# Any two $1$-forms $\alpha$, $\alpha'$ with property satisfy $\alpha = f\alpha'$ for some smooth nowhere zero function $f$?

This is a followup question to here.

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

Does it follow that any two $1$-forms $\alpha$, $\alpha'$ with this property satisfy $\alpha = f\alpha'$ for some smooth nowhere zero function $f$?

Yes. Such a 1-form is the same as a section of $N(\xi)^*$ (as in, a section of $N(\xi)^*$ canonically gives you such a 1-form and vice versa). Equivalently I'm saying "Pick a complementary line bundle; $\alpha$ is determined by its value on that line bundle." By assumption of the existence of your $\alpha$ at all (as in the previous question), the normal bundle is trivial. So sections of it are canonically identified with smooth functions, and your question becomes "Given two nonvanishing smooth functions $\alpha, \alpha'$, is there a smooth function $f$ with $\alpha = f\alpha'$? Of course, the answer is yes: $f=\alpha/\alpha'$.