# The geometry of the disk complex

Howard Masur  Department of Mathematics
University of Chicago
Chicago, Illinois 60637<EMAIL_ADDRESS>and  Saul Schleimer  Department of
Mathematics
University of Warwick
Coventry, CV4 7AL, UK<EMAIL_ADDRESS>
###### Abstract.

We give a distance estimate for the metric on the disk complex and show that
it is Gromov hyperbolic. As another application of our techniques, we find an
algorithm which computes the Hempel distance of a Heegaard splitting, up to an
error depending only on the genus.

This work is in the public domain.

###### Contents

  1. 1 Introduction
  2. 2 Background on complexes
  3. 3 Background on coarse geometry
  4. 4 Natural maps
  5. 5 Holes in general and the lower bound on distance
  6. 6 Holes for the non-orientable surface
  7. 7 Holes for the arc complex
  8. 8 Background on three-manifolds
  9. 9 Holes for the disk complex
  10. 10 Holes for the disk complex – annuli
  11. 11 Holes for the disk complex – compressible
  12. 12 Holes for the disk complex – incompressible
  13. 13 Axioms for combinatorial complexes
  14. 14 Partition and the upper bound on distance
  15. 15 Background on Teichmüller space
  16. 16 Paths for the non-orientable surface
  17. 17 Paths for the arc complex
  18. 18 Background on train tracks
  19. 19 Paths for the disk complex
  20. 20 Hyperbolicity
  21. 21 Coarsely computing Hempel distance

##  1\. Introduction

In this paper we initiate the study of the geometry of the disk complex of a
handlebody $V$. The disk complex $\mathcal{D}(V)$ has a natural simplicial
inclusion into the curve complex $\mathcal{C}(S)$ of the boundary of the
handlebody. Surprisingly, this inclusion is not a quasi-isometric embedding;
there are disks which are close in the curve complex yet very far apart in the
disk complex. As we will show, any obstruction to joining such disks via a
short path is a topologically meaningful subsurface of $S=\partial V$. We call
such subsurfaces holes. A path in the disk complex must travel into and then
out of these holes; paths in the curve complex may skip over a hole by using
the vertex representing the boundary of the subsurface. We classify the holes:

######  Theorem 1.1.

Suppose $V$ is a handlebody. If $X\subset\partial V$ is a hole for the disk
complex $\mathcal{D}(V)$ of diameter at least $61$ then:

  * •

$X$ is not an annulus.

  * •

If $X$ is compressible then there are disks $D,E$ with boundary contained in
$X$ so that the boundaries fill $X$.

  * •

If $X$ is incompressible then there is an $I$-bundle $\rho_{F}\colon T\to F$
so that $T$ is a component of $V{\smallsetminus}\partial_{v}T$ and $X$ is a
component of $\partial_{h}T$.

See Theorems 10.1, 11.6 and 12.1 for more precise statements. The $I$–bundles
appearing in the classification lead us to study the arc complex
$\mathcal{A}(F)$ of the base surface $F$. Since the $I$–bundle $T$ may be
twisted the surface $F$ may be non-orientable.

Thus, as a necessary warm-up to the difficult case of the disk complex, we
also analyze the holes for the curve complex of an non-orientable surface, as
well as the holes for the arc complex.

### Topological application

It is a long-standing open problem to decide, given a Heegaard diagram,
whether the underlying splitting surface is reducible. This question has deep
connections to the geometry, topology, and algebra of the ambient three-
manifold. For example, a resolution of this problem would give new solutions
to both the three-sphere recognition problem and the triviality problem for
three-manifold groups. The difficulty of deciding reducibility is underlined
by its connection to the Poincaré conjecture: several approaches to the
Poincaré Conjecture fell at essentially this point. See [10] for a entrance
into the literature.

One generalization of deciding reducibility is to find an algorithm that,
given a Heegaard diagram, computes the distance of the Heegaard splitting as
defined by Hempel [20]. (For example, see [5, Section 2].) The classification
of holes for the disk complex leads to a coarse answer to this question.

In every genus $g$ there is a constant $K=K(g)$ and an algorithm that, given a
Heegaard diagram, computes the distance of the Heegaard splitting with error
at most $K$.

In addition to the classification of holes, the algorithm relies on the Gromov
hyperbolicity of the curve complex [24] and the quasi-convexity of the disk
set inside of the curve complex [26]. However the algorithm does not depend on
our geometric applications of Theorem 1.1.

### Geometric application

The hyperbolicity of the curve complex and the classification of holes allows
us to prove:

The disk complex is Gromov hyperbolic.

Again, as a warm-up to the proof of Theorem 20.3 we prove that
$\mathcal{C}(F)$ and $\mathcal{A}(S)$ are hyperbolic in Corollary 6.4 and
Theorem 20.2. Note that Bestvina and Fujiwara [4] have previously dealt with
the curve complex of a non-orientable surface, following Bowditch [6].

These results cannot be deduced from the fact that $\mathcal{D}(V)$,
$\mathcal{C}(F)$, and $\mathcal{A}(S)$ can be realized as quasi-convex subsets
of $\mathcal{C}(S)$. This is because the curve complex is locally infinite. As
simple example consider the Cayley graph of $\mathbb{Z}^{2}$ with the standard
generating set. Then the cone $C(\mathbb{Z}^{2})$ of height one-half is a
Gromov hyperbolic space and $\mathbb{Z}^{2}$ is a quasi-convex subset. Another
instructive example, very much in-line with our work, is the usual embedding
of the three-valent tree $T_{3}$ into the Farey tessellation.

The proof of Theorem 20.3 requires the distance estimate Theorem 19.1: the
distance in $\mathcal{C}(F)$, $\mathcal{A}(S)$, and $\mathcal{D}(V)$ is
coarsely equal to the sum of subsurface projection distances in holes.
However, we do not use the hierarchy machine introduced in [25]. This is
because hierarchies are too flexible to respect a symmetry, such as the
involution giving a non-orientable surface, and at the same time too rigid for
the disk complex. For $\mathcal{C}(F)$ we use the highly rigid Teichmüller
geodesic machine, due to Rafi [33]. For $\mathcal{D}(V)$ we use the extremely
flexible train track machine, developed by ourselves and Mosher [27].

Theorems 19.1 and 20.3 are part of a more general framework. Namely, given a
combinatorial complex $\mathcal{G}$ we understand its geometry by classifying
the holes: the geometric obstructions lying between $\mathcal{G}$ and the
curve complex. In Sections 13 and 14 we show that any complex $\mathcal{G}$
satisfying certain axioms necessarily satisfies a distance estimate. That
hyperbolicity follows from the axioms is proven in Section 20.

Our axioms are stated in terms of a path of markings, a path in the the
combinatorial complex, and their relationship. For the disk complex the
combinatorial paths are surgery sequences of essential disks while the marking
paths are provided by train track splitting sequences; both constructions are
due to the first author and Minsky [26] (Section 18). The verification of the
axioms (Section 19) relies on our work with Mosher, analyzing train track
splitting sequences in terms of subsurface projections [27].

We do not study non-orientable surfaces directly; instead we focus on
symmetric multicurves in the double cover. This time marking paths are
provided by Teichmüller geodesics, using the fact that the symmetric Riemann
surfaces form a totally geodesic subset of Teichmüller space. The
combinatorial path is given by the systole map. We use results of Rafi [33] to
verify the axioms for the complex of symmetric curves. (See Section 16.)
Section 17 verifies the axioms for the arc complex again using Teichmüller
geodesics and the systole map. It is interesting to note that the axioms for
the arc complex can also be verified using hierarchies or, indeed, train track
splitting sequences.

The distance estimates for the marking graph and the pants graph, as given by
the first author and Minsky [25], inspired the work here, but do not fit our
framework. Indeed, neither the marking graph nor the pants graph are Gromov
hyperbolic. It is crucial here that all holes interfere; this leads to
hyperbolicity. When there are non-interfering holes, it is unclear how to
partition the marking path to obtain the distance estimate.

### Acknowledgments

We thank Jason Behrstock, Brian Bowditch, Yair Minsky, Lee Mosher, Hossein
Namazi, and Kasra Rafi for many enlightening conversations.

We thank Tao Li for pointing out that our original bound inside of Theorem
12.1 of $O(\log g(V))$ could be reduced to a constant.

##  2\. Background on complexes

We use $S_{g,b,c}$ to denote the compact connected surface of genus $g$ with
$b$ boundary components and $c$ cross-caps. If the surface is orientable we
omit the subscript $c$ and write $S_{g,b}$. The complexity of $S=S_{g,b}$ is
$\xi(S)=3g-3+b$. If the surface is closed and orientable we simply write
$S_{g}$.

###  2.1. Arcs and curves

A simple closed curve $\alpha\subset S$ is essential if $\alpha$ does not
bound a disk in $S$. The curve $\alpha$ is non-peripheral if $\alpha$ is not
isotopic to a component of $\partial S$. A simple arc $\beta\subset S$ is
proper if $\beta\cap\partial S=\partial\beta$. An isotopy of $S$ is proper if
it preserves the boundary setwise. A proper arc $\beta\subset S$ is essential
if $\beta$ is not properly isotopic into a regular neighborhood of $\partial
S$.

Define $\mathcal{C}(S)$ to be the set of isotopy classes of essential, non-
peripheral curves in $S$. Define $\mathcal{A}(S)$ to be the set of proper
isotopy classes of essential arcs. When $S=S_{0,2}$ is an annulus define
$\mathcal{A}(S)$ to be the set of essential arcs, up to isotopies fixing the
boundary pointwise. For any surface define
$\mathcal{AC}(S)=\mathcal{A}(S)\cup\mathcal{C}(S)$.

For $\alpha,\beta\in\mathcal{AC}(S)$ the geometric intersection number
$\iota(\alpha,\beta)$ is the minimum intersection possible between $\alpha$
and any $\beta^{\prime}$ equivalent to $\beta$. When $S=S_{0,2}$ we do not
count intersection points occurring on the boundary. If $\alpha$ and $\beta$
realize their geometric intersection number then $\alpha$ is tight with
respect to $\beta$. If they do not realize their geometric intersection then
we may tighten $\beta$ until they do.

Define $\Delta\subset\mathcal{AC}(S)$ to be a multicurve if for all
$\alpha,\beta\in\Delta$ we have $\iota(\alpha,\beta)=0$. Following Harvey [18]
we may impose the structure of a simplical complex on $\mathcal{AC}(S)$: the
simplices are exactly the multicurves. Also, $\mathcal{C}(S)$ and
$\mathcal{A}(S)$ naturally span sub-complexes.

Note that the curve complexes $\mathcal{C}(S_{1,1})$ and
$\mathcal{C}(S_{0,4})$ have no edges. It is useful to alter the definition in
these cases. Place edges between all vertices with geometric intersection
exactly one if $S=S_{1,1}$ or two if $S=S_{0,4}$. In both cases the result is
the Farey graph. Also, with the current definition $\mathcal{C}(S)$ is empty
if $S=S_{0,2}$. Thus for the annulus only we set
$\mathcal{AC}(S)=\mathcal{C}(S)=\mathcal{A}(S)$.

######  Definition 2.1.

For vertices $\alpha,\beta\in\mathcal{C}(S)$ define the distance
$d_{S}(\alpha,\beta)$ to be the minimum possible number of edges of a path in
the one-skeleton $\mathcal{C}^{1}(S)$ which starts at $\alpha$ and ends at
$\beta$.

Note that if $d_{S}(\alpha,\beta)\geq 3$ then $\alpha$ and $\beta$ fill the
surface $S$. We denote distance in the one-skeleton of $\mathcal{A}(S)$ and of
$\mathcal{AC}(S)$ by $d_{\mathcal{A}}$ and $d_{\mathcal{AC}}$ respectively.
Recall that the geometric intersection of a pair of curves gives an upper
bound for their distance.

######  Lemma 2.2.

Suppose that $S$ is a compact connected surface which is not an annulus. For
any $\alpha,\beta\in\mathcal{C}^{0}(S)$ with $\iota(\alpha,\beta)>0$ we have
$d_{S}(\alpha,\beta)\leq 2\log_{2}(\iota(\alpha,\beta))+2$. ∎

This form of the inequality, stated for closed orientable surfaces, may be
found in [20]. A proof in the bounded orientable case is given in [36]. The
non-orientable case is then an exercise. When $S=S_{0,2}$ an induction proves

(2.3) $d_{X}(\alpha,\beta)=1+\iota(\alpha,\beta)$

for distinct vertices $\alpha,\beta\in\mathcal{C}(X)$. See [25, Equation 2.3].

###  2.2. Subsurfaces

Suppose that $X\subset S$ is a connected compact subsurface. We say $X$ is
essential exactly when all boundary components of $X$ are essential in $S$. We
say that $\alpha\in\mathcal{AC}(S)$ cuts $X$ if all representatives of
$\alpha$ intersect $X$. If some representative is disjoint then we say
$\alpha$ misses $X$.

######  Definition 2.4.

An essential subsurface $X\subset S$ is cleanly embedded if for all components
$\delta\subset\partial X$ we have: $\delta$ is isotopic into $\partial S$ if
and only if $\delta$ is equal to a component of $\partial S$.

######  Definition 2.5.

Suppose $X,Y\subset S$ are essential subsurfaces. If $X$ is cleanly embedded
in $Y$ then we say that $X$ is nested in $Y$. If $\partial X$ cuts $Y$ and
also $\partial Y$ cuts $X$ then we say that $X$ and $Y$ overlap.

A compact connected surface $S$ is simple if $\mathcal{AC}(S)$ has finite
diameter.

######  Lemma 2.6.

Suppose $S$ is a connected compact surface. The following are equivalent:

  * •

$S$ is not simple.

  * •

The diameter of $\mathcal{AC}(S)$ is at least five.

  * •

$S$ admits an ending lamination or $S=S_{1}$ or $S_{0,2}$.

  * •

$S$ admits a pseudo-Anosov map or $S=S_{1}$ or $S_{0,2}$.

  * •

$\chi(S)<-1$ or $S=S_{1,1},S_{1},S_{0,2}$.

Lemma 4.6 of [24] shows that pseudo-Anosov maps have quasi-geodesic orbits,
when acting on the associated curve complex. A Dehn twist acting on
$\mathcal{C}(S_{0,2})$ has geodesic orbits.

Note that Lemma 2.6 is only used in this paper when $\partial S$ is non-empty.
The closed case is included for completeness.

###### Proof sketch of Lemma 2.6.

If $S$ admits a pseudo-Anosov map then the stable lamination is an ending
lamination. If $S$ admits a filling lamination then, by an argument of
Kobayashi [21], $\mathcal{AC}(S)$ has infinite diameter. (This argument is
also sketched in [24], page 124, after the statement of Proposition 4.6.)

If the diameter of $\mathcal{AC}$ is infinite then the diameter is at least
five. To finish, one may check directly that all surfaces with $\chi(S)>-2$,
other than $S_{1,1}$, $S_{1}$ and the annulus have $\mathcal{AC}(S)$ with
diameter at most four. (The difficult cases, $S_{012}$ and $S_{003}$, are
discussed by Scharlemann [35].) Alternatively, all surfaces with $\chi(S)<-1$,
and also $S_{1,1}$, admit pseudo-Anosov maps. The orientable cases follow from
Thurston’s construction [38]. Penner’s generalization [32] covers the non-
orientable cases. ∎

###  2.3. Handlebodies and disks

Let $V_{g}$ denote the handlebody of genus $g$: the three-manifold obtained by
taking a closed regular neighborhood of a polygonal, finite, connected graph
in $\mathbb{R}^{3}$. The genus of the boundary is the genus of the handlebody.
A properly embedded disk $D\subset V$ is essential if $\partial
D\subset\partial V$ is essential.

Let $\mathcal{D}(V)$ be the set of essential disks $D\subset V$, up to proper
isotopy. A subset $\Delta\subset\mathcal{D}(V)$ is a multidisk if for every
$D,E\in\Delta$ we have $\iota(\partial D,\partial E)=0$. Following McCullough
[28] we place a simplical structure on $\mathcal{D}(V)$ by taking multidisks
to be simplices. As with the curve complex, define $d_{\mathcal{D}}$ to be the
distance in the one-skeleton of $\mathcal{D}(V)$.

###  2.4. Markings

A finite subset $\mu\subset\mathcal{AC}(S)$ fills $S$ if for all
$\beta\in\mathcal{C}(S)$ there is some $\alpha\in\mu$ so that
$\iota(\alpha,\beta)>0$. For any pair of finite subsets
$\mu,\nu\subset\mathcal{AC}(S)$ we extend the intersection number:

$\iota(\mu,\nu)=\sum_{\alpha\in\mu,\beta\in\nu}\iota(\alpha,\beta).$

We say that $\mu,\nu$ are $L$–close if $\iota(\mu,\nu)\leq L$. We say that
$\mu$ is a $K$–marking if $\iota(\mu,\mu)\leq K$. For any $K,L$ we may define
$\mathcal{M}_{K,L}(S)$ to be the graph where vertices are filling $K$–markings
and edges are given by $L$–closeness.

As defined in [25] we have:

######  Definition 2.7.

A complete clean marking $\mu=\\{\alpha_{i}\\}\cup\\{\beta_{i}\\}$ consists of

  * •

A collection of base curves $\operatorname{base}(\mu)=\\{\alpha_{i}\\}$: a
maximal simplex in $\mathcal{C}(S)$.

  * •

A collection of transversal curves $\\{\beta_{i}\\}$: for each $i$ define
$X_{i}=S{\smallsetminus}\bigcup_{j\neq i}\alpha_{j}$ and take
$\beta_{i}\in\mathcal{C}(X_{i})$ to be a Farey neighbor of $\alpha_{i}$.

If $\mu$ is a complete clean marking then $\iota(\mu,\mu)\leq
2\xi(S)+6\chi(S)$. As discussed in [25] there are two kinds of elementary
moves which connected markings. There is a twist about a pants curve $\alpha$,
replacing its transversal $\beta$ by a new transversal $\beta^{\prime}$ which
is a Farey neighbor of both $\alpha$ and $\beta$. We can flip by swapping the
roles of $\alpha_{i}$ and $\beta_{i}$. (In the case of the flip move, some of
the other transversals must be cleaned.)

It follows that for any surface $S$ there are choices of $K,L$ so that
$\mathcal{M}(S)$ is non-empty and connected. We use $d_{\mathcal{M}}(\mu,\nu)$
to denote distance in the marking graph.

##  3\. Background on coarse geometry

Here we review a few ideas from coarse geometry. See [8], [12], or [15] for a
fuller discussion.

###  3.1. Quasi-isometry

Suppose $r,s,A$ are non-negative real numbers, with $A\geq 1$. If $s\leq
A\cdot r+A$ then we write $s\mathbin{\leq_{A}}r$. If $s\mathbin{\leq_{A}}r$
and $r\mathbin{\leq_{A}}s$ then we write $s\mathbin{=_{A}}r$ and call $r$ and
$s$ quasi-equal with constant $A$. We also define the cut-off function
$[r]_{c}$ where $[r]_{c}=0$ if $r<c$ and $[r]_{c}=r$ if $r\geq c$.

Suppose that $(\mathcal{X},d_{\mathcal{X}})$ and
$(\mathcal{Y},d_{\mathcal{Y}})$ are metric spaces. A relation
$f\colon\mathcal{X}\to\mathcal{Y}$ is an $A$–quasi-isometric embedding for
$A\geq 1$ if, for every $x,y\in\mathcal{X}$,

$d_{\mathcal{X}}(x,y)\mathbin{=_{A}}d_{\mathcal{Y}}(f(x),f(y)).$

The relation $f$ is a quasi-isometry, and $\mathcal{X}$ is quasi-isometric to
$\mathcal{Y}$, if $f$ is an $A$–quasi-isometric embedding and the image of $f$
is $A$–dense: the $A$–neighborhood of the image equals all of $\mathcal{Y}$.

