AA GAME OF PACKINGS
ARTHUR BARAGAR AND DANIEL LAUTZENHEISER
Abstract.
In this note, we investigate an infinite one parameter family ofcircle packings, each with a set of three mutually tangent circles. We use theseto generate an infinite set of circle packings with the
Apollonian property .That is, every circle in the packing is a member of a cluster of four mutuallytangent circles.
Let us play a game. Suppose we are given a circle, and inside that circle we havethree mutually tangent discs, two of which are tangent to the outside circle. Thechallenge is to fill in the remaining space with discs in a logical and symmetric way,as is done in Figure 1. The rules are vague, but will become clearer as we learnmore about the game. The initial setup has, modulo inversion, a one dimensionaldegree of freedom. To see this, we invert in the point of tangency of the two discsthat are tangent to the original circle. This gives us the strip version in Figure 2.In the top picture (of Figure 2), we let d be the ratio of the distance between thecenters of the two circles and the diameter of the circles. The game can be wonif d = √ n for n a positive integer (see Theorem 2.1 below). Our winning strategyfor n = 1 generates the Apollonian circle packing; for n = 2 we get a packingthat is described in [Boy74, (2.3) on p. 394], and appears in [MH91, Figure 3] and[GM10, Figure 3]; and for n = 3 we get, modulo inversion, the cross section of theSoddy sphere packing that appears in [Sod37, Figure 1]. The case n = 6 appearsin [Bar18, Figure 7] and [KN19, Figure 3]. The cases n = 10 and 14 show up in asimilar game studied by [CCS19]. We have not seen the other packings in print. Mathematics Subject Classification.
Key words and phrases.
Apollonius, Apollonian, circle packing, sphere packing, K3 surface,ample cone, lattice.L A TEXed June 2, 2020.
Figure 1.
The challenge: Given an initial configuration, fill in theremaining space so as to have a nice circle packing. a r X i v : . [ m a t h . M G ] M a y ARTHUR BARAGAR AND DANIEL LAUTZENHEISER
Figure 2.
The same configurations as in Figure 1, after invertingin a point of tangency. The packing is the image of a solid lineunder the action of the group generated by the inversions andreflections represented by the dotted lines. In this case d = 1 + √ Figure 3.
Two circle packings that have the property that everycircle is a member of a cluster of four mutually tangent circles.While Theorem 2.1 tells us that the game can be won for infinitely many d , itdoes not give us a complete strategy. Solving the problem for successive integers n = d is a combination of number theory and geometry, and has a flavor similar tothe work of Bianchi [Bia92]. Once solved, a geometric argument allows us to playa new game: Suppose we have four mutually tangent circles. Fill in the remainingspace in a nice way. Contrary to what we had imagined, the Apollonian packing isnot the only solution. See for example the packings in Figure 3. The first is a blendof the Apollonian packing with Boyd’s Example (2.3) [Boy74], and the second isa blend with a cross section of the Soddy packing. These packings have what wecall the Apollonian property : Every circle is a member of a cluster of four mutuallytangent circles. This geometric blending can be done in many ways.As in previous works, we think of circles as representing planes in the Poincar´eupper half-space model of H , a view that dates back to [Max82]. Packings thatare equivalent under inversion are thought of as different perspectives of the sameinfinite sided ideal polyhedron in H . Each perspective depends on the choice ofpoint for the point at infinity. GAME OF PACKINGS 3
There are probably many ways of winning this game. Our approach in Section 2is lattice based and is inspired by results in arithmetic geometry. For the circlepackings guaranteed by Theorem 2.1, there is a perspective so that all circles haveinteger curvature. After blending, though, this property is usually lost.
Acknowledgements.
This material is based upon work supported by the Na-tional Science Foundation under Grant No. DMS-1439786 and the Alfred P. SloanFoundation award G-2019-11406 while the first author was in residence at andthe second author was visiting the Institute for Computational and ExperimentalResearch in Mathematics in Providence, RI, during the Illustrating Mathematicsprogram. The first author wishes to thank his home institution, UNLV, for its sab-batical assistance during the Fall of 2019. The figures in this paper were producedusing McMullen’s Kleinian groups program [McM].1.
Background
The vector model of hyperbolic geometry.
