' – ye've got yer hands on him a' the while!'
Gently's gaze strayed mildly to the thunderstruck super. 'I'd like to get Hull on the wire... it may be a longish call.' He turned back to McParsons. 'You wouldn't remember what ships docked at Hull on that Tuesday... from the continent, say?'
'Fra the Continent? Och aye! There was that Porlish ship they made a' the fuss aboot aince – we ganged roon to ha' a luik at her. But concairnin' the body on yon photygraph—!'
'Thank you, Skipper,' murmured Gently distantly, 'the body on the photograph is undoubtedly your next port of call.'
They were obliging, the Hull City Police, without being able to do much more than fill in a few details. They knocked up numerous people (including constables) from the first and important hours of their slumber. No, they had no record of a man of Max's description. No, their life was not being blighted by an irruption of counterfeit hundred-dollar notes. Yes, the Polish liner _Ortory_ had broken her Danzig-New York run at Hull on the Tuesday week last. She had docked at noon and sailed again at 19.30 hours: she had discharged seventy-five crates of Russian canned salmon and picked up a Finnish trade delegation on its way to Washington. Yes, they would get on to the dock police if Gently would hang on for a while.
'So he was a Pole, was he?' brooded the super, sniffing meanly at the Navy Cut contaminating the aseptic night air of his office.
Gently shook his head. 'A Bulgar from Sofia.'
'You know that for a fact?'
'Not really... but I'm prepared to accept it as a working hypothesis.'
'How do you mean?'
'Just a hunch. I don't think someone I know could bear to tell a lie about it... provided he wasn't implicating himself.'
'And who is this someone?'
'Oh, it's a bit vague at the moment...' returned Gently evasively.
The super grunted and toyed with a retractable ball-point which seemed to be a novelty with him. 'So he was a member of this TSK... they were sending him to the States guyed up like a Yank and loaded with counterfeit... is that the angle?'
Gently nodded through his smoke.
'What was he supposed to do when he got there?'
'Oh... they'd have put him ashore quietly before the ship docked.'
'And then?'
Gently shrugged. 'Sabotage seems to be their line... he was probably going over to organize it.'
'He must have been well up in the party,' mused the super, 'it was a position of trust... what do you suppose went wrong?'
'That's something we're not likely to know.'
'A double-cross inside the party, maybe.'
'You're probably safe in saying that...'
There was a dulled, small-hour silence broken only by a scratching in the uncoupled phone and a sizzle from Gently's pipe. From the nearby harbour came the mournfully alert toot of a siren, twice repeated.
'Of course you'll get on to the Special,' muttered the super drowsily.
'Dutt's getting them for me... he was attached to them a time back.'
'They may know something... then there's the US Federal... could be something they're looking for.' The super jerked himself to attention. 'Look here... there's something that puzzles me. If this fellow was so worried about his health, why didn't he seek political asylum when he skipped the _Ortory?_ That would have been his obvious move. There was no need for all this chasing around and stowing-away aboard fishing-boats.'
Gently gave himself a little shake. 'There's the missing suitcase... if it were stuffed with hundred-dollar bills it seems a fairish reason for keeping things private.'
'But they were counterfeit!'
'He may not have known that.'
'You mean his party sent him off on this mission without telling him?'
'It would seem to square with what we know about the methods of these parties...'
The super nodded sapiently. 'But the person who swiped that suitcase must have known they were phoney, because he hasn't been passing them.'
'You can't bank on that either... the TSK weren't planning to spend them in Starmouth. What puzzles me is the way that bedroom was frisked. You don't have to tear a bedroom apart to find a suitcase...'
They were interrupted by the entry of a constable with a tray from the canteen. It bore a plate of corned-beef sandwiches and two mugs of hot coffee. Gently gladly grounded his pipe in favour of the more substantial fare – there was an almost psychic quality about corned-beef sandwiches and hot coffee at that hour of the morning. He chewed and swilled largely, and the super kept in strict step with him.
'May have hidden the stuff about the room,' mumbled the super, flipping a crumb from his moustache.
'Then why was he always carting the suitcase about with him? Everyone's agreed about that.'
'Could have been a blind.'
'Why should he bother?... the stuff would be safer by him.'
'He seems to have left it behind in the last instance, at all events,' grunted the super beefily.
'There may have been a purely incidental reason for that...'
Dutt came in, looking peeked and heavy-eyed. 'Special is going into it, sir,' he said laconically. 'I gave them a p.p. as good as I could remember and all the information we've got to date.'
'What did they say?' asked Gently, shoving him a charitable sandwich.
'Nothink, sir. They never does.'
'Did they confirm the identity of the charm?'
'Only after I'd got on to 'em, sir, and told them it was hanging up the case. You never knew such a lot for keeping their traps shut.'
Gently drank the last of his coffee and looked sadly into the empty mug before returning it to the tray. 'Maybe they don't know much... maybe they aren't going to until the day-shift turns up. Did Sergeant Dack get any results with that photograph?'
'Yessir. A lot of beautiful prints.'
'Any on record?'
'He thought there was, sir – would've sworn blind about one lot. He said they matched up with the prints of a con man who specialized in flogging licences to manufacture Starmouth Rock.'
'And did they, Dutt?'
'No, sir. They was yours.'
Gently shook his head modestly. 'You compared them with the ones out of the bedroom?'
'Yessir. No resemblance.'
'And sent a set off to town?'
'Automatic, sir.'
'Have another sandwich, Dutt.'
'Thank you, sir... this night work makes you peckish.'
The telephone scratched its gritty throat and began to emit adenoidal language. Gently picked it up and murmured kindly to it. The dock police had been roused and briefed. They had pulled in, or rather out, the two men who had been on duty at the pier where the _Ortory_ had docked on the day in question. No. 1 was applied to the line and upon invitation gave an efficient description of what occurred.
'And no civilian disembarked from the time she docked to the time you went off duty at five?' queried Gently encouragingly.
'Only one, sir, and he came down with three or four of the ship's officers... they seemed to be inspecting the cases of salmon which had been unloaded.'
'The salmon? Would that have been unloaded by the ship's crew?'
'Yes, sir, it was in this instance.'
'Down a separate gangway?'
'That's right, sir.'
'And loaded on to trucks?'
'No, sir, not directly. They built it up on a pile on the pier and it wasn't till the evening when it was taken away.'
('That's it!' whispered the super, listening on an extension, 'he bribed the sailors to get him off... they built a hollow pile for him to hide in.')
'This civilian who came to inspect the cases... when did he come ashore?'
'Just before I was relieved, sir.'
'What do you mean by "inspected"?'
'Well, sir, they appeared to be counting them... they got one or two off the top to see how many were underneath.'
'You noticed nothing unusual take place?'
'No, sir. They just did their check and then stood about talking and looking about them for a minute or two. After that they strolled up the pier to the office and went inside.'
'The civilian too?'
'Yes, sir, the civilian and the officers.'
'Can you describe the civilian?'
'Middle-aged, about five-nine, medium-build, dark, dark-eyed, slanting brows, long, straight nose, small mouth, rather harsh voice.'
'Distinguishing marks?'
'I thought he had a scar on one side of his face, sir, but I only caught a glimpse of it as he came down the gangway. The rest of the time it was turned away from me.'
'Ah!' breathed Gently and propped himself up at a better functional angle with the super's desk. 'Now... this is important... did the civilian return on board with the officers?'
'I don't know, sir. My relief came just then and I went off duty. He's in the office now, sir, if you'd like to speak to him.'
There were some confused ringing sounds at the other end and No. 2 took over. Gently repeated his question.
'Well, sir... I regret to say I didn't notice.'
'Didn't notice? Didn't the other fellow tell you there was a civilian ashore?'
'Oh yes, sir, he did. But soon after I got on the pier there was a row amongst some of the Polish seamen and it sort of took my mind off the others.'
'What sort of a row was that?'
'I don't know what it was about, sir. Half a dozen of them came ashore and started shifting some of the cases that had been unloaded. Then all of a sudden a row broke out and a couple of them started a fight. I went up and separated them, but they kept on shouting at each other and making as though they'd let fly again, so I had to stand by and keep an eye on them. In the end one of their officers came up and sent them on board again.'
'And during that little diversion the party in the pier office slipped aboard?'
'I suppose they must have done, sir... they weren't there when I checked up later.'
'So if the civilian stayed ashore you wouldn't have noticed?'
'I'm afraid not, sir... I'm very sorry...'
('Cunning lot of bastards!' interjected the super with reluctant admiration, 'you can see they're professionals!')
Gently took in a few more inches of desktop. 'Give me the other bloke again,' he said. The other bloke was given him. 'What else was going on at the pier while the _Ortory_ was there?'
'What sort of thing, sir?'
'Any loading or unloading going on?'
'There was a Swedish vessel unloading timber on the other side, sir.'
'And that meant a bit of traffic up and down the pier?'
'Quite a bit, sir. They were trucking some of it.'
'Was it going past the pile of cases from the _Ortory?'_
'Yes, sir, just behind it. Some of the trucks parked there to wait their turn.'
Gently nodded towards the slow-mantling dawn. 'And the Finnish Delegation?' he asked, 'what time did that embark?'
'Just after lunch, sir... might have been half past two.'
They sat drinking a final mug of coffee with the electric light growing thin and fey under its regulation shade. The super was looking sleepily pleased with himself, as though he felt he had a good case to go before the ratepayers, both in forgery and homicide. After all, nobody could hang Special Branch business round his neck... concern he might show, when secret agents bumped each other off on Starmouth Sands, but he was only nominally responsible...
'I suppose the bloke who did it is miles away by now,' he murmured into his coffee. 'If he shows the same ingenuity getting out of this country as he did getting into it...'
Gently shrugged slightly, but he didn't seem to be listening.
'And even if they get him I don't suppose we can make a murder rap stick...'
There was a tap on the door and the duty sergeant entered.
'Excuse me, sir,' he said to the super, 'but PC Timms has just turned in this here. It was given to him by the publican of the "Southend Smack". He changed it for a Teddy boy in his bar last night, but later on somebody tells him about some duff ones going about, so he's handed it in to be on the safe side.'
The super extended a nerveless hand. The duty sergeant placed therein a certain bill or note. And from an unexpected backyard at no great distance a cock crowed.
CHAPTER EIGHT
SUNDAY SUN FALLING steadily on the platinum beaches, on the lazy combers, on the strangely subdued streets. On the well-spaced, comely mansions of High Town. On the quaint, huddled rookeries of the Grids. On the highly-polished bonnet of a police Wolseley as it halted on the crisp gravel of Christopher Wylie's retired drive. On the more sober bonnet of PC Atkins as he knocked on the door of No. 17 Kittle Witches Grid.
'I knew he won't come to no good, that kid of Baines's,' said a frowsy matron to the newspaperman as they watched a goggle-eyed Bonce being marched away. 'I said so as soon as I saw him in that fancy get-up of his. Did you ever see such frights as they look? And then for him to be mixing with that young Wylie... I said it would be his ruination.'
'Going about the town at all hours and taking up with all sorts,' said the cook at Wylie's, relinquishing her vantage-point at the larder window, 'they should've let _me_ had the handling of Master Jeff – I'd have let him mix with riff-raff like the Baineses, _I_ would!'
'I dunno,' returned the kitchen-maid dreamily, 'I rather _liked_ him in that silly suit of his.'
The cook snorted. 'Well, you can see where it's got him now, my girl!'
In the ill-lit parlour of No. 17 John George Baines, dock labourer, sat in his shirt-sleeves staring sullenly at the _News of the World._ His wife, a bold-faced woman, was slapping together the breakfast plates at a table covered with oil-cloth and two juvenile Baineses were scuffling and screaming on the floor.
'It wouldn't have happened,' snapped Mrs Baines for the twentieth time, 'it wouldn't have happened, not if you'd kept a proper hand on him...!'
'Oh, shut your mouth, woman... it's your fault if it's anyone's.'
'You've never give him a good hiding in your life!'
'And who was it encouraged him with that bloody suit – trying to be up to His Nibs...?'
More silent was the breakfast-room in High Town. No sound fell upon the ears of Christopher Wylie, except the sobbing of his wife Cora. He stood with his back to her, staring out of the expensive oriel window, staring at his cypress and monkey-puzzle trees, his impeccable gravel drive.
'I'll get on to the chief constable,' he muttered at last, 'we'll get it straightened out, Cora... there can't be anything in it.'
'Oh, Chris... I'm so frightened... so frightened!'
'It's all a mistake... we'll get it straightened out. The lad's due for his service in October...'
Up the long High Street marched PC Atkins, the Sunday-silent High Street with its newspaper-men, milkmen and a few early-stirring visitors in holiday attire. Beside him slouched Bonce, looking neither to right nor left. Behind him frisked Nits, a chattering, excited Nits. Halfway along the High Street PC Atkins paused to address the ragged idiot. 'You run along home, m'lad, and stop making a nuisance of yourself... off with you now, off with you!' Nits backed away apprehensively while the constable's eyes were on him, but as soon as the march recommenced he was dancing along in the rear again...
The sunshine had renewed Gently's feeling of nostalgia. They had all been sunny days, on that holiday of long ago. He remembered getting sunburned and his nose peeling, and the peculiarly pungent lotion they had put on his arms to stop them blistering (though of course they did blister), and, by association the suave smell of the oiled-paper sunshades which had been fashionable about then.
'We had rooms somewhere about where we've got them now,' he confided to a bleary-eyed Dutt as they set out for headquarters. 'They used to do you awfully well in those days... I can remember having chops at breakfast.'
'Don't know as I should think so much of that, sir,' admitted Dutt honestly.
'Nonsense! You've been having these degenerate meals of bacon-and-egg too long.'
'I should think a chop sits a bit heavy on your stomach first thing, sir.'
'It's true I was only a boy, Dutt... all the same, I think I could still face one.' He plodded along silently for a space, a little frown gathered on his brow. 'We seemed to be younger in those days, Dutt...'
' _Younger_ , sir?' inquired Dutt in surprise.
'Yes, Dutt... younger.'
'Well, sir, I s'pose we was – in those days!'
But there was no smile on the face of his superior as they turned up the steps at headquarters.
The landlord of the Southend Smack was waiting patiently in the office which the super had assigned to Gently, and Copping, who had got to bed earlier than most, and was consequently his old spry self, officiously performed the introduction.
'You think you can remember the youth who changed the note?' inquired Gently dryly.
'Ho yes, sir – don't you worry about that!' replied the landlord, a red-faced beery individual called Biggers.
'You've seen him before, then?'
'Ah, I have – once or twice.'
'You know his name?'
'No. No, sir. But he's been in the bar once or twice, I can tell you that.'
'It didn't occur to you that he might be a little young to be served in a bar?'
'W'no, sir... I mean... there you are!' Biggers faltered uneasily, beginning to catch on that he wasn't Gently's blue-eyed boy. 'He _looked_ old enough, sir... couldn't be far off. You can't ask all of them to pull out their birth-certificates.'
'Was he on his own?'
'Ho yes, sir!'
'Does he always come into your bar on his own?'
'Y-yes, sir, as far as I remember.'
'How do you mean, as far as you remember?'
'Well, sir... I wouldn't like to swear he never had no one with him.'
'A woman, perhaps.'
'No, sir – no women!'
'Another youngster dressed like himself?'
'Yes, sir, that's it!'
'Dressed exactly like himself?'
'Yes, sir, exactly!'
'And younger – about a year?'
'Yes, sir... I mean...!' Biggers trailed away, realizing the trap into which he had been unceremoniously precipitated. Gently eyed him with contempt.
'This hundred-dollar bill... didn't it seem odd to you that a young fellow should have one in his possession?'
'Oh, I dunno, sir... what with the Yanks about and all...'
'And how should he have acquired it from an American?'
'Well, sir, they're master men for playing dice.'
'You thought he'd won it gambling?'
'I never really thought... that's the truth!'
'Good,' retorted Gently freezingly, 'I'm glad it's the truth, Biggers. The truth is what we are primarily interested in... let's try sticking to it, shall we? How much did you give him for it?'
'I... I give him its value.'
'How much?'
'Why, all it was worth to me...'
' _How much?_ '
Biggers halted sulkily. 'I give him a tenner... now turn round and tell me it wasn't enough, when it was a dud note in the first place!'
Gently turned his back on the sweating publican. 'Is the parade lined up?' he asked Copping.
'They're in the yard – just give me a moment.'
It was a scrupulously fair parade. Copping had wanted to impress Gently by his handling of it, and after witnessing the momentary appearance of the mailed hand lurking beneath the chief inspector's velvet glove he was glad that he had so wanted. There was something almost deceitful about Gently, he thought...
Biggers took his time in going down the line, as though wishing to display his helpful care and attention. He paused before several law-abiding youths before making his final selection. He also paused before Bonce, whose wild-eyed guilt proclaimed itself to high heaven, but the pause was a brief one and might even have been involuntary... Having done his conscientious best, he carried his findings to Gently.
'That's him... fifth from the far end... kid in the brown suit.'
Gently nodded briefly. 'And this one... the carroty-headed boy?'
'No, sir. Don't know him. Never seen him before!'
'Positive?'
'Ho yes, sir... I never forgets a face.'
The same mailed hand which Copping had so judiciously observed fell lightly on Bigger's arm and the astonished publican found himself whirled a matter of three yards in a direction not of his choosing.
'Now see here, Biggers, you've come forward voluntarily and given us some useful information, but there's not much doubt that you're sailing a bit too close to the wind. From now on there'll be an eye on you, so watch your step. Don't change any more money, American or otherwise, and if any of your customers looks a day under fifty – ask for his birth certificate. Is that clear?'
'Y-yes, sir!'
'Quite clear?'
Biggers gulped assent.
'Then get away out of here... we've finished with you – for the moment!'
A blue-bottle buzzed in a sunny pane of the office window, a casual, preoccupied buzzing which focussed and concentrated in itself a vision of all fine Sundays from time immemorial. Copping lifted the bottom of the window and let it out. It fizzed skywards in a fine frenzy of indignant release, wavered, scented a canteen dustbin and toppled down again from the height of its Homeric disdain. Copping left the window half-open.
'One at a time?' he asked.
'Yes. Shove the Baines boy into a room by himself where he can do a little quiet thinking.'
Copping nodded and went out. Gently seated himself in awful state behind the bleak steel desk with its virgin blotter, jotting-pad and desk-set. He slid open a drawer. It contained a well-thumbed copy of Moriarty's _Police_ _Law_ and some paper-clips. The drawer on the other side contained nothing but ink-stains and punch confetti.
'I wonder who the super turfed out to make room for us?' he mused to Dutt.
Copping returned, prodding Jeff before him. The Teddy boy looked a good deal less exotic in his quieter lounge-suit, but there was still plenty of swagger about him. He stared round him with a sullen defiance, his thin-lipped mouth set tight and trapped.
'Sit down,' said Gently, indicating a chair placed in front but a little to the side of the desk. Jeff sat as though he were conferring a favour. Copping took the chair on the other side and Dutt hovered respectfully in the background.
'Your full name and address?'
'You know that already—'
'Answer the inspector!' snapped Copping.
Jeff glared at him and clenched his hands. 'Jeffery Wylie, Manor House, High Town.'
'Your full name, please.'
'Jeffery... Algernon.'
Gently wrote it down on his jotter.
'Now, Wylie... you had better understand that you are here on a very serious matter, perhaps more serious than you at first supposed. You have been identified as possessing and uttering counterfeit United States currency – wait a minute!' he exclaimed, as Jeff tried to interrupt, 'You'll have plenty of opportunity to have your say – you've been identified as handling this money and we happen to know the source from which it emanated. Now what I have to say to you is this: you may be able to explain satisfactorily how you came to be in possession of that note, in which case there will be no charge made against you. But you are not obliged to give an explanation and you are not advised to if you think it may implicate you in a graver charge. At the same time, if you take the latter course I shall automatically charge you and you will be held in custody on that charge while further investigations are made. Is the situation quite plain to you?'
Jeff shuffled his feet. 'I can see you're out to get me, one way or the other...'
'We're not out to "get" anyone, Wylie, if they happen to be innocent. I'm simply warning you of where you stand. And I'd like to add to that some advice if you help us you'll be helping yourself. But it's up to you entirely. Nobody here is going to use third-degree methods.'
The Teddy boy sniffed derisively and stuck his hands into his pockets. 'I know how you get people to say what you want... I've heard what goes on.'
'Then you'd better forget what you've heard and consider your own position.'
'A fat lot of good that'll do me...'
'It'll do you more good than trying to be clever with policemen.'
'You say yourself I don't have to tell you anything.'
There was a silence during which Copping, to judge from his expression, was meditating a modified use of the third-degree methods which Gently had disowned.
'It's only his word against mine...' began Jeff at last.
Gently cocked an eyebrow. 'Whose word?'
'His – the pub-keeper's.'
'And who told you he was a publican?'
Jeff flushed. 'Isn't that what he looked like?'
'He may have looked like a publican or he may have looked like a barman. What made you think he was one and not the other?'
'I just said the first thing that came into my head, that's what I did!'
Gently nodded a mandarin nod but said nothing.
'He could have been wrong,' continued Jeff, encouraged, 'he might've just picked on me because he couldn't remember and thought you'd jump on him if he didn't find someone. He can't prove it was me.'
'I dare say other people were present...'
'There were only two of them and—' Jeff stopped abruptly, glowering.
'And they were busy playing dominoes or something?' suggested Gently helpfully.
Jeff dug deeper into his pockets. 'I won't say any more – you're trying to trap me, that's what it is! You're trying to get me to say things I don't mean...!'
'Suppose,' said Gently, beginning to draw pencil-strokes on his pad, 'suppose we go back to the beginning and try a different tack?'
'There isn't any tack to try – it wasn't me and nobody can prove it was.'
'Then you didn't change a dollar bill...'
'I never had a hundred-dollar bill in my life.' Gently's pencil paused. 'What size bill?'
Jeff bit his lip and was silent.
'He doesn't even know how to lie...' observed Copping disgustedly.
Gently finished off his stroke-pattern with aggravating deliberation. Then he felt in his pocket for the spare photograph and regarded it indifferently for a few moments. Finally he leaned across the desk and shoved it at Jeff.
'Here... take a look at this.'
Jeff unpocketed a hand to take it, but Gently was being so clumsy that he knocked it out of the Teddy boy's hand and on to the floor. Sullenly Jeff reached down and scrabbled under his chair for it.
'Was he the man who gave you the note?'
'I told you I never had one.'
'Have you ever seen this man before?'
'I saw his picture on the screen at the Marina, only it didn't have a beard.'
'But you've never seen the man?'
'No.'
Gently retrieved the photograph carefully from fingers that trembled and beckoned to Dutt.
'Take this along to the print department and see if they've got an enlargement, Dutt...'
'Print department, sir?' queried Dutt in surprise.
Gently nodded meaningly. 'And check it with the original, Dutt... it might bring out some interesting points.'
'Yessir. I get you, sir.' Dutt took the photograph gingerly by the extreme margins and went out with it. Gently picked up his pencil again and began laying out a fresh stroke-pattern. Through the open window could be heard, faint and far-off, Copping's blue-bottle or one of its mates improving the shining hour round the canteen dustbin, while more distantly sounded the hum of excursion traffic coming up the High Street. A perfect day for anything but police business...
'You see, Wylie... I'll come to the point. The note you are alleged to have had in your possession was one introduced into this country by the man on the photograph. That man, as you are aware, was murdered.'
'I never knew him – it's nothing to do with me!'
'If it's nothing to do with you then it would be a good idea to tell the truth about the note.'
'But I never had any note – it's all a lie... I keep telling you.'
Gently shook his head remorselessly. 'All you've told me to date has convinced me of the reverse. Besides, the man who identified you gave a pretty damning description when he handed in the note. That suit of yours is rather distinctive, you know. I don't suppose anybody else in Starmouth wears one excepting Baines... and I shall be questioning him in due course.'
'He's seen me before, he could have made it up.'
'He's seen you before? I thought he wasn't supposed to be known to you?'
'He _could_ have seen me before...'
'And made up the whole story about a complete stranger?' Gently hatched a few of the lines in his pattern.
Copping snorted impatiently. 'You're lying... it's too obvious. We know what you got for the note and when we picked you two up this morning you each had five-pound notes on you. What was that – a coincidence?'
'I get pocket-money!' Jeff exclaimed, 'my father isn't a labourer.'
'No, but Baines's father is. Where did _he_ get five pounds?'
'He works – he's got a job!'
'That's right – thirty bob a week as an errand boy and pays his mother a pound of it. Do you think we're fools?'
Jeff's breath came fast. 'I tip him a pound now and again...'
'And he saves it up?'
'How should I know what he does with it?'
'If you don't, nobody else does. What were you doing at ten to ten last night?'
'I... I was on the Front.'
'Alone?'
'I...'
'Answer me!' snapped Copping, 'you don't have to think if you're telling the truth. Baines was with you, wasn't he?'
'No! I mean...!'
'Yes! Of course he was. Why bother to lie? And you were skint, weren't you? You'd got rid of your precious pocket money and Baines's ten bob with it. All you'd got left was an American note – a note you'd begged, borrowed, stolen and perhaps murdered for—'
'No!'
'—and that was all there was between you and a bleak weekend. So you picked out a quiet-looking pub – one where you knew there wouldn't be many witnesses to the transaction – and slipped in and flogged the note to the publican. He wasn't offering much, was he? Less than a third of what it was worth! But you couldn't stop and argue – it might draw attention – they might ask questions you hadn't got the answers for—'
'It's a lie!' screamed Jeff, as white as a sheet, 'you're making it all up – it's all a lie!'
'Then you can prove you were somewhere else?'
'I was never near that pub!'
'Then what pub were you near?'
'I wasn't near any pub at all!'
'Is the only pub on the Front the one you weren't near?'
'I don't know... I didn't notice... I didn't go into a pub anywhere last night!'
Gently clicked his tongue. 'It's a pity about that... it might have helped you to establish an alibi that doesn't otherwise seem to be forthcoming.'
Copping repeated his snort and seemed, with flaming eyes, about to continue his verbal assault upon the shaking Teddy boy: but at that moment Dutt re-entered.
'Ah!' murmured Gently, 'did you make a comparison, Dutt?'
'Yessir.' The sergeant's eye strayed to Jeff. 'Very like, sir, at a rough check. Sergeant Dack thinks so too, sir. He's going over them proper now.'
Gently nodded and stroked off a square. 'Bring in Baines, Dutt... oh, and just a minute...'
'Yessir?'
'Take him along to the prints department first, will you?'
Dutt withdrew and Copping looked questioningly at Gently. But Gently was busy with his patterns again.
'Y-you can't go on anything Baines says,' muttered Jeff tremblingly.
'Oh? And why can't we?' barked the ferocious Copping.
'He'll say anything... you can make him say what you like.'
'If we can make him tell the truth it'll be the first time we've heard it this morning, my lad. I should button my lip, if I were you.'
Jeff licked dry lips and took the advice. There wasn't an ounce of swagger left in him. He sat sagging back in his chair, his feet at an awkward angle, his hands digging ever deeper into his pockets. Copping got up and went over to the window. The fine weather outside seemed to anger him. He studied it tigerishly for a moment, sniffed at the balmy sea air, then turned to eye the Teddy boy from between half-closed lids.
'A nice day for a picnic,' suggested Gently cautioningly.
'I was going round the links... if I'd got away early enough.'
Gently shrugged. 'Something always turns up... it's the bright day that brings forth the adder.'
But Copping sniffed and would not be comforted.
Bonce was brought in, as wild-eyed as ever, and scrubbing recently-inked fingers on the seat of his cheap trousers. Jeff pulled himself together a little at the sight of his henchman, as though conscious of a sudden that he was cutting a poor figure. Gently glanced at Dutt, who shook his head.
'Not this one, sir. Nothing like.'
'Are you sure of that?' asked Gently in surprise.
'Positive, sir.'
'Well... they're not supposed to lie! Sit down, Baines. You can wash your hands later on.'
Bonce sat down automatically in the chair indicated to him. He had an air of bereftness, as though he had lost all will of his own. His mouth was hanging a little open and his face had a boiled look. His eyes resolutely refused to focus on anything more distant than the blunt tip of his freckled nose.
Gently pondered this woebegone figure without expression.
'Robert Henry Baines of seventeen Kittle Witches Grid?'
Bonce nodded twice as though the question had operated a spring.
Gently cautioned him at some length, though it seemed doubtful if what he was saying penetrated very clearly into Bonce's shocked and bewildered mind.
'I'm going to ask you one question, Baines, and it's entirely up to you whether you answer it or not. You understand me?'
The spring was operated again. Gently paused with his pencil at one corner of his pad.
'I want you to tell me, Baines... if you assisted Wylie when, on the night of Tuesday last, he entered a rear bedroom of 52 Blantyre Road and removed from there a suitcase containing United States treasury notes.'
'Don't tell him, Bonce!' screamed Jeff, leaping to his feet, 'don't tell him, you bloody little fool!'
' _Silence_!' thundered Gently in a voice that made even Dutt wince, 'get back in your chair, Wylie!'
'But it's a lie... he'll say anything...!'
' _Get back in your chair_!'
Copping sent the Teddy boy sprawling into his seat again and held him there struggling and panting.
'Now, Baines... have you anything to answer?'
Bonce gaped and gurgled in his throat, his eyes rolling pitiably. Then the spring clicked and his head began to nod. 'I went with him... it's true... I kept watch in the alley...'
'You fool – oh, you bloody little fool!' sobbed Jeff, 'don't you understand it's murder they're after us for – don't you understand it's murder?'
There was a ripping sound as Gently's pencil crossed from one corner of the pad to the other.
The charge was made: burglary on the night of the eleventh. Jeff was in tears as he gave his statement. Of the two of them, it was Bonce who showed the better front. Having shed the intolerable load of conscious guilt he seemed to stiffen up and gain some sort of control of himself, while Jeff, on the other hand, went more and more to pieces. It was from Bonce that Gently received the more coherent picture.
They had been in 'The Feathers' late on the Tuesday evening when the prostitute Frenchy entered. She was well known to them – Jeff claimed to have slept with her and Bonce wasn't sure that Jeff hadn't – and she approached them with the information that a man-friend of hers had left in his bedroom a suitcase containing something of considerable value.
'Was she in the habit of divulging such information?' queried Gently.
Jeff stoutly denied it, but Bonce admitted one or two instances.
'And were you accustomed to act on it?'
Bonce hung his head. 'Once we did...'
Frenchy had struck a quick bargain. They would go halves in whatever the loot realized. She gave them the address, explained the situation of the bedroom and guaranteed to keep the man busy for another hour or two at least. When she left they followed her at a discreet distance and saw her meet a man resembling the one in the photograph. He had exchanged a few words with her and then signalled a taxi. The taxi had departed in the direction of the North Shore.
'Where did the taxi pick them up?' asked Gently.
'It was just outside the Marina.'
'What would have been the time?'
Bonce glanced at Jeff. 'About ten, I should think.'
'Would you know the taxi again?'
'N-no, sir, there wan't nothin' special about it.'
'From which direction did it come?'
'From the Pleasure Beach way, sir.'
The owner of the suitcase having been seen on his way, they hastened round to Blantyre Road and identified No. 52. Then they approached it by the back alley and while Bonce kept watch outside, Jeff broke into the rear bedroom.
'Weren't you taking a bit of a risk?' inquired Gently of Jeff. 'The lodger may have been out, but it's pretty certain the landlady wasn't.'
'We could see them down below,' sniffed Jeff, 'they were watching the telly.'
'The television couldn't have had much longer to go by the time you got there.'
'It's the truth, I tell you!'
'All right, all right – just answer my questions! It may have been running late on Tuesday. How long did it take you to do the job?'
'Ten minutes... quarter of an hour, perhaps.'
'No longer than that?' Gently glanced at Bonce.
'That's about it, sir.'
'But you had to hunt around for it?'
'Why should I?' sniffed Jeff, 'I knew what I was looking for... a blue suitcase with chromium locks. It was standing with the other one near the wardrobe.'
'Did you look in the other one?'
'No... I never touched it.'
'Didn't you go through the drawers or anything of that sort?'
'I tell you I didn't touch anything! I just got what I came for and went. Ask him if I aren't telling the truth.'
Bonce corroborate his leader's statement – he had returned with the blue suitcase and nothing else. They had carried it off to a quiet spot in Blantyre Gardens, forced the locks and discovered the astounding contents. Immediately there was a change of plans. Jeff decided they would tell Frenchy that they had been unable to find the suitcase – a proposition she wasn't situated to contradict – while in reality they would keep it hidden until the hue and cry had died down and then dispose of it by slow and cautious degrees. This they did, and for some reason Frenchy accepted their story without much fuss. When the murder became news and they recognized the pictures which were issued as being of Frenchy's man-friend, they had an additional incentive for keeping the stolen notes under cover. Unfortunately, their patience was soon exhausted. A financial crisis at the end of the week had slackened their caution. Surely, they had thought, there could be no harm in cashing just _one_ of that inexhaustible pile of notes... just one, to see them comfortably through the weekend...
Gently sighed at the end of the recital. 'And the rest of them, where are they now?'
Bonce swallowed and glanced again at Jeff. 'They're under the pier.'
'Which pier is that?'
'Albion Pier... there's a hole between two girders.'
'You'd better show me... Dutt!'
'Yessir?'
'Tell them to bring a car round, will you?' He returned to Bonce. 'That evening... in the bar at "The Feathers"... were all the usual crowd there?'
Bonce twisted his snub nose perplexedly. 'I – I suppose so, sir.'
'Was Artie serving at the bar?'
'Oh yes, sir.'
'That fellow who wears loud checks and lives on whisky?'
'Yes, sir.'
'Louey?'
'N-no, sir... _you_ don't often see him in the bar.'
'Peachey?'
'I think he looked in while we were talking to Frenchy...'
There was the sound of a car swinging out of the yard and Copping rose to his feet. He looked at Gently questioningly and motioned to the two youths with his head. 'Cuffs on them... just to keep on the safe side?'
Gently smiled amongst the nebulae. 'Let's be devils this morning, shall we? Let's take a risk!'
Exceeding Sunday-white lay the Albion Pier under mid-morning sun. Its two square towers, each capped with gold, notched firmly into an azure sky and its peak-roofed pavilion, home of Poppa Pickle's Pierrots, notched equally firmly into a green-and-amethyst sea. Its gates were closed. They were not to open till half past two. The brightly dressed strollers, each infected in some degree by the prevailing Sundayness, were constrained to the languid buying of ice-cream, the indifferent booking of seats or the bored contemplation of Poppa Pickle's Pierrots' pics. They didn't complain. They knew it was their lot. Being English, one was never at a loss for a moral attitude.
Even the arrival of a police car with three obvious plain-clothes men and two obvious wrong-doers didn't seriously upset the moral atmosphere, though it may have intensified it a little.
'Which end?' inquired Gently, shepherding his flock down the steps to the beach.
'This end... up here where the pier nearly touches the sand.'
They marched laboriously through soft dry sand, the cynosure for an increasing number of eyes. Dutt led the way, the Teddy boys followed, and Copping and Gently brought up the rear. Under the pier they went, where the sand was cold and grey. A forest of dank and rusty piles enclosed them in an echoing twilight.
'Up there,' snuffled Jeff, indicating a girder which nearly met the sand, 'there's another one joins it behind... it's in the gap between them.'
'Get it out,' ordered Gently to Dutt.
The gallant sergeant went down on his stomach and squirmed vigorously till he was under the girder. Then he turned on his back and began feeling in the remote obscurity beyond. He seemed to be prying there for an unconscionable length of time.
'Have you found the hole?' asked Gently, his voice echoing marinely amongst the piles.
'Yessir,' came muffledly from Dutt, 'hole's there, sir... it's what's in it I aren't sure about... couldn't get hold of me legs and pull me out, sir?'
Copping went to the rescue and a grimy Dutt renewed acquaintance with the light of day. In his arms he bore a bundle, also grimy. 'This is all there was, sir... ain't no trace of any suitcase.'
'Open it!' snapped Gently.
Copping broke the string and unwrapped the paper. There lay revealed a crumpled grey suit, a pair of two-colour shoes, shirt, socks, underclothes, suspenders and a blue bow tie.
'Sakes alive!' exclaimed Copping. 'Look at this label – Klingelschwitz – it's the same as in the boyo's suit!'
'And look at this shirt,' added Gently grimly, 'four nicely grouped stab-holes... same as in the boyo's thorax.'
A sugary thump made them all turn sharply. It was Jeff going out cold on a sand that was even colder.
CHAPTER NINE
IT WAS A hefty lunch for a hot day and Gently followed Dutt's example of shedding his jacket and rolling his sleeves up. There wasn't any frippery about it. Just straight roast beef and Yorkshire pudding, and vegetables followed by hot apple turnover with custard. But either Mrs Davis was a demon cook, or else the Starmouth ozone had really come into its own that day... there wasn't much in the way of conversation for quite some time.
'Superintendents!' muttered Gently at last, evaluating the remains of the turnover with sad resignation.
'Never alters,' agreed Dutt sympathetically, cutting an absent-minded slice.
'I can't help coming to the conclusion, Dutt...'
'Yessir?'
'... if it didn't savour of insubordination...'
'Aye, aye!' Dutt winked at his superior over a spoonful of juicy pastry. 'Don't have to say it, sir. I knows well enough what you mean.'
Gently picked up his plate and placed it at some distance from himself, as though finally to sever connections with that beguiling turnover. 'You make a pinch... you dig up some evidence... it does something to them. They're all the same, Dutt.'
'Yessir. Noticed it.'
'They suddenly turn impatient. It's an occupational disease with superintendents. At a certain stage in the proceedings they get the charge-lust. They want to charge someone. And if there's half a case against anybody it's the devil's own job to head a super off and make him be a good boy...'
'Don't we know it, sir?'
Gently drew a deep breath and pulled out his familiar sandblast. 'Of course, you have to admit it... there's enough on Baines and Wylie to make the average super sit up and howl blue murder. But at the same time, it only needs the average forensic eye. Baines isn't a liar, for instance, and Wylie's got too scared to lie. No, Dutt, no. Our super is doing himself no good by tearing the bricks apart at the Wylie's. He won't find anything, and he won't improve his standing with anyone.'
Mrs Davis brought in their cuppa, making room for the tray beside Gently. She hesitated on seeing the chief inspector's pipe on the point of being lit and then produced, from nowhere as it seemed, a capacious glass ashtray. Gently nodded a solemn acknowledgement. Mrs Davis beamed at the still-eating Dutt. 'Aren't you going down to the beach now this afternoon, Inspector?'
Gently smiled wanly and unbonneted the teapot.
'Well, sir... what do _you_ make of them clothes turning up like that?' queried Dutt when the tea was poured and Mrs Davis had retired.
'They were planted deliberately, Dutt. By the person who lifted the suitcase.'
'But how did they know where it was, sir?'
'By deduction and observation – just as we find out things.' Gently doused a match and took one or two comfortable pulls. 'Obviously... they wanted that suitcase back. Whether they still intended to use the money or not we don't know, but they feel it's important that a large consignment of it shouldn't be lying around loose... it would almost inevitably finish up in our hands. So their first move after settling with Max was to recover the suitcase and I can imagine they were a little upset to find it missing when they got to his lodgings...'
'Lord luvvus, sir – that other set of prints! I've been puzzling my loaf about them all the morning.'
'Exactly, Dutt... the first little slip our friends seem to have made. But I don't suppose they aimed to be around when those prints came to light. It was just a bit of bad luck that the suitcase had vanished into thin air...'
'So it was them who ransacked the room, sir.'
'Undoubtedly.'
'On account of he may have hidden the stuff somewhere.' 'It was a possibility they wouldn't overlook.'
Dutt gave a little chuckle. 'You're right, sir... their faces must have dropped a mile when they found the cupboard was bare!'
'A good mile, Dutt, and possibly two. It upset all their calculations. It meant they would have to hang around and look for it instead of getting to hell out of the country... and hanging around would get to be more and more dangerous as the investigation went on. At first, I imagine, they hadn't a clue about it. They may have visited the bedroom more than once and they were certainly interested to know what we found when we got there... and then, of course, they began to think it out and perhaps make some inquiries. They found out, or possibly they knew, that Max had been consorting with Frenchy... that was an obvious lead. No doubt they gave her flat a going-over. They might even have questioned her. But there was no suitcase at the flat, and all that Frenchy could tell them – even if she came clean – was of Jeff and Bonce's allegedly fruitless attempt to get the suitcase... Anyway, they got on to Jeff and Bonce somehow. It wouldn't have been too difficult if they checked up on Frenchy.'
'And then they kept them under observation, sir?'
'Just as we would have done, Dutt.'
'And last night they found out where the case was hidden – and left the clothes there for a false scent, sir?'
Gently nodded pontifically. 'A false scent for a charge-happy super.'
Dutt swallowed a mouthful of tea and looked a little dubiously at the remaining shoulder of apple turnover. 'Just one thing, sir...'
'Yes, Dutt?'
'I don't want to seem critical, sir...'
'Don't be modest, Dutt – just come to the point.'
'Well, sir, what I want to say is, how did they know we was ever going to find them clothes, let alone connect them with the Teddy boys?'
Gently nodded again and smiled around his pipe. 'That's what we want to know, isn't it, Dutt. That's going to be the clincher!'
He rose from the table and went over to Mrs Davis's telephone. The phonebook lay beside it. He flicked through it and traced down a column with a clumsy finger.
'Starmouth 75629... this is Chief Inspector Gently.' He tilted the instrument to one side so that Dutt could hear too. 'Biggers? There's something else I want to ask you, Biggers... yes, about last night.'
'Ho yes, sir?' came the publican's anxious voice from the other end.
'You told us in your statement that after you had changed the note you heard there were some counterfeit ones going about. I want to know where you obtained that information.'
'Yes, sir! Certainly, sir! It was a bloke in the bar what told me that.'
'A bloke you know?'
'Ho no, sir. Quite a stranger.'
'He was in the bar at the time of the transaction?'
'No, sir, not as I remember. The first time I noticed him there was when the young feller went out.'
'You mean he came in while the transaction was in progress?'
'Must've done, sir, 'cause he soon ups and tells me to watch my step with regard to Yank money. "Wasn't that a hundred-dollar bill?" he says. "Ah, it was," I says. "Then it's ten to one you've been had," he says, or words to that effect, "there was a sailor got copped with some this afternoon."'
'Oh did he...?' Gently exchanged a glance with Dutt.
'Yes, sir... God's honest truth!' The voice on the phone sounded panicky. 'I don't have no cause to lie, now do I—!'
'All right, Biggers... never mind the trimmings. What else did this man tell you?'
'Well, he told me I could get five years, sir, and that I ought to hand it over to the police... naturally, me just having paid ten quid...'
'We know about that. Did he say anything else?'
'No, sir... not apart from ordering a whisky. It was nearly closing-time.'
'Would you recognize him again?'
'Ho yes, sir! Like I was telling you, I never forget a face.'
'Can you describe him?'
'Well, sir... he wasn't English, that I can say.'
'Did you notice a mole on his cheek by any chance?'
'No, sir. No. But he'd got a scar running all down one side...'
Gently hung up the instrument and leaned on it ponderingly for a few moments. His eyes were fixed on Mrs Davis's flowered wallpaper, but to a watchful Dutt they seemed to be staring at something a good six feet on the other side of the wall. Then he sighed and straightened his bulky form.
'So there it is, Dutt... our clincher. And they even knew about McParsons... eh?'
Dutt shook his head ruefully. 'They must have quite an organization, sir...'
'An organization!' Gently laughed shortly. 'Well... we'd better get our own organization moving, too. Go back to headquarters, Dutt, and tell them to put a man each on the two stations and another on the bus terminus, and to warn the men on the docks to keep their eyes double-skinned. It's an even bet that our scar-faced acquaintance is well clear of Starmouth, but we can't take any risks... Then give Special a ring and let them know.'
Dutt nodded intelligently. 'And the clothes, sir...?'
'Get them sent to the lab, and the paper and string. Oh, and that cab-driver... the one who picked up Max and Frenchy on Tuesday night... see if you can get a line on him, Dutt.'
'Yessir. Do my best.'
Gently scratched a match and applied it to his pipe. 'Me, I'm going to pay a little social call in Dulford Street. I think it's time that Frenchy assisted the police by supplying the answers to one or two interesting questions.'
Dulford Street was a shabby thoroughfare adjoining the lower part of the Front. It began as though by accident where some clumsily-placed buildings had left a gap and proceeded narrowly and crookedly until it got lost in a maze of uncomely backstreets. There was a feeling of having-gone-to-seed about it, as though its original inhabitants had given it up in despair and left it to go its own way. From one end to the other it could boast of no fresh paint except the lurid red-and-cream of an odiferous fish and chip shop.
Gently eyed the assemblage moodily and applied to a new bag of peppermint creams for encouragement. Sunday was obviously an off-day in Dulford Street. The signs of life disturbing its charms were few. On the right-hand side was a frowsy little corner-shop with some newspapers in a rack at the door, and at the entry from the Front lurked a furtive and ragged figure... Nits, who had been following Gently all the way along the promenade. Gently shrugged his bulky shoulders and pushed open the clanging door of the newspaper shop.
'Chief Inspector Gently... I wonder if you can give me some information?'
It was a white-haired old lady with beaming specs and an expression of anxious affability.
'What was it you were wanting?'
'Some information, madam.'
'The newspapers is all outside... just take one, sir!'
'I want some information.' Gently raised his voice, but the only effect was to increase the old lady's look of anxiety. He pointed out of the dusty window.
'That apartment over there... do you know who lives in it?'
'Oh yes, I do! She isn't nothing to do with me!'
'Is that her permanent address or does she just make use of it?'
'Eh... eh?' The old lady peered at him as though she suspected him of having said something rude.
'Is that her permanent address?' began Gently, _fortissimo_ , then he shook his head and gave it up. 'Here, how much are these street directories?'
'They're sixpence,' retorted the old lady sharply, 'sixpence – that's what they are!'
Gently put a shilling on her rubber mat and made a noisy exit.
Frenchy's apartment, flat, or whatever other dignity it aspired to was situated above a disused fruiterer's shop. The shop itself had been anciently boarded up, but the degree of paintwork it exhibited matched evenly with that of Frenchy's door and the windows above, leaving no doubt about the contemporaneity of the decoration. Gently tried the door and found it open. It gave directly on to uncarpeted stairs which rose steeply to a narrow landing. At the top were two more doors, one with a transom light which did its best to illumine the shadow of the landing, and at this he knocked with a regular policeman's rhythm.
'Who is id...?' came Frenchy's croon.
'It's Chief Inspector Gently. All right if I come in?'
There was a creaking and scuffling, and finally the sound of shuffling footsteps. Then the door opened to display a draggle-haired Frenchy, partly-clad in a green dressing-gown. She glared at Gently.
'What are you after now?'
'I'm after you,' said Gently cheerfully, 'weren't you expecting me to call?'
Her eyes narrowed like the eyes of a cat. 'You've got nothing to pinch me for... you bloody well know it! Why can't you leave a girl alone?'
Gently tutted. 'This isn't the attitude, Frenchy. You should try to be co-operative, you know – it pays, in your profession.'
'That's none of your business and you ain't got nothing on me!'
Gently shook his head admonishingly and pressed past her into the room. It wasn't an inviting prospect. The furniture consisted of an iron bedstead, a deal table and three cheap bedroom chairs. The floor was covered with unpolished brown lino, the walls with faded paper. At the window, curtains were drawn to keep out the sun, but in spite of this the room was like a large and unventilated oven, an oven, moreover, that possessed a vigorously compounded odour, part dry rot, part cigarette smoke and part Frenchy. Gently fanned himself thoughtfully with his trilby.
'Doesn't seem a very comfortable digging for a trouper like you, Frenchy,' he observed.
'What's it got to do with you?' spat Frenchy, closing the door with a bang.
'And you're travelling light this season.' He indicated a dress and a white two-piece which hung on hangers from a hook in the wall.
'If you're going to pinch a girl for being short of clothes...!'
Gently concluded his unhurried survey with the dishevelled bed, some empty beer-bottles and a chamber-pot. 'And then again, my dear, this place is in the wrong direction...'
'Whadyermean – wrong direction?'
'It isn't in the direction the taxi took.'
'What taxi – what are you getting at?' Frenchy whisked round fiercely to confront him.
'Why... the taxi you and Max took from outside the Marina at about 10 p.m. last Tuesday. It went off towards the North Shore... that's in a diametrically opposite direction, isn't it, Frenchy?'
The sudden pallor of the blonde woman's face showed up the dark wells of her eyes like two pools, but she took a furious grip on herself. 'It's a filthy dirty lie... I didn't take no taxi! I was in "The Feathers" at ten... ask anyone who was there... ask Jeff Wylie – it was him who came away with me!' She broke off, breathing hard, crouching as though prepared to ward off a physical blow.
Gently's head wagged a measured negative and he felt in his pocket for some carelessly-folded sheets of the copy-paper. 'It won't do, Frenchy... it isn't good enough any longer. I've got a couple of statements here which tell a different story.'
'Then some b––'s been lying!' Frenchy tried to snatch the sheets out of Gently's hand.
'Nobody's been lying and you'll get a chance to read these in a couple of minutes. Now sit down like a good girl.'
Frenchy hovered a moment as though still meditating an attempt on the papers. Then she swore an atrocious oath and dumped herself down on the side of the bed, an action which endangered the decency of her sparsely-clad person. Gently turned one of the chairs back-to-front and seated himself also.
'First, I'd better have your name.'
'What's wrong with Frenchy... it suits everyone else round here.'
Gently clicked his tongue. 'Let's not be childish, Frenchy. Why make me bother the boys in Records?'
'Trust a bloody copper! So it's Meek, then. Agnes Meek.'
Gently scribbled it in his notebook. 'And where do you hail from, Agnes?'
'I was born and bred in Maida... but don't use that filthy bleeding name!'
'And when did you come up here?'
''Bout Whitsun or just before.'
'And whose idea was it?'
'Mine – who the hell's do you think it was?'
'Now Frenchy! I'm only asking a civil question.'
'And I'm telling you I came up on my own! Don't you think a girl needs a holiday?'
Gently shrugged. 'It's up to you... So you've been living at this address since Whitsun?'
'That's right.'
'And nowhere else at all?'
Frenchy swore a presumable negative.
'How did you find it? Who's your landlord?'
'Why not ask your pals up at the station – they're supposed to know every bloody thing going on round here!'
Gently sighed sadly. 'You're not being helpful, Frenchy... and I had hoped you were going to be.' He served himself a peppermint cream and chewed it sombrely for a moment. 'Well... to come to the business. I'm pinching you for conspiracy to burgle, Frenchy—'
Frenchy screeched and shot up off the bed. 'It's a frame-up, that's what it is, a filthy, stinking—!'
'Shh!' murmured Gently, 'I don't have to warn an old-stager like you.'
'They'd say anything in a jam, dirty little bastards!'
Gently handed over his sheets of copy-paper. 'In effect they said this... and there's a certain amount of evidence to back them up.'
Frenchy seized the sheets and went over to the window with them, turning her back on Gently. It didn't take her long to extract the gist of them. There was a moment when she discovered how she had been double-crossed that added three distinct new words to Gently's vocabulary.
'It's a filthy bag of lies!' she burst out at last. 'The – little liars – they're trying to pin it all on me!'
'They seem to have made a job of it, too...'
'There isn't a word of truth!'
'But there's some evidence that goes with it...'
Frenchy stormed up and down the muggy room with perspiration beading on her pasty face. 'You know what it is... You know why these pigs have said this. It's because I wouldn't go to bed with them... that's what they've wanted! They've wanted to be little men, to go to bed with a woman... they've been hanging round me ever since I came up here. But I don't go to bed with children... nobody can blame me for that!... and now they're in trouble they're trying to blame me – somebody it's easy to get in bad with the police!'
'Whoa!' interrupted Gently pacifically, 'it's no use getting out of breath, my dear. Somebody had to tell them about that suitcase and where to find it...'
'It wasn't me! I didn't know nothing about it.'
'Then who did – who else knew about it?'
'How the hell should I know? Perhaps they saw him carting it around and got the idea it was something valuable...'
'Who told you he was given to carting it around?'
'Nobody told me—!'
'And how did they know where he lodged – that he was out – that for some reason he'd left it in his room?'
'They could've watched him, couldn't they?'
'They aren't professionals, Frenchy.'
'They're sneaking little swine, that's what they are!'
She flung herself at the bed and disinterred some cigarettes from under the pillow. Gently produced a match and gave her a light, steady brown fingers against her trembling pale ones. She swallowed down the smoke as though it were nectar.
'You know, Frenchy, it isn't burglary you've got to worry about... we aren't terribly interested in that. It's the way your customer finished up on the beach the next morning that's the real headache.'
'He wasn't my customer – I never knew him!'
Gently shook his head. 'I've got another witness who saw you with him, quite independent. Do you remember having lunch at the Beachside Cafe?'
'I was never in the place!'
'And now, according to these two statements, you were the last person we know to see Max alive...'
A shudder passed through the blonde woman's body and she had to struggle to keep her hold on the jerking cigarette.
'Weight it up, Frenchy... it's a nasty position to be in.'
'But mister,' – her voice was hoarse now – 'it wasn't nothing to do with me – nothing – I'll swear to it!'
Gently shrugged and picked up his hat to fan himself again.
'I didn't have no hand in it... honest to God!'
Gently fanned himself impassively.
'I didn't – I _didn't_ – I _didn't_!' The voice was a scream now and she threw herself on her knees in a fit of anguish. 'You got to believe me... mister... you _got_ to!'
Gently nodded a single, indefinite nod and went on fanning.
'But _you've got to_ , mister!'
Gently paused at the end of a stroke. 'If,' he said, 'you _didn't_ , Frenchy, then the best thing you can do is to come clean...'
'But I can't, mister!' Her face twisted in indescribable torment.
'You can't?' Gently stared at her bleakly and recommenced his fanning.
'I can't – I _can't_! Don't you understand?'
'I understand there's a murder charge being kept on ice for someone.'
Frenchy moaned and sank in a heap on the floor. 'I didn't do it,' she babbled, 'I didn't do it... you got to believe me!'
Gently bent over and picked up the cigarette, which was making an oily mark on the dubious lino. 'Listen, Frenchy, if it's any consolation to you, I don't think you knocked off Max, and I'm not personally trying to pin it on you. But you're obviously in it up to your neck, and unless you make yourself useful to us you're going to have a pretty rough passage in court. Now what about it... suppose we do a deal?'
'I can't, mister – I daren't!'
'We'll give you protection. You've nothing to be afraid of.'
The dyed-blonde hair shook hopelessly. 'They'd get me... they always do. They don't never forget, mister.'
'Nonsense,' said Gently stoutly, 'this is England, Frenchy.'
Her haunted eyes looked up at him, hesitating. Then she gave a hysterical little laugh. 'That's what Max thought, too... he'd be safe once he got to England!'
They went down the naked stairway, Frenchy clicking her high heels, Gently clumping in the rear. She had put on her white two-piece with its red piping and split skirt, and there was almost a degree of respectability about her make-up. At the bottom she fished a key out of her handbag and locked the street door. Gently took it from her and slipped it into his pocket.
'And to save a little trouble...?'
Frenchy sniffed and tossed her head towards the corner shop. 'Mother Goffin over the way... and don't let her kid you up she's deaf.'
'I won't,' murmured Gently, 'at least, not twice in one day.'
They proceeded towards the Front, Gently feeling a trifle self-conscious beside so much window-dressing. At the corner of the street lurked Nits, his bulging eyes fixed upon them. As they drew closer he sidled out to meet them.
'Giddout of the way, you!' snapped Frenchy, angering suddenly. But Nits' attention had focused on Gently.
'You leave her alone – you leave her alone!' he piped, 'she's a good girl, you mustn't take her away!'
'Clear out!' screeched Frenchy, 'I've had enough of you hanging round me!'
Gently put his hand in his pocket for a coin, but as he did so the halfwit came flying at him with flailing arms and legs.
'You shan't take her away – you shan't – I won't let you!'
'Here, here,' said Gently, 'that's no way for a young man to behave—!'
'I'll kill you, I will, I tell you I'll kill you!'
'And I'll bleedin' kill you!' screamed Frenchy, catching Nits such a cuff across the face that he was almost cart-wheeled into the gutter. For a moment he lay there, pop-eyed and gibbering, then he sprang to his feet in a whirl of limbs and darted away down Dulford Street like a bewildered animal.
'Dirty little git!' jeered Frenchy, 'they're all the same – doesn't matter what they are. Men are all one filthy pack together!'
The super wasn't feeling his pluperfect best just then. He'd been butting his head against brick walls all day. He'd disregarded Gently, made an enemy of Christopher Wylie, been torn off a helluva strip by the chief constable, failed to find the merest trace of a suitcase full of hundred-dollar bills and, to cap it all, he was beginning to realize that he'd been wrong anyway. It was this last that really hurt. The rest he was prepared to take in his superintendental stride...
'So she won't talk!' he almost snarled, as Gently and he sluiced down canteen tea in the latter's office.
Gently shrugged woodenly. 'You can't really blame her. She's convinced she'd be signing her own death-warrant.'
'Well, if she doesn't sign it I shall – she can bank on that for a start!' yapped the super.
'Oh, I don't know...' Gently put down his cup and mopped his forehead with a handkerchief that had been seeing life. 'I've got a couple of men looking for the taxi that picked them up on Tuesday night... if we can find that, we shall be getting somewhere.'
'Now look here, Gently!' The super almost choked. 'This woman is the crux of the case. If your guessing is correct she knows everything – where he went to, who picked him up, who was after the money – she may even have been a witness to the murder, for all we know! And all you can tell me is she won't talk. That's all! They've put a scare into her, so she won't talk!'
'It isn't a small size in scares, when you come to think of it.'
'I don't care what size it was!' raved the super. 'I've got a scare up my sleeve, too, quite as big as any of theirs. We'll soon see who's got the biggest!'
Gently looked woodener than ever. 'She's got a perfect right to keep quiet. And you're overestimating your scare. There's nothing you can pin to Frenchy apart from conspiracy to burgle, and she's not such a fool that she doesn't know it.'
'Oh, she isn't, isn't she? We'll soon see about that! I'll make a pass at her with a murder charge that'll put paid to all this nonsense...!'
'No.' Gently shook his head. 'I've tried it, anyway. The position is that you _might_ get her, but they _certainly_ will. They're the ones who are holding a pistol in her back... or at least they've made her think so. No... Frenchy's our ace in the hole, and for the moment we'll have to leave her there. I've got an impression she'll be a lot more vocal when she sees certain people wearing handcuffs.'
'But how the devil are you going to get handcuffs on them when she won't talk? And the man we want – let's face it, Gently, it's the fellow with the scar who's got high jump written all over him – where will you ever lay hands on him again?'
'He was here last night,' muttered Gently obstinately.
'Last night, last night! But where is he now – today? He isn't just a criminal on the run. He's part of a powerful and ruthless organization, professionals to their fingertips.'
Gently smiled feebly. 'Even organizations are run by human beings... they're sometimes quite modest concerns when you get to grips with them. Anyway... about Frenchy. I want to ask a favour.'
The super grunted fiercely, as though indicating it wasn't his day for such things.
'I don't want her kept here... I'd like her released on bail.'
'On BAIL!!!' erupted the super, his eyes jumping open as though he had been stung.
'Yes... nothing very heavy. Just a modest little reminder.'
'But good heavens, man – bail! A woman of that character – arrested for a felony – suspected of complicity in murder – and you're asking for bail! What the devil do you think I should put on my report?'
'Just say it was at my request,' murmured Gently soothingly, 'I'll carry the can if she doesn't turn up.'
'But I'm already in bad with the CC over this business—!'
'She'll be in court. You needn't worry about that.'
The super treated Gently to several seconds of his best three-phase stare. 'All right,' he said at last, 'it's your idea, Gently. You can have her. But God help you if she's missing when we go to court. You'll have her tailed, of course?'
'Oh yes... Dutt's one of the best tails in the business. And I'd like someone to check up on the flat in Dulford Street. The rent is paid to a Mrs Goffin who keeps a newsagent's opposite... I'm just the wee-ish bit interested to know where it goes after that.'
The telephone rang and the super hooked it wearily to his ear. Gently rose to go, but the super, after a couple of exchanges, motioned for him to wait and grabbed a pencil out of his tray.
'Yes... yes... d'you mind spelling it?... yes... as in Mau-Mau... got it... you'll send his cards... right... yes... thank you!'
He hung up and pushed his desk-pad across for Gently's inspection. 'There you are – for what it's worth!'
Gently glanced at the pad and back at the super.
'The names of our playmates... Special _does_ work on a Sunday! Olaf Streifer is Scarface – he's an agent of this precious TSK Party's secret police... Maulik, it's called. Special want him in connection with some naval sabotage at Portsmouth two years ago. You seem to have got a set of his prints from somewhere, incidentally...'
Gently nodded. 'And this... Stratilesceul?'
'Stephan Stratilesceul – the lad on the slab. He wasn't known over here, but the Sûreté had records. They wanted him in connection with a similar business at Toulon... the TSK seems to have a lien on naval naughtiness.' He picked up the pad and held it up ironically. 'So now we know – and how much further does it get us?'
Gently hoisted a neutral shoulder. 'It all helps to fill in the picture... you can't know too much about a murder.'
CHAPTER TEN
IT HAD BEEN too fine.
The peerless sky which had filled the beaches yesterday had vanished overnight literally in a clap of thunder and its place was filled by low, yellow-grey cloud which drizzled warmly, as though somewhere that wonderful sun was still trying to filter its way through. Perhaps it was wise of nature. There had been havoc enough wrought by one fine Sunday. In the damp streets plastic-caped holidaymakers went about with a wonderful solicitude for their fiery backs and arms...
Now it was the cafes that came into their own. The innumerable little boxes clustered cheek by jowl all the way down Duke Street, empty and forlorn while the sun reigned, filled up now to the last tubular steel chair. After all, it wasn't an unpleasant rain... one expected it some time during the holiday. And there were worse things to be done than drinking one's coffee, smoking, writing cards and going through the newspapers...
Not that it was front-page today, their own especial murder. The super had kindly released the news of the arrest of Jeff and Bonce and the discovery of the grey suit, but in the face of fierce competition from a Cabinet re-shuffle it hadn't made the grade. It had slipped to page five. Strangely unanimous, the editors of the dailies had each come to the conclusion that the Body on the Beach wasn't going to get anywhere, and they were quietly preparing to forget the whole thing.
Like a certain superintendent, thought Gently, resting his elbows on the low wall bounding the promenade... though of course, the man had his reasons.
He hitched up his fawn raincoat and produced his pipe. He couldn't help it... this sort of weather always made him moody. To wake up and find it raining induced in him a vein of pessimism, both with himself and with society. He just wanted to turn over and go to sleep again and forget all about them...
Well... if it _would_ rain!
He lit the sizzling pipe, tossed the match on to the sand below and turned abruptly away from the melancholy sea.
Opposite him, across the carriage-way, loomed the garish tiled front of the Marina Cinema. A spare, florid-faced man with a wrinkled brow and a shock of tow hair was polishing the chrome handles of the swing-doors. Gently went across to him.
'You're at it early this morning...'
The man paused to throw him a sharp look and then went on with his polishing. 'It's got to be done some time, mate.'
'The sea air can't do them a lot of good.'
'Telling me! It plays the bloody hell with them.'
He rubbed away till he came to the top of the handle, Gently watching patiently the while. At last he straightened out and gave his cloth a shaking.
'What are you – a cop, mate?' he asked briefly.
Gently nodded sadly. 'Only I was hoping it didn't show quite so much...'
'Huh! I can always smell a cop a mile away.'
'I shouldn't have stood to windward, should I?'
The tow-haired man took a reef in his cloth and advanced to the next door-handle. 'What do you want here, anyway?'
'The usual thing. Some information.'
'And suppose I haven't got any?'
'Suppose,' said Gently smoothly, 'suppose you be a smart little ex-con and keep a civil tongue when you talk to a policeman?'
'An ex-con...! What creeping nark told you that?'
Gently smiled at a diagonal frame filled with Lollobrigida. 'You aren't the only one with a developed sense of smell...'
But he didn't get any information from the man. He didn't, or wouldn't, remember anything about people taking taxis on Tuesday night last. Yes, he would have been in the vestibule just before the last house turned out, but he was probably chatting to the cashier or one of the girls... no, there wasn't anybody else on late turn that night... no, he didn't know Frenchy or anyone like her... going straight he was, and he defied anyone to prove different.
Gently left him to his handles and plodded on down the Front, pessimism confirmed in his soul.
'The Feathers' was open, but it seemed rather a waste of electricity on such a customless morning. Its arrow was darting away with customary vigour, albeit it fizzed a little in the rain, but there were few enough strollers to be pricked into the temporary refuge of the arcade: its music drooled hollowly down empty aisles. Gently went up the steps and through the doors. Not a soul was about except the attendant, who was sweeping the floor at the far end. Through the doors of the bar, which were stood open with two chairs, could be seen a figure similarly engaged and a 'closed' notice hung rakishly on a chair-back. Obviously, they weren't expecting a rush of business.
He turned to the nearest machine and dropped a penny in the slot. It was one of the pre-war 'Stock Exchange' type and a pull on the handle yielded a brisk no-dividends. Gently tried again. He'd got quite a pocket-full of coppers. Absently he yanked the lever and watched the colourful passage of Rubber, Textiles, Railways and Gold... it seemed hard that such a well furnished wheel should come up no-dividends twice in a row. But it did. It was clever. It sorted out a solitary white from a whole rainbow of coloured, and stuck to it with an obstinate firmness.
A gigantic hand ornamented with a solitaire diamond suddenly covered the handle and its guard.
'You haven't got the knack, Inspector,' purred Louey's voice behind him, 'let an old professional show you how to beat the book!'
Gently stood back without replying and Louey pressed a coin into the slot. Then he caressed the handle with an even, almost casual pressure and the wheel drifted lazily round to a Gold segment. A second coin brought coppers cascading down the shoot.
'You see, Inspector?' Louey's gold tooth shone its message of innocent goodwill. 'It _is a_ matter of skill, after all...'
Gently shrugged and repossessed himself of his twopence. 'It needs a safe-breaker's touch... the way one tickles a combination lock.'
Louey's smile broadened. 'Some of the kids learn how to play them, though it costs them a few weeks' pocket-money. But I don't mind that... there are fifty who never learn for every one who does.'
'Sounds like an expensive accomplishment to me.'
'We have to risk our stakes, Inspector, when we're out to win something.'
Louey picked up the rest of the coins from the shoot and paid them back into the machine, one by one. They flicked up no-dividends as surely as a till flicking up no-sales.
'Skill,' purred Louey, 'you can't really call it gambling, Inspector.'
Gently quizzed the huge man's sack-like raincoat and corduroy cap. 'You were just going out?'
'My morning constitutional,' nodded Louey, 'I always take it, summer and winter.'
'Mind if I come too?'
'Delighted, Inspector! I was hoping for the opportunity of another little chat.'
He ushered Gently out, holding the door obsequiously for him. They crossed the carriage-way and turned southwards along the almost deserted Front. The rain, from being a drizzle, had now become quite steady and gusts of sea-breeze made it cut across their faces as they walked. Louey snuffed the air and looked up at the sky.
'It's set in for the day... I shall be a richer man by tomorrow night, Inspector. You remember my pussy? I expect you thought he'd got his lines crossed yesterday, but he never makes a mistake. I suppose we shan't have the pleasure of your company at the races tomorrow?'
Gently grunted. 'I follow my business... wherever it takes me.'
'Ah yes... and I see by the papers that you're making great strides. Well, well! Those two youngsters in their ridiculous suits! It must be a lesson to me to keep a tighter check on the customers in the bar...'
Gently flipped the sodden brim of his trilby. 'I still prefer your first theory, the one about a political organization.'
'You do?' Louey seemed pleasantly surprised. 'I thought you must have forgotten that, Inspector... my amateur summing-up of the case! But these new facts explode it, I'm afraid. There wouldn't seem to be much connection between Teddy boys and politics.'
'There isn't,' grunted Gently.
'Then surely we must give up my theory...?'
'We could if the Teddy boys killed Stratilesceul, but as it is they only pinched his suitcase.'
Three strides went by in silence. 'Stratilesceul?' echoed Louey, 'is that the name of the murdered man?'
'The man who skipped the _Ortory_ at Hull and was chased down here by Streifer.'
'Streifer...?' This time Gently lost count of the number of strides. 'I'm sorry, Inspector... a lot of this hasn't appeared in the papers, or if it has, I've missed it. Was it from Hull that this unfortunate man came?'
'It was.'
'And he was chased by someone?'
'By Streifer. Olaf Streifer. A member of the Maulik, the TSK secret police. It was just like in your theory, Louey... the execution of a traitor by an organization he had betrayed.'
The big man shook his head with an air of bewilderment. 'You must excuse me if I seem a little dense... I'm not so familiar with the business as yourself, Inspector. Am I to take it that the case is closed and that you have arrested this... Streifer?'
Gently didn't seem to have heard. He was poking in his pocket for a peppermint cream.
Louey gave a little laugh. 'I was saying, Inspector... has this Streifer been arrested?'
The peppermint cream was found and Gently nibbled it with deaf composure... it might have been the rain which was making him so hard of hearing. Louey shook his head again as though realizing that it was necessary to humour a Yard man. After all, he seemed to be saying, it was a privilege thus to be taken into the great man's confidence at all...
They strode on towards South Shore. The rain kept driving in from the sea. There were warm sheets of it now, really wetting, and Gently's experienced brogues were beginning to squelch. Even Louey was constrained to do up his top button, though it meant veiling the glories of a pearl tie-pin stuck in a grey silk tie but there weren't many people to see it in any case.
'Of course, it was Streifer we saw coming out of your office on Friday night,' grunted Gently at last, the peppermint cream being fairly disposed of.
'I thought we had disposed of that point, Inspector.' Louey sounded justifiably piqued. 'But it was Streifer all right, and it was your office all right.'
'Well, if you say so... but I can't imagine what he was doing there. Naturally we had a little check after you'd told us about it, but as far as we were able to discover nothing had been stolen or disturbed.' Louey turned his huge head towards Gently. 'Do you want my opinion, Inspector?'
Gently shrugged, hunched down in a leaky collar.
'My opinion is that _if_ it was Streifer and _if_ it was my office, he must have ducked in there to avoid running into your man. Doesn't that sound a reasonable explanation?'
'Very reasonable... and why did he duck out again?'
'Obviously he would have heard Peachey coming back.'
'Why wasn't he worried by the risk of meeting Peachey when he ducked in?'
'Oh, come now, Inspector, I can't work out the minute details for you...!'
'And how did he know the door was unlocked in the first place?'
'One must use one's imagination. Perhaps he took cover in the doorway, and then tried the handle...'
'Why, in fact, would he take cover at all? On Friday night he wasn't known to us, and neither was my man known to him.'
Louey chuckled softly. 'There you are, Inspector! My naïve amateur deductions don't hold water for a moment, do they? I'm afraid it's as big a mystery as ever... I would never have made a policeman.'
'One other thing,' added Gently evenly, 'how did you come to know that it was my man who saw Streifer leave your office?'
Louey's chuckle continued. 'How else could you have known about it? You admit that Streifer meant nothing to you on Friday night, so you could hardly have been making inquiries after him, Inspector...'
They had passed by the Wellesley, its wrought-iron fantasia washed and gleaming, and were approaching the weirdly incongruous skyline of the Pleasure Beach. High over all reared the Scenic Railway, a miniature Bass Rock fashioned out of painted canvas and paper mache, and under it, like a brood of Easter chicks under a hen, the gay-painted turrets and roofs of side shows, booths and the other mechanical entertainments. Harsh strains of music through the rain suggested that the Pleasure Beachers, like lesser mortals, were assuming a custom though they had it not.
Louey gestured comfortably towards the gateway. 'Rivals of mine... but they don't have a licence! Shall we stroll through?'
Gently nodded drippingly. 'I want to see the place. It's where Streifer dropped the man who was tailing him on Friday.'
'Which shows he knew his job, Inspector. Isn't this where you would come to shake off a tail?'
'I can't say I've had much experience...'
They passed under the flaunting portal with its electric jewellery. The close-packed attractions within wore a rueful look, unsupported by the crowds. Larger and more expensive pieces were frankly at a standstill – the Caterpillar had postponed its gallop, the Glee Cars their jaunting – while the smaller roundabouts and rides were operating at a profit margin which was doubtful. Booth attendants stood about in each other's stalls. They were drinking tea and staring around them morosely. The owner of the Ghost Train, for want of something to do, was riding round in his own contraption, but all its promised thrills seemed unable to raise the siege of boredom which had invested his countenance.
'Of course there's Frenchy,' brooded Gently, obstinately undiverted by all these diversions.
'Frenchy?' echoed Louey indifferently, 'is she mixed up with the business too? She took a hint the other night, Inspector. She hasn't been near the bar since then.'
'Stratilesceul was a client of hers... she went off to the North Shore with him in a taxi just before he was murdered.'
'Ah, that accounts for a rumour I heard that she had been arrested.'
'You heard such a rumour?'
'We're for ever hearing them in our business.'
'Undoubtedly... you are very well placed.'
Gently halted to inspect the front of a sideshow. It was an exhibition of methods of execution through the centuries and was advertised by some particularly lurid illustrations. He seemed to be strangely fascinated.
'And she will have given you some useful information?' suggested Louey, moving on a step impatiently.
'She knows a good deal... she'll be a devastating witness.'
'There would be some danger in it for her.'
'Danger? With police protection?'
Louey turned his back on the sideshow and busied himself with lighting a cigarette. 'If this murder was the work of an organization – and you don't seem to be in any doubt about it now – then there would be a very real danger for anyone bearing material witness. Men can be hanged, Inspector, but organizations cannot. And my feeling is that a person of Frenchy's kidney wouldn't risk too much for pure love of our excellent police force.'
Gently stooped to get a closer view of a gentleman who had been given too long a drop, with the usual top-secret result. 'You know Frenchy well?' he asked carelessly.
'I? Not apart from running her out of my bar on several occasions.'
'Dulton... Dulsome Street is where she lives.'
'Dulford Street, Inspector.'
'That's right. You've been there?'
'Not visiting Frenchy, if that's what you mean.'
'You're sure of that? Not in the last day or two?'
'Quite positive, Inspector. My tastes have never been that way inclined.'
Gently straightened up slowly. 'Odd,' he said, frowning.
'What's odd about that?'
'These two cigarette ends.' Gently felt in his pocket and produced a crumpled envelope. 'There... you see? Your blend of Russian. I found them in an ashtray in Frenchy's bedroom yesterday afternoon.'
Louey poked at them with a gigantic finger and nodded heavily. 'You're right, Inspector... it is my blend.'
'I was sure of it... I was feeling positive you'd been there.'
The grey eyes rested on his firmly, the flecked pupil seeming curiously larger than its neighbour. 'Isn't it a shame, Inspector,' purred Louey, 'I thought my cigarettes were exclusive. And now, in the commission of your duty, you've proved that someone else in Starmouth smokes them too... at least, I take it, it was in the commission of your duty?'
Gently shrugged and shoved the envelope back into his pocket.
The Scenic Railway had its shutters up, though someone was tinkering with one of the trolleys. It wasn't quite so impressive on a nearer view. Its cliffs and crags were so palpably props, its tunnels and bridges so contrived. And the rain made it look sorrier still, a great, hollow, sodden mockery. Gently took refuge in a peppermint cream as they squelched past it. If only he'd thought to bring a more reliable pair of shoes...!
'I suppose I don't have to ask you to account for your movements last Tuesday night?' he growled, as they got out on to the promenade again.
'But of course!' Louey chuckled, as though he welcomed the inquiry. 'I was having a little party in the back... Peachey, Artie, Tizer and some more of the boys. You ask them, Inspector. They'll all remember my party on Tuesday night.'
'I'm sure they will. And of course it went on till late?'
'Not terribly late. I cleared them out at two.'
'Just late enough, in fact.'
'Well... it was late enough for me.'
'And that would be your story – supposing you had to have a story?'
'Certainly, Inspector. Why should I tell any other?'
'There's no knowing what Frenchy may say.'
'She's a woman without character.'
'Or Streifer, for example.'
'Streifer?' Louey hung on to the word, as though expecting an explanation.
'And then there's your car,' continued Gently, ignoring him. 'Was that borrowed or something on the Tuesday night?'
'My car...?' This time the inquiring tone had an edge of anxiety.
'You lent it to someone – and they went up to North Shore?'
'I don't understand, Inspector. My car would have been in its lock-up in Botolph Street.'
'Even though it was seen somewhere else?'
'That would hardly be possible...'
'Then you didn't lend it to anyone?'
'No. It was never out of the garage.'
'So the people who saw it at North Shore would be liars?'
'They were certainly under a misapprehension...'
Gently flicked briskly at his over-worked trilby. 'You'll have got rid of the ring, of course... that was too much of a coincidence.'
The big features relaxed and there was a glimpse of gold. 'If you don't mind me saying it, Inspector, I think we had better consider that ring to have been an illusion.'
'I'm not subject to illusions, Louey.'
'But just once, perhaps, in a long career?'
'Not even once, and certainly not prophetically. I didn't know the TSK or its secret sign existed when I saw your ring, but I knew where I'd seen it before when it turned up a second time.'
'A trick of the memory, perhaps.'
'The police aren't much subject to them.'
'Well, shall we say rather dubious evidence?'
In a court of law it would be for the jury to decide.'
Louey laughed his low, caressing laugh. 'How we talk, Inspector... how we do. But I like these examples of your official approach to a problem. It's comforting to feel that the guardians of our law and order work so efficiently and so intelligently. As I said on a former occasion, I could only wish you had more promising material to deal with in the present instance.'
'I'll make do,' grunted Gently, 'it doesn't seem to be running out on me at the moment.'
Louey shook his head with a sort of playful sympathy. 'I respect your attitude... it's the attitude one would expect and look for in such a man as yourself. But honestly, Inspector, when one takes stock of the situation... for instance, this Streifer. What can you do about him? You can connect him with the murdered man in a dozen ways, you can show he was the most likely one to have done it – but what's all that worth when you haven't got a scrap of proof that he did it? I don't have to remind you of our careful court procedure. In some countries Streifer would be executed out-of-hand on a tenth of the evidence... and perhaps you'll allow, without too much injustice. But here you have to convince your jury. Here you are obliged to go to fanatical lengths to show proof and double proof. And you don't seem to have it, Inspector. You are faced with a planned execution, the details of which have been efficiently erased. I've no doubt that a jury would convict Streifer of something – there must be several lesser charges you could bring – but as a betting man, Inspector, I'm willing to give you ten to one they never convict him of murdering Stratilesceul.'
Louey took a farewell puff at his cigarette and seemed about to toss it away. Then he changed his mind and with a gilded smile handed it to Gently.
'Another one for your collection!'
Gently nodded and extinguished it carefully.
'The previous remarks,' continued Louey, watching him, 'supposing you have in fact arrested Streifer...?'
Gently tucked away the sodden end without replying. Louey nodded as though that were sufficient answer.
'And I don't think you will, Inspector... I don't really think you will. If he was, as you say, a member of the... what was it? A secret police? I imagine he will know his way out of a country... don't you? Especially with the assistance we must assume he will get from his organization over here.'
Gently stuck his hands in his pockets and plodded on. He seemed completely immersed in something taking place over the pale sea-horizon.
'It's wrong of me,' mused Louey, 'I shouldn't say it... but I can't help feeling a little sympathy for the man.'
'Sympathy? For a cold-blooded murderer?'
'Not a murderer, Inspector... an executioner, I think you must call him.'
'Stratilesceul's hands were tied – do you sympathize with that?'
'You're forgetting, Inspector, we also tie a man's hands for execution. If killing is the order, one may as well kill efficiently.'
'But we don't torture, Louey. Stratilesceul was burned with cigarettes.'
'Our torture is mental, Inspector... it lasts longer, and it isn't done for useful ends, such as eliciting information. No... I'm sorry. You must permit me to feel some sympathy for Streifer. He did what he did in the service of an ideal, rightly or wrongly... you really mustn't equate him with even the common hangman.'
Gently's shoulders hunched ever higher. 'He was paid, wasn't he... just like the common hangman?'
'Naturally, a labourer is worthy of his hire. But the pay wasn't his motive, you know. It wouldn't be an adequate incentive to such risk and responsibility. Your hangman is a mere assassin... you hand him his thirty pieces of silver and say Murder; we have bound your victim. And he murders, Inspector. He has your full protection. His crime is written up to humanity and he departs to spend the blood-money. Is this the way of the man you want to hang? Is this the way of any of the men you hang?'
'At least we kill only the killers...'
'Is that better than killing for an ideal?'
'It is an ideal – to protect people on their lawful occasions.'
'If only it protected them, Inspector... if only it did! But your ideal is a pathetic fallacy, I'm afraid. Of course it's wrong to say this... I understand your position. Your duty is to catch a criminal and judgment is elsewhere. But I want you to understand me when I say I feel a little sympathy with Streifer... we can talk together, Inspector. You are a man of intelligence.'
They had come to the end of the town, a straggle of houses on one hand, wasteground and the beach on the other. Louey paused as they came abreast with a decaying pill-box.
'This is as far as I go, fair weather or foul.'
Gently nodded woodenly and gave his trilby a further flick. Then he turned to face the two grey eyes which rested on him confidently, almost affectionately.
'I'm glad you made the point...' The eyes were interrogative. '... about my duty. It _is_ to catch the criminal.'
Louey's enormous head tilted backwards and forwards almost imperceptibly.
'And since I'm in betting company, Louey, I'll take you at the odds. Wasn't it ten to one you quoted?'
'At ten to one... and Louey always pays.'
'I'll have a pound on. You can open my account.'
The grey eyes flashed and the big man burst into laughter.
'You're on, Inspector... the first policeman I ever had on my books!'
Gently quizzed him expressionlessly from the depths of his comfortless collar. 'Let's hope you're lucky,' he said, 'let's hope I'm not the last.'
The lonely phone-box had a tilt in it, due to the subsidence of its sandy foundation. But it was dry inside and Gently took time off to light his pipe before getting down to business. He gave headquarters' number.
'Get me Inspector Copping.'
Copping arrived in fairly prompt switchboard time.
'Gently here... are we still entertaining Frenchy?'
'Entertaining's the word!' came Copping's disgusted voice. 'She's been yelling her head off since they brought her back... says she wants a lawyer and that we're holding her under false pretences.'
Gently grinned in a cloud of pipe-smoke. 'She's got her bail... what more does she want?'
'The cash, apparently... you seem to have pinched her at the end of the month.'
'Well... keep her nice and cosy. Has anything else come in?'
'Not a darned thing.'
'Have the lab made anything of that paper?'
'They say it's manufactured in Bristol and used for packing mattresses. I've got a man going round the stores trying to match it.'
'No prints worth having?'
'Nothing anybody's heard of.'
'You haven't traced that taxi?'
'Not so's you'd notice it.'
Gently clicked his tongue. 'It's a wet Monday all right, isn't it? Is Dutt anywhere handy?'
'He's hanging about waiting for someone to bail Frenchy.'
'I want him for a job... one of your own men will have to watch our Frenchy.'
The phone at the other end was laid down and Gently whiled away the odd moments watching two raindrops making tracks down the ebony panel in the back of the box. Then Dutt's chirpy accents saluted him from the receiver.
'Yessir? You was wanting me?'
'Yes, Dutt... I want you to do a little scouting in your old pitch in Botolph Street. There's a lock-up garage there where Louey keeps his car. You might find out if anyone noticed the car being used on Tuesday night...'
'Yessir. I think I knows the very garage you're talking about.'
'Stout fellow, Dutt. And don't forget your mac.'
'No, sir! Don't you worry!'
Gently eyed the rain-swept vista outside his box with a jaundiced stare. 'And while we're at it, Dutt, get them to send a car to pick me up at South Shore... I've had all the constitutional my constitution will stand for one wet day!'
CHAPTER ELEVEN
THERE WAS A hiatus in the proceedings and the super, excellent man, had scented it out with his keen, service-minded nostrils. Gently had come to a standstill. His case was bogging down. He had pushed it up the hill with his bulky shoulder until he was in hailing distance of the top and now, with the deceptive vision of arrests and charges dead ahead, he was stuck fast as though he had run into an invisible barrier. It was a sad sight, but not an unexpected one. The super had had a strong intuition all along that this was how it would wind up. Because he knew something about secret agents, did the super. He had come across them before in his long career and he could tell Gently, if Gently was harbouring any illusions, just how slippery these birds inevitably were...
'You see, they _plan_ their murders... that's the vital difference between them and the ordinary homicide. They know what we'll do and they take damn' good care to protect themselves.'
Gently looked up from a large-scale map and smiled with an irony which the super was unable to appreciate. 'In fact we're... "faced with a planned execution, the details of which have been efficiently erased".'
'Precisely.' The super cast him a suspicious glance. 'We may as well face it, Gently. We're not infallible. We make use of our skill and technique to the best of our ability, but the people on the other side start with an enormous advantage and if they use it intelligently then we're batting on a pretty sticky wicket.'
'I know, I've heard it once before today. We haven't got Streifer, we can't prove he did it and' – he rustled the map on his desk – 'we don't even know where it was done.'
'Well – those are the facts, Gently, and you'd better add that we've exhausted most of the chances of improving on them. Oh, I don't want to be discouraging, and I'm certainly not disparaging all the sound work you've put in getting this case into perspective, but you are scraping the bottom of the barrel now and getting precious little for it – and every hour that passes makes it less and less likely that we shall lay hands on Streifer. This isn't his first job here, you must remember. The Special have been after him before without finding hide nor hair of him and there's no reason to expect they'll be luckier this time.'
'You think I should write my report?'
The super's grizzled brows knitted in a frown. 'I'm not saying that, Gently. I'll leave you to be the judge of when you can no longer usefully continue the investigation. The point I'm making is that we should look at the thing realistically. For instance, those men of mine at the stations and the bus terminus.'
'You can have them back now,' Gently shrugged.
'And the two men you put on the taxis... they've checked and re-checked every hackney-carriage driver in town.'
Gently looked obstinate. 'That taxi must be somewhere.'
'You say it must – but your only evidence is Wylie's and Baines's statements. I wouldn't be inclined to give it too much weight if I were you.'
'They'd no reason to lie.'
'They'd every reason to lie. They wanted to make it seem that Frenchy was the principal... it could just be that she's as innocent as she says she is.'
Gently shook his head impatiently. 'Baines wasn't lying. The statements agree except where Wylie is trying to whitewash himself.'
'The fact remains that no taxi driver in town remembers the incident and nobody's got records of such a journey. Of course it's just possible that it was a taxi licensed at Norchester or Lewiston that picked them up... you know the distances, you can judge how likely it would be.'
'I'm sorry... but that taxi has got to be found.'
'Then what do you suggest – a general check-up of all the taxis in a fifty-mile radius?'
'It may come to that, though first I would like your men to re-check their re-check... it's surprising how repetition sometimes jogs people's memories.'
The super gave Gently what from meaner men would have been classed as a dirty look.
'Very well... you know your job. But remember that I've got plenty of routine work going begging when you're through with the bottom of the barrel...!'
It was a good exit line and the super duly acted upon it. Gently folded up his map with a sigh and stowed it in the drawer with the Moriarty. He didn't blame the super. He would have felt exactly the same in the great man's shoes. Police routine didn't stop because a couple of Yard men were trying to hatch a murder charge... it just became more difficult. And when the murder charge didn't look like hatching anyway, well then the Yard men started to become a nuisance about the place. The trouble was that the super hadn't got an incentive any more. He was reasonably happy with the way things had panned out. His corpse was no longer an unsolved mystery, he had pinched a small handful of auxiliaries in the case and if the principal had made tracks for a far country it wasn't through any dereliction of the super's duty... All that really concerned the super now was the propitiation of Christopher Wylie and the making of his peace with the chief constable.
Gently sighed again and unhooked his clammy raincoat. There were times when being a Central Office man wasn't all it was cracked up to be.
Accoutred for the fray, he went along to the canteen for a preliminary cup of tea. It was a quiet time there. He had the gloomy room all to himself. Behind the scenes could be heard the chink and clatter of washing-up in progress, but the only other excitement the place afforded was the distant view of someone working on a car under a tilt. Gently sauntered to the window to watch the operation while he sipped. There was something soothing about watching other people grapple with their troubles.
And then, perhaps inspired by the tea, a dreamy expression crept into his eye. He drew closer to the window. He pulled back one of the blue cotton curtains. At one stage he was even pressing his nose against the pane.
Finally he put down his cup half-finished and let himself out into the yard by the side-door.
It was an elderly car of the high-built and spacious days, and the elderly man who worked on it, though not high built, was spacious also. The dungareed rear end of him which protruded from the bonnet was particularly spacious, and so too was the language which rose in a muttered stream from somewhere in the interior. Gently hooked his fingers in the climb-proof wire fence which surrounded HQ property and conducted a leisurely survey.
'Having a spot of bother?' he inquired affably.
The stream of language faltered and a red, moon-like face disengaged itself from the oily deeps.
'Bother! Can't you hear I'm a-havin' some bother?'
'Well... it sounded like a big end gone, to say the least.'
The spacious one heaved himself upright and shored his bulk against the off-side mudguard. 'Jenny!' he observed feelingly, 'that's the bloomin' trouble – Jenny!'
'There's a woman in the case?' queried Gently, who wasn't mechanically minded.
'Woman? Naow – the Jenny! Stuck away there at the bottom till it's nearly draggin' on the ground – an' they must know it's goin' to give trouble – Jennies _allus_ give trouble!'
He waved an adjustable at Gently as though daring him to contradict, but Gently's interest had slipped to some crude white lettering just visible on the uptilted bonnet. It read: 'Henry Artichoke, Hire Car, 76 High Street.'
'This your car?' he asked casually.
''Course it's my car – who's did you think it was?' Mr Artichoke gave the vehicle a glance of mingled affection and exasperation. 'Good now as half your modern tin-lizzies – only thas like me, getting old...'
