Talk:Haar measure

Relationship to Jeffrey's prior

It might be interesting to add this to the article. 134.184.26.82 (talk) 12:34, 7 October 2020 (UTC)[reply]

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

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.

Duality

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

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

TeX

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,

f\mapsto \int _{G}f(tx)\ d\mu (x)

is a also right invariant. Thus there is unique function

\Delta

such that

\int _{G}f(tx)\ d\mu (x)=\Delta (t)\int _{G}f(x)\ d\mu (x).

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:

\int _{G}f(tx)\ d\mu _{1}(x)=\Delta (t)\int _{G}f(x)\ d\mu _{2}(x).

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

\int _{G}f(tx)\ d\mu _{1}(x)=\int _{G}f(x)\ d\Delta (t)\mu _{2}(x).

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 $\mu _{1}$ and $\mu _{2}$ be left and right invariant measures. For a set $S$ denote by $S^{-1}$ the set of inverses of elements of $S$ . Then, what you say is that there exists a positive constant $k$ such that

\mu _{1}(S^{-1})=k\,\mu _{2}(S)

for all measurable sets $S$ .

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

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

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

\mu _{1}(S^{-1})=k\,\mu _{2}(S)

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)

FWIW, agree w/Oleg, this last formula is a very natural way to write this. linas 14:11, 14 August 2006 (UTC)[reply]

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)

How to define

Q: How do you exactly define the Haar measure on an "arbitrary" (locally compact and Hausdorff) group? The uniqueness-statement does NOT define it.—Preceding unsigned comment added by 195.176.0.50 (talk)

The article defines the property of being a Haar measure and also states an existence and uniqueness result (without proof). Do you object to the use of "existence" in this statement, in the sense that without an explicit construction of Haar measure the statement of existence is meaningless?--CSTAR 19:36, 27 May 2006 (UTC)[reply]

The construction of the Haar measure is quite easy for Lie groups (confer volume form). I wonder if the article should mention it? But no, at least not until after we do the general case. -lethe ^{talk +} 19:52, 27 May 2006 (UTC)[reply]

Well by construction I meant here something more explicit, something along the lines of intuitionism or constructivism.--CSTAR 19:54, 27 May 2006 (UTC)[reply]

Oh I see. Well after checking in a book, I see that indeed the existence uses Tychanoff's theorem, so I guess it is nonconstructive. I suppose this article should mention that the general existence theorem is nonconstructive, but then go on to mention that in particular cases constructions can often be given. Like I say, the construction is easy for Lie groups, and it doesn't require choice, as far as I can tell. -lethe ^{talk +} 20:25, 27 May 2006 (UTC)[reply]

It's actually possible to give a proof w/o axiom of choice as per André Weil.--CSTAR 20:59, 27 May 2006 (UTC)[reply]

Well, I don't know about that. You brought up constructivism in the first place, but it seems irrelevant if there is a constructive existence proof. Either way, isn't the article incomplete without either a construction or a justification for the lack thereof? -lethe ^{talk +} 21:25, 27 May 2006 (UTC)[reply]

I was just trying to figure out if there was an "agenda" behind the anonymous question.

Re isn't the article incomplete without either a construction or a justification for the lack thereof? Do you mean put in a proof?--CSTAR 22:46, 27 May 2006 (UTC)[reply]

PS Given that Haar measure is unique (up to a constant) the fact that known proofs (around up to 1955) used the axiom of choice was weird. This prompted Weil (I think it was Weil) to find one which wasn't dependent on choice. --CSTAR 22:54, 27 May 2006 (UTC)[reply]

I stand corrected: It's Cartan not Weil. Googling I came up with this: [1]

Ultimately, neither Haar nor von Neumann proved the existence of invariant measures on all locally compact groups. The first one to come up with a full proof was Andŕe Weil. This proof, however, was criticized for using the Axiom of Choice in the form of Tychonoff’s Theorem. Later, Henri Cartan proved the existence of invariant measures on locally compact groups without the Axiom of Choice. Since then, several other people have also proved this theorem.

