approach. For my needs this is more than enough, since in any case I
eventually restrict attention to a small interval around $0$.

Corollary 4.11 can be formulated in a slightly more general fashion. Using the
notation
$\delta(W):=\limsup_{r\rightarrow\infty}\frac{\log\big{(}b_{W}(r)\big{)}}{r}$
for a general subset $W$ in a general metric space $X$, the following version
holds:

######  Corollary 4.13.

Let $(X,x_{0})$ be a pointed metric space, $Y,Z\subset X$ two subsets, and
$u(r)\preceq_{\infty}\varepsilon r$ for some $\varepsilon<\frac{1}{2}$. Assume
that $b_{Z}^{u}(r)=b_{Z}(r)$. If $Z\subset\mathcal{N}_{u}(Y)$, then
$\delta(Y)\geq(1-4\varepsilon)\cdot\delta(Z)$. Moreover, if $u$ is sublinear,
then $\delta(Z)=\delta(Y)$.

In particular, Corollary 4.11 holds even when the group $G$ does not have
property $(T)$. Corollary 4.11 follows from Corollary 4.13 because the fact
that $\Gamma$ is a group of isometries implies
$b_{\Gamma}^{u}(r)=b_{\Gamma}(r)$.

###### Proof of Corollary 4.13.

Consider the closest point projection $p_{Y}:Z\rightarrow Y$, denote
$y_{z}:=p_{Y}(z)$ and observe:

  1. 1.

$|y_{z}|\leq|z|+u(|z|)$.

  2. 2.

$d\big{(}y_{z},y_{z^{\prime}}\big{)}\geq
d(z,z^{\prime})-2u(\max\\{|z|,|z^{\prime}|\\})$.

The first item follows from the fact
$y_{z}\in\overline{B}\big{(}z,u(|z|)\big{)}$ and triangle inequality. The
second item follows from the quadrilateral inequality, i.e., using triangle
inequality twice along the quadrilateral
$[z,z^{\prime},y_{z^{\prime}},y_{z}]$.

The above properties allow me to use Proposition 4.10 with constant $L=1$ and
function $u^{\prime}=2u$ to get

$b_{Y}(r)\geq
b_{Z}\big{(}r-u^{\prime}(r)\big{)}/b_{Z}^{u}\Big{(}u^{\prime}\big{(}r-u^{\prime}(r)\big{)}\Big{)}$

Since I assume $b_{Z}^{u}=b_{Z}$, I can omit the superscript $u$ in the last
expression. Recalling the definition
$\delta(W)=\limsup_{r\rightarrow\infty}\frac{b_{W}(r)}{r}$, it remains to
prove:

$\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}/b_{Z}\Big{(}u^{\prime}\big{(}r-u^{\prime}(r)\big{)}\Big{)}\bigg{)}\geq(1-4\varepsilon)\cdot\delta(Z)$

The proof of this inequality involves nothing more than $\log$ rules and
arithmetic of limits:

$\begin{split}&\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}/b_{Z}\Big{(}u^{\prime}\big{(}r-u^{\prime}(r)\big{)}\Big{)}\bigg{)}\\\
&=\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\Bigg{(}\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\bigg{)}-\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}r-u^{\prime}(r)\big{)}\Big{)}\bigg{)}\Bigg{)}\\\
&\geq\limsup_{r\rightarrow\infty}\Bigg{[}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\bigg{)}-\limsup_{s\rightarrow\infty}\bigg{[}\frac{1}{s}\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}s-u^{\prime}(s)\big{)}\Big{)}\bigg{)}\bigg{]}\Bigg{]}\\\
&=\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\bigg{)}-\limsup_{s\rightarrow\infty}\frac{1}{s}\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}s-u^{\prime}(s)\big{)}\Big{)}\bigg{)}\\\
&\geq\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-2\varepsilon
r\big{)}\bigg{)}-\limsup_{s\rightarrow\infty}\frac{1}{s}\log\bigg{(}b_{Z}\Big{(}2\varepsilon
s\Big{)}\bigg{)}\\\
&=(1-2\varepsilon)\delta(Z)-2\varepsilon\delta(Z)=(1-4\varepsilon)\delta(Z)\end{split}$
(3)

Below I justify the steps in the above inequalities:

  1. 1.

First equality is by rules of $\log$.

  2. 2.

Second and third inequalities are by arithmetic of limits: let
$(a_{n})_{n},(b_{n})_{n}$ be two sequences of positive numbers, and
$A=\limsup_{n}a_{n},\ B=\limsup_{n}b_{n}$. Then
$\limsup(a_{n}-b_{n})\geq\limsup_{n}(a_{n}-B)=A-B$.

  3. 3.

Fourth inequality: $u^{\prime}(r)<2\varepsilon(r)$ for all large enough $r$.

  4. 4.

Fifth equality: definition of $\delta$.

This completes the proof in the general case, which is what is needed for the
proof of Theorem 4.3. For the more refined statement in the case $u$ is
sublinear, one has to show a bit more. From inequality 3 (specifically from
the fourth line of the inequality) it is clearly enough to prove:

  1. 1.

$\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\bigg{)}=\delta(Z)$.

  2. 2.

$\limsup_{s\rightarrow\infty}\frac{1}{s}\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}s-u^{\prime}(s)\big{)}\Big{)}\bigg{)}=0$.

Starting from the second item, indeed it holds that

$\frac{1}{s}\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}s-u^{\prime}(s)\big{)}\Big{)}\bigg{)}=\frac{u^{\prime}\big{(}s-u^{\prime}(s)\big{)}}{s}\cdot\frac{\log\bigg{(}b_{Z}\Big{(}u^{\prime}\big{(}s-u^{\prime}(s)\big{)}\Big{)}\bigg{)}}{u^{\prime}\big{(}s-u^{\prime}(s)\big{)}}$

Clearly $\limsup$ of the right factor in the above product is bounded by
$\delta(Z)$, and in particular it is uniformly bounded. On the other hand
sublinearity of $u^{\prime}$ implies that the left factor tends to $0$. I
conclude that this product tends to $0$ as $s$ tends to $\infty$.

It remains to prove
$\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\bigg{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\bigg{)}=\delta(Z)$.
In a similar fashion,

$\frac{1}{r}\log\Big{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\Big{)}=\frac{r-u^{\prime}(r)}{r}\cdot\frac{\log\Big{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\Big{)}}{r-u^{\prime}(r)}$

The left factor limits to $1$ by sublinearity of $u^{\prime}$. The right
factor is nearly the expression in the definition of $\delta(Z)$, and I want
to prove that indeed taking $\limsup$ of it equals $\delta(Z)$. _A priori_
$\\{r-u^{\prime}(r)\\}_{r\in\mathbb{R}_{>0}}$ is just a subset of
$\mathbb{R}_{>0}$, so changing variable and writing $t:=r-u^{\prime}(r)$
requires a justification. But there is no harm in assuming that $u^{\prime}$
is a non-decreasing continuous function, hence $\mathbb{R}_{\geq
R}\subset\\{r-u^{\prime}(r)\\}_{r\in\mathbb{R}_{>0}}$ for some
$R\in\mathbb{R}_{>0}$. Therefore for any sequence $r_{n}\rightarrow\infty$
there is a sequence $r_{n}^{\prime}$ with
$r_{n}=r_{n}^{\prime}-u^{\prime}(r_{n}^{\prime})$ for all large enough $n$
(note that in particular $r_{n}^{\prime}\rightarrow\infty$). In the other
direction, for every sequence $r_{n}^{\prime}\rightarrow\infty$ there is
clearly a sequence $r_{n}\rightarrow\infty$ for which
$r_{n}=r_{n}^{\prime}-u^{\prime}(r_{n}^{\prime})$. I conclude

$\limsup_{r\rightarrow\infty}\frac{\log\Big{(}b_{Z}\big{(}r-u^{\prime}(r)\big{)}\Big{)}}{r-u^{\prime}(r)}=\limsup_{r\rightarrow\infty}\frac{1}{r}\cdot\log\big{(}b_{Z}(r)\big{)}=\delta(Z)$

This completes the proof.

∎

###### Proof of Theorem 4.3.

Define $\varepsilon(G)=\frac{c^{*}(G)}{4\cdot 2\|\rho\|}$, and assume
$u(r)\preceq_{\infty}\varepsilon(G)\cdot r$. Notice that
$\varepsilon(G)<\frac{1}{2}$, and since $\delta(\Gamma)=2\|\rho\|$ Corollary
4.11 gives

$\delta(\Lambda)\geq\big{(}1-4\varepsilon(G)\big{)}\cdot
2\|\rho\|=2\|\rho\|-4\varepsilon(G)\cdot 2\|\rho\|\geq 2\|\rho\|-c^{*}(G)$

By Theorem 4.6, $\Lambda$ is a lattice. ∎

######  Remark 4.14.

The question of existence of interesting groups that coarsely cover a lattice
is a key question that arises naturally from this paper. The first question
that comes to mind is whether there exist groups that are not commensurable to
a lattice but that sublinearly, or even $\varepsilon$-linearly, cover one.
Perhaps the growth rate point of view could be used to rule out groups that
cover a lattice $\varepsilon$-linearly but not sublinearly.

##  5 Uniform Lattices

In this section I prove:

######  Theorem 5.1.

Let $G$ be a finite-centre semisimple Lie group without compact factors. Let
$\Gamma\leq G$ be a lattice, $\Lambda\leq G$ a discrete subgroup such that
$\Gamma\subset\mathcal{N}_{u}(\Lambda)$ for some sublinear function $u$. If
$\Gamma$ is uniform, then $\Lambda$ is a uniform lattice.

As in the case of lattices with property (T), uniform lattices admit the
stronger version of $\varepsilon$-linear rigidity, for any $\varepsilon<1$:

######  Theorem 5.2.

In the setting of Theorem 5.1, the conclusion holds also under the relaxed
assumption that $u(r)\preceq_{\infty}\varepsilon r$ for any $0<\varepsilon<1$.

Clearly Theorem 5.2 implies Theorem 5.1. Throughout this section, the standing
assumptions are those of Theorem 5.2.

##### Lattice Criterion.

A discrete subgroup is a uniform lattice if and only if it admits a relatively
compact fundamental domain. The criterion I use is the immediate consequence
that if $\Gamma$ is uniform and $u$ is bounded (i.e.
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$), then $\Lambda$ is a
uniform lattice.

##### Line of Proof and Use of $\varepsilon$-Linearity.

The goal is to show that the $\varepsilon$-linearity of $u$ forces
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$, i.e. that $\Gamma$
actually lies inside a bounded neighbourhood of $\Lambda$. The proof is by way
of contradiction. If there is no such $D>0$ then there are arbitrarily large
balls that do not intersect $\Lambda$. The proof goes by finding such large
$\Lambda$-free balls that are all tangent to some fixed arbitrary point $x\in
X$ (see Figure 1). The $\varepsilon$-linearity then gives rise to concentric
$\Gamma$-free balls that are arbitrarily large, contradicting the fact that
$\Gamma$ is a uniform lattice.

######  Remark 5.3.

The main difference from the non-uniform case is that for a non-uniform
lattice $\Gamma$, the space $X$ does admit arbitrarily large $\Gamma$-free
balls. This situation requires different lattice criteria and much extra work.
Still the proof for the uniform case, though essentially no more than a few
lines, lies the foundations for and presents the logic of the much more
involved case of $\mathbb{Q}$-rank $1$ lattices.

###  5.1 Notations and Terminology

The following definitions will be used repeatedly in both this section and in
Section 6. It mainly fixes terminology and notation of the geometric situation
illustrated in Figure 1.

######  Definition 5.4.

Let $H\leq G=\mathrm{Isom}(X)^{\circ}$. A set $U\subset X$ is called _$H$
-free_ if $H\cdot x_{0}\cap\mathrm{Int}(U)=\emptyset$, where $\mathrm{Int}(U)$
is the topological interior of $U$. That is, $U$ is called $H$-free if its
interior does not intersect the $H$-orbit $H\cdot x_{0}$.

######  Definition 5.5.

Denote $P_{\Lambda}(\gamma x_{0})=P_{\Lambda}(\gamma)=\lambda_{\gamma}x_{0}$.

  1. 1.

$d_{\gamma}:=d(\gamma x_{0},\lambda_{\gamma}x_{0})$.

  2. 2.

$B_{\gamma}:=B(\gamma x_{0},d_{\gamma})$. It is a $\Lambda$-free ball centred
at $\gamma x_{0}$ and tangent to $\lambda_{\gamma}x_{0}$.

  3. 3.

$x_{\gamma}^{\prime}:=\lambda_{\gamma}^{-1}\gamma x_{0}$. Notice
$|x_{\gamma}^{\prime}|=d_{\gamma}$.

  4. 4.

$B_{\gamma}^{\prime}:=\lambda_{\gamma}^{-1}B_{\gamma}=B(x_{\gamma}^{\prime},d_{\gamma})$.
It is $\Lambda$-free as a $\Lambda$-translate of the $\Lambda$-free ball
$B_{\gamma}$, and is tangent to $x_{0}$.

  5. 5.

For $s\in\mathbb{R}_{>0}$ and a ball $B=B(x,r)$, denote $sB:=B(x,sr)$, the
rescaled ball with same centre and radius $sr$.

  6. 6.

For a sequence $\gamma_{n}$, denote by
$\lambda_{n},d_{n},B_{n},B_{n}^{\prime},x_{n}^{\prime}$ the respective
$\lambda_{\gamma_{n}},d_{\gamma_{n}}$, etc.

$x_{0}$$x_{\gamma}^{\prime}$$=d_{\gamma}$$=s\cdot d_{\gamma}$$\Gamma-
free$$\Lambda-free$$\gamma x_{0}$$=d_{\gamma}$$\Lambda-
free$$\lambda_{\gamma}x_{0}$$L_{\lambda_{\gamma}^{-1}}$ Figure 1: Basic
Setting and Lemma 5.6. A $\Lambda$-free ball about $\gamma x_{0}$ of radius
$d_{\gamma}$, translated by $\lambda_{\gamma}^{-1}$ to a ball tangent to
$x_{0}$. The linear ratio between $|x_{\gamma}^{\prime}|=d_{\gamma}$ and the
$\Lambda$-free radius forces the red ball to be $\Gamma$-free.

###  5.2 Proof of Theorem 5.2

######  Lemma 5.6.

Let $x\in X$. There exists $S=S(x,u)\in(0,1)$ such that for every $s\in(0,S)$
there is $R=R(s,S)$ such that if $r>R$ and $B=B(y,r)$ is a $\Lambda$-free ball
tangent to $x$, then $sB$ is $\Gamma$-free.

In particular, the existence of arbitrarily large $\Lambda$-free balls that
are all tangent to a fixed point $x\in X$ implies the existence of arbitrarily
large $\Gamma$-free balls.

There is a slightly stronger version of Lemma 5.6 if $u$ is sublinear:

######  Lemma 5.7.

Let $G,\Gamma,\Lambda$ and $u$ be as in Theorem 5.1 (in particular, $u$ is a
sublinear function and $\Gamma\subset\mathcal{N}_{u}(\Gamma)$). For every
$x\in X$ and every $s\in(0,1)$ there exists $R=R(x,s)>0$ such that for every
$r>R$, if $B=B(y,r)$ is a $\Lambda$-free ball tangent to $x$ then $sB$ is
$\Gamma$-free.

I omit the proof of Lemma 5.7, which is a slightly simpler version of the
proof of Lemma 5.6.

###### Proof of Lemma 5.6..

The proof is more easily read if one assumes $x=x_{0}$ and $u(r)=\varepsilon
r$ so I begin with this case. Assume $B=B(y,R)$ is $\Lambda$-free for some
$y\in X,r>0$. The assumption $x=x_{0}$ gives $|y|=d(y,x_{0})=r$. For a fixed
$s\in(0,1)$, assume $sB\cap\Gamma\cdot x_{0}\neq\emptyset$. This gives rise to
an element $\gamma\in\Gamma$ such that:

  1. 1.

$|\gamma|=d(\gamma x_{0},x_{0})\leq d(y,x_{0})+d(\gamma x_{0},y)=(1+s)r$. In
particular, $d(\gamma x_{0},\Lambda\cdot x_{0})\leq\varepsilon(1+s)r$.

  2. 2.

$B\big{(}\gamma x_{0},(1-s)r\big{)}\subset B(y,r)$, so it is $\Lambda$-free.

I conclude that for such $s$ one has the inequality
$(1-s)r\leq\varepsilon(1+s)r$, i.e. $\frac{1-s}{1+s}\leq\varepsilon$. The
constant $\varepsilon$ is fixed and smaller than $1$, while $\frac{1-s}{1+s}$
limit to $1$ monotonically from below as $s>0$ tend to $0$. I conclude that
there is a segment $(0,S)\subset(0,1)$ such that for all $s\in(0,S)$, $sB$ is
$\Gamma$-free (independently of $r$).

Assume now that $x\neq x_{0}$ and $u(r)\preceq_{\infty}\varepsilon r$. As
above, if $\gamma x_{0}\in B(y,sr)$ then it is the centre of a $\Lambda$-free
ball of radius $(1-s)r$, and so $(1-s)r\leq u(|\gamma|)$. I wish to use the
$\varepsilon$-linear bound on $u$ as I did before, only this time $u$ is only
asymptotically smaller than $\varepsilon r$. To circumvent this I just need to
show that $|\gamma|$ is large enough. Indeed since $B\big{(}\gamma
x_{0},(1-s)r\big{)}$ is $\Lambda$-free it does not contain
$x_{0}\in\Lambda\cdot x_{0}$ and in particular $(1-s)r\leq d(x_{0},\gamma
x_{0})=|\gamma|$. For some $R_{1}(s)=R_{1}(s,u)$ one therefore has for all
$r>R_{1}(s)$

$(1-s)r\leq u(|\gamma|)\leq\varepsilon|\gamma|$

On the other hand $|y|\leq d(x,y)+d(x,x_{0})=r+|x|$. Consequently
$|\gamma|\leq(1+s)r+|x|$ and, for $r>R_{1}(s)$,

$(1-s)r\leq
u(|\gamma|)\leq\varepsilon|\gamma|\leq\varepsilon\big{(}(1+s)r+|x|\big{)}$

This means that $s$ for which $\Gamma\cdot x_{0}\cap B(y,sr)\neq\emptyset$
must admit, for all $r>R_{1}(s)$,

$\frac{1-s}{1+s+\frac{|x|}{r}}=\frac{(1-s)r}{(1+s)r+|x|}\leq\varepsilon<1$ (4)

The rest of the proof is just Calculus 1, and concerns with finding
$S=S(x,\varepsilon,u)\in(0,1)$ so that for any $s\in(0,S)$ there is $R(s)$
such that all $r>R(s)$ satisfy

$\varepsilon<\frac{1-S}{1+S+\frac{|x|}{r}}\leq\frac{1-s}{1+s+\frac{|x|}{r}}$
(5)

The lemma readily follows from inequalities 4,5.

