Timeline for Do the banded operators check the invariant subspace problem?
Current License: CC BY-SA 3.0
39 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Aug 1, 2013 at 8:29 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I answer some question in remark.
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| Jul 31, 2013 at 15:13 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I update the remark.
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| Jul 31, 2013 at 12:47 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add links to MSE and MO posts.
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| Jul 31, 2013 at 12:41 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add a link to a MSE post.; Post Made Community Wiki
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| Jul 31, 2013 at 10:57 | comment | added | Sebastien Palcoux | @YemonChoi: I have edited some answers about nuclearness and exactness. | |
| Jul 31, 2013 at 10:54 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add a remark and remove some confusion.
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| Jul 30, 2013 at 11:19 | history | edited | Sebastien Palcoux |
I replace the tag "reference-request" by "c-star-algebras".
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| Jul 29, 2013 at 9:47 | comment | added | Sebastien Palcoux | @YemonChoi : just a remark about your comment <<will change drastically>> : if $T \in B(H)$ is a counter-example of the ISP, and $P$ invertible, then $W^{*}(T) = W^{*}(P^{-1}TP) = B(H)$. So it's invariant under conjugation by invertible (so not at all drastically different...). Now what about $C^{*}(T) $ and $ C^{*}(P^{-1}TP)$ ? always nuclear or exact ? | |
| Jul 29, 2013 at 9:26 | comment | added | Sebastien Palcoux | To clarify this, I posted the question: : Is there an operator algebraic reformulation of the invariant subspace problem ?. | |
| Jul 29, 2013 at 6:14 | review | Close votes | |||
| Jul 29, 2013 at 12:21 | |||||
| Jul 29, 2013 at 5:55 | comment | added | Yemon Choi | +1 Taka. I do not understand why SP thinks exact C-star algebras are relevant. Note that conjugating T with some arbitrary invertible will not affect ISP but will change drastically the C-star algebra generated by the and the VN algebra generated by T | |
| Jul 28, 2013 at 23:38 | comment | added | Narutaka OZAWA | @Sébastien: first, every C*-algebra has an irreducible representation, and exact C*-algebras with tracial states is a rather restricted class. Second, Voiculescu's counterexamples are for quasidiagonal operators which do not generate exact C*-algebras, and they are not relevant to ISP (in fact they are block diagonal operators). | |
| Jul 28, 2013 at 18:07 | comment | added | Sebastien Palcoux | @YemonChoi : "vérifie" seems a better translation. To avoid any confusion, when I say that an operator $T$ checks the ISP, this means that there is a closed, non-trivial, $T$-invariant subspace. | |
| Jul 28, 2013 at 16:37 | comment | added | Yemon Choi | "check" = verifie ou safisfait? | |
| Jul 28, 2013 at 15:17 | comment | added | Bill Johnson | @Sebastier: Sorry; NTHIS (hyperinvariant). | |
| Jul 28, 2013 at 15:11 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add a second remark, and a new link.
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| Jul 28, 2013 at 14:38 | comment | added | Sebastien Palcoux | Thank you @NarutakaOZAWA ! I have replaced "thick-operator" by "banded operator". I have find your book on which section 16.3 and 16.4 is about this question. I have written your answer on MSE, here. | |
| Jul 28, 2013 at 14:36 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I remove the last question.
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| Jul 28, 2013 at 10:34 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I replace "weight" by "weighted" and I add "orthonrmal".
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| Jul 28, 2013 at 10:13 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
A add a remark about Voiculescu counter-examples, and a question.
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| Jul 28, 2013 at 9:38 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I replace "thick-diagonal" by the standard name "banded operator", and I update the last remark.
