I need to prove the following:
Let $\omega$ be a non-vanishing $1-$form on a smooth manifold $M$ such that $d\omega\wedge\omega=0.$ Show that there exists a $1-$form $\theta$ such that $d\omega=\theta\wedge\omega.$
The idea I have is write locally $\omega$ in terms of the canonical basis $$\omega=\sum_{i=1}^n a_idx_i,$$ where $n=\dim M$ and $a_i\in C^{\infty}(M).$ Then, $$d\omega=\sum_{j=1}^n\sum_{i=1}^n \frac{\partial a_i}{\partial x_j}dx_j\wedge x_i$$ and $$d\omega\wedge\omega=\sum_{k=1}^n\sum_{j=1}^n\sum_{i=1}^n a_k\frac{\partial a_i}{\partial x_j}dx_k\wedge dx_j\wedge dx_i=0.$$ I'm stuck in this part because I don't know how to justify the existence of a local $1-$form $\theta_U.$ I think that if I prove this in open subsets $U$, then it would be easy extend it to the whole manifold by using partition of unity.
How can I construct $\theta$? I have tried to obtain inspiration by studying particular examples, but not even that helps.
Thanks in advance!
Yikes! NO, don't resort to the canonical basis. (This would only work locally, anyhow.) Forget for a moment that we're talking about $d\omega$. Suppose you have a $2$-form $\phi$ so that $\phi\wedge\omega = 0$ and you want to show that $\phi = \theta\wedge\omega$ for some $1$-form $\theta$. Here's the main hint: Suppose you can extend $\omega=\omega_1$ to a basis $\omega_1,\omega_2,\dots,\omega_n$ for the $1$-forms on $M$. Write $\phi$ in terms of this basis. Can you get it now?
Now, in general, you can do this locally. Do you know a technique to turn local solutions into global ones?