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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Contact Topology
  }
    \\ {\normalsize Lectures by Peter Lambert-Cole. University of
Georgia, Spring 2022} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: 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. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-05-29 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{tuesday-january-11}{%
\section{Tuesday, January 11}\label{tuesday-january-11}}

\begin{remark}

References:

\begin{itemize}
\tightlist
\item
  \url{https://www.ma.imperial.ac.uk/~ssivek/courses/12s-math273.php}
\item
  \url{https://web.ma.utexas.edu/users/gdavtor/notes/contact_notes.pdf}
\item
  \href{https://outlookuga-my.sharepoint.com/:f:/g/personal/pbl20394_uga_edu/EjT_H9wRyAhAsZodxgXOgU0BSIYbqWav8X1jZY5v3RxqJA?e=n6dfVJ}{PLC's
  Notes}
\end{itemize}

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:

\end{remark}

\hypertarget{application-1}{%
\subsection{Application 1}\label{application-1}}

\begin{theorem}[Cerf's Theorem]

Every diffeomorphism \(f: S^3\to S^3\) extends to a diffeomorphism
\({\mathbb{B}}^4\to {\mathbb{B}}^4\).

\end{theorem}

\begin{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.

\end{remark}

\begin{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
\emph{tightness} which doesn't change under diffeomorphisms and is
fundamental to 3-manifold topology.

\begin{theorem}[Eliashberg]

There is a unique tight contact structure \(\xi_\text{std}\) on \(S^3\).

\end{theorem}

So up to isotopy, \(f\) fixes \(\xi_\text{std}\).

\end{remark}

\begin{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.

\end{remark}

\hypertarget{application-2}{%
\subsection{Application 2}\label{application-2}}

\begin{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
\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.

\end{remark}

\begin{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:

\begin{itemize}
\tightlist
\item
  Stabilization and destabilization:
\end{itemize}

\includegraphics{figures/2022-01-11_11-41-29.png}

\begin{itemize}
\tightlist
\item
  Conjugation in \(B_n\):
\end{itemize}

\includegraphics{figures/2022-01-11_11-39-52.png}

\begin{itemize}
\tightlist
\item
  Braid isotopy, which preserves braid words in \(B_n\).
\end{itemize}

\end{remark}

\begin{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

\begin{itemize}
\tightlist
\item
  \(\beta_i\in B_{n_i}\)
\item
  \(K(\beta_i) = U\)
\item
  \(n_1\geq n_2\geq \cdots \geq n_k = 1\)
\item
  \(\beta_i \to \beta_{i+1}\) is either a Markov move or an exchange
  move.
\end{itemize}

\includegraphics{figures/2022-01-11_11-46-31.png}

\end{remark}

\hypertarget{application-3}{%
\subsection{Application 3}\label{application-3}}

\begin{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
\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.

\end{remark}

\begin{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
\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.

\end{remark}

\hypertarget{contact-geometry}{%
\subsection{Contact Geometry}\label{contact-geometry}}

\begin{definition}[?]

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.

\end{definition}

\begin{example}[?]

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*}

\end{example}

\begin{exercise}[?]

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*}

\end{exercise}

\hypertarget{contact-forms-and-structures-thursday-january-13}{%
\section{Contact Forms and Structures (Thursday, January
13)}\label{contact-forms-and-structures-thursday-january-13}}

\begin{definition}[Contact form]

A \textbf{contact form} on \(Y^3\) is a 1-form \(\alpha\) with
\(\alpha \wedge d\alpha > 0\). A \textbf{contact structure} is a 2-plane
field \(\xi = \ker \alpha\) for some contact form.

\end{definition}

\begin{remark}

Forms are more rigid than structures: if \(f>0\) and \(\alpha\) is
contact, then \(f\cdot \alpha\) is also contact with
\(\ker( \alpha ) = \ker(f \alpha)\).

\end{remark}

\hypertarget{examples-of-contact-structures}{%
\subsection{Examples of Contact
Structures}\label{examples-of-contact-structures}}

\begin{example}[Standard contact structure]

On \({\mathbb{R}}^3\), a local model is \(\alpha\coloneqq\,dz- y\,dx\).

\begin{exercise}[?]

Show \(\alpha \wedge d\alpha = dz\wedge dx \wedge dy\).

\end{exercise}

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.

\end{example}

\begin{example}[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.

\end{example}

\begin{example}[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 \(\theta\).

\end{example}

\begin{example}[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.

\end{example}

\begin{example}[Overtwisted]

Set
\begin{align*}
\alpha_4 = \cos(r^2)\,dz+ \sin(r^2)\,d\theta
.\end{align*}

\begin{exercise}[?]

Compute the exterior derivative and check that this yields a contact
structure.

\end{exercise}

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 \(\pi\).

\end{example}

\begin{definition}[Contactomorphisms]

A \textbf{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).

\end{definition}

\begin{remark}

Strict contactomorphisms are more important for dynamics or geometric
applications.

\end{remark}

\begin{exercise}[?]

Prove that \(\alpha_1,\cdots, \alpha_4\) are all contactomorphic.

\end{exercise}

\begin{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:

\includegraphics{figures/2022-01-13_11-50-52.png}

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 \({\mathbf{T}} {}^{ \vee }X\).

\begin{exercise}[?]

Find out what \(\beta = \sum a_i \,dx_i\) is equal to as a section of
\({\mathbf{T}} {}^{ \vee }X\).

\end{exercise}

\end{remark}

\begin{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.

\end{remark}

\begin{exercise}[?]

Check that \({\mathbb{R}}^3 = J^1({\mathbb{R}})\) and
\({\mathbb{S}}{\mathbf{T}} {}^{ \vee }({\mathbb{R}}^2) = {\mathbb{R}}^2 \times S^1\).

\end{exercise}

\begin{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.

\end{remark}

\hypertarget{perturbing-foliation}{%
\subsection{Perturbing Foliation}\label{perturbing-foliation}}

\begin{example}[?]

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

\begin{itemize}
\tightlist
\item
  \(t>0 \implies \alpha = \,dz- y\,dx\) yields a positive contact
  structure,
\item
  \(t>0 \implies \alpha = \,dz\) is a foliation,
\item
  \(t<0 \implies \alpha = \,dz+ y\,dx\) is a negative contact structure.
\end{itemize}

\end{example}

\begin{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 \emph{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!

\begin{quote}
For a fiber bundle or vector bundle to admit an interesting foliation,
one needs a flat connection.
\end{quote}

\end{remark}

\begin{definition}[Integrability]

Any \(\xi \coloneqq\ker \alpha\) is \textbf{integrable} iff for all
vector fields \(X, Y \subseteq \xi\), their Lie bracket
\([X, Y] \subseteq \xi\).

\end{definition}

\begin{theorem}[Frobenius Integrability]

For \(\alpha\) nonvanishing on \(Y^3\), \(\ker \alpha\) is tangent to a
foliation by surfaces iff \(\alpha \wedge d\alpha = 0\).

\end{theorem}

\begin{example}[?]

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.

\end{example}

\begin{remark}

A theorem of Eliashberg and Thurston: taut foliations can be perturbed
to a (tight) positive contact structure.

\end{remark}

\hypertarget{tuesday-january-18}{%
\section{Tuesday, January 18}\label{tuesday-january-18}}

\begin{remark}

Refs:

\begin{itemize}
\tightlist
\item
  Geiges, Intro to Contact
\item
  Ozbogi-Stipsicz
\item
  Etnyre lecture notes
\item
  Massot
\item
  Sivck
\end{itemize}

\end{remark}

\begin{definition}[Standard contact structure]

For \(S^3 \subseteq {\mathbb{C}}^2\), define a form on
\({\mathbb{R}}^4\) as
\begin{align*}
\alpha \coloneqq-y_1 dx_1 + x_1 dy_1 - y_2 dx_2 + x_2 dy_2
.\end{align*}
Then then \textbf{standard contact form} on \(S^3\) is
\begin{align*}
\xi_{\text{std}} \coloneqq\ker { \left.{{\alpha}} \right|_{{S^3}} }
.\end{align*}

\end{definition}

\begin{exercise}[?]

Show that \(\alpha\) defines a contact form.

\end{exercise}

\begin{solution}

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

\begin{itemize}
\tightlist
\item
  \(d\alpha = 2(dx_1 \wedge dy_1) + 2(dx_2 + dy_2)\)
\item
  \(df = 2(x_1 dx_1 + y_1 dy_1) + 2(x_2 dx_2 + y_2 dy_2)\).
\end{itemize}

\end{solution}

\begin{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
\(\xi_p = \ker dy_1 = \mathop{\mathrm{span}}\left\{{{\partial}x_1, {\partial}y_2}\right\}\).

\includegraphics{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.

\end{remark}

\begin{definition}[Almost complex structures]

An \textbf{almost complex structure} on \(X\) is a bundle automorphism
\(J: {\mathbf{T}}X{\circlearrowleft}\) with
\(J^2 = -\operatorname{id}\).

\end{definition}

\begin{example}[?]

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*}

\end{example}

\begin{exercise}[?]

Show that \(f: {\mathbb{C}}\to {\mathbb{C}}\) is holomorphic if
\(df \circ J = J \circ df\), which corresponds to the Cauchy-Riemann
equations.

\end{exercise}

\begin{lemma}[?]

Given \(J:W\to W\), an \({\mathbb{R}}{\hbox{-}}\)subspace \(V \leq W\)
is a \({\mathbb{C}}{\hbox{-}}\)subspace iff \(J(V) = V\).

\end{lemma}

\begin{definition}[?]

The field of \(J{\hbox{-}}\)complex tangents is the hyperplane field
\begin{align*}
\xi_p \coloneqq{\mathbf{T}}S^3 \cap J({\mathbf{T}}S^3)
.\end{align*}

\end{definition}

\begin{example}[?]

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 \(\xi_{\text{std}}\).

\end{example}

\begin{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
\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.

\end{question}

\begin{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 \(J = i\),

\begin{itemize}
\tightlist
\item
  \(\delta x: dx(J {\partial}x) = dx({\partial}y) = 0\),
\item
  \({\partial}y: dx(J {\partial}y) = dx(-{\partial}x) = -1\).
\end{itemize}

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 \(\pi/2\).

\end{remark}

\begin{example}[?]

The field of complex tangencies along \(Y = f^{-1}(0)\) is the kernel of
\({ \left.{{ df(J({-})) }} \right|_{{Y}} }\).

\end{example}

\begin{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.

\end{remark}

\begin{example}[?]

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 \(S^1\).

\includegraphics{figures/2022-01-18_12-11-38.png}

\includegraphics{figures/2022-01-18_12-15-08.png}

\includegraphics{figures/2022-01-18_12-17-02.png}

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
\(S^2 = \left\{{{\left\lvert {z_1} \right\rvert}^2 + {\left\lvert {z_2} \right\rvert}^2 = 1}\right\}\):

\includegraphics{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:

\includegraphics{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.

\includegraphics{figures/2022-01-18_12-28-04.png}

\end{example}

\hypertarget{darboux-and-gromov-stability-thursday-january-20}{%
\section{Darboux and Gromov Stability (Thursday, January
20)}\label{darboux-and-gromov-stability-thursday-january-20}}

\begin{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.

\end{remark}

\begin{theorem}[Darboux]

If \((Y^3, \xi)\) is contact then for every point \(p\) there is a chart
\(U\) with coordinates \(x,y,z\) where
\(\xi = \ker (\,dz-y\,dx) = \ker (\alpha_\text{std})\).

\end{theorem}

\begin{slogan}

Locally, all contact \emph{structures} (not necessarily forms) look the
same. The mantra: local flexibility vs global rigidity.

\end{slogan}

\hypertarget{proof-of-darboux}{%
\subsection{Proof of Darboux}\label{proof-of-darboux}}

\begin{remark}

Two proofs:

\begin{itemize}
\tightlist
\item
  Geometric, due to Giroux
\item
  PDEs, which generalizes. This uses Moser's trick.
\end{itemize}

\end{remark}

\begin{proof}[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
\emph{characteristic foliation}.

\includegraphics{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
\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*}

\end{proof}

\begin{proof}[2, Moser's Trick]

By a linear change of coordinates, choose \(x,y\) along \(\xi\) to write
\(\alpha_p = \,dz\) and
\(\xi_p = \mathop{\mathrm{span}}{{\partial}x, {\partial}y}\):

\includegraphics{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.

\begin{lemma}[?]

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 \(p\).

\end{lemma}

\begin{remark}

Note that
\(\dot\alpha_t \coloneqq{\frac{\partial }{\partial t}\,} \alpha_t\), so
\begin{align*}
{\frac{\partial }{\partial t}\,} \qty{ t\alpha_1 + (1-t) \alpha_0} = \alpha_1 - \alpha_0
.\end{align*}

\end{remark}

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*}

\begin{quote}
See Prop 6.4 in Cannas da Silva.
\end{quote}

\begin{remark}

✨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*}

\end{remark}

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 \textbf{Reeb
vector field} of \(\alpha_t\), which satisfies

\begin{itemize}
\tightlist
\item
  \(\alpha_t(R_t) = 1\)
\item
  \(d\alpha_t(R_t, {-}) = 0\).
\end{itemize}

\includegraphics{figures/2022-01-20_12-11-26.png}

Then
\begin{align*}
\dot \alpha_t (R_t) = {\dot f_t \over f_t} = {\frac{\partial }{\partial t}\,} \log(f_t) \coloneqq\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.

\end{proof}

\hypertarget{gray-stability-tuesday-january-25}{%
\section{Gray Stability (Tuesday, January
25)}\label{gray-stability-tuesday-january-25}}

\begin{remark}

A homotopy of contact structures o \(Y^3\) is a smooth family
\(\left\{{\phi_t}\right\}\) of contact structures. Similarly, an
\textbf{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.

\end{remark}

\begin{lemma}[?]

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*}

\end{lemma}

\begin{proof}[?]

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
\(\mu_t = \dot \alpha_t (R_t)\).

\end{proof}

\begin{corollary}[?]

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.

\end{corollary}

\begin{proof}[?]

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.

\end{proof}

\hypertarget{legendrian-links}{%
\subsection{Legendrian Links}\label{legendrian-links}}

\begin{definition}[Legendrian and transverse knots]

Let \(Y\) be a contact 3-manifold and \(L \hookrightarrow Y\) a link.
Then \(L\) is a \textbf{Legendrian knot} iff it is everywhere tangent to
\(\xi\), so \(\alpha(L) = 0\):

\includegraphics{figures/2022-01-25_12-10-49.png}

This is a closed condition.

\(L\) is \textbf{transverse} if it is everywhere transverse to \(\xi\),
so \(\alpha(L) > 0\):

\includegraphics{figures/2022-01-25_12-14-25.png}

This is an open condition.

\end{definition}

\begin{remark}

Every Legendrian knot has a transverse pushoff (up to transverse
isotopy). Every transverse knot has a Legendrian approximation.

\end{remark}

\begin{example}[?]

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
\({ \left.{{\alpha}} \right|_{{L_3}} } = {\varepsilon}dx >0\).

\end{example}

\begin{theorem}[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 \(\alpha \coloneqq\,dz+ r^2 \,d\theta\).

\end{theorem}

\hypertarget{thursday-january-27}{%
\section{Thursday, January 27}\label{thursday-january-27}}

\begin{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 \(t\).

\end{remark}

\begin{example}[?]

\(\gamma(s) = {\left[ {x(s), y(s), z(s)} \right]}\) and
\(\xi = \ker \alpha, \alpha = \,dz- y\,dx\). Then
\(\gamma^*(\alpha) = z' \,ds- yx^1\,ds= (z' - yx')\,ds\), which is
Legendrian iff \(y=z'/x'\).

\end{example}

\begin{example}[?]

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 \(z'/x' = f'/s' = f'\).

\end{example}

\begin{remark}

There are two projections:

\begin{itemize}
\tightlist
\item
  \({\left[ {x,y,z} \right]} \to {\left[ {x,z} \right]}\), a wave front
  projection, plotted with \(y\) into the board,
\item
  \({\left[ {x,y,z} \right]} \to {\left[ {x,y} \right]}\), Lagrangian
  projection.
\end{itemize}

\end{remark}

\begin{example}[?]

Let \(\gamma(s) = {\left[ {s^2, {3\over 2}s, s^3} \right]}\), then the
two projections are as follows:

\includegraphics{figures/2022-01-27_11-25-37.png}

\end{example}

\begin{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:

\includegraphics{figures/2022-01-27_11-30-01.png}

\end{remark}

\begin{example}[?]

A front diagram of the unknot:

\includegraphics{figures/2022-01-27_11-34-42.png}

\end{example}

\begin{theorem}[?]

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:
\(\sup_{s\in I} {\left\lvert {\gamma_1(s) - \gamma_2(s)} \right\rvert} \leq {\varepsilon}\)?

\end{theorem}

\begin{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:

\includegraphics{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.

\end{remark}

\begin{definition}[The Thurston–Bennequin number]

A \textbf{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 \textbf{contact framing}. The \textbf{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.

\end{definition}

\begin{remark}

How to compute: take a pushoff and compute the linking number:

\includegraphics{figures/2022-01-27_12-07-19.png}

\end{remark}

\begin{proposition}[?]

\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.

\end{proposition}

\begin{proof}[?]

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:

\includegraphics{figures/2022-01-27_12-12-35.png}

\end{proof}

\begin{example}[?]

TB for the knots from before:

\begin{itemize}
\tightlist
\item
  The 3 unknots:

  \begin{itemize}
  \tightlist
  \item
    2 cusps, so \(-1\)
  \item
    4 cusps, so \(-2\)
  \item
    4 cusps, so \(-2\)
  \end{itemize}
\item
  The 2 trefoils:

  \begin{itemize}
  \tightlist
  \item
    \(3-{1\over 2}4 = 1\)
  \item
    \(3 - {1\over 2}6 = -6\).
  \end{itemize}
\end{itemize}

\end{example}

\begin{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.

\end{remark}

\begin{definition}[Rotation number]

The \textbf{rotation number} of \(L\) is the \emph{turning number}
\(\operatorname{rot}(L)\) in the Lagrangian projection, i.e.~how many
times a tangent vector spins after traversing the knot.

\end{definition}

\begin{example}[?]

\includegraphics{figures/2022-01-27_12-26-39.png}

It turns out that
\begin{align*}
\operatorname{rot}(L) = {1\over 2}\qty{ {\sharp}\text{down cusps} - {\sharp}\text{up cusps}}
.\end{align*}

\end{example}

\hypertarget{tuesday-february-01}{%
\section{Tuesday, February 01}\label{tuesday-february-01}}

\begin{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:

\begin{itemize}
\tightlist
\item
  Thurston-Bennequin invariant: a contact framing with respect to the
  Reeb vector field
\end{itemize}

\includegraphics{figures/2022-02-01_11-15-00.png}

\begin{itemize}
\item
  Equal to writhe minus half the number of cusps.
\item
  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.
\end{itemize}

\end{remark}

\begin{remark}

Disallowed moves:

\includegraphics{figures/2022-02-01_11-20-07.png}

Allowed moves:

\includegraphics{figures/2022-02-01_11-24-46.png}

\end{remark}

\begin{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
\((t,r) = (\mathrm{tb}(L), \operatorname{rot}(L))\)?

\end{remark}

\begin{example}[?]

For \(K\) the unknot:

\includegraphics{figures/2022-02-01_11-37-46.png}

So these numerical pairs fall into a cone:

\includegraphics{figures/2022-02-01_11-39-30.png}

\end{example}

\begin{proposition}[?]

For \(L \subseteq {\mathbb{R}}^3\) a Legendrian knot,
\begin{align*}
\mathrm{tb}(L) + \operatorname{rot}(L) \equiv 1 \operatorname{mod}2
.\end{align*}

\end{proposition}

\begin{remark}

Note that \(\chi(S) \equiv 1\operatorname{mod}2\) for \(S\) a Seifert
surface.

\end{remark}

\begin{theorem}[Bennequin-Thurston inequality]

For any Seifert surface \(S\),
\begin{align*}
\mathrm{tb}(L) + {\left\lvert { \operatorname{rot}(L) } \right\rvert} \leq -\chi(S)
.\end{align*}

\end{theorem}

\begin{remark}

This solves the geography problem: this cone contains all of the
possible pairs.

\end{remark}

\begin{theorem}[Eliashberg-Fraser]

The unknot is \textbf{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 \(L_2\).

\end{theorem}

\begin{remark}

This solves the botany problem: every red dot has exactly one
representative.

\end{remark}

\begin{remark}

Other knots are Legendrian simple, e.g.~the trefoil. A theorem of
Checkanov says the following \(5_2\) knots are not Legendrian isotopic:

\includegraphics{figures/2022-02-01_11-53-16.png}

\end{remark}

\begin{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:

\includegraphics{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.

\end{remark}

\begin{definition}[Overtwisted discs]

An \textbf{overtwisted disc} in \((Y^3, \xi)\) that is locally
contactomorphic to this local model. \(Y\) is \textbf{overtwisted} if it
contains an overtwisted disc, and is tight otherwise.

\end{definition}

\begin{theorem}[Bennequin]

\(({\mathbb{R}}^3, \xi_\text{std})\) is a tight contact structure.

\end{theorem}

\begin{theorem}[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.

\end{theorem}

\hypertarget{transverse-knots}{%
\subsection{Transverse Knots}\label{transverse-knots}}

\begin{definition}[Self-linking]

The \textbf{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
\({ \left.{{\xi}} \right|_{{S}} }\).

\end{definition}

\begin{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:

\includegraphics{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.

\end{remark}

\hypertarget{thursday-february-03}{%
\section{Thursday, February 03}\label{thursday-february-03}}

\begin{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.

\end{remark}

\begin{example}[?]

Take \(\alpha = \,dz+x\,dy- y\,dx\) and \(\Sigma = S^2\), then the
singular foliation is given by

\includegraphics{figures/2022-02-03_11-21-06.png}

\end{example}

\begin{remark}

Two possible types of singularities, the local models:

\includegraphics{figures/2022-02-03_11-24-34.png}

There are also two numerical invariants:

\begin{itemize}
\tightlist
\item
  \(e_{\pm}\): the number of positive (resp. negative) elliptics
\item
  \(h_{\pm}\): the number of positive (resp. negative) hyperbolics
\end{itemize}

A theorem
\begin{align*}
{\left\langle {c(\Sigma)},~{ \Sigma} \right\rangle} = (e_+ - h_+) - (e_- - h_-)
.\end{align*}
If \(\Sigma\) is transverse,
\(\mathrm{sl}({{\partial}}\Sigma, \Sigma) = -(e_+ - h_+) + (e_- - h_-)\).

\end{remark}

\hypertarget{local-model-1-elliptic}{%
\subsection{Local Model 1: Elliptic}\label{local-model-1-elliptic}}

\begin{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
\(\alpha(i) = x^2+y^2 = 1 > 0\).

\includegraphics{figures/2022-02-03_11-32-36.png}

Here \(\mathrm{sl} = 1\). To compute \(\mathrm{sl}\):

\begin{itemize}
\tightlist
\item
  Trivialize \({ \left.{{\xi}} \right|_{{\Sigma}} }\) to get
  \(\tau = \left\langle{e_1, e_2}\right\rangle\) a fiberwise basis for
  \(\xi\).
\item
  Let \(\tilde L\) be a pushoff in the \(e_1\) direction.
\item
  Compute \(\mathrm{sl} = \operatorname{lk}(L, \tilde L)\).
\end{itemize}

Set

\begin{itemize}
\tightlist
\item
  \(e_1 = {\partial}_x + y{\partial}_z\)
\item
  \(e_2 = {\partial}_y - x{\partial}_z\)
\item
  \(\rho = x {\partial}_x + y{\partial}_y\)
\item
  \(\theta = x{\partial}_y - y{\partial}_x\).
\end{itemize}

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

\begin{itemize}
\tightlist
\item
  \(c_1 = x\rho - y\theta + y{\partial}_z\)
\item
  \(\mkern 1.5mu\overline{\mkern-1.5muc_1\mkern-1.5mu}\mkern 1.5mu = x\rho + y{\partial}_z = \cos(\rho) + \sin(\theta) {\partial}_z\).
\end{itemize}

Example:

\begin{itemize}
\tightlist
\item
  \(\theta = 0\implies e_1 = \rho\)
\item
  \(\theta = \pi/4 \implies e_1 = {\sqrt 2\over 2}(\rho + {\partial}_z)\)
\item
  \(\theta = \pi/2 \implies e_1 = {\partial}_z\)
\end{itemize}

So here \(\operatorname{lk}(U, \tilde U) = -1\):

\includegraphics{figures/2022-02-03_11-43-46.png}

\end{remark}

\hypertarget{local-model-2-hyperbolic}{%
\subsection{Local Model 2: Hyperbolic}\label{local-model-2-hyperbolic}}

\begin{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
\(V = y{\partial}_y + dx{\partial}_x\in \ker({ \left.{{\alpha }} \right|_{{\Sigma}} })\).

\end{remark}

\begin{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
\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*}

\end{remark}

\begin{example}[?]

Note \(\chi({\mathbb{D}}) = 1\), so any vector field has a singularity?

\includegraphics{figures/2022-02-03_11-59-11.png}

\end{example}

\begin{proposition}[?]

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*}

\includegraphics{figures/2022-02-03_12-01-38.png}

\end{proposition}

\hypertarget{more-contact-geometry}{%
\subsection{More Contact Geometry}\label{more-contact-geometry}}

\begin{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

\begin{itemize}
\item
  \(\chi(\Sigma) = e({\mathbf{T}}\Sigma, V) \in H^2(\Sigma, {{\partial}}\Sigma) \cong H_0(\Sigma)\)
\item
  \(\mathrm{sl}({{\partial}}\Sigma, \Sigma) = e(\xi, V) \in H^2(\Sigma, {{\partial}}\Sigma)\)
\end{itemize}

\end{remark}

\begin{fact}

\envlist

\begin{itemize}
\tightlist
\item
  \(e_+ + e_-\) correspond to \(+1\) in \(e({\mathbf{T}}\Sigma, V)\),
\item
  \(h_+, h_-\) correspond to \(-1\) in \(e({\mathbf{T}}\Sigma, V)\).
\end{itemize}

Proof: near a zero, \(V\) determines a map \(S^1\to S^1\) and the
contribution to \(e\) is the degree of this map.

\begin{itemize}
\tightlist
\item
  \(e_+\) contributes \(-1\) to \(e(\xi, V)\), by the same computation
  of \(\mathrm{sl}(U)\) for \(U\) the unknot.
\item
  \(e_-\) contributes \(-(-1) = +1\) to \(e(\xi, V)\).
\item
  \(h_+\) contributes \(+1\) to \(e(\xi, V)\)
\item
  \(h_-\) contributes \(-(+1) = -1\) to \(e(\xi, V)\).
\end{itemize}

Proof: exercise.

\end{fact}

\begin{remark}

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:

\includegraphics{figures/2022-02-03_12-24-10.png}

The inequality follows if we can cancel every \(e_-\) with some \(h_-\).

\end{remark}

\hypertarget{tuesday-february-08}{%
\section{Tuesday, February 08}\label{tuesday-february-08}}

\begin{remark}

Topics for talks:

\begin{itemize}
\tightlist
\item
  Thom-Pontryagin
\item
  Brieskorn spheres
\item
  Milnor fibrations
\item
  Lens spaces
\end{itemize}

\end{remark}

\begin{theorem}[?]

Every closed oriented 3-manifold \(Y\) admits a (positive) contact form.

\end{theorem}

\begin{remark}

Three proofs:

\begin{itemize}
\tightlist
\item
  Lickorish-Wallace, using that \(Y\) is Dehn surgery on a link in
  \(S^3\),
\item
  Birman-Hildon, using that \(Y\) is a branched cover of \(S^3\),
\item
  Alexander, using that \(Y\) admits an open book decomposition.
\end{itemize}

\end{remark}

\begin{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.

\end{remark}

\begin{remark}

Recall:

\begin{itemize}
\tightlist
\item
  Every knot \(K\) can be \(C^0\) approximated by a transverse knot
\item
  Every link \(L\) can be \(C^0\) approximated by a transverse link
\item
  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
  \({\left[ {0, \delta, \theta} \right]}\mapsto {\left[ {q\theta, \delta, p\theta} \right]}\).
\end{itemize}

\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.

\includegraphics{figures/2022-02-08_11-57-58.png}

\end{remark}

\begin{definition}[?]

An \textbf{open book decomposition} of \(Y\) is a pair \((B, \pi)\)
where

\begin{itemize}
\tightlist
\item
  \(B\) is a link in \(Y\), called the \textbf{binding}
\item
  \(\pi: Y\setminus B\to S^1\) is a locally trivial fibration of
  relatively compact fibers \textbf{pages}
\end{itemize}

\includegraphics{figures/2022-02-08_12-09-13.png}

\end{definition}

\begin{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
\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:

\includegraphics{figures/2022-02-08_12-14-04.png}

\end{remark}

\begin{definition}[Open book decompositions supporting a contact structure]

An open book decomposition \textbf{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 \((B, \xi)\).

\end{definition}

\begin{theorem}[Thurston-Winkelnkemper]

Every open book decomposition admits a contact structure.

\end{theorem}

\begin{theorem}[Giroux]

Every \((Y^3, \xi)\) with \(Y\) closed has a supporting open book
decomposition.

\end{theorem}

\begin{proposition}[?]

If an open book decomposition supports \(\xi_1\) and \(\xi_2\), then
\(\xi_1\) is isotopic to \(\xi_2\).

\end{proposition}

\begin{proof}[?]

Two steps:

\begin{itemize}
\tightlist
\item
  Form a mapping cylinder of the monodromy map \(\phi\),
\item
  Extend over the binding, using the same idea as in Dehn surgery.
\end{itemize}

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
\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
\(\alpha \wedge d\alpha = {\varepsilon}dt \wedge d\beta_t + { \mathsf{O}} ({\varepsilon}^2)\).

