# Heegaard splittings and virtually special square complexes

Chandrika Sadanand

###### Abstract

We give a new perspective of Heegaard splittings in terms square complexes and
Guirardel’s notion of a core which allows for combinatorial measurement of the
obstruction to being a connect sum of Heegaard diagrams. A Heegaard splitting
is a decomposition of a closed orientable $3$-manifold into two isomorphic
handle bodies that have a shared boundary surface. Usually, a number of curves
on the shared boundary surface, called a Heegaard diagram, are used to
describe a Heegaard splitting. We define a larger object, the augmented
Heegaard diagram, by building on methods of Stallings and Guirardel to encode
the information of a Heegaard splitting.

Augmented Heegaard diagrams have several desirable properties: each 2-cell is
a square, they have non-positive combinatorial curvature and they are
virtually special. Restricting to manifolds that do not have $S^{1}\times
S^{2}$ as a connect summand, augmented Heegaard diagrams are tied to the
decomposition of a $3$-manifold via connect sum as described above.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Heegaard splittings and diagrams
    2. 2.2 VH complexes and special square complexes
    3. 2.3 Guirardel core
  3. 3 Heegaard splittings as Stallings maps
    1. 3.1 From Heegaard splittings to two-dimensional maps
    2. 3.2 From two-dimensional maps to Heegaard splittings
    3. 3.3 Heegaard diagrams and handle slides
    4. 3.4 A picture of Stallings maps
  4. 4 Irreducibility, immersions and square complexes
    1. 4.1 Irreducibility
    2. 4.2 Tautness
    3. 4.3 Immersions
  5. 5 Virtually special square complexes and the augmented Heegaard diagram
    1. 5.1 Non-positively curved square complexes that are virtually special
    2. 5.2 Group Actions on Trees and the Guirardel Core
    3. 5.3 Complexity of the Guirardel core and augmented Heegaard diagrams
      1. 5.3.1 Construction of augmented Heegaard diagrams
      2. 5.3.2 Area properties of the augmented Heegaard diagram
  6. 6 Future directions
  7. 7 Example: Lens spaces
    1. 7.1 Genus 1
    2. 7.2 Genus 2

##  1 Introduction

Any closed $3$-manifold $M$ is the union of two handle bodies of genus $g$
intersecting along their common boundary surface (a Heegard splitting). The
information in this decomposition is typically encoded by a set of curves on
the boundary surface called a Heegaard diagram. Stallings shows that it can
equally well be encoded by a homotopy class of maps between two $2$-complexes:
the boundary surface and the product of a bouquet of $g$ circles with it
itself. We call this map (up to homotopy) the Stallings map (see Section 3).

The handle slide group acts transitively on the set of Heegaard diagrams that
describe a particular Heegaard splitting, and $Out(F_{g})\times Out(F_{g})$
modulo a small kernel acts transitively on the Stallings maps describing a
particular Heegaard splitting. We find that these two structures are
equivalent in the following way.

Theorem 3.5 Stallings maps are in bijection with Heegaard diagrams for a given
Heegaard splitting. The handle slide group is isomorphic to $Out(F_{g})\times
Out(F_{g})$ modulo a small kernel. The bijection is equivariant with respect
to these actions and this isomorphism.

A Heegaard splitting is said to be reducible if there is an essential simple
closed curve on the splitting surface that bounds a disk in each of the handle
bodies. If a Heegaard splitting is not reducible it is said to be irreducible.
A continuous map from a surface to a product of bouquets of circles must by no
means be injective; it can exhibit “folding” and “crumpling” behaviour. In
fact, a Stallings map cannot be an embedding. The first insight of this note
is that a Stallings map for an irreducible Heegaard splitting can be
simplified (via combinatorial moves) to be a local embedding (see Section 4):

Theorem 4.12 An irreducible Heegaard splitting can be encoded by a Stallings
map which is locally an embedding.

More generally, for a Stallings map for any Heegaard splitting, as long as
$S^{2}\times S^{1}$ is not a connect summand of $M$, the map can be simplified
so that it is locally an embedding everywhere except for a minimal disjoint
union of subsurfaces. In this case, the Stallings map factors through a space,
which we call a pinched surface, resulting from collapsing these subsurfaces
to points.

Theorem 5.4 Suppose $M$ does not have $S^{2}\times S^{1}$ as a connect
summand, and has a Heegaard splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$.
Then there is a Stallings map for the Heegaard splitting that factors through
a pinched surface $\Sigma^{*}$ such that the quotient map is locally an
embedding.

Corollary Local geometric structures on the target (or codomain) $2$-complex
can be pulled back to the domain surface (or pinched suface) by the locally
injective Stallings map.

For example the codomain is naturally decomposed into squares with the link of
each vertex having cycles with at least four edges. This means the
combinatorial curvature of the codomain is at most zero. These properties can
be pulled back to the domain surface (or pinched surface) to give a non-
positively curved square complex structure. Namely the domain surface is tiled
by squares with at least four squares meeting at each vertex. In fact, more is
true about this square complex structure.

Proposition 5.1 The Stallings map can be encoded by a non-positively curved
square complex structure on the quotient of the domain $\Sigma^{*}$ that is
virtually special (in this case, this means the edges can be labelled by V
(vertical) or H (horizontal) such that each square has a pair of opposite
edges that are V, and the other pair of edges are H, and when a finite sheeted
cover is taken, the closed horizontal and vertical strips of squares satisfy
some injectivity conditions, see Section 2.2)

This vertical-horiztontal property on a surface can alternately be understood
as a half-translation structure in the case where the domain is a surface
(Namely, it is possible to make a developping mapping of the squares to
$\mathbb{R}^{2}$ to make planar charts on the complement of the vertices so
that all the transition maps between charts needed to recover the surface are
compositions of translations and rotations by $\pi$).

We now add to this square complex structure to form what we call the augmented
Heegaard diagram. This will allow for a characterization of irreducibility.

Recall that an essential simple closed curve on the splitting surface is said
to be reducing if it bounds an embedded disk in each of the handle bodies.
These disks provide an embedded copy of $S^{2}\subseteq M$. If $S^{2}\times
S^{1}$ is not a connect summand of $M$, this $S^{2}$ separates $M$ into two
connect summands. Call a Heegaard diagram simple if there is a set of disjoint
reducing curves that decompose the Heegaard splitting
$M=\mathcal{H}_{1}\sqcup\mathcal{H}_{2}$ entirely into minimal genus Heegaard
splittings of its prime connect summands. Furthermore, these reducing curves
must not intersect any of the Heegaard diagram curves. Given a Heegaard
diagram, it is difficult to know whether it is simple. One might hope that the
Heegaard diagram with the minimal number of intersections is simple, but
Osborne has given counterexamples to this [9]. Whitehead and Volodin-
Kuztnetsov-Fomenko, each conjectured that some combinatorial properties could
be used to detect whether a Heegaard diagram is simple, but their conjectures
have proven to be false, outside of specific cases of low genus [11] [12] [13]
[18].

Similarly, (assuming $S^{2}\times S^{1}$ is not a connect summand of $M$) in
Theorem 5.4, we see that the Stallings map for a Heegaard splitting sends many
curves to points, all of which are reducing. However, even after simplifying,
a Stallings map may not pinch a maximal multicurve of reducing curves,
revealing the connect sum structure of $M$. More structure is needed.

With this in mind, we construct the augmented Heegaard diagram $\mathscr{C}$
in Section 5. The augmented Heegaard diagram consists of the square complex
$\Sigma^{*}$ with some additional disks (also tiled by squares) attached. We
use the virtually special square complex structure of $\Sigma^{*}$ and
Stallings maps to construct $\mathscr{C}$, which is also virtually special. We
find that by further simplifying the Stallings map for a given Heegaard
splitting, the area of $\mathscr{C}$ can be minimized, and that in this case,
$\Sigma^{*}$ pinches a set of reducing curves that maximally decomposes $M$
with respect to connect sum.

Theorem 5.11 If the augmented Heegaard diagram $\mathscr{C}$ is of minimal
area in its $Out(F_{g})\times Out(F_{g})$ orbit, the associated Heegaard
diagram is simple.

The statement of Theorem 5.11 can be seen as a generalization and adjustment
of conjectures of Whitehead and Volodin-Kuznetsov-Fomenko, which were found to
be false [11] [12] [13] [18].

In order to constuct $\mathscr{C}$ and prove Theorem 5.11, we consider the
virtually special square complex $\Sigma^{*}$ as a pair of actions of the
fundamental group of the splitting surface $\Sigma$ on trees (see Section
5.2). Each action encodes the attachment of a handle body to the surface, and
so the information of the Heegaard splitting and $3$-manifold is contained in
the comparison of these actions.

In 2005 Vincent Guirardel introduced a kind of a “convex” core, comparing two
actions of a group on $\mathbb{R}$-trees [7]. The size of the core measures
the obstruction to the two actions being equivariant projections of a single
action on a tree. This can be interpreted as the core measures how close a
Heegaard splitting is to a connect sum of several copies of $S^{2}\times
S^{1}$. When two actions of $\pi_{1}(\Sigma)$ on a tree are exactly the same,
they encode the attachment of two handle bodies to $\Sigma$ in the same way.
This constructs a connect sum of several copies of $S^{2}\times S^{1}$. In
this case, the Guirardel core is not even full-dimensional.

More generally the Guirardel core has been used to study the orbit of cofinite
actions on an $\mathbb{R}$-tree by subgroups of $Out(F_{g})$ [2], [4], [5],
[6], [10]. This makes it particularly well-suited for the problem in this
paper, since the Stallings maps for a given Heegaard splitting form an
$Out(F_{g})\times Out(F_{g})$ orbit.

We find that the Guirardel core is indeed intimately related to the Stallings
map of a Heegaard splitting. Let $f:\Sigma\rightarrow G_{1}\times G_{2}$ be a
Stallings map encoding a Heegaard splitting (where $\Sigma$ is the splitting
surface and $G_{1}$ and $G_{2}$ are bouquets of circles). Let $S$ be the cover
of $\Sigma$ corresponding to $\ker f_{*}$. Let $T_{1}\times T_{2}$ be the
universal cover of $G_{1}\times G_{2}$ and let $\hat{f}:S\rightarrow
T_{1}\times T_{2}$ be the lift of $f$. The fundamental group of $\Sigma$ acts
on $S$ and on $T_{1}\times T_{2}$ (it acts on the latter in the form of
$f_{*}\pi_{1}(\Sigma)$). It is these actions of $\pi_{1}(\Sigma)$ on $T_{1}$
and $T_{2}$ that encode the Heegaard splitting.

${S}$${T_{1}\times T_{2}}$${\Sigma}$${G_{1}\times
G_{2}}$$\scriptstyle{\hat{f}}$$\scriptstyle{f}$

Theorem 5.7 Let $p_{1}$ and $p_{2}$ be the coordinate projections of
$T_{1}\times T_{2}$. The Guirardel core of the actions of $\pi_{1}(\Sigma)$ on
$T_{1}$ and $T_{2}$ is the smallest set $X$ such that $\hat{f}(S)\subseteq X$
and ${p_{1}}^{-1}(t)$ is connected for every $t\in T_{1}$ and
${p_{2}}^{-1}(t)$ is connected for every $t\in T_{2}$.

This leaves us with four interrelated ways to capture the information of a
Heegaard splitting and they are each in 1-1 corresponance: Heegaard diagrams,
Stallings maps, virtually special square complex structures, pairs of actions
of $\pi_{1}(\Sigma)$ on trees. There are combinatorial moves corresponding to
handle slides for Heegaard diagrams for each of the other three structures.
Just as there are many Heegaard diagrams for a single Heegaard splitting,
there are many homotopy classes of maps, many square complex structures, and
many pairs of actions of $\pi_{1}(\Sigma)$ on trees all encoding the same
Heegaard splitting.

This geometric picture leads to the question of characterizing the
relationship between properties of these structures, properties of the
Heegaard splitting and the nature of the $3$-manifold they describe. This work
is completed for genus one Heegaard splittings in Section 7, which is devoted
to Lens spaces. The tractability of special cube complexes for geometric group
theory methods led to a proof of the virtual Haken conjecture [1]. A program
exploring the higher genus cases might give a $2$-dimensional and similarly
tangible perspective on the classification of $3$-manifolds.

More specifically, one might look for algorithms to minimize the area of
$\mathscr{C}$ in terms of the core, relating properties of $\mathscr{C}$ for
irreducible 3-manifolds to properties of its JSJ-decomposition, and exploring
how to leverage methods of special square complexes to prove theorems about
$3$-manifolds via the augmented Heegaard diagram.

#### Outline

Section 2 lays out the definitions and prelimaries of Heegaard spltitings as
well as the geometric group theory tools that are used to understand Stallings
maps, such as special square complexes and the Guirardel core. Section 3
defines Stallings maps, and details their relationship to Heegaard diagrams.
Section 4 relates Stallings maps and irreducibility of Heegaard splittings,
proving Theorem 5.4. Section 5 constructs the augmented Heegaard diagram and
proves Theorem 5.11. Section 6 lists questions for future study. Section 7
works through the example of Heegaard splittings of genus one, finding their
Stallings maps, and Guirardel cores.

#### Acknowledgements

The author acknowledges Dennis Sullivan for suggesting this problem and
guiding the author through a PhD. Great thanks are due to Mark Hagen and Chris
Leininger, who listened to endless iterations of this work, providing
suggestions, feedback and support, and to Scott Taylor for reading through an
initial draft in detail and providing some corrections. Thank you to Moira
Chas for repeatedly insisting that this work should be published. Finally, the
author acknowledges Tali Pinsky, Tarik Aougab, Nissim Ranade, Cameron Crowe
and Michael Wan for helpful conversations that brought clarity.

##  2 Preliminaries

We recall some definitions and facts that will be useful in later sections.

###  2.1 Heegaard splittings and diagrams

######  Definition 2.1.

A Heegaard splitting or Heegaard decompostion of $M$ is a decomposition
$M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$, where $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ are homeomorphic $3$-manifolds that intersect in their
shared boundary, which is a surface of genus $g$. Further, each
$\mathcal{H}_{i}$ is a handle body. Namely, $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ are the result of taking a closed $3$-ball and attaching $g$
$1$-handles, or solid cylinders.

Figure 1: A handle body of genus four.

The shared boundary $\partial\mathcal{H}_{1}=\partial\mathcal{H}_{2}$ will be
called the splitting surface $\Sigma_{g}$, where $g$ is the genus.

Every compact $3$-manifold without boundary has infinitely many Heegaard
decompositions (of higher and higher genus).

######  Definition 2.2.

A Heegaard diagram describing a Heegaard splitting is a triple
$(\Sigma_{g},\mathscr{A},\mathscr{B})$, where $\mathscr{A}$ and $\mathscr{B}$
are sets of $g$ curves on a surface of genus $g$, $\Sigma_{g}$, with the
following properties.

  1. 1.

The curves in $\mathscr{A}$ represent a basis for a Lagrangian subspace of
homology (they are a maximal set of homologically non-trivial, independent,
disjoint representatives) each with null homotopic image under $\iota_{1}$,
the inclusion of $\Sigma_{g}$ into the handle body $\mathcal{H}_{1}$.

  2. 2.

the curves in $\mathscr{B}$ represent a basis for a Lagrangian subspace of
homology (they are a maximal set of homologically non-trivial, independent
disjoint representatives) each with null homotopic image under $\iota_{2}$,
the inclusion of $\Sigma_{g}$ into the handle body $\mathcal{H}_{2}$.

It is common to call the free homotopy classes in $\mathscr{A}$ blue diagram
curves and those in $\mathscr{B}$ red diagram curves. They will be depicted in
these colours in figures.

The curves in $\mathscr{A}$ dictate the attachment of $\mathcal{H}_{1}$ to
$\Sigma_{g}$ with the following procedure. Cut $\Sigma_{g}$ along the curves
of $\mathscr{A}$. This yields a sphere with $2g$ disks removed, because the
curves are homologically non-trivial and independent and because there are $g$
of them. Label all the boundary components by the curve in $\mathscr{A}$ they
were cut from. Glue $2g$ disks to this space to form a sphere. Attach a
$3$-ball to the sphere. Finally identify pairs of the glued on disks according
to the labeling on their boundary. Similarly, the curves in $\mathscr{B}$
dictate the attachment of $\mathcal{H}_{2}$ to $\Sigma_{g}$, and thus one can
reconstruct a Heegaard splitting from a diagram.

