Here's the question:
Suppose that $\omega$ is a $k$-form on an open set $U$ of $\mathbb{R}^n$ and $f:U \to \mathbb{R}$ is a $C^\infty$ function such that $f(x) \neq 0$, for all $x \in U$, and $d(f\omega)=0$. Prove that $\omega \wedge d(\omega)=0$
My attempts so far:
Differentiate the product $f\omega$, and take the wedge product with $\omega$:
$$d(f\omega)=df\wedge\omega + f d\omega=0$$ $$\Rightarrow \omega \wedge df\wedge\omega + f \omega \wedge d\omega=0$$
I see 2 cases:
- If $k$ is odd: then the product $\omega \wedge \omega$ must be zero, since if the commutation formula is used:
$$\omega \wedge \omega = (-1)^{k^2} \omega \wedge \omega = - \omega \wedge \omega$$
Then, commutating $df$ above with $\omega$ (with a sign that comes out, no problem) and dividing by $f$, which is valid since $f$ is never zero, yields the result.
- If $k$ is even: I don't really see how to extend the above argument. I'm worried I might have to toss it away and try with another tool. Or maybe I'm missing something very fundamental in here.
Any suggestion or solution is welcome. Thanks!
The result is only true (in general) when $k$ is odd. In the appropriate open subset of $\Bbb R^5$ with coordinates $(u,v,x,y,z)$, try $\omega = u(dv\wedge dx + dy\wedge dz)$.