Timeline for Asking whether there is a compact Lie group containing affine symplectic group
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2021 at 13:34 | comment | added | Denis Nardin | @AndreiSmolensky Oh yeah, sorry I was automatically making my groups locally compact (although note that usually $\mathbb{Q}/\mathbb{Z}$ is given the discrete topology, not the subspace one - exactly because we want it to be locally compact!) | |
| Nov 10, 2021 at 12:04 | comment | added | Andrei Smolensky | @DenisNardin Isn't $\mathbb{Q}/\mathbb{Z}\to S^1$ an embedding of a non-compact group into a compact one? I do not see a locally compact example, though. | |
| Nov 10, 2021 at 11:18 | history | became hot network question | |||
| Nov 10, 2021 at 11:00 | comment | added | YCor | It can't contain the symplectic group over a nonzero symplectic space. An easy reason is that in a compact Lie group, any connected abelian subgroup has the property that its centralizer has finite index in its normalizer. | |
| Nov 10, 2021 at 10:57 | history | edited | YCor | CC BY-SA 4.0 |
formatting
|
| Nov 10, 2021 at 10:49 | answer | added | Robert Bryant | timeline score: 21 | |
| Nov 10, 2021 at 8:44 | comment | added | Denis Nardin | A small note: while there is a continuous injective group homomorphism $\mathbb{R}\to T^2$, it does not "really" embeds it as a topological group, in that the topology on $\mathbb{R}$ is not the one induced by the embedding. Indeed I don't think one can embed a non-compact group into a compact group as a topological subgroup. | |
| Nov 10, 2021 at 3:17 | history | edited | En-Jui Kuo | CC BY-SA 4.0 |
added 15 characters in body
|
| S Nov 10, 2021 at 1:18 | review | First questions | |||
| Nov 10, 2021 at 7:20 | |||||
| S Nov 10, 2021 at 1:18 | history | asked | En-Jui Kuo | CC BY-SA 4.0 |