Gently nodded understandingly. 'And how's business with you these days?'
'Business? Well – I don't complain. Though I aren't saying it's like it was in the old days—!'
'Too many charas and coach-trips.'
'An' all these new-fangled cars about... still, don't run away with the idea that I'm complainin'.'
'Were you doing much last week?'
'I was out on a trip or two – can't do without me altogether, you know.'
'Last Tuesday, for instance. Did you have a trip that day?'
Mr Artichoke ruminated a moment and dashed away a raindrop which had leaked on to his oily cheek. 'Tuesday... that was the day old Hullah was buried. Yes. Yes. I had a couple of trips on the Tuesday... in the mornin' I took Sid Shorter over to see his missus at the nursing home. Then last thing they had me out to fetch an old party and her things from Norchester – that's it!'
'What time would that have been?'
'Well, I hadn't got really set down at the "Hoss-shoes"... that couldn't have been much after seven.'
'Then you went to Norchester to pick her up?'
'Her'n her things – you'd be surprised what the old gal fetched away with her!'
'Made you late, I dare say...'
'Late enough so's I didn't get into the "Hoss-shoes" again...'
'It was after ten by the time you'd got her unpacked?'
'As near to it as makes no difference... parrot she'd got too – damn' nearly had my finger as I was carting it in!'
'And where did you take her... what was her new address?'
'Oh, she was goin' to live with the Parson of St Nicholas.'
'Is that the big church?'
'No – that's St John's. St Nicholas is the one down in Lighthouse Road.'
'You mean down at South Shore?'
'That's right... the one with a herrin' stuck up for a weather-vane.'
Gently relinquished his grip on the wire fence and dived his hand into a pocket that rustled. 'The Front – was it very busy when you came back that night?'
'Huh! Usual lot of rowdies – kids, the best part on'm.'
A peppermint cream came to light and lay poised on a stubby thumb. 'Did you have any luck... like picking up an odd fare?'
Mr Artichoke raised two round eyes grown suddenly suspicious. 'Here!' he exclaimed, 'come to think of it, I don't like the side of the fence you're standing on – I don't like it at all!'
'It's the honest side, Mr Artichoke...'
'That's as may be – I don't think I like it!'
The peppermint cream went into Gently's mouth and was chewed upon thoughtfully. Mr Artichoke watched the operation indignantly, his broad face flushing a deeper shade of red. One would have thought there was something almost indecent about eating a peppermint cream.
'Now look, Mr Artichoke, I think you're in a position to help me in a rather important matter. I know you haven't got a hackney-carriage licence and that it was an offence for you to pick up a fare in the street, but if you picked up the people I think you did, then between you and me there won't be any charges... is that quite plain?'
Mr Artichoke nodded non-committally, but kept his mouth tight shut.
'Well then... did you or didn't you?'
Mr Artichoke shrugged his heavy shoulders and stared at the adjustable in his hand. 'That depends a bit on who them people was, don't it?' he remarked tentatively.
'I want you to tell me that.'
'But how am I goin' to know if they're the ones I shan't get pinched over?'
Gently returned the shrug. 'I've got a very bad memory except for criminal offences.'
Mr Artichoke brooded some more on the adjustable. 'Just suppose there were two of them – a male and a female. Is that somewhere about the mark?'
'It's right on the target.'
'And suppose this female was a blonde female – one of them there that work up this way during the season... am I still going the right way?'
Gently nodded with deliberate slowness.
'And suppose this bloke was a foreigner with a beard, dressed a bit flashy, and answering to the name of Max – and suppose they wanted taking to a house on the cliff which as far as I know has been empty for the last five years. Would I still be heading straight?'
There was the briefest of wavers in Gently's nodding and a smile little short of angelic crept over his face. 'Mr Artichoke... you've just answered the sixty-four dollar question, whether you know it or not.'
'Eh?' queried Mr Artichoke.
'The sixty-four dollar question,' repeated Gently. 'Now just stop here. Don't move. Don't go away. I'm going to have a short chat with the superintendent about his man-power problem and after that we'll make a little trip to North Shore together... who knows? We may even be lucky enough to find a tenant in that house on the cliff...'
Copping made one of the party and Bryce, at Gently's request, was added to the strength. Copping became highly indignant when he heard about Mr Artichoke's activities.
'And after all the ratepayers' money that's been spent trying to find the cabby! What's the use of issuing these licences if a lot of pirates come along and gum up our investigations for us?'
Gently clicked his tongue. 'He was only turning a slightly dishonest penny.'
'We might never have caught up with him... you admit it was pure accident.'
'Luck,' said Gently, 'you have to cultivate it in the Central Office...'
Copping snorted. 'We shouldn't have needed luck. Routine will catch a criminal if everyone is being completely honest...!'
Under Mr Artichoke's directions they proceeded north along the main Norchester road. The dreary suburbs passed by, the expensive splendours of High Town and finally the long, level, white-railed expanse of the race-course, its empty stands lifted gloomily against the rain-pale sea.
'Steady!' warned Mr Artichoke, made uneasy by the driver's reckless and newfangled technique, 'we're turning off here – if you can pull up this side of Barston!'
The driver slowed down to a dangerous thirty.
'There!' exclaimed Mr Artichoke, 'Up that little loke. There's only one house up there, so I shan't have made a mistake.'
'It's "Windy Tops"' muttered Copping, 'it belongs to one of the Thorners of Norchester.'
Gently glanced at him questioningly.
'We had some trouble with them a few years back. The Borough Engineer scheduled it as being unsafe because of cliff erosion and they made a case of it. He won the case, but there hasn't been a cliff-fall in that area from that day to this. Just mention "Windy Tops" if you want to get him in the raw.'
'It's been empty all the time?'
'Naturally. Nobody's allowed to live in it. The B.E. is just living for the day when it goes over the top.'
The narrow road skirted the northern end of the racecourse, crossed the railway line and turned abruptly left. Here the ground rose suddenly to form the first of a line of crumbling gravel cliffs and perched at the top, looking in no-wise conscious of its danger, was a small but well-architected modern house.
'Looks safe enough,' Gently murmured.
'Probably is,' grunted Copping, 'but the B.E. got rapped on the knuckles about a row of cottages that went over... he hasn't taken any chances since then.'
The road came to an end at a spacious turning place and the gate to 'Windy Tops'. Bryce was sent round to the back while Gently and Copping advanced on the front. The garden had run to seed and there was grass growing out of the crazy paving, but the house itself seemed in a fair state of preservation and Gently found himself sympathizing with the Thorners in their reluctance to abandon the place. He stooped to inspect the crazy paving.
'Someone's been up here recently all right.'
The grass had been bruised by trampling feet. But Copping was already trying the front-door handle and apparently expressing surprise at finding it locked against him. He ran an eagle eye over the front of the house and thus discovered a partly-open window which Gently had noticed as they got out of the car.
'Easy!' called Gently, 'there may be some interesting prints about.'
Copping whisked up the sash and dumped himself over the low sill. Gently followed him at a more dignified pace. It was a large room and had probably been the lounge, but it was quite empty except for some ashes of burnt paper in the grate. Copping swooped on them, sniffing like a well-trained hound.
'They're fresh!' he exclaimed, 'they haven't been there longer than a few days.'
Gently nodded and applied a speculative finger to the light switch. Pale radiance shone from an unshaded bulb.
'Every modern con... and I think I can hear a cistern hissing somewhere.'
'He's been living here!'
'Undoubtedly...'
'He might be here now!'
'There's just the remotest chance...'
The efficient Copping needed no more. He invaded the house like an unleashed jumping-cracker, pouncing from room to room, poking in cupboards, surprising the backs of doors and generally making life hectic for anything in the shape of a secret agent.
'He slept up here!' came his muffled cry from above-stairs, 'There's a mattress and some blankets... cigarette-ash... empty matchbox!'
Gently shook his head sadly and went to unlock the kitchen door. The cupboard was bare, he knew it intuitively. There had been that chance, that one chance, that Streifer had decided to lie low until the heat was off, but he had sensed it evaporating the moment he had set foot in this so-silent house. He called to Bryce.
'Any signs of life out there?'
'No sir, nobody – not even on the beach.'
'What about the garage?'
'The door's on the latch – there's nothing in there except a pair of old tyres.'
'Well, come in and give Inspector Copping a hand upstairs. You'll have to get into the loft somehow.'
Bryce came in without much enthusiasm and went up to join his superior. Gently remained below in the kitchen. There were plenty of signs there of recent occupation. On the draining-board stood a plastic cup and plate with a knife and fork, all dirty. A hot-plate was plugged in at the electric switch-point, upon it a tin kettle and nearby an aluminium teapot. In a wall-safe were a tin of condensed milk, tea, sugar, a couple of rolls... stale of course, but no staler than Saturday's rolls usually are on Monday afternoon... butter and an unopened tin of anchovies. By the wall leaned two cheap folding-stools. Under the sink stood a rusty distemper-tin containing refuse. And there were several newspapers, including Sunday's, and a pile of brown paper. Gently unfolded a _Sunday Express._ It had had a cutting taken from it. He unfolded three others. Each had cuttings taken from them.
'He's hopped it all right.' Copping came in, dusty and aggrieved. 'Bryce is up in the loft now, but he'd have hardly got up there without someone to give him a bunk... there's nothing to stand on. I'm afraid we're just too late... they always seem one jump ahead, these bastards!'
Gently pointed to the pile of brown paper. 'What do you make of that?'
Copping stared intelligently. 'Looks as though he bought a geyser or something.'
'Was that mattress upstairs a new one?'
'Brand new – and so were the blankets.'
'And what does that suggest to us... knowing what we do?'
There was a pause and then the divine spark fell. 'By glory – it's the same paper that was used to wrap the clothes!' Gently nodded approvingly. 'Used to wrap mattresses – and there's the new mattress and you can _see_ it's the same paper – it's got that crimp in it, just the same!'
'And it's had a piece torn off it... just about the same size.'
Copping's heavy features flushed with excitement. 'We've got him, then – we can tie him in! We've got proof now, good, hard, producible proof – the sort of thing juries love – material proof!'
'Just one thing, though,' murmured Gently.
'Proof!' boomed Copping, 'what more do we need?'
'We need something we haven't got right at this moment and that's the initial proof that Streifer was ever in the house at all.'
Copping faltered in his raptures. 'But good lord... it _must_ have been him!'
Gently shook an indulgent head. 'Remember that jury and keep your hands to yourself. Don't touch the paper, the taps, the dishes or anything else that's lying about. I suppose it's too late to worry about the doorknobs. As soon as Bryce is through having fun in the loft he'd better light out for your print man. It isn't likely that Streifer was too careful here... he expected to be far otherwheres when and if we ever identified the place.'
'And how right he was – how dead bloody right!'
Gently hunched his shoulders soberly. 'He's a man like you and me. People don't become magicians when they join a secret police.'
'It's enough to make you think so, the way this bloke keeps himself lost.'
A dishevelled and wash-prone Bryce was dispatched in the car and Gently, having completed his tour of the house, went out to inspect the grounds. They had nothing relevant to disclose. The tumpy wilderness which had been a lawn, the nettled and willow-herbed flower-beds, these looked as though a full five years had elapsed since a foot had trodden there, or a hand had been raised in their defence. Gently went round to the Achilles heel, the seaward side. Not more than five yards of stony land separated the house from its inevitable tumble to the beach.
'Can't last another winter,' observed Copping knowingly, 'should have gone in the January gales. It was sheer cussedness that made it hang on... there were falls everywhere except here.'
Gently approached timidly to the treacherous edge. Seventy feet below the wet sand looked dark and solid. North, south, the sullen lines of slanted combers fretted wearily, told their perpetual lie of harmlessness and non-aggression. Down by the racecourse a lonely path wore its way to the beach.
'That's it,' muttered Gently, 'they carted him down there. How far would you say we were above the Front?'
Copping did some calculations. 'Two miles, about... might be a trifle less.'
'It just about tallies... I was reckoning on two miles. They dumped him in down there in the ebb, expecting the current to pick him up and carry him right down south. It was just rank bad luck that he finished up on their doorstep again...'
'You're positive it was done in the house?'
'Quite positive... I can see the whole picture now.'
'There weren't any signs of it – no blood-stains or anything.'
Gently smiled grimly. 'Professionals, Copping. He wasn't hacked about. Didn't you notice how little blood there was on the clothes?'
'And if they'd used violence it wouldn't have shown much, not in an empty house.'
'But they didn't use any... they didn't have to. He was delivered right into their hands, unarmed and unsuspecting. My guess is that the first thing he knew about it was an automatic dug into his ribs... you don't argue with a thing like that in the first instance. By the time he'd weighed the situation up his hands were tied and he hadn't any option. It's a classic case, Copping. Only our Delilahs come a bit coarser these days.'
'Delilahs!' Copping gave a laugh. 'He must have wanted his brains tested to take up with a mare like Frenchy.'
There was a sound of two cars pulling up and they returned to the front of the house. To their surprise it was the super who came stalking up the crazy-paving. The great man had a taut look, as though primed with high enterprise, and having stalked to the end of the crazy-paving he halted smartly, straddled his legs and quizzed Gently with a sideways look in which were blended both jealousy and admiration.
'All right!' he rapped, 'you're a happy man, Gently. You know your job, and I'm just a blasted ex-infantry officer who's got shoved into a rural police force!'
Gently bowed modestly, as though disclaiming praise from such high places. The super snorted and directed his gaze at a 'Windy Tops' chimney.
'So they got him!' he jerked from the corner of his mouth, 'laid for him on information – Special and the Limehouse lot – picked him up going aboard a Polish tramp. Nearly shot two men and laid out a third. They're bringing him up now and a big-shot from Special along with them.'
There was a gratifying silence.
'You mean... _Streifer_?' gasped Copping.
The super withered him with an acetylene flicker of his eye. 'We're not after Malenkov – nor Senator McCarthy! And just to temper the general glee I may as well add that Special are not the teeniest weeniest bit interested in our lousy little homicide. They couldn't care less. What they're coming for is the whole TSK Party handed to them on a platter and if we can't produce it then they aim to make life irksome in these parts – you understand?'
Gently nodded his mandarin nod. 'I've worked with Special...'
'Then you know what's coming – and it'll be here just after tea! So get your facts marshalled, Gently. There's a top-level conference staring you in the face. Amongst other things I've had to pull the CC off a theatre party he's had planned for the last six weeks... that's one nasty-minded person who's going to be there, for a start!'
CHAPTER TWELVE
ASSUREDLY THERE WAS was an array of formidable talent lined up in the super's office on that grey August evening. It required the impression of seating accommodation from several other departments and it was sad to see so many men of such lustre crammed together like constables at a compulsory lecture. As far as sheer superiority of rank was concerned, the home team had a clear advantage. They were led by the Chief Constable of Northshire, Sir Daynes Broke, CBE, ably supported by his Assistant CC, Colonel Shotover Grout, DSO, MC, with the redoubtable flanks of Superintendent Symms and Inspector Copping. But rank, of course, wasn't everything. There was a matter of quality also, and in this respect, to judge from their attitude, the visiting team felt themselves to have the edge. They were four in number, a sort of Special commando unit. Their ranks comprised Detective Sergeants Drill and Nickman, as dour a pair of bloodhounds as ever signed reports; their lieutenant was Chief Inspector Lasher, a man who had earned the hearty dislike of a select list of international organizations. But it was their No. 1 who really set the seal on the outfit. You could feel his presence through six-inch armour plate. He was a comparatively small man with a large squarish head and blue eyes that glowed hypnotically, as though lit by the perpetual and unfaltering generation of his brain. His name was Chief Superintendent Gish and the date of his retirement had been set aside by the entire world of espionage as one for public holiday and heartfelt rejoicing.
Between these two mighty factions Gently, the lonely representative of the Central Office, felt somewhat in the character of a light skirmisher. He'd got a nuisance value, they would probably concede him that, but otherwise he was merely there as a point of reference. So he squeezed himself into a seat behind Detective Sergeant Nickman, and contented himself with issuing entirely unauthorized smoke-rings.
Chief Superintendent Gish said: 'I want to impress on everybody concerned the urgency and importance of our mission down here. We have sent you a certain amount of information already to assist you in the homicide investigation... we've got your man for you and I take it you have prepared a case against him. If you haven't, it doesn't matter because we can put him away ourselves on a certain charge of sabotage. The importance and urgency of this business lies elsewhere and it's that I want to talk about.'
He paused, not so much for comment as to drive home his conviction that comment was superfluous. Gently puffed a sly ring over Detective Sergeant Nickman's right ear. There was a general silence on all fronts.
'Very well, gentlemen,' continued Superintendent Gish, his floor confirmed. 'Now it must certainly have occurred to you, though possibly you have been unable to trace it, that Streifer has received assistance in what he did here. The circumstances of the crime as they are known to me leave no doubt about that. They are typical of the organized killing, the sort that we of the Special Branch are all too familiar with. Now that in itself is an important and urgent matter, but it becomes doubly so in the light of what I am about to tell you.
'The TSK Party came into existence shortly after the war. Officially it has no connection with the authorities on the other side of the Curtain, but I don't have to tell you that it wouldn't have thrived so long as it has done without connivance, and probably assistance, from the gentlemen over the way. It contains a strong Trotskyite element, which no doubt accounts for the nomenclature, but it pursues its aims not by assassination – though it isn't above it – but by extraordinarily well-executed sabotage.
'We first came across it in Yugoslavia. Later on it turned up in Czechoslovakia, Western Germany, France and the Suez Canal Zone. Three years ago the FBI were considerably shaken up to find it active and flourishing in the States – not just one or two agents but a complete organization, with some very dangerous contacts inside two atomic research stations. Fortunately they got on to them in time and pretty well stamped them out, though if this little affair is anything to go by the TSK still have a foot-hold over there. In the suitcase Streifer was carrying there were $1,000,000 in counterfeit bills.
'Over here our first brush with them occurred at about the same time. They suborned a couple of atomic research physicists and when the balloon went up, I regret to say that they succeeded in getting one of them out of the country. After that we had the sabotage trouble down at Portsmouth in which Streifer was identified as the agent. There was nothing else then for some time. But about a year ago, as you may remember, a rash of naval sabotage broke out from Scapa down to Plymouth and it didn't take us long to discover that the TSK were back, this time in some strength. In fact, gentlemen, they had built an organization over here, an organization similar, though perhaps not so extensive, as the one they had built in the States.
'I need hardly mention that we have left no stone unturned to get at grips with this organization. Chief Inspector Lasher and myself were assigned to the task and we have pursued every opening and lead with the not inconsiderable resources at our command. We have had some success. We have arrested and deported or imprisoned a round half-dozen of agents. But we have never been able to locate the centre, the headquarters of the organization – there are never any lines back to it. The men we arrested wouldn't talk, and the impression I received after personal interrogation was that they didn't know anyway.
'Of course, we've had theories about it. We decided early on that it was probably on the east coast. Here there's some little traffic with the other side – cargo-boats trade in and out of the ports, fishing-boats operate off-shore, liners like the _Ortory_ touch in on some pretext or another. It seemed logical to give the east coast preference. And knowing the sort of people we were up against, we didn't necessarily expect to find it in an obvious centre such as Newcastle or Hull. We felt it was much more likely to turn up in a smaller place, an innocent-seeming place... a holiday resort like Starmouth, gentlemen, with its perpetual comings and goings, its absorption with visitors, its easy-going port and fleet of fishing vessels...'
Chief Superintendent Gish dwelt fondly on his theory, as though he enjoyed its sweet reasonableness. But he had got the opposition in a raw spot. There were underground growlings from Colonel Shotover Grout, an aggressive cough from Superintendent Symms and finally the home team found its voice in an exclamation by its illustrious leader:
'But good God, man, there's nothing like that in Starmouth!'
'Indeed, Sir Daynes?' Chief Superintendent Gish looked bored.
'No, sir. Quite impossible! The Borough Police Force is one of the most efficient in the country, including the Metropolitan, and the crime figures for this town, sir, bear comparison with those of any similar town anywhere. We harbour no criminal organizations in Starmouth, political or otherwise. Starmouth is by way of being a model of a respectable popular resort.'
'Here, here!' grumbled Colonel Shotover Grout chestily. 'You are mistaken, sir, gravely mistaken.'
'I'm not prepared to say,' added Sir Daynes generously, 'that Starmouth is completely free from undesirable activity. There are features – ahem! – moral features which we would gladly see removed. But that is an evil common to this sort of town, sir, and under the present limitations forced upon the police of this country we have not the power to stamp it out, though we keep it rigidly in check. Apart from this I may safely say that Starmouth is an unusually orderly and well-policed town. I assure you that nothing of the sort you describe could establish itself here without our knowledge.'
'Quite impossible, sir!' rumbled the colonel, 'you don't know Starmouth.'
Chief Superintendent Gish let play his hypnotic blue eyes from Sir Daynes to the colonel, and back again to Sir Daynes. 'And yet you wake up one morning to find a TSK agent and saboteur lying stabbed on your beach,' he commented steelily.
'It was hardly in our province to have prevented it!' came back Sir Daynes. 'If agents and saboteurs are permitted such easy entry into this country, then responsibility for their misdeeds must lie elsewhere than with the Starmouth Borough Police.'
'I agree, Sir Daynes. My point is that the Starmouth Borough Police knew nothing of their presence until a dead body turned up.'
'And the Special Branch, sir, knew nothing of their presence until informed by the Starmouth Borough Police.'
'With some Central Office assistance.'
'Invoked in the common round of our duty.'
There was a silence-at-arms, each mighty antagonist feeling he had struck an equal blow. Chief Superintendent Gish appeared to be putting the super's desk calendar into a trance. Sir Daynes Broke was giving his best performance of an affronted nobleman. Gently, after waiting politely for the launching of some fresh assault, improved the situation by relighting his pipe and involving Detective Sergeant Nickman in a humanizing haze of Navy Cut.
'You'll have to admit,' continued Chief Superintendent Gish at length, 'that Streifer received assistance in killing Stratilesceul.'
'I admit nothing of the sort,' countered Sir Daynes warily.
'What other interpretation can be put on the facts? Is there any doubt that his hands were tied?'
'One man can tie another's hands, Superintendent.'
'He can if the other will submit to it.'
'Streifer had a gun when he was arrested. Why should he not have threatened Stratilesceul into submission?'
'Have you ever tried tying the hands of a man you are threatening with a gun, Sir Daynes?'
'He could have bludgeoned him.'
'There were no head injuries.'
'Or drugs, perhaps.'
'Where would he obtain such drugs at short notice, supposing he could have induced Stratilesceul to take them? No, Sir Daynes, it won't do. Streifer wasn't on his own. He found help here, in this town, and help for the like of Streifer can only come from one source.'
'That source, sir, need not be in Starmouth. You have offered no certain grounds for your assumption that it is in Starmouth. Since TSK agents proliferate to such an amazing extent in this country I see no reason why Streifer, having followed Stratilesceul to Starmouth, should not have summoned one of them to his assistance. He had time enough. The murder was not committed till almost a week after Stratilesceul arrived.'
'It is possible, Sir Daynes, but that is all one can say for it.'
'As possible as your own hypothesis, and a good deal more probable.'
'I beg to differ. If I thought otherwise I should not be here.'
'Then, sir, there is little doubt that you have made a fruitless journey.'
'We will defer judgment until we see the results, Sir Daynes. I do not propose to be deflected from my object.'
At this point an interruption became a diplomatic necessity and it was fortunate that Colonel Shotover Grout, who had been preparing himself with a great deal of throat work, chose the slight pause which ensued for his cue.
'I suppose we can have the fella in – question him – see what he has to say himself about the business?'
They both turned to regard the colonel with unanimous unamiability.
'I mean he's the one who knows – can't get away from that.'
Chief Superintendent Gish gestured. 'Of course he has been questioned. The results were as anticipated. You'll get nothing out of Streifer.'
'But simply as a formality, y'know—'
'This type of man never talks.'
A light of battle gleamed in Sir Daynes's eye. 'Symms!' he exclaimed, 'be good enough to have our prisoner brought in, will you?'
Superintendent Symms hesitated a moment, catching the Special Branch chief's petrifying glance.
'What are you waiting for, man?' rapped Sir Daynes, 'didn't you hear what I said?'
'I can assure you,' interrupted Chief Superintendent Gish, well below zero, 'that Streifer has been thoroughly and scientifically interrogated without the least success—'
'Superintendent Gish,' cut in Sir Daynes, 'I feel obliged to point out that Streifer is required by this authority to answer a charge of murder and that however high the Special Branch may privately rate sabotage, in the official calendar it is homicide which takes pride of place. Streifer has been brought here primarily to answer such a charge and I propose to make it forthwith. I suppose' – a sudden note of unease crept into his voice – 'I suppose a case has now been made out on which a charge can be based, Symms?'
The super looked at Copping, and Copping looked at Gently. Gently nodded and puffed some smoke at Detective Sergeant Nickman's long-suffering ear.
'Very well, then – have him brought in, Symms!'
Streifer was produced in handcuffs, presumably on the strength of his record – he certainly looked subdued enough, being prodded into the crowded office. He was a man of forty or forty-two, dark hair, dark eyes, slanting brows, a long, straight nose and a small thin-lipped mouth. He wore a well-cut suit of dark grey and had an air of refinement, almost of delicacy, about him. The only thing suggesting something else was the long, crooked scar which stretched lividly down his right cheek, beginning under the temple and trailing away at the angle of the jaw.
Colonel Shotover Grout gave a premonitory rumble. 'Cuffs, sir – take it they're absolutely necessary?'
Chief Superintendent Gish spared him a look of hypnotic pity.
'Remove them,' ordered Sir Daynes crisply. 'There would appear to be sufficient men with police training present to render the step unperilous.'
The cuffs were removed. A chair was drafted in. With a shorthand constable at his elbow Sir Daynes levelled a model charge and caution at the silent Streifer, inasmuch as he had, on the twelfth instant, with malice aforethought, stabbed to death one Stephan Stratilesceul, alias Max.
The baby being passed to Streifer, he simply shook his head.
'You don't wish to make a statement?'
'No.' His voice was harsh but not unpleasant.
'You realize the invariable penalty annexed to a conviction of homicide in this country?'
'I am not... unacquainted.'
'It is a capital offence.'
'Ah yes – England hangs.'
'Yet you still do not wish to say anything in your own defence?'
Streifer shrugged his elegant shoulders. 'Have you proof of this thing?'
'We have a very good case.'
'Enough to drop me into your pit?'
'To convict you – yes.'
'Then what should I say? Have you a confession for me to sign?'
The chief constable frowned. 'We don't do things that way. You may anticipate perfectly fair proceedings in this country. We have a case against you, but you are perfectly free to defend yourself. What you say will be equally considered with what we say in the court in which you will be tried.'
'Then I shall plead that I am innocent. What more will be necessary?'
'It will be necessary to prove it – as we shall seek to prove our contentions.'
Streifer smiled ironically and cast a deliberate glance round the assembled company. 'What pains you take! In my country we are more economical. But let me hear these contentions of yours. I have no doubt that your scrupulous system permits it.'
Sir Daynes signalled to the super, who once more communicated with Gently by the medium of Copping. Gently, however, having produced a crumpled sheet of paper, elected to pass it back to the seat of authority. The super straightened it out hastily and began to read.
'We can show that the accused, Olaf Streifer, is a member of a revolutionary party known as the TSK and that he is a member of the Maulik or secret police appertaining to that party and that previous to the present instance he has illicitly entered this country for the purpose of forwarding the aims of that party by criminal process.
'We can show that the murdered man, Stephan Stratilesceul, was also a member of the TSK party, and that he was similarly engaged in forwarding its aims.
'We can show that, on Tuesday, 5 August, Stephan Stratilesceul entered this country as a fugitive from the Polish liner _Ortory_ , which liner was at that time breaking at Hull a voyage from Danzig to New York, and that he was pursued ashore by Streifer, and that he escaped in the trawler _Harvest Sea_ , which brought him to Starmouth where he was landed on the morning of Wednesday, 6 August.
'We can show that Streifer also arrived in Starmouth, date unknown, and that he took up quarters in a deserted house known as "Windy Tops", and that he traced Stratilesceul to lodgings he had taken at 52 Blantyre Road.
'We can show that on Tuesday, 12 August, at or about 22.00 hours, Stratilesceul proceeded in a hired car to "Windy Tops" in the company of a prostitute named Agnes Meek, alias Frenchy, and that he was not again seen alive.
'We can show that his naked body, bearing four stab-wounds of which two would have been instantly fatal, as well as burns made before death, suggesting that he had been subject to torture, was washed ashore between the Albion and Wellesley Piers some time before 05.10 hours on Wednesday, 13 August, and that the time of death was estimated as being five or six hours previous, and that in the state of the tides and the offshore current then prevailing a body introduced into the sea near "Windy Tops" at or about 24.00 hours Tuesday would, with great probability, be washed ashore at the time and place at which Stratilesceul's body was washed ashore.
'We can show that following Stratilesceul's murder, burglary was committed by Streifer at 52 Blantyre Road in the hope of recovering a suitcase containing a quantity of counterfeit United States Treasury notes, but that his purpose was frustrated by a previous burglary committed by Jeffery Algernon Wylie and Robert Henry Baines on information received from Agnes Meek.
'We can show that Streifer eventually traced the suitcase to its hiding-place under the Albion Pier, that he recovered it, that he substituted for it a brown paper package containing the clothes worn by the deceased at the time of his death, and that he caused the attention of the police to be drawn to the part played by Wylie and Baines, presumably in order to mislead the investigations.' (Here the super seemed smitten by a troublesome cough and the chief constable sniffed rather pointedly.)
'We can show, finally, that the piece of brown paper used to wrap the clothes of the deceased is identical in composition with a sheet of brown paper discovered at "Windy Tops", this sheet forming part of the packing of a mattress acquired for his own use by Streifer, also discovered at "Windy Tops", and that the torn edges of the one piece match exactly the torn edges of the other piece.'
The super halted and laid down his sheet of paper.
'Excellent!' chimed in the colonel, aside. 'First-class case – magnificent phrasing!'
Chief Superintendent Gish turned his head sideways, as though he felt it unnecessary to turn it any further. 'What a pity,' he said to Gently behind, 'what a pity you couldn't have made it water-tight.'
Gently issued a quiet ring at Detective Sergeant Nickman's plastic collar-stud.
'When you've done so well... not to be able to show that Streifer was in Starmouth at the time of the murder.'
'What's that?' barked Sir Daynes, 'Not here at the time of the murder? I fail to follow you, sir, I completely fail to follow you!'
'Oh, I dare say you'll get a conviction.' The chief superintendent came back off his half-turn. 'The rest of it's so strong that it's almost bound to carry the day. But as I said, it's a pity that you have to admit a phrase like "date unknown" against the important event of Streifer's arrival in Starmouth... his defence are bound to be time-wasting and oratorical about it.'
Sir Daynes stared murder, and the chief superintendent stared it back.
'Is this a fact?' snapped the former at Gently. 'We have just heard it read,' chipped in the chief super scathingly.
Gently raised a calculating eyebrow. 'How long,' he mused, 'how long would you say it would take a man – even supposing he was a confirmed anchovy addict – to eat five average-size tins of anchovies?'
'Anchovies!' exploded Sir Daynes, 'what the devil have anchovies got to do with it, man?'
Gently shook his head. 'I was going to ask Streifer that, if he had been feeling more communicative. But there were five empty tins in his waste-bucket at "Windy Tops" and I find it difficult to believe that he consumed one whole tin each tea-time for five days together...'
'It isn't proof,' whipped in the chief super, razor-sharp.
'No, it isn't proof... just a curious example of devotion to anchovies. On the whole,' added Gently mildly, 'I was rather glad to find that a gentleman named Perkins, an employee at Starmouth Super Furnishings, was able to remember selling the mattress to a person resembling Streifer as early as Wednesday, 6 August...'
The dust died down and Sir Daynes, full of beans, returned to the problem of the reluctant Streifer.
'You have heard the case against you. I think it is plain that it requires a better answer than mere silence. In your own interest, Streifer, I advise you to be as helpful as you can.'
'In my own interest?' Streifer gave a little laugh. 'You are very kind people – very kind indeed! But what interest have I left when I am faced with this so-excellent case?'
'You will not find the police ungrateful for any assistance you may be able to give them.'
'Their gratitude would be touching. No doubt I should remember it with pleasure as I stood on your gallows.'
'If you are innocent you can do no better than tell the whole truth. You are probably aware of other charges which will be preferred if you are acquitted on this one and I can say, on certain authority, that those charges will be dropped if you give us the assistance which we know to be in your power.'
'And that would be the names of my associates in this country?'
'Their names and all the information you possess about them.'
'To turn traitor, in fact?'
'To assist the ends of justice.'
Streifer laughed again and fixed his coal-black eyes on Sir Daynes. 'Tell me,' he said, 'on this certain authority of yours – would it not be possible to forget Stratilesceul altogether if I gave this information?'
Sir Daynes jiffled impatiently, but the question pinned him down. 'No,' he admitted at last, 'that charge is irrevocable, Streifer.'
'But you could perhaps buy off the judge, or ensure that these quaint jurymen of yours returned a certain verdict?'
'Quite impossible!' rapped Sir Daynes, 'understand once and for all that such courses are not followed in this country.'
'And even if they were – even if I could be sure – even if you were to hand me a free pardon signed and sealed by your Queen herself – I would not betray the humblest comrade who marches with me towards the final liberation of mankind. That is my answer to you, the policemen. That is the only statement I wish to make. If you are just, as you claim you are just, you will take it down in writing and read it at my trial. But I have nothing more to say, excepting that.'
The silence which followed was slightly embarrassed. Sir Daynes seemed to freeze in his stern official look. Colonel Shotover Grout made rumbling noises, as though he thought the whole thing in very bad taste, and Superintendent Symms sniffed repeatedly in his superintendental way. It was the Special Branch Chief who spoke.
'You see, Sir Daynes? This is the sort of thing we are up against at every turn... you may find criminals difficult to deal with, but believe me they are child's play compared with fanatics.'
'I cannot believe he will continue in this – this obstinacy,' returned Sir Daynes, though his non-plussed tone of voice belied him, 'his life is at stake, sir. Men will attempt their defence in however desperate a situation they find themselves.'
'Not once they have become inoculated with creeds of this description,' sneered the chief super. 'They become intoxicated, Sir Daynes. They become tipsy with the most dangerous brand of aggrandizing delusion – political idealism. It means nothing for them to kill, and a triumph for them to die. We know these people. You had better let us handle them.'
Sir Daynes shook his head bewilderedly. 'I must admit that it is something new in my experience... I feel somewhat at a loss.' He glanced at the colonel. 'What is your opinion, sir?'
'Preposterous!' grumbled the colonel half-heartedly, 'unstable, sir... foreigners... unstable.'
'Then, Sir Daynes, I take it you will make no further opposition to my investigations in this town.'
Sir Daynes pursed his lips. 'If you think it is necessary it is my duty to give you every assistance.'
The chief super nodded in the comfortable consciousness of prevailed merit. 'In effect I shall be taking over the present investigations at the point where your men and Chief Inspector Gently have left off. I shall want a full report from everyone engaged on the case and in addition I intend to conduct personal interrogations to bring to light points which may not hitherto have seemed important. Inspector Gently,' – his head turned sideways again – 'I have full authority to release you and your assistant from your duties here. Later on I should like to have a private chat with you and tomorrow you will be free to return to town.'
Gently nodded his mandarin nod and slowly removed his pipe from his mouth. 'I'd like to make a point... if it isn't interrupting the proceedings too much.'
The chief super's head remained sideways in indication of his supreme patience.
'One or two side-issues have cropped up in the course of my minor activities... I would have liked another day or two to tie them up.'
'Unnecessary, Inspector Gently. They will certainly be taken care of.'
'They concern,' proceeded Gently absently, 'the organization you are interested in disbanding.'
There was a silence in the crowded room. Nine pairs of eyes focussed with one accord on the man from the Central Office.
'Of course... it's not for me to suggest the line of further investigation... I don't want to deflect the Special Branch from what it conceives to be its duty. But if they care to hold their horses for just a day or two, I feel I may be able to save them a certain amount of frustration.'
'Come to the point, man!' yapped Chief Superintendent Gish, proving for all time that his neck was fully mobile. 'What is it you're trying to say?'
'I'm trying to say,' replied Gently leisurely, 'that I'm fully aware of the identity and whereabouts of the TSK leader in this country and you could arrest him this evening... if you thought it would do you any good.'
CHAPTER THIRTEEN
THEY THOUGHT IT would do them some good for quite a long time together, did the chief super, Sir Daynes and Colonel Shotover Grout. In the first flush of enthusiasm they were for leaping into a Black Maria and descending upon Big Louey with drawn automatics and a full complement of iron-mongery. It took time and a certain amount of cold water to correct their transports. Gently was obliged to apply the latter in generous doses.
'We've got nothing on him... nothing whatever... we couldn't even take his licence away.'
'But good God, sir!' gabbled Sir Daynes, 'that ring – it's positive evidence – when he denied possession he practically declared his culpability!'
'We should never find it... he's a clever man.'
'And being able to tell you Stratilesceul's nationality when even Central Records didn't know him – it's damning, sir, absolutely damning!'
'Just his word against mine... or intelligent guessing.'
'We'd better throw a cordon round the place and raid it,' snarled the chief super, 'he'll have records – names and addresses – there'll be a short-wave transmitter somewhere.'
Gently shook his head very firmly. 'Not in Louey's place. He's far too fly. If they were ever there – which I doubt – they came out directly this Stratilesceul business got muddled.'
'But how shall we know for sure if we don't raid it?'
'We know for sure now. He would never have behaved so confidently if he'd got anything to hide.'
'There'll be something to give him away.'
'I wouldn't like to bet on it.'
'And we can't just sit around waiting for him to disappear and set up somewhere else.'
'He'll do that all the quicker if he knows you're out gunning for him.'
'I say pull him in!' erupted the colonel from his thoracic deeps. 'Confront him with the other fella – make them see the game is up!'
'I'm afraid it wouldn't have that effect, colonel... they're very old hands at this particular game.'
'But damn it, sir, we must do something!'
'Yes!' struck in Sir Daynes irritably, 'you're very good at telling us what we _can't_ do, Gently, now suppose for a change you tell us what we _can_ do?'
Gently sighed and felt about in his pockets for a peppermint cream that wasn't there.
'There's just one saving grace about this business, as far as I can see... and it's up to us to play it for all it's worth. In your previous dealings with the TSK' – he inclined his head deferentially towards the chief super – 'I think you have had to do solely with agents of the party. Is that correct?'
The chief super scowled what was presumably an affirmative.
'They were men like Streifer – men with an ideal – men who would sooner go to the gallows than give the least particle of information about the party. Now in the present instance there is a significant difference. We have here a person involved – deeply involved – who isn't a party member, who has no burning desire to liberate mankind, and who is only being prevented from giving evidence by mortal fear for her personal safety. That person is the prostitute Frenchy. She knows enough, I'm reasonably certain, to put Louey into the dock beside Streifer... perhaps somebody else too. But she's been got at. She doesn't dare testify. She's seen how Stratilesceul finished up, and no doubt she's been told that whoever she gives away, there'll always be someone left to take care of her.
'But there she is – somebody who can do our job for us. If we can only find a way to coax her to talk we shall have Louey and possibly his associates in the palm of our hand. Unfortunately it runs in a circle... we've got to pull in Louey and company before she'll talk, and before she talks we can't pull in Louey and company...'
'In fact it doesn't seem to be getting us very far, does it?' interrupted the chief super jealously.
Gently sucked a moment on his unlit pipe. 'What puzzles me is how they got her to help them in the first place,' he mused. 'I've never been able to see that quite clearly...'
'If they're terrorizing her now they could have terrorized her before.'
'I don't think so... not Frenchy. She isn't one to terrorize easily. I imagine Louey would need a corpse at his back before he could get much change out of her and the job she had to do would be better done in the spirit of co-operation than in the spirit of coercion.'
'Well then – she was paid.'
'But she didn't have any money.'
'Of course not!' snapped the chief super, 'her boyfriend would have had it.'
'She doesn't admit to any boyfriend, not even to get herself bailed.'
The chief super drew a deep and ugly breath. 'It isn't _getting_ us anywhere!' he bawled. 'Does it matter two hoots how they got her to do it? The fact is that she did do it, and precious little help it looks like being to us!'
Gently shook his head in respectful admonishment. 'It means there's a link somewhere... something we don't know about. There's a link between Louey and Frenchy, and as a result of that link Frenchy was prepared to act the decoy, without pressure and probably without payment...'
'Perhaps this Louey fella's the boyfriend himself,' suggested the colonel.
'He's too clever... and women aren't his weakness. No. It's something else.'
'I really can't see that it's important, Gently,' weighed in Sir Daynes.
'It isn't!' barked the chief super, 'we simply sit here wasting our time while the chief inspector amuses himself by...'
He broke off as a tap came at the door. It was Sergeant Dutt's homely visage that appeared.
'Begging your pardon, sir...'
'Yes? What is it?'
'It's something for Chief Inspector Gently, sir... he wanted to know directly a certain party left the premises.'
'Well, cough it up – don't stand there like a dummy!'
Dutt transferred his stolid gaze to his superior. 'It's Frenchy, sir...'
'Frenchy!' Gently rose slowly to his feet.
'I just arrived back, sir, and they tell me she was bailed aht half an hour ago.'
A faraway look stole into Gently's eye. 'And who was it, Dutt... did you get the name?'
'Yessir. It was a Mr Peach, sir.'
The faraway look lengthened till it embraced some islands of the distant Hebrides. 'Peachey!' murmured Gently, 'my old friend Peachey! I always had a feeling we should find him sewn into the lining of this case... somewhere!'
It rained still, as though it had never thought of stopping that side of Michaelmas. The picture-houses, theatres and pavilions were packed solid with moist audiences, the cafes had never had such a day, the lessees of dance-halls and amusement arcades were indulging in dreams of a late-autumn holiday at Cannes or Capri... Only the beach was having a bad time of it. Only the beach was dark and deserted and desolate to behold. Soft, unnoticed, another flood-tide crept upwards towards the hectic Front. It washed round the piles under the piers, looked up at its auld enemy, the cliffs, and made to list a few more degrees a certain post which some policemen had set up in the shingle.
But there was nobody there to see it, except a crouching halfwit. The rest of Starmouth kept tryst with their bright lights. Rain it might and rain it did, but the electric rash burned on, the music wailed, the rifles spanged, the audiences laughed and the great Till of Starmouth rang its steady chorus.
Artie in the bar was getting quite irritable with his customers, and he could afford to be. They didn't want away once they were there. And it was a gay crowd that night, on the eve of the races. Several old faces had turned up which had been missing for quite a while... it was just like it had been before that b. Inspector Gently set foot in the place, as the sporty individual observed. Even Louey seemed in a festive mood. He had been out twice in the course of the evening and each time it had been drinks all round. It was communicative, that mood of Louey's. For better or worse it affected the company in the bar. But now the clouds which had momentarily gathered about the gigantic brow had faded away, the sunshine had returned, the bar was its old happy self again...
Or it was till nine-thirty. Nine-thirty-three and a half, to be precise. At that exact moment a bulky figure in a fawn raincoat and a despairing trilby pushed through the swing-doors and looking neither to right nor left, shouldered its way across to the door opposite and disappeared again.
It was done so quickly that it might have been an optical illusion. Ferrety-face Artie had to shake his head to convince himself he wasn't seeing things. The sporty individual, halfway down his eighth Scotch, screwed up his eyes in a search for assurance that he was stone-cold sober.
'That bloke just now... it was him, washn't it?'
Artie nodded absently and moved down towards the door, as though hopeful of hearing something above the din outside.
'But whatsh he doing here... I thought Louey said it was OK?'
Artie waved him down with his hand and got still closer to the door. The whole bar held its breath in a sort of hushed watchfulness. In the comparative calm a tincased version of 'Cherry Pink' seemed to vibrate the plastic-topped tables with its singeing vehemence.
'I don't undershtand...' burbled the sporty individual, 'something's going on, Artie... I don't undershtand.'
Artie didn't either, but there wasn't very long to wait. At nine-thirty-seven, or a trifle before, the door reopened with a suddenness that nearly pinned Artie to the wall. Out waddled Peachey, red in the face. Out marched the bulky figure, his hand tucked affectionately under Peachey's arm. Again no time was wasted. Again no looks were cast to right or to left. The brief procession headed forthrightly through the swing-doors and vanished like a dream, though in this case one part of the dream was left standing in the doorway by the bar. It was Big Louey. And his gold tooth wasn't showing at all...
Outside Dutt was waiting in a police car. Peachey was bundled in and Gently gave an address to the driver which didn't sound like Headquarters. A short drive brought them to a dark and empty street where but few lamps shone islands of radiance on the gleaming pavement. Dutt alighted and stood by the door.
'Get out,' ordered Gently to Peachey.
Peachey gulped and gave a frightened look up and down the street.
'This isn't the police station! I d-demand to be taken to the police station!'
'Get out!' snapped Gently and Peachey scuttled forth like a startled rabbit. Gently followed him and after tossing a word to the driver, slammed the door resoundingly behind them. He indicated the house by which they had stopped.
'In there.'
'B-but I've g-got rights... you c-can't do this!'
Gently poked a steely finger into his plump back and Peachey forgot about his rights with great suddenness.
There was nothing alarming about the house, however. The door opened on a well-lit and comfortable-looking hall containing a hat-stand and an aspidistra on a side-table and the room into which Peachey was marshalled bore all the appurtenances of respectable boarding-house practice. Gently took off his hat and raincoat and hung them familiarly on the hat-stand.
'See if Mrs Davis has got the tea on, will you?' he said to Dutt, 'and ask her if she's got some biscuits... I like those shortbread ones we had the other night.'
Dutt departed and Gently joined Peachey in the lounge. Gently seemed in no hurry to begin business. An electric fire was glowing in the fireplace and, standing with his back to it, he slowly filled and lit his seasoned briar. Peachey watched every move with pathetic attention. Twice he seemed about to recall his flouted rights, but each time, catching Gently's mild eye, he thought better of it. The horrid ordeal ended when Dutt re-appeared bearing the tea tray. There were three cups and Peachey was even indulged with two lumps of sugar.
'And now...' mused Gently, seating himself with his teacup, 'now we can have our little chat in peace and comfort... can't we, Peachey?'
'You haven't g-got no right!' broke out the parrot-faced one unhopefully.
Gently clicked his tongue. 'No right, Peachey? Why, we're treating you like an old friend – bringing you to our nice cosy lodgings, instead of that bare old police station! Now sit yourself down on one of Mrs Davis's best chairs, and try to be a bright lad... you need to be a bright lad, don't you, Peachey?'
Peachey blinked and swallowed, then lowered himself into a chair. Gently drank a large mouthful of tea and set his cup down near the electric fire.
'You're here for a reason, Peachey. Two reasons, as a matter of fact. The unimportant reason is because there's a pack of wolves down at Headquarters who would just love to tear a little boy like you into small pieces. The important reason is that I want to talk to you off the record – no charges, no taking it down, nothing being used in evidence. Anything you tell me here is in confidence and it won't appear again till you're ready to give it in a sworn statement... you get the idea?'
Peachey's close-set eyes seemed to get closer together than ever. 'I-I'm not going to m-make a statement... I don't know nothing to make one about!'
Gently shook his head paternally. 'Don't say that, Peachey. You don't know how useful that statement's going to be. At a rough guess I should say it would make eighteen months' difference to you, besides a slimming course with the pick and shovel. You wouldn't be too handy with a pick and shovel, would you, Peachey?'
'I don't know what you're t-talking about!' Peachey gulped, his cup and saucer beginning to chatter.
'Come, come, Peachey! You're amongst friends. There's no need to be bashful. Almost any time now we're going to run you in for living on immoral earnings and I'm sure you know what that means. If you go before a beak, it'll be six months in one of our more comfortable establishments; if you go up with an indictment, it'll be two years with the pick-and-shovel boys.'
'B-but it isn't true!'
'We've got the goods, Peachey.'
'I'm a b-bookmaker's clerk – you know I am!'
'Six witnesses, Peachey, and two of them your neighbours in Sidlow Street.'
'It's a f-frame, I tell you!'
'And three past convictions, all neatly filed at Central Records... no, Peachey. You're due for a holiday. And just between us you'll be lucky if it stops there, won't you?'
The parrot-faced one put down his cup, which he was no longer in a condition to support. He made a pitiful effort to get out a cigarette, but the packet fell from his hands and its contents distributed about the floor. Dutt helped him pick them up. They got him lighted at the second attempt.
'As I was saying,' resumed Gently meditatively, 'you'll be lucky, won't you? You'll need all the goodwill that's going if you're not going to be roped in for complicity in the murder of Stephan Stratilesceul... did you know his name? At "Windy Tops"?'
He paused for artistic effect and Peachey shrank down in his chair several degrees.
'Of course, it may be that in making a statement you would incriminate yourself... there's always that to be thought about. We shall quite understand your keeping silent if you were in fact an accomplice...'
The goad was irresistible. Peachey squirmed as though it had galled him physically. 'I didn't know – I swear – it wasn't nothing to do with me!'
'Nothing to do with you? How can you, Peachey! When it was Frenchy who got him out to "Windy Tops" in the first place.'
'I tell you I didn't know... they didn't say n-nothing!'
'You mean they didn't tell you they were going to kill him?'
Peachey sucked hard on a cigarette which was coming to pieces between his lips.
'You might as well come clean, Peachey. It's off the record.'
Peachey gulped and sucked, but he had dried up again.
Gently sighed. 'Let me see if I can reconstruct it. They had a conference, didn't they? Streifer had traced Stratilesceul to his lodgings in Blantyre Road, but he was rather at a loss to know how to deal with him. It wasn't just a question of killing the man and recovering the money. Streifer could handle that well enough on his own. No – what was important about Stratilesceul was certain information he could give... with a little persuasion, perhaps... about other untrustworthy members of the TSK Party. Am I right?'
The cigarette was definitely a spent force, but Peachey kept on working at it.
'That was the problem, then – to get Stratilesceul in a place where he could be duly persuaded, and afterwards, as a mere formality, put to death. It wasn't an easy problem to solve. Stratilesceul wasn't laying himself open to being kidnapped. As far as he knew, he had shaken off the pursuit, but he was still taking precautions – like lodging in a crowded boarding-house and sticking to the frequented parts of the town. I dare say there were several plans made. The length of time it took to do the job suggests it. But they all fell through for that very simple reason – they could never get him where they could lay their hands on him.
'So we come to the final conference – Louey, Streifer and Little Peachey... because you were in on it, weren't you, Peachey? And Louey sits on a striped chair behind that very nice desk of his, thinking, thinking. At last he says to Streifer: "You're familiar with Stratilesceul's confidential record?" – And Streifer nods with that quiet little laugh of his. "Is there nothing in it that might serve our turn?" – Streifer shrugs and says: "He's fond of women." "Women!" says Louey, showing some gold, "any particular sort of women, or just women in general?" "Blonde women," says Streifer, "nice big blondes."
'At that Louey really smiles. "We've got the very thing... haven't we, Peachey?" he says, "a nice big blonde who'll do just what we ask her! Why, I dare say that if we play it right we can get friend Stephan delivered to the very door..." And what did Little Peachey say to that? He said: "Yes, Louey, of course, Louey, anything you say goes with me, Louey—'''
'I didn't know!' shrieked the tormented Peachey, 'they never said anything about killing him in front of me!'
'You didn't guess?' rapped Gently. 'You thought it was just going to be a social evening?'
'They said he'd hidden the money, that's all. They said they wanted to get him to find out what he'd done with it!'
'So you're entirely innocent – and Frenchy's entirely innocent?'
'She didn't know neither!'
'Just a couple of little lambs! And where were you when Frenchy was doing her dirty work?'
'I don't know – I was in the bar!'
'You were in the bar – then you didn't get Louey's car out of the garage?'
'No!'
'Then two witnesses we've got are liars?'
'I wasn't near the garage!'
'And you didn't pick up the reception committee and take them to "Windy Tops"?'
'... I was in the bar!'
'And you didn't wait there with them to give a hand tying up Stratilesceul?'
'I didn't – I didn't! When they'd got him in there they sent me back with Frenchy... we never knew nothing... nothing at all.'
'So it was just one big surprise when you saw it in the papers.' Gently reached down for his cup of tea and tossed it off fiercely. 'And when you found out, what did you do?'
'I didn't do nothing!' floundered Peachey, his little eyes roving from side to side as though in desperate search for escape.
'Nothing. Nothing! You knew the murderers – you'd been tricked into helping them – unless you spoke up quick you were in it along with them – and yet you did nothing. Is that your tale for the jury?'
'I ain't going before a jury!'
'Oh yes you are, Peachey, somewhere along the line.'
'But you said it wasn't evidence!'
'It will be when you've sworn it.'
'I ain't going to swear it – never – no one can make me.'
'They won't have to, Peachey. You'll do all the swearing that's necessary when you go up on a murder rap.'
'But I never did it – you know I never did it!'
'I shall feel a lot more certain when I've got a statement on paper with your signature underneath.'
Peachey shrivelled up in the chair like a punctured balloon. 'I ain't going to swear,' he whispered, 'I ain't – I ain't!'
'Then it's two years' hard at the very least.'
'I ain't going to swear, not though it was twenty.'
Gently shrugged his bulky shoulders and handed his cup to Dutt, who silently refilled it. Gently drank some and gnawed a shortbread biscuit. 'Of course, you know we've got Streifer,' he muttered casually amongst the crumbs.
'Str-Streifer?' Peachey unshrivelled a little.
Gently nodded and bit another piece.
'But Streifer is g-gone...!'
'We took the trouble to bring him back again... your grapevine can't be as good as it was.'
Peachey's small eyes fixed on the pattern of Mrs Davis's best carpet, but he made no other contribution for the moment.
'He's safe and sound,' continued Gently, 'you don't have to worry about _him_ any longer. And if a certain little bird would sing his song we could put Louey in with him. Louey in jail,' he added helpfully, 'would be just as harmless as the average mortal.'
The pattern still had Peachey fascinated.
'And with a little further assistance, Peachey – all confidential, you understand, no names published, no questions asked about how a certain individual came by his information – we could arrest and imprison or expel quite a fairish bag of unfriendly-minded persons. In fact, we could make this country a healthy place for little Peacheys to come back to after a six-month vacation... couldn't we?'
For a moment the small eyes lifted from the carpet and rested just below Gently's chin. Then they sank again, sullenly, and the dry lips bit together.
'Ah well!' sighed Gently, 'we do our best, don't we? We always do our best!' He appropriated another biscuit and crammed it into his mouth. 'Take him home, Dutt... take him to his flat in Sidlow Street. I don't suppose he wants to see Louey again tonight.'
Dutt took a step forward and Peachey looked up suddenly, his mouth dropping open.
'B-but aren't you going to p-pinch me...?'
Gently shook his head and swallowed some tea.
'B-but you've got a charge – y-you said you had!'
'Can't bother with it just now, Peachey. The local lads will see to it some time.'
'B-but it's true – you've got some witnesses!'
'You just comalongofme like the chief inspector says,' said Dutt, hoisting the parrot-faced one to his feet, 'he's done with you now... you're even getting a nice ride home. You don't want us to lock you up, do you?'
If Peachey's expression was anything to go by he did want that very thing, but neither Gently nor Dutt seemed willing to oblige. He was stood firmly in the hall while Dutt was putting on his raincoat and Gently, still ravaging amongst the biscuits, appeared to be forgetting the existence of both of them. But as Dutt reached for his hat, Gently sauntered to the lounge door.
'By the way, Peachey...'
Peachey blinked at him hopefully.
'If you were running a short-wave transmitter it would be useful to have a nice high aerial, wouldn't it?'
'T-transmitter...?'
'That's right. For sending little gossip-notes to the Continent.'
'But I don't know nothing about it!'
Gently tut-tutted and felt for a scrap of paper. 'Here we are... hot from Central Records. They released you from a stretch in '42 to go into the Services; you were trained as a radio-mechanic at Compton Bassett; radar course at Hereford in '44; demobilized as a Sergeant-Radar-mechanic in '46. Quite a distinguished career, Peachey... and of course you'd know all about building and working a simple transmitter, wouldn't you?'
Peachey gulped and tried to get some moisture on to his lips.
'And about that aerial? There aren't so many high places in Starmouth. There's the monument, but that's a bit too bare and obvious. And there's the observation tower, but that would be even worse. No... what you'd want would be something unobtrusive... something where a little private wiring wouldn't notice very much, where perhaps there was an off-season when you could do the job without interruptions. That's what you'd want, isn't it, Peachey?'
'I forgot all that... I don't remember nothing about radio!'
Gently shook his head consolingly. 'Never mind, Peachey. I dare say you will. It'll come back to you with a rush one day. Oh, and just one other thing.'
Peachey sucked in breath.
'Tell Louey I'll be in tomorrow some time to settle up a bet, will you? He'll know what I mean... just tell him that.'
Dutt hustled him out and the door closed behind them. Gently hesitated a moment till he heard the car pull away, then he returned swiftly to the lounge, uncoupled the phone, dialled a number crisply.
'Chief Inspector Gently... oh, hullo, Louey! I thought it was only fair to ring you up...'
He smiled pleasantly to himself at the note of tenseness in the voice at the other end.
'Yes, of course you have to know... with the races tomorrow too... naturally you'll be stuck if we pinch your head boy. But there's nothing to worry about, Louey... no, we came to an agreement. I've just sent him home now, as free as a bird. He's a sensible chap, Louey... knows when it's time to do a deal. We all have to play along with the police sometimes... eh? Yes... yes... Sidlow Street... yes. I'm glad it's eased your mind, Louey. Have a good day with the gee-gees tomorrow... yes... good night.'
He pressed the receiver down a moment and then dialled again.
'Gently here. Give me Copping.'
'Hullo?' came Copping's voice, 'have you had any luck? The chief super says that if you haven't—!'
'Never mind the chief super,' interrupted Gently with a grimace. 'Listen, Copping. This is vitally important. I've just sent Peach home to his flat at 27 Sidlow Street with Dutt to keep an eye on him. Now I want Dutt relieved at midnight and your best man sent to replace him. And armed, you understand? Peach has got to be guarded from now on, day and night... and heaven help the man who slips up on the assignment!'
CHAPTER FOURTEEN
STARMOUTH RACES – that colourful, moneyful, tax-free event – Starmouth Races, when a town already full to the brim began bursting at the seams. From early in the forenoon the train-loads started to emerge. By lunch-time you could hardly move on the road to the race-course, and as for getting a sit-down meal, you were lucky to pick up a couple of cheese sandwiches. But it was Starmouth Races and nobody cared. You came for the fun and the flutter and the sea-air, and if you went back skint it was all part of the outing.
They'd got a brass band from Norchester, a regular festival-winning affair. It had come out today in a fanfaronade of new grey and pink, with a man on the baton who really knew his business. Dutt was enthralled. He had always had a weakness for brass bands. When they went to town with 'Blaze Away' it touched a chord in his simple cockney heart...
'Worst day of the year!' moaned Copping to Gently, 'how can you police this lot with the men we've got? If we arrested all the dips and shysters who come up for the races it'd need a special excursion train to cart them back to town!'
The super was there, looking very spruce and commanding in his best blue with its rainbow of medal ribbons. He sharpened a glance for Gently's baggy tweed. 'I hope you know what you're doing, Gently... Gish is out for your blood if anything goes wrong.'
Gently tilted his head accommodatingly and the super passed with a sniff.
As a matter of fact, Gently was beginning to worry himself, just a little bit. The thing wasn't going to pattern at all. There had been no alarums and excursions, no rush for Sidlow Street in the quiet hours... Peachey had spent a restful night, said the report, or if not a restful one, at least a peaceful one. In the morning it was the same. The routine of 'The Feathers' had continued undisturbed. Louey had gone for his constitutional, Peachey had reported to the office, at lunchtime they had eaten together at a nearby restaurant and directly afterwards Peachey had fetched the car and driven Louey and two of the bar-regulars to the race-course. It was almost as though Louey were ignoring the situation, as though he were deliberately calling Gently's bluff. Certainly there was no anxiety in his aspect, and if Peachey was looking rather more like a boiled stuffed rabbit than usual it was hardly to be wondered at.
Gently's eye wandered through the busy crowd to the line of bookies' stands. Biggest of all flamed a great orange banner, set up on two poles, and licking across it like scarlet fire ran the legend: LOUEY ALWAYS PAYS! – Not that it was necessary, such a banner. You could hear the voice of Louey like distant thunder, over-topping crowd, band and competitors:
'FIVE TO TWO ON THE FAVOURITE... COME ON NOW... ONLY LOUEY GIVES IT... FIVE TO TWO ON THE FAVOURITE!'
His gold tooth shone, his diamond ring flashed, he loomed over the crowd like a genial Goliath. And they liked Louey. He was an institution on the race-course. Plump Peachey could hardly scribble slips out fast enough to keep pace with the money going into that gaping Gladstone.
'FIVE TO TWO ON THE FAVOURITE... TEN BOB TO WIN TWENTY-FIVE... HUNDRED TO EIGHT ON CAMBYSES... COME ON NOW, THESE ARE THE ODDS YOU'RE LOOKING FOR!'
Up beside him the sporty individual was taking signals from someone across in the stands and chalking up fresh odds on the blackboard. Down below a couple of bar-types were touting recklessly, yanking custom from the very shadow of rival stands.
'COME ON NOW... NO LIMIT... IF YOU WANT A FORTUNE COME TO LOUEY... YOU SEE MY BANNER – IT MEANS WHAT IT SAYS!... COME ALONG NOW AND DO THE INCOME-TAX COLLECTOR IN THE EYE!'
It was all so innocent, all so regular. Moral or immoral, book-making was legitimate business and watching Louey up there in all his glory tended to shake one's convictions. He looked so little like a murderous fanatic with the gallows threatening to yawn at his very feet.
But that was the situation and Gently had made sure that Louey knew where he stood. He was counterbluffing, that was all; doing what Gently would have done himself if the positions had been reversed. But counter-bluff was a temporary measure. There would be a plan behind it, a positive step. What was it cooking now, that calculating mind, when was it going to happen, and where?
Gently moved over to Dutt, who had resumed his role as Peachey's protector.
'Keep your eyes on your man,' he warned him snappily, 'he'll be easy enough to lose in a crowd like this.'
'Yessir... of course, sir. But you got to admit it's a smashing bit of brass...'
'I don't admit anything – keep your eyes on Peachey.'
Dutt clicked his heels and did as he was ordered.
Gently wandered away with a frown on his brow. He was biting Dutt's head off now! The double strain of a waiting game with Louey and a checking game with Gish was beginning to fray at his nerves. Gish wanted action. He hadn't any faith in Gently. One had a shrewd suspicion that twenty-four hours would be the limit of his patience.
A slinking figure appeared to materialize out of the worn turf in front of him and Nits' pop-eyes strained up to his own. Gently summoned up a smile for the halfwit.
'Hullo! You come to see the races too, my lad?'
Nits gibbered a moment with his invisible mouth.
'You better get over by the rails – there's a race starting in five minutes.'
'You let her come back!' piped the halfwit, 'you let her come back!'
Gently nodded gravely. Nits chittered and gabbled under his staring eyes. Then he turned to cast a glare of hatred at the towering form of Louey.
'Him – he's a very bad man – very bad!'
Gently nodded again.
'He came to see her – frighten her!' Nits hesitated and crept a little closer. 'You take him away! Yes! You take him away!' He laid a hand on Gently's sleeve.
'I'm thinking about it, Nits...'
'He's the bad one – yes! You take him away!'
Gently shrugged and slowly released his sleeve. The halfwit gabbled away furiously, darting angry glances, now at Louey, now at Gently. Gently produced a coin and offered it to him.
'Here you are... but don't go making bets with Louey.'
'Don't want it – don't want it!'
'Buy yourself an ice-cream or a pint of shrimps.'
The halfwit shook his head violently and knocked the coin out of Gently's hand. 'You take him away!' he reiterated, 'yes – you take him away!' Then he jumped backwards with a sort of frisking motion and dived away through the crowd.
There was a stir now and a general surge towards the rails. The horses had come up to the tapes and were under starter's orders. Out of a grey sky came a mild splash of sun to enliven for a moment the group of animals and riders, the brilliantly coloured shirts, the white breeches, the chestnut, grey and dun of satin flanks. Tense and nervy were the mounts, strung up and preoccupied the jockeys. A line was formed, a jumpy horse coaxed quiet and almost before one realized what was happening the tapes flew up and the field was away. Instantly a shout began to rise from the crowd, commencing near the gate and spreading right down the track. Fifty thousand pairs of eyes were each magnetized by that thundering, flying, galloping body of horse.
Out in front went the favourite, Swifty's Ghost, and following it close came Cambyses and Rockaby, the latter at a hundred to one and scarcely looked at by the punters. Three furlongs, and the field was getting lost. Six furlongs, and you could almost draw your money. Seven furlongs, and Cambyses, a big grey, was making a terrific bid and going neck-and-neck. Eight furlongs, and out of the blue came Rockaby, fairly scorching the turf, a little dun horse with a halting gallop, but moving now like a startled witch. Could Swifty's Ghost hold them? Could Cambyses maintain his challenge? – The roar of the crowd ebbed up to a fever pitch. But Rockaby drew level with a furlong to go, Rockaby slipped through with a hundred-and-fifty yards in hand, Rockaby passed the post two lengths ahead of the grey and the favourite was beaten to a place by another outsider called Watchmego. The roar died away, the roar became a buzz. They'd done it again... another race to line the bookies' pockets!
Gently hunched his shoulders and turned away from the rails, and at that precise moment things began to happen. He had only time for a confused impression; it took place like a dream. There was a crash, some angry shouting, a sound like a quantity of coins being shot on the ground, and then somebody or something struck him heavily in the back and he was lying on his face on the bruised turf.
He wasn't hurt. He got up in a hurry. All around him a crowd was milling about a centre of attraction which was otherwheres than himself. Inside this centre a dialogue for four voices was developing with great verve.
'Of course it was on purpose – I bloody saw you do it!'
'I was shoved, I tell you.'
'You can tell it to the coppers!'
'I tell you I was shoved – some bastard tripped me up!'
'Do you think we're blind?'
'Well, you don't look too bloody bright.'
'Now look here, you dirty so-and-so!'
Gently shouldered his way through. The scene enacting was self-explanatory. A bookie's stand lay on its side amid a debris of betting-slips, notes and coins, about it four angry men. Three of the men were obviously allies. The fourth, a burly gentleman in a mackinaw, appeared to be the defendant in the case.
'Police!' snapped Gently, 'you can stop that shouting. One of you tell me what's been going on here.'
The gent in the mackinaw broke off a challenge to the opposition and stared at Gently with aggressive insolence.
'Police, he says! A snouting copper! You keep your big nose out of this, mate, or it'll finish up a different shape from what it started this morning!'
'You hear him?' struck in one of the aggrieved, on his knees and trying to collect the scattered money, 'that's your man, officer – you don't have to ask! Come up and threw down my bleeding stand, he did, never as much as a word offered to him!'
'Mad!' snapped a little man with a big coloured tie, 'mad, I tell you – that's what he is!'
The gent in the mackinaw seemed about to resent this allegation when he was interrupted a second time by a new arrival. This time it was Dutt and he was propelling in front of him no less a person than Artie of the ferret face.
'I got him, sir!' panted Dutt, 'he's the one, sir – saw him wiv me own mince pies! Standing right close-up to you he was, sir, all during the race, and as soon as this lot here started he catched you a right fourpenny one and hooked it... all he didn't know was that I was watching him!'
Gently stared at the scowling bartender as though he had seen a ghost. 'Get back!' he thundered at Dutt. 'Good God, man – _don't you understand_? The whole thing's a trick to get us out of the way – get back at the double, or there may be another body on the beach tomorrow!'
The odds were still being called under the orange banner, but it wasn't Louey calling them. The slips were still being scribbled and handed out, but the man with the book wasn't Peachey. It was the sporty individual who had taken over, with one of the touts for his clerk. He welcomed Gently and Dutt derisively as they rushed up to the stand.
'Hullo-ullo! Coupla gents here getting in training for the selling-plate!'
'All right!' rasped Gently, 'where are they – where have they gone?'
'Gone, guv'nor? And who is it that's s'posed to have gone somewhere?'
Gently wasted no time. A brown hand flicked out and fifteen stone of sporty individual was picked off the stand like a pear. 'Now...! This may be fun for you, but it's murder to me, and if you don't tell me what I want to know I'll see you in dock for complicity. Where's he taken Peachey?'
'I don't know, guv, honest—!' He broke off with a yell as Gently applied pressure to his arm.
' _Where's he taken Peachey?_ '
'I don't know – we don't none of us know!'
'That's right, guv!' broke in the tout with the book, 'he just said him and Peachey had got some business to see to what he didn't want you to know about.'
'It's the truth!' shrieked the sporty individual, 'oh, my bloody arm!'
Gently threw him down against the stand, where he lay massaging his maltreated limb and moaning. 'Find Copping!' rapped Gently to Dutt, 'tell him what's happened – tell him to issue a description to all his men – send one to "The Feathers" and one to Sidlow Street – the rest fan out and search the area round the race-course. Where's Louey's car parked?' he fired at the sporty individual.
'It's over there – right by the gate!'
'Check and see if it's gone – if it has, alert all stations.'
Dutt hesitated a moment and then turned in the direction of the gate, but before he could set off an animal-like form came darting and swerving through the crowds and threw itself at Gently's feet.
'He went that way – that way! I saw him! I saw him go!'
Gently's eyes flashed. 'Which way, Nits?... which way?'
'That way!' The halfwit made a fumbling gesture towards the north end of the enclosure.
'Gorblimey!' exclaimed Dutt, 'it's "Windy Tops" again!'
Gently rounded on him. 'Forget what I've been saying – just tell Copping to bring his men up there. And when you've done that, don't wait for him... I shall probably be in need of some help!'
'Yessir!' gasped Dutt, 'yessir – I'll be there with you!'
But by that time Gently was gone.
It was a hummocky bit of paddock separating the race-course from the lane to 'Windy Tops' and Gently, past his best sprint years, found it very heavy going. At the far side was a scrubby thorn fence in which he had to find a gap. Nits, frisking along at his side, went over it like an Olympic hurdler.
'You get back, m'lad!' panted Gently, 'there'll be trouble up there!'
'You going to take him away!' chuckled Nits. 'I want to see you take him away!'
'You stop down here and you'll get a grandstand view!'
'I want to see – I want to see!'
There was no discouraging him. Gently ploughed on up the slope of the cliff. By the time he reached the gates of 'Windy Tops' he was glad of the breather offered by a pause to reconnoitre and Nits, entering into the spirit of the thing, gave up his leaping and frisking, and slid away like an eel behind the cover of some rhododendron bushes. Not a sound had come from the house. Not a vestige of life was to be seen at any of the windows. Only the front door stood half ajar, as though whoever was within didn't mean to be there for very long.
Keeping his breathing in check, Gently moved swiftly across to the threshold. Inside he could hear voices coming from somewhere at the back. Silently he worked his way down the hall towards the baize-covered door of the kitchen, which was shut, and pressed himself close to it, listening...
'No, Peachey,' came Louey's voice at its softest and silkiest, 'we don't seem able to find that money anywhere, do we?'
'B-but boss... he give me the message,' came Peachey's whine in reply.
There was the sound of a cupboard door being opened and shut, and something else moved.
'Quite empty, Peachey... not a dollar-note to be seen.'
'Boss, he t-took it with him... you don't think I'd l-lie?'
'Lie, Peachey?' Louey's laugh sounded careless and easy. 'You wouldn't lie to me, now, would you?'
'N-no, boss, of course I wouldn't!'
'And you wouldn't tell tales, Peachey, would you... not even to save your own worthless skin?'
A confused noise was Peachey's answer to this sally.
Louey's laugh came again. 'You see, Peachey, we all have our value, looked at from a certain point of view. I have mine. Streifer has his. Stratilesceul had a value too, but unfortunately for himself he lost it. And now the pressing problem of the moment, Peachey, is your value... you do see what I'm driving at?'
A strangled sound suggested that Peachey saw it very plainly.
'Yes, Peachey, I thought you would. I don't want to be unkind, you know. I'm prepared to listen to any defence you may have to offer, but it seems to me that there can't be any real doubt about the matter... doesn't it to you? Here am I, on whom the forces of liberation in this country depend, and there are you, a small and expendable unit. Now I could betray you, Peachey, and that might be wrong. But if you were to betray me, that would be a crime comparable to the crime of Judas. You understand?'
'But boss – I never – I didn't – I told them I wouldn't!'
'SILENCE!' thundered Louey's voice, stripped in a moment of its silky veneer. 'Do you think I didn't know, you miserable worm, do you think you can lift a finger without my knowing it?'
There was a pause and then he continued in his former voice: 'I like to make these matters clear. I tried to make them clear to Stratilesceul. I'm not a criminal, Peachey, in any real sense of the word. There's only one crime and that's the crime against the forces of liberation: when we, the liberators, proceed against that crime, we are guiltless of blood, we are the instruments of true justice. So I am not killing you, Peachey, from hatred or even personal considerations... I am killing you in the name of Justice, in the name of Society!'
'... No!' came Peachey's terror-stricken cry. 'Boss... you can't... you can't!'
'Oh but I can, Peachey.'
'No boss – no! It's a mistake – I never told them nothing!'
'And no more you shall!' came Louey's voice savagely, 'this is it, Peachey – this is the tool for traitors!'
Gently hurled open the door. 'Drop it!' he barked, 'drop that knife, Louey!'
The big man spun round suddenly from the sink, over which he was holding the helpless Peachey. His grey eyes were blazing with a malevolent light, strange, fey. ' _You!_ ' he articulated with a sort of hiss, '... you!'
'Yes, Louey – me. Now drop that knife and take your hands off Peach.'
'... You!' hissed Louey again, and the light in his eyes seemed to deepen.
'Stop him!' whimpered Peachey, 'oh, God, he's going to do for me!' And with the energy of despair he twisted himself out of Louey's grip and made a dive for the back door, which fortunately for him was only bolted. But Louey made no move to restrain him. His eyes remained fixed on Gently.
'Let him go!' he purred, ' _he_ won't talk... I'm not so sure now he ever would have done, are you, Chief Inspector Gently?'
'He'll talk,' retorted Gently, 'there's a limit to what you can do with a knife. Now drop it and put your hands up. It's time you started thinking of your defence.'
By way of answer Louey let the knife slide down his hand, so that now he was holding it by the tip of the blade. 'My defence, Chief Inspector Gently; you are looking at it now. Isn't it a pity? I've let a miserable parasite like Peachey escape and in his place I must execute a man of your... attainments. Isn't – it – a – pity?'
With the last four words he had reached back with his gigantic arm and was now leisurely taking aim at Gently's heart. There was no cover to dive for. There was no prospect of a quick back jump through the door. The knife was poised and on a hair-trigger, it would reach its mark long before Gently could move to evade it. And then, at the crucial split second, the knife disappeared – one instant it was flashing in Louey's hand, the next it was spirited away as though by a supernatural agency.
'You take him!' piped the delirious voice of Nits through the back door, 'ha, ha, ha! You take him – you take him!'
With a roar of anger Louey recovered himself and leaped at Gently, but it was too late. A hand that felt like a steel bar smashed into the side of his throat and he collapsed on the floor, choking and gasping, a pitiful, helpless wreck of humanity. Gently snapped handcuffs on the nerveless wrists.
'It had to come, Louey,' he said grimly, 'there has to be an end to this sort of thing.'
'Ha, ha, ha!' giggled Nits, dancing around them and brandishing Louey's knife, 'we'll take him away now – we'll take him away!'
Gently put out his hand for the knife. It was a curious weapon. The hilt and blade were one piece of steel, the former heavy, the latter relatively light and narrow. On each side of the hilt was engraved the mark of the TSK along with a number of Egyptian hieroglyphics.
'Double-edged, about three-quarters of an inch wide,' mused Gently, 'it couldn't be any other... it would have to be this one.'
Louey struggled up into a sitting position. He was still gagging for breath, his face was grey. He stared at Gently, at the knife, at the discreet links shackling his enormous wrists. 'No!' he whispered hoarsely, 'you weren't big enough... _you just weren't big_ _enough!_ '
Gently nodded sadly and slipped the knife into his pocket. 'It's you who weren't big enough, Louey... that was the mistake. We're none of us big enough... we're just very little people.'
Half the Starmouth Borough Police Force seemed to be congregating in the garden as Gently led Louey out. There was the super with Copping and three or four plain-clothes men, at least ten constables and the complete Special Branch outfit. Dutt came panting up the steps, relief showing in his face at the sight of the handcuffs and an unmarked Gently.
'You're all right then, sir – he never give you any trouble?'
Gently shrugged faintly. 'About the routine issue...'
'And Peachey, sir – you got him away safe and sound?'
'Safe and sound, Dutt... all Peachey had was a scare.'
'By thunder, Gently, you've pulled off a splendid piece of work!' exclaimed the super, striding across. 'I have to admit it – I thought you were going to fall down over this fellow. I suppose it's unnecessary to ask whether you've got the goods on him?'
'I got him red-handed... he was going to stab Peach with a TSK patent executioner's knife. I think we'll find it adds up to the weapon which was used on Stratilesceul.'
'You're an amazing fellow, Gently!' The super gazed at him with honest admiration. 'You're not an orthodox policeman, but by heaven you get the results!'
There was a cough of some penetrative power indicative of the near presence of Chief Superintendent Gish. 'I'm sure you'll forgive me for interrupting,' he observed bitterly, 'but we, at least, have still some business to transact in this affair. I take it that Chief Inspector Gently no longer has any objections to my carrying out my duty?'
Gently signified his innocence of any such desire.
'Then possibly Peach can be produced to answer a few of my irrelevant questions?'
Gently deposited Louey with Dutt and took a few steps towards the edge of the wildered garden. 'Peachey!' he called softly.
There was a rustling amongst some rhododendrons.
'Peachey... it's all right. We've got Louey under lock and key. You can come out now.'
There were further rustlings and then the parrot-faced one emerged. He was still trembling in every limb and his knees had a tendency to buckle, but the sight of so many policemen reassured him and he walked shakily over to the front of the house.
'That's the boy, Peachey... nobody's going to hurt you.'
'You got his kn-knife?' gabbled Peachey, darting a wild-eyed glance at his shackled employer.
'Yes, Peachey, we've got his knife... everything's as safe as houses. All we want now is a little information – just a little, to begin with! I suppose you're in a mood to do some talking, Peachey?'
Peachey was. He had never been so much in the mood before. Shocked to his plump core by his experiences in the house, Peachey had learned the hard way that honesty was his only hopeful policy and he was prepared to give effect to that policy in all-night sittings, if that should be required. Chief Superintendent Gish, however, was more moderate in his exactions. He was obstinately and snappily interested in but one set of facts – a short-wave transmitter and some records – and when he had obtained the address of same he departed in haste, leaving Peachey to waste his sweetness on the East Coast air.
'But you wanted a statement about the m-murder, didn't you?' asked Peachey aggrievedly, though with an anxious look at the silent Louey.
'We do, Peachey... don't you worry about that,' Gently assured him. 'We'll take you right back now and you can tell us about it over a cup of canteen tea.'
'Then there's Frenchy... she can b-back me up...'
'She hasn't been overlooked.'
'And I dare say some of the boys... it was only me what was sworn into the p-party.'
Gently nodded and urged him towards the gate. The super signed to his men and Dutt touched Louey's arm. From below them, through the scrub trees, came a murmur like a swarming of bees, a murmur that grew suddenly, became a frenzied roar. Louey stood his ground a moment. It was another race in progress.
And then there came a second sound, a rumbling, subterranean sound... like the first one and yet strangely unlike it. The roar of the crowd died down, but the second roar didn't. It seemed to be vibrating the air, the trees, the very ground itself. Yet there was nothing to see. There was nothing to account for it. It was Copping who suddenly realized what was going on.
'Run for it!' he bellowed, 'it's the house – it's going over – get the hell out of here, or we'll all be over with it!'
A sort of panic followed his words. There was a general and high-powered movement on the part of one super, one inspector, four detective sergeants, ten constables and a plump civilian in a down-hill and due south direction. This left a balance of three to be accounted for and a backward glance by Copping revealed them in a snapshot of dramatic relation which rooted him to the ground. There was Dutt, sprawling on the pavement; Gently, racing up the path; and Louey, roaring defiance from the top of the steps. And the house was already beginning to move.
'Come back!' howled Copping, 'it's on its way – come back!'
Gently pulled up short some feet from the steps. A crack was opening like magic between himself and the house.
'What are you waiting for?' roared Louey. 'Come on, Mister Chief Inspector Gently – let's die together, shall we? Let's die as though we were men – let's die as though we were more than men!'
Gently measured the distance and poised himself for the leap. Louey rattled his handcuffs in ironic invitation. Then, as though his good angel had whispered in his ear, Gently flung himself backwards instead of forwards: and at the same instant 'Windy Tops', complete in every detail, lurched out frightfully into space...
They ran to pick him up, Dutt, Copping and two uniform men. As they pulled him to safer ground another chunk of cliff dropped thundering to the beach. Down below them a raw gap loomed, large enough to put the Town Hall in. There was a curiously unnerving smell of dank and newly-revealed gravel. On the beach was piled the debris, lapping into the sea, a cloud of dust and grit still rising from it. Gently tore himself loose from his rescuers and stared down into the settling chaos.
'Not so close!' shouted Copping, 'you don't know where it's going to stop!'
But Gently remained staring from the edge of the yawning pit. Then he turned to Dutt, a curious expression on his face. 'All right... fetch him up. Use that little path over there and fetch him up.'
'Fetch him up?' echoed Dutt. 'Yessir. Of course, sir. But we'll need some picks and shovels, sir, and maybe a stretcher...'
Gently shook his head and walked away from the edge. 'Not a single shovel, Dutt... not the strap off a stretcher. Poor Louey! This is his final tragedy. He thought he was big enough to play God, but when it came to the push he couldn't even commit suicide.'
'You mean he – he's _alive?_ ' goggled Dutt.
'Yes, Dutt, and kicking too. If we'd left the door unlocked he'd have been buried in the middle of that lot, but as it is he went down on top... he's sitting there now, shaking the muck out of his ears.'
CHAPTER FIFTEEN
IT WAS GREAT stuff for the press, the double arrest of Louey and Streifer. It sent the Body On The Beach rocketing right back into the headlines. There were encomiums for Gently and encomiums for the Borough Police – it was only Chief Superintendent Gish who got a cold and cautious mention. But Gish didn't mind. They were used to it in the Special, he told everyone.
All the same, it was a pity that the best angle on the story didn't come out. It drove the reporters wild to see so much delectable copy laid for ever in the freezer. There was that transmitter under a ruined pill-box down at South Shore, for instance, with its aerial cheekily installed on the Scenic Railway... and there were the mass arrests and deportations of agents all over the country, a major operation which the British Public had an undoubted Right To Know About. But it was no use arguing. Chief Super Gish had a heart of stone. As a result of his inhuman decree the British Public were left with the vague and erroneous impression that the Body On The Beach had to do with a gang of international counterfeiters, with an element of vice thrown in by way of a gift to the Sunday papers.
BODY ON THE BEACH – VICE KING ARRESTED declared one such. BEACH MURDER TRACED TO VICE EMPIRE said a second. And it was almost pure libel – Louey only did a bit of sub-letting. After all, even revolutionary parties have to get their funds from somewhere...
But it was a London Evening that produced the really telling caption. It caused Chief Super Gish to drop dark hints about people being friendly with editors. 'Body On The Beach' ran a small by-line and then, in a triumph of Cooper Black, GENTLY DOES IT AGAIN! – with one of Gently's better press photos cut in across two columns. Of course, Gently pooh-poohed it. He folded up his copy and stuck it in his pocket with scarcely a glance. But a little later, when he thought he wasn't being watched, Copping saw him perusing that paper with more than common attention.
'We didn't waste no time, when you comes to think of it, sir,' Dutt remarked with a tinge of regret, as they stood on the bridge by the station awaiting their train. 'We comes here on the Friday night to meet a stiff what nobody don't know about and by Tuesday tea-time we got the two geezers what done it, busted up a lot of bolshies and run in a sample of ponces and Teddiesall in a long weekend, you might say.'
Gently passed him a peppermint cream and took one himself. 'We certainly haven't been too heavy on the ratepayers.'
'And me, I was just getting attached to the place, sir. I reckon it beats Sahthend hollow for some things... there's a bloke off Nelson Street as does a plaice-and-chips that knocks you backwards.'
Gently smiled at a distant tug with an orange funnel. 'Talking of plaice, there's some first-rate fishing goes on off the Albion Pier.'
'And them digs of ours, sir, they wasn't half bad neither. I reckon I could stand a week down here with the missus next Bank Holiday...'
A train-whistle sounded close at hand. Gently consulted the watch on his clumsy wrist. Beneath them an empty motor-barge came chugging past, its skipper lounging lazily by his wheel.
'But things change, Dutt... it doesn't take long to alter them. Do you know what struck me most while we were on this job?'
'No, sir. It ain't been like our other jobs, really.'
Gently took careful aim with his screwed-up peppermint cream bag and dropped it neatly on the barge-skipper's peaked cap.
'Well, Dutt, it was the donkeys.'
'The donkeys, sir?' queried Dutt.
Gently nodded and raised his hand in salute to the barge-skipper. 'They've done away with them, Dutt. There isn't one on the beach. If you'd known Starmouth when I knew Starmouth it would make you feel older... but something like that goes on all the time, doesn't it?'
# About the Author
**Alan Hunter** was born in Hoveton, Norfolk in 1922. He left school at the age of fourteen to work on his father's farm, spending his spare time sailing on the Norfolk Broads and writing nature notes for the _Eastern_ _Evening News_. He also wrote poetry, some of which was published while he was in the RAF during the Second World War. By 1950, he was running his own bookshop in Norwich and in 1955, the first of what would become a series of forty-six George Gently novels was published. He died in 2005, aged eighty-two.
# By the Same Author
The _Inspector George Gently_ series
_Gently Does It_
_Gently by the Shore_
_Gently Down the Stream_
_Landed Gently_
_Gently Through the Mill_
_Gently in the Sun_
# Copyright
Constable & Robinson Ltd
3 The Lanchesters
162 Fulham Palace Road
London W6 9ER
www.constablerobinson.com
This paperback edition published by Robinson, an imprint of Constable & Robinson Ltd, 2010
Copyright Alan Hunter 1956, 2010
The right of Alan Hunter to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988
All rights reserved. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser.
A copy of the British Library Cataloguing in Publication data is available from the British Library
ISBN: 978–1–84901–787–9
|
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"redpajama_set_name": "RedPajamaBook"
}
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Blu-ray Review: The Battery
By Adam Frazier | @ | Wednesday, September 17th, 2014 at 6:00 pm
Blu-ray l DVD l Instant
Director: Jeremy Gardner
Screenwriter: Jeremy Gardner
Cast: Jeremy Gardner, Adam Cronheim, Niels Bolle, Alana O'Brien
Not Rated | 101 Minutes
After a zombie apocalypse has overtaken the entire New England area, two former baseball players, Ben (Jeremy Gardner) and Mickey (Adam Cronheim), travel the backroads of Connecticut with no particular destination in mind.
Written and directed by Gardner, The Battery is an apocalyptic zombie flick with a focus less on monsters and mayhem and more on the small pockets of humanity left in the world.
Mickey and Ben aren't really friends, they're just two guys who met at a baseball training camp when the outbreak began. Speaking of the outbreak, it's never explained. There are no scientists or military men to explain how the apocalypse began – all we know is that the dead are up and walking around.
Mickey doesn't want to accept the harsh realities of the zombie apocalypse and longs for the creature comforts he once took for granted. Ben, on the other hand is a lawless, nomadic survivalist who insists they stay on the move.
The two are content to smoke cigarettes and drink beer in-between braining zombies with baseball bats until Mickey stumbles upon a radio transmission from a seemingly thriving community. Mickey will stop at nothing to find it, even though it's made perfectly clear that he and Ben are not welcome.
Shot in just 15 days for $6,000, Gardner's directorial debut is a sincerely funny and beautiful film. Christian Stella's sun-drenched cinematography is gorgeous, and the film's music – consisting of tracks from Rock Plaza Central, The Parlor, El Cantador, Sun Hotel, and Wise Blood – compliments the visuals perfectly. Music is a huge component of Gardner's film – it's the characters' only remaining connection to life before the outbreak.
The Battery hits Blu-ray with a 1080p high-definition widescreen transfer and 5.1 DTS HD Master Audio. As for special features, this Scream Factory release is packed:
Audio Commentary: An insightful, hilarious track featuring Ben Gardner, Adam Cronheim, and Christian Stella.
The Music of The Battery: An 11-minute featurette that looks at Chris Eaton and Don Murray's music for the film, as well as the bands and songs featured in the film.
Tools of Ignorance: The Making of The Battery: This 90-minute documentary about the making of the film features clips of Gardner's early short films, including "Bags," about killer plastic bags. The cast and crew detail their experiences making the film and Gardner explains the origins of the film and how he gathered a bunch of friends to make a feature film.
Outtake Reel: A 12-minute reel of outtakes, including bloopers and behind-the-scenes stuff.
If you're a fan of the undead (and who isn't these days?), I absolutely recommend this indie zombie flick. Scream Factory is primarily known for releasing obscure cult horror and sci-fi films from the '70s and '80s, but every now and then they're able to release a new low-budget indie flick to a large audience.
The Battery will no doubt benefit from this fantastic Scream Factory release, giving fans of flesh-eating reanimated corpses a different kind of zombie flick to digest. The Battery is currently available on Amazon.
