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\begin{document}

\title{Closing remarks
}
 \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!

diagrams for bimodules.  Shaded corresponds to $M$, unshaded to $N$.  

unit, multiplication:

\includegraphics[scale=.4]{Saturday11-30amPicture1.jpg}

Advantage:  use pictures to remember conditions.  Disadvantage:  we are secretly using other conditions, like finite index. For example, dualizability (defined pictorally below) relies on finite index:

\includegraphics[scale=.2]{Saturday11-30amPicture2.jpg}

bimodules correspond to codimension 1pictures; either points on lines or lines in the plane.  morphisms between bimodules have codimension 2 -- points in the plane.

For example, here's a picture of an $M, N$ bimodule fused wtih an $N,P$ bimodule mapping to an $M, P$ bimodule.  

\includegraphics[scale=.3]{Saturday11-30amPicture3.jpg}

3-category:

objects are codim-0 \\
1-morphisms are codimention 1 \\
2-morphisms are codimension 2 \\
3-morphisms are codimension 3 \\

Some examples of graphical notation:

\begin{itemize}
\item
\begin{tikzpicture}[baseline=0]
	\draw (0,0) circle (1cm);
\end{tikzpicture}
 Corresponds to the vacuum hilbert space

\item
\begin{tikzpicture}[baseline=0]
	\draw (0,0) circle (1cm);
	\filldraw (0,0) circle (.3mm);
	\node (0,0) [above] {$\lambda$};
\end{tikzpicture}
The hilbert space $H_\lambda$

\item
\begin{tikzpicture}[baseline=0]
	\draw (0,0) circle (1cm);
	\filldraw (-.3,0) circle (.3mm);
	\node at (-.3,0) [above] {$\lambda$};
	\filldraw (.3,0) circle (.3mm);
	\node at (.3,0) [above] {$\mu$};
\end{tikzpicture}
The hilbert space $H_\lambda \boxtimes H_\mu$.

\item The braiding:
\includegraphics[scale=.3]{Saturday11-30amPicture4.jpg}
%%%  Full braiding

\item Writing an annulus as the tensor (over a two-interval algebra) of two disks.
$$
\begin{tikzpicture}[baseline=0]
	\filldraw [fill=gray] (0,0) circle (1cm);
	\filldraw [fill=white] (0,0) circle (5mm);
\end{tikzpicture}
=
\begin{tikzpicture}[baseline=0]
	\filldraw[fill=gray] (90:5mm)--(90:1cm) arc (90:-90:1cm) -- (-90:5mm) arc (-90:90:5mm);
	\filldraw[fill=gray, xshift=-1cm] (270:5mm)--(270:1cm) arc (270:90:1cm) -- (90:5mm) arc (90:270:5mm);
	\node at (-.5,0) {$\boxtimes$};
	\draw (-.5,-1)--(-.5,-.7);
	\draw (-.5,-.5)--(-.5,-.2);
\end{tikzpicture}
$$
\end{itemize}

\begin{lemma} Neck cutting:
$$
\begin{tikzpicture}[baseline=0]
	\filldraw[fill=gray] (1,1) arc (0:-180:10mm and 5mm) -- (-1,-1) arc (-180:0:10mm and 5mm) -- (1,1);
	\draw (1,1) arc (0:180:10mm and 5mm);
	\draw[dashed] (1,-1) arc (0:180:10mm and 5mm);
\end{tikzpicture}
= \sum_{\lambda} \quad
\begin{tikzpicture}[baseline=0]
	\draw (0,1) ellipse (10mm and 5mm);
	\filldraw (0,1) circle (.5mm) node [right] {$\lambda$};
	\draw (0,-1) ellipse (10mm and 5mm);
	\filldraw (0,-1) circle (.5mm) node [right] {$\bar{\lambda}$};	
\end{tikzpicture}
$$
\end{lemma}

\begin{proof}

Finite $\mu$-index implies 

\includegraphics[scale=.3]{Saturday11-30amPicture5.jpg}

 has finite dimension as a 
$
\left( \begin{tikzpicture}[baseline=0,scale=.5]
	\draw (0,1) arc (90:270:4mm);
	\draw (0,-.2) arc (90:270:4mm);
\end{tikzpicture}
\right),
\left( \begin{tikzpicture}[baseline=0,scale=.5]
	\draw (0,1) arc (90:-90:4mm);
	\draw (0,-.2) arc (90:-90:4mm);
\end{tikzpicture}
\right)
$ 
bimodule.

