Content-Type: multipart/mixed; boundary="-------------0605021523271" This is a multi-part message in MIME format. ---------------0605021523271 Content-Type: text/plain; name="06-145.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="06-145.comments" 48 pages ---------------0605021523271 Content-Type: text/plain; name="06-145.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="06-145.keywords" Supersymmetry, KMS-functional, C*-algebra, graded derivation, entire cyclic cocycle, index, quantum field theory, noncommutative geometry, resolvent algebra ---------------0605021523271 Content-Type: application/x-tex; name="susy-final-01.05.06.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="susy-final-01.05.06.tex" %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 01.05.2006 %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[11pt]{article} \usepackage[english]{babel} \sloppy \addtolength\oddsidemargin{-2cm} \addtolength\evensidemargin{-2cm} \addtolength\textwidth{3.5cm} % \renewcommand{\baselinestretch}{1.5} \renewcommand{\baselinestretch}{1.2} \addtolength\topmargin{-3cm} \addtolength\textheight{6cm} %\addtolength\textheight{7cm} %\addtolength\topmargin{-2cm} %\addtolength\textheight{5cm} %\marginparwidth 4cm %\marginparpush 6pt %\addtolength\oddsidemargin{-1cm} %\addtolength\evensidemargin{-1cm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Special commands for this file % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeatletter \def\@begintheorem#1#2{\trivlist% \item[\hskip \labelsep{\sffamily\bfseries #2\ #1}]\itshape} \newtheorem{teo}{Theorem}[section] \newtheorem{defi}[teo]{Definition} \newtheorem{cor}[teo]{Corollary} \newtheorem{lem}[teo]{Lemma} \newtheorem{pro}[teo]{Proposition} \newtheorem{_rem}[teo]{Remark} \newtheorem{_eje}[teo]{Example} \newtheorem{claim}[teo]{Claim} 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#1,#2.{\overline{\Delta(#1,\,#2)}} \def\frak{\got} \def\wp{\got S} \def\b #1.{{\bf #1}} \def\cross#1.{\mathrel{\mathop{\times}\limits_{#1}}} \def\B{\Theta} % The revolutionary Theta notation. \def\C{\Mb{C}} \def\N{\Mb{N}} \def\P{\Mb{P}} \def\R{\Mb{R}} \def\Z{\Mb{Z}} \def\J{\Mb{J}} \def\un{\Mb{I}} \def\ff{\widehat{f}} \def\fh{\widehat{h}} \def\fk{\widehat{k}} \def\ww{\widehat{\omega}} \def\wh{\widehat} \def\wt{\widetilde} \def\ilim{\mathop{{\rm lim}}\limits_{\longrightarrow}} \def\slim{\mathop{\hbox{\rm s-lim}}} \def\cross #1.{\mathrel{\raise 3pt\hbox{$\mathop\times\limits_{#1}$}}} \def\ol #1.{\overline{#1}} \def\b #1.{{\bf #1}} \def\f #1,#2.{\mathsurround=0pt \hbox{${#1\over #2}$}\mathsurround=5pt} \def\hlf{{\f 1,2.}} \def\ker{{\rm Ker}\,} \def\span{{\rm Span}\,} \def\aut{{\rm Aut}\,} \def\dom{{\rm Dom}\,} \def\ran{{\rm Ran}\,} \def\rep{{\rm Rep}\,} \def\reg{{\rm Reg}} \def\supp{{\rm supp}\,} \def\rlf{{R(\lambda,f)}} \def\zlf{{\zeta(\lambda,f)}} \def\set #1,#2.{\left\{\,#1\;\bigm|\;#2\,\right\}} \def\maprightu #1;{\smash{\mathop{\longrightarrow}\limits^{#1}}} \def\maprightd #1;{\smash{\mathop{\longrightarrow}\limits_{#1}}} \def\maprightt #1,#2.{\mathrel{\smash{\mathop{\longrightarrow}\limits_{#1}^{#2}}}} \def\chop{\hfill\break} \def\s #1.{_{\smash{\lower2pt\hbox{\mathsurround=0pt $\scriptstyle #1$}}\mathsurround=5pt}} \def\v #1.{\mathord{\raise 3pt\hbox{\mathsurround=0pt $\mathop\vee\limits^{#1}$\mathsurround=5pt}}} % \def\bml{{\boldmath \lambda}} \newcommand{\bml}{{\mbox{\boldmath $\lambda$}}} \def\bx{{\bf x}} \def\ClifS{{{\rm Cliff}\big(\al S.(\R)\big)}} \def\ab{\allowbreak} \def\bra #1,#2.{{\left\langle #1,\,#2\right\rangle_{\al A.}}} \def\XC{\marginpar{$\longleftarrow$~{\footnotesize\bf Change}}} \def\XP#1!{\renewcommand{\baselinestretch}{.7}\marginpar{{\footnotesize #1}\hfil} \renewcommand{\baselinestretch}{1.5}} \def\XB{\marginpar{ {\footnotesize\bf Change~starts-----}\lower 11pt\hbox{\mathsurround=0pt$ \!\!\displaystyle{ \Bigg\downarrow}$\mathsurround=3pt}}} \def\XE{\marginpar{{\footnotesize\bf Change~ends-----}\raise 10pt\hbox{\mathsurround=0pt$ \!\!\displaystyle{ \Bigg\downarrow}$\mathsurround=3pt}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % End of the special commands % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{\bf Algebraic Supersymmetry: A case study} \author{ {\sc Detlev Buchholz} \\[1mm] {\footnotesize Institut f\"ur Theoretische Physik,} \\ {\footnotesize Universit\"at G\"ottingen, Friedrich-Hund-Platz 1,} \\ {\footnotesize D-37077 G\"ottingen, Germany} \\ {\footnotesize buchholz@theorie.physik.uni-goettingen.de} \\ {\footnotesize FAX: +49-551-399263} \\ \and {\sc Hendrik Grundling} \\[1mm] {\footnotesize Department of Mathematics,} \\ {\footnotesize University of New South Wales,} \\ {\footnotesize Sydney, NSW 2052, Australia.} \\ %{\footnotesize Australia.} \\ {\footnotesize hendrik@maths.unsw.edu.au} \\ {\footnotesize FAX: +61-2-93857123}} \date{\small Dedicated to Daniel Kastler on the occasion of his 80th birthday} %\date{\today{}} %\date{RUNNING TITLE: Supersymmetry and KMS functionals} \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{abstract} \noindent The treatment of supersymmetry is known to cause difficulties in the C$^*$--algebraic framework of relativistic quantum field theory; several no--go theorems indicate that super--derivations and super--KMS functionals must be quite singular objects in a C$^*$--algebraic setting. In order to clarify the situation, a simple supersymmetric chiral field theory of a free Fermi and Bose field defined on $\R$ is analyzed. It is shown that a meaningful $C^*$--version of this model can be based on the tensor product of a CAR--algebra and a novel version of a CCR--algebra, the ``resolvent algebra''. The elements of this resolvent algebra serve as mollifiers for the super--derivation. Within this model, unbounded (yet locally bounded) graded KMS--functionals are constructed and proven to be supersymmetric. From these KMS--functionals, Chern characters are obtained by generalizing formulae of Kastler and of Jaffe, Lesniewski and Osterwalder. The characters are used to define cyclic cocycles in the sense of Connes' noncommutative geometry which are ``locally entire''. \end{abstract} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction} Graded (super) derivations occur in many parts of physics: supersymmetry, BRS-constraint reduction and cyclic homology, to name a few. To adequately model these in a C*-algebra setting involves notorious domain problems. Kishimoto and Nakamura~\cite{KiNa} showed, for example, that apparently natural domain assumptions on the supersymmetry graded derivations lead to an empty theory. Similarly, supersymmetric KMS--functionals underlying the construction of cyclic cocycles as in \cite{Ka,JLO} cannot exist in the case of infinitely extended systems \cite{BuLo}. These obstructions may explain why a general C$^*$-algebraic framework for supersymmetry has not yet emerged. It thus seems worthwhile to explore representative examples in more detail in order to identify the pertinent structures. In the present article we aim to develop tools to define and analyze in a C*-algebra setting a simple but, with regard to the mathematical problems under investigation, generic supersymmetric quantum field theory. It is the model of a chiral Fermi-- and Bose--field, defined on the light ray $\R$. As the construction of this model is easily accomplished in the Wightman setting of (unbounded) quantum fields, we can concentrate here on the specific problems arising in the passage to a $C^*$--framework. Although the model has formally the structure of a tensor product of a CAR--algebra and a CCR--algebra, the adequate formulation of its $C^*$--version requires some care. It turns out that the standard Weyl algebra description of the CCR part is not suitable for the formulation of supersymmetry. We therefore introduce a more viable variant of the CCR--algebra, the resolvent algebra, which formally may be thought of as being generated by the resolvents of the underlying Bose--field. These resolvents act as mollifiers for the super--derivation and allow one to define it on a domain which is weakly dense in the underlying C$^*$--algebra in all representations of interest. The resolvents also lead to a mollified version of the fundamental relation of supersymmetry, relating the square of the super--derivation and the generator of time translations. These rather weak variants of supersymmetry turn out to be sufficient for the further analysis. Having clarified the C$^*$--algebraic formulation of supersymmetry, one has the necessary tools for the analysis of the supersymmetric KMS--functionals in this model. Again, these functionals are easily constructed in the Wightman setting. Yet, as follows from general arguments \cite{BuLo}, they cannot be extended continuously to the full underlying C$^*$--algebra. In fact, one does not have any \textit{a priori} information on their domains of definition. In the present model, the restrictions of the supersymmetric KMS--functionals to any local subalgebra of the underlying C$^*$--algebra turn out to be bounded. Thus these functionals are densely defined, but their domain of definition does not contain any non--trivial analytic elements with regard to the dynamics, as is required in the construction of cyclic cocycles given in \cite{Ka,JLO}. Nevertheless, by relying on techniques from the theory of analytic functions of several complex variables, it is possible to define cyclic cocycles in the present model as well. The restrictions of these cocycles to any fixed local subalgebra of the underlying C$^*$--algebra turn out to be entire in the sense of Connes \cite{Co}. So the present field--theoretic model allows for a satisfactory C$^*$--algebraic formulation of supersymmetry and the analysis of its consequences. There are three observations which are of interest going beyond the present model: First, a C$^*$--algebraic formulation of supersymmetry has to rely on the concept of mollifiers or, complementary, of unbounded operators affiliated with the underlying C$^*$--algebra \cite{Georg}. Second, there is growing evidence that supersymmetric KMS--functionals, although being unbounded, are locally bounded, in accordance with the heuristic considerations in \cite{BuLo}. And third, although these functionals generically do not have analytic elements in their domain of definition, they can still be used to define local versions of Connes' entire cyclic cocycles by relying on techniques from complex analysis. Based on these insights, a proper C$^*$--algebraic framework for the formulation of supersymmetry and the analysis of its consequences in quantum field theory seems within reach. We hope to return to this problem elsewhere. The plan of our paper is as follows. We will state our results in the body of the paper, and defer almost all the proofs to the appendix. In Sect.~2 we present in the Wightman framework the basic supersymmetry model which we wish to analyze; in Sect.~3 we prepare for its analysis in a C*--setting by considering algebraic mollifying relations for the quantum fields, which leads to the study of the C*--algebras generated by the resolvents of the fields. In Sect.~4 we use these tools to present the C*--algebraic framework of the model. In Sect.~5 we define (unbounded) graded KMS--functionals on the model and prove basic properties for them, including their supersymmetry invariance and their local boundedness. In Sect.~6 we use these KMS--functionals to define a Chern character formula (generalizing the construction in ~\cite{Ka,JLO}), from which we obtain a (locally) entire cyclic cocycle in the sense of Connes. This can then be taken as input to an index theory for supersymmetric quantum field theories, of the type proposed by Longo~\cite{Lo}. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The model} \label{Heur} We begin by presenting here our model in the Wightman framework, which we would like to model in a C*-algebra setting. It is the the simplest example for supersymmetry on noncompact spacetime, in that we have one dimension, one boson and one fermion. We assume chiral fields, so there is only one space-time dimension, $\R.$ The Fermi field is given by the Clifford operators $c(f)=c(f)^*,$ where $f\in\al S.(\R,\,\R)$ and \[ \big\{c(f),\,c(g)\big\}=(f,g):=\int fg\,dx\;. \] The boson field is $j(f)=j(f)^*,$ where $f\in\al S.(\R,\,\R)$ and \[ \big[j(f),\,j(g)\big]=i\sigma(f,g):=i\int fg'\,dx\;. \] The $\Z_2\hbox{--grading}$ automorphism $\gamma$ comes from the Fermi field by \[ \gamma\big(c(f)\big)=-c(f)\,,\qquad \gamma\big(j(f)\big)=j(f) \] and defines even and odd parts of the polynomial field algebra by $A_{\pm}=\big(A\pm\gamma(A)\big)\big/2\,.$ The heuristic supercharge $Q:=\int c(x)j(x)\,dx$ defines the supersymmetry generator $\delta$ as the graded derivation: \[ \delta(A):=[Q,\,A_+]+\{Q,\,A_-\} \] which satisfies $\delta(AB)=\delta(A)B+\gamma(A)\delta(B)\;.$ Note that on the generating elements of the field algebra we have: \begin{equation} \label{SusyD} \delta\big(c(f)\big)=j(f)\,,\qquad \delta\big(j(f)\big)=ic(f')\;. \end{equation} Time evolution is given by translation, i.e. \[ \alpha_t(c(f)):=c(f_t)\,\qquad\alpha_t(j(f)):=j(f_t) \] where $f_t(x):=f(x-t),$ $x\in\R\,.$ The generator of time evolution is the derivation: \begin{equation} \label{TimeD} \delta_0(c(f))=ic(f')\,,\qquad\delta_0(j(f))=ij(f')\;. \end{equation} The supersymmetry relation is valid on the field algebra: \begin{equation} \label{SuSy} \delta^2=\delta_0\;. \end{equation} Our problem is to realize this structure in a C*-algebra setting. Some problems already arise from the relation $\delta((c(f))=j(f),$ in which $\delta$ takes a bounded operator to an unbounded one. We will deal with this issue in the next section. A deeper source of problems will come from the theorems of Kishimoto and Nakamura~\cite{KiNa} which will make it hard to realize the supersymmetry relation (\ref{SuSy}) on a dense domain. % state:\chop % Let $(\al A.,\,\gamma)$ be a graded C*-algebra and $\alpha:\R\to\aut\al A.$ % a pointwise continuous action, with a generator $\delta_0$ such that % $C^\infty(\delta_0):=\bigcap\limits_{n=1}^\infty{\rm Dom}(\delta_0^n)$ % is dense. Let $\delta$ be a closable graded derivation with % $C^\infty(\delta_0)\subset{\rm Dom}(\delta)$ such that % $\delta\circ\alpha_t=\alpha_t\circ\delta$ for all $t,$ and % $\delta^2=\delta_0$ on $C^\infty(\delta_0).$ % Then $\delta$ is bounded. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{On Mollifiers and Resolvent Algebras} \label{Moll} Here we develop tools to handle the unboundedness of the range elements of $\delta\,.$ Recall that a selfadjoint operator $A$ on a Hilbert space $\al H.$ is {\it affiliated} with a C*-algebra $\al A.\subset\al B.(\al H.)$ if the resolvent ${(i\lambda\un-A)^{-1}}\in\al A.$ for some $\lambda\in\R\backslash0$ (hence for all $\lambda\in\R\backslash0).$ This notion is used by Georgescu~\cite{Georg} e.a. (and is weaker than the one used by Woronowicz~\cite{Wor1}) and it implies the usual one, i.e. that $A$ commutes with all unitaries commuting with $\al A.$ (but not conversely). % It also implies that % $\al A.$ contains all bounded continuous functions of $A.$ Observe that \[ A(i\lambda\un-A)^{-1}=\ol(i\lambda\un-A)^{-1}A.=i\lambda(i\lambda\un-A)^{-1}-\un \in\al A.\;. \] Thus the resolvent ${(i\lambda\un-A)^{-1}}=M$ acts as a ``mollifier'' for $A,$ i.e. $\ol MA.$ and $AM$ are bounded and in $\al A.,$ and $M$ is invertible such that $M^{-1}\ol MA.=A=AMM^{-1}.$ This suggests that as $AM$ and $\ol MA.$ in $\al A.$ carries the information of $A$ in bounded form, we can ``forget'' the original representation, and study the affiliated $A$ abstractly through these elements. We want to apply this idea to a representation of the bosonic fields $j(f)=j(f)^*,$ $f\in\al S.(\R)$ where \[ \big[j(f),\,j(g)\big]=i\sigma(f,\, g):=i\int fg'\, dx \] on some common dense invariant core $\al D.\subset\al H.$ of the selfadjoint fields $j(f)\,.$ It seems natural to look for mollifiers in the Weyl algebra \[ \CCR=C^*\set\exp(ij(f)),{f\in\al S.(\R)}.\,, \] (abstractly $\CCR$ is the C*-algebra generated by a set of unitaries ${\set\delta_f,{f\in\al S.(\R)}.}$ such that $\delta_f^*=\delta_{-f}$ and $\delta_f\delta_g=e^{-i\sigma(f,g)/2} \delta_{f+g}).