Geometry and arithmetic of crystallographic sphere packings

Significance

This paper studies generalizations of the classical Apollonian circle packing, a beautiful geometric fractal that has a surprising underlying integral structure. On the one hand, infinitely many such generalized objects exist, but on the other, they may, in principle, be completely classified, as they fall into, only finitely, many “families,” all in bounded dimensions.

Abstract

We introduce the notion of a “crystallographic sphere packing,” defined to be one whose limit set is that of a geometrically finite hyperbolic reflection group in one higher dimension. We exhibit an infinite family of conformally inequivalent crystallographic packings with all radii being reciprocals of integers. We then prove a result in the opposite direction: the “superintegral” ones exist only in finitely many “commensurability classes,” all in, at most, 20 dimensions.

The goal of this program is to understand the basic “nature” of the classical Apollonian gasket. Why does its integral structure exist? (Of course, it follows here from Descartes’ Kissing Circles Theorem, but is there a more fundamental, intrinsic explanation?) Are there more like it? (Around a half-dozen similarly integral circle and sphere packings were previously known, each given by an ad hoc description.) If so, how many more? Can they be classified? We develop a basic unified framework for addressing these questions, and find two surprising phenomena:

• (♡) there is indeed a whole infinite zoo of integral sphere packings, and

• (♠) up to “commensurability,” there are only finitely many Apollonian-like objects, over all dimensions.

Definition 1:

By an $S^{n-1}$-packing (or just packing) $\mathcal{P}$ of $\widehat{{\mathbb{R}}^n}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}{\mathbb{R}}^n\cup \left\{\infty \right\}$, we mean an infinite collection of oriented $\left(n-1\right)$-spheres (or co-dim-1 planes) so that: $\left(i\right)$ The interiors of spheres are disjoint, and $\left(ii\right)$ The union of the interiors of the spheres is dense in space. The bend of a sphere is the reciprocal of its (signed) radius.* To be dense but disjoint, the spheres in the packing $\mathcal{P}$ must have arbitrarily small radii, so arbitrarily large bends. If every sphere in $\mathcal{P}$ has integer bend, then we call the packing integral.

Without more structure, one can make completely arbitrary constructions of integral packings. A key property enjoyed by the classical Apollonian circle packing and connecting it to the theory of “thin groups” (see refs. 1–3) is that it arises as the limit set of a geometrically finite reflection group in hyperbolic space of one higher dimension.

Definition 2:

We call a packing $\mathcal{P}$ crystallographic if its limit set is that of some geometrically finite reflection group $\mathrm{\Gamma }<\mathrm{I}\mathrm{s}\mathrm{o}\mathrm{m}\left({\mathbb{H}}^{n+1}\right)$.

This definition is sufficiently general to encompass all previously proposed generalizations of Apollonian gaskets found in the literature, including refs. 4–10. With these two basic and general definitions in place, we may already state our first main result, confirming $\left(♡\right)$.

Theorem 3.

There exist infinitely many conformally inequivalent integral crystallographic packings.

We show in Fig. 1 but one illustrative example, whose only “obvious” symmetry is a vertical mirror image. It turns out (but may be hard to tell just from the picture) that this packing does indeed arise as the limit set of a Kleinian reflection group. The argument leading to Theorem 3 comes from constructing circle packings “modeled on” combinatorial types of convex polyhedra, as follows.

Fig. 1.

An integral crystallographic packing. (The circles are labeled with their bend.)

$\left(♡\right)$: Polyhedral Packings

Let $\mathrm{\Pi }$ be the combinatorial type of a convex polyhedron. Equivalently, $\mathrm{\Pi }$ is a 3-connected^‡ planar graph. A version of the Koebe–Andreev–Thurston Theorem^§ says that there exists a geometrization of $\mathrm{\Pi }$ (that is, a realization of its vertices in ${\mathbb{R}}^3$ with straight lines as edges and faces contained in Euclidean planes) having a midsphere (meaning, a sphere tangent to all edges). This midsphere is then also simultaneously a midsphere for the dual polyhedron $\hat{\mathrm{\Pi }}$. Fig. 2A shows the case of a cuboctahedron and its dual, the rhombic dodecahedron.

