Geodesic Envelopes in the Thurston Metric
on Teichmüller space

Assaf Bar-Natan

Department of Mathematics, Brandeis University, Waltham, MA

The Thurston metric on Teichmüller space, first introduced by W.
P. Thurston is an asymmetric metric on Teichmüller space defined
by $d_{Th}(X,Y) = \frac12 \log\sup_{\alpha}
\frac{l_{\alpha}(Y)}{l_{\alpha}(X)}$. This metric is geodesic, but
geodesics are far from unique. In this thesis, we show that in the
once-punctured torus, and in the four-times punctured sphere,
geodesics stay a uniformly-bounded distance from each other.  In other
words, we show that the width of the geodesic
envelope, $E(X,Y)$ between any pair of points $X,Y \in \mc{T}(S)$
(where $S = S_{1,1}$ or $S = S_{0,4}$) is bounded uniformly.  To do
this, we first identify extremal geodesics in $Env(X,Y)$, and show
that these correspond to stretch vectors, proving
a conjecture from <cit.>. We then compute Fenchel-Nielsen
twisting along these paths, and use these computations, along with
estimates on earthquake path lengths, to prove the main theorem.

The unit tangent sphere is the convex hull of stretch
vectors (in red). Combinatorially, this is a chamfered cube.

§ INTRODUCTION

Let $S$ be a closed orientable surface with negative Euler
characteristic, and let $\mc{T}(S)$ be its Teichmüller space. In
this paper, we study the infinitesimal geometry of the Thurston
metric, $d_{Th}$ on $\mc{T}(S)$, answering a question of Huang,
Ohshika and Papadopoulos <cit.> in the affirmative. We then
focus on two low-complexity cases, where the infinitesimal results are
used to show that Thurston geodesics in these Teichmüller spaces
have uniformly bounded width.

In Thurston's paper on the Thurston metric <cit.>, he proves
that $(\mc{T}(S), d_{Th})$ is a geodesic metric space by constructing
a special family of paths called stretch paths. For any $X
\in \mc{T}(S)$, we define the stretch vectors at $X$, $SV_X$
to be the set of $1$-jets of stretch paths starting at $X$. For
each $X$ we can also look at the unit tangent sphere, $\bm{S}_X$
which strictly contains $SV_X$. We prove:

$\bm{S}_X$ is the convex hull of $SV_X$. Moreover, $SV_X$ is
precisely the set of extreme points in $\bm{S}_X$.

Geodesics in the Thurston metric are far from unique <cit.>. We define the geodesic envelope between $X$ and $Y$
to be the collection of points along geodesics between $X$ and $Y$,
and in general this envelope has unbounded width,
denoted by $w(X,Y)$ (ie,
two geodesics between $X$ and $Y$ can grow arbitrarily far
apart from each other).

In Section <ref>, we begin by computing
estimates for the Thurston distance
between a point $X$ and its earthquake
along a simple closed curve $\alpha$, in terms of
$l_{\alpha}(X)$. We prove:

Let $\alpha$ be a simple closed curve on $S$, and let
$X \in \mc{T}(S)$. Then there exists some uniform constant
$C$ such that

\begin{align*}
d_{Th}(X,Eq_{\alpha,t}(X)) \le \log(e^{l_{\alpha}/2}t) + C
\end{align*}

We then use this result, combined with the
computations in Section <ref>
to show that:

Let $S$ be the once-punctured torus or the four-times punctured
sphere. There exists some $B>0$ such that for any $X,Y \in
\mc{T}(S)$, $w(X,Y) < B$.

Organization. We begin by going through
some background on the Thurston metric on Teichmüller
space, laminations and shearing co-ordinates, and Fenchel-Nielsen
twisting co-ordinates. Section <ref>
is dedicated to proving Theorem <ref>,
and in Section <ref>, we
prove Theorem <ref> using some computations
from subsection <ref>.

Acknowledgements. This paper is based on the
author's PhD thesis. I would like to thank my advisor, Kasra Rafi, who
has given me endless support and motivation throughout the writing of
this work.

I would also like to thank my committee members: Alexander Nabutovsky
and Giulio Tiozzo. Many thanks to Jing Tao, Yair Minsky for helpful
discussion and to Francis Bonahon, who helped me with the proof of
Lemma <ref>, and to Yi Huang, who
helped me clarify the proof of Theorem
<ref>. Special thanks to David Dumas, for
reading and commenting on earlier drafts of this paper.

Finally, I would like to acknowledge support from the Univeristy of
Toronto mathematics department, the Faculty of Arts and Sciences, the
Natural Sciences and Engineering Research Council of Canada.

§ BACKGROUND

§.§ Teichmuller space & Thurston Metric

Throughout this paper, we let $S$ be an orientable surface with no
boundary components, and negative Euler characteristic.  In
particular, this means that $S$ can be endowed with a hyperbolic

A marking on $S$ is a homeomorphism
$f:S \to X$, where $X$ is a surface endowed with a hyperbolic metric.

Two markings $f:S\to X$ and $g:S \to Y$ are equivalent
if $fg^{-1}$ is homotopic to an isometry.

We define the Teichmüller space of
$S$, denoted by $\mc{T}(S)$, to be the space
of all markings on $S$ up to equivalence.

We think of points in Teichmüller space
as a pair consisting of a metric space $X$, together with a marking
map $\vphi$.

If $\gamma$ is a curve, arc, or arc segment on $S$, then we define
$l_{\gamma}:\mc{T}(S) \to \R_{\ge 0}$ by sending $X$ to
$l_{\gamma}(X)$, the length of the geodesic representative of
$\gamma$ on $X$ relative to its endpoints.

We define the Thurston metric $d_{Th}:\mc{T}(S)\times
\mc{T}(S)\to \R_{\ge0}$

\begin{align*}
d_{Th}(X,Y) = \sup_{\alpha} \log\left(\frac{l_{\alpha}(Y)}{l_{\alpha}(X)}\right)
\end{align*}

where the supremum ranges over all simple closed curves $\alpha$
contained in $S$.

For any $X,Y\in \mc{T}(S)$,
$d_{Th}(X,Y)$ is equal to:

* $d_{Th}(X,Y) = \sup_{\alpha} \log\left(\frac{l_{\alpha}(Y)}{l_{\alpha}(X)}\right)$ (as above)

* $L(X,Y) = \inf_{f \sim id} \log(L_f)$, where $f:X\to Y$ is
a Lipschitz map homotopic to the identity, and $L_f$ is the
Lipschitz constant of $f$.

* $D(X,Y) = \inf_{f \sim id} \sup_{p\in X} \log(\|D
f_p\|)$, where $f:X\to Y$ is a homeomorphism that
is once differentiable almost everywhere.

In <cit.>, Thurston shows that
$L(X,Y) = d_{Th}(X,Y)$. It suffices to show that
$d_{Th}(X,Y) \le D(X,Y) \le L(X,Y)$. The first
inequality follows because $\frac{l_{\alpha}(Y)}{l_{\alpha}(X)}$
is always bounded above by $\sup_p \|Df_p\|$. The
latter inequality is more subtle. Trivially,
if in the definition of $L(X,Y)$,
$f$ is taken to be differentiable, then the inequality
follows by the fact that Lipschitz constants
give upper bounds for derivatives. In fact,
Thurston <cit.> explicitly
constructs a map $f$ that realizes
the infimum in $L(X,Y)$, and this map is a
homeomorphism that is differentiable almost
everywhere. Thus, the inequality holds.

For any $X,Y,Z \in \mc{T}(S)$, the Thurston metric satisfies:

* $d_{Th}(X,Y) \ge 0$

* $d_{Th}(X,Y) = 0$ if and only if $X = Y$

* $d_{Th}(X,Z) \le d_{Th}(X,Y) + d_{Th}(Y,Z)$

We use the third characterization of $d_{Th}$ in
Theorem <ref>.

* If $d_{Th}(X,Y) < 0$, then that would imply the
existence of a map $f:X\to Y$ such that $\|Df_p\| < 1$
for all $p$. In particular, this means that
$f$ is not area-preserving, which cannot happen,
as $\mathrm{Area}(X) = \mathrm{Area}(Y) = -2\pi \chi(S)$.

* If $d_{Th}(X,Y) = 0$, then there would be
a homeomorphism $f:X\to Y$ homotopic to the identity
with $\|Df_p\| = 1$ for all $p$. In particular,
$f$ must be an isometry, and $X$ and $Y$ are equivalent.

* This follows immediately from the chain rule.

The functions $d_{Th} = L = D$ define an
asymmetric complete geodesic metric on $\mc{T}(S)$ <cit.>. By
this, we mean that:

For any $X,Y,Z\in \mc{T}(S)$,

* $d_{Th}(X,Y) \ge 0$ for all $X,Y$,
with equality if and only if $X=Y$

* $d_{Th}(X,Y) \le d_{Th}(X,Z) + d_{Th}(Z,Y)$

* There exists a path $\gamma:[0,d_{Th}(X,Y)]\to \mc{T}(S)$
such that $\gamma(0)=X$, $\gamma(d_{Th}(X,Y)) = Y$, and
for any $s\le t$, $d_{Th}(\gamma(s),\gamma(t)) = t-s$.

For any simple closed curve $c$, and any $v\in T_X\mc{T}(S)$,
we define $D_v\log l_c = \frac{d}{dt}|_{t=0} \log(l_c(\alpha(t)))$,
where $\alpha(t)$ is some germ whose $1$-jet is equal to $v$. This
family of linear functionals induces the Thurston norm on

\begin{align*}
\|v\|_{Th} = \sup_{c} d\log_c(v)
\end{align*}

The Thurston metric is induced by this norm on the tangent bundle
<cit.>, and hence is a Finsler metric.

We define the unit norm sphere at $X\in \mc{T}(S)$ by:

\begin{align*}
\bm{S}_X = \{v \in T_X\mc{T}(S) : \|v\|_{Th} = 1\}
\end{align*}

Since the Thurston metric is induced by the
Thurston norm, we think of $\bm{S}_X$ as the set of tangent vectors
which arise from $1$-jets of $C^1$ geodesics starting at $X$.
By claim 1.12 and Theorem 1.11 of <cit.>,
$\bm{S}_X$ consists of all unit tangent vectors at $X$.

More details on $d_v\log_{\alpha}$ and the characterization of
the unit sphere can be found in subection <ref>
and in <cit.>.

Throughout this paper, we will work with two different co-ordinate
systems for Teichmüller space: Fenchel-Nielsen co-ordinates,
and shearing co-ordinates. We review them in this
section, following <cit.> for Fenchel-Nielsen co-ordinates, and
following <cit.> for shearing co-ordinates. For a more
generalized overview of shearing co-ordinates in the case
of a filling lamination that is not an ideal triangulation,
we refer the reader to <cit.>.

§.§ Shearing Co-ordinates & Laminations

Throughout this section, let $(X,\vphi)$
be a point in $\mc{T}(S)$. Let $\tilde X$
be the universal cover of $X$, which we will identify
with $\HH^2$. A geodesic lamination $\lambda$ on
$X$ is a closed subset of $X$ which can we decomposed
as a disjoint union of (possibly bi-infinite)
geodesics. If $(Y,\psi)$ is a different marking on $S$,
then $\psi\vphi^{-1}(\lambda)$ gives a closed collection
of disjoint arcs on $Y$, which we can turn into
geodesics by an ambient isotopy, and hence, we can think of
$\lambda$ as a lamination on
$Y$. In this manner, $\lambda$ can be thought of as a
lamination on the underlying surface $S$ without
specifying a metric.

A geodesic lamination is called complete
if its complementary components are triangles. A lamination
$\lambda$ can be lifted to a lamination $\tilde \lambda$
on $\tilde X$, so $\lambda$ is complete if and only if
$\tilde \lambda$ is a triangulation of $\HH^2$. A
lamination $\lambda$ is called chain-recurrent
if there exists a sequence of multicurves which
converge to $\lambda$ in the Hausdorff topology. An
alternate characterization of chain-recurrence is
the following:

A geodesic lamination $\lambda$ is called chain-recurrent
if for any arc segment $I$ contained in $\lambda$, and for
any $\eps > 0$, there exists a geodesic simple closed curve $\alpha$
in $S$ and an arc segment $J \subset \alpha$
such that $d_H(I,J) < \eps$, where $d_H$ is the Hausdorff

In this sense, the space of chain-recurrent laminations
is the closure of the space of multicurves,
equipped with Hausdorff convergence <cit.>.

A measured lamination is a lamination
$\lambda$ together with a measure, $\mu_{\lambda}$, defined on all
arc segments intersecting $\lambda$ transversely
whose endpoints lie on $\lambda$. We also require
$\mu$ to be invariant under homotopy that moves
the endpoints of arc segments along leaves of $\lambda$.

As an example, consider a weighted multicurve –
a disjoint union of simple closed curves $\gamma_n$ with
nonzero weights $a_n$. For any arc segment $I$,
$\mu(I) = \sum_n a_n i(\gamma_n,I)$.

The space of projective measured laminations, $\PML(S)$,
is the space of all measured laminations up to
scaling the measure by a positive real number.

We topologize $\PML(S)$ in the following manner: We
say that $\lambda_n$ converge to $\lambda$
if for every arc segment $I$ in $S$,
$\mu_{\lambda_n}(I)$ converges to $\mu_{\lambda}(I)$.

The following is useful for working with
$\PML(S)$, and appears in Chapter 1 of <cit.>:

$\PML(S)$ is compact, and moreover, is the compactification of
the space of weighted multicurves with total weight $1$.

We next explicitly describe shearing co-ordinates for
a class of laminations that will be of interest
later in this thesis.