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| Jul 28, 2013 at 8:30 | comment | added | Narutaka OZAWA | Here is the answer to the question in Remark. People often call such operators "banded operators." The C*-algebra on $\ell_2({\bf Z})$ generated by the banded operators (aka the uniform Roe algebra of $\bf Z$) is canonically isomorphic as $\ell_\infty({\bf Z})\rtimes {\bf Z}$, which is nuclear and has tracial states (coming from invariant means on $\bf Z$). So, any operator which is unitarily equivalent to a banded operator generates an exact C*-algebra having tracial states, which is not the case in general. | |
| Jul 28, 2013 at 7:41 | comment | added | Sebastien Palcoux | @BillJohnson : I don't understand, could you please enlighten me ? The reason of my misunderstanding is the following : a bilateral shift is an operator $T$ in $B(l^{2}(\mathbb{Z}))$ such that $Te_{n} = w_{n}e_{n+1}$, with $n \in \mathbb{Z}$, $(e_{n})_{n\in \mathbb{Z}}$ an orthonormal basis and $w_{n} \in \mathbb{C}$. Let $K_{n} = \langle e_{n}, e_{n+1}, e_{n+2}, ... \rangle $, then $\overline{K_{n}}$ is a closed, non-trivial, $T$-invariant subspace, right ? I trust you, so where is my mistake here ? | |
| Jul 27, 2013 at 22:03 | comment | added | Bill Johnson | It is not known whether all weighted BILATERAL shifts have NTIS, and all weighted bilateral shifts are thick-diagonal. | |
| Jul 27, 2013 at 21:19 | comment | added | Sebastien Palcoux | @JonBannon : thank you for the links containing the standard definition of "quasidiagonal" operators. As I write in my previous comments, the class of operators I defined differs, and I replace the name by thick-diagonal. These papers are about the hyperinvariant subspace problem (HISP). If (HISP) is true, then (ISP) is also true, but I don't know if the converse holds. | |
| Jul 27, 2013 at 21:07 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I improve the notation and add a remark.
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| Jul 27, 2013 at 20:29 | comment | added | Sebastien Palcoux | @YemonChoi: Now a thick-diagonal operator is not a curiosity, for two reason : firstly, it's a generalization of the weight shift operators, to be more precise, it's a finite sum of finite product of weight shift operators. However, a weight shift operator checks obviously the ISP. Secondly, I don't know yet whether or not every operators are unitary equivalent to thick-diagonal operators (i.e. thick-diagonalizable, see my MSE post). | |
| Jul 27, 2013 at 20:20 | comment | added | Sebastien Palcoux | @YemonChoi : the terminology "quasi-diagonalizable" is not standard, and not related to "quasidiagonal". The class of operators I defined is a thick generalization of the diagonal operators, in particular, it contains the shift. Now in section 5 of Jon Bannon's last link, a quasidiagonal operator is a sum of a block-diagonal and a compact operator, but the shift is not of this form. To avoid confusion, I will replace the name by "thick-diagonal". | |
| Jul 27, 2013 at 18:25 | comment | added | Jon Bannon | Regarding my last comment, the appropriate notion of "quasidiagonal" is discussed in section 5 of unf.edu/~shamid/Paper%202.pdf | |
| Jul 27, 2013 at 18:22 | comment | added | Jon Bannon | You are probably aware of this already, and I haven't checked anything at all here, but recently it has been established that the hyperinvariant subspace problem admits a reduction to a class of quasidiagonal operators: sciencedirect.com/science/article/pii/S0022123604001211 | |
| Jul 27, 2013 at 18:14 | comment | added | Yemon Choi | I think your question would be improved if you gave some justification, other than curiosity, for why you hope this restricted class of operators could suffice to test the ISP | |
| Jul 27, 2013 at 18:11 | comment | added | Yemon Choi | Is the terminology "quasi-diagonalizable" standard? and is it related to the standard terminology "quasidiagonal"? | |
| Jul 27, 2013 at 16:27 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I improve the post.
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| Jul 27, 2013 at 14:51 | comment | added | Koushik | this is a notoriously open question and your question is extremely difficult to answer | |
| Jul 27, 2013 at 12:52 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I replace "proper" by "non-trivial".
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| Jul 27, 2013 at 8:37 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add a tag and a remark.
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| Jul 27, 2013 at 8:19 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |