I have learned from Maschke's theorem that there exists an inner product on any $G$-module of a finite group $G$ that is invariant under the action of $G$. That means that we can select an orthonormal basis for the $G$-module such that $$ X(g^{-1})=X(g)^{-1}=X(g)^\dagger, $$ where $X$ is the matrix representation in that basis. If we let $\chi$ be the corresponding character, then we can write $$ \overline{\chi(g)}=\operatorname{Tr}(\overline{X(g)})=\operatorname{Tr}(\overline{X(g)}^t)=\operatorname{Tr}(X(g)^\dagger)=\operatorname{Tr}(X(g^{-1}))=\chi(g^{-1}) $$ Since a character is independent of the basis we choose, we must therefore have $\overline{\chi(g)}=\chi(g^{-1})$ whenever $\chi$ is a character of any representation of a finite group. Is this actually true, or have I overlooked something?
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1$\begingroup$ yes, it's true ${}$ $\endgroup$– user8268May 24 at 14:04
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$\begingroup$ Not sure how this is related to Maschke. Anyway, you do not need to orthogonalize $X(g)$. Just argue with the eigenvalues. $\endgroup$– Brauer SuzukiJun 5 at 4:07
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$\begingroup$ @BrauerSuzuki, I should have mentioned that, but the fact that there exists an inner product on any G-module of a finite group that is invariant under the action of $G$ is part of the proof Maschke's theorem (at least the proof I read). Can you clarify what you mean by 'argue with the eigenvalues'? $\endgroup$– LogiJun 6 at 11:09
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$\begingroup$ @Logi well, you have already accepted an answer, which argues with eigenvalues. As you can see, one does not need to orthogonalize the matrices. $\endgroup$– Brauer SuzukiJun 6 at 13:55
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$\begingroup$ @BrauerSuzuki, sure and the answer is fine. However, I do not think that the proof in the answer is any simpler than my proof. But thanks for your input. $\endgroup$– LogiJun 7 at 14:11
1 Answer
Another way to see that is as follows:
Let $G$ be a finite group, $ρ:G\rightarrow \text{GL}_n(\mathbb{C})$ a representation and $χ$ the corresponding character.
We know that for any $g\in G$, all the eigenvalues $λ_1,\dots,λ_n$ of $ρ(g)$ are roots of unity, so $|λ_i|=1,\forall i$.
Also, since $ρ(g^{-1})=ρ(g)^{-1}$, the eigenvalues of $ρ(g^{-1})$ are exactly $\frac{1}{λ_1},\dots,\frac{1}{λ_n}$.
Thus, we have: $$χ(g^{-1})=\text{Tr}\left(ρ(g^{-1})\right)=\sum\limits_{i=1}^{n}\frac{1}{λ_i}=\sum\limits_{i=1}^{n}\frac{\overline{λ_i}}{\overline{λ_i}λ_i}= \sum\limits_{i=1}^{n}\frac{\overline{λ_i}}{|λ_i|^2}=$$ $$\sum\limits_{i=1}^{n}\overline{λ_i}=\overline{\sum\limits_{i=1}^{n}λ_i}=\overline{\text{Tr}\left(ρ(g)\right)}=\overline{χ(g)}$$
So $\forall g\in G$ we have: $$\overline{\chi(g)}=\chi(g^{-1})$$
