\begin{lem}
If $\cC$ is a ribbon category and $\dim \Hom{}{V \otimes V}{V} = 1$, then any element of  $\Hom{}{V \otimes V}{V}$ is a trivalent vertex.
\end{lem}
\begin{proof}
Suppose that a positive crossing acts on this 1-dim space by the scalar $y$, rotation acts by $\lambda$, and the twist for $V$ is $\theta$.  Then the following equalities of diagrams show that $y^2 \lambda = \theta$ and $y^{-2} \lambda = \theta^{-1}$.  It follows that $\lambda = \lambda^{-1}$ and hence $\lambda= \pm 1$.  Since $\lambda$ is a cube root of unity it follows that rotation acts trivially.
\todo{add pictures}
\end{proof}


Since we will be giving more general proofs in the next section we simply illustrate this principal by sketching one example, Theorem \ref{thm:D6-coincidence}, that $\frac{1}{2} D_{6} \cong \uRep{U_{s= \exp({\frac{7}{10} 2 \pi i})}(\mathfrak{sl}_2 \oplus \mathfrak{sl}_2)}^{modularize}$ with $P \mapsto V_{(1)} \boxtimes V_{(0)}$.  

By the proofs Theorems of \ref{thm:summands} and \ref{thm:eigenvalues} we have a ribbon functor from Temperley-Lieb with $s=e^{\frac{7}{10}2\pi i}$ to $\frac{1}{2} D_{6}$ sending $V_{(1)} \mapsto P$.  Since the target category is semisimple this functor must factor through the semisimplified version of this Temperley-Lieb category, which is exactly $\uRep{U_{s= \exp({\frac{7}{10} 2 \pi i})}(\mathfrak{sl}_2)}$  Similarly there is a ribbon functor from Temperley-Lieb with $s= \exp({\frac{7}{10} 2 \pi i})$ to $\frac{1}{2} D_{6}$ sending $V_{(1)} \mapsto Q$.  

Taking the external tensor product of these two functors, we get a functor from $\uRep{U_{s= \exp({\frac{7}{10} 2 \pi i})}(\mathfrak{sl}_2 \oplus \mathfrak{sl}_2)}$ to $\frac{1}{2} D_{6}$.   By direct computation we check that this functor is dominant (that is every object in  $\frac{1}{2} D_{6}$ appears as a summand of an object in the image of the functor).  Since $\frac{1}{2} D_{6}$ is modular, and since the modularization is unique, this implies that this functor extends to an equivalence between $\frac{1}{2} D_{6} \cong \uRep{U_{s= \exp({\frac{7}{10} 2 \pi i})}(\mathfrak{sl}_2 \oplus \mathfrak{sl}_2)}^{modularize}$.

Similar arguments hold for the other cases, Theorems \ref{thm:D8-coincidence} and \ref{thm:D10-coincidence}.



Later, \todo{should we keep this?} it will be useful to have the
multiplication tables for the even parts of $\mathcal{D}(6)$ and
$\mathcal{D}(8)$, so we'll give these explicitly here.

For the even part of $\mathcal{D}(6)$, we get
\begin{equation*}
\begin{array}{l|cccccccc}
\mathcal{D}(6)  & 0 & 2       & P       & Q \\
\hline
0               & 0 & 2       & P       & Q \\
2               & 2 & 0+2+P+Q & 2+Q & 2+P  \\
P               & P & 2+Q        & 0+P   & 2 \\
Q               & Q & 2+P        & 2   & 0+Q
\end{array}
\end{equation*}

For the even part of $\mathcal{D}(8)$, we get
\begin{equation*}
\begin{array}{l|cccccccc}
\mathcal{D}(8)  & 0 & 2       & 4 & P & Q \\
\hline
0 & 0 & 2       & 4           & P & Q \\
2 & 2 & 0+2+4   & 2+4+P+Q     & 4+Q   & 4+P  \\
4 & 4 & 2+4+P+Q & 0+2+4+P+Q+4 & 2+4+P & 2+4+Q  \\
P & P & 4+Q     & 2+4+P       & 2+Q   & 0+4 \\
Q & Q & 4+P     & 2+4+Q       & 0+4   & 2+P
\end{array}
\end{equation*}

