Consider the following example of a distribution (given here):

enter image description here

I tried to draw this. If $p=(a,b,c)$ then $$ X_p = (1,0,-b), Y_p = (0,1,0)$$

Then the planes in the distribution are planes spanned by $X_p,Y_p$.

We see that the plane spanned by $X_p,Y_p$ is a plane that rotates around the vector $Y_p$ as $p$ moves along the $y$-axis.

Assume we had a surface $S$ that was tangent to all this twirling planes. Without loss of generality, assume the surface is located in $\mathbb R^3$ such that the origin is on the surface.

Then we have a plane, coincidentally parallel to the $xy$-plane, that is tangent to $S$ at $0$. In other words: the $xy$-plane is tangent to $S$.

This is as far as I can follow the explanation given in the text. But everything that follows I do not understand.

For example, only because the $xy$-plane is tangent to $S$ at $0$ it is not clear to me why $S$ would intersect the $x$-axis in a line segment (for example, $S^2$ can be tangent to the $xy$-plane an does not intersect the $x$-axis in a line segment).

But even if this was clear to me and I assume that $S$ intersects this axis in a line segment the rest of the explanation is also not clear to me: travelling along an intersection axis does not seem to contradict that the planes are twisting.

Please could someone explain this to me?


First, let us establish rigorously that the distribution $\xi$ is not integrable, to do so we will use the so-called Frobenius theorem that I recall below:

Theorem. Let $M$ be a smooth manifold and let $\xi$ be a distribution of $M$, then the two following statements are equivalent:

  • The distribution $\xi$ is integrable.

  • For all vector fields $X$ and $Y$ of $M$ such that $X,Y\in\xi$, then $[X,Y]\in\xi$.

Sketch. The proof is essentially a recursive use of the straightening theorem for vector fields around a non-singular point. A full proof can be found in the Chapter $14$ of Introduction to smooth manifolds by J. Lee. $\Box$

Going back to your case, even though $X$ and $Y$ are tangent to $\xi$ (by definition), one has $[X,Y]=\frac{\partial}{\partial z}$ which is a vector field linearly-independent from $X$ and $Y$ and hence does not belong to $\xi$. Whence the result.

Let us now shed some lights on the informal discussion you reproduced.

It helps to notice that $\xi$ is given by the kernel of $\mathrm{d}z+y\mathrm{d}x$, it is then clear that $\xi$ is invariant by translation along the $x$-axis (and the $z$-axis, for that matter). Hence, if $0\in S$ is an integral submanifold of $\xi$, then $\xi_0$ (which is the $xy$-plane ) is tangent to $S$ not only at $0$ but all along the $x$-axis (invariance of $\xi$ along this axis). Therefore, $S$ intersects the $x$-axis on a line segment. The counterexample you mentioned fails to be really one since $S^2$ is no-longer tangent to the $xy$-plane when moving along the $x$-axis from the origin.

The key point is that by invariance of $\xi$ by translation in the direction of the $x$-axis, $S$ keeps being tangent to the $xy$-plane when moving along the $x$-axis from $0$.

From this observation, the conclusion should then be really easy. A smooth surface cannot contain a line segment so that a whole open set of the $xy$-plane is in fact included in $S$. Therefore, you can travel in $S$ in the direction of the $y$-axis without leaving the $xy$-plane and $\xi$ being tangent to $S$ must be contained in $\{z=0\}$, a direct contradiction with $\xi$ twisting along the $y$-axis.


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