A proof of the Koebe-Andre'ev-Thurston theorem via flow from tangency packings
AA proof of the Koebe-Andre’ev-Thurston theorem via flowfrom tangency packings
John C. BowersDepartment of Computer ScienceJames Madison UniversityHarrisonburg, VA 22807, USA [email protected]
July 7, 2020
Abstract
Recently, Connelly and Gortler gave a novel proof of the circle packing theorem for tangencypackings by introducing a hybrid combinatorial-geometric operation, flip-and-flow, that allowstwo tangency packings whose contact graphs differ by a combinatorial edge flip to be continu-ously deformed from one to the other while maintaining tangencies across all of their commonedges. Starting from a canonical tangency circle packing with the desired number of circles afinite sequence of flip-and-flow operations may be applied to obtain a circle packing for anydesired (proper) contact graph with the same number of circles.In this paper, we extend the Connelly-Gortler method to allow circles to overlap by anglesup to π/
2. As a result, we obtain a new proof of the general Koebe-Andre’ev-Thurston the-orem for disk packings on S with overlaps and a numerical algorithm for computing them.Our development makes use of the correspondence between circles and disks on S and hyper-planes and half-spaces in the 4-dimensional Minkowski spacetime R , , which we illuminate ina preliminary section. Using this view we generalize a notion of convexity of circle polyhedrathat has recently been used to prove the global rigidity of certain circle packings. Finally, weuse this view to show that all convex circle polyhedra are infinitesimally rigid, generalizing arecent related result. The celebrated Koebe-Andre’ev-Thurston theorem (KAT) states that, given a triangulation T ofa topological sphere together with an assignment of desired overlap angles in [0 , π/
2] for eachedge of the triangulation (subject to a few mild conditions), there exists a pattern of disks on theunit sphere S , one disk for each vertex, such that the angle of overlap between two circles thatcorrespond to an edge matches the desired overlap angle for that edge. Furthermore, connectingcenters of neighboring disks along geodesic shortest paths produces a geodesic triangulation of S that is isomorphic with T , and, moreover, the disk pattern obtained by the theorem is unique up1 a r X i v : . [ m a t h . M G ] J u l o M¨obius transformations of S to itself. When all overlap angles are 0 then neighboring disks aretangent and the restricted version of this theorem is called the circle packing theorem.Numerous proofs of the circle packing theorem have appeared in the literature, for example[15, 24, 17, 3, 14, 7, 22]. Fewer proofs of the more general KAT theorem have appeared. Andre’evhad a proof [1, 2] containing errors that was later fixed by Roeder, Hubbard, and Dunbar [20].Bowers and Stephenson proved the result as a special case of their work on branched packings [9].There are also several generalizations, for instance packings with deep overlap angles of more than π/ inversive distances .When two circles overlap, their inversive distance is the cosine of their angle of overlap, but isalso defined when they do not. In this full generality much less is known. The existence of circlepatterns realizing a set of desired inversive distances corresponding to a triangulated polyhedronhas not been characterized and remains an important open problem. It is known that not allpossible assignments of inversive distances to a triangulation are realizable. Furthermore, unlike inthe tangency and overlapping cases, general inversive distance packings are not necessarily unique(as was shown in [16] and further explored in [4]), though uniqueness may be obtained with anappropriate notion of convexity for a circle packing, such as that in [5], which we generalize in thispaper. The development of new techniques seems necessary to get a handle on circle packings inthis more general setting.Recently Connelly and Gortler produced a novel proof of the circle packing theorem that obtainsa packing via a continuous motion of circles [12]. Their method starts from a canonical packing ofcircles and, through a finite sequence of moves called flip-and-flow operations, smoothly varies thestarting circle pattern into a tangency pattern with the same number of circles with any desiredcombinatorics homeomorphic to a triangulation of a sphere. Their proof is inherently algorithmicand can be numerically approximated. It is notable that their method computes both the centersand radii of the circles simultaneously. Prior algorithms for circle packing, such as [11, 18], haveeither computed all radii and then produced circle centers in a separate layout step once the radiiwhere computed or used multiple passes that alternate between an adjustment of all radii followed byan adjustment of the circle centers. Connelly and Gortler’s method is the only algorithm known tothis author that computes both centers and radii in an integrated and simultaneous way. Followingtheir approach we obtain a similar algorithm for computing shallow overlap circle packings thatalso computes circle centers and radii in an integrated way. Contributions of this paper.
In this paper we extend Connelly and Gortler’s method to obtainthe full generality of the KAT theorem. Our starting point is the ending point of their algorithm–atangency packing that has the right combinatorics, but does not yet realize desired overlap anglesbetween 0 and π/
2. Our method then uses the same flow operation as theirs without the combi-natorial flip mechanism to adjust, one by one, the overlaps across each edge of the triangulation.Considerable work is required to show that the flow exists and can be carried through to any desiredoverlap subject to the mild conditions of KAT.In order to generalize Connelly and Gortler’s result we moved the entire process from theplane onto the sphere S . Packings on the plane and on the sphere are related by stereographicprojection. Our proofs are inherently spherical and the algorithm we use to adjust the inversivedistances is carried out in situ on the sphere. We make significant use of the connection between2riangulated polyhedra in the de Sitter space and disk packings on S . This allows us to prove theinfinitesimal rigidity of all strictly convex circle polyhedra. This is a partial generalization of theresult from [6] that all c-polyhedra are rigid. This connection between circle polyhedra and de Sitterpolyhedra exists as a sort of mathematical folklore–it is known to several researchers in the fieldand occasionally plays a role in proofs (e.g. [16]), but we are not aware of any work that carefullyexplains the connection in any level of detail. As such we hope that our preliminaries section maybe found a useful introduction to this way of viewing circle packings. We remark that this view isalso useful computationally, since it provides a homogeneous representation of disks on S , whicheliminates many special cases from algorithms dealing with circle patterns and linearalizes manyproperties of circle polyhedra.An example of producing an overlap packing using the techniques of this paper is shown infigure 1. The code that produced this figure is not currently in a form fit to release publicly, butthe author is happy to make it available to any interested readers upon request. Organization of the paper.
We first give a short introduction to the analytic geometry of diskson S which identifies disks with spacelike rays through the origin of the Minkowski spacetime R , ( § § § S , remain convex, and no disk shrinks toa point or grows to encompass the entire sphere without violating a constraint. It is interestingthat our analysis here is for very general motions of circle polyhedra in which no edge’s inversivedistance is fixed, only the range of inversive distances is constrained along each edge. Our use ofthese motions to prove KAT is a significantly restricted class of this more general investigation.Next, we show that all strictly convex circle polyhedra are rigid ( §
5) and develop the necessaryrigidity theory of circle polyhedra following the same general outline as Connelly and Gortler.Though the development here is fairly standard for rigidity theory, we believe this result is ofindependent interest to the circle packing community in that it generalizes to all strictly convexcircle polyhedra the recent result that almost all circle polyhedra are infinitesimally rigid [6]. Thissection includes the development of the packing manifold, a 1-dimensional Riemannian manifold ofdisk configurations obtained by dropping one edge constraint from a convex circle polyhedra. Thepacking manifold was developed in [13, 12] for circle polyhedra where all edges are tangent savefor the dropped edge. We extend this notion to all geodesic convex circle polyhedra with shallowoverlaps.Finally, we show how to put the pieces together in order to extend Connelly and Gortler’smethod ( §
6) and prove the full Koebe-Andre’ev-Thurston method. The first part of our prooffollows the general outline of Connelly and Gortler, with the details provided by the precedingsections. This proves a restricted version of the KAT theorem for circle polyhedra we call strictlyshallow. The full generality of the theorem is then obtained by a limiting argument.We end with an intriguing example obtained by computational experimentation that shows this3dea extends beyond the KAT theorem. (a) (b)(c) (d)
Figure 1: Starting from a tangency packing (a), our proof fixes all inversive distances exceptone (represented by the gray edge) which is corrected via flow. (b) and (c) show two successiveapplications of flow along two different edges. (d) shows a final packing. The green lines are thetangent vectors to the motion of each circle’s center in the flow determined by the gray free-edge.4
Preliminaries S Our main objects of interest are disks on S . In order to parametrize disks on S we begin withthe Minkowski spacetime, R , , which is R endowed with the pseudo-Euclidean inner product ofsignature (1 , Lorentz inner product , is given by (cid:104) ( t , x , y , z ) , ( t , x , y , z ) (cid:105) , = t t − x x − y y − z z . (2.1)It is standard to treat the t = 1 subspace of R , as a model for three dimensional Euclidean space,but such a choice has certain disadvantages in our setting. Instead, we model E as the set of raysfrom the origin having a positive t -coordinate (or alternatively as the upper hemisphere of the unitsphere in R ). We parametrize this space by specifying a point on the ray ( t, x, y, z ) and thus theEuclidean point corresponds to the equivalence class of points ( λt, λx, λy, λz ) for λ >
0. To “view”a set of Euclidean points we compute the intersection of the ray with the t = 1 subspace, so that theusual E coordinates for the point represented by ( t, x, y, z ) are ( x/t, y/t, z/t ). The advantage ofviewing E as a set of rays instead of the t = 1 subspace is that compactifying E with the sphere atinfinity is natural in this view. Every ray through a point (0 , x, y, z ) represents the point at infinityin E in the direction ( x, y, z ). This extended Euclidean space , (cid:98) E is modeled by the rays in R , with t ≥
0, or equivalently with the upper t -hemisphere of the unit Euclidean 3-sphere in R , . Wecould also include those rays with negative t -coordinate, which leads to a double covering of E , inwhich every point of E also has an orientation + or − , connected by the sphere at infinity. Sucha construction leads to a model of the oriented projective space T , an oriented version of RP ,but we will not explicitly use this in this paper. See [23] for a development of oriented projectivegeometry.Let x ∈ R , . The Minkowski norm of x is the complex valued function | x | , = (cid:112) (cid:104) x , x (cid:105) , .The points for which | x | , = 0 is called the light cone . The set of points for which | x | , = − a -axis in R , called the de Sitter space , denoted d S , and the set of points for which | x | , = 1 forms a hyperboloid of two sheets which is a modelfor the hyperbolic space H . The points x ∈ R , are classified as lightlike if | x | , = 0, spacelike if | x | , <
0, and timelike if | x | , <
0. Similarly, rays, vectors, and lines through the origin of R , are classified as lightlike, timelike, or spacelike depending on whether the points along the ray, theendpoint of the vector, or the points of the line are lightlike, timelike, or spacelike. The unit sphere S . Consider the rays from the origin of R , with positive t -coordinate. Thosethat are lightlike correspond to the points of the unit sphere S in E . Those that are timelikecorrespond to points on the interior of the sphere (and thus are a model of the hyperbolic space H ). Those that are spacelike correspond to points on the exterior of S . Lorentz normals and hyperplanes.
Let N = ( a, b, c, d ) and x = ( t, x, y, z ) be two vectors in R , . The two vectors are Lorentz orthogonal , or simply orthogonal, if and only if (cid:104) N , x (cid:105) , = 0.The set Π N = { x ∈ R , : (cid:104) N , x (cid:105) , = 0 } is a hyperplane through the origin of R , with Lorentznormal , or simply normal, N . We parametrize the hyperplanes through the origin of R , bytheir Lorentz normals. We will use ( a, b, c, d ) to describe coefficients of the normal of a hyperplanethrough the origin of R , and ( t, x, y, z ) to denote a point. A hyperplane through the origin of R , is lightlike if it is tangent to the lightcone, timelike if it intersects the lightcone non-trivially, and5pacelike if it meets the lightcone only at the origin. A lightlike hyperplane has a lightlike normal,a timelike hyperplane has a spacelike normal, and a spacelike hyperplane has a timelike normal. Circles and disks on S A circle on S is given by the intersection of a hyperplane through theorigin of R , with S . Thus, we parametrize the set of circles on S by ( a, b, c, d ) where the circleis the set of points ( t, x, y, z ) ∈ S satisfying at − bx − cy − dz = 0 . (2.2)The two disks bounded by the circle ( a, b, c, d ) are the positively oriented disk at − bx − cy − dz < at − bx − cy − dz >
0. Given a disk D , we denote its boundarycircle by ∂D .Scaling the coordinates of ( a, b, c, d ) by any λ > λ < R , . Note that unlike with our model of (cid:98) E , there is no constraint on the sign of the a -coordinate.A hyperplane ( a, b, c, d ) has a real circle of intersection with S if and only if its normal isspacelike. If ( a, b, c, d ) is lightlike, then the intersection of the hyperplane ( a, b, c, d ) with S is apoint. If ( a, b, c, d ) is timelike, then the intersection is imaginary. Thus, we categorize circles anddisks of S as real, point, or imaginary depending on whether the Lorentz normal is spacelike,lightlike, or timelike. From here on we think of a 4-tuple ( a, b, c, d ) as simultaneously defining anormal vector in R , , a halfspace incident a hyperplane in R , , and a disk on S that may be real,point, or imaginary. Areas of disks on S The a -coordinate of a real disk determines a bound on the area of thedisk. When a >
0, the area of the disk is less than 2 π . When a = 0, the disk is a great circle andthus has area 2 π . When a <
0, the area of the disk is greater than 2 π . For point disks, a positive a -coordinate corresponds to zero area, while a negative a coordinate corresponds to a covering ofthe entire sphere with area 4 π . Conical caps.
Let D = ( a, b, c, d ) be a real disk of area at most 2 π . Treat ( a, b, c, d ) as a vectorin R , . Since D is real, the vector ( a, b, c, d ) is spacelike. Let D ∗ denote the ray from the origin of R , in the direction ( a, b, c, d ). This ray corresponds to a point of (cid:98) E which is finite when a > D has area less than 2 π ) and on the sphere at infinity when a = 0 (i.e. D has area 2 π ). When a > D ∗ in (cid:98) E is precisely the apex of the cone tangent to S at ∂D . When a = 0, ∂D isa great circle and the cylinder tangent to S at ∂D is parallel to the vector ( b, c, d ). Properly, thisis a cone with apex at the point of infinity in direction ( b, c, d ) tangent to S in (cid:98) E . Lorentz transformations and M¨obius transformations.
