# On SuperIntegral Kleinian Sphere Packings, Bugs, and Arithmetic Groups

Michael Kapovich<EMAIL_ADDRESS>Department of Mathematics, UC
Davis, Davis, CA 95616  and  Alex Kontorovich<EMAIL_ADDRESS>Rutgers University, New Brunswick, NJ and National Museum of Mathematics, NY,
NY

###### Abstract.

We develop the notion of a Kleinian Sphere Packing, a generalization of
“crystallographic” (Apollonian-like) sphere packings defined by Kontorovich-
Nakamura [KN19]. Unlike crystallographic packings, Kleinian packings exist in
all dimensions, as do “superintegral” such. We extend the Arithmeticity
Theorem to Kleinian packings, that is, the superintegral ones come from
${\mathbb{Q}}$-arithmetic lattices of simplest type. The same holds for more
general objects we call Kleinian Bugs, in which the spheres need not be
disjoint but can meet with dihedral angles $\pi/m$ for finitely many $m$. We
settle two questions from [KN19]: $(i)$ that the Arithmeticity Theorem is in
general false over number fields, and $(ii)$ that integral packings only arise
from non-uniform lattices.

Kapovich is supported by the NSF grant DMS-16-04241 and Simons Fellowship,
grant number 391602.

Kontorovich is supported by an NSF CAREER grant DMS-1455705, an NSF FRG grant
DMS-1463940, NSF grant DMS-1802119, a BSF grant number 2014099, and the Simons
Foundation through MoMath’s Distinguished Visiting Professorship for the
Public Dissemination of Mathematics.

##  1\. Introduction

The classical Apollonian packing in the plane, usually described by an ad hoc
construction involving inscribing tangent circles, exhibits a number of
thereafter surprising arithmetic and dynamical properties; see, e.g., [K13].
In this paper, we complete the program initiated in [KN19] to understand the
relationship between such packings and the theory of arithmetic groups in
hyperbolic space.

###  1.1. Kleinian (and Crystallographic) Packings

A sphere packing (or just “packing”) $\mathscr{P}$ of
$\mathbb{S}^{n}\cong\partial_{\infty}\mathbb{H}^{n+1}$ ($n\geq 2$) is an
infinite collection of round balls in $\mathbb{S}^{n}$ with pairwise disjoint
interiors, such that the union of the balls is dense in $\mathbb{S}^{n}$. We
identify the ideal boundary
$\mathbb{S}^{n}=\partial_{\infty}{\mathbb{H}}^{n+1}$ with the one-point
compactification of Euclidean $n$-space
$\mathbb{S}^{n}={\mathbb{R}}^{n}\cup\\{\infty\\}$. By abuse of terminology, we
will conflate the collection of balls $\mathscr{P}$ with the collection of
round spheres bounded by these balls. In view of the density condition, a
packing contains balls with arbitrarily small Euclidean radii. (Of course, the
radii are only defined once we choose an identification
$\mathbb{S}^{n}\cong\partial_{\infty}\mathbb{H}^{n+1}$, in particular, we
choose a point at $\infty$.) The bend111Note that in the theory of Kleinian
groups, “bend” more often refers to dihedral angle; but for integral sphere
packings, bend is used for inversive radii. For circles, bend is the
curvature, but for higher dimensional spheres, (Gaussian) curvature is inverse
square-radius. of a sphere is the reciprocal of its radius, with the
convention that a sphere containing $\infty$ in its interior has negative
radius and bend. A packing is integral if all its spheres have integer bends.

Next we attach to any sphere packing its “superpacking.” For an $(n-1)$-sphere
$S\subset\mathbb{S}^{n}$, denote by $R_{S}$ reflection through $S$ acting on
$\mathbb{H}^{n+1}$. Given a packing $\mathscr{P}$, let

(1) $\Gamma_{\mathscr{P}}=\left\langle R_{S}:S\in\mathscr{P}\right\rangle\ <\
\mathrm{Isom}(\mathbb{H}^{n+1})$

be the reflection group of $\mathscr{P}$, generated by reflections through the
spheres in $\mathscr{P}$. The superpacking

(2) $\widetilde{\mathscr{P}}\ :=\ \Gamma_{\mathscr{P}}\cdot\mathscr{P}$

is defined as the orbit of the packing under the action of its reflection
group, see Figure 1. A packing is superintegral if its superpacking has all
integer bends.222It turns out that this condition is strictly stronger than
integrality; that is, there exist packings, even crystallographic ones, which
are integral but not superintegral [KN19]. Note that no tangency conditions
are imposed on the spheres; indeed the circles in the packing shown in Figure
1$(a)$ are all disjoint (and this packing is superintegral).

$(a)$ $(b)$ $(c)$

Figure 1. $(a)$ A packing $\mathscr{P}$, $(b)$ its reflection group
$\Gamma_{\mathscr{P}}$, and $(c)$ its superpacking $\widetilde{\mathscr{P}}$.

As we will show, superintegrality, even in the absence of any other structure
imposed on the packing $\mathscr{P}$, is already related to
(sub)arithmeticity, as follows. Recall that a group of hyperbolic isometries
is called “$k$-arithmetic” (of simplest type, as assumed throughout) if,
possibly after conjugation, it is commensurable with the group
$O_{\mathcal{F}}(\mathfrak{o})$ of $\mathfrak{o}$-integral automorphs of a
hyperbolic333See Definition 47. quadratic form ${\mathcal{F}}$ defined over a
totally real number field $k$ with ring of integers $\mathfrak{o}$ (see, e.g.,
[VS93]). We call a Zariski-dense, discrete subgroup
$\Gamma<\mathrm{Isom}(\mathbb{H}^{n+1})$ $k$-subarithmetic if $\Gamma$ is
contained in a $k$-arithmetic lattice.

######  Theorem 3 (Subarithmeticity Theorem).

If an orbit

$\mathcal{O}=\Gamma\cdot S_{0}$

of a fixed sphere $S_{0}\subset\partial_{\infty}\mathbb{H}^{n+1}$ under a
Zariski dense subgroup $\Gamma<\mathrm{Isom}(\mathbb{H}^{n+1})$ has all
integer bends, then $\Gamma$ is ${\mathbb{Q}}$-subarithmetic. More precisely,
there exists an isotropic rational hyperbolic quadratic form ${\mathcal{F}}$
so that $\Gamma$ is contained in $O_{\mathcal{F}}({\mathbb{Z}})$.

######  Corollary 4.

In particular, if a general packing $\mathscr{P}$ happens to be superintegral,
then its reflection group $\Gamma_{\mathscr{P}}$ is necessarily
${\mathbb{Q}}$-subarithmetic.

######  Remark 5.

These conclusions need not hold for $\mathfrak{o}$-(super)integral packings,
that is, ones with all bends in a ring of integers $\mathfrak{o}$; see §1.2.

In general, there is not much more one can say about integral or superintegral
packings without assuming more structure.

######  Definition 6.

A sphere packing $\mathscr{P}$ is Kleinian if its set of limit points also
arises as the limit set of a geometrically finite444 See Definition 28. group
$\Gamma_{S}<\mathrm{Isom}(\mathbb{H}^{n+1})$. We call $\Gamma_{S}$ a symmetry
group of the packing.

In the special case that $\Gamma$ is generated by finitely many reflections in
hyperplanes, $\mathscr{P}$ is called crystallographic; such were defined and
studied in [KN19]. It is well-known and easy to see (e.g., from the Poincaré
Fundamental Polyhedron Theorem) that discrete, finitely generated, hyperbolic
reflection groups are geometrically finite, so every crystallographic packing
is also Kleinian.

The following statement is both standard and deserves to be stated explicitly.

######  Theorem 7.

Let $\Gamma_{S}<\mathrm{Isom}(\mathbb{H}^{n+1})$ be geometrically finite. Then
its domain of discontinuity is a disjoint union of open round balls (that is,
is a Kleinian sphere packing) if and only if the boundary of the convex core
$M^{*}$ of the orbifold $M=\mathbb{H}^{n+1}/\Gamma_{S}$ is totally geodesic.

An important role is played by the “supergroup” of a Kleinian packing. If
$\mathscr{P}$ has symmetry group $\Gamma_{S}$, then the supergroup
$\widetilde{\Gamma}$ is defined by

$\widetilde{\Gamma}:=\left\langle\Gamma_{S},\Gamma_{\mathscr{P}}\right\rangle.$

That is, the supergroup is the group generated by both $\Gamma_{S}$ and
$\Gamma_{\mathscr{P}}$. Note that both the symmetry group $\Gamma_{S}$ and
supergroup $\widetilde{\Gamma}$ are not uniquely determined by the packing
$\mathscr{P}$; indeed, any nontrivial normal subgroup of $\Gamma_{S}$ will
have the same limit set. A priori it is not even obvious that
$\widetilde{\Gamma}$ acts discretely, but it is in fact a lattice, acting on
$\mathbb{H}^{n+1}$ with finite covolume; see the Structure Theorem 22.

We turn our attention now to the integral and superintegral Kleinian packings.
An immediate corollary of the Subarithmeticity Theorem 3 is that, if a
Kleinian packing $\mathscr{P}$ is integral, then any symmetry group
$\Gamma_{S}$ is ${\mathbb{Q}}$-subarithmetic. If $\mathscr{P}$ is moreover
superintegral, then its supergroup $\widetilde{\Gamma}$ is itself
${\mathbb{Q}}$-arithmetic. This is because the superpacking
$\widetilde{\mathscr{P}}$ can also be given as the orbit of the packing
$\mathscr{P}$ under the action of $\widetilde{\Gamma}$, together with the
obvious fact that, if a group is subarithmetic and a lattice, then it is
arithmetic!

In the case of crystallographic packings, the main result of [KN19, Thm 18]
was the following finiteness theorem. Before stating the theorem, note that it
is shown in [KN19, Thm 3] that there exist infinitely many conformally
inequivalent superintegral crystallographic packings in certain dimensions up
to $n=18$. We say that two Kleinian packings are commensurable if (conjugates
of) their supergroups are.

######  Theorem 8 (Finiteness Theorem [KN19]).

Superintegral crystallographic packings exist in only finitely-many
dimensions, and there are finitely many in each dimension, up to
commensurability.

Indeed, if a crystallographic packing is superintegral, then its supergroup is
a ${\mathbb{Q}}$-arithmetic reflective lattice. Arithmetic reflective lattices
are known (see, e.g., the discussion in [Bel16]) to lie in finitely many
commensurability classes in finitely many dimensions, which implies Theorem 8.
Moreover, there are no ${\mathbb{Q}}$-arithmetic reflective lattices acting on
$\mathbb{H}^{n+1}$ with $n+1=20$ or $n+1\geq 22$, see [Ess96], so there are no
superintegral crystallographic packings in $n=19$ or $n\geq 21$ dimensions.
Worse yet, crystallographic packings are not yet known to exist (nevermind
integrality) in dimensions $n=14,15,16,18$ and $20$, although reflective
lattices are known in one more than these dimensions [KN19].

While the classification of commensurability types of superintegral
crystallographic packings awaits first that of arithmetic hyperbolic
reflection groups, one can completely classify superintegral Kleinian packings
in terms of arithmetic groups. The following is the first main theorem of this
paper.

######  Theorem 9 (Classification Theorem).

A hyperbolic lattice is commensurable to a supergroup of a superintegral
Kleinian packing if and only if it is a non-uniform ${\mathbb{Q}}$-arithmetic
lattice of simplest type.

So in contradistinction with the Finiteness Theorem 8 for crystallographic
packings, more general Kleinian packings exist in every dimension. Moreover,
the Classification Theorem 9 answers a question posed in [KN19] on whether
there exist superintegral crystallographic packings with cocompact
supergroups: there do not! We emphasize again that the spheres in a
superintegral Kleinian (or even just crystallographic) packing could all be
disjoint, but the supergroup $\widetilde{\Gamma}$ must have cusps. See Example
66.

######  Remark 10.

The precise role of ${\mathbb{Q}}$-isotropy of the corresponding quadratic
form leading to (super)integrality of packings is elucidated in Lemma 52,
which shows that the “covector” corresponding to the bend is itself isotropic
in the dual form.

######  Remark 11.

Note that even if a symmetry group $\Gamma_{S}$ is geometrically finite and
the convex core $M^{*}$ of $M=\mathbb{H}^{n+1}/\Gamma_{S}$ has totally
geodesic boundary (so its limit set gives rise to a Kleinian packing
$\mathscr{P}$), and if the supergroup $\widetilde{\Gamma}$ of $\mathscr{P}$ is
non-uniform and ${\mathbb{Q}}$-arithmetic of simplest type, it still need not
be the case that the packing $\mathscr{P}$ is necessarily integral, for any
conformal choice of coordinates on $\mathbb{S}^{n}$, see Example 67. But the
next theorem states that, if $M^{*}$ has only one boundary component (that is,
$\Gamma_{S}$ acts transitively on the spheres in the packing), then the
packing is necessarily superintegral.

######  Theorem 12.

$(i)$ Suppose that a supergroup $\widetilde{\Gamma}$ of a packing
$\mathscr{P}$ is a non-uniform ${\mathbb{Q}}$-arithmetic of simplest type
(i.e. is commensurable to $O_{\mathcal{F}}({\mathbb{Z}})$, where
${\mathcal{F}}$ is a rational hyperbolic quadratic form) and, moreover, a
symmetry group $\Gamma_{S}<\widetilde{\Gamma}$ acts transitively on the set of
spheres in the packing $\mathscr{P}$. Then there is always a conformal change
of coordinates such that $\mathscr{P}$ is superintegral.

$(ii)$ More generally, let $\Gamma$ be any non-uniform
${\mathbb{Q}}$-arithmetic group of the simplest type and let $S$ be a sphere
for which the reflection $R_{S}$ lies in $\Gamma$. Then the $\Gamma$-orbit of
$S$ can be made integral by a suitable conformal change of coordinates.

There are two main ingredients leading to the proof of the Classification
Theorem 9, one geometric (constructing a Kleinian packing from the Structure
Theorem 22 below), and the second arithmetic, to ensure that the packing thus
constructed is indeed superintegral; it is the second stage that forces
${\mathbb{Q}}$-arithmeticity and non-uniformity of the lattice.

A key step in the geometric argument relies on Millson’s theorem [Mi76] that
every arithmetic hyperbolic lattice of simplest type is commensurable to a
lattice $\widetilde{\Gamma}$ such that ${\mathbb{H}}^{n+1}/\widetilde{\Gamma}$
contains a nonseparating totally geodesic complete hypersurface of finite
volume.

We will also show that, in each commensurability class, there are not only
infinitely many conformally inequivalent packings (as shown for
crystallographic packings in [KN19]), but their Hausdorff dimensions can be
made arbitrarily close to maximal.

######  Theorem 13 (Abundance Theorem).

In every dimension $n\geq 2$, there exist quasiconformally inequivalent
superintegral Kleinian packings, whose limit sets have Hausdorff dimension
approaching $n$.

Note the contrast to Phillips-Sarnak’s and Doyle’s work on Schottky groups
[PS85, Theorem 5.4], [Do88], showing that their limit sets have Hausdorff
dimensions bounded strictly away from the ambient dimension $n$. Also note
that the conformally inequivalent superintegral Polyhedral Circle Packings in
$\partial_{\infty}\mathbb{H}^{3}$ constructed in [KN19, Thm 7] are all
Schottky, so their dimensions cannot approach $n=2$.

###  1.2. Packings over Number Fields

To complete the discussion of Kleinian packings (before we turn to Kleinian
“bugs”), we show that the theory breaks down over number fields, answering
another question posed in [KN19]. For a totally real number field $k$ with
ring of integers $\mathfrak{o}$, a packing is $\mathfrak{o}$-integral (resp.
$\mathfrak{o}$-superintegral) if every bend in its packing (resp.
superpacking) lies in $\mathfrak{o}$.

######  Proposition 14.

The Subarithmeticity Theorem does not hold in number fields.

Indeed, already for the simplest case of the golden ring
$k={\mathbb{Q}}(\sqrt{5})$, there exist $\mathfrak{o}$-superintegral
crystallographic packings whose supergroups are non-arithmetic. (See Example
71.) But this is not the whole story, as we also have the following.

######  Theorem 15.

For any $k$-arithmetic hyperbolic lattice of simplest type
$O_{\mathcal{F}}(\mathfrak{o})$, there is a totally real, quadratic extension
$k^{\prime}\supset k$, with ring of integers $\mathfrak{o}^{\prime}$, such
that $O_{\mathcal{F}}(\mathfrak{o})$ is commensurable to the supergroup of an
$\mathfrak{o}^{\prime}$-superintegral Kleinian packing.

Note that there is no condition on being non-uniform here, unlike Theorem 9.
Thus, even the ${\mathbb{Q}}$-anisotropic form
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-7x_{4}^{2}$ has
$\mathfrak{o}^{\prime}$-superintegral Kleinian packings, for certain quadratic
rings $\mathfrak{o}^{\prime}$. This leaves open the problem of giving a proper
formulation for how to characterize which
$\mathfrak{o}^{\prime}$-superintegral packings come from arithmetic groups.

