text:
\begin{document}

\title[The well-posedness of the Cauchy problem for the Dirac operator]{The well-posedness of the Cauchy problem for the Dirac operator on globally hyperbolic manifolds with timelike boundary}

 \thanks{Both authors are supported by the research grant { ``Geometric boundary value problems for the Dirac operator''} of the Juniorprofessurenprogramm Baden-W\"urttemberg. S.M. is partially supported within the DFG research training group GRK 1821 ``Cohomological Methods in
Geometry''.}

\subjclass[2010]{Primary 58J45; Secondary 53C50, 35L50, 35Q75}

\begin{abstract}
We consider the Dirac operator on globally hyperbolic manifolds with timelike boundary and show well-posedness of the Cauchy initial-boundary value problem coupled to MIT-boundary conditions. This is achieved by transforming the problem locally into a symmetric positive hyperbolic system, proving existence and uniqueness of weak solutions and then  using local methods developed by Lax, Phillips and Rauch, Massey to show smoothness of the solutions.
Our proof actually works for a slightly more general class of local boundary conditions.
\end{abstract}
\maketitle

\section{Introduction}

The well-posedness of the Cauchy problem for the Dirac operator on a Lorentzian manifold $(\tilde \M, g)$ is a classical problem which has been exhaustively studied in many contexts. If the underlying background is \textit{globally hyperbolic} a complete answer is known: In \cite{dimock} it was shown that a fundamental solution for the Dirac equation can obtained from a fundamental solution of the conformal wave operator via the Lichnerowicz formula. Since the Cauchy problem for the conformal wave operator is well-posed \cite{Choquet,Leray}, it follows that the Cauchy problem for the Dirac operator is also well-posed. See also \cite{bgp} for a direct treatment of Cauchy problems for Dirac operators. Even if there exists a plethora of models in physics where globally hyperbolic spacetimes has been used as background, there also exist many applications which require a manifold with non-empty boundary. Indeed, recent developments in quantum field theory focused their attention on manifolds with timelike boundary \cite{aqft,Zahn}, e.g.  anti-de Sitter spacetime \cite{Dappia1,Dappia2} and BTZ spacetime \cite{Dappia3}. Moreover, experimental setups for
studying the Casimir effect enclose (quantum) fields between walls, which may be mathematically described by introducing timelike boundaries  \cite{DappiaCas}. In these settings, the correspondence between the well-posedness of the conformal wave operator and the Dirac operator breaks down due to the boundary condition. Even if the Cauchy problem for the conformal wave operator is proved to be well-posed in a large class of boundary conditions for stationary spacetimes  \cite{Dappia},  it is still not clear how to relate this with the Cauchy problem for the Dirac operator. In fact, solvability in the Dirac case very much depends on the boundary condition, e.g. if the Dirichlet boundary condition is applied to spinors on the boundary, in general there does not exist any smooth solution to the Dirac equation.\\
The goal of this paper is to investigate the well-posedness of the Cauchy problem for the Dirac operator in globally hyperbolic manifolds with timelike boundary in the following sense:
Let $(\tilde{\M},g)$ be a globally hyperbolic spin manifold of dimension $n+1$. Globally hyperbolic as in \cite{bgp,Ge} means that the manifold admits a Cauchy surface, i.e.
a hypersurface that is hit exactly once by every inextendible timelike curve. Moreover, it can always be chosen to be smooth and spacelike, see \cite{bs}. Without further mentioning it, we always assume that our manifolds are orientable. Note that for  $n+1=4$, the case relevant for physics, such globally hyperbolic manifolds are automatically spin. Let $\N$ be a submanifold of $\tilde{\M}$ that with the induced metric is itself globally hyperbolic. Let $\tilde{\Sigma}$ be a smooth spacelike Cauchy surface of  $\tilde{\M}$. Then, $\widehat{\Sigma}\define \tilde \Sigma\cap \N$ is a spacelike Cauchy surface for $\N$. We assume that $\N$ divides $\tilde{\M}$ into two connected components. The closure of one of them we denote by $\M$ and refer to it as a \emph{globally hyperbolic manifold with timelike boundary}. On $\tilde{\M}$  we choose a
 Cauchy time function $t\colon \tilde{\M}\to \RR$.  Then $\{t^{-1}(s)\}_{s\in \RR}$ gives a foliation by Cauchy surfaces, and we set $\Sigma_s\define t^{-1}(s)\cap \mathcal{M}$.

 We always assume that we fix the spin structure on $\M$.  Let $S\M$ denote the spinor bundle over $\M$ and $S\Sigma_0$ the induced spinor bundle over $\Sigma_0$. Moreover, let $\Dir$ be the Dirac operator on $S\M$, for details see Section~\ref{preliminaries}.

  \begin{thm}\label{maintheorem}
The Cauchy problem for the Dirac operator with MIT-boundary condition on a globally hyperbolic spin manifold $\M$ with timelike boundary $\bM$ is well-posed, i.e. for any $f\in\Gamma_{cc}(S\M)$ and $h\in\Gamma_{cc}(S\Sigma_0)$   there exists a unique spacelike compact smooth solution $\psi$ to the mixed initial-boundary value problem
\begin{equation}\label{CauchyDir}
\begin{cases}{}
\Dir \psi=f   \\
\psi|_{\Sigma_0} ={h} \\
(\gamma({\n})  - \imath ) \psi|_{\bM} =0
\end{cases}
\end{equation}
 which depends continuously on the data $(f,h)$. Here the $\gamma(\n)$ denotes Clifford multiplication with ${\n}$ the outward unit normal on $\bM$ and $\Gamma_{cc}(.)$ denotes the space of sections that are compactly supported in the interior of the underlying manifold.
\end{thm}

For a subclass of stationary  spacetimes  with timelike boundary admitting a suitable timelike Killing vector field Theorem~\ref{maintheorem} was already proven in \cite{FR1}.

The \textit{MIT boundary condition}, was introduced for the first time in \cite{MIT1} in order to reproduce the confinement of quark in a finite region of space: ``Dirac waves'' are indeed reflected on the boundary. Few years later, it was used in the description of hadronic states like baryons \cite{MIT2} and mesons \cite{MIT3}. More recently, the MIT boundary condition were employed in \cite{Casimir} for computing the Casimir energy in a a three-dimensional rectangular box, in \cite{FKSY} and \cite{FR1,FR2} in order to construct an integral representation for the Dirac propagator in Kerr-Newman and Kerr Black Hole Geometry respectively, and in \cite{IR} for proving the asymptotic completeness for linear massive Dirac fields on the Schwarzschild Anti-de Sitter spacetime.
\begin{remark}\label{rem:bd}
Theorem~\ref{maintheorem} holds under more general local boundary conditions, namely replacing $\Mbc=\gamma(n)-\imath$ by any linear non-invertible map $\Mbc\colon \Gamma(S\bM) \to \Gamma(S\bM)$ with constant kernel dimension and such that $\Mbc\psi|_\bM=0$ and $\Mbc^\dagger\psi|_\bM=0$, see Remark~\ref{rem:ad} for the definition of $\Mbc^\dagger$,  both imply
\begin{equation}\label{general bound cond}
  \fiber{\psi}{\gamma(e_0)\gamma(\n)\psi}_q =0
\end{equation}
  for all $q\in \bM$, see also Remark~\ref{rem:bd2}. Here $e_0$ is the globally defined unit timelike vector field defined by~\eqref{def:e_0}.
\end{remark}

As usual,  the well-posedness of the Cauchy problem will  guarantee the existence of {\it Green operators} for the Dirac operator. In globally hyperbolic spin manifold with empty boundary, those operators play a  role in the quantization of linear field theory: Indeed they fully characterize the space of solutions of the Dirac equation \cite{DHP,dimock}, they implement the canonical anticommutation relations typical of any fermionic quantum field theory \cite{BG,BD,simo3}, and their difference, dubbed the casual propagator or Pauli-Jordan commutator,  can be used to construct quantum states \cite{FK,simo1,simo2}.

\begin{remark}The Cauchy problem~\eqref{CauchyDir} is still well-posed in a larger class of initial data. But then some compatibility condition for $f$ and $h$ on $\partial\Sigma_0$ is needed---see Remark~\ref{star}.
In the subclass of stationary spacetimes considered in \cite{FR1}, as mentioned above, this compatibility condition reduces to the one therein. Without these compatibility conditions the solution would still exist but the singularities contained in a neighborhood of $\partial\Sigma_0$ would propagate with time along lightlike geodesics.
In case, they hit again the boundary some boundary phenomena as reflection will occur.  For future work it is of course interesting to obtain  an explicit method  to  construct the  corresponding Green operators and to obtain more information on how singularities behave when hitting the boundary.
\end{remark}

Our strategy to prove the well-posedness of the Cauchy problem
is as follows: First we restrict to a  strip of finite time and prove general properties as finite propagation of speed and uniqueness in Section~\ref{sec:fps}. Using the theory of symmetric positive hyperbolic systems, see e.g. \cite{Fr2}, we investigate properties of a weak solution in Section~\ref{sec:ws}.
In Section~\ref{diffsol} we localize the problem by introducing suitable coordinates small enough that we can associate to our Dirac problem a hyperbolic system that fits into the class considered in \cite{LP, RM} and thus provides smoothness of a weak solution. Then we can use these properties to prove existence of a weak solution in a time strip.
With the help of uniqueness we can then easily glue together local solutions to obtain global ones in Section~\ref{sec:gs}.\\

\noindent \textit{Acknowledgements.}
We would like to thank Claudio Dappiaggi, Nicol\'o Drago, Felix Finster, Umberto Lupo, Oliver Petersen and Christian R\"oken for helpful discussions.

\section{Preliminaries}\label{preliminaries}

\subsection{Basis notations}
In the following we always assume that the spin structure is fixed.  We denote by $S\M$ the associated \textit{spinor bundle}, i.e. a complex vector bundle with $N\define2^{\lfloor \frac{n+1}{2}\rfloor}$-dimensional fibers, denoted by $S_p\M$ for $p\in \M$, fiberwise endowed  with the canonical scalar product on $\CC^{N}$
$$\fiber{\cdot}{\cdot}\colon S_p\M \times S_p\M \to \CC \,$$
  and with a \textit{Clifford multiplication}, i.e. a fiber-preserving map $\gamma\colon T\M\to \text{End}(S\M)$.
We denote the by $\Gamma_c(S\M)$, $\Gamma_{cc}(S\M)$, $\Gamma_{sc}(S\M)$ resp. $\Gamma(S\M)$ the spaces of compactly supported, compactly supported in the interior, spacelike compactly supported resp. smooth sections of the spinor bundle.
The \textit{(classical) Dirac operator} $\Dir \colon \Gamma(S\M) \to \Gamma(S\M)$ is defined as the composition of the connection $\nabla$ on $S\M$, obtained as a lift of the Levi-Civita connection on $T\M$, and the Clifford multiplication times $i$. Thus, in local coordinates this reads as
$$\Dir = \sum_{\mu=0}^{n} \imath \gamma(e_\mu) \nabla_{e_\mu}  $$
where $(e_\mu)_{\mu=0,\dots,n}$ is a local Lorentzian-orthonormal frame of $T\M$  and the Clifford multiplication satisfies
$ \gamma(u)\gamma(v) + \gamma(v) \gamma(u) = - 2 g(u,v)$
 for every $u,v\in T_p\M$ and $p\in\M$.

Recall that after choice of the Cauchy time function $t$ the metric on the  globally hyperbolic manifold $\tilde{\M}$ can be written as $g=g|_{\Sigma_t}-\beta^2 dt^2$ where $\beta\colon \tilde{\M}\to \RR$ is a positive smooth function. Thus, although the local orthonormal frame $e_\mu$ in general only exists locally we  have a globally defined unit timelike vector field
\begin{equation}\label{def:e_0}
e_0\define \frac{1}{\beta} \partial_t
\end{equation}   that we will use in the following. Note that in particular we have $\gamma(e_0)^{-1}=\gamma(e_0)$.

\subsection{Reformulation as a symmetric positive hyperbolic system}
In this section we will formulate the Dirac equation~\eqref{CauchyDir} as a symmetric positive hyperbolic system. For that we shortly recall the basic definition, for more details see \cite{Fr1, Fr2}.

For the following definition, let $E\to\M$ be a real or complex vector bundle with finite rank $N$ endowed with a fiberwise metric $\fiber{\cdot}{\cdot}$. Moreover, let us endow ${\Gamma_{cc}(E)}$ with $L^2$-scalar product $$\scalar{\cdot}{\cdot}_{\M}\define\int_\M \fiber{\cdot}{\cdot} \text{Vol}_\M \,,$$
where $\text{Vol}_\M$ denotes the volume element. Moreover, let $\Vert.\Vert_{L^2(\M)}\define \scalar{.}{.}_{\M}$.

\begin{definition}\label{def:symm syst}
 A linear differential operator $\oL \colon \Gamma(E) \to \Gamma(E)$ of first order is called a \emph{symmetric system} over $\M$ if
\begin{enumerate}
\item[(S)] the principal symbol $\sigma_\oL (\xi) \colon E_p \to E_p$ is Hermitian with respect to $\fiber{\cdot}{\cdot}$ for every $\xi\in T^*_p\M$ and for every $p \in \M$.
\end{enumerate}
  Additionally, we say that $\oL$ is \emph{positive} respectively \emph{hyperbolic} if it holds:
\begin{enumerate}
\item[(P)] The bilinear form $\fiber{ \,(\oL+\oL^\dagger ) \cdot}{\cdot}$ on $E_p$ is positive definite, where $\oL^\dagger$ denote the formal adjoint of $\oL$ with respect to the $L^2$-product on $\Gamma(E)$.
\item[(H)] For every future-directed timelike covector $\tau \in T_p^*\M$, the bilinear form $\fiber{\sigma_\oL (\tau) \cdot}{\cdot}$ is positive definite on $E_p$.
\end{enumerate}
\end{definition}
Let us recall that for a first-order linear operator $\oL \colon \Gamma( E) \to \Gamma( E)$ the principal symbol $\sigma_\oL\colon T^*\M \to \text{End}(E)$ can be characterized by $\oL( f u) = f \oL u + 	\sigma_\oL (d f )u$ where $u \in \Gamma(\M, E)$ and $f \in C^\infty(\M)$.
If we choose local coordinates $(t, x^1, \ldots, x^n)$ on $\M$, with $x^i$ local coordinates on $\Sigma_t$, and a local trivialization of $E$, any linear differential operator $\oL\colon \Gamma(E) \to \Gamma(E)$ of first order reads as
$$\oL\define  A_0(p) \partial_t + \sum_{j=i}^n A_j(p) \partial_{x^j} + B(p)$$
where the coefficient $A_0, A_j,B$ are $N\times N$ matrices, with $N$ being the rank of $E$, depending smoothly on $p\in\M$.
In these coordinates, Condition~(S) in Definition~\ref{def:symm syst} reduces to
$$A_0=A_0^\dagger \qquad \text{and} \qquad A_j=A_j^\dagger$$
 for $j=1,\dots, n$. Condition~(P) reads as
\begin{align}\label{def:kappa}  \kappa\define \oL+\oL^\dagger = B - \partial_t (\sqrt{g})A_0) - \sum_{j=1}^n \partial_{x^j} (\sqrt{g}A_j)>0,\end{align}
where $g$ is the absolute value of the determinant of the Lorentzian metric while condition (H) can be stated as follows: For any covector $\tau=dt + \sum_j \alpha_j dx^j$,  $$ \sigma_\oL(\tau)=A_0 + \sum_{j=1}^{N-1} \alpha_j A_j \qquad \text{is positive definite.}$$

\begin{remark}\label{Dir on halfMink}
With this definition, we can immediately notice that the Dirac operator is not a symmetric system in general. Consider for example the half Minkowski spacetime $\mathscr{M}^4\define \RR\times[0,\infty)$ endowed with the element line $$ds^2=-dt^2+dx^2+dy^2+dz^2\,.$$
In this setting the Dirac operators reads as
$$
\Dir=\imath \gamma(e_0)\partial_t+ \imath  \gamma(e_1) \partial_{x} +\imath \gamma(e_2) \partial_{y} +\imath \gamma(e_3) \partial_{z}.
$$
By straightforward computation, we obtain
\begin{itemize}
\item $\gamma (e_j)^\dagger=-\gamma(e_j)$, with $j=x,y,z$ violates condition (S);
\item $\kappa=0$ violates condition (P);
\item $\sigma_\Dir (dt)=  \gamma(e_0)$ is not positive definite, therefore condition (H) is violated.
\end{itemize}
 \end{remark}

