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\begin{document}

\title{Tomita-Takesaki Theory and the KMS condition
}
 \author{Speaker: Michael Hartglass (UC Berkeley)
 \\ Typist: James Tener
 }
 \date{\today}
 \thanks{Available online at \texttt{http://math.mit.edu/$\sim$eep/CFTworkshop}.  Please email \texttt{eep@math.mit.edu} with corrections and improvements!}

\maketitle

\begin{abstract}
Notes from the ``Conformal Field Theory and Operator Algebras workshop," August 2010, Oregon.
\end{abstract}

%%% Start typing here!

Fixing notation: $\phi$ is a faithful, normal state on a von Neumann algebra $M$.  What does that mean?

A state is a continuous linear functional $\phi:M \to \C$ such that $\phi(\id) = 1 = \norm{\phi}$.  Equivalently, $\phi(x^*x) \ge 0$ for all $x \in M$.

Faithful means that $\phi(x^*x) = 0$ if and only if $x = 0$.

Normal means that the state is continuous when $M$ is given the ultraweak topology.  That is, the topology wherre convergence is givenn by $x_\lambda \to x$ if 
\[
\sum_{i=1}^\infty \ip{x_\lambda \xi_i, \eta_i} \to \sum_{i=1}^\infty \ip{x \xi_i, \eta_i}
\]
whenever 
\[
\sum \norm{\xi_i}^2 + \norm{\eta_i}^2 < \infty.
\]

Equivalently, whenever $x_\lambda$ is an increasing net in $M$ that converges to $x$, we have $\phi(x_\lambda) \to \phi(x)$.

Put inner product on $M$ by $\ip{x,y} = \phi(y^*x)$.  This gives us a new norm, $\norm{\cdot}_2$ on $M$.  Complete $M$ with respect to this norm to get the Hilbert space $L^2(M)$.  Let $\Omega$ denote the image of $\id_M$ in $L^2(M)$.  This is called the ``vacuum vector.''

We let $M$ act on $L^2(M)$, densely defined by left multiplication.  For this action, $\Omega$ is a cyclic, separating vector for this action.  Cyclic means $M\Omega$ is dense in $L^2(M)$, and separating means $x\Omega = 0 \implies x = 0$ for all $x \in M$.

We now look at the map $S_0:M\Omega \to M\Omega$ given by $S_0(x\Omega) = x^*\Omega$.  In generral, this map is unbounded, and so cannot be extended to $L^2(M)$.  However, it is as nicely behaved as an unnbounded operator can be.  We will be especially interested in its polar decomposition.  

Can also define $F_0:M^\prime\Omega \to M^\prime\Omega$ by $F_0(x^\prime\Omega) = (x^\prime)^*\Omega$ for $x^\prime \in M^\prime$.  Recall: $M^\prime$ is the commutant of $M$, as it acts on $L^2(M)$.  i.e. $M^\prime = \{x \in B(H) : xy = yx$ for all $y \in M\}$.


Notation: if $A$ and $B$ are unbounded operators, we write $A \subset B$ if $\operatorname{dom} A \subset \operatorname{dom} B$ and $B$ agrees with $A$ when restrict to $\operatorname{dom} A$.

If $A$ is unbounded, the domain of $A^*$ is $\{\eta : \ip{A\zeta, \eta}$ is a bounded function of $\zeta\}$.  Then we have $\ip{\zeta,A^*\eta} = \ip{A\zeta,\eta}$ where the expression is defined.

\begin{fact} $S_0 \subset F_0^*$ and $F_0 \subset S_0^*$.\end{fact}
\begin{proof}[Proof of fact]Let $a \in M$, $a^\prime \in M^\prime$.  Then $\ip{S_0(a\Omega),a^\prime\Omega} = \ip{a^*\Omega,a^\prime\Omega} = \ip{(a^\prime)^*\Omega, a\Omega}$.  This is clearly a bounded functino of $a\Omega$, which implies $\operatorname{dom} S_0 \subset  \operatorname{dom} F_0^*$.  The other inclusion is proved similarly.\end{proof}

