Timeline for Topology of the "normal spectrum" of a commutative von Neumann algebra
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2012 at 19:44 | comment | added | Sergei Akbarov | This is like a dialogue between people who don't understand interlocutor's langauge. Dima, after Julia's explanation I don't need further clarifications, she clarified everything (by the way, that is why what you wrote about L^1(S) was extra). My suggestion was that if you didn't understand what I find to be wrong in your answer here, then we can discuss this in meta, like I already did once after your suggestion: tea.mathoverflow.net/discussion/1202/… This won't be mathematics, this is ethics, so don't tell me again that this is not for meta. | |
| Apr 11, 2012 at 19:25 | comment | added | Dmitri Pavlov | @Sergei: Meta is not an appropriate venue for clarifying answers (read the faq to get an idea about the purpose of the meta); such clarifications (and requests for them) belong here in the comments. This is not a discussion nor a dispute; the very fact that you seem to think otherwise indicates a problem with your attitude. | |
| Apr 11, 2012 at 8:44 | comment | added | Sergei Akbarov | @Dmitry: That was not a complaint, that was irony, I told this alredy. But if you want to continue this discussion, you can open a topic in Meta (in November you told me about this possibility mathoverflow.net/questions/80146/… , I hope, you remember this), and I promise to explain there what is still not clear. There must be judges in our dispute, otherwise it would be unsolvable. | |
| Apr 10, 2012 at 12:45 | comment | added | Dmitri Pavlov | @Sergei: Here is an alternative description of the same proof: The von Neumann algebra under consideration can be assumed to be atomic (normal characters only see the atomic part), i.e., isomorphic to the algebra of bounded functions on some set S. Its predual can be identified with L^1(S). The topological space of normal characters is the subspace of L^1(S) consisting of those elements of L^1(S) that take value 1 on some element on S and vanish elsewhere. The L^1-distance between any two such elements is 2, hence the topology is discrete. | |
| Apr 10, 2012 at 12:26 | comment | added | Dmitri Pavlov | @Sergei: Instead of asking specific questions about the parts of the proof that are unclear to you, you immediately started to complain ("Stylistically it would be more precise to say..."). I cannot but remark that such nonconstructive position can hardly help you to achieve your goals. Comments on MathOverflow are not expected to be as precise as refereed papers. Some vagueness is always expected, and you are expected to ask further questions if something is unclear to you, instead of complaining and attempting to ridicule those who are trying to help you. | |
| Apr 10, 2012 at 9:13 | comment | added | Sergei Akbarov | @Dmitri: Takesaki's III.1 is 18 pages long. In addition, the text is quite difficult for non-specialists, and one should be quite sophisticated in those tricks to see that Corollary 1.13 indeed implies what I need. For 2 days I was trying to understand your hints, and finally I had to ask another specialist, Julia Kuznetsova. Only after her explanations I understood what one can have in mind here. My opinion is that each work, including helping people, can be done punctually and agreeably for those who need help. Or carelessly and disagreeably for them. And it is Julia who did it punctually. | |
| Apr 10, 2012 at 7:01 | comment | added | Dmitri Pavlov | @Sergei: I gave you a reference to §III.1 in Takesaki's book in my first comment. Corollary III.1.13 is clearly an overkill in this case. | |
| Apr 9, 2012 at 22:35 | comment | added | Sergei Akbarov | @Dmitri: That is irony: your explanation is so vague that it is equivalent (as an explanation) to a phrase like "everything follows from the fact that the preimage of a continuous map preserves open sets". Moreover, stylistically the last phrase is more preferable, since it better expresses the aroma of absurd. :) However, Julia Kuznetsova has already gave me an exact reference, namely, Corollary 1.13 in Takesaki's chapter III. So if you don't mind, I will thank her for this. | |
| Apr 9, 2012 at 13:18 | comment | added | Dmitri Pavlov | @Sergei: I fail to see how your statement is more precise than mine. An open singleton is an open set, but not every open set is an open singleton. I also don't understand the question about the existence of the characteristic function. | |
| Apr 9, 2012 at 12:33 | comment | added | Sergei Akbarov | @Dmitri Pavlov: > Every normal character is an isolated point in the weak topology, > because the preimage of the open set (0,2) with respect to the > evaluation at the characterstic function of the corresponding > point is an open singleton. Stylistically it would be more precise to say: "...because the preimage of an open set under the action of a continuous function is always an open set". A question: why does this characteristic function exist? | |
| Apr 7, 2012 at 6:12 | comment | added | Dmitri Pavlov | @Sergei: At the bottom of the page you can find a link to the TeX source. Every normal character is an isolated point in the weak topology, because the preimage of the open set (0,2) with respect to the evaluation at the characterstic function of the corresponding point is an open singleton. | |
| Apr 6, 2012 at 8:44 | comment | added | Sergei Akbarov | Dima, and where does Takesaki prove this: "The only natural topology on the set of normal characters is the discrete topology, in particular the weak topology induced by A is discrete." | |
| Apr 6, 2012 at 6:09 | comment | added | Sergei Akbarov | Dima, apparently, I am doing something wrong, but I can't understand the text in your link because of the abundance of phrases "Math Processing Error". What should I do to look at the formulas? | |
| Apr 6, 2012 at 5:32 | comment | added | Dmitri Pavlov | The facts mentioned in the first two paragraphs one can find in Takesaki's Theory of Operator Algebras I, in particular Section III.1. The Gelfand-Neumark theorem for commutative von Neumann algebras is discussed in the last two paragraphs of the “Definitions” section in the link above. | |
| Apr 5, 2012 at 19:37 | comment | added | Sergei Akbarov | "If by a normal character you mean a normal morphism of C*-algebras" -- yes, of course. Discrete topology? So the topological explanation is trivial... That is interesting. But where is it written? By the way, in your link I did not find a mentioning of Gelfand-Neumark. | |
| Apr 5, 2012 at 18:42 | history | edited | Dmitri Pavlov | CC BY-SA 3.0 |
added 1552 characters in body
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| Apr 5, 2012 at 18:21 | history | answered | Dmitri Pavlov | CC BY-SA 3.0 |