Timeline for Easy proof of the fact that isotropic spaces are Euclidean
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 2, 2017 at 13:06 | answer | added | Chill2Macht | timeline score: 0 | |
| Jan 19, 2012 at 9:47 | answer | added | José Navarro | timeline score: 3 | |
| Oct 8, 2010 at 19:15 | vote | accept | Sergei Ivanov | ||
| Oct 6, 2010 at 17:41 | comment | added | Bill Johnson | Sorry, Greg; I did not realize that such a comment should be posted as an answer. I did it. | |
| Oct 6, 2010 at 17:40 | answer | added | Bill Johnson | timeline score: 17 | |
| Oct 6, 2010 at 10:45 | comment | added | Greg Kuperberg | @Bill: You should post your remarks as an answer to the revised question. It's an open problem for separable Banach spaces and there are inseparable counterexamples. Sweet! | |
| Oct 6, 2010 at 8:04 | comment | added | Bill Johnson | @Sergei Ivanov: The example I mentioned in the non separable setting is wrong. An example is the $\ell_p$ sum of uncountably many copies of $L_p(0,1)$, not the $\ell_p$ sum of uncountably many copies of the real line. | |
| Oct 6, 2010 at 7:28 | comment | added | Sergei Ivanov | @Bill Johnson: thanks, you answered it. I was not aware about this being an open problem. | |
| Oct 6, 2010 at 1:42 | comment | added | Bill Johnson | Not sure I understand your last question, Sergei. It is a famous problem whether a separable infinite dimensional Banach space which has a transitive isometry group must be isometrically isomorphic to a Hilbert space. Of course, if every two dimensional subspace has a transitive isometry group, then the space is a Hilbert space since then the norm satisfies the parallelogram identity. For counterexamples in the non separable setting, consider $\ell_p(A)$ with $p$ not $2$ and $A$ uncountable. | |
| Oct 6, 2010 at 1:10 | comment | added | Paul Siegel | When I was an undergraduate I read substantial portions of your book (with Burago & Burago) on metric geometry, and I remember encountering this as an exercise and getting hung up on it for about two weeks. I finally asked Ralph Spatzier who suggested more or less the exact argument that you outlined. I could barely understand the argument and went on for years with the problem of simplifying it in the back of my mind. I don't mind the more sophisticated argument anymore, but it would nevertheless be very gratifying for me personally to see this issue resolved. So thanks for the question! | |
| Oct 6, 2010 at 0:44 | answer | added | Dick Palais | timeline score: 5 | |
| Oct 5, 2010 at 23:01 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
explained what I want (hopefully) better
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| Oct 5, 2010 at 22:28 | answer | added | rpotrie | timeline score: 4 | |
| Oct 5, 2010 at 21:59 | answer | added | Greg Kuperberg | timeline score: 13 | |
| Oct 5, 2010 at 21:52 | history | asked | Sergei Ivanov | CC BY-SA 2.5 |