Describe all smooth surfaces in $\mathbb{R}^3$ with coordinates $(x,y,z)$ such that the pullback of the one-form $\theta:=dy-zdx$ is identically zero.

Question

My question is as the title states:

Describe all smooth surfaces in $\mathbb{R}^3$ with coordinates $(x,y,z)$ such that the pullback of the one-form $\theta:=dy-zdx$ is identically zero.

Now, for the distribution defined as the annhilator of $\theta$ to be involutive, there must exist some one-form $\eta:=adx+bdy+cdz$ (for some functions $a,b,c$) such that $d\theta=\theta\wedge\eta$, or \begin{align*} dx\wedge dz&=(dy-zdx)\wedge(a\,dx+b\,dy+c\,dz)\\ &=-a\,dx\wedge dy+c\, dy\wedge dz-zb\,dx\wedge dy-zc\,dx\wedge dz\\ &=-(a+zb)dx\wedge dy+c\,dy\wedge dz-zc\,dx\wedge dz, \end{align*} hence $a=-zb$, $c=0$, and $-zc=1$. But if $c=0$, then $-zc\neq 1$.

So there cannot exist any 2-dimensional manifolds for which the pullback of $\theta$ is zero. There could potentially be one dimensional manifolds, but I'm not quite sure how would I approach this case.

I feel like I'm going about this question the wrong way, since I'm getting answers I'm not expecting. Thanks in advance.

Answer 1

Even easier, note that $d\theta\wedge\theta = (dx\wedge dz)\wedge(dy-z\,dx) = -dx\wedge dy\wedge dz\ne 0$, so this distribution is not involutive. There are (even locally) no integral manifolds, as you said. (You can also see this by locally writing a putative such surface as a graph $y=\phi(x,z)$ and getting a contradiction by taking a small closed curve on the surface whose projection into the $xz$-plane encloses nonzero area.)