\subsection{The planar algebra $\pa$ is non-zero}
\label{sec:consistency}
In this section, we prove the following reassuring result.
\begin{thm}%
\label{thm:consistency}%
In the planar algebra $\pa$ described in
Definition \ref{def:pa}, the empty diagram is not equal to zero.
\end{thm}

The proof is fairly straightforward. We describe an algorithm for
evaluating any closed diagram in $\pa$, producing a number. Trivially,
the algorithm evaluates the empty diagram to $1$. We show that modifying
a closed diagram by one of the generating relations does not change the
result of the evaluation algorithm.

The algorithm we'll use actually allows quite a few choices along the
way, and the hard work will all be in showing that the answer we get
doesn't depend on these choices.\footnote{See \cite{MR1403861} and
\cite{MR2308953} for a related idea, sometimes called `confluence'.}
After that, checking that using a relation does not change the result
will be easy.

\begin{defn}[Evaluation algorithm]
\label{defn:algorithm}
This is a function from closed diagrams (no relations) to the complex
numbers.
\begin{enumerate}
\item \label{step1}%
If there are at least two $S$ boxes in the diagram, choose a pair of $S$
boxes, and an imaginary arc connecting them. (The arc should be
transverse to everything in the diagram.)

Multiply by $\frac 1 {\qi{2n-1}}$, and replace the chosen pair of $S$ boxes with a pair of $f^{(4n-4)}$ boxes, connected by $4n-4$ parallel
strands following the arc which cross above any strands of the diagram
that the arc crosses (as illustrated in Figure \ref{fig:algorithm}).
Further, for each $S$ box do the following.  Starting at the marked point walk clockwise around the box counting the number of strands you pass before you reach the point where the arc attaches. This gives two numbers; multiply the new picture by $i$ raised to the sum of these two numbers.  Restart the algorithm on the result.
\item \label{step2}%
If there is exactly one $S$ box in the diagram, evaluate as $0$.
\item \label{step3}%
If there are no $S$ boxes in the diagram, evaluate the diagram in
Temperley-Lieb at $q + q^{-1} = 2 \cos(\frac{\pi}{4n-2})$.
\end{enumerate}
\end{defn}

\begin{figure}[!ht]
\begin{align*}
\mathfig{0.17}{consistency/arc} & \mapsto \frac{i^{0+0}}{\qi{2n-1}} \mathfig{0.1}{consistency/arc-connection} \\ \\
\mathfig{0.17}{consistency/arc2} & \mapsto \frac{i^{1+0}}{\qi{2n-1}} \mathfig{0.25}{consistency/arc-connection2}
\end{align*}
\caption{The main step of the algorithm chooses an arc connecting a pair of $S$ boxes, and replaces them with Jones-Wenzl idempotents connected by $4n-4$ parallel strands that cross above the original strands of the diagram,
and inserts a factor of $\frac{1}{\qi{2n-1}}$ and a power of $i$ determined by the attachment points of the arc.}
\label{fig:algorithm}
\end{figure}

\begin{thm}
\label{thm:algorithm-well-defined}
The algorithm is well-defined, and doesn't depend on any of the choices
made.
\end{thm}
\begin{proof}
We'll prove this in five stages.
\begin{enumerate}[i]
\item \label{alg:1}%
If two applications of the algorithm use the same pairing of $S$
boxes, and the same arcs, but replace the pairs in different orders, we
get the same answer.
\item \label{alg:2}%
If we apply the algorithm to a diagram with exactly two $S$ boxes,
then we can isotope the arc connecting them without affecting the answer.
\item \label{alg:3}%
Isotoping any arc does not change the answer.
\item \label{alg:4}%
Changing the point at which an arc attaches to an $S$ box does not
change the answer.
\item \label{alg:5}%
Two applications of the algorithm which use different pairings of the $S$
boxes give the same answers.
\end{enumerate}

