# If $\chi$ is a complex-valued character of a representation of a finite group, is it always true that $\overline{\chi(g)}=\chi(g^{-1})$?

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?

• yes, it's true ${}$ May 24 at 14:04
• Not sure how this is related to Maschke. Anyway, you do not need to orthogonalize $X(g)$. Just argue with the eigenvalues. Jun 5 at 4:07
• @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'?
– Logi
Jun 6 at 11:09
• @Logi well, you have already accepted an answer, which argues with eigenvalues. As you can see, one does not need to orthogonalize the matrices. Jun 6 at 13:55
• @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.
– Logi
Jun 7 at 14:11

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: $$\overline{\chi(g)}=\chi(g^{-1})$$