Explicitly, fix $\varepsilon^{\prime}>\varepsilon$. As before, monotonic
approach of $\frac{1-s}{1+s}$ to $1$ allows to fix $S\in(0,1)$ for which
$\varepsilon<\varepsilon^{\prime}<\frac{1-s}{1+s}$ for all $s\in(0,S)$. Next
note that for any fixed $s\in(0,S)$,
$\lim_{r\rightarrow\infty}\frac{1-s}{1+s+\frac{|x|}{r}}=\frac{1-s}{1+s}$, and
that the approach in monotonically increasing with $r$. Since
$\varepsilon<\varepsilon^{\prime}$, this limit implies that for some
$R_{2}>R_{1}(S)$, all $r>R_{2}$ admit
$\varepsilon<\frac{1-S}{1+S+\frac{|x|}{r}}$. Finally notice that for any fixed
$r$ the function $\frac{1-s}{1+s+\frac{|x|}{r}}$ is again monotonically
increasing as $s$ tends to $0$ from above. Therefore inequality 5 holds for
every $s\in(0,S)$ and all $r>R_{2}(S)$ (capital $S$ is intentional and
important).

To conclude the proof, notice that if moreover $r>R_{1}(s)$ (again lowercase
$s$ is intentional and important) then inequalities 4,5 both hold. This means
that for any $s\in(0,S)$ there is $R(s):=\max\\{R_{1}(s),R_{2}(S)\\}$ such
that $r>R(s)\Rightarrow B(y,sr)$ is $\Gamma$-free. The constants
$R_{1}(s),R_{2}(S)$ have the desired dependencies, hence so does $R(s)$,
proving the lemma. ∎

######  Corollary 5.8 (Theorem 5.2).

There is a uniform bound on $\\{d_{\gamma}\\}_{\gamma\in\Gamma}$, i.e.,
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$. In particular,
$\Lambda$ is a uniform lattice.

##  6 $\mathbb{Q}$-rank 1 Lattices

In this section I prove:

######  Theorem 6.1.

Let $G$ be a real finite-centre semisimple Lie group without compact factors,
$\Gamma\leq G$ an irreducible non-uniform $\mathbb{Q}$-rank $1$ lattice,
$\Lambda\leq G$ a discrete irreducible subgroup. If
$\Gamma\subset\mathcal{N}_{u}(\Lambda)$ for some sublinear function $u$, then
$\Lambda$ is a lattice. Moreover, if
$\Gamma\not\subset\mathcal{N}_{D}(\Lambda)$ for any $D>0$, then $\Lambda$ is
also of $\mathbb{Q}$-rank $1$.

If $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$, then $\Lambda$
could be a uniform lattice. An obvious obstacle for that is if
$\Lambda\subset\mathcal{N}_{u^{\prime}}(\Gamma)$ for some sublinear function
$u^{\prime}$. This condition turns out to be sufficient for commensurability.

######  Proposition 6.2 (Proposition 6.27).

Let $G$ be a real finite-centre semisimple Lie group without compact factors,
$\Gamma\leq G$ an irreducible $\mathbb{Q}$-rank $1$ lattice, $\Lambda\leq G$ a
discrete subgroup such that $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some
$D>0$, and $\Lambda\subset\mathcal{N}_{u}(\Gamma)$ for some sublinear function
$u$. Then $\Lambda\subset\mathcal{N}_{D^{\prime}}(\Gamma)$ for some
$D^{\prime}$.

From Eskin’s and Schwartz’s results on groups at finite Hausdorff distance
(see Section 6.3), I conclude:

######  Corollary 6.3.

In the setting of Proposition 6.2 and unless $G$ is locally isomorphic to
$\mathrm{SL}_{2}(\mathbb{R})$, $\Lambda$ is commensurable to $\Gamma$.

###### Proof of Theorem 1.8.

The theorem is an immediate result of Theorem 6.1 and Corollary 6.3 ∎

###  6.1 Strategy

##### Lattice Criteria.

I use three different lattice criteria, depending on the $\mathbb{R}$-rank of
$G$ and on whether or not $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some
$D>0$. My proof is motivated by a conjecture of Margulis, that can be viewed
as an algebraic converse to the geometric structure of the compact core
described in Theorem 2.21. Over the past $30$ years this conjecture was
resolved in many major cases by Oh, Benoist and Miquel, and was recently
proved in full generality by Benoist-Miquel (see Section $1.2$ in [5] for an
overview of the milestones in solving this conjecture).

######  Theorem 6.4 (Theorem $2.16$ in [5]).

Let $G$ be a semisimple real algebraic Lie group of real rank at least $2$ and
$U$ be a non-trivial horospherical subgroup of $G$. Let $\Delta$ be a discrete
Zariski dense subgroup of $G$ that contains an indecomposable lattice
$\Delta_{U}$ of $U$. Then $\Delta$ is a non-uniform irreducible arithmetic
lattice of $G$.

See Definition 6.44 for the precise meaning of an _indecomposable_
horospherical lattice.

For $\mathbb{R}$-rank$\ 1$ groups, one has the following theorem by Kapovich
and Liu, stating that a group is geometrically finite so long as ‘most’ of its
limit points are conical. Recall $\mathcal{L}(\Delta)$ is the limit set of
$\Delta\leq\mathrm{Isom}(X)$, and $\mathcal{L}_{\mathrm{con}}(\Delta)$ is the
set of its conical limit points.

######  Theorem 6.5 (Theorem $1.5$ in [29]).

Let $X$ be a $\mathbb{R}$-rank$\ 1$ symmetric space. A discrete subgroup
$\Delta\leq\mathrm{Isom}(X)$ is geometrically infinite if and only if the set
$\mathcal{L}(\Delta)\setminus\mathcal{L}_{\mathrm{con}}(\Delta)$ of non-
conical limit points has the cardinality of the continuum.

As a direct corollary I obtain the following criterion:

######  Corollary 6.6.

Let $X$ be a $\mathbb{R}$-rank$\ 1$ symmetric space, $\Gamma\leq
G=\mathrm{Isom}(X)$ a non-uniform lattice and $\Lambda\leq G$ a discrete
subgroup. If $\mathcal{L}(\Lambda)=X(\infty)$ and
$\mathcal{L}_{\mathrm{con}}(\Gamma)\subset\mathcal{L}_{\mathrm{con}}(\Lambda)$,
then $\Lambda$ is a lattice.

###### Proof.

Since $\Gamma$ is a lattice, $\mathcal{L}(\Gamma)=X(\infty)$ and it is
geometrically finite. Theorem 6.5 implies the cardinality of
$X(\infty)\setminus\mathcal{L}_{\mathrm{con}}(\Gamma)$ is strictly smaller
than the continuum. The assumption
$\mathcal{L}_{\mathrm{con}}(\Gamma)\subset\mathcal{L}_{\mathrm{con}}(\Lambda)$
implies the same holds for $\Lambda$, and in particular that $\Lambda$ is
geometrically finite. The assumption that $\mathcal{L}(\Lambda)=X(\infty)$
implies that $\Lambda$ is geometrically finite if and only if it is a lattice.
∎

######  Corollary 6.7.

Let $X$ be a $\mathbb{R}$-rank$\ 1$ symmetric space, $\Gamma\leq
G=\mathrm{Isom}(X)$ a non-uniform lattice and $\Lambda\leq G$ a discrete
subgroup. If $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$, then
$\Lambda$ is a lattice.

###### Proof.

By definition of the limit set and of conical limit points, it is clear that
every $\Gamma$-limit point is a $\Lambda$-limit point, and every
$\Gamma$-conical limit point is also $\Lambda$-conical limit point. I conclude
from Corollary 6.6 that $\Lambda$ is a lattice. ∎

Also in higher rank the inclusion $\Gamma\subset\mathcal{N}_{D}(\Lambda)$
implies that $\Lambda$ is a lattice. This result is due to Eskin.

######  Theorem 6.8 (Eskin, see Remark 6.10 below).

Let $G$ be a real finite-centre semisimple Lie group without compact factors
and of higher rank, $\Gamma\leq G$ an irreducible non-uniform lattice,
$\Lambda\leq G$ a discrete subgroup such that
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$. Then $\Lambda$ is a
lattice. If moreover $\Lambda\subset\mathcal{N}_{D}(\Gamma)$ then $\Lambda$
and $\Gamma$ are commensurable.

Theorem 6.8 was used in the proof of quasi-isometric rigidity for higher rank
non-uniform lattices in [15] and [21]. In the (earlier) $\mathbb{R}$-rank$\ 1$
case, Schwartz [53] used an analogous statement that requires one extra
assumption.

######  Theorem 6.9 (Schwartz, see Section $10.4$ in [53] and Remark 6.10
below).

Let $G$ be a real finite-centre simple Lie group of $\mathbb{R}$-rank$\ 1$,
$\Gamma\leq G$ an irreducible non-uniform lattice, $\Lambda\leq G$ a discrete
subgroup such that both $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ and
$\Lambda\subset\mathcal{N}_{D}(\Gamma)$ for some $D>0$. Then $\Lambda$ is a
lattice. If moreover $G$ is not locally isomorphic to
$\mathrm{SL}_{2}(\mathbb{R})$, then $\Lambda$ and $\Gamma$ are commensurable.

######  Remark 6.10.

Theorem 1.6 should be viewed as a generalization of the bounded case depicted
in Theorems 6.8 and 6.9, which were known to experts in the field in the late
1990’s. Complete proofs for these statements were never given in print, and I
take the opportunity to include them here. See Section 6.3, where I also prove
Proposition 6.2. I thank Rich Schwartz and Alex Eskin for supplying me with
their arguments and allowing me to include them in this paper. I also thank my
thesis examiner Emmanuel Breuillard for encouraging me to find and make these
proofs public.

##### Line of Proof and Use of Sublinearity.

Lattices of $\mathbb{Q}$-rank $1$ admit a concrete geometric structure (see
Section 2.2). This structure is manifested in the geometry of an orbit of such
a lattice in the corresponding symmetric space $X=G/K$. One important
geometric property is the existence of a set of horoballs which the orbit of
the lattice intersects only in the bounding horospheres, and in each such
horosphere the orbit forms a (metric) cocompact lattice.

Corollary 6.7 and Theorem 6.8 reduce the proof to the case where
$\Gamma\not\subset\mathcal{N}_{D}(\Lambda)$ for any $D>0$. In that case, the
essence lies in proving the existence of horospheres in $X$ which a
$\Lambda$-orbit intersects in a cocompact lattice. This is proved purely
geometrically, using the geometric structure of $\mathbb{Q}$-rank $1$ lattices
and the sublinear constraint. Together with some control on the location of
these horospheres, I prove two strong properties:

  1. 1.

$\Lambda\cdot x_{0}$ intersects a horosphere $\mathcal{H}\subset X$ in a
cocompact lattice (Proposition 6.11).

  2. 2.

Every $\Gamma$-conical limit point is also a $\Lambda$-conical limit point
(Corollary 6.26).

The $\mathbb{R}$-rank$\ 1$ case of Theorem 6.1 follows directly from Corollary
6.6 using the second item above. The higher rank case requires a bit more,
namely to deduce from the above items that $\Lambda$ meets the hypotheses of
the Benoist-Miquel Theorem 6.4. To that end I use a well known geometric
criterion (Lemma 6.47) in order show that $\Lambda$ is Zariski dense, and a
lemma of Mostow (Lemma 6.32) to show that $\Lambda$ intersects a horospherical
subgroup in a cocompact lattice.

##### Outline for Section 6.

Section 6.2 is the core of the original mathematics of this paper. It is
devoted to proving that $\Lambda\cdot x_{0}$ intersects some horospheres in a
cocompact lattice. The proof is quite delicate and somewhat involved, and I
include a few figures and a detailed informal overview of the proof. The
figures are detailed and may take a few moments to comprehend, but I believe
they are worth the effort.

Section 6.3 deals with the case where $\Gamma\subset\mathcal{N}_{D}(\Lambda)$,
and elaborates on Schwartz’s and Eskin’s proofs of Theorem 6.8 and Theorem
6.9. Section 6.4 is devoted to the translation of the geometric results of
Section 6.2 to the algebraic language used in Theorem 6.4. Though the work is
indeed mainly one of translation, some of it is non-trivial. Finally, in
Section 6.5 I put everything together for a complete proof of Theorem 6.1.

I highly recommend the reader to have a look at the uniform case in Section 5
before reading this one.

###  6.2 A $\Lambda$-Cocompact Horosphere

Recall that $d_{\gamma}:=d(\gamma x_{0},\lambda_{\gamma}x_{0})$. In this
section I prove:

######  Proposition 6.11 (Proposition 1.9).

If $\\{d_{\gamma}\\}_{\gamma\in\Gamma}$ is unbounded, then there exists a
horosphere $\mathcal{H}$ based at $\mathcal{W}_{\mathbb{Q}}(\Gamma)$ such that
$\big{(}\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H})\big{)}\cdot x_{0}$
intersects $\mathcal{H}$ in a cocompact metric lattice. Moreover, the bounded
horoball $\mathcal{HB}$ is $\Lambda$-free.

Throughout Section 6.2 the standing assumptions are that
$\\{d_{\gamma}\\}_{\gamma}\in\Gamma$ is unbounded, and $\Gamma$ is an
irreducible $\mathbb{Q}$-rank $1$ lattice.

##### The Argument.

The proof is by chasing down the geometric implications of unbounded
$d_{\gamma}$. These implications are delicate, but similar in spirit to the
straight-forward proof for uniform lattices. The proof consists of the
following steps:

  1. 1.

Unbounded $d_{\gamma}$ results in $\Lambda$-free horoballs
$\mathcal{HB}^{\Lambda}$ tangent to $\Lambda$-orbit points. Each such horoball
is based at $\mathcal{W}_{\mathbb{Q}}(\Gamma)$, giving rise to corresponding
horoballs of $\Gamma$, denoted $\mathcal{HB}^{\Gamma}$.

  2. 2.

If $d_{\gamma}$ is large, then $\gamma x_{0}$ must lie deep inside a unique
$\Lambda$-free horoball tangent to $\lambda_{\gamma}x_{0}$. I use:

    1. (a)

A bound on the distance $d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})$.

    2. (b)

A bound on the angle
$\angle_{\lambda_{\gamma}x_{0}}([\lambda_{\gamma}x_{0},\gamma
x_{0}],[\lambda_{\gamma}x_{0},\xi))$, where $\xi\in X(\infty)$ is the base
point of a suitable $\Lambda$-free horoball tangent to
$\lambda_{\gamma}x_{0}$.

  3. 3.

There exist horospheres of $\Gamma$, say $\mathcal{H}^{\Gamma}$, such that if
$\gamma x_{0}\in\mathcal{H}^{\Gamma}$ then large $d_{\gamma}$ implies large
$\Lambda$-free areas along the bounding horosphere of some
$\mathcal{HB}^{\Lambda}$.

  4. 4.

If $\mathcal{HB}^{\Lambda}$ is (uniformly) boundedly close to some
$\Lambda$-orbit point, then I show that $\mathcal{H}^{\Lambda}$ is _almost
$\Lambda$-cocompact_, that is
$\mathcal{H}^{\Lambda}\subset\mathcal{N}_{D}(\Lambda\cdot x_{0})$ for some
universal $D=D(\Lambda)$. Together with the previous step, this yields a
uniform bound on $d_{\gamma}$ along certain horospheres of $\Gamma$.

  5. 5.

Finally I elevate the almost cocompactness to actual cocompactness and show
$\mathcal{H}^{\Lambda}\subset\mathcal{N}_{D}(\Lambda\cdot
x_{0}\cap\mathcal{H}^{\Lambda})$ for some $D>0$. This immediately elevates to
$\mathcal{H}^{\Lambda}\subset(\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H}^{\Lambda}))\cdot
x_{0}$, proving the proposition.

##### The Properties of $\Gamma$.

The geometric properties of $\Gamma$ that are used in the proof are:

  1. 1.

In higher rank, the characterization of $\mathcal{W}_{\mathbb{Q}}(\Gamma)$
using conical / non-horospherical limit points (Corollary 2.31). In
$\mathbb{R}$-rank$\ 1$, the dichotomy of limit points being either non-
horospherical or conical (Theorem 2.30).

  2. 2.

$\Gamma$-cocompactness along the horospheres of $\Gamma$.

  3. 3.

For every point $x\in X$ and $C>0$ there is a bound $K(C)$ on the number of
horospheres of $\Gamma$ that intersect $B(x,C)$ (Corollary 2.23).

####  6.2.1 $\Lambda$-Free Horoballs

I retain the notations and objects defined in Section 5.1.

######  Lemma 6.12.

There exists a $\Lambda$-free horoball tangent to $x_{0}$.

###### Proof.

Since $\\{d_{\gamma}\\}_{\gamma\in\Gamma}$ is unbounded, there are
$\gamma_{n}\in\Gamma$ with
$d_{n}=d_{\gamma_{n}}=d(\gamma_{n},\lambda_{n})\rightarrow\infty$
monotonically, where $\lambda_{n}\in\Lambda$ is a $\Lambda$-orbit point
closest to $\gamma_{n}$. Denote
$x_{n}^{\prime}=\lambda_{n}^{-1}\gamma_{n}x_{0}$,
$\eta_{n}:=[x_{0},x_{n}^{\prime}]$, and $v_{n}\in S_{x_{0}}X$ the initial
velocity vectors $v_{n}:=\dot{\eta_{n}}(0)$. The tangent space $S_{x_{0}}X$ is
compact, so up to a subsequence, $v_{n}$ converge monotonically in angle to a
direction $v\in S_{x_{0}}X$. Let $\eta$ be the unit speed geodesic ray
emanating from $x_{0}$ with initial velocity $\dot{\eta}(0)=v$. Denote
$\xi:=\eta(\infty)$ the limit point of $\eta$ in $X(\infty)$.

I claim that the horoball $\mathcal{HB}:=\cup_{t>0}B\big{(}\eta(t),t\big{)}$,
based at $\xi$ and tangent to $x_{0}$, is $\Lambda$-free. Let $t>0$ and
consider $\eta(t)$. For every $\varepsilon>0$, there is some angle
$\alpha=\alpha(t,\varepsilon)$ such that any geodesic $\eta^{\prime}$ with
$\angle_{x_{0}}(\eta,\eta^{\prime})<\alpha$ admits
$d\big{(}\eta(t),\eta^{\prime}(t)\big{)}<\varepsilon/2$. The convergence
$v_{n}\rightarrow v$ implies
$d\big{(}\eta(t),\eta_{n}(t)\big{)}<\varepsilon/2$ for all but finitely many
$n\in\mathbb{N}$. In particular,
$B\big{(}\eta(t),t\big{)}\subset\mathcal{N}_{\varepsilon}\Big{(}B\big{(}\eta_{n}(t),t\big{)}\Big{)}$
for all such $n\in\mathbb{N}$.

For a fixed $t\leq d_{n}$, it is clear from the definitions that
$B\big{(}\eta_{n}(t),t\big{)}\subset B_{n}^{\prime}=B(x_{n}^{\prime},d_{n})$.
One has $d_{n}\rightarrow\infty$, and so for a fixed $t>0$ it holds that
$t<d_{n}$ for all but finitely many $n\in\mathbb{N}$. I conclude that for any
fixed $t>0$ there is $n$ large enough such that

$B\big{(}\eta(t),t\big{)}\subset\mathcal{N}_{\varepsilon}\Big{(}B\big{(}\eta_{n}(t),t\big{)}\Big{)}\subset\mathcal{N}_{\varepsilon}B_{n}^{\prime}$

I conclude that for every $\varepsilon>0$,
$\mathcal{HB}\subset\bigcup_{n}\mathcal{N}_{\varepsilon}(B_{n}^{\prime})=\mathcal{N}_{\varepsilon}\big{(}\bigcup_{n}B_{n}^{\prime}\big{)}$.
This implies that any point in the interior of $\mathcal{HB}$ is contained in
the interior of one of the $\Lambda$-free balls $B_{n}^{\prime}$, proving
$\mathcal{HB}$ is $\Lambda$-free.