Circles in R can be repre-sented by 4-dimensional vectors, an observation that dates back to Clifford andDarboux [Boy73, Boy74]. The more modern interpretation is that they representplanes in H imbedded in a 4-dimensional Lorentz space, which in turn representcircles (or lines) on the boundary of the Poincar´e upper half-space model of H .Given a symmetric matrix J with signature (1 , R , to be the set of 4-tuples over R equipped with the negative Lorentz product u · v = u T J v . The surface x · x = 1 is a hyperboloid of two sheets. Let us distinguish a vector D with D · D > H by: H : x · x = 1 , x · D > . We define a distance on H by cosh( | AB | ) = A · B. Then H equipped with this metric is a model of H , sometimes known as the vectormodel . Equivalently, one can define V = { x ∈ R , : x · x > } and H = V / R ∗ , together with the metric defined bycosh( | AB | ) = | A · B || A || B | , where | x | = √ x · x for x ∈ V . For x · x <
0, we define | x | = i √− x · x .Planes in H are the intersection of H with hyperplanes n · x = 0 in R , . Sucha hyperplane intersects H if and only if n · n <
0. Let H n represent both thehyperplane n · x = 0 in R , and its intersection with H . The direction of n distinguishes a half space H + n = { x : n · x ≥ } , in either R , or H .The angle θ between two intersecting planes H n and H m in H is given by(1) | n || m | cos θ = n · m , ARTHUR BARAGAR AND DANIEL LAUTZENHEISER where θ is the angle in H + n ∩ H + m . If | n · m | = || n || m || , then the planes are tangentat infinity. If | n · m | > || n || m || , then the planes do not intersect, and the quantity ψ in | n || m | cosh ψ = | n · m | is the shortest hyperbolic distance between the twoplanes.The group of isometries of H is given by O + ( R ) = { T ∈ M × : T u · T v = u · v for all u , v ∈ R , , and T H = H} . Reflection in the plane H n is given by(2) R n ( x ) = x − n ( x ) = x − n · xn · n n . The group of isometries is generated by the reflections.Let ∂ H represent the boundary of H , which is a 3-sphere. It is represented by L + / R + where L + = { x ∈ R , : x · x = 0 , x · D > } . Given an E ∈ L + , let ∂ H E = ∂ H \ E R + . Then ∂ H E equipped with the metric | · | E defined by(3) | AB | E = δ A · B ( A · E )( B · E )is the Euclidean plane that is the boundary of the Poincar´e upper half-space modelof H with E the point at infinity. The quantity δ is an arbitrary scaling constant.In ∂ H E , the plane H n is represented by a circle (or line), which we denote with H n ,E (or just H n if E is understood, or sometimes just n ).The curvature (the inverse of the radius, together with a sign) of H n ,E is givenby the formula(4) n · E √ δ || n || using the metric | · | E [Bar18]. Here, || n || = − i | n | = √− n · n . By choosing asuitable orientation for n , we get the appropriate sign for the curvature.1.2. Circle packings.
The following definitions are due to or inspired by Maxwell[Max82]. We say
P ⊂ R , is a packing if for all n , n (cid:48) ∈ P , there exists a k > n · n = − k and n · n (cid:48) ≥ k . The first condition guarantees that H n is a planein H , and hence represents a circle on ∂H E . The second condition guarantees thatdistinct circles are either tangent or do not intersect. (Note that if n · n (cid:48) < − k , thenthe circles do not intersect, but their associated sides overlap.) We ignore trivialpackings of the form { n , − n } .A packing P is maximal if we cannot add a vector to P and have it still be apacking. Geometrically, this means there is no space left over where one can placeanother circle (so we sometimes use the term dense ).Given a point E ∈ ∂ H , the set P E = { H n ,E : n ∈ P} is what is traditionally thought of as a circle packing. We call P E a perspective of P . The packings in Figures 1 and 2 are different perspectives of the same packing.A packing is lattice like if P Z is a lattice in R , . GAME OF PACKINGS 5
Lattice based circle packings.