The isometries (i.e. Lorentz innerproduct preserving maps) of R , that fix the origin form a group called the Lorentz group . Thesubgroup of the Lorentz group that fix the time direction t and are orientation preserving on thespace directions x , y , and z are called the restricted Lorentz group . The elements of this groupmap the light cone to itself, or rather map S to itself. Restricted to S , the restricted Lorentzgroup is simply the 6-dimensional orientation preserving M¨obius group , M¨ob( S ), on S .6 .2 Inversive Distance The inversive distance is a M¨obius invariant measurement defined on two disks D and D of S .In our parametrization, the inversive distance d ( D , D ) is simply the negative normalized Lorentzinner product between them. d ( D , D ) = − (cid:104) D , D (cid:105) , | D | , | D | , , (2.3)When D and D are both real disks, then the normal to their defining planes are spacelike. Thuswe may normalize the coordinates of D and D so that (cid:104) D , D (cid:105) , = (cid:104) D , D (cid:105) , = −
1. In thiscase, the inversive distance between D and D simplifies to d ( D , D ) = (cid:104) D , D (cid:105) , (2.4)We will call the coordinates of a normalized real disk its de Sitter coordinates . Given any realdisk D = ( a, b, c, d ), its de Sitter coordinates are ( λa, λb, λc, λd ) where λ is the normalization factor λ = 1 / √ b + c + d − a . The absolute inversive distance , | d ( D , D ) | is sometimes useful aswell.Inverting a disk D across its boundary to obtain − D flips the sign of its inversive distancewith any other disk, since (cid:104) D , D (cid:105) , = −(cid:104)− D , D (cid:105) , .The inversive distance between two disks gives us information on how they situate relative toone another. Let D and D be two disks and C = ∂D and C = ∂D be their boundary circles.If | d ( D , D ) | > C and C are disjoint. When d ( D , D ) < −
1, one of the disks completelycontains the other on its interior. When d ( D , D ) >
1, then either the two disks are disjoint, orintersect at an annular region. When | d ( D , D ) | = 1, then C and C are tangent at a point p . If d ( D , D ) = − p or their intersection forms a crescent region that comesto a point at p . If d ( D , D ) ∈ [ − ,
1] then C and C intersect at two distinct points p and p and divide S into four lune regions, each of which is bounded by an arc of C and an arc of C .The angle of intersection θ on the interior of both D and D can be computed from the inversivedistance via cos θ = d ( D , D ) . (2.5)We call this the overlap angle of D and D . When 0 ≤ θ ≤ π/ shallow and when 0 ≤ θ < π/ strictly shallow . Note that two disks D and D have a shallow overlap if and only if d ( D , D ) ∈ [0 ,
1] and a strictly shallow overlap if andonly if d ( D , D ) ∈ (0 , M¨obius transformations and inversive distance data.
The M¨obius invariance of the inver-sive distance leads to several nice properties. First, there exists a M¨obius transformation taking apair of disks ( D , D ) to a pair ( D (cid:48) , D (cid:48) ) if and only if the inversive distances are the same betweenthe two pairs, d ( D D ) = d ( D (cid:48) , D (cid:48) ).Second, given two triples of disks ( D , D , D ) and ( D (cid:48) , D (cid:48) , D (cid:48) ) such that each triple is linearlyindependent as rays in R , , there exists a unique M¨obius transformation M taking D (cid:55)→ D (cid:48) , D (cid:55)→ D (cid:48) , and D (cid:55)→ D (cid:48) if and only if the pairwise distances match, d ( D i , D j ) = d ( D (cid:48) i , D (cid:48) j ) for i, j = 1 , , i (cid:54) = j .Finally, given two distinct disks D and D , there is a one parameter continuous differentiablesubgroup of M¨ob( S ) called a M¨obius flow that fixes both disks. More information on M¨obiusflows may be found in [4]. 7 .2.1 Planar representations of disks and inversive distances.
Any pattern of disks on S may be taken to a pattern of disks on the extended complex plane(Riemann sphere) C ∪ {∞} by stereographic projection from the North pole N . The North poleitself is mapped to ∞ . The boundary circle ∂D of a disk D either maps to a circle in C if ∂D does not contain N or to a line in C , which is a circle of infinite radius in C ∪ {∞} containing thepoint ∞ , if ∂D does contain N . Each circle in C bounds two disks in C ∪ {∞} , the first being thestandard interior disk and the second being the exterior disk containing ∞ . Topologically, both aredisks in C ∪ {∞} . A line in C (which is still a circle in C ∪ {∞} ) bounds two half-planes which aredisks in C ∪ {∞} .A key property of stereographic projection is that it preserves inversive distances between pairsof disks. Thus we may start with a set of n disks D on S , stereographically project to a set ofdisks on C ∪ {∞} and then apply a M¨obius transformation to C ∪ {∞} to obtain a new set of disk D (cid:48) in C ∪ {∞} while preserving all pairwise inversive distances.We give a partial parametrization of disks on C ∪ {∞} . Let ( x, y, r ) denote the center ( x, y ) ∈ C of a disk and r (cid:54) = 0 be its signed radius. The boundary circle of ( x, y, r ) is the circle centered at( x, y ) with radius | r | . The disk ( x, y, r ) is the usual interior of this circle when r > r <
0. Under this parametrization, the inversive distance between two disks D = ( x , y , r ) and D = ( x , y , r ) is d ( D , D ) = d − r − r r r (2.6)where d = ( x − x ) + ( y − y ) is the squared distance between their centers. By a limitingargument, the inversive distance between a finite disk D and a disk of infinite radius D may beshown to be d ( D , D ) = d r (2.7)where d here denotes the signed distance from the center ( x , y ) to the boundary line in C . Thesign of this distance is positive if ( x , y ) is not in D , negative otherwise.The inversive distance between two disks of infinite radius D and D is 1 if their boundarylines are parallel in C (and hence their boundary circles are tangent at ∞ ) or is the cosine angle ofoverlap between them θ : d ( D , D ) = cos θ . (2.8) Here we do a few quick computations with inversive distances that are used later.
Lemma 2.1.
Let b , c , and d be given. There exist two Lorentz-normalized disks with b , c , and d as the last three coordinates. Both are real, point, or imaginary depending on whether b + c + d is greater than, equal to, or less than 1.Proof. Let D = ( a, b, c, d ) be a normalized disk with the desired three last coordinates. Since D isLorentz-normalized (cid:104) D, D, (cid:105) , = − ⇒ a − b − c − d = − ⇒ a = b + c + d − → a = ±√ b + c + d −
1. Thus there are two real normalized disks
Lemma 2.2.
Let D = ( a, b, c, d ) be a normalized disk on S . Let δ ∈ R be a desired inversivedistance. Let b (cid:48) , c (cid:48) ∈ R be given. Then there exists at most two real normalized disks D (cid:48) and D (cid:48) with b and c coordinates given by b (cid:48) and c (cid:48) such that d ( D, D (cid:48) ) = d ( D, D (cid:48) ) = δ . roof. Let D = ( a, b, c, d ) and δ, b (cid:48) , c (cid:48) ∈ R be given. A normalized disk D (cid:48) with b and c coordinatesgiven by b (cid:48) and c (cid:48) and third coordinate d (cid:48) has coordinates ( (cid:112) ( b (cid:48) ) + ( c (cid:48) ) + ( d (cid:48) ) − , b (cid:48) , c (cid:48) , d (cid:48) ). Then d ( D, D (cid:48) ) = (cid:104)
D, D (cid:48) (cid:105) , = a (cid:112) ( b (cid:48) ) + ( c (cid:48) ) + ( d (cid:48) ) − − bb (cid:48) − cc (cid:48) − dd (cid:48) . Setting equal to δ and solvingfor d (cid:48) we obtain a quadratic equation in d (cid:48) , thus proving the lemma.We end with a computation in the planar representation. Lemma 2.3.
Let D and D be the interior disks of finite radius boundary circles in C ∪ {∞} . Ifthe center of D is contained in D then d ( D , D ) < (or vice versa).Proof. Since translations and rotations of C are M¨obius transformations we may, without loss ofgenerality, assume that D is centered at the origin and D is centered at a point ( x,
0) on the realaxis. Then the inversive distance is d ( D , D ) = x − r − r r r . Since D and D are interior disks, then r , r >
0. If the center of D is contained in D , then x < r ⇒ x < r ⇒ x − r − r < ⇒ d ( D , D ) <
0. The same argument holds if the center of D is contained in D . Let D and D be two disks on S not sharing the same boundary circle. The set Π of linearcombinations of D and D forms a 2-dimensional plane in R , . The intersection of this plane withd S gives a family of disks, which is called the coaxial family of D and D , denoted D ∨ D .The set of directions in R , orthogonal to both D and D spans a second 2D plane Π ⊥ throughthe origin of R , . The intersection of this plane with d S also forms a coaxial family, which we callthe orthogonal coaxial family , denoted ( D ∨ D ) ⊥ . A coaxial family and its orthogonal, takentogether, is called an Appollonian family (because such families were studied by Appollonius ofPerga). Every disk in a coaxial family meets every disk in the orthogonal family at a right angle,and thus if D ∈ D ∨ D and D (cid:48) ∈ ( D ∨ D ) ⊥ then d ( D, D (cid:48) ) = 0. Similarly, if d ( D, D ) = 0 and d ( D, D ) = 0, then D ∈ ( D ∨ D ) ⊥ .The intersection of Π with (cid:98) E is the line L connecting conical caps D ∗ and D ∗ . If one, say D isa great circle, then it is the line through the conical cap of D parallel to the normal of D . If both D and D are great circles, then Π does not meet E . Here it represents the oriented projectiveline at infinity (which is a circle) bounding the plane through the origin of E whose normal is thecross product of the normals of D and D . In this case ∂D and ∂D intersect at antipodal pointson the sphere S and the coaxial family is the set of disks whose boundary pass through these twoantipodal points. Classification of coaxial families.
Coaxial families come in three types depending on whetherthe line L intersects, is tangent to, or is disjoint from the sphere S , which we call hyperbolic , parabolic , and elliptic respectively. The orthogonal family of a hyperbolic family is elliptic (andvice versa), while the orthogonal to a parabolic family is also parabolic. The type of a coaxialfamily may be determined by measuring the absolute inversive distance between any two distinctmembers of the family. Let D and D be disks. If | d ( D , D ) | >
1, then D ∨ D is hyperbolic; if | d ( D , D ) | = 1, then parabolic; and if | d ( D , D ) | <
1, then elliptic. Thus the boundary circles of9ny two members of a hyperbolic family are disjoint, of a parabolic family are tangent, and of anelliptic family intersect at two points.
M¨obius transformations of coaxial families.
It is a standard fact of inversive geometry thatthere is a M¨obius transformation M taking one coaxial family F to another coaxial family F ifand only if F and F are the same type. Furthermore, if M takes F to F , then M also takes theorthogonal families ( F ) ⊥ to ( F ) ⊥ . This is useful because it allows us to take any coaxial familyto certain standard coaxial families that can be used to simplify computations. For example, anyhyperbolic family may be taken by a M¨obius transformation to the set of latitude disks on S .Similarly, any elliptic family may be taken to the set of longitude lines. A parabolic family makebe taken via stereographic projection to the set of lines/half-planes in C . Taking a family to somestandard view often simplifies proofs, and we make extensive use of this below. Let D = ( a , b , c , d ), D = ( a , b , c , d ), and D = ( a , b , c , d ) be three linearly independentdisks given on S . Taken as rays in R , there is a unique hyperplane Π through the origin of R , containing D , D , and D . It is an exercise to show that the Lorentz normal to this hyperplane isgiven by C ⊥ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b c d b c d b c d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a c d a c d a c d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a b d a b d a b d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a b c a b c a b c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.9)The intersection of this hyperplane with with (cid:98) E corresponds either to a plane in E togetherwith its circle at infinity, or the sphere at infinity itself in (cid:98) E . Since Π is a hyperplane throughthe origin of R , , it corresponds to a real, point, or imaginary circle depending on whether itsintersection with S is real, point, or imaginary. We call this circle the orthocircle of the triple( D , D , D ). The two disks bounded by this circle, which correspond to the intersections of thetwo half-spaces defined by Π with S we call orthodisks . The half-space in the positive directionof normal C ⊥ is the positively oriented orthodisk O + and that in the negative direction of C ⊥ isthe negatively oriented orthodisk O − . Any even permutation of the triple ( D , D , D ) definesthe same positive and negative orthodisks, whereas an odd permutation of the triple swaps thepositive for the negative.We classify triples of disks by whether their orthocircle is an imaginary, point, or real circle. Atriple ( D , D , D ) is elliptic if its orthocircle is imaginary. It is parabolic if its orthocircle is apoint. It is hyperbolic if its orthocircle is real .Now, let D be any other point of Π in R , . Since every vector from the origin to a point in Πis Lorentz-orthgonal to the normal C ⊥ we have that d ( C ⊥ , D ) = 0 and conversely if D is any disksuch that d ( C ⊥ , D ) = 0, then D is contained in Π. The intersection of Π with the d S gives usthe set of real disks meeting C ⊥ Lorentz-orthogonally. We call this set a c-plane of disks. Testingwhether a disk D is contained in the c-plane defined by a triple ( D , D , D ) follows the usual When an orthodisk is hyperbolic its interior may be viewed as a Poincar´e disk model of the hyperbolic plane,and its defining disks as hyperbolic lines. det ( O + , D ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a b c d a b c d a b c d a b c d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (2.10)determines when D is in the c-plane defined by D , D , and D . When det ( O + , D ) = 0, D isin the c-plane of ( D , D , D ). Otherwise, the sign of the determinant determines on which sideof the hyperplane Π D lies. We use this in section 3 to define the notion of convexity for circlepolyhedra. As with the triples themselves, we classify the c-planes defined by a triple as hyperbolic,parabolic, or elliptic.A result used at least as far back as Thurston’s Notes [24] is that three circles that mutuallyoverlap or are tangent have a common real orthocircle if and only if the sum of their overlap anglesis less than π . When equal to π , the three circles are coincident at a single point which correspondsto the parabolic orthocircle of the three. An elegant geometric proof of this fact may be obtainedby treating the interior of the orthocircle as the Poincar´e disk model of the hyperbolic plane. Thearcs of the three circles restricted to the interior of the orthocircle become hyperbolic lines and sincethe three circles intersect, they describe a hyperbolic triangle. The interior angles of the triangleare the angles of overlap between the circles, and thus sum to less than π . This is summarized inthe following lemma. Lemma 2.4.