###  3.2. Geodesics

Fix an interval $[u,v]\subset\mathbb{R}$. A geodesic, connecting $x$ to $y$ in
$\mathcal{X}$, is an isometric embedding $f\colon[u,v]\to\mathcal{X}$ with
$f(u)=x$ and $f(v)=y$. Often the exact choice of $f$ is unimportant and all
that matters are the endpoints $x$ and $y$. We then denote the image of $f$ by
$[x,y]\subset\mathcal{X}$.

Fix now intervals $[m,n],[p,q]\subset\mathbb{Z}$. An $A$–quasi-isometric
embedding $g\colon[m,n]\to\mathcal{X}$ is called an $A$–quasi-geodesic in
$\mathcal{X}$. A function $g\colon[m,n]\to\mathcal{X}$ is an
$A$–unparameterized quasi-geodesic in $\mathcal{X}$ if

  * •

there is an increasing function $\rho\colon[p,q]\to[m,n]$ so that
$g\circ\rho\colon[p,q]\to\mathcal{X}$ is an $A$–quasi-geodesic in
$\mathcal{X}$ and

  * •

for all $i\in[p,q-1]$,
$\operatorname{diam}_{\mathcal{X}}\left(g\left[\rho(i),\rho(i+1)\right]\right)\leq
A$.

(Compare to the definition of $(K,\delta,s)$–quasi-geodesics found in [24].)

A subset $\mathcal{Y}\subset\mathcal{X}$ is $Q$–quasi-convex if every
$\mathcal{X}$–geodesic connecting a pair of points of $\mathcal{Y}$ lies
within a $Q$–neighborhood of $\mathcal{Y}$.

###  3.3. Hyperbolicity

We now assume that $\mathcal{X}$ is a connected graph with metric induced by
giving all edges length one.

######  Definition 3.1.

The space $\mathcal{X}$ is $\delta$–hyperbolic if, for any three points
$x,y,z$ in $\mathcal{X}$ and for any geodesics $k=[x,y]$, $g=[y,z]$,
$h=[z,x]$, the triangle $ghk$ is $\delta$–slim: the $\delta$–neighborhood of
any two sides contains the third.

An important tool for this paper is the following theorem of the first author
and Minsky [24]:

######  Theorem 3.2.

The curve complex of an orientable surface is Gromov hyperbolic. ∎

For the remainder of this section we assume that $\mathcal{X}$ is
$\delta$–hyperbolic graph, $x,y,z\in\mathcal{X}$ are points, and
$k=[x,y],g=[y,z],h=[z,x]$ are geodesics.

######  Definition 3.3.

We take $\rho_{k}\colon\mathcal{X}\to k$ to be the closest points relation:

$\rho_{k}(z)=\big{\\{}w\in k\mathbin{\mid}\mbox{ for all $v\in k$,
$d_{\mathcal{X}}(z,w)\leq d_{\mathcal{X}}(z,v)$ }\big{\\}}.$

We now list several lemmas useful in the sequel.

######  Lemma 3.4.

There is a point on $g$ within distance $2\delta$ of $\rho_{k}(z)$. The same
holds for $h$. ∎

######  Lemma 3.5.

The closest points $\rho_{k}(z)$ have diameter at most $4\delta$. ∎

######  Lemma 3.6.

The diameter of $\rho_{g}(x)\cup\rho_{h}(y)\cup\rho_{k}(z)$ is at most
$6\delta$. ∎

######  Lemma 3.7.

Suppose that $z^{\prime}$ is another point in $\mathcal{X}$ so that
$d_{\mathcal{X}}(z,z^{\prime})\leq R$. Then
$d_{\mathcal{X}}(\rho_{k}(z),\rho_{k}(z^{\prime}))\leq R+6\delta.$ ∎

######  Lemma 3.8.

Suppose that $k^{\prime}$ is another geodesic in $X$ so that the endpoints of
$k^{\prime}$ are within distance $R$ of the points $x$ and $y$. Then
$d_{X}(\rho_{k}(z),\rho_{k^{\prime}}(z))\leq R+11\delta$. ∎

We now turn to a useful consequence of the Morse stability of quasi-geodesics
in hyperbolic spaces.

######  Lemma 3.9.

For every $\delta$ and $A$ there is a constant $C$ with the following
property: If $\mathcal{X}$ is $\delta$–hyperbolic and
$g\colon[0,N]\to\mathcal{X}$ is an $A$–unparameterized quasi-geodesic then for
any $m<n<p$ in $[0,N]$ we have:

$d_{\mathcal{X}}(x,y)+d_{\mathcal{X}}(y,z)<d_{\mathcal{X}}(x,z)+C$

where $x,y,z=g(m),g(n),g(p)$. ∎

###  3.4. A hyperbolicity criterion

Here we give a hyperbolicity criterion tailored to our setting. We thank Brian
Bowditch for both finding an error in our first proof of Theorem 3.11 and for
informing us of Gilman’s work [13, 14].

Suppose that $\mathcal{X}$ is a graph with all edge-lengths equal to one.
Suppose that $\gamma\colon[0,N]\to\mathcal{X}$ is a loop in $\mathcal{X}$ with
unit speed. Any pair of points $a,b\in[0,N]$ gives a chord of $\gamma$. If
$a<b$, $N/4\leq b-a$ and $N/4\leq a+(N-b)$ then the chord is $1/4$–separated.
The length of the chord is $d_{\mathcal{X}}(\gamma(a),\gamma(b))$.

Following Gilman [13, Theorem B] we have:

######  Theorem 3.10.

Suppose that $\mathcal{X}$ is a graph with all edge-lengths equal to one. Then
$\mathcal{X}$ is Gromov hyperbolic if and only if there is a constant $K$ so
that every loop $\gamma\colon[0,N]\to\mathcal{X}$ has a $1/4$–separated chord
of length at most $N/7+K$. ∎

Gilman’s proof goes via the subquadratic isoperimetric inequality. We now give
our criterion, noting that it is closely related to another paper of Gilman
[14].

######  Theorem 3.11.

Suppose that $\mathcal{X}$ is a graph with all edge-lengths equal to one. Then
$\mathcal{X}$ is Gromov hyperbolic if and only if there is a constant $M\geq
0$ and, for all unordered pairs $x,y\in\mathcal{X}^{0}$, there is a connected
subgraph $g_{x,y}$ containing $x$ and $y$ with the following properties:

  * •

(Local) If $d_{\mathcal{X}}(x,y)\leq 1$ then $g_{x,y}$ has diameter at most
$M$.

  * •

(Slim triangles) For all $x,y,z\in\mathcal{X}^{0}$ the subgraph $g_{x,y}$ is
contained in an $M$–neighborhood of $g_{y,z}\cup g_{z,x}$.

###### Proof.

Suppose that $\gamma\colon[0,N]\to\mathcal{X}$ is a loop. If $\epsilon$ is the
empty string let $I_{\epsilon}=[0,N]$. For any binary string $\omega$ let
$I_{\omega 0}$ and $I_{\omega 1}$ be the first and second half of
$I_{\omega}$. Note that if $|\omega|\geq\lceil\log_{2}N\rceil$ then
$|I_{\omega}|\leq 1$.

Fix a string $\omega$ and let $[a,b]=I_{\omega}$. Let $g_{\omega}$ be the
subgraph connecting $\gamma(a)$ to $\gamma(b)$. Note that $g_{0}=g_{1}$
because $\gamma(0)=\gamma(N)$. Also, for any binary string $\omega$ the
subgraphs $g_{\omega},g_{\omega 0},g_{\omega 1}$ form an $M$–slim triangle. If
$|\omega|\leq\lceil\log_{2}N\rceil$ then every $x\in g_{\omega}$ has some
point $b\in I_{\omega}$ so that

$d_{\mathcal{X}}(x,\gamma(b))\leq M(\lceil\log_{2}N\rceil-|\omega|)+2M.$

Since $g_{0}$ is connected there is a point $x\in g_{0}$ that lies within the
$M$–neighborhoods both of $g_{00}$ and of $g_{01}$. Pick some $b\in I_{1}$ so
that $d_{\mathcal{X}}(x,\gamma(b))$ is bounded as in the previous paragraph.
It follows that there is a point $a\in I_{0}$ so that $a,b$ are
$1/4$–separated and so that

$d_{\mathcal{X}}(\gamma(a),\gamma(b))\leq 2M\lceil\log_{2}N\rceil+2M.$

Thus there is an additive error $K$ large enough so that $\mathcal{X}$
satisfies the criterion of Theorem 3.10 and we are done. ∎

##  4\. Natural maps

There are several natural maps between the complexes and graphs defined in
Section 2. Here we review what is known about their geometric properties, and
give examples relevant to the rest of the paper.

###  4.1. Lifting, surgery, and subsurface projection

Suppose that $S$ is not simple. Choose a hyperbolic metric on the interior of
$S$ so that all ends have infinite areas. Fix a compact essential subsurface
$X\subset S$ which is not a peripheral annulus. Let $S^{X}$ be the cover of
$S$ so that $X$ lifts homeomorphically and so that
$S^{X}\mathrel{\cong}{\operatorname{interior}}(X)$. For any
$\alpha\in\mathcal{AC}(S)$ let $\alpha^{X}$ be the full preimage.

Since there is a homeomorphism between $X$ and the Gromov compactification of
$S^{X}$ in a small abuse of notation we identify $\mathcal{AC}(X)$ with the
arc and curve complex of $S^{X}$.

######  Definition 4.1.

We define the cutting relation
$\kappa_{X}\colon\mathcal{AC}(S)\to\mathcal{AC}(X)$ as follows:
$\alpha^{\prime}\in\kappa_{X}(\alpha)$ if and only if $\alpha^{\prime}$ is an
essential non-peripheral component of $\alpha^{X}$.

Note that $\alpha$ cuts $X$ if and only if $\kappa_{X}(\alpha)$ is non-empty.
Now suppose that $S$ is not an annulus.

######  Definition 4.2.

We define the surgery relation
$\sigma_{X}\colon\mathcal{AC}(S)\to\mathcal{C}(S)$ as follows:
$\alpha^{\prime}\in\sigma_{S}(\alpha)$ if and only if
$\alpha^{\prime}\in\mathcal{C}(S)$ is a boundary component of a regular
neighborhood of $\alpha\cup\partial S$.

With $S$ and $X$ as above:

######  Definition 4.3.

The subsurface projection relation
$\pi_{X}\colon\mathcal{AC}(S)\to\mathcal{C}(X)$ is defined as follows: If $X$
is not an annulus then define $\pi_{X}=\sigma_{X}\circ\kappa_{X}$. When $X$ is
an annulus $\pi_{X}=\kappa_{X}$.

If $\alpha,\beta\in\mathcal{AC}(S)$ both cut $X$ we write
$d_{X}(\alpha,\beta)=\operatorname{diam}_{X}(\pi_{X}(\alpha)\cup\pi_{X}(\beta))$.
This is the subsurface projection distance between $\alpha$ and $\beta$ in
$X$.

######  Lemma 4.4.

Suppose $\alpha,\beta\in\mathcal{AC}(S)$ are disjoint and cut $X$. Then
$\operatorname{diam}_{X}(\pi_{X}(\alpha)),d_{X}(\alpha,\beta)\leq 3$. ∎

See Lemma 2.3 of [25] and the remarks in the section Projection Bounds in
[29].

######  Corollary 4.5.

Fix $X\subset S$. Suppose that $\\{\beta_{i}\\}_{i=0}^{N}$ is a path in
$\mathcal{AC}(S)$. Suppose that $\beta_{i}$ cuts $X$ for all $i$. Then
$d_{X}(\beta_{0},\beta_{N})\leq 3N+3$. ∎

It is crucial to note that if some vertex of $\\{\beta_{i}\\}$ misses $X$ then
the projection distance $d_{X}(\beta_{0},\beta_{n})$ may be arbitrarily large
compared to $n$. Corollary 4.5 can be greatly strengthened when the path is a
geodesic [25]:

######  Theorem 4.6.

[Bounded Geodesic Image] There is constant $M_{0}$ with the following
property. Fix $X\subset S$. Suppose that $\\{\beta_{i}\\}_{i=0}^{n}$ is a
geodesic in $\mathcal{C}(S)$. Suppose that $\beta_{i}$ cuts $X$ for all $i$.
Then $d_{X}(\beta_{0},\beta_{n})\leq M_{0}$. ∎

Here is a converse for Lemma 4.4.

######  Lemma 4.7.

For every $a\in\mathbb{N}$ there is a number $b\in\mathbb{N}$ with the
following property: for any $\alpha,\beta\in\mathcal{AC}(S)$ if
$d_{X}(\alpha,\beta)\leq a$ for all $X\subset S$ then $\iota(\alpha,\beta)\leq
b$.

Corollary D of [11] gives a more precise relation between projection distance
and intersection number.

###### Proof of Lemma 4.7.

We only sketch the contrapositive: Suppose we are given a sequence of curves
$\alpha_{n},\beta_{n}$ so that $\iota(\alpha_{n},\beta_{n})$ tends to
infinity. Passing to subsequences and applying elements of the mapping class
group we may assume that $\alpha_{n}=\alpha_{0}$ for all $n$. Setting
$c_{n}=\iota(\alpha_{0},\beta_{n})$ and passing to subsequences again we may
assume that $\beta_{n}/c_{n}$ converges to $\lambda\in\mathcal{PML}(S)$, the
projectivization of Thurston’s space of measured laminations. Let $Y$ be any
connected component of the subsurface filled by $\lambda$, chosen so that
$\alpha_{0}$ cuts $Y$. Note that $\pi_{Y}(\beta_{n})$ converges to
$\lambda|_{Y}$. Again applying Kobayashi’s argument [21], the distance
$d_{Y}(\alpha_{0},\beta_{n})$ tends to infinity. ∎

###  4.2. Inclusions

We now record a well known fact:

######  Lemma 4.8.

The inclusion $\nu\colon\mathcal{C}(S)\to\mathcal{AC}(S)$ is a quasi-isometry.
The surgery map $\sigma_{S}\colon\mathcal{AC}(S)\to\mathcal{C}(S)$ is a quasi-
inverse for $\nu$.

###### Proof.

Fix $\alpha,\beta\in\mathcal{C}(S)$. Since $\nu$ is an inclusion we have
$d_{\mathcal{AC}}(\alpha,\beta)\leq d_{S}(\alpha,\beta)$. In the other
direction, let $\\{\alpha_{i}\\}_{i=0}^{N}$ be a geodesic in $\mathcal{AC}(S)$
connecting $\alpha$ to $\beta$. Since every $\alpha_{i}$ cuts $S$ we apply
Corollary 4.5 and deduce $d_{S}(\alpha,\beta)\leq 3N+3$.

Note that the composition
$\sigma_{S}\circ\nu=\operatorname{Id}|\mathcal{C}(S)$. Also, for any arc
$\alpha\in\mathcal{A}(S)$ we have
$d_{\mathcal{AC}}(\alpha,\nu(\sigma_{S}(\alpha)))=1$. Finally,
$\mathcal{C}(S)$ is $1$–dense in $\mathcal{AC}(S)$, as any arc $\gamma\subset
S$ is disjoint from the one or two curves of $\sigma_{S}(\gamma)$. ∎

Brian Bowditch raised the question, at the Newton Institute in August 2003, of
the geometric properties of the inclusion $\mathcal{A}(S)\to\mathcal{AC}(S)$.
The natural assumption, that this inclusion is again a quasi-isometric
embedding, is false. In this paper we will exactly characterize how the
inclusion distorts distance.

We now move up a dimension. Suppose that $V$ is a handlebody and $S=\partial
V$. We may take any disk $D\in\mathcal{D}(V)$ to its boundary $\partial
D\in\mathcal{C}(S)$, giving an inclusion
$\nu\colon\mathcal{D}(V)\to\mathcal{C}(S)$. It is important to distinguish the
disk complex from its image $\nu(\mathcal{D}(V))$; thus we will call the image
the disk set.

The first author and Minsky [26] have shown:

######  Theorem 4.9.

The disk set is a quasi-convex subset of the curve complex. ∎

It is natural to ask if this map is a quasi-isometric embedding. If so, the
hyperbolicity of $\mathcal{C}(V)$ immediately follows. In fact, the inclusion
again badly distorts distance and we investigate exactly how, below.

###  4.3. Markings and the mapping class group

Once the connectedness of $\mathcal{M}(S)$ is in hand, it is possible to use
local finiteness to show that $\mathcal{M}(S)$ is quasi-isometric to the
Cayley graph of the mapping class group [25].

Using subsurface projections the first author and Minsky [25] obtained a
distance estimate for the marking complex and thus for the mapping class
group.

######  Theorem 4.10.

There is a constant ${C_{0}}={C_{0}}(S)$ so that, for any $c\geq{C_{0}}$ there
is a constant $A$ with

$d_{\mathcal{M}}(\mu,\mu^{\prime})\,\,\mathbin{=_{A}}\,\,\sum[d_{X}(\mu,\mu^{\prime})]_{c}$

independent of the choice of $\mu$ and $\mu^{\prime}$. Here the sum ranges
over all essential, non-peripheral subsurfaces $X\subset S$.

This, and their similar estimate for the pants graph, is a model for the
distance estimates given below. Notice that a filling marking
$\mu\in\mathcal{M}(S)$ cuts all essential, non-peripheral subsurfaces of $S$.
It is not an accident that the sum ranges over the same set.

##  5\. Holes in general and the lower bound on distance

Suppose that $S$ is a compact connected surface. In this paper a combinatorial
complex $\mathcal{G}(S)$ will have vertices being isotopy classes of certain
multicurves in $S$. We will assume throughout that vertices of
$\mathcal{G}(S)$ are connected by edges only if there are representatives
which are disjoint. This assumption is made only to simplify the proofs — all
arguments work in the case where adjacent vertices are allowed to have
uniformly bounded intersection. In all cases $\mathcal{G}$ will be connected.
There is a natural map $\nu\colon\mathcal{G}\to\mathcal{AC}(S)$ taking a
vertex of $\mathcal{G}$ to the isotopy classes of the components. Examples in
the literature include the marking complex [25], the pants complex [9] [2],
the Hatcher-Thurston complex [19], the complex of separating curves [7], the
arc complex and the curve complexes themselves.

For any combinatorial complex $\mathcal{G}$ defined in this paper other than
the curve complex we will denote distance in the one-skeleton of $\mathcal{G}$
by $d_{\mathcal{G}}(\cdot,\cdot)$. Distance in $\mathcal{C}(S)$ will always be
denoted by $d_{S}(\cdot,\cdot)$.

###  5.1. Holes, defined

Suppose that $S$ is non-simple. Suppose that $\mathcal{G}(S)$ is a
combinatorial complex. Suppose that $X\subset S$ is an cleanly embedded
subsurface. A vertex $\alpha\in\mathcal{G}$ cuts $X$ if some component of
$\alpha$ cuts $X$.

######  Definition 5.1.

We say $X\subset S$ is a hole for $\mathcal{G}$ if every vertex of
$\mathcal{G}$ cuts $X$.

Almost equivalently, if $X$ is a hole then the subsurface projection
$\pi_{X}\colon\mathcal{G}(S)\to\mathcal{C}(X)$ never takes the empty set as a
value. Note that the entire surface $S$ is always a hole, regardless of our
choice of $\mathcal{G}$. A boundary parallel annulus cannot be cleanly
embedded (unless $S$ is also an annulus), so generally cannot be a hole. A
hole $X\subset S$ is strict if $X$ is not homeomorphic to $S$.

We now classify the holes for $\mathcal{A}(S)$.

######  Example 5.2.

Suppose that $S=S_{g,b}$ with $b>0$ and consider the arc complex
$\mathcal{A}(S)$. The holes, up to isotopy, are exactly the cleanly embedded
surfaces which contain $\partial S$. So, for example, if $S$ is planar then
only $S$ is a hole for $\mathcal{A}(S)$. The same holds for $S=S_{1,1}$. In
these cases it is an exercise to show that $\mathcal{C}(S)$ and
$\mathcal{A}(S)$ are quasi-isometric. In all other cases the arc complex
admits infinitely many holes.

