Diffgeometer1,
@Diffgeometer1@mathstodon.xyz avatar

@johncarlosbaez You are correct. I never used the fact that (\mu) is left-invariant. The proof I gave only requires that (\mu) is right-invariant.

I've seen a few books where the author starts with a left-invariant volume form on a compact Lie group (G) and then with a left (G)-action on the manifold or vector space (if one is doing representation theory and wants to define a (G)-invariant inner product), they use the same basic definition to produce a (G)-invariant structure. However, the thing that annoys me is that the proof of that (G)-invariance relies on the volume form actually being right-invariant. For a compact Lie group this is not a problem since a left-invariant volume form is necessarily right-invariant but they never mention this point.

Hall's book on Lie groups is one example that comes to mind. If one wants to construct a (G)-invariant inner product on a representation (V) of (G), he starts with an arbitrary inner product (\langle\cdot,\cdot\rangle') on (V), and then defines a (G)-invariant inner product via[\langle u,v\rangle:=\int_G\langle g\cdot u,g\cdot v\rangle' \mu_g.]However, the proof that (\langle\cdot,\cdot\rangle) is actually (G)-invariant depends on (\mu) being right-invariant but he never mentions this point which made me wonder if I was misunderstanding the construction.

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