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\author[Scott Morrison]{Scott Morrison\\ \url{http://tqft.net/} \\ joint work with Emily Peters and Noah Snyder}
\institute{UC Berkeley / Miller Institute for Basic Research}
\title[Coincidences of tensor categories]{Coincidences of tensor categories}
\date{Algebraic structures in knot theory \\ UC Riverside, November 7 2009 \\ \begin{description}\item[slides:]\url{http://tqft.net/UCR-identities} \item[article:]\url{http://tqft.net/identities}\end{description}}

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\begin{document}

\frame{\titlepage}

\begin{frame}
       \frametitle{Outline}
       \tableofcontents
\end{frame}

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\newcommand{\red}[1]{{\color{red} #1}}
\newcommand{\blue}[1]{{\color{blue} #1}}
\newcommand{\green}[1]{{\color{ForestGreen} #1}}

\section[$q$-invariants]{Quantum knot invariants}

\begin{frame}{Quantum knot invariants}
\begin{block}{}
Reshetikhin-Turaev define a polynomial knot invariant for every
\begin{itemize}
\item \red{quantum group $U_q(\mathfrak{g})$}, with $\mathfrak{g}$ a complex simple Lie algebra,
\item and \blue{irreducible representation $V$} of $U_q(\mathfrak{g})$:
\end{itemize}
\vspace{-0mm}
$$\cJ_{\red{U_q(\mathfrak{g})}, \blue{V}}(\green{K})(q).$$
\end{block}
\begin{example}
\vspace{-3mm}
$$\cJ_{\red{U_q(\mathfrak{sl}_4)}, \blue{\bigwedge^2\Complex^4}}\left(\green{\mathfig{0.1}{talks/4_1}}\right)(q) = q^{16} + q^{12} + q^{10} + q^{-10} + q^{-12} + q^{-16}.$$
\end{example}
\begin{block}{}
These invariants generalise the \textit{Jones polynomial} (\red{$SU(2)$}, \blue{$\Complex^2$}), the \textit{coloured Jones polynomials} (\blue{$\operatorname{Sym}^n \Complex^2$}), \textit{HOMFLYPT} (\red{$SU(n)$}, \blue{$\Complex^n$}) and the $2$-variable \textit{Kauffman polynomial} (\red{$SO(n)$} or \red{$Sp(2n)$}, \blue{$V^\natural$}).
\end{block}\end{frame}

\begin{frame}{}
\begin{block}{We can compute these invariants! }
A computer can calculate the universal $\cR$-matrix acting on any irreducible representation. A braid presentation of the knot tells us a sequence of matrices with entries in $\Integer[q, q^{-1}]$ to multiply, and then take trace.
\end{block}
\begin{block}{Really!}
See my \code{QuantumGroups`} package, available as part of the \code{KnotTheory`} package from \url{http://katlas.org/}.
\end{block}
\begin{example}
\code{<<KnotTheory`}
\vspace{-3mm}
\center
\code{QuantumKnotInvariant[$A_3$][Irrep[$A_3$][{0,1,0}]][Knot[4,1]] == $q^{16} + q^{12} + q^{10} + q^{-10} + q^{-12} + q^{-16}$}
\end{example}
\end{frame}

\section[identities]{Mysterious identities}

\begin{frame}{Some mysterious identities}
\begin{block}{}
Let's search for identities between these polynomials, specialising $q$ to roots of unity.
\end{block}
\begin{block}{We find lots of examples!}
\vspace{-5mm}
\begin{align*}
\J{SU(2)}{(2)}(K)(\exp(\textstyle \frac{2\pi i}{12}))
		& = 2 \\
\J{SU(2)}{(4)}(K)(\exp(\textstyle\frac{2\pi i}{20}))
		& = 2 \J{SU(2)}{(1)}(K)(\exp(\textstyle \frac{-2\pi i}{10})) \\
{\color<2->{red} \J{SU(2)}{(6)}(K)(\exp(\textstyle\frac{2\pi i}{28}))}
		& = {\color<2->{red} 2 \J{SU(4)}{(1,0,0)}(K)(\exp(\textstyle\frac{-2\pi i}{14}))} \\
\J{SU(2)}{(8)}(K)(\exp(\textstyle\frac{2\pi i}{36}))
                & = 2 \J{SO(8)}{(1,0,0,0)}(K)(-\exp(\textstyle\frac{-2 \pi i}{18})) \\
\J{SU(2)}{(12)}(K)(\exp(\textstyle\frac{2\pi i}{52})) 
		& = 2 \J{G_2}{V_{(1,0)}}(K)(\exp(\textstyle\frac{2\pi i \cdot 23}{26}))
\end{align*}
\end{block}
\begin{question}
What's going on? Is there some \emph{algebraic structure} underlying these strange identities between knot polynomials?
\end{question}
\end{frame}