###  1.3. Kleinian Bugs

To further extend the notion of a packing, we will allow spheres to meet at a
finite set of dihedral angles, as follows.

A cooriented round sphere in $\mathbb{S}^{n}$ is a round sphere together with
a choice of a nowhere vanishing normal vector field. Cooriented spheres are in
a natural bijective correspondence with cooriented hyperbolic hyperplanes in
${\mathbb{H}}^{n+1}$. A convex polyhedron, $P$, in ${\mathbb{H}}^{n+1}$ is the
intersection of a locally finite family of hyperbolic half-spaces
$H_{\alpha}^{+}$ in ${\mathbb{H}}^{n+1}$. It suffices to consider only half-
spaces $H_{\alpha}^{+}$ such that the hyperplanes $H_{\alpha}$ intersect
$\partial P$ along facets $F_{\alpha}$. We coorient these hyperplanes
$H_{\alpha}$ so that the normal vector fields point into the complementary
half-spaces $H_{\alpha}^{-}$. The ideal boundary
$\partial_{\infty}H_{\alpha}^{-}$ of the half-space $H_{\alpha}^{-}$ is a
round ball $B_{\alpha}$ bounded by the cooriented round sphere
$\partial_{\infty}H_{\alpha}$, see Figure 2.

$P$$F_{\alpha}$$H_{\alpha}$$\partial_{\infty}{\mathbb{H}}^{n+1}$$B_{\alpha}$$\partial_{\infty}{H_{\alpha}}$
Figure 2. A portion of a polyhedron $P$ having bounding facet $F_{\alpha}$
lying in the hyperplane $H_{\alpha}$. Also shown is the ideal boundary
$B_{\alpha}\subset\partial_{\infty}{\mathbb{H}}^{n+1}$ of the complementary
half-space $H_{\alpha}^{-}$, and the cooriented round sphere
$\partial_{\infty}H_{\alpha}$ bounding $B_{\alpha}$.

######  Definition 16.

A bug $\mathscr{B}$ is an infinite collection of cooriented spheres in
$\mathbb{S}^{n}\cong\partial_{\infty}\mathbb{H}^{n+1}$ containing facets of a
convex polyhedron $P=P_{\mathscr{B}}\subset{\mathbb{H}}^{n+1}$, with dihedral
angles lying in a finite subset of

$\frac{\pi}{{\mathbb{N}}}\ =\ \left\\{\frac{\pi}{m}\ :\
m\in{\mathbb{N}}\right\\},$

such that the union of round balls bounded by these spheres is dense in
$\mathbb{S}^{n}$. As before, a bug is integral if its spheres all have integer
bends.

In plainer terms (forgetting coorientations), a bug is a collection of
spheres, any pair of which is either disjoint, tangent or which intersect at
angles $\pi/m$, for finitely many $m\in{\mathbb{N}}$. See Figure 3 for an
integral bug, and Figure 4 for the corresponding hyperbolic polyhedron $P$,
which should justify our calling these objects “bugs.”

####  1.3.1. Geometric Aspects of Bugs

Before turning to any arithmetic properties of bugs, we discuss purely
geometric aspects, culminating in the Structure Theorem 22. As in (1), the
reflection group generated by reflections through the spheres in
$\mathscr{B}$, denoted $\Gamma_{\mathscr{B}}$, is discrete and has as its
fundamental domain the hyperbolic convex polyhedron $P=P_{\mathscr{B}}$
bounded by the hemispheres.

In this setting, we need a substitute for the limit set of the packing, which
will no longer necessarily contain spheres, see Figure 5(b).

######  Definition 17.

The accumulation set ${\mathcal{A}}(\mathscr{B})$ of a bug $\mathscr{B}$
consists of those points $\xi\in\partial_{\infty}{\mathbb{H}}^{n+1}$ such that
every neighborhood of $\xi$ contains infinitely many spheres in $\mathscr{B}$.

As a first (it turns out, insufficient) step, we define “geometric” bugs in
terms of symmetry groups as before:

Figure 3. An integral bug with all dihedral angles $\pi/3$. A number at the
center of a circle denotes its bend, that is, inverse radius. The thick
unlabeled blue circle has bend $-76$. The color scheme is as described in
Remark 24. Figure 4. View from $\mathbb{H}^{3}$ of the bug polyhedron $P$ from
Figure 3. The polyhedron is exterior to all solid hemispheres shown, and
interior to the one opaque hemisphere, which has bend $-76$.

$(a)$ $(b)$

Figure 5. $(a)$ The bug is Kleinian with symmetry group generated by
reflections in the thick red circles. $(b)$ The accumulation set in
$\partial_{\infty}\mathbb{H}^{n+1}$ does not in general contain spheres.

######  Definition 18.

A bug $\mathscr{B}$ is geometric if there exists a geometrically finite
subgroup $\Gamma_{S}<\mathrm{Isom}(\mathbb{H}^{n+1})$ such that

  1. (1)

the accumulation set of the bug is the limit set of $\Gamma_{S}$, and

  2. (2)

$\Gamma_{S}$ preserves the bug polyhedron $P=P_{\mathscr{B}}$.

Such a $\Gamma_{S}$ is called a symmetry group of the bug.

Recall again that a symmetry group of a bug is not unique. But its
commensurability class is uniquely determined by the bug, provided that the
accumulation set of ${\mathscr{B}}$ consists of more than two points; see
Lemma 41.

The supergroup is defined in the now-familiar way:

######  Definition 19.

Given a bug $\mathscr{B}$ and a symmetry group $\Gamma_{S}$, the supergroup

$\widetilde{\Gamma}\ :=\
\left\langle\Gamma_{S},\Gamma_{\mathscr{B}}\right\rangle\ <\
\mathrm{Isom}(\mathbb{H}^{n+1}),$

is defined as the group generated by the symmetry group $\Gamma_{S}$ and the
reflection group $\Gamma_{\mathscr{B}}$.

Naively, one could have expected that, as in the crystallographic or Kleinian
packing setting, the supergroup of a bug is a lattice; unfortunately this is
false in general (see Example 36), owing to the possible “incompatibility” of
parabolic subgroups. Recall that a discrete subgroup
$\Pi<\mathrm{Isom}({\mathbb{H}}^{n+1})$ is called parabolic555Sadly, this
classical terminology is inconsistent with the one used in the theory of
algebraic groups. if it contains a parabolic element $g$ and stabilizes the
fixed point of $g$. Such a subgroup consists of parabolic and elliptic
elements.

######  Definition 20.

An ordered pair of subgroups
$\Gamma_{1},\Gamma_{2}<\mathrm{Isom}(\mathbb{H}^{n+1})$ is cusp-compatible if,
for every maximal parabolic subgroup $\Pi_{2}<\Gamma_{2}$, there exists a
subgroup $\Pi_{1}<\Gamma_{1}$ such that the subgroup $\Pi$ generated by
$\Pi_{1},\Pi_{2}$ is virtually abelian of virtual rank $n$.

Note that cusp compatibility depends only on the commensurability classes of
$\Gamma_{1},\Gamma_{2}$.

######  Definition 21.

A bug $\mathscr{B}$ is called Kleinian if it is geometric with symmetry group
$\Gamma_{S}$, such that the ordered pair $\Gamma_{\mathscr{B}}$, $\Gamma_{S}$
is cusp-compatible.

We remark that, if $\mathscr{B}$ is not just a bug but also a packing, then
the cusp-compatibility condition is automatic, see Lemma 44.

In §2 we will prove the second main theorem of this paper, which is the
following extension of the Crystallographic Structure Theorem in [KN19, Thm
28].

######  Theorem 22 (Structure Theorem for Kleinian Bugs and Packings).

1\. If $\mathscr{B}$ is a Kleinian bug, then its supergroup
$\widetilde{\Gamma}$ is a lattice.

2\. (i) Conversely, suppose we are given a lattice
$\widetilde{\Gamma}<\mathrm{Isom}(\mathbb{H}^{n+1})$, a convex fundamental
polyhedron $D$ of $\widetilde{\Gamma}$, and a finite set $S^{\prime}$ of
elements $\gamma_{j}\in\widetilde{\Gamma}$ pairing the facets of $D$. Let
$R\subset S^{\prime}$ be a nonempty subset consisting of reflections (assuming
such exist) in some facets $F_{\alpha}\subset\partial D$, with $\alpha$ in an
indexing set $A$. As before, let $H_{\alpha}$ be the cooriented hyperbolic
hyperplanes containing the facets $F_{\alpha}$. Let $S:=S^{\prime}\setminus R$
and $\Gamma_{S}:=\left\langle S\right\rangle<\widetilde{\Gamma}$. Then the
orbit of the set of cooriented spheres in
$\\{\partial_{\infty}H_{\alpha};\alpha\in A\\}$ under the group $\Gamma_{S}$,

$\mathscr{B}\ :=\ \Gamma_{S}\cdot\\{\partial_{\infty}H_{\alpha};\alpha\in
A\\},$

is a Kleinian bug with a symmetry group $\Gamma_{S}$.

(ii) Suppose, moreover, that: (a) the facets $F_{\alpha}$, $\alpha\in A$, have
pairwise disjoint or tangent cooriented spheres $\partial_{\infty}H_{\alpha}$,
(b) the hyperplanes $H_{\alpha}$ meet the other bounding walls of $D$ either
tangentially or orthogonally (or not at all). Then $\mathscr{B}$ is in fact a
Kleinian packing, that is, if two spheres of $\mathscr{B}$ intersect, they do
so tangentially.

In particular, every lattice
$\widetilde{\Gamma}<\mathrm{Isom}(\mathbb{H}^{n+1})$ containing a reflection
yields a Kleinian bug.

######  Remark 23.

While it is not hard to show that the supergroup of a crystallographic or
Kleinian packing is a lattice, the proof of Theorem 22 is rather more
involved, see §2.

######  Remark 24.

We applied the above construction to create the bug in Figures 3–4 from the
extended Bianchi group $\widehat{Bi}(19)$666Here and throughout,
$\widehat{Bi}(D)$ denotes the extended Bianchi group, that is, the maximal
discrete subgroup of $\mathop{\hbox{Isom}}({\mathbb{H}}^{3})$ containing the
group $\mathop{\hbox{PSL}}_{2}(\mathcal{O}_{D})$, where $\mathcal{O}_{D}$ is
the ring of integers of ${\mathbb{Q}}(\sqrt{-D})$.; namely, two reflective
walls were chosen for $R\subset S^{\prime}$ from this lattice, and the bug was
created as the orbit of these two walls under the remaining generators. The
orbit of one wall was colored blue and the other red.

####  1.3.2. Arithmetic Aspects of Bugs

Turning our attention to arithmetic properties of bugs, we mimic (2) with the
following.

######  Definition 25.

The superbug $\widetilde{\mathscr{B}}$ of an (arbitrary) bug $\mathscr{B}$ is
defined as the orbit of the bug under its reflection group,
$\Gamma_{\mathscr{B}}$:

$\widetilde{\mathscr{B}}\ :=\ \Gamma_{\mathscr{B}}\cdot\mathscr{B}.$

A bug is superintegral if every sphere in its superbug has integer bend.

The Subarithmeticity Theorem 3 applies just as well to bugs, giving the
following:

######  Theorem 26.

  * •

If a bug (with no further structure imposed) is superintegral, then its
reflection group $\Gamma_{\mathscr{B}}$ is ${\mathbb{Q}}$-subarithmetic.

  * •

If a bug is Kleinian and integral, then its symmetry group $\Gamma_{S}$ is
${\mathbb{Q}}$-subarithmetic.

  * •

If a bug is Kleinian and superintegral, then its supergroup
$\widetilde{\Gamma}$ is ${\mathbb{Q}}$-arithmetic.

In the opposite direction, we already have from the Classification Theorem 9
that every non-uniform ${\mathbb{Q}}$-arithmetic lattice (of simplest type) is
commensurable to the supergroup of a superintegral Kleinian packing (and hence
bug). Similarly, we have from Theorem 12 that if a symmetry group $\Gamma_{S}$
(resp. supergroup $\widetilde{\Gamma}$) of a Kleinian bug is non-uniform and
${\mathbb{Q}}$-subarithmetic (resp. ${\mathbb{Q}}$-arithmetic), then there is
a conformal choice of coordinates for which the bug can be made integral
(resp. superintegral).

######  Remark 27.

When a Kleinian bug (or just packing) is integral but not superintegral, we
have the following curious situation: its symmetry group $\Gamma_{S}$ is a
subgroup of some integer orthogonal group $O_{{\mathcal{F}}}({\mathbb{Z}})$,
but it is also contained in a (non-arithmetic) lattice, the supergroup
$\widetilde{\Gamma}$.

Note further that there are essentially only three ways that a lattice
$\widetilde{\Gamma}<\mathrm{Isom}(\mathbb{H}^{n+1})$ (with $n\geq 2$) can be
non-arithmetic. By Selberg’s Rigidity [Sel60], after conjugation,
$\widetilde{\Gamma}$ is contained in some $O_{n+1,1}(k)$ where $k$ is a number
field with a given embedding to ${\mathbb{R}}$. For $\widetilde{\Gamma}$ to be
non-arithmetic, either:

  * $(i)$

$k$ is not totally real, or

  * $(ii)$

$k$ is totally real but for at least one non-identity embedding $\iota$, the
orthogonal group $O_{n+1,1}(\iota(k))$ is non-compact (that is, the quadratic
from is not hyperbolic), or

  * $(iii)$

$k$ is totally real and the quadratic form is hyperbolic, but the entries of
$\widetilde{\Gamma}$ as elements of $k$ have unbounded denominators.

Lattices satisfying the last condition $(iii)$ are called quasi-arithmetic by
Vinberg. The only currently known integral but non-superintegral bugs (e.g.
the crystallographic packings constructed in [KN19, Lemma 20]) all come from
this last situation (with $k={\mathbb{Q}}$). It is interesting to investigate
whether this is the only non-arithmeticity type possible for integral but not
superintegral bugs. (Note that for $\mathfrak{o}$-superintegrality, the non-
arithmeticity in Proposition 14 is of type $(ii)$; see Example 71.)

###  1.4. Outline of the paper

In §2 we prove geometric results on bugs, including the key the Structure
Theorem 22. We spend §3 discussing the arithmetic properties of bugs,
including the main Subarithmeticity Theorem 3. Of particular interest to the
reader may be §3.3 where a number of archetypal examples are constructed. We
conclude in §4 with a discussion of Hausdorff dimensions of accumulation sets
of superintegral bugs, proving the Abundance Theorem 13.

### Acknowledgements

The second-named author would like to thank Curt McMullen and Peter Sarnak for
many enlightening conversations and suggestions.

##  2\. Proof of the Structure Theorem 22

In what follows, $d$ denotes the hyperbolic distance on ${\mathbb{H}}^{n+1}$.
For a subset $A\subset{\mathbb{H}}^{n+1}$ the ideal boundary of $A$, denoted
$\partial_{\infty}A$, is the accumulation set of $A$ in the ideal boundary
sphere $\mathbb{S}^{n}=\partial_{\infty}{\mathbb{H}}^{n+1}$ of
${\mathbb{H}}^{n+1}$.

Recall that for a discrete subgroup
$\Gamma<\mathop{\hbox{Isom}}(\mathbb{H}^{n+1})$, the convex hull of the limit
set $\Lambda$ of $\Gamma$, denoted $C=C_{\Gamma}$, is the intersection of all
$\Gamma$-invariant closed convex nonempty subsets in $\mathbb{H}^{n+1}$. The
convex hull has the property that $\partial_{\infty}C=\Lambda$ unless
$\Lambda$ is a singleton (in which case $C=\emptyset$). The convex core
$M^{*}$ of the orbifold $M=\mathbb{H}^{n+1}/\Gamma$ is the quotient
$C/\Gamma$.

######  Definition 28.

A discrete subgroup $\Gamma<\mathrm{Isom}(\mathbb{H}^{n+1})$ is geometrically
finite if $\Gamma$ is virtually torsion free and there is an
$\varepsilon$-thickening of the convex core $M^{*}$ of the orbifold
$M=\mathbb{H}^{n+1}/\Gamma$ that has finite volume.

A sufficient condition for geometric finiteness is the existence of a
finitely-sided fundamental polyhedron of $\Gamma$ in ${\mathbb{H}}^{n+1}$.

######  Definition 29.

A discrete subgroup $\Gamma<\mathop{\hbox{Isom}}(\mathbb{H}^{n+1})$ is called
convex-cocompact if its limit set is not a singleton and the convex core
$M^{*}$ of the orbifold $M=\mathbb{H}^{n+1}/\Gamma$ is compact. Equivalently,
there exists a closed convex nonempty $\Gamma$-invariant subset of
$\mathbb{H}^{n+1}$ on which $\Gamma$ acts cocompactly.

Convex-cocompact subgroups have the property that

(30) $\inf\\{d(gx,x):g\in\Gamma^{*},x\in C_{\Gamma}\\}>0,$

where $\Gamma^{*}\subset\Gamma$ consists of elements of infinite order.

We refer to [Bo93, Ra06] for more background on geometrically finite and
convex-cocompact isometry groups of hyperbolic space.

