Rotational free 1-form Suppose $v = v_{\alpha}dx^{\alpha}$ is a smooth 1 form field on a smooth manifold and satisfy that
$$v \wedge dv = 0.$$
It can be represented in components as
$$v_{[\alpha,\beta}v_{\gamma]} = 0.$$
Now I want to prove that $v$ can be locally written in the form
$$v = hdf,$$
where $h$ and $f$ are scalar functions. 
I know that a one form $u$ can be locally written in the form $u = df$ if and only if
$$u_{[i,j]} = 0.$$
So $v$ can be locally written in the form $v = hdf$ if and only if
$$\left( \frac{v_i}{h} \right)_{,j} = \left( \frac{v_j}{h} \right)_{,i}.$$
And finally I arrive at 
$$hv_{i,j} - v_{i}h_{,j} = hv_{j,i} - v_{j}h_{,i}.$$
But I still can not figure out how to prove the existence of $h$ by the condition
$$v_{[\alpha,\beta}v_{\gamma]} = 0.$$
Any help is appreciated. Thanks!
 A: The condition $v \wedge dv = 0$ on $v$ implies that its kernel $$V = \{ u \in TM : v(u)=0\}$$ is an involutive distribution, meaning it is closed under the Lie bracket. To see this, let $X,Y$ be smooth vector fields in $V$ (i.e. such that $v(X) = v(Y) = 0$) and compute $$0=v\wedge dv(X,Y,Z) = v(Z) dv(X,Y) = v(Z)v([X,Y]);$$ so choosing $Z$ such that $v(Z)$ vanishes only where $v$ does we conclude that $v([X,Y])=0$. 
The Frobenius theorem tells us that an involutive distribution is tangent to a foliation by submanifolds. The idea of the proof is to construct a basis of the distribution consisting of pairwise commuting vector fields, so that the flows of these vector fields will provide natural adapted coordinates.
Once we know $V$ is tangent to a foliation by submanifolds, the local version of your claim follows: in adapted coordinates $x^i$ such that the surfaces $x^0 = \text{constant}$ are tangent to $V$, we know that $v(\partial/\partial x^i) = 0$ for $i \ne 0$; so $v$ must have component expansion $v = v_0 dx^0$. Thus $h = v_0, f = x^0$ does the job.
