Timeline for The topology of pointwise convergence with the adjoint operator on a von Neumann algebra
Current License: CC BY-SA 3.0
23 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 14, 2016 at 10:08 | comment | added | Cameron Zwarich | It turns out that my original counterexample was incorrect (for pretty silly reasons regarding misusing the definition of a real-valued measurable cardinal). The answer is affirmative for $\ell^\infty(I)$ by the Grothendieck Completeness Theorem, and it seems likely to generalize for the reasons I mention in the edited post, but it doesn't seem to follow immediately from anything else. | |
| Nov 14, 2016 at 10:06 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 13, 2016 at 11:51 | comment | added | Sergei Akbarov | And in case that this is true, I think you should write a paper, I will cite it in my work. | |
| Nov 13, 2016 at 11:47 | vote | accept | Sergei Akbarov | ||
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| Nov 13, 2016 at 11:47 | comment | added | Sergei Akbarov | Yes... Thank you, Cameron. The negative answer, actually, gives birth to another question (my last hope): if $\{u_i, i\to\infty\}$ is a net of functionals from $A_*$ such that on each totally bounded (with respect to the strong* topology) set $S\subseteq A$ they are equicontinuous and they tend to a functional $u$ on $A$ uniformly on each such $S$, will $u$ nevertheless belong to $A_*$? If this is trivial, please, add this to your answer, if not - let me know, I'll ask this question separately. | |
| Nov 13, 2016 at 9:29 | comment | added | Cameron Zwarich | Given a real-valued measurable cardinal, there is indeed a counterexample. I added the details. | |
| Nov 13, 2016 at 9:29 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 13, 2016 at 5:54 | comment | added | Cameron Zwarich | Every compact set in a TVS is bounded, the two topologies have the same bounded sets, and they agree on those bounded sets. | |
| Nov 13, 2016 at 5:49 | comment | added | Sergei Akbarov | Cameron, I think, I miss something... OK, total boundedness is the same as precompactness for both topologies. But why is the class of compact sets the same for the strong* and the σ-strong* topology? | |
| Nov 13, 2016 at 5:21 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 13, 2016 at 3:07 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 13, 2016 at 2:36 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 13, 2016 at 2:26 | comment | added | Cameron Zwarich | @SergeiAkbarov The strong$^*$ and $\sigma$-strong$^*$ topologies are both quasicomplete, i.e. every bounded and closed subset is complete. Therefore total boundedness and precompactness are equivalent notions. This follows pretty easily from the Uniform Boundedness Principle, but I'll add the details to the answer. I also think that I know how to prove this unconditionally for all von Neumann algebras, but I'll doublecheck the details first. | |
| Nov 12, 2016 at 18:35 | comment | added | Sergei Akbarov | Cameron, I don't understand this: "your question is a well-defined question about the σ-strong∗ topology". Are totaly bounded sets in $\mathcal M$ with respect to the strong∗ topology the same as totally bounded sets with respect to the σ-strong∗ topology? | |
| Nov 12, 2016 at 15:12 | comment | added | Cameron Zwarich | @MateuszWasilewski I meant that all of the $\sigma$ topologies have the same continuous linear functionals and so do all of the non-$\sigma$ topologies, so by the Krein-Smulian Theorem applied to the $\sigma$-weak topology they all have the same closed convex sets. I edited it to fix that. | |
| Nov 12, 2016 at 15:10 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 12, 2016 at 13:49 | comment | added | Mateusz Wasilewski | @CameronZwarich, could you clarify the statement about the spaces of continuous linear functionals being the same for all the topologies mentioned in your answer? For example, not every $\sigma$-weakly continuous functional is weakly continuous. | |
| Nov 12, 2016 at 12:43 | comment | added | Sergei Akbarov | OK, thank you, Cameron, I look at this. | |
| Nov 12, 2016 at 12:40 | comment | added | Cameron Zwarich | @SergeiAkbarov I added some more background info to my answer. Sorry about that. Let me know if anything else isn't clear. | |
| Nov 12, 2016 at 12:40 | history | edited | Cameron Zwarich | CC BY-SA 3.0 |
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| Nov 12, 2016 at 10:33 | comment | added | Sergei Akbarov | Cameron, could you, please, specify the terms? The $\sigma$-strong* topology, the strong* topology, what are they exactly? | |
| Nov 12, 2016 at 10:16 | history | answered | Cameron Zwarich | CC BY-SA 3.0 |