Finally, we'll also record the tensor products for $W_2 \tensor X$, for
any even vertex $X$, in $\mathcal{D}(14)$.
\begin{align*}
W_2 \tensor W_0 & = W_2 \\
W_2 \tensor W_2 & = W_0 \directSum W_2 \directSum W_4 \\
W_2 \tensor W_4 & = W_2 \directSum W_4 \directSum W_6 \\
W_2 \tensor W_6 & = W_4 \directSum W_6 \directSum W_8 \\
W_2 \tensor W_8 & = W_6 \directSum W_8 \directSum W_{10} \\
W_2 \tensor W_{10} & = W_8 \directSum W_{10} \directSum P \directSum Q \\
W_2 \tensor P & = Q \directSum W_{10} \\
W_2 \tensor Q & = P \directSum W_{10} \\
\end{align*}

\subsection{Bad footnote about FS}
This should have said "is oriented, rather than unoriented", instead of "isn't spherical".

If the planar algebra isn't spherical, there's an additional complication. Now it's possible that the projection corresponding to $V$ in the planar
algebra is anti-symmetrically self-dual.
For a self-dual object $V$, pick an isomorphism $\phi$ between $V$ and $V^*$, and write $FS(V) = \pm1$ for the $\pi$-rotation eigenvalue of this isomorphism. (This distinguishes the symmetrically self-dual and anti-symmetrically self-dual cases.)
To generalize the theorem above, one should say instead
\begin{enumerate}
\item[(2)] ... then the link invariant is $FS(V)^{\card{}}$ times a specialization of the Jones polynomial.
\item[(4)] ... then the link invariant is $FS(V)^{\card{}}$ times a specialization of the Kauffman polynomial.
\end{enumerate}
The object $V$ can't be self-dual in case (3) unless you're actually in case (2)! Moreover, in the case where you get the Jones polynomial,
the extra factor can be absorbed simply by replacing $q$ with $-q$, because the parity of the exponents in the Jones polynomial is determined by the
parity of the number of components of the link.



\subsection{No more $T_n$}
We no longer need to talk about $T_n$, because we understand that Kirby-Melvin symmetry is easy.

In an appendix of \cite{d2n}, we also looked at an "odd-box" version of
this planar algebra

\begin{defn}
We consider a skein theory with a $(k=2n+1)$ strand generator, at the
special value $q=\exp{\frac{i\pi}{k+2}}$, and relations analogous to those of Definition \ref{def:pa}:
\begin{enumerate}

\item\label{delta-T} a closed loop is equal to $2 \cos(\frac{\pi}{k+2})$,

\item\label{rotateS-T}
\inputtikz{rotateS-T}

\item\label{capS-T}
\inputtikz{capS-T}

\item\label{twoS-T}
\inputtikz{twoS-T}
\end{enumerate}
\end{defn}

A calculation analogous to that of Theorem \ref{thm:passacrossS} shows that we have the relations
\begin{equation*}
\inputtikz{pullstringoverS-T}
\qquad \text{and} \qquad \inputtikz{pullstringunderS-T}.
\end{equation*}
where $Z^- = Z^+ = (-1)^{\frac{k+1}{2}}$.


\subsection{$G_2$ conjecture}

We end with a tantalising conjecture, supported by computations for small
knots:
\begin{conj}[Identities for $n = 7$]
\begin{align*}
\restrict{\J{SU(2)}{(2n-2=12)}(K)}{q=\exp(\frac{2\pi i}{8n-4=52})} & = 2
\J{D_{2n=14}}{P}(K) \\
& = 2
\restrict{\J{G_2}{\{1,0\}\text{ ($7$-dimensional)}}(K)}{\delta = 2
\cos\left(\frac{23}{26} \cdot 2 \pi\right)} \\
\intertext{(checked for $K=0_1, 3_1, 5_1$),}
\restrict{\J{SU(2)}{(2)}(L)}{q=\exp(\frac{2\pi i}{52})} & =
\J{D_{14}}{W_2}(K) \\
& =
\restrict{\J{G_2}{\{0,2\}}(L)}{\delta =
\cos\left(\frac{23}{26} \cdot 2 \pi\right)} \\
\intertext{(checked for $L=0_1$ only!), and}
\restrict{\J{SU(2)}{(4)}(L)}{q=\exp(\frac{2\pi i}{52})} & =
\J{D_{14}}{W_4}(K) \\
& =
\restrict{\J{G_2}{\{0,1\}}(L)}{\delta =
\cos\left(\frac{23}{26} \cdot 2 \pi\right)}
\end{align*}
(checked for many knots and links, including all braid index $3$ knots with up to $10$ crossing).
\end{conj}

Alternatively, Theorem \ref{thm:G2} might be proved by finding a $Q$-system inside the $(G_2)_{\frac{1}{3}}$ planar algebra, on the representation $V_{(0,2)}$ (which corresponds to $W_2$). This would show that it's the even part of some planar algebra $\cP$. The generating projection $X$ in $\cP$ then has quantum dimension $$\dim X = \sqrt{\dim V_{(0,2)} + 1} = 2 \cos(\frac{2 \pi}{52}) < 2.$$ Further, this planar algebra is positive, and so by the classification of positive planar algebras with $\delta < 2$ must be either $A_{53}$ or $D_{14}$. Counting simple objects, we find $(G_2)_{\frac{1}{3}}$ has rank $8$, so it must be $\frac{1}{2}D_{14}$, and not $\frac{1}{2}A_{53}$.

One day, perhaps, there'll be a thorough classification of the small modular tensor categories\footnote{See \cite{MR2098028, premodular-3,MR1981895, classification-MTC} for the beginnings of this project.},
which will include a description of the isomorphisms
between examples coming from quantum groups. There would be families of such isomorphisms, for example coming from level-rank duality,
and a list of sporadic isomorphisms such as this one.

\subsection{Kirby-Melvin symmetry}
We recall that the $A_{4n-3}$ planar algebra sits as a planar subalgebra
inside $D_{2n}$. The minimal idempotents in $A_{4n-3}$ are the
Jones-Wenzl idempotents $f^{(i)}$, for $i = 0, \ldots, 4n-4$.

\begin{thm}
The Jones-Wenzl idempotents $f^{(i)}$ and $f^{(4n-4-i)}$, for $i=0,\ldots,2n-3$, are isomorphic in the planar algebra $D_{2n}$.
The Jones-Wenzl idempotent $f^{(2n-2)}$ is not simple in $D_{2n}$, but splits as the sum of two orthogonal simple idempotents, $f^{(2n-2)} = P + Q$.
\end{thm}

\begin{proof}
The second statement is immediate, since by definition of $P$ and $Q$, $P + Q = \JW{2n-2}$.

The isomorphism $\alpha: \JW{i} \rightarrow \JW{4n-4-i}$ is given by the element $S$, with $i$ strings pulled up and $4n-4-i$ strings pulled down, sandwiched between $\JW{i}$ and $\JW{4n-4-i}$; for instance, if $i=2n-3$ then $\alpha$ is

\inputtikz{SisJWisom}

Then we calculate that $\alpha \alpha^*=$

$\inputtikz{twoSsmallJW}$,
by applying relation \ref{twoS} and the partial trace relation.

Similarly, we calculate $\alpha^* \alpha= \frac{\qi{2n-1}}{\qi{4n-3-i}}  \JW{4n-4-i}$, and because we're at a special value of $q$, these coefficients are equal.  So $\alpha$ is an isomorphism from $\JW{i}$ to $\JW{4n-4-i}$.
\end{proof}

\begin{thm}[Kirby-Melvin symmetry principle]%
The coloured Jones polynomials at the root of unity $q=i \exp(\frac{\pi
i}{2r+4})$ (where the value of the Jones polynomial on unknot is $2
\cos(\frac{\pi}{r+2})$) are related as
\begin{equation*}
\Jf{SU(2)}{(k) \cup \ell}(K \cup L) = i^{(r-2k)\fr{K}} (-1)^{\sum_i \ell_i \lk{K}{L_i}} \Jf{SU(2)}{(r-k) \cup \ell}(K \cup L)
\end{equation*}
where here $K$ is a component of the link $K \cup L$, labelled by either $(k)$ or $(r-k)$, and $L$ is labelled by $\ell=(\ell_i)$.
\end{thm}

