\begin{proof}
Ideally one would prove this theorem by exhibiting explicit bases for all
the appropriate $\operatorname{Hom}$ spaces. Instead, here we'll explain
that there's a unique tensor product structure extending the tensor
product data encoded in the principal graph.

\todo{This needs to be rewritten, at the moment it's just a
copy-and-paste from an email.}

The tensor product rules for $W_j \tensor W_k$ just follow from the
tensor product rules for $A_{4n-3}$. Let's call the vertices of
$A_{4n-3}$ $V_0$ through $V_{4n-4}$. The inclusion of $A_{4n-3}$ into
$D_{2n}$ sends

\begin{equation*}
\begin{cases}
V_i \To W_i & \text{when $0 \leq i \leq 2n-3$} \\
V_{2n-2} \To P \directSum Q & \\
V_i \To W_{4n-4-i} & \text{when $2n-1 \leq i \leq 4n-4$}
\end{cases}
\end{equation*}

That is, it folds $A_{4n-3}$ in half, and splits the central vertex in
two. It's easy to prove these isomorphisms (i.e., that the images of
$V_i$ and $V_{4n-4-i}$ are the same in $D_{2n}$) using the relations for
$S$; the isomorphism is always just $S$, hooked up appropriately.

Write$i_*$ for the functor from $A_{4n_3}$ to $D_{2n}$ (ie, the map
above). It's a tensor functor, so

\begin{align*}
W_j \tensor W_k & = i_* ( V_j \tensor V_k) \\
 & i_* \left( \DirectSum_{l=\max{j-k, k-j}}^{\min{j+k, 8n-8-j-k}} V_l \right)
\end{align*}

That's just the usual tensor product rule on $A_{4n-3}$; the Steinberg
rule, but you have to remember to fold at both ends now. Because we have
to apply $i_*$ still, we end up with multiplicities of $1$ and $2$ in the
sum.

At this point we know all the tensor products amongst the $W_i$, and we
also know $P \tensor W_1$ and $Q \tensor W_1$. First look at the
associativity equation for $W_1 \tensor W_1 \tensor W_{2n-3}$, and find
that $W_1 \tensor P = W_1 \tensor Q = W_{2n-3}$. (I don't know any a
priori argument that the tensor product rules are commutative; perhaps
this is always true below index 4? But it's not hard to discover this for
$D_2n$ directly.) Now we're going to induct our way up to the tensor
product rules for $P \tensor W_k$ (and by much the same argument, $Q
\tensor W_k$, $W_k \tensor P$ and $W_k \tensor Q$).

\begin{lem}
\begin{align*}
P \tensor W_{2k+1} & = \DirectSum_{l=0}^k W_{2n-2l-3} \\
P \tensor W_{2k}     & = \begin{cases}
 P \directSum \DirectSum_{l=0}^{k-1} W_{2n-2l-4} & \text{if $k$ is even} \\
 Q \directSum \DirectSum_{l=0}^{k-1} W_{2n-2l-4} & \text{if $k$ is even}
 \end{cases}
\end{align*}
\end{lem}
\begin{proof}
By induction, and considering the associativity equation for $P \tensor
W_{m-1} W_1$.
\end{proof}
\begin{rem}
The tensor product rules for $Q \tensor W_k$, $W_k \tensor P$ and $W_k
\tensor Q$ differ from those in the lemma simply by exchanging $P$ and
$Q$.
\end{rem}

Now we know everything except the tensor products amongst the $P$s and
$Q$s.

We'll need a little lemma about ``dividing by $W_1$''.
\begin{lem}
\label{lem:dividing} Suppose $X$ is some object such that $X \tensor W_1
=  \Wodd$. Then if $n$ is even, $X$ must be one of $\Wfour, \Wfourp
\directSum P$ and $\Wfourp \directSum Q$. If $n$ is odd, $X$ must be one
of $\Wfourp, \Wfour \directSum P$ and $\Wfour \directSum Q$.
\end{lem}

Now a few calculations finish everything off. First, consider the
associativity equation for $P \tensor W_2 \tensor P$, and find that $P
\tensor Q = Q \tensor P$. Next, consider the associativity equation for
$P \tensor W_1 \tensor W_{2n-3}$, and find that
\begin{equation*}
P \tensor P \directSum P \tensor Q =
\begin{cases}
 Q \directSum \Weven & \text{if $n$ is even} \\
 P \directSum \Weven & \text{if $n$ is odd}
\end{cases}
\end{equation*}
and similarly, by considering $Q \tensor W_1 \tensor W_{2n-3}$, that
\begin{equation*}
Q \tensor Q \directSum Q \tensor P =
\begin{cases}
 P \directSum \Weven & \text{if $n$ is even} \\
 Q \directSum \Weven & \text{if $n$ is odd}
\end{cases}
\end{equation*}

Finally, $P \tensor P \tensor W_1 = P \tensor W_{2n-3} = \Wodd$, as does
$P \tensor Q \tensor W_1$, and so we can apply Lemma \ref{lem:dividing}
to both $P \tensor P$ and $P \tensor Q$. Because we know their sum, by
the calculation immediately above, only $2$ out of the $9$ possibilities
allowed by the Lemma remain. When $n$ is even,
\begin{align*}
\{ P \tensor P, P \tensor Q \} = \{ \Wfour, \Wfourp + Q \} \\
\intertext{and when $n$ is odd} \{ P \tensor P, P \tensor Q \} = \{
\Wfourp, \Wfour + P \}
\end{align*}
The same alternatives hold for $Q \tensor Q$ and $Q \tensor P$,
exchanging $P$ and $Q$ in the lines above. However, out of the four total
alternatives, only one satisfies the constraint $P \tensor Q = Q \tensor
P$, discovered above, and so we obtain the final rules appearing in the
statement of the theorem.
%%This all comes from pp1565-1567 and pp1575-1580 in Scott's notes.}
\end{proof}