Let $\lambda = \cup_{i=1}^{9g-9} \lambda_i$ be a complete geodesic
lamination consisting of $3g-3$ leaves which are simple closed curves
and $6g-6$ bi-infinite leaves.  Denote by $\mc{C}$ the closed leaves
of $\lambda$.  The leaves of $\lambda$ give an ideal triangulation of
$S$, and $\mc{C}$ form a pair-of-pants decomposition of $S$.

We wish to define a family of functions
$S_{\lambda_i}: \mc{T}(S) \to \R$, which will give a
co-ordinate system on $\mc{T}(S)$. To do this,
we first define shearing between

Let $\Delta_1$ and $\Delta_2$ be two triangles in $S\ssm \lambda$,
and let $\tilde\Delta_1$ and $\tilde\Delta_2$ be lifts of $\Delta_1$
and $\Delta_2$ to the universal cover $\tilde X = \HH^2$. Choose
some geodesic $\gamma$ separating the interiors of $\tilde \Delta_1$
and $\tilde \Delta_2$, such that $\gamma$ is the geodesic
between a vertex $v_1$ of $\tilde \Delta_1$ and to a vertex
$v_2$ of $\tilde \Delta_2$ (if $\Delta_1$ and
$\Delta_2$ are adjacent triangles on $S$, choose adjacent lifts, and
let $\gamma$ be their intersection).
We orient $\gamma$ so that the interior of $\Delta_1$ lies to
the left of $\gamma$.

An ideal triangle in $\HH^2$ has a well-defined
incenter, incircle, and three distinguished medians
along its edges, given by the intersection of
the incircle with the edges. For each triangle
$\tilde \Delta_i$, we define $m^i$ as the median of $\tilde \Delta_i$
lying on the edge separating $\gamma$ from
$int(\tilde \Delta_i)$.

Let $\vphi_i$ be the orientation-preserving
parabolic isometry of $\HH^2$ fixing $v_i$
and sending $\gamma$ to the edge of $\tilde \Delta_i$ containing
$m^i$. We set $q_i = \vphi^{-1}(m^i)$,
and define $s(\Delta_1,\Delta_2)$
to be the signed distance between $q_1$ and
$q_2$, where the sign is inherited from the
orientation of $\gamma$.

In <cit.>, it is proven that $s(\Delta_1,\Delta_2)$
is symmetric and does not depend on the geodesic $\gamma$. However,
$s(\Delta_1,\Delta_2)$ as defined above
may depend on the choice of lifts of $\Delta_1$
and $\Delta_2$, unless $\Delta_1$ and
$\Delta_2$ are adjacent.

[scale = 3]
[thick,<->] (-1.5,0) – (1.5,0);
[thick,->] (0,0) – (0,2.5);
[thick] (0.5,0) – (0.5,2.5);
[thick] (-1,0) – (-1,2.5);
[thick] (0,0) arc (0:180:0.5);
[thick] (0.5,0) arc (0:180:0.25);
[thick, dashed] (0.25,0.5) circle (0.25);
[thick, dashed] (-0.5,1) circle (0.5);

at (-0.5, 2) $\tilde\Delta_1$;
at (0.25, 2) $\tilde\Delta_2$;
at (0.1,2.5) $\tilde \gamma$;

[thick,|-|] (0,1) – (0,0.5);
[thick,->] (-0.5, 1) – (0,0.75);
[fill=white] at (-0.5,1) $s(\Delta_1,\Delta_2)$;

The set-up for computing shearing co-ordinates
between two adjacent triangles.

We are now ready to define $s_{\lambda_i}(X) = s(\Delta_1,\Delta_2)$,
where $\Delta_i$ are chosen as follows:

* If $\lambda_i$ is an infinite leaf of $\lambda$,
take $\Delta_1$ and $\Delta_2$ are the
(possibly coinciding) triangles adjacent to $\lambda_i$

* If $\lambda_i\in\mc{C}$,
we pick $\Delta_1$ and $\Delta_2$ to be
two triangles which asymptote to $\lambda_i$
from both sides of $\lambda_i$.

Let $c\in\mc{C}$, and let
$\lambda_c\subset \lambda$ be the sublamination of
$\lambda$ consisting of leaves converging to $c$.

Then $s_{c}(X)$ is well-defined
up to $\Z[\{s_{\lambda_c}\}]$.

This follows from Lemma 3.1 of <cit.>.

Choosing prescribed
lifts for every triangle in $S\ssm\lambda$,
we get a well-defined shearing co-ordinates map
$\mc{T}(S_g) \to \R^{9g-9}$ given by:

\begin{align*}
X \to \{s_l\}_{l\subset \lambda \textrm{ is a leaf}}
\end{align*}

The shearing co-ordinate map does not give an isomorphism,
as for any $\lambda$ and $X$, the collection
$\{s_{\lambda_i}(X)\}$ is linearly dependent (see
Lemma 3.2 in <cit.>). Choosing a
linearly independent subset of these gives a
homeomorphism from $\mc{T}(S_g)$ to $\R^{6g-6}$

§.§ Twisting Co-ordinates

Let $\mc{C}$ be a pair-of-pants decomposition
of a surface $S$. In the same way as in our
definition for the shearing co-ordinates,
it will be convenient to choose prescribed
lifts of every $c\in \mc{C}$ in $\tilde X$. For any
closed leaf $c\in \mc{C}$, let $P_1(c)$ and $P_2(c)$
be the (possibly same) pairs of pants
adjacent to $c$. Let
$\tilde c$ be a lift of $c$, and choose
lifts, $\tilde P_i(c)$ of $P_i(c)$ which are adjacent
to $\tilde c$ on either side of $\tilde c$.

Let $c\neq c_1$ and $c \neq c_2$ be cuffs of $P_1(c)$ and $P_2(c)$
respectively, and let $\tilde c_i$ be the lifts of $c_i$
bounding $\tilde P_i(c)$.

Let $\eta_i$ be the unique simple geodesic
segment intersecting $\tilde c_i$ with start point on $\tilde c$ and
end point on $\tilde c_i$, which is normal
to both $\tilde c$ and $\tilde c_i$. Let $p_i(c)$
denote the start-point of $\eta_i$.

We define the twist co-ordinate relative to $c$,
$\tau_c(X)$ to be the signed distance between $p_1(c)$ and $p_2(c)$,
so that the sign is positive if we turn left
to get from $p_1$ to $p_2$. Choosing different
lifts of $c$ will result in twist co-ordinates
that differ by an integer multiple of $l_c(X)$.

Additionally, it can be shown that choosing
different curves $c_i$ in the pair-of pants
will change $p_i$ by
a Gaussian-integer multiple of $l_c$, and
hence, $\tau_c(X)$ is well defined up to
an integer multiple of $l_c(X)$.

The map $FN:\mc{T}(S) \to \R^{6g-6}$
given by $X \to \{l_{c}(X),\tau_{c}(X)\}_{c\in\mc{C}}$
is a homeomorphism <cit.>, so we call
this map the Fenchel-Nielsen co-ordinate system
on $\mc{T}(S)$.

§.§ Stretch Laminations, Chain-Recurrence, and Thurston

Let $I$ be a possibly infinite closed interval.
A forward (resp. backwards)
geodesic is a map $\gamma:I\to M$
satisfying $d_{Th}(\gamma(s),\gamma(t)) = t-s$
(resp. $d_{Th}(\gamma(t),\gamma(s)) = t-s$)
for all $t\ge s$.

If $I=[a,b]$ is finite, we say that $\gamma$
starts at $\gamma(a)$ and ends at $\gamma(b)$.

Throughout this thesis, a “geodesic” means a
forward geodesic, unless otherwise stated.

Given $X,Y\in \mc{T}(S)$, we define

\begin{align*}
\Outenv(X) &=
\{Z\in \mc{T}(S) : Z \textrm{ lies on a forward geodesic
starting at } X\}\\
\In(Y) &=
\{Z\in \mc{T}(S) : Z \textrm{ lies on a backwards geodesic
starting at } Y\}
\end{align*}

For $X,Y\in \mc{T}(S)$, we define the
geodesic envelope, $\Env(X,Y) = \Outenv(X)
\cap \In(Y)$.

Given $X,Y\in\mc{T}(S)$, we can consider a sequence of multicurves
$\alpha_i$ such that $d_{Th}(X,Y) = \lim_{i\to\infty}
\frac{l_{\alpha_i}(Y)}{l_{\alpha_i}(X)}$. The space of geodesic
laminations equipped with the Hausdorff topology is compact
<cit.>, so up to subsequence, $\alpha_i$ converges to some
geodesic lamination $\lambda$. It turns out <cit.> that the
union of all Hausdorff limits of subsequences of $\{\alpha_i\}_n$ is
itself a geodesic lamination, call it $\lambda_{\{\alpha_i\}}$.
Moreover, if $\alpha'_i$ is another sequence whose length ratio
converges to $d_{Th}(X,Y)$ then any Hausdorff limit of $\{\alpha'_i\}$
is disjoint from $\lambda_{\{\alpha_i\}}$.

Thus, it makes sense to define the maximally stretched
lamination, $\Lambda(X,Y)$ as the union of all Hausdorff limits of
sequences of multicurves whose length ratio converges to

Given $X_0\in \mc{T}(S)$, and any completion of a
maximal chain-recurrent lamination $\lambda$,
there exists an analytic 1-parameter
family of metrics $X_t = \textrm{Stretch}(X_0,\lambda,t)
\subset \mc{T}(S)$ with the following properties:

* $l_{\lambda}(X_t) = e^tl_{\lambda}(X_0)$

* For $0\le s\le t$, $d_{Th}(X_s,X_t) = t-s$

* $\Lambda(X_s,X_t) = \lambda$

This family of metrics on $S$ is called the Thurston Stretch
Path associated to $X_0$ and $\lambda$. In particular, when
$\Lambda(X,Y)$ is maximal amongst all chain-recurrent laminations,
there is a unique geodesic from $X$ to $Y$, and the points along it
are precisely the points in $\Env(X,Y)$ <cit.>.

Let $X_t$ be some smooth $1$-parameter family of
surfaces in $\mc{T}(S)$.

The maps $s_{\lambda}:X_t \to \R^{9g-9}$ and
$FN:X_t \to \R^{6g-6}$ are only well-defined
up to the choices of lifts in their construction. However,
the maps $\dot s_{\lambda}:\mc{T}(S) \to \R^{9g-9}$
and $\dot{FN}:X_t \to \R^{6g-6}$ defined by
postcomposition of $s_{\lambda}$ and $FN$
by differentiation with respect to $t$ are well

Let $\lambda$ be some chain-recurrent lamination,
and let $X,Y\in \mc{T}(S)$. We

\begin{align*}
\Outenv(X,\lambda) &=
\{Z\in \Outenv(X) : \lambda \subset \Lambda(X,Z)\}\\
\In(Y, \lambda) &=
\{Z\in \In(Y) : \lambda \subset \Lambda(Y,Z)\}
\end{align*}

By this definition, $\Env(X,Y) = \Outenv(X,\Lambda(X,Y))
\cap \In(Y,\Lambda(X,Y))$. The geodesic envelope can be
thought of as a $1$-parameter family of cross-sections, where at any
time $t>0$, $\Env_t(X,Y)$ consists of all points of distance
$t$ from $X$ lying along geodesics from $X$ to $Y$.

Let $X,Y\in \mc{T}(S)$. We define the width
of the envelope $\Env(X,Y)$ as:

\begin{align*}
w(X,Y) &= \sup_{g_1,g_2 \mc{G}(X,Y)} \sup_{t\in [0,d_{Th}(X,Y)]}
d_{Th}(g_1(t), g_2(t))\\
&= \sup_t \textrm{Diam}(\textrm{Env}_t(X,Y))
\end{align*}

§ THE INFINITESIMAL ENVELOPE

The goal of this section is to understand the infinitesimal
structure of geodesic envelopes in the Thurston metric.

We begin the section by proving Theorem <ref>,
which will be one of the main ingredients in the arguments employed in
Section <ref> to prove Theorem <ref>

We continue by examining $\Env_0(X,Y)$
when $\Lambda(X,Y)$ contains a pair-of-pants decomposition of $S$. In
this case, the envelope width is determined by twist parameters along
geodesics. This will then allow us to estimate the maximal and minimal
twisting within $\Env(X,Y)$, which we will use in
Section <ref>.

§.§ Stretch Vectors in the Envelope

In this section, we study the set of tangent vectors in $T_X\mc{T}(S)$
which arise from $1$-jets of $C^1$ geodesics starting at some point
$X\in \mc{T}(S)$.  This set, called the unit norm sphere, and
denoted by $\bm{S}_X$ was shown in <cit.> to have
a combinatorial structure of a convex body with a convex
stratification whose faces come from chain-recurrent laminations
in $S$. If $v\in \bm{S}_X$ is
a $1$-jet of a stretch path $\mathrm{Stretch}(X,\lambda,t)$, we call
it a stretch vector with respect to $\lambda$, and
denote it by $v = v_{\lambda}(X)$.

The following lemma appears in various parts of <cit.>
and <cit.>. We adapt it to our language:

For any $X\in \mc{T}(S)$, $\bm{S}_X$ is a topological sphere around
$0 \in T_X\mc{T}(S)$.

Moreover, $\bm{S}_X$ is an infinite union
of convex sets, called faces, which glue together
in a combinatorial way.
In <cit.>, it is shown that there is a one-to-one
correspondence with topological chain-recurrent laminations
and faces of $\bm{S}_X$. We review their definitions
and theorems later in this section.

