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\title{Filling links and minimal surfaces in 3-manifolds}
\ctitle{Filling links and minimal surfaces in 3-manifolds}
\author{William Stagner}
\department{Mathematics}
\school{Rice University}
\degree{Doctor of Philosophy}

\committee {
        Alan Reid, Chair \\
        Edgar Odell Lovett Chair and Professor of Mathematics\and
        Christopher Leininger \\
        Professor of Mathematics\and
        Illya V. Hicks\\
        Professor of Computational and Applied Mathematics
}

\address{Houston, Texas}
\donemonth{April} \doneyear{2022} \makeindex
\begin{document}
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\chapter*{Abstract}%
\addcontentsline{toc}{chapter}{\numberline{}Abstract}%
This thesis studies this existence of filling links 3-manifolds. A link $L$ in a 3-manifold $M$ is filling in $M$ if, for any spine $G$ of $M$ disjoint from $L$, $\pi_1(G)$ injects into $\pi_1(M \setminus L )$. Conceptually, a filling link cuts through all of the topology 3-manifold. These links were first studied by Freedman-Krushkal in the concrete case of the 3-torus $M = T^3$, but they leave open the question of whether a filling link actually exists in $T^3$. We answer this question affirmatively by proving in fact that every closed, orientable 3-manifold M with fundamental group of rank 3 contains a filling link.

\chapter*{Acknowledgements}%
\addcontentsline{toc}{chapter}{\numberline{}Acknowledgements}%
I would first and foremost wish to express my deepest gratitude to my advisor, Alan Reid, for his many years of mentorship, guidance, and support. Thank you especially for your relentless encouragement during the times I struggled and for your endless patience during the times I fell short. You have taught me so much---I've accumulated a debt that I can never repay.

Of the many people at Rice who have helped me along this journey, I first wish to extend my appreciation to the other members of my thesis committee, Christopher Leininger and Illya Hicks, for their help and dedication. I also extend my sincere thanks to Shelly Harvey for her mentorship, encouragement, and her interest in the well-being of the math graduate students.

I am very grateful to have shared the last several years at Rice with so many wonderful graduate students. In particular, I'm grateful to Alan's other graduate students---Nicholas Rouse, Connor Sell, Tam Cheetham-West, Sara Edelman-Mu\~noz, and Jiayu Wan. Thank you for your friendship, as well as countless helpful conversations. I couldn't ask for a better academic family.

Many of the ideas in this thesis were clarified by discussions with Charles Ouyang and Andrea Tamburelli. Thank you for helping me believe I could write a thesis with ``minimal surfaces'' in the title. I'd like also to thank Cameron Gordon, Bob Gompf, and Daniel Allcock for first showing me the beauty of low-dimensional topology during my undergraduate studies.

I would not have been able to accomplish this without unconditional love of my parents, David and Cynthia, and the support of my friends, especially Bob and William. In particular, thank you to Kyle Crop for your unwavering dedication to our friendship. You are the best man I know.

Finally, thank you to my wife Chloe. You've done so much that I could not begin to list everything here. But above all, thank you for always believing in me.

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%     M. Freedman and V. Krushkal introduced the notion of a ``filling'' link in a 3-manifold: a link $L$ is filling in $M$ if for any spine $G$ of $M$ disjoint from $L$, $\pi_1(G)$ injects into $\pi_1(M \setminus L )$. Freedman and Krushkal show that there exist links in the $3$-torus $T^3$ that satisfy a weaker form of filling, but they leave open the question of whether $T^3$ contains an actual filling link. We answer this question affirmatively by proving in fact that every closed, orientable 3-manifold $M$ with $\rk(\pi_1(M)) = 3$ contains a filling link.

% \subsection*{Acknowledgements} The author foremost wishes to thank his Ph.D. advisor, Alan Reid, for his many years of mentorship and support, and without whom this paper would not be possible. The author would also like to thank Charles Ouyang, Andrea Tamburelli, and Zeno Huang for several helpful conversations concerning minimal surfaces. Additionally, the author would like to thank Christopher Leininger for helpful comments on an early draft of this paper.

\chapter{Introduction}

A consistent theme in geometric topology is that it is often possible to gain insight into the structure of a manifold by studying its submanifolds. In the case of a 3-manifold, the 1-dimensional submanifolds are \emph{knots} and \emph{links}. There is a long history of trying to understand the extent to which the knots and links present in $M^3$ either inform or obstruct various geometric and topological features of $M$.

M. Freedman and V. Krushkal in \cite{FK21} captured the notion of a link $L\subset M$ ``cutting through all the topology'' of $M$ by defining $L$ to be a \emph{filling link} if every spine of $M$ is $\pi_1$-injective in the link complement $M\setminus L$. While this definition is nicely intuitive, it also seems quite difficult to satisfy, since we are demanding this injectivity condition for \emph{every} spine of $M$.

Nonetheless, it is still possible to prove the existence of---or even construct---filling links in certain cases. %We define the rank of a 3-manifold $M$ to be the rank of $\pi_1(M)$. Finding filling links is trivial for 3-manifolds of rank 1, and in
We can find a filling \emph{knot} in any lens space---a 3-manifold with cyclic fundamental group formed by gluing two solid tori along their boundaries---by simply taking the core curve of one of the solid tori. More generally, we say that $M$ has rank $n$ if the smallest generating set of $\fund{M}$ contains $n$ elements. Leininger and Reid prove in the appendix to \cite{FK21} that filling links exist in 3-manifolds of rank 2. For higher rank, establishing the existence of filling links turns out to be not only difficult in general but quite subtle even for concrete examples. Evidence of this is shown in \cite{FK21}, which focuses on the 3-torus $T^3$. In \cite{FK21}, utilizing the augmentation ideal, links $L$ are constructed in $T^3$ that are ``nearly'' filling in the sense that $\pi_1(G) \to \pi_1(T^3 \setminus L)$ is injective modulo (arbitrarily large) terms in the lower central series. However, the question of whether $T^3$ contains an actual filling link is left open.

\begin{question*}
	Does the 3-torus $T^3$ contain a filling link?
\end{question*}

The goal of this thesis is to answer this question in the affirmative.

% For any given 3-manifold $M$, it is easy to produce 1-complexes that ``fill'' in this sense --- the core $C$ of a handlebody in any Heegaard splitting of $M$ is a filling 1-complex in the sense that for any spine $G$ of $M$ disjoint from $C$, $\pi_1(G)$ injects into $\pi_1(M\setminus C)$. However, the question becomes much more subtle if we begin to look for filling \emph{links} in $M$. Freedman and Krushkal's definition is motivated in part by several well-known results with the same ``filling'' flavor, such as Bing's characterization\footnote{A closed 3-manifold $M$ is diffeomorphic to $S^3$ if and only if every knot in $M$ contained in a 3-ball.} \cite{Bin58} of $S^3$ or Myers' ``disk busting curves'' theorem\footnote{Every compact 3-manifold $M$ contains a knot that intersects any essential disk or sphere in $M$.} \cite{Mye82}. These results exemplify the frequently encountered difficulty of ``upgrading'' essentially trivial statements about 1-complexes to much deeper versions about knots or links.


\begin{restatable}{thm}{mainthm}\label{thm:Main}
	Let $M$ be a closed, orientable 3-manifold of rank 3. Then $M$ contains a filling link.
\end{restatable}

\begin{cor}
	The 3-torus $T^3$ contains a filling link.
\end{cor}

 The proof \cref{thm:Main} is organized as follows: In Chapter 2, we introduce the basic objects of study---Kleinian groups, hyperbolic manifolds, and their minimal surfaces. In Chapter 3, we show that if $L$ is a link in $M$ of rank $3$ such that $M \setminus L$ admits a complete hyperbolic metric of finite volume, then either $L$ is a filling link or $\pi_1(M\setminus L)$ contains the fundamental group of a closed surface of genus $2$. In Chapter 4, we show that if the length of the shortest closed geodesic in a finite volume hyperbolic 3-manifold $N$ is sufficiently large, then $\pi_1 (N)$ cannot contain such a subgroup. In Section 4, we construct in $M$ a hyperbolic link that achieves this lower bound on the length of the shortest closed geodesic. The material in this thesis is primarily an exposition of the author's work in \cite{Sta21}.

\chapter{Preliminaries}

\section{Hyperbolic space}
The $n$-dimensional hyperbolic space $\HH^n$ is the simply-connected Riemannian $n$-manifold with constant sectional curvature $-1$. There are various models of $\HH^n$, each of which illustrates different geometric features of hyperbolic geometry. We will primarily work with the upper half-space model, which, by a slight abuse of notation, we also denote by $\HH^n$. A good reference for this section is the book by \textcite{Mar16}.

In dimension 3, the upper half-space model of $\HH^3$ is the set
$$\HH^3 = \Set*{(z,t) \given z\in \CC,\; t > 0},$$ endowed with the complete Riemannian metric with line element $$ds^2 = \frac{\abs{dz}^2 + dt^2}{t^2}.$$
The geodesics in this model are semicircles orthogonal to $\CC$ and vertical lines. The totally geodesic surfaces in \Hh---that is, embeddings of the hyperbolic plane $\HH^2\hookrightarrow\HH^3$---are precisely the hemispheres orthogonal to $\CC$ along with vertical planes.

