\subsection{The tensor category of projections of a planar algebra}
\label{sec:category}%

In this section we describe a tensor category associated to a planar
algebra, whose objects are the `projections'. This is essentially
parallel to the construction of the tensor category of bimodules over a
subfactor \cite{MR1424954}. The tensor category described here is in fact
isomorphic to that one, although we won't need to make use of this fact.
We describe the category independently here, to emphasize that it can be
constructed directly from the planar algebra, without reference to the
associated subfactor.

\begin{defn}
Given a planar algebra $P$ we construct a tensor category $\cC_P$ as follows.
\begin{itemize}
\item An object of $\cC_P$ is a projection in one of the $2n$-box algebras $P_{2n}$.
\item Given two projections $\pi_1 \in P_{2n}$ and $\pi_2 \in P_{2m}$ we define $\Hom{}{\pi_1}{\pi_2}$ to be the space $\pi_2 P_{n \rightarrow m} \pi_1$  ($P_{n \rightarrow m}$ is a convenient way of denoting $P_{n+m}$, drawn with $n$ strands going down and $m$ going up.)
\item The tensor product $\pi_1 \otimes \pi_2$ is the disjoint union of the two projections in $P_{2n+2m}$.
\item The trivial object $\id$ is the empty picture (which is a projection in $P_0$).
\item The dual $\overline{\pi}$ of a projection $\pi$ is given by rotating it 180 degrees.
\end{itemize}
\end{defn}

This category comes with a special object $X \in P_2$ which is the single strand.  Note that $X = \overline{X}$. 
We would like to be able to take direct sums in this category.  If $\pi_1$ and $\pi_2$ are orthogonal projections in the same box space $P_n$ (i.e. if $\pi_1 \pi_2 = 0 = \pi_2 \pi_1$), then their direct sum is just $\pi_1 + \pi_2$.  However, if they are not orthogonal the situation is a bit more difficult.  One solution to this problem is to replace the projections with isomorphic projections which are orthogonal.  However, this construction only makes sense on equivalence classes, so we use another construction.

\begin{defn}
Given a category $\cC$ define its matrix category $\Mat{\cC}$ as follows.
\begin{enumerate}
\item The objects of $\Mat{\cC}$ are formal direct sums of objects of $\cC$
\item A morphism of $\Mat{\cC}$ from $A_1 \oplus \ldots \oplus A_n \rightarrow B_1 \oplus \ldots \oplus B_m$ is an $m$-by-$n$ matrix whose $(i,j)$th entry is in $\Hom{\cC}{A_j}{B_i}$.
\end{enumerate}
\end{defn}

%It is easy to see that $\oplus$ is a coproduct for $\Mat{\cC}$.

If $\cC$ is a tensor category then $\Mat{\cC}$ has an obvious tensor
product (on objects, formally distribute, and on morphisms, use the
usual tensor product of matrices and the tensor product for $\cC$ on
matrix entries).  If $\cC$ is spherical then so is $\Mat{\cC}$ (where the
dual on objects is just the dual of each summand and on morphisms the dual transposes
the matrix and dualizes each matrix entry).

\begin{lem}
If $\pi_1$ and $\pi_2$ are orthogonal projections in $P_{2n}$, then $\pi_1 \oplus \pi_2 \cong \pi_1 + \pi_2$ in $\Mat{\cC_P}$.
\end{lem}
\begin{proof}
Define $f: \pi_1 \oplus \pi_2 \rightarrow \pi_1 + \pi_2$ and $g: \pi_1 + \pi_2 \rightarrow \pi_1 \oplus \pi_2$ by $f = \begin{pmatrix} \pi_1 & \pi_2 \end{pmatrix}$ and $g=\begin{pmatrix} \pi_1 \\ \pi_2 \end{pmatrix}$.  Then $f g = \id_{\pi_1 + \pi_2}$ and $g f= \id_{\pi_1 \oplus \pi_2}$.
\end{proof}

\begin{defn}
An object $\pi \in \Mat{\cC_P}$ is called minimal if $\Hom{}{\pi}{\pi}$ is $1$-dimensional.
\end{defn}
\begin{defn}
A planar algebra is called semisimple if every projection is a direct sum of minimal projections and for any pair of non-isomorphic minimal projections $\pi_1$ and $\pi_2$, we have that $\Hom{}{\pi_1}{ \pi_2} = 0$.
\end{defn}

\begin{defn}
The principal (multi-)graph of a semisimple planar algebra has as vertices the isomorphism classes of minimal projections, and there are $$\dim \Hom{}{\pi_1 \otimes X}{\pi_2} (= \dim \Hom{}{\pi_1}{\pi_2 \otimes X})$$ edges between the vertices $\pi_1 \in P_n$ and $\pi_2 \in P_m$.
\end{defn}

