LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 363, Number 9, September, Pages S -9947()539-6 Article electronically published on April, LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING REN GUO Abstract. A Euclidean (or hyperbolic) circle packing on a triangulated closed surface with prescribed inversive distance is locally determined by its cone angles. We prove this by establishing a variational principle.. Introduction.. Andreev-Thurston Theorem. In his work on constructing hyperbolic metrics on 3-manifolds, W. Thurston ([3], Chapter 3) studied a Euclidean (or hyperbolic) circle packing on a triangulated closed surface with prescribed intersection angles. Thurston s work generalizes Andreev s result of circle packing on a sphere [, ]. The special case of tangent circle packing (every intersection angle is zero) on a sphere was obtained by Koebe [6]. Suppose (Σ,T) is a triangulated closed surface so that V,E,F are sets of all vertices, edges and triangles in T. We identify vertices of T with indexes, edges of T with pairs of indexes and triangles of T with triples of indexes. This means V = {,,..., V },E = {ij i, j V } and F = { ijk i, j, k V }. Fix a vector Θ R E indexed by the set of edges E such that Θ ij [, π ]foreachij E. This vector is called a weight on (Σ,T). A circle packing on a weighted triangulated closed surface (Σ,T,Θ) is a configuration {c i,i V } of circles such that the intersection angle of c i and c j is Θ ij, which is the prescribed weight on the edge ij. To obtain a Euclidean (or hyperbolic) circle packing for (Σ,T,Θ), we start with a radius vector r =(r,r,..., r V ) R V > which assigns each vertex i V a positive number r i. A radius vector r produces a Euclidean (or hyperbolic) cone metric on the surface as follows. In Figure, consider a topological triangle ijk. One can construct a Euclidean (or hyperbolic) triangle ijp such that the edges ip, jp have lengths r i,r j respectively and the angle at p is π Θ ij. Let l ij be the length of the edge ij in the Euclidean (or hyperbolic) triangle ijp which is a function of r i,r j. In fact, in Euclidean geometry l ij = r i + r j +r ir j cos Θ ij or in hyperbolic geometry l ij =cosh (cosh r i cosh r j +cosθ ij sinh r i sinh r j ). Similarly, we obtain l jk,l ki. Received by the editors April, 9 and, in revised form, September 7, 9. Mathematics Subject Classification. Primary 5C6, 58E3. Key words and phrases. Circle packing, rigidity, variational principle. c American Mathematical Society Reverts to public domain 8 years from publication 4757

2 4758 REN GUO Θ ij p r i i Θ ki r j l ij l ki j l jk k Θ jk Figure. Andreev-Thurston circle packing As observed by Thurston [3], under the assumption that Θ ij [, π ]foreach ij E, l ij,l jk and l ki can be realized as edge lengths of a Euclidean (or hyperbolic) triangle ijk in F. Gluing all of these Euclidean (or hyperbolic) triangles in F produces a Euclidean (or hyperbolic) cone metric on (Σ,T) with possible cones at vertices of T. On this surface with a cone metric, drawing a circle centered at vertex i with radius r i for each vertex i V, we obtain a circle packing with prescribed intersection angles. At each vertex i V, the cone angle a i is the sum of all inner angles having vertex i. Thus cone angles (a,a,..., a V ) are functions of the radius vector r =(r,r,..., r V ). We try to find a radius vector when cone angles are given. The Andreev-Thurston Theorem answers the question about the existence and uniqueness of radius vectors when cone angles are given. Theorem (Andreev-Thurston). For any weighted triangulated closed surface (Σ,T,Θ) with Θ ij [, π ] for each ij E, a Euclidean (or hyperbolic) circle packing for (Σ,T,Θ) is determined by its cone angles up to Euclidean similarity (or hyperbolic isometry). Furthermore, the set of all possible cone angles form an open convex polytope in R V. For a proof, see Thurston [3], Marden-Rodin [9], Colin de Verdiére [8], He [4], Chow-Luo [9], and Stephenson []... Inversive distance. H. S. M. Coxeter [7] introduced the notion of inversive distance to describe the relationship between two circles in a Möbius plane. The notion of inversive distance generalizes the notion of an intersection angle of two circles. The inversive distance between two circles is independent of geometry, i.e., the inversive distance of two circles is the same under spherical, Euclidean and hyperbolic metrics. This generalizes the fact that the intersection angle of two circles is the same under spherical, Euclidean and hyperbolic metrics. In the Euclidean plane, consider two circles c,c with radii r,r respectively. In the relevant case of this paper, we assume that c i does not contain c j for {i, j} =

3 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4759 {, }. If the distance between their center is l, the inversive distance between c,c is given by the formula I(c,c )= l r r r r. In hyperbolic plane, consider two circles c,c with radii r,r respectively. We assume that c i does not contain c j for {i, j} = {, }. If the distance between their center is l, the inversive distance between c,c is given by the formula I(c,c )= cosh l cosh r cosh r sinh r sinh r. H 3 H C c c c c (,, ) Figure. The stereographic projection These two formulas are related by a stereographic projection. In Figure, consider the upper half space model of the hyperbolic space H 3. The hyperbolic plane H is realized as a totally geodesic plane in H 3, where it is a unit hemisphere perpendicular to the infinite boundary C. The stereographic projection is the projection starting from point (,, ), which sends objects in H into C. Let c,c be two circles in H with radii r,r and the distance between centers l. The stereographic projection produces two circles c, c on the Euclidean plane C with radii r, r and the distance between centers l. Thenwehave I(c,c )= cosh l cosh r cosh r = l r r = I( c, c ). sinh r sinh r r r We say that the stereographic projection preserves the inversive distance. When I(c,c ) (, ), the circles c,c intersect with an intersection angle arccos I(c,c ) ( π,π). When I(c,c ) [, ), the circles c,c intersect with an intersection angle arccos I(c,c ) (, π ]. When I(c,c )=, the circles c,c are tangent to each other. When I(c,c ) (, ), the circles c,c are separated..3. Inversive distance circle packing. Motivated by the application to discrete conformal mappings, Bowers-Stephenson [5] introduced inversive distance circle packing which generalizes Andreev-Thurston s intersection angle circle packing. See Stephenson [] (page 33) and Bowers-Hurdal [4] for more information. One of the

4 476 REN GUO important applications of inversive distance circle packing is the conformal mapping of the human brain; see, for example, Hurdal-Stephenson [5]. We will use the same notation as in Andreev-Thurston s intersection angle circle packing. For a triangulated closed surface (Σ,T), fix a weight I R E with I ij [, ) foreachij E. In this paper, we will not consider the case of obtuse intersection angles. (Rivin [] and Leibon [7] obtained results for certain circle patterns on surfaces allowing obtuse intersection angles.) An inversive distance circle packing for (Σ,T,I) is a configuration of circles {c i,i V } such that the inversive distance of two circles c i,c j is the prescribed number I ij for the edge ij. To obtain a Euclidean (or hyperbolic) circle packing for (Σ,T,I), we start with a which assigns each vertex i V a positive number r i. A radius vector r produces a Euclidean (or hyperbolic) cone metric on the surface as follows. Consider a topological triangle ijk F. Assign a number l ij to the edge ij E, where l ij is the distance of two circles with radii r i,r j and inversive distance I ij. Therefore l ij is a function of r i,r j. In fact, in Euclidean geometry radius vector r =(r,r,..., r V ) R V > or in hyperbolic geometry l ij = r i + r j +r ir j I ij l ij =cosh (cosh r i cosh r j + I ij sinh r i sinh r j ). Similarly, we obtain l jk,l ki. To guarantee that l ij,l jk,l ki satisfy the triangle inequality for each triangle ijk F, a vector r =(r,r,..., r V ) must satisfy certain conditions. For details, see sections. and 3.. For a radius vector r, ifl ij,l jk,l ki can be realized as edge lengths of a Euclidean (or hyperbolic) triangle for each ijk F, gluing all of these Euclidean (or hyperbolic) triangles in F produces a Euclidean (or hyperbolic) cone metric on (Σ,T) with possible cones at vertices of T. On this surface with a cone metric, drawing a circle centered at vertex i with radius r i for each vertex i V, we obtain a circle packing with prescribed inversive distance. Theconeangles(a,a,..., a V ) are functions of the radius vector r =(r,r,..., r V ). We are interested in the solutions of a radius vector when cone angles are given. Our main theorem is the local rigidity of inversive distance circle packing. Theorem. For any weighted triangulated closed surface (Σ,T,I) with I ij [, ) for each ij E, a Euclidean (or hyperbolic) inversive distance circle packing for (Σ,T,I) is locally determined by its cone angles up to Euclidean similarity (or hyperbolic isometry). The local rigidity of the Andreev-Thurston Theorem is generalized to the case of inversive distance circle packing. In our case, we are not able to obtain either the global rigidity or the description of the space of cone angles as in the Andreev- Thurston Theorem. For discussion, see sections. and 3.. In section, we prove Theorem in Euclidean geometry. In section 3, we prove Theorem in hyperbolic geometry.

