# Variation of holonomy for projective structures and an application to
drilling hyperbolic 3-manifolds

Martin Bridgeman Kenneth Bromberg Research supported by NSF grants
DMS-2005498, DMS-1564410 and the Simons Fellowship 675497Research supported by
NSF grant DMS-1906095.

###### Abstract

We bound the derivative of complex length of a geodesic under variation of the
projective structure on a closed surface in terms of the norm of the
Schwarzian in a neighborhood of the geodesic. One application is to cone-
manifold deformations of acylindrical hyperbolic 3-manifolds.

Dedicated to a great mathematician on the occasion
of his sixtieth birthday: our friend “Of course!" Francois Labourie.

##  1 Introduction

Let $X$ be a Riemann surface or, equivalently, a hyperbolic surface and
$\gamma$ a closed geodesic on $X$. A projective structure $\Sigma$ on $X$
determines a holonomy representation of $\pi_{1}(X)$. If the holonomy of
$\gamma$ is not parabolic then ${\cal L}_{\gamma}(\Sigma)$ is the complex
length of $\gamma$ which is a complex number whose imaginary part is well
defined modulo $2\pi$. We let $P(X)$ be the space of projective structures on
$X$ and $P_{\gamma}(X)$ the subspace where the holonomy of $\gamma$ is not
parabolic or the identity. Then ${\cal L}_{\gamma}$ is a smooth function on
$P_{\gamma}(X)$. The goal of this note is to gain quantitative control of the
derivative of ${\cal L}_{\gamma}$.

We begin by describing a formula for the derivative. We first identify the
universal cover $\tilde{X}$ with the upper half plane ${\mathbb{U}}$
normalized such that the imaginary axis is a lift of $\gamma$. The projective
structure $\Sigma$ determines a developing map
$f\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ which we semi-normalize so that
$f(e^{\ell}z)=e^{{\cal L}_{\gamma}(\Sigma)}f(z)$ where $\ell$ is the length of
$\gamma$ in $X$. We define the holomorphic vector ${\bf
n}=z{\frac{\partial}{\partial z}}$ on $\widehat{{\mathbb{C}}}$. Then the pull
back $f^{*}{\bf n}$ is a holomorphic vector field on ${\mathbb{U}}$ that
descends to a vector field on the annular cover $X_{\gamma}\to X$ associated
to $\gamma$. We also denote this vector field on $X_{\gamma}$ as $f^{*}{\bf
n}$. Note that $f$ is only determined up to post-composition with an element
of $\mathsf{PSL}_{2}(\mathbb{C})$ that fixes $0$ and $\infty$. Since ${\bf n}$
is invariant under any such element we have that $f^{*}{\bf n}$ is well
defined.

The tangent space $T_{\Sigma}P(X)$ is canonically identified with $Q(X)$, the
space of holomorphic quadratic differentials on $X$. The pairing of a
holomorphic vector field with a holomorphic quadratic differential $\phi$ is a
holomorphic 1-form so $f^{*}{\bf n}\cdot\phi$ is a holomorphic (and hence
closed) 1-form on $X_{\gamma}$. Our formula for the derivative of ${\cal
L}_{\gamma}$ is

###### Theorem 1.1

Given $\Sigma\in P_{\gamma}(X)$ with normalized developing map $f$ and
$\phi\in Q(X)\cong T_{\Sigma}(P_{\gamma}(X))$ we have

$d{\cal L}_{\gamma}(\phi)=-\int_{\gamma}f^{*}{\bf n}\cdot\phi.$

In the special case when $\Sigma$ is Fuchsian then ${\bf n}=f^{*}{\bf n}$ and,
as we will see below, the integral can be computed explicitly. In general, to
estimate the the integral we need quantitive control over the difference
between ${\bf n}$ and $f^{*}{\bf n}$. This is the content of our next result.

Before stating our estimates we define some norms. If $\Phi$ is a quadratic
differential on $X$ then $|\Phi|$ is an area form so its ratio with the
hyperbolic area form is a non-negative function. We let $\|\Phi(z)\|$ be this
function. It is a natural pointwise norm of $\Phi$ and we let $\|\Phi\|_{p}$
be the corresponding $L^{p}$-norms with respect to the hyperbolic area form.
We also let $\|\cdot\|$ be the hyperbolic length of vector fields. For all
norms we write $\|\cdot\|_{\gamma}$ to represent the sup norm over the curve
$\gamma$.

With these definitions we can now state our estimate.

###### Theorem 1.2

Let $\Phi$ be a quadratic differential on ${\mathbb{U}}$ and assume that an
$r$-neighborhood (in hyperbolic metric) of the geodesic $\gamma$ given by the
imaginary axis we have $\|\Phi(z)\|\leq K$ with $r<1/2$ and $K/r<1/4$. Then
there exists a locally univalent map
$f\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ such that

  * •

$\Phi=Sf$;

  * •

$f(it)\to 0$ as $t\to 0$ and $f(it)\to\infty$ as $t\to\infty$;

  * •

$\|f^{*}{\bf n}-{\bf n}\|_{\gamma}\leq\frac{9K}{r}$ and $\|f_{*}{\bf n}-{\bf
n}\|_{f\circ\gamma}\leq\frac{9K}{r}$

where $\|\cdot\|_{\gamma}$ is the supremum of the hyperbolic length of the
vector field along the curve $\gamma$ in ${\mathbb{U}}$ and
$\|\cdot\|_{f\circ\gamma}$ is the supremum of the Euclidean length of the
vector field along the curve $f\circ\gamma$ in $\mathbb{C}^{*}$.

The proof Theorem 1.2 is more complicated then one might expect. It involves
the use of Epstein surfaces and estimates in 3-dimensional hyperbolic space.

An elementary consequence of Theorem 1.2 is the following approximation for
$d\mathcal{L}_{\gamma}$.

###### Theorem 1.3

Let $\Sigma$ be a projective structure and $\gamma$ a closed geodesic of
length $\ell$ such that $\|\Phi\|\leq K$ in the $r$-neighborhood of $\gamma$
with $r\leq 1/2$ and $K/r\leq 1/4$. Then

$\left|d{\cal L}_{\gamma}(\phi)+\int_{\gamma}{\bf
n}\cdot\phi\right|\leq\frac{9K\ell}{r}\|\phi\|_{\gamma}.$

Proof: We have by Theorems 1.1 and 1.2

$\displaystyle\left|d{\cal L}_{\gamma}(\phi)+\int_{\gamma}{\bf
n}\cdot\phi\right|$ $\displaystyle\leq$ $\displaystyle\int_{\gamma}|(f^{*}{\bf
n}-{\bf n})\cdot\phi|$ $\displaystyle\leq$ $\displaystyle\|f^{*}{\bf n}-{\bf
n}\|_{\gamma}\cdot\|\phi\|_{\gamma}\int_{\gamma}\|dz\|_{{{\mathbb{H}}^{2}}}$
$\displaystyle=$ $\displaystyle\frac{9K\ell}{r}\|\phi\|_{\gamma}$

where $\|dz\|_{{{\mathbb{H}}^{2}}}$ is hyperbolic line element. $\Box$

#### Application to deformations of hyperbolic manifolds

Let $N$ be an acylindrical hyperbolizable 3-manifold with boundary $S=\partial
N$. Then for any (noded) conformal structure $Y$ in the Weil-Petersson
completion $\overline{{\mathcal{T}}(S)}$ of Teichmüller space there is a
unique geometrically finite hyperbolic structure $M_{Y}$ on $N$ with conformal
boundary $Y$. The hyperbolic structure $M_{Y}$ determines a projective
structure on $Y$ with Schwarzian quadratic differential denoted $\Phi_{Y}$.
One question that naturally arises is if the $L^{2}$-norm of $\Phi_{Y}$ is
small, does this imply that the $L^{\infty}$-norm is also small and therefore
the manifold $M_{Y}$ has almost geodesic convex core boundary. This is not the
case as the $L^{2}$-norm may be small but there are short curves where the
$L^{\infty}$-norm is large. In [BBB] we analysed this problem. We showed that
for $Y\in{\mathcal{T}}(S)$ if the $L^{2}$-norm $\|\Phi_{Y}\|_{2}$ is
sufficiently small then there is a nearby
$\hat{Y}\in\overline{{\mathcal{T}}(S)}$ (in the Weil-Petersson metric) such
that the $L^{\infty}$-norm $\|\Phi_{\hat{Y}}\|_{\infty}$ is small. This
$\hat{Y}$ will generally be a noded surface pinched along curves where the
$L^{\infty}$-norm in $Y$ was not small.

The bounds in [BBB] for nearby and small were linear in
$\|\Phi_{Y}\|_{2}^{\frac{1}{2n(S)+3}}$ where $n(S)$ is the maximum number of
disjoint geodesics, in particular for $S$ closed connected of genus $g$ then
$n(S)=3g-3$. Our application is to use our variation bound in Theorem 1.3 to
significantly improve this estimate and replace the power $\frac{1}{2n(S)+3}$
with the constant power $\frac{1}{2}$. Thus the bounds become independent of
the topology. Namely we prove:

###### Theorem 1.4

There exists $K,C>0$ such that the following holds; Let $N$ be an acylindrical
hyperbolizable manifold with boundary $S=\partial N$ and
$Y\in{\mathcal{T}}(S)$ with $\|\Phi_{Y}\|_{2}\leq K$. Then there exists a
$\hat{Y}\in\overline{{\mathcal{T}}(S)}$ with

  1. 1.

$d_{\rm WP}(Y,\hat{Y})\leq C\sqrt{\|\Phi_{Y}\|_{2}};$

  2. 2.

$\|\Phi_{\hat{Y}}\|_{\infty}\leq C\sqrt{\|\Phi_{Y}\|_{2}}.$

Our proof also holds for relative acylindrical 3-manifolds as in [BBB] but for
simplicity we will restrict to the acylindrical case.

##  2 Variation of holonomy

#### Preliminaries

We consider $X$ a Riemann surface structure on a surface $S$ and $P(X)$ the
space of projective structures on $X$. Associated to $\Sigma\in P(X)$ is the
holonomy representation
$\rho\in\operatorname{Hom}(\pi_{1}(S),\mathsf{PSL}_{2}(\mathbb{C}))$ (unique
up to conjugacy). The space $P(X)$ can be parametrized by $Q(X)$ the space of
holomorphic quadratic differentials by taking the Schwarzian derivative of its
developing map $f:{\mathbb{U}}\rightarrow\hat{\mathbb{C}}$. As $Q(X)$ is a
vector space, it is its own tangent space, and a variation of $\Sigma$ is
given by a quadratic differential $\phi\in Q(X)$. Throughout the paper we will
use $\phi$ to denote deformations of projective structures and $\Phi$ (or
$\Phi_{t}$) to denote the Schwarzian of projective structures.

The advantage of the Schwarzian $\Phi$ is that is uniquely determined by
$\Sigma$ while the holonomy representation $\rho$ and developing map $f$ are
not. Given a smooth 1-parameter family of projective structures $\Sigma_{t}$
we get a smooth family $\Phi_{t}\in Q(X)$ of Schwarzians. While the developing
maps $f_{t}\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ are not uniquely
determined we can choose them to vary smoothly and it will be convenient to do
this after making some normalizations.