Let D , D , D be three disks such that d ( D i , D j ) ∈ [0 , for i, j = 1 . . . , i (cid:54) = j andlet O + denote the positively oriented orthodisk of ( D , D , D ) . Let θ ij denote the angle of overlapbetween disks D i and D j , which is if ∂D i and ∂D j are tangent. Let θ = θ + θ + θ . Then O + is hyperbolic when θ < π , is parabolic when θ = π , and is elliptic when θ > π . M¨obius transformations of c-planes
A hyperbolic c-plane corresponds to a hyperplane thatintersects the interior of the light cone in R , and thus has a spacelike normal; a parabolic c-planecorresponds to a hyperplane tangent to the light cone with a lightlike normal; and an elliptic c-planecorresponds to a hyperplane that does not meet the interior of the light cone and has a timelikenormal. Since Lorentz transformations preserve the spacelike, lightlike, or timelikeness of vectors,this has implications for M¨obius transformations between c-planes. Given two c-planes Π and Π there exists a M¨obius transformation mapping Π bijectively to Π if and only if the two c-planesare the same type (i.e. both hyperbolic, or both parabolic, or both elliptic). Here we include a few basic facts about triples and quadruples of disks.
Parabolic triples
Let ( D , D , D ) be a parabolic triple of disks and let p be the common pointof intersection p = ∂D ∩ ∂D ∩ ∂D as in figure 2. The point p may (a) be on the interior ofthe union of the disks, (b) be on the boundary of the union of the disks next to a vanishing angle,or (c) be on the boundary of the union of the disks next to a positive angle. We now show thatif all pairwise inversive distances of the triple fall in [0 ,
1) we have case (a) and case (b) occurs ifand only if two disks are tangent and meet the third disk at an overlap of π/ a) (b) (c) Figure 2: The three cases of parabolic triples. Case (b) occur if and only if the inversive distancebetween D and both D and D is 0 and the other two are tangent at p . Case three cannot occurwith all inversive distances in [0 , Lemma 2.5.
Let D , D , and D be three disks that meet at a point p such that d ( D i , D j ) ∈ [0 , for i (cid:54) = j , i, j = 1 , , . Then either p is on the interior of D ∪ D ∪ D and no two of the disksare tangent or p is the point of tangency between two of the disks which both meet the third disk ata right angle.Proof. Stereographically project the disks to the extended complex plane (cid:98) C . Then apply a M¨obiustransformation taking p to ∞ . The result is that all three disks become lines in C as in figure 3.If it exists, let p ij denote the point of intersection between D i and D j that is not p and θ ij denotethe oriented angle of overlap between the two disks at that point (which is 0 when that point p ij = p in case (b)). Recall that the inversive distance is the angle of overlap between the disks,and stereographic projections and M¨obius transformations are conformal, so d ( D i , D j ) = cos θ ij .In case (a) the three angles are equal to the interior angle of the triangle with vertices p , p , and p and thus all inversive distances are in [0 ,
1] precisely when this triangle is non-obtuse. For case(b), without loss of generality assume that p is the point of tangency between D and D and thus p = p = ∞ . In this case θ + θ = π and thus either both angles are π/ d ( D , D ) = d ( D , D ) = 0 or one of the angles is greater than π/ p is onthe interior of D . Then the angles θ and θ are exterior angles along the same supporting lineof a triangle and therefore θ + θ > π . Thus at least one of them is greater than π/ Corollary 2.6. If D , D , and D are three disks with mutually shallow overlaps, then the sum oftheir overlaps is π . We now consider adding a fourth disk.
Lemma 2.7.
The only configurations of four disks contained in a parabolic plane that have onlyshallow overlaps are configurations comprised of two pairs of tangent disks which are mutuallyorthogonal. a) (b) (c) Figure 3: The setup for the proof of lemma 2.5. Moving the common point of the three circles to ∞ by a M¨obius transformation turns the boundary disks into half-planes. (a) and (b) are possiblewith shallow overlaps and show that the sum of the overlap angles is π . (c) is not possible withshallow overlaps. Proof.
We begin the same as in the proof of lemma 2.5 by stereographically projecting all disksto the sphere and moving the common point p to ∞ by a M¨obius transformation. The statementis now a statement about half-planes in C : the only configuration of half-planes that all havemutually shallow overlap angles are the configurations of two pairs of disjoint half-planes withparallel boundary lines that meet each other orthogonally. Let l i denote the boundary line and h i denote the half-plane corresponding to each disk D i .We first analyze the case where there is at least one tangency. Without loss of generality, assume D and D are tangent. Then l and l are parallel and by lemma 2.5 since all overlaps are shallow,both l and l meet l and l orthogonally. Then l and l are parallel so ∂D and ∂D are tangentand thus d ( D , D ) ∈ {− , } .Now for contradiction assume no tangencies occur. Since all disks overlap at a real angle eachof the lines l . . . l intersect. The intersection of the complements of h , h , and h is a triangle asin (a) of figure 3. By shallowness, this triangle is acute.Now consider any such triangle made by three mutually intersecting lines and choose an orien-tation for each line. By convention let the left-hand side of the orientated line be its half-space.Say that two of the lines agree at a corner of the triangle if there orientation is consistent along theboundary of the triangle, either counter-clockwise or clockwise. Note that when two lines agree,the oriented angle over overlap between their half-spaces equals the angle of the triangle. When thetwo orientations disagree, the angle of overlap between the half-spaces is π minus the angle formedon the interior of the triangle at each corner. We observe that either all three corners of the triangleagree or exactly two disagree with any orientation of the lines we choose.Now take the half-plane orientations induced by our disks. Because our half-plane overlapsare shallow it is not possible that two corners disagree, because a corner that disagrees with ourorientation must have an angle of at least π/ π/
2. Thus any triangle formed by any three shallow overlapping half-planes must agreewith the orientation of those half-planes, which means that the orientation of the lines must beclockwise or counter-clockwise around the triangle.13ow consider adding the fourth half-plane h to our arrangement of h , h , and h . We observethat either l does not meet any of the corners of the triangle formed by l , l , and l , or it meetsexactly one corner. In the first case any three of the lines defines a triangle and by the discussionabove, the orientations on the lines induced by h . . . h must be counter-clockwise or clockwisearound each of these triangles. However, there is no orientation of four lines in general positionthat is consistent around each of the four triangles formed by any three of the lines. Similarly,when l passes through a corner of the triangle formed by l , l , and l there are three trianglesformed, one of which is the union of the other two. But then again it is not possible to orient thelines so that the orientation is consistent across all three triangles. But if all overlaps were shallow,then any three of the half-planes must define a consistent orientation around the triangle formedby their boundary lines. Therefore this case is not possible, and we must have two pairs of tangentdisks contained in orthogonal coaxial families.As an immediate corollary we have: Corollary 2.8.
There are no five disks whose boundary circles meet at a common point such thatthe pairwise overlaps between all disks are shallow.
Coplanarity of four disks.
Let D , D , D , and D be four distinct disks and assume that thepairwise overlap angles between disks in the two triples ( D , D , D ) and ( D , D , D ) fall between0 and π/
2. When are the four disks contained in the same c-plane? Lemma 2.9 characterizes whenall four disks are coplanar.
Lemma 2.9.
Let ( D , D , D ) and ( D , D , D ) be two triples of disks such that all pairwise in-versive distances in the triples are shallow and no three of D . . . D is contained within a coaxialfamily. Let Π be the c-plane defined by the orthodisk of ( D , D , D ) . Then:1. If Π is hyperbolic and D is contained in Π then d ( D , D ) < .2. If Π is parabolic and D is contained in Π then either d ( D , D ) = d ( D , D ) = 1 and allother pairwise inversive distances are 0 or d ( D , D ) < .3. If Π is elliptic and D is contained in Π then either d ( D , D ) ≥ and D ∪ D ∪ D ∪ D = S or d ( D , D ) < . Furthermore, in the first case there does not exist any other disk that hasshallow overlaps with each of D . . . D .Proof. Case 1: In this case, let O + be the positively oriented orthodisk defining Π. Since Πis hyperbolic, this is a real disk. Assume d ( D , O + ) = 0. Since all disks D , . . . , D meet O + orthogonally, they correspond to hyperbolic half-planes in H obtained by treating the interior of O + as a Poincar´e disk model of H . This correspondence preserves overlap angles. Let h i denotethe half-plane and l i denote the boundary line corresponding to each disk D i . Let h (cid:48) i denote thereflection of h i across l i . The angle θ ij of intersection between the half-planes h i and h j is given by θ ij = arccos d ( D i , D j ) for all i, j = 1 . . . i (cid:54) = j . We now argue that it is not possible that all ofthese angles are between 0 and π/ D , D , D ) and ( D , D , D ) the lines l , l , and l form a hyperbolic triangle, as do l , l , and l . Because neither triple is contained in a coaxialfamily, neither of the triples of lines ( l , l , l ) and ( l , l , l ) meet at a common point. Therefore( l , l , l ) and ( l , l , l ) each define a hyperbolic triangle. Let T ijk denote the hyperbolic triangleformed by l i , l j , and l k . Let ˆ θ ij denote the angle between l i and l j in the triangle T ijk .14e first claim that if θ ij , θ jk , θ ki ∈ [0 , π/ T ijk either equals h i ∩ h j ∩ h k or h (cid:48) i ∩ h (cid:48) j ∩ h (cid:48) k . Indeed if T ijk ⊂ h i ∩ h j or T ijk ⊂ h (cid:48) i ∩ h (cid:48) j then ˆ θ ij = θ ij is the angle between l i and l j in T ijk . However, if one half-plane contains T ijk and the other does not, say T ijk ⊂ h i and T ijk ⊂ h (cid:48) j , then the angle of the triangle T ijk between l i and l j is ˆ θ ij = π − θ ij . In the former casewe say that the angle ˆ θ ij agrees with the orientations of h i and h j , and in the latter disagrees .Either all angles of T ijk agree or exactly two disagree with the orientations of h i , h j , h k .Since 0 ≤ θ ij ≤ π/
2, then ˆ θ ij ≤ π/ θ ij ≥ π/ θ ij + ˆ θ jk + ˆ θ ki < π by hyperbolic trigonometry and thus it is notpossible that two angles disagree with the orientation and therefore all must agree. Orient eachline l i so that its left-hand side is the half-plane h i . The property above implies that this inducedorientation is consistent (either counter-clockwise or clockwise) around the triangle T ijk .First let us assume that D and D are disjoint (i.e. d ( D , D ) (cid:54)∈ [ − , l and l donot intersect. The triangles T and T share the point p of intersection between l and l incommon, and an edge of each triangle is supported by each of l and l . This means that either T ⊂ T or vice versa, or the two triangles form a bowtie configuration on opposite sides of p .In each case, using only the fact that the orientations of the hyperplanes must agree at each cornerof the two triangles it is easily shown that either h ⊂ h or h ⊂ h . But then D ⊂ D and thus d ( D , D ) < D and D do overlap (i.e. d ( D , D ) ∈ [ − , l and l meet eitherat a hyperbolic point or at an ideal point. Then l , l , l , and l are an arrangement of four lineseach pair of which meet and no three of which meet at a common point (due to the assumption thatno three disks are contained in the same coaxial family). This arrangement defines four hyperbolictriangles by any choice of three of the lines. By the discussion above, the half-planes induce anorientation on the lines l , l , and l that is consistent around T . For contradiction assume d ( D , D ) ∈ [0 , θ ij ∈ [0 , π/
2] for i, j = 1 . . . i (cid:54) = j .This means that the half-plane of h must induce an orientation of l so that each of the triangles T , T , T , and T . However, for any placement of l no matter what orientation is chosenat least one of the triangles is not consistently oriented. Therefore it is not possible for all of theoverlap angles to be in [0 , π/ d ( D , D ) < Case 2:
This is lemma 2.7.
Case 3:
Since Π is elliptic there is a M¨obius transformation taking all of the disks in Π to theset of disks bounded by great circles on S . Such a disk is parametrized by a vector in R , with0 in the a -coordinate. Let D i = (0 , b i , c i , d i ) denote the normalized coordinates of each disk. Theinversive distance between two disks becomes: d ( D i , D j ) = − b i b j − c i c j − d i d j = − ( b i b j + c i c j + d i d j )which is simply the negative of the Euclidean inner product between the oriented normal vectorsof the planes in R supporting each disk (with the normal pointed toward the interior of thedisk). Since all great circles meet, for all i, j = 1 . . . d ( D i , D j ) ∈ [ − , d ( D i , D j ) ≥
0. Indeed, this is equivalent to claiming that the four normal vectors in E havenegative Euclidean inner product. This is possible only if the great disks cover S . Furthermore,there are no five vectors in R that all have pairwise negative Euclidean inner product.15 Circle polyhedra
We call a 3-connected planar graph P an abstract polyhedron . Let V ( P ) and E ( P ) denotethe vertex and edge sets of P . Such graphs have a well defined face set F ( P ). Let n = | V ( P ) | and m = | E ( P ) | . If each face is a triangle, then m = 3 n − P as an abstracttriangulated polyhedron . In the remainder, we will use the term polyhedron to mean triangulatedpolyhedron.Let w : E ( P ) → R be an assignment of weights to each edge of P representing desired inversivedistances. We call the pair ( P, w ) a weighted abstract polyhedron . If each w ( ij ) ∈ [0 , P, w ) shallow and strictly shallow if no 4-cycle of edges ij , jk , kl , li in E ( P ) has weight 0 foreach edge.A realization of a weighted abstract polyhedron ( P, w ) on S is a set of disks D = { D , . . . , D n } on S such that d ( D i , D j ) = w ( ij ) whenever ij ∈ E ( P ). We call the pair ( P, D ) a circle poly-hedron , or c-polyhedron . As with the abstract polyhedra, ( P, w ), we will call a circle polyhedra (strictly) shallow when it realizes a (strictly) shallow abstract polyhedron.
Proper c-polyedra.