5 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING Variational principle. Theorem is proved by establishing a variational principle. The energy function of this variational principle is derived from the derivative of the cosine law of a Euclidean (or hyperbolic) triangle. Andreev and Thurston obtained their results without using variational principles. The variational approach to circle packing was first introduced by Colin de Verdiére [8]. Since then, many works on variational principles on circle packing or circle patterns have appeared. For example, see Brägger [3], Rivin [], Leibon [7], Chow-Luo [9], Bobenko-Springborn [6], Guo [], Springborn [], Luo [8], Guo-Luo [3] and others. Applying a variational principle, Colin de Verdiére [8] proved the Andreev- Thurston Theorem in the special case of tangent circle packing, i.e., Θ ij =for each ij E. Chow-Luo [9] proved the Andreev-Thurston Theorem of intersection angle circle packing, i.e., Θ ij [, π ]foreachij E. Variational principles for polyhedral surfaces including the topic of circle packing are studied systematically in Luo [8]. Many energy functions are derived from the cosine law and its derivative. Tangent circle packing is generalized to tangent circle packing with a family of discrete curvature. For exposition of this work, see also Dai-Gu-Luo []. We follow this variational approach to prove Theorem of inversive distance circle packing, i.e., I ij [, ) foreachij E.. Circle packing in Euclidean geometry.. Space of radius vectors. As stated in the Introduction, to obtain a Euclidean cone metric, we need conditions on a radius vector. In this section we investigate the space of radius vectors for one triangle in Euclidean geometry. Consider a topological triangle ijk F. Assign a number l ij to the edge ij E, where l ij is the distance of the centers of two circles with radii r i,r j and inversive distance I ij. Therefore l ij is a function of r i,r j. In fact, in Euclidean geometry l ij = ri + r j +r ir j I ij. Similarly, we obtain l jk,l ki. Lemma 3. The set R ijk E := {(r i,r j,r k ) R 3 > l ij,l jk,l ki satisfy triangle inequality} is a connected and simply connected open subset in R 3 >. Proof. Let ρ E : R 3 > R 3 > be the map sending (r i,r j,r k )to(l ij,l jk,l ki ). We claim that ρ E is a smooth embedding. First the determinant of the Jacobian matrix of ρ E is nonzero. Next, we will show ρ E is one-to-one. This means we need to show that if the following equations have a solution (r i,r j,r k ), then the solution is unique: rj + r k +I jkr j r k = ljk, rk + r i +I kir k r i = lki, ri + rj +I ij r i r j = lij. Solving the first equation for r j,wegetr j = I jk r k ± (Ijk )r k + l jk. Since I jk > andr j >, we must have r j = I jk r k + (Ijk )r k + l jk. Hence r j is uniquely determined by r k. By the same argument r i = I ki r k + (Iki )r k + l ki.

