Given a smooth vector field $v$ on a (finite dimensional) manifold $M$, one can find the associated integral curves i.e. integral submanifolds of M such that the tangent space at any point $p\in M$ is spanned by $v_p$. A smooth vector field can be looked as a 'smooth specification of subspaces' of the tangent space at each point.

However given smooth vector fields $v_1,...,v_r$ there may not exist a regular submanifold of $M$ such that the tangent space at each point of the submanifold is spanned by these vector fields (evaluated at that point).

Frobenius theorem gives us necessary and sufficient conditions for existence of such an 'integral submanifold'.

However I am not able to visually see why integral submanifolds can not be found in general and why some conditions are indeed required on the vector fields.

1.) Is it possible to find an example of 2 vector fields in $\mathbb{R}^3$ which do not admit an integral submanifold and that this is clear just by 'looking' graphically at the vector fields.

2.) Does the issue being discussed above have anything to do with the fact (Whitney's theorem) that every $n$ dimensional manifold (satisfying some suitable conditions may be) finds an embedding into $\mathbb{R}^{2n+1}$ but not necessarily in $\mathbb{R}^{n+1}$ ? Independently of whether it is related or not, I am unable to see the need for going to $2n+1$ dimensions graphically and would appreciate if there is an intuitive way of understanding it.

Partial answers/ related comments/ references are greatly appreciated.

  • 1
    $\begingroup$ I think one good way to think about his is in terms of commutativity of flows. To have integral submanifolds, you have to get to the same place by flowing along the vector fields in different orders (at least locally). This is clearly necessary if every point is to be contained in a chart so that the vector fields of the distribution are the first coordinate vector fields in the chart. So maybe it would be fruitful to to try to visualize why two flows might not commute locally. Hope this is somewhat helpful. $\endgroup$ – Tim kinsella Nov 11 '13 at 14:37

Following example comes from John Lee's Introduction to Smooth Manifolds and answers 1) in the spirit of Tim's comment.

Consider a smooth distribution $D$ on $\mathbb{R}^3$ spanned by $X = \frac{\partial}{\partial x} + y \frac{\partial}{\partial z}$, $Y = \frac{\partial}{\partial y}$. If $N$ would be an integral manifold then tangency in $(0,0,0)$ to $X$ would imply that $N$ contains some small neighborhood containing $[(0,0,0),(x,0,0)]$ and tangency to $Y$ at $(x,0,0)$ point would imply that $N$ at $(0,0,0)$ contains some subset of $xy$ plane, which is impossible because $N$ is tangent to $xy$ plane only at $x$-axis.

Why is that? Just because $[X, Y] = -\frac{\partial}{\partial z} \neq 0$.


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