%!TEX root = ../article.tex

The theorems of this section describe how to identify an invariant coming from an object in a ribbon category as a specialization of the Jones, HOMFLYPT or Kauffman polynomials. These theorems are  well-known to the experts, and versions of them can be found in \cite{MR958895, MR1237835,MR2132671}.  Since we need explicit formulas for which specializations appear we collect the proofs here.

We then identify cases in which these theorems apply, namely $\mathcal{D}_{2n}$ for $n=2,3,4$ and $5$, and explain exactly which specializations occur.

There is a similar procedure, due to Kuperberg, for recognizing knot invariants which are specializations of the $G_2$ knot polynomial.  We apply this technique to $\mathcal{D}_{14}$.  %This proves a conjecture of Rowell's that $G_2$ at level $1/3$ is unitary \cite{MR2414692}.

The identities in this section do not follow from the knot polynomial identities in \cite{MR1134131} \cite[\S 6, Table 2]{MR980759}. (But most of those identities follow from the technique outlined in this section.)

\subsection{Recognizing a specialization of Jones, HOMFLYPT, or Kauffman}
Identifying a knot invariant as a specialization of a classical knot polynomial happens in two steps. Let's say you're looking at the knot invariant coming from an object $V$ in a ribbon category. First, you look at the direct sum decomposition of $V \tensor V$, and hope that you don't see too many summands. Theorem \ref{thm:summands} below describes how to interpret this decomposition, hopefully guaranteeing that the invariant is either trivial, or a specialization of Jones, HOMFLYPT, or Kauffman. If this proves successful, you next look at the eigenvalues of the braiding on the summands of $V \tensor V$. Theorem \ref{thm:eigenvalues} then tells you exactly which specialization you have.

\begin{thm}
\label{thm:summands}
Suppose that $V$ is a simple object in a ribbon category $\cC$ and that if $V$ is self-dual then it is symmetrically self-dual.
\begin{enumerate}
\item If $V \otimes V$ is simple, then $\dim{V}=\pm1$ and the link invariant $\J{\cC}{V} =(\dim V)^\#$ where $\#$ is the number of components of the link.
\item If $V \otimes V = \id \oplus L$ for some simple object $L$, then the link invariant $\J{\cC}{V}$ is a specialization of the Jones polynomial.
\item If $V \otimes V = L \oplus M$ for some simple objects $L$ and $M$, then the link invariant $\J{\cC}{V}$ is a specialization of HOMFLYPT.
\item If $V \otimes V = \id \oplus L \oplus M$ for some simple objects $L$ and $M$, then the link invariant $\J{\cC}{V}$ is a specialization of either the Kauffman polynomial or the Dubrovnik polynomial.
\end{enumerate}
\end{thm}


\begin{proof}
    \begin{enumerate}
    \item {\bf Trivial case} 
    
    Since the category is spherical and braided, $\End{V \otimes V} \cong \End{V \otimes V^*}$.  Hence if $V \otimes V$ is simple we must have $V \otimes V^* = \id$, so $\dim V = \pm 1$.
    Also by simplicity, $\End{V \otimes V}$ is one dimensional, and
    so, up to constants,  a crossing is equal to the identity map. Suppose now that
    $\inputtikz{Oovercrossing}=\alpha \inputtikz{Oidentity}$.  Then, Reidemeister two tells us that
    $\inputtikz{Oundercrossing}=\alpha^{-1} \inputtikz{Oidentity}$.  Capping this off shows that the twist factor is $\alpha \dim V$.  Thus the framing corrected skein relation is
    $$\inputtikz{Oovercrossing} = \dim V \inputtikz{Oidentity} = \inputtikz{Oundercrossing}.$$
    
    The equality of the two crossings lets us unlink any link, showing that the framing corrected invariant is $(\dim V)^\#$, where $\#$ is the number of components. % Hence the uncorrected invariant is $\alpha^{\text{writhe}} (\dim V)^{\#}$ where $\alpha$ is a square root of $\dim V$.

\item {\bf Jones polynomial case}

Since $\End{V\otimes V}$ is $2$-dimensional there must be a linear dependence between the crossing and the two basis diagrams of Temperley-Lieb.  (If these two Temperley-Lieb diagrams were linearly dependent, then $V\otimes V \cong \id$, contradicting the assumption).  Hence we must have a relation 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 this invariant is given by the Kauffman bracket.

\item {\bf HOMFLYPT case}

Since $\End{V\otimes V}$ is $2$-dimensional there must be a linear dependence between the two crossings and the identity (we can't use the cup-cap diagram here because $V$ is not self dual).  Hence, we have that
$$\alpha \inputtikz{Oovercrossing} + \beta \inputtikz{Oundercrossing} = \gamma \inputtikz{Oidentity}$$ for some $\alpha$, $\beta$, and $\gamma$.  If $\alpha$ or $\beta$ were zero, $\End{V\otimes V}$ would be $1$-dimensional, so we must have that $\alpha$ and $\beta$ are nonzero.  Hence we can rescale the relation so that $\alpha = w$, $\beta = -w^{-1}$, and $\gamma = z$.  Since the twist is some multiple of the single strand we can define $a$ such that the twist factor is $w^{-1}a$.  Thus we've recovered the framed HOMFLYPT skein relations.