SupposeΛ = e Z ⊕ e Z ⊕ e Z ⊕ e Z ⊂ R , and J Λ = [ e i · e j ]has integer entries. Let us fix k (usually 1 or 2) so that there exists at least one n ∈ Λ such that n · n = − k ; and a D so that D · D > D · n (cid:54) = 0 for all n ∈ Λwith n · n = − k . Such a D exists. Let us define E − k = { n ∈ Λ : n · n = − k, n · D > } . Many of the discs in E − k overlap, even if they do not intersect. The set is analogousto the Apollonian super packing [GLM + K − k = (cid:92) n ∈E − k H + n and defining the set E ∗− k = { n ∈ E − k : H n is a face of K − k } . If for all n , n (cid:48) ∈ E ∗− k , n · n (cid:48) ≥ k , then E ∗− k is a packing. If E ∈ Λ, then, witha suitable choice for δ , every circle in the perspective E ∗− k,E has integer curvature(see Equation (4)).Let O +Λ = { T ∈ O + ( R ) : T Λ = Λ } . Note that O +Λ is an arithmetic group, so it has a convex polyhedral fundamentaldomain with a finite number of faces and a finite volume. As a consequence, if E ∗− k is a packing, then the packing is maximal. We sketch the proof: Let n satisfy n · n = − k . Note that the image of n under the action of O +Λ is dense on ∂ H , sincethe fundamental domain for O +Λ has finite volume. If E ∗− k is not maximal, thenthere exists an m so that m · m = − k , m · D >
0, and H − m ⊂ K − k . By density,there exists an n ∈ E − k so that H − n ⊂ H − m . But K − k ⊂ H + n , a contradiction. Remark 1.
For a K3 surface X with Picard number ρ ≥
4, let Λ = Pic( X ) andlet D be ample. Then E − is the set of effective − X , and E ∗− is theset of irreducible − X . The cone K − is the ample cone for X .In general, K − does not yield a packing, as it may have edges. However, usinga result of Morrison [Mor84], there exist plenty of K3 surfaces where this does nothappen, including the infinite set described in Theorem 2.1.2. Playing the game
In Figure 2, we labeled the circles in our initial configuration with e , ..., e . Letus set e i · e i = −
2, so e i · e j = 2 if i (cid:54) = j and the circles are tangent (see Equation(1)). Let e · e = a , which depends on our variable d . Let J d = [ e i · e j ] = − − − a a − . ARTHUR BARAGAR AND DANIEL LAUTZENHEISER
The point of tangency between tangent circles e i and e j is e i + e j , so our point atinfinity in Figure 2 is E = e + e . Let P and P be the centers of the circles e and e , so P i = R e i ( E ) = E + 4 e i . Using Equations (4) and (3), H e ,E has diameter δ/ | P P | E = ( δ/ √ a ,so a = 4 d − . Theorem 2.1.
Let n be a positive integer and d = √ n . Let Λ n be the latticegenerated by { e , ..., e } with [ e i · e j ] = J n . Then E ∗− is a maximal circle packing.Proof. Let n , n (cid:48) ∈ E ∗− . Then n · n (cid:48) is an even integer, so we need only show n · n (cid:48) (cid:54) = 0. Suppose n · n (cid:48) = 0. Then( n − n (cid:48) ) · ( n − n (cid:48) ) = − . But Λ n does not represent − x ∈ Λ n so that x · x = − (cid:3) This theorem tells us that the game can be won if d = √ n for n a positive integer,but it does not tell us how to win the game. The strategy is to find generators for O +Λ n . Some of the reflections shown in Figure 2 are in this group for all n . Thereflection R h where h = e − e = [ − , , ,
0] is the symmetry that swaps the firstand second components (in this basis). Similarly, R v for v = e − e = [0 , , − , R s where s = e − e = [0 , − , , O +Λ n . To solve for v , we note that v · e i = 0 for i = 1, 2 and 3, so v = [ n, n, , − v · v = − n , the reflection R v is nevertheless in O +Λ n for all n . Note that R e is also in O +Λ n , so the group has a convex fundamental domain F n that includes E and is bounded by the faces H e , H h , H v , H v , and H s . For n = 1, 2, and3, this region has finite volume, so the group (cid:104) R e , R h , R v , R v , R s (cid:105) has finiteindex in O +Λ n . The groups are in fact equal (for n = 1 , , O +Λ n . This is perhaps easier todemonstrate with an example, but first some preliminaries: Let us set δ = 4, so H e ,E has unit radius, and the distance | P P | E between the centers is 2 √ n . Letus set D = e + e + e + e , which is a point a Euclidean distance √ n + 1 abovethe point Q on ∂H E in the Poincar´e model (see Figure 4), so n · D (cid:54) = 0 for all n ∈ E − (by Equation (4) and Theorem 2.1). Example (The case n = 5) . We first find O +Λ . We have a head start on thefundamental domain F , which is shown in Figure 4. We need more symmetries ofthe lattice. We begin by looking for reflections and taking them to be perpendicularto the faces we already have. Natural candidates include circles centered at the pointlabeled Q (cid:48) . The corresponding planes have normal vector a linear combination of Q (cid:48) and E . We find n = [ − , − , , Q (cid:48) and fromEquation (4), it has radius 1.It is natural then to assume that Q might be a cusp, which would suggest thatthere is a reflection in a plane that is parallel to v . Such planes have normal vectors GAME OF PACKINGS 7
Figure 4.
