Why coorientable is a dual notion of orientable?

Question

When I started to study contact geometry, I immediately met some obstacle, the notion of 'coorientable'. In Geiges' "an introduction to contact topology", he defines that the distribution $\xi$ on $M$ with codimension 1 is coorientable if $TM/\xi$ is trivial. I think it is clear, however, it is hard to understand why this definition has the name of 'coorientable'. I hope to find the dual notion of orientable in some sense and understand the connection between coorientable and orientable. Is there some intuitive way to understand the meaning of coorientable in terms of orientable?

Answer 1

$TM/\xi$ is the normal bundle to the distribution. If $M$ has a metric, you're asking to be able to put a consistent orientation on the set of vectors perpendicular to $\xi$.

In an inner product space $V$ it is reasonable to assert that $W \mapsto W^\perp$ is a duality on the subspaces of $V$, as $(W^\perp)^\perp = W$. Whence the "co".

Notice that one has $TM \cong \xi \oplus \xi^\perp \cong \xi \oplus (TM/\xi)$.

Hence if $M$ is orientable, a distribution is orientable if and only if it is co-orientable. More specifically, if $M$ is oriented, an orientation on $\xi$ determines one on $\xi^\perp$ and vice versa (by the demand that $TM \cong \xi \oplus \xi^\perp$ is an isomorphism of oriented vector bundles).

When $M$ is not orientable, the distribution $\xi$ cannot be both orientable and co-orientable. A standard example of a co-orientable distribution which is not orientable is the distribution on $\Bbb{RP}^2 \times S^1$ of tangent planes to the fibers $\Bbb{RP}^2 \times \{\theta\}$. Of course, this is not a contact structure.

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There are a number of unproved claims in this answer (for instance, in the line starting "Hence"). I suggest that if they are not obvious to you, it will be an instructive exercise in the notion of orientability of vector bundles to prove it. Whenever we use splittings as above or pass between $TM/\xi$ and $\xi^\perp$ we are using an inner product on $TM$.