\item {\bf Kauffman case}

Since $V \tensor V$ has three simple summands, its endomorphism space is $3$ dimensional. Moreover, since one of the summands is the trivial representation, one such endomorphism is the `cup-cap' diagram $\cupcap$. There must be some linear relation of the form
\begin{equation*}
p \overcrossing + q \undercrossing + r \identity + s \cupcap = 0.
\end{equation*}
The space of such relations is invariant under a $\pi/2$ rotation, and fixed under a $\pi$ rotation, so there must be a linear relation which is either a $(+1)$- or $(-1)$-eigenvector of the $\pi/2$ rotation. That is, there must be a relation of the form
\begin{equation*}
A\left( \overcrossing \pm \undercrossing\right)  = B \left(\identity \pm \cupcap\right)
\end{equation*}
If $A$ were zero, this would be a linear relation between $\identity$ and $\cupcap$, which would imply that $V \tensor V \cong \id$. Thus we can divide by $A$, and obtain either the Kauffman polynomial (Equation \eqref{eq:kauffman}) or Dubrovnik polynomial (Equation \eqref{eq:dubrovnik}) skein relation with $z = B/A$.

    \end{enumerate}
\end{proof}

This argument for the Dubrovnik polynomial is similar in spirit to Kauffman's original description in  \cite{MR958895}, and the argument for HOMFLYPT polynomial is similar to \cite[\S 4]{MR0908150}.  Similar results were also obtained in \cite{MR2132671, MR1237835}.

%\begin{cor}
%If instead $V$ is an object in a \textit{framed} braided planar algebra $\cC$, with framing number $\alpha$, then Theorem \ref{thm:summands} holds but with the corresponding framed link invariants with framing number $\alpha$. In particular, in part 3) the invariant $\J{\cC}{V}$ is a specialization of the framed HOMFLYPT invariant with $a^{-1} v = \alpha$.
%\end{cor}

We'll now need some notation for eigenvalues. Suppose $N$ appears once as a summand of $V \tensor V$, and consider the braiding as an endomorphism acting by composition on $\End{V \tensor V}$. Then the idempotent projecting onto $N \subset V \tensor V$ is an eigenvector for the braiding, and we'll write $\Reigenvalue{N}{V}$ for the corresponding eigenvalue.  The following is well-known (for example the HOMFLYPT case is essentially \cite[\S 4]{MR0908150}).

\begin{thm}
\label{thm:eigenvalues}
If one of conditions (2)-(4) of Theorem \ref{thm:summands} holds, then we can find which specialization occurs by computing eigenvalues.
    \begin{enumerate}
   \item[(2)] If $\Reigenvalue{L}{V} = \lambda$, then $\Reigenvalue{\id}{V} = -\lambda^{-3}$ and
        \begin{equation*}
        \J{\cC}{V} = \J{\SL{2}}{(1)}(a)
        \end{equation*}
    with $a=-\lambda^{2}$.
\item[(3)] If $\Reigenvalue{L}{V} = \lambda$, $\Reigenvalue{M}{V} = \mu$, and $\theta$ is the twist factor, then
        \begin{equation*}
        \J{\cC}{V} = \HOMFLY(a, z)
        \end{equation*}
    with $a=\frac{\theta}{\sqrt{-\lambda\mu}}$ and $z=\frac{\lambda+\mu}{\sqrt{-\lambda\mu}}$.
\item[(4)] If $\Reigenvalue{L}{V} = \lambda$ and $\Reigenvalue{M}{V} = \mu$, then $\lambda \mu = \pm 1$.
\begin{enumerate}
  \item
   If $\lambda \mu = -1$ then
        \begin{equation*}
        \J{\cC}{V} = \Dubrovnik(a, z)
        \end{equation*}
    with $a=\Reigenvalue{1}{V}^{-1}$ and $z=  \lambda + \mu$.
\item
 If $\lambda \mu = 1$ then
        \begin{equation*}
        \J{\cC}{V} = \Kauffman(a, z)
        \end{equation*}
    with $a=\Reigenvalue{\id}{V}^{-1}$ and $z=  \lambda + \mu$.
    \end{enumerate}
    \end{enumerate}
\end{thm}

\begin{proof}
These proofs all follow the same outline.  We consider the operator $X$ which acts on tangles with four boundary points by multiplication with a positive crossing.  We find the eigenvalues of $X$ in terms of the parameters ($a$ and/or $z$) and then solve for the parameters in terms of the eigenvalues.

    \begin{enumerate}
\item[(2)]
        The Jones skein relation for unoriented framed links is
       $$\overcrossing=i a^{\frac{1}{2}} \identity - i a^{-\frac{1}{2}} \cupcap$$
        if closed circles count for $[2]_a=(a+a^{-1})$.  
        
        The eigenvectors for $X$, multiplication by the positive crossing, are
        $$\inputtikz{JonesEigenvectors}$$
        which have eigenvalues $i a^{\frac{1}{2}}$ and $ - i a^{\frac{-3}{2}}$.

        The cup-cap picture must correspond to the summand $\id$, and so we see that if
        $\Reigenvalue{L}{V}=\lambda$,
        then
        $a=-\lambda^{2} $ and $\Reigenvalue{\id}{V} = -\lambda^{-3}$.

    \item[(3)]
        The HOMFLYPT skein relation is for oriented framed links:
\begin{equation}
w \Oovercrossing - w^{-1} \Oundercrossing = z \Oidentity,
\end{equation}
        and the characteristic equation for the operator $X$ which multiplies by the positive crossing is
        $$w x - w^{-1} x^{-1} = z \quad \Longleftrightarrow \quad  x^2- \frac{z}{w} x - \frac{1}{w^2}=0.$$

        So if $\lambda$ and $\mu$ are the eigenvalues of $X$, we have $\lambda \mu = -w^{-2}$ and
        $\lambda + \mu = \frac z w$, so that
        $$w=\frac{1}{\sqrt{-\lambda \mu}} \text{\quad and \quad} z
        = \frac{\lambda + \mu}{\sqrt{-\lambda \mu}}$$
        
        To recover $a$ we note that the twist factor is $a w^{-1}$, hence $a = w \theta$.