Nonetheless, it still possible to find a fiberwise invertible endomorphism $Q \in \Gamma(\text{End}(S\M))$ such that \underline{locally} $\oK \define Q \circ \Dir + \lambda \Id$ is a symmetric positive hyperbolic system, for a suitable $\lambda \in (0,\infty)$, and such that the Cauchy problem for $\oK$ and $\Dir$ are equivalent.

\begin{lemma}\label{Equivalence Dir and K}
Consider a globally hyperbolic spin manifold $\M$ with boundary $\bM$. Let $\Dir$ be the Dirac operator and $\lambda\in \RR$. The first order differential operator $\oK:\Gamma(S\M) \to \Gamma(S\M)$ defined by
$$\oK=-\imath \gamma(e_0)\Dir + \lambda \,
\Id$$ is a symmetric hyperbolic system for all $\lambda\in\RR$ and its Cauchy problem
\begin{equation}\label{CauchyK}
\begin{cases}{}
{\oK }\Psi=\f \in \Gamma_{c}(S\M) \\
\Psi|_{\Sigma_0} = \mathfrak{h}\in \Gamma_{c}(S\Sigma_0) \\
{\oM}\Psi|_{\bM} =0
\end{cases}
\end{equation}
is equivalent to the Cauchy problem  for the Dirac operator
\begin{equation}\label{CauchyDir2}
\begin{cases}{}
\Dir \psi=f \in \Gamma_c (S\M)  \\
\psi|_{\Sigma_0} ={h}\in \Gamma_c(S\Sigma_0) \\
\Mbc \psi|_{\bM} =0.
\end{cases}
\end{equation}
Moreover, for any compact set $\mathcal{R}\subset \M$, there exists a $\lambda$ such that $\oK$ is a symmetric positive hyperbolic system.
\end{lemma}
\begin{proof}
Since the $-\imath\gamma(e_0)\Dir$ is a symmetric hyperbolic system, see e.g. \cite[Chapter~3]{Nicolas}, also $\oK$ satisfies condition (S) in Definition~\ref{def:symm syst}. Next we show that the Cauchy problem for $\oK$ and for $\Dir$ are equivalent for any $\lambda\in \RR$. We set $\Psi=e^{-\lambda t} \psi$, which implies $\h=e^{-\lambda t} h$, and $\f=e^{-\lambda t} \gamma(e_0) f$ and obtain
\begin{align*}
\oK\Psi=\oK(e^{-\lambda t} \psi) = (-\imath\gamma(e_0)\Dir + \lambda\Id)(e^{-\lambda t}\psi) = -\imath e^{-\lambda t} \gamma(e_0) \Dir \psi = e^{-\lambda t} \gamma(e_0) f.\end{align*}
Moreover, since $e^{-\lambda t} \neq 0$ for all $t\in\RR,$ we can conclude
$$\oM \Psi|_\bM =   e^{-\lambda t} \Mbc  \psi|_{\bM} = 0 \quad \text{if and only if} \quad \Mbc\psi|_\bM =0. \qedhere
$$

Let now $\mathcal{R}\subset \M$ be compact. It remains to check condition (P). The operator $\kappa$, defined in \eqref{def:kappa}, is a zero order operator given by multiplication by smooth functions. Therefore, for any compact set $\mathcal{R}$, $\kappa|_{\Gamma(S\mathcal{R)}}$ is bounded and there exists a suitable $\lambda$ such that $\oK$ is positive definite on $\mathcal{R}$.
\end{proof}

\begin{example}\label{example2}
Let us consider the half Minkowski spacetime $\mathscr{M}^4$ and the Dirac operator
$$
\Dir=\imath \gamma(e_0)\partial_t+ \imath  \gamma(e_1) \partial_{x} +\imath \gamma(e_2) \partial_{y} +\imath \gamma(e_3) \partial_{z} \,.
$$
Here the $\gamma$-matrices are given in Remark~\ref{Dir on halfMink}.
We can see that the operator
$$\oK=-\imath \gamma(e_0)\Dir+ \lambda\Id$$
is a \emph{symmetric positive hyperbolic system} for any $\lambda>0$ on account of
\begin{gather*}
 \big(\gamma(e_0)\gamma(e_i)\big)^\dagger= \gamma(e_i)^\dagger \gamma(e_0)^\dagger = - \gamma(e_0)^\dagger\gamma(e_i)^\dagger =  \gamma(e_0)\gamma(e_i) \qquad \text{ for all $i=1,2,3$,} \\[0.75em]
\kappa= \lambda \Id_{4\times 4} {\qquad\text{and} \qquad  \sigma_\oK(dt)=\Id_{4\times4}\,.}
\end{gather*}
 \end{example}

\section{Local well-posedness of the Cauchy problem}

 Let $\Dir$ be the Dirac operator on our globally hyperbolic spin manifold with timelike boundary $\bM$.  We denote by $\T$ the \textit{time strip} given by
$$\T\define t^{-1}([0,T])$$
where $t\colon \M\to \RR$ is the chosen Cauchy time function.
 Let $\lambda$ a suitable constant in $\RR$ such that the operator $\oK\colon \Gamma(S\M) \to \Gamma(S\M)$ defined by \begin{equation}\label{eq:K}\oK=-\imath\gamma(e_0)\Dir + \lambda \,
\Id\end{equation}
is a symmetric positive hyperbolic system on
$$\R\define J^-( \mathcal{O})\cap \mathcal{T}$$
 where $\mathcal{O}$ is a compact subset of $\Sigma_T$ and $J^-(\mathcal{O})$ (resp. $J^+(\mathcal{O})$) denotes the \textit{past set} (resp. \emph{future set}), namely the set of all points that can be reached by past-directed resp. future-directed causal curves emanating from a point in $\mathcal{O}$. This is always possible by Lemma~\ref{Equivalence Dir and K} since  $\R\subset \M$ is compact. For the reason why we choose $\R$ as above compare below---especially Theorem~\ref{thm:Weak existence}.

\subsection{Uniqueness and finite propagation speed}\label{sec:fps}

In order to show existence and uniqueness of weak solutions for the Dirac Cauchy problem~\eqref{CauchyDir2}, we first shall derive the so called \textit{``energy inequality''}. This estimate has a clear physical interpretation, as we shall see in Proposition~\ref{prop:finite}: Any solution can propagate with at most speed of light.

\begin{lemma}\label{lemma: energy estimates} Let $\R$ be as above and choose the $\lambda$ in $\oK$ such that it defines a symmetric positive hyperbolic system on $\R$. Let $\Psi\in \Gamma(S\T)$ satify $\Psi|_{\Sigma_0}=0$ and $\oM\Psi|_{\bM}=0$. Then $\Psi$ satisfies the \emph{energy inequality}
\begin{equation}\label{Energy Inequality}
\|\Psi\|_{L^2(\R)}  \leq c \| \oK \Psi\|_{L^2(\R)}\,.
\end{equation}
for a constant $c>0$ independent on $\Psi$.
\end{lemma}

\begin{proof}

By Green identity we obtain
\begin{equation}\label{Green identity}
\scalar{\Psi}{ \oK \Psi }_\R - \scalar{\oK^\dagger \Psi }{ \Psi }_\R   =  \scalar{\Psi}{ \gamma(e_0)\gamma(\n) \Psi }_{\bR}
\end{equation}
where the $\gamma(\cdot)$ denotes Clifford multiplication, $\n$ is outward normal vector to $\bR$, $\oK^\dagger$  is the formal adjoint of $\oK$, and $\scalar{\cdot}{\cdot}_{\bR}$ is the induced $L^2$-product on ${\bR}$.
Subtracting $ 2 \scalar{\Psi }{ \oK \Psi}_\R$ from the latter equation, we thus obtain
\begin{equation}\label{passaggio1}
\begin{aligned}
\scalar{\Psi}{ \gamma(e_0)\gamma( \n) \Psi }_{\partial \R } - 2 \scalar{\Psi}{ \oK\Psi}_\R  &= -\scalar{\Psi }{ \oK \Psi}_\R - \scalar{\Psi}{ \oK^\dagger\Psi}_\R  \\
 &=-\scalar{\Psi}{ (\oK +\oK^\dagger )\Psi}_\R  \\
& \leq   - 2c\scalar{\Psi}{ \Psi}_\R \,,
\end{aligned}
\end{equation}
for some $c>0$ where in the last inequality we used that $\oK$ is a symmetric positive system. Next, let us decompose the boundary $\bR$ as
$$\bR=\mathcal{O} \cup \Big( \Sigma_{0}\cap J^-(\mathcal{O})\Big) \cup Y\,,$$
where $Y\define  \partial J^-(\mathcal{O})\cap \mathcal{T}$, loosely speaking $Y$ is the boundary of the light cone inside the time strip $\T$.
The boundary terms on $\Sigma_{0}\cap J^-(\mathcal{O})$ vanish by assumption on $\Psi$. Hence, we will have non zero boundary contributions only at $\mathcal{O}$ and $Y\subset\bR$. To deal with these terms, we decompose further the boundary $Y$ in
$$Y= (Y\cap \bM) \sqcup \big(Y \setminus (Y\cap \bM) \big).$$
Notice that if $\bR\cap\bM =\emptyset$ then $Y$ reduces to the boundary of the light cone $J^-(\mathcal{O})$ in the interior of $\R$. Imposing the boundary condition $\oM \Psi|_\bM=0$, we have  by condition~\eqref{general bound cond} that
$$\scalar{\Psi}{\gamma(e_0) \gamma(\n) \Psi}_\bR =0\,. $$
Therefore, the boundary terms on $Y\cap \bM$ vanish. In this way, the Green identity~\eqref{Green identity} reduces to
\[\scalar{\Psi}{ \oK \Psi }_\R - \scalar{\oK^\dagger \Psi }{ \Psi }_\R   = \scalar{\Psi}{ \gamma(e_0)\gamma( \n) \Psi }_{\mathcal{O} } + \scalar{\Psi}{ \gamma(e_0)\gamma(\n) \Psi }_{Y \setminus (Y\cap \bM) }.\]
Let us remark that the right hand side of the latter equation is non negative definite: Indeed, since  $\mathcal{O}$ is a spacelike hypersurface, $\scalar{.}{ \gamma(e_0)\gamma(\n) . }_{\mathcal{O} }$ is an inner product by \cite[Prop. 1.1]{dimock} and, thus,
$$
\scalar{\Psi}{ \gamma(e_0)\gamma(\n) \Psi }_{\mathcal{O} } >0 \,.
 $$
By continuity, also the contribution on  $Y \setminus (Y\cap \bM)$ is still positive semidefinite since  $Y \setminus (Y\cap \bM)$ is a lightlike hypersurface. Hence, plugging this consideration into Inequality~\eqref{passaggio1} we obtain
$$2\scalar{\Psi}{\lambda \Psi}_\R \leq  2 \scalar{\Psi}{ \oK\Psi}_\R.$$
Thus, by Hölder inequality for all $\Psi\in  \Gamma(S\T)$ with $\Psi|_{\Sigma_0}=0$ and $\oM\Psi|_{\partial \M}=0$ we have
\[\|\Psi\|_{L^2(\R)} \leq \lambda^{-1} \|\oK\Psi \|_{L^2(\R)}.\qedhere\]
\end{proof}

We are now ready to prove that if there exists a solution to the Cauchy problem~\eqref{CauchyK}, then it propagates with at most speed of light.

\begin{proposition}[Finite speed of propagation]\label{prop:finite}
Any solution to the Dirac Cauchy problem~\eqref{CauchyDir} propagates with at most speed of light, i.e. its support on $\T$ is inside the region
$$ \mathcal{V}\define \Big( J^+\big(\supp\, f \cap \T \big)\cup J^+(\supp h)\Big) \cap \mathcal{T},$$
see Figure~\ref{fig:finite speed}.
\end{proposition}

\begin{proof}
Consider any point $p$ outside the region $\mathcal{V}$.
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 \draw[]  (1.0,0.25) .. controls (2,1) and (6,-1) .. (9,0) ;
 \draw[]  (0.8,2) .. controls (2,2.75) and (6,0.75) .. (9,1.75);
  \draw[]  (1.0,0.25) .. controls (0.75,0.9) and (1.2,1.75) .. (0.8,2);
    \draw[dashed]  (0.9,-0.5) .. controls (1.1,-0.15) .. (1.0,0.25);
     \draw[dashed]  (0.8,2) .. controls   (0.5,2.35) .. (0.6,2.75);

  \draw[decorate, decoration=snake, segment length=5,very thick] (2,0.38)  -- ( 4,0.1)  node[midway,below,rotate=-5] {$\,\,\supp \h$}  ;

            \draw[]  plot[smooth, tension=1] coordinates { (4.5,0.8) (5,0.58) (6,0.25) (6.6,1) (6,2) (4.5, 1.5)   (4.5,0.8) };

\draw[red!80] (2,0.4) -- (0.97,1.45);
\draw[red!80] (3.99,0.1) -- (4.7,0.7);
\draw[red!80] (6.38,0.4) -- (7.4,1.47);

\draw[blue!80] (7.8,1.15) -- (7.1,-0.3);
\draw[blue!80] (7.8,1.15) -- (8.7,-0.1);

            \node  at (8,0.85) [label={$p$}] {};
            \node  at (7.8,1.15) {$\mathsmaller{\mathsmaller{\bullet}}$};
            \node  at (3.5,0.35) [label={$q$}] { };
            \node  at (3.3,0.65) {$\mathsmaller{\mathsmaller{\bullet}}$};
            \node  at (2.25,1) [label={${\mathcal{V}}$}] { };
            \node  at (7.85,-0.1) [label={$\R$}] { };
            \node  at (9.5,-0.45) [label={$\Sigma_0$}] { };
            \node  at (9.5,1.35) [label={$\Sigma_T$}] { };

    \end{tikzpicture}
    \caption{Finite propagation of speed.}
    \label{fig:finite speed}
\end{figure}
Then there exists a $\lambda$ such that $\oK$ is a symmetric  positive hyperbolic system on $\R=\mathcal{T}\cap J^-(p)$.
Let $\Psi=e^{-\lambda t}\psi$. Then, by Lemma~\ref{Equivalence Dir and K} $\Psi$ is a solution to \eqref{CauchyK} with $\f=e^{-\lambda t} \gamma(e_0)f$ and $\h= e^{-\lambda t}h$.
By Lemma~\ref{lemma: energy estimates} and $\h|_{\R\cap \Sigma_0}\equiv 0$, $\Psi$ vanishes in $\R$. Hence, $\Psi$ vanishes outside $\mathcal{V}$.
\\
The finite propagation of speed of a solution of the Dirac Cauchy problem~\eqref{CauchyDir2} then follows by Lemma~\ref{Equivalence Dir and K}.
\end{proof}

Next, we prove that if a solution exists, then it is unique.
\begin{proposition}[Uniqueness]\label{prop:unique}
Suppose there exist $\Psi,\Phi \in \Gamma(S\T)$ satisfying the same Cauchy problem~\eqref{CauchyK} for some $\f\in \Gamma_{cc}(S\M)$ and $\h\in \Gamma_{cc}(S\Sigma_0)$. Then $\Psi=\Phi$.
In particular, this also gives uniqueness of solution for the Dirac Cauchy problem~\eqref{CauchyDir2}.
\end{proposition}
\begin{proof}
Since $\Psi$ and $\Phi$ satisfy the same initial-boundary value problem~\eqref{CauchyK}, then $\Psi-\Phi\in \Gamma(S\T)$ is a solution of
$$\begin{cases}{}
\oK (\Psi-\Phi)=0  \\
(\Psi - \Phi)|_{\Sigma_0}=0 \\
\oM(\Psi-\Phi)|_{\bM} =0
\end{cases}
$$
By Proposition~\eqref{prop:finite}, the supports of $\Psi$ and $\Phi$ are contained in $\R$ for $\mathcal{O}\define \mathcal{V}\cap \Sigma_T$. Therefore, we can use Lemma~\ref{lemma: energy estimates} to conclude that $\Psi-\Phi$ is zero.

The uniqueness of the Dirac Cauchy problem~\eqref{CauchyDir2} then follows by Lemma~\ref{Equivalence Dir and K}.
\end{proof}

We conclude this section by deriving an energy inequality for the formal adjoint of $\oK$ analogous to~\eqref{Energy Inequality}. Here we restrict to the set $\Rv$  defined by
\begin{equation}\label{defRv}
\Rv\define \mathcal{T}\cap J^+(\mathcal{O}')
\end{equation}
for a compact domain $\mathcal{O}'\subset \Sigma_0$ with smooth boundary.
Using the same arguments as in the proof of Lemma~\ref{lemma: energy estimates}, we can conclude an analog energy inequality for the adjoint:
\begin{remark}\label{rem:ad}
 We denote by $\oK^\dagger$  the formal adjoint of $\oK$, i.e. the operator defined by
 \[ \scalar{\Phi}{\oK \Psi}_{\T}=\scalar{\oK^\dagger \Phi}{\Psi}_ {\T}
 \]
 for all $\Phi\in \Gamma_c(S\T)$ and $\Psi\in \Gamma_{cc}(S\T)$. Moreover, let $\oK^*$ be the adjoint of $\oK$ where $\text{dom} \oK\define \{ \Phi\in \Gamma_{c}(S\T)\ |\ \oM \Phi|_{\partial \M}=0, \Phi|_{\Sigma_0}=0\}$. By the Green identity~\eqref{Green identity} and the definition of an adjoint operator, it is
 \begin{align*} \text{dom} \oK^*&\define \Big\{\Phi\in \text{dom} \oK^\dagger_{\text{max}}\ \Big|\ \int_{(\partial \M \cap \T) \cup \Sigma_T} \fiber{\Phi}{\gamma(e_0)\gamma(n)\Psi}=0\  \forall \Psi\in \text{dom} \oK\Big\}
 \end{align*}
 where $\text{dom} \oK^\dagger_{\text{max}}$ is the maximal domain of $\oK^\dagger$.
 Note that
 $$\{\Phi\in \Gamma_c(S\T)\ |\ \Phi|_{\Sigma_T}=0,\ \int_{\partial \M \cap \T} \fiber{\Phi}{\gamma(e_0)\gamma(n)\Psi}=0\  \forall \Psi\in \text{dom} \oK\}$$ is a core of $ \oK^*$.  We set $(1-\oM^\dagger)\colon \S_q (\T\cap \partial \M)\to \S_q (\T\cap \partial \M)$ to be the orthogonal projection to
$$\{\Phi\in S_q(\T\cap \partial \M)\ |\fiber{\Phi}{\gamma(e_0)\gamma(n)\Psi}_q=0\ \forall \Psi\in  S_q(\T\cap \partial \M) \text{ with } \oM\Psi=0\}\,.$$
 We call $\oM^\dagger$ the \emph{adjoint boundary condition}, compare \cite[Sect.~7.2]{BB} for the analog notations in the elliptic case.
Using this notation, we see that
\begin{align*}\label{eq:domK}\{\Phi\in \Gamma_c(S\T)\ |\ \Phi|_{\Sigma_T}=0,\ \oM^\dagger \Phi=0 \}\end{align*} is a core of $\oK^*$.
Note that for $\oM=\gamma(n)-\imath$ being the MIT boundary condition then $\oM^\dagger=\oM$ --- see also Remark~\ref{rem:bd2}.
\end{remark}

\begin{lemma}\label{lemma: energy estimates2}
Let be $\Phi\in \Gamma(S\T)$ satisfying $\Phi|_{\Sigma_T}=0$ and $\oM^\dagger\Phi|_{\bM}=0$. Then $\Phi$ satisfies the inequality
\begin{equation}\label{Energy Inequality2}
\|\Phi\|_{L^2(\mathcal{\Rv})}  \leq C \| \oK^\dagger \Phi\|_{L^2(\mathcal{\Rv})}\,.
\end{equation}
\end{lemma}

\subsection{Weak solutions in a time strip}
\label{sec:ws}

In the following we will focus on the Cauchy problem  \eqref{CauchyK} for  $\h\equiv 0$.

\begin{remark}
Notice that choosing zero initial data is not a loss of generalities. In fact, for any Cauchy problem with any nonzero initial data $\mathfrak{h}\in \Gamma_{cc}(S\Sigma_0)$ there exists an equivalent Cauchy problem with zero initial data, namely
\begin{equation*}
\begin{cases}{}
\oK \Psi=\f  \\
\Psi|_{\Sigma_0} = \mathfrak{h} \\
\oM\Psi|_{\bM} =0
\end{cases} \, \Longleftrightarrow \quad \begin{cases}{}
\oK \tilde\Psi=\tilde{\mathfrak{f}} \\
\tilde\Psi|_{\Sigma_0} = 0  \\
\oM \tilde\Psi|_{\bM} =0
\end{cases}
\end{equation*}
{for any $\f,\tilde\f\in\Gamma_{c}(S\M)$}. Here $\tilde{\mathfrak{f}}(t,{x})\define \f(t,{x}) + \lambda\mathfrak{h}({x}) + \sum_{j=1}^n \gamma(e_0)\gamma(e_j) \nabla_{x^j} \mathfrak{h}$ and $\tilde{\Psi}(t,{x})\define \Psi(t,{x})-\mathfrak{h}({x})$. Note that $\f\in\Gamma_{cc}(S\M)$ if and only if $f\in\Gamma_{cc}(S\M)$.
\end{remark}

With the help of the Energy inequality~\eqref{Energy Inequality} we shall prove the existence of a \textit{weak solution} for the mixed initial-boundary value problem~\eqref{CauchyK} for $\oK$ as in~\eqref{eq:K} (for fixed  $\f\in\Gamma_{cc}({S\M})$ and $\h\equiv 0$).
To this end, let $\lambda$ such that $\oK^\dagger$ is a symmetric positive hyperbolic system on $\Rv=\T\cap J^-(\mathcal{O}')$ for $\mathcal{O}'\define J^-(\V\cap \Sigma_T) \cap \Sigma_0$
with
\begin{equation}\label{def:V}
\mathcal{V}\define J^+\big(\supp\,\f \cap \T \big) \cap \mathcal{T}\,.
\end{equation}
We denote by  $\fiber{\cdot}{\cdot}$ the canonical fiberwise scalar product in the spinor bundle  and by
\[\mathscr{H} \define \overline{\big({\Gamma_c(S\T)},{\scalar{.}{.}}_{\T} \big)}^{\scalar{.}{.}_{\T}} \]
 the $L^2$-completion of  $\Gamma_c(S\T)$.

\begin{definition}\label{def:Weak existence}
We call $\Psi\in\cH$ a \emph{weak solution} to the Cauchy problem~\eqref{CauchyK} with $\f\in \Gamma_{cc}(S\M)$ and $\h\equiv 0$, restricted to $\T$, if the relation
\begin{equation}\label{scalprod}
\scalar{\Phi}{\f}_{\T}=\scalar{\oK^\dagger \Phi}{\Psi}_{\T}
\end{equation}
holds for all  $\Phi \in \Gamma_{c}(S\T)$ satisfying $\oM^\dagger\Phi|_{\bM }=0$  and $\Phi|_{\Sigma_T}\equiv 0$.
\end{definition}

\begin{remark}
 Let $\Psi$ be a weak solution as above that is even smooth. Then by testing only with $\Phi\in \Gamma_{cc}(S\T)$ we immediately obtain with the Green identity that $\oK \Psi=\f$. Using this in \eqref{scalprod} with the Green identity for general $\Phi\in \Gamma_{c}(S\T)$ gives
 \[ 0= \scalar{\Phi}{\gamma(e_0)\gamma(n)\Psi}_{(\T\cap \partial\M) \cup \Sigma_0\cup \Sigma_T}. \]
 The part on $\Sigma_T$ vanishes due to $\Phi_{\Sigma_T}\equiv 0$. Moreover, using test functions $\Phi$ that have support near $\Sigma_0$ resp. $\partial \M\cap \T$ one sees as in Remark~\ref{rem:ad} that $\Psi_{\Sigma_0}=0$ resp. $\oM \Psi|_{\partial \M}=0$.
\end{remark}

\begin{remark}
 Analogously a $\Psi\in\cH$ is a weak solution to \eqref{CauchyK} with $\h\in \Gamma_{cc}(S\Sigma_0)$  if the relation
\begin{equation*}
\scalar{\Phi}{\f}_{\T}=\scalar{\oK^\dagger \Phi}{\Psi}_{\T} +\scalar{\Phi}{\gamma(e_0)\gamma(n)\h}_{\Sigma_0}
\end{equation*}
holds for all  $\Phi \in \Gamma_{c}(S\T)$ satisfying $\oM^\dagger\Phi|_{\bM }=0$  and $\Phi|_{\Sigma_T}\equiv 0$.
\end{remark}

\subsubsection{Differentiability of the solutions}\label{diffsol}

Before examining existence of a weak solution, we want to show that any weak solution is a strong solution and in particular it is smooth. Since this is a local question, we want to use the theory for hyperbolic systems on subsets of $\mathbb R^{n+1}$---in particular \cite[Section 1]{LP} that a weak solution is a (semi-)strong solution and the regularity estimates for strong solution in \cite[Theorem~3.1]{RM}. Since the definition of strong solution in the sense Lax--Phillips does not coincide with the one given by Rauch--Massey, we will denote it as semi-strong solution.
\begin{definition}\label{def:strong sol}
Let $U\subset\M$ be a compact subset in $\M$ with timelike boundary $\bM$.
We say that $\Psi\in \mathscr{H}$ is a \emph{semi-strong solution} of the initial-boundary value problem~\eqref{CauchyK} if there exists a sequence of sections $\Psi_k \in W^{1,2}(\Gamma(SU))$ such that $\oM\Psi_k=0$ on $\partial\M \cap U$ and
$$ \| \Psi_k - \Psi \|_{L^2(U)} \xrightarrow{k\to \infty} 0 \quad \text{ and } \quad \| \oK\Psi_k - \f \|_{L^2(U)} \xrightarrow{k\to \infty} 0.$$ The solution is called \emph{strong} if additionally the sequence $\Psi_k$ can chosen to be smooth.
 \end{definition}

 We concentrate on points in the boundary $p\in  \partial \M$ (the other points will even be easier since we do not have to care about boundaries) and firstly define a convenient chart as follows, compare Figure~\ref{fig:chart}.
\begin{figure}[h!]
\begin{tikzpicture}

\filldraw[fill=blue!20!white] (0.75,0.45) --(2,0.47) -- (3.25,1.2) -- (3.25,1.2) .. controls (3.45,2.7) .. (3.25,4.2) -- (3.25,4.2) --(2,4.2) -- (0.75,3.47) .. controls (0.95,1.95) .. (0.75,0.45) ;

 \draw (0,0) .. controls (4,0.1) and  (5,-0.1) .. (7,-0.2) ;
 \draw (2.5,1.5) .. controls  (6.5,1.6) and  (7.5,1.4) .. (8.5,1.3) ;
\draw[very thick] (0,0) -- (0.75,0.45);
\draw[dashed, very thick] (2,1.2) -- (2.5,1.5);
\draw[blue,dashed ] (0.75,0.45) -- (2,1.2) ;
\draw[blue, ] (0.75,0.45) -- (2,0.47) ;
\draw[blue, dashed] (2,1.2) -- (3.25,1.2) ;
\draw[blue, ] (2,0.47) -- (3.25,1.2) ;

 \draw (0,3) .. controls  (5,3.1) and  (6,2.9)  .. (7,2.8) ;
\draw (2.5,4.5)..controls (7.5,4.6) and (8.5,4.4) .. (8.5,4.3) ;
\draw[very thick] (0,3) -- (0.75,3.45);
\draw[very thick] (2,4.2) -- (2.5,4.5);
\draw[blue,] (0.75,3.45) -- (2,4.2) ;
\draw[blue, ] (0.75,3.45) -- (2,3.47) ;
\draw[blue, ] (2,4.2) -- (3.25,4.2) ;
\draw[blue, ] (2,3.47) -- (3.25,4.2) ;

 \draw  (0,0) .. controls (0.2,1.5) .. (0,3);
\draw[dashed]   (2.5,1.5) .. controls (2.7,3) .. (2.5,4.5);

