Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
Obviously this change hasn’t happened. If this change is made, then it should be Haar integral, not Haar Integral.
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 μ (“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.
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).
added pointer to:
I recently added in an explanation of why the convex hull of a compact subspace of a Banach space is compact. This is known as the Krein-Smulian theorem. While Haar integral for a (smooth) Lie-group or (possibly) pro-Lie-group could be available using easier methods, it seems that the Haar integral constructed using topological vector spaces, Banach spaces, the Krein-Milman theorem, and convex sets, cannot be made any simpler than the Krein-Smulian theorem. Also, the proof of the Peter-Weyl theorem assumes the existence of Haar integral.
There is room to streamline the writeup of that whole statement and proof, starting here.
For one, currently it suddenly says “Proof” out of the blue. What is being proven is in fact a Lemma that is used in the main proof (that deserves to be labeled such) of the main statement (that deserves to be numbered).
In its current form the whole section is unnecessarily confusing, I think.
I find the proof fairly confusing.
Here’s one thing I found confusing: the way you’ve said it, the empty set would seem to belong to S (it is G-equivariant, it is compact, it is convex, and it is a subset of the set of extensions of f). It would appear to be the minimal element under inclusion order.
Another thing that is confusing is that Zorn’s lemma, in its usual application, creates a maximal element of a poset. Here you seem to want to produce a minimal element. Are you applying Zorn’s lemma to the opposite of S under inclusion order?? How would you apply it, exactly?
Finally, there seems to be some looseness around whether whether we are using compact convex sets or weakly compact convex sets. I have a hard time bridging between (1) and (2).
Is this proof supposed to be an extract of a proof elsewhere, or is it of your own devising?
I made a more extensive edit just now adding in answers to your questions. This fixed a mistake in which I did not mention the condition of being nonempty. This must have been part of the resulting confusion in the use of Zorn’s lemma, which is indeed applied in reverse order from the typical order of inclusion of subsets. This entails the use of the lemma that a decreasing chain of compact nonempty subsets is nonempty (or alternatively, that for a compact space, a decreasing chain of nonempty closed subsets is nonempty).
For your third question, we consider compactness in the weak topology here. I’m sorry for the confusion I made here.
If it helps I would say there is not a bridge there (and I hope my edit clarified that); the Krein-Smulian theorem allows us to conclude the weak compactness of the convex hull of a weakly compact set in a Banach space (it does not allow the other conclusion).
This is something I learned form Uri Bader after asking whether there was an approach which showed the injectivity of ℝ as an object a of some category of topological vector spaces. However, the Krein-Smulian theorem concerning the weak compactness of the convex hull of a weakly compact subset of a Banach space remains a difficult theorem in this area. One theorem I saw used a reduction to the assumption of separability. In this case that could be of interest because of how a refinement of Uryssohn’s lemma produces certain partitions of unity in Radon’s theorem. Radon’s theorem is of course no harm to assume here. I hope to find out eventually if the weak topology on the Banach space C0(X)* (not to be confused with the weak* topology) hosts a Krein-Smulian theorem which is any easier, possibly by using Radon’s theorem.
Oh and I would like to add a few more thoughts:
I have added proper pointer to your source:
The \textit commands didn’t render, so I changed to what I thought you were after visually. But I’m not sure I succeeded.
Also let me say that what mildly confused me is the condition that C is compact convex in the Banach space C0(G)*, when you shouldn’t be referring to this as a Banach space at all after you change over to the weak topology. I changed the wording to make clearer what I think you meant.
But overall, it’s now a whole lot clearer – thanks.
There is still a misunderstanding: the Krein-Smulian theorem concludes compactness in the weak topology but only for a Banach space. So it uses both tolologies. Edit, oh I see, it had confused you but you saw what I meand in the comment
To be absolutely clear, the offending sentence was “The Krein-Milman theorem for the case of the textitcompact,convexsubset C of the (textitlocallyconvex) Banach-space C 0(G) * is that C is the convex hull of its extreme points.” That’s what I fixed.
Ok I see now, yes that’s definitely an error.
1 to 22 of 22