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.


1 Answer 1


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.

  • $\begingroup$ Thank you so much, this proof is very good. $\endgroup$
    – user12345
    Jun 11, 2022 at 16:40

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