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 $$d(\alpha\wedge\alpha)=(d\alpha)\wedge\alpha-\alpha \wedge(d\alpha)=0.$$

  • $\begingroup$ ah, I see when I switch $\alpha$ and $d\beta$ the sign is $+1$ not $-1$ $\endgroup$ – Luigi M Aug 23 '17 at 4:38

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