{\color{blue} Oww, my head.  I'm trying and failing to figure out the correlation between the above statement, and what I think I can prove.  Where does the $i$ come from?  why $r-2k$?  If the $i$ was $-1$, it seems like either $k$ or $r-1-k$ works in the $D_{2n}$ case, and for $T_n$ we want an always even number ($n$ odd) or an always odd number ($n$ even) which shouldn't depend on $k$.  Also, the exponent of $-1$ makes sense in the case of $D_{2n}$ but not for the $T_n$; maybe that whole exponent should be multiplied by $r$? -E}

This result first appeared as \cite[Theorem 4.20]{MR1117149}, \todo{cites for ??? and Thang Le} and we give
an elementary picture proof here, just using the $D_{2n}$ planar algebras and the $T_n$ planar algebras.
This proof, however, is restricted to the case that $r$ is not $3$ mod $4$.
The case $r \equiv 1 \pmod{4}$ uses the $D_{(r+3)/2}$ planar algebra, and the cases when $r$ is even use the $T_{r/2}$ planar algebra.

\begin{example}
See Figure \ref{KMexample}.  On an $k$-cabled component of the link, we replace $\JW{k}$ by two $S$'s, connected by $(4n-4-k)$ strands.  Now slide one of these $S$'s along until it is next to the fixed $S$, but on the other side, and replace these two $S$s, connected by $k$ strands, by $\JW{4n-4-k}$.  The difficulty is keeping track of sign changes as the $S$'s slide over and under strands.

\begin{figure}[h]
\begin{center}
\inputtikz{KMexample}
\end{center}
\caption{{\color{blue} It's hard to tell that $S$ is upside down, so I added some stars. -E} A demonstration of Kirby-Melvin symmetry for the Hopf link, at $q=i \exp(\frac{\pi
i}{6})$; we see that $\J{SU(2)}{((1),(3))} =-\J{SU(2)}{((3),(3))}$.}\label{KMexample}
\end{figure}
\end{example}

\begin{proof}[Proof of (most of) Kirby-Melvin symmetry]
Case $r \equiv 1 \pmod{4}$:
\todo{}
Case $r \equiv 0 \pmod{4}$:
Case $r \equiv 2 \pmod{4}$:
\end{proof}


\begin{example}
We compute the second colored jones polynomial of the trefoil
\end{example}



More specifically, choose a link diagram in which any critical points -- ie crossings, cups and caps -- occur at different heights.  Then the link is built of cross sections on which most strands are straight, and either one strand has a max or min, or two strands cross:

\inputtikz{Crossing} , \quad
\inputtikz{Cap} , \quad
\inputtikz{Cup}

The axioms of a braided tensor category let us interpret a crossing as a map $V \otimes V \rightarrow V \otimes V$, a cup as a map $\id \rightarrow V \otimes V$, and a cap as a map $V \otimes V \rightarrow \id$ (and a straight strand as $\text{id}:V \rightarrow V$), and so the link diagram gives a composition of maps $\id \rightarrow V \otimes V \rightarrow \cdots \rightarrow V \otimes V \rightarrow \id$. By Schur's lemma, this map is some number times the identity -- this number is meant to be our invariant.  See Figure \ref{trefoileg} for an example.

\begin{figure}[h]
\center{\inputtikz{Trefoil}}
\caption{\todo{This diagram should have a label $V$ somewhere on it}Read from the bottom up in a braided tensor category, this knot diagram gives a map
$iId
\stackrel{\cup}{\rightarrow}
V \otimes V
\stackrel{\text{id} \otimes \text{id} \otimes \cup}{\longrightarrow}
V \otimes V \otimes V \otimes V
\stackrel{\text{id} \otimes X \otimes \text{id}}{\longrightarrow}
V \otimes V \otimes V \otimes V
\stackrel{\text{id} \otimes X \otimes \text{id}}{\longrightarrow}
V \otimes V \otimes V \otimes V
\stackrel{\text{id} \otimes X \otimes \text{id}}{\longrightarrow}
V \otimes V \otimes V \otimes V
\stackrel{\cap \otimes \text{id} \otimes \text{id}}{\longrightarrow}
V \otimes V
\stackrel{\cap}{\rightarrow}
\id$,
in other words a number.\todo{what sort of number? -S}}\label{trefoileg}
\end{figure}

