# Tuesday, January 11 ::: {.remark} References: - - - [PLC's Notes](https://outlookuga-my.sharepoint.com/:f:/g/personal/pbl20394_uga_edu/EjT_H9wRyAhAsZodxgXOgU0BSIYbqWav8X1jZY5v3RxqJA?e=n6dfVJ) Emphasis for the course: applications to low-dimensional topology, lots of examples, and ways to construct contact structures. The first application is critical to 4-manifold theory: ::: ## Application 1 ::: {.theorem title="Cerf's Theorem"} Every diffeomorphism \( f: S^3\to S^3 \) extends to a diffeomorphism \( {\mathbb{B}}^4\to {\mathbb{B}}^4 \). ::: ::: {.remark} This isn't true in all dimensions! This is essentially what makes Kirby calculus on 4-manifolds possible without needing to track certain attaching data. ::: ::: {.remark} There is a standard contact structure on \( S^3 \): regard \( {\mathbb{C}}^2 \cong {\mathbb{R}}^4 \) and suppose \( f:S^3\to S^3 \). There is an intrinsic property of contact structures called *tightness* which doesn't change under diffeomorphisms and is fundamental to 3-manifold topology. ::: {.theorem title="Eliashberg"} There is a unique tight contact structure \( \xi_\text{std} \) on \( S^3 \). ::: So up to isotopy, \( f \) fixes \( \xi_\text{std} \). ::: ::: {.remark} A useful idea: tiling by holomorphic discs. This involves taking \( S^1 \) and foliating the bounded disc by geodesics -- by the magic of elliptic PDEs, this is unobstructed and can be continued throughout the disc just using convexity near the boundary. In higher dimensions: \( {\mathbb{B}}^4 \) is foliated by a 2-dimensional family of holomorphic discs. ::: ## Application 2 ::: {.remark} Another application: monotonic simplification (?) of the unknot. Given a knot \( K \hookrightarrow S^3 \), a theorem of Alexander says \( K \) can be braided about the \( z{\hbox{-}} \)axis, which can be described by a word \( w\in B_n \), the braid group \[ B_n = \left\{{ \sigma_1,\cdots, \sigma_{n-1} {~\mathrel{\Big\vert}~}[\sigma_i, \sigma_j] = 1 \,{\left\lvert {i-j} \right\rvert}\geq 2,\, \sigma_i \sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i \sigma_{i+1}\, i=1,\cdots, n-2 }\right\} .\] This captures positive vs negative braiding on nearby strands, commuting of strands that are far apart, and the Reidemeister 3 move. Write \( K = K(\beta) \) for \( \beta \) a braid for the braid closure. ::: ::: {.remark} Markov's theorem: if \( K = K(\beta_1), K(\beta_2) \) where \( \beta_1 \in B_n \) and \( \beta_2\in B_m \) with \( m,n \) not necessarily equal, then there is a sequence of Markov moves \( \beta_1 \) to \( \beta_2 \). The moves are: - Stabilization and destabilization: ![](figures/2022-01-11_11-41-29.png) ```{=html} ``` - Conjugation in \( B_n \): ```{=html} ``` ![](figures/2022-01-11_11-39-52.png) - Braid isotopy, which preserves braid words in \( B_n \). ::: ::: {.remark} A theorem of Birman-Menasco: if \( K(\beta) = U \) is the unknot for \( \beta \in B_n \), then there is a sequence of braids \( \left\{{\beta_i}\right\}_{i\leq k} \) with \( \beta_k = 1 \in B_1 \) such that - \( \beta_i\in B_{n_i} \) - \( K(\beta_i) = U \) - \( n_1\geq n_2\geq \cdots \geq n_k = 1 \) - \( \beta_i \to \beta_{i+1} \) is either a Markov move or an exchange move. ![](figures/2022-01-11_11-46-31.png) ::: ## Application 3 ::: {.remark} Genus bounds. A theorem due to Thurston-Eliashberg: if \( \xi \) is either a taut foliation or a tight contact structure on a 3-manifold \( Y \) and \( \Sigma \neq S^2 \) is an embedded orientable surface in \( Y \), then there is an Euler class \( e(\xi) \in H^2(Y) \). Then \[ {\left\lvert { {\left\langle { e(\xi)},~{ \Sigma } \right\rangle} } \right\rvert} \leq -g(\Sigma) ,\] which after juggling signs is a lower bound on the genus of any embedded surface. ::: ::: {.remark} Taut foliations: the basic example is \( F\times S^1 \) for \( F \) a surface. The foliation carries a co-orientation, and the tangencies at critical points of an embedded surface will have tangent planes tangent to the foliation, so one can compare the co-orientation to the outward normal of the surface to see if they agree or disagree and obtain a sign at each critical point. Write \( c_\pm \) for the number of positive/negative elliptics and \( h_\pm \) for the hyperbolics. Then \[ \chi = (e_+ + e_-) - (h_+ + h_-) ,\] by Poincaré-Hopf. On the other hand, \( {\left\langle {e(\xi)},~{\Sigma} \right\rangle} = (e_+ - h_+) - (e_- - h_-) \), so adding this yields \[ {\left\langle {e(\xi)},~{\Sigma} \right\rangle} +\chi = 2(e_+ - h_+) \leq 0 .\] Isotope the surface to cancel critical points in pairs to get rid of caps/cups so that only saddles remain. ::: ## Contact Geometry ::: {.definition title="?"} A contract structure on \( Y^{2n+1} \) is a hyperplane field (a codimension 1 subbundle of the tangent bundle) \( \xi = \ker \alpha \) such that \( \alpha \wedge (d\alpha){ {}^{ \scriptscriptstyle\wedge^{n} } } > 0 \) is a positive volume form. ::: ::: {.example title="?"} On \( {\mathbb{R}}^3 \), \[ \alpha = dz - ydx \implies d\alpha = -dy \wedge dx = dx\wedge dy ,\] so \[ \alpha \wedge d\alpha = (dz-ydx)\wedge(dx\wedge dy) = dz\wedge dx\wedge dy = dx\wedge dy\wedge dz .\] ::: ::: {.exercise title="?"} On \( {\mathbb{R}}^5 \), set \( \alpha = dz-y_1 dx_1 - y_2 dx_2 \). Check that \[ \alpha\wedge (d\alpha)^2 = 2(dz \wedge dx_1 \wedge dy_1 \wedge dx_2 \wedge dy_2) .\] ::: # Contact Forms and Structures (Thursday, January 13) ::: {.definition title="Contact form"} A **contact form** on \( Y^3 \) is a 1-form \( \alpha \) with \( \alpha \wedge d\alpha > 0 \). A **contact structure** is a 2-plane field \( \xi = \ker \alpha \) for some contact form. ::: ::: {.remark} Forms are more rigid than structures: if \( f>0 \) and \( \alpha \) is contact, then \( f\cdot \alpha \) is also contact with \( \ker( \alpha ) = \ker(f \alpha) \). ::: ## Examples of Contact Structures ::: {.example title="Standard contact structure"} On \( {\mathbb{R}}^3 \), a local model is \( \alpha\coloneqq\,dz- y\,dx \). ::: {.exercise title="?"} Show \( \alpha \wedge d\alpha = dz\wedge dx \wedge dy \). ::: Write \( \xi = \mathop{\mathrm{span}}_{\mathbb{R}}({\partial}y, y {\partial}z + {\partial}x) \), which yields planes with a corkscrew twisting. Verify this by writing \( \alpha = 0 \implies {\frac{\partial z}{\partial x}\,} = y \), so the slope depends on the \( y{\hbox{-}} \)coordinate. ::: ::: {.example title="Rotation of the standard structure"} On \( {\mathbb{R}}^3 \), take \( \alpha_2 \coloneqq dz + x\,dy \) and check \( \alpha_2 \wedge d \alpha_2 = dz \wedge dx \wedge dy \). This is a rigid rotation by \( \pi/2 \) of the previous \( \alpha \), so doesn't change the essential geometry. ::: ::: {.example title="Radially symmetric contact structure"} Again on \( {\mathbb{R}}^3 \), take \( \alpha_3 = dz + {1\over 2}r^2 d\theta \). Check that \( d \alpha_3 = r\,dr\wedge \,d\theta \) and \( \alpha_3 \wedge d \alpha_3 = r\,dz\wedge \,dr\wedge \,d\theta \). Then \( \xi = \mathop{\mathrm{span}}_{\mathbb{R}}({\partial}r, {1\over 2} r^2 {\partial}z + {\partial}\theta ) \). Note that as \( r\to \infty \), the slope of these planes goes to infinity, but doesn't depend on \( z \) or \( \theta \). ::: ::: {.example title="Rectangular version of radially symmetric structure"} Set \( \alpha_4 = \,dz+ {1\over 2}(x\,dy- y\,dx) \), then this is equal to \( \alpha_3 \) in rectangular coordinates. ::: ::: {.example title="Overtwisted"} Set \[ \alpha_4 = \cos(r^2)\,dz+ \sin(r^2)\,d\theta .\] ::: {.exercise title="?"} Compute the exterior derivative and check that this yields a contact structure. ::: Now note that \[ \alpha = 0 \implies {\frac{\partial z}{\partial \theta}\,} = -{\sin(r^2)\over \cos(r^2)} = -\tan(r^2) ,\] which is periodic in \( r \). So a fixed plane does infinitely many barrel rolls along a ray at a constant angle \( \theta_0 \). This is far too twisty -- to see the twisting, consider the graph of \( (r, \tan(r^2)) \) and note that it flips over completely at odd multiples of \( \pi/2 \). In the previous examples, the total twist for \( r\in (-\infty, \infty) \) was less than \( \pi \). ::: ::: {.definition title="Contactomorphisms"} A **contactomorphism** is a diffeomorphism \[ \psi: (Y_1^3, \xi_1) \to (Y_2^3,\xi_2) \] such that \( \phi_*(\xi_1) = \xi_2 \) (tangent vectors push forward). A strict contactomorphism is a diffeomorphism \[ \phi: (Y_1^3, \ker \alpha_1) \to (Y_2^3, \ker \alpha_2) .