% use options % '[beamer]' for a digital projector % '[trans]' for an overhead projector % '[handout]' for 4-up printed notes \documentclass[beamer]{beamer} % change talk_preamble if you want to modify the slide theme, colours, and settings for trans and handout modes. \newcommand{\pathtotrunk}{../../} \input{\pathtotrunk talks/talk_preamble.tex} %\setbeameroption{previous slide on second screen=right} \author[Scott Morrison]{Scott Morrison \\ \texttt{http://tqft.net/} \\ joint work with Stephen Bigelow, Emily Peters and Noah Snyder} \institute{Microsoft Station Q / UC Santa Barbara} \title{Extended Haagerup exists!} \date{Modular tensor categories, Indiana, March 22 2009 \\ \url{http://tqft.net/indiana-EH}} \usepackage{multimedia} \begin{document} \frame{\titlepage} \begin{frame} \frametitle{Outline} \tableofcontents \end{frame} \beamertemplatetransparentcovered \mode{\setbeamercolor{block title}{bg=green!40!black}} \beamersetuncovermixins {\opaqueness<1->{60}} {} \section{Subfactors for native speakers of $\tensor$-categories} \subsection{Subfactors and $\tensor$-categories} \begin{frame} \frametitle{Subfactors and $\tensor$-categories}% \begin{itemize} \item Extremal finite-index $II_1$ subfactors correspond to unitary spherical $\tensor_A, \tensor_B$-categories with a chosen generator $\leftidx{_A}{X}{_B}$. \begin{itemize} \item (objects are bimodules for $A \subset B$, and $X$ is the `regular' bimodule $\leftidx{_A}{B}{_B}$) \end{itemize} \item The `index' $[A\!:\!B]$ of the subfactor is $(\qdim X)^2$. \item The `principal graph' encodes the tensor products $V \tensor_A X$ and $V \tensor_B X^*$. \item The `even part' is the subcategory of $A\!-\!A$ objects (alternatively of $B\!-\!B$ objects). \item The double of the even part is a modular tensor category. \begin{itemize} \item (it doesn't matter which even part you take, because they're Morita equivalent) \item (these MTCs may be `exotic', that is, don't come from quantum groups \href{http://arxiv.org/abs/0710.5761}{[Hong, Rowell, Wang])} \end{itemize} \end{itemize} \end{frame} \subsection{Planar algebras: pictures for subfactors} \begin{frame} \frametitle{Planar algebras: pictures for subfactors} \begin{tabular}{c|c} spherical tensor categories & subfactor planar algebras \\ \hline\hline $\mathfig{0.2}{category-tangle}$ & $\mathfig{0.2}{subfactor-tangle}$ \\ oriented edges & unoriented edges, with two shadings \\ an edge label for each object & edges only labelled by $X$ \\ many $\operatorname{Hom}$-spaces & just $\mathcal{P}_{k,\pm}$ \end{tabular} \begin{itemize} \item To produce the planar algebra, restrict your objects to $X \tensor X^* \tensor X \tensor \cdots $. \item To recover the $\tensor_A, \tensor_B$-category, form the idempotent completion. \end{itemize} \end{frame} \AtBeginSection[] { \begin{frame} \frametitle{Outline} \tableofcontents[currentsection] \end{frame} } \section{Haagerup's classification up to index $3+\sqrt{3}$} \begin{frame} \beamersetuncovermixins {\opaqueness<1->{0}} {} \frametitle{Haagerup's list} \begin{itemize} \item<1-> In 1994 Haagerup classified possible principal graphs for subfactors with index less than $3+\sqrt{3}$: \only<1|handout:0>{ \begin{itemize} \item $\mathfig{0.15}{graphs/Haagerup}, \mathfig{0.225}{graphs/EH}, \mathfig{0.3}{graphs/EEH}, \ldots$, \item $\mathfig{0.3}{graphs/HA}$, \item \vspace{0.25cm} $\mathfig{0.15}{graphs/hexagon}, \mathfig{0.225}{graphs/Ehexagon}, \ldots.