If the area of each disk of a circle polyhedron is strictly less than 2 π we sayit is strictly proper and if we relax this so that one disk may have area 2 π we call it proper . Geodesic triangulations.
Let ( P, D ) be a proper shallow c-polyhedron. Let G be a graphdrawing on S given by placing a vertex v i at the center of each disk D i ∈ D and for each edge ij ∈ E ( P ) drawing the shortest path great circle arc on S from v i to v j . Since ( P, D ) is properand shallow, the centers of two disks D i and D j connected by an edge ij cannot be antipodal, andtherefore the shortest path is well defined. Furthermore, by elementary geometry it can be shownthat this path lies entirely within D i ∪ D j . We call G the induced geodesic graph of ( P, D ).If no two arcs of G self intersect and no vertex of G lies on the interior of an arc it is not incidentto in P , then G is a triangulation of S . In this case we say that ( P, D ) has a induced geodesictriangulation G and call ( P, D ) a geodesic circle polyhedron . The Koebe-Andre’ev-Thruston theorem.
We may now state the Koebe-Andre’ev-Thurstontheorem. Our statement is similar in presentation to that in [8].
Theorem 3.1 (Koebe-Andre’ev-Thurston) . Let ( P, w ) be a shallow abstract triangulated polyhe-dron, with P not a tetrahedron, and let the following conditions hold.1. If e , e , e form a closed loop of edges with (cid:80) i =1 arccos w ( e i ) ≥ π , then e , e , and e bounda face of P .2. If e , e , e , and e form a closed loop of edges with (cid:80) i =1 arccos w ( e i ) = 2 π then e , e , e ,and e bound the union of two faces of P .Then there exists a geodesic c-polyhedron ( P, D ) realizing ( P, w ) . In section 6 we give a novel proof of this theorem generalizing the proof of the tangency casefrom [12]. 16 .2 Convex circle polyhedra
Let ( P, D ) be a triangulated circle polyhedron and let ( D i , D j , D k ) be the disks corresponding toa face ijk ∈ F ( P ). Let O + be the positively oriented orthodisk of ( D i , D j , D k ). We say that O + separates ( P, D ) if and only if there exist disks D + and D − in D such that det ( D + , O + ) > det ( D − , O + ) <
0. A face separates ( P, D ) precisely when its orthodisks do. ( P, D ) is convex ifand only if no face separates ( P, D ). Interpreted in R , this is the usual separation of the raysof D by the unique hyperplane containing the rays D i , D j , and D k . Under this interpretation aconvex circle polyhedron is a convex polyhedral solid cone with apex at the origin in R , . The configuration space of circle polyhedra.
Given an abstract triangulated polyhedron P a configuration of P is a set of n disks D , one for each vertex in P . We think of D as a point in R n and call it the configuration space of P . Let D ( t ) be a continuous path in R n on some interval[ a, b ] such that D ( a ) is strictly convex and D ( b ) is strictly non-convex. Both strict convexity andstrict non-convexity are open conditions determined by the signs of determinants of each face in P with all other vertices in P (see ( 2.10)). Then there is an open ball around D ( a ) in R n such thatall configurations in the ball are strictly convex and all disks are real and distinct and an open ballaround D ( b ) in which all configurations are strictly non-convex and all disks are real and distinct.Thus, there must be some intermediate time a < t (cid:48) < b at which either a disk D i contracts to apoint disk, or two neighboring disks coincide, or ( P, D ( t (cid:48) )) is convex, but not strictly so. In thethird case, the associated polyhedron in S is no longer strictly convex, which occurs when twofaces become co-planar. Thus in ( P, D ( t (cid:48) )) we have two faces ijk and kjl incident an edge jk suchthat all four disks D i , D j , D k , and D l are coplanar. From this discussion we immediately have: Lemma 3.2.
Let ( P, D ( t )) be a path in R n defined for a (possibly bi-infinite) interval t ∈ I suchthat there is a time t (cid:48) ∈ I where ( P, D ( t (cid:48) )) is strictly convex. If for all t ∈ I we have that1. D i ( t ) is a real disk for all i ∈ V ( P ) ,2. D i ( t ) (cid:54) = D j ( t ) for all ij ∈ E ( P ) , and3. if ijk, kjl ∈ F ( P ) are faces sharing a common edge jk then D i ( t ) , D j ( t ) , D k ( t ) , and D l ( t ) are not co-planar,then ( P, D ( t )) is strictly convex for all t ∈ I . We call a path ( P, D ( t )) a motion of the circle polyhedron ( P, D ( t )). Note that a motion doesnot necessarily preserve inversive distances. We investigate the properties of motions in section 4after deriving some basic properties of circle polyhedra in the next section. Conical cap polyhedra.
Let ( P, D ) be a triangulated circle polyhedron on S , none of whichhas area greater than or equal to 2 π (equivalently, the a coordinate of each disk is greater than 0).Construct a Euclidean polyhedron by placing placing each vertex i ∈ V ( P ) at the conical cap D ∗ i and realizing each edge ij ∈ E ( P ) as the line segment D ∗ i D ∗ j . The resulting triangulated Euclideanpolyhedron is the (Euclidean) conical cap polyhedron of ( P, D ). The following lemma relatesthe convexity of the cap polyhedron with that of ( P, D ). Lemma 3.3.
Let ( P, D ) be a strictly proper triangulated circle polyhedron on S , and let (cid:98) P denoteits conical cap polyhedron in E . Then ( P, D ) is a convex circle polyhedron if and only if (cid:98) P is aconvex Euclidean polyhedron. roof. Let D i = ( a i , b i , c i , d i ), D j = ( a j , b j , c j , d j ) and D k = ( a k , b k , c k , d k ) be the disks of D corre-sponding to a face ijk ∈ F ( P ). Since no disk has 2 π area or more, a i > a j >
0, and a k >
0. Thuswe may scale the coefficients of each disk to obtain new coordinates: D i = (1 , b i /a i , c i /a i , d i /a i ), D j = (1 , b j /a j , c j /a j , d j /a j ), and D k = (1 , b k /a k , c k /a k , d k /a k ) without flipping the orientation ofany disks. Let O + be the positively oriented orthodisk of ijk and let D = (1 , b/a, c/a, d/a ) beanother disk in ( P, D ) also scaled so that the first coordinate is 1. From equation 2.10 we have that det ( O + , D ) = 1 aa i a j a k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b/a c/a d/a b i /a i c i /a i d i /a i b j /a j c j /a j d j /a j b k /a k c k /a k d k /a k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (3.1)The determinant on the right hand side of (3.1) is the standard determinant whose sign is usedto determine whether the point ( b/a, c/a, d/a ) lies on the positively oriented side (determinant ispositive), negatively oriented side (determinant is negative), or inside (determinant is zero) the ori-ented Euclidean plane meeting D ∗ i , D ∗ j , and D ∗ k . Since a, a i , a j , a k >
0, the sign of this determinantis the same as det ( O + , D ). Thus if D (cid:48) is another disk in D , D ∗ and ( D (cid:48) ) ∗ lie on the same side ofthe plane meeting D ∗ i , D ∗ j , and D ∗ k if and only if det ( O + , D ) and det ( O + , D (cid:48) ) have the same sign.Therefore no plane supporting any face of the cap polyhedron separates any other vertices of thepolyhedron, which is thus convex. Extending to large disks.
Lemma 3.3 shows that if all disks of a strictly convex circle polyhe-dron ( P, D ) have area less than 2 π , then the resulting cap polyhedron (cid:98) P is convex. We may extendthe notion of conical caps to disks that have area greater than 2 π . Indeed for such a disk D we willsimply take its conical cap to be the conical cap of the disk opposite it along its boundary ∂D . Inother words D ∗ = ( − D ) ∗ . Now, consider the determinant used in (3.1) and a triangle ijk ∈ F ( P ).Let O + denote the orthodisk of ( D i , D j , D k ). Let Π be the hyperplane passing through the originof R , containing the rays D ∗ i , D ∗ j , D ∗ k . This corresponds to a plane in E passing through theconical cap points that are the intersections of the rays D ∗ i , D ∗ j , D ∗ k with E . Now let D and D (cid:48) be two other disks in D . By convexity, the rays D ∗ and ( D (cid:48) ) ∗ lie on the same side of Π. If bothhave area less than 2 π , then both have a positive a -coordinate, and thus the corresponding conicalcap points in E lie on the same side of the Euclidean plane corresponding to Π. However, if oneof the disks, say D has area greater than 2 π , then its a -coordinate is negative. Thus by equation(3.1) the determinant on the right hand side of the equation has a flipped sign, which means that in E , the conical cap points corresponding to D ∗ and ( D (cid:48) ) ∗ lie on opposite sides of Π. Furthermore, D and − D lie on opposite sides of Π and − D lies on the same side of Π as the conical cap in E (since it is obtained by a positive scaling factor). Therefore D ∗ and its corresponding conical cappoint lie on opposite sides of Π. From this discussion we see that the cap polyhedron for a strictlyconvex circle polyhedron which has some disks of area greater than 2 π will not be convex. This isprecisely because the conical caps for disks with area greater than 2 π lie on the “wrong side” of theplane supporting any face of the polyhedron. However, this wrong side property turns out to bewell behaved. We now use it to recover a convex Euclidean fan for every vertex of the polyhedronthat allows us to use 3D Euclidean geometry to investigate the geometry of the rays in R , . Thisconstruction will be crucial to our proof of infinitesimal rigidity of convex circle polyhedra. The convex Euclidean triangle fan for a vertex.
Consider some vertex 0 and its neighbors1 . . . k given in counter clockwise order. Let v i denote the conical cap point of D i . The triangles18 v v , v v v , . . . , v v k v form a triangle fan in the cap polyhedron around v . When all disks havearea less than 2 π , this fan is convex and lies on a convex polyhedral cone with apex v . However,if D . . . D k are a mix of disks with area less than 2 π and area greater than 2 π , then some of theconical caps are on the wrong side of the planes supporting the triangles and those planes separatethe points v . . . v k . However, let D i be a disk with area greater than 2 π . Let Π be any hyperplanepassing through v and the origin of R , that does not contain v i . Reflecting v i through v toobtain v (cid:48) i = v − ( v i − v ) gives a point on the opposite side of Π. Now consider some triangle of thefan v v i v i +1 and two other vertices v j and v k of the fan. Let Π i denote the hyperplane supportingthe rays D ∗ , D ∗ i , and D ∗ i +1 . If v j and v k correspond to disks D j and D k that have area strictly lessthan 2 π , then v j and v k lie on the same side of Π i as D ∗ j and D ∗ k . However, if one has area greaterthan 2 π , say D i , then v i lies on the wrong side. In this case replace v i with v (cid:48) i its reflection through v . Let v (cid:48) , . . . , v (cid:48) k denote the resulting points (which are reflected whenever a point is “wrong” andare left alone otherwise). We claim that the resulting triangle fan v v (cid:48) v (cid:48) , v v (cid:48) v (cid:48) , . . . v v (cid:48) k v (cid:48) formsa triangle fan on a convex polyhedral cone. Indeed, by construction the plane Π i in E supportinga face v v (cid:48) i v (cid:48) i +1 not only contains v , v (cid:48) i and v (cid:48) i +1 , but also contains v i and v i +1 . Furthermore, thecorresponding hyperplane containing Π i and the origin of R , contains D ∗ , D ∗ i , and D ∗ i +1 . Byconvexity the normal rays for all other disks in D lie on the same side of Π i since 0 i ( i + 1) ∈ F ( P ).But each v (cid:48) i lies on the same side of Π i as its corresponding disk D i . Thus no two vertices of thefan are separated by Π i . Therefore the fan is a convex triangle fan on a convex polyhedral cone.We call this convex triangle fan the convex Euclidean triangle fan for vertex We now derive some properties of circle polyhedra we use later.
Lemma 3.4.
Let ( P, D ) be a strictly shallow circle polyhedron and ij ∈ E ( P ) . Then the centers of D i and D j are not antipodal.Proof. Antipodal disks either have the same boundary circle or disjoint boundary circles. Since( P, D ) is shallow the boundary circles of neighboring disks must either be tangent or intersect.Thus the only possible antipodal case is when two neighboring disks share the same boundarycircle. Assume for contradiction that there is an edge ij ∈ E ( P ) such that ∂D i = ∂D j . If D i = D j then d ( D i , D j ) = − P, D ) is strictly shallow. Suppose then that D i = − D j .Let ijk ∈ F ( P ) be a face containing the edge ij . We may, by a M¨obius transformation of S ,take D i and D j to the two hemispheres whose boundary is the equator ( D i = (0 , , ,
1) and D j = (0 , , , − D k = ( a k , b k , c k , d k ). The inversive distances from D k to D i and D j are given by d ( D i , D k ) = − d k and d ( D j , D k ) = d k . Thus if d k (cid:54) = 0 one of theserepresents a non-shallow overlap, a contradiction. Thus d k = 0 and D k meets both disks D i and D j orthogonally. Now consider the other triangle jil ∈ F ( P ) incident the edge ij . By the sameargument D l must overlap both D i and D j orthogonally. But then there is a 4-cycle of edges kj , jl , li , ik that all have inversive distance 0, contradicting that ( P, D ) is strictly shallow.Because no two neighboring disks are antipodal, the shortest path between their centers is welldefined. Corollary 3.5.
Let ( P, D ) be a strictly shallow circle polyhedron and ij ∈ E ( P ) . There is a uniqueshortest geodesic path on S between the centers of D i and D j . emma 3.6. Let ( D , D , D ) be a triple of disks with shallow overlaps. Let s be the geodesicshortest path connecting the centers of D and D . Then either1. D is disjoint from s ,2. the boundary of D is the great circle containing s , or3. D and D are tangent at a point p on s and D is orthogonal to both D and D and tangentto the great circle containing s at p .Proof. Assume that D is not disjoint from s and ∂D is not the great circle containing s . Weprove that D is tangent to s at a point p that is also the point of tangency between D and D .First, let us make several elementary remarks about disks and inversive distances. Remark 1:
Let D and D (cid:48) be disks such that d ( D, D (cid:48) ) ∈ [ − ,
1] and let p be one of the intersectionpoints (or sole point of tangency) between the boundary circles ∂D and ∂D (cid:48) . The spherical radiusarc from the center of D to p contains a point of the interior of D (cid:48) if and only if the overlap between D and D (cid:48) is greater than π/
2, or equivalently if d ( D, D (cid:48) ) < Remark 2:
From remark 1 it follows immediately that if d ( D, D (cid:48) ) ≥ D is notcontained within D (cid:48) and vice versa. Remark 3: If D and D (cid:48) are two disks that do not contain each other’s centers and D (cid:48)(cid:48) ⊂ D (cid:48) isa disk with the same center as D (cid:48) but a strictly smaller radius, then d ( D, D (cid:48) ) < d ( D, D (cid:48)(cid:48) ).By remarks 1 and 2, D cannot contain the endpoints of s on its interior. Thus it is eithertangent to s at a point p on its interior, or contains an interior segment s (cid:48) of s .In the first case, we show that D and D are tangent to each other at p . Indeed, assumenot. Then p is on the interior of either D or D , say D . Let r be the radius arc of D alongthe segment s . By construction, D is tangent to r at p , which is interior to D . But then theradius arc between the center of D and the intersection points of ∂D with ∂D is readily shownto intersect the interior of D . Then by remark 1, D and D do not have a shallow overlap, acontradiction. Thus D and D are tangent at p . The great circle containing s meets both D and D orthogonally and thus, since D is tangent to this great circle, it must be in the orthogonalcoaxial family ( D ∨ D ) ⊥ .Now assume that D contains a proper sub-segment s (cid:48) of s . In this case, by remark 3, we canshrink D while keeping its radius fixed until it is tangent to s at a point p to obtain a disk D (cid:48) . Byremark 3 d ( D , D (cid:48) ) > d ( D , D ). But then by the paragraph above, D and D must be tangentat p and D (cid:48) must meet them both orthogonally, so d ( D , D (cid:48) ) = 0. Therefore d ( D , D ) < P, D ), let ∆ ijk denote the geodesic triangle correspondingto each face ijk ∈ F ( P ). Let F ( i ) denote the faces incident i . The link of i is the union of thegeodesic triangles incident i , lnk( i ) = ∪ ijk ∈ F ( i ) ∆ ijk . The link of vertex i always contains its disk D i in a shallow geodesic circle polyhedron. Lemma 3.7.
Let ( P, D ) be a strictly shallow geodesic circle polyhedron. Then for each vertex i ∈ V ( P ) D i ∩ lnk ( i ) = D i .Proof. Consider one of the geodesic triangles ∆ ijk making up the link of i . Let s denote the geodesicarc on the boundary of ∆ ijk corresponding to jk ∈ E ( P ). By lemma 3.6 D i does not cross s orcontain it on its interior (it may contain it entirely, if D i is a great disk, be tangent to it, or bedisjoint from it). On the other hand, D i does contain portions of the arcs of ∆ ijk corresponding to20he other two edges ik and jk on its interior. Thus the boundary of lnk( i ) may be tangent to D i but does not cross into its interior. Furthermore, since ( P, D ) is geodesic, the boundary of lnk( i ) isa simple spherical polygon. Thus D i is contained within lnk( i ).Given that the link of any vertex contains the disk for a proper shallow circle polyhedron, we cannow put bounds on the pairwise inversive distances between any two disks that are not neighborsin P . Lemma 3.8.
Let ( P, D ) be a shallow geodesic circle polyhedron. Then for all non-edge pairs i, j ∈ V ( P ) d ( D i , D j ) ≥ with equality if and only if there exists vertices l, k ∈ V ( P ) such that ilk ∈ F ( P ) , klj ∈ F ( P ) , and d ( D i , D l ) = d ( D l , D j ) = d ( D j , D k ) = d ( D k , D i ) = 0 and d ( D k , D l ) =1 (and therefore ( P, D ) is not strictly shallow).Proof. Let i and j be two distinct vertices of P that are not connected by an edge. Let T bethe induced geodesic triangulation. Let L i and L j denote links of i and j in T . Since T is atriangulation and ij (cid:54)∈ E ( P ), L i and L j are either disjoint, meet at a single vertex, or share aboundary edge between two vertices, say l and k where ilk denotes the face of L i and klj denotesthe face of L j incident along the edge lk . By lemma 3.7, D i ⊂ L i and D j ⊂ L j . Thus if L i and L j are disjoint or meet at a vertex then D i and D j are disjoint. If L i and L j meet along anedge kl then it is possible that D i and D j be tangent. This occurs only if D i is tangent to theboundary of L i and D j is tangent to the boundary of L j . But by lemma 3.6 this occurs if and onlyif d ( D i , D l ) = d ( D l , D j ) = d ( D j , D k ) = d ( D k , D i ) = 0 and d ( D k , D l ) = 1. Recall that a motion ( P, D ( t )) is a path through the configuration space R n (see sec. 3.1). In thissection we investigate the properties of motions that maintain a set of soft constraints defined onthe edges of a polyhedron P . Let an abstract triangulated polyhedron P be given and let D be arealization of the vertex set V ( P ) as a set of disks on S . An unconstrained motion of D definedon an interval I is a continuous family of realizations D ( t ) defined for t ∈ I such that D (0) = D .Each disk D i ∈ D has a continuous motion in D ( t ) given by a continuous path D i ( t ) in R , .If, for each ij ∈ E ( P ) and all t ∈ I the inversive distance across the edge between D i and D j stays in the interval [0 , shallow-constrained . If further, at no pointin the motion does a 4-cycle of edges all have inversive distance 0, we say that the motion is strictlyshallow-constrained . Note that the inversive distances along each edge are constrained to theinterval but are free to change within that interval. We now show that strictly shallow-constrained motions of disk patterns that start as geodesic circlepolyhedra, remain geodesic circle polyhedra throughout the motion.
Theorem 4.1.
Let ( P, D ( t )) be a strictly shallow-constrained motion of a disk pattern defined for t ∈ I . If there exists a t (cid:48) ∈ I such that ( P, D ( t (cid:48) )) is a geodesic circle polyhedron, then ( P, D ( t )) isa geodesic circle polyhedron for all t ∈ I .Proof. For each t ∈ I , let G ( t ) be the induced geodesic graph for ( P, D ( t )) and let n denote thenumber of vertices. Let v t = [ x ( t ) y ( t ) z ( t ) . . . x n ( t ) y n ( t ) z n ( t )] Tt be the vector of coordinates21f the vertex positions p i = ( x i , y i , z i ) of the center p i of each disk D i on S in E . The space of allpossible v t is a compact connected subset of R n which is called the configuration space of G . Let C denote the configuration space of G .Since D ( t ) is continuous, G ( t ) is a continuous family of graph drawings and v t traces out a pathin C . Notice that if G ( t ) is a triangulation, then there is an open neighborhood U of v t where all u ∈ U correspond to triangulations of S . Similarly, if G ( t ) has self-intersecting edges that intersectstrictly on their interior, then there is a neighborhood of G ( t ) in C in which all nearby drawingsare strictly self-intersecting. Additionally, if each edge length is non-zero, then there is an openneighborhood of G ( t ) in C in which all edge lengths are non-zero.Therefore, if there exists at any time t a graph G ( t ) that is not a triangulation of S there mustbe a time 0 < t (cid:48) ≤ t at which G ( t (cid:48) ) either has an edge p i p j shrink to zero length, or a vertex p i intersect an edge p j p k for some triangle ijk ∈ F ( G ). In the first case, the inversive distance d ( D i , D j ) is either −
1, if D i = D j , or is less than −
1, if D i ⊂ D j or D j ⊂ D i . This contradictsthat the motion is shallow-constrained. On the other hand, if p i intersects p j p k , then the disk D i intersects the geodesic shortest path segment between the centers of D j and D k . But thiscontradicts lemma 3.6 (note that D i cannot be the great disk whose boundary ∂D i contains p j p k and have its center on p j p k so condition 1 of lemma 3.6 is not applicable). Similar to the previous section, we show that strictly shallow-constrained motions of disk patternsthat start out as both geodesic and strictly convex circle polyhedra stay strictly convex circlepolyhedra throughout the motion.
Theorem 4.2.
Let ( P, D ( t )) be a strictly shallow-constrained motion of a disk pattern defined for t ∈ [0 , T ) for some T ∈ R ∪ {∞} . Let n > be the number of vertices in P . If ( P, D (0)) is astrictly convex geodesic circle polyhedron, then ( P, D ( t )) is a strictly convex circle polyhedron forall t ∈ [0 , T ) .Proof. We first note that for every ij ∈ E ( P ) D i ( t ) (cid:54) = D j ( t ) at any time t ∈ [0 , T ). To the contrary,if D i ( t ) = D j ( t ) then d ( D i ( t ) , D j ( t )) = − t (cid:48) ∈ [0 , T ) ( P, D ( t (cid:48) )) is not strictly convex. Bylemma 3.2 there must be a time t (cid:48)(cid:48) ∈ [0 , T ) where ( P, D ( t (cid:48)(cid:48) )) has two neighboring faces ijk and kjl such that D i ( t (cid:48)(cid:48) ), D j ( t (cid:48)(cid:48) ), D k ( t (cid:48)(cid:48) ), and D l ( t (cid:48)(cid:48) ) are coplanar.Let Π denote the c-plane containing D i ( t (cid:48)(cid:48) ), D j ( t (cid:48)(cid:48) ), D k ( t (cid:48)(cid:48) ), and D l ( t (cid:48)(cid:48) ). By lemma 2.9, Πis not hyperbolic. If it is parabolic, then the four cycle of edges ij , jl , lk , ki all have inversivedistance 0, contradicting the strictly shallow assumption of the hypothesis. If it is elliptic, thenthe union D i ( t (cid:48)(cid:48) ) ∪ D j ( t (cid:48)(cid:48) ) ∪ D k ( t (cid:48)(cid:48) ) ∪ D l ( t (cid:48)(cid:48) ) covers the sphere S and any other disk D m ( t (cid:48)(cid:48) ) hasa negative inversive distance to at least one of D i ( t (cid:48)(cid:48) ), D j ( t (cid:48)(cid:48) ), D k ( t (cid:48)(cid:48) ), or D l ( t (cid:48)(cid:48) ). Without lossof generality assume that d ( D i ( t (cid:48)(cid:48) ) , D m ( t (cid:48)(cid:48) )) <
0. If im ∈ E ( P ) this contradicts the shallownessassumption. Suppose im (cid:54)∈ E ( P ). By theorem 4.1 ( P, D ( t (cid:48)(cid:48) )) is geodesic. But then by lemma 3.8, d ( D i ( t (cid:48)(cid:48) ) , D m ( t (cid:48)(cid:48) )) >
1, a contradiction.
In this section we show that the extra conditions in the statement of the Koebe-Andre’ev-Thurstontheorem (conditions 1 and 2) allow us to avoid motions of disk patterns in which a disk approaches22 point-disk (either by its area approaching 0 or 4 π ). One of our main analysis tools comes from thefollowing observation. Suppose we reparametrize each disk by the coordinates of its center ( x, y, z )on the sphere S and its spherical radius ρ ∈ [0 , π ], with ρ = 0 , π denoting a disk whose boundaryhas degenerated into a single point. Then the configuration of one disk is a subset of R which lieswithin the region defined by the bounds − ≤ x, y, z ≤ ≤ ρ ≤ π . A parametrization of n disks similarly falls within a bounded region of R n . Consider any motion D ( t ) of disks underthis parametrization defined for t ∈ [0 , T ) for some T ∈ R + ∪ {∞} . The motion traces a path inthe configuration space R n , which by the discussion above falls within a bounded region. Then bythe Bolzano-Weierstrass theorem, there exists a convergent subsequence. We now ask, when is itpossible that as t → T , the radius of some disk D i ( t ) approaches zero?In general, we can by a M¨obius flow collect all disk boundary circles down to a point. Simplyselect two antipodal points neither of which is contained on the boundary of any disk and thencompute the standard M¨obius flow out from one antipodal point towards the other. All points ofthe sphere collect at the sink pole and thus every disk approaches either 0 or 4 π area. To avoid thissituation we must additionally pin-down three disks. Let ijk ∈ F ( P ). We call a motion ( P, D ( t )) a pinned motion with respect to ijk if D i ( t ), D j ( t ), and D k ( t ) are constant throughout the motion. Lemma 4.3.
Let ( P, D ( t )) be a strictly shallow-constrained motion of a circle polyhedron for t ∈ [0 , T ) for some T ∈ R + ∪{∞} with ( P, D ( t )) geodesic. Suppose some disk boundary ∂D i vanishes tothe a point as t → T (but at no earlier time). Let V denote the maximal set of vertices with vanishingboundary circles that are edge-connected to i and V (cid:48) denote the set of vertices with boundaries thatdo not contract but are connected by an edge to a vertex in V . Then | V (cid:48) | ≤ .Furthermore:1. If | V (cid:48) | = 4 , then the vertices form a 4-cycle of edges that limit to inversive distance 0 alongeach edge and the edges do not bound two neighboring faces of P .2. If | V (cid:48) | = 3 , then the vertices form a 3-cycle of edges that limit to inversive distance 0 alongeach edge and it is not possible that all three are pinned. Furthermore, the sum of the overlapangles of the 3-cycle converges to π .3. If | V (cid:48) | ≤ , then V ∪ V (cid:48) = V ( P ) and it is not possible that three disks are pinned in ( P, D ( t )) .Proof. Suppose the boundary of some disk D ( t ) vanishes as t → T . By the Bolzano-Weierstrasstheorem there is a convergent subsequence of ( P, D ( t )). Let D (cid:48) i denote the disk that D i ( t ) convergesto in this subsequence. Call this the convergent set D . By theorem 4.1, ( P, D ( t )) is geodesic for all t ∈ [0 , T ). Inversive Distance Convergence Property (IDCP).
By strict shallowness, for every edge ij ∈ E ( P ), d ( D i ( t ) , D j ( t )) ∈ (0 ,
1] and thus d ( D (cid:48) i , D (cid:48) j ) ∈ [0 , i, j ∈ V ( P ), d ( D i ( t ) , D j ( t )) > d ( D (cid:48) i , D (cid:48) j ) ≥
1. Let V be the maximaledge-connected set (to the vanishing disk D ( t )) of vertices whose boundary circles vanish and V (cid:48) bethe vertices connected by an edge in P to a vertex in V with boundary circles that do not vanish.Every disk in V converges to a single point p in the convergent set. Distinct disk property.
We first claim that it is not the case that D (cid:48) i = D (cid:48) j for any distinct i, j ∈ V (cid:48) . Assume not. The radii of D (cid:48) i and D (cid:48) j are bounded away from 0. Then the inversivedistance is d ( D (cid:48) i , D (cid:48) j ) = − roof that | V (cid:48) | < . We next claim that | V (cid:48) | <
5. For contradiction assume that V (cid:48) contains atleast five vertices. Let D (cid:48) , D (cid:48) , D (cid:48) , D (cid:48) , and D (cid:48) denote the disks corresponding to these vertices.By the discussion above these five disks are distinct. Since each of them is the convergence disk ofa disk overlapping some disk in V and everything in V converges to a point p , the boundary circles ∂D (cid:48) , . . . , ∂D (cid:48) all contain p . Now, by theorem 4.1, ( P, D ( t )) is geodesic for all t ∈ [0 , T ). Then bycorollary 2.8 at least two of the disks must have a non-shallow overlap, which contradicts (IDCP). Analysis of | V (cid:48) | = 4 . If | V (cid:48) | = 4, the four disks D (cid:48) , D (cid:48) , D (cid:48) , and D (cid:48) meet at a point. Bylemma 2.5, in order to not contradict (IDCP), two pairs of these disks are tangent and the pairs aremutually orthogonal. Without loss of generality assume that ( D (cid:48) , D (cid:48) ) are tangent and ( D (cid:48) , D (cid:48) )are tangent, and the two pairs are mutually orthogonal.Then D (cid:48) , D (cid:48) , D (cid:48) , D (cid:48) is a cycle of disks each of which is (cyclically) orthogonal to the next.Thus at all t ∈ [0 , T ) at least one of the overlap angles for the edges 12, 13, 24, and 34 is greaterthan 0 but in the limit equal 0. Assume that the cycle of edges (12 , , ,
41) bounds two adjacentfaces of P . Then all of the other vertices must be in V and all of the disks save D through D limitto the point common to D (cid:48) , D (cid:48) , D (cid:48) , D (cid:48) . But then the geodesic triangulation induced by ( P, D ( t ))approaches the configuration of geodesic triangles p p p , p p p , p p p , p p p where p i denotes thecenter of D (cid:48) i and p is the point corresponding to the collapsed disks in V . But the outer boundaryof this configuration is a 4-cycle ( p , p , p , p ) and thus not a triangulation of S . But then forsufficiently large t , the geodesic graph induced by ( P, D ( t )) is not a triangulation. This contradictstheorem 4.1. Analysis of | V (cid:48) | = 3 . In this case the three disks D (cid:48) , D (cid:48) , D (cid:48) must all have pairwise inversivedistances in [0 ,
1] and thus 123 is a cycle of edges in P . By lemma 2.5, the sum of the overlap angles is π . Now suppose all three disks are pinned: D ( t ) = D (cid:48) , D ( t ) = D (cid:48) , and D ( t ) = D (cid:48) are constant.By strict shallowness, none of the three are tangent at p , and thus any small disk containing p willoverlap at least one of the disks by more than π/
2. Thus any of the disks corresponding to verticesin P will overlap at least one of them by more than π/
2. But any of the disks in V converge to smalldisks containing p , and thus at some time t (cid:48) ∈ [0 , T ) there is a disk D i for some i ∈ V overlappingone of the disks, say D ( t (cid:48) ) such that d ( D i ( t (cid:48) ) , D ( t (cid:48) )) < Analysis of | V (cid:48) | ≤ . In this case, since V (cid:48) is the boundary of a connected set of vertices V on atriangulated polyhedron and | V (cid:48) | ≤ P is either in V or in V (cid:48) . Then if any three disks are pinned at least one of the disks in V is pinned and thus it cannotvanish, a contradiction.We now show that the extra conditions from the Koebe-Andre’ev-Thurston theorem (in thelimit), coupled with three disks whose radii are bounded away from zero guarantees that no diskvanishes for any strictly shallow motion of a disk pattern that starts out geodesic and has threepinned disks. Theorem 4.4.
Let ( P, D ( t )) be a strictly shallow motion of a disk pattern for t ∈ [0 , T ) for some T ∈ R + ∪ {∞} such that ( P, D (0)) is geodesic. Suppose further that for one face ijk the disks D i ( t ) , D j ( t ) , and D k ( t ) are constant. Finally, letting d ( e ( t )) denote the inversive distance between thedisks corresponding to the endpoints of an edge e ∈ E ( P ) at time t ∈ [0 , T ) , suppose that: . If e , e , e form a closed loop of edges such that at some t ∈ [0 , T ]lim t → t (cid:88) k =1 arccos d ( e k ( t )) ≥ π, then e , e , and e bound a face of P .2. If e , e , e , and e form a closed loop of edges such that at some t ∈ [0 , T ] , lim t → T (cid:88) i =1 arccos d ( e i ) = 2 π then e , e , e , and e bound the union of two faces of P .Then no disk in D ( t ) vanishes as t → T .Proof. Since ( P, D (0)) is geodesic, by theorem 4.1, ( P, D ( t )) is geodesic for all t ∈ [0 , T ).Suppose now that some disk’s radius vanishes and let V be a maximally edge-connected set ofdisks whose radius vanishes. Let V (cid:48) denote the vertices of V ( P ) \ V connected by an edge in E ( P )to a vertex in V . By lemma 4.3, | V (cid:48) | ≤ | V (cid:48) | ≤
2, then it is not possible that three disks are pinned, a contradiction.Suppose that | V (cid:48) | = 3. Then by lemma 4.3 the vertices of V (cid:48) are connected by a three cycleof edges e , e , and e whose angle sum approaches π . Then by hypothesis e , e , and e bounda face of P . Then every vertex not in V (cid:48) must be in V . But none of the vertices in V can beconstant, since their boundary circles contract to a point. But then the vertices in V (cid:48) must bepinned contradiction lemma 4.3.Finally, suppose that | V (cid:48) | = 4. Then by lemma 4.3 there is a 4-cycle of edges e , e , e , and e connecting the vertices of E ( P ) whose inversive distances all approach 0 and this 4-cycle does notbound two neighboring faces of P . But by hypothesis, these edges must bound two neighboringfaces, a contradiction. We now develop the infinitesimal rigidity of strictly convex circle packings. This section generalizesthe results of [6] in the context of convex circle polyhedra.
Let ( P, D ) be a triangulated circle polyhedron with n vertices. Since P is triangulated it has m = 3 n − D as a configuration in the configuration space R n : D = [ a b c d . . . a n b n c n d n ] T ∈ R n . Where the coordinates of each disk D i = ( a i , b i , c i , d i ) are normalized to lie on the de Sitter sphere.Then d ( D i , D j ) = −(cid:104) D i , D j (cid:105) , . Define a measurement function f : R n → R n − , which measuresthe negative length of each edge ij ∈ E ( P ) as well as half the squared Minkowski norm of each25ertex. Thus f has 3 n − f ij indexed by the edges of P and n measurements f i indexed by the vertices of P . The entries of f are defined by f ij ( D ) = − d ( D i , D j ) = (cid:104) D i , D j (cid:105) , (5.1)and f i ( D ) = (1 / (cid:104) D i , D i (cid:105) , . (5.2)Let J denote the Jacobian matrix of f . J is called the rigidity matrix of ( P, D ). Each rowcorresponds to either an edge ij or a vertex i of P . The columns represent the 4 n coordinates ofthe configuration D . The row corresponding to the ij edge of J is zero everywhere except at theentries corresponding to the coordinates of D i = ( a i , b i , c i , d i ) and D j = ( a j , b j , c j , d j ). The entriesof the row are . . . a i b i c i d i . . . a j b j c j d j . . .J ij = ( . . . . . . a j − b j − c j − d j . . . . . . a i − b i − c i − d i . . . . . . ) . (5.3)(Notationally, in the equation above, the first line gives the column labels and the second line thedefinition of the row J ij .) The row corresponding to the vertex i of J is similarly zero everywhereexcept at the entries corresponding to the coordinates of D i : . . . a i b i c i d i . . .J i = ( . . . . . . a i − b i − c i − d i . . . . . . ) . (5.4)By construction, the null space of J corresponds to the space of infinitesimal motions of the diskset D that maintain the inversive distances on each edge ij ∈ E ( P ) and maintain that each vertex i ∈ V ( P ) remains on the de Sitter sphere. Since the Lorentz transformations map d S to itselfand maintain the Minkowski inner product, they form a 6-dimensional family of trivial motions.Thus the dimension of the null space of J must be at least 6, corresponding to the ininitessimalLorentz transformations. By the rank-nullity theorem J has rank at most 4 n −
6. If the rank of J equals 4 n −
6, then the only motions of ( P, D ) must be Lorentz transformations, and thereforetrivial. In this case we say that ( P, D ) is infinitesimally rigid . Otherwise, we say that ( P, D ) isinfinitesimally flexible. Proof of infinfinitesimal rigidity of strictly shallow convex geodesic circle polyhedra.
We now prove that strictly convex geodesic circle polyhedra are infinitesimally rigid. We firstremind the reader of the Cauchy index lemma, which have at this point become a standard toolfor analyzing polyhedra. We state Cauchy’s index lemma in two forms: the weak Cauchy indexlemma, which was used by Cauchy to prove the global rigidity of convex Euclidean polyhedra; anda stronger form, which implies the weak.Let G be a graph drawn on a topological sphere with arcs representing edges such that no twoarcs intersect and no vertex touches an edge it is not incident to. Call this a topologically planargraph drawing . The orientation on the sphere imposes a cyclic ordering of the edges incidentaround each vertex. Label each edge of G with a sign +, − , or 0. Around any vertex of G in orderwe may list the signs in cyclic order (ignoring 0) and count the number of sign changes in the list.Call the number of sign changes around a vertex v the index of v , denoted ind( v ). Since the signchanges are a rotation around each vertex, the index of any vertex is even. Cauchy’s weak indexlemma follows. 26 emma 5.1 (Weak Cauchy index lemma) . Let G be a topologically planar graph drawing. Labeleach edge of G with + , − , or such that at least one edge of G is not labeled 0. Let ind ( v ) denotethe index of each vertex v ∈ V ( G ) . Then there exists at least one vertex v incident to an edgelabeled + or − such that ind ( v ) ∈ { , } . The strong form of the lemma involves the sum of the indices and easily implies the weak form.
Lemma 5.2 (Strong Cauchy index lemma) . Let G be a geodesic graph drawing with n vertices.Label each edge of G with + , − , or such that at least one edge of G is not labeled 0. Let ind ( v ) denote the index of each vertex v ∈ V ( G ) . Let n (cid:48) denote the number of vertices incident an edgelabeled + or − . Let s = (cid:80) v ∈ V ( G ) ind ( v ) . Then s ≤ n (cid:48) − . We now establish the following lemma, which implies the infinitesimal rigidity of strictly convexcircle polyhedra.
Lemma 5.3.
Let ( P, D ) be a strictly convex geodesic circle polyhedron with P not a tetrahedron.Let J be its rigidity matrix. Then the rank of J T is n − .Proof. By the rank-nullity theorem, rank J T = 4 n − J T v = has no non-trivialsolutions. We prove this by contradiction. Assume J T v = and v (cid:54) = . The vector v has an entryfor each edge ij ∈ E ( P ), which we denote by ω ij , and an entry for each vertex i ∈ V ( P ), which wedenote by ω i .Consider the four rows of J T corresponding to a vertex i ∈ V ( P ). The condition that J T v = restricted to the four coordinate rows of vertex i in J T is equivalent to the vector equation = ω i ( a i , − b i , − c i , − d i ) T + (cid:88) ij ∈ E ( P ) ω ij ( a j , − b j , − c j , − d j ) T . (5.5)We first remark that from this formulation we see immediately that it is not possible for ω ij = 0for all ij ∈ E ( P ) and ω i (cid:54) = 0 since the coordinates of a disk on the de Sitter sphere cannot all be0. Obviously, if (5.5) is satisfied, then we may reflect through the spatial coordinates to obtain = ω i ( a i , b i , c i , d i ) T + (cid:88) ij ∈ E ( P ) ω ij ( a j , b j , c j , d j ) T . (5.6)Now, label each edge ij ∈ E ( P ) with a +, a − , or a 0 depending on whether ω ij > ω ij <
0, or ω ij = 0. Since at least one edge corresponds to a non-zero entry ω ij in v , there must be an edge oflabeled with either a + or a − . Then by the weak Cauchy index lemma there is a vertex 0 ∈ V ( P )such that the counterclockwise order of labeled edges (those that are either + or − ) around 0 hasat most two sign changes.Assume first that there are two sign changes. Let 1 , . . . , k ∈ V ( P ) denote the labeled neighborsof 0 in order indexed so that edge 01 is the labeled with a +, 0 k is labeled with a − , and i is thelargest index such that 0 i is labeled with a +. Without loss of generality assume that no disk hasarea equal to 2 π since we may apply a nearby M¨obius transformation to ensure this property, andby construction the M¨obius transformations are the trivial motions and thus do not change therank of our Jacobian.Denote the vertices of the cap polyhedron triangle fan for vertex 0 by v , v , . . . , v k and let v , v (cid:48) , . . . , v (cid:48) k denote the vertices of the corresponding convex Euclidean triangle fan ((as defined in27ection 3.2). Since v , v (cid:48) , . . . , v (cid:48) k is convex, there is a Euclidean plane Π through v that separates v (cid:48) , . . . , v (cid:48) i from v (cid:48) i +1 , . . . , v (cid:48) k and contains none of them on its interior.Let Π (cid:48) denote the hyperplane through the origin of R , containing Π. By construction, if D j has area less than 2 π , then v j lies on the same side of Π as v (cid:48) j and v (cid:48) j lies on the same side of Π (cid:48) asthe ray D ∗ j . On the other hand if D j has area greater than π then v j and v (cid:48) j are on opposite sidesof Π and the ray D ∗ j and v (cid:48) j lie on the same side of Π (cid:48) (see section 3.2). Thus Π (cid:48) separates the rayscorresponding to the disks with + signs from those with − signs. Now, scale each of D ∗ . . . D ∗ k bythe appropriate value ω j . We have that D ∗ j lies on the same side of Π (cid:48) as ω j D ∗ j when ω j > (cid:48) separates the vectors with positive ω j from those withnegative ω j , this scaling moves all scaled vectors to the same side of Π (cid:48) . But then all the scaledvectors lie on the same side of a hyperplane through the origin and therefore their sum cannot be . Thus (5.6) is not satisfiable with two sign changes.Essentially the same argument holds when the labels all have the same sign. Here we select Πto be a plane through v such that all other v (cid:48) i lie on the same side. Then all the values are eitherall scaled by a positive scalar or all scaled by a negative scalar. In either case the same argumentshows that (5.6) cannot be satisfied. The lemma follows.From this lemma a version of the Legendre-Cauchy-Dehn lemma for strictly convex circle poly-hedra follows immediately. Corollary 5.4.
