Timeline for ${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?
Current License: CC BY-SA 3.0
24 events
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| Aug 3, 2014 at 5:37 | comment | added | Sebastien Palcoux | @NikWeaver: I've posted it as an answer (perhaps a better place than in the original post). | |
| Aug 3, 2014 at 2:56 | comment | added | Nik Weaver | @SébastienPalcoux: if you've answered the infinite dimensional question, perhaps you could say so in an edit to your original post, to give the answer some visibility. | |
| Aug 2, 2014 at 21:14 | comment | added | Sebastien Palcoux | Infinite tensor product of states and your answer. | |
| Aug 2, 2014 at 17:42 | comment | added | Sebastien Palcoux | It seems there are counterexamples for the generalized question by using $(\mathcal{A} \otimes I_2)^{\otimes \infty}$ and $(\mathcal{B}\otimes I_2)^{\otimes \infty}$. So we need to generalize more the question by replacing "finite dimensional" by "hyperfinite". | |
| Aug 1, 2014 at 19:28 | comment | added | Nik Weaver | Honestly? I think you should spend a couple of days working on it yourself before opening a new post. Otherwise it feels like you are abusing the community's goodwill to have people do your thinking for you. Just my opinion. | |
| Aug 1, 2014 at 19:14 | comment | added | Sebastien Palcoux | @NikWeaver: if you've no objection, I will open a new post for this generalized question. | |
| Aug 1, 2014 at 18:56 | comment | added | Nik Weaver | Oh you're right, I take it back. | |
| Aug 1, 2014 at 18:53 | comment | added | Sebastien Palcoux | @NikWeaver: yes but this nontrivial intersection could be also finite dimensional, else $\mathcal{M}$ and $\mathcal{N}$ themselves answer the generalized question, but do they exist? | |
| Aug 1, 2014 at 18:38 | comment | added | Nik Weaver | Sorry, I pressed enter before I was finished ... I was saying, now replace with $\mathcal{A}\otimes\mathcal{M}$ and $\mathcal{B}\otimes\mathcal{N}$ where $\mathcal{M}$ and $\mathcal{N}$ are $II_1$ factors with trivial intersection whose commutants have nontrivial intersection. | |
| Aug 1, 2014 at 18:36 | comment | added | Sebastien Palcoux | @NikWeaver: yes precisely, for having an infinite dimensional intersection by modifying your example, we need to go to $\mathcal{A} \otimes I_{\infty}$ which have commutant $\mathcal{A}' \otimes B(H)$, which is not ${\rm II}_1$ but ${\rm II}_{\infty}$. | |
| Aug 1, 2014 at 18:30 | comment | added | Nik Weaver | @SébastienPalcoux: I think the same idea works though. In my example we go to $\mathcal{A} \otimes I_2$ and $\mathcal{B}\otimes I_2$ which have commutants $\mathcal{A}' \otimes M_2$ and $\mathcal{B}$ | |
| Aug 1, 2014 at 18:27 | comment | added | Sebastien Palcoux | @NikWeaver: Is it also trivially false if we enlarge the question, replacing "$\mathcal{A}' \cap \mathcal{B}' = \mathbb{C}$" by "$\mathcal{A}' \cap \mathcal{B}'$ finite dimensional" ? In your example, $\mathcal{A^{(2)}}' \cap \mathcal{B^{(2)}}' \simeq M_2(\mathbb{C})$ which is finite dimensional. | |
| Aug 1, 2014 at 18:23 | comment | added | Sebastien Palcoux | @YemonChoi: OK thank you, the next time, I will ask. Now, $M_2(\mathcal{A}') \cap M_2(\mathcal{B}') \neq \mathbb{C}$, i.e. they "don't intersect" in the scalar. | |
| Aug 1, 2014 at 18:19 | comment | added | Yemon Choi | @SébastienPalcoux moreover, I feel it is more courteous to ASK someone who answers your question if they have made minor errors or typos. | |
| Aug 1, 2014 at 18:17 | comment | added | Yemon Choi | @SébastienPalcoux you changed "now intersect" to "don't intersect". That is not fixing a typo. | |
| Aug 1, 2014 at 18:09 | history | edited | Nik Weaver | CC BY-SA 3.0 |
added 129 characters in body
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| Aug 1, 2014 at 18:01 | comment | added | Sebastien Palcoux | @YemonChoi: Is it fashion to keep an answer with obvious typos? | |
| Aug 1, 2014 at 17:56 | comment | added | Yemon Choi | However I agree with part of Sebastien's edit, namely that you probably mean M2 of A', etc | |
| Aug 1, 2014 at 17:55 | comment | added | Nik Weaver | hey, I got the check mark, that's all that matters | |
| Aug 1, 2014 at 17:54 | comment | added | Yemon Choi | I rolled back Sebastien's edit, which seemed to change completely the meaning of what Nik Weaver wrote | |
| Aug 1, 2014 at 17:53 | history | rollback | Yemon Choi |
Rollback to Revision 1
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| Aug 1, 2014 at 17:50 | vote | accept | Sebastien Palcoux | ||
| Aug 1, 2014 at 17:49 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
typos
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| Aug 1, 2014 at 17:17 | history | answered | Nik Weaver | CC BY-SA 3.0 |