\chapter{A quantum energy inequality involving local modular data} \cite{Much2022} \begin{equation*} \innerp{\psi}{\energydensity(f)\psi} \ge - \epsilon - \norm{\smash[b]{\Delta}_{\smash[t]{\sharp}}^{-1/2} \ft{g}_{\lambda}(K_{\raisebox{5pt}{\footnotesize$\sharp$}}) \energydensity(f) \fockvaccum} \end{equation*} \section{Misc} \todo{Put this somwhere else.} A \emph{Lorentz transform} is a linear automorphism of Minkowski spacetime which preserves the Lorentz bilinear form. Lorentz transforms are usually represented by (real) $4 \times 4$ matrices, with respect to the standard basis. the \emph{Lorentz group} $\FullLorentzGroup$. \begin{equation*} \FullPoincareGroup = \RR^4 \ltimes \FullLorentzGroup \end{equation*} The relativistic transformation law for one-particle states is given by \begin{equation*} \parens[\big]{U(a,\Lambda) \psi}(p) = e^{ia \cdot p} \psi(\Lambda^{-1} p), \quad \psi \in \hilb{H}, (a,\Lambda) \in \ProperOrthochronousPoincareGroup. \end{equation*} The mapping $(a,\Lambda) \mapsto U(a,\Lambda)$ is a (irreducible) unitary representation of the proper orthochronous Poincaré group $\ProperOrthochronousPoincareGroup$ on the one-particle Hilbert space. By applying any of the second quantization functors we obtain a representation on the multi-particle state space. \begin{equation*} \parens[\big]{U(a,\Lambda) \psi}{}_n(p_1,\ldots,p_n) = e^{ia \cdot (p_1 + \cdots + p_n)} \psi_n(\Lambda^{-1} p_1, \ldots, \Lambda^{-1} p_n), \end{equation*} Poincaré covariance \begin{equation} \label{equation:poincare-covariance-local-algebras} U(g) \localalg{\spacetimeregion{O}} U(g)^* = \localalg{g\spacetimeregion{O}} \qquad g \in \ProperOrthochronousPoincareGroup \end{equation} \begin{definition}{Von Neumann Algebra of Local Observables}{} \begin{equation*} \localalg{\spacetimeregion{O}} = \braces{b(\varphi(f)) \mid b, f \in \realschwartz{M}, \supp f \subset \spacetimeregion{O}}'' \end{equation*} \end{definition} \section{Basic Concepts of Modular Theory} \index{modular!theory} If $\hilb{H}$ is a Hilbert space we shall denote the $C^*$-algebra of all bounded linear operators on $\hilb{H}$ by $B(\hilb{H})$. \begin{definition}{Cyclic and Separating Vectors}{} Suppose $\hilb{H}$ is a Hilbert space and $\mathcal{A}$ is a $C^*$-subalgebra of $B(\hilb{H})$. A vector $\Omega \in \hilb{H}$ is called \begin{itemize} \item \emph{cyclic}\index{cyclic vector} for $\mathcal{A}$ if the vector set $\mathcal{A} \Omega$ is dense in $\hilb{H}$. \item \emph{separating}\index{separating vector} for $\mathcal{A}$ if the map $A \mapsto A \Omega$ from $\mathcal{A}$ into $\hilb{H}$ is injective. \end{itemize} \end{definition} Occasionally, a vector that is both cyclic and separating is called \emph{standard}\index{standard vector}. Recall that the commutant of a set $\mathcal{S} \subset B(\hilb{H})$ of operators is defined as the set of all operators $T \in B(\hilb{H})$ which commute with all operators $S$ in $\mathcal{S}$. We shall denote the commutant of $\mathcal{S}$ by $\mathcal{S}'$.\nomenclature{$\mathcal{A}'$}{commutant of $\mathcal{A}$} \begin{proposition}{}{cyclic-separating} \begin{enumerate}[label=(\roman*),nosep,leftmargin=*,widest=ii] \item A vector is cyclic for $\mathcal{A}$ if and only if it is separating for $\mathcal{A}'$. \item If $\vNa{M}$ is a von Neumann algebra, then a vector is cyclic and separating for $\vNa{M}$ if and only if it is cyclic and separating for $\vNa{M}'$. \end{enumerate} \end{proposition} \begin{proof} \todo{xxx} The second assertion directly follows from the first and the fact that $\vNa{M}'' = \vNa{M}$. \end{proof} If $\Omega$ is separating for $\mathcal{A}$, then every element of $\mathcal{A}\Omega$ is of the form $A\Omega$ with a unique $A \in \mathcal{A}$. This allows us to define an (anti-linear) operator $S_0$ in $\hilb{H}$ with domain $\mathcal{A}\Omega$ by \begin{equation} \label{equation:definition-s0} \quad S_0 A\Omega \defequal S_0 A^*\Omega \qquad A \in \mathcal{A}. \end{equation} The operator $S_0$ is densely defined if and only if $\Omega$ is cyclic for $\mathcal{A}$. Since the $*$-operation on $\mathcal{A}$ is involutive, the range of $S_0$ coincides with its domain. \begin{lemma}{}{} If $\Omega$ is a cyclic and separating vector for a von Neumann algebra $\mathcal{A}$, then the operator $S_0$ defined by~\eqref{equation:definition-s0} is closable. \end{lemma} \begin{proof} By \cref{proposition:cyclic-separating}, $\Omega$ is also cyclic and separating for the commutant $\vNa{A}'$. Hence we may, analogously to $S_0$, define another anti-linear operator $F_0$ in $\hilb{H}$ with dense domain $\mathcal{A}' \Omega$ by \begin{equation*} \quad F_0 B\Omega \defequal F_0 B^*\Omega \qquad B \in \mathcal{A'}. \end{equation*} By definition of $S_0$ and $F_0$ we have for every $A \in \mathcal{A}$ and $B \in \mathcal{A}'$ \begin{equation*} \innerp{S_0 A \Omega}{B \Omega} = \innerp{\Omega}{AB \Omega} = \innerp{\Omega}{BA \Omega} = \innerp{F_0 B\Omega}{A \Omega}. \end{equation*} This adjoint identity establishes that $S_0 \subset F_0^*$. (The \enquote{twisted} appearance of the identity is correct, since it involves anti-linear operators on both sides.) The Hilbert adjoint $F_0^*$ of $F_0$ is closed. Hence, we have shown that $S_0$ has a closed extension, and this implies that $S_0$ is closable. \end{proof} \begin{definition}{Tomita operator}{} Suppose $\Omega$ is a cyclic and separating vector for a von Neumann algebra $\mathcal{A}$. The closure $S = \operatorclosure{S_0}$ of the operator $S_0$ defined on $\mathcal{A}\Omega$ by $S_0 A\Omega = S_0 A^*\Omega$ for $A \in \mathcal{A}$ is called the \emph{Tomita operator}\index{Tomita operator}\index{operator!Tomita}\nomenclature{$S$}{Tomita operator} for the pair $(\mathcal{A},\Omega)$. \end{definition} It is a well-known fact that closed operators can be decomposed in a similar fashion to the polar coordinate representation $z = e^{i\arg z} \abs{z}$ of a complex number. We state the theorem in its somewhat uncommon variant for anti-linear operators, as this is our only use case. \begin{theorem}{Polar Decomposition for Anti-Linear Closed Operators}{polar-decomposition} \index{polar decomposition} Let $T$ be an arbitrary closed anti-linear operator in a Hilbert space $\hilb{H}$. Then there exist a positive selfadjoint linear operator $\abs{T}$ and a partial anti-linear isometry $U$ such that \begin{equation*} T = U \abs{T} \qquad \bracks[\big]{\text{in particular, $\Domain{T} = \Domain{\abs{T}}$}}. \end{equation*} The operators $U$ and $\abs{T}$ are uniquely determined given the additional conditions \begin{equation*} \ker\abs{T} = \ker T \qquad (\ker U)^\perp = (\ker T)^\perp \qquad \ran U = \overline{\ran T}. \end{equation*} \end{theorem} Proofs of this statement are contained in~\cite{ReedSimon1} and~\cite{Schmüdgen2012}. When we speak of \emph{the} polar composition we tacitly assume that the additional conditions ensuring uniqueness are satisfied. Now we are able to introduce the fundamental objects of modular theory. \begin{definition}{Modular Conjugation, Modular Operator}{} Suppose $\vNa{M}$ is a von