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\begin{document}

\title[Modularity of the rep.~cat.~of a conformal net, II]{Modularity of the category of representation of a conformal net, II
}
 \author{Speaker:  Marcel Bischoff
  \\ 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!

Outline:
\begin{enumerate}
\item Introduction
\item Two interval inclusions
\item Modularity
\end{enumerate}

\begin{goal}
Let $\cA$ be a completely rational conformal net.  Orit showed the first few of these:
\begin{enumerate}
\item {\bf Semisimplicity:}  Every seperable non-degenerate rep is completely reducible.
\item  The number of unitary equiv. classes of irreducible reps is finite
\item  {\bf Finite statistics:} Every separable irreducible representation has finite statistical dimension
\item  {\bf Modularity:}  $\Rep_f(\cA)$ has a monoid structure with simple unit and duals (conjugates) and a maximally non-degenerate braiding, thus is modular.
\end{enumerate}
\end{goal}

\section{Introduction}
Assume
$\cA$ is a completely rational conformal net, i.e.
$$ \cI \ni I \longmapsto \cA(I) \subset B(H_0)$$
with $H_0$ the vacuum Hilbert space, $\Omega \in H_0$ the vacuum vector, $U \acton H_0$ unitary 
positive energy representation of 
$PSU(1,1)$. These data fullfil some axioms (Corbett) plus the  additional assumption
of \emph{complete rationality}:
 \begin{enumerate}
 \item strong additivity
 \item split property
 \item finite $\mu_2$ index
 \end{enumerate}

Recall a representation of $\cA$ is a collection of reps $\{ \pi_I\}_{I\in\cI}$ with $\pi_I: \cA(I) \rightarrow B(H)$
which are compatible. If $H$ separable (then we call $\pi$ a seperable representation), 
for all $I \in \cI$ there is $\rho \simeq \pi$ (we also write $\rho\in[\pi]$;
the equivalence class  $[\pi]$ is called sector) on $H_0$ with 
$\rho_{I'}=id_{A(I')}$. Thus the representation acts trivial outside $I$. $\rho$ then is called {\em localized in $I$}.
One has a monoidal structure, given by composition of localized endomorphism (Yoh showed relation to Connes fusion).


Conjugates:  Let $\pi \simeq \rho$ be a separable non-degenerate representation localized in $I$.  
Let $P$, $Q$ be two other intervals.  

Let $r_Q\in PSU_\pm(1,1)$ reflection associated to the intervall $Q$, cf:

\includegraphics[scale=.3]{Saturday9amPicture1.jpg}

Then we can define another representation by
$$\bar{\rho_I}(x)= J_P \rho_{r_Q (I)}(J_Q x J_Q) J_P$$
where $J_P$ is the modular conjugation for the algbra $\cA(P)$. i.e.
 $J_P \cA(P) J_P=\cA(P)'$. Remember that we have Bisognano-Wichman property, telling us that
$J_P x J_P=U(r_P) x U(r_P)^*$ holds, where $U$ is now the extended (anti) unitary 
representation of $PSU_\pm(1,1)$, i.e. $J_P$ acts geometrically by a reflection. This ensures the above 
formular is well defined. 

It turns out the equivalence class $[\bar{\rho}_I]$ does not depend on $P$, $Q$.



\begin{thm}
If $\pi$ is separable and irreducible with finite statistical dimension, 
then there exists a conjugate representation $\bar{\pi}$.  
If $\pi$ is M\"{o}bius covariant, then also $\bar{\pi}$. In particular
if $\rho \in [\pi]$ like above then $\bar{\rho} \in [\bar{\pi}]$
\end{thm}
So the conjugate representation
is given by the above formular up to some choice in the unitary equivalence class.

\section{Two interval inclusions}
We begin with some fact from subfactor theory
\begin{fact}
Let $N\subset M$ be an inclusion of type III factors, which is irreducible (ie $N' \cap M = \C \id$) 
and has finite index: $[M:N] \leq \infty$.
We assume we have a canonical endomorphism $\gamma:M \hookrightarrow N $,  $\gamma(x)=J_N J_M x J_M J_N$ for $x \in M$. 
Then are equivalent:
\begin{enumerate}
	\item $\sigma \in \End(N):  \sigma \prec \gamma|_N$, i.e. there is $U \in N$ such that $U \sigma(x) = \gamma(x)U$
	\item There is $\psi \in M$ such that $\psi x = \sigma(x) \psi$ for all $x \in N$.
\end{enumerate}
\end{fact}

This we want to apply to the two intervall inclusion $\cA(E) \subset \hat{\cA}(E):=\cA(E')'$

\includegraphics[scale=.3]{Saturday9amPicture2.jpg}

with the canonical endomorphism $\gamma_E:  \hat{\cA}(E) \hookrightarrow {\cA}(E)$.

