Note: These are notes live-tex’d from a graduate course in Contact Topology taught by Peter Lambert-Cole at the University of Georgia in Spring 2022. As such, any errors or inaccuracies are almost certainly my own.
Last updated: 2022-05-29
References:
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:
Every diffeomorphism \(f: S^3\to S^3\) extends to a diffeomorphism \({\mathbb{B}}^4\to {\mathbb{B}}^4\).
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.
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.
There is a unique tight contact structure \(\xi_\text{std}\) on \(S^3\).
So up to isotopy, \(f\) fixes \(\xi_\text{std}\).
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.
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 \begin{align*} 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\} .\end{align*} 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.
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:
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
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 \begin{align*} {\left\lvert { {\left\langle { e(\xi)},~{ \Sigma } \right\rangle} } \right\rvert} \leq -g(\Sigma) ,\end{align*} which after juggling signs is a lower bound on the genus of any embedded surface.
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 \begin{align*} \chi = (e_+ + e_-) - (h_+ + h_-) ,\end{align*} by Poincaré-Hopf. On the other hand, \({\left\langle {e(\xi)},~{\Sigma} \right\rangle} = (e_+ - h_+) - (e_- - h_-)\), so adding this yields \begin{align*} {\left\langle {e(\xi)},~{\Sigma} \right\rangle} +\chi = 2(e_+ - h_+) \leq 0 .\end{align*} Isotope the surface to cancel critical points in pairs to get rid of caps/cups so that only saddles remain.
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.
On \({\mathbb{R}}^3\), \begin{align*} \alpha = dz - ydx \implies d\alpha = -dy \wedge dx = dx\wedge dy ,\end{align*} so \begin{align*} \alpha \wedge d\alpha = (dz-ydx)\wedge(dx\wedge dy) = dz\wedge dx\wedge dy = dx\wedge dy\wedge dz .\end{align*}
On \({\mathbb{R}}^5\), set \(\alpha = dz-y_1 dx_1 - y_2 dx_2\). Check that \begin{align*} \alpha\wedge (d\alpha)^2 = 2(dz \wedge dx_1 \wedge dy_1 \wedge dx_2 \wedge dy_2) .\end{align*}
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.
Forms are more rigid than structures: if \(f>0\) and \(\alpha\) is contact, then \(f\cdot \alpha\) is also contact with \(\ker( \alpha ) = \ker(f \alpha)\).
On \({\mathbb{R}}^3\), a local model is \(\alpha\mathrel{\vcenter{:}}=\,dz- y\,dx\).
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.
On \({\mathbb{R}}^3\), take \(\alpha_2 \mathrel{\vcenter{:}}= dz + x\,dy\) 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.
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 \(\theta\).
Set \(\alpha_4 = \,dz+ {1\over 2}(x\,dy- y\,dx)\), then this is equal to \(\alpha_3\) in rectangular coordinates.
Set \begin{align*} \alpha_4 = \cos(r^2)\,dz+ \sin(r^2)\,d\theta .\end{align*}
Compute the exterior derivative and check that this yields a contact structure.
Now note that \begin{align*} \alpha = 0 \implies {\frac{\partial z}{\partial \theta}\,} = -{\sin(r^2)\over \cos(r^2)} = -\tan(r^2) ,\end{align*} 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 \(\pi\).
A contactomorphism is a diffeomorphism \begin{align*} \psi: (Y_1^3, \xi_1) \to (Y_2^3,\xi_2) \end{align*} such that \(\phi_*(\xi_1) = \xi_2\) (tangent vectors push forward).
A strict contactomorphism is a diffeomorphism \begin{align*} \phi: (Y_1^3, \ker \alpha_1) \to (Y_2^3, \ker \alpha_2) .\end{align*} such that \(\phi^*( \alpha_2) = \alpha_1\) (forms pull back).
Strict contactomorphisms are more important for dynamics or geometric applications.
Prove that \(\alpha_1,\cdots, \alpha_4\) are all contactomorphic.
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:
So we can write \begin{align*} \lambda = \sum y_i \,dx_i \in \Omega^1({\mathbf{T}} {}^{ \vee }X) ,\end{align*} regarding the \(y_i\) as functions on \({\mathbf{T}} {}^{ \vee }X\) and \(\,dx_i\) as 1-forms on \({\mathbf{T}} {}^{ \vee }X\).
Find out what \(\beta = \sum a_i \,dx_i\) is equal to as a section of \({\mathbf{T}} {}^{ \vee }X\).
To get a contact manifold of dimension \(2n+1\), consider the 1-jet space \(J^1(X) \mathrel{\vcenter{:}}= T {}^{ \vee }X \times {\mathbb{R}}\). 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 \(\alpha \mathrel{\vcenter{:}}=-{ \left.{{\lambda}} \right|_{{{\mathbb{S}}{\mathbf{T}} {}^{ \vee }X}} }\) is contact.
Check that \({\mathbb{R}}^3 = J^1({\mathbb{R}})\) and \({\mathbb{S}}{\mathbf{T}} {}^{ \vee }({\mathbb{R}}^2) = {\mathbb{R}}^2 \times S^1\).
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.
Define \begin{align*} \alpha_t = \,dz- ty\,dx\qquad t\in {\mathbb{R}} \end{align*} to get a 1-parameter family of 1-forms. Check that \(\alpha_t \wedge d \alpha_t = t(\,dz\wedge \,dx\wedge \,dy)\). Consider \(t\in (-{\varepsilon}, {\varepsilon})\):
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.
Any \(\xi \mathrel{\vcenter{:}}=\ker \alpha\) is integrable iff for all vector fields \(X, Y \subseteq \xi\), their Lie bracket \([X, Y] \subseteq \xi\).
For \(\alpha\) nonvanishing on \(Y^3\), \(\ker \alpha\) is tangent to a foliation by surfaces iff \(\alpha \wedge d\alpha = 0\).
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.
A theorem of Eliashberg and Thurston: taut foliations can be perturbed to a (tight) positive contact structure.
Refs:
For \(S^3 \subseteq {\mathbb{C}}^2\), define a form on \({\mathbb{R}}^4\) as \begin{align*} \alpha \mathrel{\vcenter{:}}=-y_1 dx_1 + x_1 dy_1 - y_2 dx_2 + x_2 dy_2 .\end{align*} Then then standard contact form on \(S^3\) is \begin{align*} \xi_{\text{std}} \mathrel{\vcenter{:}}=\ker { \left.{{\alpha}} \right|_{{S^3}} } .\end{align*}
Show that \(\alpha\) defines a contact form.
Write \(f = x_1^2 + y_1^2 + x_2^2 + y_2^2\), then \begin{align*} { \left.{{ \alpha}} \right|_{{S^3}} } \wedge { \left.{{ d\alpha}} \right|_{{S^3}} } > 0 \iff df\wedge d \alpha \wedge d \alpha > 0 .\end{align*}
Check that
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 \(\xi_p = \ker dy_1 = \mathop{\mathrm{span}}\left\{{{\partial}x_1, {\partial}y_2}\right\}\).
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.
An almost complex structure on \(X\) is a bundle automorphism \(J: {\mathbf{T}}X{\circlearrowleft}\) with \(J^2 = -\operatorname{id}\).
For \(X = {\mathbb{C}}^2\), take \begin{align*} {\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 .\end{align*}
Show that \(f: {\mathbb{C}}\to {\mathbb{C}}\) is holomorphic if \(df \circ J = J \circ df\), which corresponds to the Cauchy-Riemann equations.
Given \(J:W\to W\), an \({\mathbb{R}}{\hbox{-}}\)subspace \(V \leq W\) is a \({\mathbb{C}}{\hbox{-}}\)subspace iff \(J(V) = V\).
The field of \(J{\hbox{-}}\)complex tangents is the hyperplane field \begin{align*} \xi_p \mathrel{\vcenter{:}}={\mathbf{T}}S^3 \cap J({\mathbf{T}}S^3) .\end{align*}
Consider \({\mathbf{T}}_p S^3\) for \(p={\left[ {1,0,0,0} \right]}\), then \begin{align*} 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\} ,\end{align*} so \(\xi_p = \mathop{\mathrm{span}}{ {\partial}x_1, {\partial}y_2}\) is the intersection and coincides \(\xi_{\text{std}}\).
