Timeline for The isometry group of a product of two Riemannian manifolds
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 17, 2021 at 16:13 | history | edited | Ivan Solonenko | CC BY-SA 4.0 |
added 557 characters in body
|
| Jul 16, 2021 at 11:00 | comment | added | V. Rogov | (upd: of course $H_M \subset I(M)$ and $H_N \subset I(N)$ are not normal, unless $M$ itself is a Lie group | |
| Jul 15, 2021 at 14:40 | comment | added | V. Rogov | (Is it important that $M$ is of non-negative curvature and $N$ is flat? I'd suspect, the same must hold whenever $M$ and $N$ have curvature of different signs) | |
| Jul 15, 2021 at 14:39 | comment | added | V. Rogov | There is also the following perspective: assume that $M$ and $N$ are homogenous and denote by $H_M \trianglelefteq I(M)$ (resp. $H_N \trianglelefteq I(N)$) the isotropy group (a stabiliser of a point). Then $I(M) \times I(N)$ acts on $M\times N$ transitively and thus $I(M \times N) / I(M) \times I(N)$ maps to $H_{M\times N}$. In fact, it further projects to $H_{M \times N} / H_M \times H_N$, and I think this is an isomorphism. What you are actually proving is that if $M$ and $N$ have different curvature, $H_{M \times N} =H_N \times H_M$, right? | |
| Jul 15, 2021 at 11:46 | history | answered | Ivan Solonenko | CC BY-SA 4.0 |