######  Definition 5.3.

If $X$ is a hole and if $\pi_{X}(\mathcal{G})\subset\mathcal{C}(X)$ has
diameter at least $R$ we say that the hole $X$ has diameter at least $R$.

######  Example 5.4.

Continuing the example above: Since the mapping class group acts on the arc
complex, all non-simple holes for $\mathcal{A}(S)$ have infinite diameter.

Suppose now that $X,X^{\prime}\subset S$ are disjoint holes for $\mathcal{G}$.
In the presence of symmetry there can be a relationship between
$\pi_{X}|\mathcal{G}$ and $\pi_{X^{\prime}}|\mathcal{G}$ as follows:

######  Definition 5.5.

Suppose that $X,X^{\prime}$ are holes for $\mathcal{G}$, both of infinite
diameter. Then $X$ and $X^{\prime}$ are paired if there is a homeomorphism
$\tau\colon X\to X^{\prime}$ and a constant $L_{4}$ so that

$d_{X^{\prime}}(\pi_{X^{\prime}}(\gamma),\tau(\pi_{X}(\gamma)))\leq L_{4}$

for every $\gamma\in\mathcal{G}$. Furthermore, if $Y\subset X$ is a hole then
$\tau$ pairs $Y$ with $Y^{\prime}=\tau(Y)$. Lastly, pairing is required to be
symmetric; if $\tau$ pairs $X$ with $X^{\prime}$ then $\tau^{-1}$ pairs
$X^{\prime}$ with $X$.

######  Definition 5.6.

Two holes $X$ and $Y$ interfere if either $X\cap Y\neq\emptyset$ or $X$ is
paired with $X^{\prime}$ and $X^{\prime}\cap Y\neq\emptyset$.

Examples arise in the symmetric arc complex and in the discussion of twisted
$I$–bundles inside of a handlebody.

###  5.2. Projection to holes is coarsely Lipschitz

The following lemma is used repeatedly throughout the paper:

######  Lemma 5.7.

Suppose that $\mathcal{G}(S)$ is a combinatorial complex. Suppose that $X$ is
a hole for $\mathcal{G}$. Then for any $\alpha,\beta\in\mathcal{G}$ we have

$d_{X}(\alpha,\beta)\leq 3+3\cdot d_{\mathcal{G}}(\alpha,\beta).$

The additive error is required only when $\alpha=\beta$.

###### Proof.

This follows directly from Corollary 4.5 and our assumption that vertices of
$\mathcal{G}$ connected by an edge represent disjoint multicurves. ∎

###  5.3. Infinite diameter holes

We may now state a first answer to Bowditch’s question.

######  Lemma 5.8.

Suppose that $\mathcal{G}(S)$ is a combinatorial complex. Suppose that there
is a strict hole $X\subset S$ having infinite diameter. Then
$\nu\colon\mathcal{G}\to\mathcal{AC}(S)$ is not a quasi-isometric embedding. ∎

This lemma and Example 5.2 completely determines when the inclusion of
$\mathcal{A}(S)$ into $\mathcal{AC}(S)$ is a quasi-isometric embedding. It
quickly becomes clear that the set of holes tightly constrains the intrinsic
geometry of a combinatorial complex.

######  Lemma 5.9.

Suppose that $\mathcal{G}(S)$ is a combinatorial complex invariant under the
natural action of $\mathcal{MCG}(S)$. Then every non-simple hole for
$\mathcal{G}$ has infinite diameter. Furthermore, if $X,Y\subset S$ are
disjoint non-simple holes for $\mathcal{G}$ then there is a quasi-isometric
embedding of $\mathbb{Z}^{2}$ into $\mathcal{G}$. ∎

We will not use Lemmas 5.8 or 5.9 and so omit the proofs. Instead our interest
lies in proving the far more powerful distance estimate (Theorems 5.10 and
13.1) for $\mathcal{G}(S)$.

###  5.4. A lower bound on distance

Here we see that the sum of projection distances in holes gives a lower bound
for distance.

######  Theorem 5.10.

Fix $S$, a compact connected non-simple surface. Suppose that $\mathcal{G}(S)$
is a combinatorial complex. Then there is a constant ${C_{0}}$ so that for all
$c\geq{C_{0}}$ there is a constant $A$ satisfying

$\sum[d_{X}(\alpha,\beta)]_{c}\mathbin{\leq_{A}}d_{\mathcal{G}}(\alpha,\beta).$

Here $\alpha,\beta\in\mathcal{G}$ and the sum is taken over all holes $X$ for
the complex $\mathcal{G}$. ∎

The proof follows the proof of Theorems 6.10 and 6.12 of [25], practically
word for word. The only changes necessary are to

  * •

replace the sum over all subsurfaces by the sum over all holes,

  * •

replace Lemma 2.5 of [25], which records how markings differing by an
elementary move project to an essential subsurface, by Lemma 5.7 of this
paper, which records how $\mathcal{G}$ projects to a hole.

One major goal of this paper is to give criteria sufficient obtain the reverse
inequality; Theorem 13.1.

##  6\. Holes for the non-orientable surface

Fix $F$ a compact, connected, and non-orientable surface. Let $S$ be the
orientation double cover with covering map $\rho_{F}\colon S\to F$. Let
$\tau\colon S\to S$ be the associated involution; so for all $x\in S$,
$\rho_{F}(x)=\rho_{F}(\tau(x))$.

######  Definition 6.1.

A multicurve $\gamma\subset\mathcal{AC}(S)$ is symmetric if
$\tau(\gamma)\cap\gamma=\emptyset$ or $\tau(\gamma)=\gamma$. A multicurve
$\gamma$ is invariant if there is a curve or arc $\gamma^{\prime}\subset F$ so
that $\gamma=\rho_{F}^{-1}(\gamma^{\prime})$. The same definitions holds for
subsurfaces $X\subset S$.

######  Definition 6.2.

The invariant complex $\mathcal{C}^{\tau}(S)$ is the simplicial complex with
vertex set being isotopy classes of invariant multicurves. There is a
$k$–simplex for every collection of $k+1$ distinct isotopy classes having
pairwise disjoint representatives.

Notice that $\mathcal{C}^{\tau}(S)$ is simplicially isomorphic to
$\mathcal{C}(F)$. There is also a natural map
$\nu\colon\mathcal{C}^{\tau}(S)\to\mathcal{C}(S)$. We will prove:

######  Lemma 6.3.

$\nu\colon\mathcal{C}^{\tau}(S)\to\mathcal{C}(S)$ is a quasi-isometric
embedding.

It thus follows from the hyperbolicity of $\mathcal{C}(S)$ that:

######  Corollary 6.4 ([4]).

$\mathcal{C}(F)$ is Gromov hyperbolic. ∎

We begin the proof of Lemma 6.3: since $\nu$ sends adjacent vertices to
adjacent edges we have

(6.5) $d_{S}(\alpha,\beta)\leq d_{\mathcal{C}^{\tau}}(\alpha,\beta),$

as long as $\alpha$ and $\beta$ are distinct in $\mathcal{C}^{\tau}(S)$. In
fact, since the surface $S$ itself is a hole for $\mathcal{C}^{\tau}(S)$ we
may deduce a slightly weaker lower bound from Lemma 5.7 or indeed from Theorem
5.10.

The other half of the proof of Lemma 6.3 consists of showing that $S$ is the
only hole for $\mathcal{C}^{\tau}(S)$ with large diameter. After a discussion
of Teichmüller geodesics we will prove:

There is a constant $K$ with the following property: Suppose that
$\alpha,\beta$ are invariant multicurves in $S$. Suppose that $X\subset S$ is
an essential subsurface where $d_{X}(\alpha,\beta)>K$. Then $X$ is symmetric.

From this it follows that:

######  Corollary 6.6.

With $K$ as in Lemma 16.4: If $X\subset S$ is a hole for
$\mathcal{C}^{\tau}(S)$ with diameter greater than $K$ then $X=S$.

###### Proof.

Suppose that $X\subset S$ is a strict subsurface, cleanly embedded. Suppose
that $\operatorname{diam}_{X}(\mathcal{C}^{\tau}(S))>K$. Thus $X$ is
symmetric. It follows that $\partial X{\smallsetminus}\partial S$ is also
symmetric. Since $\partial X$ does not cut $X$ deduce that $X$ is not a hole
for $\mathcal{C}^{\tau}(S)$. ∎

This corollary, together with the upper bound (Theorem 13.1), proves Lemma
6.3.

##  7\. Holes for the arc complex

Here we generalize the definition of the arc complex and classify its holes.

######  Definition 7.1.

Suppose that $S$ is a non-simple surface with boundary. Let $\Delta$ be a non-
empty collection of components of $\partial S$. The arc complex
$\mathcal{A}(S,\Delta)$ is the subcomplex of $\mathcal{A}(S)$ spanned by
essential arcs $\alpha\subset S$ with $\partial\alpha\subset\Delta$.

Note that $\mathcal{A}(S,\partial S)$ and $\mathcal{A}(S)$ are identical.

######  Lemma 7.2.

Suppose $X\subset S$ is cleanly embedded. Then $X$ is a hole for
$\mathcal{A}(S,\Delta)$ if and only if $\Delta\subset\partial X$. ∎

This follows directly from the definition of a hole. We now have an straight-
forward observation:

######  Lemma 7.3.

If $X,Y\subset S$ are holes for $\mathcal{A}(S,\Delta)$ then $X\cap
Y\neq\emptyset$. ∎

The proof follows immediately from Lemma 7.2. Lemma 5.9 indicates that Lemma
7.3 is essential to proving that $\mathcal{A}(S,\Delta)$ is Gromov hyperbolic.

In order to prove the upper bound theorem for $\mathcal{A}$ we will use pants
decompositions of the surface $S$. In an attempt to avoid complications in the
non-orientable case we must carefully lift to the orientation cover.

Suppose that $F$ is non-simple, non-orientable, and has non-empty boundary.
Let $\rho_{F}\colon S\to F$ be the orientation double cover and let
$\tau\colon S\to S$ be the induced involution. Fix
$\Delta^{\prime}\subset\partial F$ and let
$\Delta=\rho_{F}^{-1}(\Delta^{\prime})$.

######  Definition 7.4.

We define $\mathcal{A}^{\tau}(S,\Delta)$ to be the invariant arc complex:
vertices are invariant multi-arcs and simplices arise from disjointness.

Again, $\mathcal{A}^{\tau}(S,\Delta)$ is simplicially isomorphic to
$\mathcal{A}(F,\Delta^{\prime})$. If $X\cap\tau(X)=\emptyset$ and
$\Delta\subset X\cup\tau(X)$ then the subsurfaces $X$ and $\tau(X)$ are paired
holes, as in Definition 5.5. Notice as well that all non-simple symmetric
holes $X\subset S$ for $\mathcal{A}^{\tau}(S,\Delta)$ have infinite diameter.

Unlike $\mathcal{A}(F,\Delta^{\prime})$ the complex
$\mathcal{A}^{\tau}(S,\Delta)$ may have disjoint holes. Nonetheless, we have:

######  Lemma 7.5.

Any two non-simple holes for $\mathcal{A}^{\tau}(S,\Delta)$ interfere.

###### Proof.

Suppose that $X,Y$ are holes for the $\tau$–invariant arc complex,
$\mathcal{A}^{\tau}(S,\Delta)$. It follows from Lemma 16.4 that $X$ is
symmetric with $\Delta\subset X\cup\tau(X)$. The same holds for $Y$. Thus $Y$
must cut either $X$ or $\tau(X)$. ∎

##  8\. Background on three-manifolds

Before discussing the holes in the disk complex, we record a few facts about
handlebodies and $I$–bundles.

Fix $M$ a compact connected irreducible three-manifold. Recall that $M$ is
irreducible if every embedded two-sphere in $M$ bounds a three-ball. Recall
that if $N$ is a closed submanifold of $M$ then $\operatorname{fr}(N)$, the
frontier of $N$ in $M$, is the closure of $\partial N{\smallsetminus}\partial
M$.

###  8.1. Compressions

Suppose that $F$ is a surface embedded in $M$. Then $F$ is compressible if
there is a disk $B$ embedded in $M$ with $B\cap\partial M=\emptyset$, $B\cap
F=\partial B$, and $\partial B$ essential in $F$. Any such disk $B$ is called
a compression of $F$.

In this situation form a new surface $F^{\prime}$ as follows: Let $N$ be a
closed regular neighborhood of $B$. First remove from $F$ the annulus $N\cap
F$. Now form $F^{\prime}$ by gluing on both disk components of $\partial
N{\smallsetminus}F$. We say that $F^{\prime}$ is obtained by compressing $F$
along $B$. If no such disk exists we say $F$ is incompressible.

######  Definition 8.1.

A properly embedded surface $F$ is boundary compressible if there is a disk
$B$ embedded in $M$ with

  * •

${\operatorname{interior}}(B)\cap\partial M=\emptyset$,

  * •

$\partial B$ is a union of connected arcs $\alpha$ and $\beta$,

  * •

$\alpha\cap\beta=\partial\alpha=\partial\beta$,

  * •

$B\cap F=\alpha$ and $\alpha$ is properly embedded in $F$,

  * •

$B\cap\partial M=\beta$, and

  * •

$\beta$ is essential in $\partial M{\smallsetminus}\partial F$.

A disk, like $B$, with boundary partitioned into two arcs is called a bigon.
Note that this definition of boundary compression is slightly weaker than some
found in the literature; the arc $\alpha$ is often required to be essential in
$F$. We do not require this additional property because, for us, $F$ will
usually be a properly embedded disk in a handlebody.

Just as for compressing disks we may boundary compress $F$ along $B$ to obtain
a new surface $F^{\prime}$: Let $N$ be a closed regular neighborhood of $B$.
First remove from $F$ the rectangle $N\cap F$. Now form $F^{\prime}$ by gluing
on both bigon components of $\operatorname{fr}(N){\smallsetminus}F$. Again,
$F^{\prime}$ is obtained by boundary compressing $F$ along $B$. Note that the
relevant boundary components of $F$ and $F^{\prime}$ cobound a pair of pants
embedded in $\partial M$. If no boundary compression exists then $F$ is
boundary incompressible.

######  Remark 8.2.

Recall that any surface $F$ properly embedded in a handlebody $V_{g}$, $g\geq
2$, is either compressible or boundary compressible.

Suppose now that $F$ is properly embedded in $M$ and $\Gamma$ is a multicurve
in $\partial M$.

######  Remark 8.3.

Suppose that $F^{\prime}$ is obtained by a boundary compression of $F$
performed in the complement of $\Gamma$. Suppose that $F^{\prime}=F_{1}\cap
F_{2}$ is disconnected and each $F_{i}$ cuts $\Gamma$. Then $\iota(\partial
F_{i},\Gamma)<\iota(\partial F,\Gamma)$ for $i=1,2$.

It is often useful to restrict our attention to boundary compressions meeting
a single subsurface of $\partial M$. So suppose that $X\subset\partial M$ is
an essential subsurface. Suppose that $\partial F$ is tight with respect to
$\partial X$. Suppose $B$ is a boundary compression of $F$. If $B\cap\partial
M\subset X$ we say that $F$ is boundary compressible into $X$.

######  Lemma 8.4.

Suppose that $M$ is irreducible. Fix $X$ a connected essential subsurface of
$\partial M$. Let $F\subset M$ be a properly embedded, incompressible surface.
Suppose that $\partial X$ and $\partial F$ are tight and that $X$ compresses
in $M$. Then either:

  * •

$F\cap X=\emptyset$,

  * •

$F$ is boundary compressible into $X$, or

  * •

$F$ is a disk with $\partial F\subset X$.

###### Proof.

Suppose that $X$ is compressible via a disk $E$. Isotope $E$ to make $\partial
E$ tight with respect to $\partial F$. This can be done while maintaining
$\partial E\subset X$ because $\partial F$ and $\partial X$ are tight. Since
$M$ is irreducible and $F$ is incompressible we may isotope $E$, rel
$\partial$, to remove all simple closed curves of $F\cap E$. If $F\cap E$ is
non-empty then an outermost bigon of $E$ gives the desired boundary
compression lying in $X$.

Suppose instead that $F\cap E=\emptyset$ but $F$ does cut $X$. Let
$\delta\subset X$ be a simple arc meeting each of $F$ and $E$ in exactly one
endpoint. Let $N$ be a closed regular neighborhood of $\delta\cup E$. Note
that $\operatorname{fr}(N){\smallsetminus}F$ has three components. One is a
properly embedded disk parallel to $E$ and the other two $B,B^{\prime}$ are
bigons attached to $F$. At least one of these, say $B^{\prime}$ is trivial in
the sense that $B^{\prime}\cap\partial M$ is a trivial arc embedded in
$\partial M{\smallsetminus}\partial F$. If $B$ is non-trivial then $B$
provides the desired boundary compression.

Suppose that $B$ is also trivial. It follows that $\partial E$ and one
component $\gamma\subset\partial F$ cobound an annulus $A\subset X$. So
$D=A\cup E$ is a disk with $(D,\partial D)\subset(M,F)$. As $\partial
D=\gamma$ and $F$ is incompressible and $M$ is irreducible deduce that $F$ is
isotopic to $E$. ∎

###  8.2. Band sums

A band sum is the inverse operation to boundary compression: Fix a pair of
disjoint properly embedded surfaces $F_{1},F_{2}\subset M$. Let
$F^{\prime}=F_{1}\cup F_{2}$. Fix a simple arc $\delta\subset\partial M$ so
that $\delta$ meets each of $F_{1}$ and $F_{2}$ in exactly one point of
$\partial\delta$. Let $N\subset M$ be a closed regular neighborhood of
$\delta$. Form a new surface by adding to $F^{\prime}{\smallsetminus}N$ the
rectangle component of $\operatorname{fr}(N){\smallsetminus}F^{\prime}$. The
surface $F$ obtained is the result of band summing $F_{1}$ to $F_{2}$ along
$\delta$. Note that $F$ has a boundary compression dual to $\delta$ yielding
$F^{\prime}$: that is, there is a boundary compression $B$ for $F$ so that
$\delta\cap B$ is a single point and compressing $F$ along $B$ gives
$F^{\prime}$.

###  8.3. Handlebodies and I-bundles

Recall that handlebodies are irreducible.

Suppose that $F$ is a compact connected surface with at least one boundary
component. Let $T$ be the orientation $I$–bundle over $F$. If $F$ is
orientable then $T\mathrel{\cong}F\times I$. If $F$ is not orientable then $T$
is the unique $I$–bundle over $F$ with orientable total space. We call $T$ the
$I$–bundle and $F$ the base space. Let $\rho_{F}\colon T\to F$ be the
associated bundle map. Note that $T$ is homeomorphic to a handlebody.

If $A\subset T$ is a union of fibers of the map $\rho_{F}$ then $A$ is
vertical with respect to $T$. In particular take
$\partial_{v}T=\rho_{F}^{-1}(\partial F)$ to be the vertical boundary of $T$.
Take $\partial_{h}T$ to be the union of the boundaries of all of the fibers:
this is the horizontal boundary of $T$. Note that $\partial_{h}T$ is always
incompressible in $T$ while $\partial_{v}T$ is incompressible in $T$ as long
as $F$ is not homeomorphic to a disk.

Note that, as $|\partial_{v}T|\geq 1$, any vertical surface in $T$ can be
boundary compressed. However no vertical surface in $T$ may be boundary
compressed into $\partial_{h}T$.