∎

######  Lemma 6.13.

Suppose $\mathcal{HB}$ is a $\Lambda$-free horoball, based at some point
$\xi\in X(\infty)$. Then $\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$.

###### Proof.

For any geodesic $\eta$ with limit $\xi$, the size $d(x_{0},\gamma x_{0})$ of
the $\Gamma$-orbit points $\gamma x_{0}$ that lie boundedly close to $\eta$
grows linearly in the distance to any fixed horosphere based at $\xi$, and in
particular to $\mathcal{H}=\partial\mathcal{HB}$. The sublinear constraint
$d(\gamma x_{0},\lambda_{\gamma}x_{0})\leq u(|\gamma|)$ together with the fact
that $\mathcal{HB}$ is $\Lambda$-free imply that the size of such $\gamma$ is
bounded. In $\mathbb{R}$-rank$\ 1$ every limit point is either conical or in
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$, proving the lemma in this case. For higher
rank, the above argument actually shows more: it shows that a point
$\xi^{\prime}\in\mathcal{N}_{\frac{\pi}{2}}(\xi)$ is not conical, because
every geodesic with limit $\xi^{\prime}\in\mathcal{N}_{\frac{\pi}{2}}(\xi)$
entres $\mathcal{HB}$ at a linear rate (Lemma 2.32). Hattori’s
characterization of $\mathcal{W}_{\mathbb{Q}}(\Gamma)$ (Corollary 2.31)
implies $\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$.

∎

######  Definition 6.14.

Given a $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$, Lemma 6.13 gives
rise to a horoball of $\Gamma$ based at the same point at $X(\infty)$. Call
this the _horoball corresponding to $\mathcal{HB}^{\Lambda}$_, and denote it
by $\mathcal{HB}^{\Gamma}$. The corresponding horosphere is denoted
$\mathcal{H}^{\Gamma}$.

######  Remark 6.15.

In the course of my work I had had a few conversations with Omri Solan
regarding the penetration of geodesics into $\Lambda$-free horoballs. Assuming
$\Lambda\subset\mathcal{N}_{u}(\Gamma)$ implies that $\Lambda$ preserves
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$ (see Lemma 6.29). This is the case in the
motivating setting where $\Lambda$ is an abstract finitely generated group
that is SBE to $\Gamma$, see Claim 3.26 in Chapter 3. In the case of
$SL_{2}(\mathbb{R})$ Omri suggested to use the action of $\Lambda$ on the
Bruhat-Tits tree of $SL_{2}(\mathbb{Q}_{p})$ (for all primes $p$) and the
classification of these elements into elliptic and hyperbolic elements
(separately for each $p$) in order to deduce that $\Lambda$ actually lies in
$SL_{2}(\mathbb{Z})$. We did not pursue that path nor its possible
generalization to the $SL_{n}$ case and general Bruhat-Tits buildings.

####  6.2.2 A $\Gamma$-orbit point Lying Deep Inside a $\Lambda$-Free Horoball

I established the existence of $\Lambda$-free horoballs. It may seem odd that
the first step in proving $\Lambda\cdot x_{0}$ is ‘almost everywhere’ is
proving the existence of $\Lambda$-free regions. But this fits perfectly well
with the algebraic statement that non-uniform lattices must admit unipotent
elements (see Proposition $5.3.1$ in [38]). The goal of this section is to
obtain some control on the location of the $\Lambda$-free horoballs, in order
to conclude that some $\gamma x_{0}$ lies deep inside
$\mathcal{HB}^{\Lambda}$. This results in yet more $\Lambda$-free regions,
found on the bounding horosphere.

I need one property of sublinear functions. I thank Panos Papazoglou for
noticing a mistake in the original formulation.

######  Lemma 6.16.

Let $u$ be a sublinear function, $f,g:\mathbb{R}_{\geq
0}\longrightarrow\mathbb{R}_{>0}$ two positive monotone functions with
$\lim_{x\rightarrow\infty}f(x)+g(x)=\infty$. If for all large enough $x$ it
holds that $f(x)\leq u\big{(}f(x)+g(x)\big{)}$, then for every $1<s$ and all
large enough $x$ it holds that $f(x)\leq u\big{(}s\cdot g(x)\big{)}$. In
particular $f(x)\leq u^{\prime}\big{(}g(x)\big{)}$ for some sublinear function
$u^{\prime}$.

###### Proof.

Assume as one may that $u$ is non-decreasing. By definition of sublinearity
$\lim_{x\rightarrow\infty}\frac{u\big{(}f(x)+g(x)\big{)}}{f(x)+g(x)}=0$, so by
hypothesis $\lim_{x\rightarrow\infty}\frac{f(x)}{f(x)+g(x)}=0$. This means
that for every $\varepsilon>0$ one has $f(x)\leq\varepsilon\cdot g(x)$ for all
large enough $x$, resulting in

$f(x)\leq u\big{(}f(x)+g(x)\big{)}\leq u\big{(}(1+\varepsilon)\cdot
g(x)\big{)}$

Notice that for a fixed $s>0$, the function $u^{\prime}(x)=u(sx)$ is
sublinear, as

$\lim_{x\rightarrow\infty}\frac{u(sx)}{x}=\lim_{x\rightarrow\infty}s\cdot\frac{u(sx)}{sx}=0$

∎

######  Lemma 6.17.

Let $C>0$. There is $L=L(C)$ such that if $\mathcal{HB}^{\Lambda}$ is any
$\Lambda$-free horoball tangent to a point $x\in B(x_{0},C)$ then
$d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})\leq L$. Moreover, there is a
sublinear function $u^{\prime}$ such that:

$L(C)\leq\begin{cases}u^{\prime}(C)&\text{if
}\mathcal{HB}^{\Gamma}\subset\mathcal{HB}^{\Lambda}\\\ C&\text{if
}\mathcal{HB}^{\Lambda}\subset\mathcal{HB}^{\Gamma}\end{cases}$

###### Proof.

If $\mathcal{HB}^{\Lambda}\subset\mathcal{HB}^{\Gamma}$, then clearly
$d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})\leq C$, simply because
$\mathcal{HB}^{\Gamma}$ is $\Gamma$-free and in particular cannot contain
$x_{0}$. Therefore $\mathcal{H}^{\Gamma}$ must separate
$\mathcal{H}^{\Lambda}$ from $x_{0}$ and in particular
$d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})\leq C$.

Assume that $\mathcal{HB}^{\Gamma}\subset\mathcal{HB}^{\Lambda}$, and denote
$l=d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})$. The horoball
$\mathcal{HB}^{\Gamma}$ is a horoball of $\Gamma$, hence $\Gamma\cdot x_{0}$
is $D_{\Gamma}$-cocompact along $\mathcal{H}^{\Gamma}$ and there is an element
$\gamma\in\Gamma$ with $|\gamma|\leq C+l+D_{\Gamma}$ and $\gamma
x_{0}\in\mathcal{H}^{\Gamma}$. Since $\mathcal{HB}^{\Lambda}$ is
$\Lambda$-free one has

$l\leq d(\gamma x_{0},\lambda_{\gamma}x_{0})\leq u(|\gamma|)\leq
u(C+l+D_{\Gamma})$

$C,D_{\Gamma}$ are fixed, so this inequality can only occur for boundedly
small $l$, say $l<L^{\prime}(C)$ ($D_{\Gamma}$ is a universal constant and may
be ignored). Consult figure 2 for a geometric visualization of this situation.

It remains to show that $L^{\prime}(C)$ is indeed sublinear in $C$. First
define $L(C)$ to be the minimal $L$ that bounds the distance
$d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})$ for all possible
$\mathcal{HB}^{\Lambda}$ tangent to a point $x\in B(x_{0},C)$. This is indeed
a minimum, since by Corollary 2.23 there are only finitely many horoballs of
$\Gamma$ intersecting $B\big{(}x_{0},C+L^{\prime}(C)\big{)}$. For every $C$
there is thus a horoball $\mathcal{HB}^{\Gamma}_{C}$ and an element
$\gamma\in\Gamma$ such that $\gamma x_{0}\in\mathcal{H}^{\Gamma}$,
$d(\mathcal{H}^{\Lambda}_{C},\mathcal{H}^{\Gamma}_{C})=L(C)$ and $|\gamma|\leq
C+L(C)+D_{\Gamma}$. The fact that $\mathcal{HB}^{\Lambda}_{C}$ is
$\Lambda$-free implies

$L(C)=d(\mathcal{H}^{\Lambda}_{C},\mathcal{H}^{\Gamma}_{C})\leq
u(|\gamma|)\leq u\big{(}C+L(C)+D_{\Gamma}\big{)}$

Lemma 6.16 implies that $L(C)\leq u^{\prime}(C)$ for some sublinear function
$u^{\prime}$. ∎

$Rad=C$$x_{0}$$\Lambda-free\ \mathcal{HB}^{\Lambda}$$\xi$$\Gamma-free\
\mathcal{HB}^{\Gamma}$$\leq L(C)$$\gamma
x_{0}$$P_{\mathcal{H}^{\Gamma}}(x_{0})$$\leq L(C)+D_{\Gamma}$ Figure 2: Lemma
6.17. A $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$ intersects a ball of
radius $C$ about $x_{0}$. The associated $\Gamma$-free horoball
$\mathcal{HB}^{\Gamma}$ is boundedly close, essentially due to the uniform
cocompactness of $\Gamma\cdot x_{0}$ along the $\Gamma$ horospheres.

The following is an immediate corollary, apparent already in the above proof.

######  Corollary 6.18.

For every $C>0$ there is a bound $K=K(C)$ and a fixed set
$\xi_{1},\xi_{2},\dots\xi_{K}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)\subset
X(\infty)$ so that every $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$
which is tangent to some point $x\in B(x_{0},C)$ is based at $\xi_{i}$ for
some $i\in\\{1,\dots,K\\}$. In particular, for any specific point
$x\in\mathcal{N}_{C}(\Lambda\cdot x_{0})$ there are at most $K$ $\Lambda$-free
horoballs tangent to $x$.

###### Proof.

Let $\mathcal{HB}^{\Lambda}$ be a horoball tangent to a point $x\in
B(x_{0},C)$. Lemma 6.17 bounds
$d(\mathcal{HB}^{\Lambda},\mathcal{HB}^{\Gamma})$ by $L(C)$, hence
$\mathcal{HB}^{\Gamma}$ is tangent to a point $x^{\prime}\in
B\big{(}x_{0},C+L(C)\big{)}$. By Corollary 2.23 there are only finitely many
possibilities for such $\mathcal{HB}^{\Gamma}$. In particular there are
finitely many base-points for these horoballs, say
$\xi_{1},\xi_{2},\dots,\xi_{K(C)}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$.
Finally, recall that a horoball is determined by a base point and a point
$x\in X$ tangent to it, so the last statement of the corollary holds for any
$x\in B(x_{0},C)$. But the property in question is $\Lambda$-invariant so the
same holds for any point $x\in\Lambda\cdot
B(x_{0},C)=\mathcal{N}_{C}(\Lambda\cdot x_{0})$. ∎

The bound on $d(\mathcal{HB}^{\Lambda},\mathcal{HB}^{\Gamma})$ given by Lemma
6.17 further strengthen the relation between $\mathcal{HB}^{\Lambda}$ and
$\mathcal{HB}^{\Gamma}$. The ultimate goal is to show that the
$\mathcal{HB}^{\Lambda}$-s play the role of the $\Gamma$-horoballs in the
geometric structure of $\mathbb{Q}$-rank $1$ lattices, namely to show that
$\Lambda\cdot x_{0}$ is cocompact on the $\mathcal{H}^{\Lambda}$-s. This
requires to actually find $\Lambda$-orbit points somewhere in $X$, and not
just $\Lambda$-free regions as was done up to now. As one might suspect, these
points arise as $\lambda_{\gamma}x_{0}$ corresponding to points $\gamma
x_{0}\in\mathcal{H}^{\Gamma}$, which exist in abundance since $\Gamma\cdot
x_{0}\cap\mathcal{H}^{\Gamma}$ is a cocompact lattice in
$\mathcal{H}^{\Gamma}$.

The hope is that a $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$ tangent to
$\lambda_{\gamma}x_{0}$ would correspond to a horoball of $\Gamma$ tangent to
$\gamma x_{0}$. This would have forced all the $\lambda_{\gamma}$ to actually
lie on the same bounding horosphere, and $\\{\lambda_{\gamma}x_{0}\mid\gamma
x_{0}\in\mathcal{H}^{\Gamma}\\}$ would then be a cocompact lattice in
$\mathcal{H}^{\Lambda}$. This hope turns out to be more or less true, but it
requires some work. The goal of the rest of this section is to establish a
relation between a $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$ tangent to
$\lambda_{\gamma}x_{0}$ and $\gamma x_{0}$. I start with some notations.

######  Definition 6.19.

In light of Corollary 6.18, there is a finite number $N$ of $\Lambda$-free
horoballs tangent to $x_{0}$. Denote:

  1. 1.

$\\{\mathcal{HB}^{\Lambda}_{i}\\}_{1=1}^{N}$ are the $\Lambda$-free horoballs
tangent to $x_{0}$.

  2. 2.

$\xi^{i}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$ is the base point of
$\mathcal{HB}^{\Lambda}_{i}$.

  3. 3.

$v^{i}\in S_{x_{0}}X$ is the unit tangent vector in the direction $\xi^{i}$.

  4. 4.

$\eta^{i}:=[x_{0},\xi_{i})$ is the unit speed geodesic ray emanating from
$x_{0}$ with limit $\xi^{i}$. In particular $v^{i}=\frac{d}{dt}\eta^{i}(0)$.

  5. 5.

$\mathcal{HB}^{\Gamma}_{i}$ is the horoball of $\Gamma$ that corresponds to
$\mathcal{HB}^{\Lambda}_{i}$, based at $\xi^{i}$.

  6. 6.

$\mathcal{HB}^{\Lambda}_{\lambda,i},\xi^{i}_{\lambda},\eta^{i}_{\lambda}$ are
the respective $\lambda$-translates of the objects above. For example,
$\mathcal{HB}^{\Lambda}_{\lambda,i}:=\lambda\cdot\mathcal{HB}^{\Lambda}_{i}$.

  7. 7.

$\mathcal{H}$ decorated by the proper indices denotes the horosphere bounding
$\mathcal{HB}$, the horoball with respective indices, e.g.
$\mathcal{H}^{\Lambda}_{i}:=\partial\mathcal{HB}^{\Lambda}_{i}$.

  8. 8.

For an angle $\alpha>0$ and a tangent vector $v_{0}\in S_{x}X$, define

    1. (a)

The _$\alpha$ -sector of $v$ in $S_{x}X$_ is the set $\\{v\in S_{x}X\mid
v\in\mathcal{N}_{\alpha}(v_{0})\\}$. Recall that the metric on $S_{x}X$ is the
angular metric.

    2. (b)

The _$\alpha$ -sector of $v$ in $X$_ are all points $y\in X$ for which the
tangent vector at $0$ of the unit speed geodesic $[x,y]$ lies in the
$\alpha$-sector of $v$ in $SxX$.

######  Lemma 6.20.

For every angle $\alpha\in(0,\frac{\pi}{2})$ there exists $D=D(\alpha)$ such
that if $d_{\gamma}>D$ then for some $i\in\\{1,\dots,N\\}$, $\gamma x_{0}$
lies inside the $\alpha$-sector of $v^{i}_{\lambda_{\gamma}}$ at
$\lambda_{\gamma}x_{0}$. Furthermore whenever $\alpha$ is uniformly small
enough, there is a unique such $i=i(\gamma)$, independent of $\alpha$.

###### Proof.

Translation by the isometry $\lambda_{\gamma}^{-1}$ preserves angles and
distances, so it is enough to prove that there is an $i$ for which
$x^{\prime}_{\gamma}:=\lambda^{-1}\gamma x_{0}$ lies inside the
$\alpha$-sector of $v^{i}$, and that this $i$ is unique if $\alpha$ is
uniformly small. Assume towards contradiction that there is
$\alpha\in(0,\frac{\pi}{2})$ and a sequence $\gamma_{n}\in\Gamma$,
$\lambda_{n}:=\lambda_{\gamma_{n}}\in\Lambda$ with $d_{\gamma_{n}}$ unbounded,
and $x^{\prime}_{n}:=\lambda_{n}^{-1}\gamma_{n}x_{0}$ not lying in the union
of the $\alpha$-sectors of $v^{i}$. By perhaps taking smaller $\alpha$ I may
assume all the $\alpha$-sectors of the $v^{i}$ in $S_{x_{0}}X$ are pairwise
disjoint. This can be done because there are only finitely many $v^{i}$.

Compactness of $S_{x_{0}}X$ allows me to take a converging subsequence
$v^{\prime}_{n}:=\dot{[x_{0},x^{\prime}_{n}]}$, with limit direction
$v^{\prime}$. Denote by $\eta^{\prime}$ the geodesic ray emanating from
$x_{0}$ with initial velocity $v^{\prime}$. The exact same argument of Lemma
6.12 proves that $\eta^{\prime}(\infty)$ is the base point of a $\Lambda$-free
horoball tangent to $x_{0}$. But this means $v^{\prime}=v^{i}$ for some
$i\in\\{1,\dots,N\\}$, contradicting the fact that all $v^{\prime}_{n}$ lie
outside the $\alpha$-sectors of the $v^{i}$. This proves that there is a bound
$D=D(\alpha)$ such that if $d_{\gamma}>D$ then $x_{\gamma}^{\prime}$ lies
within the $\alpha$-sector of some $v^{i}$.

The proof clearly shows that whenever $\alpha$ is small enough so that the
$\alpha$-sectors of the $v^{i}$ are disjoint, $x_{\gamma}^{\prime}$ lies in
the $\alpha$-sector of a unique $v^{i}$ as soon as $d_{\gamma}>D(\alpha)$.

∎

######  Remark 6.21.

In the proof of Lemma 6.12 I used compactness of $S_{x_{0}}X$ to induce a
converging subsequence of directions. Lemma 6.20 actually shows that the fact
there are finitely many $\Lambda$-free horoballs tangent to $x_{0}$ implies _a
posteriori_ that there was not much choice in the process - all directions
$[x_{0},x_{\gamma}^{\prime}]$ must fall into one of the finitely many
directions $v^{i}$.

Next, I want to control the actual location of certain points with respect to
the horoballs of interest, and not just the angles. This turns out to be a
more difficult of a task than one might suspect, since control on angles does
not immediately give control on distances.

Recall that large $\Lambda$-free balls near $x_{0}$ imply large concentric
$\Gamma$-free balls. The precise quantities and bounds are given by Lemma 5.6
(one can use Lemma 5.7 to obtain a slightly cleaner statement).

######  Proposition 6.22.

Let $S\in(0,1)$ be the constant given by Lemma 5.6, and let $s\in(0,S)$. There
is a bound $D=D(s)$ such that $d_{\gamma}>D$ implies that $\gamma x_{0}$ lies
$sd_{\gamma}$ deep in $\mathcal{HB}^{\Lambda}_{\lambda_{\gamma}}$.

###### Proof.

The proof is a bit delicate but very similar to that of Lemma 5.6. In essence,
I use the $\Gamma$-free balls near $x_{0}$ to produce a $\Gamma$-free
cylinder, which would force a certain geodesic not to cross a horosphere of
$\Gamma$, i.e. force it to stay inside a $\Gamma$-free horoball.

