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Suppose $\omega$ is a nonvanishing $1$-form on a $3$-manifold $M$. It's known that $\ker\omega$ is an integral distribution iff $\omega\wedge d\omega=0$.

I'm trying to understand this, but I don't get why $\ker\omega$ integrable implies $\omega\wedge d\omega=0$.

I worked out so far: suppose $\omega\wedge d\omega=0$. Let $X,Y$ be vector fields in $\ker\omega$. Then $$ d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y]) $$ so $\omega([X,Y])=-d\omega(X,Y)$. Since $\omega$ is nonvanishing, we can find $Z$ such that $\omega(Z)\neq 0$. Then $$ 0=(\omega\wedge d\omega)(X,Y,Z)=c\omega(Z)d\omega(X,Y) $$ for some $c\neq 0$. Thus $d\omega(X,Y)=0$, so $\omega([X,Y])=0$, so $[X,Y]\in\ker\omega$. Then $\ker\omega$ is involutive, so by Frobenius, it is integrable.

On the other hand, if $\ker\omega$ is integrable, then the annihilator ideal $I(\ker\omega)$ is closed under $d$. Clearly $\omega\in I(\ker\omega)$, so by assumption $d\omega\in I(\ker\omega)$. I'm trying to show $\omega\wedge d\omega(X,Y,Z)=0$ for any three vector fields, but it looks like a dead end. Is it possible to assume $X,Y\in\ker\omega$ without loss of generality some how? Because that would make the calculation work out.

Clara
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1 Answers1

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Yes, you can assume $X,Y\in\ker(\omega)$. Since $M$ is three dimensional $L=\bigwedge^3TM$ is a line bundle over $M$, and $\omega\wedge d\omega$ is naturally defined on (local) sections of this line bundle. Take a local basis $(X,Y)$ of vector fields parallel to $\ker(\omega)$, and a smooth local extension $Z$ of some vector $Z_m\in T_mM\setminus\ker(\omega_m)$, then $Z\wedge X\wedge Y$ is a local basis of the line bundle $L$, and evaluating $\omega\wedge d\omega$ on it yields zero (through your calcuation), so that this form is actually locally zero, and hence globally zero.

Olivier Bégassat
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