Neumann algebra acting on a Hilbert space $\hilb{H}$, and suppose $\Omega \in \hilb{H}$ is a cyclic and separating vector for $\vNa{M}$. Let $S$ be the Tomita operator for $(\vNa{M},\Omega)$ and let \begin{equation*} S = J \Delta^{1/2} \end{equation*} be its polar decomposition. The anti-unitary operator $J$ is called \emph{modular conjugation}\index{modular!conjugation}\nomenclature{$J$}{modular conjugation}. The positive selfadjoint operator $\Delta$ is called \emph{modular operator}\index{modular!operator}\index{operator!modular}\nomenclature{$\Delta$}{modular operator}. The pair $(J,\Delta)$ is said to be the \emph{modular data}\index{modular!data}\index{modular!objects} associated to the pair $(\vNa{M},\Omega)$. \end{definition} \todo{clarify why $J$ is anti-unitary} \begin{definition}{Modular Group}{} Adopt the notation of the foregoing definition. The mapping $\RR \ni t \mapsto \Delta^{it}$ is called the \emph{modular group}\index{modular!group} associated to $(\vNa{M},\Omega)$. \end{definition} The modular group is a strongly continuous one-parameter unitary group on $\hilb{H}$. \newpage \begin{proposition}{}{modular-data-unitary} Suppose $\vNa{M}$ is a von Neumann algebra acting on a Hilbert space $\hilb{H}$. Let $U$ be a unitary operator on $\hilb{H}$. Then $U\vNa{M}U^*$ is a von Neumann algebra on $\hilb{H}$. Suppose further that $\Omega \in \hilb{H}$ is a cyclic and separating vector for $\vNa{M}$. Then $U \Omega$ is cyclic and separating for $U\vNa{M}U^*$. Let $(J,\Delta)$ be the modular data associated to $(\vNa{M},\Omega)$. Then $(UJU^*,U{\Delta}U^*)$ is the modular data associated to $(U\vNa{M}U^*,U\Omega)$. \end{proposition} \begin{proof} To prove the first assertion, consider any $A \in (U\vNa{M}U^*)''$. By the double commutant theorem, it suffices to show that $A \in U\vNa{M}U^*$. As $\vNa{M}$ is a von Neumann algebra, this is equivalent to $U^*\! AU \in \vNa{M}''$, again by the double commutant theorem. Let $B \in \vNa{M}'$. It is easy to check that $UBU^* \in (U\vNa{M}U^*)'$. By assumption, $A$ lies in the commutant of $(U\vNa{M}U^*)'$. Thus we find that $[U^*\! AU,B] = U^* [A,UBU^*] U = 0$, as desired. The set of vectors $U\vNa{M}U^* U\Omega = U\vNa{M}\Omega$ is dense in $\hilb{H}$, since it is the image of $\vNa{M} \Omega$ under the homeomorphism $U$. Thus, the vector $U\Omega$ is cyclic for $U\vNa{M}U^*$. Let us show that it is also separating. Suppose $A$ is in $\vNa{M}$ and $UAU^*U\Omega = UA\Omega = 0$ Since unitaries are injective, $A\Omega = 0$. Now $A=0$ follows from the assumption that $\Omega$ is separating for $\vNa{M}$. We have shown that the mapping $UAU^*U\Omega = UA\Omega$ from $U\vNa{M}U^* \to \hilb{H}$ is injective. Let $S = \overline{S_0}$ be the Tomita operator associated to $(\vNa{M},\Omega)$, and let $S' = \overline{S'_0}$ be the Tomita operator associated to $(U\vNa{M}U^*,U\Omega)$. Then we have \begin{equation*} (S'_0 U) A \Omega = S'_0 (U A U^*) U \Omega = (U A^* U^*) U \Omega = U A^* \Omega = U S_0 A \Omega \end{equation*} for all $A \in \vNa{M}$. Consequently, $S'_0 = U S_0 U^*$ as operators with domain $U\vNa{M}\Omega$. Taking the closure, we obtain $S' = U S U^*$. We can write this as $S' = UJU^* U\Delta^{1/2} U^*$, where $S = J \Delta^{1/2}$ is the polar decomposition of the Tomita operator. It is straightforward to check that $UJU^*$ is anti-unitary and $U\Delta^{1/2} U^*$ is positive selfadjoint, and satisfy the additional condition of \cref{theorem:polar-decomposition}. The uniqueness of the polar decomposition implies that $UJU^*$ is the modular conjugation and $U\Delta U^*$ is the modular conjugation associated to the pair $(U\vNa{M}U^*,U\Omega)$. \end{proof} \newpage Finally, let us outline how modular theory enters into algebraic quantum field theory. \begin{theorem}{Reeh-Schlieder Theorem}{reeh-schlieder} \todo{spell it out} \end{theorem} By Reeh-Schlieder (\cref{theorem:reeh-schlieder}), the vacuum $\Omega$ is cyclic and separating for $\localalg{\spacetimeregion{O}}$. Thus, modular theory \section{The Geometric Action of the Modular Operator Associated With a Wedge Domain} \begin{definition}{Right and Left Wedge, General Wedges}{} The \emph{right wedge}\index{wedge!right}\nomenclature[WR]{$\rightwedge$}{right wedge} and \emph{left wedge}\index{wedge!left}\nomenclature[WL]{$\leftwedge$}{left wedge} in Minkowski space $M$ are the open subsets \begin{equation*} \rightwedge \defequal \braces[\big]{x \in M \vcentcolon x^1 > \abs{x^0}} \quad \text{and} \quad \leftwedge \defequal \braces[\big]{x \in M \vcentcolon x^1 < -\abs{x^0}}. \end{equation*} We say that a spacetime region $W \subset M$ is a \emph{wedge}\index{wedge} if there exists an element $g$ of the Poincaré group such that $W = g \rightwedge$. \end{definition} Instead of the right wedge, we could just as well have used the left wedge to define the notion of a general wedge, since they are transformed into each other by space inversion. \begin{lemma}{}{general-wedge-from-right-wedge} If a spacetime region $W$ is a wedge, then there exists an element $g$ of the proper orthochronous Poincaré group such that $W = g \rightwedge$. \end{lemma} \begin{proof} \todo{xxx} \end{proof} In the standard representation of the Lorentz group, the boost (or velocity transformation) along the $x^1$-axis with rapidity $2 \pi t$ is given by the matrix\footnote{ This matrix depends on the choice of metric signature. Ours is $(+,-,-,-)$. For $(-,+,+,+)$, use \begin{equation*} \Lambda(t) = \begin{pmatrix} \phantom{-}\cosh(2 \pi @ t) & -\sinh(2 \pi @ t) & \; 0 \; & \; 0 \; \\ -\sinh(2 \pi @ t) & \phantom{-}\cosh(2 \pi @ t) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}. \end{equation*} } \begin{equation*} \Lambda(t) = \begin{pmatrix} \cosh(2 \pi @ t) & \sinh(2 \pi @ t) & \; 0 \; & \; 0 \; \\ \sinh(2 \pi @ t) & \cosh(2 \pi @ t) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \end{equation*} The following proposition shows that $t \mapsto \Lambda(t)$ is a one-parameter subgroup of the stabilizer group of the right wedge with respect to the action of the Lorentz group on subsets of Minkowski space. \begin{proposition}{}{} \begin{enumerate}[label=(\roman*),nosep,leftmargin=*,widest=ii] \item $\Lambda(s + t) = \Lambda(s) \Lambda(t)$ for all $s,t \in \RR$. \item $\Lambda(t) \rightwedge = \rightwedge$ for all $t \in \RR$. \end{enumerate} \end{proposition} \begin{proof} The first property can be verified by direct computation. Let us prove the second. By definition, the image $\Lambda(t) x$ of a vector $x \in M$ lies in $\rightwedge$ if and only if \begin{equation*} x^0 \sinh(2 \pi t) + x^1 \cosh(2 \pi t) > \abs{x^0 \cosh(2 \pi t) + x^1 \sinh(2 \pi t)}, \end{equation*} or equivalently \begin{equation*} x^0 \parens[\big]{\sinh(2 \pi t) \mp \cosh(2 \pi t)} + x^1 \parens[\big]{\cosh(2 \pi t) \mp \sinh(2 \pi t)} > 0 \end{equation*} for both sign choices. Using the definitions of the hyperbolic sine and cosine, this may be further simplified to $(x^1 \mp x^0) e^{2 \pi t} > 0$, which holds if and only if $x^1 > \abs{x^0}$, since the exponential is always positive. So we have