6 476 REN GUO After substituting r i,r j in the third equation, we need to show it has a unique solution for r k. Consider the function f(t) =ri (t)+rj (t)+i ij r i (t)r j (t) lij which is obtained by replacing r k by t for simplicity of notation. Now f (t) =(r i +I ij r j )r i(t)+(r j +I ij r i )r j(t). Also, we have r i(t) (Iki = I ki + )t. (I ki )t + lki If I ki [, ], then each term in r i (t) isnonpositive.thusr i (t) <. If I ki (, ), then r i(t) < I ki + (I ki )t (I ki )t = I ki + I ki <. By the same argument, we have r j (t) <. Since r i+i ij r j > andr j +I ij r i >, then f (t) < holds. Therefore f(t) = has a unique solution. To determine the image ρ E (R 3 >), we consider the image of the boundary of R 3. If r i =, then r k = l ki,r j = l ij. Thus ljk = rj + rk +I jk r j r k = lij + lki +I jk l ij l ki. On the other hand, if r i >, then r k <l ki,r j <l ij. We have ljk <lij + lki +I jk l ij l ki. Therefore the image ρ E (R 3 >) is the subset of vectors (l ij,l jk,l ki ) satisfying ljk <lij + lki +I jk l ij l ki, lki <ljk + lij +I ki l jk l ij, lij <lki + ljk +I ij l ki l jk. In fact, ρ E (R 3 >) is a cone bounded by three surfaces. In R 3, the intersection of any of the two surfaces is a straight line. The three lines of intersection are l ij =andl jk = l ki, l jk =andl ki = l ij, l ki =andl ij = l jk. Let L be the subset of R 3 > formed by vectors (l ij,l jk,l ki ) satisfying the triangle inequality. Therefore L is a cone bounded by three planes. The three lines of intersection of the three planes are the same as that of the boundary surfaces of ρ E (R 3 >). We claim that ρ E (R 3 > ) L is a cone. connected. In fact, if I ij [, ], then in R 3 > we have Therefore it is connected and simply {(l ij,l jk,l ki ) lij <lki + ljk +I ij l ki l jk } {(l ij,l jk,l ki ) l ij <l ki + l jk }. If I ij (, ), then in R 3 > we have {(l ij,l jk,l ki ) lij <lki + ljk +I ij l ki l jk } {(l ij,l jk,l ki ) l ij <l ki + l jk }. To determine the shape of ρ E (R 3 >) L, there are four cases to consider.

7 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4763 If I ij,i jk,i ki (, ), then ρ E (R 3 >) L= L. If I ij [, ],I jk,i ki (, ), then ρ E (R 3 >) L is a cone bounded by one surface and two planes. If I ij,i jk [, ],I ki (, ), then ρ E (R 3 > ) L is a cone bounded by two surfaces and one plane. If I ij,i jk,i ki [, ], then ρ E (R 3 >) L= ρ E (R 3 >). To finish the proof of the lemma, by definition, R ijk E ρ E is a smooth embedding, R ijk E = ρ E (ρ E(R 3 >) L). Since is connected and simply connected. In the construction of the energy function, we will use a natural variable u = (u,u,..., u V ), where u i = ln(r i ). Let τ E : R 3 R 3 > be the map sending (u i,u j,u k )to(r i,r j,r k ). Since τ E is a diffeomorphism, we have Corollary 4. The set U ijk E := τe (Rijk E ) is a connected and simply connected open subset in R 3. To prove Theorem, we construct a concave energy function on U ijk E for each triangle ijk F. We obtain the local rigidity of inversive distance circle packing from this energy function. We cannot prove the global rigidity due to the fact that U ijk E is not always convex in this general setting. In fact, l ij,l jk,l ki satisfying the triangle inequality is equivalent to () (l ij + l jk + l ki )(l jk + l ki l ij )(l jk + l ij l ki )(l ij + l jk l ki ) >. Substituting the functions l ij,l jk,l ki in terms of u i,u j,u k into inequality (), we see that U ijk E is defined by the inequality () I jk e u i + I ki e u j + I ij e u k + (I jki ki + I ij ) e u i+u j + (I kii ij + I jk ) e u j+u k + (I iji jk + I ki ) e u k+u i >. For example, when I ij = I jk = I ki =, the inequality () determines a subset which is not convex. In the Andreev-Thurston Theorem, the space of cone angles is determined by a continuity argument. Since we cannot prove the global rigidity, we are not able to determine the space of cone angles in the case of inversive distance circle packing... Energy function. To prove Theorem, we will construct an energy function on the space of radius vectors. First, we construct an energy function on the space for one triangle. To simplify notation, consider the triangle 3 F and the map UE 3 R 3 E L R 3 >, (u,u,u 3 ) (r,r,r 3 ) (l,l,l 3 ) (α,α,α 3 ), where r i = e u i for i =,, 3. For three fixed positive numbers I,I 3,I 3, we have li = rj + r k +I jkr j r k for {i, j, k} = {,, 3}. For simplicity of notation, we set (l,l,l 3 )=(l 3,l 3,l ). Also, α,α,α 3 are inner angles of a Euclidean triangle with edge lengths l,l,l 3. By the cosine law, cos α i = l i +l j +l k l j l k for {i, j, k} = {,, 3}. Therefore α,α,α 3 are functions in terms of u,u,u 3. Lemma 5. The Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 is symmetric.