As in Lemma 6.20 it is only required to show that
$x^{\prime}=\lambda_{\gamma}^{-1}\gamma x_{0}$ is $sd_{\gamma}$ deep inside
$\mathcal{HB}^{\Lambda}_{i(\gamma)}$. I start with proving that
$x^{\prime}\in\mathcal{HB}^{\Gamma}_{i(\gamma)}$. I learned the hard way that
even this is not a triviality. Recall the notation $B_{\gamma}=B(\gamma
x_{0},d_{\gamma})$. The ball
$B^{\prime}_{\gamma}=\lambda_{\gamma}^{-1}B_{\gamma}$ is a $\Lambda$-free ball
of radius $d_{\gamma}$ about $x^{\prime}=\lambda_{\gamma}^{-1}\gamma x_{0}$.
Denote by $x^{\prime}_{t}$ the point at time $t$ along the unit speed geodesic
$\eta^{\prime}:=[x_{0},x^{\prime}]$. It holds that $|x_{t}^{\prime}|=t$ and,
for $t\leq d_{\gamma}$, $x_{t}^{\prime}$ is the centre of a $\Lambda$-free
ball of radius $t$ tangent to $x_{0}$. The constant $s$ is fixed and by Lemma
5.6 there is $T^{\prime}=T^{\prime}(s)$ such that if $t>T^{\prime}$, the ball
$sB\big{(}x^{\prime}_{t},t\big{)}$ is $\Gamma$-free.

The next goal is to show that $x_{T}^{\prime}\in\mathcal{HB}^{\Gamma}$ for
some adequate $T$. For any time $T>0$, let $\alpha=\alpha(\varepsilon,T)$ be
the angle for which $d\big{(}\eta(T),\eta^{i(\gamma)}(T)\big{)}<\varepsilon$
for every $\eta$ in the $\alpha$-sector of $v^{i(\gamma)}$. By perhaps taking
smaller $\alpha$ I may assume that $\alpha$ is uniformly small as stated in
Lemma 6.20. Let $D(\alpha)$ be the bound given by Lemma 6.20 guaranteeing

$D(\alpha)<d_{\gamma}\Rightarrow
d\big{(}x^{\prime}_{T},\eta^{i(\gamma)}(T)\big{)}<\varepsilon$

For my needs in this lemma $\varepsilon$ may as well be chosen to be $1$. I
now choose a specific time $T$ for which I want $x_{T}^{\prime}$ and
$\eta^{i(\gamma)(T)}$ to be close. There are only finitely many $\Lambda$-free
horoballs $\\{\mathcal{HB}^{\Lambda}_{i}\\}_{i\in\\{1,\dots,N\\}}$ tangent to
$x_{0}$, giving rise to a uniform bound
$L=\max_{i\in\\{1,\dots,N\\}}\\{d(\mathcal{H}^{\Lambda}_{i},\mathcal{H}^{\Gamma}_{i})\\}$
on the distance
$d(\mathcal{H}^{\Lambda}_{i(\gamma)},\mathcal{H}^{\Gamma}_{i(\gamma)})$. Fix
$T$ to be any time in the open interval
$(T^{\prime}+L+\varepsilon,d_{\gamma})$. The fact that $L+\varepsilon<T$
implies that $\eta^{i(\gamma)}(T)$ lies at least $\varepsilon$-deep inside
$\mathcal{HB}^{\Gamma}_{i(\gamma)}$, and therefore
$\eta^{\prime}(T)\in\mathcal{HB}^{\Gamma}_{i(\gamma)}$. Recall that any point
on $\mathcal{H}^{\Gamma}$ is $D_{\Gamma}$-close to a point $\gamma
x_{0}\in\mathcal{H}^{\Gamma}$. By perhaps enlarging $T$ and shrinking $\alpha$
if necessary, I may assume that $D_{\Gamma}<sT$. Thus for all $T<t\leq
d_{\gamma}$, $x_{t}^{\prime}$ is the centre of a $\Gamma$-free ball of radius
$st>sT>D_{\Gamma}$, hence $\\{x^{\prime}_{t}\\}_{T\leq t\leq d_{\gamma}}$ does
not cross a horosphere of $\Gamma$. Since
$x_{T}^{\prime}\in\mathcal{HB}^{\Gamma}_{i(\gamma)}$, this implies that
$x_{t}^{\prime}$ stays in $\mathcal{HB}^{\Gamma}_{i(\gamma)}$ for all $T<t\leq
d_{\gamma}$. In particular
$x^{\prime}_{d_{\gamma}}=x^{\prime}\in\mathcal{HB}^{\Gamma}_{i(\gamma)}$.

To get the result of the proposition, recall that
$sB_{\gamma}^{\prime}=B(x^{\prime},sd_{\gamma})$ is $\Gamma$-free, so
$x^{\prime}$ must be at distance at least $sd_{\gamma}-D_{\Gamma}$ from any
horosphere of $\Gamma$, and in particular from
$\mathcal{H}^{\Gamma}_{i(\gamma)}$. In terms of Busemann functions, this means
that $b_{\eta^{i(\gamma)}}(x^{\prime})\leq-sd_{\gamma}+D_{\Gamma}$ whenever
one can find such $T^{\prime}+L+\varepsilon<T<d_{\gamma}$. Since
$\mathcal{HB}^{\Lambda}_{i(\gamma)}$ is tangent to $x_{0}$, the corresponding
horoball $\mathcal{HB}^{\Gamma}_{i(\gamma)}$ lies inside it, and so
$x^{\prime}$ lies $(sd_{\gamma}-D_{\Gamma})$-deep inside
$\mathcal{HB}^{\Lambda}_{i(\gamma)}$. A close look at the argument yields the
desired bound $D=D(s)$ such that the above holds whenever $d_{\gamma}>D(s)$.
To help the reader take this closer look, I reiterate the choice of constants
and their dependencies as they appear in the proof:

  1. 1.

Fix $\varepsilon=1$.

  2. 2.

Let $T^{\prime}=T^{\prime}(s)$ the constant from Lemma 5.6 and
$L=\max_{i\in\\{1,\dots,N\\}}\\{d(\mathcal{HB}^{\Lambda}_{i},\mathcal{HB}^{\Gamma}_{i})\\}$.

  3. 3.

Fix $T>T^{\prime}+L+1$.

  4. 4.

Fix $\alpha=\alpha(1,T)$.

  5. 5.

Fix $D(s)=\max\\{D(\alpha),T+1\\}$.

I remark, for the reader worried about the $D_{\Gamma}$ which appears in the
final bound but not in the statement, that (a) $D_{\Gamma}$ is a fixed
universal constant and may as well be ignored, and (b) the discrepancy can be
formally corrected by taking a slightly larger $s<s^{\prime}$ to begin with
and as a result perhaps enlarging the bound $D$ for $d_{\gamma}$). Also note
that $L=L(\Lambda)$ is a universal constant.

∎

####  6.2.3 Intersection of $\Lambda$-Free Regions and the Existence of a
$\Lambda$-Cocompact Horosphere

In this section I find $\Lambda$-orbit points that lie close to the bounding
horosphere of a $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$. In order to
find such points I need to make sure $\mathcal{HB}^{\Lambda}$ is not contained
inside a much larger $\Lambda$-free horoball. I introduce the following
definition.

######  Definition 6.23.

A $\Lambda$-free horoball $\mathcal{HB}^{\Lambda}$ is called _maximal_ if it
is tangent to a point $x=\lambda x_{0}\in\Lambda\cdot x_{0}$. It is called
_$\varepsilon$ -almost maximal_ if $d(\Lambda\cdot
x_{0},\mathcal{H}^{\Lambda})<\varepsilon$.

######  Remark 6.24.

It may happen that a discrete group admits free but not _maximally free_
horoballs - see discussion in section $4$ of [56]. In any case it is clear
that any $\Lambda$-free horoball can be ‘blown-up’ to an $\varepsilon$-almost
maximal $\Lambda$-free horoball, for every $\varepsilon>0$. Moreover, every
two $\varepsilon$-almost maximal horoballs based at the same point
$\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$ lie at distance at most $\varepsilon$
of one another. For my needs any fixed $\varepsilon$ would suffice, and I fix
$\varepsilon=1$.

######  Lemma 6.25.

There is $D_{\Lambda}>0$ such that if $\mathcal{HB}^{\Lambda}$ is $1$-almost
maximal $\Lambda$-free horoball then
$\mathcal{H}^{\Lambda}\subset\mathcal{N}_{D_{\Lambda}}(\Lambda\cdot x_{0})$,
i.e. $d(x,\Lambda\cdot x_{0})\leq D_{\Lambda}$ for all
$x\in\mathcal{H}^{\Lambda}$.

Notice that Lemma 6.25 does not state $\Lambda\cdot x_{0}$ even intersects
$\mathcal{H}^{\Lambda}$.

###### Proof.

I start with a short sketch of the proof. Consider a $1$-maximal horoball and
a point $x$ on its bounding horosphere with $d(x,\Lambda\cdot x_{0})=D$. One
may translate this situation to $x_{0}$, which results in a $\Lambda$-free
horoball $\mathcal{HB}^{\Lambda}$ intersecting the (closed) $D$-ball about
$x_{0}$ at a point $w$ with $B(w,D)$ $\Lambda$-free.

The proof differs depending on whether
$\mathcal{HB}^{\Gamma}\subset\mathcal{HB}^{\Lambda}$ or the other way round,
since I use the bounds from 6.17:

  1. 1.

If $\mathcal{HB}^{\Gamma}\subset\mathcal{HB}^{\Lambda}$, there is a sublinear
bound on $d(\mathcal{HB}^{\Lambda},\mathcal{HB}^{\Gamma})$, which readily
yields a bound on $D$.

  2. 2.

if $\mathcal{HB}^{\Lambda}\subset\mathcal{HB}^{\Gamma}$ there is a bound on
$d(x_{0},\mathcal{HB}^{\Gamma})$ that is independent of $D$. So there are only
finitely many possibilities for $\mathcal{HB}^{\Gamma}$, independent of $D$.
Hence there are only finitely many possible base points for
$\mathcal{HB}^{\Gamma}$. These in turn correspond to possible base points for
such $\mathcal{HB}^{\Lambda}$, and this finiteness yields a bound on the
distance $d(\mathcal{HB}^{\Gamma},\mathcal{HB}^{\Lambda})<L$ that is
independent of $D$. The rest of the proof is quite routine.

Let $\mathcal{HB}^{\Lambda}$ be a $1$-almost maximal $\Lambda$-free horoball.
By definition there is $\lambda\in\Lambda$ and $z\in\mathcal{H}^{\Lambda}$
such that $d(\lambda x_{0},z)<1$. Fix $D>0$. I show that if there is some
$z^{\prime}\in\mathcal{H}^{\Lambda}$ for which $d(z^{\prime},\Lambda\cdot
x_{0})\geq D$, then $D$ must be uniformly small. Exactly how small will be set
in the course of the proof.

Fix $D>1$ and assume that there is $z^{\prime}\in\mathcal{H}^{\Lambda}$ with
$d(z,\Lambda\cdot x_{0})\geq D$. Up to sliding $z^{\prime}$ along
$\mathcal{H}^{\Lambda}$, the continuity of the function $x\mapsto
d(x,\Lambda\cdot x_{0})$ together with Intermediate Value Theorem allows to
assume that $d(z^{\prime},\Lambda\cdot x_{0})=D$. Let
$\lambda^{\prime}\in\Lambda$ be the element for which
$d(z^{\prime},\lambda^{\prime}x_{0})=D$. Translating by $\lambda^{\prime-1}$
yields

  1. 1.

A $\Lambda$-free horoball
$\mathcal{HB}^{\Lambda}_{0}:=\lambda^{\prime-1}\mathcal{HB}^{\Lambda}$.

  2. 2.

A point $w:=\lambda^{\prime-1}z^{\prime}\in\mathcal{H}^{\Lambda}_{0}$ for
which $|w|=d(w,x_{0})=d(w,\Lambda\cdot x_{0})=D$.

Assume first that
$\mathcal{HB}^{\Gamma}_{0}\subset\mathcal{HB}^{\Lambda}_{0}$. By Lemma 6.17
there is a sublinear function $u^{\prime}$ such that
$d(\mathcal{H}^{\Gamma}_{0},\mathcal{H}^{\Lambda}_{0})\leq u^{\prime}(D)$.
This yields a point $\gamma x_{0}\in\mathcal{H}^{\Gamma}_{0}$ for which
$d(w,\gamma x_{0})\leq u^{\prime}(D)+D_{\Gamma}$. Thus $|\gamma x_{0}|\leq
D+u^{\prime}(D)+D_{\Gamma}$ and the reverse triangle inequality gives

$D-\big{(}u^{\prime}(D)+D_{\Gamma}\big{)}\leq
d(w,\lambda_{\gamma}x_{0})-d(w,\gamma x_{0})<d(\gamma
x_{0},\lambda_{\gamma}x_{0})$

Together with the bound $d(\gamma x_{0},\lambda_{\gamma}x_{0})\leq u(|\gamma
x_{0}|)$ and rearranging, one obtains

$D\leq u\big{(}D+u^{\prime}(D)+D_{\Gamma}\big{)}+u^{\prime}(D)+D_{\Gamma}$

The right hand side is clearly a sublinear function in $D$, hence this
inequality may hold only for boundedly small $D$, say $D<D_{1}$. I conclude
that $\mathcal{HB}^{\Gamma}_{0}\subset\mathcal{HB}^{\Lambda}_{0}$ may occur
only when $D<D_{1}$. Notice that $D_{1}$ depends only on $u$ and $u^{\prime}$,
and not on $\mathcal{HB}^{\Lambda}$.

Assume next that $\mathcal{HB}^{\Lambda}_{0}\subset\mathcal{HB}^{\Gamma}_{0}$,
and that the containment is strict. Since $x_{0}\in\Gamma\cdot x_{0}$, the
geodesic $\tau:=[x_{0},w]$ is of length $D$ and intersects
$\mathcal{H}^{\Gamma}_{0}$. Denote by $t_{0}\in[0,D)$ the time in which $\tau$
intersects $\mathcal{H}^{\Gamma}_{0}$, and let
$w^{\prime}:=\tau(t_{0})\in\mathcal{H}^{\Gamma}_{0}$ be the intersection
point. In particular $|w^{\prime}|=t_{0}$. It is clear that
$B(w^{\prime},t_{0})$ is $\Lambda$-free, as a subset of the ball $B(w,D)$.
Again there is $\gamma x_{0}\in
B(w^{\prime},D_{\Gamma})\cap\mathcal{H}^{\Gamma}_{0}$ and so $|\gamma
x_{0}|\leq t_{0}+D_{\Gamma}$. By reverse triangle inequality

$t_{0}-D_{\Gamma}\leq d(w^{\prime},\lambda_{\gamma}x_{0})-d(w^{\prime},\gamma
x_{0})\leq d(\gamma x_{0},\lambda_{\gamma}x_{0})$

and the sublinear constraint gives $t_{0}-D_{\Gamma}\leq u(t_{0}+D_{\Gamma})$.
This can only happen for boundedly small $t_{0}$, say $t_{0}<T$. I conclude
that if $\mathcal{HB}^{\Lambda}_{0}\subset\mathcal{HB}^{\Gamma}_{0}$, then
$\mathcal{HB}^{\Gamma}_{0}$ is a horoball of $\Gamma$ tangent to some point
$y\in B(x_{0},T)$.

By Corollary 2.23 there are finitely many horoballs of $\Gamma$ tangent to
points in $B(x_{0},T)$. In particular there is a finite set
$\\{\xi_{1}^{\prime},\dots,\xi_{K}^{\prime}\\}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$
of possible base points for $\mathcal{HB}^{\Gamma}_{0}$. This set depends only
on $T$, and since the choice of $T$ was completely independent of $D$, the set
of possible base points is independent of $D$ as well. Let
$\widetilde{\mathcal{HB}^{\Gamma}_{i}}$ be the horoball of $\Gamma$ based at
$\xi_{i}^{\prime}$.

I can now bound the distance
$d(\mathcal{HB}^{\Gamma}_{0},\mathcal{HB}^{\Lambda}_{0})$. Let $1\leq i\leq K$
be an index for which there is a $\Lambda$-free horoball based at
$\xi_{i}^{\prime}$ that is contained in
$\widetilde{\mathcal{HB}^{\Gamma}_{i}}$. There is thus some $1$-almost-maximal
$\Lambda$-free horoball based at $\xi^{\prime}_{i}$. Fix an arbitrary such
$1$-almost-maximal $\Lambda$-free horoball
$\widetilde{\mathcal{HB}^{\Lambda}_{i}}$ for each such $i$, and let
$L_{i}:=d(\widetilde{\mathcal{HB}^{\Lambda}_{i}},\widetilde{\mathcal{HB}^{\Gamma}_{i}})$.
Finally, define $L:=\max\\{L_{i}\\}+1$ among such $i$. As stated in Remark
6.24,
$d(\mathcal{HB}^{\Lambda}_{0},\widetilde{\mathcal{HB}^{\Lambda}_{i}})\leq 1$
for some $i$, therefore
$d(\mathcal{HB}^{\Gamma}_{0},\mathcal{HB}^{\Lambda}_{0})\leq L$.

Recall $|w|=D$ and $B(w,D)$ is $\Lambda$-free. It holds that
$d(w,\mathcal{H}^{\Gamma}_{0})\leq L$, and so there is $\gamma
x_{0}\in\mathcal{H}^{\Gamma}_{0}$ for which $d(w,\gamma x_{0})\leq
L+D_{\Gamma}$. In particular $|\gamma x_{0}|\leq D+L+D_{\Gamma}$ (in fact it
is clear that $|\gamma x_{0}|\leq T+D_{\Gamma}$, but this won’t be necessary).
Reverse triangle inequality gives

$D-(L+D_{\Gamma})\leq d(w,\lambda_{\gamma}x_{0})-d(w,\gamma x_{0})\leq
d(\gamma x_{0},\lambda_{\gamma}x_{0})$

and from the sublinear constraint I conclude $D-(L+D_{\Gamma})\leq
u(D+L+D_{\Gamma})$. Since $L,D_{\Gamma}$ are fixed constants independent of
$D$, this can only hold for boundedly small $D$, say $D<D_{2}$. In particular,
one gets a uniform bound $D_{\Lambda}:=\max\\{D_{1},D_{2}\\}$ such that
$x\in\mathcal{H}^{\Lambda}\Rightarrow d(x,\Lambda\cdot x_{0})<D_{\Lambda}$.

∎

$w$$x_{0}$$\begin{array}[]{l}\Lambda-free\\\ \
B(w,D)\end{array}$$\tau(t_{0})$$\tau$$\Lambda-free\
\mathcal{HB}^{\Lambda}{}$$\xi$$\Gamma-free\ \mathcal{HB}^{\Gamma}{}$$\gamma
x_{0}$ Figure 3: Lemma 6.25, case
$\mathcal{HB}^{\Lambda}\subset\mathcal{HB}^{\Gamma}$. The red horosphere of
$\Gamma$ is trapped between $x_{0}$ and $\mathcal{H}^{\Lambda}$, and is at
distance $t_{0}$ from $x_{0}$. A $\Gamma$-orbit point on the red horosphere
close to $x_{0}$ allows to use sublinearity to get a bound on $t_{0}$.

######  Corollary 6.26.

Every $\Gamma$-conical limit point is a $\Lambda$-conical limit point.