######  Definition 2.3.

Two Heegaard diagrams $(\mathscr{A},\mathscr{B})$ and
$(\mathscr{A}^{\prime},\mathscr{B}^{\prime})$ on a surface of genus $g$ differ
by a handle slide if $\mathscr{B}=\mathscr{B}^{\prime}$, and
$\mathscr{A}\setminus\\{\gamma\\}=\mathscr{A}^{\prime}\setminus\\{\gamma^{\prime}\\}$
such that for some $\delta\in\mathscr{A}$, $\delta$, $\gamma$, and
$\gamma^{\prime}$ bound a pair of pants. The roles of $\mathscr{A}$ and
$\mathscr{B}$ may be reversed as well.

######  Theorem 2.4.

(Reidemeister-Singer [15] [16]) Given two Heegaard diagrams for a Heegaard
decomposition, one can be obtained from the other by performing a finite
sequence of handle slides.

###  2.2 VH complexes and special square complexes

Here, the definitions from “VH complexes, towers and subgroups of $F\times F$”
by Bridson and Wise [3] as well as “Special Cube complexes” by Haglund and
Wise [8] are introduced.

######  Definition 2.5.

A combinatorial map of CW complexes is a continuous map between two CW
complexes such that each open cell is sent homeomorphically onto an open cell.
A CW complex is called combinatorial if all its attaching maps are
combinatorial. Finally a square complex is a combinatorial CW complex of
dimension two, where the top dimensional cells are attached along loops that
are four edges long.

######  Definition 2.6.

A square complex is VH if the edges can be partitioned into two sets: vertical
(V) and horizontal (H) so that the attaching map of each $2$-cell alternates
between vertical and horizontal edges.

Let $X$ be a VH complex. Let $X_{H}$ be the $0$-skeleton $X^{(0)}$ with the
horizontal edges attached and let $X_{V}$ be the $0$-skeleton $X^{(0)}$ with
the vertical edges attached.

######  Definition 2.7.

The connected components of the complement $X\setminus X_{H}$ are called
horizontal corridors. Similarly, the connected components of the complement
$X\setminus X_{V}$ are called the vertical corridors. Consider how the
horizontal corridors must attach to $X_{H}$ to reconstruct $X$ (and similarly
for the vertical corridors and $X_{V}$). A VH complex is clean if the
attaching maps of the horizontal (and respectively vertical) corridors to
$X_{H}$ (and respectively $X_{V}$) are injective.

VH complexes are the predecessors of special cube complexes which are cube
complexes that don’t exhibit certain pathological behaviours. We recall the
following definition, but ask the reader to see [8] for the details.

######  Definition 2.8.

A square complex is special if:

  1. 1.

Each hyperplane embeds.

  2. 2.

No hyperplane directly self osculates.

  3. 3.

No hyperplane indirectly self osculates.

A square complex is virtually special if it has a finite-sheeted cover that is
special.

See [8] for definitions of hyperplane and direct and indirect self osculation.

###  2.3 Guirardel core

Below, the definitions of “Core and intersection number for group actions on
trees” by Guirardel [7] are introduced. The reader is encouraged to consult
this paper as a reference for more details.

Begin with two actions of a group $G$ on a trees $T_{1}$ and $T_{2}$, or
equivalently an action of $G$ on $T_{1}\times T_{2}$. In this paper we will be
looking at the special case where $T_{1}=T_{2}=T$ and it is an infinite tree
where every vertex is of the same finite valence, and edges are of length 1.
Below, we give Vincent Guirardel’s definitions for this case [7]

######  Definition 2.9.

A direction $\delta$ is a connected component of $T\setminus\\{x\\}$ for some
point $x$ in $T$. A quadrant $Q=\delta_{1}\times\delta_{2}$ is the cartesian
product of two directions (and therefore is a subset of $T\times T$).

######  Definition 2.10.

Let $(t_{1},t_{2})\in T\times T$. A quadrant $Q=\delta_{1}\times\delta_{2}$ is
heavy if there exists a sequence $\\{\gamma_{n}\\}\subseteq G$ such that

  * •

$\gamma_{n}\cdot(t_{1},t_{2})\in Q$ for all $n$,

  * •

$d_{\delta_{1}}(t_{1},\gamma_{n}\cdot t_{1})\rightarrow\infty$ as
$n\rightarrow\infty$ (where $d_{\delta_{1}}(t_{1},\gamma_{n}\cdot t_{1})$ is
the distance between $t_{1}$ and $\gamma_{n}\cdot t_{1}$ in $\delta_{1}$) and

  * •

$d_{\delta_{2}}(t_{2},\gamma_{n}\cdot t_{2})\rightarrow\infty$ as
$n\rightarrow\infty$ (where $d_{\delta_{2}}(t_{2},\gamma_{n}\cdot t_{2})$ is
the distance between $t_{2}$ and $\gamma_{n}\cdot t_{2}$ in $\delta_{2}$).

$Q$ is light if it is not heavy.

Now we can define the core, which is used to compare the pair of actions on
$T$.

######  Definition 2.11.

The core $C$ of $T\times T$ with respect to the action of $G$ is the
complement of the union of light quadrants in $T\times T$. The intersection
number of the two actions is the co-volume of the action of $G$ on $C$. In
other words, the intersection number is the number of 2-cells in $C/G$, or the
area of $C/G$.

The larger the intersection number, the larger the obstruction to finding a
third tree $T_{3}$ with an action of $\pi_{1}(\Sigma_{g})$ and surjective
equivariant maps onto $T_{1}$ and $T_{2}$. In the case where $G$ acts on
$T_{1}$ and $T_{2}$ in the same way, $C/\pi_{1}(\Sigma_{g})$ has zero area. It
is common for the area to be infinite, and we will confront this fact in
Section 5.

We finish by stating a characterization of the core that will be helpful later
on.

######  Theorem 2.12 (Proposition 5.1 in [7]).

Let $G$ act on two trees $T_{1}$ and $T_{2}$ such that $C\neq\emptyset$. Let
$F\subseteq T_{1}\times T_{2}$ be a non-empty closed connected $G$-invariant
subset with connected fibres. Then $F$ contains $C$. Moreover, $C$ is the
intersection of all such sets $F$.

##  3 Heegaard splittings as Stallings maps

###  3.1 From Heegaard splittings to two-dimensional maps

Given a Heegaard splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$, with a
splitting surface
$\Sigma_{g}=\partial\mathcal{H}_{1}=\partial\mathcal{H}_{2}$, there are two
inclusion maps $\iota_{i}:\Sigma_{g}\longrightarrow\mathcal{H}_{i}$. Knowing
these two maps is sufficient to reconstruct $M$. The two maps can be expressed
as a single product map
$\iota_{1}\times\iota_{2}:\Sigma_{g}\longrightarrow\mathcal{H}_{1}\times\mathcal{H}_{2}$.

Now, we simplify by considering this map up to homotopy, and considering the
spaces up to homotopy equivalence. We post-compose $\iota_{1}$ and $\iota_{2}$
by a deformation retraction that takes $\mathcal{H}_{i}$ to a graph of genus
$g$, $\vee^{g}S^{1}$. Let $\iota^{\prime}_{1}$ and $\iota^{\prime}_{2}$ be the
respective compositions. This gives a map from a surface to a product of
graphs.

$f=\iota^{\prime}_{1}\times\iota^{\prime}_{2}:\Sigma_{g}\longrightarrow
G_{1}\times G_{2}$

There was some choice in the construcion of $f$, above, and to account for
this, we consider all possible outcomes together as a class of maps
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$.

######  Definition 3.1.

Let $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ be a Heegaard splitting. Let
$\Sigma_{g}$ be the splitting surface and let $\iota_{i}$ be in the inclusion
maps $\Sigma_{g}\rightarrow\mathcal{H}_{i}$ for $i=1,2$. The set
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ is defined as
$\\{f:\Sigma_{g}\rightarrow G_{1}\times G_{2}\\}$, where $G_{i}$ are graphs of
genus $g$ and each coordinate $f_{i}$ is homotopic to $\iota_{i}$, up to
homotopy equivalence of the target, $\mathcal{H}_{i}$.

We say that a map $f\in\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$
_encodes_ or describes the Heegaard splitting
$M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$.

Note that if $f\in\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ and
$f^{\prime}$ is homotopic to $f$, $f^{\prime}$ must also be in
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$. Therefore
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ can be partitioned into
homotopy classes.

These maps are truly two-dimensional in the sense that both their domain and
target spaces are two-dimensional and they display properties such as folding,
and branch covering behaviour (locally homotopic to the holomorphic map
$z\mapsto z^{n}$).

######  Remark 3.2.

Note that $\iota^{\prime}_{i}$ has the following properties:

  1. 1.

The homology kernel of $\iota^{\prime}_{i}$ has rank $g$ and has a basis with
representatives that are simple, closed, non-intersecting curves.

  2. 2.

These representatives have null homotopic image (because being null-homologous
and null-homotopic are equivalent in a graph).

  3. 3.

The image of the homology under ${\iota^{\prime}_{i}}_{*}$ must be rank $g$ as
well, meaning that ${\iota^{\prime}_{i}}_{*}$ is surjective on homology. This
implies that the map induced on fundamental groups must also be surjective,
since the codomain is a graph.

The first property implies the other two above. These properties are not
changed up to homotopy nor homotopy equivalence, and therefore any map in
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ will bear these properties
in each of its coordinates.

###  3.2 From two-dimensional maps to Heegaard splittings

Begin with a map $f$ from a surface of genus $g$ to the product of two
bouquets of $g$ circles, whose coordinates’ homology kernel is of rank $g$,
and each have a basis with representatives that are simple, closed, non-
intersecting curves (property (1) from Remark 3.2 above). One may construct a
Heegaard splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$. This is done simply
by taking the two sets of $g$ curves representing a basis for the homology
kernels as a Heegaard diagram. Indeed, $f$ describes
$M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ uniquely.

######  Proposition 3.3.

Consider a map $f:\Sigma_{g}\rightarrow G_{1}\times G_{2}$, where $G_{i}$ are
graphs of genus g. Suppose $f=f_{1}\times f_{2}$ and the homology kernels of
${f_{i}}_{*}$ with $i=1,2$ have bases that can each be represented by a set of
$g$ disjoint simple closed curves. Then there exists a unique Heegaard
splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ such that
$f\in\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$.

###### Proof.

This proof considers one coordinate of the map $f$ at a time. We show that
there is a unique inclusion of $\Sigma_{g}$ into a handle body
$\mathcal{H}_{1}$ as its boundary that is equivalent to $f_{1}$ up to homotopy
and homotopy equivalence of the target. The proof for $f_{2}$ follows
similarly.

##### Existence

Consider $f_{1}:\Sigma_{g}\rightarrow G_{1}$, the first coordinate of $f$.
Recall that property (1) implies properties (2) and (3). Properties (1) and
(2) guarantee the existence of $g$ simple, closed, non-intersecting curves
with null homotopic image (which represent a basis of the homology kernel of
$f_{1}$). Therefore, the map $f_{1}$ factors through a space constructed by
gluing $g$ disks to $\Sigma_{g}$ along these curves. We note that this space
has non-trivial $\pi_{2}$, but that the target $G_{1}$ does not, and so
$f_{1}$ must in fact factor through inclusion to the handle body
$\mathcal{H}_{1}$ obtained by gluing a $3$-ball along the $\pi_{2}$ class of
the previous space. See Figure 2 below. Note that the map $d$ (in Figure 2)
induces an isomorphism on homology and inherits the property (3) from $f_{1}$
($\pi_{1}$-surjectiveness).

${\Sigma_{g}\includegraphics[width=113.81102pt]{surfacegenus4.pdf}}$${\includegraphics[width=56.9055pt]{graphgenus4.pdf}G_{1}}$${\mathcal{H}_{1}\includegraphics[width=113.81102pt]{handlebodycells.pdf}}$$\scriptstyle{f_{1}}$$\scriptstyle{d}$
Figure 2: The map $f_{1}$ must factor through a handle body $\mathcal{H}_{1}$.

It remains to show that the map $d$ is part of a homotopy equivalence, and
also the uniqueness of $\mathcal{H}_{1}$ with this property.

To show that $d$ is part of a homotopy equivalence, consider the following
diagram.

${A:=\vee^{g}S^{1}}$${\mathcal{H}_{1}}$${G_{1}}$${\vee^{g}S^{1}=:B}$$\scriptstyle{h}$$\scriptstyle{d}$$\scriptstyle{j}$$\scriptstyle{e}$

The diagram shows homotopy equivalences $h$ and $j$ between $\vee^{g}S^{1}$
and $\mathcal{H}_{1}$ and a graph of genus $g$ respectively. It suffices to
show that $j\circ d\circ h$ is part of a homotopy equivalence. To this end, a
map, $e$ will be constructed that forms a homotopy equivalence together with
$j\circ d\circ h$. The labels $A$ and $B$ help clarify discussion about the
map $e$.

The fundamental groups of these spaces will be discussed assuming the chosen
base point of both copies of $\vee^{g}S^{1}$ is the wedge point.

Consider a copy of $S^{1}$ in $B$. It represents a generator of the
fundamental group, $[\alpha]$. The map $d$ inherits $\pi_{1}$-surjectiveness
from $f_{1}$, and so there is a preimage of $[\alpha]$ by $j\circ d\circ h$.
Call this preimage $[a]$. Pick some $a\in[a]$. Now we construct $e$ by sending
$\alpha$ to the image of $a$ linearly, and sending each copy of $S^{1}$ in $B$
similarly to the representative chosen by the procedure outlined in this
paragraph.

To show that $e$ and $j\circ d\circ h$ are homotopy inverses, consider a copy
of $S^{1}$ in $B$. Consider it as a based loop $\beta$. By construction,
$(j\circ d\circ h)\circ e(\beta)$ and $\beta$ are base-point homotopic.
Turning attention to $e\circ(j\circ d\circ h)$, consider a copy of $S^{1}$ in
$A$. Consider it as a loop, $\gamma$, that represents an element of
$\pi_{1}(A)$. Note that the construction of $e$ guarantees that both
$[\gamma]$ and its image by $e\circ(j\circ d\circ h)$ have the same image in
$\pi_{1}(A)$.

Simultaneously applying these based homotopies, we get a homotopy between
identity and $(j\circ d\circ h)\circ e$ and $e\circ(j\circ d\circ h)$
respectively.

##### Uniqueness

Uniqueness follows rather quickly. Suppose there were inclusions into two
handle bodies $\mathcal{H}_{1}$ and $\widetilde{\mathcal{H}_{1}}$ that were
equivalent to $f_{1}$. These are shown below with $d$ and $\tilde{d}$ both
homotopy equivalences.

${\widetilde{\mathcal{H}_{1}}}$${\Sigma_{g}}$${G_{1}}$${\mathcal{H}_{1}}$$\scriptstyle{\tilde{d}}$$\scriptstyle{f_{1}}$$\scriptstyle{d}$

One can simply compose $d$ and $\tilde{d}$ to show that in fact the two
inclusions are equivalent (up to homotopy and homotopy equivalence of the
target). Thus each coordinate of $f$ encodes the attachment of a handle body
to $\Sigma_{g}$ in a unique way, encoding a unique Heegaard splitting. ∎

From this proof, it is clear that the Heegaard splitting
$M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ can be reconstructed from any map in
$\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$. This justifies the
terminology of Definition 3.1 (that a map
$f\in\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ _encodes_ the Heegaard
splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$).

We summarize the situation with the following corollary.

######  Corollary 3.4.

A map $f:\Sigma_{g}\rightarrow G_{1}\times G_{2}$ encodes a Heegaard splitting
iff it satisfies property (1) from Remark 3.2, Section 3.1.

###  3.3 Heegaard diagrams and handle slides

#### Heegard diagrams and Stallings maps

In order to study $\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ more
systematically and relate it to Heegaard diagrams, we focus on the subset
$\mathcal{B}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}\subseteq\mathcal{A}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$
where $G_{1}$ and $G_{2}$ are bouquets of $g$ circles. Let
$\overline{\mathcal{B}}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ denote the
homotopy classes of maps in $\mathcal{B}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$.
Next, consider the equivalence relation on
$\overline{\mathcal{B}}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ generated by
$[f]\sim[f^{\prime}]$ if and only if there exists $f\in[f]$,
$f^{\prime}\in[f^{\prime}]$ and $k$ such that $k\circ f=f^{\prime}$, where
$k:\vee^{g}S^{1}\rightarrow\vee^{g}S^{1}$ is a homeomorphism. Finally, let
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}:=\overline{\mathcal{B}}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}/\sim$.

