Timeline for Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?
Current License: CC BY-SA 3.0
10 events
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| Mar 11, 2013 at 7:53 | vote | accept | archipelago | ||
| Mar 10, 2013 at 15:47 | comment | added | Todd Trimble | @archipelago: if you look at example 4.5 in the first paper cited in Benjamin's answer (page 14), you will see an explicit example. Particularly, let $H$ be a non-trivial complete group, and let $G$ be the product of $\aleph_1$ copies of $H$. The subgroup of $G$ consisting of tuples where at most countably many components are non-identity elements furnishes an explicit counterexamople. | |
| Mar 10, 2013 at 13:53 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Mar 10, 2013 at 13:49 | comment | added | Benjamin Steinberg | Not all groups are G_\delta complete. The paper later shows that such groups are not Dieudonne complete and esoteric examples can be found by googling. Locally compact groups and Polish groups are G_'\delta complete if I understood correctly. | |
| Mar 10, 2013 at 13:46 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Mar 10, 2013 at 13:36 | vote | accept | archipelago | ||
| Mar 10, 2013 at 13:36 | |||||
| Mar 8, 2013 at 11:44 | comment | added | Benjamin Steinberg | The OP asks 2 questions. One involves metric spaces and topological groups and the other Lie groups and manifolds. The first question is answered in the paper I linked. | |
| Mar 8, 2013 at 11:42 | comment | added | Benjamin Steinberg | By the topological category I meant topological groups not Lie groups. So as a topological group we just expect a metric space not a Riemannian manifold. | |
| Mar 8, 2013 at 6:36 | comment | added | YCor | By "completely answered in the topological category" means that a certain analogous question is answered. In the setting here, it shows that any Lie group (with an arbitrary number of components) is isometry group of a complete metric space (btw I don't really call this the "topological category"). But it does not say if it can be chosen as a Riemannian manifold. | |
| Mar 8, 2013 at 3:31 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |