# On train track splitting sequences

Howard Masur  Department of Mathematics
University of Chicago
Chicago, IL 60607, USA<EMAIL_ADDRESS>,  Lee Mosher  Department of
Mathematics
Rutgers-Newark, State University of New Jersey
Newark, NJ 07102, USA<EMAIL_ADDRESS>and  Saul Schleimer  Department of
Mathematics
University of Warwick
Coventry, CV4 7AL, UK<EMAIL_ADDRESS>
###### Abstract.

We show that the subsurface projection of a train track splitting sequence is
an unparameterized quasi-geodesic in the curve complex of the subsurface. For
the proof we introduce induced tracks, efficient position, and wide curves.

This result is an important step in the proof that the disk complex is Gromov
hyperbolic. As another application we show that train track sliding and
splitting sequences give quasi-geodesics in the train track graph,
generalizing a result of Hamenstädt [Invent. Math.].

Masur and Mosher are grateful to the National Science Foundation for its
suport.

This work is in the public domain.

##  1\. Introduction

Train track splitting sequences are an integral part of the study of surface
diffeomorphisms [19, 15, 14]. There are links between splitting sequences of
tracks and the curves they carry, on the one hand, and Teichmüller geodesics
and measured foliations on the other. Of particular importance, in light of
the hierarchy machine [12] and the closely related control of Teichmüller
geodesics [16, 17], is to understand how a splitting sequence interacts with
subsurface projection. We give a detailed structure theorem (Theorem 5.3) that
explains this interaction. Our main application is the following result,
needed for Masur and Schleimer’s work on the disk complex.

For any surface $S$ with $\xi(S)\geq 1$ there is a constant $Q=Q(S)$ with the
following property: For any sliding and splitting sequence
$\\{\tau_{i}\\}_{i=0}^{N}$ of birecurrent train tracks in $S$ and for any
essential subsurface $X\subset S$ if $\pi_{X}(\tau_{N})\neq\emptyset$ then the
sequence $\\{\pi_{X}(\tau_{i})\\}_{i=0}^{N}$ is a $Q$–unparameterized quasi-
geodesic in the curve complex $\mathcal{C}(X)$.

As another application of Theorem 5.3 we generalize, via a very different
proof, a result of Hamenstädt [8, Corollary 3]:

For any surface $S$ with $\xi(S)\geq 1$ there is a constant $Q=Q(S)$ with the
following property: If $\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding and splitting
sequence in the train track graph $\mathcal{T}(S)$, injective on slide
subsequences, then $\\{\tau_{i}\\}$ is a $Q$–quasi-geodesic.

For the proof of Theorem 5.3 we introduce induced tracks, efficient position,
and wide curves. For any essential subsurface $X\subset S$ and track
$\tau\subset S$ there is an induced track $\tau|X$. Induced tracks generalize
the notion of subsurface projection of curves. Efficient position of a curve
with respect to a track $\tau$ is a simultaneous generalization of curves
carried by $\tau$ and curves dual to $\tau$ (called hitting $\tau$ efficiently
in [15]). Efficient position of $\partial X$ allows us to pin down the
location of the induced track $\tau|X$. Wide curves are our combinatorial
analogue of curves of definite modulus in a Riemann surface. The structure
theorem (5.3) then implies Theorem 5.5: this, together with subsurface
projection, controls the motion of a splitting sequence through the complex of
curves $\mathcal{C}(X)$.

Theorem 6.2, our second application of the structure theorem, is a direct
consequence of Theorem 6.1, stated in terms of the marking graph. Theorem 6.1
requires a delicate induction proof conceptually similar to the hierarchy
machine developed in [12]. We do not deduce Theorem 6.1 directly from the
results of [12]; in particular it is not known if splitting sequences fellow-
travel resolutions of hierarchies.

### Acknowledgments

We thank Yair Minsky for enlightening conversations.

##  2\. Background

We provide the definitions needed for Theorem 5.3 and its corollaries.

###  2.1. Coarse geometry

Suppose that $Q\geq 1$ is a real number. For real numbers $r,s$ we write
$r\leq_{Q}s$ if $r\leq Qs+Q$ and say that $r$ is quasi-bounded by $s$. We
write $r=_{Q}s$ if $r\leq_{Q}s$ and $s\leq_{Q}r$; this is called a quasi-
equality.

For a metric space $(\mathcal{X},d_{\mathcal{X}})$ and finite diameter subsets
$A,B\subset\mathcal{X}$ define
$d_{\mathcal{X}}(A,B)=\operatorname{diam}_{\mathcal{X}}(A\cup B)$. Following
Gromov [6], a relation $f\colon\mathcal{X}\to\mathcal{Y}$ of metric spaces is
a $Q$–quasi-isometric embedding if for all $x,y\in\mathcal{X}$ we have
$d_{\mathcal{X}}(x,y)=_{Q}d_{\mathcal{Y}}(f(x),f(y))$. (Here
$f(x)\subset\mathcal{Y}$ is the set of points related to $x$.) If,
additionally, the $Q$–neighborhood of $f(\mathcal{X})$ equals $\mathcal{Y}$
then $f$ is a $Q$–quasi-isometry and $\mathcal{X}$ and $\mathcal{Y}$ are
quasi-isometric.

If $[m,n]$ is an interval in $\mathbb{Z}$ and $f\colon[m,n]\to\mathcal{Y}$ is
a quasi-isometric embedding then $f$ is a $Q$–quasi-geodesic. Now suppose that
$Q>1$ is a real number, $[m,n]$ and $[p,q]$ are intervals in $\mathbb{Z}$, and
$f\colon[m,n]\to\mathcal{Y}$ is a relation. Then $f$ is a $Q$–unparameterized
quasi-geodesic if there is a strictly increasing function
$\rho\colon[p,q]\to[m,n]$ so that $f\circ\rho$ is a $Q$–quasi-geodesic and for
all $i\in[p,q-1]$ the diameter of
$f\left(\big{[}\rho(i),\rho(i+1)\big{]}\right)$ is at most $Q$.

###  2.2. Surfaces, arcs, and curves

Let $S=S_{g,n}$ be a compact, connected, orientable surface of genus $g$ with
$n$ boundary components. The complexity of $S$ is $\xi(S)=3g-3+n$. A curve in
$S$ is an embedding of the circle into $S$. An arc in $S$ is a proper
embedding of the interval $[0,1]$ into $S$. A curve or arc $\alpha\subset S$
is trivial if $\alpha$ separates $S$ and one component of
$S{\smallsetminus}\alpha$ is a disk; otherwise $\alpha$ is essential. A curve
$\alpha$ is peripheral if $\alpha$ separates $S$ and one component of
$S{\smallsetminus}\alpha$ is an annulus; otherwise $\alpha$ is non-peripheral.
A connected subsurface $X\subset S$ is essential if every component of
$\partial X$ is essential in $S$ and $X$ is neither a pair of pants
($S_{0,3}$) nor a peripheral annulus (the core curve is peripheral). Note that
$X$ inherits an orientation from $S$. This, in turn, induces an orientation on
$\partial X$ so that $X$ is to the left of $\partial X$.

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 in $S$. Let
$\mathcal{AC}(S)=\mathcal{C}(S)\cup\mathcal{A}(S)$. If
$\alpha,\beta\in\mathcal{AC}(S)$ then the geometric intersection number [5] of
$\alpha$ and $\beta$ is

$i(\alpha,\beta)=\min\big{\\{}|a\cap b|:a\in\alpha,b\in\beta\big{\\}}.$

A finite subset $\Delta\subset\mathcal{AC}(S)$ is a multicurve if
$i(\alpha,\beta)=0$ for all $\alpha,\beta\in\Delta$.

If $T\subset S$ is a subsurface with $\partial T$ a union of smooth arcs,
meeting perpendicularly at their endpoints, then define

$\operatorname{index}(T)=\chi(T)-\frac{c^{+}(T)}{4}+\frac{c^{-}(T)}{4}$

where $c^{\pm}(T)$ is the number of outward (inward) corners of $\partial T$.
Note that index is additive: $\operatorname{index}(T\cup
T^{\prime})=\operatorname{index}(T)+\operatorname{index}(T^{\prime})$ as long
as the interiors of $T,T^{\prime}$ are disjoint.

###  2.3. Train tracks

For detailed discussion of train tracks see [19, 15, 14]. A pretrack
$\tau\subset S$ is a properly embedded graph in $S$ with additional structure.
The vertices of $\tau$ are called switches; every switch $x$ is equipped with
a tangent $v_{x}\in T_{x}^{1}S$. We require every switch to have valence
three; higher valence is dealt with in [15]. The edges of $\tau$ are called
branches. All branches are smoothly embedded in $S$. All branches incident to
a fixed switch $x$ have derivative $\pm v_{x}$ at $x$.

An immersion $\rho\colon\mathbb{R}\to S$ is a train-route (or simply a route)
if

  * •

$\rho(\mathbb{R})\subset\tau$ and

  * •

$\rho(n)$ is a switch if and only if $n\in\mathbb{Z}$.

The restriction $\rho|[0,\infty)$ is a half-route. If $\rho$ factors through
$\mathbb{R}/m\mathbb{Z}$ then $\rho$ is a train-loop. We require, for every
branch $b$, a train-route travelling along $b$.

For each branch $b$ and point $p\in b$, a component $b^{\prime}$ of
$b{\smallsetminus}\\{p\\}$ is a half-branch. Two half-branches
$b^{\prime},b^{\prime\prime}\subset b$ are equivalent if $b^{\prime}\cap
b^{\prime\prime}$ is again a half-branch. Every switch divides the three
incident half-branches into a pair of small half-branches on one side and a
single large half-branch on the other. A branch $b$ is large (small) if both
of its half-branches are large (small); if $b$ has one large and one small
half-branch then $b$ is called mixed.

Let $\mathcal{B}=\mathcal{B}(\tau)$ be the set of branches of $\tau$. A
function $w\colon\mathcal{B}\to\mathbb{R}_{\geq 0}$ is a transverse measure on
$\tau$ if $w$ satisfies the switch conditions: for every switch $x\in\tau$ we
have $w(a)+w(b)=w(c)$, where $a^{\prime},b^{\prime}$ are the small half-
branches and $c^{\prime}$ is the large half-branch meeting $x$. Let $P(\tau)$
be the projectivization of the cone of transverse measures; define $V(\tau)$
to be the vertices of the polyhedron $P(\tau)$.

$\begin{array}[]{c}\includegraphics[width=10cm]{split}\\\ \\\ \\\
\includegraphics[width=10cm]{slide}\end{array}$ Figure 1. Top: A large branch
admits a right, central, and left splitting. Bottom: a mixed branch admits a
slide.

We may split a pretrack along a large branch or slide it along a mixed branch;
see Figure 1. (Slides are called shifts in [15].) The inverse of a split or
slide is called a fold. Note that the inverse of a slide may be obtained via a
slide followed by an isotopy.

$\begin{array}[]{c}\includegraphics[height=3.5cm]{tie2}\end{array}$ Figure 2.
The local model for $N(\tau)$ near a switch, with horizontal and vertical
boundary in the correct orientation. The dotted lines are ties.

Suppose that $\tau\subset S$ is a pretrack. Let $N=N(\tau)\subset S$ be a tie
neighborhood of $\tau$: so $N$ is a union of rectangles
$\\{R_{b}\mathbin{\mid}b\in\mathcal{B}\\}$ foliated by vertical intervals (the
ties). At a switch, the upper and lower thirds of the vertical side of the
large rectangle are identified with the vertical side of the small rectangles,
as shown in Figure 2. Since $N$ is a union of rectangles it follows that
$\operatorname{index}(N)=0$. The horizontal boundary $\partial_{h}N$ is the
union of $\partial_{h}R_{b}$, for $b\in\mathcal{B}$, while the vertical
boundary is $\partial_{v}N={\overline{\partial
N{\smallsetminus}\partial_{h}N}}$.

Let $N=N(\tau)$ be a tie neighborhood. Let $T$ be a complementary region of
$\tau$: a component of the closure of $S{\smallsetminus}N$. Define the
horizontal and vertical boundary of $T$ to be $\partial_{h}T=\partial
T\cap\partial_{h}N$ and $\partial_{v}T=\partial T\cap\partial_{v}N$. Note that
all corners of $T$ are outward, so
$\operatorname{index}(T)=\chi(T)-\frac{1}{4}|\partial\partial_{h}T|$.

Suppose $\tau\subset S$ is a pretrack. The subsurface filled by $\tau$ is the
union of $N$ with all complementary regions $T$ of $\tau$ that are disks or
peripheral annuli.

######  Definition 2.1.

Suppose that $\tau\subset S$ is a pretrack and $N=N(\tau)$. We say that $\tau$
is a train track if $\tau$ is compact, every component of $\partial N$ has at
least one corner and every complementary region $T$ of $\tau$ has negative
index.

######  Definition 2.2.

In a sliding and splitting sequence $\\{\tau_{i}\\}$ of train tracks each
$\tau_{i+1}$ is obtained from $\tau_{i}$ by a slide or a split.

###  2.4. Carrying, duality, and efficient position

Suppose that $\tau\subset S$ is a train track. If $\sigma$ is also a track,
contained in $N=N(\tau)$ and transverse to the ties, then we write
$\sigma\prec\tau$ and say that $\sigma$ is carried by $\tau$. For example, if
$\tau$ is a fold of $\sigma$ then $\sigma$ is carried by $\tau$.

A properly embedded arc or curve $\beta\subset N$ is carried by $\tau$ if
$\beta$ is transverse to the ties and
$\partial\beta\cap\partial_{h}N=\emptyset$. Thus if $\beta$ is carried then
$\partial\beta\subset\partial_{v}N$. Again we write $\beta\prec\tau$ for
carried arcs and curves.

######  Definition 2.3.

Suppose that $\alpha\subset S$ is a properly embedded arc or curve. Then
$\alpha$ is in efficient position with respect to $\tau$, denoted
$\alpha\dashv\tau$, if

  * •

every component of $\alpha\cap N$ is a tie or is carried by $\tau$ and

  * •

every region $T\subset S{\smallsetminus}(N\cup\alpha)$ has negative index or
is a rectangle.

Suppose that $\alpha\dashv\tau$. If $\alpha\subset N$ then $\alpha$ is
carried, $\alpha\prec\tau$. If no component of $\alpha\cap N$ is carried then
$\alpha$ is dual to $\tau$ and we write $\alpha\pitchfork\tau$. If
$\Delta\subset\mathcal{AC}(S)$ is a multicurve then we write
$\Delta\prec\tau$, $\Delta\pitchfork\tau$, or $\Delta\dashv\tau$ if all
elements of $\Delta$ are disjointly and simultaneously carried, dual, or in
efficient position.

######  Remark 2.4.

Our notion of duality is called hitting efficiently by Penner and Harer [15,
page 19]. Note that $\alpha\pitchfork\tau$ if and only if $\alpha$ is carried
by some extension of the dual track $\tau^{*}$, also defined in [15].
Likewise, if $\alpha\dashv\tau$ and $\alpha\cap N$ consists of carried arcs
then $\alpha$ is carried by some extension of $\tau$.

An index argument proves:

######  Lemma 2.5.

If $\alpha$ is a properly embedded curve or arc in efficient position with
respect to a train track $\tau\subset S$ then $\alpha$ is essential and non-
peripheral in $S$. ∎

One of the goals of this paper is to prove the converse of Lemma 2.5; this is
done in Theorem 4.1. Following Lemma 2.5 we may define
$\mathcal{C}(\tau)=\\{\alpha\mathbin{\mid}\alpha\prec\tau\\}$ and
$\mathcal{C}^{*}(\tau)=\\{\alpha\mathbin{\mid}\alpha\pitchfork\tau\\}$. Notice
that if $\sigma\prec\tau$ is a track then
$\mathcal{C}(\sigma)\subset\mathcal{C}(\tau)$ and
$\mathcal{C}^{*}(\tau)\subset\mathcal{C}^{*}(\sigma)$.

A branch $b\in\mathcal{B}(\tau)$ is recurrent if there is some
$\alpha\prec\tau$ that meets $R_{b}$. The track $\tau$ is recurrent if every
branch is recurrent. Transverse recurrence is defined by replacing carrying by
duality [15, page 20]. The track $\tau$ is birecurrent if $\tau$ is recurrent
and transversely recurrent [15, Section 1.3]. In a slight departure from
Penner and Harer’s terminology [15, page 27] we will call a birecurrent track
$\tau$ complete if all complementary regions have index $-1/2$. (When
$S=S_{1,1}$ there is, instead, a single complementary region with index $-1$.)