Our definitions here are particularly simple because we work in the
context of unshaded planar algebras. A slight variation works for a
shaded planar algebra as well. \footnote{
For a shaded planar algebra, we construct a $2$-category with two
objects `shaded' and `unshaded'. The $1$-morphisms from $A$ to $B$
are now the projections in the planar algebra, where $A$ is the shading at the marked point, and $B$ is the shading opposite the marked point. The $2$-morphisms are defined the same way as the morphisms were above. This $2$-category is equivalent to the $2$-category of $N,M$ bimodules for the type $II_1$ subfactor $N \subset M$ associated to the shaded planar algebra, as described in \cite{MR1424954}. }

\subsubsection{The minimal projections of $\pa$}

Now we can use the definitions of the previous section to explain why we call $\pa$ the $D_{2n}$ planar algebra.

\begin{thm}
The planar algebra $D_{2n}$ is semi-simple, with minimal projections $f^{(k)}$ for $k=0,
\ldots, 2n-3$ along with $P$ and $Q$ defined by
\begin{align*}
P & = \frac{1}{2}\left(f^{(2n-2)} + S\right) \\
\intertext{and}
Q & = \frac{1}{2}\left(f^{(2n-2)} - S\right).
\end{align*}
The principal graph is the Dynkin diagram $D_{2n}$.
\begin{equation*}
\mathfig{0.5}{graphs/d2n}
\end{equation*}
\end{thm}
\begin{proof} Observe that $\JW{2n-2} \cdot S = S$ (as the identity has weight $1$ in $\JW{2n-2}$ and all non-identity Temperley-Lieb pictures have product $0$ with $S$) and $S^2=\JW{2n-2}$. We see that $P$ and $Q$ are projections.
Let $\mathcal{M} = \{f^{(1)}, \ldots f^{(2n-3)}, P, Q\}$.  By Lemmas \ref{LemmasA1} and \ref{LemmasA2} (below) every projection in $\mathcal{M}$ is minimal. Lemma \ref{lem:no-homs} says there are no nonzero morphisms between different elements of $\mathcal{M}$. By Lemmas \ref{LemmasB1}, \ref{LemmasB2}, and \ref{LemmasB3}, we see that for each $Y \in \mathcal{M}$, the projection $Y \tensor f^{(1)}$ is isomorphic to a direct sum of projections in $\mathcal{M}$.  Thus, because every projection is a summand of $\id \in P_n$ for some $n$, every minimal projection is in $\mathcal{M}$.  Finally from Lemmas \ref{LemmasB1}, \ref{LemmasB2}, and \ref{LemmasB3} we read off that the principal graph for our planar algebra is the Dynkin diagram $D_{2n}$.
\end{proof}

\begin{rem}
Since $S^* = S$, all the projections are self-adjoint. The projections
$f^{(k)}$ are all self-dual. The projections $P$ and $Q$ are self-dual
when $n$ is odd, and when $n$ is even, $\overline{P} = Q$ and
$\overline{Q} = P$. These facts follow immediately from the definitions,
and the rotation relation (\ref{rotateS}).
\end{rem}

\begin{lem} \label{LemmasA1}
The Jones-Wenzl idempotents $f^{(k)}$ for $k=0, \ldots, f^{(2n-3)}$ are minimal.
\end{lem}

\begin{rem}
The minimality of the empty diagram, $f^{(0)}$, is exactly the fact that any closed diagram evaluates to a multiple of the empty diagram; that is, $\dim{\pa_0}=1$.
\end{rem}

\begin{proof}
The space $\Hom{}{f^{(i)}} {f^{(i)}}$ consists of all diagrams obtained by filling in the empty ellipse in the following diagram.

