How to determine the foliation of an involutive distribution?

Question

The question is as follows:

Consider the distribution $D$ in $\mathbb{R}^3$ generated by the vector fields:

$$ X_1 = \frac{\partial}{\partial x} + \cos x \cos y \frac{\partial}{\partial z}, ~~~~ ~~~~ ~~~ X_2 = \frac{\partial}{\partial y} - \sin x \sin y \frac{\partial}{\partial z}$$ Check that $D$ is involutive and determine the foliation $F$ that integrates it.

$\textbf{Some attempt:}$

We have \begin{align}[X_1 , X_2] &= [\frac{\partial}{\partial x} + \cos x \cos y \frac{\partial}{\partial z} , \frac{\partial}{\partial y} - \sin x \sin y \frac{\partial}{\partial z}] \\&= - \cos x \sin y \frac{\partial}{\partial z} + \sin y \cos x \frac{\partial}{\partial z} \\&= 0 \end{align} So it is involutive.

And we know that since $D$ is involutive, therefore, is integrable. And we know that the integral manifolds of an integrable distribution define a foliation of our manifold.

So how can we find the foliation in our example parcticaly?

Can someone help me to understand this?

Thanks!

Answer 1

Differential forms win again. Note that $D$ is given by $\omega = 0$, where $$\omega = dz-\cos x\cos y\,dx + \sin x\sin y\,dy.$$ As it happens, $d\omega = 0$ [in general, integrability will be couched in the statement that $d\omega \wedge\omega = 0$], so $\omega$ is exact. Indeed, $\omega = df$, where $f(x,y,z)= z - \sin x\cos y$. Thus the foliation is given by the level sets $f(x,y,z) = z-\sin x\cos y = c$ as $c$ varies over $\Bbb R$.

Answer 2

Hint: Since $[X_1,X_2]=0$, the foliation is defined by $\phi_\circ\psi_t{t'}(u)$ where $\phi_t$ is the flow of $X_1$ and $\psi_t$ the flow of $X_2$, $u(x,y,z)$ and $t,t'\in\mathbb{R}$.