Part 1. Let $\mathscr{B}$ be a Kleinian bug and let
$P\subset{\mathbb{H}}^{n+1}$ be the corresponding convex polyhedron. We wish
to show that the supergroup $\widetilde{\Gamma}$ is a lattice.

If $\Phi\subset{\mathbb{H}}^{n+1}$ is a convex fundamental domain of an
arbitrary discrete subgroup $\Gamma<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$,
then the $\Gamma$-orbit of the relative interior of $\partial_{\infty}\Phi$
(with respect to $\mathbb{S}^{n}$) is dense in the domain of discontinuity of
$\Gamma$ in $\mathbb{S}^{n}$. Since $\mathscr{B}$ is a bug, for $\Phi=P$, this
relative interior of $\partial_{\infty}P$ is empty and, hence the limit set of
$\Gamma_{\mathscr{B}}$ is equal to the entire sphere $\mathbb{S}^{n}$.

The ideal boundary
$\partial_{\infty}P\subset\partial_{\infty}{\mathbb{H}}^{n+1}=\mathbb{S}^{n}$
is the disjoint union

(31) $\partial_{\infty}P={\mathcal{A}}(P)\sqcup{\mathcal{V}}(P),$

where each point of ${\mathcal{V}}(P)$ is isolated in the accumulation set
${\mathcal{A}}(P):={\mathcal{A}}(\mathscr{B})$ of $\mathscr{B}$; elements of
${\mathcal{V}}(P)$ are the “ideal vertices” of $P$. For each
$\eta\in{\mathcal{V}}(P)$ and every horoball
${\mathrm{B}}\subset{\mathbb{H}}^{n}$ centered at $\eta$, the intersection
$P\cap{\mathrm{B}}$ has finite volume.

The convex polyhedron $P=P_{\mathscr{B}}$ is the fundamental domain of the
reflection group $\Gamma_{\mathscr{B}}<\mathrm{Isom}({\mathbb{H}}^{n+1})$
generated by isometric reflections in facets of $P$, if and only if each
dihedral angle of $P$ belongs to $\frac{\pi}{{\mathbb{N}}}$, see [Ma88, Ra06].

We start the proof of Part 1 with two auxiliary lemmata.

For each nonempty closed convex subset $C\subset{\mathbb{H}}^{n+1}$ we have
the nearest-point projection $\pi_{C}:{\mathbb{H}}^{n+1}\to C$. This
projection is a continuous (actually, $1$-Lipschitz) map.

######  Lemma 32.

Suppose that $C$ is a closed convex subset of ${\mathbb{H}}^{n+1}$ and
$C\subset A$, where $A$ is a closed subset of ${\mathbb{H}}^{n+1}$ such that
$\partial_{\infty}C=\partial_{\infty}A$. Then the restriction of $\pi_{C}$ to
$A$ is a proper map, i.e. preimages of compact subsets of $C$ are compact.

###### Proof.

Assume that this is not the case. Then there exist sequences $y_{i}\in C$ and
$x_{i}\in A$ such that $\pi(x_{i})=y_{i}$, $\lim y_{i}=y\in C$, while $\lim
x_{i}=\xi\in\partial_{\infty}A=\partial_{\infty}C$. Let $\rho=y\xi$ denote the
geodesic ray in ${\mathbb{H}}^{n+1}$ emanating from $y$ and asymptotic to
$\xi$. By convexity of $C$, $\rho$ is contained in $C$. Since $\lim
x_{i}=\xi$, the angles $\angle x_{i}y\xi$ converge to zero as $i\to\infty$,
since $x_{i}\to\xi$. If follows that the points $z_{i}:=\pi_{\rho}(x_{i})$
satisfy

$\lim_{i\to\infty}\frac{d(x_{i},z_{i})}{d(x_{i},y_{i})}=\lim_{i\to\infty}\frac{d(x_{i},z_{i})}{d(x_{i},y)}=0.$

This is a contradiction since $z_{i}\in C$ and $y_{i}$ is the nearest-point
projection of $x_{i}$ to $C$. ∎

######  Lemma 33.

Suppose that $\Gamma^{\prime}\lhd\Gamma<\mathrm{Isom}(\mathbb{H}^{n+1})\cong
PO(n+1,1)$, where $\Gamma^{\prime}$ is a Zariski dense discrete subgroup. Then
$\Gamma$ is also discrete.777The same holds for subgroups of arbitrary
algebraic Lie groups with discrete center.

###### Proof.

Let $\gamma_{i}\in\Gamma$ be a sequence converging to $1\in PO(n+1,1)$. Then
for every $\gamma\in\Gamma^{\prime}$ we have

$\lim_{i\to\infty}\gamma_{i}\gamma\gamma_{i}^{-1}=\gamma.$

Since $g_{i}:=\gamma_{i}\gamma\gamma_{i}^{-1}\in\Gamma^{\prime}$ and the
latter is discrete, it follows that $g_{i}=\gamma$ for all sufficiently large
$i$ (depending on $\gamma$). Since $\Gamma<PO(n+1,1)$ is Zariski dense, it
contains a Zariski dense Schottky subgroup $\Sigma$ of finite rank $r>1$ with
free generators $\sigma_{1},...\sigma_{r}$. Hence there exists $i_{0}$ such
that for all $i\geq i_{0}$ we have that

$\gamma_{i}\sigma_{k}\gamma_{i}^{-1}=\sigma_{k},\quad k=1,...,r.$

By the Zariski density of $\Sigma$, we obtain that $\gamma_{i}$ belongs to the
center of $PO(n+1,1)$ for all $i\geq i_{0}$. Since $PO(n+1,1)$ is centerless,
$\gamma_{i}=1$ for all $i\geq i_{0}$. Thus, $\Gamma$ is discrete. ∎

Recall that
$\Gamma_{1}:=\Gamma_{\mathscr{B}}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ is
the discrete reflection group with the convex fundamental polyhedron
$P\subset{\mathbb{H}}^{n+1}$. As we noted earlier, the limit set of
$\Gamma_{\mathscr{B}}$ is the entire sphere at infinity
$\mathbb{S}^{n}=\partial_{\infty}{\mathbb{H}}^{n+1}$; in particular,
$\Gamma_{\mathscr{B}}$ is Zariski dense in $PO(n+1,1)$. Since $\mathscr{B}$ is
a bug, the polyhedron $P$ has infinitely many faces, equivalently,
${\mathcal{A}}(P)$ is nonempty. Thus, if we put an accumulation point in
${\mathcal{A}}(P)$ at infinity (in the upper half-space model), then the
spheres of $\mathscr{B}$ are cooriented so that they bound round balls in
${\mathbb{R}}^{n}$. In this setting, the density condition simply means that
the union of these round balls is dense in ${\mathbb{R}}^{n}$.

The group $\Gamma_{2}:=\Gamma_{S}$ is a symmetry group of the bug
$\mathscr{B}$, i.e. a group of isometries of ${\mathbb{H}}^{n+1}$ preserving
$P$. Then supergroup of the bug $\mathscr{B}$ is the subgroup
$\widetilde{\Gamma}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ generated by
$\Gamma_{1},\Gamma_{2}$.

Since $\Gamma_{2}$ preserves $P$, it normalizes $\Gamma_{1}$; hence,
$\Gamma_{1}$ is a normal subgroup of $\widetilde{\Gamma}$. In view of Lemma 33
and Zariski density of $\Gamma_{1}$, the subgroup
$\widetilde{\Gamma}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ is discrete. Our
goal is to show that $\widetilde{\Gamma}$ is not just Zariski dense and
discrete, but is also a lattice. We build up to the general case from some
initial assumptions which simplify the exposition.

######  Theorem 34.

If, in addition to the above assumptions, $\Gamma_{2}$ is convex-cocompact and
${\mathcal{A}}(P)=\partial_{\infty}P$ (i.e. ${\mathcal{V}}(P)=\emptyset$),
then the supergroup $\widetilde{\Gamma}$ is a uniform lattice in
$\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$.

###### Proof.

Since ${\mathbb{H}}^{n+1}/\widetilde{\Gamma}$ is homeomorphic (actually,
isometric) to $P/\Gamma_{2}$, it suffices to show that the quotient
$P/\Gamma_{2}$ is compact. Since $\Gamma_{2}$ is convex-cocompact, it acts
cocompactly on the convex hull $C$ of its limit set
$\Lambda=\partial_{\infty}P$. Let $K\subset C$ be a compact fundamental set of
the action of $\Gamma_{2}$ on $C$, i.e. $\Gamma_{2}\cdot K=C$. Then, by Lemma
32 (applied with $A=P$), the preimage $K^{\prime}:=\pi^{-1}_{C}(K)\cap P$ is
also compact. Since $\pi_{C}$ is equivariant with respect to the action of
$\Gamma_{2}$, it follows that $K^{\prime}$ is a compact fundamental set of the
action of $\Gamma_{2}$ on $P$. Hence, $P/\Gamma_{2}$ is compact and
$\widetilde{\Gamma}$ is a uniform lattice. ∎

Next we relax the assumption that the ideal boundary $\partial_{\infty}P$
coincides with the accumulation set ${\mathcal{A}}(P)$ (that is, we allow
${\mathcal{V}}(P)$ in (31) to be nonempty), while keeping convex-cocompactness
of $\Gamma_{2}$. We let $\mu_{n+1}$ denote the Margulis constant of
${\mathbb{H}}^{n+1}$ (see e.g. [Ka] or [Ra06]).

######  Theorem 35.

If $\Gamma_{2}$ is convex-cocompact, then the supergroup $\widetilde{\Gamma}$
is a lattice in $\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$.

###### Proof.

Let $C=C_{\Gamma_{2}}$ again denote the closed convex hull of the limit set of
$\Gamma_{2}$.

Each ideal vertex $\eta_{j}\in{\mathcal{V}}(P),j\in J$, represents a finite
volume cusp of the orbifold ${\mathcal{O}}={\mathbb{H}}^{n+1}/\Gamma_{1}\cong
P$; its stabilizer $\Gamma_{1,j}$ in $\Gamma_{1}$ is generated by reflections
in the faces of $P$ asymptotic to $\eta_{j}$. We will be using certain open
horoballs ${\mathrm{B}}_{\eta_{j}}\subset\mathbb{H}^{n+1}$ centered at the
points $\eta_{j}$. We call such a horoball
${\mathrm{B}}\subset\mathbb{H}^{n+1}$ an $\epsilon$-Margulis horoball, where
$0<\epsilon\leq\mu_{n+1}$, if it is the maximal horoball such that for each
$x\in{\mathrm{B}}$, there exists a parabolic element $g\in\Gamma_{1,j}$
satisfying

$d(x,g(x))<\epsilon.$

The Margulis lemma implies that Margulis horoballs ${\mathrm{B}}_{\eta_{j}}$
are pairwise disjoint. In view of (30), we can choose $\epsilon$ such that
each $\epsilon$-Margulis horoball is disjoint from $C$.888This is not really
necessary but provides a clearer picture of the situation. From now on, we fix
a collection ${\mathrm{B}}_{\eta_{j}},j\in J$, of such $\epsilon$-Margulis
horoballs. This collection is necessarily $\Gamma_{2}$-invariant. Since each
intersection ${\mathrm{B}}_{\eta_{j}}\cap P$ has finite volume and
${\mathbb{H}}^{n+1}/\widetilde{\Gamma}$ has the same volume as $P/\Gamma_{2}$,
in order to prove that $\widetilde{\Gamma}$ is a lattice it suffices to show
that the quotient

$(P\ \setminus\ \bigcup_{j\in J}{\mathrm{B}}_{\eta_{j}})/\Gamma_{2}$

is compact. (This will also force finiteness of the number of
$\Gamma_{2}$-orbits of the horoballs ${\mathrm{B}}_{\eta_{j}}$.) Compactness
of the quotient again follows by applying Lemma 32 to the set

$A:=P\ \setminus\ \bigcup_{j\in J}{\mathrm{B}}_{\eta_{j}}$

as in the proof of Theorem 34. ∎

Lastly, we would like to relax the convex-cocompactness assumption on
$\Gamma_{2}$, replacing it with geometric finiteness. Unfortunately, in
general, the subgroup
$\widetilde{\Gamma}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ will not be a
lattice, already in the case $n=3$. This can happen even if the limit set
${\mathcal{A}}(P)$ is Zariski-dense in $\mathbb{S}^{n}$.

Figure 6. A Kleinian bug which is not cusp-compatible.

######  Example 36.

A concrete example is the bug consisting of circles of unit diameter centered
at the set ${\mathbb{Z}}[i]\cup(\tfrac{1}{2}+\tfrac{1}{2}i+{\mathbb{Z}}[i]),$
that is, the Gaussian integers and their translates by the vector
$\tfrac{1}{2}+\tfrac{1}{2}i$. The group $\Gamma_{1}$ is generated by
reflections in these circles. The group $\Gamma_{2}$ is generated by
reflections in the lines $\\{\Re(z)=0\\}$ and $\\{\Re(z)=1\\}$. The situation
is illustrated in Figure 6. The accumulation point is $\infty$, and the group
$\widetilde{\Gamma}=\left\langle\Gamma_{1},\Gamma_{2}\right\rangle$ is a non-
lattice. On the other hand, if we take $\Gamma^{\prime}_{2}$ generated by the
reflections in the lines $\\{\Re(z)=0\\}$, $\\{\Re(z)=1\\}$, $\\{\Im(z)=0\\}$,
$\\{\Im(z)=1\\}$, then the group
$\widetilde{\Gamma}^{\prime}=\left\langle\Gamma_{1},\Gamma^{\prime}_{2}\right\rangle$
is a lattice.

######  Example 37.

More generally, let $\Gamma_{2}$ be a geometrically finite subgroup of
$\mathop{\hbox{Isom}}({\mathbb{H}}^{3})$ which has at least one rank-one cusp.
Then (by either using the Andreev–Thurston theorem, see [T81, §§13.6–13.7], or
Brooks’s theorem, see [Br86] or [Ka]) there exists a quasiconformal
deformation $\Gamma^{\prime}_{2}$ of $\Gamma_{2}$ and a reflection group
$\Gamma_{1}$ such that the fundamental domain of $\Gamma_{1}$ is invariant
under $\Gamma^{\prime}_{2}$ and its accumulation set equals the limit set of
$\Gamma^{\prime}_{2}$, but the groups $\Gamma_{1}$, $\Gamma^{\prime}_{2}$ are
not cusp-compatible.

A necessary condition for $\widetilde{\Gamma}$ to be a lattice is that the
$\widetilde{\Gamma}$-stabilizer of each parabolic fixed point of $\Gamma_{2}$
is virtually abelian of virtual rank $n$. (Recall that a group $G$ is a
virtually abelian if it contains a free abelian subgroup of finite index. The
virtual rank of $G$ is defined to be the rank of any maximal free abelian
subgroup of $G$. By Bieberbach’s Theorem, a discrete parabolic subgroup of
$\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ is necessarily virtually abelian of
virtual rank $\leq n$.) We thus arrive at the following definition of cusp-
compatibility, for the pair of discrete groups
$\Gamma_{1}=\Gamma_{\mathscr{P}}$, $\Gamma_{2}=\Gamma_{S}$, repeated from
Definition 20.

######  Definition 38.

The subgroups $\Gamma_{1},\Gamma_{2}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$
are cusp-compatible if for every maximal parabolic subgroup
$\Pi_{2}<\Gamma_{2}$, there exists a subgroup $\Pi_{1}<\Gamma_{1}$ such that
the subgroup $\Pi$ generated by $\Pi_{1},\Pi_{2}$ is virtually abelian of
virtual rank $n$.

Observe that the subgroup $\Pi_{1}$ in this definition necessarily fixes the
limit set (the single fixed point at infinity)
$\\{\lambda\\}\subset\mathbb{S}^{n}$ of $\Pi_{2}$. Thus, $\Pi$ fixes $\lambda$
as well. In view of the discreteness of $\widetilde{\Gamma}$ (and, hence, of
$\Pi$), we can (and will) as well assume that $\Pi_{1}$ is the full stabilizer
of $\lambda$ in $\Gamma_{1}$. (The group $\Pi_{1}$ might be finite.) Since
$\Gamma_{1}$ is normal in $\widetilde{\Gamma}$, the subgroup $\Pi_{1}$ is
normal in $\Pi$ and $\Pi$ splits as the semidirect product
$\Pi_{1}\rtimes\Pi_{2}$.

######  Lemma 39.

1\. The group $\Pi_{1}$ is generated by reflections in the facets of $P$
asymptotic to $\lambda$ and, hence, has a fundamental domain $P_{\lambda}$ in
${\mathbb{H}}^{n+1}$ equal to the intersection of all half-spaces defined by
facets of $P$ asymptotic to $\lambda$.

3\. The fundamental domain $P_{\lambda}$ is invariant under the action of
$\Pi_{2}$.

###### Proof.

1\. Let $s_{\alpha}$ denote the isometric reflections in the facets
$F_{\alpha}$ of $P$ generating the reflection group $\Gamma_{1}$. Every
element $\gamma\in\Gamma_{1}$ is represented by a reduced word $w$ in the
generators $s_{\alpha}$. Suppose that $w$ has the form $u\cdot s_{\alpha}\cdot
v$, where $v$ is the product of reflections in the facets of $P_{\lambda}$,
while $s_{\alpha}$ is the reflection in a facet $F_{\alpha}$ not asymptototic
to $\lambda$. Then $s_{\alpha}v(P)$ is contained in the half-space
$H^{-}_{\alpha}$ whose closure in
${\mathbb{H}}^{n+1}\cup\partial_{\infty}{\mathbb{H}}^{n+1}$ does not contain
$\lambda$. It follows that $u\cdot s_{\alpha}\cdot v(P)$ is also contained in
$H^{-}_{\alpha}$ and, hence, $\gamma$ cannot possibly fix $\lambda$.

2\. Since $P$ is preserved by $\Gamma_{2}$, and, hence, by its subgroup
$\Pi_{2}$, and $\Pi_{2}$ fixes $\lambda$, the elements of $\Pi_{2}$ send
facets of $P$ asymptotic to $\lambda$ to facets of $P$ asymptotic to
$\lambda$. ∎

The following lemma provides a list of equivalent algebraic and geometric
characterizations of cusp-compatibility in the context of the pair of groups
$\Gamma_{1}=\Gamma_{\mathscr{P}}$, $\Gamma_{2}=\Gamma_{S}$:

######  Lemma 40.

The following are equivalent for subgroups $\Pi_{1},\Pi_{2}$ of
$\Gamma_{1},\Gamma_{2}$ as above:

  1. (1)

The subgroup $\Pi$ generated by $\Pi_{1},\Pi_{2}$ is virtually abelian of
virtual rank $n$.

  2. (2)

Virtual ranks of $\Pi_{1},\Pi_{2}$ add up to $n$.

  3. (3)

$\Pi$ acts cocompactly on $\partial_{\infty}{\mathbb{H}}^{n+1}\ \setminus\
\\{\lambda\\}$.

  4. (4)

$\Pi_{2}$ acts cocompactly on the intersection
$P_{\lambda}\cap\partial{\mathrm{B}}$, for every horoball
${\mathrm{B}}\subset{\mathbb{H}}^{n+1}$ centered at $\lambda$.

  5. (5)

$\Pi_{2}$ acts cocompactly on the intersection $P\cap\partial{\mathrm{B}}$,
for every horoball ${\mathrm{B}}\subset{\mathbb{H}}^{n+1}$ centered at
$\lambda$.

  6. (6)

$\Pi_{2}$ acts with finite covolume on the intersection
$P_{\lambda}\cap{\mathrm{B}}$, for every horoball
${\mathrm{B}}\subset{\mathbb{H}}^{n+1}$ centered at $\lambda$.

  7. (7)

$\Pi_{2}$ acts with finite covolume on the intersection $P\cap{\mathrm{B}}$,
for some horoball ${\mathrm{B}}\subset{\mathbb{H}}^{n+1}$ centered at
$\lambda$.

###### Proof.

The equivalence (1) $\Leftrightarrow$ (2) follows from the semidirect product
decomposition $\Pi=\Pi_{1}\rtimes\Pi_{2}$. The equivalence (1)
$\Leftrightarrow$ (3) follows from the fact that a discrete subgroup of
${\mathbb{R}}^{n}$ is a uniform lattice if and only if this subgroup has rank
$n$.

To prove the equivalence (3) $\Leftrightarrow$ (4) observe that, by Lemma 39,
$P_{\lambda}\cap\partial{\mathrm{B}}$ is the fundamental domain for the action
of $\Pi_{1}$ on the horosphere $\partial{\mathrm{B}}$ and this intersection is
invariant under $\Pi_{2}$. Therefore, cocompactness of the action of $\Pi$ on
the horosphere is equivalent to the cocompactness of the action of $\Pi_{2}$
on $P_{\lambda}\cap\partial{\mathrm{B}}$.

The proofs of equivalences (3) $\Leftrightarrow$ (5) $\Leftrightarrow$
(6)$\Leftrightarrow$ (7) are similar to the proof of (3) $\Leftrightarrow$ (4)
and are left to the reader. ∎

Subgroups $\Pi_{1},\Pi_{2}$ are called compatible if they satisfy one of the
equivalent conditions in this lemma. Note that since $P\subset P_{\lambda}$,
compactness of $(P_{\lambda}\cap\partial{\mathrm{B}})/\Pi_{2}$, implies that
only finitely many $\Pi_{2}$-orbits of faces of $P$ might intersect
${\mathrm{B}}$. Moreover, the compatibility of $\Pi_{1},\Pi_{2}$ (in the form
of the 3rd condition) implies that any sequence of faces of $P$ converging to
$\lambda$ (possibly outside of a horoball ${\mathrm{B}}$) is contained in
finitely many $\Pi_{2}$-orbits of faces.

######  Lemma 41.

Suppose that ${\mathcal{A}}(P)$ has cardinality $\geq 2$. Then any two
symmetry groups of $\mathscr{B}$ are commensurable and, in particular, the
cusp-compatibility of the subgroups $\Gamma_{1},\Gamma_{2}$ depends only on
$P$.

###### Proof.

Suppose that
$\Gamma_{2},\Gamma^{\prime}_{2}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ are
geometrically finite subgroups preserving $P$ and having ${\mathcal{A}}(P)$ as
their limit sets. Since $|{\mathcal{A}}(P)|\geq 2$, by the definition of
geometric finiteness, the groups $\Gamma_{2},\Gamma^{\prime}_{2}$ act with
finite covolume on the $\epsilon$-neighborhood $Y$ of the convex hull of
${\mathcal{A}}(P)$ (which is their common limit set). Since both
$\Gamma_{2},\Gamma^{\prime}_{2}$ preserve the polyhedron $P$, they generate a
discrete subgroup $\Gamma_{3}<\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$,
which, therefore, acts properly discontinuously, with finite covolume on $Y$.
It follows that $|\Gamma_{3}:\Gamma_{2}|<\infty$,
$|\Gamma_{3}:\Gamma^{\prime}_{2}|<\infty$, which implies commensurability of
$\Gamma_{2},\Gamma^{\prime}_{2}$. Therefore, the pair
$(\Gamma_{1},\Gamma_{2})$ is cusp-compatible if and only if
$(\Gamma_{1},\Gamma^{\prime}_{2})$ is. ∎

######  Remark 42.

Example 36 shows that, when $|{\mathcal{A}}(P)|=1$, cusp-compatibility depends
not only on $P$ but also on $\Gamma_{2}$.

######  Theorem 43.

If $\Gamma_{2}$ is geometrically finite and the pair $\Gamma_{1},\Gamma_{2}$
is cusp-compatible, then the subgroup $\widetilde{\Gamma}$ generated by
$\Gamma_{1},\Gamma_{2}$ is a lattice in
$\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$.

###### Proof.

Similarly to the proof of Theorem 35, we define open horoballs
${\mathrm{B}}_{\eta_{j}},j\in J$. Since each $\eta_{j}$ does not belong to
${\mathcal{A}}(P)$, it follows that all $\eta_{j}$’s lie in the discontinuity
domain of the group $\Gamma_{2}$. This, together with geometric finiteness of
$\Gamma_{2}$ implies that there exists a collection
$\\{\bar{{\mathrm{B}}}_{\lambda_{i}};i\in I\\}$ of closed (Margulis) horoballs
centered at parabolic fixed points of $\Gamma_{2}$, such that:

  1. (1)

The collection of horoballs is $\Gamma_{2}$-invariant.

  2. (2)

The horoballs are pairwise disjoint and disjoint from the horoballs
$\bar{{\mathrm{B}}}_{\eta_{j}}$.

  3. (3)

Each horoball $\bar{{\mathrm{B}}}_{\lambda_{i}}$ intersects only those faces
of $P$ which are asymptotic to $\lambda_{i}$.

Set

$A:=P\ \setminus\ \left(\bigcup_{j\in
J}{\mathrm{B}}_{\eta_{j}}\cup\bigcup_{i\in
I}{\mathrm{B}}_{\lambda_{i}}\right).$

There are two cases to consider. First, suppose that
${\mathcal{A}}(P)=\Lambda(\Gamma_{2})$ is a singleton $\\{\lambda\\}$. Then
$\\{{\mathrm{B}}_{\lambda_{i}}:i\in I\\}$ consists of a single horoball
${\mathrm{B}}$. Consider the nearest-point projection
$\pi_{\bar{{\mathrm{B}}}}:{\mathbb{H}}^{n+1}\to\bar{{\mathrm{B}}}$. For
$x\notin\bar{{\mathrm{B}}}$, this projection is given by the unique
intersection point of the geodesic ray $x\lambda$ with the horosphere
$\partial{\mathrm{B}}$. As in the proof of Theorem 35, the restriction of
$\pi_{\bar{{\mathrm{B}}}}$ to $A$ is a proper map. The image
$\pi_{\bar{{\mathrm{B}}}}(A)$ is contained in $\partial{\mathrm{B}}\cap P$: By
the convexity of $P$, for each $x\in A$ the ray $x\lambda$ is contained in
$P$, hence, the intersection point $x\lambda\cap\partial{\mathrm{B}}$ belongs
to $\partial{\mathrm{B}}\cap P$.

Since the pair $\Gamma_{1},\Gamma_{2}$ is cusp-compatible, by Lemma 40(5), the
action of $\Gamma_{2}$ on $P\cap\partial{\mathrm{B}}$ is cocompact. Combining
this with the fact that the map
$\pi_{\bar{{\mathrm{B}}}}:A\to\bar{{\mathrm{B}}}$ is proper and its image is
contained in $\partial{\mathrm{B}}\cap P$, we conclude that the quotient
$A/\Gamma_{2}$ is compact. Since $(P\cap{\mathrm{B}})/\Gamma_{2}$ has finite
volume and the volume of each cusp $P\cap{\mathrm{B}}_{\eta_{j}}$ is finite,
we conclude that $P/\Gamma_{2}$ has finite volume and, therefore,
$\widetilde{\Gamma}$ is a lattice.

We now consider the “generic” case, when $\Gamma_{2}$ has at least two limit
points. As before, let $C=C_{\Gamma_{2}}$ denote the closed convex hull in
${\mathbb{H}}^{n+1}$ of the limit set $\Lambda=\Lambda(\Gamma_{2})$. Since
$\Lambda={\mathcal{A}}(P)\subset\partial_{\infty}P$, convexity of $P$ implies
that $C\subset P$. We let $C^{\prime}\subset C$ be the complement to the union
of open horoballs ${\mathrm{B}}_{\lambda_{i}},i\in I$. The group $\Gamma_{2}$
acts cocompactly on $C^{\prime}$ since it is geometrically finite. As in the
proof of Theorem 35, the restriction of $\pi_{C}$ to $A$ is a proper map and
$(A\cap\pi_{C}^{-1}(C^{\prime}))/\Gamma_{2}$ is compact. Lastly, Lemma 40 (and
the discussion following it) implies that for every horoball
${\mathrm{B}}_{\lambda_{i}},i\in I$, its $\Gamma_{2}$-stabilizer
$\Pi_{\lambda_{i}}$ acts with finite covolume on

$P\cap\pi_{C}^{-1}({\mathrm{B}}_{\lambda_{i}}).$

Hence, $P/\Gamma_{2}$ has finite volume and, since
$Vol(P/\Gamma_{2})=Vol({\mathbb{H}}^{n+1}/\widetilde{\Gamma})$, it follows
that $\widetilde{\Gamma}$ is a lattice. ∎

The next lemma establishes the cusp-compatibility condition in the the special
case of bugs which are packings.

######  Lemma 44.

If $\mathscr{B}$ is a Kleinian packing then the cusp-compatibility condition
always holds.

###### Proof.

Since $\mathscr{B}$ is a Kleinian packing, the polyhedron $P$ is the
intersection of half-spaces $H^{+}_{\alpha}$, which are bounded by pairwise
disjoint hyperbolic hyperplanes $H_{\alpha}$. In particular,
$\partial_{\infty}P$ has no isolated points and, hence, equals the
accumulation set ${\mathcal{A}}(P)$, which is the limit set $\Lambda$ of
$\Gamma_{S}$, and $P=C_{\Gamma_{S}}$, the closed convex hull of $\Lambda$.
Since $\Gamma_{S}$ is geometrically finite, $C_{\Gamma_{S}}/\Gamma_{S}$ has
finite volume. But $C_{\Gamma_{S}}/\Gamma_{S}=P/\Gamma_{S}$ is naturally
isometric to ${\mathbb{H}}^{n+1}/\widetilde{\Gamma}$. We conclude that
$\widetilde{\Gamma}$ is a lattice, which is equivalent to the cusp-
compatibility condition. ∎

Part 2.

Proof of Part 2(i). Step 1. For each defining hyperplane $H_{\alpha}$ of $D$,
$\alpha\notin A$, we let $H_{\alpha}^{+}\subset{\mathbb{H}}^{n+1}$ denote the
closed half-space bounded by $H_{\alpha}$ and containing $D$. We define
$D_{S}$ as the intersection of these half-spaces,

$D_{S}=\bigcap_{\alpha\notin A}H_{\alpha}^{+}.$

Each facet of $D_{S}$ is contained in one of the facets of $D$ which is not of
the form $F_{\alpha},\alpha\in A$. And conversely, each facet of $D$ not of
the form $F_{\alpha},\alpha\in A$, is contained in one of the facets of
$D_{S}$. The generators $\gamma\in S$ still pair the facets of $D_{S}$. We
leave it to the reader to check that, since $D$ is a fundamental polyhedron of
$\Gamma$, the conditions of the Poincaré’s Fundamental Domain Theorem (see
[Ra06]) still hold for $D_{S}$ and the face-pairing transformations $\gamma\in
S$, and, hence, $D_{S}$ is a fundamental polyhedron of $\Gamma_{S}$. Since, by
the construction, $D_{S}$ is finitely-sided, the group $\Gamma_{S}$ is
geometrically finite with the fundamental domain $D_{S}$.

Step 2. We let ${\mathcal{R}}$ denote the collection of all
$\widetilde{\Gamma}$-images of the hyperplanes $H_{\alpha},\alpha\in A$. Since
$D$ is a fundamental domain of $\widetilde{\Gamma}$, none of the hyperplanes
in ${\mathcal{R}}$ intersects the interior of $D$. The hyperplanes in
${\mathcal{R}}$ define a partition of ${\mathbb{H}}^{n+1}$ into (convex)
connected components; the closure of one of these components, denoted $P$,
contains $D$. We let $\Gamma_{R}<\widetilde{\Gamma}$ denote the subgroup
generated by reflections in the hyperplanes $H\in{\mathcal{R}}$. Thus,
$\Gamma_{R}$ is a nontrivial normal subgroup in the lattice
$\widetilde{\Gamma}$; hence, the limit set of $\Gamma_{R}$ is the entire
sphere $\mathbb{S}^{n}$. The polyhedron $P$ is a fundamental domain of
$\Gamma_{R}$ and $\Gamma_{R}$ is generated by reflections in those hyperplanes
$H\in{\mathcal{R}}$ for which $H\cap P$ are facets of $P$. Since
$\Gamma_{R}<\widetilde{\Gamma}$ and the latter is a lattice, it follows that
the dihedral angles of $P$ come from a finite subset of
$\frac{\pi}{{\mathbb{N}}}$. We conclude that $P$ defines a bug $\mathscr{B}$.

Step 3. We next verify that $\Gamma_{S}$ is a symmetry group of the bug
$\mathscr{B}$. We first check that the generators $\gamma\in S$ of
$\Gamma_{S}$ preserve the polyhedron $P$. It suffices to show that each
$\gamma$ sends facets of $P$ to facets of $P$. The element $\gamma$ pairs a
facet $F_{\gamma}$ of $D$ to a facet $F^{\prime}_{\gamma}$ of $D$. Let
$G_{\beta}$ be a facet of $P$ and $H_{\beta}\in{\mathcal{R}}$ be the
hyperbolic hyperplane containing $G_{\beta}$. We pick a generic base-point $o$
in the interior of $D$ and a generic point $x$ on the facet $G_{\beta}$. Then,
by the convexity of $P$ and since $o\in D\subset P$, the geodesic segment $ox$
is disjoint from all the hyperplanes $H\in{\mathcal{R}}$ except for the point
$x\in G_{\beta}$. Similarly, the geodesic segment $o\gamma(o)$ is contained in
$D\cup\gamma(D)$ and crosses their intersection at an interior point of
$F^{\prime}_{\gamma}$ (since $o$ was chosen generically). It follows that the
segment $o\gamma(o)$ is also disjoint from all the hyperplanes in
${\mathcal{R}}$. Thus, the union

$o\gamma(o)\cup\gamma(o)\gamma(x)$

is a path connecting $o$ to $\gamma(x)$ and disjoint from all the hyperplanes
in ${\mathcal{R}}$ except for the point $\gamma(x)\in\gamma H_{\beta}$. Hence,
$\gamma(x)$ lies in a facet of $P$. Thus, we verified that the generators
$\gamma\in S$ have the property that they send facets of $P$ to facets of $P$
and, moreover, respect their coorientation: The half-space $H^{+}_{\alpha}$
determined by a facet $F_{\alpha}$ and containing $o$, maps to the half-space
$H^{+}_{\alpha^{\prime}}$ containing $o$. Therefore, the entire group
$\Gamma_{S}$ preserves $P$.