    \item[(4)]
       For the Dubrovnik or Kauffman skein relation we have
       $$ \overcrossing \pm \undercrossing = z\left(\identity\pm \cupcap \right).$$  Multiplying by the crossing we see that, 
       $$\inputtikz{crossingsquared} \pm \identity = z \left(\overcrossing \pm a^{-1} \cupcap\right).$$  Subtracting $a^{-1}$ times the first equation from the second and rearranging slightly we see that the characteristic equation for the crossing operator is $(x-a^{-1})(x^2-z x\pm 1)=0$.  Hence the eigenvalues are $a^{-1}$, $\lambda$, and $\mu$, where $\lambda+\mu = z$ and $\lambda \mu = \pm 1$.  Since $a$ is the twist factor it is the inverse of the eigenvalue corresponding to $\id$ (compare with case (2)).
           
    \end{enumerate}
\end{proof}

\subsection{Knot polynomial identities for $\mathcal{D}_{4}$, $\mathcal{D}_{6}$, $\mathcal{D}_{8}$ and $\mathcal{D}_{10}$}

We state four theorems, give two lemmas, and then give rather pedestrian proofs of the theorems.  Snazzier proofs appear in Section \ref{sec:coincidences}, as special cases of Theorem \ref{thm:identities-geq}.  In each of these theorems, we relate two quantum knot invariants via an intermediate knot invariant coming from $\mathcal{D}_{2n}$.  You can think of these results as purely about quantum knot invariants, although the proofs certainly use $\mathcal{D}_{2n}$.

\begin{thm}[Identities for $n=2$]
\label{thm:identities-2}
\begin{align*}
\restrict{\J{\SL{2}}{(2)}(K)}{q=\exp(\frac{2\pi i}{12})} & = 2 \J{\mathcal{D}_4}{P}(K) \\
& = 2
\end{align*}
\end{thm}

%\todo{Say something about linking number mod 3, for two component links?}

\begin{thm}[Identities for $n=3$]
\label{thm:identities-3}
\begin{align*}
\restrict{\J{\SL{2}}{(4)}(K)}{q=\exp(\frac{2\pi i}{20})} & = 2 \J{\mathcal{D}_6}{P}(K) \\ & = 2 \restrict{\J{\SL{2}}{(1)}(K)}{q=\exp(- \frac{2\pi i}{10})}
\end{align*}
\end{thm}

\begin{thm}[Identities for $n=4$]
\label{thm:identities-4}
\begin{align*}
\restrict{\J{\SL{2}}{(6)}(K)}{q=\exp(\frac{2\pi i}{28})} & = 2 \J{\mathcal{D}_8}{P}(K) \\
                                     & = 2 \HOMFLY(K)(\exp(2\pi i\frac{3}{14}), \exp(\frac{2\pi i}{14}) - \exp(-\frac{2\pi i}{14})) \\
                & = 2 \HOMFLY(K)(\exp(2\pi i\frac{5}{7}), \exp(- \frac{2\pi i}{14}) - \exp(\frac{2\pi i}{14}))
\end{align*}
\end{thm}
\begin{rem}
This isn't just any specialization of the HOMFLYPT polynomial:
\begin{align*}
\HOMFLY(K)&(\exp(2\pi i\frac{3}{14}), \exp(\frac{2\pi i}{14}) - \exp(-\frac{2\pi i}{14})) \\
    & = \restrict{\HOMFLY(K)(q^3,q-q^{-1})}{q=\exp(\frac{2\pi i}{14})} \\
    & =  \restrict{\J{\SL{3}}{(1,0)}(K)}{q=\exp(\frac{2\pi i}{14})} \\
\intertext{and}
\HOMFLY(K)&(\exp(2\pi i\frac{5}{7}), \exp(- \frac{2\pi i}{14}) - \exp(\frac{2\pi i}{14}))    \\
    & = \restrict{\HOMFLY(K)(q^4,q-q^{-1})}{q=\exp(-\frac{2\pi i}{14})} \\
    & =  \restrict{\J{\SL{4}}{(1,0,0)}(K)}{q=\exp(-\frac{2\pi i}{14})} \\
    & =  - \restrict{\J{\SL{4}}{(1,0,0)}(K)}{q=-\exp(-\frac{2\pi i}{14})}
\end{align*}
(The last identity here follows from the fact that every exponent of $q$ in $\J{\SL{4}}{(1,0,0)}(K)$ is odd. We've included this form here to foreshadow \S \ref{sec:applications} where we'll give an independent proof of this theorem, and in which this particular value of $q=-\exp(-\frac{2\pi i}{14})$ will spontaneously appear.)
\end{rem}

\begin{thm}[Identities for $n=5$]
\label{thm:identities-5}
\begin{align*}
\restrict{\J{\SL{2}}{(8)}(K)}{q=\exp(\frac{2\pi i}{36})} & = 2 \J{\mathcal{D}_{10}}{P}(K) \\
        & = 2 \Dubrovnik(K) (\exp(2 \pi i \frac{4}{36}), \exp(2 \pi i \frac{2}{36}) + \exp(2 \pi i \frac{16}{36}))
\end{align*}
\end{thm}
\begin{rem}
Again, this isn't just any specialization of the Dubrovnik polynomial:
\begin{align*}
\Dubrovnik(K) (\exp(2 \pi i \frac{4}{36}),& \exp(2 \pi i \frac{2}{36}) + \exp(2 \pi i \frac{16}{36})) \\
& = \restrict{\Dubrovnik(K) (q^7,q-q^{-1})}{q=-\exp (\frac{-2 \pi i}{18})}\\
& = \restrict{\J{\SO{8}}{(1,0,0,0)}(K)}{q=-\exp(\frac{-2\pi i}{18})}.
\end{align*}
\end{rem}

