# Frobenius Theorem

Say a manifold M has 3 vector fields S,T and R whose Lie brackets satisfy the equations $[S,T]=R$, $[R,S]=T$ and $[T,R]=S$

Then I suppose the following properties hold for M,

• There exists a metric on M whose Killing Fields are $S$,$T$ and $R$

• There exists a foliation of M with manifolds on which $SO(3)$ has a transitive action.

There are many possible loose ends in the above statements, like the metric can be pseudo-Riemannian (surely Schwarzschild Metric is an example which satisfies the above) and if M is a $4$-manifold then the foliation is probably only by 2-spheres.

I guess this is an application/special case of the Frobenius Theorem or its dual.

I would like to know what is the precise statement along these lines and its proof (reference) and if there is some general framework in which this fits in. (like for some arbitrary group instead of just $SO(3)$)

-
Minor correction: Should be $SU(2)$, not $SO(3)$. It is the simply connected group which will act here. – David Speyer Jul 7 '10 at 14:14
@David I had the Schwarzschild space-time in mind which being a real manifold can't have a $SU(2)$ action on it. It is foliated precisely by $S^2$s on which $SO(3)$ acts. Am I confusing something? – Anirbit Jul 8 '10 at 5:09
The point is if you have an action of Lie algebra then it gives a (local) action of the Lie group. You can glue a global action if you fields are complete, but this will be in general an action of SIMPLY CONNECTED Lie group (and $SO(3)$ is not s.c. and its cobver is $SU(2)=S^3$). – Anton Petrunin Jul 10 '10 at 16:32
Since you had Schwarzshchild in mind, for SO(3), you may be interested in Szenthe, "On the global geometry of spherically symmetric space-times" Math. Proc. Cambridge Philos. Soc., 2004, 137, 741-754 – Willie Wong Jul 11 '10 at 11:04
Why should the foliation be by 4-spheres? I can certainly come up with 4-manifolds with an SU(2) or SO(3) action with three-dimensional orbits. – José Figueroa-O'Farrill Jul 11 '10 at 13:14

I assume that $M$ is compact. [More generally you may assume that the vector fields are complete i.e. they have infinite integral curves.]
All diffeomorphism obtained by integrating your vector fields give an $S^3$-action on your manifold --- your assumption is just a reformulation in terms of Lie algebra.
@Anirbit (1) instead of compactness you may assume that your vector fields are complete (see my answer) --- that is to rule out case when your manifold is an open set in a big manifold with an $S^3$-action. (2) Yes (3) There is no canonical metric, you may start with any, pass to an average and get an invariant one --- it is possible if the group is compact. (4) Essentially you need a link between Lie algebra and Lie groups... --- I will think of a good book... – Anton Petrunin Jul 8 '10 at 13:02