\] such that \( \phi^*( \alpha_2) = \alpha_1 \) (forms pull back). ::: ::: {.remark} Strict contactomorphisms are more important for dynamics or geometric applications. ::: ::: {.exercise title="?"} Prove that \( \alpha_1,\cdots, \alpha_4 \) are all contactomorphic. ::: ::: {.remark} Recall that \( X \) has a cotangent bundle \( {\mathbf{T}} {}^{ \vee }X \xrightarrow{\pi} X \) of dimension \( 2 \dim X \). There is a canonical 1-form \( \lambda \in \Omega^1({\mathbf{T}} {}^{ \vee }X) \), i.e. a section of \( T {}^{ \vee }(T {}^{ \vee }X) \). Given any smooth section \( \beta\in \Gamma({\mathbf{T}} {}^{ \vee }X_{/ {X}} ) \) there is a unique 1-form \( \lambda \) on \( {\mathbf{T}} {}^{ \vee }X \) such that \( \beta^*( \lambda) = \beta \), regarding \( \beta \) as a smooth map on the left and a 1-form on the right. In local coordinates \( (x_1,\cdots, x_n) \) on \( X \), write \( y_i = dx_i \) on the fiber of \( {\mathbf{T}} {}^{ \vee }X \). Why this works: the fibers are collections of covectors, so if \( x_i \) are horizontal coordinates there is a dual vertical coordinate in the fiber: ```{=html} ``` ![](figures/2022-01-13_11-50-52.png) So we can write \[ \lambda = \sum y_i \,dx_i \in \Omega^1({\mathbf{T}} {}^{ \vee }X) ,\] regarding the \( y_i \) as functions on \( {\mathbf{T}} {}^{ \vee }X \) and \( \,dx_i \) as 1-forms on \( {\mathbf{T}} {}^{ \vee }X \). ::: {.exercise title="?"} Find out what \( \beta = \sum a_i \,dx_i \) is equal to as a section of \( {\mathbf{T}} {}^{ \vee }X \). ::: ::: ::: {.remark} To get a contact manifold of dimension \( 2n+1 \), consider the 1-jet space \( J^1(X) \coloneqq T {}^{ \vee }X \times {\mathbb{R}} \). Write the coordinates as \( (x,y)\in {\mathbf{T}} {}^{ \vee }X \) and \( z\in {\mathbb{R}} \) and define \( \alpha = \,dz- \lambda \), the claim is that this is contact. For dimension \( 2n-1 \), choose a cometric on \( X \) and take \( {\mathbb{S}}{\mathbf{T}} {}^{ \vee }X \) the unit cotangent bundle of unit-length covectors. Then \( \alpha \coloneqq-{ \left.{{\lambda}} \right|_{{{\mathbb{S}}{\mathbf{T}} {}^{ \vee }X}} } \) is contact. ::: ::: {.exercise title="?"} Check that \( {\mathbb{R}}^3 = J^1({\mathbb{R}}) \) and \( {\mathbb{S}}{\mathbf{T}} {}^{ \vee }({\mathbb{R}}^2) = {\mathbb{R}}^2 \times S^1 \). ::: ::: {.remark} A neat theorem: the contact geometry of \( {\mathbb{S}}{\mathbf{T}} {}^{ \vee }{\mathbb{R}}^3 \) is a perfect knot invariant. This involves assigning to knots unique Legendrian submanifolds. ::: ## Perturbing Foliation ::: {.example title="?"} Define \[ \alpha_t = \,dz- ty\,dx\qquad t\in {\mathbb{R}} \] to get a 1-parameter family of 1-forms. Check that \( \alpha_t \wedge d \alpha_t = t(\,dz\wedge \,dx\wedge \,dy) \). Consider \( t\in (-{\varepsilon}, {\varepsilon}) \): - \( t>0 \implies \alpha = \,dz- y\,dx \) yields a positive contact structure, - \( t>0 \implies \alpha = \,dz \) is a foliation, - \( t<0 \implies \alpha = \,dz+ y\,dx \) is a negative contact structure. ::: ::: {.remark} What is a (codimension \( r \)) foliation on an \( n{\hbox{-}} \)manifold? A local diffeomorphism \( U\cong {\mathbb{R}}^n \times {\mathbb{R}}^{n-r} \) with *leaves* \( {\operatorname{pt}}\times{\mathbb{R}}^{n-r} \). For example, \( {\mathbb{R}}^3\cong {\mathbb{R}}\times{\mathbb{R}}^2 \) with coordinates \( t \) and \( (x, y) \). We're leaving out a lot about how many derivatives one needs here! > For a fiber bundle or vector bundle to admit an interesting foliation, one needs a flat connection. ::: ::: {.definition title="Integrability"} Any \( \xi \coloneqq\ker \alpha \) is **integrable** iff for all vector fields \( X, Y \subseteq \xi \), their Lie bracket \( [X, Y] \subseteq \xi \). ::: ::: {.theorem title="Frobenius Integrability"} For \( \alpha \) nonvanishing on \( Y^3 \), \( \ker \alpha \) is tangent to a foliation by surfaces iff \( \alpha \wedge d\alpha = 0 \). ::: ::: {.example title="?"} Consider \( \alpha = \,dz- y\,dx \), so \( \ker \alpha = \mathop{\mathrm{span}}_{\mathbb{R}}\left\{{{\partial}y, y{\partial}z + {\partial}x}\right\} \) which bracket to \( {\partial}z \not \in \ker \alpha \). This yields a non-integrable contact structure. On the other hand, for \( \alpha = \,dz \), \( \ker \alpha = \mathop{\mathrm{span}}_{\mathbb{R}}\left\{{{\partial}x, {\partial}y}\right\} \) which bracket to zero. So this yields a foliation. ::: ::: {.remark} A theorem of Eliashberg and Thurston: taut foliations can be perturbed to a (tight) positive contact structure. ::: # Tuesday, January 18 ::: {.remark} Refs: - Geiges, Intro to Contact - Ozbogi-Stipsicz - Etnyre lecture notes - Massot - Sivck ::: ::: {.definition title="Standard contact structure"} For \( S^3 \subseteq {\mathbb{C}}^2 \), define a form on \( {\mathbb{R}}^4 \) as \[ \alpha \coloneqq-y_1 dx_1 + x_1 dy_1 - y_2 dx_2 + x_2 dy_2 .\] Then then **standard contact form** on \( S^3 \) is \[ \xi_{\text{std}} \coloneqq\ker { \left.{{\alpha}} \right|_{{S^3}} } .\] ::: ::: {.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 \[ { \left.{{ \alpha}} \right|_{{S^3}} } \wedge { \left.{{ d\alpha}} \right|_{{S^3}} } > 0 \iff df\wedge d \alpha \wedge d \alpha > 0 .\] 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 \( p= {\left[ {1,0,0,0} \right]} \subseteq S^3 \), \( {\mathbf{T}}_pS^3 = \mathop{\mathrm{span}}\left\{{{\partial}y_1, {\partial}x_2, {\partial}y_2}\right\} \). and \( \alpha_p = -0 dx_1 + 1dy_1 -0 dx_2 + 0dy_2 = dy_1 \) and \( \xi_p = \ker dy_1 = \mathop{\mathrm{span}}\left\{{{\partial}x_1, {\partial}y_2}\right\} \). ```{=html} ``` ![](figures/2022-01-18_11-29-05.png) Then \( \xi_p \leq {\mathbf{T}}_p {\mathbb{C}}^2 = \mathop{\mathrm{span}}\left\{{{\partial}x_1, {\partial}y_1, {\partial}x_2, {\partial}y_2}\right\} \cong {\mathbb{C}}^4 \) is a distinguished complex line. ::: ::: {.definition title="Almost complex structures"} An **almost complex structure** on \( X \) is a bundle automorphism \( J: {\mathbf{T}}X{\circlearrowleft} \) with \( J^2 = -\operatorname{id} \). ::: ::: {.example title="?"} For \( X = {\mathbb{C}}^2 \), take \[ {\partial}x_1 &\mapsto {\partial}y_1 \\ {\partial}y_1 &\mapsto - {\partial}x_1 \\ {\partial}x_2 &\mapsto {\partial}y_2 \\ {\partial}y_2 &\mapsto -{\partial}x_2 .\] ::: ::: {.exercise title="?"} Show that \( f: {\mathbb{C}}\to {\mathbb{C}} \) is holomorphic if \( df \circ J = J \circ df \), which corresponds to the Cauchy-Riemann equations. ::: ::: {.lemma title="?"} Given \( J:W\to W \), an \( {\mathbb{R}}{\hbox{-}} \)subspace \( V \leq W \) is a \( {\mathbb{C}}{\hbox{-}} \)subspace iff \( J(V) = V \). ::: ::: {.definition title="?"} The field of \( J{\hbox{-}} \)complex tangents is the hyperplane field \[ \xi_p \coloneqq{\mathbf{T}}S^3 \cap J({\mathbf{T}}S^3) .\] ::: ::: {.example title="?"} Consider \( {\mathbf{T}}_p S^3 \) for \( p={\left[ {1,0,0,0} \right]} \), then \[ J(\mathop{\mathrm{span}}\left\{{{\partial}y_1, {\partial}x_1, {\partial}y_2}\right\}) = \mathop{\mathrm{span}}\left\{{-{\partial}x_1, {\partial}y_2, -{\partial}x_2}\right\} ,\] so \( \xi_p = \mathop{\mathrm{span}}{ {\partial}x_1, {\partial}y_2} \) is the intersection and coincides \( \xi_{\text{std}} \). ::: ::: {.question} Where does \( \alpha \) come from? Let \( \rho = \sum x_i {\partial}x_i + \sum y_i {\partial}y_i \) be the radial vector field, so \( \rho = {1\over 2}\operatorname{grad}{\left[ {\sum x_i^2 + \sum y_i^2} \right]} \). Setting \( \omega \coloneqq\bigwedge dx_i \wedge \bigwedge dy_i \), then \( \alpha = \iota_p\omega \coloneqq\omega(p, {-}) \) is the interior product of \( \omega \). Then \[ \alpha = dx_1 \wedge dy_1 (x_1 {\partial}x_1 + y_1 {\partial}y_1 + \cdots ) + \cdots = x_1 dy_1 -y_1 dx_1 + \cdots .\] So the contact form comes from pairing the symplectic form against a radial vector field. ::: ::: {.remark} Recall \( f \coloneqq\sum x_i^2 + \sum y_i^2 \) satisfies \( df = 2\sum x_i dx_i + 2\sum y_i dy_i \). Note that \( J \) acts on 1-forms by \( J^*(dx)({-}) = dx(J({-})) \). For \( J = i \), - \( \delta x: dx(J {\partial}x) = dx({\partial}y) = 0 \), - \( {\partial}y: dx(J {\partial}y) = dx(-{\partial}x) = -1 \). 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 \( Y = f^{-1}(0) \) is the kernel of \( { \left.{{ df(J({-})) }} \right|_{{Y}} } \). ::: ::: {.remark} Methods of getting contact structures: for a vector field \( X \), being contact comes from \( {\mathcal{L}}_X \omega = \omega \). For functions \( f:{\mathbb{C}}^2 \to {\mathbb{R}} \), being contact comes from \( \alpha = d^{\mathbb{C}}f \) 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 \[ {\mathbb{C}}^2\setminus\left\{{0}\right\}&\to {\mathbb{CP}}^1 \cong S^2 \\ {\left[ {z, w} \right]} &\mapsto {\left[ {z: w} \right]} ,\] which restricts to a map \( S^3\to S^2 \) defining the Hopf fibration. If \( L \) is a complex line through 0, then \( L \cap S^3 \) is a Hopf fiber that is homeomorphic to \( S^1 \). ```{=html} ``` ![](figures/2022-01-18_12-11-38.png) ```{=html} ``` ![](figures/2022-01-18_12-15-08.png) ```{=html} ``` ![](figures/2022-01-18_12-17-02.png) Take \[ {\mathbb{C}}^2 &\to {\mathbb{R}}^2 \\ (z_1, z_2)&\mapsto ({\left\lvert {z_1} \right\rvert}, {\left\lvert {z_2} \right\rvert}) .\] Consider the image of \( S^2 = \left\{{{\left\lvert {z_1} \right\rvert}^2 + {\left\lvert {z_2} \right\rvert}^2 = 1}\right\} \): ```{=html} ``` ![](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: ```{=html} ``` ![](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. ```{=html} ``` ![](figures/2022-01-18_12-28-04.png) ::: # 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 \( \xi = \ker (\,dz-y\,dx) = \ker (\alpha_\text{std}) \). ::: ::: {.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 \( {\mathbf{T}}_p S \pitchfork\xi_p \). This produces a set of curves in \( S \) which are tangent to \( \xi_p \) everywhere, called the *characteristic foliation*. ```{=html} ``` ![](figures/2022-01-20_11-24-33.png) Then \( { \left.{{\alpha}} \right|_{{S}} } = \,dz \), 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 \( t\in (-{\varepsilon}, {\varepsilon}) \) around \( p \) along \( X \). Choose a vector field \( {\partial}t \) which is transverse to \( S \) and contained in \( \xi \). Then \( \alpha({\partial}t) = 0 \), 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 \( \xi_p = \mathop{\mathrm{span}}{{\partial}x, {\partial}y} \): ```{=html} ``` ![](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 \( \alpha_0 \simeq\alpha_1 \) 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 \( \dot\alpha_t \coloneqq{\frac{\partial }{\partial t}\,} \alpha_t \), so \[ {\frac{\partial }{\partial t}\,} \qty{ t\alpha_1 + (1-t) \alpha_0} = \alpha_1 - \alpha_0 .\] ::: 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 \( {\frac{\partial }{\partial t}\,} \) of both sides here to get \[ \Phi_t^*( \dot \alpha_t + {\mathcal{L}}_{V_t} \alpha_t ) = \dot f_t \alpha_0 .\] > See Prop 6.4 in Cannas da Silva. ::: {.remark} ✨Cartan's magic formula✨: \[ {\mathcal{L}}_V(\alpha) = d(\iota_V \alpha) + \iota_V(d\alpha) ,\] so \[ {\mathcal{L}}_{V_t}(\alpha_t) = d(\alpha_t(V_t)) + d\alpha_t(V_t, {-}) = 0 + d\alpha_t(V_t, {-}) .\] ::: We can thus write this equation as \[ \Phi_t^*(\dot \alpha_t + d\alpha_t(V_t, {-})) = \dot f_t \alpha_0 = \dot f_t \qty{\Phi_t^*(\alpha_t) \over f_t } .\] Applying \( (\Phi_t^*)^{-1} \) yields \[ \dot \alpha_t + d\alpha_t(V_t, {-})= {\dot f_t \over f_t }\alpha_t .\] Now try to solve this for \( V_t \). Let \( R_t \) be the **Reeb vector field** of \( \alpha_t \), which satisfies - \( \alpha_t(R_t) = 1 \) - \( d\alpha_t(R_t, {-}) = 0 \). ```{=html} ``` ![](figures/2022-01-20_12-11-26.png) Then \[ \dot \alpha_t (R_t) = {\dot f_t \over f_t} = {\frac{\partial }{\partial t}\,} \log(f_t) \coloneqq\mu_t ,\] so \( \dot\alpha_t(R_t) \) determines \( f_t \) by first integrating and exponentiating. We now need to solve \[ { \left.{{d\alpha_t(V_t, {-})}} \right|_{{\xi_t}} } = { \left.{{\mu_t \alpha_t - \dot \alpha_t}} \right|_{{\xi_t}} } .\] 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. ::: # Gray Stability (Tuesday, January 25) ::: {.remark} A homotopy of contact structures o \( Y^3 \) is a smooth family \( \left\{{\phi_t}\right\} \) of contact structures. Similarly, an **isotopy** of structures such that \( \left\{{D\phi_t(\xi_0)}\right\} \) for an isotopy \( \phi_t: Y\to Y \) with \( \phi_0 = \operatorname{id} \). 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, \[ {\frac{\partial }{\partial t}\,} \phi^*_t(\alpha_t) \mathrel{\Big|}_{t=t_0} = \phi_{t_0}^*(\dot \alpha_{t_0} + {\mathcal{L}}_{X_{t_0}} \alpha_{t_0} ) .\] ::: ::: {.proof title="?"} Write \[ \phi_x^*(\alpha_y) = {\frac{\partial }{\partial x}\,} ? + {\frac{\partial }{\partial y}\,}? = \phi_{x_0}^* {\mathcal{L}}_{X_0} {\mathcal{L}}_X \alpha_{y_0} + \phi_{x_0}^* \alpha_y ,\] and proceed similarly to the proof of Darboux's theorem. Pick \( \left\{{\phi_t}\right\} \) 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 \( \lambda_t: Y\to {\mathbb{R}}_+ \). What does \( \phi_t \) need to look like? Differentiate in \( t \): \[ \phi^*_{t_0}(\cdot \alpha_{t_0} + {\mathcal{L}}_{V_{t_0}} \alpha_{t_0} ) = \dot \lambda_t \alpha_0 = \dot \lambda_t \qty{ \phi^*_{t_0} (\alpha_t) \over \lambda_t } .\] Apply \( (\phi^*_{t_0})^{-1} \): \[ \cdot\alpha_t + {\mathcal{L}}_{V_t}\alpha_t = \mu_t \alpha_t\qquad \mu_t = (\phi^*_{t_0})^{-1}(\dot\lambda_t \over \lambda_t) .\] 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 \( S^3 \subseteq {\mathbb{C}}^2 \) that is transverse to the radial vector field. Then \[ \alpha = x_1 dy_1 - y_1 dx_1 + x_2 dy_2 - y_2 dx_2\mathrel{\Big|}_y \] defines the standard tight contact structure. ::: ::: {.proof title="?"} Write \( Y \subseteq {\mathbb{R}}\times S^3 \) in coordinates \( (f(x), x) \) as the graph of a function \( f: S^3\to {\mathbb{R}} \). Take an isotopy \( Y_t = (tf(x), x) \subseteq {\mathbb{R}}\times S^3 \) to get a family of contact forms where \( \alpha_0 = \alpha_\text{std} \) 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 \( L \hookrightarrow Y \) a link. Then \( L \) is a **Legendrian knot** iff it is everywhere tangent to \( \xi \), so \( \alpha(L) = 0 \): ```{=html} ``` ![](figures/2022-01-25_12-10-49.png) This is a closed condition. \( L \) is **transverse** if it is everywhere transverse to \( \xi \), so \( \alpha(L) > 0 \): ```{=html} ``` ![](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 \( {\mathbb{R}}^3 \) and \( \alpha_\text{std}= dz-ydx \), then the \( y{\hbox{-}} \)axis \( L_1 \coloneqq\left\{{{\left[ {0,t,0} \right]}}\right\} \) is Legendrian. Similarly the \( x{\hbox{-}} \)axis \( L_2 \) is Legendrian, checking that \( {\mathbf{T}}L_2 = \mathop{\mathrm{span}}\left\{{{\left[ {1,0,0} \right]}}\right\} \). However the slight pushoff \( L_3 \coloneqq\left\{{{\left[ {t, -{\varepsilon}, 0} \right]}}\right\} \) is transverse since \( { \left.{{\alpha}} \right|_{{L_3}} } = {\varepsilon}dx >0 \). ::: ::: {.theorem title="Neighborhood theorem, Darboux for Legendrian/transverse knots"} Every Legendrian has a neighborhood contactomorphic to the zero section in \( J_1 S^1 = {\mathbf{T}}S^1 \times{\mathbb{R}} \). Every transverse has a neighborhood contactomorphic to the \( z{\hbox{-}} \)axis in \( {\mathbb{R}}\times S^1 \) with \( \alpha \coloneqq\,dz+ r^2 \,d\theta \). ::: # Thursday, January 27 ::: {.remark} Goal: classify Legendrian knots up to (Legendrian) isotopy. Recall a knot \( \gamma: S^1 \hookrightarrow Y \) satisfies \( \gamma^*(\alpha) = 0 \), and a Legendrian isotopy is a 1-parameter family \( \gamma_t \) which are Legendrian for all \( t \). ::: ::: {.example title="?"} \( \gamma(s) = {\left[ {x(s), y(s), z(s)} \right]} \) and \( \xi = \ker \alpha, \alpha = \,dz- y\,dx \). Then \( \gamma^*(\alpha) = z' \,ds- yx^1\,ds= (z' - yx')\,ds \), which is Legendrian iff \( y=z'/x' \). ::: ::: {.example title="?"} Let \( f:{\mathbb{R}}\to {\mathbb{R}} \) and take the 1-jet \( \gamma(s) = {\left[ {s, f'(s), f(s)} \right]} \) of the graph of \( f \) -- this is like the graph of the 1st order Taylor expansion. This is Legendrian since \( s'=1 \) implies \( z'/x' = f'/s' = f' \). ::: ::: {.remark} There are two projections: - \( {\left[ {x,y,z} \right]} \to {\left[ {x,z} \right]} \), a wave front projection, plotted with \( y \) into the board, - \( {\left[ {x,y,z} \right]} \to {\left[ {x,y} \right]} \), Lagrangian projection. ::: ::: {.example title="?"