The fundamental domain for O +Λ .that are linear combinations of v and Q . We find s = [ − , − , , O +Λ . The circle it generates has curvature √
5, is tangent to v at Q so has center on h , and using Equation (3) (applied to its center R s ( E )) weverify that it is the one shown in Figure 4.The resulting region has finite volume, so is a candidate for the fundamentaldomain of O +Λ . Our region has two cusps: E and Q . We choose our normalvectors to be primitive , meaning they have coprime coordinates. The faces at E include one with normal vector e , which has norm e · e = −
2, while the facesat Q have normal vectors with norms − − −
10, and −
40. Thus, there cannotbe a symmetry that sends Q to E . If O +Λ has a symmetry not generated by thefaces of our region, then there must exist a symmetry of this region, and since itfixes E , it must be a Euclidean symmetry of Figure 4. Clearly no such symmetryexists, so our region is a fundamental domain for O +Λ , and hence O +Λ = (cid:104) R e , R h , R v , R v , R s , R s , R s (cid:105) . Thus, E − = O +Λ ( e ), since e is the only face with norm −
2. (Note that, if avector n ∈ Λ n has norm − O +Λ n ,then it must be a face of the fundamental domain, since R n ∈ O +Λ n .) We setΓ = (cid:104) R h , R v , R v , R s , R s , R s (cid:105) (we removed R e ). Then E ∗− = Γ ( e ).The reflections R s and R s generalize to all odd n . Let Q (cid:48) = [4 − n, − n, , s = ( Q (cid:48) − E ) /
2. Then R s ∈ O +Λ n and on ∂ H E , is inversion in a circle ofradius 1 centered at Q (cid:48) . We use Q = [ − n , − n , ,
1] to generate s = nQ − v .Then R s ∈ O +Λ n and on ∂ H E is inversion in a circle with curvature √ n that istangent to v at Q .Something similar works for even n as well. In this case Q = [1 − n, − n, , Q (cid:48) = [2 − n/ , − n/ , , n ≡ s = Q (cid:48) − E , whichhas R s ∈ O +Λ n and on ∂ H E is inversion in a circle of radius √ Q (cid:48) .(This finishes the case n = 6.) For n ≡ s = ( Q − E ) /
2. Then R s ∈ O +Λ n , and on ∂ H E is inversion in a circle of radius one centered at Q . (Thisfinishes the case n = 4.) Then we can get s = ( n/ Q (cid:48) − v , which generates areflection in O +Λ n corresponding to a circle of curvature (cid:112) n/
2. However, sometimes
ARTHUR BARAGAR AND DANIEL LAUTZENHEISER
Figure 5.