######  Lemma 2.6.

Suppose that $\sigma\subset S$ is a birecurrent track. Then
$\mathcal{C}^{*}(\sigma)$ has infinite diameter inside of $\mathcal{C}(S)$.

###### Proof.

Let $\tau$ be a complete track extending $\sigma$ [15, Corollary 1.4.2].
Section 3.4 of [15] and a dimension count gives a lamination
$\lambda\pitchfork\tau$ so that $i(\lambda,\alpha)\neq 0$ for all
$\alpha\in\mathcal{C}(S)$. Now an argument of Kobayashi [10], refined by Luo
[11, page 124], implies that
$\mathcal{C}^{*}(\tau)\subset\mathcal{C}^{*}(\sigma)$ has infinite diameter. ∎

###  2.5. Vertex cycles

When $\alpha\prec\tau$ is a curve there is a transverse measure $w_{\alpha}$
defined by taking $w_{\alpha}(b)=|\alpha\cap t|$ where $t$ is any tie of the
rectangle $R_{b}$. Conversely, for any integral transverse measure $w$ there
is a multicurve $\alpha_{w}$ — take $w(b)$–many horizontal arcs in $R_{b}$ and
glue endpoints as dictated by the switch conditions.

Note that if $v\in V(\tau)$ then there is a minimal integral measure $w$
projecting to $v$. Since $v$ is an extreme point of $P(\tau)$ deduce that
$\alpha_{w}$ is an embedded curve. We call $\alpha_{w}$ a vertex cycle of
$\tau$ and henceforth use $V(\tau)$ to denote the set of vertex cycles.

###  2.6. Wide curves

Let $N=N(\tau)$ be a tie neighborhood.

######  Definition 2.7.

A multicurve $\Delta\dashv\tau$ is wide if there is an orientation of the
components of $\Delta$ so that

  * •

for every $b\in\mathcal{B}(\tau)$, all arcs of $\Delta\cap R_{b}$ are to the
right of each other (see the top of Figure 3) and

  * •

for every complementary region $T$ of $\tau$, all arcs of $\Delta\cap T$ are
to the right of each other (see the bottom of Figure 3).

$\begin{array}[]{c}\includegraphics[height=2.2cm]{right}\\\ \\\
\includegraphics[height=3.5cm]{rightdual}\end{array}$ Figure 3. Above: Arcs of
$\Delta\cap R_{b}$ to the right of each other. The vertical dotted line are
ties; the heavy horizontal lines are arcs of $\partial_{h}N$. Below: Arcs
meeting the complementary region $T$, all to the right of each other.

It follows from the definition that if $\Delta\dashv\tau$ is wide then for any
branch $b\in\mathcal{B}(\tau)$ the intersection $\Delta\cap R_{b}$ has at most
two components.

######  Lemma 2.8.

Every vertex cycle $\alpha\in V(\tau)$ is wide.

For an even more precise characterization of vertex cycles see Lemma 3.11.3 of
[14]. The proof below recalls a surgery technique used in the sequel.

###### Proof of Lemma 2.8.

We prove the contrapositive. Suppose that $\alpha$ is not wide. Orient
$\alpha$. There are three cases.

Suppose there is a branch $b\subset\tau$ and an oriented tie $t\subset R_{b}$
where $x$ and $y$ are consecutive (along $t$) points of $\alpha\cap t$ so that
the signs of intersection at $x$ and $y$ are equal. Let $[x,y]$ be the subarc
of $t$ bounded by $x$ and $y$. Surger $\alpha$ along $[x,y]$ to form curves
$\beta,\gamma\prec\tau$. See Figure 4. Thus $w_{\alpha}=w_{\beta}+w_{\gamma}$
and $\alpha$ is not a vertex cycle.

2pt $t$ [l] at 149.1 173.6 $\alpha$ [l] at 291.6 145.5 $\beta$ [l] at 291.6
127 $\gamma$ [l] at 291.6 37.1

$\begin{array}[]{c}\includegraphics[width=5.5cm]{surger}\end{array}$

Figure 4. Surgery when adjacent intersections have the same sign.

Suppose instead that $x,y,z$ are consecutive (along $t$) points of $\alpha\cap
t$ with alternating sign. In this case there is again a surgery along $[x,z]$
producing curves $\beta$ and $\gamma$. See Figure 5 for one of the possible
arrangements of $\alpha$, $\beta$ and $\gamma$. Again
$w_{\alpha}=w_{\beta}+w_{\gamma}$ is a non-trivial sum and $\alpha$ is not a
vertex cycle.

2pt $t$ [l] at 149.1 264.0 $\alpha$ [l] at 291.6 235.9 $\beta$ [l] at 291.6
217.3 $\gamma$ [l] at 291.6 163.6

$\begin{array}[]{c}\includegraphics[width=5.5cm]{surger2}\end{array}$

Figure 5. Surgery when the three intersections have alternating sign.

In the remaining case $w_{\alpha}(a)\leq 2$ for all $a\in\mathcal{B}$ and
there are branches $b,c\in\mathcal{B}$ where the arcs of $\alpha\cap R_{b}$
are to the right of each other while the arcs of $\alpha\cap R_{c}$ are to the
left of each other. See Figure 6 for the two ways $\alpha$ may be carried by
$\tau$.

$\begin{array}[]{c}\includegraphics[height=2.5cm]{disagree2}\\\
\includegraphics[height=2.5cm]{disagree1}\end{array}$ Figure 6. The curve
$\alpha$ meets $R_{b}$ and $R_{c}$ twice.

If $\alpha$ is carried as in the top line of Figure 6 then surger $\alpha$ in
both rectangles $R_{b}$ and $R_{c}$ as was done in Figure 4. This shows that
$w_{\alpha}$ is a non-trivial sum and so $\alpha$ is not a vertex cycle. Now
suppose that $\alpha$ is carried as in the bottom line of Figure 6. Note that
the closure of $\alpha{\smallsetminus}(R_{b}\cup R_{c})$ is a union of four
arcs. Two of these, $\beta^{\prime}$ and $\gamma^{\prime}$, meet both $R_{b}$
and $R_{c}$. Since $S$ is orientable no tie-preserving isotopy of $N$ throws
$\beta^{\prime}$ onto $\gamma^{\prime}$. Let
$\alpha^{\prime}\cup\alpha^{\prime\prime}=\alpha{\smallsetminus}(\beta^{\prime}\cup\gamma^{\prime})$.
Create an embedded curve $\beta$ by taking two parallel copies of
$\beta^{\prime}$ and joining them to $\alpha^{\prime}$ and
$\alpha^{\prime\prime}$. Similarly create $\gamma$ by joining two parallel
copies of $\gamma^{\prime}$ to the arcs $\alpha^{\prime}$ and
$\alpha^{\prime\prime}$. It follows that $w_{\beta}\neq w_{\gamma}$. Since
$2w_{\alpha}=w_{\beta}+w_{\gamma}$ again $\alpha$ is not a vertex cycle. ∎

###  2.7. Combing

Suppose that $\alpha\prec\tau$ has $w_{\alpha}(b)\leq 1$ for every branch
$b\subset\tau$. Orient and transversely orient $\alpha$ to agree with the
orientation of the surface $S$. We think of the orientation as pointing in the
$x$–direction and the transverse orientation pointing in the $y$–direction. A
half-branch $b\subset\tau{\smallsetminus}\alpha$, sharing a switch with
$\alpha$, twists to the right if any train-route through $b$ locally has
positive slope. Otherwise $b$ twists to the left. If all branches on one side
of $\alpha$ twist to the right then that side of $\alpha$ has a right combing,
and similarly for a left combing. See Figure 8 for an example where both sides
are combed to the left.

###  2.8. Curve complexes and subsurface projection

For more information on the curve complex see [11, 12]. Impose a simplicial
structure on $\mathcal{AC}(S)$ where $\Delta\subset\mathcal{AC}(S)$ is a
simplex if and only if $\Delta$ is a multicurve. The complex of curves
$\mathcal{C}(S)$ and the arc complex $\mathcal{A}(S)$ are the subcomplexes
spanned by curves and arcs, respectively. Note that if the complexity $\xi(S)$
is at least two then $\mathcal{C}(S)$ is connected [9, Proposition 2]. For
surfaces of lower complexity we alter the simplicial structure on
$\mathcal{C}(S)$.

Define the Farey tessellation $\mathcal{F}$ to have vertex set
$\mathbb{Q}\cup\\{\infty\\}$. A collection of slopes
$\Delta\subset\mathcal{F}$ spans a simplex if $ps-rq=\pm 1$ for all
$p/q,r/s\in\Delta$. If $S=S_{1,1}$ or $S_{0,4}$ we take
$\mathcal{C}(S)=\mathcal{F}$; that is, there is an edge between curves that
intersect in exactly one point (for $S_{1,1}$) or two points (for $S_{0,4}$).
Note that for surfaces with $\xi(S)\geq 1$ the inclusion of $\mathcal{C}(S)$
into $\mathcal{AC}(S)$ is a quasi-isometry.

Suppose now that $X\mathrel{\cong}S_{0,2}$ is an annulus. Define
$\mathcal{A}(X)$ to be the set of all essential arcs in $X$, up to isotopy
fixing $\partial X$ pointwise. For $\alpha,\beta\in\mathcal{A}(X)$ define

$i(\alpha,\beta)=\min\big{\\{}|(a\cap b){\smallsetminus}\partial
X|:a\in\alpha,b\in\beta\big{\\}}.$

As usual, multicurves give simplices for $\mathcal{A}(X)$.

If $\alpha,\beta$ are vertices of $\mathcal{C}(S)$, $\mathcal{A}(S)$, or
$\mathcal{AC}(S)$ then define $d_{S}(\alpha,\beta)$ to be the minimal number
of edges in a path, in the one-skeleton, connecting $\alpha$ to $\beta$; the
containing complex will be clear from context. Note that if $\alpha$ $\beta$
are distinct arcs of $\mathcal{A}(X)$, when $X$ is an annulus, then
$d_{X}(\alpha,\beta)=1+i(\alpha,\beta)$ [12, Equation 2.3].

As usual, suppose that $\xi(S)\geq 1$. Fix an essential subsurface $X\subset
S$ with $\xi(X)<\xi(S)$. We suppose that $X$ is either a non-peripheral
annulus or a surface of complexity at least one. (The case of an essential
annulus inside of $S_{1}$ is not relevant here.) Following [12], we will
define the subsurface projection relation
$\pi_{X}\colon\mathcal{AC}(S)\to\mathcal{AC}(X)$. Let $S^{X}$ be the cover of
$S$ corresponding to the inclusion $\pi_{1}(X)<\pi_{1}(S)$. The surface
$S^{X}$ is not compact; however, there is a canonical (up to isotopy)
homeomorphism between $X$ and the Gromov compactification of $S^{X}$. This
identifies the arc and curve complexes of $X$ and $S^{X}$. Fix
$\alpha\in\mathcal{AC}(S)$. Let $\alpha^{X}$ be the preimage of $\alpha$ in
$S^{X}$.

If $\alpha^{X}$ contains a non-peripheral curve in $S^{X}$ then
$\pi_{X}(\alpha)=\\{\alpha\\}$. Otherwise, place every essential arc of
$\alpha^{X}$ into the set $\pi_{X}(\alpha)$. If neither obtains then
$\pi_{X}(\alpha)=\emptyset$. If $\pi_{X}(\alpha)=\emptyset$ then we say that
$\alpha$ misses $X$. If $\pi_{X}(\alpha)\neq\emptyset$ then $\alpha$ cuts $X$.

Suppose that $\alpha,\beta\in\mathcal{C}(S)$. If $\pi_{X}(\alpha)$ and
$\pi_{X}(\beta)$ are nonempty define

$d_{X}(\alpha,\beta)=\operatorname{diam}_{X}\big{(}\pi_{X}(\alpha)\cup\pi_{X}(\beta)\big{)}.$

Likewise define the distance $d_{X}(A,B)$ between finite sets
$A,B\subset\mathcal{C}(S)$. When $\tau$ is a track, we use the shorthand
$\pi_{X}(\tau)$ for the set $\pi_{X}(V(\tau))$. If $\sigma$ is also a track,
we write $d_{X}(\tau,\sigma)$ for the distance
$d_{X}(\pi_{X}(\tau),\pi_{X}(\sigma))$.

We end with a lemma connecting the subsurface projection of carried (or dual)
curves to the behavior of wide curves.

######  Lemma 2.9.

Suppose that $X\subset S$ is an essential surface and $\tau$ is a track. If
some $\alpha\prec\tau$ ($\alpha\pitchfork\tau$) cuts $X$ then there is a
vertex cycle $\beta\prec\tau$ (wide dual $\beta\pitchfork\tau$) cutting $X$.

###### Proof.

Some multiple of $\alpha\prec\tau$ is a sum of vertices: $m\cdot
w_{\alpha}=\sum n_{i}w_{i}$ where $w_{i}$ is the integral transverse measure
for the vertex $\beta_{i}\in V(\tau)$. Via a sequence of tie-preserving
isotopies of $N$ we may arrange for all of the $\beta_{i}$ to realize their
geometric intersection with each other. Note that there is an isotopy
representative of $\alpha$ contained inside of a small neighborhood of the
union $B=\cup\beta_{i}$.

To prove the contrapositive, suppose that none of the $\beta_{i}$ cut $X$. It
follows that $X$ may be isotoped in $S$ to be disjoint from $B$. Thus $\alpha$
misses $X$, as desired. A similar discussion applies when
$\alpha\pitchfork\tau$. ∎

##  3\. Induced tracks

Suppose that $\tau\subset S$ is a train track. Suppose that $X\subset S$ is an
essential subsurface with $\xi(X)<\xi(S)$. Let $S^{X}$ be the corresponding
cover of $S$. Let $\tau^{X}$ be the preimage of $\tau$ in $S^{X}$; note that
the pretrack $\tau^{X}$ satisfies all of the axioms of a train track except
compactness.

Define $\mathcal{AC}(\tau^{X})$ to be the set of essential arcs and essential,
non-peripheral curves properly embedded in the Gromov compactification of
$S^{X}$ with interior a train-route or train-loop carried by $\tau^{X}$. A bit
of caution is required here — inessential arcs and peripheral curves may be
carried by $\tau^{X}$ but these are not admitted into
$\mathcal{AC}(\tau^{X})$. Define
$\mathcal{A}(\tau^{X}),\mathcal{C}(\tau^{X})\subset\mathcal{AC}(\tau^{X})$ to
be the subsets of arcs and curves respectively. Define
$\mathcal{AC}^{*}(\tau^{X})$ to be the set of dual essential arcs and dual
essential, non-peripheral curves, up to isotopies fixing $\tau^{X}$ setwise.

###  3.1. Induced tracks for non-annuli

If $X$ is not an annulus define $\tau|X$, the induced track, to be the union
of the branches of $\tau^{X}$ crossed by an element of
$\mathcal{C}(\tau^{X})$.

######  Lemma 3.1.

If $X$ is not an annulus then the induced track $\tau|X$ is compact.

###### Proof.

Note train-routes in $\tau^{X}$ that are mapped properly to $S^{X}$ are
uniform quasi-geodesics in $S^{X}$ [14, Proposition 3.3.3]. Thus there is a
compact core $X^{\prime}\subset S^{X}$, homeomorphic to $X$, so that any route
meeting $S^{X}{\smallsetminus}X^{\prime}$ has one endpoint on the Gromov
boundary of $S^{X}$. It follows that $\tau|X\subset X^{\prime}$. ∎

Note that $\tau|X$ may not be a train track: $N=N(\tau|X)$ may have smooth
boundary components and complementary regions with non-negative index.
However, since all complementary regions of $\tau^{X}$ have negative index it
follows that if a complementary region $T$ of $\tau|X$ has non-negative index
then $T$ is a peripheral annulus meeting a smooth component of $\partial N$.

