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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, but they are not M¨obius 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 visiting Institut de Math´ematiques de Toulouse, Universit´ePaul Sabatier (Toulouse III), he would like to thank it for its hospitality. He also would like to thank the China Scholarship Council for financial support.
2. Proof of the theorem The proof of our theorem uses a well-known infinitesimal flexible Euclidean poly- hedron and the Pogorelov map which preserves the relative distances between two points in the configurations in different geometries. We first give a rapid prelimi- nary. Let hx, yi = −x0y0 + x1y1 + x2y2 + x3y3 be the symmetric 2-form in the Min- 4 skowski space R1, recall that the hyperbolic space is 3 4 2 (2.1) H = {x ∈ R1| kxk = −1, x0 > 0} 4 with the induced Riemannian metric on it, which is a hyperboloid in R1. The totally geodesic planes in H3 are the intersections between H3 and hyperplanes in R4 which pass through the origin. 3 3 3 3 Let B be the unit ball in R , then, there is a projective map pH : H → B given by ρ((x0, x1, x2, x3)) = (x1, x2, x3)/x0, which is a homeomorphism and which maps geodesic in H3 into geodesic in R3. This map is the projective model (Klein model) of the hyperbolic space. −1 3 3 A hyperideal hyperbolic polyhedron is the image of pH : Q ∩ B → H , where Q is a Euclidean polyhedron in R3 such that all vertices of Q lie out of B3 and all edges of Q intersect with B3. For a point A in R3 −B3, consider the space A⊥ of the NON-RIGIDITY OF SPHERICAL INVERSIVE DISTANCE CIRCLE PACKINGS 3
4 −1 −1 points in R1 which are orthogonal to pH (A) in the symmetric 2-form. pH (A) is a 3 ∗ ⊥ 3 hyperbolic plane in H . Then take A = pH(A ) ∩ B , so it is a hyperbolic plane in the Klein model of the 3-dimensional hyperbolic space B3. Thus its boundary is 3 2 ∗ a round circle CA in ∂B = S . A is called the hyperbolic plane dual to A. Bya simple argument, the planes dual to two vertices A and B of Q don’t intersect. The length of an edge of a hyperideal hyperbolic polyhedron is defined as the distance between the dual planes. The de Sitter space can be defined as 3 4 2 (2.2) S1 = {x ∈ R1| kxk =1} 4 with the induced Lorentzian metric on it, which is a one-sheeted hyperboloid in R1. 3 3 The totally geodesic planes in S1 are the intersections of S1 with the hyperplanes in R4 which pass through the origin. Let S3 R4 2 (2.3) 1,+ = {x ∈ 1| kxk =1, x0 > 0} be the upper de Sitter space. S3 R3 3 As for the hyperbolic space, there is a projective map p : 1,+ → − B given by ρ((x0, x1, x2, x3)) = (x1, x2, x3)/x0, which is a homeomorphism and which maps S3 R3 geodesic in 1,+ into geodesic in . S3 3 In the projective model of 1,+, a geodesic maybe pass through B , and if it is the case, then the geodesic is time-like. If a geodesic does not pass through the closure of B3, then this geodesic is space-like. For more details on distances in the de Sitter space, see [5]: for two points x and S3 y in 1,+, if the geodesic [x, y] is a time-like geodesic, then the distance d between them is the negative number d such that cosh(d) = hx, yi; if the geodesic [x, y] is a space-like geodesic, the distance d between them is the unique number in i[0, π] such that cosh(d)= hx, yi. There is a duality between points in the de Sitter space and oriented hyperplanes in the hyperbolic 3-space: consider the projective model of the upper de Sitter space R3 −B3, when A lies in R3 −B3, then the hyperplane A∗ constructed above viewed as in hyperbolic 3-space is the dual of A. When A and B are two points in R3 − B3, such that the Euclidean line L connecting A to B passes through B3, then the de Sitter distance between A and B is essentially the hyperbolic distance between the two planes A∗ and B∗: let l be the distance between A∗ and B∗ in the hyperbolic space, then l = −d. It is also essentially the inversive distance between the two circles CA and CB, where CA and ∗ ∗ 2 CB are the ideal boundaries of the planes A and B in S = ∂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−t which have the same combinatorics as Q and the same corresponding edge lengths. Proof. We first recall Sch¨onhardt’s twisted octahedron (see [8] and [2]): let ABC be 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 the image of ABC under a screw motion with axis L and rotation angle π/2. Con- sider a polyhedron Q bounded by triangles ABC, A0B0C0, ABC0, A0BC, AB0C, 4 JIMINGMAANDJEAN-MARCSCHLENKER
A0B0C, AB0C0, and A0BC0. The polyhedron Q is combinatorially isomorphic to an octahedron, and has three edges with dihedral angles bigger than π: the edges AB0, BC0, and CA0, see Figure 1.
