We give a counterexample of Bowers-Stephenson's conjecture in the spherical
case: spherical inversive distance circle packings are not determined by their
inversive distances.Comment: 6 pages, one pictur

A.V. Pogorelov

E. Schönhardt

F. Luo

I. Izmestiev

J.-M. Schlenker

Jean-Marc Schlenker

Jiming Ma

P. Bowers

English

arXiv

Crossref

Discrete & Computational Geometry

Non-rigidity of Spherical Inversive Distance Circle Packings

Ma, Jiming

SCHLENKER, Jean-Marc

Open Repository and Bibliography - Luxembourg

arXiv:1105.1469v1  [math.DG]  7 May 2011NON-RIGIDITY OF SPHERICAL INVERSIVE DISTANCECIRCLE PACKINGSJIMING MA AND JEAN-MARC SCHLENKERAbstract. We give a counterexample of Bowers-Stephenson’s conjecture inthe spherical case: spherical inversive distance circle packings are not deter-mined by their inversive distances.1. IntroductionIn this note we study inversive distance circle packing metrics on a surface F .1.1. Polyhedral surface. Given a triangulated closed orientable surface F , a Eu-clidean (resp. spherical or hyperbolic) polyhedral surface is a map l : E → R+,where E is the set of all edges of the triangulation, such that when e1, e2 and e3are the three edges of a triangle, then l(e1) + l(e2) > l(e3) (it is also required thatl(e1) + l(e2) + l(e3) < 2pi in the spherical case). From this l, there is a polyhedralmetric in F such that the restriction of the metric to each triangle is isometricto a triangle in E2 (resp. S2 or H2) and the length of an edge e is given by l(e).For instance, the boundary of a generic convex polyhedron in E3 (resp. S3 or H3)admits a natural polyhedral metric.The discrete curvature k of a polyhedral surface is the map k : V → R, whereV is the set of all vertices of the triangulation, and for a vertex v ∈ V , k(v) =2pi −∑mi=1 θi, where θi are the angles at the vertex v.1.2. Inversive distance circle packings. The notion of inversive distance circlepacking was introduced by Bowers-Stephenson in [1], it is a generalization of An-dreev and Thurston’s circle packings on a surface, where two circles may intersector not. We just give the definition of the spherical inversive distance circle packing,for Euclidean and hyperbolic cases, see [1] and [3] for more detailed discussions.For two circles C1 and C2 centered at v1, v2 of radii r1 and r2 in S2, so that v1and v2 are of distance l apart, the inversive distance I = I(C1, C2) between them is(1.1) I =cos(l)− cos(r1) cos(r2)sin(r1) sin(r2).When viewed B3 as the Klein model of H3, the inversive distance is essentiallythe hyperbolic distance(or the intersection angle) between the two totally geodesicplanes in H3 with Ci as their ideal boundaries. When those planes intersect, theDate: May 3, 2011.2000 Mathematics Subject Classification. 52C25, 53B30.Key words and phrases. inversive distance circle packing, rigidity.The first author was supported in part by NSFC 10901038 and Shanghai NSF 10ZR1403600.The second author was supported in part by ANR programs ETTT, ANR-09-BLAN-0116-01,2009-13, and ACG, ANR-10-BLAN-0105, 2010-14.12 JIMING MA AND JEAN-MARC SCHLENKERinversive distance is the cos of their angle, and if they’re disjoint, it is the cosh oftheir distance.In a triangulated surface F , a spherical inversive distance circle packing isgiven as follows: fix a vector I ∈ [−1,∞)E, called the inversive distance vec-tor. For any r ∈ (0,∞)V , called the radius vector, define the edge length byl(e) =√r(u)2 + r(v)2 + 2r(u)r(v)I(e) for an edge e with u and v as its end points.If for any triangle with e1, e2 and e3 as its three edges, we have l(e1)+ l(e2) > l(e3)and l(e1) + l(e2) + l(e3) < 2pi, then the edge length function l : E → R