Timeline for Is there an operator algebraic reformulation of the invariant subspace problem?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2020 at 18:52 | comment | added | LSpice | Radjavi and Rosenthal - Invariant subspaces. | |
| Oct 30, 2013 at 10:01 | vote | accept | Sebastien Palcoux | ||
| Jul 29, 2013 at 15:47 | comment | added | Sebastien Palcoux | So if $T$ is a counter-example of the ISP, $\overline{\mathbb{C}[T]}^{wot}$ is a reductive and non-selfadjoint algebra. For the converse, if $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ has no non-trivial invariant subspace, then it's reductive and also non-selfadjoint algebra (because if $\mathcal{A}$ selfadjoint, then it's irreducible and so $\mathcal{A} = B(H)$, contradiction with the commutativity of $\mathcal{A}$). So the question is : does $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ has no non-trivial invariant subspace, implies $T$ a counter-example of the ISP ? | |
| Jul 29, 2013 at 15:21 | comment | added | Sebastien Palcoux | In particular, if $T$ is a counter-example of the ISP, then $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ is reductive because its invariant subspace are $\{0 \}$ and $H$ (obviously reducing the algebra). So if the RAP is true, $\mathcal{A}$ is self-adjoint, in particular, $T^{*} \in \mathcal{A}$, but $\mathcal{A} \subset \{ T\}'$ and $T$ is non-normal, so contradiction, and ISP is true. | |
| Jul 29, 2013 at 15:15 | comment | added | Sebastien Palcoux | Thank you @MikeJury. A counter-example $T \in B(H) $ of the ISP is of course non-normal. If $R\in B(H)$ is normal, there is a theorem due to Sarason : $\overline{\mathbb{C}[R]}^{wot}$ is a $∗$-algebra if and only if $R$ is reductive, i.e. every closed invariant subspace $K \subset H$ of $R$, is a reducing subspace. If I'm not mistaken, the RAP generalizes this theorem to every operator. | |
| Jul 29, 2013 at 13:55 | history | answered | Mike Jury | CC BY-SA 3.0 |