3

Let $M$ be a $(2n+1)$-manifold and $\xi$ a smooth hyperplane field. It's easy to see that for each point $p$ exists a neighborhood $U$ of $p$ such that $\xi_{|U} = \ker \alpha_{U}$, where $\alpha_U$ is a smooth 1-form. I have read that if $M$ and $\xi$ are both oriented, we can find $\alpha$ $\textbf{globally}$ defined such that $\xi = \ker \alpha$ in all $M$. I can't see why, I mean, if $M$ is oriented by a volume form $\omega \in \Omega^{2n+1}(M)$, without supposing any kind of regularity in $\xi$ I don't know how to construct this globally defined 1-form. This exercise is proposed in $\textit{Introductory Lectures on Contact Geometry}$ by John B. Etnyre.

EDITED: In this post have a nice proof of this statement. I am going to introduce another proof based on it without using Riemannian metrics. When we say that $\xi$ is orientable we refer to $TM/\xi$ is the trivial bundle (by definition). We will say that $\xi$ is $\textit{corientable}$. In fact, since $\xi$ is a field of hyperplanes the previous quotient is a line bundle, that is, a 1-dimensional bundle.

Suposse $\xi$ is coorientable. If $TM/\xi$ is trivial, there exists a $\textbf{global}$ section throughout $TM/\xi$ and therefore a global section $\overline{\alpha}$ in $(TM/\xi)^*$. Consider $\pi : TM \to (TM/\xi)^*$ and the pullback $\alpha := \pi^*\overline{\alpha}$. In fact, this $\alpha$ satisfies $\ker \alpha = \xi$ by definition, and it's globally defined. Therefore we have proved that

$$ \xi \text{ coorientable} \Longrightarrow \alpha \text{ globally defined} $$

Reciprocally, assume $\alpha$ globally defining $\xi = \ker \alpha$ throughout $M$. Then this $\alpha$ induces a non-zero section in $TM/\xi$, that's mean, $TM/\xi$ is trivial.

JulianDoyle
  • 387
  • 1
  • 12
  • Here is an answer using the fact that $\xi$ is coorientable and an auxilliary riemannian metric. It says that if $\xi$ is coorientable, as $TM/\xi$ is isomorphic to $\xi^{\perp}$ by the auxilliary riemannian metric, $\xi^{\perp}$ is trivial and the result follows. – Didier Oct 28 '20 at 08:44
  • But, $M$ is orientable if and only if $TM/\xi$ is trivial? – JulianDoyle Oct 28 '20 at 09:04
  • No, $\xi$ is co-orientable if and only if $TM/\xi$ is trivial. – Didier Oct 28 '20 at 09:19
  • Nice! I can refine the proof. Thank you very much! I'm going to add it with details without using Riemannian metrics. – JulianDoyle Oct 28 '20 at 09:28

1 Answers1

2

Your exercice does not use the usual co-orientability property of globally defined contact $1$-form, but use the orientability of $M$ and $\xi$ instead. Here is an idea. $M$ is orientable, that means that $\Lambda^{2n+1}TM$ is a trivial line bundle. $\xi$ is orientable, so $\Lambda^{2n}\xi$ is a trivial line bundle. Given a riemannian metric $g$ on $M$, you can define $\xi^{\perp}$ as the $g$-orthogonal supplementary of $\xi$ in $TM$. Thus, you can show that

\begin{align} \Lambda^{2n+1}TM = \Lambda^{2n}\xi\otimes \Lambda^1 \xi^{\perp} \end{align} And triviality of the two first line bundles implies the triviality of $\Lambda^1\xi^{\perp}$ as a line bundle. Thus you can find a non-vanishing section of $\Lambda^1\xi^{\perp}$, that is a non-vanishing $1$-form in $\Gamma(\Lambda^1\xi^{\perp})$, which will be a globally defined contact $1$-form.

Didier
  • 20,820
  • Nice answer! I will edit the question commenting another proof using the ``standard´´ definition for $\xi$, that is, co-orientability, which is equivalent. – JulianDoyle Oct 28 '20 at 09:50
  • 1
    Oh! You can use the same idea without a Riemannian metric, isn't it? You can simply define $\eta$ like the direct sum decomposition of $TM = \xi \oplus \eta$. And the same plot gives us the answer. Is this correct? – JulianDoyle Oct 28 '20 at 10:04
  • Yes, a riemannian metric just permits to chose a "natural" supplement in the context, but it is not necessary. – Didier Oct 28 '20 at 10:11