For the proofs of these statements, we'll need to know how $P \tensor P$ decomposes in each $\mathcal{D}_{2n}$. The following formula was proved in \cite{MR1145672}.
\begin{equation}
\label{eq:P-squared}%
P \tensor P \iso \begin{cases}
Q \directSum \DirectSum_{l=0}^{\frac{n-4}{2}} f^{(4l+2)} & \text{when $n$ is even} \\
P \directSum \DirectSum_{l=0}^{\frac{n-3}{2}} f^{(4l)}   & \text{when $n$ is odd}
\end{cases}
\end{equation}
In particular,
\begin{align*}
P \tensor P & \iso Q                                        & \text{in $\mathcal{D}_{4}$,} \displaybreak[1] \\
P \tensor P & \iso P \directSum f^{(0)}                     & \text{in $\mathcal{D}_{6}$,} \displaybreak[1] \\
P \tensor P & \iso Q \directSum f^{(2)}                     & \text{in $\mathcal{D}_{8}$, and} \displaybreak[1] \\
P \tensor P & \iso P \directSum f^{(0)} \directSum f^{(4)}  & \text{in $\mathcal{D}_{10}$.}
\end{align*}

Further, we'll need a lemma calculating the eigenvalues of the braiding.
\begin{lem} \label{lem:eigenvalues}
Suppose $X$ is an idempotent in the set $\{ f^{(2)}, f^{(6)}, \ldots, f^{(2n-6)}, Q \}$ if $n$ is even, or $X \in \{ f^{(0)}, f^{(4)}, \ldots, f^{(2n-6)}, P \}$ if $n$ is odd. Then the eigenvalues for the braiding in $\mathcal{D}_{2n}$ are 
\begin{equation*}
R_{X \subset P \tensor P} =  (-1)^k q^{k(k+1)-2n(n-1)}
\end{equation*}
where $2k$ is the number of strands in the idempotent $X$.
\end{lem}


\begin{proof}
The endomorphism space for $P \tensor P$ is spanned by the projections onto the direct summands described above in Equation \eqref{eq:P-squared},
and thus by the diagrams
\begin{equation*}
\inputtikz{eigenvalues/basis}.
\end{equation*}

We calculate

\begin{equation*}
\inputtikz{eigenvalues/twist} =  \inputtikz{eigenvalues/middle} = \inputtikz{eigenvalues/flip}.
\end{equation*}
Here there are negative half-twists on $2n-2$ strands below the top $P$s, and a positive half-twist on $2k$ strands above $X$. The $2n-2-k$ strands
connecting the two $P$s each have a negative kink. 

A positive half-twist on $\ell$ strands adjacent to an ``uncappable" element, such as a minimal projection, gives a factor of $(i s)^{\ell(\ell-1)/2}$, a negative half-twist on  $\ell$ strands adjacent to an uncappable element gives a factor of $(- i s^{-1})^{\ell(\ell-1)/2}$, and a negative kink gives a factor of $-i s^{-3}$. Remembering $q=s^2$, this shows that

\begin{equation*}
\inputtikz{eigenvalues/flip} =  (-1)^k q^{k(k+1)-2n(n-1)}\inputtikz{eigenvalues/untwist}
\end{equation*}

Thus
\begin{equation*}
R_{X \subset P \tensor P} = (-1)^k q^{k(k+1)-2n(n-1)}. \qedhere
\end{equation*}

%
%Along with the fact that $P = f^{(2n-2)} P$, this then implies the lemma:
%\scott{on LHS here, draw trivalent graphs}\begin{align*}
%\mathfig{0.2}{eigenvalues/thick-twist} & = q^{-n(2n-2)} \mathfig{0.2}{eigenvalues/twist} \\
%                                       & = q^{-n(2n-2)} R^{PP}_{f^{(2k)}} \mathfig{0.2}{eigenvalues/untwist} \\
%                                       & = R^{PP}_{f^{(2k)}} \mathfig{0.2}{eigenvalues/untwist}.
%\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:identities-2}]
In $\mathcal{D}_4$, $P \otimes P \cong Q$, so part one of Theorem \ref{thm:summands} applies.  Furthermore $\dim P = 1$, so the unframed invariant for the object $P$ in $\mathcal{D}_4$ is trivial. The first equation is just Theorem \ref{thm:half}.
\end{proof}


The same argument yields a previously known identity \cite{MR1134131}. Consider Temperley-Lieb at $q=\exp({\frac{2 \pi i}{6}})$, and notice that $\JW{1} \otimes \JW{1} \cong \JW{0}$ and $\dim \JW{1} = 1$. Thus $\restrict{\J{\SL{2}}{(1)}(K)}{q=\exp(\frac{2\pi i}{6})} = 1$.

