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Let $G$ be a compact group with Haar measure $m$, then $m$ is left-invariant, in the sense that $\int_{G} f(x) \ dm(x) = \int_{G} f(s^{-1}x) \ dm(x)$ for all $s\in G$ and for all $f\in C(G)$, and $m$ is also right-invariant.

Why does $m$ satisfies the relation $\int_{G} f(x) \ dm(x) = \int_{G} f(x^{-1}) \ dm(x)$ for all $f\in C(G)$?

My idea was to show that $f\mapsto m(\tilde{f})$ with $\tilde{f}(x) = f(x^{-1})$ is also a haar measure on $G$ and then the equality would follow from the uniqueness of the haar measure. But I don't know how to show the left and right invariance.

Amirhossein Haddadian
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Jonas Stange
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  • Isn't more natural to work on the level of measures? If $A\mapsto m(A)$ is a left-invariant measure, then $A\mapsto m(A^{-1})$ is necessarily right invariant. As you seem to assume that $G$ is unimodular, the desired follows by the usual approximation of measurable functions by simple functions. – WoolierThanThou Dec 17 '19 at 22:34
  • Have you tried writing down what left invariance of $f\mapsto m(\tilde{f})$ would mean? – Eric Wofsey Dec 17 '19 at 22:39
  • The theorem I'm referring to is from Rudins 'Functional analysis' and he is not working with unimodular groups, so I think there has to be another way to get to the equality of the integrals – Jonas Stange Dec 17 '19 at 22:41
  • @Eric Wofsey Left invariance would mean, that $\int_{G} f(s x^{-1}) \ dm(x) = \int_{G} f(x^{-1}) \ dm(x)$ holds true for all $s\in G$ – Jonas Stange Dec 17 '19 at 22:43
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    @WoolierThanThou $G$ is a assumed to be a compact group and compact groups are automatically unimodular. If $\Delta$ is the modular function of the Haar measure then for a compact group $G$, $\Delta(G)$ is a compact subgroup of $(0,\infty)$ under multiplication and hence must be ${1}$. – Rhys Steele Dec 17 '19 at 22:43

1 Answers1

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So, I think I figured it out. Let $f\in C(G)$ and $\tilde{f}(x) = f(x^{-1})$ for $x\in G$. Then $\tilde{f}\in C(G)$ and we get $\int_{G} (L_{s}\tilde{f})(x) dm(x) = \int_{G} \tilde{f}(sx) dm(x) = \int_{G} \tilde{f}(x) dm(x) = \int_{G} f(x^{-1}) dm(x)$, where I used the Left-Invariance of $m$ in the second equality. Then $f \mapsto m(\tilde{f})$ is also a Haar Measure and the uniqueness gives $m(\tilde{f}) = m(f)$. Does anybody see any mistakes?

Jonas Stange
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    This answer has the right idea but things could be written more clearly. What your computation checks is that the measure $\tilde{m}$ is right-invariant since $\int_G L_s \tilde{f}(x) dm(x) = \int_G \tilde{f}(sx) dm(x) = \int_G f(xs^{-1}) d\tilde{m}(x)$. This is fine since a compact group is unimodular so that $\tilde{m}$ being right invariant implies that it is (a scalar multiple of) the Haar measure $m$. In fact, the result of this question is true for a group $G$ iff that group is unimodular so you should expect to explicitly use the fact that $m$ is both left and right invariant somewhere. – Rhys Steele Dec 18 '19 at 14:11
  • But shouldn't it be $\int_{G} \tilde{f}(sx)dm(x) = \int_{G} f((sx)^{-1})dm(x) = \int_{G} f(sx)d\tilde{m}(x)$ or am I missing something? – Jonas Stange Dec 18 '19 at 14:24
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    Only the $x$ gets inverted in changing from $dm(x)$ to $d\tilde{m}(x)$ in the integral so you have $\int_G f(sx) d\tilde{m}(x) = \int_G f(sx^{-1}) dm(x)$ – Rhys Steele Dec 18 '19 at 14:28
  • Oh yeah, I think I should have it now. Thanks a lot. – Jonas Stange Dec 18 '19 at 14:36