Fig. 2.

Geometrization and cluster/cocluster pair for $\mathrm{\Pi }=$ cuboctahedron with dual $\hat{\mathrm{\Pi }}=$ rhombic dodecahedron. (A) $\mathrm{\Pi }$, $\hat{\mathrm{\Pi }}$, and their midsphere. (B) An orthogonal cluster and cocluster pair with nerves $\mathrm{\Pi }$ and $\hat{\mathrm{\Pi }}$.

Stereographically projecting to $\widehat{{\mathbb{R}}^2}$, we obtain a cluster (just meaning, a finite collection) $\mathcal{C}$ of circles, whose nerve (that is, tangency graph) is isomorphic to $\mathrm{\Pi }$, and a cocluster, $\hat{\mathcal{C}}$, with nerve $\hat{\mathrm{\Pi }}$, which meets $\mathcal{C}$ orthogonally. Again, the example of the cuboctahedron is shown in Fig. 2B.

Definition 4:

The orbit $\mathcal{P}=\mathcal{P}\left(\mathrm{\Pi }\right)=\mathrm{\Gamma }\cdot \mathcal{C}$ of the cluster $\mathcal{C}$ under the group $\mathrm{\Gamma }=⟨\hat{\mathcal{C}}⟩$ generated by reflections through the cocluster $\hat{\mathcal{C}}$ is said to be modeled on the polyhedron $\mathrm{\Pi }$.

Lemma 5.

An orbit modeled on a polyhedron is a crystallographic packing.

See Fig. 3 for a packing modeled on the cuboctahedron. Such packings are unique up to conformal/anticonformal maps by Mostow rigidity, but Möbius transformations do not generally preserve arithmetic.

Fig. 3.

A packing modeled on the cuboctahedron, shown with cocluster.

Definition 6:

We call a polyhedron $\mathrm{\Pi }$ integral if there exists some packing modeled on $\mathrm{\Pi }$, which is integral.

It is not hard to see that the cuboctahedron is indeed integral, as is the tetrahedron, which corresponds to the classical Apollonian gasket. It is a fundamental problem to classify all integral polyhedra.

Let us point out some basic difficulties with this problem. First of all, it is nontrivial to determine whether, given a particular polyhedron, there exists some packing modeled on it that is integral. Indeed, Koebe–Andreev–Thurston geometrization is an infinite limiting process, and how is one to know whether 3.9999 is really 4? To the rescue is Mostow rigidity again, which implies that one can always find cluster/cocluster configurations with all centers and radii algebraic. This means that after computing enough decimal places, one can guess what the nearby algebraic values might be and then rigorously verify whether the guess gives the correct tangency data. This algorithm works for small examples, but once $\mathrm{\Pi }$ is sufficiently complicated, it may take a very long time for the guessing process to halt. (The code for this algorithm is available at math.rutgers.edu/$\sim$alexk/crystallographic.)

Despite these difficulties, we are able to show the following toward $\left(♡\right)$.

Theorem 7.

Infinitely many polyhedra are integral and give rise to infinitely many conformally inequivalent integral polyhedral packings.

This of course implies Theorem 3. We stress the conformal inequivalence here because it turns out that infinitely many polyhedra give rise to the same crystallographic packing; so, the first part of Theorem 7, that infinitely many polyhedra are integral, does not by itself imply Theorem 3. To explain the main ideas in the proof, we need some more notation.

Returning to the general setting of crystallographic packings, recall that $\mathcal{P}$ is assumed to arise as the limit set of a discrete group $\mathrm{\Gamma }$; we call the latter a symmetry group of $\mathcal{P}$.