We end this section with:

######  Lemma 8.5.

Suppose that $F$ is a compact, connected surface with $\partial
F\neq\emptyset$. Let $\rho_{F}\colon T\to F$ be the orientation $I$–bundle
over $F$. Let $X$ be a component of $\partial_{h}T$. Let $D\subset T$ be a
properly embedded disk. If

  * •

$\partial D$ is essential in $\partial T$,

  * •

$\partial D$ and $\partial X$ are tight, and

  * •

$D$ cannot be boundary compressed into $X$

then $D$ may be properly isotoped to be vertical with respect to $T$. ∎

##  9\. Holes for the disk complex

Here we begin to classify the holes for the disk complex, a more difficult
analysis than that of the arc complex. To fix notation let $V$ be a
handlebody. Let $S=S_{g}=\partial V$. Recall that there is a natural inclusion
$\nu\colon\mathcal{D}(V)\to\mathcal{C}(S)$.

######  Remark 9.1.

The notion of a hole $X\subset\partial V$ for $\mathcal{D}(V)$ may be phrased
in several different ways:

  * •

every essential disk $D\subset V$ cuts the surface $X$,

  * •

$\overline{S{\smallsetminus}X}$ is incompressible in $V$, or

  * •

$X$ is disk-busting in $V$.

The classification of holes $X\subset S$ for $\mathcal{D}(V)$ breaks roughly
into three cases: either $X$ is an annulus, is compressible in $V$, or is
incompressible in $V$. In each case we obtain a result:

Suppose $X$ is a hole for $\mathcal{D}(V)$ and $X$ is an annulus. Then the
diameter of $X$ is at most $5$.

Suppose $X$ is a compressible hole for $\mathcal{D}(V)$ with diameter at least
$15$. Then there are a pair of essential disks $D,E\subset V$ so that

  * •

$\partial D,\partial E\subset X$ and

  * •

$\partial D$ and $\partial E$ fill $X$.

Suppose $X$ is an incompressible hole for $\mathcal{D}(V)$ with diameter at
least $61$. Then there is an $I$–bundle $\rho_{F}\colon T\to F$ embedded in
$V$ so that

  * •

$\partial_{h}T\subset S$,

  * •

$X$ is isotopic in $S$ to a component of $\partial_{h}T$,

  * •

some component of $\partial_{v}T$ is boundary parallel into $S$,

  * •

$F$ supports a pseudo-Anosov map.

As a corollary of these theorems we have:

######  Corollary 9.2.

If $X$ is hole for $\mathcal{D}(V)$ with diameter at least $61$ then $X$ has
infinite diameter.

###### Proof.

If $X$ is a hole with diameter at least $61$ then either Theorem 11.6 or
Theorem 12.1 applies.

If $X$ is compressible then Dehn twists, in opposite directions, about the
given disks $D$ and $E$ yields an automorphism $f\colon V\to V$ so that $f|X$
is pseudo-Anosov. This follows from Thurston’s construction [38]. By Lemma 2.6
the hole $X$ has infinite diameter.

If $X$ is incompressible then $X\subset\partial_{h}T$ where $\rho_{F}\colon
T\to F$ is the given $I$–bundle. Let $f\colon F\to F$ be the given pseudo-
Anosov map. So $g$, the suspension of $f$, gives a automorphism of $V$. Again
it follows that the hole $X$ has infinite diameter. ∎

Applying Lemma 5.8 we find another corollary:

######  Theorem 9.3.

If $S=\partial V$ contains a strict hole with diameter at least $61$ then the
inclusion $\nu\colon\mathcal{D}(V)\to\mathcal{C}(S)$ is not a quasi-isometric
embedding. ∎

##  10\. Holes for the disk complex – annuli

The proof of Theorem 10.1 occupies the rest of this section. This proof shares
many features with the proofs of Theorems 11.6 and 12.1. However, the
exceptional definition of $\mathcal{C}(S_{0,2})$ prevents a unified approach.
Fix $V$, a handlebody.

######  Theorem 10.1.

Suppose $X$ is a hole for $\mathcal{D}(V)$ and $X$ is an annulus. Then the
diameter of $X$ is at most $5$.

We begin with:

######  Claim.

For all $D\in\mathcal{D}(V)$, $|D\cap X|\geq 2$.

###### Proof.

Since $X$ is a hole, every disk cuts $X$. Since $X$ is an annulus, let
$\alpha$ be a core curve for $X$. If $|D\cap X|=1$, then we may band sum
parallel copies of $D$ along an subarc of $\alpha$. The resulting disk misses
$\alpha$, a contradiction. ∎

Assume, to obtain a contradiction, that $X$ has diameter at least $6$. Suppose
that $D\in\mathcal{D}(V)$ is a disk chosen to minimize $D\cap X$. Among all
disks $E\in\mathcal{D}(V)$ with $d_{X}(D,E)\geq 3$ choose one which minimizes
$|D\cap E|$. Isotope $D$ and $E$ to make the boundaries tight and also tight
with respect to $\partial X$. Tightening triples of curves is not canonical;
nonetheless there is a tightening so that $S{\smallsetminus}(\partial
D\cup\partial E\cup X)$ contains no triangles. See Figure 1.

$\begin{array}[]{ccc}\includegraphics[height=3.5cm]{badtriangles}&&\includegraphics[height=3.5cm]{goodtriangles}\\\
\end{array}$ Figure 1. Triangles outside of $X$ (see the left side) can be
moved in (see the right side). This decreases the number of points of $D\cap
E\cap(S{\smallsetminus}X)$.

After this tightening we have:

######  Claim.

Every arc of $\partial D\cap X$ meets every arc of $\partial E\cap X$ at least
once.

###### Proof.

Fix components arcs $\alpha\subset D\cap X$ and $\beta\subset E\cap X$. Let
$\alpha^{\prime},\beta^{\prime}$ be the corresponding arcs in $S^{X}$ the
annular cover of $S$ corresponding to $X$. After the tightening we find that

$|\alpha\cap\beta|\geq|\alpha^{\prime}\cap\beta^{\prime}|-1.$

Since $d_{X}(D,E)\geq 3$ Equation 2.3 implies that
$|\alpha^{\prime}\cap\beta^{\prime}|\geq 2$. Thus $|\alpha\cap\beta|\geq 1$,
as desired. ∎

######  Claim.

There is an outermost bigon $B\subset E{\smallsetminus}D$ with the following
properties:

  * •

$\partial B=\alpha\cup\beta$ where $\alpha=B\cap D$, $\beta=\partial
B{\smallsetminus}\alpha\subset\partial E$,

  * •

$\partial\alpha=\partial\beta\subset X$, and

  * •

$|\beta\cap X|=2$.

Furthermore, $|D\cap X|=2$.

See the lower right of Figure 2 for a picture.

###### Proof.

Consider the intersection of $D$ and $E$, thought of as a collection of arcs
and curves in $E$. Any simple closed curve component of $D\cap E$ can be
removed by an isotopy of $E$, fixed on the boundary. (This follows from the
irreducibility of $V$ and an innermost disk argument.) Since we have assumed
that $|D\cap E|$ is minimal it follows that there are no simple closed curves
in $D\cap E$.

So consider any outermost bigon $B\subset E{\smallsetminus}D$. Let
$\alpha=B\cap D$. Let $\beta=\partial B{\smallsetminus}\alpha=B\cap\partial
V$. Note that $\beta$ cannot completely contain a component of $E\cap X$ as
this would contradict either the fact that $B$ is outermost or the claim that
every arc of $E\cap X$ meets some arc of $D\cap X$. Using this observation,
Figure 2 lists the possible ways for $B$ to lie inside of $E$.

2pt $E$ [bl] at 22 2 $\alpha$ [tl] at 78 30

$\begin{array}[]{cc}\includegraphics[height=3.2cm]{possiblealphas1}\end{array}$

Figure 2. The arc $\alpha$ cuts a bigon $B$ off of $E$. The darker part of
$\partial E$ are the arcs of $E\cap X$. Either $\beta$ is disjoint from $X$,
$\beta$ is contained in $X$, $\beta$ meets $X$ in a single subarc, or $\beta$
meets $X$ in two subarcs.

Let $D^{\prime}$ and $D^{\prime\prime}$ be the two essential disks obtained by
boundary compressing $D$ along the bigon $B$. Suppose $\alpha$ is as shown in
one of the first three pictures of Figure 2. It follows that either
$D^{\prime}$ or $D^{\prime\prime}$ has, after tightening, smaller intersection
with $X$ than $D$ does, a contradiction. We deduce that $\alpha$ is as
pictured in lower right of Figure 2.

Boundary compressing $D$ along $B$ still gives disks
$D^{\prime},D^{\prime\prime}\in\mathcal{D}(V)$. As these cannot have smaller
intersection with $X$ we deduce that $|D\cap X|\leq 2$ and the claim holds. ∎

Using the same notation as in the proof above, let $B$ be an outermost bigon
of $E{\smallsetminus}D$. We now study how $\alpha\subset\partial B$ lies
inside of $D$.

######  Claim.

The arc $\alpha\subset D$ connects distinct components of $D\cap X$.

###### Proof.

Suppose not. Then there is a bigon $C\subset D{\smallsetminus}\alpha$ with
$\partial C=\alpha\cup\gamma$ and $\gamma\subset\partial D\cap X$. The disk
$C\cup B$ is essential and intersects $X$ at most once after tightening,
contradicting our first claim. ∎

We finish the proof of Theorem 10.1 by noting that $D\cup B$ is homeomorphic
to $\Upsilon\times I$ where $\Upsilon$ is the simplicial tree with three edges
and three leaves. We may choose the homeomorphism so that $(D\cup B)\cap
X=\Upsilon\times\partial I$. It follows that we may properly isotope $D\cup B$
until $(D\cup B)\cap X$ is a pair of arcs. Recall that $D^{\prime}$ and
$D^{\prime\prime}$ are the disks obtained by boundary compressing $D$ along
$B$. It follows that one of $D^{\prime}$ or $D^{\prime\prime}$ (or both) meets
$X$ in at most a single arc, contradicting our first claim. ∎

##  11\. Holes for the disk complex – compressible

The proof of Theorem 11.6 occupies the second half of this section.

###  11.1. Compression sequences of essential disks

Fix a multicurve $\Gamma\subset S=\partial V$. Fix also an essential disk
$D\subset V$. Properly isotope $D$ to make $\partial D$ tight with respect to
$\Gamma$.

If $D\cap\Gamma\neq\emptyset$ we may define:

######  Definition 11.1.

A compression sequence $\\{\Delta_{k}\\}_{k=1}^{n}$ starting at $D$ has
$\Delta_{1}=\\{D\\}$ and $\Delta_{k+1}$ is obtained from $\Delta_{k}$ via a
boundary compression, disjoint from $\Gamma$, and tightening. Note that
$\Delta_{k}$ is a collection of exactly $k$ pairwise disjoint disks properly
embedded in $V$. We further require, for $k\leq n$, that every disk of
$\Delta_{k}$ meets some component of $\Gamma$. We call a compression sequence
maximal if either

  * •

no disk of $\Delta_{n}$ can be boundary compressed into
$S{\smallsetminus}\Gamma$ or

  * •

there is a component $Z\subset S{\smallsetminus}\Gamma$ and a boundary
compression of $\Delta_{n}$ into $S{\smallsetminus}\Gamma$ yielding an
essential disk $E$ with $\partial E\subset Z$.

We say that such maximal sequences end essentially or end in $Z$,
respectively.

All compression sequences must end, by Remark 8.3. Given a maximal sequence we
may relate the various disks in the sequence as follows:

######  Definition 11.2.

Fix $X$, a component of $S{\smallsetminus}\Gamma$. Fix $D_{k}\in\Delta_{k}$. A
disjointness pair for $D_{k}$ is an ordered pair $(\alpha,\beta)$ of essential
arcs in $X$ where

  * •

$\alpha\subset D_{k}\cap X$,

  * •

$\beta\subset\Delta_{n}\cap X$, and

  * •

$d_{\mathcal{A}}(\alpha,\beta)\leq 1$.

If $\alpha\neq\alpha^{\prime}$ then the two disjointness pairs
$(\alpha,\beta)$ and $(\alpha^{\prime},\beta)$ are distinct, even if $\alpha$
is properly isotopic to $\alpha^{\prime}$. A similar remark holds for the
second coordinate.

The following lemma controls how subsurface projection distance changes in
maximal sequences.

######  Lemma 11.3.

Fix a multicurve $\Gamma\subset S$. Suppose that $D$ cuts $\Gamma$ and choose
a maximal sequence starting at $D$. Fix any component $X\subset
S{\smallsetminus}\Gamma$. Fix any disk $D_{k}\in\Delta_{k}$. Then either
$D_{k}\in\Delta_{n}$ or there are four distinct disjointness pairs
$\\{(\alpha_{i},\beta_{i})\\}_{i=1}^{4}$ for $D_{k}$ in $X$ where each of the
arcs $\\{\alpha_{i}\\}$ appears as the first coordinate of at most two pairs.

###### Proof.

We induct on $n-k$. If $D_{k}$ is contained in $\Delta_{n}$ there is nothing
to prove. If $D_{k}$ is contained in $\Delta_{k+1}$ we are done by induction.
Thus we may assume that $D_{k}$ is the disk of $\Delta_{k}$ which is boundary
compressed at stage $k$. Let $D_{k+1},D_{k+1}^{\prime}\in\Delta_{k+1}$ be the
two disks obtained after boundary compressing $D_{k}$ along the bigon $B$. See
Figure 3 for a picture of the pair of pants cobounded by $\partial D_{k}$ and
$\partial D_{k+1}\cup\partial D_{k+1}^{\prime}$.

2pt $\delta$ [bl] at 82 55.5 $D_{k}$ [bl] at 184 95 $D_{k+1}$ [bl] at 40 49
$D^{\prime}_{k+1}$ [bl] at 148.5 47.5 $\Gamma$ [l] at 120 28

$\begin{array}[]{c}\includegraphics[height=3.5cm]{pants}\end{array}$

Figure 3. All arcs connecting $D_{k}$ to itself or to $D_{k+1}\cup
D^{\prime}_{k+1}$ are arcs of $\Gamma\cap P$. The boundary compressing arc
$B\cup S$ meets $D_{k}$ twice and is parallel to the vertical arcs of
$\Gamma\cap P$.

Let $\delta$ be a band sum arc dual to $B$ (the dotted arc in Figure 3). We
may assume that $|\Gamma\cap\delta|$ is minimal over all arcs dual to $B$. It
follows that the band sum of $D_{k+1}$ with $D_{k+1}^{\prime}$ along $\delta$
is tight, without any isotopy. (This is where we use the fact that $B$ is a
boundary compression in the complement of $\Gamma$, as opposed to being a
general boundary compression of $D_{k}$ in $V$.)

There are now three possibilities: neither, one, or both points of
$\partial\delta$ are contained in $X$.

First suppose that $X\cap\partial\delta=\emptyset$. Then every arc of
$D_{k+1}\cap X$ is parallel to an arc of $D_{k}\cap X$, and similarly for
$D_{k+1}^{\prime}$. If $D_{k+1}$ and $D_{k+1}^{\prime}$ are both components of
$\Delta_{n}$ then choose any arcs $\beta,\beta^{\prime}$ of $D_{k+1}\cap X$
and of $D_{k+1}^{\prime}\cap X$. Let $\alpha,\alpha^{\prime}$ be the parallel
components of $D_{k}\cap X$. The four disjointness pairs are then
$(\alpha,\beta)$, $(\alpha,\beta^{\prime})$, $(\alpha^{\prime},\beta)$,
$(\alpha^{\prime},\beta^{\prime})$. Suppose instead that $D_{k+1}$ is not a
component of $\Delta_{n}$. Then $D_{k}$ inherits four disjointness pairs from
$D_{k+1}$.

Second suppose that exactly one endpoint $x\in\partial\delta$ meets $X$. Let
$\gamma\subset D_{k+1}$ be the component of $D_{k+1}\cap X$ containing $x$.
Let $X^{\prime}$ be the component of $X\cap P$ that contains $x$ and let
$\alpha,\alpha^{\prime}$ be the two components of $D_{k}\cap X^{\prime}$. Let
$\beta$ be any arc of $D_{k+1}^{\prime}\cap X$.

If $D_{k+1}\mathbin{\notin}\Delta_{n}$ and $\gamma$ is not the first
coordinate of one of $D_{k+1}$’s four pairs then $D_{k}$ inherits disjointness
pairs from $D_{k+1}$. If $D_{k+1}^{\prime}\mathbin{\notin}\Delta_{n}$ then
$D_{k}$ inherits disjointness pairs from $D_{k+1}^{\prime}$.

Thus we may assume that both $D_{k+1}$ and $D_{k+1}^{\prime}$ are in
$\Delta_{n}$ or that only $D_{k+1}^{\prime}\in\Delta_{n}$ while $\gamma$
appears as the first arc of disjointness pair for $D_{k+1}$. In case of the
former the required disjointness pairs are $(\alpha,\beta)$,
$(\alpha^{\prime},\beta)$, $(\alpha,\gamma)$, and $(\alpha^{\prime},\gamma)$.
In case of the latter we do not know if $\gamma$ is allowed to appear as the
second coordinate of a pair. However we are given four disjointness pairs for
$D_{k+1}$ and are told that $\gamma$ appears as the first coordinate of at
most two of these pairs. Hence the other two pairs are inherited by $D_{k}$.
The pairs $(\alpha,\beta)$ and $(\alpha^{\prime},\beta)$ give the desired
conclusion.

Third suppose that the endpoints of $\delta$ meet $\gamma\subset D_{k+1}$ and
$\gamma^{\prime}\subset D_{k+1}^{\prime}$. Let $X^{\prime}$ be a component of
$X\cap P$ containing $\gamma$. Let $\alpha$ and $\alpha^{\prime}$ be the two
arcs of $D_{k}\cap X^{\prime}$. Suppose both $D_{k+1}$ and $D_{k+1}^{\prime}$
lie in $\Delta_{n}$. Then the desired pairs are $(\alpha,\gamma)$,
$(\alpha^{\prime},\gamma)$, $(\alpha,\gamma^{\prime})$, and
$(\alpha^{\prime},\gamma^{\prime})$. If $D_{k+1}^{\prime}\in\Delta_{n}$ while
$D_{k+1}$ is not then $D_{k}$ inherits two pairs from $D_{k+1}$. We add to
these the pairs $(\alpha,\gamma^{\prime})$, and
$(\alpha^{\prime},\gamma^{\prime})$. If neither disk lies in $\Delta_{n}$ then
$D_{k}$ inherits two pairs from each disk and the proof is complete. ∎

Given a disk $D\in\mathcal{D}(V)$ and a hole $X\subset S$ our Lemma 11.3
allows us to adapt $D$ to $X$.

######  Lemma 11.4.

Fix a hole $X\subset S$ for $\mathcal{D}(V)$. For any disk
$D\in\mathcal{D}(V)$ there is a disk $D^{\prime}$ with the following
properties:

  * •

$\partial X$ and $\partial D^{\prime}$ are tight.

  * •

If $X$ is incompressible then $D^{\prime}$ is not boundary compressible into
$X$ and $d_{\mathcal{A}}(D,D^{\prime})\leq 3$.

  * •

If $X$ is compressible then $\partial D^{\prime}\subset X$ and
$d_{\mathcal{AC}}(D,D^{\prime})\leq 3$.

Here $\mathcal{A}=\mathcal{A}(X)$ and $\mathcal{AC}=\mathcal{AC}(X)$.

###### Proof.

If $\partial D\subset X$ then the lemma is trivial. So assume, by Remark 9.1,
that $D$ cuts $\partial X$. Choose a maximal sequence with respect to
$\partial X$ starting at $D$.

Suppose that the sequence is non-trivial ($n>1$). By Lemma 11.3 there is a
disk $E\in\Delta_{n}$ so that $D\cap X$ and $E\cap X$ contain disjoint arcs.

If the sequence ends essentially then choose $D^{\prime}=E$ and the lemma is
proved. If the sequence ends in $X$ then there is a boundary compression of
$\Delta_{n}$, disjoint from $\partial X$, yielding the desired disk
$D^{\prime}$ with $\partial D^{\prime}\subset X$. Since $E\cap
D^{\prime}=\emptyset$ we again obtain the desired bound.