$ \end{itemize}} \only<2|handout:0>{ \begin{itemize} \item $\mathfig{0.15}{graphs/Haagerup-green}, \mathfig{0.225}{graphs/EH}, \mathfig{0.3}{graphs/EEH}, \ldots$, \item $\mathfig{0.3}{graphs/HA-green}$, \item \vspace{0.25cm} $\mathfig{0.15}{graphs/hexagon}, \mathfig{0.225}{graphs/Ehexagon}, \ldots.$ \end{itemize}} \only<3|handout:0>{ \begin{itemize} \item $\mathfig{0.15}{graphs/Haagerup-green}, \mathfig{0.225}{graphs/EH}\color{red}{, \mathfig{0.3}{graphs/EEH-red}, \ldots}$, \item $\mathfig{0.3}{graphs/HA-green}$, \item \vspace{0.25cm} \color{red}{$\mathfig{0.15}{graphs/hexagon-red}, \mathfig{0.225}{graphs/Ehexagon-red}, \ldots$}. \end{itemize}} \only<4>{ \begin{itemize} \item $\mathfig{0.15}{graphs/Haagerup-green}, \mathfig{0.225}{graphs/EH-blue}\color{red}{, \mathfig{0.3}{graphs/EEH-red}, \ldots}$, \item $\mathfig{0.3}{graphs/HA-green}$, \item \vspace{0.25cm} \color{red}{$\mathfig{0.15}{graphs/hexagon-red}, \mathfig{0.225}{graphs/Ehexagon-red}, \ldots$}. \end{itemize}} \item<2-> Haagerup and \href{http://dx.doi.org/10.1007/s002200050574}{Asaeda \& Haagerup (1999)} constructed two of these possibilities. \item<3-> \href{http://www.springerlink.com/content/c4885q02dflfttwm/}{Bisch (1998)} and \href{http://arxiv.org/abs/0711.4144}{Asaeda \& Yasuda (2007)} ruled out infinite families. \item<4-> Today, we construct the missing example (`extended Haagerup'), and complete the classification. \end{itemize} \end{frame} \begin{frame} \begin{itemize} \item Asaeda \& Haagerup's constructions find a biunitary flat connection, using Ocneanu's paragroup formalism. \item \href{http://www.ams.org/mathscinet-getitem?mr=MR1633929}{Ikeda (1998)} did numerical calculations suggesting that the extended Haagerup subfactor existed ``up to $10^{-9}$''. \item Our techniques are very different. Although we made heavy use of a computer while searching for a solution, verifying the answer only requires a computer to multiply large matrices, using exact arithmetic. \end{itemize} \end{frame} \section{Constructing planar algebras} \subsection{Annular and quadratic tangles} \begin{frame} \begin{project} Now that we know about planar algebras, we can give \emph{skein theory} constructions of subfactors. \end{project} \begin{itemize} \item \href{http://math.berkeley.edu/~vfr}{Jones' work} on \emph{annular} and \emph{quadratric} tangles gives you information about the generators and relations to expect. \item \href{http://arxiv.org/abs/0902.1294}{Peters [math/0902.1294]} has implemented this approach to give a new construction of the Haagerup subfactor. Our work follows that model closely. \end{itemize} \end{frame} \begin{frame} \frametitle{Annular Temperley-Lieb} Every planar algebra $\mathcal{P}$ is a representation of the \emph{annular Temperley-Lieb} category. Every representation breaks up as a sum of representations generated by \emph{lowest weight vectors} $T_{k,\lambda} \in \mathcal{P}_k$ satisfying \begin{align*} \inputtikz{Generator/Cap1} = \inputtikz{Generator/Cap2} & = \cdots = 0, & \mathfig{0.19}{Generator/Rotate} & = \lambda T \end{align*} for some $k \geq 0$ and $\lambda \in \sqrt[k]{1}$. Thus the planar algebra is generated by