Strictly convex circle polyhedra are infinitesimally rigid.
Remark.
The above result on the infinitesimal rigidity of strictly convex circle polyhedra mayalso be obtained by applying a theorem from the unpublished manuscript in [21] to the connectionbetween circle polyhedra and de Sitter polyhedra.
Let ( P, D ) be a convex circle polyhedron with n vertices such that every disk D i = ( a i , b i , c i , d i ) isgiven in de Sitter coordinates and let J be its (4 n − × n rigidity matrix. J has rank 4 n − ?? . The nullspace of J is 6-dimensional, corresponding to the dimension of the Lorentzgroup on R , (equivalently the M¨obius group on S ). As in [12], we want to mod out the M¨obiusgroup by the coordinates of three disks. We do this by fixing one face ijk ∈ F ( P ) and removingthe b i , c i , d i , b j , and c j columns and one of the columns corresponding to the spacial coordinatesof D k . The resulting matrix J ijk is a square (4 n − × (4 n −
6) matrix called the square rigiditymatrix for the face ijk ∈ F ( P ). We now show that this matrix has full rank. Lemma 5.5.
The rank of the square rigidity matrix J ijk is n − .Proof. We begin by pinning b i , c i , d i , b j , and c j . By lemma 2.2 any continuous motion fixing thesefive coordinates must fix the two disks D i and D j , and therefore all 8 coordinates of the two disks.In standard rigidity theory parlance, this is called pinning . Fixing two disks determines a uniqueM¨obius flow of the third disk D k . The flow must vary at least one of the space coordinates of D k . Pinning this column stops the flow, since the derivative of the flow must be zero at this pointand all other coordinates are determined by these 6 choices. Without loss of generality, assumethis is the b k column. Now, add 6 rows to J , one for each of the pinned coordinates, which is 0everywhere except with a 1 in the pinned coordinate column. We call the resulting 4 n × n matrix¯ J the pinned rigidity matrix . 28ssume there is a non-trivial v such that ¯ J v = 0. By the discussion above, v must have 0’s in allof its rows corresponding to the coordinates of vertices i , j , and k and is therefore not the derivativeof a M¨obius transformation. But by construction, J v = 0 as well and thus v is in the kernel of J . But the kernel of J is corresponds exactly to the derivatives of the M¨obius transformations on D , a contradiction. Therefore ¯ J has rank 4 n . Then every column of ¯ J is linearly independent ofthe others and in particular the columns corresponding to the 4 n − J and the corresponding columns must also be linearly independent in J . Thus if we delete the 6columns corresponding to the pinned coordinates from J we obtain a 4 n − × n − In [12], Connelly and Gortler describe an algorithm for computing a tangency packing of a givenabstract triangulated polyhedron using a finite number of operations they call a flip-and-flow op-eration. Suppose P and P (cid:48) are abstract triangulated polyhedra that combinatorially differ by asingle edge flip operation and ( P, D ) is a tangency packing of disks D in the plane. Let P − denotethe common vertices, edges, and faces of P and P (cid:48) . P − has the same number of vertices, one feweredge, and two fewer faces than P and P (cid:48) . The tangency packing ( P, D ) is infinitesimally rigid.The removal of one edge gives rise to a single non-trivial infinitesimal motion and the configurationspace of disks realizing tangencies along the edges of P − is a 1-dimensional manifold they call thepacking manifold. Moving along this manifold gives rise to a continuous motion D ( t ) of the diskswith D (0) = D which maintains tangencies along every edge of P − . Meanwhile as t increases, theinversive distance between the disks corresponding to the endpoints of the edge e ∈ E ( P ) /E ( P − )is initially equal to 1 and strictly increases while the inversive distance between the two disks cor-responding to the endpoints of the edge e (cid:48) ∈ E ( P (cid:48) ) /E ( P − ) is initially greater than one and strictlydecreases. The motion continues until the inversive distance along e (cid:48) becomes 1 at some finite time t . At that point ( P (cid:48) , D ( t )) is a tangency packing. Thus any combinatorial flip can be realized bya continuous motion that keeps all neighboring circles tangent except across the edge that needs toflip.Any two triangulations with the same number of vertices are connected by a finite sequence ofcombinatorial flip operations. Thus, if there exists any canonical triangulation that can be shownto have a circle packing, then all triangulations with the same number of vertices are reachablevia the Connelly-Gortler flip-and-flow operation. In fact, such canonical circle packings are easyto come by. Select any three mutually tangent disks in the plane. These disks define an intersticeregion and there is a unique disk on the interior of this interstice whose boundary is tangent tothe other three. Thus inductively a circle packing with n disks whose contact graph is an abstracttriangulated polyhedron may be obtained by starting with any three mutually tangent disks andrepeatedly filling in interstices inductively n − S may be obtained via stereographic projection onto the sphere such that nodisk has area greater than 2 π . The conical cap polyhedron for P is known as its Koebe polyhedron and is known to be convex, which by lemma 3.3 implies ( P, D ) is convex in our sense. Furthermore,the geodesic arc between to neighboring disks stays entirely within their union and passes throughthe point of tangency. Thus, because the disks are each disjoint, the induced geodesic graph is ageodesic triangulation. In our terminology, the tangency circle packing theorem is: Theorem 6.1 (Tangency Circle Packing Theorem (TCPT)) . Let ( P, w ) be a weighted abstracttriangulated polyhedron where w ( ij ) = 1 for all ij ∈ E ( P ) . Then there exists a geodesic, strictlyconvex circle polyhedron ( P, D ) realizing ( P, w ) . From this starting point we now show that the flow used by Connelly and Gortler to obtaintheir proof of TCPT may be extended past tangency to bring disks into shallow overlaps. Given anabstract shallow-weighted triangulated polyhedron (
P, w ) our starting point is the tangency packingwith P as its contact graph guaranteed by TCPT. From there we apply a series of flow operationsto correct, edge-by-edge, the inversive distances from tangency to whatever the desired inversivedistance is in ( P, w ). If (
P, w ) is strictly shallow, then we can correct each edge by a single flow.When (
P, w ) is not strictly shallow, but has some desired inversive distances of 0, we require a morenuanced limiting argument.
We first prove a version of our main theorem restricted to strictly shallow polyhedra. In the nextsection we extend this result to handle non-strictly shallow polyhedra. The key difference is thatstrictly shallow polyhedra that are strictly convex maintain strict convexity under certain motionsdue to theorem 4.2 and our main argument in this section requires strict convexity in order to applythe infinitesimal rigidity results from the last section.
Let ( P, D ) be a strictly shallow, strictly convex circle polyhedron. Let J q denote the square rigiditymatrix of ( P, D ) with respect to some face q ∈ F ( P ). By lemma 5.5 the rank of J q is 4 n −
6. Now,consider removing one of the edge constraint rows imposed by P for some edge e = ij ∈ E ( P ).The result is to remove the row of J q corresponding to e . Denote P with the edge e removed by P − e and J q with the e row removed by J − e . Since J q has full rank, so does J − e . Call J − e the free-edge rigidity matrix of ( P, D ) with respect to q and e . We have that: Lemma 6.2.
The free-edge rigidity matrix J − e of a convex circle polyhedron ( P, D ) with respectto a face q and edge e is non-singular. Finally, as in [12], we note that the matrix J q has a useful block form. Let i , j , and k denotethe vertices of the pinned face q . Each of the edge rows of J q corresponding to the edges ij , jk , and ki have non-zero entries only in the columns of J q that correspond to the coordinates of the disks D i , D j , D k that were not already removed from J to obtain J q . This is also the case for the rowscorresponding to the vertices i , j , and k designed to keep D i , D j , and D k in the de Sitter sphere.For D i , D j , and D k , six of their coordinate columns were removed from the rigidity matrix J toobtain J q . Simultaneously removing the remaining six coordinate columns along with the edge rowsfor ij , jk , and ki and the vertex rows for i , j , and k from J q leaves us with a square matrix of size30 n −
12. Its edge rows correspond to the edges of P that have at least one endpoint not in { i, j, k } .Its vertex rows and columns correspond to the vertices of P that are not i , j , or k . Call this the unmarked rigidity matrix and denote it by J u . From the block form and previous lemma wehave that: Lemma 6.3.
The unmarked rigidity matrix J u of a strictly convex circle polyhedron ( P, D ) withrespect to a face q and edge ij is non-singular. The configuration space ( P, D ) with three disks D i , D j , and D k fixed is R n − .Now, select an edge e ∈ E ( P ) that is not ij , jk , or ki . Denote by S − eP the subset of theconfiguration space that represent all configurations ( P, D (cid:48) ) that are strictly shallow strictly convexgeodesic circle polyhedra and for every edge e (cid:48) ∈ E ( P ) that is not e , the inversive distance betweenthe disks corresponding to the endpoints of e (cid:48) in ( P, D (cid:48) ) is the same as the inversive distance betweenthe endpoints of e (cid:48) in ( P, D ). In other words, ( P, D ) and ( P, D (cid:48) ) agree in inversive distances on allbut one edge. We enumerate several conditions on ( P, D (cid:48) ) that follow this construction. Because( P, D (cid:48) ) is strictly shallow and geodesic, by lemma 3.8 non-adjacent disks are disjoint.We remark that the conditions for a point to be in S − eP are open conditions on the n − S − eP . Lemma 6.4. S − eP is a 1-dimensional smooth manifold.Proof. The argument here is essentially Connelly and Gortler’s with modified details. Let ( P, D ) bea circle polyhedron corresponding to a point p ∈ S − eP and consider a sufficiently small neighborhood U of p in S − eP . Every configuration ( P, D (cid:48) ) corresponding to a point of U has a radius for each ofthe free disks in D (cid:48) in (0 , π ) and all non-edge inversive distances are greater than 1. Locally weneed only consider the 3 n −
10 inversive distances corresponding to the edges of P that are not thefree-edge e and are not incident the pinned face ijk and the constraints keeping each of the n − D (cid:48) constrained to the de Sitter sphere. The partial derivatives of these constraints are the4 n −
13 rows of J u corresponding to the edges that are not e , which by lemma 6.3 is non-singular.The result then follows by the implicit function theorem. Let (
P, w ) be an abstract weighted triangulated polyhedron. Let ( P, D ) be a circle polyhedron withthe same abstract polyhedron P . We call ( P, D ) w -bounded if d ( D i , D j ) ≥ w ( ij ) for every edge ij ∈ E ( P ).We now use the Connelly-Gortler flow to adjust inversive distances. Unlike their use, we donot need to make combinatorial changes to our polyhedron, since our starting point is a tangencypacking with the correct combinatorics. We simply adjust the inversive distances along one freeedge at a time, keeping the inversive distance across all other edges fixed.Assume we have a desired set of weights w we want to see realized as inversive distances for eachedge in P . Throughout our use of flows we maintain the invariant that our polyhedron ( P, D ) is w -bounded. In the remainder let d ( e ) denote the inversive distance across an edge e in ( P, D ), d ( e ( t ))denote the inversive distance across the edge in ( P, D ( t )), and w ( e ) denote the desired inversivedistance. 31e begin with a w -bounded polyhedron ( P, D ). Select an edge e for which d ( e ) > w ( e ). Weapply a motion called a flow (defined below) to ( P, D ) that strictly decreases the inversive distancealong e while maintaining the inversive distance across every other edge. In order to mod out theM¨obius transformations from this motion, we pin three disks corresponding to a face of ijk ∈ P .This choice necessarily fixes the inversive distances across the edges ij , jk , and ki . Therefore wechoose ijk so that e is not one of its edges.With three disks pinned, the configuration of D is described by a point in R n − . Let J u be theunmarked rigidity matrix of ( P, D ) with respect to ijk . Since J u is non-singular, its inverse exists.Let D (cid:48) := J − u v e where v e is the vector whose entries are all 0 except that the entry correspondingto the edge e is -1. D (cid:48) defines a velocity field on the n − P, D ) that keeps alledge inversive distances fixed save the free edge e , which has an inversive distance that decreasesat a constant rate. Since J u is non-singular at ( P, D ) it is non-singular over a Zariski open subset U of R n − and thus defines a smooth velocity field over U including ( P, D ) that we use to set upa system of ordinary differential equations (ODEs).Starting at ( P, D ) and integrating forward in time obtains a maximal trajectory ( P, D ( t )) forsome time interval [0 , T ) where T ∈ R ∪ {∞} , which we call the flow trajectory . By the standardtheory of ODEs this trajectory will either leave any compact set in U or continue for infinite time.Since the inversive distance d ( e ( t )) decreases at a constant rate while all other edge inversivedistances are fixed, at some finite time t (cid:48) ∈ [0 , T ) we will either achieve an inversive distance of w ( e ) along the edge e in ( P, D ( t (cid:48) )) or T is finite and ( P, D ( T )) (cid:54)∈ S − eP . In the former case, weflow along the motion until from t = 0 to t (cid:48) to obtain a new polyhedron ( P, D ( t (cid:48) )) at which point d ( e ( t (cid:48) )) = w ( e ). In the latter case ( P, D ( T )) is either not strictly shallow, not strictly convex, orsome disk has degenerated to a point. We now add conditions necessary to ensure that this lattercase does not occur. Lemma 6.5.