shown that \begin{equation} \label{equation:image-right-wedge} \Lambda(t) x \in \rightwedge \iff x \in \rightwedge. \end{equation} This implies that $\Lambda(t)\rightwedge \subset \rightwedge$ for all $t \in \RR$. Conversely, given an arbitrary vector $y \in \rightwedge$, we have to find $x \in \rightwedge$ such that $\Lambda(t) x = y$. Consider $x = \Lambda(-t) y$. Clearly, $x \in \rightwedge$, because of $y \in \rightwedge$ and~\eqref{equation:image-right-wedge}. Now it follows from $\Lambda(-t) = \Lambda(t)^{-1}$ that in fact $\Lambda(t) x = y$. \end{proof} \begin{theorem}{Bisognano-Wichmann Theorem \textmd{\cite{Bisognano1975}}}{} For the theory of a free scalar field in Minkowski spacetime, let $\spacetimeregion{O} \mapsto \localalg{\spacetimeregion{O}}$ be the net of von Neumann algebras of local observables. If $(J,\Delta)$ is the modular data associated to the algebra $\localalg{\rightwedge}$ of the right wedge and the vacuum $\Omega$, then \begin{equation*} J = \Theta \cdot U\parens[\big]{0, R_{23}(\pi)} \qquad \Delta^{it} = U\parens[\big]{0,\Lambda(t)}, \end{equation*} where $U$ is the theory's unitary representation of the proper orthochronous Poincaré group. \end{theorem} \todo{give definition of $\Theta$ and $R_{23}$} Note that above statement is for the right wedge only. Let us investigate how the modular group changes, if we consider another wedge region. By \cref{lemma:general-wedge-from-right-wedge} any wedge $W$ can be obtained as $W = g\rightwedge$, where $g$ is a proper orthochronous Poincaré transformation. The covariance property~\eqref{equation:poincare-covariance-local-algebras} of $\vNa{R}$ implies \begin{equation*} \localalg{W} = U(g) \localalg{\rightwedge} U(g)^*. \end{equation*} The vacuum $\Omega$ is Poincaré invariant: \begin{equation*} U(g) \Omega = \Omega. \end{equation*} We write $(J_W,\Delta_W)$ for the modular data associated to $(\localalg{W},\Omega)$. By \cref{proposition:modular-data-unitary} \begin{equation*} J_W = U(g) J U(g)^* \qquad \Delta_W = U(g) \Delta U(g)^* \end{equation*} Recall that the modular group $\Delta_W^{it}$ is defined by means of functional calculus. This raises the following problem: given a selfadjoint operator $A$, a unitary operator $U$ and a suitable function $f$ we want to express $f(UAU^*)$ in terms of $f(A)$, if possible. Note that two different functional calculi are at play here, the former is for $UAU^*$ and the latter for $A$. Simple functions such as polynomials suggest $f(UAU^*) = Uf(A)U^*$. That this is generally true is the statement of the following Lemma. \begin{lemma}{}{functional-calclus-unitary-trafo} Suppose that $A$ is a selfadjoint operator on a Hilbert space $\hilb{H}$, with spectral measure $E_A$. Suppose $U$ is an unitary operator on $\hilb{H}$, and let $E_{U\! @AU^*}$ denote the spectral measure of the (selfadjoint) operator $UAU^*$. Then we have $U E_A U^* = E_{U\! @AU^*}$, and \begin{equation*} U f(A) U^* = f(U\! @@AU^*) \end{equation*} for all Borel functions $f : \RR \to \CC$. \end{lemma} \question{Ist diese Aussage korrekt? Ist mein Beweis richtig? Geht der auch einfacher?