We define the \emph{conformal boundary} $\partial\HH^n$ as the set of equivalence classes of oriented geodesics, where we consider two oriented geodesics $\alpha,\; \beta$ to be equivalent if $$\limsup_{t\to\infty}\abs*{\alpha(t)-\beta(t)}<\infty.$$ In the upper half-space model of $\Hh$, we identify the conformal boundary $\partial\Hh$ with the Riemann sphere $\hat \CC = \CC \cup \{\infty\}$.

Every orientation-preserving isometry of $\Hh$ uniquely extends to a conformal automorphism of $\hat \CC$. Conversely, every conformal automorphism of $\hat\CC$ determines an orientation-preserving isometry via the Poincar\'e extension. The conformal automorphisms of $\hat \CC$ are precisely the linear fractional transformations of the form $$z \mapsto \frac{az + b}{cz + d},\quad ad-bc=1,$$ so we identify $\operatorname{Isom}^+(\HH^3) \cong \PSL_2(\CC)$. We will often represent an isometry as an actual matrix, suppressing the sign ambiguity that arises from lifting the projection $\SL_2(\CC) \to \PSL_2(\CC)$. We resolve this ambiguity for the trace of an isometry $\gamma\in \PSL_2(\CC)$ by instead working with $\abs{\tr\gamma}$ and $(\tr\gamma)^2$, which are well-defined.

There are three types of isometries of $\Hh$, classified by their fixed points. Viewing these isometries as matrices, the three types are conveniently determined by their traces. Specifically, let $\g \in \pslc$ be nontrivial.

\begin{itemize}
	\item \g is \emph{elliptic} if \g has a fixed point in $\Hh$. Equivalently, $\tr\g\in \mathbb{R}$ and $\abs{\tr\g}<2$.
	\item \g is \emph{parabolic} if \g has no fixed points in \Hh but exactly one fixed point in \ph. Equivalently, $\tr\g = \pm 2$.
	\item \g is \emph{loxodromic} if \g has no fixed points in \Hh but exactly two fixed points in \ph. Equivalently, either $\tr\g \not\in \mathbb{R}$ or $\tr\g \in \mathbb{R}$ with $\abs{\tr\g}>2$.
\end{itemize}

The two fixed points of a loxodromic $\gamma$ are connected by a geodesic axis in \Hh along which \g translates in a screw motion. This is captured by the \emph{complex translation length} $$\ell(\gamma) = \ell_0(\gamma) + \theta(\gamma),$$ where $\ell_0(\gamma) > 0$ is the distance that \g translates along its axis, and $\theta(\gamma)$ is the angle of the resulting rotation. The axis of $\gamma$ projects to a closed geodesic in $\HH/\Gamma$ of length $\ell_0(\g)$, which is of smallest length in its free homotopy class. The free homotopy classes corresponding to a parabolic or elliptic isometries contain arbitrarily short loops.

\subsection{Discrete groups of isometries}

A \emph{Kleinian group} is a discrete subgroup of $\isom$. A Kleinian group \G acts properly discontinuously on \Hh: for any closed ball $B\subset \Hh$, the set $\Set{\g \in \G \given \g(B)\cap B \neq \emptyset}$ is finite. In this case, the quotient $\Hh/\G$ inherits the structure of a complete hyperbolic 3-orbifold.

The torsion elements of a Kleinian group are precisely the elliptic isometries. On the other hand, elliptic elements are, by definition, the isometries with fixed points in \Hh. Hence, torsion-free Kleinian groups act freely on \Hh. We will consider only torsion-free Kleinian groups, so $\HH^3/\Gamma$ is a complete hyperbolic 3-manifold.

A subgroup of isometries $\Gamma\subset \isom$ is \emph{elementary} if its action on $\Hh\cup \hat\CC$ has a finite orbit, otherwise $\Gamma$ is \emph{nonelementary}. It is a classical result of \textcite{For29} that a Kleinian group is elementary if and only if it is either finite or contains a finite-index free abelian group of rank 1 or 2.

The maximal elementary subgroups of a torsion-free Kleinian group \G are the stabilizers of points $\zeta\in \hat\CC$. If $\zeta$ is the fixed point of a parabolic, then either $\stab\zeta$ is infinite cyclic, generated by a single parabolic, or $\stab(\zeta)$ is free abelian of rank 2, generated by a pair of distinct parabolics. In these two cases, we call $\zeta$ a \emph{rank one cusp} or \emph{rank 2 cusp} of \G, respectively. We will often refer to a rank 2 cusp as simply a \emph{cusp} of \G. Two cusps $\zeta$ and $\xi$ are called \emph{equivalent} if $\zeta=\gamma\cdot \xi$ for some $\gamma\in\G$.

\subsection{The conformal boundary}

Let \G be a Kleinian group. The \emph{limit set} of \G, denoted $\Lambda(\Gamma)$, is the set of limit points of $\Gamma$ in $\hat\CC$. In particular, $\Lambda(\Gamma)$ is closed, as it is the closure of the set of loxodromic or parabolic fixed points of \G. The complementary open set $\OG = \hat\CC \setminus \LG$ is called the \emph{regular set} of \G. It is the largest open subset of $\hat\CC$ on which \G acts properly discontinuously.

By enlarging the action of \G to include $\OG$, we define the quotient space
$$\MG = \parens[\big]{\Hh\cup \OG}/\G.$$
With our standing assumption that \G is torsion-free, the quotient \MG is an oriented manifold, called the \emph{Kleinian manifold of $\Gamma$}, with (possibly empty) boundary $\PMG = \OG/\G$, called the \emph{conformal boundary} of $\Gamma$. Note that the complete hyperbolic manifold $ \HH^3 / \Gamma $ is diffeomorphic to the interior $\operatorname{int}(\M(\Gamma))$. In particular, $\fund\MG \cong \Gamma$.

The geometry and topology of the conformal boundary of the Kleinian manifold $\MG$ is intimately related to that of its interior. This was first explored in the pioneering work of \textcite{Ahl64}, which established a fundamental result:

\begin{thm}[\cite{Ahl64}]
	If \G is a finitely generated, torsion-free Kleinian group, then $\PMG$ is the union of finitely many Riemann surfaces with at most finitely many punctures.
\end{thm}
\subsection{Fundamental domains}

A \emph{fundamental domain} of a Kleinian group \G is an open subset $\FF\subset \Hh$ such that
\begin{itemize}
	\item $\displaystyle \bigcup_{\g \in \G} \gamma\cdot \overbracket[0.65pt][0pt]{\mathcal{F}}= \Hh$
	\item $\displaystyle \FF \cap \g\cdot \FF = \emptyset \text{ for $g\neq \text{id}$}$
	\item $\partial \overbracket[0.65pt][0pt]{\mathcal{F}}$ has measure zero.
\end{itemize}

A Kleinian group is said to have \emph{finite covolume} if it has a fundamental domain of finite hyperbolic volume. Similarly, a Kleinain group said to be \emph{cocompact} if it has a fundamental domain with compact closure in $\Hh$. In both cases, $\OG=\emptyset$, so $\MG\cong \Hh/\G$. However, when \G is non-cocompact of finite volume, the ends of any fundamental domain $\mathcal F$ limit onto a finite set of ideal points $\zeta\in \CC$, which are rank two cusps of \G. A neighborhood of such a cusp projects to an end of $\MG$ isometric to $T^2 \times [0,\infty)$, called a \emph{solid cusp torus}. As such, we call a noncompact hyperbolic manifold of finite volume a \emph{cusped} manifold.

A Kleinian group is called \emph{geometrically finite} if it possesses a fundamental region that is the interior of finite-sided hyperbolic polyhedron. Finite covolume Kleinian groups are geometrically finite, but geometrically finite Kleinian groups may have infinite covolume, in general. However, finite volume and geometrically finite Kleinian groups share several similar topological features.