Let ( P, w ) be a strictly shallow weighted abstract triangulated polyhedron, not a tetra-hedron, and let ( P, D ) be a strictly shallow strictly convex geodesic circle polyhedron that is w -bounded. Let e be an edge such that d ( e ) > w ( e ) and ijk be a pinned face that is not bounded by e . Suppose further that if e , e , e form a closed loop of edges such that (cid:80) k =1 arccos w ( e k ) ≥ π then e , e , and e bound a face of P .Then at a finite t (cid:48) the flow reaches a point of the trajectory such that d ( e ( t (cid:48) )) = w ( e ) and forall ≤ t (cid:48)(cid:48) ≤ t (cid:48) ( P, D ( t (cid:48)(cid:48) )) is in S − eP . In other words, if we have the first condition from the Koebe-Andre’ev-Thurston theorem toour desired weight function w , then our flow trajectory can be used to correct any edge not yetmatching our desired weight. (The second condition from the Koebe-Andre’ev-Thurston theoremon 4-cycles of edges does not apply in this case because our polyhedra are strictly shallow. For afour-cycle of edges to have overlaps summing to 2 π when all overlap angles are between 0 and π/ π/ Proof.
Let [0 , T ] be an interval of the flow trajectory in which d ( e ( t )) ≥ w ( e ) for all t ∈ [0 , T ]. Westart strictly shallow, and we maintain all inversive distances except along e , which itself remainsstrictly shallow (since it is decreasing and d ( e ( t )) > w ( e ) > P, D ( t )) is strictlyshallow for all [0 , T ]. Then by theorems 4.1 & 4.2, the trajectory remains geodesic and strictlyconvex.Then, by our extra condition that closed loops of three edges having angle sums of ≥ π mustbound a face, and the fact that the only inversive distance that is changing is along e , condition 132f theorem 4.4 is maintained for t ∈ [0 , T ], and therefore no disk vanishes throughout the motion.(Condition 2 is maintained vacuously because our polyhedra and desired weights w are strictlyshallow.)Then ( P, D ( t )) is in S e − P for every finite interval maintaining d ( e ( t )) ≥ w ( e ). But the d ( e ( t ))is decreasing at a constant rate as t increases. Therefore there must be a finite time t (cid:48) such that d ( e ( t (cid:48) )) = w ( e ).We now prove the Koebe-Andre’ev-Thurston theorem for strictly shallow circle polyhedra. Inthe next section, we extend this to the full Koebe-Andre’ev-Thurston theorem. Theorem 6.6 (Strictly shallow Koebe-Andre’ev-Thurston) . Let ( P, w ) be a strictly shallow abstracttriangulated polyhedron, not a tetrahedron, and let the following condition hold. If e , e , e form aclosed loop of edges with (cid:80) i =1 arccos w ( e i ) ≥ π , then e , e , and e bound a face of P . Then thereexists a geodesic circle polyhedron ( P, D ) realizing ( P, w ) .Proof. Start with the geodesic, strictly convex tangency circle polyhedron ( P, D ) with the com-binatorics P guaranteed by the tangency circle packing theorem. Since all inversive distances in( P, D ) are 1, ( P, D ) is w -bounded.We now produce a finite sequence of polyhedra ( P, D ) , ( P, D , . . . , ( P, D k ) such that ( P, D ) =( P, D ) and ( P, D k ) realizes ( P, w ) and each of which agrees with (
P, w ) on more edge than theprevious polyhedron in the sequence. Let e be an edge of ( P, D i ) such d ( e ) > w ( e ). By ourhypothesis (which is maintained inductively), by lemma 6.5 we may apply the flow ( P, D i ( t )) tocontinuously decrease d ( e ( t )) until it equals w ( e ), thus obtaining the next polyhedron ( P, D i +1 )which agrees with w on one more edge (namely e ). In the worst case, we must apply this operation | E ( P ) | times to correct one edge at a time until they all match w . Remark.
As pointed out in a personal correspondence by Steven Gortler it is likely that theingredients above may be used to flow all disks from tangency to their desired inversive distancesnearly simultaneously. First select a single triangle of P and flow to obtain the desired inversivedistances along its edges. Then fix this triangle and flow the remaining disks. We leave the detailsof this approach to future work. Proof of Koebe-Andre’ev-Thurston.
We now use a limiting argument to obtain the proof of the fullKoebe-Andre’ev-Thurston theorem. Let (
P, w ) be an abstract weighted triangulated polyhedronthat is shallow, but not strictly shallow and satisfies the extra conditions of KAT. Then thereis some 4-cycle of edges ij , jl , lk , ki bounding faces ijk and kjl such that w ( ij ) = w ( jl ) = w ( lk ) = w ( ki ) = 0. Let w : E ( P ) → R + be the weight function defined by w ( ij ) = w ( ki ) = 0, w ( jk ) = w ( jk ), and w ( e ) = 1 for all other edges. Then ( P, w ) is a strictly shallow weight functionand by theorem 6.6 we obtain a strictly shallow strictly convex geodesic realization ( P, D ). Byconstruction, this realization now has the correct desired inversive distances along the edges of theface ijk ∈ F ( P ).Now, for any fixed (cid:15) > w (cid:15) denote the function such that w (cid:15) ( ij ) = w (cid:15) ( ki ) = 0, w (cid:15) ( jk ) = 1and for all other edges e , w (cid:15) ( e ) = max { (cid:15), w ( e ) } . Consider the sequence of weighted abstractpolyhedra ( P, w ) , ( P, w / ) , ( P, w / ) , . . . . Starting with ( P, D ) we produce a sequence of circlepolyhedra ( P, D ) , ( P, D / ) , ( P, D / ) , . . . as follows. Each ( P, D / i +1 ) is obtained from the pre-vious ( P, D / i ) by pinning the face ijk and flowing across each of the remaining edges that are33ot yet at their correct inversive distance. Thus, between each consecutive pair of polyhedra inthe sequence, we have a continuous motion of disks which is geodesic, strictly shallow, and strictlyconvex. Throughout the motion each inversive distance is (non-strictly) decreasing. Furthermore,for any edge e such that w ( e ) >
0, after finitely many polyhedra in the sequence the edge e achieves w ( e ) as its inversive distance which is then held constant throughout the rest of the motion. Forthe edges where w ( e ) = 0 (that are not ij or ki ) the inversive distances approach 0 as the motioncontinues.By the Bolzano-Weierstrass theorem, there is a convergent subsequence of this motion thatconverges to a set of disks D (cid:48) v for each v ∈ V ( P ). If we require that ( P, w ) satisfies conditions 1and 2 of the Koebe-Andre’ev-Thurston theorem, then our motion satisfies conditions 1 and 2 oftheorem 4.4 and thus no disk vanishes. Therefore every disk D (cid:48) v is a real disk and for every edge uv ∈ E ( P ) we must have that d ( D (cid:48) u , D (cid:48) v ) = w ( uv ).This completes the proof of theorem 3.1. This paper presents a new proof of the general Koebe-Andre’ev-Thurston theorem which extends theideas of Connelly and Gortler’s proof in the tangency case to handle circle packings with overlaps.Still open are the problems of existence (unsolved) and uniqueness (solved only for non-unitaryconvex circle polyhedra) of circle packings with mixes of overlapping and non-overlapping circles.Unlike the overlap packings of the Koebe-Andre’ev-Thurston theorem, such polyhedra need notbe convex nor geodesic. That said, however, the flow trajectory can, in fact, be used to producepackings with a mix of overlaps and non-overlaps, both for convex geodesic circle polyhedra andcircle polyhedra that are neither convex nor geodesic. An example is shown in figure 4. This isan intriguing experimental observation and we hope that the present work may serve as a furtherlaunching off point in the effort to characterize the existence and uniqueness of circle packings onthe sphere.Also, in this paper we have made extensive use of two related notions: geodesy and convexity.The Ma-Schlenker counter-examples to the global rigidity of general inversive distance circle pack-ings give rise to a family of examples (explored in [4]) that are non-convex and have realizationsthat are geodesic, as well as realizations that are non-geodesic. Contrary to such examples, our in-vestigation in the present paper has shown that in some cases geodesic circle polyhedra and convexcircle polyhedra are intimately related. In some sense this seems to connect to the all of a convexEuclidean polyhedra are visible to any point on its interior, whereas some points on the interiorof a non-convex Euclidean polyhedra may see all of the polyhedra, but there always exist somepoints on the interior for which some of the polyhedron is obstructed. We leave the reader with onefurther intriguing question: is it the case that a circle polyhedron is convex if and only if all M¨obiustransformations of the polyhedron are geodesic? This would align with both the Ma-Schlenkerexamples and much of the development in this paper, but remains an open question.
Acknowledgements.
We especially thank Philip Bowers for a great many helpful conversations,challenges, and encouragement during the development of the present work. We also very muchthank Bob Connelly for the invitation to Cornell in the fall of 2019 that led the author to begindeveloping code for experimenting with flip-and-flow and lead eventually to this paper. We thankKen Stephenson for pointing out a Bolzano-Weierstrass based proof of the Ring Lemma for tan-34 a) (b)(c) (d)
Figure 4: Our proof guarantees that we can move along the flow trajectory to obtain any desiredshallow packing, but do not rule out using the flow to obtain packings with non-shallow inversivedistances. This figure shows the application of a flow to obtain a circle polyhedron with inversivedistances that are shallow and inversive distances that are larger than 1. (a) is a configurationobtained numerically by the the flow. The gray edge represents our free edge and the green vectorsfrom the center of each circle represent the vector of the motion that is maintaining all otherinversive distances. In (b), (c), and (d) we continue to apply the flow on the same edge. In (b), thepolyhedron is still convex and geodesic. In (c), the polyhedron has ceased being convex (we wouldneed to flip the gray edge at an intermediate point between (b) and (c) to maintain convexity.In (d) the polyhedron is both non-convex and non-geodesic as the center of one circle has movedthrough the edge of one of its link triangles.gency packings which informed our analysis of disk motions. We also thank Steven Gortler, TonyNixon, and Bernd Schulze for helpful conversations and guidance. Finally, we thank both BrownUniversity’s Institute for Computational and Experimental Research Mathematics (ICERM) and35he American Institute of Mathematics (AIM) and the organizers for hosting meetings on circlepacking and rigidity theory that indirectly led to the development of this paper.
References [1] E.M. Andre’ev. On convex polyhedra in Lobachevski spaces.
Mat. Sbornik , 81(123):445–478,1970.[2] E.M. Andre’ev. On convex polyhedra of finite volume in Lobachevski spaces.
Mat. Sbornik ,83(125):256–260, 1970.[3] Alan F. Beardon and Kenneth Stephenson. The uniformization theorem for circle packings.
Indiana Univ. Math. J. , 39:1383–1425, 1990.[4] John C. Bowers and Philip L. Bowers. Ma–Schlenker c-octahedra in the 2-sphere.
Discrete &Computational Geometry , Sep 2017.[5] John C. Bowers, Philip L. Bowers, and Kevin Pratt. Rigidity of circle polyhedra in the 2-sphereand of hyperideal polyhedra in hyperbolic 3-space.
Trans. Amer. Math. Soc. , 2018.[6] John C. Bowers, Philip L. Bowers, and Kevin Pratt. Almost all circle polyhedra are rigid.
Geometriae Dedicata , 203(1):337–346, 2019.[7] Philip L. Bowers. The upper Perron method for labelled complexes with applications to circlepackings.
Proc. Camb. Phil. Soc. , 114:321–345, 1993.[8] Philip L. Bowers. Combinatorics encoding geometry: The legacy of Bill Thurston in the storyof one theorem. In K. Ohshika and A. Papadopoulos, editors,
In the tradition of Thurston:geometry and topology . Springer, 2020.[9] Philip L. Bowers and Kenneth Stephenson. A branched Andreev-Thurston theorem for circlepackings of the sphere.
Proc. London Math. Soc. (3) , 73:185–215, 1996.[10] Philip L. Bowers and Kenneth Stephenson. Uniformizing dessins and Bely˘ı maps via circlepacking.
Memoirs of the AMS , 170(805), 2004.[11] Charles R Collins and Kenneth Stephenson. A circle packing algorithm.
Computational Ge-ometry , 25(3):233–256, 2003.[12] Robert Connelly and Steven J. Gortler. Packing disks by flipping and flowing, 2019.[13] Robert Connelly, Steven J Gortler, and Louis Theran. Rigidity for sticky discs.
Proceedings ofthe Royal Society A , 475(2222):20180773, 2019.[14] Yves Colin de Verdi`ere. Une principe variationnel pour les empilements de cercles.
InventionesMathematicae , 104:655–669, 1991.[15] Paul Koebe. Kontaktprobleme der konformen Abbildung.
Ber. S¨achs. Akad. Wiss. Leipzig,Math.-Phys. Kl. , 88:141–164, 1936. 3616] Jiming Ma and Jean-Marc Schlenker. Non-rigidity of spherical inversive distance circle pack-ings.
Discrete & Computational Geometry , 47(3):610–617, February 2012.[17] Al Marden and Burt Rodin. On Thurston’s formulation and proof of Andreev’s theorem. In
Computational Methods and Function Theory, Proceeding, Valparaiso 1989 , pages 103–115.Springer-Verlag, Berlin, Heidelberg, New York, Tokyo, 1990. Lecture Notes in Mathematics,Vol. 1435.[18] Gerald L Orick, Kenneth Stephenson, and Charles Collins. A linearized circle packing algo-rithm.
Computational Geometry , 64:13–29, 2017.[19] Igor Rivin. Euclidean structures on simplicial surfaces and hyperbolic volume.
Ann. of Math. ,139(3):553, May 1994.[20] Roland K.W. Roeder, John H. Hubbard, and William D. Dunbar. Andreev’s theorem onhyperbolic polyhedra.
Annales de l’Institut Fourier , 57(3):825–882, 2007.[21] Franco V. Saliola and Walter Whiteley. Some notes on the equivalence of first-order rigidityin various geometries, 2007.[22] Kenneth Stephenson.
Introduction to Circle Packing: the Theory of Discrete Analytic Func-tions . Camb. Univ. Press, New York, 2005. (ISBN 0-521-82356-0, QA640.7.S74).[23] Jorge Stolfi. Oriented projective geometry. In