\section[MTCs]{Modular $\tensor$-categories}

\begin{frame}{Algebraic structure: modular tensor categories}
\begin{block}{}
At a root of unity, the representation theory of a quantum group \emph{truncates} to a \textbf{modular $\tensor$-category} with finitely many objects.
\end{block}
\begin{example}[$SU(3)$ at `level $3$', $q=\exp(\textstyle \frac{2 \pi i}{12} )$]
$$\roundframe{\mathfig{0.4}{talks/SU3}} \scalebox{3}{$\rightsquigarrow$} \roundframe{\mathfig{0.3}{talks/SU3-level3}}$$
\end{example}
\end{frame}

\begin{frame}{Braided tensor categories are like finite groups}
\begin{block}{}
\[
\begin{array}{ccc}
\mathfig{0.27}{talks/SU3-auto} & \mathfig{0.24}{talks/SU3-sub} & \mathfig{0.12}{talks/SU3-quotient} \\
\text{automorphisms} & \text{sub-categories} & \text{quotients}
\end{array}
\]
\end{block}

\begin{block}{}
\begin{itemize}
\item Not all automorphisms come from `group-like' sub-categories.
\item Not all quotients are `modular', or even $\tensor$.
\end{itemize}
\end{block}

\begin{block}{}
These algebraic operations explain identities between the corresponding knot invariants.
\end{block}
\end{frame}

\subsection{De-equivariantisation}

\begin{frame}
\begin{block}{}
We'd like to prove 
\vspace{-2mm}
$$\J{SU(2)}{(6)}(K)(\exp(\textstyle\frac{2\pi i}{28}))
		= 2 \J{SU(4)}{(1,0,0)}(K)(\exp(\textstyle\frac{-2\pi i}{14})).$$
\end{block}
\begin{block}{}
On right hand side, we look at the modular tensor category $SU(2)$ at $q=\exp(\frac{2 \pi i}{28})$. This has $12$ objects, so we call it $\blue{SU(2)_{11}}$ (`$SU(2)$ at level $11$').
\only<1>{$$\mathfig{0.6}{coincidences/SU2-plain}$$}
\only<2>{$$\mathfig{0.6}{coincidences/SU2-Z2}$$}
\only<3->{$$\mathfig{0.6}{coincidences/SU2-all}$$}
\visible<2->{which has subcategories
\vspace{-3mm}
$$\green{\Integer / 2 \Integer} \visible<3->{\subset \blue{SO(3)_6}} \subset \blue{SU(2)_{11}}$$}
\vspace{-5mm}
\end{block}
\visible<4>{
\begin{block}{}
Here's the subcategory $SO(3)_6$.
\vspace{-3mm}
$$\blue{SO(3)_6} = \mathfig{0.36}{coincidences/SO3}$$
\vspace{-3mm}
\end{block}
}
\end{frame}

\begin{frame}
\begin{block}{}
Take the quotient $SO(3)_6 / 2$; it's a new modular tensor category.
\vspace{-4mm}
$$\blue{SO(3)_6/2} = \mathfig{0.18}{coincidences/evenD6}$$
\vspace{-5mm}
\end{block}
\begin{block}{}
Quotients of braided $\tensor$-categories are usually called `de-equivariantisations'.
\end{block}
\begin{block}{}
To match conventions between $SU$ and $SO$,  replace $q$ with $q^2$. The object $(6)$ splits into two pieces, $P$ and $Q$, with the same knot invariants.
\end{block}
\begin{cor}
$$\J{\blue{SU(2)_{11}}}{\red{(6)}}(\green{K})(\exp(\textstyle \frac{2 \pi i}{28})) = 2 \J{\blue{SO(3)_6/2}}{\red P}(\green{K})(\exp(\textstyle \frac{2 \pi i}{14})).$$
\end{cor}
\end{frame}

