%!TEX root = ../article.tex

In the previous section we found identities between knot polynomials coming from (a priori) different ribbon categories.  In Section \ref{sec:ribbonfunctors} we showed that these identities must come from unexpected functors between these ribbon categories.  In this section we explain how these coincidences of tensor categories follow from general theory.  One should think of the results of this section as quantum analogs of small coincidences in group theory, such as $\operatorname{Alt}_5 \cong \mathbf{PSL}_2(\mathbb{F}_5)$.

There are three important sources of unexpected equivalences (or autoequivalences) between ribbon categories coming from quantum groups: coincidences of small Dynkin diagrams, (deequivariantization related to) generalized Kirby-Melvin symmetry, and level-rank duality.   

There are sometimes coincidences between Dynkin diagrams in different families. For instance, the Dynkin diagrams $A_3$ and $D_3$ are equal, from which it follows that $\SL{4} \iso \SO{6}$ and the associated categories of representations of quantum groups are equivalent too.

Kirby-Melvin symmetry relates link invariants coming from different objects in the same category, when that category has an invertible object.  
Under certain auspicious conditions, one can go further and deequivariantize by the invertible object.

Level-rank duality is a collection of equivalences relating $SU(n)_k$ with $SU(k)_n$, and relating $SO(n)_k$ with $SO(k)_n$,
where $SU(n)_k$ or $SO(n)_k$ refers to the semisimplified representation category of the rank-$n$ quantum group, at a carefully chosen root of unity which depends on the `level' $k$. In some sense, level-rank duality is more natural in the context of $U(n)$ and $O(n)$, and new difficulties arise formulating level-rank duality for the quantum groups $SU(n)$ and $SO(n)$. We give, in Theorem \ref{thm:SO3-duality}, a precise statement for $SO$ level-rank duality with $n=3$ and $k$ even.

We will discuss each of these three sources of unexpected equivalences in the following sections, and then use them to prove the following results:

\begin{thm}
\label{thm:D6-coincidence}
There is an equivalence of modular tensor categories $$\frac{1}{2} \mathcal{D}_{6} \cong \uRep{U_{s = \exp\left(\frac{7}{10} 2 \pi i\right)}(\SL{2} \oplus \SL{2}})^{modularize},$$ sending $P \mapsto V_{(1)} \boxtimes V_{(0)}$.
\end{thm}

\begin{thm}
\label{thm:D8-coincidence}
There is an equivalence of modular tensor categories $$\frac{1}{2} \mathcal{D}_{8} \cong \uRep{U_{s=\exp \left(\frac{5}{14} 2 \pi i\right)}(\SL{4})}^{modularize},$$ sending $P \mapsto V_{(100)}$.
\end{thm}

\begin{thm}
\label{thm:D10-coincidence}
The modular tensor category $\frac{1}{2} \mathcal{D}_{10}$ has an order $3$ automorphism, fixing $\JW{0}, \JW{4}$ and $\JW{6}$, and permuting $$\baselinetrick{\xymatrix@C-8mm@R-3mm{P \ar@{|->}@/^/[rr] & & Q \ar@{|->}@/^/[dl] \\ & f^{(2)} \ar@{|->}@/^/[ul]&}}.$$
\end{thm}

Finally we note that there are other coincidences of small tensor categories that do not follow from these general techniques.  In particular it would be very interesting to better explain the coincidences involving $G_2$.

\subsection{Dynkin diagram coincidences and quantum groups}

The definition of the quantum group and its ribbon category of representations depend only on the Dynkin diagram itself.  For the quantum group and its tensor category this is obvious from the presentation by generators and relations.  For the braiding and the ribbon structure this follows from the independence of choice of decomposition of the longest word in the Weyl group in the multiplicative formula for the R-matrix.

In particular, every coincidence between Dynkin diagrams lifts to a statement about the quantum groups.  We will use that $D_2 = A_1 \times A_1$, that $D_3 = A_3$, and that $D_4$ has triality symmetry. 

The reason these coincidences are useful is that they give two different diagrammatic presentations of the same ribbon category.  For example, the fact that $B_1 = A_1$ tells you that the even part of Temperley-Lieb can be described using the Dubrovnik category, which we used implicity in Section \ref{sec:recognizeD}. The only coincidence we don't use is $B_2 = C_2$. Since $B_2$ is the Dynkin diagram for $\SO{5}$, there is no relationship via level-rank duality with the $\mathcal{D}_{2n}$ planar algebras.

\subsection{Kirby-Melvin symmetry}\label{sec:KM}
Kirby-Melvin symmetry relates link invariants from one
representation of a quantum group to link invariants coming from another
representation which is symmetric to it under a
symmetry of the affine Weyl chamber.  This symmetry principle was proved
in type $A_1$ by Kirby and Melvin \cite{MR1117149}, in type $A_n$ by Kohno and 
Takata \cite{MR1227008}, and for a general quantum group by Le \cite{MR1749439}.  There is another proof in the type $A$ case, using conformal inclusions, due to Xu \cite{MR2124554}.
We give a diagrammatic  proof which generalizes this result to tensor categories which might not come from quantum groups.

Suppose that $\cC$ is a semi-simple ribbon category and that $X$ is an object which is invertible in the sense that $X \otimes X^* \cong \id$.  Kirby-Melvin symmetry relates link invariants coming from a simple object $A$ to invariants coming from the (automatically simple) object $A\otimes X$.  

The key observation is that, for any simple $A$, the objects $A
\otimes X$ and $X \otimes A$ are simple (since $\Hom{}{A \otimes X}{ A \otimes X}
= \Hom{}{A \otimes X \otimes X^*}{ A}$), so the Hom space between them is one dimensional.  Thus the over-crossing and under-crossing must be scalar multiples.  Define $c_A$ by the following formula,
$$\inputtikz{labelledundercrossing} = c_A \inputtikz{labelledovercrossing}.$$

Note that $c_A^{-1} \dim A \dim X = S_{XA}$ where $S$ is the $S$-matrix.  Using the formula for the square of the crossing in terms of the ribbon element, we see that $c_A = \frac{\theta_A \theta_X}{\theta_{A\otimes X}}$.



\begin{thm}
\label{thm:ourKM}
Let $\cC$ be a semi-simple ribbon category, $A$ be a simple object in $\cC$, $X$ be a simple invertible object, and $L$ a link with $\# L$ components.  Then, $$\J{\cC}{A \otimes X}(L) = \J{\cC}{A}(L) \J{\cC}{X}(L) = (\dim X)^{\# L} \J{\cC}{A}(L).$$
\end{thm}
\begin{proof}
First look at the framed version of the knot invariants.  The framed $A \otimes X$ invariant comes from cabling $L$ and labeling one of the two cables $A$ and the other one $X$.  We unlink the link labeled $A$ from the link labeled $X$ by successively changing crossings where $X$ goes under $A$ to crossings where $X$ goes over $A$.  Each crossing in the original link gives rise to two crossings between the $X$-labelled link and the $A$-labelled link, and exactly one of these crossings needs to be switched.  Furthermore, the sign of the crossing that needs to be switched is the same as the sign of the original crossing.  See the following diagram for what happens at each positive crossing.

$$\inputtikz{labelledovercrossingAX} = \inputtikz{labelleddoubleovercrossingzoom} = c_A^{-1}\inputtikz{labelleddoubleovercrossingzoomswitched}$$

Hence, unlinking the $X$-labelled link from the $A$-labelled link picks up a factor of $c_A^{-writhe}$.  At this point, the link labelled by $A$ lies completely behind the link labelled by $X$, and we can compute their invariants separately. Thus,
$$\theta_{A \otimes X}^{writhe} \J{\cC}{A \otimes X}(L) = c_A^{-writhe} \theta_A^{writhe} \J{\cC}{A}(L) \theta_X^{writhe} \J{\cC}{X}(L).$$
Rearranging terms and writing $c_A$ in terms of twist factors, we see that $\J{\cC}{A \otimes X}(L) = \J{\cC}{A}(L) \J{\cC}{X}(L)$.  The final equation follows from Theorem \ref{thm:summands}.
\end{proof}


Note that $\dim{X}$ above has to be $1$ or $-1$, since $\dim{X}=\dim{X^*}$ and 
$X \otimes X^*=1$.

Suppose that you have a finite ribbon category whose fusion graph is symmetric.  Take $X$ to be any projection which is symmetric in the fusion graph with $\id$.  Then it is easy to see that its Frobenius-Perron dimension $\dim_{FP}(X) = 1$, and thus that $X$ is invertible.  Hence, any time the fusion graph has a symmetry so do the knot invariants.