\begin{proof}[Proof of Theorem \ref{thm:identities-3}]
In $\mathcal{D}_{6}$, we have that $P \tensor P \iso P \directSum f^{(0)}$, so part two of Theorem \ref{thm:summands} applies, and we know $\J{\mathcal{D}_6}{P}(K)$ is some specialization of the Jones polynomial. Using Lemma \ref{lem:eigenvalues}, we compute the two eigenvalues as
$$\Reigenvalue{f^{(0)}}{P}  = \exp({\frac{2 \pi i}{20}})^{-12} = -(\exp({2 \pi i \frac{-3}{10}}))^{-3}$$
and
$$\Reigenvalue{P}{P}  = \exp({\frac{2 \pi i}{20}})^{-6}  = \exp({2 \pi i\frac{-3}{10}})$$
which is consistent with $\Reigenvalue{P}{P}=\lambda$ and $\Reigenvalue{f^{(0)}}{P}=-\lambda^{-3}$.  So, we conclude that $a=-\lambda^2=-\exp({-6\frac{2 \pi i}{10}}) = \exp({-\frac{2 \pi i}{10}})$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:identities-4}]
In much the same way, for $\mathcal{D}_8$ we have $P \tensor P \iso Q \directSum f^{(2)}$, so $\J{\mathcal{D}_8}{P}(K)$ is some
specialization of the HOMFLYPT polynomial. The eigenvalues are
\begin{align*}
\lambda = \Reigenvalue{f^{(2)}}{P} & =  \exp( 2\pi i \frac{10}{14})  \\
\intertext{and}
\mu = \Reigenvalue{Q}{P} & =  \exp(2\pi i\frac{1}{14}),
\end{align*}
so $\frac{1}{\sqrt{-\lambda \mu}} = \pm \exp\left(2\pi i \frac{-2}{14}\right)$.
The twist factor is $\theta=\exp\left(2 \pi i \frac{-2}{14}\right)$, and so we get $\J{\mathcal{D}_8}{P}(K) = \HOMFLY(K)(a,z)$ with either
$$a= \exp(2 \pi i \frac{5}{7}),\, z=\exp(-2 \pi i \frac{1}{14}) - \exp(2 \pi i \frac{1}{14})$$
(taking the `positive' square root) or
$$a= \exp(2 \pi i \frac{3}{14}),\, z=\exp(2 \pi i \frac{1}{14}) - \exp(- 2 \pi i \frac{1}{14})$$
(taking the other).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:identities-5}]
Again, in $\mathcal{D}_{10}$ we have $P \tensor P \iso P \directSum f^{(0)} \directSum f^{(4)}$, so $\J{\mathcal{D}_{10}}{P}(K)$ is a specialization of either the Kauffman polynomial or the Dubrovnik polynomial.
The eigenvalues are
\begin{align*}
a^{-1} &			 = \Reigenvalue{f^{(0)}}{P} 
%			   = q^{-2(n-1)n} 
%                          = \exp(2 \pi i \frac{-40}{36}) 
                          = \exp(2 \pi i \frac{-1}{9}), \\
\lambda & = \Reigenvalue{f^{(4)}}{P} 
%			 = q^{6-2(n-1)n} 
%                          = \exp(2 \pi i \frac{-34}{36}) 
                          = \exp(2 \pi i \frac{1}{18}) \\
\intertext{and}
\mu & = \Reigenvalue{P}{P}
%		    = q^{(n-1)n-2(n-1)n} 
%                 = \exp(2 \pi i \frac{-20}{36}) 
                    = \exp(2 \pi i \frac{4}{9})
\end{align*}

Now we apply Theorem \ref{thm:eigenvalues} (4) to these; we see that $\Reigenvalue{f^{(4)}}{P}\Reigenvalue{P}{P} = -1$, so we're in the Dubrovnik case. 
We read off $z =  \exp(2 \pi i \frac{1}{18}) + \exp(2 \pi i \frac{4}{9})$.

We've now shown that
\begin{align*}
\restrict{\J{\SL{2}}{(8)}(K)}{q=\exp(\frac{2\pi i}{36})} & = 2 \J{\mathcal{D}_{10}}{P}(K) \\
        & = 2 \Dubrovnik(K) (\exp(2 \pi i \frac{4}{36}), \exp(2 \pi i \frac{2}{36}) + \exp(2 \pi i \frac{16}{36})).
\end{align*}

To get the last identity, we note that
$$\restrict{(q^7,q-q^{-1})}{q=-\exp (\frac{-2 \pi i}{18})} = (\exp(2 \pi i \frac{4}{36}), \exp(2 \pi i \frac{16}{36}) + \exp(2 \pi i \frac{2}{36}))$$
and use the specialization appearing in Equation \eqref{eq:dubrovnik-D}.
\end{proof}

We remark that when $2n \geq 12$, Equation \eqref{eq:P-squared} shows that $P \tensor P$ has at least three summands which are not isomorphic to $f^{(0)}$,
and thus Theorem \ref{thm:summands} does not apply.

\subsection{Recognizing specializations of the $G_2$ knot invariant}

If $V$ is an object in a ribbon category such that $V \otimes V \cong \id \oplus V \oplus A \oplus B$ then it is reasonable to guess that the knot invariants coming from $V$ are specializations of the $G_2$ knot polynomial.  In particular $\mathcal{D}_{14}$ might be related to $G_2$, since in $\mathcal{D}_{14}$ we have $P \tensor P \iso f^{(0)} \directSum f^{(4)} \directSum f^{(8)} \directSum P$ by Equation \eqref{eq:P-squared}.  In this section we prove such a relationship using results of Kuperberg \cite{MR1265145}.  Applying Kuperberg's theorem requires some direct but tedious calculations.

In work in progress, Snyder has shown that, outside of a few small exceptions, all nontrivial knot invariants coming from tensor categories with $V \otimes V \cong \id \oplus V \oplus A \oplus B$ come from the $G_2$ link invariant, which would obviate the need for these calculations.  (The ``nontrivial" assumption in the last sentence is crucial as the standard representation of the symmetric group $S_n$, or more generally the standard object in Deligne's category $S_t$, also satisfies  $V \otimes V \cong \id \oplus V \oplus A \oplus B$.) 