Step 4. Since, by the construction, $\widetilde{\Gamma}$ is a lattice
generated by $\Gamma_{R}$ and $\Gamma_{S}$, the group $\Gamma_{S}$ acts on the
set of facets of $P$ with finitely many orbits. Thus, if $\xi$ is an
accumulation point of the bug $\mathscr{B}$, there is a facet $G_{\alpha}$ of
$P$ and an infinite sequence $\gamma_{i}\in\Gamma_{S}$ such that
$\gamma_{i}(G_{\alpha})$ converges to $\xi$. In other words,
${\mathcal{A}}(\mathscr{B})$ is contained in the limit set of $\Gamma_{S}$.
The opposite inclusion follows from the fact that $\Gamma_{S}$ preserves the
polyhedron $P$. We conclude, therefore, that $\Gamma_{S}$ is a symmetry group
of $\mathscr{B}$ and $\mathscr{B}$ is geometric with
$\Gamma_{R}=\Gamma_{\mathscr{B}}$.

Step 5. Lastly, the cusp-compatibility of the pair of groups
$\Gamma_{R},\Gamma_{S}$ follows from the fact that they generate a lattice
$\widetilde{\Gamma}$.

Thus, we proved that $\mathscr{B}$ is a Kleinian bug, which concludes the
proof of Part 2(i) of Theorem 22.

Proof of Part 2(ii). We continue with the notation from the proof of Part
2(i). However, before starting the actual proof of (ii) we will have a
discussion related to combinatorics of convex fundamental polyhedra. Each
facet $F$ of $\partial D$ is paired with another facet $F^{\prime}$ by a
unique generator $\gamma=\gamma_{F,F^{\prime}}\in S^{\prime}$;
$\gamma(F)=F^{\prime}$, where, possibly, $F=F^{\prime}$.

######  Remark 45.

Following Ratcliffe in [Ra06], we require that if a generator $\gamma\in
S^{\prime}$ preserves a facet $F$ of $D$, then it fixes $F$ pointwise, i.e. is
a reflection in $F$. To achieve this, one performs, if necessary, a
subdivision of geometric facets of $D$. We refer to [Ra06] for details.
Accordingly, if $F_{1},F_{2}$ are distinct facets of the same geometric facet
of $\partial D$ and $\gamma_{i}:F_{i}\to F^{\prime}_{i},i=1,2$, pair these
facets, and it happens that $\gamma_{1}=\gamma_{2}$ are equal as elements of
$\widetilde{\Gamma}$, then we still regard $\gamma_{1},\gamma_{2}$ as distinct
elements of $S^{\prime}$.

A ridge of a convex polyhedron in ${\mathbb{H}}^{n+1}$ is the
$n-1$-dimensional intersection of two facets.

The pseudogroup ${\mathcal{G}}$. The pair $(D,S^{\prime})$ defines a
pseudogroup ${\mathcal{G}}$ acting on $\partial D$, which we discuss below.
Each generator $\gamma\in S^{\prime}$ is an element of ${\mathcal{G}}$; it is
a partially defined map between the two facets of $D$ paired as $\gamma:F\to
F^{\prime}$. Then the unique generator sending $F^{\prime}\to F$ is
$\gamma^{-1}$. A composition $\gamma_{2}\circ\gamma_{1}$ of two generators

$\gamma_{1}:F_{1}\to F_{1}^{\prime},\gamma_{2}:F_{2}\to F_{2}^{\prime}$

is admissible if $F_{1}^{\prime}\cap F_{2}$ is a common ridge of these facets.
A (possibly empty) word

$w=\gamma_{l}\circ...\circ\gamma_{1},\gamma_{i}:F_{i}\to
F^{\prime}_{i},i=1,...,l,$

is admissible if each consecutive composition in it is admissible. Thus, each
admissible word defines a map from one ridge $E$ to another ridge
$E^{\prime}$, the domain and the range of $w$ (unless $l=1$ in which case the
domain and the range are facets). Here $E$ is a boundary ridge of $F_{1}$ and
$E^{\prime}$ is a boundary ridge of $E_{l}^{\prime}$. The pseudogroup
${\mathcal{G}}$ then consists of admissible compositions of the generators.
Note that each admissible word is necessarily a reduced word in the alphabet
$S^{\prime}$: $\gamma_{i+1}\neq\gamma^{-1}_{i}$ for each $i$.

The pseudogroup ${\mathcal{G}}$ defines an equivalence relation
$\sim_{{\mathcal{G}}}$ on $D$: $x\sim_{{\mathcal{G}}}y$ ($x$ is
${\mathcal{G}}$-equivalent to $y$) if and only if there exists an element
$\gamma\in{\mathcal{G}}$ sending $x$ to $y$. This equivalence relation is the
one obtained by saturating the non-reflexive and non-transitive relation given
by

$x\sim y,x\in F,y\in F^{\prime},y=\gamma_{F,F^{\prime}}(x).$

An important fact, coming from the assumption that $D$ is a fundamental
polyhedron of $\Gamma$ is that the natural projection of quotient spaces

$D/_{\sim_{\mathcal{G}}}~{}\to{\mathbb{H}}^{n+1}/\widetilde{\Gamma}$

is a homeomorphism; this implies that two points in $D$ are
${\mathcal{G}}$-equivalent if and only if they are
$\widetilde{\Gamma}$-equivalent, i.e. belong to the same
$\widetilde{\Gamma}$-orbit.

Ridge-chains and cycles. Suppose that $E=E_{1}=F_{1}\cap F_{0}^{\prime}$ is a
ridge of $D$. Let $\gamma_{1}\in S^{\prime}$ denote the generator pairing the
facet $F_{1}$ to another facet, $F_{1}^{\prime}$; this yields a new ridge
$E_{2}:=\gamma_{1}(E_{1})=F_{1}^{\prime}\cap F_{2}$. Then let $\gamma_{2}\in
S^{\prime}$ be the generator pairing $F_{2}$ to a facet $F_{2}^{\prime}$. The
composition $\gamma_{2}\circ\gamma_{1}$ is admissible (by the construction).
This composition process continues (uniquely), until we return to the original
ridge so that

$\gamma_{k}:E_{k}\to
E_{1},\quad\hbox{and}\quad\gamma_{k}(F_{k})=F^{\prime}_{k}=F_{0}^{\prime}.$

(The process has to terminate since $D$ has only finitely many faces; in the
case of polyhedra with infinitely many faces, such termination is a
consequence of one of the axioms of fundamental polyhedra.) There is an
important caveat regarding this definition that applies in the special case
when the first return to the initial ridge yields $F^{\prime}_{k}=F_{1}$
instead of $F^{\prime}_{k}=F^{\prime}_{0}$; we discuss this in Remark 46
below.

The finite sequence

$c_{E}=(\gamma_{1},\gamma_{2},...,\gamma_{k})$

is called a ridge-cycle; it corresponds to the word

$w_{c_{E}}=\gamma_{k}\circ...\circ\gamma_{1},$

which is an admissible composition. Its subwords

$w_{c_{E,E^{\prime}}}=\gamma_{l}\circ...\circ\gamma_{1},\quad l\leq k,$

correspond to ridge-chains

$c_{E,E^{\prime}}=(\gamma_{1},\gamma_{2},...,\gamma_{l}),$

where $E^{\prime}=E_{l+1}=\gamma_{l}(E_{l})$. (The notation is slightly
ambiguous since the chain $c_{E,E^{\prime}}$ is not uniquely determined by
$E,E^{\prime}$; the same, of course, applies to the notation $c_{E}$.) The
element $\gamma_{c_{E,E^{\prime}}}\in\Gamma$ corresponding to the composition
$w_{c_{E,E^{\prime}}}$ sends $E_{1}$ to $E_{k+1}$. We will refer to
$w_{c_{E,E^{\prime}}}$ as the word of the ridge-chain $c_{E,E^{\prime}}$.

We let $\theta_{i}$ denote the interior dihedral angles of $D$ along the
ridges $E_{i}$; the sum

$\theta_{c_{E,E^{\prime}}}=\sum_{i=1}^{l}\theta_{i}$

is the total angle of the ridge-chain $c_{E,E^{\prime}}$.

######  Remark 46.

1\. In the sequence of facets given by a ridge-chain, we could have
$F_{i}^{\prime}=F_{i}$; this happens when $\gamma_{i}$ is the reflection in
the facet $F_{i}$. Accordingly, in the case $F_{i}^{\prime}=F_{i}$ we will
have $\gamma_{i}(E_{i})=E_{i+1}=E_{i}$.

2\. In this situation (i.e. $F_{i}^{\prime}=F_{i}$), the word $w_{c_{E}}$
contains a prefix subword which is a “palindrome”

$u=(\gamma_{i-1}\circ...\circ\gamma_{1})^{-1}\circ\gamma_{i}\circ(\gamma_{i-1}\circ...\circ\gamma_{1}).$

The word $u$ represents an element of $\Gamma$ sending $F_{1}$ back to itself
and the ridge $E$ back to itself. In this case, of course, $u$ does not send
$F_{1}$ to $F_{0}^{\prime}$ as required by the definition of a ridge-cycle.
Thus, the actual cycle-word $w$ will be longer than $u$ and will equal $v\circ
u$, where $v$ is another palindromic composition, starting with a face-pairing
$F^{\prime}_{0}\to F_{0}$. As a simple example of this situation, one can take
the case when both $F_{0}^{\prime},F_{1}$ are reflective facets with the
corresponding generating reflections $\gamma_{0},\gamma_{1}$ respectively.
Then $w=\gamma_{0}\circ\gamma_{1}$.

3\. Each ridge $E$ of $D$ defines exactly two ridge-cycles, which differ by
swapping the facets $F_{1},F_{0}^{\prime}$, reversing the order in the
sequence $(\gamma_{i})$ and inverting the generators in the cycle. The total
angle is, of course, independent of which of the two ridge-cycles is used.

Conversely, suppose we are given an admissible composition

$w=\gamma_{l}\circ...\circ\gamma_{1}$

representing $\gamma\in\widetilde{\Gamma}$ sending a ridge
$E=F_{0}^{\prime}\cap F_{1}$ to a ridge $E^{\prime}$. This alone, however,
does not guarantee that corresponding sequence of generators is a chain since
for some $i$, $1\leq i<k$, we may have that
$\gamma_{i}\circ...\circ\gamma_{1}$ sends $E$ to itself and
$\gamma_{i}:F_{i}\to F^{\prime}_{i}=F^{\prime}_{0}$. Taking the minimal $i$
with this property we obtain a ridge-cycle $c_{E}$ and the corresponding word
$w_{c_{E}}$. Hence, we decompose $w$ as

$w=w^{\prime}\circ(w_{c_{E}})^{t},$

where $w^{\prime}$ is a subword in $w_{c_{E}}$ which represents an element
$\gamma^{\prime}$, $\gamma^{\prime}(E)=E^{\prime}$. In particular,
$w^{\prime}$ is the word of a ridge-chain $c_{E,E^{\prime}}$.

The fact that $D$ is a convex fundamental polyhedron of $\widetilde{\Gamma}$
implies that for every ridge $E$, the total angle $\theta_{E}$ is of the form
$\frac{2\pi}{m}$, and $\gamma_{c_{E}}$ has order $m$. Suppose now that the
ridge-cycle $c_{E}$ is such that the consecutive facets $F_{i},F^{\prime}_{i}$
are equal (this happens when $\gamma_{i}$ is a reflection). In this case the
angle $\theta_{i}$ appears twice in the sum defining the total angle
$\theta_{E}$. In particular, if $\theta_{1}=\theta_{k}=\frac{\pi}{2}$ and
$\theta_{s}=\frac{\pi}{2}$ for some $1\leq s\leq k$, and
$\gamma_{s},\gamma_{k}$ are reflections, then both $\theta_{1},\theta_{s}$
contribute twice to the total angle and, thus, there are exactly two
possibilities for the ridge-cycle $c_{E}$:

1\. $k=2,s=1$, $E=E_{1}=E_{2}$, $\theta_{c_{E}}=\pi$, and the ridge-cycle is
as in the simple example in Remark 46(2) above. In particular, the reflections
$\gamma_{1},\gamma_{2}$ commute and $\gamma_{i}(H_{j})=H_{j}$, $j=1,2$ is
taken modulo $2$.

2\. $k=4,s=2$, $E_{1}\neq E_{2}$, $\theta_{c_{E}}=2\pi$, and, up to inversion,

$w_{c_{E}}=\tau_{0}\circ\gamma_{1}^{-1}\circ\tau_{2}\circ\gamma_{1}$

where $\tau_{2}=\gamma_{2}$ is the reflection in the facet $F_{2}$, while
$\tau_{0}=\gamma_{4}$ is the reflection in the facet $F_{0}^{\prime}=F_{4}$.
The composition $\gamma_{c_{E}}$ then has order $1$, i.e. the word $w_{c_{E}}$
represents the neutral element of the group $\widetilde{\Gamma}$. In
particular, in this case,

$\tau_{2}=\gamma_{1}\circ\tau_{0}\circ\gamma_{1}^{-1}.$

In both cases 1 and 2, there are exactly four images of $D$ (one of which is
$D$ itself) under the elements of $\widetilde{\Gamma}$, sharing the ridge $E$,
all with right dihedral angles at $E$. The hyperplanes $H_{0},H_{1}$ bounding
these images and containing the faces $F^{\prime}_{0},F_{1}$ are orthogonal to
each other and divide the hyperbolic space in four quadrants, each containing
one of the above images of $D$. The stabilizer of $E$ in $\widetilde{\Gamma}$
preserves both hyperplanes.

We are now ready for the proof of Part 2(ii). Suppose that the bug
$\mathscr{B}$ is not a packing and two walls in ${\mathcal{R}}$ have nonempty
transversal intersection in the hyperbolic space; the intersection necessarily
has codimension $2$. Since $D$ is a fundamental polyhedron of
$\widetilde{\Gamma}$, this intersection comes from $D$ in the following sense:
There exists a pair of ridges $E,E^{\prime}$ (possibly equal!) formed,
respectively, by pairs of facets $F_{1},F^{\prime}_{0}$ (the ridge $E$) and
$F_{s},F^{\prime}_{s}$ (the ridge $E^{\prime}$). The facets
$F^{\prime}_{0},F^{\prime}_{s}$ are contained in hyperplanes
$H=H_{0},H^{\prime}=H_{s}\in{\mathcal{R}}$ fixed by reflections
$\tau_{0},\tau_{s}$. There exists an element $g\in\widetilde{\Gamma}$ which
carries $E$ to $E^{\prime}$ and sends $H$ to a hyperplane $g(H)$ which meets
$H^{\prime}$ orthogonally along $E^{\prime}$. In particular, $E,E^{\prime}$
are in the same $\widetilde{\Gamma}$-orbit. It follows that there exists a
chain $c_{E,E^{\prime}}$ corresponding to a word $w_{E,E^{\prime}}$
representing an element $\gamma=\gamma_{E,E^{\prime}}$ which sends $E$ to
$E^{\prime}$. The element $\gamma$ need not be equal to $g$, but
$g=\gamma^{\prime}\circ\gamma$, where $\gamma^{\prime}\in\widetilde{\Gamma}$
is an elliptic isometry preserving $E^{\prime}$. Since $H,H^{\prime}$ are both
in ${\mathcal{R}}$, the dihedral angles of $D$ along $E,E^{\prime}$ are both
right angles. Thus, the discussion above regarding the ridge-chain
$w_{E,E^{\prime}}$ applies. In particular, both $H,g(H)$ define facets of $D$.
Recall that by the assumptions of Part 2(ii), the hyperplanes in
${\mathcal{R}}$ defining facets of $D$ are pairwise disjoint, which implies
that $g(H)=H^{\prime}$, a contradiction.

This concludes the proof of Theorem 22. ∎

##  3\. (Sub)Arithmeticity

In this section, we prove the various (sub)arithmeticity theorems. We will be
repeatedly using the Lorentzian model of hyperbolic space $\mathbb{H}^{n+1}$
and the corresponding parameterization of round spheres in
$\mathbb{S}^{n}=\partial_{\infty}\mathbb{H}^{n+1}$ by unit vectors of the
associated quadratic form.

######  Definition 47.

A quadratic form ${\mathcal{F}}$ over a totally real number field $k$ is
called hyperbolic if it has signature $(n+1,1)$ in the identity embedding
$k\to{\mathbb{R}}$, and is definite in all others embeddings. A quadratic
space is an $(n+2)$-dimensional real vector space $V$ together with a real
quadratic form ${\mathcal{F}}$ on $V$ defining a bilinear from
$\left\langle\cdot,\cdot\right\rangle$ in the usual way. A quadratic space is
said to be hyperbolic if ${\mathcal{F}}$ has signature $(n+1,1)$.

