# $\alpha \wedge d\alpha=0$ on 3 manifold for a $1$-form $\alpha$

Is it true that, for any $1$-form on a $3$-mfld $\alpha$, $\alpha \wedge d\alpha=0$?

I'd argue like $$0=d(0)=d(\alpha \wedge \alpha)= 2d(\alpha)\wedge \alpha$$

since for any $1$-form $\alpha \wedge \alpha=0$, but then there are counterexample like $\alpha= dz+xdy$ on $\Bbb R^3$ s.t. $\alpha \wedge d\alpha\neq0$

Where am I making a mistake?

For $1$-forms $\alpha$ and $\beta$, $$d(\alpha\wedge\beta)=(d\alpha)\wedge\beta-\alpha\wedge(d\beta)$$ so that $$d(\alpha\wedge\alpha)=(d\alpha)\wedge\alpha-\alpha \wedge(d\alpha)=0.$$
• ah, I see when I switch $\alpha$ and $d\beta$ the sign is $+1$ not $-1$ – Luigi M Aug 23 '17 at 4:38