\draw[dashed,blue,]  (0.75,0.45) .. controls (0.95,1.95) .. (0.75,3.47);
\draw[dashed,blue,]  (2,0.47) .. controls (2.2,1.97) .. (2,3.47);
\draw[blue,]  (2,0.47) .. controls (2.2,1.97) .. (2,3.47);
\draw[dashed,blue,]  (3.25,1.2) .. controls (3.45,2.7) .. (3.25,4.2);

            \node  at (1.6,0.95) {$\mathsmaller{{\bullet}}$};
            \node  at (1.6,0.70) {$p$};
            \node  at (1.6,3.95) {$\mathsmaller{{\bullet}}$};
            \node  at (2.1,3.95) {$\varrho(t)$};
\draw[dashed,black, thick]  (1.6,0.95)  .. controls (1.8,2.45) .. (1.6,3.95);
\draw[black, very thick,->]  (1.6,0.95)  -- (1.7,1.55);
            \node  at (1.4,1.3) {$\bf v$};

            \node  at (6,0.5) {$\Sigma_p$};
            \node  at (6,3.5) {$\Sigma_{\varrho(t)}$};

    \draw[dashed,blue, ,->]  (-0.7,1) .. controls (0.75,1) .. (1.39,0.65) ;
     \node  at (-1.5,1.1) {$\widehat{B}_\epsilon(p)$};

    \draw[dashed,blue, ,->]  (-0.7,4) .. controls (0.75,4) .. (1.39,3.65) ;
     \node  at (-1.5,4.1) {$\widehat{B}_\epsilon(\varrho(t))$};

            \node  at (-0.4,0.5) {$\widehat{\Sigma}_c$};
    \draw[black,thick ,->]  (-0.3,0.5)  .. controls (0,0.55) .. (0.4,0.3) ;

            \node  at (-0.5,3.3) {$\widehat{\Sigma}_{\gamma(t)}$};
   \draw[black, thick,->]  (-0.3,3.4)  .. controls (0,3.45) .. (0.4,3.3) ;

    \end{tikzpicture}
    \caption{Fermi coordinates on each Cauchy surface.}
    \label{fig:chart}
\end{figure}
 Let ${\Sigma}_p$ be the Cauchy surface of $\M$ to which $p$ belongs to. For $p\in \partial \M$ let $\widehat{\Sigma}_p \define \Sigma_p\cap \partial \M$ be the corresponding Cauchy surface in the boundary. Let $\varrho\colon [0,\epsilon]\to \partial \M$ be the timelike geodesic in $\partial \M$ starting at $p$ with velocity ${\bf v} \in T_p\bM$ where ${\bf v}$ is a normalized, future-directed, timelike vector perpendicular to $\widehat{\Sigma}_p$. \label{def:U_p}
 Let $\widehat{B}_\epsilon (\varrho(t))$ be the $\epsilon$-ball in $\hat{\Sigma}_{\varrho(t)}$ around $\varrho(t)$. On this balls we choose geodesic normal coordinates $\widehat{\kappa}_t\colon B_\epsilon^{n-1}(0)\to \widehat{B}_\epsilon (\varrho(t))$. Moreover, inside each $\Sigma_{\varrho(t)}$ we choose Fermi coordinates with base $\widehat{B}_\epsilon (\varrho(t))$
  and thus get a chart in $\Sigma_{\varrho(t)}$ around $\varrho(t)$ as
\begin{align*}
\widetilde\kappa_t\colon B_\epsilon^{n-1}(0)\times [0,\epsilon]\subset \mathbb R^{n}&\to U_{\epsilon} (\widehat{B}_\epsilon (\varrho(t)))\define \{ q\in \Sigma_{\varrho(t)}\ |\ \mathrm{dist}_{\Sigma_{\varrho(t)}}(q, \widehat{B}_\epsilon (\varrho(t))\leq \epsilon\}
\\
 (y,z) & \mapsto \exp^{\perp, \Sigma_{\varrho(t)}}_{\widehat{\kappa}_t(y)}(z)
\end{align*}
where $\exp^{\perp, \Sigma_{\varrho(t)}}_{\widehat{\kappa}_t(y)}(z)$ is the exponential map in $\Sigma_{\varrho(t)}$ starting at $\widehat{\kappa}_t(y)$ with velocity perpendicular to $\widehat{\Sigma}_{\varrho(t)}=\partial \Sigma_{\varrho(t)}$ pointing in the interior and magnitude $z$. For more details on the geometric maps compare \cite{GS}.

   Putting all this together we obtain a chart
  \begin{align*}
   \kappa_{p} \colon [0,\epsilon] \times B_\epsilon^{n-1}(0)\times [0,\epsilon] \subset \mathbb R^{n+1}&\to U_p\define \bigcup_{t\in [0,\epsilon]} U_\epsilon (\widehat{B}_\epsilon (\varrho(t))) \subset \M\\
   (t,y,\bar{z})&\mapsto \widetilde\kappa_t(y,\bar{z}).
  \end{align*}
    Note that sections of the spinor bundle $SU_p$ are now simply vector-valued functions $U_p\to \mathbb C^N$ where $N$ is the rank of the spinor bundle.\\
  In these coordinates our Cauchy problem will take the form as in \cite{LP, RM}. To see this, let us first consider the model case of `half' of the Minkowski space.

 \begin{example}\label{ex:mink}
  Let $\tilde{\mathscr{M}}^n$ be the Minkowski space with coordinates  $x=(x^0, \ldots, x^n)$. We set $t\define x^0$, $y=(x^1,\ldots, x^{n-1})$ and $z\define x^n$. For $|a|<1$ the hypersurface $\N_a\define \{z=at\}$ is timelike and $\mathscr M^n_a=\{ z\geq at\}$ is a globally hyperbolic manifold with a timelike boundary. We set $\tilde{z}\define z-at$ and use $(t,y,\tilde{z})$ as new coordinates on $\mathscr M^n_a$.
  Then, together with $\gamma(e_0)=\gamma(e_0)^{-1}$, we have \begin{align*}
  -\i \gamma(e_0)\Dir &= \partial_t +\sum_{j=1}^{n-1} \gamma (e_0)\gamma (e_j)\partial_{x^j}+\gamma ({e_0})\gamma ({e_n})\partial_{z}\\
  &=\big(1-a\gamma ({e_0})\gamma ({e_n})\big)\partial_t +  \sum_{j=1}^{n-1} \gamma ({e_0})\gamma ({e_j})\partial_{x^j}+\gamma ({e_0})\gamma ({e_n}){\partial_{\tilde{z}}}.
  \end{align*}
  Since $a<1$, the coefficient in front of $\partial_t$ is invertible.
  Thus, $\widehat{\oK}\define (1-a\gamma ({e_0})\gamma ({e_n}))^{-1}\oK$ with $\oK=-\imath\gamma(e_0)\Dir+\lambda \mathrm{Id}$, as in Example~\ref{example2}, is just given by
  \begin{equation}\label{eq:Pform}
   \widehat{\oK}= \partial_t + \underbrace{\sum_{j=1}^{n-1} A_j(x)\partial_{x^j} + A_{\tilde{z}}(x) \partial_{\tilde{z}} + B(x)}_{=: -\oH}.
  \end{equation}
  Since $A_{\tilde{z}}(x)=\big(1-a\gamma ({e_0})\gamma ({e_n})\big)^{-1}\gamma ({e_0})\gamma ({e_n})$ is in particular nonsingular on $\partial \mathscr M^n_a$ and since $\ker\oM|_q$ varies smoothly with $q\in\partial \mathscr M_a^n$, after restricting to some cube in $\mathscr M^n_a$ we are exactly in the situation considered in \cite{LP, RM}.
 \end{example}

 \begin{corollary}\label{cor_Up}
  For the Dirac operator $\Dir$ on a globally hyperbolic spin manifold with timelike boundary $\bM$ and $p\in \partial \M$, there is a sufficiently small $\epsilon >0$ such that in the coordinates $\kappa_p$ from above there is an invertible operator $\oE\colon \Gamma(U_p, \mathbb C^N)\to \Gamma(U_p, \mathbb C^N)$ such that $\oE\oK$ has the form~\eqref{eq:Pform} with $A_{\bar{z}}$ nonsingular on the boundary. In particular, any weak solution of the Cauchy problem~\eqref{CauchyK} gives rise to a weak solution to the Cauchy problem
    \begin{equation}\label{CauchyK_2}
\left\{\begin{matrix}
\widehat{\oK}\define \oE\oK \Psi&=\f \in \Gamma_{c}(SU_p) \\
\Psi|_{V_p}& = \mathfrak{h}\in \Gamma_{c}(SV_p) \\
\oM\Psi|_{\bM} & =0 \qquad\qquad\quad
\end{matrix} \right.
\end{equation}
 on $U_p$ and vice versa. Here $V_p\define U_p\cap t^{-1}(0)$ and weak solution of \eqref{CauchyK_2} is defined analogously as in Definition~\ref{def:Weak existence}.
 \end{corollary}

 \begin{proof}
  By the choice of the coordinates,  the Dirac operator will look in $p$ exactly  as for the Minkowski space computed in Example~\ref{ex:mink}. Note that the role of $\N_a$ is taken by the tangent plane of $\partial \M\hookrightarrow \M$ in $p$.
  Since everything is continuous, we can find a sufficiently small $\epsilon>0$ such that there is an invertible linear map $\oE \colon \Gamma(U_p, \mathbb C^N)\to \Gamma(U_p, \mathbb C^N)$ with $\oE|_{S_p\M}=-\i\big(1-a\gamma(e_0)\gamma(e_{{n}})\big)^{-1}\gamma(e_0)$ such that $\oE\oK$ has the required form.

  Moreover, we obtain a weak solution of \eqref{CauchyK_2} as required by taking the weak solution of \eqref{CauchyK} where the right handside is given by $(\oE^{-1}\f, \h, 0)$.
 \end{proof}

\begin{lemma}[Local strong solution]\label{lem:strong}
A weak solution $\Psi$ of the Cauchy problem~\eqref{CauchyK_2} is locally a strong solution.
\end{lemma}

\begin{proof}
 The last Corollary tells us that we can apply \cite[Section 2]{LP} in order to obtain the existence of a semi-strong solution, i.e. there is a sequence of continuous sections $\Phi_k\in W^{1,2}(U_p, \mathbb C^N)$ with $\oM\Phi_k|_{\partial \M\cap U_p}=0$ and $\Vert \Phi_k-\Psi\Vert_{L^2(U_p)}\to 0$ and $\Vert\widehat{\oK}\Phi_k-\f\Vert_{L^2(U_p)}\to 0$. It remains to argue, that we can approximate the $\Phi_k$  by smooth $\Psi_k$ still fulfilling the boundary condition and the convergences from above.

 This can be achieved using the standard theory of Sobolev spaces. We refer to \cite{Evans} for more details. First, choose $u_i\in \Gamma(U_p, \mathbb C^N)$, $i=1,\ldots, r$,  such that for each $q\in \partial \M\cap U_p$ they form a basis of $\ker \oM|_q$ and are linearly independent in all $q\in U_p$. Since $\M$ depends smoothly on the base point and has constant rank this is always possible.
Choose  $u_j\in \Gamma(U_p, \mathbb C^N)$, $j=r+1,\ldots, N$ such that $u_1(q), \ldots, u_N(q)$ is a basis of $\mathbb C^N$ at each $q\in U_p$. A section $\Phi\in W^{1,2}(U_p, \mathbb C^N)$ can now be expressed as $\Phi=\sum_{i=1}^N a_i u_i$ for $a_i\colon U_p\to \mathbb C$. We denote by $\Phi^+$ the part of $\Phi$ spanned by $u_1$ to $u_r$ and set $\Phi^-\define\Phi-\Phi^+$. Using the $a_i$ as the new coordinates, we have $\Psi_k^+\in W^{1,2}(U_p, \mathbb C^r)$ and
$\Psi_k^-\in W^{1,2}(U_p, \mathbb C^{N-r})$. Thus, there is a sequence $\Psi_{k,j}^+\in \Gamma(U_p, \mathbb C^r)$ that converges to $\Psi_k^+$ in $W^{1,2}$ and analogously a smooth sequence $\Psi_{k,j}^-$ converging to $\Psi_k^-$ in $W^{1,2}$. Moreover, by definition $\mathtt{tr}\, \Phi^+_k= 0$, where $\mathtt{tr}$ is the trace map $W^{1,2}(U_p, \mathbb C^r)\to L^2 (U_p\cap \partial \M, \mathbb C^r)$. Thus, $\Phi_{k,j}^+$ can be chosen to be zero on $U_p\cap \partial \M$. Thus, $\Phi_{k,j}=\Phi_{k,j}^++\Phi_{k,j}^-$, where we use the embeddings $\mathbb C^r\hookrightarrow \mathbb C^N$ and  $\mathbb C^r\hookrightarrow \mathbb C^{N-r}$ from above, are smooth sections fulfilling $\oM\Phi_{k,j}|_{U_p\cap \partial \M}=0$. Choosing a diagonal sequence we obtain smooth $\Psi_k$ approximating $\Psi$ as in Definition~\ref{def:strong sol}.
\end{proof}

Next we want to see whether the strong solution is actually smooth. Just assuming  $\f\in \Gamma_c(S\M)$ and $\h\in \Gamma_c(S\Sigma_0)$ (and not as in Theorem~\ref{maintheorem} compactly supported in the interior) is not sufficient to guarantee
that the solution of the Cauchy problem~\eqref{CauchyK} is smooth.
\begin{example}
Let $\mathscr M^n_a$ be the half Minkowski spacetime as described in Example~\ref{ex:mink} and consider the Cauchy problem~\eqref{CauchyK_2}. Assume that $\oM$ does not depend on $t$, that is (e.g.) true for MIT boundary conditions. In local coordinates the Cauchy problem reads as
\[
\begin{cases}{}
\oK \Psi= (\partial_t - \oH) \Psi=\f  \\
\Psi|_{t=0} = \mathfrak{h} \\
\oM\Psi|_{\tilde{z}=0}= 0.
\end{cases}
\]
Suppose that $\Psi$ is $k$-differentiable. Then it satisfies
$$ 0 = \partial_t^k \big( \oM \Psi|_{\tilde{z}=0} \big)|_{t=0} =  \big( \oM \partial_t^k \Psi|_{\tilde{z}=0} \big)|_{t=0} = \big( \oM \oH^k \Psi(t)|_{\tilde z=0} \big)|_{t=0} =  \oM \big(\oH^k \h\big)|_{\tilde z=0} \,. $$
Therefore, any initial data have to satisfy a compatibility condition.
 \end{example}

\begin{remark}\label{star}
 Let us remark that in the previous example we massively used that $\oM$ and $\oH$  does not depend on $t$. In the more general case of $\oK$ for a Dirac operator on a globally hyperbolic manifold in the coordinates on $U_p$ defined on page~\pageref{def:U_p} this is in general not the case.   A general compatibility condition on $\f$ and $\h$ can be obtained by setting
$$ \h_k=
\sum_{j=0}^{k-1}
\frac{(k-1)!}{j! (k-1-j)!}
\big( \partial^j_t  \oH \big)|_{V_p} \h_{k-1-j}+ \partial_t^k  \f|_{V_p}
$$
with $\h_0=\h$, $V_p=U_p\cap \Sigma_0$ and imposing that the data $\h\in \Gamma_c(S\Sigma_0)$ and $\f\in\Gamma(S\M$) satisfy
$$ \sum_{j=1}^k
 \frac{k!}{(k-j)!}
( \partial_t^j \oM )|_{V_p}\h_{j-1} =0
$$
for all $k\geq 1$. Here $\oH$ has the form~\eqref{eq:Pform}.
Translating this back for our Dirac Cauchy problem ~\eqref{CauchyDir2} in the Hamiltonian form where $\Ham\define \partial_t-\imath\gamma(e_0)\Dir$
$$(\partial_t - \Ham )\psi = -\imath \gamma(e_0) f$$
the compatibility condition for $h\in\Gamma_c(S\Sigma_0)$ and $\f\in\Gamma(S\M)$ reduces to
$$ \sum_{j=1}^k
\frac{(k)!}{j! (k-j)!}
\Big( \partial_t^j \Mbc \Big)\Big|_{\partial\Sigma_0} h_{k-1} =0$$
for all $k\geq 1$ where
$$ h_k=
\sum_{j=0}^{k-1}
\frac{(k-1)!}{j! (k-1-j)!}
 (\partial^j_t \Ham)|_{\partial\Sigma_0} \, h_{k-1-j} + \partial_t^p \big(-\imath \gamma(e_0)f)|_{\partial\Sigma_0}
$$
with $h_0=h$.
\end{remark}
The claim in Remark~\ref{star} follows by localizing a solution in any set $U_p$ defined as above and then applying \cite[Theorem 3.1]{RM}.