8 4764 REN GUO Proof. Due to the cosine law of a Euclidean triangle, α,α,α 3 are functions of l,l,l 3. The following formula can be proved by direct calculation. For {i, j, k} = {,, 3}, dα dα = l l cosα 3 cos α cos α 3 cosα dl dl. sin α dα i l j l k 3 l 3 cos α cos α dl 3 Next, by the construction l i = r j + r k +I jkr j r k, for {i, j, k} = {,, 3}, where I jk is a constant. Differentiating the two sides of the equality, we obtain dl i = l i + r j r k dr j + l i + r k r j dr k. l i r j l i r k Finally, r i = e u i implies r i du i = dr i,fori =,, 3. Combining the three relations, we have, for {i, j, k} = {,, 3}, dα dα = l l cosα 3 cos α cos α 3 cosα dα sin α i l j l k 3 l 3 cos α cos α l + r r3 l + r3 r l r l r 3 l + r r3 l + r3 r r l r l r r du du 3 l3 + r r l3 + r r r 3 du 3 l 3 r l 3 r We write the above formula as dα dα = dα 3 N sin α i l j l k du du du 3.. To show that the Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 is symmetric is equivalent to showing that the matrix N is symmetric. Bythecosinelaw,wehave l l + l l3 l3 + l l l 4N = l + l l3 l l + l3 l l l3 + l l l + l3 l l3 l3 l + r r3 l + r3 r l + r r 3 l + r 3 r. l3 + r r l3 + r r

9 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4765 To simplify notation, let a := l,b := l,c := l 3,x := r r 3 a,y := r 3 r b,z :=. Note that ax + by + cz =. Thus a a+ b c c+ a b +x x 4N = a + b c b b+ c a y +y c + a b b+ c a c +z z ) =. r r c ( a (a+b c)y+(c+a b)z c a b ax (c+a b)z b a c+ax+(a+b c)y c a b+by+(b+c a)z b (b+c a)z+(a+b c)x a b c by (a+b c)x b c a cz (b+c a)y a b c+cz+(c+a b)x c (c+a b)x+(b+c a)y This matrix is symmetric due to the equality ax + by + cz =. Glickenstein [] gave a new proof of the above lemma. The new proof puts the result in a general context and reveals its geometric meaning. Lemma 6. The Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 has one zero eigenvalue with associated eigenvector (,, ) and two negative eigenvalues. Proof. Using the notation in Lemma 5, we need to show that the matrix 4N has one zero eigenvalue and two positive eigenvalues. Since 4N is symmetric and the sum of entries in each row is zero, we can write 4N as B B 3 B 3 B B 3 B 3 B B, B B B B where B,B,B 3 are functions of a, b, c, x, y, z. The characteristic equation of 4N is λ(λ +(B + B + B 3 )λ +3(B B + B B 3 + B 3 B )) =. There is a zero eigenvalue, and the associated eigenvector is (,, ). We claim that B + B + B 3 < andb B + B B 3 + B 3 B >. These two inequalities imply that the matrix 4N has two positive eigenvalues. In the following we verify the two inequalities. First, B + B + B 3 < trace of 4N = (B + B + B 3 ) > a + b + c +(b c)x +(c a)y +(a b)z > l + l + l3 +(l l3) r r3 l l + l + l3 +(l 3 l ) r 3 r l +(l l ) r r l 3 > + l l+l l 3 l 4 l3 4 l r l + l l+l l 3 l 4 l3 4 3 l r l + l 3l +l3l l 4 l 4 3 l r l 3 > : l + l + l3 +Ξ r +Ξ r +Ξ 3 r3 >. For {i, j, k} = {,, 3}, sincel i = r j + r k +I jkr j r k and I ij [, ), we have l i >r j. Without loss of generality, we may assume l l l 3. Therefore Ξ and Ξ 3.

10 4766 REN GUO Case. If Ξ, then l + l + l3 +Ξ r +Ξ r +Ξ 3 r3 l + l + l 3 +Ξ 3 r 3 >l + l + l 3 +Ξ 3 l = l + l + l3 + l 3 l + l 3 l l4 l4 l l l = l ((l l )l +l l 3 + l l 3) > due to the assumption l l. Case. If Ξ <, then l + l + l3 +Ξ r +Ξ r +Ξ 3 r3 >l + l + l 3 +Ξ r +Ξ 3 r 3 >l + l + l 3 +Ξ l 3 +Ξ 3 l = l + l + l 3 + l l + l l 3 l 4 l 4 3 l l 3 = l (l l +l l 3 +l 3l l 4 l 4 l 4 3) l3 + l 3l + l3l l 4 l 4 l l l = l (l + l + l 3 )(l + l l 3 )(l 3 + l l )(l + l 3 l ) > due to the triangle inequality. Thus B + B + B 3 < holds. To prove B B + B B 3 + B 3 B >, we substitute to obtain B =4N 3 = a b c by (a + b c)x, B =4N 3 = b c a cz (b + c a)y, B 3 =4N = c a b ax (c + a b)z B B + B B 3 + B 3 B =ab +bc +ac a b c +(b c 3a +6ab +4ac)xy +(c a 3b +6bc +4ab)yz +(a b 3c +6ac +4bc)zx +a(a + b c)x +b(b + c a)y +c(c + a b)z +(a + b + c)(ax + by + cz) =(ab +bc +ac a b c )(xy + yz + zx +) +(ax + by + cz)((a + b c)x +(b + c a)y +(c + a b)z) +(a + b + c)(ax + by + cz) =(ab +bc +ac a b c )(xy + yz + zx +) due to the fact that ax + by + cz =.

11 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4767 Since ab +bc +ac a b c =l l +l l 3 +l 3 l l4 l4 l4 3 =(l + l + l 3 )(l + l l 3 )(l 3 + l l )(l + l 3 l ) >, we have B B + B B 3 + B 3 B > xy + yz + zx +> (3) (r r)(r 3 r)l +(r r3)(r r)l +(r3 r)(r r3)l 3 + ll l 3 >. To verify inequality (3), first note that, since li = r j + r k +I jkr j r k and I jk [, ), we have li r j + r k for {i, j, k} = {,, 3}. Since the coefficient of l on the left hand side of (3) satisfies (r r)(r 3 r)+l l 3 (r r)(r 3 r)+(r + r3)(r + r) =rr +rr 3 >, by replacing l by r + r 3, we see that the left hand side of (3) is not less than (4) (r r )(r 3 r )(r + r 3)+(r r 3)(r r )l +(r 3 r )(r r 3)l 3 +(r + r 3)l l 3. Since the coefficient of l in (4) satisfies (r r3)(r r)+(r + r3)l 3 (r r3)(r r)+(r + r3)(r + r) =rr +rr 3 >, by replacing l by r + r 3, we see that (4) is not less than (5) (r r)(r 3 r)(r + r3)+(r r3)(r r)(r + r3) +(r3 r)(r r3)l 3 +(r + r3)(r + r3)l 3. Since the coefficient of l 3 in (8) is positive, by replacing l 3 by r + r, we see that (5)isnotlessthan (6) (r r)(r 3 r)(r + r3)+(r r3)(r r)(r + r3) +(r3 r)(r r3)(r + r)+(r + r3)(r + r3)(r + r). Since (6) is equal to 8r r r 3,weseethatB B + B B 3 + B 3 B >. For the special case of intersection angle circle packing, Lemma 6 was proved in Chow-Luo [9] (Lemma 3.). In that case, by a geometric reasoning due to Thurston [3], the angle α i is a decreasing function of r i and an increasing function of r j for j i. That means α i r i < and α i r j > forj i. Therefore the diagonal entries of the matrix N in Lemma 5 are positive and the off-diagonal entries of N are negative. By a theorem in linear algebra, this property implies Lemma 6. But Chow-Luo s proof does not work in the general case of inversive distance circle packing, because it is not always true that the angle α i is a decreasing function of r i andanincreasing

