\documentclass[11pt, letterpaper]{amsart} \usepackage{amsthm, amssymb} \usepackage{graphicx,tikz} \usepackage[all]{xy} \xyoption{knot} \usetikzlibrary{decorations.pathreplacing} \usetikzlibrary{arrows} \setlength{\parindent}{0pt} \setlength{\parskip}{\baselineskip} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{thm}{Theorem}[section] \newtheorem*{claim}{Claim} \newtheorem*{lemma*}{Lemma} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{cor}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{prop}[theorem]{Proposition} \theoremstyle{definition} \newtheorem*{definition}{Definition} \newtheorem*{defn}{Definition} \newtheorem*{example}{Example} \newtheorem*{cexample}{Counter-example} \newtheorem*{remark}{Remark} \newtheorem*{homework}{Homework} \newtheorem*{question}{Question} \newtheorem*{answer}{Answer} \newtheorem*{idea}{Idea} \newtheorem*{recall}{Recall} \newtheorem*{note}{Note} \newtheorem*{fact}{Fact} \newtheorem*{goal}{Goal} \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cC}{\mathcal{C}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cH}{\mathcal{H}} \newcommand{\cI}{\mathcal{I}} \newcommand{\cQ}{\mathcal{Q}} \newcommand{\cU}{\mathcal{U}} \newcommand{\cX}{\mathcal{X}} \newcommand{\cY}{\mathcal{Y}} \newcommand{\g}{\mathfrak{g}} \newcommand{\h}{\mathfrak{h}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\id}{\mathbf{1}} \newcommand{\directSum}{\oplus} \newcommand{\DirectSum}{\bigoplus} \newcommand{\tensor}{\otimes} \newcommand{\Tensor}{\bigotimes} \newcommand{\acton}{\curvearrowright} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\Rep}{\operatorname{Rep}} \newcommand{\norm}[1]{\| #1 \|} \newcommand{\abs}[1]{\left| #1 \right|} \newcommand{\ip}[1]{\langle #1 \rangle} \newcommand{\su}[1]{\mathfrak{su}(#1)} \newcommand{\bT}{\mathbb{T}} \newcommand{\Trot}{\bT_{rot}} \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}