\begin{equation*}
\mathfig{0.14}{projections/fi}
\end{equation*}

We want to show that any such diagram which is non-zero is equal to a multiple of the diagram gotten by inserting the identity into the empty ellipse. By Corollary \ref{cor:atmost1S}, we need only consider diagrams with $0$ or $1$ $S$ boxes. First consider inserting any Temperley-Lieb diagram.  Since any cap applied to a Jones-Wenzl is zero, the ellipse must contain no cups or caps, hence it is a multiple of the identity.  Now consider any diagram with exactly one $S$.  Since $S$ has $4n-4$ strands, and $2i \leq 4n-6$, any such diagram must cap off the $S$, hence it vanishes.
\end{proof}


\begin{lem} \label{LemmasA2}
The projections $P = \frac{1}{2}\left(f^{(2n-2)} + S\right)$ and $Q = \frac{1}{2}\left(f^{(2n-2)} - S\right)$ are minimal.
\end{lem}
\begin{proof}
The two proofs are identical, so we do the $P$ case.  The space $\Hom{}{P} {P}$ consists of all ways of filling in the following diagram.

\begin{equation*}
\mathfig{0.13}{projections/PP}
\end{equation*}

We want to show that any such diagram which is non-zero is equal to a multiple of the diagram with the identity inserted.  Again we use Corollary \ref{cor:atmost1S}. First consider any Temperley-Lieb diagram drawn there. Since any cap applied to $P$ is zero, the diagram must have no cups or caps, hence it is a multiple of the identity.  Now consider any diagram with exactly one $S$.  Since $S$ has $4n-4$ strands, any such diagram which does not cap off $S$ must be (up to rotation) the following diagram.

\begin{equation*}
\mathfig{0.13}{projections/PPS}
\end{equation*}

Since $PS=P$, this diagram is a multiple of the diagram with the identity inserted.

\end{proof}

\begin{lem}
\label{lem:no-homs}
If $A$ and $B$ are two distinct projections from the set $$\{f^{(0)}, f^{(1)}, \ldots, f^{(2n-3)}, P, Q\}$$ then
$\Hom{}{A}{B} = 0$.
\end{lem}
\begin{proof}
Suppose $A$ and $B$ are distinct Jones-Wenzl projections.  Any morphism between them with exactly one $S$ must cap off the $S$, and so is $0$.  Any morphism between them in Temperley-Lieb must cap off either $A$ or $B$ and so is zero.  If $A$ is a Jones-Wenzl projection, while $B$ is $P$ or $Q$, exactly the same argument holds.

If $A=P$ and $B=Q$, we see that the morphism space is spanned by Temperley-Lieb diagrams and the diagram with a single $S$ box. Changing basis, the morphism space is spanned by non-identity Temperley-Lieb diagrams, along with $P$ and $Q$. Non-identity Temperley-Lieb diagrams are all zero as morphisms, because they result in attaching a cap to both $P$ and $Q$. The elements $P$ and $Q$ are themselves zero as morphisms from $P$ to $Q$, because $PQ=QP=0$.
\end{proof}

\begin{lem} \label{LemmasB1}
The projection $f^{(k)} \tensor f^{(1)}$ is isomorphic to $f^{(k-1)} \directSum f^{(k+1)}$ for $k = 1,\ldots, 2n-4$.
\end{lem}
\begin{proof}
This is a well known result about Temperley-Lieb.   The explicit isomorphisms are

\begin{align*}
\begin{pmatrix}
\frac{\qi{m}}{\qi{m+1}} \cdot \mathfig{0.1}{projections/capk_fX_f} \\ \mathfig{0.1}{projections/id_fX_f}
\end{pmatrix}: &\mathfig{0.1}{projections/fk1} \to f^{(k-1)} \directSum f^{(k+1)} \\
\intertext{and}
\begin{pmatrix}
\mathfig{0.1}{projections/cupk_f_fX}  & \mathfig{0.1}{projections/id_f_fX}
\end{pmatrix}: &  f^{(k-1)} \directSum f^{(k+1)} \to \mathfig{0.1}{projections/fk1}
.\end{align*}
The fact that these are inverses to each other is exactly Wenzl's relation \eqref{WenzlsRelation}.
\end{proof}

\begin{lem} \label{LemmasB2}
The projection $f^{(2n-3)} \tensor f^{(1)}$ is isomorphic to $f^{(2n-4)} \directSum P \directSum Q$.
\end{lem}
\begin{proof}
The explicit isomorphisms are:

\begin{align*}
\begin{pmatrix}
\frac{\qi{m}}{\qi{m+1}} \cdot \mathfig{0.1}{projections/capn_fX_f} \\ \mathfig{0.1}{projections/id_fX_P} \\ \mathfig{0.1}{projections/id_fX_Q}
\end{pmatrix}: \mathfig{0.1}{projections/fn1} \to f^{(2n-4)} \directSum P \directSum Q \\
\intertext{and}
\begin{pmatrix}
\mathfig{0.1}{projections/cupn_f_fX} & \mathfig{0.1}{projections/id_P_fX} & \mathfig{0.1}{projections/id_Q_fX}
\end{pmatrix}: f^{(2n-4)} \directSum P \directSum Q \to \mathfig{0.1}{projections/fn1}.
\end{align*}
The fact that these are inverses to each other follows from Wenzl's relation and the fact that $P$ and $Q$ absorb Jones-Wenzl idempotents (ie, $f^{(2n-3)} \cdot P=P$ and $f^{(2n-3)} \cdot Q=Q$).
\end{proof}

\begin{lem} \label{LemmasB3}
$P \tensor f^{(1)} \iso f^{(2n-3)}$ and $Q \tensor f^{(1)} \iso f^{(2n-3)}$.
\end{lem}

\begin{proof}
We claim that the maps
$$\qi{2} \cdot
\begin{tikzpicture}[baseline]
    \node at (.2,.5) (mid) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};
    \node at (0,-.5) (bottom) [rectangle,draw]{$ \JW{2n-3} $};
    \draw (mid.-40)--(bottom.35);
    \draw (mid.-20) arc (-180:0:.3cm) -- ++(90:.5cm);
    \draw (mid.-160)--(bottom.145);
     \node at (mid.-90) [below] {...};
\end{tikzpicture}
: P \tensor f^{(1)} \rightarrow f^{(2n-3)}$$
and
$$
\begin{tikzpicture}[baseline]
    \node at (0,.5) (top) [rectangle,draw] {$ \JW{2n-3} $};
    \node at (.2,-.5) (mid) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};
    \draw (top.-35)--(mid.40);
    \draw (top.-145)--(mid.160);
    \draw (mid.20) arc (180:0:.3cm) -- ++(-90:.5cm);
    \node at (mid.90) [above] {...};
\end{tikzpicture}
:f^{(2n-3)} \rightarrow P \tensor f^{(1)}$$
are isomorphisms inverse to each other.  To check this, we need to verify
$$\qi{2} \cdot
\begin{tikzpicture}[baseline]
    \node at (0,1) (top) [rectangle,draw] {$ \JW{2n-3} $};
    \node at (.2,0) (mid) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};
    \node at (0,-1) (bottom) [rectangle,draw]{$ \JW{2n-3} $};
    \draw (top.-35)--(mid.40);
    \draw (top.-145)--(mid.160);
    \draw (mid.-40)--(bottom.35);
    \draw (mid.20) arc (180:0:.3cm) -- ++(-90:.5cm);
    \draw (mid.-20) arc (-180:0:.3cm) -- ++(90:.5cm);
    \draw (mid.-160)--(bottom.145);
    \node at (mid.90) [above] {...};
     \node at (mid.-90) [below] {...};
\end{tikzpicture}
= \JW{2n-3} \text{ \qquad and \qquad }
\qi{2} \cdot
\begin{tikzpicture}[baseline]
    \node at (0,0) (top) [rectangle,draw] {$ \JW{2n-3} $};
    \node at (.2,-1) (mid) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};

    \draw (top.-35)--(mid.40);
    \draw (top.-145)--(mid.160);
     \draw (mid.20) arc (180:0:.3cm) -- ++(-90:.5cm);

    \node at (mid.90) [above] {...};

    \node at (.2,1) (toptop) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};
    \draw (top.35)--(toptop.-40);
    \draw (top.145)--(toptop.-160);
     \draw (toptop.-20) arc (-180:0:.3cm) -- ++(90:.5cm);

    \node at (toptop.-90) [below] {...};
\end{tikzpicture}
= P \tensor \JW{1}.$$

The first equality is straightforward:  capping $P=\frac{1}{2}(\JW{2n-2} + S)$ on the right side kills its $S$ component, and then the equality follows from the partial trace relation and the observation that  $\frac{\qi{2n-1}}{\qi{2n-2}}=\frac{2}{\qi{2}}$. The second will take a bit more work to establish, but it's not hard. We first observe that $P \cdot \JW{2n-3} = P$, then expand both $P$s as $\frac{1}{2} (\JW{2n-2} + S)$. Thus
\begin{align*}
\inputtikz{PEnP} \\
\inputtikz{PEnPExpanded},
\intertext{and applying Wenzl's relation to the first term and the two-S relation to the fourth term yields}
& = \inputtikz{PEnPWenzl}. \\
\intertext{Because $\qi{k}=\qi{4n-2-k}$ we can cancel the second and fifth terms; and using $\frac{\qi{2n-1}}{\qi{2n-2}}=\frac{2}{\qi{2}}$ again we get }
\inputtikz{PEnPcancel}.
\end{align*}