defines aspherical polyhedral metric on F , which is called the spherical inversive distancecircle packing metric with inversive distance I.The geometric meaning is that in F with this polyhedral metric, if we drawcircles with radii r at the vertices V , then the inversive distance of two circles atthe end points of an edge e is the given number I(e).It was conjectured by Bowers and Stephenson [1] that inversive distance circlepackings have a global rigidity property: an inversive distance circle packing isdetermined by its combinatoric, inversive distance vector and discrete curvature atthe vertices. Luo [3] proved Bowers-Stephenson’s conjecture in the hyperbolic andEuclidean cases. In this note, we give a counterexample in the spherical case:Theorem 2.4. There is a triangulation of S2 and two spherical inversive dis-tance circle packings with the same inversive distance and discrete curvature, butthey are not Mo¨bius equivalent.The example we construct actually have zero discrete curvature at all vertices,so they are inversive distance circle patterns on the (non-singular) sphere.Acknowledgements: This work was done when the first author was visitingInstitut de Mathe´matiques de Toulouse, Universite´ Paul Sabatier (Toulouse III),he would like to thank it for its hospitality. He also would like to thank the ChinaScholarship Council for financial support.2. Proof of the theoremThe proof of our theorem uses a well-known infinitesimal flexible Euclidean poly-hedron and the Pogorelov map which preserves the relative distances between twopoints in the configurations in different geometries. We first give a rapid prelimi-nary.Let 〈x, y〉 = −x0y0 + x1y1 + x2y2 + x3y3 be the symmetric 2-form in the Min-skowski space R41, recall that the hyperbolic space is(2.1) H3 = {x ∈ R41| ‖x‖2 = −1, x0 > 0}with the induced Riemannian metric on it, which is a hyperboloid in R41. Thetotally geodesic planes in H3 are the intersections between H3 and hyperplanes inR4 which pass through the origin.Let B3 be the unit ball in R3, then, there is a projective map pH : H3 → B3given by ρ((x0, x1, x2, x3)) = (x1, x2, x3)/x0, which is a homeomorphism and whichmaps geodesic in H3 into geodesic in R3. This map is the projective model (Kleinmodel) of the hyperbolic space.A hyperideal hyperbolic polyhedron is the image of p−1H: Q ∩ B3 → H3, whereQ is a Euclidean polyhedron in R3 such that all vertices of Q lie out of B3 and alledges of Q intersect with B3. For a point A in R3−B3, consider the space A⊥ of theNON-RIGIDITY OF SPHERICAL INVERSIVE DISTANCE CIRCLE PACKINGS 3points in R41 which are orthogonal to p−1H(A) in the symmetric 2-form. p−1H(A) is ahyperbolic plane in H3. Then take A∗ = pH(A⊥) ∩ B3, so it is a hyperbolic planein the Klein model of the 3-dimensional hyperbolic space B3. Thus its boundary isa round circle CA in ∂B3 = S2. A∗ is called the hyperbolic plane dual to A. By asimple argument, the planes dual to two vertices A and B of Q don’t intersect. Thelength of an edge of a hyperideal hyperbolic polyhedron is defined as the distancebetween the dual planes.The de Sitter space can be defined as(2.2) S31 = {x ∈ R41| ‖x‖2 = 1}with the induced Lorentzian metric on it, which is a one-sheeted hyperboloid in R41.The totally geodesic planes in S31 are the intersections of S31 with the hyperplanesin R4 which pass through the origin. Let(2.3) S31,+ = {x ∈ R41| ‖x‖2 = 1, x0 > 0}be the upper de Sitter space.As for the hyperbolic space, there is a projective map p : S31,+ → R3 −B3 givenby ρ((x0, x1, x2, x3)) = (x1, x2, x3)/x0, which is a homeomorphism and which mapsgeodesic in S31,+ into geodesic in R3.In the projective model of S31,+, a