Corollary

Assume that $(W, S)$ is irreducible. Then: a) If $B_M$ is non-degenerate, the W-module $E$ is absolutely simple. b) If $B_M$ is degenerate, the W-module $E$ is not semi-simple.

In case a), Prop. 7 shows that $E$ is simple, hence also absolutely simple ($§ 2$, no. 1, Prop. 1).

In case b), $E^0 \neq 0$, $E \neq E^0$ (since $B_M \neq 0$), and Prop. 7 shows that $E^0$ has no complement stable under $W$; thus, the W-module $E$ is not semi-simple.