} Let \( \gamma(s) = {\left[ {s^2, {3\over 2}s, s^3} \right]} \), then the two projections are as follows: ```{=html} ``` ![](figures/2022-01-27_11-25-37.png) ::: ::: {.remark} The front projection uniquely determines \( L \), since the \( y \) coordinate can be recovered as \( y=z'/x' \). So for example, there is no ambiguity about crossing order: the more negatively sloped line in a diagram is the over-crossing: ```{=html} ``` ![](figures/2022-01-27_11-30-01.png) ::: ::: {.example title="?"} A front diagram of the unknot: ```{=html} ``` ![](figures/2022-01-27_11-34-42.png) ::: ::: {.theorem title="?"} Every knot \( K \hookrightarrow{\mathbb{R}}^3 \) can be \( C^0 \) approximation by a Legendrian knot \( L \). Idea: zigzags in an \( {\varepsilon} \) tube in the knot diagram, which will be Legendrian. How to measure: \( \sup_{s\in I} {\left\lvert {\gamma_1(s) - \gamma_2(s)} \right\rvert} \leq {\varepsilon} \)? ::: ::: {.remark} Note that \( \mathsf{Lie}({\operatorname{SO}}_3) \coloneqq{\mathbf{T}}_e({\operatorname{SO}}_3) = {\mathfrak{su}}_2 \), spanned by roll, pitch, and yaw generators: ```{=html} ``` ![](figures/2022-01-27_11-48-58.png) So measuring the number of rotations along each generator after traversing \( L \) in a full loop yields integer invariants. ::: ::: {.definition title="The Thurston–Bennequin number"} A **framing** of a knot \( K \) is a trivialization of its normal bundle, so an identification of \( \nu(K) \cong S^1\times {\mathbb{D}}^2 \). The potential framings are in \( \pi_1({\operatorname{SO}}_2) \cong \pi_1(S^1) \cong {\mathbb{Z}} \), since a single vector field normal (?) to the knot determines the framing by completing to an orthonormal basis. The Reeb vector field is never tangent to a Legendrian knot, so this determines a framing called the **contact framing**. The **Thurston--Bennequin number** is the different between the 0-framing and the contact framing. The 0-framing comes from a Seifert surface. This is an invariant of Legendrian knots, since Legendrian isotopy transports frames. Note that adding zigzags adds cusps, and thus decreases this number. ::: ::: {.remark} How to compute: take a pushoff and compute the linking number: ```{=html} ``` ![](figures/2022-01-27_12-07-19.png) ::: ::: {.proposition title="?"} \[ \mathrm{tb}(L) = w(L) - {1\over 2}C(L) ,\] where \( w(L) \) is the writhe and \( C(L) \) is the number of cusps. ::: ::: {.proof title="?"} The linking number is \( {1\over 2}(c_+(L) - c_-(L)) \), half of the signed number of crossings. ```{=html} ``` Here all 4 crossing have the same sign: ![](figures/2022-01-27_12-12-35.png) ::: ::: {.example title="?"} TB for the knots from before: - The 3 unknots: - 2 cusps, so \( -1 \) - 4 cusps, so \( -2 \) - 4 cusps, so \( -2 \) - The 2 trefoils: - \( 3-{1\over 2}4 = 1 \) - \( 3 - {1\over 2}6 = -6 \). ::: ::: {.remark} Since adding zigzags decreases \( \mathrm{tb} \), define \( \mathrm{TB} \) to be the max over all Legendrian representatives of \( K \). This distinguishes mirror knots. In fact \( \mathrm{tb}(L) \leq 2g_3(L) - 1 \) (the Bennequin bound), involving the 3-genus. ::: ::: {.definition title="Rotation number"} The **rotation number** of \( L \) is the *turning number* \( \operatorname{rot}(L) \) in the Lagrangian projection, i.e. how many times a tangent vector spins after traversing the knot. ::: ::: {.example title="?"} ```{=html} ``` ![](figures/2022-01-27_12-26-39.png) It turns out that \[ \operatorname{rot}(L) = {1\over 2}\qty{ {\sharp}\text{down cusps} - {\sharp}\text{up cusps}} .\] ::: # Tuesday, February 01 ::: {.remark} Last time: front diagrams \( {\left[ {x,y,z} \right]}\mapsto {\left[ {x,z} \right]} \), 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 ```{=html} ``` ![](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: ```{=html} ``` ![](figures/2022-02-01_11-20-07.png) Allowed moves: ```{=html} ``` ![](figures/2022-02-01_11-24-46.png) ::: ::: {.remark} Geography problem: given a smooth knot \( K \), which pairs \( (t, r) \in {\mathbb{Z}}^2 \) are realized as \( (\mathrm{tb}(L), \operatorname{rot}(L)) \) for \( L \) a Legendrian representative of \( K \)? Botany problem: given \( (t, r) \in {\mathbb{Z}}^2 \), how many inequivalent \( L \) representing \( K \) realize \( (t,r) = (\mathrm{tb}(L), \operatorname{rot}(L)) \)? ::: ::: {.example title="?"} For \( K \) the unknot: ```{=html} ``` ![](figures/2022-02-01_11-37-46.png) So these numerical pairs fall into a cone: ```{=html} ``` ![](figures/2022-02-01_11-39-30.png) ::: ::: {.proposition title="?"} For \( L \subseteq {\mathbb{R}}^3 \) a Legendrian knot, \[ \mathrm{tb}(L) + \operatorname{rot}(L) \equiv 1 \operatorname{mod}2 .\] ::: ::: {.remark} Note that \( \chi(S) \equiv 1\operatorname{mod}2 \) for \( S \) a Seifert surface. ::: ::: {.theorem title="Bennequin-Thurston inequality"} For any Seifert surface \( S \), \[ \mathrm{tb}(L) + {\left\lvert { \operatorname{rot}(L) } \right\rvert} \leq -\chi(S) .\] ::: ::: {.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 \( \operatorname{rot}(L_1) = \operatorname{rot}(L_2) \), 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: ```{=html} ``` ![](figures/2022-02-01_11-53-16.png) ::: ::: {.remark} This all depended on the standard contact form. Consider instead the overtwisted disc: take \( {\mathbb{R}}^3 \) with \( \alpha = \cos(r)\,dz+ \sin(r) \,d\theta \). Take the curve \( {\left[ {r,\theta, z} \right]} = \gamma(t) \coloneqq{\left[ {1, t, 0} \right]} \), a copy of \( S^1 \) in the \( x,y{\hbox{-}} \)plane. Then \( \gamma' = {\left[ {0,1,0} \right]} \), and at \( \theta=\pi, \alpha = \cos(\pi)\,dz+ \sin(\pi) \,d\theta= -\,dz \), but at \( r=0 \) \( \alpha = \,dz \), so traversing a ray from \( 0 \) to \( -1 \) in the \( x,y{\hbox{-}} \)plane forces the contact plane to flip: ```{=html} ``` ![](figures/2022-02-01_12-02-08.png) One can check that \( \mathrm{tb} \) is given my \( \operatorname{lk}(L, L') = 0 \) where \( L' \) is a pushoff of \( L \), and can be made totally disjoint from \( L \) in this case by moving in the \( z{\hbox{-}} \)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"} \( ({\mathbb{R}}^3, \xi_\text{std}) \) 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 \( \operatorname{lk}(T, T') \) for \( T' \) a pushoff of \( T \) determined by a trivialization of \( { \left.{{\xi}} \right|_{{S}} } \). ::: ::: {.remark} In this case, \( \xi \) restricts to an \( {\mathbb{R}}^2 \) bundle over \( \Sigma \), which is trivial since \( \Sigma \) is closed with boundary and \( e(\xi) \in H^2(S) = 0 \). To see this, use \( H^2(S) \cong H_0(S, {{\partial}}S) = 0 \) by Lefschetz duality. This yields a section of the frame bundle over \( S \), which gives a pushoff direction along the first basis vector: ```{=html} ``` ![](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 \( \pi_1(S) \to {\mathbb{Z}} \), which factors through \( \pi_1(S)^{\operatorname{ab}}= H_1(S) \). The difference in surfaces is measured by \( {\left\langle {e(S)},~{ \Sigma_1 {\textstyle\coprod}_T \Sigma_2 } \right\rangle} \), which is a glued surface. ::: # 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 \( {{\partial}}\Sigma \) transverse to \( \xi \). Let \( F \) be the characteristic foliation, the singular foliation on \( \Sigma \) induced by \( { \left.{{\xi}} \right|_{{\Sigma}} } \). Equivalently, if \( \xi = \ker \alpha \), consider the 1-form \( { \left.{{\alpha}} \right|_{{\Sigma}} } \). Generically, \( { \left.{{\ker \alpha}} \right|_{{{\mathbf{T}}\Sigma}} } \) 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 ```{=html} ``` ![](figures/2022-02-03_11-21-06.png) ::: ::: {.remark} Two possible types of singularities, the local models: ```{=html} ``` ![](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 \[ {\left\langle {c(\Sigma)},~{ \Sigma} \right\rangle} = (e_+ - h_+) - (e_- - h_-) .\] If \( \Sigma \) is transverse, \( \mathrm{sl}({{\partial}}\Sigma, \Sigma) = -(e_+ - h_+) + (e_- - h_-) \). ::: ## Local Model 1: Elliptic ::: {.remark} \( \sigma \) is the \( x,y{\hbox{-}} \)plane and \( \xi = \ker (\,dz+ x\,dy- y\,dx) \) with \( { \left.