Notice that $g$ has the effect of permuting the circles of $\vee^{g}S^{1}$ or
reversing the orientation of some of the circles of $\vee^{g}S^{1}$.
Therefore, elements in the quotient, $B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$
broadly keep track of which part of $\Sigma_{g}$ wrap around a circle of
$\vee^{g}S^{1}$ without recording which circle is which, or the direction in
which the surface wraps around it. It turns out that this information is
equivalent to a Heegaard diagram.

######  Theorem 3.5.

There is a bijection $\mathscr{S}:\\{\text{Heegaard diagrams for
}M=\mathcal{H}_{1}\cup\mathcal{H}_{2}\\}\rightarrow
B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$. Any map that is a representative of
$\mathscr{S}(x)$ is called a Stallings map for the Heegaard diagram $x$.

###### Proof.

We define a bijection
$\mathscr{S}^{-1}:B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}\rightarrow\\{\text{Heegaard
diagrams for }M=\mathcal{H}_{1}\cup\mathcal{H}_{2}\\}$ as follows.

##### Defining $\mathscr{S}^{-1}$

Begin with $[f]\in B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ and pick any
representative $f\in[f]$. Let $f_{1}$ and $f_{2}$ be the two coordinates of
$f$ (in other words, $f=f_{1}\times f_{2}$). By Proposition 3.3, one can think
of the first coordinate $f_{1}$, as inclusion of $\Sigma_{g}$ into a handle
body $\mathcal{H}_{1}$ as its boundary, followed by a homotopy equivalence
$f_{1}^{\prime}:\mathcal{H}_{1}\rightarrow\vee^{g}S^{1}$. Let
$h:\vee^{g}S^{1}\rightarrow\mathcal{H}_{1}$ be the other part of this homotopy
equivalence. There are many Heegaard diagrams for the attachment of
$\mathcal{H}_{1}$ to $\Sigma_{g}$ by this inclusion. However, $[f]$ determines
one of them canonically using the following idea, and this canonical diagram
is taken to be $\mathscr{S}^{-1}([f])$.

Let $b\in\vee^{g}S^{1}$ be the bouquet point. Pick distinct points
$s_{1},\ldots,s_{g}$ in $(\vee^{g}S^{1})$, such that $s_{i}\neq b$, and there
is one point on each circle. Pick generators $\\{\sigma_{i}\\}$ for
$\pi_{1}(\vee^{g}S^{1})$ such that the circle that contains $s_{i}$ is a
representative for $\sigma_{i}$. The idea is that the points $\\{s_{i}\\}$
determine disks in $\mathcal{H}_{1}$, whose boundaries form the desired
Heegaard diagram. The details of this process are laid out below.

Choose smooth structures on $\mathcal{H}_{1}$ and on the components of
$\vee^{g}S^{1}\setminus\\{b\\}$. Then modify $f_{1}$ by a homotopy so that it
is smooth with respect to these structures (using Whitney’s approximation
theorem) and so that it is transverse to the points $s_{i}$ (using Thom
transversality). Now $\\{f_{1}^{-1}(s_{i})\\}$ is a collection of simple
closed curves. Label each curve in $f_{1}^{-1}(s_{i})$ by $\sigma_{i}$. This
way, one can associate an intersection word in the letters $\\{\sigma_{i}\\}$
for curves on $\Sigma_{g}$.

Suppose there is an $i$ such that $f_{1}^{-1}(s_{i})$ consists of at least two
simple closed curves $\gamma$ and $\delta$. If there is an oriented simple
curve arc $\alpha$ in $\Sigma_{g}\setminus\\{f_{1}^{-1}(s_{i})\\}$
intersecting $\gamma$ and $\delta$ whose intersection word is
$\sigma_{i}\sigma_{i}^{-1}$, then a neighbourhood of
$\alpha\cup\gamma\cup\delta$ is a pair of pants $\mathcal{P}$. One boundary
component of the pants is homotopic to $\gamma$, one is homotopic to $\delta$,
and the third is homotopic to a concatenation of the form
$\gamma\cdot\alpha\cdot\delta\cdot\alpha^{-1}$ (with appropriate choices of
orientation for $\gamma$ and $\delta$). In this case, $f_{1}$ can be modified
by a homotopy supported on $\mathcal{P}$ so that $f_{1}^{-1}(\sigma_{i})$ is a
single curve tracing $\gamma\cdot\alpha\cdot\delta\cdot\alpha^{-1}$. Each such
modification reduces the number of curves in $\\{f_{1}^{-1}(s_{i})\\}$, and so
after a point, there are no more arcs travelling between two curves in
$\Sigma_{g}$ with intersection word $\sigma_{i}\sigma_{i}^{-1}$ for all $i$.
Next, modify $f_{1}$ by another homotopy to remove any null homotopic curves
in $\\{f_{1}^{-1}(s_{i})\\}$ (this is possible because $\vee^{g}S^{1}$ is
aspherical). Both these homotopies can be realized by modifying $f_{1}$ by a
homotopy to that it is locally a Morse function, and then pushing parts of the
surface above or below the height value $s_{i}$.

$s_{i}$$s_{i}$ Figure 3: The coordinates of the Stallings map locally as a
Morse function.

After these modifications, $\\{f_{1}^{-1}(s_{i})\\}$ is a set of disjoint
essential simple closed curves, and there are no cancelling homotopic pairs.
This means that these curves fall into at most $3g-3$ homotopy classes.

In the proof of Proposition 3.3, we see that $f_{1}$ factors through a map
$d:\mathcal{H}_{1}\rightarrow\vee^{g}S^{1}$, and $d$ is part of a homotopy
equivalence, together with a map $h:\vee^{g}S^{1}\rightarrow\mathcal{H}_{1}$.
This implies that the curves $\\{f_{1}^{-1}(s_{i})\\}$ are null homotopic when
included into $\mathcal{H}_{1}$. Therefore they bound (immersed) disks in
$\mathcal{H}_{1}$. In fact, it is possible to find embedded disks that these
curves bound. This will be proved in Proposition 4.4 and does not depend on
any additional hypotheses.

Consider $h(\vee^{g}S^{1})$. Each circle must be sent to a curve in
$\mathcal{H}_{1}$ that intersects these disks in some pattern. Modifying $h$
by a homotopy so that these intersections are transverse, ${d}_{*}\circ
h_{*}(\sigma_{i})$ can be read off as the intersection word of the $h$-image
of the $i^{\text{th}}$ circle with the disks. Since $d$ and $h$ form a
homotopy equivalence, ${d}_{*}\circ h_{*}(\sigma_{i})=\sigma_{i}$. So, the
$i^{\text{th}}$ circle must have intersection word that reduces to
$\sigma_{i}$ in the free group $\pi_{1}(\vee^{g}S^{1})$. However, since it is
not possible for an arc to run between two distinct curves with intersection
subword $\sigma_{i}\sigma_{i}^{-1}$, the only possibility is that $h$ can be
modified by a homotopy so that each circle in $h(\vee^{g}S^{1})$ intersects
the corresponding disk exactly once. This implies that $f^{-1}(s_{i})$
contains a non-separating curve. If $f^{-1}(s_{i})$ contained a second curve
as well for some $i$, then there would be an arc $\gamma$ intersecting these
two distinct curves with intersection word $\sigma_{i}\sigma_{i}^{-1}$. This
is a contradiction. Therefore $f^{-1}(s_{i})$ consists of a single non-
separating curve.

Thus, there are exactly $g$ disjoint non-separating simple closed curves in
$\\{f_{1}^{-1}(s_{i})\\}$. Because these $g$ curves are null homotopic when
included into $\mathcal{H}_{1}$, are homologically independent, and we know
${f_{1}}_{*}$ is $\pi_{1}$-surjective; from Proposition 3.3, we know they must
form a Heegaard diagram for the attachment of $\mathcal{H}_{1}$ to
$\Sigma_{g}$.

To show that this half Heegaard diagram is unique, consider that if there was
another half Heegaard diagram obtained in the way above, the curves of this
diagram would be the preimages of $s_{1},\ldots s_{g}$ by a map
$\tilde{f}_{1}$ that is homotopic to $f_{1}$. Label the curves of this second
diagram $\sigma_{1},\ldots\sigma_{g}$. Since the two diagrams are not the
same, there must be an $i$ such that the $\sigma_{i}$ two curves are not
homotopic. Consider the homotopy between $f_{1}$ and $\tilde{f}_{1}$. The
preimage of $s_{i}$ by this homotopy sweeps out a surface immersed in
$\Sigma_{g}$, bounded by the two $\sigma_{i}$ curves. This implies that the
two $\sigma_{i}$ curves are homologous. Two non-separating curves are
homologous if and only if they are homotopic. This is a contradiction.
Therefore the half Heegaard diagram obtained above is unique.

In the same way, the second half of the Heegaard diagram is uniquely
determined by $f_{2}$. We let this Heegaard diagram be
$\mathscr{S}^{-1}([f])$.

##### Checking $\mathscr{S}^{-1}$ is surjective

Given a Heegaard diagram for $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$, a map
$f:\Sigma_{g}\rightarrow\vee^{g}S^{1}\times\vee^{g}S^{1}$ that is sent to this
Heegaard diagram by $\mathscr{S}^{-1}$ can be constructed. To construct the
first coordinate $f_{1}$, attach disks and a $3$-ball to $\Sigma_{g}$
according to the blue Heegaard diagram curves to obtain $\mathcal{H}_{1}$.
Then take a smooth structure on $\mathcal{H}_{1}$, and choose a tubular
neighbourhood of the disks (each neighbourhood homeomorphic to
$\mathbb{D}\times(-1,1)$ with $\mathbb{D}\times\\{0\\}$ being the disk around
which the neighbourhood is chosen).

Let $f_{1}$ send each disk to a distinct point in $\\{s_{1},s_{2},\ldots
s_{g}\\}$. The tubular neighbourhood around each disk is mapped to the
interval of $\vee^{g}S^{1}\setminus\\{b\\}$ containing the corresponding
$s_{i}$ by forgetting the first coordinate of $\mathbb{D}\times(-1,1)$, and
applying a homeomorphism from $(-1,1)$ to the desired interval. The complement
of the tubular neighbourhoods is mapped to the bouquet point $b$. The second
coordinate, $f_{2}$ is constructed similarly.

Note that $\\{f_{j}^{-1}(s_{i})\\}_{i,j}$ are the diagram curves and
$\mathscr{S}^{-1}([f])$ is the Heegaard diagram we began with. the second
coordinate $f_{2}$ is constructed similarly.

##### Checking $\mathscr{S}^{-1}$ is injective

Suppose $[f]$ and $[f^{\prime}]$ are in
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ and that
$\mathscr{S}^{-1}([f])=\mathscr{S}^{-1}([f^{\prime}])$. Let $f\in[f]$ and
$f^{\prime}\in[f^{\prime}]$. Then after modifying $f$ and $f^{\prime}$ by a
homotopy if necessary, $\\{(f^{\prime})^{-1}(s_{i})\\}_{i}$ is a set of $g$
simple closed curves that homotopic to the $g$ curves
$\\{f^{-1}(s_{i})\\}_{i}$ for all $i$. After possibly composing $f$ with a
homeomorphism $k:\vee^{g}S^{1}\rightarrow\vee^{g}S^{1}$, we may assume that
for each $i$, the curve $(f^{\prime})^{-1}(s_{i})$ is homotopic to
$f^{-1}(s_{i})$. Modifying $f$ further by homotopy, we may assume
$f^{-1}(s_{i})=(f^{\prime})^{-1}(s_{i})$ for all $i$.

Since $\vee^{g}S^{1}\setminus\cup_{i}\\{s_{i}\\}$ is contractible, $f$ can be
modified by a homotopy that is supported on the complement of
$\cup_{i}f^{-1}(s_{i})$ so that the resulting map sends every point outside a
tubular neighbourhood of $\cup_{i}f^{-1}(s_{i})$ to the bouquet point $b$. The
components of the tubular neighbourhood of $\cup_{i}f^{-1}(s_{i})$ are each
homeomorphic to $S^{1}\times(-1,1)$, and the modified $f$ sends the tubular
neighbourhood of $f^{-1}(s_{i})$ to the maximal interval in $\vee^{g}S^{1}$
containing $s_{i}$. The map $f^{\prime}$ can be modified by a homotopy to have
the same properties.

The modified maps $f$ and $f^{\prime}$ now agree on the complement of a
tubular neighbourhood of $\cup_{i}f^{-1}(s_{i})$. It remains to show that they
agree up to homotopy on each component of the tubular neighbourhood of
$\cup_{i}f^{-1}(s_{i})$. This amounts to showing that the path
$\\{\pi\\}\times(-1,1)$ in the tubular neighbourhood of $f^{-1}(s_{i})$ is
mapped by $f$ and $f^{\prime}$ to wrap around the $i^{\text{th}}$ circle in
$\vee^{g}S^{1}$ in the same direction. However, this can be guaranteed by an
appropriate choice of $k$ earlier. Therefore $[f]=[f^{\prime}]$. ∎

#### Handle slides and Stallings maps

Given any two Heegaard diagrams for $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$,
there is a finite sequence of handle slides that can be applied to one to
obtain the other. Now that we know that The set of Stallings maps
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ is in bijection with the set of
Heegaard diagrams for $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$, one might wonder
about the relationship between these Stallings maps and how it relates to the
notion of handle slide.

The group $Out(F_{g})\times Out(F_{g})$ acts on
$\mathcal{B}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ by the action $(a,b)\cdot
f_{1}\times f_{2}=a\circ f_{1}\times b\circ f_{2}$ where $a$ and $b$ denote
elements in $Out(F_{g})$ as well as maps
$\vee^{g}S^{1}\rightarrow\vee^{g}S^{1}$ inducing these elements on the
fundamental group (these maps must be part of homotopy equivalences). Note
that up to homotopy, there is only one choice of the maps $a$ and $b$. One
choice is made for each element of $Out(F_{g})$ so that this action is well
defined.

Note that any map that is part of a homotopy equivalence of $\vee^{g}S^{1}$ to
itself determines an element of $Out(F_{g})$ by its action on the fundamental
group. Recall that any two maps in
$\mathcal{B}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ differ by a homotopy
equivalence of each of the two coordinates of
$\vee^{g}S^{1}\times\vee^{g}S^{1}$. Therefore, given
$f,f^{\prime}\in\mathcal{B}_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$, there is an
element $(a,b)$ of $Out(F_{g})\times Out(F_{g})$ such that $(a,b)\cdot f$ is
homotopic to $f^{\prime}$. This induces a transitive action on the quotient
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$. The action is well defined because
given $a\in Out(F_{g})$, for every homeomorphism
$k:\vee^{g}S^{1}\rightarrow\vee^{g}S^{1}$, there is another homeomorphism
$j:\vee^{g}S^{1}\rightarrow\vee^{g}S^{1}$ such that the map $a\circ f_{1}$ is
homotopic to $j\circ(a\circ k\circ f_{1})$. Namely, $j$ can be chosen to be a
homeomorphism in the homotopy class of $a\circ k^{-1}\circ a^{-1}$ (recall $a$
is part of a homotopy equivalence, so there must be a homeomorphism in this
class).

Since $B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$ is in bijection with the set of
Heegaard diagrams for $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ (Theorem 3.5),
the action of $Out(F_{g})\times Out(F_{g})$ must translate into an action by
handle slides. Let $K^{\prime}$ be the normal closure of the elements of
$Out(F_{g})$ that permute the generators or replace a generator with its
inverse. Then let $K_{1}\leq Out(F_{g})\times Out(F_{g})$ be the subgroup
$K^{\prime}\times\mathds{1}$ and let $K_{2}\leq Out(F_{g})\times Out(F_{g})$
be the subgroup $\mathds{1}\times K^{\prime}$. Then $K:=K_{1}\cap K_{2}$ is
the kernel of the action of $Out(F_{g})\times Out(F_{g})$ on
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$.

######  Proposition 3.6.

The group $Out(F_{g})\times Out(F_{g})/K$ is isomorphic to $H\times H$ where
$H$ is the handle slide group acting on a handle body of genus g. The actions
of $Out(F_{g})\times Out(F_{g})/K$ on $B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$
and $H\times H$ on the set of Heegaard diagrams of a Heegaard splitting of
genus $g$ are equivariant with respect to the bijection $\mathscr{S}$ via this
isomorphism.

###### Proof.

The handle slide group $H$ is generated by individual handle slides. Beginning
with a generator $(a,b)$ of $Out(F_{g})\times Out(F_{g})$, we would like to
show that the map $\mathscr{S}([f])\mapsto\mathscr{S}((a,b)\cdot[f])$ acts on
the set of Heegaard diagrams in one of two ways on each set of $g$ curves. It
either as the identity in $H$, or it acts as a single handle slide. We will
see that this defines a homomorphism $Out(F_{g})\times Out(F_{g})\rightarrow
H\times H$, and show that its kernel is $K$. As with most proofs in this
paper, we prove this statement for one coordinate, and the second coordinate
follows similarly.

Let $j$ be an automorphism of the free group $F_{g}$. Then $j$ induces a
homotopy equivalence of the bouquet of $g$ circles to itself. In the case
where $j$ is an inner automorphism, the homotopy equivalence is homotopic to
identity. If not, $j$ represents an element of $Out(F_{g})$. The group
$Out(F_{g})$ is generated by automorphisms that send a single generator to its
inverse, and act on the other generators as identity, as well as automorphisms
that send a generator to the product of two generators and act on the other
generators as identity. The first type fixes every element of
$B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$, and therefore the Heegaard diagram
obtained is unchanged as well. If $j$ is of the second type, its action on
$f_{1}$ can be realized gradually using a path $p(t)$ in the outer space
$\mathcal{O}_{g}$ such that $p(0)\circ f_{1}=f_{1}$ and $p(1)\circ
f_{1}=j\circ f_{1}$. Intuitively, $p(t)$ folds one circle around another as
shown in the comic strip in Figure 4.

Figure 4: A generator of $Out(F_{g})$ is shown inducing a handle slide

At each moment in time $t$, Let $Y(t)$ be the target space of $p(t)\circ
f_{1}$. Choose points $\\{s_{i}\\}$ one at the midpoint of each edge of
$Y(t)$. Take the preimage of these points and simplify by a homotopy as in the
proof of Proposition 3.5, so that there is a single curve in the preimage of
each $s_{i}$. When $t\neq 0$ or $1$, one connected component $P$ of
$Y(t)\setminus\cup_{i}\\{s_{i}\\}$ contains a single trivalent vertex. The
preimage of $P$ must be a subset of $\Sigma_{g}$ bounded by three curves in
the preimage of $\\{s_{i}\\}$. This surface cannot have positive genus because
$p(t)$ is a homotopy equivalence. If the surface had positive genus, there
would be two intersecting degree one homology classes that have trivial image
by $p(t)\circ f_{1}$. This contradicts Property (1) of Stallings maps.
Therefore, $P$ must be a pair of pants. When $t=0$ or $1$, the preimage of
$\\{s_{i}\\}$ are Heegaard diagrams (as in Proposition 3.5) and one of the
cuffs of the pair of pants $P$ disappears. This is exactly the combinatorial
move known as a handle slide.

This defines a homomorphism $\varphi:Out(F_{g})\rightarrow H$. A composition
of generators of $Out(F_{g})$ may act by permuting the generators. These
compositions must have trivial image by $\varphi$ since the set of midpoints
of edges is unchanged by a permutation of the copies of $S^{1}$. Thus the
kernel of $\varphi$ is $K^{\prime}$. Then $\varphi\times\varphi$ induces an
isomorphism between the quotient $Out(F_{g})\times Out(F_{g})/K$ and $H\times
H$, as desired. ∎

###  3.4 A picture of Stallings maps

We would like to study the nature of these maps. It is natural to ask whether
they can be homeomorphisms (thinking of these as the nicest possible map
between two-dimensional spaces). We will see in the final section about Lens
spaces that there is one Heegaard splitting that can be encoded as a
homeomorphism between tori (the genus one Heegard splitting of $S^{3}$).
However, outside the genus one case, this is not possible, since the target
space is no longer a manifold.

It is valuable to spend a moment understanding the topology of a cartesian
product of two bouquets of $g$ circles. There are three types of points:
manifold points (which have neighbourhoods homeomorphic to discs), book-like
points (which have neighbourhoods homeomorphic to several disks intersecting
in a curve), and the singular point (which is the product of the two wedge
points). All together this forms a non-positively curved square complex with
one vertex, $2g$ edges and $g^{2}$ squares. It is comprised of many one-square
tori that intersect pairwise in an edge or a vertex. The link of the vertex is
homeomorphic to the complete bipartite graph $K_{2g,2g}$. Note that this graph
is not planar, and so $(\vee^{g}S^{1})\times(\vee^{g}S^{1})$ cannot be
embedded in $\mathbb{R}^{3}$.

Figure 5: topology of product neighbourhoods of $G_{1}\times G_{2}$. The
bottom left neighbourhood cannot be embedded in $\mathbb{R}^{3}$, due to
additional square “tiles” that are not shown here. The ellipses indicate that
these are missing.

##  4 Irreducibility, immersions and square complexes

We will now focus on the case where the Heegaard splitting is irreducible.
Here, is it possible to pull our maps tight in some sense, removing folding
and crumpling, and this gives an even more concrete picture of the maps,
allowing understanding of the reducible case through decomposition by connect
sum. Some set up is needed to formalize and prove this result. We begin with
some definitions

######  Definition 4.1.

Consider an embedded curve $\gamma$ on $\Sigma_{g}$. Take an open tubular
neighbourhood $T$ of $\gamma$ of width small enough that $T$ is homeomorphic
to an open cylinder. Consider a tubular neighbourhood $U$ of $\gamma$ that is
contained in $T$. Removing the image of $\gamma$ from $U$ leaves two connected
components. One of these components will be called $A_{U}$ and the other will
be called $B_{U}$. Now we define an equivalence relation on the set of tubular
neighbourhoods of $\gamma$ contained in $T$. Any two components of (possibly
distinct) deleted tubular neighbourhoods are equivalent if and only if their
intersection is non-empty. Notice that there are two equivalence classes and
$A_{U}$ and $B_{U}$ are in different classes for any given $U$. Call these two
classes the two sides of $\gamma$.

A set intersects a side of  $\gamma$ if it intersects all the sets in this
equivalence class. Consider an arbitrary constant curvature metric on
$\Sigma_{g}$. Let $U_{\epsilon}$ be the tubular neighbourhood of $\gamma$ of
width $\epsilon$ (with $\epsilon$ small enough so that $U_{\epsilon}$ is
contained in $T$). Then the number of connected components of the intersection
of a set $X$ with a side of $\gamma$ is defined as the limit of the number of
connected components of $X\cap A_{U_{\epsilon}}$ as $\epsilon\rightarrow 0$
(respectively $B_{U_{\epsilon}}$, depending on which side).

A wave move is a combinatorial change that one can make to a Heegaard diagram
without changing the $3$-manifold it describes. The proof of this is
originally written by Volodin, Kuznetzov and Fomenko in 1974 [19], but is
explained below.

######  Definition 4.2.

Given a half Heegaard diagram on the surface $\Sigma_{g}$, a wave move
consists of the following steps.

  1. 1.

Pick a simple arc $\gamma$ that begins and ends on a diagram curve $\delta$,
does not intersect any other diagram curves, and intersects only one side of
$\delta$ in two components;

  2. 2.

Thicken the union of $\delta$ and $\gamma$ in $\Sigma_{g}$ to obtain a pair of
pants (see Figure 6);

  3. 3.

Replace $\delta$ with another cuff of this pair of pants (if it creates a new
Heegaard diagram).

Figure 6: The diagram curve $\delta$ is shown in red and the arc $\gamma$ is
shown in yellow. Representatives of the free homotopy classes $[\alpha]$ and
$[\beta]$, shown in two ways: (i) as a union of segments of $\delta$ and
$\gamma$, and (ii) as the cuffs of a pair of pants constructed by $\delta$ and
$\gamma$.

Call the two curves forming the other cuffs of the pair of pants $\alpha$ and
$\beta$. We shall see in the proof of the next theorem that either replacing
$\delta$ with $\alpha$ or $\delta$ with $\beta$ always results in another
Heegaard diagram. Further, we shall see that the new Heegaard diagram
describes the same Heegaard decomposition.

######  Theorem 4.3 (Volodin, Kuznetzov, Fomenko [19]).

Applying a wave move to a half Heegaard diagram does not change the attachment
of a handle body described.

###### Proof.

To show that one of these swaps must give a new Heegaard diagram for the same
Heegaard decomposition, two cases must be explored combinatorially.

##### Case 1: $\alpha$ or $\beta$ is isotopic to a diagram curve.

If $\alpha$ is isotopic to another curve in the original half Heegaard
diagram, then $\beta$ must not be, since $\delta$ must be homologically
independent from the other curves. Then, by a single handle slide, $\delta$
can be replaced by $\beta$ to give a new Heegaard diagram, describing the same
attachment of a handlebody to $\Sigma_{g}$. Similarly, if $\beta$ is isotopic
to a curve in the original half Heegaard diagram, then $\delta$ can be
replaced with $\alpha$.

##### Case 2: Neither $\alpha$ nor $\beta$ is isotopic to a diagram curve

Assume neither $\alpha$ nor $\beta$ is isotopic to a curve in the original
half Heegaard diagram. In this case, cut $\Sigma_{g}$ along all the diagram
curves, and label them to recall the identifications needed to recover
$\Sigma_{g}$. This is a sphere with $2g$ discs removed. Note that, by the way
it is defined, $\gamma$ does not intersect any diagram curves. Therefore, the
graph given by the union of $\delta$ and the $\gamma$ can be embedded in the
sphere with $2g$ discs removed. By thickening this graph in this space, the
pair of pants bounded by $\delta$, $\alpha$, and $\beta$ are embedded as well.

Figure 7: The embedding of the pair of pants introduced in Figure 6.

Either $\alpha$ or $\beta$ must separate one copy of $\delta$ from the other.
Without loss of generality let it be $\beta$. The aim is to show that $\delta$
can be replaced by $\beta$ to give a new Heegaard diagram associated to the
same attachment of a handle body.

This will be shown by constructing a sequence of embedded pairs of pants that
dictate a sequence of handle slides that ultimately make the desired
replacement.

Cut along $\beta$. This produces two connected components. Pick either one of
these to work with. Call this space $S$. We will embed our pairs of pants that
dictate handle slides in this space. Identify as many of the diagram curves as
possible to each other. Each identification creates a handle with a diagram
curve wrapped around it.

Figure 8: $S$ is shown before and after identifications.

In general $S$ is now a surface of genus $p$ with $n+2$ boundary components
($n$ boundary components besides $\delta$ and $\beta$). Further, $S$ has a
diagram curve around each handle, and isotopic to each of the boundary
components except for $\beta$. Using the classification of surfaces, the
surface $S$ and the diagram curves are homeomorphic to the surface and curves
shown in Figure 9. Consider the following pair of pants decomposition by
curves: $\delta,\gamma_{1}\ldots,\gamma_{2p},\ldots,\gamma_{2p+n}=\beta$.

Figure 9: A pair of pants decomposition of $S$.

By a handle slide, the problematic curve can be replaced by $\gamma_{1}$ to
give a new Heegaard diagram associated to the same Heegaard decomposition. A
second handle slide allows $\gamma_{1}$ to be replaced by $\gamma_{2}$. Next
$\gamma_{2}$ is replaced with $\gamma_{3}$, and so on until
$\gamma_{2p+(n-1)}$ is replaced with $\beta$, as desired.

∎

The author rediscovered the wave move for herself during her graduate studies,
and found that it is incredibly useful! It is the key to almost every proof in
this paper.

###  4.1 Irreducibility

Recall that Heegaard splitting is _irreducible_ if no essential simple closed
curve on the splitting surface bounds a disk in $\mathcal{H}_{1}$ and in
$\mathcal{H}_{2}$.

######  Proposition 4.4.

Let $\mathcal{H}$ be a handle body and let $\Sigma$ be the surface that is its
boundary. Let $\gamma$ be an essential simple closed curve on $\Sigma$, that
is nullhomotopic in $\mathcal{H}$. Then $\gamma$ bounds an embedded disk in
$\mathcal{H}$.

###### Proof.

Choose half a Heegaard diagram, that describes the attachment of $\mathcal{H}$
to $\Sigma$. Since $\gamma$ is nullhomotopic in $\mathcal{H}$, the cyclic word
produced by its intersection with the diagram curves must reduce to identity
in the free group on $g$ generators. Thus, the arcs of $\gamma$ that run
between diagram curves dictate a number of wave moves that change the diagram
so that it does not intersect $\gamma$ at all, while still describing the
attachment of $\mathcal{H}$ to $\Sigma$.

Cut $\mathcal{H}$ along the embedded disks bounded by diagram curves. Note
that $\gamma$ is not cut since it does not intersect any diagram curves. The
remaining space is homotopic to a closed $3$-ball with $2g$ disks removed on
the boundary. $\gamma$ is an embedded curve on the boundary of this $3$-ball.
It is clear that $\gamma$ bounds an embedded disk in this space. Identifying
pairs of disks to recover $\mathcal{H}$, we see that $\gamma$ bounds an
embedded disk in $\mathcal{H}$. ∎

######  Remark 4.5.

Proposition 4.4 shows that irreducibility of a Heegaard splitting is
equivalent to the following: no essential simple closed curve in $\Sigma$ is
nullhomotopic in both $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$. This property
is much more easily verified than the definition, and will be used extensively
in what follows.

######  Definition 4.6.

Let $S$ be a closed orientable surface and let $G_{1}$, $G_{2}$ be graphs. A
Stallings map $f:S\longrightarrow G_{1}\times G_{2}$ is said to be irreducible
if it encodes an irreducible Heegaard decomposition.

######  Remark 4.7.

Proposition 4.4 implies that if $f$ is an irreducible map, then
$\text{ker}f_{*}$ has no non-trivial element with a representative that is a
simple closed curve.

######  Notation 1.

The traditional picture of a Heegaard diagram is of red and blue curves.
Following this tradition, we will refer to the two sets of curves as though
one set is red and the other is blue.

######  Proposition 4.8.

Let $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ be an irreducible Heegaard
decomposition. The complement of the curves of any associated Heegaard diagram
in $\Sigma_{g}$ is homeomorphic to a finite union of discs. In other words,
the Heegaard diagram fills the surface.

###### Proof.

Consider the handle body $\mathcal{H}_{i}$ and the $g$ curves of a Heegaard
diagram that encode its attachment to $\Sigma_{g}$. Notice that the complement
of these curves in $\Sigma_{g}$ is sent to a null-homotopic subset of
$\mathcal{H}_{i}$ by $\iota_{i}$.

Now suppose the proposition were false. That is, suppose that there was a
Heegaard diagram for $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ that has a
complement with a connected component that is not a disk. This implies that
there is an essential simple closed curve in the complement of the diagram.
Since it does not intersect any blue curves, it must have null-homotopic image
by $\iota_{1}$ and since it does not intersect any red curves, it must have
null-homotopic images by $\iota_{2}$. This contradicts the irreducibility of
of $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$. ∎

######  Notation 2.

The discs described above can be thought of as _polygons_ with alternating red
and blue edges made from segments of Heegaard diagram curves.

###  4.2 Tautness

######  Definition 4.9.

A Heegaard diagram is taut if every connected component of the complement
intersects a side of a diagram curve in at most one connected component.

A region of $\Sigma_{g}$ is shown in Figure 10, with two complementary
polygonal components of two different Heegaard diagrams. The first picture is
not taut. The open polygon intersects the same side of a curve in two
connected components. The second picture is taut. Here, while the open polygon
does intersect the sides of a curve in two connected components, they are the
opposite sides.

Figure 10: Two regions of Heegaard diagrams. The left is not taut, while the
right is taut.

######  Theorem 4.10.

Every Heegaard decomposition has at least one associated taut Heegaard
diagram.

###### Proof.

The proof is constructive, beginning with any Heegaard diagram and modifying
it until a taut Heegaard diagram is achieved.

If this Heegaard diagram is already taut, we are finished.

If not, there exists a connected component of the complement that intersects
some diagram curve twice on the same side. Such incidences will be called
problems. Notice that multiple problems may occur on a single complementary
component, or along a single curve.

Given a problem, without loss of generality, let the problematic curve be red
and call it $\delta$. Consider a curve segment $\gamma$ with one end point on
each of the two segments shared by the problematic curve and problematic
component, and with interior inside the problematic component. Such a segment
is shown in Figure 11 for the problem first seen in Figure 10.

Figure 11: An orange segment that intersects the same side of a diagram curve
in two connected components.

Thicken the union of $\delta$ and $\gamma$ in $\Sigma_{g}$ to obtain a pair a
pants. One cuff of the pair of pants is isotopic to $\delta$, call the other
two cuffs $\alpha$ and $\beta$. Replace the problematic curve $\delta$ with
$\alpha$ or $\beta$ to produce a new Heegaard diagram. This Heegaard diagram
describes the same Heegaard splitting by Theorem 4.3, since we have just
applied a wave move.

This swap strictly reduces the number of intersections between red and blue
curves.

Figure 12 includes the minimal number of blue arcs possible so that the
complementary region containing $\gamma$ intersects a side of $\delta$ in two
connected components . The elimination of an intersection point is seen in
either case.

$\delta$$\alpha$$\beta$$\gamma$ Figure 12: Applying wave move to the red half
of the Heegaard diagram, seen in the pair of pants. Figure 13: A neighbourhood
of the problematic curve $\delta$ before and after the wave move, this time
with a few more blue arcs.

Repeat this process as long as there is a problem, each time reducing the
intersections between red and blue curves by at least one. The number of
intersections is finite and bounded below by zero. Therefore, this process
must end with a taut configuration. ∎

Applying Theorem 4.10 to irreducible Heegaard splittings, where every Heegaard
diagram is filling (Proposition 4.8) gives the following result immediately.

######  Theorem 4.11.

Every irreducible Heegaard splitting has a taut and filling Heegaard diagram.

###  4.3 Immersions

The goal of this section is to prove the following theorem.

######  Theorem 4.12.

There is a locally injective map
$f:\Sigma_{g}\longrightarrow(\vee^{g}S^{1})\times(\vee^{g}S^{1})$ encoding
every irreducible Heegaard decomposition of genus greater than zero.

This theorem is a corollary of the following lemma, which we will prove in
detail.

######  Lemma 4.13.

The Stallings map corresponding to a taut and filling Heegaard diagram without
bigons is locally injective.

Taking Lemma 4.13 to be true, all Heegaard diagrams of irreducible Heegaard
splittings are filling by Proposition 4.8, and we have seen in Theorem 4.11
that there is a taut diagram for each irreducible decomposition. Therefore
Theorem 4.12 follows.

To prove Lemma 4.13, consider a taut, filling Heegaard diagram without bigons
for some Heegaard decomposition $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$. Let
the curves describing the attachment of $\mathcal{H}_{1}$ to $\Sigma_{g}$ be
blue. Similarly, let the curves describing the attachment of $\mathcal{H}_{2}$
to $\Sigma_{g}$ be red.

This diagram will be used to give a particular map
$f:\Sigma_{g}\longrightarrow(\vee^{g}S^{1})\times(\vee^{g}S^{1})$. We will see
that $f$ is locally injective and then the proof of Proposition 4.13, will be
reduced to checking that $f$ describes $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$.

##### Defining $f$:

In the target, $(\vee^{g}S^{1})\times(\vee^{g}S^{1})$, consider each
coordinate $(\vee^{g}S^{1})$. For simplicity (and foreshadowing), call one the
blue coordinate and the other the red coordinate. After removing the wedge
point of $\vee^{g}S^{1}$, $g$ open intervals remain. Orient these intervals
arbitrarily. In the blue copy of $\vee^{g}S^{1}$, Pick $g$ points, one at the
midpoint of each of these intervals. Call these blue points. Pick a bijection
between the set of blue points and the set of blue diagram curves. Repeat the
process on the red coordinate, and call the chosen points red points. Now
orient the diagram curves on $\Sigma_{g}$ arbitrarily.

Figure 14: A region of $\Sigma_{g}$ is shown with orientations chosen on the
Heegaard diagram curves. The two coordinates of
$(\vee^{g}S^{1})\times(\vee^{g}S^{1})$ are shown with blue and red points
chosen as well as orientations on the edges.

Recalling the definition of “side,” there is now a right side (and left side)
of each curve using these orientations. In the bouquet of circles, there is
also a right side of each of the points in the intervals picked out in the
preceding paragraph.

Due to Proposition 4.8, this Heegaard diagram induces a cell decomposition
structure on $\Sigma_{g}$. Every vertex in this cell decomposition is
quadrivalent. Consider the dual cell decomposition. Every face in the dual
decomposition must have four edges (it is a cell decomposition of squares). We
metrize $\Sigma_{g}$ so that these dual $2$-cells are squares with the diagram
curves intersecting the midpoint of the edges of each square. Each square has
one blue curve and one red curve running through it, cutting it precisely into
quarters. The two diagram curves that run through a given square are
associated to two points: one in the blue coordinate $\vee^{g}S^{1}$ and one
in the red coordinate $\vee^{g}S^{1}$. We would like to construct a map
$f:\Sigma_{g}\longrightarrow(\vee^{g}S^{1})\times(\vee^{g}S^{1})$. The idea is
that $f$ sends the interior of each square homeomorphically to the product of
maximal intervals which contain these two points. The interior of the squares
are mapped by $f$ such that the right side of the red curve is sent to the
right side of the red point and the right side of the blue curve is sent to
the right side of the blue point.

Figure 15: The definition of $f$ on the interior of a square.

Now we say precisely how to set this up. Consider $f$ as two coordinates
$f=f_{1}\times f_{2}$, where $f_{i}:\Sigma_{g}\longrightarrow\vee^{g}S^{1}$.
Each of the $1$-cells in the square cell decomposition of $\Sigma_{g}$
intersects exactly one diagram curve in one point. If this diagram curve is
blue, let $f_{1}$ send the open $1$-cell to the maximal interval subset of
$\vee^{g}S^{1}$ that contains the blue point assigned to the diagram curve.
This should be done in such a way so that the half interval that intersects
the right side of the diagram curve is sent to the right of the blue point.
Let the boundary of this $1$-cell be sent to the wedge point. If the diagram
curve is red, let $f_{2}$ send the open $1$-cell to the maximal interval
subset of $\vee^{g}S^{1}$ that contains the red point assigned to the diagram
curve, with the boundary of the $1$-cell being sent to the wedge point. Now
consider a closed $2$-cell in the square cell decomposition of $\Sigma_{g}$.
$f_{1}$ and $f_{2}$ are defined on alternating boundary edges of this cell. To
define $f_{1}$, we consider the square as $[0,1]\times[0,1]$, where $f_{1}$ is
defined on $[0,1]\times\\{0\\}$ and on $[0,1]\times\\{1\\}$. The rest of the
square is sent by $f_{1}$ to a straight-line homotopy between the
definition(s) of $f_{1}$ on $[0,1]\times\\{0\\}$ and $[0,1]\times\\{1\\}$. The
map $f_{2}$ is defined on the closed $2$-cell similarly. The map $f$ is
defined this way on all of $\Sigma_{g}$ and it is by definition continuous. It
is also clear that $f$ restricted to the interior of each cell is a
homeomorphism onto a cell of the square complex structure of
$(\vee^{g}S^{1})\times(\vee^{g}S^{1})$. Therefore $f$ is a combinatorial map
of square complexes (see Section 2.2 for definitions).

In order to prove Proposition 4.13, we must check that $f$ is locally
injective and that it describes the Heegaard decomposition. In order to do
this, we need two lemmas.

######  Lemma 4.14.

$f$ is locally injective.

###### Proof.

In order to prove that $f$ is an immersion, or locally injective, it suffices
to show that $f$ is locally injective at the vertices of $\Sigma_{g}$. This is
because $f$ is a combinatorial map of square complexes (See [8]).

Consider a vertex, $v$, of the square complex structure on $\Sigma_{g}$. Note
that it is contained in the interior of a single polygon of the dual cell
decomposition induced by the Heegaard diagram curves (since the squares are
dual to this decomposition). Consider a small disk neighbourhood $D$ of $v$
and the intersection of the cells of the square complex structure with $D$.
Recall that $f$ sends $1$-cells homeomorphically to $1$-cells, and $2$-cells
homeomorphically to $2$-cells. Therefore, in order for $f$ to be locally
injective at $v$, it is sufficient that $f(D\cap\sigma)\cap
f(D\cap\delta)=\emptyset$ where $\sigma,\delta$ are cells in $\Sigma_{g}$ of
the same dimension.

Suppose two $1$-cells $\sigma$ and $\delta$ meeting at $v$ have
$f(D\cap\sigma)\cap f(D\cap\delta)\neq\emptyset$. It follows that
$f(\sigma)=f(\delta)$. By the definition of $f_{1}$ and $f_{2}$ on the
$1$-skeleton of $\Sigma_{g}$, this means that $\sigma$ and $\delta$ intersect
the same Heegaard diagram curve. (Recall that each $1$-cell intersects exactly
one diagram curve.) This means that the polygon dual to the vertex $v$ has two
edges that are intervals on a single diagram curve. The hypothesis that the
images of the intersection with $D$ intersect implies that one side of this
diagram curve intersects the open polygon in two connected components, one
along each of these edges. Thus the Heegaard diagram is not taut. This is a
contradiction.

Suppose there are two $2$-cells $\sigma$ and $\delta$, that meet at $v$ and
$f(\sigma)=f(\delta)$. Suppose further that $f(D\cap\sigma)\cap
f(D\cap\delta)\neq\emptyset$. There must be two $1$-cells $e_{1}$ and $e_{2}$
that lie consecutively in the boundary of $\sigma$ that intersect $D$. The
edges $e_{1}$ and $e_{2}$ must be sent to consecutive $1$-cells in the
boundary of $f(\sigma)$, incident to $f(v)$. This is because $f$ is a
continuous combinatorial map of square complexes. Similarly, There are two
$1$-cells, $e_{3}$ and $e_{4}$ that lie consecutively in the boundary of
$\delta$ that must be sent to two consecutive edges in the boundary of
$f(\delta)$ incident to $f(v)$. The set $\\{e_{i}\\}$ must contain at least
three elements. If it contained 2 or fewer elements, it would imply that only
two squares meet at $v$ (or that $v$ does not have a surface neighbourhood).
If only two squares meet at $v$, then the Heegaard diagram dual to the square
complex structure contains a bigon. This contradicts the assumptions of our
setting, see Lemma 4.13.

Since $f(\delta)=f(\sigma)$, it is possible to find three or four connected
components of $D\cap(\cup_{i}e_{i})\setminus\\{v\\}$ that are sent by $f$ to
two distinct connected components of edges in $f(D)\setminus f(v)$. We have
already shown this is not possible earlier in this proof, as this means that
there are two distinct $1$-cells $e_{k}$ and $e_{j}$ meeting at $v$ that have
$f(D\cap e_{j})\cap f(D\cap e_{k})\neq\emptyset$.

Since $f$ is a combinatorial map, these are the only pathologies that would
prevent $f$ from being injective on $D$. Therefore $f$ is locally injective at
the vertices of $\Sigma_{g}$ and therefore an immersion. ∎

We are now ready to prove Proposition 4.13. The remaining work consists of
checking that $f$ describes the given Heegaard decomposition.

###### Proof of Lemma 4.13.

Begin with a taut filling Heegaard diagram on a surface of genus greater than
zero. Construct the locally injective map
$f:\Sigma_{g}\rightarrow(\vee^{g}S^{1})\times(\vee^{g}S^{1})$ as above. It
remains to show that $f$ describes the Heegaard decomposition we began with.

Recall that $f$ describes $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ if, up to
homotopy equivalence of the domain, and product homotopy equivalence of the
target,
$f=\iota_{1}\times\iota_{2}:\Sigma_{g}\longrightarrow\mathcal{H}_{1}\times\mathcal{H}_{2}$,
where $\iota_{i}$ are the inclusion maps from the separating surface to
$\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ respectively. First, we check that
$f_{1}=\iota_{1}$ up to homotopy equivalence of the target space. Consider the
fibres of $f_{1}$. In what follows, refer to the example shown in Figure 16.
Recall that $f_{1}$ sends the interior of each square in $\Sigma_{g}$ to a
maximal interval in $\vee^{g}S^{1}$. The blue curve running through the square
is sent to the blue point in this interval. In fact, the full pre-image of
each blue point in $\Sigma_{g}$ is a blue diagram curve. The full pre-image of
any other point in the same maximal interval as a blue point is an isotopic
curve. It may help to think about these statements on one closed square at a
time first, and then put them together. The full pre-image of the wedge point
is graph on $\Sigma_{g}$ given by the $1$-skeleton of the square complex
structure, with the edges that intersect blue diagram curves removed. This can
be seen by thinking about each square first. The pre-image of the wedge point
in each closed square is the boundary edges that do not intersect a blue
diagram curve. Notice that all the fibres except the pre-image of the wedge
point are topological circles. All this follows from the definition of $f$.

Figure 16: A genus two example of one coordinate of $f$.

Consider including $\Sigma_{g}$ into a larger space $\mathcal{H}$, defined as
follows. Remove the $f_{1}$ pre-image of the wedge point from $\Sigma_{g}$.
Each connected component of the remaining space is made up of the squares
through which a given blue diagram curve runs. It is a cylinder of squares of
“height” $1$. There are $g$ such cylinders remaining, and they have a product
structure of $(0,1)\times S^{1}$ induced by the fibres of $f$. Include these
cylinders into solid cylinders, $(0,1)\times D^{2}$ so that the $S^{1}$ fibres
are included as boundaries to the $D^{2}$ fibres of the solid cylinder. The
idea is to glue $\Sigma_{g}$ back together, and glue the solid cylinders back
together accordingly. We now formalize this idea as follows. First glue two
disks to each solid cylinder, so that it is $[0,1]\times D^{2}$ instead of
$(0,1)\times D^{2}$. Parametrize the $D^{2}$ coordinate of these cylinders as
a unit disk using polar coordinates $(r,\theta)$, $0\leq r\leq 1$. Now glue
$\Sigma_{g}$ back together along the pre-image of the wedge point. If two
boundary disks ($\\{0\\}\times D^{2}$ or $\\{1\\}\times D^{2}$) are identified
along an arc on their $S^{1}$ boundary, each point in this arc belongs to a
ray in each of the two disks. Identify these rays. Call the resulting space
$\mathcal{H}$. Note that the centres of all the boundary disks are identified
in this process. This is because the pre-image of the wedge point is
connected. If it were not connected, then a collection of Heegaard diagram
curves would separate the connected components. The union of these curves
would be separating on $\Sigma_{g}$ and this contradicts the Heegaard diagram
curves being homologically independent.

######  Claim 1.

$\mathcal{H}$ is a handle body and there is a homeomorphism
$\phi:\mathcal{H}\longrightarrow\mathcal{H}_{1}$ so that the inclusion
$\iota:\Sigma_{g}\hookrightarrow\mathcal{H}$ is equal to $\phi\circ\iota_{1}$.

Recall that in order to complete the proof of Proposition 4.13, we wanted to
show that $f_{1}=\iota_{1}$ up to homotopy equivalence. If Claim 1 is true,
the proof is completed by noting that $\vee^{g}S^{1}$ is a deformation retract
of $\mathcal{H}$. The deformation retraction is realized by taking each point
on each disk in the solid cylinders $(r,\theta)$, and continuously shrinking
the radius r to $0$. Call this deformation retraction $d$. Each cylinder
becomes an interval, and the way the solid cylinders were glued together leads
to these intervals being glued together into a bouquet of circles. This can be
seen by considering the glueing locus under $d$, and noting that it is all
sent to a single point in the end. Then note that $f_{1}=d\circ\iota$ because
we set up the product structure on the hollow cylinders so that each $S^{1}$
fibre was the $f_{1}$ pre-image of a point in $\vee^{g}S^{1}$ (not the wedge
point). So we have $f_{1}=d\circ\iota=d\circ\phi\circ\iota_{1}$, and
$d\circ\phi$ is a homotopy equivalence as desired (since the composition of a
homeomorphism and deformation retraction gives a homotopy equivalence together
with inclusion composed with a homeomorphism). This completes the proof of
Lemma 4.13. ∎

Now we prove Claim 1.

