# Showing that a non-degenerate invariant bilinear form on a irreducible representation is either symmetric or anti-symmetric

Let $$H \subset GL(n,\mathbb{C})$$ be a group, acting irreducible on $$V=\mathbb{C}^n$$ and let $$F$$ be a non-degenerate bilinear form on $$V=\mathbb{C}^n$$, which is invariant under $$H$$. I now want to show, that $$F$$ has to be either symmetric or anti-symmetric. In fact i found a similar question here, but i dont completely understand the proof.

What we have:
We have the bilinear form $$F: V \times V \to \mathbb{C}$$ satisfying $$F(hx,hy)=F(x,y)$$ for all $$h \in H$$ and $$x,y \in V$$. Irreducibility of $$H$$ implies that this form is non-degenerate. As in said question, one can consider the mapping $$\varphi: V \to V^{*}$$, $$v \mapsto F(v, - )$$. But i dont understand, how this is used in the further proof. It would be awesome, if someone could exxplain this a little bit to me.

Here's an alternate proof. Given a non-degenerate $$G$$-invariant bilinear form $$F$$, define $$F_s(x,y) = \frac{1}{2}( F(x,y) + F(y,x))$$ and $$F_a(x,y) = \frac{1}{2}(F(x,y)-F(y,x))$$. So, $$F_s$$ is the symmetric part of $$F$$ and $$F_a$$ is the antisymmetric part.
Because $$F$$ is $$G$$-invariant, it follows easily that both $$F_s$$ and $$F_a$$ are $$G$$-invariant so that we obtain $$G$$-equivariant maps $$\phi_a,\phi_s:V\rightarrow V^\ast$$ defined by$$\phi_s(x) = F_s(x,\cdot)$$ and $$\phi_a(x) = F_a(x,\cdot)$$. Because the representations are irreducible, each of $$\phi_a,\phi_s$$ is either an isomorphism or the zero map.
If they're both the zero map, then this implies that both $$F_s$$ and $$F_a$$ are zero, which then implies $$F = F_s + F_a$$ is the zero form, contradicting non-degeneracy. If exactly one is the zero map, then $$F = F_s$$ or $$F= F_a$$, so we get the result you want.
Thus, there is one case left to consider: both $$\phi_a$$ and $$\phi_s$$ are isomorphisms. If this happens, consider the composition $$\phi_a^{-1}\circ \phi_s:V\rightarrow V$$, which is a $$G$$-equivariant isomorphism of $$V$$. By Schur's Lemma, this composition is a non-zero multiple $$\lambda$$ of the identity map.
Said another way, $$\phi_s = \lambda \phi_a$$ for some complex number $$\lambda$$. That is, for any $$v\in V$$, we have $$F_s(v,\cdot) = \lambda F_a(v,\cdot)$$. Or, to say it even simpler, $$F_s = \lambda F_a$$. This implies that $$F_s$$ is anti-symmetric, which then implies $$F_s = 0$$, giving a contradiction.