{{\alpha }} \right|_{{\Sigma }} }= x\,dy- y\,dx \). Set \( V: x{\partial}_x + y{\partial}_y \) and \( L' = \left\langle{x{\partial}_y - y{\partial}_x}\right\rangle \), and \( \alpha(i) = x^2+y^2 = 1 > 0 \). ```{=html} ``` ![](figures/2022-02-03_11-32-36.png) Here \( \mathrm{sl} = 1 \). To compute \( \mathrm{sl} \): - Trivialize \( { \left.{{\xi}} \right|_{{\Sigma}} } \) to get \( \tau = \left\langle{e_1, e_2}\right\rangle \) a fiberwise basis for \( \xi \). - Let \( \tilde L \) be a pushoff in the \( e_1 \) direction. - Compute \( \mathrm{sl} = \operatorname{lk}(L, \tilde L) \). Set - \( e_1 = {\partial}_x + y{\partial}_z \) - \( e_2 = {\partial}_y - x{\partial}_z \) - \( \rho = x {\partial}_x + y{\partial}_y \) - \( \theta = x{\partial}_y - y{\partial}_x \). Then \[ x\rho - y\theta = x(x{\partial}_x + y{\partial}_y) - y(-y {\partial}_x + x{\partial}_y) = (x^2+y^2)\,dx .\] Then - \( c_1 = x\rho - y\theta + y{\partial}_z \) - \( \mkern 1.5mu\overline{\mkern-1.5muc_1\mkern-1.5mu}\mkern 1.5mu = x\rho + y{\partial}_z = \cos(\rho) + \sin(\theta) {\partial}_z \). Example: - \( \theta = 0\implies e_1 = \rho \) - \( \theta = \pi/4 \implies e_1 = {\sqrt 2\over 2}(\rho + {\partial}_z) \) - \( \theta = \pi/2 \implies e_1 = {\partial}_z \) So here \( \operatorname{lk}(U, \tilde U) = -1 \): ```{=html} ``` ![](figures/2022-02-03_11-43-46.png) ::: ## Local Model 2: Hyperbolic ::: {.remark} Here \( \xi \) is the \( x,y{\hbox{-}} \)plane, so \( \xi = \ker (\,dz+ 2x\,dy+ y\,dx) \) with \( { \left.{{\alpha}} \right|_{{\Sigma}} } = 2x\,dy+ y\,dx \) and \( V = y{\partial}_y + dx{\partial}_x\in \ker({ \left.{{\alpha }} \right|_{{\Sigma}} }) \). ::: ::: {.remark} The Euler class of a real vector bundle \( E \xrightarrow{\pi} X \) is the obstruction to finding a nonvanishing section \( s \) of \( E \), given by \( e(E) \in H^k(X) \). It is Poincare dual to \( [s^{-1}(0)] \in H_{n-k}(X, {{\partial}}X) \). For the tangent bundle, \( e({\mathbf{T}}X)\in H^{n}(X) \), and \[ {\left\langle {e({{\mathbf{T}}X} )},~{[X]} \right\rangle} = \chi(X) .\] Since a section of \( {\mathbf{T}}X \) is a vector field, \( e({\mathbf{T}}X) \) is an obstruction to finding a nonvanishing vector field. If \( {{\partial}}X \neq \emptyset \) and \( t \) is a section of \( { \left.{{E}} \right|_{{{{\partial}}X}} } \), there is a relative Euler class \( e(E, t)\in H^k(X, {{\partial}}X) \cong H_{n-k}(X) \). Similarly, \[ {\left\langle {e({\mathbf{T}}X, t)},~{[X]} \right\rangle} = \chi(X) .\] ::: ::: {.example title="?"} Note \( \chi({\mathbb{D}}) = 1 \), so any vector field has a singularity? ```{=html} ``` ![](figures/2022-02-03_11-59-11.png) ::: ::: {.proposition title="?"} The total class is the sum of the relative obstructions. If \( \sigma = \Sigma_1 { \displaystyle\coprod_{{{\partial}}} } \Sigma_2 \) and \( \tau \) is a nonvanishing section of \( { \left.{{\Sigma}} \right|_{{{{\partial}}\Sigma_1}} } = { \left.{{\Sigma}} \right|_{{{{\partial}}\Sigma_2}} } \), then \[ c(E) = e({ \left.{{E}} \right|_{{\Sigma_1}} }, \tau) + c({ \left.{{E}} \right|_{{\Sigma_2}} }, \tau) .\] ```{=html} ``` ![](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 \( V \in \xi \cap{\mathbf{T}}\Sigma \). We can assume \( V \) is outward-pointing along \( {{\partial}}\Sigma \). Check that - \( \chi(\Sigma) = e({\mathbf{T}}\Sigma, V) \in H^2(\Sigma, {{\partial}}\Sigma) \cong H_0(\Sigma) \) - \( \mathrm{sl}({{\partial}}\Sigma, \Sigma) = e(\xi, V) \in H^2(\Sigma, {{\partial}}\Sigma) \) ::: ::: {.fact} ```{=tex} \envlist ``` - \( e_+ + e_- \) correspond to \( +1 \) in \( e({\mathbf{T}}\Sigma, V) \), - \( h_+, h_- \) correspond to \( -1 \) in \( e({\mathbf{T}}\Sigma, V) \). 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: ```{=html} ``` ![](figures/2022-02-03_12-24-10.png) The inequality follows if we can cancel every \( e_- \) with some \( h_- \). ::: # Tuesday, February 08 ::: {.remark} Topics for talks: - Thom-Pontryagin - Brieskorn spheres - Milnor fibrations - Lens spaces ::: ::: {.theorem title="?"} Every closed oriented 3-manifold \( Y \) admits a (positive) contact form. ::: ::: {.remark} Three proofs: - Lickorish-Wallace, using that \( Y \) is Dehn surgery on a link in \( S^3 \), - Birman-Hildon, using that \( Y \) is a branched cover of \( S^3 \), - Alexander, using that \( Y \) admits an open book decomposition. ::: ::: {.remark} Dehn surgery for slope \( p/q \): for \( K \hookrightarrow S^3 \), cut out \( \nu(K) \cong S^1\times {\mathbb{D}} \) and re-glue by a map \( {{\partial}}(S^1\times {\mathbb{D}}) \to {{\partial}}\nu(K) \) such that \( [\left\{{0}\right\} \times {{\partial}}{\mathbb{D}}] = p[m] + q[\ell] \in H^1({{\partial}}\nu (K) ) \). Use that \( \nu(K) \cong S^1\times {\mathbb{D}} \) and \( {{\partial}}\nu(K) \cong S^1\times S^1 = T^2 \). Idea: wrapped \( p \) times longitudinally, \( q \) times around the meridian. ::: ::: {.remark} Recall: - Every knot \( K \) can be \( C^0 \) approximated by a transverse knot - Every link \( L \) can be \( C^0 \) approximated by a transverse link - Neighborhood theorem: for every transverse knot \( K \), there is a \( w(K) \) and a contactomorphism to a standard model: \( S^1\times {\mathbb{D}} \) in coordinates \( (\phi, r, \theta) \) with \( 0\leq r\leq \delta \) and \( \alpha = d\phi + r^2\,d\theta \). Re-gluing corresponds to the map \( {\left[ {0, \delta, \theta} \right]}\mapsto {\left[ {q\theta, \delta, p\theta} \right]} \). \[ {\left[ {0, \delta, \mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu} \right]} &\mapsto {\left[ {q\mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu, \delta, p\mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu} \right]} \\ {\left[ {\mkern 1.5mu\overline{\mkern-1.5mu\pi\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu} \right]} &\mapsto {\left[ {\phi,r,\theta} \right]} .\] If \( p, q \) are coprime there exist \( m,n \) with \( pm-qn = 1 \). So define \[ \psi: {\left[ {\mkern 1.5mu\overline{\mkern-1.5mu\pi\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mur\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu} \right]} &\mapsto {\left[ {\phi,r,\theta} \right]} ,\] so \[ \psi^*(\alpha) = d(\alpha\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu+ m\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu) + r^2d(p\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu+ n\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu) = (q+r^2p)d\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu+ (m+r^2n)d\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu .\] We want \( \alpha = h_1(r) d\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu+ h_2(r) d\mkern 1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu \) to be contact and satisfy \( (h_1, h_2) = (r^2, 1) \) near \( r=0 \) and \( (q+r^2 p, m+r^2 n) \) near \( r=\delta \). This requires \[ d\alpha = h_1' \,dr\wedge d\mkern 1.5mu\overline{\mkern-1.5mu\phi \mkern-1.5mu}\mkern 1.5mu+ h_2' \,dr\wedge d\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu= (h_2 h_1' - h_1 h_2') dr \wedge d\mkern 1.5mu\overline{\mkern-1.5mu\theta \mkern-1.5mu}\mkern 1.5mu\wedge d\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu ,\] which happens iff \[ \operatorname{det} \begin{bmatrix} h_2 & h_2' \\ h_1 & h_1' \end{bmatrix} > 0 .\] Think of \( {\left[ {h_2, h_1} \right]} \) as a path with tangent vector \( {\left[ {h_2', h_1'} \right]} \). This requires moving counterclockwise. ```{=html} ``` ![](figures/2022-02-08_11-57-58.png) ::: ::: {.definition title="?"