# Need help understanding this example of a distribution

Consider the following example of a distribution (given 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.