\begin{corollary}[Local smooth solution]\label{smooth small}
Let $U_p$ as above. Then a strong solution  for the Cauchy problem~\eqref{CauchyK} is smooth if and only if $\h\in \Gamma_c(U_p, \mathbb C^N)$ and $\f\in\Gamma(U_p, \mathbb C^N$) satisfy
\begin{equation}\label{compatibility cond}
 \sum_{j=1}^k
 \frac{k!}{(k-j)!}
( \partial_t^j \oM )|_{\partial\Sigma_0}\, \h_{k-1} =0\,.
\end{equation}
\end{corollary}

\begin{remark}\label{rem:compcond}
 If we choose an initial data $\h$ resp. $\f$ with compact support in the interior of $\Sigma_0$ resp. $\M$, the compatibility condition is automatically satisfied. Actually for $\f$ it is enough to be zero in a neighborhood of $\partial \Sigma_0\subset \M$.
 \end{remark}

\begin{corollary}\label{smooth_on_T}
 A weak solution of the Cauchy problem~\eqref{CauchyK_2} for $\f\in \Gamma_{cc}(S\M)$ and $\h\in \Gamma_{cc}(S\Sigma_0)$ in the time strip $\mathcal{T}$ provided by Theorem~\ref{thm:Weak existence} is smooth. In particular, there is a smooth solution of the Dirac Cauchy problem~\eqref{CauchyDir} in $\mathcal{T}$.
\end{corollary}

\begin{proof}
 First let $p\in \partial \M\cap \Sigma_{\hat{t}}$ for some $\hat{t}\in
[0,T]$. Let $\varrho\colon [0,\hat{t}]\to \partial M$ be a timelike curve
with $\varrho(0)\in \Sigma_0$ and $p=\varrho(\hat{t})$.
We fix $\epsilon>0$ such that
we have Fermi coordinates on a 'cube' $U_{\varrho(t)}$ around $\varrho(t)$
as in Section~\ref{diffsol} for all $t\in [0,\hat{t}]$ and such that Corollary~\ref{cor_Up} holds for those cubes. This is always possible since  the image of
$\varrho$ is compact and everything depends smoothly on the basepoints.

For $U_{\varrho(t)}$ we know that the compatibility condition~
\eqref{compatibility cond} is fulfilled by assumption. Thus, Corollary~
\ref{smooth small} tells us that the weak solution $\Psi$ is smooth in
$U_{\varrho(0)}$ and that for every $a\in [0,\epsilon]$ the function
$\h_a\define \Psi|_{U_{\varrho(0)}\cap \Sigma_{a}}$ as new initial data
$\h$ together with the original $\f$ still fulfill the compatibility
condition. Moreover, $\Psi|_{U_{\varrho(a)}}$ is still a weak solution to
the initial data $(\h_a, \f)$ on $U_{\varrho(a)}$. Thus, we can again use
Corollary~\ref{smooth small} where $\Sigma_a$ now takes the role of $\Sigma_0$. Iterating this procedure, we obtain smoothness on
all $U_{\varrho(t)}$ for $t\in [0,\hat{t}]$, i.e. in particular in $p$.

For $p\in \M\setminus \partial \M$ we choose a timelike curve $\varrho
\colon [0,\hat{t}]\to \M\setminus \partial \M$  with $\varrho(0)\in
\Sigma_0$ and $p=\varrho(T)$ and proceed as before. It is even easier since we can just use geodesic normal coordinates in the Cauchy surfaces around each $\rho(t)$.

The existence of smooth solutions to the Dirac Cauchy problem~\eqref{CauchyDir2} then follows by Lemma~\ref{Equivalence Dir and K}.
\end{proof}

\begin{remark}\label{rem:bd2}
In view of Remark~\ref{rem:bd}, we want to comment on the assumption on the boundary condition that we have used up to here. The energy inequalities~\eqref{Energy Inequality} and~\eqref{Energy Inequality2} need that $\oM\Psi|_{\bM}=0$ and $\oM^\dagger\Psi|_{\bM}=0$ both imply
  $$\fiber{\Psi}{\gamma(e_0)\gamma(\n)\Psi}_q =0$$
  for all $q\in\bM$. Moreover, in order to apply \cite{LP} in Lemma~\ref{lem:strong} and \cite{RM} in Corollary~\ref{smooth small}, we additionally use that $\ker\oM|_q$ is nonempty and varies smoothly with $q\in\bM$. \\
  The properties collected above are valid for the MIT bag boundary condition $\Mbc\define (\gamma(\n) -\imath)$ (and analogously for $\Mbc\define (\gamma(\n) +\imath)$). Indeed, on account of
\begin{align*}
\fiber{\Psi }{\gamma(e_0) \gamma(\n) \Psi}_q = \fiber{\gamma(e_0) \Psi }{ \gamma(\n) \Psi}_q &=\fiber{- \gamma(\n) \gamma(e_0)  \Psi }{ \Psi}_q =\\ &= \fiber{\gamma(e_0) \gamma(\n) \Psi }{ \Psi}_q = \overline{\fiber{\Psi }{\gamma(e_0) \gamma(\n) \Psi}_q}
\end{align*}
we obtain $  \fiber{\Psi }{\gamma(e_0) \gamma(\n) \Psi}_q \in \RR\,.$
By using MIT boundary conditions, we have
\begin{align*}
\fiber{\Psi }{\gamma(e_0) \gamma( n )\Psi}_q &= \fiber{\Psi}{ \gamma(e_0) \imath \Psi}_q=\fiber{\gamma(e_0)\Psi}{  \imath \Psi}_q=\\
&=\fiber{- \gamma(e_0) \imath\Psi}{  \Psi}_q=\fiber{-\gamma(e_0) \gamma(\n) \Psi}{  \Psi}_q=-\overline{\fiber{\Psi}{ \gamma(e_0) \gamma( n )\Psi}_q }\,,
\end{align*}
which implies $\fiber{\Psi }{\gamma(e_0) \gamma(\n) \Psi}_q=0$ for $\oM\Psi=0$.
The rest follows since $\oM=\oM^\dagger$ as we will see in the following: Note that this in particular implies that $\Ham$ as in Remark~\ref{star} restricted to a fixed Cauchy surface $\Sigma_t$ is essentially self-adjoint. First we rewrite the boundary condition $\oM\Psi=0$ as
$$P_+\Psi:=\frac{1}{2}(\Id+\imath\gamma(n))\Psi=0$$
 and we set $P_- \Psi:=\frac{1}{2}(\Id-\imath \gamma(n))\Psi=0$. It is easy to see that
$$P_++P_-=\Id \qquad P_\pm^2=P_\pm \qquad \text{and}\qquad P_+P_-=P_-P_+$$
and in particular that
$\fiber{P_+\Psi }{P_-\Phi}_q=0$ .
This implies that
$P_+\Psi=0$ is tantamount to set $\Psi=P_- \tilde{\Psi}$ for some $\tilde\Psi$.
Thus,  for $\Psi$ verifying $\oM\Psi=0$, the condition $\fiber{\Phi }{\gamma(e_0) \gamma( n )\Psi}_q =0$ holds if and only if for any $\tilde{\Psi}$ it is satisfied
\begin{align*}
0&=\fiber{\Phi }{\gamma(e_0) \gamma( n )P_- \tilde\Psi}_q=\fiber{\gamma(e_0)\Phi }{( \gamma( n )+\imath) \tilde\Psi}_q=
\\& =\fiber{(- \gamma( n )-\imath)\gamma(e_0)\Psi }{ \tilde\Phi}_q=\fiber{\gamma(e_0)( \gamma( n )-\imath)\Phi }{ \tilde\Psi}_q\,,
\end{align*}
which implies $ \oM\Phi=(\gamma( n )-\imath)\Phi =0\,.$

Another example is the chirality operator $\Mbc \define (\text{Id}-\gamma(n) \mathcal{G})$ where $\mathcal{G}$ is the
restriction to $\partial \M$ of an endomorphism-field of $S\M$ which is involutive,
unitary, parallel and anti-commuting with the Clifford multiplication on $\M$, \cite[Section 1.5]{Gin}.
\end{remark}

\subsubsection{Existence of the weak solution}

Up to now we have seen in Corollary~\ref{smooth_on_T}, Proposition~\ref{prop:finite} and Proposition~\ref{prop:unique}, that if there exists a weak solution $\Psi$ of the Cauchy problem~\eqref{CauchyK} then it is smooth, unique and it vanishes outside $\mathcal{V}$ defined by~\ref{def:V}. It just remains to prove existence.

\begin{thm}[Weak existence]\label{thm:Weak existence}
There exists a unique weak solution $\Psi\in\cH$ to the Cauchy problem~\eqref{CauchyK} with $\f\in \Gamma_{cc}(S\M)$ and $\h\equiv 0$, restricted to $\T$.
\end{thm}

\begin{proof}
As a consequence of the above, the integrands in~\eqref{scalprod} are potentially non zero only in $\Rv \subset \T$ of the form~\eqref{defRv} for $\mathcal O'\define J^-(\V\cap \Sigma_T) \cap \Sigma_0$. Therefore, \eqref{scalprod} can be rewritten as
$$ \scalar{\Phi}{\f}_{\Rv}=\scalar{\oK^\dagger \Phi}{\Psi}_{\Rv}.
 $$
 On account of Lemma~\ref{lemma: energy estimates2} and using arguments similar to Proposition~\ref{prop:unique}, we can notice that the  kernel of the operator $\oK^\dagger$ acting on $$\text{dom}\,  \oK^\dagger\define \{\Phi\in \Gamma_c(S\T)\ |\ \Phi|_{\Sigma_T}=0, \Mbc^\dagger \Phi|_{\partial \M}=0\}$$ is trivial.  Let now $\ell\colon \oK^\dagger(\text{dom}\,  \oK^\dagger)	\to \CC$ be the linear functional defined by
$$\ell(\Theta)=\scalar{\Phi}{\f}_{\Rv}$$
where $\Phi$ satisfies $\oK^\dagger \Phi=\Theta$. By the energy inequality~\eqref{Energy Inequality2}, $\ell$ is bounded:
\begin{align*}\ell(\Theta)=\scalar{\Phi}{\f}_{\Rv} &\leq \|\f\|_{L^2(\Rv)} \, \|\Phi\|_{L^2(\Rv)} \\& \leq \lambda^{-1} \|\f\|_{L^2(\Rv)} \|\oK^\dagger \Phi\|_{L^2(\Rv)} = \lambda^{-1} \|\f\|_{L^2(\Rv)} \|\Theta\|_{L^2(\Rv)} ,\end{align*}
where in the first inequality we used Cauchy-Schwartz inequality.
Then $\ell$ can be extended to a continuous functional defined on the $L^2$-completion of $\oK^\dagger(\text{dom} \, \oK^\dagger)$ denoted by  $\mathscr{H}\subset \mathcal{H}$. Finally, by Riesz representation theorem, there exists a unique element $\Psi\in \mathcal{H}$ such that
$$ \ell({\Theta}) =  \scalar{\Theta}{\Psi}_{\Rv} \,. $$
for all $\Theta\in \oK^\dagger(\text{dom})$. Thus, we obtain
$$ \scalar{\Phi}{\f}_{\Rv}= \ell({\Theta})=\scalar{\Theta}{\Psi}_{\Rv} =\scalar{\oK^\dagger \Phi}{\Psi}_{\Rv}$$
for all $\Phi\in \text{dom} \oK^\dagger$. This concludes our proof.
\end{proof}

\section{Global well-posedness of Cauchy problem}
\label{sec:gs}
Up to now we obtained a weak solution in a time strip and showed that it is actually smooth if the initial data are compactly supported in the interior (or more generally fulfill the compatibility condition~\eqref{compatibility cond}, compare Remark~\ref{rem:compcond}).
We can finally put everything together to  obtain global well-posedness of the Cauchy problem~\eqref{CauchyDir}.

\begin{proof}[Proof of Theorem~\ref{maintheorem}]
Fix $h\in\Gamma_{cc}(S\Sigma_0)$. By Corollary~\ref{smooth small}, for any $T\in [0,\infty)$ there exists a smooth solution $\psi_T$ to the Dirac Cauchy problem~\eqref{CauchyDir} in the time strip $\T_T\define t^{-1}([0,T])$. Now consider $T_1,T_2 \in [0,\infty)$ with $T_2 > T_1$. By uniqueness of solution, see Proposition~\ref{prop:finite}, we have $\psi_{T_2}|_{\T_{T_1}}=\psi_{T_1}$.  Hence, we can glue everything together to obtain a smooth solution for all $\T\geq 0$. A similar arguments holds for negative time. \\
Since $h\in\Gamma_{cc}(S\Sigma_0)$, $f\in \Gamma_{cc}(S\M)$ it follows by the finite propagation of speed, see Proposition~\ref{prop:finite}, that the solution is spacelike compact. For the continuous dependency on the initial data, see the remark below.
\end{proof}

