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Nov 1, 2021 at 21:20 comment added KSackel To clarify, the quotient is therefore just in bijection with primitive elements in $\mathbb{Z}^3$ (given by the image of $\gamma$).
Nov 1, 2021 at 20:53 comment added KSackel Because contact isotopy generates contactomorphisms, we have that up to contact isotopy, tight contact structures up to isotopy are in bijection with $MCG(T^3)/\mathrm{Stab}(\xi_0)$. On the one hand, $MCG(T^3) = GL_3(\mathbb{Z})$. On the other hand, given $\xi_0$, $T^3$ comes with a canonical generator $\gamma \in H_1(T^3)$ coming from representing the unique class nulhomologous with respect to the Stein filling, and $\mathrm{Stab}(\xi_0)$ consists precisely of those elements fixing this class, i.e. $\mathrm{Stab}(\xi_0) = \mathbb{Z}^2 \rtimes \mathrm{GL}_2(\mathbb{Z})$.
Nov 1, 2021 at 19:34 comment added magicker72 These are all stated up to contactomorphism. Up to contact isotopy, there are infinitely distinct many tight contact structures that are all contactomorphic to the standard Stein fillable one.
Oct 30, 2021 at 5:10 history answered KSackel CC BY-SA 4.0