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\author[Noah Snyder]{Noah Snyder\\ \texttt{http://math.columbia.edu/$\sim$nsnyder} \\ joint work with Scott Morrison and Emily Peters}
\institute{Indiana University}
\title{Knot polynomial identities and coincidences of tensor categories}

\begin{document}

\frame{\titlepage}

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

\beamertemplatetransparentcovered 

\mode<beamer>{\setbeamercolor{block title}{bg=green!40!black}}

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\section{Tensor categories}

\AtBeginSection[]
{
   \begin{frame}<beamer>
       \frametitle{Outline}
       \tableofcontents[currentsection]
   \end{frame}
}

\subsection{Tensor categories}

\begin{frame}{Tensor categories}
\begin{block}{Structures}
\begin{itemize}
\item Abelian category
\item Tensor product of representations $V \otimes W$
\item Trivial representation $\mathbf{1}$
\item Associator $\omega_{X,Y,Z}: (X \otimes Y) \otimes Z \rightarrow X \otimes (Y \otimes Z)$
\item $u_X: \mathbf{1} \otimes X \rightarrow X$
\item etc.
\end{itemize}
\end{block}

These should satisfy some properties, for example you can associate $X \otimes Y \otimes Z \otimes W$ in different ways, you can look at $\mathbf{1} \otimes X \otimes Y$, etc.
\end{frame}

\begin{frame}{Fusion categories}
The category of finite dimensional vector spaces has certain properties which make it easier to work with.  We concentrate on tensor categories which satisfy these properties.

\begin{defn}
A \emph{fusion category} is tensor category $\mathcal{C}$ which satisfies:
\begin{itemize}
\item For every object $X$ there exists a dual $X^*$.
\item All Hom spaces are finite dimensional.
\item $\mathcal{C}$ is semisimple.
\item $\mathcal{C}$ has finitely many simple objects, up to isomorphism.
\end{itemize}
\end{defn}

\end{frame}

\begin{frame}{Why study fusion categories?}
\begin{block}{}
Fusion categories come up in many settings:
\begin{itemize}
\item von Neumann algebras
\item Knot polynomials
\item $3$-dimensional TQFTs (Reshetikhin-Turaev and Turaev-Viro)
\item Representation theory
\item Solid state physics
\item Quantum information theory
\end{itemize}
\end{block}

\end{frame}

\begin{frame}{Examples of fusion categories}

\begin{block}{Trivial example}
The category of vector spaces $\mathrm{Vec}$.  Only one simple object $\mathbf{1}$.
\end{block}

\begin{block}{Representations of $G$}
The category of representations of a finite group $\mathrm{Rep}(G)$
\end{block}

\begin{block}{$G$-graded vector spaces}
The category of $G$-graded vector spaces $\mathrm{Vec}_G$.  Here the simple objects are indexed by elements of $G$ and the tensor product is given by the group structure.
\end{block}
\end{frame}

\begin{frame}{Twisting by $3$-cocycles}
One way that fusion categories get trickier is that a lot of information is encoded in the associator.  For example, we can look at $G$-graded vector spaces with a nontrivial associator.

\begin{block}{Associator}
$$\omega_{\alpha, \beta, \gamma}: V_{\alpha \beta \gamma}= (V_\alpha \otimes V_\beta) \otimes V_\gamma \rightarrow V_\alpha \otimes (V_\beta \otimes V_\gamma) = V_{\alpha \beta \gamma}$$ assigns a scalar to every triple $(\alpha, \beta, \gamma)$.  
\end{block}

\begin{block}{}
\begin{itemize}
\item Compatibility = $\omega$ is a $3$-cocycle
\item $\mathrm{Vec}(G, \omega)$ up to equivalence only depends on $\omega \in H^3(G, k^\times)$.
\end{itemize}
\end{block}
\end{frame}

\begin{frame}{Diagram categories}
Where else do tensor categories come from?

\begin{block}{Diagram categories}
Objects: Boundaries of diagrams

Morphisms: (Linear comb. of) diagrams with fixed boundaries

Composition: Vertical stacking

Tensor product: Horizontal disjoint union
\end{block}

\begin{block}{Temperley-Lieb}
$$\inputtikz{TLexample} \in \{\mathrm{Hom}(\bullet^{\otimes 3} \rightarrow \bullet) \}$$
\end{block}
\end{frame}

\begin{frame}{An example}
\begin{block}{The Fibonacci category}
This fusion category has two simple objects $X$ and $1$.  

