Timeline for Spaces that are finitely covered by manifolds
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2017 at 6:27 | vote | accept | Jens Reinhold | ||
| Jan 31, 2017 at 6:10 | comment | added | მამუკა ჯიბლაძე | Yes, thanks. In other words this is also the intersection of all conjugate subgroups, of which there are finitely many. Probably this is actually largest normal subgroup contained in the subgroup. | |
| Jan 31, 2017 at 6:02 | comment | added | Ian Agol | @მამუკაჯიბლაძე: This is true more generally for a finite-index subgroup of a group. The group acts on cosets of a finite index subgroup as permutations. The kernel of this permutation action is a normal subgroup contained in the finite-index subgroup. | |
| Jan 31, 2017 at 6:02 | comment | added | Jens Reinhold | Great, that makes sense! I feel enlightened - thanks. | |
| Jan 31, 2017 at 6:01 | comment | added | Ian Agol | @JensReinhold: The point is that since $Y$ (and X) are finite complexes, their fundamental groups are torsion-free. So the automorphisms of $\tilde{M}$ I described cannot have fixed points, hence $N$ is a manifold, not an orbifold. There are some details here, but I think that's where finiteness of Y is used. | |
| Jan 31, 2017 at 5:57 | comment | added | Jens Reinhold | @Ian Agol: Thanks, that helped. But I still do not see where this uses that $Y$ is a finite complex. At some point this has to enter, as described in my question: Namely we have that $M \times BG$, for $G$ any finite group, is finitely covered by $M \times EG \simeq M$, but it is NOT homotopy equivalent to a manifold (not even homotopy equivalent to a finite complex). | |
| Jan 31, 2017 at 5:57 | comment | added | მამუკა ჯიბლაძე | Does this imply that any finite index subgroup of a hyperbolic group contains a finite index normal subgroup? If yes, is there some canonically defined one? | |
| Jan 31, 2017 at 5:51 | comment | added | Ian Agol | @JensReinhold: The point is that any homotopy equivalence of a hyperbolic manifold is homotopy equivalent to an isometry. If $M$ is hyperbolic, the finite-sheeted cover $X\to Y$ might not be regular, but we can pass to a finite cover so that $\tilde{X}\to Y$ is regular, and still homotopy equivalent to a hyperbolic manifold $\tilde{M}$. Then the covering translations of $\tilde{X}\to Y$ give homotopy equivalences of $\tilde{M}$, which in turn is realized by a finite group of isometries. The quotient is the desired manifold $N$. | |
| Jan 31, 2017 at 5:47 | history | edited | Ian Agol | CC BY-SA 3.0 |
added 660 characters in body
|
| Jan 31, 2017 at 5:45 | comment | added | Jens Reinhold | I am aware of Mostow's theorem which says that every homotopy equivalence of a hyperbolic manifold is homotopic to an isometry. But how exactly does that imply what I want? The analogue of Mostow's result for $S^1$ is obvious, but nevertheless I do not see how this implies that every finite complex that is finitely covered by some $X \simeq S^1$ has to be itself homotopy equivalent to $S^1$. | |
| Jan 31, 2017 at 5:23 | vote | accept | Jens Reinhold | ||
| Jan 31, 2017 at 5:23 | |||||
| Jan 31, 2017 at 4:21 | history | answered | Ian Agol | CC BY-SA 3.0 |