# Do we have $\rho(g^*)=\rho(g)^*$ for a group representation $\rho$

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.

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 $$G$$.