In the following, by a {\em trivalent vertex} we mean a rotationally invariant map $V \tensor V \to V$ for some symmetrically self-dual object $V$. By a {\em tree} we mean a trivalent graph without cycles (allowing disjoint components).

\begin{thm}[{\cite[Theorem 2.1]{MR1265145}}] 
\label{thm:kuperberg}%
Suppose we have a symmetrically self-dual object $V$ and a trivalent vertex in a ribbon category $\cC$, such that trees with $5$ or fewer boundary points form a basis for the spaces $\Inv{\cC}{V^{\otimes k}}$ for $k \leq 5$. Then the link invariant $\J{\cC}{V}$ is a specialization of the $G_2$ link invariant for some $q$.%such that $\dim V = q^{10}+q^8+q^2+1+q^{-2}+q^{-8}+q^{-10}$. \noah{Figure out which q?}
\end{thm}

\begin{rem}
The trivalent vertex in $\cC$ is some scalar multiple of the trivalent vertex in the $G_2$ spider.  Note that the $G_2$ link invariant is the same at $q$ and $-q$ since all the relations only depend on $q^2$.  
\end{rem}

\begin{lem} \label{lem:yucky}
Suppose that $\cC$ is a pivotal tensor category with a trivalent vertex such that trees form a basis of $\Inv{}{V^{\otimes k}}$ for $k \leq 3$.  Then
\begin{enumerate}
\item trees are linearly independent in $\Inv{}{V^{\otimes 4}}$ if and only if \begin{equation}-2 b^4 d^5 + b^4 d^6 - 2 b^3 d^4 t + (b^2 d^4 - b^2 d^6) t^2 \neq 0,\end{equation}
\item trees are linearly independent in $\Inv{}{V^{\otimes 5}}$ if and only if
\begin{align*}
& b^{20} \left(d^{15}-10 d^{13}-5 d^{12}+65 d^{11}-62 d^{10}\right) \\
& +5 b^{19} t\left(d^{14}+d^{13}-7 d^{12}-d^{11}+10 d^{10}\right) & \displaybreak[1]  \\
& -5 b^{18} t^2\left(d^{15}-10 d^{13}-3d^{12}+55 d^{11}-61 d^{10}\right) & \\
& -5 b^{17} t^3\left(6 d^{14}+7 d^{13}-40 d^{12}-41 d^{11}+83 d^{10}\right) & \displaybreak[1]  \\
& +5 b^{16} t^4\left(2 d^{15}+3 d^{14}-15 d^{13}-17 d^{12}+72d^{11}-68 d^{10}\right) & \\
& +b^{15} t^5\left(2 d^{15}+60 d^{14}+60 d^{13}-405 d^{12}-485 d^{11}+930 d^{10}\right) & \displaybreak[1]  \\
& -5 b^{14} t^6\left(3 d^{15}+12 d^{14}-8 d^{13}-64 d^{12}+3 d^{11}+71 d^{10}\right) & \\
& -5 b^{13} t^7\left(5 d^{14}+5 d^{13}-44 d^{12}-50 d^{11}+96 d^{10}\right) & \displaybreak[1]  \\
& +5 b^{12} t^8\left(3 d^{15}+12 d^{14}-6 d^{13}-70 d^{12}-17 d^{11}+112d^{10}\right)& \\
& -5 b^{11} t^8\left(2 d^{15}+6 d^{14}-5 d^{13}-29 d^{12}+4 d^{11}+45 d^{10}\right)& \displaybreak[1]  \\
& +b^{10} t^{10}\left(2 d^{15}+5 d^{14}-5 d^{13}-20 d^{12}+10 d^{11}+33 d^{10}\right) &\\ & \neq 0
\end{align*}
\end{enumerate}

Where $d$, $b$ and $t$ are defined by
\begin{align*}
\mathfig{0.04}{G2/loop} &= d,\\
\mathfig{0.04}{G2/bigon} &=  b \;\;
\begin{tikzpicture}[baseline=-.1] \draw (0,-1)--(0,1); \end{tikzpicture} \\
\mathfig{0.1}{G2/triangle} &= t \mathfig{0.08}{G2/vertex}.
\end{align*}
\end{lem}
\begin{proof}
Compute the matrix of inner products between trees.  Each of these inner products can be calculated using only the relations for removing circles, bigons, and triangles.  If the determinant of this matrix is nonzero then the trees are linearly independent.
\end{proof}

For $\mathcal{D}_{2n}$ the single strand corresponds to $P$,  and the trivalent vertex is
$$\inputtikz{trivalentD14},$$ which is rotationally invariant, because $P$ is invariant under $180$-degree rotation.


In order to apply Lemma \ref{lem:yucky} we must compute the values of $b$ and $t$ in $\frac{1}{2} \mathcal{D}_{14}$.  In order to do so we simplify the expression for the trivalent vertex.

\begin{lem}
$$\inputtikz{trivalentD14}=\inputtikz{trivalentfewerPs}$$
\end{lem}
\begin{proof}
Expand $\JW{12} = P + Q$ and use the fact that $P \otimes Q$, $Q \otimes P$, and $Q \otimes Q$ do not have nonzero maps to $P$.
\end{proof}

\begin{lem}
In $\frac{1}{2} \mathcal{D}_{14}$, using the above trivalent vertex, we have that $d$ is the root of $x^6 - 3x^5 -6 x^4 +4 x^3 + 5x^2- x-1$ which is approximately $4.14811$, $b$ is the root of $x^6-12x^5-499x^4-2760x^3-397x^2+276x-1$ which is approximately $0.00364276 $ and $t$ is the root of $x^6 +136 x^5+5072x^4+53866x^3+13132x^2+721x+1$ which is approximately $-0.00142366 $.
\end{lem}
\begin{proof}
The formula for $d$ is just the dimension of $P$.