If $X$ gives a Kirby-Melvin symmetry, then if you're lucky you can set $X \cong \id$ using the deequivariantization procedure.  Furthermore, even if you can't deequivariantize immediately (for example, if $\dim X \neq 1$) you might still be able to modify the category $\cC$ is some mild way (changing the braiding or changing the pivotal structure, neither of which changes the link invariant significantly) and then be able to deequivariantize.  We give three examples of this:

Consider $\Rep{U_{q=-\exp \left(-2 \pi i \frac{1}{10} \right)}(\SL{2})}$.  The representation $V_3$ is invertible and thus gives a Kirby-Melvin symmetry.  We can make this monoidal category into a ribbon category in many ways: first we can choose $s = q^{\frac{1}{2}}$ in two different ways; second we can choose either the usual pivotal structure or the unimodal one.  For each of these four choices we check each of the conditions needed to define the deequivariantization $\cC//V_3$ (transparency, dimension $1$, and twist factor $1$).

\begin{figure}[!ht]
\begin{center}
\begin{tabular}{c|c|c|c}
&$V_3$ transparent & $\dim V_3$ & $\theta_{V_3}$\\
\hline
$\Rep{U_{s=\exp\left(2 \pi i \frac{1}{5} \right)}(\SL{2})}$&Yes&-1&1\\
\hline
$\Rep{U_{s=\exp\left(2 \pi i \frac{7}{10} \right)}(\SL{2})}$&No&-1&-1\\
\hline
$\uRep{U_{s=\exp\left(2 \pi i \frac{1}{5} \right)}(\SL{2})}$&No&1&-1\\
\hline
$\uRep{U_{s=\exp\left(2 \pi i \frac{7}{10} \right)}(\SL{2})}$& Yes&1&1
\end{tabular}
\end{center}
\end{figure}

Let $\rRep U_q(\mathfrak{g})$ denote the full subcategory of representations whose highest weights are in the root lattice.  (Notice that this ribbon category only depends on $q$, not on a choice of $s=q^{\frac {1}{L}}$.  Furthermore, it does not depend on the choice of ribbon element.)

\begin{lem} \label{lem:KM-SL2}
$\rRep U_{q=-\exp \left(-2 \pi i \frac{1}{10} \right)}(\SL{2}) \cong \uRep{U_{s=\exp\left(2 \pi i \frac{7}{10} \right)}(\SL{2})^{modularize}}$.
\end{lem}

\begin{proof}
We restrict the deequivariantization $$\cF: \uRep{U_{s=\exp\left(2 \pi i \frac{7}{10} \right)}(\SL{2})} \rightarrow \uRep{U_{s=\exp\left(2 \pi i \frac{7}{10} \right)}(\SL{2})} // V_3$$ to $\rRep$.  Since $\otimes V_3$ acts freely on the isomorphism classes of simple objects and since every orbit contains exactly one object in $\rRep$ the restriction of this functor is an equivalence by Lemma \ref{lem:free-quotient}.
\end{proof}

We will need two similar results, for $\Rep U_{q=-\exp \left(-2 \pi i \frac{1}{10}\right)}(\SL{2} \oplus \SL{2})$ and for $\Rep{U_{q=-\exp \left(-\frac{2\pi i}{14}\right)}(\SL{4})}$.

In $\Rep U_{q=-\exp \left(-2 \pi i \frac{1}{10}\right)}(\SL{2} \oplus \SL{2})$ we can consider the root representations, those of the form $V_a \boxtimes V_b$ with both $a$ and $b$ even, as well as the vector representations, those $V_a \boxtimes V_b$ with $a+b$ even. We call these the vector representations because they become the vector representations under the identification $\SL{2} \directSum \SL{2} \iso \SO{4}$.

\begin{lem} \label{lem:KM-SO4}
\begin{align*}
& \rRep U_{q=-\exp \left(-2 \pi i \frac{1}{10}\right)}(\SL{2} \directSum \SL{2}) \\
& \qquad \cong \vRep U_{q=-\exp \left(-2 \pi i \frac{1}{10} \right)}(\SL{2} \directSum \SL{2}) // V_3 \boxtimes V_3  \\
& \qquad \cong \uRep{U_{s=\exp \left(2 \pi i \frac{7}{10} \right)}(\SL{2} \directSum \SL{2})}^{modularize}.
\end{align*}
\end{lem}
\begin{proof}
We make the abbreviations
\begin{align*}
\cR & = \rRep U_{q=-\exp \left(-2 \pi i \frac{1}{10}\right)}(\SL{2} \directSum \SL{2}) \\
\cV & = \vRep U_{q=-\exp \left(-2 \pi i \frac{1}{10} \right)}(\SL{2} \directSum \SL{2}) \\
\cU & = \uRep{U_{s=\exp \left(2 \pi i \frac{7}{10} \right)}(\SL{2} \directSum \SL{2})}.
\end{align*}
It is easy to check that $\cR$ and $\cV$ are not affected by either the choice of $s$ (recall in this situation $s$ is a square root of $q$, required for the definition of the braiding), or changing between the usual and the unimodal pivotal structures.
Thus we have inclusions $$\cR \subset \cV \subset \cU.$$
The invertible objects in $\cU$ are the representations  $V_0 \boxtimes V_0, V_0 \boxtimes V_3, V_3 \boxtimes V_0$ and $V_3 \boxtimes V_3$. For any choice of $s$ and either pivotal structure, $V_3 \boxtimes V_3$ is transparent. The representations $V_0 \boxtimes V_3$ and $V_3 \boxtimes V_0$ are transparent only with $s=\exp \left(2 \pi i \frac{7}{10} \right)$ and the unimodal pivotal structure. Under tensor product, the invertible objects form the group $\Integer/2 \Integer \times \Integer/2 \Integer$.
The invertible objects in $\cV$ are $V_0 \boxtimes V_0$ and $V_3 \boxtimes V_3$, forming the group $\Integer/2 \Integer$.

We have
\begin{align*}
(V_a \boxtimes V_b) \tensor (V_0 \boxtimes V_3) & \iso V_a \boxtimes V_{3-b} \\
(V_a \boxtimes V_b) \tensor (V_3 \boxtimes V_0) & \iso V_{3-a} \boxtimes V_b \\
(V_a \boxtimes V_b) \tensor (V_3 \boxtimes V_3) & \iso V_{3-a} \boxtimes V_{3-b},
\end{align*}
and so see that the action of the group of invertible objects is free.
Each $\Integer/2 \Integer \times \Integer/2 \Integer$ orbit on $\cU$ contains exactly one object from $\cR$, and each $\Integer/2 \Integer$ orbit on $\cV$ contains exactly one object from $\cR$. See Figure \ref{fig:SO4-quotients}.
\begin{figure}[!ht]
\begin{center}
\begin{tabular}{cc}
$\mathfig{0.35}{coincidences/so4-4fold-quotient}$ &
$\mathfig{0.35}{coincidences/so4-vector-quotient}$ \\
(a) & (b)
\end{tabular}
\end{center}
\caption{(a) The $4$-fold quotient of the Weyl alcove and (b) the $2$-fold quotient of the Weyl alcove, with vector representations marked. Lemma \ref{lem:KM-SO4} identifies the two resulting $4$-object categories.}
\label{fig:SO4-quotients}
\end{figure}

Thus both equivalences in this lemma are deequivariantizations, by applying Lemma \ref{lem:free-quotient} to the inclusions $\cR \subset \cU$ and $\cV \subset \cU$.
\end{proof}

In $\Rep{U_{q=-\exp \left(-\frac{2\pi i}{14}\right)}(\SL{4})}$ we can again consider two subcategories, the root representations and the vector representations. The root representations of $\SL{4}$ are those whose highest weight is an $\Natural$-linear combination of $(2,-1,0), (-1,2,-1), (0,-1,2)$ in $\Natural^3$. They form an index $4$ sublattice of the weight lattice. The Weyl alcove for $\SL{4}$ at a $14$-th root of unity consists of those weights $(a,b,c) \in \Natural^3$ with $a+b+c \leq 3$, and so the relevant root representations are $V_{(000)}, V_{(101)}, V_{(210)}, V_{(012)}$ and $V_{(020)}$. The vector representations $\vRep U_{q=-\exp \left(-\frac{2\pi i}{14}\right)}(\SL{4})$ are those that become vector representations under the identification $\SL{4} \iso \SO{6}$ (this is $A_3 = D_3$), namely those $V_{(abc)}$ with $a+c$ even. These form an index $2$ sublattice of the weight lattice, containing the root lattice. Both sublattices are illustrated in Figure \ref{fig:SL4-lattices}; hopefully having these diagrams in mind will ease later arguments.