We will be interested in the complex length of an element
$\gamma\in\pi_{1}(S)$ that represents a closed geodesic for the hyperbolic
structure on $X$. After fixing $\gamma$, we identify ${\mathbb{U}}$ with
$\tilde{X}$ so that the deck action of $\gamma$ on ${\mathbb{U}}$ is given by
$\gamma(z)=e^{\ell}z$ where $\ell$ is the length of the geodesic
representative of $\gamma$. Under the covering map ${\mathbb{U}}\to X$ the
imaginary axis is taken to this geodesic. Our second normalization is to
choose the developing map $f$ and corresponding holonomy representation $\rho$
such that $\rho(\gamma)(z)=e^{\cal L}z$. Then ${\cal L}$ is the complex length
${\cal L}_{\gamma}(\Sigma)$. Note that this is only well defined modulo $2\pi
i$.

Extending these normalizations to the 1-parameter family $\Sigma_{t}$ we get a
smooth family of developing maps $f_{t}$ and holonomy representations with
$\rho(\gamma)(z)=e^{{\cal L}_{t}}z$. Note that while ${\cal L}$ is only well
defined modulo $2\pi i$, after this choice is made there is a unique choice of
${\cal L}_{t}$ that makes the path continuous. In particular the time zero
derivative $\dot{\cal L}$ of ${\cal L}_{t}$ is well defined.

We also let $v$ be the vector field on ${\mathbb{U}}$ such that $f_{*}v(z)$ is
the tangent vector of the path $f_{t}(z)$ at $t=0$. The vector field $v$ is
not equivariant under the deck action but a standard computation gives

$v-\beta_{*}v=f^{*}\dot{\rho}(\beta)$

for all $\beta\in\pi_{1}(S)$ where
$\dot{\rho}(\beta)\in\mathfrak{sl}_{2}\mathbb{C}$ is the time zero of the
$\rho_{t}(\beta)$. As $\rho_{t}(\gamma)(z)=e^{{\cal L}_{t}}z$ we have that
$\dot{\rho}(\gamma)=\dot{\cal L}z{\frac{\partial}{\partial z}}$ so

$v-\gamma_{*}v=f^{*}\left(\dot{\cal L}z{\frac{\partial}{\partial z}}\right).$
(2.1)

If $U$ is an open neighborhood in ${\mathbb{U}}$ where $f_{t}$ is injective
then $(U,f_{t})$ is a projective chart for $\Sigma_{t}$. In the chart $(U,f)$
the vector field $v$ is represented as $g{\frac{\partial}{\partial z}}$ where
$g$ is a holomorphic function. In this chart the Schwarzian derivative of $v$
is $g_{zzz}dz^{2}$. This is a quadratic differential and a computation gives
that it is $\phi$ the time zero derivative of the path $\Phi_{t}$.

#### Model deformations

We are now ready to prove Theorem 1.1. Here’s the outline:

  * •

We first construct a model deformation $\phi_{\lambda}$ on the projective
structure on the annulus $X_{\gamma}$.

  * •

The model deformation will be very explicit so that we can directly calculate
the line integral of $f^{*}{\bf n}\cdot\phi_{\lambda}$ over $\gamma$.

  * •

We then find a specific $\lambda$ so that the 1-form $f^{*}{\bf
n}\cdot\phi-f^{*}{\bf n}\cdot\phi_{\lambda}$ is exact. Then the line integrals
over both 1-forms will be equal so our previous calculation will give the
theorem.

A minor issue with this outline is that the model deformation will only be
defined on a sub-annulus of $X_{\gamma}$. We begin with this difficulty.

The set $f^{-1}(\\{0,\infty\\})$ will be discrete in $\widehat{{\mathbb{C}}}$
and $\gamma$-invariant so it will descend to a discrete set in $X_{\gamma}$.
Let $\gamma^{\prime}$ be a smooth curve, homotopic to $\gamma$, that misses
this set and let $A$ be an annular neighborhood of $\gamma^{\prime}$ that is
also disjoint from this set. Then $\tilde{A}$, the pre-image of $A$ in
${\mathbb{U}}$, we also be disjoint from $f^{-1}(\\{0,\infty\\})$. As
$\tilde{A}$ is simply connected we can choose a well defined function $\log f$
on $\tilde{A}$ and define the vector field $v_{\lambda}$ on $\tilde{A}$ by
$f_{*}v(z)=\lambda f(z)\log f(z){\frac{\partial}{\partial z}}$. We let
$\phi_{\lambda}$ be the Schwarzian of $v_{\lambda}$ and differentiating
$\lambda z\log z$ three times we see that
$\phi_{\lambda}=f^{*}\left(-\frac{\lambda}{z^{2}}dz^{2}\right)$. Then
$\phi_{\lambda}$ is our model deformation.

Note the choice of $\log$ defines ${\cal L}$ as a complex number rather than
just a number modulo $2\pi$. We will use this in the rest of the proof. Next
we compute the line integral of $f^{*}{\bf n}\cdot\phi_{\lambda}$ over
$\gamma^{\prime}$:

###### Lemma 2.1

$\int_{\gamma^{\prime}}f^{*}{\bf n}\cdot\phi_{\lambda}=-\lambda{\cal L}$

Proof: Let $\sigma\colon[0,1]\to{\mathbb{U}}$ be a smooth path that projects
to $\gamma^{\prime}$ in $X_{\gamma}$. In particular
$\sigma(1)=\gamma(\sigma(0))$ and $f(\sigma(1))=f(\gamma(\sigma(0))=e^{{\cal
L}}f(\sigma(0))$. Then

$\displaystyle\int_{\gamma^{\prime}}f^{*}{\bf n}\cdot\phi_{\lambda}$
$\displaystyle=$
$\displaystyle\int_{\sigma}f^{*}\left(z{\frac{\partial}{\partial
z}}\cdot-\frac{\lambda}{z^{2}}dz^{2}\right)$ $\displaystyle=$
$\displaystyle-\lambda\int_{f\circ\sigma}\frac{dz}{z}$ $\displaystyle=$
$\displaystyle-\lambda(\log f(\sigma(1))-\log f(\sigma(0))$ $\displaystyle=$
$\displaystyle-\lambda{\cal L}.$

$\Box$

Next we give a criteria for the form $f^{*}{\bf n}\cdot\phi$ to be exact.

###### Lemma 2.2

Let $v$ be a $\gamma$-equivariant holomorphic vector field on $\tilde{A}$ with
Schwarzian $\phi$. Then $f^{*}{\bf n}\cdot\phi$ is exact on $A$.

Proof: The vector field $v$ is a section of the holomorphic tangent bundle
$T_{\mathbb{C}}X_{\gamma}$ so $v_{z}$ is a function, $v_{zz}$ is a 1-form,
$v_{zzz}=\phi$ is a quadratic differential. Therefore $f^{*}{\bf n}\cdot
v_{zz}-v_{z}$ is a holomorphic function on $X_{\gamma}$ and we’ll show that it
is a primitive of $f^{*}{\bf n}\cdot\phi$.

We can do this calculation in a chart. Namely choose an open neighborhood $U$
in ${\mathbb{U}}$ such $f$ is injective on $U$. Then as above $(U,f)$ is a
chart for the projective structure and in this chart $v$ has the form
$g{\frac{\partial}{\partial z}}$ where $g$ is a holomorphic function on $f(U)$
and $\phi$ is $g_{zzz}dz^{2}$. It follows that on this chart $v_{z}$ is
$g_{z}$, $v_{zz}$ is $g_{zz}dz$ and $\phi=v_{zzz}$ is $g_{zzz}dz^{2}$. As the
derivative of $zg_{zz}-g_{z}$ is $g_{zzz}$ this shows that $f^{*}{\bf n}\cdot
v_{zz}-v_{z}$ is a primitive for $f^{*}{\bf n}\cdot\phi$ as claimed. $\Box$

Proof of Theorem 1.1: Since $f^{*}{\bf n}\cdot\phi$ is holomorphic it is
closed and we have

$\int_{\gamma}f^{*}{\bf n}\cdot\phi=\int_{\gamma^{\prime}}f^{*}{\bf
n}\cdot\phi.$

Next we calculate to see that

$v_{\lambda}-\gamma_{*}v_{\lambda}=f^{*}\left(\lambda{\cal
L}z{\frac{\partial}{\partial z}}\right)$

so if $\lambda=\dot{\cal L}/{\cal L}$ then by (2.1) we have that
$v-v_{\dot{\cal L}/{\cal L}}$ is $\gamma$-invariant. Therefore by Lemma 2.2
the 1-form $f^{*}{\bf n}\cdot(\phi-\phi_{\dot{\cal L}/{\cal L}})$ is exact and

$\int_{\gamma^{\prime}}f^{*}{\bf n}\cdot\phi=\int_{\gamma^{\prime}}f^{*}{\bf
n}\cdot\phi_{\dot{\cal L}/{\cal L}}=-\dot{\cal L}$

where the last equality comes from Lemma 2.1. Combining the first and last
equality gives the theorem. $\Box$

We conclude this section with a simple formula for the line integral when the
projective structure is Fuchsian. For this, rather than represent the annulus
as a quotient of the upper half plane ${\mathbb{U}}$ for this calculation it
is convenient to represent the annulus as an explicit subset of $\mathbb{C}$.
Namely, let

$A_{\ell}=\left\\{z\in\mathbb{C}\ |\
e^{\frac{-\pi^{2}}{\ell}}<|z|<e^{\frac{\pi^{2}}{\ell}}\right\\}.$

This annulus is conformally equivalent $X_{\gamma}$ when
$\ell=\ell_{\gamma}(X)$ and the circle $|z|=1$ is the closed geodesic of
length $\ell$ in $A_{\ell}$. For this representation of the annulus the vector
field ${\bf n}$ is written as ${\bf n}=\frac{2\pi
i}{\ell}z{\frac{\partial}{\partial z}}$ and we observe that ${\bf n}$ extends
to a holomorphic vector field on all of $\widehat{{\mathbb{C}}}$ that is $0$
at $z=0$ and $z=\infty$.

To decompose the quadratic differential $\phi$ on $A_{\ell}$ we let $D^{+}$
and $D^{-}$ be the disks in $\widehat{{\mathbb{C}}}$ with
$|z|>e^{\frac{-\pi^{2}}{\ell}}$ and $|z|<e^{\frac{\pi^{2}}{\ell}}$,
respectively. We recall that any holomorphic function $\psi$ on $A_{\ell}$ can
be written as $\psi_{+}+\psi_{0}+\psi_{-}$ where $\psi_{+}$ and $\psi_{-}$
extend to holomorphic functions on $D^{+}$ and $D^{-}$ that are zero at
$z=\infty$ and $z=0$, respectively, and $\psi_{0}$ is constant. Then $\phi$
can be written as $\phi=\frac{\psi}{z^{2}}dz^{2}$ so the decomposition of
$\psi$ gives

$\phi=\phi_{+}+\phi_{0}+\phi_{-}$ (2.2)

where $\phi_{+}$ and $\phi_{-}$ extend to holomorphic quadratic differentials
on $D^{+}$ and $D^{-}$ with simple poles at $z=\infty$ and $z=0$,
respectively, and $\phi_{0}$ is a constant multiple of $\frac{dz^{2}}{z^{2}}$.