The definition of $\tau|X$ implies that $\tau|X$ is recurrent. Carrying,
duality, efficient position and wideness with respect to an induced track are
defined as in Section 2.4. Define $\mathcal{C}(\tau|X)\subset\mathcal{C}(X)$,
the subset of curves carried by $\tau|X$. Note that
$\mathcal{C}(\tau|X)=\mathcal{C}(\tau^{X})$. Define
$\mathcal{AC}^{*}(\tau|X)\subset\mathcal{AC}(X)$ to be the subset of arcs and
curves dual to $\tau|X$. Note that
$\mathcal{AC}^{*}(\tau|X)\supset\mathcal{AC}^{*}(\tau^{X})$.

Now, $\tau|X$ fails to be transversely recurrent exactly when it carries a
peripheral curve. We say that a branch $b\subset\tau|X$ is transversely
recurrent with respect to arcs and curves if there is
$\alpha\in\mathcal{AC}^{*}(\tau|X)$ meeting $b$. Then $\tau|X$ is transversely
recurrent with respect to arcs and curves if every branch $b$ is.

######  Lemma 3.2.

Suppose that $\tau$ is transversely recurrent in $S$. Then $\tau|X$ is
transversely recurrent with respect to arcs and curves in $X$. Furthermore:
suppose that $\tau|X$ is transversely recurrent with respect to arcs and
curves in $X$. If $\sigma\subset\tau|X$ is a train track then $\sigma$ is
transversely recurrent in $X$.

###### Proof.

The first claim follows from the definitions. An index argument proves the
second claim. ∎

Here is our second surgery argument.

######  Lemma 3.3.

Suppose that $\tau$ is a track and $X\subset S$ is an essential subsurface,
yet not an annulus. For every $\alpha\in\mathcal{A}(\tau^{X})$ at least one of
the following holds:

  * •

There is an arc $\beta\in\mathcal{A}(\tau^{X})$ so that $\beta$ is wide and
$i(\alpha,\beta)=0$.

  * •

There is a curve $\gamma\in\mathcal{C}(\tau|X)$ so that $i(\alpha,\gamma)\leq
2$.

The statement also holds replacing $\mathcal{A},\mathcal{C}$ by
$\mathcal{A}^{*},\mathcal{C}^{*}$.

###### Proof.

The proof is modelled on that of Lemma 2.8. If $\alpha\prec\tau^{X}$ is wide
we are done. If not, as $\alpha$ is a quasi-geodesic [14, Proposition 3.3.3],
orient $\alpha$ so that $\alpha$ is wide outside of a compact core for
$S^{X}$. Now we induct on the total number of arcs of intersection between
$\alpha$ and rectangles $R_{b}\subset N(\tau^{X})$ meeting the compact core.

Let $t$ be a tie of $R_{b}$. Orient $t$. Suppose that $x,y$ are consecutive
(along $t$) points of $\alpha\cap t$. Suppose that the sign of intersection at
$x$ equals the sign at $y$. Let $[x,y]$ be the subarc of $t$ bounded by $x$
and $y$. As in Lemma 2.8 surger $\alpha$ along $[x,y]$ to form an arc
$\beta^{\prime}$ and a curve $\gamma$. See Figure 4 with $\beta^{\prime}$
substituted for $\beta$.

Note that $\gamma$ is essential in $S^{X}$, by an index argument. If $\gamma$
is non-peripheral then the second conclusion holds. So suppose that $\gamma$
is peripheral. Then $\alpha$ is obtained by Dehn twisting $\beta^{\prime}$
about $\gamma$. So $\beta^{\prime}$ is properly isotopic to $\alpha$ and has
smaller intersection with $R_{b}$; thus we are done by induction.

Suppose instead that $x,y,z$ are consecutive (along $t$) points of $\alpha\cap
t$, with alternating sign. Surger $\alpha$ along $[x,z]$ to form an arc
$\beta^{\prime}$ and a curve $\gamma$. See Figure 5, with $\beta^{\prime}$
substituted for $\beta$, for one of the possible arrangements of $\alpha$,
$\beta^{\prime}$, and $\gamma$. Again $\gamma$ is essential. If $\gamma$ is
non-peripheral then the second conclusion holds and we are done. If $\gamma$
is peripheral then, as $\alpha$ and $\beta^{\prime}$ differ by a half-twist
about $\gamma$, we find that $\beta^{\prime}$ is properly isotopic to
$\alpha$. Since $\beta^{\prime}$ has smaller intersection with $R_{b}$ we are
done by induction.

All that remains is the case that $\alpha$ meets every rectangle $R_{b}$ in
most a pair of arcs of opposite orientation. For every branch $b$ where
$\alpha$ meets $R_{b}$ twice, choose a subarc $t_{b}$ of a tie in $R_{b}$ so
that $\alpha\cap t_{b}=\partial t_{b}$. We call $t_{b}$ a chord for $\alpha$.
For every $t_{b}$ there is a subarc $\alpha_{b}\subset\alpha$ so that
$\partial t_{b}=\partial\alpha_{b}$. A chord $t_{b}$ is innermost if there is
no chord $t_{c}$ with $\alpha_{c}$ strictly contained in $\alpha_{b}$. Let
$t_{b}$ be the first innermost chord. Let $\alpha^{\prime}$ be the component
of $\alpha{\smallsetminus}\alpha_{b}$ before $\alpha_{b}$. Build a route
$\beta$ by taking two copies of $\alpha^{\prime}$ and joining them to
$\alpha_{b}$. Note that $\beta\cap R_{c}$ is a single arc or a pair of arcs
exactly as $\alpha_{b}$ or $\alpha^{\prime}$ meets $R_{c}$. Thus $\beta$ is
wide. Also, $\beta$ is essential: otherwise $t_{b}\cup\alpha_{b}$ bounds a
disk with index one-half, a contradiction. By construction $i(\alpha,\beta)=0$
and Lemma 3.3 is proved. ∎

###  3.2. Induced tracks for annuli

Suppose that $X\subset S$ is an annulus. Define $\tau|X$ to be the union of
branches $b\subset\tau^{X}$ so that some element of $\mathcal{A}(\tau^{X})$
travels along $b$. (Note that $\tau|X$, if nonempty, is not compact.) Define
$\mathcal{A}(\tau|X)=\mathcal{A}(\tau^{X})$ and also the duals
$\mathcal{A}^{*}(\tau|X)\supset\mathcal{A}^{*}(\tau^{X})$.

Define $V(\tau|X)$ in $\mathcal{A}(\tau|X)$ to be the set of wide carried
arcs. Define $V^{*}(\tau|X)$ dually.

######  Lemma 3.4.

Suppose that $X\subset S$ is an essential annulus. If
$\mathcal{A}^{(*)}(\tau|X)$ is nonempty then $V^{(*)}(\tau|X)$ is nonempty.
Let $N=N(\tau|X)$. If $\gamma\dashv\tau|X$ is a wide essential arc then
$\gamma$ meets each rectangle of $N$ and each region of
$S^{X}{\smallsetminus}N$ in at most a single arc.

For example, if $\gamma\prec\tau|X$ is a wide essential arc then $\gamma$
embeds into $\tau|X$.

###### Proof of Lemma 3.4.

We prove the second conclusion; the first is similar. Suppose that $R$ is
either a rectangle or region so that $\gamma\cap R$ is a pair of arcs to the
right of each other. Let $\delta$ be an arc properly embedded in
$R{\smallsetminus}\gamma$ so that $\delta\cap\gamma=\partial\delta$. Let
$\gamma^{\prime}$ be the component of $\gamma{\smallsetminus}\partial\delta$
so that $\partial\gamma^{\prime}=\partial\delta$. If
$\gamma^{\prime}\cup\delta$ bounds a disk in $S^{X}$ then this disk has index
one-half and we contradict efficient position. If $\gamma^{\prime}\cup\delta$
bounds an annulus then $\gamma$ was not essential, another contradiction. ∎

Suppose that $\alpha$ is the core curve of the annulus $X$.

######  Lemma 3.5.

If $\alpha$ is not carried by $\tau|X$ then $V(\tau|X)=\mathcal{A}(\tau|X)$.
If $\alpha$ is not dual to $\tau|X$ then
$V^{*}(\tau|X)=\mathcal{A}^{*}(\tau|X)$. ∎

######  Lemma 3.6.

Suppose that $\alpha\prec\tau|X$. One side of $\alpha$ is combed if and only
if both sides are combed in the same direction if and only if some isotopy
representative of $\alpha$ is dual to $\tau|X$. ∎

##  4\. Finding efficient position

After discussing the various sources of non-uniqueness we prove in Theorem 4.1
that efficient position exists

Let $N=N(\tau)$; suppose that $\alpha\dashv\tau$. A rectangle $T\subset
S{\smallsetminus}(N\cup\alpha)$ is vertical if $\partial T$ has a pair of
opposite sides meeting $\alpha$ and $\partial_{v}N$ respectively. Define
horizontal rectangles similarly. Figure 7 depicts the two kinds of rectangle
swap.

$\begin{array}[]{c}\includegraphics[height=3.5cm]{rect_{s}wap2}\end{array}$
Figure 7. Pieces of $N$ are shown, with vertical and horizontal boundary in
the correct orientation; the dotted lines are ties. The left and right
pictures show a vertical and horizontal rectangle swap, respectively.

Now suppose that $\alpha\prec\tau$, every rectangle $R_{b}\subset N$ meets
$\alpha$ in at most a single arc, and one side of $\alpha$ is combed. Let $A$
be a small regular neighborhood of $\alpha$. Then an annulus swap interchanges
$\alpha$ and the component of $\partial A$ on the combed side. See Figure 8.

$\begin{array}[]{c}\includegraphics[height=3.5cm]{ann_{s}wap}\end{array}$
Figure 8. Both sides of $\alpha$ are combed (to the left). Thus both boundary
components of the annulus shown are dual to $\tau$ and both differ from the
carried core curve by an annulus swap.

######  Theorem 4.1.

Suppose that $\xi(S)\geq 1$ and $\tau\subset S$ is a birecurrent train track.
Suppose that $\Delta\subset\mathcal{AC}(S)$ is a multicurve. Then efficient
position for $\Delta$ with respect to $\tau$ exists and is unique up to
rectangle swaps, annulus swaps, and isotopies of $S$ preserving the foliation
of $N(\tau)$ by ties.

######  Remark 4.2.

When $S=S_{1}$ is a torus, Lemma 14 of [7] proves the existence of efficient
position for curves with respect to Reebless bigon tracks. Uniqueness of
efficient position follows from a slight generalization of Section 4.1 using
bigon swaps.

###  4.1. Uniqueness of efficient position

Suppose that $\alpha$ and $\beta$ are isotopic curves and in efficient
position with respect to $\tau$. We induct on $i(\alpha,\beta)$. For the base
case suppose that $|\alpha\cap\beta|=0$. Then $\alpha$ and $\beta$ cobound an
annulus $A\subset S$ so that $\partial A$ has no corners [3, Lemma 2.4]. Since
$N=N(\tau)$ is a union of rectangles the intersection $A\cap N$ is also a
union of rectangles. Thus $\operatorname{index}(N\cap A)=0$. By the hypothesis
of efficient position any region $T\subset{\overline{A{\smallsetminus}N}}$ has
non-positive index and has all corners outwards. By the additivity of index it
follows that $\operatorname{index}(T)=0$. It follows that each region $T$ is
either an annulus without corners or a rectangle.

Suppose that some region $T$ is an annulus without corners. Then we must have
$T=A$. For if $\partial T$ meets $\partial N$ then $\partial N$ has a
component without corners, contrary to assumption. Since $T=A$ it follows that
$\alpha$ and $\beta$ are isotopic in the complement of $N$ and we are done.

So we may assume that all regions of $A{\smallsetminus}N$ are rectangles. (In
particular, $A\cap N\neq\emptyset$.) Note that if a region $R$ is a horizontal
rectangle then there is no obstruction to doing a rectangle swap across $R$.
After doing all such swaps we may assume that $A{\smallsetminus}N$ contains no
horizontal rectangles.

We now abuse terminology slightly by assuming that the position of $N$
determines that of $\tau$. So if $A$ contains vertical rectangles then there
are switches of $\tau$ contained in $A$. This implies that $A$ contains half-
branches of $\tau$. Let $b^{\prime}$ be a half-branch in $A$, meeting
$\partial A$. If $b^{\prime}$ is large then there is a vertical rectangle swap
removing three half-branches from $A$. After doing all such swaps we may
assume that any such $b^{\prime}$ is small. If $R$ is a vertical rectangle
meeting $\partial A$ then $R$ has two horizontal sides. If neither of these
meets a switch on its interior then again there is a swap removing three half-
branches from $A$.

After doing all such swaps if there are still vertical rectangles in $A$ then
we proceed as follows: every vertical rectangle must have a horizontal side
that properly contains the horizontal side of another vertical rectangle. (For
example, in Figure 8 number the rectangles above the core curve
$R_{0},R_{1},R_{2}$ from left to right. Note that the left horizontal side of
$R_{i}$ strictly contains the right horizontal side of $R_{i-1}$.) It follows
that the union of these vertical rectangles gives an annulus swap which we
perform. Thus, we are reduced to the situation where $A$ contains no
horizontal or vertical rectangles.

If $A\subset N$ then $\alpha$ and $\beta$ are both carried. For any tie
$t\subset N$, any component $t^{\prime}\subset t\cap A$ is an essential arc in
$A$. (To see this, suppose that $t^{\prime}$ is inessential. Let $B\subset A$
be the bigon cobounded by $t^{\prime}$ and $\alpha^{\prime}\subset\alpha$,
say. Since $\alpha$ is carried, $\alpha^{\prime}$ is transverse to the ties.
We define a continuous involution on $\alpha^{\prime}$; for every tie $s$ and
for every component $s^{\prime}\subset s\cap B$ transpose the endpoints of
$s^{\prime}$. As this involution is fixed point free, we have reached a
contradiction.) It follows that $A$ is foliated by subarcs of ties and we are
done.

There is one remaining possibility in the base case of our induction: $A\cap
N\neq\emptyset$, $A\not\subset N$, and $A$ contains no switches of $\tau$.
Thus every region of $A\cap N$ and of $A{\smallsetminus}N$ is a rectangle
meeting both $\alpha$ and $\beta$. Any region $R$ of $A\cap N$ is foliated by
(subarcs of) ties and, as above, all ties meet $R$ essentially. Thus $R$ gives
a parallelism between (carried arcs) ties of $\alpha$ and $\beta$. It follows
that $A$ gives an isotopy between $\alpha$ and $\beta$, sending ties to ties.
This completes the proof of uniqueness when $|\alpha\cap\beta|=0$.

For the induction step assume $|\alpha\cap\beta|>0$. Since $\alpha$ is
isotopic to $\beta$ the Bigon Criterion [3, Lemma 2.5], [4, Proposition 1.3]
implies that there is a disk $B\subset S$ with exactly two outward corners $x$
and $y$ so that $B\cap(\alpha\cup\beta)=\partial B$. Suppose that $x$ is a
dual intersection: an intersection of a tie of $\alpha$ and a carried arc of
$\beta$. See Figure 9.

2pt $\alpha$ [bl] at 228.4 120.4 $\beta$ [l] at 291.6 20.6

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

Figure 9. The left corner is a dual intersection, between a tie of $\alpha$
and a carried arc of $\beta$. If the half-route $\rho$ exits through $\alpha$
or $\beta$ then a bigon or non-trivial trigon is created.

Let $\alpha^{\prime}=\alpha\cap B$ and $\beta^{\prime}=\beta\cap B$. Orient
$\beta^{\prime}$ away from $x$. Let $z\in\alpha^{\prime}$ be immediately
adjacent to $x$. Without loss of generality we may assume that $z$ is to the
left of $\beta^{\prime}$, near $x$. Let $\rho$ be the half-route starting at
$z$, initially agreeing with $\beta$, and turning left at every switch. If
$\rho\subset B$ then eventually $\rho$ repeats a branch $b$ in the same
direction; it follows that there is a curve $\gamma\prec\tau$ contained in $B$
contradicting Lemma 2.5. However, if $\rho$ exits $B$ through
$\alpha^{\prime}$ ($\beta^{\prime}$) then we contradict efficient position of
$\alpha$ ($\beta$).

