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\begin{document}

\title{Transport formula, Connes fusion with the vector representation
}
 \author{Speaker:  Andr\'{e} Henriques
  \\ 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!

Recall a formula from Scott's talk, of the form
$$\phi \circ \phi = \sum c \phi \circ \phi$$ with indices of the $\phi$s switched around.

Here's a finite-dimensional analogy:

We need 5 irreps of $SU(N)$, $A,B,C,U,V$.  $U$ and $V$ are miniscule, meaning $V_{[k]} = \wedge^k \C^N$ for some $k$.  This means $\Hom(- \otimes V,  -)$ is at most one dimensional (recall Pieri rule from earlier today).

Once and for all, fix a basis vector $\phi_{X,Y}^V$ of $\Hom(X \otimes V,  Y)$.

Consider 

\includegraphics[scale=.3]{Friday11-30amPicture1.jpg}

It is an element in $$\Hom(A \tensor V \tensor U, C) = \Hom(A \tensor U \tensor V, C) $$ (this uses the fact that the cat of $SU(N)$ reps has a symmetric braiding).  And $$ \Hom(A \tensor U \tensor V, C) = \DirectSum_{B'} \Hom(A \tensor U, B')\tensor \Hom (B' \tensor V, C)$$
Assume our index set contains only those $B'$ which make a non-zero contribution.

Say our original element is 
$\phi_{CB}^U \circ \phi_{BA}^V$; elements in the direct sum are of the form $\phi_{B' A}^V \tensor \phi_{C B'}^V$.  

If I'm given elements $v \in V$, $u \in U$, I have 
$$ \phi_{CB}^U(u) \phi_{BA}^V(v) = \sum_{B'} c_{BB'}  \phi_{CB'}^V(v) \phi_{B'A}^U(u)
$$
which is an equation in $\Hom(A,C)$.

We now do the same thing for primary fields.  We have $H_i, H_j, H_k$ and instead of $U$, $V$ we'll have $C^\infty$ functions to $U$ or $V$.

\includegraphics[scale=.3]{Friday11-30amPicture2.jpg}

Given $a \in C^\infty(S^1,U)$ and $b \in C^ \infty (S^1, V)$ with disjoint support (and disjoint from $\{ 1 \}$), we get $$\phi_{ik}^U(a) \phi_{kj}^V(b) = \sum_{k'} c_{kk'} \phi_{ik'}^V(b) \phi_{k' j}^U(a).$$

What's important is that we have complete control over which $c_{k k'}$ are and aren't zero, since we have a formula of the form
$$c_{k k'} = \frac{\prod \Gamma(\cdots )\prod \Gamma(\cdots )}{\prod \Gamma(\cdots )\prod \Gamma(\cdots )}$$

Now, if $a$ and $b$ were delta functions, this would be a solution of the KZ-equation.  As it is, it's a smeared solution to the KZ-equation.

Goal:  we want to understand $H_\Box \boxtimes H_f$; we expect it to be the direct sum over $g$ obtained by adding one box to $f$.  This isn't going to be quite enough; as $H_\Box$ doesn't generated the fusion ring, we also need to understand
$H_{[k]} \boxtimes H_f$.  This is done with similar techniques and is twice as technical; Wasserman does it, but we won't.

$H_\Box \boxtimes H_f$ is a completion of $\Hom_{\tilde{L_{I'}G}}(H_0, H_\Box) \tensor H_0 \tensor \Hom_{\tilde{L_{I}G}}(H_0, H_f)$.  This (minus the $H_0$) is what Yoh talked about.  It's also a completion of 
$$\Hom_{\tilde{L_{I'}G}}(H_0, H_\Box) \tensor H_f$$ -- this is the nonsymmetric version of Connes fusion.  This is where we're going to work.

Given $a \in L^2(I,V)$ ( $V=\C^N=V_\Box$))
, you get $a_{\Box 0} =  \phi_{\Box 0}^\Box(a)   \in \Hom_{\tilde{L_{I'}G}}(H_0,H_\Box)$

Given $h \in \tilde{L_I G}$ you get $\pi_\Box(h) \in \Hom_{\tilde{L_{I'} G}}(H_\Box, H_\Box)$

Elements of the form $x=\sum_n \pi_\Box(h^n) a_{\Box \Box}^n$ are dense in $Hom(H_n, H_\Box)$.

Goal:   We simplify our lives and look just at $x=a_{\Box 0}$, $y \in H_f$.  We want to compute the norm of $x \tensor y \in H_\Box \boxtimes H_f$. 

By definition, 
\begin{align*}
\norm{x \tensor y}^2 & = \ip{x^* x y, y} \\
& = \ip{\pi_f(x^* x)y,y}
\end{align*}

(Note this middle term doesn't really make sense, but we make sense of it by: )
$x^* x \in \Hom_{\cA(I')}(H_0, H_0) = \cA(I)$ by Haag duality.  This $\cA(I)$ acts everywhere

\begin{thm} Technical theorem (transport formula) [W]:

$\pi_f(a^*_{\Box 0} a_{\Box 0}) 
\sum_{g=f+\Box \text{ permissible}} \lambda_g a_{gf}^* a_{gf}
$
with all $\lambda_g >0$.  
\end{thm}

Showing that $\lambda_g \geq 0$ isn't bad; showing they're not $=0$ is the hard part.

\begin{proof}
Sketch:

$f=0$ is easy, equality on both sides.

What about $f=\Box$?  $\pi_\Box (a_{\Box 0}^* a_{\Box 0})$ takes $H_\Box$ back to itself.  Let's precompose now with a map from the trivial to the box rep:  $b_{\Box 0}$.

First, we rewrite it with $a^*$ and $a$ separated.  By the braiding we have the second and third equalities.  

\includegraphics[scale=.5]{Friday11-30amPicture3.jpg} and 

\includegraphics[scale=.3]{Friday11-30amPicture4.jpg}


This is true for all $b$.  {\color{blue} By denseness it follows more generality}.  Non-zeroness follows from non-zeroness of braiding things, which we already knew.
\end{proof}

Now back to our computation:

\begin{align*}
\norm{x \tensor y}^2 & = \ip{x^* x y, y} \\
& = \ip{\pi_f(x^* x)y,y} \\
& = \sum_{g=f+\Box} \lambda_g \ip{a_{gf}^* a_{gf} y, y} \\
& = \sum_{g=f+\Box}  \ip{ \sqrt{\lambda_g} a_{gf} y, \sqrt{\lambda_g} a_{gf}  y} \\
\end{align*}

\begin{conclusion}  $H_\Box \boxtimes H_f \rightarrow \DirectSum_{g=f+\Box} H_g$ given by $x \tensor y \mapsto \DirectSum \sqrt{\lambda_g}a_{gf}y$ is an isometry; it's also surjective, since $\lambda_g > 0$, so therefore is an isomorphism.  
\end{conclusion}


\end{document}
