We let $Q$ denote the standard hyperbolic quadratic form with half-Hessian:

(48) $Q\ =\ \begin{pmatrix}&&-\tfrac{1}{2}\\\ &I&\\\
-\tfrac{1}{2}&&\end{pmatrix}.$

###  3.1. Inversive coordinates

Before embarking on the proofs, we recall the very convenient (in this
context) inversive coordinate system (see, e.g., [K17, LMW02, Wil82]). For
convenience, we work here with the standard hyperbolic quadratic form $Q$, but
the entire discussion applies to all quadratic forms after an appropriate
change of coordinates.

To a cooriented round sphere $S$ in the boundary
$\mathbb{S}^{n}=\partial_{\infty}\mathbb{H}^{n+1}={\mathbb{R}}^{n}\cup\\{\infty\\}$
having center $z=(x_{1},\dots,x_{n})$ and signed radius $r$, we associate the
column vector

$v_{S}\ :=\ (\tfrac{1}{r},\tfrac{z}{r},\tfrac{1}{\hat{r}})^{t}\ \in\
{\mathbb{R}}^{n+2}.$

Here $\hat{r}$ is the co-radius, defined to be the signed radius of the image
of $S$ under reflection through the unit sphere; more concretely,

(49) $\hat{r}={r\over|z|^{2}-r^{2}}.$

When $S$ is a hyperplane, the inversive coordinates are those obtained from a
limit of spheres. That is, $1/r=0$, $1/\hat{r}$ is half the distance from the
hyperplane to the origin, and $z/r$ is the unit normal to the cooriented
hyperplane.

Rewriting (49) as

$|\tfrac{z}{r}|^{2}-\tfrac{1}{r}\tfrac{1}{\hat{r}}=1,$

we see that $Q(v_{S})=1$.

The Möbius group $Mob_{n}$ (the group of Möbius transformations of
$\mathbb{S}^{n}$) acts on the space $Sph(n)$ of (cooriented) round spheres and
the group $O_{Q}$ of automorphisms of the form $Q$ acts on the 1-sheeted
hyperboloid $\\{Q(v)=1\\}$.

######  Lemma 50.

The map $j:S\mapsto v_{S}$ is equivariant with respect to the actions of
$Mob_{n}$ and $O_{Q}$. More precisely, $j$ conjugates the Möbius action on the
space of cooriented spheres $S\in Sph(n)$ to the Lorentzian action on the
inversive coordinate vectors $v_{S}\in\\{Q_{S}=1\\}$.

###### Proof.

The proof is essentially contained in the proof of [Iv92, Theorem 7.5].
Iversen constructs an equivariant map $\iota$ of $\mathbb{S}^{n}$ to the
projectivization of the conic $\\{Q=0\\}$. In the proof he verifies that

$\iota R_{S}\iota^{-1}=\tau_{j(S)},$

where $R_{S}$ is inversion in $S\in Sph(n)$ and $\tau_{v}$ is the Lorentzian
reflection in $v^{\perp}$, the Lorentzian orthogonal complement to the vector
$v$. To prove equivariance of $j$ it remains to observe that for every $g\in
Mob_{n}$, $S\in Sph(n)$ and $v_{S}=j(S)$,

$\iota\sigma_{gS}\iota^{-1}=\iota gR_{S}g^{-1}\iota^{-1}=g\iota
R_{S}\iota^{-1}g^{-1}=g\tau_{v_{S}}g^{-1}=\tau_{gv_{S}}.$

Hence, $j$ sends the sphere $gS$ to $gv_{S}$. ∎

######  Remark 51.

Iversen uses the description of spheres $S$ in ${\mathbb{R}}^{n}$ by the
quadratic equations

$b\left\langle x,x\right\rangle-2\left\langle x,f\right\rangle+a=0,$

Assuming the normalization $|f|^{2}-ab=1$, the inversive coordinates then
become (for $r<\infty$)

$(b,f,a).$

The natural invariant, the negative of the cosine of the angle between the
spheres, is given by

$\left\langle f_{1},f_{2}\right\rangle-\frac{1}{2}(a_{1}b_{2}+b_{1}a_{2}),$

see [Iv92, sect. I.8], i.e. the Lorentzian inner product for vectors in
$\\{Q=-1\\}$. In the case when $S$ is a cooriented hyperplane, $b=0$, $f$ is
the unit vector normal to $S$ and $a=\hat{r}$ is the coradius. This gives an
alternative proof of the lemma.

Fixing one sheet of the two-sheeted hyperboloid $\\{Q=-1\\}$ as a model of
$\mathbb{H}^{n+1}$, the original sphere $S$ corresponds to the boundary at
infinity of the intersection with $\\{Q=-1\\}$ of the plane $Q$-orthogonal to
$v_{S}$. Then under the isomorphism $j_{*}$ induced by $j$,
$\mathrm{Isom}(\mathbb{H}^{n+1})\cong O^{+}_{Q}({\mathbb{R}})$, where
$O_{Q}^{+}$ is the “orthochronous” subgroup of $O_{Q}$ which preserves the two
sheets of $Q=-1$ (rather than allowing them to interchange).

The pair $({\mathbb{R}}^{n+2},Q)$ defines a hyperbolic quadratic space. Write
$({\mathbb{R}}^{n+2})^{*}$ for the dual vector space, and $Q^{*}$ for the
induced dual form. The key observation elucidating the role of isotropic
vectors in the study of the arithmetic of sphere packings is the following.

######  Lemma 52.

The “bend” covector $\mathbf{b}=(0,\dots,0,-2)\in({\mathbb{R}}^{n+2})^{*}$ is
isotropic,

(53) $Q^{*}(\mathbf{b})=0$

and captures the bend of a sphere $S$ with inversive coordinates
$v_{S}=(\frac{1}{r},\frac{z}{r},\frac{1}{\hat{r}})^{t}$. That is,

(54) $\mathbf{b}(v_{S})=\frac{1}{r}.$

Similarly, the co-bend covector $\widehat{\mathbf{b}}=(-2,0,\dots,0)$ is also
isotropic, and has $\widehat{\mathbf{b}}(v_{S})=\frac{1}{\hat{r}}$. In the
dual inner product $\left\langle\cdot,\cdot\right\rangle_{*}$ defined by
$Q^{*}$, we have

(55) $\left\langle\mathbf{b},\widehat{\mathbf{b}}\right\rangle_{*}=-2.$

###### Proof.

Direct and elementary computation.∎

We can also identify the hyperbolic space ${\mathbb{H}}^{n+1}$ with a
component of the two-sheeted hyperboloid $\\{{\mathcal{F}}=-1\\}$, where
$(V,{\mathcal{F}})$ is another real hyperbolic quadratic space of the same
dimension $n+2$. A convenient way to choose coordinates here is as follows.
Let $\left\langle\cdot,\cdot\right\rangle_{\mathcal{F}}$ denote the bilinear
form on $V$ corresponding to ${\mathcal{F}}$, and let ${\mathcal{F}}^{*}$ be
the dual form on the dual space $V^{*}$.

######  Lemma 56.

The dual space $V^{*}$ admits an orthogonal splitting

(57) $V^{*}=V_{1}^{*}\oplus V_{2}^{*}$

so that $\dim V_{1}^{*}=2$ with

(58)
$\bigg{.}{\mathcal{F}}^{*}\bigg{|}_{V_{1}^{*}}=\left(\begin{array}[]{cc}0&-2\\\
-2&0\end{array}\right),$

and ${\mathcal{F}}^{*}$ restricts to a definite form on the second factor.

###### Proof.

Take two linearly independent light-like covectors $\alpha_{0},\alpha_{n+1}\in
V^{*}$ (i.e.
${\mathcal{F}}^{*}(\alpha_{0})={\mathcal{F}}^{*}(\alpha_{n+1})=0$), and
rescale $\alpha_{n+1}$ to ensure that
$\left\langle\alpha_{0},\alpha_{n+1}\right\rangle_{{\mathcal{F}}^{*}}=-2$, as
in (55). Let $V_{1}^{*}$ be the span of these vectors. Defining $V_{2}^{*}$ to
be the orthogonal complement spanned by an arbitrary orthonormal system
$\alpha_{1},\dots,\alpha_{n}$ gives the required splitting. ∎

We next transition to the number-theoretic discussion.

######  Addendum 59.

Suppose that ${\mathcal{F}}$ is a hyperbolic quadratic form on an
$(n+2)$-dimensional real vector space $V$ and there is a basis of $V$ with
respect to which the form is a hyperbolic over a (totally real) field $k$.
Then:

1\. There is field extension $k^{\prime}\supset k$ of degree at most two with
ring of integers $\mathfrak{o}^{\prime}$, and a choice of covectors
$\alpha_{0},\dots,\alpha_{n+1}$ so that the splitting (57) is realized with
$V_{1}^{*}$ spanned by $\alpha_{0},\alpha_{n+1}$ and
$V_{2}^{*}=\left\langle\alpha_{1},\dots,\alpha_{n}\right\rangle$, where
$\alpha_{0}$ can be chosen to be defined over $\mathfrak{o}^{\prime}$.

2\. If $k={\mathbb{Q}}$ and ${\mathcal{F}}$ is isotropic over ${\mathbb{Q}}$,
then we can take $k^{\prime}=k={\mathbb{Q}}$ and
$\mathfrak{o}^{\prime}={\mathbb{Z}}$. Moreover, $V$ contains a rational unit
vector $u$ (of ${\mathcal{F}}$).

###### Proof.

1\. For a general field $k$, we can find an isotropic covector $\alpha_{0}$
over a suitable quadratic extension $k^{\prime}$.999As an aside, recall
Godement’s compactness criterion, that $O_{\mathcal{F}}(\mathfrak{o})$ is non-
uniform if and only if ${\mathcal{F}}$ is isotropic over $k$, see [BHC62]; and
furthermore, if this is the case, then $k={\mathbb{Q}}$. Indeed, if
${\mathcal{F}}$ represents $0$ nontrivially, then so does every Galois
conjugate ${\mathcal{F}}^{\sigma}$. But ${\mathcal{F}}^{\sigma}$ is definite
since ${\mathcal{F}}$ is hyperbolic, so there can be no other Galois
conjugates, and $k={\mathbb{Q}}$. We may clear denominators to ensure that
$\alpha_{0}$ is defined over $\mathfrak{o}^{\prime}$. Then we proceed as in
the proof of Lemma 56, finding a second isotropic vector $\alpha_{n+1}$ (again
over $k^{\prime}$) and rescaling to ensure (58). Lastly, choose an orthonormal
basis $\alpha_{1},\dots,\alpha_{n}$ for the orthogonal complement $V_{2}^{*}$
of $V_{1}^{*}$.

2\. Since ${\mathcal{F}}$ is isotropic over ${\mathbb{Q}}$, so is the dual
form ${\mathcal{F}}^{*}$. We then let $\alpha_{0}$, $\alpha_{n+1}$ denote
linearly independent rational isotropic vectors of ${\mathcal{F}}^{*}$. As
before, by rescaling we may assume that $\alpha_{0}$ is an integer vector and
(58) is satisfied. In order to find a rational unit vector, we similarly find
two rational isotropic vectors $v,w\in V$ such that $\left\langle
v,w\right\rangle_{\mathcal{F}}=1/2$. Then the vector $u:=v+w$ has
$\left\langle u,u\right\rangle_{\mathcal{F}}=1$, as desired. ∎

###  3.2. Constructing (Super)Integral Kleinian Bugs/Packings

###### Proof of Theorem 12.

We first prove part $(ii)$. Let $\Gamma$ be a non-uniform
${\mathbb{Q}}$-arithmetic group of simplest type, commensurable to
$O_{\mathcal{F}}({\mathbb{Z}})$, where ${\mathcal{F}}$ is a hyperbolic
quadratic form defined over ${\mathbb{Q}}$ (with respect to some basis for
$V$). Let $R_{S}\in\Gamma$ be a reflection with respect to a rational vector
$u=w_{S}$:

$R_{S}:x\mapsto I-2{\left\langle x,w_{S}\right\rangle\over\left\langle
w_{S},w_{S}\right\rangle}w_{S}$

Hence, $a=\left\langle w_{S},w_{S}\right\rangle$ is rational. Dividing
${\mathcal{F}}$ by $a$ we obtain a new rational hyperbolic form
$a^{-1}{\mathcal{F}}$, with respect to which $w_{S}$ is a unit vector. Then we
choose coordinates $\alpha_{0},\dots,\alpha_{n+1}$ for $V^{*}$ as in Addendum
59, adapted to the form $a^{-1}{\mathcal{F}}$.

Since $\Gamma$ is commensurable to $O_{\mathcal{F}}({\mathbb{Z}})$, all the
vectors in the $\Gamma$-orbit of the vector $u$ have uniformly bounded
denominators, hence, their bends $\alpha_{0}(\Gamma u)$ also have uniformly
bounded denominators. Under the correspondence, $\gamma(S)\mapsto\gamma(u)$
defined in §3.1 (with respect to the form $a^{-1}{\mathcal{F}}$), the radius
of the sphere $\gamma(S)$ equals $\alpha_{0}(u)$. Hence, by a suitable
rescaling in the Euclidean space ${\mathbb{R}}^{n}$ (equivalently, rescaling
the choice of $\alpha_{0}$ to clear the denominators), all the bends can be
made integral, as claimed.

The proof of part $(i)$ is now clear. By rescaling the form ${\mathcal{F}}$,
we get a new form rational hyperbolic form $a^{-1}{\mathcal{F}}$ such that the
reflection $R_{S}\in\Gamma$ is given by a unit (with respect to
$a^{-1}{\mathcal{F}}$) vector $w_{S}$. The group $\widetilde{\Gamma}$, of
course, preserves the new form. Since, by the assumption, the action of the
symmetry group $\Gamma_{S}<\widetilde{\Gamma}$ on the packing $\mathscr{P}$ is
transitive, so is the action of $\widetilde{\Gamma}$ on the superpacking
$\widetilde{\mathscr{P}}$. Thus, by Part (ii), after a suitable rescaling, the
superpacking $\widetilde{\mathscr{P}}$ becomes integral. ∎

We now prepare for the proof of Theorem 15. Before getting to questions of
(super)integrality, we construct the necessary packings as follows.

######  Proposition 60.

Let $O_{{\mathcal{F}}}(\mathfrak{o})<\mathrm{Isom}(\mathbb{H}^{n+1})$ be a
$k$-arithmetic lattice of simplest type. Then there is a sequence of
conformally inequivalent Kleinian packings $\mathscr{P}_{j}$ such that
$O_{{\mathcal{F}}}(\mathfrak{o})$ is commensurable to a supergroup
$\widetilde{\Gamma}_{j}$ of $\mathscr{P}_{j}$ which acts transitively on
$\mathscr{P}_{j}$.

###### Proof.

While this result is stated in arithmetic terms, most of the proof is non-
arithmetic.

Geometric setup. Suppose that $\Gamma<G=\mathrm{Isom}({\mathbb{H}}^{n+1})$ is
a torsion-free lattice such that the hyperbolic manifold
$M={\mathbb{H}}^{n+1}/\Gamma$ contains a properly embedded complete connected
nonseparating totally geodesic hypersurface of finite $n$-dimensional volume
$N$. The hypersurface $N$ lifts to a hyperplane $H$ in ${\mathbb{H}}^{n+1}$.
We will assume that the reflection $R_{H}$ in $H$ normalizes the group
$\Gamma$ and let $\widetilde{\Gamma}$ denote the subgroup of $G$ generated by
$R_{H}$ and $\Gamma$. The group $\widetilde{\Gamma}$ contains $\Gamma$ as an
index 2 subgroup; hence, $\widetilde{\Gamma}$ is again a lattice. Since
$R_{H}$ normalizes $\Gamma$, it descends to an isometric reflection
$\sigma:M\to M$ fixing $N$ pointwise.

######  Remark 61.

Abundance of examples of this type (comping from arithmetic groups) was first
established by John Millson in [Mi76]. More precisely, he proved that every
$k$-arithmetic lattice $O_{{\mathcal{F}}}(\mathfrak{o})$ in $O({n+1,1})$ of
simplest type (where $\mathfrak{o}$ is the ring of integers of the field $k$
and ${\mathcal{F}}$ is a hyperbolic quadratic form over $k$) is commensurable
to a group $\widetilde{\Gamma}$ as above.

Our goal is define a sequence of Kleinian packings $\mathscr{P}_{j}$ with the
supergroups $\widetilde{\Gamma}_{j}$ commensurable to $\widetilde{\Gamma}$ and
acting transitively on $\mathscr{P}_{j}$.

