# Visualizing Frobenius Theorem

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.

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 $[X, Y] = -\frac{\partial}{\partial z} \neq 0$.