###### Proof of Claim 1.

Recall how $\mathcal{H}_{1}$ is attached to $\Sigma_{g}$: a disk is attached
along each blue diagram curve on $\Sigma_{g}$, and then a $3$-ball is attached
to the resulting $2$-complex. We claim that the method of attaching
$\mathcal{H}$ is equivalent to this. Consider, in the construction of
$\mathcal{H}$, the disk $\\{\frac{1}{2}\\}\times D^{2}$ in each solid
cylinder. The $S^{1}$ boundary of such a disk is a blue diagram curve. So both
constructions attach disks along the blue diagram curves. Let $A$ be the space
remaining when we remove these disks and $\Sigma_{g}$ from $\mathcal{H}$. We
will prove that $A$ is homeomorphic to a $3$-ball that is attached to the rest
of $\mathcal{H}$ in the same way that the $3$-ball is attached in the
construction of $\mathcal{H}_{1}$.

The middle disk $\\{\frac{1}{2}\\}\times D^{2}$ cuts each solid cylinder in
half. Each half is a cone on a disk given by the “bottom” and “sides” of the
cylinder. This is shown below.

Figure 17: The cone structure of half of a solid cylinder.

The two cones share the disk $\\{\frac{1}{2}\\}\times D^{2}$. Note that the
cone structure restricted to $\\{0\\}\times D^{2}$ and $\\{1\\}\times D^{2}$
is compatible with the polar coordinates chosen on these disks. Recall that
when the solid cylinders are glued together to form $\mathcal{H}$, rays of one
of these disks are identified with rays of another. This implies that $A$ can
be expressed as a cone relating to the cone structure we have placed on the
half cylinders. Cut $\Sigma_{g}$ along the blue diagram curves (cut the
cylinders in half). Glue in disks along all boundary components. This is
simply $S^{2}$. $A$ is the interior of the cone on this space. That is, an
open $3$-ball. In this way, it is seen that $\mathcal{H}$ and
$\mathcal{H}_{1}$ are attached to $\Sigma_{g}$ in the same way, completing the
proof. ∎

######  Remark 4.15.

It is natural to ask whether Theorem 4.12 has a converse. Namely if $f$ being
locally injective implies that the Heegaard splitting it encodes is
irreducible. Osborne has provided examples that show that this is not the case
[14]. Figure 18 depicts one such diagram for a genus two Heegaard splitting
for a lens space with fundamental group $\mathbb{Z}_{13}$. The diagram is taut
and filling, and therefore the map encoding it is locally injective, however
it must be reducible since the genus is greater than one.

Figure 18: Heegaard diagram for a lens space with fundamental group
$\mathbb{Z}_{13}$

##  5 Virtually special square complexes and the augmented Heegaard diagram

We have seen the emergence of square complex structures as we study Stallings
maps. We now discuss the properties of these square complexes, and interpret
them further in the language of geometric group theory.

###  5.1 Non-positively curved square complexes that are virtually special

Note that the cartesian product of two graphs naturally has the structure of a
non-positively curved square complex. A square complex is non-positively
curved if cycles in the link of every vertex have length at least four. The
precise definitions and terminology about square complex structures can be
found in [8].

When $f$ is locally injective, a non-positively curved square complex
structure can be pulled back to the surface $\Sigma_{g}$, giving $\Sigma_{g}$
a tiling by squares where at least four squares meet at each vertex. The
hyperplanes of this square complex structure (the curves dual to the square
tiling) are precisely the Heegaard diagram curves. This is seen in the
constructive proof of Theorem 4.12 and Proposition 4.13. Therefore the
following properties of the square complex structure are immediate.

  * •

Hyperplanes do not self intersect (the hyperplanes are Heegaard diagram
curves, and these do not self intersect)

  * •

The square complex is a VH-complex (V stands for vertical and H stands for
horizontal. The blue curves of the Heegaard diagram are the vertical
hyperplanes, and the red curves are the horizontal hyperplanes, see Definition
2.6).

  * •

Hyperplanes are two-sided (because all tubular neighbourhoods of simple closed
curves on $\Sigma_{g}$ are orientable)

  * •

Hyperplanes do not directly self-osculate (since the Heegaard diagram is
taut). Therefore the VH-complex is clean.

The definitions of clean VH-complexes can be found breifly in Section 2.2. For
more detail, see [3]. Note that the hyperplanes may indirectly self-osculate
and they may also interosculate, so the square complex structure is not
special and so, while $f$ is a local embedding, it is not a local isometry.

######  Proposition 5.1.

The square complex structure on $\Sigma_{g}$ dual to a taut Heegaard diagram
is virtually special.

###### Proof.

This is a corollary of Theorem 5.7 in [8], which states that any compact VH-
complex with clean horizontal hyperplanes is virtually special. The square
complex structure here has both horizontal and vertical hyperplanes being
clean. The surface $\Sigma_{g}$ is certainly compact. Thus, we have the
desired result. ∎

The connection between the square complex structure and geometric group theory
methods can be used to understand $f$ better (see Section 5.2).

To end the section, we state a generalized version of Theorem 4.12 for all
Heegaard splittings.

######  Definition 5.2.

A pinched surface $\mathcal{P}$ is given by the quotient $S/\sim$ of a surface
by an equivalence relation of the following form. Given a finite set of
compact connected subsurfaces of $S$ with boundary, $x\sim y\iff x=y$ or $x$
and $y$ are in the same component of the chosen subsurfaces. We call the
images of the non-trivial equivalence classes pinching points

We distinguish pinched surfaces from singular surfaces with nodes, since a
subsurface that has more boundary components than an annulus can be pinched to
a point in the quotient in Definition 5.2, leading to multiple connected
compact surfaces identified at a single point in the quotient.

######  Definition 5.3.

The graph of a pinched surface $S/\sim$ is given as follows. Let $X$ be the
union of the finite set of subsurfaces of $S$ in the definition of $\sim$.
There is one vertex for each connected component of $X$, as well as one vertex
for each connected component of $S\setminus X$. There is an edge connecting
two vertices if the regions of $S$ they represent share a boundary.

Note that this graph is bipartite. A connected component of $X$ cannot share a
boundary with another connected component of $X$ since they are disjoint
compact subsurfaces. Therefore, the corresponding vertices cannot have an edge
connecting them. If two connected components of the complement of $X$ shared a
boundary, this would mean that the boundary curve is a connected component of
$X$. However, $X$ is a union of subsurfaces, so this is not possible.
Therefore the corresponding vertices cannot have an edge connecting them.

######  Theorem 5.4.

Suppose that $M$ does not have $S^{2}\times S^{1}$ as a connect summand, and
has Heegaard splitting $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$. Then there is a
Stallings map $f$ for this splitting that factors through a quotient to a
pinched surface $\Sigma^{*}$. The pinched surface $\Sigma^{*}$ has a square
complex structure such that the the quotient map is a combinatorial map of
square complexes that is locally injective. Further, the Heegaard splitting
can be recovered from the square complex structure on $\Sigma^{*}$.

###### Proof.

By Theorem 4.10, a taut Heegaard diagram for
$M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ can be chosen. Then let $f$ be a
Stallings map descibing it. Next, take a the union of some tubular
neighbourhoods of the curves. Any component of the complement must have null
homotopic image by $f$. The pinched surface $\Sigma^{*}$ is constructed by
making each non-simply connected component of the complement a single
equivalence class, and making any other point a singleton equivalence class,
and then taking the quotient. Up to homotopy, $f$ must factor through
$\Sigma^{*}$.

The graph of $\Sigma^{*}$ cannot have any cycles. A cycle consists of a set of
vertices and edges arranged in a circle. Since the graph is bipartite, if
there was a cycle, the vertices must alternate in whether they correspond to a
subsurface of $\Sigma_{g}$ that is pinched, or not. In particular, this
implies that the cycle cannot consist exclusively of vertices that correspond
to subsurfaces of $\Sigma_{g}$ that are not pinched. Let
$A\subseteq\Sigma_{g}$ be a subsurface that is pinched, and whose
corresponding vertex is part of the cycle. Any essential simple closed curve
$\gamma$ on $A$ is a reducing curve. Further, it is non separating in
$\Sigma_{g}$ since a path can be found from one side of $\gamma$ to the other
that projects to a path that goes around the cycle in the graph of
$\Sigma^{*}$. By Proposition 4.4, the reducing curve $\gamma$ bounds an
embedded disk in $\mathcal{H}_{1}$ and another disk in $\mathcal{H}_{2}$. This
forms a copy of $S^{2}\subseteq M$ that is not separating (since $\gamma$ is
not separating). This implies that $S^{2}\times S^{1}$ is a connect summand of
$M$, a contradiction.

Note that a closed surface subset of $\Sigma^{*}$ cannot be homeomorphic to a
sphere. This could only occur when in $\Sigma_{g}$, a connected component $C$
of the tubular neighbourhood of the Heegaard diagram has genus 0. Then in the
quotient, every complementary surface region bordering the image of $C$ is
simply connected, forming a sphere. However, then each diagram curve in $C$
would be separating on the sphere in $\Sigma^{*}$, and therefore separating in
$\Sigma_{g}$ (since the graph of $\Sigma^{*}$ is a tree). This is a
contradiction.

Further, there are no open surface subsets of $\Sigma^{*}$ whose closure is a
pseudo-manifold. If this were the case, there would be a connected component
$C$ of the tubular neighbourhood of the Heegaard diagram with two boundary
curves that bound the same complementary region. In this case, any one of
those boundary curves is a reducing, non-separating curve on $\Sigma_{g}$, a
contradiction (again, because it implies that $S^{2}\times S^{1}$ is a connect
summand of $M$).

Therefore $\Sigma^{*}$ consists of several closed surfaces of positive genus
identified along points in a tree like fashion (more than two may be
identified at a point).

Consider the image of the Heegaard diagram curves in $\Sigma^{*}$. If a
diagram curve passed from one surface subset of $\Sigma^{*}$ to another, it
must have passed through one of the pinching points. This is not possible by
the definition of $\Sigma^{*}$. If a number of curves are homologous in one of
the surface subspaces, they must have been homologous in $\Sigma_{g}$. If any
two quotient curves intersect in $\Sigma^{*}$, the original two curves must
have intersected in $\Sigma_{g}$. Finally if a quotient diagram curve
$\overline{\gamma}$ were to separate the surface subspace it lies in, the
original curve $\gamma$ must have been separating in $\Sigma_{g}$. This is
because the complement of every subsurface of $\Sigma_{g}$ used in the
definition of $\Sigma^{*}$ is not connected (since the graph of $\Sigma^{*}$
is a tree, removing any vertex leaves a disconnected space). Therefore there
cannot be any curve from one side of $\gamma$ to the other side, since it
would project to a cycle in the graph of $\Sigma^{*}$.

Therefore the image of the Heegaard diagram curves on each surface subspace of
$\Sigma^{*}$ are disjoint, as well as homologically non-trivial and
independent. If any surface subset of $\Sigma^{*}$ contains fewer red or blue
curves than its genus, then $\Sigma_{g}$ had fewer red or blue curves
respectively than its genus. If a surface subset of $\Sigma^{*}$ had more blue
(respectively red) curves than its genus, since they do not intersect each
other, one must be separating on the subsurface, which in turn implies that it
is separating on $\Sigma_{g}$, since the graph of $\Sigma^{*}$ is a tree. This
is a contradiction. Therefore the quotient of the diagram curves form a
Heegaard diagram on each surface subset of $\Sigma^{*}$. This Heegaard diagram
must be taut and filling because the diagram on $\Sigma_{g}$ was taut, and the
quotient procedure described precisely takes each component of the complement
that was not homeomorphic to a disk, and makes it so. Therefore the Lemma 4.13
can be applied to each surface subset to show that the quotient map is
homotopic to a locally injective map.

Note that any simple closed curves on $\Sigma_{g}$ in the pre-image of a point
of $\Sigma^{*}$ are reducing curves (that is, they are null homotopic in
$\mathcal{H}_{1}$ and in $\mathcal{H}_{2}$). Thus if
$\Sigma_{g}\neq\Sigma^{*}$, $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ is
reducible and the restriction of $f$ to the surface subspaces of $\Sigma^{*}$
are encodings of Heegaaard splittings of connect summands.

The square complex structure on each closed surface subset of non-positive
Euler characteristic is dual to the Heegaard diagram on the subsurface, so it
contains all the information of the Heegaard splitting. ∎

In the case where $S^{2}\times S^{1}$ is a connect summand, the portion of
$\Sigma^{*}$ corresponding to $S^{2}\times S^{1}$ may not be a surface
subspace, but rather a pseudo manifold of genus 0, on which $f$ cannot be
locally injective. In the cases where $\Sigma_{g}=\Sigma^{*}$ , $f$ itself is
locally injective and $\Sigma_{g}$ inherits a square complex structure so that
$f$ is a combinatorial map of square complexes.

There have been several proposed methods to combinatorially reduce or simplify
a Heegaard diagram to a normal form from which the topology of the 3-manifold
is more readily recognized ( for example the Whitehead conjecture [20] and
Volodin-Kuznetsov-Fomenko conjecture [19] for $S^{3}$). Each is true in
certain cases, but not generally [12]. At this point, it is natural to ask
whether Theorem 5.4 gives one such method to recognize the connect sum
structure of the 3-manifold. However, Remark 4.15 describes an example of the
failure of taut Stallings maps to pinch every reducing curve. Nevertheless,
the structure of Stallings maps suggest that there is something beyond the
surface that can be manipulated and simplified. We expose and explore this in
the remainder of the paper, where we will use the following definition.

######  Definition 5.5.

A Heegaard diagram is simple if there is a maximal disjoint set of separating
reducing curves on $\Sigma_{g}$ that do not intersect the Heegaard diagram
(these reducing curves must induce the prime decomposition of the
$3$-manifold, possibly with additional $S^{3}$ connect summands if the
Heegaard splitting is stabilized). A Heegaard diagram is not simple if every
maximal disjoint set of separating reducing curves on $\Sigma_{g}$ intersects
the Heegaard diagram. A Stallings map describing a simple Heegaard diagram is
also called simple.

Note that if a Heegaard diagram for the $3$-manifold $M$ is simple and $f$ is
a Stallings map for the Heegaard diagram, the surface subspaces of
$\Sigma^{*}$ each form minimal genus ($\geq 1$) Heegaard diagrams for the
prime connect summands of $M$.

###  5.2 Group Actions on Trees and the Guirardel Core

Let $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ be a $3$-manifold with a Heegaard
decomposition. Let $f$ be a Stallings map for this Heegaard splitting. By the
proof of Theorem 4.10, we may assume that $f$ is taut after possibly replacing
it with another Stallings map in its $Out(F_{g})\times Out(F_{g})$ orbit. In
this section, we uniformly assume that $S^{2}\times S^{1}$ is not a connect
summand of $M$. By Theorem 5.4, $f$ may be modified by a homotopy so that it
factors through a pinched surface $\Sigma^{*}$ with a square complex
structure, and $f$ is a combinatorial map of square complexes that is locally
injective on each surface subspace of $\Sigma^{*}$. The Heegaard diagram
corresponding to $f$ is taut but not necessarily filling. This is the setting
for the analysis that follows.

Consider the following commutative diagram of covering spaces, where $T_{1}$
and $T_{2}$ are each a copy of the universal cover of $\vee^{g}S^{1}$, $S$ is
the cover of $\Sigma_{g}$ corresponding to the kernel of
$f_{*}:\pi_{1}(\Sigma_{g})\rightarrow\pi_{1}((\vee^{g}S^{1})\times(\vee^{g}S^{1}))$,
and $\hat{f}$ is the corresponding lift of $f$.

${S}$${T_{1}\times
T_{2}}$${\Sigma_{g}}$${(\vee^{g}S^{1})\times(\vee^{g}S^{1})}$$\scriptstyle{\hat{f}}$$\scriptstyle{\pi_{1}(\Sigma_{g})}$$\scriptstyle{\pi_{1}(\Sigma_{g})\times\pi_{1}(\Sigma_{g})\geq\Delta_{\pi_{1}(\Sigma_{g})}}$$\scriptstyle{f}$

Consider the action of $\pi_{1}(\Sigma_{g})$ on this diagram. The group
$\pi_{1}(\Sigma_{g})$ acts on $S$ by deck transformations (with kernel equal
to $\ker(f_{*})$). It also acts on $T_{i}$ by deck transformations. This is
because $T_{i}$ is the universal cover of $\vee^{g}S^{1}$ with the free group
on $g$ generators $F_{g}$, acting as deck transformations. The inclusion of
$\Sigma_{g}$ as the boundary of a handle body is $\pi_{1}$-surjective, and
therefore $f_{i*}\colon\pi_{1}(\Sigma_{g})\rightarrow F_{g}$ is surjective as
well. Thus each element $\alpha\in\pi_{1}(\Sigma_{g})$ acts by the deck
transformation $f_{i*}(\alpha)$ on $T_{i}$. This creates an action of
$\pi_{1}(\Sigma_{g})\times\pi_{1}(\Sigma_{g})$ on $T_{1}\times T_{2}$, and the
diagonal subgroup $\Delta_{\pi_{1}(\Sigma_{g})}$ (elements of the form
$(x,x)$) gives an action of $\pi_{1}(\Sigma_{g})$ on $T_{1}\times T_{2}$. The
map $\hat{f}$ is equivariant with respect to the actions of
$\pi_{1}(\Sigma_{g})$ on $S$ and on $T_{1}\times T_{2}$.

We would like to understand the diagonal action of $\pi_{1}(\Sigma_{g})$ on
$T_{1}\times T_{2}$ because this is where the information of the $3$-manifold
$M$ is hidden. The Guirardel core (Definition 2.11, and defined with more
details in [7]) provides a way to understand the compatibility of the two
coordinates of this action. The larger the volume of the core (a subset of
$T_{1}\times T_{2}$), the larger the obstruction to finding a third tree
$T_{3}$ with an action of $\pi_{1}(\Sigma_{g})$ and surjective equivariant
maps onto $T_{1}$ and $T_{2}$.

For instance, if $\pi_{1}(\Sigma_{g})$ acts on $T_{1}$ and $T_{2}$ in the same
way, the Guirardel core has zero volume and every simple closed curve in
$\Sigma_{g}$ is reducing. The corresponding Heegaard diagram is the connect
sum of several genus one diagrams for $S^{2}\times S^{1}$. This diagram
completely reveals the connect sum structure of this particular $3$-manifold.

The precise relationship between the Guirardel core and Heegaard
decompositions can be elucidated using Stallings maps, as we shall see in the
following theorem. Recall that the Guirardel core is the subset of
$T_{1}\times T_{2}$ given by removing all the light quadrants (see Definition
2.11, which recounts the definition in [7]). Let $\hat{f}_{1}$ and
$\hat{f}_{2}$ be the coordinates of $\hat{f}$.

######  Lemma 5.6.

A quadrant $\delta_{1}\times\delta_{2}$ of $T_{1}\times T_{2}$ is heavy iff it
intersects $\hat{f}(S)$.

###### Proof.

Let $\hat{f}_{1}$ and $\hat{f}_{2}$ be the coordinates of $\hat{f}$. Suppose
$\delta_{1}\times\delta_{2}\cap\hat{f}(S)=\emptyset$. Using any point in
$\hat{f}(S)$ as the base point, its orbit must be contained in $\hat{f}(S)$
because the map $\hat{f}$ is equivariant with respect to the group action. For
this reason, there can be no sequence of elements in the orbit of the base
point that is contained in $\delta_{1}\times\delta_{2}$. Therefore
$\delta_{1}\times\delta_{2}$ must not be heavy (and therefore must be light).

Conversely, suppose $\delta_{1}\times\delta_{2}\cap\hat{f}(S)\neq\emptyset$.
This is equivalent to
$\hat{f}_{1}^{-1}(\delta_{1})\cap\hat{f}_{2}^{-1}(\delta_{2})\neq\emptyset$.
We show that $\delta_{1}\times\delta_{2}$ must be heavy. Consider the covers
$S_{1}$ and $S_{2}$ of $\Sigma_{g}$ corresponding to the subgroups
$\ker{f_{1}}_{*}$ and $\ker{f_{2}}_{*}$ of $\pi_{1}(\Sigma_{g})$ respectively.
These covers are planar surfaces that are the boundaries of the universal
covers of the handle bodies $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$
respectively. In fact, $\hat{f}_{1}$ can be interpreted as the projection from
$S$ to $S_{1}$ followed by the inclusion of $S_{1}$ into the universal cover
of $\mathcal{H}_{1}$, followed by a deformation retraction $d_{1}$ of this
space to $T_{1}$ (a lift of the deformation retraction of $\mathcal{H}_{1}$ to
$\vee^{g}S^{1}$). The blue Heegaard diagram curves lift to separating curves
in $S_{1}$ and the red Heegaard diagram curves lift to separating curves in
$S_{2}$. This is because on $\Sigma_{g}$ they are the pre-images of midpoints
of edges in $\vee^{g}S^{1}$ by $f_{1}$ and $f_{2}$ respectively, and so in
$S_{i}$, we may think of them as the pre-images of the mid points of edges of
$T_{i}$ by the deformation retraction $d_{i}$ alluded to above. Recall that
these curves are taut but not filling.

To understand how a path $\gamma$ in $\hat{S}$ is sent to a path in $T_{1}$ by
$\hat{f}_{1}$, we may project $\gamma$ to $S_{1}$ and observe the algebraic
intersection with blue diagram curves. (Recall that the blue diagram curves
are the pre-images of the midpoints of the edges in $T_{1}$). The sum of the
absolute value of the algebraic intersection numbers of $\gamma$ with each of
the blue diagram curves is the distance between the end points of
$\hat{f}_{1}(\gamma)$ in $T_{1}$.

Let $x\in\hat{f}_{1}^{-1}(\delta_{1})\cap\hat{f}_{2}^{-1}(\delta_{2})$. If $x$
is on (the lift of a) Heegaard diagram curve, it can be perturbed slightly to
be off all diagram curves while remaining in
$x\in\hat{f}_{1}^{-1}(\delta_{1})\cap\hat{f}_{2}^{-1}(\delta_{2})$ (since this
set is open and the curves are not full-dimensional sets).

From here, we seek a sequence $\\{\alpha_{i}\cdot x\\}_{i}$ where
$\alpha_{i}\in\pi_{1}(\Sigma_{g})$, such that the sequences of distances
$\\{d_{T_{1}}(x,\alpha_{i}\cdot x)\\}_{i}$ and $\\{d_{T_{2}}(x,\alpha_{i}\cdot
x)\\}_{i}$ are both unbounded. We construct a sequence by constructing a path
in $S$ beginning at $x$ that does not backtrack significantly when its image
is projected to each tree. The backtracking in the $T_{i}$ is monitored by
algebraic intersection with blue diagram curves when projected to $S_{i}$.

Since $x$ is not on a lifted Heegaard diagram curve, it must be in the
interior of a connected component of the complement in $S$. Call these
complementary components. Since $f$ is locally injective, this complementary
component cannot have a problem, and therefore, at most two edges of the
boundary are sent to the boundary of $\delta_{1}\times\delta_{2}$ and they
cannot both be the same colour. The complementary component must have at least
four edges, so there must be a way for a path beginning at $x$ to exit the
complementary component via a vertex (transversally crossing a red and blue
hyperplane simultaneously) without exiting
$\hat{f}_{1}^{-1}(\delta_{1})\cap\hat{f}_{2}^{-1}(\delta_{2})$.

The path has now entered a new complementary component in $S$. We now extend
the path further, taking care not to back track at each step. This new
complementary component must also not have a problem. Therefore, as long as
the path does not exit this complementary component by crossing the same two
edges it crossed upon entry, the path’s image will not back track in $T_{1}$
and in $T_{2}$. The complementary component has least four edges, and so
again, it is possible to exit this complementary component via a vertex
(simultaneously transversally crossing a blue and red hyperplane) without
backtracking in $T_{1}$ and in $T_{2}$.

We continue in this way, extending the path. Each time the path crosses
diagonally over a vertex (simultaneously transversally crossing a blue and red
diagram curve), its image goes one unit deeper in $T_{1}$ and $T_{2}$.

Let $N$ be the number of complementary components in $\Sigma_{g}$. We extend
the path from $x$ as above until it visits $N+10$ complementary components. At
this point, we extend the path by a segment connecting it to the the nearest
point in the $\pi_{1}(\Sigma_{g})$ orbit of $x$. This is at most $N$
complementary components away. The completed path now connects $x$ to a point
$\alpha_{1}\cdot x$ for some $\alpha_{1}\in\pi_{1}(\Sigma_{g})$. The image of
the path moves distance $N+10$ units deeper into $\delta_{1}$ and
$\delta_{2}$, and then backtracks at most $N$ units. In total then,
$d_{T_{j}}(x,\alpha_{1}\cdot x)\geq 10$ for $j=1,2$.

We find $\alpha_{i+1}\cdot x$ by extending our path from $\alpha_{i}\cdot x$
in the same way as above: first taking care to move $N+10$ unit deeper into
both trees, and then proceeding to the nearest point in the orbit of $x$. This
constructs a sequence $\\{\alpha_{i}\\}$ so that $d_{T_{j}}(x,\alpha_{i}\cdot
x)\geq 10i$, for both $j=1,2$. These sequences of distances are unbounded as
desired. Therefore, by definition, $\delta_{1}\times\delta_{2}$ is heavy. ∎

Lemma 5.6 implies that the Guirardel core contains $\hat{f}(S)$, but it also
cannot be much larger than $\hat{f}(S)$ (because every quadrant not
intersecting $\hat{f}(S)$ is light).

In fact, the Guirardel core is the minimal (with respect to inclusion), non-
empty $\pi_{1}(\Sigma_{g})$–invariant closed connected set such that the
fibres of the projections to $T_{1}$ and $T_{2}$ are connected (see [7],
section 5). The theorem below then follows and characterizes the relationship
between $f$ and the core.

######  Theorem 5.7.

Let $f:\Sigma_{g}\rightarrow(\vee^{g}S^{1})\times(\vee^{g}S^{1})$ be a locally
injective Stallings map describing a Heegaard decomposition. The Guirardel
core of the actions of $\pi_{1}(\Sigma)$ on $T_{1}$ and $T_{2}$ is the
smallest set $X$ such that $\hat{f}(S)\subseteq X$ and ${p_{1}}^{-1}(t)$ is
connected for every $t\in T_{1}$ and ${p_{2}}^{-1}(t)$ is connected for every
$t\in T_{2}$.

###### Proof.

Let $X$ be the the smallest set $X$ such that $\hat{f}(S)\subseteq X$ and
${p_{1}}^{-1}(t)$ is connected for every $t\in T_{1}$ and ${p_{2}}^{-1}(t)$ is
connected for every $t\in T_{2}$. The core contains $\hat{f}(S)$ and the
fibres of $p_{1}$ and $p_{2}$ are connected in the core, so the core must
contain $X$. On the other hand, $X$ is a non-empty,
$\pi_{1}(\Sigma_{g})$–invariant, closed, connected set, and the projections to
$T_{1}$ and $T_{2}$ are connected by definition, so the core must be contained
in $X$. This implies that $X$ is the Guirardel core. ∎

###  5.3 Complexity of the Guirardel core and augmented Heegaard diagrams

Let $f$ be a taut Stallings map describing a Heegaard splitting for a
$3$-manifold $M$. We continue to assume that $S^{2}\times S^{1}$ is not a
connect summand of $M$. Recall that $f$ factors through a pinched surface
$\Sigma^{*}$. Let the factor map be
$g:\Sigma^{*}\rightarrow\vee^{g}S^{1}\times\vee^{g}S^{1}$. Let
$\hat{\Sigma}^{*}$ be the cover of $\Sigma^{*}$ that corresponds to the kernel
of $g$. Then $\hat{f}$ factors through $\hat{\Sigma}^{*}$ with factor map
$\hat{g}:\hat{\Sigma}^{*}\rightarrow T_{1}\times T_{2}$.

${S}$${T_{1}\times
T_{2}}$${\hat{\Sigma}^{*}}$${\Sigma_{g}}$${(\vee^{g}S^{1})\times(\vee^{g}S^{1})}$${\Sigma^{*}}$$\scriptstyle{\hat{f}}$$\scriptstyle{\hat{g}}$$\scriptstyle{f}$$\scriptstyle{g}$

Given a Heegaard splitting, all choices of Stallings maps $[f]$ encoding it
have the same $\pi_{1}$-kernel, $f_{*}$, so the topology of the cover $S$ is
fixed. As we pass through the $\text{Out}F_{g}\times\text{Out}F_{g}$ orbit of
$f$, passing through all the maps $\Sigma_{g}\rightarrow\vee^{g}S^{1}$ that
describe the splitting, the square complex structure on $\hat{\Sigma}^{*}$
changes and $\hat{f}$ changes. If there is a reducing curve in $\Sigma_{g}$,
it lifts to $S$. The disk its image bounds in $T_{1}\times T_{2}$ is in the
Guirardel core, because of the coordinate-wise convex condition. A longer
reducing curve suggests more squares in the core needed to form a disk that
this curve bounds. This leads to the following question.

######  Question 5.8.

Does the smallest (in some sense) Guirardel core in the Out$F_{g}\times$
Out$F_{g}$ orbit give an $f$ that is simple (pinches a maximal set of reducing
curves, see Definition 5.5)?

If the answer to this question is positive, we would have a new
characterization for Heegaard diagrams that are connect sums of irreducible
diagrams. Then working with Guirardel cores, it may be possible to find a
combinatorial algorithm to move within the Out$F_{n}\times$ Out$F_{n}$ orbit
to find these diagrams.

One attempt to compare sizes of Guirardel cores, comes from looking at their
quotients. Given a Stallings map $f$ encoding a Heegaard splitting, let
$\mathscr{G}_{f}$ be the quotient of the Guirardel core (of the action of
$\pi_{1}(\Sigma_{g})$ on $T_{1}$ and $T_{2}$) by the diagonal action of
$\pi_{1}(\Sigma_{g})$. The area of $\mathscr{G}_{f}$, in terms of number of
squares, could be used as a measure of the size of the Guirardel core. However
even $\mathscr{G}_{f}$ is likely infinite in many cases. Nevertheless, the
following construction gives a partial answer to Question 5.8. We look at a
certain subset of the core whose quotient does have finite area.

####  5.3.1 Construction of augmented Heegaard diagrams

Begin with a taut Stallings map for a Heegaard splitting for a $3$-manifold
$M$ that does not have $S^{2}\times S^{1}$ as a connect summand. Consider an
embedded disk in $T_{1}\times T_{2}$ whose boundary is a subset of
$\hat{f}(S)$, and when pulled back to $S$, and then projected to $\Sigma_{g}$,
the boundary is a simple closed curve $\gamma$. Note that such a disk is in
the Guirardel core due to the convexity of fibres. We would like to focus on
this part of the Guirardel core.

Take the pinched surface $\hat{f}$ factors through, $\hat{\Sigma}^{*}$, and
attach an isomorphic disk (tiled with squares in the same way) along the
pullback of its boundary. The quotient map
$\hat{g}:\hat{\Sigma}^{*}\rightarrow T_{1}\times T_{2}$ can be extended to
this disk by sending it to the isomorphic disk bounded by $\hat{g}(\gamma)$ in
$T_{1}\times T_{2}$. The unextended map $\hat{g}$ is locally injective. If the
extended $\hat{g}$ is no longer locally injective, we do not want to include
the disk. In this way, the augmented Heegaard diagram is built by attaching
disks that do not disrupt the local injectivity. A precise definition will
follow shortly.

Given a taut Stallings map $f$ for a Heegaard splitting of $M$, the boundary
components of the subsurfaces that are pinched by $f$ provide a set of
disjoint reducing curves. We may use the two disks each of these reducing
curves bounds (one in each handle body) to decompose $M$ by connect sum. The
connect summands come with Heegaard splittings of lower geni (that add to give
the original genus). In this way, we look at the closed surface subspaces of
$\Sigma^{*}$ as Heegaard splittings of connect summands of $M$.

We may ask whether each of these are reducible. If $f$ restricted to one of
the surface subspaces describes a reducible Heegaard splitting, then by
Haken’s lemma, there must be a reducing curve on the splitting surface. Since
$S^{2}\times S^{1}$ is not a connect summand, this curve must be separating,
and thus we may use the two disks it bounds (one in each handle body) to
decompose $M$ further by connect sum. We continue in this way until we have a
set of disjoint reducing simple closed curves on $\Sigma$ such that the
connect components of the complement represent irreducible Heegaard splittings
of the prime connect summands of $M$.

