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 $SO(3)$)

  • $\begingroup$ Minor correction: Should be $SU(2)$, not $SO(3)$. It is the simply connected group which will act here. $\endgroup$ – David E Speyer Jul 7 '10 at 14:14
  • $\begingroup$ @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? $\endgroup$ – Anirbit Jul 8 '10 at 5:09
  • $\begingroup$ 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 $SU(2)=S^3$). $\endgroup$ – Anton Petrunin Jul 10 '10 at 16:32
  • $\begingroup$ 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 $\endgroup$ – Willie Wong Jul 11 '10 at 11:04
  • $\begingroup$ 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. $\endgroup$ – 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.

For any compact Lie group acting smoothly on a manifold there is a invariant Riemannian metric --- this can be constructed as an average of a given metric by the group action.

  • $\begingroup$ Thanks for your answer. For which argument are you needing compactness? (The Schwarzschild space-time is non-compact but it shows this structure) Are you saying that given a set of vector fields satisfy the Lie-Algebra of a group G one can integrate them to get an action of G on the manifold? Further for the averaging process what is the canonical metric one starts with? In the generic situation the construction begins with only a differential manifold being given. Any reference you can give would be helpful which say shows some examples of computation of the group from the vector fields. $\endgroup$ – Anirbit Jul 8 '10 at 5:07
  • $\begingroup$ @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... $\endgroup$ – Anton Petrunin Jul 8 '10 at 13:02
  • $\begingroup$ Where do you need completeness of the geodesics? Atleast in Schwarzschild space-time the geodesics are not complete (almost by definition of what it means to have a Black-Hole) and the above structure fits through. I am aware that given a Lie Algebra one can exponentiate it to recover a simply-connected Lie Group corresponding to it. What I am not clear of is how to get an action for this group on the manifold and then see that its orbits can foliate the manifold. Anyway waiting for your book reference. $\endgroup$ – Anirbit Jul 8 '10 at 16:52
  • $\begingroup$ @Anirbit, Sorry, I did not read your comment correctly and said something irrelevant... I do NOT need "completeness of the geodesics" --- I need "completeness of the fields". A field is complete if it defines a global flow; i.e. one can start at any point and go along integral curve to the field for arbitrary time. I'm sorry I could not find a nice book --- the problem is I did not ready study the subject --- ask someone else OR if you know how to get from Lie algebra to its Lie group then try to mimic the same argument for your vector fields and you will get what you want... $\endgroup$ – Anton Petrunin Jul 10 '10 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.