It follows that the corner $x$ either lies in $S{\smallsetminus}N$ or is the
intersection of carried arcs of $\alpha$ and $\beta$. The same holds for $y$.
In either case we cut off of $B$ a small neighborhood of $x$ and of $y$: when
the corner lies in $N$ we use a subarc of a tie to do the cutting. The result
$B^{\prime}$ is a rectangle with the components of $\partial_{h}B$ contained
in $\alpha$ and $\beta$ respectively. As $\operatorname{index}(B^{\prime})=0$
the argument given in the case of an annulus gives a sequence of rectangle
swaps moving $\alpha$ across $B$. This reduces $|\alpha\cap\beta|$ by two and
so completes the induction step.

The proof when $\alpha$ and $\beta$ are arcs follows the above but omitting
any mention of annulus swaps.

Finally, suppose that $\Delta,\Gamma$ are isotopic multicurves, both in
efficient position. We may isotope $\Gamma$ to $\Delta$, as above, being
careful to always use innermost bigons. This completes the proof that
efficient position is unique.

We end this subsection with a useful corollary:

######  Corollary 4.3.

Suppose that $\Gamma\subset\mathcal{AC}(S)$ is a finite collection of arcs and
curves in efficient position. Then we may perform a sequence of rectangle
swaps to realize the pairwise geometric intersection numbers.

###### Proof.

Let $\Gamma=\\{\gamma_{i}\\}_{i=1}^{k}$. By induction, the curves of
$\Gamma^{\prime}=\Gamma{\smallsetminus}\\{\gamma_{k}\\}$ realize their
pairwise geometric intersection numbers. If $\gamma_{k}$ meets some
$\gamma_{i}\in\Gamma^{\prime}$ non-minimally then by the Bigon Criterion [5,
page 46] there is an innermost bigon between $\gamma_{k}$ and some
$\gamma_{j}\in\Gamma^{\prime}$. We now may reduce the intersection number
following the proof of uniqueness of efficient position. ∎

###  4.2. Existence of efficient position

Our hypotheses are weaker, and thus our discussion is more detailed, but the
heart of the matter is inspired by [11, pages 122-123].

We may assume that $\tau$ fills $S$; for if not we replace $S$ by the
subsurface filled by $\tau$. Since $\tau$ is transversely recurrent for any
$\epsilon,L>0$ there is a finite area hyperbolic metric on the interior of $S$
and an isotopy of $\tau$ so that: every branch of $\tau$ has length at least
$L$ and every train-route $\rho\prec\tau$ has geodesic curvature less than
$\epsilon$ at every point [15, Theorem 1.4.3].

Let $\tau^{\mathbb{H}}$ be the lift of $\tau$ to $\mathbb{H}=\mathbb{H}^{2}$,
the universal cover of $S$. Every train-route $\rho\prec\tau^{\mathbb{H}}$
cuts $\mathbb{H}$ into a pair $H^{\pm}(\rho)$ of open half-planes. Fix a route
$\rho\prec\tau^{\mathbb{H}}$ and a half-branch
$b^{\prime}\subset\tau^{\mathbb{H}}$ so that there is some $n\in\mathbb{Z}$
with $b^{\prime}\cap\rho=\rho(n)$. We say the branch $b$ is rising or falling
with respect to $\rho$ as the large half-branch at the switch $\rho(n)$ is
contained in $\rho|[n,\infty)$ or contained in $\rho|(-\infty,n]$.

######  Claim 4.4.

For any route $\rho\prec\tau^{\mathbb{H}}$ one side of $\rho$ has infinitely
many rising branches while the other side has infinitely many falling
branches.

###### Proof.

Note that there are infinitely many half-branches on both sides of $\rho$: if
not then $\partial_{h}N(\tau)$ would have a component without corners,
contrary to assumption. Suppose that there are only finitely many rising
branches along $\rho$. Then there is a curve $\gamma\prec\tau$ so that
$w_{\gamma}(b)\leq 1$ for every branch $b$ and so that the two sides of
$\gamma$ are combed in opposite directions. Thus $\tau$ is not recurrent, a
contradiction. The same contradiction is obtained if there are only finitely
many falling branches along $\rho$. ∎

######  Claim 4.5.

For any route $\rho\prec\tau^{\mathbb{H}}$ and for any family of half-routes
$\\{\beta_{n}\\}$ if $\beta_{n}\cap\rho=\rho(n)$ then
$\lim_{n\to\infty}\beta_{n}(\infty)=\rho(\infty)$.

###### Proof.

Let $x=\rho(\infty)\in\partial_{\infty}\mathbb{H}$. Consider the subsequence
$\\{\beta_{n}\\}$ where the first branch of each $\beta_{n}$ is falling. Let
$P_{n}=\rho|(-\infty,n]\cup\beta_{n}$, oriented away from $\rho(-\infty)$.
Note that $P_{n}(\infty)=\beta_{n}(\infty)$. Recall that $\rho$ and $P_{n}$
are both uniformly close to geodesics [15, pages 61–62]. Thus
$P_{n}(\infty)\to x$ as $n\to\infty$.

Now consider the subsequence $\\{\beta_{n}\\}$ where the first branch of each
$\beta_{n}$ is rising. Let $P_{n}=\beta_{n}\cup\rho|[n,\infty)$ oriented
towards $x$; so $P_{n}(\infty)=x$ for all $n$. Note that
$P(-\infty)=\beta_{n}(\infty)$. Since all complementary regions of
$\tau^{\mathbb{H}}$ have negative index none of the $P_{n}$ may cross each
other. It follows that either the $P_{n}$ exit compact subsets of
$\mathbb{H}$, and we are done, or the $P_{n}$ converge [15, Theorem 1.5.4] to
$P$, a train-route with $P(\infty)=x$. Since $P$ does not cross any $P_{n}$
deduce that $P$ and $\rho$ are disjoint. But this contradicts [14, Corollary
3.3.4]: train-routes that share an endpoint must share a half-route. ∎

Given distinct points $x,y,z\in S^{1}=\partial_{\infty}\mathbb{H}$, arranged
counterclockwise, let $(y,z)$ be the component of
$S^{1}{\smallsetminus}\\{y,z\\}$ that does not contain $x$. Let $[y,z]$ be the
closure of $(y,z)$. Thus $x\in(z,y)$, $(y,z)\cap[z,y]=\emptyset$, and
$(y,z)\cup[z,y]=S^{1}$.

######  Claim 4.6.

For any distinct $z,y\in S^{1}$ there is a train-route $\rho$ so that one of
the intervals $\partial_{\infty}H^{\pm}(\rho)$ is contained in $(z,y)$.

###### Proof.

The endpoints of train-routes are dense in
$S^{1}=\partial_{\infty}\mathbb{H}$. Fix $x\in(z,y)$ so that $x$ is the
endpoint of a train-route $\gamma$. Since there are infinitely many rising
branches along $\gamma$ (Claim 4.4) the claim follows from the rising case of
Claim 4.5. ∎

Let $H_{x,y}\subset\mathbb{H}$ be the convex hull of
$(x,y)\subset\partial_{\infty}\mathbb{H}$. Let $\mathcal{H}_{x,y}$ be the
union of all open half-planes $H(\rho)$ so that
$\partial_{\infty}H(\rho)\subset(x,y)$. Since train-routes have geodesic
curvature less than $\epsilon$ at every point:

######  Claim 4.7.

The union $\mathcal{H}_{x,y}$ is contained in an $\delta$–neighborhood of
$H_{x,y}$, where $\delta$ may be taken as small as desired by choosing
appropriate $\epsilon,L$. ∎

A set $X\subset\mathbb{H}$ is $\epsilon^{\prime}$–convex if every pair of
points in $X$ can be connected by a path in $X$ which has geodesic curvature
less than $\epsilon^{\prime}$ at every point.

######  Claim 4.8.

$\mathbb{H}{\smallsetminus}\mathcal{H}_{x,y}$ is closed and
$\epsilon^{\prime}$–convex, where $\epsilon^{\prime}$ may be taken as small as
desired by choosing appropriate $\epsilon,L$.

###### Proof.

This is proved in detail on pages 122-123 of [11]. ∎

######  Claim 4.9.

The point $x$ is an accumulation point of
$\partial(\mathbb{H}{\smallsetminus}\mathcal{H}_{x,y})$.

###### Proof.

Pick a sequence of subintervals $(x_{n},y_{n})\subset(x,y)$ so that
$x_{n},y_{n}\to x$ as $n\to\infty$. By Claim 4.6 for every $n$ there is a
route $\rho_{n}$ and a half-plane $H_{n}=H(\rho_{n})$ so that
$\partial_{\infty}H_{n}\subset(x_{n},y_{n})$. It follows that
$H_{n}\subset\mathcal{H}_{x,y}$. Let $r_{n}$ be any bi-infinite geodesic
perpendicular to $\partial H_{x,y}$ and meeting $H_{n}$. Thus $r_{n}\to x$ as
$n\to\infty$. By Claim 4.7 the intersection
$r_{n}\cap\partial(\mathbb{H}{\smallsetminus}\mathcal{H}_{x,y})$ is nonempty,
and we are done. ∎

The next lemma is not needed for the proof of Theorem 4.1: we state it and
give the proof in order to introduce necessary techniques and terminology.

######  Lemma 4.10.

For any non-parabolic point $x\subset S^{1}$ there is a sequence of train-
routes $\\{\rho_{n}\\}$ with associated half-planes $\\{H(\rho_{n})\\}$
forming a neighborhood basis for $x$.

###### Proof.

Let $y,z$ be arbitrary points of $S^{1}$ so that $x,y,z$ are ordered
counterclockwise. It suffices to construct a train-route separating $x$ from
$(y,z)$.

First assume that $x$ is the endpoint of a route $\rho$. Claim 4.4 implies
that there are infinitely many rising branches $\\{a_{m}\\}$ on one side of
$\rho$ and infinitely many falling branches $\\{c_{n}\\}$ on the other side.
Run half-routes $\alpha_{m}$ and $\gamma_{n}$ through $a_{m}$ and $c_{n}$; so
each half-route meets $\rho$ in a single switch. By Claim 4.5 the endpoints
converge $\alpha_{m}(\infty),\gamma_{n}(\infty)\to x$. Thus sufficiently large
$m,n$ give a train-route

$\alpha_{m}\cup\rho|_{[m,n]}\cup\gamma_{n}$

that separates $x$ from $(y,z)$, as desired.

For the general case consider

$\mathcal{K}=\mathbb{H}{\smallsetminus}(\mathcal{H}_{z,x}\cup\mathcal{H}_{x,y}).$

Note that $x$ is an accumulation point of $\mathcal{K}$ (by Claim 4.9 and
because $\mathcal{H}_{z,x}$ cannot contain points of
$\partial(\mathbb{H}{\smallsetminus}\mathcal{H}_{x,y})$). Fix any basepoint
$w\in\mathcal{K}$. By Claim 4.8 $\mathcal{K}$ is $\epsilon^{\prime}$–convex.
Thus there is a path $r\subset\mathcal{K}$ from $w$ to $x$ which has geodesic
curvature less than $\epsilon^{\prime}$ at every point. Since $x$ is not a
parabolic point the projection of $r$ to $S$ recurs to the thick part of $S$;
thus $r$ meets infinitely many branches $\\{b_{n}\\}$ of $\tau^{\mathbb{H}}$.

Suppose that $b$ is a branch of $\tau^{\mathbb{H}}$ lying in $\mathcal{K}$. If
the two sides of $b$ meet $\mathcal{H}_{z,x}$ and $\mathcal{H}_{x,y}$ then $b$
is a bridge of $\mathcal{K}$. If the sides of $b$ meet neither
$\mathcal{H}_{z,x}$ nor $\mathcal{H}_{x,y}$ then $b$ is an interior branch of
$\mathcal{K}$. If exactly one side of $b$ lies in $\mathcal{K}$ then $b$ is a
boundary branch. If both sides lie in $\mathcal{H}_{z,x}$ (or both sides lie
in $\mathcal{H}_{x,y}$) then $b$ is an exterior branch. See Figure 10 below.

By convexity the path $r$ is disjoint from the exterior branches. After a
small isotopy the path $r$ is also disjoint from the boundary branches, meets
the interior branches transversely, and still has geodesic curvature less than
$\epsilon^{\prime}$ at every point.

Now, if $r$ travels along a bridge then there are routes $\rho^{\pm}$ cutting
off half-planes $H^{\pm}$ lying in $\mathcal{H}_{z,x}$ and $\mathcal{H}_{x,y}$
respectively. Then either $x$ is the endpoint of a train-route or a cut and
paste of $\rho^{\pm}$ gives the desired route $\Gamma$ separating $x$ from
$(y,z)$. In either case we are done.

So suppose that $r$ only meets interior branches $\\{b_{n}\\}$ of
$\mathcal{K}$. Let $\gamma_{n}$ be any train-route travelling along $b_{n}$.
If any of the $\gamma_{n}$ land at $x$ we are done, as above. Supposing not:
Fixing orientations and passing to a subsequence we may assume that
$\gamma_{n}(\infty)\to x$ as $n\to\infty$. There are now two cases: suppose
that for infinitely many $n$ we find that $\gamma_{n}(-\infty)\in[y,z]$. Then
passing to a further subsequence we have that $\gamma_{n}\to\Gamma$ where
$\Gamma(\infty)=x$ [15, Theorem 1.5.4]; thus $x$ is the endpoint of a train-
route and we are done as above. The other possibility is that for some
sufficiently large $n$ both endpoints $\gamma_{n}(\pm\infty)$ lie in $(z,y)$.
Since $b_{n}$ is an interior branch $\gamma_{n}$ separates $x$ from $(y,z)$
and Lemma 4.10 is proved. ∎

###  4.3. Finding invariant efficient position

Fix $\alpha\in\mathcal{C}(S)$. (The case where $\alpha\in\mathcal{A}(S)$ is
dealt with at the end.) Let $\alpha^{\prime}$ be a component of the lift of
$\alpha$ to the universal cover $\mathbb{H}$. Let $\pi_{1}(\alpha)$ be the
cyclic subgroup (of the deck group) preserving $\alpha^{\prime}$. Let
$\\{x,y\\}=\partial_{\infty}\alpha^{\prime}\subset S^{1}$. We take

$\mathcal{K}=\mathbb{H}{\smallsetminus}(\mathcal{H}_{y,x}\cup\mathcal{H}_{x,y}).$

By construction $\mathcal{K}$ is $\pi_{1}(\alpha)$–invariant. By Claims 4.8
and 4.9 the set $\mathcal{K}$ is closed, $\epsilon^{\prime}$–convex, and has
$\\{x,y\\}\subset\partial_{\infty}\mathcal{K}$. By Lemma 4.10 the only non-
parabolic points of $\partial_{\infty}\mathcal{K}$ are $x$ and $y$. As in the
proof of Lemma 4.10 we find a bi-infinite path $r\subset\mathcal{K}$
connecting $y$ to $x$, with geodesic curvature less than $\epsilon^{\prime}$
at every point. See Figure 10.

2pt $y$ [b] at 146 292.1 $x$ [t] at 146 0 $\mathcal{H}_{x,y}$ [B] at 217.8
116.4 $\mathcal{H}_{y,x}$ [B] at 59.7 167.1

$\begin{array}[]{c}\includegraphics[height=7cm]{K}\end{array}$

Figure 10. The path $r$ runs from $y$ to $x$. To simplify the figure, no
interior branches are shown.

Let $H(r)$ be the open half-plane to the right of $r$. If we remove the union

$\bigcup_{g\in\pi_{1}(\alpha)}g\cdot H(r)$

from $\mathbb{H}$ then, as with Claim 4.8, what remains is closed and
$\epsilon^{\prime\prime}$–convex for some small $\epsilon^{\prime\prime}$. It
follows that we may homotope the path $r$ to become a
$\pi_{1}(\alpha)$–invariant smooth path, contained in $\mathcal{K}$ and
transverse to the interior branches, and avoiding the exterior branches of
$\mathcal{K}$. A further equivariant isotopy ensures that $r$ also avoids the
boundary branches of $\mathcal{K}$. Orient $r$ from $y$ to $x$.

######  Remark 4.11.

Suppose that $\gamma\prec\tau^{\mathbb{H}}$ is a train-route that separates
$x$ from $y$. Note that if there exists a non-identity element
$g\in\pi_{1}(\alpha)$ so that $\gamma$ and $g\cdot\gamma$ meet then $r$ is
carried by $\tau^{\mathbb{H}}$, thus $\alpha\prec\tau$, and we are done. We
will henceforth assume that train-routes separating $x$ from $y$ are disjoint
from their non-trivial translates.