} An **open book decomposition** of \( Y \) is a pair \( (B, \pi) \) where - \( B \) is a link in \( Y \), called the **binding** - \( \pi: Y\setminus B\to S^1 \) is a locally trivial fibration of relatively compact fibers **pages** ```{=html} ``` ![](figures/2022-02-08_12-09-13.png) ::: ::: {.remark} An open book decomposition is determined by its monodromy map \( \phi: \Sigma_0\to \Sigma_0 \), which determines a class \( [\phi] \in {\operatorname{MCG}}(\Sigma_0) \). Form \[ Y\setminus\nu(B) \cong {\Sigma \times I \over \phi(x) \times\left\{{0}\right\} \sim x\times\left\{{1}\right\}} ,\] which is a glued cylinder: ```{=html} ``` ![](figures/2022-02-08_12-14-04.png) ::: ::: {.definition title="Open book decompositions supporting a contact structure"} An open book decomposition **supports** a contact structure \( \xi \) iff there exists a contact form \( \alpha \) such that \( d\alpha \) is an area form on each page and \( B \) is a transverse link in \( (B, \xi) \). ::: ::: {.theorem title="Thurston-Winkelnkemper"} Every open book decomposition admits a contact structure. ::: ::: {.theorem title="Giroux"} Every \( (Y^3, \xi) \) with \( Y \) closed has a supporting open book decomposition. ::: ::: {.proposition title="?"} If an open book decomposition supports \( \xi_1 \) and \( \xi_2 \), then \( \xi_1 \) is isotopic to \( \xi_2 \). ::: ::: {.proof title="?"} Two steps: - Form a mapping cylinder of the monodromy map \( \phi \), - Extend over the binding, using the same idea as in Dehn surgery. Choose an area form \( \omega \) on \( \Sigma \) and a primitive \( \beta \) with \( d\beta = \omega \). Let \( \beta_1 \coloneqq\phi^*\beta \) and \( \beta_0 = \beta \), then set \[ \beta_t = t\beta_1 + (1-t)\beta_0 .\] This yields a 1-form on \( \Sigma\times I \) that extends to the mapping cylinder. Moreover \( d\beta_t = td\beta_1 + (1-t)d\beta_0 \) is an area form on \( \sigma\times \left\{{t}\right\} \) and \( \alpha = \,dt+ {\varepsilon}\beta_t \) is a contact form for small \( {\varepsilon}> 0 \). Then \( d\alpha = {\varepsilon}d\beta_t + {\varepsilon}\,dt\wedge \dot{\beta}_t \) and \( \alpha \wedge d\alpha = {\varepsilon}dt \wedge d\beta_t + { \mathsf{O}} ({\varepsilon}^2) \). ::: # Tuesday, February 15 Missed due to orthodontic appointment! Please send me notes. :) # Thursday, February 17 ::: {.remark} Let \( \Sigma \subseteq (Y^3, \xi) \). - Characteristic foliation: \( F = \xi \cap{\mathbf{T}}\Sigma \), complicated but necessary - Dividing set: a multicurve, simpler ::: ::: {.theorem title="?"} If \( \Sigma \) is convex with a dividing set \( \Gamma \) and \( F \) is any foliation divided by \( \Gamma \), there is a \( C^0{\hbox{-}} \)small isotopy \( \phi_t \) wt - \( \phi_0(\Sigma) = \Sigma, \phi_t(\Gamma) = \Gamma \) - \( \phi_t(\Sigma) \) is convex for all \( t \in [0, 1] \) - The characteristic foliation of \( \phi_1(\Sigma) \) is \( F \). ::: ::: {.remark} Idea: dividing sets give ways to detect overtwisted contact structures. ::: ::: {.remark} If \( \Sigma = S^2 \) and \( {\sharp}\Gamma \geq 2 \), then \( (Y, \xi) \) is overtwisted. Recall that an overtwisted disc is an embedded \( D^2 \) with Legendrian boundary such that \( \mathrm{tb}({{\partial}}D) = 0 \) and \( \mathrm{tw}(\xi, {{\partial}}D) \). ```{=html} ``` ![](figures/2022-02-17_11-24-46.png) Spheres can have exactly one dividing component. ::: ::: {.exercise title="?"} Generalize to an arbitrary number of components \( {\sharp}\Gamma = n \). ::: ::: {.remark} Same if \( \Sigma \neq S^2 \) and \( \Gamma \) contains a contractible curve. Contrapositively, if \( (Y, \xi) \) is tight, then either - \( \Sigma = S^2 \) and \( \Gamma \) is connected, or - \( \Sigma \neq S^2 \) and \( \Gamma \) has no contractible components. ::: ::: {.exercise title="?"} Consider tight contact structures on \( S^3 \). Choose Darboux \( B^3 \) neighborhoods at the ends, and note the interior is \( S^2\times [0, 1] \): ```{=html} ``` ![](figures/2022-02-17_11-37-35.png) The \( S^3\times \left\{{t_0}\right\} \) slices can be perturbed to be complex. So there is only one tight contact structure on \( S^3 \). ::: ::: {.remark} What can \( F \) look like on an \( S^2 \) in a tight \( (Y,\xi) \)? \( F \) can be perturbed to be Morse-Smale. - There are a finite number of elliptic/hyperbolic singularities - There are nondegenerate periodic orbits, either attracting or repelling - There are no saddle-saddle arcs - The limit sets are singularities or periodic orbits Dimension 3: strange attractors! Two types of limit sets: - \( \omega \) limit sets: \( x\in Y \) where there exists a sequence \( \left\{{t_1 < \cdots }\right\} \) with \( \phi(t_k)\to x \). - \( \alpha \) limit sets: \( x\in Y \) where there exists a sequence \( \left\{{t_1 > \cdots }\right\} \) with \( \phi(t_k)\to x \). ::: ::: {.remark} For \( S^2 \), take \( S^+ \) with an outward pointing vector field. ```{=html} ``` ![](figures/2022-02-17_11-50-05.png) There are no periodic orbits since \( (Y, \xi) \) is tight. The only limit sets are singular points. \( \chi(D) = 1 = {\sharp}e - {\sharp}h \). Stable manifold of \( h \): \( {\operatorname{Stab}}_h \) are \( x\in D^2 \) such that there exists a flow like with \( \phi(0) = x \) and \( \phi(t) \to h \) Form a 1-complex \( \displaystyle\bigcup_h { \operatorname{cl}} _X({\operatorname{Stab}}_h) \) -- this contains no cycles, thus this is a tree, and the dividing set is a neighborhood of the tree. ```{=html} ``` ![](figures/2022-02-17_11-58-12.png) ::: ::: {.proposition title="?"} If \( F \) on \( \Sigma \) is Morse-Smale, then it admits dividing curves. ::: ::: {.proof title="?"} Let \( G = \displaystyle\bigcup_h { \operatorname{cl}} ({\operatorname{Stab}}_h) \cup\displaystyle\bigcup_e e_t \) along with all of the repelling periodic orbits. Then \( \Gamma = {{\partial}}\nu(G) \) divides \( F \). ::: ::: {.theorem title="?"} If \( \Sigma \) is orientable, then there is a \( C^\infty \) small perturbation of \( F \) such that it is Morse-Smale. ::: ::: {.proposition title="?"} Every oriented \( \Sigma \subseteq (Y, \xi) \) can be perturbed to be convex. ::: ::: {.proof title="?"} Near \( \Sigma \), \( \alpha = \beta_t + \alpha_t \,dt \) and \( \beta_0 \) define \( F \). By Peixoto there exists \( \tilde \beta_t \) such that\( \tilde \beta_t \) defines a Morse-Smale \( F \). For \( {\left\lVert {\beta - \tilde\beta} \right\rVert}_{C^\infty} \ll {\varepsilon} \), \( \tilde \alpha = \tilde \beta_t + \alpha_t \,dt \) is contact. Then \( \alpha_s = s\tilde\alpha + (1-s)\alpha \) is a path of contact forms, so by Gray stability there is an isotopy \( \phi_s \) such that \( \phi_s^*(\alpha_s) = \lambda_s \alpha \) and we can take \( \phi_1(\Sigma) \) to be our surface. ::: ::: {.proposition title="?"} If \( (\Sigma, \tilde F) \) admits dividing curves, then it is convex. ::: # Thursday, February 24 ::: {.remark} Last time: there is a unique tight contact structure on \( S^3 \), using the existence of a contact structure on \( S^3\times I \). Next: tight contact structures on - \( T^2\times I \) - \( S^1\times {\mathbb{D}}^2 \) - \( L(p, q) \) - \( T^3 \) Given dividing sets of \( \Gamma_0, \Gamma_1 \in T^2\times I \), how can contact structures vary in a family. Tightness implies no contractible components in \( \Gamma \), so \( \Gamma \) consists of \( 2n \) embedded curves of slow \( p/q \). So the dividing set is governed by two parameters. ::: ::: {.remark} The only change to the dividing set in a generic family can be: - Retrograde saddle-saddle, yielding by pass moves. ```{=html} ``` ![](figures/2022-02-24_11-26-51.png) ::: ::: {.proposition} Given any contact structure on \( \Sigma \times I \) with dividing sets \( \Gamma_0, \Gamma_1 \), \( \xi \) is determined by a finite number of bypass moves. ::: ::: {.proof title="?"} Diagrams? \( \vdots \) ::: ::: {.remark} Given \( \Gamma_0 \) with slope \( p/q \) and \( \Gamma_1 \) with slope \( r/s \), form a Farey graph: ```{=html} ``` ![](figures/2022-02-24_12-00-24.png) \( \vdots \) ::: ::: {.proposition title="Legendrian Darboux"} If \( L \) is a Legendrian knot in \( M \), then a neighborhood of \( L \) is contactomorphic to a neighborhood of a zero section in \( J(S^1) \cong {\mathbb{R}}\times {\mathbf{T}} {}^{ \vee }S^1 \cong S^1\times {\mathbb{R}}^2 \).. ::: ::: {.remark} Write this in coordinates as \( (z, (x, y)) \), so \( \alpha = \,dz-y\,dx \) with \( x\in {\mathbb{R}}/{\mathbb{Z}} \). Then \( v(L) = \left\{{y^2+z^2\leq {\varepsilon}}\right\} \), \( y=r\cos\theta, z=r\sin\theta \). \( T^2 = \left\{{x, \theta}\right\}, { \left.