Talk:Haar measure

The article says:

It turns out that there is, up to a multiplicative constant, only one translation invariant measure on X which is finite on all compact sets.

This can't be quite right. On R, for example, Borel measure and Lebesgue measure both have this property. Perhaps it's correct if "measure" is replaced with "complete measure". --Zundark, 2002 Mar 6

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You're mistaken. "X" was defined as the sigma algebra generated by the compact sets. Lebesgue measure has a more extensive domain than that; Borel measure does not.

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I think some of the facts on Pontryagin duality really belong here.

Charles Matthews 08:20, 17 May 2004 (UTC)

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TeX isn't working. Argh!

CSTAR 15:24, 6 Jun 2004 (UTC)

Question about this article

Let me quote the last section from the article

The modular function

Here we show how right and left Haar measures are related. Note that the left translate of a right Haar measure (or integral) is a Haar measure (or integral): More precisely, if μ is a right Haar measure,

<math>f \mapsto \int_G f(t x) \ d\mu(x) <math>

is a also right invariant. Thus there is unique function <math>\Delta<math> such that

<math>\int_G f(t x) \ d \mu(x) = \Delta(t) \int_G f(x) \ d \mu(x) .<math>

From what I see, μ is the right measure. If you say that the left and the right measures are related, where is the left measure in here? And which is the relation? Oleg Alexandrov 00:06, 15 Jan 2005 (UTC)

This defines the modular function. The left translate of a right invariant Haar measure is also a right invariant Haar measure. Thus the modular function. However having defined the modular function, the modular function also serves to relate right and left Haar measures.CSTAR 00:35, 15 Jan 2005 (UTC)

I still don't get it. OK, it is my fault I guess; I know what an invariant measure is, but I don't know what a modular function is. What I really wanted, is something of the form:

<math>\int_G f(t x) \ d \mu_1(x) = \Delta(t) \int_G f(x) \ d \mu_2(x) .<math>

with μ1 being the left measure, μ2 being the right measure. In other words, I want some explicit formula showing how they relate. The above argument you wrote is probably correct, but it is hard to interpret (for me). Oleg Alexandrov 01:55, 15 Jan 2005 (UTC)

Nope that won't work. What you wrote says

<math>\int_G f(t x) \ d \mu_1(x) = \int_G f(x) \ d \Delta(t) \mu_2(x) .<math>

which would say the left translate of a right Haar measure is a multiple of a left Haar measure. However, the left translate of a right Haar measure is also a right Haar measure so what you're looking for is an assertion that a right Haar measure is a multiple of a left Haar measure, e.g. is itself left Haar. That's only true for unimodular groups. CSTAR 06:20, 15 Jan 2005 (UTC)

I was aware that precisely this would not work. OK, let me try one last time. Is there a way (any way) to establish an explicit equation between the left and the right measures, containing both and μ2? If you say "No", I will give up. Oleg Alexandrov 17:52, 15 Jan 2005 (UTC)

Basically taking the inverse on G interchanges left and right. So μ1(g) is proportional to μ2(g^-1), if you understand the notation ...

Charles Matthews 19:15, 15 Jan 2005 (UTC)

Good. OK, all I wanted is a formula in the The modular function of Haar measure article which contains both μ1(g) and μ2(g^-1). But I guess that is not achievable. Oleg Alexandrov 19:55, 15 Jan 2005 (UTC)

I don't know what I was thinking when I wrote the above. I understand what you say, and this is indeed neat! Basically, let <math>\mu_1<math> and <math>\mu_2<math> be left and right invariant measures. For a set <math>S<math> denote by <math>S^{-1}<math> the set of inverses of elements of <math>S<math>. Then, what you say is that there exists a positive constant <math>k<math> such that

<math>\mu_1(S^{-1})=k\,\mu_2(S)<math>

for all measurable sets <math>S<math>.

I will have this written in the article sometime. Oleg Alexandrov 21:56, 16 Jan 2005 (UTC)

I thought that fact was already in it. If μ is a left Haar measure then

<math> \int f(x^{-1}) d \mu(x) <math>

is a right invariant integral. Maybe it should be made more explicit by letting f be an indicator function. CSTAR 22:53, 16 Jan 2005 (UTC)

I think the equation <math>\mu_1(S^{-1})=k\,\mu_2(S)<math> could be the simplest thing to write. Also, the sentence "Here we show how right and left Haar measures are related." should go from the last section in the section where the actual equation containing both the right measure and the left measure is. That sentence is misleading in the last section, because that section talks not about the relation between the left and the right measures, but rather about the relation between a right measure and its left translate. Oleg Alexandrov 00:59, 17 Jan 2005 (UTC)

The theorem

I forgot to say in my change to the Haar measure article that I removed the statement of the theorem because it shows up twice. Oleg Alexandrov 04:54, 17 Jan 2005 (UTC)