Let $b$ be any interior branch of $\mathcal{K}$ and let $\gamma$ be any train-
route travelling along $b$. Since $b$ is interior, $\gamma$ must separate $x$
from $y$. Orient $\gamma$ from $\mathcal{H}_{y,x}$ to $\mathcal{H}_{x,y}$.
(Thus if $\gamma$ and $r$ meet once then the tangent vectors to $r$ and
$\gamma$, in that order, form a positive frame.) The orientation of $\gamma$
gives an orientation to $b$. Moreover, as $b$ is an interior branch a cut and
paste argument shows that the orientation on $b$ is independent of our choice
of $\gamma$. Orient all interior branches in this fashion and note that these
orientations agree at interior switches.

We say that $p\in r\cap b$ has positive or negative sign as the tangent
vectors to $r$ and $b$ (in that order) form a positive or negative frame.
Suppose that there are ${\sf N}\in\mathbb{N}$ orbits of points of negative
sign, under the action of $\pi_{1}(\alpha)$. We now induct on ${\sf N}$.

Suppose that ${\sf N}$ is zero. Any bigon between $r$ and a train-route is
contained in $\mathcal{K}$ and so contributes one point of positive and one
point of negative sign. So if there are no points of negative sign then there
are no bigons and $r$ is in efficient position with respect to
$\tau^{\mathbb{H}}$. Recall that $r$ is $\pi_{1}(\alpha)$–invariant. So
$\beta\subset S$, the image of $r$ under the universal covering map, is an
immersed curve in $S$ homotopic to $\alpha$. If $\beta$ is embedded then we
are done. If not then the Bigon Criterion for immersed curves [18] implies
that $\beta$ must have either a monogon or a bigon of self-intersection. If
$\beta$ has a monogon $B$ of self-intersection then, since index is additive,
$\tau$ must be disjoint from $B$. Thus we can homotope $\beta$ to remove $B$
while fixing $\tau$ pointwise. If $\beta$ has a bigon $B$ of self-intersection
then, as in the proof of uniqueness in Section 4.1, we may remove $B$ via a
sequence of rectangle swaps. After removing all monogons and bigons of self-
intersection the curve $\beta$ is embedded and in efficient position.

2pt $r$ [l] at 98.9 19.1 $r$ [l] at 315.7 19.1 $\gamma_{L}$ [l] at 148.1 183.2
$\gamma_{R}$ [l] at 365.9 56.7

$\begin{array}[]{c}\includegraphics[height=4cm]{bigons}\end{array}$

Figure 11. Left: The lowest point shown has negative sign. The paths $r$ and
$\gamma_{L}$ form a bigon. Right: The corresponding figure for $\gamma_{R}$.

Suppose that ${\sf N}$ is positive. Let $b$ be a branch with a point $p\in
r\cap b$ of negative sign. Let $\gamma_{R}$ ($\gamma_{L}$) be the half-route
starting at $p$, travelling in the direction of $b$, and thereafter turning
only right (left). Each of $\gamma_{R}$ and $\gamma_{L}$ must have at least
one bigon with $r$, as their points at infinity lie in $(x,y)$. There are now
two (essentially identical) cases:

  * •

There is a bigon $B$ between $r$ and $\gamma_{R}$, to the right of $r$, so
that the corners of $B$ appear in the same order along $r$ and $\gamma_{R}$.

  * •

There is a bigon $B$ between $r$ and $\gamma_{L}$, to the right of $r$, so
that the corners of $B$ appear in opposite order along $r$ and $\gamma_{L}$.

See Figure 11. If neither case holds then any half-route ending at $p$ must
originate in $(x,y)$, contradicting the fact that $p$ has negative sign. Note
that, by Remark 4.11, $\gamma_{R}$ ($\gamma_{L}$) is disjoint from its non-
trivial $\pi_{1}(\alpha)$ translates. It follows that the bigon $B$ is also
disjoint from its non-trivial translates. Finally, we may equivariantly
isotope $r$ across $\pi_{1}(\alpha)\cdot B$. Since the arc of
$\gamma_{R}\cap\partial B$ (respectively $\gamma_{L}\cap\partial B$) is combed
outside of $B$ this isotopy reduces ${\sf N}$ by at least one. (Again, see
Figure 11.) This completes the proof of Theorem 4.1 when $\alpha$ is a curve.

###  4.4. Efficient position for arcs and multicurves

Now suppose that $\alpha\subset S$ is an essential arc. Let $\alpha^{\prime}$
be a lift of $\alpha$ to $\mathbb{H}$, the universal cover of $S$. Note that
$\\{x,y\\}=\partial_{\infty}\alpha^{\prime}$ is a pair of parabolic points.
Construct
$\mathcal{K}=\mathbb{H}{\smallsetminus}(\mathcal{H}_{y,x}\cup\mathcal{H}_{x,y})$
as before. The proof now proceeds as above, omitting any mention of
$\pi_{1}(\alpha)$, equivariance, or annulus swaps.

Finally suppose that $\Delta$ is a multicurve. As shown in Section 4.2 we may
isotope, individually, every $\alpha\in\Delta$ into efficient position. By
Corollary 4.3 all $\alpha\in\Delta$ may be realized disjointly in efficient
position. This completes the proof of Theorem 4.1. ∎

##  5\. The structure theorem

###  5.1. Bounding diameter

We now bound the diameters of the sets of wide arcs and curves carried by the
induced track.

######  Lemma 5.1.

Suppose that $\tau$ is a birecurrent track and $X\subset S$ is an essential
annulus. If $\tau|X\neq\emptyset$ then the diameter of $V(\tau|X)\cup
V^{*}(\tau|X)$ inside of $\mathcal{A}(X)$ is at most eight.

###### Proof.

In the proof we use $V,V^{*}$ to represent $V(\tau|X)$ and $V^{*}(\tau|X)$.
Since $\tau|X\neq\emptyset$ it follows that $\mathcal{A}(\tau|X)$ is nonempty.
The first conclusion of Lemma 3.4 now implies that $V$ is nonempty.

######  Claim.

$V^{*}\neq\emptyset$.

###### Proof.

By Lemma 2.6 there is a dual curve $\beta\in\mathcal{C}^{*}(\tau)$ so that
$i(\alpha,\beta)>0$. Thus there is a lift $\beta^{\prime}\subset S^{X}$ with
closure an essential arc. Since $\tau|X\subset\tau^{X}$ it follows that
$\beta^{\prime}\in\mathcal{A}^{*}(\tau|X)$. The first conclusion of Lemma 3.4
now implies that $V^{*}$ is nonempty. ∎

######  Claim.

If $\beta\in V$ and $\gamma\in V^{*}$ then $i(\beta,\gamma)\leq 3$.

###### Proof.

Suppose that $i(\beta,\gamma)=n\geq 4$. Let $\\{\gamma_{i}\\}_{i=1}^{n-1}$ be
the components of $\gamma{\smallsetminus}\beta$ with both endpoints on
$\beta$. Let $R_{i}$ be the components of
$S^{X}{\smallsetminus}(\beta\cup\gamma)$ with compact closure. We arrange
matters so that opposite sides of $R_{i}$ are on $\gamma_{i}$ and
$\gamma_{i+1}$. Let $R$ be the union of the $R_{i}$. Since
$\operatorname{index}(R)=0$ every region $T$ of the closure of
$R{\smallsetminus}N(\tau|X)$ also has index zero and so is a rectangle. If $T$
meets both $\gamma_{i}$ and $\gamma_{i+1}$ then $\gamma$ was not wide, a
contradiction. As $n-1\geq 3$ any region $T$ meeting $\gamma_{2}$ is a compact
rectangle component of the closure of $S^{X}{\smallsetminus}N(\tau|X)$. An
index argument implies that $\tau^{X}$ and thus $\tau\subset S$ has a
complementary region with non-negative index, a contradiction. ∎

Since $V,V^{*}$ are nonempty it follows that $\operatorname{diam}(V\cup
V^{*})\leq 8$. ∎

Now suppose that $X$ is not an annulus. Prompted by Lemma 2.8 we define

$W(\tau^{X})=\left\\{\alpha\in\mathcal{AC}\big{(}\tau^{X}\big{)}\mathbin{\mid}\mbox{$\alpha$
is wide}\right\\}.$

Define $W^{*}(\tau^{X})$ similarly, replacing $\mathcal{AC}(\tau^{X})$ by
$\mathcal{AC}^{*}(\tau^{X})$.

######  Lemma 5.2.

There is a constant ${\sf K_{1}}={\sf K_{1}}(S)$ with the following property.
Suppose that $\tau$ is a track and $X\subset S$ is an essential subsurface
(not an annulus) with $\pi_{X}(\tau)\neq\emptyset$. Then the diameter of
$W(\tau^{X})\cup W^{*}(\tau^{X})$ inside of $\mathcal{AC}(X)$ is at most ${\sf
K_{1}}$. Furthermore if, after isotoping $\partial X$ into efficient position,
the induced orientation on $\partial X$ is not wide then either
$\mathcal{C}(\tau|X)$ or $\mathcal{C}^{*}(\tau|X)$ has diameter at most two in
$\mathcal{C}(X)$.

###### Proof.

In the proof we use $W,W^{*},\mathcal{AC},\mathcal{AC}^{*}$ to represent
$W(\tau^{X})$ and so on. Since $\pi_{X}(\tau)\neq\emptyset$ there is some
vertex cycle $\alpha\in V(\tau)$ so that $\alpha$ cuts $X$. Since $\alpha$ is
wide (Lemma 2.8) there is a lift $\alpha^{\prime}\prec\tau^{X}$ which is also
wide; deduce that $W$ is nonempty.

######  Claim.

$W^{*}\neq\emptyset$.

###### Proof.

By Lemma 2.6 there is a dual curve $\alpha\in\mathcal{C}^{*}(\tau)$ cutting
$X$. By Lemma 2.9 there is a wide dual $\beta$ that also cuts $X$. Thus there
is a lift $\beta^{\prime}\subset S^{X}$ with closure an essential wide arc or
wide essential, non-peripheral curve. So $\beta^{\prime}\in W^{*}$ as desired.
∎

Now isotope $\partial X$ into efficient position. Let $X^{\prime}$ be the
compact component of the preimage of $X$ under the covering map $S^{X}\to S$.
Note that $\partial X^{\prime}$ is in efficient position with respect to
$\tau^{X}$. Note that the covering map $S^{X}\to S$ induces a homeomorphism
between $X^{\prime}$ and $X$. Let $N^{X}=N(\tau^{X})\subset S^{X}$ be the
preimage of $N=N(\tau)$. Let $N^{\prime}=X^{\prime}\cap N^{X}$. Again, the
covering map induces a homeomorphism between $N^{\prime}$ and $N\cap X$.

Suppose that $\partial X$, with its induced orientation, is not wide. If
$\partial X$ fails to be wide in $S{\smallsetminus}N(\tau)$ then there is a
properly embedded, essential arc $\gamma\subset X$ disjoint from $N(\tau)$.
Lift $\gamma$ to $\gamma^{\prime}\subset X^{\prime}$. Adjoin to
$\gamma^{\prime}$ geodesic rays in $S^{X}{\smallsetminus}X^{\prime}$ to obtain
an essential, properly embedded arc $\gamma^{\prime\prime}\subset S^{X}$. Note
that $i(\gamma^{\prime\prime},\alpha)=0$ for every $\alpha\in\mathcal{AC}$;
only intersections in $X^{\prime}$ contribute to geometric intersection number
as computed in $S^{X}$. This implies that
$\operatorname{diam}_{X}(\mathcal{C}(\tau|X))\leq 2$. Furthermore,
$i(\gamma^{\prime\prime},\beta)\leq 2$ for every $\beta\in W^{*}$. This gives
the desired diameter bound for $W\cup W^{*}$.

If, instead, $\partial X$ fails to be wide in $N(\tau)$ then there is a
properly embedded, essential arc $\gamma\subset X$ that is a subarc of a tie.
Again, lift and extend to an essential arc $\gamma^{\prime\prime}\subset
S^{X}$ so that $i(\gamma^{\prime\prime},\beta)=0$ for any
$\beta\in\mathcal{AC}^{*}$. This implies that
$\operatorname{diam}_{X}(\mathcal{C}^{*}(\tau|X))\leq 2$. We also have
$i(\gamma^{\prime\prime},\alpha)\leq 2$ for any $\alpha\in W$. Again the
diameter is bounded.

Now suppose that $\partial X$ is wide. Thus, for every
$b\in\mathcal{B}(\tau)$, the rectangle $R_{b}$ meets $\partial X$ in at most a
pair of arcs. It follows that $N\cap X$, and thus $N^{\prime}$, is a union of
at most $2|\mathcal{B}(\tau)|$ subrectangles of the form
$R^{\prime},R^{\prime\prime}\subset R_{b}$. Suppose that $\alpha\in W$ and
$\beta\in W^{*}$. Then $\alpha$ and $\beta$ each meet a subrectangle
$R^{\prime}$ in at most two arcs. Thus $\alpha$ and $\beta$ intersect in at
most four points inside of $R^{\prime}$. Thus $i(\alpha,\beta)\leq
8|\mathcal{B}(\tau)|$ and Lemma 5.2 is proved. ∎

###  5.2. Accessible intervals

Suppose that $\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding and splitting sequence of
birecurrent train tracks. Suppose $X\subset S$ is an essential subsurface, yet
not an annulus, with $\xi(X)<\xi(S)$. Define

$\displaystyle
m_{X}=\min\big{\\{}i\in[0,N]\mathbin{\mid}\operatorname{diam}_{X}(\mathcal{C}^{*}(\tau_{i}|X))\geq
3\big{\\}}$ and $\displaystyle
n_{X}=\max\big{\\{}i\in[0,N]\mathbin{\mid}\operatorname{diam}_{X}(\mathcal{C}(\tau_{i}|X))\geq
3\big{\\}}.$

If either $m_{X}$ or $n_{X}$ is undefined or if $n_{X}<m_{X}$ then $I_{X}$,
the accessible interval is empty. Otherwise, $I_{X}=[m_{X},n_{X}]$.

If $X\subset S$ is an annulus, then $I_{X}$ is defined by replacing
$\mathcal{C}$ by $\mathcal{A}$ and increasing the lower bound on diameter from
3 to 9. We may now state the structure theorem:

######  Theorem 5.3.

For any surface $S$ with $\xi(S)\geq 1$ there is a constant ${\sf K_{0}}={\sf
K_{0}}(S)$ with the following property: Suppose that
$\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding and splitting sequence of birecurrent
train tracks in $S$ and suppose that $X\subset S$ is an essential subsurface.

  * •

For every $[a,b]\subset[0,N]$ if $[a,b]\cap I_{X}=\emptyset$ and
$\pi_{X}(\tau_{b})\neq\emptyset$ then $d_{X}(\tau_{a},\tau_{b})\leq{\sf
K_{0}}$.

Suppose $i\in I_{X}$. If $X$ is an annulus:

  * •

The core curve $\alpha$ is carried by and wide in $\tau_{i}$.

  * •

Both sides of $\alpha$ are combed in the induced track $\tau_{i}|X$.

  * •

If $i+1\in I_{X}$ then $\tau_{i+1}|X$ is obtained by taking subtracks, slides,
or at most a pair of splittings of $\tau_{i}|X$.

If $X$ is not an annulus:

  * •

When in efficient position $\partial X$ is wide with respect to $\tau_{i}$.

  * •

The track $\tau_{i}|X$ is birecurrent and fills $X$.

  * •

If $i+1\in I_{X}$ then $\tau_{i+1}|X$ is either a subtrack, a slide, or a
split of $\tau_{i}|X$.

###### Proof.

Fix an interval $[a,b]\subset[0,N]$. Note that $\tau_{b}\prec\tau_{a}$ and so
$\tau_{b}^{X}\prec\tau_{a}^{X}$. Thus
$\mathcal{AC}(\tau_{b}^{X})\subset\mathcal{AC}(\tau_{a}^{X})$ while
$\mathcal{AC}^{*}(\tau_{a}^{X})\subset\mathcal{AC}^{*}(\tau_{b}^{X})$.

