Kernel of Cotangent Vector Equals Tangent Space to Submanifold

Question

Given a $1$-form $\omega=dx+ydz$ on $\mathbb{R}^3$, does there exist a $2$-dimensional submanifold $S\subset \mathbb{R}^3$ such that $$\ker(\omega_p)=T_pS$$ for all $p\in S$?

This is a homework question on which I have been stuck for quite a while. How should one determine the kernel of a covector $\omega_p$ given a general $1$-form $\omega$? I know that if $S$ is the preimage of a regular value of some smooth map $F:\mathbb{R}^3\to \mathbb{R}$, then $T_pS=\ker(T_pF)$. So in this case can I conclude that $$\ker(\omega_p)=T_pS\quad \forall p\in S\iff \omega_p=T_pF\quad \forall p\in S$$ or is the above not so obvious/false?

If $\omega=df$ (i.e. when it is exact) though, we do know that $\omega_p$ is the map $$T_pf:T_p\mathbb{R}^3\to T_{f(p)}\mathbb{R}$$

But now I'm totally lost since $S$ need not be the preimage of a regular value of $f$. I suspect that the answer to the original problem is no when $\omega$ is not exact. Can anyone give a hint on how to solve this problem? Any help would be appreciated.

Answer 1

To give some insight on what is to follow, let us notice that $\omega$ is a contact form on $\mathbb{R}^3$ so that any integral submanifold of $\ker(\omega)$ has dimension at most $1$. I am basically going to prove this result.

First, notice that $\mathrm{d}\omega=\mathrm{d}y\wedge\mathrm{d}z$ is non-degenerate (as a bilinear form) on $\ker(\omega)$, indeed one has: $$\omega\wedge\mathrm{d}\omega=\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\neq 0.$$ Notice that if $S$ is an integral submanifold of $\ker(\omega)$ and $p\in S$, then $\omega_{\vert S}=0$ and $\mathrm{d}\omega_{\vert S}=0$, meaning that $T_pS$ is an isotropic subvector space of the symplectic vector space $(\ker(\omega_p),\mathrm{d}{\omega_p}_{\vert\ker(\omega_p)})$, so that: $$\dim T_pS\leqslant\dim(T_pS)^\perp=2-\dim T_pS,$$ and $T_pS$ has at most dimension $1$ and so does $S$.

This proof is just a matter of linear algebra, there is nothing more to it. Notice that it is crucial to know that the bilinear form is non-degenerate to compute the dimension of the orthogonal.

Reminders. Here are the relevant definitions.

Let $V$ be a finite-dimensional vector space over $\mathbb{R}$ and let $B\colon V\times V\rightarrow\mathbb{R}$ be a bilinear form.

Definitions. Let $E$ be a subvector space of $V$, then $E^\perp:=\{v\in V:\forall w\in E,B(v,w)=0\}$, $E$ is said to be isotropic whenever $E\subset E^\perp$ and $B$ is non-degenerate when $\ker(B):=V^\perp=\{0\}$.

The main result used is that when $B$ is non-degenerate, then $\dim E^\perp=\dim V-\dim E$.

Let us define the linear map $u\colon V\rightarrow E^*$ by $u(v)=B(v,\cdot)$, its kernel is given by $E^\perp$ and its image is $E^*$. Indeed, since $B$ is non-degenerate and $V$ is finite-dimensional, $v\in V\rightarrow B(v,\cdot)\in V^*$ is an isomorphism. Whence the result, using the rank-nullity theorem.

A symplectic vector space is a vector space endowed with a non-degenerate skew-symmetric bilinear form.

Answer 2

Here's a more basic argument. Locally, any smooth surface $S\subset\Bbb R^3$ is in fact of the form $f=0$ for some smooth function $f$ (having $0$ a regular value). Thus, locally, we would have to have $df = \lambda\omega$ for some nonvanishing function $\lambda$. So, can we locally have $df = \lambda(dx+y\,dz)$?

There are various ways to proceed, but perhaps the easiest is to note that this tells us that $$\frac{\partial f}{\partial x} = \lambda, \quad \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial z} = \lambda y.$$ In particular, $0=\dfrac{\partial^2f}{\partial x\partial y} = \dfrac{\partial^2f}{\partial y\partial x}=\dfrac{\partial\lambda}{\partial y}$ and so $$0=\frac{\partial^2f}{\partial z\partial y}= \frac{\partial^2f}{\partial y\partial z} = \lambda + y\frac{\partial \lambda}{\partial y} = \lambda.$$ This is an obvious contradiction.