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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?

Thanks!

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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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