To see that this is an invariant of links, we must check Reidemeister moves one, two and three.  Reidemeister two and three are, basically, the axioms of a braided tensor category \eep{but they have funny names like ``quasitriangularity,'' I think}.  However, Reidemeister one fails, but all is not lost.  Since $V$ is a simple object in our category, the diagrams
$$ \inputtikz{ROne} \text{\qquad and \qquad } \inputtikz{otherROne}$$
are in $\Hom{}{V}{V}$, and are therefore multiples of the identity.   Say the first is $\alpha \id$, then the second must be $\alpha^{-1} \id$.\footnote{To see this, call the second one $\beta \id$ for now.  By drawing a circle and applying a Reidemeister two move, we create a circle with both kinds of kink in it.  Pulling both of these out, we find that $\alpha \beta=1$.} Therefore, we've constructed not a link invariant, but a framed link invariant.  Call this  $\tilde{\mathcal{J}}_{\cC,V}$. We can turn this into a genuine link invariant:

\begin{defn}
Let $\J{\cC}{V}(L)=\alpha^{w(L)} \tilde{\mathcal{J}}_{\cC,V}(L)$\eep{maybe minus writhe ...}, where $w(L)$ is the writhe of the framed link $L$ and $\tilde{J}_{\cC,V}(L)$ is as above.
\end{defn}

\begin{thm}
$\J{\cC}{V}(L)$ is an invariant of links.
\end{thm}

\subsubsection{The colored Jones polynomial}
An example ... \todo{}

{\color{green}
We need to talk about orientations all through here! It doesn't matter for colored Jones, but it will for P and Q. -S
}

\todo{Look up some references for colored Jones, and see what conventions people use?}

The $n$-colored Jones polynomial is constructed this way, in the category $\mathcal{A}_{\infty}$ with the simple object $\JW{n}$.  The Temperley-Lieb planar algebra, at $q$ not a root of unity, has this structure, and is a simple place to do calculations.  The cup map becomes
\inputtikz{JWcup}
 and cap is
\inputtikz{JWcap}
and a crossing is
\inputtikz{JWcrossing},
 with each crossing of strands resolved in the usual Kauffman way.  So for example, to compute the third colored Jones polynomial of the Hopf link, we evaluate

 \inputtikz{CJHopf}\todo{some of these f's should be upside down, following orientations!}

This picture can be simplified by remembering that $\JW{n}$ is an idempotent, and therefore we can push all the $\JW{n}$s around on each component until they're all next to each other, and replace them by a single $\JW{n}$.  Therefore we can compute the $n$-colored Jones polynomial by $n$-cabling our link (ie, replacing all strands by $n$ parallel strands) and putting one $\JW{n}$ on each component.

\todo{Can we repeat this picture, with fewer f's?}

\subsection{$\frac{1}{2}\mathcal{D}_{14}$ is a specialization of $G_2$}

If $X$ is an object in a ribbon category such that $X \otimes X \cong \id \oplus X \oplus A \oplus B$ then it is reasonable to guess that the category you are looking at might be equivalent to a tensor category coming from $G_2$ with $X$ playing the role of the $7$-dimensional representation.  Thus one might guess from Equation \eqref{eq:P-squared} that $D_{14}$ might be related to $G_2$, since in $D_{14}$ we have $P \tensor P \iso f^{(0)} \directSum f^{(4)} \directSum f^{(8))} \directSum P$.  In this section we prove such a result using results of Kuperberg \cite{MR1265145}.  Applying Kuperberg's result requires some direct but tedious calculations.  In work in progress, Snyder has shown that, outside of a few small exceptions, all braided non-symmetric tensor categories with $X \otimes X \cong \id \oplus X \oplus A \oplus B$ come from quantum $G_2$, which would obviate the need for these calculations.

In this section we will be using two semisimple versions of the representation theory of $G_2$ at a root of unity. The first, $U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}}$ is Kuperberg's diagram category $U_q(\mathfrak{g}_2)\text{-spider}$ modulo negligible morphisms. The second, $U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}$, 
is the category of tilting modules of Lusztig's integral form of $U_q(\mathfrak{g}_2)$, again modulo negligible morphisms. Usually, the integral form is defined over $\Integer[q,q^{-1}]$, but we will need to enlarge the ground ring to $\Integer[q,q^{-1},(q+q^{-1})^{-1}]$; this is harmless, as we'll only ever consider roots of unity `larger' than $i$.