$ Unfortunately this is not possible because: \begin{pro} \label{CCRX} The Weyl algebra $\CCR$ contains no nonzero element $M$ such that $j(f)M$ is bounded for some $f\in\al S.(\R)\backslash0\,.$ Thus $\CCR$ contains no mollifier for any nonzero $j(f),$ and $j(f)$ is not affiliated with $\CCR\,.$ \end{pro} \begin{beweis} Assume that $M\in\CCR$ is nonzero such that $j(f)M$ is bounded for some nonzero $f\in\al S.(\R)\,.$ Let $U(t):=\exp(itj(f)),$ and denote the spectral resolution of $j(f)$ by $j(f)={\int\lambda\,dP(\lambda)},$ then \begin{eqnarray*} \left\|{(U(t)-\un)M}\right\| &=& \bigg\|\int(e^{it\lambda}-1)dP(\lambda)M\bigg\| \\[1mm] &=& |t|\bigg\|\int{(e^{it\lambda}-1)\over t\lambda}\,dP(\lambda) \int\lambda'\,dP(\lambda')M\bigg\| \\[1mm] &\leq& C|t|\|j(f)M\|\longrightarrow 0 \end{eqnarray*} as $t\to0,$ where we used the bound ${|{e^{ix}-1\over x}|}0\,. \end{equation} We define our {\bf resolvent algebra} $\al R.(X,\,\sigma)$ as the abstract C*-algebra generated by $\pi_S(\al R._0),$ i.e. we factor $\al R._0$ by $\ker\pi_S$ and complete w.r.t. the operator norm of $\pi_S\,.$ We state some elementary properties of $\al R.(X,\,\sigma)\,.$ \begin{teo} \label{Relemen} Let $(X,\,\sigma)$ be a given nondegenerate symplectic space, and define $\al R.(X,\,\sigma)$ as above. Then for all $\lambda,\,\mu\in\R\backslash 0$ and $f,\, g\in X\backslash 0$ we have: \begin{itemize} \item[(i)] $[\rlf,\,R(\mu,f)]=0\,.$ Substitute $\mu=-\lambda$ to see that $\rlf$ is normal. \item[(ii)] $\big\|\rlf\big\|=|\lambda|^{-1}\,.$ \item[(iii)] $\rlf$ is analytic in $\lambda\,.$ Explicitly, the series expansion: \[ \rlf=\sum_{n=0}^\infty(\lambda_0-\lambda)^n\,R(\lambda_0,f)^{n+1}i^n, \qquad\lambda,\;\lambda_0\not=0\qquad\qquad \hbox{(Von Neumann series)} \] converges in norm whenever $|\lambda_0-\lambda|<|\lambda_0|\;.$ \item[(iv)] $R(\lambda,tf)$ is norm continuous in $t\in\R\backslash 0\,.$ \item[(v)] $\rlf R(\mu,g)^2\rlf= R(\mu,g)\rlf^2 R(\mu,g)\,.$ \item[(vi)] Let $T\in{\rm Sp}(X,\sigma)$ be a symplectic transformation. then $\alpha\big(\rlf\big):=R(\lambda,Tf)$ defines a unique automorphism $\alpha\in\aut\al R.(X,\,\sigma)\,.$ \end{itemize} \end{teo} Note that the von Neumann series for $\rlf$ converges for any $z\in\C$ with $|z-\lambda_0|<|\lambda_0|,$ i.e. on a disk which stays off the real line. Using different $\lambda_0's$ we can thus define $R(z,f)$ for any complex $z$ not on the real line. and deduce the properties in the definition for these from the series. Thus we obtain also resolvents $R(z,f)$ for complex $z$ in $\al R.(X,\,\sigma)\,.$ Any operator family $R_{\lambda}$ satisfying the resolvent equation (\ref{Resolv}) is called by Hille a pseudo-resolvent (cf. p215 in~\cite{Yos}), and for such a family we know (cf. Theorem~1 p216 in~\cite{Yos}) that: \begin{itemize} \item{} All $R_\lambda$ have a common range and a common null space. \item{} A pseudo resolvent $R_{\lambda}$ is the resolvent for an operator $B$ iff $\ker R_\lambda=\{0\}\,,$ and in this case $\dom B=\ran R_\lambda$ for all $\lambda\,.$ \end{itemize} Thus we define: \begin{defi} A {\bf regular representation} $\pi\in\rep\al R.(X,\,\sigma)$ is a Hilbert space representation such that \[ \ker\pi\big(R(1,f)\big)=\{0\}\qquad\forall\;f\in \al S.(\R,\R)\backslash 0\;. \] We denote the collection of regular representations by $\reg\,.$ \end{defi} Obviously many regular representations are known, e.g. $\pi_S$ and the Fock representation. Given a $\pi\in\rep\al R.(X,\,\sigma)$ with $\ker\pi\big(R(1,f)\big)=\{0\},$ we can define a field operator by \[ j_\pi(f):=i\un-\pi\big(R(1,f)\big)^{-1}\; \] with domain $\dom j_\pi(f)=\ran\pi\big(R(1,f)\big)\,.$ Thus for $\pi\in\reg,$ all the field operators $j_\pi(f),$ $f\in \al S.(\R,\R)$ are defined, and we have the resolvents $\pi(\rlf)=(i\lambda\un-j_\pi(f))^{-1}\,.$ \begin{teo} \label{RegThm} Let $\al R.(X,\,\sigma)$ be as above, and let $\pi\in\rep\al R.(X,\,\sigma)$ satisfy $\ker\pi\big(R(1,f)\big)=\{0\}= \ker\pi\big(R(1,h)\big)$ for given $f,\; h\in X.$ Then \begin{itemize} \item[(i)] $j_\pi(f)$ is selfadjoint, and $\pi(\rlf)\dom j_\pi(h) \subseteq\dom j_\pi(h)\,.$ \item[(ii)] $\lim\limits_{\lambda\to\infty}i\lambda\pi(\rlf)\psi=\psi$ for all $\psi\in\al H._\pi,$ \item[(iii)] $\lim\limits_{s\to 0}i\pi(R(1,sf))\psi=\psi$ for all $\psi\in\al H._\pi.$ \item[(iv)] The space $\al D.:={\pi\big(R(1,f)R(1,h)\big)\al H._\pi}$ is a joint dense domain for $j_\pi(f)$ and $j_\pi(h)$ and we have: $[j_\pi(f),\,j_\pi(h)]=i\sigma(f,h)$ on $\al D.,$ \item[(v)] $j_\pi(\lambda f+h)=\lambda j_\pi(f)+j_\pi(h)$ for all $\lambda\in\R$ on $\al D.,$ \item[(vi)] $j_\pi(f)\pi(\rlf)=\pi(\rlf)j_\pi(f)=i\lambda\pi(\rlf)-\un$ on $\dom j_\pi(f),$ \item[(vii)] $\big[j_\pi(f),\pi(R(\lambda,h))\big]=i\sigma(h,f)\pi(R(\lambda,h)^2)$ on $\dom j_\pi(f),$ \item[(viii)] Denote $W(f):=\exp(ij_\pi(f))\,,$ then \begin{eqnarray*} W(f)W(h) &=& e^{i\sigma(f,h)}W(h)W(f) \\[1mm] W(f)\pi\big(R(\lambda,h)\big)W(f)^* &=& \pi\big(R(\lambda+i\sigma(f,h),\,h)\big)\,. \end{eqnarray*} Moreover $W(f)\al D.\subseteq\al D.\supseteq W(h)\al D.,$ hence $\al D.:={\pi\big(R(1,f)R(1,h)\big)\al H._\pi}$ is a common core for $j_\pi(f)$ and $j_\pi(h)\,.$ \end{itemize} \end{teo} A distinguished regular representation of $\al R.(X,\,\sigma)$ is of course the defining strongly regular representation $\pi_S.$ By definition $\al R.(X,\,\sigma)$ is faithfully represented in it, and moreover, there is a common dense invariant domain $\al D._0$ for all the field operators $j\s\pi_S.(f),$ $f\in X.$ This domain can be enlarged to a dense invariant domain $\al D._S$ for both the resolvents and the fields simply by applying all polynomials in $j\s\pi_S.(f)$ and $\pi_S(R(\lambda,f))$ to $\al D._0,$ which makes sense, because from (i) above all resolvents preserve the joint domain $\mathop{\bigcap}\set \dom j\s\pi_S.(f),{f\in X}..$ Thus we can form the *-algebra of (unbounded) operators \[ \al E._0:=*\hbox{--alg}\set{ j\s\pi_S.(f),\; \pi_S(R(\lambda,f))},f\in X,\;\lambda\in\R\backslash 0. \] on $\al D._S\,.$ Then $\al E._0$ contains of course the *-algebra $\pi_S(\al R._0)$ generated by resolvents alone, which is dense in $\al R.(X,\,\sigma).$ We will need these *-algebras $\al E._0\supset\pi_S(\al R._0)$ below, and will generally not indicate the faithful representation $\pi_S$ w.r.t. which they are defined. Note that for any strongly regular state $\omega,$ its cyclic GNS-vector is in the domain of all $j\s\pi_\omega.(f)\,,$ hence $\omega$ extends to define a functional on $\al E._0\,.$ Thus we give a meaning to all expressions of the form \begin{eqnarray*} &&\!\! \omega\big(j(f_1)\cdots j(f_n)R(\lambda_1,g_1)\cdots)R(\lambda_k,g_k)\big) \\[1mm] &&\qquad :=\left(\Omega\s\omega.,\,j\s\pi_\omega.(f_1)\cdots j\s\pi_\omega.(f_n) \pi_\omega\big(R(\lambda_1,g_1)\cdots)R(\lambda_k,g_k)\big)\Omega\s\omega. \right) \end{eqnarray*} as above. A very important class of states on $\CCRX$ are the quasifree states, which we will need below. They are given by \[ \omega(\delta_f) = \exp\big(-\hlf\langle f | f \rangle_\omega\big) , \quad f \in X, \] where $\langle \, \cdot \, | \, \cdot \, \rangle_\omega $ is a (possibly semi--definite) scalar product on the complex linear space $X + i X$ satisfying $$ \langle f | g \rangle_\omega- \langle g | f \rangle_\omega = i \sigma(f,g), \quad f,g \in X. $$ Any quasifree state is also regular in the strong sense. By a routine computation one can represent the expectation values of products of Weyl operators in a quasifree state in the form $$ \omega(\delta_{f_1} \cdots \delta_{f_n}) = \exp\Big(- \sum_{k 0$ \begin{eqnarray} & & \omega(R(\lambda_1,f_1) \cdots R(\lambda_n,f_n)) \nonumber\\[1mm] & & = (-i)^n \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, e^{-{\sum}_k t_k \lambda_k} \omega(\delta\s t_1 f_1. \cdots \delta\s t_n f_n.) \nonumber\\[1mm] \label{QFresolvent} & & = (-i)^n \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, \exp\Big(-\sum_k t_k \lambda_k - \sum_{k 0$. The relation (\ref{QFresolvent}) should be regarded as the definition of quasifree states on the resolvent algebra. In our calculations below, we will frequently need the following differentiablility of quasifree states: \begin{pro} \label{MasterLemma} Let $\omega$ be a quasifree state as above, and let $x_k\in\R \mapsto f_k(x_k) \in X$, $k = 1, \ldots n$ be paths in $X$ for which the functions $ x_k,x_l \longmapsto \langle f_k( x_k) | f_l( x_l) \rangle_\omega, \quad k,l = 1, \dots n,$ are smooth. Then \begin{eqnarray*} &&\!\!\!\!\! \frac{\partial}{\partial x_r} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) \\[1mm] &=& - \sum_{k=1}^{r-1} \frac{\partial }{\partial x_r}\Big(\big\langle f_k(x_k) | f_r(x_r) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r\partial\lambda_k} % \frac{\partial}{\partial \lambda_k} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] &&\qquad-\hlf\frac{\partial }{\partial x_r}\Big(\big\langle f_r(x_r) | f_r(x_r) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r^2} % \frac{\partial}{\partial \lambda_r} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] &&\qquad - \sum_{k=r+1}^n \frac{\partial }{\partial x_r}\Big( \big\langle f_r(x_r) | f_k(x_k) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r\partial \lambda_k} % \frac{\partial}{\partial \lambda_k} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) \end{eqnarray*} % \begin{eqnarray*} % &&\!\!\!\!\! \frac{\partial}{\partial x_r} \, % \omega\Big(R(\mu_1, f_1(x_1)) \cdots R(\mu_n, f_1(x_n))\Big) % \\[1mm] % &=& - \sum_{k=1}^n % \frac{\partial }{\partial x_r}\Big(\big\langle f_r(x_r) | % f_k(x_k) \big\rangle_\omega + % \big\langle f_k(x_k) | % f_r(x_r) % \big\rangle_\omega \Big) % \, \frac{\partial}{\partial \mu_r} \frac{\partial}{\partial \mu_k} % \, \omega\Big(R(\mu_1, f_1(x_1)) \cdots R(\mu_n, f_1(x_n))\Big). % \end{eqnarray*} and all the partial derivatives involved in this formula exist. \end{pro} \section{C*-algebra formulation of Supersymmetry.} Here we want to write our model of Section~\ref{Heur} in a C*-algebra framework. However, to motivate our choices made below, let us recall a theorem of Kishimoto and Nakamura~\cite{KiNa}: \begin{teo} Let $\al A.$ be a C*-algebra with $\Z_2\hbox{--grading}$ $\gamma,$ let $\alpha:\R\to\aut\al A.$ be a pointwise continuous action with generator $\delta_0$ having a smooth domain $C^\infty(\delta_0):= \bigcap\limits_{n=1}^\infty\dom(\delta_0^n)\,.$ Let $\delta$ be a closable graded derivation with $\dom(\delta)\supset C^\infty(\delta_0),$ $\delta\circ\alpha_t =\alpha_t\circ\delta$ for all $t,$ and $\delta^2=\delta_0$ on $ C^\infty(\delta_0).$ Then $\delta$ is bounded. \end{teo} Thus it will be hard to obtain the supersymmetry relation on natural dense domains. For the fermion field, let $\al H.=L^2(\R)$ and define ${\rm CAR}( \al H.)$ in Araki's self-dual form (cf.~\cite{Ar}) as follows. On $\al K.:=\al H.\oplus\al H.$ define an antiunitary involution $\Gamma$ by $\Gamma(h_1\oplus h_2):=\ol h._2\oplus\ol h._1\;.$ Then ${\rm CAR}(\al H.)$ is the unique simple C*--algebra with generators $\set\Phi(k),k\in{\al K.}.$ such that $k\to\Phi(k)$ is antilinear, $\Phi(k)^*=\Phi(\Gamma k)\,,$ and \[ \left\{\Phi(k_1),\,\Phi(k_2)^*\right\}=(k_1,\,k_2)\un\;,\qquad k_i\in\al K.\;. \] The correspondence with the heuristic creators and annihilators of fermions is given by $\Phi(h_1\oplus h_2)=a(h_1)+a^*(\ol h._2)\;,$ where \[ a(h)=\int a(x)\,\ol h(x).\,dx\;,\qquad a^*(h)=\int a^*(x)\, h(x)\, dx\;. \] To obtain the Clifford operators $c(f)=c(f)^*,$ $f\in\al S.(\R,\,\R)$ we take $c(f):=\Phi(f\oplus f)/\sqrt{2}\,,$ in which case we have $c(f)=c(f)^*$ and $\{c(f),\,c(g)\}=(f,g)=\int fg\,dx\,.$ Let $\ClifS:=C^*\set c(f),{f\in\al S.(\R)}.$ and notice that $\wt{c}(f):=i\Phi(f\oplus -f)/\sqrt{2}$ also satisfies the Clifford relations, hence generates another copy of $\ClifS$ in ${\rm CAR}(\al H.)\,,$ and together these two Clifford algebras generate all of ${\rm CAR}(\al H.)\,.$ In fact, since the $c(f)$ and $\wt c(g)$ anticommute, we have that ${\rm CAR}(\al H.) \cong{{\rm Cliff}\big(\al S.(\R)\oplus\al S.(\R)\big)}\,.$ Conversely, if we are given a real pre-Hilbert space $X$ with complexified completion $Y$ and a projection $P$ and antiunitary involution $\Gamma$ such that $\Gamma P\Gamma=\un-P$ and these preserve $X,$ then we have an isomorphism ${\rm Cliff}(X)\cong{\rm CAR}(Y)$ given by $\Phi(x)=\big(c(Px)-ic(\Gamma Px) +c(P\Gamma x)+ic((\un-P)x)\big)/\sqrt{2}\,.$ %% % [Can't see why the C*-alg generated by $\{c(f);f\in\al S.\}$ is % ${\rm CAR}(\al H.)$- not true in one dim, when $a=\sigma_x+i\sigma_y$ % and $c=(a+a^*)/\sqrt{2}=\sigma_x\sqrt{2}$]\chop For the bosonic part we take the resolvent algebra $\al R.(\al S.(\R),\sigma)$ where $\sigma(f,g):=\int fg'\,dx\,,$ and so the full C*-algebra in which we want to define our model is \[ \al A.:=\ClifS\otimes\al R.(\al S.(\R),\sigma) \] where the tensor norm is unique because the CAR-algebra is nuclear. The grading automorphism $\gamma$ is the identity on $\al R.(\al S.(\R),\sigma)\,,$ and $\gamma(\Phi(k))=-\Phi(k)$ for all $k$ on the CAR-part. Next, we want to define on some suitable domain in $\al A.$ the supersymmetry graded derivation $\delta$ corresponding to the relations~(\ref{SusyD}). First, considering $\delta\big(j(f)\big)=ic(f'),$ since $\delta$ is a derivation on the bosonic part, it is natural to define \begin{equation} \label{DerR} \delta(\rlf):=ic(f')\,\rlf^2\quad\in\al A.\,. \end{equation} However due to the unbounded rhs of $\delta\big(c(f)\big)=j(f)$ we cannot define $\delta$ directly on the $c(f),$ so we need to multiply by mollifiers. Define \begin{eqnarray} \label{DefZ} \zeta(f)&:=&c(f)R(1,f)\,, \\[1mm] \label{DerZ} \hbox{then}\quad\qquad \delta(\zeta(f))&:=&i R(1 ,f)-\un+ic(f)c(f')R(1,f)^2\quad\in\al A. \end{eqnarray} where we made use of the graded derivation property, the relations~(\ref{SusyD}) and $j(f)R(\lambda,f)=i\lambda R(\lambda,f)-\un\,.$ Next, we would like to extend $\delta$ as a graded derivation to the *-algebra generated by these basic objects: \[ \al D._S:=\hbox{*--alg} \set{\un,\;\rlf,\,\;\zeta(f)},{\lambda\in\R\backslash 0,\; f\in\al S.(\R)\backslash 0}.\subset\al A.\;. \] Observe that $\al D._S$ is not norm-dense in $\al A.,$ however due to~Theorem~\ref{RegThm}(ii) applied to $R(\lambda,f)c(f)= \zeta(f/\lambda),$ it will be strong operator dense in $\al A.$ in any regular representation. Note that $\delta$ does not preserve $\al D._S,$ it takes its image in the norm dense *-algebra \[ \al A._0:=\hbox{*--alg} \set{\un,\;\rlf,\,\;c(f)},{\lambda\in\R\backslash 0,\; f\in\al S.(\R)\backslash 0}.\subset\al A.\;. \] To see that $\delta$ extends as a graded derivation to $\al D._S,$ we proceed as follows. Let $\pi_0$ be any representation of % ${\rm CAR}(\al H.),$ $\ClifS$ then $\pi_0\otimes\pi_S$ is a faithful representation of $\al A.$ and there is a common dense invariant domain $\al D.:=\al H.\s\pi_0. \otimes\al D._S$ for all $\pi_0(c(f))\otimes\un,$ $\un\otimes j\s\pi_S.(f)$ and $\un\otimes \pi_S(\rlf)$ where $\al D._S$ denotes the domain of $\al E._0$ defined at the end of Section~\ref{Moll}. (Henceforth we will not indicate tensoring by $\un$ nor the representations $\pi_0,\;\pi_S$ when the context makes clear what is meant). Let \[ \al E.:=*\hbox{--algebra}\set{c(f),\;j(f),\;\rlf},{f\in\al S.(\R),\; \lambda\in\R\backslash 0}.\supset\al A._0 \] so we have the *-algebras $\al R._0\subset\al E._0\subset\al E.$ on $\al D.\,.$ Define on the generating elements of $\al E.$ a map $\overline{\delta}$, setting \begin{eqnarray*} \overline{\delta}(j(f)) &=& i c(f^\prime), \\[1mm] \overline{\delta}(R(\lambda,f)) &=& i c(f^\prime) R(\lambda,f)^2, \\[1mm] \overline{\delta}(c(f)) &=& j(f). \end{eqnarray*} We will see that this map extends to a graded derivation on $\al E.$. For the proof it suffices to show that $\overline{\delta}$ is linear and satisfies the graded Leibniz rule on any finite polynomial involving operators $j(f)$, $R(\lambda,f)$ and $c(f)$, i.e. in each instance only a finite number of test functions $f$ and real parameters $\lambda$ are involved. We will take advantage of this fact as follows. Let $X_s \subset \al S.(\R)$ be any \textit{finite}--dimensional subspace and consider the subalgebra $\al E.(X_s) \subset \al E.$ generated by the elements $j(f)$, $R(\lambda,f)$ and $c(f)$ with $\lambda \in \R\backslash 0$, $f \in X_s$. We extend $X_s$ to a space $ {X_s}^\prime \subset\al S.