# Coorientation of contact structures

When reading about contact geometry one quickly encounters the notion of a cooriented contact structure/form. But I do not seem to be able to find a definition of "coorientation".

In some places they define a cooriented contact structure as one induced by a globally defined 1-form, but in other places (such as wikipedia) they note that coorientation implies the global definition of the contact form.

Is there a general notion of coorientation on manifolds or distributions (on manifolds)? Is this related to the general notion of orientation of a manifold?

Let $$M$$ be a manifold and let $$\xi$$ be a subvector bundle of $$TM$$, then $$\xi$$ is coorientable if, and only, if the vector bundle $$TM/\xi$$ is trivial.

Proposition. Let $$\xi$$ be a field of hyperplanes of $$TM$$, then $$\xi$$ is coorientable if, and only, if $$\xi$$ is the kernel of a differential form of degree $$1$$ of $$M$$.

Proof. Let $$g$$ be an auxiliary Riemannian metric on $$M$$, then $$TM/\xi=\xi^\perp$$.

If $$\xi$$ is coorientable, then $$\xi^\perp$$ admits a section $$X$$ and $$\alpha:=g(X,\cdot)\in\Omega^1(M)$$ satisfies $$\ker(\alpha)=\xi$$.

Conversely, if $$\xi=\ker(\alpha)$$ with $$\alpha\in\Omega^1(M)$$, then one can pick a unit vector field such that $$\alpha(X)>0$$, this gives a global section of $$\xi^\perp$$ and $$\xi$$ is coorientable. $$\square$$

Therefore, the two given definitions of a cooriented contact structure are equivalent!

For further information, I recommend checking An introduction to contact topology by Hansjörg Geiges.

• I had indeed also found this definition but also thought it was rather an implication of some god-given definition (I was rather thinking that the "co" came from the complement than from the quotient, but it does make a lot of sense). Also thanks for the proof of the equivalence. Sep 9, 2018 at 0:33