\begin{figure}[!ht]
\begin{center}
%(a) $\mathfig{0.6}{coincidences/sl4-plain}$
\begin{tabular}{cc}
$\mathfig{0.45}{coincidences/sl4-vector-labels}$ &
$\mathfig{0.45}{coincidences/sl4-root-labels}$ \\
(a) & (b)
\end{tabular}
\end{center}
\caption{The $\SL{4}$ Weyl alcove at a $14$-th root of unity, showing (a) the vector representation sublattice and (b) the root representation sublattice.}
\label{fig:SL4-lattices}
\end{figure}

\begin{lem} \label{lem:KM-SO6}
\begin{align*}
\rRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) & \cong \vRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) // V_{(030)}  \\
	& \cong \uRep{U_{s=\exp \left(2 \pi i \frac{5}{14}\right)}(\SL{4})}^{modularize}.
\end{align*}
\end{lem}
\begin{proof}
We make the abbreviations
\begin{align*}
\cR & = \rRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) \\
\cV & = \vRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) \\
\cU & = \uRep{U_{s=\exp \left(2 \pi i \frac{5}{14}\right)}(\SL{4})}.
\end{align*}
It is easy to check that $\cR$ and $\cV$ are not affected by either the choice of $s$ (recall in this situation $s$ is a $4$-th root of $q$, required for the definition of the braiding), or any variation of pivotal structure. Thus we have inclusions $$\cR \subset \cV \subset \cU.$$
The invertible objects in $\cU$ are the representations  $V_{(000)}, V_{(300)}, V_{(030)}$ and $V_{(003)}$. For any choice of $s$ and pivotal structure, $V_{(030)}$ is transparent. The representations $V_{(300)}$ and $V_{(003)}$ are transparent only with $s=\exp \left(2 \pi i \frac{5}{14}\right)$ and the unimodal pivotal structure. Under tensor product, the invertible objects form the group $\Integer/4 \Integer$.
The invertible objects in $\cV$ are $V_{000}$ and $V_{030}$, forming the group $\Integer/2 \Integer$.

The action of the group of invertible objects is free, and shown in Figure \ref{fig:SL4-action}. Each $\Integer/4 \Integer$ orbit on $\cU$ contains exactly one object from $\cR$, and each $\Integer/2 \Integer$ orbit on $\cV$ contains exactly one object from $\cR$. See Figure \ref{fig:SL4-quotients}.

\begin{figure}[!ht]
\begin{tabular}{ccc}
$\mathfig{0.3}{coincidences/sl4-300}$ &
$\mathfig{0.3}{coincidences/sl4-030}$ &
$\mathfig{0.3}{coincidences/sl4-003}$
\\
(a) & (b) & (c)
\end{tabular}
\caption{The action of tensor product with an invertible object. (a) $- \tensor V_{(300)}$ and (c) $- \tensor V_{(003)}$ act by orientation reversing isometries, while (b) $- \tensor V_{(030)}$ acts by a $\pi$ rotation.}
\label{fig:SL4-action}
\end{figure}

\begin{figure}[!ht]
\begin{center}
\begin{tabular}{cc}
$\mathfig{0.45}{coincidences/sl4-4fold-quotient}$ &
$\mathfig{0.45}{coincidences/sl4-vector-quotient}$ \\
(a) & (b)
\end{tabular}
\end{center}
\caption{(a) The $4$-fold quotient of the Weyl alcove and (b) the $2$-fold quotient of the Weyl alcove, with vector representations marked.  Lemma \ref{lem:KM-SO6} identifies the two resulting $5$-object categories.}
\label{fig:SL4-quotients}
\end{figure}

Thus both equivalences in the Lemma are de-equivariantizations, by applying Lemma \ref{lem:free-quotient} to the inclusions $\cR \subset \cU$ and $\cV \subset \cU$.
\end{proof}


Finally, the usual statement in the literature of generalized Kirby-Melvin symmetry involves changing the label of only one component on the link.  This can be proved in a completely analogous way to the result above.  We recall the statement here.

\begin{thm}
\label{thm:KM}
Let $J_{A_1, \ldots, A_k}(L)$ be the value of a framed link $L$ (with components $L_1, \ldots, L_k$), 
labeled by simple objects $A_1, \ldots, A_k$.  Suppose now that $A_1$ is replaced by 
$A_1 \otimes X$ (with $X$ invertible).  Then 
$$J_{A_1 \otimes X, A_2, \ldots, A_k}(L) = 
	\dim{X}  \cdot c_X^{\operatorname{writhe}(L_1)} \cdot \prod_{i=1,\ldots,k} c_{A_i}^{\operatorname{linking}(L_1', L_i)} \cdot J_{A_1, \ldots, A_k}(L)
	$$
where $L_1'$ is a copy of $L_1$ running parallel to $L_1$ in the 
blackboard framing.  
\end{thm}

\subsection{Level-rank duality}
%Nice statements of Schur-Weyl duality for the orthogonal and symplectic groups can be found in \cite{0712.0944} and \cite{MR2342000} respectively, with proofs in the original papers of Brauer and Brown \cite{MR1503378,MR0072122,MR0075931}.
Level-rank duality is a collection of ideas saying that the semisimplified representation theory of a quantum group at a certain root of unity is related to that of a different quantum group, at a (potentially) different root of unity. The rank of a quantum group in this setting is dimension of its natural representation (i.e. the $n$ in $\SO{n}$ or $\SL{n}$).  The level describes the root of unity.  The name ``level" comes from the connection between quantum groups at roots of unity and projective representations of loop groups at a fixed level.  Here the relationship between the root of unity and the level is given  by the formula
$$k  = \frac{l}{2 D} - \check{h}$$
where $l$ is the order of the root of unity, $D$ is the lacing number of the quantum group, and $\check{h}$ is the dual Coxeter number. See Figure \ref{fig:quantum-group-data} for the values for each simple Lie algebra. Notice that not all roots of unity come from loop groups under this correspondence. 

\begin{figure}[!ht]
\begin{center}
\begin{tabular}{cccccc}
type & Lie group & rank  & lacing number $D$ & dual Coxeter number $\check{h}$ & L \\
\hline
$A_n$ & $\SL{n+1}$ & $n$ & $1$ & $n+1$ & $n+1$ \\
$B_{\text{$n$ even}}$ & $\SO{2n+1}$ & $n$ & $2$ & $2n-1$ & 1\\
$B_{\text{$n$ odd}}$ & $\SO{2n+1}$ & $n$ & $2$ & $2n-1$ & 2\\
$C_n$ & $\SP{2n}$ & $n$ & $2$ & $n+1$ & $1$ \\
$D_{\text{$n$ even}}$ & $\SO{2n}$ & $n$ & $1$ & $2n-2$ & 2 \\
$D_{\text{$n$ odd}}$ & $\SO{2n}$ & $n$ & $1$ & $2n-2$ & 4 \\
$E_n$ & $E_{6|7|8}$ & $6, 7, 8$ & $1$ & $12,18,30$ & $3,2,1$ \\
$F_4$ & $F_4$ & $4$ & $2$ & $9$ & $1$ \\
$G_2$ & $G_2$ & $2$ & $3$ & $4$ & $1$
\end{tabular}
\end{center}
\caption{Combinatorial data for the simple Lie algebras.}
\label{fig:quantum-group-data}
\end{figure}

Nonetheless there are versions of level-rank duality for quantum groups at roots of unity not corresponding to loop groups.  In this context what the ``level" measures is which quantum symmetrizers vanish, while the rank measures which quantum antisymmetrizers vanish.  At the level of combinatorics, the rank gives the bound on the number of rows in Young diagrams, while the level gives a bound on the number of columns, and duality is realized by reflecting Young diagrams thus interchanging the roles of rank and level.

We want statements of level-rank duality that give equivalences of braided tensor categories.  In order to get such precise statements several technicalities appear.  First, level-rank duality concerns $SO$, not $Spin$, so we only look at the vector representations.  Second, there is a subtle relationship between the roots of unity you need to pick on each side of the equivalence.  In particular, if the root of unity on the left side is of the form $\exp(\frac{2 \pi i}{m})$ then the root of unity on the right side typically will not be of that form.  Finally, level-rank duality is most natural as a statement about $U$ and $O$, not about $SU$ and $SO$.  Getting statements about $SU$ and $SO$ requires considering modularizations.  (It may seem surprising that this is even possible, since we know that $\Rep(U_{\zeta(\ell)}(\SO{n}))$ is already a modular tensor category \cite[Theorem 6]{MR2286123}. When we restrict to the subcategory of vector representations, however, we lose modularity.)