We use the alternate description of the trivalent vertex to reduce the calculation of $b$ and $t$ to a calculation in Temperley-Lieb  which we do using the formulas of  \cite{MR1280463}.

\begin{align*}
\mathfig{0.1}{G2/bigon} &= \inputtikz{bigonD14} = \inputtikz{bigonD14simplified} \displaybreak[1] \\
&=\inputtikz{bigonD14furthersimplified}= b' P 
\end{align*}

where $b'$ is the coefficient for removing bigons labelled with $\JW{12}$ in Temperley-Lieb.

The calculation for $t$ is similar: we replace each trivalent vertex with a trivalent vertex with a $P$ on the outside and $\JW{12}$s in the middle.  Then we reduce the inner triangle in Temperley-Lieb.
\end{proof}

\begin{thm}
\label{thm:G2-links}%
For $\ell = -3$ or $10$,
\begin{align*}\restrict{\J{\SL{2}}{(12)}(K)}{q=\exp(\frac{2\pi i}{52})} & = 2 \J{\mathcal{D}_{14}}{P}(K) \\
& = 2\restrict{\J{G_2}{V_{(10)}}(K)}{q=\exp{2\pi i \frac{\ell}{26}}} \\
\end{align*}
\end{thm}
\begin{proof}
Since $\dim \Inv{}{P^{\otimes 0}} = \dim \Inv{}{P^{\otimes 2}} = \dim \Inv{}{P^{\otimes 3}} = 1$ and $\dim \Inv{}{P} = 0$, trees form a basis of $\Inv{}{P^{\otimes k}}$ for $k \leq 3$.  By Lemma \ref{lem:yucky} we see that trees are linearly independent in $\Inv{}{P^{\otimes 4}}$ and $\Inv{}{P^{\otimes 5}}$.  A dimension count shows that trees form a basis for these spaces.   Now we apply Theorem \ref{thm:kuperberg} to see that the theorem holds for some $q$. 
We need to normalize the $\mathcal{D}_{14}$ trivalent vertex for $P$ before it satisfies the $G_2$ relations, specifically multiplying it by the largest real root of $x^{12} - 645 x^{10} -10928 x^8 - 32454 x^6 - 4752 x^4 + 2 x^2 +1$. The quantities $b$ and $t$ are both homogeneous of degree $2$ with respect to scaling the trivalent vertex, so they are both multiplied by the square of this quantity. We now solve the equations
\begin{align*}
d & = q^{10} + q^{8} + q^2 + 1 + q^{-2} + q^{-8} + q^{-10}  \\
b & = -\left(q^6 + q^4 + q^2 + q^{-2} + q^{-4} + q^{-6}\right) \\
t & = q^4 + 1 +q^{-4}
\end{align*}  
and find that they have a four solutions, $q= \exp{2\pi i \frac{\ell}{26}}$ with $\ell=\pm 3, \pm 10$. Not all of these give the correct twist factor, however. The twist factor for $P$ is $\exp(2 \pi i \frac{-10}{26})$, while the twist factor for the representation $V_{(10)}$ of $G_2$ is $q^{12}$; these only agree for $\ell = -3$ or $10$. Since the identity holds for some $q$, and the knot invariant only depends on $q^2$, the identity must hold for each of these values.
\end{proof}


\subsection{Ribbon functors} \label{sec:ribbonfunctors}
The proofs of Theorems \ref{thm:summands} , \ref{thm:eigenvalues}, and \ref{thm:kuperberg} actually construct ribbon functors from a certain diagrammatic category to the ribbon category $\cC$.  Combining this functor with the description of quantum group categories by diagrams in \cite{MR1710999, MR1854694} and \cite{MR1182414} one could prove the coincidences described in the introduction (that is, Theorems \ref{thm:D6-coincidence}, \ref{thm:D8-coincidence} and \ref{thm:D10-coincidence}).  To do this we need the following lemma.

\begin{lem}
Suppose that $\cC$ is a ribbon category such that $\cC^{ss}$ is premodular, that $\cD$ is a pseudo-unitary modular category, and that $\cF$ is a dominant ribbon functor $\cC \rightarrow \cD$. Then $\cD \cong \cC^{ss\ modularize}$.
\end{lem}
\begin{proof}
Since $\cD$ is pseudo-unitary the functor must factor through the semisimplification, and thus  the result follows from the uniqueness of modularization.
\end{proof}

In our cases $\cD_{2n}$ is the target category, and is certainly unitary and modular.  The source category is a category of diagrams (coming from Temperley-Lieb, Kauffman/Dubrovnik, HOMFLYPT, or the $G_2$ spider).   Dominance of the functor is a simple calculation in the fusion ring of $\mathcal{D}_{2n}$.  If $q$ is a large enough root of unity, then the semisimplification of that diagram category has been proven to be pre-modular for each of these cases \cite{MR1854694, MR2132671, MR1710999} except the $G_2$ spider.  Hence the argument of the last subsection does not yet give a proof of the $G_2$ coincidence.  We give a completely different proof in the next subsection.