A fundamental domain for a subgroup of O +Λ and itsinversion in a circle centered at Q .one can do better, as in the case n = 12, where we can let s = (3 Q − v ) /
2, whichcorresponds to a circle with curvature √ / Example (The case n = 7) . We proceed as described above to find s = [ − , − , , s = [ − , − , ,
6] (see Figure 5). We find a couple more reflections that fillin the remaining gap: s = [ − , − , ,
15] and s = [ − , − , , s · s = −
2, which suggests that the cusps at P and E might be symmetric. Weinvert in the point Q and get the second region in Figure 5, which appears to haverotational symmetry about the center Q = R s ( Q ) of s (in ∂ H Q ). A rotationby π about a line with endpoints A and B in ∂ H has the equation φ A,B ( x ) = 2(( A · x ) B + ( B · x ) A ) A · B − x . The rotation φ Q ,Q is in O +Λ , and E ∗− = Γ ( e ) whereΓ = (cid:104) R h , R v , R v , R s , φ Q ,Q (cid:105) On ∂ H E , the map φ A,B is represented by the M¨obius transformation(5) γ = (cid:20) ˜ A + ˜ B − A ˜ B − ( ˜ A + ˜ B ) (cid:21) , where ˜ A and ˜ B are the complex numbers that represent A and B in ∂ H E ∼ C .Another useful formula gives the distance x from a point A to a line H n (so n · E = 0): x = δA · n √− n · n A · E .
So, for example, if we think of e as the x -axis and v as the y -axis, then thecoordinates ( A x , A y ) of a point A ∈ ∂ H E is( A x , A y ) = (cid:18) A · v √ nA · E , A · e A · E (cid:19) . A plane with normal vector n is represented by a circle with radius given by Equa-tion (4) and center (cid:18) n · v √ n n · E , n · e n · E (cid:19) . GAME OF PACKINGS 9
Figure 6.
The fundamental domain for O Λ . The dotted circlerepresents a map that is inversion in that circle composed withrotation by π about its center.The point Q generalizes for n ≡ Q = [ − n + 25 , − n + 9 , n + 10 , n − φ Q ,Q ∈ O +Λ n . Geomet-rically, Q is the center of the rectangle formed by h , v , s and s when Q is thepoint at infinity. Example (The case n = 10) . The group O Λ is generated by the usual reflections,the reflection n = [ − , − , , − R P ( x ) = 2 P · x P · P P − x for P = [6 , , − , − P · P = 16, so P is a point in the hyperbolicspace. This map is the − H centered at P . As an action on ∂ H E , it is thecomposition of inversion in a circle and rotation by π about its center. The centeris − R P ( E ), and Equation (4) gives i times the circle’s curvature (see Figure 6). Example (The case n = 11) . Every good game has its boss levels, and 11 is oneof them. Chasing reflections is a never ending pursuit that leads one into the cusp Q = [ − , − , ,
1] (see Figure 7 and also Equation (6)). Inverting in this pointreveals the rotational symmetry about Q ; Q = R h ( Q ) = [ − , − , , Q = [ − , − , , h + s and Q . These threerotations, together with the usual reflections, generate O +Λ . Example (The case n = 15) . There is a point P = [ − , − , , ∈ H such that − R P ∈ O +Λ , but since P · e = 0, it is not a symmetry of E ∗− . We compose with R e to get a rotation by π whose endpoints are the irrational points Q and Q inTable 1. Together with φ Q ,Q (see Equation (6)) and the usual reflections, thesegenerate the packing. Example (The case n = 21) . This is the first group that includes a glide transla-tion: T = − − − − − − − − . McMullen’s code does not allow for a map like this. This packing can be generated by asubgroup of index two that is generated by just reflections, though we also edited his code.
Figure 7.
A never ending set of reflections that generate thetiling for n = 11, converging on the cusp Q = [ − , − , ,
1] (left).On the right, the same image inverted in Q , illuminating therotational symmetry about Q , Q and Q . Figure 8.
The tiling with n = 21 and the eigenvector A atinfinity. The map T reflects the central circle along a vertical axisand dilates it by a factor of λ , giving the large circle e boundingthe picture on the left. The figure on the right represents variousreflective symmetries of the packing.This was found by finding a cusp similar to Q and guessing that there shouldbe a symmetry that sends one to the other. The group is generated by the usualreflections including R s and R s , the map T , and the reflection R s where s =[ − , − , , T are 1, − λ = 9 + 4 √
5, and λ − . Let A bethe eigenvector associated to λ . A perspective with A the point at infinity is shownin Figure 8.A hyperbolic translation has an eigenvalue λ >
1, and the rest are λ − and 1with multiplicity 2. Let A and B be the eigenvectors associated to λ and λ − , GAME OF PACKINGS 11
Figure 9.