(\R) $ by adding to the elements of $X_s$ also their \textit{first} derivatives. Picking in ${X_s}^\prime$ some (finite) orthonormal basis $\{h_n\}$ with regard to the scalar product $( \cdot , \, \cdot ),$ we have for any $f \in X_s$ the ``completeness relations'' $ \sum_n \, ( h_n , f) \, h_n = f, \ \sum_n \, ( h_n , f^\prime ) \, h_n = f^\prime $. Next, we define an operator $Q_s \in \al E.$, setting $$ Q_s = {\sum}_n c(h_n) j(h_n). $$ As $Q_s$ is of fermionic (odd) type, we can consistently define with the help of it a graded derivation $\overline{\delta}_s$ on $\al E.$, setting for even and odd elements $E_\pm \in \al E.$, respectively, $$ \overline{\delta}_s(E_+) = [Q_s, E_+] = Q_s E_+ - E_+ Q_s, \quad \overline{\delta}_s(E_-) = \{Q_s,E_-\} = Q_s E_- + E_- Q_s. $$ Computing the action of $\overline{\delta}_s$ on the even elements $j(f)$, $R(\lambda,f)$ and odd elements $c(f)$, where $\lambda \in \R\backslash 0$, $f \in X_s$, we obtain from the basic relations in $\al E.$ by some elementary algebraic manipulations. \begin{eqnarray*} \overline{\delta}_s(j(f)) &=& i {\sum}_n \, c(h_n)\, ( h_n , f^\prime ) = i c(f^\prime)\,, \\[1mm] \overline{\delta}_s(R(\lambda,f)) &=& i {\sum}_n \, c(h_n) \,( h_n , f^\prime) \, R(\lambda,f)^2 = i c(f^\prime) R(\lambda,f)^2, \\[1mm] \overline{\delta}_s(c(f)) &=& {\sum}_n \, ( h_n,\, f)\, j(h_n) = j(f). \end{eqnarray*} Thus we conclude that the action of $\overline{\delta}$ on the generating elements of $\al E.(X_s)$ coincides with the action of the graded derivation $\overline{\delta}_s$. As the choice of the subspace $X_s$ was arbitrary, it follows that $\overline{\delta}$ extends to a graded derivation on the whole polynomial algebra $\al E.$. The final step consists in showing that the action of $\delta$ on the generating elements $R(\lambda,f)$, $\zeta(f)$ of $\al D._S$ coincides with the action of the graded derivation $\overline{\delta}$. But this follows immediately from the relations given above. Thus $\delta$ extends to a graded derivation with domain $\al D._S\,$ and range in $\al A._0$. Uniqueness is clear from the graded derivation property, so we have proven: \begin{teo} \label{DwellDf} There is a unique graded derivation $\delta:\al D._S\to\al A.$ satisfying relations (\ref{DerR}) and (\ref{DerZ}). \end{teo} Next we need to define the time evolution derivation $\delta_0$ in this C*-setting. From the equations (\ref{TimeD}) this suggest that we define on $\al E.$ a *-derivation $\overline{\delta}_0$ satisfying: \begin{eqnarray*} % \label{TimeE} \overline{\delta}_0(j(f)) &=& i j(f^\prime), \\[1mm] \overline{\delta}_0(R(\lambda,f)) &=& i R(\lambda,f) j(f^\prime) R(\lambda,f), \\[1mm] \overline{\delta}_0(c(f)) &=& i \, c(f^\prime) \end{eqnarray*} and then proceed to the corresponding mollified relations in $\al A..$ For the proof that $\overline{\delta}_0$ extends to a % (skew symmetric) *-derivation on $\al E.,$ we proceed as in the discussion of the superderivation: We pick any finite dimensional subspace $X_s \subset \al S.(\R),$ consider the corresponding subalgebra $\al E.(X_s) \subset \al E.$ and choose in the extended space ${X_s}^\prime \subset\al S.(\R)$, containing the elements of $X_s$ and their first derivatives, some orthonormal basis $\{h_n\}$. In addition to the completeness relations mentioned above we will also make use of $ \sum_n \, (h_n, f ) \, h_n^\prime = f^\prime$ for $f \in X_s$. We consider now the symmetric operator in $\al E.$ $$ H_s = \frac{1}{2} \, {\sum}_n \big\{ i c({h_n}^\prime) c(h_n) + j(h_n) j(h_n) \big\}. $$ Putting $$ \overline{\delta}_{0 \, s}(\,\cdot \,) = [H_s, \, \cdot \,], $$ it induces a *--derivation on $\al E.$. Its action on the generating elements of $\al E.(X_s)$ can easily be computed: \begin{eqnarray*} \overline{\delta}_{0 \, s}(j(f)) &=& {\sum}_n \, j(h_n) \, i (h_n , f^\prime ) = i \, j(f^\prime), \\[1mm] \overline{\delta}_{0 \, s}(R(\lambda,f)) &=& {\sum}_n \, i (h_n , f^\prime )\, R(\lambda,f) j(h_n) R(\lambda,f) = i \, R(\lambda,f) j(f^\prime) R(\lambda,f), \\[1mm] \overline{\delta}_{0 \, s}(c(f)) &=& \frac{1}{2} \, {\sum}_n \, i \big\{ ( h_n , f )\, c({h_n}^\prime) + (h_n , f^\prime )\, c(h_n) \big\} = i \, c(f^\prime). \end{eqnarray*} Thus we conclude as in the preceding discussion that the action of $\overline{\delta}_0$ on the generating elements of $\al E.(X_s)$ coincides with the action of the derivation $\overline{\delta}_{0 \, s}$. As $X_s$ was arbitrary, it follows that $\overline{\delta}_0$ extends to a derivation on the whole polynomial algebra $\al E.$. The supersymmetry relation $\overline{\delta}^2 = \overline{\delta}_0$ can now be verified on the generating elements of $\al E.$ and thus holds on the whole algebra $\al E..$ The question now is how one should define the time evolution $\delta_0$ and the square $\delta^2$ on the C*-algebra $\al A.$ from the unbounded versions in $\al E..$ Since $\delta:\al D._S\to\al A._0,$ its square $\delta^2$ does not make sense on $\al D._S.$ Note however that for every $A\in\al A._0$ there is a monomial $M\in\al D._S$ of resolvents $\rlf$ such that \[ AM\in\al D._S\ni MA\,. \] (By Theorem~\ref{RegThm}(ii) we know that in regular representations we can let these mollifiers $M$ go to $\un$ in the strong operator topology.) \begin{defi} \label{deltasquared} For each $A\in \al D._S$ let $M_A\in \al D._S$ be a monomial of resolvents $\rlf$ such that $M_A\delta(A)\in \al D._S.$ Define \[ M_A\delta^2(A):=\delta\big(M_A\delta(A)\big)-\delta(M_A)\delta(A)\in\al A._0\;. \] \end{defi} Note that this definition coincides with $M_A\overline\delta^2(A)$ in $\al E.,$ however the definition above involves only bounded quantities, so it can be defined independently in the C*-setting on $\al D._S\,.$ Of course we then have the mollified SUSY--relations $M_A\delta^2(A)=M_A\overline{\delta}_0(A)$ for all $A\in\al D._S$ from the unbounded SUSY relation in $\al E.\,.$ This is not however acceptable for a bounded SUSY--relation until we have demonstrated the connection of $M_A\overline{\delta}_0(A)$ with the time evolution. The time evolution $\alpha:\R\to\aut\al A.$ is just translation, as this is a chiral theory \[ \alpha_t(c(f)):=c(f_t)\,,\quad\alpha_t\left(\rlf\right) =R(\lambda,f_t)\;. \] The desired connection \begin{equation} \label{timeconn} M_A\overline{\delta}_0(A)=-i{d\over dt} M_A\,\alpha_t(A)\Big|_0 \end{equation} exists only in specific regular representations on suitable domains, and for these one will then have supersymmetry. In many applications, one only needs the supersymmetry weakly, i.e. \[ \omega(BM_A\overline{\delta}_0(A)C)=-i{d\over dt} \omega(BM_A\,\alpha_t(A)C)\Big|_0=\omega(BM_A\delta^2(A)C) \] for $A,B,C$ in a suitable domain and $\omega$ a distinguished functional. We will verify this relation explicitly below for the functionals used in our constructions. \section{ Graded KMS--functionals.} Graded KMS--functionals are used in supersymmetric theories to calculate cyclic cocycles~\cite{JLO, Ka}, and here we want to develop this theory in the current context for our simple supersymmetric model as a first application of it. \begin{defi} \label{KMSfunDef} Let $\al A.$ be a unital C*-algebra with a grading automorphism $\gamma\in\aut\al A.,$ $\gamma^2=\iota,$ and a (pointwise continuous) action $\alpha:\R\to \aut\al A.$ such that $\alpha_t\circ\gamma =\gamma\circ\alpha_t$ for all $t\,.$ Then a {\bf graded KMS--functional} is a (possibly unbounded) functional $\varphi$ on $\al A.$ such that \begin{itemize} \item[(i)] $\dom\varphi$ is a unital dense *--subalgebra of $\al A.$ such that \[ \gamma(\dom\varphi)\subseteq\dom\varphi\supseteq\alpha_t(\dom\varphi) \quad\forall\,t, \] \item[(ii)] For all $A,\,B\in\dom\varphi$ there is a continuous complex function $F\s{A,B}.:S\to\C$ on the strip $S:=\R+i{[0,1]}$ which is analytic on the interior of $S$ and satisfying on the boundary: \begin{eqnarray*} F\s{A,B}.(t)&=& \varphi\left(A\,\alpha_t(B)\right)\quad\forall\,t \\[1mm] F\s{A,B}.(t+i)&=&\varphi\left(\alpha_t(B)\gamma(A)\right)\quad\forall\,t\in\R\,. \end{eqnarray*} \item[(iii)] For $A,\,B\in\dom\varphi$ we have \[ \left|F\s{A,B}.(t+is)\right|0$ but are independent of $n.$ Thus condition ~(\ref{EntCon}) holds for $\tau,$ i.e. \[ \lim_{n\to\infty}n^{1/2}\|\tau_n\|_*^{1/n}=0\;. \] \end{teo} Using this, we can now prove that: \begin{teo} \label{TCycCoc} The sequence $\wt\tau:=(\tau_0,0,-\tau_2,0,\tau_4,0,-\tau_6,\ldots)\in \al C.(\al D._k)$ defines an entire cyclic cocycle for each $k>0\,,$ i.e. \[ (b\tau_{n-1})(a_0,\ldots,a_n)=(B\tau_{n+1})(a_0,\ldots,a_n) \,,\qquad n=1,3,5,\ldots \] and the entire analyticity condition holds. \end{teo} It is possible to have taken the choice $\partial=b-B$ for the cyclic coboundary operator above;- this is in fact done in \cite{JO1}, and would have led to the cyclic cocycle ${(\tau_0,0,\tau_2,0,\tau_4,0,\ldots)}$ instead of $\wt\tau$ above. Note that whilst we have obtained entire cyclic cocycles on each compact set $[-k,\,k]$ these do not define an entire cyclic cocycle on $\al D._{\rm comp}:=\hbox{*--alg}\set{\un,\,R(1,f),\,\zeta(f)}, {\rm supp}(f)\;\hbox{is compact}.\subset\al D._S$ because one can choose a sequence $\{a_0,\,a_1,\ldots\}$ with $a_j\in\al D._{k_j}$ where $k_j$ grows sufficiently fast so that through the dependencies of the constants $A$ and $B$ in Theorem~\ref{CoBd} on $k_j$ the entire analytic condition fails. One expects to use an inductive limit argument, to define an index on $\al D._{\rm comp}$ from the indices on the $\al D._k\,.$ From this cyclic cocycle we can calculate an index for this quantum field theory, but its physical significance is presently unclear, though one would expect it to remain stable under deformations. This type of index is discussed in more detail in Longo~\cite{Lo}. \section{Conclusions} In this paper we have explored how supersymmetric quantum fields can be treated in a C*-algebra setting, avoiding the obstructions found by Kishimoto and Nakamura~\cite{KiNa} and by Buchholz and Longo \cite{BuLo}. We did this in detail for a simple one--dimensional model. In order to establish a reasonable domain of definition for the super--derivation, we found it necessary to analyze a notion of ``mollifiers'' for the quantum fields and to introduce a corresponding C*-algebra, the resolvent algebra. The full algebra $\al A.$ defining the model can then be taken as the tensor product of this resolvent algebra and the familiar CAR--algebra. The super--derivation is defined on a subalgebra which is weakly dense in $\al A.$ in all representations of physical interest; alternatively, one can define it on a norm dense subalgebra of $\al A.$ with range in a *--algebra ${\cal E}$ of bounded and unbounded operators which are affiliated with $\al A.$ in the sense of \cite{Georg}. Similarly, the basic supersymmetry relation can either be formulated in a mollified form on some weakly dense domain or, alternatively, as a relation between maps which have been extended to the *--algebra ${\cal E}$. These findings reveal some basic features of supersymmetric quantum field theories which have to be taken into account in a general C*-framework covering such theories. The tools developed here should also be useful in other areas of quantum field theory where one needs to use graded derivations, e.g. in BRS--constraint theory. We also exhibited in the present model graded KMS--functionals for arbitrary positive temperatures which are supersymmetric. In accordance with the general results in \cite{BuLo}, these functionals are unbounded. Yet their restrictions to any local subalgebra of the underlying C*-algebra $\al A.$ are bounded. It is an interesting question whether these functionals are also locally normal with respect to the vacuum representation of the theory, as one would heuristically expect. The KMS--functionals were then employed to define cyclic cocycles. In view of the fact that the domain of definition of these functionals does not contain analytic elements with regard to the time evolution, the strategy outlined in \cite{Ka,JLO} could not be applied here. That these cocycles can be constructed, nevertheless, is due to the fact that the functionals inherit sufficiently strong analyticity properties from the KMS--condition which allow one to perform the necessary complex integrations. Moreover, the resulting cocycles are entire on all local algebras. These functionals may thus be taken as an input for a quantum index theory as suggested by Longo~\cite{Lo}. Such an index should be stable under deformations, and one can easily think of possible deformations of our model, e.g. deform the supersymmetry generator $Q$ by an appropriate function $M\,:$ \begin{eqnarray*} Q_M&:=&\int M(x)\,j(x)\,c(x)\,dx \\[1mm] \hbox{so}\qquad \delta_M(c(f))&=&j(Mf),\qquad \delta_M(j(f))=c(Mf') \\[1mm] \hbox{thus:}\qquad \delta_M^2&=:&\delta_{0M} \end{eqnarray*} defines the new generator for time evolution. \section{Appendix} \subsection*{Proof of Theorem~\ref{Relemen}} (i) By (\ref{Resolv}) we have that $i(\mu-\lambda)\rlf R(\mu,f)= \rlf-R(\mu,f)=-\big(R(\mu,f)-\rlf\big)=i(\mu-\lambda)R(\mu,f)\rlf\,,$ i.e. ${[\rlf,\,R(\mu,f)]}=0\,.$\chop (ii) The Fock representation is a subrepresentation of $\pi_S$, and the resolvents of the fields $\varphi(f)$ give the Fock representation induced on $\al R.(X,\sigma),$ i.e. $\pi(\rlf)=(i\lambda-\varphi(f))^{-1}\,.$ Since this is nonzero, $\rlf\not=0$ for all nonzero $f$ and $\lambda\,.$ Now by $\rlf^*=R(-\lambda,f)$ we get \[ 2|\lambda|\|\rlf\|^2=\|2\lambda\rlf\rlf^*\|=\|\rlf-\rlf^*\| \leq 2\|\rlf\|\,. \] Thus by $\rlf\not=0$ we find $\|\rlf\|\leq 1/|\lambda|\,.$ Now \[ \|\rlf\|\geq\|\pi(\rlf)\|=\|(i\lambda-\varphi(f))^{-1}\| =\sup_{t\in\sigma(\varphi(f))}\Big|{1\over i\lambda-t}\Big|= {1\over|\lambda|} \] using the fact that the spectrum $\sigma(\varphi(f))=\R\,.$ Thus $\|\rlf\|=1/|\lambda|\,.$\chop (iii) Rearrange equation~(\ref{Resolv}) to get: \[ \rlf\big(\un-i(\lambda_0-\lambda)R(\lambda_0,f)\big)=R(\lambda_0,f)\,. % \|\rlf-R(\mu,f)\|=|\lambda-\mu|\cdot\|\rlf R(\mu,f)\| % \leq|\lambda-\mu|\big/|\lambda\mu| \] Now by (ii) above, if $\big|\lambda_0-\lambda\big|<\big|\lambda_0\big|$ then ${\big\|i(\lambda_0-\lambda)R(\lambda_0,f)\big\|}<1,$ and hence ${\big(\un-i(\lambda_0-\lambda)R(\lambda_0,f)\big)^{-1}}$ exists, and is given by a norm convergent power series in ${i(\lambda_0-\lambda)R(\lambda_0,f)}\,.$ That is, we have that \[ \rlf=R(\lambda_0,f)\big(\un-i(\lambda_0-\lambda)R(\lambda_0,f)\big)^{-1} =\sum_{n=0}^\infty(\lambda_0-\lambda)^n\,R(\lambda_0,f)^{n+1}i^n \] when $\big|\lambda_0-\lambda\big|<\big|\lambda_0\big|,$ as claimed. \chop % from which it is clear that $\rlf$ is norm continuous in $\lambda$ % for nonzero $\lambda\,.$ We also get from equation~(\ref{Resolv}) % that % \begin{eqnarray} % & & % \left\|\big(R(\lambda+h,f)-\rlf)\big)\big/h+iR(\lambda+h,f)\rlf\right\|=0 % \nonumber\\[1mm] % \label{R-ODE} % \hbox{so} & & {d\over d\lambda}\rlf=-i\rlf^2 % \end{eqnarray} % and hence $\rlf$ is smooth in $\lambda,$ determined by the ODE~(\ref{R-ODE}), % and one value of $\rlf,$ $\lambda\not=0$ (making use of equation~(\ref{Rinvol}) % to pass through $\lambda=0\,).$ % Consider the series $S(\lambda):={\sum\limits_{n=0}^\infty i^n(\lambda_0-\lambda)^n % R(\lambda_0,f)^{n+1}}$ for $\lambda\not=0\not=\lambda_0.$ It converges in norm % for $|\lambda_0-\lambda|<|\lambda_0|$ and % \[ % {d\over d\lambda}S({\lambda})=\sum_{n=1}^\infty i^nn(\lambda_0-\lambda)^{n-1} % R(\lambda_0,f)^{n+2}=iS(\lambda)^2\,. % \] % For the last inequality we used spectral theory, and the fact that % $S(\lambda)=h(R(\lambda_0,f))$ where $h(t):=t\big/[1-(\lambda-\lambda_0)it]\,.$ % Then we have ${d\over d\lambda}h(t)=ih(t)^2$ so that from the series % expansions we get ${d\over d\lambda}S({\lambda})=iS(\lambda)^2\,.$ % Since $S(\lambda_0)=R(\lambda_0,f),$ we conclude that for % $|\lambda-\lambda_0|<|\lambda_0|$ we have % \[ % \rlf=S(\lambda)=\sum_{n=0}^\infty(\lambda_0-\lambda)^ni^nT(\lambda_0,f)^{n+1}\,. % \] (iv) From equation~(\ref{Rhomog}) we get $R(\lambda,tf)={1\over\lambda} R(1,{t\over\lambda}f)={1\over t}R({\lambda\over t},f)$ so \begin{eqnarray*} R(\lambda,sf) &-& R(\lambda,tf)=\f 1,s.R(\f\lambda,s.,f)- \f 1,t.R(\f\lambda,t.,f) \\[1mm] &=& \f 1,s.\left(R(\f\lambda,s.,f)-R(\f\lambda,t.,f)\right) +\left(\f 1,s.-\f 1,t.\right)R(\f\lambda,t.,f) \\[1mm] &=&\f i\lambda,s.\left(\f 1,t.-\f 1,s.\right)R(\f\lambda,s.,f) R(\f\lambda,t.,f)+\left(\f 1,s.-\f 1,t.\right)R(\f\lambda,t.,f)\,. \end{eqnarray*} Thus $\left\|R(\lambda,sf) - R(\lambda,tf)\right\|\leq\left|\f 1,s. -\f 1,t.\right|\,2\left|\f t,\lambda.\right|$ from which continuity away from zero is clear.\chop (v) This follows directly from equation~(\ref{Rccr}) by interchanging $\lambda$ and $f$ with $\mu$ and $g$ resp.\chop (vi) Recall that we have a faithful (strongly regular) representation $\pi_S$ of $\al R.(X,\sigma)$ which is an extension of a regular representation of the Weyl algebra $\CCRX$ such that $\pi\s S.\left(\CCRX\right)'' \supset\pi\s S.\left(\al R.(X,\sigma)\right)$ and such that each $\pi_S(\rlf)$ is the resolvent of the generator $j\s\pi_S.(f)$ of the one-parameter group $t\to\pi_S(\delta\s tf.).$ Let $T\in{\rm Sp}(X,\sigma)$ then it defines an automorphism of $\CCRX$ by $\alpha_T(\delta_f)= \delta\s Tf.$ which preserves the set of strongly regular states, and in fact defines a bijection on the set of strongly regular states by $\omega\to\omega\circ\alpha_T.$ Now $\pi_S$ is the direct sum of the GNS--representations of all the strongly regular states, and hence $\pi_S\circ\alpha_T$ is just $\pi_S$ where its direct summands have been permuted. Such a permutation of direct summands can be done by conjugation of a unitary, thus $\pi_S$ is unitarily equivalent to $\pi_S\circ\alpha_T,$ and so we can extend $\alpha_T$ by unitary conjugation to $\pi\s S.\left(\CCRX\right)''.$ By equation~(\ref{Laplace1}) we get that $\alpha_T\left(\pi_S(\rlf)\right)=\pi_S(R(\lambda,Tf)),$ and hence $\alpha_T$ preserves $\al R.(X,\sigma),$ so defines an automorphism on it. % the map $\rlf\to R(\lambda,Tf)$ % is a permutation of the generating elements % $\set{\rlf},\lambda\in\R\backslash 0,\;f\in X\backslash 0.$ % of $\al R._0$ % which obviously preserves the relations % (\ref{Rinvol}) to (\ref{Rsum}), hence defines a *--automorphism of % $\al R._0\,.$ Since $T$ is invertible and linear it maps strongly regular % representations to strongly regular ones. So composition with $\alpha$ % preserves the set of strongly regular representations, and hence % $\pi\s S.\circ\alpha$ is unitarily equivalent to $\pi\s S..$ % Thus $\alpha$ preserves $\ker\pi\s S.$ and hence defines an automorphism % of $\al R.(X,\sigma)\,.$ \subsection*{Proof of Theorem~\ref{RegThm}} (i) Observe that by Theorem 1 p216 of Yosida~\cite{Yos}, we deduce from ${\ker\pi(R(1,f))}=\{0\}$ that $\pi(\rlf)$ is the resolvent of $j_\pi(f),$ i.e. we have now for all $\lambda\not=0$ that $j_\pi(f)=i\lambda\un-\pi(\rlf))^{-1}\,.$ Then \begin{eqnarray*} j_\pi(tf) &=& i\un-\pi(R(1,tf))^{-1}=i\un-t\pi(R(\f 1,t.,f))^{-1} \\[1mm] &=& t\left(i\f 1,t.{\un}-\pi\big(R(\f 1,t.,f)\big)^{-1}\right) =t\,j_\pi(f)\,. \end{eqnarray*} Thus \begin{eqnarray*} j_\pi(f)^* &=& \left(i\un-\pi(R(1,f))^{-1}\right)^* \supseteq -i\un-\left(\pi(R(1,f))^{-1}\right)^*\\[1mm] &=&-i\un-\pi(R(1,f)^*)^{-1} =-i\un-\pi(R(-1,f))^{-1} \\[1mm] &=&-i\un+\pi(R(1,-f))^{-1} = -j_\pi(-f)=j_\pi(f) \end{eqnarray*} and hence $j_\pi(f)$ is symmetric. To see that it is selfadjoint note that: \[ \ran\left(j_\pi(f)\pm i\un\right)=\ran\left(-\pi\big(R(\pm 1,f)\big)^{-1}\right) =\dom\left(\pi\big(R(\pm 1,f)\big)\right)=\al H._\pi \] hence the deficiency spaces $\left(\ran\left(j_\pi(f)\pm i\un\right)\right)^\perp =\{0\}$ and so $j_\pi(f)$ is selfadjoint.\chop For the domain claim, recall that $\dom j_\pi(f)=\ran\pi\big(R(1,f)\big)\,.$ So \begin{eqnarray*} \pi(\rlf)\dom j_\pi(h)&=&\pi(\rlf)\pi\big(R(1,h)\big)\al H._\pi \\[1mm] &=&\pi\left(R(1,h)\rlf+i\sigma(f,h)R(1,h)\rlf^2R(1,h)\right)\al H._\pi \\[1mm] &\subseteq&\pi\left(R(1,h)\right)\al H._\pi=\dom j_\pi(h). \end{eqnarray*} (ii) Let $j_\pi(f)=\int\lambda dP(\lambda)$ be the spectral resolution of $j_\pi(f).$ Then $\pi(R(\mu,f))=\int{1\over i\mu-\lambda}dP(\lambda)$ hence \[ i\mu\pi(R(\mu,f))\psi=\int{i\mu\over i\mu-\lambda}\,dP(\lambda)\psi\quad\forall\;\psi \in\al H._\pi\,. \] Since $\left|{i\mu\over i\mu-\lambda}\right|<1$ (for $\mu\in\R\backslash 0)$ the integrand is dominated by $1$ which is an $L^1\hbox{--function}$ with respect to $dP(\lambda),$ and as we have pointwise that $\lim\limits_{\mu\to\infty}{i\mu\over i\mu-\lambda}=1,$ we can apply the dominated convergence theorem to get that \[ \lim_{\mu\to\infty}i\mu\pi(R(\mu,f))\psi=\int dP(\lambda)\psi=\psi\,. \] (iii) $i\pi(R(1,sf))\psi=\int{i\over i-s\lambda}\,dP(\lambda)\psi\to\psi$ as $s\to0$ by the same argument as in (ii)\,.\chop (iv) Let $\al D.:={\pi\big(R(1,f)R(1,h)\big)\al H._\pi},$ then by definition $\al D.\subseteq\ran\pi(R(1,f))=\dom j_\pi(f)\,.$ Moreover ${\pi\big(R(1,f)R(1,h)\big)\al H._\pi} = {\pi\big(R(1,h)[R(1,f)+i\sigma(f,h)R(1,f)^2R(1,h)]\big)\al H._\pi}\subseteq\ran\pi(R(1,h))= \dom j_\pi(h),$ i.e. $\al D.\subseteq\dom j_\pi(f)\cap\dom j_\pi(h)\,.$ That $\al D.$ is dense, follows from (iii) of this theorem, using \[ \lim_{s\to 0}\lim_{t\to 0}\pi\big(R(1,sf)R(1,th)\psi=-\psi \] for all $\psi\in\al H._\pi\,,$ as well as $sR(1,sf)=R(1/s,\,f)$ and the fact mentioned before (cf. Theorem~1 p216 in~\cite{Yos}) that all $\pi(\rlf)$ have the same range for $f$ fixed.\chop Let $\psi\in\al D.,$ i.e. $\psi=\pi\big(R(1,f)R(1,h)\big)\varphi$ for some $\varphi\in\al H._\pi\,.$ Then \begin{eqnarray*} & &\pi\big(R(1,h)R(1,f)\big)\big[j_\pi(f),j_\pi(h)\big]\psi \\[1mm] & &\qquad\quad=\pi\big(R(1,h)R(1,f)\big)\big[\pi(R(1,f))^{-1},\pi(R(1,h))^{-1}\big] \pi\big(R(1,f)R(1,h)\big)\varphi \\[1mm] & & = \pi\big(R(1,f)R(1,h)-R(1,h)R(1,f)\big)\varphi = i\sigma(f,h)\pi\big(R(1,h)R(1,f)^2R(1,h)\big)\varphi \\[1mm] & & = i\sigma(f,h)\pi\big(R(1,h)R(1,f)\big)\psi\;. \end{eqnarray*} Since $\ker\pi\big(R(1,h)R(1,f)\big)=\{0\}$ it follows that $\big[j_\pi(f),j_\pi(h)\big]=i\sigma(f,h)$ on $\al D.\,.$\chop (v) From Equation~(\ref{Rhomog}) we have that \[ \pi(\rlf)=(1\lambda-j_\pi(f))^{-1}=\f 1,\lambda.\pi\big(R(1,\f 1,\lambda.f)\big) =\f 1,\lambda.\cdot\left(i{\un}-j_\pi(\f 1,\lambda.f)\right)^{-1} \] and hence that $j_\pi(f)=\lambda \,j_\pi(\f 1,\lambda.f),$ i.e. $j_\pi(\lambda f)=\lambda j_\pi(f)$ for all $\lambda\in\R\backslash 0\,.$ In equation~(\ref{Rsum}): \[ \pi\left(\rlf R(\mu,g)\right)= \pi\left(R(\lambda+\mu,\,f+g)[\rlf+R(\mu,g) +i\sigma(f,g)\rlf^2R(\mu,g)]\right) \] multiply on the left by $i(\lambda+\mu)\un-j_\pi(f+g)$ and apply to ${(i\mu\un-j_\pi(g))}{(i\lambda\un-j_\pi(f))}\psi,$ $\psi\in\al D.$ to get \[ \big(i(\lambda+\mu)\un-j_\pi(f+g)\big)\psi =\left((i\mu\un-j_\pi(g))+(i\lambda\un-j_\pi(f))\right)\psi \] making use of $\left[(i\mu\un-j_\pi(g)), (i\lambda\un-j_\pi(f))\right]\psi=i\sigma(g,f)\psi\,.$ Thus $j_\pi(f+g)=j_\pi(f)+j_\pi(g)$ on $\al D..$\chop (vi) From the spectral resolution for $j_\pi(f)$ we have trivially that on $\dom j_\pi(f)$ \[ j_\pi(f)\pi(R(\mu,f))=\pi(R(\mu,f))j_\pi(f)=\int{\lambda\over i\mu\un-\lambda}dP(\lambda) =i\mu\pi(R(\mu,f))-\un\,. \] (vii) Let $\psi\in\dom j_\pi(f)=\ran\pi(\rlf),$ i.e. $\psi=\pi(\rlf)\varphi$ for some $\varphi\in\al H._\pi.$ Then \begin{eqnarray*} \pi(\rlf)\big[j_\pi(f),\,\pi(R(\lambda,g))\big]\psi &=& \pi(\rlf)\big[j_\pi(f),\,\pi(R(\lambda,g))\big]\pi(\rlf)\varphi \\[1mm] &=& \pi\left(\big[R(\lambda,g),\rlf\big]\right)\varphi =i\sigma(g,f)\pi\left(\rlf R(\lambda,g)^2\rlf\right)\varphi \\[1mm] &=& i\sigma(g,f)\pi\left(\rlf R(\lambda,g)^2\right)\psi\;. \end{eqnarray*} Since $\ker\pi(\rlf)=\{0\}\;,$ it follows that \[ \big[j_\pi(f),\,\pi(R(\lambda,g))\big]= i\sigma(g,f)\pi\left( R(\lambda,g)^2\right) \] on $\dom j_\pi(f)\,.$\chop (viii) We first prove the second equality. Let $\psi,\,\varphi\in\wt\al D.:={\rm Span}\set{\chi\s[-a,a].\big(j_\pi(f)\big)\al H._\pi},a>0.$ which is a dense subspace. Since ${\left\|j_\pi(f)\restriction\chi\s[-a,a].\big(j_\pi(f)\big)\al H._\pi\right\|} \leq a\,,$ we can use the exponential series, i.e. \[ W(f)\psi:=\exp\big(ij_\pi(f)\big)\psi =\sum_{n=0}^\infty{\big(ij_\pi(f)\big)^n\over n!}\psi \qquad\forall\;\psi\in\wt\al D.\;. \] By the usual rearrangement of series we then have \[ \left(\varphi,\,W(f)\pi\big(R(\lambda,h)\big)W(f)^*\psi\right)= \sum_{n=0}^\infty{1\over n!}\left(\varphi,\,\big({\rm ad}\,ij_\pi(f)\big)^n\big(\pi(R(\lambda,h)) \big)\psi\right) \] for all $\varphi,\,\psi\in\wt\al D.\,.$ Using part (vii) we have \begin{eqnarray*} \big({\rm ad}\,ij_\pi(f)\big)\big(\pi(R(\lambda,h)^k)\big) &=&k\,\sigma(f,h)\pi\big(R(\lambda,h)^{k+1}\big) \\[1mm] \hbox{thus}\qquad\qquad\big({\rm ad}\,ij_\pi(f)\big)^n\big(\pi(R(\lambda,h)) &=& n!\,\sigma(f,h)^n\pi\big(R(\lambda,h)^{n+1}\big) \\[1mm] \hbox{so}\qquad\qquad \left(\varphi,\,W(f)\pi\big(R(\lambda,h)\big)W(f)^*\psi\right) &=& \sum_{n=0}^\infty\sigma(f,h)^n\left(\varphi,\,\pi(R(\lambda,h)^{n+1}) \psi\right) \\[1mm] &=& \left(\varphi,\,\pi(R(\lambda+i\sigma(f,h),\,h)) \psi\right) \end{eqnarray*} where we made use of the Von Neumann series (Theorem~\ref{Relemen}(iii)) in the last step. Since the operators involved are bounded and $\wt\al D.$ is dense, it follows that \begin{equation} \label{AdWR} W(f)\pi\big(R(\lambda,h)\big)W(f)^*=\pi(R(\lambda+i\sigma(f,h),\,h))\;. \end{equation} To prove the first equation, let us write $W(h)$ in terms of resolvents. Note that $\lim\limits_{n\to\infty}(1-it/n)^{-n}=e^{it},$ $t\in\R$ and so by the bound: $\sup\limits_{t\in\R}\left|(1-it/n)^{-n}\right|=\sup\limits_{t\in\R}\left(1+t^2/n^2\right)^{-n} =1,$ it follows from spectral theory (cf. Theorem~VIII.5(d), p262 in \cite{RS1}) that \[ W(h)=e^{ij_\pi(h)}=\lim_{n\to\infty}\left(1-ij_\pi(h)/n\right)^{-n} =\lim_{n\to\infty}\pi\left(iR(1,-h/n)\right)^n \] in strong operator topology. Apply equation~(\ref{AdWR}) to this to get \begin{eqnarray*} W(f)W(h)W(f)^* &=& \slim_{n\to\infty}\pi\left(iR(1+i\sigma(f,-\f h,n.),-\f h,n.) \right)^n \\[1mm] &=& \slim_{n\to\infty}\left(\un-i\big(\sigma(f,h)+j_\pi(h)\big)\big/n\right)^{-n} \\[1mm] &=&\exp[i\sigma(f,h)+ij_\pi(h)]=e^{i\sigma(f,h)}W(h) \end{eqnarray*} as required. Now \[ W(f)\al D.=W(f)\pi\big(\rlf R(\mu,h)\big)\al H._\pi =\pi\big(\rlf R(\mu+i\sigma(f,h),h)\big)\al H._\pi=\al D. \] hence we conclude that $\al D.$ is a core for $j_\pi(f)$ (cf. Theorem~VIII.11, p269 in \cite{RS1}). \subsection*{Proof of Proposition~\ref{MasterLemma}} % (I) \XP Simple minded proof (needing fewer assumptions).! Recall Equation~(\ref{QFresolvent}) \begin{eqnarray*} & & \omega(R(\lambda_1,f_1) \cdots R(\lambda_n,f_n)) \nonumber\\[1mm] & & = (-i)^n \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, \exp\Big(-\sum_k t_k \lambda_k - \sum_{k0$ for all $k.$ Thus we can use the dominated convergence theorem for derivatives (cf. Theorem~2.7 in~\cite{Foll}) to conclude that ${\omega(R(\lambda_1,f_1) \cdots R(\lambda_n,f_n))}$ is differentiable in all $x_i\hbox{--variables,}$ and the partial derivatives can be taken into the integral to give via (\ref{partiallambda}): \begin{eqnarray*} &&\!\!\!\!\! \frac{\partial}{\partial x_r} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) \\[1mm] &=& - \sum_{k=1}^{r-1} \frac{\partial }{\partial x_r}\Big(\big\langle f_k(x_k) | f_r(x_r) \big\rangle_\omega \Big) \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, {\partial^2\over\partial \lambda_r\partial\lambda_k}\,F(\bx) \\[1mm] &&\qquad-\hlf\frac{\partial }{\partial x_r}\Big(\big\langle f_r(x_r) | f_r(x_r) \big\rangle_\omega \Big) \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, {\partial^2\over\partial \lambda_r^2}\,F(\bx) \\[1mm] % \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] &&\qquad - \sum_{k=r+1}^n \frac{\partial }{\partial x_r}\Big( \big\langle f_r(x_r) | f_k(x_k) \big\rangle_\omega \Big) \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, {\partial^2\over\partial \lambda_r\partial\lambda_k}\,F(\bx)\,. % \frac{\partial}{\partial \lambda_k} \end{eqnarray*} We need to argue that we can take the partial derivatives w.r.t. the $\lambda_i\hbox{'s}$ through the integrals above. Now if we have that $\langle f_r(x_r) | f_r(x_r) \rangle_\omega=0,$ then each integrand factorises into a $t_r\hbox{--dependent}$ and a $t_r\hbox{--independent}$ part. Then the integrand w.r.t. the $t_r\hbox{--variable}$ is of the form $t_r^k\exp\big(-t_r\lambda_r\big)$ for $k=0,\,1,\,2\,,$ and so we get explicitly from the Laplace transforms that we can take $\partial\big/\partial\lambda_r$ through the $t_r\hbox{--integral.}$ This takes care of the part of the integral corresponding to those variables $t_r$ for which $\langle f_r(x_r) | f_r(x_r) \rangle_\omega=0\,.$ The remaining factor $\wt{F}(\bx)$ of $F(\bx)$ depends only on variables $t_r$ for which we have that $\langle f_r(x_r) | f_r(x_r) \rangle_\omega>0\,.$ Then ${\left|{\partial^2\over\partial \lambda_r\partial\lambda_k}\,\wt{F}(\bx)\right|} \leq {t_rt_k\exp\Big(-\hlf \sum_l \, t_l^2 \langle f_l | f_l \rangle_\omega\Big)}$ which is integrable w.r.t. the remaining variables, and likewise we also get a dominating function for the first derivatives. Thus by dominated convergence (uniformly in the $\lambda\hbox{--variables}$) we can take the partial derivatives in $\lambda_i$ through the integral in in the remaining variables. Thus we get \begin{eqnarray*} &&\!\!\!\!\! \frac{\partial}{\partial x_r} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) \\[1mm] &=& - \sum_{k=1}^{r-1} \frac{\partial }{\partial x_r}\Big(\big\langle f_k(x_k) | f_r(x_r) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r\partial\lambda_k} % \frac{\partial}{\partial \lambda_k} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] &&\qquad-\hlf\frac{\partial }{\partial x_r}\Big(\big\langle f_r(x_r) | f_r(x_r) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r^2} % \frac{\partial}{\partial \lambda_r} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] &&\qquad - \sum_{k=r+1}^n \frac{\partial }{\partial x_r}\Big( \big\langle f_r(x_r) | f_k(x_k) \big\rangle_\omega \Big) \, \frac{\partial^2}{\partial \lambda_r\partial \lambda_k} % \frac{\partial}{\partial \lambda_k} \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) \end{eqnarray*} %%%######### % \chop % (II) % \XP Original proof (needing $X=$Schwartz space, distributions)! % Recall Equation~(\ref{QFresolvent}) % \begin{eqnarray*} % & & \omega(R(\lambda_1,f_1) \cdots R(\lambda_n,f_n)) \nonumber\\[1mm] % & & = (-i)^n % \int_0^\infty \! dt_1 \dots \int_0^\infty \! dt_n \, % \exp\Big(-\sum_k t_k \lambda_k % - \sum_{k 0$, $k = 1, \dots n$. % Clearly, this function is analytic with respect to all $ M_{kl}$ % on the domain $\mbox{Re} M > 0$ and continuous at the % boundary. Moreover, one has on $\mbox{Re} M > 0$ % \begin{eqnarray*} % && \frac{\partial}{\partial M_{i_1 j_1}} \dots % \frac{\partial}{\partial M_{i_m j_m}} \, F(M;\bml) \\[1mm] % && = (-1)^m \int_0^\infty \! dt_1 \dots % \int_0^\infty \! dt_n \, t_{i_1} t_{j_1} \cdots % t_{i_m} t_{j_m} \, \exp\Big(-\sum_k t_k \lambda_k % - \sum_{k,l} t_k t_l \, M_{kl}\Big) \\[1mm] % && = (-1)^m \frac{\partial}{\partial \lambda_{i_1}} % \frac{\partial}{\partial \lambda_{j_1}} \dots % \frac{\partial}{\partial \lambda_{i_m}} \frac{\partial}{\partial \lambda_{j_m}} % \, F(M; \bml), % \end{eqnarray*} % and these equalities extend by continuity to $\mbox{Re} M \geq % 0$, ie all derivatives with respect to $M_{ij}$ are defined at the % boundary ``from the inside''. % Next, we consider for $\bx = (x_1, \dots x_p)$ % varying in some open set of $\R^p$ smooth matrix valued functions % $\bx \mapsto M(\bx)$ with $\mbox{Re} M(\bx) > 0$. Then % by the chain rule, % \begin{eqnarray} % && \frac{\partial}{\partial x_r} F(M(\bx); \bml) % = \sum_{i,k} \frac{\partial M_{ik}(\bx)}{\partial x_r} \, % \frac{\partial F(M; \bml) }{\partial M_{ik}} % \bigg|_{M = M(\bx)} \nonumber\\[1mm] % \label{partialF} % && = - \sum_{i,k} \frac{\partial M_{ik}(\bx)}{\partial x_r} % \, \frac{\partial}{\partial \lambda_i} \frac{\partial}{\partial \lambda_k} % F(M; \bml) \bigg|_{M = M(\bx)}. % \end{eqnarray} % Repeating this computation % one finds that $\bx \mapsto F(M(\bx); \bml) $ % is smooth and that there are similar equations for the higher partial % derivatives. These equations can be extended in the sense of % distributions by a deformation argument % to paths $\bx \mapsto M(\bx)$ with % $\mbox{Re} M(\bx) \geq 0 $. % But, in view of the second equality and its generalizations, % this distributional equality then is also valid in % the sense of ordinary functions. % So we conclude that for any smooth path $\bx \mapsto M(\bx)$ % with $\mbox{Re} M(\bx) \geq 0 $ the function % $\bx \mapsto F(M(\bx); \bml) $ is smooth and % satisfy relations of the preceding type for all % its partial derivatives. % % We put together now the preceding results in order to establish % in quasifree states % the smoothness of products of the resolvents with regard to the test functions. % Recall that we consider paths % $x_k \mapsto f_k(x_k) \in X$, $k = 1, \dots n$, % such that the functions % $ x_k, x_l \longmapsto \langle f_k(x_k) | f_l(x_l) % \rangle_\omega, % \quad k,l = 1, \dots n,$ % are smooth. Setting % \[ % M_{kl} (\bx) = \big\langle f_k(x_k) | f_l( x_l) \big\rangle_\omega, \quad % k < l\,; \qquad\quad % M_{ll} (\bx) = \big\langle f_l(x_l) | f_l(x_l) % \big\rangle_\omega/2, \qquad\quad % M_{kl} (\bx) = 0, \quad k > l, % \] % where $\bx = (x_1, \dots x_n)$, % we obtain a smooth matrix valued function $\bx \mapsto M(\bx)$ which % satisfies $\mbox{Re} M(\bx) \geq 0\,.