elements of this form. \end{frame} \begin{frame} If $T$ is a lowest weight vector in $\mathcal{P}_5$, then we write $\mathcal{ATL}_{+1}(T)$ for the ``annular consequences'' of $T$ in $\mathcal{P}_{6}$. \begin{align*} \begin{tikzpicture}[ATLsix] \clip (0,0) circle (5cm); \draw[shaded] (162:6cm) -- (162:5cm) .. controls (162:3cm) and (126:3cm) .. (126:5cm) -- (126:6cm); \draw[shaded] (0,0) -- (90:6cm)--(54:6cm)--(0,0); \draw[shaded] (0,0) -- (-18:6cm)--(18:6cm)--(0,0); \draw[shaded] (0,0) -- (-90:6cm)--(-54:6cm)--(0,0); \draw[shaded] (0,0) -- (-162:6cm)--(-126:6cm)--(0,0); \node at (0,0) [Tbox] (T) {$T$}; \node at (T.180) [left] {$\star$}; \node at (180:5cm) [right] {$\star$}; \draw [ultra thick] (0,0) circle (5cm); \end{tikzpicture} \; , \; \begin{tikzpicture}[ATLsix] \clip (0,0) circle (5cm); \draw[shaded] (0,0)--(162:6cm) -- (126:6cm) -- (126:5cm) .. controls (126:3cm) and (90:3cm) .. (90:5cm) -- (90:6cm) -- (54:6cm) --(0,0); \draw[shaded] (0,0) -- (-18:6cm)--(18:6cm)--(0,0); \draw[shaded] (0,0) -- (-90:6cm)--(-54:6cm)--(0,0); \draw[shaded] (0,0) -- (-162:6cm)--(-126:6cm)--(0,0); \node at (0,0) [Tbox] (T) {$T$}; \node at (T.180) [left] {$\star$}; \node at (180:5cm) [right] {$\star$}; \draw [ultra thick] (0,0) circle (5cm); \end{tikzpicture} \; , \ldots , \; \begin{tikzpicture}[ATLsix, rotate=72] \clip (0,0) circle (5cm); \draw[shaded] (0,0)--(162:6cm) -- (126:6cm) -- (126:5cm) .. controls (126:3cm) and (90:3cm) .. (90:5cm) -- (90:6cm) -- (54:6cm) --(0,0); \draw[shaded] (0,0) -- (-18:6cm)--(18:6cm)--(0,0); \draw[shaded] (0,0) -- (-90:6cm)--(-54:6cm)--(0,0); \draw[shaded] (0,0) -- (-162:6cm)--(-126:6cm)--(0,0); \node at (0,0) [Tbox] (T) {$T$}; \node at (T.-108) [below] {$\star$}; \node at (108:5cm) [right] {$\star$}; \draw [ultra thick] (0,0) circle (5cm); \end{tikzpicture} \end{align*} \begin{lem} If $\delta > 2$, the annular consequences $\mathcal{ATL}_{+k}(T)$ of a lowest weight vector $T$ are all linearly independent. \end{lem} \end{frame} \begin{frame} \frametitle{Quadratic tangles} Further, by counting the dimensions of the $\mathcal{ATL}$ representations, we can expect relations between certain \emph{quadratic tangles}. \begin{example}[Extended Haagerup] $$\mathcal{P} \iso V_{0,\delta} \directSum V_{8,-1} \directSum V_{10,\bullet} \directSum \cdots$$ If $T$ generates $V_{8,-1}$, then \begin{itemize} \item $\mathfig{0.2}{quadratic/small-T2-8} \in P_8$ is a linear combination of $T$ and $\mathcal{TL}$, \item there no lowest weight vector in $P_{9}$ , so $\mathfig{0.2}{quadratic/small-T2-7}$ must be a linear combination of $\mathcal{ATL}_{+1}(T)$ and $\mathcal{TL}$, \item there's only one lowest weight vector in $P_{10}$, so some linear combination of $\mathfig{0.2}{quadratic/small-T2-6}$ and $\mathfig{0.2}{quadratic/small-T2-6t}$ must lie in $\mathcal{ATL}_{+2}(T) \directSum \mathcal{TL}$. \end{itemize} \end{example} \end{frame} \subsection{Existence} \begin{frame} \frametitle{Existence} \begin{question} Consider a planar algebra $\mathcal{P}$ generated by an element $T$ satisfying relations like these. Is it the extended Haagerup planar algebra? \end{question} \begin{itemize} \item In fact, if we can show that $\mathcal{P}$ is a \emph{subfactor planar algebra} with the correct index (the largest root of $x^3 - 8 x^2 + 17 x - 5$, $\sim 4.3772...$), then Haagerup's classification guarantees it must have the desired principal graph. \item It's probably also possible to compute the principal graph directly from the skein theory. \end{itemize} \end{frame} \begin{frame} \frametitle{When is $\mathcal{P}$ a subfactor planar algebra?} \begin{itemize} \item Do we know the relations are consistent? (i.e., is $\mathcal{P} \neq 0$?) \item Is the planar algebra unitary? \item Is $\dim\mathcal{P}_0 = 1$? That is, can we evaluate every closed diagram using the relations? \end{itemize} To answer the first two questions, we can try to find an element $T$ inside a larger unitary planar algebra. Fortunately, there's an obvious place to look: \begin{thm} Every subfactor planar algebra $\mathcal{P}$ is a subalgebra of the \emph{graph planar algebra} of the principal graph $\Gamma(\mathcal{P})$. \end{thm} The \emph{evaluation problem} requires some clever skein theory. \end{frame} \begin{frame} \frametitle{The graph planar algebra} \begin{defn} The \emph{graph planar algebra} $GPA(\Gamma)$ of a bipartite graph $\Gamma$ has \begin{description} \item[spaces] $GPA(\Gamma)_k = \Complex\left\{\text{length $2k$ loops on $\Gamma$}\right\}$ and \item[$\mathcal{TL}$ action] described in Jones', \href{}{``\emph{The planar algebra of a bipartite graph}''}, in terms of the Perron-Frobenius eigenvector. \end{description} \end{defn} The graph planar algebra is always \emph{unitary} and \emph{spherical}. It is not a subfactor planar algebra since $$\dim(GPA(\Gamma)_0) = \card{\text{even vertices}} > 1.$$ \begin{example} $\dim(GPA(\Gamma(\mathcal{H}))_4) = 375$, $\dim(GPA(\Gamma(\mathcal{EH}))_{8}) = 148475$ \end{example} \end{frame} \subsection{Construction} \begin{frame} \frametitle{Construction} We now start solving equations in $GPA(\Gamma(\mathcal{EH}))_{16}$. \begin{itemize} \item Considerations from Jones' Quadratic Tangles tells us the rotational eigenvalue of the generator must be $-1$. \item We cut down $GPA(\Gamma(\mathcal{EH}))_{16}$ to a $19$ dimensional space using the linear relations $$ \inputtikz{Generator/Cap1} = \inputtikz{Generator/Cap2} = \cdots = 0 \, , \, \mathfig{0.19}{Generator/Rotate} = - T . $$ \end{itemize} \end{frame} \begin{frame} \begin{itemize} \item The generator must satisfy $T^2 = f^{(16)}$ --- this gives $148475$ quadratic equations! \item We look at specially chosen subset of these and find a solution $T$ by ad-hoc methods. \end{itemize} \end{frame} \begin{frame} \label{frame:construction-last} At this point you can forget everything I told you about finding $T$! \begin{itemize} \item We can write down an explicit description of $T$ as an element of the $148475d$ vector space. \goto{frame:explicit-generator}{See the generator!} \item It's tedious but straightforward to check by hand that $T$ is a lowest weight vector with rotational eigenvalue $-1$. \item Certainly the subalgebra generated by $T$ is both nontrivial and unitary. \item We still need to see that all closed diagrams can be evaluated, that is, $\mathcal{P}(T)_0 = \Complex$. \end{itemize} \end{frame} \section{Skein theory for extended Haagerup} \subsection{Quadratic relations} \begin{frame} \frametitle{Quadratic relations} We have a candidate generator; let's find the particular quadratic relations it satisfies. First, with the aid of a computer (doing exact arithmetic!) we calculate six moments. \begin{align*} \tr{T^2} & = [9] \sim 24.66097...