######  Claim.

If $[a,b]\cap I_{X}=\emptyset$ and $\pi_{X}(\tau_{b})\neq\emptyset$ then
$d_{X}(\tau_{a},\tau_{b})\leq{\sf K_{0}}$.

###### Proof.

Fix, for the duration of the claim, a vertex cycle $\beta\in V(\tau_{b})$ so
that $\beta$ cuts $X$. Since $\beta$ is also carried by $\tau_{a}$ there is,
by Lemma 2.9, a vertex cycle $\alpha\in V(\tau_{a})$ cutting $X$. Pick
$\alpha^{\prime}\in\pi_{X}(\alpha)$ and $\beta^{\prime}\in\pi_{X}(\beta)$.
Note Lemma 2.8 implies that $\alpha^{\prime}$ is wide in $\tau_{a}^{X}$ while
$\beta^{\prime}$ is wide in $\tau_{b}^{X}$. The proof divides into cases
depending on the relative positions of $a,b,m_{X}$ and $n_{X}$.

######  Case I.

Suppose $n_{X}<a$ or $n_{X}$ is undefined.

Note that $\beta^{\prime}\prec\tau_{a}^{X}$. If $X$ is an annulus then since
$a\mathbin{\notin}I_{X}$ the diameter of $\mathcal{A}(\tau_{a}|X)$ is at most
eight; thus $d_{X}(\alpha,\beta)\leq 8$ and we are done.

Suppose that $X$ is not an annulus. If $\beta^{\prime}$ is an arc then Lemma
3.3 gives two cases: we may replace $\beta^{\prime}$ by $\gamma$ which is
either a wide arc in $\tau_{a}^{X}$ or is an essential non-peripheral curve in
$\tau_{a}|X$. (If $\beta^{\prime}$ is a curve then let
$\gamma=\beta^{\prime}$.) In either case Lemma 3.3 ensures that
$i(\gamma,\beta^{\prime})\leq 2$ and so $d_{X}(\gamma,\beta^{\prime})\leq 4$.
If $\gamma$ is an arc then both $\alpha^{\prime}$ and $\gamma$ are wide so
Lemma 5.2 gives $d_{X}(\alpha,\beta)\leq{\sf K_{1}}+4$. If $\gamma$ is a curve
pick any $\delta\in V(\tau_{a}|X)$. Then Lemma 5.2 implies that
$d_{X}(\alpha^{\prime},\delta)\leq{\sf K_{1}}$. Also, $a\mathbin{\notin}I_{X}$
implies that $d_{X}(\delta,\gamma)\leq 2$. Thus $d_{X}(\alpha,\beta)\leq{\sf
K_{1}}+6$.

######  Case II.

Suppose $b<m_{X}$ or $m_{X}$ is undefined.

If $X$ is an annulus, then Lemma 5.1 gives wide duals $\alpha^{*}\in
V^{*}(\tau_{a}|X)$ and $\beta^{*}\in V^{*}(\tau_{b}|X)$ so that
$d_{X}(\alpha^{\prime},\alpha^{*}),d_{X}(\beta^{\prime},\beta^{*})\leq 8$. It
follows that the arc $\alpha^{*}$ also lies in $\mathcal{A}^{*}(\tau_{b}|X)$.
Since $b\mathbin{\notin}I_{X}$ we have $d_{X}(\alpha^{*},\beta^{*})\leq 8$.
Thus $d_{X}(\alpha,\beta)\leq 24$, as desired.

If $X$ is not an annulus, then by Lemma 5.2 there is a wide dual
$\alpha^{*}\in W^{*}(\tau_{a}^{X})$ so that
$d_{X}(\alpha^{\prime},\alpha^{*})\leq{\sf K_{1}}$. Again, $\alpha^{*}$ is
also an element of $\mathcal{AC}^{*}(\tau_{b}^{X})$ but may not be wide there.
If $\alpha^{*}$ is an arc then Lemma 3.3 gives two cases: we may replace
$\alpha^{*}$ by $\gamma^{*}$ which is either a wide dual arc to $\tau_{b}^{X}$
or is an essential non-peripheral dual curve to $\tau_{b}^{X}$. (If
$\alpha^{*}$ is a curve then let $\gamma^{*}=\alpha^{*}$.) So
$i(\alpha^{*},\gamma^{*})\leq 2$ and thus $d_{X}(\alpha^{*},\gamma^{*})\leq
4$. If $\gamma^{*}$ is a wide dual arc then Lemma 5.2 implies that
$d_{X}(\gamma^{*},\beta^{\prime})\leq{\sf K_{1}}$ and so
$d_{X}(\alpha,\beta)\leq 2{\sf K_{1}}+4$. If $\gamma^{*}$ is a dual curve
then, as $b\mathbin{\notin}I_{X}$, any dual wide curve $\delta^{*}\in
V^{*}(\tau_{b}|X)$ has $d_{X}(\gamma^{*},\delta^{*})\leq 2$. Again, Lemma 5.2
implies that $d_{X}(\delta^{*},\beta^{\prime})\leq{\sf K_{1}}$ and so
$d_{X}(\alpha,\beta)\leq 2{\sf K_{1}}+6$.

######  Case III.

Suppose $a\leq n_{X}<c<m_{X}\leq b$.

The first two cases bound $d_{X}(\tau_{c},\tau_{b})$ and
$d_{X}(\tau_{a},\tau_{c})$; thus we are done by the triangle inequality.

######  Case IV.

Suppose $a\leq n_{X}<m_{X}\leq b$ and $m_{X}=n_{X}+1$.

Let $c=n_{X}$ and $d=m_{X}$. The first two cases bound
$d_{X}(\tau_{d},\tau_{b})$ and $d_{X}(\tau_{a},\tau_{c})$. Since $V(\tau_{c})$
and $V(\tau_{d})$ have bounded intersection $d_{X}(\tau_{c},\tau_{d})$ is also
bounded and the claim is proved. ∎

Now fix $i\in I_{X}$.

######  Claim.

If $X$ is an annulus:

  * •

The core curve $\alpha$ is carried by and is wide in $\tau_{i}$.

  * •

Both sides of $\alpha$ are combed in the induced track $\tau_{i}|X$.

  * •

If $i+1\in I_{X}$ then $\tau_{i+1}|X$ is obtained by taking subtracks, slides,
or at most a pair of splittings of $\tau_{i}|X$.

###### Proof.

Since $i\in I_{X}$, both $\mathcal{A}(\tau_{i}|X)$ and
$\mathcal{A}^{*}(\tau_{i}|X)$ have diameter at least nine. From Lemma 5.1
deduce that the inclusions $V\subset\mathcal{A}$ and
$V^{*}\subset\mathcal{A}^{*}$ are strict. Thus by Lemma 3.5 the core curve
$\alpha$ is both carried by and dual to $\tau_{i}|X$. The second statement now
follows from Lemma 3.6. Thus at least one side of $\alpha$ is combed in
$\tau_{i}^{X}$. Projecting from $S^{X}$ to $S$ we find that
$\alpha\prec\tau_{i}$. If $\alpha$ is not wide in $\tau_{i}$ then we deduce
that neither side of $\alpha$ is combed in $\tau^{X}$, a contradiction.

Suppose that $\tau_{i}$ slides to $\tau_{i+1}$. Then, up to isotopy,
$\tau_{i+1}$ slides to $\tau_{i}$. Since slides do not kill essential arcs it
follows that $\tau_{i+1}|X$ is obtained from $\tau_{i}|X$ by an at most
countable collection of slides.

Now suppose that $\tau_{i+1}$ is obtained by splitting $\tau_{i}$ along a
large branch $b$. Thus $\tau_{i+1}^{X}$ is obtained from $\tau_{i}^{X}$ by
splitting all of the countably many lifts of $b$. Every essential arc carried
by $\tau_{i+1}|X$ is also carried by $\tau_{i}^{X}$. Let
$\tau^{\prime}\subset\tau_{i}^{X}$ be the union of these essential routes. It
follows that $\tau_{i+1}|X$ is obtained from $\tau^{\prime}$ by splitting
along lifts of $b$ that are also large branches of $\tau^{\prime}$. Since both
sides of $\alpha$ are combed in $\tau_{i}|X$ the same is true in
$\tau^{\prime}$ and so any component of $\tau^{\prime}{\smallsetminus}\alpha$
is a tree without large branches. The track $\tau^{\prime}$ therefore has only
finitely many large branches, all contained in $\alpha$. Since $\alpha$ is
wide in $\tau_{i}$ there are at most two preimages of the large branch $b$
contained in $\alpha\subset S^{X}$. Thus $\tau_{i+1}|X$ is obtained from
$\tau^{\prime}$ by at most two splittings. This proves the claim. (See Figure
12 for pictures of how $\alpha$ may be carried by $\tau_{i}$ and how splitting
$b$ effects $\tau_{i}|X$.) ∎

$\begin{array}[]{cc}\includegraphics[height=3cm]{widelarge00}&\includegraphics[height=3cm]{widelarge01}\\\
\\\
\includegraphics[height=3cm]{widelarge10}&\includegraphics[height=3cm]{widelarge11}\end{array}$
Figure 12. Four of the possible ways for an oriented, carried, wide curve
$\alpha$ to meet a large rectangle $R_{b}$ of $N(\tau)$. Note that when $X$ is
an annulus and $\alpha$ is the core curve the upper left picture implies that
neither side of $\alpha$ is combed in $\tau|X$. Splitting the upper right
deletes zero or two components of $\tau_{i}|X{\smallsetminus}\alpha$. In the
bottom row, only the left splitting is possible when $i,i+1\in I_{X}$. On the
bottom left one component of $\tau_{i}|X{\smallsetminus}\alpha$ is deleted and
$\tau_{i}|X$ is split once. On the bottom right $\tau_{i}|X$ is split twice.

######  Claim.

Suppose $X$ is not an annulus.

  * •

When in efficient position $\partial X$ is wide with respect to $\tau_{i}$.

  * •

The track $\tau_{i}|X$ is birecurrent and fills $X$.

  * •

If $i+1\in I_{X}$ then $\tau_{i+1}|X$ is either a subtrack, a slide, or a
split of $\tau_{i}|X$.

###### Proof.

Since $i\in I_{X}$ the second conclusion of Lemma 5.2 implies that $\partial
X$ is wide. The induced track $\tau_{i}|X$ carries a pair of curves at
distance at least three, so fills $X$. Also, $\tau_{i}|X$ is recurrent by
definition. For any branch $b^{\prime}\in\tau_{i}|X\subset S^{X}$, let
$b\subset\tau$ be the image in $S$. Since $\tau$ is transversely recurrent
there is a dual curve $\beta$ meeting $b$. Lifting $\beta$ to a curve or arc
$\beta^{\prime}\subset S^{X}$ gives a dual to $\tau|X$ meeting $b^{\prime}$.
Thus $\tau|X$ is transversely recurrent with respect to arcs and curves, as
defined in Section 3.1

Now, if $\tau_{i}$ slides to $\tau_{i+1}$ then, as in the annulus case,
$\tau_{i}|X$ slides to $\tau_{i+1}|X$. Suppose instead that $\tau_{i}$ splits
to $\tau_{i+1}$ along the branch $b\in\mathcal{B}(\tau_{i})$. Thus
$\tau_{i}|X$ splits (or isotopes) to a track $\tau^{\prime}$ so that
$\tau_{i+1}|X$ is a subtrack. Let $R_{b}\subset N(\tau_{i})$ be the rectangle
corresponding to the branch $b$. Isotope $\partial X$ into efficient position
with respect to $N(\tau_{i})$ and recall that $X$ is to the left of $\partial
X$. Note that, by an isotopy, we may arrange for all curves in
$\mathcal{C}(\tau_{i}|X)$ to be disjoint from $\partial X$. Let $\beta\subset
R_{b}$ be a central splitting arc: a carried arc completely contained in
$R_{b}$.

If $\beta\cap X$ is empty then $\tau_{i}|X$ is identical to $\tau_{i+1}|X$. If
$\beta\subset X$ then $\tau_{i+1}|X$ is either a subtrack or a splitting of
$\tau_{i}|X$, depending how the carried curves of $\tau_{i}|X$ meet the lift
of $R_{b}$. In all other cases $\tau_{i+1}|X$ is a subtrack of $\tau_{i}|X$.
See Figure 12 for some of the ways carried subarcs of $\partial X$ may meet
$R_{b}$. If $\partial X\cap R_{b}$ contains a tie then $\tau_{i}|X$ is
identical to $\tau_{i+1}|X$. This completes the proof of the claim. ∎

Thus Theorem 5.3 is proved. ∎

We now rephrase a result of Masur and Minsky using the refinement procedure of
Penner and Harer [15, page 122].

######  Theorem 5.4.

[13, Theorem 1.3] For any surface $S$ with $\xi(S)\geq 1$ there is a constant
$Q=Q(S)$ with the following property: For any sliding and splitting sequence
$\\{\tau_{i}\\}_{i=0}^{N}$ of birecurrent train tracks in $S$ the sequence
$\\{V(\tau_{i})\\}_{i=0}^{N}$ forms a $Q$–unparameterized quasi-geodesic in
$\mathcal{C}(S)$.

Theorem 5.3 implies that the same result holds after subsurface projection.

######  Theorem 5.5.

For any surface $S$ with $\xi(S)\geq 1$ there is a constant $Q=Q(S)$ with the
following property: For any sliding and splitting sequence
$\\{\tau_{i}\\}_{i=0}^{N}$ of birecurrent train tracks in $S$ and for any
essential subsurface $X\subset S$ if $\pi_{X}(\tau_{N})\neq\emptyset$ then the
sequence $\\{\pi_{X}(\tau_{i})\\}_{i=0}^{N}$ is a $Q$–unparameterized quasi-
geodesic in $\mathcal{C}(X)$.

###### Proof.

By the first conclusion of Theorem 5.3 we may restrict attention to the
subinterval $[p,q]=I_{X}\subset[0,N]$.

Fix any vertex $\alpha\in V(\tau_{q}|X)$. Note $\alpha$ is carried by
$\tau_{i}|X$ for all $i\leq q$. So define $\sigma_{i}\subset\tau_{i}|X$ to be
the minimal pretrack carrying $\alpha$. Since $\sigma_{i}$ does not carry any
peripheral curves $\sigma_{i}$ is a train track. Note that $\sigma_{i}$ is
recurrent by definition and transversely recurrent by Lemma 3.2. Applying
Theorem 5.3, for all $i\in[p,q-1]$ the track $\sigma_{i+1}$ is a slide, a
split, or identical to the track $\sigma_{i}$.

Theorem 5.4 implies the sequence $\\{V(\sigma_{i})\\}$ is a
$Q$–unparameterized quasi-geodesic in $\mathcal{C}(X)$. Note that
$d_{X}(\sigma_{i},\tau_{i}|X)$ is uniformly bounded because $\sigma_{i}$ is a
subtrack.

Since $\alpha\prec\tau_{i}|X$ the curve $\alpha$ is also carried by
$\tau_{i}$. By Lemma 2.9 there is a vertex cycle $\beta_{i}\prec\tau_{i}$ that
cuts $X$. Since $\beta_{i}$ is wide (Lemma 2.8) any element
$\beta_{i}^{\prime}\in\pi_{X}(\beta)$ is carried by and wide in
$\tau_{i}^{X}$. It follows that $\beta_{i}^{\prime}$ and the vertex cycles of
$\tau_{i}|X$ have bounded intersection. Thus $d_{X}(\tau_{i},\tau_{i}|X)$ is
uniformly bounded and we are done. ∎

##  6\. Further applications of the structure theorem

We now turn to Theorems 6.1 and 6.2; both are slight generalizations of a
result of Hamenstädt [8, Corollary 3]. Our proofs, however, rely on Theorem
5.3 and are quite different from the proof found in [8].

###  6.1. The marking and train track graphs

Suppose that $S$ is not an annulus. A finite subset
$\mu\subset\mathcal{AC}(S)$ fills $S$ if for all $\beta\in\mathcal{C}(S)$
there is a $\gamma\in\mu$ so that $i(\beta,\gamma)\neq 0$. If
$\mu,\nu\subset\mathcal{AC}(S)$ then we define

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

Also, let $i(\mu)=i(\mu,\mu)$ be the self-intersection number. A set $\mu$ is
a $k$–marking if $\mu$ fills $S$ and $i(\mu)\leq k$. Two sets $\mu,\nu$ are
$\ell$–close if $i(\mu,\nu)\leq\ell$.