Definition 8:

Given a packing $\mathcal{P}$ with symmetry group $\mathrm{\Gamma }$, we define its supergroup, $\tilde{\mathrm{\Gamma }}$, to be the group generated by $\mathrm{\Gamma }$ itself, plus reflections through all spheres in $\mathcal{P}$. Abusing notation, we may write this as

$$\tilde{\mathrm{\Gamma }}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}⟨\mathrm{\Gamma },\mathcal{P}⟩\hspace{0.17em}<\hspace{0.17em}\mathrm{I}\mathrm{s}\mathrm{o}\mathrm{m}\left({\mathbb{H}}^{n+1}\right).$$

In the case of a polyhedral packing $\mathcal{P}=\mathcal{P}\left(\mathrm{\Pi }\right)$, the supergroup is simply the group generated by reflections in both the cluster and cocluster, $\tilde{\mathrm{\Gamma }}=\hspace{0.17em}⟨\hat{\mathcal{C}},\mathcal{C}⟩$.

Definition 9:

The superpacking, $\tilde{\mathcal{P}}$, of $\mathcal{P}$ with symmetry group $\mathrm{\Gamma }$ is the orbit of $\mathcal{P}$ under its supergroup, that is,

$$\tilde{\mathcal{P}}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}\tilde{\mathrm{\Gamma }}\cdot \mathcal{P}.$$

Note that the superpacking is not a packing by our definition as the sphere interiors are no longer disjoint.^¶

Definition 10:

We call a packing $\mathcal{P}$ superintegral if every bend in its superpacking $\tilde{\mathcal{P}}$ is an integer. (Note that an unrelated notion of “superintegrality” is defined in ref. 12, section 8.)

Remark 11.

While different symmetry groups $\mathrm{\Gamma }$ lead to different (but commensurate) supergroups $\tilde{\mathrm{\Gamma }}$, the superpackings are universal, the same for all choices of $\mathrm{\Gamma }$.

Returning to polyhedral packings, we say that a polyhedron is superintegral if some packing modeled on it is. To prove Theorem 7, we actually prove the following stronger statement.

Theorem 12.

Infinitely many polyhedra are superintegral and give rise to infinitely many conformally inequivalent superintegral crystallographic packings.

Although every previously known integral packing was also superintegral, we discover that the latter is a strictly stronger condition.

Lemma 13.

There exist infinitely many conformally inequivalent crystallographic packings that are integral but not superintegral.

Remark 14:

Just one example of an integral but not superintegral polyhedron is the hexagonal pyramid. See also Remark 21.

To prove Theorem 12, we define certain operations on “seed” polyhedra, which we call “growths,” including doubling the seed along a vertex or a face, and observe that, while these generally wreak havoc on the resulting packings $\mathcal{P}$, so $\mathcal{P}\left(growth\right)$ and $\mathcal{P}\left(seed\right)$ are usually conformally inequivalent, the superpackings $\tilde{\mathcal{P}}$ are essentially preserved, in fact

$$\tilde{\mathcal{P}}\left(growth\right)\hspace{0.17em}\subset \hspace{0.17em}\tilde{\mathcal{P}}\left(seed\right).$$

In particular, if a polyhedron is superintegral, then all of its growths are also superintegral, and hence integral. This proves Theorem 12 and hence Theorem 3.

$\left(♠\right)$: Classifying Superintegral Crystallographic Packings

Toward the opposite general problem of classifying integral and superintegral crystallographic packings, we make two basic observations. The first, having nothing to do with integrality, shows that the entire theory of crystallographic packings is “low”-dimensional.

Theorem 15.

Crystallographic packings can only exist in dimensions $n<996$.

To prove this, we need the following.

Lemma 16.

The supergroup $\tilde{\mathrm{\Gamma }}$ of a crystallographic packing $\mathcal{P}$ with symmetry group $\mathrm{\Gamma }$ is a lattice, that is, it acts on ${\mathbb{H}}^{n+1}$ with finite covolume.

