Timeline for Reflexive norm-closed subalgebras of $B(X)$
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| S Jun 14, 2021 at 18:01 | history | bounty ended | CommunityBot | ||
| S Jun 14, 2021 at 18:01 | history | notice removed | CommunityBot | ||
| Jun 6, 2021 at 17:18 | history | edited | Onur Oktay | CC BY-SA 4.0 |
edited title
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| Jun 6, 2021 at 17:11 | comment | added | Onur Oktay | @YemonChoi Thank you for the notice, I edited as suggested. | |
| Jun 6, 2021 at 17:05 | history | edited | Onur Oktay | CC BY-SA 4.0 |
$A$ is a Banach (norm-closed) subalgebra of $B(X)$
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| Jun 6, 2021 at 17:01 | comment | added | Yemon Choi | Since this now has a bounty, can you please edit your question to clarify whether A is supposed to be a norm-closed subalgebra of B(X)? This ambiguity seems to have created some confusion in comments. | |
| S Jun 6, 2021 at 16:58 | history | bounty started | Onur Oktay | ||
| S Jun 6, 2021 at 16:58 | history | notice added | Onur Oktay | Draw attention | |
| May 8, 2021 at 11:35 | history | edited | Onur Oktay | CC BY-SA 4.0 |
added 246 characters in body
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| May 7, 2021 at 1:58 | comment | added | Onur Oktay | Attaching a unit to an algebra would create an ideal of codimension 1. Thus, even if $A$ has no finite codimensional maximal left ideals, $A$ with unit attached does not have the same property. | |
| May 7, 2021 at 1:54 | comment | added | Onur Oktay | @TomaszKania Any Banach algebra, which is a dual Banach space, with a separately weak* continuous multiplication can be embedded into $B(X)$ for some reflexive Banach space $X$ via a weak* continuous isometry [Corollary 3.8 in doi:10.4064/sm178-3-3]. When $A$ is reflexive itseft, then $A$ is trivially a subalgebra of $B(A)$. What Hilbert-Schmidt algebra misses is, indeed, the identity. | |
| May 6, 2021 at 19:14 | comment | added | Tomasz Kania | @NikWeaver, when you talk about Hilbert-Schmidt operators, it is not quite an example for 2 and 3 since most likely Onur wants a closed subalgebra of $B(X)$. | |
| May 6, 2021 at 18:37 | comment | added | Onur Oktay | @NikWeaver $\ell^p$ ($1<p<\infty$) with pointwise operations are reflexive Banach algebras. One may attach a unit without losing reflexivity. | |
| May 6, 2021 at 18:35 | comment | added | Onur Oktay | @NikWeaver Sorry for the mistake. I must have gotten mixed up with the link Tomasz Kania directed me to. $K(l^2)$ is amenable, has a bounded approximate identity, but no identity for sure. Could you open up what you meant by "first identify a Banach algebra with the three properties"? Will this lead to an existence or a non-existence? | |
| May 6, 2021 at 18:24 | comment | added | Nik Weaver | Are there even any infinite dimensional unital Banach algebras which are reflexive as Banach spaces? | |
| May 6, 2021 at 18:19 | comment | added | Nik Weaver | Mmm, I don't think $K(l^2)$ has a unit. One approach to constructing an example would be to first identify a Banach algebra with the three properties, then try to find a matricial norm structure extending the original norm which makes the product completely contractive. Then by general theory it will be (completely) isometrically isomorphic to a subalgebra of $B(H)$. | |
| May 6, 2021 at 16:44 | comment | added | Onur Oktay | @NikWeaver Professor Weaver, I am so glad to see your reply. There are relatively common examples when one drops one of the 3 conditions above. If A is a unital Banach algebra, which is also a Hilbert space, then it is finite dimensional (e.g., doi.org/10.1090/S0002-9904-1963-11035-6) For another example, the algebra $K(l^2)$ of compact operators is not reflexive, but has the other 2 properties. | |
| May 6, 2021 at 16:43 | comment | added | Onur Oktay | @TomaszKania Special thanks you for your reply, for I am aware of your level of knowledge and expertise in the field, as anybody would who's even remotely familiar with this field. My amateur attempts to produce reflexive operator algebras with the 3rd property from reflexive subspaces did not get me so far. | |
| May 6, 2021 at 15:36 | comment | added | Nik Weaver | I guess the algebra of Hilbert-Schmidt operators on $l^2$ satisfies 2 and 3, but of course it isn't unital. | |
| May 6, 2021 at 12:14 | comment | added | Tomasz Kania | Possibly relevant: mathoverflow.net/questions/299311/reflexive-operator-algebra | |
| May 6, 2021 at 11:20 | history | asked | Onur Oktay | CC BY-SA 4.0 |