Suppose we have a complex irreducible representation $(V, \phi)$ of a finite group $G$. (Where $\phi : G \rightarrow \text{Aut}_{\mathbb{C}}(V)$). Suppose $V$ it is isomorphic to its dual $V^*$, and that $G$ acts on $V^*$ by $(g\alpha)(v) = \alpha(g^{-1}v). $ Then there exists a $G$-equivariant vector space isomorphism $f : V \tilde{\longrightarrow} V^*$. There must be a homomorphism $\psi : G \rightarrow \text{Aut}_{\mathbb{C}}(V^*)$ such that $(V^*,\psi)$ is another representation of $G$. Is there a way to link $\phi$ and $\psi$ to each other? E.g. via $f$?
My goal is to show that for all $g \in G$, $\chi_V(g) = \overline{\chi_V(g)}$, by the way.
Best Answer
This is by definition of the $G$-equivariant map. We have, for all $g \in G, v \in V$:
$$ f(g\cdot v) = g \cdot f(v). $$ (See: https://en.wikipedia.org/wiki/Representation_theory#Equivariant_maps_and_isomorphisms.) You should note however, that in this case, the sets $V$ and $V^*$ aren't just any old $G$-sets. The dot stands not just for some action, but for the action given by the representations $\phi, \psi$, respectively. Knowing this allows you to write, for all $g \in G$:
$$ f\circ \phi (g)(v) = f(\phi(g)(v)) = f(g\cdot v) = g \cdot f(v) = \psi(g)(f(v)) = \psi(g) \circ f (v), $$
so $f\circ \phi(g) = \psi(g) \circ f$. Now as $f$ is in this case bijective, we can rewrite this as $ \psi(g) = f\circ \phi(g)\circ f^{-1}$. As similar linear maps have the same trace, it now follows that $$\chi_V(g) = \text{tr}(\phi(g)) = \text{tr}(f \circ \phi(g) \circ f^{-1}) =\text{tr}(\psi(g))= \chi_{V^*}(g). $$ Moreover, it's a fact of complex linear algebra that tr$(f^{-1}) = \overline{\text{tr}(f)}$.
(See: https://groupprops.subwiki.org/wiki/Trace_of_inverse_is_complex_conjugate_of_trace.)
Another general result is that $\chi_{V^*}(g) = \overline{\chi_V(g)}$. By the above it now quickly follows that $$\chi_{V}(g) = \chi_{V^*}(g) = \overline{\chi_V(g)}. $$