$ % Plugging this matrix into (\ref{gauss}) and making % use of relation (\ref{QFresolvent}), we see that the function % $$ % x_1 \dots x_n \longmapsto % \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) % $$ % is smooth. Moreover, with the help of relation (\ref{partialF}) we % can compute its partial derivatives, giving % \begin{eqnarray*} % &&\!\!\!\!\! \frac{\partial}{\partial x_r} \, % \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) % \\[1mm] % &=& - \sum_{k=1}^{r-1} % \frac{\partial }{\partial x_r}\Big(\big\langle f_k(x_k) | % f_r(x_r) \big\rangle_\omega \Big) % \, \frac{\partial^2}{\partial \lambda_r\partial\lambda_k} % % \frac{\partial}{\partial \lambda_k} % \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] % &&\qquad-\hlf\frac{\partial }{\partial x_r}\Big(\big\langle f_r(x_r) | % f_r(x_r) \big\rangle_\omega \Big) % \, \frac{\partial^2}{\partial \lambda_r^2} % \frac{\partial}{\partial \lambda_r} % \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big)\\[1mm] % &&\qquad - \sum_{k=r+1}^n % \frac{\partial }{\partial x_r}\Big( % \big\langle f_r(x_r) | % f_k(x_k) % \big\rangle_\omega \Big) % \, \frac{\partial^2}{\partial \lambda_r\partial \lambda_k} % % \frac{\partial}{\partial \lambda_k} % \, \omega\Big(R(\lambda_1, f_1(x_1)) \cdots R(\lambda_n, f_1(x_n))\Big) % \end{eqnarray*} \subsection*{Proof of Proposition~\ref{KMSbasics}} To prove this theorem, we first need to establish the following lemma. \begin{lem} \label{AnalFn} For the strip $S:=\R+i[0,1[\subset\C,$ let $F: S\to\C$ be a continuous function, analytic on the interior of $S,$ which satisfies for some $C>0$ and $\lambda\in\C,$ $|\lambda|=1$ the conditions: \begin{eqnarray*} \big|F(t+is)\big| &\leq& C(1+|t|)^N\quad\forall\,t\in\R,\;s\in(0,1) \\[1mm] \hbox{and}\qquad\qquad F(t+i)&=&\lambda\,F(t)\quad\forall\,t\in\R\;. \end{eqnarray*} Then $F=0$ if $\lambda\not=1$ and $F=\hbox{constant}$ if $\lambda=1\,.$ \end{lem} \begin{beweis} Note that $\C$ is covered by the strips $S_n:= S+in.$ We define $G:\C\to\C$ by $G(z):=\lambda^nF(z-in)$ whenever $z\in S_n$ This is consistent, because on the joining lines $\R+in=S_n\cap S_{n-1}$ we have that $G(t+in)=\lambda^nF(t)=\lambda^{n-1}F(t+i)\,.$ By the continuity of $F$ on $S_0= S,$ $G$ is continuous. Now $G$ is analytic on the interior of each $S_n$ and continuous on the boundary, i.e. continuous on the lines $\R+in$ and analytic on either side of them. So it follows from a well-known theorem of analytic continuation that $G$ is analytic on the lines $\R+in$ (cf.~\cite{Ne} p183) hence entire. Moreover $G(z+i)=\lambda G(z)$ for all $z\,.$ Let $\Gamma$ be a closed anticlockwise circle of radius $R$ centered at the fixed point $z_0\,.$ If $z\in \Gamma\cap S_n$ then \begin{eqnarray*} \big|G(z)\big|&=& \big|\lambda^nF(z-in)\big|=\big|F(z-in)\big| \\[1mm] &\leq&C\big(1+|{\rm Re}(z-in)|\big)^N=C\big(1+|{\rm Re}(z)|\big)^N \\[1mm] &\leq&C\big(1+\big(R+|{\rm Re}(z_0)\big)|\big)^N \leq C\big(1+R+|z_0|\big)^N \end{eqnarray*} which is independent of $n,$ i.e. $|G(z)|\leq C\big(1+R+|z_0|\big)^N$ for all $z\in\Gamma\,.$ Applying this to the Cauchy integral formula: \begin{eqnarray*} G^{(k)}(z_0)&=&{k!\over 2\pi i}\int_\gamma{G(z)\over(z-z_0)^{k+1}}\,dz\qquad\quad \hbox{we find:} \\[1mm] \left|G^{(k)}(z_0)\right|&\leq&k!R\sup_{z\in\Gamma}{\big|G(z)\big|\over R^{k+1}} \leq k!\,C\,{\big(1+R+|z_0|\big)^N\over R^k}\;. \end{eqnarray*} When $k>N+1$ this goes to zero as $R\to\infty,$ hence $G^{(k)}(z_0)=0$ for all $k>N+1\,.$ This is true for all $z_0\in\C,$ so $G$ is a polynomial. However only a constant polynomial can satisfy $G(z+i)=\lambda G(z)$ (or else it has infinitely many zeroes), hence $G$ is a constant, and if $\lambda\not=1,$ the only possible constant is zero. \end{beweis} (i) The KMS-condition for $A=\un$ reads $F\s\un,B.(t)=\varphi\big(\alpha_t(B)\big) =F\s\un,B.(t+i)\,.$ Thus by lemma~\ref{AnalFn} it follows that $F\s\un,B.$ is constant, hence that $\varphi\big(\alpha_t(B)\big)$ is independent of $t\,.$\chop (ii) Let $\gamma(A)=-A\in\dom\varphi\,,$ then $F\s A,\un.(t+i)=\varphi(\gamma(A))=-\varphi(A)=-F\s A,\un.(t)$ so by lemma~\ref{AnalFn} we have $0=F\s A,\un.=\varphi(A)\,.$ For any $B\in\dom\varphi$ decompose $B=B_++B_-$ into $\gamma\hbox{-even}$ and odd parts then we get $\varphi(B)=\varphi(B_+)=\varphi(\gamma(B))\,.$\chop Finally, let a functional $\varphi$ satisfy the graded KMS-condition on a set $Y\subset\dom\varphi\,.$ Let $A,\,B\in\span Y,$ i.e. $A=\sum\limits_i\lambda_iA_i\,,$ $B=\sum\limits_j\mu_jB_j$ for $A_i,\,B_j\in Y\,,$ and $\lambda_i,\,\mu_j\in\C\,.$ Then \[ F\s A,B.(t):=\varphi\big(A\alpha_t(B)\big)=\sum_{i,j}\lambda_i\mu_j\, \varphi\big(A_i\alpha_t(B_j)\big)= \sum_{i,j}\lambda_i\mu_j\,F\s A_i,B_j.(t)\,. \] Since $\varphi$ is $\gamma\hbox{--KMS}$ on $Y$ the $F\s A_i,B_j.$ are $\gamma\hbox{--KMS}$ functions. Thus $F\s A,B.$ is continous on $S,$ analytic on its interior, and \begin{eqnarray*} F\s A,B.(t+i)&=&\sum_{i,j}\lambda_i\mu_j\,F\s A_i,B_j.(t+i) =\sum_{i,j}\lambda_i\mu_j\, \varphi\big(\alpha_t(B_j)\gamma(A_i)\big) \\[1mm] &=& \varphi\big(\alpha_t(B)\gamma(A)\big) \;,\\[1mm] \left|F\s A,B.(t+is)\right|&\leq&\sum_{i,j}|\lambda_i\mu_j|\, \left|F\s A_i,B_j.(t+is)\right| \\[1mm] &\leq&\sum_{i,j}|\lambda_i\mu_j|\,C_{ij}(1+|t|)^{N_{ij}} \leq C(1+|t|)^N \end{eqnarray*} where $C:=\sum\limits_{i,j}|\lambda_i\mu_j|\,C_{ij}$ and $N:=\mathop{\rm max}\limits_{ij}(N_{ij})\,.$ So $\varphi$ is $\gamma\hbox{--KMS}$ on $\span Y\,.$ \subsection*{Proof of Proposition~\ref{KMStensor}} Since $\dom\psi$ and $\al B.$ are dense *--algebras, it follows that $\dom\varphi:={\span \set C\otimes B,C\in\dom\psi,\;B\in{\al B.}.}$ is a dense *-algebra of $\al A.=\al C.\otimes\al B.,$ and that it is invariant w.r.t. $\gamma\otimes\iota$ and $\sigma\otimes\beta\,.$ Thus by the last part of Proposition~\ref{KMSbasics} it suffices to verify the KMS--property on $\set C\otimes B,C\in\dom\psi,\;B\in{\al B.}.\,.$ Let $A_i:=C_i\otimes B_i\,,$ $i=1,2$ for $C_i\in\dom\psi\subset\al C.,$ and $B_i\in\al B.\,.$ Consider for $t\in\R$ the function \begin{eqnarray*} F\s A_1,A_2.(t)&:=& \varphi\left(A_1(\sigma\otimes\beta)_t(A_2)\right) \\[1mm] &=&\varphi\big((C_1\otimes B_1)(\sigma_t(C_2)\otimes\beta_t(B_2))\big) \\[1mm] &=&\psi\big(C_1\,\sigma_t(C_2)\big)\,\omega\big(B_1\,\beta_t(B_2)\big) =F^\psi\s C_1,C_2.(t)\,F^\omega\s B_1,B_2.(t) \end{eqnarray*} where $F^\psi\s C_1,C_2.$ and $F^\omega\s B_1,B_2.$ are the KMS--functions of $\psi$ and $\omega$ resp.. Thus, using their analytic properties, it follows that $F\s A_1,A_2.$ extends to an function on the strip $S=\R+i[0,1]\,,$ analytic on its interior and continuous on the boundary, given by $F\s A_1,A_2.(z)=F^\psi\s C_1,C_2.(z)\,F^\omega\s B_1,B_2.(z)\,.$ Moreover \begin{eqnarray*} F\s A_1,A_2.(t+i)&=&F^\psi\s C_1,C_2.(t+i)\,F^\omega\s B_1,B_2.(t+i) =\psi\big(\sigma_t(C_2)\,\gamma(C_1)\big)\,\omega\big(\beta_t(B_2)\,B_1\big) \\[1mm] &=& \varphi\left(\sigma_t(C_2)\gamma(C_1)\otimes\beta_t(B_2)\,B_1\right) =\varphi\left((\sigma\otimes\beta)_t(A_2)(\gamma\otimes\iota)(A_1)\right) \end{eqnarray*} and thus $F\s A_1,A_2.$ will be a $(\gamma\otimes\iota)\hbox{--KMS}$ function if the tempered growth property also holds. We have ${\big|F^\psi\s C_1,C_2.(t+is)\big|}\leq K(1+|t|)^N$ for a constant $K$ and $N\in\N$ depending on $C_i\,.$ Now as $\omega$ is a state we have from the KMS--property that ${\big|F^\omega\s B_1,B_2.(t+is)\big|}\leq{\|B_1\|\,\|B_2\|}$ for $s=0,\,1\,.$ So by the maximum modulus principle (apply it after first mapping $S$ to the unit disk by the Schwartz mapping principle) it follows that ${\big|F^\omega\s B_1,B_2.(z)\big|}\leq{\|B_1\|\,\|B_2\|}$ for all $z\in S\,.$ Thus for $t\in\R,$ $s\in[0,1]$ we have \[ \big|F\s A_1,A_2.(t+is)\big|\leq \|B_1\|\,\|B_2\|\, K(1+|t|)^N \] and so the tempered growth property holds for $F\s A_1,A_2.\,.$ Thus $\varphi$ is a $(\gamma\otimes\iota)\hbox{--KMS}$ functional. \subsection*{Proof of Theorem~\ref{KMSresolv}} (i) To prove there is a quasi--free state on $\CCR$ defined by $\omega(\delta_f):=\exp[-s(f,f)/2]\,,$ $f\in\al S.(\R,\R)$ where ${s(f,f)}:={\int{p\over 1-e^{-p}}\left|\widehat{f}(p)\right|^2dp}\,,$ it suffices to show that ${|\sigma(f,h)|^2}\leq{4\,s(f,f)\,s(h,h)}$ by~\cite{MaVe}. \begin{eqnarray*} |\sigma(f,h)|^2&=&\Big|\int ip\,\widehat{f}(p)\,\ol\widehat{h}(p).\,dp\Big|^2 \leq \Big(\int\left|p\,\widehat{f}(p)\,\widehat{h}(p)\right|dp\Big)^2 \\[1mm] &=&\Big(2\int_0^\infty p \left|\widehat{f}(p)\,\widehat{h}(p)\right|dp\Big)^2 \qquad\quad\hbox{as $\big|p\widehat{f}(p)\,\widehat{h}(p)\big|$ is even by $\widehat{f}(-p)=\ol\widehat{f}(p).$}\\[1mm] &\leq&4\int_0^\infty p\big|\widehat{f}(p)\big|^2dp\, \int_0^\infty k\big|\widehat{h}(k)\big|^2dk\qquad\qquad\hbox{by Cauchy--Schwartz}\\[1mm] &<&4\int_0^\infty{ p\over 1-e^{-p}}\big|\widehat{f}(p)\big|^2dp\, \int_0^\infty{ k\over 1-e^{-k}}\big|\widehat{h}(k)\big|^2dk\quad\qquad \hbox{as $p<{ p\over 1-e^{-p}}$ for $p>0$} \\[1mm] &<&4\int_{-\infty}^\infty{ p\over 1-e^{-p}}\big|\widehat{f}(p)\big|^2dp\, \int_{-\infty}^\infty{ k\over 1-e^{-k}}\big|\widehat{h}(k)\big|^2dk\quad\qquad \hbox{as $0<{ p\over 1-e^{-p}}$ for all $p$} \\[1mm] &=&4\,s(f,f)\,s(h,h) \end{eqnarray*} (ii) Next we need to show that this quasi--free state $\omega$ on $\CCR$ extends to a KMS--functional on $\pi_\omega(\CCR)'',$ and hence on $\al R.(\al S.(\R),\sigma).$ We first prove that $\omega$ is KMS on the *--algebra $\Delta_c % ${\Delta(\al S.(\R),\sigma)} :={\span\set\delta_f,{f\in\al S.(\R),\;\supp \widehat{f}\;\hbox{compact}}.}\,.$ Let $\delta_f,\,\delta_h\in\Delta_c$ and consider \begin{eqnarray*} F_1(t)&:=&\omega\big(\delta\s f.\alpha_t(\delta\s h.)\big)= e^{-i\sigma(f,h_t)/2}\omega(\delta\s f+h_t.) \\[1mm] &=&\exp\big[-\f i,2.\,\sigma(f,h_t)-\f 1,2.\,s(f+h_t,f+h_t)\big]\\[1mm] &=&\exp\Big[\f 1,2.\int\Big(-p\,\widehat{f}\,\ol\widehat{h_t}. -{p\over 1-e^{-p}} \big|\widehat{f}+\widehat{h_t}\big|^2\Big)\,dp\Big] \\[1mm] &=&\exp\Big[\f 1,2.\int p \Big({|\widehat{f}|^2+|\widehat{h}|^2\over e^{-p}-1} -{2-e^{-p}\over 1-e^{-p}}\,e^{ipt}\widehat{f}\,\ol\widehat{h}. -{e^{-ipt}\over 1-e^{-p}}\,\ol\widehat{f}.\,\widehat{h}\Big)\,dp\Big] \end{eqnarray*} where we used $\widehat{h_t}(p)=e^{-ipt}\widehat{h}(p)\,.$ Now put $K:={\exp\Big[\hlf\int p{|\widehat{f}|^2+|\widehat{h}|^2\over e^{-p}-1}\,dp\Big]\,,}$ and substitute $p\to-p$ in the last integral, using $\widehat{f}\,\ol\widehat{h}.(-p) =\ol\widehat{f}.\,\widehat{h}(p)$ to get: \[ F_1(t)=K\,\exp\Big[-\int{pe^{ipt}\over 1-e^{-p}}\,\widehat{f}\,\ol\widehat{h}.(p)\,dp\Big]\;. \] By a similar calculation we find $\omega\big(\alpha_t(\delta\s h.)\delta\s f.\big)=F_1(t+i)\,$ and this suggests that we define the KMS--function for $z\in S=\R+i[0,1]$ by: \begin{equation} \label{KMSdeltas} F(z):=K\,\exp\Big[-\int{pe^{ipz}\over 1-e^{-p}}\,\widehat{f}\,\ol\widehat{h}.(p)\,dp\Big]\;. \end{equation} Let $z=x+iy\in S=\R+i[0,1],$ then we know that the integral exists for $y=0,\,1\,.$ For $y\in(0,1)$ the function $\left|{p\,e^{ipz}\over 1-e^{-p}}\right|={p\,e^{-py}\over 1-e^{-p}}$ is bounded, so the integral exists for all $z\in S,$ hence the definition~(\ref{KMSdeltas}) makes sense for $F\,.$ It is however not clear that $F$ is analytic. However, recall that by assumption $\supp \wh{f}$ and $\supp \wh{h}$ are compact, then it follows from the dominated convergence theorem that $F$ is continuous on $S$ and as the differential w.r.t. $z$ of the integrand is continuous in $p,$ it also implies that $F$ is analytic on the interior of $S.$ Thus $F$ is the KMS--function for $\omega,$ hence by the last part of Proposition~\ref{KMStensor}, $\omega$ is a KMS--state on $\Delta_c\,.$ Next, we prove that $\pi_\omega(\Delta_c)$ is strong operator dense in $\pi_\omega(\CCR)''\,.$ It suffices to show that for each $f\in\al S.(\R)$ there is a sequence $\{f_n\}\subset\al S.(\R)$ with Fourier transforms of compact support, such that ${\omega\big(\delta_g\,\delta\s f_n.\,\delta_h\big)}\to {\omega\big(\delta_g\,\delta\s f.\,\delta_h\big)}$ for all $g,\,h\in\al S.(\R)$ as $n\to\infty\,.$ Fix an $f\in\al S.(\R)$ and define $f_n\in\al S.(\R)$ by its Fourier transform $\widehat{f_n}=K_n\cdot\widehat{f}$ where each $K_n:\R\to[0,1]$ is a smooth bump function of compact support which is $1$ on ${[-n,n]}$ and zero on ${\R\backslash[-n-1,\,n+1]}\,.$ Then \begin{eqnarray*} \omega\big(\delta_g\,\delta\s f_n.\,\delta_h\big) &=& e^{-i\sigma(g+f_n,\,f_n+h)/2} \omega\big(\delta\s g+f_n+h.\big) \\[1mm] &=& \exp\left[-\f i, 2.\,\sigma(g+f_n,\,f_n+h)-\hlf\,s\big( g+f_n+h,\, g+f_n+h\big)\right] \\[1mm] &=& \exp\Big[\hlf\int\Big(-p(\wh{g}+\wh{f_n})(\ol\wh{f_n}.+\ol\wh{h}.) -{p\over 1-e^{-p}}\big|\wh{g}+\wh{f_n}+\wh{h}\big|^2\Big)\,dp\Big] \\[1mm] &\maprightt\infty,n.& \exp\Big[\hlf\int\Big(-p(\wh{g}+\wh{f})(\ol\wh{f}.+\ol\wh{h}.) -{p\over 1-e^{-p}}\big|\wh{g}+\wh{f}+\wh{h}\big|^2\Big)\,dp\Big] \\[1mm] &=& \omega\big(\delta_g\,\delta\s f.\,\delta_h\big) \end{eqnarray*} using dominated convergence as $f$ is $L^1\,.$ Next, we want to use the strong operator denseness of $\pi_\omega(\Delta_c)$ in $\pi_\omega(\CCR)''$ to show that $\omega$ is KMS on all of $\pi_\omega(\CCR)''\,.$ Let $A,\,B\in\pi_\omega(\CCR)''$ be selfadjoint, then by the Kaplansky density theorem (cf. Theorem~5,3.5 p329 in~\cite{KaRiI}) it follows that there are sequences $\{A_n\},\;\{B_n\}\subset\pi_\omega(\Delta_c)$ of selfadjoint elements such that $\|A_n\|\leq\|A\|,$ $\|B_n\|\leq\|B\|$ and $A_n\to A,$ $B_n\to B,$ in strong operator topology. Now $\omega\circ\alpha_t=\omega$ because $s(f_t,\,f_t)=s(f,f),$ and thus there is an implementing unitary $U_t\in\al B.(\al H._\omega)$ such that ${U_t\pi_\omega(D)U_t^*}=\pi_\omega(\alpha_t(D))$ for $D\in\CCR$ and $U_t\Omega_\omega=\Omega_\omega\,.$ Abbreviate the KMS--functions $F_n(z):=F\s A_n,B_n.(z)$ of $\omega,$ then for all $t\in\R\,:$ \begin{eqnarray*} F_n(t)-F_k(t)&=&\omega\big(A_n\,\alpha_t(B_n)\big)- \omega\big(A_k\,\alpha_t(B_k)\big)\\[1mm] &=&\big(A_n\Omega_\omega,\,U_tB_n\Omega_\omega\big) -\big(A_k\Omega_\omega,\,U_tB_k\Omega_\omega\big) \\[1mm] &=&\big(U_t^*A_n\Omega_\omega,\,(B_n-B_k)\Omega_\omega\big) +\big(U_t^*(A_n-A_k)\Omega_\omega,\,B_k\Omega_\omega\big) \\[1mm] \hbox{Thus}\qquad \left|F_n(t)-F_k(t)\right| &\leq& \big\|(B_n-B_k)\Omega_\omega\|\cdot\|A_n\|+ \big\|(A_n-A_k)\Omega_\omega\|\cdot\|B_k\|\\[1mm] &\leq& \big\|(B_n-B_k)\Omega_\omega\|\cdot\|A\|+ \big\|(A_n-A_k)\Omega_\omega\|\cdot\|B\| \end{eqnarray*} and this is independent of $t.$ Similarly \[ \left|F_n(t+i)-F_k(t+i)\right| \leq\big\|(A_n-A_k)\Omega_\omega\|\cdot\|B\|+ \big\|(B_n-B_k)\Omega_\omega\|\cdot\|A\|\;. \] Now $F_n-F_k$ is a analytic function on the strip $S,$ so by combining The Riemann mapping theorem with the maximum modulus principle we have that $|F_n-F_k|$ takes its maximum on the boundary of $S\,.$ Thus by the inequalities above, $F_n-F_k$ converges uniformly to zero, hence the uniform limit $F:=\lim\limits_nF_n$ exists and is analytic and bounded on $S.$ Since ${\omega\big(A_n\alpha_t(B_n)\big)}={\big(A_n\Omega_\omega,\,U_tB_n\Omega_\omega\big)} \to{\big(A\Omega_\omega,\,U_tB\Omega_\omega\big)} ={\omega\big(A\alpha_t(B)\big)}$ it follows that $F$ is the KMS--function $F\s A,B.$ of $\omega.$ Thus $\omega$ is KMS on the selfadjoint part, hence on all of $\pi_\omega(\CCR)''\,.$ Since $\omega$ is quasifree, it is strongly regular, hence the resolvents of the generators of the one-parameter groups $t\to\pi_\omega(\delta\s tf.)$ will provide a representation of $\al R.(\al S.