I think Hewitt and Stromberg has the constructive (i.e. non-choice) proof. --CSTAR 23:18, 27 May 2006 (UTC)[reply]

They use Tychonoff on page 6. 94.21.49.195 (talk) 23:30, 1 April 2009 (UTC)[reply]

For Lie groups everything is quite explicit. Here is a quote from "Noncommutative harmonic analysis" by Michael Taylor (Sect.0.2):

The term "Haar measure" in this context is a slight misnomer. Haar solved the more delicate problem of showing that every locally compact topological group has a locally finite left invariant Baire measure. The construction given above for Lie groups is quite simple and was known long before Haar's work.

Boris Tsirelson (talk) 19:38, 23 March 2009 (UTC)[reply]

From the indentations I would infer that you are replying to the first question in the subsection, but your comment seems to be directed at another question or remark. --CSTAR (talk) 21:20, 23 March 2009 (UTC)[reply]

Sorry... Nothing really important, just a note about the history. Somewhat related to "The construction of the Haar measure is quite easy for Lie groups" above (by Lethe). Boris Tsirelson (talk) 21:29, 23 March 2009 (UTC)[reply]

Borel sets

The article currently states early on that:

Let G be a locally compact topological group. In this article, the σ-algebra generated by all compact subsets of G is called the Borel algebra.[1]

with a footnote that an anon elaborated on:

[1] We follow the conventions of Halmos' textbook. Many authors instead use the term Borel algebra to denote the σ-algebra generated by the open sets.

I don't see that any part of the rest of the article depends on whether the sigma algebra was defined using open or closed sets. (If there is such content, it should be clarified locally). Thus, I propose eliminating the footnote, and replacing the sentence by:

Let G be a locally compact topological group, and let its Borel algebra be the σ-algebra corresponding to the topology of the group.

This avoids confusion over an extraneous, unrelated point. linas 14:23, 14 August 2006 (UTC)[reply]

I think the footnote is very important and we should not remove it. —Preceding unsigned comment added by 201.231.28.30 (talk • contribs)

I agree with linas that we should remove this extraneous detail. --MarSch (talk) 11:09, 17 December 2007 (UTC)[reply]

It seems very odd and rather confusing to use these uncommon conventions on Borel sets. Does any reasonably modern measure theory book use the term "Borel sets" in this way? I don't think so: Bogachev, Fremlin, Folland, Royden-Fitzpatrick, Rudin certainly don't. It is also inaccurate to say that Halmos considers the σ-algebra generated by the compact sets: he uses the term Borel sets for the σ-ring generated by the compact sets. Dottyhill (talk) 08:34, 2 October 2012 (UTC)[reply]

More on Borel sets

The article defines Haar measure on the sigma algebra generated by compact sets. However, some classic books like Rudin's "Fourier Analysis on Groups" consider Haar measure over the sigma algebra generated by all open sets. Rudin's book does not prove explicitly the existance of such a measure. Is there a mistake on Rudin's book or Haar measure can be defined over the sigma algebra generated by all open sets? I think this is a very important matter since it has consequences on the mesurability of the continuos functions. Can someone provide a clarification about this matter and maybe an example that shows that one needs to restrict the construction to the sigma algebra generated by compact sets or a reference to a book where the measure is constructed over the sigma algebra generated by all open sets? That would be a great contribution to this article! —Preceding unsigned comment added by 201.231.28.30 (talk • contribs)

No, Rudin's book is correct, but so is the WP article. The statement given in the article is exactly as given in Halmos' book "Measure Theory" (now admittedly klunky, but authoritative and completely accessible). Haar measure can also be approached via Radon measures on locally compact spaces; these are non-negative functionals on the space of continuous functions of compact support. There is a unique (up to a scalar) non-zero left invariant Radon measure on a locally compact group. Now Radon measures can be "extended" in a canonical way to the σ-algebra of Borel sets (generated by open sets). See Rudin's "Real and Complex Analysis 2.14" (Big Rudin 1966) for a precise statement and the proof. The delicate point here is uniqueness, which requires somewhat different regularity properties for the measures in the statements.

Most of these technical distinctions are immaterial for metrizable separable locally compact groups or compact groups. This would be massive technical overkill for the article to elaborate further, I think.

--CSTAR 20:32, 11 September 2006 (UTC)[reply]

The theorem is stated incorrectly.

