Suppose (G) is a compact Lie group and (M) is a manifold with a left action by (G). If (f\in C^\infty(M)), then one can use the ``averaging trick" to produce a function (F\in C^\infty(M)) which is invariant under (G), that is, (F(g\cdot p)=F(p)) for all (p\in M), (g\in G). Explicitly,[F(p):=\int_G f(g\cdot p)\mu_g.]where (\mu) is a left-invariant volume form on (G) which is normalized, i.e.,[\int_G\mu=1.]It seems to me that the proof of this fact relies on another fact: for (G) compact, a left-invariant volume form on (G) is also RIGHT-invariant as well. This is how the proof goes (I think): for (p\in M), let (f_p:G\rightarrow \mathbb{R}) be defined by (f_p(g):=f(g\cdot p)). Then[F(p)=\int_Gf_p \mu.] Then for (x\in G),[F(x\cdot p)=\int_G f_{x\cdot p}\mu=\int_G f_p(gx)\mu_g][=\int_G (R_x^\ast f_p)\mu=\int_G (R^\ast_xf_p)R^\ast_x\mu=\int_G R_x^\ast(f_p\mu)][=\int_Gf_p\mu=F(p)]where we have used the fact that (\mu), in addition to being left-invariant (by hypothesis), is also right-invariant when (G) is compact, that is, (R^\ast_x\mu=\mu). This also shows that (R_x) is an orientation preserving diffeomorphism on (G) which explains the last equality. So again, in proving that the definition of (F) produces a (G)-invariant function on (M), does one need to use the fact that left-invariant (\mu) is necessarily right-invariant for (G) compact?
