*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.
*

dzackgarza@gmail.com

Last updated: 2022-05-29

References:

- https://www.ma.imperial.ac.uk/~ssivek/courses/12s-math273.php
- https://web.ma.utexas.edu/users/gdavtor/notes/contact_notes.pdf
- PLC’s Notes

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:

- Stabilization and destabilization:

- Conjugation in \(B_n\):

- Braid isotopy, which preserves braid words in \(B_n\).

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.

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 \(t\in (-{\varepsilon}, {\varepsilon})\):

- \(t>0 \implies \alpha = \,dz- y\,dx\) 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.

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:

- Geiges, Intro to Contact
- Ozbogi-Stipsicz
- Etnyre lecture notes
- Massot
- Sivck

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

- \(d\alpha = 2(dx_1 \wedge dy_1) + 2(dx_2 + dy_2)\)
- \(df = 2(x_1 dx_1 + y_1 dy_1) + 2(x_2 dx_2 + y_2 dy_2)\).

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 \(J = i\),

- \(\delta x: dx(J {\partial}x) = dx({\partial}y) = 0\),
- \({\partial}y: dx(J {\partial}y) = dx(-{\partial}x) = -1\).

So \(J^*(dx) = -dy\), and \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: