Timeline for Simple $C^*$ algebras whose all commutator elements have scalar square
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 11, 2018 at 15:16 | vote | accept | Ali Taghavi | ||
| May 11, 2018 at 10:26 | comment | added | YCor | You're right, and one can thus conclude. I finally posted an answer based on the better-known von Neumann's bicommutant theorem. | |
| May 11, 2018 at 10:07 | answer | added | YCor | timeline score: 5 | |
| May 11, 2018 at 9:13 | comment | added | Ali Taghavi | @YCor But topological irreduciblity is equivalent to the algebraic irreducibiity. | |
| May 10, 2018 at 9:31 | comment | added | YCor | Yes I know, this is why my comment is rather a sort of expectation. Still, topological simplicity implies the existence of a (topologically) irreducible faithful Hilbert module. Then one would need a substitute for the Jacobson density theorem, it does not sound hopeless to me. | |
| May 10, 2018 at 9:13 | comment | added | Ali Taghavi | @YCor I learn a lot from your great comment. however the topological simplicity is not equivalent to the algebraic one. I know a non unital example: The algebra of compact operators is a simple $C^*$ algebra but it has dense ideal of finite rank operator or trace class operatores or Hilbert Schmidt operators. | |
| May 9, 2018 at 10:17 | comment | added | YCor | Such an algebra satisfies the identity $[[X,Y]^2,Z]$ and hence is a PI-algebra (Polynomial Identity). Simplicity probably implies that it's primitive (a little doubt, because of the topological vs purely algebraic setup). In encyclopediaofmath.org/index.php/PI-algebra it is said that a primitive PI-algebra is a matrix algebra over a skew-field, which would entail a negative answer to your question. | |
| May 9, 2018 at 9:24 | history | edited | YCor |
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| May 9, 2018 at 9:03 | comment | added | Ali Taghavi | @Vincent Yes I mean complex $C^*$ algebra. In particular I wonder whether there is an infinite dimensional one? | |
| May 9, 2018 at 8:44 | comment | added | Vincent | Well, over $\mathbb{R}$ the quaternions qualify, but if you complexify them you'll get $M_2(\mathbb{C})$ which you already ruled out. | |
| May 9, 2018 at 8:28 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 5 characters in body
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| May 9, 2018 at 8:07 | history | asked | Ali Taghavi | CC BY-SA 4.0 |