Tags: Adam Cronheim, Alana O'Brien, Jeremy Gardner, Niels Bolle, Scream Factory, The Battery, Zombies
Groot Meets Sepultura In 'Groots Bloody Groots' (Video)
Comic Review: Translucid #6
• Blu-ray Review: Road Games (Collector's Edition)
• Blu-ray Review: The Blob (1988) Collector's Edition
• Blu-ray Review: The Devil Rides Out
• Blu-ray Review: Vice Squad (Collector's Edition)
• Movie Review: Bliss
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,604
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using System;
using System.Linq;
using System.Windows.Controls;
using System.Windows.Controls.Primitives;
using PMKS;
using System.Windows;
using System.Windows.Data;
using System.Windows.Media;
using System.Windows.Shapes;
using Point = System.Windows.Point;
namespace PMKS_Silverlight_App
{
public abstract class DisplayVectorBaseShape : Path
{
#region Fields
protected readonly double yOffset;
protected readonly double xOffset;
protected readonly double factor;
#endregion
#region Dependency Properties
public static readonly DependencyProperty StartProperty
= DependencyProperty.Register("Start",
typeof(double[]), typeof(DisplayVectorBaseShape),
new PropertyMetadata(null, OnTimeChanged));
public double[] Start
{
get { return (double[])GetValue(StartProperty); }
set { SetValue(StartProperty, value); }
}
public static readonly DependencyProperty EndProperty
= DependencyProperty.Register("End",
typeof(double[]), typeof(DisplayVectorBaseShape),
new PropertyMetadata(null, OnTimeChanged));
public double[] End
{
get { return (double[])GetValue(EndProperty); }
set { SetValue(EndProperty, value); }
}
#endregion
protected DisplayVectorBaseShape(double factor, double strokeThickness, double xOffset, double yOffset)
{
Height = Width = DisplayConstants.UnCroppedDimension;
this.xOffset = xOffset;
this.yOffset = yOffset;
this.factor = factor;
Data = new LineGeometry();
StrokeThickness = strokeThickness;
}
protected static void OnTimeChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
{
var vector = ((DisplayVectorBaseShape)d);
if (vector.End == null || vector.Start == null || vector.Start.Contains(double.NaN)
|| vector.End.Contains(double.NaN)) return;
var xStart = vector.Start[0] + vector.xOffset;
var yStart = vector.Start[1] + vector.yOffset;
((LineGeometry)vector.Data).StartPoint = new Point(xStart, yStart);
((LineGeometry)vector.Data).EndPoint = new Point(xStart + vector.factor * vector.End[0], yStart + vector.factor * vector.End[1]);
}
public void ClearBindings()
{
ClearValue(StartProperty);
ClearValue(EndProperty);
ClearValue(OpacityProperty);
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,247
|
\section{Introduction}\label{S1}
In this paper, we investigate the interior and boundary pointwise regularity for viscosity solutions of fully nonlinear uniformly elliptic equations
\begin{equation}\label{FNE}
F(D^2u,Du,u,x)=f~~\mbox{in}~\Omega
\end{equation}
and its corresponding Dirichlet problems respectively. Here, $\Omega\subset R^n$ is a bonded domain and $F$ is a real fully nonlinear operator defined in $S^n\times R^n\times R\times \bar\Omega$, where $S^n$ denotes the set of $n\times n$ symmetric matrices.
For studying fully nonlinear equations, some structure condition is necessary. We mainly assume the following structure condition: for any $M,N\in S^n$, $p,q\in R^n$ and $s,t\in R$ with $|s|,|t|\leq K$, we have for $a.e.~x\in \bar \Omega$,
\begin{equation}\label{SC2}
\begin{aligned}
&M^-(M-N,\lambda,\Lambda)-\mu|p-q|(|p|+|q|)-b(x)|p-q|-c(x)\omega_0(K,|s-t|)\\
&\leq F(M,p,s,x)-F(N,q,t,x)\leq \\
&M^+(M-N,\lambda,\Lambda)+\mu|p-q|(|p|+|q|)+b(x)|p-q|+c(x)\omega_0(K,|s-t|),\\
\end{aligned}
\end{equation}
where $0<\lambda\leq \Lambda$, $\mu$ are nonnegative constants and $b,c$ are nonnegative functions. The $M^-,M^+$ denote the Pucci operators (see \Cref{d-Sf}). The $\omega_0$ is a modulus of continuity depending on $K$, i.e., for any $K>0$, $\omega_0(K,\cdot)$ is a nondecreasing function and $\omega_0(K,s)\rightarrow 0$ as $s\rightarrow 0$. In this paper, we assume that all functions are measurable in $x$. We will drop ``$a.e.$'' in the following arguments and use ``for any'' instead.
This structure condition allows equations to have quadratic growth in the gradient and the following is a typical example:
\begin{equation}\label{e.linear-1}
a^{ij}(x)u_{ij}(x)+\mu^{ij}(x)u_i(x)u_j(x)+b^i(x)u_i(x)+c(x)h(u)=f(x),
\end{equation}
where the Einstein summation convention is used (similarly hereinafter), i.e., repeated indices means summation.
Quadratic growth in the gradient is also called natural growth which means that the equation is invariant under nonlinear transformation. For instance, if $u$ is a solution of \cref{FNE} and $v=T(u)$ where $T\in C^2$ and $T'>0$, $v$ is a solution of some equation satisfying the structure condition \cref{SC2}. Equations with quadratic growth in the gradient have been studied to some extent. Ladyzhenskaya and Ural'tseva obtained a priori estimates and proved the existence of classical solutions for quasilinear elliptic equations (see \cite[Theorem III]{MR0149075} and \cite[Chapter 4 and Chapter 6]{MR0244627}). Trudinger \cite{MR701522} extended these results to fully nonlinear elliptic equations. For regularity theory, Sirakov \cite{MR2592289} proved the interior and boundary $C^{\alpha}$ regularity for viscosity solutions of fully nonlinear elliptic equations. Recently, Nornberg \cite{MR3980853} obtained the $C^{1,\alpha}$ regularity. For more works on the equations with quadratic growth in the gradient, we refer to the references in \cite{MR3980853} and \cite{MR2592289}.
We also consider equations under the following special structure condition:
\begin{equation}\label{SC1}
\begin{aligned}
&M^-(M-N,\lambda,\Lambda)-b(x)|p-q|-c(x)|s-t|\\
&\leq F(M,p,s,x)-F(N,q,t,x)\leq\\
&M^+(M-N,\lambda,\Lambda)+b(x)|p-q|+c(x)|s-t|.
\end{aligned}
\end{equation}
In this case, \cref{FNE} is the natural generalization of the linear uniformly elliptic equations in nondivergence form
\begin{equation}\label{e.linear}
a^{ij}(x)u_{ij}(x)+b^{i}(x)u_i(x)+c(x)u(x)=f(x)~~\mbox{in}~~ \Omega.
\end{equation}
With the special structure condition \cref{SC1}, we can obtain various explicit estimates for solutions.
In this paper, we aim to develop the interior and boundary pointwise regularity systematically for viscosity solutions under structure conditions \cref{SC2} and \cref{SC1} (see \Cref{MR} for details). In particular, part of results are new even for the Laplace equation. We make minimum requirements on the coefficients and prescribed data to obtain the corresponding regularity. In other words, our results are optimal (even for the linear elliptic equations). Moreover, all proofs are fairly simple (compared with previous results). This is one of our goal: make the regularity theory easy.
To study fully nonlinear elliptic equations, one may assume first that the solution is a classical solution and then obtain a priori estimates. Based on a priori estimates, the existence of a classical solution can be proved by the method of continuity. The benefit is that we can differentiate the equation directly. This method has its limitations as well. It usually relies on higher smoothness assumptions on the operator, the solution and the domain etc. (e.g. \cite[Chapter 9]{MR1351007} and \cite{MR701522}). In addition, it often obtains the global (or local) estimates rather than pointwise estimates. Moreover, the proofs are relatively complicated compared with the proofs in this paper.
Another way to attack an equation is proving the existence of a solution in some weak sense first and then obtaining the regularity later. Viscosity solution is a kind of weak solution which is introduced by Crandall and Lions \cite{MR690039} (see also \cite{MR503721,MR597451}) and suitable for elliptic equations in nondivergence form, especially for fully nonlinear elliptic equations. The related theories of existence, uniqueness and regularity have been studied extensively. (see \cite{MR1351007,MR1118699,MR2084272} and the references therein).
Among various regularity results for fully nonlinear elliptic equations, pointwise regularity occupies an important position. Caffarelli \cite{MR1005611} (see also \cite{MR1351007}) proved the interior pointwise $C^{1,\alpha}$ and $C^{2,\alpha}$ regularity. Recently, Lian and Zhang \cite{MR4088470} obtained the boundary pointwise $C^{1,\alpha}$ and $C^{2,\alpha}$ regularity. For equations with lower terms, Savin \cite{MR2334822} proved the interior pointwise $C^{2,\alpha}$ regularity for small solutions without the usual assumption that $F$ is convex or concave in $M$. Silvestre and Sirakov \cite{MR3246039} proved the boundary pointwise $C^{1,\alpha}$ and $C^{2,\alpha}$ regularity on flat boundaries for equations depending the gradient. For equations with quadratic growth in the gradient, Sirakov \cite{MR2592289} proved the interior and boundary pointwise $C^{\alpha}$ regularity. The interior pointwise $C^{1,\alpha}$ regularity and boundary pointwise $C^{1,\alpha}$ regularity on flat domains were obtained by Nornberg \cite{MR3980853}. We point out that the pointwise regularity also attract a lot of attention for other types of equation, such as parabolic equations (see \cite{MR1135923,MR1139064,MR1151267}) and the Monge-Amp\`{e}re equation (see \cite{MR2983006}).
Essentially, the behavior of a solution near a point is determined by the coefficients and the prescribed data near the same point. The pointwise regularity shows clearly how the coefficients and the prescribed data influence the behavior of the solution. In fact, pointwise regularity gives deeper understanding of the relation between the solution and the coefficients and the prescribed data (see also the comment at the beginning of \Cref{C2a-mu}). Moreover, the classical local and global regularity can be obtained directly from the pointwise regularity. It is worth to point out that the assumptions for pointwise regularity could be weaker than that for local and global regularity. For example, for the boundary pointwise $C^{k,\alpha}$ regularity, the boundary $\partial \Omega$ maybe not a graph of some function locally, which is necessary for usual boundary regularity since one need to flatten the boundary by a transformation
The perturbation and compactness techniques are adopted in this paper. The perturbation technique is motivated originally from \cite{MR1351007} and the application to boundary regularity is inspired by \cite{MR3780142} and \cite{MR4088470}. The compactness technique has been inspired by \cite{MR3246039} and \cite{Wang_Regularity}. As stated in \cite{Wang_Regularity}, the advantages of compactness technique are that we don't need to solve an equation and use its solution to approximate the origin solution and we don't need the difference of these two solutions to satisfy some equation. In fact, our proofs in this paper don't depend on any solvability. The regularity for a complicated equation \cref{FNE} can be deduced from that of basic equations (see \Crefrange{l-3modin1}{l-32}) by relatively simple proofs. This demonstrates the great power of the perturbation and compactness techniques. Moreover, the regularity results are usually sharp based on this two techniques.
This paper is organized as follows. In \Cref{MR}, we state our main results and make some comments. In particular, we will compare them with previous related results. \Cref{P1} is devoted to prepare some necessary preliminaries. Especially, we present the compactness and the closedness for a family of viscosity solutions. The pointwise regularity will be proved in the subsequent sections. We obtain the interior $C^{1,\alpha}$ regularity, $C^{2,\alpha}$ regularity and $C^{k,\alpha}$ regularity ($k\geq 3$) in \Cref{In-C1a-mu}, \Cref{In-C2a-mu} and \Cref{In-Cka-mu} respectively. The corresponding boundary regularity are proved in \Cref{C1a-mu} to \Cref{Cka-mu}. \Cref{C1C2Ck} gives proofs of the interior $C^{2}$ regularity and boundary $C^1$ regularity. Finally, we derive the interior $C^{k,\mathrm{lnL}}$ ($k\geq 2$) and boundary $C^{1,\mathrm{lnL}}$ regularity in \Cref{CLlnL}.
In the second half of this section, we introduce some notions and notations. They will make the statements of our main results and the proofs concise and readable, which is one of our goals. Our paper treats the regularity of solutions in H\"{o}lder spaces and we adopt the classical definitions of
H\"{o}lder spaces. For a bounded domain $\Omega$ and $k\geq 0$, $C^{k}(\Omega)$ denotes the space of functions who have continuous derivatives up to order $k$ in $\Omega$. The $C^{k}(\bar{\Omega})$ contains functions in $C^{k}(\Omega)$ whose derivatives of order $l$ ($0\leq l\leq k$) have continuous extensions to $\bar{\Omega}$. For $f\in C^{k}(\bar{\Omega})$, define its norm
\begin{equation*}
\|f\|_{C^{k}(\bar{\Omega})}=\sum_{i=0}^{k} \|D^{i}f\|_{L^{\infty}(\Omega)}.
\end{equation*}
In addition, we use $C^{k,\alpha}(\bar{\Omega})$ ($0<\alpha\leq 1$) to denote the space of functions in $C^{k}(\bar{\Omega})$ whose $k$-th order derivatives satisfies for some constant $C$
\begin{equation*}
\sup_{x,y\in \Omega,x\neq y}\frac{|D^kf(x)-D^kf(y)|}{|x-y|^{\alpha}} \leq C.
\end{equation*}
Then we associate the space with the following norm
\begin{equation*}
\|f\|_{C^{k,\alpha}(\bar{\Omega})}=\|f\|_{C^{k}(\bar{\Omega})}+\sup_{x,y\in \Omega,x\neq y}\frac{|D^kf(x)-D^kf(y)|}{|x-y|^{\alpha}}.
\end{equation*}
If
\begin{equation*}
\sup_{x,y\in \Omega,x\neq y}\frac{|D^kf(x)-D^kf(y)|}{|x-y|\big|\ln\min(|x-y|,1/2)\big|} \leq C,
\end{equation*}
we would call $f\in C^{k,\mathrm{lnL}}(\bar{\Omega})$, the so called ``ln-Lipschitz'' spaces and set
\begin{equation*}
\|f\|_{C^{k,\mathrm{lnL}}(\bar{\Omega})}=\|f\|_{C^{k}(\bar{\Omega})}+\sup_{x,y\in \Omega,x\neq y}\frac{|D^kf(x)-D^kf(y)|}{|x-y|\big|\ln\min(|x-y|,1/2)\big|}.
\end{equation*}
Moreover, we say $f\in C^{k,\alpha}(\Omega)$ ($f\in C^{k,\mathrm{lnL}}(\Omega)$) if $f\in C^{k,\alpha}(\bar{\Omega}')$ ($f\in C^{k,\mathrm{lnL}}(\bar{\Omega}')$) for any subdomain $\Omega'\subset\subset \Omega$.
In this paper, we mainly consider the pointwise regularity and we need to introduce the definition of pointwise $C^{k,\alpha}$ for a function.
\begin{definition}\label{d-f}
Let $\Omega\subset R^n$ be a bounded set (may be not a domain), $f:\Omega\rightarrow R$ be a function and $\omega$ be a modulus of continuity. We say that $f$ is $C^{k,\omega}$ ($k\geq 0$) at $x_0\in \Omega$ or $f\in C^{k, \omega}(x_0)$ if there exist a $k$-th order polynomial $P_{x_0}$ and $r_0>0$ such that
\begin{equation}\label{m-holder}
|f(x)-P_{x_0}(x)|\leq |x-x_0|^{k}\omega(|x-x_0|),~~\forall~x\in \Omega\cap B_{r_0}(x_0).
\end{equation}
There may exist multiple $P_{x_0}$ (e.g. $\Omega=B_1\cap R^{n-1}$). Then we take $P_{x_0}$ with
\begin{equation*}
\|P_{x_0}\|=\min \left\{\|P\|\big | \cref{m-holder} ~\mbox{holds with}~P\right\},
\end{equation*}
where $\|P\|=\sum_{m=0}^{k}|D^m P(x_0)|$. Define
\begin{equation*}
D^mf(x_0)=D^mP_{x_0}(x_0)~~\mbox{and}~~\|f\|_{C^{k}(x_0)}=\|P_{x_0}\|.
\end{equation*}
If $\omega(r)=Kr^{\alpha}$ ($0<\alpha\leq 1$) for some constant $K$, we call that $f$ is $C^{k,\alpha}$ at $x_0\in \Omega$ or $f\in C^{k, \alpha}(x_0)$. Then define
\begin{equation*}
[f]_{C^{k,\alpha}(x_0)}=\min \left\{K\big | \cref{m-holder} ~\mbox{holds with}~P_{x_0}~\mbox{and}~K\right\}
\end{equation*}
and
\begin{equation*}
\|f\|_{C^{k, \alpha}(x_0)}=\|f\|_{C^{k}(x_0)}+[f]_{C^{k, \alpha}(x_0)}.
\end{equation*}
If $f\in C^{k, \alpha}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{k,\alpha}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{k}(x)}+\sup_{x\in \Omega} [f]_{C^{k, \alpha}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{k,\alpha}(\bar{\Omega})$.
If $\omega$ is a Dini function, i.e.,
\begin{equation}\label{e.1.Dini}
I_{\omega}:=\int_{0}^{r_0}\frac{\omega(r)}{r} dr<\infty,
\end{equation}
we say that $f$ is $C^{k,\mathrm{Dini}}$ at $x_0$ or $f\in C^{k,\mathrm{Dini}}(x_0)$. Similarly, we define
\begin{equation*}
[f]_{C^{k,\mathrm{Dini}}(x_0)}=\min \left\{I_{\omega}+\omega(r_0) \big | \cref{m-holder} ~\mbox{holds with}~P_{x_0}~\mbox{and}~\omega\right\}
\end{equation*}
and
\begin{equation*}
\|f\|_{C^{k, \mathrm{Dini}}(x_0)}=\|f\|_{C^{k}(x_0)}+[f]_{C^{k, \mathrm{Dini}}(x_0)}.
\end{equation*}
If $f\in C^{k,\mathrm{Dini}}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{k,\mathrm{Dini}}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{k}(x)}+\sup_{x\in \Omega} [f]_{C^{k, \mathrm{Dini}}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{k,\mathrm{Dini}}(\bar{\Omega})$.
If $\omega(r)=Kr|\ln r|$ for some constant $K$, we call that $f$ is $C^{k,\mathrm{lnL}}$ at $x_0$ or $f\in C^{k,\mathrm{lnL}}(x_0)$. Then define
\begin{equation*}
[f]_{C^{k,\mathrm{lnL}}(x_0)}=\min \left\{K \big | \cref{m-holder} ~\mbox{holds with}~P_{x_0}~\mbox{and}~K\right\}
\end{equation*}
and
\begin{equation*}
\|f\|_{C^{k, \mathrm{lnL}}(x_0)}=\|f\|_{C^{k}(x_0)}+[f]_{C^{k, \mathrm{lnL}}(x_0)}.
\end{equation*}
If $f\in C^{k, \mathrm{lnL}}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{k,\mathrm{lnL}}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{k}(x)}+\sup_{x\in \Omega} [f]_{C^{k,\mathrm{lnL}}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{k,\mathrm{lnL}}(\bar{\Omega})$.
If $\omega$ is only a modulus of continuity rather than a Dini function,
we just say that $f$ is $C^{k}$ at $x_0$ or $f\in C^{k}(x_0)$. If $f\in C^{k}(x)$ for any $x\in \Omega$ with the same $\omega$ and $r_0$, and
\begin{equation*}
\|f\|_{C^{k}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{k}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{k}(\bar{\Omega})$.
Finally, if $f\in C^{k,\alpha}(\bar{\Omega}')$ for any $\Omega'\subset\subset \Omega$, we call $f\in C^{k,\alpha}(\Omega)$. Similarly, we can define $f\in C^{k,\mathrm{Dini}}(\Omega)$, $f\in C^{k,\mathrm{lnL}}(\Omega)$ and $f\in C^{k}(\Omega)$.
\end{definition}
\begin{remark}\label{r-1.1}
If $\Omega$ is a bounded smooth domain, the definitions of $C^{k,\alpha}(\bar{\Omega})$ etc. are equivalent to the above usual classical definitions. In this paper, we mainly treat the cases that $\Omega$ is a bounded domain or $\Omega$ is part of the boundary of a domain.
\end{remark}
Next, we define some other types of continuity.
\begin{definition}\label{d-2}
Let $\Omega$, $f$ and $\omega$ be as in \Cref{d-f}. We say that $f$ is $C^{-1,\omega}$ at $x_0$ or $f\in C^{-1,\omega}(x_0)$ if there exists $r_0>0$ such that
\begin{equation}\label{e.c-1}
\|f\|_{L^n(\bar{\Omega}\cap B_r(x_0) )}\leq \omega(r), ~\forall ~0<r<r_0.
\end{equation}
If $\omega(r)=Kr^{\alpha}$ ($0<\alpha\leq 1$) for some constant $K$, we call that $f$ is $C^{-1,\alpha}$ at $x_0\in \Omega$ or $f\in C^{-1, \alpha}(x_0)$. Then define
\begin{equation*}
\|f\|_{C^{-1,\alpha}(x_0)}=\min \left\{K \big | \cref{e.c-1} ~\mbox{holds with}~K\right\}.
\end{equation*}
If $f\in C^{-1, \alpha}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{-1,\alpha}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{-1,\alpha}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{-1,\alpha}(\bar{\Omega})$.
If $\omega$ is a Dini function, we say that $f$ is $C^{-1,\mathrm{Dini}}$ at $x_0$ or $f\in C^{-1,\mathrm{Dini}}(x_0)$. Similarly, we define
\begin{equation*}
\|f\|_{C^{-1,\mathrm{Dini}}(x_0)}=\min \left\{I_{\omega} \big | \cref{e.c-1} ~\mbox{holds with}~\omega\right\}.
\end{equation*}
If $f\in C^{-1,\mathrm{Dini}}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{-1,\mathrm{Dini}}(x)}<+\infty,
\end{equation*}
we say that $f\in C^{-1,\mathrm{Dini}}(\bar{\Omega})$.
If $\omega(r)=Kr|\ln r|$ for some constant $K$, we call that $f$ is $C^{-1,\mathrm{lnL}}$ at $x_0$ or $f\in C^{-1,\mathrm{lnL}}(x_0)$. Then define
\begin{equation*}
\|f\|_{C^{-1,\mathrm{lnL}}(x_0)}=\min \left\{K \big | \cref{e.c-1} ~\mbox{holds with}~K\right\}.
\end{equation*}
If $f\in C^{k, \mathrm{lnL}}(x)$ for any $x\in \Omega$ with the same $r_0$ and
\begin{equation*}
\|f\|_{C^{-1,\mathrm{lnL}}(\bar{\Omega})}:= \sup_{x\in \Omega} \|f\|_{C^{-1,\mathrm{lnL}}(x_0)}<+\infty,
\end{equation*}
we say that $f\in C^{-1,\mathrm{lnL}}(\bar{\Omega})$.
If $\omega$ is only a modulus of continuity rather than a Dini function,
we just say that $f$ is $C^{-1}$ at $x_0$ or $f\in C^{-1}(x_0)$. We also define
\begin{equation*}
\|f\|_{C^{-1}(0)}=\|f\|_{L^n(\Omega\cap B_{r_0})}.
\end{equation*}
If $f\in C^{-1}(x)$ for any $x\in \Omega$ with the same $\omega$ and $r_0$, we say that $f\in C^{-1}(\bar{\Omega})$ and define
\begin{equation*}
\|f\|_{C^{-1}(\bar{\Omega})}=\|f\|_{L^n(\Omega)}.
\end{equation*}
\end{definition}
\begin{remark}\label{r-5}
In this paper, for $f\in C^{k,\mathrm{Dini}}$ ($k\geq -1$), we always use $\omega_f$ to denote its corresponding Dini function.
\end{remark}
Since the boundary pointwise regularity is also considered, we give the definitions of the pointwise geometric conditions on the domain. Usually, if we say that $\partial \Omega$ is $C^{k,\alpha}$ near $x_0\in \partial \Omega$, it means that $\partial \Omega\cap B_{r_0}(x_0)$ (for some $r_0>0$) can be represented as a graph of a $C^{k,\alpha}$ function. Here, we use a more general pointwise definition for the smoothness of the boundary, which is adopted from \cite{MR4088470}.
\begin{definition}\label{d-re} Let $\Omega$ be a bounded domain, $x_0\in \Gamma\subset \partial \Omega$ and $\omega$ be a modulus of continuity. We say that $\partial \Omega$ (or $\Gamma$) is $C^{k,\omega}$ ($k\geq 1$) at $x_0$ or $\partial \Omega\in C^{k,\omega}(x_0)$ (or $\Gamma\in C^{k,\omega}(x_0)$) if there exist a constant $r_0>0$, a coordinate system $\{x_1,...,x_n \}$ (isometric to the original coordinate system) and a $k$-th order polynomial $P(x')$ with $P(0)=0$ and $DP(0)=0$ such that $x_0=0$ in this coordinate system,
\begin{equation}\label{e-re}
B_{r_0} \cap \{(x',x_n)\big |x_n>P(x')+|x'|^k\omega(|x'|)\} \subset B_{r_0}\cap \Omega
\end{equation}
and
\begin{equation}\label{e-re2}
B_{r_0} \cap \{(x',x_n)\big |x_n<P(x')-|x'|^k\omega(|x'|)\} \subset B_{r_0}\cap \Omega^c.
\end{equation}
Then, define
\begin{equation*}
\|\partial \Omega\|_{C^{k}(x_0)}=\|\Gamma\|_{C^{k}(x_0)}=\|P\|=\sum_{m=0}^{k}|D^m P(0)|.
\end{equation*}
If $\omega(r)=Kr^{\alpha}$ ($0<\alpha\leq 1$) for some constant $K$, we call that $\Gamma$ is $C^{k,\alpha}$ at $x_0$ or $\Gamma\in C^{k,\alpha}(x_0)$. Then define
\begin{equation*}
[\Gamma]_{C^{k,\alpha}(x_0)}=\min \left\{K\big | \cref{e-re} ~\mbox{and}~ \cref{e-re2}~\mbox{hold with}~P~\mbox{and}~K\right\}
\end{equation*}
and
\begin{equation*}
\|\Gamma\|_{C^{k, \alpha}(x_0)}=\|\Gamma\|_{C^{k}(x_0)}+[\Gamma]_{C^{k, \alpha}(x_0)}.
\end{equation*}
If $\Gamma\in C^{k, \alpha}(x)$ for any $x\in \Gamma$ with the same $r_0$ and
\begin{equation*}
\|\Gamma\|_{C^{k,\alpha}}:= \sup_{x\in \Gamma} \|\Gamma\|_{C^{k}(x)}+\sup_{x\in \Gamma}~[\Gamma]_{C^{k, \alpha}(x)}<+\infty,
\end{equation*}
we say that $\Gamma\in C^{k,\alpha}$.
If $\omega$ is a Dini function, we say that $\Gamma$ is $C^{k,\mathrm{Dini}}$ at $x_0$ or $\Gamma\in C^{k,\mathrm{Dini}}(x_0)$. Similarly, we define
\begin{equation*}
[\Gamma]_{C^{k,\mathrm{Dini}}(x_0)}=\min \left\{I_{\omega}+\omega(r_0)\big | \cref{e-re} ~\mbox{and}~ \cref{e-re2}~\mbox{hold with}~P~\mbox{and}~\omega\right\}
\end{equation*}
and
\begin{equation*}
\|\Gamma\|_{C^{k, \mathrm{Dini}}(x_0)}=\|\Gamma\|_{C^{k}(x_0)}+[\Gamma]_{C^{k, \mathrm{Dini}}(x_0)},
\end{equation*}
where
\begin{equation*}
I_{\omega}=\int_{0}^{r_0} \frac{\omega(r)}{r} dr.
\end{equation*}
If $\Gamma\in C^{k,\mathrm{Dini}}(x)$ for any $x\in \Gamma$ with the same $r_0$ and
\begin{equation*}
\|\Gamma\|_{C^{k,\mathrm{Dini}}}:= \sup_{x\in \Gamma} \|\Gamma\|_{C^{k}(x)}+\sup_{x\in \Gamma}~ [\Gamma]_{C^{k, \mathrm{Dini}}(x)}<+\infty,
\end{equation*}
we say that $\Gamma\in C^{k,\mathrm{Dini}}$.
If $\omega(r)=Kr|\ln \min(r,1/2)|$ for some constant $K$, we call that $\Gamma$ is $C^{k,\mathrm{lnL}}$ at $x_0$ or $\Gamma\in C^{k,\mathrm{lnL}}(x_0)$. Then define
\begin{equation*}
[\Gamma]_{C^{k,\mathrm{lnL}}(x_0)}=\min \left\{K\big | \cref{e-re} ~\mbox{and}~ \cref{e-re2}~\mbox{hold with}~P~\mbox{and}~K\right\}
\end{equation*}
and
\begin{equation*}
\|\Gamma\|_{C^{k, \mathrm{lnL}}(x_0)}=\|\Gamma\|_{C^{k}(x_0)}+[\Gamma]_{C^{k, \mathrm{lnL}}(x_0)}.
\end{equation*}
If $\Gamma\in C^{k, \mathrm{lnL}}(x)$ for any $x\in \Gamma$ with the same $r_0$ and
\begin{equation*}
\|\Gamma\|_{C^{k,\mathrm{lnL}}}:= \sup_{x\in \Gamma} \|\Gamma\|_{C^{k}(x)}+\sup_{x\in \Gamma} ~[\Gamma]_{C^{k,\mathrm{lnL}}(x)}<+\infty,
\end{equation*}
we say that $\Gamma\in C^{k,\mathrm{lnL}}$.
If $\omega$ is only a modulus of continuity rather than a Dini function,
we may simply say that $\Gamma$ is $C^{k}$ at $x_0$ or $\Gamma\in C^{k}(x_0)$. If $\Gamma\in C^{k}(x)$ for any $x\in \Gamma$ with the same $\omega$ and $r_0$ and
\begin{equation*}
\|\Gamma\|_{C^{k}}:= \sup_{x\in \Gamma} \|\Gamma\|_{C^{k}(x)}<+\infty,
\end{equation*}
we say that $\Gamma\in C^{k}$.
\end{definition}
\begin{remark}\label{r-12}
One feature of this definition is that $\partial \Omega$ may not be represented as a graph of a function near $x_0$. For example, let
\begin{equation*}
\Omega=B_1\cap \left\{(x',x_n)\big | x_n>|x'|^2/2\right\}\backslash \left\{(x',x_n)\big | x_n=|x'|^2/2+|x'|^4, |x|\leq 1/2 \right\}.
\end{equation*}
Then $\partial \Omega$ is $C^{2,\alpha}$ at $0$ for any $0<\alpha\leq 1$ by the definition.
\end{remark}
In addition, we define the oscillation of $\partial \Omega$, which is also adopted from \cite{MR4088470}.
\begin{definition}\label{d.osc}
Let $\Omega$ be a bounded domain and $x_0\in \partial \Omega$. Given $r_0>0$ and $\nu\in R^n$ with $|\nu|=1$, set
\begin{equation*}
A(\nu)=\left\{r\big|\{x\in B(x_0,r_0)\big|~(x-x_0)\cdot \nu >r\}\subset \Omega\right\}
\end{equation*}
and
\begin{equation*}
B(\nu)= \left\{r\big|\{x\in B(x_0,r_0)\big|~(x-x_0)\cdot \nu <-r\}\subset \Omega^c \right\}
\end{equation*}
If $A(\nu)$ and $B(\nu)$ are not empty, define
\begin{equation*}
\underset{B_{r_0},\nu}{\mathrm{osc}}~\partial\Omega=\inf_{r\in A(\nu)} r +\inf_{r\in B(\nu)} r
\end{equation*}
and
\begin{equation*}
\underset{B_{r_0}}{\mathrm{osc}}~\partial\Omega=\inf_{|\nu|=1} \underset{B_{r_0},\nu}{\mathrm{osc}}~\partial\Omega.
\end{equation*}
\end{definition}
\begin{remark}\label{r-0}
If we study the boundary regularity in later sections, we always assume that $\underset{B_{r}}{\mathrm{osc}}~\partial\Omega$ exists for any $r>0$ and
\begin{equation*}
\underset{B_{r}}{\mathrm{osc}}~\partial\Omega= \underset{B_{r},e_n}{\mathrm{osc}}~\partial\Omega=\underset{x\in \partial \Omega\cap B_r}{\sup} x_n -\underset{x\in \partial \Omega\cap B_r}{\inf} x_n.
\end{equation*}
\end{remark}
In the following, we introduce some notions with respect to viscosity solutions, which are standard (see \cite{MR1376656}, \cite{MR1351007} and \cite{MR1118699}).
\begin{definition}\label{d-viscoF}
We say that $u\in C(\Omega)$ is an $L^p$-viscosity ($p>n/2$) subsolution (resp., supersolution) of \cref{FNE} if
\begin{equation*}
\begin{aligned}
ess~\underset{y\to x}{\lim\sup}\left(F(D^2\varphi(y),D\varphi(y),u(y),y)-f(y)\right)\geq 0\\
\left(\mathrm{resp.},~ess~\underset{y\to x}{\lim\inf}\left(F(D^2\varphi(y),D\varphi(y),u(y),y)-f(y)\right)\leq 0\right)
\end{aligned}
\end{equation*}
provided that for $\varphi\in W^{2,p}(\Omega)$, $u-\varphi$ attains its local maximum (resp., minimum) at $x\in\Omega$.
We call $u\in C(\Omega)$ an $L^p$-viscosity solution of \cref{FNE} if it is both an $L^p$-viscosity subsolution and supersolution of \cref{FNE}.
\end{definition}
\begin{remark}\label{r-14}
In this paper, we deal with the $L^n$-viscosity solution. If $\varphi\in W^{2,p}(\Omega)$ is replaced by $\varphi\in C^2(\Omega)$, we arrive at the definition of $C$-viscosity solution. If all functions are continuous in their variables, these two notions of viscosity solutions are equivalent.
\end{remark}
We introduce the Pucci class, which is basic for fully nonlinear elliptic equations.
\begin{definition}\label{d-Sf}
For $M\in S^n$, denote its eigenvalues by $\lambda_i$ ($1\leq i\leq n$) and set
\begin{equation*}
M^+(M,\lambda,\Lambda)=\Lambda\left(\sum_{\lambda_i>0}\lambda_i\right)
+\lambda\left(\sum_{\lambda_i<0}\lambda_i\right),
M^-(M,\lambda,\Lambda)=\lambda\left(\sum_{\lambda_i>0}\lambda_i\right)
+\Lambda\left(\sum_{\lambda_i<0}\lambda_i\right).
\end{equation*}
Then we can define the Pucci class as follows. Let $b\in L^p(\Omega) (p>n)$ and $f\in L^n(\Omega)$. We say that $u\in \underline{S}(\lambda,\Lambda,\mu,b,f)$ if $u$ is an $L^n$-viscosity subsolution of
\begin{equation}\label{SC2Sf}
M^+(D^2u,\lambda,\Lambda)+\mu|Du|^2+b|Du|= f.
\end{equation}
Similarly, we denote $u\in \bar{S}(\lambda,\Lambda,\mu,b,f)$ if $u$ is an $L^n$-viscosity supersolution of
\begin{equation}\label{SC2Sf-}
M^-(D^2u,\lambda,\Lambda)-\mu|Du|^2-b|Du|= f.
\end{equation}
We also define
\begin{equation*}
S^*(\lambda,\Lambda,\mu,b,f)=\underline{S}(\lambda,\Lambda,\mu,b,-|f|)\cap \bar{S}(\lambda,\Lambda,\mu,b,|f|).
\end{equation*}
We will denote $\underline{S}(\lambda,\Lambda,\mu,b,f)$ ($\bar{S}(\lambda,\Lambda,\mu,b,f)$, $S^*(\lambda,\Lambda,\mu,b,f)$) by $\underline{S}(\mu,b,f)$ ($\bar{S}(\mu,b,f)$, $S^*(\mu,b,f)$) and $M^+(M,\lambda,\Lambda)$ ($M^-(M,\lambda,\Lambda)$) by $M^+(M)$ ($M^-(M)$) when the choice of $\lambda,\Lambda$ is understood.
\end{definition}
We use the Einstein summation convention throughout this paper, i.e., repeated indices are implicitly summed over. Symbols in this paper are standard and are listed below.
\begin{notation}\label{no1.1}
\begin{enumerate}~~\\
\item $\{e_i\}^{n}_{i=1}$: the standard basis of $R^n$, i.e., $e_i=(0,...0,\underset{i^{th}}{1},0,...0)$.
\item $x'=(x_1,x_2,...,x_{n-1})$ and $x=(x_1,...,x_n)=(x',x_n)$ .
\item $S^{n}$: the set of $n\times n$ symmetric matrices and $\|A\|=$ the spectral radius of $A$ for any $A\in S^{n}$.
\item $|x|:=\left(\sum_{i=1}^{n} x_i^2\right)^{1/2}$ for $x\in R^n$.
\item $R^n_+=\{x\in R^n\big|x_n>0\}$.
\item $B_r(x_0)=B(x_0,r)=\{x\in R^{n}\big| |x-x_0|<r\}$, $B_r=B_r(0)$, $B_r^+(x_0)=B_r(x_0)\cap R^n_+$ and $B_r^+=B^+_r(0)$.
\item $T_r(x_0)\ =\{(x',0)\in R^{n}\big| |x'-x_0'|<r\}$ \mbox{ and } $T_r=T_r(0)$.
\item $A^c$: the complement of $A$ and $\bar A $: the closure of $A$, where $ A\subset R^n$.
\item $\mathrm{diam}(A)$: the diameter of $A$ and $\mathrm{dist}(A,B)$: the distance between $A$ and $B$, where $ A,B\subset R^n$.
\item $a^+$: $\max(a,0)$, the positive part of $a$; $a^-$: $\max(-a,0)$, the negative part of $a$ for $a\in R$.
\item $\Omega_r=\Omega\cap B_r$ and $(\partial\Omega)_r=\partial\Omega\cap B_r$.
\item $\varphi _i=D_i \varphi=\partial \varphi/\partial x _{i}$, $\varphi _{ij}=D_{ij}\varphi =\partial ^{2}\varphi/\partial x_{i}\partial x_{j}$ and we also use similar notations for higher order derivatives.
\item $D^0\varphi =\varphi$, $D \varphi=(\varphi_1 ,...,\varphi_{n} )$ and $D^2 \varphi =\left(\varphi _{ij}\right)_{n\times n}$ etc. $D_{x'} \varphi=(\varphi_1 ,...,\varphi_{n-1} )$ and $D_{x'}^2 \varphi =\left(\varphi _{ij}\right)_{(n-1)\times (n-1)}$ etc.
\item $|D^k\varphi |= \left(\sum_{1\leq i_1,\cdots,i_k\leq n}\varphi_{i_1,\cdots,i_k}^2\right)^{1/2}$ for $k\geq 1$. For $k=2$, we also write $\|D^2\varphi\|$ since it is a matrix.
\end{enumerate}
\end{notation}
\section{Main results}\label{MR}
In this section, we list a series of pointwise regularity for viscosity solutions, which will be proved in later sections. There exists a constant $0<\bar{\alpha}\leq 1$ depending only on $n,\lambda$ and $\Lambda$ (see \Crefrange{l-3modin1}{l-32}) such that the theorems in this section hold.
For any fully nonlinear operator $F$ appearing in this paper, we always assume that $F$ satisfies \cref{SC2} and $F(0,0,0)=0$ if $F$ is independent of $x$ unless otherwise stated. If we study the interior pointwise regularity, we always consider the equation in $B_1$ and study the regularity at the center $0$; if we study the boundary pointwise regularity, we always assume that $0\in \partial \Omega$ and study the boundary regularity at $0$. In both cases, we always assume that $r_0=1$ in \Cref{d-f}, \Cref{d-2} and \Cref{d-re}. When we say that $\partial \Omega$ is $C^{k,\alpha}$ at $0$, it always indicates that \cref{e-re} and \cref{e-re2} hold with $P(0)=0$ and $DP(0)=0$.
Our main results are stated as follows.
\subsection{The $C^{k,\alpha}$ regularity}
In addition to the structure condition, it is necessary to make assumptions on the oscillation of $F$ in $x$. Suppose that we consider an equation in a domain $\Omega$. Let $x_0\in \bar{\Omega}$ be the point where we intend to study the behavior of the solution. For the $C^{1,\alpha}$ regularity, we introduce the following to estimate the oscillation of $F$ in $x$ near $x_0$. There exist a fully nonlinear operator $F_{x_0}$ and $r_0>0$ such that for any $M\in S^n$ and $x\in \bar\Omega\cap B_{r_0}(x_0)$
\begin{equation}\label{e.C1a.beta}
|F(M,0,0,x)-F_{x_0}(M)|\leq \beta_1(x,x_0) \omega_1(\|M\|)+\gamma_1(x,x_0) ,
\end{equation}
where $\beta_1,\gamma_1\geq 0$ with $\beta_1(x,x)\equiv \gamma_1(x,x) \equiv 0$ ; $\omega_1\geq 0$ is nondecreasing and there exists a constant $K_0$ such that for any $0<r<1$,
\begin{equation}\label{e.omega-1}
r\omega_1\left(r^{-1}\|M\|\right)\leq K_0\omega_1(\|M\|).
\end{equation}
We also consider a special case that $F$ satisfies the following:
\begin{equation}\label{e.C1a.beta-2}
|F(M,0,0,x)-F_{x_0}(M)|\leq \beta_1(x,x_0) \|M\|+\gamma_1(x,x_0).
\end{equation}
In this case, we can obtain explicit estimates for solutions.
If we study pointwise regularity, we always assume that $x_0=0$ and $r_0=1$ in \cref{e.C1a.beta} and \cref{e.C1a.beta-2} unless otherwise stated. For simplicity, we denote $\beta_1(x)=\beta_1(x,0)$ and $\gamma_1(x)=\gamma_1(x,0)$. We remark here that if $F$ is continuous in $x$, we may use $F(M,0,0,x_0)$ to measure the oscillation of $F$ in $x$ in \cref{e.C1a.beta} instead of $F_{x_0}(M)$.
Clearly, the linear equation \cref{e.linear} satisfies \cref{e.C1a.beta-2} for $\beta_1(x,x_0)=|a^{ij}(x)-a^{ij}_{x_0}|$ and $\gamma_1\equiv 0$, where $a^{ij}_{x_0}$ is a constant matrix with eigenvalues lying in $[\lambda,\Lambda]$.
First, we give the interior pointwise $C^{1,\alpha}$ regularity.
\begin{theorem}\label{t-C1a-i}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation}\label{e.2.1}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation}
Suppose that $F$ satisfies \cref{e.C1a.beta} with $\|\beta_1\|_{C^{-1,1}(0)}\leq \delta_0$ and $\gamma_1\in C^{-1,\alpha}(0)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,\alpha$ and $\omega_1$. Assume that $b\in L^{p}(B_1)$ $(p=n/(1-\alpha))$, $c\in C^{-1,\alpha}(0)$ and $f\in C^{-1,\alpha}(0)$.
Then $u\in C^{1,\alpha}(0)$, i.e., there exists a linear polynomial $P$ such that
\begin{equation}\label{e.C1a-1-i-mu}
|u(x)-P(x)|\leq C |x|^{1+\alpha}, ~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C1a-2-i-mu}
|Du(0)|\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,\alpha,\mu,\|b\|_{L^p(B_1)}, \|c\|_{C^{-1,\alpha}(0)},\omega_0,\|\gamma_1\|_{C^{-1,\alpha}(0)}$, $\omega_1,\|f\|_{C^{-1,\alpha}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1a-1-i}
|u(x)-P(x)|\leq C |x|^{1+\alpha}\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{-1,\alpha}(0)}+\|\gamma_1\|_{C^{-1,\alpha}(0)}\right), ~\forall ~x\in B_{1},
\end{equation}
and
\begin{equation}\label{e.C1a-2-i}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }( B_1)}+\|f\|_{C^{-1,\alpha}(0)}+\|\gamma_1\|_{C^{-1,\alpha}(0)}\right),
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,\|b\|_{L^p(B_1)}$ and $\|c\|_{C^{-1,\alpha}(0)}$.
\end{theorem}
\begin{remark}\label{r-1}
The $C^{1,\alpha}$ estimate is also called Cordes-Nirenberg estimate. Caffarelli \cite{MR1005611,MR1351007} proved the interior $C^{1,\alpha}$ regularity for fully nonlinear elliptic equations without lower order terms. Swiech \cite{MR1606359} obtained interior $C^{1,\alpha}$ regularity for $\mu=0$, $b,c\in L^{\infty}$ and less general $\omega_0$. Recently, Nornberg \cite{MR3980853} proved $C^{1,\alpha}$ regularity for equations in the general form. However, he proved the regularity for some $\alpha<\bar{\alpha}$ and $c\in L^{\infty}$. In addition, it requires the solvability for some equation and its proof is relatively complicated.
\end{remark}
\begin{remark}\label{r-2}
The condition $b\in L^p$ ($p=n/(1-\alpha)$) can be replaced by the following: there exists $p_0>n$ and a constant $K_0$ such that $\|b\|_{L^{p_0}(B_r)}\leq K_0 r^{\alpha-1+n/p_0}$ for any $0<r<1$. The reason that we require $b\in L^p$ for some $p>n$ is the dependence on the closedness (\Cref{l-35}) and the H\"{o}lder regularity (\Cref{l-3Ho}) for viscosity solutions.
\end{remark}
\begin{remark}\label{r-3}
If $\bar{\alpha}\leq \alpha<1$, \Cref{t-C1a-i} can also be proved by a similar proof provided an additional assumption that $F$ and $F_{x_0}$ are convex in $M$. Similarly, interior and boundary pointwise $C^{k,\alpha}$ ($\bar{\alpha}\leq \alpha<1,k\geq 1$) can be obtained by strengthening the assumption on $F$ correspondingly.
\end{remark}
\begin{remark}\label{r-2.4}
Since we study fully nonlinear elliptic equations in general form, the statement of the theorem seems a little complicated. In fact, the assumption includes mainly three parts (similarly hereinafter): the structure condition \cref{SC2}, the oscillation of $F$ in $x$ \cref{e.C1a.beta} and the prescribed data appearing in the equation \cref{e.2.1} (i.e., $f$). The \cref{SC2} describes the dependence of $F$ on $M,p$ and $s$, and \cref{e.C1a.beta} describes the dependence of $F$ on $x$. Both of them are necessary for regularity of fully nonlinear elliptic equations. Hence, we need to make necessary assumptions on the parameters appearing in \cref{SC2} (i.e., $\mu,b,c$ and $\omega_0$) and \cref{e.C1a.beta} (i.e., $\beta_1$ and $\omega_1$). For boundary regularity, the prescribed data also contain $g$ and $\partial \Omega$ (see boundary regularity below).
\end{remark}
Next, we intend to present the interior pointwise $C^{2,\alpha}$ regularity. For the $C^{1,\alpha}$ ($0<\alpha<1$) or lower regularity, assuming $b\in L^p$ and $c\in L^n$ is appropriate in \cref{SC2}. For higher regularity (including the $C^{1,\mathrm{lnL}}$ or higher regularity), we assume that \cref{SC2} holds with $b,c\in L^{\infty}$ without loss of generality. Equivalently, for higher regularity, we always assume that $b\equiv b_0$ and $c\equiv c_0 $ for some positive constants $b_0,c_0$ in \cref{SC2}.
In addition, for the $C^{2,\alpha}$ regularity, we require that $\omega_0$ satisfies the following homogeneous condition: there exists a constant $K_0>0$ such that
\begin{equation}\label{e.omega0}
\omega_0(\cdot,r s)\leq K_0r^{\alpha}\omega_0(\cdot,s), ~\forall ~0<r<1,s>0.
\end{equation}
If we consider \cref{e.linear-1} with $h(u)=u^q$, \cref{e.omega0} indicates that $q\geq \alpha$ is necessary for $C^{2,\alpha}$ regularity. This can be shown by constructing a simple example.
Similar to the $C^{1,\alpha}$ regularity, we use the following to estimate the oscillation of $F$ in $x$ near $x_0\in\bar{\Omega}$. There exists $r_0>0$ such that for any $(M,p,s)\in S^n\times R^n\times R$ and $x\in \bar\Omega\cap B_{r_0}(x_0)$,
\begin{equation}\label{e.C2a-KF}
|F(M,p,s,x)-F_{x_0}(M,p,s)|\leq \beta_2(x,x_0)\omega_2(\|M\|,|p|,|s|),
\end{equation}
where $F_{x_0}$ satisfies the same structure condition as $F$ and $\beta_2\geq 0$ with $\beta_2(x,x)\equiv 0$. Here, $\omega_2\geq 0$ is nondecreasing in each variable and there exists a positive constant $K_0$ such that for any $0<r<1$,
\begin{equation}\label{e.omega}
r\omega_2\left(r^{-1}\|M\|,|p|,|s|\right)\leq K_0\omega_2(\|M\|,|p|,|s|).
\end{equation}
We remark here that if $F$ satisfies \cref{e.C2a-KF}, it satisfies \cref{e.C1a.beta} with $\omega_2(\cdot,0,0)$. For the $C^{1,\alpha}$ regularity, the assumption on $\beta_1$ is different from that on $\gamma_1$ and hence we adopt \cref{e.C1a.beta}. For the $C^{2,\alpha}$ regularity, \cref{e.C2a-KF} is enough.
This paper also considers the following special case for which we can obtain explicit estimates:
\begin{equation}\label{e.C2a-KF-0}
|F(M,p,s,x)-F_{x_0}(M,p,s)|\leq \beta_2(x,x_0) \left(\|M\|+|p|+|s|\right)+\gamma_2(x,x_0),
\end{equation}
where $\gamma_2\geq 0$ with $\gamma_2(x,x)\equiv 0$.
Similar to the $C^{1,\alpha}$ regularity, if we study pointwise regularity, we always assume that $x_0=0$ and $r_0=1$. For simplicity, we denote $\beta_2(x,0)$ by $\beta_2(x)$ and $\gamma_2(x,0)$ by $\gamma_2(x)$. If $F$ is continuous in $x$, we may use $F(M,p,s,x_0)$ to measure the oscillation of $F$ in $x$ in \cref{e.C2a-KF} instead of $F_{x_0}(M,p,s)$.
For the linear equation \cref{e.linear}, the corresponding operator is $F(M,p,s,x)=a^{ij}(x)M_{ij}+b^{i}(x)p_i+c(x)s$. Let $F_{x_0}(M,p,s)=a^{ij}_{x_0}M_{ij}+b^{i}_{x_0}p_i+c_{x_0}s$, where $a^{ij}_{x_0}$ is a constant matrix with eigenvalues lying in $[\lambda,\Lambda]$; $b^{i}_{x_0}$ is a constant vector and $c_{x_0}$ is a constant. Then
\begin{equation*}
|F(M,p,s,x)-F_{x_0}(M,p,s)|\leq \beta_2(x,x_0)\omega_2(\|M\|,|p|,|s|),
\end{equation*}
where $\beta_2(x,x_0)=\max \left(|a^{ij}(x)-a^{ij}_{x_0}|,|b^{i}(x)-b^{i}_{x_0}|,
|c(x)-c_{x_0}|\right)$ and $\omega_2(\|M\|,|p|,|s|)=\|M\|+|p|+|s|$.
Now, we state the interior pointwise $C^{2,\alpha}$ regularity.
\begin{theorem}\label{t-C2a-i}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\alpha}(0)$. Assume that $\omega_0$ satisfies \cref{e.omega0} and $f\in C^{\alpha}(0)$.
Then $u\in C^{2,\alpha}(0)$, i.e., there exists a quadratic polynomial $P$ such that
\begin{equation}\label{e.C2a-1-i-mu}
|u(x)-P(x)|\leq C |x|^{2+\alpha}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{e.C2a-3-i-mu}
|F(D^2P,DP(x),P(x),x)-f(0)|\leq C|x|^{\alpha}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C2a-2-i-mu}
|Du(0)|+\|D^2u(0)\|\leq C,
\end{equation}
where $C$ depends only on $n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,
\|\beta_2\|_{C^{\alpha}(0)},\omega_2,\|f\|_{C^{\alpha}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\alpha}(0)$, we have the following explicit estimates
\begin{equation}\label{e.C2a-1-i}
|u(x)-P(x)|\leq C |x|^{2+\alpha}\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\alpha}(0)}
+\|\gamma_2\|_{C^{\alpha}(0)}\right), ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{e.C2a-3-i}
|F(D^2P,DP(x),P(x),x)-f(0)|\leq C|x|^{\alpha}
\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\alpha}(0)}
+\|\gamma_2\|_{C^{\alpha}(0)}\right), ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C2a-2-i}
|Du(0)|+\|D^2u(0)\|\leq C\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\alpha}(0)}
+\|\gamma_2\|_{C^{\alpha}(0)}\right),
\end{equation}
where $C$ depends only on $n,\lambda, \Lambda,\alpha, b_0,c_0$ and $\|\beta_2\|_{C^{\alpha}(0)}$.
\end{theorem}
\begin{remark}\label{r-2.1}
This is the classical $C^{2,\alpha}$ regularity (Schauder estimate), which has been studied extensively for linear or nonlinear elliptic equations. With respect to pointwise regularity for fully nonlinear elliptic equations, it was first proved by Caffarelli \cite{MR1005611,MR1351007}. Savin \cite{MR2334822} extended to equations with lower order terms. It is interesting that the convexity of $F$ can be removed by assuming that $\|u\|_{L^{\infty}}$ is small enough and $F\in C^{1}$ (see also \cite[Section 4]{MR2928094}). In regards to equations with quadratic growth in gradient, Trudinger \cite{MR701522} had obtained a priori estimates under less general conditions. As far as we know, there is no $C^{2,\alpha}$ regularity result for equations with quadratic growth in gradient.
\end{remark}
\begin{remark}\label{r-2.5}
In \Cref{t-C2a-i}, the assumptions on $\beta_2$ and $f$ can be replaced with the following weaker ones (see the proof in \Cref{In-C2a-mu}):
\begin{equation*}
\|\beta_2\|_{L^n(B_r)}\leq Cr^{1+\alpha}~\mbox{and}~
\|f-f_0\|_{L^n(B_r)}\leq Cr^{1+\alpha},~\forall ~0<r<1,
\end{equation*}
where $f_0$ is a constant. To make the statement clear, we work under the assumptions in \Cref{t-C2a-i}. Similar comments can be made for other regularity.
\end{remark}
\begin{remark}\label{r-2.12}
For the linear equation \cref{e.linear}, if the coefficients $a^{ij},b^{i}$ and $c$ is $C^{\alpha}$ at $0$, $\beta_2\in C^{\alpha}(0)$. Then \Cref{t-C2a-i} reduces to the classical Schauder estimates for linear equations.
\end{remark}
\begin{remark}\label{r-22}
If $F(0,0,0,x)\equiv 0$, we can take $\gamma_2\equiv 0$ in \cref{e.C2a-KF-0} to estimate the oscillation of $F$ in $x$. Then we obtain estimates similar to \crefrange{e.C2a-1-i}{e.C2a-2-i} without ``$\gamma_2$'' involved in the righthand. This is a natural generalization of Schauder estimates for linear equations.
\end{remark}
\begin{remark}\label{r-2.17}
The requirement that $F$ is convex in $M$ originates from the model problem \Cref{l-3modin2}.
\end{remark}
In this paper, we also obtain higher pointwise regularity. We require that $\omega_0$ satisfies the following necessary homogeneous condition: for some constant $K_0>0$,
\begin{equation}\label{e.omega0-2}
\omega_0(\cdot,rs)\leq K_0r\omega_0(\cdot,s), ~\forall ~r>0,s>0.
\end{equation}
For the equation \cref{e.linear-1}, if $h\in C^{1}(R)$, \cref{e.omega0-2} holds.
Before stating the regularity result, we introduce the following definition to estimate the oscillation of $F$ in $x$.
\begin{definition}\label{d-FP}
Let $k\geq 0$, $\Omega$ be a bounded domain and $F:S^n\times R^n\times R\times \bar{\Omega}\rightarrow R$. We use the following to estimate the oscillation of $F$ in $x$ near $x_0\in \bar{\Omega}$. There exists $r_0>0$ such that for any $(M,p,s)\in S^n\times R^n\times R$ and $x\in \bar\Omega\cap B_{r_0}(x_0)$,
\begin{equation}\label{holder}
|F(M,p,s,x)-F_{x_0}(M,p,s,x)|\leq \beta_3(x,x_0)\omega_3(\|M\|,|p|,|s|),
\end{equation}
where $F_{x_0} \in C^{k,\bar{\alpha}}(S^n\times R^n\times R\times \bar{\Omega})$ and satisfies the same structure condition as $F$; $\beta_3(x,x_0)=|x-x_0|^k\omega(|x-x_0|)$ and $\omega$ is a modulus of continuity. Here, $\omega_3\geq 0$ is nondecreasing in each variable and there exists a positive constant $K_0$ such that for any $0<r<1$,
\begin{equation}\label{e.omega-3}
r\omega_3\left(r^{-1}\|M\|,|p|,|s|\right)\leq K_0\omega_3(\|M\|,|p|,|s|).
\end{equation}
For pointwise regularity, we always assume that $x_0=0$ and $r_0=1$. For simplicity, we also denote $\beta_3(x,0)$ by $\beta_3(x)$.
If $\omega(r)=Kr^{\alpha}$ ($0<\alpha< \bar{\alpha}$) for some constant $K$, we call that $F$ is $C^{k,\alpha}$ at $x_0$ or $F\in C^{k, \alpha}(x_0)$. Then define
\begin{equation*}
\|\beta_3(\cdot,x_0)\|_{C^{k,\alpha}(x_0)}=\min \left\{K\big | \cref{holder} ~\mbox{holds with}~K\right\}
\end{equation*}
and denote
\begin{equation*}
\omega_4(r)=\|F_{x_0}\|_{C^{k,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{\Omega})},~\forall ~r>0,
\end{equation*}
where $\textbf{B}_r:=\left\{(M,p,s)\big| \|M\|+|p|+|s|<r\right\}$. If $\bar{\alpha}\leq \alpha\leq 1$, we require that $F_{x_0}\in C^{k+1,\bar\alpha}$ and set
\begin{equation*}\label{e.cka.F}
\omega_4(r)=\|F_0\|_{C^{k+1,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{\Omega})},~\forall ~r>0.
\end{equation*}
If $F\in C^{k, \alpha}(x)$ for any $x\in \bar{\Omega}$ with the same $r_0,\omega_3$ and $\omega_4$, and
\begin{equation*}
\|\beta_3\|_{C^{k,\alpha}(\bar{\Omega})}:=\sup_{x_0\in \Omega} \|\beta_3(\cdot,x_0)\|_{C^{k,\alpha}(x_0)}<+\infty,
\end{equation*}
we say that $F\in C^{k, \alpha}(\bar{\Omega})$.
Similarly, we can define $F\in C^{k,\mathrm{Dini}}$ and $F\in C^{k,\mathrm{lnL}}$. If $\omega$ is only a modulus of continuity rather than a Dini function, we may say that $F\in C^{k}$.
\end{definition}
Take the linear equation \cref{e.linear} for example again. If $a^{ij},b^{i}$ and $c$ are $C^{k,\alpha}$ at $x_0$, we can take $F_{x_0}$ as
\begin{equation*}
F_{x_0}(M,p,s,x)=a_{x_0}^{ij}(x)M_{ij}+b_{x_0}^{i}(x)p_i+c_{x_0}(x)s,
\end{equation*}
where $a_{x_0}^{ij},b_{x_0}^{i},c_{x_0}$ are the $k$-th order Taylor polynomials of $a^{ij},b^{i},c$ at $x_0$ respectively. Then $F_{x_0} \in C^{\infty}(S^n\times R^n\times R\times \bar{\Omega})$ and $F$ is $C^{k,\alpha}$ at $x_0$. Roughly speaking, if the coefficients are $C^{k,\alpha}$ at $x_0$, the operator is $C^{k,\alpha}$ at $x_0$.
The following is the interior pointwise $C^{k,\alpha}$ ($k\geq 3$) regularity.
\begin{theorem}\label{t-Cka-i}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F\in C^{k-2,\alpha}(0) (k\geq 3)$ is convex in $M$. Assume that $\omega_0$ satisfies \cref{e.omega0-2} and $f\in C^{k-2,\alpha}(0)$.
Then $u\in C^{k,\alpha}(0)$, i.e., there exists a $k$-th order polynomial $P$ such that
\begin{equation}\label{e.Cka-1-i}
|u(x)-P(x)|\leq C |x|^{k+\alpha}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{e.Cka-3-i}
|F(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C|x|^{k-2+\alpha}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.Cka-2-i}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where $N_f$ is the $(k-2)$-th order Taylor polynomial of $f$ at $0$, i.e.,
\begin{equation*}
N_f(x)=f(0)+f_{i}(0)x_{i}+\cdots+f_{i_1\cdots i_{k-2}}(0)x_{i_1}\cdots x_{i_{k-2}}/(k-2)!
\end{equation*}
and $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0,\omega_0,\|\beta_3\|_{C^{k-2,\alpha}(0)},\omega_3,\omega_4$, $\|f\|_{C^{k-2,\alpha}}(0)$ and $\|u\|_{L^{\infty}(B_1)}$.
\end{theorem}
\begin{remark}\label{r-2.10}
For higher regularity, there is no explicit estimates since the coefficients of the linearized equation depend on the solution (see \Cref{In-l-62}). But for linear equation \cref{e.linear}, we have the following explicit estimates instead of \crefrange{e.Cka-1-i}{e.Cka-2-i}:
\begin{equation*}
|u(x)-P(x)|\leq C |x|^{k+\alpha}\left(\|u\|_{L^{\infty }(B_1)}
+\|f\|_{C^{k-2,\alpha}(0)}\right), ~~\forall ~x\in B_{1},
\end{equation*}
\begin{equation*}
|F(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C|x|^{k-2+\alpha}\left(\|u\|_{L^{\infty }(B_1)}
+\|f\|_{C^{k-2,\alpha}(0)}\right), ~~\forall ~x\in B_{1}
\end{equation*}
and
\begin{equation*}
|Du(0)|+\cdots+|D^ku(0)|\leq C\left(\|u\|_{L^{\infty }(B_1)}
+\|f\|_{C^{k-2,\alpha}(0)}\right),
\end{equation*}
where $C$ depends only on $k,n,\lambda,\Lambda,\alpha,\|a^{ij}\|_{C^{k-2,\alpha}(0)},\|b^i\|_{C^{k-2,\alpha}(0)}$ and $\|c\|_{C^{k-2,\alpha}(0)}$.
\end{remark}
\begin{remark}\label{r-2.8}
The observation \cref{e.Cka-3-i} is key for higher regularity and we refer to the proof in \Cref{In-Cka-mu} for details.
\end{remark}
\begin{remark}\label{r-2.6}
For higher pointwise regularity, our results are new even for equations without lower order terms.
\end{remark}
Now, we arrive at presenting the boundary regularity. For the boundary pointwise $C^{1,\alpha}$ regularity, we have
\begin{theorem}\label{t-C1a-mu}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $b\in L^{p}(\Omega\cap B_1)$ $(p=n/(1-\alpha))$, $f\in C^{-1,\alpha}(0)$, $\partial \Omega\in C^{1,\alpha}(0)$ and $g\in C^{1,\alpha}(0)$.
Then $u\in C^{1,\alpha}(0)$, i.e., there exists a linear polynomial $P$ such that
\begin{equation}\label{e.C1a-1-mu}
|u(x)-P(x)|\leq C |x|^{1+\alpha}, ~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.C1a-3-mu}
D_{x'}u(0)=D_{x'}g(0)
\end{equation}
and
\begin{equation}\label{e.C1a-2-mu}
|Du(0)|\leq C ,
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,\mu,\|b\|_{L^p(\Omega\cap B_1)}, \|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)}, \|f\|_{C^{-1,\alpha}(0)}$, $\|g\|_{C^{1,\alpha}(0)}$ and $\|u\|_{L^{\infty }(\Omega\cap B_1)}$.
In particular, if $\mu=0$, we have the following explicit estimates
\begin{equation}\label{e.C1a-1}
|u(x)-P(x)|\leq C |x|^{1+\alpha}\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{-1,\alpha}(0)}+\|g\|_{C^{1,\alpha}(0)}\right), ~\forall ~x\in \Omega\cap B_{1}
\end{equation}
and
\begin{equation}\label{e.C1a-2}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{-1,\alpha}(0)}+\|g\|_{C^{1,\alpha}(0)}\right),
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,\|b\|_{L^p(\Omega\cap B_1)}$ and $\|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)}$.
\end{theorem}
\begin{remark}\label{r-21}
In \cite{MR2853528}, Ma and Wang introduced a definition of pointwise $C^{1,\alpha}$ for the boundary, which is similar to \Cref{d-re}. Then they proved (for $\mu=b=0$) the boundary pointwise $C^{1,\alpha}$ regularity for some $\alpha<\bar{\alpha}$ since the Harnack inequality was used. Huang, Zhai and Zhou \cite{MR3933752} extended this result to equations with unbounded $b$.
Silvestre and Sirakov \cite{MR3246039} proved (for $\mu =0$ and $b\in L^{\infty}$) the boundary $C^{1,\alpha}$ regularity for any $\alpha<\bar{\alpha}$. However, it is not a pointwise regularity and requires that $\partial \Omega\in C^2$ since flattening the boundary is carried out in the proof.
For equations with quadratic growth in the gradient, Nornberg \cite{MR3980853} obtained the boundary $C^{1,\alpha}$ regularity for some $\alpha<\bar{\alpha}$. It is not either a pointwise regularity and $\partial \Omega\in C^{1,1}$ is needed. In addition, its proof doesn't apply to the Pucci class.
Wang \cite{MR1139064} gave a definition of pointwise $C^{1,\alpha}$ for the boundary for parabolic equations, which is similar to this paper. Then he proved the corresponding pointwise boundary $C^{1,\alpha}$ regularity for $\mu=b=0$.
The first boundary pointwise $C^{1,\alpha}$ regularity for any $0<\alpha<\bar{\alpha}$ is given in \cite{MR4088470} (without lower terms). In this paper, we extend to the general equations. We remark here that our proof is simpler than that of above results.
\end{remark}
\begin{remark}\label{r-24}
If $u$ satisfies $F(D^2u,Du,u,x)=f$, by the structure condition \cref{SC2},
\begin{equation*}
u\in S^*(\lambda,\Lambda,\mu,b,|f|+|F(0,0,0,\cdot)|
+c\omega_0(\|u\|_{L^{\infty}(\Omega\cap B_1)},\|u\|_{L^{\infty}(\Omega\cap B_1)})).
\end{equation*}
Then we can obtain the corresponding boundary pointwise $C^{1,\alpha}$ regularity. In fact, in this case, we can obtain the boundary pointwise $C^{1,\alpha}$ regularity for any $0<\alpha<1$ if we assume that $c\in C^{-1,\alpha}(0)$ additionally.
\end{remark}
\begin{remark}\label{r-2.14}
Since we treat the Pucci class, the method of solving an auxiliary equation to approximate the solution (e.g. \cite[Section 3.2]{MR3980853}) is invalid. The method of compactness can avoid the solvability.
\end{remark}
Let $\gamma_1$ be as in \cref{e.C1a.beta}. If $\gamma_1(\cdot, x_0)\in C^{-1,\alpha}(x_0)$ for any $x_0\in \Omega$ with the same $r_0$ (see \Cref{d-2}) and $\|\gamma_1\|_{C^{-1,\alpha}(\bar{\Omega})}:= \sup_{x_0\in \Omega} \|\gamma_1(\cdot, x_0)\|_{C^{-1,\alpha}(x_0)}<+\infty$, we say that $\gamma_1\in C^{-1,\alpha}(\bar{\Omega})$. We also use similar notations for $\|\beta_2\|_{C^{\alpha}(\bar{\Omega})}$, $\|\gamma_2\|_{C^{\alpha}(\bar{\Omega})}$ in the following.
By combining the interior and boundary regularity with standard covering arguments, we have the following local and global $C^{1,\alpha}$ regularity, where we use the same $r_0$ in \Cref{d-f}, \Cref{d-2} and \Cref{d-re} (similar hereinafter). Its proof is standard and we omit it.
\begin{corollary}\label{t-C1a-global}
Let $0<\alpha<\bar{\alpha}$, $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} for any $x_0\in \Omega\cup \Gamma$, $\beta_1(x,x_0)\leq \delta_0$ for any $x\in B_{r_0}(x_0)\cap \Omega$ and $\gamma_1\in C^{-1,\alpha}(\bar{\Omega})$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,\alpha$ and $\omega_1$. Assume that $b\in L^{p}(\Omega)$ $(p=n/(1-\alpha))$, $c\in C^{-1,\alpha}(\bar{\Omega})$, $f\in C^{-1,\alpha}(\bar{\Omega})$, $\Gamma\in C^{1,\alpha}$ and $g\in C^{1,\alpha}(\bar{\Gamma})$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{1,\alpha}(\bar{\Omega}')$ and
\begin{equation}\label{e.C1a-1-i-mu-global}
\|u\|_{C^{1,\alpha}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,\alpha,r_0,\mu,\|b\|_{L^p(\Omega)}, \|c\|_{C^{-1,\alpha}(\bar{\Omega})},\omega_0$, $\|\gamma_1\|_{C^{-1,\alpha}(\bar{\Omega})}$, $\omega_1$, $\|\Gamma\|_{C^{1,\alpha}}$, $\Omega'$, $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$, $ \|f\|_{C^{-1,\alpha}(\bar{\Omega})},\|g\|_{C^{1,\alpha}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1a-1-i-global}
\|u\|_{C^{1,\alpha}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}+\|f\|_{C^{-1,\alpha}(\bar{\Omega})}+\|\gamma_1\|_{C^{-1,\alpha}(\bar{\Omega})}
+\|g\|_{C^{1,\alpha}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,\alpha,r_0,\|b\|_{L^p(\Omega)}, \|c\|_{C^{-1,\alpha}(\bar{\Omega})}$, $\|\Gamma\|_{C^{1,\alpha}}$, $\Omega'$ and $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$.
If $\Gamma=\emptyset$, we obtain the interior $C^{1,\alpha}$ estimates analogous to \cref{e.C1a-1-i-mu-global} and \cref{e.C1a-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{1,\alpha}$ estimates analogous to \cref{e.C1a-1-i-mu-global} and \cref{e.C1a-1-i-global} on $\bar{\Omega}$.
\end{corollary}
\begin{remark}\label{r-2.11}
If $F$ is uniformly continuous in $x$, we may use $F(M,0,0,x_0)$ to measure the oscillation of $F$ in $x$ in \cref{e.C1a.beta} instead of $F_{x_0}(M)$. Then there exists $r_0>0$ such that
\begin{equation*}
\beta_1(x,x_0)\leq \delta_0,~\forall ~x,x_0\in\Omega\cup \Gamma~\mbox{with}~|x-x_0|\leq r_0.
\end{equation*}
Hence, the $C^{1,\alpha}$ regularity holds for equations with continuous coefficients.
\end{remark}
For the boundary pointwise $C^{2,\alpha}$ regularity, we have
\begin{theorem}\label{t-C2a}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\alpha}(0)$. Assume that $\omega_0$ satisfies \cref{e.omega0}, $f\in C^{\alpha}(0)$, $\partial \Omega\in C^{2,\alpha}(0)$ and $g\in C^{2,\alpha}(0)$.
Then $u\in C^{2,\alpha}(0)$, i.e., there exists a quadratic polynomial $P$ such that
\begin{equation}\label{e.C2a-1-mu}
|u(x)-P(x)|\leq C |x|^{2+\alpha}, ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.C2a-3-mu}
|F(D^2P,DP(x),P(x),x)-f(0)|\leq C|x|^{\alpha}, ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.C2a-4-mu}
D^l_{x'}u(x',P_0(x')):=D^l_{x'}P(x',P_0(x'))=D^l_{x'}g(x',P_0(x'))~\mbox{at}~0,~\forall ~0\leq l\leq 2
\end{equation}
and
\begin{equation}\label{e.C2a-2-mu}
|Du(0)|+\|D^2u(0)\|\leq C,
\end{equation}
where $P_0$ denotes the polynomial in \cref{e-re} and \cref{e-re2} and $C$ depends only on $n, \lambda, \Lambda,\alpha,\mu$, $b_0$, $c_0$, $\omega_0$, $\|\beta_2\|_{C^{\alpha}(0)}$, $\omega_2$, $\|\partial \Omega\cap B_1\|_{C^{2,\alpha}(0)}, \|f\|_{C^{\alpha}(0)},\|g\|_{C^{2,\alpha}(0)}$ and $\|u\|_{L^{\infty }(\Omega\cap B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\alpha}(0)$, we have the following explicit estimates
\begin{equation}\label{e.C2a-1}
|u(x)-P(x)|\leq C |x|^{2+\alpha}\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}
+\|f\|_{C^{\alpha}(0)}+\|\gamma_2\|_{C^{\alpha}(0)}+\|g\|_{C^{2,\alpha}(0)}\right), ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.C2a-3}
\begin{aligned}
&|F(D^2P,DP(x),P(x),x)-f(0)|\\
&\leq C|x|^{\alpha}\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{\alpha}(0)}
+\|\gamma_2\|_{C^{\alpha}(0)}+\|g\|_{C^{2,\alpha}(0)}\right),
~~\forall ~x\in \Omega\cap B_{1}
\end{aligned}
\end{equation}
and
\begin{equation}\label{e.C2a-2}
|Du(0)|+\|D^2u(0)\|\leq C\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}
+\|f\|_{C^{\alpha}(0)}+\|\gamma_2\|_{C^{\alpha}(0)}+\|g\|_{C^{2,\alpha}(0)}\right),
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,b_0,c_0$,$\|\beta_2\|_{C^{\alpha}(0)}$ and $\|\partial \Omega\cap B_1\|_{C^{2,\alpha}(0)}$.
\end{theorem}
\begin{remark}\label{r-2.9}
The boundary pointwise $C^{2,\alpha}$ regularity was first proved in \cite{MR4088470} for equations without lower terms.
\end{remark}
\begin{remark}\label{r-2.7}
We don't require that $F$ is convex in $M$, which is different from the interior $C^{2,\alpha}$ regularity. This can be tracked to that the model problem doesn't need this condition (see \Cref{l-32}).
\end{remark}
Similar to $C^{1,\alpha}$ regularity, we have the following corollary.
\begin{corollary}\label{t-C2a-global}
Let $0<\alpha<\bar{\alpha}$, $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\alpha}(\bar\Omega)$. Assume that $\omega_0$ satisfies \cref{e.omega0}, $f\in C^{\alpha}(\bar\Omega)$, $\Gamma\in C^{2,\alpha}$ and $g\in C^{2,\alpha}(\bar{\Gamma})$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{2,\alpha}(\bar{\Omega}')$ and
\begin{equation}\label{e.C2a-1-i-mu-global}
\|u\|_{C^{2,\alpha}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,r_0,\mu,b_0,c_0,\omega_0,\|\beta_2\|_{C^{\alpha}(\bar\Omega)},\omega_2$,
$\|\Gamma\|_{C^{2,\alpha}}$, $\Omega'$, $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$, $\|f\|_{C^{\alpha}(\bar\Omega)}$, $\|g\|_{C^{2,\alpha}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\alpha}(\bar\Omega)$, we have the following explicit estimates
\begin{equation}\label{e.C2a-1-i-global}
\|u\|_{C^{2,\alpha}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}+\|f\|_{C^{\alpha}(\bar\Omega)}+
\|\gamma_2\|_{C^{\alpha}(\bar\Omega)}+\|g\|_{C^{2,\alpha}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,\alpha,r_0,b_0,c_0,\|\beta_2\|_{C^{\alpha}(\bar\Omega)}$, $\|\Gamma\|_{C^{2,\alpha}}$, $\Omega'$ and $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$.
If $\Gamma=\emptyset$, we obtain the interior $C^{2,\alpha}$ estimates analogous to \cref{e.C2a-1-i-mu-global} and \cref{e.C2a-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{2,\alpha}$ estimates analogous to \cref{e.C2a-1-i-mu-global} and \cref{e.C2a-1-i-global} on $\bar{\Omega}$.
\end{corollary}
For the boundary pointwise $C^{k,\alpha}$ regularity, we have
\begin{theorem}\label{t-Cka}
Let $0<\alpha<\bar{\alpha}$ and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,\alpha}(0) (k\geq 3)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,\alpha}(0)$, $\partial \Omega\in C^{k,\alpha}(0)$ and $g\in C^{k,\alpha}(0)$.
Then $u\in C^{k,\alpha}(0)$, i.e., there exists a $k$-th order polynomial $P$ such that
\begin{equation}\label{e.Cka-1}
|u(x)-P(x)|\leq C |x|^{k+\alpha}, ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.Cka-3}
|F(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C|x|^{k-2+\alpha}, ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.Cka-5}
D^l_{x'}u(x',P_0(x'))=D^l_{x'}g(x',P_0(x'))~\mbox{at}~0,~\forall ~0\leq l\leq k
\end{equation}
and
\begin{equation}\label{e.Cka-2}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where $P_0$ denotes the polynomial in \cref{e-re} and \cref{e-re2} and $N_f$ is the $(k-2)$-th order Taylor polynomial of $f$ at $0$; in addition, $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0$, $\omega_0$, $\|\beta_3\|_{C^{k-2,\alpha}(0)}$, $\omega_3,\omega_4$, $\|\partial \Omega\cap B_1\|_{C^{k,\alpha}(0)}$, $\|f\|_{C^{k-2,\alpha}}(0)$, $\|g\|_{C^{k,\alpha}}(0)$ and $\|u\|_{L^{\infty}(\Omega\cap B_1)}$.
\end{theorem}
\begin{remark}\label{r-4}
The relation \cref{e.Cka-5} is key for the proof of boundary pointwise $C^{k,\alpha}$ regularity. See the proof in \Cref{Cka-mu} for details.
\end{remark}
\begin{remark}\label{r-23}
This boundary pointwise regularity is new even for the Laplace equation.
\end{remark}
Similar to $C^{1,\alpha}$ and $C^{2,\alpha}$ regularity, we have the following corollary.
\begin{corollary}\label{t-Cka-global}
Let $0<\alpha<\bar{\alpha}$, $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,\alpha}(\bar \Omega)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,\alpha}(\bar\Omega)$, $\Gamma\in C^{k,\alpha}$ and $g\in C^{k,\alpha}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{k,\alpha}(\bar{\Omega}')$ and
\begin{equation}\label{e.Cka-1-i-mu-global}
\|u\|_{C^{k,\alpha}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,r_0,\mu,b_0,c_0,\|\beta_3\|_{C^{k-2,\alpha}(\bar\Omega)},\omega_3,
\omega_4$, $\|\Gamma\|_{C^{k,\alpha}}$, $\|f\|_{C^{k-2,\alpha}(\bar\Omega)}$,
$\|g\|_{C^{k,\alpha}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{k,\alpha}$ estimates analogous to \cref{e.Cka-1-i-mu-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{k,\alpha}$ estimates analogous to \cref{e.Cka-1-i-mu-global} on $\bar{\Omega}$.
\end{corollary}
\subsection{The $C^{k}$ regularity}
Besides the classical $C^{k,\alpha}$ regularity ($0<\alpha<1$), we also obtain other types of pointwise regularity. It is well known that even for the Poisson equation, $f\in C^0$ (i.e., $\alpha=0$) doesn't imply the $C^2$ regularity of the solution and $f\in C^{0,1}$ (i.e., $\alpha=1$) doesn't imply the $C^{2,1}$ regularity of the solution.
However, it is also well known that if $f$ is Dini continuous, the $C^2$ regularity holds; if $f\in C^{0,1}$, one can obtain $C^{2,\mathrm{lnL}}$ regularity (the so-called ``ln-Lipschitz'' regularity). Based on the same idea to the $C^{k,\alpha}$ regularity, we can derive systematically these two kinds of regularity. Their proofs are similar to that of $C^{k,\alpha}$ regularity (with minor modifications) and we only give the details of proofs for the interior pointwise $C^2$ regularity, the boundary pointwise $C^1$ regularity, the interior pointwise $C^{k,\mathrm{lnL}}$ ($k\geq 2$) regularity and the boundary pointwise $C^{1,\mathrm{lnL}}$ regularity. For the convenience of readers and citation, we itemize the regularity results as follows.
\begin{theorem}\label{t-C1-i}
Let $p>n$ and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} with $\|\beta_1\|_{C^{-1,1}(0)}\leq \delta_0$ and $\gamma_1\in C^{-1,\mathrm{Dini}}(0)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,p$ and $\omega_1$. Assume that $b\in L^{p}(B_1)$, $c\in C^{-1,\mathrm{Dini}}(0)$ and $f\in C^{-1,\mathrm{Dini}}(0)$
Then $u\in C^{1}(0)$, i.e., there exist a linear polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.C1-1-i-mu}
|u(x)-P(x)|\leq |x|\omega_u (|x|), ~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C1-2-i-mu}
|Du(0)|\leq C,
\end{equation}
where for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\begin{aligned}
&\omega_u(r)=C\left(r^{\tilde{\alpha}}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}
+r^{\tilde{\alpha}}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\tilde{\alpha}}}d\rho
\right), \\ &\tilde{\alpha}=\min(\bar{\alpha}/2,1-n/p),~ \tilde{\omega}=\max(\omega_c,\omega_{\gamma_1},\omega_f);
\end{aligned}
\end{equation*}
$\eta$ and $C$ depend only on $n,\lambda,\Lambda,p,\mu,\|b\|_{L^p(B_1)}, \|c\|_{C^{-1,\mathrm{Dini}}(0)}$, $\omega_0$, $\|\gamma_1\|_{C^{-1,\mathrm{Dini}}(0)}$,
$\omega_1$, $\|f\|_{C^{-1,\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1-1-i}
|u(x)-P(x)|\leq |x|\omega_u (|x|)\left(\|u\|_{L^{\infty }( B_1)}
+\|f\|_{C^{-1,\mathrm{Dini}}(0)}+\|\gamma_1\|_{C^{-1,\mathrm{Dini}}(0)}\right), ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C1-2-i}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{-1,\mathrm{Dini}}(0)}
+\|\gamma_1\|_{C^{-1,\mathrm{Dini}}(0)}\right),
\end{equation}
where $\eta$ and $C$ depend only on $n,\lambda,\Lambda,p,\|b\|_{L^p(B_1)}$ and $\|c\|_{C^{-1,\mathrm{Dini}}(0)}$.
\end{theorem}
For $C^2$ regularity, we assume that $\omega_0$ can be written as:
\begin{equation}\label{e.omega0-c-1}
\omega_0(K,s)=\hat\omega_0(K)\omega_{\omega_0}(s), ~\forall ~K,s>0,
\end{equation}
where $\hat{\omega}_0$ is nondecreasing and $\omega_{\omega_0}$ is a Dini function. For example, the following equation satisfies \cref{e.omega0-c-1}
\begin{equation*}
\Delta u=\frac{1}{\ln^2|u|+1}
\end{equation*}
and the $C^{2}$ regularity holds for it.
\begin{theorem}\label{t-C2-i}
Let $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\mathrm{Dini}}(0)$. Assume that $\omega_0$ satisfies \cref{e.omega0-c-1} and $f\in C^{\mathrm{Dini}}(0)$.
Then $u\in C^{2}(0)$, i.e., there exist a quadratic polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.C2-1-i-mu}
|u(x)-P(x)|\leq |x|^2\omega_u(|x|), ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{e.C2-3-i-mu}
|F(D^2P,DP(x),P(x),x)-f(0)|\leq C\omega_{\beta_2}(|x|), ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.C2-2-i-mu}
|Du(0)|+\|D^2u(0)\|\leq C,
\end{equation}
where for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\begin{aligned}
\omega_u(r)&=C\left(r^{\bar{\alpha}/2}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}
+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
\right),~ \tilde{\omega}=\max(\omega_{\omega_0},\omega_{\beta_2},\omega_f);
\end{aligned}
\end{equation*}
$\eta$ and $C$ depend only on $n,\lambda,\Lambda,\mu,b_0,c_0,\omega_0,
\|\beta_2\|_{C^{\mathrm{Dini}}(0)},\omega_{\beta_2}, \omega_{2},
\|f\|_{C^{\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\mathrm{Dini}}(0)$, we have the following explicit estimates
\begin{equation}\label{e.C2-1-i}
|u(x)-P(x)|\leq |x|^2\omega_u(|x|)\left(\|u\|_{L^{\infty }(B_1)}
+\|f\|_{C^{\mathrm{Dini}}(0)}
+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}\right), ~\forall ~x\in B_1,
\end{equation}
\begin{equation}\label{e.C2-3-i}
\begin{aligned}
|&F(D^2P,DP(x),P(x),x)-f(0)|\\
&\leq C\omega_{\beta_2}(|x|)\left(\|u\|_{L^{\infty}(B_1)}
+\|f\|_{C^{\mathrm{Dini}}(0)}
+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}\right),~\forall ~x\in B_{1}
\end{aligned}
\end{equation}
and
\begin{equation}\label{e.C2-2-i}
|Du(0)|+\|D^2u(0)\|\leq C\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\mathrm{Dini}}(0)}
+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}\right),
\end{equation}
where $\eta$ and $C$ depend only on $n,\lambda,\Lambda,b_0,c_0,\|\beta_2\|_{C^{\mathrm{Dini}}(0)}$ and $\omega_{\beta_2}$.
\end{theorem}
\begin{remark}\label{r-2.16}
The $C^2$ regularity has been investigated extensively. For Poisson equation, it can be derived by the integral representation. For linear equations, Burch \cite{MR521856} obtained the $C^2$ regularity for weak solutions. Sperner \cite{MR650493} proved the $C^2$ regularity for Dirichlet problems. Wang \cite{MR2273802} deduced explicit formula for the modulus of continuity of $D^2u$ for $C^2$ solutions by a simple method. For fully nonlinear elliptic equations, Kovats \cite{MR1629506} first obtained the $C^2$ regularity under Dini conditions. Later, it was extended to boundary by Zou and Chen \cite{MR1924273}. Wang \cite{MR2273802} also proved the $C^2$ regularity for fully nonlinear elliptic equations with additional assumption $F\in C^{1,1}$. Our result is more general than above results and the proof is simple.
\end{remark}
\begin{remark}\label{r-2.13}
In fact, $\bar{\alpha}/2$ can be replaced by any $0<\alpha_0<\bar{\alpha}$ in the expression of
$\omega_u$ (see the proof in \Cref{C1C2Ck} for details). Similar remarks can be made for other $C^k$ ($k\geq 2$) regularity.
\end{remark}
\begin{remark}\label{r-2.15}
Note that the constant $C$ depends not only $\|\beta_2\|_{C^{\mathrm{Dini}}(0)}$ but also $\omega_{\beta_2}$. The reason is that $C^2$ regularity is a critical case for $\beta_2$ and the normalization procedure have to depend on $\omega_{\beta_2}$ (see the normalization procedure in \Cref{C1C2Ck}).
\end{remark}
\begin{theorem}\label{t-Ck-i}
Let $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F\in C^{k-2,\mathrm{Dini}}(0)$ $(k\geq 3)$ is convex in $M$. Assume that $\omega$ satisfies \cref{e.omega0-2} and $f\in C^{k-2,\mathrm{Dini}}(0)$.
Then $u\in C^{k}(0)$, i.e., there exist a $k$-th order polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.Ck-i-1}
|u(x)-P(x)|\leq |x|^k\omega_u(|x|), ~~\forall ~x\in B_1,
\end{equation}
\begin{equation}\label{e.Ck-i-3}
|F(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C|x|^{k-2}\omega_{\beta_3}(|x|), ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.Ck2-i-2}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where $N_f$ is the $(k-2)$-th order Taylor polynomial of $f$; for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\begin{aligned}
\omega_u(r)&=C\left(r^{\bar{\alpha}/2}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
\right),~ \tilde{\omega}=\max(\omega_{\beta_3},\omega_f);
\end{aligned}
\end{equation*}
$\eta$ and $C$ depend only on $k,n,\lambda,\Lambda,\mu,b_0,c_0,\omega_0,\|\beta_3\|_{C^{k-2,\mathrm{Dini}}(0)},
\omega_3,\omega_4$, $\|f\|_{C^{k-2,\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$.
\end{theorem}
For the boundary pointwise $C^{1}$, $C^{2}$ and $C^{k}$ regularity, we have
\begin{theorem}\label{t-C1}
Let $u$ satisfy
\begin{equation}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation}
Suppose that $b\in L^{p}(\Omega\cap B_1)(p>n)$, $f\in C^{-1,\mathrm{Dini}}(0)$, $\partial\Omega\in C^{1,\mathrm{Dini}}(0)$ and $g\in C^{1,\mathrm{Dini}}(0)$.
Then $u$ is $C^{1}$ at $0$, i.e., there exist a linear polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.C1-1-mu}
|u(x)-P(x)|\leq |x|\omega_u(|x|), ~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.C1-3-mu}
D_{x'}u(0)=D_{x'}g(0)
\end{equation}
and
\begin{equation}\label{e.C1-2-mu}
|Du(0)|\leq C ,
\end{equation}
where for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\begin{aligned}
& \omega_u(r)=C\left(r^{\tilde{\alpha}}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}
+r^{\tilde{\alpha}}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\tilde{\alpha}}}d\rho
\right) \\ &\tilde{\alpha}=\min(\bar{\alpha}/2,1-n/p), \tilde{\omega}=\max(\omega_f, \omega_\Omega, \omega_g);
\end{aligned}
\end{equation*}
$\eta$ and $C$ depend only on $n, \lambda, \Lambda,\mu,p,\|b\|_{L^p(\Omega\cap B_1)},\|\partial \Omega\cap B_1\|_{C^{1,\mathrm{Dini}}(0)}, \omega_{\Omega}$, $\|f\|_{C^{-1,\mathrm{Dini}}(0)}$, $\|g\|_{C^{1,\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty }(\Omega\cap B_1)}$.
In particular, if $\mu=0$, we have the following explicit estimates
\begin{equation}\label{e.C1-1}
|u(x)-P(x)|\leq |x|\omega_u (|x|)\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{-1,\mathrm{Dini}}(0)}+\|g\|_{C^{1,\mathrm{Dini}}(0)}\right), ~~\forall ~x\in \Omega\cap B_{1}
\end{equation}
and
\begin{equation}\label{e.C1-2}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{-1,\mathrm{Dini}}(0)}
+\|g\|_{C^{1,\mathrm{Dini}}(0)}\right),
\end{equation}
where $\eta$ and $C$ depend only on $n, \lambda, \Lambda,p,\|b\|_{L^p(\Omega\cap B_1)},\|\partial \Omega\cap B_1\|_{C^{1,\mathrm{Dini}}(0)}$ and $\omega_{\Omega}$.
\end{theorem}
\begin{remark}\label{r-2.18}
Ma and Wang \cite{MR2853528} have proved the boundary pointwise $C^1$ regularity with $\mu=b=0$. Huang, Zhai and Zhou \cite{MR3933752} obtained the boundary pointwise $C^1$ regularity for linear equations with $b\in C^{-1,\mathrm{Dini}}(0)$.
\end{remark}
\begin{remark}\label{r-2.19}
Note that the constant $C$ depends not only
$\|\partial \Omega\cap B_1\|_{C^{1,\mathrm{Dini}}(0)}$ but also $\omega_{\Omega}$. The reason is similar to the $C^2$ regularity for $\beta_2$ (see the proof in \Cref{C1C2Ck}).
\end{remark}
\begin{theorem}\label{t-C2}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\mathrm{Dini}}(0)$. Assume that $\omega_0$ satisfies \cref{e.omega0-c-1}, $f\in C^{\mathrm{Dini}}(0)$, $\partial\Omega\in C^{2,\mathrm{Dini}}(0)$ and $g\in C^{2,\mathrm{Dini}}(0)$.
Then $u\in C^{2}(0)$, i.e., there exists a quadratic polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.C2-1-mu}
|u(x)-P(x)|\leq |x|^2\omega_u(|x|), ~~\forall ~x\in \Omega \cap B_{1},
\end{equation}
\begin{equation}\label{e.C2-3-mu}
|F(D^2P,DP(x),P(x),x)-f(0)|\leq C\omega_{\beta_2}(|x|), ~~\forall ~x\in \Omega \cap B_{1},
\end{equation}
\begin{equation}\label{e.C2-4-mu}
D^l_{x'}u(x',P_0(x'))=D^l_{x'}g(x',P_0(x'))~\mbox{at}~0,~\forall ~0\leq l\leq 2
\end{equation}
and
\begin{equation}\label{e.C2-2-mu}
|Du(0)|+\|D^2u(0)\|\leq C,
\end{equation}
where for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\omega_u(r)=C\left(r^{\bar{\alpha}/2}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}
+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
\right),~ \tilde\omega(r)=\max(\omega_{\beta_2}, \omega_f, \omega_\Omega, \omega_g);
\end{equation*}
the $P_0$ denotes the polynomial in \cref{e-re} and \cref{e-re2}; $\eta$ and $C$ depend only on $n,\lambda,\Lambda,\mu$, $b_0$, $c_0,\omega_0,
\|\beta_2\|_{C^{\mathrm{Dini}}(0)},\omega_{\beta_2},\omega_{2}$,
$\|\partial \Omega\cap B_1\|_{C^{2,\mathrm{Dini}}(0)}$, $\|f\|_{C^{\mathrm{Dini}}(0)}$,
$\|g\|_{C^{2,\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty}(\Omega\cap B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\mathrm{Dini}}(0)$, we have the following explicit estimates
\begin{equation}\label{e.C2-1}
\begin{aligned}
|&u(x)-P(x)|\\
&\leq |x|^2\omega_u(|x|)\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{\mathrm{Dini}}(0)}+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}
+\|g\|_{C^{2,\mathrm{Dini}}(0)}\right), ~~\forall ~x\in \Omega\cap B_1,
\end{aligned}
\end{equation}
\begin{equation}\label{e.C2-3}
\begin{aligned}
|&F(D^2P,DP(x),P(x),x)-f(0)|\\
&\leq C\omega_{\beta_2}(|x|)\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{\mathrm{Dini}}(0)}+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}
+\|g\|_{C^{2,\mathrm{Dini}}(0)}\right), ~~\forall ~x\in \Omega \cap B_{1}
\end{aligned}
\end{equation}
and
\begin{equation}\label{e.C2-2}
|Du(0)|+\|D^2u(0)\|\leq C\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{C^{\mathrm{Dini}}(0)}+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}
+\|g\|_{C^{2,\mathrm{Dini}}(0)}\right),
\end{equation}
where $\eta$ and $C$ depend only on $n,\lambda,\Lambda,b_0,c_0,\|\beta_2\|_{C^{\mathrm{Dini}}(0)}$
$\omega_{\beta_2}$ and $\|\partial \Omega\cap B_1\|_{C^{2,\mathrm{Dini}}(0)}$.
\end{theorem}
\begin{theorem}\label{t-Ck}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial\Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,\mathrm{Dini}}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,\mathrm{Dini}}(0)$, $\partial\Omega\in C^{k,\mathrm{Dini}}(0)$ and $g\in C^{k,\mathrm{Dini}}(0)$.
Then $u\in C^{k}(0)$, i.e., there exist a $k$-th order polynomial $P$ and a modulus of continuity $\omega_u$ such that
\begin{equation}\label{e.Ck-1}
|u(x)-P(x)|\leq |x|^k\omega_u(|x|), ~~\forall ~x\in \Omega\cap B_{1},
\end{equation}
\begin{equation}\label{e.Ck-3}
|F(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C|x|^{k-2}\omega_{\beta_2}(|x|), ~~\forall ~x\in \Omega \cap B_{1},
\end{equation}
\begin{equation}\label{e.Ck-5}
D^l_{x'}u(x',P_0(x'))=D^l_{x'}g(x',P_0(x'))~\mbox{at}~0,~\forall ~0\leq l\leq k
\end{equation}
and
\begin{equation}\label{e.Ck2-2}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where for $0<r<\eta$ $(\omega_u(r)\equiv C$ for $\eta\leq r<1)$,
\begin{equation*}
\omega_u(r)=C\left(r^{\bar{\alpha}/2}+\int_{0}^{r/\eta}\frac{\tilde\omega(\rho)d\rho}{\rho}
+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
\right),~ \tilde\omega(r)=\max(\omega_{\beta_3},\omega_f, \omega_\Omega, \omega_g);
\end{equation*}
the $N_f$ is the $(k-2)$-th order Taylor polynomial of $f$; the $P_0$ denotes the polynomial in \cref{e-re} and \cref{e-re2}; $\eta$ and $C$ depend only on $k,n,\lambda,\Lambda,\mu,b_0,c_0,\omega_0,
\|\beta_3\|_{C^{k-2,\mathrm{Dini}}(0)}$, $\omega_3$, $\omega_4$,
$\|\partial \Omega\cap B_1\|_{C^{k,\mathrm{Dini}}(0)}$,
$\|f\|_{C^{k-2,\mathrm{Dini}}(0)},\|g\|_{C^{k,\mathrm{Dini}}(0)}$ and
$\|u\|_{L^{\infty}(\Omega\cap B_1)}$.
\end{theorem}
Similar to $C^{k,\alpha}$ regularity, by combining the interior and boundary regularity, we have the local and global $C^{k}$ regularity. For the completeness and the convention of citation, we list them as follows.
\begin{corollary}\label{t-C1-global}
Let $p>n$ and $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} for any $x_0\in \Omega\cup \Gamma$, $\beta_1(x,x_0)\leq \delta_0$ for any $x\in B_{r_0}(x_0)\cap \Omega$ and $\gamma_1\in C^{-1,\mathrm{Dini}}(\bar{\Omega})$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,p$ and $\omega_1$. Assume that $b\in L^{p}(\Omega)$, $c\in C^{-1,\mathrm{Dini}}(\bar\Omega)$, $f\in C^{-1,\mathrm{Dini}}(\bar\Omega)$, $\Gamma\in C^{1,\mathrm{Dini}}$ and $g\in C^{1,\mathrm{Dini}}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{1}(\bar{\Omega}')$ and
\begin{equation}\label{e.C1-1-i-mu-global}
\|u\|_{C^{1}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,p,r_0,\mu,\|b\|_{L^p(\Omega)},
\|c\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})},\omega_0,
\|\Gamma\|_{C^{1,\mathrm{Dini}}}$, $\omega_{\Omega}$, $\Omega'$, $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$, $\|f\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})},
\|g\|_{C^{1,\mathrm{Dini}}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1-1-i-global}
\|u\|_{C^{1}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}
+\|f\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})}
+\|\gamma_1\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})}
+\|g\|_{C^{1,\mathrm{Dini}}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,p,r_0,\|b\|_{L^p(\Omega)},
\|c\|_{C^{-1,\mathrm{Dini}}(\bar{\Omega})}$, $\|\Gamma\|_{C^{1,\mathrm{Dini}}}$, $\omega_{\Omega}$, $\Omega'$ and $\mathrm{dist}(\Omega',\partial \Omega\backslash \Gamma)$.
If $\Gamma=\emptyset$, we obtain the interior $C^{1}$ estimates analogous to \cref{e.C1-1-i-mu-global} and \cref{e.C1-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{1}$ estimates analogous to \cref{e.C1-1-i-mu-global} and \cref{e.C1-1-i-global} on $\bar{\Omega}$.
\end{corollary}
\begin{corollary}\label{t-C2-global}
Let $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF} with $\beta_2\in C^{\mathrm{Dini}}(\bar{\Omega})$. Assume that $\omega_0$ satisfies \cref{e.omega0-c-1}, $f\in C^{\mathrm{Dini}}(\bar{\Omega})$, $\Gamma\in C^{2,\mathrm{Dini}}$ and $g\in C^{2,\mathrm{Dini}}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{2}(\bar{\Omega}')$ and
\begin{equation}\label{e.C2-1-i-mu-global}
\|u\|_{C^{2}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,r_0,\mu,b_0,c_0,\omega_0$,
$\|\beta_2\|_{C^{\mathrm{Dini}}(\bar{\Omega})}$, $\omega_{\beta_2}$, $\omega_2$, $\|\Gamma\|_{C^{2,\mathrm{Dini}}}$, $\|f\|_{C^{\mathrm{Dini}}(\bar{\Omega})}$,
$\|g\|_{C^{2,\mathrm{Dini}}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0} with $\gamma_2\in C^{\mathrm{Dini}}(\bar\Gamma)$, we have the following explicit estimates
\begin{equation}\label{e.C2-1-i-global}
\|u\|_{C^{2}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}+\|f\|_{C^{\mathrm{Dini}}(\bar{\Omega})}
+\|\gamma_2\|_{C^{\mathrm{Dini}}(\bar{\Omega})}
+\|g\|_{C^{2,\mathrm{Dini}}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,r_0,\mu,b_0,c_0$, $\|\beta_2\|_{C^{\mathrm{Dini}}(\bar{\Omega})}$, $\omega_{\beta_2},\omega_2$ and $\|\Gamma\|_{C^{2,\mathrm{Dini}}}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{2}$ estimates analogous to \cref{e.C2-1-i-mu-global} and \cref{e.C2-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{2}$ estimates analogous to \cref{e.C2-1-i-mu-global} and \cref{e.C2-1-i-global} on $\bar{\Omega}$.
\end{corollary}
\begin{corollary}\label{t-Ck-global}
Let $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,\mathrm{Dini}}(\bar \Omega)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,\mathrm{Dini}}(\bar\Omega)$, $\Gamma\in C^{k,\mathrm{Dini}}$ and $g\in C^{k,\mathrm{Dini}}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{k}(\bar{\Omega}')$ and
\begin{equation}\label{e.Ck-1-i-mu-global}
\|u\|_{C^{k}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,r_0,\mu,b_0,c_0,
\|\beta_3\|_{C^{k-2,\mathrm{Dini}}(\bar\Omega)},\omega_3,\omega_4$,
$\|\Gamma\|_{C^{k,\mathrm{Dini}}}$, $\|f\|_{C^{k-2,\mathrm{Dini}}(\bar\Omega)}$,
$\|g\|_{C^{k,\mathrm{Dini}}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{k}$ estimates analogous to \cref{e.Ck-1-i-mu-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{k}$ estimates analogous to \cref{e.Ck-1-i-mu-global} on $\bar{\Omega}$.
\end{corollary}
\subsection{The $C^{k,\mathrm{lnL}}$ regularity}
In this subsection, we state the so-called ``ln-Lipschitz'' regularity. The first is the $C^{0,\mathrm{lnL}}$ regularity.
\begin{theorem}\label{t-C0ln-i}
Let $p>n$ and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} with $\|\beta_1\|_{C^{-1,1}(0)}\leq \delta_0$ and $\gamma_1\in L^{n}(B_1)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,p$ and $\omega_1$. Assume that $b\in L^{p}(B_1)$, $c\in L^{n}(B_1)$ and $f\in L^{n}(B_1)$.
Then $u\in C^{0,\mathrm{lnL}}(0)$, i.e.,
\begin{equation}\label{e.C0ln-1-i-mu}
|u(x)-u(0)|\leq C|x|\big|\ln |x|\big|, ~\forall ~x\in B_{1/2},
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,p,\mu,\|b\|_{L^p(B_1)}, \|c\|_{L^{n}(B_1)},\omega_0,\|\gamma_1\|_{L^{n}(B_1)}$, $\omega_1,\|f\|_{L^{n}(B_1)}$ and $\|u\|_{L^{\infty }(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimate
\begin{equation}\label{e.C0ln-1-i}
|u(x)-u(0)|\leq C |x|\big|\ln |x|\big|\left(\|u\|_{L^{\infty }(B_1)}+\|f\|_{L^{n}(B_1)}+\|\gamma_1\|_{L^{n}(B_1)}\right), ~\forall ~x\in B_{1/2},
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,p,\|b\|_{L^p(B_1)}$ and $\|c\|_{L^{n}(B_1)}$.
\end{theorem}
\begin{remark}\label{r-2.20}
Usually, one obtains the $C^{\alpha}$ regularity under the condition $f\in L^n$. \Cref{t-C0ln-i} shows a stronger regularity (note that $C^{0,\mathrm{lnL}}$ is embedded into $C^{\alpha}$ for any $0<\alpha<1$). For the Poisson equation, $f\in L^n$ implies $u\in W^{2,n}$, which is the ``almost'' $C^{0,1}$. The $C^{0,\mathrm{lnL}}$ can be regarded as another kind of ``almost'' $C^{0,1}$.
\end{remark}
\begin{theorem}\label{t-C1ln-i}
Let $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C1a.beta} with $\|\beta_1\|_{L^{n}(B_r)}\leq \delta_0 r/|\ln r|$ for any $0<r<1$ and $\gamma_1\in L^{\infty}(B_1)$, where $\delta_0$ depends only on $n,\lambda,\Lambda$ and $\omega_1$. Assume that $f\in L^{\infty}(B_1)$.
Then $u$ is $C^{1,\mathrm{lnL}}$ at $0$, i.e., there exists a linear polynomial $P$ such that
\begin{equation}\label{e.C1ln-1-i-mu}
|u(x)-P(x)|\leq C |x|^2\big|\ln |x|\big|, ~~\forall ~x\in B_{1/2}
\end{equation}
and
\begin{equation}\label{e.C1ln-2-i-mu}
|Du(0)|\leq C ,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,\mu,b_0, c_0,\omega_0,
\|\gamma_1\|_{L^{\infty}(B_1)},\omega_1$, $\|f\|_{L^{\infty}(B_1)}$ and
$\|u\|_{L^{\infty }(B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1ln-1-i}
|u(x)-P(x)|\leq C |x|^2\big|\ln |x|\big|\left(\|u\|_{L^{\infty }( B_1)}
+\|f\|_{L^{\infty}(B_1)}+\|\gamma_1\|_{L^{\infty}(B_1)}\right), ~~\forall ~x\in B_{1/2}
\end{equation}
and
\begin{equation}\label{e.C1ln-2-i}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }(B_1)}
+\|f\|_{L^{\infty}(B_1)}+\|\gamma_1\|_{L^{\infty}(B_1)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,b_0$ and $c_0$.
\end{theorem}
\begin{remark}\label{r-2.3}
It is well known that $f\in L^{\infty}$ doesn't imply $u\in C^{1,1}$. \Cref{t-C1ln-i} can be regarded as a kind of ``almost'' $C^{1,1}$ regularity. Interestingly, if $f$ and $\beta$ satisfy the same conditions as in \Cref{t-C1ln-i}, the solution $u$ belongs to $W^{2,\mathrm{BMO}}$ (see \cite{MR1978880}), which is another kind of ``almost'' $C^{1,1}$ regularity.
\end{remark}
\begin{theorem}\label{t-Ckln-i}
Let $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Suppose that $F\in C^{k-2,1}(0)$ $(k\geq 2)$ is convex in $M$. Assume that $\omega_0$ satisfies \cref{e.omega0-2} and $f\in C^{k-2,1}(0)$.
Then $u\in C^{k,\mathrm{lnL}}(0)$, i.e., there exists a $k$-th polynomial $P$ such that
\begin{equation}\label{e.Ckln-1-i}
|u(x)-P(x)|\leq C |x|^{k+1}\big|\ln|x|\big|, ~~\forall ~x\in B_{1/2},
\end{equation}
\begin{equation}\label{e.Ckln-3-i}
|F(D^2P(x),DP(x),P(x),x)|\leq C|x|^{k-1}, ~~\forall ~x\in B_{1/2}
\end{equation}
and
\begin{equation}\label{e.Ckln-2-i}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,\mu, b_0,c_0,\omega_0,
\|\beta_3\|_{C^{k-2,1}(0)},\omega_3,\omega_4$, $\|f\|_{C^{k-2,1}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$.
\end{theorem}
\begin{remark}\label{r-2.22}
In \cite{MR2273802}, Wang proved the $C^{2,\mathrm{lnL}}$ regularity for classical solutions with $F\in C^{1,1}$.
\end{remark}
We also obtain the boundary pointwise $C^{0,\mathrm{lnL}}$, $C^{1,\mathrm{lnL}}$ and $C^{k,\mathrm{lnL}}$ regularity:
\begin{theorem}\label{t-C0ln-mu}
Let $p>n$ and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $b\in L^{p}(\Omega\cap B_1)$, $f\in L^{n}(\Omega\cap B_1)$, $\underset{B_{r}}{\mathrm{osc}}~\partial\Omega\leq \delta_0r/|\ln r|$ for any $0<r<1$ and $g\in C^{0,1}(0)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda$ and $p$.
Then $u\in C^{0,\mathrm{lnL}}(0)$, i.e.,
\begin{equation}\label{e.C0ln-1-mu}
|u(x)-u(0)|\leq C |x|\big|\ln |x|\big|, ~\forall ~x\in \Omega\cap B_{1/2},
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda,p,\mu,\|b\|_{L^p(\Omega\cap B_1)}, \|f\|_{L^{n}(\Omega\cap B_1)}$, $\|g\|_{C^{0,1}(0)}$ and $\|u\|_{L^{\infty }(\Omega\cap B_1)}$.
In particular, if $\mu=0$, we have the following explicit estimate
\begin{equation}\label{e.C0ln-1}
|u(x)-u(0)|\leq C |x|\big|\ln |x|\big|\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{L^{n}(\Omega\cap B_1)}+\|g\|_{C^{0,1}(0)}\right), ~\forall ~x\in \Omega\cap B_{1/2},
\end{equation}
where $C$ depends only on $n,\lambda, \Lambda,p$ and $\|b\|_{L^p(\Omega\cap B_1)}$.
\end{theorem}
\begin{remark}\label{r-2.23}
To the best of our knowledge, this regularity is new even for the Laplace equation.
\end{remark}
\begin{remark}\label{r-2.21}
Note that the condition on $\partial \Omega$ for $C^{0,\mathrm{lnL}}$ regularity is similar to the condition on $\beta$ for $C^{1,\mathrm{lnL}}$ regularity. For $C^{k}$ regularity in last subsection, there is also a similarity between them. We make a summary here.
The $\beta$ can be regarded as the oscillation of the coefficient of second order term of the equation and the $C^2$ regularity is a critical case for $\beta$. If we intend to obtain higher regularity than $C^2$ (e.g. $C^{2}$, $C^{k,\alpha}$ ($k\geq 2$)), we need that $\beta$ has a decay (e.g. $\beta\in C^{Dini}$, $\beta\in C^{k-2,\alpha}$). Then by a normalization procedure, we can assume that $\beta$ is small. If we intend to obtain lower regularity than $C^2$ (e.g. $C^{0,\mathrm{lnL}}$, $C^{1,\alpha}$), that $\beta$ has a decay is not necessary. Then we can not make $\beta$ small by normalization and we must make the assumption that $\beta$ is small in the theorem (e.g. $\|\beta\|_{L^n(B_r)}\leq \delta_0r/|\ln r|$, $\|\beta\|_{L^n(B_r)}\leq \delta_0 r$ where $\delta_0$ is a small constant).
For $\partial \Omega$, we have a similar explanation. In the proof of boundary regularity, the key is to estimate $x_n$ on $\partial \Omega$ (see the proof in \Cref{C1a-mu}). Hence, $C^1$ regularity is critical for $\partial \Omega$ in some sense. If we intend to obtain higher regularity than $C^1$ (e.g. $C^{1}$, $C^{k,\alpha}$ ($k\geq 1$)), we need that $\mathrm{osc}~\partial \Omega$ has a decay (e.g. $\partial \Omega\in C^{1,Dini}$, $\partial \Omega\in C^{k,\alpha}$). Then by a normalization procedure, we can assume that $\mathrm{osc}~\partial \Omega$ is small. If we intend to obtain lower regularity than $C^1$ (e.g. $C^{0,\mathrm{lnL}}$), that $\mathrm{osc}~\partial \Omega$ has a decay is not necessary. Then we can not make $\beta$ small by normalization and we must make the assumption that $\mathrm{osc}~\partial \Omega$ is small in the theorem (e.g. $\underset{B_{r}}{\mathrm{osc}}~\partial\Omega\leq \delta_0r/|\ln r|$ where $\delta_0$ is a small constant).
\end{remark}
\begin{theorem}\label{t-C1ln}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial\Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} with $\|\beta_1\|_{L^{n}(B_r)}\leq \delta_0 r/|\ln r|$ for any $0<r<1$ and $\gamma_1\in L^{\infty}(\Omega\cap B_1)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda$ and $\omega_1$. Assume that $f\in L^{\infty}(\Omega\cap B_1)$, $\partial\Omega\in C^{1,1} (0)$ and $g\in C^{1,1}(0)$.
Then $u$ is $C^{1,\mathrm{lnL}}$ at $0$, i.e., there exists a linear polynomial $P$ such that
\begin{equation}\label{e.C1ln-1-mu}
|u(x)-P(x)|\leq C |x|^2\big|\ln |x|\big|, ~~\forall ~x\in \Omega\cap B_{1/2}
\end{equation}
and
\begin{equation}\label{e.C1ln-2-mu}
|Du(0)|\leq C ,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,\mu,b_0, c_0,\omega_0,
\|\gamma_1\|_{L^{\infty}(\Omega\cap B_1)}, \omega_1$,
$\|\partial \Omega\cap B_1\|_{C^{1,1}(0)}$, $\|f\|_{L^{\infty}(\Omega\cap B_1)}$, $\|g\|_{C^{1,1}(0)}$ and $\|u\|_{L^{\infty }(\Omega\cap B_1)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimates
\begin{equation}\label{e.C1ln-1}
\begin{aligned}
|&u(x)-P(x)|\\
&\leq C |x|^2\big|\ln |x|\big|\left(\|u\|_{L^{\infty }(\Omega\cap B_1)}
+\|f\|_{L^{\infty}(\Omega\cap B_1)}+\|\gamma_1\|_{L^{\infty}(\Omega\cap B_1)}
+\|g\|_{C^{1,1}(0)}\right), ~~\forall ~x\in \Omega\cap B_{1/2}
\end{aligned}
\end{equation}
and
\begin{equation}\label{e.C1ln-2}
|Du(0)|\leq C \left(\|u\|_{L^{\infty }(\Omega\cap B_1)}+\|f\|_{L^{\infty}(\Omega\cap B_1)}
+\|\gamma_1\|_{L^{\infty}(\Omega\cap B_1)}+\|g\|_{C^{1,1}(0)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,b_0, c_0$ and
$\|\partial \Omega\cap B_1\|_{C^{1,1}(0)}$.
\end{theorem}
\begin{theorem}\label{t-Ckln}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,1}(0)$ $(k\geq 2)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,1}(0)$, $\partial\Omega\in C^{k,1} (0)$ and $g\in C^{k,1}(0)$.
Then $u\in C^{k,\mathrm{lnL}}(0)$, i.e., there exists a $k$-th polynomial $P$ such that
\begin{equation}\label{e.Ckln-1}
|u(x)-P(x)|\leq C |x|^{k+1}\big|\ln|x|\big|, ~~\forall ~x\in \Omega\cap B_{1/2},
\end{equation}
\begin{equation}\label{e.Ckln-3}
|F(D^2P(x),DP(x),P(x),x)|\leq C|x|^{k-1}, ~~\forall ~x\in \Omega\cap B_{1/2}
\end{equation}
and
\begin{equation}\label{e.Ckln-2}
|Du(0)|+\cdots+|D^ku(0)|\leq C,
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,\mu, b_0,c_0,\omega_0,\|\beta_3\|_{C^{k-2,1}(0)},
\omega_3,\omega_4$, $\|\partial \Omega\cap B_1\|_{C^{k,1}(0)}$, $\|f\|_{C^{k-2,1}(0)}$,
$\|g\|_{C^{k,1}(0)}$ and $\|u\|_{L^{\infty}(\Omega\cap B_1)}$.
\end{theorem}
Similar to $C^{k,\alpha}$ regularity, by combining the interior and boundary regularity, we have the local and global $C^{k,\mathrm{lnL}}$ regularity. For the completeness and the convention of citation, we list them as follows.
\begin{corollary}\label{t-C0ln-global}
Let $p>n$, $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} for any $x_0\in \Omega\cup \Gamma$, $\beta_1(x,x_0)\leq \delta_0$ for any $x\in B_{r_0}(x_0)\cap \Omega$, $\gamma_1\in L^{n}(\Omega)$ and $\underset{B_{r}(x_0)}{\mathrm{osc}}~\partial\Omega\leq \delta_0r/|\ln r|$ for any $x_0\in \Gamma$ and $0<r<r_0$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda,p$ and $\omega_1$. Assume that $b\in L^{p}(\Omega)$, $c\in L^{n}(\Omega)$, $f\in L^{n}(\Omega)$ and $g\in C^{0,1}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{0,\mathrm{lnL}}(\bar{\Omega}')$ and
\begin{equation}\label{e.C0ln-1-i-mu-global}
\|u\|_{C^{0,\mathrm{lnL}}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,p,r_0,\mu,\|b\|_{L^p(\Omega)}, \|c\|_{L^{n}(\Omega)},\omega_0,\|\gamma_1\|_{L^{n}(\Omega)},\omega_1$, $\|f\|_{L^{n}(\Omega)}$, $\|g\|_{C^{0,1}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimate
\begin{equation}\label{e.C0ln-1-i-global}
\|u\|_{C^{0,\mathrm{lnL}}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}+\|f\|_{L^{n}(\Omega)}+\|g\|_{C^{0,1}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,p,r_0,\|b\|_{L^p(\Omega)}$ and $\|c\|_{L^{n}(\Omega)}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{0,\mathrm{lnL}}$ estimates analogous to \cref{e.C0ln-1-i-mu-global} and \cref{e.C0ln-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{0,\mathrm{lnL}}$ estimates analogous to \cref{e.C0ln-1-i-mu-global} and \cref{e.C0ln-1-i-global} on $\bar{\Omega}$.
\end{corollary}
\begin{corollary}\label{t-C1ln-global}
Let $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F$ satisfies \cref{e.C1a.beta} for any $x_0\in \Omega\cup \Gamma$, $\beta_1(x,x_0)\leq \delta_0/|\ln|x-x_0||$ for any $x\in B_{r_0}(x_0)\cap \Omega$ and $\gamma_1\in L^{\infty}(\Omega)$, where $0<\delta_0<1$ depends only on $n,\lambda,\Lambda$ and $\omega_1$. Assume that $f\in L^{\infty}(\Omega)$, $\Gamma\in C^{1,1}$ and $g\in C^{1,1}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{1,\mathrm{lnL}}(\bar{\Omega}')$ and
\begin{equation}\label{e.C1ln-1-i-mu-global}
\|u\|_{C^{1,\mathrm{lnL}}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,r_0,\mu,b_0,c_0,\omega_0,
\|\gamma_1\|_{L^{\infty}(\Omega)}, \|\Gamma\|_{C^{1,1}}, \|f\|_{L^{\infty}(\Omega)},\|g\|_{C^{1,1}(\bar\Gamma)}$ and $\|u\|_{L^{\infty }(\Omega)}$.
In particular, if $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}, we have the following explicit estimate
\begin{equation}\label{e.C1ln-1-i-global}
\|u\|_{C^{1,\mathrm{lnL}}(\bar{\Omega}')}\leq C \left(\|u\|_{L^{\infty }(\Omega)}+\|f\|_{L^{\infty}(\Omega)}+\|g\|_{C^{1,1}(\bar\Gamma)}\right),
\end{equation}
where $C$ depends only on $n,\lambda,\Lambda,r_0,b_0,c_0$ and $\|\Gamma\|_{C^{1,1}}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{1,\mathrm{lnL}}$ estimates analogous to \cref{e.C1ln-1-i-mu-global} and \cref{e.C1ln-1-i-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{1,\mathrm{lnL}}$ estimates analogous to \cref{e.C1ln-1-i-mu-global} and \cref{e.C1ln-1-i-global} on $\bar{\Omega}$.
\end{corollary}
\begin{corollary}\label{t-Ckln-global}
Let $\Gamma\subset \partial \Omega $ be relatively open (maybe empty) and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega;\\
&u=g&& ~~\mbox{on}~~\Gamma.
\end{aligned}\right.
\end{equation*}
Suppose that $F\in C^{k-2,1}(\bar \Omega)$ $(k\geq 2)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Assume that $f\in C^{k-2,1}(\bar\Omega)$, $\Gamma\in C^{k,1}$ and $g\in C^{k,1}(\bar\Gamma)$.
Then for any $\Omega'\subset\subset \Omega\cup \Gamma$, we have $u\in C^{k,\mathrm{lnL}}(\bar{\Omega}')$ and
\begin{equation}\label{e.Ckln-1-i-mu-global}
\|u\|_{C^{k,\mathrm{lnL}}(\bar{\Omega}')}\leq C,
\end{equation}
where $C$ depends only on $n,\lambda, \Lambda,r_0,\mu, b_0,c_0,\omega_0,
\|\beta_3\|_{C^{k-2,1}(\bar\Omega)},\omega_3,\omega_4$, $\|\Gamma\|_{C^{k,1}}$,
$\|f\|_{C^{k-2,1}(\bar\Omega)}$, $\|g\|_{C^{k,1}(\bar{\Gamma})}$ and $\|u\|_{L^{\infty}(\Omega)}$.
If $\Gamma=\emptyset$, we obtain the interior $C^{k,\mathrm{lnL}}$ estimates analogous to \cref{e.Ckln-1-i-mu-global} on any $\bar{\Omega}'\subset\subset \Omega$; if $\Gamma=\partial \Omega$, we obtain the global $C^{k,\mathrm{lnL}}$ estimates analogous to \cref{e.Ckln-1-i-mu-global} on $\bar{\Omega}$.
\end{corollary}
\subsection{Some comments}
Here, we try to explain the key idea used in this paper. Consider the following linear equations for example.
\begin{equation}\label{e.linear-2}
\left\{\begin{aligned}
&a^{ij}u_{ij}+b^{i}u_i+cu=f&&~~\mbox{in}~~ \Omega\cap B_1;\\
&u=g&& ~~\mbox{on}~~\partial \Omega\cap B_1,
\end{aligned}\right.
\end{equation}
where $0\in \partial \Omega$ and we study the regularity near $0$.
The main idea is perturbation, which can be tracked to \cite{MR1005611}. Roughly speaking, if we know the regularity for harmonic functions, the regularity for \cref{e.linear-2} can be obtained by a perturbation argument. In the usual perturbation technique, the coefficients $a^{ij}$ and the righthand $f$ are regarded as a perturbation of $\delta_{ij}$ (the Kronecker symbol) and $0$. In this paper, we also regard the coefficients $b^i,c_i$, the boundary value $g$ and the curved boundary $\partial \Omega\cap B_1$ as the perturbation of $0,0,0$ and a hyperplane respectively.
More precisely, take the boundary $C^{1,\alpha}$ regularity for instance. The proof contains mainly two steps. First, if \cref{e.linear-2} is quite close to (e.g., $\|a^{ij}-\delta_{ij}\|_{L^{\infty}}\leq \delta$, $\|b\|_{L^{\infty}}\leq \delta$, $\|c\|_{L^{\infty}}\leq \delta$, $\|f\|_{L^{\infty}}\leq \delta$, $\|g\|_{L^{\infty}}\leq \delta$ and $\mathrm{osc}_{B_1}\partial \Omega\leq \delta$ for some small enough $\delta$)
\begin{equation}\label{e.2.2}
\left\{\begin{aligned}
&\Delta u=0&&~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1,
\end{aligned}\right.
\end{equation}
then the solution can be approximated by a linear polynomial in $\Omega\cap B_{\eta}$ for some $0<\eta<1$. This can be proved by the method of compactness which has been inspired by \cite{MR3246039} and \cite{Wang_Regularity}. Indeed, if the conclusion is false, we will have a sequence of problems in the form of \cref{e.linear-2} whose coefficients and prescribed data converge to that of \cref{e.2.2}. If the corresponding sequence of solutions are compact (e.g. by the uniform H\"{o}lder continuity \Cref{l-3Ho}), there exists a subsequence of solutions converging to some function $u$. Combining with the closedness result (e.g. by \Cref{l-35}), $u$ is a solution of \cref{e.2.2}. Then $u$ can be approximated by a linear polynomial, which will lead to a contradiction.
Second, by a scaling argument, we have a sequence of estimates in $\Omega\cap B_{\eta^m} (m\geq 1)$, which implies the boundary $C^{1,\alpha}$ regularity. This step only uses the scaling invariance of equations. Usually, we need to make necessary normalization such that the scaling argument can be carried out.
We point out that Nornberg \cite{MR3980853} also regarded $b^i,c$ as perturbation for $C^{1,\alpha}$ regularity. However, we deal with this in a more delicate way especially for higher regularity. Silvestre and Sirakov \cite{MR3246039} also obtained the boundary pointwise $C^{2,\alpha}$ regularity for flat boundaries. But they first established the regularity for the equation $\Delta u+b_0^iu_i=0$ where $b_0$ is constant vector. Then they proved $C^{2,\alpha}$ regularity for general equations $a^{ij} u_{ij}+b^i(x)u_i=f$ by a perturbation argument. On the contrast, we obtain the pointwise $C^{2,\alpha}$ regularity directly based on the regularity of $\Delta u=0$.
\section{Preliminaries}\label{P1}
In this section, we gather some preliminary results which will be used for proving our main regularity in following sections. Since we use the technique of perturbation and compactness, we need to prepare some compactness results for solutions. There exist two constants $0<\bar{\alpha}<1$ and $\bar{C}$ (fixed throughout this paper) depending only on $n,\lambda$ and $\Lambda$ such that the following four lemmas hold.
The first result is the interior $C^{1,\alpha}$ regularity, which is almost a direct result of the interior H\"{o}lder regularity for the Pucci class and Jensen's uniqueness theorem (see \cite[Corollary 5.7]{MR1351007}).
\begin{lemma}\label{l-3modin1}
Let $u$ satisfy $F(D^2u)=0$ in $B_1$. Then $u\in C^{1,\bar{\alpha}}(\bar{B}_{1/2})$ and
\begin{equation}\label{e.l3modin1-1}
\|u\|_{C^{1,\bar{\alpha}}(\bar{B}_{1/2})}\leq \bar{C}\|u\|_{L^{\infty}(B_1)}.
\end{equation}
In particular, $u\in C^{1,\bar\alpha}(0)$, i.e., there exists a linear polynomial $P$ such that
\begin{equation}\label{e.3.1}
|u(x)-P(x)|\leq \bar{C} |x|^{1+\bar{\alpha}}\|u\|_{L^{\infty }(B_1)}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{e.3.2}
|Du(0)|\leq \bar{C}\|u\|_{L^{\infty }(B_1)}.
\end{equation}
\end{lemma}
Next lemma concerns the interior $C^{2,\alpha}$ regularity, which was first proved independently by Evans \cite{MR649348} and Krylov \cite{MR661144} for $C^2$ solutions. The proof for viscosity solutions was given by Cabr\'{e} and Caffarelli ( see \cite{MR1391943} and \cite[Theorem 6.6]{MR1351007}).
\begin{lemma}\label{l-3modin2}
Let $F$ be convex in $M$ and $u$ satisfy $F(D^2u)=0$ in $B_1$. Then $u\in C^{2,\bar{\alpha}}(\bar{B}_{1/2})$ and
\begin{equation}\label{e.l3modin2-1}
\|u\|_{C^{2,\bar{\alpha}}(\bar{B}_{1/2})}\leq \bar{C}\|u\|_{L^{\infty}(B_1)}.
\end{equation}
In particular, $u\in C^{2,\bar\alpha}(0)$, i.e., there exists a quadratic polynomial $P$ such that
\begin{equation}\label{e.3.3}
|u(x)-P(x)|\leq \bar{C} |x|^{2+\bar{\alpha}}\|u\|_{L^{\infty }(B_1)}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{e.3.5}
F(D^2u(0))=0
\end{equation}
and
\begin{equation}\label{e.3.4}
|Du(0)|+\|D^2u(0)\|\leq \bar{C}\|u\|_{L^{\infty }(B_1)}.
\end{equation}
\end{lemma}
The following is the boundary $C^{1,\alpha}$ regularity, which was first proved by Krylov \cite{MR688919} for $C^2$ solutions. Later, it was simplified by Caffarelli (see \cite[Theorem 9.31]{MR1814364} and \cite[Theorem 4.28]{MR787227}), which is applicable to the Pucci class.
\begin{lemma}\label{l-31}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&u\in S(\lambda,\Lambda,0)&& ~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1.
\end{aligned}\right.
\end{equation*}
Then $u\in C^{1,\bar{\alpha}}(0)$ and for some constant $a$,
\begin{equation}\label{e.l31-1}
|u(x)-ax_n|\leq \bar{C} |x|^{1+\bar{\alpha}}\|u\|_{L^{\infty }(B_1^+)}, ~~\forall ~x\in B_{1}^+
\end{equation}
and
\begin{equation}\label{e.l31-2}
|a|\leq \bar{C}\|u\|_{L^{\infty }(B_1^+)}.
\end{equation}
\end{lemma}
Finally, we present the boundary $C^{2,\alpha}$ regularity. It was first proved for $C^2$ solutions by Krylov \cite{MR688919}. For viscosity solutions, it was proved by Silvestre and Sirakov \cite[Lemma 4.1]{MR3246039} by combining the $C^{1,\alpha}$ regularity and the technique of difference quotient.
\begin{lemma}\label{l-32}
Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u)=0&& ~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1.
\end{aligned}\right.
\end{equation*}
Then $u\in C^{2,\bar{\alpha}}(0)$ and for some constants $a$ and $b_{in} (1\leq i \leq n)$,
\begin{equation}\label{e.l32-1}
|u(x)-ax_n-\frac{1}{2}b_{in}x_ix_n|\leq \bar{C} |x|^{2+\bar{\alpha}}\|u\|_{L^{\infty }(B_1^+)}, ~~\forall ~x\in B_{1}^+,
\end{equation}
\begin{equation}\label{e.l32-2}
F(b_{in})=0
\end{equation}
and
\begin{equation}\label{e.l32-3}
|a|+|b_{in}|\leq \bar{C}\|u\|_{L^{\infty }(B_1^+)}.
\end{equation}
\end{lemma}
\begin{remark}\label{r-31}
In \cref{e.l32-1}, the Einstein summation convention is used, i.e., $b_{in}x_ix_n$ denotes the sum $\sum_{i=1}^{n}b_{in}x_ix_n$ (similarly hereinafter). In \cref{e.l32-2}, $b_{in}$ denotes the corresponding Hessian matrix $a_{ij}$ of $b_{in}x_ix_n$ (similarly hereinafter) whose elements are
\begin{equation*}
a_{ij}=\left\{\begin{aligned}
&0&&i<n,j<n;\\
&\frac{b_{in}}{2}&&i<n,j=n;\\
&b_{nn}&&i=n,j=n.
\end{aligned}
\right.
\end{equation*}
\end{remark}
\begin{remark}\label{r-33}
We remark here that Harnack inequality is a common key element for above regularity results. From Harnack inequality, the interior H\"{o}lder regularity for the Pucci class can be easily derived. Since the difference quotient of a solution belongs to the Pucci class, we have the interior $C^{1,\alpha}$ regularity (\Cref{l-3modin1}). For the interior $C^{2,\alpha}$ regularity (\Cref{l-3modin2}), the key is that $u_{ee}$ (the second derivative of $u$ along $e\in R^n$) is a subsolution and then the weak Harnack inequality is applicable for $\sup u_{ee} -u_{ee}$. By combing Harnack inequality and a proper barrier, the boundary $C^{1,\alpha}$ regularity (\Cref{l-31}) follows. With regards to the boundary $C^{2,\alpha}$ regularity (\Cref{l-32}), note that $u_i$ ($1\leq i\leq n-1$) belongs to the Pucci class and then apply \Cref{l-31} to $u_i$ to conclude that $u_i\in C^{1,\alpha}$. Then the boundary $C^{2,\alpha}$ regularity can be obtained.
\end{remark}
\begin{remark}\label{r-32}
If we study linear uniformly elliptic equations in nondivergence form, equations in \Cref{l-3modin1} and \Cref{l-3modin2} will reduce to $\Delta u=0$ in $B_1$ and equations in \Cref{l-31} and \Cref{l-32} will reduce to
\begin{equation*}
\left\{\begin{aligned}
&\Delta u=0&& ~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1.
\end{aligned}\right.
\end{equation*}
The corresponding results of \Crefrange{l-3modin1}{l-32} for Laplace equation are well known basic facts. Through a perturbation argument as in this paper, we can obtain the interior and boundary pointwise $C^{k,\alpha}$ regularity for any $0<\alpha<1$ and $k\geq 1$. This makes the pointwise regularity theory for linear equations quite easy.
\end{remark}
Since we use compactness method, we need the following closedness result for viscosity solutions ( see \cite[Proposition 2.3]{MR3980853}). It states that the limit of a sequence of solutions is a solution of some equation as well.
\begin{lemma}\label{l-35}
Let $F$ and $\{F_m\}_{m\geq 1}$ be fully nonlinear operators. Assume that $b\in L^p (\Omega) (p>n)$, $f,f_m\in L^n(\Omega)$. Let $u_m\in C(\Omega)$ be $L^n$-viscosity subsolutions (or supersolutions) of
\begin{equation*}
F_m(D^2u_m,Du_m,u_m,x)= f_m~~\mbox{in}~~\Omega.
\end{equation*}
Suppose that $u_m\rightarrow u$ in $L^{\infty}_{\mathrm{loc}}(\Omega)$ as $m\rightarrow \infty$ and for any ball $B\subset \subset \Omega$ and $\varphi \in W^{2,n}(B)$, $\|(G_m-G)^+\|_{L^n(B)}\left(~\mbox{or}~\|(G_m-G)^-\|_{L^n(B)}\right)\rightarrow 0$ as $m\rightarrow \infty$ where
\begin{equation*}\label{FkCg}
G_m(x):=F_m(D^2\varphi,D\varphi,u_m,x)-f_m(x)~~\mbox{and}~~G(x):=F(D^2\varphi,D\varphi,u,x)-f(x).
\end{equation*}
Then $u$ is an $L^n$-viscosity subsolution (or supersolution) of
\begin{equation}\label{FkC}
F(D^2u,Du,u,x)= f~~\mbox{in}~~\Omega.
\end{equation}
If $F$ and $f$ are continuous in $x$, it is enough to take $\varphi \in C^2(\bar B)$, in which case $u$ is a $C$-viscosity subsolution (or supersolution) of \cref{FkC}.
\end{lemma}
\begin{remark}\label{r-3.2}
In this paper, we always use the case that $F$ and $f$ are continuous in $x$. Hence, we always take $\varphi \in C^2(\bar B)$ in the proofs.
\end{remark}
\begin{remark}\label{r-34}
If all functions concerned in \Cref{l-35} are continuous in $x$ and we consider $C$-viscosity solutions, the closedness result is easily to establish by the definition of viscosity solution and assuming $\|(G_m-G)^+\|_{L^{\infty}(B)}(~\mbox{or}~\|(G_m-G)^-\|_{L^{\infty}(B)})\rightarrow 0$.
\end{remark}
\Cref{l-35} requires that $u_m$ converges uniformly. The following lemma (see \cite[Theorem 2]{MR3980853}) states the interior H\"{o}lder regularity, which provides necessary compactness for solutions and then guarantee the uniform convergence. Moreover, it also contains the boundary pointwise H\"{o}lder regularity, which will be used in the normalization procedure for boundary pointwise $C^{1,\alpha}$ regularity (see the proof in \Cref{C1a-mu}).
\begin{lemma}\label{l-3Ho}
Let $u$ satisfy
\begin{equation*}
u\in S^*(\lambda,\Lambda,\mu,b,f) ~~\mbox{in}~~\Omega.
\end{equation*}
Then there exists $0<\alpha<1$ depending only on $n,\lambda,\Lambda,p$ and $\|b\|_{L^p(\Omega)}$ such that $u\in C^{\alpha}(\bar\Omega')$ for any $\Omega'\subset\subset\Omega$ and
\begin{equation*}
\|u\|_{C^{\alpha}(\bar\Omega')}\leq C,
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda,p,\mu,\|b\|_{L^p(\Omega)},\|f\|_{L^n(\Omega)}$, $\Omega'$, $\mathrm{dist}(\Omega',\partial \Omega)$ and $\|u\|_{L^{\infty}(\Omega)}$.
In addition, suppose that $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $\Omega$ satisfies the exterior cone condition at $0\in \partial \Omega$ and $g\in C^{\alpha_1}(0)$ for some $0<\alpha_1<1$. Then there exists $0<\alpha\leq \alpha_1$ depending only on $n,\lambda,\Lambda,p,\|b\|_{L^p(\Omega)}$ and the cone such that $u\in C^{\alpha}(0)$ and
\begin{equation*}
|u(x)-u(0)|\leq C|x|^{\alpha}, ~\forall ~x\in \Omega_1,
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda,p,\mu,\|b\|_{L^p(\Omega_1)},\|f\|_{L^n(\Omega_1)}$, $\|u\|_{L^{\infty}(\Omega_1)}$ and the cone.
\end{lemma}
We also need the compactness of solutions up to the boundary for the boundary regularity. In the following, we prepare the compactness up to the boundary following the idea of \cite{MR4088470}. First, we introduce the Alexandrov-Bakel'man-Pucci maximum principle (see \cite[Proposition 2.5]{MR3980853}).
\begin{lemma}\label{l-3ABP}
Let $u$ satisfy
\begin{equation*}
M^+(D^2u,\lambda,\Lambda)+\mu|Du|^2+b|Du|\geq f~~\mbox{in}~~\Omega,
\end{equation*}
where $b\in L^p(\Omega)$ $(p>n)$ and $f\in L^n(\Omega)$. Suppose that
\begin{equation*}
\mu \|f^-\|_{L^n(\Omega)}\cdot\mathrm{diam}(\Omega)\leq \delta,
\end{equation*}
where $\delta>0$ depends only on
$n,\lambda,\Lambda,p,\mathrm{diam}(\Omega)$ and $\|b\|_{L^p(\Omega)}$. Then
\begin{equation*}
\max_{\bar{\Omega}}u\leq \max_{\partial \Omega}u+C \|f^-\|_{L^n(\Omega)},
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda,p,\mathrm{diam}(\Omega)$ and $\|b\|_{L^p(\Omega)}$.
\end{lemma}
The following lemma provides a uniform estimate for solutions, which is a kind of ``equicontinuity'' up to the boundary in some sense.
\begin{lemma}\label{l-33}
Let $0<\delta<1$ be as in \Cref{l-3ABP} and $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1.
\end{aligned}\right.
\end{equation*}
Suppose that $\|u\|_{L^{\infty}(\Omega_{1})}\leq 1$, $\mu\leq \delta/C_0$, $\|b\|_{L^{p}(\Omega_1)}\leq\delta$, $\|f\|_{L^{n}(\Omega_1)}\leq\delta$, $\|g\|_{L^{\infty}((\partial \Omega)_1)}\leq \delta$
and $\underset{B_1}{\mathrm{osc}}~\partial\Omega \leq \delta$, where $C_0$ (to be specified later) depends only on $n,\lambda$ and $\Lambda$.
Then
\begin{equation*}
\|u\|_{L^{\infty}(\Omega_{\delta})}\leq C\delta,
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda$ and $p$.
\end{lemma}
\proof Set $B^{+}=B^{+}_{1}-\delta e_n $, $T=T_1-\delta e_n$ and then $\partial \Omega\cap B_{1/4}\subset B^+$. Take
\begin{equation*}
v(x)=C\big((1-\delta)^{-\alpha}-|x+e_n|^{-\alpha}\big).
\end{equation*}
Then by choosing proper constants $C$ and $\alpha$ (depending only on $n,\lambda$ and $\Lambda$), we have $v(-\delta e_n)=0$ and
\begin{equation*}
\left\{\begin{aligned}
&M^{+}(D^2v,\lambda,\Lambda)\leq 0 &&\mbox{in}~~B^+; \\
&v\geq 0 && \mbox{in}~~ B^+;\\
&v\geq 1 &&\mbox{on}~~\partial B^+\backslash T.
\end{aligned}
\right.
\end{equation*}
Let $w=u-v$ and then $w$ satisfies
\begin{equation*}
\left\{
\begin{aligned}
&w\in \underline{S}(\lambda,\Lambda,2\mu,b,\tilde{f}) &&\mbox{in}~~ \Omega \cap B^+; \\
&w\leq g &&\mbox{on}~~\partial \Omega \cap \bar{B}^+;\\
&w\leq 0 &&\mbox{on}~~\partial B^+\cap \Omega,
\end{aligned}
\right.
\end{equation*}
where $\tilde{f}=-|f|-2\mu|Dv|^2-b|Dv|$.
From the definition of $v$,
\begin{equation*}
\|v\|_{L^{\infty}(B_{4\delta}^+-\delta e_n)}
= \|v-v(-\delta e_n)\|_{L^{\infty}(B_{4\delta}^+-\delta e_n)}
\leq C\delta,
\end{equation*}
where $C$ depends only on $n,\lambda$ and $\Lambda$. For $w$, note that
\begin{equation*}
\|\tilde{f}\|_{L^n(\Omega_1)}\leq \|f\|_{L^n(\Omega_1)}+\|2\mu |Dv|^2\|_{L^n(\Omega_1)}
+\|b|Dv|\|_{L^n(\Omega_1)}\leq C_0/2,
\end{equation*}
where $C_0$ depends only on $n,\lambda$ and $\Lambda$. Then
\begin{equation*}
2\mu \|\tilde{f}\|_{L^n(\Omega_1)}\leq \delta.
\end{equation*}
By applying the Alexandrov-Bakel'man-Pucci maximum principle to $w$ (\Cref{l-3ABP}), we have
\begin{equation*}
\begin{aligned}
\sup_{\Omega \cap B^+} w&
\leq\|g\|_{L^{\infty }(\partial \Omega \cap B^+)}+C\left(\|f\|_{L^n}+
\mu\|Dv\|^2_{L^{\infty}}+\|b\|_{L^n} \|Dv\|_{L^{\infty}}\right)
\leq C\delta,
\end{aligned}
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda$ and $p$. Thus,
\begin{equation*}
\sup_{\Omega_{\delta}} u\leq \sup_{\Omega\cap (B_{4\delta}^+-\delta e_n)} u \leq \|v\|_{L^{\infty}(B_{4\delta}^+-\delta e_n)}+\sup_{\Omega \cap B^+} w\leq C\delta.
\end{equation*}
The proof for
\begin{equation*}
\inf_{\Omega_{\delta}} u \geq -C\delta
\end{equation*}
is similar and we omit it here. Therefore, the proof is completed.\qed~\\
\begin{remark}\label{r-3.1}
If $\Omega$ satisfies the exterior cone condition at every boundary point (or $\partial \Omega\cap B_1\in C^{0,1}$), we have the equicontinuity up to the boundary by \Cref{l-3Ho}. However, it is not assumed in this paper since we study the pointwise regularity.
\end{remark}
\begin{remark}\label{r-3.3}
The method of compactness only requires some kind of compactness for solutions. It can be proved by constructing an appropriate barrier (i.e., $v$ in the proof). That is, the compactness is easy to prepare. Instead, the method of solving an equation to approximate the solution is more complicated. For example, it needs the smoothness of the boundary if one intend to attack the boundary regularity by solving an auxiliary equation.
\end{remark}
Based on above lemma, the following corollary follows easily.
\begin{corollary}\label{c-31}
For any $0<r<1$ and $\varepsilon>0$, there exists $\delta>0$ (depending only on $n,\lambda,\Lambda,p,r$ and $\varepsilon$) such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $\|b\|_{L^{p}(\Omega_1)}\leq \delta$, $\|f\|_{L^{n}(\Omega_1)}\leq \delta$, $\|g\|_{L^{\infty}((\partial \Omega)_1)}\leq \delta$
and $\underset{B_1}{\mathrm{osc}}~\partial\Omega \leq \delta$, then
\begin{equation*}
\|u\|_{L^{\infty}(\Omega \cap B(x_0,\delta))}\leq \varepsilon, ~~\forall~x_0\in \partial \Omega\cap B_r.
\end{equation*}
\end{corollary}
Next, we prove the equicontinuity of the solutions, which provides the necessary compactness.
\begin{lemma}\label{l-34}
For any $\Omega'\subset\subset \bar{\Omega}\cap B_1$ and $\varepsilon>0$, there exists $\delta>0$ (depending only on $n,\lambda,\Lambda,p,\varepsilon$ and $\Omega'$) such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $\|b\|_{L^{p}(\Omega_1)}\leq \delta$, $\|f\|_{L^{n}(\Omega_1)}\leq \delta$, $\|g\|_{L^{\infty}((\partial \Omega)_1)}\leq \delta$
and $\underset{B_1}{\mathrm{osc}}~\partial\Omega \leq \delta$, then for any $x,y\in \Omega'$ with $|x-y|\leq \delta$, we have
\begin{equation*}
|u(x)-u(y)|\leq \varepsilon.
\end{equation*}
\end{lemma}
\proof By \Cref{c-31}, for any $\varepsilon>0$, there exists $\delta_1>0$ depending only on $n,\lambda,\Lambda,p,\varepsilon$ and $\Omega'$ such that for any $x,y\in \Omega'$ with $\mathrm{dist}(x,\partial \Omega)\leq \delta_1$ and $|x-y|\leq \delta_1$, we have
\begin{equation}\label{e.l34-2.1}
|u(x)-u(y)|\leq |u(x)|+|u(y)|\leq \varepsilon.
\end{equation}
If $\mathrm{dist}(x,\partial \Omega)> \delta_1$, by the interior H\"{o}lder estimate (\Cref{l-3Ho}),
\begin{equation}\label{e.l34-2.2}
|u(x)-u(y)|\leq C\frac{|x-y|^{\alpha}}{\delta_1^{\alpha}},
\end{equation}
where $C$ and $0<\alpha<1$ depend only on $n,\lambda,\Lambda,\mu$ and $p$. Take $\delta$ small enough such that
\begin{equation*}
C\frac{\delta^{\alpha}}{\delta_1^{\alpha}}\leq \varepsilon.
\end{equation*}
Then by combining \cref{e.l34-2.1} and \cref{e.l34-2.2}, the conclusion follows.~\qed~\\
\section{Interior $C^{1,\alpha}$ regularity}\label{In-C1a-mu}
In this section, we give the proof of the interior pointwise $C^{1,\alpha}$ regularity \Cref{t-C1a-i}. First, we show that the solution in \Cref{t-C1a-i} can be approximated by a linear function in a small scale provided that the coefficients and the prescribed data are small enough.
\begin{lemma}\label{In-l-C1a-mu}
Suppose that $F$ satisfies \cref{e.C1a.beta}. For any $0<\alpha<\bar{\alpha}$, there exists $\delta>0$ depending only on $n,\lambda,\Lambda,\alpha$ and $\omega_1$ such that if $u$ satisfies
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1
\end{equation*}
with $\|u\|_{L^{\infty}(B_1)}\leq 1$, $\mu\leq \delta$, $\|b\|_{L^{p}(B_1)}\leq \delta$, $\|c\|_{L^{n}(B_1)}\leq \delta$, $\omega_0(1,1)\leq 1$, $\|\beta_1\|_{L^n(B_1)}\leq \delta$, $\|\gamma_1\|_{L^n(B_1)}\leq \delta$ and $\|f\|_{L^{n}(B_1)}\leq \delta$, then there exists a linear polynomial $P$ such that
\begin{equation}\label{In-e.l4.0-mu}
\|u-P\|_{L^{\infty}(B_{\eta})}\leq \eta^{1+\alpha}
\end{equation}
and
\begin{equation}\label{In-e.14.2-mu}
|DP(0)|\leq \bar{C},
\end{equation}
where $\eta$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
\end{lemma}
\begin{remark}\label{r-4.2}
In fact, we can obtain a little stronger result: there exists $0<\eta_0<1$ such that for any $0<\eta<\eta_0$, \cref{In-e.l4.0-mu} holds (see the proof below).
\end{remark}
\begin{remark}\label{In-r-41-mu}
\Cref{In-l-C1a-mu} is sometimes called the ``key lemma'' (or ``key step'') in the proof of regularity. Usually, one uses a solution of a homogenous equation to approximate $u$ (e.g. \cite[Lemma 8.2]{MR1351007} and \cite[Lemma 3.7]{MR3980853}). Although the compactness method is applied there, they need the solvability of some equation, which is avoided in our proof.
\end{remark}
\proof We prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<\bar{\alpha}$, $\omega_1$ and sequences of $F_m,u_m, f_m$ satisfying
\begin{equation*}
F_m(D^2u_m, Du_m, u_m,x)=f_m ~~\mbox{in}~~ B_1.
\end{equation*}
In addition, $F_m$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\mu_m,b_m,c_m$ and $\omega_m$. Moreover, $F_m$ satisfy \cref{e.C1a.beta} with $F_{0m}$, $\beta_m,\gamma_m$ and $\omega_1$. Furthermore, $\|u_m\|_{L^{\infty}(B_1)}\leq 1$, $\|\beta_m\|_{L^n(B_1)}\leq 1/m$, $\|\gamma_m\|_{L^{n}(B_1)}\leq 1/m$, $\mu_m\leq 1/m$, $\|b_m\|_{L^{p}(B_1)}\leq 1/m$, $\|c_m\|_{L^{n}(B_1)}\leq 1/m$, $\omega_m(1,1)\leq 1$ and $\|f_m\|_{L^{n}(B_1)}\leq 1/m$. Finally, we have
\begin{equation}\label{In-e.lC1a.1-mu}
\|u_m-P\|_{L^{\infty}(B_{\eta})}> \eta^{1+\alpha}, \forall~P~\mbox{with}~|DP(0)|\leq \bar{C},
\end{equation}
where $0<\eta<1$ is taken small such that
\begin{equation}\label{In-e.lC1a.2-mu}
\bar{C}\eta^{\bar{\alpha}-\alpha}<1/2.
\end{equation}
Note that $u_m$ are uniformly bounded (i.e., $\|u_m\|_{L^{\infty}(B_1)}\leq 1$) and equicontinuous (see \Cref{l-3Ho}) in any $B\subset\subset B_1$. Hence, there exist a subsequence (denoted by $u_m$ again) and $u:B_1\rightarrow R$ such that $u_m\rightarrow u$ in $L^{\infty}_{\mathrm{loc}}(B_1)$. In addition, since $F_{0m}(0)=0$ and $F_{0m}$ are Lipschitz continuous in $M$ with a uniform Lipschitz constant depending only on $n,\lambda$ and $\Lambda$, there exist a subsequence (denoted by $F_{0m}$ again) and $F:S^n\rightarrow R$ such that $F_{0m}\rightarrow F$ in $L^{\infty}_{\mathrm{loc}}(S^n)$. Furthermore, for any ball $B\subset\subset B_1$ and $\varphi\in C^{2}(\bar{B})$, let $G_m=F_m(D^2\varphi,D\varphi,u_m,x)-f_m(x)$ and $G=F(D^2\varphi)$. Then,
\begin{equation*}\label{In-gkC1a-mu}
\begin{aligned}
\|G_m-G\|_{L^n(B)}=&\|F_m(D^2\varphi,D\varphi,u_m,x)-F(D^2\varphi)-f_m\|_{L^n(B)}\\
=&\|F_m(D^2\varphi,D\varphi,u_m,x)-F_m(D^2\varphi,0,0,x)+F_m(D^2\varphi,0,0,x)\\
&-F_{0m}(D^2\varphi)+F_{0m}(D^2\varphi)-F(D^2\varphi)-f_m\|_{L^n(B)}\\
\leq& \|\mu_m|D\varphi|^2\|_{L^n(B)}+\|b_m|D\varphi|\|_{L^n(B)}
+\|c_m\omega_m(1,|u_m|)\|_{L^n(B)}\\
&+\|\beta_m\omega_1(\|D^2\varphi\|_{L^{\infty}(B)})+\gamma_m\|_{L^n(B)}\\
&+\|F_{0m}(D^2\varphi)-F(D^2\varphi)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
\leq&\mu_m \||D\varphi|^2\|_{L^n(B)}
+\|b_m\|_{L^p(B)}\|D\varphi\|_{L^{\frac{np}{p-n}}(B)}+\|c_m\|_{L^n(B)}\omega_m(1,1)\\
&+\|\beta_m\|_{L^n(B)} \omega_1(\|D^2\varphi\|_{L^{\infty}(B)})+\|\gamma_m\|_{L^n(B)} \\
&+\|F_{0m}(D^2\varphi)-F(D^2\varphi)\|_{L^n(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$.
By \Cref{l-35}, $u$ is a viscosity solution of
\begin{equation*}
F(D^2u)=0 ~~\mbox{in}~~ B_1.
\end{equation*}
By \Cref{l-3modin1}, there exists a linear polynomial $P$ such that
\begin{equation*}
|u(x)-P(x)|\leq \bar{C} |x|^{1+\bar{\alpha}}, ~~\forall ~x\in B_{1}
\end{equation*}
and
\begin{equation*}
|DP(0)|\leq \bar{C}.
\end{equation*}
Hence, by noting \cref{In-e.lC1a.2-mu}, we have
\begin{equation}\label{In-e.lC1a.3-mu}
\|u-P\|_{L^{\infty}(B_{\eta})}\leq \eta^{1+\alpha}/2.
\end{equation}
From \cref{In-e.lC1a.1-mu},
\begin{equation*}
\|u_m-P\|_{L^{\infty}(B_{\eta})}> \eta^{1+\alpha}.
\end{equation*}
Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-P\|_{L^{\infty}(B_{\eta})}\geq \eta^{1+\alpha},
\end{equation*}
which contradicts with \cref{In-e.lC1a.3-mu}. ~\qed~\\
Now, we can prove the interior $C^{1,\alpha}$ regularity.
\noindent\textbf{Proof of \Cref{t-C1a-i}.} We make some normalization first such that \Cref{In-l-C1a-mu} can be applied. Let $\delta$ be the constant as in \Cref{In-l-C1a-mu} and we take $\delta_0$ ($\delta_0$ is from the assumptions of \Cref{t-C1a-i}) small enough such that $K_0^2\delta_0\leq \delta$. We can assume that
\begin{equation}\label{In-e.tC1a-ass-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(B_1)}\leq 1,\\
&\mu\leq \frac{\delta}{8C_0^2},~\|b\|_{L^p(B_1)}\leq \frac{\delta}{8C_0},
~\|c\|_{C^{-1,\alpha}(0)}\leq \frac{\delta}{4},~\omega_0(1+C_0,C_0)\leq1,\\
&\|\beta_1\|_{C^{-1,1}(0)}\leq \frac{\delta}{K_0},~\|\gamma_1\|_{C^{-1,\alpha}(0)}\leq \frac{\delta}{4}~~\mbox{and}~~\|f\|_{C^{-1,\alpha}(0)}\leq \delta,
\end{aligned}
\end{equation}
where $C_0$ is a constant (depending only on $n,\lambda,\Lambda$ and $\alpha$) to be specified later.
Otherwise, we consider for some $0<\rho<1$,
\begin{equation}\label{e.4.1}
\bar{u}(y)=\frac{u(x)-u(0)}{\rho^{\alpha_0}},
\end{equation}
where $y=x/\rho$ and $0<\alpha_0<1$ is a H\"{o}lder exponent (depending only on $n,\lambda,\Lambda$ and $\|b\|_{L^p(B_1)}$) such that $u\in C^{2\alpha_0}(0)$ (by \Cref{l-3Ho}). Then $\bar{u}$ satisfies
\begin{equation*}
\bar{F}(D^2\bar{u}, D\bar{u}, \bar{u},y)=\bar{f} ~~\mbox{in}~~ B_1,
\end{equation*}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}\label{In-e.tC1a-n1-mu}
\begin{aligned}
&\bar{F}(M,p,s,y)
=\rho^{2-\alpha_0}F(\rho^{\alpha_0-2}M,\rho^{\alpha_0-1}p,\rho^{\alpha_0}s+u(0),x)
~~\mbox{and}~~\bar{f}(y)=\rho^{2-\alpha_0}f(x).
\end{aligned}
\end{equation*}
It is easy to check that $\bar F$ satisfies the structure condition \cref{SC2} with $\lambda$, $\Lambda$, $\bar{\mu}$, $\bar{b}$, $\bar{c}$ and $\bar{\omega}_0$, where
\begin{equation*}
\begin{aligned}
&\bar{\mu}=\rho^{\alpha_0}\mu,~\bar{b}(y)=\rho b(x),
~\bar{c}(y)=\rho c(x) ~~\mbox{and}~~
\bar{\omega}_0(\cdot,\cdot)=\rho^{1-\alpha_0}\omega_0(\cdot+|u(0)|,\cdot).
\end{aligned}
\end{equation*}
In addition, we take $\bar{F}_0(M)=\rho^{2-\alpha_0}F_0(\rho^{\alpha_0-2}M)$ for any $M\in S^n$, where $F_0$ is from \cref{e.C1a.beta}. Then $\bar{F}_0$ is uniformly elliptic with constants $\lambda$ and $\Lambda$, and
\begin{equation*}
\begin{aligned}
|\bar{F}&(M,0,0,y)-\bar{F}_0(M)|\\
=&\rho^{2-\alpha_0}|F(\rho^{\alpha_0-2}M,0,u(0),x)
-F_0(\rho^{\alpha_0-2}M)|\\
=&\rho^{2-\alpha_0}|F(\rho^{\alpha_0-2}M,0,u(0),x)
-F(\rho^{\alpha_0-2}M,0,0,x)\\
&+F(\rho^{\alpha_0-2}M,0,0,x)-
F_0(\rho^{\alpha_0-2}M)|\\
\leq & \rho^{2-\alpha_0}c(x)\omega_0(|u(0)|,|u(0)|)
+K_0\beta_1(x)\omega_1(\|M\|)+\rho^{2-\alpha_0}\gamma_1(x)\\
:=&\bar\beta_1(y)\omega_1(\|M\|)+\bar\gamma_1(y),
\end{aligned}
\end{equation*}
where $\bar\beta_1(y)=K_0\beta_1(x)$ and $\bar\gamma_1(y)=\rho^{2-\alpha_0}\gamma_1(x)+\rho^{2-\alpha_0}c(x)\omega_0(|u(0)|,|u(0)|)$.
From above arguments,
\begin{equation*}
\begin{aligned}
&\|\bar{u}\|_{L^{\infty}(B_1)}\leq \frac{1}{\rho^{\alpha_0}}|x|^{2\alpha_0}[u]_{C^{2\alpha_0}(0)}
\leq \rho^{\alpha_0}[u]_{C^{2\alpha_0}(0)},\\
&\bar\mu=\rho^{\alpha_0}\mu,\\
&\|\bar{b}\|_{L^{p}(B_1)}=\rho^{1-\frac{n}{p}}\|b\|_{L^{p}(B_{\rho})}\leq \rho^{\alpha}\|b\|_{L^{p}(B_1)},\\
&\|\bar{c}\|_{C^{-1,\alpha}(0)}\leq \rho^{\alpha}\|c\|_{C^{-1,\alpha}(0)},\\
& \bar{\omega}_0(1+C_0,1)=\rho^{1-\alpha_0}\omega_0(1+C_0+|u(0)|,1),\\
&\|\bar\beta_1\|_{C^{-1,1}(0)}\leq K_0\|\beta_1\|_{C^{-1,1}(0)}\leq \frac{\delta}{K_0},\\
&\|\bar{\gamma}_1\|_{C^{-1,\alpha}(0)}
\leq\rho^{1-\alpha_0+\alpha}(\|\gamma_1\|_{C^{-1,\alpha}(0)}
+\omega_0(|u(0)|,|u(0)|)\|c\|_{C^{-1,\alpha}(0)}),\\
&\|\bar{f}\|_{C^{-1,\alpha}(0)}
\leq\rho^{1-\alpha_0+\alpha}\|f\|_{C^{-1,\alpha}(0)}.
\end{aligned}
\end{equation*}
Therefore, by taking $\rho$ small enough (depending only on $n,\lambda,\Lambda,\alpha, \mu, \|b\|_{L^{p}(B_1)}$, $\|c\|_{C^{-1,\alpha}(0)}$, $\omega_0$,
$\|\gamma_1\|_{C^{-1,\alpha}(0)}$, $\omega_1$, $\|f\|_{C^{-1,\alpha}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$), the assumptions in \cref{In-e.tC1a-ass-mu} for $\bar{u}$ can be guaranteed. Then the regularity of $u$ can be derived from that of $\bar{u}$. Hence, without loss of generality, we assume that \cref{In-e.tC1a-ass-mu} holds for $u$.
Now, we prove that $u$ is $C^{1,\alpha}$ at $0$ and we only need to prove the following. There exist a sequence of linear polynomials $P_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{In-e.tC1a-3-mu}
\|u-P_m\|_{L^{\infty }(B _{\eta^{m}})}\leq \eta ^{m(1+\alpha )}
\end{equation}
and
\begin{equation}\label{In-e.tC1a-4-mu}
|P_m(0)-P_{m-1}(0)|+\eta^m|DP_m(0)-DP_{m-1}(0)|\leq \bar{C}\eta^{m(1+\alpha)},
\end{equation}
where $\eta$, depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{In-l-C1a-mu} .
It is standard that \cref{In-e.tC1a-3-mu} and \cref{In-e.tC1a-4-mu} imply \cref{e.C1a-1-i-mu} and \cref{e.C1a-2-i-mu}. Indeed, assume that $P_m$ has the form $P_m(x)=a_m+b_{m,i}x_i$ ($1\leq i\leq n$ and the Einstein summation convention is used). Then \cref{In-e.tC1a-4-mu} reads
\begin{equation*}
|a_m-a_{m-1}|+\eta^m|b_{m,i}-b_{m-1,i}|\leq \bar{C}\eta ^{m(1+\alpha)},~\forall ~m\geq 1.
\end{equation*}
Thus, there exist constants $a$ and $b_i$ ($1\leq i\leq n$) such that ,
\begin{equation*}
|a_m-a|+\eta^m|b_{m,i}-b_{i}|\leq C\eta^{m(1+\alpha)}~(\forall ~m\geq 1)~~\mbox{and}~~|a|+|b_i|\leq C,
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda$ and $\alpha$. Let $P(x)=a+b_ix_i$ and then \cref{e.C1a-2-i-mu} holds. In addition, for any $x\in B_1$, there exists $m\geq 0$ such that $\eta^{m+1}\leq |x|< \eta^{m}$. Hence,
\begin{equation*}
\begin{aligned}
|u(x)-P(x)|\leq & |u(x)-P_m(x)|+|P_m(x)-P(x)|\\
\leq &\eta^{m(1+\alpha )}+|a_m-a|+\eta^m|b_{m,i}-b_{i}|\\
\leq & C\eta^{m(1+\alpha )}\leq C\eta^{(m+1)(1+\alpha )}\leq C|x|^{1+\alpha}.
\end{aligned}
\end{equation*}
That is, \cref{e.C1a-1-i-mu} holds. Similar arguments can be made for other pointwise regularity and we will omit them in the proofs.
Now, we prove \cref{In-e.tC1a-3-mu} and \cref{In-e.tC1a-4-mu} by induction. For $m=0$, by setting $P_0=P_{-1}\equiv 0$, the conclusion holds clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{In-e.tC1a-v1-mu}
v(y)=\frac{u(x)-P_{m_0}(x)}{r^{1+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{In-e.C1as-F-mu}
\tilde{F}(D^2v, Dv, v,y)=\tilde{f} ~~\mbox{in}~~ B_1,
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{1}{r^{\alpha-1}}F(r^{\alpha-1}M,r^{\alpha}p+DP_{m_0}(x),r^{1+\alpha}s
+P_{m_0}(x),x),\tilde{f}(y)=\frac{f(x)}{r^{\alpha-1}}.
\end{aligned}
\end{equation*}
In addition, we define $\tilde{F}_0(M)=r^{1-\alpha}F_0(r^{\alpha-1}M)$ for $M\in S^n$.
In the following, we show that \cref{In-e.C1as-F-mu} satisfies the assumptions of \Cref{In-l-C1a-mu}. First, it is easy to verify that (note that we have assumed that $F_0(0)=0$)
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(B_1)}\leq 1, ~(\mathrm{by}~ \cref{In-e.tC1a-3-mu} ~\mathrm{and}~ \cref{In-e.tC1a-v1-mu})\\
&\|\tilde{f}\|_{L^{n}(B_1)}= r^{-\alpha}\|f\|_{L^{n}(B_r)}\leq \|f\|_{C^{-1,\alpha}(0)}\leq \delta, ~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu})\\
&\tilde{F}_0(0)=r^{1-\alpha}F_0(0)=0.
\end{aligned}
\end{equation*}
By \cref{In-e.tC1a-4-mu}, there exists a constant $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$ such that $|P_{m}(0)|+|DP_m(0)|\leq C_0$ ($\forall~0\leq m\leq m_0$). Then it is easy to verify that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu}$, $\tilde{b}$, $\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation}\label{In-e.tC1a-n2-mu}
\begin{aligned}
\tilde{\mu}=r^{1+\alpha}\mu,~\tilde{b}(y)=rb(x)+2C_0r\mu,~
\tilde{c}(y)=r^{1-\alpha}c(x),~
\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{aligned}
\end{equation}
Hence,
\begin{equation*}
\begin{aligned}
&\tilde{\mu}= r^{1+\alpha}\mu\leq \delta, ~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu} ~\mathrm{and}~ \cref{In-e.tC1a-n2-mu})\\
&\|\tilde{b}\|_{L^{p}(B_1)}\leq r^{1-\frac{n}{p}}\|b\|_{L^{p}(B_r)}+2C_0|B_1|^{1/p}r\mu\leq r^{\alpha}\|b\|_{L^{p}(B_1)}+4C_0\mu \leq \delta,~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu} ~\mathrm{and}~ \cref{In-e.tC1a-n2-mu})\\
&\|\tilde{c}\|_{L^n(B_1)}=r^{-\alpha}\|c\|_{L^n(B_r)}\leq \|c\|_{C^{-1,\alpha}(0)}\leq \delta,~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu} ~\mathrm{and}~ \cref{In-e.tC1a-n2-mu})\\
&\tilde{\omega}_0(1,1)=\omega_0(1+C_0,1)\leq \delta. ~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu}~\mathrm{and}~ \cref{In-e.tC1a-n2-mu})\\
\end{aligned}
\end{equation*}
Finally, we compute the oscillation of $\tilde{F}$ in $y$:
\begin{equation*}\label{In-e.tC1a-n2-mu-2}
\begin{aligned}
|\tilde{F}&(M,0,0,y)-\tilde{F}_0(M)|\\
=&r^{1-\alpha}|F(r^{\alpha-1}M,DP_{m_0},P_{m_0}(x),x)
-F_0(r^{\alpha-1}M)|\\
=&r^{1-\alpha}|F(r^{\alpha-1}M,DP_{m_0},P_{m_0}(x),x)
-F(r^{\alpha-1}M,0,0,x)\\
&+F(r^{\alpha-1}M,0,0,x)-
F_0(r^{\alpha-1}M)|\\
\leq & r^{1-\alpha} (C_0^2\mu+C_0 b(x)+c(x)\omega_0(C_0,C_0))
+K_0\beta_1(x)\omega_1(\|M\|)+r^{1-\alpha}\gamma_1(x)\\
:=&\tilde{\beta}_1(y)\omega_1(\|M\|)+\tilde{\gamma}_1(y),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_1(y)=K_0\beta_1(x),
\tilde{\gamma}_1(y)=r^{1-\alpha}\gamma_1(x)+ r^{1-\alpha} (C_0^2\mu+C_0 b(x)+c(x)\omega_0(C_0,C_0)).\\
\end{equation*}
Then
\begin{equation*}\label{In-e.tC1a-n22-mu}
\begin{aligned}
\|\tilde{\beta}_1\|_{L^n(B_1)}=&K_0r^{-1}\|\beta_1\|_{L^n(B_r)}
\leq K_0 \|\beta_1\|_{C^{-1,1}(0)}\leq \delta, ~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu}~\mathrm{and}~ \cref{In-e.tC1a-n2-mu}),\\
\|\tilde{\gamma}_1\|_{L^n(B_1)}\leq& r^{-\alpha}\|\gamma_1\|_{L^n(B_r)}+
C_0^2|B_1|^{1/n}r^{1-\alpha}\mu+C_0|B_1|^{\alpha/n}\|b\|_{L^{p}(B_r)}\\
&+\omega_0(C_0,C_0)r^{-\alpha}\|c\|_{L^n(B_r)}\\
\leq & \|\gamma_1\|_{C^{-1,\alpha}(0)}+2C_0^2\mu
+2C_0\|b\|_{L^{p}(B_1)}+\omega_0(C_0,C_0)\|c\|_{C^{-1,\alpha}(0)}\\
\leq& \delta.~(\mathrm{by}~ \cref{In-e.tC1a-ass-mu} ~\mathrm{and}~ \cref{In-e.tC1a-n2-mu})\\
\end{aligned}
\end{equation*}
Therefore, \cref{In-e.C1as-F-mu} satisfies the assumptions of \Cref{In-l-C1a-mu} and there exists a linear polynomial $\tilde{P}$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{P}\|_{L^{\infty }(B _{\eta})}&\leq \eta ^{1+\alpha}
\end{aligned}
\end{equation*}
and
\begin{equation*}
|D\tilde{P}(0)|\leq \bar{C}.
\end{equation*}
Let $P_{m_0+1}(x)=P_{m_0}(x)+r^{1+\alpha}\tilde{P}(y)$. Then \cref{In-e.tC1a-4-mu} holds for $m_0+1$. Recalling \cref{In-e.tC1a-v1-mu}, we have
\begin{equation*}
\begin{aligned}
&\|u-P_{m_0+1}\|_{L^{\infty}(B_{\eta^{m_0+1}})}\\
&= \|u-P_{m_0}-r^{1+\alpha}\tilde{P}\|_{L^{\infty}(B_{\eta r})}\\
&= \|r^{1+\alpha}v-r^{1+\alpha}\tilde{P}\|_{L^{\infty}(B_{\eta})}\\
&\leq r^{1+\alpha}\eta^{1+\alpha}=\eta^{(m_0+1)(1+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{In-e.tC1a-3-mu} holds for $m=m_0+1$. By induction, the proof is completed.
Finally, we consider the special case, i.e., $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}. Set
\begin{equation*}
K=\|u\|_{L^{\infty}(B_1)}
+\delta^{-1}\left(\|f\|_{C^{-1,\alpha}(0)}+4\|\gamma\|_{C^{-1,\alpha}(0)}\right)
\end{equation*}
and define for $0<\rho<1$
\begin{equation*}
\bar{u}(y)=\frac{u(x)-u(0)}{K},
\end{equation*}
where $y=x/\rho$. By taking $\rho$ small enough (depending only on $n,\lambda,\Lambda,\alpha, \|b\|_{L^{p}(B_1)}$ and $\|c\|_{C^{-1,\alpha}(0)}$), the assumptions \cref{In-e.tC1a-ass-mu} can be guaranteed. Then, by the same proof, we have the explicit estimates \cref{e.C1a-1-i} and \cref{e.C1a-2-i}. \qed~\\
\begin{remark}\label{r-4.3}
The proof is also called ``scaling argument'' in the regularity theory. Essentially, it is just a sequence of repetition of \Cref{In-l-C1a-mu} in different scales.
\end{remark}
\begin{remark}\label{r-4.4}
Above shows clearly that our proof is simple. Moreover, the result is optimal (it means that we don't impose any additional conditions on the operator, coefficients and the prescribed data) since the proof depends only on the scaling property of the equation.
\end{remark}
\begin{remark}\label{r-4.7}
Taking the transformation form \cref{e.4.1} is motivated by \cite{MR3980853} (see the definition of $\tilde{u}$ in Claim 3.2), which can be tracked to \cite{MR1139064} (see Section 1.3 there).
\end{remark}
\section{Interior $C^{2,\alpha}$ regularity}\label{In-C2a-mu}
In this section, we prove the interior pointwise $C^{2,\alpha}$ regularity \Cref{t-C2a-i}. For simplicity, we call a function $Q:R^n\rightarrow R$ a $k$-form ($k\geq 1$) if $Q$ can be written as
\begin{equation*}
Q(x)=\frac{1}{k!}\cdot\sum_{1\leq i_1,...,i_{k}\leq n}a_{i_1\cdots i_{k}}x_{i_1}\cdots x_{i_{k}},
\end{equation*}
where $a_{i_1\cdots i_{k}}$ are constants. In addition, we define
\begin{equation*}
\|Q\|=\sum_{1\leq i_1,...,i_{k}\leq n}|a_{i_1\cdots i_{k}}|.
\end{equation*}
The following lemma is the ``key step'' for interior pointwise $C^{2,\alpha}$ regularity, which is similar to \Cref{In-l-C1a-mu}.
\begin{lemma}\label{In-l-C2a-mu}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF}. For any $0<\alpha<\bar{\alpha}$, there exists $\delta>0$ depending only on $n,\lambda,\Lambda,\alpha$ and $\omega_2$ such that if $\omega_0$ satisfies \cref{e.omega0} and $u$ satisfies
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1
\end{equation*}
with $\|u\|_{L^{\infty}(B_1)}\leq 1$, $u(0)=0$, $Du(0)=0$, $\mu,b_0,c_0\leq \delta$, $\omega_0(1,1)\leq 1$, $\|\beta_2\|_{L^{\infty}(B_1)}\leq \delta$ and $\|f\|_{L^{\infty}(B_1)}\leq \delta$, then there exists a $2$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(B_{\eta})}\leq \eta^{2+\alpha},
\end{equation*}
\begin{equation*}
F_0(D^2Q,0,0)=0
\end{equation*}
and
\begin{equation*}
\|Q\|\leq \bar{C}+1,
\end{equation*}
where $\eta$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
\end{lemma}
\begin{remark}\label{r-4.1}
Note that by \Cref{t-C1a-i}, $u\in C^{1,\alpha}(0)$. Hence, $Du(0)$ is well defined.
\end{remark}
\begin{remark}\label{r-4.6}
Of course, we can use $L^n$ norm rather $L^{\infty}$ norm to measure $\beta_2$ and $f$. Then we can obtain the pointwise $C^{2,\alpha}$ regularity under the assumptions presented in \Cref{r-2.5}.
\end{remark}
\begin{remark}\label{r-4.5}
If we only intend to derive that there exists a second order polynomial $P$ such that
\begin{equation*}
\|u-P\|_{L^{\infty}(B_{\eta})}\leq \eta^{2+\alpha},
\end{equation*}
it is enough to make the same assumptions as that of \Cref{In-l-C1a-mu}. It indicates that we can approximate the solution in some scale only if the coefficients and the prescribed data are small in ``soft'' norms. The requirements that $\beta_2\in C^{\alpha}(0)$ and $f\in C^{\alpha}(0)$ are only used in the ``scaling argument'' (see \Cref{In-t-C2as-mu} below).
\end{remark}
\proof Similar to \Cref{In-l-C1a-mu}, we prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<\bar{\alpha},\omega_2$ and sequences of $F_m,u_m,f_m$ satisfying
\begin{equation*}
F_m(D^2u_m, Du_m, u_m,x)=f_m ~~\mbox{in}~~ B_1.
\end{equation*}
In addition, $F_m$ are convex in $M$ and satisfy the structure condition \cref{SC2} with $\lambda,\Lambda$, $\mu_m$, $b_m$, $c_m$ and $\omega_m$. Moreover, $F_m$ satisfy \cref{e.C2a-KF} with $F_{0m},\beta_m$ and $\omega_2$. Furthermore, $\|u_m\|_{L^{\infty}(B_1)}\leq 1$, $u_m(0)=0$, $Du_m(0)=0$, $\mu_m\leq 1/m$, $b_m\leq 1/m$, $c_m\leq 1/m$, $\|\beta_{m}\|_{L^{\infty}(B_1)}\leq 1/m$ and $\|f_m\|_{L^{\infty}(B_1)}\leq 1/m$. Finally, for any $2$-form $Q$ satisfying $\|Q\|\leq \bar{C}+1$ and
\begin{equation}\label{In-e.lC2a-3-mu}
F_{0m}(D^2Q,0,0)=0,
\end{equation}
we have
\begin{equation}\label{In-e.lC2a-1-mu}
\|u_m-Q\|_{L^{\infty}(B_{\eta})}> \eta^{2+\alpha},
\end{equation}
where $0<\eta<1$ is taken small such that
\begin{equation}\label{In-e.lC2a-2-mu}
\bar{C}\eta^{\bar{\alpha}-\alpha}<1/2.
\end{equation}
As before, $u_m$ are uniformly bounded and equicontinuous and then there exist a subsequence (denoted by $u_m$ again) and $u:B_1\rightarrow R$ such that $u_m\rightarrow u$ in $L^{\infty}_{\mathrm{loc}}(B_1)$. In addition, since $F_{0m}(0,0,0)=0$ and $F_{0m}$ are Lipschitz continuous in $M$ with a uniform Lipschitz constant, there exist a subsequence (denoted by $F_{0m}$ again) and $F:S^n\rightarrow R$ such that $F_{0m}(\cdot,0,0)\rightarrow F$ in $L^{\infty}_{\mathrm{loc}}(S^n)$. Furthermore, for any ball $B\subset\subset B_1$ and $\varphi\in C^2(\bar{B})$, let $r=\|D^2\varphi\|_{L^{\infty}(B)}+\|D\varphi\|_{L^{\infty}(B)}+1$, $G_m=F_m(D^2\varphi,D\varphi,u_m,x)-f_m$ and $G=F(D^2\varphi)$. Then,
\begin{equation*}\label{In-gkC2a-mu}
\begin{aligned}
\|G&_m-G\|_{L^n(B)}\\
\leq& \|F_m(D^2\varphi,D\varphi,u_m,x)-F(D^2\varphi)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
=& \|F_m(D^2\varphi,D\varphi,u_m,x)-F_{0m}(D^2\varphi,D\varphi,u_m)
+F_{0m}(D^2\varphi,D\varphi,u_m)\\
&-F_{0m}(D^2\varphi,0,0)+F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
\leq& \|\beta_{m}\omega_2(r,r,r)+r^2\mu_m+rb_m+c_m\omega_m(1,1)\|_{L^n(B)}\\
&+\|F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^{n}(B)}+\|f_m\|_{L^n(B)}\\
\leq& \omega_2(r,r,r)\|\beta_m\|_{L^{n}(B)}
+C(r^2\mu_m+rb_m+c_m)\\
&+\|F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^{n}(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$.
By \Cref{l-35}, $u$ is a viscosity solution of
\begin{equation*}
F(D^2u)=0 ~~\mbox{in}~~ B_1.
\end{equation*}
From the $C^{1,\alpha}$ estimate for $u_m$ (see \Cref{t-C1a-i}) and noting $u_m(0)=0$ and $Du_m(0)=0$, we have
\begin{equation*}
|u_m(x)|\leq C|x|^{1+\alpha}, ~~~~\forall~x\in B_{1},
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda,\alpha$ and $\omega_2$. Since $u_m$ converges to $u$,
\begin{equation*}
|u(x)|\leq C|x|^{1+\alpha}, ~~~~\forall~x\in B_{1},
\end{equation*}
which implies $u(0)=0$ and $Du(0)=0$.
By \Cref{l-3modin2}, there exists a $2$-form $\tilde{Q}$ such that
\begin{equation}\label{In-e.c2a-3-mu}
|u(x)-\tilde{Q}(x)|\leq \bar{C} |x|^{2+\bar{\alpha}}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation*}
F(D^2\tilde{Q})=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}.
\end{equation*}
Combining \cref{In-e.lC2a-2-mu} and \cref{In-e.c2a-3-mu}, we have
\begin{equation}\label{In-e.lC2a-3-mu}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta})}\leq \eta^{2+\alpha}/2.
