I am reading a representation theory book, and it says:
For every representation $\vartheta$ of a finite group $G$ and for every unitary representation of an arbitrary group $G$ $$\chi(s^{-1}) = \bar{\chi}(s)$$ the bar denotes complex conjugation.
I know for the unitary representation, we have: $$\chi(s^{-1})=tr(D(s^{-1}))=tr(D(s)^{-1})=tr(D^*(s))=\bar{\chi}(s)$$
My questions are:
- How to prove the case for "non-"unitary representation?
- Why the book says "finite group" and "arbitrary group"? Does the property of being "unitary" make that difference? Why?
Thanks!
The point is that any representation of a finite group can be made into a unitary representation by redefining the inner product. Indeed, if $(v_1,v_2)$ is the original inner product on $V$, and $D$ is a representation of the finite group $G$ on $V$, then $$ \left<v_1 , v_2\right> := \frac{1}{|G|} \sum_{s \in G} (D(s) v_1, D(s) v_2)$$
is also an inner product on $V$, and moreover, the representation $D$ is orthogonal with respect to this new inner product $\left< v_1, v_2 \right>$.
So to prove that $\chi(s^{-1}) = \bar\chi(s)$ for the representation $D$ of the finite group $G$, you should write the matrices $D(s)$ using a basis of vectors that are orthogonal with respect to the new inner product $\left< v_1, v_2 \right>$. Written in this basis, the $D(s)$'s are unitary matrices, so the $D(s)^{-1} = D^\dagger(s)$ step in your proof holds, and the result ${\rm tr}(D(s^{-1})) = {\rm tr}(D(s))^\star$ is valid. And since the traces of the matrices are invariant under changes of basis, you can deduce that ${\rm tr}(D(s^{-1})) = {\rm tr}(D(s))^\star$ is true for all choices of basis for $V$.