## Criteria for Involutive Subbundles

Preliminaries: Let $M$ be a smooth manifold with tangent bundle $TM$. A vector subbundle $VM$ of $TM$ is called involutive if the section space $\Gamma(VM)$ of $VM$ is closed under the Lie bracket of $\Gamma(TM)$ or in other words if $[X,Y] \in \Gamma(VM)$ for all $X,Y \in \Gamma(VM)$.

On the other side the Lie bracket of two vector field can be expressed entirely by the flow transformations of the fields, that is we have:

$$[X,Y] = \frac{1}{2}\frac{\partial^2}{\partial_t^2} |_{t=0}(Fl^Y _{-t}\circ Fl^X _{-t}\circ Fl^Y _{t}\circ Fl^X _{t})$$

where $Fl^X$ and $Fl^Y$ are the flow transformations of $X$ and $Y$ respectively.

Now the question is, can (and if yes, how) we decide whether or not o subbundle of $TM$ is involutive entirely in terms of flow transformations?

If somehow we want to proof that the bracket is closed on a subbundle but we know very little about it but instead know much about the associated flow transformations.

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The subbundle is involutive if and only if the image of any given $x \in M$ under all of the flow transformations is an immersed smooth submanifold of $M$ with dimension equal to the rank of the subbundle.

ADDED: One direction follows by the Frobenius theorem. The other direction is even easier.

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If you have a set of vector fields whose flows form a group which is closed in the topology of $C^1$ convergence on compact sets, then you can compose and take limits, so you get bracket closure. If you have constant orbit rank, these vector fields span a subbundle of the tangent bundle.

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 you nean "span an involution"? Because "span a subbundle" was already one of my assumptions. – Mark.Neuhaus Mar 22 2012 at 9:36 Mark: I think Ben meant generate (=bracket-generate) a subbundle, as opposed to span (=linearly span). – Claudio Gorodski Mar 22 2012 at 14:02

I think the definitive work on that has been done by Hector Sussmann (Orbits of families of vector fields and integrability of distributions. Trans. Amer. Math. Soc. 180 (1973), 171–188, freely available here). Let $D$ be a smooth subbundle of $TM$ and $G$ the pseudo-group of local diffeomorphims generated by the flows of smooth sections of $D$. Sussmann proved there is a smallest $G$-invariant smooth distribution $\bar D$ containing $D$, which moreover is involutive and admits maximal integral manifolds through any point. It is important to note that the dimension of $\bar D_x$ will not be constant for $x\in M$. We have $\bar D=D$ if and only if $D$ is $G$-invariant.

Sussmann paper indeed studies a more general situation, appropriate for control theory, and contains many more very interesting results about the complete integrability of distributions. In fact I would recommend it as a "classic".

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 Although I agree with your characterization of Sussmann's work, the question is about a much simpler situation, namely a subbundle, which has a fixed constant rank, say, $r$. This means that in a neighborhood of each point in $M$, there are $r$ nonzero linearly independent vector fields spanning the subbundle. The involutivity assumption implies that the Lie bracket of any pair of these vector fields lies in their span. With these assumption the classical Frobenius theorem already tells you exactly what the flow looks like in a neighorhood of that point. – Deane Yang Mar 22 2012 at 22:54