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\begin{document}

\title{Modularity of the category of representations of a conformal net (Part 1) 
}
 \author{Speaker:  Orit Davidovich
  \\ Typist: Owen Gwilliam
   }
 \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!

\begin{recall} A {\em conformal net} is a cosheaf of von Neumann algebras on $S^1$ that satisfies properties expected of the algebra of local observables which is covariant with respect to the $PSU(1,1)$-action.\footnote{Don't take this definition too seriously. It may not literally be a cosheaf, just a precosheaf, and we don't allow ourselves to evaluate on the whole circle.}\end{recall}

\begin{example} The vacuum conformal net associated to a loop group. \end{example}

A representation of a conformal net is a compatible family $\{ \pi_I \, : \, I \subset S^1\}$.

\begin{example} Every positive energy representation of the vacuum conformal net associated to a loop group. \end{example}

We will now make extra assumptions on our conformal nets:
\begin{enumerate}

\item separability;

\item split property;

\item strong additivity.

\end{enumerate}

For loop groups, (1) is obvious, and Wassermann proves (2) and (3) in his paper.

There are two equivalent forms of the split property:
\begin{itemize}

\item if $\bar{I} \cap \bar{J} = \emptyset$, then $\cA(I) \otimes \cA(J) \overset{\cong}{\rightarrow} \cA(I) \vee \cA(J) \subset B(H_0)$

\item if $\bar{I} \subset J$, then there exists a type I von Neumann algebra N such than $\cA(I) \subset N \subset \cA(J)$.

\end{itemize}

We want to consider separable representations (i.e., the Hilbert space is separable) and nondegenerate representations (for any interval $I$ and $x \in \cA(I)$, if $\pi(x) \zeta = 0 \Rightarrow \zeta =0$.

Remark: The identity in $\cA(I)$ may not go to the identity in a representation $\pi$. All we know is that it goes to a projection that commutes with $\pi(\cA(I))$.

\begin{claim}
$\pi$ is nondegenerate if and only if $\pi_I(1) = \id_\cH$.
\end{claim}

Let $Rep_S(\cA)$ denote the category of nondegenerate separable representations of $\cA$.

\begin{claim} 
If the $\mu_2$-index is finite, then $Rep_S(\cA)$ is a modular tensor category!
\end{claim}

{\bf Goal:} Show $Rep_S(\cA)$ is semisimple. That is it has finitely many irreducible representations and every representation is completely reducible.

We'll assume the finite index condition from hereon.

\begin{defn} 
A representation $\pi$ of $\cA$ is {\em localized} at $I \subset S^1$ if
\begin{itemize}

\item $\pi$ is defined on $\cH_0$ (always the vacuum Hilbert space -- the defining representation of $\cA$)

\item $\pi_{I'} = \id$

\end{itemize}
\end{defn}

\begin{proposition} 
For $\pi \in Rep_S(\cA)$, $\pi$ is unitarily equivalent to a representation localized in any interval $I \subset S^1$.
\end{proposition}

\begin{proof} 
We'll need the following facts about Type $III_1$ factors:
\begin{enumerate}

\item they are simple as algebras;

\item every representation on a separable Hilbert space is also continuous with respect to the strong operator topology;

\item hence $\cA(I) \hookrightarrow \pi_I(\cA(I))$, and

\item $\pi_1(\cA(I)$ is a von Neumann algebra so $\cA(I) \cong \pi_I(\cA(I))$.

\end{enumerate}

Yoh told us that any two representations of a type $III$ factor are unitarily equivalent. Thus there exists a unitary map $u \in Hom(\cH_0, \cH)$ such that $u \pi_{I_0'}(x) = xu$ for all $x \in \cA(I_0')$.