Assume now that the sequence is trivial ($n=1$). Then take $E=D\in\Delta_{n}$
and the proof is identical to that of the previous paragraph. ∎

######  Remark 11.5.

Lemma 11.4 is unexpected: after all, any pair of curves in $\mathcal{C}(X)$
can be connected by a sequence of band sums. Thus arbitrary band sums can
change the subsurface projection to $X$. However, the sequences of band sums
arising in Lemma 11.4 are very special. Firstly they do not cross $\partial X$
and secondly they are “tree-like” due to the fact every arc in $D$ is
separating.

When $D$ is replaced by a surface with genus then Lemma 11.4 does not hold in
general; this is a fundamental observation due to Kobayashi [21] (see also
[17]). Namazi points out that even if $D$ is only replaced by a planar surface
Lemma 11.4 does not hold in general.

###  11.2. Proving the theorem

We now prove:

######  Theorem 11.6.

Suppose $X$ is a compressible hole for $\mathcal{D}(V)$ with diameter at least
$15$. Then there are a pair of essential disks $D,E\subset V$ so that

  * •

$\partial D,\partial E\subset X$ and

  * •

$\partial D$ and $\partial E$ fill $X$.

###### Proof.

Choose disks $D^{\prime}$ and $E^{\prime}$ in $\mathcal{D}(V)$ so that
$d_{X}(D^{\prime},E^{\prime})\geq 15$. By Lemma 11.4 there are disks $D$ and
$E$ so that $\partial D,\partial E\subset X$, $d_{X}(D^{\prime},D)\leq 6$, and
$d_{X}(E^{\prime},E)\leq 6$. It follows from the triangle inequality that
$d_{X}(D,E)\geq 3$. ∎

##  12\. Holes for the disk complex – incompressible

This section classifies incompressible holes for the disk complex.

######  Theorem 12.1.

Suppose $X$ is an incompressible hole for $\mathcal{D}(V)$ with diameter at
least $61$. Then there is an $I$–bundle $\rho_{F}\colon T\to F$ embedded in
$V$ so that

  * •

$\partial_{h}T\subset\partial V$,

  * •

$X$ is a component of $\partial_{h}T$,

  * •

some component of $\partial_{v}T$ is boundary parallel into $\partial V$,

  * •

$F$ supports a pseudo-Anosov map.

Here is a short plan of the proof: We are given $X$, an incompressible hole
for $\mathcal{D}(V)$. Following Lemma 11.4 we may assume that $D,E$ are
essential disks, without boundary compressions into $X$ or
$S{\smallsetminus}X$, with $d_{X}(D,E)>43$. Examine the intersection pattern
of $D$ and $E$ to find two families of rectangles $\mathcal{R}$ and
$\mathcal{Q}$. The intersection pattern of these rectangles in $V$ will
determine the desired $I$–bundle $T$. The third conclusion of the theorem
follows from standard facts about primitive annuli. The fourth requires
another application of Lemma 11.4 as well as Lemma 2.6.

###  12.1. Diagonals of polygons

To understand the intersection pattern of $D$ and $E$ we discuss diagonals of
polygons. Let $D$ be a $2n$ sided regular polygon. Label the sides of $D$ with
the letters $X$ and $Y$ in alternating fashion. Any side labeled $X$ (or $Y$)
will be called an $X$ side (or $Y$ side).

######  Definition 12.2.

An arc $\gamma$ properly embedded in $D$ is a diagonal if the points of
$\partial\gamma$ lie in the interiors of distinct sides of $D$. If $\gamma$
and $\gamma^{\prime}$ are diagonals for $D$ which together meet three
different sides then $\gamma$ and $\gamma^{\prime}$ are non-parallel.

######  Lemma 12.3.

Suppose that $\Gamma\subset D$ is a collection of pairwise disjoint non-
parallel diagonals. Then there is an $X$ side of $D$ meeting at most eight
diagonals of $\Gamma$.

###### Proof.

A counting argument shows that $|\Gamma|\leq 4n-3$. If every $X$ side meets at
least nine non-parallel diagonals then $|\Gamma|\geq\frac{9}{2}n>4n-3$, a
contradiction. ∎

###  12.2. Improving disks

Suppose now that $X$ is an incompressible hole for $\mathcal{D}(V)$ with
diameter at least $61$. Note that, by Theorem 10.1, $X$ is not an annulus. Let
$Y=\overline{S{\smallsetminus}X}$.

Choose disks $D^{\prime}$ and $E^{\prime}$ in $V$ so that
$d_{X}(D^{\prime},E^{\prime})\geq 61$. By Lemma 11.4 there are a pair of disks
$D$ and $E$ so that both are essential in $V$, cannot be boundary compressed
into $X$ or into $Y$, and so that $d_{\mathcal{A}(X)}(D^{\prime},D)\leq 3$ and
$d_{\mathcal{A}(X)}(E^{\prime},E)\leq 3$. Thus $d_{X}(D^{\prime},D)\leq 9$ and
$d_{X}(E^{\prime},E)\leq 9$ (Lemma 5.7). By the triangle inequality
$d_{X}(D,E)\geq 61-18=43$.

Recall, as well, that $\partial D$ and $\partial E$ are tight with respect to
$\partial X$. We may further assume that $\partial D$ and $\partial E$ are
tight with respect to each other. Also, minimize the quantities
$|X\cap(\partial D\cap\partial E)|$ and $|D\cap E|$ while keeping everything
tight. In particular, there are no triangle components of $\partial
V{\smallsetminus}(D\cup E\cup\partial X)$. Now consider $D$ and $E$ to be
even-sided polygons, with vertices being the points $\partial D\cap\partial X$
and $\partial E\cap\partial X$ respectively. Let $\Gamma=D\cap E$. See Figure
4 for one a priori possible collection $\Gamma\subset D$.

$\begin{array}[]{c}\includegraphics[height=3.5cm]{rectanglewithbadarcs}\end{array}$
Figure 4. In fact, $\Gamma\subset D$ cannot contain simple closed curves or
non-diagonals.

From our assumptions and the irreducibility of $V$ it follows that $\Gamma$
contains no simple closed curves. Suppose now that there is a
$\gamma\subset\Gamma$ so that, in $D$, both endpoints of $\gamma$ lie in the
same side of $D$. Then there is an outermost such arc, say
$\gamma^{\prime}\subset\Gamma$, cutting a bigon $B$ out of $D$. It follows
that $B$ is a boundary compression of $E$ which is disjoint from $\partial X$.
But this contradicts the construction of $E$. We deduce that all arcs of
$\Gamma$ are diagonals for $D$ and, via a similar argument, for $E$.

Let $\alpha\subset D\cap X$ be an $X$ side of $D$ meeting at most eight
distinct types of diagonal of $\Gamma$. Choose $\beta\subset E\cap X$
similarly. As $d_{X}(D,E)\geq 43$ we have that $d_{X}(\alpha,\beta)\geq
43-6=37$.

Now break each of $\alpha$ and $\beta$ into at most eight subarcs
$\\{\alpha_{i}\\}$ and $\\{\beta_{j}\\}$ so that each subarc meets all of the
diagonals of fixed type and only of that type. Let $R_{i}\subset D$ be the
rectangle with upper boundary $\alpha_{i}$ and containing all of the diagonals
meeting $\alpha_{i}$. Let $\alpha_{i}^{\prime}$ be the lower boundary of
$R_{i}$. Define $Q_{j}$ and $\beta_{j}^{\prime}$ similarly. See Figure 5 for a
picture of $R_{i}$.

2pt $R_{i}$ [l] at 92 61 $\alpha_{i}$ [bl] at 93 146 $\alpha_{i}^{\prime}$
[tr] at 63 3

$\begin{array}[]{c}\includegraphics[height=3.5cm]{rectangle}\end{array}$

Figure 5. The rectangle $R_{i}\subset D$ is surrounded by the dotted line. The
arc $\alpha_{i}$ in $\partial D\cap X$ is indicated. In general the arc
$\alpha^{\prime}_{i}$ may lie in $X$ or in $Y$.

Call an arc $\alpha_{i}$ large if there is an arc $\beta_{j}$ so that
$|\alpha_{i}\cap\beta_{j}|\geq 3$. We use the same notation for $\beta_{j}$.
Let $\Theta$ be the union of all of the large $\alpha_{i}$ and $\beta_{j}$.
Thus $\Theta$ is a four-valent graph in $X$. Let $\Theta^{\prime}$ be the
union of the corresponding large $\alpha_{i}^{\prime}$ and
$\beta_{i}^{\prime}$.

######  Claim 12.4.

The graph $\Theta$ is non-empty.

###### Proof.

If $\Theta=\emptyset$, then all $\alpha_{i}$ are small. It follows that
$|\alpha\cap\beta|\leq 128$ and thus $d_{X}(\alpha,\beta)\leq 16$, by Lemma
2.2. As $d_{X}(\alpha,\beta)\geq 37$ this is a contradiction. ∎

Let $Z\subset\partial V$ be a small regular neighborhood of $\Theta$ and
define $Z^{\prime}$ similarly.

######  Claim 12.5.

No component of $\Theta$ or of $\Theta^{\prime}$ is contained in a disk
$D\subset\partial V$. No component of $\Theta$ or of $\Theta^{\prime}$ is
contained in an annulus $A\subset\partial V$ that is peripheral in $X$.

###### Proof.

For a contradiction suppose that $W$ is a component of $Z$ contained in a
disk. Then there is some pair $\alpha_{i},\beta_{j}$ having a bigon in
$\partial V$. This contradicts the tightness of $\partial D$ and $\partial E$.
The same holds for $Z^{\prime}$.

Suppose now that some component $W$ is contained in an annulus $A$, peripheral
in $X$. Thus $W$ fills $A$. Suppose that $\alpha_{i}$ and $\beta_{j}$ are
large and contained in $W$. By the classification of arcs in $A$ we deduce
that either $\alpha_{i}$ and $\beta_{j}$ form a bigon in $A$ or $\partial X$,
$\alpha_{i}$ and $\beta_{j}$ form a triangle. Either conclusion gives a
contradiction. ∎

######  Claim 12.6.

The graph $\Theta$ fills $X$.

###### Proof.

Suppose not. Fix attention on any component $W\subset Z$. Since $\Theta$ does
not fill, the previous claim implies that there is a component
$\gamma\subset\partial W$ that is essential and non-peripheral in $X$. Note
that any large $\alpha_{i}$ meets $\partial W$ in at most two points, while
any small $\alpha_{i}$ meets $\partial W$ in at most $32$ points. Thus
$|\alpha\cap\partial W|\leq 256$ and the same holds for $\beta$. Thus
$d_{X}(\alpha,\beta)\leq 36$ by the triangle inequality. As
$d_{X}(\alpha,\beta)\geq 37$ this is a contradiction. ∎

The previous two claims imply:

######  Claim 12.7.

The graph $\Theta$ is connected. ∎

There are now two possibilities: either $\Theta\cap\Theta^{\prime}$ is empty
or not. In the first case set $\Sigma=\Theta$ and in the second set
$\Sigma=\Theta\cup\Theta^{\prime}$. By the claims above, $\Sigma$ is connected
and fills $X$. Let $\mathcal{R}=\\{R_{i}\\}$ and $\mathcal{Q}=\\{Q_{j}\\}$ be
the collections of large rectangles.

###  12.3. Building the I-bundle

We are given $\Sigma$, $\mathcal{R}$ and $\mathcal{Q}$ as above. Note that
$\mathcal{R}\cup\mathcal{Q}$ is an $I$–bundle and $\Sigma$ is the component of
its horizontal boundary meeting $X$. See Figure 6 for a simple case.

2pt $R_{i}$ [l] at 312 173 $Q_{j}$ [l] at 388 371

$\begin{array}[]{c}\includegraphics[height=4.5cm]{rectangles}\end{array}$

Figure 6. $\mathcal{R}\cup\mathcal{Q}$ is an $I$–bundle: all arcs of
intersection are parallel.

Let $T_{0}$ be a regular neighborhood of $\mathcal{R}\cup\mathcal{Q}$, taken
in $V$. Again $T_{0}$ has the structure of an $I$–bundle. Note that
$\partial_{h}T_{0}\subset\partial V$, $\partial_{h}T_{0}\cap X$ is a component
of $\partial_{h}T_{0}$, and this component fills $X$ due to Claim 12.6. We
will enlarge $T_{0}$ to obtain the correct $I$–bundle in $V$.

Begin by enumerating all annuli $\\{A_{i}\\}\subset\partial_{v}T_{0}$ with the
property that some component of $\partial A_{i}$ is inessential in $\partial
V$. Suppose that we have built the $I$–bundle $T_{i}$ and are now considering
the annulus $A=A_{i}$. Let $\gamma\cup\gamma^{\prime}=\partial
A\subset\partial V$ with $\gamma$ inessential in $\partial V$. Let
$B\subset\partial V$ be the disk which $\gamma$ bounds. By induction we assume
that no component of $\partial_{h}T_{i}$ is contained in a disk embedded in
$\partial V$ (the base case holds by Claim 12.5). It follows that $B\cap
T_{i}=\partial B=\gamma$. Thus $B\cup A$ is isotopic, rel $\gamma^{\prime}$,
to be a properly embedded disk $B^{\prime}\subset V$. As $\gamma^{\prime}$
lies in $X$ or $Y$, both incompressible, $\gamma^{\prime}$ must bound a disk
$C\subset\partial V$. Note that $C\cap T_{i}=\partial C=\gamma^{\prime}$,
again using the induction hypothesis.

It follows that $B\cup A\cup C$ is an embedded two-sphere in $V$. As $V$ is a
handlebody $V$ is irreducible. Thus $B\cup A\cup C$ bounds a three-ball
$U_{i}$ in $V$. Choose a homeomorphism $U_{i}\mathrel{\cong}B\times I$ so that
$B$ is identified with $B\times\\{0\\}$, $C$ is identified with
$B\times\\{1\\}$, and $A$ is identified with $\partial B\times I$. We form
$T_{i+1}=T_{i}\cup U_{i}$ and note that $T_{i+1}$ still has the structure of
an $I$–bundle. Recalling that $A=A_{i}$ we have
$\partial_{v}T_{i+1}=\partial_{v}T_{i}{\smallsetminus}A_{i}$. Also
$\partial_{h}T_{i+1}=\partial_{h}T_{i}\cup(B\cup C)\subset\partial V$. It
follows that no component of $\partial_{h}T_{i+1}$ is contained in a disk
embedded in $\partial V$. Similarly, $\partial_{h}T_{i+1}\cap X$ is a
component of $\partial_{h}T_{i+1}$ and this component fills $X$.

After dealing with all of the annuli $\\{A_{i}\\}$ in this fashion we are left
with an $I$–bundle $T$. Now all components of $\partial\partial_{v}T$ [sic]
are essential in $\partial V$. All of these lying in $X$ are peripheral in
$X$. This is because they are disjoint from $\Sigma\subset\partial_{h}T$,
which fills $X$, by induction. It follows that the component of
$\partial_{h}T$ containing $\Sigma$ is isotopic to $X$.

This finishes the construction of the promised $I$–bundle $T$ and demonstrates
the first two conclusions of Theorem 12.1. For future use we record:

######  Remark 12.8.

Every curve of $\partial\partial_{v}T=\partial\partial_{h}T$ is essential in
$S=\partial V$.

###  12.4. A vertical annulus parallel into the boundary

Here we obtain the third conclusion of Theorem 12.1: at least one component of
$\partial_{v}T$ is boundary parallel in $\partial V$.

Fix $T$ an $I$–bundle with the incompressible hole $X$ a component of
$\partial_{h}T$.

######  Claim 12.9.

All components of $\partial_{v}T$ are incompressible in $V$.

###### Proof.

Suppose that $A\subset\partial_{v}T$ was compressible. By Remark 12.8 we may
compress $A$ to obtain a pair of essential disks $B$ and $C$. Note that
$\partial B$ is isotopic into the complement of $\partial_{h}T$. So
$\overline{S{\smallsetminus}X}$ is compressible, contradicting Remark 9.1. ∎

######  Claim 12.10.

Some component of $\partial_{v}T$ is boundary parallel.

###### Proof.

Since $\partial_{v}T$ is incompressible (Claim 12.9) by Remark 8.2, we find
that $\partial_{v}T$ is boundary compressible in $V$. Let $B$ be a boundary
compression for $\partial_{v}T$. Let $A$ be the component of $\partial_{v}T$
meeting $B$. Let $\alpha$ denote the arc $A\cap B$.

The arc $\alpha$ is either essential or inessential in $A$. Suppose $\alpha$
is inessential in $A$. Then $\alpha$ cuts a bigon, $C$, out of $A$. Since $B$
was a boundary compression the disk $D=B\cup C$ is essential in $V$. Since $B$
meets $\partial_{v}T$ in a single arc, either $D\subset T$ or
$D\subset\overline{V{\smallsetminus}T}$. The former implies that
$\partial_{h}T$ is compressible and the latter that $X$ is not a hole. Either
gives a contradiction.

It follows that $\alpha$ is essential in $A$. Now carefully boundary compress
$A$: Let $N$ be the closure of a regular neighborhood of $B$, taken in
$V{\smallsetminus}A$. Let $A^{\prime}$ be the closure of $A{\smallsetminus}N$
(so $A^{\prime}$ is a rectangle). Let $B^{\prime}\cup B^{\prime\prime}$ be the
closure of $\operatorname{fr}(N){\smallsetminus}A$. Both $B^{\prime}$ and
$B^{\prime\prime}$ are bigons, parallel to $B$. Form $D=A^{\prime}\cup
B^{\prime}\cup B^{\prime\prime}$: a properly embedded disk in $V$. If $D$ is
essential then, as above, either $D\subset T$ or
$D\subset\overline{V{\smallsetminus}T}$. Again, either gives a contradiction.

It follows that $D$ is inessential in $V$. Thus $D$ cuts a closed three-ball
$U$ out of $V$. There are two final cases: either $N\subset U$ or $N\cap
U=B^{\prime}\cup B^{\prime\prime}$. If $U$ contains $N$ then $U$ contains $A$.
Thus $\partial A$ is contained in the disk $U\cap\partial V$. This contradicts
Remark 12.8. Deduce instead that $W=U\cup N$ is a solid torus with meridional
disk $B$. Thus $W$ gives a parallelism between $A$ and the annulus $\partial
V\cap\partial W$, as desired. ∎

######  Remark 12.11.

Similar considerations prove that the multicurve

$\\{\partial A\mathbin{\mid}\mbox{$A$ is a boundary parallel component of
$\partial_{v}T$}\\}$

is disk-busting for $V$.

###  12.5. Finding a pseudo-Anosov map

Here we prove that the base surface $F$ of the $I$–bundle $T$ admits a pseudo-
Anosov map.

As in Section 12.2, pick essential disks $D^{\prime}$ and $E^{\prime}$ in $V$
so that $d_{X}(D^{\prime},E^{\prime})\geq 61$. Lemma 11.4 provides disks $D$
and $E$ which cannot be boundary compressed into $X$ or into
$\overline{S{\smallsetminus}X}$ – thus $D$ and $E$ cannot be boundary
compressed into $\partial_{h}T$. Also, as above, $d_{X}(D,E)\geq 61-18=43$.

After isotoping $D$ to minimize intersection with $\partial_{v}T$ it must be
the case that all components of $D\cap\partial_{v}T$ are essential arcs in
$\partial_{v}T$. By Lemma 8.5 we conclude that $D$ may be isotoped in $V$ so
that $D\cap T$ is vertical in $T$. The same holds of $E$. Choose $A$ and $B$,
components of $D\cap T$ and $E\cap T$. Each are vertical rectangles. Since
$\operatorname{diam}_{X}(\pi_{X}(D))\leq 3$ (Lemma 4.4) we now have
$d_{X}(A,B)\geq 43-6=37$.

We now begin to work in the base surface $F$. Recall that $\rho_{F}\colon T\to
F$ is an $I$–bundle. Take $\alpha=\rho_{F}(A)$ and $\beta=\rho_{F}(B)$. Note
that the natural map $\mathcal{C}(F)\to\mathcal{C}(X)$, defined by taking a
curve to its lift, is distance non-increasing (see Equation 6.5). Thus
$d_{F}(\alpha,\beta)\geq 37$. By Theorem 10.1 the surface $F$ cannot be an
annulus. Thus, by Lemma 2.6 the subsurface $F$ supports a pseudo-Anosov map
and we are done.

###  12.6. Corollaries

We now deal with the possibility of disjoint holes for the disk complex.

######  Lemma 12.12.

Suppose that $X$ is a large incompressible hole for $\mathcal{D}(V)$ supported
by the $I$–bundle $\rho_{F}\colon T\to F$. Let
$Y=\partial_{h}T{\smallsetminus}X$. Let
$\tau\colon\partial_{h}T\to\partial_{h}T$ be the involution switching the ends
of the $I$–fibres. Suppose that $D\in\mathcal{D}(V)$ is an essential disk.

  * •

If $F$ is orientable then $d_{\mathcal{A}(F)}(D\cap X,D\cap Y)\leq 6$.

  * •

If $F$ is non-orientable then $d_{X}(D,\mathcal{C}^{\tau}(X))\leq 3$.

###### Proof.

By Lemma 11.4 there is a disk $D^{\prime}\subset V$ which is tight with
respect to $\partial_{h}T$ and which cannot be boundary compressed into
$\partial_{h}T$ (or into the complement). Also, for any component
$Z\subset\partial_{h}T$ we have $d_{\mathcal{A}(Z)}(D,D^{\prime})\leq 3$.

Properly isotope $D^{\prime}$ to minimize $D^{\prime}\cap\partial_{v}T$. Then
$D^{\prime}\cap\partial_{v}T$ is properly isotopic, in $\partial_{v}T$, to a
collection of vertical arcs. Let $E\subset D^{\prime}\cap T$ be a component.
Lemma 8.5 implies that $E$ is vertical in $T$, after an isotopy of
$D^{\prime}$ preserving $\partial_{h}T$ setwise. Since $E$ is vertical, the
arcs $E\cap\partial_{h}T\subset D^{\prime}$ are $\tau$–invariant. The
conclusion follows. ∎

Recall Lemma 7.3: all holes for the arc complex intersect. This cannot hold
for the disk complex. For example if $\rho_{F}\colon T\to F$ is an $I$–bundle
over an orientable surface then take $V=T$ and notice that both components of
$\partial_{h}T$ are holes for $\mathcal{D}(V)$. However, by the first
conclusion of Lemma 12.12, $X$ and $Y$ are paired holes, in the sense of
Definition 5.5. So, as with the invariant arc complex (Lemma 7.5), all holes
for the disk complex interfere:

######  Lemma 12.13.

Suppose that $X,Z\subset\partial V$ are large holes for $\mathcal{D}(V)$. If
$X\cap Z=\emptyset$ then there is an $I$–bundle $T\mathrel{\cong}F\times I$ in
$V$ so that $\partial_{h}T=X\cup Y$ and $Y\cap Z\neq\emptyset$.

###### Proof.

Suppose that $X\cap Z=\emptyset$. It follows from Remark 9.1 that both $X$ and
$Z$ are incompressible. Let $\rho_{F}\colon T\to F$ be the $I$–bundle in $V$
with $X\subset\partial_{h}T$, as provided by Theorem 12.1. We also have a
component $A\subset\partial_{v}T$ so that $A$ is boundary parallel. Let $U$ be
the solid torus component of $V{\smallsetminus}A$. Note that $Z$ cannot be
contained in $\partial U{\smallsetminus}A$ because $Z$ is not an annulus
(Theorem 10.1).

Let $\alpha=\rho_{F}(A)$. Choose any essential arc $\delta\subset F$ with both
endpoints in $\alpha\subset\partial F$. It follows that
$\rho_{F}^{-1}(\delta)$, together with two meridional disks of $U$, forms an
essential disk $D$ in $V$. Let $W=\partial_{h}T\cup(U{\smallsetminus}A)$ and
note that $\partial D\subset W$.

If $F$ is non-orientable then $Z\cap W=\emptyset$ and we have a contradiction.
Deduce that $F$ is orientable. Now, if $Z$ misses $Y$ then $Z$ misses $W$ and
we again have a contradiction. It follows that $Z$ cuts $Y$ and we are done. ∎

##  13\. Axioms for combinatorial complexes

The goal of this section and the next is to prove, inductively, an upper bound
on distance in a combinatorial complex $\mathcal{G}(S)=\mathcal{G}$. This
section presents our axioms on $\mathcal{G}$: sufficient hypotheses for
Theorem 13.1. The axioms, apart from Axiom 13.2, are quite general. Axiom 13.2
is necessary to prove hyperbolicity and greatly simplifies the recursive
construction in Section 14.

######  Theorem 13.1.

Fix $S$ a compact connected non-simple surface. Suppose that
$\mathcal{G}=\mathcal{G}(S)$ is a combinatorial complex satisfying the axioms
of Section 13. Let $X$ be a hole for $\mathcal{G}$ and suppose that
$\alpha_{X},\beta_{X}\in\mathcal{G}$ are contained in $X$. For any constant
$c>0$ there is a constant $A$ satisfying:

$d_{\mathcal{G}}(\alpha_{X},\beta_{X})\mathbin{\leq_{A}}\sum[d_{Y}(\alpha_{X},\beta_{X})]_{c}$

where the sum is taken over all holes $Y\subseteq X$ for $\mathcal{G}$.

The proof of the upper bound is more difficult than that of the lower bound,
Theorem 5.10. This is because naturally occurring paths in $\mathcal{G}$
between $\alpha_{X}$ and $\beta_{X}$ may waste time in non-holes. The first
example of this is the path in $\mathcal{C}(S)$ obtained by taking the short
curves along a Teichmüller geodesic. The Teichmüller geodesic may spend time
rearranging the geometry of a subsurface. Then the systole path in the curve
complex must be much longer than the curve complex distance between the
endpoints.

In Sections 16, 17, 19 we will verify these axioms for the curve complex of a
non-orientable surface, the arc complex, and the disk complex.

###  13.1. The axioms

Suppose that $\mathcal{G}=\mathcal{G}(S)$ is a combinatorial complex. We begin
with the axiom required for hyperbolicity.

######  Axiom 13.2 (Holes interfere).

All large holes for $\mathcal{G}$ interfere, as given in Definition 5.6.

Fix vertices $\alpha_{X},\beta_{X}\in\mathcal{G}$, both contained in a hole
$X$. We are given $\Lambda=\\{\mu_{n}\\}_{n=0}^{N}$, a path of markings in
$X$.

######  Axiom 13.3 (Marking path).

We require:

  1. (1)

The support of $\mu_{n+1}$ is contained inside the support of $\mu_{n}$.

  2. (2)

For any subsurface $Y\subseteq X$, if $\pi_{Y}(\mu_{k})\neq\emptyset$ then for
all $n\leq k$ the map $n\mapsto\pi_{Y}(\mu_{n})$ is an unparameterized quasi-
geodesic with constants depending only on $\mathcal{G}$.

The second condition is crucial and often technically difficult to obtain.

We are given, for every essential subsurface $Y\subset X$, a perhaps empty
interval $J_{Y}\subset[0,N]$ with the following properties.

######  Axiom 13.4 (Accessibility).

The interval for $X$ is $J_{X}=[0,N]$. There is a constant ${B_{3}}$ so that

  1. (1)

If $m\in J_{Y}$ then $Y$ is contained in the support of $\mu_{m}$.

  2. (2)

If $m\in J_{Y}$ then $\iota(\partial Y,\mu_{m})<{B_{3}}$.

  3. (3)

If $[m,n]\cap J_{Y}=\emptyset$ then $d_{Y}(\mu_{m},\mu_{n})<{B_{3}}$.

There is a combinatorial path
$\Gamma=\\{\gamma_{i}\\}_{i=0}^{K}\subset\mathcal{G}$ starting with
$\alpha_{X}$ ending with $\beta_{X}$ and each $\gamma_{i}$ is contained in
$X$. There is a strictly increasing reindexing function $r\colon[0,K]\to[0,N]$
with $r(0)=0$ and $r(K)=N$.

######  Axiom 13.5 (Combinatorial).

There is a constant ${C_{2}}$ so that:

  * •

$d_{Y}(\gamma_{i},\mu_{r(i)})<{C_{2}}$, for every $i\in[0,K]$ and every hole
$Y\subset X$,

  * •

$d_{\mathcal{G}}(\gamma_{i},\gamma_{i+1})<{C_{2}}$, for every $i\in[0,K-1]$.

######  Axiom 13.6 (Replacement).

There is a constant ${C_{4}}$ so that:

  1. (1)

If $Y\subset X$ is a hole and $r(i)\in J_{Y}$ then there is a vertex
$\gamma^{\prime}\in\mathcal{G}$ so that $\gamma^{\prime}$ is contained in $Y$
and $d_{\mathcal{G}}(\gamma_{i},\gamma^{\prime})<{C_{4}}$.

  2. (2)

If $Z\subset X$ is a non-hole and $r(i)\in J_{Z}$ then there is a vertex
$\gamma^{\prime}\in\mathcal{G}$ so that
$d_{\mathcal{G}}(\gamma_{i},\gamma^{\prime})<{C_{4}}$ and so that
$\gamma^{\prime}$ is contained in $Z$ or in $X{\smallsetminus}Z$.

There is one axiom left: the axiom for straight intervals. This is given in
the next subsection.

###  13.2. Inductive, electric, shortcut and straight intervals

We describe subintervals that arise in the partitioning of $[0,K]$. As
discussed carefully in Section 13.3, we will choose a lower threshold
${L_{1}}(Y)$ for every essential $Y\subset X$ and a general upper threshold,
${L_{2}}$.

######  Definition 13.7.

Suppose that $[i,j]\subset[0,K]$ is a subinterval of the combinatorial path.
Then $[i,j]$ is an inductive interval associated to a hole $Y\subsetneq X$ if

  * •

$r([i,j])\subset J_{Y}$ (for paired $Y$ we require $r([i,j])\subset J_{Y}\cap
J_{Y^{\prime}}$) and

  * •

$d_{Y}(\gamma_{i},\gamma_{j})\geq{L_{1}}(Y)$.

When $X$ is the only relevant hole we have a simpler definition:

######  Definition 13.8.

Suppose that $[i,j]\subset[0,K]$ is a subinterval of the combinatorial path.
Then $[i,j]$ is an electric interval if $d_{Y}(\gamma_{i},\gamma_{j})<{L_{2}}$
for all holes $Y\subsetneq X$.

Electric intervals will be further partitioned into shortcut and straight
intervals.

######  Definition 13.9.

Suppose that $[p,q]\subset[0,K]$ is a subinterval of the combinatorial path.
Then $[p,q]$ is a shortcut if

  * •

$d_{Y}(\gamma_{p},\gamma_{q})<{L_{2}}$ for all holes $Y$, including $X$
itself, and

  * •

there is a non-hole $Z\subset X$ so that $r([p,q])\subset J_{Z}$.

######  Definition 13.10.

Suppose that $[p,q]\subset[0,K]$ is a subinterval of the combinatorial path
and is contained in an electric interval $[i,j]$. Then $[p,q]$ is a straight
interval if $d_{Y}(\mu_{r(p)},\mu_{r(q)})<{L_{2}}$ for all non-holes $Y$.

Our final axiom is:

######  Axiom 13.11 (Straight).

There is a constant $A$ depending only on $X$ and $\mathcal{G}$ so that for
every straight interval $[p,q]$:

$d_{\mathcal{G}}(\gamma_{p},\gamma_{q})\mathbin{\leq_{A}}d_{X}(\gamma_{p},\gamma_{q})$

###  13.3. Deductions from the axioms

Axiom 13.3 and Lemma 3.9 imply that the reverse triangle inequality holds for
projections of marking paths.

######  Lemma 13.12.

There is a constant ${C_{1}}$ so that

$d_{Y}(\mu_{m},\mu_{n})+d_{Y}(\mu_{n},\mu_{p})<d_{Y}(\mu_{m},\mu_{p})+{C_{1}}$

for every essential $Y\subset X$ and for every $m<n<p$ in $[0,N]$. ∎

We record three simple consequences of Axiom 13.4.

######  Lemma 13.13.

There is a constant ${C_{3}}$, depending only on ${B_{3}}$, with the follow
properties:

  * (i)

If $Y$ is strictly nested in $Z$ and $m\in J_{Y}$ then $d_{Z}(\partial
Y,\mu_{m})\leq{C_{3}}$.

  * (ii)

If $Y$ is strictly nested in $Z$ then for any $m,n\in J_{Y}$,
$d_{Z}(\mu_{m},\mu_{n})<{C_{3}}$.

  * (iii)

If $Y$ and $Z$ overlap then for any $m,n\in J_{Y}\cap J_{Z}$ we have
$d_{Y}(\mu_{m},\mu_{n}),d_{Z}(\mu_{m},\mu_{n})<{C_{3}}$.

###### Proof.

We first prove conclusion (i): Since $Y$ is strictly nested in $Z$ and since
$Y$ is contained in the support of $\mu_{m}$ (part (1) of Axiom 13.4), both
$\partial Y$ and $\mu_{m}$ cut $Z$. By Axiom 13.4, part (2), we have that
$\iota(\partial Y,\mu_{m})\leq{B_{3}}$. It follows that $\iota(\partial
Y,\pi_{Z}(\mu_{m}))\leq 2{B_{3}}$. By Lemma 2.2 we deduce that $d_{Z}(\partial
Y,\mu_{m})\leq 2\log_{2}{B_{3}}+3$. We take ${C_{3}}$ larger than this right
hand side.

Conclusion (ii) follows from a pair of applications of conclusion (i) and the
triangle inequality.

For conclusion (iii): As in (ii), to bound $d_{Z}(\mu_{m},\mu_{n})$ it
suffices to note that $\partial Y$ cuts $Z$ and that $\partial Y$ has bounded
intersection with both of $\mu_{m},\mu_{n}$. ∎

We now have all of the constants ${C_{1}},{C_{3}},{C_{2}},{C_{4}}$ in hand.
Recall that $L_{4}$ is the pairing constant of Definition 5.5 and that $M_{0}$
is the constant of 4.6. We must choose a lower threshold ${L_{1}}(Y)$ for
every essential $Y\subset X$. We must also choose the general upper threshold,
${L_{2}}$ and general lower threshold ${L_{0}}$. We require, for all essential
$Z,Y$ in $X$, with $\xi(Z)<\xi(Y)\leq\xi(X)$:

(13.14) $\displaystyle{L_{0}}>{C_{3}}+2{C_{2}}+2L_{4}$ (13.15)
$\displaystyle{L_{2}}>{L_{1}}(X)+2L_{4}+6{C_{1}}+2{C_{2}}+14{C_{3}}+10$
(13.16) $\displaystyle{L_{1}}(Y)>M_{0}+2{C_{3}}+4{C_{2}}+2L_{4}+{L_{0}}$
(13.17) $\displaystyle{L_{1}}(X)>{L_{1}}(Z)+2{C_{3}}+4{C_{2}}+4L_{4}$

##  14\. Partition and the upper bound on distance

In this section we prove Theorem 13.1 by induction on $\xi(X)$. The first
stage of the proof is to describe the inductive partition: we partition the
given interval $[0,K]$ into inductive and electric intervals. The inductive
partition is closely linked with the hierarchy machine [25] and with the
notion of antichains introduced in [34].

We next give the electric partition: each electric interval is divided into
straight and shortcut intervals. Note that the electric partition also gives
the base case of the induction. We finally bound
$d_{\mathcal{G}}(\alpha_{X},\beta_{X})$ from above by combining the
contributions from the various intervals.

###  14.1. Inductive partition

We begin by identifying the relevant surfaces for the construction of the
partition. We are given a hole $X$ for $\mathcal{G}$ and vertices
$\alpha_{X},\beta_{X}\in\mathcal{G}$ contained in $X$. Define

$B_{X}=\\{Y\subsetneq X\mathbin{\mid}\mbox{$Y$ is a hole
and~{}}d_{Y}(\alpha_{X},\beta_{X})\geq{L_{1}}(X)\\}.$

For any subinterval $[i,j]\subset[0,K]$ define

$B_{X}(i,j)=\\{Y\in
B_{X}\mathbin{\mid}d_{Y}(\gamma_{i},\gamma_{j})\geq{L_{1}}(X)\\}.$

We now partition $[0,K]$ into inductive and electric intervals. Begin with the
partition of one part $\mathcal{P}_{X}=\\{[0,K]\\}$. Recursively
$\mathcal{P}_{X}$ is a partition of $[0,K]$ consisting of intervals which are
either inductive, electric, or undetermined. Suppose that
$[i,j]\in\mathcal{P}_{X}$ is undetermined.

######  Claim.

If $B_{X}(i,j)$ is empty then $[i,j]$ is electric.

###### Proof.

Since $B_{X}(i,j)$ is empty, every hole $Y\subsetneq X$ has either
$d_{Y}(\gamma_{i},\gamma_{j})<{L_{1}}(X)$ or $Y\mathbin{\notin}B_{X}$. In the
former case, as ${L_{1}}(X)<{L_{2}}$, we are done.

So suppose the latter holds. Now, by the reverse triangle inequality (Lemma
13.12),

$d_{Y}(\mu_{r(i)},\mu_{r(j)})<d_{Y}(\mu_{0},\mu_{N})+2{C_{1}}.$

Since $r(0)=0$ and $r(K)=N$ we find:

$d_{Y}(\gamma_{i},\gamma_{j})<d_{Y}(\alpha_{X},\beta_{X})+2{C_{1}}+4{C_{2}}.$

Deduce that

$d_{Y}(\gamma_{i},\gamma_{j})<{L_{1}}(X)+2{C_{1}}+4{C_{2}}<{L_{2}}.$

This completes the proof. ∎

Thus if $B_{X}(i,j)$ is empty then $[i,j]\in\mathcal{P}_{X}$ is determined to
be electric. Proceed on to the next undetermined element. Suppose instead that
$B_{X}(i,j)$ is non-empty. Pick a hole $Y\in B_{X}(i,j)$ so that $Y$ has
maximal $\xi(Y)$ amongst the elements of $B_{X}(i,j)$

Let $p,q\in[i,j]$ be the first and last indices, respectively, so that
$r(p),r(q)\in J_{Y}$. (If $Y$ is paired with $Y^{\prime}$ then we take the
first and last indices that, after reindexing, lie inside of $J_{Y}\cap
J_{Y^{\prime}}$.)

######  Claim.

The indices $p,q$ are well-defined.

###### Proof.

By assumption $d_{Y}(\gamma_{i},\gamma_{j})\geq{L_{1}}(X)$. By Equation 13.14,

${L_{1}}(X)>{C_{3}}+2{C_{2}}.$

We deduce from Axiom 13.4 and Axiom 13.5 that $J_{Y}\cap r([i,j])$ is non-
empty. Thus, if $Y$ is not paired, the indices $p,q$ are well-defined.

Suppose instead that $Y$ is paired with $Y^{\prime}$. Recall that measurements
made in $Y$ and $Y^{\prime}$ differ by at most the pairing constant $L_{4}$
given in Definition 5.5. By (13.16),

${L_{1}}(X)>{C_{3}}+2{C_{2}}+2L_{4}.$

We deduce again from Axiom 13.4 that $J_{Y^{\prime}}\cap r([i,j])$ is non-
empty.

