Proposition 1

Let G be a group and $\rho$ an irreducible linear representation of G on a vector space V; assume that there exists an element g of G such that $\rho(g)$ is a pseudo-reflection.

(i) Every endomorphism of V commuting with $\rho(G)$ is a homothety, and $\rho$ is absolutely irreducible.

(ii) Assume that V is finite dimensional. Let B be a non-zero bilinear form on V invariant under $\rho(G)$. Then B is non-degenerate, either symmetric or skew-symmetric, and every bilinear form on V invariant under $\rho(G)$ is proportional to B.

Let u be an endomorphism of V commuting with $\rho(G)$. Let g be an element of G such that $\rho(g)$ is a pseudo-reflection and let D be the image of $1 - \rho(g)$. Since D is of dimension 1 and $u(D) \subset D$, there exists $\alpha$ in K such that $u - \alpha . 1$ vanishes on D; the kernel N of $u - \alpha . 1$ is then a vector subspace of V invariant under $\rho(G)$ and is non-zero as it contains D; since $\rho$ is irreducible, $N = V$ and $u = \alpha . 1$. The second part of (i) follows from the first by Algebra, Chap. VIII, § 13, no. 4, Cor. of Prop. 5.

Let N (resp. $N'$) be the subspace of V consisting of those x such that $B(x, y) = 0$ (resp. $B(y, x) = 0$) for all y in V; since B is invariant under $\rho(G)$, the subspaces N and $N'$ of V are stable under $\rho(G)$ and distinct from V since $B \neq 0$. Since $\rho$ is irreducible, $N = N' = 0$ and B is non-degenerate.

Since V is finite dimensional, every bilinear form on V is given by the formula

for some endomorphism u of V. If $B'$ is invariant under $\rho(G)$, the endomorphism u commutes with $\rho(G)$. Indeed, let $x, y$ be in V and let g be in G; since B and $B'$ are invariant under $\rho(G)$, we have hence $u(\rho(g)(x)) = \rho(g)(u(x))$ since B is non-degenerate. By (i), there exists $\alpha$ in K with $u = \alpha.1$, so $B' = \alpha.B$.

In particular, we can apply this to the bilinear form $B'(x, y) = B(y, x)$; then $B(y, x) = \alpha.B(x, y) = \alpha^2.B(y, x)$ for all $x, y$ in V, and since B is nonzero we have $\alpha^2 = 1$, hence $\alpha = 1$ or $\alpha = -1$. Thus, B is either symmetric or skew-symmetric.