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Let $M$ be a closed $3$-manifold, and let $\xi$ be a $2$-dimensional subbundle of $TM$. There is a nowhere zero $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$. How do I deduce the following for integrable $\xi$?

  1. We can write $d\alpha = \omega \wedge \alpha$ for some $1$-form $\omega$.
  2. Any two choices $\omega$, $\omega'$ satisfying this equation have $\omega' - \omega = g\alpha$ for some smooth function $g$.
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Part $1$ follows immediately from (one formulation of) Frobenius' theorem. Specifically, if $\xi$ is integrable, then $\alpha \wedge d\alpha = 0$. Since $\alpha$ is a $1$-form, that implies (take local coordinates, for example) that $d\alpha = \alpha \wedge \omega$ for some $\omega\in \Omega^1(M)$.

For part $2$, suppose $d\alpha = \alpha\wedge \omega'$. Then $\alpha \wedge (\omega - \omega') = 0$, forcing $\omega - \omega' = f\alpha$ for some $f\in C^\infty(M)$ (again, local coordinates might clarify the last equality).

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