{{\alpha}} \right|_{{T^2}} } = y\,d\theta-y\,dx= {\varepsilon}\cos\theta (\,d\theta- \,dx) \). Unwrap: ```{=html} ``` ![](figures/2022-02-24_12-15-41.png) Note that \( \,d\alpha> 0 \) at \( \pi/2 \) and \( \,d\alpha< 0 \) at \( 3\pi/2 \). Idea: given two unrelated surfaces with their own foliations, how do they interact at the boundary? Dividing sets on each can be extended into the annulus, and this reduces to a combinatorial problem of how to connected arcs: ```{=html} ``` ![](figures/2022-02-24_12-26-17.png) ::: # Tuesday, March 15 > See ::: {.remark} Last time: classifying tight contact structures on \( T^3 \). Some contact structure: \[ \xi_n = \ker( \cos(2\pi n z) \,dx- \sin(2\pi n z)\,dy) .\] Realize \( T^3 \) as a cube with faces glued, then moving in the \( z \) direction twists \( n \) times as you traverse the cube. We can reduce this to \( \xi_1 \) using \( {\left[ {x,y,z} \right]}\mapsto {\left[ {x,y,nz} \right]} \). ::: ::: {.remark} Goal: classify tight contact structures on lens spaces \( L_{p, q} = T^2\times I/\sim \). We can discretize the contact structure on \( \Sigma\times I \) into a finite number of *bypass moves* on the dividing sets. The basic move: ```{=html} ``` ![](figures/2022-03-15_11-30-20.png) ::: ::: {.definition title="Basic slice"} A **basic slice** is a contact structure on \( T^2 \) such that - \( T^2\times\left\{{0}\right\} \) is convex with 2 dividing curves of slope 0 - \( T^2\times\left\{{1}\right\} \) is convex with 2 dividing curves of slope -1 - \( \xi \) is tight - \( \xi \) is minimally twisting, so if \( T^2 \subseteq T^2\times I \) is convex then \( {\mathrm{slope}}(r) \in [-1, 0] \). ::: ::: {.proposition title="?"} There are exactly 2 basic slices. Both embed in \( (T^3, \xi_1) = \ker(\cos(2\pi z)\,dx- \sin(2\pi z)\,dy) = T^2\times I/\sim \), and are given by - \( (T_2\times [0, 1/8], \xi_1) \) - \( (T_2\times [1/2, 5/8], \xi_1) \) ::: ::: {.proof title="?"} Step 1: There are at most 2 basic slices. Reduce to \( S^1\times D^2 \) by removing a convex annulus. Note that \( T^2\times I\setminus(S^1\times I) \cong S^1\times I^2 \cong S^1\times D^2 \). ```{=html} ``` ![](figures/2022-03-15_11-40-54.png) Since the boundary is convex, we can make the foliations on both of the ruling curves of slope \( \infty \). > ? Take an annulus \( A \) with some condition on \( {{\partial}}A \), perturb to be convex? Something contradicts the "minimally twisting" assumption, involving these pics: ```{=html} ``` ![](figures/2022-03-15_11-51-21.png) Smooth corners? > ? ::: ::: {.definition title="Relative Euler class"} Let \( (M,\xi) \) be a contact 3-manifold with \( { \left.{{\xi}} \right|_{{{{\partial}}M}} } \) trivial. Let \( s \) be a nonvanishing section of \( { \left.{{\xi}} \right|_{{{{\partial}}M}} } \), then the **relative Euler class** \( e(\xi, s) \in H^2(M, {{\partial}}M;{\mathbb{Z}}) \cong H_1(M) \) (by Lefschetz duality) is the dual of the vanishing set of an extension of \( s \) to a section of \( \xi \) on \( M \). ::: ::: {.remark} In this case \( \dim s^{-1}= \dim M - \dim \xi \). ::: ::: {.lemma title="?"} If \( \Sigma \hookrightarrow(M,\xi) \) is a properly embedded convex surface and \( s \) is a section of \( { \left.{{\xi}} \right|_{{{{\partial}}M}} } \) that is tangent to \( {{\partial}}\Sigma \) with the correct orientation, then \[ {\left\langle {e(\xi, s)},~{\Sigma} \right\rangle} = \chi(\Sigma_+) - \chi(\Sigma_-) .\] where \( {\left\langle {{-}},~{{-}} \right\rangle}: H^2(M, {{\partial}}M;{\mathbb{Z}}) \times H_2(M, {{\partial}}M; {\mathbb{Z}}) \to {\mathbb{Z}} \). ::: ::: {.remark} Note \( H_2(T^2\times I, {{\partial}}; {\mathbb{Z}}) = \left\langle{[\alpha\times I], [\beta\times I]}\right\rangle \) where \( H_2(T^2) = \left\langle{\alpha, \beta}\right\rangle \). ::: # Tuesday, March 22 ## Farey Graphs ::: {.remark} Build a graph on the hyperbolic plane in the Poincare disc model: ```{=html} ``` ![](figures/2022-03-22_11-20-24.png) Here every midpoint corresponds to adding numerators and denominators respectively. Associate slopes: - \( 0/1 \leadsto 1\alpha + 0 \beta \) - \( 1/0 \leadsto 0\alpha + 1 \beta \) - \( 1/1 \leadsto 1\alpha + 1 \beta \) Any pair of these is a \( {\mathbb{Z}}{\hbox{-}} \)basis for \( H^1(T^2; {\mathbb{Z}})\cong {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{2} } } \). Use \( {\operatorname{SL}}_2({\mathbb{Z}}) \hookrightarrow{\operatorname{PSL}}_2({\mathbb{C}}) \) to realize any change of basis as an isometry of \( {\mathfrak{h}} \). This makes the interior/exterior of any tile isometric to the full upper/lower half-disc. ::: ::: {.remark} Basic moves: bypasses ```{=html} ``` ![](figures/2022-03-22_11-29-28.png) The first case corresponds to slopes \( r\in (-\infty , -1) \) and the second to \( r\in (-1, -1/2) \). Idea: the resulting dividing set is locally constant in perturbations of \( r \), provided one doesn't cross the endpoints of the curve for the bypass move. This produces a continued fraction defined inductively by \( r_0 = {\left\lfloor -{p\over q} \right\rfloor} \), writing \( -{p\over q} = r_0 -{1\over {p'/q'}} = - {q'\over p'} \) with \( -p/q < -p'/q' < -1 \) and thus \( 0 < -{p\over q} - r_0 < 1 \), so set \( r_1 = {\left\lfloor -{p'\over q'} \right\rfloor} \). This yields \[ -{p\over q} = r_0 - {1\over r_1 - {1\over r_2 - \cdots}} = [r_0, r_1,\cdots, r_m] ,\] which terminates in finitely many steps since \( p/q \) is rational. Note that \( r_i \leq -1 \implies {\left\lfloor r_i \right\rfloor}\leq -2 \). ::: ::: {.proposition title="?"} If \( r=-p/q = [r_0, \cdots, r_k] \) in a continued fraction expansion and \( s=a/b \) is the first point connected to \( p/q \) while moving counterclockwise from \( 0/1 \) on the Farey graph, then \( -a/b = [r_0, \cdots, r_{k}+1] \). ::: ::: {.remark} This gives the minimal graph path from \( p/q \) back to \( 0/1 \) by jumping the maximal distance along the circle to \( a/b \). Noting that \( [r_1, \cdots, r_{k-1}, -1] = [r_1, \cdots, r_{k-1} + 1] \) which is a shorter continued fraction. ::: ::: {.example title="?"} Let \( p/q = 53/17 \), then - \( r_0 = -4 = -68/17 \) - \( r_1 = -2 = -30/15 \) - \( r_2 = -2 = -26/13 \) - \( r_3 = -2 = \cdots \) So this yields \( [-4, -2, \cdots_7, -2, -3] \). ::: ::: {.remark} Idea: decompose \( p/q = [r_0, \cdots, r_k] \) surgery into integer surgeries on a link with \( k \) components. ::: # Tuesday, March 29 ::: {.remark} Goal: classification of tight contact structures on lens spaces. Lens spaces: \( L_{p, q} = S^3/C_p \) where the action is \( {\left[ {z_1, z_2} \right]} \mapsto {\left[ {e^{2\pi i\over p}, e^{2\pi iq \over p}} \right]} \) which has order \( p \). Note \( L_{p, q} \cong L_{p, q'} \) when \( q\equiv q'\operatorname{mod}p \), so we can assume \( -p 0 \). Write \( \phi = \sum x_i^2 + \sum y_i^2 \), then \( S^3 = \phi^{-1}(1) \). - Letting \( \rho = \sum x_i {\partial}x_i + \sum y_i {\partial}y_i = {1\over 2}\operatorname{grad}\phi \) in the standard metric yields a contact form \( \alpha = \omega(\rho, {-}) \). - Since \( { \left.{{\omega}} \right|_{{\xi_\text{std}}} } > 0 \), this yields an area form on contact planes. - There is also a complex structure \( J: {\mathbf{T}}_p {\mathbb{C}}^2 \to {\mathbf{T}}_p {\mathbb{C}}^2 \) where \( J({\partial}x_i) = {\partial}y_i \) and \( J({\partial}y_i) = - {\partial}x_i \) with a compatibility \( g(x, y) = \omega(x, Jy) \). ::: ::: {.definition title="Fillings"} A complex symplectic manifold \( (X^4, \omega, J) \) is a filling of \( (Y^3, \xi) \) if \( Y = {{\partial}}X \), - **Stein filling**: \( (X^4, J) \) is a Stein manifold, and \( \xi = {\mathbf{T}}Y \cap J({\mathbf{T}}Y) \). - **Strong filling**: if there is an outward pointing (Liouville) vector field \( \rho \) with \( {\mathcal{L}}_\rho \omega = \omega \) with \( \xi \ker (\omega(\rho, {-})) \) (which is always contact). Note \( {\mathcal{L}}_\rho \omega = d(\iota_\rho \omega) + \iota_\rho(d\omega) \) where the 2nd term vanishes for a symplectic form. - **Weak filling**: \( { \left.{{\omega}} \right|_{{\xi}} } > 0 \). > Note that we aren't defining what "Stein" means here. ::: ::: {.theorem title="?"} There are strict implications - Stein \( \implies \) - Strong \( \implies \) - Weak \( \implies \) - Tight. Note that the last implication is the harder part of the theorem. ::: ::: {.problem title="?"