\subsection{First consequences of the relations}
Recall from the introduction that we are considering the planar algebra $\pa$ generated by a single box $S$ with $4n-4$ strands, with $q=\exp(\frac{\pi i}{4n-2})$, modulo the following relations.
\begin{enumerate}

\item  A closed circle is equal to $[2]_q = (q+q^{-1}) = 2 \cos(\frac{\pi}{4n-2})$ times the empty diagram.

\item\inputtikz{rotateS}

\item\inputtikz{capS}

\item\inputtikz{twoS}

\end{enumerate}

\begin{rem}
Relation \eqref{delta} fixes the index $\qi{2}^2$ of the planar algebra as a `special value' as in \S \ref{sec:special-TL} of the form $\qi{2} = 2 \cos(\frac{\pi}{k+2})$.
Note that usually at special values, one imposes a further relation, that the corresponding Jones-Wenzl idempotent $f^{(k)}$ is zero, in order that the planar algebra be positive definite.
As it turns out, we \emph{don't} need to impose this relation by hand; it will follow, in Theorem \ref{easyconsequences}, from the other relations.

According to the philosophy of \cite{MR1929335, quadratic} any planar algebra is generated by boxes which satisfy ``annular relations"  like \eqref{rotateS} and \eqref{capS}, while particularly nice planar algebras require in addition only ``quadratic relations" which involve two boxes.  Our quadratic relation \eqref{twoS}, in which the two $S$ boxes are not connected, is unusually strong and makes many of our subsequent arguments possible. Notice that this relation also implies relations with a pair of $S$ boxes connected by an arbitrary number of strands.
\end{rem}

We record for future use some easy consequences of the relations of Definition \ref{def:pa}.

\begin{thm}\label{easyconsequences}
The following relations hold in $\pa$.
\begin{enumerate}
\item
\inputtikz{StimesS}

(Here $2n-2$ strands connect the two $S$ boxes on the left hand side.)
\item
\inputtikz{otherScaps}

\item\label{JWzero}
\inputtikz{JWzerostatement}

\item\label{enoughcappings}
 For $T, T'  \in \pa_{4n-3}$, if

$$
\inputtikz{enoughcappings}
$$

then $T=T'$.  More generally, if $T,T' \in \pa_m$ for $m \geq {4n-3}$, and $4n-4$ consecutive cappings of $T$ and $T'$ are equal, then $T=T'$.

\end{enumerate}
\end{thm}

\begin{proof}
\begin{enumerate}

\item This follows from taking a partial trace (that is, connecting top right strings to bottom right strings) of the diagrams of the two-$S$ relation (\ref{twoS}), and applying the partial trace relation from Equation \eqref{eq:PartialTraceRelation}.

\item This is a straightforward application of the rotation relation (\ref{rotateS}) and the capping relation (\ref{capS}).

\item Using Wenzl's relation (Equation \eqref{WenzlsRelation}) we calculate

\begin{align*}
\inputtikz{JWzeroWenzl}
\intertext{then replace $\qi{4n-3}$ and $\qi{4n-4}$ by $1$ and $\qi{2}$, and apply the two $S$ relation thrice, obtaining}
\inputtikz{JWzeroputinS}
\intertext{We can then use the two-$S$ relation on the middle two $S$ boxes of the second picture, and apply the partial trace relation (Equation \eqref{eq:PartialTraceRelation}) to the resulting $\JW{4n-4}$. We thus see}
\qi{4n-3}
\begin{tikzpicture}[baseline]
    \clip (-1,-1.3) rectangle (1,1.3);
    \node at (0,0) (box) [rectangle,draw] {$\hspace{.1cm} \JW{4n -3} \hspace{.1cm} $};
    \draw (box.-50)-- ++(-90:1cm);
    \draw (box.-90) -- ++(-90:1cm);
    \draw (box.-150)--++(-90:1cm);
    \node at (-.25,-.6) {...};
    \draw (box.-30) -- ++(-90:1cm);
    \draw (box.50)-- ++(90:1cm);
    \draw (box.90) -- ++(90:1cm);
    \draw (box.150)--++(90:1cm);
    \node at (-.25,.6) {...};
    \draw (box.30) -- ++(90:1cm);
\end{tikzpicture} \inputtikz{JWzerotakeoutS}
\inputtikz{JWzerosimplify}
\end{align*}


\item Thanks to Stephen Bigelow for pointing out this fact.

On the one hand, $\JW{4n-3}$ is a weighted sum of Temperley-Lieb pictures, with the weight of $\id$ being $1$:   $$\JW{4n-3} = \id + \sum_{P \in \TL_{4n-3}, P \neq \id} \alpha_P \cdot P;$$

On the other hand, $\JW{4n-3}=0$.  Therefore $$\id  = \id - \JW{4n-3} = \sum_{P \in \TL_{4n-3},  P \neq \id} -\alpha_P \cdot P.$$
If $P \in \TL_{4n-3}$ and $P \neq \id$, then $P$ has a cap somewhere along the boundary, so it follows from our hypotheses that $P T = P T'$, and therefore $$T = \left ( \sum_{P \in \TL_{4n-3},  P \neq \id} -\alpha_P \cdot P\right) T =  \left( \sum_{P \in \TL_{4n-3},  P \neq \id} -\alpha_P \cdot P \right) T' = T' .$$
\end{enumerate}

\end{proof}
