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\begin{document}

\title{Connes fusion
}
 \author{Speaker:  Yoh Tanimoto \\ Typist: Emily Peters
 }
 \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!

The plan is to relate Connes fusion and endomorphisms.

In this talk, $M$ is always a type {III} factor.  For our purposes, it suffices to have the following property:
\begin{fact}
Any representation of $M$ on a separable Hilbert space, is implemented by a unitary operator.  Another way of saying this is that any two representations are equivalent.
\end{fact}

\begin{defn}
An $(M,M)$ bimodule is a Hilbert space $X$ with commuting actions of $M$ and $M^{op}$.
\end{defn}

\begin{defn}
An endomorphism of $M$ is a unital *-homomorphism of $M$ into $M$.
\end{defn}

\begin{example}
$L^2(M)$ is a trivial bimodule.  For $x, y \in M$ and $\xi \in L^2(M)$, $x \cdot \xi \cdot y$.
\end{example}

\begin{example}
If $\rho$ is an endomorphism of $M$, then it also acts on $L^2(M)$.  We define $\rho(x) \cdot \xi \cdot y=\rho(x)JY^*J\xi$.  Call this bimodule $X_\rho$.
\end{example}

\begin{prop}
Any bimodule is unitarily equivalent to some $X_\rho$.
\end{prop}

\begin{proof}
From the first fact, as representation of $M^{op}$, $X$ and $L^2(M)$ are equivalent.  We may assume that $X=L^2(M)$ as $M^{op}$-modules.  The action of $M$ commutes with $M^{op}$.  $(M^{op})'=M$; the image of $M$ is $M$.
\end{proof}

\begin{prop}
$X_{\rho_1} \simeq X_{\rho_2}$ iff there is a $u \in \cU(M)$ such that $u \rho_1(x)u^*=\rho_2(x)$.
\end{prop}

\begin{proof}
 in one direction, $u$ commutes with $M^{op}$.  In the other, let $u$ implement the equivalence.  $u$ must commute with $M^{op}=M'$, so $u \in M$.
\end{proof}

\begin{tabular}{c|c|c|c}
 & direct sum & subobject & \\
 \hline
 bimodule & $X\directSum Y$ & invariant subspace & fusion \\
 \hline
 endomorphisms & $P_1 \perp P_2$, $P_1+P_2=I$,  & $P \in M$, $[P,\rho(x)]=0$,  & composition \\
 & $v_I: P_i \simeq I$:   & $V:P \simeq I$.  $V\rho(x)V^*$. & $\rho_2 \circ \rho_1$.  \\
 & $V_1 \rho_1(x)V_1^* + V_2 \rho_2(x)V_2^*$ & & \\
\end{tabular}

Let $X$, $Y$ be bimodules.  $\cX=\Hom(L^2(M)_M, X_M)$ and $\cY=\Hom({}_M L^2(M), {}_M Y)$.

We consider $\cX \tensor \cY$ with an inner product,
$\langle x_1 \tensor y_1, x_2 \tensor y_2 \rangle = \langle x_2^*x_1 y_2^* y_1 \Omega, \Omega \rangle
$
Here we use $x_2^* x_1 \in M$ and $y_2^* y_1 \in M^{op}$.  (this is because $x \in \Hom(L^2(M)_M, X_M)$ and $x^* \in \Hom (X_M,L^2(M)_M)$ gives us $x^*x \in \Hom(L^2(M)_M, L^2(M)_M)$ ie $x^* x \in M$.)

\begin{lemma} The form thus defined on $\cX \otimes \cY$ is an inner product
\end{lemma}

\begin{proof}
Show positive definiteness.

Let $z=\sum_i x_i \otimes y_i$; then $\langle z, z \rangle = \sum_{i,j} \langle x_i^* x_j y_i^* y_j \Omega, \Omega \rangle$

Now $x = (x_i^* x_j) \in M_n(M)$; rewrite it as 
$$x=
\left(
\begin{array}{c}
x_1^* \\
x_2^* \\
\vdots \\
x_n^*
\end{array}
\right)
\cdot
\left(
\begin{array}{cccc}
x_1 & x_2 & \ldots & x_n
\end{array}
\right)$$
We can write $x=a^* a$ where $a \in M_n(M)$.  Similarly for $y$, $y=b^*b$.

So, $\langle z, z \rangle = \sum_{i,j} \langle x_i^* x_j y_i^* y_j \Omega, \Omega \rangle 
= \sum_{i,j} \sum_{p,q} \langle
a_{pi}^*a_{pj} b_{qi}^* b_{qj} \Omega, \Omega \rangle$

Now by orthogonality, all of these commute and so
$
\sum_{i,j} \sum_{p,q} \langle
a_{pi}^*a_{pj} b_{qi}^* b_{qj} \Omega, \Omega \rangle
=
\sum_{p,q} \sum_{i,j} \langle
a_{pj}  b_{qj} \Omega, a_{qi} b_{qi}\Omega \rangle
=\sum_{p,q} \norm{\sum_j a_{pj} b_{qj} \Omega}^2 \geq 0
$
\end{proof}

We define on $\cX \tensor \cY$ actions of $M$, $M^{op}$ by $a, b \in M$ by $a \cdot x \tensor y \cdot b=ax \tensor J b^* J y$

\begin{prop}
These actions are well-defined.
\end{prop}

\begin{defn}
call the completion of $\cX \tensor \cY$ the {\em fusion} of $X$ and $Y$, $X \boxtimes Y$.
\end{defn}

\begin{theorem}
Let $\rho_1$, $\rho_2$ be endomorphisms of $M$.  Then $X_{\rho_1} \boxtimes X_{\rho_2} \simeq X_{\rho_2 \circ \rho_1}$.
\end{theorem}

\begin{proof}
The operator 
$$ V: x \tensor y \mapsto \rho_2(x) y \Omega
$$
 is a unitary.  Remains to show that it's an intertwiner:
 \begin{align*}
 V \cdot a \cdot x\tensor y \cdot b \\
 = V \rho_1(a) x \tensor Jb^*Jy  \\
 = \rho_2(\rho_1(a)x)JB^*Jy\Omega \\
 = \rho_2 \rho_1(a) \rho_2(x) JB^*Jy \Omega \\
 =\rho_2 \rho_1(a) J b^* J \rho_2(x)y \Omega \\
 =\rho_2 \rho_1(x) Jb^*JV x\tensor y
 \end{align*}
 
\end{proof}

\begin{corollary}
$X_{\rho_1} \boxtimes (X_{\rho_2} \boxtimes X_{\rho_3}) \simeq X_{\rho_3 \rho_2 \rho_1} \simeq (X_{\rho_1} \boxtimes X_{\rho_2}) \boxtimes X_{\rho_3}$
\end{corollary}

\end{document}