Removing the interiors of the (finitely many) solid cusp tori in a finite volume hyperbolic 3-manifold $\MG$ yields a compact submanifold $\Mog\subset \MG$ such that $\fund{\Mog} \to \fund{\MG}\cong \G$ is an isomorphism. It is a classical result of \textcite{Mar74} that a similar ``almost compact'' property characterizes geometrically finite Kleinian groups. To formulate this, it will be necessary to also remove cusp neighborhoods of rank one cusps of a geometrically finite $\G$. These neighborhoods project to an end of $\MG$ isometric to $(S^1\times \mathbb{R})\times [0,\infty)$, called a \emph{solid cusp cylinder}. 

\begin{thm}[\textcite{Mar74}]\label{geom-finite-criterion}
	A nonelementary, torsion-free Kleinian group \G is geometrically finite if and only if 
	\begin{enumerate}[(i)]
	\item \G has at most finitely many equivalence classes of rank one and rank two cusps;
	\item The punctures on \PMG are arranged in pairs such that each pair corresponds two the two ends of a solid cusp cylinder; and,
	\item Removing from \MG the interiors of all solid cusp cylinders and solid cusp tori results in a compact manifold \Mog.
\end{enumerate}
\end{thm}

\subsection{Kleinian subgroup dichotomy}
To understand the subgroups of a Kleinian group $\G$ that are \emph{not} geometrically finite, we first record two fundamental results. A 3-manifold $M$ is said to be \emph{topologically tame} if $M$ is homeomorphic to the interior of a compact 3-manifold.
\begin{thm}[Tameness Theorem, Agol \cite{Ago04}, Calegari--Gabai \cite{CG06}]\label{thm:tameness}
	If \G is a torsion-free, finitely generated Kleinian group, then $\Hh/\G$ is topologically tame.
\end{thm}
\begin{thm}[Canary Covering Theorem \cite{Can96}]\label{thm:covering}
	Let $\hat N$ be a topologically tame hyperbolic 3-manifold which covers another hyperbolic 3-manifold $N$ by a local isometry $p\colon \hat N\to N$. If $\hat E$ is a geometrically infinite end of $\hat N$, then either
\begin{enumerate}[(a)]
	\item $\hat E$ has a neighborhood $\hat U$ such that $p$ is finite-to-one on $\hat U$, or
	\item $N$ has finite volume and has a finite cover $N'$ which fibers over the circle such that if $N_S$ denotes the cover of $N'$ associated to the fiber subgroup, then $\hat N$ is finitely covered by $N_S$.
\end{enumerate}
\end{thm}

Combining \cref{thm:tameness} and \cref{thm:covering} yields a remarkable dichotomy of the finitely generated subgroups of $\G$ (see, e.g. \cite[Corollary 8.1]{Can08}).
\begin{thm}[Subgroup Tameness Theorem]\label{thm:subgroup-tameness}
	Let $N = \Hh/\G$ be a finite volume hyperbolic 3-manifold and $\Gamma'$ be a finitely generated subgroup of $\G$. Then either
	\begin{enumerate}
		\item $\Gamma$ is a virtual surface fiber group, i.e. there exists a finite cover $\widetilde M \to M\setminus L$ and a surface bundle $\Sigma \hookrightarrow \widetilde M \twoheadrightarrow S^1$ with $\Gamma = \pi_1(\Sigma)$, or
		\item $\Gamma$ is geometrically finite.
	\end{enumerate}
\end{thm}
\section{Minimal surfaces}
A standard reference for Riemannian geometry is \cite{dCar92}.
\subsection{Curvature}
Let $M$ be a Riemannian manifold with Levi-Civita connection $\nabla$. We denote the space of smooth vector fields on $M$ by $\mathfrak{X}(M)$. Recall the \emph{Riemann curvature tensor} of $M$ is the map
\begin{align*}
	R\colon \XM\times \XM\times\XM&\to\XM\\
	(X,Y,Z)&\mapsto R(X,Y)Z
\end{align*}
defined by
\begin{equation*}
	R(X,Y)Z = [\nabla_X,\nabla_Y]Z - \nabla_{[X,Y]}Z.
\end{equation*}
Given a point $p\in M$ and two-dimensional subspace $\sigma\subset T_p(M)$ with basis $u,v$, we define the \emph{sectional curvature of $\sigma$ at $p$} by $$K(\sigma) = K(u,v) = \frac{\ang{R(u,v)u,v}}{\abs{u\wedge v}^2}.$$ This quantity does not depend on the choice of basis $\{u,v\}$.

If $M$ is a surface, then of course $T_p(M)$ is itself two-dimensional. The sectional curvature then defines a smooth function $K:M\to \mathbb R$, called the \emph{Gaussian curvature} of $M$, by $K(p) = K(T_p(M))$.

\subsection{The Gauss equation}\label{sec:gauss-equation}
Let $S\subset M$ be an oriented hypersurface in $M$. The Riemannian metric on $M$ induces a Riemannian metric on $S$ so that the inclusion $S\to M$ is an isometric embedding. The Levi-Civita connection on $S$, denoted $\nabla^S$, corresponding to this metric is simply given by projecting the ambient connection $\nabla$ onto the tangent space $TS$.

In addition to projecting $\nabla$ onto $TS$, we can capture the extrinsic geometry of $S$ by projecting onto $TS^\perp$. To be precise, let $\nu \in T_pS^\perp$ be the unit normal vector determined by the orientations of $S$ and $M$. Given vector fields $X,Y\in \mathfrak X(S)$, choose local extensions $\overline X, \overline Y$ to a tubular neighborhood of $S$. We define the \emph{second fundamental form} $\Pi\colon T_p(S) \times T_p(S) \to \mathbb R$ by
$$ \Pi(X,Y) = \ang{\nabla_{\overline X}\overline Y, \nu}.$$ It is a routine calculation that $\Pi$ is a symmetric, bilinear form. In particular, $\Pi(X,Y)$ at $p$ only depends on $X(p)$ and $Y(p)$.

The \emph{shape operator} $\mathscr{S}_\nu$ is the unique linear map $T_pS \to T_pS$ satisfying $$\ang{\mathscr{S}_\nu(X),Y} = \Pi(X,Y).$$ Since the second fundamental form $\Pi$ is symmetric, we see that the shape operator is self-adjoint.

Now let us consider the case that $M$ is a Riemannian 3-manifold and $S\subset M$ an oriented surface. Since the shape operator $\mathscr{S}_\nu$ is self-adjoint, we can choose an orthonormal basis of eigenvectors $\{e_1,e_2\}$ of $T_pS$. The corresponding real eigenvalues $\lambda_1(p),\lambda_2(p)$ are called the \emph{principal curvatures} of $S$ in $M$ at $p$.

Now we are able to state the Gauss equation.

\begin{thm}[Gauss Equation]
	Let $K_M$ and $K_S$ be the sectional curvature operators defined with respect to the Riemann curvature tensors on $M$ and $S$, respectively. Then
	\begin{equation}
		K_S(T_p S) - K_M(T_p S) = \lambda_1(p)\lambda_2(p).
	\end{equation}
	Or more succinctly,
	\begin{equation}\label{eqn:gauss}
		K_S - K_M = \lambda_1\lambda_2.
	\end{equation}
\end{thm}

\subsection{Minimal surfaces} Let $S\subset M$ be an oriented surface in a Riemannian 3-manifold $M$. The trace of the shape operator, called the \emph{mean curvature} of $S$, is a measure of the extrinsic curvature of $S$ relative to $M$. We say that $S$ is a \emph{minimal surface} if its mean curvature is identically zero. The critical property that follows from this restriction on the extrinsic curvature is variational: minimal surfaces are precisely the critical point of the area functional. A surface that is a local minimum of the area functional is called \emph{stable}.

One of the essential tools in the study of minimal surfaces is the monotonicity formula, which gives local control on the area growth of a minimal surface inside the neighborhood of a point. There are several versions of the monotonicity formula, but we focus on the statement for hyperbolic manifolds, first described by \textcite{And82}.

\begin{thm}[Monotonicity Formula \protect{\cite[Theorem 1]{And82}}]\label{monotonicity-formula}
	Let $M$ be a hyperbolic 3-manifold and $S\subset M$ a minimal surface. Fix a point $x\in S$. Then the function $$r\mapsto \frac{\area(S\cap B(x,r))}{2\pi(\cosh r-1)}$$ is monotonic increasing.
\end{thm}

The quantity $2\pi(\cosh r-1)$ is the area of a totally geodesic disk of radius $r$, so $$\lim_{r\to0^+} \frac{\area(S\cap B(x,r))}{2\pi(\cosh r-1)} = 1.$$ Applying \cref{monotonicity-formula}, we obtain an area estimate central to our argument.

\begin{lem}\label{monotonicity}
	Let $M$ be a hyperbolic 3-manifold and let $B(x,r)$ be the open ball of radius $r \leq \operatorname{injrad}_M(x)$ centered at $x$. If $S$ is a minimal surface passing through $x$, then 
	\begin{equation}
		\area(S\cap B(x,r)) \geq 2\pi(\cosh r - 1).
	\end{equation}
\end{lem}
\begin{rmk}
	We can understand \cref{monotonicity} as saying that if a minimal surface passes through the center of a hyperbolic ball, then the area of the surface inside that ball is at least as large as the area of an equatorial disk. 
\end{rmk}



\chapter{Filling links and surface subgroups}
\section{Filling links}

Given a closed 3-manifold $M$, we say that a 1-complex $G$ embedded in $M$ \emph{carries $\fund{M}$} if the inclusion $G\hookrightarrow M$ induces a surjection $\fund G \twoheadrightarrow \fund{M}$. Such a 1-complex always exists---e.g., the 1-skeleton of a triangulation of $M$ as well as the core of a handlebody in a Heegaard splitting of $M$ both carry $\fund M$.

