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?


1 Answer 1


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – NDewolf
    Sep 9, 2018 at 0:33

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