We found the papers  \cite{MR1710999} (on the $SU$ case) %Blanchet
and \cite{MR1854694} (on the $SO$ and $Sp$ cases) % Beliakova-Blanchet
to be exceedingly useful, and we'll give statements and proofs that closely follow their methods. %We'll restrict attention to the $SO$ cases of level-rank duality, although we're admittedly tempted to explain everything, since the full story is remarkably interesting and apparently obscure in the literature! Additionally, we'll pay particular attention below to the `low rank' cases, of $SO(3)$, $SO(4)$ and $SO(6)$, which do not fit immediately into the usual families of simple Lie algebras, although essentially no new ideas are required to treat these.
Level-rank duality for $SO(3)-SO(4)$ appears in the paper \cite{MR2469528}, where it is used to prove Tutte's golden identity for the chromatic polynomial.  For our particular case of level-rank duality involving $\mathfrak{so}_3$ and the $\mathcal{D}_{2m}$ subfactor planar algebra, see the more physically minded \cite{MR1279547}.
For some more background on level-rank duality, see \cite{MR1103065, MR1063959, MR1710999, MR2124554} for the $SU$ cases, \cite{MR1414470} for level-rank duality at the level of $3$-manifold invariants and \cite{MR1601870} for loop groups. 

%Before we begin, we'll explain precisely what we mean by $SO(n)_k$. This should be the subcategory of vector representations inside the semisimple representation category of $U_q(\SO{n})$ for some root of unity $q$. We take only the vector representations; the full representation category corresponds to $\operatorname{Spin}(n)$. Looking at the dual Coxeter numbers from Figure \ref{fig:quantum-group-data}, we expect to be working at a primitive $\ell_{n,k}$ root of unity, $q = \zeta(\ell_{n,k})$ where

%It's tempting to  define $$SO(n)_k = \vRep(U_{\zeta(\ell_{n,k})}(\SO{n})),$$ and then hope to prove that $SO(n)_k \iso SO(k)_n$.  However, it turns out this is insufficient, for two reasons. First, in order to get statements of the form $SO(n)_k \iso SO(k)_n$, at the level of braided tensor categories, we actually need to choose \textit{different} primitive roots on either side. That is, it's impossible to uniformly specify which primitive root of unity to use.  To get around this, we'll no longer write $SO(n)_k$, but spell out the particular specialization we intend in each case.
%Second, even if we choose roots of unity carefully, in general these categories are simply not equivalent, even at the level of the fusion ring! To correct this, we need to modularize both sides by a certain $\Integer/2\Integer$ subcategory. It may seem surprising that this is even possible, since we know that $\Rep(U_{\zeta(\ell)}(\SO{n}))$ is already a modular tensor category \cite[Theorem 6]{MR2286123}. When we restrict to the subcategory of vector representations, however, we lose modularity; $\vRep(U_{\zeta(\ell_{n,k})}(\SO{n}))$ contains a symmetric subcategory of transparent dimension $1$ objects, namely the trivial representation $V_0$ and $V_{k e_1}$, the representation with highest weight $k$ times that of the standard representation. 

As explained by Beliakova and Blanchet, level-rank duality is easiest to understand in a diagrammatic setting, where it says that $U(n)_k \iso U(k)_n$ and $O(n)_k \iso O(k)_n$, with $U$ and $O$ being interpreted as categories of tangles modulo either the HOMFLYPT or Dubrovnik relations. The equivalences come from almost trivial symmetries of the relations. %Making the connection with quantum groups introduces many difficulties; in particular instead of a relationship between $SU(n)_k$ and $SU(k)_n$, or between $SO(n)_k$ and $SO(k)_n$, we nearly always need to look at modularizations of these categories. Even then, the general case for arbitrary $n$ and $k$ with a common divisor is complicated. We give a precise statement for $SO(3)_k$-$SO(k)_3$ level-rank duality when $k$ is even, as Theorem \ref{thm:SO3-duality}.
The reason this modularization is necessary is that to recover $SO$ from $O$, we need to quotient out by the determinant representation. Thus, to translate an equivalence $O(n)_k \iso O(k)_n$ into something like $SO(n)_k \iso SO(k)_n$, we find that in each category there is the `shadow' of the determinant representation in the other category, which we still need to quotient out. See Figure \ref{fig:LR-schematic} for a schematic diagram illustrating this.

\begin{figure}[!ht]
$$\xymatrix{
&O(n)_k=O(k)_n \ar[ddl]_{\mathrm{det}_n \cong \id} \ar[ddr]^{\mathrm{det}_k \cong \id} & \\
&& \\
SO(n)_k \ar[ddr]_{\mathrm{det}_k \cong \id} && SO(k)_n \ar[ddl]^{\mathrm{det}_n \cong \id}\\
&&\\
&SO(n)_k^{\operatorname{mod}} \cong SO(k)_n^{\operatorname{mod}}&
}$$
\caption{A schematic description of $SO$ level-rank duality, suppressing the details of the actual roots of unity appearing.}
\label{fig:LR-schematic}
\end{figure}

Here is the precise statement of level-rank duality which we will be using.
Define \begin{equation*}
\ell_{n,k} = \begin{cases} 2 (n+k-2) & \text{if $n$ and $k$ are even} \\ 4(n+k-2) & \text{if $n$ is odd and $k$ is even}
\\ n+k-2 & \text{if $n$ is even and $k$ is odd}%\\ 2(n+k-2) & \text{if $n$ and $k$ are odd.}
\end{cases}
\end{equation*}

\begin{thm}[SO level-rank duality]
\label{thm:DLR}
Suppose $n, k \geq 3$ are not both odd. Suppose $q_1$ is a primitive root of unity with order $\ell_{n,k}$. Choose $q_2$ so that
\begin{equation*}
-1 = \begin{cases}
q_1 q_2 & \text{if $n$ and $k$ are both even} \\
q_1^2 q_2 & \text{if $n$ is odd and $k$ is even} \\
q_1 q_2^2 & \text{if $n$ is even and $k$ is odd} \\
%q_1^2 q_2^2 & \text{if $n$ and $k$ are both odd.}
\end{cases}
\end{equation*}
As ribbon categories, there is an equivalence
\begin{equation*}
\vRep(U_{q=q_1}(\SO{n}))//V_{k e_1} \iso \vRep(U_{q=q_2}(\SO{k}))//V_{n e_1}.
\end{equation*}
\end{thm}
\begin{rem}
When both $n$ and $k$ are odd, there is some form of level-rank duality in terms of the Dubrovnik skein relation,  pursued in  \cite{MR1854694} where it is called the $B^{n,-k}$ case.  However it does not seem possible to express this case purely in terms of quantum groups.  %This is essentially because when we look at an `allowed' root of unity on one side, we always find a `disallowed' root of unity appearing on the other.  It may be the case that this level-rank duality has a nice statement in terms of the orthosymplectic supergroup $OSP(1|2m)$.
\end{rem}

\begin{rem}
Notice that the order of $q_2$ is always $\ell_{k,n}$. When $n$ and $k$ are both even then the roots of unity on both sides come from loop groups.  However, when $n$ or $k$ is odd the roots of unity are not the ones coming from loop groups.
%\sout{On the left hand side, we have the simplest interpretation of $SO(n)_k$: $q_1$ being a root of unity of order $\ell_{n,k}$ corresponds exactly to `level $k$'. Notice that the order of $q_2$ is always $\ell_{k,n}$ \todo{double check}. Thus when $n$ and $k$ are both even, the right hand side can be interpreted as $SO(k)_n$, but only with a particular choice of root of unity relative to the choice of $q_1$. On the other hand, when either $n$ or $k$ is odd the root of unity on one side of the equivalence does not correspond to an integer level. In all cases, the value of $\ell'$ (in the notation of \cite{MR2286123}) on the right hand side is $n+k-2$. Thus when $k$ is even the wall of the Weyl alcove appearing on the right side is perpendicular to a long root (since the lacing number is $D=1$ so $D \mid \ell'$), while when $k$ is odd the wall is perpendicular to a short root (since then $D=2$ so $D \nmid \ell'$).|}
\end{rem}


%\begin{thm}[SO level-rank duality]
%\label{thm:DLR}
%Suppose $n, k \geq 3$ and $k$ is even. As ribbon categories,
%\begin{equation*}
%\vRep(U_{q=\zeta(\ell_{n,k})}(\SO{n}))//V_{k e_1} \iso \vRep(U_{q=-\zeta(\ell_{k,n})^{-1}}(\SO{k}))//V_{n e_1}.
%\end{equation*}
%\end{thm}
%\begin{rem}
%On the left hand side, we have the simplest interpretation of $SO(n)_k$.

