self-dual if it is isomorphic to its dual.

When $G = V_4$, it is not hard to see that if $M = (V_1, V_2; a,
b)$ is a projective-free representation of $V_4$, then $M^* =
(V_2^*, V_1^*; a^*, b^*)$.

\[prop:dual\] Even dimensional indecomposable representations of $V_4$ are self-dual.

A non-trivial indecomposable representation of $V_4$ is even dimensional if and only if it is self-dual.

If $M$ is a non-trivial odd dimensional representation of $V_4$, then we know that $M \cong \Omega^l(k)$ for some $l \ne
0$. Then $M^* \cong (\Omega^l(k))^* \cong \Omega^{-l}(k)$. In particular, $M^* \ncong M$.

Proofs
======

In this section we give the proofs of the lemmas and propositions that were used in the classification theorem and applications.

Proof of proposition \[prop1\]
------------------------------

Let $M$ be a projective-free $V_4$ module. Then we have we have $ab(M) = 0$, it follows that $UM$ is included in $U^V_4$. Remarkably one can show that if $M$ is additionally not trivial representation and $M$ is indecomposable then $UM$ is actually equal $M^V_4$, the $V_4$ invariant submodule of $M$. Consider short exact sequence of $V_4$ modules $$0 {\ensuremath{\rightarrow}}UM {\ensuremath{\rightarrow}}M {\ensuremath{\rightarrow}}M/UM {\ensuremath{\rightarrow}}0.$$ Let $\pi: M {\ensuremath{\rightarrow}}UM$ be a vector space retraction of the inclusion 