\end{proof}

\hypertarget{tuesday-february-15}{%
\section{Tuesday, February 15}\label{tuesday-february-15}}

Missed due to orthodontic appointment! Please send me notes. :)

\hypertarget{thursday-february-17}{%
\section{Thursday, February 17}\label{thursday-february-17}}

\begin{remark}

Let \(\Sigma \subseteq (Y^3, \xi)\).

\begin{itemize}
\tightlist
\item
  Characteristic foliation: \(F = \xi \cap{\mathbf{T}}\Sigma\),
  complicated but necessary
\item
  Dividing set: a multicurve, simpler
\end{itemize}

\end{remark}

\begin{theorem}[?]

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

\begin{itemize}
\tightlist
\item
  \(\phi_0(\Sigma) = \Sigma, \phi_t(\Gamma) = \Gamma\)
\item
  \(\phi_t(\Sigma)\) is convex for all \(t \in [0, 1]\)
\item
  The characteristic foliation of \(\phi_1(\Sigma)\) is \(F\).
\end{itemize}

\end{theorem}

\begin{remark}

Idea: dividing sets give ways to detect overtwisted contact structures.

\end{remark}

\begin{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
\(\mathrm{tw}(\xi, {{\partial}}D)\).

\includegraphics{figures/2022-02-17_11-24-46.png}

Spheres can have exactly one dividing component.

\end{remark}

\begin{exercise}[?]

Generalize to an arbitrary number of components \({\sharp}\Gamma = n\).

\end{exercise}

\begin{remark}

Same if \(\Sigma \neq S^2\) and \(\Gamma\) contains a contractible
curve. Contrapositively, if \((Y, \xi)\) is tight, then either

\begin{itemize}
\tightlist
\item
  \(\Sigma = S^2\) and \(\Gamma\) is connected, or
\item
  \(\Sigma \neq S^2\) and \(\Gamma\) has no contractible components.
\end{itemize}

\end{remark}

\begin{exercise}[?]

Consider tight contact structures on \(S^3\). Choose Darboux \(B^3\)
neighborhoods at the ends, and note the interior is
\(S^2\times [0, 1]\):

\includegraphics{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 \(S^3\).

\end{exercise}

\begin{remark}

What can \(F\) look like on an \(S^2\) in a tight \((Y,\xi)\)? \(F\) can
be perturbed to be Morse-Smale.

\begin{itemize}
\tightlist
\item
  There are a finite number of elliptic/hyperbolic singularities
\item
  There are nondegenerate periodic orbits, either attracting or
  repelling
\item
  There are no saddle-saddle arcs
\item
  The limit sets are singularities or periodic orbits
\end{itemize}

Dimension 3: strange attractors! Two types of limit sets:

\begin{itemize}
\tightlist
\item
  \(\omega\) limit sets: \(x\in Y\) where there exists a sequence
  \(\left\{{t_1 < \cdots }\right\}\) with \(\phi(t_k)\to x\).
\item
  \(\alpha\) limit sets: \(x\in Y\) where there exists a sequence
  \(\left\{{t_1 > \cdots }\right\}\) with \(\phi(t_k)\to x\).
\end{itemize}

\end{remark}

\begin{remark}

For \(S^2\), take \(S^+\) with an outward pointing vector field.

\includegraphics{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.

\includegraphics{figures/2022-02-17_11-58-12.png}

\end{remark}

\begin{proposition}[?]

If \(F\) on \(\Sigma\) is Morse-Smale, then it admits dividing curves.

\end{proposition}

\begin{proof}[?]

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 \(F\).

\end{proof}

\begin{theorem}[?]

If \(\Sigma\) is orientable, then there is a \(C^\infty\) small
perturbation of \(F\) such that it is Morse-Smale.

\end{theorem}

\begin{proposition}[?]

Every oriented \(\Sigma \subseteq (Y, \xi)\) can be perturbed to be
convex.

\end{proposition}

\begin{proof}[?]

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.

\end{proof}

\begin{proposition}[?]

If \((\Sigma, \tilde F)\) admits dividing curves, then it is convex.

\end{proposition}

\hypertarget{thursday-february-24}{%
\section{Thursday, February 24}\label{thursday-february-24}}

\begin{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

\begin{itemize}
\tightlist
\item
  \(T^2\times I\)
\item
  \(S^1\times {\mathbb{D}}^2\)
\item
  \(L(p, q)\)
\item
  \(T^3\)
\end{itemize}

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.

\end{remark}

\begin{remark}

The only change to the dividing set in a generic family can be:

\begin{itemize}
\item
  Retrograde saddle-saddle, yielding by pass moves.
\end{itemize}

\includegraphics{figures/2022-02-24_11-26-51.png}

\end{remark}

\begin{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.

\end{proposition}

\begin{proof}[?]

Diagrams?

\(\vdots\)

\end{proof}

\begin{remark}

Given \(\Gamma_0\) with slope \(p/q\) and \(\Gamma_1\) with slope
\(r/s\), form a Farey graph:

\includegraphics{figures/2022-02-24_12-00-24.png}

\(\vdots\)

\end{remark}

\begin{proposition}[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
\(J(S^1) \cong {\mathbb{R}}\times {\mathbf{T}} {}^{ \vee }S^1 \cong S^1\times {\mathbb{R}}^2\)..

\end{proposition}

\begin{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:

\includegraphics{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:

\includegraphics{figures/2022-02-24_12-26-17.png}

\end{remark}

\hypertarget{tuesday-march-15}{%
\section{Tuesday, March 15}\label{tuesday-march-15}}

\begin{quote}
See \url{https://arxiv.org/pdf/math/9910127.pdf}
\end{quote}

\begin{remark}

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
\({\left[ {x,y,z} \right]}\mapsto {\left[ {x,y,nz} \right]}\).

\end{remark}

\begin{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 \emph{bypass moves} on the
dividing sets. The basic move:

\includegraphics{figures/2022-03-15_11-30-20.png}

\end{remark}

\begin{definition}[Basic slice]

A \textbf{basic slice} is a contact structure on \(T^2\) such that

\begin{itemize}
\tightlist
\item
  \(T^2\times\left\{{0}\right\}\) is convex with 2 dividing curves of
  slope 0
\item
  \(T^2\times\left\{{1}\right\}\) is convex with 2 dividing curves of
  slope -1
\item
  \(\xi\) is tight
\item
  \(\xi\) is minimally twisting, so if \(T^2 \subseteq T^2\times I\) is
  convex then \({\mathrm{slope}}(r) \in [-1, 0]\).
\end{itemize}

\end{definition}

\begin{proposition}[?]

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

\begin{itemize}
\tightlist
\item
  \((T_2\times [0, 1/8], \xi_1)\)
\item
  \((T_2\times [1/2, 5/8], \xi_1)\)
\end{itemize}

\end{proposition}

\begin{proof}[?]

Step 1: There are at most 2 basic slices. Reduce to \(S^1\times D^2\) by
removing a convex annulus. Note that
\(T^2\times I\setminus(S^1\times I) \cong S^1\times I^2 \cong S^1\times D^2\).

\includegraphics{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 \(\infty\).

\begin{quote}
?
\end{quote}

Take an annulus \(A\) with some condition on \({{\partial}}A\), perturb
to be convex? Something contradicts the ``minimally twisting''
assumption, involving these pics:

\includegraphics{figures/2022-03-15_11-51-21.png}

Smooth corners?

\begin{quote}
?
\end{quote}

\end{proof}

\begin{definition}[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 \textbf{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 \(M\).

\end{definition}

\begin{remark}

In this case \(\dim s^{-1}= \dim M - \dim \xi\).

\end{remark}

\begin{lemma}[?]

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
\({\left\langle {{-}},~{{-}} \right\rangle}: H^2(M, {{\partial}}M;{\mathbb{Z}}) \times H_2(M, {{\partial}}M; {\mathbb{Z}}) \to {\mathbb{Z}}\).

\end{lemma}

\begin{remark}

Note
\(H_2(T^2\times I, {{\partial}}; {\mathbb{Z}}) = \left\langle{[\alpha\times I], [\beta\times I]}\right\rangle\)
where \(H_2(T^2) = \left\langle{\alpha, \beta}\right\rangle\).

\end{remark}

\hypertarget{tuesday-march-22}{%
\section{Tuesday, March 22}\label{tuesday-march-22}}

\hypertarget{farey-graphs}{%
\subsection{Farey Graphs}\label{farey-graphs}}

\begin{remark}

Build a graph on the hyperbolic plane in the Poincare disc model:

\includegraphics{figures/2022-03-22_11-20-24.png}

Here every midpoint corresponds to adding numerators and denominators
respectively.

Associate slopes:

\begin{itemize}
\tightlist
\item
  \(0/1 \leadsto 1\alpha + 0 \beta\)
\item
  \(1/0 \leadsto 0\alpha + 1 \beta\)
\item
  \(1/1 \leadsto 1\alpha + 1 \beta\)
\end{itemize}

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.

\end{remark}

\begin{remark}

Basic moves: bypasses

\includegraphics{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
\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 \(r_i \leq -1 \implies {\left\lfloor r_i \right\rfloor}\leq -2\).

\end{remark}

\begin{proposition}[?]

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
\(-a/b = [r_0, \cdots, r_{k}+1]\).

\end{proposition}

\begin{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.

\end{remark}

\begin{example}[?]

Let \(p/q = 53/17\), then

\begin{itemize}
\tightlist
\item
  \(r_0 = -4 = -68/17\)
\item
  \(r_1 = -2 = -30/15\)
\item
  \(r_2 = -2 = -26/13\)
\item
  \(r_3 = -2 = \cdots\)
\end{itemize}

So this yields \([-4, -2, \cdots_7, -2, -3]\).

\end{example}

\begin{remark}

Idea: decompose \(p/q = [r_0, \cdots, r_k]\) surgery into integer
surgeries on a link with \(k\) components.

\end{remark}

\hypertarget{tuesday-march-29}{%
\section{Tuesday, March 29}\label{tuesday-march-29}}

\begin{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 \(-p<q\leq 0\), so
\(p/q < -1\).

Some examples:

\begin{itemize}
\tightlist
\item
  \(L_{1, q} \cong S^3\)
\item
  \(L_{2, 1} \cong {\mathbb{RP}}^3\cong {\operatorname{SO}}_3({\mathbb{R}})\).
\end{itemize}

There is a genus 1 Heegaard splitting. The double branched cover of a
2-bridge link is a lens space:

\includegraphics{figures/2022-03-29_11-31-57.png}

All lens spaces can be generated by genus 1 Heegaard splittings?

\end{remark}

\begin{remark}

\(-p/q\) 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.

\end{remark}

\begin{remark}

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.:

\includegraphics{figures/2022-03-29_12-11-25.png}

Stabilize up to \(r_K+1\) on each Legendrian knot. Fact: yields a Stein
fillable thing, implies tight contact structure.

\end{remark}

\begin{remark}

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:

\begin{itemize}
\tightlist
\item
  \((3, 0)\)
\item
  \((2, 1)\)
\item
  \((1, 2)\)
\item
  \((0, 3)\)
\end{itemize}

So there are \(\prod_{1\leq i\leq k} (-r_k - 1)\) tight contact
structures on \(-p/q = {\left[ {r_0, \cdots, r_k} \right]}\).

\end{remark}

\hypertarget{tuesday-april-05}{%
\section{Tuesday, April 05}\label{tuesday-april-05}}

\hypertarget{symplectic-fillings}{%
\subsection{Symplectic Fillings}\label{symplectic-fillings}}

\begin{example}[Properties of the standard contact structure on $S^3$]

Consider \((S^3, \xi_\text{std}) \subseteq {\mathbb{C}}^2\); some things
that are true:

\begin{itemize}
\tightlist
\item
  There is a symplectic form
  \(\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2\) where \(d\omega = 0\)
  and \(\omega \wedge \omega =2 d\mathrm{Vol} > 0\). Write
  \(\phi = \sum x_i^2 + \sum y_i^2\), then \(S^3 = \phi^{-1}(1)\).
\item
  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
  \(\alpha = \omega(\rho, {-})\).
\item
  Since \({ \left.{{\omega}} \right|_{{\xi_\text{std}}} } > 0\), this
  yields an area form on contact planes.
\item
  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
  \(g(x, y) = \omega(x, Jy)\).
\end{itemize}

\end{example}

\begin{definition}[Fillings]

A complex symplectic manifold \((X^4, \omega, J)\) is a filling of
\((Y^3, \xi)\) if \(Y = {{\partial}}X\),

\begin{itemize}
\tightlist
\item
  \textbf{Stein filling}: \((X^4, J)\) is a Stein manifold, and
  \(\xi = {\mathbf{T}}Y \cap J({\mathbf{T}}Y)\).
\item
  \textbf{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.
\item
  \textbf{Weak filling}: \({ \left.{{\omega}} \right|_{{\xi}} } > 0\).
\end{itemize}

\begin{quote}
Note that we aren't defining what ``Stein'' means here.
\end{quote}

\end{definition}

\begin{theorem}[?]

There are strict implications

\begin{itemize}
\tightlist
\item
  Stein \(\implies\)
\item
  Strong \(\implies\)
\item
  Weak \(\implies\)
\item
  Tight.
\end{itemize}

Note that the last implication is the harder part of the theorem.

\end{theorem}

\begin{problem}[?]

Given \((Y, \xi)\), classify all fillings.

\end{problem}

\begin{example}[?]

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 \(T^2 \times {\mathbb{B}}^2\).

\end{example}

\begin{example}[?]

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
\(c_1(T_X, J)\).

\end{example}

\begin{example}[?]

Brieskorn spheres are examples of fillings, related to Milnor fibers.
For \(p,q,r \geq 2\), define
\begin{align*}
\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\}
.\end{align*}

In this case, we have:

\includegraphics{figures/2022-04-05_11-58-26.png}

Note that \({\varepsilon}= 0\) yields a singular variety, while
\({\varepsilon}>0\) small yields a smooth manifold.

\end{example}

\begin{exercise}[?]

Show \(\Sigma_{p,q,r}\) is the \(r{\hbox{-}}\)fold cyclic branched cover
of \(S^3\) over the torus knot \(T_{p, q}\).

\end{exercise}

\begin{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
\({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}z = {1\over 2}\qty{{\partial}x + i {\partial}y}\).

\end{remark}

\begin{exercise}[?]

Let \(f(z) = {\left\lvert {z} \right\rvert}^2\) and check

\begin{itemize}
\tightlist
\item
  \({\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f = {\partial}(z d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) = dz \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = -2i (dx \wedge dy)\).
\item
  \(d = {\partial}+ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\)
\end{itemize}

Practicing this type of change of variables is important!

\end{exercise}

\begin{definition}[Levi forms and plurisubharmonicity]

Let \(\phi: X\to {\mathbb{R}}\) for \(X\) a complex manifold, then the
\textbf{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
\textbf{plurisubharmonic} if \({\mathcal{L}}\phi\) is positive
semidefinite at every point.

\end{definition}

\begin{example}[?]

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.

\end{example}

\begin{proposition}[?]

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.

\end{proposition}

\begin{example}[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.

\end{example}

\hypertarget{thursday-april-21}{%
\section{Thursday, April 21}\label{thursday-april-21}}

\begin{quote}
Note: student talks in previous weeks!
\end{quote}

\begin{remark}

Possible topics for the remainder of the class:

\begin{itemize}
\tightlist
\item
  Open book decompositions
\item
  Every \((Y^3, \xi)\) is homotopic to a contact structure.
\item
  Seifert fibered spaces
\end{itemize}

\end{remark}

\hypertarget{seifert-fibered-spaces}{%
\subsection{Seifert Fibered Spaces}\label{seifert-fibered-spaces}}

\begin{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 \emph{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.

\end{remark}

\begin{definition}[Seifert fibered spaces]

A \textbf{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

\begin{itemize}
\tightlist
\item
  \(Y\setminus\cup_i\nu(L_i) = (\Sigma\setminus\left\{{{\operatorname{pt}}_1,\cdots, {\operatorname{pt}}_n}\right\}) \times S^1\)
\item
  \(\nu L_i = S^1\times {\mathbb{D}}^2\) is glued in by \(p_i/q_i\) Dehn
  surgery.
\end{itemize}

\end{definition}

\begin{example}[?]

\(L(p, q)\) is \(-p/q\) surgery on \(S^1\), or by a slam-dunk move:

\includegraphics{figures/2022-04-21_11-46-26.png}

\(\Sigma(p,q,r)\):

\includegraphics{figures/2022-04-21_11-48-50.png}

\end{example}

\begin{exercise}[?]

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.

\end{exercise}

\begin{exercise}[?]

Prove that \(\Sigma(p,q,r)\) comes from the plumbing diagram for the
Milnor fibration using Kirby calculus.

\end{exercise}

\hypertarget{tuesday-april-26}{%
\section{Tuesday, April 26}\label{tuesday-april-26}}

\begin{remark}

Recall that \(\operatorname{PHS}^3 = \Sigma(2,3,5)\) has a
Stein-fillable (and hence tight) contact structure.

\end{remark}

\begin{theorem}[?]

The negative \(-\Sigma(2,3,5)\) admits no tight contact structures.

\end{theorem}

\begin{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
\({{\partial}}X = T^2\cup T^2\cup T^2\):

\includegraphics{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
\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
\(S^1\times {\mathbb{D}}^2\):

\includegraphics{figures/2022-04-26_11-24-08.png}

\end{remark}

\begin{exercise}[?]

Show via Kirby calculus:

\includegraphics{figures/2022-04-26_11-25-37.png}

\end{exercise}

\begin{lemma}[?]

There exist Legendrian representatives \(F_2, F_3\) with twisting
numbers \(m_2, m_3 = -1\).

\end{lemma}

\begin{proof}[?]

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.

\end{proof}

\addsec{ToDos}
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\printbibliography[title=Bibliography]


\end{document}