The hypersurface $N$ does not separate $M$ and, hence, defines a nontrivial
element $\xi$ of $H^{1}(M)$, which is Poincaré-dual to the locally-finite
fundamental class of the hypersurface $N$. Since $\sigma$ fixes $N$ pointwise,
$\sigma^{*}(\xi)=-\xi$. The class $\xi$ defines a homomorphism

$\phi:\Gamma\to H_{1}(M)\to{\mathbb{Z}}.$

Let $p:\hat{M}\to M$ denote the infinite cyclic covering corresponding to the
kernel $Ker(\phi)$ of $\phi$. Since $\sigma^{*}(\xi)=-\xi$, $Ker(\phi)$ is
$R_{H}$-invariant; hence $\sigma$ lifts to a reflection
$\tau:\hat{M}\to\hat{M}$ fixing pointwise one of the components $N_{0}$ of the
preimage of $N$ in $\hat{M}$. We let $D$ denote a component of $p^{-1}(M-N)$
whose boundary contains $N_{0}$. Then $\bar{D}$ is a fundamental domain of the
action of the deck-transformation group
${\mathbb{Z}}=\left\langle\theta\right\rangle$ of the regular covering
$p:\hat{M}\to M$. Furthermore, for each $j\in{\mathbb{N}}$,

$D_{j}:=\bigcup_{-j\leq i\leq j}\theta^{i}(\bar{D})=\bigcup_{0\leq i\leq
j}\theta^{i}(\bar{D})\cup\tau(\bigcup_{0\leq i\leq j}\theta^{i}(\bar{D}))$

is a fundamental domain for the index $2j$ subgroup in ${\mathbb{Z}}$. Each
domain $D_{j}$ is $\tau$-invariant, has finite volume ($2j$ times the volume
of $M$) and two boundary component, both totally geodesic in $\hat{M}$ and
isometric to $N$ via the restriction of the covering map $p$.

######  Remark 62.

For future reference, we record the following obvious properties of the
domains $D_{j}$:

1\. $cl(D_{j})\subset D_{j+1}$ for each $j$.

2\. $\bigcup_{j\geq 1}D_{j}=\hat{M}$.

Figure 7. Infinite cyclic covering101010Many thanks to Jules Flin for this
drawing.

The fundamental group $\Gamma_{j}$ of $D_{j}$ embeds in $\pi_{1}(\hat{M})$
(since $D_{j}$ has totally-geodesic boundary) and, hence, $\pi_{1}(M)$. Since
$\pi_{1}(M)$ is isomorphic to the lattice $\Gamma$, the group $\Gamma_{j}$
will be identified with a subgroup (again denoted $\Gamma_{j}$) of $\Gamma$.
Then the preimage of $D_{j}$ in ${\mathbb{H}}^{n+1}$ under the covering map
$q:{\mathbb{H}}^{n+1}\to\hat{M}$ contains a (unique) $\Gamma_{j}$-invariant
component $P_{j}=\tilde{D}_{j}$. Since $\partial N$ is totally-geodesic,
$P_{j}$ is a convex domain with totally-geodesic boundary. By the
construction, each boundary component of $P_{j}$ is fixed by a reflection in
$\widetilde{\Gamma}$ conjugate to $R_{H}$. Thus, the collection of hyperplanes
bounding $P_{j}$ defines a sphere packing $\mathscr{P}_{j}$ with the
reflection group $\Gamma_{R,j}$, generated by reflections in the walls
bounding $P_{j}$. The group $\Gamma_{j}$ acts on $P_{j}$ with quotient of
finite volume (isometric to $D_{j}$); in particular, the limit set
$\Lambda_{j}$ of $\Gamma_{j}$ equals $\partial_{\infty}P_{j}$ and $P_{j}$ is
the closed convex hull of this limit set, $P_{j}=C_{\Gamma_{j}}$. We conclude
that $\Gamma_{j}$ is geometrically finite. Thus, $\mathscr{P}_{j}$ is a
Kleinian packing.

The supergroup of the packing generated by $\Gamma_{R,j}$ and $\Gamma_{j}$,
however, acts on $\mathscr{P}_{j}$ with two orbits: These two orbits
correspond to the two connected components of $\partial D_{j}$. To fix this
problem, we note that the reflection $\tau$ discussed above swaps these
boundary components. Hence, we lift $\tau$ to a reflection $\tilde{\tau}$ in
${\mathbb{H}}^{n+1}$ preserving $P_{j}$ and let $\widetilde{\Gamma}_{j}$
denote the subgroup of $\mathop{\hbox{Isom}}({\mathbb{H}}^{n+1})$ generated by
$\Gamma_{j}$ and $\tilde{\tau}$. The group $\widetilde{\Gamma}_{j}$, which is
an index 2 extension of $\Gamma_{j}$, is still a symmetry group of the packing
$\mathscr{P}_{j}$ and acts transitively on the packing.

We are almost done with the proof of the proposition. It remains to show that
the packings $\mathscr{P}_{j}$ are conformally inequivalent for different
$j$’s; equivalently, we claim that the convex subsets
$P_{j}\subset{\mathbb{H}}^{n+1}$ are pairwise non-isometric.

To distinguish the sets $P_{j}$, we define the following invariant:

$\rho_{j}=\sup_{\bar{x}\in D_{j}}d(\bar{x},\partial D_{j})=\sup_{{x}\in
P_{j}}d({x},\partial P_{j}).$

Here $d({x},\partial P_{j})$ denotes the minimal distance from $x$ to the
points of $\partial P_{j}$, similar for $d(\bar{x},\partial D_{j})$. Since
each $P_{j}$ has finite volume, $\rho_{j}$ is finite for each $j$. Then Remark
62(1) implies that the sequence $\rho_{j}$ is strictly increasing with $j$
and, hence $P_{j}$’s are pairwise non-isometric. ∎

######  Remark 63.

1\. In §4 we will give a different argument for conformal inequivalence of
packings (possibly after passing to a subsequence) using Hausdorff dimensions
of the limit sets of the groups $\Gamma_{j}$.

2\. Two Kleinian packings are conformally equivalent if and only if they are
quasiconformally (more precisely, quasisymmetrically) equivalent, provided
that $n\geq 2$, see [Fr06, BKM09]. Hence, we obtain infinitely many
quasiconformally inequivalent packings in every dimension $n\geq 2$.

###### Proof of Theorem 15.

We are given $O_{\mathcal{F}}(\mathfrak{o})$, a $k$-arithmetic hyperbolic
lattice of simplest type, with $\mathfrak{o}$ the ring of integers of $k$. In
Proposition 60, we constructed a Kleinian packing $\mathscr{P}=\Gamma_{S}\cdot
S_{0}$ with a symmetry group $\Gamma_{S}$ which acts transitively on the
spheres in the packing, and an arithmetic super-symmetry group
$\widetilde{\Gamma}$ (commensurable to $O_{\mathcal{F}}(\mathfrak{o})$). The
group $\widetilde{\Gamma}$ contains a reflection $R$ through the sphere
$S_{0}$. As in the proof of Theorem 12, the normal vector $w\in V$ to $S_{0}$
has all coordinates in $k$. By Addendum 59, after a conformal change of
coordinates on ${\mathbb{R}}^{n}$, the “bend” covector $\alpha_{0}$ is defined
over $\mathfrak{o}^{\prime}$, the ring of integers of a quadratic extension
$k^{\prime}$ of $k$. The supergroup $\widetilde{\Gamma}$ of the packing is
commensurable to $O_{\mathcal{F}}(\mathfrak{o})$, and so the orbit of $w$
under the supergroup is defined entirely over $k$; it is only when we measure
the bends using the covector $\alpha_{0}$ (after a suitable rescaling) that we
obtain elements of $\mathfrak{o}^{\prime}$. Regardless, the superpacking has
all bends in $\mathfrak{o}^{\prime}$, as claimed. ∎

###### Proof of the Classification Theorem 9.

The same proof gives the forward direction of the Classification Theorem 9;
indeed, if ${\mathcal{F}}$ is defined over ${\mathbb{Q}}$ and isotropic, then
by the same argument as above, the packing constructed in Proposition 60 is
superintegral by Part $(i)$ of Theorem 12. The backwards direction is a direct
consequence of the Subarithmeticity Theorem 3, which we turn to now. We will
give two proofs, one that is basically identical to the proof of [KN19, Thm
19], and another using somewhat different ideas. ∎

###### First Proof of Theorem 3.

Let $Q$ be the standard quadratic form of the signature $(n+1,1)$ as in
equation (48), $\Gamma<G=O_{Q}^{+}({\mathbb{R}})$ be a discrete, Zariski dense
subgroup acting on the inversive coordinates $v_{S_{0}}$ of a sphere $S_{0}$,
so that the bends, that is, first entries, in the orbit $\mathcal{O}=\Gamma
v_{S_{0}}$ are all integers. The action of $\Gamma$ is on the left (on column
vectors) and involves all the entries of $v_{S_{0}}$; we conjugate it to a
right action just on (row vectors of) bends, as follows.

By the Zariski density of $\Gamma$, the orbit $\mathcal{O}$ is also Zariski
dense in the one-sheeted hyperboloid $Q=1$, and hence contains $n+2$ linearly
independent vectors
$\\{v_{1}=v_{S_{0}},v_{2},\dots,v_{n+2}\\}\subset\mathcal{O}$. These vectors
provide a coordinate system for the dual vector space
$({\mathbb{R}}^{n+2})^{*}$ so that if one applies these coordinates to the
covector $\alpha_{0}$ (that is, the first coordinate in ${\mathbb{R}}^{n+2}$),
one gets integers. Moreover, for each $\gamma\in\Gamma$, the pairing of
vectors and covectors $\left\langle\gamma
v_{i},\alpha_{0}\right\rangle=\left\langle
v_{i},{}^{*}\gamma^{-1}\alpha_{0}\right\rangle$ is also an integer. Hence, the
vectors $v_{i}$ evaluate to integers on the $\Gamma$-orbit of the covector
$\alpha_{0}$ in $({\mathbb{R}}^{n+2})^{*}$. Hence $\Gamma$ preserves the
finite index sublattice $L$ generated by the $\Gamma$-orbit of $\alpha_{0}$ in
$({\mathbb{Z}}^{n+2})^{*}$; here $({\mathbb{Z}}^{n+2})^{*}$ consists of
covectors with integer coordinates with respect to the coordinate system given
by $v_{1},...,v_{n+2}$. Thus $\Gamma<G^{L}$ is a subgroup of $G^{L}=\\{g\in
O_{Q}^{+}({\mathbb{R}}):gL=L\\}$. The latter group is easily seen to be (and
is sometimes taken to be the very definition of) a congruence subgroup. Note
also that the co-vector $\alpha_{0}$ is rational with respect to the integral
structure given by $L$ and isotropic, see Lemma 52. It follows that $G^{L}$,
regarded as a lattice acting on the dual vector space $V^{*}$, is non-uniform,
i.e. contains a unipotent element. Then $G^{L}$ itself, regarded as a subgroup
of $G$, contains a unipotent element, and, hence, is non-uniform. To conclude:
$\Gamma$ is a subgroup of a non-uniform ${\mathbb{Q}}$-arithmetic hyperbolic
group of simplest type, as claimed. ∎

###### Second Proof of Theorem 3.

Let $\Gamma<O_{Q}^{+}({\mathbb{R}})$ be a discrete, Zariski dense subgroup
acting on the inversive coordinates $v_{S_{0}}$ of a sphere $S_{0}$, so that
the bends, that is, first entries, in the orbit $\mathcal{O}=\Gamma v_{S_{0}}$
are all integers. The first coordinate on ${\mathbb{R}}^{n+2}$ is a (nonzero)
linear functional $\alpha$ on the real vector space $V={\mathbb{R}}^{n+2}$.
The key is the following general lemma:

######  Lemma 64.

Let $V$ be a finite-dimensional real vector space, $\Gamma<GL(V)$ an
irreducible subgroup, i.e. a subgroup which has no proper invariant subspaces.
Let $v\in V,\alpha\in V^{*}$ be nonzero vectors with the property that
$\alpha(gv)\in{\mathbb{Z}}$ for all $g\in\Gamma$. Define the
${\mathbb{Z}}$-submodule $L$ in $V$ generated by the orbit $\Gamma\cdot v$.
Then $L$ is a free ${\mathbb{Z}}$-module of rank equal to the dimension of
$V$.

###### Proof.

The group $\Gamma$ obviously acts on $L$ by automorphisms and $\alpha$ still
takes only integer values on $L$. The irreducibility of $\Gamma$ implies that
$L$ spans $V$ as a real vector space. It remains to prove that $L$ is a
discrete subgroup of $V$ regarded as an abelian Lie group. Let $\bar{L}$
denote the closure of $L$ in the classical topology on $V$ and let
$W:=\bar{L}_{0}$ be the identity component of this Lie subgroup of $V$; this
component is a (real) linear subspace in $V$. The group $\Gamma$ preserves
this subspace. In view of irreducibility of $\Gamma$, the subspace $W$ is
either $\\{0\\}$ or the entire $V$. However, $\alpha\neq 0$ still takes only
integer values on $W$, hence, $W\neq V$ and we conclude that $W=\\{0\\}$, i.e.
$L$ is a discrete subgroup of $V$. ∎

We apply this lemma in our setting. The submodule $L\subset
V={\mathbb{R}}^{n+2}$ defines an integral structure on $V$. We claim that the
quadratic form $Q$ is rational with respect to this integral structure. Since
$G_{\mathbb{Z}}$, the set of integer points in $G=O_{Q}$, contains $\Gamma$,
it is Zariski dense in $G$, hence, is an (arithmetic) lattice. It follows from
Exercise 4 in [WM15, Sect. 5A] that $G$ is defined over ${\mathbb{Q}}$. We
claim that the form $Q$ is also defined over ${\mathbb{Q}}$. The proof is the
same as the one of Exercise 4 in [WM15, Sect. 5A]: Consider the vector space
$U$ of all quadratic forms on $V$. In view of Zariski density of $\Gamma$ in
$G$, there is a unique $\Gamma$-invariant line in $U\otimes{\mathbb{C}}$.
Since $\Gamma$ consists of integer matrices (with respect to the integer
structure on $V$ defined by $L$), for every Galois automorphism $\sigma$ of
${\mathbb{C}}$, for every $q\in U$, we have

$\gamma^{*}(q)^{\sigma}=\gamma^{*}(q^{\sigma}),\gamma\in\Gamma.$

Therefore, since $Q$ is $\Gamma$-invariant, so are the forms
$Q^{\sigma},\sigma\in Gal({\mathbb{C}})$. Thus, the forms $Q^{\sigma}$ belong
to the line ${\mathbb{C}}Q$, i.e. for every $\sigma\in Gal({\mathbb{Q}})$
there exists $z\in{\mathbb{C}}^{\times}$ such that

$Q^{\sigma}=zQ.$

Lastly, $Q(v)=1$ and $v\in L$, hence,

$z=zQ(v)=Q^{\sigma}(v)=Q^{\sigma}(v^{\sigma})=(Q(v))^{\sigma}=1.$

It follows that $Q$ itself is a rational form with respect to the rational
structure on $V$ defined by the lattice $L$.

Lastly, the argument that $\Gamma$ is contained in a nonuniform lattice is the
same as in the first proof: The covector $\alpha_{0}$ is isotropic and
rational with respect to the integral structure defined by the lattice $L$. ∎

###  3.3. Caveats and Examples

We collect here some examples that illustrate various caveats given in the
introduction to the main theorems. We begin with an explicit example of a
superintegral Kleinian packing which is not crystallographic.

$(a)$ $(b)$

Figure 8. $(a)$ A fundamental domain for the extended Bianchi group
$\widehat{Bi}(23)$, and $(b)$ a superintegral Kleinian packing attached to it.

######  Example 65.

The extended Bianchi group $\widehat{Bi}(23)$ (see footnote 6) is not
reflective (see [BM13]). Indeed, applying Vinberg’s algorithm [Vin72] to the
quadratic form $f=-2xy+2z^{2}+2zw+\frac{23+1}{2}w^{2}$ shows that the subgroup
of $O_{f}({\mathbb{Z}})$ generated by all reflections has infinite index in
$O_{f}({\mathbb{Z}})$. One can give a fundamental domain in $\mathbb{H}^{3}$
for $O_{f}({\mathbb{Z}})$ as shown in Figure 8$(a)$; here the blue walls act
by reflections, and there is a pair of commuting unipotent elements, one
identifying the green walls, and another identifying the orange ones. By Part
2(ii) of the Structure Theorem 22, we can construct a superintegral packing
from this fundamental polyhedron; see Figure 8$(b)$. So this is a
superintegral packing which is Kleinian but not crystallographic.111111Note
that this particular packing appeared previously in work of Stange [St18] and
Martin [Ma19] but was not recognized there as being dense (and hence wasn’t
considered a “packing” by our definition), due to the significant distances
between disjoint circles. The general theory given here makes this density
apparent.

As mentioned above Remark 10, a superintegral Kleinian packing can have all
spheres disjoint (as observed already in [KN19]); but its supergroup must
still be non-uniform and ${\mathbb{Q}}$-arithmetic. One such is the following.

1$m=30$234567891011

$(a)$ $(b)$

Figure 9. $(a)$ The Coxeter diagram for $\widehat{Bi}(30)$, and $(b)$ a
superintegral packing with all spheres disjoint.