###### Proof.

Let $\xi\in X(\infty)$ be a $\Gamma$-conical limit point. Let
$\eta:\mathbb{R}_{\geq 0}\rightarrow X$ be a geodesic with $\eta(\infty)=\xi$.
By definition there is a bound $D>0$ and sequences
$t_{n}\rightarrow\infty,\gamma_{n}\in\Gamma$ such that
$d\big{(}\gamma_{n}x_{0},\eta(t_{n})\big{)}<D$. Consider the corresponding
$\lambda_{n}:=\lambda_{\gamma_{n}}$ and $\lambda_{n}x_{0}$. If $d_{n}$ is
uniformly bounded, then $\xi$ is $\Lambda$-conical by definition. Otherwise,
assume $d_{n}$ is monotonically increasing to $\infty$. For some fixed
$s\in(0,1)$ it holds that for all but finitely many $n\in\mathbb{N}$,
$\gamma_{n}x_{0}$ is $sd_{n}$ deep inside
$\mathcal{HB}^{\Lambda}_{n}:=\mathcal{HB}^{\Lambda}_{\lambda_{n},i(\gamma_{n})}$.
I assume $d_{n}$ is large enough so that $sd_{n}>D$, and in particular
$\eta(t_{n})\in\mathcal{HB}^{\Lambda}_{n}$. Let
$\xi_{n}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$ be the respective base points of
$\mathcal{HB}^{\Lambda}_{n}$. The point $\xi$ is $\Gamma$-conical, and by
Theorem 2.29 $\frac{\pi}{2}\leq d(\xi,\mathcal{W}_{\mathbb{Q}}(\Gamma))\leq
d(\xi,\xi_{n})$.

The proof differs depending on whether the above inequality is strict or not
for any $n\in\mathbb{N}$. Assume first that for some $m\in\mathbb{N}$,
$d(\xi,\xi_{m})=\frac{\pi}{2}$. By item $2$ of Lemma 2.32,
$d\big{(}\mathcal{H}^{\Lambda}_{m},\eta(t)\big{)}$ is uniformly bounded, i.e.,
there is $C>0$ such that for every $t>0$ there is
$x_{t}\in\mathcal{H}^{\Lambda}_{m}$ for which
$d\big{(}x_{t},\eta(t)\big{)}<C$. By Lemma 6.25, $d(x_{t},\Lambda\cdot
x_{0})<D_{\Lambda}$, hence $d\big{(}\eta(t_{n}),x_{t_{n}}\big{)}\leq
C+D_{\Lambda}$. This means that $\xi$ is $\Lambda$-conical.

Otherwise, for all $n\in\mathbb{N}$ it holds that
$\frac{\pi}{2}<d(\xi,\xi_{n})$. The fact that
$\eta(t_{n})\in\mathcal{HB}^{\Lambda}_{n}$ together with Lemma 2.32 implies
that at some later time the geodesic ray $\eta$ leaves
$\mathcal{HB}^{\Lambda}_{n}$. Thus there is $s_{n}>t_{n}$ for which
$\eta(s_{n})\in\mathcal{H}^{\Lambda}_{n}$. Since $\mathcal{HB}^{\Lambda}_{n}$
are maximal $\Lambda$-free horoballs, Lemma 6.25 gives rise to points
$\lambda_{n}x_{0}$ such that $d\big{(}\lambda_{n}x_{0},\eta(s_{n})\big{)}\leq
D_{\Lambda}$. This renders $\xi$ as a $\Lambda$-conical limit point, as
wanted.

∎

###### Proof of Proposition 6.11.

The strategy is as follows. For
$\mathcal{HB}^{\Lambda}=\mathcal{HB}^{\Lambda}_{{\lambda_{\gamma}},i(\gamma)}$,
one uses Proposition 6.22 to get that
$\mathcal{HB}^{\Gamma}\subset\mathcal{HB}^{\Lambda}$ and that the distance
$d(\mathcal{H}^{\Lambda},\mathcal{H}^{\Gamma})$ is large with $d_{\gamma}$.
The horosphere $\mathcal{H}^{\Gamma}$ admits a $\Gamma$-cocompact metric
lattice, and so the projections of these metric lattice points onto
$\mathcal{H}^{\Lambda}$ form a cocompact metric lattice in
$\mathcal{H}^{\Lambda}$. It remains to show that for each
$\gamma^{\prime}x_{0}\in\Gamma\cdot x_{0}\cap\mathcal{H}^{\Gamma}$, the
corresponding $\lambda^{\prime}=\lambda_{\gamma^{\prime}}$ indeed lies on the
same $\mathcal{H}^{\Lambda}$ and boundedly close to the projection
$P_{\mathcal{H}^{\Lambda}}(\gamma^{\prime}x_{0})$. This is done by putting
together all the geometric facts obtained up to this point, specifically Lemma
6.25. One delicate fact that will be of use is that two maximal $\Lambda$-free
horoballs that are based at the same point must be equal, because none of them
can contain a $\Lambda$-orbit point while on the other hand both bounding
horospheres intersect $\Lambda\cdot x_{0}$.

Fix $s>0$ for which Proposition 6.22 yields a corresponding bound $D(s)$, and
let $\gamma\in\Gamma$ such that
$sd_{\gamma}>2\cdot\big{(}D_{\Lambda}+D(s)\big{)}$. Consider the (maximal)
$\Lambda$-free horoball $\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$
based at $\xi^{i(\gamma)}_{\lambda}$. I show that $\Lambda\cdot
x_{0}\cap\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ is a cocompact
metric lattice in $\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$. I keep
the subscript notation because the proof is a game between
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ and another
$\Lambda$-free horoball.

Let $\mathcal{HB}^{\Gamma}_{\lambda_{\gamma},i(\gamma)}$ be the
$\Gamma$-horoball corresponding to
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$. I can conclude that
$\mathcal{HB}^{\Gamma}_{\lambda_{\gamma},i(\gamma)}\subset\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$,
because the choice of $d_{\gamma}>D(s)$ guarantees $\gamma x_{0}$ is
$sd_{\gamma}$ deep inside
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$. In particular
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ is not $\Gamma$-free.
Moreover, it holds that
$L:=d(\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)},\mathcal{H}^{\Gamma}_{\lambda_{\gamma},i(\gamma)})\geq
sd_{\gamma}$. Let $\gamma^{\prime}\in\Gamma$ be any element in the cocompact
metric lattice $\Gamma\cdot
x_{0}\cap\mathcal{H}^{\Gamma}_{\lambda_{\gamma},i(\gamma)}$, and consider two
associated points: (a) $\lambda^{\prime}x_{0}=\lambda_{\gamma^{\prime}}x_{0}$
and (b) the projection of $\gamma^{\prime}x_{0}$ on
$\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$, denoted
$p_{\gamma}^{\prime}:=P_{\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}}(\gamma^{\prime}x_{0})\in\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$.
The horoball $\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ is a
maximal $\Lambda$-free horoball so it is also $1$-almost maximal, hence
$d(p_{\gamma}^{\prime},\Lambda\cdot x_{0})\leq D_{\Lambda}$ and the following
holds:

$sd_{\gamma}\leq L\leq d_{\gamma^{\prime}}\leq
d(\gamma^{\prime}x_{0},p_{\gamma}^{\prime})+d(p_{\gamma}^{\prime},\Lambda\cdot
x_{0})\leq L+D_{\Lambda}$ (6)

Consider $\xi^{i(\gamma^{\prime})}_{\lambda^{\prime}}$, and assume towards
contradiction that
$\xi^{i(\gamma^{\prime})}_{\lambda_{\gamma^{\prime}}}\neq\xi^{i(\gamma)}_{\lambda_{\gamma}}$.
Both points lie in $\mathcal{W}_{\mathbb{Q}}(\Gamma)$ and therefore must be at
Tits distance $\pi$ of each other. Therefore the fact that
$\gamma^{\prime}x_{0}$ lies in
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ implies that the
geodesic $[\gamma^{\prime}x_{0},\xi^{i(\gamma^{\prime})}]$ leaves
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ at some point
$z\in\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$.

The fact that $D(s)\leq sd_{\gamma}\leq d_{\gamma^{\prime}}$ implies that
$\gamma^{\prime}x_{0}$ lies $s^{2}d_{\gamma}$ deep inside
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}$.
Therefore the point $z$ also lies at least $s^{2}d_{\gamma}$ deep inside
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}$, and
therefore $z$ is the centre of a $\Lambda$-free horoball of radius at least
$s^{2}d_{\gamma}$.

By choice of $d_{\gamma}$ the point $z$ therefore admits a $2D_{\Lambda}$
neighbourhood that is $\Lambda$-free. But $z$ lies on
$\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$, a maximal horosphere of
$\Lambda$, contradicting Lemma 6.25. I conclude that
$\xi^{i(\gamma^{\prime})}_{\lambda_{\gamma^{\prime}}}=\xi^{i(\gamma)}_{\lambda_{\gamma}}$,
so $\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ and
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}$ are
two $\Lambda$-free horoballs that are tangent to a $\Lambda\cdot x_{0}$ point
and based at the same point at $\infty$. This implies
$\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}=\mathcal{HB}^{\Lambda}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}$,
and in particular $\lambda_{\gamma^{\prime}}x_{0}\in\mathcal{H}^{\Lambda}$.
Finally, it is clearly seen from Inequality 6 that

$d(\lambda^{\prime}x_{0},p_{\gamma}^{\prime})\leq
d(\lambda^{\prime}x_{0},\gamma^{\prime}x_{0})+d(\gamma^{\prime}x_{0},p_{\gamma}^{\prime})\leq
d_{\gamma^{\prime}}+L\leq L+D_{\Lambda}+L$

The element $\gamma^{\prime}x_{0}\in\Gamma\cdot
x_{0}\cap\mathcal{H}^{\Gamma}_{\lambda_{\gamma},i(\gamma)}$ was as arbitrary
element, and the above argument shows that the corresponding $\Lambda$-orbit
points satisfy:

  1. 1.

$\lambda^{\prime}x_{0}$ all lie on
$\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$.

  2. 2.

Each $p_{\gamma}^{\prime}$ is $2L+D_{\Lambda}$ close to the point
$\lambda^{\prime}x_{0}$.

This shows that the cocompact metric lattice
$\\{p_{\gamma}^{\prime}\mid\gamma^{\prime}x_{0}\in\mathcal{H}^{\Gamma}_{\lambda_{\gamma},i(\gamma)}\\}$
lies in a bounded neighbourhood of the set of points $\Lambda\cdot
x_{0}\cap\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$, proving that
$\Lambda\cdot x_{0}\cap\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$ is
a cocompact metric lattice in
$\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$. Lemma 2.25 elevates this
to

$\big{(}\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)})\big{)}\cdot
x_{0}\cap\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$

being a cocompact metric lattice in
$\mathcal{H}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$, completing the proof. ∎

$\gamma
x_{0}$$d_{\gamma}$$\gamma^{\prime}x_{0}$$\lambda_{\gamma^{\prime}}x_{0}$$\mathcal{HB}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}^{\Lambda}$$z^{\prime}:=\pi_{\mathcal{HB}_{\lambda_{\gamma^{\prime}},i(\gamma^{\prime})}^{\Lambda}}(\gamma^{\prime}x_{0})$$\overbrace{}^{D_{1}}_{1}$$\xi_{\lambda_{\gamma^{\prime}}}^{i(\gamma^{\prime})}$$\xi_{\lambda_{\gamma}}^{i(\gamma)}$$\geq\frac{1}{2}d_{\gamma}$$\mathbf{z}$$\geq\frac{1}{2}d_{\gamma^{\prime}}$$\begin{array}[]{l}\Lambda-
free\ ball\\\ \
B\left(z,\frac{1}{2}d_{\gamma}\right)\end{array}$$\mathcal{HB}_{\lambda_{\gamma},i(\gamma)}^{\Lambda}$$\mathcal{HB}_{\lambda_{\gamma},i(\gamma)}^{\Gamma}$$\lambda_{\gamma}x_{0}$
Figure 4: Proposition 6.11. Assuming towards contradiction that
$\xi^{i(\gamma)}_{\lambda_{\gamma}}\neq\xi^{i(\gamma^{\prime})}_{\lambda_{\gamma^{\prime}}}$
results in a point $z\in\mathcal{HB}^{\Lambda}_{\lambda_{\gamma},i(\gamma)}$
(blue coloured and bold faced in the bottom part of the figure) admitting a
large $\Lambda$-free neighbourhood, contradicting almost cocompactness.

###  6.3 The Bounded Case

Proposition 6.11 is enough in order to prove Theorem 6.1 in the case that
$\Gamma$ does not lie in a bounded neighbourhood of $\Lambda$. The case where
$\Gamma$ and $\Lambda$ lie at bounded Hausdorff distance, i.e. where
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ and
$\Lambda\subset\mathcal{N}_{D}(\Gamma)$ for $D>0$, arose naturally in the
context of quasi-isometric completeness of non-uniform lattices. The precise
statements are given in Theorems 6.8 and 6.9 above. I present their proofs in
Section 6.3.2 below.

The notable difference between Theorem 6.8 and Theorem 6.9 is that for higher
rank groups, the inclusion $\Lambda\subset\mathcal{N}_{D}(\Gamma)$ is only
required to prove commensurability. In view of Corollary 6.7, this allows me
to omit that assumption from Theorem 1.6. Notice also that for groups with
property (T) the result easily follows from the (much more recent) result by
Leuzinger in Theorem 4.7.

In the context of commensurability in the sublinear setting, I can only prove
a limited result, Namely that $\Lambda$ is commensurable to $\Gamma$ if
$\Gamma$ is an irreducible $\mathbb{Q}$-rank $1$ lattice and both
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ and
$\Lambda\subset\mathcal{N}_{u}(\Gamma)$ for some constant $D>0$ and a
sublinear function $u$. This is done via a reduction to the bounded case.

######  Proposition 6.27.

Let $G$ be a real finite-centre semisimple Lie group without compact factors,
$\Gamma\leq G$ an irreducible lattice of $\mathbb{Q}$-rank $1$, $\Lambda\leq
G$ a discrete subgroup, and $u$ a sublinear function. If
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ for some $D>0$ and
$\Lambda\subset\mathcal{N}_{u}(\Gamma)$, then actually
$\Lambda\subset\mathcal{N}_{D^{\prime}}(\Gamma)$ for some $D^{\prime}>0$.
Moreover, if $G$ is of $\mathbb{R}$-rank$\ 1$, the conclusion holds under the
relaxed assumption that $u(r)\preceq_{\infty}\varepsilon r$ for some
$\varepsilon<1$.

######  Remark 6.28.

While the setting of Proposition 6.2 is indeed rather limited, the situation
that both $\Gamma\subset\mathcal{N}_{u}(\Lambda)$ and
$\Lambda\subset\mathcal{N}_{u}(\Gamma)$ arises naturally from the motivating
example of SBE-rigidity in Theorem 3.7. Notice however that Theorem 3.7 is not
known for groups $G$ that admit $\mathbb{R}$-rank$\ 1$ factors, which is the
only setting for which I can prove Proposition 6.2.

####  6.3.1 A Reduction

I start with the proof of Proposition 6.27. The first step is to establish the
fact that $\Lambda$ must preserve $\mathcal{W}_{\mathbb{Q}}(\Gamma)$.

######  Lemma 6.29.

Let $G$ be a real finite-centre semisimple Lie group without compact factors,
$\Gamma\leq G$ an irreducible non-uniform lattice of $\mathbb{Q}$-rank $1$,
$\Lambda\leq G$ a discrete subgroup. Assume that
$\Gamma\subset\mathcal{N}_{u}(\Lambda)$ and that
$\Lambda\subset\mathcal{N}_{u^{\prime}}(\Gamma)$ for sublinear functions
$u,u^{\prime}$. Then
$\Lambda\cdot\mathcal{W}_{\mathbb{Q}}(\Gamma)\subset\mathcal{W}_{\mathbb{Q}}(\Gamma)$.
Moreover, if $G$ is of $\mathbb{R}$-rank$\ 1$, the conclusion holds under the
relaxed assumption that $u^{\prime}(r)\preceq_{\infty}\varepsilon r$ for some
$\varepsilon<1$.

###### Proof.

The proof is similar to the argument of Lemma 6.13, and uses the linear
penetration rate of a geodesic into a horoball. Let
$\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$, and let $\mathcal{H}^{\Gamma}$ be a
horosphere bounding a $\Gamma$-free horoball $\mathcal{HB}^{\Gamma}$ with
$\mathcal{H}^{\Gamma}\cap\Gamma\cdot x_{0}$ a metric lattice in
$\mathcal{H}^{\Gamma}$. Assume first that $u^{\prime}$ is sublinear. Since
$\mathcal{HB}^{\Gamma}$ is $\Gamma$-free and
$\Lambda\subset\mathcal{N}_{u^{\prime}}(\Gamma)$, I can conclude that
$\Lambda\cdot
x_{0}\cap\mathcal{HB}^{\Gamma}\subset\mathcal{N}_{u^{\prime}}(\mathcal{H}^{\Gamma})$.
Recall (Lemma 2.32) that every geodesic ray $\eta$ with limit point
$\xi^{\prime}\in\mathcal{N}_{\frac{\pi}{2}}(\xi)$ penetrates
$\mathcal{HB}^{\Gamma}$ at linear rate. Therefore for every such geodesic ray
$\eta$ and every sublinear function $v$ there is $R=R(\eta,v)>0$ for which
$\mathcal{N}_{v}(\eta_{\restriction_{r>R}})$ is $\Lambda$-free.

On the other hand, let $\lambda\in\Lambda$, and assume towards contradiction
that $\lambda\xi\notin\mathcal{W}_{\mathbb{Q}}(\Gamma)$. Then by Proposition
2.31 there is a $\Gamma$-conical limit point
$\xi^{\prime}\in\mathcal{N}_{\frac{\pi}{2}}(\lambda\xi)$. The hypothesis that
$\Gamma\subset\mathcal{N}_{u}(\Lambda)$ then implies that for every $R>0$,
$\mathcal{N}_{u}(\eta_{\restriction{r>R}})\cap\Lambda\cdot
x_{0}\neq\emptyset$. Translating by $\lambda^{-1}$ yields a contradiction to
the previous paragraph. I conclude that
$\lambda\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$.

I now modify the argument to include $u^{\prime}(r)\preceq_{\infty}\varepsilon
r$ when $G$ is of $\mathbb{R}$-rank$\ 1$. In this case, the only point
$\xi^{\prime}\in\mathcal{N}_{\frac{\pi}{2}}(\lambda\xi)$ is $\lambda\xi$
itself. Therefore by the same argument as above, the assumption that
$\lambda\xi\notin\mathcal{W}_{\mathbb{Q}}(\Gamma)$ implies that the
$u$-sublinear neighbourhood of every geodesic ray with limit point $\xi$
intersects $\Lambda\cdot x_{0}$. I.e., for every $\eta$ with limit point $\xi$
and every $R>0$ it holds that
$\mathcal{N}_{u}(\eta_{\restriction_{r>R}})\cap\Lambda\cdot
x_{0}\neq\emptyset$. On the other hand, every such geodesic penetrates
$\mathcal{HB}^{\Gamma}$ at $1$-linear rate. This amounts to the following
fact: if $v^{\prime}(r)=\varepsilon r$ for some $\varepsilon\in(0,1)$, then
for some $R>0$, the set
$\mathcal{N}_{u}(\eta_{\restriction_{r>R}})\cap\mathcal{N}_{v^{\prime}}(\mathcal{H}^{\Gamma})=\emptyset$.
This is a contradiction to $\Lambda\subset\mathcal{N}_{u^{\prime}}(\Gamma)$. ∎

###### Proof of Proposition 6.27.

Assume towards contradiction that there is a sequence $\lambda_{n}$ such that
$d(\lambda_{n}x_{0},\Gamma\cdot x_{0})>n$. Recall that $\Gamma\cdot x_{0}$ is
a cocompact metric lattice in the compact core of $\Gamma$. This implies that
there is a number $D^{\prime}>0$ such that any $\lambda\in\Lambda$ for which
$\lambda x_{0}\notin\mathcal{N}_{D^{\prime}}(\Gamma\cdot x_{0})$ must lie at
least $\frac{1}{2}D^{\prime}$-deep inside a horoball of $\Gamma$. I can assume
that for all $n\in\mathbb{N}$ there are corresponding horoballs of $\Gamma$,
which I denote $\mathcal{HB}^{\Gamma}_{n}$, for which $\lambda_{n}\cdot
x_{0}\in\mathcal{HB}^{\Gamma}_{n}$. The fact that
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ then implies that
$\mathcal{N}_{D}(\Lambda\cdot x_{0})$ covers a cocompact metric lattice in
$\mathcal{H}^{\Gamma}_{n}$, namely the metric lattice $\Gamma\cdot
x_{0}\cap\mathcal{H}^{\Gamma}_{n}$. In the terminology of Section 6.2,
$\mathcal{H}^{\Gamma}_{n}$ is almost $\Lambda$-cocompact, or $D$-almost
$\Lambda$-cocompact.