Pick $\pi_i$ an irreducible separable representation with finite index, $\rho_i \in [\pi_i]$ localized on $I_1$.  

Then exist a conjugate $\bar{\pi}_i$ and we pick $\bar{\rho}_i \in [\bar{\pi}_i]$ localized in $I_2$.

There exist a up to constant unique intertwiner (think of co-evaluation map) $R_i \in \Hom(\id,\rho_i \bar{\rho}_i) \in \cA(E)$, i.e. $R_i(x)=\rho_i(\bar{\rho}_i(x))R_i$.  

Thus using $\sigma=\rho_i\bar\rho_i$ in the above fact we get 
$\rho_i \bar{\rho}_i \prec \lambda_E=\gamma_E |_{\cA(E)}$.
On the lefthand side we can even take a sum over mutually non-equivalent representations with finite index $\Gamma_f$
and the inequivality still holds:
$$\DirectSum_{i \in \Gamma_f} \rho_i \bar{\rho}_i \prec \lambda_E=\gamma_E |_{\cA(E)}$$
because the endomorphism are mutually inequivalent.
It turns out by some further arguments:
$$\DirectSum_{i \in \Gamma_f} \rho_i \bar{\rho}_i \simeq \lambda_E=\gamma_E |_{\cA(E)}$$

Taking the index on both sides one can conclude:
$$\sum_{\Gamma_f} d(\rho_i)^2 = [\hat{\cA}(E): \cA(E)]=\mu_2$$

We will use another fact from subfactor theory
\begin{fact} Let
$\gamma(x)=\sum_iU_i \sigma_i(x) U_i^*$ for $x \in N$ with $\sigma_i$ irreducible,
$U_i$ partial isometries, such that $\sum_{i} U_i^* U_i=\id$, $U_j U_i^*=\delta_{ij} \id$.
Then every $x \in M$ is of the form $x=\sum x_i \psi_i$ for unique $x_i \in N$.
\end{fact}

So, for each $x \in \hat{\cA}(E)$ we have a decomposition $x=\sum_{i \in \Gamma_f} x_i R_i$ with unique $x_i \in \cA(E)$.
Thus every element of the bigger factor can be written as elements of the smaller subfactor 
and intertwiner $\lbrace R_i\rbrace$:
$$ 
	\hat{\cA}(E)=\cA(E) \vee \{ R_i \}'
$$

The two-intervall inclusion is connected to the intertwiner $R_i$, thus connected to the representation theory of the net. 

\section{Modularity}

\begin{prop}
  Every irreducible seperable representation of $\cA$ has finite statistical dimension.
\end{prop}

\begin{proof}
Sketch:  Let $\rho, \rho' \in [\pi] $ be localized in the two components of $E$ respectivly and 
$u\in \Hom(\rho, \rho') \subset \hat{\cA}(E)$ their intertwiner. 
By the last fact we can uniquely write $u$ as 
$u=\sum u_i R_i$. Then exist an $i$ such that $u_i \neq 0$ and a short calculation shows that $u_i \in \Hom(\rho_i\rho, id)$, i.e.
there exist an non trivial intertwiner $\rho_i\rho$ with the vacuum representation for some $i$.
Duality
implies the existence of a non-trivial intertwiner between $\rho$ and $\bar\rho_i$ given essentially by:
$$
\xymatrix{
	\rho  \ar[rr]^{coev_{\bar \rho_i}\otimes 1} && \bar\rho_i \rho_i \rho \ar[rr]^{1\otimes u_i} && 
	\bar\rho_i
}
$$
and because $\rho, \bar\rho_i$ both 
are irreducible this means $\rho \simeq \bar{\rho_i}$.
\end{proof}

Next:  what's the braiding in this category?
Braiding is given by a bijective morphism
$\epsilon(\rho, \eta) \in \Hom(\rho \eta, \eta \rho)$ satisfying some identities. 

