Timeline for Orbits of Lie Algebra Actions
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 7 at 22:18 | comment | added | Ali Taghavi | @RobertBryant is the question obvious in the particular case :foliation of TM by verticql space? | |
| Feb 3, 2012 at 17:06 | comment | added | Nicola Ciccoli | About the last lines in this remark there is a very nice blog post by Nguyen Tien Zung which summarizes some known results (and showing that in many cases tha type of foliation coming from a $\mathbb R^n$-action is quite limited. zung.zetamu.net/2012/01/rn-actions-on-n-dimensional-manifolds | |
| Jan 17, 2012 at 7:42 | answer | added | Nicola Ciccoli | timeline score: 3 | |
| Jan 16, 2012 at 0:23 | comment | added | Robert Bryant | Besides the obvious conditions that the subbundle $D\subset TM$ of vectors tangent to the leaves be trivial and integrable, I doubt that there is much you can say. If you want the action to be complete, each leaf of $\mathcal{F}$ will have to have the topology of a discrete quotient of a fixed Lie group $G$, and, for example, that will rule out leaves having the topology of the $7$-sphere, even though it is parallelizable. It's probably not even that easy to say when an integrable $2$-plane field $D\subset TM$ that is smoothly trivial can be spanned by two commuting vector fields $X$ and $Y$. | |
| Jan 15, 2012 at 22:55 | comment | added | Zouhair | Yes, I want the action to be free, so $\rm{dim}\;\mathfrak{g}$=dimension of the foliation $<\infty$. The Lie algebra of fields tangent to the foliation is infinite-dimensional. | |
| Jan 15, 2012 at 22:42 | history | edited | Zouhair | CC BY-SA 3.0 |
added 35 characters in body
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| Jan 15, 2012 at 21:51 | comment | added | Mariano Suárez-Álvarez | (I guess, though, you want the action of $\mathfrak g$ to be free, as in the first sentence?) | |
| Jan 15, 2012 at 21:36 | comment | added | Mariano Suárez-Álvarez | If you let $\mathfrak g$ be the Lie algebra of fields tangent to the foliation and $\rho$ the inclusion, you always have that the foliation is the set of $\mathfrak g$-orbits, no? | |
| Jan 15, 2012 at 21:08 | history | edited | Zouhair | CC BY-SA 3.0 |
edited body
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| Jan 15, 2012 at 20:34 | history | asked | Zouhair | CC BY-SA 3.0 |