We further analyze the structure of $\bm{S}_X$, and
prove Theorem <ref>,
answering Conjecture 1.12 of <cit.> in the affirmative:

The set of stretch vectors in $\bm{S}_X$ with respect to completions
of maximal chain-recurrent laminations is precisely the set of
extreme points in $\bm{S}_X$

This result not only allows us to characterize
faces in $\bm{S}_X$ using stretch paths, but
also allows us to find extreme points in
the infinitesimal envelope from $X$
to $Y$. These extreme points can later be used to
give upper and lower bounds on the twisting width
between any two geodesics from $X$ to $Y$, not
just stretch lines starting at $X$, as
computed in Section <ref>.

Throughout this section, if $\Lambda$ is some lamination, we fix the
following notation:

* We denote by $\Lambda^{CR}$ the largest sublamination of
$\Lambda$ which is chain-recurrent (see Section <ref>).

* If $\Lambda$ is chain-recurrent, we denote by
$CR(\Lambda)$ to be the set of chain-recurrent laminations
containing $\Lambda$ that are maximal with respect to inclusion.
Abusing notation, we write $CR(\emptyset)$ to denote
the set of all chain-recurrent laminations that are maximal
with respect to inclusion.

* If $\Lambda$ is chain-recurrent or is empty, we define:

\begin{align*}
=\{\mu\in CR(\Lambda) : \forall \nu \in CR(\Lambda),
\mu \subset \nu \Rightarrow \mu = \nu\}
\end{align*}

* If $\Lambda$ is chain-recurrent or empty, we denote by
$M(\Lambda)$ to be the set of completions of laminations in

* If $\Lambda$ is chain-recurrent or empty, and
$X\in \mc{T}(S)$, we define
$SV_X(\Lambda) = \{v_{\mu}(X):\mu \in M(\Lambda)\}$.  We abuse
notation and write $SV_X = SV_X(\emptyset) = \{v_{\mu}(X): \mu\in

Using this notation, Theorem <ref>
says that the set of extreme points of $\bm{S}_X$
is precisely $SV_X(\emptyset)$. We prove some lemmas and state some
facts about the above sets and spaces.

If $\Lambda$ is chain-recurrent or empty, $CR(\Lambda)$
is compact in the Hausdorff topology.

The space of all geodesic laminations on a surface is
compact <cit.>, so it suffices to show that
the set of chain-recurrent laminations containing
$\Lambda$ is closed. Note that containment of $\Lambda$ is a closed
condition, since laminations are closed subsets of $S$. Let
$\Lambda_n$ be a converging sequence of chain-recurrent laminations.
If $\alpha_n^k$ is a sequence of simple closed multicurves
converging in the Hausdorff topology to $\Lambda_n$, then
$\alpha_n^n$ is a sequence of simple closed curves converging to

The following lemma follows from the proof of Theorem 8.5 in
<cit.>, and also appears as Corollary 2.3 in <cit.>
and in the discussion following Remark 2.9 in <cit.>.

If $\Lambda$ is chain-recurrent or empty,
and if $\lambda\in MCR(\Lambda)$, then
for any two completions $\lambda_1$ and
$\lambda_2$ of $\lambda$, we have that
for any $X\in \mc{T}(S)$, and $t \ge 0$,
and in particular,
$v_{\lambda_1}(X) = v_{\lambda_2}(X)$

§.§ The case for $\Env_0(X,Y)$

Let $X,Y \in \mc{T}(S)$ be two points in Teichmuller space, and let
$\mc{G}(X,Y)$ denote the set of all  geodesics parametrized by arc
length from $X$ to $Y$. By definition, $\cup \mc{G}(X,Y) = \Env(X,Y)$.
If $g\in \mc{G}(X,Y)$, we define the $1$-jet of $g$ by
$v_g(X) = \frac{d}{dt}|_{t=0} g(t)$. Note that $v_g(X)\in T_X
\mc{T}(S)$ is a unit tangent vector.

The infinitesimal envelope, $\Env_0(X,Y)\subset
T_X\mc{T}(S)$ is defined by:

\begin{align*}
\Env_0(X,Y) = \{v_g(X) : g\in \mc{G}(X,Y)\}
\end{align*}

For any $X,Y\in \mc{T}(S)$, we denote by $\Lambda(X,Y)$ the maximally
stretched lamination between $X$ and $Y$. In this subsection, we prove
the following:

For any $X,Y\in \mc{T}(S)$, $\Env_0(X,Y)$ is the convex hull of:
$SV_X(\Lambda(X,Y))$. Moreover,
$SV_X(\Lambda(X,Y))$ is precisely the set of extremal vectors
in $\Env_0(X,Y)$.

As an example for Theorem <ref>,
consider the genus $2$ surface, $S_2$, and let
$\lambda = \alpha \cup \beta \cup \gamma$ be a pair-of-pants
decomposition consiting of non-separating curves.
Let $X \in \mc{T}(S_2)$ be arbitrary.
Working in Fenchel-Nielsen co-ordinates with respect
to $\alpha,\beta,\gamma$, $\Outenv(X,\lambda)$
can be thought of as a $3$-dimensional cone lying
in $\R^{6}$. In particular, every $Z$ in $\Outenv(X,\lambda)$
of distance $t$ from $X$ has the same length co-ordinates,
and the only interesting co-ordinates are the twists.

If $Y\in \Outenv(X,\lambda)$, we
can consider the $3$-dimensional projection of
$\Env_0(X,Y)$ to the tangent subspace of $T_X\mc{T}(S_2)$
spanned by the directions corresponding to
twist co-ordinates. Theorem <ref>
then says that the extremal vectors in
this projection must be stretch vectors with
respect to laminations in $M(\lambda)$.

Using the formulas developed in subsection <ref>,
we can plot all $1$-jets of stretch paths
emenating from $X$ and maximally-stretching $\lambda$. This
is Figure <ref>.

The projection of all stretch vectors emenating from
$X$ and maximally stretching $\lambda$. Combinatorially,
this is a chamfered cube.

The red dots in the above picture are stretch vectors
corresponding to the $32$ chain-recurrent completions
of $\lambda$. The rest of the points are
stretch paths corresponding to non chain-recurrent
laminations. The eight red vertices at the “corners” of
the projected infinitesimal envelope correspond
to the completions $\Lambda$ of $\lambda$ that have the property
that for any pair of curves in $\lambda$, there exists a leaf
$l \subset \Lambda$ asymptotic to both. There are
precisely $8$ such laminations,
corresponding to the $2^3$ possible directions
in which leaves can asymptotically twist around
$\alpha,\beta$, and $\gamma$.
that have leaves

We will split the proof of Theorem <ref> up into three
main lemmas:

For any $X,Y\in \mc{T}(S)$, if
$\Lambda(X,Y)$ has finitely-many completions,
then $Env_0(X,Y)$ is the convex hull of

We will use this lemma to show the general case:

For any $X,Y\in \mc{T}(S)$, $Env_0(X,Y)$ is the convex hull of:

Finally, we show:

Let $\lambda \in M(\Lambda(X,Y))$,
and let $v_{\lambda} \in SV_X(\Lambda(X,Y))$ be
a convex combination:
$v_{\lambda} = \sum_i a_i v_{\lambda_i}$, where
$\lambda_i \in M(\Lambda(X,Y))$. Then
$v_{\lambda_i} = v$ for all $i$.

Before we prove the lemmas, we will prove a helpful technical lemma
about oriented foliations and their $1$-jets. When we say
oriented foliation, we mean a foliation
on some manufold $M$ such that every leaf carries with it an
orientation, and that these orientations vary continuously along any
path transverse to the foliation.

Let $\mc{F}_1,\mc{F}_2,\hdots$ be smooth oriented foliations defined
on some open domain with smooth (possibly empty) boundary $0\in
U \subset \R^k$. For each oriented foliation, we can construct a
vector field of unit-length vectors tangent to the foliation. Such
a vecor field can be constructed by taking the associated line field
of the foliation and assigning a direction at every point using the
orientation of the foliation. We will refer to the flow along the
vector field associated to the foliation by the `flow along the

Let $U^{\eps} = \{y\in U: d(y,\del y) >\eps\}$
and let $f_i:[0,\eps)\times U^{\eps} \to \R^k$
be defined by setting $f_i(t,x)$ to be the flow for time $t$ along
$\mc{F}_i$ starting from $x$. Taking $\eps$ sufficiently small,
we can always ensure that $0 \in U^{\eps}$, and that
all of the flows $f_i$ are defined at $x=0$.

We prove:

Let $\alpha:[0,T) \to U$ be a path differentiable
at $0$, and such that
$\alpha(0) = 0$. Assume that for every
$t< T$, there exist
$t_1,\hdots,t_n\ge 0$ and $i_1,\hdots,i_n$ such that
$t = \sum_i t_i$ and $\alpha(t)
= f_{i_1}(t_1,f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0)))))$.
Furthermore, assume that the non-negative span,
$\Span_{\ge 0}(\{f_i\}_i)$ is
a closed subset of $\R^N$. Then $\alpha'(0) \in \Span_{\ge
0}(\{\mc{F}_i'(0)\}_{i}) = \Span_{\ge 0}(\{\frac{\del}{\del t}f_i

The picture of the set-up in <ref>

In order to estimate $\alpha'(0)$, we first note that
as $t$ gets really small, so must the $t_i$'s. Thus,
for small $t$, we can take a first-order expansion of
$\alpha(t) = f_{i_1}(t_1,f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0)))))$ when $t_1$ is small to get:

\begin{align*}
\alpha(t) &= f_{i_1}(0,f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0))))) \\
&\qquad   + t_1 \frac{\del f_{i_1}}{\del t}(0, f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0)))))
+ o(t_1^2)\\
&= f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0))))
+ t_1 \frac{\del f_{i_1}}{\del t}(0, f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0)))))
+ o(t_1^2)
\end{align*}

Where we understand $\frac{\del f_{i_1}}{\del t}$ to be the derivative
of $f_{i_1}$ with respect to the first co-ordinate.

Expanding the $f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0))))$ in terms
of $t_2$, we get:

\begin{align} \label{eq:alpha_est1}
\alpha(t) &=
+ t_2 \frac{\del f_{i_2}}{\del t}(0, f_{i_3}(t_3,(\hdots,(f_{i_n}(t_n,0))))) \\
&\quad + t_1 \frac{\del f_{i_1}}{\del t}(0,
f_{i_2}(t_2,(\hdots,(f_{i_n}(t_n,0))))) + o(t^2)
\end{align}

If we continue to expand in this manner, we

\begin{align} \label{eq:alpha_est2}
\alpha(t) &= f_{i_n}(t_n,0)
+\sum_{j=1}^{n-1} t_j \frac{\del f_{i_j}}{\del t}
(0,f_{i_{j+1}}(t_{j+1},(\hdots,(f_{i_n}(t_n,0))))) + o(t^2)\\
&= t_n \frac{\del f_{i_n}}{\del t}(0,0)
+\sum_{j=1}^{n-1} t_j \frac{\del f_{i_j}}{\del t} (0,f_{i_{j+1}}(t_{j+1},(\hdots,(f_{i_n}(t_n,0))))) + o(t^2)\\
\end{align}

We denote $v_j = \frac{\del}{\del t} f_j (0,0)$,
and notice that $f_j(t_j,0) = t_j v_j + o(t^2)$.

\begin{align*}
f_{i_{n-1}}(t_{n-1},f_{i_n}(t_n,0)) &= f_{i_{n-1}}(t_{n-1}, t_n
v_{i_n} + o(t^2))\\
&= f_{i_{n-1}}(t_{n-1},0) + D_0 f_{i_{n-1}}(t_{n-1},0) \cdot (t_n
+ o(t^2))\\
&= t_{i_{n-1}}v_{n-1} + t_n D_0 f_{i_{n-1}}(t_{n-1},0)\cdot
v_{i_n} + o(t^2)\\
&= t_{i_{n-1}}v_{n-1} + t_n D_0 (t_{n-1} v_{i_{n-1}} + o(t^2))
\cdot v_{i_n}
+ o(t^2)\\
&= t_{n-1}v_{i_{n-1}} + t_n t_{n-1} v_{i_{n-1}} + o(t^2) \\
&= t_{n-1}v_{i_{n-1}} + o(t^2)
\end{align*}

Where we used the fact that all functions are smooth and
hence have bounded derivatives in a neighbourhood
of $0$. Note that we can continue this computation
to replace the composition terms in
<ref> with simpler, linear terms:

\begin{align} \label{eq:alpha_est3}
\alpha'(0) &= t_n v_{i_n} +\sum_{j=1}^{n-1}
t_j \frac{\del f_{i_j}}{\del t}
(0,t_{j+1} v_{i_{j+1}} + o(t^2)) + o(t^2)\\
&= t_n v_n + \sum_{j=1}^{n-1} t_j \left(v_{i_j} + D_0 f_{i_j}(0,0) \cdot
(t_{j+1} v_{i_{j+1}} + o(t^2))\right) + o(t)\\
&= \sum_{j=1}^n t_j v_{i_j} + o(t^2)
\end{align}

Since $W = \Span_{\ge 0}(\{v_i\}_i)$ is closed,
we also get that for any $t > 0$, we have
that $d(\alpha(t), W) = o(t^2)$. This means
that $\alpha'(0)$ is in $W$.

Assume that $\Lambda(X,Y)$
has finitely-many completions,
and let $v \in \Env_0(X,Y)$. Then $v$ is a convex
combination of vectors in $SV_X(\Lambda(X,Y))$.