A blend of the n = 7 packing and the Apollonian packingrespectively. Then the corresponding map on C is τ λ,A,B ( z ) = ( λ ˜ A − ˜ B ) z + (1 − λ ) ˜ A ˜ B ( λ − z + ˜ A − λ ˜ B .
For the above case ( n = 21), rather than come up with a representation for theglide reflection T , let us note that the eigenvector associated to 1 is s , so thecomposition S = R s ◦ T is an orientation preserving map that is in our group. Itis the composition of a hyperbolic translation with rotation by π about its line oftranslation, and is therefore represented by σ ( z ) = − τ − λ,A,B ( z ).3. Gluing and slicing
Let us play a new game.
Find a maximal circle packing with the propertythat every circle in the packing is a member of a cluster of four mutually tangentcircles. Of course, the Apollonian circle packing satisfies this property, so thechallenge is to be different.Let us begin, for example, with the fundamental domain for n = 7 shown inFigure 6, and let us move v to the left one unit (let us call that new line v (cid:48) ). Thisgives us a new group Γ (cid:48) = (cid:104) R v (cid:48) , R h , R v , R s , φ Q ,Q (cid:105) and Γ (cid:48) ( e ) is the packingshown in Figure 9. This packing clearly has a cluster of four mutually tangentcircles, and since Γ (cid:48) acts transitively on the packing (it is the orbit of a singleelement), every circle is a member of a cluster of four mutually tangent circles. Wesay that the packing has the Apollonian property .We should think of this as gluing two compatible fundamental domains together.(A similar process in described in [CCS19].) To the left of the fundamental domainfor n = 7, we glued along the plane H v a reflected version of our fundamental do-main for the Apollonian packing. Since the faces that intersect H v are compatible,the new fundamental domain generates a circle packing. We can do this for any n ,giving us the following result: Theorem 3.1.
There exists an infinite number of maximal circle packings with theApollonian property.Proof.
There is one minor detail we should address: For a given n , we do not knowwhether Γ n acts transitively on E ∗− , nor whether this is the case for infinitely many n . However, suppose the fundamental domain F n has a face other than e whosenorm is also −
2. Then the reflection through that face is in O +Λ n and hence byadding that reflection to Γ n , we can get a different subset of E − that is a maximalcircle packing. By doing this for all faces of F n except for e , we get a group thatacts transitively on the new packing. Thus, when we replace R v with R v (cid:48) for thismodified packing, we get a packing with the Apollonian property. (cid:3) Figure 10.
A blend of the n = 5 and n = 7 packings, glued onthe face v , and on the face v . Figure 11.
The fundamental domain for O +Λ / , the set E ∗− , anda packing that is a subset of E − .We can glue on the face v as well, when the two fundamental domains arecompatible. For example, in Figure 10, we have glued the fundamental domains for n = 5 and n = 7 together in two different ways, giving us two different packings. Remark 2.
Gluing is a geometric process, so it is no surprise that the integralcurvature property is lost when two fundamental domains are glued together. Butnot always, so let us suggest a new rule/game: Find an infinite set of maximalcircle packings that have both the Apollonian property and the integral curvatureproperty.3.2.
Closing the gap?
Is there a strategy to create infinitely many packings wherethe gap d is between 1 and √
2? For example, if we let n = 3 /
2, then J n still hasinteger entries so we can investigate E ∗− . A fundamental domain for O +Λ / is shownin Figure 11, and because it includes a face that is not perpendicular to H e , theset E ∗− is not a packing. However, if we reflect our fundamental domain acrossthat face and glue the two domains together, we get a subgroup of index two in thesymmetries of E ∗− that generates a packing.We have not played this game long enough to know if there is a strategy thatgives us packings where the gap d converges to 1.3.3. Filling in ghost circles.
When a packing has a ghost circle (e.g. n = 7 inFigure 13), we can fill it in or reflect across it (see Figure 12). This correspondsto slicing off a portion of the fundamental domain and including the resulting new GAME OF PACKINGS 13
Figure 12.
