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My question is

Why is the Haar measure on a Lie Group unique upto scalar multiple?

I know how to show it for $\mathbb R^n$ because there I have countable many open balls that form a base and the measure of the unit ball around 0 gives the constant scalar.

How to show for general Lie Group?

Angry_Math_Person
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  • Calm down. The left and right Haar measures on a locally compact topological group are unique. See here for a proof. – Bumblebee Nov 27 '20 at 20:08
  • What you said certainly is true. But I wanted a proof specifically for Lie Groups. The point is to show existence of Haar measure on Lie Group you can do it in a much easier way than for a general Topological group. I wanted to know if there's a way to show uniqueness easily too. – Angry_Math_Person Nov 27 '20 at 20:13
  • Choosing a coordinate chart around the identity, then the approach you use in $\mathbb{R}^n$ should work fine on a general Lie group. – Kajelad Nov 28 '20 at 06:37
  • @Angry_Math_Person A locally compact is far from being a "general topological group". Many interesting usual topological groups are not locally compact (infinite-dimensional normed spaces, many automorphism groups...) – YCor Nov 28 '20 at 16:13

1 Answers1

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Let $\mu,\nu$ be two left Haar measures on a locally compact group. Then $\mu+\nu$ is a left Haar measure, and $\mu$ has a density $f$ with respect to it. Then $f$ is left-invariant, so is constant. Hence $\mu$ is a scalar multiple of $\mu+\nu$.

I don't really see how to make it simpler, even with additional assumptions?

Abdelmalek Abdesselam
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YCor
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  • Nice argument. I don't remember seeing that in textbook treatments. – Abdelmalek Abdesselam Nov 30 '20 at 20:22
  • I am missing something I believe. What is the reason that $\mu$ has a density $f$ with respect to $\mu + \nu$? And why doesn't the same reason not tell you more directly that it has a density w.r.t $\nu$? – Vincent Nov 30 '20 at 20:34
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    Because $\mu\ll\mu+\nu$, so $\mu$ has a density with respect to $\mu+\nu$ by the [Radon-Nikodym theorem] (I should have assumed the group $\sigma$-compact to ensure that the measure are $\sigma$-finite, i.e., for Lie groups, that the number of components is countable. In general we don't know a priori that $\mu\ll\nu$. (Recall that $\mu_1\ll\mu_2$ means that $\mu_2(A)=0$ implies $\mu_1(A)=0$: one says that $\mu_1$ is absolutely continuous with respect to $\mu_2$). – YCor Dec 01 '20 at 03:05
  • @YCor Thanks! That clarifies it! – Vincent Dec 01 '20 at 10:07
  • Why is the density f left invariant? – Angry_Math_Person Dec 02 '20 at 12:49
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    It's straightforward that it's left-invariant modulo measure zero ($gf-f$ is a.e. zero for each given $g$). Some argument to show it's indeed constant modulo measure zero should indeed be added. – YCor Dec 02 '20 at 13:15
  • @YCor btw I asked a question about that last step here. – Carlos Esparza Mar 01 '22 at 18:54