Define $\rho_I (x) = u \pi_I(x) u^*$ for all $x \in \cA(I)$. By construction $\rho$ is localized at $I_0$ and is equivalent to $\pi$.
\end{proof}


\begin{note} $\rho$ localized at $I_0$ insures that $\rho(\cA(I)) \subset \cA(I)$ for every $I \supset I_0$. \end{note}

\begin{proof} 
Haag duality says $\cA(I) = \cA(I')'$. Thus $\rho_I(\cA(I))$ commutes with $\cA(I') = \rho_{I'} (\cA(I'))$. Use locality in $\cA$ and regularity. 
\end{proof}

Dimension: $\pi$ an irreducible in $Rep_S(\cA)$. Inclusion of type $III_1$ factors implies $\pi(\cA(I)) \subset \pi(\cA(I'))'$. The {\em in dex} $[\pi(\cA(I'))': \pi(\cA(I))] \in [1, \infty]$.

\begin{example}
$M$ a factor then $M \hookrightarrow Mat_k(M)$ by diagonal matrices, and $[Mat_k(M): M] = k^2$.
\end{example}

\begin{proposition}
For $\pi$ an irrep, the index does not depend on the interval $I$.
\end{proposition}

Heuristic idea behind proof: the index should only depend on the equivalence class, so since $\pi \cong \rho$ localized on any interval $I$, then the index should not depend on the interval $I$.

\begin{remark} Corbett assumed that representations were conformally invariant, but if this index is finite, then the representation is conformally invariant (though we might not know this {\em a priori}.
\end{remark}

\begin{defn} 
For an irrep $\pi$, the {\em statistical dimension} $d(\pi)$ is the square root of the index.
\end{defn}

\begin{thm}[Longo] 
The statistical dimension and the quantum dimension agree.
\end{thm}

Thus the index is the square of the quantum dimension, so if you prefer the categorical notion of quantum dimension, you can stick to that.

\begin{remark} if the index equals one, then a representation is invertible. That is, you can find another representation that tensors with it to give the unit.
\end{remark}

Now let's consider the inclusion of two disjoint intervals $I$ and $J$. Let $E = I \cup J$. Then $\cA(E) = \cA(I) \vee \cA(J) \subset B(\cH_0)$. We have inclusion of type $III_1$ factors $\cA(E) \subset \cA(E')'$.

\begin{proposition} The index $[ \cA(E')' : \cA(E)]$ does not depend on $E$. \end{proposition}

\begin{definition} $\mu_2(\cA)$ is this index.\end{definition}

\begin{claim} Given the extra assumptions on our conformal net $\cA$, if $\mu_2 (\cA) < \infty$, then $\mu_2(\cA) $ gives an upper bound on the number of isomorphism classes of irreducible representations in $Rep_S(\cA)$.\end{claim}

\begin{remark} $\mu_2 = \sum_i d(\pi_i)^2$ where $i$ runs over isomorphism classes of irreps. (Note that $d(\pi_i) \geq 1$.) \end{remark}

{\bf Complete reducibility:}

We want to make the universal enveloping $C^*$ algebra of our cosheaf. There is a subalgebra $C^*_S(\cA)$ related to separability.

If $\pi$ is a separable nondegenerate rep of $C^*(\pi)$ on $\cH$, then we can decompose $\cH = \int_X^\oplus \cH_x \, d\nu(x)$ and $\pi = \int_X^\oplus \pi_X\, d\nu(x)$.

Let $R \subset \pi(C^*(\cA))''$ act diagonally on $\cH$. There are two important possibilities: $R$ is the center, and $R$ is a maximal abelian

Here's the strategy to disintegrate the representation:

\begin{enumerate}

\item use the central decomposition

\item show all the $\pi_x$ are type $I$

\item the $\pi_x$ are multiples of an irreducible

\item outside the null set, $x \neq x'$ then $\pi_x$ and $\pi_{x'}$ are not multiples of the same irrep

\item there are only finitely many irreps so $|X|$ is finite and thus our representation decomposes into a sum

\end{enumerate}

\end{document}
