There are many choices in the procedure above, and so there are several sets
of curves one may end with. For any given possibility, we may construct an
augmented Heegaard diagram.

######  Definition 5.9.

Given a taut Stallings map $f$ for a Heegaard splitting, and a maximal set of
disjoint reducing curves $\\{\gamma_{1},\ldots\gamma_{n}\\}$, constructed as
above (namely, so that they include the curves bounding the subsurfaces
pinched by $f$), we construct the augmented Heegaard diagram by gluing disks
to $\Sigma^{*}$ along each $\gamma_{i}$. Beginning with the pinched surface
$\Sigma^{*}$, for each $\gamma_{i}$, consider its lift to $\hat{\Sigma}^{*}$,
$\hat{\gamma}_{i}$ (recall $\hat{\Sigma}^{*}$ is the pinched surface $\hat{f}$
factors through). The image $\hat{g}(\hat{\gamma}_{i})$ in $T_{1}\times T_{2}$
bounds a disk $D$. We attach an isometric copy of $D$ to $\Sigma^{*}$ along
$\gamma_{i}$ and extend the map $g$ to send this copy to the projected image
of $D$ in $\vee^{g}S^{1}\times\vee^{g}S^{1}$. By modifying $\gamma_{i}$ by a
homotopy, and removing some peripheral squares of $D$, it is possible to
choose $D$ so that the extended map $g$ remains locally injective. The
resulting square complex is an augmented Heegaard diagram $\mathscr{C}$.

Notice that we now have the following commutative diagram, with
$\hat{\mathscr{C}}$ being the cover of $\mathscr{C}$ corresponding to $\ker
g_{*}$.

${S}$${T_{1}\times
T_{2}}$${\hat{\mathscr{C}}}$${\Sigma_{g}}$${(\vee^{g}S^{1})\times(\vee^{g}S^{1})}$${\mathscr{C}}$$\scriptstyle{\hat{f}}$$\scriptstyle{\hat{g}}$$\scriptstyle{f}$$\scriptstyle{g}$

The purpose of only attaching disks that maintain the local injectivity of
$\hat{g}$ is that this ensures that $\hat{\mathscr{C}}$ and $\mathscr{C}$ are
clean $VH$-complexes. Note that $\mathscr{C}$ is compact, and therefore
virtually special.

######  Remark 5.10.

Let $M$ be a $3$-manifold that does not have $S^{1}\times S^{2}$ as a connect
summand. A Heegaard decomposition of $M$ is irreducible if and only if it has
an augmented Heegaard diagram that is a surface.

It is also notable that one could define the augmented Heegaard diagram for a
Stallings map $f$ that is not taut. In this case, $f$ would still factor
through a pinched surface, however, the factor map would not be locally
injective. In this case, the image of $\hat{f}$ is may not be contained in the
Guirardel core. However, one could still attach disks along reducing curves as
in Definition 5.9 to obtain an augmented Heegaard diagram.

####  5.3.2 Area properties of the augmented Heegaard diagram

The fact that one may take irreducible Heegaard diagrams and then take their
connect sum to form a Heegaard diagram for the connect sum of the
$3$-manifolds they represent implies that certainly there exist Heegaard
diagrams that are simple, as in Definition 5.5. Equivalently, there exist
Stallings maps that pinch a maximal reducing multicurve. However, as mentioned
earlier, it is difficult to detect from the combinatorics of a Heegaard
diagram whether it is simple. The following theorem relates the area (number
of squares) of an augmented Heegaard diagram to whether it is simple or not.

######  Theorem 5.11.

Let $M=\mathcal{H}_{1}\cup\mathcal{H}_{2}$ be a Heegaard splitting of a
$3$-manifold $M$ that does not have $S^{2}\times S^{1}$ as a connect summand.
In the space of Stallings maps, $B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$, each
has several augmented Heegaard diagrams. Of all of these, the Stallings map
with the smallest-area augmented Heegaard diagram $\mathscr{C}$, has
$\mathscr{C}=\Sigma^{*}$. In other words, a Stallings map with the smallest-
area augmented Heegaard diagram is simple.

We will see in the proof below that a Stallings map with the smallest-area
augmented Heegaard diagram must also be taut. To make a Stallings map taut is
equivalent to simplifying a Heegaard diagram with wave moves along curves that
do not intersect the Heegaard diagram. While it had been conjectured by
Whitehead and Volodin-Kuznetso-Fomenko that this may be enough to simplify a
Heegaard diagram (of $S^{3}$), it has been shown that this is not enough [11]
[12] [13] [18]. Theorem 5.11 quantifiably describes how to continue to
simplify all the way.

Overall, we see that asking for a Stallings map to be taut ties it to the
Guirardel core (Theorem 5.7). In order for this Stallings map to be simple,
the Guirardel core must have a sense of minimal complexity, which is made
precise by minimizing the area over all possible augmented Heegaard diagrams
(Theorem 5.11).

###### Proof.

Assume that $S^{2}\times S^{1}$ is not a connect summand of $M$. Let $[f]\in
B_{M,\mathcal{H}_{1},\mathcal{H}_{2}}$. Suppose that
$\mathscr{C}\neq\Sigma^{*}$. We claim that by applying a sequence of handle
slides, the area of $\mathscr{C}$ can be reduced. Specifically, we reduce the
area by applying wave moves to pinch curves that previously had a disk
attached along them. The idea is that this eliminates the need to attach any
disks to $\Sigma^{*}$ to obtain $\mathscr{C}$.

Consider $\gamma$, a simple closed curve in $\Sigma^{*}$, along which a disk
is attached in $\mathscr{C}$. Let $\hat{\gamma}$ be a lift of $\gamma$ to $S$.
Then $\hat{f}(\hat{\gamma})$ bounds a disk in $T_{1}\times T_{2}$.

Consider the intersection word of $\gamma$ with the Heegaard diagram
corresponding to $[f]$. Let $b_{1},\ldots b_{n}$ be the blue letters that
appear and let $r_{1},\ldots r_{m}$ be the red letters that appear in this
intersection word. (In other words, $b_{1},\ldots b_{n}$ are the blue curves
$\gamma$ intersects and $r_{1},\ldots r_{m}$ are the red curves $\gamma$
intersects). The cyclic intersection word of $\gamma$ can be reduced to the
trivial word using the relation that $r$’s and $b$’s commute. This implies
that each intersection noted by $b_{i}$ is paired with another noted
$b_{i}^{-1}$ with which it will cancel in the reduction process (respectively
for $r$’s). Each cancelling pair bounds a hyperplane in the disk bounded by
$\hat{f}(\hat{\gamma})$. Further, two hyperplanes in the disk intersect if and
only if their end points on $\hat{f}(\hat{\gamma})$ are linked.

$\hat{f}(\hat{\gamma})$ Figure 19: Disk in $T_{1}\times T_{2}$ with
$\hat{f}(\hat{\gamma})$ as boundary.

Since $\hat{f}(\hat{\gamma})$ is null homotopic in $T_{1}\times T_{2}$, there
exists an $i$ such that a subsegment of $\gamma$ provides a path from $b_{i}$
to itself intersecting the same side twice without intersecting any other blue
curves (corresponding to the place in the intersection word where cyclic
reduction can begin). In other words, a wave move can be applied to $b_{i}$
along a subsegment of $\gamma$. Note that this segment of $\gamma$ may
intersect red curves, and so after applying the wave move, new squares may
emerge on $\Sigma^{*}$ corresponding to the intersection of $b_{i}$ and these
red curves, see Figure 20.

$\gamma$$\gamma$$r_{1}$$b_{2}$$r_{1}$$b_{1}$$b_{2}$$r_{1}$$b_{2}$$r_{1}$$b_{1}$$b_{1}$$b_{2}$$\gamma$$r_{1}$$b_{1}$$b_{2}$$r_{1}$$b_{2}$$b_{1}$$r_{1}$$b_{2}$Portion
of original Heegaard diagram. Disk tiled by squares attached along $\gamma$ in
$\mathscr{C}$. Heegaard diagram after wave move applied to $b_{1}$ along
segment of $\gamma$. Disk attached along $\gamma$ in $\mathscr{C}$ now has
fewer squares. Several wave moves applied so that Heegaard diagram no longer
intersects $\gamma$. Here, $\gamma$ is pinched in $\mathscr{C}$ and does not
bound a disk of positive area. Figure 20: Before and after several wave moves
have been applied. Heegaard diagram curves have been oriented arbitrarily so
that intersection with $\gamma$ can be seen more easily. Note that in this
case, one new intersection is created in the Heegaard diagram, between $r_{1}$
an $b_{1}$.

Continue in this way, applying wave moves until $\gamma$ does not intersect
any Heegaard diagram curves. At this point $\gamma$ is pinched in
$\Sigma^{*}$. Further, any new squares added to $\Sigma^{*}$ that remain after
this process correspond to pairs of Heegaard diagram curves whose intersection
points with $\gamma$ are linked. Thus the new squares are in bijection with a
subset of the squares in the disk $\gamma$ bounded in $\mathscr{C}$ before the
modifications, see Figure 20. (While this is not always the case, there are
instances where strictly fewer new squares are created. One such example is
seen in Figure 20)

We claim that the segments of the Heegaard diagram curves that were deleted in
these wave moves must have had some intersections with other diagram curves.
Thus, the total number of squares in $\mathscr{C}$ has been reduced by this
move. This is proven as follows:

Begin by assuming the deleted arcs of Heegaard diagram curves had no
intersections. Orient $\gamma$ and choose a base point away from any
intersections with diagram curves. Take closed tubular neighbourhoods of the
arcs of $b_{1},\ldots b_{n}$ and $r_{1},\ldots,r_{m}$ that are deleted in the
process above. Call them $\tau_{1}\ldots\tau_{n}$ and $\nu_{1},\ldots\nu_{m}$
respectively. Choose these neighbourhoods so that they do not intersect the
base point of $\gamma$, and so that the boundaries of $\tau_{1}\ldots\tau_{n}$
and $\nu_{1},\ldots\nu_{m}$ that run alongside $b_{1},\ldots b_{n}$ and
$r_{1},\ldots,r_{m}$ do not intersect any other curves in the original
Heegaard diagram. This is possible because of our assumption that the arcs of
$b_{1},\ldots b_{n}$ and $r_{1},\ldots,r_{m}$ that are deleted do not have any
intersections.

Consider a new curve $\eta$ given by beginning at the base point of $\gamma$
and following $\gamma$, until the first boundary of $\tau_{1}\ldots\tau_{n}$
or $\nu_{1},\ldots\nu_{m}$ is intersected. Then follow that boundary until it
intersects $\gamma$ again. At this point follow $\gamma$ again until another
boundary of $\tau_{1}\ldots\tau_{n}$ and $\nu_{1},\ldots\nu_{m}$ is
intersected. Then follow this boundary until it returns to $\gamma$. Continue
in this way, see Figure 21.

The process ends if $\eta$ returns to the base point of $\gamma$. There are
only finitely many boundary arcs of $\tau_{1}\ldots\tau_{n}$ and
$\nu_{1},\ldots\nu_{m}$, and segments of $\gamma$ between these. If the
process does not end, $\eta$ must eventually intersect itself somewhere other
than the base point. This is not possible because, for each arc in $\eta$,
there is a unique arc that could possibly precede it. Therefore, the process
ends and $\eta$ is a simple closed curve.

Further, we have assumed that the boundaries of $\tau_{1}\ldots\tau_{n}$ and
$\nu_{1},\ldots\nu_{m}$ do not intersect any Heegaard diagram curves, and so
$\eta$ does not intersect any of the curves of the original Heegaard diagram.
Therefore, before the modifications, $\eta$ is a curve that must be pinched in
$\Sigma^{*}$, or bound a disk in $\Sigma_{g}$.

Depending on the choice of base point on $\gamma$, there are many choices for
$\eta$. As long as the deleted arcs of $b_{1},\ldots b_{n}$ and
$r_{1},\ldots,r_{m}$ are not all on the same side of $\gamma$, it is possible
to choose the base point so that $\eta$ intersects $\gamma$ transversally, see
Figure 21. If $\eta$ is pinched, this is a contradiction since the pinched
curves and the curves along which disks are glued in to form $\mathscr{C}$ are
disjoint. If $\eta$ bounds a disk in $\Sigma_{g}$, this disk must not contain
any Heegard diagram curves. However, by construction there are Heegaard
diagram curves on either side of $\eta$, a contradiction.

Figure 21: By choosing the base point between two deleted arcs of Heegaard
diagram curves that are on opposite sides of $\gamma$, $\eta$ intersects
$\gamma$ transversally (figure on the left). Otherwise, this may not be the
case (figure on the right).

In the case where the deleted arcs of $b_{1},\ldots b_{n}$ and
$r_{1},\ldots,r_{m}$ are all on the same side of $\gamma$, the assumption that
they have no intersections imply that $\gamma$ bounds a disk on this side in
$\Sigma^{*}$ (before attaching disks to form $\mathscr{C}$!). This is because,
by definition of $\Sigma^{*}$, the Heegaard diagrams on each surface subspace
of $\Sigma^{*}$ are filling. This is a contradiction because $\gamma$ must be
an essential curve in $\Sigma^{*}$.

In either case, the deleted arcs of $b_{1},\ldots b_{n}$ and
$r_{1},\ldots,r_{m}$ must have had some intersections. This translates to
squares being deleted in $\Sigma^{*}$ when the wave moves are applied. Looking
to $\mathscr{C}$, we see that some squares may be added, but these are in
bijection with a subset of the squares that are deleted. The above now
guarantees that this is a proper subset, and so, applying these wave moves
reduces the area of $\mathscr{C}$. ∎

######  Corollary 5.12.

Given a $3$-manifold $M$ that does not have $S^{1}\times S^{2}$ as a connect
summand, the minimum-area augmented Heegaard diagram over all possible
augmented Heegaard diagrams describing $M$ corresponds to a Heegaard splitting
that is not stabilized.

###### Proof.

Given a Heegaard splitting of $M$, the minimum-area augmented Heegaard diagram
$\mathscr{C}$ for this splitting is equal to $\Sigma^{*}$ and corresponds to a
simple Heegaard diagram. If the Heegaard splitting was stabilized, one of the
surface subspaces $X$ of $\Sigma^{*}$ must be a surface of genus one that
forms an augmented Heegaard diagram for $S^{3}$. Consider collapsing this
subsurface to a point to make a new pinched surface. This new pinched surface
must be an augmented Heegaard diagram for a Heegaard splitting of $M$ as well,
since it has the same set of connect summands that are not $S^{3}$. However,
the area of the new augmented Heegaard diagram must be strictly less than the
original augmented Heegaard diagram. This implies that the original augmented
Heegaard diagram was not of minimum area across all possible augmented
Heegaard diagrams describing $M$, a contradiction. Thus the original Heegaard
splitting must have been stabilized.

∎

##  6 Future directions

While the previous section gives a partial answer to Question 5.8, it would be
great if one could find a more explicit description of augmented Heegaard
diagrams, as complexity of the Guirardel core, as opposed to using the
Stallings map. This would allow results and methods for Guirardel cores to be
applied to Heegaard splittings more readily.

This paper connects several different structures: Heegaard diagrams, Stallings
maps, virtually special square complexes, Guirardel cores of a surface group
acting on a pair of trees. One natural question is how the properties of the
$3$-manifold all these structures describe is manifested.

######  Question 6.1.

What methods and theorems from the study of Guirardel cores and virtually
special square complexes be used to prove statements about $3$-manifolds,
using the connection provided by Theorems 5.4, 5.7 and 5.11

##  7 Example: Lens spaces

###  7.1 Genus 1

Consider genus 1 Heegaard splittings. These are lens spaces. They are all
prime, and $S^{2}\times S^{1}$ is the only lens space that is reducible. This
is because any essential simple closed curve on the two-dimensional torus does
not separate the surface. Therefore the existence of a curve that bounds a
disk in both handle bodies implies the existence of a two-dimensional sphere
embedded in the lens space that does not separate. It is well-known that an
embedded, non-separating $2$-sphere implies that $S^{2}\times S^{1}$ is a
connect summand of the manifold. This is a contradiction, as we have excluded
$S^{2}\times S^{1}$ in our hypothesis and lens spaces are prime. Therefore,