I first prove that every horoball of $\Gamma$ contains a $\Lambda$-free
horoball (this is of course immediate if
$\Lambda\subset\mathcal{N}_{C}(\Gamma)$ for some $C>0$). Assume towards
contradiction that there is a horoball $\mathcal{HB}^{\Gamma}$ of $\Gamma$
that does not contain a $\Lambda$-free horoball. Denote
$\mathcal{H}^{\Gamma}:=\partial\mathcal{HB}^{\Gamma}$. In the notations of the
previous paragraph, I can assume without loss of generality that
$\mathcal{HB}^{\Gamma}=\mathcal{HB}^{\Gamma}_{n}$ for all $n\in\mathbb{N}$.
Denote by $\xi$ the base point of $\mathcal{HB}^{\Gamma}$, fix some arbitrary
$x\in\mathcal{H}^{\Gamma}$ and consider the geodesic ray $\eta:=[x,\xi)$. The
constraint that $\Lambda\subset\mathcal{N}_{u}(\Gamma)$ implies that for every
$R>0$ there is some $L>0$ for which the ball $B\big{(}\eta(L+t),R\big{)}$ is
$\Lambda$-free, for all $t\geq 0$. In particular, for all large enough
$n\in\mathbb{N}$ (depending on $R$), the horosphere
$\mathcal{H}(\xi,\lambda_{n}x_{0})$ that is parallel to $\mathcal{H}^{\Gamma}$
and that passes through $\lambda_{n}x_{0}$ contains a point that is the centre
of $\Lambda$-free ball of radius $R$. This property is $\Lambda$-invariant, as
well as the fact that $\mathcal{HB}^{\Gamma}$ is based at
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$. In particular, these two properties hold
for the horoballs
$\mathcal{HB}_{n}:=\lambda_{n}^{-1}\cdot\mathcal{HB}^{\Gamma}$, whose
respective base points I denote
$\xi_{n}:=\lambda_{n}^{-1}\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$.

Fix $R=D+2D_{\Gamma}$ (recall that $D_{\Gamma}$ is such that every horosphere
$\mathcal{H}$ of $\Gamma$ admits
$\mathcal{H}\subset\mathcal{N}_{D_{\Gamma}}(\Gamma\cdot
x_{0}\cap\mathcal{H})$). Let $L=L(D+2D_{\Gamma})$ be the corresponding bound
from the previous paragraph. For every $n>L$ the horoball $\mathcal{HB}_{n}$
has bounding horosphere $\mathcal{H}_{n}$ that admits a point
$z_{n}\in\mathcal{H}_{n}$ for which $B(z_{n},D+2D_{\Gamma})$ is
$\Lambda$-free. Moreover, the same is true for every horosphere that is
parallel to $\mathcal{H}_{n}$ which lies inside $\mathcal{HB}_{n}$. Since
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$, this means that every horosphere that
lies inside $\mathcal{HB}_{n}$ admits a point that is the centre of a
$\Gamma$-free ball of radius $2D_{\Gamma}$. I conclude that none of those
horospheres could be the horosphere of $\Gamma$ corresponding to the parabolic
limit point $\xi_{n}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$. Since
$x_{0}\in\mathcal{H}_{n}$ it must therefore be that $\mathcal{H}_{n}$ is a
horosphere of $\Gamma$. But this contradicts the fact that
$z_{n}\in\mathcal{H}_{n}$ and $B(z_{n},2D_{\Gamma})$ is $\Gamma$-free. This
shows that no horoball of $\Gamma$ contains a sequence of $\Lambda$-orbit
points that lie deeper and deeper in that horoball. Put differently, it shows
that every horoball of $\Gamma$ contains a $\Lambda$-free horoball.

I remark that the above argument shows something a bit stronger, which I will
not use but which I find illuminating. It proves that as soon as $d(\lambda
x_{0},\Gamma\cdot x_{0})$ is uniformly large enough, say more than $M$, then
$\lambda x_{0}$ must lie on a $(D+2D_{\Gamma}$)-almost $\Lambda$-cocompact
horosphere parallel to $\mathcal{H}^{\Gamma}$, where $\mathcal{H}^{\Gamma}$ is
the bounding horosphere of any horoball of $\Gamma$ in which $\lambda x_{0}$
lies (recall that it must lie in at least one such horoball). On the other
hand if $d(\lambda x_{0},\Gamma\cdot x_{0})<M$, then since every point in the
$\Gamma$-orbit lies on a horosphere of $\Gamma$ one concludes that $\lambda
x_{0}$ lies on a horosphere $\mathcal{H}$ based at
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$ that is $(M+D)$-almost $\Lambda$-cocompact.

I can now assume that every $\mathcal{HB}^{\Gamma}_{n}$ contains a
$\Lambda$-free horoball. In particular it contains a $1$-almost maximal
$\Lambda$-free horoball $\mathcal{HB}^{\Lambda}_{n}$ (see Definition 6.23). By
definition there is a point $\lambda_{n}^{\prime}x_{0}$ that is at distance at
most $1$ from $\mathcal{H}^{\Lambda}_{n}=\partial\mathcal{HB}^{\Lambda}_{n}$.
Up to enlarging $d(\lambda_{n}x_{0},\Gamma\cdot x_{0})$ or decreasing it by at
most $1$, I can assume $\lambda_{n}=\lambda_{n}^{\prime}$ to begin with.
Consider $\mathcal{HB}_{n}:=\lambda_{n}^{-1}\cdot\mathcal{HB}^{\Lambda}_{n}$
with $\mathcal{H}_{n}=\partial\mathcal{HB}_{n}$. This is a sequence of
horoballs, each of which contains a $\Lambda$-free horoball at depth at most
$1$, based at corresponding parabolic limit points
$\xi_{n}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$, and tangent to points that are
at distance at most $1$ from $x_{0}$, i.e., $\mathcal{H}_{n}\cap
B(x_{0},1)\neq\emptyset$.

Since $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ I conclude that each of the
$\mathcal{HB}_{n}$ contain a horoball of depth at most $D+1$ that is
$\Gamma$-free. Therefore the horoball of $\Gamma$ that is based at $\xi_{n}$
must have its bounding horosphere intersecting $B(x_{0},D+2)$. By Corollary
2.23 there are only finitely such horoballs. I conclude that there are
finitely many points
$\xi^{\prime}_{1},\dots,\xi^{\prime}_{K}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$
such that for every $n\in\mathbb{N}$ there is $i(n)\in\\{1,\dots,K\\}$ with
$\xi_{n}=\xi^{\prime}_{i(n)}$. From the Pigeonhole Principle there is some
$\xi^{\prime}\in\\{\xi^{\prime}_{1}\dots,\xi^{\prime}_{K}\\}$ for which
$\xi_{n}=\xi^{\prime}$ for infinitely many $n\in\mathbb{N}$. Passing to a
subsequence I assume that this is the case for all $n\in\mathbb{N}$.

To begin with the $\mathcal{HB}^{\Gamma}_{n}$ are horoballs of $\Gamma$, and
therefore as in the first case the bounding horospheres
$\mathcal{H}^{\Gamma}_{n}$ are $D+2D_{\Gamma}$-almost $\Lambda$-cocompact.
This is a $\Lambda$-invariant property and therefore the same holds for the
$\lambda_{n}^{-1}$ translate of it. These are the horospheres which are based
at $\xi^{\prime}$ and lie outside $\mathcal{HB}_{n}$ at distance
$d(\lambda_{n}x_{0},\Gamma\cdot x_{0})>n-1$ from $\mathcal{H}_{n}$. They form
a sequence of outer and outer horospheres based at the same point at
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$, all of which are $D+2D_{\Gamma}$-almost
$\Lambda$-cocompact. This is a contradiction, since the union of such
horospheres intersect every horoball of $X$, contradicting the existence of
$\Lambda$-free horoballs. Formally, take some
$\zeta\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$ different from $\xi^{\prime}$.
Since both $\xi^{\prime}$ and $\zeta$ lie in
$\mathcal{W}_{\mathbb{Q}}(\Gamma)$, they admit $d_{T}(\zeta,\xi^{\prime})=\pi$
and there is a geodesic $\eta$ with $\eta(-\infty)=\xi^{\prime}$ and
$\eta(\infty)=\zeta$. Let $\mathcal{HB}^{\Gamma}_{\zeta}$ be the horoball of
$\Gamma$ that is based at $\zeta$. By the first step of this proof, every such
horoball must contain a $\Lambda$-free horoball
$\mathcal{HB}^{\Lambda}_{\zeta}$. Therefore there is some $T>0$ such that for
all $t>T$ the point $\eta(t)$ lies $2(D+2D_{\Gamma})$ deep in
$\mathcal{HB}^{\Lambda}_{\zeta}$. I conclude that for all $t>T$,
$B\big{(}\eta(t),2D\big{)}$ is $\Lambda$-free. On the other hand, for
arbitrarily large $t$ it holds that the horosphere based at $\xi^{\prime}$ and
tangent to $\eta(t)$ is $D+2D_{\Gamma}$-almost $\Lambda$-cocompact, and in
particular $d\big{(}\eta(t),\Lambda\cdot x_{0}\big{)}<D+2D_{\Gamma}$, a
contradiction.

I conclude that $\Lambda\subset\mathcal{N}_{D^{\prime}}(\Gamma)$ for some
$D^{\prime}>0$, as claimed. ∎

###### Proof of Theorem 1.8.

The theorem follows immediately from Proposition 6.27 together with Theorems
6.8 and 6.9. ∎

####  6.3.2 The Arguments of Schwartz and Eskin

##### The $\mathbb{R}$-rank$\ 1$ case.

The statement of Theorem 6.9 is a slight modification of Schwartz’s original
formulation. His framework leads to a discrete subgroup $\Delta\leq G$ such
that:

  1. 1.

Every element of $\Delta$ _quasi-preserves_ the compact core of the lattice
$\Gamma$. Namely, each element of $\Delta$ is an isometry of $X$ that
preserves $\mathcal{W}_{\mathbb{Q}}(\Gamma)$ and that maps every horosphere of
$\Gamma$ to within the $D=D(\Delta)$ neighbourhood of some other horosphere of
$\Gamma$.

  2. 2.

It holds that $\Gamma\subset\mathcal{N}_{D}(\Delta)$.

From these two properties Schwartz is able to deduce that $\Delta$ has finite
covolume, i.e. that $\Delta$ is a lattice in $G$. Here is a sketch of his
argument, which works whenever $\Gamma$ is a $\mathbb{Q}$-rank $1$ lattice.

######  Theorem 6.30.

In the setting described above, $\Delta$ is a lattice in $G$.

###### Proof sketch.

Consider $X_{0}^{\prime}:=\bigcup_{g\in\Delta}g\cdot X_{0}$, where $X_{0}$ is
the compact core of $\Gamma$. This space serves as a ‘compact core’ for
$\Delta$: the fact that $\Delta$ quasi-preserves $X_{0}$ implies that
$X_{0}^{\prime}\subset\mathcal{N}_{D}(X_{0})$. It is a $\Delta$-invariant
space, and therefore one gets an isometric action of $\Delta$ on
$X_{0}^{\prime}$. This action is cocompact: the reason is that $\Gamma$ acts
cocompactly on $X_{0}$, and $\Gamma\subset\mathcal{N}_{D}(\Delta)$. Formally,
every point in $X_{0}^{\prime}$ is $D$-close to a point in $X_{0}$. Every
point in $X_{0}$ is $D_{\Gamma}$-close to a point in $\Gamma\cdot x_{0}$.
Every point in $\Gamma\cdot x_{0}$ is $D$-close to a point in $\Delta\cdot
x_{0}$. Therefore the ball of radius $2D+D_{\Gamma}$ contains a fundamental
domain for the action of $\Delta$ on $X_{0}^{\prime}$.

It remains to see that the action of $\Delta$ on $X\setminus X_{0}^{\prime}$
is of finite covolume. As a result of the cocompact action of $\Delta$ on
$X_{0}^{\prime}$, there is $B:=B(x_{0},R)$ so that
$X_{0}^{\prime}\subset\Delta\cdot B$. $X_{0}^{\prime}$ is the complement of a
union of horoballs, which one may call _horoballs of $\Delta$_, with bounding
_horospheres of $\Lambda$_. The fact that $\Gamma$ is of $\mathbb{Q}$-rank $1$
means that the horoballs of $\Gamma$ are disjoint, and therefore those of
$\Lambda$ are almost disjoint: there is some $C>0$ such that for every
horosphere $\mathcal{H}$ of $\Lambda$ and every point $x\in\mathcal{H}$,
$d(x,X_{0}^{\prime})<C$. Up to enlarging the radius of $B$ by $C$, I can
assume that $\mathcal{H}\subset\Delta\cdot B$ for every horosphere
$\mathcal{H}$ of $\Lambda$.

Each horoball of $\Lambda$ is based at $\mathcal{W}_{\mathbb{Q}}(\Gamma)$, and
each lies uniformly boundedly close to the corresponding horoballs of
$\Gamma$. From Corollary 2.23 one therefore sees that there are finitely many
horoballs of $\Delta$ that intersect $B$. Denote them by
$\mathcal{HB}_{1},\dots,\mathcal{HB}_{N}$, their bounding horospheres by
$\mathcal{H}_{i}=\partial\mathcal{HB}_{i}$, and their intersection with $B$ by
$B_{i}:=B\cap\mathcal{HB}_{i}$. Let also
$\xi_{i}\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$ denote the base point of each
$\mathcal{HB}_{i}$. Each $B_{i}$ is pre-compact and therefore the projection
of each $B_{i}$ on $\mathcal{H}_{i}$ is pre-compact as well (this is a
consequence e.g. of the results of Heintze-Im hof recalled in Remark 2.6). Let
$D_{i}\subset\mathcal{H}_{i}$ be a compact set that contains this projection,
i.e. $P_{\mathcal{H}_{i}}(B_{i})\subset D_{i}\subset\mathcal{H}_{i}$. In
particular $B\cap\mathcal{H}_{i}\subset D_{i}$.

Observe now that for every horoball $\mathcal{HB}$ of $\Delta$, with bounding
horosphere $\mathcal{H}=\partial\mathcal{HB}$, the $\Delta$-orbit of every
point $x\in\mathcal{H}$ intersects some $D_{i}$. First notice that for
$x\in\mathcal{H}$ the choice of $B$ implies that the $\Delta$-orbit of $x$
must intersect $B$, say $gx\in B$. In particular $g\mathcal{H}\cap
B\neq\emptyset$, and since $g\mathcal{HB}$ is a horoball of $\Delta$ then by
definition $g\mathcal{H}=\mathcal{H}_{i}$ for some $i\in\\{1,\dots,N\\}$. One
concludes that indeed $gx\subset D_{i}$.

Moreover, let $y\in X$ is any point that lies inside a horoball $\mathcal{HB}$
of $\Delta$, and $x=P_{\mathcal{H}}(y)$ its projection on the bounding
horosphere $\mathcal{H}=\partial\mathcal{HB}$. By the previous paragraph there
is some $g\in\Delta$ and $i\in\\{1,\dots,N\\}$ for which $gx\in D_{i}$, and
therefore it is clear that $gy$ lies on a geodesic emanating from $D_{i}$ to
$\xi_{i}$.

Finally, define $\mathrm{Cone}(D_{i})$ to be the set of all geodesic rays that
emanate from $D_{i}$ and with limit point $\xi_{i}$. The previous paragraph
proves that $\bigcup_{i=1}^{N}\mathrm{Cone}(D_{i})$ contains a fundamental
domain for the action of $\Delta$ on $X\setminus X_{0}^{\prime}$. Moreover,
the fact that $D_{i}\subset\mathcal{H}_{i}$ is compact readily implies that
each $\mathrm{Cone}(D_{i})$ has finite volume, and so this fundamental domain
is of finite volume.

To conclude, $B\cup\big{(}\bigcup_{i=1}^{N}\mathrm{Cone}(D_{i})\big{)}$ is a
set of finite volume and it contains a fundamental domain for the
$\Delta$-action on $X$, as claimed. The proof of commensurability of $\Delta$
and $\Gamma$ is given in full in [53].

∎

There is one essential difference between Theorem 6.9 and Theorem 6.30, namely
the assumption that $\Lambda\subset\mathcal{N}_{D}(\Gamma)$ rather than that
$\Lambda$ quasi-preserves the compact core of $\Gamma$. In Schwartz’s work,
the fact that $\Delta\subset\mathcal{N}_{D}(\Gamma)$ is not relevant (even
though it easily follows from the construction of his embedding of $\Delta$ in
$G$). He only uses the two properties described above, namely the quasi-
preservation of $X_{0}$ and $\Gamma\subset\mathcal{N}_{D}(\Delta)$.

The assumption that $\Lambda$ quasi-preserves the compact core of $\Gamma$
does not feel appropriate in the context of sublinear rigidity, while the
metric condition $\Lambda\subset\mathcal{N}_{D}(\Gamma)$ seems much more
natural. It is a stronger condition as I now show. By Lemma 6.29,
$\Lambda\cdot\mathcal{W}_{\mathbb{Q}}(\Gamma)\subset\mathcal{W}_{\mathbb{Q}}(\Gamma)$.
Let $\mathcal{H}^{\Gamma}_{1}$ be a horosphere of $\Gamma$, based at
$\xi\in\mathcal{W}_{\mathbb{Q}}(\Gamma)$, and let $\gamma
x_{0}\in\mathcal{H}^{\Gamma}_{1}$ be some point on the metric lattice of
$\Gamma\cdot x_{0}$ on $\mathcal{H}^{\Gamma}_{1}$. There is an element
$\lambda\in\Lambda$ such that $d(\lambda x_{0},\gamma x_{0})<D$. Moreover,
since $\Lambda\subset\mathcal{N}_{D}(\Gamma)$ one knows that the parallel
horoball that lies $D$-deep inside $\mathcal{HB}^{\Gamma}_{1}$ is
$\Lambda$-free.