\end{equation}
Since $F_{0m}(D^2\tilde{Q},0,0)\rightarrow F(D^2\tilde{Q})=0$, there exist $t_m$ with $t_m\rightarrow 0$ and $|t_m|\leq 1$ (for $m$ large) such that
\begin{equation*}
F_{0m}(D^2\tilde{Q}+t_m\delta_{nn},0,0)=0,
\end{equation*}
where $\delta_{nn}$ denotes a matrix $a_{ij}$ whose elements are all $0$ except $a_{nn}=1$ (similarly hereinafter). Let $Q_m(x)=\tilde{Q}(x)+t_mx_n^2/2$ be a $2$-form and $\|Q_m\|\leq \bar{C}+1$.
Hence, \cref{In-e.lC2a-1-mu} holds for $Q_m$. That is,
\begin{equation*}
\|u_m-Q_m\|_{L^{\infty}(B_{\eta})}> \eta^{2+\alpha}.
\end{equation*}
Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta})}\geq \eta^{2+\alpha},
\end{equation*}
which contradicts with \cref{In-e.lC2a-3-mu}. ~\qed~\\
\begin{remark}\label{r-2.2}
Usually, \Cref{In-l-C2a-mu} is proved by solving an auxiliary solution $v$ and considering the equation satisfied by $u-v$ (see \cite[Lemma 7.9]{MR1351007}). As pointed out in \cite[Chapte 1.3]{Wang_Regularity}, the benefits of the method of compactness are that we don't need the solvability and $u-v$ satisfying some equation is not necessary.
\end{remark}
Next, we show the interior pointwise $C^{2,\alpha}$ regularity under very ``strong'' conditions.
\begin{theorem}\label{In-t-C2as-mu}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF}. Let $0<\alpha <\bar{\alpha}$, $\omega_0$ satisfy \cref{e.omega0} and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Assume that
\begin{equation}\label{In-e.C2as-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(B_1)}\leq 1,~ u(0)=0,~ Du(0)=0, \\
&\mu\leq \frac{\delta_1}{4C_0},~ b_0\leq \frac{\delta_1}{2},~ c_0\leq \frac{\delta_1}{K_0},~
\omega_0(1+C_0,1)\leq 1,\\
&|\beta_2(x)|\leq \delta_1|x|^{\alpha}~~\mbox{and}~~|f(x)|\leq \delta_1|x|^{\alpha}, ~~\forall ~x\in B_1,\\
\end{aligned}
\end{equation}
where $\delta_1$ depends only on $n,\lambda,\Lambda,\alpha,\omega_0$ and $\omega_2$, and $C_0$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
Then $u\in C^{2,\alpha}(0)$, i.e., there exists a $2$-form $Q$ such that
\begin{equation}\label{In-e.C2as-1-mu}
|u(x)-Q(x)|\leq C |x|^{2+\alpha}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{In-e.C2as-3-mu}
F_0(D^2Q,0,0)=0
\end{equation}
and
\begin{equation}\label{In-e.C2as-2-mu}
\|Q\| \leq C,
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda$ and $\alpha$.
\end{theorem}
\begin{remark}\label{r-5.1}
Similar to the proof of $C^{1,\alpha}$ regularity, \Cref{In-t-C2as-mu} is the ``scaling argument'', which is just the consequence of a sequence of scaling versions of \Cref{In-l-C2a-mu}.
\end{remark}
\proof To prove that $u$ is $C^{2,\alpha}$ at $0$, we only need to prove the following. There exist a sequence of $2$-forms $Q_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{In-e.C2as-4-mu}
\|u-Q_m\|_{L^{\infty }(B _{\eta^{m}})}\leq \eta ^{m(2+\alpha )},
\end{equation}
\begin{equation}\label{In-e.C2as-5-mu}
F_0(D^2Q_m,0,0)=0
\end{equation}
and
\begin{equation}\label{In-e.C2as-6-mu}
\|Q_m-Q_{m-1}\|\leq (\bar{C}+1)\eta ^{(m-1)\alpha},
\end{equation}
where $\eta$ depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{In-l-C2a-mu}.
We prove above conclusion by induction. For $m=0$, by setting $Q_0=Q_{-1}\equiv 0$, \crefrange{In-e.C2as-4-mu}{In-e.C2as-6-mu} hold clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{In-e.C2as-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^{2+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{In-e.C2as-F-mu}
\tilde{F}(D^2v,Dv,v,y)=\tilde{f} ~~\mbox{in}~~ B_1,
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{1}{r^{\alpha}}F(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p
+DQ_{m_0}(x),r^{2+\alpha}s+Q_{m_0}(x),x),\\
&\tilde{f}(y)=\frac{f(x)}{r^{\alpha}}.\\
\end{aligned}
\end{equation*}
In addition, define
\begin{equation*}
\tilde{F}_0(M,p,s)=\frac{1}{r^{\alpha}}F_0(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p
,r^{2+\alpha}s).
\end{equation*}
In the following, we show that \cref{In-e.C2as-F-mu} satisfies the assumptions of \Cref{In-l-C2a-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(B_1)}\leq 1, v(0)=0, Dv(0)=0,
~~(\mathrm{by}~ \cref{In-e.C2as-be-mu},~ \cref{In-e.C2as-4-mu} ~\mbox{and}~ \cref{In-e.C2as-v-mu})\\
&\|\tilde{f}\|_{L^{\infty}(B_1)}=\frac{\|f\|_{L^{\infty}(B_r)}}{r^{\alpha}}\leq \delta_1,
~(\mathrm{by}~ \cref{In-e.C2as-be-mu})\\
&\tilde{F}_0(0,0,0)=r^{-\alpha}F_0(D^2Q_{m_0},0,0)=0.~(\mathrm{by}~ \cref{In-e.C2as-5-mu})\\
\end{aligned}
\end{equation*}
By \cref{In-e.C2as-6-mu}, we can choose a constant $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$ such that $\|Q_m\|\leq C_0$ ($\forall~0\leq m\leq m_0$). Then $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde\omega_0$, where (note that $\omega_0$ satisfies \cref{e.omega0})
\begin{equation*}
\tilde{\mu}=r^{2+\alpha}\mu,~~\tilde{b}=rb_0+2C_0r\mu,~~\tilde{c}=K_0 r^{\alpha+\alpha^2}c_0
~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence, $\tilde{\omega}_0$ satisfies \cref{e.omega0} and from \cref{In-e.C2as-be-mu},
\begin{equation*}
\begin{aligned}
\tilde{\mu}\leq \mu\leq \delta_1,~\tilde{b}\leq b_0+2C_0\mu\leq \delta_1,~
\tilde{c}\leq K_0 c_0\leq \delta_1~~\mbox{and}~~\tilde{\omega}_0(1,1)\leq 1.
\end{aligned}
\end{equation*}
Finally, by combining \cref{SC2}, \cref{e.C2a-KF} and \cref{In-e.C2as-be-mu}, we have
\begin{equation*}
\begin{aligned}
|\tilde{F}&(M,p,s,y)-\tilde{F}_0(M,p,s)|\\
=&r^{-\alpha}|F(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p+DQ_{m_0}(x),
r^{2+\alpha}s+Q_{m_0}(x),x)\\
&-F_0(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p,r^{2+\alpha}s)|\\
\leq& r^{-\alpha}|F(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p
+DQ_{m_0}(x),r^{2+\alpha}s+Q_{m_0}(x),x)\\
&-F(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p,r^{2+\alpha}s,x)|\\
&+r^{-\alpha}|F(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p,r^{2+\alpha}s,x)
-F_0(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p,r^{2+\alpha}s)|\\
\leq &r^{-\alpha}\Big(2C_0r^{1+\alpha}\mu |p||x|+ C_0^2\mu|x|^2+C_0 b_0|x|+K_0 c_0\omega_0(|s|+C_0,C_0)|x|^{2\alpha}\\
&+\beta_2(x)\omega_2(\|M\|+C_0,|p|,|s|)\Big)\\
\leq& \delta_1\omega_2(\|M\|+C_0,|p|,|s|)+\delta_1|p|+\delta_1C_0
+\delta_1\omega_0(|s|+C_0,C_0)\\
:=& \tilde{\beta}_2(y)\tilde{\omega}_2(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_2(y)\equiv \delta_1$ and $\tilde{\omega}_2(\|M\|,|p|,|s|)=
\omega_2(\|M\|+C_0,|p|,|s|)+|p|+C_0+\omega_0(|s|+C_0,C_0)$. Then $\tilde{\omega}_2$ satisfies \cref{e.omega} and
\begin{equation*}
\|\tilde{\beta}_2\|_{L^{\infty}(B_1)}\leq \delta_1.
\end{equation*}
Choose $\delta_1$ small enough (depending only on $n,\lambda,\Lambda,\alpha,\omega_0$ and $\omega_2$) such that \Cref{In-l-C2a-mu} holds for $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. Since \cref{In-e.C2as-F-mu} satisfies the assumptions of \Cref{In-l-C2a-mu}, there exists a $2$-form $\tilde{Q}(y)$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(B_{\eta})}&\leq \eta ^{2+\alpha},
\end{aligned}
\end{equation*}
\begin{equation*}
\tilde{F}_0(D^2\tilde{Q},0,0)=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}+1.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+r^{\alpha}\tilde{Q}(x)$. Then \cref{In-e.C2as-5-mu} and \cref{In-e.C2as-6-mu} hold for $m_0+1$. Recalling \cref{In-e.C2as-v-mu}, we have
\begin{equation*}
\begin{aligned}
\|u-Q_{m_0+1}(x)\|_{L^{\infty}(B_{\eta^{m_0+1}})}&= \|u-Q_{m_0}(x)-r^{\alpha}\tilde{Q}(x)\|_{L^{\infty}(B_{\eta r})}\\
&= \|r^{2+\alpha}v-r^{2+\alpha}\tilde{Q}(y)\|_{L^{\infty}(B_{\eta})}\\
&\leq r^{2+\alpha}\eta^{2+\alpha}=\eta^{(m_0+1)(2+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{In-e.C2as-4-mu} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Now, we can give the~\\
\noindent\textbf{Proof of \Cref{t-C2a-i}.} In fact, \Cref{In-t-C2as-mu} has contained the essential ingredients for the $C^{2,\alpha}$ regularity. In some sense, the following proof is just a normalization procedure that makes the assumptions in \Cref{In-t-C2as-mu} satisfied. We prove \Cref{t-C2a-i} in two cases.
\textbf{Case 1:} the general case, i.e., $F$ satisfies \cref{SC2} and \cref{e.C2a-KF}. Throughout the proof for this case, $C$ always denotes a constant depending only on $n, \lambda,\Lambda,\alpha,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\alpha}(0)}$, $\omega_2$, $\|f\|_{C^{\alpha}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$.
Let $F_1(M,p,s,x)=F(M,p,s,x)-f(0)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar B_1$. Then $u_1$ satisfies
\begin{equation*}
F_1(D^2u,Du,u,x)=f_1~~\mbox{in}~~B_1,
\end{equation*}
where $f_1(x)=f(x)-f(0)$. Hence,
\begin{equation*}
|f_1(x)|\leq C|x|^{\alpha}, ~~\forall ~x\in B_1.
\end{equation*}
Set $u_1(x)=u(x)-u(0)-Du(0)\cdot x$ and $F_2(M,p,s,x)=F_1(M,p+Du(0),s+u(0)+Du(0)\cdot x,x)$. Then $u_1$ satisfies
\begin{equation*}
F_2(D^2u_1,Du_1,u_1,x)=f_1~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_1(0)=0, Du_1(0)=0.
\end{equation*}
Next, take $u_2(x)=u_1(x)-\tau x_n^2$ and $F_3(M,p,s,x)=F_2(M+2\tau\delta_{nn},p+2\tau x_n,s+\tau x_n^2,x)$ for $\tau\in R$. Then $u_2$ satisfies
\begin{equation*}
F_3(D^2u_2,Du_2,u_2,x)=f_1~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_2(0)=0, Du_2(0)=0.
\end{equation*}
Define the fully nonlinear operators $F_{10},F_{20}$ and $F_{30}$ in a similar way as $F_1,F_2$ and $F_3$ (only replacing $F$ by $F_0$ and setting $x=0$ in the expression). By the structure condition, there exists $\tau\in R$ such that $F_{30}(0,0,0)=0$ and
\begin{equation}\label{In-e.tC2a-2-mu}
\begin{aligned}
|\tau|&\leq |F_{20}(0,0,0)|/\lambda\leq |F_0(0,Du(0),u(0))-f(0)|/\lambda\leq C.
\end{aligned}
\end{equation}
For $\rho\in R$, define $y=x/\rho$, $u_3(y)=u_2(x)/\rho$, $F_4(M,p,s,y)=\rho F_3(M/\rho, p, \rho s,x)$ and $F_{40}(M,p,s)=\rho F_{30}(M/\rho, p,\rho s)$. Then $u_3$ satisfies
\begin{equation}\label{In-F5-mu}
F_4(D^2u_3,Du_3,u_3,y)=f_2~~\mbox{in}~~B_1,
\end{equation}
where $f_2(y)=\rho f_1(x)$.
Now, we can check that \cref{In-F5-mu} satisfies the conditions of \Cref{In-t-C2as-mu} by choosing a proper $\rho$. First, it can be verified easily that
\begin{equation*}
\begin{aligned}
u_3(0)=0 ,Du_3(0)=0~\mbox{and}~|f_2(y)|&= \rho|f_1(x)|\leq C\rho^{1+\alpha}|y|^{\alpha}~\mbox{in}~B_1.
\end{aligned}
\end{equation*}
Next, by the interior $C^{1,\alpha}$ regularity for $u$,
\begin{equation*}
\|u_3\|_{L^{\infty}(B_1)}\leq \rho^{-1} \|u_2\|_{L^{\infty}(B_{\rho})}
\leq \rho^{-1}\left( \|u_1\|_{L^{\infty}(B_{\rho})}+C\rho^{2}\right)
\leq \rho^{-1}\left(C\rho^{1+\alpha}+C\rho^{2}\right)\leq C\rho^{\alpha}.
\end{equation*}
Furthermore, $F_{40}(0,0,0)=\rho F_{30}(0,0,0)=0$, and $F_{40}$ and $F_{30}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde\omega_0$, where
\begin{equation*}
\tilde{\mu}= \rho\mu,~~
\tilde{b}= \rho b_0+C\rho\mu,~~\tilde{c}= \rho^{1/2} c_0~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\rho^{1/2} \omega_0(\cdot+C,\cdot).
\end{equation*}
Then $\tilde{\omega}_0$ satisfies \cref{e.omega0}.
Finally, we check the oscillation of $F_4$ in $y$.
\begin{equation*}
\begin{aligned}
|F&_4(M,p,s,y)-F_{40}(M,p,s)|\\
=&\rho\Big(F(\rho^{-1}M
+2\tau\delta_{nn},p+2\tau x_n+Du_1(0),\rho s+\tau x_n^2+u_1(0)+Du_1(0)\cdot x,x)\\
&-F_0(\rho^{-1}M+2\tau\delta_{nn},p+Du_1(0),\rho s+u_1(0))\Big)\\
\leq &\rho\Big(F(\rho^{-1}M+2\tau\delta_{nn},p+2\tau x_n+Du_1(0),
\rho s+\tau x_n^2+u_1(0)+Du_1(0)\cdot x,x)\\
&-F(\rho^{-1}M+2\tau\delta_{nn},p+Du_1(0),\rho s+u_1(0),x)\Big)\\
&+\rho\Big(F(\rho^{-1}M+2\tau\delta_{nn},p+Du_1(0),\rho s+u_1(0),x)\\
&-F_0(\rho^{-1}M+2\tau\delta_{nn},p+Du_1(0),\rho s+u_1(0))\Big)\\
\leq& \rho\beta_2(x)\omega_2(\rho^{-1}\|M\|+C,|p|+C,\rho|s|+C)+C\rho\mu(2|p|+C|x|+C)|x|\\
&+C\rho b_0|x|+K_0\rho c_0\omega_0(|s|+C,C)|x|^{\alpha}\\
\leq& \rho^{\alpha}\Big(C\omega_2\Big(\|M\|+C,|p|+C,|s|+C\Big)
+C|p|+C+C\omega_0(|s|+C,C)\Big)|y|^{\alpha}\\
:= &\tilde{\beta}_2(y)\tilde{\omega}_2\Big(\|M\|,|p|,|s|\Big),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_2(y)=C\rho^{\alpha}|y|^{\alpha}$,
\begin{equation*}
\tilde{\omega}_2(\|M\|,|p|,|s|)=\omega_2\left(\|M\|+C,|p|+C,|s|+C\right)
+C|p|+C+C\omega_0(|s|+C,C).
\end{equation*}
Clearly, $\tilde{\omega}_2$ satisfies \cref{e.omega}.
Take $\delta_1$ small enough such that \Cref{In-t-C2as-mu} holds with $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $n, \lambda,\Lambda,\alpha,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\alpha}(0)}$, $\omega_2$, $\|f\|_{C^{\alpha}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$) such that
\begin{equation*}
\begin{aligned}
&\|u_3\|_{L^{\infty }(B_1)}\leq 1,~\tilde\mu \leq \frac{\delta_1}{4C_0},~
\tilde{b}\leq \frac{\delta_1}{2},~ \tilde c\leq \frac{\delta_1}{K_0},~
\tilde{\omega}_0(1+C_0,1)\leq 1,\\
&|\tilde\beta_2(y)|\leq \delta_1|y|^{\alpha}~~\mbox{and}~~|f_2(y)|\leq \delta_1|y|^{\alpha}, ~~\forall ~y\in B_1,
\end{aligned}
\end{equation*}
where $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{In-t-C2as-mu}. Therefore, the assumptions in \Cref{In-t-C2as-mu} are satisfied for \cref{In-F5-mu}. By \Cref{In-t-C2as-mu}, $u_3$ and hence $u$ is $C^{2,\alpha}$ at $0$, and the estimates \crefrange{e.C2a-1-i-mu}{e.C2a-2-i-mu} hold.
\textbf{Case 2:} $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0}. Let $K=\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\alpha}(0)}+\|\gamma_2\|_{C^{\alpha}(0)}$ and $u_1=u/K$. Then $u_1$ satisfies
\begin{equation}\label{e.5.1}
F_1(D^2u_1,Du_1,u_1,x)=f_1~~\mbox{in}~~B_1,
\end{equation}
where $F_1(M,p,s,x)=F(KM,Kp,Ks,x)/K$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar B_1$ and $f_1=f/K$.
Obviously,
\begin{equation*}
\|u_1\|_{L^{\infty }(B_1)}\leq 1~~\mbox{and}~~\|f_1\|_{C^{\alpha}(0)}\leq 1.
\end{equation*}
Also, $F_1$ satisfies the structure condition \cref{SC1} with the same $\lambda,\Lambda,b_0$ and $c_0$. In addition, define $F_{10}(M,p,s)=F_0(KM,Kp,Ks)/K$. Then,
\begin{equation*}
\begin{aligned}
|F_1(M,p,s,x)-F_{10}(M,p,s)|= & K^{-1}|F(KM,Kp,Ks,x)-F_0(KM,Kp,Ks)|\\
\leq & \beta_2(x)\left(\|M\|+|p|+|s|\right)+\tilde\gamma_2(x),
\end{aligned}
\end{equation*}
where $\tilde\gamma_2(x)=K^{-1}\gamma_2(x)$ and hence $\|\tilde{\gamma}_2\|_{C^{\alpha}(0)}\leq 1$.
Apply \textbf{Case 1} to \cref{e.5.1}, we obtain that $u_1$ and hence $u$ is $C^{2,\alpha}$ at $0$ and the estimates \crefrange{e.C2a-1-i}{e.C2a-2-i} hold. \qed~\\
\begin{remark}\label{r-6-2}
In fact, from above proof, we obtain that there exists a quadratic polynomial $P$ such that
\begin{equation*}
F_0(D^2P,0,0)=f(0).
\end{equation*}
Since $F$ satisfies \cref{e.C2a-KF},
\begin{equation*}
\begin{aligned}
|F&(D^2P,DP(x),P(x),x)-f(0)|\\
\leq& |F(D^2P,DP(x),P(x),x)-F_0(D^2,DP(x),P(x))|\\
&+|F_0(D^2,DP(x),P(x))-F_0(D^2,0,0)|
+|F_0(D^2,0,0)-f(0)|\\
\leq &C |x|^{\alpha}, ~~\forall ~x\in B_{1}.
\end{aligned}
\end{equation*}
That is, \cref{e.C2a-3-i-mu} holds.
\end{remark}
\section{Interior $C^{k,\alpha}$ regularity}\label{In-Cka-mu}
In this section, we prove the interior pointwise $C^{k,\alpha}$ regularity for $k\geq 3$. First, we prove interior local $C^{k,\alpha}$ regularity (similar to \Crefrange{l-3modin1}{l-32}).
\begin{lemma}\label{In-l-62}
Suppose that $F_0\in C^{k-2,\bar\alpha}(S^n\times R^n\times R\times \bar{B}_1)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Let $u$ satisfy
\begin{equation*}
F_0(D^2u,Du,u,x)=0 ~~\mbox{in}~~B_1.
\end{equation*}
Then $u\in C^{k,\alpha}(\bar{B}_{1/2})$ for any $0<\alpha<\bar{\alpha}$ and
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}_{1/2})}\leq C_k,
\end{equation*}
where $C_k$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,\omega_4$ and $\|u\|_{L^{\infty }(B_1)}$.
In particular, $u\in C^{k,\alpha}(0)$ and there exists a $k$-th order polynomial $P$ such that
\begin{equation}\label{In-e.l62-1}
|u(x)-P(x)|\leq C_k |x|^{k+\alpha}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{In-e.l62-2}
|F_0(D^2P(x),DP(x),P(x),x)|\leq C_k|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{In-e.l62-3}
\|P\|\leq C_k.
\end{equation}
\end{lemma}
\proof Since $F_0\in C^{k-2,\bar{\alpha}}$, for any $x_0,x\in B_1$, $(M,p,s)\in S^n\times R^n\times R$,
\begin{equation*}
\begin{aligned}
&|F_0(M,p,s,x)-F_0(M,p,s,x_0)|\\
&\leq |F_{0,x_i}(M,p,s,\xi)||x_i-x_{0i}|\\
&\leq (|F_{0,x_i}(M,p,s,\xi)-F_{x_i}(0,0,0,0)|+|F_{0,x_i}(0,0,0,0)|)|x_i-x_{0i}|\\
&\leq C\left(\|M\|^{\bar\alpha}+|p|^{\bar\alpha}+|s|^{\bar\alpha}+1\right)|x-x_0|\\
&:=\tilde{\beta}_2(x,x_0)\tilde{\omega}_2(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_2(x,x_0)=|x-x_0|,~\tilde{\omega}_2(\|M\|,|p|,|s|)
=C\left(\|M\|^{\bar\alpha}+|p|^{\bar\alpha}+|s|^{\bar\alpha}+1\right)
\end{equation*}
and $C$ depends only on $\omega_4$. Here,
\begin{equation*}
\omega_4(r):=\|F_0\|_{C^{k-2,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{B}_1)},~\forall ~r>0.
\end{equation*}
Moreover, $\omega_0$ satisfies \cref{e.omega0} and $\tilde{\omega}_2$ satisfies \cref{e.omega}. By \Cref{t-C2a-i}, $u\in C^{2,\alpha}(x_0)$ for any $0<\alpha<\bar{\alpha}$ and $x_0\in \bar{B}_{3/4}$. Hence, $u\in C^{2,\alpha}(\bar B_{3/4})$ and
\begin{equation*}
\|u\|_{C^{2,\alpha}(\bar{B}_{3/4})}\leq C,
\end{equation*}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,\omega_4$ and $\|u\|_{L^{\infty }(B_1)}$. Unless stated otherwise, $C$ always has the same dependence in the following proof.
Now, we show that $u\in C^{3,\alpha}$, which can be proved by the standard technique of difference quotient. Let $h>0$ be small and $1\leq l\leq n$. Take the difference quotient on both sides of the equation and we have
\begin{equation*}
a_{ij}(\Delta_l^h u)_{ij}=-G ~~~~\mbox{in}~~ B_{5/8},
\end{equation*}
where
\begin{equation*}
\begin{aligned}
&\Delta_l^h u(x)=\left(u(x+he_l)-u(x)\right)/h,\\
&a_{ij}(x)=\int_{0}^{1}F_{0,ij}(\xi)dr,\\
&G(x)=\int_{0}^{1}F_{0,p_i}(\xi)\Delta_l^h u_i+F_{0,s}(\xi)\Delta_l^h u+F_{0,x_l}(\xi)dr\\
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\xi=r\left(D^2u(x+he_l),Du(x+he_l),u(x+he_l),x+he_l\right)
+(1-r)\left(D^2u(x),Du(x),u(x),x\right).
\end{aligned}
\end{equation*}
Thus, $a_{ij}$ is uniformly elliptic with ellipticity constants $\lambda$ and $\Lambda$. Moreover, it is easy to check that $\|a_{ij}\|_{C^{\alpha^2}(\bar{B}_{5/8})}\leq C$ and $\|G\|_{C^{\alpha^2}(\bar{B}_{5/8})}\leq C$.
By the Schauder's estimate for linear equations, we have $\Delta_l^h u \in C^{2,\alpha^2}(\bar{B}_{1/2})$. Hence, $u_l \in C^{2,\alpha^2}(\bar{B}_{1/2})$ and
\begin{equation*}
\|u_l\|_{C^{2,\alpha^2}(\bar{B}_{1/2})}\leq C,
\end{equation*}
which implies $u\in C^{3,\alpha^2}$ and
\begin{equation*}
\|u\|_{C^{3,\alpha^2}(\bar{B}_{1/2})}\leq C.
\end{equation*}
Since $u\in C^{3,\alpha^2}$, for $1\leq l \leq n$, $u_l$ satisfies
\begin{equation}\label{In-uksat}
a_{ij}(u_l)_{ij}=-G,
\end{equation}
where
\begin{equation}\label{In-aijuk}
a_{ij}(x)=F_{0,ij}(D^2u,Du,u,x)
\end{equation}
and
\begin{equation}\label{In-Guk}
G(x)=F_{0,p_i}(D^2u,Du,u,x)u_{il}+F_{0,s}(D^2u,Du,u,x)u_l+F_{0,x_l}(D^2u,Du,u,x).
\end{equation}
Then it can be seen that $a_{ij}\in C^{\alpha}(\bar{B}_{1/2})$ and $G\in C^{\alpha}(\bar{B}_{1/2})$. By the Schauder's estimate for linear equations again, $u_l\in C^{2,\alpha}(\bar{B}_{1/4})$. Then $u\in C^{3,\alpha}(\bar{B}_{1/4})$ and
\begin{equation*}
\|u\|_{C^{3,\alpha}(\bar{B}_{1/4})}\leq C.
\end{equation*}
Since $F_0\in C^{k-2,\alpha}$, by considering \crefrange{In-uksat}{In-Guk} iteratively, we obtain
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}_{2^{-k}})}\leq C.
\end{equation*}
By a standard covering argument, $u\in C^{k,\alpha}(\bar{B}_{1/2})$ and
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}_{1/2})}\leq C.
\end{equation*}
In particular, $u\in C^{k,\alpha}(0)$ and there exists a $k$-th order polynomial $P$ such that \cref{In-e.l62-1} and \cref{In-e.l62-3} hold. From \cref{In-e.l62-1}, for any $0\leq l\leq k$,
\begin{equation*}\label{In-e.Q-mu}
|D^lu(x)-D^lP(x)|\leq C |x|^{k-l+\alpha}, ~~\forall ~x\in B_{1}.
\end{equation*}
Thus,
\begin{equation*}
\begin{aligned}
&|F_0(D^2P(x),DP(x),P(x),x)|\\
&=|F_0(D^2P(x),DP(x),P(x),x)-F_0(D^2u(x),Du(x),u(x),x)|\\
&=\bigg|\int_{0}^{1} F_{0,ij}(\xi)(P_{ij}-u_{ij})+F_{0,p_i}(\xi)(P_{i}-u_{i})
+F_{0,s}(\xi)(P-u)dr \bigg|\\
&\leq C|x|^{k-2+\alpha},
\end{aligned}
\end{equation*}
where $\xi=r\left(D^2P(x),DP(x),P(x),x\right)+(1-r)\left(D^2u(x),Du(x),u(x),x\right)$. Then
\begin{equation*}
D^{l}_x \Big(F_0(D^2P(x),DP(x),P(x),x)\Big)\Big|_{x=0}=0,~\forall ~0\leq l\leq k-2.
\end{equation*}
Combining with $F_0\in C^{k-2,\bar{\alpha}}$, \cref{In-e.l62-2} holds. \qed~\\
\begin{remark}\label{r-6.1}
In above proof, we don't assume that $F_0(0,0,0,x_0)=0$. Thus, \Cref{t-C2a-i} can not be applied directly. However, this assumption is not essential and we can consider $F_0(M,p,s,x)-F_0(0,0,0,x_0)$.
\end{remark}
\begin{remark}\label{r-6.2}
If we consider the linear equation \cref{e.linear}, the coefficients are independent of $u$. Then we obtain
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}_{1/2})}\leq C_k\|u\|_{L^{\infty }(B_1)},
\end{equation*}
where $C_k$ depends only on $k,n,\lambda,\Lambda,\alpha,
\|a^{ij}\|_{C^{k-2,\alpha}(\bar{B}_1)},\|b^{i}\|_{C^{k-2,\alpha}(\bar{B}_1)}$ and $\|c\|_{C^{k-2,\alpha}(\bar{B}_1)}$.
Based on this estimate, we can obtain the higher order pointwise regularity with explicit estimates as illustrated in \Cref{r-2.10}.
\end{remark}
In the following, we prove the interior pointwise $C^{k,\alpha}(k\geq 3)$ regularity \Cref{t-Cka-i} by induction. For $k=3$, $F\in C^{1,\alpha}(0)$ and $F(M,p,s,0)\equiv F_0(M,p,s,0)$ for any $(M,p,s)\in S^n\times R^n\times R$ (by \cref{holder}). Hence,
\begin{equation*}
\begin{aligned}
&|F(M,p,s,x)-F_0(M,p,s,0)|\\
&\leq |F(M,p,s,x)-F_0(M,p,s,x)|+|F_0(M,p,s,x)-F_0(M,p,s,0)|\\
&\leq \|\beta_3\|_{C^{1,\alpha}(0)}\omega_3(\|M\|,|p|,|s|)|x|^{1+\alpha}
+|D_xF_{0}(M,p,s,\xi)||x|\\
&\leq C\left(\omega_3(\|M\|,|p|,|s|)+\|M\|^{\alpha}+|p|^{\alpha}+|s|^{\alpha}+1\right)|x|\\
&:=\tilde\omega_3(\|M\|,|p|,|s|)|x|.
\end{aligned}
\end{equation*}
Clearly, $\tilde\omega_3$ satisfies \cref{e.omega}. Then, from the interior pointwise $C^{2,\alpha}$ regularity \Cref{t-C2a-i}, $u\in C^{2,\alpha}(0)$. Hence, we can assume (and do assume throughout this section) that the $C^{k-1,\alpha}(0)$ regularity holds if $F\in C^{k-2,\alpha}(0)$ and we need to derive the $C^{k,\alpha}(0)$ regularity.
The following lemma is a higher order counterpart of \Cref{In-l-C1a-mu} and \Cref{In-l-C2a-mu}.
\begin{lemma}\label{In-l-Cka-mu}
Let $0<\alpha<\bar{\alpha}$. Suppose that $F\in C^{k-2,\alpha}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Then there exists $\delta>0$ depending only on $k,n,\lambda,\Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that if $u$ satisfies
\begin{equation*}
F(D^2u,Du,u,x)=f ~~\mbox{in}~~B_1
\end{equation*}
with $\|u\|_{L^{\infty}(B_1)}\leq 1$, $u(0)=0,Du(0)=0,\cdots,D^{k-1}u(0)=0$, $\mu,b_0,c_0\leq 1$, $\|\beta_3\|_{C^{k-2,\alpha}(0)}\leq \delta$ and $\|f\|_{C^{k-3,\alpha}(0)}\leq \delta$, then there exists a $k$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(B_{\eta})}\leq \eta^{k+\alpha},
\end{equation*}
\begin{equation*}
|F_0(D^2Q(x),DQ(x),Q(x),x)|\leq C |x|^{k-2+\bar\alpha}, ~\forall ~x\in B_1,
\end{equation*}
and
\begin{equation*}
\|Q\|\leq C,
\end{equation*}
where $C$ and $\eta$ depend only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
\end{lemma}
\begin{remark}\label{r-6-1}
Note that we have assumed that the interior pointwise $C^{k-1,\alpha}$ regularity holds by induction. Hence, $u(0),Du(0),\cdots,D^{k-1}u(0)$ are well defined.
\end{remark}
\proof Throughout this proof, $C$ always denotes a constant depending only on $k,n,\lambda, \Lambda$, $\alpha$, $\omega_0$, $\omega_3$ and $\omega_4$. As before, we prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<\bar{\alpha},\omega_0,\omega_3,\omega_4$ and sequences of $F_m$, $u_m,f_m$ satisfying
\begin{equation*}
F_m(D^2u_m,Du_m,u_m,x)=f_m ~~\mbox{in}~~B_1.
\end{equation*}
In addition, $F_m$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\mu\equiv 1,b\equiv 1,c\equiv 1$ and $\omega_0$. Moreover, $F_m$ are $C^{k-2,\alpha}$ at $0$ with $F_{0m}$, $\beta_m$ and $\omega_3$, and
\begin{equation*}
\|F_{0m}\|_{C^{k-2,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{B}_1)}\leq \omega_4(r),~\forall ~r>0.
\end{equation*}
Furthermore, $\|u_m\|_{L^{\infty}(B_1)}\leq 1$, $u(0)=0, Du_m(0)=0,\dots,D^{k-1}u_m(0)=0$, $\|\beta_{m}\|_{C^{k-2,\alpha}(0)}\leq 1/m$ and $\|f_m\|_{C^{k-3,\alpha}(0)}\leq 1/m$. Finally, for any $k$-form $Q$ satisfying $\|Q\|\leq C_k$ and
\begin{equation}\label{In-e.cka.Q-mu}
|F_{0m}(D^2Q(x),DQ(x),Q(x),x)|\leq C_k |x|^{k-2+\bar\alpha}, ~\forall ~x\in B_1,
\end{equation}
we have
\begin{equation}\label{In-e.lCka.1-mu}
\|u_m-Q\|_{L^{\infty}(B_{\eta})}> \eta^{k+\alpha},
\end{equation}
where $C_k$ is to be specified later and $0<\eta<1$ is taken small such that
\begin{equation}\label{In-e.lCka.2-mu}
C_k\eta^{\alpha_1-\alpha}<1/2
\end{equation}
for $\alpha_1=(\bar{\alpha}+\alpha)/2$.
As before, $u_m$ are uniformly bounded and equicontinuous. Hence, there exists some function $u$ such that $u_m\rightarrow u$ in $L^{\infty}_{\mathrm{loc}}(B_1)$. Since $\|F_{0m}\|_{C^{k-2,\bar{\alpha}}(A)}$ are uniformly bounded for any $A\subset\subset S^n\times R^n\times R\times \bar{B}_1$, there exist a subsequence (denoted by $F_{0m}$ again) and $F_0\in C^{k-2,\bar{\alpha}}(S^n\times R^n\times R\times \bar{B}_1)$ such that $F_{0m}\rightarrow F_0$ in $C^{k-2,\alpha_1}$ in any compact subset of $S^n\times R^n\times R\times \bar{B}_1$ and $\|F_0\|_{C^{k-2,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{B}_1)}\leq \omega_4(r)$. Moreover, it is easy to verify that $F_0$ satisfies the structure condition \cref{SC2} with $\lambda,\Lambda,\mu\equiv 1,b\equiv 1,c\equiv 1$ and $\omega_0$. Furthermore, for any ball $B\subset\subset B_1$ and $\varphi\in C^2(\bar{B})$, let $r=\|D^2\varphi\|_{L^{\infty}(B)}+\|D\varphi\|_{L^{\infty}(B)}+1$, $G_m=F_m(D^2\varphi,D\varphi,u_m,x)-f_m$ and $G(x)=F_0(D^2\varphi,D\varphi,u,x)$. Then
\begin{equation*}
\begin{aligned}
&\|G_m-G\|_{L^n(B)}\\
&\leq \|F_m(D^2\varphi,D\varphi,u_m,x)-F_{0m}(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}\\
&+\|F_{0m}(D^2\varphi,D\varphi,u_m,x)-F_0(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
&\leq \omega_3(r,r,r)\|\beta_{m}\|_{L^n(B)}
+\|F_{0m}(D^2\varphi,D\varphi,u_m,x)-F_0(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$. From \Cref{l-35}, $u$ satisfies
\begin{equation*}
F_0(D^2u,Du,u,x)=0 ~~\mbox{in}~~B_{1}.
\end{equation*}
By the $C^{k-1,\alpha}$ estimate for $u_m$ and noting $u_m(0)=0$, $Du_m(0)=0$, $\cdots$, $D^{k-1}u_m(0)=0$, we have
\begin{equation*}
|u_m(x)|\leq C|x|^{k-1+\alpha} ,~~~~\forall~x\in B_1.
\end{equation*}
Since $u_m$ converges to $u$,
\begin{equation*}
|u(x)|\leq C|x|^{k-1+\alpha} ,~~~~\forall~x\in B_1,
\end{equation*}
which implies $u(0)=0$, $Du(0)=0$, $\cdots$, $D^{k-1}u(0)=0$. Combining with \Cref{In-l-62}, there exists a $k$-form $\tilde{Q}$ such that
\begin{equation}\label{In-e.cka-5}
|u(x)-\tilde{Q}(x)|\leq C_1 |x|^{k+\alpha_1}, ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{In-e.cka-4}
|F_0(D^2\tilde{Q}(x),D\tilde{Q}(x),\tilde{Q}(x),x)|\leq C_1|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation*}
\|\tilde{Q}\|\leq C_1,
\end{equation*}
where $C_1$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Now, we try to construct a sequence of $k$-forms $\tilde{Q}_m$ such that \cref{In-e.cka.Q-mu} holds for $\tilde{Q}_m$ and $\tilde{Q}_m\rightarrow \tilde{Q}$ as $m\rightarrow \infty$. Let $Q_m$ be $k$-forms ($m\geq 1$) with $\|Q_m\|\leq 1$ to be specified later and $\tilde{Q}_m=Q_m+\tilde{Q}$. Denote $F_{0m}(D^2\tilde{Q}_m(x),D\tilde{Q}_m(x),\tilde{Q}_m(x),x)$ by $G_m(x)$. Since $G_m\in C^{k-2,\bar\alpha}(\bar{B}_{1/2})$, \cref{In-e.cka.Q-mu} holds for $\tilde{Q}_m$ if we show that $D^iG_m(0)=0$ for any $1\leq i\leq k-2$.
Note that $u_m(0)=0,Du_m(0)=0,\cdots,D^{k-1}u_m(0)=0$ and from the interior pointwise $C^{k-1,\alpha}$ regularity \Cref{t-Cka-i} (i.e., $P\equiv N_f\equiv 0$ in \cref{e.Cka-3-i}),
\begin{equation*}
|F_{0m}(0,0,0,x)|\leq C|x|^{k-3+\alpha}.
\end{equation*}
We remark here that if $k=3$, we should use the interior pointwise $C^{2,\alpha}$ regularity \Cref{t-C2a-i} with $P\equiv f(0)\equiv 0$ in \cref{e.C2a-3-i-mu}. Thus,
\begin{equation*}
\begin{aligned}
&|G_m(x)|\\
&\leq |F_{0m}(D^2\tilde{Q}_m(x),D\tilde{Q}_m(x),\tilde{Q}_m(x),x)-F_{0m}(0,0,0,x)|
+|F_{0m}(0,0,0,x)|\\
&=|\int_{0}^{1} F_{0m,ij}(\xi)\tilde{Q}_{m,ij}+F_{0m,p_i}(\xi)\tilde{Q}_{m,i}+F_{0m,s}(\xi)\tilde{Q}_mdr|
+|F_{0m}(0,0,0,x)|\\
&\leq C|x|^{k-3+\alpha},
\end{aligned}
\end{equation*}
where $\xi=r\left(D^2\tilde{Q}_m(x),D\tilde{Q}_m(x),\tilde{Q}_m(x),x\right)
+(1-r)\left(0,0,0,x\right)$.
Hence, to verify \cref{In-e.cka.Q-mu} for $\tilde{Q}_m$, we only need to prove
\begin{equation*}
D^{k-2}G_m(0)=D^{k-2}\left( F_{0m}(D^2\tilde{Q}_m(x),D\tilde{Q}_m(x),\tilde{Q}_m(x),x)\right)\bigg|_{x=0}=0.
\end{equation*}
Indeed,
\begin{equation*}
\begin{aligned}
D^{k-2}G_m(0)&=F_{0m,ij}(0)D^{k-2}\tilde{Q}_{m,ij}+(D^{k-2}_{x}F_{0m})(0)\\
&=F_{0m,ij}(0)D^{k-2}\tilde{Q}_{ij}+(D^{k-2}_{x}F_{0m})(0)+F_{0m,ij}(0)D^{k-2}Q_{m,ij},
\end{aligned}
\end{equation*}
where $D^{k-2}_{x}$ means that we take $(k-2)$-th order derivatives only with respect to $x$. Since
\begin{equation*}
\begin{aligned}
F&_{0m,ij}(0)D^{k-2}\tilde{Q}_{ij}+(D^{k-2}_{x}F_{0m})(0)\rightarrow F_{0,ij}(0)D^{k-2}\tilde{Q}_{ij}+(D^{k-2}_{x}F_{0})(0)\\
&=D^{k-2}\left(F_0(D^2\tilde{Q}(x),D\tilde{Q}(x),\tilde{Q}(x),x)\right)\bigg |_{x=0}=0~(\mbox{by}~ \cref{In-e.cka-4})
\end{aligned}
\end{equation*}
and $\lambda\leq F_{0m,ij}(0)\leq \Lambda$, we can choose proper $Q_m$ such that $D^{k-2}G_m(0)=0$ for any $m\geq 1$ and $\|Q_m\|\rightarrow 0$ as $m\rightarrow \infty$. Hence, there exists a constant $C_k(\geq C_1+1)$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that $\|\tilde{Q}_m\|\leq C_k$ for any $m\geq 1$ and
\begin{equation*}
\begin{aligned}
|G_m(x)|=|F_{0m}(D^2\tilde{Q}_m(x),D\tilde{Q}_m(x),\tilde{Q}_m(x),x)|\leq C_k|x|^{k-2+\alpha}.
\end{aligned}
\end{equation*}
Thus, \cref{In-e.lCka.1-mu} holds for $\tilde{Q}_m$. Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta})}\geq \eta^{k+\alpha}.
\end{equation*}
However, by \cref{In-e.lCka.2-mu} and \cref{In-e.cka-5}, we have
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta})}\leq \eta^{k+\alpha}/2,
\end{equation*}
which is a contradiction. ~\qed~\\
Now, similar to the $C^{2,\alpha}$ regularity, we prove the interior pointwise $C^{k,\alpha}$ regularity under strong conditions.
\begin{theorem}\label{In-t-Ckas-mu}
Let $0<\alpha <\bar{\alpha}$. Suppose that $F\in C^{k-2,\alpha}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Let $u$ satisfy
\begin{equation*}
F(D^2u,Du,u,x)=f ~~\mbox{in}~~B_1.
\end{equation*}
Assume that
\begin{equation}\label{In-e.tCkas-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(B_1)}\leq 1,u(0)=0,Du(0)=0,\dots,D^{k-1}u(0)=0,\\
&\mu\leq \frac{1}{4C_0},b_0\leq \frac{1}{2},c_0\leq \frac{1}{K_0},\\
&|\beta_3(x)|\leq \delta_1|x|^{k-2+\alpha}~~\mbox{and}~~|f(x)|\leq \delta_1|x|^{k-2+\alpha}, ~~\forall ~x\in B_1,\\
\end{aligned}
\end{equation}
where $\delta_1\leq \delta$ ($\delta$ is as in \Cref{In-l-Cka-mu}) and $C_0$ depend only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Then $u\in C^{k,\alpha}(0)$ and there exists a $k$-form $Q$ such that
\begin{equation*}\label{In-e.tCkas-1-mu}
|u(x)-Q(x)|\leq C |x|^{k+\alpha}, ~~\forall ~x\in B_{1},
\end{equation*}
\begin{equation*}
|F_0(D^2Q(x),DQ(x),Q(x),x)|\leq C |x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}
\end{equation*}
and
\begin{equation*}\label{In-e.tCkas-2-mu}
\|Q\|\leq C,
\end{equation*}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
\end{theorem}
\proof As before, to prove that $u$ is $C^{k,\alpha}$ at $0$, we only need to prove the following. There exist a sequence of $k$-forms $Q_m$ ($m\geq 0$) such that for all $m\geq 1$,
\begin{equation}\label{In-e.tCkas-6-mu}
\|u-Q_m\|_{L^{\infty }(B _{\eta^{m}})}\leq \eta ^{m(k+\alpha )},
\end{equation}
\begin{equation}\label{In-e.tCkas-9-mu}
|F_0(D^2Q_m(x),DQ_m(x),Q_m(x),x)|\leq \tilde{C}|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}
\end{equation}
and
\begin{equation}\label{In-e.tCkas-7-mu}
\|Q_m-Q_{m-1}\|\leq \tilde{C}\eta ^{(m-1)\alpha},
\end{equation}
where $\tilde{C}$ and $\eta$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
We prove the above by induction. For $m=1$, by \Cref{In-l-Cka-mu}, there exists a $k$-form $Q_1$ such that \crefrange{In-e.tCkas-6-mu}{In-e.tCkas-7-mu} hold for some $C_1$ and $\eta_1$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ where $Q_0\equiv 0$. Take $\tilde{C}\geq C_1,\eta\leq \eta_1$ and then the conclusion holds for $m=1$. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{In-e.tCkas-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^{k+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{In-e.Ckas-F-mu}
\tilde{F}(D^2v,Dv,v,y)=\tilde{f} ~~\mbox{in}~~ B_1,
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{F(r^{k-2+\alpha}M+D^2Q_{m_0}(x),r^{k-1+\alpha}p+DQ_{m_0}(x),
r^{k+\alpha}s+Q_{m_0}(x),x)}{r^{k-2+\alpha}},\\
&\tilde{f}(y)=\frac{f(x)}{r^{k-2+\alpha}},
\end{aligned}
\end{equation*}
In addition, define $\tilde{F}_0$ in a similar way to the definition of $\tilde{F}$ (only replacing $F$ by $F_0$).
In the following, we show that \cref{In-e.Ckas-F-mu} satisfies the assumptions of \Cref{In-l-Cka-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(B_1)}\leq 1, v(0)=0,\cdots,D^{k-1}v(0)=0,~(\mathrm{by}~ \cref{In-e.tCkas-be-mu},~\cref{In-e.tCkas-6-mu} ~\mbox{and}~ \cref{In-e.tCkas-v-mu})\\
&|\tilde{f}(y)|\leq r^{-(k-2+\alpha)}|f(x)|\leq \delta_1|y|^{k-2+\alpha}, ~\forall ~y\in B_1. ~(\mathrm{by}~\cref{In-e.tCkas-be-mu})\\
\end{aligned}
\end{equation*}
By \cref{In-e.tCkas-7-mu}, there exists a constant $C_{0}$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that $\|Q_m\|\leq C_{0}$ ($\forall~0\leq m\leq m_0$). It is easy to check that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where (note that $\omega_0$ satisfies \cref{e.omega0-2})
\begin{equation*}
\tilde{\mu}=r^{k+\alpha}\mu,~~\tilde{b}= rb_0+2 C_0 r\mu ,~~\tilde{c}= K_0 r^{2} c_0~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence, $\tilde{\omega}_0$ satisfies \cref{e.omega0-2} and from \cref{In-e.tCkas-be-mu},
\begin{equation*}
\begin{aligned}
&\tilde{\mu}\leq \mu\leq 1,~\tilde{b}\leq b_0+2C_0\mu\leq 1,~\tilde{c}\leq c_0\leq 1.
\end{aligned}
\end{equation*}
In addition, for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation}\label{e.6.1}
\begin{aligned}
|&\tilde{F}(M,p,s,y)-\tilde{F}_0(M,p,s,y)|\\
&=r^{-(k-2+\alpha)}\bigg(F(r^{k-2+\alpha}M+D^2Q_{m_0}(x),r^{k-1+\alpha}p+DQ_{m_0}(x),
r^{k+\alpha}s+Q_{m_0}(x),x)\\
&-F_0(r^{k-2+\alpha}M+D^2Q_{m_0}(x),r^{k-1+\alpha}p+DQ_{m_0}(x),
r^{k+\alpha}s+Q_{m_0}(x),x)\bigg)\\
&\leq r^{-(k-2+\alpha)}\beta_3(x)\omega_3(r^{k-2+\alpha}\|M\|+C_0r^{k-2},
r^{k-1+\alpha}|p|+C_0r^{k-1},r^{k+\alpha}|s|+C_0r^{k})\\
&\leq \delta_1\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)|y|^{k-2+\alpha}\\
&:=\tilde{\beta}_3(y)\tilde{\omega}_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation}
where $\tilde{\beta}_3(y)=\delta_1|y|^{k-2+\alpha}$ and $\tilde{\omega}_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)$. Hence, $\tilde{\omega}_3$ satisfies \cref{e.omega-3} and
\begin{equation*}
\|\tilde{\beta}_3\|_{C^{k-2,\alpha}(0)}\leq \delta_1.
\end{equation*}
Finally, we show that
\begin{equation}\label{In-e.cka-6}
\|\tilde{F}_0\|_{C^{k-2,\bar\alpha}(\bar{\textbf{B}}_\rho\times \bar{B}_1)}\leq \tilde{\omega}_4(\rho),~\forall ~\rho>0,
\end{equation}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$. First, note that
\begin{equation*}
\begin{aligned}
&\tilde{F}_0(M,p,s,y)\\
&=\tilde{F}_0(M,p,s,y)-\tilde{F}_0(0,0,0,y)+\tilde{F}_0(0,0,0,y)\\
&=\int_{0}^{1} F_{0,ij}(\xi)M_{ij}+rF_{0,p_i}(\xi)p_i+r^2F_{0,s}(\xi)sd\rho
+\frac{F_0(D^2Q_{m_0}(x),DQ_{m_0}(x),Q_{m_0}(x),x)}{r^{k-2+\alpha}},\\
\end{aligned}
\end{equation*}
where $\xi=\rho\left(r^{k-2+\alpha}M,r^{k-1+\alpha}p,r^{k+\alpha}s,x\right)
+\left(D^2Q_{m_0}(x),DQ_{m_0}(x),Q_{m_0}(x),x\right)$. Hence, by combining with \cref{In-e.tCkas-9-mu}, $\|\tilde{F}_0\|_{C^{k-3}(\bar{\textbf{B}}_\rho\times \bar B_1)}$ has a uniform bound. Next, from the definition of $\tilde{F}_0$, if any $(k-2)$-th derivative of $\tilde{F}_0$ involves one derivative with respect to $M,p$ or $s$, it is bounded; if we take $(k-2)$-th derivative with respect to $y$, (note that $D^{k-2}F_0\in C^{\bar\alpha}$ and \cref{In-e.tCkas-9-mu})
\begin{equation*}
\begin{aligned}
D^{k-2}_{y}&\tilde{F}_0(M,p,s,y)\\
=&D^{k-2}_{y}\left(\tilde{F}_0(M,p,s,y)-\tilde{F}_0(0,0,0,y)\right)
+D^{k-2}_{y}\tilde{F}_0(0,0,0,y)\\
=&r^{-\alpha}D^{k-2}_{x}\Big(F_0(r^{k-2+\alpha}M+D^2Q_{m_0}(x),
r^{k-1+\alpha}p+DQ_{m_0}(x),r^{k+\alpha}s+Q_{m_0}(x),x)\\
&-F_0(D^2Q_{m_0}(x),DQ_{m_0}(x),Q_{m_0}(x),x)\Big)\\
&+r^{-\alpha}D^{k-2}_{x}F_0(D^2Q_{m_0}(x),DQ_{m_0}(x),Q_{m_0}(x),x)\\
\leq& C,
\end{aligned}
\end{equation*}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Similarly, $[\tilde{F}_0]_{C^{k-2,\bar\alpha}(\bar{\textbf{B}}_\rho\times \bar B_1)}$ is also bounded. Therefore, \cref{In-e.cka-6} holds.
Choose $\delta_1$ small enough (depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$) such that \Cref{In-l-Cka-mu} holds for $\tilde\omega_0,\tilde\omega_3,\tilde\omega_4$ and $\delta_1$. Since \cref{In-e.Ckas-F-mu} satisfies the assumptions of \Cref{In-l-Cka-mu}, there exist a $k$-form $\tilde{Q}$ and constants $\tilde{C}\geq C_1$ and $\eta\leq \eta_1$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(B_{\eta})}&\leq \eta^{k+\alpha},
\end{aligned}
\end{equation*}
\begin{equation}\label{e.6.2}
|\tilde F_0(D^2\tilde{Q}(y),D\tilde{Q}(y),\tilde{Q}(y),y)|\leq \tilde{C}|y|^{k-2+\bar\alpha},
~~\forall ~y\in B_{1}
\end{equation}
and
\begin{equation*}
\|\tilde{Q}\|\leq \tilde{C}.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+r^{\alpha}\tilde{Q}(x)$. Then \cref{In-e.tCkas-7-mu} holds for $m_0+1$ clearly. Next, by rescaling back, \cref{e.6.2} reads
\begin{equation*}
|F_0(D^2Q_{m_0+1}(x),DQ_{m_0+1}(x),Q_{m_0+1}(x),x)|
\leq \tilde{C}|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{r},
\end{equation*}
which implies
$D^{k-2} \left(F_0(D^2Q_{m_0+1}(x),DQ_{m_0+1}(x),Q_{m_0+1}(x),x)\right)=0$ at $0$. Combining with $F_0\in C^{k,\bar{\alpha}}$, above inequality also holds in $B_1$, i.e., \cref{In-e.tCkas-9-mu} holds for $m_0+1$. Recalling \cref{In-e.tCkas-v-mu}, we have
\begin{equation*}
\begin{aligned}
\|u-Q_{m_0+1}(x)\|_{L^{\infty}(B_{\eta^{m_0+1}})}&= \|u-Q_{m_0}(x)-r^{\alpha}\tilde{Q}(x)\|_{L^{\infty}(B_{\eta r})}\\
&= \|r^{k+\alpha}v-r^{k+\alpha}\tilde{Q}(y)\|_{L^{\infty}(B_{\eta})}\\
&\leq r^{k+\alpha}\eta^{k+\alpha}=\eta^{(m_0+1)(k+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{In-e.tCkas-6-mu} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Next, we give the~\\
\noindent\textbf{Proof of \Cref{t-Cka-i}.} As before, in the following proof, we just make necessary normalization to satisfy the assumptions of \Cref{In-t-Ckas-mu}. Throughout this proof, $C$ always denotes a constant depending only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,\alpha}(0)},$ $\omega_3,\omega_4, \|f\|_{C^{k-2,\alpha}}(0)$ and $\|u\|_{L^{\infty}(B_1)}$.
Let $F_1(M,p,s,x)=F(M,p,s,x)-N_f(x)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar{B}_1$ where $N_f(x)=f(0)+f_{i}(0)x_{i}+\cdots+f_{i_1\cdots i_{k-2}}(0)x_{i_1}\cdots x_{i_{k-2}}/(k-2)!$. Then $u$ satisfies
\begin{equation*}
F_1(D^2u,Du,u,x)=f_1 ~~\mbox{in}~~B_1,
\end{equation*}
where $f_1(x)=f(x)-N_f(x)$. Thus,
\begin{equation*}
|f_1(x)|\leq [f]_{C^{k-2,\alpha}(0)}|x|^{k-2+\alpha}\leq C|x|^{k-2+\alpha}, ~~\forall ~x\in B_1.
\end{equation*}
Note that $u\in C^{k-1,\alpha}(0)$ and define
\begin{equation*}
N_u(x)=u(0)+u_i(0)x_i+\cdots+u_{i_1\cdots i_{k-1}}(0)x_{i_1}\cdots x_{ i_{k-1}}/(k-1)!.
\end{equation*}
Set $u_1(x)=u(x)-N_u(x)$ and $F_2(M,p,s,x)=F_2(M+D^2N_u(x),p+DN_u(x),s+N_u(x),x)$. Then $u_1$ satisfies
\begin{equation*}
F_2(D^2u_1,Du_1,u_1,x)=f_1 ~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_1(0)=0,Du_1(0)=0,\dots,D^{k-1}u_1(0)=0.
\end{equation*}
Next, take $y=x/\rho$ and $u_2(y)=u_1(x)/\rho^2$, where $0<\rho<1$ is a constant to be specified later. Then $u_2$ satisfies
\begin{equation}\label{In-F6-k-mu}
F_3(D^2u_2,Du_2,u_2,y)=f_2~~\mbox{in}~~B_1,
\end{equation}
where
\begin{equation*}
\begin{aligned}
F_3(M,p,s,y)=F_2(M,\rho p,\rho^2s,x)~\mbox{and}~f_2(y)=f_1(x).
\end{aligned}
\end{equation*}
Finally, define the fully nonlinear operator $F_{30}$ in the same way as $F_3$ (only replacing $F$ by $F_0$).
Now, we try to choose a proper $\rho$ such that \cref{In-F6-k-mu} satisfies the assumptions of \Cref{In-t-Ckas-mu}. First, $u_2(0)=0,Du_2(0)=0,\dots,D^{k-1}u_2(0)=0$ clearly. Combining with the $C^{k-1,\alpha}$ regularity at $0$, we have
\begin{equation*}
\begin{aligned}
\|u_2\|_{L^{\infty}(B_1)}&= \|u_1\|_{L^{\infty}(B_\rho)}/\rho^2
\leq C\rho^{k-3+\alpha}.
\end{aligned}
\end{equation*}
Next,
\begin{equation*}
|f_2(y)|=|f_1(x)|\leq C|x|^{k-2+\alpha}=C\rho^{k-2+\alpha}|y|^{k-2+\alpha},~\forall ~y\in B_1,
\end{equation*}
It is easy to verify that $F_3$ and $F_{30}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde\mu,\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= \rho^2\mu,~~\tilde{b}= \rho b_0+C\rho\mu,~~\tilde{c}= K_0\rho^2c_0~~\mbox{and}~~
\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C,\cdot).
\end{equation*}
Clearly, $\tilde{\omega}_0$ satisfies \cref{e.omega0-2}.
Finally, we show $F_3\in C^{k-2,\alpha}(0)$. Indeed,
\begin{equation*}
\begin{aligned}
&|F_3(M,p,s,y)-F_{30}(M,p,s,y)|\\
&=|F(M+D^2N_u,\rho p+DN_u,\rho^2s+N_u,x)-F_{0}(M+D^2N_u,\rho p+DN_u,\rho^2s+N_u,x)|\\
&\leq \beta_3(x)\omega_3(\|M\|+C,|p|+C,|s|+C)\\
&\leq C\rho^{k-2+\alpha}\omega_3(\|M\|+C,|p|+C,|s|+C)|y|^{k-2+\alpha}\\
&:=\tilde{\beta}_3(y)\tilde\omega_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_3(y)=C\rho^{k-2+\alpha}|y|^{k-2+\alpha},~
\tilde\omega_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C,|p|+C,|s|+C)
\end{equation*}
and $\tilde\omega_3$ satisfies \cref{e.omega-3}. Moreover,
\begin{equation*}
\|F_{30}\|_{C^{k,\bar\alpha}(\bar{\textbf{B}}_r\times \bar{B}_1)}\leq \tilde{\omega}_4(r),~\forall ~r>0,
\end{equation*}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,\alpha}(0)},$ $\omega_3$, $\omega_4$, $\|f\|_{C^{k-2,\alpha}}(0)$ and $\|u\|_{L^{\infty}(B_1)}$.
Take $\delta_1$ small enough such that \Cref{In-t-Ckas-mu} holds for $\tilde{\omega}_0$, $\tilde{\omega}_3,\tilde{\omega}_4$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0$, $c_0,\omega_0$, $\omega_3,\omega_4, \|f\|_{C^{k-2,\alpha}}(0)$ and $\|u\|_{L^{\infty}(B_1)}$) such that the conditions of \Cref{In-t-Ckas-mu} are satisfied. Then $u_2$ and hence $u$ is $C^{k,\alpha}$ at $0$, and the estimates \crefrange{e.Cka-1-i}{e.Cka-2-i} hold. \qed~\\
\begin{remark}\label{r-6-2}
In fact, from above proof, we obtain that there exists a $k$-th order polynomial $P$ such that
\begin{equation*}
|F_0(D^2P(x),DP(x),P(x),x)-N_f(x)|\leq C |x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}.
\end{equation*}
Note that $F$ is $C^{k-2,\alpha}$ at $0$, \cref{e.Cka-3-i} holds for $F$.
\end{remark}
\section{Boundary $C^{1,\alpha}$ regularity}\label{C1a-mu}
In the next three sections, we give the detailed proofs of boundary pointwise regularity. In this section, we prove the boundary $C^{1,\alpha}$ regularity for the Pucci class. Similar to the interior $C^{1,\alpha}$ regularity, we first prove that the solution in \Cref{t-C1a-mu} can be approximated by a linear function at some scale provided that the coefficients and the prescribed are small enough (recall that $\Omega_1=\Omega\cap B_1$ and $(\partial\Omega)_1=\partial\Omega\cap B_1$).
\begin{lemma}\label{l-C1a-mu}
For any $0<\alpha<\bar{\alpha}$, there exists $\delta>0$ depending only on $n,\lambda,\Lambda$ and $\alpha$ such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&u\in S^*(\lambda,\Lambda,\mu,b,f)&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $\mu\leq \delta$, $\|b\|_{L^{p}(\Omega_1)}\leq \delta$ $(p=n/(1-\alpha))$, $\|f\|_{L^{n}(\Omega_1)}\leq \delta$, $\|g\|_{L^{\infty}((\partial \Omega)_1)}\leq \delta$
and $\underset{B_1}{\mathrm{osc}}~\partial\Omega \leq \delta$, then there exists a constant $a$ such that
\begin{equation*}\label{e.l4.0-mu}
\|u-ax_n\|_{L^{\infty}(\Omega_{\eta})}\leq \eta^{1+\alpha}
\end{equation*}
and
\begin{equation*}\label{e.14.2-mu}
|a|\leq \bar{C},
\end{equation*}
where $\eta$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
\end{lemma}
\begin{remark}\label{r-7-1}
Since we treat the Pucci class, the method of approximating the solution by solving some equation (e.g. \cite{MR3980853}) is unreliable. Instead, \Cref{l-C1a-mu} can be proved by the method of compactness (e.g. \cite{MR3246039,MR4088470}) or the method of combing Harnack inequality and a proper barrier (e.g. \cite{MR2853528,MR1139064}). In fact, our proof of this lemma is inspired directly by \cite{MR4088470}, which can be tracked to \cite{MR3246039} and \cite[Chapter 4]{Wang_Regularity}.
Note that we can prove the boundary pointwise $C^{1,\alpha}$ for the Pucci class and for any $\alpha<\bar{\alpha}$. This shows a firm evidence of the benefits of the method of compactness.
\end{remark}
\proof As before, we prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<\bar{\alpha}$ and sequences of $u_m,\mu_m,b_m,f_m,g_m$ and $\Omega_m$ satisfying
\begin{equation*}
\left\{\begin{aligned}
&u_m\in S^*(\lambda,\Lambda,\mu_m,b_m,f_m)&& ~~\mbox{in}~~\Omega_m\cap B_1;\\
&u_m=g_m&& ~~\mbox{on}~~\partial \Omega_m\cap B_1
\end{aligned}\right.
\end{equation*}
with $\|u_m\|_{L^{\infty}(\Omega_m\cap B_1)}\leq 1$, $\mu_m\leq 1/m$, $\|b_m\|_{L^{p}(\Omega_m\cap B_1)}\leq 1/m$, $\|f_m\|_{L^{n}(\Omega_m\cap B_1)}\leq 1/m$, $\|g_m\|_{L^{\infty}(\partial \Omega_m\cap B_1)}\leq 1/m$
and $\underset{B_1}{\mathrm{osc}}~\partial\Omega_m\leq 1/m$ such that
\begin{equation}\label{e.lC1a.1-mu}
\|u_m-ax_n\|_{L^{\infty}(\Omega_{m}\cap B_{\eta})}> \eta^{1+\alpha}, \forall~|a|\leq \bar{C},
\end{equation}
where $0<\eta<1$ is taken small such that
\begin{equation}\label{e.lC1a.2-mu}
\bar{C}\eta^{\bar{\alpha}-\alpha}<1/2.
\end{equation}
Note that $u_m$ are uniformly bounded. In addition, by \Cref{l-34}, $u_m$ are equicontinuous up to the boundary. More precisely, for any $\Omega'\subset\subset B_1^+\cup T_1$, $\varepsilon>0$, there exist $\delta>0$ and $m_0$ such that for any $m\geq m_0$ and $x,y\in \Omega'\cap \bar{\Omega}_m$ with $|x-y|<\delta$, we have $|u(x)-u(y)|\leq \varepsilon$. Hence, by the Arzel\`{a}-Ascoli theorem, there exists a subsequence (denoted by $u_m$ again) and $u:B_1^+\cup T_1\rightarrow R$ such that $u_m$ converges uniformly to $u$ on compact subsets of $B_1^+\cup T_1$.
Furthermore, for any $B\subset\subset B_1^+$ and $\varphi\in C^2(\bar{B})$, let $G_m=M^+(D^2\varphi)+\mu_m|D\varphi|^2+b_m|D\varphi|+|f_m|$ and $G=M^+(D^2\varphi)$. Then,
\begin{equation*}\label{gkC1a-mu}
\begin{aligned}
\|G_m-G\|_{L^n(B)}&=\|\mu_m|D\varphi|^2+b_m|D\varphi|+|f_m|\|_{L^n(B)}\\
&\leq \mu_m \||D\varphi|^2\|_{L^n(B)}+\|b_m\|_{L^p(B)}\|D\varphi\|_{L^{\frac{np}{p-n}}(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$. By \Cref{l-35}, $u\in \underline{S}(\lambda,\Lambda,0)$ in $B_{1}^+$. Similarly, $u\in \bar{S}(\lambda,\Lambda,0)$ and hence $u\in S(\lambda,\Lambda,0)$.
Next, we show that $u$ is continuous up to $T_1$ and $u=0$ there. For any $x_0\in T_1$ and $\varepsilon>0$, let $\delta>0$ be small to be specified later and $\tilde{x}\in B^+(x_0,\delta)\subset B_1^+$. Since $u_m$ converges to $u$ uniformly, there exists $m_0$ such that for any $m\geq m_0$ and $x\in B^+(x_0,\delta)\cap \bar{\Omega}_m$, we have
\begin{equation*}
|u_m(x)-u(x)|\leq \varepsilon/2.
\end{equation*}
Take $m$ large enough such that $\tilde{x}\in \Omega_m$ and $\|g_m\|_{L^{\infty}(\partial \Omega_m\cap B_1)}\leq \varepsilon/4$. Note that $u_m\in C(\bar{\Omega}_m\cap B_1)$ and we can take $\delta$ small such that $|u_m(\tilde{x})|\leq \varepsilon/2$. Hence,
\begin{equation*}
|u(\tilde{x})|=|u(\tilde{x})-u_m(\tilde{x})+u_m(\tilde{x})|\leq |u(\tilde{x})-u_m(\tilde{x})|+|u_m(\tilde{x})|\leq \varepsilon.
\end{equation*}
Therefore, $u$ is continuous up to $T_1$ and $u\equiv 0$ on $T_1$.
Above arguments show that
\begin{equation*}
\left\{\begin{aligned}
&u\in S(\lambda,\Lambda,0)&& ~~\mbox{in}~~B_{1}^+;\\
&u=0&& ~~\mbox{on}~~T_{1}.
\end{aligned}\right.
\end{equation*}
By \Cref{l-31}, there exists a constant $\bar{a}$ such that
\begin{equation*}
|u(x)-\bar{a}x_n|\leq \bar{C} |x|^{1+\bar{\alpha}}, ~~\forall ~x\in B_{1}^+
\end{equation*}
and
\begin{equation*}
|\bar{a}|\leq \bar{C}.
\end{equation*}
Combining with \cref{e.lC1a.2-mu}, we have
\begin{equation}\label{e.lC1a.3-mu}
\|u-\bar{a}x_n\|_{L^{\infty}(B_{\eta}^+)}\leq \eta^{1+\alpha}/2.
\end{equation}
However, from \cref{e.lC1a.1-mu},
\begin{equation*}
\|u_m-\bar{a}x_n\|_{L^{\infty}(\Omega_m\cap B_{\eta})}> \eta^{1+\alpha}.
\end{equation*}
Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-\bar{a}x_n\|_{L^{\infty}(B_{\eta}^+)}\geq \eta^{1+\alpha},
\end{equation*}
which contradicts with \cref{e.lC1a.3-mu}. ~\qed~\\
Now, we can prove the boundary pointwise $C^{1,\alpha}$ regularity.
\noindent\textbf{Proof of \Cref{t-C1a-mu}.} Since $\partial \Omega\in C^{1,\alpha}(0)$, we have
\begin{equation}\label{e.tC1a-1-mu}
|x_n|\leq \|\partial \Omega\|_{C^{1,\alpha}(0)}|x'|^{1+\alpha}, ~\forall ~x\in (\partial \Omega)_1.
\end{equation}
We assume that $g(0)=0$ and $Dg(0)=0$. Otherwise, we may consider $v(x)=u(x)-g(0)-Dg(0)\cdot x$. Then the regularity of $u$ follows easily from that of $v$. Since $g\in C^{1,\alpha}(0)$, we have
\begin{equation}\label{e.tC1a-2-mu}
|g(x)| \leq [g]_{C^{1,\alpha}(0)} |x|^{1+\alpha},~\forall~x\in (\partial \Omega)_1.
\end{equation}
Let $\delta$ be as in \Cref{l-C1a-mu}, which depends only on $n,\lambda,\Lambda$ and $\alpha$. Without loss of generality, we assume that
\begin{equation}\label{e.tC1a-ass-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(\Omega_1)}\leq 1,~~\mu\leq \frac{\delta}{6C_0^2},~~\|b\|_{L^p(\Omega_1)}\leq \frac{\delta}{3C_0},\\
&\|f\|_{C^{-1,\alpha}(0)}\leq \frac{\delta}{3},~~[g]_{C^{1,\alpha}(0)}\leq \frac{\delta}{2}~~\mbox{and}~~\|\partial \Omega\|_{C^{1,\alpha}(0)}\leq \frac{\delta}{2C_0},\\
\end{aligned}
\end{equation}
where $C_0>1$ is a constant (depending only on $n,\lambda,\Lambda$ and $\alpha$) to be specified later. Otherwise, note that $\Omega$ satisfies the exterior cone condition at $0$ and we may consider for $0<\rho<1$,
\begin{equation}\label{e.c1a-v}
\bar{u}(y)=\frac{u(x)}{\rho^{\alpha_0}},
\end{equation}
where $y=x/\rho$ and $0<\alpha_0<1$ is a H\"{o}lder exponent (depending only on $n,\lambda,\Lambda,\|b\|_{L^p(B_1)}$ and $\|\partial \Omega\|_{C^{1,\alpha}(0)}$) such that $u\in C^{2\alpha_0}(0)$ (by \Cref{l-3Ho}).
Then we have
\begin{equation*}
\left\{\begin{aligned}
&\bar{u}\in S^{*}(\lambda,\Lambda,\bar{\mu},\bar{b},\bar{f})&&
~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&\bar{u}=\bar{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation*}
where
\begin{equation*}\label{e.tC1a-n1-mu}
\bar{\mu}=\rho^{\alpha_0}\mu,~~\bar{b}(y)=\rho b(x),~~\bar{f}(y)=\rho^{2-\alpha_0}f(x),
~~\bar{g}(y)=\rho^{-\alpha_0}g(x)~~\mbox{and}
~~\tilde{\Omega}=\rho^{-1}\Omega.
\end{equation*}
Hence,
\begin{equation*}
\begin{aligned}
&\|\bar{u}\|_{L^{\infty}(\tilde{\Omega}_1)}\leq \frac{1}{\rho^{\alpha_0}}|x|^{2\alpha_0}[u]_{C^{2\alpha_0}(0)}
\leq \rho^{\alpha_0}[u]_{C^{2\alpha_0}(0)},\\
&\bar{\mu}=\rho^{\alpha_0}\mu,\\
&\|\bar{b}\|_{L^{p}(\tilde{\Omega}_1)}=\rho^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_\rho)}\leq \rho^{\alpha}\|b\|_{L^{p}(\Omega_1)},\\
&\|\bar{f}\|_{C^{-1,\alpha}(0)}\leq\rho^{1-\alpha_0+\alpha}\|f\|_{C^{-1,\alpha}(0)},\\
&[\bar{g}]_{C^{1,\alpha}(0)}\leq \rho^{1-\alpha_0+\alpha}[g]_{C^{1,\alpha}(0)},\\
&\|\partial \tilde{\Omega}\|_{C^{1,\alpha}(0)}\leq \rho^{\alpha}\|\partial \Omega\|_{C^{1,\alpha}(0)}.
\end{aligned}
\end{equation*}
By choosing $\rho$ small enough (depending only on $n,\lambda,\Lambda,\alpha,\mu, \|b\|_{L^{p}(\Omega_1)},\|\partial \Omega\|_{C^{1,\alpha}(0)}$, $\|f\|_{C^{-1,\alpha}(0)},[g]_{C^{1,\alpha}(0)}$ and $\|u\|_{L^{\infty}(\Omega_1)}$), the assumptions \cref{e.tC1a-ass-mu} for $\bar{u}$ can be guaranteed. Hence, we can make the assumption \cref{e.tC1a-ass-mu} for $u$ without loss of generality.
Now, we prove that $u$ is $C^{1,\alpha}$ at $0$ and we only need to prove the following. There exists a sequence of constants $a_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{e.tC1a-3-mu}
\|u-a_mx_n\|_{L^{\infty }(\Omega _{\eta^{m}})}\leq \eta ^{m(1+\alpha )}
\end{equation}
and
\begin{equation}\label{e.tC1a-4-mu}
|a_m-a_{m-1}|\leq \bar{C}\eta ^{m\alpha},
\end{equation}
where $\eta$, depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{l-C1a-mu} .
We prove the above by induction. For $m=0$, by setting $a_0=a_{-1}=0$, the conclusion holds clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{e.tC1a-v1-mu}
v(y)=\frac{u(x)-a_{m_0}x_n}{r^{1+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&v\in S^*(\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{f})&& ~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&v=\tilde{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation*}
where
\begin{equation}\label{e.tC1a-n2-mu}
\begin{aligned}
&\tilde{\mu}=2r^{1+\alpha}\mu,~\tilde{b}(y)=rb(x),~
\tilde{f}(y)=\frac{|f(x)|+b(x)|a_{m_0}|+2\mu|a_{m_0}|^2}{r^{\alpha-1}},\\
&\tilde{g}(y)=\frac{g(x)-a_{m_0}x_n}{r^{1+\alpha}}~~\mbox{and}~~
\tilde{\Omega}=\frac{\Omega}{r}.
\end{aligned}
\end{equation}
By \cref{e.tC1a-4-mu}, there exists a constant $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$ such that $|a_{m}|\leq C_0$ ($\forall~0\leq m\leq m_0$). Then it is easy to verify that
\begin{equation}\label{e.tC1a-n22-mu}
\begin{aligned}
&\|v\|_{L^{\infty}(\tilde{\Omega}_1)}\leq 1, ~(\mathrm{by}~ \cref{e.tC1a-3-mu} ~\mathrm{and}~ \cref{e.tC1a-v1-mu})\\
&\tilde\mu= 2r^{1+\alpha}\mu\leq \delta, ~(\mathrm{by}~ \cref{e.tC1a-ass-mu} ~\mathrm{and}~ \cref{e.tC1a-n2-mu})\\
& \|\tilde{b}\|_{L^{p}(\tilde{\Omega}_1)}=r^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_r)}\leq r^{\alpha}\|b\|_{L^{p}(\Omega_1)}\leq \delta,~(\mathrm{by}~ \cref{e.tC1a-ass-mu}~\mathrm{and}~ \cref{e.tC1a-n2-mu})\\
&\|\tilde{f}\|_{L^{n}(\tilde{\Omega}_1)}\leq\frac{\|f\|_{L^{n}(\Omega_r)}}{r^{\alpha}}
+\frac{C_0}{r^{\alpha}}r^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_r)}+\frac{2 C_0^2 \mu}{r^{\alpha-1}}\\
&\leq \|f\|_{C^{-1,\alpha}(0)}+C_0\|b\|_{L^{p}(\Omega_1)}+2\mu C_0^2\leq \delta, ~(\mathrm{by}~ \cref{e.tC1a-ass-mu}~\mathrm{and}~ \cref{e.tC1a-n2-mu})\\
&\|\tilde{g}\|_{L^{\infty}((\partial \tilde{\Omega})_1)}\leq \frac{1}{r^{1+\alpha}}\left([g]_{C^{1,\alpha}(0)}r^{1+\alpha}+C_0\|\partial \Omega\|_{C^{1,\alpha}(0)}r^{1+\alpha}\right)\\
&\leq \delta, ~(\mathrm{by}~\crefrange{e.tC1a-1-mu}{e.tC1a-ass-mu} ~\mathrm{and}~ \cref{e.tC1a-n2-mu})\\
&\underset{B_1}{\mathrm{osc}}~\partial\tilde{\Omega}=
\frac{1}{r}\underset{B_r}{\mathrm{osc}}~\partial\Omega \leq \|\partial \Omega\|_{C^{1,\alpha}(0)} r^{\alpha} \leq \delta. ~(\mathrm{by}~ \cref{e.tC1a-1-mu}~\mathrm{and}~ \cref{e.tC1a-ass-mu})
\end{aligned}
\end{equation}
Hence, by \Cref{l-C1a-mu}, there exists a constant $\tilde{a}$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{a}y_n\|_{L^{\infty }(\tilde{\Omega} _{\eta})}&\leq \eta ^{1+\alpha}
\end{aligned}
\end{equation*}
and
\begin{equation*}
|\tilde{a}|\leq \bar{C}.
\end{equation*}
Let $a_{m_0+1}=a_{m_0}+r^{\alpha}\tilde{a}$. Then \cref{e.tC1a-4-mu} holds for $m_0+1$. Recalling \cref{e.tC1a-v1-mu}, we have
\begin{equation*}
\begin{aligned}
&\|u-a_{m_0+1}x_n\|_{L^{\infty}(\Omega_{\eta^{m_0+1}})}\\
&= \|u-a_{m_0}x_n-r^{\alpha}\tilde{a}x_n\|_{L^{\infty}(\Omega_{\eta r})}\\
&= \|r^{1+\alpha}v-r^{1+\alpha}\tilde{a}y_n\|_{L^{\infty}(\tilde{\Omega}_{\eta})}\\
&\leq r^{1+\alpha}\eta^{1+\alpha}=\eta^{(m_0+1)(1+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{e.tC1a-3-mu} holds for $m=m_0+1$. By induction, the proof is completed.
For the special case $\mu=0$, set
\begin{equation*}
K=\|u\|_{L^{\infty}(\Omega_1)}+\delta^{-1}\left(3\|f\|_{C^{-1,\alpha}(0)}
+2[g]_{C^{1,\alpha}(0)}\right)
\end{equation*}
and define for $0<\rho<1$
\begin{equation*}
\bar{u}(y)=u(x)/K,
\end{equation*}
where $y=x/\rho$. Then by taking $\rho$ small enough (depending only on $n,\lambda,\Lambda,\alpha, \|b\|_{L^{p}(B_1)}$ and $\|\partial \Omega\|_{C^{1,\alpha}(0)}$), \cref{e.tC1a-ass-mu} can be guaranteed. Hence, for $\mu=0$, we have the explicit estimates \cref{e.C1a-1} and \cref{e.C1a-2}.\qed~\\
\begin{remark}\label{r-42-mu}
From above proof, it shows clearly that the assumption $\partial \Omega \in C^{1,\alpha}(0)$ is used to estimate $x_n$ on $\partial \Omega$ (see \cref{e.tC1a-n22-mu} for the estimate on $g$). This observation is originated from \cite{MR3780142} and pointed out in \cite{MR4088470}. It is also key to higher regularity in following sections.
\end{remark}
\section{Boundary $C^{2,\alpha}$ regularity}\label{C2a-mu}
In this section, we prove the boundary pointwise $C^{2,\alpha}$ regularity. The key is that if $u(0)=0$ and $Du(0)=0$, the boundary $C^{2,\alpha}$ regularity holds even if $\partial \Omega\in C^{1,\alpha}(0)$.
In fact, we can say a little more about above idea. Take the linear equation \cref{e.linear} for example. We have the following observation. If $u(0)=0$, the assumption for $c$ can be reduced one order; if $u(0)=0$ and $Du(0)=0$, the assumption for $c$ can be reduced two order, and the assumptions for $b^i$, $\partial \Omega$ can be reduced one order; if $u(0)=0$, $Du(0)=0$ and $D^2u(0)=0$, the assumption for $c$, the assumptions for of $b^i$, $\partial \Omega$ and the assumption for $a^{ij}$ can be reduced three order, two order and one order respectively. For instance, if we intend to derive the $C^{3,\alpha}(0)$ regularity and we know $u(0)=0$, $Du(0)=0$ and $D^2u(0)=0$ beforehand, then $c\in L^n$, $b^i\in L^{p} (p=n/(1-\alpha))$, $\partial \Omega\in C^{1,\alpha}$ and $a^{ij}\in C^{\alpha}$ are enough (see also \Cref{r-9.1} below).
This observation can be used to prove the regularity of free boundary problems, which will be dealt in a separate work. It is hard to be observed by studying the local or global regularity.
For boundary pointwise regularity, we call $Q:R^n\rightarrow R$ a $k$-form ($k\geq 1$) if $Q$ can be written as
\begin{equation*}
Q(x)=\frac{1}{k!}\cdot \sum_{1\leq i_1,...,i_{k-1}\leq n}a_{i_1\cdots i_{k-1}}x_{i_1}\cdots x_{i_{k-1}}x_n,
\end{equation*}
where $a_{i_1\cdots i_{k-1}}$ are constants. In addition, we define
\begin{equation*}
\|Q\|=\sum_{1\leq i_1,...,i_{k-1}\leq n}|a_{i_1\cdots i_{k-1}}|.
\end{equation*}
Note that this definition is different from that in \Cref{In-C2a-mu}. For simplicity, we use the same symbol $Q$.
The following lemma is similar to \Cref{l-C1a-mu}, but without the term $x_n$ in the estimate.
\begin{lemma}\label{l-C2a-mu}
Suppose that $F$ satisfies \cref{e.C2a-KF}. For any $0<\alpha<\bar{\alpha}$, there exists $\delta>0$ depending only on $n,\lambda,\Lambda,\alpha$ and $\omega_2$ such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $u(0)=0$, $Du(0)=0$, $\mu,b_0,c_0\leq \delta$, $\omega_0(1,1)\leq 1$, $\|\beta_2\|_{L^{\infty}(\Omega_1)}\leq \delta$, $\|f\|_{L^{\infty}(\Omega_1)}\leq \delta$, $\|g\|_{C^{1,\alpha}(0)}\leq \delta$ and $\|\partial \Omega\|_{C^{1,\alpha}(0)} \leq \delta$, then there exists a $2$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(\Omega_{\eta})}\leq \eta^{2+\alpha},
\end{equation*}
\begin{equation*}
F_0(D^2Q,0,0)=0
\end{equation*}
and
\begin{equation*}
\|Q\|\leq \bar{C}+1,
\end{equation*}
where $\eta$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
\end{lemma}
\proof As before, we prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<
\bar{\alpha},\omega_2$ and sequences of $F_m,u_m,f_m,g_m,\Omega_m$ satisfying
\begin{equation*}
\left\{\begin{aligned}
&F_m(D^2u_m,Du_m,u_m,x)=f_m&& ~~\mbox{in}~~\Omega_m\cap B_1;\\
&u_m=g_m&& ~~\mbox{on}~~\partial \Omega_m\cap B_1.
\end{aligned}\right.
\end{equation*}
In addition, $F$ satisfies the structure condition \cref{SC2} with $\lambda,\Lambda,\mu_m,b_m,c_m$ and $\omega_m$. Moreover, $F_m$ satisfy \cref{e.C2a-KF} with $F_{0m},\beta_m$ and $\omega_2$. Furthermore, $\|u_m\|_{L^{\infty}(\Omega_m\cap B_1)}\leq 1$, $u_m(0)=0$, $Du_m(0)=0$, $\mu_m\leq 1/m$, $b_m\leq 1/m$, $c_m\leq 1/m$, $\|\beta_{m}\|_{L^{\infty}(\Omega_m\cap B_1)}\leq 1/m$, $\|f_m\|_{L^{\infty}(\Omega_m\cap B_1)}\leq 1/m$, $\|g_m\|_{C^{1,\alpha}(0)}\leq 1/m$ and $\|\partial \Omega_m\|_{C^{1,\alpha}(0)} \leq 1/m$. Finally, for any $2$-form $Q$ satisfying $\|Q\|\leq \bar{C}+1$ and
\begin{equation}\label{e.lC2a-3-mu}
F_{0m}(D^2Q,0,0)=0,
\end{equation}
we have
\begin{equation}\label{e.lC2a-1-mu}
\|u_m-Q\|_{L^{\infty}(\Omega_{m}\cap B_{\eta})}> \eta^{2+\alpha},
\end{equation}
where $0<\eta<1$ is taken small such that
\begin{equation}\label{e.lC2a-2-mu}
\bar{C}\eta^{\bar{\alpha}-\alpha}<1/2.
\end{equation}
As before, $u_m$ are uniformly bounded and equicontinuous and then there exist a subsequence (denoted by $u_m$ again) and $u:B_1^+\cup T_1\rightarrow R$ such that $u_m\rightarrow u$ uniformly for any $\Omega'\subset\subset B_1^+\cup T_1$. In addition, there exist a subsequence of $F_{0m}$ (denoted by $F_{0m}$ again) and $F:S^n\rightarrow R$ such that $F_{0m}(\cdot,0,0)\rightarrow F$ uniformly on compact subsets of $S^n$.
Furthermore, for any $B\subset\subset B_1^+$ and $\varphi\in C^2(\bar{B})$, let $r=\|D^2\varphi\|_{L^{\infty}(B)}+\|D\varphi\|_{L^{\infty}(B)}+1$, $G_m=F_m(D^2\varphi,D\varphi,u_m,x)-f_m$ and $G=F(D^2\varphi)$. Then,
\begin{equation*}\label{gkC2a-mu}
\begin{aligned}
\|G&_m-G\|_{L^n(B)}\\
\leq&\|F_m(D^2\varphi,D\varphi,u_m,x)-F_{0m}(D^2\varphi,D\varphi,u_m)
+F_{0m}(D^2\varphi,D\varphi,u_m)\\
&-F_{0m}(D^2\varphi,0,0)+F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
\leq&\|\beta_{m}\omega_2(r,r,r)+\mu_m|D\varphi|^2+b_m|D\varphi|
+c_m\|\omega_m(1,|u_m|)\|_{L^n(B)}\\
&+\|F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^{n}(B)}+\|f_m\|_{L^n(B)}\\
\leq& \omega_2(r,r,r)\|\beta_m\|_{L^{n}(B)}
+C(r^2\mu_m+rb_m+c_m)\\
&+\|F_{0m}(D^2\varphi,0,0)-F(D^2\varphi)\|_{L^{n}(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$. By \Cref{l-35}, $u$ is a viscosity solution of
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u)=0&& ~~\mbox{in}~~B_{1}^+;\\
&u=0&& ~~\mbox{on}~~T_{1}.
\end{aligned}\right.
\end{equation*}
Note that $u_m\in S^*(\lambda,\Lambda,\mu_m,b_m,|f_m|+c_m+|F_m(0,0,0,\cdot)|)$. By the boundary $C^{1,\alpha}$ estimate for $u_m$ (see \Cref{t-C1a-mu}) and noting $u_m(0)=0$ and $Du_m(0)=0$, we have
\begin{equation*}
\|u_m\|_{L^{\infty }(\Omega_m\cap B_r)}\leq Cr^{1+\alpha}, ~~~~\forall~0<r<1,
\end{equation*}
where $C$ depends only on $n,\lambda,\Lambda,\alpha$ and $\omega_2$. Since $u_m$ converges to $u$ uniformly,
\begin{equation*}
\|u\|_{L^{\infty }(B_r^+)}\leq Cr^{1+\alpha}, ~~~~\forall~0<r<1,
\end{equation*}
Hence, $u(0)=0$ and $Du(0)=0$.
By \Cref{l-32}, there exists a $2$-form $\tilde{Q}$ such that
\begin{equation}\label{e.c2a-3-mu}
|u(x)-\tilde{Q}(x)|\leq \bar{C} |x|^{2+\bar{\alpha}}, ~~\forall ~x\in B_{1}^+,
\end{equation}
\begin{equation*}
F(D^2\tilde{Q})=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}.
\end{equation*}
Combining \cref{e.lC2a-2-mu} and \cref{e.c2a-3-mu}, we have
\begin{equation}\label{e.lC2a-3-mu}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta}^+)}\leq \eta^{2+\alpha}/2.
\end{equation}
Similar to the interior $C^{2,\alpha}$ regularity (see \Cref{In-l-C2a-mu} and its proof), there exist a sequence of constants $t_m$ with $t_m\rightarrow 0$ and $|t_m|\leq 1$ (for $m$ large) such that
\begin{equation*}
\begin{aligned}
F_{0m}(D^2Q_m,0,0)=0,
\end{aligned}
\end{equation*}
where $Q_m(x)=\tilde{Q}(x)+t_mx_n^2/2$.
