# Frobenius theorem with lesser regularity

Clearly, if one is given a $C^1$ sub-bundle $V$ of the tangent space of a smooth manifold $M$, wheather $V$ comes from a $C^2$ foliation of the manifold is decided by the conditions of the Frobenius theorem.

Of course one can define a $C^1$ foliation of a $C^0$ sub-bundle $V$ as well. However the conditions of the Frobenius theorem are not applicable anymore since $V$ is not differentiable hence there is no good? notion of Lie product.

Does anyone know of an analog of the Frobenius theorem for $C^0$ sub-bundles, or any result in this direction?

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It might be possible to salvage the involutivity condition when working with differential forms. The annihilator of a $C^0$ sub-bundle will consist of $C^0$ forms, whose derivatives can be interpreted as distributional currents. Currents in general are not multiplicative, so don't form an algebra. However, they may be considered as a differential module over $C^k$ functions, with appropriate $k$. Unfortunately, I don't know whether such conditions are sufficient the standard Frobenious result to hold. – Igor Khavkine Sep 5 '12 at 8:46
I just found the following on MO that is basically the same question: mathoverflow.net/questions/12266/… Has I seen that question I would have not asked this one. Thanks for the answers though. – The Common Crane Sep 5 '12 at 20:31

Not an answer, but a pointer to the difficulties:

If you restrict yourself to just one dimensional distributions, you immediately see a problem.

Let $V$ be a continuous (non-vanishing) vector field on $\mathbb{R}^n$, $n > 1$, it clearly generates a $C^0$ sub-bundle of the tangent bundle. By Peano's existence theorem we have that $V$ has at least one integral curve passing through any initial point. On the other hand, contrasted against Picard's existence theorem which requires Lipschitz regularity of the vector field (or in your case, $C^1$ suffices), this integral curve is not guaranteed to be unique.

Why are curves important? Frobenius in $C^1$ can be interpreted in the "integral form" as the following: To be sure that the distribution integrates to a submanifold, if you travel infinitesimally first in the $V_1$ direction, then in the $V_2$ direction, and compare with traveling first in the $V_2$ direction, then in the $V_1$ direction, the difference should be in a direction that is inside the subbundle. (In other words, you cannot move off the manifold by moving inside the manifold.)

So to have any hope of having a Frobenius like statement you need to overcome this problem of having way too many integral curves. This, in fact, also causes problems for the existence of a foliation!

An example:

Consider the vector field $V$ on $\mathbb{R}^2$ given by $$V(x,y) = (1, \sqrt{|y|})$$ Integral curves through $(0,0)$ include $$\gamma_{0}(t) = (t,0)\qquad \gamma_{+}(t) = (t,t^2 / 4) \qquad \gamma_{-}(t) = (t,-t^2/4)$$ for positive time $t$, and only $$\gamma(t) = (t,0)$$ is admissible for negative time $t$. (In fact, one can also transition from $\gamma_0$ to a suitably translated $\gamma_{\pm}$ at any $t \geq 0$.) This shows that every integral curve of $V$ must become tangent to the $x$ axis in finite negative time. Hence it is impossible to obtain a foliation to this non-vanishing, continuous vector field.

Thus you see that the issue of regularity occurs when just one dimension is considered, much before the statement of Frobenius theorem, which is a compatibility condition on multiple dimensions, comes into play.

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