\subsection{Level-rank duality}

\begin{frame}{Level-rank duality: ``$SO(n)_m \iso SO(m)_n$''}
Level-rank duality is tricky! The correct statement is
\begin{thm}
With $n$ odd, $q$ a $4(n+m-2)$-th root of unity,
$$\restrict{SO(n)}{q}/2 \iso \restrict{SO(m)}{-q^{-1}}/2.$$
\end{thm}

\begin{block}{}
Translating to levels, this is $SO(n)_m/2 \iso SO(m)_n/2$, but \emph{not} at the obvious root of unity!
\end{block}

\begin{block}{}
The quotients are by $V_{m e_1}$ and $V_{n e_1}$, the highest multiples of the standard representation.
\end{block}
\end{frame}

\begin{frame}[plain]
\begin{example}[$(SO(3)_6)/2 \iso (SO(6)_3)/2$]
$$\roundframe{$\mathfig{0.22}{coincidences/evenD6-2}$} \xrightarrow{\qquad\iso\qquad} \roundframe{$\mathfig{0.38}{coincidences/sl4-D6-labels}$}$$
Here we show $SO(6)_3$ as the `vector' subset of $Spin(6)_4 \iso SU(4)_3$.
\end{example}
\begin{cor}
\vspace{-4mm}
$$ \J{SO(3)/2}{P}(K)(\exp(\textstyle \frac{2 \pi i}{14})) = \J{SO(6)/2}{(200)}(K)(-\exp(\textstyle\frac{-2 \pi i}{14}))$$
\end{cor}
\end{frame}

\subsection{Kirby-Melvin symmetry}

\begin{frame}{Kirby-Melvin symmetry}
\begin{block}{}
\begin{tabular}{m{6.5cm}c}
`Include up' to all of $SU(4)$. There's a ``Kirby-Melvin symmetry''  given by \mbox{$-\otimes(300)$}, interchanging $(200)$ and $(100)$.
&
$$\roundframe{\mathfig{0.28}{coincidences/sl4-KM}}$$
\end{tabular}
\end{block}

\begin{block}{}
Kirby-Melvin symmetries aren't quite `quotients' unless we change the pivotal structure. Knot invariants may change by a sign.
\end{block}

\begin{cor}{}
\vspace{-4mm}
$$\J{SU(4)}{(200)}(K)(-\exp(\textstyle \frac{-2 \pi i}{14}) = -\J{SU(4)}{(100)}(K)(-\exp(\textstyle \frac{-2 \pi i}{14}).$$
\end{cor}
\end{frame}

\section{Conclusion}
\subsection{Putting it all together}
\begin{frame}{Putting it all together}
\begin{thm}
$\restrict{\J{SU(2)}{(6)}(K)}{q=\exp(\frac{2\pi i}{28})} = 2 \restrict{\J{SU(4)}{(1,0,0)}(K)}{q=\exp(-\frac{2\pi i}{14})}$
\end{thm}
\begin{proof}
\vspace{-3mm}
\begin{align}
\J{SU(2)}{(6)}(K)(e^{\frac{2\pi i}{28}}) & =  \J{SO(3)_6}{(6)}(K)(e^{\frac{2\pi i}{14}}) \tag{sub-category} \\
 & = 2 \J{SO(3)_6/2}{P}(K)(e^{\frac{2\pi i}{14}}) \tag{quotient} \\
 & = 2 \J{SO(6)_3/2}{2e_3}(K)(-e^{-\frac{2\pi i}{14}}) \tag{level-rank} \\
 & = 2 \J{SU(4)}{2e_1}(K)(-e^{-\frac{2\pi i}{14}}) \tag{$D_3=A_3$} \\
 & = - 2 \J{SU(4)}{e_1}(K)(-e^{-\frac{2\pi i}{14}}) \tag{Kirby-Melvin} \\
 & =  2 \J{SU(4)}{e_1}(K)(e^{-\frac{2\pi i}{14}}) \tag{parity}
\end{align}
\vspace{-5mm}
\noqed
\end{proof}
\end{frame}

\subsection{Thank you!}
\begin{frame}{Conclusion}
\begin{block}{}
It's fun to explain strange identities between knot polynomials by understanding algebraic relationships between the underlying modular tensor categories.
\end{block}
\begin{block}{Read our paper \url{http://tqft.net/identities} for}
\begin{itemize}
\item all the coincidences and automorphisms related to $SO(3)_m/2$,
\item a nice summary of level-rank duality, especially for $SO(3)$,
\item the best description of Kirby-Melvin symmetry in the literature,
\item many more pretty pictures!
\end{itemize}
\end{block}

\hfill Thank you!
\end{frame}

\end{document}
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