Suppose now, for a contradiction, that $J_{Y}\cap J_{Y^{\prime}}\cap r([i,j])$
is empty. Define

$h=\max\\{\ell\in[i,j]\mathbin{\mid}r(\ell)\in J_{Y}\\},\quad
k=\min\\{\ell\in[i,j]\mathbin{\mid}r(\ell)\in J_{Y^{\prime}}\\}$

Without loss of generality we may assume that $h<k$. It follows that
$d_{Y^{\prime}}(\gamma_{i},\gamma_{h})<{C_{3}}+2{C_{2}}$. Thus
$d_{Y}(\gamma_{i},\gamma_{h})<{C_{3}}+2{C_{2}}+2L_{4}$. Similarly,
$d_{Y}(\gamma_{h},\gamma_{j})<{C_{3}}+2{C_{2}}$. Deduce

$d_{Y}(\gamma_{i},\gamma_{j})<2{C_{3}}+4{C_{2}}+2L_{4}<{L_{1}}(X),$

the last inequality by (13.16). This is a contradiction to the assumption. ∎

######  Claim.

The interval $[p,q]$ is inductive for $Y$.

###### Proof.

We must check that $d_{Y}(\gamma_{p},\gamma_{q})\geq{L_{1}}(Y)$. Suppose first
that $Y$ is not paired. Then by the definition of $p,q$, (2) of Axiom 13.4,
and the triangle inequality we have

$d_{Y}(\mu_{r(i)},\mu_{r(j)})\leq d_{Y}(\mu_{r(p)},\mu_{r(q)})+2{C_{3}}.$

Thus by Axiom 13.5,

$d_{Y}(\gamma_{i},\gamma_{j})\leq
d_{Y}(\gamma_{p},\gamma_{q})+2{C_{3}}+4{C_{2}}.$

Since by (13.17),

${L_{1}}(Y)+2{C_{3}}+4{C_{2}}<{L_{1}}(X)\leq d_{Y}(\gamma_{i},\gamma_{j})$

we are done.

When $Y$ is paired the proof is similar but we must use the slightly stronger
inequality ${L_{1}}(Y)+2{C_{3}}+4{C_{2}}+4L_{4}<{L_{1}}(X)$. ∎

Thus, when $B_{X}(i,j)$ is non-empty we may find a hole $Y$ and indices $p,q$
as above. In this situation, we subdivide the element
$[i,j]\in\mathcal{P}_{X}$ into the elements $[i,p-1]$, $[p,q]$, and $[q+1,j]$.
(The first or third intervals, or both, may be empty.) The interval
$[p,q]\in\mathcal{P}_{X}$ is determined to be inductive and associated to $Y$.
Proceed on to the next undetermined element. This completes the construction
of $\mathcal{P}_{X}$.

As a bit of notation, if $[i,j]\in\mathcal{P}_{X}$ is associated to $Y\subset
X$ we will sometimes write $I_{Y}=[i,j]$.

###  14.2. Properties of the inductive partition

######  Lemma 14.1.

Suppose that $Y,Z$ are holes and $I_{Z}$ is an inductive element of
$\mathcal{P}_{X}$ associated to $Z$. Suppose that $r(I_{Z})\subset J_{Y}$ (or
$r(I_{Z})\subset J_{Y}\cap J_{Y^{\prime}}$, if $Y$ is paired). Then

  * •

$Z$ is nested in $Y$ or

  * •

$Z$ and $Z^{\prime}$ are paired and $Z^{\prime}$ is nested in $Y$.

###### Proof.

Let $I_{Z}=[i,j]$. Suppose first that $Y$ is strictly nested in $Z$. Then by
(ii) of Lemma 13.13, $d_{Z}(\mu_{r(i)},\mu_{r(j)})<{C_{3}}$. Then by Axiom
13.5

$d_{Z}(\gamma_{i},\gamma_{j})<{C_{3}}+2{C_{2}}<{L_{1}}(Z),$

a contradiction. We reach the same contradiction if $Y$ and $Z$ overlap using
(iii) of Lemma 13.13.

Now, if $Z$ and $Y$ are disjoint then there are two cases: Suppose first that
$Y$ is paired with $Y^{\prime}$. Since all holes interfere, $Y^{\prime}$ and
$Z$ must meet. In this case we are done, just as in the previous paragraph.

Suppose now that $Z$ is paired with $Z^{\prime}$. Since all holes interfere,
$Z^{\prime}$ and $Y$ must meet. If $Z^{\prime}$ is nested in $Y$ then we are
done. If $Y$ is strictly nested in $Z^{\prime}$ then, as $r([i,j])\subset
J_{Y}$, we find that as above by Axioms 13.5 and (ii) of Lemma 13.13 that

$d_{Z^{\prime}}(\gamma_{i},\gamma_{j})<{C_{3}}+2{C_{2}}$

and so $d_{Z}(\gamma_{i},\gamma_{j})<{C_{3}}+2{C_{2}}+2L_{4}<{L_{1}}(Z)$, a
contradiction. We reach the same contradiction if $Y$ and $Z^{\prime}$
overlap. ∎

######  Proposition 14.2.

Suppose $Y\subsetneq X$ is a hole for $\mathcal{G}$.

  1. (1)

If $Y$ is associated to an inductive interval $I_{Y}\in\mathcal{P}_{X}$ and
$Y$ is paired with $Y^{\prime}$ then $Y^{\prime}$ is not associated to any
inductive interval in $\mathcal{P}_{X}$.

  2. (2)

There is at most one inductive interval $I_{Y}\in\mathcal{P}_{X}$ associated
to $Y$.

  3. (3)

There are at most two holes $Z$ and $W$, distinct from $Y$ (and from
$Y^{\prime}$, if $Y$ is paired) such that

     * •

there are inductive intervals $I_{Z}=[h,i]$ and $I_{W}=[j,k]$ and

     * •

$d_{Y}(\gamma_{h},\gamma_{i}),d_{Y}(\gamma_{j},\gamma_{k})\geq{L_{0}}$.

######  Remark 14.3.

It follows that for any hole $Y$ there are at most three inductive intervals
in the partition $\mathcal{P}_{X}$ where $Y$ has projection distance greater
than ${L_{0}}$.

###### Proof of Proposition 14.2.

To prove the first claim: Suppose that $I_{Y}=[p,q]$ and
$I_{Y^{\prime}}=[p^{\prime},q^{\prime}]$ with $q<p^{\prime}$. It follows that
$[r(p),r(q^{\prime})]\subset J_{Y}\cap J_{Y^{\prime}}$. If $q+1=p^{\prime}$
then the partition would have chosen a larger inductive interval for one of
$Y$ or $Y^{\prime}$. It must be the case that there is an inductive interval
$I_{Z}\subset[q+1,p^{\prime}-1]$ for some hole $Z$, distinct from $Y$ and
$Y^{\prime}$, with $\xi(Z)\geq\xi(Y)$. However, by Lemma 14.1 we find that $Z$
is nested in $Y$ or in $Y^{\prime}$. It follows that $Z=Y$ or $Y$, a
contradiction.

The second statement is essentially similar.

Finally suppose that $Z$ and $W$ are the first and last holes, if any,
satisfying the hypotheses of the third claim. Since
$d_{Y}(\gamma_{h},\gamma_{i})\geq{L_{0}}$ we find by Axiom 13.5 that

$d_{Y}(\mu_{r(h)},\mu_{r(i)})\geq{L_{0}}-2{C_{2}}.$

By (13.14), ${L_{0}}-2{C_{2}}>{C_{3}}$ so that

$J_{Y}\cap r(I_{Z})\neq\emptyset.$

If $Y$ is paired then, again by (13.14) we have
${L_{0}}>{C_{3}}+2{C_{2}}+2L_{4}$, we also find that $J_{Y^{\prime}}\cap
r(I_{Z})\neq\emptyset$. Symmetrically, $J_{Y}\cap r(I_{W})$ (and
$J_{Y^{\prime}}\cap r(I_{W})$) are also non-empty.

It follows that the interval between $I_{Z}$ and $I_{W}$, after reindexing, is
contained in $J_{Y}$ (and $J_{Y^{\prime}}$, if $Y$ is paired). Thus for any
inductive interval $I_{V}=[p,q]$ between $I_{Z}$ and $I_{W}$ the associated
hole $V$ is nested in $Y$ (or $V^{\prime}$ is nested in $Y$), by Lemma 14.1.
If $V=Y$ or $V=Y^{\prime}$ there is nothing to prove. Suppose instead that $V$
(or $V^{\prime}$) is strictly nested in $Y$. It follows that

$d_{Y}(\gamma_{p},\gamma_{q})<{C_{3}}+2{C_{2}}<{L_{0}}.$

Thus there are no inductive intervals between $I_{Z}$ and $I_{W}$ satisfying
the hypotheses of the third claim. ∎

The following lemma and proposition bound the number of inductive intervals.
The discussion here is very similar to (and in fact inspired) the antichains
defined in [34, Section 5]. Our situation is complicated by the presence of
non-holes and interfering holes.

######  Lemma 14.4.

Suppose that $X,\alpha_{X},\beta_{X}$ are given, as above. For any
$\ell\geq(3\cdot{L_{2}})^{\xi(X)}$, if $\\{Y_{i}\\}_{i=1}^{\ell}$ is a
collection of distinct strict sub-holes of $X$ each having
$d_{Y_{i}}(\alpha_{X},\beta_{X})\geq{L_{1}}(X)$ then there is a hole
$Z\subseteq X$ such that $d_{Z}(\alpha_{X},\beta_{X})\geq{L_{2}}-1$ and $Z$
contains at least ${L_{2}}$ of the $Y_{i}$. Furthermore, for at least
${L_{2}}-4({C_{1}}+{C_{3}}+2{C_{3}}+2)$ of these $Y_{i}$ we find that
$J_{Y_{i}}\subsetneq J_{Z}$. (If $Z$ is paired then $J_{Y_{i}}\subsetneq
J_{Z}\cap J_{Z^{\prime}}$.) Each of these $Y_{i}$ is disjoint from a distinct
vertex $\eta_{i}\in[\pi_{Z}(\alpha_{X}),\pi_{Z}(\beta_{X})]$.

###### Proof.

Let $g_{X}$ be a geodesic in $\mathcal{C}(X)$ joining $\alpha_{X},\beta_{X}$.
By the Bounded Geodesic Image Theorem (Theorem 4.6), since ${L_{1}}(X)>M_{0}$,
for every $Y_{i}$ there is a vertex $\omega_{i}\in g_{X}$ such that
$Y_{i}\subset X{\smallsetminus}\omega_{i}$. Thus $d_{X}(\omega_{i},\partial
Y_{i})\leq 1$. If there are at least ${L_{2}}$ distinct $\omega_{i}$,
associated to distinct $Y_{i}$, then
$d_{X}(\alpha_{X},\beta_{X})\geq{L_{2}}-1$. In this situation we take $Z=X$.
Since $J_{X}=[0,N]$ we are done.

Thus assume there do not exist at least ${L_{2}}$ distinct $\omega_{i}$. Then
there is some fixed $\omega$ among these $\omega_{i}$ such that at least
$\frac{\ell}{{L_{2}}}\geq 3(3\cdot{L_{2}})^{\xi(X)-1}$ of the $Y_{i}$ satisfy

$Y_{i}\subset(X{\smallsetminus}\omega).$

Thus one component, call it $W$, of $X{\smallsetminus}\omega$ contains at
least $(3\cdot{L_{2}})^{\xi(X)-1}$ of the $Y_{i}$. Let $g_{W}$ be a geodesic
in $\mathcal{C}(W)$ joining $\alpha_{W}=\pi_{W}(\alpha_{X})$ and
$\beta_{W}=\pi_{W}(\beta_{X})$. Notice that

$d_{Y_{i}}(\alpha_{W},\beta_{W})\geq d_{Y_{i}}(\alpha_{X},\beta_{X})-8$

because we are projecting to nested subsurfaces. This follows for example from
Lemma 4.4. Hence $d_{Y_{i}}(\alpha_{W},\beta_{W})\geq{L_{1}}(W)$.

Again apply Theorem 4.6. Since ${L_{1}}(W)>M_{0}$, for every remaining $Y_{i}$
there is a vertex $\eta_{i}\in g_{W}$ such that

$Y_{i}\subset(W{\smallsetminus}\eta_{i})$

If there are at least ${L_{2}}$ distinct $\eta_{i}$ then we take $Z=W$.
Otherwise we repeat the argument. Since the complexity of each successive
subsurface is decreasing by at least $1$, we must eventually find the desired
$Z$ containing at least ${L_{2}}$ of the $Y_{i}$, each disjoint from distinct
vertices of $g_{Z}$.

So suppose that there are at least ${L_{2}}$ distinct $\eta_{i}$ associated to
distinct $Y_{i}$ and we have taken $Z=W$. Now we must find at least
${L_{2}}-4({C_{1}}+{C_{3}}+2{C_{3}}+2)$ of these $Y_{i}$ where
$J_{Y_{i}}\subsetneq J_{Z}$.

To this end we focus attention on a small subset
$\\{Y^{j}\\}_{j=1}^{5}\subset\\{Y_{i}\\}$. Let $\eta_{j}$ be the vertex of
$g_{Z}=g_{W}$ associated to $Y^{j}$. We choose these $Y^{j}$ so that

  * •

the $\eta_{j}$ are arranged along $g_{Z}$ in order of index and

  * •

$d_{Z}(\eta_{j},\eta_{j+1})>{C_{1}}+{C_{3}}+2{C_{3}}+2$, for $j=1,2,3,4$.

This is possible by (13.15) because

${L_{2}}>4({C_{1}}+{C_{3}}+2{C_{3}}).$

Set $J_{j}=J_{Y^{j}}$ and pick any indices $m_{j}\in J_{j}$. (If $Z$ is paired
then $Y^{j}$ is as well and we pick $m_{j}\in J_{Y^{j}}\cap
J_{(Y^{j})^{\prime}}$.) We use $\mu(m_{j})$ to denote $\mu_{m_{j}}$. Since
$\partial Y^{j}$ is disjoint from $\eta_{j}$, Axiom 13.4 and Lemma 2.2 imply

(14.5) $d_{Z}(\mu(m_{j}),\eta_{j})\leq{C_{3}}+1.$

Since the sequence $\pi_{Z}(\mu_{n})$ satisfies the reverse triangle
inequality (Lemma 13.12), it follows that the $m_{j}$ appear in $[0,N]$ in
order agreeing with their index. The triangle inequality implies that

$d_{Z}(\mu(m_{1}),\mu(m_{2}))>{C_{3}}.$

Thus Axiom 13.4 implies that $J_{Z}\cap[m_{1},m_{2}]$ is non-empty. Similarly,
$J_{Z}\cap[m_{4},m_{5}]$ is non-empty. It follows that $[m_{2},m_{4}]\subset
J_{Z}$. (If $Z$ is paired then, after applying the symmetry $\tau$ to $g_{Z}$,
the same argument proves $[m_{2},m_{4}]\subset J_{Z^{\prime}}$.)

Notice that $J_{2}\cap J_{3}=\emptyset$. For if $m\in J_{2}\cap J_{3}$ then by
(14.5) both $d_{Z}(\mu_{m},\eta_{2})$ and $d_{Z}(\mu_{m},\eta_{3})$ are
bounded by ${C_{3}}+1$. It follows that

$d_{Z}(\eta_{2},\eta_{3})<2{C_{3}}+2,$

a contradiction. Similarly $J_{3}\cap J_{4}=\emptyset$. We deduce that
$J_{3}\subsetneq[m_{2},m_{4}]\subset J_{Z}$. (If $Z$ is paired $J_{3}\subset
J_{Z}\cap J_{Z^{\prime}}$.) Finally, there are at least

${L_{2}}-4({C_{1}}+{C_{3}}+2{C_{3}}+2)$

possible $Y_{i}$’s which satisfy the hypothesis on $Y^{3}$. This completes the
proof. ∎

Define

$\mathcal{P}_{\text{ind}}=\\{I\in\mathcal{P}_{X}\mathbin{\mid}\mbox{ $I$ is
inductive}\\}.$

######  Proposition 14.6.

The number of inductive intervals is a lower bound for the projection distance
in $X$:

$d_{X}(\alpha_{X},\beta_{X})\geq\frac{|\mathcal{P}_{\text{ind}}|}{2(3\cdot{L_{2}})^{\xi(X)-1}+1}-1.$

###### Proof.

Suppose, for a contradiction, that the conclusion fails. Let $g_{X}$ be a
geodesic in $\mathcal{C}(X)$ connecting $\alpha_{X}$ to $\beta_{X}$. Then, as
in the proof of Lemma 14.4, there is a vertex $\omega$ of $g_{X}$ and a
component $W\subset X{\smallsetminus}\omega$ where at least
$(3\cdot{L_{2}})^{\xi(X)-1}$ of the inductive intervals in $I_{X}$ have
associated surfaces, $Y_{i}$, contained in $W$.

Since $\xi(X)-1\geq\xi(W)$ we may apply Lemma 14.4 inside of $W$. So we find a
surface $Z\subseteq W\subsetneq X$ so that

  * •

$Z$ contains at least ${L_{2}}$ of the $Y_{i}$,

  * •

$d_{Z}(\alpha_{X},\beta_{X})\geq{L_{2}}$, and

  * •

there are at least ${L_{2}}-4({C_{1}}+{C_{3}}+2{C_{3}}+2)$ of the $Y_{i}$
where $J_{Y_{i}}\subsetneq J_{Z}$.

Since $Y_{i}\subsetneq Z$ and $Y_{i}$ is a hole, $Z$ is also a hole. Since
${L_{2}}>{L_{1}}(X)$ it follows that $Z\in B_{X}$. Let
$\mathcal{Y}=\\{Y_{i}\\}$ be the set of $Y_{i}$ satisfying the third bullet.
Let $Y^{1}\in\mathcal{Y}$ and $\eta_{1}\in g_{Z}$ satisfy $\partial
Y^{1}\cap\eta_{1}=\emptyset$ and $\eta_{1}$ is the first such. Choose $Y^{2}$
and $\eta_{2}$ similarly, so that $\eta_{2}$ is the last such. By Lemma 14.4

(14.7) $d_{Z}(\eta_{1},\eta_{2})\geq L_{2}-4({C_{1}}+{C_{3}}+2{C_{3}}+2).$

Let $p=\min I_{Y^{1}}$ and $q=\max I_{Y^{2}}$. Note that $[p,q]\subset J_{Z}$.
(If $Z$ is paired with $Z^{\prime}$ then $[p,q]\subset J_{Z}\cap
J_{Z^{\prime}}$.) Again by (1) of Axiom 13.4, and Lemma 2.2,

$d_{Z}(\mu_{r(p)},\partial Y^{1})<{C_{3}}.$

It follows that

$d_{Z}(\mu_{r(p)},\eta_{1})\leq{C_{3}}+1$

and the same bound applies to $d_{Z}(\mu_{r(q)},\eta_{2})$. Combined with
(14.7) we find that

$d_{Z}(\mu_{r(p)},\mu_{r(q)})\geq{L_{2}}-4{C_{1}}-4{C_{3}}-10{C_{3}}-10.$

By the reverse triangle inequality (Lemma 13.12), for any $p^{\prime}\leq
p,q\leq q^{\prime}$,

$d_{Z}(\mu_{r(p^{\prime})},\mu_{r(q^{\prime})})\geq{L_{2}}-6{C_{1}}-4{C_{3}}-10{C_{3}}-10.$

Finally by Axiom 13.5 and the above inequality we have

$d_{Z}(\gamma_{p^{\prime}},\gamma_{q^{\prime}})\geq{L_{2}}-6{C_{1}}-4{C_{3}}-10{C_{3}}-10-2{C_{2}}.$

By (13.15) the right-hand side is greater than ${L_{1}}(X)+2L_{4}$ so we
deduce that $Z\in B_{X}(p^{\prime},q^{\prime})$, for any such
$p^{\prime},q^{\prime}$. (When $Z$ is paired deduce also that $Z^{\prime}\in
B_{X}(p^{\prime},q^{\prime})$.)

