# Thursday, February 03 :::{.remark} Last time: self-linking of transverse knots. Today: surfaces with transverse boundary. Let $\Sigma$ be a surface embedded in $(Y, \xi)$ with $\bd \Sigma$ transverse to $\xi$. Let $F$ be the characteristic foliation, the singular foliation on $\Sigma$ induced by $\ro{\xi}{\Sigma}$. Equivalently, if $\xi = \ker \alpha$, consider the 1-form $\ro{\alpha}{\Sigma}$. Generically, $\ro{\ker \alpha}{\T\Sigma}$ is 1-dimensional except at finitely many points where $\alpha_p = 0$, i.e. $\xi$ is tangent to $\Sigma$. This line field integrates to a singular foliation. Recall that $\mathrm{sl}(L)$ is the self-linking number. ::: :::{.example title="?"} Take $\alpha = \dz +x\dy - y\dx$ and $\Sigma = S^2$, then the singular foliation is given by ![](figures/2022-02-03_11-21-06.png) ::: :::{.remark} Two possible types of singularities, the local models: ![](figures/2022-02-03_11-24-34.png) There are also two numerical invariants: - $e_{\pm}$: the number of positive (resp. negative) elliptics - $h_{\pm}$: the number of positive (resp. negative) hyperbolics A theorem \[ \inp{c(\Sigma)}{ \Sigma} = (e_+ - h_+) - (e_- - h_-) .\] If $\Sigma$ is transverse, $\mathrm{sl}(\bd\Sigma, \Sigma) = -(e_+ - h_+) + (e_- - h_-)$. ::: ## Local Model 1: Elliptic :::{.remark} $\sigma$ is the $x,y\dash$plane and $\xi = \ker (\dz + x\dy - y\dx)$ with $\ro \alpha \Sigma = x\dy - y\dx$. Set $V: x\del_x + y\del_y$ and $L' = \gens{x\del_y - y\del_x}$, and $\alpha(i) = x^2+y^2 = 1 > 0$. ![](figures/2022-02-03_11-32-36.png) Here $\mathrm{sl} = 1$. To compute $\mathrm{sl}$: - Trivialize $\ro{\xi}{\Sigma}$ to get $\tau = \gens{e_1, e_2}$ a fiberwise basis for $\xi$. - Let $\tilde L$ be a pushoff in the $e_1$ direction. - Compute $\mathrm{sl} = \lk(L, \tilde L)$. Set - $e_1 = \del_x + y\del_z$ - $e_2 = \del_y - x\del_z$ - $\rho = x \del_x + y\del_y$ - $\theta = x\del_y - y\del_x$. Then \[ x\rho - y\theta = x(x\del_x + y\del_y) - y(-y \del_x + x\del_y) = (x^2+y^2)\dx .\] Then - $c_1 = x\rho - y\theta + y\del_z$ - $\bar{c_1} = x\rho + y\del_z = \cos(\rho) + \sin(\theta) \del_z$. Example: - $\theta = 0\implies e_1 = \rho$ - $\theta = \pi/4 \implies e_1 = {\sqrt 2\over 2}(\rho + \del_z)$ - $\theta = \pi/2 \implies e_1 = \del_z$ So here $\lk(U, \tilde U) = -1$: ![](figures/2022-02-03_11-43-46.png) ::: ## Local Model 2: Hyperbolic :::{.remark} Here $\xi$ is the $x,y\dash$plane, so $\xi = \ker (\dz + 2x\dy + y\dx)$ with $\ro{\alpha}{\Sigma} = 2x\dy + y\dx$ and $V = y\del_y + dx\del_x\in \ker(\ro \alpha \Sigma)$. ::: :::{.remark} The Euler class of a real vector bundle $E \mapsvia{\pi} X$ is the obstruction to finding a nonvanishing section $s$ of $E$, given by $e(E) \in H^k(X)$. It is Poincare dual to $[s\inv(0)] \in H_{n-k}(X, \bd X)$. For the tangent bundle, $e(\T X)\in H^{n}(X)$, and \[ \inp{e(\TX)}{[X]} = \chi(X) .\] Since a section of $\T X$ is a vector field, $e(\T X)$ is an obstruction to finding a nonvanishing vector field. If $\bd X \neq \emptyset$ and $t$ is a section of $\ro{E}{\bd X}$, there is a relative Euler class $e(E, t)\in H^k(X, \bd X) \cong H_{n-k}(X)$. Similarly, \[ \inp{e(\T X, t)}{[X]} = \chi(X) .\] ::: :::{.example title="?"} Note $\chi(\DD) = 1$, so any vector field has a singularity? ![](figures/2022-02-03_11-59-11.png) ::: :::{.proposition title="?"} The total class is the sum of the relative obstructions. If $\sigma = \Sigma_1 \glue{\bd} \Sigma_2$ and $\tau$ is a nonvanishing section of $\ro{\Sigma}{\bd \Sigma_1} = \ro{\Sigma}{\bd\Sigma_2}$, then \[ c(E) = e(\ro{E}{\Sigma_1}, \tau) + c(\ro E {\Sigma_2}, \tau) .\] ![](figures/2022-02-03_12-01-38.png) ::: ## More Contact Geometry :::{.remark} Let $\Sigma$ have transverse boundary with characteristic foliation $F$, and let $V$ be the vector field directing $F$, so $V \in \xi \intersect \T \Sigma$. We can assume $V$ is outward-pointing along $\bd \Sigma$. Check that - $\chi(\Sigma) = e(\T\Sigma, V) \in H^2(\Sigma, \bd \Sigma) \cong H_0(\Sigma)$ - $\mathrm{sl}(\bd\Sigma, \Sigma) = e(\xi, V) \in H^2(\Sigma, \bd \Sigma)$ ::: :::{.fact} \envlist - $e_+ + e_-$ correspond to $+1$ in $e(\T\Sigma, V)$, - $h_+, h_-$ correspond to $-1$ in $e(\T\Sigma, V)$. Proof: near a zero, $V$ determines a map $S^1\to S^1$ and the contribution to $e$ is the degree of this map. - $e_+$ contributes $-1$ to $e(\xi, V)$, by the same computation of $\mathrm{sl}(U)$ for $U$ the unknot. - $e_-$ contributes $-(-1) = +1$ to $e(\xi, V)$. - $h_+$ contributes $+1$ to $e(\xi, V)$ - $h_-$ contributes $-(+1) = -1$ to $e(\xi, V)$. Proof: exercise. ::: :::{.remark} Bennequin inequality: \[ \mathrm{sl}(T, \Sigma) \leq -\chi(\Sigma) \implies e_+ + h_+ + e_- + h_- \leq -(e_+ + e_- - h_+ - h_-) \iff e_- \leq h_- .\] Try to cancel in pairs: ![](figures/2022-02-03_12-24-10.png) The inequality follows if we can cancel every $e_-$ with some $h_-$. :::