Timeline for Is this a submanifold?
Current License: CC BY-SA 4.0
10 events
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| Mar 9, 2019 at 23:15 | comment | added | L.F. Cavenaghi | @AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem. | |
| Mar 9, 2019 at 22:29 | comment | added | Ali Taghavi | You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case. | |
| Mar 9, 2019 at 22:18 | comment | added | L.F. Cavenaghi | @AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has? | |
| Mar 9, 2019 at 21:45 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 9, 2019 at 21:44 | comment | added | Ali Taghavi | @ArunDebray yes thank you but $\tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction! | |
| Mar 9, 2019 at 21:42 | comment | added | Arun Debray | "every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)\in\tilde S$ is homeomorphic to $\mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold. | |
| Mar 9, 2019 at 21:21 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 9, 2019 at 21:07 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 9, 2019 at 20:54 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 9, 2019 at 20:36 | history | answered | Ali Taghavi | CC BY-SA 4.0 |