Let $v(t):[0,1] \to \mc{T}(S)$ be a geodesic path parametrized by
arc length whose $1$-jet at $0$ is $v$. Since $v\in \Env_0(X,Y)$, we
can freely assume that $v(t)$ lies in $\Env(X,Y)$ for any
sufficiently small $t$.

Working in co-ordinates $\vphi:\mc{T}(S_g) \to \R^{6g-6}$, for each
completion $\lambda$ of $\Lambda(X,Y)$, we get a foliation of
$\R^{6g-6}$ defined by the Thurston stretch line
corresponding to $\lambda$. This foliation also
comes with a natural flow direction by
taking the forward corresponding to the stretch
path defined by $\lambda$.

Note that when $\lambda$ contains a pair-of-pants decomposition,
shearing co-ordinates are smoothly related to Fenchel-Nielsen
co-ordinates on $\mc{T}(S_g)$ by the computations done in
subsection <ref>. More generally, by Theorem A of
<cit.> and computations in <cit.>,
we have that length functions of simple closed
curves are smooth in the shearing co-ordinates. Since
Teichmuller space is locally parametrized by length
functions of $6g-6$ simple closed curves, it follows
that we can think of shearing co-ordinates not just
as topological co-ordinates on $\mc{T}(S_g)$,
but as smooth co-ordinates as well. Stretch lines are smooth in
shearing co-ordinates, since they can be realized as rays starting at the origin in the shearing co-ordinates corresponding to the lamination defining the stretch path <cit.>. Thus, stretch lines are also smooth in Fenchel-Nielsen co-ordinates. In
particular, it follows that if $\lambda$ is a completion of a
maximal chain-recurrent lamination containing $\Lambda(X,Y)$, we get
a smooth oriented foliation (which we will call $\mc{F}_{\lambda}$)
of $\R^{6g-6}$ under any co-ordinates on Teichmüller space.

By Thurston's construction of geodesics using concatenation of
stretch lines, it follows that for each $t$, $v(t)$ can be expressed
as a concatenation of flows along the foliations $\mc{F}_{\lambda}$.
Moreover, a careful reading of Theorem 8.5 of <cit.> actually
says that we can choose these laminations to lie in

By Lemma <ref>, using the fact
that $SV_X(\Lambda(X,Y))$ is finite and hence $\Span_{\ge
0}(SV_X(\Lambda(X,Y)))$ is closed, we get that $v\in \Span_{\ge
0}(SV_X(\Lambda(X,Y)))$ and we write $v = \sum_{i=1}^{N} a_i
v_{\lambda_i}$, where $\lambda_i \in M(\Lambda(X,Y))$.

Next, we show that $v$ is a convex linear combination of
We write $v = \sum_i a_i v_{\lambda_i}$,
and consider the path:

\begin{align*}
\beta(t) = f_1(k a_1t,f_2(ka_2t,(\hdots,(f_n(ka_nt,0)))))
\end{align*}

Where $k$ is chosen such that $k\sum a_i = 1$, and
where we denote $f_i(t,X) = \textrm{Stretch}(X,\lambda_i,t)$.
Note that $\beta$ is a length-parametrized geodesic, since
$\Lambda(X,Y)$ is maximally stretched along it. By the
same computations in Lemma <ref>, we get
that $\beta'(0) = k v$. In particular,
since $v$ and $\beta'(0)$ are unit-length vectors,
it follows that $k = 1$, and $\sum_i a_i = 1$.

We now treat the cases when $\Lambda(X,Y)$
has infinitely many completions.

If $\Lambda$ is chain-recurrent or
empty, then $\Span_{\ge 0}(SV_X(\Lambda))$
is closed.

We will show that for $\lambda_n \in MCR(\Lambda)$, if
$v_{\lambda_n} \in SV_X(\Lambda)$ converge to some $v$, then $v$ is
in the convex hull of $SV_X(\Lambda)$. By <ref>, up
to subsequence, there exists some $\lambda \in CR(\Lambda)$ such
that $\lambda_n \to \lambda$ in the Hausdorff topology.

By maximality of $\lambda_n$,
$S\ssm \lambda_n$ has at most finitely-many
completions to a triangulation, and
since $\lambda$ is a Hausdorff limit of $\lambda_n$, it shares this
property. Let $\lambda^1,\hdots,\lambda^k \in M(\lambda)$ be the
finitely-many completions of maximal chain-recurrent laminations
containing $\lambda$.

Since the complementary components of $\lambda_n$
stabilize close to the complementary components of
$\lambda$, it follows that for any sufficiently large $n$,
there exist completions of $\lambda_n^i \in M(\lambda_n)$
which Hausdorff converge to $\lambda^i$. We can do
this, for example, by adding in leaves
into the complementary components of $\lambda_n^i$
which converge to the added leaves of $\lambda^i$.

For any $t$, we consider the family
$S_n(t) = \mathrm{Stretch}(X, \lambda_n^i,t)$. Since
$S_n(t)$ are smooth geodesics, it follows that
they have uniformly bounded first derivatives. By
the Arzela-Ascoli theorem, up to subsequence, $S_n(t)$
converge to a continuous path, $S_{\infty}(t)$ in
$\mc{T}(S)$, starting at $X$. Moreover, $S_{\infty}$
is differentiable at $0$, and has derivative
equal to $v$. Note that
because $\lambda_n^i$ converge to $\lambda^i$,
and $\lambda \subset \lambda^i$,
it follows that $\lambda \subset \Lambda(X,S_{\infty}(t))$,
and so $\Lambda(X,S_{\infty}(t))$ has
finitely-many completions.

By <ref>,
we have that $\Env_0(X, S_{\infty}(t))$
is the convex hull of $SV_X(\Lambda(X,S_{\infty}(t)))$,
which is closed.
Since this is true for all $t$, a similar
Taylor series argument as in the previous
lemmas above shows that,
in fact, $v = \frac{d}{dt}|_{t=0} S_{\infty}(t)$
lies in the convex hull of $SV_X(\Lambda(X,S_{\infty}(t)))$,
as desired.

The proof of Lemma <ref>
follows in the same way as the proof of
Lemma <ref>,
where we now use Lemma <ref>
to meet the conditions of
Lemma <ref>.

The reason we split Lemma <ref>
into two lemmas is a bit of a subtlety. In the argument
above, we showed that the non-negative
span of $SV_X(\Lambda(X,Y))$ is closed. A priori,
the entire proof of Lemma <ref>
could have been skipped if we knew that $SV_X(\Lambda(X,Y))$
was closed.

To argue this, let $v_{\lambda_n}$ be a
converging sequence of stretch vectors
converging to some $v$. One should show that the Hausdorff
limit of $\lambda_n \in MCR(\Lambda(X,Y))$ is a
lamination $\lambda \in MCR(\Lambda(X,Y))$, and then use
the fact that stretch vectors change continuously
in their defining laminations to get that
$v = v_{\lambda}$. It is easy to see that
$\lambda$ must be chain-recurrent, and we must show that it
is maximal amongst all chain-recurrent laminations,

Showing that $MCR(\Lambda(X,Y))$ is
closed is non-trivial, and may in fact be false. It would follow
from Remark 2.9 of <cit.>, that says that the complementary
components of a maximal chain-recurrent lamination consists of ideal
triangles, once-punctured monogons, or once-punctured bigons.

The argument is that Remark 2.9 of <cit.> tells us that
the complementary components of $\lambda$ must also
be triangles or once-punctured monogons and bigons, and
hence $\lambda$ is maximal. However, a proof of the remark
is not furnished anywhere in literature, and it is not clear
if it is even true. It is possible that $\lambda_n$
each have a complementary component that is an ideal
square, and that in the limit, $\lambda$
has an ideal square as a complementary component, and one
of the diagonal leaves can be added to $\lambda$
preserving chain-recurrence.

It remains to show that $SV_X(\Lambda(X,Y))$
consists of extremal vectors in $\Env_0(X,Y)$.
This follows immediately from the proof of
Theorem 5.2 in <cit.>. We
provide the main argument, but refer the
reader to details in <cit.>.
We prove Lemma <ref>:

Let $\Lambda$ be chain-recurrent or empty,
and let $\lambda \in M(\Lambda)$,
and let $v_{\lambda} \in SV_X(\Lambda)$ be
a convex combination:
$v_{\lambda} = \sum_i a_i v_{\lambda_i}$, where
$\lambda_i \in M(\Lambda)$. Then
$v_{\lambda_i} = v_{\lambda}$ for all $i$.

For a lamination $\mu$, denote by $\mu^{CR}$ the maximal
chain-recurrent sublamination of $\mu$. To prove the lemma, suffices
to show that $\lambda_i^{CR} = \lambda^{CR}$.

Suppose not, and let $\alpha_n$ be a sequence of
simple closed multi-curves Hausdorff-converging to
$\lambda^{CR}$. Denoting the curve complex
of $S$ by $\mathcal{C}(S)$, and fixing $X\in \mc{T}(S)$,
we define a map from
$T\mc{T}(S) \times \mathcal{C}(S)$ by sending
$v,\alpha$ to $D_v \log l_{\alpha}$.
This map is linear in the first agument, and is continuous in the
second argument, where we think of $\mathcal{C}(S)$ as a space
endowed with the Hausdorff topology on $X$.

By Thurston's construction of $\mathrm{Stretch}(X,\lambda,t)$, it
follows that if $\alpha_n$ Hausdorff converge to $\lambda^{CR}$,
then $\lim_{n\to \infty} D_{v_{\lambda}} \log(l_{\alpha_n}) = 1$.
We argue that for any $i$,
$\lim_{n\to \infty} D_{v_{\lambda_i}} \log(l_{\alpha_n}) < 1$,
and the contradiction follows.

Indeed, if $\lim_{n\to \infty} D_{v_{\lambda_i}} \log(l_{\alpha_n})
= c \ge 1$, then there would exist some weights $w_n$ on
$\alpha_n$ and some measured lamination
$\lambda^m$ whose support contains $\lambda^{CR}$,
for which $\alpha_n$ converge to $\lambda^m$ in
measure. Moreover, for this measured lamination,
we would have $D_{v_{\lambda_i}} \log(l_{\lambda^m}) = c$.
This means that $\lambda^{CR}$ is maximally stretched
along $v_{\lambda_i}$, a contradiction.

§.§ Proving Theorem <ref>

We now prove Theorem <ref>:

Let $v$ be an extremal point in $\bm{S}_X$. By Remark
1.12 and Theorem 1.11 of <cit.>,
$v$ must lie in $\Env_0(X, Y)$
for some $Y \in \mc{T}(S)$. In particular, it must be
extremal in $\Env_0(X, Y)$, and hence,
by Theorem <ref>, $v \in SV_X$.
Conversely, by Lemma <ref>,
any stretch vector must be extremal.

§.§ Computing Twisting from Shearing

We compute the twist parameters along Thurston geodesics defined by
laminations that are completions of pair of pants decompositions of
$S$. A large portion of this work was independently done in
<cit.>, but we leave it here for the sake of exposition.
We begin by reviewing some notation that will help us
later on.

§.§ Introduction and Notation

Let $\mc{C} = \{c_1,\hdots, c_{N}\}$ be a pair-of-pants
decomposition of $S$, and let $\lambda$ be a chain-recurrent
completion of $\mc{C}$ to a triangulation. Let $X\in \mc{T}(S)$, and let
$X_t$ be the Thurston geodesic $\textrm{Stretch}(X,\lambda,t)$.  To
compute the twisting along $X_t$, we must compute
$\tau_{c_i}(t)$ as a function of
$s_{\lambda}$. This can be done from the definition, but depends on
the topological type of $\lambda$. Let $c\in \mc{C}$ be some curve, and
$\tilde c$ a lift of it. Let $P_1$, $P_2$ be its adjacent
pairs-of-pants, with lifts $\tilde P_i$ adjacent to $\tilde c$.

Let $p_1(t)$ and $q_2(t)$ be defined as in Section <ref>
be the distinguished points used to compute $s_c(X_t)$ and
$\tau_c(X_t)$, where we choose lifts carefully so that $p_i$ and $q_i$
lie on the same lift $\tilde c$ of $c$.

We define $\Delta_{P_i,c,\lambda}(t)$ as the signed distance from
$q_i(t)$ to $p_i(t)$, where the sign is positive if to get from
$q_i(t)$ to $p_i(t)$, one has to turn left from the perspective of
$\tilde P_i$. The functions $\Delta_{P_i,c,\lambda}(t)$ are actually
intrinsic to $P_i$, and are well-defined up to a choice of the lifts
of all of the curves in $\mc{C}$. In particular,
$\Delta_{P_1,c,\lambda}(t)+\Delta_{P_2,c,\lambda}(t) = s_c(X_t)
- \tau_c(X_t)$, so we devote the rest of this section to computing
$\Delta_{P,c,\lambda}(t)$, where $P$ is a pair of pants in $S\ssm
\mc{C}$, $c\in \mc{C}$ and $\lambda$ is a triangulation of $P$.

A geodesic triangulation of a hyperbolic pair of pants $P$
is a decomposition of $P$ into two ideal triangles,
whose edges are glued together.

Let $P$ be a hyperbolic pair of pants, then there are
exactly $32$ geodesic triangulations of $P$ up to
boundary-preserving isomorphism of $P$.

This fact is stated in the beginning of section 3.3 of <cit.>,
and follows from the discussion in section 2.6 of <cit.>.

In fact, let $P$ be a pair of pants with
cuff curves $\gamma_1$, $\gamma_2$, and $\gamma_3$.
The topological type of a geodesic triangulation of $P$ is
precisely characterized by the following information

* The twist direction at every cuff $\gamma_i$ of $P$
(there are two possibilities per cuff, so eight options in total)

* The number of leaves converging to
$\gamma_1, \gamma_2, \gamma_3$, with
possibilities $(2,2,2), (1,1,4), (1,4,1)$ and $(4,1,1)$.

