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!
Best Answer
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.