For a compact group $G$ and any finite-dimensional complex representation $(\pi,V)$, how to show that $V$ admits an invariant inner product?

Question

To demonstrate that an inner product $\langle\ ,\ \rangle$ on $V$ is invariant, one need to check $$\langle\pi(g)v,\pi(g)w\rangle=\langle v,w\rangle.$$ For an arbitrary inner product $\langle\langle\ ,\ \rangle\rangle$, consider the following inner product: $$\langle v,w\rangle=\int_G\langle\langle\pi(g)v,\pi(g)w\rangle\rangle\textrm{d}g.$$ Now I try to show that $\langle\pi(h)v,\pi(h)w\rangle=\langle v,w\rangle$ as $$\begin{aligned} \langle\pi(h)v,\pi(h)w\rangle&=\int_G\langle\langle\pi(g)\pi(h) v,\pi(g)\pi(h) w\rangle\rangle\textrm{d}g & \\ &= \int_G\langle\langle\pi(gh) v,\pi(gh) w\rangle\rangle\textrm{d}g & \\ &= \cdots= \langle v,w\rangle,& \\ \end{aligned}$$ but I find it hard to check that $$\int_G\langle\langle\pi(gh) v,\pi(gh) w\rangle\rangle\textrm{d}g=\int_G\langle\langle\pi(g)v,\pi(g)w\rangle\rangle\textrm{d}g.$$ Could you please provide me with some hints?

Answer 1

From Haar measure, we have $\textrm{d}g=\textrm{d}(gh)$. Then it follows that $$∫_G⟨⟨π(gh)v,π(gh)w⟩⟩\textrm{d}g=\int_G⟨⟨π(g)v,π(g)w⟩⟩\textrm{d}(gh^{-1})=∫_G⟨⟨π(g)v,π(g)w⟩⟩\textrm{d}g$$ by $g\mapsto gh^{-1}$.