######  Example 66.

Consider the extended Bianchi group $\widetilde{\Gamma}=\widehat{Bi}(30)$.
Equivalently, $\widetilde{\Gamma}\cong O_{\mathcal{F}}({\mathbb{Z}})$ is the
integer orthogonal group preserving the form
${\mathcal{F}}=-xy+z^{2}+30w^{2}$. This group is reflective (see [BM13]), and
applying Vinberg’s algorithm [Vin72] gives the Coxeter diagram121212The
diagram has nodes for each facet, and facets that are orthogonal are not
connected; the dihedral angle $\pi/3$ is denoted by a single line, angle
$\pi/4$ is a double line, and angle $\pi/\infty=0$ is a thick solid line.
Nodes of separated facets are connected with a dotted line.  shown in Figure
9. The node marked “8” is totally isolated from the others, being either
orthogonal or some given distance apart from the other generating reflective
walls. (So are nodes “9”, “10”, and “11” but we choose to use “8”.) Dropping
this wall from the generators and letting the remaining generators act on it
by reflections (as in the Structure Theorem 22) gives the superintegral
crystallographic packing shown in Figure 9(b).

As observed already in [KN19], if a packing cannot be realized as the orbit of
a single sphere (that is, there is no symmetry group for which the action on
the spheres in the packing is transitive), then the resulting packing (or bug)
need not be superintegral (or even integral); cf. Remark 11. Here is an
explicit example of a non-integral bug with non-uniform,
${\mathbb{Q}}$-arithmetic supergroup.

######  Example 67.

_Let $\widetilde{\Gamma}$ be the extended Bianchi group
$\widetilde{\Gamma}=\widehat{Bi}(6)$. Equivalently, $\widetilde{\Gamma}\cong
O_{\mathcal{F}}({\mathbb{Z}})$ is isomorphic to the integer orthogonal group
preserving the form ${\mathcal{F}}=-xy+z^{2}+6w^{2}$. That it is reflective
(generated by reflections) is essentially due to Bianchi [Bia92]; see also
[BM13]. Applying Vinberg’s algorithm to this group produces the following
normal vectors _

$({0,0,-1,0})^{t},({1,0,1,0})^{t},({0,0,0,-1})^{t},$
$({6,0,0,1})^{t},({-1,1,0,0})^{t},({2,2,0,1})^{t}.$

_Making a (choice of) change of variables from ${\mathcal{F}}$ to the
universal form $Q$ in (48), these correspond to spheres with the following
(realization of) inversive coordinates:_

$v_{1}=(0,-1,0,0)^{t},\ v_{2}=(0,1,0,1)^{t},\ v_{3}=(0,0,-1,0)^{t},\ $ (68)
$v_{4}=(0,0,1,\sqrt{6})^{t},\ v_{5}=(1,0,0,-1)^{t},\
v_{6}=(\sqrt{2},0,\sqrt{3},\sqrt{2})^{t}.$

_SeeFigure 10(a) for the corresponding spheres (as circles in
${\mathbb{R}}^{2}$). Writing $V$ for the $4\times 6$ matrix whose columns are
$v_{j}$, we can compute the $6\times 6$ Gramian of all $Q$-inner products:_

$V^{t}QV\ =\ {\mathcal{G}}=\left(\begin{array}[]{cccccc}1&-1&0&0&0&0\\\
-1&1&0&0&-\frac{1}{2}&-\frac{1}{\sqrt{2}}\\\ 0&0&1&-1&0&-\sqrt{3}\\\
0&0&-1&1&-\sqrt{\frac{3}{2}}&0\\\ 0&-\frac{1}{2}&0&-\sqrt{\frac{3}{2}}&1&0\\\
0&-\frac{1}{\sqrt{2}}&-\sqrt{3}&0&0&1\\\ \end{array}\right).$

_Equivalently, $\widetilde{\Gamma}$ has the Coxeter diagram given in Figure
10(b). A different realization of $v_{j}$ from (68) will of course have
different inversive coordinates $V$ but the Gramian and Coxeter diagram are
invariants. _

123456

$(a)$ $(b)$

Figure 10. $(a)$ The spheres with inversive coordinates (68) in Example 67.
$(b)$ Their Coxeter diagram.

Now we construct the bug. Write $R_{j}$ for the Möbius action of reflection
through sphere $v_{j}$, that is,

$R_{j}=I_{4\times 4}+2v_{j}\cdot v_{j}^{t}\cdot Q.$

We apply Part 2(i) of the Structure Theorem 22 with
$S^{\prime}=\\{R_{1},\dots,R_{6}\\}$, $\left\langle
S^{\prime}\right\rangle=\widetilde{\Gamma}$, and take $R=\\{R_{3},R_{6}\\}$ so
that $S=S^{\prime}\setminus R$. The bug we obtain is then the orbit

(69) $\mathscr{B}=\left\langle S\right\rangle\cdot\\{v_{3},v_{6}\\},$

as shown in Figure 11. This particular realization of the bug is evidently
non-integral, but we have not yet ruled out that there cannot be some other
conformally equivalent realization of this bug which is integral. (Indeed,
there exist realizations of the classical Apollonian packing that are non-
integral.)

Figure 11. A non-integral bug; the bend of a circle is shown at its center.

######  Lemma 70.

There does not exist a conformal realization of the bug in (69) that is
integral.

###### Proof.

We mimic the first proof of Theorem 3 by looking at a right “bends” action.
This time, we will do it using an over-determined system of equations, leading
to an irrational linear relation among the bends, which is realization-
independent (and hence the bug cannot ever be integral).

To begin, notice that the spheres

$v_{3},~{}v_{6},~{}~{}R_{2}\cdot v_{6},~{}~{}R_{1}\circ R_{2}\cdot
v_{6},~{}~{}\hbox{and}~{}~{}R_{5}\circ R_{1}\circ R_{2}\cdot v_{6}$

are all in the bug $\mathscr{B}$. Write $W$ for the $4\times 5$ matrix whose
columns are the inversive coordinates of these spheres. The kernel of $W$,

$\mathcal{K}\ :=\ \ker(W)\ :=\ \\{g\in Mat_{5\times 5}({\mathbb{R}}):Wg=0\\},$

is independent of the realization of the bug. Indeed, if
$g\in\mathrm{Isom}(\mathbb{H}^{3})$ is any isometry and we move the whole bug
by left-acting by $g$, then the inversive coordinates matrix $W$ changes to
$gW$, leaving the kernel $\mathcal{K}$ invariant. Notice that the kernel
contains, e.g.,

$K\ =\ \left(\begin{array}[]{ccccc}-\sqrt{3}&0&0&0&0\\\ 1&0&0&0&0\\\
-3&0&0&0&0\\\ 0&0&0&0&0\\\ 1&0&0&0&0\\\ \end{array}\right)\in\mathcal{K}.$

So in any realization of this bug $\mathscr{B}$, if the spheres in $W$ have
bends, resp., $a$, $b$, $c$, $d$, and $e$, say, then

$0\ =\ (a,b,c,d,e)\cdot K\cdot(1,0,0,0,0)^{t}\ =\ -\sqrt{3}a+b-3c+e.$

Suppose this is the case with bends $a,\dots,e$ all integers; then we must
have $a=0$. Note that this applies to not only one particular realization of
these five spheres, but to any such; in particular, the entire orbit of these
five spheres under the symmetry group $\Gamma=\left\langle S\right\rangle$ has
the bend $a=0$. But then $\Gamma$ satisfies an extra polynomial equation, and
is not Zariski dense, which is a contradiction. ∎

Lastly, we prove Proposition 14 by exhibiting the following.

Figure 12. Icosahedral packing

######  Example 71.

In [KN19], a procedure was given for construction a packing
$\mathscr{P}=\mathscr{P}(\Pi)$ modeled on a (convexly-realizable combinatorial
type of a) polyhedron $\Pi$. In the case of $\Pi$ being the icosahedron, the
resulting packing shown in Figure 12. In can be shown that the entire
superpacking can be made to have bends in $\mathfrak{o}={\mathbb{Z}}[\phi]$,
where $\phi={1+\sqrt{5}\over 2}$ is the golden mean. But Vinberg’s criterion
applied to the “superGramian” (in the nomenclature of [KN19]) shows that the
supergroup is non-arithmetic, because the quadratic form it defines (over
${\mathbb{Q}}(\phi)$) fails to become definite under the Galois conjugate
embedding. For details of this computation, see
http://math.rutgers.edu/~alexk/maths/Icosahedron.nb.

##  4\. Hausdorff dimensions of limit sets of packings

In Proposition 60 and its application to (the forward direction of) the
Classification Theorem 9, we showed how to construct superintegral Kleinian
packings from ${\mathbb{Q}}$-arithmetic non-uniform lattices. We now extend
this construction to show that for infinitely many $j$’s the Hausdorff
dimensions of the limit sets of the groups $\Gamma_{j}$’s constructed in the
proof of Proposition 60 are pairwise distinct and, moreover, the sequence of
Hausdorff dimensions converges to the maximal Hausdorff dimension, $n$. This
will follow from the following theorem where $\dim$ stands for the Hausdorff
dimension, which implies Theorem 13. We continue with the notation introduced
in the proof of Proposition 60.

######  Theorem 72.

$\lim_{j\to\infty}\dim(\Lambda(\Gamma_{j}))=n$ and for all $j$,
$\dim(\Lambda(\Gamma_{j}))<n$.

Note first that the discrete groups $\Gamma_{j}$ contain geometrically finite
subgroups which are conjugates of $\pi_{1}(N)$. In particular,

$\delta(\Gamma_{j})\geq\dim(\Lambda(\pi_{1}(N)))=n-1\geq\frac{n}{2},$

provided that $n\geq 2$.

We next discuss a relation between the Hausdorff dimension of the limit set,
the critical exponent and the bottom of the spectrum of the Laplacian.

For a complete connected Riemannian manifold $M$ let $\lambda(M)$ denote
bottom of the $L^{2}$-spectrum of the Laplacian of $M$. This number can be
computed via Rayleigh quotients:

$\lambda=\inf\frac{\int_{\Omega}|\nabla u|^{2}}{\int_{\Omega}u^{2}}$

where the infimum is taken over all smooth compactly supported functions
(called “test” or “trial” functions) $u\in C^{\infty}_{c}(M)$ and
$\Omega:=\\{u>0\\}$.

######  Theorem 73 (Semicontinuity of $\lambda$).

$\lambda(M)$ is upper semicontinuous with respect to the topology of smooth
Gromov–Hausdorff convergence:

$M_{i}\to M\Rightarrow\lim\inf_{i\to\infty}\lambda(M_{i})\leq\lambda(M).$

###### Proof.

This is clear using the Rayleigh quotient definition: Every test-function on
$M$ is the $C^{1}$-limit of a sequence of test-functions on $M_{i}$’s. ∎

######  Remark 74.

1\. The same theorem (and proof) applies to Riemannian orbifolds.

2\. The bottom of the spectrum is not continuous with respect to the topology
of smooth Gromov–Hausdorff convergence. For instance, for $n\geq 3$ let $M$ be
a hyperbolic $n+1$-manifold of finite volume, let $M_{i}\to M$ be a profinite
sequence of (finite) covers of $M$. Then the manifolds $M_{i}$ converge to the
hyperbolic $n$-space, $\lambda(M_{i})=0$, while
$\lambda({\mathbb{H}}^{n+1})=n^{2}/4$.

Given a discrete subgroup $\Gamma<\mathrm{Isom}({\mathbb{H}}^{n+1})$, let
$\delta(\Gamma)$ denote the critical exponent of $\Gamma$ (see e.g. [Ni89]). A
discrete subgroup $\Gamma<\mathrm{Isom}({\mathbb{H}}^{n+1})$ is called
nonelementary if its limit set consists of more than two points. If $\Gamma$
is geometrically finite and nonelementary, then $\delta(\Gamma)$ equals the
Hausdorff dimension of the limit set of $\Gamma$, see [Pat76, Su84, Tu84,
Ni89]. We will need the Elstrod–Patterson–Sullivan formula (see e.g. [Ni89]),
relating, for a discrete subgroup $\Gamma<\mathrm{Isom}({\mathbb{H}}^{n+1})$,
the critical exponent $\delta=\delta(\Gamma)$ and the bottom of the spectrum
$\lambda=\lambda({\mathbb{H}}^{n+1}/\Gamma)$:

######  Theorem 75.

$\lambda=\left(\frac{n}{2}\right)^{2},\hbox{~{}~{}if~{}~{}}\delta\leq\frac{n}{2},$
$\lambda=\delta(n-\delta),\hbox{~{}~{}if~{}~{}}\delta\geq\frac{n}{2}.$

######  Definition 76.

A sequence of closed subgroups $\Gamma_{i}$ of a Lie group $G$ is said to
converge to a closed subgroup $\Gamma<G$ geometrically or in Chabauty topology
if the following two conditions are met:

1\. For every $\gamma\in\Gamma$ there exists a sequence
$\gamma_{i}\in\Gamma_{i}$ which converges to $\gamma$.

2\. If a sequence $\gamma_{i}\in\Gamma_{i}$ subconverges to $\gamma\in G$,
then $\gamma\in\Gamma$.

Suppose now that $X$ is a complete connected Riemannian manifold, $G$ is the
isometry group of $X$. Then $G$ (equipped with the compact-open topology) is a
Lie group. Fix a base-point $x\in X$. Consider a sequence of subgroups
$\Gamma_{i}<G$ and a subgroup $\Gamma<G$ and quotient manifolds/orbifolds
$M_{i}=X/\Gamma_{i},M=X/\Gamma$. Let $\bar{x}_{i},\bar{x}$ denote the
projections of $x$ to $M_{i}$, $M$ respectively.

The following theorem was proven in [BP92] in the case when $X$ is the
hyperbolic space (which will suffice for us), but the same proof works for any
complete connected Riemannian manifold.

######  Theorem 77 (See [BP92]).

A sequence of discrete subgroups $\Gamma_{i}<G$ converges geometrically to a
discrete subgroup $\Gamma<G$ if and only if the sequence of pointed Riemannian
manifolds/orbifolds $(M_{i},\bar{x}_{i})$ converges to $(M,\bar{x})$ in the
smooth Gromov–Hausdorff topology.

######  Corollary 78 (Semicontinuity of $\lambda$ and $\delta$).

(a) Suppose that $\Gamma_{i}$ is a sequence of discrete (nonelementary)
subgroups of $\mathrm{Isom}({\mathbb{H}}^{n+1})$ converging to a discrete
subgroup $\Gamma<\mathrm{Isom}({\mathbb{H}}^{n+1})$. Then

$\lim\inf_{i\to\infty}\lambda({\mathbb{H}}^{n+1}/\Gamma_{i})\leq\lambda({\mathbb{H}}^{n+1}/\Gamma).$

(b) Suppose, in addition, that
$\inf\\{\delta(\Gamma),\delta(\Gamma_{i}),i\in{\mathbb{N}}\\}\geq n/2$. Then

(79) $\lim\inf_{i\to\infty}\delta(\Gamma_{i})\geq\delta(\Gamma).$

In particular, if $\delta(\Gamma_{i})\leq\delta(\Gamma)$ for all $i$ (e.g. if
$\Gamma_{i}<\Gamma$) then

$\lim_{i\to\infty}\delta(\Gamma_{i})=\delta(\Gamma).$

###### Proof.

The first inequality is a direct corollary of the two theorems above. The
second inequality follows from the relation of $\delta$ and $\lambda$:

$\lambda=\delta(n-\delta)\hbox{~{}~{}if~{}~{}}\delta\geq\frac{n}{2},$

see Theorem 75. ∎

Lastly, we need the following theorem due to T. Roblin [Ro05] (see also R.
Brooks [Br85]):

######  Theorem 80.

Let $\Gamma<\mathrm{Isom}({\mathbb{H}}^{n+1})$ be a geometrically
finite131313More generally, a group of divergence type, i.e. a group whose
Poincaré series diverges at the critical exponent. subgroup and
$\hat{\Gamma}<\Gamma$ a normal subgroup with amenable quotient
$\Gamma/\hat{\Gamma}$. Then $\delta(\hat{\Gamma})=\delta(\Gamma)$.

Thus, if $\Gamma$ is a lattice and $\hat{\Gamma}<\Gamma$ is a normal subgroup
with cyclic quotient, then $\delta(\hat{\Gamma})=n$.

We can now finish the proof of Theorem 72. Let $\Gamma_{j}$ be the discrete
groups as in the theorem.

######  Lemma 81.

The sequence $\Gamma_{j}$ geometrically converges to $\hat{\Gamma}$.

###### Proof.

This follows from the fact that the sequence of domains $D_{j}$ exhausts
$\hat{M}$ (see Remark 62), which ensures Gromov–Hausdorff convergence of the
corresponding hyperbolic manifolds, hence, geometric convergence of discrete
subgroups. ∎

Thus, we conclude that

$\lim_{j\to\infty}\dim(\Lambda(\Gamma_{j}))=\dim(\Lambda(\Gamma))=n.$

On the other hand, $\delta(\Gamma_{j})<n$ since each $\Gamma_{j}$ is
geometrically finite and its limit set is a proper subset of $\mathbb{S}^{n}$,
(Sullivan [Su84] and Tukia [Tu84], independently). This concludes the proof of
Theorem 72 and, hence, of Theorem 13. ∎

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