We first sketch a proof of this lemma. Let $\mathrm{\Gamma }$ be a symmetry group for $\mathcal{P}$; then, it is assumed to be geometrically finite (recall that this means some uniform thickening of the convex core of $\mathrm{\Gamma }$ has finite volume). Since $\mathrm{\Gamma }$ is a reflection group, it has an essentially unique fundamental polyhedron $\mathcal{F}:\hspace{0.17em}=\hspace{0.17em}\mathrm{\Gamma }\backslash {\mathbb{H}}^{n+1}$. The domain of discontinuity $\mathrm{\Omega }$ of $\mathrm{\Gamma }$ (that is, the complement in $d{\mathbb{H}}^{n+1}$ of its limit set ${\mathrm{\Lambda }}_{\mathrm{\Gamma }}$) is the union of disjoint open geometric balls, since the limit set ${\mathrm{\Lambda }}_{\mathrm{\Gamma }}$ is assumed to coincide with the set of limit points of $\mathcal{P}$. The quotient $\mathrm{\Omega }/\mathrm{\Gamma }$ is then a disjoint union of finitely many open ends. For each end, we develop the domain under the $\mathrm{\Gamma }$-action and fill an open ball, the boundary of which is then an (unoriented) sphere in $\mathcal{P}$. A geodesic hemisphere above such a ball is a frontier of the flare, cutting the walls it meets of $\mathcal{F}$ either tangentially or at right angles (for, otherwise, the spheres in $\mathcal{P}$ would overlap). Hence, when we form the supergroup $\tilde{\mathrm{\Gamma }}$ by adjoining to $\mathrm{\Gamma }$ reflections through all of the spheres in $\mathcal{P}$, we obtain a discrete action, and moreover the original domain of discontinuity $\mathrm{\Omega }$ has been entirely cut out, rendering $\tilde{\mathrm{\Gamma }}$ a lattice.

Returning to Theorem 15, Vinberg (14) and Prokhorov (15) showed that hyperbolic reflection lattices can only exist in dimensions $n<996$, and hence crystallographic packings are similarly bounded in dimension, proving the theorem. (The number 996 is not expected to be sharp.)

Next, we show that not only is the dimension bounded, but if we assume superintegrality, then (up to commensurability) there are only finitely many Apollonian-like objects.

Definition 17:

Two crystallographic packings are said to be commensurate if their supergroups are.

Theorem 18.

There are only finitely many commensurability classes of superintegral crystallographic packings, all of dimension $n\le 20$.

To prove this theorem, we show the following.

Theorem 19.

If $\mathcal{P}$ is a superintegral crystallographic packing, then its supergroup $\tilde{\mathrm{\Gamma }}$ is arithmetic.^#

In fact, to conclude arithmeticity, it is sufficient that the orbit under the supergroup $\tilde{\mathrm{\Gamma }}$ of a single sphere $S\in \mathcal{P}$ has all integer bends. Let us sketch a proof. To a (positively oriented) sphere $S$ of center $\mathbf{\text{z}}=\left(z_1,\dots z_n\right)$ and radius $r$, we attach the “inversive coordinates”

$${\mathbf{\text{v}}}_S\hspace{0.17em}≔\hspace{0.17em}\left(\hat{b},b,b\mathbf{\text{z}}\right).$$

Here, $b=1/r$ is the bend, and $\hat{b}=1/\hat{r}$ is the cobend, that is, the reciprocal of the coradius, the latter defined as the radius of the sphere after inversion through the unit sphere; see the discussion in, e.g., refs. 17 and 18. The vector ${\mathbf{\text{v}}}_S$ lies on a one-sheeted hyperboloid $Q=-1$, where $Q$ is the (universal) “discriminant” form,

$$Q=\left(\begin{array}{cc}\hfill & \hfill \frac{1}{2}\hfill \\ \frac{1}{2}\hfill & \hfill \hfill \\ \hfill & \hfill \hfill & \hfill -I_{n-1}\end{array}\right).$$

