# Darboux and Gromov Stability (Thursday, January 20) :::{.remark} Almost-complex structures: weaker than an actual complex structure, but not necessarily integrable. Useful for studying pseudoholomorphic curves. A necessary and sufficient condition for integrability: the Nijenhuis tensor $N_J = 0$ iff $J$ is integrable. In real dimension 2, all $J$ are integrable. ::: :::{.theorem title="Darboux"} If $(Y^3, \xi)$ is contact then for every point $p$ there is a chart $U$ with coordinates $x,y,z$ where $\xi = \ker (\dz -y\dx) = \ker (\alpha_\std)$. ::: :::{.slogan} Locally, all contact *structures* (not necessarily forms) look the same. The mantra: local flexibility vs global rigidity. ::: ## Proof of Darboux :::{.remark} Two proofs: - Geometric, due to Giroux - PDEs, which generalizes. This uses Moser's trick. ::: :::{.proof title="1"} Locally write $\xi = \ker \alpha$ with $\alpha \wedge d \alpha > 0$. Pick a contact plane $\xi_p$ and let $S$ be a transverse surface, so $\T_p S \transverse \xi_p$. This produces a set of curves in $S$ which are tangent to $\xi_p$ everywhere, called the *characteristic foliation*. ![](figures/2022-01-20_11-24-33.png) Then $\ro{\alpha}{S} = \dz$, which is a 1-form that is nonvanishing near $p$ and is locally integrable. Sending $\alpha \to X$ a vector field along $S$ yields a set of integral curves tracing out the characteristic foliation. This yields an $x$ direction and a $z$ direction on $S$ by flowing $t\in (-\eps, \eps)$ around $p$ along $X$. Choose a vector field $\del t$ which is transverse to $S$ and contained in $\xi$. Then $\alpha(\del t) = 0$, so we can write \[ \alpha = f\dx + g\dz + h\dt = f\dx + g\dz .\] Since $g(p) = 1$, replace $\alpha$ with ${1\over g}\alpha$ which is positive near $p$ and doesn't change the contact structure $\xi$. So write \[ \alpha = f\dx + \dz \implies \alpha \wedge d \alpha = \alpha \wedge \qty{ f_t \dy\wedge \dx + f_z \dz \wedge \dx } = -f_t \dx \wedge \dt \wedge \dz > 0 ,\] meaning $f_t <0$ and we can set $y = f(x,z,t)$. This yields \[ \alpha = \dz + f\dx = \dz - y\dx .\] ::: :::{.proof title="2, Moser's Trick"} By a linear change of coordinates, choose $x,y$ along $\xi$ to write $\alpha_p = \dz$ and $\xi_p = \spanof{\del x, \del y}$: ![](figures/2022-01-20_11-43-26.png) Write $(\alpha_0)_p$ for the original form and $\alpha_1 = \dz - y\dx$ the standard form, then the claim is that $\alpha_0 \homotopic \alpha_1$ through a path of contact forms. :::{.lemma title="?"} In a neighborhood of $p$, there is a family $\alpha_t$ for $t\in [0, 1]$. To obtain this, interpolate: \[ d\alpha_t = t d\alpha_1 + (1-t) d\alpha_0 \implies \alpha_t \wedge d\alpha_t = t^2 \alpha_1 \wedge d\alpha_1 + t(1-t) (\alpha_0 \wedge d\alpha_1 + \alpha_1 \wedge d\alpha_0) + (1-t)^2 \alpha_0 \wedge d\alpha_0 .\] The first and last terms are positive since the $\alpha_i$ are contact. For the middle term, $\alpha_0 = \alpha_1$ near $p$, so by continuity this is positive in some neighborhood of $p$. ::: :::{.remark} Note that $\dot\alpha_t \da \dd{}{t} \alpha_t$, so \[ \dd{}{t} \qty{ t\alpha_1 + (1-t) \alpha_0} = \alpha_1 - \alpha_0 .\] ::: We'll assume that there is a time-dependent vector field $V_t \in \xi_t$ with flow $\Phi_t$ such that $(\Phi_t)_*(\xi_t) = \xi_0$. We'll also require $\xi_t = \ker \alpha_t$, so this is a contactomorphism for each $t$. The goal is to show $(\Phi_1)_*(\xi_0) = \xi_1$, or equivalently $\Phi_t^* \alpha_t = f_t \alpha_0$ with $f_t > 0$. Take $\dd{}{t}$ of both sides here to get \[ \Phi_t^*( \dot \alpha_t + \mcl_{V_t} \alpha_t ) = \dot f_t \alpha_0 .\] > See Prop 6.4 in Cannas da Silva. :::{.remark} ✨Cartan's magic formula✨: \[ \mcl_V(\alpha) = d(\iota_V \alpha) + \iota_V(d\alpha) ,\] so \[ \mcl_{V_t}(\alpha_t) = d(\alpha_t(V_t)) + d\alpha_t(V_t, \wait) = 0 + d\alpha_t(V_t, \wait) .\] ::: We can thus write this equation as \[ \Phi_t^*(\dot \alpha_t + d\alpha_t(V_t, \wait)) = \dot f_t \alpha_0 = \dot f_t \qty{\Phi_t^*(\alpha_t) \over f_t } .\] Applying $(\Phi_t^*)\inv$ yields \[ \dot \alpha_t + d\alpha_t(V_t, \wait)= {\dot f_t \over f_t }\alpha_t .\] Now try to solve this for $V_t$. Let $R_t$ be the **Reeb vector field** of $\alpha_t$, which satisfies - $\alpha_t(R_t) = 1$ - $d\alpha_t(R_t, \wait) = 0$. ![](figures/2022-01-20_12-11-26.png) Then \[ \dot \alpha_t (R_t) = {\dot f_t \over f_t} = \dd{}{t} \log(f_t) \da \mu_t ,\] so $\dot\alpha_t(R_t)$ determines $f_t$ by first integrating and exponentiating. We now need to solve \[ \ro {d\alpha_t(V_t, \wait)}{\xi_t} = \ro{\mu_t \alpha_t - \dot \alpha_t}{\xi_t} .\] Since \( d \alpha_t \) is a volume form on $\xi_t$, it identifies vector fields in $\xi_t$ with 1-forms on $\xi_t$ using the happy coincidence that $n=2$ so $1\mapsto n-1 = 1$. So $V_t$ is uniquely determined by the solution to the above equation. :::