\begin{proof}[Stage \ref{alg:1}]
Switching the order of two pairs of $S$ boxes produces Temperley-Lieb
diagrams that differ only where the corresponding two arcs intersected;
there we see one set of $4n-4$ parallel strands passing either over or
under the other set of $4n-4$ parallel strands. However, by Lemma
\ref{lem:f-crossings}, these are the same in Temperley-Lieb at $q + q^{-1} =
2 \cos(\frac{\pi}{4n-2})$. Thus after Step \ref{step3} of the evaluation
algorithm we get the same result.
\noqed\end{proof}
\begin{proof}[Stage \ref{alg:2}]
This follows easily from the fact that Temperley-Lieb is braided, and the final statement in
Lemma \ref{lem:f-twist}.
\noqed\end{proof}
\begin{proof}[Stage \ref{alg:3}]
In order to isotope an arbitrary arc, we make use of Stage \ref{alg:1} to
arrange that this arc corresponds to the final pair of $S$ boxes chosen.
Stage \ref{alg:2} then allows us to move the arc.
\noqed\end{proof}
\begin{proof}[Stage \ref{alg:4}]
Changing the point of attachment of an arc by one step clockwise results in a
Temperley-Lieb diagram at Step \ref{step3} which differs just by a factor
of $i$, according to the first part of Lemma \ref{lem:f-twist}. See the second part of Figure \ref{fig:algorithm}, which illustrates exactly this situation. This
exactly cancels with the factor of $i$ put in by hand by the algorithm.  Furthermore, moving the point of attachment across the marked point does not change the diagram, but does multiply it by a factor of
$i^{4n-4}=1$.
\noqed\end{proof}
\begin{proof}[Stage \ref{alg:5}]
We induct on the number of $S$ boxes in the diagram. If there are fewer
than $3$ $S$ boxes, there is no choice in the pairing. If there are
exactly $3$ $S$ boxes, the evaluation is automatically $0$.

Otherwise, consider two possible first choices of a pair of $S$ boxes.
Suppose one choice involves boxes which we'll call $A$ and $B$, while the
other involves boxes $C$ and $D$. There are two cases depending on
whether the sets $\{A,B\}$ and $\{C,D\}$ are disjoint, or have one common
element, say $D=A$. If the sets are disjoint, we (making use of the
inductive hypothesis), continue the algorithm which first removes $A$ and
$B$ by next removing $C$ and $D$, and continue the algorithm which first
removes $C$ and $D$ by next removing $A$ and $B$. The argument given in
Stage \ref{alg:1} shows that the final results are the same.
Alternatively, if the sets overlap, say with $A=D$, we choose some fourth $S$ box, say
$E$. After removing $A$ and $B$, we remove $C$ and $E$, while after
removing $A$ and $C$ we remove $B$ and $E$, and in each case we then
finish the algorithm making the same choices in either application. The
resulting Temperley-Lieb diagrams which we finally evaluate in Step
\ref{step3} differ exactly by the two sides of the identity in Lemma
\ref{lem:2f}. (More accurately, in the case that the arcs connecting
these pairs of $S$ boxes cross strands in the original diagram, the
resulting Temperley-Lieb diagrams differ by the two sides of that
equation sandwiched between some fixed Temperley-Lieb diagram; see
the remark following Lemma \ref{lem:2f}.)
\noqed\end{proof}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:consistency}]
We just need to check that modifying a closed diagram by one of the
relations from Definition \ref{def:pa} does not change the answer.

\paragraph{Relation \eqref{delta}}
Make some set of choices for running the
algorithm, choosing arcs that avoid the disc in which the relation is
being applied. The set of choices is trivially valid both before and
after applying the relation. Once we reach Step \ref{step3} of the
algorithm, the Temperley-Lieb diagrams differ only by the relation, which
we know holds in Temperley-Lieb!

\paragraph{Relation \eqref{rotateS}}%
Run the algorithm, choosing the $S$ we want to rotate as one of the first
pair of $S$ boxes, using the same arc both before and after rotating the
$S$. The algorithm gives answers differing just by a factor of $i$,
agreeing with the relation. See Figure \ref{fig:rotateS}.

\paragraph{Relation \eqref{capS}}%
If there's exactly one $S$ box,
the algorithm gives zero anyway. If there's at least two $S$ boxes, choose the $S$ with a cap on it as a member of one of the pairs. Once
we reach Step \ref{step3}, the $S$ with a cap on it will have been
replaced with an $f^{(4n-4)}$ with a cap on one end, which gives $0$ in
Temperley-Lieb.

\paragraph{Relation \eqref{twoS}}%
When running the algorithm, on the diagram with more $S$ boxes, ensure that the pair of $S$ boxes affected
by the relation are chosen as a pair in Step \ref{step1}, with an arc
compatible with the desired application of the relation.

\begin{figure}[!ht]
\begin{align*}
\mathfig{0.14}{consistency/arc} & \mapsto  \frac{1}{\qi{2n-1}} \mathfig{0.09}{consistency/arc-connection} \\
-i \mathfig{0.14}{consistency/rotation} & \mapsto  \frac{i(-i)}{\qi{2n-1}} \mathfig{0.22}{consistency/rotation-connection}
\end{align*}
\caption{The left hand sides are related by the rotation relation \ref{rotateS}. The algorithm gives the same answer in both cases.}
\label{fig:rotateS}
\end{figure}

\end{proof}