\subsection{A partial braiding} \label{sec:braiding}

Recall the definition of a crossing given in \S \ref{ssTL}.  This still defines an element of $\pa$ and, away from $S$ boxes, diagrams related by a framed three-dimensional isotopy are equal in the planar algebra.  However, one needs to be careful manipulating these crossings and $S$ boxes at the same time.

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

\item\label{pullunder}
$\inputtikz{pullstringunderS}$

\end{enumerate}
\end{thm}

\begin{proof}

\begin{enumerate}
\item
This is a straightforward consequence of part \ref{enoughcappings} of
Theorem \ref{easyconsequences}. Capping at any position from $2$ to
$4n-3$ in either of the above pictures gives zero; the first capping of
the lefthand picture is

\begin{align*}
\begin{tikzpicture}[baseline]
    \clip (-1.1,-1.4) rectangle (1.1,1.4);
    \node (S) at (0,0) [circle, draw] {$S$};
    \filldraw (S.180) circle (.5mm);
    \draw[rounded corners=2mm] (S.135) -- ++(90:.5cm) -- ++(180:.5cm) -- ++ (-90:1.7cm) -- ++(0:1.6cm) -- ++(90:3cm);
    \draw (S.90) node [above] {...};
    \draw (S.45) -- ++(90:15mm);
\end{tikzpicture} & =i S, \\
\intertext{and the first capping of the righthand picture is}
\inputtikz{cappullstringoverS1}
& = iq^{3/2} \inputtikz{cappullstringoverS2} \\
& = iq^{3/2} (iq^{1/2})^{4n-5} S \\
& = i S.
\end{align*}
Thus these two pictures are equal.

\item This is essentially identical to the previous argument, except that the factor picked up by resolving the crossings of the second picture is $$-iq^{-3/2} (-iq^{-1/2})^{4n-5} =- i;$$ hence the minus sign in the relation.
\end{enumerate}
\end{proof}

\begin{rem}
Upon reading this paper, one might hope that all subfactor planar algebras are braided, or partially braided.  Unfortunately this is far from being the case.  For the representation theory of the annular Temperley-Lieb category for $\qi{2} >2$, set out in \cite{MR1659204} and in the language of planar algebras in \cite{MR1929335}, implies that one cannot pull strands across lowest weight generators, even up to a multiple.  To see this, resolve all crossings in either of the equations in Theorem \ref{thm:passacrossS}; such an identity would give a linear dependence between ``annular consequences" of the generator.  For the other $\qi{2} < 2$ examples, namely $E_6$ and $E_8$,  \cite{math/0903.0144} shows that Equation \eqref{pullover} holds, but not Equation \eqref{pullunder}, even up to a coefficient.  The $\qi{2} = 2$ cases remain interesting.

%The $\delta < 2$ planar algebras are ``quantum subgroups" of $U_q(\mathfrak{su}_2)$ (in the sense of Ocneanu \cite{Ocneanu} or Kirillov-Ostrik \cite{KO}).  It would also be interesting to describe which quantum subgroups of other quantum groups satisfy relations like those in Theorem \ref{thm:passacrossS}.
\end{rem}

\begin{cor}
\label{cor:atmost1S}
Any diagram in $\pa$ is equal to a sum of diagrams involving at most one
$S$.
\end{cor}

\begin{proof}
When a diagram has more than one $S$, use the above relations to move one
of the $S$'s next to another one, then apply relation (\ref{twoS}) of
Definition \ref{def:pa} to replace the two $S$'s with a Jones-Wenzl
idempotent.  Resolve all the crossings and proceed inductively.
\end{proof}

\begin{cor}
\label{cor:evaluation} Every closed diagram is a multiple of the empty
diagram.
\end{cor}
\begin{proof}
By the previous corollary, a closed diagram can be written in terms of
closed diagrams with at most one $S$. If a closed diagram has exactly one
$S$, it must be zero, because the $S$ must have a cap attached to it
somewhere. If a closed diagram has no $S$'s, it can be rewritten as a
multiple of the empty diagram using Relation \ref{delta}, which allows us
to remove closed loops.
\end{proof}

\begin{cor}
\label{cor:spherical} The planar algebra $\pa$ is spherical.
\end{cor}
\begin{proof}
A braiding always suffices to show that a planar algebra is spherical; even
though there are signs when a strand passes underneath an $S$, we can
check that $\pa$ is spherical simply by passing strands above everything
else in the diagram.
\begin{equation*}
\mathfig{0.1}{spherical/1} = \mathfig{0.1}{spherical/2} =
\mathfig{0.1}{spherical/3}
\end{equation*}
\end{proof}

%We record some relations among annular consequences in case anyone ever needs them.
%
%\begin{fact}\label{acrelations}
%The following relations among annular consequences hold:
%$$A_{4n-3}(S)=-\qi{4n-4}A_{4n-4}(S) - \qi{4n-5} A_{4n-5}(S) - \cdots - \qi{2} A_2(S) - \qi{1} A_1(S);$$
%$$A_{4n-2}(S)=-\qi{4n-4} A_1(S) - \qi{4n-5} A_2(S) - \cdots - \qi{2} A_{4n-5} (S) - \qi{1} A_{4n-4}(S).$$
%\end{fact}
%
%This fact is proved by grabbing $(4n-3)$ strands and replacing them with $1-\JW{4n-3}$, written as a sum of non-identity Temperley-Lieb diagrams, and knowing the coefficients of these diagrams in $\JW{4n-3}$ (see for example [Morrison, ???]).