Timeline for Bi-annihilator of a subspace of the dual of an infinite-dimensional vector space
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2020 at 1:30 | comment | added | Xiaosong Peng | Maybe you can see Proposition 1.3.5 of the book "Hopf algebras" by David E. Radford | |
| Aug 16, 2020 at 21:58 | history | became hot network question | |||
| Aug 16, 2020 at 15:02 | comment | added | lefuneste | Dear LSpice, thank you for your explanations. I know nothing about topological vector spaces but I appreciate your point that if I did I would have immediately realized that the question is quite easy, whereas in reality I spent much time coming up with a solution. | |
| Aug 16, 2020 at 15:00 | comment | added | LSpice | For example, your surprising true fact is the statement that all subspace are closed in the weak topology (with respect to the full algebraic dual). | |
| Aug 16, 2020 at 14:54 | comment | added | LSpice | I think @user131781's point is less that the problem can't be stated, or even answered, without topology, and more that it is probably most useful to think of the problem in topological terms even if one is not required to do so (for example, this reveals 'heuristically' that the answer must be no, even if one is left to, as you and I did, construct a specific non-closed subspace). | |
| Aug 16, 2020 at 14:29 | comment | added | lefuneste | @user131781 Of course we can get by without topology: already two answers show this, just 17 minutes after the question was posted! | |
| Aug 16, 2020 at 14:26 | answer | added | lefuneste | timeline score: 3 | |
| Aug 16, 2020 at 14:12 | answer | added | LSpice | timeline score: 6 | |
| Aug 16, 2020 at 14:12 | comment | added | user131781 | In the infinite dimensional case, you can’t get by without topology, in this case the weak topologies $\sigma(V,V^\ast)$ and $\sigma(V^\ast,V)$. It is well known that the bipolars of subspaces of either of your spaces are precisely their closures for the corresponding topologies. So if they are not closed, then your claims will fail. | |
| Aug 16, 2020 at 14:00 | comment | added | LSpice | A maybe-more accessible question that keeps it all on the dual side: since $((\Gamma^\diamond)^\circ)^\diamond$ equals $\Gamma^\diamond$, we can ask: if $\Gamma$ is contained in $\Lambda$ and $\Gamma^\diamond$ equals $\Lambda^\diamond$, then under what circumstances does $\Gamma$ equal $\Lambda$? | |
| Aug 16, 2020 at 13:52 | history | asked | lefuneste | CC BY-SA 4.0 |