MTW in Exercise 4.4 calls a 1-form $A_\alpha$ a rotation free 1-form if $$\textbf{A}\wedge\textbf{dA}=0.$$ And claims that all such 1-forms may be written as $$\textbf{A}=\phi\,\textbf{d}\psi$$ for scalar $\phi$ and $\psi$. (Assume the space to be Minkowski or of simple enough topology)
Can you help me prove?
Let $D$ be the kernel of $A$, eg. $D=\ker A=\{v\in TM:\ A(v)=0\}$.
Frobenius' theorem tells us that $D$ is integrable iff for any $v,w\in D$ we have $dA(v,w)=0$.
If we have $A\wedge dA=0$, then for any $v,w\in D$ and $u\notin D$ we have $$ 0=(A\wedge dA)(u,v,w)=\frac{1}{2}\left( A(u)dA(v,w)+A(v)dA(w,u)+A(w)dA(u,v) \right)=\frac{1}{2}A(u)dA(v,w), $$ since by assumption $A(u)\neq 0$, it implies that $dA(v,w)=0$, and since, $v,w$ are arbitrary, as long as they are in $D$, it implies that $dA|_D=0$. By Frobenius' theorem then the distribution $D$ is integrable, and since its dimension is $n-1$, there is a local foliation of $M$ by hypersurfaces $\Sigma_r$ ($r$ is some parameter) such that $T_x\Sigma_r=D_x$, and since $A$ annihilates these spaces, $A$ is a normal form to the foliation $\Sigma$, and since the leaves are $n-1$-dimensional, the set of all normal forms is one dimensional (at each point), so all normal forms $N$ are proportional to $A$, eg. $A=\phi N$ for some function $\phi$
But then, we can locally specify the foliation by the level set of some function $\psi$, implying that for some $N$ normal form we have $N=d\psi$, which then gives $$A=\phi d\psi$$ as was given.