###### Lemma 2.3

Let $\phi$ be a holomorphic quadratic differential on $X_{\gamma}$. Then

$\int_{\gamma}{\bf n}\cdot\phi=\int_{\gamma}{\bf n}\cdot\phi_{0}$

and

$\left|\int_{\gamma}{\bf n}\cdot\phi\right|=\ell\|\phi_{0}\|_{\infty}$

Proof: In the annulus $A_{\gamma}$ the line integral along $\gamma$ is the
line integral over the circle $|z|=1$.

Note that ${\bf n}\cdot\phi_{\pm}$ extends to a 1-form on $D^{\pm}$ so

$\int_{|z|=1}{\bf n}\cdot\phi_{\pm}=0$

since by Cauchy’s Theorem for any line integral of a closed curve over a
holomorphic 1-form on a simply connected region is zero. Therefore

$\int_{|z|=1}{\bf n}\cdot\phi=\int_{|z|=1}{\bf n}\cdot\phi_{+}+{\bf
n}\cdot\phi_{0}+{\bf n}\cdot\phi_{-}=\int_{|z|=1}{\bf n}\cdot\phi_{0}.$

We have the covering map ${\mathbb{U}}\rightarrow A_{\ell}$ given by
$z\rightarrow z^{\frac{2\pi i}{\ell}}$. Thus pushing forward the hyperbolic
metric on ${\mathbb{U}}$ we have that the hyperbolic metric $g_{A_{\ell}}$ on
$A_{\ell}$ is

$g_{A_{\ell}}=\left(\frac{\ell}{2\pi|z|\cos\left(\frac{\ell}{2\pi}\log|z|\right)}\right)^{2}|dz|^{2}.$

As $\phi_{0}=\frac{\psi_{0}}{z^{2}}dz^{2}$ we therefore have

$\|\phi_{0}\|_{\infty}=\sup_{z}\frac{|\phi_{0}(z)|}{g_{A_{\ell}}(z)}=\frac{4\pi^{2}|\psi_{0}|}{\ell^{2}}.$

Thus

$\int_{|z|=1}{\bf n}\cdot\phi_{0}=\frac{2\pi
i\psi_{0}}{\ell}\int_{|z|=1}\frac{dz}{z}=\frac{-4\pi^{2}\psi_{0}}{\ell}$

and

$\left|\int_{|z|=1}{\bf n}\cdot\phi_{0}\right|=\ell\|\phi_{0}\|_{\infty}.$

$\Box$

##  3 Derivative bounds on univalent maps

Next we prove Theorem 1.2. The proof here is considerably more involved. We
begin by reducing it to a more explicit statement.

###### Theorem 3.1

Let $\Phi$ be a quadratic differential on ${\mathbb{U}}$ and assume that on an
$r$-neighborhood of the imaginary axis (in the hyperbolic metric) we have
$\|\Phi(z)\|\leq K$ with $r<1/2$ and $K/r<1/4$. Then there exists a locally
univalent map $f\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ such that

  * •

$Sf=\Phi$;

  * •

$f(it)\to 0$ as $t\to 0$ and $f(it)\to\infty$ as $t\to\infty$;

  * •

$f(i)=i$, $|f^{\prime}(i)-1|\leq\frac{9K}{r}$ and
$|\frac{1}{f^{\prime}(i)}-1|\leq\frac{9K}{r}$.

We now see how this implies Theorem 1.3:

Proof of Theorem 1.2 assuming Theorem 3.1: Given $ie^{t}\in{\mathbb{U}}$ with
$t\in\mathbb{R}$ we let $\Phi^{t}$ be the pull back of $\Phi$ by the isometry
$\gamma_{t}(z)=e^{t}z$. As $\gamma_{t}$ preserves the imaginary axis we still
have that $\|\Phi^{t}(z)\|\leq K$ for $z$ in an $r$-neighborhood of the axis.
Therefore by Theorem 3.1 we have a locally univalent map
$f^{t}\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ satisfying the 3 bullets.

We now let $f=f^{0}$. Since $Sf^{t}=\Phi^{t}$ we have that
$S(f^{t}\circ\gamma_{-t})=\Phi$. Since $f$ and $f^{t}\circ\gamma_{-t}$ have
the same Schwarzian they differ by post-composition with an element of
$\mathsf{PSL}_{2}(\mathbb{C})$. Since these two maps have the same behavior as
$it\to 0$ and $it\to\infty$ this element of $\mathsf{PSL}_{2}(\mathbb{C})$
must be of the form $z\mapsto e^{\lambda}z$. We note that ${\bf n}$ is
invariant under these maps (which includes the maps $\gamma_{t}$). In
particular this implies that

$\left|\frac{1}{(f^{t})^{\prime}(i)}i-i\right|=\|(f^{t})^{*}({\bf n}(i))-{\bf
n}(i)\|_{g_{{{\mathbb{H}}^{2}}}}=\|f^{*}({\bf n}(f(it)))-{\bf
n}(it)\|_{g_{{{\mathbb{H}}^{2}}}}$

so the inequality (3) gives

$\|f^{*}({\bf n}(f(it))-{\bf n}(it)\|_{{{\mathbb{H}}^{2}}}\leq\frac{9K}{r}.$

Similarly if $g_{euc}$ is the Euclidean metric on $\mathbb{C}^{*}$ then

$|(f^{t})^{\prime}(i)i-i|=|(f^{t})_{*}({\bf n}(i))-{\bf n}(i)|=\|f_{*}({\bf
n}(it))-{\bf n}(f(it))\|_{g_{euc}}$

$\Box$

Here is a brief outline of the proof of Theorem 3.1:

  * •

Given the quadratic differential $\Phi$ on ${\mathbb{U}}$ there is an
immersion $f_{0}\colon{\mathbb{U}}\to{{\mathbb{H}}^{3}}$ such that composition
of $f_{0}$ with the hyperbolic Gauss map is a map
$f\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ with $Sf=\Phi$.

  * •

The surface $f_{0}\colon{\mathbb{U}}\to{{\mathbb{H}}^{3}}$ is the Epstein
surface for $\Phi$. In [Eps], Epstein gives formulas for the metric and shape
operator of this surface in terms of the hyperbolic metric on ${\mathbb{U}}$
and $\Phi$.

  * •

We will use Epstein’s formulas to show that the curve $t\mapsto f_{0}(ie^{t})$
is nearly unit speed and has small curvature. This will imply that $f_{0}(it)$
limits to distinct points as $t\to 0$ and $t\to\infty$. We then normalize so
that these limiting points are $0$ and $\infty$.

  * •

The proof is then completed by a calculation of the hyperbolic Gauss map using
the Minkowski model for hyperbolic space.

Before starting the proof of Theorem 3.1 we review the necessary facts about
Epstein surfaces. These surfaces are defined for any conformal metric on
${\mathbb{U}}$ and a holomorphic quadratic differential $\Phi$. Here we will
restrict to the hyperbolic metric.

The projective second fundamental form for the hyperbolic metric
$g_{{\mathbb{H}}^{2}}$ is

${\rm II}=\Phi+\bar{\Phi}+g_{{\mathbb{H}}^{2}}.$

The projective shape operator is given by the formula

$g_{{\mathbb{H}}^{2}}(\hat{B}v,w)={\rm II}(v,w).$

We then define a dual pair by

$g=\frac{1}{4}\left({\operatorname{id}}+\hat{B}\right)^{*}g_{{{\mathbb{H}}^{2}}}\qquad
B=\left({\operatorname{id}}+\hat{B}\right)^{-1}\left({\operatorname{id}}-\hat{B}\right).$
(3.3)

By inverting it follows also that

$g_{{{\mathbb{H}}^{2}}}=\left({\operatorname{id}}+B\right)^{*}g\qquad\hat{B}=\left({\operatorname{id}}+B\right)^{-1}\left({\operatorname{id}}-B\right).$
(3.4)

We will also use two maps on the unit tangent bundle
$T^{1}{{\mathbb{H}}^{3}}$. First we have the projection $\pi\colon
T^{1}{{\mathbb{H}}^{3}}\to{{\mathbb{H}}^{3}}$. We also have the hyperbolic
Gauss map $\mathfrak{g}:T^{1}{{\mathbb{H}}^{3}}\to\widehat{{\mathbb{C}}}$
which takes each unit tangent vector to the limit of the geodesic ray starting
from the vector.

######  Theorem 3.2 (Epstein, [Eps])

Given any holomorphic quadratic differential $\Phi$ on ${\mathbb{U}}$ there
exists a smooth map $\hat{f}\colon{\mathbb{U}}\to T^{1}{{\mathbb{H}}^{3}}$
with

  1. 1.

Where $B$ is non-singular the map $f_{0}=\pi\circ\hat{f}$ is smooth,
$g=f_{0}^{*}g_{{{\mathbb{H}}^{3}}}$ and $B$ is the shape operator for the
immersed surface.

  2. 2.

The map $f=\mathfrak{g}\circ\hat{f}$ is locally univalent with $Sf=\Phi$.

  3. 3.

The eigenvalues of $\hat{B}$ are $1\pm 2\|\Phi(z)\|$.

###  3.1 Geodesic curvature

The path $\gamma(t)=ie^{t}$ is a unit speed parameterization of the imaginary
axis. We will now begin the computation of the curvature of
$\alpha=f_{0}\circ\gamma$.

We begin with a few preliminaries. Let $\hat{B}_{0}$ be the traceless part of
$\hat{B}$. Then ${\rm II}_{0}(X,Y)=g_{{{\mathbb{H}}^{2}}}(\hat{B}_{0}X,Y)$ is
the traceless part of ${\rm II}$. We then have
$\hat{B}={\operatorname{id}}+\hat{B}_{0}$ and ${\rm II}_{0}=\Phi+\bar{\Phi}$.
We also need to relate the Riemannian connection $\hat{\nabla}$ for
$g_{{{\mathbb{H}}^{2}}}$ and the Riemannian connection $\nabla$ for $g$. A
simple calculation (see [KS, Lemma 5.2]) gives

$({\operatorname{id}}+\hat{B})\nabla_{X}Y=\hat{\nabla}_{X}(({\operatorname{id}}+\hat{B})Y).$

###### Lemma 3.3

Define the holomorphic function $h$ by

$h=\Phi({\bf n},{\bf n}).$

Then

$\left({\operatorname{id}}+\hat{B}\right)\nabla_{\dot{\gamma}}\dot{\gamma}=4\operatorname{Re}dh(\dot{\gamma})\bar{\bf
n}\quad\mbox{and}\quad\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}=|dh(\dot{\gamma})|.$

Proof: We work in the complexified tangent bundle of ${\mathbb{U}}$. We want
to compute $\nabla_{\dot{\gamma}}{\dot{\gamma}}$ where $\dot{\gamma}$ is the
tangent vector to the path $\gamma$. As
$\left({\operatorname{id}}+\hat{B}\right)\nabla=\hat{\nabla}\left({\operatorname{id}}+\hat{B}\right)$
then

$\displaystyle\left({\operatorname{id}}+\hat{B}\right)\nabla_{\dot{\gamma}}\dot{\gamma}$
$\displaystyle=$
$\displaystyle\hat{\nabla}_{\dot{\gamma}}\left({\operatorname{id}}+\hat{B}\right)\dot{\gamma}$
$\displaystyle=$ $\displaystyle
2\hat{\nabla}_{\dot{\gamma}}\dot{\gamma}+\hat{\nabla}_{\dot{\gamma}}\hat{B}_{0}\dot{\gamma}$
$\displaystyle=$
$\displaystyle\hat{\nabla}_{\dot{\gamma}}\hat{B}_{0}\dot{\gamma}$

since $\hat{\nabla}_{\dot{\gamma}}\dot{\gamma}=0$ as $\gamma$ is a geodesic in
$g_{{{\mathbb{H}}^{2}}}$.