A 1-complex $G$ that carries $\fund{M}$ is called a \emph{spine} of $M$ if the first betti number $\beta_1(G)$ satisfies $\beta_1(G) \leq \beta_1(G')$ for all $G'\hookrightarrow M$ that carry $\fund{M}$. We say that  link $L\subset M$ is \emph{filling} in $M$ if for any spine $G$ of $M$ disjoint from $L$, $\pi_1(G)$ injects into $\pi_1(M \setminus L)$.%We consider two spines to be equivalent if they are related by a sequence of edge contractions and isotopies. Hence we see that a link $L$ is filling if and only if for any minimal Heegaard splitting $M = H_1\cup H_2$ with $L$ disjoint from one of the two Heegaard handlebodies $H_i$, the inclusion $H_i\to M\setminus L$ induces an injection $\pi_1(H_i)\hookrightarrow\pi_1(M \setminus L)$.

\begin{figure*}[h]
	\centering
	\includegraphics[width=0.4\linewidth]{spine.png}
	\caption{Standard spine for the 3-torus $T^3$.}
\end{figure*}

The main result of this section is the following algebraic alternative to filling. A \emph{hyperbolic link} is a link $L$ in a 3-manifold $M$ such that $M\setminus L$ admits a complete hyperbolic metric of finite volume.

\begin{prop}\label{prop:alternative}
	Let $M$ be a closed, orientable 3-manifold of rank $n\geq 3$, and let $L\subset M$ be a hyperbolic link with at least $n+1$ components. Then either $L$ is filling or $\pi_1(M\setminus L)$ contains surface subgroup of genus $1<g<n$.
\end{prop}

\begin{cor}\label{cor:alternative}
	Let $M$ be a closed, orientable 3-manifold of rank $3$, and let $L\subset M$ be a hyperbolic link with at least 4 components. Then either $L$ is filling or $\pi_1(M\setminus L)$ contains a genus 2 surface subgroup.
\end{cor}

\begin{rmk}
	As described in \cite[Appendix]{FK21}, it is always possible to find a hyperbolic link in a compact, orientable 3-manifold with any prescribed number of components by appealing to \cite{Mye82}.
\end{rmk}

\section{Proof of \cref*{prop:alternative}}
The proof of \cref{prop:alternative} will comprise the remainder of this chapter. Let $M$ be a closed, orientable 3-manifold of rank $n\geq 3$. Fix a hyperbolic link $L\subset M$ with at least $n+1$ components and an embedding of a spine $i \colon G \hookrightarrow M \setminus L$. Denote the image of the induced map on fundamental groups by $\Gamma = i_*(\pi_1 G) \subset \pi_1(M\setminus L)$.

Since $M\setminus L$ is hyperbolic, \cref{thm:subgroup-tameness} asserts that either:
\begin{itemize}
	\item $\Gamma$ is a virtual surface fiber group, or
	\item $\Gamma$ is geometrically finite.
\end{itemize}
We prove \cref{prop:alternative} by considering these two cases in turn.

\subsection{Virtual fiber case}
Assume $\Gamma$ is a virtual surface fiber group, so there exists a finite cover $\widetilde M \to M\setminus L$ and a surface bundle $\Sigma \hookrightarrow \widetilde M \twoheadrightarrow S^1$ with $\Gamma = \pi_1(\Sigma)$. Since $\widetilde M$ is non-compact, the fiber surface $\Sigma$ is also non-compact. In particular, $\Gamma = \pi_1(\Sigma)$ is free of some rank $k$. Then $\pi_1(\widetilde M)$ decomposes as $$\fund{\widetilde M} \cong F(k) \rtimes \mathbb Z.$$
However, since $L$ has at least $n+1$ components, we have $$b_1(\widetilde M)\geq b_1(M\setminus L) \geq n+1,$$ so the rank $k$ of \G satisfies $k = n$, and $\Gamma \cong F(n) \cong \pi_1(G)$.
It follows from the Hopfian property of free groups that $\fund{G} \to \fund{M\setminus L}$ is injective.

Hence we see that if $L$ is \emph{not} filling, then there exists a spine $G \hookrightarrow M \setminus L$ such that the corresponding subgroup $\Gamma$ is geometrically finite. Moreover, by the argument from the preceding paragraph, we need only to consider geometrically finite subgroups $\Gamma$ that are not free.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Geometrically finite case}

Now suppose $\Gamma$ is geometrically finite and non-free. If $\Gamma$ contains parabolics, denote by $\M_0(\Gamma)$ the manifold obtained by removing from \MG the interiors of all solid cusp cylinders and solid cusp tori. When there is no risk of ambiguity, we write more simply $\M_0 = \M_0(\Gamma)$. Since $\G$ is geometrically finite, Marden's criterion (\cref{geom-finite-criterion}) asserts that $\M_0$ is compact.

Since $b_1(\Gamma) \leq n$, half-lives-half-dies (see e.g., \cite[Lemma 3.5]{Hat}) implies the total genus of $\partial \M_0$ is at most $n$. We now analyze the possibilities for $\partial \M_0$.

\emph{$\partial \M_0$ is a union of incompressible tori.} By the hyperbolization theorem, either $\M_0\cong T^2\times I$, or $\operatorname{int}(\M_0) = \HH^3/\Gamma$ has finite volume. Since $\rk(T^2\times I) = 2 < n = \rk(\Gamma)$, we conclude that $\operatorname{int}(\M_0) = \HH^3/\Gamma$ has finite volume. Then the inclusion $\Gamma \subset \pi_1(M\setminus L)$ induces a finite-sheeted covering $\HH^3/\Gamma \to M\setminus L$. In particular, $\Hh/\G$ has at least as many cusps as $M\setminus L$. But recall that $M\setminus L$ has at least $n+1$ cusps, while $\HH^3/\Gamma$ has at most $n$ cusps, so we reach a contradiction.
%To see that the case $\M_0\cong T^2\times I$ is impossible, note that the cusp tori in $\partial \M_0$ correspond to parabolic subgroups belonging to distinct conjugacy classes. In particular, a loop on a cusp torus in $\M_0$ cannot be freely homotopic into a different cusp torus. But in $\M_0\cong T^2\times I$, each simple loop in one of the boundary tori cobounds an annulus with another simple loop in the opposite boundary component.

\emph{$\partial \M_0$ contains an incompressible component of genus $1<g<n$.} We have immediately that $\Gamma$, and hence also $\pi_1(M\setminus L)$, contains a surface subgroup of genus $1<g<n$, as desired.

\emph{$\partial \M_0$ is incompressible of genus $n$.} We first compute the betti numbers of $\M_0$. Since $\M_0$ is connected with non-empty boundary, we immediately have $b_0(\M_0) = 1$ and $b_3(\M_0) = 0$. To compute $b_1$, first note that $b_1(\M_0) \geq b_1(\partial \M_0) = n$ by half-lives-half-dies. On the other hand, $\Gamma \cong \pi_1(\M_0)$ is rank at most $n$, so $b_1(\M_0) \leq n$. Hence, $b_1(\M_0) = n$.

Now recall that $$\chi(\M_0) = \frac{1}{2}\chi(\partial \M_0) = 1-n,$$ which follows from an elementary application of the Mayer--Vietoris sequence. From this we obtain $b_2(\M_0) = 0$.

Suppose now $\Gamma = \langle \gamma_1, \gamma_2, \dots,\gamma_n \rangle$. Since $b_1(\Gamma) = n$, the images of $\gamma_1, \gamma_2, \dots,\gamma_n$ in $H_1(\Gamma; \mathbb Q)$ are linearly independent over $\mathbb Q$. This together with $H_2(\Gamma; \mathbb Q) = 0$ allows us to apply \cite[Theorem 7.4]{Sta65}, which states that $\gamma_1, \gamma_2, \dots,\gamma_n$ freely generate a subgroup of $\Gamma$, which of course is equal to $\Gamma$. But this yields a contradiction, since $\Gamma$ contains a non-free subgroup induced by $\partial \M_0$, which was assumed to be incompressible of genus $n$.

\emph{$\partial \M_0$ contains a compressible component.} Let $D\subset \partial \M_0$ be a compressing disk and $Y$ the manifold obtained by cutting along $D$. If $D$ is non-separating, then $$\Gamma \cong \pi_1 (Y) * \ZZ$$ with $\rk(\pi_1 (Y))=n-1>1$. On the other hand, if $D$ is separating with components $Y'$, $Y''$, then by Grushko's theorem,  we have $$\Gamma \cong \pi_1(Y') * \pi_1(Y''),$$ such that $\rk(\pi_1 Y'), \rk(\pi_1Y'') < n.$ Since $\rk(\Gamma)>3$, we see that $\pi_1 Y'$ or $\pi_1 Y''$ (or both) must have rank greater than one.

Thus, whether $D$ is separating or non-separating, in both cases we see that $\Gamma$ decomposes as $\Gamma \cong \Gamma' *\Gamma''$, for some subgroup $\Gamma'$ of rank $k$, $1<k<n$. We now reason inductively via the following lemma. Applying \cref{lem:induction} to $\Gamma'$ completes the proof of \cref{prop:alternative}.

\begin{lem}\label{lem:induction}
	Suppose $\Gamma$ decomposes as a free product $\Gamma \cong \Gamma'*\Gamma''$, where $\Gamma'<\Gamma$ is a subgroup of rank $k$, $1< k< n$. Then either $\Gamma'$ is free or $\Gamma'$ contains a surface subgroup of genus $g$, $1<g\leq k$.
\end{lem}
\begin{proof}
	We begin with $k=2$, so $\Gamma'$ is a two-generator subgroup of the Kleinian group $\fund{M\setminus L}$. We use the classical description of such subgroups by \textcite[][Theorem VI.4.1]{JS79}, yielding the following possibilities for $\Gamma'$:
	\begin{enumerate}[(1)]
		\item $\Gamma'$ has finite index in $\pi_1(M\setminus L)$, or
		\item $\Gamma'$ is free of rank 2, or
		\item $\Gamma'$ is free-abelian of rank 2.
	\end{enumerate}
	
	Clearly case (1) is impossible since $b_1(\pi_1(M \setminus L)) \geq 4$.
	