\\ \tr{T^3} & = 0 \\ \tr{T^4} & = [9] \\ \tr{\rho^{1/2}(T)^2} & = -[9] \\ \tr{\rho^{1/2}(T)^3} & = -i\frac{[18]}{\sqrt{[8][10]}} \sim -15.29004i \\ \tr{\rho^{1/2}(T)^4} & = \frac{1}{5}\left(46 \lambda ^4-2 \lambda ^2-94\right) \sim 34.1409... \end{align*} with $\lambda$ the root of $x^6 +2 x^4 -3x^2 -5$ which is approximately $1.12867i$. \end{frame} \begin{frame} Bessel's inequality now gives the desired relations \begin{align*} \scalebox{0.25}{\inputtikz{InnerProducts/T-arc}} & = \scalebox{0.85}{$i \frac{\sqrt{[8][10]}}{[9]}$} \scalebox{0.25}{\inputtikz{InnerProducts/T2-7}} \\ \scalebox{0.25}{\inputtikz{InnerProducts/T-2arcs}} & = \scalebox{0.85}{$\frac{\lambda^3}{\delta}$} \scalebox{0.25}{\inputtikz{InnerProducts/T2-6s}} - \scalebox{0.75}{$\frac{4 \lambda ^5+7 \lambda ^3-5 \lambda }{10}$} \scalebox{0.25}{\inputtikz{InnerProducts/T2-6t}} \end{align*} %X= i sqrt([8][10])/[9] = 0.9992 i %Y=Root[-125 - 229 #1^2 - 46 #1^4 + 5 #1^6 &, 5], Z= - Root[-125 - 5812 #1^2 - 6896 #1^4 + 1600 #1^6 &, 4] Note that the first equation only holds with the given shading. For the Haagerup and extended Haagerup graph, when we write $\mathfig{0.2}{quadratic/small-T2-7}$ in terms of $\mathcal{ATL}_{+1}(T)$ a certain coefficient is `unexpectedly' zero. \end{frame} \begin{frame} Now substitute the first equation into the last term of the second, and expand the Jones-Wenzl idempotents. \begin{align*} \mathfig{0.25}{jellyfish/left} & = \mathfig{0.25}{jellyfish/right0} + \mathfig{0.25}{jellyfish/right1} + \\ & \qquad \qquad + \mathfig{0.25}{jellyfish/right2} + \mathfig{0.25}{jellyfish/right3} \end{align*} This relation lets us `pull a generator through a pair of strands'. This increases the number of generators in the diagram, so it isn't immediately obvious how this helps evaluate closed diagrams! \end{frame} \subsection{The jellyfish algorithm} \begin{frame} \begin{thm} Stephen's \emph{jellyfish algorithm} shows these relations suffice to evaluate arbitrary closed diagrams. %\begin{center}\movie{$\mathfig{0.7}{jellyfish/echeng-jellyfish-lake-palau}$}{jellyfish-lake.mov}\end{center} $$\mathfig{0.7}{jellyfish/echeng-jellyfish-lake-palau}$$ \end{thm} \end{frame} \begin{frame} \beamersetuncovermixins {\opaqueness<1->{0}} {} Begin with arbitrary planar network of $T$s. \only<1|handout:0>{ $$\mathfig{0.65}{jellyfish/network}$$}\only<2>{ $$\mathfig{0.65}{jellyfish/network-paths}$$ } \only<3|handout:0>{$$\mathfig{0.65}{jellyfish/network-paths-2}$$} \only<4|handout:0>{$$\mathfig{0.65}{jellyfish/network-paths-3}$$} \only<5-|handout:0>{$$\mathfig{0.65}{jellyfish/network-paths-4}$$}\uncover<2->{ Now float each generator to the surface, using the relation. } \end{frame} \begin{frame} The diagram now looks like a polygon with some diagonals, labelled by the numbers of strands connecting generators. $$\mathfig{0.3}{jellyfish/curved-diagonals} = \mathfig{0.3}{jellyfish/diagonals}$$ \vspace{-4mm} \begin{itemize}\item Each such polygon has a corner, and the generator there is connected to one of its neighbours by at least $8$ edges. \item Use $T^2 = f^{(16)}$ to reduce the number of generators, and recursively evaluate the entire diagram. \end{itemize} \end{frame} \begin{frame} \frametitle{Thank you!