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 \(\omega \mathrel{\vcenter{:}}=\bigwedge dx_i \wedge \bigwedge dy_i\), then \(\alpha = \iota_p\omega \mathrel{\vcenter{:}}=\omega(p, {-})\) is the interior product of \(\omega\). Then \begin{align*} \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 .\end{align*} So the contact form comes from pairing the symplectic form against a radial vector field.
Recall \(f \mathrel{\vcenter{:}}=\sum x_i^2 + \sum y_i^2\) 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 \(J = i\),
So \(J^*(dx) = -dy\), and \begin{align*} J^*(df) = 2x_1 (-dy_1) + 2y_1 (dx_1) + 2x_2 (-dy_2) + 2y_2 (dx_2) = -2 \alpha .\end{align*}
Thus \(J^*(df)\) is a rotation of \(df\) by \(\pi/2\).
The field of complex tangencies along \(Y = f^{-1}(0)\) is the kernel of \({ \left.{{ df(J({-})) }} \right|_{{Y}} }\).
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.
The standard contact structure is orthogonal to the Hopf fibration: define a map \begin{align*} {\mathbb{C}}^2\setminus\left\{{0}\right\}&\to {\mathbb{CP}}^1 \cong S^2 \\ {\left[ {z, w} \right]} &\mapsto {\left[ {z: w} \right]} ,\end{align*} 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 \(S^1\).
Take \begin{align*} {\mathbb{C}}^2 &\to {\mathbb{R}}^2 \\ (z_1, z_2)&\mapsto ({\left\lvert {z_1} \right\rvert}, {\left\lvert {z_2} \right\rvert}) .\end{align*} Consider the image of \(S^2 = \left\{{{\left\lvert {z_1} \right\rvert}^2 + {\left\lvert {z_2} \right\rvert}^2 = 1}\right\}\):
The preimage is \(S^1\times S^1\). This can be realized as a tetrahedron with sides identified:
There are Hopf fibers on the ends, and undergo a \(\pi/2\) twist as you move through the tetrahedron.
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.
If \((Y^3, \xi)\) is contact then for every point \(p\) there is a chart \(U\) with coordinates \(x,y,z\) where \(\xi = \ker (\,dz-y\,dx) = \ker (\alpha_\text{std})\).
Locally, all contact structures (not necessarily forms) look the same. The mantra: local flexibility vs global rigidity.
Two proofs:
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.
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 \begin{align*} \alpha = f\,dx+ g\,dz+ h\,dt= f\,dx+ g\,dz .\end{align*} 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 \begin{align*} \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 ,\end{align*} meaning \(f_t <0\) and we can set \(y = f(x,z,t)\). This yields \begin{align*} \alpha = \,dz+ f\,dx= \,dz- y\,dx .\end{align*}
By a linear change of coordinates, choose \(x,y\) along \(\xi\) to write \(\alpha_p = \,dz\) and \(\xi_p = \mathop{\mathrm{span}}{{\partial}x, {\partial}y}\):
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.
In a neighborhood of \(p\), there is a family \(\alpha_t\) for \(t\in [0, 1]\).
To obtain this, interpolate: \begin{align*} 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 .\end{align*} 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 \(p\).
Note that \(\dot\alpha_t \mathrel{\vcenter{:}}={\frac{\partial }{\partial t}\,} \alpha_t\), so \begin{align*} {\frac{\partial }{\partial t}\,} \qty{ t\alpha_1 + (1-t) \alpha_0} = \alpha_1 - \alpha_0 .\end{align*}
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 \begin{align*} \Phi_t^*( \dot \alpha_t + {\mathcal{L}}_{V_t} \alpha_t ) = \dot f_t \alpha_0 .\end{align*}
See Prop 6.4 in Cannas da Silva.
✨Cartan’s magic formula✨: \begin{align*} {\mathcal{L}}_V(\alpha) = d(\iota_V \alpha) + \iota_V(d\alpha) ,\end{align*} so \begin{align*} {\mathcal{L}}_{V_t}(\alpha_t) = d(\alpha_t(V_t)) + d\alpha_t(V_t, {-}) = 0 + d\alpha_t(V_t, {-}) .\end{align*}
We can thus write this equation as \begin{align*} \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 } .\end{align*} Applying \((\Phi_t^*)^{-1}\) yields \begin{align*} \dot \alpha_t + d\alpha_t(V_t, {-})= {\dot f_t \over f_t }\alpha_t .\end{align*} Now try to solve this for \(V_t\). Let \(R_t\) be the Reeb vector field of \(\alpha_t\), which satisfies
Then \begin{align*} \dot \alpha_t (R_t) = {\dot f_t \over f_t} = {\frac{\partial }{\partial t}\,} \log(f_t) \mathrel{\vcenter{:}}=\mu_t ,\end{align*} so \(\dot\alpha_t(R_t)\) determines \(f_t\) by first integrating and exponentiating.
We now need to solve \begin{align*} { \left.{{d\alpha_t(V_t, {-})}} \right|_{{\xi_t}} } = { \left.{{\mu_t \alpha_t - \dot \alpha_t}} \right|_{{\xi_t}} } .\end{align*} 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.
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.
For \(\phi_t\) an isotopy generated by the flow of \(X_t\) and \(\alpha_t\) a family of 1-forms, \begin{align*} {\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} ) .\end{align*}
Write \begin{align*} \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 ,\end{align*} 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 \(t\): \begin{align*} \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 } .\end{align*} Apply \((\phi^*_{t_0})^{-1}\): \begin{align*} \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) .\end{align*} 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: \begin{align*} \dot \alpha_t + d\alpha_t(V_t) + \iota_{V_t} d\alpha_t = \mu_t \alpha_t ,\end{align*} and \(d\alpha_t(V_t) = 0\), so \begin{align*} \iota_{V_t}d\alpha_t = \mu_t \alpha_t - \dot \alpha_t .\end{align*} Plug in the Reeb vector field \(R_t\), then \(\alpha_t(R_t) = 0\) so \(\mu_t = \dot \alpha_t (R_t)\).
Let \(Y\) be n \(S^3 \subseteq {\mathbb{C}}^2\) that is transverse to the radial vector field. Then \begin{align*} \alpha = x_1 dy_1 - y_1 dx_1 + x_2 dy_2 - y_2 dx_2\mathrel{\Big|}_y \end{align*} defines the standard tight contact structure.
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.
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 \(\alpha(L) = 0\):
This is a closed condition.
\(L\) is transverse if it is everywhere transverse to \(\xi\), so \(\alpha(L) > 0\):
This is an open condition.
Every Legendrian knot has a transverse pushoff (up to transverse isotopy). Every transverse knot has a Legendrian approximation.
Take \({\mathbb{R}}^3\) and \(\alpha_\text{std}= dz-ydx\), then the \(y{\hbox{-}}\)axis \(L_1 \mathrel{\vcenter{:}}=\left\{{{\left[ {0,t,0} \right]}}\right\}\) 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 \(L_3 \mathrel{\vcenter{:}}=\left\{{{\left[ {t, -{\varepsilon}, 0} \right]}}\right\}\) is transverse since \({ \left.{{\alpha}} \right|_{{L_3}} } = {\varepsilon}dx >0\).
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 \(\alpha \mathrel{\vcenter{:}}=\,dz+ r^2 \,d\theta\).
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 \(t\).
\(\gamma(s) = {\left[ {x(s), y(s), z(s)} \right]}\) and \(\xi = \ker \alpha, \alpha = \,dz- y\,dx\). Then \(\gamma^*(\alpha) = z' \,ds- yx^1\,ds= (z' - yx')\,ds\), which is Legendrian iff \(y=z'/x'\).
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 \(z'/x' = f'/s' = f'\).
There are two projections:
Let \(\gamma(s) = {\left[ {s^2, {3\over 2}s, s^3} \right]}\), then the two projections are as follows:
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:
A front diagram of the unknot:
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: \(\sup_{s\in I} {\left\lvert {\gamma_1(s) - \gamma_2(s)} \right\rvert} \leq {\varepsilon}\)?
Note that \(\mathsf{Lie}({\operatorname{SO}}_3) \mathrel{\vcenter{:}}={\mathbf{T}}_e({\operatorname{SO}}_3) = {\mathfrak{su}}_2\), spanned by roll, pitch, and yaw generators:
So measuring the number of rotations along each generator after traversing \(L\) in a full loop yields integer invariants.
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.
How to compute: take a pushoff and compute the linking number:
\begin{align*} \mathrm{tb}(L) = w(L) - {1\over 2}C(L) ,\end{align*} where \(w(L)\) is the writhe and \(C(L)\) is the number of cusps.
The linking number is \({1\over 2}(c_+(L) - c_-(L))\), half of the signed number of crossings.
Here all 4 crossing have the same sign:
TB for the knots from before:
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.
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.
It turns out that \begin{align*} \operatorname{rot}(L) = {1\over 2}\qty{ {\sharp}\text{down cusps} - {\sharp}\text{up cusps}} .\end{align*}
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:
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.
Disallowed moves:
Allowed moves:
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 \((t,r) = (\mathrm{tb}(L), \operatorname{rot}(L))\)?
For \(K\) the unknot:
So these numerical pairs fall into a cone:
For \(L \subseteq {\mathbb{R}}^3\) a Legendrian knot, \begin{align*} \mathrm{tb}(L) + \operatorname{rot}(L) \equiv 1 \operatorname{mod}2 .\end{align*}
Note that \(\chi(S) \equiv 1\operatorname{mod}2\) for \(S\) a Seifert surface.
For any Seifert surface \(S\), \begin{align*} \mathrm{tb}(L) + {\left\lvert { \operatorname{rot}(L) } \right\rvert} \leq -\chi(S) .\end{align*}
This solves the geography problem: this cone contains all of the possible pairs.
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 \(L_2\).
This solves the botany problem: every red dot has exactly one representative.
Other knots are Legendrian simple, e.g. the trefoil. A theorem of Checkanov says the following \(5_2\) knots are not Legendrian isotopic:
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 \({\left[ {r,\theta, z} \right]} = \gamma(t) \mathrel{\vcenter{:}}={\left[ {1, t, 0} \right]}\), 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:
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.
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.
\(({\mathbb{R}}^3, \xi_\text{std})\) is a tight contact structure.
For every closed oriented \(Y^3\), every homotopy class of 2-plane fields on \(Y\) contains a unique (up to isotopy) overtwisted contact structure.
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 \({ \left.{{\xi}} \right|_{{S}} }\).
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:
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.
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.
Take \(\alpha = \,dz+x\,dy- y\,dx\) and \(\Sigma = S^2\), then the singular foliation is given by
Two possible types of singularities, the local models:
There are also two numerical invariants:
A theorem \begin{align*} {\left\langle {c(\Sigma)},~{ \Sigma} \right\rangle} = (e_+ - h_+) - (e_- - h_-) .\end{align*} If \(\Sigma\) is transverse, \(\mathrm{sl}({{\partial}}\Sigma, \Sigma) = -(e_+ - h_+) + (e_- - h_-)\).
\(\sigma\) 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 \(\alpha(i) = x^2+y^2 = 1 > 0\).
Here \(\mathrm{sl} = 1\). To compute \(\mathrm{sl}\):
Set
Then \begin{align*} x\rho - y\theta = x(x{\partial}_x + y{\partial}_y) - y(-y {\partial}_x + x{\partial}_y) = (x^2+y^2)\,dx .\end{align*}
Then
Example:
So here \(\operatorname{lk}(U, \tilde U) = -1\):
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 \(V = y{\partial}_y + dx{\partial}_x\in \ker({ \left.{{\alpha }} \right|_{{\Sigma}} })\).
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 \begin{align*} {\left\langle {e({{\mathbf{T}}X} )},~{[X]} \right\rangle} = \chi(X) .\end{align*} 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, \begin{align*} {\left\langle {e({\mathbf{T}}X, t)},~{[X]} \right\rangle} = \chi(X) .\end{align*}
Note \(\chi({\mathbb{D}}) = 1\), so any vector field has a singularity?
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 \begin{align*} c(E) = e({ \left.{{E}} \right|_{{\Sigma_1}} }, \tau) + c({ \left.{{E}} \right|_{{\Sigma_2}} }, \tau) .\end{align*}
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
\(\chi(\Sigma) = e({\mathbf{T}}\Sigma, V) \in H^2(\Sigma, {{\partial}}\Sigma) \cong H_0(\Sigma)\)
\(\mathrm{sl}({{\partial}}\Sigma, \Sigma) = e(\xi, V) \in H^2(\Sigma, {{\partial}}\Sigma)\)
Proof: near a zero, \(V\) determines a map \(S^1\to S^1\) and the contribution to \(e\) is the degree of this map.
Proof: exercise.
Bennequin inequality: \begin{align*} \mathrm{sl}(T, \Sigma) \leq -\chi(\Sigma) \implies e_+ + h_+ + e_- + h_- \leq -(e_+ + e_- - h_+ - h_-) \iff e_- \leq h_- .\end{align*} Try to cancel in pairs:
The inequality follows if we can cancel every \(e_-\) with some \(h_-\).
Topics for talks:
Every closed oriented 3-manifold \(Y\) admits a (positive) contact form.
Three proofs:
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.
Recall:
\begin{align*} {\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]} .\end{align*} If \(p, q\) are coprime there exist \(m,n\) with \(pm-qn = 1\). So define \begin{align*} \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]} ,\end{align*} so \begin{align*} \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 .\end{align*} 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 \begin{align*} 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 ,\end{align*} which happens iff \begin{align*} \operatorname{det} \begin{bmatrix} h_2 & h_2' \\ h_1 & h_1' \end{bmatrix} > 0 .\end{align*}
Think of \({\left[ {h_2, h_1} \right]}\) as a path with tangent vector \({\left[ {h_2', h_1'} \right]}\). This requires moving counterclockwise.
An open book decomposition of \(Y\) is a pair \((B, \pi)\) where
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 \begin{align*} Y\setminus\nu(B) \cong {\Sigma \times I \over \phi(x) \times\left\{{0}\right\} \sim x\times\left\{{1}\right\}} ,\end{align*} which is a glued cylinder:
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 \((B, \xi)\).
Every open book decomposition admits a contact structure.
Every \((Y^3, \xi)\) with \(Y\) closed has a supporting open book decomposition.
If an open book decomposition supports \(\xi_1\) and \(\xi_2\), then \(\xi_1\) is isotopic to \(\xi_2\).
Two steps:
Choose an area form \(\omega\) on \(\Sigma\) and a primitive \(\beta\) with \(d\beta = \omega\). Let \(\beta_1 \mathrel{\vcenter{:}}=\phi^*\beta\) and \(\beta_0 = \beta\), then set \begin{align*} \beta_t = t\beta_1 + (1-t)\beta_0 .\end{align*} 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 \(\alpha \wedge d\alpha = {\varepsilon}dt \wedge d\beta_t + { \mathsf{O}} ({\varepsilon}^2)\).
Missed due to orthodontic appointment! Please send me notes. :)
Let \(\Sigma \subseteq (Y^3, \xi)\).
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
Idea: dividing sets give ways to detect overtwisted contact structures.
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 \(\mathrm{tw}(\xi, {{\partial}}D)\).
Spheres can have exactly one dividing component.
Generalize to an arbitrary number of components \({\sharp}\Gamma = n\).
Same if \(\Sigma \neq S^2\) and \(\Gamma\) contains a contractible curve. Contrapositively, if \((Y, \xi)\) is tight, then either
Consider tight contact structures on \(S^3\). Choose Darboux \(B^3\) neighborhoods at the ends, and note the interior is \(S^2\times [0, 1]\):
The \(S^3\times \left\{{t_0}\right\}\) slices can be perturbed to be complex. So there is only one tight contact structure on \(S^3\).
What can \(F\) look like on an \(S^2\) in a tight \((Y,\xi)\)? \(F\) can be perturbed to be Morse-Smale.
Dimension 3: strange attractors! Two types of limit sets:
For \(S^2\), take \(S^+\) with an outward pointing vector field.
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.
If \(F\) on \(\Sigma\) is Morse-Smale, then it admits dividing curves.
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 \(F\).
If \(\Sigma\) is orientable, then there is a \(C^\infty\) small perturbation of \(F\) such that it is Morse-Smale.
Every oriented \(\Sigma \subseteq (Y, \xi)\) can be perturbed to be convex.
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.
If \((\Sigma, \tilde F)\) admits dividing curves, then it is convex.
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
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.
The only change to the dividing set in a generic family can be:
Retrograde saddle-saddle, yielding by pass moves.
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.
Diagrams?