Furthermore, $X \otimes X \cong X \oplus \mathbf{1}$.
\end{block}

\begin{block}{Relations}
\begin{itemize}
\item $\mathfig{0.06}{G2/I.pdf} + \frac{1}{\tau} \cupcap = \identity$
\item $\mathfig{0.03}{G2/bigon.pdf} =  \mathfig{0.002}{G2/strand.pdf}$
\item $\mathfig{0.04}{G2/loop.pdf} = \tau$
\end{itemize}
\end{block}

Where $\tau = \frac{1+\sqrt{5}}{2}$.
\end{frame}


\begin{frame}{Braidings}
\begin{block}{What is a braiding?}
Often we have that $V \otimes W \cong W \otimes V$.

Even better we might have $\sigma_{V,W}: V \otimes W \rightarrow W \otimes V$.

This could satisfy the relations of the symmetric group or the braid group.
\end{block}

\begin{block}{The braiding on the Fibonacci category}
$$\inputtikz{overcrossing} \mapsto \zeta \inputtikz{identity} + \zeta^{-1} \inputtikz{cupcap}.$$

Where $\zeta = e^{\frac{3}{10} 2 \pi i}$ is a  certain $10$th root of unity.
\end{block}

If the braiding is nondegenerate, we call the category modular.

\end{frame}

\begin{frame}
\begin{block}{Examples of coincidences in finite group theory:}
\begin{itemize}
\item $\mathrm{PSL}_2(\mathbb{F}_5) \cong A_5$
\item $S_6$ has an outer automorphism
\end{itemize}
\end{block}

\begin{block}{Goal of this talk}
Find coincidences of modular categories that involve the even part of the subfactor planar algebra $D_{2n}$.  (This is a modular category related to $\mathfrak{so}_3$.)
\end{block}

\begin{block}{Methods}
\begin{itemize}
\item Hands-on technique for recognizing some important families of tensor categories by looking at $X \otimes X$.
\item Theoretical explanation of coincidences using Level-Rank duality, Kirby-Melvin symmetry, and coincidences of Dynkin diagrams.
\end{itemize}
\end{block}
\end{frame}

\begin{frame}
\begin{block}{Why should you care?}
Classifying modular categories is a long way off, so why should you care about coincidences?  Because they have applications!
\begin{itemize}
\item These coincidences give strange identities between knot polynomials.
\item One coincidence is $\frac{1}{2} D_{14} \cong (G_2)_{\frac {1}{3}}$.  Since the former is known to be unitary, this coincidence answers a question of Rowell.
\end{itemize}
\end{block}
\end{frame}

\subsection{The quantum subgroup $D_{2n}$ and its structure}

\begin{frame}
\begin{block}{What is $D_{2n}$?}
\begin{itemize}
\item Start with the braided category $TL_{q+q^{-1}}$ for $q=e^{\frac{\pi i}{4n-2}}$.
\item ``Modularize," by setting $f^{(4n-3)} \cong \id$
\item This quotient isn't braided, instead it's a spin-modular category.
\item But the even part $\frac{1}{2}D_{2n}$ is modular.
\item $\frac{1}{2}D_{2n}$ is the modularization of $U_q(\mathfrak{so}_3) = \frac{1}{2} TL_{q+q^{-1}}$.
\item $D_{2n}$ is related to the Dynkin diagram $D_{2n}$, but it is \emph{not} related to $SO_{4n}$.
\end{itemize}
\end{block}
\end{frame}

\begin{frame}
\begin{block}{Building $\frac{1}{2}D_8$}
We start with 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}


\begin{frame}{Skein theory description of $D_{2n}$}
Fix $q=\exp(\frac{\pi i}{4n-2})$.
Let $\pa$ be the planar algebra generated by a single ``box" $S$ with $4n-4$ strands, modulo the following relations.
\begin{itemize}
\item\label{delta} A closed circle is equal to $[2]_q = (q+q^{-1}) = 2 \cos(\frac{\pi}{4n-2})$ times the empty diagram.

\item\label{rotateS} Rotation relation:
\inputtikz{rotateS}

\item\label{capS} Capping relation:
\inputtikz{capS}

\end{itemize}
\end{frame}

\begin{frame}
\begin{itemize}
\item\label{twoS} Two $S$ relation:
\inputtikz{twoS}
\end{itemize}
\end{frame}

\begin{frame}
$D_{2n}$ is not braided.  However it has a ``braiding up to sign."

\begin{thm}\label{thm:passacrossS}
You can isotope a strand above an $S$ box, but isotoping a strand below
an $S$ box introduces a factor of $-1$.
\begin{enumerate}
\item
$\inputtikz{pullstringoverSsmaller}$

\item
$\inputtikz{pullstringunderSsmaller}$

\end{enumerate}
\end{thm}

Since diagrams in $\frac{1}{2}D_{2n}$ have an even number of strands, $\frac{1}{2} D_{2n}$ is braided.

\end{frame}

\begin{frame}
\begin{thm}
The planar algebra $D_{2n}$ is semi-simple, with minimal projections $f^{(k)}$ for $k=0,
\ldots, 2n-3$ along with $P$ and $Q$ defined by
\begin{align*}
P & = \frac{1}{2}\left(f^{(2n-2)} + S\right) \\
\intertext{and}
Q & = \frac{1}{2}\left(f^{(2n-2)} - S\right).
\end{align*}
The principal graph is the Dynkin diagram $D_{2n}$.
\begin{equation*}
\mathfig{0.5}{graphs/d2n}
\end{equation*}
\end{thm}
\end{frame}

\subsection{Coincidences involving $D_{2n}$}

\begin{frame}
\begin{block}{Coincidences involving $D_{2n}$}
\begin{itemize}
\item $\frac{1}{2} D_{4} \cong \mathbb{Z}/3$ (but an unusual braiding!) with $P \mapsto \chi_{e^{\frac{2 \pi i}{3}}}$.
\item $\frac{1}{2} D_{6} \cong U_{e^{-\frac{2\pi i}{10}}}(\mathfrak{sl}_2 \times \mathfrak{sl}_2)^{unimodularize}$ with $P \mapsto V_{(1)} \boxtimes V_{(0)}$.
\item $\frac{1}{2} D_{8} \cong U_{e^{-\frac{2 \pi i}{14}}}(\mathfrak{sl}_4)^{unimodularize}$ with $P \mapsto V_{(100)}$.
\item $\frac{1}{2} D_{10}$ has an automorphism: $\xymatrix@C-8mm@R-3mm{P \ar@{|->}@/^/[rr] & & Q \ar@{|->}@/^/[dl] \\ & f^{(2)} \ar@{|->}@/^/[ul]&}$.
\item $\frac{1}{2} D_{14} \cong U_{e^{2 \pi i\frac{23}{26}}}(\mathfrak{g}_2)$ with $P \mapsto V_{(10)}$.
\end{itemize}
\end{block}
(Technical point: on the righthand side we've changed the pivotal structure and modularized, furthermore we've been \emph{very careful} about choosing the right root of $q$ in the definition of the braiding.)
\end{frame}

\section{Recognizing braided categories via $X \tensor X$}
\subsection{$P \tensor P$}

\begin{frame}
In $D_{2n}$ we have the following decomposition for $P \otimes P$ (Izumi).

\newcommand{\Wfour}{\DirectSum_{l=0}^{\frac{n-3}{2}} f^{(4l)}}
\newcommand{\Wfourp}{\DirectSum_{l=0}^{\frac{n-4}{2}} f^{(4l+2)}}

\begin{itemize}
\item If $n$ is even: $P \otimes P \cong Q \directSum \Wfourp$
\item If $n$ is odd: $P \otimes P \cong P \directSum \Wfour$
\end{itemize}

In particular,
\begin{itemize}
\item In $D_4$, $P \otimes P \cong Q$
\item In $D_6$, $P \otimes P \cong \id \oplus P$
\item In $D_8$, $P \otimes P \cong f^{(2)} \oplus Q$
\item In $D_{10}$, $P \otimes P \cong \id \oplus f^{(4)} \oplus P$
\item In $D_{14}$, $P \otimes P \cong \id \oplus f^{(4)} \oplus f^{(8)} \oplus P$
\end{itemize}
\end{frame}

