# Thursday, February 24

:::{.remark}
Last time: there is a unique tight contact structure on $S^3$, using the existence of a contact structure on $S^3\times I$.
Next: tight contact structures on

- $T^2\times I$
- $S^1\times \DD^2$
- $L(p, q)$
- $T^3$

Given dividing sets of $\Gamma_0, \Gamma_1 \in T^2\times I$, how can contact structures vary in a family.
Tightness implies no contractible components in $\Gamma$, so $\Gamma$ consists of $2n$ embedded curves of slow $p/q$.
So the dividing set is governed by two parameters.
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:::{.remark}
The only change to the dividing set in a generic family can be:

- Retrograde saddle-saddle, yielding by pass moves.

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![](figures/2022-02-24_11-26-51.png)

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:::{.proposition}
Given any contact structure on $\Sigma \times I$ with dividing sets $\Gamma_0, \Gamma_1$, $\xi$ is determined by a finite number of bypass moves.
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:::{.proof title="?"}
Diagrams?

$\vdots$

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:::{.remark}
Given $\Gamma_0$ with slope $p/q$ and $\Gamma_1$ with slope $r/s$, form a Farey graph:

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![](figures/2022-02-24_12-00-24.png)


$\vdots$

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:::{.proposition title="Legendrian Darboux"}
If $L$ is a Legendrian knot in $M$, then a neighborhood of $L$ is contactomorphic to a neighborhood of a zero section in $J(S^1) \cong \RR\times \T\dual S^1 \cong S^1\times \RR^2$..
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:::{.remark}
Write this in coordinates as $(z, (x, y))$, so $\alpha = \dz -y\dx$ with $x\in \RR/\ZZ$.
Then $v(L) = \ts{y^2+z^2\leq \eps}$, $y=r\cos\theta, z=r\sin\theta$.
$T^2 = \ts{x, \theta}, \ro{\alpha}{T^2} = y\dtheta -y\dx = \eps\cos\theta (\dtheta - \dx)$.
Unwrap:

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![](figures/2022-02-24_12-15-41.png)

Note that $\dalpha > 0$ at $\pi/2$ and $\dalpha < 0$ at $3\pi/2$.
Idea: given two unrelated surfaces with their own foliations, how do they interact at the boundary?
Dividing sets on each can be extended into the annulus, and this reduces to a combinatorial problem of how to connected arcs:

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![](figures/2022-02-24_12-26-17.png)

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