Timeline for Subgroups of free abelian groups are free: a topological proof?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2009 at 1:21 | comment | added | HJRW | I agree completely, Igor - it's clearly the geometry, rather than the topology, that's important. | |
| Nov 9, 2009 at 1:07 | comment | added | Tom Church | It is hard to be sure when talking about assumptions like this that are always unspoken, but I think you could prove Bieberbach without this. Thinking about the proof, the following statement seems attainable: if G is a lattice in Isom(R^n), then the translational part of G is finite index. This subgroup, of course, is a discrete subgroup of R^n, so by Tyler's argument above it's free abelian. This wouldn't yield another proof of the result, it would just say that Bieberbach's theorem is independent of it. | |
| Nov 8, 2009 at 22:58 | comment | added | Andy Putman | True. However, spaces of curvature 0 have fundamental groups that are virtually free abelian, which isn't quite good enough (spaces of curvature -infty, ie graphs, have fundamental groups that are free on the nose). Moreover, the proof that spaces of zero curvature have virtually free abelian fundamental groups (which is part of Bieberbach's theorem) is quite a bit deeper than the fact that graphs have free fundamental groups. I'd have to think about it a bit, but it wouldn't surprise me if the fact that subgroups of free abelian groups are free abelian is already used in the proof. | |
| Nov 8, 2009 at 19:32 | comment | added | Igor Belegradek | I think it is the geometry that plays a role, not the topology. Locally graphs are spaces of curvature minus infinity, while tori are spaces of curvature zero. Both curvature conditions are therefore inherited by the covering spaces. | |
| Nov 8, 2009 at 16:02 | history | answered | HJRW | CC BY-SA 2.5 |