Let $I_{V}$ be the first inductive interval chosen by the procedure with the
property that $I_{V}\cap[p,q]\neq\emptyset$. Note that, since $I_{Y^{1}}$ and
$I_{Y^{2}}$ will also be chosen, $I_{V}\subset[p,q]$. Let
$p^{\prime},q^{\prime}$ be the indices so that $V$ is chosen from
$B_{X}(p^{\prime},q^{\prime})$. Thus $p^{\prime}\leq p$ and $q\leq
q^{\prime}$. However, since $I_{V}\subset[p,q]\subset J_{Z}$, Lemma 14.1
implies that $V$ is strictly nested in $Z$. (When pairing occurs we may find
instead that $V\subset Z^{\prime}$ or $V^{\prime}\subset Z$.) Thus
$\xi(Z)>\xi(V)$ and we find that $Z$ would be chosen from
$B_{X}(p^{\prime},q^{\prime})$, instead of $V$. This is a contradiction. ∎

###  14.3. Electric partition

The goal of this subsection is to prove:

######  Proposition 14.8.

There is a constant $A$ depending only on $\xi(X)$, so that: if
$[i,j]\subset[0,K]$ is a electric interval then

$d_{\mathcal{G}}(\gamma_{i},\gamma_{j})\mathbin{\leq_{A}}d_{X}(\gamma_{i},\gamma_{j}).$

We begin by building a partition of $[i,j]$ into straight and shortcut
intervals. Define

$C_{X}=\\{Y\subsetneq X\mathbin{\mid}\mbox{$Y$ is a non-hole and
}d_{Y}(\mu_{r(i)},\mu_{r(j)})\geq{L_{1}}(X)\\}.$

We also define, for all $[p,q]\subset[i,j]$

$C_{X}(p,q)=\\{Y\in C_{X}\mathbin{\mid}J_{Y}\cap[r(p),r(q)]\neq\emptyset\\}.$

Our recursion starts with the partition of one part,
$\mathcal{P}(i,j)=\\{[i,j]\\}$. Recursively $\mathcal{P}(i,j)$ is a partition
of $[i,j]$ into shortcut, straight, or undetermined intervals. Suppose that
$[p,q]\in\mathcal{P}(i,j)$ is undetermined.

######  Claim.

If $C_{X}(p,q)$ is empty then $[p,q]$ is straight.

###### Proof.

We show the contrapositive. Suppose that $Y$ is a non-hole with
$d_{Y}(\mu_{r(p)},\mu_{r(q)})\geq{L_{2}}$. Since ${L_{2}}>{C_{3}}$, Axiom 13.4
implies that $J_{Y}\cap[r(p),r(q)]$ is non-empty. Also, the reverse triangle
inequality (Lemma 13.12) gives:

$d_{Y}(\mu_{r(p)},\mu_{r(q)})<d_{Y}(\mu_{r(i)},\mu_{r(j)})+2{C_{1}}.$

Since ${L_{2}}>{L_{1}}(X)+2{C_{1}}$, we find that $Y\in C_{X}$. It follows
that $Y\in C_{X}(p,q)$. ∎

So when $C_{X}(p,q)$ is empty the interval $[p,q]$ is determined to be
straight. Proceed onto the next undetermined element of $\mathcal{P}(i,j)$.
Now suppose that $C_{X}(p,q)$ is non-empty. Then we choose any $Y\in
C_{X}(p,q)$ so that $Y$ has maximal $\xi(Y)$ amongst the elements of
$C_{X}(p,q)$. Notice that by the accessibility requirement that
$J_{Y}\cap[r(p),r(q)]$ is non-empty.

There are two cases. If $J_{Y}\cap r([p,q])$ is empty then let
$p^{\prime}\in[p,q]$ be the largest integer so that $r(p^{\prime})<\min
J_{Y}$. Note that $p^{\prime}$ is well-defined. Now divide the interval
$[p,q]$ into the two undetermined intervals $[p,p^{\prime}]$,
$[p^{\prime}+1,q]$. In this situation we say $Y$ is associated to a shortcut
of length one and we add the element $[p^{\prime}+\frac{1}{2}]$ to
$\mathcal{P}(i,j)$.

Next suppose that $J_{Y}\cap r([p,q])$ is non-empty. Let
$p^{\prime},q^{\prime}\in[p,q]$ be the first and last indices, respectively,
so that $r(p^{\prime}),r(q^{\prime})\in J_{Y}$. (Note that it is possible to
have $p^{\prime}=q^{\prime}$.) Partition
$[p,q]=[p,p^{\prime}-1]\cup[p^{\prime},q^{\prime}]\cup[q^{\prime}+1,q]$. The
first and third parts are undetermined; either may be empty. This completes
the recursive construction of the partition.

Define

$\mathcal{P}_{\text{short}}=\\{I\in\mathcal{P}(i,j)\mathbin{\mid}\mbox{$I$ is
a shortcut}\\}$

and

$\mathcal{P}_{\text{str}}=\\{I\in\mathcal{P}(i,j)\mathbin{\mid}\mbox{$I$ is
straight}\\}.$

######  Proposition 14.9.

With $\mathcal{P}(i,j)$ as defined above,

$d_{X}(\gamma_{i},\gamma_{j})\geq\frac{|\mathcal{P}_{\text{short}}|}{2(3\cdot{L_{2}})^{\xi(X)-1}+1}-1.$

###### Proof.

The proof is identical to that of Proposition 14.6 with the caveat that in
Lemma 14.4 we must use the markings $\mu_{r(i)}$ and $\mu_{r(j)}$ instead of
the endpoints $\gamma_{i}$ and $\gamma_{j}$. ∎

Now we “electrify” every shortcut interval using Theorem 13.1 recursively.

######  Lemma 14.10.

There is a constant ${L_{3}}={L_{3}}(X,\mathcal{G})$, so that for every
shortcut interval $[p,q]$ we have
$d_{\mathcal{G}}(\gamma_{p},\gamma_{q})<{L_{3}}$.

###### Proof.

As $[p,q]$ is a shortcut we are given a non-hole $Z\subset X$ so that
$r([p,q])\subset J_{Z}$. Let $Y=X{\smallsetminus}Z$. Thus Axiom 13.6 gives
vertices $\gamma_{p}^{\prime},\gamma_{q}^{\prime}$ of $\mathcal{G}$ lying in
$Y$ or in $Z$, so that
$d_{\mathcal{G}}(\gamma_{p},\gamma_{p}^{\prime}),d_{\mathcal{G}}(\gamma_{q},\gamma_{q}^{\prime})\leq{C_{4}}$.

If one of $\gamma_{p}^{\prime},\gamma_{q}^{\prime}$ lies in $Y$ while the
other lies in $Z$ then

$d_{\mathcal{G}}(\gamma_{p},\gamma_{q})<2{C_{4}}+1.$

If both lie in $Z$ then, as $Z$ is a non-hole, there is a vertex
$\delta\in\mathcal{G}(S)$ disjoint from both of $\gamma_{p}^{\prime}$ and
$\gamma_{q}^{\prime}$ and we have

$d_{\mathcal{G}}(\gamma_{p},\gamma_{q})<2{C_{4}}+2.$

If both lie in $Y$ then there are two cases. If $Y$ is not a hole for
$\mathcal{G}(S)$ then we are done as in the previous case. If $Y$ is a hole
then by the definition of shortcut interval, Lemma 5.7, and the triangle
inequality we have

$d_{W}(\gamma_{p}^{\prime},\gamma_{q}^{\prime})<6+6{C_{4}}+{L_{2}}$

for all holes $W\subset Y$. Notice that $Y$ is strictly contained in $X$. Thus
we may inductively apply Theorem 13.1 with $c=6+6{C_{4}}+{L_{2}}$. We deduce
that all terms on the right-hand side of the distance estimate vanish and thus
$d_{\mathcal{G}}(\gamma_{p}^{\prime},\gamma_{q}^{\prime})$ is bounded by a
constant depending only on $X$ and $\mathcal{G}$. The same then holds for
$d_{\mathcal{G}}(\gamma_{p},\gamma_{q})$ and we are done. ∎

We are now equipped to give:

###### Proof of Proposition 14.8.

Suppose that $\mathcal{P}(i,j)$ is the given partition of the electric
interval $[i,j]$ into straight and shortcut subintervals. As a bit of
notation, if $[p,q]=I\in\mathcal{P}(i,j)$, we take
$d_{\mathcal{G}}(I)=d_{\mathcal{G}}(\gamma_{p},\gamma_{q})$ and
$d_{X}(I)=d_{X}(\gamma_{p},\gamma_{q})$. Applying Axiom 13.5 we have

(14.11) $\displaystyle d_{\mathcal{G}}(\gamma_{i},\gamma_{j})$
$\displaystyle\leq\sum_{I\in\mathcal{P}_{\text{str}}}d_{\mathcal{G}}(I)+\sum_{I\in\mathcal{P}_{\text{short}}}d_{\mathcal{G}}(I)+{C_{2}}|\mathcal{P}(i,j)|$

The last term arises from connecting left endpoints of intervals with right
endpoints. We must bound the three terms on the right.

We begin with the third; recall that
$|\mathcal{P}(i,j)|=|\mathcal{P}_{\text{short}}|+|\mathcal{P}_{\text{str}}|$,
that $|\mathcal{P}_{\text{str}}|\leq|\mathcal{P}_{\text{short}}|+1$, and that
$|\mathcal{P}_{\text{short}}|\mathbin{\leq_{A}}d_{X}(\gamma_{i},\gamma_{j})$.
The second inequality follows from the construction of the partition while the
last is implied by Proposition 14.9. Thus the third term of Equation 14.11 is
quasi-bounded above by $d_{X}(\gamma_{i},\gamma_{j})$.

By Lemma 14.10, the second term of Equation 14.11 at most
${L_{3}}|\mathcal{P}_{\text{short}}|$. Finally, by Axiom 13.11, for all
$I\in\mathcal{P}_{\text{str}}$ we have

$d_{\mathcal{G}}(I)\mathbin{\leq_{A}}d_{X}(I),$

Also, it follows from the reverse triangle inequality (Lemma 13.12) that

$\sum_{I\in\mathcal{P}_{\text{str}}}d_{X}(I)\leq
d_{X}(\gamma_{i},\gamma_{j})+(2{C_{1}}+2{C_{2}})|\mathcal{P}_{\text{str}}|+2{C_{2}}.$

We deduce that $\sum_{I\in\mathcal{P}_{\text{str}}}d_{\mathcal{G}}(I)$ is also
quasi-bounded above by $d_{X}(\gamma_{i},\gamma_{j})$. Thus for a somewhat
larger value of $A$ we find

$d_{\mathcal{G}}(\gamma_{i},\gamma_{j})\mathbin{\leq_{A}}d_{X}(\gamma_{i},\gamma_{j}).$

This completes the proof. ∎

###  14.4. The upper bound

We will need:

######  Proposition 14.12.

For any $c>0$ there is a constant $A$ with the following property. Suppose
that $[i,j]=I_{Y}$ is an inductive interval in $\mathcal{P}_{X}$. Then we
have:

$d_{\mathcal{G}}(\gamma_{i},\gamma_{j})\mathbin{\leq_{A}}\sum_{Z}[d_{Z}(\gamma_{i},\gamma_{j})]_{c}$

where $Z$ ranges over all holes for $\mathcal{G}$ contained in $X$.

###### Proof.

Axiom 13.6 gives vertices $\gamma^{\prime}_{i}$,
$\gamma^{\prime}_{j}\in\mathcal{G}$, contained in $Y$, so that
$d_{\mathcal{G}}(\gamma_{i},\gamma^{\prime}_{i})\leq{C_{4}}$ and the same
holds for $j$. Since projection to holes is coarsely Lipschitz (Lemma 5.7) for
any hole $Z$ we have $d_{Z}(\gamma_{i},\gamma^{\prime}_{i})\leq 3+3{C_{3}}$.

Fix any $c>0$. Now, since

$\displaystyle d_{\mathcal{G}}(\gamma_{i},\gamma_{j})$ $\displaystyle\leq
d_{\mathcal{G}}(\gamma^{\prime}_{i},\gamma^{\prime}_{j})+2{C_{3}}$

to find the required constant $A$ it suffices to bound
$d_{\mathcal{G}}(\gamma^{\prime}_{i},\gamma^{\prime}_{j})$. Let
$c^{\prime}=c+6{C_{3}}+6$. Since $Y\subsetneq X$, induction gives us a
constant $A$ so that

$\displaystyle d_{\mathcal{G}}(\gamma^{\prime}_{i},\gamma^{\prime}_{j})$
$\displaystyle\mathbin{\leq_{A}}\sum_{Z}[d_{Z}(\gamma^{\prime}_{i},\gamma^{\prime}_{j})]_{c^{\prime}}$
$\displaystyle\leq\sum_{Z}[d_{Z}(\gamma_{i},\gamma_{j})+6{C_{3}}+6]_{c^{\prime}}$
$\displaystyle<(6{C_{3}}+6)N+\sum_{Z}[d_{Z}(\gamma_{i},\gamma_{j})]_{c}$

where $N$ is the number of non-zero terms in the final sum. Also, the sum
ranges over sub-holes of $Y$. We may take $A$ somewhat larger to deal with the
term $(6{C_{3}}+6)N$ and include all holes $Z\subset X$ to find

$\displaystyle d_{\mathcal{G}}(\gamma_{i},\gamma_{j})$
$\displaystyle\mathbin{\leq_{A}}\sum_{Z}[d_{Z}(\gamma_{i},\gamma_{j})]_{c}$

where the sum is over all holes $Z\subset X$. ∎

###  14.5. Finishing the proof

Now we may finish the proof of Theorem 13.1. Fix any constant $c\geq 0$.
Suppose that $X$, $\alpha_{X}$, $\beta_{X}$ are given as above. Suppose that
$\Gamma=\\{\gamma_{i}\\}_{i=0}^{K}$ is the given combinatorial path and
$\mathcal{P}_{X}$ is the partition of $[0,K]$ into inductive and electric
intervals. So we have:

(14.13) $\displaystyle d_{\mathcal{G}}(\alpha_{X},\beta_{X})$
$\displaystyle\leq\sum_{I\in\mathcal{P}_{\text{ind}}}d_{\mathcal{G}}(I)+\sum_{I\in\mathcal{P}_{\text{ele}}}d_{\mathcal{G}}(I)+{C_{2}}|\mathcal{P}_{X}|$

Again, the last term arises from adjacent right and left endpoints of
different intervals.

We must bound the terms on the right-hand side; begin by noticing that
$|\mathcal{P}_{X}|=|\mathcal{P}_{\text{ind}}|+|\mathcal{P}_{\text{ele}}|$,
$|\mathcal{P}_{\text{ele}}|\leq|\mathcal{P}_{\text{ind}}|+1$ and
$|\mathcal{P}_{\text{ind}}|\mathbin{\leq_{A}}d_{X}(\alpha_{X},\beta_{X})$. The
second inequality follows from the way the partition is constructed and the
last follows from Proposition 14.6. Thus the third term of Equation 14.13 is
quasi-bounded above by $d_{X}(\alpha_{X},\beta_{X})$.

Next consider the second term of Equation 14.13:

$\displaystyle\sum_{I\in\mathcal{P}_{\text{ele}}}d_{\mathcal{G}}(I)$
$\displaystyle\mathbin{\leq_{A}}\sum_{I\in\mathcal{P}_{\text{ele}}}d_{X}(I)$
$\displaystyle\leq
d_{X}(\alpha_{X},\beta_{X})+(2{C_{1}}+2{C_{2}})|\mathcal{P}_{\text{ele}}|+2{C_{2}}$

with the first inequality following from Proposition 14.8 and the second from
the reverse triangle inequality (Lemma 13.12).

Finally we bound the first term of Equation 14.13. Let $c^{\prime}=c+{L_{0}}$.
Thus,

$\displaystyle\sum_{I\in\mathcal{P}_{\text{ind}}}d_{\mathcal{G}}(I)$
$\displaystyle\leq\sum_{I_{Y}\in\mathcal{P}_{\text{ind}}}\left(A^{\prime}_{Y}\left(\sum_{Z\subsetneq
Y}[d_{Z}(I_{Y})]_{c^{\prime}}\right)+A^{\prime}_{Y}\right)$ $\displaystyle\leq
A^{\prime\prime}\left(\sum_{I\in\mathcal{P}_{\text{ind}}}\sum_{Z\subsetneq
X}[d_{Z}(I)]_{c^{\prime}}\right)+A^{\prime\prime}\cdot|\mathcal{P}_{\text{ind}}|$
$\displaystyle\leq A^{\prime\prime}\left(\sum_{Z\subsetneq
X}\sum_{I\in\mathcal{P}_{\text{ind}}}[d_{Z}(I)]_{c^{\prime}}\right)+A^{\prime\prime}\cdot|\mathcal{P}_{\text{ind}}|$

Here $A^{\prime}_{Y}$ and the first inequality are given by Proposition 14.12.
Also $A^{\prime\prime}=\max\\{A^{\prime}_{Y}\mathbin{\mid}Y\subsetneq X\\}$.
In the last line, each sum of the form
$\sum_{I\in\mathcal{P}_{\text{ind}}}[d_{Z}(I)]_{c^{\prime}}$ has at most three
terms, by Remark 14.3 and the fact that $c^{\prime}>{L_{0}}$. For the moment,
fix a hole $Z$ and any three elements
$I,I^{\prime},I^{\prime\prime}\in\mathcal{P}_{\text{ind}}$.

By the reverse triangle inequality (Lemma 13.12) we find that

$d_{Z}(I)+d_{Z}(I^{\prime})+d_{Z}(I^{\prime\prime})<d_{Z}(\alpha_{X},\beta_{X})+6{C_{1}}+8{C_{2}}$

which in turn is less than $d_{Z}(\alpha_{X},\beta_{X})+{L_{0}}$.

It follows that

$[d_{Z}(I)]_{c^{\prime}}+[d_{Z}(I^{\prime})]_{c^{\prime}}+[d_{Z}(I^{\prime\prime})]_{c^{\prime}}<[d_{Z}(\alpha_{X},\beta_{X})]_{c}+{L_{0}}.$

Thus,

$\displaystyle\sum_{Z\subsetneq
X}\sum_{I\in\mathcal{P}_{\text{ind}}}[d_{Z}(I)]_{c^{\prime}}$
$\displaystyle\leq{L_{0}}\cdot N+\sum_{Z\subsetneq
X}[d_{Z}(\alpha_{X},\beta_{X})]_{c}$

where $N$ is the number of non-zero terms in the final sum. Also, the sum
ranges over all holes $Z\subsetneq X$.

Combining the above inequalities, and increasing $A$ once again, implies that

$d_{\mathcal{G}}(\alpha_{X},\beta_{X})\mathbin{\leq_{A}}\sum_{Z}[d_{Z}(\alpha_{X},\beta_{X})]_{c}$

where the sum ranges over all holes $Z\subseteq X$. This completes the proof
of Theorem 13.1. ∎

##  15\. Background on Teichmüller space

Our goal in Sections 16, 17 and 19 will be to verify the axioms stated in
Section 13 for the complex of curves of a non-orientable surface, for the arc
complex, and for the disk complex. Here we give the necessary background on
Teichmüller space.

Fix now a surface $S=S_{g,n}$ of genus $g$ with $n$ punctures. Two conformal
structures on $S$ are equivalent, written $\Sigma\sim\Sigma^{\prime}$, if
there is a conformal map $f\colon\Sigma\to\Sigma^{\prime}$ which is isotopic
to the identity. Let $\mathcal{T}=\mathcal{T}(S)$ be the Teichmüller space of
$S$; the set of equivalence classes of conformal structures $\Sigma$ on $S$.

Define the Teichmüller metric by,

$d_{\mathcal{T}}(\Sigma,\Sigma^{\prime})=\inf_{f}\left\\{\frac{1}{2}\log
K(f)\right\\}$