} Given \( (Y, \xi) \), classify all fillings. ::: ::: {.example title="?"} Consider \( (T^3, \xi_n) \) -- if \( n=1 \), this is Stein fillable, and for \( n\geq 2 \) these are weakly fillable but not strongly fillable. In this case, all of the filling manifolds are \( T^2 \times {\mathbb{B}}^2 \). ::: ::: {.example title="?"} For lens spaces \( L_{p, q} \) all tight contact structures are Stein fillable with the same smooth filling. Take the linear plumbing \( X \) of copies of \( S^2 \) corresponding to \( -p/q = {\left[ {r_1, r_2, \cdots, r_k} \right]} \) as a continued fraction expansion. They're distinguished by Chern classes \( c_1(T_X, J) \). ::: ::: {.example title="?"} Brieskorn spheres are examples of fillings, related to Milnor fibers. For \( p,q,r \geq 2 \), define \[ \Sigma(p,q,r) \coloneqq\left\{{F_{p,q,r}(x,y,z) = x^p + y^q + z^r = {\varepsilon}}\right\} \cap S^5 \subseteq {\mathbb{C}}^3 = \mathop{\mathrm{span}}_{\mathbb{C}}\left\{{x,y,z}\right\} .\] In this case, we have: ```{=html} ``` ![](figures/2022-04-05_11-58-26.png) Note that \( {\varepsilon}= 0 \) yields a singular variety, while \( {\varepsilon}>0 \) small yields a smooth manifold. ::: ::: {.exercise title="?"} Show \( \Sigma_{p,q,r} \) is the \( r{\hbox{-}} \)fold cyclic branched cover of \( S^3 \) over the torus knot \( T_{p, q} \). ::: ::: {.remark} Let \( J: {\mathbf{T}}X\to {\mathbf{T}}X \) with \( J^2 = -\operatorname{id} \), so the eigenvalues are \( \pm i \). So consider complexifying to \( {\mathbf{T}}_{\mathbb{C}}X \coloneqq{\mathbf{T}}X \otimes_{\mathbb{R}}{\mathbb{C}} \), so e.g. \( {\partial}x_k \mapsto (a_k + i b_k) {\partial}x_k \). This splits into positive (holomorphic) and negative (antiholomorphic) eigenspaces \( {\mathbf{T}}^{1, 0}_{\mathbb{C}}X \oplus {\mathbf{T}}^{0, 1}_{\mathbb{C}}X \). Take a change of basis \( {\left[ {x_1, y_1, x_2, y_2} \right]} \mapsto {\left[ {z_1, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_1, z_2, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_2} \right]} \) which yields \( {\partial}z = {1\over 2}\qty{{\partial}x - i {\partial}y} \) and \( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}z = {1\over 2}\qty{{\partial}x + i {\partial}y} \). ::: ::: {.exercise title="?"} Let \( f(z) = {\left\lvert {z} \right\rvert}^2 \) and check - \( {\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f = {\partial}(z d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) = dz \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = -2i (dx \wedge dy) \). - \( d = {\partial}+ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu} \) Practicing this type of change of variables is important! ::: ::: {.definition title="Levi forms and plurisubharmonicity"} Let \( \phi: X\to {\mathbb{R}} \) for \( X \) a complex manifold, then the **Levi form** of \( \phi \) is \[ {\mathcal{L}}\phi = {\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\phi = \sum_{i, j} {{\partial}^2 \over {\partial}z_j { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_k} dz_j \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_k ,\] generalizing the Hessian. The function \( \phi \) is **plurisubharmonic** if \( {\mathcal{L}}\phi \) is positive semidefinite at every point. ::: ::: {.example title="?"} Consider \( \phi: {\mathbb{C}}\to {\mathbb{R}} \), then \[ {\mathcal{L}}\phi &= {\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\phi \\ &= 2\qty{{1\over 2}\qty{\phi_x + i\phi_y}}\\ &= {1\over 2}\qty{ {1\over 2}\qty{\phi_{xx} - \phi_{xy}} + {1\over 2}\qty{ \phi_{yx} - i \phi_{yy}}} \\ &= {1\over 4}\qty{\phi_{xx} + \phi_{yy}}\\ &= {1\over 4}\Delta\phi ,\] so plurisubharmonic implies positive Laplacian. Note that in 1 dimension, \( \Delta f = 0 \implies f'' = 0 \), so \( (x, f(x)) \) is a straight line. In higher dimensions, \( f''>0 \) forces convexity, so secant lines are under the straight lines, hence the "sub" in subharmonic. ::: ::: {.proposition title="?"} If \( \phi: X\to {\mathbb{R}} \) is plurisubharmonic and \( 0 \) is a regular value, then \( (\phi^{-1}(0), \xi) \) (where \( \xi \) is its complex tangencies) forms a contact structure and the sub-level set \( \phi^{-1}(-\infty, 0] \) is a Stein filling. ::: ::: {.example title="A basic example of a plurisubharmonic function"} The radical function \( \phi: {\mathbb{C}}^3\to {\mathbb{R}} \) where \( \phi(z_1,z_2,z_3) = \sum {\left\lvert {z_i} \right\rvert}^2 \) is plurisubharmonic, as is its restriction to any submanifold of \( {\mathbb{C}}^3 \), including any filling of \( \Sigma_{p,q,r} \). Hard theorem: any Stein manifold and any Stein filling essentially comes from this construction. ::: # Thursday, April 21 > Note: student talks in previous weeks! ::: {.remark} Possible topics for the remainder of the class: - Open book decompositions - Every \( (Y^3, \xi) \) is homotopic to a contact structure. - Seifert fibered spaces ::: ## Seifert Fibered Spaces ::: {.remark} Brieskorn spheres \( \Sigma(p,r,q) \coloneqq\left\{{x^p + y^q + z^r = 0}\right\} \cap S_{\varepsilon}^5 \subseteq {\mathbb{C}}^3 \) are 3-manifolds *foliated* by \( S^1 \). Note that \( S_1\to X \to S^2 \) for \( X = S^3 \) or \( L(p, q) \) are actual fibrations. Idea: a foliation by \( F \)'s is a decomposition \( X = {\textstyle\coprod}F \to B \) which is a fibration with ramification in some fibers. ::: ::: {.definition title="Seifert fibered spaces"} A **Seifert fibered space** associated to \( (\Sigma, (p_1/q_1, \cdots, p_n/q_n)) \) with \( p_i/q_i\in {\mathbb{Q}} \) and \( \Sigma \) an orbifold surface is a 3-manifold \( Y \) and knots \( L_1,\cdots, L_n \) with neighborhoods \( \nu L_i \) such that - \( Y\setminus\cup_i\nu(L_i) = (\Sigma\setminus\left\{{{\operatorname{pt}}_1,\cdots, {\operatorname{pt}}_n}\right\}) \times S^1 \) - \( \nu L_i = S^1\times {\mathbb{D}}^2 \) is glued in by \( p_i/q_i \) Dehn surgery. ::: ::: {.example title="?"} \( L(p, q) \) is \( -p/q \) surgery on \( S^1 \), or by a slam-dunk move: ```{=html} ``` ![](figures/2022-04-21_11-46-26.png) \( \Sigma(p,q,r) \): ```{=html} ``` ![](figures/2022-04-21_11-48-50.png) ::: ::: {.exercise title="?"} Show that for \( \Sigma(p,q,r) \), removing the axes in \( {\mathbb{C}}^3 \) yields a trivial fibration by copies of \( S^1 \) over \( S^2\setminus{{\operatorname{pt}}_1,{\operatorname{pt}}_2,{\operatorname{pt}}_3} \) and check the surgery slopes. ::: ::: {.exercise title="?"} Prove that \( \Sigma(p,q,r) \) comes from the plumbing diagram for the Milnor fibration using Kirby calculus. ::: # Tuesday, April 26 ::: {.remark} Recall that \( \operatorname{PHS}^3 = \Sigma(2,3,5) \) has a Stein-fillable (and hence tight) contact structure. ::: ::: {.theorem title="?"} The negative \( -\Sigma(2,3,5) \) admits no tight contact structures. ::: ::: {.remark} Let \( S=S^3\setminus\left\{{ {\operatorname{pt}}_1,{\operatorname{pt}}_2,{\operatorname{pt}}_3 }\right\} \) be a pair of pants and consider \( X = S\times S^1 \). Note \( {{\partial}}X = T^2\cup T^2\cup T^2 \): ```{=html} ``` ![](figures/2022-04-26_11-20-51.png) Note \( -\Sigma(2,3,5) = \Sigma(2,-3,-5) \), since \( \operatorname{PHS}^3 \) is \( -1 \) surgery on the trefoil. Glue in 3 solid torii by \[ A_1 = { \begin{bmatrix} {2} & {-1} \\ {1} & {0} \end{bmatrix} }, \quad A_2 = { \begin{bmatrix} {3} & {1} \\ {-1} & {0} \end{bmatrix} }, \quad A_3 = { \begin{bmatrix} {5} & {1} \\ {-1} & {0} \end{bmatrix} } .\] acting on \( {\left[ {m, \lambda} \right]} \) in \( S^1\times {\mathbb{D}}^2 \): ```{=html} ``` ![](figures/2022-04-26_11-24-08.png) ::: ::: {.exercise title="?"} Show via Kirby calculus: ```{=html} ``` ![](figures/2022-04-26_11-25-37.png) ::: ::: {.lemma title="?"} There exist Legendrian representatives \( F_2, F_3 \) with twisting numbers \( m_2, m_3 = -1 \). ::: ::: {.proof title="?"} Idea: by stabilization, we can assume \( m_2,m_3 < 0 \), and the claim is that we can destabilize them back up to \( -1 \) simultaneously using bypass moves. Reduce to studying dividing sets on \( T^2\times I \) or \( S^1\times {\mathbb{D}}^2 \). Check that the dividing set has slope \( -1/2 \), which implies that there is an overtwisted disc. Reduce to \( 2\cdot 3\cdot 5 = 30 \) cases, check that an overtwisted disc can be found in each case. :::