%Since $k$ is even, the highest power of $2$ dividing $\ell_{k,n} = 2(n+k-2)$ is $1$ when $n$ is odd and at least $2$ when $n$ is even. Thus $-\zeta(\ell_{k,n})^{-1}$ has order $n+k-2$ when $n$ is odd, and order $2(n+k-2)$ when $n$ is even. So when $n$ is even the right hand side is a version of $SO(k)_n$ with a nonstandard primitive root of unity, while when $n$ is odd the right hand side is a stranger beast: it is a quotient of $SO(k)$ at a root of unity that does not correspond to an integer level.

%However, the root of unity at which we've specialized $SO(k)$ on the right hand side may not correspond to level $n$ at all. In particular, if  $\ell'$ is the order of $\zeta_{\ell_{k,n}}^{n+k-3}$, then we have
%\begin{equation*}
%\frac{\ell_{k,n}}{\ell'} = 
%\begin{cases}
%1 & \text{if $n = k \pmod{2}$,} \\
%2 & \text{if $n$ is odd and $k$ is even or if $n$ is even and $k = 1 \pmod{4}$, or} \\
%4 & \text{if $n$ is even and $k = 3 \pmod{4}$.}
%\end{cases}
%\end{equation*}
%\end{rem}

%Further,
%\begin{thm}
%\label{thm:negate-q}
%As pivotal monoidal categories,
%\begin{equation*}
%\vRep(U_q(\SO{2m})) \iso \vRep(U_{-q}(\SO{2m}))
%\end{equation*}
%via a map which sends the braiding on the standard representation to its negative (and thus this isn't an isomorphism of ribbon categories).
%\end{thm}

%Combining these, we can say
%\begin{cor}
%\label{cor:DLR}
%\begin{equation*}
%\vRep(U_{\zeta_{\ell_{n,k}}}(\SO{n}))//V_{k e_1} \iso
%	\begin{cases}
%		\vRep(U_{\zeta_{\ell_{k,n}}^{n+k-3}}(\SO{k}))//V_{n e_1} & \text{if $n = k \pmod{2}$} \\
%		\vRep(U_{\zeta_{\ell_{k,n}}^{-1}}(\SO{k}))//V_{n e_1} & \text{if $n \neq k \pmod{2}$}		
%	\end{cases}
%\end{equation*}
%\end{cor}
%Notice that on the right side the specialization is always a primitive $\ell_{k,n}$ root of unity, so it's reasonable to consider these isomorphisms as a realization of the ``$SO(n)_k \iso SO(k)_n$'' idea. Note however, that this isomorphism negates the braiding when $n$ and $k$ have different parities.

%In the remainder of the paper, we'll only need the cases $n=3$, $k=4, 6$ or $8$, so we're particularly interested in the case $n$ odd, $k$ even, where $\ell_{k,n}  = 2 \ell_{n,k}$, so we have
%\begin{cor}
%When $n$ is odd and $k$ is even, 
%\begin{equation*}
%\vRep(U_{\zeta_{\ell_{n,k}}}(\SO{n}))//V_{k e_1} \iso
%		\vRep(U_{\zeta_{\ell_{n,k}}^{-2}}(\SO{k}))//V_{n e_1}.	
%\end{equation*}
%\end{cor}

\begin{proof}
We begin by defining a diagrammatic category $\cO(t,w)$, and then seeing that a certain $\Integer/2\Integer \times \Integer/2\Integer$ quotient can be realised via two steps of deequivariantization in two different ways. In the first way, after the initial deequivariantization we obtain a category equivalent to $\vRep(U_{q_1}(\SO{n}))$, while in the second way we obtain a category equivalent to $\vRep(U_{q_2}(\SO{k}))$ instead. The second steps of deequivariantizations give the categories in the statement above; since both are the modular quotient of $\cO(t, w)$ for a certain $t$ and $w$, they are equivalent.

\begin{defn}
The category $\widetilde{\cO}(t,w)$ is the idempotent completion of the BMW category (the quotient of the tangle category by the Dubrovnik skein relations) with
\begin{align*}
a & = w^{t-1} \\
z & = w-w^{-1}.
\end{align*}
The category $\cO(t,w)$ is the quotient of $\widetilde{\cO}(t,w)$ by all negligible morphisms.
\end{defn}


Now define $w_{n,k}$ by
\begin{equation*}
w_{n,k}  = \begin{cases} q_1 & \text{if $n$ is even} \\
q_1^2 & \text {if $n$ is odd.}
\end{cases} 
\end{equation*}
Note the $w_{n,k}$ is a root of unity of order $2(n+k-2)$ when $k$ is even and of order $n+k-2$ with $k$ is odd. The hypotheses of the theorem then ensure that
\begin{equation*}
-w_{n,k}^{-1}  = \begin{cases}q_2 & \text{if $k$ is even} \\
q_2^2 & \text{if $k$ is odd.}\end{cases}
\end{equation*}

\begin{lem}
\label{lem:Oiso}
For $n,k \in \Natural$ and not both odd, the categories $\cO(n,w_{n,k})$ and $\cO(k,-w_{n,k}^{-1})$ are equivalent.
\end{lem}
\begin{proof}
In $\cO(k, -w_{n,k}^{-1})$, $z = -w_{n,k}^{-1}+w_{n,k}$, which is the same value of $z$ as appears in $\cO(n,w_{n,k})$. Similarly, in $\cO(k, -w_{n,k}^{-1})$, we have 
\begin{align*}
a & = (-w_{n,k}^{-1})^{k-1} \\
   & = \begin{cases}
   	 	-w_{n,k}^{1-k} = w_{n,k}^{n+k-2+1-k} & \text{if $k$ is even} \\
		w_{n,k}^{1-k} = w_{n,k}^{n+k-2+1-k} & \text{if $k$ is odd}
	 \end{cases} \\
   & = w_{n,k}^{n-1}
\end{align*}
and so the same values of $a$ appear in both categories; thus they actually have exactly the same definition.
\end{proof}

\begin{lem}
When $n \in \Natural$, the category $\cO(n,w)$ has a transparent object with quantum dimension $1$, which we'll call $\det{}_{n}$. Further, if $w=w_{n,k}$, there is another such object $\det{}_{k}$ coming from $\cO(k,-w^{-1})$ via the equivalence of the previous lemma. These transparent objects form the group $\Integer/2\Integer \times \Integer/2\Integer = \{\id, \det{}_{n}, \det{}_{k}, \det{}_{n} \tensor \det{}_{k}\}$ under tensor product.
\end{lem}
\begin{proof}
See \cite[Lemmas 4.1.ii and 4.3]{MR1854694}.
\end{proof}

%Now define
%\begin{equation*}
%q_{n,k} =
%\begin{cases}
%%	\zeta(2(n+k-2)) & \text{if $n$ is odd and $k$ is odd} \\
%	\zeta(4(n+k-2)) & \text{if $n$ is odd and $k$ is even} \\
%%	\zeta(n+k-2) & \text{if $n$ is even and $k$ is odd} \\
%	\zeta(2(n+k-2)) & \text{if $n$ is even and $k$ is even} \\
%\end{cases}
%\end{equation*}

Write $\ell(q)$ for the order of a root of unity $q$, and define 
\begin{equation*}
\ell'(q) = \begin{cases} \ell(q) & \text{if $2 \nmid \ell(q)$} \\ \ell(q)/2 & \text{if $2 \mid \ell(q)$.}\end{cases}
\end{equation*}

\begin{lem}
We can identify the deequivariantizations as
\begin{align*}
\vRep(U_{q}(\SO{n})) & \iso
	\begin{cases}
		\cO(n,q)//\det{}_{n} & \text{if $n$ is even} \\
		\cO(n,q^2)//\det{}_{n} & \text{if $n$ is odd}
	\end{cases}
\end{align*}
for any $q$, as long as if $q$ is a root of unity, when $n$ is even, $\ell'(q) \geq n-2$, and when $n$ is odd, $\ell'(q) \geq 2(n-2)$ when $2 \mid \ell'(q)$ and $\ell'(q) > n-1$ when $2 \nmid \ell'(q)$.