We are now in the position to discuss the stability of the Cauchy problem. Since the proof is independent on the presence of the boundary and it does rely mostly on functional analytic techniques, we shall omit it and we refer to \cite[Section 5]{Ba} for further details.
\begin{proposition}\label{stability}
Consider a globally hyperbolic spacetime $\M$ with boundary $\bM$ and denote with $S\M$ the spinor bundle over $\M$. Moreover,  let
$\Gamma_{0}(S\Sigma_0)\times \Gamma_{0}(S\M)$ denote the space of data satisfying the compatibility condition~\eqref{compatibility cond}. Then the map
$$ \Gamma_{0}(S\Sigma_0) \times \Gamma_0(S\M) \to  \Gamma_{sc}(S\M)$$
 which assign to $  (\h,\f) $ a solution $\Psi$ to the Cauchy problem~\eqref{CauchyK} is continuous.
\end{proposition}

A byproduct of the well-posedness of the Cauchy problem is the existence of Green operators:

\begin{proposition}\label{Green}
The Dirac operator is Green hyperbolic, i.e. there exist linear  maps, dubbed \emph{advanced/retarded Green operator},
$\Green^\pm\colon \Gamma_{cc}(S\M)  \to \Gamma_{sc}(S\M)$ satisfying
\begin{enumerate}[(i)]
\item $\Green^\pm \circ \Dir\,  f = \Dir\circ \Green^\pm f=f$ for all $ f \in \Gamma_{cc}(S\M)$;
\item  $\supp(\Green^\pm f ) \subset J^\pm (\supp f )$ for all $f \in \Gamma_{cc} (S\M)$,
\end{enumerate}
where $J^\pm$ denote the causal future (+) and past (-).
\end{proposition}

\begin{proof}
Let $f\in \Gamma_{cc}(S\M)$ and choose $t_0\in \RR$ such that $\supp f \subset J^+( \Sigma_{t_0})$.
By Theorem~\ref{maintheorem}, there exists a unique solution $\psi(f)$ to the Cauchy problem
$$
\begin{cases}
\Dir \psi =  f\\
\psi|_{\Sigma_0} = 0 \\
\Mbc\Psi|_{\bM} =0.
\end{cases}
$$
For $f\in \Gamma_{cc}(S\M)$ we set
$\Green^+f\define\psi$ and notice that $\Dir\circ\Green^+f= \Dir \psi = f$. Note that by the finite speed of propagation, cf. Proposition~\ref{prop:finite}, $\Green^+f\in \Gamma_{sc}(S\M)$. Moreover, $\Green^+ \circ \Dir \psi= \Green^+f=\psi$ which finishes the proof of (i). By Proposition~\ref{prop:finite}, we obtain $\supp\Green^+f \subset J^+(f)$ and this conclude the proof of (ii). \\
The existence of the retarded Green operator $\Green^-$ is proven analogously.
\end{proof}

\vspace{0.5cm}


TITLE: Is there a name for this quaternion-like algebra?
QUESTION [1 upvotes]: $\bullet\ \textbf{The Algebra in question}$
So if you take the ring of integers appended with the cube root of unity, $R=\mathbb{Z}[\zeta_3]$, it seems like you can construct this weird non-commutative algebra, $H=R[u,v]$, with the properties
$$u^2+u+1=v^2+v+1=0\quad\text{and}\quad vu = \zeta_3uv.$$
It would seem at first that $1,u,v,$ and $uv$ form the minimal basis of $H$ over $R$. But with some calculation,
$$v^2u=-vu-u=-\zeta_3uv-u\quad\text{and}$$
$$v^2u=\zeta_3^2uv^2=-\zeta_3uv^2-uv^2=\zeta_3uv+\zeta_3u+uv+u,$$
one can show that
$$(-2\zeta_3-1)uv=(\zeta_3+2)v$$
and therefore that $uv=\zeta_3v\in\text{span}_R(1,u,v)$. Or something similar. So $[H:R]=3$?
$\bullet\ \text{Motivation}$
I was trying to construct something like the quaternions but with a norm of $a^3+b^3+c^3$ instead of $a^2+b^2+c^2+d^2$. I'd started out with $u^3=v^3=1$ but switched to aforementioned equations to reduce the basis. And indeed the construction seems successful. If we take $\phi_i:u,v\rightsquigarrow \zeta_3^iu,\zeta_3^iv$ then we get for $\alpha=a+bu+cv$ that
$$\alpha\cdot\phi_1(\alpha)\cdot\phi_2(\alpha)=a^3+b^3+c^3.$$
I'd suspect that this isn't exactly a multaplicative norm but it's interesting enough to investigate.
$\bullet\ \text{Question}$
So what even do you call a structure like this? The search "noncommutative ring extension" doesn't yield anything helpful -- at least, nothing helpful that I can make sense of. I've been reading about Banach and Clifford algebras with no luck. I've also read Conway's quaternion/octonion book but don't have it available at the moment.

REPLY [0 votes]: You want $H$ to be an $R$-algebra generated by $u$ and $v$ over $R$, as an algebra. Assuming $H$ is associative, it is a quotient of the algebra $\bar H:=R\langle u,v\rangle/I$, where $R\langle u,v\rangle$ is the ring of polynomials in two non-commuting variables $u$ and $v$ with coefficients in $R$, and $I$ is the two-sided ideal generated by your relations;
$$I:=(u^2+u+1,v^2+v+1,vu-\zeta uv).$$
For future convencience set $\varepsilon:=1-\zeta\in R$.
Observation 1: In $\bar{H}$ we have $\varepsilon u=\varepsilon v=\zeta^2\varepsilon uv$.
Proof.  In the quotient $\bar H=R\langle u,v\rangle/I$ we have
\begin{eqnarray*}
u+uv&=&u(1+v)=-uv^2=-\zeta v^2u&=&\zeta (1+v)u=\zeta u+\zeta vu&=&\zeta u+\zeta^2uv,\\
v+uv&=&(1+u)v=-u^2v=-\zeta vu^2&=&\zeta v(1+u)=\zeta v+\zeta vu&=&\zeta v+\zeta^2uv,
\end{eqnarray*}
and comparing these two shows that
$$
(1-\zeta)u=-(1-\zeta^2)uv=(1-\zeta)v.
$$
Note that $-(1-\zeta^2)=\zeta^2(1-\zeta)$, so this already shows that $\varepsilon u=\zeta^2\varepsilon uv=\varepsilon v$.$\hspace{10pt}\square$
Observation 2: In $\bar{H}$ we have $3=0$ and $\varepsilon u=\varepsilon v=\varepsilon uv=\varepsilon$.
From the identities $\varepsilon u=\zeta^2\varepsilon uv=\varepsilon v$ it follows that
\begin{eqnarray*}
\varepsilon u
&=&\zeta^2\varepsilon uv
=\zeta\varepsilon vu
&=&\zeta\varepsilon u^2
&=&-\zeta\varepsilon(u+1),\\
\varepsilon v
&=&\zeta^2\varepsilon uv
&=&\zeta^2\varepsilon v^2
&=&-\zeta^2\varepsilon(v+1),
\end{eqnarray*}
which shows that
$$
\zeta\varepsilon=-(1+\zeta)\varepsilon u=\zeta^2\varepsilon u
\qquad\text{ and }\qquad
\zeta^2\varepsilon=-(1+\zeta^2)\varepsilon v=\zeta\varepsilon v,
$$
and hence that $\varepsilon u=\zeta^2\varepsilon$ and $\varepsilon v=\zeta\varepsilon$. Then
$$\zeta\varepsilon^2
=\zeta(1-\zeta)\varepsilon
=\zeta\varepsilon-\zeta^2\varepsilon
=\varepsilon v-\varepsilon u=0,$$
because we saw in observation 1 that $\varepsilon u=\varepsilon v$. As $\zeta\varepsilon^2=-3\zeta^2$ it follows that also $3=0$ in $\bar{H}$. Moreover, this shows that $\zeta\varepsilon=\varepsilon$ and so $\varepsilon uv=\zeta^2\varepsilon uv=\varepsilon u=\varepsilon v=\zeta\varepsilon=\varepsilon$.$\hspace{10pt}\square$

Now we can reframe the question; we have seen that $\{1,\zeta,u,v,uv\}$ is a $\Bbb{Z}$-basis for $\bar{H}$, but multiplication in $\bar{H}$ is significantly simpler on the $\Bbb{Z}$-basis $\{1,1-\zeta,1-u,1-v,(1-u)(1-v)\}$.
Set $x:=1-u$ and $y:=1-v$. Observation 2 shows that $3,\varepsilon(u-1),\varepsilon(v-1),\varepsilon(uv-1)\in I$, and so $\varepsilon x,\varepsilon y,\varepsilon xy\in I$. Because $3\in I$ we see that $\bar{H}$ is an $R/3R$-algebra, where
$$R/3R\cong\Bbb{Z}[X]/(X^2+X+1,3)\cong\Bbb{F}_3[X]/(X^2+X+1)=\Bbb{F}_3[X]/(1-X)^2.$$
This shows that $R/3R\cong\Bbb{F}_3[\varepsilon]$ where $\varepsilon^2=0$. Of course also $u^2+u+1,v^2+v+1\in I$, where
$$u^2+u+1=(1-u)^2=x^2
\qquad\text{ and }\qquad
v^2+v+1=(1-v)^2=y^2,$$
and so $x^2,y^2\in I$. A routine check shows that $yx-xy+\varepsilon\in I$, so $\bar{H}/(\varepsilon)$ is commutative; in fact
$$\bar{H}/(\varepsilon)\cong\Bbb{F}_3[x,y]/(x^2,y^2).$$
This is a commutative ring of $3^4=81$ elements, and $H$ all its quotients are isomorphic to
$$\Bbb{F}_3[x,y]/(x^2,y^2),\qquad \Bbb{F}_3[x,y]/(x,y)^2,\qquad\Bbb{F}_3[x]/(x^2),\qquad\Bbb{F}_3,\qquad\text{ or }\qquad 0.$$
That is to say, if $\varepsilon=0$ in $H$, or equivalently if $\zeta=1$ in $H$, then $H$ is isomorphic to one of these five commutative rings. Otherwise, if $\varepsilon\neq0$ in $H$ then $H$ is isomorphic to a quotient of $\bar{H}$ by a two-sided ideal not containing $\varepsilon$. It turns out that every non-zero two-sided ideal of $\bar{H}$ contains $\varepsilon$:
Observation 3: If $J\subset\bar{H}$ is a nonzero two-sided ideal then $\varepsilon\in J$.
Proof. Let $J\subset\bar{H}$ be a two-sided ideal not containing $\varepsilon$ and let $h\in J$. Let $a,b,c,d,e\in\Bbb{F}_3$ such that
$$h=a+b\varepsilon+cx+dy+exy.$$
Then clearly $\varepsilon h=a\varepsilon\in J$, so $a=0$. Because $xy-yx=\varepsilon$ in $\bar{H}$ we see that $J$ contains
$$xh-hx=(d+e)\varepsilon
\qquad\text{ and }\qquad
hy-yh=(c+e)\varepsilon,$$
so $c+e=0$ and $d+e=0$. Then $h=b\varepsilon+c(x+y-xy)$ and so
$$h(1+y)(x-y)=(b\varepsilon+c(x+y))(x-y)=c(yx-xy)=-c\varepsilon,$$
hence also $c=0$. Then $h=b\varepsilon\in J$ and so $h=0$, which shows that $J=0$.$\hspace{10pt}\square$
In conclusion, there is only one such non-commutative algebra, up to isomorphism, and it is the finite ring
$$\Bbb{F}_3[\varepsilon]\langle x,y\rangle/(x^2,y^2,\varepsilon x,\varepsilon y,yx-xy+\varepsilon,),$$
of order $3^5=243$ in which $\varepsilon x=\varepsilon y=\varepsilon xy=0$.


text:
\begin{document}

\title{Asymmetric Hopfield neural network and \\Twisted tetrahedron equation}

\author{\rule{0pt}{7mm} Dmitry V. Talalaev\thanks{dtalalaev@yandex.ru}\\
{\small\it
Moscow State University, Faculty of Mechanics and Mathematics}\\
{\small\it 119991 Moscow, Russian Federation}}

\date{}

\maketitle

\abstract{
We generalize the approach of \cite{T1} for the case of the Hopfield neural network in the recall stage on a triangular lattice with isotropic weights. It appears that some properties of this model, in particular the probability of passing a trajectory in time dynamics, obeys the Gibbs distribution with a partition function having a vertex realization. Moreover the corresponding weight matrix satisfies the TTE - some deformation of the Zamolodchikov tetrahedron equation, the latter playing the role analogous to the Yang-Baxter equation in 3-dimensional statistical models.
 }

\vskip 5mm
~\\
\tableofcontents
\vskip 2cm
\section{Introduction}
Models of artificial and natural neural networks for a long time have been shown to be related to the integrable models in lattice statistical physics. The main emphasis of this work is on some new kind of relation between the Ising model and the Hopfield model of associative memory.

\subsection*{Acknowledgments}
The work was partially supported by the RFBR grant 17-01-00366.

\subsection{Hopfield model}
The Hopfield network \cite{Hop} is one of the fundamental models of content-addressable associative memory system. This model have demonstrated some very interesting collective properties of the neural network behavior and is used in artificial neural networks as like as in neurophysiology \cite{Tso} in research of memory capacities, memory retrieval process, short-term plasticity, working memory properties and many other questions. The model is represented by the complete graph with $N$ vertexes (neurons) with a connectivity matrix $W_{ij}$ characterizing the conductivity of the synapse between $i$-th and $j$-th neurons. At each time the system is characterized by its neurons states $\{x_i\},i=1,\ldots,N$ $x_i=\pm 1.$ Our interest is focused on the network which undergoes the synchronous evolution in discrete time. In the deterministic version the state of $i$-th neuron at the next step $x'_i$ is determined by the formula:
\bea
x'_i=\left\{
\begin{array}{ccc}
1 & \mbox{if}  & \sum_{j} W_{ij}x_j >t_i \\
-1 & \mbox{if} &  \sum_{j} W_{ij}x_j < t_i \\
x_i & \mbox{else.} &
\end{array}\right.
\eea
We principally deal with the probabilistic Hopfield model in which the transition is performed with the probability specified by the Fermi sigmoid function
\bea
\label{prob}
P(x',x)=\prod_{i}(1+e^{-\beta x'_i(\sum_j W_{ij}x_j-t_i)})^{-1}.\nn
\eea
For the threshold level $t_i=0$ this expression can be rewritten as
\bea
P(x',x)= {e^{-\frac \beta 2 \sum_{ij}W_{ij}x'_i x_j}}/ {\sum_{x''}e^{-\frac \beta 2 \sum_{ij}W_{ij}x''_i x_j}}.\nn
\eea
The similarity of such an expression with the partition function of the Ising model was remarked in \cite{Hop}. We make this resemblance more formal in the next section.

We need to say some words about the learning rules. Following the Hebbian paradigm the learning of the Hopfield network on the set of $n$ patterns $\{\epsilon^1,\ldots,\epsilon^n\}$ is given by the formula
\bea
W_{ij}=\frac 1 n \sum_{k}\epsilon_i^k \epsilon_j^k.\nn
\eea
In what follows we start with the learned network with isotropic weight function.
The work of the network in the recall stage consists of iterative time dynamics defined by the transformation probability (\ref{prob}). The principal question in this domain is the basins of attraction for such a process.