we must compute the dimension of some $\Hom$ spaces.  We use  frobenius reciprocity (and secretly dualizability) and see

\includegraphics[scale=.4]{Saturday11-30amPicture6.jpg}

\end{proof}

A diagrammatic argument for modularity:

\begin{proof}[Proof that $\Rep_f(\cA)$ is modular:]

Let $k$ be a transparent object, ie an object such that the positive and negative braidings are equal for all $\lambda$,
ie

\includegraphics[scale=.3]{Saturday11-30amPicture7.jpg}

(note we can rewrite this by saying the full twist is the identity.)
Think of $-z$ direction as time.  As a movie, this is just $k$ moving in a full circle once around $\lambda$.  

As a consequence, we get 

\includegraphics[scale=.3]{Saturday11-30amPicture8.jpg}

This can be rewritten using the neck cutting lemma; then we flatten this and get $k$ taking a trip around this hole.

\includegraphics[scale=.3]{Saturday11-30amPicture9.jpg}


Now let's redraw this in 3D again:  

\includegraphics[scale=.3]{Saturday11-30amPicture10.jpg}

or

\includegraphics[scale=.4]{Saturday11-30amPicture11.jpg}

(1)=(2) is an equality of maps from 
\begin{tikzpicture}
	\filldraw [fill=gray] (0,0) circle (1cm);
	\filldraw [fill=white] (0,0) circle (5mm);
	\filldraw (0,.75) circle (.5mm);
\end{tikzpicture}
to
\begin{tikzpicture}
	\filldraw [fill=gray] (0,0) circle (1cm);
	\filldraw [fill=white] (0,0) circle (5mm);
	\filldraw (0,-.75) circle (.5mm);
\end{tikzpicture}

(1) is equivariants w.r.t.~
\begin{tikzpicture}
	\filldraw [fill=gray] (0,0) circle (1cm);
	\filldraw [fill=white] (0,0) circle (5mm);
	\draw[ ultra thick] (160:1cm) arc (160:200:1cm);
	\draw[ ultra thick] (160:5mm) arc (160:200:5mm);	
\end{tikzpicture}
because \\
\includegraphics[scale=.3]{Saturday11-30amPicture12.jpg}

(2) is equivariant w.r.t.~\begin{tikzpicture}
	\filldraw [fill=gray] (0,0) circle (1cm);
	\filldraw [fill=white] (0,0) circle (5mm);
	\draw[ ultra thick] (20:1cm) arc (20:-20:1cm);
	\draw[ ultra thick] (20:5mm) arc (20:-20:5mm);	
\end{tikzpicture}
for the same reason.

The map for (1)=(2) can therefore be fused with 
\begin{tikzpicture}
	\filldraw [fill=gray] (20:5mm) arc (20:-20:5mm) -- (-20:10mm) arc (-20:20:10mm) -- (20:5mm);
	\filldraw [fill=gray] (160:5mm) arc (160:200:5mm) -- (200:10mm) arc (200:160:10mm) -- (160:5mm);	
\end{tikzpicture}
to get a map 

\includegraphics[scale=.5]{Saturday11-30amPicture13.jpg}

This looks like a torus, but is really two disks with tubes connecting them.  (Audience member:  shouldn't the tubes be going under the disks, in that second picture?  Andr\'{e}:  yes, but I don't want to redraw them!)

The map $\phi_1$ respects $\DirectSum_{\lambda}$; $\phi_2$ respects $\DirectSum_{\mu}$.

Therefore $\phi=\phi_1=\phi_2$ restricts to a map

\includegraphics[scale=.3]{Saturday11-30amPicture14.jpg}

For this map to be equivariant w.r.t.~the action of the algebras, we need $k=0$.

\end{proof}


\end{document}
































