# Gray Stability (Tuesday, January 25)

:::{.remark}
A homotopy of contact structures o $Y^3$ is a smooth family $\ts{\phi_t}$ of contact structures.
Similarly, an **isotopy** of structures such that $\ts{D\phi_t(\xi_0)}$ for an isotopy $\phi_t: Y\to Y$ with $\phi_0 = \id$.
If $Y^3$ is closed then every homotopy of contact structures is an isotopy.
Theorem: contact structures mod isotopy is discrete, which critically uses closedness.
:::

:::{.lemma title="?"}
For $\phi_t$ an isotopy generated by the flow of $X_t$ and $\alpha_t$ a family of 1-forms,
\[
\dd{}{t} \phi^*_t(\alpha_t) \mid_{t=t_0} = \phi_{t_0}^*(\dot \alpha_{t_0} + \mcl_{X_{t_0}} \alpha_{t_0} )
.\]
:::

:::{.proof title="?"}
Write
\[
\phi_x^*(\alpha_y) = \dd{}{x} ? + \dd{}{y}? =  \phi_{x_0}^* \mcl_{X_0} \mcl_X \alpha_{y_0} + \phi_{x_0}^* \alpha_y
,\]
and proceed similarly to the proof of Darboux's theorem.

Pick $\ts{\phi_t}$ a homotopy, one can choose $\alpha_t$ with $\xi_t = \ker \alpha_t$ for all $t$.
Apply Moser's trick: assume there exists a $\phi_t$ with $\phi_t^*(\alpha_t) = \lambda_t \alpha_0$ and try to find $v_t$ generating it, where $\lambda_t: Y\to \RR_+$.
What does $\phi_t$ need to look like?
Differentiate in $t$:
\[
\phi^*_{t_0}(\cdot \alpha_{t_0} + \mcl_{V_{t_0}} \alpha_{t_0} ) = \dot \lambda_t \alpha_0 = \dot \lambda_t \qty{ \phi^*_{t_0} (\alpha_t) \over \lambda_t }
.\]
Apply $(\phi^*_{t_0})\inv$:
\[
\cdot\alpha_t + \mcl_{V_t}\alpha_t = \mu_t \alpha_t\qquad \mu_t = (\phi^*_{t_0})\inv(\dot\lambda_t \over \lambda_t)
.\]
Use that $V_t$ is always tangent to the contact structure, so $V_t \in \xi_t$, to assume $\alpha_t(V_t) =0$.
Apply Cartan:
\[
\dot \alpha_t + d\alpha_t(V_t) + \iota_{V_t} d\alpha_t = \mu_t \alpha_t
,\]
and $d\alpha_t(V_t) = 0$, so
\[
\iota_{V_t}d\alpha_t = \mu_t \alpha_t - \dot \alpha_t
.\]
Plug in the Reeb vector field $R_t$, then $\alpha_t(R_t) = 0$ so $\mu_t = \dot \alpha_t (R_t)$.

:::

:::{.corollary title="?"}
Let $Y$ be n $S^3 \subseteq \CC^2$ that is transverse to the radial vector field.
Then 
\[
\alpha = x_1 dy_1 - y_1 dx_1 + x_2 dy_2 - y_2 dx_2\mid_y
\]
defines the standard tight contact structure.
:::

:::{.proof title="?"}
Write $Y \subseteq \RR\times S^3$ in coordinates $(f(x), x)$ as the graph of a function $f: S^3\to \RR$.
Take an isotopy $Y_t = (tf(x), x) \subseteq \RR\times S^3$ to get a family of contact forms where $\alpha_0 = \alpha_\std$ and $\alpha_1$ is some unknown form.
By Gray stability, the contact structures are isotopic.
:::

## Legendrian Links

:::{.definition title="Legendrian and transverse knots"}
Let $Y$ be a contact 3-manifold and $L \injects Y$ a link.
Then $L$ is a **Legendrian knot** iff it is everywhere tangent to $\xi$, so $\alpha(L) = 0$:

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![](figures/2022-01-25_12-10-49.png)

This is a closed condition.


$L$ is **transverse** if it is everywhere transverse to $\xi$, so $\alpha(L) > 0$:

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![](figures/2022-01-25_12-14-25.png)

This is an open condition.
:::

:::{.remark}
Every Legendrian knot has a transverse pushoff (up to transverse isotopy).
Every transverse knot has a Legendrian approximation.
:::

:::{.example title="?"}
Take $\RR^3$ and $\alpha_\std = dz-ydx$, then the $y\dash$axis $L_1 \da\ts{\tv{0,t,0}}$ is Legendrian.
Similarly the $x\dash$axis $L_2$ is Legendrian, checking that $\T L_2 = \spanof\ts{\tv{1,0,0}}$.
However the slight pushoff $L_3 \da \ts{\tv{t, -\eps, 0}}$ is transverse since $\ro{\alpha}{L_3} = \eps dx >0$.
:::

:::{.theorem title="Neighborhood theorem, Darboux for Legendrian/transverse knots"}
Every Legendrian has a neighborhood contactomorphic to the zero section in $J_1 S^1 = \T S^1 \cross \RR$.
Every transverse has a neighborhood contactomorphic to the $z\dash$axis in $\RR\times S^1$ with $\alpha \da \dz + r^2 \dtheta$.
:::