Brian Bi

Remark: In the proof of Theorem 4.6.2, a claim is made that if $$B_1, B_2$$ are two $$G$$-invariant positive definite Hermitian forms on $$V$$, then $$B_1(v, w) = B_2(Av, w)$$ for some homomorphism $$A : V \to V$$. To see why, first observe that by nondegeneracy, $$B_1(v, w) = \varphi_v(w) = B_2(v', w)$$ for some unique $$\varphi_v \in V^*, v' \in V$$. Define $$A$$ such that $$Av$$ is that unique vector $$v'$$ for each $$v \in V$$. By linearity of $$B_1, B_2$$ in the first argument, it follows that $$A$$ is $$\mathbb{C}$$-linear. Furthermore, for all $$g \in G$$, we have $$B_1(gv, w) = B_1(v, g^{-1}w) = B_2(Av, g^{-1}w) = B_2(gAv, w)$$, therefore $$Agv = gAv$$. Therefore $$A$$ is a homomorphism of $$\mathbb{C}[G]$$-modules, as claimed.

It is also claimed that we can start with any positive definite form $$B$$ and the form $$\overline{B}$$ constructed from $$B$$ will be Hermitian. It's not obvious to me how $$\overline{B}$$ becomes Hermitian if $$B$$ wasn't already Hermitian, but this doesn't matter, because it is easy to construct $$B$$ to be Hermitian in the first place (and in fact this is already implied by positive definiteness).