Timeline for Is every nontrivial idempotent in the Cuntz algebra, a commutator element?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 23, 2018 at 23:55 | comment | added | long trail | Oops! The statement that if $p \oplus 0_{k-1}$ and $q \oplus 0_{k-1}$ are M-vN equivalent in $M_k(A)$, then $p$ and $q$ are M-vN equivalent in $A$ is True. I forgot that if $[p]_0=[q]_0$, then there exists $k$ such that $p \oplus 1_{k-1}$ and $q \oplus 1_{k-1}$ are M-vN equivalent in $M_k(A)$. This implies that $p \oplus 1_{k-1} \oplus 0_k$ and $q \oplus 1_{k-1} \oplus 0_k$ are unitarily equivalent in $M_{2k}(A)$. (All this can be found in Sec 2.2 and 3.1 of Rordam, Larsen, and Laustsen.) I believe the algebra still works to show that $q$ a lin comb of commutators implies $p$ the same. | |
| May 17, 2018 at 20:18 | comment | added | Ali Taghavi | please see my comment to the following answer in MSE. I think that there are some contradictory situation. what is my mistake math.stackexchange.com/questions/2785117/… | |
| May 17, 2018 at 17:46 | comment | added | long trail | It is possible, that's one of the details about the semigroup $V(A)$ of projections that can't be overlooked. However, you can check (the details are a little long in notation, but mechanical) that if two projections $p,q $ in a $C^*$-algebra are stably similar, in the sense that $P = p \oplus 0_{k-1}$ and $Q = q \oplus 0_{k-1}$ are conjugates $P = UQU^*$ via some unitary $U$ in $M_k(A)$, then $p$ is a linear combination of commutator elements if and only if $q$ is such an element. It's a little lengthy to check but I'm fairly certain it's just a matrix multiplication problem. | |
| May 17, 2018 at 15:39 | vote | accept | Ali Taghavi | ||
| May 17, 2018 at 4:59 | comment | added | Ali Taghavi | Thank you. It is not a comment but is a very helpful answer. Just a question is not possible that two projections $e,f$ are not Murray Von Neuman equivalent but $e\oplus 0\simeq f\oplus 0$? | |
| May 16, 2018 at 14:55 | history | edited | long trail | CC BY-SA 4.0 |
commutator element --> linear combination of commutator elements in first paragraph
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| May 16, 2018 at 14:25 | history | edited | long trail | CC BY-SA 4.0 |
(edited commutator bracket to get the order right)
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| May 16, 2018 at 14:19 | history | answered | long trail | CC BY-SA 4.0 |