# Is there a nowhere $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$?

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 $\xi$?

• You mean a nowhere zero $1$-form? – Patrick Da Silva Feb 26 '16 at 20:18

This is true if and only if the normal bundle to $\xi$ is trivial. In one direction, if $N(\xi) = TM/\xi$ is trivial, the bundle map $TM \to N(\xi) \to \Bbb R$ defines a 1-form $\alpha$ as desired (recalling that $T^*M = \text{Hom}(TM,\Bbb R)$ as vector bundles). On the other hand, given such an $\alpha$, put a Riemannian metric on $M$ and consider the vector field $\alpha^\sharp$; by definition, $g(\alpha^\sharp, \cdot) = \alpha(\cdot)$, and in particular $\alpha(\alpha^\sharp) \neq 0$. So $\alpha^\sharp$ is a nonvanishing vector field, always transverse to $\xi$, hence defines a trivialization of $N(\xi)$.
(This assumption - that the normal bundle is trivial - is usually called $\xi$ being "co-orientable".)