Differential form and wedge product Let $M$ be a differentiable manifold and $w$ a one-differential form on $M$ such that $dw \wedge w$ = $0$ and $w(p)$ is not equal to $0$ for any point $p$ in $M$.
How to show that there exist a one-differential form $v$ on $M$ such that $dw = v \wedge w$ ?
 A: As Ted Shifrin notes in the comments, this holds for any $2$-form $\eta$ on $M$ with $\eta \wedge \omega=0$.
Let $U$ be a coordinate neighborhood for $M$. Denote by $\omega_1$ the restriction of $\omega$ to $U$.
Since $\omega_1$ is nowhere-vanishing, we can find $1$-forms $\omega_2,\ldots,\omega_n$ on $U$ such that for every $p\in U$ the family $\big(\omega_1(p),\ldots, \omega_n(p)\big)$ forms a basis of the cotangent space of $M$ at $p$. Thus, there are smooth functions $c_{i,j}\colon U \rightarrow\mathbb{R}$ such that on $U$ we have
$$\eta=\sum_{i<j}c_{i,j}\ \omega_i\wedge\omega_j.$$
Now, the equality $\eta \wedge \omega=0$ and the alternating property of the wedge product together imply that on $U$ we have
$$0=\eta\wedge \omega_1= \sum_{i<j}c_{i,j}\ \omega_i\wedge\omega_j\wedge \omega_1= \sum_{1<i<j}c_{i,j}\ \omega_i\wedge\omega_j\wedge \omega_1.$$
Since the family of $3$-forms $\big(\omega_i\wedge\omega_j\wedge \omega_1\big)_{1<i<j}$ is linearly independent, it follows that $c_{i,j}=0$ for $1<i<j$. Thus, on $U$ we have
$$\eta=\sum_{j}c_{1,j}\ \omega_1\wedge \omega_j=\omega_1\wedge\sum_{j}c_{1,j}\ \omega_j.$$
Finally, set $v_U\colon=-\sum_j c_{1,j}\ \omega_j$.
By using a partition of unity subordinate to all coordinate neighborhoods for $M$ (the $U$'s), one then glues together the local (smooth) $1$-forms (the $v_U$'s) to obtain a global (smooth) $1$-form $v$ on $M$ such that $\eta=v\wedge \omega$.
