To me, this seems like a true statement, but I'm not sure on the best way to prove it. If $\chi$ is the character for a representation $\rho$ such that $\rho(g)=\rho_g$, then the associating representation for $\overline{\chi}$ would be $\overline{\rho}$ such that $\overline{\rho}(g)=\overline{\rho_g}$.

$\overline{\rho}$ is definitely a valid representation, since it's a group homomorphism, and the character $\overline{\chi}$ of this representation would be how we defined it. Is this all there really is to it? Is there an easier way to think about it?



This is not the correct way to think about this. I assume that $\rho$ is a representation of a finite group. Then notice that $\rho(g)^n$ is the identity matrix, where $n$ is the order of $g$. That means the eigenvalues of $\rho(g)$ must be $n$-th roots of unity, so if $\lambda$ is an eigenvalue, $\lambda^{-1}=\bar{\lambda}$. Note the above also means $\rho(g)$ is unitary, so that $\overline{\rho(g)}=\rho(g^{-1})$. Therefore $\chi(g^{-1})=\overline{\chi(g)}$.

For any $(\rho,V)$, you also get a representation $(\rho^*,V^*)$ on the dual space $V^*$, by $(\rho^*(g)\varphi)(v)=\varphi(\rho(g^{-1})v)=\varphi(\rho(g)^{-1}v)$. Putting this all together, you see that $\chi_{\rho^*}=\bar{\chi_\rho}$.


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