Timeline for Centralizers in C*-algebra
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 9, 2012 at 9:51 | answer | added | Martin Argerami | timeline score: 6 | |
| Jan 6, 2012 at 22:14 | comment | added | Jesse Peterson | mathoverflow.net/questions/85092/… | |
| Jan 6, 2012 at 21:28 | comment | added | Jesse Peterson | @Spelas: Since you are interested in properties for general $C^*$-algebras I'll ask the von Neumann algebra question in a separate post. | |
| Jan 6, 2012 at 20:25 | comment | added | spelas | I am interested in the situation of general $C^*$-algebras, and some "nontrivial" correspondence between $a$ and $b$. Since the correspondence in $B(H)$ is quite "strong", I expected that something nonobvious can be said also in general $C^*$-algebras. | |
| Jan 5, 2012 at 18:29 | history | edited | Jesse Peterson |
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| Jan 5, 2012 at 18:27 | comment | added | Jesse Peterson | @Yemon: This doesn't work for all von Neumann algebras. For instance if $\mathbb F_2 = \langle a, b \rangle$ is the free group, $A = \{ a \}''$, $B = \{ a^2 \}''$. Then it is known that $A' \cap L\mathbb F_2 = B' \cap L\mathbb F_2$. In fact, my first guess would be that this property actually characterizes $B(H)$. | |
| Jan 5, 2012 at 18:24 | comment | added | Jesse Peterson | @Spelas: Since you want to apply Borel functional calculus this is really a question about von Neumann algebras rather than $C^*$-algebras. It might be more insightful to edit the question to ask which von Neumann algebras $N$ have the property that for all commutative von Neumann subalgebras $A, B \subset N$ we have $A' \cap N = B' \cap N \implies A = B$. | |
| Jan 5, 2012 at 18:10 | comment | added | Yemon Choi | @spelas: if that is the particular question you are interested in, why don't you edit the question? Moreover, if you have a particular class of noncommutative $C^\ast$ algebras in mind, you should specify which ones you mean. (Note that if $A$ is a von Neumann algebra then everything works as in $B(H)$.) | |
| Jan 5, 2012 at 9:26 | comment | added | spelas | Since the condition trivially holds if A is commutative, we exclude that case. In the noncommutative case, does it hold the same correspondence as in $B(H)$? | |
| Jan 4, 2012 at 16:31 | comment | added | Jesse Peterson | What type of general statement are you looking for? Have you considered the case when $A$ is commutative? | |
| Jan 4, 2012 at 12:35 | comment | added | spelas | Yes, I am assuming that a and b are self-adjoint. Sorry for the mistake and thank you for pointing it out. | |
| Jan 4, 2012 at 12:33 | history | edited | spelas | CC BY-SA 3.0 |
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| Jan 4, 2012 at 6:36 | comment | added | Yemon Choi | Certainly your condition implies that $a$ lies in the double commutant of $b$, and that $b$ lies in the double commutant of $a$, but off the top of my head I don't see how to say more without using some assumptions such as normality. | |
| Jan 4, 2012 at 6:32 | comment | added | Yemon Choi | Regarding your last comment: are you assuming $a$ is normal? (Otherwise take $a$ to be quasinilpotent.) | |
| Jan 4, 2012 at 6:24 | history | edited | Alain Valette | CC BY-SA 3.0 |
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| Jan 4, 2012 at 6:20 | history | asked | spelas | CC BY-SA 3.0 |