The theorem says that $\mu (B)>0$ for any nonempty Borel set $B$ . What if B is a singleton set? This is ridiculous. 128.208.1.224 (talk) 21:26, 4 March 2009 (UTC)[reply]

The theorem says that

\mu (B)>0

for any open nonempty Borel set

B

. If a singleton set is open, then indeed it must have positive measure, and the group is discrete. JackSchmidt (talk) 22:50, 4 March 2009 (UTC)[reply]

Harr measure on unitary matrices

Could someone explicitly discuss the Harr measure on unitary matrices? It is very important in physics (especially for SU(N) in QFT and unitary transformations in quantum information theory). I haven't found anywhere it is explained well. Njerseyguy (talk) 16:14, 15 July 2009 (UTC)[reply]

Existence of the Haar measure

In section "Existence and uniqueness of the left Haar measure" it is said that

<<It turns out that there is, up to a positive multiplicative constant, only one left-translation-invariant countably additive regular measure μ on the Borel subsets of G such that μ(U) > 0 for any open non-empty Borel set U. Such a measure is called a left Haar measure>>

While in section "Uses" it is stated that

<<A frequently used technique for proving the existence of a Haar measure on a locally compact group G is showing the existence of a left invariant Radon measure on G.>>

Why bother to "use a technique for proving the existence of a Haar measure" if it always exists? Where am I missing the point? —Preceding unsigned comment added by 88.1.123.97 (talk) 17:03, 6 November 2009 (UTC)[reply]

Split

There is a split and an expand tag. The section as it stands cannot be split because a split article would not be viable. The expand tag seems reasonable so I have removed the split tag. Op47 (talk) 15:09, 6 April 2012 (UTC)[reply]

Suggested need for expansion of section 'Haar integral'

Can someone say what expansion they would like to see in this section? The statements under Haar integral seem quite clear. If the question is 'How to define the Haar integral', or another such question, then this is properly a question about the definition of the 'Lebesgue integral' with respect to a given measure. Also there are links to 'Lebesgue integration' AND 'indicator functions' here.

Further, with the 4 examples of Haar measures beneath this heading, it seems that there is enough meat in this section for someone who longs for some computational examples of Haar integrals. 99.33.255.116 (talk) 22:42, 2 June 2013 (UTC)[reply]

Open submanifolds

The measure for open submanifolds is stated to be dⁿx/J(x). But shoulden't it be $d^{n}x/|\det(J(x))|$ instead? --Jobu0101 (talk) 09:22, 14 December 2016 (UTC)[reply]

Existence of Haar measure: regular content?

Quote from the section about existence:

The function

{\displaystyle \mu _{A}}

is additive on disjoint compact subsets of

{\displaystyle G}

, which implies that it is a regular content.

I understand that this is a content. I don't see, however, how this is a regular content. Additivity for disjoint compact sets is part of the definition of content. But in order to get a Radon measure from it, we need a regular content. I couldn't find anywhere a convincing proof that the so-constructed content is regular. Could someone who understands this please expand upon that claim? --138.246.2.201 (talk) 12:13, 28 July 2017 (UTC)[reply]

An unstated change of hyperbola in the middle of a paragraph

The section titled Examples contains this paragaph:

"* The unit hyperbola $\{(x,y):x^{2}-y^{2}=1,x>0\}$ can be taken as a group under multiplication defined as with split-complex numbers $z=x+yj,\ \ j^{2}=+1.$ The usual area measure in the crescent $C=\{(x,y):|y|<x,\ \ x^{2}-y^{2}<1\}$ serves to define hyperbolic angle as the area of its hyperbolic sector. The Haar measure of the unit hyperbola is generated by the hyperbolic angle of segments on the hyperbola. For instance, a measure of one unit is given by the segment running from (1,1) to (e,1/e), where e is Euler's number. Hyperbolic angle has been exploited in mathematical physics with rapidity standing in for classical velocity."

Unfortunately the points (1,1) and (e,1/e) do not lie on the stated hyperbola, but instead on the hyperbola given by

{(x,y) in R² | xy = 1}.

I hope that someone familiar with this example can fix this.

In addition, this article has many places where it could be phrased a lot more clearly.50.205.142.50 (talk) 22:00, 8 June 2020 (UTC)[reply]