To compute this term we note $\dot{\gamma}={\bf n}+\bar{\bf n}$. Therefore,
since ${\rm II}_{0}$ is symmetric we have

$\displaystyle g_{{{\mathbb{H}}^{2}}}\left(\hat{B}_{0}\dot{\gamma},{\bf
n}\right)$ $\displaystyle=$ $\displaystyle{\rm II}_{0}({\bf n}+\bar{\bf
n},{\bf n})$ $\displaystyle=$ $\displaystyle\Phi({\bf n},{\bf n})=h.$

Using the compatibility of the metric with the Riemannian connection and that
${\bf n}$ is parallel along $\gamma$ we have

$dh(\dot{\gamma})=g_{{{\mathbb{H}}^{2}}}\left(\hat{\nabla}_{\dot{\gamma}}\left(\hat{B}_{0}\dot{\gamma}\right),{\bf
n}\right).$

A similar calculation gives

$d\bar{h}(\dot{\gamma})=g_{{{\mathbb{H}}^{2}}}\left(\hat{\nabla}_{\dot{\gamma}}\left(\hat{B}_{0}\dot{\gamma}\right),\bar{\bf
n}\right).$

Let

$v=2dh(\dot{\gamma})\bar{\bf n}+2{d\bar{h}(\dot{\gamma})}{\bf n}.$

Note that we are extending $g_{{\mathbb{H}}^{2}}$ to be $\mathbb{C}$-linear on
$T{\mathbb{U}}\otimes\mathbb{C}$ and therefore

$g_{{{\mathbb{H}}^{2}}}({\bf n},{\bf n})=g_{{{\mathbb{H}}^{2}}}(\bar{\bf
n},\bar{\bf n})=0\quad\mbox{and}\quad g_{{\mathbb{H}}^{2}}({\bf n},\bar{\bf
n})=1/2.$

It follows that

$g_{{{\mathbb{H}}^{2}}}(v,{\bf n})=dh(\dot{\gamma})\quad\mbox{and}\quad
g_{{{\mathbb{H}}^{2}}}(v,\bar{\bf n})=d\bar{h}(\dot{\gamma}).$

As ${\bf n}$ and $\bar{\bf n}$ span the tangent space this implies

$\left({\operatorname{id}}+\hat{B}\right)\nabla_{\dot{\gamma}}\dot{\gamma}=\hat{\nabla}_{\dot{\gamma}}\hat{B}_{0}(\dot{\gamma})=v.$

Since $\|v\|_{g_{{{\mathbb{H}}^{2}}}}=2|dh|$ this gives

$\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}=\frac{1}{2}\|\left({\operatorname{id}}+\hat{B}\right)\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g_{{{\mathbb{H}}^{2}}}}=\frac{1}{2}\|\hat{\nabla}_{\dot{\gamma}}\hat{B}_{0}\dot{\gamma}\|_{g_{{{\mathbb{H}}^{2}}}}=|dh(\dot{\gamma})|.$

$\Box$

Next we use the Cauchy integral formula to bound $dh$.

###### Lemma 3.4

Assume that $r\leq 1/2$ and $K<1$. If $\|\Phi(z)\|\leq K$ for $z$ in the
$r$-neighborhood of $\gamma$ then the geodesic curvature $\kappa_{\gamma}$ of
$\gamma$ in the metric $g$ satisfies

$\kappa_{\gamma}\leq\frac{5K}{4r(1-K)^{2}}.$

Proof: We have

$\kappa_{\gamma}=\frac{\|(\nabla_{\dot{\gamma}}\dot{\gamma})^{\perp}\|_{g}}{\|\dot{\gamma}\|_{g}^{2}}\leq\frac{\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}}{\|\dot{\gamma}\|_{g}^{2}}$

where $(\nabla_{\dot{\gamma}}\dot{\gamma})^{\perp}$ is the component of
$\nabla_{\dot{\gamma}}\dot{\gamma}$ perpendicular to $\dot{\gamma}$.

To bound $\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}$ we will work in the disk
model for ${{\mathbb{H}}^{2}}$ with $\gamma$ the geodesic following the real
axis and we will bound the curvature at the origin. With this normalization we
have ${\bf n}=\frac{1-z^{2}}{2}{\frac{\partial}{\partial z}}$ and therefore

$h(z)=\frac{\Phi(z)(1-z^{2})^{2}}{4}.$

(Here we are not distinguishing between $\Phi$ as quadratic differential and
$\Phi$ as holomorphic function.) At zero,
$\dot{\gamma}=\frac{1}{2}{\partial_{x}}$ so by Lemma 3.3 we have

$\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}=|dh(\dot{\gamma})|=|h_{x}(0)|/2=|h_{z}(0)|/2$

where the last equality uses that $h$ is holomorphic.

We now bound $|h_{z}(0)|$ using the Cauchy integral formula. Given that
$\|\Phi(z)\|\leq K$ in the $r$-neighborhood of the origin (in the hyperbolic
metric) we have

$|h(z)|\leq\frac{1}{4}|1-z^{2}|^{2}|\Phi(z)|\leq\frac{|1-z^{2}|^{2}}{(1-|z|^{2})^{2}}K$

when $|z|\leq R=\tanh(r/2)$. Therefore

$\displaystyle|h_{z}(0)|$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\left|\int_{|z|=R}\frac{h(z)}{z^{2}}dz\right|$
$\displaystyle\leq$ $\displaystyle\frac{KR}{2\pi
R^{2}(1-R^{2})^{2}}\int_{0}^{2\pi}\left|1-R^{2}e^{2i\theta}\right|^{2}d\theta$
$\displaystyle=$ $\displaystyle\frac{K}{2\pi
R(1-R^{2})^{2}}\int_{0}^{2\pi}\left(1-2R^{2}\cos(2\theta)+R^{4}\right)d\theta$
$\displaystyle=$ $\displaystyle\frac{K(1+R^{4})}{R(1-R^{2})^{2}}.$

If $r\leq 1/2$ then, since $\tanh(r/2)\leq r/2$, we have $R\leq 1/4$. As the
derivative of $\tanh^{-1}R$ is $1/(1-R^{2})$ when $R\leq 1/4$ we have
$\tanh^{-1}R\leq 16R/15$ and $\tanh(r/2)\geq 15r/32$. Combining this with our
estimate on $|h_{z}(0)|$ we have

$|h_{z}(0)|\leq\frac{32}{15r}\cdot\frac{\left(1+(1/4)^{4}\right)K}{\left(1-(1/4)^{2}\right)^{2}}\leq\frac{5K}{2r}$

and

$\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}\leq\frac{5K}{4r}.$

We now obtain a lower bound on $\|\dot{\gamma}\|_{g}$. For this we recall that
by Theorem 3.2 the eigenvalues of $\hat{B}$ at $z\in{{\mathbb{H}}^{2}}$ are
$1\pm 2\|\Phi(z)\|$. If $z\in\gamma$ then $\|\Phi(z)\|\leq K$ giving

$\|\dot{\gamma}\|_{g}=\frac{1}{2}\|\left({\operatorname{id}}+\hat{B}\right)\dot{\gamma}\|_{g_{{{\mathbb{H}}^{2}}}}\geq
1-K.$

Therefore

$\kappa_{\gamma}\leq\frac{\|\nabla_{\dot{\gamma}}\dot{\gamma}\|_{g}}{\|\dot{\gamma}\|_{g}^{2}}<\frac{5K}{4r(1-K)^{2}}.$

$\Box$

We can now bound the curvature of $\alpha$ in ${{\mathbb{H}}^{3}}$.

###### Lemma 3.5

Assume that $r\leq 1/2$ and $K<1$. Then if $\|Sf(z)\|\leq K$ for all $z$ in
the $r$-neighborhood of $\gamma$ then the geodesic curvature $\kappa_{\alpha}$
of $\alpha=f_{0}\circ\gamma$ in ${{\mathbb{H}}^{3}}$ satisfies

$\kappa_{\alpha}\leq\frac{3K}{2r(1-K)^{2}}.$

Proof: To bound the geodesic curvature of the curve $\alpha=f_{0}\circ\gamma$
we again need to bound the covariant derivative
$\bar{\nabla}_{\dot{\alpha}}\dot{\alpha}$. This will have a component tangent
to the immersed surface which is
$(f_{0})_{*}\nabla_{\dot{\gamma}}\dot{\gamma}$ plus an orthogonal component
whose length is $\|B(\dot{\gamma})\|_{g}$. This last norm is bounded by the
product of the maximal eigenvalue of $B$ and $\|\dot{\gamma}\|_{g}$. As
$B=({\operatorname{id}}+\hat{B})^{-1}({\operatorname{id}}-\hat{B})$ then
eigenvalues of $B$ at a point $f(z)$ in the immersed surface are
$-\frac{\|\Phi(z)\|}{\|\Phi(z)\|\pm 1}$. If $z$ is in $\gamma$ we have
$\|\Phi(z)\|\leq K$ so the maximum eigenvalue is bounded above by $K/(1-K)$.
It follows that

$\kappa_{\alpha}\leq\sqrt{\kappa_{\gamma}^{2}+\left(\frac{K}{1-K}\right)^{2}}\leq\frac{K}{r(1-K)^{2}}\sqrt{(5/4)^{2}+(1/2)^{2}}\leq\frac{3K}{2r(1-K)^{2}}.$

$\Box$

###  3.2 Curves with geodesic curvature $\kappa_{g}<1$

It is well know that curves in ${{\mathbb{H}}^{n}}$ with geodesic curvature
$\leq\kappa<1$ are quasi-geodesics. In particular, a bi-infinite path with
curvature $\leq\kappa$ will limit to distinct endpoints. We need a
quantatative version of this statement.

###### Lemma 3.6

Let $\alpha\colon\mathbb{R}\to{{\mathbb{H}}^{3}}$ be smooth, bi-infinite curve
with curvature at most $\kappa<1$. Normalize $\alpha$ in the upper half space
model of ${{\mathbb{H}}^{3}}$ so that $\alpha(0)=(0,0,1)$ and
$\alpha^{\prime}(0)=(0,0,\lambda)$ with $\lambda>0$. Then there are distinct
points $z_{-},z_{+}\in\widehat{{\mathbb{C}}}$ with
$\underset{t\to\pm\infty}{\lim}\alpha(t)=z_{\pm}$ with

$|z_{\pm}|^{\mp
1}\leq\frac{1}{\kappa}\left(1-\sqrt{1-\kappa^{2}}\right)\leq\kappa.$

Let $P_{t}$ be the hyperbolic plane orthogonal to $\alpha$ at $\alpha(t)$.
Then $\partial P_{t}\to z_{\pm}$ at $t\to\pm\infty$.