The principal interplay of the Hopfield model and the Ising one is due to the energy argument. It turns our that for the symmetric weight function the expression
\bea
E=-\frac 1 2 \sum_{i,j} W_{ij} x_i x_j -\sum_i t_i x_i\nn
\eea
plays the role of the energy in the sense that in time dynamics it either lower or stays the same. This observation allows to analyze the asymptotic behavior of the Hopfield model. The stable points for the Hopfield time dynamics are hence the states of local minimum energy for the Ising model on the same lattice. In this paper we explore another relationship between two such models: the Hopfield model on a two-dimensional lattice and the Ising model on the 3-dimensional lattice, making one of the spacial directions relevant to the time evolution.

\subsection{Time evolution and the Ising model}
Let us consider the Hopfield model on the triangular lattice (figure \ref{pic-lattice}) colored by 3 colors in $\Z/3\Z.$ The weights are nontrivial for neighbors with appropriate colors: for the neuron with the color $c$ we choose only such influencing neurons whose colors are $(c-1) ~mod ~3.$ This network is not symmetric despite the most traditional normalization.

\begin{figure}[h!]
\center
\includegraphics[width=80mm]{tri-lattice}
\caption{Triangular lattice}
\label{pic-lattice}
\end{figure}

This lattice corresponds to the projection of the cubic lattice to the plane $i+j+k=0$. The vertexes with different colors represent the planes $i+j+k=c~ mod~ 3.$ The temporal behavior of this lattice is equivalent to the 3 independent 3-dimensional cubic lattices. We then consider the one of these 3 lattices.

The conditional probability that the model passes throw the states with free initialization data is:
\bea
P=\prod_{i+j+k=a}^b\frac {e^{-\beta s_{ijk}(s_{ijk-1}+s_{i-1 jk}+s_{i j-1k})}}{ch(\beta (s_{ijk-1}+s_{i-1 jk}+s_{i j-1k}))}\nn
\eea
\begin{Lem}
Let $\sigma_2$ be the second symmetric polynomial on 3 variables $s_1,~s_2$ and $s_3$ taking values $\pm 1.$ Then
\bea
\label{eq1}
ch(\beta(s_1+s_2+s_3))=ch^3\beta+sh^2\beta ch \beta \sigma_2.
\eea
\end{Lem}
Demonstration is straightforward.
\\
Moreover we have:
\bea
\label{eq2}
e^{\gamma \sigma_2}=(ch^3\gamma +sh^3 \gamma)+(ch^2 \gamma sh \gamma+ch \gamma sh^2 \gamma)\sigma_2.
\eea
Expressions (\ref{eq1}) and (\ref{eq2}) turn out to be proportional if
\bea
\label{eq3}
th^2 \beta=th \gamma/(1- th \gamma+th^2 \gamma)
\eea
With $\gamma$ solving the equation (\ref{eq3}) the conditional probability for the Hopfield model is identical to the Gibbs distribution for the anisotropic Ising model on cubic lattice with additional diagonals with specially chosen anisotropy parameters and with border terms
\bea
P(s)=\frac 1 Z \prod_{i,j,k} e^{-\beta s_{ijk}(s_{i-1 j k}+s_{i j-1 k}+s_{i j k-1})-\gamma(s_{i-1 j k}s_{i j-1 k}+
s_{i j-1 k} s_{i j k-1}+s_{i j k-1} s_{i-1 j k})}.\nn
\eea
\begin{figure}[h!]
\center
\includegraphics[width=70mm]{Cube2}
\caption{Ising model lattice}
\label{edges}
\end{figure}

The main problems stated in the Hopfield theory are related with stationary points of this dynamics and with correlations. For example the correlation length in the Ising model and its particular dependence on both parameters $\beta$ and $\lambda$ is related with such questions as cluster dimension and time evolution.

\subsection{3D Ising model and TTE}
Let us remind the main steps of the construction of \cite{T1}.

\paragraph{Dual lattice}
The isotropic Ising model is described by the Hamiltonian
\bea
H(\sigma)=\sum_{d(i,j)=1} \sigma_i \sigma_j\nn
\eea
where $i\in\Lambda$ is a periodic 3d lattice, $\sigma_i$ is the spin variable associated to the $i$-th vertex. $d(i,j)$ is the standard (Manhattan) metric on the cubic lattice.
The partition function is defined by the Gibbs rule:
\bea
\label{part}
Z(t)=\sum_{(\sigma)}\exp(t H(\sigma))
\eea
where the sum is taken over the space of all spin configurations.

To work with the dual lattice we introduce the variables $s_{ij}=\sigma_i \sigma_j$ on edges and associate the space $\C^2$ to each edge of $\Lambda.$ Then we consider the dual lattice $\Lambda^*$ those vertices are 3-cubes of $\Lambda$, the edges of $\Lambda^*$ are 2-faces of $\Lambda.$ We associate a vector space $V_f=(\C^2)^{\otimes 4}\simeq \C^{16}$ to each edge of the dual lattice $\Lambda^*.$

\paragraph{Recursion on the moduli of solutions for the $n$-simplex equation}
In \cite{KST} we define a moduli space $\mathfrak{S}_n(X)$ of solutions for the set theoretical $n$-simplex equation in correspondences with the underlying set $X$. In \cite{T1} we constructed a recursion
\bea
\tau: \mathfrak{S}_n(X)\to \mathfrak{S}_{n+1}(X^{2n})\nn
\eea
assigning a set of colors of 1-edges of a 2-face to this 2-face. We then applied such a recursion to the solution for the  Yang-Baxter equation given by the matrix
\bea
R_M=\left(\begin{array}{cccc}
1 & 0 & 0 & 1\\
0 & 1 & 1 & 0\\
0 & 1 & 1 & 0\\
1 & 0 & 0 & 1
\end{array}\right).
\eea
\begin{Rem}
This matrix encodes the conditions on edge-spins for the Ising model.
\end{Rem}
Then $\tau(R)$ produces an auxiliary weight matrix $W_0$ for the 3D Ising model.

\paragraph{Counting edges of the 3-cubes}
First let us make some remarks on the notations in hypercube combinatorics.
The $n$-cube subfaces are denoted by sequences $(a_1,\ldots,a_n)$ where $a_i$ are either $1$ or $0$ or $*$. The number of stars is the dimension of a subface.
Then define a sequence
\bea
\zeta=(0,1,\ldots,).\nn
\eea
\begin{Def}
An $(n-1)$-subface of an $n$-face is called incoming if the $i$-th $*$ is replaced by $\zeta_i,$ otherwise it is called outgoing.
\end{Def}
\begin{Rem}
We also introduce a lexicographic order on subfaces. For example the standard $2$-cube has $4$-edges, 2 incoming and 2 outgoing, we denoted their orders in the second line of the table.
\begin{table}[h!]
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
0*& *1 & 1*& *0\\
\hline
$i_1$ & $i_2$ & $o_1$ & $o_2$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\end{Rem}

We then choose a cocycle on 3-cube colorings determined by the following choice of 3 edges on each 3-cube of a 4-cube given by the  tables
\begin{table}[h!]
\caption{Left side of the TE}
\label{tab-LTE}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
0***& *1** & **0* & ***1\\
\hline
01*0 & 011* & 0*00 & 00*1 \\
000* & *100 & 110* & *111 \\
0*11 & 11*1 & *001 & 1*01 \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{table}[h!]
\caption{Right side of the TE}
\label{tab-RTE}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
***0 & **1* & *0** & 1***\\
\hline
1*00 & 0*10 & 001* & 100* \\
00*0 & 111* & *000 & 1*11 \\
*110 & *011 & 10*1 & 11*0 \\
\hline
\end{tabular}
\end{center}
\end{table}
The picture \ref{pic-edges} illustrates the table \ref{tab-LTE}.
\begin{figure}[h!]
\center
\includegraphics[width=80mm]{edges}
\caption{Chosen subgraph in the 4-cube}
\label{pic-edges}
\end{figure}
The picture \ref{pic-torus} visualizes the same subset taking in mind the genus 1 embedding of the 4-cube. \begin{figure}[h!]
\center
\includegraphics[width=80mm]{torus}
\caption{Torus embedding}
\label{pic-torus}
\end{figure}
\begin{Rem}
The essential property of the subgraphs defined by tables \ref{tab-LTE} and \ref{tab-RTE} is that they are exchanged by the indexes transformation $1 \leftrightarrow 4, 2\leftrightarrow 3.$
\end{Rem}

\paragraph{Twisted tetrahedron equation}
The chosen subgraphs properties can be encoded by the weight matrix $W(i,j,k)=W_0\times \exp(t\sum_{u=1}^6 \sigma_{\varphi(u)})$ where the choice of edges is given by the tables above.
We demonstrated in \cite{T1} the following
\begin{Lem}
The matrix $W(i,j+1,k+2)$ gives a weight matrix for the 3D Ising model for each choice of parameters $i,j$ and $k.$ This means that the partition function $Z(t)$ can be obtained as a product
\bea
Z(t)=\prod_{(\alpha,\beta,\gamma)\in\Lambda^*} W_{\alpha\beta\gamma}(i,j,k)
\eea
over the dual lattice $\Lambda^*.$
\end{Lem}

\begin{Th}
\label{th-1}
The weight matrix $W(i,j,k)$ satisfies the equation
\bea
&&W_{653}^A(1,2,3) W_{642}^A(1,2,4) W_{541}^A(1,3,4) W_{321}^A(2,3,4)\nn\\
&&=
W_{356}(2,3,4) W_{246}(1,3,4) W_{145}(1,2,4) W_{123}(1,2,3)
\label{TTE}
\eea
where the upper index $A$ means the conjugation with a linear operator $A$ in each tensor component symbolizing the basis change in $V_f$ exchanging the 1-edges of a 2-face as follows $i_1 \leftrightarrow o_2, i_2\leftrightarrow o_1.$
\end{Th}
\begin{Rem}
We call the equation \ref{TTE} the {\bf twisted tetrahedron equation}.
The importance of the result of theorem \ref{th-1} is that the equation \ref{TTE} is closely resembles the Zamolodchikov tetrahedron equation \cite{Zam} by the following transformation:
\bea
&&{\color{red}P_{16} P_{25}} W_{123}^{{\color{red}A}}(1,2,3) W_{145}^{{\color{red}A}}(1,2,4) W_{246}^{{\color{red}A}}(1,3,4) W_{356}^{{\color{red}A}}(2,3,4){\color{red}P_{16} P_{25}} \nn\\
&&=
W_{356}(2,3,4) W_{246}(1,3,4) W_{145}(1,2,4) W_{123}(1,2,3).\nn
\label{TTE2}
\eea
where $P_{ij}$ are just transpositions in the pair of $i$-th and $j$-th spaces.

The Zamolodchikov tetrahedron equation manifests the integrable properties of the 3D statistical model in virtue of the main result of \cite{T2}. This results is aimed to construct a commutative family including the transfermatrix of the model further targeting in application of the Bethe ansatz technique in an analogous way to the 2-dimensional integrable statistical models \cite{Bel}.
\end{Rem}

\section{Hopfield vertex model}
\subsection{Diagonal combinatorics}
We reproduce the scheme of \cite{T1} in the context of the Ising model on the lattice of figure \ref{edges}.
Let us attach a spin variable to each diagonal by the same rule as for the edge-spins - by the product of the vertex-spins in its ends.
\begin{figure}[h!]
\center
\includegraphics[width=70mm]{diagonals}
\caption{Diagonal choice}
\label{fig-diagonals}
\end{figure}
\begin{Lem}
For each 2-face of a 3-cube the value of the diagonal spin is related with the edge-spins by the formula
\bea
s=s_{i_1}\times s_{o_2}=s_{i_2}\times s_{o_1}.\nn
\eea
\end{Lem}
The demonstration is via the figure \ref{fig-diagonals}.

The choice of the diagonals on the left and right hand sides of the tetrahedron equation are given by the figures \ref{fig-diag1} and \ref{fig-diag2}.
\begin{figure}[h!]
\center
\includegraphics[width=80mm]{diag1}
\caption{Diagonals on LHS}
\label{fig-diag1}
\end{figure}
\begin{figure}[h!]
\center
\includegraphics[width=80mm]{diag2}
\caption{Diagonals on RHS}
\label{fig-diag2}
\end{figure}

On both figures we singled out the chosen diagonals. We should comment the meaning of these figures. The 3-cubes of the 4-cube are realized as the horizontal and vertical bands. They are marked by the appropriate codes. We denoted the diagonals which are not embedded into the torus by curved blue lines. We need also to note that some of curved lines and embedded diagonals are taken twice.

\begin{Lem}
The chosen diagonals on the left and right hand sides are exchanged by the indexes transformation $1\leftrightarrow 4,~2\leftrightarrow 3.$
\end{Lem}
The demonstration is straightforward.

\subsection{TTE}
\begin{Def}
Let us define the weight matrix $W_h(i,j,k)$ which is constructed with a help of the 3D Ising model weight matrix $W(i,j,k)$ as follows
\bea
W_h(i,j,k)=W(i,j,k)\times \exp(\gamma\sum_f \sigma_{i_1} \sigma_{o_2}).
\eea
Here the sum is over 2-faces of a 3-cube, and the spins variables correspond to the edges of a 2-face.
\end{Def}

\begin{Th}\label{th-2}
The weight matrix $W_h(i,j,k)$ satisfies the twisted tetrahedron equation
\bea
&&W_{653}^A(1,2,3) W_{642}^A(1,2,4) W_{541}^A(1,3,4) W_{321}^A(2,3,4)\nn\\
&&=
W_{356}(2,3,4) W_{246}(1,3,4) W_{145}(1,2,4) W_{123}(1,2,3).
\label{TTE1}
\eea
where the upper index $A$ has the same meaning as in theorem \ref{th-1}.
\end{Th}
The proof  reproduces the proof of theorem 1 of \cite{T1}. It ensues the same observation that the configuration of chosen graph on the left hand side transforms to the graph on the right hand side by the index transformation $1\leftrightarrow 4,~2\leftrightarrow 3$ which can be encoded by the specified matrix notations.

\subsection{Conclusion}
Our main hope concerns the possibility of the Bethe ansatz method application to the description of the critical behavior of the Hopfield neural network. This could be fruitful in such a technique as the simulated annealing in neural networks \cite{An}.

config: arxiv


TITLE: A Necessary Condition for Measurable Sets
QUESTION [0 upvotes]: Suppose $E \subset \mathbb{R}^n$ is measurable. I want to show that for any subset $A \subset E$ we have
\begin{align*}
m(E) = m_*(A) + m^*(E \setminus A)
\end{align*}
where $m_*(A) = \sup\{m(\mathcal{C}) : \mathcal{C} \subset A \text{ is closed }\}$ and $m^*(A) = \inf \{m(\mathcal{O}): \mathcal{O}\supset A \text{ is open}\}$. My definition of measuribility is that for any $\epsilon > 0$ there exists an open set $\mathcal{U} \supset E$ such that $\lambda^*(\mathcal{U} \setminus E) \leq \epsilon$.
I've solved it for the case $m(E) = 0$, so we can assume $E$ has nonzero measure. Any hints or help would be appreciated. Thanks!
Edit: Fixing a typo. $A$ is suppose to be an arbitrary subset of $E$ not $\mathbb{R}^n$.