(\R),\sigma),$ where it is defined by Equation~(\ref{QFresolvent}) via spectral theory. \subsection*{Proof of Theorem~\ref{ThetaProp}} For $z\in S=\R+i[0,1]$ consider \begin{eqnarray} G(z)&:=& \lim_{\varepsilon\to 0^+}\Big( \int_{-\infty}^{-\varepsilon} +\int_{\varepsilon}^\infty\Big){e^{ipz}\over 1-e^{-p}}\, \,{\widehat{f}(p)}\,\ol\hat{g}(p).\,dp \nonumber\\[1mm] &=&\lim_{\varepsilon\to 0^+}\int_\varepsilon^\infty \Big({e^{ipz}\over 1-e^{-p}}\,{\widehat{f}(p)}\,\ol\hat{g}(p). +{e^{-ipz}\over 1-e^{p}}\,\ol{\widehat{f}(p)}.\,{\hat{g}(p)} \Big)\,dp\qquad\hbox{by $\widehat{f}(-p)=\ol\widehat{f}(p).$}\nonumber\\[1mm] &=&\int_0^\infty \Big[\Big(e^{ipz}+{e^{-p}e^{ipz}\over 1-e^{-p}}\Big) \,{\widehat{f}(p)}\,\ol\hat{g}(p). - {e^{-p}e^{-ipz}\over 1-e^{-p}}\,\ol{\widehat{f}(p)}.\,{\hat{g}(p)} \Big]\,dp \nonumber \\[1mm] &=&(g,F(D)f) +\int_0^\infty dp\int dx\int dy\,f(x)\,g(y){e^{-p}\over 1-e^{-p}} \Big(e^{ipz}e^{ip(y-x)}-e^{-ipz}e^{ip(x-y)}\Big)\nonumber\\[1mm] \label{GIntegr} &=&(g,F(D)f) +2i\int_0^\infty dp\int dx\int dy\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x) \end{eqnarray} where $D=id/dx$ and $F(p):=e^{ipz}\chi\s[0,\infty).(p)\,,$ and we used the fact that the Fourier transform diagonalises $D\,.$ To use Fubini to rearrange these integrals, we need to show that the integrand is integrable. % and this will also produce estimates to prove part~(ii). We need to separate the low $p$ from the high $p$ behaviour in the last integral. For the low $p$ behaviour, consider the integral \[ \int_0^1 dp\int dx\int dy\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\;. \] Rearrange the integrand to $f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}}\,p(z+y-x) \,{\sin p(z+y-x)\over p(z+y-x)}$ and observe that $p(z+y-x)\in S=\R+i[0,1]\,$ since $p\in[0,1]\,.$ Now $H(z):=\sin(z)\big/z$ is analytic in $S,$ so $|H(z)|$ takes its maximum on the boundary. On $\R,$ $|H(x)|\leq 1,$ and for $\R+i$ we have \[ \big|H(x+i)\big|^2=\left|{\sin(x+i)\over x+i}\right|^2 ={\cosh^21-\cos^2x\over x^2+1}\leq\cosh^2 1 \] and thus for the integrand \begin{equation} \label{LowIngrand} \left|f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}}\,p(z+y-x) \,{\sin p(z+y-x)\over p(z+y-x)}\right| \leq \left|f(x)\,g(y)\right|\,{e^{-p}\over 1-e^{-p}}\,p|z+y-x|\cosh 1 \end{equation} which is clearly integrable because $f$ and $g$ are Schwartz functions so take care of the linear factor in $x$ and $y\,.$ For the high $p$ behaviour, consider the remaining part of the integral~(\ref{GIntegr}), i.e. \[ \int_1^\infty dp\int dx\int dy\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\;. % =\int_1^\infty dp\,{e^{-p}\over 1-e^{-p}}\Big[e^{ipz}\ol\wh{f}(p).\, % \wh{g}(p)-e^{-ipz}\wh{f}(p)\,\ol\wh{g}(p).\Big] \] Now for $z=t+is\in S,$ $t\in\R,$ $s\in[0,1]$ we have \begin{eqnarray*} \left|f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\right| &\leq& \hlf\left|f(x)\,g(y)\right|\,{e^{-p}\over 1-e^{-p}} \,\left|e^{i p(z+y-x)}-e^{-i p(z+y-x)}\right|\\[1mm] \leq \hlf\left|f(x)\,g(y)\right|\,{e^{-p}\over 1-e^{-p}} \,\left|e^{-ps}+e^{ps}\right| &\leq& \left|f(x)\,g(y)\right|\,{ e^{p(s-1)}\over 1-e^{-p}} % \leq \left|f(x)\,g(y)\right|\,{1\over 1-e}\,. \end{eqnarray*} This is integrable for $s\in[0,1),$ but not for $s=1\,.$ However, we will see below that $G$ is continuous on $S,$ and this will be enough. Thus the integrand in (\ref{GIntegr}) is integrable for $z\in\R+i[0,1),$ and we can apply the Fubini theorem to rearrange the order of integrals, and we get: \begin{equation} \label{GFubini} G(z)=(g,F(D)f) +2i\int dx\int dy\,f(x)\,g(y)\,\int_0^\infty dp\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\,. \end{equation} To prove (i), let $z=0,$ so $G(0)=\theta(f,g)$ and $F(p)=\chi\s[0,\infty).(p)$ hence $P:=F(D)=2\pi\times\hbox{projection}$ onto positive spectrum of $D\,.$ So \begin{eqnarray*} G(0)=\theta(f,g)&=& (g,Pf) +2i\int dx\int dy\,f(x)\,g(y)\,\int_0^\infty dp\,{e^{-p}\over 1-e^{-p}} \,\sin p(y-x) \\[1mm] &=&(g,Pf)-2i\int dx\int dy\,f(x)\,g(y)\,(y-x)\,\int_0^\infty dp\, \ln( 1-e^{-p})\,\cos p(y-x)\\[1mm] &=&\big(g,(P+T)f) \end{eqnarray*} through an integration by parts. Consider the operator \[ \big(Tf\big)(x):=2i\int dy\,f(y)\,(x-y)\,\int_0^\infty dp\, \ln( 1-e^{-p})\,\cos p(x-y)=:\int dy\,f(y)\,K(x-y) \] which is obviously a kernel operator with kernel $K.$ Due to the factor ${(x-y)}$ in $K,$ $T$ is unbounded. Note however that \[ \left|{\partial^n\over\partial t^n}\,\ln(1-e^{-p})\,\cos pt\right| \leq\big|p^n\ln(1-e^{-p})\big| \] which is integrable in $p.$ [To see this, note that for $02$ we have $\big|p^n\ln(1-e^{-p})\big|=p^n\big(e^{-p}+e^{-2p}/2+e^{-3p}/3+\cdots\big) \leq p^n\,e^{-p}\big(1+e^{-p}/2+e^{-2p}/3+\cdots\big) \leq p^n\,e^{-p}\sum_{k=0}^\infty(\hlf)^k\leq 2 p^n\,e^{-p}$ which is integrable.] Thus by dominated convergence the function $t\to{\int_0^\infty dp\, \ln( 1-e^{-p})\,\cos pt}$ is is smooth. Thus the kernel $K$ of $T$ is smooth. If $J$ is a compact interval then $P_JTP_J$ has kernel $\chi\s J.(x)K(x-y)\chi\s J.(y)$ which is smooth and bounded on $J\,.$ Thus $P_JTP_J$ is trace--class by Theorem~1 p128 in Lang~\cite{La}. We also have an explicit proof that $P_JTP_J$ is trace--class below in the proof of Theorem~\ref{LocBd}. Selfadjointness now follows from the fact that $\theta(f,f)$ is real by its formula. This proves (i). To prove (ii), Note that we already proved above that $G(z)$ is well-defined for $z\in\R+i[0,1)\,.$ To prove that it is well defined on all of $S,$ it is only necessary to prove integrability for the high $p$ part of the integral. For this \begin{eqnarray} &&\Big|2i\int_1^\infty dp\int dx\int dy\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\Big|\nonumber \\[1mm] &=&\Big|\int_1^\infty dp\,{e^{-p}\over 1-e^{-p}} \Big[e^{ipz}\wh{f}(p)\,\ol\wh{g}(p).-e^{-ipz}\ol\wh{f}(p).\,\wh{g}(p) \Big]\Big|\nonumber \\[1mm] &\leq&\int_1^\infty dp\,{e^{-p}\over 1-e^{-p}} \big(e^{-ps}+e^{ps}\big)\,\big|\wh{f}(p)\,\wh{g}(p)\big|\nonumber \\[1mm] \label{HighpIneq} &\leq&\int_1^\infty dp\,{2\over 1-e^{-p}} \big|\wh{f}(p)\,\wh{g}(p)\big|\leq{2\over 1-e^{-1}}\|f\|\,\|g\| \end{eqnarray} where $z=t+is\in S,$ and so $G(z)$ is well-defined for $z\in S\,.$ To establish the stated inequality for $z=s+it\in S,$ consider Equation~(\ref{GFubini}). Now $|F(p)|=e^{-sp}\chi\s[0,\infty).(p)$ implies $\|F\|=1,$ so $\|F(D\|=1$ and hence ${\big|(g,F(D)f, )\big|}\leq{\|f\|\,\|g\|}\,.$ The high $p$ part of the integral in (\ref{GFubini}) has an estimate (\ref{HighpIneq}), so for the low $p$ integral we have for its integrand the inequality (\ref{LowIngrand}) above, so that \begin{eqnarray*} &&\!\!\!\!\Big|\int_0^1 dp\int dx\int dy\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}} \,\sin p(z+y-x)\Big| \\[1mm] &\leq&\cosh1\cdot\int dx\int dy\,\big|f(x)\,g(y)\big|\,\big(1+|t|+|x-y|\big)\, \int_0^1 dp\,{p\,e^{-p}\over 1-e^{-p}} \\[1mm] &=& C+|t|E \end{eqnarray*} for some finite constants $C$ and $E.$ Combining this with (\ref{HighpIneq}) and the estimate for ${\big|(g,\,F(D)f)\big|},$ we obtain ${\big|G(t+is)\big|}\leq A+B|t|$ for constants $A,\,B$ as desired. It remains to prove that $G$ is continuous on $S$ and analytic in its interior. Now \[ \Big|{d\over dz}\,f(x)\,g(y)\,{e^{-p}\over 1-e^{-p}}\,\sin p(z+y-x)\Big| \leq\big|f(x)\,g(y)\big|\,{p\,e^{-p}\over 1-e^{-p}}\,\hlf(e^{-sp}+e^{sp}) \] which is $L^1$ for all $s\in(0,1)\,.$ Thus the last integral in (\ref{GFubini}) is analytic on the interior of $S\,.$ That ${(g,F(D)f)}$ is analytic in $z$ follows from spectral theory, hence $G$ is analytic on the interior of $S\,.$ For continuity on $S,$ we already have that ${(g,F(D)f)}$ is continuous in $z,$ and by inequalities (\ref{HighpIneq}) and (\ref{LowIngrand}) we can get $L^1$ estimates to ensure that $G$ is continuous on $S.$ \subsection*{Proof of Theorem~\ref{QFCAR}} We now show that quasi--free functional $\psi$ with two point functional $\theta$ is a graded KMS--functional on $\ClifS\,.$ Its domain $\dom\psi=\hbox{*-alg}\set c(f), {f\in\al S.(\R)}.$ is clearly a unital dense *-algebra of $\ClifS$ which is invariant w.r.t. both the grading $\gamma$ and the time evolution $\alpha_t\,,$ so part (i) of Definition~\ref{KMSfunDef} is satisfied. For the KMS-condition (ii), it suffices to check it for the monomials $c(f_1)\cdots c(f_k)\,.$ Let $A=c(f_1)\cdots c(f_k)$ and $B=c(g_1)\cdots c(g_m)\,$ where $k+m=2n\,.$ Then from (\ref{QFodd}) and (\ref{QFeven}) we get \begin{eqnarray} F\s A,B.(t)&:=&\psi\big(A\,\alpha_t(B)\big)= \psi(c(f_1)\cdots c(f_{k})c(T_tg_1)\cdots c(T_tg_m)\nonumber\\[1mm] \label{QFtheta} &=& (-1)^{\left({n\atop 2}\right)}\sum_P(-1)^P\prod_{j=1}^n \theta\big(h\s P(j).,\,h\s P(n+j).\big) \end{eqnarray} where $h_1=f_1,\ldots,h_k=f_k,\,h_{k+1}=T_tg_1,\ldots,h_{2n}=T_tg_m$ and $T_tf:=f_t$ is translation by $t\,.$ Since $P(j)n\,.$} \] Since by the formula for $\theta$ we have $\theta\big(T_tf,T_tg\big)= \theta\big(f,\,g\big)$ the last two types are the same and constant in $t.$ For the first type, we get by definition functions $G(t)={\theta\big(f,T_tg\big)}$ as in Theorem~\ref{ThetaProp}(ii), which we therefore know extend analytically to the strip $S\,.$ Thus since Equation~(\ref{QFtheta}) expresses $F\s A,B.(t)$ as a polynomial of constant functions and functions of the form $G\,,$ it follows that $F\s A,B.(t)$ extends to a continuous function on $S$ which is analytic on its interior. Now \begin{eqnarray} G(t+i)&=& \lim_{\varepsilon\to 0^+}\Big( \int_{-\infty}^{-\varepsilon} +\int_{\varepsilon}^\infty\Big){e^{ip(t+i)}\over 1-e^{-p}}\, \,{\widehat{f}(p)}\,\ol\hat{g}(p).\,dp \nonumber\\[1mm] &=& \lim_{\varepsilon\to 0^+}\Big( \int_{-\infty}^{-\varepsilon} +\int_{\varepsilon}^\infty\Big){e^{ipt}\over e^{p}-1}\, \,{\widehat{f}(p)}\,\ol\hat{g}(p).\,dp \nonumber\\[1mm] &=& \lim_{\varepsilon\to 0^+}\Big( \int_{-\infty}^{-\varepsilon} +\int_{\varepsilon}^\infty\Big){e^{-ipt}\over e^{-p}-1}\, \,{\widehat{f}(-p)}\,\ol\hat{g}(-p).\,dp \nonumber\\[1mm] &=&- \lim_{\varepsilon\to 0^+}\Big( \int_{-\infty}^{-\varepsilon} +\int_{\varepsilon}^\infty\Big){e^{-ipt}\over 1- e^{-p}}\, \,\ol{\widehat{f}(p)}.\,\hat{g}(p)\,dp \nonumber\\[1mm] \label{GKMStheta} &=&-\theta(T_tg,f)=\psi\big(\alpha_t(c(g))\gamma(c(f))\big) \end{eqnarray} which is the graded KMS-condition for $F\s c(f),c(g).(t)= \psi\big(c(f)\alpha_t(c(g))\big)\,.$ The terms $\theta(T_tg,f)$ are exactly the ones which occur in the corresponding expression~(\ref{QFtheta}) for ${\psi\big(\alpha_t(B)\,\gamma(A)\big)}$ so the graded KMS-condition for $F\s A,B.$ follows from the one for $G,$ Equation~(\ref{GKMStheta}). It remains to prove the growth condition~(iii) of Definition~\ref{KMSfunDef}. We already have \[ \big|G(t+is)\big|\leq a+b|t|\quad\hbox{for}\;t\in\R,\; s\in[0,1] \] by Theorem~\ref{ThetaProp}(ii). So from formula~(\ref{GKMStheta}) we get that for $t+is\in S:$ \[ \big|F\s A,B.(t+is)\big|\leq (a_1+b_1|t|)\cdots(a_n+b_n|t|)\leq C\,(1+|t|)^n \] for suitable constants $a_i,\,b_i$ and $C\,.$ Thus $\psi$ is a graded KMS-functional. \subsection*{Proof of Theorem~\ref{KMSfSUSY}} By construction $\dom\varphi$ contains the *-algebra generated by all $c(f),\;f\in\al S.(\R)$ as well as $\al R.(\al S.(\R),\sigma)$ and so it will certainly contain the *-algebra generated by $\un$ and all $c(f),\;\rlf\,,$ which is $\al A._0\,.$ So (i) is trivially true. Next, for (ii), we need to prove the SUSY-invariance of $\varphi\,,$ and for this, we need the following lemma. \begin{lem} \label{PhijR} For all $g,\;f_i\in\al S.(\R)\backslash 0$ and $\lambda_i\in\R\backslash 0$ we have % \[ % \varphi\big(j(f_0)\,R(\lambda_1,f_1)\cdots R(\lambda_n,f_n)\big) % =\sum_{k=1}^n s(f_k,f_0)\cdot %\varphi\big(j(f_0)\,j(f_k)\big)\cdot % \varphi\left(R(\lambda_1,f_1)\cdots R(\lambda_k,f_k)^2 % \cdots R(\lambda_n,f_n)\right) % \] \begin{eqnarray*} &&\!\!\!\!\!\!\!\!\!\! \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,j(g)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &=&\sum_{k=1}^n s(f_k,g)\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_k,f_k)^2\cdots R(\lambda_n,f_n) \,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &&+\sum_{k=n+1}^m s(g,f_k)\; \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_k,f_k)^2\cdots R(\lambda_m,f_m)\Big) \end{eqnarray*} where $\varphi$ is a strongly regular state on $\al R.(\al S.(\R),\sigma)$ so these expressions make sense on $\al E._0\,.$ \end{lem} \begin{beweis} Recall by Theorem~\ref{KMSresolv} that $\varphi$ is a quasi--free state on $\CCR$ defined by $\varphi(\delta_f):=\exp[-s(f,f)/2]\,,$ $f\in\al S.(\R,\R)$ where $s$ is given in Equation~(\ref{KMSCCR}). Since the maps $t,r\to s(rf,tg)$ are smooth, we can apply Proposition~\ref{MasterLemma} to $\varphi$ w.r.t. the maps $t\to tf\,.$ From the two relations ${d\over dt}R(\lambda,tf)=j(f)R(\lambda,tf)^2$ and $\lim\limits_{t\to 0}\lambda i\pi_\varphi(R(\lambda,tf))\psi=\psi,$ we get \begin{eqnarray*} &&\!\!\!\!\!\!\!\!\!\! \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,j(g)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &=&-\mu^2\,\lim_{t\to 0}{\partial\over \partial t}\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\mu,tg)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &=& -\mu^2\,\lim_{t\to 0}\bigg\{ -\sum_{k=1}^n{d\over dt}\,s(f_k,tg)\, {\partial^2\over \partial \mu\partial\lambda_k}\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\mu,tg)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &&-\hlf\,{d\over d t}\, s(tg,tg)\;{\partial^2\over \partial \mu^2}\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\mu,tg)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &&-\sum_{k=n+1}^m{d\over dt}\, s(tg,f_k)\;{\partial^2\over \partial \mu\partial\lambda_k}\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\mu,tg)\,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \bigg\} \\[1mm] &&\qquad\qquad\qquad\hbox{(By Proposition~\ref{MasterLemma})} \\[1mm] &=&\sum_{k=1}^n s(f_k,g)\, \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_k,f_k)^2\cdots R(\lambda_n,f_n) \,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_m,f_m)\Big) \\[1mm] &&+\sum_{k=n+1}^m s(g,f_k)\; \varphi\Big(R(\lambda_1,f_1)\cdots R(\lambda_n,f_n) \,R(\lambda_{n+1},f_{n+1})\cdots R(\lambda_k,f_k)^2\cdots R(\lambda_m,f_m)\Big) \end{eqnarray*} where we used ${d\over d\lambda}\rlf=-i\rlf^2\,.$ % \begin{eqnarray*} % &&\!\!\!\!\!\varphi\big(j(f_0)\,R(\lambda_1,f_1)\cdots R(\lambda_n,f_n)\big) % = -\lambda_0^2\,\lim_{t\to 0}{\partial\over \partial t}\, % \varphi\big(R(\lambda_0,tf_0)\,R(\lambda_1,f_1)\cdots R(\lambda_n,f_n)\big) % \\[1mm] % &=& -\lambda_0^2\,\lim_{t\to 0}\Big\{-\hlf\,{\partial\over \partial t}\, % s(tf_0,tf_0)\;{\partial^2\over \partial \lambda_0^2}\, % \varphi\big(R(\lambda_0,tf_0)\,R(\lambda_1,f_1)\cdots R(\lambda_n,f_n)\big) % \\[1mm] % &&\qquad-\sum_{k=1}^n{\partial\over \partial t}\, % s(f_k,tf_0)\;{\partial^2\over \partial \lambda_0\lambda_k}\, % \varphi\big(R(\lambda_0,tf_0)\,R(\lambda_1,f_1)\cdots R(\lambda_n,f_n)\big) % \Big\} % \\[1mm] % &&\qquad\qquad\qquad\hbox{(By Proposition~\ref{MasterLemma})} \\[1mm] % &=&-\lambda_0^2\,\lim_{t\to 0}\sum_{k=1}^n % s(f_k,f_0)\; % \varphi\big(R(\lambda_0,tf_0)^2\,R(\lambda_1,f_1)\cdots % R(\lambda_k,f_k)^2\cdots R(\lambda_n,f_n)\big) % \\[1mm] % &=&\sum_{k=1}^n % s(f_k,f_0)\; % \varphi\big(R(\lambda_1,f_1)\cdots % R(\lambda_k,f_k)^2\cdots R(\lambda_n,f_n)\big) % \end{eqnarray*} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % %% THE WEYL RELATION PROOF % % We will abbreviate the notation % $s(f,f)=:s(f)$ below. So, using the Weyl relation: % \begin{eqnarray} % &&\!\!\!\!\! % \varphi\left(j(f_0)\,\delta\s t_1f_1.\cdots\delta\s t_kf_k.\right) % =-i{d\over dt_0}\,\varphi\left(\delta\s t_0f_0.\cdots\delta\s t_kf_k.\right) % \Big|_{t_0=0} % \nonumber\\[1mm] % &=&-i{d\over dt_0}\, % \exp\Big(\hlf \sum_{0\leq is\,.