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 semisimple pivotal tensor category $\cC$, such that trees with $5$ or fewer boundary points form a basis for the spaces $\Inv{\cC}{V^{\otimes k}}$ for $k \leq 5$. There is a pivotal functor $\cF: U_q(\mathfrak{g}_2)\text{-spider} \rightarrow \cC$, which sends the single strand to $V$ and the trivalent vertex to the trivalent vertex.  If $\cC$ is a ribbon category, then $\cF$ is a ribbon functor.
\end{thm}

\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 \mathfig{0.04}{G2/strand},\\
\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}

\begin{lem}
If $q$ is a root of unity, then $\left(U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}}\right)^{modularize} \cong U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}.$
\end{lem}
\begin{proof}
We first describe the action of the restricted integral form of the quantum group $U_q(\mathfrak{g}_2)$ on $$V_{(1,0)} = \Integer[q,q^{-1}]\{v_{1,0}, v_{-1,1},v_{2,-1},v_{0,0},v_{-2,1},v_{1,-1},v_{-1,0}\}$$ via
\begin{equation*}
\mathfig{0.65}{G2/standard-rep}
\end{equation*}
where the red arrows (pointing left and down-right) indicate the actions of the generators $F_1$ and $F_2$, the blue arrows (pointing right and up-left) indicate the actions of the generators $E_1$ and $F_1$, and $K_1$ and $K_2$ act by $q^3$ to the power of the first and second labels of each weight vector. The arrows with $[2]$ next to them indicate that the corresponding generator takes a weight vector to $[2] = q+q^{-1}$ times another. The higher divided powers for $E_2$ and $F_2$ are all zero, and the higher divided powers for $E_1$ and $F_1$ are zero except that $E_1^{(2)}v_{-2,1} = v_{2,-1}$ and $F_1^{(2)}v_{2,-1}=v_{-2,1}$.
Thus for example, $K_2 v_{-2,1} = q^3 v_{-2,1}$ and $E_1 v_{-2,1} = (q+q^{-1})v_{0,0}$.

To describe a trivalent vertex in $U_q(\mathfrak{g}_2)\text{-tilting}$, we need to specify a map from $V_{(1,0)} \rightarrow V_{(1,0)}^{\otimes 2}$, which we describe by its action on the highest weight vector:
\begin{equation*}
\operatorname{tri}(v_{1,0}) = (q+q^{-1})\left(-q^6 v_{1,0} \tensor v_{0,0} + q^4 v_{-1,1} \tensor v_{2,-1} - q v_{2,-1} \tensor v_{-1,1} + v_{0,0} \tensor v_{1,0}\right)
\end{equation*}

Similarly, we have maps $\operatorname{cup} : V_{(0,0)} \rightarrow V_{(1,0)}^{\otimes 2}$ and $\operatorname{cap} : V_{(1,0)}^{\otimes 2} \rightarrow V_{(0,0)}$ described by
\begin{align*}
\operatorname{cup}(v_{0,0}) & = q^{10} v_{1,0} \tensor v_{-1,0} - q^9 v_{-1,1} \tensor v_{1,-1} + q^6 v_{2,-1} \tensor v_{-2,1} \\
						& - q^6 (q+q^{-1}) v_{0,0} \tensor v_{0,0} + q^4 v_{-2,1} \tensor v_{2,-1} - q v_{1,-1} \tensor v_{-1,1} + v_{-1,0} \tensor v_{1,0} \\
\end{align*}
and
\begin{align*}
\operatorname{cap}(v_{1,0} \tensor v_{-1,0}) & = v_{0,0} &
\operatorname{cap}(v_{-1,1} \tensor v_{1,-1} ) & = - q^{-1} v_{0,0} \\
\operatorname{cap}(v_{2,-1} \tensor v_{-2,1}) & = q^{-4} v_{0,0} &
\operatorname{cap}(v_{0,0} \tensor v_{0,0}) & = -q^{-6} (q+q^{-1})^{-1} v_{0,0} \\
\operatorname{cap}(v_{-2,1} \tensor v_{2,-1}) & = q^{-6} v_{0,0} &
\operatorname{cap}(v_{1,-1} \tensor v_{-1,1}) & = -q^{-9} v_{0,0} \\
\operatorname{cap}(v_{-1,0} \tensor v_{1,0}) &= q^{-10} v_{0,0}.
\end{align*}
Setting $q=1$ we see the formulas for $\operatorname{cup}$ and $\operatorname{cap}$ are symmetric, so $V_{(1,0)}$ is symmetrically self-dual.
We check that $\operatorname{tri}$ really is a trivalent vertex, by calculating that its rotational eigenvalue is $+1$, by comparing the following two morphisms.
\begin{equation*}
\mathfig{0.16}{G2/rotate-vertex} = \mathfig{0.08}{G2/vertex2}
\end{equation*}