In particular when $q=q_1$ we obtain
\begin{align*}
\cO(n,w_{n,k})//\det{}_{n} & \iso \vRep(U_{q_1}(\SO{n})) \displaybreak[1] \\
\intertext{and further,}
\cO(n,w_{n,k})//\det{}_{k} & \iso	\vRep(U_{q_2}(\SO{k})) \end{align*}
Moreover, in $\cO(n,w_{n,k})//\det{}_{n}$, we have $\det{}_{k} \iso V_{k e_1}$ and in $\cO(n,w_{n,k})//\det{}_{k}$, we have $\det{}_{n} \iso V_{n e_1}$.
\end{lem}
\begin{proof}
The first equivalence follows from the main results of \cite{MR2132671}. We give a quick sketch of their argument.  The fact that the eigenvalues of the $R$-matrix acting on the standard representation of $\SO{n}$ are $q^{-2n+2}, -q^{-2}$ and $q^2$ when $n$ is odd, or $q^{-n+1}, -q^{-1}$ and $q$ when $n$ is even ensures that this is a functor from $\widetilde{\cO}(n,q^2)$ or $\widetilde{\cO}(n,q)$, by Theorems \ref{thm:summands}, \ref{thm:eigenvalues} and \S \ref{sec:ribbonfunctors}.  One then checks that the functor factors through the deequivariantization.  Finally, by computing dimensions of Hom-spaces one concludes that the functor must kill all negligibles and must be surjective.

One can check that $\ell'(q_1) = n+k-2$ when $n$ is even or $2(n+k-2)$ when $n$ is odd, and so the required inequalities always hold for $\SO{n}$. 

The last equivalence follows from the first and Lemma \ref{lem:Oiso},:
\begin{align*}
\cO(n,w_{n,k})//\det{}_{k} & \iso \cO(k,-w_{n,k}^{-1})//\det{}_{k} \displaybreak[1] \\
					& \iso \vRep(U_{q_2}(\SO{k})) \displaybreak[1]
\end{align*}
Here we check that
$\ell'(q_2)= n+k-2$ when $k$ is even or $2(n+k-2)$ when $k$ is odd, satisfying the inequalities for $\SO{k}$.
\end{proof}

The proof of the theorem is now immediate; we write $\cO(n,w_{n,k})//\{\det{}_n,\det{}_k\}$ in two different ways, obtaining
\begin{align*}
\cO(n,w_{n,k})//\{\det{}_n,\det{}_k\} & = \cO(n,w_{n,k})// \det{}_n //\det{}_k \\
						  & \iso 	\vRep(U_{q_1}(\SO{n})) // V_{k e_1} \displaybreak[1] \\
\intertext{and}
\cO(n,w_{n,k})//\{\det{}_n,\det{}_k\} & = \cO(n,w_{n,k})// \det{}_k //\det{}_n \displaybreak[1] \\
						  & \iso \vRep(U_{q_2}(\SO{k})) // V_{n e_1}. \qedhere
\end{align*}
\end{proof}
\begin{rem}
One can easily verify an essential condition, that the twist factor for $V_{ke_1}$ inside $\vRep(U_{\zeta(\ell_{n,k})}(\SO{n}))$ is $+1$, from the formulas for the twist factor given in \S \ref{sec:quantumgroups}.
\end{rem}


Finally, we specialize to the case $n=3$, where the $\mathcal{D}_{2m}$ planar algebras appear. 

\begin{thm}[$SO(3)$-$SO(k)$ level-rank duality]
\label{thm:SO3-duality}
Suppose $k \geq 4$ is even. There is an equivalence of ribbon categories
\begin{equation*}
\frac{1}{2} \mathcal{D}_{k+2} \iso \vRep(U_{q=-\exp\left(-\frac{2 \pi i}{2k+2}\right)}(\SO{k}))//V_{3 e_1}
\end{equation*}
sending the tensor generator $W_2$ of $\frac{1}{2} \mathcal{D}_{k+2}$ to $V_{2 e_1}$ and $P$ to $V_{2e_{\frac{k}{2}-1}}$.
\end{thm}

This follows immediately, from the description in \S \ref{sec:d2n} of the even part of $\mathcal{D}_{2n}$ as $\frac{1}{2}\mathcal{D}_{2n} \cong \vRep(U_{q=\exp(\frac{2\pi i}{8n-4})}(\SO{3})^{modularize}$, and the general case of level-rank duality.

\subsection{Applications}
\label{sec:applications}%

\subsubsection{Knot invariants}
Combining $SO(3)$-$SO(k)$ level-rank duality for even $k \geq 8$ with Kirby-Melvin symmetry, we obtain the following knot polynomial identities:
\begin{thm}[Identities for $n \geq 3$]
\label{thm:identities-geq}
For all knots $K$,
\begin{align}
\restrict{\J{\SL{2}}{(2n-2)}(K)}{q=\exp(\frac{2\pi i}{8n-4})} & = 2 \J{\mathcal{D}_{2n}}{P}(K) \notag \displaybreak[1]  \\
                                                   & = 2 \restrict{\J{\SO{2n-2}}{2e_{n-2}}(K)}{q=-\exp(-\frac{2\pi i}{4n-2})} \notag \displaybreak[1] \\
                                                   & = (-1)^{1+\ceil{\frac{n}{2}}} 2 \restrict{\J{\SO{2n-2}}{e_{n-2}}(K)}{q=-\exp(-\frac{2\pi i}{4n-2})}
                                                   \label{eq:identities-geq-5} \displaybreak[1] \\
\intertext{and for all links $L$,}                                               
\restrict{\J{\SL{2}}{(2)}(L)}{q=\exp(\frac{2\pi i}{8n-4})}    & = \J{\mathcal{D}_{2n}}{W_2}(L) \notag \displaybreak[1] \\
                                                   & = \restrict{\J{\SO{2n-2}}{2e_1}(L)}{q=-\exp(-\frac{2\pi i}{4n-2})} \notag \displaybreak[1] \\
                                                   & =  \restrict{\J{\SO{2n-2}}{e_1}(L)}{q=-\exp(-\frac{2\pi i}{4n-2})}. 
\end{align}
(The representation of $\SO{2n-2}$ with highest weight $e_{n-1}$ is one of the spinor representations.)
%\todo{The $\pm$ above might come from linking numbers in Kirby-Melvin symmetry. It should be $+1$ for knots.}
\end{thm}

\begin{proof}
The first two identities are immediate applications of Theorems \ref{thm:half} and \ref{thm:SO3-duality}. For the next identity, we use the statement of Kirby-Melvin symmetry in Theorem \ref{thm:ourKM}, with $A = V_{2e_{n-2}}$ and $X = V_{3e_{n-2}}$. %Using the quantum Racah rule, we compute that $X^* = X$ when $n$ is odd, while $X^* =  V_{3e_{n-1}}$ when $n$ is even. In either case, 
We calculate that $\dim{X}=(-1)^{1+\ceil{\frac{n}{2}}}$ by the following trick.  At $q=\exp(\frac{2\pi i}{4n-2})$, this dimension must be $+1$, since it is the dimension of an invertible object in a unitary tensor category.  At $q=\exp(-\frac{2\pi i}{4n-2})$ it is the same, since quantum dimensions are invariant under $q \mapsto q^{-1}$, and finally we can calculate the sign at $q=-\exp(-\frac{2\pi i}{4n-2})$ by checking the parity of the exponents in the Weyl dimension formula. This implies that $X \tensor X^* \iso V_0$.  Using the Racah rule, we find $A \tensor X =  V_{e_{n-2}}$. 
%The twist factor for $X$ is 
%\begin{align*}
%c_0 & = q^{\frac{3}{4}(n-1)(2n-1)} \\
%      & = \exp\left(2\pi i \frac{2n-2}{4n-2} \frac{3}{4} (n-1)(2n-1) \right) \\ 
%       & = \exp\left(2\pi i \frac{3}{4} (n-1)^2 \right) \\
%       & = \begin{cases} 1 & \text{if $n$ is odd, and} \\ -i & \text{if $n$ is even.}\end{cases}
%\end{align*}
%The twist factor for $A_1 \tensor X$ is $c_1 = q^{\frac{1}{4}(n-1)(2n-3)} = \exp\left(2\pi i \frac{n-1}{2n-1} \frac{1}{4} (n-1)(2n-3) \right)$ which is just yucky; presumably I'm misunderstanding how this works --- presumably this factor magically goes away in the framing-independent case.

Now we do the same computation again with $A = V_{2e_1}$ and $X=V_{3e_1}$.  This case is simpler since $\dim V_{3e_1} = 1$.

\end{proof}

\begin{rem}
We found keeping all the details of this theorem straight very difficult, and we'd encourage you to wonder if we eventually got it right. We had some help, however, in the form of computer computations. You too can readily check the details of this theorem on small knots and links, assuming you have access to \MMA. Download and install the \package{KnotTheory} package from \url{http://katlas.org}. This includes with it the \package{QuantumGroups} package written by Morrison, which, although rather poorly documented, provides the function \code{QuantumKnotInvariant}. This function can in principle compute any knot invariant coming from an irreducible  representation of a quantum group, but in practice runs into time and memory constraints quickly.  The explicit commands for checking small cases of the above theorem are included as a \MMA \ notebook \code{aux/check.nb} with the \code{arXiv} source of this paper.