If $\gamma$ is a cuff curve of $P$,
and $\lambda$ is a triangulation of $P$, then we say that

* $\lambda$ is $3$-symmetric if $\lambda$ has precisely
two leaves converging to $\gamma$

* $\lambda$ is $2$-symmetric around $\gamma$
if $\lambda$ has precisely four leaves converging
to $\gamma$

* $\lambda$ is asymmetric around $\gamma$
if $\lambda$ has precisely one leaf converging
to $\gamma$

Asymmetric, $2$-symmetric, and $3$-symmetric laminations $\lambda$. The curve $\gamma$ is the bottom one

We compute $\Delta_{c,P,\lambda}$ in each of these
cases. We will first fix some notation in all of the following

Let $\gamma_i$ be the three boundary components of
$P$, and we will always assume that $c = \gamma_1$.
Let $\gamma_{ij}$ be the leaf of $\lambda$ which is
asymptotic to both $\gamma_i$ and $\gamma_j$.  For ease of notation
later on, we define $l_i = l_{\gamma_i}(X_0)$, and $s_i
= s_{\gamma_i}(X_0)$, $s_{ji} = s_{ij} = s_{\gamma_{ij}}(X_0)$. Note
that in the different possibilities for $\lambda$, not all of
the variables $s_{ij}$ have meaning. For example,
if $\lambda$ is $2$-symmetric, $\gamma_{23}$
is replaced by $\gamma_{22}$.

§.§ $\lambda$ is $3$-symmetric around $c$

In this subsection, we compute $\Delta_{P,c,\lambda}$, where
$\lambda$ is a $3$-symmetric triangulation which twists left
at every cuff. We will compute $\Delta_{P,c,\lambda}(X_0)$
in terms of the shearing co-ordinates of $X_0$.

We fix an identification of $\tilde S$ with the
upper-half plane such that $\tilde \gamma_1$ is the line
$\{(0,t):t\in\R\}$, oriented so that $P$ lies to the left of this

We know that $\gamma_{13}$ is asymptotic to
$\gamma_1$, and twists left, meaning that there is a
lift of it of the form $\{(x,t):t\in \R\}$, for some
$x<0$. Similarly, $\gamma_{12}$ has a lift of the form
$\{(y,t):t\in \R\}$, oriented so that $(0,0)$
lies to the right of them.

Up to rescaling the identification
above, we can assume that $y = x+1$. This will allow
us to explicitly compute $x$ later on. In our choice
of specified lifts for the shearing co-ordinates,
we choose these lifts of $\gamma_{ij}$ with this
orientation. This choice
will not matter after differentiating, as a
different choice of lifts will only change
the answer by an additive constant.

The deck action of $[\gamma_1]^{-1}$ on $\HH^2$
sends the vertical lines $\{(a,t):t\in \R\}$ to
$\{(e^{-l_1}a,t):t\in \R\}$, and
in particular, must send one lift of
$\gamma_{ij}$ to another one. Drawing two
lifts of $\gamma_{23}$, one from $(x,0)$ to $(x+1,0)$
and one from $(x+1,0)$ to $(e^{-l_1}x,0)$,
we obtain lifts of the complementary triangles
of $\lambda$.

(/e - )/2

[thick,->] (0,0) – (0,3);
[thick,<->] (0.25,0) – (-4.3,0);

[ForestGreen] at (,3.2) $\tilde \gamma_{12}$;
[OrangeRed] at (,3.2) $\tilde \gamma_{13}$;
at (0,3.2) $\tilde \gamma_{1}$;

[thick, ForestGreen] (, 0) arc (0:180:1/3);
[thick, NavyBlue] (, 0) arc (0:180:7/24);
[thick, ForestGreen] (, 0) arc (0:180:13/48);
[thick, OrangeRed] (-2/3, 0) arc (0:180:1/6);

[thick, OrangeRed] (/e, 0) arc (0:180:1/2);
[thick, NavyBlue] (/e, 0) arc (0:180:3/8);
[thick, OrangeRed] (/e, 0) arc (0:180:5/16);
[thick, ForestGreen] (/e-1, 0) arc (0:180:(/e-1-)/2);

[thick, ->] (, 0) arc (0:180:1/4) node[anchor=north, pos=0.5] $\gamma_2$;
[thick, ->] (/e, 0) arc (0:180:1/4) node[anchor=north, pos=0.5] $\gamma_3$;

[thick, OrangeRed, ->] (,0) – (,3);
[thick, ForestGreen,->] (,0) – (,3);
[thick, NavyBlue] (, 0) arc (180:0:(- )/2);
[thick, BurntOrange] ((+ )/2,- ) circle ((-)/2);

in 1,2,3,4

[thick, OrangeRed,->] (/(e^),0) – (/(e^),3);
[thick,ForestGreen,->] (/(e^),0) – (/(e^),3);
[thick,NavyBlue] (/e^, 0) arc (180:0:(/e^- /e^)/2);
[thick,NavyBlue] (/e^(-1), 0) arc (180:0:(/e^- /e^(-1))/2);
[thick,BurntOrange, ->] ((/(e^) + /(e^))/2,/(e^) - /(e^)) circle ((/(e^)-/(e^))/2);
[thick,BurntOrange, ->] ((/(e^) + /(e^(-1)))/2,-/(e^(-1)) + /(e^)) circle ((/(e^(-1))-/(e^))/2);

at (,-0.2) $x$;
at (,-0.2) $x+1$;
at (/e,-0.2) $e^{-l_1}x$;
at ((+)/2,(-)/2 + 0.2) $q$;
(,0) circle (0.02);
(,0) circle (0.02);
(/e,0) circle (0.02);
((+)/2,(-)/2) circle (0.02);

at ((+ )/2, 2.5) $\Delta_1$;
at ((+ /e)/2, 2.5) $\Delta_2$;

[thick,|-|] (,2*) – (,2*) node[fill=white, anchor=center, pos=0.5, rotate=90] $s_{12}$;
[thick,|-|] (/e,2*) – (/e,1/e) node[fill=white, anchor=center, pos=0.5,rotate=90] $s_{13}$;

Lifting a $3$-symmetric lamination

In particular, one can readily see from this picture
that $s_{12} + s_{13} = -l_1$, and in general,
$|s_{ij} + s_{jk}| = l_j$, where the sign is
positive if $\gamma_{ij}$ and $\gamma_{jk}$ twist
to the right at $\gamma_j$, and is negative
otherwise. This is a special case of Lemma 3.2 in

By definition, the shearing co-ordinates of $\gamma_{12}$ is the
difference between the medians along $\tilde \gamma_{12}$ of the two
triangles drawn ($\Delta_1$ and $\Delta_2$). This gives us the
(Euclidean) radius of the incircle on the right, which tells us that
$e^{-l_1}x - (x+1) = e^{s_{12}}$.

In particular,

\begin{align*}
x = \frac{1 + e^{s_{12}}}{e^{-l_1} - 1}
\end{align*}

The next step is to find the endpoints of $\tilde \gamma_2$
in terms of $x$. Since $\gamma_{12}$ is
asymptotic to $\gamma_2$, it follows that one of
the endpoints of $\gamma_2$ is $x+1$. The
second endpoint can be computed by taking the
limit $[\gamma_2]^n \circ x$, where $[\gamma_2]$
is the deck transformation corresponding to
$\gamma_2$. Call this second endpoint $p^*$. We
now wish to compute $p^*$ in terms of $x$ and the
shearing co-ordinates.

To do this, we apply the isometry $\vphi(z) = \frac{x-z}{z-(x+1)}$,
which will send the lift of $\gamma_2$ at
$[p^*, x+1]$ to an arc $[\vphi(p^*), \infty]$. This
isometry also sends the arc $[x,x+1]$ to the arc $[0,\infty]$,
and we can now use shearing co-ordinates to find that its
first translate under $[\gamma_2]$ must lie on $[w,\infty]$
Where $w = e^{s_{23}} + e^{s_{23}}e^{s_{12}}$ (see picture):

(-4/e + 3)/2
*(1 + 2*-2*)̊

[thick,->] (0,0) – (0,3);
[thick,<->] (-0.5,0) – (4,0);

[thick, ->] (3.2,0) – (3.2,3);
at (3.2, 3.2) $\tilde \gamma_{2}$;

[NavyBlue] at (,3.2) $\tilde \gamma_{23}$;
[thick, NavyBlue, ->] (,0) – (,3);
[thick, OrangeRed] (, 0) arc (180:0:(- )/2);
[thick, BurntOrange] ((+ )/2,- ) circle ((-)/2);
[ForestGreen] at (,3.2) $\tilde \gamma_{12}$;
[thick, ForestGreen,->] (,0) – (,3);

[thick, NavyBlue, ->] (,0) – (,3);
[thick, OrangeRed] (, 0) arc (180:0:(- )/2);
[thick, BurntOrange] ((+ )/2,- ) circle ((-)/2);

+ 0.5
[thick, ForestGreen, ->] (,0) – (,3);
[thick, OrangeRed] (, 0) arc (180:0:(- )/2);
[thick, BurntOrange] ((+ )/2,- ) circle ((-)/2);

+ 0.3
[thick, NavyBlue, ->] (,0) – (,3);
[thick, OrangeRed] (, 0) arc (180:0:0.3/2);
[thick, BurntOrange] ((+ )/2,0.3) circle (0.3/2);

+ 0.2
[thick, ForestGreen, ->] (,0) – (,3);
[thick, OrangeRed] (, 0) arc (180:0:0.2/2);
[thick, BurntOrange] ((+ )/2,0.2) circle (0.2/2);

[thick] (+0.3,1.5) circle (0.005);
[thick] (+0.4,1.5) circle (0.005);
[thick] (+0.5,1.5) circle (0.005);

at (,-0.2) $0$;
(,0) circle (0.02);
at (,-0.2) $e^{s_{23}}$;
(,0) circle (0.02);
at (,-0.2) $w$;
(,0) circle (0.02);

at (-0.2,1.6) $i$;
(,1.6) circle (0.02);

[thick,|-|] (,0.9) – (,1.6) node[fill=white, anchor=center, pos=0.5, rotate=90] $s_{23}$;
[thick,|-|] (,-) – (,0.9) node[fill=white, anchor=center, pos=0.5, rotate=90] $s_{12}$;

at ((+ )/2, 2.5) $\Delta_1$;
at ((+ /e)/2, 2.5) $\Delta_2$;

[thick,|-|] (,2*) – (,2*) node[fill=white, anchor=center, pos=0.5, rotate=90] $s_{12}$;
[thick,|-|] (/e,2*) – (/e,2*/e) node[fill=white, anchor=center, pos=0.5,rotate=90] $s_{13}$;

The lift of the $3$-symmetric lamination after applying the isometry $\vphi$

Since the translation length of the aciton of $[\gamma_2]$
is precisely $l_2$, it follows that we must have that
$\vphi(p^*) - w = e^{l_2}(\vphi(p^*))$, or, solving for
$\vphi(p^*)$ and plugging in $\vphi^{-1}(z) = x + \frac{z}{z+1}$

\begin{align*}
p^* = x + \left(\frac{e^{s_{23}} + e^{-l_2}}{e^{s_{23}}+1}\right)
\end{align*}

All that remains is to compute the intersection point of
$[0,\infty]$ and the orthogeodesic between $[p^*, x+1]$ and
$[0,\infty]$. In other words, we must find the point
$p$ as described at the start of this section.

Under our lift, the incircle of $\Delta_1$ intersects
$\tilde \gamma_{13}$ at $i$, meaning that the horocycle connecting
this intersection point to $\tilde \gamma_1$ must also
intersect $\tilde \gamma_1$ at $i$. Thus,
$\Delta_{P,c,\lambda}(X_0) = \log(-ip)$.

Let $C_1$ be a circle whose center lies on the $x$-axis intersects the
$x$-axis at $(a,0)$ and $(b,0)$ with $0<a<b$. Let $C_2$
be a circle centered at $0$ intersecting $C_1$ at right angles.
Then $C_1$ has radius $\sqrt{ab}$

The triangle $(0,0), (0,\frac{a+b}{2}), C_1\cap C_2$ is a
right triangle with hypothenuse $\frac{a+b}{2}$, and other whose
other edges have lengths equal to $\frac{b-a}{2}$ and to
the radius of $C_2$. The claim follows.

By the above claim, it follows that $p = i\sqrt{(x+1)p^*}$,
and since $q(t)$ is located at $i$ by the choice
of normalization, we get that:

\begin{align} \label{eq:3-symmetric-delta}
\Delta_{P,c,\lambda}(X_0)
= \frac12\log\left((x+1)\left(x+\frac{e^{s_{23}}
+ e^{-l_2}}{e^{s_{23}}+1}\right)\right)
\end{align}

where $x = \frac{1 + e^{s_{12}}}{e^{-l_1} - 1}$, and
$s_{ij}$ are the shearing co-ordinates of $X_0$, as described

Recall that these shearing co-ordinates $\{s_{ij}\}$
satisfy the system of equations $s_{ij} + s_{jk} = - l_j$,
as we assumed that the arcs $\gamma_{ij}$ all twist
left around the cuffs. In particular, by solving this system
of equations, we can compute $\Delta_{P,c,\lambda}(X_0)$
as an explicit function of $l_{\gamma_1},l_{\gamma_2}$, and

A similar computation also holds when the twist directions
of $\lambda$ around the $\gamma_i$'s are arbitrary. However,
in this case, every appearance of $l_i$ is replaced
with $-l_i$ if $\lambda$ twists to the right around
$\gamma_i$, and the twisting is in the opposite direction
if $\lambda$ twists to thr right around $\gamma_1$.

For example, if $\lambda$ twists to
the right around $\gamma_1$, then we choose a normalization
and lifts of $\gamma_{13}$ and $\gamma_{12}$
that are vertical lines intersecting the $x$-axis at $x$ and $x+1$.
The next lift of $\gamma_{13}$ is then the vertical
line at $e^{l_1}x$, and not at $e^{-l_1}x$. The
computation of $x, p^*$, and therefore
$\Delta_{\mc{P},c,\lambda}$ follow in the same way,
but with $l_1$ replaced with $-l_1$, and the signed
distance from $p$ to $q$ picking up a negative sign.

If $\lambda$ twists to the right around $\gamma_2$,
then in Figure <ref>, $\tilde \gamma_2$
would be drawn starting at $x+1$, and going to the
right of $\tilde \gamma_{12}$. After passing through
the same mobius transformation, we arrive at a reflected
picture in place of Figure <ref>,
where the deck transformation
pushing lifts of $\gamma_{23}$ towards
$\tilde \gamma_2$ acts in the opposite direction.
Thus, we would get $\vphi(p^*)-w = e^{-l_2}\vphi(p^*)$,
and the rest follows.

Similar computations can be done for all permutations
of twisting directions of $\lambda$ around the $\gamma_i$'s,

\begin{align}\label{eq:3-symmetric}
\eps_1\frac12\log\left((x+1)\left(x+\frac{e^{s_{23}}
+ e^{-\eps_2l_2}}{e^{s_{23}}+1}\right)\right)
\end{align}

* $\eps_i = 1$ if $\lambda$ twists to the
left around $l_i$, and $\eps_i = -1$ otherwise.

* $x=x(t) = \frac{1 + e^{s_{12}}}{e^{-\eps_1l_1} - 1}$,

* $s_{ij}$ are the shearing co-ordinates, which satisfy
$s_{ij} = \frac12(\eps_kl_k - \eps_il_i - \eps_jl_j)$ for $k\neq
i \neq j \neq k$

The formula above for $\Delta_{P,\gamma_1,\lambda}$
does not appear to be symmetric under replacing the labels of
$\gamma_2$ and $\gamma_3$, and indeed
there is no reason for it to be, as we made the
choice to compute $p$ using $\gamma_2$.
However, one may verify that choosing to compute
$p$ using $\gamma_3$ is analogous to replacing
$x$ with $x+1$, and replacing $x+1$ with $e^{-l_1}x$, and
rescaling by a factor of $e^{-l_1}x-(x+1)$. This yields
a difference of $l_1/2$ in the computation, as expected.

§.§ $\lambda$ is $2$-symmetric around $c$

In this section, we compute $\Delta_{P,c,\lambda}$, where
$\lambda$ is a $2$-symmetric or asymmetric triangulation with
respect to $c$, which twists left
at every cuff. We will compute $\Delta_{P,c,\lambda}(X_0)$
in terms of the shearing co-ordinates of $X_0$.

Let $\gamma_i$ be the three boundary components of
$P$, and label $c = \gamma_1$. Let $\gamma_{ij}$ be the
leaf of $\lambda$ which is asymptotic to both $\gamma_i$ and
$\gamma_j$. Note that not all combinations of $i$ and $j$
yield a viable $\gamma_{ij}$, as in the case
that $\lambda$ is $2$-symmetric, $\gamma_{23}$ does
not exist as no leaf is aysmptotic to both $\gamma_2$
and $\gamma_3$. In this case,

We define $l_i = l_{\gamma_i}(X_0)$,
and $s_i = s_{\gamma_i}(X_0)$, $s_{ij} = s_{\gamma_{ij}}(X_0)$.

We follow a similar computation as in the section above to get that:

\begin{align}\label{eq:2-symmetric}
\Delta_{P,c,\lambda}(X_0) = \eps_1\frac12\log\left((x+1)(x+e^{-\eps_2l_2})\right)
\end{align}

\begin{align*}
x = \frac{1 + e^{s_{12}} + e^{s_{12}+s_{11}}
+ e^{s_{12}+s_{11}+s_{13}}}{e^{-\eps_1l_1} - 1}
\end{align*}

* $\eps_i = 1$ if $\lambda$ twists to the
left around $l_i$, and $\eps_i = -1$ otherwise.

* $s_{ij}$ are the shearing co-ordinates, which satisfy:

\begin{align*}
s_{11} &= \frac12(-\eps_1l_1 + \eps_2l_2 + \eps_3l_3)\\
s_{12} &= -\eps_2 l_2\\
s_{13} &= -\eps_3 l_3
\end{align*}

§.§ $\lambda$ is asymmetric around $c$

In this section, we compute $\Delta_{P,c,\lambda}$, where
$\lambda$ is a $2$-symmetric triangulation with
respect to $c$, which twists left
at every cuff.

Let $\gamma_i$ be the three boundary components of
$P$, and label $c = \gamma_1$ as before. Let $\gamma_{ij}$ be
defined as before, and assume that $\lambda$ is
$2$-symmetric around $\gamma_2$. When computing
$\Delta_{P,c,\lambda}$, we will use an orthogeodesic
from $\tilde \gamma_2$ to $\tilde \gamma_1$. We
define $l_i$, $s_{ij}$ as before, noting that in
this case, the only co-ordinates are
$s_{12}$, $s_{23}$, and $s_{22}$.

We follow a similar computation as in the section above to get that,

\begin{align} \label{eq:a-symmetric}
\Delta_{P,c,\lambda}(X_0) = \eps_1\frac12\log\left((x+1)\left(x+
\frac{e^{s_{22}} + e^{s_{22}+s_{23}} + e^{2s_{22} + s_{23}} + e^{-\eps_2l_2}}
{e^{s_{22}} + e^{s_{22} + s_{23}} + e^{2s_{22} + s_{23}} + 1}
\right)\right)
\end{align}

* $\eps_i = 1$ if $\lambda$ twists to the
left around $l_i$, and $\eps_i = -1$ otherwise.

* $x = \frac{1}{e^{-\eps_1l_1} - 1}$

* $s_{ij}$ are the shearing co-ordinates, which satisfy:

\begin{align*}
s_{22} &= \frac12(-\eps_2 l_2 + \eps_1 l_1 + \eps_3 l_3)\\
s_{12} &= -\eps_1 l_1\\
s_{23} &= -\eps_3 l_3
\end{align*}

§.§ Twist Widths Between Stretch Paths

We have seen how twisting co-ordinates change along stretch paths
defined by any lamination which is a completion of a pair-of-pants
decomposition of a surface $S$. We now end this subsection
with an explicit formula for the twisting along
stretch paths.

Consider the foliation of $\mc{T}(S)$ given by stretch paths
corresponding to $\lambda$. For any $Y\in \mc{T}(S)$, we can define
the negative-time stretch path from $Y$ by setting
$\mathrm{Stretch}(Y, \lambda, -t)$ to be the unique point $X \in
\mc{T}(S)$ such that $\mathrm{Stretch}(X, \lambda, t) = Y$,
and $d_{Th}(\mathrm{Stretch}(Y,\lambda,-t),Y) = t$. Denote
$Y^{\lambda}_{-t} = \mathrm{Stretch}(Y,\lambda,-t)$. For $X\in
\mc{T}(S)$, we denote $X^{\lambda}_t = \mathrm{Stretch}(X,\lambda,t)$,

Fix a pair of pants decomposition $\mc{P}$, and some
curve $c\in \mc{P}$. Let $P_1$ and $P_2$ be the (possibly non-distinct)
pairs-of-pants adjacent to $c$. We employ the
shorthand: $\Delta_{\lambda}^i = \Delta_{P_i,c,\lambda}$,
where $\Delta_{P_i,c,\lambda}$ was defined
in Section <ref>

Note that for any $X\in \mc{T}(S)$, we have:

\begin{align} \label{formula:twist-and-shear}
\tau_c(X) = s_c(X) - \Delta_{\lambda}^1 - \Delta_{\lambda}^2
\end{align}

For any $X,Y\in \mc{T}(S)$, we have:

\begin{align*}
\tau_c(X_t^{\lambda}) &=  \tau_c(X)e^t +
\Delta_{\lambda}^1(0)e^t
+\Delta_{\lambda}^2(0)e^t- \Delta_{\lambda}^1(t)
\tau_c(Y_{-t}^{\lambda}) &= \tau_c(Y)e^{-t} +
\Delta_{\lambda}^1(0)e^{-t}
+\Delta_{\lambda}^2(0)e^{-t}- \Delta_{\lambda}^1(-t)
\end{align*}

Where $\Delta_{\lambda}^i(s)$ is $\Delta_{\lambda}^i$, where
every shearing co-ordinate is multiplied by a factor
of $e^s$ for positive or negative $s$.

Note that $s_c(X^{\lambda}_t) = s_c(X)e^{t}$, and similarly
$s_c(Y^{\lambda}_{-t}) = s_c(Y)e^{-t}$. This
is because $c$ is contained in $\lambda$, and in the shearing
co-ordinates $(s_{\lambda})$ on Teichmüller
space, stretch lines are exponential scalings.

For any $t$, and plugging in $s_c(X_t) = e^t s_c(X)$ to
Formula <ref>,
we have $\tau_c(X_t) = s_c(X)e^t - \Delta_{\lambda}^1(t) -
\Delta_{\lambda}^2(t)$. To find $s_c(X)$,
we set $t=0$ and rearrange, giving us
$s_c(X) = \tau_c(X) + \Delta_{\lambda}^1(0) + \Delta_{\lambda}^2(0)$.
We then get:

\begin{align*}
\tau_c(X_t^{\lambda}) &= \tau_c(X)e^t+
\Delta_{\lambda}^1(0)e^t
+\Delta_{\lambda}^2(0)e^t- \Delta_{\lambda}^1(t)
\end{align*}

A similar computation follows for $Y_{-t}^{\lambda}$.

If $\nu$ is another completion of
$\mc{P}$, we define the twist width
$\Delta \tau_c(\lambda,\nu,t)$ by
$\tau_c(X_t^{\lambda}) - \tau_c(X_t^{\nu})$, and write:

\begin{align} \label{eq:twist-width}
\Delta \tau_c(\lambda,\nu,t)
&= e^t\Delta_{\lambda}^1(0) -\Delta_{\lambda}^1(t)\\
&\quad + e^t\Delta_{\lambda}^2(0) -\Delta_{\lambda}^2(t)\\
&\quad - \left(e^t\Delta_{\nu}^1(0) -\Delta_{\nu}^1(t)\right)\\
&\quad - \left(e^t\Delta_{\nu}^2(0) -\Delta_{\nu}^2(t)\right)
\end{align}

In order to estimate the twist width, it suffices to estimate each of
the terms above separately.

Observe that by the computations
in the previous section, $\Delta_{\lambda}^i(t)$ is
always of the form $\frac12\log(R_{\mc{P},\lambda^i}(t))$,
where $R_{\mc{P},\lambda}^i(t)$ is some rational polynomial
over the set $\{e^{l_{\alpha}(t)}\}_{\alpha\in \mc{P}}$.
Thus, if $\alpha\in \mc{P}$, then we have that
$l_{\alpha}(t) = l_{\alpha}(X_0)e^t$, and so, for any real $k$,
we compute:

\begin{align*}
&= \frac12 e^t \log(e^{kl_{\alpha}(X_0)}R_{\mc{P},\lambda}^i(0))
- \frac12 \log(e^{kl_{\alpha}(t)}R_{\mc{P},\lambda}^i(t))\\
&= \frac12 e^t\left(kl_{\alpha}(X_0) + \log(R_{\mc{P},\lambda}^i(0))\right)
- \frac12 \left(kl_{\alpha}(t) + \log(R_{\mc{P},\lambda}^i(t))\right)\\
&= \frac12 e^t \log(R_{\mc{P},\lambda}^i(0))
- \frac12 e^t \log(R_{\mc{P},\lambda}^i(0))\\
&=e^t\Delta_{\lambda}^i(0) -
\Delta_{\lambda}^i(t)
\end{align*}

Where $t$ can also hold negative values. Thus, we are left with:

When computing twist widths, we are free to mutliply and divide
$R_{\mc{P},\lambda}^i(t)$ by any factor of the form
$e^{l_{\alpha}(t)}$, where $\alpha\in \mc{P}$.

§ BOUNDED WIDTH

In this section, we prove Theorem <ref>.
To prove this theorem, we first use Theorem <ref>
to show that it suffices to compute the distance between
points in the envelope that lie on stretch paths. We then
examine the lengths of simple closed curves
in the maximally-stretched lamination. We use twisting
computations from subection <ref>
to find upper and lower boundes of the twisting around it.
When the curve is long, we
use an earthquake around it to bound the
distance in the envelope. When the curve is short, we
resort to other estimates of the Thurston distance.

Throughout this section, we will use coarse estimate
notation as in <cit.>. We introduce it here.

Given two quantities (or functions), $A$ and $B$,
we write $A \pl B$ if there exists a constant $C$
independent of $A$ and $B$ such that $A \le B+C$.

Similarly, we write $A \tl B$ if there exists a constant $C$
independent of $A$ and $B$ such that $A \le BC$.
We write $A \pe B$ (resp. $A \te B$) if
$A \pl B$ and $B \pl A$ (resp. $A \tl B$ and
$B \tl A$)

The relations $\tl, \te, \pl,$ and $\pe$ do not
behave as regular equalities and inequalities. For example,
if $A \pe B$ and $C \pe D$, it is not necessarily
true that $\frac{A}{C} \pe \frac{B}{D}$.

