Timeline for Commutative algebras whose bidual is commutative
Current License: CC BY-SA 3.0
13 events
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| Jun 9, 2014 at 22:16 | comment | added | ACL | Had you had a look at the paper On the second conjugate space of a Banach algebra as an algebra, projecteuclid.org/euclid.pjm/1103037121, by Paul Civin and Bertram Yood. They seem to give a fairly detailed study of the question. In particular, they give conditions under which the analogue of this algebra (in the Banach category) is not commutative. (It is almost never.) | |
| Jan 30, 2014 at 19:17 | comment | added | YCor | I think that if one works without AC, it would be completely insane to denote by $\prod k$ a product of spaces each being isomorphic to $k$ (with no prescribed isomorphisms). | |
| Jan 30, 2014 at 17:39 | history | edited | Gro-Tsen | CC BY-SA 3.0 |
note the terminology "Arens multiplication"
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| Jan 30, 2014 at 17:39 | comment | added | Jeremy Rickard | @YvesCornulier: I think that Johannes' point was that I wrote about a direct sum of one-dimensional spaces, without specifying that they came with simultaneous choices of isomorphisms with $k$. In that case, it's possible without choice that the quotient of the direct product by the direct sum (not the direct product itself) is zero, which would make my argument fail. But it's OK if I use a family of one-dimensional spaces furnished with isomorphisms with $k$. | |
| Jan 30, 2014 at 15:43 | comment | added | YCor | @Johannes: I don't understand: there are plenty of explicit nonzero elements in $\prod_{i\in I}k$ for any infinite set $I$, e.g., there is the diagonal embedding of $k$; there is the inclusion of $\bigoplus k$ which also contains plenty of explicit elements. | |
| Jan 30, 2014 at 15:25 | history | edited | Gro-Tsen | CC BY-SA 3.0 |
answer if k noetherian integral domain and A finite over k
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| Jan 30, 2014 at 9:01 | comment | added | Jeremy Rickard | Sorry, I mean not both $V$ and $U/V$ can be reflexive. | |
| Jan 30, 2014 at 8:48 | comment | added | Jeremy Rickard | @JohannesHahn: OK, I meant each copy of $k$ to have a specified generator. In particular, if $U$ is the space of sequences of elements of the ground field, and $V$ the subspace of finite sequences, then not both $U$ and $U/V$ can be reflexive. Unless I'm being stupid? | |
| Jan 29, 2014 at 20:21 | comment | added | Johannes Hahn | @JeremyRickard Without some form of AC, $\prod k$ might be zero itself. | |
| Jan 29, 2014 at 19:05 | comment | added | Jeremy Rickard | @JohannesHahn: I don't think it's possible for every vector space to be reflexive. If $V=\bigoplus k$ is a countable direct sum of one-dimensional spaces, and $V$ is reflexive, then the dual of $W=\prod k/\bigoplus k$ is zero, so $W$ is not reflexive. I'm not sure this invalidates your point, though. | |
| Jan 29, 2014 at 18:15 | answer | added | Qiaochu Yuan | timeline score: 3 | |
| Jan 29, 2014 at 16:30 | comment | added | Johannes Hahn | If I remember correctly, without AC it is possible that every vector space is reflexive that is $V\cong DD(V)$ via the canonical homomorphism. A choice-free counterexample would therefore have to be other a ring that is not a ring (and might not exist at all for all I know ...) | |
| Jan 29, 2014 at 16:15 | history | asked | Gro-Tsen | CC BY-SA 3.0 |