	If $\Gamma'$ is free of rank 2 as in case (2), then we are done.
	
	Now we consider case (3). Let $\phi\colon \pi_1(M\setminus L) \to \pi_1(M)$ be the map on $\pi_1$ induced by trivially filling each component of $L$. Recall that $\ker(\phi)$ is normally generated by a collection of meridians $\{\mu_1,\dots,\mu_r\}$, where each $\mu_i$ is contained in a peripheral subgroup $P_i$ corresponding to a component of $L$. Since $M\setminus L$ is atoroidal, the rank 2 abelian subgroup $\Gamma' \cong \ZZ\times \ZZ$ is conjugate into some peripheral subgroup $P_j$ of $\pi_1(M\setminus L)$. After choosing a framing $P_j = \langle \mu_j, \lambda_j \rangle$, we see that $\rk (\phi(P_j)) = \rk(\langle \phi(\lambda_j)\rangle) \leq 1$. In particular, $\rk(\phi(\Gamma')) \leq 1$, so $\rk(\phi(\Gamma)) \leq n-1$. But recall that $G$ is a spine of $M$, so the composition $$ \pi_1(G) \xrightarrow{i_*} \pi_1(M\setminus L) \xrightarrow{\phi} \pi_1(M)$$ is surjective by definition. In particular, $$\rk(\phi(i_*(G))) = \rk(\phi(\Gamma)) = n,$$ which is a contradiction. Thus, we exclude the possibility of case (3) and conclude the proof for $k=2$.

	If $n=3$, then $k=2$ is the only case to consider in the lemma, so we are done. Hence, assume now that $n>3$. We proceed by induction on $k$.
	
	Suppose $\Gamma\cong \Gamma'*\Gamma''$, where $\rk(\Gamma')=k$, $2<k<n$. Then $\Gamma'$ is geometrically finite, so we conduct a similar case analysis of $\partial\M_0(\Gamma')$, which has total genus at most $k$.
	
	\emph{$\partial \M_0(\Gamma')$ is a union of incompressible tori.} This situation is impossible by the same argument we applied to $\partial\Mog$. That is, if $\partial\Mogg$ consists of incompressible tori, then either $\mathcal{M}_0(\Gamma') \cong T^2\times I$ or $\operatorname{int}(\Mogg) = \HH^3/\Gamma'$ has finite volume. But again, $\rk(T^2\times I) = 2 < k = \rk(\Gamma')$, so the first case is impossible. We conclude that $\HH^3/\Gamma'$ has finite volume, and hence, $\HH^3/\Gamma' \to M\setminus L$ is a finite covering. But $M\setminus L$ has at least $n+1$ cusps, while $\HH^3/\Gamma'$ has at most $k<n$ cusps, a contradiction.

	\emph{$\partial \M_0(\Gamma')$ contains an incompressible component of genus $g$, $1<g\leq k$.} Immediately $\Gamma'$ contains a surface subgroup of the required genus.
	
	\emph{$\partial \M_0(\Gamma')$ contains a compressible component.} Applying the same argument as with \G, we obtain a splitting $\Gamma' \cong \Lambda * \Lambda'$, where, without loss of generality, $\rk(\Lambda)=\ell$, $1<\ell<k$. But then by induction, $\Lambda$ is either free or contains a surface subgroup of genus $1<g\leq \ell<k$, and therefore $\Gamma'$ is either free or contains a surface subgroup of genus $1<g<k$.
\end{proof}
\chapter{Minimal surfaces in hyperbolic 3-manifolds}
\section{Genus bounds of essential surfaces}\label{section:min-surfaces}

% We now use tools from minimal surface theory to translate the algebraic obstruction of Proposition~\ref{prop:alternative} into a geometric one. First we recall some basic facts about minimal surfaces.


%Recall that there is a one-to-one correspondence between conjugacy classes of surface subgroups of a hyperbolic 3-manifold $M$ and homotopy classes of closed $\pi_1$-injective surfaces immersed in $M$.
There is an extensive body of work constructing stable minimal surfaces which are area-minimizing in their homotopy classes, based on the seminal work of Schoen and Yau \cite{SY79}, Sacks and Uhlenbeck \cites{SU81,SU82}, and many others \cites{MY82, HS88}. The following result of \textcite{HW17} allows us to find least-area representatives in cusped hyperbolic 3-manifolds.
%Recall that a closed, orientable surface $S$ immersed in a compact 3-manifold $M$ is called \emph{essential} if $S$ is $\pi_1$-injective and non-boundary parallel.
\begin{prop}[\protect{\cite[Theorem 1.1]{HW17}}]\label{thm:least-area-exist}
	Let $S$ be a closed orientable surface of genus at least two, which is immersed in a cusped hyperbolic 3-manifold $M^3$. If $S$ is $\pi_1$-injective, then $S$  is homotopic to an immersed least-area minimal surface in $M$.
\end{prop}
%Using this theorem we can reason about surface subgroups by passing to a least-area representative in the corresponding homotopy class.
The next lemma supplies a lower bound for the genus of a least-area surface in terms of its area. Originally attributed to an unpublished work of Uhlenbeck, it is well known to differential geometers.

\begin{lem}\label{lem:genus-area-bound}
	A closed minimal surface $S$ of genus $g$ immersed in a hyperbolic 3-manifold $M$ satisfies
	\begin{equation}
		g \geq \frac{\area(S)}{4\pi} + 1.
	\end{equation}
\end{lem}
\begin{proof}
	Let $\lambda_1,\lambda_2$ be the principal curvatures of $S$. Since $S$ is minimal, $$\tr(\mathscr{S}_\nu) = \lambda_1 + \lambda_2 = 0.$$ Then by \cref{eqn:gauss}, the Gaussian curvature of $S$ satisfies $$K_S =K_M + \lambda_1\lambda_2 = -1 - \lambda_1^2 \leq -1.$$ We now apply Gauss--Bonnet,
	\begin{equation*}
		2\pi\chi(S) = \int_S K_S\ dA \leq \int_S -1\ dA = -\area(S).
	\end{equation*}
	Thus we obtain $2\pi\chi(S) = 2\pi(2-2g) \leq -\area(S)$, so $$g\geq \frac{\area(S)}{4\pi} + 1,$$ as desired.
\end{proof}

% \begin{cor}
%   Let $S$ be a closed, incompressible, least-area surface of genus $g$ immersed in a hyperbolic 3-manifold. If $\area(S) > 4\pi$, then $g \geq 3$.
% \end{cor}

\section{Systoles and area bounds}

Now we relate the area of minimal surfaces to the geometry of the ambient 3-manifold. Let $M$ be a hyperbolic 3-manifold and $S$ be a closed surface of genus $g>1$ with an immersion $i\colon S \looparrowright M$. A simple closed curve $\alpha \subset S$ is called an \emph{accidental parabolic} if $i_*(\alpha) \in \pi_1(M)$ is a parabolic element of $\fund{M}$.

\begin{lem}\label{prop:area-systole-bound}
	Let $M$ be a complete hyperbolic 3-manifold and $S$ a closed, stable minimal surface immersed in $M$ with no accidental parabolics. Then the area of $S$ satisfies
	\begin{equation}
		\area(S) \geq 2\pi \left(\cosh\left(\frac{\sys(M)}{2}\right) - 1\right),
	\end{equation}
	where $\sys(M)$ denotes the length of the shortest closed geodesic in $M$.
\end{lem}

\begin{proof}
	Since $\pi_1(M)$ is LERF (see e.g. \cite[Chapter 5.2, H.11]{AFW15} for details), we pass to a finite cover $\tilde M$ in which $S$ embeds. Let $p\colon \HH^3 \to \tilde M$ be a covering map, and let $\tilde{S}$ be a lift of $S$ to $\HH^3$. Take any point $x\in \tilde{S}$ and let $\tilde{S}_0$ be a connected component of $S \cap B(x, \sys(M)/2)$ containing $x$. We first claim that $\left.p\right|_{\tilde{S}_0}$ is injective. To see this, suppose $x_0, x_1 \in \tilde{S}_0$ are distinct with $p(x_0) = p(x_1)$, and let $\gamma\in \operatorname{Isom}^+(\HH^3)$ be the covering transformation with $\gamma(x_0) = x_1$. Since $\tilde{S}_0$ is connected, let $\alpha\colon [0,1] \to  \tilde{S}_0$ be a path connecting $x_0$ to $x_1$. Then the loop $p\circ \alpha$ lies in the free homotopy class $\left[\gamma\right]\in \pi_1(\tilde M)$. In particular, $p\circ \alpha$ is essential in $\tilde M$, and by incompressibility, also essential in $S$. Now homotope $\alpha$ rel endpoints to the unique geodesic arc $\alpha'$ connecting $x_0$ and $x_1$. Composing with $p$ yields a homotopy in $\tilde M$ from $p\circ \alpha$ to $p\circ \alpha'$, which has length $\ell_0(p\circ \alpha') = d_{\HH_3}(x_0, x_1) < \sys(\tilde M)$. Thus $p\circ \alpha$ is an accidental parabolic in $S$, a contradiction.
	