Define $k_{0}=\max_{\tau}i(V(\tau))$, where $\tau$ ranges over tracks with
vertex cycles $V(\tau)$ filling $S$. Define
$\ell_{0}=\max_{\tau,\sigma}i(V(\tau),V(\sigma))$, where $\sigma$ ranges over
tracks obtained from $\tau$ by a single splitting. Referring to [12] for the
necessary definitions, we define $k_{1}=\max_{\mu}i(\mu)$, where $\mu$ ranges
over complete clean markings of $S$. Define
$\ell_{1}=\max_{\mu,\nu}i(\mu,\nu)$, where $\nu$ ranges over markings obtained
from $\mu$ by a single elementary move. Define
$\ell_{2}=\max_{\tau}\min_{\mu}i(V(\tau),\mu)$.

Note that there are only finitely many tracks $\tau$ and finitely many
complete clean markings $\mu$, up to the action of $\mathcal{MCG}(S)$. An
index argument bounds $|\mathcal{B}(\tau)|$ and so bounds the number of
splittings of $\tau$. Lemma 2.4 of [12] bounds the number of elementary moves
for $\mu$. Thus the quantities $k_{0},k_{1},\ell_{0},\ell_{1}$ are well-
defined. An upper bound for $\ell_{2}$ can be obtained by surgering $V(\tau)$
to obtain a complete clean marking: see the discussion preceding Lemma 6.1 in
[1]. Now define $k=\max\\{k_{0},k_{1}\\}$ and
$\ell=\max\\{\ell_{0},\ell_{1},\ell_{2}\\}$. Define $\mathcal{M}(S)$ to be the
marking graph: the vertices are $k$–markings and the edges are given by
$\ell$–closeness. (When $S$ is an annulus we take
$\mathcal{M}(S)=\mathcal{A}(S)$. Recall that $\mathcal{A}(S)$ is quasi-
isometric to $\mathcal{MCG}(S,\partial)\cong\mathbb{Z}$.)

That $\mathcal{M}(S)$ is connected now follows from the discussion at the
beginning of [12, Section 6.4]. Accordingly, define
$d_{\mathcal{M}(S)}(\mu,\nu)$ to be the length of the shortest edge-path
between the markings $\mu$ and $\nu$.

Since the above definitions are stated in terms of geometric intersection
number, the mapping class group $\mathcal{MCG}(S)$ acts via isometry on
$\mathcal{M}(S)$. Counting the appropriate set of ribbon graphs proves that
the action has finitely many orbits of vertices and edges. The Alexander
method [4, Section 2.4] proves that vertex stabilizers are finite and hence
the action is proper. It now follows from the Milnor-Švarc Lemma [2,
Proposition I.8.19] that any Cayley graph for $\mathcal{MCG}(S)$ is quasi-
isometric to $\mathcal{M}(S)$.

Define $\mathcal{T}(S)$, the train track graph as follows: vertices are
isotopy classes of birecurrent train tracks $\tau\subset S$ so that $V(\tau)$
fills $S$. Connect vertices $\tau$ and $\sigma$ by an edge exactly when
$\sigma$ is a slide or split of $\tau$. Let
$d_{\mathcal{T}(S)}(\tau,\upsilon)$ be the minimal number of edges in a path
in $\mathcal{T}(S)$ connecting $\tau$ to $\upsilon$, if such a path exists.
Note that the map $\tau\mapsto V(\tau)$ from $\mathcal{T}(S)$ to
$\mathcal{M}(S)$ sends edges to edges (or to vertices) and thus is distance
non-increasing. For further discussion of graphs tightly related to
$\mathcal{T}(S)$ see [8].

We adopt the following notations. If $\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding
and splitting sequence in $\mathcal{T}(S)$ and $I=[p,q]\subset[0,N]$ is a
subinterval then $|I|=q-p$ and
$d_{\mathcal{T}(S)}(I)=d_{\mathcal{T}(S)}(\tau_{p},\tau_{q})$. If
$\tau,\sigma\in\mathcal{T}(S)$ then define

$d_{\mathcal{M}(X)}(\tau,\sigma)=d_{\mathcal{M}(X)}\big{(}V(\tau|X),V(\sigma|X)\big{)}.$

Also take $d_{\mathcal{M}(X)}(I)=d_{\mathcal{M}(X)}(\tau_{p},\tau_{q})$.

######  Theorem 6.1.

For any surface $S$ with $\xi(S)\geq 1$ there is a constant $Q=Q(S)$ with the
following property: Suppose that $\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding and
splitting sequence in $\mathcal{T}(S)$. Then the sequence
$\\{V(\tau_{i})\\}_{i=0}^{N}$, as parameterized by splittings, is a $Q$–quasi-
geodesic in the marking graph.

Our final generalization of [8, Corollary 3] follows from Theorem 6.1:

######  Theorem 6.2.

For any surface $S$ with $\xi(S)\geq 1$ there is a constant $Q=Q(S)$ with the
following property: If $\\{\tau_{i}\\}_{i=0}^{N}$ is a sliding and splitting
sequence in $\mathcal{T}(S)$, injective on slide subsequences, then
$\\{\tau_{i}\\}$ is a $Q$–quasi-geodesic.

Notice that here, unlike Theorem 6.1, the parameterization is by index. We
require:

######  Lemma 6.3.

There is a constant $A=A(S)$ so that if $\\{\tau_{i}\\}_{i=1}^{N}$ is an
injective sliding sequence in $\mathcal{T}(S)$ then $N+1\leq A$. ∎

###### Proof of Theorem 6.2.

Let $\\{\tau_{i}\\}$ be the given sliding and splitting sequence in
$\mathcal{T}(S)$. Let $I=[p,q]\subset[0,N]$ be a subinterval. Note that
$d_{\mathcal{T}(S)}(I)\leq|I|$ because $\\{\tau_{i}\\}$ is an edge-path in
$\mathcal{T}(S)$.

Define $\mathcal{S}(I)$ to be the set of indices $r\in I$ where $r+1\in I$ and
$\tau_{r+1}$ is a splitting of $\tau_{r}$. Thus $|I|\leq_{A}|\mathcal{S}(I)|$,
where $A$ is the constant of Lemma 6.3. Now, Theorem 6.1 implies
$|\mathcal{S}(I)|\leq_{Q}d_{\mathcal{M}(S)}(I)$. Finally, since $\tau\mapsto
V(\tau)$ is distance non-increasing we have $d_{\mathcal{M}(S)}(I)\leq
d_{\mathcal{T}(S)}(I)$. Deduce that $|I|\leq_{Q}d_{\mathcal{M}(S)}(I)$, for a
somewhat larger value of $Q$. ∎

######  Remark 6.4.

Note that we have not used the connectedness of $\mathcal{T}(S)$, an issue
that appears to be difficult to approach combinatorially. For a proof of
connectedness see [8, Corollary 3.7].

###  6.2. Hyperbolicity and the distance estimate

To prove Theorem 6.1 we will need:

######  Theorem 6.5.

[11, Theorem 1.1] For every connected compact orientable surface $X$ there is
a constant $\delta_{X}$ so that $\mathcal{C}(X)$ is $\delta_{X}$–hyperbolic.

An important consequence of the Morse Lemma [2, Theorem III.H.1.7] is a
reverse triangle inequality.

######  Lemma 6.6.

For every $\delta$ and $Q$, there is a constant ${\sf R_{0}}={\sf
R_{0}}(\delta,Q)$ with the following property: For any $\delta$–hyperbolic
space $\mathcal{X}$, for any $Q$–unparameterized quasi-geodesic
$f\colon[m,n]\to\mathcal{X}$, and for any $a,b,c\in[m,n]$, if $a\leq b\leq c$
then

$d_{\mathcal{X}}(\alpha,\beta)+d_{\mathcal{X}}(\beta,\gamma)\leq
d_{\mathcal{X}}(\alpha,\gamma)+{\sf R_{0}}$

where $\alpha,\beta,\gamma=f(a),f(b),f(c)$. ∎

We now take ${\sf R_{0}}={\sf R_{0}}(\delta,Q(S))$ where
$\delta=\max\\{\delta_{X}\mathbin{\mid}X\subset S\\}$, as provided by Theorem
6.5, and $Q(S)$ is the constant of Theorem 5.5. The next central result needed
is the distance estimate for $\mathcal{M}(S)$. Let $[\cdot]_{C}$ be the cut-
off function:

$[x]_{C}\,=\,\left\\{\begin{array}[]{cc}0,&\mbox{if $x<C$}\\\ x,&\mbox{if
$x\geq C$}\end{array}\right\\}.$

We may now state the distance estimate:

######  Theorem 6.7.

[12, Theorem 6.12] For any surface $S$, there is a constant ${\sf C}(S)$ so
that for every $C\geq{\sf C}(S)$ there is an ${\sf E}\geq 1$ so that for all
$\mu,\nu\in\mathcal{M}(S)$

$d_{\mathcal{M}(S)}(\mu,\nu)\,\,=_{\sf E}\,\,\sum[d_{X}(\mu,\nu)]_{C}$

where the sum ranges over essential subsurfaces $X\subset S$.

###  6.3. Marking distance

Suppose that $\\{\tau_{i}\\}_{i=0}^{N}\subset\mathcal{T}(S)$ is a sliding and
splitting sequence. Let $V_{i}=V(\tau_{i})$ be the set of vertex cycles of
$\tau_{i}$. As $i(V_{i})\leq k_{0}$ and $i(V_{i},V_{i+1})\leq\ell_{0}$ the map
$i\mapsto V_{i}$ gives rise to an edge-path in $\mathcal{M}(S)$.

Suppose that $[p,q]\subset[0,N]$. Let $\mathcal{S}_{X}(p,q)$ be the set of
indices $r\in[p,q-1]$ so that $\tau_{r+1}|X$ is a splitting of $\tau_{r}|X$.
(When $X$ is an annulus $\tau_{r+1}|X$ may also differ from $\tau_{r}|X$ by a
pair of splits.) Since slides do not effect $P(\tau)$ [15, Proposition 2.2.2]
the distance in $\mathcal{M}(S)$ between $V_{p}$ and $V_{q}$ is at most
$|\mathcal{S}_{S}(p,q)|$.

######  Remark 6.8.

We do not place indices $r$ onto $S_{X}$ where $\tau_{r+1}|X$ is a subtrack of
$\tau_{r}|X$; the number of such indices is bounded by a constant depending
only on $X$.

In the other direction, we must bound the number of splittings between
$\tau_{p}$ and $\tau_{q}$ in terms of the marking distance between $V_{p}$ and
$V_{q}$. This will be done inductively. As a notational matter set
$I_{S}=[0,N]$. When $X\subset S$ is essential take $V(\tau|X)$ to be the
vertex cycles of the induced track. Recall that $I_{X}\subset I_{S}$, defined
in Section 5.2, is the accessible interval for $X\subset S$. If
$I=[m,n]\subset[0,N]$ is an interval then define
$\mathcal{S}_{X}(I)=\mathcal{S}_{X}(m,n)$,
$d_{X}(I)=d_{X}(\tau_{m},\tau_{n})$, and so on.

######  Proposition 6.9.

Suppose that $X\subset S$ is an essential subsurface and $J_{X}\subset I_{X}$
is a subinterval. There is a constant $A=A(X)$, independent of the sequence
$\\{\tau_{i}\\}$, so that
$|\mathcal{S}_{X}(J_{X})|\leq_{A}d_{\mathcal{M}(X)}(J_{X}).$

The rest of this section is devoted to the proof of Proposition 6.9, from
which Theorem 6.1 follows.

###  6.4. Inductive and straight intervals

We fix two thresholds ${\sf T_{0}},{\sf T_{1}}$ so that:

$\displaystyle\max\big{\\{}6{\sf N_{1}}+2{\sf N_{2}}+2{\sf K_{0}}(X)+2,2{\sf
R_{0}},M_{2}(X),{\sf C}(X)\big{\\}}$ $\displaystyle\leq{\sf T_{0}}(X)$
$\displaystyle\max\big{\\{}{\sf T_{0}}(X)+2{\sf R_{0}},{\sf B_{0}}{\sf
N_{2}}\big{\\}}$ $\displaystyle\leq{\sf T_{1}}(X)$

Here ${\sf N_{1}}$ is an upper bound for $d_{Y}(\alpha,\beta)$ where $Y\subset
S$ is any essential subsurface, $\tau$ is a track, and $\alpha$ and $\beta$
are wide with respect to $\tau$. The constant ${\sf B_{0}}$ is an upper bound
for the number of branches in any induced track. The constant ${\sf N_{2}}$ is
an upper bound for the distance (in any subsurface projection) between the
vertices of $\tau$ (or $\tau|X$) and the vertices of a single splitting or
subtrack of $\tau$. Also, $M_{2}(X)$ is the constant provided by Lemma 6.1 of
[12].

Recall that the interval $J_{X}\subset I_{X}$ is given in Proposition 6.9.

######  Definition 6.10.

Suppose that $Y\subset X$ is an essential subsurface with $\xi(Y)<\xi(X)$. If
$d_{Y}(J_{X})\geq{\sf T_{0}}(X)$ then we call $Y$ an inductive subsurface of
$X$ and take $J_{Y}=I_{Y}\cap J_{X}$ as the associated inductive subinterval
of $J_{X}$. If $d_{Y}(J_{X})<{\sf T_{0}}(X)$ then we set $J_{Y}=\emptyset$.

Suppose $I$ is a subinterval of $J_{X}$. Define $\operatorname{diam}_{Y}(I)$
to be the diameter, in $\mathcal{AC}(Y)$ of the union $\cup_{i\in
I}\pi_{Y}(\tau_{i})$.

######  Definition 6.11.

A subinterval $I\subset J_{X}$ is a straight subinterval for $X$ if for all
essential subsurfaces $Y\subset X$, with $\xi(Y)<\xi(X)$, we have
$\operatorname{diam}_{Y}(I)\leq{\sf T_{1}}(X)$.

######  Lemma 6.12.

If $I\subset J_{X}$ is disjoint from all inductive subintervals of $J_{X}$
then $I$ is straight for $X$.

###### Proof.

Fix an essential $Y\subset X$ with $\xi(Y)<\xi(X)$. It suffices to show, for
every subinterval $J\subset I$, that $d_{Y}(J)\leq{\sf T_{1}}(X)$.

If $J\cap I_{Y}=\emptyset$ then Theorem 5.3 implies $d_{Y}(J)\leq{\sf K_{0}}$.
Suppose that $J$ meets $I_{Y}$; thus $J_{Y}=\emptyset$ by hypothesis and so
$Y$ is not inductive. It follows that $d_{Y}(I)<{\sf T_{0}}(X)$. By Lemma 6.6
we have $d_{Y}(J)<{\sf T_{0}}(X)+2{\sf R_{0}}$. ∎

######  Lemma 6.13.

There is a constant $A=A(X)$, independent of $\\{\tau_{i}\\}$, so that if
$I\subset J_{X}$ is straight then $|\mathcal{S}_{X}(I)|\leq_{A}d_{X}(I)$.

###### Proof.

If $X$ is an annulus then, by Theorem 5.3, for every $r\in I$ the core curve
$\alpha\subset X$ is carried by and wide in $\tau_{r}$. It follows that the
number of switches in $\alpha\subset\tau_{r}|X$ is bounded by some constant
$K=K(S)$. Let $q=\max I$ and pick any $\beta\in V(\tau_{q}|X)$. As in the
proof of Theorem 5.5 let $\sigma_{r}\subset\tau_{r}|X$ be the minimal subtrack
carrying $\beta$. Thus $\sigma_{r}$ has either exactly four branches and two
switches, or is an embedded arc. It follows that every $K^{2}/4$ consecutive
splittings in $S_{X}(I)$ induces at least one splitting in the sequence of
tracks $\\{\sigma_{r}\\}$. Therefore the singleton sets $V(\sigma_{r})$ form a
quasi-geodesic in $\mathcal{A}(X)$. Since $V(\sigma_{r})\subset V(\tau_{r}|X)$
the proof is complete when $X$ is an annulus.