In these coordinates,

$$\tilde{\mathrm{\Gamma }}<O_Q\left(\mathbb{R}\right)$$ [1]

is a right action by Möbius transformations on the row vector ${\mathbf{\text{v}}}_S$. Since $\tilde{\mathrm{\Gamma }}$ is a lattice, it is essentially (up to finite index components) Zariski dense in $O_Q$; hence, the orbit $\mathcal{O}={\mathbf{\text{v}}}_S\cdot \tilde{\mathrm{\Gamma }}$ of $S$ is essentially Zariski dense in the quadric $Q=-1$. There is then a choice of cluster ${\mathcal{C}}_S\subset \mathcal{O}$ of $n+2$ spheres whose matrix $\mathcal{V}$ of inversive coordinates has (full) rank $n+2$. Make such a choice arbitrarily. This cluster $\mathcal{V}$ has a Gram matrix of inversive products,

$$\mathcal{G}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}\mathcal{V}\cdot Q\cdot {\mathcal{V}}^{†},$$ [2]

which is invertible (also has rank $n+2$). Let

$$\mathcal{F}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}{\mathcal{G}}^{-1}$$

be its inverse, which also induces a quadratic form having signature $\left(1,n+1\right)$. Then $\tilde{\mathrm{\Gamma }}$ is conjugate to a “bends” group,

$$\tilde{\mathcal{A}}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}\mathcal{V}\cdot \tilde{\mathrm{\Gamma }}\cdot {\mathcal{V}}^{-1}\hspace{0.17em}<\hspace{0.17em}{O}_{\mathcal{F}}\left(\mathbb{R}\right),$$

which now acts on the left on the (second) column vector of bends $\mathbf{\text{b}}=\mathcal{V}\cdot {\left(\mathrm{0,1,0},\dots ,0\right)}^{†}$ in $\mathcal{V}$; this vector $\mathbf{\text{b}}$ lies on the cone $\mathcal{F}=0$, and $\tilde{\mathcal{A}}$ is a lattice in $O_{\mathcal{F}}\left(\mathbb{R}\right)$. Though a priori real valued, we claim that $\mathcal{F}$ is in fact rational. Indeed, by assumption, the $\tilde{\mathcal{A}}$-orbit

$$\mathcal{B}=\tilde{\mathcal{A}}\cdot \mathbf{\text{b}}$$

lies in ${\mathbb{Z}}^{n+2}\cap \left\{\mathcal{F}=0\right\}$ and is Zariski dense in the cone. However, a quadratic form having a Zariski dense set of integer points $\mathcal{B}$ on the cone $\mathcal{F}=0$ is easily seen to be rational, as claimed. Next, we observe that, since $\tilde{\mathcal{A}}$ is a linear action, it in fact preserves a full-rank $\mathbb{Z}$-lattice $\mathrm{\Lambda }$. However, the group

$$O_{\mathcal{F}}^{\mathrm{\Lambda }}=\left\{g\in O_{\mathcal{F}}\left(\mathbb{R}\right):g\mathrm{\Lambda }=\mathrm{\Lambda }\right\}$$

is easily seen to be congruence and contains $\tilde{\mathcal{A}}$. Hence, $\tilde{\mathcal{A}}$ is arithmetic, as is its conjugate $\tilde{\mathrm{\Gamma }}$. This proves Theorem 19.

Returning to Theorem 18, this now follows the amazing fact (19–22) that there are only finitely many commensurability classes of arithmetic reflection groups. The ones defined over $\mathbb{Q}$, as supergroups of superintegral packings are (see the proof of Theorem 19), all have dimension $n+1\le 21$ (see ref. 23); this proves Theorem 18.

It turns out that superintegrality is a necessary condition in Theorem 19, and mere integrality is insufficient. Indeed, we discover the following.

Lemma 20.

There exist infinitely many conformally inequivalent integral (but of course not superintegral) packings whose supergroups are nonarithmetic.

Remark 21:

The supergroup of the hexagonal pyramid is nonarithmetic; see also Remark 14.

Remark 22:

Note also that there is no contradiction with Theorem 12 (and Theorem 3), as the packings constructed there fall into finitely many commensurability classes.

Given these finiteness results, the complete classification of superintegral crystallographic packings will then rely on understanding to what extent a converse of Theorem 19 may be true.