\noop{
\begin{mma}
\begin{inm}<<KnotTheory`\end{inm}
\begin{printm}
Loading KnotTheory` version of January 20, 2009. \\
Read more at \url{http://katlas.org/wiki/KnotTheory}.
\end{printm}
\begin{inm}
\\
CheckIdentity1[$n\_$][$K\_$] := \\
\quad Chop[N[ \\
\quad\quad    QuantumKnotInvariant[$A_1$, Irrep[$A_1$][$\{2n-2\}$]] \\
\quad\quad\quad [$K$][Exp[($2\pi I$)/($8n-4$)]] \\
\quad\quad    - 2 $(-1)^{1+\operatorname{Ceiling}[n/2]}$ QuantumKnotInvariant[$D_{n-1}$, Irrep[$D_{n-1}$][UnitVector[$n-1$,$n-1$]]] \\
\quad\quad\quad [$K$][-Exp[-(($2\pi I$)/($4n-2$))]] \\
\quad ]]
\end{inm}
\begin{inm}
CheckIdentity1[$5$] /@ {Knot[$0$,$1$], Knot[$3$,$1$], Knot[$5$,$1$]}
\end{inm}
\begin{outm}
$\{0,0,0\}$
\end{outm}
\begin{inm}
Table[CheckIdentity1[$n$][Knot[$5$,$2$]], $\{n, 5, 6\}$]
\end{inm}
\begin{outm}
$\{0,0\}$
\end{outm}
\begin{inm}
CheckIdentity1[$7$][Knot[$3$,$1$]]
\end{inm}
\begin{outm}
$0$
\end{outm}
\begin{inm}
\\
CheckIdentity2[$n\_$][$K\_$] := \\
\quad Chop[N[ \\
\quad\quad    QuantumKnotInvariant[$A_1$, Irrep[$A_1$][$\{2\}$]] \\
\quad\quad\quad [$K$][Exp[($2\pi I$)/($8n-4$)]] \\
\quad\quad    - QuantumKnotInvariant[$D_{n-1}$, Irrep[$D_{n-1}$][UnitVector[$n-1$,$1$]]] \\
\quad\quad\quad [$K$][-Exp[-(($2\pi I$)/($4n-2$))]] \\
\quad ]]
\end{inm}
\begin{inm}
CheckIdentity2[$5$] /@ {Knot[$0$,$1$], Knot[$3$,$1$], Knot[$5$,$1$]}
\end{inm}
\begin{outm}
$\{0,0,0\}$
\end{outm}
\begin{inm}
Table[CheckIdentity1[$n$][Knot[$5$,$2$]], $\{n, 5, 6\}$]
\end{inm}
\begin{outm}
$\{0,0\}$
\end{outm}
\end{mma}
}
\end{rem}

Note that the $n=5$ case of Equation \eqref{eq:identities-geq-5} in Theorem \ref{thm:identities-geq} reproduces the statement of Theorem \ref{thm:identities-5}.

The $n=3$ and $n=4$ cases of Theorem \ref{thm:identities-geq} also reproduce previous results. 
The Lie algebra $\SO{2n-2}$ has Dynkin diagram $D_{n-1}$, with the spinor representations
corresponding to the two extreme vertices. At $n=4$, $D_{n-1}$ becomes
the Dynkin diagram $A_3$, and the spinor representations become the
standard and dual representations; this explains Theorem
\ref{thm:identities-4}. See Theorem \ref{thm:D8-coincidence} and Figure \ref{fig:SO6} for a full explanation. 

At $n=3$, $D_{n-1}$ becomes $A_1 \times A_1$, and
the spinor representations become $(\text{standard}) \boxtimes (\text{trivial})$
and $(\text{trivial}) \boxtimes (\text{standard})$, giving the case
described in Theorem \ref{thm:identities-3}. See Theorem \ref{thm:D6-coincidence} and Figure \ref{fig:SO4} for a full explanation.

\subsubsection{Coincidences}

\begin{proof}[Proof of Theorem \ref{thm:D6-coincidence}]
We want to construct an equivalence $$\frac{1}{2} \mathcal{D}_{6} \cong \uRep{U_{s = \exp\left(\frac{7}{10} 2 \pi i\right)}(\SL{2} \oplus \SL{2})}^{modularize},$$ sending $P \mapsto V_{(1)} \boxtimes V_{(0)}$.

First, we recall that the $k=4$ case of $SO(3)$ level-rank duality (Theorem \ref{thm:SO3-duality}) gave us the equivalence  
\begin{align*}
\frac{1}{2} \mathcal{D}_{6} & \iso \vRep U_{q=\exp( \frac{2 \pi i}{20})} (\SO{3})^{modularize} \\
& \iso \vRep U_{q=-\exp \left( -\frac{2 \pi i}{10}\right)}(\SO{4}) // V_{3e_1} 
\end{align*}
Since the Dynkin diagrams $D_2$  and $A_1 \times A_1$ 
coincide we can replace $\SO{4}$ by $\SL{2} \oplus \SL{2}$. The representation $V_{3e_1}$ of $\SO{4}$ is sent to $V_3 \boxtimes V_3$, so we have 
\begin{align*}
\frac{1}{2} \mathcal{D}_{6} & \iso \vRep U_{q=-\exp \left( -\frac{2\pi i}{10} \right)}(\SL{2} \oplus \SL{2}) // V_3 \boxtimes V_3
\end{align*}
Next, by Lemma \ref{lem:KM-SO4} we can replace this $2$-fold quotient of the vector representations with a $4$-fold quotient of the entire representations category of $\SL{2} \oplus \SL{2}$, as long as we carefully choose $s$ and the unimodal pivotal structure. Figure \ref{fig:SO4} shows the identification between the objects of $\frac{1}{2} \mathcal{D}_6$ and the corresponding objects in the $4$-fold quotient.
\end{proof}
\begin{figure}[!ht]
$$
\mathfig{0.5}{coincidences/SO4-chamber}
$$
\caption{
The simple objects of $\frac{1}{2} \mathcal{D}_6$ may be identified with the representatives of the vector representations in the quotient $\Rep U_{q=-\exp \left( -\frac{2 \pi i}{10}\right)}(\SO{4}) // V_{3e_1}$, via level-rank duality. We can replace $\SO{4}$ here with $\SL{2} \directSum \SL{2}$. The object $V_{3e_1}$ becomes $V_3 \boxtimes V_3$. The circles above indicate the vector representations, labelled by the corresponding objects of $\frac{1}{2}\mathcal{D}_6$. Next we can apply Lemma \ref{lem:KM-SO4} to realize $\frac{1}{2} \mathcal{D}_6$ as the modularization of $\uRep{U_{s = \exp\left(\frac{7}{10} 2 \pi i\right)}(\SL{2} \oplus \SL{2})}$. In this modularization, we quotient out the four corner vertices. Note that $P$ is sent to $V_1 \boxtimes V_0$, and in particular the knot invariant coming from $P$ recovers a specialization of the Jones polynomial.
}
\label{fig:SO4}
\end{figure}
\begin{rem}
Theorem \ref{thm:identities-3} is now an immediate corollary.
\end{rem}


\begin{rem}
The coincidence of Dynkin diagrams $D_2 = A_1 \times A_1$ also implies that the $D_2$ specialization of the Dubrovnik polynomial is equal to the square of the Jones polynomial:
$$\Dubrovnik(K)(q^3,q-q^{-1}) = J(K)(q)^2.$$
This was proved by Lickorish \cite[Theorem 3]{MR966143}, without using quantum groups.
\end{rem}

\begin{cor}
Looking at the object $\JW{2} \in \frac{1}{2} \mathcal{D}_6$, we have
\begin{equation*}
\restrict{\J{\SL{2}}{(2)}(K)}{q=\exp(\frac{2\pi i}{20})}  = \restrict{\J{\SL{2}}{(1)}(K)^2}{q=\exp(-\frac{2\pi i}{10})}.
\end{equation*}
This identity is closely related to Tutte's golden identity, c.f. \cite{MR2469528}.
\end{cor}

\begin{proof}[Proof of Theorem \ref{thm:D8-coincidence}]
We want to construct an equivalence
\begin{align*}
\frac{1}{2} \mathcal{D}_{8} & \cong \vRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SO{6}) // (V_{3e_1}) \\
			   & \cong \uRep{U_{s=\exp \left(\frac{5}{14} 2 \pi i\right)}(\SL{4})}^{modularize},
\end{align*}
sending $P$ to a spinor representation of $\SO{6}$ and to $V_{(100)}$, the standard representation of $\SL{4}$.

