\subsection{A basis for the planar algebra}%
\label{sec:basis}%
In this section we present an explicit basis for the
planar algebra, and use this to show that the generators and relations
presentation from Definition \ref{def:pa} really does result in a
positive definite planar algebra.

Each vector space $\pa_m$ of the planar algebra also appears as a
$\operatorname{Hom}$ space of the corresponding tensor category of
projections, specifically as $\Hom{}{\id}{X^{\tensor m}}$. We'll use a standard approach for describing bases for
semisimple tensor categories, based on tree-diagrams.

For each triple of self-adjoint minimal projections $p,q,r$, we need to fix an
orthogonal basis for $\Hom{}{\id}{p \tensor q \tensor r}$. Call these bases $\{v_\lambda\}_{\lambda \in \mathcal{B}(p,q,r)}$.
If we take the adjoint of $v_\lambda \in \Hom{}{\id}{p \tensor q \tensor r}$, we get
$v_\lambda^* \in \Hom{}{p \tensor q \tensor r}{\id}$.

In fact, we'll only need to do this when one of the the three projections $p, q$ and $r$ is just $X$. In these cases, we've already implicitly described
the $\operatorname{Hom}$ spaces in Lemmas \ref{LemmasB1}, \ref{LemmasB2} and \ref{LemmasB3}.

We can now interpret certain planar trivalent graphs as notations for elements
of the planar algebra. The graphs have oriented edges labelled by projections, but where we allow
reversing the orientation and replacing the projection with its dual.\footnote{We'll leave off orientations on edges labelled by $X$, since it's self dual. In fact, all the minimal projections are self dual, except for $P$ and $Q$ when $n$ is even, in which case $\overline{P} = Q$ and $\overline{Q}=P$.} The graphs have vertices labelled by elements of the sets $\mathcal{B}(p,q,r)$
described above (where $p, q$ and $r$ are the projections on the edges
\emph{leaving} the vertex).  If $\card{\mathcal{B}(p,q,r)}=1$ we may leave off the label at that vertex.

To produce an element of the planar algebra from such a graph, we simply
replace each edge labelled by a projection $p$ in $\pa_{2m}$ with $m$
parallel strands, with the projection $p$ drawn across them, and each
trivalent vertex labelled by $\lambda$ with the element $v_\lambda \in
\Hom{}{\id}{p \tensor q \tensor r}$.

As a first example,
\begin{defn}
We call the norm of the element $v_\lambda$, with $\lambda \in
\mathcal{B}(p,q,r)$, the theta-symbol:

\begin{equation*}
\theta(p,q,r;\lambda) := \mathfig{0.2}{basis/theta}
\end{equation*}
\end{defn}

\begin{defn}
Fix a list of minimal projections $(p_i)_{0\leq i\leq k+1}$, called the
boundary. A tree diagram for this boundary is a trivalent graph of the form:
\begin{equation*}
\mathfig{0.75}{basis/tree}
\end{equation*}

It is labelled by
\begin{itemize}
\item another list of minimal projections $(q_i)_{1\leq i \leq k-1}$
such that $q_i$ is a summand of $q_{i-1} \tensor p_i$ for each $1\leq i
\leq k$, or equivalently that $\mathcal{B}(\overline{q_i}, q_{i-1}, p_i) \neq
\eset$ (here we make the identifications $q_0 = p_0$ and $q_k =
\overline{p_{k+1}}$),
\item and for each $1\leq i\leq k$, a choice of orthogonal basis vector
$v_{\lambda_i}$, with $\lambda_i \in \mathcal{B}(\overline{q_i}, q_{i-1}, p_i)$.
\end{itemize}
\end{defn}

\begin{thm}
\label{thm:identity-tree}
The $k$-strand identity can be written as a sum of tree diagrams.
(We'll assume there are no multiple edges in the principal graph for the exposition here; otherwise, we need to remember labels at vertices.)
Let $\Gamma_{k-1}$ be the set of length $k-1$ paths on the principal graph starting at $X$. (Thus if $\gamma \in \Gamma_{k-1}$, $\gamma_0 = X$ and the endpoint of the path is $\gamma_{k-1}$.)
Then
\begin{equation*}
\mathfig{0.18}{basis/identity} = \sum_{\gamma \in \Gamma_{k-1}} \prod_{i=0}^{k-2} \frac{\tr{\gamma_{i+1}}}{\theta(\overline{\gamma_{i}},\gamma_{i+1},X))} \mathfig{0.18}{basis/identity-term}
\end{equation*}
\end{thm}
\begin{proof}
We induct on $k$.

When $k=1$, the result is trivially true; the only path in $\Gamma_0$ is the constant path, with $\gamma(0)=X$, and there's no coefficient.

To prove the result for $k+1$, we replace the first $k$ strands on the left, obtaining
\begin{equation*}
\mathfig{0.18}{basis/identity_1} = \sum_{\gamma \in P_k} \prod_{i=0}^{k-2} \frac{\tr{\gamma_{i+1}}}{\theta(\overline{\gamma_{i}},\gamma_{i+1},X)} \mathfig{0.18}{basis/identity-term_1}
\end{equation*}
and then use the identity
\begin{equation*}
\mathfig{0.075}{basis/gamma_1} = \sum_{\substack{\text{$\gamma_k$ adjacent}\\\text{to $\gamma_{k-1}$}}} \frac{\tr{\gamma_k}}{\theta(\overline{\gamma_{k-1}},\gamma_k,X)} \mathfig{0.075}{basis/gamma-term}
\end{equation*}
(which certainly holds with some coefficients, by the definition of the principal graph, and with these particular coefficients by multiplying in turn both sides by each of the terms on the right)
to obtain the desired result.
\end{proof}

\begin{thm}
The tree diagrams with boundary labelled entirely by $X$ give a positive orthogonal basis for the
invariant space $\Hom{}{\id}{X^{\tensor n}}$.
\end{thm}
\begin{rem}
Actually, the tree diagrams with boundary $(p_i)$ give an orthogonal basis for the invariant space
$\Hom{}{\id}{\Tensor_i p_i}$, but we won't prove that here. We'd need to exhibit explicit bases for all the triple invariant spaces in order
to check positivity, and a slightly stronger version of Theorem \ref{thm:identity-tree}.
\end{rem}
\begin{proof}
To see that the tree diagrams are all orthogonal is just part of the
standard machinery of semisimple tensor categories --- make repeated use
of the formulas
\begin{align*}
\mathfig{0.08}{basis/tr-p} & = \tr{p} \\
\mathfig{0.12}{basis/open-theta} & = \delta_{p=q} \delta_{\mu=\lambda^*} \frac{\theta(\overline{p},r,s;\lambda)}{\tr{p}} \mathfig{0.032}{basis/strand}
\end{align*}
where $\lambda \in \mathcal{B}(\overline{p},r,s)$ and $\mu \in \mathcal{B}(\overline{s},\overline{r},q)$.
(In fact, this proves that the tree diagrams are orthogonal for arbitrary boundaries).

The norm of a tree diagram is a ratio of theta symbols and traces of
projections.
The trace of $f^{(k)}$ is $\qi{k+1}$, and $\tr{P} = \tr{Q} = \frac{\qi{2n-2}}{2}$, and these quantities are all positive at our value of $q$. Further, the theta symbols with one edge labelled by $X$ are all easy to calculate (recall the relevant one-dimensional bases for the $\operatorname{Hom}$ spaces were described in Lemmas \ref{LemmasB1}, \ref{LemmasB2} and \ref{LemmasB3}), and
in fact are just equal to traces of these same projections:
\begin{align*}
\theta(f^{(k-1)},f^{(k)},X) & = \tr{f^{(k)}} \\
\theta(f^{(2n-3)},P,X) & = \tr{P} \\
\theta(f^{(2n-3)},Q,X) & = \tr{Q} \\
\end{align*}
Since these are all positive, the norms of tree diagrams are positive.

To see that the tree diagrams span, we make use of Theorem \ref{thm:identity-tree}, and Lemma \ref{lem:no-homs}.
Take an arbitrary open diagram $D$ with $k$ boundary points, and write it as $D \cdot \id_k$. Apply Theorem \ref{thm:identity-tree} to $\id_k$, and
observe that all terms indexed by paths not ending at $f^{(0)}$ are zero, by Lemma \ref{lem:no-homs} (here we think of $D$ as having an extra boundary point labeled by $f^{(0)}$, so we get a map from $f^{(0)}$ to the endpoint of the path).
In the remaining terms, we have the disjoint union (after erasing the innermost edge labeled by $f^{(0)}$) of a closed diagram
and a tree diagram. Since all closed diagrams can be evaluated, by Corollary \ref{cor:evaluation}, we see we have rewritten an arbitrary diagram
as a linear combination of tree diagrams.
\end{proof}

\begin{cor}
The planar algebra given by generators and relations in Definition
\ref{def:pa} is positive definite.
\end{cor}

Therefore, $\pa$ is indeed the subfactor planar algebra with principal graph $D_{2n}$.