	Now we have $\left.p\right|_{\tilde{S}_0}\colon \tilde{S}_0 \to S$ is an isometric embedding, so in particular, $\area(S) \geq \area(\tilde{S}_0)$. Then applying \cref{monotonicity} yields
	$$\area(S) \geq \area(\tilde{S}_0) \geq  2\pi\left(\cosh\left(\frac{\sys(\tilde M)}{2}\right) - 1\right) \geq 2\pi\left(\cosh\left(\frac{\sys(M)}{2}\right) - 1\right),$$
	since $\sys(\tilde M) \geq \sys(M)$.
\end{proof}

The previous proposition in essence says that, as long as $M$ does not have short geodesics, the stable minimal surfaces (without accidental parabolics) in $M$ must be sufficiently complex. We now give a similar systole bound for hyperbolic 3-manifolds that contain an immersed genus 2 surface with accidental parabolics.

\begin{lem}\label{lem:accidental-parabolic}
	Let $M$ be a finite volume hyperbolic 3-manifold and $S$ a closed, $\pi_1$-injective surface of genus $2$ immersed in $M$. If $S$ has an accidental parabolic, then $\sys(M) \leq 2\arccosh(9/2) = 4.369...$.
\end{lem}

To prove this, we will utilize a result of \textcite{AR00} which provides a systole bound induced by the presence of incompressible surfaces with cusps.

\begin{lem}[\protect{\cite[Theorem 5.2]{AR00}}]\label{adams-reid-result}
	Let $M$ be a compact 3-manifold with nonempty torus boundaries and hyperbolic interior, such that it contains a properly embedded or properly immersed incompressible boundary-incompressible surface $S_{g,p}$, with genus $g$ and $p$ punctures, $p\geq 1$.
	\begin{enumerate}[(i)]
		\item If $p\geq 2$, then $\begin{aligned}[t]
	\sys(M)\leq \max\big\{3.525..., \min\{ 2\arccosh ((12g+5p-13)/2),\\4\arccosh((6g-6+3p)/p)\}\big\}.
\end{aligned}$
		\item If $p=1$, then $\sys(M) \leq 3.525\dots$ for $g=1$ and $\sys(M)\leq 2\arccosh(6g-5)$ for $g\geq 2$.
	\end{enumerate}
\end{lem}
\begin{proof}[Proof of \cref{lem:accidental-parabolic}]
	Since $\pi_1(M)$ is LERF (see e.g. \cite[Chapter 5.2, H.11]{AFW15} for details), we pass to a finite cover $\tilde M$ in which $S$ embeds. Additionally, we truncate $\tilde M$ by removing cusp neighborhoods disjoint from $S$, and so we may take $\tilde M$ to be compact with toroidal boundary. Now take a simple closed curve in $S$ representing the accidental parabolic. By the annulus theorem, this curve cobounds an annulus with another simple closed curve in $\partial \tilde M$. Now we compress $S$ to $\partial \tilde M$ along this annulus. The result is either two once-punctured tori or a single twice-punctured torus, depending on whether the compressing annulus separates $S$. 
	
	In the first case, the once-punctured tori are boundary incompressible, otherwise cutting along a boundary compressing disk yields an essential annulus, which is impossible since $\tilde M$ is hyperbolic. Now consider the second case, so $\tilde M$ contains a properly embedded twice-punctured torus. If this surface is boundary compressible, then again cutting along a compressing disk yields either a once-punctured torus and essential annulus, which is impossible, or a single once-punctured torus, which by the above must be boundary incompressible. Hence $\tilde M$ contains either a once or twice-punctured torus which is incompressible and boundary incompressible. With this, we apply \cref{adams-reid-result} to obtain
	\begin{equation*}
		\sys(\tilde M) \leq \max\{2 \arccosh 3, \min\{2\arccosh(9/2), 4\arccosh(3) \} \} = 2\arccosh(9/2).
	\end{equation*}
	But since $\tilde M$ covers $M$, we have $\sys(M) \leq \sys(\tilde M) \leq 2\arccosh(9/2)$, a contradiction.
\end{proof}

We are now able to prove the main result of this section.

\begin{prop}\label{thm:systole-genus-bound}
	Let $M$ be a finite volume hyperbolic 3-manifold and $S$ a closed, $\pi_1$-injective surface of genus $g>1$ immersed in $M$. If $\sys(M) > 2\arccosh(9/2) = 4.369...$, then $g \geq 3$.
\end{prop}

\begin{proof}
	First, suppose that $S$ has an accidental parabolic. The systole assumption $\sys(M) > 2\arccosh(9/2)$ implies $g\neq 2$ by \cref{lem:accidental-parabolic}, so $g\geq 3$.
	
	Now assume $S$ has no accidental parabolics. By \cref{thm:least-area-exist}, we may, after a homotopy, assume $S$ is least-area. Then applying \cref{prop:area-systole-bound} with $\sys(M) > 2\arccosh(9/2)$ yields $\area(S) \geq 7\pi$. Immediately by \cref{lem:genus-area-bound}, we have $g\geq 3$, as desired.
\end{proof}

\chapter{A universal congruence link}


\section{Principal congruence manifolds}

Denote by $\mathcal O_d$ the ring of integers of the number field $\mathbb Q(\sqrt{-d})$ for $d>0$ a square-free integer. This gives rise to a family of finite-covolume Kleinian groups $\Gamma_d = \PSL_2(\mathcal O_d)$ called the \emph{Bianchi groups}. Note that $\mathcal O_1 = \ZZ[i]$, so for $d=1$ we will simply write $\Gamma_1 = \PSL_2(\ZZ[i])$. The Bianchi groups arise as the natural 3-dimensional analogue of the modular group $\PSL_2(\ZZ)$: the non-cocompact arithmetic Kleinian groups are precisely the Kleinian groups \G that, perhaps after a conjugation in $\pslc$, are commensurable with some $\Gamma_d$; that is, $\G\cap \Gamma_d$ has finite index in both \G and $\Gamma_d$ (see \cite[Chapters 8 and 9]{MR03} for background).

To an ideal $I < \mathcal O_d$, there is an associated \emph{principal congruence subgroup} $$\Gamma(I) = \ker \left\{ \PSL_2(\mathcal O_d) \to \PSL_2(\mathcal O_d/I) \right\}.$$ The quotient $\HH^3/\Gamma(I)$ is called a \emph{principal congruence manifold}. The next proposition utilizes the algebra of principal congruence subgroups to get good control over the geometry of their manifold quotients.

\begin{prop}\label{prop:cong-systole-bound}
	Let $\langle \alpha \rangle < \mathcal O_d$ be a principal ideal. If $\lvert \alpha \rvert > \sqrt{11}$, then $$\sys(\HH^3/\Gamma\big(\langle \alpha \rangle\big) > 2\arccosh(9/2).$$
\end{prop}

\begin{proof}
	Let $\gamma\in \Gamma(I)$ be loxodromic, and write $$ \gamma = \pm\begin{pmatrix}1+a & b \\ c & 1 + d \end{pmatrix}, \quad a,b,c,d \in I.$$
	Since $\det \gamma = 1$, we have
	$$
		\pm\tr \gamma = 2 + (bc - ad) \equiv 2 \pmod {I^2}.
	$$
	Hence, for a principal ideal $I = \langle \alpha \rangle$, we may write $\pm \tr \gamma = z\alpha^2 + 2$ for some $z\in \mathcal O_d$. Then the reverse triangle inequality yields
	\begin{equation}
		\lvert \tr\gamma\rvert \geq \big\lvert \lvert z\rvert\lvert \alpha\rvert^2 - 2\big\rvert > \lvert11-2\rvert = 9.
	\end{equation}
	Now recall that the complex translation length $\ell(\gamma) = \ell_0(\gamma) + i\theta(\gamma)$ satisfies $$\tr\gamma = 2\cosh\left(\ell(\gamma)/2\right).$$ Hence we obtain $$\ell_0(\gamma) \geq 2\arccosh(\lvert \tr\gamma\rvert/2) > 2\arccosh(9/2),$$ as desired.
\end{proof}