Proof: Let $H_{\delta}$ be a convex region in ${{\mathbb{H}}^{3}}$ bounded by
a plane of constant curvature $\delta<1$. If $\alpha^{\prime}(0)$ is tangent
to $\partial H_{\delta}$ and $\delta>\kappa$ then in neighborhood of zero the
only intersection of $\alpha$ with $H_{\delta}$ will be at $\alpha(0)$. Now
suppose there is some $t_{0}$ with, say, $t_{0}>0$ such that $\alpha(t_{0})\in
H_{\delta}$. Then by compactness of the interval $[0,t_{0}]$ there is a
minimum $\epsilon>0$ such that $\alpha$ restricted $[0,t_{0}]$ is contained in
the $\epsilon$-neighborhood of $H_{\delta}$. This will be a convex region
$H_{\delta^{\prime}}$ bounded by a plane of constant curvature
$\delta^{\prime}>\delta$ and $\alpha$ will be tangent to the boundary of this
region. However, all of $\alpha$ restricted to $[0,t_{0}]$ will be contained
in $H_{\delta^{\prime}}$ contradicting that at the point of tangency the
intersection of $\alpha$ and $H_{\delta^{\prime}}$ will be an isolated point.
This implies that $\alpha(0)$ is the unique point where $\alpha$ intersects
$H_{\delta}$ and $\alpha$ is disjoint from the interior of $H_{\delta}$.

Now take the union of the interior of all regions $H_{\delta}$ with
$\delta>\kappa$ that are tangent to $\alpha^{\prime}(0)$ and let $\mathcal{K}$
be its complement. Then the image of $\alpha$ will be contained in
$\mathcal{K}$.

Figure 1: In this 2-dimensional figure the union of the $H_{\delta}$ is the
shaded region. In ${{\mathbb{H}}^{3}}$ one rotates this region about the
vertical axis. The relationship between the angle $\theta$ and a $\kappa$ is
given by $\kappa=\sin(\theta)$. One then computes that $\mathcal{K}_{-}$ is a
disk of radius $\frac{1}{\kappa}\left(1-\sqrt{1-\kappa^{2}}\right)$.

The intersection of the closure $\mathcal{K}$ with
$\widehat{{\mathbb{C}}}=\partial{{\mathbb{H}}^{3}}$ will be two regions
$\mathcal{K}_{-}$ and $\mathcal{K}_{+}$ with the accumulation set of
$\alpha(t)$ as $t\to\pm\infty$ contained in $\mathcal{K}_{\pm}$. Then
$\mathcal{K}_{-}$ is the disk
$|z|\leq\frac{1}{\kappa}\left(1-\sqrt{1-\kappa^{2}}\right)$ (see Figure 1) and
by symmetry $\mathcal{K}_{+}$ is the region with
$1/|z|\leq\frac{1}{\kappa}\left(1-\sqrt{1-\kappa^{2}}\right)$ (including
$\infty\in\widehat{{\mathbb{C}}}$).

Let $z_{-}$ be a point in the accumulation set of $\alpha(t)$ as $t\to-\infty$
and let $b_{-}$ be a Buseman function based at $z_{-}$. We then observe that
the angle between $-\alpha^{\prime}(t)$ and the gradient $\nabla b_{-}$ is
$\leq\theta$. It is enough to do this calculation when $t=0$. We then let
$\mathfrak{h}$ be the horosphere based at $z_{-}$ that goes through
$\alpha(0)$. Then (along $\mathfrak{h}$) the gradient $\nabla b_{-}$ is the
inward pointing normal vector field to $\mathfrak{h}$. Then angle will be
greatest when $z_{-}$ is in $\partial\mathcal{K}_{-}$ and a direct computation
gives that the angle in this case is $\theta$. The bound on the angle implies
that $b_{-}(\alpha(t))\to\infty$ as $t\in-\infty$ and therefore $\gamma(t)\to
z_{-}$ as $t\to-\infty$.

A similar argument shows that $\alpha(t)\to z_{+}$ as $t\to\infty$ for some
$z_{+}\in\mathcal{K}_{+}$. Let $\beta$ be the geodesic in ${{\mathbb{H}}^{3}}$
with endpoints $z_{-}$ and $z_{+}$.

Let $z$ be a point in $\partial P_{t}$ and let $\mathfrak{h}_{z}$ be the
horosphere based at $z$ that goes through $\alpha(t)$ (and is therefore
tangent to $\alpha^{\prime}(t)$). Let $\mathfrak{h}_{z,\beta}$ be the
horosphere based at $z$ that is tangent to $\beta$. We claim that the
hyperbolic distance between these two horospsheres is
$\leq\tanh^{-1}(\kappa)$. For this we can assume $t=0$ and let $\mathcal{H}$
be the convex hull of the regions $\mathcal{K}_{\pm}$. Then
$P_{0}\cap\mathcal{H}$ is a disk $D$ of radius $\tanh^{-1}(\kappa)$ and
$\beta$ will be contained in $\mathcal{H}$ and intersect $D$. Now let
$\mathfrak{h}_{\pm}$ be the two horospheres based at $z$ that are tangent to
$\mathcal{H}$. The distance between $\mathfrak{h}_{z}$ and each of the
$\mathfrak{h}_{\pm}$ will be $\tanh^{-1}(\kappa)$ and $\mathfrak{h}_{z,\beta}$
will lie between the $\mathfrak{h}_{\pm}$. This implies the distance bound.

We now adjust the picture and so that $z_{-}=0$ and $z_{+}=\infty$ in the
upper half space model for ${{\mathbb{H}}^{3}}$. Assume that $t_{i}\to-\infty$
and let $z_{i}$ be points in $\partial P_{t_{i}}$. We’ll show that $z_{i}\to
0$. Assume not. Then, after passing to a subsequence we can assume that
$|z_{i}|$ is bounded below away from zero. Then for each $i$ we apply the
isometry $z\mapsto z/|z_{i}|$ to the points $\alpha(t_{i})$ and the
horospheres $\mathfrak{h}_{z_{i}}$ and $\mathfrak{h}_{z_{i},\beta}$ to get new
objects $\bar{\alpha}(t_{i})$, $\bar{\mathfrak{h}}_{z_{i}}$ and
$\bar{\mathfrak{h}}_{z_{i},\beta}$. This isometry fixes $\beta$ and as
$|z_{i}|$ is bounded below we still have that $\bar{\alpha}(t_{i})\to
0\in\widehat{{\mathbb{C}}}$. Then horospheres
$\bar{\mathfrak{h}}_{z_{i},\beta}$ will have Euclidean radius $1$ while the
Euclidean radius of the $\bar{\mathfrak{h}}_{z_{i}}$ will go to infinity,
contradicting that the distance between these horospheres is bounded by
$\tanh^{-1}(\kappa)$. Therefore $|z_{i}|\to 0$, proving the last claim. $\Box$

###  3.3 Gauss Map

If $S$ is an immersed surface in ${{\mathbb{H}}^{3}}$ let

$n\colon S\to T^{1}{{\mathbb{H}}^{n}}$

be the lift to the unit tangent bundle. The Gauss map for $S$ is

$\mathfrak{g}_{S}\colon S\to\widehat{{\mathbb{C}}}$

with $\mathfrak{g}_{S}=\mathfrak{g}\circ n$. The following lemma gives the
derivative of $\mathfrak{g}_{S}$.

###### Lemma 3.7

Let $S$ be an oriented, immersed surface in ${{\mathbb{H}}^{3}}$ with
$\hat{S}$ its lift to $T^{1}{{\mathbb{H}}^{3}}$. Normalize so that $p=(0,1)\in
S$ the lift of this point to $\hat{S}$ is the vector ${\partial_{y}}$. Let
$G\colon T_{p}S\to T_{i}\mathbb{C}$ be the linear map with
$G({\partial_{x}})={\partial_{x}}$ and $G({\partial_{t}})={\partial_{y}}$ and
let $\mathfrak{g}_{S}\colon S\to\widehat{{\mathbb{C}}}$ be the Gauss map. Then

$(\mathfrak{g}_{S})_{*}(p)=G\circ({\operatorname{id}}+B)$

where $B\colon T_{p}S\to T_{p}S$ is the shape operator.

The proof is a straightforward calculation that is simplest to do in the
Minkowski model for ${{\mathbb{H}}^{n}}$.

Let $\langle,\rangle$ be the inner product on $\mathbb{R}^{n,1}$. Then

${{\mathbb{H}}^{n}}=\\{x\in\mathbb{R}^{n,1}\ |\ \langle x,x\rangle=-1\\}$

with tangent space

$T_{x}{{\mathbb{H}}^{n}}=\\{v\in\mathbb{R}^{n,1}\ |\ \langle x,v\rangle=0\\}.$

Then the restriction of $\langle,\rangle$ to each $T_{x}{{\mathbb{H}}^{n}}$ is
positive definite and gives a metric of constant sectional curvature equal to
$-1$. This is a model for hyperbolic space.

In this model the sphere at infinity for ${{\mathbb{H}}^{n}}$ is the
projectivized light cone which we identify with the sphere unit sphere in the
plane $x_{n+1}=1$. In this model there is a very simple formula for the Gauss
map.

######  Lemma 3.8 (Bryant, [Bry])

For $x\in{{\mathbb{H}}^{n}}$ and $v\in T^{1}_{x}{{\mathbb{H}}^{n}}$ the
hyperbolic Gauss map is given by

$\mathfrak{g}(x,v)=\frac{x+v}{\left\langle x+v,(0,\dots,0,1)\right\rangle}.$

Proof: Note that this formula is clear when $x=(0,\dots,0,1)$. Then general
case the follows from equivariance. $\Box$

The formula for the Riemannian connection on ${{\mathbb{H}}^{n}}$ is also very
simple. The Minkowski connection $\hat{\nabla}$ on $\mathbb{R}^{3,1}$ is flat
with $\hat{\nabla}_{X}Y=X(Y)$. Thus if $\nabla$ is the Riemannian connection
on ${{\mathbb{H}}^{3}}$ then by compatibility we have

$X(Y)=\nabla_{X}Y+\langle X(Y),N\rangle N.$

where $N$ is the normal to ${{\mathbb{H}}^{3}}$ in $\mathbb{R}^{3,1}$. We note
that $N:{{\mathbb{H}}^{3}}\rightarrow\mathbb{R}^{3,1}$ is given by $N(x)=x$.

From this formula for $\nabla$ we can also calculate the shape operator $B$.
We have

$BX=\nabla_{X}(n)=X(n)-\langle X(n),N\rangle N.$

As $N(p)=p$ then $X(N)=X$ giving

$\langle X(n),N\rangle=-\langle n,X(N)\rangle=-\langle n,X\rangle=0$

so

$BX=X(n).$

Note that tangent spaces in $\mathbb{R}^{3,1}\cong\mathbb{R}^{4}$ are
canonically identified and if $p=(0,0,0,1)$ then the linear map $G\colon
T_{p}S\to T_{\mathfrak{g}_{S}(p)}\partial{{\mathbb{H}}^{3}}$ (from Lemma 3.7)
is the identity map. Therefore, we need to show that

$(\mathfrak{g}_{S})_{*}(p)={\operatorname{id}}+B.$

With these preliminaries done we can now prove the lemma.

Proof of Lemma 3.7: By Lemma 3.8 we have

$\mathfrak{g}_{S}(x)=\frac{x+n(x)}{\langle x+n(x),(0,0,0,1)\rangle}.$

Note that for $v\in T_{p}S$ the derivative of $x\mapsto x+n(x)$ in the
direction of $v$ is $v+Bv$ by the calculation above. As $\langle
v+Bv,(0,0,0,1)\rangle=0$ this implies that at $p$ the derivative of the
denominator above is zero which gives that

$(\mathfrak{g}_{S})_{*}(p)v=v+Bv=({\operatorname{id}}+B)v$

as claimed. $\Box$

Using the above, we prove the following bound on the derivative of $f$.

###### Lemma 3.9

Let $f:{\mathbb{U}}\rightarrow\hat{\mathbb{C}}$ be a locally univalent map
with Epstein maps $f_{0}\colon{\mathbb{U}}\rightarrow{{\mathbb{H}}^{3}}$ and
$\hat{f}_{0}\colon{\mathbb{U}}\to T^{1}{{\mathbb{H}}^{3}}$ normalized as
follows:

  1. 1.

$f_{0}(i)=(0,1)$;

  2. 2.

$\hat{f}_{0}(i)={\partial_{y}}$;

  3. 3.

$(f_{0})_{*}(i){\partial_{y}}=\lambda{\partial_{t}}$ with
$\lambda=\|{\partial_{y}}\|_{g}$.

Then

$|f^{\prime}(i)|=1\qquad\mbox{and}\qquad|f^{\prime}(i)-1|\leq 2\|\Phi(i)\|$

where $\Phi=Sf$ is the Schwarzian.

Proof: Let $P\subset T_{(0,1)}{{\mathbb{H}}^{3}}$ be the image of
$T_{i}{\mathbb{U}}$ under the map $(f_{0})_{*}(i)$. We then have the following
sequence of isometries:

$(T_{i}{\mathbb{U}},g_{{\mathbb{H}}^{2}})\overset{{\operatorname{id}}+B}{\relbar\joinrel\relbar\joinrel\relbar\joinrel\longrightarrow}(T_{i}{\mathbb{U}},g)\overset{(f_{0})_{*}(i)}{\relbar\joinrel\relbar\joinrel\longrightarrow}(P,g_{{\mathbb{H}}^{3}})\overset{G}{\relbar\joinrel\relbar\joinrel\longrightarrow}(T_{i}{\mathbb{U}},g_{{{\mathbb{H}}^{2}}})$

and by Lemma 3.7 this composition is the map $f_{*}(i)$. In particular
$f_{*}(i)$ is an isometry of $T_{i}(U)$ to itself so $|f^{\prime}(i)|=1$.

As all the above maps are isometries we can do the computation in any of the
metrics. For convenience we will do it in the $g$-metric. Note that
$|f^{\prime}(i)-1|$ is the distance between the vectors
$f^{\prime}(i){\partial_{x}}$ and ${\partial_{x}}$ in $T_{i}{\mathbb{U}}$ or,
since $f$ is conformal, the distance between $f^{\prime}(i){\partial_{y}}$ and
${\partial_{y}}$ with the hyperbolic metric. In the $g$-metric this is the
distance between $(\operatorname{id}+B){\partial_{y}}$ and
$\frac{{\partial_{y}}}{\lambda}$ (by our normalization (3)). That is

$\displaystyle|f^{\prime}(i)-1|$ $\displaystyle=$
$\displaystyle\left\|(\operatorname{id}+B){\partial_{y}}-\frac{{\partial_{y}}}{\lambda}\right\|_{g}$
$\displaystyle=$
$\displaystyle\left\|(\lambda-1)\frac{{\partial_{y}}}{\lambda}+B{\partial_{y}}\right\|_{g}$
$\displaystyle\leq$
$\displaystyle|\lambda-1|\cdot\left\|\frac{{\partial_{y}}}{\lambda}\right\|_{g}+\|B{\partial_{y}}\|_{g}$
$\displaystyle=$ $\displaystyle|\lambda-1|+\|B{\partial_{y}}\|_{g}.$

As $(\operatorname{id}+B)^{-1}=\frac{1}{2}(\operatorname{id}+\hat{B})$ and by
Theorem 3.2 the eigenvalues of $\hat{B}$ are $1\pm 2\|Sf(i)\|$ we have

$\lambda=\|{\partial_{y}}\|_{g}=\frac{1}{2}\|(\operatorname{id}+\hat{B}){\partial_{y}}\|_{g_{{{\mathbb{H}}^{2}}}}\leq
1+\|Sf(i)\|.$

Therefore

$|\lambda-1|\leq\|Sf(i)\|.$

Also $({\operatorname{id}}+\hat{B})B=({\operatorname{id}}-\hat{B})$ giving

$\|B{\partial_{y}}\|_{g}=\frac{1}{2}\|(\operatorname{id}+\hat{B})B{\partial_{y}}\|_{g_{{{\mathbb{H}}^{2}}}}=\frac{1}{2}\|(\operatorname{id}-\hat{B}){\partial_{y}}\|_{g_{{{\mathbb{H}}^{2}}}}=\|Sf(i)\|.$

Combining these bounds, we obtain the result. $\Box$

###  3.4 Proof of Theorem 3.1:

Let $f^{1}\colon{\mathbb{U}}\to\widehat{{\mathbb{C}}}$ be the map given by
Lemma 3.9 and $f^{1}_{0}\colon U\to\widehat{{\mathbb{C}}}$ the associated
Epstein map. By our assumptions $r<1/2$ and $K/r<1/4$ so $K<1/8$. Then Lemma
3.5 gives that

$\kappa_{\alpha}<\frac{2K}{r}<1$

where $\kappa_{\alpha}$ is the curvature of $\alpha=f_{0}^{1}\circ\gamma$ in
${{\mathbb{H}}^{3}}$. The normalization (3) in Lemma 3.9 and the bound on
$\kappa_{\alpha}$ allows us to apply Lemma 3.6. Therefore we have
$z_{\pm}\in\widehat{{\mathbb{C}}}$ such that
$\underset{t\to\pm\infty}{\lim}f^{1}_{0}(it)=z_{\pm}$ and if $P_{t}$ are the
planes perpendicular to $\alpha$ at $\alpha(t)$ then
$\underset{t\to\pm\infty}{\lim}P_{t}=z_{\pm}$. Thus as $f^{1}(it)\in\partial
P_{t}\cap\widehat{{\mathbb{C}}}$ then
$\underset{t\to\pm\infty}{\lim}f^{1}(it)=z_{\pm}$.

We let $f=m\circ f^{1}$ where

$m(z)=i\cdot\frac{i-z^{+}}{i-z^{-}}\cdot\frac{z-z^{-}}{z-z^{+}}.$

Then $m(z^{-})=0,m(z^{+})=\infty$ and $m(i)=i$. Therefore $f$ has the desired
normalization and we are left to bound the derivative at $i$ which will follow
from Lemma 3.9 if we can bound the derivative of $m$ at $i$. Computing we have

$|m^{\prime}(i)-1|=\left|\frac{z^{-}}{i-z^{-}}-\frac{i/z^{+}}{i/z^{+}-1}\right|\leq\frac{2\kappa_{\alpha}}{1-\kappa_{\alpha}}.$

For the reciprocal we have

$\left|\frac{1}{m^{\prime}(i)}-1\right|=\left|\frac{(i-z_{-})^{2}}{i/z_{+}(z_{-}/z_{+}-1)}-\frac{z_{-}}{i}\right|\leq\frac{\kappa_{\alpha}(1+\kappa_{\alpha})^{2}}{1-\kappa_{\alpha}^{2}}+\kappa_{\alpha}=\frac{2\kappa_{\alpha}}{1-\kappa_{\alpha}}.$

As $\kappa_{\alpha}<2K/r$ and by our assumption that $K/r<1/4$ then
$\kappa_{\alpha}<1/2$. Combining these we get that

$\left|m^{\prime}(i)^{\pm
1}-1\right|<\frac{2\kappa_{\alpha}}{1-\kappa_{\alpha}}<\frac{8K}{r}.$

By Lemma 3.9 we have $|(f^{1})^{\prime}(i)|=1$ and
$|(f^{1})^{\prime}(i)-1|<2K$ giving

$\displaystyle\left|f^{\prime}(i)^{\pm 1}-1\right|$ $\displaystyle=$
$\displaystyle\left|((f^{1})^{\prime}(i)m^{\prime}(i))^{\pm 1}-1\right|$
$\displaystyle<$ $\displaystyle\left|(f^{1})^{\prime}(i)\right|^{\pm
1}\cdot\left|m^{\prime}(i)^{\pm 1}-1\right|+\left|(f^{1})^{\prime}(i)^{\mp
1}-1\right|$ $\displaystyle<$ $\displaystyle\frac{8K}{r}+2K<\frac{9K}{r}.$

$\Box$

##  4 Application to hyperbolic three-manifolds

We now prove our application to deformations of hyperbolic 3-manifolds given
in Theorem 1.4 which we now restate.

Theoreom 1.4: There exists $K,C>0$ such that the following holds; Let $N$ be
an acylindrical hyperbolizable manifold with boundary $S=\partial N$ and
$Y\in{\mathcal{T}}(S)$ with $\|\Phi_{Y}\|_{2}\leq K$. Then there exists a
$\hat{Y}\in\overline{{\mathcal{T}}(S)}$ with

  1. 1.

$d_{\rm WP}(Y,\hat{Y})\leq C\sqrt{\|\Phi_{Y}\|_{2}};$

  2. 2.

$\|\Phi_{\hat{Y}}\|_{\infty}\leq C\sqrt{\|\Phi_{Y}\|_{2}}.$

The above result actually holds for relative acylindrical 3-manifolds as in
[BBB]. For simplicity we are restricting to the acylindrical setting here. At
most points the proof is the same as in [BBB]. However, at one key point we
will improve the estimate. We will restrict to proving that improved estimate
and allow the reader to refer to [BBB] for the remainder of the proof.

Let ${\cal C}$ be a collection of essential simple closed curves on $S$ and
$\hat{N}$ the 3-manifold obtained from removing the curves in ${\cal C}$ from
the level surface $S\times\\{1/2\\}$ in a collar neighborhood $S\times[0,1]$
of $N$. Given a $Y\in{\mathcal{T}}(S)$ there is a unique hyperbolic structure
$\hat{M}_{Y}$ on $\hat{N}$ with conformal boundary $Y$. Again there is a
projective structure on $Y$ with Schwarzian $\hat{\Phi}_{Y}$. Key to our
bounds is choosing ${\cal C}$ such that in the complement of the standard
collars of ${\cal C}$ we have bounds on the pointwise norm of
$\hat{\Phi}_{Y}$. It is at this stage that we improve our estimate.