$ % In the last step we used: % \[ % (z-S)^{-1} - (z-S_0)^{-1} = (z-S_0)^{-1} Q (z-S)^{-1} % \] So from (\ref{DiffRes}) we conclude that $E((s,\infty)) - E_0((s,\infty))$ is a trace class operator. Taking into account that $ E_0((s,\infty)) (2S_0 -1) = 0$ we obtain \begin{equation} \label{EStrace} E((s,\infty)) (2S -1) = \big(E((s,\infty)) - E_0((s,\infty))\big) (2S -1) + E_0((s,\infty)) \, 2Q, \end{equation} showing that $E((s,\infty)) (2S -1)$ and hence {\it a fortiori} $E((\mbox{\footnotesize $\frac{1}{2}$}, \infty))\, (2S -1)$ is trace class. A similar argument establishes that $E((-\infty,-\hlf)) (2S +1)$ is trace class, and hence $E([-\hlf,\hlf]^c)\cdot(2|S|-\un)$ is trace class. From Equation~(\ref{EStrace}) we get that \begin{eqnarray} \left\|E((s,\infty)) (2S -1)\right\|_1 &\leq& \left\|\big(E((s,\infty)) - E_0((s,\infty))\big) (2S -1)\right\|_1 +\left\|E_0((s,\infty)) \, 2Q\right\|_1 \nonumber\\[1mm] \label{Etraceineq} &\leq&\left\|\big(E((s,\infty)) - E_0((s,\infty))\big) (2S -1)\right\|_1 +2\|Q\|_1 \end{eqnarray} since all the terms in (\ref{EStrace}) are trace class, and $\|AQ\|_1\leq\|A\|\cdot\|Q\|_1$ for $A$ bounded. Let $P_0$ be the projection onto the eigenspace of $S_0$ with eigenvalue $\hlf$ (since $S_0=\hlf\Bigl({0\atop I}{I\atop 0}\Bigr)$ this is just the space of even functions w.r.t. the decomposition associated with $P\,).$ Then by substituting \[ (z-S_0)^{-1}=(z-\hlf)^{-1}P_0+(z+\hlf)^{-1}(\un-P_0) \qquad\hbox{and}\qquad(z-S)^{-1}(2S-\un)=(2z-1)(z-S)^{-1}-2\un \] into (\ref{DiffRes})$\times(2S-\un)$ we get that \begin{eqnarray*} &&\!\!\!\!\!\!\! \big[E((s,\infty)) - E_0((s,\infty))\big](2S-\un) \\[1mm] &=& \f 1,2\pi i. \Big[P_0Q\,\int_{C} \! dz\,(z-\hlf)^{-1}\, \Big(\f 2z-1,z-S.-2\Big) +(\un-P_0)Q\,\int_{C} \! dz\,(z+\hlf)^{-1}\,\Big(\f 2z-1,z-S.-2\Big) \Big] \\[1mm] &=&\f 1,2\pi i.P_0Q\,\int_{C} \! dz\,2 \,(z-S)^{-1}-2P_0Q +\f 1,2\pi i.(\un-P_0)Q\,\int_{C} \! dz\, {\Big(\f 2z-1,z+1/2.\Big)\over z-S} \\[1mm] &=&P_0\,Q\,2E((s,\infty))-2P_0\,Q+(\un-P_0)Q\,f(S) \end{eqnarray*} where $f(z):=\f 2z-1,z+1/2.\,\chi\s H.(z)$ and $H:=\set z\in\C, {{\rm Re}(z)> s}.\,.$ Now $\|f(S)\|\leq\|f\restriction(s,\infty)\|_\infty=2\,,$ and so \[ \Big\|\big[E((s,\infty)) - E_0((s,\infty))\big](2S-\un)\Big\|_1 \leq6\|Q\|_1\leq a\|P_JTP_J\|_1 \] for a constant $a>0\,,$ where we obtain the last inequality $\|Q\|_1\leq\hbox{const.}\|P_JTP_J\|_1$ from the decomposition in (\ref{RQstart}) from which $PQP=-iB=-iP(P_JTP_J)(\un-P),$ $(\un-P)QP=P(P_JTP_J)P$ etc. Thus by (\ref{Etraceineq}) we get \[ \|E((\hlf,\infty))(2S-1)\|_1=\|E(\hlf,\infty)E(s,\infty)(2S-1)\|_1 \leq\|E(s,\infty)(2S-1)\|_1\leq a\|P_JTP_J\|_1\,. \] A similar argument establishes that $\|E(-\infty,-\hlf)(2S+1)\|_1 \leq a\|P_JTP_J\|_1$ and hence that $ \left\|E([-\hlf,\hlf]^c)\cdot(2|S|-\un)\right\|_1\leq b\left\|P_JTP_J\right\|_1\, $ for a positive constant $b\,.$ \chop (ii) Recall that $S$ has a purely discrete spectrum in $[-\hlf,\hlf]^c\,,$ so choose an orthonormal system $\set{ e_j\in\al K.},j\in\J\subseteq\N.$ of eigenvectors of $S,$ corresponding to eigenvalues $s_j\in[-\hlf,\hlf]^c$ and exhausting these eigenspaces (some $s_j$ will coincide for higher multiplicities). Let $E_j$ be the one--dimensional orthogonal projection onto $e_j$, $j \in \J$, and let $\mbox{\boldmath $\lambda$} = (\lambda_1, \lambda_2, \dots ) \in \bigoplus_{j \in \J} \, \{-1,1\}$. We define on ${\cal K}$ the unitaries $$V(\mbox{\boldmath $\lambda$}) := E([-\hlf,\hlf]) + \sum_{j \in \J} \lambda_j E_j\,.$$ Since $V(\mbox{\boldmath $\lambda$})$ commutes with $\Gamma$, these unitaries induce an action $\gamma:\bigoplus_{j \in \J} \, \{-1,1\} \to\aut{\rm CAR}(\al K.)$ given by $$ \gamma_{\mbox{\boldmath $\scriptstyle\lambda$}} (\Phi(f)) := \Phi(V(\mbox{\boldmath $\lambda$}) f). $$ Since $V(\mbox{\boldmath $\lambda$})^2=\un\,,$ we can decompose ${\rm CAR}(\al K.)$ into odd and even parts w.r.t. each $\gamma_{\mbox{\boldmath $\scriptstyle\lambda$}}\,.$ Moreover, since $V(\mbox{\boldmath $\lambda$})$ commutes with $S$ we have that $ \psi \circ \gamma_{\mbox{\boldmath $\scriptstyle\lambda$}} = \psi $ and so $\psi$ must vanish on the odd part of ${\rm CAR}_0$ with respect to each $\gamma_{\mbox{\boldmath $\scriptstyle\lambda$}}\,.$ Let $\al C.\subset{\rm CAR}_0$ be the *--algebra generated by $\set{\Phi( e_j)},j\in\J.\cup\set{\Phi(f)},{f\in E\big([-\hlf,\hlf]\big) \al K.}.\,,$ then $\al C.$ is mapped to itself by all $\gamma_{\mbox{\boldmath $\scriptstyle\lambda$}}\,.$ Since the two-point functional $\theta$ is bounded on $L^2(J),$ it suffices to calculate the norm of $\psi$ on $\al C.,$ and in fact on the intersection of all the even parts of $\al C.$ with respect to $\gamma_{\mbox{\boldmath $\scriptstyle\lambda$}}\,,$ and we denote this *-algebra by $\varepsilon(\al C.)\,.$ It is produced by a projection $\varepsilon$ which we can consider as the projection defined on $\al C.$ by averaging over the action of $\gamma_{\mbox{\boldmath $\scriptstyle\lambda$}}$ on $\al C..$ Since for each $A\in\al C.$ only a finite number of $j\hbox{'s}$ are involved, these averages will again be in the *-algebra $\al C..$ Now we only need to consider monomials in the $\Phi( e_j)$ and $\Phi( e_j)^*$ which are even in each index $j\,.$ In a given monomial $\Phi( e_{j_1})\cdots\Phi( e_{j_n})\in\varepsilon(\al C.)$ if we collect all (even number of) terms with the same $j$ together, we can then simplify it with the relations $2 \, \Phi(e_j)^2 = \langle \Gamma e_j | e_j \rangle 1 $ and $\big[\Phi(e_j)^*\Phi(e_j)\big]^2=\Phi(e_j)^*\Phi(e_j)-\f 1,4.\big|(\Gamma e_j, e_j)\big|^2\un\,.$ Thus $\varepsilon(\al C.)$ is generated by the two-dimensional abelian *-algebras $\al C._j:=\hbox{*--alg}\{ \Phi(e_j)^*\Phi(e_j)\}$ and $\al C._0:=\hbox{*--alg}\set\Phi(f),{f\in E\big([-\hlf,\hlf] \big)\al K.}.\,.$ Since for $i\not=j$ we have $$ \varepsilon\left(\big[\Phi(e_i)^*\Phi(e_i),\,\Phi(e_j)^*\Phi(e_j)\big]\right) =\varepsilon\left(\ol\big(\Gamma e_i,e_j\big).\Phi(e_j)\Phi(e_i)+ \big(\Gamma e_i,e_j\big)\Phi(e_j)^*\Phi(e_i)^*\right)=0 $$ it follows that all the $\al C._i$ commute, and in fact we have the (incomplete) tensor product decomposition $$ \varepsilon({\cal C}) = \bigotimes_{j \in \{0\} \cup \J} \, {\cal C}_j. $$ Moreover, $\psi$ is a product functional on this tensor product. Hence its norm, if it exists, is given by $$\| \psi \| = \prod_{j \in \{0\} \cup \J} \| \psi_j \|,$$ where $\psi_j$ denotes the restriction of $\psi$ to $ {\cal C}_j$. Now $\psi_0$ is by construction a state on ${\cal C}_0$ because $\hlf+S$ is positive on ${E\big([-\hlf,\hlf] \big)\al K.},$ hence the two-point function is positive and so $\| \psi_0 \| = 1$. Since for $a,b \in \C$ $$ \psi(a1+b\,\Phi(e_j)^* \Phi(e_j)) = a + b \, (\mbox{\footnotesize $\frac{1}{2}$} + s_j) $$ and $$ \| a1+b\,\Phi(e_j)^* \Phi(e_j) \| = \max\{|a|,|a+b|\} $$ one obtains $ \| \psi_j \| = 2 |s_j| $, $j \in \J$. Thus since $2|s_j|>1,$ a necessary and sufficient condition for the existence of $\| \psi \|$ is $\sum_{j \in \J} (2 |s_j| -1) < \infty$. However, this is guaranteed by part (i) \end{beweis} Using this lemma, we can now prove: \begin{lem} \label{eT1norm} We have $$\big\|\varphi\restriction\al A._0(J)\big\| \leq\exp\big(b\|P_JTP_J\|_1\big)$$ where $b$ is a positive constant (independent of $J$) and $\|\cdot\|_1$ denotes the trace--norm. \end{lem} \begin{beweis} recall from Equation~(\ref{psiProd}) that \[ \big\|\varphi\restriction\al A._0(J)\big\| = \prod_{j \in \{0\} \cup \Mb{J}} \| \psi_j \| =\prod_{j \in \Mb{J}}2|s_j|=\prod_{j \in \Mb{J}}(1+t_j) <\infty \] where $t_j:=2|s_j|-1\,.$ Now $\ln(1+x)\leq x$ for $x\geq 0\,,$ so \[ \ln\prod_{j =1}^N(1+t_j)=\sum_{j=1}^N\ln(1+t_j)\leq\sum_{j=1}^Nt_j \qquad\hbox{hence}\qquad\prod_{j \in \Mb{J}}(1+t_j)\leq \exp\Big(\sum_{j\in\J}t_j\Big) \] Now $\sum\limits_{j\in\J}t_j=\sum\limits_{j\in\J}(2|s_j|-1)=\big\| E([-\hlf,\hlf]^c)\cdot(2|S|-\un)\big\|_1 \leq b\left\|P_JTP_J\right\|_1$ for a constant $b>0$ by Lemma~\ref{RomeWork}(i). Combining these claims prove the lemma \end{beweis} To conclude the proof of the theorem, we need to estimate $\left\|P_JTP_J\right\|_1\,.$ Recall from Theorem \ref{ThetaProp} that \begin{eqnarray} \big(P_JTP_Jf\big)(x)&:=&2i\,\chi\s J.(x)\int dy\,f(y)\,(x-y)\,\int_0^\infty dp\, \ln( 1-e^{-p})\,\cos p(x-y)\,\chi\s J.(y) \nonumber \\[1mm] &=& i\,\chi\s J.(x)\int dy\,f(y)\,(x-y)\,\int_{-\infty}^\infty dp\, \ln( 1-e^{-|p|})\,e^{ip(x-y)}\,\chi\s J.(y) \nonumber \\[1mm] \label{PTPcomm} &=&{\rm const.}\Big(\chi\s J.(X)\,\Big[X,\,\ln(1-e^{-|P|})\Big]\, \chi\s J.(X)\,f\Big)(x) \end{eqnarray} where $X$ is the multiplication operator $(Xf)(x)=xf(x),$ and $P$ is as usual $i{d\over dx}\,,$ and the constant incorporates the $2\pi$ factors from the Fourier transforms. Now $D:=-\ln(1-e^{-|P|})$ is a positive operator, so write trivially $ \chi\s J.(X)\,X\,D\,\chi\s J.(X)=\big(\chi\s J.(X)\,X\,D^{1/2}\big) \big(D^{1/2}\,\chi\s J.(X)\big)\,,$ then we show that both factors are Hilbert--Schmidt. Now \begin{eqnarray*} \big(\chi\s J.(X)\,X\,D^{1/2}f\big)(x)&=&\int dy\,K(x,y)\qquad \qquad\hbox{where:}\\[1mm] K(x,y)&=&\chi\s J.(x)\,x\,\int dp\,\big[-\ln( 1-e^{-|p|})\big]^{1/2}\, e^{ip(x-y)} \\[1mm] \hbox{so}\qquad \big\|\chi\s J.(X)\,X\,D^{1/2}\big\|_2^2&=& \int dx\,dy\; \big|K(x,y)\big|^2 %\\[1mm] =\int_Jdx\, x^2\,\int dp\,\big[-\ln( 1-e^{-|p|})\big] \\[1mm] &\leq& {\rm const.}|J|^3\,, \end{eqnarray*} using the integrability of $\ln( 1-e^{-|p|})\,.$ Likewise, we get \[ \big\|(D^{1/2}\,\chi\s J.(X)\big\|_2^2= \int_Jdx\,\int dp\,\big[-\ln( 1-e^{-|p|})\big]\leq {\rm const.}|J|\,. \] Thus $\chi\s J.(X)\,X\,D\,\chi\s J.(X)$ is trace class, and as $P_JTP_J$ is, so is ${\chi\s J.(X)\,D\,X\,\chi\s J.(X)}\,.$ For their trace norms we find \[ \big\|\chi\s J.(X)\,X\,D\,\chi\s J.(X)\big\|_1 \leq\big\|\chi\s J.(X)\,X\,D^{1/2}\big\|_2\cdot \big\|(D^{1/2}\,\chi\s J.(X)\big\|_2 \leq{\rm const.}|J|^2 \] and likewise $\big\|\chi\s J.(X)\,D\,X\, \chi\s J.(X)\big\|_1\leq{\rm const.}|J|^2\,.$ Thus by (\ref{PTPcomm}) we get that \[ \big\|P_JTP_J\big\|_1\leq {\rm const.}\big\|\chi\s J.(X)\,X\,D\,\chi\s J.(X)\big\|_1 +{\rm const.}\big\|\chi\s J.(X)\,D\,X\, \chi\s J.(X)\big\|_1\leq K|J|^2 \] for a constant $K$ (independent of $J$). Now from Lemma~\ref{eT1norm} we get the claim of the theorem, i.e. that \[ \big\|\varphi\restriction\al A._0(J)\big\| \leq\exp\big(K\,|J|^2\big)\,. \] \subsection*{Proof of Theorem~\ref{CoBd}} % Let $a_i\in\al D._c$ where $1=0,\ldots,\,n\,.$ Thus there is some Fix a compact interval $J=[-k,\,k]$ and let $a_i\in\al A._0(J)$ for all $i\,,$ then as $\alpha_t(a_i)\in\al A._0(J+t)\,$ we have $\alpha\s t_0.(a_0)\cdots\alpha\s t_n.(a_n)\in \al A._0([-M,M])$ where $M:=k+\sup\limits_i|t_i|\,.$ Thus by Theorem~\ref{LocBd} we get \begin{eqnarray*} &&\!\!\!\!\!\!\!\! \Big|\varphi\big(\alpha\s t_0.(a_0)\cdots\alpha\s t_n.(a_n)\big)\Big| \leq e^{4KM^2}\,\|a_0\|\cdots\|a_n\|\;. \\[1mm] \hbox{Now}\qquad &&M^2= k^2+\sup_i t_i^2+2k\,\sup_i|t_i| \leq k^2+\sum_it_i^2+2k\,\sup_i(1+t_i^2) \\[1mm] &&\leq k^2+\sum_it_i^2+2k\Big(1+\sum_it_i^2)\Big) =k^2+2k+(1+2k)\,\sum_it_i^2 \\[1mm] \hbox{hence:}\qquad &&\!\!\!\!\!\!\!\! \Big|\varphi\big(\alpha\s t_0.(a_0)\cdots\alpha\s t_n.(a_n)\big)\Big| \leq A\,\exp\Big(B\,\sum_it_i^2\Big)\,\|a_0\|\cdots\|a_n\|\;. \end{eqnarray*} for suitable constants $A$ and $B$ depending only on $k$ (but not on $n$). Now let $t_0= 0,\ab\;t_1=s_1,\ab\;t_2=s_1+s_2,\ldots,\,t_n= s_1+\cdots+s_n\,,$ and define for all $s_i\in\R$: \[ F\s a_o,\ldots a_n.(s_1,\ldots,\,s_n):= \exp\Big(-B\hbox{$\sum\limits_{k=1}^n$}(s_1+\cdots+s_k)^2\Big)\cdot \varphi\Big(a\s 0.\,\alpha\s s_1.(a_1)\cdots\alpha\s s_1+\cdots+s_n.(a_n)\Big)\,. \] Then we have $\big|F\s a_o,\ldots a_n.(s_1,\ldots,\,s_n)\big| \leq A\,\|a_0\|\cdots\|a_n\|\;,$ and by the KMS--property of $\varphi,$ the function $F\s a_o,\ldots a_n.$ can be analytically continued in each variable $s_j$ into the strip $S_j:=\{z_j\in\C^n\,\big|\,{\rm Im}\,z_j\in[0,1]\},$ keeping the other variables real. This produces functions $F^{(j)}\s a_o,\ldots a_n.$ analytic in the flat tubes $T^j:=\R^{n-1}\times S_j\,,$ and hence by using the Flat Tube Theorem~\ref{FTT} inductively, we obtain an analytic continuation of $F\s a_o,\ldots a_n.$ into the tube $\al T._n:=\R^n+i\Sigma_n$ where $\Sigma_n:= \set{\bf s}\in\R^n,0\leq s_i\;\forall\,i,\;\;s_1+\cdots+s_n\leq1.\,,$ coinciding with all $F^{(j)}\s a_o,\ldots a_n.$ on $T^j\,.$ We want to obtain a bound for this analytic function $F\,.$ We start by finding bounds for the $F^{(j)}\,.$ Let $G(s_1,\ldots,\,s_n):={\varphi\Big(a\s 0.\,\alpha\s s_1.(a_1) \cdots\alpha\s s_1+\cdots+s_n.(a_n)\Big)}$ which has analytic extensions to each $T^j\,,$ and by the definition of KMS--functionals we know that $\big|G(s_1,\ldots,\,s_j+ir_j,\ldots,\,s_n)\big|\leq C(1+|s_j|)^N$ where $C$ and $N$ are independent of $s_j$ and $r_j\in[0,1]\,.$ Now \begin{eqnarray} &&\!\!\!\!\!\! F\s a_o,\ldots a_n.({s_1,\ldots,\,s_j+ir_j,\ldots,\,s_n})= \nonumber\\[1mm] \label{Festimate} &&G({ s_1,\ldots,\,s_j+ir_j,\ldots,\,s_n})\,\exp\big[Br_j^2(n+1-j)- B\hbox{$\sum\limits_{k=1}^n$}(s_1+\cdots+s_k)^2 +i\theta\big] \end{eqnarray} where $\theta$ is real. Thus from the exponential damping factor in $s_j$ we conclude that $F^{(j)}\s a_o,\ldots a_n.$ is bounded. By the maximum modulus principle (applied after first mapping $S_j$ to a unit disk by the Schwartz mapping principle), the bound of $|F^{(j)}\s a_o,\ldots a_n.|$ is attained on the boundary of $S_j$ (this also follows from the Phragmen Lindel\"of theorem, cf. p138 in~\cite{Conw}). So on the real part of the boundary of $S_j$ we have already from above that $ \big|F^{(j)}\s a_o,\ldots a_n. (s_1,\ldots,\,s_n)\big| \leq A\,\|a_0\|\cdots\|a_n\| $ and by the KMS--condition and translation invariance of $\varphi$ we have on the other part \begin{eqnarray*} &&\big|G(s_1,\ldots,\,s_j+i,\ldots,\,s_n)\big|= \Big|\varphi\Big(\alpha\s s_1+\cdots s_j.(a_j)\cdots \alpha\s s_1+\cdots+ s_n.(a_n) a\s 0.\,\alpha\s s_1.(a_1)\cdots\alpha\s s_1+\cdots+s_{j-1}. (a_{j-1})\Big)\Big| \\[1mm] &&\qquad\qquad\qquad \leq A\,\exp\Big(B\hbox{$\sum\limits_{k=1}^n$}(s_1+\cdots+s_k)^2\Big)\cdot \|a_0\|\cdots\|a_n\|\qquad\hbox{hence by (\ref{Festimate}):} \\[1mm] &&\big|F^{(j)}\s a_o,\ldots a_n.(s_1,\ldots,s_j+i,\ldots,\,s_n)\big| \leq A\,e^{Bn}\,\|a_0\|\cdots\|a_n\|\qquad\hbox{and thus as $e^{Bn}>1\,,$} \\[1mm] &&\big|F^{(j)}\s a_o,\ldots a_n.(s_1,\ldots,z_j,\ldots,\,s_n)\big|\leq A\,e^{Bn}\,\|a_0\|\cdots\|a_n\|=:C\qquad\hbox{for all $z_j\in S_j\,.$} \end{eqnarray*} Now define $H_\alpha(z_1,\ldots,\,z_n):=[F\s a_o,\ldots a_n.(z_1,\ldots,\,z_n) -e^{i\alpha}C]^{-1}$ where $\alpha\in[0,2\pi]\,$ for $(z_1,\ldots,\,z_n) \in \al T._n\,.$ Then by the estimates above for $|F^{(j)}\s a_o,\ldots a_n.|\,,$ each map $z_j\to H_\alpha(s_1,\ldots,z_j,\ldots,\,s_n)$ is analytic on the strip $S_j\,,$ and thus by the Flat Tube Theorem~\ref{FTT}, $H_\alpha$ has a unique extension as an analytic function to $\al T._n\,,$ and hence cannot have any singularities in $\al T._n\,,$ i.e. $F\s a_o,\ldots a_n.(z_1,\ldots,\,z_n) \not=e^{i\alpha}C$ for all $\alpha\,.$ By continuity of $F\,,$ the image set $F\s a_o,\ldots a_n.(\al T._n)$ must be connected. By assumption, this set has some points inside the circle $|z|=C\,,$ hence the entire image set is inside the circle $|z|=C\,,$ i.e. \[ \big|F\s a_o,\ldots a_n.(z_1,\ldots,\,z_n)\big|\leq A\,e^{Bn}\,\|a_0\|\cdots\|a_n\| \qquad\forall\;(z_1,\ldots,\,z_n)\in\al T._n\quad \hbox{and}\quad a_i\in\al A._0(J)\,. \] Consider now the Chern character formula~(\ref{ChCa}): \begin{eqnarray*} \tau_n(a_0,\ldots,a_n)&:=& i^{\epsilon_n}\int_{\sigma_n} \varphi\Big(a_0\,\alpha_{is_1}\big(\delta\gamma(a_1)\big)\, \alpha_{is_2}\big(\delta(a_2)\big)\,\alpha_{is_3}\big(\delta\gamma(a_3)\big) \cdots \\[1mm] & &\qquad\qquad\cdots\alpha_{is_n}\big(\delta\gamma^n(a_n)\big)\Big)\, ds_1\cdots ds_n\,,\qquad a_i\in\al D._c\,, \\[1mm] &=& i^{\epsilon_n} \int_{\Sigma_n}\varphi\big(b_0\,\alpha_{ir_1}(b_1)\cdots\alpha_{ir_1+\cdots+ ir_n}(b_n)\big) \,dr_1\cdots dr_n \end{eqnarray*} where we made a change of variables $s_1=r_1,\;s_2=r_1+r_2,\ldots,s_n=r_1+\cdots+r_n$ and substitutions $a_0=b_0,$ $b_1=\delta\gamma(a_1),\ldots, b_n=\delta\gamma^n(a_n)$ as in Section~\ref{JLOsection}, making use of the Flat Tube Theorem. (Note that $b_i\in\al A._0(J)\;\forall\,i$ for some $J\,.)$ In fact, from the uniqueness part of the extensions to $\al T._n$ we have that on $\al T._n$ \[ \varphi\big(b_0\,\alpha_{z_1}(b_1)\cdots\alpha_{z_1+\cdots+ z_n}(b_n)\big) =\exp\big[B\sum_{k=1}^n(z_1+\cdots+ z_n)^2\big]\cdot F\s b_o,\ldots b_n.(z_1,\ldots,\,z_n) \] and so for $(z_1,\ldots,\,z_n)=i(r_1,\ldots,\,r_n)\in i\Sigma_n$ we have \begin{eqnarray*} \Big|\varphi\big(b_0\,\alpha_{ir_1}(b_1)\cdots\alpha_{ir_1+\cdots+ ir_n}(b_n)\big) \Big| &\leq& \exp\Big[-B\sum_{k=1}^n(r_1+\cdots+ r_n)^2+Bn\Big] \,A\,\|b_0\|\cdots\|b_n\| \\[1mm] \hbox{hence:}\qquad \big|\tau_n(a_0,\ldots,a_n)\big| &\leq& {A\over n!}\,e^{Bn}\,\|b_0\|\cdots\|b_n\|\leq{A\over n!}\,e^{Bn}\, \|a_0\|_*\cdots\|a_n\|_* \end{eqnarray*} where we used first, that the volume of $|\Sigma_n|=1/n!\,,$ and second, that $\|b_j\|\leq\|a_j\|_*$ because $b_j=\delta\gamma(a_j)=-\gamma\delta(a_j)$ for $j>0\,.$ Thus $\|\tau_n\|_*\leq A\,e^{Bn}\big/n!$ and hence it is clear that $\lim\limits_{n\to\infty}n^{1/2}\|\tau\|_*^{1/n}\leq e^B\,\lim\limits_{n\to\infty}n^{1/2}(A/n!)^{1/n}=0$ by Stirling's formula, which concludes the proof. \subsection*{Proof of Theorem~\ref{TCycCoc}} By Theorem~\ref{CoBd} we already have the entireness condition for $\widetilde{\tau}$ so it is only necessary to prove the cocycle condition for $a_i\in\al D._c\,:$ \[ (b{\tau}_{n-1})(a_0,\ldots,a_n)=(B{\tau}_{n+1})(a_0,\ldots,a_n) \,,\qquad n=1,3,5,,\ldots \] with $b$ and $B$ given by Equations (\ref{Defb}) and (\ref{DefB}). We will roughly follow the technique used in \cite{JLOK}, but due to the different analytic properties of our model, we will need to go explicitly through the steps. In order to manipulate the expressions involved with Equation~(\ref{CocEq}), we need the results in the following Lemma. \begin{lem} \label{CocLem} Let $b_i\in\al A._0\,,$ then: \begin{itemize} \item[(i)] $\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_n}.(b_n)\Big) =\varphi\Big(\gamma(b_n)\,\alpha\s{i(1-s_n)}.(b_0)\,\alpha\s{i(1-s_n+s_1)}.(b_1) \cdots\alpha\s{i(1-s_n+s_{n-1})}.(b_{n-1})\Big)$ for all $(s_1,\ldots,\,s_n)\in\sigma_n\;.$ \item[(ii)] $\displaystyle{\int_{\sigma_n}\varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_n}.(b_n)\Big)\,ds_1\cdots ds_n= \int_{\sigma_n}\varphi\Big(\gamma(b_n)\,\alpha\s{is_1}.(b_0) \cdots\alpha\s{is_n}.(b_{n-1})\Big)}\,ds_1\cdots ds_n$ \item[(iii)] The functions $(t_1,\ldots,\,t_n)\to \varphi\Big(b_0\,\alpha\s{t_1}.(b_1)\cdots\delta\big(\alpha\s t_k.(b_k)\big) \cdots\alpha\s{t_n}.(b_n)\Big)$ and\chop $(t_1,\ldots,\,t_n)\to \varphi\Big(b_0\,\alpha\s{t_1}.(b_1)\cdots\gamma\big(\alpha\s t_k.(b_k)\big) \cdots\alpha\s{s_n}.(b_n)\Big)$ both have analytic continuations to $\R^n+i\sigma_n\,,$ and for these we have\chop $\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\delta\big(\alpha\s is_k.(b_k)\big) \cdots\alpha\s{is_n}.(b_n)\Big) =\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s is_k.\big(\delta(b_k)\big) \cdots\alpha\s{is_n}.(b_n)\Big)$ and \chop $\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\gamma\big(\alpha\s is_k.(b_k)\big) \cdots\alpha\s{is_n}.(b_n)\Big) =\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s is_k.\big(\gamma(b_k)\big) \cdots\alpha\s{is_n}.(b_n)\Big)\,.$ \item[(iv)] For $j=2,\ldots,\,n$ we have: \begin{eqnarray} &&\!\!\!\! \int_{\sigma_{n+1}}{\partial\over\partial s_j} \varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big)\,ds_1\cdots ds_{n+1} \nonumber\\[1mm] &&=\int_{\sigma_n}\Big[\varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s is_j.(b_jb_{j+1})\cdots\alpha\s{is_n}.(b_{n+1})\Big) \nonumber\\[1mm] \label{ByParts1} &&\qquad\qquad- \varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s is_{j-1}.(b_{j-1}b_j)\cdots\alpha\s{is_n}.(b_{n+1})\Big)\Big] \,ds_1\cdots ds_n \\[1mm] &&\!\!\!\! \int_{\sigma_{n+1}}{\partial\over\partial s_1} \varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big)\,ds_1\cdots ds_{n+1} \nonumber\\[1mm] \label{ByParts2} &&=\int_{\sigma_n}\Big[ % \nonumber\\[1mm] % \label{ByParts2} % &&\qquad\qquad+ \varphi\Big(b_0\,\alpha\s{is_1}.(b_1b_2) \,\alpha\s is_{2}.(b_3)\cdots\alpha\s{is_n}.(b_{n+1})\Big) -\varphi\Big(b_0\,b_1\,\alpha\s{is_1}.(b_2) \cdots\alpha\s{is_n}.(b_{n+1})\Big)\Big] \,ds_1\cdots ds_n\qquad\qquad{} \\[1mm] &&\!\!\!\! \int_{\sigma_{n+1}}{\partial\over\partial s_{n+1}} \varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big)\,ds_1\cdots ds_{n+1} \nonumber\\[1mm] \label{ByParts3} &&=\int_{\sigma_n}\Big[\varphi\Big(\gamma(b_{n+1})\,b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_n}.(b_{n})\Big) % \nonumber\\[1mm] % \label{ByParts3} % &&\qquad\qquad- -\varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_n}.(b_nb_{n+1})\Big)\Big] \,ds_1\cdots ds_n \end{eqnarray} \end{itemize} \end{lem} \begin{beweis} (i) Recall that the left hand side is defined by the analytic extension of the function ${F\s t_1\cdots,t_n.(b_0,\cdots,\,b_n)}:= {\varphi\big(b_0\,\alpha\s t_1.(b_1)\cdots \alpha\s t_n.(b_n)\big)}$ to the tube $\R^n+i\sigma_n$ by the KMS--condition and Flat Tube theorem, so $\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_n}.(b_n)\Big) :={F\s is_1\cdots,is_n.(b_0,\cdots,\,b_n)}\,.$ By the invariance ${\varphi\circ\alpha_t}=\varphi$ we have \begin{eqnarray*} F\s t_1\cdots,t_n.(b_0,\cdots,\,b_n)&=&\varphi\big(\alpha_t(b_0)\,\alpha\s t+t_1.(b_1) \cdots\alpha\s t+t_n.(b_n)\big)\\[1mm] &=&F\s t,\,t+t_1\cdots,\,t+t_n.(\un,\,b_0,\cdots,\,b_n)= F\s t+t_1\cdots,\,t+t_n.(\alpha_t(b_0),\,b_1,\cdots,\,b_n)\,. \end{eqnarray*} The latter function has an analytic continuation in the variables ${(t+t_1,\ldots,\,t+t_n)}$ to $\R^n+i\sigma_n$ and from the former function it also has an analytic extension in $t$ to the strip $\R+i[0,1]\,.$ Thus by the flat tube theorem we get a unique analytic extension to all of $\R^{n+1}+i\sigma\s n+1.\,.$ Put $t_j=is_j$ where ${\bf s}\in\sigma_n$ and $t=i(1-s_n)\,,$ then \[ F\s is_1\cdots,is_n.(b_0,\cdots,\,b_n) =F\s i(1-s_n),\,i(1-s_n+s_1),\cdots,\,i(1-s_n+s_{n-1}),\,i. (\un,\,b_0,\cdots,\,b_n)\; \] which is justified because we have that the variables \[ {\bf r}=(r_1,\cdots,r_n):= (1-s_n,\,1-s_n+s_1,\ldots,\,1-s_n+s_{n-1})\in\sigma_n\,. \] Now the function ${F\s ir_1,\cdots,\,ir_n,\,i. (\un,\,b_0,\cdots,\,b_n)}$ is obtained from the analytic extension of ${F\s t_1,\cdots,\,t_n,\,v. (\un,\,b_0,\cdots,\,b_n)}\,,$ $t_i,\;v\in\R$ to $\R^{n+1}+i\sigma\s n+1.\,.$ By the KMS--condition: \begin{eqnarray*} &&\!\!\!\!\! F\s t_1,\cdots,\,t_n,\,i.(\un,\,b_0,\cdots,\,b_n) = \varphi\Big(b_n\,\gamma\big(\alpha\s{t_1}.(b_0)\cdots\alpha\s{t_n}.(b_{n-1})\big)\Big) \\[1mm] && = \varphi\Big(\gamma(b_n)\,\alpha\s{t_1}.(b_0)\cdots\alpha\s{t_n}.(b_{n-1})\Big) =F\s t_1,\cdots,\,t_n.(\gamma(b_n),\,b_0,\cdots,\,b_{n-1}) \end{eqnarray*} Thus by uniqueness of the analytic continuations we have \[ F\s is_1\cdots,is_n.(b_0,\cdots,\,b_n)=F\s ir_1,\cdots,\,ir_n,\,i. (\un,\,b_0,\cdots,\,b_n)=F\s ir_1,\cdots,\,ir_n.(\gamma(b_n),\,b_0,\cdots,\,b_{n-1}) \] which is the statement (i) of the lemma.\chop (ii) By part (i) we have: \begin{eqnarray*} &&\!\!\!\!\! \int_{\sigma_n}\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_n}.(b_n)\Big) \,ds_1\cdots ds_n \\[1mm] &&=\int_{\sigma_n}\varphi\Big(\gamma(b_n)\,\alpha\s{i(1-s_n)}.(b_0)\, \alpha\s{i(1-s_n+s_1)}.(b_1) \cdots\alpha\s{i(1-s_n+s_{n-1})}.(b_{n-1})\Big) \,ds_1\cdots ds_n \\[1mm] &&=\int_{\sigma_n}\varphi\Big(\gamma(b_n)\,\alpha\s{ir_1}.(b_0)\, \cdots\alpha\s{ir_n}).(b_{n-1})\Big)\,dr_1\cdots dr_n \end{eqnarray*} making use of the change of variables ${\bf s}\to{\bf r}$ above (with Jacobian $=1),$ and the fact that ${\bf r}\in\sigma_n$ iff ${\bf s}\in\sigma_n\,.$\chop (iii) Since $\varphi\Big(b_0\,\alpha\s{t_1}.(b_1)\cdots\delta\big(\alpha\s t_k.(b_k)\big) \cdots\alpha\s{t_n}.(b_n)\Big)=\varphi\Big(b_0\,\alpha\s{t_1}.(b_1)\cdots \alpha\s t_k.\big(\delta(b_k)\big) \cdots\alpha\s{t_n}.(b_n)\Big)$ and the latter obviously has an analytic extension to $\R^n+\sigma_n$ the claim follows. Likewise for the other one. \chop (iv) For $2\leq j\leq n$ we have \begin{eqnarray*} &&\!\!\!\!\! \int_{s_{j-1}}^{s_{j+1}}{\partial\over\partial s_j} \varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big) \,ds_j \\[1mm] &&=\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_{j-1}}.(b_{j-1}) \,\alpha\s{is_{j+1}}.(b_{j})\,\alpha\s{is_{j+1}}.(b_{j+1}) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big) \\[1mm] &&\quad-\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_{j-1}}.(b_{j-1}) \,\alpha\s{is_{j-1}}.(b_{j})\,\alpha\s{is_{j+1}}.(b_{j+1}) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big) \\[1mm] &&=\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_{j-1}}.(b_{j-1}) \,\alpha\s{is_{j+1}}.(b_{j}b_{j+1}) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big) \\[1mm] &&\quad-\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots\alpha\s{is_{j-1}}.(b_{j-1} b_{j})\,\alpha\s{is_{j+1}}.(b_{j+1}) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big) \end{eqnarray*} from which equation~(\ref{ByParts1}) follows by a change of label of the integration variables. For equation~(\ref{ByParts2}) we substitute $j=1,\;s\s j-1.=0$ into the last equation. For equation~(\ref{ByParts3}) we substitute $j=n+1,\;s\s j+1.=1$ into the last equation to get \begin{eqnarray*} &&\!\!\!\!\! \int_{\sigma_{n+1}}{\partial\over\partial s_{n+1}} \varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_{n+1}}.(b_{n+1})\Big)\,ds_1\cdots ds_{n+1} \\[1mm] &&=\int_{\sigma_{n}}\Big[\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots \alpha\s{is_{n}}.(b_{n})\,\alpha\s{i}.(b_{n+1})\Big) -\varphi\Big(b_0\,\alpha\s{is_1}.(b_1)\cdots \alpha\s{is_{n}}.(b_{n}b_{n+1})\Big)\Big]\,ds_1\cdots ds_{n} \\[1mm] &&=\int_{\sigma_n}\Big[\varphi\Big(\gamma(b_{n+1})\,b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_n}.(b_{n})\Big) -\varphi\Big(b_0\,\alpha\s{is_1}.(b_1) \cdots\alpha\s{is_n}.(b_nb_{n+1})\Big)\Big] \,ds_1\cdots ds_n \end{eqnarray*} making use of part (i) for the KMS--condition. \end{beweis} Let us begin with the right hand side of our desired Equation~(\ref{CocEq}). From the definition~(\ref{DefB}) we have for $a_i\in\al D._c\,$ via $\delta(\un)=0$ that: \begin{eqnarray*} &&\!\!\!\!\! \big(B{\tau}_{n+1}\big)(a_0,\ldots,\,a_n) =i^{\epsilon_{n+1}}\int_{\sigma_{n+1}}ds_1\cdots ds_{n+1}\,\Big[ \varphi\Big(\alpha\s is_1.(\delta\gamma a_0)\cdots\alpha\s is_{n+1}. (\delta\gamma^{n+1} a_n)\Big) \\[1mm] &&\quad +\sum_{j=1}^n(-1)^{nj}\,\varphi\Big(\alpha\s is_1.(\delta \gamma^2 a_{n+1-j})\cdots \alpha\s is_j.(\delta\gamma^{j+1} a_n)\,\alpha\s is_{j+1}.(\delta\gamma^{j+1} a_0) \cdots\alpha\s is_{n+1}.(\delta\gamma^{n+1} a_{n-j})\Big)\Big]\,. \end{eqnarray*} We can now use Lemma~\ref{CocLem}(ii) in all the terms on the right hand side to bring the factor with $a_0$ to the front: \begin{eqnarray*} &&\!\!\!\!\! \big(B{\tau}_{n+1}\big)(a_0,\ldots,\,a_n) =i^{\epsilon_{n+1}}\int_{\sigma_{n+1}}ds_1\cdots ds_{n+1}\,\Big[ \varphi\Big(\delta\gamma( a_0)\,\alpha\s is_1.(\delta\gamma^2 a_1) \cdots \\[1mm] &&\quad\cdots\alpha\s is_{n}. (\delta\gamma^{n+1} a_n)\,\alpha\s is_{n+1}.(\un)\Big) +\sum_{j=1}^n(-1)^{nj}\,\varphi\Big(\delta\gamma^{j+1} (a_0)\, \alpha\s is_1.(\delta\gamma^{j+2} a_{1})\cdots \\[1mm] &&\qquad\cdots \alpha\s is_{n-j}.(\delta\gamma^{n+1} a_{n-j})\,\alpha\s is_{n-j+1}.(\un)\, \alpha\s is_{n-j+2}.(\gamma\delta\gamma^2 a_{n+1-j}) \cdots\alpha\s is_{n+1}.(\gamma\delta\gamma^{j+1} a_{n})\Big)\Big] \\[1mm] &&\hbox{now substitute $\varphi\to\varphi\circ\gamma^{j+1}$ in the last term, and use $\gamma\circ\delta=-\delta\circ\gamma$ and Lemma~\ref{CocLem}(iii):}\\[1mm] &&=i^{\epsilon_{n+1}}\int_{\sigma_{n+1}}ds_1\cdots ds_{n+1}\,\Big[(-1)^{n+1} \varphi\Big(\delta( a_0)\,\alpha\s is_1.(\delta\gamma a_1) \cdots\alpha\s is_{n}. (\delta\gamma^{n} a_n)\,\alpha\s is_{n+1}.(\un)\Big) \\[1mm] &&\quad +\sum_{j=1}^n(-1)^{nj}\,\varphi\Big((-1)^{j+1}\delta(a_0)\,(-1)^{j+1} \alpha\s is_1.(\delta\gamma a_{1})\cdots \\[1mm] &&\qquad\cdots(-1)^{j+1} \alpha\s is_{n-j}.(\delta\gamma^{n-j} a_{n-j})\,\alpha\s is_{n-j+1}.(\un)\, (-1)^{j}\alpha\s is_{n-j+2}.(\delta\gamma^j a_{n+1-j}) \cdots(-1)^j\alpha\s is_{n+1}.(\delta\gamma a_{n})\Big)\Big] \\[1mm] &&=i^{\epsilon_{n+1}}(-1)^{n+1}\int_{\sigma_{n+1}}ds_1\cdots ds_{n+1}\,\Big[ \varphi\Big(\delta( a_0)\,\alpha\s is_1.(\delta\gamma a_1) \cdots\alpha\s is_{n}. (\delta\gamma^{n} a_n)\,\alpha\s is_{n+1}.(\un)\Big) \\[1mm] &&\quad +\sum_{j=1}^n\varphi\Big(\delta(a_0)\, \alpha\s is_1.(\delta\gamma a_{1})\cdots \\[1mm] &&\qquad\cdots \alpha\s is_{n-j}.(\delta\gamma^{n-j} a_{n-j})\,\alpha\s is_{n-j+1}.(\un)\, \alpha\s is_{n-j+2}.(\delta\gamma^j a_{n+1-j}) \cdots\alpha\s is_{n+1}.(\delta\gamma a_{n})\Big)\Big]\,. \end{eqnarray*} Now recall that $\wt\tau:=(\tau_0,0,-\tau_2,0,\tau_4,\ldots)\in \al C.(\al D._c),$ and hence we may assume that $n$ is odd in the preceding expression (if $n$ is even, $B\wt\tau_{n+1}=0\,).$ Thus \begin{eqnarray*} &&\!\!\!\!\! \big(B{\tau}_{n+1}\big)(a_0,\ldots,\,a_n) =\int_{\sigma_{n+1}}ds_1\cdots ds_{n+1}\,\Big[ \sum_{j=0}^n\varphi\Big(\delta(a_0)\, \alpha\s is_1.(\delta\gamma a_{1})\cdots \\[1mm] &&\qquad\cdots \alpha\s is_{n-j}.(\delta\gamma^{n-j} a_{n-j})\,\alpha\s is_{n-j+1}.(\un)\, \alpha\s is_{n-j+2}.(\delta\gamma^{n+1-j} a_{n+1-j}) \cdots\alpha\s is_{n+1}.(\delta\gamma^n a_{n})\Big)\Big]\,. \end{eqnarray*} Since $\alpha\s is_{n-j+1}.(\un)=\un\,,$ we can do the integrals w.r.t. $s_{n-j+1},$ and so using $0\leq s_1\leq s_2\leq\cdots\leq s_{n+1}\leq 1$ and a relabelling of variables, we get \begin{eqnarray} &&\!\!\!\!\! \big(B{\tau}_{n+1}\big)(a_0,\ldots,\,a_n) =\int_{\sigma_{n}}ds_1\cdots ds_{n}\,\Big[ \sum_{j=1}^{n-1}(s\s n-j+1.-s\s n-j.)\,\varphi\Big(\delta(a_0)\, \alpha\s is_1.(\delta\gamma a_{1})\cdots \nonumber\\[1mm] &&\qquad\qquad\qquad\cdots \alpha\s is_{n-j}.(\delta\gamma^{n-j} a_{n-j})\, \alpha\s is_{n-j+1}.(\delta\gamma^{n+1-j} a_{n+1-j}) \cdots\alpha\s is_{n}.(\delta\gamma^n a_{n})\Big) \nonumber \\[1mm] &&\qquad\qquad\qquad\qquad\qquad +(s_1+1-s_n)\,\varphi\Big(\delta(a_0)\,\alpha\s is_1.(\delta\gamma a_{1})\cdots \alpha\s is_{n}.(\delta\gamma^n a_{n})\Big) \Big] \nonumber \\[1mm] \label{BtFinal} &&=\int_{\sigma_{n}}ds_1\cdots ds_{n}\, \varphi\Big(\delta(a_0)\,\alpha\s is_1.(\delta\gamma a_{1})\cdots \alpha\s is_{n}.(\delta\gamma^n a_{n})\Big)\,. \end{eqnarray} Next, we turn our attention to the left hand side of our desired Equation~(\ref{CocEq}). Observe first that we have \begin{eqnarray*} \tau_n\big(\gamma a_0,\ldots,\,\gamma a_n\big) &=&(-1)^n\tau_n\big( a_0,\ldots,\, a_n\big)\qquad\quad \hbox{because:} \\[1mm] \varphi\left(\gamma(a_0)\,\alpha\s is_1.(\delta\gamma(\gamma a_1))\cdots \alpha\s is_n.(\delta\gamma^n(\gamma a_1))\right) &=& (-1)^n\,\varphi\left(a_0\,\alpha\s is_1.(\delta\gamma a_1)\cdots \alpha\s is_n.(\delta\gamma^na_n)\right) \end{eqnarray*} since $\delta\circ\gamma=-\gamma\circ\delta,$ $\varphi\circ\gamma=\varphi$ and by Lemma~\ref{CocLem}(iii). Thus $\wt{\tau}\circ\gamma=\wt{\tau},$ and so we have $\wt{a}=\gamma a$ in definition~(\ref{Defb}). An application of definition~(\ref{Defb}) to the left hand side of Equation~(\ref{CocEq}) yields: \begin{eqnarray} &&\big(b\,\tau_{n-1}\big)(a_0,\ldots,\,a_n)= i^{\epsilon_{n-1}}\int_{\sigma_{n-1}}ds_1\cdots ds_{n-1}\Big[ \sum_{j=0}^{n-1}(-1)^j\varphi\Big(a_0\,\alpha\s is_1.(\delta\gamma a_1) \cdots \alpha\s is_j.(\delta\gamma^j(a_j\,a_{j+1}))\cdots \nonumber \\[1mm] \label{LHSCocEq} &&\qquad\qquad\cdots\alpha\s is_{n-1}.(\delta\gamma^{n-1}a_n)\Big) +(-1)^{n}\,\varphi\left((\gamma a_n)\,a_0\, \alpha\s is_1.(\delta\gamma a_1)\cdots \alpha\s is_{n-1}.(\delta\gamma^{n-1}a_{n-1}) \right)\Big]\,. \end{eqnarray} We examine the terms in this sum more closely: \begin{eqnarray*} \hbox{$j=0$:} && \int_{\sigma_{n-1}}ds_1\cdots ds_{n-1}\,\varphi\Big(a_0\,a_1\,\alpha\s is_1.(\delta\gamma a_2) \cdots \alpha\s is_{n-1}.(\delta\gamma^{n-1}a_n)\Big) \\[1mm] \hbox{$j=1$:} && -\int_{\sigma_{n-1}}ds_1\cdots ds_{n-1}\,\varphi\Big(a_0\,\alpha\s is_1.(\delta\gamma (a_1a_2))\,\alpha\s is_2.(\delta a_3)\cdots \alpha\s is_{n-1}.(\delta \gamma^{n-1}a_n)\Big) \\[1mm] &&=-\int_{\sigma_{n-1}}ds_1\cdots ds_{n-1}\,\varphi\Big(a_0\,\alpha\s is_1. \big((\delta\gamma(a_1)\,\gamma a_2+a_1\,\delta\gamma a_2\big)\,\alpha\s is_2. (\delta a_3)\cdots \alpha\s is_{n-1}.(\delta \gamma^{n-1}a_n)\Big) \\[1mm] &&=-\int_{\sigma_{n-1}}ds_1\cdots ds_{n-1}\,\Big[\varphi\Big(a_0\,\alpha\s is_1. \big((\delta\gamma(a_1)\,\gamma a_2\big)\,\alpha\s is_2. (\delta a_3)\cdots \alpha\s is_{n-1}.(\delta \gamma^{n-1}a_n)\Big) \\[1mm] &&\qquad\qquad +\; \varphi\Big(a_0\,a_1\,\alpha\s is_1. (\delta\gamma a_2)\,\alpha\s is_2. (\delta a_3)\cdots \alpha\s is_{n-1}.(\delta \gamma^{n-1}a_n)\Big)\Big] \\[1mm] &&-\int_{\sigma_{n}}ds_1\cdots ds_{n}\,{\partial\over\partial s_1} \,\varphi\Big(a_0\,\alpha\s is_1.(a_1)\,\alpha\s is_2.(\delta\gamma a_2)\, \alpha\s is_3.(\delta a_3)\cdots\alpha\s is_{n}.(\delta\gamma^{n+1}a_n)\Big) \end{eqnarray*} where we made use of Equation~(\ref{ByParts2}) in the last step. Notice that we get a cancellation between the middle term and the $j=0$ term in the sum. For $1