We will now narrow our focus to a particular principal congruence group $\Gamma(\langle 3 + 2i\rangle) < \PSL_2(\mathcal \ZZ[i])$. It was shown in \cite{BGR19b} that in fact $\HH^3/\Gamma(\langle 3 + 2i\rangle)$ is the complement of a link $L_0$ in $S^3$. Recent work of \textcites{DOR22} produced the diagram of $L_0$ in \cref{fig:link}. For now, note $\lvert 3 + 2i\rvert = \sqrt{13} > \sqrt{11}$.
\begin{figure*}[h]
	\centering
	\includegraphics[width=\linewidth]{link_figure.png}
	\caption[Principal congruence link $L_0$, $I =\ang{3+2i}$]{Principal congruence link $L_0$, $I =\ang{3+2i}$ \cites[Figure 6]{BGR19}{DOR22}
	\label{fig:link}}
\end{figure*}
%so Proposition~\ref{prop:cong-systole-bound} and Theorem~\ref{thm:systole-genus-bound} imply the following.
\begin{cor}\label{cor:no-subgroup}
	The link group $\pi_1(S^3\setminus L_0)$ and all its finite index subgroups do not contain a surface subgroup of genus 2.
\end{cor}
\begin{proof}
	Let $\Gamma<\pi_1(S^3\setminus L_0)$ have finite index, and suppose $\Gamma$ contains a genus 2 surface subgroup. We then obtain an immersed, $pi_1$-injective genus 2 surface in $\Hh/\Gamma$. Now \cref{thm:systole-genus-bound} asserts $\sys(\Hh/\Gamma)\leq 2\arccosh(9/2)$. But since systole is non-decreasing in finite covers, we have $$\sys(\fund{S^3\setminus L_0})\leq \sys(\Hh/\Gamma) \leq 2\arccosh(9/2).$$ On the other hand, by \cref{prop:cong-systole-bound} and the discussion preceding the corollary, we have $\sys(\fund{S^3\setminus L_0}) > 2\arccosh(9/2)$, a contradiction. 
\end{proof}

\Cref{cor:no-subgroup} in particular implies that there is no incompressible genus 2 surface immersed in $S^3\setminus L_0$. To our knowledge, this is the first known hyperbolic link in $S^3$ with a known obstruction to immersed incompressible surfaces of a given genus. In fact, by taking finite cyclic covers corresponding to unknotted components of $L_0$, we obtain an infinite family of links with this property.

\begin{cor}
	There exist infinitely many hyperbolic links $L\subset S^3$ such that $S^3\setminus L$ does not contain an immersed, closed, orientable, incompressible surface of genus 2.
\end{cor}

\begin{question*}
	Does there exist a hyperbolic knot $K\subset S^3$ such that $S^3\setminus K$ does not contain an immersed, closed, orientable, incompressible surface of genus 2?
\end{question*}


\section{Universal Links}
It is a classical theorem of \textcite{Ale20} that every closed orientable smooth $n$-manifold is a branched cover of $S^n$. The branch locus in Alexander's theorem is in general only a codimension 2 CW-complex, not a submanifold. However, in the 3-dimensional setting, \textcite{Hil74}, \textcite{Hir74}, and \textcite{Mon74} independently proved that there always exists a branched cover $M^3 \to S^3$ whose branch locus is a \emph{knot}. Incidentally, this is very similar to the present consideration of filling subsets: it is easy to construct a filling complex (e.g. the core of a handlebody in a Heegaard splitting), but it is much more difficult to upgrade such a complex to be a submanifold.

Thurston extended this idea in an unpublished manuscript \cite{Thu82} by constructing a link $L$ in $S^3$ with the remarkable property that \emph{every} closed, orientable 3-manifold is a branched covering of $S^3$ with branch set precisely $L$.

\begin{defn}
	A link $L$ in $S^3$ is called \emph{universal} if every closed, orientable 3-manifold is a branched cover of $S^3$ with branch set $L$.
\end{defn}

The goal of this section is to show that the principal congruence link $L_0$ is universal. Known universal links include the Whitehead link \cite{HLM83}, Borromean rings \cite{HLM83}, and non-torus 2-bridge links \cite{HLM85}. Note that this implies the figure-eight knot is universal, a result which we will use later. In fact, it is an open question whether all hyperbolic links in the 3-sphere are universal,\footnote{In fact, the only known links known \emph{not} to be universal are iterated cables of torus links.} but it is difficult in practice to determine for any given link. The most straightforward strategy is to bootstrap off of other known universal links by utilizing the following basic observation.

\begin{lem}\label{lem:sublink-universal}
	If a link $L\subset S^3$ contains a sublink $L'$ that is universal, then $L$ is universal.
\end{lem}

\begin{proof}
	Let $p\colon M \to S^3$ be a covering branched over $L' \subset S^3$. Recall that $p$ is determined by its monodromy $m\colon \pi_1(S^3 \setminus L') \to S_n$ (see, e.g. \cite[Chapter 10]{Rol76}). We can trivially extend the monodromy to $m\colon \pi_1(S^3 \setminus L) \to S_n$ by mapping each meridian of $L \setminus L'$ to the identity permutation and hence obtain a cover $M\to S^3$ branched over $L$.
\end{proof}

% \begin{lem}\label{lem:branched-universal}
%   Let $p\colon S^3 \to S^3$ be a covering branched over a link $L$ and let $L'$ be a link disjoint from $L$. If $p^{-1}(L')$ is universal, then $L \cup L'$ is universal.
% \end{lem}

% \begin{proof}
%   If $q \colon M \to S^3$ is a covering branched over $p^{-1}(L')$, then $p\circ q \colon M \to S^3$ is a covering branched over $L\cup L'$.
% \end{proof}

We now seek to find a universal sublink of $L_0$. This is complicated by the sheer complexity of the diagram in \cref{fig:link}. Unfortunately, the computational tools of \cite{DOR22} used to construct the diagram are unable to recover the rich symmetry that arithmetic manifolds tend to exhibit. We give two computational methods to circumvent this problem: a geometric approach using SnapPy~\cite{SnapPy} and an algebraic one using Magma~\cite{Magma}.
%The Magma computation is rigorous and verifiably correct but not as straightforward.

\begin{lem}\label{lem:fig-eight-component}
	The principal congruence link $L_0$ has the figure-eight knot as a component.
\end{lem}

\begin{proof}[Proof using SnapPy]
	We use the triangulation \verb|pSL_3_plus_2_sqrt_minus_1.tri| given in \cite{BGR19a} of $S^3 \setminus L_0$. We claim that the component of $L_0$ corresponding to cusp 20 of this triangulation is the figure-eight knot. To see this, we trivially Dehn fill every other component sequentially, retriangulating after every step.
	\begin{quote}
		\begin{verbatim}
In[1]: M = Manifold("pSL_3_plus_2_sqrt_minus_1.tri")

In[2]: for _ in range(20):
         M.dehn_fill((1,0),0)
         M = M.filled_triangulation()
         # fill the first 20 cusps

In[3]: for _ in range(M.num_cusps() - 1):
         M.dehn_fill((1,0),-1)
         M = M.filled_triangulation()
         # fill the last 21 cusps

In[4]: N = Manifold("4_1")
       # figure-eight knot

In[5]: M.is_isometric_to(N)
Out[5]: True
		\end{verbatim}
	\end{quote}
	Note that filling each component one at a time and calling \verb|M.filled_triangulation| afterwards is essential to avoid floating-point errors.
\end{proof}

\begin{proof}[Proof using Magma]
	Here we take a group-theoretic approach to Dehn filling and make extensive use of the methods and subroutines given in \cite{BGR19a}. We use the following presentation of $\PSL_2(\ZZ[i])$ given in \cite{Swa71}: $$ \PSL_2(\ZZ[i]) = \langle a,\ell,t,u \mid \ell^2=(t\ell)^2=(u\ell)^2=(a\ell)^2=a^2=(ta)^3=(ua\ell)^3=1, [t,u] = 1 \rangle.$$ The generators correspond to matrices
	\begin{equation*}
		a = \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix},\quad \ell = \begin{pmatrix} -i & 0 \\ 0 & i\end{pmatrix},\quad
		t = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix},\quad u = \begin{pmatrix} 1 & i \\ 0 & 1\end{pmatrix}.
	\end{equation*}
	Since $\Gamma(\langle 3 + 2i\rangle)$ is isomorphic to a link group $\pi_1(S^3 \setminus L_0)$, we have that $\Gamma(\langle 3 + 2i\rangle)$ is normally generated by the meridians of $L_0$. Using the method of \cite[Section 13]{BGR19a}, we compute a complete set $P$ of meridians given in \cref{table:meridians}. Each cusp of $\Gamma(\langle 3 + 2i\rangle)$ corresponds to precisely one meridian in $P$.
	
	To simplify the computation, we note that since $\PSL_2(\ZZ[i])$ has a single cusp, the peripheral subgroups of $\Gamma(\langle 3 + 2i\rangle)$ are all mutually conjugate in $\PSL_2(\ZZ[i])$; in fact, a computation using the methods of \cite{BGR19a} shows that they are all conjugate to the subgroup $\langle t^{13}, t^{-5}u\rangle$. Hence $\Gamma(\langle 3 + 2i\rangle)$ is the normal closure of $\langle t^{13}, t^{-5}u\rangle$ in $\PSL_2(\ZZ[i])$. 
	