Let $\lambda^{\prime}\in\Lambda$ be an arbitrary element of $\Lambda$, and
consider $\lambda^{\prime}\cdot\mathcal{H}^{\Gamma}_{1}$. The last statement
in the previous paragraph is $\Lambda$-invariant, and so the horoball that
lies $D$-deep inside $\lambda^{\prime}\cdot\mathcal{HB}^{\Gamma}_{1}$ is
$\Lambda$-free. The fact that $\Gamma\subset\mathcal{N}_{D}(\Lambda)$ then
implies that the parallel horoball that lies $2D$ deep inside
$\lambda^{\prime}\mathcal{HB}^{\Gamma}_{1}$ is $\Gamma$-free. Let
$\mathcal{H}^{\Gamma}_{2}$ be the horosphere of $\Gamma$ that is based at
$\lambda^{\prime}\xi$. The last statement amounts to saying that
$\mathcal{H}^{\Gamma}_{2}$ lies at most $2D$-deep inside
$\lambda^{\prime}\mathcal{HB}^{\Gamma}_{1}$. On the other hand, one has
$d(\lambda^{\prime}\lambda
x_{0},\lambda^{\prime}\mathcal{H}^{\Gamma}_{1})=d(\lambda
x_{0},\mathcal{H}^{\Gamma}_{1})\leq D$, so there is a $\Lambda$-orbit point
that lies within $D$ of $\lambda^{\prime}\mathcal{H}^{\Gamma}_{1}$. The
parallel horoball that lies $D$-deep inside $\mathcal{HB}^{\Gamma}_{2}$ must
also be $\Lambda$-free, so I conclude that $\mathcal{H}^{\Gamma}_{2}$ must be
contained in the parallel horoball to
$\lambda^{\prime}\mathcal{HB}^{\Gamma}_{1}$ which contains it and that is at
distance $D$ from it. I conclude that
$d(\lambda^{\prime}\mathcal{H}^{\Gamma}_{1},\mathcal{H}^{\Gamma}_{2})\leq 2D$,
and so that $\Lambda$ quasi-preserves $X_{0}$.

######  Remark 6.31.

It is interesting to note that Schwartz’s arguments are similar in spirit to
my arguments in Section 6.2. In fact, one could also prove Theorem 6.9 using
the same type of arguments that appear repeatedly in section 6.2, namely by
moving $\Lambda$-free horoballs around the space, specifically the proof of
Proposition 6.27. I do not present this alternative proof here.

##### Higher rank.

Eskin’s proof is ergodic, and based on results of Mozes [41] and Shah [55]. I
produce it here without the necessary preliminaries, which are standard.

###### Proof of Theorem 6.8.

To prove that $\Lambda$ is a lattice amounts to finding a finite non-zero
$G$-invariant measure on $\Lambda\backslash G$. By Theorem $2$ in [41], if
$P\leq G$ is a parabolic subgroup then every $P$-invariant measure on
$\Lambda\backslash G$ is automatically $G$-invariant. Fix a minimal parabolic
subgroup $P\leq G$ and let $\mu_{0}$ be some fixed probability measure on
$\Lambda\backslash G$. Since $P$ is amenable it admits a tempered Følner
sequence $F_{n}\subset P$, and one can average $\mu_{0}$ along each $F_{n}$ to
get a sequence of probability measures $\mu_{n}$. The weak* compactness of the
unit ball in the space of measures on $\Lambda\backslash G$ implies that there
exists a weak* limit $\mu$ of the $\mu_{n}$. The measure $\mu$ is
automatically a finite $P$-invariant measure. It remains to show that $\mu$ is
not the zero measure. To see this it is enough to show that for some compact
set $C_{\Lambda}\subset\Lambda\backslash G$ and some $\Lambda
g=x\in\Lambda\backslash G$, one has

$0<\liminf_{n}\frac{1}{|F_{n}|}\int_{F_{n}}{\mathbbm{1}_{C_{\Lambda}}(xp^{-1})dp}$
(7)

Fix some compact neighbourhood $C_{\Gamma}\subset\Gamma\backslash G$ of the
trivial coset $\Gamma e$. The hypothesis
$\Gamma\subset\mathcal{N}_{D}(\Lambda)$ implies that there is a corresponding
compact neighbourhood $C_{\Lambda}\subset\Lambda\backslash G$ of the trivial
coset $\Lambda e$ such that for any $p\in P$, it holds that $\Gamma gp^{-1}\in
C_{\Gamma}\Rightarrow\Lambda gp^{-1}\in C_{\Lambda}$ (simply take
$C_{\Lambda}$ to be the $D+1$-blowup of $C_{\Gamma}$). The action of $P$ on
$\Gamma\backslash G$ is uniquely ergodic, therefore

$0<\mu_{\Gamma}(C_{\Gamma})=\lim_{n}\frac{1}{|F_{n}|}\int_{F_{n}}{\mathbbm{1}_{C_{\Gamma}}(\Gamma
p^{-1})dp}$

where $\mu_{\Gamma}$ denotes the natural $G$-invariant measure on
$\Gamma\backslash G$. The defining property of $C_{\Lambda}$ ensures that
Inequality (7) is satisfied, implying that $\mu$ is a non-zero $P$-invariant
probability measure on $\Lambda\backslash G$. I conclude that $\mu$ is also
$G$-invariant, and that $\Lambda$ is a lattice. If moreover
$\Lambda\subset\mathcal{N}_{D}(\Gamma)$, one may use Shah’s Corollary [55] to
conclude that $\Lambda$ is commensurable to $\Gamma$. ∎

###  6.4 Translating Geometry into Algebra

The goal of this section is to prove that the results of Section 6.2 imply
that $\Lambda$ satisfies the hypotheses of the Benoist-Miquel criterion
Theorem 6.4. Namely, that $\Lambda$ is Zariski dense, and that it intersects a
horospherical subgroup in a cocompact indecomposable lattice. These are
algebraic properties, and the proof that $\Lambda$ satisfies them is in
essence just a translation of the geometric results of Section 6.2 to an
algebraic language. The geometric data given by Section 6.2 is that for some
horosphere $\mathcal{H}$ bounding a $\Lambda$-free horoball,
$\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H})\cdot x_{0}$ intersects
$\mathcal{H}$ in a cocompact metric lattice (Proposition 6.11), and that the
set of $\Lambda$-conical limit points contains the set of $\Gamma$-conical
limit points (Corollary 6.26). Note that since $K$ is compact the former
implies that $\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H})$ is a uniform lattice
in $\mathrm{Stab}_{G}(\mathcal{H})$.

####  6.4.1 A Horospherical Lattice

I assume that $\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H})$ is a lattice in
$\mathrm{Stab}_{G}\mathcal{H}$, and I want to show that $\Lambda$ intersects a
horospherical subgroup $U$ of $G$ in a lattice. This step requires quite a bit
of algebraic background, which I give below in full. In short, the first goal
is to show that $\mathrm{Stab}_{G}(\mathcal{H})$ admits a subgroup
$U\leq\mathrm{Stab}_{G}(\mathcal{H})$ that is a horospherical subgroup of $G$.
A lemma of Mostow (Lemma 6.32 below) allows to conclude that $\Lambda$
intersects $U$ in a lattice.

######  Lemma 6.32 (Lemma $3.9$ in [39]).

Let H be a Lie group having no compact connected normal semisimple non-trivial
Lie subgroups, and let $N$ be the maximal connected nilpotent normal Lie
subgroup of $H$. Let $\Gamma\leq H$ be a lattice. Then $N/N\cap\Gamma$ is
compact.

######  Remark 6.33.

In the original statement Mostow uses the term ‘analytic group’, which I
replaced here with ‘connected Lie subgroup’. This appears to be Mostow’s
definition of an analytic group. See e.g. Section $10$, Chapter $1$ in [32].
In Chevalley’s _Theory of Lie Groups_ , he defines a _Lie group_ as a locally
connected topological group whose identity component is an analytic group
(Definition $1$, Section $8$, Chapter $4$ in [12]), and proves (Theorem $1$,
Section $4$, Chapter $4$ therein) a 1-1 correspondence between analytic
subgroups of an analytic group and Lie subalgebras of the corresponding Lie
algebra.

Lemma 6.32 lays the rationale for the rest of this section. Explicitly, I
prove that $\mathrm{Stab}_{G}(\mathcal{H})$ admits a subgroup that is a
horospherical subgroup $U$ of $G$ (Corollary 6.36), and that $U$ is maximal
connected nilpotent normal Lie subgroup of $\mathrm{Stab}_{G}(\mathcal{H})$
(Corollary 6.42).

In order to use Lemma 6.32, I show that the horospherical subgroup $N_{\xi}$
is a maximal normal nilpotent connected Lie subgroup of
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$, and that
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ admits no compact normal factors.
This requires to establish the structure of
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$.

######  Definition 6.34.

In the notation $h_{\xi}^{t}=\exp(tX)$ and
$A_{\xi}=\exp\big{(}Z(X)\cap\mathfrak{p}\big{)}$ of Proposition 2.8, define
$A_{\xi}^{\perp}$ to be the codimension-$1$ submanifold of $A_{\xi}$ that is
orthogonal to $\\{h_{\xi}(t)\\}_{t\in\mathbb{R}}$ (with respect to the Killing
form in the Lie algebra).

######  Claim 6.35.

Every element $a\in A_{\xi}^{\perp}$ stabilizes
$\mathcal{H}=\mathcal{H}(x_{0},\xi)$.

###### Proof.

An element in $A_{\xi}$ is an element that maps $x_{0}$ to a point on a flat
$F\subset X$ that contains the geodesic ray $[x_{0},\xi)$. If $a\in
A_{\xi}^{\perp}$, then the geodesic $[x_{0},ax_{0}]$ is orthogonal to
$[x_{0},\xi)$, and lies in $F$. From Euclidean geometry and structure of
horospheres in Euclidean spaces, it is clear that
$ax_{0}\in\mathcal{H}(x_{0},\xi)$. Since $a\in G_{\xi}$, this means
$a\mathcal{H}=\mathcal{H}(ax_{0},\xi)=\mathcal{H}(x_{0},\xi)=\mathcal{H}$. ∎

######  Corollary 6.36.

Let $\mathcal{H}$ be a horosphere based at $\xi$. Then
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}=(K_{\xi}A_{\xi}^{\perp})^{\circ}N_{\xi}$,
and in particular it contains a horospherical subgroup of $G$. Moreover,
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ is normal in
$\mathrm{Stab}_{G}(\xi)^{\circ}$ and acts transitively on $\mathcal{H}$.

###### Proof.

Clearly $(K_{\xi}A_{\xi}^{\perp})^{\circ}N_{\xi}$ is a codimension-$1$
subgroup of $\mathrm{Stab}_{G}(\xi)^{\circ}$. Since
$\mathrm{Stab}_{G}(\mathcal{H})\neq\mathrm{Stab}_{G}(\xi)$ (e.g.
$h_{\xi}^{t}\notin\mathrm{Stab}_{G}(\mathcal{H})$ for $t\neq 0$), it is enough
to show that
$(K_{\xi}A_{\xi}^{\perp})^{\circ}N_{\xi}\leq\mathrm{Stab}_{G}(\mathcal{H})$.
Let $kan\in(K_{\xi}A_{\xi}^{\perp})^{\circ}N_{\xi}$. It fixes $\xi$, so it is
enough to show that $kanx_{0}\in\mathcal{H}$. Since $k\in K_{\xi}$ and
$kx_{0}=x_{0}$, it stabilizes $\mathcal{H}$. From Claim 6.35
$a\in\mathrm{Stab}_{G}(\mathcal{H})$. So it remains to check that $N_{\xi}$
stabilizes $\mathcal{H}$, but this is more or less the definition: fixing a
base point $x_{0}$, the horospheres based at $\xi$ are parameterized by
$\mathbb{R}$. Denote them by $\\{\mathcal{H}_{t}\\}_{t\in\mathbb{R}}$, where
$\mathcal{H}=\mathcal{H}_{0}$. In this parameterization, any element $g\in
G_{\xi}$ acts on $\\{\mathcal{H}_{t}\\}_{t\in\mathbb{R}}$ by translation. I
can thus define for $g\in\mathrm{Stab}_{G}(\xi)$ the real number $l(g)$ to be
that number for which $g\mathcal{H}_{t}=\mathcal{H}_{t+l(g)}$. Clearly
$l\big{(}h_{\xi}(t)\big{)}=t$. The element $n$ fixes $\xi$, so one has

$h_{\xi}^{-t}nh_{\xi}^{t}\mathcal{H}_{0}=h_{\xi}^{-t}\mathcal{H}_{t+l(n)}=\mathcal{H}_{t+l(n)-t}=\mathcal{H}_{l(n)}$

The fact that $n\in Ker(T_{\xi})$, i.e. that
$\lim_{t\rightarrow\infty}h_{\xi}^{-t}nh_{\xi}^{t}=e_{G}$ readily implies that
necessarily $l(n)=0$. I conclude that
$(K_{\xi}A_{\xi}^{\perp})^{\circ}N_{\xi}=\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$,
as wanted.

Next recall that $\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ acts transitively on
$X$. Let $x,y\in\mathcal{H}$, and consider
$g\in\mathrm{Stab}_{G}(\xi)^{\circ}$ with $gx=y$. Writing an element $g\in
G_{\xi}$ as $ka_{t}a_{\perp}n\in K_{\xi}h_{\xi}^{t}A_{\xi}^{\perp}N_{\xi}$,
the argument above shows that
$kh_{\xi}^{t}a_{\perp}n\mathcal{H}_{0}=\mathcal{H}_{0}$ if and only if $t=0$,
i.e., if and only if $g\in\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$.

Finally, let $g\in\mathrm{Stab}_{G}(\xi)$ and
$h\in\mathrm{Stab}_{G}(\mathcal{H})$. By the discussion above
$h\cdot\mathcal{H}_{t}=\mathcal{H}_{t}$ for all $t\in\mathbb{R}$. Clearly
$-l(g)=l(g^{-1})$, and therefore

$ghg^{-1}\mathcal{H}_{0}=gh\mathcal{H}_{-l(g)}=g\cdot\mathcal{H}_{-l(g)}=\mathcal{H}_{0}$

Therefore $\mathrm{Stab}_{G}(\mathcal{H})$ is normal in
$\mathrm{Stab}_{G}{\xi}$, and the same is true for the respective identity
components. ∎

######  Corollary 6.37.

$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ is a connected Lie group with no
connected compact normal semisimple non-trivial Lie subgroups.

###### Proof.

Every compact subgroup of $G$ fixes a point. Let $H\leq G$ be some closed
subgroup. It is standard to note that a normal $N\leq H$ that fixes a point
$x\in X$ must fix every point in the orbit $H\cdot x$: $hnh^{-1}hx=hx$. Since
$H=\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ acts transitively on $\mathcal{H}$,
it shows that a normal compact subgroup of
$\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$ fixes every point in $\mathcal{H}$.
An isometry fixing a horosphere pointwise while fixing its base point is
clearly the identity, proving the claim. ∎

The following fact is well known but I could not find it in the literature.

######  Corollary 6.38.

A horosphere in $X$ is not convex.

###### Proof.

Let $\mathcal{H}^{\prime}$ be some horosphere in $X$, with base point
$\zeta\in X(\infty)$, and assume towards contradiction that it is convex. Fix
$x\in\mathcal{H}^{\prime}$ and $a_{t}^{\prime}$ the one parameter subgroup
with $\eta^{\prime}(\infty)=a_{t}^{\prime}x$, and denote
$\mathcal{H}^{\prime}_{t}=\mathcal{H}(a_{t}^{\prime}x,\zeta)$. Let $e_{G}\neq
n\in N_{\zeta}$ ($N_{\zeta}$ defined with respect to $a_{t}^{\prime}$ in a
corresponding Langlands decomposition), and consider the curve
$\eta_{n}^{\prime}(t):=a_{t}^{\prime}nx$. I claim that this is a geodesic. On
the one hand, the fact that $\mathcal{H}^{\prime}$ is convex implies that the
geodesic segment $[x,nx]$ is contained in $\mathcal{H}^{\prime}$. Therefore
$a_{t}^{\prime}[x,nx]=[a_{t}^{\prime}x,a_{t}^{\prime}nx]\subset\mathcal{H}^{\prime}_{t}$.
More generally it is clear that because
$a_{t}^{\prime}\mathcal{H}^{\prime}_{s}=\mathcal{H}^{\prime}_{s+t}$ it holds
that $\mathcal{H}^{\prime}_{t}$ is convex for every $t$ as soon as it is
convex for some $t$.

On the other hand, for every point $y\in[x,nx]$,
$d(y,\mathcal{H}^{\prime}_{t})=t$, and more generally for any
$y\in[a_{s}^{\prime}nx,a_{s}^{\prime}x]$ it holds that
$d(y,\mathcal{H}_{t}^{\prime})=|s-t|$. In particular this is true for
$\eta_{n}^{\prime}(t)=y_{t}:=a_{t}^{\prime}nx$. I get that
$d\big{(}\eta_{n}^{\prime}(t),\eta_{n}^{\prime}(s)\big{)}=|s-t|$. Therefore
$\eta_{n}^{\prime}$ is a geodesic (to be pedantic one has to show that
$\eta_{n}^{\prime}$ is a continuous curve, which is a result of the fact that
$a_{t}^{\prime}$ is a one parameter subgroup of isometries). Clearly

$d\big{(}\eta_{n}^{\prime}(t),\eta^{\prime}(t)\big{)}=d(a_{t}^{\prime}nx,a_{t}^{\prime}x)=d(nx,x)$

and therefore $\eta_{n}^{\prime}$ is at uniformly bounded distance to
$\eta^{\prime}$. This bounds $d(\eta_{n},\eta_{n}^{\prime})$ as bi-infinite
geodesics, i.e. for all $t\in\mathbb{R}$, not just as infinite rays. The Flat
Strip Theorem (Theorem $2.13$, Chapter $2.2$ in [11]), then implies that the
geodesics $\eta_{n},\eta_{n}^{\prime}$ bound a flat strip: an isometric copy
of $\mathbb{R}\times[0,l]$ (where $l=d(x,nx)$).

Up to now I did not use the fact that $n\in N_{\xi}$, only that the point $nx$
lies on a geodesic that is contained in
$\mathcal{H}^{\prime}=\mathcal{H}^{\prime}_{0}$. Therefore the entire bi-
infinite geodesic that is determined by $[x,nx]$ lies on a $2$-dimensional
flat $F$ that contains $\eta^{\prime}$. The two elements $n,a_{t}^{\prime}$
therefore admit $nx,a_{t}^{\prime}x\in F$. It is a fact that two such elements
must commute. I can conclude therefore that $[n,a_{t}^{\prime}]=e_{G}$, which
contradicts the fact that that $n\in N_{\zeta}=Ker(T_{\zeta})$. ∎

######  Lemma 6.39 (Theorem $11.13$ in [52]).

Let $N$ be a connected real Lie group. Then $\mathrm{Lie}(N)$ is a nilpotent
Lie algebra if and only if $N$ is a nilpotent Lie group.

######  Proposition 6.40 (Proposition $13$, Section $4$, Chapter $1$ in [7]).

In the notation of Proposition 2.8, $\mathfrak{n}_{\xi}=\mathrm{Lie}(N_{\xi})$
is a maximal nilpotent ideal in $\mathfrak{g}_{\xi}=\mathrm{Lie}(G_{\xi})$.

######  Remark 6.41.

  1. 1.

The presentation of $\mathfrak{n}_{\xi}$ in [8] is given by means of the root
space decomposition of $\mathrm{Stab}_{G}(\xi)$, that appears in Proposition
$2.17.13$ in [19].

  2. 2.

