$UM \hookrightarrow M$. Define a $V_4$ action on the vector space $M/UM \oplus M$ as follows: $$\begin{aligned}
a(x, y) := (0, ax) \\
b(x, y) := (0, by).\end{aligned}$$ Then it is easy to verify that the map $x \mapsto (x, \pi(x))$ establishes an isomorphism of $V_4$ modules between $M$ and $M/UM
\oplus UM$. Thus $M$ is determined by the vector spaces $M/UM$ and $UM$ and the linear maps $a, b: M/UM {\ensuremath{\rightarrow}}M$. This data amounts to giving a special representation of $Q$.

In the other direction, suppose $[V_1, V_2; \psi_1, \psi_2]$ is a special representation of $Q$. Define a $V_4$ action on the vector space $V_1 \oplus V_2$ by setting $a(x, y):=(0, \psi_1(x))$ and $b(x, y):= (0, \psi_2(x))$. This is easily shown to be a projective free $V_4$ module.

We leave it as an exercise to the reader to verify that the recipes are inverses to each other.

It is also clear that these recipes respect direct sum of representations. Thus the indecomposables are also in 1-1 correspondence.

Proof of lemma \[lemma1\]
-------------------------

Suppose $M$ contains a copy of $\Omega^{l}(k)$, for some $l \ge 1$. $$\xymatrix@=2em{
    &  \overset{g_0}{\bullet} \ar[dr]_b & & \overset{g_1}{\bullet} \ar[dl]^a \ar[dr] &  & \overset{g_2}{\bullet} \ar[dl]  & \bullet  