Question about differential form $\omega = y dx + dz$ is a differential form in $\mathbb{R}^3$, then what is ${\rm ker}(\omega)$? Is ${\rm ker}(\omega)$ integrable? Can you teach me about this question in details? Many thanks!
 A: In this context, I presume that $\ker(\omega)$ denotes the subbundle of $T \mathbb{R}^3$ with fibre $\ker(\omega)_p := \ker(\omega_p)$ at each $p \in \mathbb{R}^3$, where $\ker(\omega_p)$ is the kernel of the functional $\omega_p : T_p \mathbb{R}^3 \to \mathbb{R}$. How does this functional look like? Well, setting $x^1 = x$, $x^2 = y$, $x^3 = z$ for notational convenience, the standard basis for $T_p \mathbb{R}^3$ is
$$
 \left\{ \left.\frac{\partial}{\partial x^1}\right|_p, \left.\frac{\partial}{\partial x^2}\right|_p, \left.\frac{\partial}{\partial x^3}\right|_p\right\},
$$
which satisfy
$$
 dx^i_p \left( \left.\frac{\partial}{\partial x^j}\right|_p \right) = \begin{cases} 1 &\text{if $i=j$,}\\ 0 &\text{if $i \neq j$.} \end{cases}
$$
Now, in general, $\omega = x^2 dx^1 + dx^3$, so that at a given $p = (p^1,p^2,p^3) \in \mathbb{R}^3$,
$$
 \omega_p = p^2 dx^1_p + dx^3_p.
$$
Hence, for $v \in T_p \mathbb{R}^3$,
$$
 \omega_p(v) = (p^2 dx^1_p + dx^3_p)\left( v^1 \left.\frac{\partial}{\partial x^1}\right|_p + v^2 \left.\frac{\partial}{\partial x^2}\right|_p + v^3 \left.\frac{\partial}{\partial x^3}\right|_p \right) = \;?
$$
In particular, then, what is $\ker(\omega)_p := \ker(\omega_p)$?
Next, let's turn to the question of integrability. Recall that a subbundle $E$ of $T \mathbb{R}^3$ is integrable if and only if for any sections $v$, $w$ of $E$, the Lie bracket $[v,w]$ of $v$ and $w$ is also a section of $E$, where
$$
 [v,w] = \left[v^1 \frac{\partial}{\partial x^1} + v^2 \frac{\partial}{\partial x^2} + v^3 \frac{\partial}{\partial x^3}, w^1 \frac{\partial}{\partial x^1} + w^2 \frac{\partial}{\partial x^2} + w^3 \frac{\partial}{\partial x^3} \right]\\ = \sum_{i=1}^3 \sum_{j=1}^3 \left(v^j \frac{\partial w^i}{\partial x^j} - w^j \frac{\partial v^i}{\partial x^j} \right) \frac{\partial}{\partial x^i}.
$$ 
Now, in the case of $\ker(\omega)$, by construction, by construction,
$$
 \Gamma(\ker(\omega)) = \{\xi \in \Gamma(T \mathbb{R}^3) \mid \omega(\xi) = 0\},
$$
i.e., the sections of $\ker(\omega)$ are precisely the vector fields $\xi$ such that $\omega(\xi) = 0$, where, by essentially the same computation as above,
$$
 \omega(\xi) = (x^2 dx^1 + dx^3)\left( \xi^1 \frac{\partial}{\partial x^1} + \xi^2 \frac{\partial}{\partial x^2} + \xi^3 \frac{\partial}{\partial x^3} \right) = \; ?
$$
Hence, what you need to do is use this computation to algebraically characterise the vector fields $\xi$ such that $\omega(\xi) = 0$, and then use this to show that for any vector fields $\xi$ and $\eta$ with $\omega(\xi) = 0$ and $\omega(\eta) = 0$, $\omega([\xi,\eta]) = 0$.
A: $\omega$ is nowhere zero so its kernel at each point is 2 dimensional. Let us find 2 vector fields that span this kernel. If we take a vector $v=a\partial_x+b\partial_y+c\partial_z$  then $\omega(v)=ay+c.$ Hence $v\in Ker(\omega)$ iff $c=-ay$. So we can take as a basis for the kernel of $\omega$ say $v_1=\partial_x-y\partial_z$, $v_2=\partial_y$. 
As for integrability of the distribution $Ker(\omega)$, this is equivalent to the vanishing of $\omega\wedge d\omega$. But the latter is $dx\wedge dy\wedge dz$, ie nowhere vanishing, hence the distribution is not integrable.  
