Timeline for geometry and connected sum of aspherical closed manifolds
Current License: CC BY-SA 4.0
6 events
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| Feb 17, 2022 at 23:18 | comment | added | Vitali Kapovitch | sorry, I should have used a different letter for the subgroup in the second part of the above comment. So, $G$ is a connected Lie group and $H$ is a compact subgroup (not necessarily maximal compact $K$). Then still $G/H$ has at most two ends by the argument I gave. | |
| Feb 17, 2022 at 23:05 | comment | added | Vitali Kapovitch | @OtisChodosh Sorry, I didn't think it through very carefully but what I had in mind was this. A connected lie group $G$ topologically is $\mathbb R^n\times K$ where $K$ is maximal compact subgroup. This has 1 end if $n>1$ and two ends if $n=1$. If $K$ is a compact subgroup of $G$ then $G/K$ will have no more than 2 ends. I have not thought through the case if $K$ is not compact (it should be true too) but it does not arise for Riemannian homogeneous spaces which I think is what the OP is interested in. | |
| Feb 17, 2022 at 22:41 | comment | added | Otis Chodosh | @VitaliKapovitch interesting, thanks. Can you explain why a homogeneous space has at most two ends (I assume this is a very basic question)? | |
| Feb 17, 2022 at 22:39 | comment | added | Vitali Kapovitch | @OtisChodosh that link does answer the question. The fundamental group of $M_1\#M_2$ is a free product and has infinitely many ends. Hence so does the universal cover of $M_1\#M_2$. On the other hand the universal cover of a locally homogeneous space is homogeneous and hence has at most two ends. | |
| Feb 17, 2022 at 22:11 | comment | added | Otis Chodosh | I am not sure if this answers your question, but mathoverflow.net/questions/46874/… might be relevant. | |
| Feb 17, 2022 at 22:08 | history | asked | Ian Gershon Teixeira | CC BY-SA 4.0 |