} \begin{proof} For each regular value $\lambda \in \rho(A)$ let \begin{equation*} R_A(\lambda) = (A-\lambda)^{-1} \end{equation*} denote the resolvent operator of $A$. This proof is based on Stone's Formula \todo{reference}, which relates the resolvent to the spectral projections of $A$: If $E_A$ is the spectral measure of $A$ and $\alpha < \beta$ are real numbers, then \begin{equation*} \stronglim_{\varepsilon \downarrow 0} \frac{1}{\pi i} \int_{\alpha}^{\beta} \bracks{R_A(\lambda + i \varepsilon) - R_A(\lambda - i \varepsilon)} d\lambda = E_A \parens[\big]{\bracks{\alpha,\beta}} + E_A \parens[\big]{\parens{\alpha,\beta}} \end{equation*} Recall that a spectral measure is countably additive. As a consequence, \begin{equation*} \stronglim_{\alpha \uparrow a} \stronglim_{\beta \downarrow b} \stronglim_{\varepsilon \downarrow 0} \frac{1}{2\pi i} \int_{\alpha}^{\beta} \bracks{R_A(\lambda + i \varepsilon) - R_A(\lambda - i \varepsilon)} d\lambda = E_A \parens[\big]{\bracks{a,b}} \end{equation*} for all $a \in \RR \cup \braces{-\infty}$, $b \in \RR \cup \braces{\infty}$. Observe that $\rho(A) = \rho(U\! @AU^*)$ and that for each (common) regular value $\lambda$ we have \begin{equation*} R_{U\! @AU^*}(\lambda) = U R_A(\lambda) @ U^*\!. \end{equation*} Since conjugation with an unitary commutes with the strong operator limit, we obtain \begin{equation*} E_{U\! @AU^*} \parens[\big]{\bracks{a,b}} = U E_A \parens[\big]{\bracks{a,b}} U^* \end{equation*} for all $a,b \in \RR$. The collection $\mathcal{A}$ of all subsets $S$ of $\RR$ such that $E_{U\! @AU^*} \parens[\big]{S} = U E_A \parens[\big]{S} U^*$ is a $\sigma$-algebra on $\RR$. We have shown that all closed intervals belong to $\mathcal{A}$. It is well known that the Borel-$\sigma$-algebra $\mathcal{B}$ of $\RR$ is generated by the closed intervals. Hence, $\mathcal{B} \subset \mathcal{A}$. This shows that the spectral measures $U E_A U^*$ and $E_{U\! @AU^*}$ coincide. \end{proof} \begin{equation*} \Delta_W^{it} = U(g) \Delta^{it} U(g)^* = U(g) U\parens[\big]{0,\Lambda(t)} U(g)^* = U\parens[\big]{g(0,\Lambda(t))g^{-1}} \end{equation*} Recall that Stones Theorem \todo{add reference} states that every strongly continuous one-parameter unitary group is of the form $t \mapsto e^{itK}$ with a uniquely determined selfadjoint operator $K$, which is called \emph{infinitesimal generator} of the group. \begin{definition}{Modular Hamiltonian}{} The infinitesimal generator of the modular group associated to a spacetime region $\spacetimeregion{O}$ is called the \emph{modular Hamiltonian}\index{modular!Hamiltonian}\nomenclature{$K_{\spacetimeregion{O}}$}{modular Hamiltonian for $\spacetimeregion{O}$} for said region, and denoted $K_{\spacetimeregion{O}}$. \end{definition} In other words, $K_{\spacetimeregion{O}}$ is the unique selfadjoint operator such that $\Delta_{\spacetimeregion{O}}^{it} = e^{itK_{\spacetimeregion{O}}}$ for all $t \in \RR$. \begin{proposition}{}{} The modular Hamiltonian for the right wedge is given by $d \Gamma(A)$, where \begin{equation*} A\psi(p) = - \frac{2\pi}{i} \parens[\big]{\partial_0 \psi(p) \, p^1 + \partial_1 \psi(p) \, p^0} \end{equation*} \end{proposition} \section{Complex Lorentz Transformations} \subsection{Analytic Continuation of the Space-Time Translation Group} \subsection{Complex Lorentz Boosts} \begin{lemma}{}{} Suppose $A$ is a selfadjoint operator on some Hilbert space $\hilb{H}$. For all complex numbers $z$ define a closed normal operator $V(z) = e^{izA}$ by means of functional calculus. Let $g$ be a xxx function. Then the range of the bounded operator $g(A)$ is contained in the domain of $V(z)$ for all $z$, and \begin{equation*} V(z) g(A) = \int e^{iz \lambda} g(\lambda) dE_A(\lambda). \end{equation*} \end{lemma} \subsection{A Convolution Theorem for Vector-Valued Tempered Distributions} \blockcquote{Bisognano1975}{% The extension to vector-valued tempered distributions is trivial. } \chapterbib \cleardoublepage % vim: syntax=mytex