The idea how to define $\epsilon$ is to transport $\rho$ and $\eta$ in disjoint 
regions (so they commute), exchange the order, and than transport back. This does not 
depend one the explicit choice of the regions. One could for example transport $\eta$ to the left
or to the right, this gives in particular two (a priori) inequivalent choices.

So let $\rho$, $\eta$ be localized in some intervalls, cf

\includegraphics[scale=.3]{Saturday9amPicture4.jpg}
 
Let $\eta_{L/R} \in[\eta]$ be to equivalent representations localized
left and right from $\rho$, respectively and $T_{L/R} \in \Hom(\eta, \eta_{L/R})$ 
intertwiners. Note that $\rho\eta_{R/L} = \eta_{R/L}\rho$.

Define $\epsilon(\rho, \eta)$
$$
\epsilon(\rho,\eta) 
\equiv
\begin{xy}
	(-5,17)*{\eta};
	(5,17)*{\rho};
	(-5,-17)*{\rho};
	(5,-17)*{\eta};
	(5,10)*{}; (5,15)*{} **\dir{-}; 
	(-5,10)*{}; (-5,15)*{} **\dir{-}; 
	\vtwist~{(-5,10)}{(5,10)}{(-5,-10)}{(5,-10)}; 
	(5,-10)*{}; (5,-15)*{} **\dir{-}; 
	(-5,-10)*{}; (-5,-15)*{} **\dir{-};
	p*\frm{-};
\end{xy} 
:=
\begin{xy} 
	(-5,17)*{\eta};
	(5,17)*{\rho};
	(-5,-17)*{\rho};
	(5,-17)*{\eta};
	(-5,10)*{\scriptstyle T_\mathrm{L}^\ast}; 
	(5,5)*{}; (5,15)*{} **\dir{-};
	(-5,5)*{}; (-5,7)*{} **\dir{-}; 
 (-5,10)*\xycircle(2.65,2.65){-}="1_y"; (-5,15)*{} **\dir{-}; 
	(5,-10)*{\scriptstyle T_\mathrm{L}}; 
	(-5,-5)*{}; (-5,-15)*{} **\dir{-};
	(5,-5)*{}; (5,-7)*{} **\dir{-}; 
	(-5,5)*{}; (5,-5) **\dir{-};
	(5,5)*{}; (-5,-5) **\dir{-};
 (5,-10)*\xycircle(2.65,2.65){-}="1_x"; (5,-15)*{} **\dir{-};
\end{xy}
= T_\mathrm{L}^\ast \rho(T_\mathrm{L}) 
$$
Then 
$$\epsilon(\eta,\rho)^*
\equiv
\begin{xy}
	(-5,17)*{\eta};
	(5,17)*{\rho};
	(-5,-17)*{\rho};
	(5,-17)*{\eta};
	(5,10)*{}; (5,15)*{} **\dir{-}; 
	(-5,10)*{}; (-5,15)*{} **\dir{-}; 
	\vcross~{(-5,10)}{(5,10)}{(-5,-10)}{(5,-10)}; 
	(5,-10)*{}; (5,-15)*{} **\dir{-}; 
	(-5,-10)*{}; (-5,-15)*{} **\dir{-};
	p*\frm{-};
\end{xy} 
:=
\begin{xy} 
	(-5,17)*{\eta};
	(5,17)*{\rho};
	(-5,-17)*{\rho};
	(5,-17)*{\eta};
	(-5,10)*{\scriptstyle T_\mathrm{R}^\ast}; 
	(5,5)*{}; (5,15)*{} **\dir{-};
	(-5,5)*{}; (-5,7)*{} **\dir{-}; 
 (-5,10)*\xycircle(2.65,2.65){-}="1_y"; (-5,15)*{} **\dir{-}; 
	(5,-10)*{\scriptstyle T_\mathrm{R}}; 
	(-5,-5)*{}; (-5,-15)*{} **\dir{-};
	(5,-5)*{}; (5,-7)*{} **\dir{-}; 
	(-5,5)*{}; (5,-5) **\dir{-};
	(5,5)*{}; (-5,-5) **\dir{-};
 (5,-10)*\xycircle(2.65,2.65){-}="1_x"; (5,-15)*{} **\dir{-};
\end{xy}
= T_\mathrm{R}^\ast \rho(T_\mathrm{R}) 
$$
 thus is given by the other choice.