} \begin{center}\movie{\placefig{0.8}{jellyfish/echeng-jellyfish-lake-palau}}{jellyfish-lake.mov}\end{center} \end{frame} \appendix \subsection{An explicit generator} \begin{frame} \label{frame:explicit-generator} To specify the generator $T$ we need to give its value on every one of the $148475$ loops of length $16$. \begin{equation*} T(\gamma) = r \cdot \sigma(\gamma) \cdot p_{\widehat{\gamma}} \cdot \frac{1}{d_{\gamma_1}} \cdot \prod_{i=1}^{16} \frac{1}{\sqrt{d_{\gamma_i}}}. \end{equation*} \begin{itemize} \item $r=-1843700 + 5847375 d^2 - 1614050 d^4$ \item If $\gamma$ is a path on the principal graph, produce a sequence $\widehat{\gamma} \in \{0,1,2\}^8$ so that if $\gamma_{2i-1}$ is in the $j$-th arm of the principal graph, then $\widehat{\gamma}_i = j$. \item Let $\sigma(\gamma)$ to be $-1$ raised to the number of times the vertices $v_0, w_0, z_i$ and $a_i$ appear in $\gamma$. \end{itemize} We still need to specify $3^8 = 6561$ values for $p_{\widehat{\gamma}}$. \end{frame} \begin{frame}[shrink=0.8] Fix $\lambda = \sqrt(2-d^2)$. Define $21$ elements of $\Integer[\lambda]$ \begin{align*} p_{00000001} & = -2 \lambda ^4-\lambda ^2+9 & p_{00000011} & = -\lambda ^5-\lambda ^3+3 \lambda \displaybreak[1] \\ p_{00000101} & = 2 \lambda ^4+\lambda ^2-9 & p_{00000111} & = 1 \displaybreak[1] \\ p_{00001001} & = \lambda ^5-\lambda ^3-3 \lambda & p_{00001011} & = \lambda ^3-1 \displaybreak[1] \\ p_{00010001} & = 2 \lambda ^4+\lambda ^2-9 & p_{00010011} & = \lambda ^5-\lambda ^4+\lambda ^2-3 \lambda +4 \displaybreak[1] \\ p_{00010101} & = \lambda ^4-2 \lambda ^2+1& p_{00010111} & = 1-\lambda ^4 \displaybreak[1] \\ p_{00011011} & = \lambda ^4-\lambda ^2-3 & p_{00100101} & = 5-2 \lambda ^4 \displaybreak[1] \\ p_{00100111} & = \lambda ^2+1 & p_{00101011} & = -\lambda ^5-\lambda ^3+\lambda +1 \displaybreak[1] \\ p_{00101101} & = \lambda ^5-\lambda& p_{00110011} & = 2 \lambda ^5+5 \lambda ^3+4 \lambda \displaybreak[1] \\ p_{00110111} & = -\lambda ^5-2 \lambda ^3-4 \lambda ^2-\lambda -5 & p_{01010101} & = -4 \lambda ^4+3 \lambda ^2+7 \displaybreak[1] \\ p_{01010111} & = \lambda ^4+\lambda ^2 & p_{01011011} & = \lambda ^4-2 \lambda ^2-4 \displaybreak[1] \\ p_{01110111} & = \lambda ^4+6 \lambda ^2+6 & & \end{align*} %Notice that all these elements lie in the $\Integer$-lattice generated by $p_{00000001}$, $p_{00000011}$, $p_{00000111}$, $p_{00001011}$, $p_{00010111}$ and $p_{00101011}$. (You can %verify this using the function \code{VerifyZ$\lambda$Lattice} in the \MMA notebook.) \end{frame} \begin{frame} Extend these definitions to every $p_w$ for $w \in \{0,1\}^8$ by the rules \begin{align*} p_{abcdefgh} & = - p_{bcdefgha} \\ p_{abcdefgh} & = \overline{p}_{ahgfedcb} \\ \intertext{and} p_{00000000} & = 0 \\ p_{abcd1111} & = 0. \end{align*} Note that we have to check these rules are well-defined. For example, one can get from $p_{00110011}$ to $p_{01100110}$ either by rotating, or by reversing. \end{frame} \begin{frame} Further extend these definitions to every $p_w$ for $w \in \{0,1,2\}^8$ by the rules \begin{align*} p_{x 0 y} + p_{x 1 y} + p_{x 2 y} = 0. \end{align*} \return{frame:construction-last}{Return to the talk...} \end{frame} \end{document} % ----------------------------------------------------------------