\(\vdots\)
Given \(\Gamma_0\) with slope \(p/q\) and \(\Gamma_1\) with slope \(r/s\), form a Farey graph:
\(\vdots\)
If \(L\) is a Legendrian knot in \(M\), then a neighborhood of \(L\) is contactomorphic to a neighborhood of a zero section in \(J(S^1) \cong {\mathbb{R}}\times {\mathbf{T}} {}^{ \vee }S^1 \cong S^1\times {\mathbb{R}}^2\)..
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:
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:
Last time: classifying tight contact structures on \(T^3\). Some contact structure: \begin{align*} \xi_n = \ker( \cos(2\pi n z) \,dx- \sin(2\pi n z)\,dy) .\end{align*} 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 \({\left[ {x,y,z} \right]}\mapsto {\left[ {x,y,nz} \right]}\).
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:
A basic slice is a contact structure on \(T^2\) such that
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
Step 1: There are at most 2 basic slices. Reduce to \(S^1\times D^2\) by removing a convex annulus. Note that \(T^2\times I\setminus(S^1\times I) \cong S^1\times I^2 \cong S^1\times D^2\).
Since the boundary is convex, we can make the foliations on both of the ruling curves of slope \(\infty\).
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Take an annulus \(A\) with some condition on \({{\partial}}A\), perturb to be convex? Something contradicts the “minimally twisting” assumption, involving these pics:
Smooth corners?
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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 \(M\).
In this case \(\dim s^{-1}= \dim M - \dim \xi\).
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 \begin{align*} {\left\langle {e(\xi, s)},~{\Sigma} \right\rangle} = \chi(\Sigma_+) - \chi(\Sigma_-) .\end{align*} where \({\left\langle {{-}},~{{-}} \right\rangle}: H^2(M, {{\partial}}M;{\mathbb{Z}}) \times H_2(M, {{\partial}}M; {\mathbb{Z}}) \to {\mathbb{Z}}\).
Note \(H_2(T^2\times I, {{\partial}}; {\mathbb{Z}}) = \left\langle{[\alpha\times I], [\beta\times I]}\right\rangle\) where \(H_2(T^2) = \left\langle{\alpha, \beta}\right\rangle\).
Build a graph on the hyperbolic plane in the Poincare disc model:
Here every midpoint corresponds to adding numerators and denominators respectively.
Associate slopes:
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.
Basic moves: bypasses
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 \begin{align*} -{p\over q} = r_0 - {1\over r_1 - {1\over r_2 - \cdots}} = [r_0, r_1,\cdots, r_m] ,\end{align*} which terminates in finitely many steps since \(p/q\) is rational. Note that \(r_i \leq -1 \implies {\left\lfloor r_i \right\rfloor}\leq -2\).
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 \(-a/b = [r_0, \cdots, r_{k}+1]\).
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.
Let \(p/q = 53/17\), then
So this yields \([-4, -2, \cdots_7, -2, -3]\).
Idea: decompose \(p/q = [r_0, \cdots, r_k]\) surgery into integer surgeries on a link with \(k\) components.
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 \(-p<q\leq 0\), so \(p/q < -1\).
Some examples:
There is a genus 1 Heegaard splitting. The double branched cover of a 2-bridge link is a lens space:
All lens spaces can be generated by genus 1 Heegaard splittings?
\(-p/q\) Dehn surgery is equivalent to a sequence of linked unknots with numbers \(r_1,\cdots, r_k\). When can this be done in a way that preserves the contact structure? Idea: Legendrian surgery, which removes a Legendrian knot and reglues.
Let \(L\) be Legendrian and \(\nu(L)\) is a standard neighborhood (so standard contact structure). Then \({{\partial}}\nu(L) \cong T^2\) is convex with 2 dividing curves, where “slope” is the contact framing. For \({\left[ {\theta, x, y} \right]} \in S^1\times {\mathbb{R}}^2\), set \(\alpha = \,dx+ y\,d\theta\). Then \({\left[ {\theta, 0, 0} \right]}\) is Legendrian. When can we extend \(\xi\) uniquely across surgery \(S^1\times {\mathbb{D}}^2\)? Need to attach handles along integer framing (choice of integer in \(\pi_1 {\operatorname{SO}}_2({\mathbb{R}}) \cong {\mathbb{Z}}\) corresponding to trivializing the normal bundle \(\nu (K)\) in an embedding). Need good surgery slopes: \(\left\{{n}\right\}_{n\in {\mathbb{Z}}} \cap\left\{{1\over k}\right\}_{k\in {\mathbb{Z}}} = \left\{{\pm 1}\right\}\), relative to the tb-framing. So \(\mathrm{tb}-1\) is the best framing.:
Stabilize up to \(r_K+1\) on each Legendrian knot. Fact: yields a Stein fillable thing, implies tight contact structure.