Since $\dim \Inv{}{V_{(1,0)}^{\otimes 0}} = \dim \Inv{}{V_{(1,0)}^{\otimes 2}} = \dim \Inv{}{V_{(1,0)}^{\otimes 3}} = 1$ and $\dim \Inv{}{V_{(1,0)}} = 0$, trees form a basis of $\Inv{}{V_{(1,0)}^{\otimes k}}$ for $k \leq 3$.
We can calculate values of $d$, $b$, and $t$ in $U_q(\mathfrak{g}_2)\text{-tilting}$ by comparing the morphisms on by sides of the equations
\begin{align*}
d & = \mathfig{0.07}{G2/cup-cap} \\
b \mathfig{0.08}{G2/strand2} & = \mathfig{0.12}{G2/bigon-vertex} \\
t \mathfig{0.07}{G2/vertex2} & = \mathfig{0.1}{G2/triangle-vertex},
\end{align*}
obtaining
\begin{align*}
d & = q^{10} + q^8 + q^2 + 1 + q^{-2} + q^{-8} + q^{-10} \\
b & = -q^6 - q^4 - q^2 - q^{-2} - q^{-4} - q^{-6} \\
t & = q^4 + 1 + q^{-4}.
\end{align*}
These are, unsuprisingly, exactly the coefficients appearing in the definition of the spider; we remark however that when Kuperberg discovered the $G_2$ spider, he didn't need to do this calculation!

Now, by Lemma \ref{lem:yucky} trees are independent in $\Inv{}{V^{\otimes 4}}$ and $\Inv{}{V^{\otimes 5}}$.  A dimension count shows that they must form a basis there.  Hence, by Theorem \ref{thm:kuperberg}, there is a functor from the diagrammatic category to the category of tilting modules.  As with all pivotal functors this gives a functor between semisimplifications $\cF: U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}} \rightarrow U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}$.  Finally, since $V_{(1,0)}$ is a tensor generator for the category of tilting modules (this is an easy induction using the Racah formula for tensor products \cite{MR2286123}), and since $U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}$ is modular \cite[Theorem 6]{MR2286123}, this functor is a modularization.
\end{proof}

\begin{rem}
Why did we need to invert $q+q^{-1}$? You can see from our formula for $\operatorname{cap}$ that something goes wrong if $q+q^{-1}=0$; in fact, the Weyl representation $V_{(1,0)}$ of the usual integral form is not even self-dual at $q+q^{-1} = 0$ and so is not a tilting module. The spider is still well behaved when $q+q^{-1}=0$ but presumably has nothing to do with the quantum group.
\end{rem}

\begin{rem}
In all likelihood $U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}}\cong U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}$, without modularizing.  However, proving this would require some detailed calculations. Whether these categories are equivalent before we even semisimplify is a very interesting question!
\end{rem}

Our next goal is to show that $\frac{1}{2} D_{14}$ is also equivalent to $\left(U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}}\right)^{modularize}$ at $q=e^{2\pi i\frac{23}{26}}$.  Here $P$ corresponds to the single strand, 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 $\frac{1}{2} D_{14}$.  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 maps to $P$.
\end{proof}

\begin{lem}
In $\frac{1}{2} D_{14}$ we have that $d$ is largest root of $x^6 - 6x^5 -24 x^4 +32 x^3 + 80x^2-32 x-64$, approximately $8.29623$, $b$ is the smallest positive root of $x^6-22x^5+115x^4-48x^3-37x^2+17x-1$, approximately $0.070448$ and $t$ is the smallest (in magnitude) negative root of $x^6 -304 x^5-3008624x^4+42227282x^3-3190558x^2+13349x+1$, approximately $-0.0000736155$.
\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 did with a computer, but which appears in \cite{MR1280463}).