\subsection{Recognizing $\mathcal{D}_{2n}$ modular categories}\label{sec:recognizeD}

Earlier in this section we found knot polynomial identities and coincidences of modular tensor categories by observing that $P^{\otimes 2}$ broke up in some particular way.  In this section we work in the reverse direction.  The category $\frac{1}{2}\mathcal{D}_{2n}$ has a small object $\JW{2}$ and $\JW{2} \tensor \JW{2} \cong \id \oplus \JW{2} \oplus \JW{4}$.  If we are to have a coincidence of modular tensor categories $\mathcal{D}_{2n} \cong \cC$ then there must be an object in $\cC$ which breaks up the same way.  Using the characterization of the Kauffman and Dubrovnik categories above we can prove that  $\mathcal{D}_{2n} \cong \cC$ by producing this object.  In the following theorem, we use this technique to show $\frac{1}{2} \mathcal{D}_{14} \cong \Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}$, for $\ell = -3$ or $10$, sending $P \mapsto V_{(10)}$. It's also possible to prove Theorems \ref{thm:D6-coincidence}, \ref{thm:D8-coincidence} and \ref{thm:D10-coincidence} by this technique, although we don't do this.

\begin{thm}\label{thm:G2}
There is an equivalence of modular tensor categories 
$$\Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}\cong \frac{1}{2} \mathcal{D}_{14},$$ 
where $\ell = -3$ or $10$,
sending $\JW{2} \mapsto V_{(02)}$.  Under this equivalence we also have $P \mapsto V_{(10)}$.
\end{thm}
\begin{proof}
Using the Racah rule for tensor products in $\Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}$ we see that $V_{(02)}^{\otimes 2} \cong \id \oplus V_{(01)} \oplus V_{(02)} $.  

The eigenvalues for the square of a crossing can be read off from twist factors: $$R_{X \subset Y \tensor Y}^2 = \theta_X \theta_Y^{-2}.$$
The twist factors for the representations $V_{(00)}, V_{(01)}$ and $V_{(02)}$ are $1, q^{24}$ and $q^{60}$ respectively, so the corresponding eigenvalues for the crossing are $q^{-60}, \sigma_1 q^{-48}$ and $\sigma_2 q^{-30}$ for some signs $\sigma_1$ and $\sigma_2$. We thus compute, whether we are in the Kauffman or Dubrovnik settings, that  $a=q^{60}$ and $z=\sigma_1 q^{-48} + \sigma_2 q^{-30}$.

If we are in the Kauffman setting, we must have $\sigma_1 \sigma_2 q^{-78} = 1$, so $\sigma_1 = \sigma_2$. We now see the dimension formula $d=\frac{a+a^{-1}}{z} - 1$ can not be equal to 
$\dim(\JW{2})=[3]_{q=\exp(\frac{2 \pi i}{52})}$ for any choice of $\sigma_1, \sigma_2$.

Hence we must be in the Dubrovnik setting where we have $\sigma_1 = -\sigma_2$
and $d=\frac{a-a^{-1}}{z} + 1$.
Now the dimensions match up exactly when $\sigma_1 =-1$ and $\sigma_2 = 1$.

By \S \ref{sec:ribbonfunctors} and Theorems \ref{thm:summands} and \ref{thm:eigenvalues} we have a functor from the Dubrovnik category with $a= \exp(2 \pi i \frac{1}{13})$ and $z=\exp(2 \pi i \frac{1}{26}) - \exp(2 \pi i \frac{-1}{26})$ to $\Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}$.   Since the target category is pseudo-unitary \cite{MR2414692}, this functor factors through the semi\-simpli\-fi\-cat\-ion of the diagram category, which is the premodular category $\Rep{U_{q=\exp(2 \pi i \frac{1}{52})}(\SO{3})}$.  Since the target is modular \cite{MR2286123} and the functor is dominant (a straightforward calculation via the Racah rule in the Grothendieck group of $\Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}$) this functor induces an equivalence between the modularization of $\Rep{U_{q=\exp(2 \pi i \frac{1}{52})}(\SO{3})}$, which is nothing but $\frac{1}{2} \mathcal{D}_{14}$, and $\Rep{U_{\exp({2 \pi i\frac{\ell}{26}})}(\mathfrak{g}_2)}$.

The correspondence between simples shown in Figure \ref{fig:g2-weyl-chamber}, can be computed inductively.  Begin with the observation that $\JW{2}$ is sent to $V_{(02)}$ by construction; after that, everything else is determined by working out the tensor product rules in both categories.
\end{proof}

\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 $q=\pm \exp({2\pi i\frac{-3}{26}})$, and the
correspondence with the even vertices of $\mathcal{D}_{14}$.}
\label{fig:g2-weyl-chamber}
\end{figure}

Note that $q=\pm \exp({2\pi i\frac{-3}{26}})$ corresponds to the fractional level $\frac{1}{3}$ of $G_2$ (see \cite{MR2286123}), which has previously been conjectured to be unitary \cite{MR2414692}. This theorem proves that conjecture.

Finally, we note that the same method gives an equivalence between 
$\Rep{U_{\pm \exp({2 \pi i\frac{-3}{28}})}(\mathfrak{g}_2)}$
and the subcategory of $\Rep{U_{\exp({2 \pi i\frac{1}{28}})}(\SP{6})}$ generated by the representation $V_{(012)}$, sending the  representation $V_{(12)}$ of $\mathfrak{g}_2$ to $V_{(012)}$. On the $\mathfrak{g}_2$ side, we have $V_{(12)}^{\tensor 2} \iso V_{(00)} \directSum V_{(01)} \directSum V_{(02)}$ with corresponding eigenvalues $1, \exp(2\pi i \frac{3}{14})$ and $-\exp(2\pi i \frac{4}{14})$. On the $\SP{6}$ side we have $V_{(012)}^{\tensor 2} \iso V_{(000)} \directSum V_{(010)} \directSum V_{(200)}$ with corresponding eigenvalues $1, \exp(2\pi i \frac{3}{14})$ and $-\exp(2\pi i \frac{4}{14})$. Thus both categories, which are each modular, are the modularization of the semisimplification of the Kauffman category at $a=1, z=\exp(2\pi i \frac{3}{14}) - \exp(2 \pi i \frac{4}{14})$. This proves the conjecture of \cite{MR2414692} that $G_2$ at level $\frac{2}{3}$ is also unitary.