There are \(-r_0-1\) ways to perform \(-r_0 - 2\) stabilizations. E.g. for \(-r_0-2 = 3\), break into positive and negative stabilizations:
So there are \(\prod_{1\leq i\leq k} (-r_k - 1)\) tight contact structures on \(-p/q = {\left[ {r_0, \cdots, r_k} \right]}\).
Consider \((S^3, \xi_\text{std}) \subseteq {\mathbb{C}}^2\); some things that are true:
A complex symplectic manifold \((X^4, \omega, J)\) is a filling of \((Y^3, \xi)\) if \(Y = {{\partial}}X\),
Note that we aren’t defining what “Stein” means here.
There are strict implications
Note that the last implication is the harder part of the theorem.
Given \((Y, \xi)\), classify all fillings.
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 \(T^2 \times {\mathbb{B}}^2\).
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 \(c_1(T_X, J)\).
Brieskorn spheres are examples of fillings, related to Milnor fibers. For \(p,q,r \geq 2\), define \begin{align*} \Sigma(p,q,r) \mathrel{\vcenter{:}}=\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\} .\end{align*}
In this case, we have:
Note that \({\varepsilon}= 0\) yields a singular variety, while \({\varepsilon}>0\) small yields a smooth manifold.
Show \(\Sigma_{p,q,r}\) is the \(r{\hbox{-}}\)fold cyclic branched cover of \(S^3\) over the torus knot \(T_{p, q}\).
Let \(J: {\mathbf{T}}X\to {\mathbf{T}}X\) with \(J^2 = -\operatorname{id}\), so the eigenvalues are \(\pm i\). So consider complexifying to \({\mathbf{T}}_{\mathbb{C}}X \mathrel{\vcenter{:}}={\mathbf{T}}X \otimes_{\mathbb{R}}{\mathbb{C}}\), 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 \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}z = {1\over 2}\qty{{\partial}x + i {\partial}y}\).
Let \(f(z) = {\left\lvert {z} \right\rvert}^2\) and check
Practicing this type of change of variables is important!
Let \(\phi: X\to {\mathbb{R}}\) for \(X\) a complex manifold, then the Levi form of \(\phi\) is \begin{align*} {\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 ,\end{align*} generalizing the Hessian. The function \(\phi\) is plurisubharmonic if \({\mathcal{L}}\phi\) is positive semidefinite at every point.
Consider \(\phi: {\mathbb{C}}\to {\mathbb{R}}\), then \begin{align*} {\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 ,\end{align*} 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.
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.
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.
Note: student talks in previous weeks!
Possible topics for the remainder of the class:
Brieskorn spheres \(\Sigma(p,r,q) \mathrel{\vcenter{:}}=\left\{{x^p + y^q + z^r = 0}\right\} \cap S_{\varepsilon}^5 \subseteq {\mathbb{C}}^3\) 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.
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
\(L(p, q)\) is \(-p/q\) surgery on \(S^1\), or by a slam-dunk move:
\(\Sigma(p,q,r)\):
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.
Prove that \(\Sigma(p,q,r)\) comes from the plumbing diagram for the Milnor fibration using Kirby calculus.
Recall that \(\operatorname{PHS}^3 = \Sigma(2,3,5)\) has a Stein-fillable (and hence tight) contact structure.
The negative \(-\Sigma(2,3,5)\) admits no tight contact structures.
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 \({{\partial}}X = T^2\cup T^2\cup T^2\):
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 \begin{align*} 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} } .\end{align*} acting on \({\left[ {m, \lambda} \right]}\) in \(S^1\times {\mathbb{D}}^2\):
Show via Kirby calculus:
There exist Legendrian representatives \(F_2, F_3\) with twisting numbers \(m_2, m_3 = -1\).
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.