At this point, we use part \ref{enoughcappings} of Theorem \ref{easyconsequences} to show

$$\inputtikz{JWEnS}.$$

Start at the top left of the pictures, and take the first $4n-4$ cappings in the counterclockwise direction.  Most of these give zero immediately, and the three we are left to check are

\begin{align*}
\inputtikz{JWEnStopcap}, \\
\inputtikz{JWEnSbottomcap} \\
\intertext{and}
\inputtikz{JWS}.
\end{align*}

The first two of these follows from the partial trace relation (Equation \eqref{eq:PartialTraceRelation}) and $\JW{2n-3} \cdot S =S$, and the third follows from $\JW{2n-2} \cdot S =S$.  Therefore, we conclude
$$
\qi{2} \cdot
\begin{tikzpicture}[baseline]
    \node at (0,0) (top) [rectangle,draw] {$ \JW{2n-3} $};
    \node at (.2,-1) (mid) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};

    \draw (top.-35)--(mid.40);
    \draw (top.-145)--(mid.160);
     \draw (mid.20) arc (180:0:.3cm) -- ++(-90:.5cm);

    \node at (mid.90) [above] {...};

    \node at (.2,1) (toptop) [rectangle, draw] {$ \hspace{.7cm} P \hspace{.7cm} $};
    \draw (top.35)--(toptop.-40);
    \draw (top.145)--(toptop.-160);
     \draw (toptop.-20) arc (-180:0:.3cm) -- ++(90:.5cm);

    \node at (toptop.-90) [below] {...};
\end{tikzpicture}
=
\frac{\qi{2}}{4} \cdot \frac{2}{\qi{2}}
\left(
\begin{tikzpicture}[baseline]
    \clip (-1,-1.3) rectangle (1,1.3);
    \node at (0,0) (box) [rectangle,draw] {$\JW{2n-2}$};

    \draw (box.-145)--++(-90:1cm);
    \node at (-.1,-.6) {...};
    \draw (box.-35) -- ++(-90:1cm);
    \draw (box.-60) -- ++(-90:1cm);

    \draw (box.35) -- ++(90:1cm);
    \draw (box.60) -- ++(90:1cm);
    \node at (-.1,.6) {...};
    \draw (box.145) -- ++(90:1cm);

    \draw (.9,1.5)--(.9,-1.5);
\end{tikzpicture}
+
\begin{tikzpicture}[baseline]
    \clip (-.8,-1.3) rectangle (1,1.3);
    \node at (0,0) (box) [ellipse,draw] {$ \hspace{.15 cm} S \hspace{.15cm} $};
    \filldraw (box.180) circle (.5mm);

    \draw (box.-145)--++(-90:1cm);
    \node at (-.1,-.6) {...};
    \draw (box.-35) -- ++(-90:1cm);

    \draw (box.35) -- ++(90:1cm);
    \node at (-.1,.6) {...};
    \draw (box.145) -- ++(90:1cm);

    \draw (.9,1.5)--(.9,-1.5);
\end{tikzpicture}
\right)
=
\begin{tikzpicture}[baseline]
    \clip (-1,-1.3) rectangle (1,1.3);
    \node at (0,0) (box) [rectangle,draw] {$\hspace{3mm} P \hspace{3mm}$};

    \draw (box.-145)--++(-90:1cm);
    \node at (-.1,-.6) {...};
    \draw (box.-35) -- ++(-90:1cm);
    \draw (box.-60) -- ++(-90:1cm);

    \draw (box.35) -- ++(90:1cm);
    \draw (box.60) -- ++(90:1cm);
    \node at (-.1,.6) {...};
    \draw (box.145) -- ++(90:1cm);

    \draw (.9,1.5)--(.9,-1.5);
\end{tikzpicture},
$$
which is what we wanted to show.
\end{proof}

\newcommand{\Wfour}{\DirectSum_{l=0}^{\frac{n-3}{2}} f^{(4l)}}
\newcommand{\Wfourp}{\DirectSum_{l=0}^{\frac{n-4}{2}} f^{(4l+2)}}

%\todo{add enough to this theorem to prove that $D_\text{odd}$ doesn't exist}

\subsubsection{The tensor product decompositions} We do not prove the formulas that follow, and they are not essential to this paper.  Nevertheless, we include the full tensor product table of $D_{2n}$ for the sake of making this description of $D_{2n}$ as complete as possible. Partial
tensor product tables appears in \cite[\S 3.5]{MR1145672} and \cite[\S 7]{MR1936496}.

