Timeline for Finite codimensional subvector space of $C^{*}$ algebras which contains no invertible elements
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2015 at 13:47 | vote | accept | Ali Taghavi | ||
| Mar 3, 2015 at 18:41 | comment | added | Ali Taghavi | @RasmusBentmann please see the above comment. | |
| Mar 3, 2015 at 18:40 | comment | added | Ali Taghavi | @MannyReyes Please see the answer to this post (For $n(M_{n}(\mathbb{C}))$. | |
| Mar 2, 2015 at 10:56 | answer | added | Hannes Thiel | timeline score: 6 | |
| Feb 10, 2015 at 18:02 | comment | added | Ali Taghavi | @MannyReyes please see the above comment. | |
| Feb 10, 2015 at 18:02 | comment | added | Ali Taghavi | @RasmusBentmann I receive an interesting link in the comment to this post mathoverflow.net/questions/196182/… | |
| Feb 10, 2015 at 17:20 | comment | added | Ali Taghavi | @MannyReyes As I wrote above I realize that my argument is incorrect. Any way I think that if $n(M_{n}(\mathbb{C})$ does not go to infinity, the problem about $B(H)$ is not so obvious. | |
| Feb 10, 2015 at 17:14 | comment | added | Ali Taghavi | My argument was based on this incomplete idea:let $n-1$ be the maximum rank of matrices in our vector space. So we assume that we have a matrice in the form $I_{n-1}\oplus 0$.Now my error was that I imagined for all other matrices either the last columns or the last array must vanish. | |
| Feb 10, 2015 at 17:13 | comment | added | Rasmus | No, I don't have a counterexample. | |
| Feb 10, 2015 at 17:11 | comment | added | Ali Taghavi | @RasmusBentmann In fact $n(A)=1$ is necessary and sufficient condition for exsistence of a character.(As a consequence of Gleason Zelazko theoerm. Now I realize that may be my argument of $n(M_{n}(\mathbb{C})$ is not complete. Do you have a counter example? | |
| Feb 10, 2015 at 8:31 | comment | added | Rasmus | One small observation is that $n(A)=1$ if $A$ has a character. How did you get $n(M_n(\mathbb C))\geq n$? | |
| Feb 9, 2015 at 10:36 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 9, 2015 at 8:51 | comment | added | Ali Taghavi | So it would be interesting to find (and classify) some simple algebra with finite $n(A)$ which is certainly different from "1". | |
| Feb 9, 2015 at 8:43 | comment | added | Ali Taghavi | @MannyReyes Yes. As you said, $n(A)=1$ does not implies commutativity. But one can show that $n(A)=1$ can not occur for simple $C^{*}$ algebras. This a consequence of Han Banach theorem and Gleason-Kahane-Zelako theorem | |
| Feb 7, 2015 at 20:37 | comment | added | Manny Reyes | this is a good point about $B(H)$, as long as you truly know that $n(M_n(\mathbb{C}) = n$. (I see that it's $\leq n$, but not quite equality.) Regarding $n(A) = 1$, note that for any (not necessarily commutative) unital C*-algebra $B$ one has $n(B \times \mathbb{C}) = 1$. | |
| Feb 7, 2015 at 19:41 | comment | added | Ali Taghavi | So the next question would be: "what is the interpretation of finitness of $n(A)$ or $n(A)=1$. Does the later implies commutativity? Thanks again for your comments. | |
| Feb 7, 2015 at 19:39 | comment | added | Ali Taghavi | @MannyReyes according to your comment, I think the answer is negative for $B(H)$ since each $M_{n}(\mathbb{C}$ is embeded in $B(H)$. On the other hand this codimension goes to infinity as $n\to \infty$. Am I right? | |
| Feb 7, 2015 at 19:20 | comment | added | Ali Taghavi | @MannyReyes thank you for you comment and your revision. No, I do not know how to try this question for $B(H)$. There is no any determinant type function. | |
| Feb 7, 2015 at 19:08 | comment | added | Manny Reyes | Have you considered your question for $A = B(H)$, with $H$ an infinite-dimensional Hilbert space? | |
| Feb 7, 2015 at 19:04 | history | edited | Manny Reyes | CC BY-SA 3.0 |
Added three words to make the question match the title and to avoid a trivial answer (the zero subspace).
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| Feb 7, 2015 at 18:30 | history | edited | Ali Taghavi |
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| Feb 7, 2015 at 18:14 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 7, 2015 at 18:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 7, 2015 at 17:57 | history | asked | Ali Taghavi | CC BY-SA 3.0 |