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I am reading the book Representations of Compact Lie Groups. On page 79, in the proof of Theorem 4.6, it is said that $b: V \times \overline{W} \to \mathbb{C}$ is $G$-invariant. We have \begin{align} b: V \times \overline{W} & \to \mathbb{C} \\ (v, w) & \mapsto \int_G \overline{<g \alpha, v>}<g \beta, w>dg. \end{align} We need to show that \begin{align} b(gv,gw) = b(v,w). \end{align} That is \begin{align} \int_G \overline{<g \alpha, gv>}<g \beta, gw>dg = \int_G \overline{<g \alpha, v>}<g \beta, w>dg. \end{align} But I don't know how to prove the above identity. Any help will be greatly appreciated! Thank you very much.

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What you wrote is not what you need to show. Given $$b(v, w) = \int_G \overline{ \langle g\alpha , v\rangle } \langle g\beta, w\rangle dg$$ you want $\langle h v, hw\rangle = \langle v, w\rangle $ for all $h\in G$, which is $$\int_G \overline{ \langle g\alpha , hv\rangle } \langle g\beta, hw\rangle dg= \int_G \overline{ \langle g\alpha , v\rangle } \langle g\beta, w\rangle dg$$ Using $\langle g\alpha , hv\rangle = \langle h^{-1}g\alpha, v\rangle$ and similar for $hw$, together with the fact that $dg$ is left-invariant, $$\begin{split} b(hv, hw) &= \int_G \overline{ \langle g\alpha , hv\rangle } \langle g\beta, hw\rangle dg \\ &= \int_G \overline{ \langle h^{-1}g\alpha , v\rangle } \langle h^{-1}g\beta, w\rangle dg \\ &= \int_G \overline{ \langle h^{-1} g\alpha , v\rangle } \langle h^{-1} g\beta, w\rangle d(h^{-1}g) \\ &= \int_G \overline{ \langle g\alpha , v\rangle } \langle g\beta, w\rangle dg\\ &= b(v, w). \end{split}$$

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