Timeline for Polynomial Vector Fields on the 3-Sphere
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2011 at 17:00 | history | edited | Daniel Fleisher | CC BY-SA 3.0 |
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| Aug 7, 2011 at 22:08 | comment | added | Daniel Fleisher | Yes, that's pretty much what I want. The fact that I haven't found anything resembling this makes me somewhat pessimistic. | |
| Aug 7, 2011 at 21:36 | answer | added | José Figueroa-O'Farrill | timeline score: 5 | |
| Aug 7, 2011 at 21:21 | comment | added | André Henriques | I think that I now understand what your question really is. You have the Lie algebra of polynomial vector fields on $S^3$, which I defined in my answer. You'd like to know if there is a way of describing it in a "concrete" way (i.e., in the way the Witt algebra is usually defined: by formulas). So your question could be formulated as: "could someone provide a nice basis of the Lie algebra of polynomial vector fields on $S^3$ in which the Lie bracket can be described by a nice closed formula?". | |
| Aug 7, 2011 at 20:52 | history | edited | Daniel Fleisher | CC BY-SA 3.0 |
added 159 characters in body; edited title
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| Aug 7, 2011 at 20:40 | answer | added | André Henriques | timeline score: 5 | |
| Aug 7, 2011 at 20:35 | history | edited | Daniel Fleisher | CC BY-SA 3.0 |
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| Aug 7, 2011 at 20:28 | answer | added | John Pardon | timeline score: 2 | |
| Aug 7, 2011 at 20:22 | comment | added | Daniel Fleisher | Let me redact: what is the the diffeomorphism group of $S^3$? | |
| Aug 7, 2011 at 20:19 | comment | added | Ryan Budney | Regarding question (1), it's not clear what your question is. Presumably you aren't interested in just a definition. Your question (1.0) on the other hand is quite simple and seems to have been completely answered in the comments. | |
| Aug 7, 2011 at 20:14 | comment | added | Daniel Fleisher | Yes, this is actually why I'm interested in $S^3$, actually. The problem is that choosing a basis for the Lie algebra and left-translating them is somehow insufficient. I'm having difficulty exactly why. Take $S^1$ - it's Lie algebra is trivial. You left-translate a basis and you don't get the Witt algebra. | |
| Aug 7, 2011 at 20:10 | comment | added | Francesco Polizzi | In fact, $S^3$ is a Lie group. One can find three linearly-independent and nonvanishing vector fields, trivializing $TS^3$ and forming a basis for the corresponding Lie algebra, see en.wikipedia.org/wiki/3-sphere | |
| Aug 7, 2011 at 20:08 | comment | added | Daniel Fleisher | I suspected [something like] this. That's why I'm interested in $C^{\infty}$ maps $S^3\rightarrow S^3(\cong\mathbb{R}^3\cup\{\infty\})$. Thanks for the bundle clarification. Very clear. | |
| Aug 7, 2011 at 19:58 | comment | added | Alain Valette | The tangent bundle of $S^3$ is trivial (view $S^3$ as the group of unit quaternions, and use the group law to trivialize the bundle) so the smooth sections are $C^\infty$ maps $S^3\rightarrow\mathbb{R}^3$. | |
| Aug 7, 2011 at 19:30 | history | asked | Daniel Fleisher | CC BY-SA 3.0 |