\subsection{Recognizing categories by $X \otimes X$}
\begin{frame}
Suppose that $X$ is an object in a unimodal braided semisimple tensor category $\cC$.
\begin{block}{}
\begin{tabular}{c|c}
if $X \tensor X \iso$ & then $\cC$ takes a braided functor from \\
\hline
$A$ & cyclic group category\\
$1 \oplus A$ & Temperley-Lieb category\\
$A \oplus B$ & HOMFLY skein category\\
$1 \directSum A \oplus B$ & Kauffman or Dubrovnik skein categories\\
$1 \directSum X \directSum A \oplus B$ & $U_q(G_2)$ (if you're lucky...)
\end{tabular}
\end{block}

Furthermore, the eigenvalues of the braiding determine the parameters in the right column.
\end{frame}

\begin{frame}
\begin{proof}
These results follow from standard skein theory arguments.  For example suppose $X \otimes X \cong 1 \oplus A$. 
Since $\End{X\otimes X}$ is $2$-dimensional there must be a linear dependence  of the form $$\inputtikz{overcrossing}=A \inputtikz{identity} + B \inputtikz{cupcap}.$$
Following Kauffman, rotate this equation, glue them together and apply Reidemeister $2$ to see that $B=A^{-1}$ and $A^2+A^{-2} = \dim V$.  Hence there's a braided functor from Temperley-Lieb.

The other cases are only a little more complicated.
\end{proof}

Technical comments: Notice that $TL$ is related to $U_q(\mathfrak{sl}_2)$ by changing the pivotal structure.  Furthermore note that in the above argument $A$ could be either square root of $A^2$.  Finally the functor need not be full (=surjective on morphisms).
\end{frame}

\subsection{More on $G_2$}

\begin{frame}{More on the $G_2$ case.}

\begin{defn}
A trivalent vertex is a rotationally symmetric map $X \otimes X \rightarrow X$.  A tree is a trivalent graph without cycles (but allowing disjoint components).
\end{defn}

\begin{thm}[Kuperberg]
Suppose we have a trivalent vertex in a semisimple pivotal tensor category $\cC$, such that the graphs
\begin{equation*}
\mathfig{0.04}{G2/lollipop} \qquad \mathfig{0.04}{G2/bigon} \qquad \mathfig{0.1}{G2/triangle} \qquad \mathfig{0.08}{G2/square} \qquad \mathfig{0.1}{G2/pentagon}
\end{equation*}
are each equal to linear combinations of trees. Then there is a faithful pivotal functor $Rep(U_q(G_2)) \to \cC$ for some $q \in \Complex$.
\end{thm}
\end{frame}

\begin{frame}{Simplifying Kuperberg's conditions}
\emph{Generically} having a braiding implies a pentagon-bursting relation.

\begin{equation*}
\mathfig{0.2}{G2/vertex-crossing-1}=\mathfig{0.2}{G2/vertex-crossing-2},
\end{equation*} 

Expand the crossings and there will be exactly one pentagon.  This gives an easier way to find a pentagon-bursting relation.
\end{frame}

%\begin{frame}
%\begin{lem}
%\label{lem:pentagon}
%Suppose we have a trivalent vertex in a \emph{braided} pivotal tensor category $\cC$, such that the graphs
%\begin{equation*}
%\mathfig{0.02}{G2/lollipop} \qquad \mathfig{0.02}{G2/bigon} \qquad \mathfig{0.05}{G2/triangle} \qquad \mathfig{0.04}{G2/square}
%\end{equation*}
%are each equal to linear combinations of trees. Further suppose that the braiding is of the form 
%\begin{equation*}
%\mathfig{0.06}{G2/crossing} = x \mathfig{0.06}{G2/vert}+ y \mathfig{0.06}{G2/horiz} + z \mathfig{0.04}{G2/I} + w \mathfig{0.075}{G2/H},
%\end{equation*}
%with at least one of $z$ and $w$ nonzero. Then \begin{equation*}
%\mathfig{0.05}{G2/pentagon}
%\end{equation*}
%is a linear combination of trees.
%\end{lem}
%\end{frame}