Hence, \cref{e.lC2a-1-mu} holds for $Q_m$, i.e.,
\begin{equation*}
\|u_m-Q_m\|_{L^{\infty}(\Omega_m\cap B_{\eta})}> \eta^{2+\alpha}.
\end{equation*}
Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta}^+)}\geq \eta^{2+\alpha},
\end{equation*}
which contradicts with \cref{e.lC2a-3-mu}. ~\qed~\\
The following is a special (also essential) result for the boundary pointwise $C^{2,\alpha}$ regularity. The key is that if $u(0)=0$ and $Du(0)=0$, the $C^{2,\alpha}$ regularity holds even if $\partial \Omega\in C^{1,\alpha}(0)$. This was first observed in \cite{MR4088470}.
\begin{theorem}\label{t-C2as-mu}
Let $0<\alpha <\bar{\alpha}$ and $F$ satisfy \cref{e.C2a-KF}. Suppose that $\omega_0$ satisfies \cref{e.omega0} and $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1.
\end{aligned}\right.
\end{equation*}
Assume that
\begin{equation}\label{e.C2as-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(\Omega_1)}\leq 1,~ u(0)=0,~ Du(0)=0, \\
&\mu\leq \frac{\delta_1}{4C_0},~ b_0\leq \frac{\delta_1}{2},~ c_0\leq \frac{\delta_1}{K_0},~
\omega_0(1+C_0,1)\leq 1,\\
&|\beta_2(x)|\leq \delta_1|x|^{\alpha},~|f(x)|\leq \delta_1|x|^{\alpha}, ~~\forall ~x\in \Omega_1,\\
&|g(x)|\leq \frac{\delta_1}{2}|x|^{2+\alpha}, ~~\forall ~x\in (\partial \Omega)_1~~\mbox{and}~~\|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)} \leq \frac{\delta_1}{2C_0},\\
\end{aligned}
\end{equation}
where $\delta_1$ depends only on $n,\lambda,\Lambda,\alpha,\omega_0$ and $\omega_2$, and $C_0$ depends only on $n,\lambda,\Lambda$ and $\alpha$.
Then $u\in C^{2,\alpha}(0)$, i.e., there exists a $2$-form $Q$ such that
\begin{equation}\label{e.C2as-1-mu}
|u(x)-Q(x)|\leq C |x|^{2+\alpha}, ~~\forall ~x\in \Omega_{1},
\end{equation}
\begin{equation}\label{e.C2as-3-mu}
F_0(D^2Q,0,0)=0
\end{equation}
and
\begin{equation}\label{e.C2as-2-mu}
\|Q\| \leq C,
\end{equation}
where $C$ depends only on $n, \lambda, \Lambda$ and $\alpha$.
\end{theorem}
\proof To prove that $u$ is $C^{2,\alpha}$ at $0$, we only need to prove the following. There exist a sequence of $2$-forms $Q_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{e.C2as-4-mu}
\|u-Q_m\|_{L^{\infty }(\Omega_{\eta^{m}})}\leq \eta ^{m(2+\alpha )},
\end{equation}
\begin{equation}\label{e.C2as-5-mu}
F_0(D^2Q_m,0,0)=0
\end{equation}
and
\begin{equation}\label{e.C2as-6-mu}
\|Q_m-Q_{m-1}\|\leq (\bar{C}+1)\eta ^{(m-1)\alpha},
\end{equation}
where $\eta$ depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{l-C2a-mu}.
We prove the above by induction. For $m=0$, by setting $Q_0=Q_{-1}\equiv 0$, \crefrange{e.C2as-4-mu}{e.C2as-6-mu} hold clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{e.C2as-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^{2+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{e.C2as-F-mu}
\left\{\begin{aligned}
&\tilde{F}(D^2v,Dv,v,y)=\tilde{f}&& ~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&v=\tilde{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar{\tilde{\Omega}}_1$,
\begin{equation*}\label{e.c2a.b}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{1}{r^{\alpha}}
F(r^{\alpha}M+D^2Q_{m_0}(x),r^{1+\alpha}p+DQ_{m_0}(x),r^{2+\alpha}s
+Q_{m_0}(x),x),\\
&\tilde{f}(y)=\frac{f(x)}{r^{\alpha}},~~\tilde{g}(y)=\frac{g(x)-Q_{m_0}(x)}{r^{2+\alpha}}
~~\mbox{and}~~\tilde{\Omega}=\frac{\Omega}{r}.\\
\end{aligned}
\end{equation*}
In addition, define
\begin{equation*}
\tilde{F}_0(M,p,s)=\frac{1}{r^{\alpha}}F_0(r^{\alpha}M+D^2Q_{m_0},r^{1+\alpha}p
,r^{2+\alpha}s).
\end{equation*}
In the following, we show that \cref{e.C2as-F-mu} satisfies the assumptions of \Cref{l-C2a-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(\tilde{\Omega}\cap B_1)}\leq 1, v(0)=0, Dv(0)=0,
~~(\mathrm{by}~ \cref{e.C2as-be-mu},~\cref{e.C2as-4-mu}~\mbox{and}~ \cref{e.C2as-v-mu})\\
&\|\tilde{f}\|_{L^{\infty}(\tilde{\Omega}\cap B_1)}=\frac{\|f\|_{L^{\infty}(\Omega\cap B_r)}}{r^{\alpha}}\leq \delta_1,
~(\mathrm{by}~\cref{e.C2as-be-mu})\\
&\|\partial \tilde{\Omega}\cap B_1\|_{C^{1,\alpha}(0)} \leq r^{\alpha}\|\partial\Omega\cap B_1\|_{C^{1,\alpha}(0)}\leq \delta_1,
~(\mathrm{by}~\cref{e.C2as-be-mu})\\
&\tilde{F}_0(0,0,0)=r^{-\alpha}F_0(D^2Q_{m_0},0,0)=0.~(\mathrm{by}~\cref{e.C2as-5-mu})
\end{aligned}
\end{equation*}
By \cref{e.C2as-6-mu}, there exists a constant $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$ such that $\|Q_m\|\leq C_0$ ($\forall~0\leq m\leq m_0$). For any $0<\rho<1$ (note that $Q_{m_0}$ is a $2$-form),
\begin{equation*}
\|\tilde{g}\|_{L^{\infty}(\partial \tilde{\Omega}\cap B_{\rho})} \leq \frac{1}{r^{2+\alpha}}\left(\frac{\delta _1}{2}(\rho r)^{2+\alpha}+
C_0\cdot \frac{\delta_1}{2C_0}(\rho r)^{2+\alpha}\right)\leq \delta_1 \rho^{2+\alpha}.
~~(\mathrm{by}~\cref{e.C2as-be-mu})
\end{equation*}
Hence,
\begin{equation*}
\|\tilde{g}\|_{C^{1,\alpha}(0)}\leq \delta_1.
\end{equation*}
It is easy to verify that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= r^{2+\alpha}\mu,~~\tilde{b}= rb_0+2 C_0 r\mu ,~~\tilde{c}= K_0 r^{\alpha+\alpha^2}c_0~~
\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence, $\tilde{\omega}_0$ satisfies \cref{e.omega0} and from \cref{e.C2as-be-mu},
\begin{equation*}
\tilde{\mu}\leq \delta_1,~
\tilde{b}\leq \delta_1, ~\tilde{c}\leq \delta_1
~~\mbox{and}~~\tilde{\omega}_0(1,1)\leq 1.
\end{equation*}
Similar to the interior $C^{2,\alpha}$ regularity, by combining \cref{SC2}, \cref{e.C2a-KF} and \cref{e.C2as-be-mu}, we have
\begin{equation*}
\begin{aligned}
|&\tilde{F}(M,p,s,y)-\tilde{F}_0(M,p,s)|\\
&\leq r^{-\alpha}\Big(2C_0r^{1+\alpha}\mu |p||x|+ C_0^2\mu|x|^2+C_0 b_0|x|+K_0 c_0\omega_0(|s|+C_0,C_0)|x|^{2\alpha}\\
&+\beta_2(x)\omega_2(\|M\|+C_0,|p|,|s|)\Big)\\
&\leq \delta_1\omega_2(\|M\|+C_0,|p|,|s|)+\delta_1|p|
+\delta_1C_0+\delta_1\omega_0(|s|+C_0,C_0)\\
&:= \tilde{\beta}_2(y)\tilde{\omega}_2(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_2(y)\equiv \delta_1$ and $\tilde{\omega}_2(\|M\|,|p|,|s|)=
\omega_2(\|M\|+C_0,|p|,|s|)+|p|+C_0+\omega_0(|s|+C_0,C_0)$. Then $\tilde{\omega}_2$ satisfies \cref{e.omega} and
\begin{equation*}
\|\tilde{\beta}_2\|_{L^{\infty}(\tilde{\Omega}_1)}\leq \delta_1.
\end{equation*}
Choose $\delta_1$ small enough (depending only on $n,\lambda,\Lambda,\alpha,\omega_0$ and $\omega_2$) such that \Cref{l-C2a-mu} holds for $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. Since \cref{e.C2as-F-mu} satisfies the assumptions of \Cref{l-C2a-mu}, there exists a $2$-form $\tilde{Q}(y)$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(\tilde{\Omega} _{\eta})}&\leq \eta ^{2+\alpha},
\end{aligned}
\end{equation*}
\begin{equation*}
\tilde{F}_0(D^2\tilde{Q},0,0)=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}+1.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+r^{\alpha}\tilde{Q}(x)$. Then \cref{e.C2as-5-mu} and \cref{e.C2as-6-mu} hold for $m_0+1$. Recalling \cref{e.C2as-v-mu}, we have
\begin{equation*}
\begin{aligned}
&\|u-Q_{m_0+1}(x)\|_{L^{\infty}(\Omega_{\eta^{m_0+1}})}\\
&= \|u-Q_{m_0}(x)-r^{\alpha}\tilde{Q}(x)\|_{L^{\infty}(\Omega_{\eta r})}\\
&= \|r^{2+\alpha}v-r^{2+\alpha}\tilde{Q}(y)\|_{L^{\infty}(\tilde{\Omega}_{\eta})}\\
&\leq r^{2+\alpha}\eta^{2+\alpha}=\eta^{(m_0+1)(2+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{e.C2as-4-mu} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Now, we give the~\\
\noindent\textbf{Proof of \Cref{t-C2a}.} In fact, as before, the following proof is mere a normalization procedure in some sense. We prove the theorem in two cases.
\textbf{Case 1:} the general case, i.e., $F$ satisfies \cref{SC2} and \cref{e.C2a-KF}. Throughout the proof for this case, $C$ always denotes a constant depending only on $n, \lambda,\Lambda,\alpha,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\alpha}(0)}$, $\omega_2$, $\|\partial \Omega\|_{C^{2,\alpha}(0)}, \|f\|_{C^{\alpha}(0)},
\|g\|_{C^{2,\alpha}(0)}$ and $\|u\|_{L^{\infty }(\Omega_1)}$.
Let $F_1(M,p,s,x)=F(M,p,s,x)-f(0)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar\Omega_1$. Then $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_1(D^2u,Du,u,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $f_1(x)=f(x)-f(0)$. Thus,
\begin{equation*}
|f_1(x)|\leq C|x|^{\alpha}, ~~\forall ~x\in \Omega_1.
\end{equation*}
Next, set $u_1(x)=u(x)-N_{g}(x)$ and $F_2(M,p,s,x)=F_1(M+D^2N_{g}(x),p+DN_{g}(x),s+N_{g}(x),x)$ where $N_{g}(x)=g(0)+Dg(0)\cdot x+x^TD^2g(0)x/2$. Then $u_1$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_2(D^2u_1,Du_1,u_1,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_1=g_1&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_1(x)=g(x)-N_{g}(x)$. Hence,
\begin{equation*}
|g_1(x)|\leq C|x|^{2+\alpha}, ~~\forall ~x\in (\partial \Omega)_1
\end{equation*}
and
\begin{equation*}
\begin{aligned}
|F_2(0,0,0,x)|=&\left|F\left(D^2N_{g}(x),DN_{g}(x),N_{g},x\right)-f(0)\right|\\
=&\big|F\left(D^2N_{g}(x),DN_{g}(x),N_{g}(x),x\right)-F_0(D^2N_{g}(x),DN_{g}(x),N_{g}(x))\\
&+F_0(D^2N_{g}(x),DN_{g}(x),N_{g})-F_0(0,0,0)-f(0)\big|\\
\leq& \beta_2(x)\omega_2(C,C,C)+C+|f(0)|\leq C.
\end{aligned}
\end{equation*}
Note that
\begin{equation*}
u_1\in S^{*}(\lambda,\Lambda,\mu,\hat{b},|f_1|
+c_0\omega_0(\|u\|_{L^{\infty}(\Omega_1)}+\|g\|_{C^{2,\alpha}(0)},u_1)+|F_2(0,0,0,\cdot)|).
\end{equation*}
where $\hat{b}=b_0+2\mu\|g\|_{C^{2,\alpha}(0)}$. By \Cref{t-C1a-mu}, $u_1\in C^{1,\alpha}(0)$, $Du_1(0)=(D_{x'}u_1(0),(u_1)_n(0))=(D_{x'}g_1(0),(u_1)_n(0))=(0,...,0,(u_1)_n(0))$ and
\begin{equation}\label{e.tC2a-1-mu}
\begin{aligned}
|(u_1)_n(0)| &\leq C.
\end{aligned}
\end{equation}
Assume that $\Omega$ satisfies \cref{e-re} and \cref{e-re2} with $P(x')=x'^{T}Ax'/2$ for some $A\in S^{n-1}$. Define $u_2(x)=u_1(x)-N_u(x)$ and $F_3(M,p,s,x)=F_2(M+D^2N_u(x),p+DN_u(x),s+N_u(x),x)$ where $N_u(x)=(u_1)_n(0)\left(x_n-x'^{T}Ax'/2\right)$. Then $u_2$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_3(D^2u_2,Du_2,u_2,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_2=g_2&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_2=g_1-N_u(x)$. Moreover, $u_2(0)=0$, $Du_2(0)=0$ and
\begin{equation*}
|g_2(x)|\leq |g_1(x)|+|N_u(x)|\leq C|x|^{2+\alpha}, ~~\forall ~x\in (\partial \Omega)_1.
\end{equation*}
We remark here that this step shows clearly where the condition $\partial \Omega\in C^{2,\alpha}(0)$ is used. If $\partial \Omega \in C^{2,\alpha}(0)$, we can assume that $Du_2(0)=0$ and keep that $g_2$ has a decay of order $2+\alpha$ near $0$.
Next, take $u_3(x)=u_2(x)-\tau x_n^2$ and $F_4(M,p,s,x)=F_3(M+2\tau\delta_{nn},p+2\tau x_n,s+\tau x_n^2,x)$. Then $u_3$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_4(D^2u_3,Du_3,u_3,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_3=g_3&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_3=g_2-\tau x_n^2$. Similar to the previous step, $u_3(0)=0$, $Du_3(0)=0$ and
\begin{equation*}
|g_3(x)|\leq C|x|^{2+\alpha}, ~~\forall ~x\in (\partial \Omega)_1.
\end{equation*}
Define the fully nonlinear operators $F_{10},F_{20},F_{30}$ and $F_{40}$ in a similar way as $F_1,F_2,F_3$ and $F_4$ (only replacing $F$ by $F_0$ and setting $x=0$ in the expression). By the structure condition, there exists $\tau\in R$ such that $F_{40}(0,0,0)=0$ and
\begin{equation}\label{e.tC2a-2-mu}
\begin{aligned}
|\tau|&\leq |F_{30}(0,0,0)|/\lambda\leq C.
\end{aligned}
\end{equation}
Finally, let $y=x/\rho$, $u_4(y)=u_3(x)/\rho$, $F_5(M,p,s,y)=\rho F_4(M/\rho, p,\rho s,x)$ and $F_{50}(M,p,s)=\rho F_{40}(M/\rho, p,\rho s)$. Then $u_4$ satisfies
\begin{equation}\label{F5-mu}
\left\{\begin{aligned}
&F_5(D^2u_4,Du_4,u_4,y)=f_2&& ~~\mbox{in}~~\tilde{\Omega}_1;\\
&u_4=g_4&& ~~\mbox{on}~~(\partial \tilde\Omega)_1,
\end{aligned}\right.
\end{equation}
where $f_2(y)=\rho f_1(x)$, $g_4(y)=g_3(x)/\rho$ and $\tilde{\Omega}=\Omega/\rho$.
Now, we can check that \cref{F5-mu} satisfies the conditions of \Cref{t-C2as-mu} by choosing a proper $\rho$. First, it can be checked easily that
\begin{equation*}
\begin{aligned}
u_4(0)&=0 ,Du_4(0)=0,|f_2(y)|= \rho|f_1(x)|\leq C\rho^{1+\alpha}|y|^{\alpha},~~\forall ~y\in\tilde{\Omega}_1,\\
|g_4(y)|&= \rho^{-1}|g_3(x)|\leq \rho^{-1}(|g_2(x)|+|\tau|x_n^2)
\leq C\rho^{-1} \left(|x|^{2+\alpha}+|x|^4\right)\\
&\leq C\rho^{1+\alpha}|y|^{2+\alpha},~~\forall ~y\in\partial \tilde{\Omega}_1.\\
\|\partial \tilde{\Omega}\cap &B_1\|_{C^{1,\alpha}(0)}\leq \rho\|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)} \leq \rho\|\partial \Omega\cap B_1\|_{C^{2,\alpha}(0)}\leq C\rho.
\end{aligned}
\end{equation*}
Next, by the boundary $C^{1,\alpha}$ regularity,
\begin{equation*}
\|u_4\|_{L^{\infty}(\tilde{\Omega}_1)}= \frac{\|u_3\|_{L^{\infty}(\Omega_\rho)}}{\rho}
\leq \frac{1}{\rho} \left(\|u_2\|_{L^{\infty}(\Omega_\rho)}+|\tau|\rho^2\right)
\leq \frac{C}{\rho}\left(\rho^{1+\alpha}+\rho^2\right)\leq C\rho^{\alpha}.
\end{equation*}
Furthermore, $F_{50}(0,0,0)=0$, and $F_5$ and $F_{50}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= \rho\mu,\tilde{b}= \rho b_0+C\rho\mu,\tilde{c}= \rho^{1/2}c_0
~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\rho^{1/2}\omega_0(\cdot+C,\cdot).
\end{equation*}
Finally, we check the oscillation of $F_5$ in $y$. Define $N(x)=N_g(x)+N_u(x)+\tau x_n^2$, which is a quadratic polynomial and we have
\begin{equation*}
\|N\|_{C^{2}(\bar{\Omega}_1)}\leq C.
\end{equation*}
Similar to the interior $C^{2,\alpha}$ regularity,
\begin{equation*}
\begin{aligned}
|F&_5(M,p,s,y)-F_{50}(M,p,s)|\\
=&\rho\Big(F(\rho^{-1}M+D^2N,p+DN,\rho s+N,x)
-F_0(\rho^{-1}M+D^2N,p+DN(0),\rho s+N(0),0)\Big)\\
\leq& \rho\beta_2(x)\omega_2(\rho^{-1}\|M\|+C,
|p|+C,\rho|s|+C)+C\rho\mu(|p|+C)|x|\\
&+C\rho b_0|x|+\rho c_0\omega_0(|s|+C,\rho|x|)\\
\leq& \left(C\rho^{\alpha}\omega_2\left(\|M\|+C,|p|+C,|s|+C\right)+C\rho^{2}|p|
+C\rho^{2}+C\rho^{1+\alpha}\omega_0(|s|+C,1)\right)|y|^{\alpha}\\
:= &\tilde{\beta}_2(y)\tilde{\omega}_2\left(\|M\|,|p|,|s|\right),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_2(y)=C\rho^{\alpha}|y|^{\alpha}$,
\begin{equation*}
\tilde{\omega}_2(\|M\|,|p|,|s|)=\omega_2\left(\|M\|+C,|p|+C,|s|+C\right)
+C|p|+C+C\omega_0(|s|+C,C)
\end{equation*}
and $\tilde{\omega}_2$ satisfies \cref{e.omega}.
Take $\delta_1$ small enough such that \Cref{t-C2as-mu} holds with $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $n, \lambda,\Lambda,\alpha,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\alpha}(0)}$, $\omega_2$, $\|\partial \Omega\|_{C^{2,\alpha}(0)}, \|f\|_{C^{\alpha}(0)},
\|g\|_{C^{2,\alpha}(0)}$ and $\|u\|_{L^{\infty }(\Omega_1)}$) such that
\begin{equation*}
\begin{aligned}
&\|u_4\|_{L^{\infty }(\tilde{\Omega}_1)}\leq 1,\tilde\mu \leq \frac{\delta_1}{4C_0},\tilde{b}\leq \frac{\delta_1}{2}, \tilde c\leq \frac{\delta_1}{K_0},\tilde{\omega}_0(1+C_0,1)\leq 1,\\
&|\tilde\beta_2(y)|\leq \delta_1|y|^{\alpha},|f_2(y)|\leq \delta_1|y|^{\alpha}, ~~\forall ~y\in \tilde\Omega_1,\\
&|g_4(y)|\leq \frac{\delta_1}{2}|y|^{2+\alpha}, ~~\forall ~y\in (\partial \tilde\Omega)_1
~~\mbox{and}~~\|\partial \tilde\Omega\cap B_1\|_{C^{1,\alpha}(0)} \leq \frac{\delta_1}{2C_0},
\end{aligned}
\end{equation*}
where $C_0$ depending only on $n,\lambda,\Lambda$ and $\alpha$, is as in \Cref{t-C2as-mu}. Therefore, the assumptions in \Cref{t-C2as-mu} are satisfied for \cref{F5-mu}. By \Cref{t-C2as-mu}, $u_4$ and hence $u$ is $C^{2,\alpha}$ at $0$, and the estimates \crefrange{e.C2a-1-mu}{e.C2a-2-mu} hold.
\textbf{Case 2:} $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0}. Let $K=\|u\|_{L^{\infty }(\Omega_1)}+\|f\|_{C^{\alpha}(0)}+\|g\|_{C^{2,\alpha}(0)}
+\|\gamma_2\|_{C^{\alpha}(0)}$ and $u_1=u/K$. Then $u_1$ satisfies
\begin{equation}\label{e.8.1}
\left\{\begin{aligned}
&F_1(D^2u_1,Du_1,u_1,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_1=g_1&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation}
where $F_1(M,p,s,x)=F(KM,Kp,Ks,x)/K$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar\Omega_1$, $f_1=f/K$ and $g_1=g/K$.
Clearly,
\begin{equation*}
\|u_1\|_{L^{\infty }(\Omega_1)}\leq 1,~\|f_1\|_{C^{\alpha}(0)}\leq 1~~\mbox{and}~~ \|g_1\|_{C^{2,\alpha}(0)}\leq 1.
\end{equation*}
In addition, $F_1$ satisfies the structure condition \cref{SC1} with the same $\lambda,\Lambda,b_0$ and $c_0$. In addition, define $F_{10}(M,p,s)=F_0(KM,Kp,Ks)/K$. Then,
\begin{equation*}
\begin{aligned}
|F_1(M,p,s,x)-F_{10}(M,p,s)|= & K^{-1}|F(KM,Kp,Ks,x)-F_0(KM,Kp,Ks)|\\
\leq & \beta_2(x)\left(\|M\|+|p|+|s|\right)+\tilde\gamma_2(x),
\end{aligned}
\end{equation*}
where $\tilde\gamma_2(x)=K^{-1}\gamma_2(x)$ and thus $\|\tilde{\gamma}_2\|_{C^{\alpha}(0)}\leq 1$.
Apply \textbf{Case 1} to \cref{e.8.1}, we obtain that $u_1$ and hence $u$ is $C^{2,\alpha}$ at $0$, and the estimates \crefrange{e.C2a-1}{e.C2a-2} hold. \qed~\\
\section{Boundary $C^{k,\alpha}$ regularity}\label{Cka-mu}
In this section, we prove the boundary pointwise $C^{k,\alpha}$ regularity for $k\geq 3$. First, we establish the model problem (similar to \Crefrange{l-3modin1}{l-32}) for boundary $C^{k,\alpha}$ regularity. The following lemma states the local $C^{2,\alpha}$ regularity up to the boundary which combines the interior pointwise regularity and the boundary pointwise regularity. The proof is standard and we adopt it from \cite{MR3246039}.
\begin{lemma}\label{l-61}
Let $0<\alpha<\bar{\alpha}$. Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF} for any $x_0\in \bar{B}_1^+$ with $\beta_2\in C^{\alpha}(\bar{B}_1^+)$. Assume that $\omega_0$ satisfies \cref{e.omega0} and $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=0&& ~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1.
\end{aligned}\right.
\end{equation*}
Then $u\in C^{2,\alpha}(\bar{B}^+_{1/2})$ and
\begin{equation*}
\|u\|_{C^{2,\alpha}(\bar{B}^+_{1/2})}\leq C,
\end{equation*}
where $C$ depends only on $n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,
\|\beta_2\|_{C^{\alpha}(\bar{B}_1^+)},\omega_2$ and $\|u\|_{L^{\infty }(B_1^+)}$.
\end{lemma}
\proof We only need to prove that given $x\in \bar{B}^+_{1/2}$, there exists a quadratic polynomial $P_x$ such that for any $y\in \bar{B}^+_{1/2}$,
\begin{equation}\label{e.l61-3}
|u(y)-P_x(y)|\leq C|x-y|^{2+\alpha},
\end{equation}
where $C$ depends only on $n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,\omega_2,
\|\beta_2\|_{C^{\alpha}(\bar{B}_1^+)}$ and $\|u\|_{L^{\infty }(B_1^+)}$. Throughout this proof, $C$ always denotes a constant having the same dependence.
Let $\tilde{x}=(x',0)$. By the boundary $C^{2,\alpha}$ regularity \Cref{t-C2a}, there exists a quadratic polynomial $P_{\tilde{x}}$ such that
\begin{equation*}
|u(y)-P_{\tilde{x}}(y)|\leq C|y-\tilde{x}|^{2+\alpha}, ~\forall ~y\in \bar{B}_{1}^{+}.
\end{equation*}
Set $v(y)=u(y)-P_{\tilde{x}}(y)$ and $v$ satisfies
\begin{equation}\label{e.Cka-4}
\tilde{F}(D^2v(y),Dv(y),v(y),y)=0~~~~\mathrm{in}~B_{x_n}(x),
\end{equation}
where $\tilde{F}(M,p,s,y)=F(M+D^2P_{\tilde{x}}(y),p+DP_{\tilde{x}}(y),s+P_{\tilde{x}}(y),y)$. Then \cref{e.Cka-4} satisfies the conditions of the interior $C^{2,\alpha}$ regularity (see \Cref{t-C2a-i}). Hence, there exists a quadratic polynomial $P$ such that
\begin{equation*}
\begin{aligned}
&|v(y)-P(y)|\leq C\|v\|_{L^\infty(B_{x_n}(x))}\frac{|y-x|^{2+\alpha}}{|x_n|^{2+\alpha}}
\leq C |y-x|^{2+\alpha}~~\mathrm{in}~B_{x_n/2}(x),\\
&|P(x)|=|v(x)|=|u(x)-P_{\tilde{x}}(x)|\leq C|x_n|^{2+\alpha},\\
&|D P(x)|\leq C\frac{\|v\|_{L^\infty(B_{x_n}(x))}}{|x_n|}\leq C|x_n|^{1+\alpha},\\
&|D^2P(x)|\leq C\frac{\|v\|_{L^\infty(B_{x_n}(x))}}{x_n^2}\leq C|x_n|^{\alpha}.
\end{aligned}
\end{equation*}
Let $P_x=P_{\tilde{x}}+P$. If $|y-x|<x_n/2$, we have
\begin{equation*}
|u(y)-P_x(y)|=|v(y)-P(y)|\leq C|y-x|^{2+\alpha}.
\end{equation*}
If $|y-x|>x_n/2$, we have
\begin{equation*}
\begin{aligned}
|u(y)-P_x(y)|&\leq |u(y)-P_{\tilde{x}}(y)|+|P(y)|\\
&\leq C|y-\tilde{x}|^{2+\alpha}+|P(x)+DP(x)\cdot (y-x)+(y-x)^TD^2P(x)(y-x)|\\
&\leq C|y-\tilde{x}|^{2+\alpha}+C(x_n^{2+\alpha}+|y-x|x_n^{1+\alpha}+|y-x|^2x_n^{\alpha})\\
&\leq C|y-x|^{2+\alpha}.
\end{aligned}
\end{equation*}~\qed \\
Now, we establish the model problem for boundary $C^{k,\alpha}$ regularity.
\begin{lemma}\label{l-62}
Suppose that $F_0\in C^{k-2,\bar\alpha}(S^n\times R^n\times R\times \bar{B}_1^+)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F_0(D^2u,Du,u,x)=0&& ~~\mbox{in}~~B_1^+;\\
&u=0&& ~~\mbox{on}~~T_1.
\end{aligned}\right.
\end{equation*}
Then $u\in C^{k,\alpha}(\bar{B}^+_{1/2})$ for any $0<\alpha<\bar{\alpha}$ and
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}^+_{1/2})}\leq C_k,
\end{equation*}
where $C_k$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,\omega_4$ and $\|u\|_{L^{\infty }(B_1^+)}$.
In particular, $u\in C^{k,\alpha}(0)$ and there exists a $k$-th order polynomial $P=\sum_{m=1}^{k} Q_m$ where $Q_m(1\leq m\leq k)$ are $m$-forms, such that
\begin{equation}\label{e.l62-1}
|u(x)-P(x)|\leq C_k |x|^{k+\alpha}, ~~\forall ~x\in B_{1}^+,
\end{equation}
\begin{equation}\label{e.l62-2}
|F_0(D^2P(x),DP(x),P(x),x)|\leq C_k|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}^+
\end{equation}
and
\begin{equation}\label{e.l62-3}
\sum_{m=1}^{k}\|Q_m\|\leq C_k.
\end{equation}
\end{lemma}
\proof The proof is similar to \Cref{In-l-62}. Since $F_0\in C^{k-2,\bar{\alpha}}$, for any $x_0,x\in B_1\cup T_1$, $(M,p,s)\in S^n\times R^n\times R$,
\begin{equation*}
\begin{aligned}
&|F_0(M,p,s,x)-F_0(M,p,s,x_0)|\\
&\leq |F_{0,x_i}(M,p,s,\xi)||x_i-x_{0i}|\\
&\leq (|F_{0,x_i}(M,p,s,\xi)-F_{0,x_i}(0,0,0,0)|+|F_{0,x_i}(0,0,0,0)|)|x_i-x_{0i}|\\
&\leq C\left(\|M\|^{\bar\alpha}+|p|^{\bar\alpha}+|s|^{\bar\alpha}+1\right)|x-x_0|\\
&:=\tilde{\beta}_2(x,x_0)\tilde{\omega}_2(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_2(x,x_0)=|x-x_0|,~\tilde{\omega}_2(\|M\|,|p|,|s|)
=C\left(\|M\|^{\bar\alpha}+|p|^{\bar\alpha}+|s|^{\bar\alpha}+1\right),
\end{equation*}
and $C$ depends only on $\omega_4$. Moreover, $\omega_0$ satisfies \cref{e.omega0} and $\tilde{\omega}_2$ satisfies \cref{e.omega}. By \Cref{l-61}, $u\in C^{2,\alpha}(\bar{B}_{3/4}^+)$ for any $0<\alpha<\bar{\alpha}$ and
\begin{equation*}
\|u\|_{C^{2,\alpha}(\bar{B}_{3/4}^+)}\leq C,
\end{equation*}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu,b_0,c_0,\omega_0,\omega_4$ and $\|u\|_{L^{\infty }(B_1^+)}$. Unless stated otherwise, $C$ always has the same dependence in the following proof.
Now, we show that $u\in C^{3,\alpha}$, which can be proved by the standard technique of difference quotient. The only difference from the interior case is that we can't take the difference quotient along $e_n$. Let $h>0$ be small and $1\leq l\leq n-1$. By taking the difference quotient along $e_l$ on both sides of the equation and applying the Schauder estimates for linear equations, we obtain (similar to \Cref{In-l-62}) that $u_l \in C^{2,\alpha^2}(\bar B_{1/2}^+)$ and
\begin{equation*}
\|u_l\|_{C^{2,\alpha^2}(\bar{B}_{1/2}^+)}\leq C.
\end{equation*}
It follows that $u_{il}\in C^{1,\alpha^2}$ ($i+l<2n$). Combining with $F_0(D^2u,Du,u,x)=0$ and the implicit function theorem, $u_{nn}\in C^{1,\alpha^2}$. Thus, $u\in C^{3,\alpha^2}$ and
\begin{equation*}
\|u\|_{C^{3,\alpha^2}(\bar B_{1/2}^+)}\leq C.
\end{equation*}
Since $u\in C^{3,\alpha^2}$, for $1\leq l <n$, $u_l$ satisfies
\begin{equation}\label{uksat}
a_{ij}(u_l)_{ij}=-G,
\end{equation}
where
\begin{equation}\label{aijuk}
a_{ij}(x)=F_{0,ij}(D^2u,Du,u,x)
\end{equation}
and
\begin{equation}\label{Guk}
G(x)=F_{0,p_i}(D^2u,Du,u,x)u_{il}+F_{0,s}(D^2u,Du,u,x)u_l+F_{0,x_l}(D^2u,Du,u,x).
\end{equation}
Then it can be seen that $a_{ij}\in C^{\alpha}(\bar{B}^+_{1/2})$ and $G\in C^{\alpha}(\bar{B}^+_{1/2})$. By the Schauder's estimates for linear equations, $u_l\in C^{2,\alpha}$. Then $u\in C^{3,\alpha}$ and
\begin{equation*}
\|u\|_{C^{3,\alpha}(\bar{B}_{1/4}^+)}\leq C.
\end{equation*}
Since $F_0\in C^{k-2,\bar\alpha}$, by considering \crefrange{uksat}{Guk} iteratively and carrying out a covering argument, we obtain
\begin{equation*}
\|u\|_{C^{k,\alpha}(\bar{B}^+_{1/2})}\leq C.
\end{equation*}
In particular, $u\in C^{k,\alpha}(0)$ and there exists a $k$-th order polynomial $P$ such that \cref{e.l62-1} holds. Note that $u(0)=0$, $u_i(0)=0~(1\leq i<n)$, $u_{ij}(0)=0 ~(1\leq i,j<n)$ etc. Hence, $P$ has the form $P=\sum_{m=1}^{k} Q_m$ where $Q_m$ are $m$-forms. Moreover, \cref{e.l62-3} holds. Finally, similar to \Cref{In-l-62},
\begin{equation*}
\begin{aligned}
&|F_0(D^2P(x),DP(x),P(x),x)|\\
&=|F_0(D^2P(x),DP(x),P(x),x)-F_0(D^2u(x),Du(x),u(x),x)|\\
&=|\int_{0}^{1} F_{0,ij}(\xi)(P_{ij}-u_{ij})+F_{0,p_i}(\xi)(P_{i}-u_{i})
+F_{0,s}(\xi)(P-u)dr|\\
&\leq C|x|^{k-2+\alpha},
\end{aligned}
\end{equation*}
where $\xi=r\left(D^2P(x),DP(x),P(x),x\right)
+(1-r)\left(D^2u(x),Du(x),u(x),x\right)$. Then
\begin{equation*}
D^{l}_x \Big(F_0(D^2P(x),DP(x),P(x),x)\Big)\Big|_{x=0}=0,~\forall ~0\leq l\leq k-2.
\end{equation*}
Combining with $F_0\in C^{k-2,\bar{\alpha}}$, \cref{e.l62-2} holds. \qed~\\
In the following, we prove the boundary pointwise $C^{k,\alpha}(k\geq 3)$ regularity \Cref{t-Cka} by induction. For $k=3$, $F\in C^{1,\alpha}(0)$ and by \cref{holder}, $F(M,p,s,0)\equiv F_0(M,p,s,0)$ for any $(M,p,s)\in S^n\times R^n\times R$. Hence,
\begin{equation*}
\begin{aligned}
&|F(M,p,s,x)-F(M,p,s,0)|\\
&\leq |F(M,p,s,x)-F_0(M,p,s,x)|+|F_0(M,p,s,x)-F_0(M,p,s,0)|\\
&\leq \|\beta_3\|_{C^{1,\alpha}(0)}\omega_3(\|M\|,|p|,|s|)|x|^{1+\alpha}
+|D_xF_{0}(M,p,s,\xi)||x|\\
&\leq C\left(\omega_3(\|M\|,|p|,|s|)+\|M\|^{\alpha}+|p|^{\alpha}+|s|^{\alpha}+1\right)|x|\\
&:=\tilde\omega_3(\|M\|,|p|,|s|)|x|.
\end{aligned}
\end{equation*}
In addition, $\tilde\omega_3$ satisfies \cref{e.omega}. Then, from the boundary pointwise
$C^{2,\alpha}$ regularity \Cref{t-C2a}, $u\in C^{2,\alpha}(0)$. Hence, we can assume that the boundary $C^{k-1,\alpha}(0)$ regularity holds if $F\in C^{k-2,\alpha}(0)$ and we need to prove the boundary $C^{k,\alpha}(0)$ regularity.
The following lemma is a higher order counterpart of \Cref{l-C1a-mu} and \Cref{l-C2a-mu}.
\begin{lemma}\label{l-Cka-mu}
Let $0<\alpha<\bar{\alpha}$. Suppose that $F\in C^{k-2,\alpha}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Then there exists $\delta>0$ depending only on $k,n,\lambda,\Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $u(0)=0,Du(0)=0,\cdots,D^{k-1}u(0)=0$, $\mu,b_0,c_0\leq 1$, $\|\beta_3\|_{C^{k-2,\alpha}(0)}\leq \delta$, $\|f\|_{C^{k-3,\alpha}(0)}\leq \delta$, $\|g\|_{C^{k-1,\alpha}(0)}\leq \delta$ and $\|\partial \Omega\|_{C^{1,\alpha}(0)} \leq \delta$, then there exists a $k$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(\Omega_{\eta})}\leq \eta^{k+\alpha},
\end{equation*}
\begin{equation*}
|F_0(D^2Q(x),DQ(x),Q(x),x)|\leq C |x|^{k-2+\bar\alpha},~\forall ~x\in \Omega_1
\end{equation*}
and
\begin{equation*}\label{e.16.2-mu}
\|Q\|\leq C,
\end{equation*}
where $C$ and $\eta$ depend only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
\end{lemma}
\begin{remark}\label{r-9.2}
Note that we only assume that $\partial \Omega\in C^{1,\alpha}(0)$. Clearly, the $C^{1,\alpha}(0)$ regularity holds. Since $u(0)=0$ and $Du(0)=0$, from the boundary pointwise $C^{2,\alpha}$ regularity (see last section), $u\in C^{2,\alpha}(0)$. By induction, we can assume that $u\in C^{k-1,\alpha}(0)$. In fact, for $\beta_3$, we have similar consideration and assuming $\beta_3\in C^{\alpha}(0)$ is enough.
\end{remark}
\proof Throughout this proof, $C$ always denotes a constant depending only on $k,n,\lambda, \Lambda$, $\alpha$, $\omega_0$, $\omega_3$ and $\omega_4$. As before, we prove the lemma by contradiction. Suppose that the lemma is false. Then there exist $0<\alpha<\bar{\alpha},\omega_0,\omega_3,\omega_4$ and sequences of $F_m$, $u_m,f_m,g_m,\Omega_m$ satisfying
\begin{equation*}
\left\{\begin{aligned}
&F_m(D^2u_m,Du_m,u_m,x)=f_m&& ~~\mbox{in}~~\Omega_m\cap B_1;\\
&u_m=g_m&& ~~\mbox{on}~~\partial \Omega_m\cap B_1.
\end{aligned}\right.
\end{equation*}
In addition, $F_m$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\mu\equiv 1,b\equiv 1,c\equiv 1$ and $\omega_0$. Moreover, $F_m$ is $C^{k-2,\alpha}$ at $0$ with $F_{0m}$, $\beta_m$ and $\omega_3$, and
\begin{equation*}
\|F_{0m}\|_{C^{k-2,\bar{\alpha}}(\bar{\textbf{B}}_r\times \overline{\Omega_m\cap B_1})}\leq \omega_4(r),~\forall ~r>0.
\end{equation*}
Furthermore, $\|u_m\|_{L^{\infty}(\Omega_m\cap B_1)}\leq 1$, $u(0)=0, Du_m(0)=0,\dots,D^{k-1}u_m(0)=0$, $\|\beta_{m}\|_{C^{k-2,\alpha}(0)}\leq 1/m$, $\|f_{m}\|_{C^{k-3,\alpha}(0)}\leq 1/m$, $\|g_{m}\|_{C^{k-1,\alpha}(0)}\leq 1/m$ and $\|\partial \Omega_m\|_{C^{1,\alpha}(0)} \leq 1/m$. Finally, for any $k$-form $Q$ satisfying $|Q|\leq C_k$ and
\begin{equation}\label{e.cka.Q-mu}
|F_{0m}(D^2Q(x),DQ(x),Q(x),x)|\leq C_k |x|^{k-2+\bar\alpha},~\forall ~x\in \Omega_m\cap B_1
\end{equation}
we have
\begin{equation}\label{e.lCka.1-mu}
\|u_m-Q\|_{L^{\infty}(\Omega_{m}\cap B_{\eta})}> \eta^{k+\alpha},
\end{equation}
where $C_k$ is to be specified later and $0<\eta<1$ is taken small such that
\begin{equation}\label{e.lCka.2-mu}
C_k\eta^{\alpha_1-\alpha}<1/2
\end{equation}
for $\alpha_1=(\bar{\alpha}+\alpha)/2$.
As before, $u_m$ are uniformly bounded and equicontinuous, and there exists $u:B_1^+\cup T_1\rightarrow R$ such that $u_m$ converges uniformly to $u$ on compact subsets of $ B_1^+\cup T_1$.
In addition, there exists $F_0\in C^{k-2,\bar{\alpha}}(S^n\times R^n\times R\times \bar{B}_1^+)$ such that $F_{0m}\rightarrow F_0$ in $C^{k-2,\alpha_1}$ in any compact subset of $S^n\times R^n\times R\times \bar{B}_1^+$ and $\|F_0\|_{C^{k-2,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{B}_1^+)}\leq \omega_4(r)$. Moreover, it is easy to verify that $F_0$ satisfies the structure condition \cref{SC2} with $\lambda,\Lambda,\mu\equiv 1,b\equiv 1,c\equiv 1$ and $\omega_0$. Furthermore, for any $B\subset\subset B_1^+$ and $\varphi\in C^2(\bar{B})$, let $r=\|D^2\varphi\|_{L^{\infty}(B)}+\|D\varphi\|_{L^{\infty}(B)}+1$, $G_m=F_m(D^2\varphi,D\varphi,u_m,x)-f_m$ and $G(x)=F_0(D^2\varphi,D\varphi,u,x)$. Then
\begin{equation*}
\begin{aligned}
&\|G_m-G\|_{L^n(B)}\\
&\leq \|F_m(D^2\varphi,D\varphi,u_m,x)-F_{0m}(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}\\
&+\|F_{0m}(D^2\varphi,D\varphi,u_m,x)-F_0(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}+\|f_m\|_{L^n(B)}\\
&\leq \omega_3(r,r,r)\|\beta_{m}\|_{L^n(B)}
+\|F_{0m}(D^2\varphi,D\varphi,u_m,x)-F_0(D^2\varphi,D\varphi,u_m,x)\|_{L^n(B)}+\|f_m\|_{L^n(B)}.
\end{aligned}
\end{equation*}
Thus, $\|G_m-G\|_{L^n(B)}\rightarrow 0$ as $m\rightarrow \infty$. Hence, from \Cref{l-35}, $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_0(D^2u,Du,u,x)=0&& ~~\mbox{in}~~B_{1}^+;\\
&u=0&& ~~\mbox{on}~~T_{1}.
\end{aligned}\right.
\end{equation*}
By the $C^{k-1,\alpha}$ estimate for $u_m$ and noting $u_m(0)=0$, $Du_m(0)=0$, $\cdots$, $D^{k-1}u_m(0)=0$, we have
\begin{equation*}
\|u_m\|_{L^{\infty }(\Omega_m\cap B_r)}\leq Cr^{k-1+\alpha} ,~~~~\forall~0<r<1.
\end{equation*}
Since $u_m$ converges to $u$ uniformly,
\begin{equation*}
\|u\|_{L^{\infty }(B_r^+)}\leq Cr^{k-1+\alpha}, ~~~~\forall~0<r<1.
\end{equation*}
Hence, $u(0)=0$, $Du(0)=0$, $\cdots$, $D^{k-1}u(0)=0$. Combining with \Cref{l-62}, there exists a $k$-form $\tilde{Q}$ such that
\begin{equation}\label{e.cka-5}
|u(x)-\tilde{Q}(x)|\leq C_1 |x|^{k+\alpha_1}, ~~\forall ~x\in B_{1}^+,
\end{equation}
\begin{equation}\label{e.cka-4}
|F_0(D^2\tilde{Q}(x),D\tilde{Q}(x),\tilde{Q}(x),x)|\leq C_1|x|^{k-2+\bar\alpha}, ~~\forall ~x\in B_{1}^+
\end{equation}
and
\begin{equation*}
\|\tilde{Q}\|\leq C_1,
\end{equation*}
where $C_1$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Similar to \Cref{In-l-Cka-mu}, we can construct a sequence of $k$-forms $\tilde{Q}_m$ such that \cref{e.cka.Q-mu} holds for $\tilde{Q}_m$ and $\tilde{Q}_m\rightarrow \tilde{Q}$ as $m\rightarrow \infty$. Thus, \cref{e.lCka.1-mu} holds for $\tilde{Q}_m$. Let $m\rightarrow \infty$, we have
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta}^+)}\geq \eta^{k+\alpha},
\end{equation*}
However, by \cref{e.lCka.2-mu} and \cref{e.cka-5},
\begin{equation*}
\|u-\tilde{Q}\|_{L^{\infty}(B_{\eta}^+)}\leq \eta^{k+\alpha}/2,
\end{equation*}
which is a contradiction. ~\qed~\\
The following is a boundary pointwise $C^{k,\alpha}$ regularity. The key is that if $u(0)=0$, $Du(0)=0$, $\cdots$ $D^{k-1}u(0)=0$, the boundary pointwise $C^{k,\alpha}$ regularity holds even if $\partial \Omega\in C^{1,\alpha}(0)$.
\begin{theorem}\label{t-Ckas-mu}
Let $0<\alpha <\bar{\alpha}$. Suppose that $F\in C^{k-2,\alpha}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Let $u$ satisfy
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1.
\end{aligned}\right.
\end{equation*}
Assume that
\begin{equation}\label{e.tCkas-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(\Omega_1)}\leq 1,~ u(0)=0,~Du(0)=0,\dots,D^{k-1}u(0)=0,\\
&\mu\leq \frac{1}{4C_0},~b_0\leq \frac{1}{2},~c_0\leq 1,\\
&|\beta_3(x)|\leq \frac{\delta_1}{K_0}|x|^{k-2+\alpha}, ~|f(x)|\leq \delta_1|x|^{k-2+\alpha}, ~~\forall ~x\in \Omega_1,\\
&|g(x)|\leq \frac{\delta_1}{2}|x|^{k+\alpha}, ~~\forall ~x\in (\partial \Omega)_1
~~\mbox{and}~~\|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)} \leq \frac{\delta_1}{2C_0},\\
\end{aligned}
\end{equation}
where $\delta_1\leq \delta$ ($\delta$ is as in \Cref{l-Cka-mu}) and $C_0$ depend only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Then $u\in C^{k,\alpha}(0)$ and there exists a $k$-form $Q$ such that
\begin{equation}\label{e.tCkas-1-mu}
|u(x)-Q(x)|\leq C |x|^{k+\alpha}, ~~\forall ~x\in \Omega_{1},
\end{equation}
\begin{equation*}
|F_0(D^2Q(x),DQ(x),Q(x),x)|\leq C |x|^{k-2+\bar\alpha}, ~~\forall ~x\in \Omega_{1}
\end{equation*}
and
\begin{equation}\label{e.tCkas-2-mu}
\|Q\|\leq C,
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
\end{theorem}
\proof As before, to prove that $u$ is $C^{k,\alpha}$ at $0$, we only need to prove the following. There exist a sequence of $k$-forms $Q_m$ ($m\geq 0$) such that for all $m\geq 1$,
\begin{equation}\label{e.tCkas-6-mu}
\|u-Q_m\|_{L^{\infty }(\Omega _{\eta^{m}})}\leq \eta ^{m(k+\alpha )},
\end{equation}
\begin{equation}\label{e.tCkas-9-mu}
|F_0(D^2Q_m(x),DQ_m(x),Q_m(x),x)|\leq \tilde{C}|x|^{k-2+\alpha}, ~~\forall ~x\in \Omega_{1}
\end{equation}
and
\begin{equation}\label{e.tCkas-7-mu}
\|Q_m-Q_{m-1}\|\leq \tilde{C}\eta ^{(m-1)\alpha},
\end{equation}
where $\tilde{C}$ and $\eta$ depend only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
We prove the above by induction. For $m=1$, by \Cref{l-Cka-mu}, there exists a $k$-form $Q_1$ such that \crefrange{e.tCkas-6-mu}{e.tCkas-7-mu} hold for some $C_1$ and $\eta_1$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ where $Q_0\equiv 0$. Take $\tilde{C}\geq C_1, \eta\leq \eta_1$ and then the conclusion holds for $m=1$. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{e.tCkas-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^{k+\alpha}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{e.Ckas-F-mu}
\left\{\begin{aligned}
&\tilde{F}(D^2v,Dv,v,y)=\tilde{f}&& ~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&v=\tilde{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar{\tilde\Omega}_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{F(r^{k-2+\alpha}M+D^2Q_{m_0}(x),r^{k-1+\alpha}p+DQ_{m_0}(x),
r^{k+\alpha}s+Q_{m_0}(x),x)}{r^{k-2+\alpha}},\\
&\tilde{f}(y)=\frac{f(x)}{r^{k-2+\alpha}},~
\tilde{g}(y)=\frac{g(x)-Q_{m_0}(x)}{r^{k+\alpha}}~~\mbox{and}~~\tilde{\Omega}=\frac{\Omega}{r}.\\
\end{aligned}
\end{equation*}
In addition, define $\tilde{F}_0$ in a similar way to the definition of $\tilde{F}$ (only replacing $F$ by $F_0$).
In the following, we show that \cref{e.Ckas-F-mu} satisfies the assumptions of \Cref{l-Cka-mu}. First, by \cref{e.tCkas-7-mu}, there exists a constant $C_{0}$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that $\|Q_m\|\leq C_{0}$ ($\forall~0\leq m\leq m_0$). Then it is easy to verify that (note that $Q_{m_0}$ is a $k$-form)
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(\tilde{\Omega}_1)}\leq 1, v(0)=0,\cdots,D^{k-1}v(0)=0,~(\mathrm{by}~ \cref{e.tCkas-be-mu},~ \cref{e.tCkas-6-mu} ~\mbox{and}~\cref{e.tCkas-v-mu})\\
&|\tilde{f}(y)|\leq r^{-(k-2+\alpha)}|f(x)|\leq \delta_1|y|^{k-2+\alpha}, ~\forall ~y\in \tilde{\Omega}_1,~(\mathrm{by}~ \cref{e.tCkas-be-mu})\\
&|\tilde{g}(y)|\leq \frac{1}{r^{k+\alpha}}\left(\frac{\delta_1( r|y|)^{k+\alpha}}{2}+
\frac{C_{0}\delta_1(r|y|)^{k+\alpha}}{2C_0}\right)\leq \delta_1|y|^{k+\alpha},~\forall ~y\in (\partial\tilde{\Omega})_1,~(\mathrm{by}~ \cref{e.tCkas-be-mu})\\
&\|\partial \tilde{\Omega}\cap B_1\|_{C^{1,\alpha}(0)}\leq r^{\alpha}\|\partial\Omega\cap B_1\|_{C^{1,\alpha}(0)}\leq \delta_1.~(\mathrm{by}~ \cref{e.tCkas-be-mu})
\end{aligned}
\end{equation*}
Next, it is easy to check that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}=r^{k+\alpha}\mu,~~\tilde{b}= rb_0+2 C_0 r\mu ,~~\tilde{c}= K_0 r^{2}c_0~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence, $\tilde{\omega}_0$ satisfies \cref{e.omega0-2} and from \cref{e.tCkas-be-mu},
\begin{equation*}
\begin{aligned}
&\tilde{\mu}\leq \mu\leq 1,~\tilde{b}\leq b_0+2C_0\mu\leq 1,~\tilde{c}\leq c_0\leq 1.
\end{aligned}
\end{equation*}
In addition, for $(M,p,s,y)\in S^n\times R^n\times R\times \bar{\tilde{\Omega}}_1$,
\begin{equation}\label{e.9.1}
\begin{aligned}
|&\tilde{F}(M,p,s,y)-\tilde{F}_0(M,p,s,y)|\\
&\leq r^{-(k-2+\alpha)}\beta_3(x)\omega_3(r^{k-2+\alpha}\|M\|+C_0r^{k-2},
r^{k-1+\alpha}|p|+C_0r^{k-1},r^{k+\alpha}|s|+C_0r^{k})\\
&\leq \delta_1\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)|y|^{k-2+\alpha}\\
&:=\tilde{\beta}_3(y)\tilde{\omega}_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation}
where $\tilde{\beta}_3(y)=\delta_1|y|^{k-2+\alpha}$ and $\tilde{\omega}_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)$. Hence, $\tilde{\omega}_3$ satisfies \cref{e.omega-3} and
\begin{equation*}
\|\tilde{\beta}_3\|_{C^{k-2,\alpha}(0)}\leq \delta_1.
\end{equation*}
Finally, with the aid of \cref{e.tCkas-9-mu}, we can show that (similar to the interior $C^{k,\alpha}$ regularity)
\begin{equation*}
\|\tilde{F}_0\|_{C^{k-2,\bar\alpha}(\bar{\textbf{B}}_\rho\times \bar{\tilde\Omega}_1)}\leq \tilde{\omega}_4(\rho),~\forall ~\rho>0,
\end{equation*}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$.
Choose $\delta_1$ small enough (depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$) such that \Cref{l-Cka-mu} holds for $\tilde\omega_0,\tilde\omega_3,\tilde\omega_4$ and $\delta_1$. Since \cref{e.Ckas-F-mu} satisfies the assumptions of \Cref{l-Cka-mu}, there exist a $k$-form $\tilde{Q}$, constants $\tilde{C}\geq C_1$ and $\eta\leq \eta_1$ depending only on $k,n,\lambda, \Lambda,\alpha,\omega_0,\omega_3$ and $\omega_4$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(\tilde{\Omega} _{\eta})}&\leq \eta^{k+\alpha},
\end{aligned}
\end{equation*}
\begin{equation*}
|\tilde F_0(D^2\tilde{Q}(y),D\tilde{Q}(y),\tilde{Q}(y),y)|\leq \tilde{C}|y|^{k-2+\bar\alpha}
, ~~\forall ~y\in \tilde{\Omega}_{1}
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \tilde{C}.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+r^{\alpha}\tilde{Q}(x)$. Then \cref{e.tCkas-9-mu} and \cref{e.tCkas-7-mu} hold for $m_0+1$. Recalling \cref{e.tCkas-v-mu}, we have
\begin{equation*}
\begin{aligned}
\|u-Q_{m_0+1}(x)\|_{L^{\infty}(\Omega_{\eta^{m_0+1}})}&= \|u-Q_{m_0}(x)-r^{\alpha}\tilde{Q}(x)\|_{L^{\infty}(\Omega_{\eta r})}\\
&= \|r^{k+\alpha}v-r^{k+\alpha}\tilde{Q}(y)\|_{L^{\infty}(\tilde{\Omega}_{\eta})}\\
&\leq r^{k+\alpha}\eta^{k+\alpha}=\eta^{(m_0+1)(k+\alpha)}.
\end{aligned}
\end{equation*}
Hence, \cref{e.tCkas-6-mu} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
\begin{remark}\label{r-9.1}
Roughly speaking, in above proof, the condition on $\beta_3$ is only used in \cref{e.9.1}. Suppose that $\omega_3$ satisfies for some constant $K_0$ (e.g., $\omega_3(\|M\|,|p|,|s|)=\|M\|+|p|+|s|$)
\begin{equation*}
r\omega_3\left(r^{-1}\|M\|,r^{-1}|p|,r^{-1}|s|\right)\leq K_0\omega_3(\|M\|,|p|,|s|)~\forall ~r>0.
\end{equation*}
Then we can obtain \cref{e.9.1} by only assuming $\beta_3\in C^{\alpha}$.
Furthermore, if we consider the linear equation \cref{e.linear} and assume that $u(0)=0,Du(0)=0,\dots,D^{k-1}u(0)=0$, we can prove the $C^{k,\alpha}(0)$ regularity under the condition that $a^{ij}\in C^{\alpha}(0)$, $b^i\in L^{p} (p=n/(1-\alpha))$, $c\in L^n$ and $\partial \Omega\in C^{1,\alpha}(0)$. The proof is similar to the above. This corresponds to the comment at the beginning of \Cref{C2a-mu}.
\end{remark}
Now, we give the~\\
\noindent\textbf{Proof of \Cref{t-Cka}.} As before, in the following proof, we just make necessary normalization to satisfy the conditions of \Cref{t-Ckas-mu}. Throughout this proof, $C$ always denotes a constant depending only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,\alpha}(0)},$ $\omega_3,\omega_4,\|\partial \Omega\|_{C^{k,\alpha}(0)}$, $\|f\|_{C^{k-2,\alpha}}(0)$, $\|g\|_{C^{k,\alpha}}(0)$ and $\|u\|_{L^{\infty}(\Omega_1)}$.
Let $F_1(M,p,s,x)=F(M,p,s,x)-N_f(x)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar{\Omega}_1$ where $N_f(x)=f(0)+f_{i}(0)x_{i}+\cdots+f_{i_1\cdots i_{k-2}}(0)x_{i_1}\cdots x_{i_{k-2}}/(k-2)!$. Then $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_1(D^2u,Du,u,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $f_1(x)=f(x)-N_f(x)$. Thus,
\begin{equation*}
|f_1(x)|\leq [f]_{C^{k-2,\alpha}(0)}|x|^{k-2+\alpha}\leq C|x|^{k-2+\alpha}, ~~\forall ~x\in \Omega_1.
\end{equation*}
Next, set $u_1(x)=u(x)-N_g(x)$ and $F_2(M,p,s,x)=F_1(M+D^2N_g(x),p+DN_g(x),s+N_g(x),x)$ where $N_g(x)=g(0)+g_{i}(0)x_{i}+\cdots+g_{i_1\cdots i_k}(0)x_{i_1}\cdots x_{i_k}/k!$. Then $u_1$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_2(D^2u_1,Du_1,u_1,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_1=g_1&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_1(x)=g(x)-N_g(x)$. Hence,
\begin{equation}\label{e.cka-1}
|g_1(x)|\leq [g]_{C^{k,\alpha}(0)}|x|^{k+\alpha}\leq C|x|^{k+\alpha}, ~~\forall ~x\in (\partial \Omega)_1.
\end{equation}
Assume that $\Omega$ satisfies \cref{e-re} and \cref{e-re2} with $P$ satisfying $P(0)=0$ and $DP(0)=0$. Note that $u_1\in C^{k-1,\alpha}(0)$ and define
\begin{equation*}
N_u(x)=\left((u_1)_n(0)+(u_1)_{in}(0)x_{i}+\cdots+(u_1)_{i_1\cdots i_{k-2}n}(0)x_{i_1}\cdots x_{i_{k-2}}\right)\left(x_n-P(x')\right).
\end{equation*}
Combining with \cref{e.Cka-5} and \cref{e.cka-1}, we have
\begin{equation*}
D^l N_u(0)=D^l u_1(0),~\forall ~0\leq l\leq k-1.
\end{equation*}
Let $u_2(x)=u_1(x)-N_u(x)$ and $F_3(M,p,s,x)=F_2(M+D^2N_u(x),p+DN_u(x),s+N_u(x),x)$. Then $u_2$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_3(D^2u_2,Du_2,u_2,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_2=g_2&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
and
\begin{equation*}
u_2(0)=0,~Du_2(0)=0,\dots,D^{k-1}u_2(0)=0,
\end{equation*}
where $g_2=g_1-N_u(x)$. By the boundary $C^{k-1,\alpha}$ regularity,
\begin{equation}\label{e.tCka-1-mu}
\begin{aligned}
|Du_1(0)|+\cdots+|D^{k-1}u_1(0)| \leq C.
\end{aligned}
\end{equation}
Hence,
\begin{equation*}
\begin{aligned}
|g_2(x)|& \leq |g_1(x)|+|N_u(x)|\\
&\leq C\left(1+|(u_1)_n(0)|+\cdots+|(u_1)_{i_1\cdots i_{k-2}n}(0)|\right)|x|^{k+\alpha}\\
&\leq C|x|^{k+\alpha}, ~~\forall ~x\in (\partial \Omega)_1.
\end{aligned}
\end{equation*}
Finally, take $y=x/\rho$ and $u_3(y)=u_2(x)/\rho^2$, where $0<\rho<1$ is a constant to be specified later. Then $u_3$ satisfies
\begin{equation}\label{F6-k-mu}
\left\{\begin{aligned}
&F_4(D^2u_3,Du_3,u_3,y)=f_2&& ~~\mbox{in}~~\tilde{\Omega}_1;\\
&u_3=g_3&& ~~\mbox{on}~~(\partial \tilde{\Omega})_1,
\end{aligned}\right.
\end{equation}
where
\begin{equation*}
\begin{aligned}
&F_4(M,p,s,y)=F_3\left(M,\rho p,\rho^2s,\rho y\right),\\
&f_2(y)=f_1(x),~ g_3(y)=\frac{g_2(x)}{\rho^2} ~~\mbox{and}~~ \tilde{\Omega}=\frac{\Omega}{\rho}.
\end{aligned}
\end{equation*}
Define the fully nonlinear operator $F_{40}$ in the same way as $F_4$ (only replacing $F$ by $F_0$).
Now, we try to choose a proper $\rho$ such that \cref{F6-k-mu} satisfies the conditions of \Cref{t-Ckas-mu}. Let $N(x)=N_g(x)+N_u(x)$. First, $u_3(0)=0,Du_3(0)=0,\dots,D^{k-1}u_3(0)=0$ clearly. Combining with the $C^{k-1,\alpha}$ regularity at $0$, we have
\begin{equation*}
\begin{aligned}
\|u_3\|_{L^{\infty}(\tilde{\Omega}_1)}&= \|u_2\|_{L^{\infty}(\Omega_\rho)}/\rho^2
\leq C\rho^{k-3+\alpha}.
\end{aligned}
\end{equation*}
Next,
\begin{equation*}
\begin{aligned}
&|f_2(y)|=|f_1(x)|\leq C|x|^{k-2+\alpha}=C\rho^{k-2+\alpha}|y|^{k-2+\alpha}~\mbox{in}~\tilde{\Omega}_1,\\
&|g_3(y)|=|g_2(x)|/\rho^2\leq C|x|^{k+\alpha}/\rho^2\leq C\rho^{k-2+\alpha}|y|^{k+\alpha}
~\mbox{on}~\partial \tilde{\Omega}_1,\\
&\|\partial \tilde{\Omega}\cap B_1\|_{C^{1,\alpha}(0)}\leq \rho\|\partial \Omega\cap B_1\|_{C^{1,\alpha}(0)} \leq \rho\|\partial \Omega\cap B_1\|_{C^{k,\alpha}(0)}\leq C\rho.
\end{aligned}
\end{equation*}
Furthermore, $F_4$ and $F_{40}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde\mu,\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= \rho^2\mu,~~\tilde{b}= \rho b_0+2C\rho\mu,~~\tilde{c}= \rho^2c_0~~\mbox{and}~~
\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C,\cdot).
\end{equation*}
Finally, we show that $F_4\in C^{k-2,\alpha}(0)$. Indeed,
\begin{equation*}
\begin{aligned}
&|F_4(M,p,s,y)-F_{40}(M,p,s,y)|\\
&=|F(M+D^2N,\rho p+DN,\rho^2s+N,x)-F_{0}(M+D^2N,\rho p+DN,\rho^2s+N,x)|\\
&\leq C\rho^{k-2+\alpha}\omega_3(\|M\|+C,|p|+C,|s|+C)|y|^{k-2+\alpha}\\
&:=\tilde{\beta}_3(y)\tilde\omega_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_3(y)=C\rho^{k-2+\alpha}|y|^{k-2+\alpha},~
\tilde\omega_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C,|p|+C,|s|+C)
\end{equation*}
and $\tilde\omega_3$ satisfies \cref{e.omega-3}. In addition,
\begin{equation*}
\|F_{40}\|_{C^{k,\alpha}(\bar{\textbf{B}}_r\times \bar{\tilde\Omega}_1)}\leq \tilde{\omega}_4(r),~\forall ~r>0,
\end{equation*}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,\alpha}(0)},$ $\omega_3,\omega_4$, $\|\partial \Omega\|_{C^{k,\alpha}(0)}$, $\|f\|_{C^{k-2,\alpha}}(0), \|g\|_{C^{k,\alpha}}(0)$ and $\|u\|_{L^{\infty}(\Omega_1)}$.
Take $\delta_1$ small enough such that \Cref{t-Ckas-mu} holds for $\tilde{\omega}_0$, $\tilde{\omega}_3,\tilde{\omega}_4$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $k,n,\lambda, \Lambda,\alpha,\mu, b_0$, $c_0,\omega_0$, $\omega_3,\omega_4,\|\partial \Omega\|_{C^{k,\alpha}(0)}, \|f\|_{C^{k-2,\alpha}}(0), \|g\|_{C^{k,\alpha}}(0)$ and $\|u\|_{L^{\infty}(\Omega_1)}$) such that the assumptions of \Cref{t-Ckas-mu} are satisfied. Then $u_3$ and hence $u$ is $C^{k,\alpha}$ at $0$, and the estimates \crefrange{e.Cka-1}{e.Cka-2} hold. \qed~\\
\section{The $C^{k}$ regularity}\label{C1C2Ck}
With minor modifications, the pointwise $C^{k}$ and $C^{k,\mathrm{lnL}}$ regularity can be proved in a similar way to the pointwise $C^{k,\alpha}$ regularity. In this paper, we only give detailed proofs for several results and other regularity can be proved similarly. In this section, we prove the interior pointwise $C^{2}$ regularity and the boundary pointwise $C^{1}$ regularity under the corresponding Dini conditions on the coefficients and the prescribed data.
For $C^{k}$ regularity, we can use a similar ``key step'' as the $C^{k,\alpha}$ regularity. The following is the ``key step'' for interior $C^{2}$ regularity, whose proof is almost the same as that of \Cref{In-l-C2a-mu} and we omit it.
\begin{lemma}\label{In-l-C2-mu}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF}. There exists $\delta>0$ depending only on $n,\lambda,\Lambda,\alpha$ and $\omega_2$ such that if $u$ satisfies
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1
\end{equation*}
with $\|u\|_{L^{\infty}(B_1)}\leq 1$, $u(0)=0$, $Du(0)=0$, $\mu,b_0,c_0\leq \delta$, $\omega_0(1,1)\leq 1$, $\|\beta_2\|_{L^{\infty}(B_1)}\leq \delta$ and $\|f\|_{L^{\infty}(B_1)}\leq \delta$, then there exists a $2$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(B_{\eta})}\leq \eta^{2+\bar\alpha/2},
\end{equation*}
\begin{equation*}
F_0(D^2Q,0,0)=0
\end{equation*}
and
\begin{equation*}
\|Q\|\leq \bar{C}+1,
\end{equation*}
where $\eta$ depends only on $n,\lambda$ and $\Lambda$.
\end{lemma}
Now, we give the ``scaling argument'' of interior $C^2$ regularity.
\begin{theorem}\label{In-t-C2s-mu}
Suppose that $F$ is convex in $M$ and satisfies \cref{e.C2a-KF}. Let $\omega_0$ satisfy \cref{e.omega0-c-1} and $u$ satisfy
\begin{equation*}
F(D^2u, Du, u,x)=f ~~\mbox{in}~~ B_1.
\end{equation*}
Assume that
\begin{equation}\label{In-e.C2s-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(B_1)}\leq 1,~ u(0)=0,~Du(0)=0, \\
&\mu\leq \frac{\delta_1}{4C_0},~ b_0\leq \frac{\delta_1}{2},~ c_0\leq \frac{\delta_1}{\omega_{\omega_0}(C_0)+1},~\hat\omega_0(1+C_0)\leq 1,\\
&|\beta_2(x)|\leq \delta_1\omega_{\beta}(|x|),~|f(x)|\leq \delta_1\omega_f(|x|), ~~\forall ~x\in B_1\\
&\tilde\omega(1)\leq 1~~\mbox{and}~~\int_{0}^{1}\frac{\tilde\omega(\rho)}{\rho}d\rho\leq 1
~~(\tilde\omega=\max(\omega_{\omega_0},\omega_{\beta},\omega_f)),
\end{aligned}
\end{equation}
where $\delta_1$ depends only on $n,\lambda,\Lambda,\omega_0$ and $\omega_2$, and $C_0$ depends only on $n,\lambda$ and $\Lambda$.
Then $u\in C^{2}(0)$, i.e., there exist a $2$-form $Q$ and a modulus of continuity $\omega_{u}$ such that
\begin{equation}\label{In-e.C2s-1-mu}
|u(x)-Q(x)|\leq |x|^{2}\omega_{u}(|x|), ~~\forall ~x\in B_{1},
\end{equation}
\begin{equation}\label{In-e.C2s-3-mu}
F_0(D^2Q,0,0)=0
\end{equation}
and
\begin{equation}\label{In-e.C2s-2-mu}
\|Q\| \leq C,
\end{equation}
where for $0<r<\eta^2$ $(\omega_u(r)\equiv C$ for $\eta^2\leq r<1)$,
\begin{equation*}
\omega_u(r)=C\left(r^{\bar{\alpha}/2}+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
+\int_{0}^{r/\eta^2}\frac{\tilde\omega(\rho)}{\rho}d\rho\right);
\end{equation*}
$\eta$ and $C$ depend only on $n,\lambda$ and $\Lambda$.
\end{theorem}
\begin{remark}\label{r-10.2}
In fact, $\bar{\alpha}/2$ can be replaced by any $0<\alpha_0<\bar{\alpha}$ in the expression of
$\omega_u$ since we can define
\begin{equation*}
A_m=\max(\tilde\omega(\eta^{m}),\eta^{\alpha_0} A_{m-1})
\end{equation*}
in the proof below.
\end{remark}
\proof To prove that $u$ is $C^{2}$ at $0$, we only need to prove the following. There exist a sequence of $2$-forms $Q_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{In-e.C2s-4-mu}
\|u-Q_m\|_{L^{\infty }(B_{\eta^{m}})}\leq \eta ^{2m}A_m,
\end{equation}
\begin{equation}\label{In-e.C2s-7-mu}
F_0(D^2Q_m,0,0)=0
\end{equation}
and
\begin{equation}\label{In-e.C2s-6-mu}
\|Q_m-Q_{m-1}\|\leq (\bar{C}+1)A_{m-1},
\end{equation}
where
\begin{equation}\label{In-e.tC2-Ak}
A_{-1}=A_0=1,~A_m=\max(\tilde\omega(\eta^{m}),\eta^{\bar\alpha/2} A_{m-1})
(m\geq 1)
\end{equation}
and $\eta$ depends only on $n,\lambda$ and $\Lambda$.
Before proving \crefrange{In-e.C2s-4-mu}{In-e.C2s-6-mu}, we show that they imply \crefrange{In-e.C2s-1-mu}{In-e.C2s-2-mu}. Indeed, by \cref{In-e.tC2-Ak} and noting that $\tilde{\omega}$ is a Dini function, $A_m$ is nonincreasing to $0$ as $m\rightarrow \infty$ and $\sum_{m=1}^{\infty} A_m<\infty$. In addition, for any $m\geq 0$,
\begin{equation*}
\sum_{i=m}^{\infty} A_i\leq \sum_{i=m}^{\infty}\tilde\omega(\eta^i)+\eta^{\bar\alpha/2}\sum_{i=m}^{\infty} A_i+A_{m-1},
\end{equation*}
which implies
\begin{equation}\label{e.10.1}
\begin{aligned}
\sum_{i=m}^{\infty} A_i&\leq \frac{1}{1-\eta^{\bar\alpha/2}}\left(\sum_{i=m}^{\infty}\tilde\omega(\eta^i)+A_{m-1}\right)\\
&=\frac{1}{1-\eta^{\bar\alpha/2}}\left(\sum_{i=m}^{\infty}\frac{\tilde\omega(\eta^i)
\left(\eta^{i-1}-\eta^i\right)}{\eta^{i-1}-\eta^i}+A_{m-1}\right)\\
&=\frac{1}{\left(1-\eta^{\bar\alpha/2}\right)\left(1-\eta\right)}\left(\sum_{i=m}^{\infty}
\frac{\tilde\omega(\eta^i)\left(\eta^{i-1}-\eta^i\right)}{\eta^{i-1}}+A_{m-1}\right)\\
&\leq \frac{1}{\left(1-\eta^{\bar\alpha/2}\right)\left(1-\eta\right)}\left(\int_{0}^{\eta^{m-1}}
\frac{\tilde\omega(\rho)}{\rho}d\rho+A_{m-1}\right).
\end{aligned}
\end{equation}
Furthermore, for $m\geq 1$,
\begin{equation}\label{e.10.3}
\begin{aligned}
A_m\leq &\tilde{\omega}(\eta^{m})+\eta^{\bar{\alpha}/2}A_{m-1}\\
\leq & \sum_{i=1}^{m}\eta^{(m-i)\bar{\alpha}/2}\tilde{\omega}(\eta^{i})
+\eta^{m\bar{\alpha}/2}\\
=& \eta^{m\bar{\alpha}/2}
\sum_{i=1}^{m}\frac{\tilde{\omega}(\eta^{i})}{\eta^{i\bar{\alpha}/2}}
+\eta^{m\bar{\alpha}/2}\\
\leq & C\eta^{m\bar{\alpha}/2}\sum_{i=1}^{m}
\frac{\tilde{\omega}(\eta^{i})}{\eta^{(i-1)\bar{\alpha}/2}\eta^{i-1}}(\eta^{i-1}-\eta^{i})
+\eta^{m\bar{\alpha}/2}\\
\leq & C\eta^{m\bar{\alpha}/2}\int_{\eta^{m}}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho+\eta^{m\bar{\alpha}/2},
\end{aligned}
\end{equation}
where $C$ depends only on $n,\lambda$ and $\Lambda$.
Assume that $Q_m$ has the form $Q_m(x)=a_{m,ij}x_ix_j/2$ ($1\leq i,j\leq n$ and the Einstein summation convention is used). Then \cref{In-e.C2s-6-mu} reads
\begin{equation*}
|a_{m,ij}-a_{m-1,ij}|\leq A_{m-1},~\forall ~m\geq 1,~1\leq i,j\leq n.
\end{equation*}
Since $\sum_{m=1}^{\infty} A_m<\infty$, there exists constants $a_{ij}$ ($1\leq i,j\leq n$) such that
\begin{equation}\label{e.10.2}
|a_{m,ij}-a_{ij}|\leq C\sum_{i=m}^{\infty}A_i~(\forall ~m\geq 1)~~\mbox{and}~~
|a_{ij}|\leq C,
\end{equation}
where $C$ depends only on $n,\lambda$ and $\Lambda$. Let $Q(x)=a_{ij}x_ix_j/2$ and then \cref{In-e.C2s-3-mu} and \cref{In-e.C2s-2-mu} hold.
In addition, for any $x\in B_{\eta^2}$, there exists $m\geq 2$ such that $\eta^{m+1}\leq |x|< \eta^{m}$. Then
\begin{equation*}
\begin{aligned}
|u(x&)-Q(x)|\\
\leq& |u(x)-Q_m(x)|+|Q_m(x)-Q(x)|\\
\leq & \eta^{2m}A_m+\eta^{2m}C\sum_{i=m}^{\infty}A_i ~(\mbox{by}~\cref{In-e.C2s-4-mu}~\mbox{and}~\cref{e.10.2})\\
\leq & C\eta^{2m}\left(A_m+A_{m-1}
+\int_{0}^{\eta^{m-1}}\frac{\tilde\omega(\rho)}{\rho}d\rho\right)
~(\mbox{by}~\cref{e.10.1})\\
\leq & C\eta^{2m}\left(A_{m-1}
+\int_{0}^{\eta^{m-1}}\frac{\tilde\omega(\rho)}{\rho}d\rho\right)\\
\leq &C\eta^{2m}\left(\eta^{(m-1)\bar{\alpha}/2}\int_{\eta^{m-1}}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho+\eta^{(m-1)\bar{\alpha}/2}
+\int_{0}^{\eta^{m-1}}\frac{\tilde\omega(\rho)}{\rho}d\rho\right) ~(\mbox{by}~\cref{e.10.3})\\
\leq &C|x|^2\left(|x|^{\bar{\alpha}/2}+|x|^{\bar{\alpha}/2}\int_{|x|}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
+\int_{0}^{|x|/\eta^2}\frac{\tilde\omega(\rho)}{\rho}d\rho\right).
\end{aligned}
\end{equation*}
For $0<r<\eta^2$, define
\begin{equation*}
\omega_u(r)=C\left(r^{\bar{\alpha}/2}+r^{\bar{\alpha}/2}\int_{r}^{1} \frac{\tilde{\omega}(\rho)}{\rho^{1+\bar{\alpha}/2}}d\rho
+\int_{0}^{r/\eta^2}\frac{\tilde\omega(\rho)}{\rho}d\rho\right).
\end{equation*}
Then $\omega_u$ is a modulus of continuity and \cref{In-e.C2s-1-mu} holds.
Now, we prove \crefrange{In-e.C2s-4-mu}{In-e.C2s-6-mu} by induction. For $m=0$, by setting $Q_0=Q_{-1}\equiv 0$, \crefrange{In-e.C2s-4-mu}{In-e.C2s-6-mu} hold clearly.
Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{In-e.C2s-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^2A_{m_0}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{In-e.C2s-F-mu}
\tilde{F}(D^2v,Dv,v,y)=\tilde{f} ~~\mbox{in}~~ B_1,
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{1}{A_{m_0}}F(A_{m_0}M+D^2Q_{m_0},rA_{m_0}p+DQ_{m_0}(x),
r^2A_{m_0}s+Q_{m_0}(x),x),\\
&\tilde{f}(y)=\frac{f(x)}{A_{m_0}}.\\
\end{aligned}
\end{equation*}
In addition, define
\begin{equation*}
\tilde{F}_0(M,p,s)=\frac{1}{A_{m_0}}F_0(A_{m_0}M+D^2Q_{m_0},rA_{m_0}p
,r^{2}A_{m_0}s).
\end{equation*}
In the following, we show that \cref{In-e.C2s-F-mu} satisfies the assumptions of \Cref{In-l-C2-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(B_1)}\leq 1, v(0)=0, Dv(0)=0,
~(\mathrm{by}~\cref{In-e.C2s-be-mu},~\cref{In-e.C2s-4-mu} ~\mbox{and}~\cref{In-e.C2s-v-mu})\\
&\|\tilde{f}\|_{L^{\infty}(B_1)}=\frac{\|f\|_{L^{\infty}(B_r)}}{A_{m_0}}\leq \delta_1.
~(\mathrm{by}~ \cref{In-e.C2s-be-mu})\\
&\tilde{F}_0(0,0,0)=\frac{F_0(D^2Q_{m_0},0,0)}{A_{m_0}}=0.~(\mathrm{by}~\cref{In-e.C2s-7-mu})
\end{aligned}
\end{equation*}
By \cref{In-e.C2s-6-mu}, we can choose a constant $C_0$ depending only on $n,\lambda$ and $\Lambda$ such that $\|Q_{m}\|\leq C_0$ ($\forall~0\leq m\leq m_0$). Then $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}=r^2A_{m_0}\mu,~\tilde{b}=rb_0+2C_0r\mu,
~\tilde{c}=c_0~~\mbox{and}~~
\tilde{\omega}_0(K,s)=A_{m_0}^{-1}\omega_0(K+C_0,r^2A_{m_0}s),~\forall ~K,s >0.
\end{equation*}
Hence, from \cref{In-e.C2s-be-mu},
\begin{equation*}
\begin{aligned}
\tilde{\mu}\leq \mu\leq \delta_1,~\tilde{b}\leq b_0+2C_0\mu\leq \delta_1,~
\tilde{c}\leq c_0\leq \delta_1
\end{aligned}
\end{equation*}
and (note that $\omega_0$ satisfies \cref{e.omega0-c-1})
\begin{equation*}
\tilde{\omega}_0(1,1)=A_{m_0}^{-1}\omega_0(1+C_0,r^2A_{m_0})
=\hat\omega_0(1+C_0)\omega_{\omega_0}(r^2A_{m_0})/A_{m_0}\leq 1.
\end{equation*}
Moreover, $\tilde{\omega}_0$ satisfies \cref{e.omega0-c-1} with
\begin{equation*}
\hat{\tilde{\omega}}_0(K)=\hat{\omega}_0(K+C_0)~~\mbox{and}~~
\omega_{\tilde{\omega}_0}(s)=A_{m_0}^{-1}\omega_{\omega_0}(r^2A_{m_0}s),~\forall ~K,s>0.
\end{equation*}
Finally, take $\eta$ small enough such that
\begin{equation*}
\eta C_0\leq 1.
\end{equation*}
By combining with the structure condition \cref{SC2}, \cref{e.C2a-KF} and \cref{In-e.C2s-be-mu}, we have for $(M,p,s,y)\in S^n\times R\times R\times \bar B_1$ (note that $r^{\bar\alpha/2}=\eta^{m_0\bar\alpha/2}\leq A_{m_0}$),
\begin{equation*}
\begin{aligned}
|&\tilde{F}(M,p,s,y)-\tilde{F}(M,p,s,0)|\\
&\leq\frac{\beta_2(x)\omega_2(\|M\|+C_0,|p|,|s|)
+2C_0 r^2A_{m_0}\mu|p|+ C_0^2r^2\mu+C_0rb_0
+c_0\hat\omega_0(|s|+C_0)\omega_{\omega_0}(C_0 r^2)}{A_{m_0}}\\
&\leq \delta_1\omega_2(\|M\|+C_0,|p|,|s|)+\delta_1|p|+\delta_1C_0
+\delta_1\hat\omega_0(|s|+C_0)\\
&:=\tilde{\beta}_2(y)\tilde{\omega}_2(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_2(y)\equiv\delta_1$ and $\tilde{\omega}_2(\|M\|,|p|,|s|)=
\omega_2(\|M\|+C_0,|p|,|s|)+|p|+C_0+\hat\omega_0(|s|+C_0)$. Then $\tilde{\omega}_2$ satisfies \cref{e.omega} and
\begin{equation*}
\|\tilde{\beta}_2\|_{L^{\infty}(B_1)}\leq \delta_1.
\end{equation*}
Choose $\delta_1$ small enough (depending only on $n,\lambda,\Lambda,\omega_0$ and $\omega_2$) such that \Cref{In-l-C2-mu} holds for $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. Since \cref{In-e.C2s-F-mu} satisfies the assumptions of \Cref{In-l-C2-mu}, there exists a $2$-form $\tilde{Q}(y)$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(B_{\eta})}&\leq \eta ^{2+\bar{\alpha}/2},
\end{aligned}
\end{equation*}
\begin{equation*}
\tilde{F}_0(D^2\tilde{Q},0,0)=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}+1.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+A_{m_0}\tilde{Q}(x)$. Then \cref{In-e.C2s-7-mu} and \cref{In-e.C2s-6-mu} hold for $m_0+1$. Recalling \cref{In-e.C2s-v-mu}, we have (note that $\eta^{\bar{\alpha}/2}A_{m_0}\leq A_{m_0+1}$ by the definition of $A_{m_0}$)
\begin{equation*}
\begin{aligned}
&\|u-Q_{m_0+1}(x)\|_{L^{\infty}(B_{\eta^{m_0+1}})}\\
&= \|u-Q_{m_0}(x)-A_{m_0}\tilde{Q}(x)\|_{L^{\infty}(B_{\eta r})}\\
&= \|r^2A_{m_0}v-r^2A_{m_0}\tilde{Q}(y)\|_{L^{\infty}(B_{\eta})}\\
&\leq r^2A_{m_0}\eta^{2+\bar{\alpha}/2}\leq\eta^{2(m_0+1)}A_{m_0+1}.
\end{aligned}
\end{equation*}
Hence, \cref{In-e.C2s-4-mu} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Now, we give the~\\
\noindent\textbf{Proof of \Cref{t-C2-i}.} As before, we prove the theorem in two cases.
\textbf{Case 1:} the general case, i.e., $F$ satisfies \cref{SC2} and \cref{e.C2a-KF}. Let $\omega=\max(\omega_{\omega_0},\omega_{\beta},\omega_f)$. Throughout the proof for this case, $C$ always denotes a constant depending only on $n, \lambda,\Lambda,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\mathrm{Dini}}(0)}$, $\omega_{\beta_2}$, $\omega_2$, $\|f\|_{C^{\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$. Note that $\omega_{\beta_2}$ is a Dini function and there exists $0<\rho_1<1$ such that $\omega_{\beta_2}(\rho_1)\leq \delta_0$ where $\delta_0$ is as in \Cref{t-C1a-i} with $\alpha=\bar{\alpha}/2$ there. Then $\|\beta_2\|_{L^{\infty }(B_{\rho_1})}\leq \delta_0$. By \Cref{t-C1a-i}, $u\in C^{1,\bar\alpha/2}(0)$ and
\begin{equation*}
\|u\|_{ C^{1,\bar\alpha/2}(0)}\leq C.
\end{equation*}
Let $F_1(M,p,s,x)=F(M,p,s,x)-f(0)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar B_1$. Then $u_1$ satisfies
\begin{equation*}
F_1(D^2u,Du,u,x)=f_1~~\mbox{in}~~B_1,
\end{equation*}
where $f_1(x)=f(x)-f(0)$. Hence,
\begin{equation*}
|f_1(x)|\leq \omega_f(|x|), ~~\forall ~x\in B_1.
\end{equation*}
Set $u_1(x)=u(x)-u(0)-Du(0)\cdot x$ and $F_2(M,p,s,x)=F_1(M,p+Du(0),s+u(0)+Du(0)\cdot x,x)$. Then $u_1$ satisfies
\begin{equation*}
F_2(D^2u_1,Du_1,u_1,x)=f_1~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_1(0)=0, Du_1(0)=0.
\end{equation*}
Next, take $u_2(x)=u_1(x)-\tau x_n^2$ and $F_3(M,p,s,x)=F_2(M+2\tau\delta_{nn},p+2\tau x_n,s+\tau x_n^2,x)$ for $\tau\in R$. Then $u_2$ satisfies
\begin{equation*}
F_3(D^2u_2,Du_2,u_2,x)=f_1~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_2(0)=0, Du_2(0)=0.
\end{equation*}
As before, define the fully nonlinear operators $F_{10},F_{20}$ and $F_{30}$ similarly. By the structure condition, there exists $\tau\in R$ such that $F_{30}(0,0,0)=0$ and
\begin{equation}\label{In-e.tC2a-2-mu}
\begin{aligned}
|\tau|&\leq |F_{20}(0,0,0)|/\lambda\leq |F_0(0,Du(0),u(0))-f(0)|/\lambda\leq C.
\end{aligned}
\end{equation}
For $\rho\leq \rho_1$, define $y=x/\rho$, $u_3(y)=u_2(x)/\rho$, $F_4(M,p,s,y)=\rho F_3(M/\rho, p, \rho s,x)$ and $F_{40}(M,p,s)=\rho F_{30}(M/\rho, p,\rho s)$. Then $u_3$ satisfies
\begin{equation}\label{In-C2-F5-mu}
F_4(D^2u_3,Du_3,u_3,y)=f_2~~\mbox{in}~~B_1,
\end{equation}
where $f_2(y)=\rho f_1(x)$.
Now, we can check that \cref{In-C2-F5-mu} satisfies the conditions of \Cref{In-t-C2s-mu} by choosing a proper $\rho$. First, it can be verified easily that
\begin{equation*}
\begin{aligned}
u_3(0)=0 ,Du_3(0)=0
\end{aligned}
\end{equation*}
and
\begin{equation*}
|f_2(y)|= \rho|f_1(x)|\leq \rho\omega_f(\rho|y|):=\rho \omega_{f_2}(|y|)~\mbox{in}~B_1.
\end{equation*}
Next, by the interior $C^{1,\bar\alpha/2}$ regularity for $u$,
\begin{equation*}
\|u_3\|_{L^{\infty}(B_1)}\leq \rho^{-1} \|u_2\|_{L^{\infty}(B_{\rho})}
\leq \rho^{-1}\left( \|u_1\|_{L^{\infty}(B_{\rho})}+C\rho^{2}\right)
\leq \rho^{-1}\left(C\rho^{1+\bar\alpha/2}+C\rho^{2}\right)\leq C\rho^{\bar\alpha/2}.
\end{equation*}
Furthermore, $F_{40}(0,0,0)=\rho F_{30}(0,0,0)=0$, and $F_4$ and $F_{40}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde\omega_0$, where
\begin{equation*}
\tilde{\mu}= \rho\mu,~\tilde{b}= \rho b_0+C\rho\mu,~\tilde{c}= \rho^{1/2}
c_0~~\mbox{and}~~
\tilde{\omega}_0(K,s)=\rho^{1/2}\omega_0(K+C,\rho s), ~\forall ~K,s>0.
\end{equation*}
Thus, $\tilde{\omega}_0$ satisfies \cref{e.omega0-c-1} with
\begin{equation*}
\hat{\tilde\omega}_0(K)=\rho^{1/4}\hat{\omega}_0(K)~~\mbox{and}~~
\omega_{\tilde{\omega}_0}(s)=\rho^{1/4}\omega_{\omega_0}(\rho s), ~\forall ~K,s>0.
\end{equation*}
Finally, we check the oscillation of $F_4$ in $y$.
\begin{equation*}
\begin{aligned}
|F&_4(M,p,s,y)-F_{40}(M,p,s)|\\
=&\rho\Big(F(\rho^{-1}M+2\tau\delta_{nn},p+2\tau x_n+Du_1(0),
\rho s+\tau x_n^2+u_1(0)+Du_1(0)\cdot x,x)\\
&-F_0(\rho^{-1}M+2\tau\delta_{nn},p+Du_1(0),\rho s+u_1(0))\Big)\\
\leq& \rho\beta_2(x)\omega_2(\rho^{-1}\|M\|+C,|p|+C,\rho|s|+C)+C\rho\mu(2|p|+C|x|+C)|x|\\
&+C\rho b_0|x|+\rho c_0\omega_0(|s|+C,C|x|)\\
\leq& C\omega_2(\|M\|+C,|p|+C,\rho|s|+C)\omega_{\beta_2}(|x|)+C(|p|+C)|x|\\
&+C\rho|x|+\rho\hat\omega_0(|s|+C)\omega_{\omega_0}(C|x|)\\
:= & \tilde{\beta}_2(y)\tilde{\omega}_2\left(\|M\|,|p|,|s|\right),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
&\tilde{\beta}_2(y)=CC_{\rho}\omega_{\tilde{\beta}_2}(|y|),~
\omega_{\tilde{\beta}_2}(|y|)=\frac{\omega_{\beta_2}(\rho|y|)}{2C_{\rho}}
+\rho\omega_{\omega_0}(C\rho|y|),~
C_{\rho}=\int_{0}^{3\rho}\frac{\omega_{\beta_2}( r)}{r}dr~~\mbox{and}\\
&\tilde{\omega}_2\left(\|M\|,|p|,|s|\right)=\omega_2\left(\|M\|+C,|p|+C,|s|+C\right)
+C|p|+C+\hat\omega_0(|s|+C).
\end{aligned}
\end{equation*}
Clearly, $\tilde{\omega}_2$ satisfies \cref{e.omega}. In addition, $C_{\rho}\rightarrow 0$ as $\rho\rightarrow 0$ and
\begin{equation*}
\int_{0}^{1}\frac{\omega_{\tilde{\beta}_2}(r)}{r}dr
=\frac{1}{2C_{\rho}}\int_{0}^{\rho}\frac{\omega_{\beta_2}(r)}{r}dr
+\rho\int_{0}^{C\rho}\frac{\omega_{\omega_0}(r)}{r}dr.
\end{equation*}
Moreover,
\begin{equation*}
\frac{1}{2}+\rho\int_{0}^{3C\rho}\frac{\omega_{\omega_0}(r)}{r}dr\geq \int_{0}^{3}\frac{\omega_{\tilde{\beta}_2}(r)}{r}dr\geq \int_{1}^{3}\frac{\omega_{\tilde{\beta}_2}(r)}{r}dr\geq \omega_{\tilde{\beta}_2}(1)\ln 3.
\end{equation*}
Take $\delta_1$ small enough such that \Cref{In-t-Cklns-mu} holds with $\tilde{\omega}_0,\tilde{\omega}_2$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $n, \lambda,\Lambda,\mu,b_0,c_0,\omega_0$, $\|\beta_2\|_{C^{\mathrm{Dini}}(0)}$, $\omega_{\beta_2}$, $\omega_2$, $\|f\|_{C^{\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty }(B_1)}$) such that
\begin{equation*}
\begin{aligned}
&\|u_3\|_{L^{\infty }(B_1)}\leq 1,~\tilde\mu \leq \frac{\delta_1}{4C_0},~
\tilde{b}\leq \frac{\delta_1}{2},~
\tilde c\leq \frac{\delta_1}{\omega_{\tilde\omega_0}(C_0)+1},~
\hat{\tilde{\omega}}_0(1+C_0)\leq 1,\\
&|\tilde\beta_2(y)|\leq \delta_1\omega_{\tilde{\beta}_2}(|y|),~|f_2(y)|\leq \delta_1\omega_{f_2}(|y|), ~~\forall ~y\in B_1,\\
&\tilde\omega(1)\leq 1~~\mbox{and}~~\int_{0}^{1}\frac{\tilde\omega(\rho)}{\rho}d\rho\leq 1
~~(\tilde\omega=\max(\omega_{\tilde{\omega}_0},\omega_{\tilde\beta_2},\omega_{f_2})),
\end{aligned}
\end{equation*}
where $C_0$ depending only on $n,\lambda$ and $\Lambda$, is as in \Cref{In-t-C2s-mu}. Therefore, the assumptions in \Cref{In-t-C2s-mu} are satisfied for \cref{In-C2-F5-mu}. By \Cref{In-t-C2s-mu}, $u_3$ and hence $u$ is $C^{2}$ at $0$, and the estimates \crefrange{e.C2-1-i-mu}{e.C2-2-i-mu} hold.
\textbf{Case 2:} $F$ satisfies \cref{SC1} and \cref{e.C2a-KF-0}. Let $K=\|u\|_{L^{\infty }(B_1)}+\|f\|_{C^{\mathrm{Dini}}(0)}+\|\gamma_2\|_{C^{\mathrm{Dini}}(0)}$ and $u_1=u/K$. Then $u_1$ satisfies
\begin{equation}\label{e.10.4}
F_1(D^2u_1,Du_1,u_1,x)=f_1~~\mbox{in}~~B_1,
\end{equation}
where $F_1(M,p,s,x)=F(KM,Kp,Ks,x)/K$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar B_1$ and $f_1=f/K$.
Thus,
\begin{equation*}
\|u_1\|_{L^{\infty }(B_1)}\leq 1~~\mbox{and}~~ \|f_1\|_{C^{\mathrm{Dini}}(0)}\leq 1.
\end{equation*}
In addition, define $F_{10}(M,p,s)=F_0(KM,Kp,Ks)/K$. Then $F_1$ and $F_{10}$ satisfy the structure condition \cref{SC1} with the same $\lambda,\Lambda,b_0$ and $c_0$. Moreover, $F_1$ satisfies \cref{e.C1a.beta-2} with $\beta_1(x)=\beta_2(x)$ and $\gamma_1(x)=\gamma_2(x)/K$. Hence, $\|\gamma_1\|_{C^{\mathrm{Dini}}(0)}\leq 1$.
Apply \textbf{Case 1} to \cref{e.10.4}, we obtain that $u_1$ and hence $u$ is $C^2$ at $0$, and the estimates \crefrange{e.C2-1-i}{e.C2-2-i} hold. \qed~\\
In the second half of this section, we prove the boundary $C^{1}$ regularity.
\noindent\textbf{Proof of \Cref{t-C1}.} As before, we assume that $g(0)=0$ and $Dg(0)=0$. Let $\tilde\alpha=\min(\bar{\alpha}/2,1-n/p)$ and $\delta$ be as in \Cref{l-C1a-mu}, which depends only on $n,\lambda,\Lambda$ and $p$. We also assume that
\begin{equation}\label{e.tC1-ass}
\begin{aligned}
&\|u\|_{L^{\infty}(\Omega_1)}\leq 1,~\mu\leq \frac{\delta}{6C_0^2},~\|b\|_{L^p(\Omega_1)}\leq \frac{\delta}{3C_0},\\
&\|f\|_{L^n(\Omega_r)}\leq \frac{\delta}{3}\omega_f(r), ~\forall ~0<r<1,\\
&\|g\|_{L^{\infty}((\partial\Omega)_r)}\leq \frac{\delta}{2}r\omega_g(r), ~\forall ~0<r<1,\\
&\underset{B_{r}}{\mathrm{osc}}~\partial\Omega\leq \frac{\delta}{2C_0}r\omega_{\Omega}(r), ~\forall ~0<r<1,\\
&\tilde\omega(1)\leq 1~~\mbox{and}~~\int_{0}^{1}\frac{\tilde\omega(\rho)}{\rho}d\rho\leq1
~(\tilde\omega=\max(\omega_f,\omega_{g},\omega_{\Omega})),
\end{aligned}
\end{equation}
where $C_0>1$ is a constant (depending only on $n,\lambda,\Lambda$ and $p$) to be specified later.
Otherwise, note that $\Omega$ satisfies the exterior cone condition at $0$ and we may consider for $0<\rho<1$,
\begin{equation}\label{e.c1a-v}
\bar{u}(y)=\frac{u(x)}{\rho^{\alpha_0}},
\end{equation}
where $y=x/\rho$ and $0<\alpha_0<1$ is a H\"{o}lder exponent (depending only on $n,\lambda,\Lambda,\|b\|_{L^p(B_1)}$ and $\|\partial \Omega\|_{C^{1,\mathrm{Dini}}(0)}$) such that $u\in C^{2\alpha_0}(0)$ (by \Cref{l-3Ho}).
Then we have
\begin{equation*}
\left\{\begin{aligned}
&\bar{u}\in S^{*}(\lambda,\Lambda,\bar{\mu},\bar{b},\bar{f})&&
~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&\bar{u}=\bar{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation*}
where
\begin{equation*}\label{e.tC1-n1-mu}
\bar{\mu}=\rho^{\alpha_0}\mu,~~\bar{b}(y)=\rho b(x),~~\bar{f}(y)=\rho^{2-\alpha_0}f(x),
~~\bar{g}(y)=\rho^{-\alpha_0}g(x)~~\mbox{and}
~~\tilde{\Omega}=\rho^{-1}\Omega.
\end{equation*}
Hence,
\begin{equation*}
\begin{aligned}
&\|\bar{u}\|_{L^{\infty}(\tilde{\Omega}_1)}\leq \frac{1}{\rho^{\alpha_0}}|x|^{2\alpha_0}[u]_{C^{2\alpha_0}(0)}
\leq \rho^{\alpha_0}[u]_{C^{2\alpha_0}(0)},\\
&\bar{\mu}=\rho^{\alpha_0}\mu,\\
&\|\bar{b}\|_{L^{p}(\tilde{\Omega}_1)}=\rho^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_\rho)}\leq \rho^{\tilde\alpha}\|b\|_{L^{p}(\Omega_1)},\\
&\|\bar{f}\|_{L^{n}(\tilde{\Omega}_r)}=\rho^{1-\alpha_0}\|f\|_{L^{n}(\Omega_{\rho r})}
\leq \rho^{1-\alpha_0}\omega_f(\rho r):=\rho^{1-\alpha_0}\omega_{\bar{f}}(r),\\
&\|\bar{g}\|_{L^{\infty}((\partial\tilde{\Omega})_r)}
=\rho^{-\alpha_0}\|g\|_{L^{\infty}((\partial\Omega)_{\rho r})}\leq
\rho^{1-\alpha_0}r\omega_g(\rho r):=\rho^{1-\alpha_0}r\omega_{\bar{g}}(r),\\
&\underset{B_{r}}{\mathrm{osc}}~\partial\tilde{\Omega}
=\rho^{-1}\underset{B_{\rho r}}{\mathrm{osc}}~\partial\Omega
\leq \omega_{\Omega}(\rho r)=C_{\rho}\frac{\omega_{\Omega}(\rho r)}{C_{\rho}}:=C_{\rho}\omega_{\tilde\Omega}(r),
\end{aligned}
\end{equation*}
where
\begin{equation*}
C_{\rho}=\int_{0}^{3\rho}\frac{\omega( r)}{r}dr.
\end{equation*}
By choosing $\rho$ small enough (depending only on $n,\lambda,\Lambda,p,\mu,\|b\|_{L^{p}(\Omega_1)},\|\partial \Omega\|_{C^{1,\mathrm{Dini}}(0)}$, $\omega_{\Omega}$, $\|f\|_{C^{-1,\mathrm{Dini}}(0)}$, $[g]_{C^{1,\mathrm{Dini}}(0)}$ and $\|u\|_{L^{\infty}(\Omega_1)}$), the assumptions \cref{e.tC1-ass} for $\bar{u}$ can be guaranteed. Hence, we can make the assumption \cref{e.tC1-ass} for $u$ without loss of generality.
Now we prove that $u$ is $C^{1}$ at $0$ and we only need to prove the following. There exists a sequence $a_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{e.tC1-3}
\|u-a_mx_n\|_{L^{\infty }(\Omega _{\eta^{m}})}\leq \eta^m A_m
\end{equation}
and
\begin{equation}\label{e.tC1-4}
|a_m-a_{m-1}|\leq \bar{C} A_{m-1},
\end{equation}
where
\begin{equation}\label{e.tC1-Ak}
A_{-1}=A_0=1,A_m=\max(\tilde\omega(\eta^{m}),\eta^{\tilde\alpha} A_{m-1}) (m\geq 1)
\end{equation}
and $\eta$, depending only on $n,\lambda$ and $\Lambda$, is as in \Cref{l-C1a-mu} (take $\alpha=\tilde{\alpha}$ there).
We prove the above by induction. For $m=0$, by setting $a_0=a_{-1}=0$, the conclusion holds clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{e.tC1-v1}
v(y)=\frac{u(x)-a_{m_0}x_n}{rA_{m_0}}.
\end{equation}
Then $v$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&v\in S^*(\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{f})&& ~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&v=\tilde{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation*}
where
\begin{equation*}
\begin{aligned}
&\tilde{\mu}=2rA_{m_0}\mu,~\tilde{b}(y)=rb(x),~
\tilde{f}(y)=\frac{r|f(x)|+rb(x)|a_{m_0}|+2r\mu|a_{m_0}|^2}{A_{m_0}},\\
&\tilde{g}(y)=\frac{g(x)-a_{m_0}x_n}{rA_{m_0}}
~~\mbox{and}~~ \tilde{\Omega}=\frac{\Omega}{r}.
\end{aligned}
\end{equation*}
From the definition of $A_{m_0}$ (see \cref{e.tC1-Ak}),
\begin{equation}\label{e.C1-4}
r^{\tilde{\alpha}}=\eta^{m_0\tilde{\alpha}}\leq A_{m_0}\leq 1.
\end{equation}
By \cref{e.tC1-4}, there exists a constant $C_0$ depending only on $n,\lambda,\Lambda$ and $p$ such that $|a_{m}|\leq C_0$ ($\forall~0\leq m\leq m_0$). Then it is easy to verify that
\begin{equation}\label{e.tC1-esti}
\begin{aligned}
& \|v\|_{L^{\infty}(\tilde{\Omega}_1)}\leq 1, ~(\mathrm{by}~ \cref{e.tC1-3} ~\mathrm{and}~ \cref{e.tC1-v1})\\
&\tilde\mu\leq 2rA_{m_0}\mu\leq \delta, ~(\mathrm{by}~ \cref{e.tC1-ass})\\
&\|\tilde{b}\|_{L^{p}(\tilde{\Omega}_1)}=r^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_r)}\leq \delta,~(\mathrm{by}~ \cref{e.tC1-ass})\\
&\|\tilde{f}\|_{L^{n}(\tilde{\Omega}_1)}\leq A_{m_0}^{-1}\left(\|f\|_{L^{n}(\Omega_r)}
+C_0r^{1-\frac{n}{p}}\|b\|_{L^{p}(\Omega_r)}+2C_0^2r\mu\right)\\
&~~~~\leq A_{m_0}^{-1}\left(\frac{\delta\tilde\omega(r)}{3}
+\frac{\delta r^{\tilde{\alpha}}}{3}+\frac{\delta r}{3}\right)\\
&~~~~\leq \delta, ~(\mathrm{by} ~\cref{e.tC1-ass}~\mathrm{and}~\cref{e.tC1-Ak})\\
&\|\tilde{g}\|_{L^{\infty}(\partial \tilde{\Omega}\cap B_1)}\leq \frac{1}{rA_{m_0}}\left(\frac{\delta r\tilde\omega(r)}{2}+\frac{C_0\delta r\tilde\omega(r)}{2C_0}\right)\leq \delta, ~(\mathrm{by}~ \cref{e.tC1-ass})\\
&\underset{B_1}{\mathrm{osc}}~\partial\tilde{\Omega}=
\frac{1}{r}\underset{B_r}{\mathrm{osc}}~\partial\Omega \leq \delta\tilde\omega(r) \leq \delta ~(\mathrm{by}~ \cref{e.tC1-ass}).
\end{aligned}
\end{equation}
By \Cref{l-C1a-mu} (take $\alpha=\tilde{\alpha}$ there), there exists a constant $\tilde{a}$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{a}y_n\|_{L^{\infty }(\tilde{\Omega} _{\eta})}&\leq \eta ^{1+\tilde{\alpha}}
\end{aligned}
\end{equation*}
and
\begin{equation*}
|\tilde{a}|\leq \bar{C}.
\end{equation*}
Let $a_{m_0+1}=a_{m_0}+A_{m_0}\tilde{a}$. Then \cref{e.tC1-4} holds for $m_0+1$. Recalling \cref{e.tC1-Ak} and \cref{e.tC1-v1}, we have
\begin{equation*}
\begin{aligned}
&\|u-a_{m_0+1}x_n\|_{L^{\infty}(\Omega_{\eta^{m_0+1}})}\\
&= \|u-a_{m_0}x_n-A_{m_0}\tilde{a}x_n\|_{L^{\infty}(\Omega_{\eta r})}\\
&= \|rA_{m_0}v-rA_{m_0}\tilde{a}y_n\|_{L^{\infty}(\tilde{\Omega}_{\eta})}\\
&\leq rA_{m_0}\eta^{1+\tilde{\alpha}}\leq\eta^{m_0+1}A_{m_0+1}.
\end{aligned}
\end{equation*}
Hence, \cref{e.tC1-3} holds for $m=m_0+1$. By induction, the proof is completed.
For the special case $\mu=0$, set
\begin{equation*}
K=\|u\|_{L^{\infty}(\Omega_1)}+\delta^{-1}\left(3\|f\|_{C^{-1,\mathrm{Dini}}(0)}
+2\|g\|_{C^{1,\mathrm{Dini}}(0)}\right)
\end{equation*}
and define for $0<\rho<1$
\begin{equation*}
\bar{u}(y)=u(x)/K,
\end{equation*}
where $y=x/\rho$. Then by taking $\rho$ small enough (depending only on $n, \lambda, \Lambda,\mu$, $p,\|b\|_{L^p(\Omega_1)}$ and $\|\partial \Omega\|_{C^{1,\mathrm{Dini}}(0)}$), \cref{e.tC1-ass} can be guaranteed. Hence, for $\mu=0$, we have the explicit estimates \cref{e.C1-1} and \cref{e.C1-2}.\qed~\\
\section{The $C^{k,\mathrm{lnL}}$ regularity}\label{CLlnL}
In this section, we prove the interior $C^{k,\mathrm{lnL}}$ ($k\geq 2$) regularity and the boundary $C^{1,\mathrm{lnL}}$ regularity. The following lemma is the ``key step'' for the interior $C^{k,\mathrm{lnL}}$ regularity and we omit its proof.
\begin{lemma}\label{In-l-Ckln-mu}
Suppose that $F\in C^{k-2,1}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Then there exists $\delta>0$ depending only on $k,n,\lambda,\Lambda,\omega_0,\omega_3$ and $\omega_4$ such that if $u$ satisfies
\begin{equation*}
F(D^2u,Du,u,x)=f ~~\mbox{in}~~B_1
\end{equation*}
with $\|u\|_{L^{\infty}(B_1)}\leq 1$, $u(0)=0,Du(0)=0,\cdots,D^{k}u(0)=0$, $\mu,b_0,c_0\leq 1$, $\|\beta_3\|_{C^{k-2,1}(0)}\leq \delta$ and $\|f\|_{C^{k-2,1}(0)}\leq \delta$, then there exists a $(k+1)$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(B_{\eta})}\leq \eta^{k+1},
\end{equation*}
\begin{equation*}
|F_0(D^2Q(x),DQ(x),Q(x),x)|\leq C |x|^{k-1+\bar{\alpha}},~\forall ~x\in B_1
\end{equation*}
and
\begin{equation*}
\|Q\|\leq C,
\end{equation*}
where $C$ and $\eta$ depend only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$.
\end{lemma}
\begin{theorem}\label{In-t-Cklns-mu}
Suppose that $F\in C^{k-2,1}(0)$ is convex in $M$ and $\omega_0$ satisfies \cref{e.omega0-2}. Let $u$ satisfy
\begin{equation*}
F(D^2u,Du,u,x)=f ~~\mbox{in}~~B_1.
\end{equation*}
Assume that
\begin{equation}\label{In-e.tCklns-be-mu}
\begin{aligned}
&\|u\|_{L^{\infty}(B_1)}\leq 1,~u(0)=0,~Du(0)=0,\cdots,D^ku(0)=0,\\
&\mu\leq \frac{1}{4C_0},~b_0\leq\frac{1}{2},~c_0\leq \frac{1}{K_0},\\
&|\beta_3(x)|\leq \delta_1|x|^{k-1}~~\mbox{and}~~|f(x)|\leq \delta_1|x|^{k-1}, ~~\forall ~x\in B_1,\\
\end{aligned}
\end{equation}
where $\delta_1\leq \delta$ ($\delta$ is as in \Cref{In-l-Ckln-mu}) and $C_0$ depend only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$.
Then $u\in C^{k,\mathrm{lnL}}(0)$ and
\begin{equation}\label{In-e.tCklns-1-mu}
|u(x)|\leq C |x|^{k+1}|\ln|x||, ~~\forall ~x\in B_{1/2},
\end{equation}
where $C$ depends only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$.
\end{theorem}
\proof To prove that $u$ is $C^{k,\mathrm{lnL}}$ at $0$, we only need to prove the following. There exist a sequence of $(k+1)$-forms $Q_m$ ($m\geq 0$ and $Q_0\equiv 0$) such that for all $m\geq 1$,
\begin{equation}\label{In-e.tCklns-6-mu}
\|u-Q_m\|_{L^{\infty }(\Omega _{\eta^{m}})}\leq \eta ^{m(k+1)},
\end{equation}
\begin{equation}\label{In-e.tCklns-9-mu}
|F_0(D^2Q_m(x),DQ_m(x),Q_m(x),x)|\leq \tilde{C}|x|^{k-1+\bar{\alpha}},~\forall ~x\in B_1
\end{equation}
and
\begin{equation}\label{In-e.tCklns-7-mu}
\|Q_m-Q_{m-1}\|\leq \tilde{C},
\end{equation}
where $\tilde{C}$ and $\eta$ depends only on $n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$.
Before proving \crefrange{In-e.tCklns-6-mu}{In-e.tCklns-7-mu}, we show that they imply \cref{In-e.tCklns-1-mu}. Indeed, by \cref{In-e.tCklns-7-mu},
\begin{equation*}
\|Q_m\|\leq m\tilde{C}\leq \tilde{C}|\ln \eta^m|.
\end{equation*}
Thus, by combing with \cref{In-e.tCklns-6-mu}, we have
\begin{equation*}
\|u\|_{L^{\infty }(\Omega_{\eta^{m}})}\leq \|Q_m\|\eta^{m(k+1)}+\eta ^{m(k+1)}
\leq(\tilde{C}+1)\eta^{m(k+1)}\cdot |\ln \eta^m|,
\end{equation*}
which implies \cref{In-e.tCklns-1-mu}.
Now, we prove \crefrange{In-e.tCklns-6-mu}{In-e.tCklns-7-mu} by induction. For $m=1$, by \Cref{In-l-Ckln-mu}, there exists a $(k+1)$-form $Q_1$ such that \crefrange{In-e.tCklns-6-mu}{In-e.tCklns-9-mu} hold for some $C_1$ and $\eta_1$ depending only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$ where $Q_0\equiv 0$. Take $\tilde{C}\geq C_1$, $\eta\leq \eta_1$ and then the conclusion holds for $m=1$. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{In-e.tCklns-v-mu}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^{k+1}}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{In-e.Cklns-F-mu}
\tilde{F}(D^2v,Dv,v,y)=\tilde{f} ~~\mbox{in}~~B_1,
\end{equation}
where for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=\frac{1}{r^{k-1}}F(r^{k-1}M+D^2Q_{m_0}(x),r^kp+DQ_{m_0}(x),
r^{k+1}s+Q_{m_0}(x),x),\\
&\tilde{f}(y)=\frac{f(x)}{r^{k-1}},
\end{aligned}
\end{equation*}
In addition, define $\tilde{F}_0$ in a similar way to the definition of $\tilde{F}$ (only replacing $F$ by $F_0$).
In the following, we show that \cref{In-e.Cklns-F-mu} satisfies the assumptions of \Cref{In-l-Ckln-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(B_1)}\leq 1,~ v(0)=0,~Dv(0)=0,\cdots,D^2v(0)=0,~(\mathrm{by}~ \cref{In-e.tCklns-be-mu},~\cref{In-e.tCklns-6-mu}~\mbox{and}~\cref{In-e.tCklns-v-mu})\\
&|\tilde{f}(y)|\leq r^{-(k-1)}|f(x)|\leq \delta_1|y|^{k-1}, ~\forall ~y\in B_1. ~(\mathrm{by}~ \cref{In-e.tCklns-be-mu})
\end{aligned}
\end{equation*}
By \cref{In-e.tCklns-7-mu}, there exists a constant $C_{0}$ depending only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$ such that $\|Q_m\|\leq mC_{0}$ ($\forall~0\leq m\leq m_0$). It is easy to check that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where (note that $\eta^m m\leq 1$, $\forall~ m\geq 0$ and $\omega_0$ satisfies \cref{e.omega0-2})
\begin{equation*}
\tilde{\mu}=r^{k+1}\mu,~~\tilde{b}= rb_0+2m_0C_0 r\mu ,~~\tilde{c}= K_0r^{2}c_0~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence, $\tilde{\omega}_0$ satisfies \cref{e.omega0-2} and from \cref{In-e.tCklns-be-mu},
\begin{equation*}
\begin{aligned}
&\tilde{\mu}\leq \mu\leq 1,~\tilde{b}\leq b_0+2C_0\mu\leq 1,~\tilde{c}\leq c_0\leq 1.
\end{aligned}
\end{equation*}
In addition, by \cref{In-e.tCklns-be-mu}, for $(M,p,s,y)\in S^n\times R^n\times R\times \bar B_1$,
\begin{equation*}
\begin{aligned}
|\tilde{F}(M,p,s,y)-\tilde{F}_0(M,p,s,y)|&\leq \delta_1\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)|y|^{k-1}\\
&:=\tilde{\beta}_3(y)\tilde{\omega}_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where $\tilde{\beta}_3(y)=\delta_1|y|^{k-1}$ and $\tilde{\omega}_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C_0,|p|+C_0,|s|+C_0)$. Hence, $\tilde{\omega}_3$ satisfies \cref{e.omega-3} and
\begin{equation*}
\|\tilde{\beta}_3\|_{C^{k-2,1}(0)}\leq \delta_1.
\end{equation*}
Finally, with the aid of \cref{In-e.tCklns-9-mu}, we can show that (similar to the interior $C^{k,\alpha}$ regularity)
\begin{equation}\label{In-e.Ckln-6}
\|\tilde{F}_0\|_{C^{k-1,\bar{\alpha}}(\bar{\textbf{B}}_{\rho}\times \bar{B}_1)}\leq \tilde{\omega}_4(\rho),~\forall ~\rho>0,
\end{equation}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$.
Choose $\delta_1$ small enough (depending only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$) such that \Cref{In-l-Ckln-mu} holds for $\tilde\omega_0,\tilde\omega_3,\tilde\omega_4$ and $\delta_1$. Since \cref{In-e.Cklns-F-mu} satisfies the assumptions of \Cref{In-l-Ckln-mu}, there exist a $(k+1)$-form $\tilde{Q}$ and constants $\tilde{C}\geq C_1$ and $\eta\leq \eta_1$ depending only on $k,n,\lambda, \Lambda,\omega_0,\omega_3$ and $\omega_4$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(B_{\eta})}&\leq \eta^{k+1},
\end{aligned}
\end{equation*}
\begin{equation*}
|\tilde F_0(D^2\tilde{Q}(y),D\tilde{Q}(y),\tilde{Q}(y),y)|\leq \tilde{C} |y|^{k-1+\bar{\alpha}},
~\forall ~y\in B_1
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \tilde{C}.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+\tilde{Q}(x)$. Then \cref{In-e.tCklns-9-mu} and \cref{In-e.tCklns-7-mu} hold for $k_0+1$. Recalling \cref{In-e.tCklns-v-mu}, we have
\begin{equation*}
\begin{aligned}
\|u-Q_{m_0+1}(x)\|_{L^{\infty}(B_{\eta^{m_0+1}})}&= \|u-Q_{m_0}(x)-\tilde{Q}(x)\|_{L^{\infty}(B_{\eta r})}\\
&= \|r^{k+1}v-r^{k+1}\tilde{Q}(y)\|_{L^{\infty}(B_{\eta})}\\
&\leq r^{k+1}\eta^{k+1}=\eta^{(m_0+1){k+1}}.
\end{aligned}
\end{equation*}
Hence, \cref{In-e.tCklns-6-mu} hold for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Now, we give the~\\
\noindent\textbf{Proof of \Cref{t-Ckln-i}.} As before, in the following proof, we just make necessary normalization to satisfy the conditions of \Cref{In-t-Cklns-mu}. Throughout this proof, $C$ always denotes a constant depending only on $k,n,\lambda, \Lambda,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,1}(0)}$, $\omega_3,\omega_4, \|f\|_{C^{k-2,1}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$.
Let $F_1(M,p,s,x)=F(M,p,s,x)-N_f(x)$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar{B}_1$ where $N_f(x)=f(0)+f_{i}(0)x_{i}+\cdots+f_{i_1\cdots i_{k-2}}(0)x_{i_1}\cdots x_{i_{k-2}}/(k-2)!$. Then $u$ satisfies
\begin{equation*}
F_1(D^2u,Du,u,x)=f_1 ~~\mbox{in}~~B_1,
\end{equation*}
where $f_1(x)=f(x)-N_f(x)$. Thus,
\begin{equation*}
|f_1(x)|\leq [f]_{C^{k-2,1}(0)}|x|^{k-1}\leq C|x|, ~~\forall ~x\in B_1.
\end{equation*}
Note that $u\in C^{k,\bar\alpha/2}(0)$ and define
\begin{equation*}
N_u(x)=u(0)+u_i(0)x_i+\cdots+u_{i_1\cdots i_{k}}(0)x_{i_1}\cdots x_{ i_{k}}/k!.
\end{equation*}
Set $u_1(x)=u(x)-N_u(x)$ and $F_2(M,p,s,x)=F_2(M+D^2N_u(x),p+DN_u(x),s+N_u(x),x)$. Then $u_1$ satisfies
\begin{equation*}
F_2(D^2u_1,Du_1,u_1,x)=f_1 ~~\mbox{in}~~B_1
\end{equation*}
and
\begin{equation*}
u_1(0)=0,~Du_1(0)=0, \cdots, D^ku_1(0)=0.
\end{equation*}
Next, take $y=x/\rho$ and $u_2(y)=u_1(x)/\rho^2$, where $0<\rho<1$ is a constant to be specified later. Then $u_2$ satisfies
\begin{equation}\label{In-Ckln-F6-k-mu}
F_3(D^2u_2,Du_2,u_2,y)=f_2~~\mbox{in}~~B_1,
\end{equation}
where
\begin{equation*}
\begin{aligned}
&F_3(M,p,s,y)=F_2\left(M,\rho p,\rho^2s,\rho y\right)~~\mbox{and}~~f_2(y)=f_1(x).
\end{aligned}
\end{equation*}
Finally, define the fully nonlinear operator $F_{30}$ in the same way as $F_3$ (only replacing $F$ by $F_0$).
Now, we try to choose a proper $\rho$ such that \cref{In-Ckln-F6-k-mu} satisfies the conditions of \Cref{In-t-Cklns-mu}. First, $u_2(0)=0,Du_2(0)=0,\cdots,D^ku_2(0)=0$ clearly. Combining with the $C^{k,\bar\alpha/2}$ regularity for $u_2$, we have
\begin{equation*}
\begin{aligned}
\|u_3\|_{L^{\infty}(B_1)}&= \|u_2\|_{L^{\infty}(B_\rho)}/\rho^2
\leq C\rho^{k-2+\bar\alpha/2}.
\end{aligned}
\end{equation*}
Next,
\begin{equation*}
|f_2(y)|=|f_1(x)|\leq C|x|^{k-1}=C\rho^{k-1}|y|^{k-1},~\forall ~y\in B_1,
\end{equation*}
It is easy to verify that $F_3$ and $F_{30}$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde\mu,\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= \rho^2\mu,~\tilde{b}= \rho b_0+C\rho\mu,~\tilde{c}= K_0\rho^2c_0~~\mbox{and}~~
\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C,\cdot).
\end{equation*}
In addition, $\tilde{\omega}_0$ satisfies \cref{e.omega0-2}.
Finally, we show $F_3\in C^{k-2,1}(0)$. Indeed,
\begin{equation*}
\begin{aligned}
&|F_3(M,p,s,y)-F_{30}(M,p,s,y)|\\
&=|F(M+D^2N_u,\rho p+DN_u,\rho^2s+N_u,x)-F_{0}(M+D^2N_u,\rho p+DN_u,\rho^2s+N_u,x)|\\
&\leq C\rho^{k-1}\omega_3(\|M\|+C,|p|+C,|s|+C)|y|^{k-1}\\
&:=\tilde{\beta}_3(y)\tilde\omega_3(\|M\|,|p|,|s|),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_3(y)=C\rho^{k-1}|y|^{k-1},~
\tilde\omega_3(\|M\|,|p|,|s|)=\omega_3(\|M\|+C,|p|+C,|s|+C)
\end{equation*}
and $\tilde\omega_3$ satisfies \cref{e.omega-3}. Moreover,
\begin{equation*}
\|F_{30}\|_{C^{k-1,\bar{\alpha}}(\bar{\textbf{B}}_r\times \bar{B}_1)}\leq \tilde{\omega}_4(r),~\forall ~r>0,
\end{equation*}
where $\tilde{\omega}_4$ depends only on $k,n,\lambda, \Lambda,\mu, b_0,c_0,\omega_0$, $\|\beta_3\|_{C^{k-2,1}(0)}$, $\omega_3,\omega_4$, $\|f\|_{C^{k-2,1}(0)}$, and $\|u\|_{L^{\infty}(B_1)}$.
Take $\delta_1$ small enough such that \Cref{In-t-Cklns-mu} holds for $\tilde{\omega}_0$, $\tilde{\omega}_3,\tilde{\omega}_4$ and $\delta_1$. From above arguments, we can choose $\rho$ small enough (depending only on $k,n,\lambda, \Lambda,\mu, b_0,c_0$, $\omega_0$, $\|\beta_3\|_{C^{k-2,1}(0)}$, $\omega_3,\omega_4, \|f\|_{C^{k-2,1}(0)}$ and $\|u\|_{L^{\infty}(B_1)}$) such that the conditions of \Cref{In-t-Cklns-mu} are satisfied. Then $u_3$ and hence $u$ is $C^{k,\mathrm{lnL}}$ at $0$, and the estimates \crefrange{e.Ckln-1-i}{e.Ckln-2-i} hold. \qed~\\
In the following, we give the proof of the boundary $C^{1,\mathrm{lnL}}$ regularity. The following is the corresponding ``key step'' and we omit its proof.
\begin{lemma}\label{l-C1ln-mu}
Suppose that $F$ satisfies \cref{e.C1a.beta}. There exists $\delta>0$ depending only on $n,\lambda,\Lambda$ and $\omega_1$ such that if $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1
\end{aligned}\right.
\end{equation*}
with $\|u\|_{L^{\infty}(\Omega_1)}\leq 1$, $u(0)=0$, $Du(0)=0$, $\mu,b_0,c_0\leq \delta$, $\omega_0(1,1)\leq 1$, $\|\beta_1\|_{L^{\infty}(\Omega_1)}\leq \delta$, $\|f\|_{L^{\infty}(\Omega_1)}\leq \delta$, $\|\partial \Omega\|_{C^{1,1}(0)} \leq \delta$ and $\|g\|_{C^{1,1}(0)} \leq \delta$, then there exists a $2$-form $Q$ such that
\begin{equation*}
\|u-Q\|_{L^{\infty}(\Omega_{\eta})}\leq \eta^{2},
\end{equation*}
\begin{equation*}
F_0(D^2Q,0,0)=0
\end{equation*}
and
\begin{equation*}
\|Q\|\leq \bar{C}+1,
\end{equation*}
where $\eta$ depends only on $n,\lambda$ and $\Lambda$.
\end{lemma}
\begin{theorem}\label{t-C1lns}
Suppose that $F$ satisfies \cref{e.C1a.beta} and $u$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F(D^2u,Du,u,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u=g&& ~~\mbox{on}~~(\partial \Omega)_1.
\end{aligned}\right.
\end{equation*}
Assume that
\begin{equation}\label{e.C1ln-ubeg}
\begin{aligned}
&\|u\|_{L^{\infty}(\Omega_1)}\leq 1,~ u(0)=0,~Du(0)=0,\\
&\mu\leq \frac{\delta_1}{4C_0^2},~ b_0\leq \frac{\delta_1}{4C_0},~
c_0\leq \frac{\delta_1}{4},~\omega_0(1+C_0,C_0)\leq1,\\
&|\beta_1(x)|\leq \frac{\delta_1}{|\ln |x||}, ~\forall ~x\in \Omega_1,~\|\gamma_1\|_{L^{\infty}(\Omega_1)}\leq \frac{\delta_1}{4}, \\ &\|f\|_{L^{\infty}(\Omega_1)}\leq \delta_1,~
|g(x)|\leq \frac{\delta_1}{2}|x|^2,~\forall ~x\in(\partial \Omega)_1~~\mbox{and}~~ \|\partial \Omega\cap B_1\|_{C^{1,1}(0)} \leq \frac{\delta_1}{2C_0},\\
\end{aligned}
\end{equation}
where $\delta_1$ depends only on $n,\lambda,\Lambda$ and $\omega_1$, and $C_0$ depends only on $n,\lambda$ and $\Lambda$.
Then $u$ is $C^{1,\mathrm{lnL}}$ at $0$ and
\begin{equation*}
|u(x)|\leq C |x|\big|\ln |x|\big|, ~~\forall ~x\in \Omega_{1/2},
\end{equation*}
where $C$ depends only on $n,\lambda$ and $\Lambda$.
\end{theorem}
\proof To prove that $u$ is $C^{1,\mathrm{lnL}}$ at $0$, we only need to prove the following. There exist a sequence of $2$-forms $Q_m$ ($m\geq -1$) such that for all $m\geq 0$,
\begin{equation}\label{e.C1ln-u}
\|u-Q_m\|_{L^{\infty }(\Omega_{\eta^{m}})}\leq \eta ^{2m},
\end{equation}
\begin{equation}\label{e.C1lns-F}
F_0(D^2Q_m)=0
\end{equation}
and
\begin{equation}\label{e.C1ln-bk}
\|Q_m-Q_{m-1}\|\leq \bar{C}+1,
\end{equation}
where $\eta$ depending only on $n,\lambda$ and $\Lambda$, is as in \Cref{l-C1ln-mu} .
Now we prove \crefrange{e.C1ln-u}{e.C1ln-bk} by induction. For $m=0$, by setting $Q_0=Q_{-1}\equiv 0$, the conclusion holds clearly. Suppose that the conclusion holds for $m=m_0$. We need to prove that the conclusion holds for $m=m_0+1$.
Let $r=\eta ^{m_{0}}$, $y=x/r$ and
\begin{equation}\label{e.C1ln-v1}
v(y)=\frac{u(x)-Q_{m_0}(x)}{r^2}.
\end{equation}
Then $v$ satisfies
\begin{equation}\label{e.C1lns-F-mu}
\left\{\begin{aligned}
&\tilde{F}(D^2v,Dv,v,y)=\tilde{f}&& ~~\mbox{in}~~\tilde{\Omega}\cap B_1;\\
&v=\tilde{g}&& ~~\mbox{on}~~\partial \tilde{\Omega}\cap B_1,
\end{aligned}\right.
\end{equation}
where for $(M,p,s,x)\in S^n\times R^n\times R\times \bar{\tilde{\Omega}}_1$,
\begin{equation*}
\begin{aligned}
&\tilde{F}(M,p,s,y)=F(M+D^2Q_{m_0},rp+DQ_{m_0}(x),r^2s+Q_{m_0}(x),x),\\
&\tilde{f}(y)=f(x),~\tilde{g}(y)=\frac{g(x)-Q_{m_0}(x)}{r^2}~~\mbox{and}~~\tilde{\Omega}=\frac{\Omega}{r}.
\end{aligned}
\end{equation*}
In addition, define
\begin{equation*}
\tilde{F}_0(M)=F_0(M+D^2Q_{m_0}).
\end{equation*}
In the following, we show that \cref{e.C1lns-F-mu} satisfies the assumptions of \Cref{l-C1ln-mu}. First, it is easy to verify that
\begin{equation*}
\begin{aligned}
&\|v\|_{L^{\infty}(\tilde{\Omega}_1)}\leq 1,~ v(0)=0,~ Dv(0)=0,
~(\mathrm{by}~\cref{e.C1ln-ubeg},~ \cref{e.C1ln-u} ~\mbox{and}~ \cref{e.C1ln-v1})\\
&\|\tilde{f}\|_{L^{\infty}(\tilde{\Omega}_1)}=\|f\|_{L^{\infty}(\Omega_r)}\leq \delta_1,
~(\mathrm{by}~\cref{e.C1ln-ubeg})\\
&\|\partial \tilde{\Omega}\cap B_1\|_{C^{1,1}(0)} \leq r\|\partial\Omega\cap B_1\|_{C^{1,1}(0)}\leq \delta_1,
~(\mathrm{by}~\cref{e.C1ln-ubeg})\\
&\tilde{F}_0(0)=F_0(D^2Q_{m_0})=0.~(\mathrm{by}~ \cref{e.C1lns-F})
\end{aligned}
\end{equation*}
In addition, for any $0<\rho<1$ (note that $Q_{m_0}$ is a $2$-form),
\begin{equation*}
\|\tilde{g}\|_{L^{\infty}(\partial \tilde{\Omega}\cap B_{\rho})} \leq \frac{1}{r^{2}}\left(\frac{\delta _1}{2}(\rho r)^{2}+
C_0\cdot \frac{\delta_1}{2C_0}(\rho r)^{2}\right)\leq \delta_1 \rho^{2}.
~~(\mathrm{by}~\cref{e.C1ln-ubeg})
\end{equation*}
Hence,
\begin{equation*}
\|\tilde{g}\|_{C^{1,1}(0)}\leq \delta_1.
\end{equation*}
By \cref{e.C1ln-bk}, there exists a constant $C_0$ depending only on $n,\lambda$ and $\Lambda$ such that $\|Q_m\|\leq mC_0$ ($\forall~0\leq m\leq m_0$). It is easy to verify that $\tilde{F}$ and $\tilde{F}_0$ satisfy the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where (note that $\eta^m m\leq 1$, $\forall~ m\geq 0$)
\begin{equation*}
\tilde{\mu}=r^2\mu,~\tilde{b}= rb_0+2m_0C_0r^2\mu,~\tilde{c}= c_0
~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\omega_0(\cdot+C_0,\cdot).
\end{equation*}
Hence,
\begin{equation*}
\begin{aligned}
&\tilde{\mu}\leq \mu\leq \delta_1,~\tilde{b}\leq b_0+2C_0\mu\leq \delta_1
~~\mbox{and}~~\tilde{c}\leq c_0\leq \delta_1.
\end{aligned}
\end{equation*}
Moreover, by combining the structure condition \cref{SC1}, \cref{e.C2a-KF} and \cref{e.C1ln-ubeg}, for $(M,y)\in S^n\times \bar{\tilde{\Omega}}_1$,
\begin{equation*}
\begin{aligned}
&|\tilde{F}(M,0,0,y)-\tilde{F}_0(M)|\\
&= |F(M+D^2Q_{m_0},DQ_{m_0}(x),Q_{m_0}(x),x)-F_0(M+D^2Q_{m_0})|\\
&\leq |F(M+D^2Q_{m_0},DQ_{m_0}(x),Q_{m_0}(x),x)-F(M+D^2Q_{m_0},0,0,x)|\\
&~~~~+|F(M+D^2Q_{m_0},0,0,x)-F_0(M+D^2Q_{m_0})|\\
&\leq C_0^2\mu+C_0b_0+c_0\omega_0(C_0,C_0)+\beta_1(x)\omega_1(\|M\|+m_0C_0)+\gamma_1(x)\\
&\leq\delta_1\omega_1(\|M\|+C_0)+\delta_1\\
&:=\tilde{\beta}_1(y)\tilde{\omega}_1(\|M\|)+\tilde{\gamma}_1(y),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_1(y)\equiv \delta_1,~
\tilde{\gamma}_1(y)\equiv \delta_1~~\mbox{and}~~\tilde{\omega}_1(\|M\|)=\omega_1(\|M\|+C_0).
\end{equation*}
Choose $\delta_1$ small enough (depending only on $n,\lambda,\Lambda$ and $\omega_1$) such that \Cref{l-C1ln-mu} holds for $\tilde\omega_0,\tilde\omega_1$ and $\delta_1$. Since \cref{e.C1lns-F-mu} satisfies the assumptions of \Cref{l-C1ln-mu}, there exists a $2$-form $\tilde{Q}(y)$ such that
\begin{equation*}
\begin{aligned}
\|v-\tilde{Q}\|_{L^{\infty }(\tilde{\Omega}_{\eta})}&\leq \eta^2,
\end{aligned}
\end{equation*}
\begin{equation*}
\tilde{F}_0(D^2\tilde{Q})=0
\end{equation*}
and
\begin{equation*}
\|\tilde{Q}\|\leq \bar{C}+1.
\end{equation*}
Let $Q_{m_0+1}(x)=Q_{m_0}(x)+\tilde{Q}(x)$. Then \cref{e.C1lns-F} and \cref{e.C1ln-bk} hold for $m_0+1$. Recalling \cref{e.C1ln-v1}, we have
\begin{equation*}
\begin{aligned}
&\|u-Q_{m_0+1}(x)\|_{L^{\infty}(\Omega_{\eta^{m_0+1}})}\\
&= \|u-Q_{m_0}(x)-\tilde{Q}(x)\|_{L^{\infty}(\Omega_{\eta r})}\\
&= \|r^{2}v-r^{2}\tilde{Q}(y)\|_{L^{\infty}(\tilde{\Omega}_{\eta})}\\
&\leq r^{2}\eta^{2}= \eta^{2(m_0+1)}.
\end{aligned}
\end{equation*}
Hence, \cref{e.C1ln-u} holds for $m=m_0+1$. By induction, the proof is completed.\qed~\\
Now, we give the ~\\
\noindent\textbf{Proof of \Cref{t-C1ln}.} As before, we prove the theorem in two cases.
\textbf{Case 1:} the general case, i.e., $F$ satisfies \cref{SC2} and \cref{e.C1a.beta}. Throughout the proof for this case, $C$ always denotes a constant depending only on $n,\lambda,\Lambda,\mu,b_0, c_0,\omega_0$, $\omega_1$, $\|\partial \Omega\|_{C^{1,1}(0)}$, $\|f\|_{L^{\infty}(\Omega_1)}$, $\|g\|_{C^{1,1}(0)}$ and $\|u\|_{L^{\infty }(\Omega_1)}$.
Let $u_1(x)=u(x)-N_{g}(x)$ and $F_1(M,p,s,x)=F(M,p+DN_{g}(x),s+N_{g}(x),x)$ where $N_{g}(x)=g(0)+Dg(0)\cdot x$. Then $u_1$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_1(D^2u_1,Du_1,u_1,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u_1=g_1&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_1(x)=g(x)-N_{g}(x)$. Hence,
\begin{equation*}
|g_1(x)|\leq C|x|^{2}, ~~\forall ~x\in (\partial \Omega)_1
\end{equation*}
and
\begin{equation*}
\begin{aligned}
|F_1(0,0,0,x)|=\left|F\left(0,DN_{g}(x),N_{g},x\right)\right|\leq C.
\end{aligned}
\end{equation*}
Note that
\begin{equation*}
u_1\in S^{*}(\lambda,\Lambda,\mu,\hat{b},|f|
+c_0\omega_0(\|u\|_{L^{\infty}(\Omega_1)}+\|g\|_{C^{1,1}(0)},u_1)+|F_1(0,0,0,\cdot)|).
\end{equation*}
where $\hat{b}=b_0+2\mu\|g\|_{C^{1,1}(0)}$. By \Cref{t-C1a-mu}, $u_1\in C^{1,\alpha}(0)$, $Du_1(0)=(0,...,0,(u_1)_n(0))$ and
\begin{equation*}
\begin{aligned}
|(u_1)_n(0)| &\leq C.
\end{aligned}
\end{equation*}
Define $u_2(x)=u_1(x)-N_u(x)$ and $F_2(M,p,s,x)=F_1(M,p+DN_u(x),s+N_u(x),x)$ where $N_u(x)=(u_1)_n(0)x_n$. Then $u_2$ satisfies
\begin{equation*}
\left\{\begin{aligned}
&F_2(D^2u_2,Du_2,u_2,x)=f&& ~~\mbox{in}~~\Omega_1;\\
&u_2=g_2&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation*}
where $g_2=g_1-N_u(x)$. Moreover, $u_2(0)=0$ and $Du_2(0)=0$. Since $\partial \Omega$ is $C^{1,1}$ at $0$,
\begin{equation*}
|g_2(x)|\leq |g_1(x)|+|N_u(x)|\leq C|x|^{2}, ~~\forall ~x\in (\partial \Omega)_1.
\end{equation*}
Finally, let $y=x/\rho$, $u_3(y)=u_2(x)/\rho$ and $F_3(M,p,s,y)=\rho F_2(M/\rho, p,\rho s,x)$. Then $u_3$ satisfies
\begin{equation}\label{F4-Cln-mu}
\left\{\begin{aligned}
&F_3(D^2u_3,Du_3,u_3,y)=f_1&& ~~\mbox{in}~~\tilde{\Omega}_1;\\
&u_3=g_3&& ~~\mbox{on}~~(\partial \tilde\Omega)_1,
\end{aligned}\right.
\end{equation}
where $f_1(y)=\rho f(x)$, $g_3(y)=g_2(x)/\rho$ and $\tilde{\Omega}=\Omega/\rho$. Finally, define the fully nonlinear operator $F_{30}(M):=\rho F_0(\rho ^{-1}M)$.
Now, we can check that \cref{F4-Cln-mu} satisfies the conditions of \Cref{t-C1lns} by choosing a proper $\rho$. First, it can be checked easily that
\begin{equation*}
\begin{aligned}
&u_3(0)=0 ,~Du_3(0)=0,~|f_1(y)|= \rho|f(x)|\leq C\rho,~\forall ~y\in\tilde{\Omega}_1,\\
&|g_3(y)|= \frac{|g_2(x)|}{\rho}\leq C\rho |y|^2,~\forall ~y\in\partial \tilde{\Omega}_1,\\
&\|\partial \tilde{\Omega}\cap B_1\|_{C^{1,\alpha}(0)}\leq \rho\|\partial \Omega\cap B_1\|_{C^{1,1}(0)}.
\end{aligned}
\end{equation*}
Next, by the boundary $C^{1,\bar\alpha/2}$ regularity for $u_1$,
\begin{equation*}
\|u_3\|_{L^{\infty}(\tilde{\Omega}_1)}= \frac{\|u_2\|_{L^{\infty}(\Omega_\rho)}}{\rho}
\leq C\rho^{\bar\alpha/2}.
\end{equation*}
It can be checked that $F_4$ satisfies the structure condition \cref{SC2} with $\lambda,\Lambda,\tilde{\mu},\tilde{b},\tilde{c}$ and $\tilde{\omega}_0$, where
\begin{equation*}
\tilde{\mu}= \rho\mu,~\tilde{b}= \rho b_0+C\rho\mu,~\tilde{c}= \rho^{1/2}c_0
~~\mbox{and}~~\tilde{\omega}_0(\cdot,\cdot)=\rho^{1/2}\omega_0(\cdot+C,\cdot).
\end{equation*}
Finally, we check the oscillation of $F_3$ in $y$. Define $N(x)=N_g(x)+N_u(x)$, which is a linear polynomial and we have
\begin{equation*}
\|N\|_{C^{1}(\bar{\Omega}_1)}\leq C.
\end{equation*}
We compute
\begin{equation*}
\begin{aligned}
|F&_4(M,0,0,y)-F_{40}(M)|\\
=&|\rho F(\frac{M}{\rho},DN,N,x)-\rho F_0(\frac{M}{\rho})|\\
\leq & |\rho F(\frac{M}{\rho},DN,N,x)-\rho F(\frac{M}{\rho},0,0,x)|
+|\rho F(\frac{M}{\rho},0,0,x)-\rho F_0(\frac{M}{\rho})|\\
\leq& \rho(C^2\mu+Cb_0+c_0\omega_0(C,C))
+K_0\beta_1(x)\omega_1(\|M\|)+\rho\gamma_1(x)\\
:= &\tilde{\beta}_1(y)\omega_1(\|M\|)+\tilde{\gamma}_1(y),
\end{aligned}
\end{equation*}
where
\begin{equation*}
\tilde{\beta}_1(y)=K_0\beta_1(x)\leq \delta_0 K_0/\ln |y|~~\mbox{and}~~
\tilde{\gamma}_1=\rho\gamma_1(x)+\rho(C^2\mu+Cb_0+c_0\omega_0(C_0,C_0)).
\end{equation*}
Take $\delta_1$ small enough such that \Cref{t-C1lns} holds with $\tilde{\omega}_0,\omega_1$ and $\delta_1$. From above arguments, we take $\delta_0$ satisfying $K_0\delta_0\leq \delta_1$ and $\rho$ small enough (depending only on $n,\lambda,\Lambda,\mu,b_0, c_0,\omega_0$, $\omega_1$, $\|\partial \Omega\|_{C^{1,1}(0)}$, $\|f\|_{L^{\infty}(\Omega_1)},\|g\|_{C^{1,1}(0)}$ and $\|u\|_{L^{\infty }(\Omega_1)}$) such that
\begin{equation*}
\begin{aligned}
&\|u_3\|_{L^{\infty}(\Omega_1)}\leq 1,~ \tilde\mu\leq \frac{\delta_1}{4C_0^2},~ \tilde b\leq \frac{\delta_1}{4C_0},~ \tilde c\leq \frac{\delta_1}{4},~\tilde\omega_0(1+C_0,C_0)\leq1, \\ &|\tilde\beta_1(y)|\leq \frac{\delta_1}{|\ln |y||}, ~\forall ~x\in \Omega_1,~
\|\tilde\gamma_1\|_{L^{\infty}(\Omega_1)}\leq \frac{\delta_1}{4},~ \|f_1\|_{L^{\infty}(\Omega_1)}\leq \delta_1,\\
&|g_2(x)|\leq \frac{\delta_1}{2}|x|^2, ~\forall ~x\in(\partial \Omega)_1~~\mbox{and}~~\|\partial \tilde\Omega\cap B_1\|_{C^{1,1}(0)} \leq \frac{\delta_1}{2C_0},\\
\end{aligned}
\end{equation*}
where $C_0$ depending only on $n,\lambda$ and $\Lambda$, is as in \Cref{t-C1lns}. Therefore, the assumptions in \Cref{t-C1lns} are satisfied. By \Cref{t-C1lns}, $u_3$ and hence $u$ is $C^{1,\mathrm{lnL}}$ at $0$, and the estimates \cref{e.C1ln-1} and \cref{e.C1ln-2} hold.
\textbf{Case 2:} $F$ satisfies \cref{SC1} and \cref{e.C1a.beta-2}. Let $K=\|u\|_{L^{\infty }(\Omega_1)}+\|f\|_{L^{\infty}(\Omega_1)}+\|\gamma_1\|_{L^{\infty}(\Omega_1)}
+\|g\|_{C^{1,1}(0)}$ and $u_1=u/K$. Then $u_1$ satisfies
\begin{equation}\label{e.11.1}
\left\{\begin{aligned}
&F_1(D^2u_1,Du_1,u_1,x)=f_1&& ~~\mbox{in}~~\Omega_1;\\
&u_1=g_1&& ~~\mbox{on}~~(\partial \Omega)_1,
\end{aligned}\right.
\end{equation}
where $F_1(M,p,s,x)=F(KM,Kp,Ks,x)/K$ for $(M,p,s,x)\in S^n\times R^n\times R\times \bar\Omega_1$, $f_1=f/K$ and $g_1=g/K$. Then
\begin{equation*}
\|u_1\|_{L^{\infty }(\Omega_1)}\leq 1,~\|f_1\|_{L^{\infty }(\Omega_1)}\leq 1 ~~\mbox{and}~~\|g_1\|_{C^{1,1}(0)}\leq 1.
\end{equation*}
In addition, define $F_{10}(M,p,s)=F_0(KM)/K$. Then $F_1$ and $F_{10}$ satisfy the structure condition \cref{SC1} with the same $\lambda,\Lambda,b_0$ and $c_0$. Moreover, $F_1$ satisfies \cref{e.C1a.beta-2} with $\beta_1(x)=\beta_2(x)$ and $\gamma_1(x)=\gamma_2(x)/K$. Hence, $\|\gamma_1\|_{L^{\infty }(\Omega_1)}\leq 1$.
Apply \textbf{Case 1} to \cref{e.11.1}, we obtain that $u_1$ and hence $u$ is $C^{1,\mathrm{lnL}}$ at $0$, and the estimates \cref{e.C1ln-1} and \cref{e.C1ln-2} hold. \qed~\\
\bibliographystyle{model4-names}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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Спаська коса — довга серпоподібна коса, що обмежує Спаське урочище на півдні. До виникнення міста Миколаєва коса називалася «Жабурна».
Назва
Стара назва коси пов'язана з тим, що навколо неї, особливо в затоні, що утворився, росло багато тину, нитчатки, що українською називалося «жабуриння». Назва відома до заснування Миколаєва і згадується в документі; дано запорожцями, що ловили рибу на берегах Бузького лиману на орендних умовах. Згодом ця коса була названа Спаською на ім'я Спаського урочища, яке вона обмежувала з півдня. На плані міста Миколаєва назва «Спаська коса» вперше з'являється у 1890 році, однак в той же час вона вказана у лоціях 1851 і 1867 років.
Галерея
Примітки
географія Миколаєва
Коси України
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
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HomeThe Frivolist4 Powerful LGBTQ Exhibits to Catch This Spring
4 Powerful LGBTQ Exhibits to Catch This Spring
By Mikey Rox
Punch up your post-brunch weekend activities with these exhibits of cultural and LGBTQ significance.
1. Superfine! Art Fair, New York, N.Y.
Superfine! NYC moves this year's everyman art fair, May 1 to 5, from Manhattan's Meatpacking District to SoHo with a particular emphasis on queer culture, including Brooklyn artist Adam Chuck's Call Me By Your Preferred Pronoun special project. Commissioned by the fair's directors, the exhibit's Polaroid-style paintings evoke the continued narrative of the Oscar-nominated film's Oliver and Elio, reimagined with the broader LGBTQ community in mind. Other not-to-miss exhibit highlights include works from contemporary Provincetown artists Andrew Moncrief and Thom Jackson, cloaked-identity Andy Blank's $200-or-less "f*&%ing awesome" pieces made with museum-quality materials (a ground-level Banksy-esque opportunity, if you buy into the manufactured hype), and tapestry-based mixed media paintings that dabble in the occult from Chris Roberts-Antieau. Tickets are available at superfinenyc.eventbrite.com and include complimentary beer or wine for fairgoers.
2. Rise Up: Stonewall and the LGBTQ Rights Movement, Washington, D.C.
While history tries to whitewash what actually happened in New York's Greenwich Village in June 1969, "Rise Up: Stonewall and the LGBTQ Rights Movement" will set the record straight through powerful artifacts, images and historical print publications from those landmark moments in LGBTQ liberation. Housed at D.C.'s Newseum, the exhibit will explore key moments of gay rights history, including the 1978 assassination of Harvey Milk; the AIDS crisis; U.S. Rep. Barney Frank's public coming out in 1987; hate-crime legislation efforts; "Don't ask, don't tell"; and the fight for marriage equality. The program will feature journalists, authors, politicians and other newsmakers who have led the fight for equality, through Dec. 31, after which the exhibit will travel nationally.
3. Queer Miami: A Fascinating Look at the History of LGBTQ Communities in Miami, Miami, Fla.
Revisit Miami's queer past through this exciting exhibit of artifacts, photographs and archival footage, curated by local historian Julio Capó Jr., which chronicles more than 100 years of the area's queer communities' will to survive, thrive and rise above discrimination, isolation and violence. An expert on the intersection of gender and sexuality throughout history, Capó, who is an associate professor of history at the University of Massachusetts, Amherst, also is the author of Welcome to Fairyland: Queer Miami Before 1940, which has received six honors, including several awards from the Florida Historical Society, Florida Book Awards and Southern Historical Association. The exhibition will run until Sept.1 at HistoryMiami Museum.
4. Queer California: Untold Stories, Oakland, Calif.
With a concentrated focus on transgender communities, people of color and women, Queer California: Untold Stories at the Oakland Museum of California, through Aug.11, explores the lesser-told history of minorities within a minority through rarely seen artifacts, archival documents, photographs, costumes and ephemera such as 'zines, stickers and flyers. Powerful examples of social activism through contemporary artwork and historical materials align important milestones in LGBTQ culture, showcasing a diversity of queer identities, civil rights and resistance to oppression. OMCA includes representations of queer bodies and material with sexual content in this exhibition to affirm the ways they have been central to the experience of LGBTQ individuals and communities.
Mikey Rox is an award-winning journalist and LGBT lifestyle expert whose work has been published in more than 100 outlets across the world. He spends his time writing from the beach with his dog Jaxon. Connect with Mikey on Instagram @mikeyrox
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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\section{INTRODUCTION}
\label{sec:intro}
To date, numerous adaptive optics surveys have attempted to directly image young extrasolar planets in the near infrared\cite{Biller07, Laf07a, Mas05}. While these surveys have placed strong statistical limits on extrasolar planet distributions, no planet has yet been imaged around a main sequence or pre-main sequence star. Clearly, directly imaging planets is a nontrivial undertaking. It is important to weigh the survey instrument capability against theoretical predictions as to planet properties and distribution (e.g., from evolutionary models, radial velocity studies, etc.) and design a study which maximizes the chance of detecting a planet while also providing a scientifically interesting null result. In this paper, we will consider how best to maximize scientific output for a 50 night planet finding survey with the Near Infrared Coronagraphic Imager (NICI) at Gemini South.
NICI combines a number of techniques to attenuate
starlight and suppress superspeckles: 1) coronagraphic imaging, 2)
dual channel imaging for Spectral Differential Imaging (SDI)\cite{Mar05,Biller07} and 3)
operation in a fixed Cassegrain rotator mode for Angular Differential
Imaging (ADI) a.k.a. roll subtraction\cite{Liu04,Mar06}. While coronagraphic imaging, SDI, and ADI have all been used individually in large planet-finding
surveys, this is the first time they will all be used together in a large survey. Thus, it is important to carefully consider how to best carry out an observing campaign with NICI.
Both SDI and ADI techniques seek to separate real objects from speckles. SDI achieves this by exploiting a spectral feature in the desired target (previous implementations utilize the 1.6 $\mu$m methane absorption feature robustly observed in substellar objects cooler than 1400 K). Images are taken simultaneously both within and outside the chosen absorption feature. Due to the simultaneity of the observations, the star and the coherent speckle pattern are largely identical in both filters, while any faint companion with that absorption feature is bright in one filter and faint in the other. Subtracting the two images thus removes the starlight and speckle patterns while a real companion with the chosen absorption feature remains in the image. In other words, the off-absorption band image acts as an ideal reference point spread function (henceforth PSF) for the absorption band image. Utilizing a signature spectral feature of substellar objects greatly reduces the number of false positives detected (e.g. a background object, while real, will drop out of the SDI
subtraction since it will not have methane absorption).
In previous surveys\cite{Biller07}, SDI was used at two discrete rotator angles for each object -- a real object on the sky will modulate with a change of rotator angle; however, speckles which are instrumental phenomena will not. Thus, by subtracting data taken at multiple roll angles, speckles are further attenuated and any real object in the frame will display a characteristic jump in roll angle.
ADI employs a similar strategy to build a high quality reference PSF to remove speckles. Instead of observing at discrete roll angles, ADI leaves the rotator off and allows the telescope optics to rotate on the sky (this works best at Cassegrain focus). In a sequence of images taken at different parallactic angles, a real companion will track on the sky with the parallactic angle, while speckles will move randomly. From a series of images, a reference PSF can be constructed for and subtracted from each individual image, attenuating quasi-static speckle structure. Combining both SDI and ADI techniques thus allows an even greater degree of speckle supression.
SDI is the base mode of the NICI instrument. Each exposure produces two frames, one from each of the two channels. Usually, each channel is operated with a different filter, one off the absorption feature (usually on the blue side), and one on the feature (usually redward of the first filter). Whether SDI is combined with ADI or used at discrete roll angles is the choice of the observer, and is not necessarily a straightforward choice. For more details regarding the NICI instrument itself, see Chun et al.\cite{Chun08}, this volume.
To develop the most appropriate observing strategy for the NICI science campaign, we must first
determine a number of quantities (some inherent to the objects observed, some properties of the
NICI instrument itself), including:
1) the sky coverage and relative performance of the ADI plus SDI and SDI discrete roll angles observation modes (covered in Section 2)
2) the inherent luminosity of young planets and the required $\Delta$mag to image them (covered in Section 3.1)
3) the expected methane break strength for cool substellar objects and young planets, which will constrain how well our spectral difference works and how much of a $\Delta$mag boost
we can get from it (also covered in Section 3.2)
4) the achieved contrast of the NICI instrument as a function of separation from the primary star (relative $H$ band vs. $K$ band contrast discussed in Section 3.3, $H$ band contrast discussed in detail in Artigau et al.\cite{Art08} in this volume)
5) the achieved boost in contrast compared to the standard AO due to the spectral subtraction or due to azimuthal differential imaging (discussed in Artigau et al.\cite{Art08} in this volume)
6) the throughput of the NICI instrument at various wavelengths
In this document, we will therefore broadly discuss two major trades for the NICI campaign: operating with rotator off and on (ADI plus SDI vs. SDI discrete roll angles, respectively), and $H$ band vs. $K$ band performance.
\section{Rotator On vs. Rotator Off}
\label{sec:rotonoff}
Previous direct detection surveys for planets have either utilized the multi-wavelength Spectral Differential Imaging technique (SDI) with one or more fixed rotator angles\cite{Biller07}
or utilized the Azimuthal Differential Imaging technique (ADI) at a single wavelength with the telescope rotator turned off\cite{Laf07a}. The NICI instrument is the first to combine both capabilities. Combining both techniques thus allows an even greater degree of speckle supression. However, while any observation can utilize the narrow-band methane filters to do SDI, ADI is not necessarily possible on all objects. To successfully perform ADI on an object, two conditions must be satisfied:
1) During the observing period, the target must rotate on the sky by at least 20$^{\circ}$ (corresponding to a rotation on the chip of 2 pixels at 1"). In ADI mode, a sequence of images is taken while the object moves through a variety of parallactic angles. For each image in the sequence, a reference PSF is built from the remaining images in the sequence. If the target has not rotated through a large enough angle, no images in the series will be appropriate for using as a reference PSF, because a companion object will not move sufficiently to prevent self-subtraction.
2) At the same time, the target must not rotate too much. For a given exposure time (e.g. 1 minute), if sky rotation is too much, an object on the sky at separation of greater than 2" will appear blurred in each image in the series (referred to hereafter as "shear"), also leading to lost companion flux.
Thus, only objects with enough total sky rotation ($>$20$^{\circ}$) but not too
high a rate of sky rotation (thus causing shear), are appropriate for rotator-off observations. Below, we discuss necessary criteria for rotator-off observations, the amount of sky accessible to rotator-off vs. rotator-on observations, and compare rotator-on vs. rotator-off performance.
\subsection{Analytical Considerations}
\label{sec:rotonoffanalytical}
For any ADI observation, we need to balance having sufficient sky rotation to construct a reference PSF over a multi-image sequence vs. the acceptable shear per image. To constrain available sky rotation as a function of declination and hour angle, we plot maps on the sky (declination vs. hour angle) of rotation on the sky in an hour's observation in the left panel of Fig.~\ref{fig:rotairmass}. Hour angle plotted on the X axis of this figure is the hour angle at the beginning of the observation. (We have not plotted rotation over transit since parallactic angle changes so rapidly over transit that ADI observations are impossible due to shear -- see below.) For an object separated by 1" from the parent star, a sky rotation of 20$^{\circ}$ corresponds to a rotation on the chip of two pixels. With typical AO PSF FWHM$\sim$4 pix, this corresponds to about 1/2 FWHM of the AO PSF and is the minimum rotation necessary to ensure that a good reference PSF can be made. Thus, it is clear that only select portions of the sky are available to ADI at any given point in the night (although usable ADI time can be doubled by observing
both before and after transit, omitting the period of highest sky rotation over transit). In contrast, most of the observable sky is available for SDI at discrete roll angles (see the right panel of Fig.~\ref{fig:rotairmass}, assuming the only constraint on whether SDI can be performed is having an airmass less than 2).
\begin{figure}
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=3.2in]{rotmapha2.eps} &
\includegraphics[width=3.2in]{ammapha2.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:rotairmass} Left: Rotation map on the sky (hour angle vs. declination) in an hour's observation at Gemini South. Hour angle plotted on the X axis of this figure is the hour angle at the beginning of the observation. Contours are given at sky rotations of 60, 40, 30, 20, 15, 10, and 5$^{\circ}$. Targets with sky rotation of greater than 20$^{\circ}$ in an hour are observable with the ADI technique. Right: Airmass map on the sky for an hour's observation at Gemini South. Targets with airmass less than 2 are observable with the SDI technique at fixed rotator angles.
}
\end{figure}
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.2in]{shear2.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:shear1} Map of shear as a function of sky position (hour angle vs. declination) in a 1 minute exposure at Gemini South for an object 1" from the target. Contours are shown at shear levels of .1, .2, .5, 1, 2, and 5 pixels.
}
\end{figure}
Too much sky rotation can lead to blurring, or "shear", of a companion object image over one exposure. A map of shear at 1" from the primary in a 1 minute exposure is shown in Fig.~\ref{fig:shear1}. Thus, observing an object at -35 declination starting right after a transit will provide $>$30$^{\circ}$ of sky rotation, but also up to two pixels worth of shear for an object at 1" in a 1 minute exposure. Unfortunately, the portion of the sky that produces the greatest sky rotation will also produce the greatest shear, further reducing the part of the sky accessible with ADI.
It is unclear intuitively how much shear is too much. To estimate the efficiency loss due to shear, we used a simple geometric model. We assume a circular PSF, which will be elongated by shear. The total energy in the PSF stays the same, but the peak flux decreases as PSF area increase.
We model this elongated shape in a straightforward manner, assuming that the area of the PSF increases from a circular shape:
\begin{equation}
A_{noshear} = \pi r_{PSF}^2
\end{equation}
to the sheared final PSF (modeled as a rectangular sheared segment, with semi-circle edges):
\begin{equation}
A_{shear} = \pi r_{PSF}^2 + 2r_{PSF}*s
\end{equation}
where s is the total shear in pixels.
Thus, the peak flux F in the object is decreased by:
\begin{equation}
\frac{F_{shear}}{F_{noshear}} = \frac{A_{shear}}{A_{noshear}}
\end{equation}
and the sensitivity loss $\eta$ is given by:
\begin{equation}
\eta = 1 - \frac{F_{shear}}{F_{noshear}}
\end{equation}
Shear in pixel vs. efficiency loss is presented in Fig.~\ref{fig:shear2}. Even for shears as much as 2 or 3 pixels, efficiency losses are less than 50\%, which makes highly sheared observations non-ideal, but still scientifically useful. One of the advantages of the NICI coronagraph compared to standard AO is the ability to have longer base exposure times because the coronagraph prevents saturation at the core of the target. In a situation where rotation would cause high shear, one can reduce base exposure time to reduce shear, but this negates the capability of the coronagraph to perform deep observations. Choosing a shorter base exposure time to reduce shear will also result in decreased depth of observation per image, an increase in telescope overhead, and additional read noise.
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.2in]{shear3_PSF3.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:shear2} Shear in pixel vs. sensitivity loss. Even up to 2 pixel shear leads to less than 50\% sensitivity loss.
Initial and sheared PSF shapes are shown; the sheared PSF is modeled
as a rectangle (width given by the total shear in pixels) with two semicircular caps.
}
\end{figure}
\subsection{On Sky Case Study}
\label{sec:rotonoffcasestudy}
Combining ADI with SDI theoretically allows for even more improvement in quasi-static speckle suppression compared to the SDI technique with discrete rotator angles. To quantify the relative performance of each technique, we compare results from two very similar datasets, one observed with ADI plus SDI, the other observed with SDI at discrete rotator angles.
On March 28, 2008, TWA7 was observed with NICI at Gemini South for 61 minutes prior to transit with the rotator off and 73 after transit with the rotator at two different discrete rotator angles (half at a rotator angle of 0$^{\circ}$, half at a rotator angle of 35$^{\circ}$). TWA7 (10 42 30.1 -33 40 16.2, J2000) is an M1 star, 8-10 Myr old, 55 $\pm$16 pc from Earth (assuming TW Hya membership \cite{ZS04}), with a background object at a separation of 3"\cite{Neu00}
(2.5" separation at discovery), and 9.4 magnitudes fainter than the primary in $H$. A base exposure time of 30 seconds was used for each dataset. A 4\% filter centered on 1.58 $\mu$m was used in NICI's blue channel, and a 4\% filter centered on 1.65 $\mu$m was used in NICI's red channel. TWA7 rotated on the sky by 75$^{\circ}$ during the rotator off ADI plus SDI observation. Thus, many images were available to construct an ideal reference PSF, but significant blurring was also observed in TWA7B (3" separation).
Each frame was flat-fielded and bad pixel corrected. The red frame from each red/blue frame image set was then scaled to the Airy pattern of the blue image, derotated, and aligned with its corresponding blue frame. A custom shift and subtract algorithm was used to align each red/blue frame image set on the speckle pattern. After precision alignment, the red frame was subtracted from the blue frame. For the SDI at discrete rotator angle dataset, all subtracted frames at each rotator angle were then median combined and the final 35 degree image was subtracted from the final 0 degree image.
For the ADI plus SDI dataset, a reference PSF was constructed for each subtracted image from the rest of the subtracted images in the sequence. The LOCI algorithm\cite{Laf07b,Art08} was used to construct the reference PSF. All frames which did not rotate on the sky by at least 1/2 FWHM at 1" were discarded when building the reference PSF. Final ADI plus SDI and SDI with fixed rotator angle images are presented in Fig.~\ref{fig:rotonoffimage}.
Contrast curves for each of these cases were generated by calculating the standard deviation within a 7 pixel wide annulus at each radial distance from the star. Magnitudes of contrast were scaled according to the unsaturated PSF of TWA7B in these images. Contrast curves are presented in Fig.~\ref{fig:rotonoffcontrast}. For this dataset, ADI plus SDI achieves an extra magnitude of contrast compared to SDI at discrete rotator angles. This is somewhat due to the fact that this is a particularly optimal ADI observation within 3", and a somewhat sub-optimal SDI observation.
Simulated objects were also added into the ADI plus SDI datasets. Three simulated objects were added at separations of .35", .7", and 1", with $\Delta$mag = 9.5, 10.5, and 11.5 mag, respectively. These simulated objects are scaled images of TWA7B. A very strong methane break was assumed, thus, these objects appear only the blue filter. The objects were all detected at the 10-20 $\sigma$ level.
It is important to note that, while the ADI plus SDI dataset was well optimized, the SDI at discrete rotator angles dataset is somewhat sub-optimal. While the base exposure time for the ADI plus SDI dataset had to be limited to 30 seconds to reduce image shear, a longer base exposure time would have been optimal for the SDI at discrete rotator angle dataset. This would have allowed for deeper imaging, as well as a reduction in read noise and telescope overhead. In addition, the NICI detector suffers somewhat from checkerboard pattern noise, which is well removed in the ADI plus SDI dataset. However, as opposed to other implentations of SDI, only one dither position was used in the SDI discrete rotator angle dataset. This means that greater pattern noise remains after data reduction, resulting in a higher noise floor compared to the ADI plus SDI dataset. Finally, it is possible to get 75$^{\circ}$ of sky rotation only in a very small part of the sky; less sky rotation means less well constructed reference PSFs in the inner arcsec of the image. In other word, the inner arcsec in this ADI plus SDI dataset looks especially clean compared to more typical conditions. The noise floor encountered in the ADI plus SDI dataset at separations greater than 1" is likely due to read noise -- this dataset was taken with 4 nondestructive reads (NDR). Boosting the number of nondestructive will help bring down this limiting read noise floor. Effort is currently being put into increasing the number of NDRs possible with NICI -- by the start of the NICI campaign, the NICI instrument will be able to be run with an arbitrary
number of NDRs, thus significantly reducing the operational read noise floor.
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.4in]{rotonoff_masked_annot.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:rotonoffimage} Left: Final ADI plus SDI reduced image, Right: Final SDI with rotator fixed image. The inner .3" (under the coronagraphic mask) has been masked out in these images. Images are 4.7''$\times$6''. The TWA7B background object is apparent at a separation of 3" in these images. Each frame was flat-fielded and bad pixel corrected. The red frame from each red/blue frame image set was then scaled to the Airy pattern of the blue image, derotated, and aligned with its corresponding blue frame. A custom shift and subtract algorithm was used to align each red/blue frame image set on the speckle pattern. After precision alignment, the red frame was subtracted from the blue frame. For the SDI at discrete rotator angle dataset, all subtracted frames at each rotator angle were then median combined and the final 35 degree image was subtracted from the final 0 degree image. For the ADI plus SDI dataset, a reference PSF was constructed for each subtracted image from the rest of the subtracted images in the sequence. The LOCI algorithm\cite{Laf07b,Art08} was used to construct the reference PSF. All frames which did not rotate on the sky by at least 1/2 FWHM at 1" were discarded when building the reference PSF. Simulated objects were also added into the ADI plus SDI datasets. Three simulated objects were added at separations of .35", .7", and 1", with $\Delta$mag = 9.5, 10.5, and 11.5 mag, respectively. Note that the SDI with rotator fixed final image clearly has a higher noise floor than the ADI plus SDI final image; this is at least partly due to the fact that only one dither position was used with the SDI rotator fixed data.
}
\end{figure}
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.4in]{contrast_rotonoff3.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:rotonoffcontrast} $H$ band contrast curves for the ADI plus SDI and SDI at discrete rotator angles data. Contrast curves for each of these cases were generated by calculating the standard deviation within a 7 pixel wide annulus at each radial distance from the star. Magnitudes of contrast were scaled according to the unsaturated PSF of TWA7B in these images. For this dataset, ADI plus SDI achieves an extra magnitude of contrast compared to SDI at discrete rotator angles. This is somewhat due to the fact that this is a particularly optimal ADI observation within 3", and a somewhat sub-optimal SDI observation. The three simulated objects are overplotted here as diamond points.
}
\end{figure}
\section{$H$ band vs. $K$ band performance}
\label{sec:HK}
Previous near-infrared direct detection searches for methane-rich planets have favored the $H$ band methane break over the $K$ band\cite{Biller07, Laf07a}. However, $K$ band also can provide a valuable search space, as well as follow-up capabilities. In this section, we investigate whether $H$ or $K$ provide more fertile ground for planet searches (e.g. inherent contrasts and methane break strengths in each band), as well as compare $H$ and $K$ band NICI performance.
\subsection{$H$ vs. $K$ necessary contrasts}
\label{sec:HKcontrast}
We used recent young planet models\cite{For08} to constrain the inherent expected luminosity of young planets, and thus the contrasts necessary to detect them. Planets around main sequence stars vs. planetary mass objects around very low mass stars and
brown dwarfs likely have
disparate formation mechanisms; the former likely form via core accretion or disk instability in a circumstellar disk, while the latter form (most likely) in a manner analogous to stars. Thus, different initial
conditions must be taken into account when modeling each population. Unfortunately, until
recently, all available evolutionary models for planetary mass objects used "hot start" initial conditions --
starting from a very high initial entropy state. This choice is likely appropriate for brown dwarfs, but
not for true planets forming around a star (which require a lower entropy initial condition.) Recently, Fortney et al. (accepted)\cite{For08} have modeled spectra for young planets using these more realistic
initial conditions.
In order to evaluate $H$ vs. $K$ band performance in advance of the NICI science campaign,
we calculated the necessary $\Delta$mag to image planets of varying masses in the $H$ and $K$ bands. Star-planet contrasts as a function of age for 1-10 M$_{Jup}$ planet models (core accretion on left, hot start on right, models from Fortney et al. accepted\cite{For08}) around main sequence K2 and M0 stars are plotted in Figs.~\ref{fig:K2} and ~\ref{fig:M0}. (We used the simplifying assumption of main sequence primaries -- in reality, especially for the M0 primary, this assumption is probably not quite true.) In general, much higher contrasts ($\Delta$mag=12-16 vs. $\Delta$mag=8-12) are required to
image core accretion planets vs. hot start objects. For objects of these ages and masses, $K$ band
provides a more favorable $\Delta$mag than $H$. In addition, these model planets are not especially overluminous in $K$ at very young ages. Previous surveys have operated on the assumption that young objects are especially bright compared to older ($>$300 Myr) objects. These new models suggest this brightness bump at early ages is considerably less than expected. Thus, upcoming direct planet surveys perhaps do not
need to give very young objects ultra-high priorities. Additionally, a viable strategy in light of these faint models would be to go very deep on fewer objects -- according to these predictions, a shallow survey is not likely to produce
any detections or meaningful limits on planet distribution.
\begin{figure}
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=3in]{deltamag_K2.eps} &
\includegraphics[width=3in]{deltamag_K2_hotstart.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:K2} Required Contrasts to image planets of a range of masses for a K2 main sequence primary. Left: Fortney et al. 2008 core accretion models. Right: Fortney et al. 2008 hot start models.}
\end{figure}
\subsection{$H$ vs. $K$ methane break strength}
\label{sec:HKbreak}
Prominent methane absorption features appear both in the $H$ band
(1.6 $\mu$m) and $K$ band (2.12 $\mu$m) spectra of T dwarfs
(T$_{eff}$ $<$ 1200 K). Representative
L and T dwarf spectra\cite{Cus05}
in both the $H$ and $K$ band are presented in Fig.~\ref{fig:spectra}.
While previous
methane imagers such as the SDI imager at the VLT and MMT have utilized
the $H$ band methane absorption feature, it is important to evaluate
the strength of both the $H$ and $K$ band features to determine which feature
best optimizes the performance of the NICI methane imager.
To determine the relative strengths of the $H$ band and $K$ band methane
absorption features we define a number of methane spectral indices of the form:
\begin{equation}
\frac{Flux_{unabsorbed}}{Flux_{absorbed}} =
\frac{\int^{\lambda_2}_{\lambda_1} S_{\lambda} F({\lambda}) d\lambda}
{\int^{\lambda_4}_{\lambda_3} S_{\lambda} F({\lambda}) d\lambda}
\end{equation}
where S is the object flux and F is the filter transmission at wavelength
$\lambda$. Spectral indices were calculated from
spectra of 56 L dwarfs and 35 T dwarfs\cite{Kna04}.
Spectra for these objects were obtained from
Sandy Leggett's L and T dwarf archive\footnote[2]{http://www.jach.hawaii.edu/$\sim$skl/LTdata.html}. Spectral index as a function of spectral type
is presented in Fig.~\ref{fig:filters}. (Although synthetic model spectra are available in Fortney et al. 2008, we did not calculate spectral indices for these model spectra, since methane opacities for the $H$ and $K$ bandheads are not well known. The dwarf L and T spectra used for this analysis are not exact analogues to young planets, but should have similar methane break strengths.) The standard deviation per spectral
type for these spectral indices range from $\sim$0.05 for late L's to
$\sim$1.0 for mid to late T's. In other words, the standard deviation of these
indices are fairly small and indices can be used to derive an accurate spectral
type for T dwarfs.
\begin{figure}
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=3in]{deltamag_M0.eps} &
\includegraphics[width=3in]{deltamag_M0_hotstart.eps} \\
\end{tabular}
\end{center}
\caption{\label{fig:M0} Required Contrasts to image planets of a range of masses for a M0 main sequence primary. Left: Fortney et al. 2008 core accretion models. Right: Fortney et al. 2008 hot start models.}
\end{figure}
Two $H$ band indices were calculated. The first of these,
$H$(1.5625 $\mu$m - 1.5875 $\mu$m)/(1.6125 $\mu$m-1.6375 $\mu$m),
is the index calculated from the filters currently implemented in the
VLT and MMT SDI imager. SDI filter transmission curves were convolved into these calculations. The second $H$ band index,
$H$(1.56 $\mu$m - 1.60 $\mu$m)/(1.635 $\mu$m-1.675 $\mu$m), is the methane
spectral index defined by Geballe et al. 2002\cite{Geb02}. In this case,
a tophat function was used for the filter transmission. One $K$ band index was also calculated -- K(2.09 $\mu$m - 2.15 $\mu$m)/
(2.19 $\mu$m - 2.25 $\mu$m).
A tophat function was used for the filter transmission. For mid to late T dwarfs, the $K$ band
methane break is considerably stronger than the $H$ band break -- thus, future direct detection planet finding surveys should
strongly consider incorporating $K$ band methane break filters.
These indices only take into account the spectrum of a cool exoplanet
companion. In the real differential imaging case, we must also take into
account the spectrum of the parent star. In general, the star's spectrum
will appear ``blue'' to any IR methane imager -- at these wavelengths,
the IR starlight is from the Rayleigh-Jeans tail of its blackbody spectrum.
This blue appearance will be slightly stronger in the
K band compared to the $H$ band, but in general, the blue color of the star
will be small for either band ($<$0.2 mag).
\begin{figure}
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=2in]{Cushing11_SDI.eps} &
\includegraphics[width=2in]{Cushing11_Geballe.eps} \\
\includegraphics[width=2in]{Cushing13_K.eps} & \\
\end{tabular}
\end{center}
\caption[] { \label{fig:spectra} $H$ band (top) and $K$ band (bottom)
spectra for a variety of L and T dwarfs. Filter bandpasses are overlaid
for all three filter systems used to calculate spectral indices.
Methane absorption features
appear strongly in both the $H$ and $K$ band spectra.
Spectra from Cushing et al. 2005\cite{Cus05}.
}
\end{figure}
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.2in]{ind_comp.eps}
\end{tabular}
\end{center}
\caption[] { \label{fig:filters} Methane spectral indices for 2 $H$ band
and 1 $K$ band filter systems.
We plot numerical spectral types on the x-axis; a numerical
type of 8 corresponds to a L8 spectral type, a numerical type of 16
corresponds to a T6 spectral type, etc. Spectral indices were calculated
from spectra of 56 L dwarfs and 35 T dwarfs\cite{Kna04}.
Note that the $K$ band Methane break is
stronger than the $H$ band Methane break for comparable filter systems.
}
\end{figure}
\subsection{$H$ band vs. $K$ band on sky performance}
To quantify the achieved contrast in $H$ vs. $K$ band, we compare results from two similar datasets, one taken in $H$ band, one in $K$. We observed HD 129642 for 15 minutes in the $K$ band (using Brackett $\gamma$ in one channel and $K$ continuum in the other) and 30 minutes in the $H$ band (using a 1\% filter centered on 1.58 $\mu$m in NICI's blue channel, and a 1\% filter centered on 1.65 $\mu$m in NICI's red channel). HD 129642 is a K2 star 28.6 pc distant surrounded by 5 companion candidates, one of which is likely bound\cite{Egg07}. A base exposure time of 30 seconds was used. The data were reduced according to procedures described in Section 2.2, and also in Artigau et al.\cite{Art08}, this volume. Contrast curves were generated using the standard deviation in annular regions 7 pixels in diameter and scaling magnitudes according to the known companions in the field. $H$ and $K$ contrast curves are presented in Fig.~\ref{fig:HKcontrast}. Inside of 1", the $H$ band provides up to .4 mag better contrast than K (as expected because of the deeper exposure). Otherwise contrast in these two bands appears to be roughly similar and limited by read noise beyond 1'' (8 NDRs were used with this dataset; boosting the number of NDRs will significantly decrease the read noise floor). A more in depth comparison of $H$ vs. $K$ performance is planned for upcoming commissioning runs.
\begin{figure}
\begin{center}
\begin{tabular}{c}
\includegraphics[width=3.0in]{HKcontrast.eps}
\end{tabular}
\end{center}
\caption[] { \label{fig:HKcontrast} Contrast curves for $H$ and $K$ bands. Contrast curves for each of these cases were generated by calculating the standard deviation within a 7 pixel wide annulus at each radial distance from the star. Magnitudes of contrast were scaled according to the unsaturated PSFs of companion candidates\cite{Egg07}. Inside of 1", the $H$ band provides up to .4 mag better contrast than K, but otherwise contrast in these two bands appears to be similar.