Question 23.

Given an arithmetic reflection group, is it commensurate with the supergroup of a superintegral crystallographic packing?

We will say that an arithmetic group “supports” a packing if the answer to the above is YES. We have investigated this question in some special cases and found the following positive results.

Theorem 24.

The answer to Question 23 is YES for all nonuniform lattices over $\mathbb{Q}$ in dimension $n=2$. Namely, every reflective (that is, commensurate to a reflection group) Bianchi group supports a superintegral crystallographic packing.

In higher dimensions, we are also able to show the following.

Theorem 25.

The answer is YES for certain lattices in dimensions $n\le 13,$ and $n=\mathrm{17,18}$; that is, superintegral crystallographic packings exist in all these dimensions.

Before saying more about these theorems, let us point out that we suspect that the answer may be NO in general.

Remark 26:

At present, we do not know of a single superintegral (or even integral) packing whose supergroup is cocompact. In dimension $n=2$, the integral orthogonal groups preserving the form $x_1^2+x_2^2+x_3^2-d{x}_4^2$ are cocompact and reflective only when the coefficient $d=7$ or 15, see ref. 24. We suspect, but do not know how to prove, that neither of these reflection groups support crystallographic packings. See Remark 29.

Remark 27:

Taking, e.g., $\mathfrak{o}=\mathbb{Z}\left[\phi \right]$ the ring of the golden mean, we can construct $\mathfrak{o}$-superintegral packings (that is, with all bends in $\mathfrak{o}$), and having supergroup the right-angled dodecahedron (which is arithmetic and co-compact). It is an interesting problem to extend our theory to packings with bends in integer rings. (And more generally to complex hyperbolic space, $\mathrm{S}\mathrm{U}\left(n,1\right)$, etc.)

Theorems 24 and 25 follow from our Structure Theorem:

Theorem 28 (Structure Theorem for Crystallographic Packings).

Let $\tilde{\mathcal{C}}$ be a set of walls (that is, spheres), the reflections through which generate a hyperbolic lattice, and orient these walls so that the fundamental domain is the intersection of their exteriors. Assume that $\tilde{\mathcal{C}}$ decomposes into a cluster/cocluster pair:

$$\tilde{\mathcal{C}}=\mathcal{C}⨆\hat{\mathcal{C}}$$ [3]

so that

• •Any pair of spheres in $\mathcal{C}$ is either disjoint or tangent, and
• •Any sphere in $\mathcal{C}$ is either disjoint, tangent, or orthogonal to any in $\hat{\mathcal{C}}$.

Let $\mathrm{\Gamma }\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}⟨\hat{\mathcal{C}}⟩$ be the (thin) group generated by reflections through the cocluster. Then the cluster orbit under this group, $\mathcal{P}\hspace{0.17em}:\hspace{0.17em}=\hspace{0.17em}\mathrm{\Gamma }\cdot \mathcal{C}$, is a crystallographic packing.

Conversely, every crystallographic packing arises in this way.

The converse direction follows from our proof of Lemma 16, and the forward direction uses similar ideas. Hence, answering Question 23 for a given reflection lattice is equivalent to finding a decomposition as in Eq. 3, or proving that one cannot exist.

Remark 29:

In the case of the cocompact forms in Remark 26, we are not yet able, after some effort, to find a reflective subgroup (or prove it does not exist) with a suitable decomposition of the form Eq. 3.

Returning to Theorem 24, our proof of this result relies on the complete classification by Belolipetsky and Mcleod (25) of reflective Bianchi groups. For example, the Bianchi group ${\mathrm{P}\mathrm{S}\mathrm{L}}_2\left(\mathbb{Z}\left[\sqrt{-6}\right]\right)$ is commensurate to a maximal reflection group having the Coxeter diagram illustrated in Fig. 4. (we follow Vinberg’s convention for the labeling, as indicated there.)

Fig. 4.

