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The general Frobenius Theorem stating that

Let $u_1,\dots,u_k$ be $k$ smooth linearly independent vector field on $M$. Let $$ W=\operatorname{Span}(u_1,\cdots,u_k) $$ Then $[u_i,u_j]\in W$ for any $i,j$ if and only if there exist foliation by $k$ dimension hypersurface tangent to $M$.

To my understanding,

there exist foliation by $k$ dimension hypersurface tangent to $M$.

means there is a cooridnate $(w_1,\cdots,w_{n-k},x_1,\cdots,x_k)$ such that $$ u_i=\frac{\partial}{\partial x_i} $$

I know the proof of the special case where $k=2$. To prove the general case, there is a hint saying using induction on $k$. However, I am not clear how to commit the induction. I tried to consider $[u_{k-1},u_k]$, but we only have $$ [u_{k-1},u_k]\in\operatorname{Span}(u_1,\cdots,u_k) $$ rather than $$ [u_{k-1},u_k]\in\operatorname{Span}(u_{k-1},u_k) $$ So I cannot perform the simpler case. Could anyone help?

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    $\begingroup$ The case k=1 follows from an existence and uniqueness theorem for ODE's. The step to k=2 from the proof you know should be the same as the on from k-1 to k in the induction. $\endgroup$ – MBN May 11 '14 at 18:56
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    $\begingroup$ Although I think this is a good and interesting question, I don't think it belongs on this site (math.SE would be better, clearly). $\endgroup$ – Danu May 11 '14 at 23:42
  • $\begingroup$ @MBN, sorry I don't get your words. Could you explain a little more? $\endgroup$ – hxhxhx88 May 12 '14 at 13:41
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Here is a way of thinking about Frobenius theorem that might help to understand the theorem and its proof.

The condition that one has a set of $k$ linearly independent complete vector fields on $M$ closed by the Lie bracket of vector fields means that one has a Lie subalgebra of the Lie algebra of vector fields of $M$.

One can, then, invoke Picard-Lindelöf's theorem (on existence and uniqueness of solutions for ODE's) and use the flow of each vector field to provide an action of the associated Lie group on $M$. This action induces a foliation on $M$, the orbits of this action are the leaves of the foliation.

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  • $\begingroup$ Yes, I can imagine such picture, but still not sure how to prove rigorously. Anyway, thanks:) $\endgroup$ – hxhxhx88 May 16 '14 at 19:28
  • $\begingroup$ This is a, rigorous, proof of it. It is not hard to work the details of it. $\endgroup$ – R.S. May 16 '14 at 19:36

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