# Tuesday, February 01 :::{.remark} Last time: front diagrams $\tv{x,y,z}\mapsto \tv{x,z}$, where $\alpha = \dz -y\dx$ forces $y=\ds/\dx$ can be recovered as the slope in the projection. Note that we can also recover crossing information from the Legendrian condition, since $y$ always points into the board, so more negative slopes go on top. Some invariants: - Thurston-Bennequin invariant: a contact framing with respect to the Reeb vector field ![](figures/2022-02-01_11-15-00.png) - Equal to writhe minus half the number of cusps. - Rotation numbers: Turning number of $L$ with respect to $\xi$, after fixing a trivialization of $\xi$. Equal to ${1\over 2}(D-U)$, the number of down/up cusps respectively. ::: :::{.remark} Disallowed moves: ![](figures/2022-02-01_11-20-07.png) Allowed moves: ![](figures/2022-02-01_11-24-46.png) ::: :::{.remark} Geography problem: given a smooth knot $K$, which pairs $(t, r) \in \ZZ^2$ are realized as $(\mathrm{tb}(L), \rot(L))$ for $L$ a Legendrian representative of $K$? Botany problem: given $(t, r) \in \ZZ^2$, how many inequivalent $L$ representing $K$ realize $(t,r) = (\mathrm{tb}(L), \rot(L))$? ::: :::{.example title="?"} For $K$ the unknot: ![](figures/2022-02-01_11-37-46.png) So these numerical pairs fall into a cone: ![](figures/2022-02-01_11-39-30.png) ::: :::{.proposition title="?"} For $L \subseteq \RR^3$ a Legendrian knot, \[ \mathrm{tb}(L) + \rot(L) \equiv 1 \mod 2 .\] ::: :::{.remark} Note that $\chi(S) \equiv 1\mod 2$ for $S$ a Seifert surface. ::: :::{.theorem title="Bennequin-Thurston inequality"} For any Seifert surface $S$, \[ \mathrm{tb}(L) + \abs{ \rot(L) } \leq -\chi(S) .\] ::: :::{.remark} This solves the geography problem: this cone contains all of the possible pairs. ::: :::{.theorem title="Eliashberg-Fraser"} The unknot is **Legendrian simple**: if $\mathrm{tb}(L_1) = \mathrm{tb}(L_2)$ and $\rot(L_1) = \rot(L_2)$, then $L_1$ is isotopic to $L_2$. ::: :::{.remark} This solves the botany problem: every red dot has exactly one representative. ::: :::{.remark} Other knots are Legendrian simple, e.g. the trefoil. A theorem of Checkanov says the following $5_2$ knots are not Legendrian isotopic: ![](figures/2022-02-01_11-53-16.png) ::: :::{.remark} This all depended on the standard contact form. Consider instead the overtwisted disc: take $\RR^3$ with $\alpha = \cos(r)\dz + \sin(r) \dtheta$. Take the curve $\tv{r,\theta, z} = \gamma(t) \da \tv{1, t, 0}$, a copy of $S^1$ in the $x,y\dash$plane. Then $\gamma' = \tv{0,1,0}$, and at $\theta=\pi, \alpha = \cos(\pi)\dz + \sin(\pi) \dtheta = -\dz$, but at $r=0$ $\alpha = \dz$, so traversing a ray from $0$ to $-1$ in the $x,y\dash$plane forces the contact plane to flip: ![](figures/2022-02-01_12-02-08.png) One can check that $\mathrm{tb}$ is given my $\lk(L, L') = 0$ where $L'$ is a pushoff of $L$, and can be made totally disjoint from $L$ in this case by moving in the $z\dash$plane. ::: :::{.definition title="Overtwisted discs"} An **overtwisted disc** in $(Y^3, \xi)$ that is locally contactomorphic to this local model. $Y$ is **overtwisted** if it contains an overtwisted disc, and is tight otherwise. ::: :::{.theorem title="Bennequin"} $(\RR^3, \xi_\std)$ is a tight contact structure. ::: :::{.theorem title="Eliashberg"} For every closed oriented $Y^3$, every homotopy class of 2-plane fields on $Y$ contains a unique (up to isotopy) overtwisted contact structure. ::: ## Transverse Knots :::{.definition title="Self-linking"} The **self-linking number** $\mathrm{sl}(T, S)$ of a transverse knot rel a Seifert surface $S$ is $\lk(T, T')$ for $T'$ a pushoff of $T$ determined by a trivialization of $\ro{\xi}{S}$. ::: :::{.remark} In this case, $\xi$ restricts to an $\RR^2$ bundle over $\Sigma$, which is trivial since $\Sigma$ is closed with boundary and $e(\xi) \in H^2(S) = 0$. To see this, use $H^2(S) \cong H_0(S, \bd S) = 0$ by Lefschetz duality. This yields a section of the frame bundle over $S$, which gives a pushoff direction along the first basis vector: ![](figures/2022-02-01_12-27-33.png) This turns out to be well-defined: it's independent of the surface $S$ chosen and the trivialization of $\xi$. The difference of two trivializations gives a map $\pi_1(S) \to \ZZ$, which factors through $\pi_1(S)^\ab = H_1(S)$. The difference in surfaces is measured by $\inp{e(S)}{ \Sigma_1 \disjoint_T \Sigma_2 }$, which is a glued surface. :::