geodesic maybe pass through B3, and if it isthe case, then the geodesic is time-like. If a geodesic does not pass through theclosure of B3, then this geodesic is space-like.For more details on distances in the de Sitter space, see [5]: for two points x andy in S31,+, if the geodesic [x, y] is a time-like geodesic, then the distance d betweenthem is the negative number d such that cosh(d) = 〈x, y〉; if the geodesic [x, y] isa space-like geodesic, the distance d between them is the unique number in i[0, pi]such that cosh(d) = 〈x, y〉.There is a duality between points in the de Sitter space and oriented hyperplanesin the hyperbolic 3-space: consider the projective model of the upper de Sitter spaceR3−B3, when A lies in R3−B3, then the hyperplane A∗ constructed above viewedas in hyperbolic 3-space is the dual of A.When A and B are two points in R3 − B3, such that the Euclidean line Lconnecting A to B passes through B3, then the de Sitter distance between A andB is essentially the hyperbolic distance between the two planes A∗ and B∗: let lbe the distance between A∗ and B∗ in the hyperbolic space, then l = −d. It is alsoessentially the inversive distance between the two circles CA and CB, where CA andCB are the ideal boundaries of the planes A∗and B∗ in S2 = ∂B3.Lemma 2.1. There is a Euclidean polyhedron Q such that(1) all vertices of Q lie out of B3,(2) all edges of Q intersect with B3,(3) in any neighborhood of Q, there are two Euclidean polyhedra Qt and Q−twhich have the same combinatorics as Q and the same corresponding edgelengths.Proof. We first recall Scho¨nhardt’s twisted octahedron (see [8] and [2]): let ABCbe an equilateral triangle in R3, and let L be a line that passes through the cen-ter of ABC and it is orthogonal to the plane of the triangle. Let A0B0C0 be theimage of ABC under a screw motion with axis L and rotation angle pi/2. Con-sider a polyhedron Q bounded by triangles ABC, A0B0C0, ABC0, A0BC, AB0C,4 JIMING MA AND JEAN-MARC SCHLENKERA0B0C, AB0C0, and A0BC0. The polyhedron Q is combinatorially isomorphic toan octahedron, and has three edges with dihedral angles bigger than pi: the edgesAB0, BC0, and CA0, see Figure 1.BCB0C0A0AFigure 1. Scho¨nhardt’s twisted octahedronThis polyhedron Q is infinitesimal flexible (see [2]): there are vectors ηA0 , ηB0 ,ηC0 in R3 such that the polyhedron Qt with the vertices A, B, C, A0 + tηA0 , B0 +tηB0 , C0 + tηC0 , which is a small deformation of Q, is a non-trivial infinitesimallyisometric deformation of Q, where ηA0 is a vector orthogonal to the plane A0BC ofnorm 1 and pointing out from Q, and similarly for ηB0 and ηC0 . Then by a directcalculation(or see Lemma 4.1 of [2]), the pairs of corresponding edges of Qt andQ−t have the same lengths, for 0 < t small enough.Let O be the center of the polyhedron Q, i.e, O lies in the line L and its distancesto the planes ABC and A0B0C0 are both equal to h > 0. Let a be the edge lengthof the equilateral triangle ABC, a simple calculation shows that conditions (1) and(2) of the lemma are equivalent to(1) h2 + a2/3 > 1,(2) h2 + a2/12 < 1,(3) a2/3 < 1.So, we can assign h = 1/2 and a a little bigger than 3/2, and the lemma follows.Pogorelov [4] has found remarkable maps from S3× S3 and H3×H3 to R3 ×R3,see also [5], [6], [7] for other forms of these maps and their infinitesimal versions.What we really need is the first four properties of the following proposition, westate it along with other properties for the sake of future reference.Proposition 2.2. There exists a map Φ : S31,+ × S31,+ → R3 × R3 such that:(1) Φ is a homeomorphism from S31,+ × S31,+ to its image in R3 × R3,(2) the restriction of Φ to the diagonal △ ⊂ S31,+ × S31,+ is the projective mapp(its image is in the diagonal △′ ⊂ R3 × R3),(3) let α be a time-orientation preserving isometry of S31, and x ∈ S31,+ withα(x) ∈ S31,+, we have Φ(x, α(x)) = (y, y′) in R3, then there is a Euclideanisometry β such that for all x with α(x) ∈ S31,+, we have y′ = β(y),(4) if [x, y] and [x′, y′] are two time-like geodesics of the same length in S31,+, and if p1, p2 are the projections of R3 × R3 on the two factors, then[p1 ◦ Φ ◦ (x, x′), p1 ◦ Φ ◦ (y, y′)] and [p2 ◦ Φ ◦ (x, x′), p2 ◦ Φ ◦ (y, y′)] aregeodesics of the same length in R3,NON-RIGIDITY OF SPHERICAL INVERSIVE DISTANCE CIRCLE PACKINGS 5(5) if g1, g2 : [0, 1]→ S31,+ are space-like geodesic segments parametrized at thesame speed, then p1 ◦Φ ◦ (g1, g2) and p2 ◦Φ ◦ (g1, g2) are geodesic segmentsparametrized at the same speed,(6) there exists a point x0 = p−1(0) ∈ S31,+ such that, for each 2-plane Π ⊂ S31,+containing x0,(2.4) ∀x ∈ Π, ∀y ∈ S31,+, p1 ◦ Φ(x, y) ∈ p(Π).The proof of this proposition can be obtained by following those given by Pogorelov’sbook for the hyperbolic space or the sphere. More precisely, it is straightforwardto adapt the proof of §3 Lemmas 1-4 and §4 Theorems 1-2 in Chapter V of [4].Or from Section 6 of [9]: in Proposition 6.3 and 6.4 of [9], we should replacef(a, b) = (a2− b2)2−8(a2+ b2−2)(in hyperbolic case and 1 > a, b ≥ 0) to g(a, b) =−(a2 − b2)2 + 8(a2 + b2 − 2) (in de Sitter case and 1 < a, b). Note that g(a, b) =−(a2 − b2)2 + 8(a2 + b2 − 2) is not always positive for 1 < a, b, but this is true for−4 < a2−b2 < 4, so {(ξ, η) ∈ (R3−B3)×(R3−B3)|−4 < |ξ|2−|η|2 < 4} ⊂ Im(Φ),which is an open neighborhood of the diagonal of (R3−B3)× (R3−B3) ⊂ R3×R3.For the proof of (3) of Proposition 2.2, we just recall that for a time-orientationpreserving isometry α of S31, in the matrix presentation A4×4 of it, the (1, 1)-entryof A is positive, and then (3) of Proposition 2.2 follows from arguments similar toProposition 6.5 of [9].For the proof of (4) of Proposition 2.2, we need the transitivity of the time-orientation subgroup of Iso(S31) on the space of time-like geodesic segments of afixed length, which can be see from the duality between the de sitter space and thehyperbolic space. From this, we have an time-orientation isometry α, such thatα([x, y]) = [x′, y′], and then (4) follows from (3). (5) is similar.The map Φ in the Minkowski coordinate is given as follows: let (x, y) ∈ S31,+ ×S31,+, where x = (x0, x1, x2, x3) and y = (y0, y1, y2, y3), then(2.5) Φ(x, y) = 2((x1, x2, x3), (y1, y2, y3))/(x0 + y0) ∈ R3 × R3.The converse Φ−1 : R3 × R3 → S31,+ × S31,+ is given by(2.6)((ξ1, ξ2, ξ3), (η1, η2, η3))→ (ρ((4−|η|2+|ξ|2, ξ1, ξ2, ξ3)), ρ((4−|ξ|2+|η|2, η1, η2, η3))),where ρ is the linear normalization such that ρ((4−|η|2+ |ξ|2, ξ1, ξ2, ξ3)) and ρ((4−|ξ|2 + |η|2, η1, η2, η3)) lie in the hyperboloid S31,+.Remark 2.3. The Pogorelov’s maps in [4] and [9] are a little different, i.e, up to themultiple constant 2, we choose the one similar to [9], which is convenient for us.Theorem 2.4. There is a triangulation of S2 and two spherical inversive distancecircle packings with the same inversive distance and discrete curvature, but they arenot Mo¨bius equivalent.Proof. The triangulation of S2 is given from the boundary of the Euclidean octa-hedron in Lemma 2.1. From Lemma 2.1, we have two Euclidean polyhedra whichhave the same edge lengths, but they are not congruent, say Qt and Q−t for a fixedt > 0 small enough, which are very near to Q. Denoted the vertices of Q by vi andthe corresponding vertices of Qt(Q−t) by vit(vi−t). Note that (vi, vi) ∈ Im(Φ) byProposition 2.2 (2), and from Proposition 2.2 (1), we can assume (vit, vi−t) ∈ Im(Φ).Since Φ−1 ◦ (vi, vi) give us a polyhedron in S31,+ such that each of the edges aretime-like. Then we use Proposition 2.2 (4), p1◦Φ−1◦(vit, vi−t) and p2◦Φ−1◦(vit, vi−t)6 JIMING MA AND JEAN-MARC SCHLENKERgive us two polyhedra, say Pt and P−t, in S31,+ such that each of the edges are time-like and which have the same corresponding edge lengths. For each vertex of Pt,we have a circle in S2, which is the ideal boundary of the hyperbolic plane dualto the vertex. But recall that the de Sitter length here is essentially the inversivedistance of the circles corresponding to two ideal vertices of the hyperideal hyper-bolic polyhedra. So, we have two spherical inversive distance circle packing metrics,they induced the same standard spherical metric in S2, thus they have the samediscrete curvature zero. These two spherical inversive distance circle packing arenot Mo¨bious equivalent can be seen also from the Pogorelov map.Corollary 2.5. There is a hyperideal polyhedron P such that each face of it is atriangle and in any neighborhood of P , there are two hyperideal polyhedra Pt andP−t which have the same combinatorics and the corresponding edges of them havethe same length.Proof. Now from Lemma 2.1, we have two Euclidean polyhedra Qt and Q−t, whichhave the same edge length, but they are not congruent. Then we use Proposition2.2, we get two polyhedra in S31,+ such that each of the edges are time-like.Such polyhedra in S31,+ can be viewed as hyperideal hyperbolic polyhedra, andthe distance in the de Sitter geometry is just the distance of the circles correspondingto two ideal vertices of the hyperideal hyperbolic polyhedra, which is the edge lengthof the hyperideal hyperbolic polyhedra. Remark 2.6. Our polyhedra above are not convex, a similar phenomena appears inconvex hyperbolic polyhedra, but, some of the faces are not triangle, see Theorem2’ of [6].References1. P. Bowers and K. Stephenson, Uniformizing dessins and Belyi (maps via circle packing),Mem. Amer. Math. Soc. 170(2004), no. 805.2. I. Izmestiev, Examples of infinitesimally flexible 3-dimensional hyperbolic cone-manifolds, toappear in J. Math. Soc. Japan.3. F. Luo, Rigidity of polyhedral surfaces, III, arXiv: math.GT/1010.3284.4. A. V. Pogorelov, Extrinsic geometry of convex surfaces, Translations of Mathematical Mono-graphs Vol. 35. American Mathematical Society, Providence, RI, 1973.5. Jean-Marc Schlenker, Me´triques sur les polye`dres hyperboliques convexes, Journal of Differ-ential Geometry, 48, no.2 (1998), 323-405.6. Jean-Marc Schlenker, Dihedral angles of convex polyhedra, Discrete and Computational Ge-ometry, 23, no.3 (2000), 409-417.7. Jean-Marc Schlenker, A rigidity criterion for non-convex polyhedra, Discrete and Computa-tional Geometry, 33, no.2(2005), 207-221.8. Erich Scho¨nhardt, Uber die Zerlegung von Dreieckspolyedern in Tetraeder, Math. Ann., 98,no.1(1928), 309-312.9. Javier Virto, On the characterization of polyhedra in hyperbolic 3-Space, arXiv:math.MG/1006.4445.School of Mathematical Sciences, Fudan University, Shanghai, China, 200433E-mail address: majiming@fudan.edu.cnInstitut de Mathe´matiques de Toulouse, UMR CNRS 5219, Universite´ Paul Sabatier,118 route de Narbonne, 31062 Toulouse Cedex 9, FranceE-mail address: schlenker@math.univ-toulouse.fr

Non-rigidity of spherical inversive distance circle packings

6 pages, one pictureInternational audienceWe give a counterexample of Bowers-Stephenson's conjecture in the spherical case: spherical inversive distance circle packings are not determined by their inversive distances

Non-rigidity of spherical inversive distance circle packings

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