The Coxeter diagram for the reflective subgroup of the maximal discrete extension of the Bianchi group ${\mathrm{P}\mathrm{S}\mathrm{L}}_2\left(\mathbb{Z}\left[\sqrt{-6}\right]\right)$. Vertices denote reflecting walls, and two vertices are connected by: $\left(i\right)$ a dotted line, if the walls’ interiors are disjoint; $\left(ii\right)$ a thick line, if the walls meet at a cusp; $\left(iii\right)\hspace{0.17em}m-2$ thin lines, if the walls meet at dihedral angle $\pi /m$; or $\left(iv\right)$ no line, if the walls intersect orthogonally.

One realization of the Coxeter diagram in Fig. 4 is given by reflecting walls (circles) illustrated in Fig. 5A, with the same labeling. (The reader may check that the angles of intersection are as claimed in the Coxeter diagram.) The reader may now also verify that the decomposition of labeled walls as:

$$\mathcal{C}=\left\{1\right\},\hat{\mathcal{C}}=\left\{\mathrm{2,3},\dots ,6\right\},$$ [4]

satisfies the assumptions of Theorem 28 and hence gives rise to a crystallographic packing by taking the orbit of $\mathcal{C}$ under the group of reflections through $\hat{\mathcal{C}}$. The resulting packing is shown in Fig. 5B, which is the familiar cuboctahedral packing in disguise. (Compare with Fig. 3.)

Fig. 5.

(A) The reflecting walls in Fig. 4. (B) The packing resulting from the orbit generated by $\hat{\mathcal{C}}$ on the cluster $\mathcal{C}$ in the decomposition Eq. 4.

All but one of the (finite list of) reflective Bianchi groups can be similarly verified via the Structure Theorem to support superintegral packings; see Figs. 1 and 2 in ref. 25. The only case in which the decomposition Eq. 3 is not straightforward from studying this Coxeter data is the Bianchi group on the Eisenstein integers (that is, adjoining the cube root of unity). It turns out, in this case, that the Coxeter diagram in the literature has a minor mistake that can be traced to an early paper of Shaiheev (26); it has propagated in the literature ever since. The issue comes from the execution of Vinberg’s algorithm for reflection subgroups, which, for the Eisenstein integers, has extra stabilizers due to the larger group of units. The true diagram is

and the Eisenstein Bianchi group has a subgroup with Coxeter diagram

This last diagram supports a decomposition as in Eq. 3 by taking either $\mathcal{C}=\left\{1\right\}$ or $\mathcal{C}=\left\{3\right\}$. We are thus finished sketching the only nonimmediate case of Theorem 24.

Remark 30:

In fact, it turns out that all previously known integral circle packings (and many new ones) arise in this way as limit sets of thin subgroups of reflective Bianchi groups.^∥

To prove Theorem 25, we apply the Structure Theorem to (manipulations of) certain other Coxeter diagrams, e.g., Vinberg’s diagrams (27) in dimensions $n+1\le 14$ for the reflective subgroup of the integer orthogonal group preserving the form $-2x_0^2+x_1^2+\cdots +x_{n+1}^2$. In dimensions $n+1=\mathrm{15,16,17}$ and 21, arithmetic reflection groups are known, but currently not crystallographic packings.

Integral but Nonsuperintegral Packings

Let us say more about what happens in Remarks 14 and 21. When $\mathrm{\Pi }$ is the hexagonal pyramid, its supergroup $\tilde{\mathrm{\Gamma }}=\hspace{0.17em}⟨\mathcal{C},\hat{\mathcal{C}}⟩$ can be computed to have (symmetric) Gram matrix (see Eq. 2)

