Background: Suppose $E=TM$ is the tangent bundle to some differentiable manifold $M^n$. If we specify some subbundle $D\subset TM$ (distribution of $k$-planes) then there are two natural situations that arise. We may have that at some point $p$ there is an immersed submanifold $N^k\subset M^n$ passing through $p$ such that for all $q\in N$, $T_q N=D_q$. If this is true then we say $D$ is a integrable at $p$. Otherwise it is non-intregrable.

Frobenius Integrability states that every involutive distribution is completely integrable, i.e. $M$ is foliated by integral manifolds. One way of phrasing involutivity is to require that any lie bracket of vector fields lying in $D$ stay in $D$. There is another version, which is the one that I prefer, using differential forms. Note that any $k$-distribution is cut out locally by $n-k$ independent 1-forms $\theta_1,\ldots,\theta_{n-k}$. Call $\Theta$ the ideal generated by these, then we say $\Theta$ is involutive if $d\Theta\subset \Theta$,i.e. it is a differential ideal.

Situation: This last version generalizes to arbitrary vector bundles (with connection) easily. Suppose $E$ is a rank $N>n$ vector bundle over $M^n$. Then I can specify any sub-vector-bundle $D\subset E$ by $$D_x:=\lbrace v\in E_x | \theta_1(v)=\ldots=\theta_{n-k}(v)=0\rbrace $$

for some collection of 1-forms, i.e. sections of the dual bundle $E^*$. I can extend the above definition of involutive distribution to this subbundle in the obvious way. Let me take this as a definition of integrable subbundle.

Question: If the geometric concept associated to an integrable subbundle of $TM$ is a foliation, what is the geometric concept associated to an integrable subbundle of $E$?

I am only beginning to dig into exterior differential systems, so any well articulated answer/exposition is appreciated.

EDIT: The differential one gets from a connection actually just lands in vector valued forms, i.e. the Twisted de Rham complex. So the complex one would like to get (which looks like a Koszul complex possibly) is not obtained in a canonical way with a connection. Other differential operators would be needed.

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    $\begingroup$ Thanks to the experienced commenters below I see that the "obvious way" is not so obvious! Posting this has helped me realize that. $\endgroup$ – Justin Curry Oct 31 '10 at 16:05
  • $\begingroup$ Please allow me to add the Lie-Algebroid tag. I guess it fits in well here. $\endgroup$ – Stefan Waldmann Mar 19 '11 at 10:41

Note that for the tangent bundle you do NOT need a connection to define the exterior derivative on the dual bundle---the cotangent bundle. If you follow your prescription in the case that $E$ is the tangent bundle, using some arbitrary connection to define the derivatives of the $1-$forms, then note that, first of all these derivatives are not "in the right place", and moreover what you get will depend on the choice of connection, so the "definition" that you are hinting at for a generalized notion of involutive is really not well-defined.

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    $\begingroup$ In particular, it is not at all clear what "in the obvious way" means in the question. I suggest that the questioner try to work out the details of what "involutive" is supposed to mean for a vector bundle. As far as I know, there is no such concept. $\endgroup$ – Deane Yang Oct 30 '10 at 20:35
  • $\begingroup$ Ah, part of my question was whether the notion of integrable or involutive had been worked out for arbitrary vector bundles. $\endgroup$ – Justin Curry Oct 30 '10 at 21:01
  • $\begingroup$ @Justin Curry: I don't believe so. Part of the reason is that it is not at all clear how to generalize the notion of integrability for an arbitrary vector bundle. (Actually, "worked out" is a somewhat misleading way of putting it: it makes it sound like there must be a natural generalization of those notions for subbundles of bundles more general than the tangent bundle, while what Deane Yang and I have been trying to say is that we do not believe this is so.) $\endgroup$ – Dick Palais Oct 30 '10 at 22:22
  • $\begingroup$ Justin, in your question, you say "I can extend the above definition of involutive distribution to this subbundle in the obvious way. Let me take this as a definition of integrable subbundle." So aren't you claiming that you do have a definition of an involutive or integrable distribution for a vector bundle? $\endgroup$ – Deane Yang Oct 30 '10 at 23:52

As it was already pointed out, on a bare vector bundle there is no intrinsic notion of "integrability". However, things change when you pass to a Lie algebroid: In this case the vector bundle $E$ is equipped with two extra things: a Lie bracket for the smooth secitons $\Gamma^\infty(E)$ and an "anchor" map $\rho\colon E \longrightarrow TM$ which is a bundle map such that it yields a Leibniz rule for the two Lie brackets involveld: the one on $E$ and the canonical one on $TM$.

The tangent bundle itself is clearly an example for a Lie algebroid and so is a bundle of Lie algebras etc. Yet another example comes from Poisson geometry: any Poisson tensor on $M$ makes $T^*$ a Lie algebroid.

Now there are many notions of integrabilty associated to this structure: first, the image of the anchor map $\rho$ is a involutive distribution inside $TM$, now of course the rank may vary. Nevertheless there is a good notion of integrability (Stefan-Sussman). But you can also ask yourself whether a Lie algebroid "integrates" to something called a Lie groupoid in a similar way as a Lie algebra integrates to its simply-connected connected Lie group. This is a by far more complicated question which has ultimately been solved by Crainic and Fernandes only a couple of years ago.

Yet another instance where we encounter "integrability" for a general vector bundle is the generalized geometry: here you start with a vector bundle $E$ which is equipped with the extra structure of what is called a Courant algebroid. This is a maximally indefinite fibre metric, a bracket for the sections, and an achor map. Now the axioms are rather funny (weird) but $E = TM \oplus T^*M$ provides an example. Inside a Courant algebroid you can ask for maximally isotropic subbundles $L \subseteq E$ which are closed under the bracket ("integrable"!) These are called Dirac structures and play an important role in many recent applications in geometric mechanics of constraint systems, generalized geometries a la Hitchin, Poisson geometry etc.

The literature is waste and a little googling in the arXive will provide more than you want :)

  • $\begingroup$ I ended up finding out a little about Lie Algebroids, but thanks for posting. Also I am familiar with some of the Marsden work on Dirac structures, but never thought of it in an integrable context. Thanks! $\endgroup$ – Justin Curry Mar 23 '11 at 15:15

By now you've probably realized that if $E$ is just a bare vector bundle without additional structure (like a connection) then your $d\theta$ makes no sense for a section of $E^*$,

But to emphasize: in this "pure" vector bundle setting there cannot be an analogous notion of integrability of subbundles, since any two subbundles of the same dimension are locally isomorphic. So subbundles have no local structure besides the dimension while distributions of $k$-planes have plenty of local invariants.

  • $\begingroup$ I realized now that you actually demanded the existence of a connection in the original question, sorry my answer is a little out of context. $\endgroup$ – Michael Bächtold Mar 19 '11 at 10:54

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