Timeline for Symmetries of contractable subsets of $\Bbb R^n$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 14, 2022 at 19:12 | comment | added | Nick S | @M.Winter As I was saying, stupid mistake :) | |
| May 14, 2022 at 19:11 | comment | added | M. Winter | @NickS The symmetries of the half-sphere fix an axis. | |
| May 14, 2022 at 19:09 | comment | added | Nick S | I'm probably making a stupid mistake, but isn't half sphere in $\mathbb R^3$ such an example? | |
| May 14, 2022 at 17:41 | vote | accept | M. Winter | ||
| May 14, 2022 at 17:41 | history | edited | M. Winter | CC BY-SA 4.0 |
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| May 14, 2022 at 17:41 | answer | added | M. Winter | timeline score: 6 | |
| May 14, 2022 at 14:45 | history | edited | M. Winter | CC BY-SA 4.0 |
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| May 14, 2022 at 14:37 | history | edited | M. Winter | CC BY-SA 4.0 |
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| May 13, 2022 at 17:06 | history | edited | M. Winter | CC BY-SA 4.0 |
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| May 12, 2022 at 16:20 | comment | added | LSpice | That seems only to argue that every $g$ has a fixed point, not that all of $G$ does; but @YCor's citation does the job. I guess the latter is a special case of the Bruhat–Tits fixed-point theorem, which I should have thought of since I recently asked a question about it. Thanks! | |
| May 12, 2022 at 16:08 | comment | added | YCor | @LSpice a result of Jung (1905) is that every nonempty bounded subset of a Hilbert space is contained in a unique ball of minimal radius. In particular the center of this ball is fixed by the isometry group. | |
| May 12, 2022 at 16:07 | comment | added | M. Winter | @LSpice I considered this folklore, but I have no quick argument at hand. My vague intuition is that if $g\in G$ has not fixed point then it sends $x\mapsto Ax + b$ with $A$ fixing the orthogonal complement of $b$. So no orbit of $G$ can be compact. | |
| May 12, 2022 at 15:52 | comment | added | LSpice | I'm sure it's obvious to anyone who might be in a position to answer this question, but why does the compactness of $K$ imply that $G = \operatorname{Isom}(\mathbb R^n, K)$ has a fixed point on $\mathbb R^n$? I could imagine buying it if we had some sort of appropriate compactness for $G$; do we? | |
| May 12, 2022 at 15:43 | answer | added | Andronick Arutyunov | timeline score: -2 | |
| May 12, 2022 at 15:07 | history | asked | M. Winter | CC BY-SA 4.0 |