Timeline for Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2013 at 15:51 | history | bounty ended | Sebastien Palcoux | ||
| Sep 27, 2013 at 8:09 | comment | added | Sebastien Palcoux | A generic way for tweaking a question, is to improve it by excluding the counter-examples. Here we can improve the definition of "see the ISP" by : $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T') $ and $T' \ne v_{1} T v^{*}_{1} + v_{2} T v^{*}_{2}$ (with $v_{1}$, $v_{2}$ as in your answer), then : "$T$ is an ISP counter-example" $\Leftrightarrow$ "$T'$ is an ISP counter-example". Is this what you thought? Else what do you suggest ? | |
| Sep 26, 2013 at 20:11 | comment | added | Leonel Robert | @SébastienPalcoux: I guess so. Though I'm not sure that the question cannot be made more interesting by tweaking it a little. | |
| Sep 26, 2013 at 13:09 | comment | added | Sebastien Palcoux | Thank you Leonel ! This map is continuous for the main topologies of operators algebras (norm-topology, weak-topology, strong-topology...), and this argument runs also without a $\star$-structure (on the algebra), so that it shows that no category of operator algebras see the ISP. Is it right ? | |
| Sep 26, 2013 at 12:56 | vote | accept | Sebastien Palcoux | ||
| Sep 26, 2013 at 4:27 | history | answered | Leonel Robert | CC BY-SA 3.0 |