\begin{align*}
\mathfig{0.1}{G2/bigon} &= \inputtikz{bigonD14} = \inputtikz{bigonD14simplified} \\
&=\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}%
We have the following equivalences of ribbon categories, at $q=e^{2\pi i\frac{23}{26}}$.
$$\frac{1}{2}D_{14} \cong \left(U_q(\mathfrak{g}_2)\text{-spider}^{\text{ss}}\right)^{modularize} \cong U_q(\mathfrak{g}_2)\text{-tilting}^{\text{ss}}.$$
See Figure \ref{fig:g2-weyl-chamber} for the correspondence between objects.
\end{thm}
\begin{proof}
We use Theorem \ref{thm:kuperberg} to construct a functor $\cF: U_q(\mathfrak{g}_2)\text{-spider} \rightarrow D_{14}$ sending the single strand to $P$. Since $D_{14}$ is semisimple this factors through the semisimplification.  Because $P$ is a tensor generator in $D_{14}$ and $D_{14}$ is modular, this functor is a modularization.  Since all modularizations are equivalent the theorem follow from the existence of $\cF$.

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 space.  Hence $\cF$ exists by Theorem \ref{thm:kuperberg}.

The correspondence between simples shown in Figure \ref{fig:g2-weyl-chamber}, can be computed inductively.  Begin with the observation that $P$ is sent to $V_{(1,0)}$ by construction; after that, everything else is determined by working out the tensor product rules in both categories.  To compute tensor products in a truncation of $G_2$ we use the convenient `Racah formula' (or `Steinberg rule') given in \cite{MR2286123}.  For example, $V_{(1,0)} \otimes V_{(1,0)} \cong \id \oplus V_{(1,0)} \oplus V_{(0,1)} \oplus V_{(2,0)}$ while $P \otimes P \cong \id \oplus P \oplus W_4 \oplus W_8$, and by computing dimensions we see that $V_{(0,1)}$ goes to $W_4$ while $V_{(2,0)}$ goes to $W_8$ (and not vice-versa).
\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 $q=e^{2\pi i\frac{23}{26}}$, and the
correspondence with the even vertices of $D_{14}$.}
\label{fig:g2-weyl-chamber}
\end{figure}


Note that $q=e^{2\pi i\frac{23}{26}}$ 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.

Computer calculations suggest that $(G_2)_{\frac{2}{3}}$ is equivalent to a subcategory of $(C_3)_3$.  We expect that the techniques in this section will carry over to that example.

\begin{cor}
\label{cor:G2-identities}
This isomorphism of categories immediately gives several identities between knot polynomials, analogous to those of Theorems  \ref{thm:identities-2}, \ref{thm:identities-3}, \ref{thm:identities-4} and \ref{thm:identities-5}:
\begin{align*}
\restrict{\J{\SL{2}}{(2)}(K)}{q=\exp(\frac{2\pi i}{52})} & =\restrict{\J{G_2}{V_{(0,2)}}(K)}{q=\exp{2\pi i \frac{23}{26}}}\\
\restrict{\J{\SL{2}}{(12)}(K)}{q=\exp(\frac{2\pi i}{52})} & = 2 \J{D_{14}}{P}(K)\\
& = 2\restrict{\J{G_2}{V_{(1,0)}}(K)}{q=\exp{2\pi i \frac{23}{26}}} \\
\intertext{and also}
\restrict{\J{\SL{2}}{(12)}(K)}{q=\exp(\frac{2\pi i}{52})} & = 2 \J{D_{14}}{Q}(K) \\
& = 2\restrict{\J{G_2}{V_{(2,1)}}(K)}{q=\exp{2\pi i \frac{23}{26}}}
\end{align*}
Further identities relating $\J{\SL{2}}{(4)}(K)$, $\J{\SL{2}}{(6)}(K)$, $\J{\SL{2}}{(8)}(K)$ and $\J{\SL{2}}{(10)}(K)$ at $q=\exp(\frac{2\pi i}{52})$ with evaluations of various $G_2$ knot polynomials at $q=\exp{2\pi i \frac{23}{26}}$ can be read off from Figure \ref{fig:g2-weyl-chamber}.
\end{cor}