Timeline for vanishing of vector field in infinite dimensions
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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Jul 8, 2013 at 5:08 | comment | added | Dan Lee | To answer my own question, I suppose that the Schauder fixed point theorem (being an infinite dimensional analog of the Brouwer fixed point theorem) provides some understanding of the case of the unit ball. | |
Jul 8, 2013 at 4:46 | history | edited | Dan Lee | CC BY-SA 3.0 |
added 126 characters in body
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Jul 8, 2013 at 4:43 | comment | added | Dan Lee | Given how embarrassingly wrong my original premise was, it's a bit awkward to ask this, but I still wonder if something interesting can be said about the unit ball in Banach space. | |
Jul 8, 2013 at 4:19 | vote | accept | Dan Lee | ||
Jul 7, 2013 at 11:00 | review | Close votes | |||
Jul 7, 2013 at 16:50 | |||||
Jul 7, 2013 at 10:40 | comment | added | Ryan Budney | Dan, what you claim as a "simple fact" is not true. Take for example $S^1 \times [0,1]$, this has an inward-pointing everywhere non-zero vector field. The Poincare-Hopf index theorem is what tells you when you have to have a zero, and that's given in terms of the Euler characteristic of the manifold. | |
Jul 7, 2013 at 10:21 | answer | added | Vivek Shende | timeline score: 5 | |
Jul 7, 2013 at 0:35 | comment | added | Dan Lee | I know that the answer can't be super simple since there is no compactness, but maybe there are hypotheses that make it work. (For example, something like the Palais-Smale condition.) Even one specific situation in the literature where such an argument was used would be interesting to me. | |
Jul 6, 2013 at 20:58 | comment | added | Matthias Ludewig | This depends on the topology though. In general, closed balls can be compact if the model space is not Banach. | |
Jul 6, 2013 at 20:52 | comment | added | Ben McKay | I would bet against because compactness isn't available. Closed balls aren't compact. | |
Jul 6, 2013 at 20:41 | history | asked | Dan Lee | CC BY-SA 3.0 |