	We finish by using Magma to kill all but one meridian, resulting in the figure-eight knot group.
	\begin{quote}
		\begin{verbatim}
Bianchi<a,l,t,u> := Group<a,l,t,u|a^2,l^2,(t*l)^2,(u*l)^2,
                         (a*l)^2,(t*a)^3,(u*a*l)^3,(t,u)>;

G := NormalClosure(Bianchi, sub<Bianchi | t^13, t^-5*u>);
G := Rewrite(Bianchi, G : Simplify := false);
// G = Gamma(<3 + 2i>)

Q := quo< G | P[1..13], P[15..42] >;
Q := ReduceGenerators(Q);
// Filling all but cusp 14
// P is the set of meridians in Table 1

H<a,b> := Group<a,b | a^-1*b*a*b^-1*a*b=b*a^-1*b*a>;
// Figure-eight knot group

SearchForIsomorphism(Q,H,7);
// true
		\end{verbatim}
	\end{quote}
	\vspace{-1em}
\end{proof}

Since the figure-eight knot is universal \cite{HLM85}, \cref{lem:sublink-universal} and \cref{lem:fig-eight-component} immediately imply the main result of this section.

\begin{prop}\label{prop:cong-universal}
	The principal congruence link $L_0$ is universal.
\end{prop}

% \begin{proof}
% Consider the sublink $L'_0\subset L_0$ of Figure~\ref{fig:sublink}. By Lemma~\ref{lem:sublink-universal}, it suffices to show that $L'_0$ is universal. It can be verified either by hand or with the program SnapPy~\cite{SnapPy} that $L'_0$ is isotopic to the link $C = C_1\cup C_2\cup C_3\cup C_4$ of Figure ***. The universality of the link $C$ follows from an argument in \cite{HLM85}, which we summarize briefly.

% Consider the irregular 3-fold cover $p\colon S^3 \to S^3$ branched over the unlink $C_1 \cup C_2$ with monodromy mapping the two meridians to transpositions $(12)$ and $(23)$, respectively. The preimage $p^{-1}(C_3 \cup C_4) = A \cup B$ is shown in Figure ***. Now let $q\colon S^3 \to S^3$ be the 3-fold cyclic covering branched over $A$. It is easy to see that $q^{-1}(B)$ is the Borromean rings, which is universal by \cite{HLM83}. The result now follows by applying Lemma~\ref{lem:branched-universal} to the branched covering $q\circ p$.
% \end{proof}
% The argument in *** that was used to show the universality of the Whitehead link and Borromean rings in fact proves along the way that $C_4$ is also universal. It is not explicitly stated and may not be immediately apparent, so for convenience we sketch the argument here.

% Let $M$ be a closed, orientable 3-manifold. By *** there is a 3-fold branched cover $p: M \to S^3$ with branch set a link $L$ corresponding to a monodromy representation $m: \pi_1(S^3 \setminus L) \to S_3$ which maps each meridian of $L$ to one of the three transpositions $(12),(13),(23)$. Fixing a diagram of $L$, we may think of this monodromy as a 3-coloring of the strands of $L$. This 3-coloring determines the representation and hence the branched covering (up to isomorphism).

% Consider the \emph{Montesinos move} illustrated in Figure~\ref{montesinos-move}. This results in a new link $L'$ and branched cover whose monodromy is identical to $m$ outside of a ball. Since the 3-fold cover of a ball branched over two unknotted, unlinked arcs is again a ball, the effect of the Montesinos move on $M$ is to remove a ball and then glue it back in differently. Hence the topological type of $M$ is unchanged and we obtain a 3-fold cover $M \to S^3$ branched over $L'$.

% \begin{figure}[h]
%   \begin{center}
%   \includegraphics[scale=0.5]{figures/montesinos_move.pdf}
%   \end{center}
%   \caption{The Montesinos move.\label{fig:montesinos-move}}
% \end{figure}

% Now consider a diagram expressing $L$ as the closure of some braid. After applying Montesinos moves to switch crossings types if necessary, we use Montesinos moves as in Figure~*** to replace each crossing with a circle. We then Modify each circle as in Figure~*** to obtain the link $L'$ in Figure~***.
% \begin{figure}[ht]
%   \centering
%   \includegraphics[scale=0.75]{figures/sublink.pdf}
%   \caption{The 4-component sublink $L'_0 \subset L_0$.\label{fig:sublink}}
% \end{figure}
% \scalebox{0.5}[0.5]{
% \input{smooth.tex}}
\section{Proof of \cref{thm:Main}}
We recall our main theorem from the introduction.
\mainthm*
The theorem follows immediately by combining the following proposition with \cref{cor:alternative}.

\begin{prop}\label{prop:genus-2-small-existence}
	Let $M$ be a closed, orientable 3-manifold. Then there exists a hyperbolic link $L\subset M$ with at least 4 components such that $\pi_1(M\setminus L)$ does not contain a genus 2 surface subgroup.
\end{prop}
\begin{proof}
	By \cref{prop:cong-universal}, there exists a covering $p\colon M\to S^3$ branched over $L_0$. Since $L_0$ is a $42$-component link, certainly $L = p^{-1}(L_0)$ has at least 4 components. Now $p$ restricts to an unbranched finite-sheeted covering $p \colon M\setminus L \to S^3 \setminus L_0$, so $\pi_1(M\setminus L) \subset \pi_1(S^3 \setminus L_0)$. Hence $\pi_1(M\setminus L)$ does not contain a genus 2 surface subgroup by \cref{cor:no-subgroup}.
\end{proof}

\begin{table}
	% \renewcommand{\arraystretch}{0.9}
	\doublespacing
	%\caption[The set $P$ of meridians of $L_0$]{The set $P$ of meridians of $L_0$. Here $h = uau^{-2}au^{-1}ta$.\label{table:meridians}}
	\begin{tabular}{l | l}
		$ t^{-5}u $                                 & $ ht^8ut^{-5}uh^{-1} $                   \\
		$ h^2t^8ut^{-5}ut^{-5}uh^{-2} $             & $ at^8ut^{-5}ut^{-5}ua^{-1} $            \\
		$ hat^8ut^{-5}ut^{-5}ua^{-1}h^{-1} $        & $ h^2at^8ut^{-5}ua^{-1}h^{-2} $          \\
		$ tat^8ut^{-5}ut^{-5}ua^{-1}t^{-1} $        & $ t^2at^8ut^{-5}ua^{-1}t^{-2} $          \\
		$ t^3at^8ut^{-5}ua^{-1}t^{-3} $             & $ t^4at^8ut^{-5}ua^{-1}t^{-4} $          \\
		$ t^5at^8ut^{-5}ua^{-1}t^{-5} $             & $ t^6at^8ut^{-5}ua^{-1}t^{-6} $          \\
		$ t^{-1}at^{-5}ua^{-1}t $                   & $ t^{-2}at^8ut^{-5}ut^{-5}ua^{-1}t^2 $   \\
		$ t^{-3}at^8ut^{-5}ut^{-5}ua^{-1}t^3 $      & $ t^{-4}at^8ut^{-5}ut^{-5}ua^{-1}t^4 $   \\
		$ t^{-5}at^8ut^{-5}ua^{-1}t^5 $             & $ t^{-6}at^8ut^{-5}ua^{-1}t^6 $          \\
		$ that^8ut^{-5}ut^{-5}ua^{-1}h^{-1}t^{-1} $ & $ t^2hat^8ut^{-5}ua^{-1}h^{-1}t^{-2} $   \\
		$ t^3hat^8ut^{-5}ua^{-1}h^{-1}t^{-3} $      & $ t^4hat^8ut^{-5}ua^{-1}h^{-1}t^{-4} $   \\
		$ t^5hat^8ut^{-5}ua^{-1}h^{-1}t^{-5} $      & $ t^6hat^8ut^{-5}ua^{-1}h^{-1}t^{-6} $   \\
		$ t^{-1}hat^8ut^{-5}ua^{-1}h^{-1}t $        & $ t^{-2}hat^8ut^{-5}ua^{-1}h^{-1}t^2 $   \\
		$ t^{-3}hat^8ut^{-5}ua^{-1}h^{-1}t^3 $      & $ t^{-4}hat^8ut^{-5}ua^{-1}h^{-1}t^4 $   \\
		$ t^{-5}hat^8ut^{-5}ua^{-1}h^{-1}t^5 $      & $ t^{-6}hat^8ut^{-5}ua^{-1}h^{-1}t^6 $   \\
		$ th^2at^8ut^{-5}ua^{-1}h^{-2}t^{-1} $      & $ t^2h^2at^8ut^{-5}ua^{-1}h^{-2}t^{-2} $ \\
		$ t^3h^2at^8ut^{-5}ua^{-1}h^{-2}t^{-3} $    & $ t^4h^2at^8ut^{-5}ua^{-1}h^{-2}t^{-4} $ \\
		$ t^5h^2at^8ut^{-5}ua^{-1}h^{-2}t^{-5} $    & $ t^6h^2at^8ut^{-5}ua^{-1}h^{-2}t^{-6} $ \\
		$ t^{-1}h^2at^8ut^{-5}ua^{-1}h^{-2}t $      & $ t^{-2}h^2at^8ut^{-5}ua^{-1}h^{-2}t^2 $ \\
		$ t^{-3}h^2at^8ut^{-5}ua^{-1}h^{-2}t^3 $    & $ t^{-4}h^2at^8ut^{-5}ua^{-1}h^{-2}t^4 $ \\
		$ t^{-5}h^2at^8ut^{-5}ua^{-1}h^{-2}t^5 $    & $ t^{-6}h^2at^8ut^{-5}ua^{-1}h^{-2}t^6 $ \\
		\hline
	\end{tabular}
	\centering
	\caption[The set $P$ of meridians of $L_0$]{The set $P$ of meridians of $L_0$. Here $h = uau^{-2}au^{-1}ta$.}
	\label{table:meridians}
\end{table}

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