§.§ Twisting and Earthquakes

In this section, we prove the following, which is of interest
of its own right:

Let $\alpha$ be a simple closed curve on $S$, and let
$X \in \mc{T}(S)$. Then:

\begin{align*}
d_{Th}(X,Eq_{\alpha,t}(X)) \pl \log(e^{l_{\alpha}/2}t)
\end{align*}

The following lemma will be useful in the proof of
Proposition <ref>:

Let $\eps > 0$, and $X\in \mc{T}(S)$ be given. Let $\alpha,\beta$ be
geodesic arc segments, and assume that $\beta$ is  contained in the
$\eps$-thick part of $X$, $X_{\ge \eps}$. Then

\begin{align*}
i(\alpha,\beta) \le \frac{4l_{\alpha}(X)l_{\beta}(X)}{\eps^2}
\end{align*}

Lemma <ref> has appeared in literature in
the setting of flat structures <cit.>, and in various other
contexts for extremal lengths <cit.>. We prove it here in the
setting of hyperbolic lengths and geometric intersection numbers. In
particular, we get:

If $\alpha,\beta$ are any closed curves contained in $X_{\ge \eps}$,
then $i(\alpha,\beta) \tl_{\eps} l_{\alpha}(X)l_{\beta}(X)$

We prove Lemma <ref>:

Let $w$ be the subarc of $\alpha$ of length $\eps/2$ which
maximizes $|w\cap \beta|$. Since $\beta$ is contained
$X_{\ge \eps}$, it follows that every
arc segment of $\beta \ssm (\beta \cap w)$ must be of
length at least $\eps/2$, otherwise we could find an essential loop of
length $\eps$ in $X_{\ge \eps}$ ($\alpha$ and $\beta$ are geodesic
segments and hence in minimal position). Denote $|w\cap \beta| = N$,
and by the above observation, we get that $N \le \frac{l_{\beta}}{\eps/2}$.
Additionally, since $w$ was chosen to maximize $|w\cap \beta|$, it follows
that $\frac{l_{\alpha}}{\eps/2} \le N$, and the proof
follows by stringing together these inequalities.

To prove Proposition <ref>, we will use the coarse
estimates of the Thurston metric from <cit.> using a short

Let $X\in \mc{T}(S_{g,n})$. A short marking on $X$ is
a collection of $N = 3g-g + n$ simple closed curves, $\{\beta_i,
\beta_i'\}_{i=1}^{N}$ obtained as follows:

* Define $\beta_1$ to be the shortest curve on $X$. Then
inducively, define $\beta_i$ to be the shortest curve on $X$
disjoint from $\beta_j$ for all $j < i$.

* Define $\beta_i'$ to be the shortest curve which
intersects $\beta_i$ transversely.

We call $\beta_i$ and $\beta_i'$ duals to each other, and
we denote $\bar \beta_i = \beta_i'$ and $\bar\beta_i' = \beta_i$.

We can estimate intersection numbers of curves with a short marking
in a similar way to Proposition 3.1 of <cit.>. By gaining
control of the number of intersections between a curve $\alpha$ and the
curves in the short marking, we can gain control
of the length of each curve in the short marking after earthquaking
along $\alpha$. We prove:

Let $\mu_X$ be a short marking on $X$, and let $\alpha$ be a simple
closed curve. For any $\beta\in \mu_X$, we have:

\begin{align*}
i(\alpha,\beta)/l_{\beta}(X) \tl e^{l_{\alpha}/2}
\end{align*}

We split the proof into two cases, depending on the length
of $\beta$. We denote $l_{\beta} = l_{\beta}(X)$, and
always assume that $i(\alpha,\beta) \ge 1$, otherwise
the lemma follows trivially.

$l_{\beta} < \frac1e$: In this case, $\beta$
has a collar of radius $2\sinh^{-1}(1/\sinh(l_{\beta}/2))$. By
estimating $\sinh(x) \le 2x$ for $x < 1$, and applying the
inequality $\sinh^{-1}(x) \ge \log(x)$, we
obtain a lower-bound on the width of the collar around $\beta$:

For every intersection of $\alpha$ and $\beta$, $\alpha$
must enter and exit the collar of $\beta$, meaning that $i(\alpha,\beta) \le \frac{l_{\alpha}}{2\log(1/l_{\beta})}$
Since $\alpha$ intersects $\beta$, it follows in particular that
we must have $l_{\alpha} \ge 2\log(1/l_{\beta})$, or,
$l_{\beta} \ge e^{-l_{\alpha}/2}$. Thus,

\begin{align*}
\frac{i(\alpha,\beta)}{l_{\beta}} \le \frac{l_{\alpha}}{l_{\beta}2\log(1/l_{\beta})}
\end{align*}

The function $x\log(1/x)$ is increasing on $[0,\frac1e]$, meaning that
$2l_{\beta}\log(1/l_{\beta}) \ge e^{-l_{\alpha}/2}l_{\alpha}l_{\alpha}$, and
plugging this in, we get the appropriate estimate.

$\frac1e \le l_{\beta}$: By Propositon 3.1 of
<cit.>, there exists some $C$ such that $l_{\alpha} \ge
C\sum_{\beta\in \mu_X} i(\alpha,\beta)l_{\bar\beta}$. In particular,
we get $i(\alpha,\beta) \le C\frac{l_{\alpha}}{l_{\bar\beta}}$, so
$\frac{i(\alpha,\beta)}{l_{\beta}} \le
\frac{l_{\alpha}}{l_{\beta}l_{\bar\beta}}$.

Let $k(S) < \frac1e$ be such that if $l_{\bar\beta} \le k(S)$, then $\bar\beta$ has a
collar of width at least $B(S)$, where $B(S)$ is the Bers constant of $S$. In particular,
if $l_{\bar\beta} \le k(S)$, then this means that $\bar\beta$ is a curve in the short
marking that was a part of a short pair-of-pants decomposition.

If $l_{\bar\beta} \ge k(S)$, then $\frac{i(\alpha,\beta)}{l_{\beta}} \le C\frac{e}{K(S)} l_{\alpha}
\tl e^{l_{\alpha}/2}$.

We now assume that $\frac1e \le l_{\beta}$ and that $l_{\bar\beta} \le k(S)$. In this case,
$\bar\beta$ is part of a short pair-of-pants decomposition, and
we can use the collar estimates from before to get that $\bar\beta$ has a collar
of width at least $2\log(1/l_{\bar\beta})$, so $l_{\beta} \ge 2\log(1/l_{\bar\beta})$.
If $\alpha$ intersects $\bar\beta$, then we must have $l_{\alpha} \ge 2\log(1/l_{\bar\beta})$,
and, in a similar fashion to the case where $l_{\beta} < \frac1e$, we get:

\begin{align*}
\frac{i(\alpha,\beta)}{l_{\beta}} \le \frac{l_{\alpha}}{l_{\beta}l_{\bar\beta}}
\le \frac{l_{\alpha}}{l_{\bar\beta}2\log(1/l_{\bar\beta})} \le e^{l_{\alpha}/2}
\end{align*}

We are left with the case that $\alpha$ does not intersect $\bar\beta$. Let
$U(\bar\beta)$ be a collar around $\bar\beta$ whose
boundary components have length at least $\frac1e$. By construction
of the dual, $\beta$ must be entirely contained in $X_{\ge \frac1e} \cup U(\bar\beta)$. Let
$\beta_1 = \beta \cap X_{\ge \frac1e}$. Note that since $\alpha$ does not intersect
$\bar\beta$, it follows that $\alpha$ is disjoint from $U(\bar\beta)$, and
so $i(\alpha,\beta) = i(\alpha,\beta_1)$. By Lemma <ref>,
$i(\alpha,\beta_1) \tl l_{\alpha}l_{\beta_1} \tl l_{\alpha}l_{\beta}$. Thus,
$\frac{i(\alpha,\beta)}{l_{\beta}} \tl e^{l_{\alpha}/2}$.

(Theorem E in <cit.>)
For any $X,Y\in \mc{T}(S_{g,n})$, we have:

\begin{align*}
d_{Th}(X,Y) \pe \max_{\beta\in \mu_X} \log \frac{l_{\beta}(Y)}{l_{\beta}(X)}
\end{align*}

We now prove Proposition <ref>:

Let $\mu_X$ be a short marking on $X$. For
any $\beta \in \mu_X$, if $\beta$ is disjoint from $\alpha$, then
$l_{\beta}(Eq_{\alpha,t}(X)) = l_{\beta}(X)$, and so
$\log \frac{l_{\beta}(Eq_{\alpha,t}(X))}{l_{\beta}(X)} = 0$. Otherwise,
we can bound the length $l_{\beta}(Eq_{\alpha,t}(X))$ by:

\begin{align*}
l_{\beta}(Eq_{\alpha,t}(X)) \le l_{\beta} + t i(\alpha,\beta)
\end{align*}

In particular, by Theorem <ref>,

\begin{align*}
d_{Th}(X,Eq_{\alpha,t}(X)) \pe \max_{\beta\in \mu_X} \log \left(1 + t\frac{i(\alpha,\beta)}{l_{\beta}(X)}\right)
\end{align*}

Applying Proposition <ref>, we get that
$\frac{i(\alpha,\beta)}{l_{\beta}} \tl e^{l_{\alpha}/2}$, meaning
that $d_{Th}(X,Eq_{\alpha,t}(X)) \pl \log(e^{l_{\alpha}/2}t)$, as desired.

§.§ Bounded Envelopes in $\mc{T}(S)$

In this section, we prove that geodesic envelopes have uniformly
bounded width in $\mc{T}(S_{1,1})$
and in $\mc{T}(S_{0,4})$. We first show:

Let $S$ be the once-punctured torus or the four-times
punctured sphere. There exists some $B = B(S)\in \R$ such that for
any $X,Y \in \mc{T}(S)$, we have that if $\Lambda(X,Y)$ is a simple
closed curve, then $w(X,Y) < B$.

From the lemma, we immediately obtain:

Let $S$ be the once-punctured torus or the four-times punctured
sphere. There exists some $B\in \R$ such that for any $X,Y \in
\mc{T}(S)$, $w(X,Y) < B$.

Let $X,Y \in \mc{T}(S)$ be arbitrary, and let
$\Lambda(X,Y)$ be the maximally-stretched lamination
from $X$ to $Y$. For the four-times punctured sphere,
by Proposition 26 of
<cit.>, $\Lambda(X,Y)$ is either maximal,
in which case there is a unique geodesic from
$X$ to $Y$, or $\Lambda(X,Y)$
contains a simple closed curve, $\alpha$.
For the once-punctured torus, the same result
holds by Theorem 1.1 of <cit.>. Now,
notice that geodesic from $X$ to $Y$ must be contained
in $\Outenv(X,\alpha) \cap In(Y,\alpha)$, which, by
Lemma <ref>, has width bounded by $B$.

The rest of this section is dedicated to proving
Lemma <ref>. We first fix some
notation. Fix $Y\in \mc{T}(S)$
and a simple closed curve $\alpha$.

* Denote by $\lambda^{L}$ and $\lambda^R$ be the two
maximal chain-recurrent laminations containing $\alpha$, twisting
to the left and right (respectively) around $\alpha$
(see Figure <ref>)

* If $S = S_{1,1}$, define $l_0 = l_{\alpha}(Y)/2$,
if $S = S_{0,4}$, define $l_0 = l_{\alpha}(Y)/4$.
Define $l_s = l_0e^s$ for any real $s$.

* For $Y\in \mc{T}(S)$,
we define $Y_{-t}^{L}$ as the unique point in $\mc{T}(S_{0,4})$
such that $\Lambda(Y_{-t}^{L},Y) = \lambda^L$,
and $d_{Th}(Y_{-t}^L,Y) = t$. Define
$Y_{-t}^R$ in the same manner.

* Let $P_1, P_2$ be the two (possibly non-distinct) pairs of
pants on opposite sides of $\alpha$. We define
$\Delta_i^L(t) = \Delta_{P_i,\alpha,\lambda^L}(t)$
and $\Delta_i^R(t) = \Delta_{P_i,\alpha,\lambda^R}(t)$
(see Section <ref>)

Definition of $\lambda^R$ (right) and $\lambda^L$ (left).
The blue curve is $\alpha$

Employing the above notation, and following the
computations in Lemma <ref>, we
compute the relative twisting
in either case of $S$:

Let $\alpha,l_0,l_s,Y_{-t}^{\pm}\in \mc{T}(S)$ be defined as above,

\begin{align}\label{eq:four-punctured-twist-width}
\mathrm{Twist}_{\alpha}(Y_{-t}^{+},Y_{-t}^{-})
= 4e^{-t}\log\coth(l_0/2)-
\end{align}

We first treat the case where $S = S_{0,4}$. In this
case, we use the computations from
Lemma <ref> to get:

\begin{align*}
\mathrm{Twist}_{\alpha}(Y_{-t}^L, Y_{-t}^R)
= e^{-t}(\Delta_1^L(0) + \Delta_2^L(0))
+ e^{-t}(\Delta_1^R(0) + \Delta_2^R(0))\\
- (\Delta_1^L(-t) + \Delta_2^L(-t))
- (\Delta_1^R(-t) + \Delta_2^R(-t))
\end{align*}

Using Example <ref> with $l_2 = l_3 = 0$,
and $s_{11} = -2l_s$, we get
$x = 2\frac{1+e^{-l_s}}{e^{-2l_s}-1}$, and so:

\begin{align*}
\Delta_1^L = \Delta_2^L
&= \frac12\log((x+1)^2)\\
&= \log\left(\frac{e^{-2l_s}+1+2e^{-l_s}}{1-e^{-2l_s}}\right)\\
&= \log\coth(l_s)
\end{align*}

where in the last line we used Fact <ref>.
and $\Delta_1^R = \Delta_2^R =
and the result follows.