The first step is the $k=6$ special case of Theorem \ref{thm:SO3-duality} on level-rank duality. The second step uses the coincidence of Dynkin diagrams $D_3 = A_3$ to obtain
$$\vRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SO{6} // (V_{3e_1}) \cong
	\vRep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) // V_{(030)}$$
after which Lemma \ref{lem:KM-SO6} gives the desired result.
For more details see Figure \ref{fig:SO6}.
\end{proof}
\begin{rem}
Theorem \ref{thm:identities-4} is now an immediate corollary.
\end{rem}

\begin{figure}[!pht] 
\begin{equation*}
\mathfig{0.8}{coincidences/sl4-modularisation-domain}
\end{equation*}

\caption{
We can realise $\frac{1}{2} \mathcal{D}_{8}$ as the vector representations in the $2$-fold quotient $\Rep U_{q=-\exp \left(-\frac{2 \pi i}{14} \right)}(\SL{4}) // (V_{(030)})$, via level-rank duality and the $A_3 = D_3$ coincidence of Dynkin diagrams. The figure shows a fundamental domain for the $2$-fold quotient. The objects of $\frac{1}{2} \mathcal{D}_{8}$ are shown circled (with fainter circles in the other domain showing their other representatives). Now we can apply Lemma \ref{lem:KM-SO6}, and instead identify these vector representations with representations in the $4$-fold quotient $\uRep{U_{s=\exp \left(2 \pi i \frac{5}{14}\right)}(\SL{4})}^{modularize}$ of the unimodal representation theory of $\SL{4}$, at a particular choice of $s$. These identifications are shown as arrows. Note that $P$ is sent to $V_{(100)}$, the standard representation of $\SL{4}$. In particular, the knot invariant coming from $P$ matches up with a specialization of the HOMFLYPT polynomial.
}
\label{fig:SO6}
\end{figure}


\begin{proof}[Proof of Theorem \ref{thm:D10-coincidence}]
We want to show that $\frac{1}{2} \mathcal{D}_{10}$ has an order $3$ automorphism: $$\baselinetrick{\xymatrix@C-8mm@R-3mm{P \ar@{|->}@/^/[rr] & & Q \ar@{|->}@/^/[dl] \\ & f^{(2)} \ar@{|->}@/^/[ul]&}}.$$

\newcommand{\isoto}{\xrightarrow{\iso}}
Again, we first apply the $k=8$ special case of level-rank duality (Theorem \ref{thm:SO3-duality})  to see there is a functor $$\mathcal{L} :\frac{1}{2}\mathcal{D}_{10} \isoto \vRep U_{q=-\exp\left(-\frac{2\pi i}{18}\right)}(\SO{8}) // V_{(3000)}$$ with $\JW{2}$ corresponding to $V_{(1000)}$ and $P$ to $V_{(0002)}$.
In an exactly analogous manner as in Lemmas \ref{lem:KM-SO4} and \ref{lem:KM-SO6}, we can identify this two-fold quotient of the vector representations of $\SO{8}$ with a four-fold quotient of all the representations in the Weyl alcove. That is, there is a functor
\begin{multline*} \mathcal{K} : \vRep U_{q=-\exp\left(-\frac{2\pi i}{18}\right)}(\SO{8}) // V_{(3000)} \isoto \\
    \uRep U_{q=-\exp\left(-\frac{2\pi i}{18}\right)}(\SO{8}) // (V_{(3000)}, V_{(0030)}, V_{(0003)}).
\end{multline*}

The triality automorphism of the Dynkin diagram $D_4$ gives an automorphism $T$ of this category.  A direct computation shows that $T$ induces a cyclic permutation of  $P$, $Q$, and $\JW{2}$ in $\frac{1}{2} \mathcal{D}_{10}$.
For example,
\begin{align*}
\mathcal{L}^{-1}(\mathcal{K}^{-1}(T(\mathcal{K}(\mathcal{L}(\JW{2}))))) & = \mathcal{L}^{-1}(\mathcal{K}^{-1}(T(V_{(1000)}))) \\
	        & = \mathcal{L}^{-1}(\mathcal{K}^{-1}(V_{(0001)})) \\
	        & = \mathcal{L}^{-1}(\mathcal{K}^{-1}(V_{(0001)} \tensor V_{(0003)})) \\
	        & = \mathcal{L}^{-1}(\mathcal{K}^{-1}(V_{(0002)})) \\
	        & = \mathcal{L}^{-1}(V_{(0002)}) \\
	        & = P
\end{align*}

See Figures \ref{fig:SO8} and \ref{fig:SO8-4fold} for more details. It may be that this automorphism of $\frac{1}{2}\mathcal{D}_{10}$ is related to the exceptional modular invariant associated to $\SL{2}$ at level $16$ described in \cite{MR1105431}.
\end{proof}

\begin{figure}[!pht]
\begin{equation*}
\mathfig{0.8}{coincidences/SO8-chamber}
\end{equation*}
\caption{ We can realise $\frac{1}{2} \mathcal{D}_{10}$ as the vector representations in the $2$-fold quotient $\vRep U_{q=-\exp \left(-\frac{2 \pi i}{18} \right)}(\SO{8}) // (V_{(3000)})$, via level-rank duality.
The Weyl alcove for $\SO{8}$ at $q=-\exp\left(-\frac{2\pi i}{18}\right)$ consists of those $V_{(abcd)}$ such that $a+2b+c+d \leq 3$.  In particular, $b=0$ or $b=1$.  So we draw this alcove as two tetrahedra, the $V_{\star 0 \star \star}$ tetrahedron, and the $V_{\star1 \star \star}$ tetrahedron.  The vector representations are those $V_{(abcd)}$ with $c+d$ even.
We show a fundamental domain for the modularization involution $\tensor V_{(3000)}$, which acts on the $V_{\star0 \star \star}$ tetrahedron by $\pi$ rotation about the line joining $\frac{3}{2}000$ and $00\frac{3}{2}\frac{3}{2}$ and on the $V_{\star1 \star \star}$ tetrahedron by $\pi$ rotation about the line joining $\frac{1}{2}100$ and $01\frac{1}{2}\frac{1}{2}$.
The tensor category of $\frac{1}{2}\mathcal{D}_{10}$ is equivalent to the tensor subcategory of this modularization consisting of images of vector representations, with the equivalence sending $\JW{0} \mapsto V_{(0000)}$, $\JW{2} \mapsto V_{(1000)}$, $\JW{4} \mapsto V_{(0011)}$, $\JW{6} \mapsto V_{(0100)}$,  $P \mapsto V_{(0020)}$ and $Q \mapsto V_{(1020)}$.
The blue arrows shows the action of the triality automorphism $V_{(abcd)} \mapsto V_{(cbda)}$ for $\SO{8}$ on the image of $\frac{1}{2}\mathcal{D}_{10}$. This action is computed via the equivalence with the $4$-fold quotient of all representations, shown in Figure \ref{fig:SO8-4fold}. Notice that under this automorphism $P$ is sent to the standard representation.  In particular, the knot invariant coming from $P$ matches up with a specialization of the Kauffman polynomial.
} \label{fig:SO8}
\end{figure}

\begin{figure}[!htb]
\begin{equation*}
\mathfig{0.6}{coincidences/SO8-4fold-quotient}
\end{equation*}
\caption{The action of the $D_4$ triality automorphism on the four-fold quotient
$\Rep U_{q=-\exp \left(-\frac{2 \pi i}{18} \right)}(\SO{8}) // (V_{(3000)}, V_{(0030)}, V_{(0003)}).$}
\label{fig:SO8-4fold}
\end{figure}


%Our last application combines level-rank duality with triality.

%\begin{thm}
%$\J{\mathcal{D}_{10}}{P}(K) = \J{\mathcal{D}_{10}}{W_2}(K)$
%\end{thm}
%\begin{proof}
%Use the $n=5$ case of Theorem \ref{thm:identities-geq} and triality for $SO(8)$:
%\begin{align*}
%\J{\mathcal{D}_{10}}{P}(K) & = \restrict{\J{SO(8)}{e_4}(K)}{q=\exp(\frac{2\pi i}{18})} \\
%                             & = \restrict{\J{SO(8)}{e_1}(K)}{q=\exp(\frac{2\pi i}{18})} \\
%                            & = \J{\mathcal{D}_{10}}{W_2}(K) \qedhere
%\end{align*}
%\end{proof}