}
\end{figure}
\section{Conclusions}
Careful scheduling is necessary to use ADI with SDI, since both shear per image and total rotation on the sky need to be taken into account. There will always be parts of the sky where ADI is completely unfeasible. In areas of the sky with enough target rotation but not too high a rate of rotation, ADI plus SDI does seem to provide considerable enhancements in speckle suppression over SDI at discrete rotation angles, although part of the difference for this particular comparison dataset may have resulted from the fact that the SDI dataset at discrete rotation angles was non-optimal.
New, more realistic "cold start" models for young planets less than 300 Myr in age\cite{For08} suggest planets are fainter than originally predicted; thus, somewhat higher contrasts will be necessary to detect them (i.e., 12-16 mag vs. 8-12). Both theoretically necessary contrasts and predicted methane break strengths appear to be optimized in $K$ band relative to $H$ band. However, $H$ band contrast performance appears to be somewhat better than $K$ band inside 1". Additionally, it is important to consider total NICI instrument throughput in $H$ and $K$ bands when choosing between these, which we have not discussed in this document.
\acknowledgments
Support for this work was provided by NASA through Hubble Fellowship grant HF-1204.01-A awarded by the Space Telescope Science Institute, which is operated by the Association of Universities for Research in Astronomy, Inc., for NASA, under contract NAS 5-26555.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,763
|
#ifndef _SYS__CALLOUT_H
#define _SYS__CALLOUT_H
#include <sys/queue.h>
struct lock_object;
LIST_HEAD(callout_list, callout);
SLIST_HEAD(callout_slist, callout);
TAILQ_HEAD(callout_tailq, callout);
struct callout {
union {
LIST_ENTRY(callout) le;
SLIST_ENTRY(callout) sle;
TAILQ_ENTRY(callout) tqe;
} c_links;
sbintime_t c_time; /* ticks to the event */
sbintime_t c_precision; /* delta allowed wrt opt */
void *c_arg; /* function argument */
void (*c_func)(void *); /* function to call */
struct lock_object *c_lock; /* lock to handle */
short c_flags; /* User State */
short c_iflags; /* Internal State */
volatile int c_cpu; /* CPU we're scheduled on */
};
#endif
|
{
"redpajama_set_name": "RedPajamaGithub"
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{"url":"https:\/\/www.physicsforums.com\/threads\/do-electrons-have-mass.12863\/","text":"# Do Electrons have MASS?\n\n1. Jan 19, 2004\n\n### timejim\n\nWhirling around Atoms are the Electrons. I was wondering if they have MASS? If so, it would seem there tremendous rotational velocity would cause them to fly away from the nucleus but they don't, so I kinda think maybe they do not have mass, only some form of energy. Can you help answer this for me?\n\n2. Jan 19, 2004\n\n### suyver\n\nYes, electrons have mass. A single electron weighs about 9.109534 x 10$^{-31}$ kg.\n\nThe reason they don't 'fly away': quantum mechanics.\n\n3. Jan 20, 2004\n\n### ZapperZ\n\nStaff Emeritus\nThere is a very simple experiment often done in college level intro physics labs that measure the e\/m ratio of electrons. Here, the path of electrons coming off a hot cathode are bent by an external uniform magnetic field (typically generated by a helmholtz coil). As one increases the potential between the anode and the cathode (thus imparting more kinetic energy to the electrons), one sees that the path of the electrons are bent less and less. An object with a mass that is travelling faster will require a greater force to bend its path by the same amount, or given a constant bending force (as in this experiment), an object with mass that is moving faster will be bent less. This essentially is the same principle applied in a mass spectrometer!\n\nThis experiment would have been useless, or give weird results, if electrons do not have a mass.\n\nZz.\n\n4. Jan 20, 2004\n\n### GCT\n\nYes electrons have mass. Most chemical phenomena are based on this fact. For your latter question, remember that E=mc^2.\n\n5. Jan 20, 2004\n\n### FZ+\n\nThe great evil of the Bohr model strikes again!\n\nElectrons do not whirl around atoms. (or nuclei, as I think you meant to say) They exist as an uncertain \"electron cloud\".\n\nIn general, massless particles travel at c. The electron does not travel at c, so it has rest mass.\n\n6. Jan 23, 2004\n\n### nautica\n\nYou are correct about the \"uncertainty principle\" in that we can not be certain where an electron will be at any particular cloud, but I believe that it is still up for debate as to wether it is a cloud or a cloud due to the whirling of the electrons around the nucleus.\n\nI guess this would depend on wether the electon was considered to be a particle or a wave????\n\nNautica\n\n7. Jan 23, 2004\n\nStaff Emeritus\nEven if the electron is considered a particle, it does not have a well defined position, trajectory, or momentum, due to uncertainty. So the whirling concept brings in false assumptions - at least to my mind.\n\n8. Jan 23, 2004\n\n### nautica\n\nI will agree with that. So are you saying the \"cloud\" has mass and there is not actually a point mass.\n\nNautica\n\n9. Jan 23, 2004\n\nThe electron is a point mass, but when it's in a \"cloud\" it's uncertain where this mass is, and the more you know where it is the less you know when it was there. Why this is important is because it's not just that our instruments suck it\u2019s that the information isn't even there to \"know.\"\n\nLast edited: Jan 23, 2004\n10. Feb 6, 2004\n\n### GRQC\n\nThe \"clouds\", or orbitals, are solutions to Schrodinger's equation (not directly an uncertainty principle issue). We're quite certain they (orbitals) exist, and that they represent a probability density of finding the electron. The notion of a \"chunk\" of matter moving around is classical and fall apart very quickly in the quantum regime.\n\nIf you consider the shape of p or d orbitals, there's no way the electron can be \"whirling\". Furthermore (to kill the whirling idea altogether), ask yourself what happens if a charge accelerates.\n\n11. Feb 6, 2004\n\n### Nereid\n\nStaff Emeritus\norbitals - quintessential QM\n\nVery good GRQC, we should use the concept of orbitals - as in chemical bonds etc - more often in explanations of QM, the nature of the electron, etc.\n\nLast edited: Feb 7, 2004\n12. Mar 11, 2007\n\n### raman\n\nIt's all clear today about the electrondynamics. however i can not stop to think that how did physisits or chemists measure the mass of an electron. and if this was a measurement what was the reference and how much error is acounted for?\n\n13. Mar 11, 2007\n\n### Staff: Mentor\n\nLast edited: Mar 11, 2007\n14. Mar 11, 2007\n\n### Staff: Mentor\n\nI think you left out a few words there:\n\n\"The electron's charge to mass ratio was known (Thomson)...\"\n\n15. Mar 11, 2007\n\n### Staff: Mentor\n\nYeah - I left out some words while changing the order of the sentence.\n\nThank jt - I added the crucial words.\n\n16. Apr 3, 2007\n\n### neu\n\nThis is true, but how would you describe the \"motion\" of a free electron? It would be fine to describe it as a particle with velocity as well as a wave which is a solution to the schrodinger eq with no potential\n\n17. Apr 3, 2007\n\n### Manchot\n\nEven if they behaved classically, they still wouldn't \"fly away,\" as long as they had a low enough energy. After all, planets most certainly have mass and also orbit the sun.\n\n18. Apr 11, 2007\n\n### aku2590\n\nCathode Rays\n\nHey...i really dont get this part of dischage tubes....a potential difference is applied and the 2 electrodes are cathode and anode...now if electrons are produced and emitted at the cathode...shouldnt they always hit the anode...since they would need to hit the anode to complete the circuit?? :S Cuz for cathode rays in TV's they go through the anode...and i dont understand how this works :S....??? Any ideas anyone please??\n\n19. Apr 11, 2007\n\n### christianjb\n\nAn electron in any orbital isn't going anywhere, because an orbital is a stationary state solution to the S.E.\n\nHowever, the probability density does 'swirl' in practice because the wavefunction is a linear combination over all orbitals. That allows for interference between the phases of differing orbitals creating time dependent regions of low and high probability.\n\nThe 'swirls' are not much like planetary orbits, but there is movement.\n\n------------------\n\nThis applet should convince you that particle wavefunctions do swirl in a potential well.\n\nEdit: This applet is even better- it's for the H-atom","date":"2016-09-30 01:37:06","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.48350557684898376, \"perplexity\": 1287.2080765048242}, \"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-2016-40\/segments\/1474738661974.78\/warc\/CC-MAIN-20160924173741-00166-ip-10-143-35-109.ec2.internal.warc.gz\"}"}
| null | null |
{"url":"https:\/\/motls.blogspot.com\/2013\/10\/2013-nobel-prize-in-physics-englert.html?showComment=1381299459015&m=1","text":"## Tuesday, October 08, 2013\n\n### 2013 Nobel prize in physics: Englert, Higgs\n\nCongratulations to Fran\u00e7ois Englert (left) and Peter Higgs (right)!\n\nToday's Nobel prize in physics has been the most anticipated Nobel prize in any discipline for decades, especially because of the July 4th, 2012 discovery of the God particle, the most important finding in experimental particle physics in at least 30 years.\n\nWell, I am sure it's fair to say that the Higgs boson discovery was more novel, original, and groundbreaking than the discovery of the top quark in the mid 1990s. Some people might argue that the Higgs boson discovery dwarfs the discovery of the massive gauge bosons in the early 1980s, too. Another way to see how the discovery is important: The word \"Higgs\" appears in 469 articles on this blog \u2013 and it is not primarily a Higgs blog in any sense. ;-)\n\nAt any rate, people have known about a not-yet-rewarded breakthrough that visibly surpassed all other breakthroughs, a breakthrough that connects the stories of the heroes of theory and the heroes of experiments in the traditional way which is why it made sense that the 2013 Nobel prize in physics could be linked to this development.\n\nThe Belgian and British physicists finally won the prize:\n\u201cfor the theoretical discovery of a mechanism that contributes to our understanding of the origin of mass of subatomic particles, and which recently was confirmed through the discovery of the predicted fundamental particle, by the ATLAS and CMS experiments at CERN\u2019s Large Hadron Collider.\u201d\nSee the advanced information about the prize on the Nobel Prize server (PDF) or a popular presentation (PDF).\n\nPeter Higgs was guaranteed to be rewarded \u2013 he was the only original discoverer who realized and emphasized the existence of the extra particle, one that is now known to have the mass $$126\\GeV\/c^2$$. But the history of the Higgs mechanism and related ideas is somewhat complicated so the precise choice of the winners was seen as a controversial topic by some \u2013 but the story of the Higgs mechanism isn't quite an exception. In particular, Robert Brout (died 2011) and Fran\u00e7ois Englert wrote a paper about the Higgs mechanism (spontaneous gauge symmetry breaking) before Peter Higgs did (and he did it independently). See some extra words about the history of the Higgs mechanism.\n\nAt 11:40 am, I took the following screenshot of our Higgs Nobel poll because I needed 20 seconds to send the results to Carl XVI Gustaf, the monarch needed 3 minutes to run to the other side of his castle and call the Nobel prize committee, and some additional seconds were needed to finalize the verdict. In total, 5 minutes were needed.\n\nThe original announcement was scheduled for 11:45 am or later and 60 minutes were indeed added because Carl XVI Gustaf's dog ate the homework with the list of winners at 11:44 am and it had to be resent.\n\nThe 11:40 am screenshot looked like this:\n\nOut of the 209 votes collected among some of the TRF readers, the following contributors to the Higgs et al. mechanism were supported:\n1. Peter Higgs: 43%\n2. Fran\u00e7ois Englert: 15%\n3. Jeffrey Goldstone: 12%\n4. Philip Anderson (1977): 8%\n5. Rolf Heuer (CERN): 6%\n6. Tom Kibble: 5%\nThe remaining candidates, Guralnik, Weinberg, Gianotti, Nambu, Incandela, and Hagen received 7,5,4,4,2,1 votes, respectively. Extra 1+1+1 votes for Higgs, Nambu, and Goldstone between 11:45 and 12:45 didn't change anything important. At most 3 winners may be picked according to the dynamite inventor's rules. While I agree with Sean Carroll that the limit \"3\" is arbitrary and artificial, I can't overlook that he misses a more important point, namely that some rules have to exist that prevent inflation of the Nobel prize winners and that preserve their relatively exclusive status (well, relatively, because I have personally talked to 28 Nobel prize winners so far; one winner whom I talked to 10+ times, a very modest man, died 3 weeks ago, RIP David).\nCERN live: See CERN Higgs Nobel webcast where CERN bosses comment on the prize \u2013 unless you are late\nFor that reason, prizes to large experimental teams sound undesirable to me. Prizes for their spokespersons are problematic as well because these spokespeople are largely political animals and a scientific Nobel prize shouldn't degenerate into a political one, into an inferior award similar to the Nobel Peace Prize or Nobel Literature Prize (the Economics Memorial Nobel Prize is an intermediate case). In my eyes, the Nobel prize in physics (and chemistry and medicine) are rare examples of prizes that haven't become ridiculous tools of politics or mediocrity and I believe that the committees behind these awards are expected to preserve this special status of the scientific Nobel prizes.\n\nAt any rate, the poll suggested Higgs, Englert, and perhaps Goldstone \u2013 and this, either with or without Goldstone, happened to be my choice I've been talking about for a year (see my public comment posted yesterday at viXra including Goldstone). Fran\u00e7ois Englert has been my main \"horse\" whom I was promoting as a winner (not only because of his extensive and interesting work on supersymmetry, supergravity, strings, and branes); click at the link. During the announcement, Englert was called from Stockholm (they couldn't reach Higgs but he warned about his Tuesday invisibility in advance) and asked what are the remaining open issues once the Standard Model is completed. His answer was 1) whether there is broken SUSY at the LHC scale, 2) dark matter, 3) quantum gravity. Our man. ;-) He said it \"wasn't very unpleasant\" to have won the prize.\n\nSo I agreed with the TRF readers which was quite a reason for Carl XVI Gustaf and pals to make the same choice (well, without Jeffrey Goldstone, sorry), after the extra hour they spent by fighting against the contrarians.\n\nCongratulations to the winners for their well-deserved prize \u2013 if they care about the prize. ;-)\n\nIf you're asking how it's possible that this long text was already published at 12:45 pm, well, I admit that it had been ready for hours before. I had two versions of this text, almost identical ones, that only differ by having Goldstone among the winners. ;-) (Dennis Overbye needed just an hour for a story.) But I was sure that the experimenters wouldn't be included etc. It's not because I don't admire or value the work of the experimenters but because their victory has been the result of heavily collective efforts and the Nobel prize was simply not created to reward large groups. (Joe Incandela, the CMS boss, wisely said after the announcement that for them, the prize is the discovery.)\n\n1. Congratulations to the winners! The prediction and discovery of the Higgs Boson is the culmination of many years of painstaking effort and ingenuity. But there is one thing I have not yet understood (maybe because I missed something in your posts): Were the winners selected by the poll on the TRF? Does this mean that Lubos Motl was actually involved in the decision making?\n\n2. Dear TheDOC,\n\nagreed! As you may perhaps guess, I am bound by commitments of confidentiality in the case that the answer is Yes.\n\nBest wishes\nLubos\n\n3. If you had said 'yes', I would have said, \"Wow! The presence of people like you on the committee has reinstated my belief in the Nobel Physics prize. I can now brag to the world that I know people who decided the Nobel Prize!!! That is soooo awesome!!!!\"\n\nOf course, this is purely a hypothetical scenario since you have kept completely silent on the matter. :D\n\n4. Exactly, you may at most silently wow.\n\nI tried not to deny any such rumors because no denials could erase doubts, anyway. ;-)\n\n5. The best Android Volume Booster https:\/\/play.google.com\/store\/search?q=best%20volume%20booster%20162&c=apps\n\n6. Experimental high energy physics has reached the point where a Nobel, which is supposed to reward the physicist for something extraordinary and also for his (her) general contributions to physics, cannot be given. 3000 +3000 physicists ! only the spokesmen (women)would be grossly unfair. And CERN, as represented by its director even more so .\n\nMaybe, to encourage collective work in science a benevolent billionaire could setup a prize to be awarded to groups which discover something extraordinary. The group would get the prestige and the students would be invaluable in experimental work ( they do shifts :) ).\n\n7. Something that is always so tantalising about this blog is how you can never quite be sure if Lubos is being satirical.\nHe is always so plausible, yet the harder you want to believe the more that little doubt nags away, that all may not be quite as it seems. The uncertainty seems quite inescapable.\n\n8. Perhaps but I said elsewhere, one may also justify the omission of CERN differently: judge them in the same league as the theorists. CERN did the same discovery as the fathers of the mechanism but 48 years later.\n\nThey should get a prize for something they find first, at least something that good theorists are not virtually certain about in advance.\n\n9. My proposal is because the experimental work is not in the same grouping for a Nobel: independent work etc.\n\nNevertheless, if experimental physicists are not encouraged to gather together and try to validate predictions then theory becomes a mathematical game.\n\nIn some sense future generations may look at this as one experiment, its proposal by theorists (~7 of them independently?) 48 years ago and its final validation by 6000+ experimentalists in 2012..\n\n10. It seems to me that Nobel prizes for original scientific ideas that result in major discoveries. These ideas can be mathematical or experimental but they need to be original. Nobel prizes are not given for carrying out, however skilfully, work that could equally well be carried out by other well trained persons. I find it hard to believe that there were 6000 significant original ideas involved in the experimental work at CERN that resulted in the discovery of the Higgs boson.\n\n11. This year's Nobel prizes provide even greater evidence of the Zionist conspiracy than usual.\n\nTwo of the Nobel prizes winners in medicine (Rothman and Schekman) are Jewish\n\nhttp:\/\/www.timesofisrael.com\/two-americans-german-win-nobel-medicine-prize\/\n\nTheir main competitors, who did not get them this time are even worse: they are actually Israelis.\n\nOf the two physics laureates, Englert is not only Jewish but is also a professor at Tel Aviv University:\n\nhttp:\/\/www.jpost.com\/International\/Tel-Aviv-U-affiliated-prof-and-Holocaust-survivor-shares-Nobel-for-physics-328171\n\nAnd finally Chemistry. All three winners Martin Karplus, Michael Levitt, Arieh Warshel are Jewish and the last two are Israeli citizens:\n\nhttp:\/\/www.jpost.com\/Jewish-World\/Jewish-Features\/Israeli-scientists-awarded-Nobel-Prize-in-Chemistry-328246\n\nWhat happened to the goys? Can't they do science any more. I am reminded of the following passage from Stanislaw Ulam's \"Adventures of a Mathematician\" (Johnny is John von Neumann):\n\nI told Banach about an expression Johnny had once used in conversation with me in Princeton before stating some non-\u00adJewish mathematician's result, \"Die Goim haben den folgenden Satzbewiesen\" (The goys have proved the following theorem). Banach, who was pure goy, thought it was one of the funniest sayings he had ever heard. He was enchanted by its implication that if the goys could do it, Johnny and I ought to be able to do it better. Johnny did not invent this joke, but he liked it and we started using it.\n\n12. That is why I am proposing that in order to encourage young physicists to do experimental high energy physics, or astrophysics or many of the disciplines that require group work of many people, there should be a different prize, recognizing the dedication and good work performed by the group as a whole.\n\nThe way it is going young physicists will be choosing subjects that would allow them to shine by themselves. That would be the end of experimental high energy physics ( at least until robotics advances enough to replace 6000 physicists) and string theory will never be validated and will remain a mathematical proposition.\n\nLet us take one experiment which has 3000 physicists signing the Higgs paper. Maybe half of them are graduate students whose contribution to the Higgs paper has been shifts upon shifts of data taking. The original bit in their thesis will be the study of a paraicular channel for a particular interaction and there the can show their ingenuity. Or if their talents are computational the design of a specific trigger for a specific interaction etc. None of these are Nobel level original researches, as is true for the grand majority of thesis written in physics and not only. One adds a small stone on the edifice being validated.\n\nThere are some \"original\" experimental proposals : search for axions, or search for monopoles etc. If found then the experimentalist who fueled the interest on the subject to the rest of the group will be eligible, as was Rubbia with the W and Z when they were a gleam in theorists eyes and he moved the earth to get the collider going so they could find them.\n\nThe present collider was a collaborative effort of the HEP community, no strong personalities behind the decision but careful weighted arguments for a discovery machine.\n\nI am just making the observation that as time goes on it will be hard to find brilliant graduate students and post docs to enter the HEP experimental field, and then theories will become just mathematical guesses.\n\n13. I quite agree although I don't the necessity of calling it \"the Nobel Prize\". There are quite many other very respectable prizes and it would certainly be a good idea to have one especially for groups of experimentalists.\n\nOn the other hand, I am getting a little fed up of people referring to \"mathematical games\". There are indeed \"mathematical games\" and sometimes they involve quite beautiful mathematics (to those who believe that beautiful mathematics must derive from physics I recommend J.H. Conway's \"On numbers and games\" or several other books by the same author) but most mathematics is not a \"game\".\n\nThose who this in this way of mathematics (and I include among them) remind me of Newton's letter to Halley (of the comet):\n\n\"Mathematicians, that find out, settle and do all the business, must content themselves with being nothing but dry calculators and drudges, and another that does nothing but pretend and grasp at all the things must carry away all the inventions as well of those that were to follow him as of those that went before\u2026\u201d\n\n14. I want to add here a second way how an experimental high energy physicist might become eligible for the Noble as is, at this day and age.\n\nBy pursuing the same analysis over many experiments after having discovered an anomaly in one of experiments he\/sh took part and thus establishing an unexplained\/unexpected up to then effect.\n\nFor example there exists the soft photon anomaly http:\/\/indico.cern.ch\/getFile.py\/access?resId=1&materialId=slides&confId=93025 which has not been satisfactorily settled as far as I know, within the known theories. Suppose , and this is just for an example, it turns out that it is a clear signal of the string structure of the fundamental particles. Then the experimentalists that patiently pursued and established the signal over many experiments would be eligible, in my opinion.\n\n15. Right but so far it looks too vague for a \"discovery\", right? What could it be? Why haven't we heard about it later?\n\n16. \"The Journey IS the Reward\"\n\n-- Chinese Proverb\n\n\"The Journey is better than the End\"\n\n-- Cervantes\n\nThe above quote is used by John Wooden (legendary UCLA basketball coach aka \"Wizard of Westwood\") in his <a href=\"http:\/\/www.ted.com\/talks\/john_wooden_on_the_difference_between_winning_and_success.html>TED talk<\/a> (\"Difference between Success & Winning\")\n\n17. Professor Higgs stole the prize from the author of the theory of Superunification\nhttp:\/\/leonov-higgsnot.blogspot.ru\/2013\/10\/professor-higgs-stole-prize-from-author.html\nhttp:\/\/decelerator.blogspot.ru\/2012\/07\/well-be-at-higgs-seminar-in-melbourne.html?showComment=1381499379623#c1353513545228568291\nThe nature of the formation of mass was first discovered by Russian scientist Vladimir Leonov in 1996, and not by Professor Higgs:\nLeonov V. S. Quantum Energetics. Volume 1. Theory of Superunification. Cambridge International Science Publishing, 2010, 745 pages.\nhttp:\/\/www.cisp-publishing.com\/acatalog\/info_54.html\nChapter 3. Unification of electromagnetism and gravitation (pages 167-261)\n3.5.1. Formation of mass (pages 218-219)\nTo understand the nature of gravitation and gravity it is necessary to understand the nature of the mass of the particle (solid). In the classic theory of gravitation, the mass of the particle (solid) is used as a measure of gravity and inertia. Einstein added that the mass is the measure of curvature of space-time. Now, the theory of Superunification shows that the spherical deformation of quantised space-time is the measure of mass. Thus, this shows that the mass is a non-independent secondary formation in the quantised space-time, does not represent an isolated system (thingin-itself) and is an open quantum mechanics systems, linked permanently with the quantised medium as its bunch of the energy of spherical deformation of the medium. In fact, the classic mass typical of physics dissolved in the quantised space-time as the measure of matter which in the region of the microworld of the elementary particles simply does not actually exist. In reality, there is only the spherically deformed local region of the quantised space-time whose deformation energy (3.56) determines the particle mass. Therefore, the movement of the particle with the mass in the superelastic quantised medium is the wave transfer of the energy of spherical deformation of the medium governed by the effect of the principle of corpuscular-wave dualism. The Superunification theory makes it possible to derive equations describing the mass m by the vector of spherical deformation D (3.43) of the quantised space-time. The Gauss theory determines unambiguously the mass by the flow of deformation vector (3.43) penetrating through the closed surface S around the particle [12]:\nEquation of the mass (3.85)\nEquation (3.85) treats the mass of the particle (solid) as the parameter of spherical deformation of quantised space-time. Remove the spherical deformation from the quantised medium and the mass disappears. This is observed in annihilation of the positron and the electron when the energy W of spherical deformation of the particles is released and transfers to the electromagnetic energy of radiation of gamma quanta [13]:\n(3.87)\nEquation (3.87) determines the equivalence of the mass and energy of deformation of the quantised space-time.\nThe theory of Superunification has experimental support:\n3.5.3. Simple quantum mechanics effects (pages 224-227)\n10.9.1. Results of the tests of a quantum engine for generating thrust without the ejection of reactive mass (pages 685-689)\nhttp:\/\/theoryofsuperunification-leonov.blogspot.ru\/2011_05_01_archive.html\nVideo: The tests 2009 of a quantum pulsed engine for generating thrust without the ejection of reactive mass. http:\/\/theoryofsuperunification-leonov.blogspot.ru\/2011\/07\/video-tests-2009-of-quantum-pulsed.html\n\nThe Higgs boson has no experimental verification. This is a major scientific fraud, deception and falsification.\nLeonov. Higgs boson does not exist in nature:\nhttp:\/\/www.scientificamerican.com\/article.cfm?id=higgs-cern-lhc-discovery\nhttp:\/\/www.theguardian.com\/science\/blog\/2012\/jul\/04\/higgs-boson-universe-peter-higgs\nhttp:\/\/leonov-higgsnot.blogspot.ru\/2012\/07\/higgs-boson-does-not-exist-in-nature.html\n\n18. It seems to me that this forum has already exceeded its quota of unrecognised geniuses, so I suggest taking your fascinating theory elsewhere.\n(There is a blog which is, I think, just perfect for this sort of stuff. Search the Internet for \"Not even wrong\" .)\n\n19. Or may use the viXra ;)\n\n20. If I\u2019m not mistaken Higgs is the only physicist name that\nhas been verbalised?\n\nE.g. you often hear the term \u2018Higgsing a theory\u2019\u2026","date":"2021-10-24 04:03:33","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 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\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5405583381652832, \"perplexity\": 1627.024278145106}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"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-2021-43\/segments\/1634323585837.82\/warc\/CC-MAIN-20211024015104-20211024045104-00665.warc.gz\"}"}
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Work For Frederick Community College!
Founded in 1957, FCC is one of 16 community colleges in Maryland and is an accredited, public, two-year, degree-granting institution. More than 17,000 people take credit and noncredit classes each year. The day student population is more traditional—younger, many students out of high school. Many working adults with family, perhaps like you, take classes at night, on weekends, or through non-traditional formats such as online, TV courses, credit by exam, independent study, Weekend College, or portfolio development where you earn credits for your work and life experiences. You'll find our faculty caring and outstanding.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 5,406
|
package com.google.cloud.teleport.v2.utils;
import static com.google.common.truth.Truth.assertThat;
import org.apache.avro.Schema;
import org.apache.beam.vendor.guava.v26_0_jre.com.google.common.base.Supplier;
import org.junit.Test;
import org.junit.runner.RunWith;
import org.junit.runners.JUnit4;
/** Tests for {@link SerializableSchemaSupplier}. */
@RunWith(JUnit4.class)
public class SerializableSchemaSupplierTest {
private static final String AVRO_SCHEMA_STRING =
"{\n"
+ " \"type\" : \"record\",\n"
+ " \"name\" : \"test_file\",\n"
+ " \"namespace\" : \"com.test\",\n"
+ " \"fields\" : [\n"
+ " {\n"
+ " \"name\": \"id\",\n"
+ " \"type\": \"string\"\n"
+ " },\n"
+ " {\n"
+ " \"name\": \"price\",\n"
+ " \"type\": \"double\"\n"
+ " }\n"
+ " ]\n"
+ "}";
@Test
public void schemaSupplierProvidesValidSchema() {
Schema expectedSchema = new Schema.Parser().parse(AVRO_SCHEMA_STRING);
Supplier<Schema> schemaSupplier = SerializableSchemaSupplier.of(expectedSchema);
assertThat(schemaSupplier.get()).isEqualTo(expectedSchema);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,805
|
Q: Facebook likes button in winform application i want to implement Facebook likes button in winform application. I am not able to find any document or example to solve my requirement. Is this Possible to LIKES page from winform application?
A: My guess is you have to create a browser control on your winform application that shows a page that hosts your like button.
http://msdn.microsoft.com/en-us/library/aa752041%28v=vs.85%29.aspx
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,254
|
Who loves paying fees? Not you, that's who. That's why we're slashing them starting immediately! When you buy BTC, ETH, MTL, or any of the 16 (and growing) currencies available on Metal Pay, you'll only pay 1%. How much will that save you? Here's what you get for a hundred bucks.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,742
|
Class ShellView
End Class
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,358
|
package edu.umd.cs.findbugs.classfile.engine.bcel;
import org.apache.bcel.generic.MethodGen;
import edu.umd.cs.findbugs.ba.heap.LoadAnalysis;
import edu.umd.cs.findbugs.ba.heap.LoadDataflow;
import edu.umd.cs.findbugs.classfile.CheckedAnalysisException;
import edu.umd.cs.findbugs.classfile.IAnalysisCache;
import edu.umd.cs.findbugs.classfile.MethodDescriptor;
/**
* Analysis engine to produce LoadDataflow objects for analyzed methods.
*
* @author David Hovemeyer
*/
public class LoadDataflowFactory extends AnalysisFactory<LoadDataflow> {
/**
* Constructor.
*/
public LoadDataflowFactory() {
super("field load analysis", LoadDataflow.class);
}
/*
* (non-Javadoc)
*
* @see
* edu.umd.cs.findbugs.classfile.IAnalysisEngine#analyze(edu.umd.cs.findbugs
* .classfile.IAnalysisCache, java.lang.Object)
*/
public LoadDataflow analyze(IAnalysisCache analysisCache, MethodDescriptor descriptor) throws CheckedAnalysisException {
MethodGen methodGen = getMethodGen(analysisCache, descriptor);
if (methodGen == null)
return null;
LoadAnalysis analysis = new LoadAnalysis(getDepthFirstSearch(analysisCache, descriptor), getConstantPoolGen(
analysisCache, descriptor.getClassDescriptor()));
LoadDataflow dataflow = new LoadDataflow(getCFG(analysisCache, descriptor), analysis);
dataflow.execute();
return dataflow;
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,206
|
Compromise v. Prudence
Articles, Senior Fellows / June 7, 2012 / Peter Wehner
In his book The Years of Lyndon Johnson: The Passage of Power, Robert Caro–in the context of the civil rights struggle–writes this:
Johnson refused to compromise. In public, in answer to a press conference question about the possibility of one, he said, "I am in favor of passing it [the bill] in the Senate exactly in its present form." In private, talking to legislative leaders, he had a more pungent phrase. "There will be no wheels and no deals." There was, as always, a political calculation behind his stance. "I knew," he was to tell Doris Goodwin, "that if I didn't get out in front on this issue, [the liberals] would get me…. I had to produce a civil rights bill that was even stronger than the one they'd have gotten if Kennedy had lived." And there was, as always, something more than calculation. Assuring Richard Goodwin there would be "no compromises on civil rights; I'm not going to bend an inch," he added, "In the Senate [as Leader] I did the best I could. But I had to be careful…. But I always vowed that if I ever had the power I'd make sure every Negro had the same chance as every white man. Now I have it. And I'm going to use it."
The issue of compromise is an important one in politics, and there is much to be said on its behalf. "In the Constitutional Convention, the spirit of compromise reigned in grace and glory," Catherine Drinker Bowen wrote in Miracle at Philadelphia. "As Washington presided, it sat on his shoulder like the dove. Men rise to speak and one sees them struggle with the bias of birthright, locality, statehood…. One sees them change their minds, fight against pride and when the moment comes, admit their error."
Some conservatives seem instinctively hostile to comprise in principle, as if it is inherently a sign of weakness, of lack of commitment and resolve, and that it inevitably leads to bad outcomes. As a "constitutional conservative," I dissent from this attitude.
It's worth noting that two of the most impressive figures in American history, James Madison and Abraham Lincoln, showed the ability to compromise at key moments. It was Lincoln, as a young man, who said, "The true rule, in determining to embrace, or reject any thing, is not whether it have any evil in it; but whether it has more of evil, than of good. There are few things wholly evil, or wholly good. Almost everything, especially of governmental policy, is an inseparable compound of the two; so that our best judgment of the preponderance between them is continually demanded." The Lincoln biographer William Lee Miller, building on this point, added, "[M]any reflective moralists, and most serious politicians, including Abraham Lincoln, perceive … that good and evil come mixed and that the moral life most of the time (not quite all of the time) consists of making discriminate judgments, judgments at the margins, discernments of less and more…"
At the same time, there are those who speak as if compromise is itself, in principle, a moral good. But that approach is also flawed and potentially dangerous. Compromise for its own sake can set back the cause of justice. In the wrong hands, in weak hands, it can produce pernicious results. The point is that compromise can't be judged in the abstract; it can only be assessed in particular circumstances. It takes wisdom and statesmanship to discern when to hold firm (on fundamental principles) and when to give ground (on tactics and secondary issues).
Which brings me back to LBJ. Most modern-day liberals who excoriated conservatives for being "rigid" and opposing "compromise" on matters having to do with the budget and raising the debt ceiling–the pseudo-sophisticated putdown is that they are nihilists–would (rightly) celebrate President Johnson's refusal to compromise on civil rights. Which may get us somewhat closer to the heart of the matter.
The word compromise is something of a Rorschach test. Those who hold a liberal worldview often consider conservatives who fight hard for their cause to be inflexible and unreasonable, just as those who hold a conservative worldview often consider liberals who fight hard for their cause to be inflexible and unreasonable. What determines whether we judge a politician to be a profile in courage or a profile in intransigence almost always depends on whether we're sympathetic to the cause they are championing.
As a general matter, then, compromise is neither a moral good nor a moral evil; it's contextual. And it's why prudence, not compromise, is rightly considered to be among the highest of all the political virtues.
Peter Wehner is a senior fellow at the Ethics and Public Policy Center.
TAGS: #Articles
Five Reasons Why Romney is the Favorite
Hate and Forgiveness
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 416
|
package org.apache.joshua.decoder;
import static java.util.Arrays.asList;
import static java.util.Collections.emptyList;
import static java.util.Collections.emptyMap;
import static org.apache.joshua.decoder.hypergraph.ViterbiExtractor.getViterbiFeatures;
import static org.apache.joshua.decoder.hypergraph.ViterbiExtractor.getViterbiString;
import static org.apache.joshua.decoder.hypergraph.ViterbiExtractor.getViterbiWordAlignmentList;
import static org.apache.joshua.util.FormatUtils.removeSentenceMarkers;
import java.util.List;
import org.apache.joshua.decoder.ff.FeatureFunction;
import org.apache.joshua.decoder.hypergraph.HyperGraph;
import org.apache.joshua.decoder.hypergraph.KBestExtractor.DerivationState;
import org.apache.joshua.decoder.segment_file.Sentence;
/**
* This factory provides methods to create StructuredTranslation objects
* from either Viterbi derivations or KBest derivations.
*
* @author fhieber
*/
public class StructuredTranslationFactory {
/**
* Returns a StructuredTranslation instance from the Viterbi derivation.
*
* @param sourceSentence the source sentence
* @param hypergraph the hypergraph object
* @param featureFunctions the list of active feature functions
* @return A StructuredTranslation object representing the Viterbi derivation.
*/
public static StructuredTranslation fromViterbiDerivation(
final Sentence sourceSentence,
final HyperGraph hypergraph,
final List<FeatureFunction> featureFunctions) {
final long startTime = System.currentTimeMillis();
final String translationString = removeSentenceMarkers(getViterbiString(hypergraph));
return new StructuredTranslation(
sourceSentence,
translationString,
extractTranslationTokens(translationString),
extractTranslationScore(hypergraph),
getViterbiWordAlignmentList(hypergraph),
getViterbiFeatures(hypergraph, featureFunctions, sourceSentence).getMap(),
(System.currentTimeMillis() - startTime) / 1000.0f);
}
/**
* Returns a StructuredTranslation from an empty decoder output
* @param sourceSentence the source sentence
* @return a StructuredTranslation object
*/
public static StructuredTranslation fromEmptyOutput(final Sentence sourceSentence) {
return new StructuredTranslation(
sourceSentence, "", emptyList(), 0, emptyList(), emptyMap(), 0f);
}
/**
* Returns a StructuredTranslation instance from a KBest DerivationState.
* @param sourceSentence Sentence object representing the source.
* @param derivationState the KBest DerivationState.
* @return A StructuredTranslation object representing the derivation encoded by derivationState.
*/
public static StructuredTranslation fromKBestDerivation(
final Sentence sourceSentence,
final DerivationState derivationState) {
final long startTime = System.currentTimeMillis();
final String translationString = removeSentenceMarkers(derivationState.getHypothesis());
return new StructuredTranslation(
sourceSentence,
translationString,
extractTranslationTokens(translationString),
derivationState.getModelCost(),
derivationState.getWordAlignmentList(),
derivationState.getFeatures().getMap(),
(System.currentTimeMillis() - startTime) / 1000.0f);
}
private static float extractTranslationScore(final HyperGraph hypergraph) {
if (hypergraph == null) {
return 0;
} else {
return hypergraph.goalNode.getScore();
}
}
private static List<String> extractTranslationTokens(final String translationString) {
if (translationString.isEmpty()) {
return emptyList();
} else {
return asList(translationString.split("\\s+"));
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,088
|
Q: Magento-- add to cart-- url change In My site when i click addtocart button it returns the following url:
http://mywebsite.com/ajax/index/add/uenc/a...cC5uZXQvYXBwbGlhbmNlcy5odG1s/product/83/isAjax/1
But i need the specific url which is below:
http://mywebsite.com/checkout/cart/add/uenc/a...cC5uZXQvYXBwbGlhbmNlcy5odG1s/product/83/isAjax/1
can i know where to edit this?
A: Of course you can change it in all your theme files.
However, if you want to change all add to cart behavior site wide you may want to extend the CartController and forward the add action to your own ajax module.
An example:
app/code/local/MyCompany/Ajax/controllers/CartController.php:
<?php
require_once 'Mage/Checkout/controllers/CartController.php';
class MyCompany_Ajax_CartController extends Mage_Checkout_CartController {
public function addAction() {
$this->_forward('add', 'index', 'ajax', $this->getRequest()->getParams());
}
}
app/code/local/MyCompany/Ajax/etc/config.xml:
<?xml version="1.0"?>
<config>
[...]
<frontend>
<routers>
<checkout>
<args>
<modules>
<mycompany_ajax before="Mage_Checkout">MyCompany_Ajax</mycompany_ajax>
</modules>
</args>
</checkout>
</routers>
</frontend>
</config>
Another way is to add an observer that listens to the controller_action_predispatch_checkout_cart_add event and forward there.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,012
|
This module is an opportunity for you to devise, develop and complete a substantial piece of independent research at Master's level, in dissertation form, on a legal topic of your choice linked to your study on the BPTC.
You will work under the guidance of a project supervisor, using the knowledge and understanding you have gained on your Qualifying Law Degree and on the LLM in Bar Professional Training programme.
The LLM Project module focuses on the development of skills in three key areas: problem identification and analysis, research and information gathering and the written presentation of results. You will develop your ability to undertake legal research, addressing complex and current issues in your subject area. You will also engage critically with research material and analyse current skills and practice. The LLM Project module will also develop your skill in effectively formulating, communicating and presenting written argument to both a specialist and non-specialist audience.
You will learn through a combination of tutor-directed and independent learning practice. The module will contain enhanced taught elements of research skills to support you in your learning.
In the initial stage of the module your focus will be on tutor-directed learning via a module dedicated e-learning site (eLP), where you will be provided with detailed learning materials and resources, including guided reading and self-evaluative tasks, to help you develop the research and writing skills that are needed for you to undertake a project. During the induction period you will receive instruction on how to structure and conduct research for your dissertation, as well as how to develop your legal writing and referencing skills. In addition, you will have the opportunity to receive formative feedback on short writing tasks as you progress through the year.
Independent learning will focus on you identifying the reading and research required to complete your dissertation. Additionally, you will be allocated a project supervisor for your project. Your summative assessment for the module is your finished research project and here too you will be provided with feedback on your work. Thus, through such independent study, research and feedback you can also develop your learning through reflective practice.
The LLM is supported by the appointment of an individual project supervisor to guide you through a research-oriented and research enquiry based learning process. You are encouraged to maintain regular contact with your supervisor, with whom you will discuss your research topic and methodology to help clarify your understanding, analysis and presentation. In semester one, you will be required to submit a project proposal and ethics Form. Your proposal will identify a title, the underlying aims and structure of your work, any preliminary sources and a timetable for its completion the following year. Once submitted, your supervisor will provide you with formative feedback and form a view on whether your proposal and ethics form are Competent/Not Yet Competent and what you might do to improve your work. Failure to submit a project proposal by the stated deadline in semester one will prevent you from progressing on the LLM route. As you proceed through the year, you will receive written feedback from your project supervisor on draft work as you work towards completion of your project in semester 3.
The University is well placed to support you in learning and research with an excellent library and teaching facilities, access to on-line legal databases, resources and appropriate software. Programme support will be provided through the e-learning portal and resources through the University Library. This will provide you with tutored guidance on legal skills, research and assessment in this module. The Programme Leader can be contacted in addition to your supervisor in respect of any problems.
The Programme Administration and Student Progress teams are responsible for the non-academic administration of the module, such as receiving your completed project, returning your marked project and recording your marks. They and the project supervisor will contact you throughout the duration of your module with important dates and other issues.
• Developed critical knowledge and understanding of your chosen subject, informed by current practice, advanced scholarship and research, including a critical awareness and evaluation of current issues and developments in the subject.
• Developed an ability to formulate a research proposal which enables you to engage in critical discourse in relation to your chosen area of study, to formulate balanced judgements if information is ambiguous or incomplete and communicate clearly to both a specialist and a non-specialist audience.
• Developed an ability to conceptualise, theorise and reflect on your own practice and the practice of others and undertake research which addresses complex issues and which advances understanding in your chosen area of study.
• You will be able to devise, develop and complete a substantial independent research project, including selecting an appropriate subject area, conducting research and making editorial choices and decisions.
Summative assessment: The summative assessment on this module is your submission of a 8,000 to 10,000 word research project.
There are opportunities for formative assessment throughout the module including advice on the suitability of your proposed topic, on your research techniques, on whether your project is sufficiently analytical and well-structured and written and complies with University ethics requirements. There is initial legal research instruction, with linked resources, which provide self-test questions and exercises on legal research and writing. Thereafter you will receive feedback from your supervisor on draft work as you work towards producing your dissertation.
The marking criteria used to assess the project are outlined and explained when you embark on the module to ensure you can understand what is expected of you and how you will be judged on your performance. Once your project has been marked, you will be provided with a written feedback sheet on how well it met with the assessment criteria.
The aim of this module is to enable you to enhance your Bar qualification, your knowledge, your research skills and your employability through the completion of a supervised independent research project on a specialist topic of your choosing. You will integrate your understanding of topics and skills learnt throughout the Bar Professional Training Course (BPTC), gaining an internationally recognised Masters qualification.
As part of this module, you will critically engage with research outputs as part of your research-rich learning, drawing from Northumbria's extensive on-line databases and library facilities. You will further develop the necessary skills to conduct effective legal research, essential and integral skills to successful practice at the Bar. The final assessed research project will consolidate your understanding of a linked area of study on the BPTC and develop your skills of critical analysis.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,465
|
\section{Introduction}
Bose-Einstein condensates generated in trapped
atom experiments constitute mesoscopic quantum objects that are well
localized in space and exhibit quantum properties such as interference
and phase coherence. They represent a highly controllable and
potentially rich template for addressing fundamental questions
related to quantum measurement and configuring precision measurement
devices \cite{bhongale}. Here we study the unique problem of transporting such
finite-sized BEC's on rough surfaces, a situation inevitably encountered with
BEC's generated on chips. There the roughness originates from the
fluctuations of the surface fields \cite{kruger}.
In a broader context, transport of superfluids through
disordered potentials has always been a topic of theoretical intrigue
ever since the first experimental observations concerning super-flow
of Helium-4 through porous media \cite{superflow}. Huang and Meng
first proposed a hard-sphere Bose gas in a random potential as a model
describing the static properties of this scenario \cite{huang}. Even
though mostly qualitative, this work was very insightful and provided
definite clues to the connection between boson localization and
condensate depletion. Subsequently, using a quantum hydrodynamic approach, Giorgini \emph{et al.} \cite{giorgini}
looked at the disorder induced phonon damping in a Bose superfluid. However,
not much else was said about dynamics. In fact, after almost two decades, questions
related to the transport of superfluids through disorder have hardly
been touched mostly due to the lack of proper experimental
setups. However, very recent, highly precise experiments on trapped
atoms \cite{randy,randy2,demarco} have begun to expose puzzling new
aspects of such transport through disorder. A sharp reduction in
dissipation rate beyond the Landau critical velocity \cite{landau} is
observed. Further, the experiments also indicate a power-law
dependance of the dissipation rate on the interaction strength. Finally, when we view these
disordered BEC experiments as well localized
macroscopic quantum objects sliding on rough surfaces, it touches
upon another interesting research area with open questions related to
the origins of non-linear friction laws \cite{friction}.
Trapped atomic systems constitute a promising new platform whereby the
many-body physics of various disorder-induced phenomena may be
understood. Contrary to condensed matter, these atomic
systems are intrinsically pure and allow for controlled insertion of
disorder by, {\it e.g.}, modulating the lattice potential
\cite{massimo}, or creating an optical speckle pattern
\cite{randy,aspect}. Experimentalists are even able to control, to a
good extent, the correlation properties of these disorder
potentials.
The significance of this type of extreme control needs no further
discussion if we simply note that, for the first time, experiments are
able to provide explicit access to the localized wavefunction, thus
allowing direct evidence of Anderson localization, an exquisite
phenomenon responsible for the complete disappearance of metallic
conductivity \cite{massimo,aspect,anderson}. Such spectacular
developments, with no surprise, give compelling reasons to
theoretically investigate the disordered Bose system in the light of
the new techniques of probing made available by ultra-cold
atoms. Especially, with the possibility of tuning the inter-particle
interactions via techniques of Feshbach resonance \cite{feshbach}, ultra-cold atom
experiments provide a systematic way to investigate the role of
interactions in the localization phenomenon; an aspect that has
generated much debate in the field. Numerous transport experiments can
be performed, requiring new insights due to the finite spatial extent
of these trapped-atom condensates.
In addition, excitations in a finite BEC are fundamentally different
in symmetry from those in the translationally invariant counterpart.
Unfortunately, at present there is no well established
theoretical framework for describing such non-equilibrium hydrodynamic
scenarios. Experimental observations \cite{randy,randy2,demarco}
continue to demand a clear understanding of the physical mechanism.
Recent numerical simulations based on the Gross-Pitaevskii equation
\cite{albert}, confirm some of these observations and provide some
hints towards the possible physical mechanism, but a more detailed
theoretical model is missing. As will be shown below, our model provides a
microscopic underpinning of the experimental and numerical
results. This allows us to determine exponents governing the power law
dependence of the dissipation on the scattering length, giving a
deeper understanding of the interplay between disorder and
interactions. Also, as will be evident from our discussion, such exponents
critically depend on dimensionality.
In this paper, we present a theory for determining the transport
properties of a trapped BEC flowing through a disordered potential,
similar to that used in the experiments of
Refs.~\cite{randy,randy2,aspect}. In the past, similar interest in
transport of localized wavepackets, for example in microwaves
\cite{tiggelen}, has been limited to time scales of the order of
the diffusion time $\tau$. While our proposed framework may also be applied
to this regime as well, we intend to focus on the specific
experimental situation discussed in Refs.~\cite{randy,randy2}.
Here, the dissipative transport is achieved by generating a center-of-mass
mode of the trapped condensate and allowing it to propagate through an
optical speckle potential (see Fig.~\ref{schematic}). The damping of this mode
is measured as a function of interaction strength and the
center-of-mass velocity.
The strength of the disorder is such that the period of oscillations $T \ll \tau$.
\section{Collective Modes (Excitations) of a Trapped BEC}
A tightly confined BEC can be made to oscillate as a whole, for
instance, by abruptly shifting the trap by a small distance.
If the displacement is small, these excitations can be easily
described by linearizing the Gross-Pitaevskii (GP) equation
\begin{eqnarray}
i\hbar\frac{\partial}{\partial t} \psi_0({\bf r},t)=\left[-\frac{\hbar^2\nabla^2}{2m}+V_{{\rm ext}}({\bf r})+\lambda |\psi_0({\bf r},t)|^2\right]\psi_0({\bf r},t),&&\nonumber
\end{eqnarray}
where $V_{{\rm ext}}({\bf r})$ is the trapping potential, and the
interaction parameter $\lambda$, in terms of the $s$-wave atom-atom
scattering length, $a$, is $\lambda=4\pi\hbar^2a/m$. The set of
Bogoliubov equations that follow provide the eigenenergies,
$\Lambda_i$, of the elementary excitations. As is apparent from the above GP equation, interactions play an important role in defining these excitations.
\begin{figure}[h]
\includegraphics[scale=.37]{disorder-schematic.eps}
\caption{(Color online) Sketch of the experimental scenario in question. The zigzag
line corresponds to the speckle potential that is superimposed
onto the smooth harmonic potential. The condensate (shown in blue)
is shifted to one edge of the trap and released at $t=0$,
resulting in dipole oscillations.}
\label{schematic}
\end{figure}
Among other types of collective oscillations exhibited by the trapped
condensates, of special significance is the dipole mode. It
corresponds to the center of mass motion of the condensate which due
to the harmonic confinement has the same frequency as the trap and is
unaltered by interactions. In general, for 3D confinement, there exists
a whole set of low energy modes that can be obtained analytically by
solving the coupled hydrodynamic equations (derived from the GP
equation above) for the velocity, ${\bf v}({\bf r},t)$, and the
density, $n({\bf r},t)$:
$\partial n/\partial
t+{\rm div}({\bf v}n)=0$ and
$m\partial {\bf v}/\partial t +\nabla
\left\{m{\bf v}^2/2+V_{{\rm ext}}+\lambda n\right\}=0$.
These equations allow for
solutions, where the condensate is oscillating as a whole
\cite{stringari}, with frequencies given by
$\omega_e=\omega_{{\rm ho}}\sqrt{2n_r^2+2n_r\ell+3n_r+\ell}$,
assuming $V_{{\rm ext}}({\bf r})$ is harmonic with frequency
$\omega_{{\rm ho}}$. The dipole mode is the one with quantum numbers
$(n_r=0,\ell=1)$. In the absence of any other potential (disordered
or not), the harmonic trap is special, leaving the center-of-mass
motion completely decoupled from the relative degrees of freedom even
for large amplitudes. However this feature is immediately lost as soon
as any other external potential, {\it e.g.} disorder, is turned on,
resulting in coupling of different modes, eventually leading to the
complete decay of a well defined oscillating mode.
\section{Key Concepts} Our model consists of two main ingredients, \emph{disorder}
and \emph{inhomogeneity}, which we describe below. \\
\subsection{Disorder} For treating disorder within an analytic
approach, among several well established techniques in the literature
\cite{replica,methods}, we find it convenient to use the \emph{replica} method.
The disorder potential $U_{{\rm d}}({\bf x})$ is described
in terms of a Gaussian distribution function with a strength $\gamma$
such that
$P[U_{{\rm d}}]=\exp [-1/(2\gamma^2)\int {\rm d}{\bf
x}{\rm d}{\bf x}'U_{{\rm d}}({\bf x})K^{-1}({\bf x}-{\bf
x}') U_{{\rm d}}({\bf x}')]$
and $K$ describes the spatial
correlations \cite{replica}. Such procedure automatically allows for a
freedom in choosing the actual disorder potential, the details of
which are quickly erased by multiple scattering. To further simplify
our investigation, we consider a white noise correlation function
$K({\bf x})\sim \delta({\bf x})$. Thus,
$\langle U_{{\rm d}}({\bf
x})\rangle_{{\rm dis}}=0$ and $\langle U_{{\rm d}}({\bf
x})U_{{\rm d}}({\bf x}')\rangle_{{\rm dis}}=\gamma^2\delta({\bf
x}-{\bf x}')$,
where $\langle\cdot\cdot\cdot\rangle_{{\rm dis}}$ means averaging
over disorder. However, the \emph{trick} of using the \emph{replica}
method lies in following a particular procedure for taking the
sequence of averages (quantum followed by disorder), essential for
calculating observables.
While the \emph{replica trick} has been typically used in relation to
Anderson localization of non-interacting electrons,
we adopt it in the context of ultra-cold
interacting atoms. A brief outline of this technique is provided in
the Appendix. Aside from technical aspects,
the \emph{replica trick} allows for a systematic diagrammatic
perturbation expansion within which all transport
properties can be calculated, at least in principle.\\
\subsection{Inhomogeneity} Next we use local density
approximation (LDA) \cite{stringari} to account for the inhomogeneous density
distribution of the trapped condensate. This is achieved by defining a position
dependent chemical potential,
\begin{eqnarray}
\mu\rightarrow \mu({\bf
r})&=&\mu-V_{{\rm trap}}({\bf r}),\label{lda1}
\end{eqnarray}
clearly justified in the Thomas-Fermi limit \cite{stringari}, which is typically always the case
for trapped-atom BEC experiments. Thus, the density of the stationary
condensate in the absence of disorder is given by
\begin{eqnarray}
\rho({\bf r})=\mu({\bf r})/\lambda\label{lda2}
\end{eqnarray}
Further, we assume that as the condensate undergoes dipole
oscillations, on average the shape of the condensate is retained,
meaning $\rho({\bf r}+{\bf r_{\rm cm}}(t))=\rho({\bf r})$. This is
an extremely good assumption, it stems from the rigidity associated with
the Bose condensate.
This greatly simplifies the
problem by allowing us to define all the relevant quantities locally,
for example we define the local Bogoliubov excitation energies \cite{ripka}
\begin{eqnarray}
\Lambda_{{\bf k}}[{\bf r}]=\sqrt{e_k^2-\lambda^2\rho({\bf r})^2}\label{bdglocal}
\end{eqnarray}
with $e_k=\hbar^2k^2/2m+\lambda\rho({\bf r})$. Also, from the linear
part of the above dispersion relation, we can define a local sound
velocity\\
\begin{eqnarray}
c[{\bf r}]=\sqrt{\lambda\rho({\bf r})/m}.\nonumber
\end{eqnarray}
\begin{figure}[ht]
\includegraphics[scale=.42]{distances_def.eps}
\caption{(Color online) Sketch showing the different relevant distances defined
in the text. Dark (light) blue color represent spatial regions of
the condensate where the local sound velocity is greater (lesser)
than the center-of-mass velocity. Graphic arrows indicate excitation moving
along the $z$ direction.}
\label{distances}
\end{figure}
\section{The Model} The total Hamiltonian for the problem can be
written as the sum
\begin{eqnarray}
\hat{H}&=&\underbrace{\hat{H}_{{\rm free}}+\hat{H}_{{\rm int}}}+\hat{V}_{{\rm trap}}+\hat{U}_{{\rm d}}\nonumber\\
&=&\hspace{0.74cm}\underbrace{\hat{H}_{{\rm eff}}\hspace{0.45cm}+\hat{V}_{{\rm trap}}}+\underbrace{\hat{U}_{{\rm d}}}\nonumber\\
&&\hspace{1.4cm}{\rm LDA}\hspace{1cm}{\rm replica},\nonumber
\end{eqnarray}
where $\hat{H}_{{\rm free}}$ is for the free non-interacting atoms
and $\hat{H}_{{\rm int}}$ is due to the atom-atom interaction. As
indicated above, we first write an effective Hamiltonian for the
bosons in the absence of any potentials by treating
$\hat{H}_{{\rm int}}$ at the mean field level. Diagonalizing
$\hat{H}_{{\rm eff}}$, the lowest eigenvalue corresponds to the
condensate mode at temperatures $T<T_{{\rm BEC}}$. The excitations
above the ground state are nothing but the Bogoliubov modes described
above and are orthogonal to the condensate mode. Any other external
potential will in general couple these different eigenmodes. However,
as mentioned earlier, the trapping potential is typically smooth and
LDA holds, drastically simplifying the calculation. The only piece that
remains to be included is the potential, $\hat{U}_{{\rm d}}$. At any
spatial location, the disorder tends to deplete the condensate by
populating the Bogoliubov modes. However, this does not happen if the
condensate is at rest, with zero center-of-mass velocity. Due to the superfluid
property of the BEC, Landau criterion holds and there is a non-zero
critical velocity below which no excitations can be generated and the
nonequilibrium state will persist forever. This is precisely where
the finite size of the condensate plays an important role. At any
given center-of-mass velocity $v$, on the contrary, there is always a finite
region of the condensate, shown by the light blue color in
Fig.~\ref{distances}, where $v>c[{\bf r}]$ thus satisfying Landau
criterion for excitations and providing a channel for the energy to be
transferred from the center-of-mass motion to the Bogoliubov
modes. In essence, this is the physical mechanism that provides the
frictional force responsible for slowing down the condensate center-of-mass
motion.\\
\textbf{Remark:}
\emph{Before we proceed to calculations, we need to take
into account the fact that, previously, excitations were calculated in
the rest frame of the condensate. Thus, if we wish to continue
calculating everything else in the same frame, it implies having to
use a time dependent disorder potential.}\\
This is certainly a
non-trivial complication and cannot be simply neglected. Fortunately,
LDA turns out to be perfectly suited for such a situation. It
allows us to work in the lab frame instead, hence static disorder. Now, from Eq.~\eqref{bdglocal}, the locally defined
Bogoliubov modes are eigenstates of momentum and hence remain
unaltered under a Galilean frame transformation, only their energies are
shifted according to:
\begin{eqnarray}
\Lambda_{\bf
k}\rightarrow\Lambda_{\bf k}^{{\rm lab}}&=&\Lambda_{\bf k}-\hbar{\bf
k}\cdot {\bf v}.\nonumber
\end{eqnarray}
Thus, we have gathered all the ingredients for conducting the transport
calculations via the diagrammatics of the \emph{replica trick}. The above
model is quite general and allows for calculating observables in a
broad range of scenarios by including appropriate orders in the
perturbation expansion. For example, if the time scale of interest is
larger than the Heisenberg time (time scale defined by the trapping
potential), one needs to include appropriate particle-hole diagrams to
describe the diffusive dynamics. On the other hand, if the dynamics
occurs on the trap scale, then depending on the disorder strength,
appropriate single particle diagrams may be sufficient. The only point
one needs to remember is that the bare propagators of the replica
theory are those for the quasiparticles representing atoms dressed by
interactions with the corresponding Green's function given by~\cite{ripka,mahan}:
\begin{eqnarray}
\mathscr{G}({\bf k},i\omega_n)[{\bf r}]=\frac{1}{\Lambda_{{\bf k}}^{{\rm lab}}[{\bf r}]-i\omega_n}+\frac{1}{\Lambda_{{\bf k}}^{{\rm lab}}[{\bf r}]+i\omega_n},\nonumber
\end{eqnarray}
where $\omega_n$ is the bosonic Matsubara frequency.
\section{Experiment} As elaborated in the introduction, our
model is motivated by recent experiments performed on cold trapped
atomic condensates. Hence, we illustrate the applicability of
the model by considering the actual experimental situation of
Refs.~\cite{randy,randy2}: $^7{\rm Li}$ atoms trapped in an axial
harmonic trap with $\omega_z=2\pi\times 5.5\,\, {\rm Hz}$, aspect
ratio $\alpha= \omega_{\perp}/\omega_z \approx 46$. The $s$-wave
scattering length is tuned via a Feshbach resonance to about
$a=25\,a_{{\rm B}}$, resulting in an axial condensate with the size
represented by the Thomas-Fermi (TF) radius, which in units of the
axial oscillator length $\ell_z$, is given by $R\approx 13.6
\,\ell_z$. These parameters conform with the experimental data for
damping of the BEC dipole mode used in the plot of
Fig.~\ref{oscillations}. The disorder potential is produced using an
optical speckle with gaussian correlation:
\begin{eqnarray}
\langle
U_{{\rm d}}({\bf x})U_{{\rm d}}({\bf
x}')\rangle_{{\rm dis}}&=&V_{{\rm d}}^2\exp\left[-\frac{2({\bf x}-{\bf
x}')^2}{\sigma^2}\right].\label{disorderfunction}
\end{eqnarray}
Further, while no data on temperature is available, from the
observations presented, it is safe to assume that the experiment is
conducted at temperatures well below $T_{{\rm BEC}}$ and thus a $T=0$
theoretical prescription should suffice.
The TF density of the stationary condensate in cylindrical
coordinates follows from Eqs.~\eqref{lda1} and \eqref{lda2}
\begin{eqnarray}
\rho(r_{\perp}
,z)&=&\frac{m\omega_z^2(R^2-\alpha^2r_{\perp}^2-z^2)}{2\lambda},\nonumber
\end{eqnarray}
with the normalization fixed by the TF axial size $R$ via
\begin{eqnarray}
N=4\pi\int_0^{R/\alpha} \int_{0}^{R[r_{\perp}]}\rho(r_{\perp},z){\rm d}z\,\,r_{\perp}{\rm d}r_{\perp},\label{normalization}
\end{eqnarray}
with the different distances indicated in
Fig.~\ref{distances}. For convenience, we define the dimensionless
ratio $\xi[r_{\perp}]=v/c[r_{\perp}]$, where $c[r_{\perp}]$ is the
speed of sound at the coordinate $\{r_{\perp},z=0\}$. We also use the
notation, $c\equiv c[0]$ and $\xi\equiv\xi[0]$. We point out that even
though the aspect ratio is large, the chemical potential $\mu\sim{\cal
O}(\hbar\omega_{\perp})$, hence a full 3D treatment is
required. Finally, we observe that the oscillation data shown in
Fig.~\ref{oscillations} indicates
dynamics over time scales much shorter than the diffusion time
dictated by the disorder. This greatly reduces the manipulations and it
suffices to calculate only the single particle diagrams.
The imaginary part of the selfenergy, obtained from the dressed
propagator of the non-interacting quasiparticles, immediately provides
the transition rate $W({\bf k}\rightarrow {\bf k}',{\bf r})$ at the
spatial point ${\bf r}$ (for details refer to the Appendix). However, we alert the reader that one
typically never measures the single particle propagator in an
experiment and we still need to connect to the damping of the
classical dipole mode. We achieve this by recalling that the energy
per condensate particle is nothing but the chemical potential
$\mu({\bf r})$. Thus, the total energy transfer rate out of the ${\bf
k=0}$ condensate mode is precisely
\begin{eqnarray}
\Gamma=\int_{\Omega} {\rm d}{\bf r}\int_{k>\kappa}{\rm d}{\bf k}\,\,\mu({\bf r})W(0\rightarrow {\bf k},{\bf r}),\label{energylossrate}
\end{eqnarray}
where $\kappa$ is some infrared cutoff imposed by the trapping
potential and $\Omega$ is the condensate volume. The above integral
can be approximated and has a nice analytic form if excitations are
allowed only along the axial (z) direction. This constraint can be
understood to arise from the high aspect ratio $\alpha$ of the
trapping potential. Even at large $k$, where this approximation may
fail, we expect the density of states to be diminished enough to
accrue any appreciable error. With this we arrive at the final result
in terms of dimensionless quantities represented by
$(\bar{\,\,\,\,})$(physical quantities scaled in units of the axial
harmonic trap: lengths in units of $\ell_z=\sqrt{\hbar/m\omega_z}$ and
energies in $\hbar\omega_z$)
\begin{eqnarray}
\bar{\Gamma}[\xi]=\frac{\bar{\gamma}^2\bar{R}^4}{4\sqrt{2}\pi\bar{a}}\int_0^1 t^{2}\,\,
{\cal I}[\xi/\sqrt{t},\xi_{{\rm min}}/\sqrt{t}]\,\,{\rm d}t, \label{main1}
\end{eqnarray}
where the upper (lower) integration limit corresponds to the center
(radial edge) of the condensate, $\xi_{{\rm min}}$ is related to
$\kappa$, $\xi_{{\rm min}}=\hbar\kappa/2mc$ and the function ${\cal
I}$ is defined by Eqs.~\eqref{boundaryfunction} and \eqref{function} of Appendix.
\section{Results and Discussion} To begin with, we first look at the
plot in Fig.~\ref{oscillations}(a) that shows the velocity dependence
of the function ${\cal I}$ for points along the long axis of the
condensate. The sharp discontinuity near the speed of sound $c$ is a
clear indication of the origin of distinct regimes of damping as the
BEC flows through the disordered medium. Essentially, for velocities
significantly smaller than the speed of sound $c$, excitations are
only allowed in a small shell near the surface of the condensate. This
region tends to grow inwards, eventually allowing excitations
everywhere in the condensate, as the center-of-mass velocity
reaches $c$. However, increasing the velocity beyond this critical
value, the excited volume remains fixed while the Bogoliubov density
of states keeps falling, resulting in net decay of the
damping rate with velocity.
The above expression for $\bar{\Gamma}$ represents the first of the two main
results of this article. It conveniently leads us to a dynamical
equation for the condensate motion, if we picture the BEC oscillating
in the random potential as a rigid body whose hydrodynamic
oscillations are damped as the kinetic energy is transferred to the
internal excitations. Such damped oscillatory motion of the BEC center-of-mass
coordinate can easily be described by an equation for the peak
velocity of the condensate motion in terms of $\xi_{{\rm peak}}$
\begin{eqnarray}
\frac{\partial\xi_{{\rm peak}}}{\partial t}= -\frac{2}{N\bar{R}^2 \xi_{{\rm peak}}}\tilde{\Gamma}[\xi_{{\rm peak}}]=-\beta \xi_{{\rm peak}},
\label{main2}
\end{eqnarray}
where the energy dissipation rate is averaged over one oscillation
period, represented by $\tilde{\Gamma}=(1/T)\int_0^T\Gamma {\rm d}t$.
The above equation for $\xi_{{\rm peak}}$ represents the second main
result of this article. These results immediately provide important
asymptotic exponents governing the dependence of the velocity-damping rate
$\beta$ on the scattering length. For this, all that is required is to
remember that the total number of atoms $N$ is fixed, automatically
implying $R\sim a^{1/5}$ from the normalization condition Eq.~
\eqref{normalization}; thus, $c\sim a^{1/5}$ follows as well. Now, from the
functional form given by Eq.~\eqref{function},
\begin{eqnarray}
\beta \sim \left\{
\begin{array}{l}
a^{-6/5} \,\,\,\,\forall\,\,\xi\ll 1 \Rightarrow {\cal I}\sim 1/a\\
a^0 \,\,\,\,\,\,\,\,\,\,\,\,\,\forall\,\,\xi\gg 1
\end{array}
\right.
\end{eqnarray}
The latter limit is easily confirmed from the plot,
Fig.~10, in Ref. \cite{randy2}. While, data for the former limit
is not present in the same figure, available data points for $\xi <1$
show a clear trend towards the predicted limiting behavior. Instead,
we indirectly verify this limit by extracting the damping rate as a
function of $\xi$ from the data shown in Fig.~\ref{oscillations}(b)
which can then be written as a function of scattering length, $a$, by
fixing $v$ in Eq.~\eqref{main1}. We again find very good agreement,
confirming the exponent corresponding to the former limit as well.
\begin{figure}[t]
\includegraphics[scale=.43]{testfancy_final_v2.eps}
\caption{(Color online) (a) Damping function along the long axis of the condensate
plotted as a function of $\xi$. (b) Thick solid curve:
theoretical prediction for the peak center-of-mass displacement of the
condensate as a function of time for $\bar{\gamma}=0.5$, filled
circles connected with thin line are experimental data points for
the center-of-mass position as the condensate oscillates in the trap
\cite{randy2}. Square marker indicates the point where $\xi_{{\rm peak}}=1$.}
\label{oscillations}
\end{figure}
The only parameter that remains to be estimated is the
$\delta$-correlated disorder strength $\gamma$. We remind the reader
that, in the actual experiment, the disorder potential, Eq.~
\eqref{disorderfunction}, has a finite correlation length $\sigma$.
Such spatial correlations are indeed outside the scope of the
current LDA based model. Therefore, $\gamma$ in our model maybe
considered as a free parameter representing the renormalized value of
the disorder strength.
Our result, the solution of Eq.~\eqref{main2} plotted in Fig.~\ref{oscillations}(b), is in excellent
quantitative agreement with the experimental data, indicating the
validity of a delta-correlated ansatz for the speckle potential. The
proposed model is also able to capture all the damping time scales via
only coupling to the Bogoliubov excitations. Thus, we identify the
most dominant source of dissipation, sufficient to describe transport
in such ultra-cold trapped-atom experiments.
Finally, we shall comment on the strength of the disorder. For this,
let us use the value of $\bar{\gamma}$ estimated in
Fig.~\ref{oscillations}(b) to find the magnitude of disorder-induced
condensate depletion given via the method of Ref.~\cite{huang}
\begin{eqnarray}
\rho_{{\rm dep}}(0)&=&\frac{m^2\gamma^2}{8\pi^{3/2}\hbar^4}\sqrt{\frac{\rho(0)}{a}}. \nonumber
\end{eqnarray}
We find $\rho_{{\rm dep}}(0)/\rho(0)=.002$, implying
that the condensate density remains almost unchanged after inclusion
of speckle potential and thus justifying the assumption of weak disorder,
permitting perturbative treatment up to second order in disorder strength.
Furthermore, it
also points to an important aspect about the disorder potential used
in the experiment~\cite{randy2}. There, even though the disorder
strength seems to be large, given by $V_{{\rm d}}\approx 0.6\,\mu$,
our theoretical model points to a critical interplay between disorder
correlations and interactions, resulting in a dressed disorder with a
small renormalized strength $\gamma$.
\section{Conclusion}
To summarize, we have given a theory for describing the transport
properties of \emph{finite-size} atomic condensates flowing through
disorder potentials. We emphasize the intricate role played by the
inhomogeneous character of the condensate, resulting in a completely
different form of the damping function, ${\cal I}$, above and below
the critical velocity, $c$. The validity of our model is established
via comparison with the experimental observations of damped
hydrodynamic dipole oscillations of the condensate \cite{randy2}. The
predicted power-law dependance of the damping on the interaction parameter is
in excellent agreement with experimental observations. Our
results, while based on a white noise disorder model, show extremely
good agreement with experimental data, thereby, indicating the
insignificant role of finite correlations in determining the
dissipation time scales. While a full theory was not necessary for
the aspects considered here, an improved quantitative picture can be
realized by including finite disorder-correlations, for example, via
a coherent potential approximation \cite{cpadig}. Such procedure leads
to an effective, renormalized disorder vertex, whose value can then be
compared with the $\gamma$ obtained here. However, these calculations
are beyond the current model and would be considered in a future work.
Furthermore, although the effects of Anderson localization turn out not to be relevant
for these experiments\cite{randy,randy2}, we predict they could be if the center-of-mass
motion is very slow and therefore the physics is dominated by diffusion.
These effects could be incorporated into our calculations by including appropriate
diagrams corresponding to particle-hole processes. Finally, the analysis presented here can be easily translated in the language of
friction indicating the manifestation of non-linear friction on mesoscopic quantum objects.
\begin{acknowledgments}
We are extremely
grateful to Randy Hulet, Scott Pollack, and Dan Dries for invaluable
discussions relating to their experimental findings on the dissipative
transport. We acknowledge the financial support from the W. M. Keck
Foundation and NSF. C.J.B. and P.K. also acknowledge the financial
support from ARO Award W911NF-07-1-0464 with the funds from the DARPA
OLE Program.
\end{acknowledgments}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
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Q: Show country specific website in search We have a website which we want to replicate and put on country specific domain names.
How can we ensure only these country specific domain names show up in searches?
i.e. - ensure that http://getbunnybox.nl shows up for Dutch searchers instead of https://getbunnybox.com.
Is this possible?
A: HTML5 provides the alternate link type which, together with the hreflang attribute, can be used for translations:
If the alternate keyword is used with the hreflang attribute, and that attribute's value differs from the root element's language, it indicates that the referenced document is a translation.
The hreflang (as well as the lang) attribute takes a BCP 47 language tag as value, which can have a region subtag.
So if you have a document in Dutch and you want to target the Netherlands, you can use the language tag nl-NL (and nl if the country doesn't matter, and nl-BE if you want to target Dutch speakers in Belgium, etc.).
From each page of getbunnybox.com, you could link to each corresponding page of getbunnybox.nl (and vice versa) with a link element:
<!-- on http://getbunnybox.com/hello -->
<link href="http://getbunnybox.nl/hallo" rel="alternate" hreflang="nl-NL" />
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"redpajama_set_name": "RedPajamaStackExchange"
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Savasana, as many know, means corpse pose in Sanskrit (sava= corpse, asana=pose). It is the pose most commonly used to end an asana practice. It is "intended" to be the ultimate relaxation pose, from which we are meant to arise feeling rejuvenated in mind, body and spirit. Yet, how many of you out there have put your bodies into savasana and felt anything but relaxed and mindful? I definitely relate to this and know of many of my students who can as well.
So my question is what is savasana about really? What is more important: the position of the body or the alignment of the mind with the body?
I have tried…and tried…and tried some more to place my body exactly as various teachers have described to me. I understand they are trying to align my body for optimal energy flow through my body, but what if this results in tension, mental and physical to hold a position? For far too long I have felt that I was trying to squeeze my body into a cookie cutter shape called "savasana"…why? For what purpose?
Now what? That at least is the question that had been buzzing through my mind. Something had to change for me to find savasana, the pose of ultimate relaxation.
Then, of course, as I am ending a class and guiding to my students into savasana it hit me like a freight train. Who says your savasana pose has to look any way in particular. The goal and purpose of the pose is relaxation, both internal and external. It is a time for the body to take what has been infused into it during the practice and do something productive with that information. So perhaps it is time to let go of our "traditional" definitions of savasana; isn't that what we are supposed to be doing anyway – letting go?
So next time you find yourself "setting up" savasana and your body and mind are fighting it – try something different. Shake out your limbs…wiggle your bum…sway your shoulders…roll your head back and forth a little. See where body ends up. Then breath and notice. Notice where you are still holding…holding onto energy, tightness, emotion, stuff…and breath into that space that distracts. And as you exhale, imagine that distraction deflating as a sail with no wind.
Be Curious. Be Adventurous. It is your body your practice!
Let me know what you find on your little adventure with savasana! #InSearchof Savasana !
And always remember, every time you step on your mat it will be a new adventure!
Wandering Spirit Yoga Adventures for 2014: Look Out World Here Come the Yogis!
Happy Belated New Year! I can't believe it is 2014 already…where did 2013 go!
I remember as a small kid my parents telling me, warning me, to slow down because times just seems to speed up as we grow up. Of course I didn't understand that then, but with each passing year it makes more and more sense. It also makes me more aware of how I spend my time and making the most of my time. Sooooooo… look out 2014, WSYT has some fabulous adventures planned for this upcoming year!
First up this month is the WINTER ESCAPE TO PARADISE in Cat Island Bahamas. Actually, I am headed down to the Bahamas in two days from now to set up for the arrival of the crew this coming Saturday! (Apparently, I planned this well as another cold front rolls into North Carolina just as I leave!) I am pumped to be bringing another crew down home to the island on which I grew up. It is hard to explain what makes this island so magical…but I will try for you.
As I sit at the water's edge in Fernandez Bay with the waves tickling my feet I feel at peace. There is an energy here that calms my internal dialogue and allows me to sit still, internally and externally.
I can sit quietly for hours watching the soft ripples atop the water as a breeze tiptoes by. My endless to-do lists no longer matter. I do not need to be anywhere other than exactly where I am in that moment.
It is a feeling, a focus, a practice I strive for when I am not here. This feeling of complete calm, quiet, contentment in serene stillness is a gift the island gives. And this is the gift I look to share with others.
So now that I have you ready to jump on a plane down to Cat Island…keep your eyes peeled as I will be posting our August Cat Island trip dates in short order!
Next up: SUN, SEA, SAIL & YOGA! Yes, that is right, WSYT is setting sail for a week of cruising around the Leeward Islands the second week of June.
Never sailed…don't worry! You have option: I can teach you to sail or you can just kick back and enjoy the ride. If you are worried about motion sickness…got that covered too. We will be cruising around on a 40' catamaran which is much more stable than a traditional mono hull sailboat.
My personal favorite is meditating to the sound of the ocean while watching the sun set! I have been cruising around these islands for over a decade; there are some great adventures to be had around these islands as well as some great nooks to get our yoga on!
Remember, WSYT is all about taking your yoga off the beaten path!
RISHIKESH: RUSTIC, REAL, & RASA!
Lila will be teaching a daily class introducing Tantric Philosophy and Rasa Yoga. A two hour yoga asana practice will follow the discussion time to dive deeper into your asana yoga practice.
Get your yoga on Indian style with optional afternoon or evening practice at a local yoga hall. Sample and learn more about how yoga is still being taught in India!
And then there is white water rafting, hiking, local spice markets and the Taj Mahal to visit! Don't worry, I have built in 'wandering' time into the schedule for you to get lost and soak up the culture on your own. And I promise you this will be an adventure in and of itself with many stories to share. And to top it all off we will be there during Diwali, India's Festival of Lights, as well as the Rishikesh's International Yoga and Music Festival!
Yikes, I get so excited as I write about the trips WSYT has looking forward into this year. Believe it or not but we are already looking in another trip down to Ecuador to do Yoga and Service, so keep your eyes peeled for that trip to be posted up on the website soon!
So check out this year's trips and tell your friends about them! Help spread the word that Wandering Spirit Yoga Travel is here to help you do something different with your yoga practice!
The holiday season is here once again! Christmas is now less than a week away! Another chance for families to gather together and celebrate, eat and be merry. I LOVE getting together with my family.
There is the chaos of bringing multiple branches of family together under one roof. There is the family politics to navigate. There is order to keep among the kids, as well as cooking, cleaning and holiday cheer on top! It is CHOAS, not to mention limited personal time or space.
Really, don't get me wrong; I really do love holiday gatherings with my family. I just need a way to find calm amidst all the chaos in order to keep my holiday cheer going. Of course I turn to my practice for a little self care. I have found that taking 10 minutes out of my day to practice a few poses along with some breathing and short meditation supercharges me with holiday cheer (cheesy, yes I know). I could even do them in bed!
I practice this in bed right when I get up because it is the most efficient time for me and a great way to start my day. But by no means is this a mandatory time or place to bring this practice… just an alternative if you can't find a space for a yoga mat anywhere.
Begin by lying on your back and exhale your right knee up to your chest. Interlace the fingers all the way to the webbing and place this grip on your shin just below your knee joint.
Keep the back of your head grounded as you shrug your chin to your chest and look down the centre line of your body. Press the tops of your shoulders away from your ears, shoulder blades grounded beneath and elbows hugging into your waistline.
Take a deep breath in through your nose, filling your lungs up as much as you can. On the exhale, also through the nose, begin to pull your knee joint down towards your armpit. Stop when you run out of breath. Maintain the compression you have gained as you take another breath (tuck your chin in even tighter to the chest as you inhale).
Again on the exhale continue to pull the knee cap down towards the armpit. I like to imagine I might one day nestle my knee cap into my arm pit one day. Your leg should not get stuck on the top of the rib cage; it should be grazing the outside edge of the right side of the rib cage. Repeat inhale chin in, exhale pull the knee in for 3-5 rounds. Release right let down and repeat on the left leg.
Bring both legs up to the chest and wrap your arms around your legs to grab opposite elbows (if this is not possible then grabbing forearms or wrists works as well). Keep back of head grounded and chin shrugged into chest.
Take a full breath in and on the exhaled begin to hug your legs straight down into your chest and torso for the complete exhale. Hold your legs where ever you have hugged them into as you take another deep breath in tucking chin in tighter (fabulous stretch for neck from base of the skull to tops of the shoulders).
On the next exhale continue to hug your legs into your torso. Inhale maintain the compression in hips, exhale continue pulling. Do this for 3-5 rounds of breathing.
Slowly release legs to the floor and enjoy the sensation of blood rushing back into the hip joints to flush out the crud.
Note: It is pretty normal for most, especially when we are not warmed up, to feel a pinching sensation in the hips. Don't be frightened of the sensation. Use your breath to try and work through it slowly. However, if you ever feel a piercing pain back off. Sensation good, pain bad.
Lie on your back, knees bent with feet on the ground hip width distance apart. Check your feet are parallel to one another and heels as close to your sit-bones as physically possible.
Press the back of your head gently towards the ground, chin off of chest. Place your arms alongside your torso, palms facing down, shoulder blades ground and tops of shoulders pressing gently away from your ears.
Feel your feet and arms touching the ground; take a breath in and on the exhale press through your feet and arms to lift the hips and pubic bone up off the ground until thighs approximately parallel to the ground. Your bum will likely engage but do not let it be the driving action to lifting the hips and pubic bone. You do not want to tuck your tailbone.
Check that your head remains gently pressing down to the ground beneath it, chin off the chest, and tops of shoulders continue to press away from your ears. Again in this pose you should again feel a great stretch in the back of the neck. Your shoulder blades should also still be ground so you feel an opening sensation in the front of the chest.
Keep your hips, knees and ankle joints all in one line. Hug in your inner thighs, press down with the inner edges of feet your fee and engage your quadriceps to help with maintaining the alignment here.
Remember not to over use your gluteus muscles (bum muscles) otherwise you may end up tucking your tailbone. Instead focus extending your tailbone away from your head, stretching and decompressing the spine.
When you find your edge (where you feel your legs working and neck/should/chest stretching and a lot of space to fill your lungs with air) hold the pose for 30-60 seconds. Stay connected with your breath. The moment that you start holding your breath, your muscles will no longer want to work for you.
Slowly roll the spine back down to the ground and extend the legs out. Take several slow deep inhales and exhales before moving on to the next pose.
Bend your knees and place feet flat on the ground. On an exhale, lift your knees so they stack above the hip joints and lift you feet so your ankles are in line with your knee joints with shins parallel to the ground.
Bring your right hand to the outside of the left knee joint. On an exhale, slowly guide your legs over to the right hand side of your body until your right leg is resting on the floor with the left leg stacked on top of it.
If your left hip is not stacked above the right then shuffle your hips to the left so you can comfortably line your left hip atop your right.
Take your right hand and place it on top of the outside left knee joint. Inhale your left hand straight up towards the ceiling and on the exhale slowly lower it down to the ground on the left hand side of your body. You should end up with your finger tips in line with the top of your shoulder or below the shoulder joint, not above the shoulder joint.
Gently turn your gaze over to your left hand. If you feel discomfort in the neck as you look to the left, bring the back of your head slightly more towards the right before turning the head to look to the left.
Now focus on taking full lung breaths in and on the exhales relax your left shoulder down and use your right hand to gently press the left knee down. Hold on the inhale and work deeper on the exhale. Complete at least 3-5 rounds of breathing.
To release and move to the other side: bring your gaze back to center. Bring your right hand beneath your right knee joint and on an inhale slowly guide your legs back to center. Then place your left hand to the outside of the right knee and, with your exhale, slowly guide your legs to the left and repeat above.
I hope this little mini practice is helpful to you during your holiday celebrations. Share these poses with your family and friends and help spread holiday calmness as well as cheer!
The practice of hatha yoga focuses on strengthening the mind as well as the physical body. Even though both areas are important, mental strengthening is often forgotten in this physically focused practice. How does it happen? How can you strengthen your mind as you strengthen your body?
Think about the last time you fell out of a pose.
What was it that caused you to fall?
It's likely that when you began your yoga practice, you fell out of poses often. You lacked the physical strength or flexibility to successfully hold a pose. But you practiced… over and over and over again. You built up strength and flexibility one muscle strand at a time. Through the practice of returning to the mat, over and over again, you began the process of strengthening your mind, whether you were aware of it or not.
Now, poses are getting easier, yet still you find yourself falling down now and then. The cause is likely your mental focus. The moment you go from thinking about engaging your inner thigh muscles in a crescent lunge to thinking about your weekend shopping list… BOOM! you are on your bum again.
In order to work deeper into a pose, you must focus your mind pointedly on what you are doing in that moment. (I have found in my own practice that I am often unaware of my meandering mind until I fall on my bum.) The mind is not so different from the body in this sense. The more you practice honing your mind to be in the present moment, the easier it will become to stay away from meandering thoughts. Your mind will grow stronger as you nurture and grow your mental focus.
Through mental persistence combined with your physical practice, you will continue to strengthen your mind. Yes, there will be times when you will get frustrated with your practice… it happens to everyone, even teachers. Instead of giving up, you can choose to come back to the mat. The practice of returning to the mat, even after a rough practice, will nurture and strengthen your mental persistence. You will always have proverbial "wrenches" thrown into your practice, but your continual return to the mat will strengthen both you mentally and physically. That is how life works and life will reflect itself in your practice.
Each return to your practice will bring a new lesson, a new way to grow, heal or strengthen your body and mind.
What strategies do you use to focus your mind when you are on the mat?
I am a hot yogi at heart. Hot yoga is what started my love affair with yoga and it is the first yoga I was trained to teach.
Since becoming a teacher I have come to appreciate a love many different lineages, styles and flavors of yoga out there. There seems to be a never ending evolution of yoga flavors these days which both excites and concerns me, but I diverge (perhaps a future post).
When I sat down and thought about what to write for this week I felt compelled to write about what it was with hot yoga that I connected with where so many other yoga classes had fallen short with me.
My practice did begin with the Bikram series and so I start there. What hooked me, and what I believe hooked so many others – is the focused direction of how to get into and out of the poses and where I should be feeling things during the poses. There wasn't the infusion of spiritual philosophy to confuse directions of how to get into and out of poses that I had always found so distracting and annoying.
I also did not feel judged by the teachers in any way. For what I could or couldn't do in the poses and whether I held any spiritual belief in yogic philosophy (although now it is funny to me how much I am enjoying learning about yogic philosophy… I simply wasn't ready then).
It was a pure asana practice. I felt physically challenged. I felt that I had a lot to learn about the practice, and through practice I learned a lot about myself, whether I liked or not.
I am sure there are some of you saying that there are plenty of other lineages that have detailed asana direction, so why HOT yoga. There are a few ways to answer that.
Firstly, I had been sampling so many flavors of yoga for the previous 5-6 years and just getting really frustrated. When I found what I felt to be a good fit for me, I ran with it. It has been 8+ years now that I have followed this yoga path and in the past year I have finally been ready to venture out and see, sample and practice what else is out there in the yoga world, and I am have fun doing it now.
Another reason the heat made sense to me is that I came into the yoga with a lot of tightness from many years of playing sports and the heat helped me to open up and get into my areas of injury and pain. This allowed me to rehab my body from the inside out. It has not always been the most pleasant process, but it has been the most effective and the least invasive.
Lastly, I love the heat because I sweat. I don't mean just sweating in the hot yoga room, but I am just generally a sweaty person. Even when I took unheated classes, I sweat way more than the average person… so in the hot room I just blended in. It was great! No more feeling embarrassed about my profuse sweating, because everyone else was also standing in their own puddles.
So, now that you know my story of attraction to yoga… stop and take a moment to think about your story. What is it that really got you hooked into yoga? Share your stories in the comments below or on the WSYT Facebook page so others, teachers and students alike, may learn from them!
Welcome to the Wandering Spirit Yoga Travel Blog! When I sat down to think about what I wanted this blog to be about it was pretty easy… ADVENTURES! Every day is an adventure if you stop and think about it.
Over the years of learning about yoga asana, meditation and pranayama, I have come to the realization that these practices are really no different to the adventures of traveling to a new faraway country. No matter how well you plan ahead with your practice or how good your intentions are, there is always an element of the unknown until you get there. When you arrive… proceed forward with curiosity, an open mind… and work with whatever is presented to you! This philosophy holds true throughout my practice and my travels.
With this in mind I look forward to bringing you all things yoga, travel, adventure and fun such as: information about exciting places you can take your practice around the globe; yoga dorkiness on poses, practice, breath, science+yoga, fun yoga factoids to boggle your friends with and other relevant and interesting tidbits I find along the way that will contribute to your life's journey!
This is a whole new adventure for me and I am excited to have your company along the way. I have been working in the travel industry for the past ten years designing, coordinating and running international adventure programs for other people.
My love affair with yoga began in 2001, and I have been teaching since 2008. Now, I am taking my passion for yoga and my experience in the adventure travel industry and blending them together. My goal is to share my passion with all of you out there looking to do something a little different and a little crazy with your vacation, while still bringing your asana, pranayama, and meditation practice with you.
I look forward to have you join me on this journey to take yoga off the beaten path!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,724
|
package test.others;
import test.test.ITest;
import test.test.ITestHandle;
class Node {
private Node next;
public Node getNext() {
return next;
}
public void hello() {
}
public void setNext(Node next) {
this.next = next;
}
}
public class TestLinkAndArrayList {
public static void main(String[] args) {
int size = 2999 * 10000;
testArrayList(size);
// testLinkedList(size);
}
static void testArrayList(int size) {
final Node[] ns = new Node[size];
for (int i = 0; i < ns.length; i++) {
ns[i] = new Node();
}
ITestHandle.doTest(new ITest() {
@Override
public void test(int i1) throws Exception {
ns[i1].hello();
}
}, size, "TestArrayList");
}
static void testLinkedList(int size) {
Node n = new Node();
final Node r = n;
for (int i = 0; i < size; i++) {
Node t = new Node();
n.setNext(t);
n = t;
}
ITestHandle.doTest(new ITest() {
@Override
public void test(int i1) throws Exception {
Node t = r;
for (; ; ) {
t.hello();
t = t.getNext();
if (t == null) {
break;
}
}
}
}, 1, "TestLinkedList");
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,076
|
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