# If $d(f\omega)=0$, then $\omega \wedge d(\omega)=0$

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!

• Would it perhaps be more natural to wedge with $\omega$ on the other side? – Chappers Dec 19 '15 at 18:10
• Is $U$ really an open subset of $\mathbb{R}$ (of dimension 1, no exponent)? Because if so there aren't many $k$-forms on $U$ with $k > 1$... – Najib Idrissi Dec 19 '15 at 18:10
• @NajibIdrissi Whoops, huge typo. It should be $\mathbb{R}^n$. Thanks! – Jorge Arce Garro Dec 19 '15 at 18:12
• @Chappers: Perhaps, though the commutation formula can be used and the same results are obtained. Right? – Jorge Arce Garro Dec 19 '15 at 18:16

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 $\omega = u(dv\wedge dx + dy\wedge dz)$.
• That's awesome. Then an $f$ that works for the exercise is $f(u,v,x,y,z)=1/u$, and the appropiate open subset $U$ is anyone that doesn't intersect the hyperplane where $u=0$. Thank you very much :) – Jorge Arce Garro Dec 19 '15 at 19:51
• Another counterexample, this one is even lame. Take $U$ to be $\mathbb{R}$ without zero. On $\mathbb{R}$ with coordinate $x$, take the 0-form $\omega = x$. If $f(x)=1/x$, then $d(f\omega)=d(1)=0$, but $\omega \wedge d\omega = x \wedge dx \neq 0$. – Jorge Arce Garro Dec 19 '15 at 20:05
• I don't usually think about $0$-forms for these, but you're right, of course. I would not write the last wedge (just $x\,dx$). – Ted Shifrin Dec 19 '15 at 20:10