# Tuesday, January 18 :::{.remark} Refs: - Geiges, Intro to Contact - Ozbogi-Stipsicz - Etnyre lecture notes - Massot - Sivck ::: :::{.definition title="Standard contact structure"} For $S^3 \subseteq \CC^2$, define a form on $\RR^4$ as \[ \alpha \da -y_1 dx_1 + x_1 dy_1 - y_2 dx_2 + x_2 dy_2 .\] Then then **standard contact form** on $S^3$ is \[ \xi_{\std} \da \ker \ro{\alpha}{S^3} .\] ::: :::{.exercise title="?"} Show that $\alpha$ defines a contact form. ::: :::{.solution} Write $f = x_1^2 + y_1^2 + x_2^2 + y_2^2$, then \[ \ro{ \alpha}{S^3} \wedge \ro{ d\alpha}{S^3} > 0 \iff df\wedge d \alpha \wedge d \alpha > 0 .\] Check that - $d\alpha = 2(dx_1 \wedge dy_1) + 2(dx_2 + dy_2)$ - $df = 2(x_1 dx_1 + y_1 dy_1) + 2(x_2 dx_2 + y_2 dy_2)$. ::: :::{.remark} Note that at $p= \tv{1,0,0,0} \subseteq S^3$, $\T_pS^3 = \spanof\ts{\del y_1, \del x_2, \del y_2}$. and $\alpha_p = -0 dx_1 + 1dy_1 -0 dx_2 + 0dy_2 = dy_1$ and $\xi_p = \ker dy_1 = \spanof\ts{\del x_1, \del y_2}$. ![](figures/2022-01-18_11-29-05.png) Then $\xi_p \leq \T_p \CC^2 = \spanof\ts{\del x_1, \del y_1, \del x_2, \del y_2} \cong \CC^4$ is a distinguished complex line. ::: :::{.definition title="Almost complex structures"} An **almost complex structure** on $X$ is a bundle automorphism $J: \T X\selfmap$ with $J^2 = -\id$. ::: :::{.example title="?"} For $X = \CC^2$, take \[ \del x_1 &\mapsto \del y_1 \\ \del y_1 &\mapsto - \del x_1 \\ \del x_2 &\mapsto \del y_2 \\ \del y_2 &\mapsto -\del x_2 .\] ::: :::{.exercise title="?"} Show that $f: \CC\to \CC$ is holomorphic if $df \circ J = J \circ df$, which corresponds to the Cauchy-Riemann equations. ::: :::{.lemma title="?"} Given $J:W\to W$, an $\RR\dash$subspace $V \leq W$ is a $\CC\dash$subspace iff $J(V) = V$. ::: :::{.definition title="?"} The field of $J\dash$complex tangents is the hyperplane field \[ \xi_p \da \T S^3 \intersect J(\T S^3) .\] ::: :::{.example title="?"} Consider $\T_p S^3$ for $p=\tv{1,0,0,0}$, then \[ J(\spanof\ts{\del y_1, \del x_1, \del y_2}) = \spanof\ts{-\del x_1, \del y_2, -\del x_2} ,\] so $\xi_p = \spanof{ \del x_1, \del y_2}$ is the intersection and coincides $\xi_{\std}$. ::: :::{.question} Where does $\alpha$ come from? Let $\rho = \sum x_i \del x_i + \sum y_i \del y_i$ be the radial vector field, so $\rho = {1\over 2}\grad\tv{\sum x_i^2 + \sum y_i^2}$. Setting $\omega \da \Wedge dx_i \wedge \Wedge dy_i$, then $\alpha = \iota_p\omega \da \omega(p, \wait)$ is the interior product of $\omega$. Then \[ \alpha = dx_1 \wedge dy_1 (x_1 \del x_1 + y_1 \del y_1 + \cdots ) + \cdots = x_1 dy_1 -y_1 dx_1 + \cdots .\] So the contact form comes from pairing the symplectic form against a radial vector field. ::: :::{.remark} Recall $f \da \sum x_i^2 + \sum y_i^2$ satisfies $df = 2\sum x_i dx_i + 2\sum y_i dy_i$. Note that $J$ acts on 1-forms by $J^*(dx)(\wait) = dx(J(\wait))$. For $J = i$, - $\delta x: dx(J \del x) = dx(\del y) = 0$, - $\del y: dx(J \del y) = dx(-\del x) = -1$. So $J^*(dx) = -dy$, and \[ J^*(df) = 2x_1 (-dy_1) + 2y_1 (dx_1) + 2x_2 (-dy_2) + 2y_2 (dx_2) = -2 \alpha .\] Thus $J^*(df)$ is a rotation of $df$ by $\pi/2$. ::: :::{.example title="?"} The field of complex tangencies along $Y = f\inv (0)$ is the kernel of $\ro{ df(J(\wait)) } {Y}$. ::: :::{.remark} Methods of getting contact structures: for a vector field $X$, being contact comes from $\mcl_X \omega = \omega$. For functions $f:\CC^2 \to \RR$, being contact comes from $\alpha = d^\CC f$ being contact. See strictly plurisubharmonic functions and Levi pseudoconvex subspaces. ::: :::{.example title="?"} The standard contact structure is orthogonal to the Hopf fibration: define a map \[ \CC^2\smz &\to \CP^1 \cong S^2 \\ \tv{z, w} &\mapsto \tv{z: w} ,\] which restricts to a map $S^3\to S^2$ defining the Hopf fibration. If $L$ is a complex line through 0, then $L \intersect S^3$ is a Hopf fiber that is homeomorphic to $S^1$. ![](figures/2022-01-18_12-11-38.png) ![](figures/2022-01-18_12-15-08.png) ![](figures/2022-01-18_12-17-02.png) Take \[ \CC^2 &\to \RR^2 \\ (z_1, z_2)&\mapsto (\abs{z_1}, \abs{z_2}) .\] Consider the image of $S^2 = \ts{\abs{z_1}^2 + \abs{z_2}^2 = 1}$: ![](figures/2022-01-18_12-20-49.png) The preimage is $S^1\times S^1$. This can be realized as a tetrahedron with sides identified: ![](figures/2022-01-18_12-24-27.png) There are Hopf fibers on the ends, and undergo a $\pi/2$ twist as you move through the tetrahedron. ![](figures/2022-01-18_12-28-04.png) :::