3

Let $\pi:G\rightarrow \text{GL}(V)$ be a finite-dimnesional complex representation of finite group $ G $. If there is a non-zero $ G $-invariant symmetric bilinear form on $ V $, then the symmetric square $\text{Sym}^2(V)$ must contain the trivial subrepresentaion. I don't know how to get the trivial subrepresentation. I try to use the universal property of tensor product to get a linear function $\text{Sym}^2(V)\rightarrow\mathbb{C}$ which is invariant under the $ G $-action, but I dont't know what to do next. Any help or hit would be highly appreciated!

Kevin
  • 423
  • 1
    Show that there's a nondegenerate $G$-invariant pairing $\text{Sym}^2(V) \otimes S^2(V^{\ast}) \to \mathbb{C}$. $S^2(V^{\ast})$ is the space of symmetric bilinear forms on $V$. – Qiaochu Yuan Sep 24 '22 at 02:27

1 Answers1

0

Let $b: V \times V \rightarrow \mathbb{C}$ be a $G$-invariant symmetric bilinear form on $V$.

  1. Define $f: S^2(V) \rightarrow \mathbb{C}$ by $f(vw) = b(v,w)$ for all $v, w \in V$. (Such a map exists since $b$ is symmetric bilinear.)

  2. Check that $f$ is a surjective homomorphism of $\mathbb{C}[G]$-modules. (In other words, $G$-invariant.)

  3. Conclude that $S^2(V)$ has a trivial submodule.

Mikko Korhonen
  • 26,156