Given a nowhere zero vector field $Z$, does there exist a one-form $\gamma$ such that $\gamma(Z) = 1$?

Question

Take $M$ a smooth manifold, and $Z$ a vector field on $M$ such that $Z(p)\neq0$ for all $p\in M$. Is there a one form $\gamma \in \Omega^1(M)$ such that $\gamma(Z)=1$? I started to work locally, but couldn't work it out.

Answer 1

Here's a sketch of a solution along the lines you're looking for, @KlaasVanderBroeck.

In some neighborhood $U$ of each point $p\in M$, we can choose local coordinates $(x^1,\dots,x^n)$ so that $Z = \partial/\partial x^1$. (This is sometimes called the Flowbox Theorem.) Cover $M$ with such neighborhoods $U_\alpha$, with $x_\alpha\colon U_\alpha\to \Bbb R^n$ the charts as described.

Choose a partition of unity $\phi_i$ subordinate to the open covering $\{U_\alpha\}$, and consider the $1$-form $$\gamma = \sum_i \phi_i dx^1_{\alpha(i)},$$ where $\phi_i$ is supported in $U_{\alpha(i)}$. I leave it to you to check that $\gamma$ satisfies your conditions.

Answer 2

First note that $\langle Z\rangle$ defines a subbundle of $TM$. By choosing a Riemannian metric on $M$, we can write $TM = \langle Z\rangle \oplus E$, so any $V \in \Gamma(TM)$ can be written uniquely as $fZ + e$ where $f \in C^{\infty}(M)$ and $e \in \Gamma(E)$. Now define $\gamma(V) = \gamma(fZ+e) = f$. Then $\gamma$ is a smooth one-form on $M$ and $\gamma(Z) = \gamma(1Z+0) = 1$.