nonsimplerecurrence.tex
\subsection{Non-Simple Loop Recurrences}\label{s:nonsimple}

As mentioned in the previous section, multiplication by simple loops does not provide sufficient recurrence relations to reduce any trace diagram to its simplest pieces. In the rank three case, for example, multiplication by $\Tr{\xb_1\xb_2^{-1}\xb_3}$ requires a non-simple loop:
$$\tikz[scale=1.2]{ \draw[trivalent] (0,0)to[bend left=80]node[small matrix]{\xb_1}(0,1)to[bend left=80]node[small matrix]{\xb_2}(0,0) (0,0)to[bend right=20](.5,-.2)to[bend right=80]node[small matrix]{\xb_3}(.5,1.2)to[bend right=20](0,1) (.5,-.2)to[bend right=20](1,-.4)to[bend right=80](1,1.4)to[bend right=20](.5,1.2); \draw[blue,trivalent] (-.1,-.1)to[bend left=80]node[small matrix]{\xb_1}(0,1.1)to[bend left=80]node[small matrix,pos=.55]{\xb_2}(.2,0) to[bend right=20](.5,-.1)to[bend right=80]node[small matrix]{\xb_3}(.4,1.1)to[bend left=40](1.1,1.5) to[bend left=80](1.1,-.5)to[bend left=30](.5,-.3)to[bend left=20](-.1,-.1); }$$
While Corollary \ref{c:simplerecurrence} can be applied to any simple loop in a trace diagram, it is not sufficient to compute the value of an arbitrary diagram. The simplest case where it fails is the barbell'' depicted in Figure \ref{fig:barbellcase}. There is no simple loop recurrence, since subtracting one from either loop produces a non-admissible diagram. However, the non-simple loop depicted on the right does provide a recurrence.
\begin{figure}[h]
\tikz[trivalent,every node/.style={basiclabel}]{
\draw(0,-.1)circle(.3)(0,1.1)circle(.3)(0,.2)to node[auto]{$2$}(0,.8);
\node at(.5,.1){$1$};\node at(.5,1.3){$1$};}
\tikz[trivalent,every node/.style={basiclabel}]{
\draw(0,-.1)circle(.3)(0,1.1)circle(.3)(0,.2)to(0,.8);
\foreach\xa in{1,-1}{
\draw[blue,xscale=\xa](0,-.6)to[out=180,in=-90](-.5,-.1)to[wavyup](-.2,.5)to[wavyup](-.5,1.1)to[out=90,in=180](0,1.6);
}
\node[blue,rightlabel]at(.6,1.3){$\gamma$};
}.
\caption{Trace diagram with no simple recurrence (left). The non-simple loop which can be used to produce a recurrence (right).}\label{fig:barbellcase}
\end{figure}

This section describes the process for generating recurrences in such cases. The main difference is in the gluing process: the vertices must be glued in a specific order, and the coefficients at vertices depend on changes to the labels made at prior vertices. For example, in the product
$$\tikz[trivalent,scale=1.3]{ \coordinate(vxa)at(0,.45) edge[bend left]node[rightlabel,pos=1]{c}(-.4,1) edge[bend right]node[leftlabel,pos=1]{b}(.4,1) edge[]node[basiclabel,anchor=north,pos=1]{a}(0,0); \draw[blue](-.2,0)to[wavyup](-.6,1)node[leftlabel]{1} (.2,0)to[wavyup](.6,1)node[rightlabel]{1};} ,$$
applying Proposition \ref{l:vertexglue} on the left will change the values of $a$ and $c$ at the vertex. Applying the same identity on the right of the vertex requires using the new values.

%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{The Barbell}
We begin with a special case. Define
$$\tilde\chi_{a,c}^b({\bf X},{\bf Y})= \tikz[trivalent]{ \draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9); \node[basiclabel]at(.5,.1){a};\node[basiclabel]at(.5,1.6){c}; \node[rightlabel]at(0,.5){b}; \node[small matrix]at(-.4,-.3){\bf X};\node[small matrix]at(-.4,1.3){\bf Y};}$$
When applied to the simple loops in the diagram, \eqref{eq:loopone} implies
\begin{align*}
\Tr{\bf X} \cdot \tilde\chi_{a,c}^b &=
\tilde\chi_{a+1,c}^b - \nFus ba{a-1}a{a-1} \cdot \tilde\chi_{a-1,c}^b \\
\Tr{\bf Y} \cdot \tilde\chi_{a,c}^b &=
\tilde\chi_{a,c+1}^b - \nFus bc{c-1}c{c-1} \cdot \tilde\chi_{a,c-1}^b.
\end{align*}
These equations can be rearranged to provide recurrences for the $\bf X$ and $\bf Z$ loops. The third recurrence corresponds to the multiplication by $\Tr{XY}$:

