Existence of a foliation on $\mathbb{R}^3$

Let $f \in C^\infty\left(\mathbb{R}^3\right)$. Let $D$ be the distribution given, at every $p \in \mathbb{R}^3$, by \begin{equation*} D_p = \text{span}\Bigg\{ \left.\frac{\partial}{\partial x}\right|_p , \left.\left(\frac{\partial}{\partial y}+ f \frac{\partial}{\partial z}\right)\right|_p\Bigg\} \subset T_p \mathbb{R}^3 \, . \end{equation*} Give sufficient and necessarily conditions for the existence of a foliation on $\mathbb{R}^3$ such that for all $p \in \mathbb{R}^3$, the tangent space to the leaf through $p$ is $D_p$.

By the Global Frobenius Theorem, there exists a bijection between foliations and involutive distributions. Thus if $D$ is an involutive distribution, there exists a corresponding foliation. Let us work out the condition when $D$ is an involutive distribution. For this, we require that for all $X,Y \in \Gamma\left(D\right)$, we have $\left[X,Y\right] \in \Gamma\left(D\right)$. Let $g \in C^\infty\left(\mathbb{R}^3\right)$ be a test function.

\begin{equation*} \begin{split} \left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}+f\frac{\partial}{\partial z}\right]\left(g\right) &=\frac{\partial^2g}{\partial x \partial y} + \frac{\partial f}{\partial x}\frac{\partial g}{\partial z}+ f \frac{\partial^2 g}{\partial x \partial z} - \frac{\partial^2 g}{\partial y \partial x} - f \frac{\partial^2g}{\partial z\partial x} \\ &= \frac{\partial f}{\partial x}\frac{\partial g}{\partial z} \, . \end{split} \end{equation*} Since $g$ was a test function, we conclude that \begin{equation*} \left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}+f\frac{\partial}{\partial z}\right] = \frac{\partial f}{\partial x}\frac{\partial }{\partial z} \, . \end{equation*}

If $D$ is an involutive distribution, this result should be a linear combination of the basis elements but that can clearly never be the case. The only case that this can happen is if $f \equiv 0$. But that looks too easy to me, so I'm wondering if I'm correct. Thanks in advance!

Solution 1:

Your solution is almost correct, at least your computations are. You should have conclude that $\partial_xf\equiv 0$, so that $f=f(y,z)$ is a function of $y$ and $z$ only.

Perhaps another way to conclude would have been to use the differential form version of Frobenius theorem, stating that $D=\ker\alpha$ is integrable if, and only if, $\alpha\wedge\operatorname{d}\alpha=0$. In your case, $\alpha$ can be taken to be $\operatorname{d}z-f\operatorname{d}y$, so that $\alpha\wedge\operatorname{d}\alpha=-\operatorname{d}f\wedge\operatorname{d}y\wedge\operatorname{d}z=-\partial_xf\operatorname{d}x\wedge\operatorname{d}y\wedge\operatorname{d}z$, and the same conclusion follows.