$$\mathcal{G}=\left(\begin{array}{cccccccccccccc}-1& 1& 1& 1& 1& 1& 1& 0& 0& 0& 0& 0& 0& \frac{2}{\sqrt{3}}\\ \cdot & -1& 1& 5& 7& 5& 1& 0& 2\sqrt{3}& 4\sqrt{3}& 4\sqrt{3}& 2\sqrt{3}& 0& 0\\ \cdot & \cdot & -1& 1& 5& 7& 5& 0& 0& 2\sqrt{3}& 4\sqrt{3}& 4\sqrt{3}& 2\sqrt{3}& 0\\ \cdot & \cdot & \cdot & -1& 1& 5& 7& 2\sqrt{3}& 0& 0& 2\sqrt{3}& 4\sqrt{3}& 4\sqrt{3}& 0\\ \cdot & \cdot & \cdot & \cdot & -1& 1& 5& 4\sqrt{3}& 2\sqrt{3}& 0& 0& 2\sqrt{3}& 4\sqrt{3}& 0\\ \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 4\sqrt{3}& 4\sqrt{3}& 2\sqrt{3}& 0& 0& 2\sqrt{3}& 0\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 2\sqrt{3}& 4\sqrt{3}& 4\sqrt{3}& 2\sqrt{3}& 0& 0& 0\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 5& 7& 5& 1& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 5& 7& 5& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 5& 7& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 5& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1& 1\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & -1\end{array}\right).$$ [5]

Vinberg’s Arithmeticity Criterion (28) (see also ref. 29, Theorem 3.1) says in this context that $\tilde{\mathrm{\Gamma }}$ is arithmetic if and only if cyclic products of $2\mathcal{G}$ are always integers. This is almost the case for Eq. 5, except for the entry $\frac{2}{\sqrt{3}}$ in the top right; hence $\tilde{\mathrm{\Gamma }}$ is nonarithmetic (see Lemma 20). However, it is nearly so; indeed, $\tilde{\mathrm{\Gamma }}$, viewed as a subgroup of $O_Q\left(\mathbb{R}\right)$ (see Eq. 1), can be conjugated to lie in $O_Q\left(\mathbb{Z}\left[\frac{1}{3}\right]\right)$ with unbounded denominators in its entries. The latter group is a perfectly nice $S$-arithmetic lattice in the product $O_Q\left(\mathbb{R}\right)\times O_Q\left({\mathbb{Q}}_3\right)$, but $\tilde{\mathrm{\Gamma }}$ is already a lattice on projection to the first factor, $O_Q\left(\mathbb{R}\right)$. This too implies that $\tilde{\mathrm{\Gamma }}$ is nonarithmetic and, in this sense, is reminiscent of constructions of nonarithmetic groups by Deligne and Mostow (30). It is interesting to understand whether all integral but nonsuperintegral packings arise this way.

Local–Global Principles

We conclude with a discussion of whether local–global principles hold for bends of crystallographic circle ($n=2$) packings. (For higher-dimensional sphere packings, this problem becomes easier; see, e.g., ref. 31.) As explained in ref. 1 for the case of the classical Apollonian packing, the “asymptotic” local–global principle is proved in ref. 32. This method was extended in the thesis of Zhang (33) to show the same statement for packings modeled on the octahedron. Most recently, Fuchs, Stage, and Zhang (34) showed that the Bourgain–Kontorovich method extends to the following context:

Theorem 31.

Let $\mathcal{P}$ be a packing with symmetry group $\mathrm{\Gamma }$ and let $C\in \mathcal{P}$. Assume that there is a circle $C\prime \in \mathcal{P}$ tangent to $C$ so that the stabilizer of $C\prime$ in $\mathrm{\Gamma }$ is a congruence (Fuchsian) group. Then the orbit $\mathrm{\Gamma }\cdot C$ satisfies an asymptotic local–global principle.

The assumption of the existence of such a companion circle $C\prime$ is a generalization of Sarnak’s observation (35) in the classical Apollonian case that such leads to certain shifted binary quadratic forms representing bends in the orbit. We show that this condition is both satisfied and not satisfied infinitely often.

Theorem 32.

The assumptions (and hence conclusions) of Theorem 31 are satisfied for infinitely many conformally inequivalent superintegral crystallographic packings. The same statement holds with “are satisfied” replaced by “are not satisfied.”

Thus, even the asymptotic local–global problem remains open in this generality. Lacking evidence against the local–global principle in all of these examples, we conjecture that it does indeed hold.