Timeline for Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 11, 2015 at 23:39 | comment | added | Will Sawin | @Igor Rivin that it is a manifold is precisely the thing I don't really know how to check. | |
| Aug 11, 2015 at 22:40 | comment | added | Igor Rivin | Look at my latest edit... By the way, the first thing that needs to be checked is that your construction actually gives a manifold (this is not an issue in low dimensions), which means checking that the link of a singular point is always an actual (as opposed to a rational homology) sphere. | |
| Aug 11, 2015 at 20:28 | comment | added | Igor Rivin | Fair enough. How do you know that once you go to the fiber, this problem goes away? | |
| Aug 11, 2015 at 20:14 | comment | added | Will Sawin | @IgorRivin For instance because $\mathbb Z^n \rtimes S_n$ is not generated by elements with fixed points, because there is a homomorphism to $\mathbb Z$ where you sum up the $n$ elements of $S^n$ and everything with a fixed point is in the kernel. | |
| Aug 11, 2015 at 20:11 | comment | added | Igor Rivin | @WillSawin why does it have a nontrivial fundamental group? | |
| Aug 11, 2015 at 19:51 | comment | added | Will Sawin | @Igor Rivin otherwise the quotient has nontrivial fundamental group and thus is not a simplex. | |
| Aug 11, 2015 at 18:59 | comment | added | Igor Rivin | Why do you want to go to the fiber? | |
| Aug 11, 2015 at 10:02 | comment | added | arivero | Thanks, it is more clear now :-) At least it could work for S^3; I will try to visualize the general case. | |
| Aug 11, 2015 at 4:00 | history | answered | Will Sawin | CC BY-SA 3.0 |