Timeline for Infinite-dimensional classical Lie algebras
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2018 at 14:53 | comment | added | LSpice | @RobertFurber, interesting; I didn't know that about TVS theory. Sorry for my inapposite remark. | |
| Feb 26, 2018 at 0:14 | comment | added | Robert Furber | @LSpice In my previous comment I was only requiring the metric to define a vector space topology, not to be translation invariant (this is the usual convention in topological vector space theory). Your last parenthesized statement is not correct -- $F^\infty$ is actually complete in the unique uniformity defined by the direct sum topology (reference: Bourbaki's Topological Vector Spaces, III.21 Corollary 2). What my previous comment is saying, in this context, is that this uniformity is not metrizable. | |
| Feb 21, 2018 at 14:25 | comment | added | LSpice | @RobertFurber, reasoning about topologies on subspaces means that you want to use a translation-invariant metric, right? Then it seems to me that the notion of completeness depends only on the uniform structure, not the specific metric; and that the uniform structure in turn depends only on the topology and the group structure. (That is, "cannot be complete in any metric" really just means "isn't complete", I think.) | |
| Feb 21, 2018 at 10:23 | history | edited | Thomas Rot | CC BY-SA 3.0 |
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| Feb 19, 2018 at 7:40 | comment | added | Robert Furber | For what it's worth, $F^\infty$ cannot be complete in any metric, by the Baire category theorem. (It is the union of countably many finite-dimensional subspaces, which are closed sets of empty interior). | |
| Feb 18, 2018 at 20:24 | comment | added | Denis Nardin | That is, the crucial fact is that $\mathrm{hocolim}_n \mathrm{Isom}(V,\mathbb{R}^n)=*$ for all inner product spaces $V$. | |
| Feb 18, 2018 at 20:18 | comment | added | Denis Nardin | For what is worth, I think the reason why groups like $O(\infty)$ show up in homotopy theory is that $\mathbb{R}^\infty$ really should be thought of as an ind-space, not as a space (precisely, as an ind-object in the topological category of finite dimensional inner product spaces and isometric embeddings, and in fact the terminal object of the ind-category). | |
| Feb 18, 2018 at 19:03 | comment | added | Thomas Rot | @DenisNardin: That is a very helpful comment. I always assumed this to be true without checking, but it is obviously false. There is another group lurking around on $F^\infty$ which is the space of invertible operators which differ of the identity by a finite rank operator. Let me think about it and see if I can say something correct in the next couple of days. | |
| Feb 18, 2018 at 18:56 | history | edited | Thomas Rot | CC BY-SA 3.0 |
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| Feb 18, 2018 at 18:42 | comment | added | Denis Nardin | Actually there are invertible linear operators on $\mathbb{R}^\infty$ not in $O(∞)$ (think about permutation operators for a permutation of $\mathbb{N}$ with infinite support). Dunno about the homotopy type. | |
| Feb 18, 2018 at 14:37 | history | edited | Thomas Rot | CC BY-SA 3.0 |
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| Feb 18, 2018 at 14:32 | history | answered | Thomas Rot | CC BY-SA 3.0 |