\section{Sources of coincidences}
\begin{frame}
\begin{block}{Level-rank duality}
Relates $(\mathfrak{so}_3)_k$ to $(\mathfrak{so}_k)_3$.
\end{block}

\begin{block}{Kirby-Melvin symmetry}
Generalized Kirby-Melvin symmetry relates $A \mapsto A \otimes B$ when $\dim B = 1$.  
\end{block}

\begin{block}{Coincidences of small Dynkin diagrams}
$D_2 = A_1 \times A_1$, $D_3 = A_3$, and $D_4$ has a `triality' automorphism.
\end{block}
\end{frame}

\subsection{Level-Rank duality}
\begin{frame}{Level-Rank duality a la Beliakova-Blanchet}
\begin{itemize}
\item The BMW (or ``Kauffman skein'') category depends on parameters $q$ and $m$.
\begin{equation*}
\overcrossing - \undercrossing = i(q-q^{-1})\left(\identity - \cupcap\right)
\end{equation*}
\begin{align*}
\positivetwist & = -i q^{m-1} \verticalstrand & \negativetwist & = i q^{1-m} \verticalstrand.
\end{align*}
\item It has a manifest symmetry. If \begin{equation*}
q_{n,k} = \begin{cases}\exp\left(\frac{2\pi i}{n+k-2}\right) & \text{when $k$ is odd, or} \\ \exp\left(\frac{2\pi i}{2(n+k-2)}\right) & \text{when $k$ is even.}\end{cases} \end{equation*}
then switching between $m=n$, $q=q_{n,k}$ and $m=k$, $q=-q_{n,k}^{-1}$ doesn't change the relations.
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item You should think of this symmetry as relating ``$U_{q_{n,k}}(O_n)$'' and ``$U_{-q_{n,k}^{-1}}(O_k)$,''
\item except that the usual quantum group is really $U_q(\mathrm{Spin})$.
\item To get from O to SO you de-equivariantize, and to get from Spin to SO you restrict to the vector representations.
\item Level-rank duality relates de-equivariantizations of the vector reps of the usual $\mathfrak{so}$ quantum groups.
\end{itemize}
\end{frame}

\begin{frame}
$$\xymatrix{&BMW_{n, q_{n,k}}=BMW_{k,-q_{n,k}^{-1}} \ar[ddl]^{\mathrm{det}_n \cong \id} \ar[ddr]^{\mathrm{det}_k \cong \id} & \\
&& \\
U_{q_{n,k}}^{vect}(\mathfrak{so}_n) \ar[ddr]^{\mathrm{det}_k \cong \id} &&U_{-q_{n,k}^{-1}}^{vect}(\mathfrak{so}_k) \ar[ddl]^{\mathrm{det}_n \cong \id}\\
&&\\
&U_{q_{n,k}}^{vect}(\mathfrak{so}_n)^{mod} \cong U_{-q_{n,k}^{-1}}^{vect}(\mathfrak{so}_k)^{mod}&
}$$
\end{frame}

\subsection{Kirby-Melvin symmetry}
\begin{frame}{Kirby-Melvin symmetry}
\begin{itemize}
\item Generalized Kirby-Melvin symmetry relates $A$ and $A \otimes B$ (for all $A$) when $\dim B = 1$.  
\item Replacing $A$ with $A \otimes B$ modifies the braiding in an understandable way (because the overcrossing and undercrossing labelled by $B$ differ by a scalar).
\item In many cases generalized Kirby-Melvin symmetry says that $A$ and $A\otimes B$ have the same image in the ``unimodularized" category (where we first change the pivotal structure and then modularize).  This is only true if you're very careful about the choice of root of $q$ used in defining the braiding.
\end{itemize}
\end{frame}

\subsection{Coincidences of Dynkin diagrams}
\begin{frame}{Coincidences of Dynkin diagrams}
\begin{itemize}
\item $D_2 = A_1 \times A_1$, hence $U_q(\mathfrak{so}_4) \cong U_q(\mathfrak{sl}_2) \boxtimes U_q(\mathfrak{sl}_2)$.
\item $D_3 = A_3$, hence $U_q(\mathfrak{so}_6) \cong U_q(\mathfrak{sl}_4)$.
\item $D_4$ has a triality automorphism, hence $U_q(\mathfrak{so}_8)$ has a triality automorphism.
\end{itemize}
\end{frame}

\section{Examples and applications}

\subsection{$G_2$ and unitarity}
\begin{frame}{Unitary of $(G_2)_{\frac{1}{3}}$}
\begin{block}{}
Rowell showed that $(G_2)_{\frac{1}{3}}$ and $(G_2)_{\frac{2}{3}}$ are pseudo-unitary, and asked if they were unitary.  
\end{block}

\begin{block}{}
But $D_{14} \cong (G_2)_{\frac{1}{3}}$, and the former is unitary.
\end{block}

\begin{block}{}
We expect that similar techniques would show that $(G_2)_{\frac{2}{3}}$ agrees with a subcategory of $(\mathrm{Sp}_6)_3$, and thus is unitary.
\end{block}
\end{frame}

\begin{frame}
\begin{figure}[!ht]
\begin{equation*}
\mathfig{0.3}{coincidences/g2-weyl-chamber}
\end{equation*}
\caption{The positive Weyl chamber for $G_2$, showing the surviving irreducible representations in the semisimple quotient at $q=e^{2\pi i\frac{23}{26}}$, and the
correspondence with the even vertices of $\mathcal{D}_{14}$.}
\label{fig:g2-weyl-chamber}
\end{figure}
\end{frame}

\subsection{Knot polynomial identities}

\begin{frame}{Knot polynomial identities}
\begin{block}{Knot invariants from $P$}
\begin{itemize}
\item The usual Reshetikhin-Turaev technique gives knot invariants coming from $P$.
\item Any closed diagram involving a single $S$ is zero.
\item If $K$ is a knot (but not a link!) then $$\frac{1}{2} \J{SU(2)}{(2n-2)}(K)(e^{\frac{2\pi i}{8n-4}}) =  \J{D_{2n}}{P}(K).$$
\item So coincidences involving $D_{2n}$ give identities relating colored Jones polynomials to other knot polynomials.
\end{itemize}
\end{block}
\end{frame}

\begin{frame}
\begin{thm}
\begin{align*}
\restrict{\J{SU(2)}{(2)}(K)}{q=e^{\frac{2\pi i}{12}}} & = 2 \J{D_4}{P}(K)  \\
& = 2 \\
\restrict{\J{SU(2)}{(4)}(K)}{q=e^{\frac{2\pi i}{20}}} & = 2 \J{D_6}{P}(K)  \\ & = 2 \restrict{\J{SU(2)}{(1)}(K)}{q=\exp(- \frac{2\pi i}{10})} \\
\restrict{\J{SU(2)}{(6)}(K)}{q=e^{\frac{2\pi i}{28}}} & = 2 \J{D_8}{P}(K)  \\
                & = 2 \restrict{\HOMFLY(K)(q^4 , q-q^{-1})}{q=e^{\frac{-2\pi i}{14}}} \\
\restrict{\J{SU(2)}{(8)}(K)}{q=e^{\frac{2\pi i}{36}}} & = 2 \J{D_{10}}{P}(K) \\
                & = 2 \restrict{\Kauffman(K) (-iq^7,i(q-q^{-1}))}{q=-e^{\frac{-2 \pi i}{18}}} \\
\end{align*}
\end{thm}
\end{frame}


\begin{frame}{The $D_8$-$\mathrm{SL}_4$ coincidence}
$$\mathfig{0.45}{coincidences/sl4-modularisation-domain}$$
\end{frame}

\begin{frame}
\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}
\begin{align}
\J{SU(2)}{(6)}(K)(e^{\frac{2\pi i}{28}}) & = 2 \J{D_8}{P}(K) \notag\\
 & = 2 \J{SO(6)}{2e_3}(K)(-e^{-\frac{2\pi i}{14}}) \tag{LR} \\
 & = 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{KM} \\
 & =  2 \J{SU(4)}{e_1}(K)(e^{-\frac{2\pi i}{14}}) \tag{parity}
\end{align}
\noqed
\end{proof}
\end{frame}



\end{document}
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