Positive vs negative Dehn twist monoids Given a bounded surface $S$ and a mapping class $h$, construct the open book 3-manifold $(S,h)$. If $h$ lies in the positive monoid of the mapping class group, then $(S,h)$ supports a tight contact structure; if $h$ lies in the negative monoid of the mapping class group, then $(S,h)$ supports an overtwisted contact structure. My question is why there are different conclusions. The difference between a positive or negative Dehn twist seems to be just one of orientation, so it's not clear why they should have such different effects. Is there something fundamentally "handed" about contact structures that makes the difference? Or is there a bigger difference between the two monoids that I'm missing?
[This question is motivated by the fabulous article, "Contact Geometry and the Mapping Class Group," by Joan Licata, in the April 2022 issues of the Notices of the AMS.]
 A: A general discussion may help clarify these points. There's a lot more to say (e.g. how the positive monoid gives Weinstein fillings, or how negative Dehn twists explicitly give overtwisted disks), but hopefully this is a start.

(1) Orientations in contact geometry
The most common definition of a contact manifold is as a pair $(Y^{2n+1},\xi)$ where $Y$ is an odd-dimensional manifold and $\xi \leq TY$ is a maximally non-integrable hyperplane distribution. If $\xi = \ker \alpha$ locally, then maximal non-integrability is the condition that $\alpha \wedge d\alpha^n \neq 0$. Many times, contact geometers restrict to the setting of cooriented contact manifolds, i.e. an orientation of $TY/\xi$. In such a case, $\alpha$ may be defined globally (in such a way that the corresponding trivialization of $TY/\xi$ matches its chosen orientation), and $\alpha \wedge d\alpha^n$. So in fact, cooriented contact manifolds always come with natural orientations. If $n$ is odd (e.g. $n=1$, meaning $2n+1 = 3$), then notice that negating $\alpha$ preserves the orientation of the contact manifold. This actually means that in dimensions $4n+3$, the contact manifold is canonically oriented even if the contact structure is not cooriented!
Here's a little digression, if you like thinking naturally without picking a contact form. Any hyperplane distribution $\xi$ determines a Lie bracket pairing $$\wedge^2\xi \rightarrow TY/\xi$$ (note that the full Lie bracket involves sections, but the image in $TY/\xi$ is well-defined as a bundle map) and maximal nonintegrability means that this pairing is non-degenerate.
Remark: The Frobenius Theorem is the statement we have a foliation if and only if this Lie bracket map is zero; hence we see the meaning of maximal non-integrability.
Remark 2: If you've chosen $\alpha$, yielding a map $TY/\xi \rightarrow \mathbb{R}$, then the composition $\wedge^2\xi \rightarrow \mathbb{R}$ is just $-d\alpha|_{\xi}$.
This non-degeneracy is equivalent to the $n$-fold wedge product $$\wedge^{2n}\xi \rightarrow \otimes^n(TY/\xi)$$ being an isomorphism of line bundles, or equivalently, the contact structure affords an isomorphism $$\wedge^{2n}\xi^* \cong \otimes^n(TY/\xi)^*$$ Using this isomorphism, we find $$\wedge^{2n+1}(T^*Y) = (TY/\xi)^* \otimes \wedge^{2n}\xi^* \cong \otimes^{n+1}(TY/\xi)^*$$ An orientation may be thought of as a trivialization (up to positive scale) of $\wedge^{2n+1}(T^*Y)$, which in turn is just a trivialization (up to positive scale) of $\otimes^{n+1}(TY/\xi)^*$. Since $\otimes^2 \ell$ is always canonically oriented for a line bundle, we see that if $n$ is odd, we get a canonical orientation, and if $n$ is even, an orientation is equivalent to a coorientation of the contact structure.

(2) Orientations and supporting open books
Part of the trickery is that the notion of a positive or negative Dehn twist depends precisely on the orientation of the surface $S$. A positive Dehn twist on $S$ is a negative Dehn twist on $\overline{S}$, and vice versa. For that reason, the positive monoid on $S$ is canonically identified with the negative monoid on $\overline{S}$.
In 3-dimensional contact geometry, the Thurston-Winkelnkemper construction takes as input the data of $(S^2,h)$ and constructs a contact manifold on the corresponding open book. However, when one really wants to say this precisely, one usually notes the following important point: $S$ comes with an orientation, and the output contact structure comes with a coorientation. It is also worth noting that if the input is $(\overline{S},h)$, the result is not just the same contact structure with the reverse coorientation! (Indeed, reversing coorientation preserves tightness, whereas the whole point of the story is that negative Dehn twists yield overtwistedness.)
Let me say a little about how the construction works. The idea is to first form the mapping torus $M_h = (S \times [0,1])/((x,1) \sim (h(x),0))$. Fix an exact area form $\omega = d\lambda$ on $S$ such that on an open collar neighborhood of the boundary $$\mathrm{Op}(\partial S) \cong (-\epsilon,0] \times \partial S,$$ we have $\lambda = e^t\lambda_0$ where $t$ is the coordinate on $(-\epsilon,0]$ and $\lambda_0$ a nowhere zero form on the boundary circles $\partial S$. There is a technicality, which is that we require that $h^*\lambda = \lambda$ itself, but this can be arranged up to isotopy of $h$. Let $\theta$ be the coordinate along the $[0,1]$ direction of $M_h$. Then we have a natural contact form $$\alpha := d\theta - \lambda$$ on $M_h$. The open book is constructed by gluing in a copy of $D^2 \times \partial S$ along the boundary $S^1 \times \partial S \cong \partial M_h$ via a filling procedure which can be made standard (since we know what $\alpha$ looks like near $\partial M_h$). If instead we used $-\omega = -d\lambda$, then $M_h$ is given the contact form $d\theta+\lambda$, which is not the same as $-\alpha$, and in particular, the construction does not simply reverse the coorientation. You can think instead that it both reverses the coorientation but also negates the $\theta$ direction independently. The filling procedure for gluing in $D^2 \times \partial S$ takes into account this orientation data along the boundary as well.
The reason that this construction does not take opposite orientations to the same contact structure with opposite coorientations may be thought of in the following way. On $M_h$, we are using the orientation which descends from the product orientation on $S \times [0,1]$. Reversing the orientation on $S$ thus reverses the orientation on $M_h$, but orientation in dimension 3 is invariant of the choice of coorientation, so this construction really is producing a different contact structure!
Adding to the digression of Section (1), more generally, the Thurston-Winkelnkemper construction takes as input an exact symplectic manifold $(S,\omega=d\lambda)$ with convex boundary (and such that the monodromy satisfies $h^*\lambda = \lambda$). We see that handedness still matters in the same way: we simultaneously reverse the coorientation but also negate $\theta$, and so the corresponding constructions are not the same under the involution of primitive $\lambda \mapsto -\lambda$. In terms of orientations, on $M_h$, we get that the $\theta$ direction parametrizes $TY/\xi$, and projection along $\partial_{\theta}$ identifies $TY$ with $\xi$. You can then compare this with the contact-form-invariant discussion of orientations above.