\begin{proposition}\label{prop:barbellrecurrence}
$$\tikz[trivalent]{ \foreach\xa in{1,-1}{ \draw[blue,xscale=\xa](0,-.9)to[out=180,in=-90](-.6,-.3)to[wavyup](-.2,.5)to[wavyup](-.6,1.3)to[out=90,in=180](0,1.9);} \draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9); \node[basiclabel]at(.2,-.3){a};\node[basiclabel]at(.2,1.3){c}; \node[coordinate,pin={[basiclabel,pin edge={darkgray}]right:b}]at(0,.8){}; \node[small matrix]at(-.4,-.3){\bf X};\node[small matrix]at(-.4,1.3){\bf Y}; \node[blue,rightlabel]at(.5,1.6){1}; } = \sum_{\substack{a'\in\iadm{a,1}\\ b'\in\iadm{b,1}\cap\iadm{a,a'}\\ c'\in\iadm{c,1}\cap\iadm{c,b'}\\ b''\in\iadm{b',1}\cap\iadm{a,a'}\cap\iadm{c,c'}}} \mfs_{a'}(1,a)\mfs_{c'}(1,c)\nFus aa{a'}b{b'}\nFus cc{c'}b{b'}\nFus{a'}a{a'}{b'}{b''}\nFus{c'}c{c'}{b'}{b''} \tikz[trivalent]{ \draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9); \node[basiclabel]at(.5,.1){a'};\node[basiclabel]at(.5,1.6){c'}; \node[rightlabel]at(0,.5){b''}; \node[small matrix]at(-.4,-.3){\bf X};\node[small matrix]at(-.4,1.3){\bf Y};}.$$
\begin{proof}
The computation is a straightforward application of Lemma \ref{l:vertexglue} and the bubble identity \eqref{eq:bubbleidentity}. For simplicity, the matrices are not shown on the diagram; nothing changes if they are included.
\begin{align*}
\tikz[trivalent]{
\foreach\xa in{1,-1}{
\draw[blue,xscale=\xa](0,-.9)to[out=180,in=-90](-.6,-.3)to[wavyup](-.2,.5)to[wavyup](-.6,1.3)to[out=90,in=180](0,1.9);}
\draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9);
\node[basiclabel]at(.15,-.25){$a$};\node[basiclabel]at(.15,1.35){$c$};
\node[coordinate,pin={[basiclabel,pin edge={darkgray}]right:$b$}]at(0,.8){};
\node[blue,rightlabel]at(.5,1.6){$1$};
}
=
\hspace{-.5cm}
\hspace{-.5cm}
&    \mfs_{b'}(1,b)\Fus aa{a'}b{b'}\mfs_{b'}(1,b)\Fus cc{c'}b{b'}\mfb_{b'}(1,b)
\tikz[trivalent]{
\draw[blue](0,-.9)to[out=0,in=-90](.6,-.3)to[wavyup](.2,.5)to[wavyup](.6,1.3)to[out=90,in=0](0,1.9)
(0,-.9)to[out=180,in=-90](-.6,-.5)to[out=90,in=-135](-.4,-.3)
(0,1.9)to[out=180,in=90](-.6,1.5)to[out=-90,in=135](-.4,1.3);
\draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9);
\node[basiclabel]at(.15,-.25){$a$};\node[basiclabel]at(.15,1.35){$c$};
\node[leftlabel]at(-.2,.15){$a'$};\node[leftlabel]at(-.3,.9){$c'$};
\node[coordinate,pin={[basiclabel,pin edge={darkgray}]right:$b'$}]at(0,.8){};
\node[blue,rightlabel]at(.5,1.6){$1$};
} \\
=
\hspace{-1cm}
\hspace{-1cm}
& \Fus aa{a'}b{b'}\Fus cc{c'}b{b'}\mfb_{b'}(1,b)
\mfs_{b''}(1,b')\Fus {a'}a{a'}{b'}{b''}\mfs_{b''}(1,b')\Fus{c'}c{c'}{b'}{b''}\mfb_{b''}(1,b')
\tikz[trivalent]{
\draw[blue]
(0,-.9)to[out=180,in=-90](-.6,-.5)to[out=90,in=-135](-.4,-.3)
(0,-.9)to[out=0,in=-90](.6,-.5)to[out=90,in=-45](.4,-.3)
(0,1.9)to[out=180,in=90](-.6,1.5)to[out=-90,in=135](-.4,1.3)
(0,1.9)to[out=0,in=90](.6,1.5)to[out=-90,in=45](.4,1.3);
\draw(0,-.3)circle(.4)(0,1.3)circle(.4)(0,.1)to(0,.9);
\node[toplabel]at(0,-.7){$a$};\node[bottomlabel]at(0,1.7){$c$};
\node[leftlabel]at(-.2,.15){$a'$};\node[leftlabel]at(-.3,.9){$c'$};
\node[rightlabel]at(.3,.1){$a'$};\node[rightlabel]at(.3,1.1){$c'$};
\node[rightlabel]at(0,.5){$b''$};
\node[blue,rightlabel]at(.6,-1){$1$};
\node[blue,rightlabel]at(.5,1.6){$1$};
} \\
%    =
%    \hspace{-1cm}
%    \hspace{-1cm}
%    & \Fus aa{a'}b{b'}\Fus cc{c'}b{b'}\mfb_{b'}(1,b)
%        \Fus {a'}a{a'}{b'}{b''}\Fus{c'}c{c'}{b'}{b''}\mfb_{b''}(1,b')
%        \mfs_{a'}(1,a)\mfb_{a'}(1,a)\mfs_{c'}(1,c)\mfb_{c'}(1,c)
%    \tikz[trivalent,every node/.style={basiclabel}]{
%        \draw(0,-.1)circle(.3)(0,1.1)circle(.3)(0,.2)to node[auto]{$b''$}(0,.8);
%        \node at(.5,.1){$a'$};\node at(.5,1.3){$c'$};
%    } \\
=
\hspace{-1cm}
\hspace{-1cm}
&    \mfs_{a'}(1,a)\mfs_{c'}(1,c)\nFus aa{a'}b{b'}\nFus cc{c'}b{b'}\nFus{a'}a{a'}{b'}{b''}\nFus{c'}c{c'}{b'}{b''}
\tikz[trivalent,every node/.style={basiclabel}]{
\draw(0,-.1)circle(.3)(0,1.1)circle(.3)(0,.2)to node[auto]{$b''$}(0,.8);
\node at(.5,.1){$a'$};\node at(.5,1.3){$c'$};
}
\end{align*}
\end{proof}
\end{proposition}

The explicit computation of these functions is shown in Table \ref{t:barbell} with $x=\mathrm{tr}(\xb)$, $y=\mathrm{tr}(\mathbf{Y})$, and $z=\mathrm{tr}(\xb\mathbf{Y})$. We omit the cases where $b=0$, since they can be obtained as a product of rank one central functions.

{\renewcommand\arraystretch{1.5}
\begin{table}[ht]
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
{\bf rank} & $\tilde\chi_{a,c}^b$ & formula\\
\hline\hline
$\bf 4$ & $\tilde\chi_{1,1}^2$ & $z-\frac{x y}{2}$\\
\hline
$\bf 5$ & $\tilde\chi_{2,1}^2$ & $x z-\frac{x^2 y}{2}$\\
& $\tilde\chi_{1,2}^2$ & $y z-\frac{x y^2}{2}$\\
\hline
$\bf 6$ & $\tilde\chi_{3,1}^2$ & $x^2 z-\frac{x^3 y}{2} + \frac{x y}{3}-\frac{2 z}{3}$ \\
& $\tilde\chi_{2,2}^2$ & $x y z-\frac{1}{2} x^2 y^2$ \\
& $\tilde\chi_{1,3}^2$ & $y^2 z-\frac{x y^3}{2} + \frac{x y}{3}-\frac{2 z}{3}$ \\
\hline
$\bf 7$ & \emph{4 cases} & $\cdots$ \\
\hline
$\bf 8$ & $\tilde\chi_{2,2}^4$ & $z^2-x y z+\frac{x^2 y^2}{6}+\frac{x^2}{3}+\frac{y^2}{3}-\frac{4}{3}$ \\
& \emph{5 more cases} & $\cdots$ \\
\hline
\end{tabular}
\medskip
\caption{Barbell Functions.}\label{t:barbell}
\end{center}
\end{table}
}

%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{The General Case}

Other non-simple recurrence formulas may be similarly derived. In every case, the coefficients will be signed summations over the normalized fusion coefficients, with the sign depending on the particular topology and orientation of the diagram.