For a complete hyperbolic surface $X$ let $X^{<\epsilon}$ be the set of points
of injectivity radius $<\epsilon$. Recall that by the collar lemma ([Bus])
there is an $\epsilon_{2}>0$ such that $X^{<\epsilon_{2}}$ is a collection of
standard collars about simple closed geodesics of length $<2\epsilon_{2}$ and
cusps. Explicitly we choose $\epsilon_{2}=\sinh^{-1}(1)$, the Margulis
constant. If $\gamma$ is closed geodesic of length $<2\epsilon_{2}$ we let
$U_{\gamma}$ be this collar and for a collection of curves ${\cal C}$ we let
$U_{{\cal C}}$ be the union of these standard collars. We can also choose
$\overline{\epsilon}_{2}>0$ so that for any point in
$X^{<\overline{\epsilon}_{2}}$ the disk of radius $\operatorname{inj}_{X}(z)$
is contained in $X^{<\epsilon_{2}}$. Again, by elementary calculation, we can
choose $\overline{\epsilon}_{2}=\sinh^{-1}(1/\sqrt{3})$. We denote by
$\hat{U}_{\gamma}$ and $\hat{U}_{{\cal C}}$ the corresponding sub-annuli of
$U_{\gamma}$ and $U_{{\cal C}}$. We note that if
$\ell_{\gamma}(X)<2\overline{\epsilon}_{2}$ then
$d(\hat{U}_{\gamma},X-U_{\gamma})\geq\overline{\epsilon}_{2}$.

We can now state our improved estimate:

###### Theorem 4.1

There exists constants $C_{0},C_{1}$ such that if $\|\Phi_{Y}\|_{2}\leq C_{0}$
then exists a collection of curves ${\cal C}$ such that $\ell_{\cal C}(Y)\leq
2\|\Phi_{Y}\|_{2}$ and

$\left\|\hat{\Phi}_{Y}(z)\right\|\leq C_{1}\sqrt{\|\Phi_{Y}\|}$

for $z\in Y-\hat{U}_{\cal C}$.

The proof of Theorem 1.4 follows by combining Theorem 4.1 and [BBB, Theorem
3.5]. We now briefly outline this and refer the reader to [BBB] for further
details.

Proof of Theorem 1.4 using Theorem 4.1: In [BBB, Theorem 3.5] we prove that we
can find curves $\mathcal{C}$ such that

  1. 1.

$d_{\rm WP}(Y,\hat{Y})\leq C\sqrt{\ell_{\mathcal{C}}(Y)};$

  2. 2.

$\|\Phi_{\hat{Y}}\|_{\infty}\leq C\sqrt{\ell_{\mathcal{C}}(Y)}.$

In [BBB] we were only able to find $\mathcal{C}$ such that
$\ell_{c}(Y)\leq\|\Phi_{Y}\|_{2}^{\frac{2}{2n(S)+3}}$ for all
$c\in\mathcal{C}$ which gave the genus dependent bound
$\ell_{\mathcal{C}}(Y)\leq n(S)\|\Phi_{Y}\|_{2}^{\frac{2}{2n(S)+3}}$ on the
total length of $\mathcal{C}$. Substituting this bound gave us our original
result in [BBB].

To obtain the new bounds in Theorem 1.4 we apply [BBB, Theorem 3.5] with the
improved bound $\ell_{\mathcal{C}}(Y)\leq 2\|\Phi_{Y}\|_{2}$ from Theorem 4.1.
$\Box$

###  4.1 $L^{2}$ and $L^{\infty}$-norms for quadratic differentials

In order to prove our estimate, we will first need some results relating the
$L^{2}$ and $L^{\infty}$-norms for holomorphic quadratic differentials.

We have the following which bounds the pointwise norm in terms of the
$L_{2}$-norm of $\phi$.

###### Lemma 4.2

Given $\Phi\in Q(X)$ we have:

  * •

(Teo, [Teo]) If $z\in X^{\geq\epsilon}$ then

$\|\Phi(z)\|\leq C(\epsilon)\|\Phi\|_{2}$

where $C(x)=\frac{4\pi}{3}(1-\operatorname{sech}^{6}(x/2))^{-1/2}$.

  * •

(Bridgeman-Wu, [BW]) If $z\in\hat{U}_{\gamma}$ where $\gamma$ is a closed
geodesic of length $<2\overline{\epsilon}_{2}$ then

$\|\Phi(z)\|\leq\frac{\left.\|\Phi\right|_{U_{\gamma}}\|_{2}}{\sqrt{\operatorname{inj}_{X}(z)}}.$

Proof: The first statement is the main result of [Teo]. The second statement
doesn’t appear explicitly in [BW] but is a simple consequence of it. By [BW,
Proposition 3.3, part 4] we have for $z\in\hat{U}_{\gamma}$

$\|\Phi(z)\|\leq
G(\operatorname{inj}_{X}(z))\left.\|\Phi\right|_{U_{\gamma}}\|_{2}$

for some explicit function $G$. Then by direct computation in [BW, Equation
3.13] we prove that $G(t)\leq 1/\sqrt{t}$ for $t\leq\epsilon_{2}$. The result
then follows. $\Box$

For a sufficiently short closed geodesic $\gamma$ on $X$ we can lift $\Phi\in
Q(X)$ to the annular cover $X_{\gamma}$. The covering map is injective on
$U_{\gamma}$ in $X_{\gamma}$ so on $U_{\gamma}$ we have our annular
decomposition given in (2.2)

$\Phi=\Phi^{\gamma}_{-}+\Phi^{\gamma}_{0}+\Phi^{\gamma}_{+}.$

We will need the following $L^{\infty}$-bound in the thin part. In [Wol, Lemma
11]) Wolpert proved a bound which is qualitatively the same but we will need
the following quantitative version which we derive independently.

###### Lemma 4.3

Let $z\in\hat{U}_{\gamma}$. Then

$\|\Phi(z)\|\leq\|\Phi^{\gamma}_{0}\|_{\infty}+2C(\overline{\epsilon}_{2})\left.\|\Phi\right|_{U_{\gamma}}\|_{2}$

Proof: This is again essentially also contained in [BW]. We have on
$\hat{U}_{\gamma}$ the splitting of $\Phi$ into

$\Phi=\Phi^{\gamma}_{0}+\Phi^{\gamma}_{+}+\Phi^{\gamma}_{-}.$

Therefore

$\|\Phi(z)\|\leq\|\Phi^{\gamma}_{0}(z)\|+\|\Phi^{\gamma}_{+}(z)\|+\|\Phi^{\gamma}_{-}(z)\|.$

We need to bound the norms $\|\Phi^{\gamma}_{0}(z)\|$,
$\|\Phi^{\gamma}_{+}(z)\|$, and $\|\Phi^{\gamma}_{-}(z)\|$. By definition
$\|\Phi^{\gamma}_{0}(z)\|\leq\|\Phi^{\gamma}_{0}\|_{\infty}$. By [BW,
Proposition 3.3, part 3]

$\|\Phi^{\gamma}_{\pm}(z)\|\leq
C(\overline{\epsilon}_{2})\left.\|\Phi\right|_{U_{\gamma}}\|_{2}.$

Thus it follows that

$\|\Phi(z)\|\leq\|\Phi^{\gamma}_{0}\|_{\infty}+2C(\overline{\epsilon}_{2})\left.\|\Phi\right|_{U_{\gamma}}\|_{2}$

$\Box$

Our next estimate is a statement about holomorphic quadratic differentials
that we will use to choose the curves $\mathcal{C}$.

###### Lemma 4.4

Let $\Phi\in Q(X)$ with $\|\Phi\|_{2}<1/2$. Let $\mathcal{C}$ be the
collection of simple curves $\gamma$ such that $L_{\gamma}(X)\leq
2\overline{\epsilon}_{2}$ and
$\left.\|\Phi\right|_{\hat{U}_{c}}\|_{\infty}\geq\sqrt{\|\Phi\|_{2}}$. Then

$L_{\mathcal{C}}(X)\leq 2\|\Phi\|_{2}.$

Furthermore for $z\in X-\hat{U}_{\mathcal{C}}$ then
$\|\Phi(z)\|\leq\sqrt{\|\Phi\|_{2}}$.

Proof: For each $\gamma\in{\cal C}$ choose $z_{\gamma}\in\hat{U}_{\gamma}$
with $\|\Phi(z_{\gamma})\|\geq\sqrt{\|\Phi\|_{2}}$. Squaring the second bound
from Lemma 4.2 gives the bound

$\|\Phi\|_{2}\leq\|\Phi(z_{\gamma})\|^{2}\leq\frac{\left\|\left.\Phi\right|_{U_{\gamma}}\right\|^{2}_{2}}{{\operatorname{inj}_{X}(z_{\gamma})}}.$

Using that $\operatorname{inj}_{X}(z_{\gamma})\geq\ell_{\gamma}(X)/2$ and
summing gives

$\frac{\|\Phi\|_{2}}{2}\sum_{\gamma\in\mathcal{C}}\ell_{\gamma}(X)\leq\sum_{\gamma\in\mathcal{C}}\left.\|\Phi\right|_{U_{\gamma}}\|^{2}_{2}\leq\|\Phi\|^{2}_{2}$

which rearranges to give

$\ell_{{\cal C}}(X)=\sum_{\gamma\in\mathcal{C}}\ell_{\gamma}(X)\leq
2\|\Phi\|_{2}.$

By the first bound in Lemma 4.2 for $z\in X^{\geq\overline{\epsilon}_{2}}$ we
have as $C(\overline{\epsilon}_{2})\leq 1.1$

$\|\phi(z)\|\leq
C(\overline{\epsilon}_{2})\|\Phi\|_{2}\leq\sqrt{\|\Phi\|_{2}}$

where the second inequality uses that $\|\Phi\|_{2}\leq 1/2$. If $z\in
X^{<\overline{\epsilon}_{2}}$ but $z\not\in\hat{U}_{\gamma}$ for some
$\gamma\in{\cal C}$ we have

$\|\Phi(z)\|\leq\sqrt{\|\Phi\|_{2}}$

be the definition of ${\cal C}$. These two inequalities give the second bound.
$\Box$

###  4.2 Drilling

The proof of Theorem 4.1 uses the deformation theory of hyperbolic cone-
manifolds developed by Hodgson and Kerkhoff in [HK] and generalized by the
second author in [Brm, Bro]. In particular if ${\cal C}$ is sufficiently short
in $M_{Y}$ (or in $Y$) then there will be a one-parameter family of hyperbolic
cone-manifolds $M_{t}$ where the complement of the singular locus is
homeomorphic $\hat{N}$, the conformal boundary of $M_{t}$ is $Y$ and the cone
angle is $t$. The $M_{t}$ will also determine projective structures on $Y$
with Schwarzian quadratic differentials $\Phi_{t}$ and derivative
$\phi_{t}=\dot{\Phi}_{t}$. Then when $t=2\pi$ we have $M_{Y}=M_{2\pi}$,
$\Phi_{Y}=\Phi_{2\pi}$ and when $t=0$ we have $\hat{M}_{Y}=M_{0}$.