There are two main objects in the literature that are referred to as the
_nilpotent radical_ or the _nilradical_ of a Lie algebra. These are: (a) the
maximal nilpotent ideal of the Lie algebra, and (b) the intersection of the
kernels of all irreducible finite-dimensional representations. Proposition
$13$ in Section $4$ of Chapter $9$ in [7] shows that in the case of Lie
algebras of parabolic Lie groups, these notions coincide.

######  Corollary 6.42.

$N_{\xi}$ is a maximal connected nilpotent normal Lie subgroup of the identity
connected component $\mathrm{Stab}_{G}(\mathcal{H})^{\circ}$.

###### Proof.

Lemma 6.39 implies $N_{\xi}$ is nilpotent. Since
$\mathrm{Stab}_{G}{\mathcal{H}}\vartriangleleft\mathrm{Stab}_{G}(\xi)$, every
normal subgroup of $\mathrm{Stab}_{G}(\mathcal{H})$ containing $N_{\xi}$ is in
fact a normal subgroup of $\mathrm{Stab}_{G}(\xi)$, still containing
$N_{\xi}$. It remains to prove maximality of $N_{\xi}$ among all connected
nilpotent normal Lie subgroups of $\mathrm{Stab}_{G}(\xi)$. Any such subgroup
$N^{\prime}\vartriangleleft\mathrm{Stab}_{G}(\xi)$ gives rise to an ideal
$\mathfrak{n}^{\prime}$ of
$\mathfrak{g}_{\xi}=\mathrm{Lie}\big{(}\mathrm{Stab}_{G}(\xi)\big{)}$, and by
Lemma 6.39 it is a nilpotent ideal. Therefore by Proposition 6.40 it is
contained in $\mathfrak{n}_{\xi}=\mathrm{Lie}(N_{\xi})$, implying that
$N^{\prime}\leq N_{\xi}$.

∎

######  Corollary 6.43.

A lattice in $\mathrm{Stab}_{G}(\mathcal{H})$ intersects the horospherical
subgroup $N_{\xi}$ in a lattice.

###### Proof.

Corollaries 6.37 and 6.42 imply that the pair
$N_{\xi}\vartriangleleft\mathrm{Stab}_{G}(\mathcal{H})$ satisfy the hypotheses
of Mostow’s Lemma 6.32. ∎

####  6.4.2 Indecomposable Horospherical Lattices

It is shown in [5] that if a horospherical lattice is contained in a Zariski
dense discrete subgroup, then the indecomposability condition is equivalent to
the irreducibility of the ambient group. The latter is imposed on $\Lambda$ as
a hypothesis in Theorem 6.1. The precise definitions and statements are as
follows.

######  Definition 6.44 (Definition $2.14$ in [5]).

For a semisimple real algebraic Lie group $G$ and $U$ a horospherical subgroup
of $G$, let $\Delta_{U}$ be a lattice in $U$.

  1. 1.

$\Delta_{U}$ is _irreducible_ if for any proper normal subgroup $N$ of
$G^{\circ}$, one has $\Delta_{U}\cap N=\\{e\\}$.

  2. 2.

$\Delta_{U}$ is _indecomposable_ if one cannot write $G^{\circ}$ as a product
$G^{\circ}=N^{\prime}N^{\prime\prime}$ of two proper normal subgroups
$N^{\prime},N^{\prime\prime}\vartriangleleft G$ with finite intersection such
that the group

$\Delta_{U}^{\prime}:=(\Delta_{U}\cap N^{\prime})(\Delta_{U}\cap
N^{\prime\prime})$

has finite index in $\Delta_{U}$.

######  Definition 6.45 (See Section $2.4.1$ in [5]).

Let $G$ be a semisimple real algebraic Lie group. A discrete subgroup
$\Lambda\leq G$ is said to be _irreducible_ if, for all proper normal
subgroups $N\vartriangleleft G$, the intersection $\Lambda\cap N$ is finite.

######  Lemma 6.46 (Lemma $4.3$ in [5]).

Let $G$ be a semisimple real algebraic Lie group, $U\subset G$ a non-trivial
horospherical subgroup, and $\Delta_{U}\leq U$ a lattice of $U$ which is
contained in a discrete Zariski dense subgroup $\Delta$ of $G$. Then the
following are equivalent:

  1. 1.

$\Delta$ is irreducible.

  2. 2.

$\Delta_{U}$ is irreducible.

  3. 3.

$\Delta_{U}$ is indecomposable.

####  6.4.3 Zariski Density

The last requirement is for $\Lambda$ to be Zariski dense. I use a geometric
criterion which is well known to experts.

######  Lemma 6.47 (Proposition $2$ in [30]).

Let $X$ be a symmetric space of noncompact type, $G=\mathrm{Isom}(X)^{\circ}$.
A subgroup $\Delta\leq G$ is Zariski dense if and only if:

  1. 1.

The group $\Delta$ does not globally fix a point in $X(\infty)$, i.e.
$\Delta\not\leq\mathrm{Stab}_{G}(\zeta)$ for any $\zeta\in X(\infty)$.

  2. 2.

The identity component of the Zariski closure of $\Delta$ does not leave
invariant any proper totally geodesic submanifold in $X$.

In the proof I use several facts - mostly algebraic, and two geometric. I
warmly thank Elyasheev Leibtag for his help and erudition in algebraic groups.
The first property I need is very basic.

######  Lemma 6.48.

Let $\Delta\leq G$ be a discrete subgroup, and let $H\leq G$ be the Zariski
closure of $\Delta$. Then $\Delta\cap H^{\circ}$ is of finite index in
$\Delta$.

###### Proof.

The subgroup $H^{\circ}$ is normal and of finite index in $H$. ∎

The following fact is probably known to experts. It appears in a recent work
by Bader and Leibtag [2].

######  Lemma 6.49 (Lemma $3.9$ in [2]).

Let $k$ be a field, $\mathbf{G}$ a connected $k$ algebraic group, $P\leq
G=\mathbf{G}(\mathbb{R})$ a parabolic subgroup. Then the centre of $G$
contains the centre of $P$.

Still on the algebraic side, I need a Theorem of Dani, generalizing the Borel
Density Theorem.

######  Theorem 6.50 (See [14]).

Let $\mathbf{S}$ be a real solvable algebraic group. If
$S=\mathbf{S}(\mathbb{R})$ is $\mathbb{R}$-split, then every lattice
$\Gamma_{S}\leq S$ is Zariski dense.

######  Remark 6.51.

It is a fact (see Theorem $15.4$ and Section $18$ in [6]) that:

  1. 1.

Every unipotent group over $\mathbb{R}$ is $\mathbb{R}$-split.

  2. 2.

For a field $k$ of characteristic $0$, a solvable linear algebraic $k$-group
is $k$-split if and only if its maximal torus is $k$-split.

Finally I need two geometric facts. The first is a characterization
determining when does a unipotent element belongs to $N_{\zeta}$ for some
$\zeta\in X(\infty)$.

######  Proposition 6.52 (Proposition $4.1.8$ in [19]).

Let $X$ be a symmetric space of noncompact type and of higher rank, $n\in
G=\mathrm{Isom}(X)^{\circ}$ a unipotent element, and $\zeta\in X(\infty)$. The
following are equivalent:

  1. 1.

For $N_{\zeta}$ as in Proposition 2.8, $n\in N_{\zeta}$.

  2. 2.

For some geodesic ray $\eta$ with $\eta(\infty)=\zeta$ it holds that
$\lim_{t\rightarrow\infty}d\big{(}n\eta(t),\eta(t)\big{)}=0$.

  3. 3.

For every geodesic ray $\eta$ with $\eta(\infty)=\zeta$ it holds that
$\lim_{t\rightarrow\infty}d\big{(}n\eta(t),\eta(t)\big{)}=0$.

The last property I need is a characterization of the displacement function
for unipotent elements.

######  Proposition 6.53 (See proof of Proposition $3.4$ in [4]).

Let $X$ be a symmetric space of noncompact type, $\zeta\in X(\infty)$ some
point and $n\in N_{\zeta}$ an element of the unipotent radical of
$\mathrm{Stab}_{G}(\zeta)$. The displacement function $x\mapsto d(nx,x)$ is
constant on horospheres based at $\zeta$, and for every $\varepsilon>0$ there
is a horoball $\mathcal{HB}_{\varepsilon}$ based at $\zeta$ such that
$d(nx,x)<\varepsilon$ for every $x\in\mathcal{HB}_{\varepsilon}$.

######  Corollary 6.54.

Assume that:

  1. 1.

$\big{(}\Lambda\cap\mathrm{Stab}_{G}(\mathcal{H})\big{)}\cdot x_{0}$ is a
cocompact metric lattice in a horosphere $\mathcal{H}\subset X$ bounding a
$\Lambda$-free horoball.

  2. 2.

Every $\Gamma$-conical limit point is a $\Lambda$-conical limit point.

Then $\Lambda$ is Zariski dense.

###### Proof.

I show the criteria of Lemma 6.47 are met, starting with
$\Lambda\not\leq\mathrm{Stab}_{G}(\zeta)$ for any $\zeta\in X(\infty)$. To
this end, I first prove that $\Lambda\cdot x_{0}$ is not contained in any
bounded neighbourhood of any horosphere $\mathcal{H}^{\prime}$. Let
$\xi^{\prime}$ be the base point of $\mathcal{H}^{\prime}$. By Hattori’s Lemma
2.32 (and Remark 2.33), it is enough to find a $\Lambda$-conical limit point
$\zeta^{\prime}$ with $d_{T}(\xi^{\prime},\zeta^{\prime})\neq\frac{\pi}{2}$.
Take some $\zeta^{\prime\prime}\in X(\infty)$ at Tits distance $\pi$ of
$\xi^{\prime}$, i.e. take a flat $F$ on which $\xi^{\prime}$ lies and let
$\zeta^{\prime\prime}$ be the antipodal point to $\xi^{\prime}$ in $F$. Fix
$\varepsilon=\frac{\pi}{4}$. By Proposition 2.4, there are neighbourhoods of
the cone topology $U,V\subset X(\infty)$ of
$\xi^{\prime},\zeta^{\prime\prime}$ (respectively) so that every point
$\zeta^{\prime}\in V$ admits $d_{T}(\xi^{\prime},\zeta^{\prime})\geq
d_{T}(\xi^{\prime},\zeta^{\prime\prime})-\frac{\pi}{4}=\frac{3}{4}\pi$. Recall
that the set of $\Gamma$-conical limit points is dense (in the cone topology),
so the second hypothesis implies there is indeed a $\Lambda$-conical limit
point in $V$ and therefore at Tits distance different (in this case larger)
than $\frac{\pi}{2}$ from $\xi^{\prime}$. I conclude that $\Lambda\cdot x_{0}$
is not contained in any bounded metric neighbourhood of any horosphere of $X$.

Assume towards contradiction that $\Lambda\leq\mathrm{Stab}_{G}(\zeta)$. I
show that this forces $\Lambda\cap N_{\zeta}\neq\emptyset$. By Proposition
6.52 it is enough to find a unipotent element $\lambda\in\Lambda$ and a
geodesic $\eta$ with $\eta(\infty)=\zeta$ such that
$\lim_{t\rightarrow\infty}d\big{(}\lambda\eta(t),\eta(t)\big{)}=0$. Let $F$ be
a maximal flat with $\xi,\zeta\in F(\infty)$, $x\in F$ some point and
$X,Y\in\mathfrak{a}\leq\mathfrak{p}$ two vectors such that $\exp(tY)=\eta(t)$
for the unit speed geodesic $\eta=[x,\zeta)$, and $\exp(tX)=\eta^{\prime}(t)$
for the unit speed geodesic $\eta^{\prime}=[x,\xi)$ (where
$\mathfrak{a}\leq\mathfrak{p}$ a maximal abelian subalgebra in a suitable
Cartan decomposition $\mathfrak{g}=\mathfrak{p}\oplus\mathfrak{k}$). Let
$\mathrm{Stab}_{G}(\xi)=K_{\xi}A_{\xi}N_{\xi}$ be the decomposition described
in Proposition 2.8 with respect to $X$ (notice that $N_{\xi}$ does not depend
on choice of $X$, see item $3$ of Proposition $2.17.7$ in [19]).

The assumption that $\Lambda\leq\mathrm{Stab}_{G}(\zeta)$ implies that for any
$\lambda\in\Lambda$ the distance $d\big{(}\lambda\eta(t),\eta(t)\big{)}$
either tends to $0$ as $t\rightarrow\infty$ or is uniformly bounded for
$t\in\mathbb{R}$. In the latter case there is some constant $c>0$ for which
$d\big{(}\lambda\eta(t),\eta(t)\big{)}=c$ for all $t\in\mathbb{R}$. As in the
proof of Corollary 6.38, the Flat Strip Theorem (Theorem $2.13$, Chapter $2.2$
in [11]) implies that $\lambda$ and $a_{t}:=\exp(tY)$ commute.

From the first hypothesis of the statement and Mostow’s result (Corollary
6.43) I know that $\Lambda\cap N_{\xi}$ is a cocompact lattice in $N_{\xi}$
(attention to subscripts). Therefore $\Lambda\cap N_{\xi}$ is Zariski dense in
$N_{\xi}$ (Theorem 6.50). Moreover, since commuting with an element is an
algebraic property, an element $g\in G$ that commutes with $\Lambda\cap
N_{\xi}$ must also commute with its Zariski closure, namely with $N_{\xi}$.
This means that if $a_{t}$ commutes with all $\Lambda\cap N_{\xi}$ then it
commutes with $N_{\xi}$, i.e. $a_{t}n=na_{t}$ for all $t\in\mathbb{R}$ and all
$n\in N_{\xi}$. I know that $a_{t}\in A_{\xi}$ commutes with both $K_{\xi}$
and $A_{\xi}$ (Proposition 2.8) therefore if $a_{t}$ also commutes with
$N_{\xi}$ then $a_{t}$ lies in the centre of $\mathrm{Stab}_{G}(\xi)$. This
means that $a_{t}$ is central in $G$ (Lemma 6.49). For a group $G$ with
compact centre this cannot happen, so there is indeed some unipotent element
$\lambda\in\Lambda\cap N_{\xi}$ for which
$\lim_{t\rightarrow\infty}d\big{(}\lambda\eta(t),\eta(t)\big{)}=0$. I conclude
from Proposition 6.52 that $\Lambda\cap N_{\zeta}\neq\emptyset$.

The first paragraph of the proof implies in particular that $\Lambda\cdot
x_{0}$ does not lie in any bounded neighbourhood of a horosphere
$\mathcal{H}^{\prime}$ based at $\zeta$. The assumption that
$\Lambda\subset\mathrm{Stab}_{G}(\zeta)$ implies that every
$\lambda\in\Lambda$ acts by translation on the filtration
$\\{\mathcal{H}^{\prime}_{t}\\}_{t\in\mathbb{R}}$ by horospheres based at
$\zeta$. Therefore as soon as $\Lambda\cdot x_{0}\not\subset\mathcal{H}_{t}$
for some $t\in\mathbb{R}$ one concludes that $\zeta$ is a horospherical limit
point of $\Lambda$, i.e. that every horoball based at $\zeta$ intersects the
orbit $\Lambda\cdot x_{0}$.

By Proposition 6.53 it holds that for a unipotent element $g\in N_{\zeta}$ the
displacement function $x\mapsto d(gx,x)$ depends only on the horosphere
$\mathcal{H}^{\prime}_{t}$ in which $x$ lies and that, for
$x_{t}\in\mathcal{H}^{\prime}_{t}$ it holds that
$\lim_{t\rightarrow\infty}d(gx_{t},x_{t})=0$ (up to reorienting the filtration
$t\in\mathbb{R}$ so that $\eta(t)\in\mathcal{H}^{\prime}_{t})$. For a non-
trivial element $\lambda_{\zeta}\in\Lambda\cap N_{\zeta}$ the previous
paragraph therefore yields a sequence of elements $\lambda_{n}\in\Lambda$ such
that
$\lim_{n\rightarrow\infty}d(\lambda_{\zeta}\lambda_{n}x_{0},\lambda_{n}x_{0})=0$,
contradicting the discreteness of $\Lambda$. I conclude that
$\Lambda\not\leq\mathrm{Stab}_{G}(\zeta)$ for every $\zeta\in X(\infty)$.

Assume that $H:=\big{(}\overline{\Lambda}^{Z}\big{)}^{\circ}$, the identity
connected component of the Zariski closure of $\Lambda$, stabilizes a totally
geodesic submanifold $Y\subset X$. By Lemma 6.48, $\Lambda_{0}:=\Lambda\cap H$
is of finite index in $\Lambda$, therefore
$\Lambda_{0}\cap\mathrm{Stab}_{G}(\mathcal{H})$ is also a cocompact lattice in
$\mathrm{Stab}_{G}(\mathcal{H})$. The fact that
$\big{(}\Lambda_{0}\cap\mathrm{Stab}_{G}(\mathcal{H})\big{)}\cdot x_{0}$ is a
cocompact metric lattice in $\mathcal{H}$ readily implies that
$\big{(}\Lambda_{0}\cap\mathrm{Stab}_{G}(\mathcal{H})\big{)}\cdot y$ is a
cocompact metric lattice in $\mathcal{H}_{y}=\mathcal{H}(y,\xi)$. This goes to
show that there is no loss of generality in assuming $x_{0}\in\mathcal{H}\cap
Y$. It follows that $\Lambda_{0}\cap\mathrm{Stab}_{G}(\mathcal{H})\cdot
x_{0}\subset Y\cap\mathcal{H}$, and therefore
$\mathcal{H}\subset\mathcal{N}_{D}(Y)$ for some $D>0$. A horosphere is a
codimension-$1$ submanifold, implying that $Y$ is either all of $X$ or of
codimension-$1$. The latter forces $Y=\mathcal{H}$, which is impossible since
$\mathcal{H}$ is not totally geodesic ($\mathcal{H}$ is not convex, see
Corollary 6.38). I conclude that $H$ does not stabilize any totally geodesic
proper submanifold, and hence that $\Lambda$ is Zariski dense. ∎

###  6.5 Proof of Theorem 6.1

I now complete the proof of the main sublinear rigidity theorem for
$\mathbb{Q}$-rank $1$ lattices.

###### Proof of Theorem 6.1.

If $\\{d_{\gamma}\\}_{\gamma\in\Gamma}$ is bounded, then $\Lambda$ is a
lattice by Corollary 6.6 or Theorem 6.8, depending on the $\mathbb{R}$-rank of
$G$.

If $\\{d_{\gamma}\\}_{\gamma\in\Gamma}$ is unbounded, then Proposition 6.11
and Corollary 6.26 both hold. In $\mathbb{R}$-rank$\ 1$ the proof again
follows immediately from Corollary 6.6 using Lemma 2.17 and Corollary 6.26.

In higher rank, the results of Section 6.4 allows one to conclude that
$\Lambda$ is an irreducible, discrete, Zariski dense subgroup that contains a
horospherical lattice. By Theorem 6.4, this renders $\Lambda$ a lattice. It is
a $\mathbb{Q}$-rank $1$ lattice as a result of Theorem 2.21. ∎

######  Remark 6.55.

The sublinear nature of the hypothesis in Theorem 1.6 induces coarse metric
constraints. A horospherical lattice on the other hand is a very precise
object. It is not clear how to produce unipotent elements in $\Lambda$, or
even general elements that preserve some horosphere. The proof above produces
a whole lattice of unipotent elements in $\Lambda$ (this is Corollary 6.43);
it is also the only proof that I know which produces even a single unipotent
element in $\Lambda$.

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