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Reading a paper I came through an argument proving the following:

Let be given a smooth action of $\mathbb{R}^n$ on a manifold $M$, such that it is infinitesimally free and its orbits are pairwise diffeomorphic. If the orbits of this action are the fibers of a surjective submersion $\pi:M\to P$ then $\pi$ is a locally trivial fiber bundle whose standard fiber is $\mathbb{T}^k\times\mathbb{R}^{n-k}$.

Is this just an easy exercise? or is a special case of some other result? and in such a case is there a useful reference?
I am asking this question in order to have a proper way to refer to this result.

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If you don't get an answer here, you might consider asking at mathoverflow.net . –  Qiaochu Yuan May 6 '11 at 21:17
    
Dear Qiaochu Yuan, thanks for the attention. Probably just after your comment, Ryan Budney posted his answer. Thanks to him, I find the confirm of the argument in the paper, and that it is not special case of some other result but just a combination of standard constructions. –  Giuseppe Tortorella May 7 '11 at 5:32
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That the action is infinitesimally free tells you the orbits are diffeomorphic to $\mathbb R^n / G$ for various discrete subgroups $G$ of $\mathbb R^n$. This is just general nonsense with Lie group actions. And $\mathbb R^n/G$ is diffeomorphic to $\mathbb T^k \times \mathbb R^{n-k}$, that's pretty much just linear algebra -- take a basis for $G$ as a $\mathbb Z$-module, and extend that to a basis for $\mathbb R^n$, that gives you your diffeomorphism. In this argument $k$ need not be constant.

The last thing you're interested in is showing $k$ is constant and that the orbit types are constant. This apparently is the reason for the assumption of the bundle. The main thing this seems to be telling you is that the orbits are closed. Because if the orbit is closed you can construct a transversal and apply the above argument. This tells your your manifold $M$ fibers over something with fiber is $\mathbb T^k \times \mathbb R^{n-k}$, and that something would be $P$.

So I don't think there's a real reference for this result, it's just a standard putting-together of basic results about group actions. Take a look at Conlon's book on manifolds -- he gives some similar (but not quite the same) types of arguments concerning actions, commuting vector fields and such.

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"I don't think there's a real reference for this result, it's just a standard putting-together of basic results about group actions." Dear Ryan Budney. Thank you for the attention. I just started learning analysis on Smooth manifolds. Again the proof you sketched help me to consolidate the comprehension of the argument exposed in the paper. P.S.:Thank also for the suggestion of Conlon, I like it very much. –  Giuseppe Tortorella May 7 '11 at 5:24
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