Using the methods of this paper, one could prove that these tensor product formulas hold by producing explicit bases for all the appropriate $\operatorname{Hom}$ spaces in the tensor category of projections.  However, this method would not show that these formulas are the only extension of the data encoded in the principal graph.

Much of this is proved in \cite{MR1145672}, except for the formula for $f^{(j)} \tensor f^{(k)}$ when $2n-2 \leq j+k \leq 4n-4$ in Equation \eqref{eq:wjwk-large}
and the formula for $P \tensor f^{(2k+1)}$ and $Q \tensor f^{(2k+1)}$ in Equation \eqref{eq:pqw-odd}. Nonetheless, the methods of \cite{MR1145672} readily extend to give the remaining formulas. With the same exceptions, along with Equations \eqref{eq:pw-even} and \eqref{eq:qw-even},
these are proved in \cite{MR1936496}, by quite different methods.
Further, \cite{MR1145672} proves there is no associative tensor product
extending the tensor product data encoded in the principal graphs $D_{2n+1 \geq 5}$ with an odd number of vertices.

\begin{thm}
The tensor product structure is commutative, and described by the following isomorphisms.

When $j+k < 2n-2$
\begin{equation*}
f^{(j)} \tensor f^{(k)}  \iso \DirectSum_{l=\frac{\abs{j-k}}{2}}^{\frac{j+k}{2}} f^{(2l)}
\end{equation*}
and when $2n-2 \leq j+k \leq 4n-4$
\begin{equation}
\label{eq:wjwk-large}
f^{(j)} \tensor f^{(k)}  \iso \begin{cases}
 \left(\displaystyle \DirectSum_{l=\frac{\abs{j-k}}{2}}^{2n-3-\frac{j+k}{2}} f^{(2l)}\right) \directSum \left(\displaystyle \DirectSum_{l=2n-2-\frac{j+k}{2}}^{n-2} 2 f^{(2l)} \right) \directSum P \directSum Q  & \text{if $j+k$ is even} \\
 \left(\displaystyle \DirectSum_{l=\frac{\abs{j-k}}{2}}^{2n-3-\frac{j+k}{2}} f^{(2l)}\right) \directSum \left(\displaystyle \DirectSum_{l=2n-2-\frac{j+k}{2}}^{n-\frac{3}{2}} 2 f^{(2l)} \right) & \text{if $j+k$ is odd.}
\end{cases}
\end{equation}
Moreover
\begin{equation}
\label{eq:pqw-odd}%
P \tensor f^{(2k+1)} \iso Q \tensor f^{(2k+1)}
                    \iso \DirectSum_{l=0}^k f^{(2n-2l-3)}
\end{equation}
\begin{align}
\label{eq:pw-even}
P \tensor f^{(2k)}     & \iso \begin{cases}
 P \directSum \DirectSum_{l=0}^{k-1} f^{(2n-2l-4)} & \text{if $k$ is even} \\
 Q \directSum \DirectSum_{l=0}^{k-1} f^{(2n-2l-4)} & \text{if $k$ is odd}
 \end{cases} \\
\label{eq:qw-even}
Q \tensor f^{(2k)}     & \iso \begin{cases}
 Q \directSum \DirectSum_{l=0}^{k-1} f^{(2n-2l-4)} & \text{if $k$ is even} \\
 P \directSum \DirectSum_{l=0}^{k-1} f^{(2n-2l-4)} & \text{if $k$ is odd}
 \end{cases} \\
\intertext{and when $n$ is even}
\notag P \tensor P & \iso Q \directSum \Wfourp   \\
\notag Q \tensor Q & \iso P \directSum \Wfourp  \\
\notag P \tensor Q & \iso Q \tensor P \iso \Wfour \\
\intertext{and when $n$ is odd}
\notag P \tensor P & \iso P \directSum \Wfour \\
\notag Q \tensor Q & \iso Q \directSum \Wfour \\
\notag P \tensor Q & \iso Q \tensor P \iso  \Wfourp.
\end{align}
\end{thm}