Since $\Omega$ is cyclic and separating for $M$, it is also cyclic and separating for $M^\prime$, so $S_0$ and $F_0$ are both densely defined.  By a standard result in operator theory, both $S_0$ and $F_0$ are closable.  That is, when we take the closure of the graph of $S_0$ or $F_0$, it remains the graph of a linear operator.  These new operators have the continuity property that if $x_n \to x$ and $S_0(x_n)$ converges, then $S_0(x_n) \to S_0(x)$.

\begin{theorem}
Let $S$ and $F$ be the closures of $S_0$ and $F_0$, respectively.  Then $S = F^*$ and $F = S^*$.
\end{theorem}

Since $S$ and $F$ are losed and densely defined, so we have a polar decomposition $S = J\Delta^{\frac12}.$
Recall that $S$ is conjugate linear, so $J$ is conjugate linear.  In fact, $J$ is an isometry.  In general, $\Delta$ will be unbounded.

\begin{claim}
$J\Delta^\frac12J = \Delta^{-\frac12}$
\end{claim}
\begin{proof}
$S=S^{-1}$, so $S^{-1} = (J\Delta^{\frac12})^{-1} = \Delta^{-\frac12}J$.   Rearranging proves the claim.
\end{proof}

We now want to show $JMJ = M^\prime$.  We begin with some preliminary results.

\begin{lemma}
With $\phi$ as before, and $\psi \in M_*$ such that $\abs{\psi(y^*x)}^2 \le \phi(x^*x)\phi(y^*y)$.  Then $S = F^*$ and $F = S^*$ and given $\lambda > 0$ then there exists $a\in M$ with $\norm{a} < \frac12$ such that $\psi(x) = \lambda\phi(ax) + \lambda^{-1} \phi(xa)$.
\end{lemma}

This is a sort of non-commutative Radon-Nikodym derivative.

\begin{lemma}
Let $\lambda$ be as before.  Given $a^\prime \in M^\prime$, theere is an $a \in M$ such that $a\Omega \in \operatorname{dom}(F)$ and $a^\prime \Omega = (\lambda S + \lambda^{-1}F)a\Omega$.
\end{lemma}
\begin{proof}[Sketch of proof]
From basic facts about the GNS construction, $\phi(x) = \ip{x\Omega,\Omega}$.  We then have
\[
\ip{x\Omega,a^\prime\Omega} = \lambda\ip{ax\Omega,\Omega} + \lambda^{-1}\ip{xa\Omega,\Omega}
\]
assuming $\norm{a} < 1$.  Once we check that everything is in the right domain, we get that 
\[
\lambda\ip{ax\Omega,\Omega} + \lambda^{-1}\ip{xa\Omega,\Omega} = \lambda \ip{x \Omega, Sa \Omega} + \lambda^{-1} \ip{a\Omega, S(x\Omega)}.
\]
The last expression is $\lambda^{-1}\ip{x\Omega,F(a\Omega)}$.
\end{proof}

\begin{lemma}
Let $\lambda$, $a$ and $a^\prime$ be as before.  Then if $xi, \eta \in \operatorname{dom}(F) \cap \operatorname{dom}(S)$, we have $\lambda\ip{SaS\xi,\eta} + \lambda^{-1}\ip{FaF\xi,\eta} = \ip{a^\prime\xi,\eta}$.
\end{lemma}
\begin{proof}
\[
F(0) = \int_{-\infty}^\infty \frac{F(it + \frac12) + F(it - \frac12)}{2\cosh(\pi t)} dt
\]
provided $f$ is bounded, holomorphic on the strip $-\frac12 \le \operatorname{Re}(z) \le \frac12$.
\end{proof}

\begin{proposition}
Let $\lambda$, $a$ and $a^\prime$ be as before.  Then
\[
a = \int_{-\infty}^\infty \lambda^{2it} \frac{\Delta^{it} J a^\prime J \Delta^{-it}}{\cosh(\pi t)} dt
\]
\end{proposition}
\begin{proof}[Proof idea]
$\lambda^{2it} etc.$
extend holomorphically, apply prev lemma
\end{proof}