Motivated by the Lemma 4.4 above, we will fix ${\cal C}$ to be the collection
of closed geodesics $\gamma$ on $Y$ with $\ell_{\gamma}(Y)\leq
2\overline{\epsilon}_{2}$ and
$\|\left.\Phi_{Y}\right|_{\hat{U}_{\gamma}}\|_{\infty}\geq\sqrt{\|\Phi_{Y}\|_{2}}$.
Then by the above $\ell_{\mathcal{C}}(Y)\leq 2\|\Phi_{Y}\|_{2}$. We first will
need to apply the following theorem.

######  Theorem 4.5 (Bridgeman-Bromberg, [BB])

There exists an $L_{0}>0$ and $c_{\rm drill}>0$ such that the following holds.
Let $M$ be a conformally compact hyperbolic 3-manifold and $\mathcal{C}$ a
collection of simple closed geodesics in $M$ each of length $\leq L_{0}$. Let
$M_{t}$ be the unique hyperbolic cone-manifold with cone angle $t\in(0,2\pi]$
about $\mathcal{C}$ and $\Sigma_{t}$ the projective structure on the boundary.
If $\Phi_{t}$ is the Schwarzian of the uniformization map for $\Sigma_{t}$ and
$\phi_{t}=\dot{\Phi}_{t}$ then

$\|\phi_{t}\|_{2}\leq c_{\rm drill}\sqrt{L_{\mathcal{C}}(M)}.$

By Bers inequality, $L_{\mathcal{C}}(M)\leq 2\ell_{\mathcal{C}}(Y)\leq
4\|\Phi_{Y}\|_{2}$. Thus we obtain

###### Theorem 4.6

If $N$ is an acylindrical hyperbolic manifold with $\|\Phi_{Y}\|_{2}<L_{0}/4$
then there exists a family of curves $\mathcal{C}$ such that
$\ell_{\mathcal{C}}(Y)\leq 2\|\Phi_{Y}\|_{2}$ and

$\|\phi_{t}\|_{2}\leq 2c_{drill}\sqrt{\|\Phi_{Y}\|_{2}}$

for $t\in(0,2\pi]$.

We now promote the $L^{2}$-bound to a pointwise bound in the complement of the
collars of $\mathcal{C}$. We first need the following lemma.

###### Lemma 4.7

There exists a universal constants $D>0$ such that if
$\|\Phi_{Y}\|_{2}<L_{0}/4$ and $\|\Phi_{t}(z)\|\leq 1/36$ for all $z\in
Y-\hat{U}_{{\cal C}}$ then for $z\in Y-\hat{U}_{{\cal C}}$ we have

$\|\phi_{t}(z)\|\leq D\sqrt{\|\Phi_{Y}\|_{2}}.$

Proof: We have the $L^{2}$-bound $\|\phi_{t}\|_{2}\leq
2c_{drill}\sqrt{\|\Phi_{Y}\|_{2}}$. We need to convert this to a pointwise
bound.

For $z\in Y^{\geq\overline{\epsilon}_{2}}$ we have

$\|\phi_{t}(z)\|\leq C(\overline{\epsilon}_{2})\|\phi_{t}\|_{2}\leq
2c_{drill}C(\overline{\epsilon}_{2})\sqrt{\|\Phi_{Y}\|_{2}}.$

Now assume $\ell_{\gamma}(Y)\leq 2\overline{\epsilon}_{2}$ but
$\gamma\not\in{\cal C}$. For $z\in\hat{U}_{\gamma}$ by Lemma 4.3 we have

$\|\phi_{t}(z)\|\leq\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}+2C(\overline{\epsilon}_{2})\|\phi_{t}\|_{2}\leq\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}+4c_{drill}C(\overline{\epsilon}_{2})\sqrt{\|\Phi_{Y}\|_{2}}.$

We let $D_{0}=4c_{drill}C(\overline{\epsilon}_{2})$. Since the above holds for
all $z\in\hat{U}_{\gamma}$ we also have

$\|\phi_{t}\|_{\gamma}\leq\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}+D_{0}\sqrt{\|\Phi_{Y}\|_{2}}.$
(4.5)

To finish the proof we need to bound
$\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}$ by a constant
multiple of $\sqrt{\|\Phi_{Y}\|_{2}}$. By definition, for any geodesic $\beta$
with $\ell_{\beta}(Y)\leq 2\overline{\epsilon}_{2}$ then
$d(\hat{U}_{\beta},Y-U_{\beta})\geq\overline{\epsilon}_{2}$. Therefore
$d(\gamma,\hat{U}_{\cal C})\geq 2\overline{\epsilon}_{2}>1/2$ and the $1/2$
neighborhood of $\gamma$ is in $Y-\hat{U}_{{\cal C}}$. By our assumptions
$\|\Phi_{t}(z)\|\leq 1/36$ for $z\in Y-\hat{U}_{{\cal C}}$ and therefore by
Theorem 1.3 we have

$\left|\frac{\dot{\cal
L}_{\gamma}(t)}{\ell_{\gamma}(Y)}+\frac{1}{\ell_{\gamma}(Y)}\int_{\gamma}{\bf
n}\cdot\phi_{t}\right|\leq\frac{9\cdot(1/36)}{1/2}\|\phi_{t}\|_{\gamma}=\frac{1}{2}\|\phi_{t}\|_{\gamma}.$

Applying Lemma 2.3 to the integral and rearranging and then applying (4.5)
gives

$\left|\frac{\dot{\cal
L}_{\gamma}(t)}{\ell_{\gamma}(Y)}\right|\geq\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}-\frac{1}{2}\|\phi_{t}\|_{\gamma}\geq\frac{1}{2}\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}-\frac{D_{0}}{2}\sqrt{\|\Phi_{Y}\|_{2}}.$
(4.6)

Next we bound the ratio on the left. In [Bro], the second author analysed the
change in length of geodesics under cone deformations. We now apply a number
of results from [Bro]. By [Bro, Equation 4.6] we have

$|\dot{\cal L}_{\gamma}(t)|\leq 4L_{{\cal C}}(t)L_{\gamma}(t).$ (4.7)

We need to bound the two terms on the right. By [Bro, Proposition 4.1] we have
that $L_{\cal C}(t)\leq L_{\cal C}(2\pi)$ for $t\in(0,2\pi]$. When $t=2\pi$
the manifold is non-singular so we can apply the Bers inequality to see that
$L_{\cal C}(2\pi)\leq 2\ell_{\cal C}(Y)$. Since
$|\dot{L}_{\gamma}(t)|\leq|\dot{\cal L}_{\gamma}(t)|$ we can combine (4.7)
with the bound on $L_{\cal C}(t)$ to get

$\left|\frac{\dot{L}_{\gamma}(t)}{L_{\gamma}(t)}\right|\leq\left|\frac{\dot{{\cal
L}}_{\gamma}(t)}{L_{\gamma}(t)}\right|\leq 8\ell_{\cal C}(Y)$

which integrates to give

$L_{\gamma}(t)\leq L_{\gamma}(2\pi)e^{8\pi\ell_{\cal C}(Y)}\leq
2\ell_{\gamma}(Y)e^{16\pi\|\Phi_{Y}\|_{2}}.$

Here the second inequality comes from applying the Bers inequality to $\gamma$
in $M=M_{2\pi}$ and the bound $\ell_{\cal C}(Y)\leq 2\|\Phi_{Y}\|_{2}$. As we
have assumed a universal bound $\|\Phi_{Y}\|_{2}\leq L_{0}/4$ these two
inequalities give

$\left|\frac{\dot{{\cal L}}_{\gamma}(t)}{\ell_{\gamma}(Y)}\right|\leq
16e^{16\pi\|\Phi_{Y}\|_{2}}\ell_{\cal C}(Y)\leq D_{1}\|\phi_{Y}\|_{2}.$ (4.8)

where $D_{1}=16e^{4\pi L_{0}}$ is a universal constant. Combining (4.6) and
(4.8) gives

$\left\|\left(\phi_{t}\right)^{\gamma}_{0}\right\|_{\infty}\leq(D_{0}+2D_{1})\sqrt{\|\Phi_{Y}\|_{2}}.$

Finally by equation 4.5 we have for $z\in\hat{U}_{\gamma}$

$\|\phi_{t}(z)\|\leq D\sqrt{\|\Phi_{Y}\|_{2}}$

where $D=2D_{0}+2D_{1}=32e^{4\pi L_{0}}+8c_{drill}C(\overline{\epsilon_{2}})$.
$\Box$

Finally we can now prove Theorem 4.1 which we restate.

Theorem 4.1 There exists constants $C_{0},C_{1}$ such that if
$\|\Phi_{Y}\|_{2}\leq C_{0}$ then exists a collection of curves ${\cal C}$
such that for each $\gamma\in{\cal C}$ we have $\ell_{\cal C}(Y)\leq
2\|\Phi_{Y}\|_{2}$ and

$\left\|\hat{\Phi}_{Y}(z)\right\|\leq C_{1}\sqrt{\|\Phi_{Y}\|}$

for $z\in Y-\hat{U}_{\cal C}$.

Proof: We let $C_{0}=\min\\{(1/64)^{2},1/(128\pi D)^{2},L_{0}/4\\}$ and
$C_{1}=2\pi D+1$. If $\|\Phi_{Y}\|_{2}<C_{0}$ then by Lemma 4.4 for $z\in
Y-\hat{U}_{\mathcal{C}}$ we have

$\|\Phi_{Y}(z)\|\leq\sqrt{\|\Phi_{Y}\|_{2}}\leq\frac{1}{64}.$

We now show that $\|\Phi_{t}(z)\|\leq 1/36$ for $z\in Y-\hat{U}_{\mathcal{C}}$
and $t\in(0,2\pi]$. If not, then by continuity there is a $t_{0}>0$ be such
that $\|\Phi_{t}(z)\|=1/36$ and $\|\Phi_{t}(z)\|<1/36$ for $t\in(t_{0},2\pi]$.
For $t\geq t_{0}$ as $\|\Phi_{Y}\|_{2}<L_{0}/4$ by Lemma 4.7 above, we have if
$z\in Y-\hat{U}_{\mathcal{C}}$ then $\|\phi_{t}(z)\|\leq
D\sqrt{\|\Phi_{Y}\|_{2}}$. Therefore integrating we have

$\frac{1}{64}\leq\|\Phi_{t_{0}}(z)-\Phi_{Y}(z)\|\leq\int_{t_{0}}^{2\pi}\|\phi_{t}(z)\|dt<2\pi
D\sqrt{\|\Phi_{Y}\|_{2}}\leq\frac{2\pi D}{128\pi D}\leq\frac{1}{64}.$

This gives our contradiction. Thus we can apply Lemma 4.7 and integrate to get

$\|\hat{\Phi}_{Y}(z)-\Phi_{Y}(z)\|\leq\int_{0}^{2\pi}\|\phi_{t}(z)\|dt\leq
2\pi D\sqrt{\|\Phi_{Y}\|_{2}}.$

Thus

$\|\hat{\Phi}_{Y}(z)\|\leq\|\Phi_{Y}(z)\|+2\pi
D\sqrt{\|\Phi_{Y}\|_{2}}\leq(2\pi D+1)\sqrt{\|\Phi_{Y}\|_{2}}.$

$\Box$

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