B 0
C0
A 0
C
A B
Figure 1. Sch¨onhardt’s twisted octahedron
This polyhedron Q is infinitesimal flexible (see [2]): there are vectors ηA0 , ηB0 , 3 0 0 ηC0 in R such that the polyhedron Qt with the vertices A, B, C, A + tηA0 , B + 0 tηB0 , C + tηC0 , which is a small deformation of Q, is a non-trivial infinitesimally 0 isometric deformation of Q, where ηA0 is a vector orthogonal to the plane A BC of norm 1 and pointing out from Q, and similarly for ηB0 and ηC0 . Then by a direct calculation(or see Lemma 4.1 of [2]), the pairs of corresponding edges of Qt and Q−t have the same lengths, for 0
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, we state it along with other properties for the sake of future reference. S3 S3 R3 R3 Proposition 2.2. There exists a map Φ: 1,+ × 1,+ → × such that: S3 S3 R3 R3 (1) Φ is a homeomorphism from 1,+ × 1,+ to its image in × , S3 S3 (2) the restriction of Φ to the diagonal △ ⊂ 1,+ × 1,+ is the projective map p(its image is in the diagonal △′ ⊂ R3 × R3), S3 S3 (3) let α be a time-orientation preserving isometry of 1, and x ∈ 1,+ with S3 ′ R3 α(x) ∈ 1,+, we have Φ(x, α(x)) = (y,y ) in , then there is a Euclidean S3 ′ isometry β such that for all x with α(x) ∈ 1,+, we have y = β(y), ′ ′ S3 (4) if [x, y] and [x ,y ] are two time-like geodesics of the same length in 1,+ 3 3 , and if p1,p2 are the projections of R × R on the two factors, then ′ ′ ′ ′ [p1 ◦ Φ ◦ (x, x ),p1 ◦ Φ ◦ (y,y )] and [p2 ◦ Φ ◦ (x, x ),p2 ◦ Φ ◦ (y,y )] are geodesics of the same length in R3, NON-RIGIDITY OF SPHERICAL INVERSIVE DISTANCE CIRCLE PACKINGS 5
S3 (5) if g1,g2 : [0, 1] → 1,+ are space-like geodesic segments parametrized at the same speed, then p1 ◦ Φ ◦ (g1,g2) and p2 ◦ Φ ◦ (g1,g2) are geodesic segments parametrized at the same speed, 0 −1 S3 S3 (6) there exists a point x = p (0) ∈ 1,+ such that, for each 2-plane Π ⊂ 1,+ containing x0, S3 (2.4) ∀x ∈ Π, ∀y ∈ 1,+, p1 ◦ Φ(x, y) ∈ p(Π). The proof of this proposition can be obtained by following those given by Pogorelov’s book for the hyperbolic space or the sphere. More precisely, it is straightforward to 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 replace f(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
S3 give us two polyhedra, say Pt and P−t, in 1,+ 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 dual to the vertex. But recall that the de Sitter length here is essentially the inversive distance 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 same discrete curvature zero. These two spherical inversive distance circle packing are not M¨obious 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 a triangle and in any neighborhood of P , there are two hyperideal polyhedra Pt and P−t which have the same combinatorics and the corresponding edges of them have the same length.
Proof. Now from Lemma 2.1, we have two Euclidean polyhedra Qt and Q−t, which have the same edge length, but they are not congruent. Then we use Proposition S3 2.2, we get two polyhedra in 1,+ such that each of the edges are time-like. S3 Such polyhedra in 1,+ can be viewed as hyperideal hyperbolic polyhedra, and the distance in the de Sitter geometry is just the distance of the circles corresponding to two ideal vertices of the hyperideal hyperbolic polyhedra, which is the edge length of the hyperideal hyperbolic polyhedra. Remark 2.6. Our polyhedra above are not convex, a similar phenomena appears in convex hyperbolic polyhedra, but, some of the faces are not triangle, see Theorem 2’ of [6]. References
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School of Mathematical Sciences, Fudan University, Shanghai, China, 200433 E-mail address: [email protected]
Institut de Mathematiques´ de Toulouse, UMR CNRS 5219, Universite´ Paul Sabatier, 118 route de Narbonne, 31062 Toulouse Cedex 9, France E-mail address: [email protected]