Two variations on the n = 7 packing.face as a member of the new packing, or reflecting across it. In Figure 5 (right),this corresponds to cutting the region in half with a line through Q perpendicularto h . 4. Appendix
In Figures 13, 14 and 15, we present strip versions of the circle packings for n = 1through 26.In Tables 1, 2 and 3 we give the generators for Γ n for n ≤
26. For all n , the setof generators for Γ n include the first four generators in Table 1, namely R h , R v , R v , and R s . The packing is Γ n ( e ). Though Γ n acts transitively on the packingfor n ≤
26, we have no reason to believe that it does so in general.In Tables 1, 2 and 3, the types are:(1) Reflections R n where n · n <
0, which in C correspond to inversion in acircle centered at the given point and with the given radius r .(2) Rotations φ A,B by π about a line in H with endpoints A and B , where A · A = B · B = 0; see Equation (5) for the corresponding M¨obius map.(3) Inversion − R P through a point P in H (so P · P > C isinversion in the given circle composed with rotation by π about its center.The coordinates in C correspond to the choice of e as the real axis, v as theimaginary axis, and P as the point i . References [Bar18] Arthur Baragar,
Higher dimensional Apollonian packings, revisited , Geom. Dedicata (2018), 137–161, DOI 10.1007/s10711-017-0280-7. MR3820499[Bia92] Luigi Bianchi,
Sui gruppi di sostituzioni lineari con coefficienti appartenentia corpi quadratici immaginarˆı , Math. Ann. (1892), no. 3, 332–412, DOI10.1007/BF01443558 (Italian). MR1510727[Boy73] David W. Boyd, The osculatory packing of a three dimensional sphere , Canadian J.Math. (1973), 303–322, DOI 10.4153/CJM-1973-030-5. MR320897[Boy74] , A new class of infinite sphere packings , Pacific J. Math. (1974), 383–398.MR350626[CCS19] Debra Chait, Alisa Cui, and Zachary Stier, A taxonomy of crystallographic spherepackings (2019), to appear, available at arXiv:1903.03563v1 . Figure 13.
The packings for n = 1 through 9. GAME OF PACKINGS 15
Figure 14.
The packings for n = 10 through 18. n Type in Λ n in C All n R h h = [ − , , , y = 1 R v v = [ n, n, , − x = 0 R v v = [0 , , − , x = √ nR s s = [0 , − , ,
0] 0, r = 24 R s s = [ − , − , ,
1] 2 + i , r = 15 R s s = [ − , − , , √ r = 1 R s s = [ − , − , , √ + i , r = √ R s s = [ − , − , , √ r = √ R s s = [ − , − , , √ r = 1 φ Q ,Q Q = [ − , − , , √ iQ = [ − , − , , √ i R s s = [ − , − , ,
1] 2 √ i , r = 1 R s s = [ − , − , , √ , r = √ R s s = [ − , − , , √ i , r = √ R s s = [ − , − , ,
1] 3, r = 1 R s s = [ − , − , − , − + i , r = R s s = [ − , − , ,
1] 2 + i , r = R s s = [ − , − , , √ r = √ − R P P = [6 , , − , − √ + i , r = √ φ Q ,Q Q = [ − , − , , √ i Q = [ − , − , , √
11 + iφ Q ,Q Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ i R s s = [ − , − , , √
12 + i , r = 1 R s s = [ − , − , , √ , r = √ R s s = [ − , − , , √ r = 1 R s s = [ − , − , , √ i , r = √ − R P P = [ − , − , , √ i , r = R s s = [ − , − , , √ r = √ R s s = [ − , − , , √ , r = √ R s s = [ − , − , , √ + i , r = √ R s s = [ − , − , , √ + i , r = √ φ Q ,Q Q = [ − , − , , √
15 + iQ = [ − , − , , √ i φ Q ,Q Q , Q = [ − ± √ , − , , −√ − (3 ± √ i R s s = [ − , − , ,
1] 4 + i , r = 1 R s s = [ − , − , , r = R s s = [ − , − , , + i , r = − R P P = [ − , − , , i , r = Table 1.
