Timeline for Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2013 at 14:53 | comment | added | Todd Trimble | Why is this the answer that was accepted? Not that this doesn't have good information, but it's speculative and restricted to the Lie group case. By contrast, Benjamin gave a solid reference for a complete characterization for the topological case. | |
| Mar 10, 2013 at 13:36 | vote | accept | archipelago | ||
| Mar 11, 2013 at 7:53 | |||||
| Mar 10, 2013 at 13:36 | vote | accept | archipelago | ||
| Mar 10, 2013 at 13:36 | |||||
| Mar 8, 2013 at 1:03 | comment | added | Ryan Budney | I think there are perhaps better ways to introduce handedness into the manifold than the "global" way I did above. I imagine you could put a little bit of curvature in the fibre-cross-base directions in a handed way. That would be more natural and generic. | |
| Mar 8, 2013 at 0:34 | comment | added | Ryan Budney | I was trying to introduce a "handedness" to the bundle. The fiber having symmetry group $\mathbb Z_2$ was too simple. | |
| Mar 8, 2013 at 0:32 | history | edited | Ryan Budney | CC BY-SA 3.0 |
fiber symmetry group was too small
|
| Mar 7, 2013 at 23:48 | comment | added | André Henriques | Sorry, I fail to understand how your "thickened moebius band" construction cuts down the isometry group from $O(2)$ to $SO(2)$. | |
| Mar 7, 2013 at 23:13 | comment | added | Ryan Budney | I think that's manageable. I haven't given it as much thought as I should have but I've edited in a sketch of how to address your concern. | |
| Mar 7, 2013 at 23:12 | history | edited | Ryan Budney | CC BY-SA 3.0 |
added 747 characters in body
|
| Mar 7, 2013 at 22:30 | comment | added | André Henriques | Ryan: can you make the last step of your construction explicit in the case $G=S^1$? | |
| Mar 7, 2013 at 20:46 | comment | added | YCor | Any connected Lie group admits a left-invariant Riemannian metric (compact allows to find a bi-invariant one but this is probably useless here). | |
| Mar 7, 2013 at 19:44 | comment | added | duetosymmetry | Does this only apply to Lie groups with semisimple Lie algebra? <=> the Killing form is nondegenerate. Otherwise the manifold is not Riemannian, correct? | |
| Mar 7, 2013 at 19:31 | history | answered | Ryan Budney | CC BY-SA 3.0 |