\begin{theorem}
$JMJ = M^\prime$ and $\Delta^{it} M \Delta^{-it} M$ for all $t \in \R$.
\end{theorem}
\begin{proof}[Proof idea]
Take a unitary $u \in M^\prime$.  Then $a = u^*au$.  Pull $u$ under the integral given above, and note that the Fourier transforms of $u^*\Delta^{it}Ja^\prime J\Delta^{-it}u$ and $\Delta^{it} Ja^\prime J \Delta^{-it}$ are equal, so the operators are equal.  Plugging in $t = 0$ impliess $JM^\prime J \subseteq M$.  By symmetry $JMJ \subseteq M^\prime$, which proves the theorem.  We used the fact that an operator commuting with every unitary in $M^\prime$ must commute with everything in $M^\prime$, and is therefore in $M$ (by the double commutant theorem).
\end{proof}

\begin{example}
$M = M_2(\C)$, and $\phi(x) = \operatorname{tr}(ax)$ where $a = \begin{bmatrix}\mu_1 & 0\\ 0 &\mu_2\end{bmatrix}$ for $\mu_i > 0$ and $\mu_1 + \mu_2 = 1$.  In this example, $J(e_{ij}) = \sqrt{\frac{\mu_i}{\mu_j}} e_{ji}$ (extended by conjugate linearity) and $\Delta(e_{ij}) = \frac{\mu_i}{\mu_j} e_{ij}$ (extended linearly).
\end{example}

\begin{example}
Here we show an example where $S$ is unbounded. Consider 
\[
M = \bigotimes_{i=1}^\infty M_2(\C).
\]
Our state is $\phi = \bigotimes_{i=1}^\infty \phi_i$, where $\phi_i(x) = \operatorname{tr}(a_ix)$ as above with $\mu_{1,i} \to 0$ as $i \to \infty$ (and thus $\mu_{2,i} \to 1$).
\end{example}

Back to the $M_2(\C)$ example.  Fix $x,y \in M_2(\C)$.  Define $f(z)$ on the strip $0 \le \operatorname{Im}z \le 1$ by $f(z) = \ip{\Delta^{-iz}y\Omega,x\Omega}$.  Then, for $t \in \R$, $f(t) = \phi(\sigma_t^\phi(x)y)$ and $f(t+i) = \phi(y\sigma_t^\phi(x))$. Here, $\sigma_t^\phi(x) := \Delta^{it}x\Delta{-it}$.  

\begin{theorem}[KMS condition]
Define $f(z) = \ip{\Delta^{-iz}x\Omega,y\Omega}$ for $x,y \in M$ fixed.  Then $f(t) = \phi(\sigma_t^\phi(x)y)$ and $f(t+i) = \phi(y\sigma_t^\phi(x))$.  If $\alpha_t$ is a strongly continuous 1 parameter group of automorphisms of $M$ satisfying $\phi \circ \alpha_t = \phi$ and there exists a function $G$, holomorphic in the strip, such that $G(t + i) = \phi(y\sigma_t^\phi(x))$ and $G(t) = \phi(\sigma_t^\phi(x)y)$, then $\alpha_t = \sigma_t^\phi$ for all $t \in \R$.
\end{theorem} 

\begin{corollary}
The following are equivalent:
\begin{enumerate}
\item $\phi(ax) = \phi(xa)$ for all $x \in M$
\item $\sigma_t^\phi(a) = a$ for all $t \in \R$.
\end{enumerate}
\end{corollary}
\begin{proof}
(1) $\implies$ (2).  If $x \in M$, then 
\[
\ip{x^*\Omega,a\Omega} = \ip{\Omega,xa\Omega} = \ip{\Omega,ax\Omega} = \ip{a^*\Omega,x\Omega} = \ip{S(a\Omega),x\Omega}.
\]
This implies $a\Omega \in \operatorname{dom}S^*$ and $S^*(a\Omega) = a^*\Omega$.  Sine $S^* = \Delta^\frac12 J$, $a^*\Omega$ has to be fixed by $\Delta$.  Hence $a\Omega$ is fixed by $\Delta$ and thus $\Delta^{it}$.\\

(2) $\implies$ (1).  We have $\phi(\sigma_t^\phi(x)a) = \phi(\sigma_t^\phi(xa)) = \phi(xa)$.  This implies $f(z)$ is constant along the real axis, which implies that it is constant everywhere in the strip $0 \le \operatorname{Im}(z) \le 1$.  Plugging in $t=0$ we get $\phi(xa) = \phi(ax)$.
\end{proof}


\end{document}