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Assume that $G\leq GL_n$ is a symmetric group (i.e. if $g\in G$ then $g^*\in G$ and $G$ is Zariski closed) and $K$ be a maximal compact subgroup of $G$. Say $\rho:G\rightarrow GL(V)$ is a representation of the group $G$. I denote the action of $g\in G$ on $v\in V$ by $g\cdot v$. Say $\langle \_,\_\rangle$ is a $K$-invariant Hermitian inner product on $V$. Does it hold that $$ \langle v,g\cdot w\rangle=\langle g^*\cdot v,w\rangle? $$ Or equivalently, do we have $\rho(g)^*=\rho(g^*)$?

I assume that since $\langle\_,\_\rangle$ is $K$-invariant, that should be the case but I could not prove it. Thanks for your help.

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It turns out that $\langle v,g\cdot w\rangle=\langle g^*\cdot v,w\rangle$ holds since the Zariski closure $\overline{K}$ of $K$ equals $G$.

Proof : Since $\langle\_,\_\rangle$ is $K$ invariant, we have $\rho(K)\subseteq U(V)$ where $U(V)$ denotes the unitary group on $V$. Thus, the equation $$ \langle v,g\cdot w\rangle=\langle g^*\cdot v,w\rangle$$ holds for $g\in K$. On the other hand, as $\langle\_,\_\rangle$ is Hermitian, this equation is a polynomial in the entries of $g$ (I assume by convention that the Hermitian inner product is anti-linear in the first coordinate). Since $\overline{K}=G$ and the equation holds for $K$, it also holds for $G$.

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