Next, we consider the case where $S = S_{1,1}$.
Using Example <ref>
with $l_2 = l_1 = 2l_s$ and $l_3 = 0$, and
getting $s_{12} = -2l_s$, and $s_{23} = \frac12(l_1-l_2-l_3) = 0$.
This gives $x = \frac{1+e^{-2l_s}}{e^{-2l_s}-1}$,
and we compute:

\begin{align*}
x+1 &= \frac{2e^{-2l_s}}{e^{-2l_s}-1}\\
&= \frac12 \frac{(1+e^{-2l_s})^2}{e^{-2l_s}-1}
\end{align*}

Thus, after applying Fact <ref>,
we get:

\begin{align*}
\Delta_1^L = \log \left(\coth(l_s)\right)
\end{align*}

$\Delta_2^L$, and $\Delta_i^R$ are similarly computed,
giving the result.

Parts of the above lemma also follows from formulas 18 and 19 in
<cit.>, with slight modifications to treat negative $t$.
However, it is a reaffirming sanity check that the computations
preformed in Section <ref> agree with those
done in <cit.>

Let $S$ be the once-punctured torus or
the four-times punctured sphere, and let
$\alpha$ be a simple closed curve in $S$.
Then there exists a constant $\eps_0$
such that for any $X \in \mc{T}(S)$,
if $l_{\alpha}(X) < \eps_0$, then
$\alpha$ is the systole of $X$.

Any pair of distinct simple closed curves
on $S$ must intersect. Therefore, if
$\alpha$ is sufficiently short, then
by the collar lemma, any other simple closed
curve in $S$ must be at least the length
of the Bers constant of $S$. In particular,
this implies that $\alpha$ is the shortest
curve on $S$.

We henceforth write $\eps_S$ to be the constant
from Lemma <ref>.

Let $S = S_{1,1}$ or $S = S_{0,4}$,
and let $X\in \mc{T}(S)$. Let
$Y \in \mc{T}(S)$ be obtained from $X$ by
twisting along a simple closed curve $\alpha$
Let $\beta = \bar\alpha$ be the dual curve
of $\alpha$ as in Definition <ref>.
If $l_{\alpha}(X) < \eps_S$, then

§.§ The Thin Part of $\mc{T}(S)$

In this subsection, we show that
$d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is uniformly
bounded if $l_0e^{-t}$ is small.

Let $\eps = \min(\eps_{S_{0,4}},\eps_{S_{1,1}},\log(2))$, and assume
that $l_0 \le \eps$. Furthermore,
assume that $Y\in \mc{T}(S)$ for $S = S_{1,1}$ or
$S = S_{0,4}$. Then $d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is bounded

Using Theorem E of <cit.>, it suffices to check that
$\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}$ is uniformly
bounded, where $\beta$ is the dual curve to $\alpha$. Recall that
$l_{-t} = l_0e^{-t} \pe l_{\alpha}(Y_{-t}^L) = l_{\alpha}(Y_{-t}^R)$.
By the collar lemma, and Proposition 3.1 of <cit.>, we know
that $l_{\beta}(Y_{-t}^L) \te \log(1/l_{-t}) \tg \log(1/\eps)$.
Since $Y_{-t}^R$ is obtained from $Y_{-t}^L$ by twisting along
$\alpha$ for time $\mathrm{Twist}_{\alpha}(Y_{-t}^L, Y_{-t}^R)$, we
can estimate the length ratio of $\beta$ by:

\begin{align*}
\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}
&\le \frac{l_{\beta}(Y_{-t}^L) + l_{\alpha}(Y_{-t}^L)
&\tl 1+ \frac{l_0e^{-t}}{\log(1/\eps)} 4\left(e^{-t}
\log\coth(l_0) + \log\coth(l_0e^{-t})\right)
\end{align*}

Where we used the fact that $l_{\alpha}(Y_{-t}^L)
\le 4 l_0e^s$.

We remark that by our choice of $\eps$, we have that
$\frac{1}{l_0e^{-t}} \ge \log\coth(l_0e^{-t}) > 0$, meaning that:

\begin{align*}
\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}
&\tl 1+ 4\frac{l_0e^{-t}}{\log(1/\eps)}
\left(\frac{e^{-t}}{l_0} + \frac{e^{t}}{l_0}\right)
=1 + \frac{1}{\log(1/\eps)} (e^{-2t} + 1)
\end{align*}

Which is uniformly bounded in $l_0$ and in $t$.

Assume that $\eps \le l_0\le 2$, and that $l_0e^{-t} < \eps$ from
the previous lemma. Then $d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is bounded

As in the previous lemma, we must estimate

\begin{align*}
\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}
&\tl 1+ \frac{l_0e^{-t}}{\log(1/\eps)} 4\left(e^{-t}
\log\coth(l_0) + \log\coth(l_0e^{-t})\right)
\end{align*}

The $\log\coth(l_0)$ term is bounded uniformly,
so it suffices to observe that
$l_0e^{-t} \log\coth(l_0e^{-t})$ is
bounded uniformly.

Assume that $2 \ge l_0$, and that $l_0e^{-t} < \eps$ from the
previous lemmas. Then $d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is bounded

Let $t_0 = \log(l_0)$, and set $s = t-t_0$. Note
that $t_0 \ge \log(2) > 0$ and that $s\in (-\log(\eps),\infty)$
is bounded from below. With this notation,
and the same argumens as before, we have:

\begin{align*}
\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}
&\tl 1+ 4\frac{e^{-s}}{s}
\|\left(e^{-s-t_0} \log\coth(e^{t_0})
\end{align*}

As $t_0$ ranges from $0$ to $\infty$,
we have that $\frac{1}{e^{t_0}} > \log\coth(e^{t_0})> 0$,
and similarly $\frac{1}{e^{-s}} > \log\coth(e^{-s}) > 0$,

\begin{align*}
\frac{l_{\beta}(Y_{-t}^R)}{l_{\beta}(Y_{-t}^L)}
&\tl 1+ 4\frac{e^{-s}}{s}\left(e^{-s}+e^s)\right)
\end{align*}

Which is bounded uniformly in $s$, since
$s$ is bounded from below by $-\log(\eps)$.

Together, the previous three lemmas tell us:

Let $\eps = \min(\eps_{S_{1,1}},\eps_{S_{0,4}},\log(2))$,
and assume that $l_0e^{-t} \le \eps$. Then
$d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is bounded uniformly.

§.§ The Thick Part of $\mc{T}(S)$

In this subsection, we show that
$d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is uniformly bounded
when $l_0e^{-t}$ is large.

In the setting of the above lemmas, if $1 < l_0e^{-t}$, then
$d_{Th}(Y_{-t}^L, Y_{-t}^R)$ is uniformly bounded.

We first treat the case when $S = S_{1,1}$
is the once-punctured torus. We recall
that $l_{\alpha}(Y_{-t}^L) = l_{\alpha}(Y_{-t}^R)
= 2 l_0$, and by Proposition <ref>,
it suffices to verify that for $l_0e^{-t} >1$, we
have that

\begin{align*}
e^{l_0e^{-t}} \mathrm{Twist}_{\alpha}(Y_{-t}^L,Y_{-t}^R)
&= 4e^{l_0e^{-t}} (e^{-t}\log\coth(l_0)
&\le 4e^{l_0e^{-t}} e^{-t}\log\coth(l_0)
+4e^{l_0e^{-t}} \log\coth(l_0e^{-t})
\end{align*}

is bounded uniformly. Note that
$1 < \coth(l_0) < \coth(l_0e^{-t})$,
so it suffices to observe that $e^{2l_0e^{-t}}\log\coth(l_0e^{-t})$
is bounded uniformly in $t$ and $l_0$, as long
as $l_0 e^{-t} > 1$.

The case when $S = S_{0,4}$ is similar; we must verify
that $e^{2l_0e^{-t}}
\mathrm{Twist}_{\alpha}(Y_{-t}^L,Y_{-t}^R)$ is uniformly bounded.

It is a striking coincidence that the twisting
along a simple closed curve is decaying
at least as fast as $e^{-2l_0e^{-t}}$ in the
case of the once-punctured torus and the four-times punctured
sphere. In higher complexity surfaces, this is not
generally the case. In fact, the factor of $\frac12$ is
crucial in Proposition <ref>,
since $e^{(2+\eps)l_0e^{-t}}\log\coth(l_0e^{-t})$
is not uniformly bounded for any $\eps > 0$.

In fact, similar computations can be performed to estimate
$d_{Th}(Y_{-t}^R, Y_{-t}^L)$, giving:

$d_{Th}(Y_{-t}^L, Y_{-t}^R)$ and $d_{Th}(Y_{-t}^R, Y_{-t}^L)$ are
uniformly bounded.

We've seen the cases for when $l_0e^{-t} < \eps$
and when $l_0e^{-t} >1$.
then let $Y_0^R$ (resp. $Y_0^L$) be the point on $Y_{-t}^R$
(resp. $Y_0^L$) for which $l_{\alpha}(Y_0^L) = 1$.

Let $t$ be such that
$\eps \le l_0e^{-t} \le 1$, by the triangle inequality,
\le 2(1-\eps) + d_{Th}(Y_0^R,Y_0^L)$
which is uniformly bounded. A similar
argument can be shown for

§.§ Geodesic Envelopes

Before proving Lemma <ref>, we
first analyze the large-scale structure of
$Env(X,Y)$ for $X, Y \in \mc{T}(S)$, where $S = S_{1,1}$
or $S = S_{0,4}$. Let $\alpha$ be a
simple closed curve in $S$, and let
$Y, Y_{-t}^L$ and $Y_{-t}^R$ be defined as before

The set $In(Y, \alpha)$ is a closed
set. Moreover, in Fenchel-Nielsen
co-ordinates, we can write:

\begin{align*}
In(Y, \alpha)
= \{X\in \mc{T}(S) : \tau_{\alpha}(Y_{-t}^L) \le
\tau_{\alpha}(X)\le \tau_{\alpha}(Y_{-t}^R),
l_{\alpha}(X) = l_{\alpha}(Y)e^{-t} \textrm{for some $t$}\}
\end{align*}

Where $\tau_{\alpha}(Y_{-t}^R)$
and $\tau_{\alpha}(Y_{-t}^L)$
are functions of only
$l_{\alpha}(Y)$ and $t$.

Denote the second set by $A(Y)$.  Note that $Y_{-t}^R$ and
$Y_{-t}^L$ are two smooth paths in $\mc{T}(S)$, which intersect at
$Y$. In particular, they split $\mc{T}(S)$ into four complementary
components (this is because stretch paths do not accumulate inside
$\mc{T}(S)$), one of which is $A(Y)$.

If $X \in A(Y)$, then consider $\mathrm{Stretch}(X,\lambda^L,t)$,
and note that this path must $Y_{-t}^R$ at $X'$.
The contatenation of $\mathrm{Stretch}(X,\lambda^L,t)$
from $X$ to $X'$, and the path $Y_{-t}^R$ from
$X'$ to $Y$ shows that $X \in In(Y,\alpha)$.

[scale = 2]

Conversely, assume that $X \in In(Y,\alpha)$ but not in
$A(Y)$. Without loss of generality,
assume that $\tau_{\alpha}(X) > \tau_{\alpha}(Y_{-t}^R)$.
Let $\gamma(t)$ be the path from
$X$ to $Y$ which maximally stretches $\alpha$ along it.
By Theorem <ref>, it follows that
\ge \tau_{\alpha}(\mathrm{Stretch}(X,\lambda^R,t))$
for any $t$, and so $\gamma(t)$ cannot reach
$Y$ (by construction of $Y_{-t}^R$).

Let $S  =S_{1,1}$ or $S = S_{0,4}$.
For any $X,Y \in \mc{T}(S)$, we have:

* $Env(X,Y)$ is a unique geodesic from $X$ to $Y$ or,

* $Env(X,Y)$ is a combinatorial
quadrilateral whose edges are stretch paths

If we are not in the first case, then
$\Lambda(X,Y)$ must be a simple closed curve, $\alpha$.
Notice that $Env(X,Y) = In(Y,\alpha) \cap
\Outenv(X,\alpha)$. From Theorem <ref>,
we see that $\Outenv(\Lambda(X,Y))$ is a
cone starting at $X$ and bounded by
$\mathrm{Stretch}(X,\lambda^L,t)$ and
$\mathrm{Stretch}(X, \lambda^R,t)$. The intersection
of two cones in $\mc{T}(S)$ is a combinatorial

This characterization allows us to prove
Lemma <ref>

Let $X, Y \in \mc{T}(S)$ be such that
$\Lambda(X,Y) = \alpha$. For any $Z,Z' \in Env(X,Y)$,
if $l_{\alpha}(Z) = l_{\alpha}(Y)$, then
$d_{Th}(Z,Z')$ is bounded uniformly. In particular,
$w(X,Y)$ is uniformly bounded.

We must have that
$\mathrm{Twist}_{\alpha}(Z,Z') \le
\mathrm{Twist}_{\alpha}(Y_{-t}^L,Y_{-t}^R)$, so
using the exact same
computations as in the previous subsection,
$d_{Th}(Z,Z') \le d_{Th}(Y_{-t}^L,Y_{-t}^R)$.
The theorem then follows from
Lemma <ref>