Let $V$ be a vector space over $\mathbb{C}$ of finite dimension $n$, $G$ is a finite group and $T:G\rightarrow GL(V)$ its irreducible representation that sends each $g$ into $T_g$. Let $E:V^{\bigoplus 2}\rightarrow \mathbb{C}$ be an arbitrary hermitian form on $V$. It is easy to verify $H$ that sends a pair of vectors $(x,y)$ into $\sum_{g\in G} E(T_g x,T_g y)$ is a $G$-invariant hermitian form, i.e. each $T_g$ respects $H$. It means that an invariant hermitian form always exists.

I am trying to prove that each $G$-invariant hermitian form $H_i$ is $\alpha_i H$ for some $\alpha_i \in \mathbb{C}$. Could you give me any hints or help to develop the following idea?

There is a widely known result that is called Schur's lemma: each endomorphism of irreducible complex representation is scalar (has a form $x\rightarrow \lambda x$). It is easy to deduce the following lemma:
If $T_1$ and $T_2$ - isomorphic irreducible representations then each isomorphism $\phi$ has a form $x \phi_0$ for some $x\in \mathbb{C}$ and fixed isomorphism $\phi_0$.
Is it possible to solve my problem using this lemma? Or somehow otherwise?


I think it is most useful here to interpret any Hermitian form$~h$, which I'm supposing conjugate-linear in its first argument, as associated to a conjugate linear mapping $f_h:V\to V^*$ defined by $f_h(v)=h(v,\cdotp)\in V^*$. Now a first thing to check is that $H$, which is nonzero because $H(v,v)$ is an average over positive values, makes $f_H$ invertible: the kernel of $f_H$ is easily seen to be a $G$-invariant subspace of $V$ (becuase $H$ is $G$-invariant) and not being the whole space it can only be $\{0\}$, and then $f_H$ is invertible because $\dim V=\dim V^*$.

Now if $H_1$ is another $G$-invariant Hermitian form, the composition $(f_H)^{-1}\circ f_{H_1}$ is a $\Bbb C$-linear map $V\to V$ that can be checked to intertwine with the $G$-action. This allows you to apply Schur's lemma.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy