Timeline for Subalgebras of $B(H)$ consisting of all operators leaving a given finite dimensional space invariant
Current License: CC BY-SA 4.0
12 events
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| Sep 13, 2021 at 13:15 | comment | added | Ali Taghavi | @NarutakaOZAWA Yes I can see the reason now. One define a morphism from $A$ to $L(V)$ by restriction. But the jacobson radical of L(V) is zero. its premimage is the ideal I mentioned. But what about K-groups? | |
| Sep 13, 2021 at 12:24 | comment | added | Ali Taghavi | @NarutakaOZAWA Thank you for your interesting comment. I guess that the maximal left ideal your are indicating to is the left ideal of all operatores vanishing at V. | |
| Sep 13, 2021 at 7:34 | comment | added | Ali Taghavi | @NarutakaOZAWA I was aware of this fact for commutative Banach algebra(May be a corollary ir an exercize in Rudin Book) but what is the statement for noncommutative case? BTW what can be said about k groups of this algebra) | |
| Sep 12, 2021 at 23:08 | comment | added | Narutaka OZAWA | Whenever nontrivial, your $A$ is not semi-simple and hence cannot be a C*-algebra. | |
| Sep 12, 2021 at 22:41 | comment | added | Ali Taghavi | @MarkSapir But if we allow H to be a finite dim space the answer to the first question is negative for ex" dim H=2 and dim V=1 . We get the three dimensional algebra isomorphic to upper triangle matrices. It can not become a C^* algebra. | |
| Sep 12, 2021 at 22:38 | comment | added | Ali Taghavi | @MarkSapir I do not care this case since the answer is obvious. | |
| Sep 12, 2021 at 22:38 | comment | added | user160032 | @MarkSapir Then $A= B(H)$ which is an obvious $C^*$-algebra? | |
| Sep 12, 2021 at 21:36 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 12, 2021 at 21:27 | history | edited | Ali Taghavi |
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| Sep 12, 2021 at 20:24 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 12, 2021 at 20:10 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 12, 2021 at 20:04 | history | asked | Ali Taghavi | CC BY-SA 4.0 |