We assume for the rest of the proof that $X$ is not an annulus. The map
$i\mapsto V(\tau_{i}|X)$, taking tracks to their vertex cycles, is generally
not injective. (For example, see [15, Proposition 2.2.2].) However:

######  Claim 6.14.

There is a constant ${\sf N_{0}}={\sf N_{0}}(X)$, independent of
$\\{\tau_{i}\\}$, so that if $V(\tau_{r}|X)=V(\tau_{s}|X)$ then
$|\mathcal{S}_{X}(r,s)|\leq{\sf N_{0}}$.

###### Proof.

Let $\mu=V(\tau_{r}|X)$. Our hypothesis on $\tau_{s}|X$ and induction proves
that $V(\tau_{t}|X)=\mu$ for all $t\in[r,s]$. Recurrence and uniqueness of
carrying [14, Proposition 3.7.3] implies that $\tau_{t+1}|X$ is a split or a
slide of $\tau_{t}|X$, and not a subtrack, for all $t\in[r,s-1]$.

If $t\in[r,s]$ and $b\in\mathcal{B}(\tau_{t}|X)$ then define
$w_{\mu}(b)=\sum_{\alpha\in\mu}w_{\alpha}(b)$. Let

$M(t)=\big{(}w_{\mu}(b):\mbox{$b$ is a large branch of $\tau_{t}|X$}\big{)}$

be the sequence of given numbers, arranged in non-increasing order. Note that
if $\tau_{t+1}|X$ is a slide of $\tau_{t}|X$ then $M(t+1)=M(t)$. However, if
$t\in\mathcal{S}_{X}(r,s)$ then the recurrence of $\tau_{t}|X$ implies that
$M(t+1)<M(t)$, in lexicographic order. Since there are only finitely many
possibilities for an induced track $\tau|X$, up to the action of
$\mathcal{MCG}(X)$, the claim follows. ∎

Notice that if $V(\tau_{i+1}|X)\neq V(\tau_{i}|X)$ then $V(\tau_{i+1}|X)\neq
V(\tau_{j}|X)$ for $j\leq i$. This is because $P(\tau_{k+1}|X)\subset
P(\tau_{k}|X)$ for all $k$. Using $C=1+\max\\{{\sf C}(X),{\sf T_{1}}(X)\\}$ as
the cut-off in Theorem 6.7 gives some constant of quasi-equality, say ${\sf
E}$. Define ${\sf R_{1}}={\sf E}+1$.

Suppose that $[p,q]=I$, the straight subinterval of $J_{X}$ given by Lemma
6.13. We define a function $\rho\colon[0,M]\to I$ as follows: let $\rho(0)=p$
and let $\rho(n+1)$ be the smallest element in $[\rho(n),q]$ with
$d_{\mathcal{M}(X)}(\tau_{\rho(n)},\tau_{\rho(n+1)})={\sf R_{1}}$. (If
$\rho(n+1)$ is undefined then take $M=n+1$ and $\rho(M)=q$.) Let $B(\mu)$ be
the ball of radius ${\sf R_{1}}$ about the marking $\mu\in\mathcal{M}(X)$.
Define

${\sf V}=\max\big{\\{}|B(\mu)|:\mu\in\mathcal{M}(X)\big{\\}}.$

Deduce from Claim 6.14 and the remark immediately following that, for all
$n\in[0,M-1]$,

$\displaystyle|\mathcal{S}_{X}(\rho(n),\rho(n+1))|\leq{\sf N_{0}}{\sf V}.$
Thus $\displaystyle|\mathcal{S}_{X}(I)|\leq{\sf N_{0}}{\sf V}\cdot M.$

So to prove Lemma 6.13 it suffices to bound $M$ from above in terms of
$d_{X}(I)$.

######  Claim 6.15.

Fix $n\in[0,M-2]$. Let $\tau,\sigma=\tau_{\rho(n)},\tau_{\rho(n+1)}$. Then
$d_{X}(\tau,\sigma)\geq{\sf R_{0}}+1$.

###### Proof.

We use Theorem 6.7. Note that $d_{\mathcal{M}(X)}(\tau,\sigma)={\sf R_{1}}$.
Since ${\sf R_{1}}$ is greater than the additive error there is at least one
non-vanishing term in the sum $\sum_{Y\subset X}[d_{Y}(\tau,\sigma)]_{C}$.

However, since $[\rho(n),\rho(n+1)]\subset[p,q]$ and $[p,q]=I$ is straight we
have $d_{Y}(\tau,\sigma)\leq{\sf T_{1}}(X)$ for all $Y\subset X$ with
$\xi(Y)<\xi(X)$. Thus $d_{X}(\tau,\sigma)$ is the only term of the sum greater
than the cut-off $C$. Since $C>{\sf T_{1}}(X)\geq{\sf R_{0}}$, we have
$d_{X}(\tau,\sigma)\geq{\sf R_{0}}+1$ and the claim is proved. ∎

Thus we have

$\displaystyle d_{X}(I)$ $\displaystyle\geq-(M-1)\cdot{\sf
R_{0}}+\sum_{n=0}^{M-1}d_{X}(\tau_{\rho(n)},\tau_{\rho(n+1)})$
$\displaystyle\geq M-1+d_{X}(\tau_{\rho(M-1)},\tau_{\rho(M)})$
$\displaystyle\geq M-1$

where the first and second lines follow from Lemma 6.6 and Claim 6.15
respectively. This completes the proof of Lemma 6.13. ∎

######  Lemma 6.16.

There is a constant $A=A(X)$ with the following property: Suppose that
$J_{Y}\subset J_{X}$ is an inductive interval. Suppose that $I\subset J_{Y}$
is a straight subinterval for $X$. Then $|\mathcal{S}_{X}(I)|\leq A$.

###### Proof.

Let $[p,q]=I$. Applying Theorem 5.3, as $p\in J_{Y}\subset I_{Y}$, the
multicurve $\partial Y$ is wide with respect to $\tau_{p}$. It follows that
$\partial Y$ is also wide with respect to $\tau_{p}^{X}$. Note that the curves
of $V(\tau_{p}|X)$ are also wide with respect to $\tau_{p}^{X}$. Repeating
this discussion for $q$, and then applying Lemma 5.2 and the triangle
inequality gives a uniform bound for $d_{X}(\tau_{p},\tau_{q})$. The lemma now
follows from Lemma 6.13. ∎

###  6.5. Long and short intervals

######  Definition 6.17.

A straight subinterval $I$ for $X$ is short if $d_{X}(I)\leq 4{\sf R_{0}}$.
Otherwise $I$ is long.

By Lemma 6.13, if $I$ is a short straight interval then $|\mathcal{S}_{X}(I)|$
is uniformly bounded by a constant depending only on $X$.

Let ${\sf Ind}$ be the set of inductive subsurfaces $Y\subset X$. Define ${\sf
Ind}^{\prime}={\sf Ind}\cup\\{X\\}$. Note that every maximal subinterval of
$J_{X}{\smallsetminus}\cup_{Y\in{\sf Ind}}J_{Y}$ is straight, by Lemma 6.12.
We partition these maximal subintervals into the sets ${\sf Long}$ and ${\sf
Short}$ as the given interval is long or short respectively.

######  Lemma 6.18.

There is a constant $A=A(X)$, independent of $\\{\tau_{i}\\}$, so that

$\sum_{I\in{\sf Long}}|\mathcal{S}_{X}(I)|\leq_{A}d_{X}(J_{X}).$

###### Proof.

From Lemma 6.13 we deduce

$\sum_{I\in{\sf Long}}|\mathcal{S}_{X}(I)|\leq_{A}|{\sf Long}|+\sum_{I\in{\sf
Long}}d_{X}(I)$

where the first term on the right arises from addition of additive errors. By
the definition of a long straight interval and from Lemma 6.6 deduce

$4{\sf R_{0}}|{\sf Long}|\leq\sum_{I\in{\sf Long}}d_{X}(I)\leq
d_{X}(J_{X})+2{\sf R_{0}}|{\sf Long}|.$

Thus $2{\sf R_{0}}|{\sf Long}|\leq d_{X}(J_{X})$. These inequalities combine
to prove the lemma, for a somewhat larger value of $A=A(X)$. ∎

######  Lemma 6.19.

There is a constant $A=A(X)$, independent of $\\{\tau_{i}\\}$, so that

$\sum_{I\in{\sf Short}}|\mathcal{S}_{X}(I)|\leq_{A}|{\sf Ind}^{\prime}|.$

###### Proof.

By Lemma 6.13 the number of splittings in any short straight interval is a
priori bounded (depending only on $X$). Since $|{\sf Short}|\leq|{\sf
Ind}^{\prime}|$ the lemma follows. ∎

######  Lemma 6.20.

If $Z\in{\sf Ind}$ then

$\operatorname{card}\big{\\{}Y\in{\sf Ind}\mathbin{\mid}Z\subset
Y,\,\xi(Z)<\xi(Y)\big{\\}}\leq 2(\xi(X)-\xi(Z)-1).$

This follows from and is strictly weaker than Theorem 4.7 and Lemma 6.1 of
[12]. We give a proof, using our structure theorem, to extract the necessary
lower bound for ${\sf T_{0}}(X)$.

###### Proof of Lemma 6.20.

Suppose that $U\in{\sf Ind}$ contains $Z$. Suppose $J_{Z}=[p,q]$ and
$J_{X}=[m,n]$. Thus $\partial Z$ is wide with respect to $\tau_{p}$. So
$d_{U}(\tau_{p},\partial Z)\leq{\sf N_{1}}$, by the definition of ${\sf
N_{1}}$, and the same holds at the index $q$. Thus
$d_{U}(\tau_{p},\tau_{q})\leq 2{\sf N_{1}}$. The subsurface $U$ precedes or
succeeds $Z$ if $d_{U}(\tau_{m},\tau_{p})$ or $d_{U}(\tau_{q},\tau_{n})$,
respectively, is greater than or equal to $2{\sf N_{1}}+{\sf N_{2}}+{\sf
K_{0}}(X)+1$. Note that $U$ must precede or succeed $Z$ (or both) as otherwise
$d_{U}(\tau_{m},\tau_{n})<6{\sf N_{1}}+2{\sf N_{2}}+2{\sf K_{0}}(X)+2\leq{\sf
T_{0}}(X)$, a contradiction.

It now suffices to consider subsurfaces $U$ and $V$ that both succeed and both
contain $Z$. If $\max J_{U}\leq\max J_{V}$ then $U\subset V$. For, if not,
$\partial V$ cuts $U$ while missing $Z$. Since $\partial V$ is wide at the
index $r=\max J_{V}$ we deduce that

$\displaystyle d_{U}(\tau_{q},\tau_{n})\leq{}$ $\displaystyle
d_{U}(\tau_{q},\partial Z)+d_{U}(\partial Z,\partial V)+d_{U}(\partial
V,\tau_{r})+$
$\displaystyle+d_{U}(\tau_{r},\tau_{r+1})+d_{U}(\tau_{r+1},\tau_{n})$
$\displaystyle\leq{}$ $\displaystyle 2{\sf N_{1}}+{\sf N_{2}}+{\sf
K_{0}}(X)+1$

and this is a contradiction. Thus the surfaces in ${\sf Ind}$ that strictly
contain $Z$, and succeed $Z$, are nested. ∎

######  Definition 6.21.

Assign an index $r\in\mathcal{S}_{X}(J_{X})$ to a subsurface $Y\subset X$ if
$Y\in{\sf Ind}^{\prime}$, $r\in J_{Y}$, $\tau_{r+1}|Y$ is a splitting of
$\tau_{r}|Y$ and there is no subsurface $Z\subset Y,\xi(Z)<\xi(Y)$ with those
three properties.

######  Lemma 6.22.

There is a constant $A=A(X)$, independent of $\\{\tau_{i}\\}$, so that the
number of splittings contained in inductive intervals is quasi-bounded by
$|{\sf Ind}|+\sum_{Y\in{\sf Ind}}d_{Y}(J_{X})$.

###### Proof.

Fix $Y\in{\sf Ind}$. Consider an index $r\in J_{Y}$ that is assigned to $X$.
Let $I\subset J_{Y}$ be the maximal interval containing $r$ so that all
indices in $\mathcal{S}_{X}(I)$ are assigned to $X$. We now show that $I$ is
straight. Let $Z$ be any essential subsurface of $X$ with $\xi(Z)<\xi(X)$ and
let $[r,s]=J\subset I$ be any subinterval. If $J\cap I_{Z}=\emptyset$ then
Theorem 5.3 implies that $d_{Z}(J)\leq{\sf K_{0}}$. If $J$ meets $I_{Z}$ then,
as no splittings of $J$ are assigned to $Z$ we deduce that $\tau_{s}|Z$ is
obtained from $\tau_{r}|Z$ by sliding and taking subtracks only. Thus
$d_{Z}(J)\leq{\sf B_{0}}{\sf N_{2}}\leq{\sf T_{1}}(X)$, as desired.

By Lemma 6.16 we find that $|\mathcal{S}_{X}(I)|$ is bounded. It follows that
the number of splittings in the inductive intervals is quasi-bounded by
$\sum_{Y\in{\sf Ind}}|\mathcal{S}_{Y}(J_{Y})|$.

By induction, Proposition 6.9 gives

$|\mathcal{S}_{Y}(J_{Y})|\leq_{A}d_{\mathcal{M}(Y)}(J_{Y}).$

Taking a cutoff of $C=1+\max\\{{\sf C}(Y),{\sf T_{0}}(X)+2{\sf R_{0}}\\}$ and
applying the distance estimate Theorem 6.7 we have a quasi-inequality

$d_{\mathcal{M}(Y)}(J_{Y})\,\leq_{\sf E}\sum_{Z\subset Y}[d_{Z}(J_{Y})]_{C}.$

Since $d_{Z}(J_{Y})\leq d_{Z}(J_{X})+2{\sf R_{0}}$ for all $Z\subset Y$, it
follows that non-zero terms in the sum only arise for subsurfaces in ${\sf
Ind}^{\prime}(Y)=\\{Z\in{\sf Ind}^{\prime}\mathbin{\mid}Z\subset Y\\}$. Since
$2{\sf R_{0}}\leq{\sf T_{0}}(X)$ we have $[d_{Z}(J_{Y})]_{C}\leq 2\cdot
d_{Z}(J_{X})$. Making $A=A(X)$ larger if necessary we have

$\displaystyle|\mathcal{S}_{Y}(J_{Y})|$ $\displaystyle\leq_{A}\sum_{Z\in{\sf
Ind}^{\prime}(Y)}d_{Z}(J_{X}).$ Thus $\displaystyle\sum_{Y\in{\sf
Ind}}|\mathcal{S}_{Y}(J_{Y})|$ $\displaystyle\leq_{A}|{\sf
Ind}|+\sum_{Y\in{\sf Ind}}\sum_{Z\in{\sf Ind}^{\prime}(Y)}d_{Z}(J_{X})$
$\displaystyle\leq_{A}|{\sf Ind}|+\sum_{Y\in{\sf Ind}}d_{Y}(J_{X})$

where the final quasi-inequality follows from Lemma 6.20, taking $A$ larger as
necessary. Note that the term $|{\sf Ind}|$ on the middle line arises by
adding additive errors. This proves Lemma 6.22. ∎

Since every index in $\mathcal{S}_{X}(J_{X})$ is either in a long or short
straight interval or in an inductive interval, from Lemmas 6.18, 6.19, and
6.22 and increasing $A$ slightly, we have:

$|\mathcal{S}_{X}(J_{X})|\leq_{A}d_{X}(J_{X})+|{\sf
Ind}^{\prime}|+\sum_{Y\in{\sf Ind}}d_{Y}(J_{X}).$

Note that $|{\sf Ind}^{\prime}|\leq_{A}d_{\mathcal{M}(X)}(J_{X})$; this
follows from the hierarchy machine (in particular Lemma 6.2 and Theorem 6.10
of [12]) and because ${\sf T_{0}}(X)\geq M_{2}(X)$, the constant of Lemma 6.1
in [12]. Finally,

$\sum_{Y\in{\sf Ind}^{\prime}}d_{Y}(J_{X})\leq_{A}d_{\mathcal{M}(X)}(J_{X})$

follows from the distance estimate (Theorem 6.7) and because ${\sf
T_{0}}(X)\geq{\sf C}(X)$. This completes the proof of Proposition 6.9 and thus
the proof of Theorem 6.1. ∎

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