Note: $T_\mathrm{L/R}^\ast\rho(T_\mathrm{L/R})$  is indeed  
$$\xymatrix{
	\rho \eta \ar[rr]^{1\otimes T_{L/R}} && \rho\eta_{N/L} = \eta_{N/L}\rho
	\ar[rr]^{T_{L/R}^\ast \otimes 1} && 
	\bar\rho_i
}
$$ using that the categorical tensorproduct $\rho\eta\equiv \rho\otimes \eta$ is the composition of localized 
endomorphism.



\begin{defn}
$\rho$ and $\eta$ have {\em trivial monodromy} if $\epsilon(\rho,\eta)=\epsilon(\eta,\rho)^*$
or equivalently $\epsilon_M(\rho,\eta):=\epsilon(\rho,\eta) \epsilon(\eta, \rho)=\id$, i.e.
$$
\begin{xy}
	\vtwist~{(-5,10)}{(5,10)}{(-5,0)}{(5,0)}; 
	\vtwist~{(-5,0)}{(5,0)}{(-5,-10)}{(5,-10)};
\end{xy}
=
\begin{xy}
	(-5,-10)*{}; (-5,10)*{} **\dir{-};
	(5,-10)*{}; (5,10)*{} **\dir{-};
\end{xy} 
$$

\end{defn}
Note that $\epsilon_M([\rho],[\eta]) = \epsilon_M(\rho,\eta)$ is well-defined, i.e. the monodromy just depends
on sectors and not on the representations itself.

\begin{defn}
$\pi$ separable, non-degenerate representation of $\cA$ is called {\em finite} if one 
of the following equivalent conditions holds
\begin{itemize}
\item $\pi$ is a finite direct sum of irreps.
\item $\pi$ has finite statistical dimension
\item $\pi(C^*(\cA))'$ is finite.
\end{itemize}
\end{defn}

Let $\Rep_f(\cA)$ be the category of all finite reps.

\begin{defn}
$\rho$ is called degenerate with respect to braiding if $\epsilon_M(\rho,\eta)=1$ for all $\eta\in\Rep_f(\cA)$.

The center $Z_2(\Rep_f)$ is the set of degenerate w.r.t.~braiding reps.
\end{defn}

Note: in a modular category $\cC$, $Z_2(\cC)$ is trivial, i.e sums of $\id$.
This is the most non-trivial fact to check. 

We use two ingredients:
\begin{description}
\item[Criterion for degeneracy]  $\epsilon_M(\rho,\eta)=1$ iff $\rho(T)=T$ for $T \in \Hom(\eta_L,\eta_R)$.\\
\begin{proof} $\epsilon_M(\rho_\eta)\equiv T_L^* \rho(T_L T_R^*) T_R=1$ iff $\rho(T_L T_R^*) = T_L T_R^*$.
The statement follows, realizing  $T_L T_R^*$ equals $T^*$ up to some constant: 

\includegraphics[scale=.3]{Saturday9amPicture7.jpg}

\end{proof}

\item[Criterion for triviality of a representation]  If $\rho$ act trivially on $\hat{\cA}(E)$ then $\rho \simeq N \cdot id$, thus 
trivial.
\end{description}

\begin{thm}
$Z_2(\Rep_f \cA)$ is trivial thus $\Rep_f \cA$ is modular.
\end{thm}

\begin{proof}
  $\pi \in Z_2(\Rep_f(\cA))$ and $\rho \in [\pi]$ localized as above and 
$E$ the union of intervalls left and right from the localization intervall of $\rho$. $\rho \in Z_2$ implies $\rho(T)=\id$
	for all possible charge transporters $T$ from left to the right using the first criterion.
  
  We have seen that the big factor $\hat{\cA}(E)$ is generated by the small $\cA(E)$ and 
  the intertwinner $R_i$, this turns out to be equivalent with $\hat{\cA}(E)$ generated by 
  $\cA(E)$ and interwiner $T_i$ which transport $\eta=\rho_i$ from left to right, i.e.
  $$\hat{\cA}(E)=\cA(E) \vee \{ R_i\}=\cA(E) \vee \{ T_i\}$$
  
  By definition $\rho$ acts trivially on $\cA(E)$, but also on all charge transporters $T_i$ thus on $\hat{\cA}(E)$.  
  But this is the second criteria which implies triviality of $\rho$ thus $\pi$.
  Thus the center is trivial.
 \end{proof}

\end{document}
































