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1. I have changed the name to Haar Integral – if that’s ok – since the perspective I have added to the article leaves Haar measure as a consequence of Haar integral, and not the other way around.

edeany@umich.edu

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeFeb 23rd 2020

Obviously this change hasn’t happened. If this change is made, then it should be Haar integral, not Haar Integral.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 24th 2020

These are some major edits. In general, it’s a good idea for the author making such major edits to document them here so that we can discuss them, rather than leaving it to others to hunt them down through red-and-green, which can be at times a chore.

For example, a decision was made to erase the positivity condition on Radon measures. Why was this done? Notice that in doing so, the definition of $\mu$ (“in the usual sense of measure theory”) as a certain supremum doesn’t seem quite right.

Not a huge change, but for some reason, Haar’s name is capitalized, but Hausdorff’s has been changed to lower-case. Why change an earlier author’s decision?

Is the lesser-known proof in the “Analogy” section documented somewhere? I think it’s an interesting question, just what is the dependence on choice principles in constructing Haar measures (it would be surprising to me if it were really needed, since uniqueness up to scalar multiple suggests a certain canonicity).

I’m not immediately tuning in to the remark on bar constructions, but maybe I haven’t thought hard enough. Are there also degeneracy maps floating around?

I may get in there at some point to prettify the formatting. Only in recent months have I begun to appreciate the considerable virtues of TikZ.

• CommentRowNumber4.
• CommentAuthorDean
• CommentTimeFeb 24th 2020
• (edited Feb 24th 2020)
>>In general, it's a good idea for the author making such major edits to document them here so that we can discuss them, rather than leaving it to others to hunt them down through red-and-green, which can be at times a chore.

Ok. I'll do that next time.

>>For example, a decision was made to erase the positivity condition on Radon measures. Why was this done? Notice that in doing so, the definition of $\mu$ ("in the usual sense of measure theory") as a certain supremum doesn't seem quite right.

This is a good point. I should have been more attentive to the discrepancy in the article. It is natural to want the Riesz representation theorem to apply, so I changed it to require positivity again.

Then again, does uniqueness hold even without the positivity assumption?

The lesser known proof was shown by Uri Bader in response to a mathover flow question of mine. That is here: https://mathoverflow.net/questions/351091/existence-and-uniqueness-of-haar-measure-on-compacta-a-cohomological-approach

He mentioned I could email him if I wanted to know more, so I sent him a few questions about whether it is known already.

>>I think it's an interesting question, just what is the dependence on choice principles in constructing Haar measures (it would be surprising to me if it were really needed, since uniqueness up to scalar multiple suggests a certain canonicity).

This and the standard proof use the axiom of choice, but there is a constructive version: "a simplified and constructive proof of the existence and uniqueness of haar measure" by E.M. Alfsen.
• CommentRowNumber5.
• CommentAuthorDean
• CommentTimeFeb 24th 2020
• (edited Feb 24th 2020)
P.S. Under convolution, $C(G)$ is a unitless algebra. The monad at hand is therefore unitless, and the bar resolution of such a unitless monad has only faces and no degeneracies. Hence the scare-quotes. I clarified this in the most recent edit.

I had the same question, as to why this is a unitless monad, and how to fix it. Of course, there is a canonical way of adding units to an $\mathbb{R}[G]$-algebra: take product with $\mathbb{R}[G]$. This adds the degeneracy maps you were thinking of. On further contemplation, this is exactly the role of the dirac delta function. So adding dirac delta functions $\delta_g$ gives the resolution degeneracy maps.
• CommentRowNumber6.
• CommentAuthorDean
• CommentTimeFeb 24th 2020
If you end up fixing this up with your Tex knowledge, you might wish to make a table out of my analogies in the analogies section. I tried to do this, but I got slowed down learning the notation and I kind of gave up.
• CommentRowNumber7.
• CommentAuthorDean
• CommentTimeFeb 25th 2020
It looks like if we switch $G$-Ban to a category where $G$ acts by isometries (of course, this is more natural), then we get positivity in the proof! The proof is essentially the same as before, except that the resulting extension to Y has norm $1$. From this, positivity follows. I'll fix it shortly.
• CommentRowNumber8.
• CommentAuthorGuest
• CommentTimeMar 3rd 2020
Update: I have fixed the proof of existence of Haar integral; now the constructed integral satisfies positivity. It turns out it is as simple as working in the category of norm preserving $G$-actions.

Secondly, I have added a section connecting this page to the page on extensive and intensive quantities. Haar integral is a nice choice of extensive quantity, namely the unique $G$-invariant norm preserving one.
• CommentRowNumber9.
• CommentAuthorDean
• CommentTimeMar 3rd 2020

The article and its proofs are designed around the Haar integral instead of the Haar measure. This potentially more streamlined approach avoids heavy use of measure theory (or, I suppose one could say that it puts all the measure theory into the Riesz representation theorem). Considering this, I am attempting to rename the article Haar integral (last time it didn’t work for some reason).

• CommentRowNumber10.
• CommentAuthorDean
• CommentTimeMar 3rd 2020

My apologies, I had capitalized integral.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeApr 8th 2021