The generators of Γ n . GAME OF PACKINGS 17 n Type in Λ n in C R s s = [ − , − , , √ r = 1 R s s = [ − , − , , √ + i , r = √ φ Q ,Q Q = [ − , − , , √ i Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ − i φ Q ,Q Q = [ − , − , , √ i R s s = [ − , − , , √ r = √ R s s = [ − , − , , √ , r = √ φ Q ,Q Q = [ −
12 + √ , − − √ , , √ −√ i Q = [ − − √ , −
12 + √ , , √ √ i − R P P = [ − , − , , √ i , r = φ Q ,Q Q = [ − , − , , √ i Q = [ − , − , , √
19 + iφ Q ,Q Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ i R s s = [ − , − , , √
20 + i , r = 1 R s s = [ − , − , , √ , r = √ R s s = [ − , − , , √ r = 1 R s s = [ − , − , , √ + i , r = R s s = [ − , − , , √ + i , r = √ R s s = [ − , − , , √ + i , r = √ R s [ − , − , , √ r = 1 R s [ − , − , , √ , r = √ R s [ − , − , , √ , r = √ (see text) T σ R s s = [ − , − , , √ r = √ − R P P = [ − , − , , √ i , r = φ Q ,Q Q , Q = [ − ± √ , − , , √ ± i √ φ Q ,Q Q = [ − , − , , √
23 + iQ = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ i Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ − i φ Q ,Q Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ − i Table 2.
The generators of Γ n , continued. Figure 15.
The packings for n = 19 through 26. [GLM +
06] Ronald L. Graham, Jeffrey C. Lagarias, Colin L. Mallows, Allan R. Wilks, andCatherine H. Yan,
Apollonian circle packings: geometry and group theory. II. Super-Apollonian group and integral packings , Discrete Comput. Geom. (2006), no. 1,1–36, DOI 10.1007/s00454-005-1195-x. MR2183489 GAME OF PACKINGS 19 n Type in Λ n in C R s s = [ − , − , , √
24 + i , r = 1 R s s = [ − , − , , √ , r = √ R s s = [ − , − , , √ + i , r = √ φ Q ,Q Q , Q = [ − ± √ , − , , √ ± √ i φ Q ,Q Q , Q = [ − ± √ , − , ± √ , √ ∓ √ ± √ i R s s = [ − , − , ,
1] 5, r = 1 R s s = [ − , − , , + i , r = − R P P = [ − , − , , i , r = − R P P = [ − , − , ,
2] 4 + i , r = − R P P = [ − , − , ,
1] 2 + i , r = R s s = [ − , − , , √ r = √ φ Q ,Q Q = [ − , − , , √ i Q = [ − , − , , √ i φ Q ,Q Q = [ − , − , , √ − i φ Q ,Q Q = [ − , − , , √ i φ Q ,Q Q , Q = [ − ∓ √ , − ± √ , , √ ±√ i Table 3.
The generators of Γ n , continued. [GM10] Gerhard Guettler and Colin Mallows, A generalization of Apollonian packingof circles , J. Comb. (2010), no. 1, [ISSN 1097-959X on cover], 1–27, DOI10.4310/JOC.2010.v1.n1.a1. MR2675919[KN19] Alex Kontorovich and Kei Nakamura, Geometry and arithmetic of crystallographicsphere packings , Proc. Natl. Acad. Sci. USA (2019), no. 2, 436–441, DOI10.1073/pnas.1721104116. MR3904690[MH91] S. S. Manna and H. J. Herrmann,
Precise determination of the fractal dimensions ofApollonian packing and space-filling bearings , J. Phys. A (1991), no. 9, L481–L490.MR1117857[Max82] George Maxwell, Sphere packings and hyperbolic reflection groups , J. Algebra (1982), no. 1, 78–97, DOI 10.1016/0021-8693(82)90318-0. MR679972[McM] Curtis T. McMullen, Kleinian groups , http://people.math.harvard.edu/~ctm/programs/index.html .[Mor84] D. R. Morrison, On K surfaces with large Picard number , Invent. Math. (1984),no. 1, 105–121, DOI 10.1007/BF01403093. MR728142[Sod37] Frederick Soddy, The bowl of integers and the hexlet , Nature (January 9, 1937),77–79, DOI 10.1038/139077a0.
Department of Mathematical Sciences, University of Nevada, Las Vegas, NV 89154-4020
E-mail address : [email protected] Eastern Sierra College Center, 4090 W. Line Street, Bishop, CA 93514-7306
E-mail address ::