REPLY [0 votes]: I think there is a typo in your statement, it probably should read:
$$m(E)=m_*(A\cap E)+m^*(E\setminus A)$$
I will assume you have already proven that $m^*$ coincides with the outer measure defined in the usual sense. First note that, by monotonicity and sub-aditivity:
$$m_*(A\cap E)+m^*(E\setminus A)\le m^*(A\cap E)+m^*(E\setminus A)\le(E)$$
Now, if we knew that $m_*(G\cup F)\ge m_*(G) + m_*(F)$ for any subsets $F,G\subset \mathbb{R}^n$, then it would follow that:
$$m_*(A\cap E)+m^*(E\setminus A)\ge m_*(A \cap E)+m_*(E\setminus A)\ge m_*((A\cap E)\cup (E\setminus A))\ge m_*(E)$$
But, since $E$ is measurable, so is $E^c$, hence given $\varepsilon >0$ there exists an open set $U \supset E^c$ such that $\lambda(U\setminus (E^c))\le \varepsilon$, this is $\lambda(E \cap U)\le \varepsilon$. But $U^c$ is closed and $E\subset U^c$, hence:
$$m_*(E)\le \lambda(E)=\lambda(E\cap U)+\lambda(E\setminus U) \le \varepsilon + \lambda (E\setminus U)\le \varepsilon + \lambda (U^c)\le \varepsilon + m_*(E)$$
Since $\varepsilon>0$ is arbitrary, it follows that $m_*(E)=\lambda(E)$.
We have therefore reduce the original problem to proving that $m_*(G\cup F)\ge m_*(G) + m_*(F)$ for any subsets $F,G\subset \mathbb{R}^n$. I leave this to you, you can also find the proof in Halmos' Measure Theory, in the section on inner measures.


TITLE: If $\int {yy''dx} = 3xy$ is it possible to find y'?
QUESTION [4 upvotes]: So one of the people I know on discord gave me this problem to solve for fun.
I've tried using integration by parts on the left side and it didn't do anything.
The integral transforms into the ODE $yy''- (3xy)'=0$ but it's non-linear and I have no idea how to solve that. Would a series solution work?
Notice that I'm looking for y' not y so maybe there's some trick there?
Any help is appreciated. Thank you for reading.
Here, $y'$ and $y''$ denote the first and second order derivatives of $y$ w.r.t. $x$.

REPLY [0 votes]: There is impossible to directly integration of both sides , you have to face directly to solve $yy''-(3xy)'=0$ .
$yy''=3xy'+3y$
$y''=\dfrac{3xy'}{y}+3$
you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=416


TITLE: Groups acting on groups
QUESTION [2 upvotes]: When a group acts on a set, many doors open up. The orbits partition the set, the stabilizers are subgroups, the size of an orbit is the index of the stabilizer, the # of orbits is the average # of fixed points, etc. But when a finite group acts on a finite-dimensional complex vector space, the additional structure reveals even more information about the action. For example, the representation decomposes into irreducible subrepresentations that can be described in a character table.
I would suspect that if a group $G$ acts on a group $H$ in a sensible way, then one can derive additional information beyond the possibilities of a group acting on a set, just like in representation theory. By "sensible way" I mean that the action satisfies the additional property $g * (h h') = ( g * h) (g * h')$ for all $g \in G$ and $h, h' \in H$, which gives a group homomorphism $G \to \mathrm{Aut}(H)$. One example is $G$ acting on itself by conjugation. If $H$ is an abelian group, then 2021 Wikipedia would call this sort of action (together with $H$) a $G$-module, although I think there are different uses of that term.
One consequence of the new property is that
$$g * 1_H = g * (1_H 1_H) = (g * 1_H)(g * 1_H) \in H \text, $$
so $g * 1_H = 1_H$ for all $g \in G$; $1_H$ is always a fixed point of the action.

What are some properties of groups acting on groups that are not always true of groups acting on sets?

Imposing too many constraints can make the phenomenon bland. If $g * (h h') = (g * h)h'$ for all $g \in G$ and $h, h' \in H$, then $g * h = g * (1_H h) = (g * 1_H)h = 1_H h = h$ for all $h \in H$, so the action is trivial. This is related to Notions of groups acting on groups, although that question relinquishes the assumption that $g * (h h') = (g * h) (g * h')$.

REPLY [2 votes]: If a group $G$ acts on another group $H$ (by automorphisms), then a distinguishing condition takes place, namely:
$$\operatorname{Fix}(g)\le H, \forall g\in G \tag 1$$
where $\operatorname{Fix}(g):=\{h\in H\mid \phi_g(h)=h\}$, being $\phi\colon G\longrightarrow \operatorname{Aut}(H)$ the action. In fact, for every $g\in G$, $\phi_g(1_H)=1_H$, whence $1_H\in\operatorname{Fix}(g)\ne\emptyset$; moreover, by definition, $\operatorname{Fix}(g)\subseteq H$; finally, for every $g\in G$ and $h_1,h_2\in\operatorname{Fix}(g)$, $\phi_g(h_1h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2)^{-1}=$ $h_1h_2^{-1}$, whence $h_1h_2^{-1}\in\operatorname{Fix}(g)$.
If $G$ and $H$ are both finite, say $G=\{g_1,\dots,g_{|G|}\}$ and $H=\{h_1,\dots,h_{|H|}\}$, then the usual equation:
$$\sum_{i=1}^{|H|}|\operatorname{Stab}(h_i)|=\sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 2$$
and the usual condition (by Burnside's Lemma):
$$|G|\mid \sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 3$$
give new opportunities, according to the fact that -by Lagrange's Theorem- the summands in the RHS of $(2)$ and in $(3)$ must all divide $|H|$.
As an entry test for this new setting, let's take $G=C_p$ and $H=C_q$, where $p$ and $q$ are distinct primes. If a nontrivial homomorphism exists, then $|\operatorname{Fix}(g_{\bar j})|=1$ and $|\operatorname{Stab}(h_{\bar i})|=1$, for some $\bar j\in \{1,\dots,p\}$, $\bar i\in\{1,\dots,q\}$. Then $(2)$ and $(3)$ yield:
$$[\space k+(q-k)p=l+(p-l)q\Longrightarrow k(p-1)=l(q-1)\space]\space\wedge\space [\space p\mid pq-l(q-1)\space ] \tag 4$$
for some $1\le k\le q$ and $1\le l\le p$. Now:

if $p>q$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l \Longrightarrow$ $l=p\Longrightarrow$ $k(p-1)=p(q-1)\Longrightarrow$ $k=\frac{p}{p-1}(q-1)>q-1\Longrightarrow$ $k=q$; but $(k,l)=(q,p)$ is not a solution of $(4)$-1st term of the "$\wedge$": so, for $p>q$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$;
if $p<q$ and $p\nmid q-1$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l$, and we fall back into the previous case.

Therefore, if $p\nmid q-1$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$. This hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(C_q)$. (Incidentally, that for $p\mid q-1$ there is actually a nontrivial homomorphism $\phi$, it is shown e.g. here.)
As one more application, let's now assume $G=C_p$ and $|H|=q^2$, still with $p,q$ distinct primes. If a nontrivial homomorphism exists, then $(2)$ and $(3)$ yield:
$$[\space k+(q^2-k)p=l_0+l_1q+(p-l_0-l_1)q^2\space]\space\wedge\space [\space p\mid k\space ] \tag 5$$
for some $1\le k\le q^2$ and $1\le l_0+l_1\le p$. Since $p\mid k\Longrightarrow p\le k\le q^2$, for $p>q^2$ there are no nontrivial homomorphisms $\phi$. This suffices to give a different proof to, e.g., this question (and the conclusion doesn't change with $\Bbb Z_2\times \Bbb Z_2$ in place of $\Bbb Z_4$). Again, this hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(H)$.


TITLE: Expected number of non-empty regions
QUESTION [3 upvotes]: Consider $d$ dimensional space cut by $n$ hyperplanes in general position, each one of which goes through the origin. The number of distinct regions created is known to be:
$$2\sum_{i=0}^{d-1} {n -1 \choose i}.$$
Let us assume the hyperplanes are created uniformly at random.  Now we sample $k$ points uniformly at random on the surface of the unit sphere centred at the origin.  How many distinct regions, on average will have at least one point in it?

REPLY [3 votes]: Here is an answer for $d=2$ and general $n.$ Perhaps it simplifies, but I don't immediately see how. The approach might be more effective for  $d \gt 2$ than integrating over a high dimensional simplex.
$$\frac{n}{2^k}\sum_0^k\binom{k}{\ell}\left(\frac{\ell}{n+\ell-1}+\frac{k-\ell}{n+k-\ell-1}\right).  $$  I don't immediately see how to simplify that. The reasoning seems valid and agrees with the answer for $n=2.$
One peculiarity of the problem is that we have regions whose sizes come from some distribution, but we have two of each size. Filtering out that feature makes the case $d=2$, which seems pretty challenging, at least by the approach I outlined in my first answer, much easier to solve.
So, as before, $n$ hyperplanes in general position through the origin in $\mathbb{R}^d$ cut the surface of the unit sphere, each creating two hemispheres and collectively $2R$ regions. Each hemisphere cut into $R$ regions, one of each size. In this version choose a hemisphere (it matters not which), randomly choose $k$ points in that hemisphere, and ask:

What is the expected number of regions with at least one point?

For $d=2$ this translates to: We have a unit interval and pick $n-1$ points splitting it into $n$ subintervals whose sizes come from a Poisson distribution. Then we pick $k$ sample points in the same interval and wonder how many subintervals are hit. Now forget the initial distribution and simply say that first we pick $n+k-1$ points and then randomly pick $n-1$ of them to be dividers and the remaining $k$ to be test points. Suddenly we have stars and bars (note that that reference switches the use of $n,k$) available to us.  If $x_i$ is the number of test points in region $i$ then we have an ordered sum of non-negative integers $\sum_1^nx_i=k.$ The number of such sums is $\binom{k+n-1}{n-1}.$ The number with all the $x_i \gt 0$ is $\binom{k-1}{n-1}$ and the number with exactly $j$ of the $x_i$ positive is $\binom{n}{j}\binom{k-1}{j-1}.$ So the expected number of non-empty regions is $$E(n,k)=\frac{\sum_1^n j\binom{n}{j}\binom{k-1}{j-1}}{\binom{k+n-1}{n-1}}=\frac{nk}{n+k-1}.$$  Comments:

I'll leave out the ugly details of how I summed that with the assurance that calculations agree.
The final answer is so simple that I imagine there is an elegant derivation which escaped me.
Evidently $E(n,k)=E(k,n),$ the answer is unchanged by switching $k$ and $n$. In hindsight, that is clear from the stars and bars.

Here is the reasoning for the formula claimed at the top:
Go back to picking $k$ test points distributed over the entire sphere. Pick any one of the hyperplanes. Then the $k$ points are split into $\ell$ in one half and $k-\ell$ in the other. The chance of any particular split is $2^{-k}\binom{k}{\ell}.$ It follows that the expected number of regions is $$\frac{n}{2^k}\sum_0^k\binom{k}{\ell}\left(\frac{\ell}{n+\ell-1}+\frac{k-\ell}{n+k-\ell-1}\right).  $$


text:
\section{Conclusion and Future Work}
\label{sec:concl}

This paper extended the spectral evolution
model~\citep{kunegis-semodel-2013} to estimate link prediction in
complex networks. It characterizes the evolution of a network in terms
of the evolution of its spectrum, i.e., the eigenvalues of its
adjacency matrix. On the one hand, the proposed approach uses the
Rayleigh quotient for fast computation of eigenvalue
approximations. On the other hand, the estimation of new edges
formation is cast as the problem of predicting eigenvalues'
trajectories. In particular, linear and quadratic regression methods
showcase the proposed approach.  It has been validated that in the
growth of some Twitter mention networks, the eigenvectors remain
largely constant, while their spectrum grows irregularly. Since
computing the spectral decomposition at every time step for each
network is a computationally expensive task, the Rayleigh quotient
proved to be a feasible tool to efficiently calculating the
approximated eigenvalues trajectories. It also avoids the exchange in
the location of eigenvectors seen in stability tests.

It was further observed that the prediction performance of some
methods improve by modifying the train-test ratio in such a way that
the size of the train set increases. Especially, the performance of
the linear and quadratic regression methods are enhanced because the
ratio modifications include more available information for the
prediction. Also, the execution time reduction introduced by the
Rayleigh quotient is noteworthy. In fact, the analysis using this
technique is twice as fast, in comparison to other common techniques,
for most of the Twitter mention networks studied here. For
experimentation, this was crucial because accessing large networks
became an option. Based on the extensive experimental exploration
summarized in this manuscript, the authors strongly believe that
proposed approach has the potential to become a significant addition
to the spectral evolution model approach.

For future research, the use of matrix representations other than the
adjacency matrix need to be researched. In general, it is known that
eigenvalues and eigenvectors are most meaningful when used to
understand a natural operator or a natural quadratic form, such as the
Laplacian matrix. In applications, it would be interesting to apply
the approach to other networks such as gene co-expression networks,
where genes represent vertices and there is a link between two genes
$g_1$ and $g_2$ if there is a positive relation in their expression,
i.e., $g_1$ and $g_2$ use to express at the same
time~\citep{ruan-coexp-2010,stuart-coexp-2003a}. In this case, new
relations in the gene co-expression can be predicted through the
extended spectral evolution model. Finally, considering different
growth methods in the analysis in order to capture more information
from the spectrum of the networks remains an important research
direction.


TITLE: Question about Hermitian Symmetric matrix
QUESTION [1 upvotes]: Assume that $A$ is a Hermitian symmetric n x n matrix with complex entries having all of its eigenvalues lying inside the interval, (-1, 1). Is $A^3 + Id$ always a positive definite matrix? My hunch says yes, but I am not sure how to prove it. Any help would be greatly appreciated.

REPLY [2 votes]: Hint: What are the eigenvalues of $A^3+I$?


TITLE: Set of n variables($A_1,A_2,... A_n$) are linearly independent. Prove that ($A_1^t,A_2^t,... A_n^t$) is also linearly independent
QUESTION [0 upvotes]: Suppose we have n square matrices $A_1,A_2...A_n$ that are linearly independent.Prove that the set $A_1^t,A_2^t..A_n^t$ are also linearly independent

REPLY [3 votes]: Suppose they are not linearly independent so that there exist scalars $\alpha_1,\ldots,\alpha_n$ (not all $0$) such that
$$\alpha_1A_1^t+\cdots+\alpha_nA_n^t=0.$$
What happens if you apply the transpose operation to each side of the equation?
Edit: If $t$ denotes a natural power and not transpose, then the statement is false. The set $$\left\{\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}\right\}$$ is a counterexample in the case of $t=2$.