12 4768 REN GUO function of r j for j i. For example, when l =,l =,l 3 =3,r = r = r 3 =, the matrix N is with eigenvalue, 7 4, 7 4. This shows that α r <, while α r 3 > for some value of l i s and r i s. Since, by Corollary 4, the space UE 3 of vectors (u,u,u 3 ) is connected and simply connected, Lemma 5 and Lemma 6 imply Corollary 7. The differential -form 3 i= α idu i is closed. For any c UE 3, the integration w(u,u,u 3 )= (u,u,u 3 ) 3 c i= α idu i is a concave function on UE 3 and satisfies, for i =,, 3, w (7) = α i. u i Proof of Theorem in Euclidean geometry. Let s prove the local rigidity of inversive distance circle packing in Euclidean geometry. For a weighted triangulated closed surface (Σ,T,I) with I ij [, ) foreachij E, let U E be the open subset of R V formed by the vectors u =(u,u,..., u V ) satisfying (u i,u j,u k ) U ijk E whenever ijk F. For a radius vector r =(r,r,..., r V ), we may rescale it to get cr, while the cone angles remain the same. Under the rescaling, u changes to u+ln(c)(,,..., ). Therefore we only consider the intersection U E P where P is the hyperplane defined by V i= u i =. Let a i be the cone angle at vertex i under the Euclidean cone metric. Then V i= a i = the sum of all inner angles = π F. Therefore the space of possible cone angles is contained in the intersection R V > Q, whereq is the hyperplane defined by V i= a i = π F. We have a map ξ : U E P R V > Q. By Corollary 7, for each triangle ijk F there is a function w(u i,u j,u k ). Define a function W : U E R by W (u,u,..., u V )= 9 ijk F w(u i,u j,u k ), where the sum is over all triangles in F. By (7), W u i equals the sum of inner angles having vertex i, i.e., the cone angle at vertex i. Therefore, when restricted on U E P, the gradient W is the map ξ : U E P R V > Q. We claim that the restriction of W on U E P is strictly concave. By the definition of W, we get the relation between the Hessian matrixes H(W )= H(w(u i,u j,u k )), ijk F where we think of w(u i,u j,u k ) as a function on U E.

13 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4769 For any x U E P, xh(w )x T = ijk F = ijk F xh(w(u i,u j,u k ))x T (x i,x j,x k )H(w(u i,u j,u k ))(x i,x j,x k ) T, since H(w(u i,u j,u k )) is negative semi-definite for each ijk F. Thus xh(w )x T = implies (x i,x j,x k )H(w(u i,u j,u k ))(x i,x j,x k ) T =foreach ijk. By Lemma 6, (x i,x j,x k ) is an eigenvector of zero eigenvalue. Therefore (x i,x j,x k )=c ijk (,, ) for some constant c ijk. If two triangles share a vertex, then the corresponding constants c ijk of the two triangles must be the same. In fact, consider the two triangles ijk and ij k which share the vertex i. For ijk, we have (x i,x j,x k )=c ijk (,, ); in particular, x i = c ijk. Also, for ij k,wehave (x i,x j,x k )=c ij k (,, ); in particular, x i = c ij k. Hence c ijk = x i = c ij k. Since the surface is connected, the corresponding constants c ijk of any triangles are the same. Thus c ijk must be a constant c independent of the triangle ijk. Therefore x = c(,,..., ). Since x P, i.e, x i =, we have c =. Thus x = (,,..., ). This shows that the Hessian H(W )isnegativedefinitewhen restricted on U E P. The local rigidity of inversive distance circle packing in Euclidean geometry follows from the following lemma in analysis Lemma 8. If X is an open set in R n and the Hessian matrix of f is positive definite for all x X, then the gradient f : X R n is a local diffeomorphism. 3. Circle packing in hyperbolic geometry 3.. Space of radius vectors. As stated in the Introduction, to obtain a hyperbolic cone metric, we need conditions on a radius vector. In this section we investigate the space of radius vectors for one triangle in hyperbolic geometry. Consider a topological triangle ijk F. Assign a number l ij to the edge ij E, where l ij is the distance of the centers of two circles with radii r i,r j and inversive distance I ij. Therefore l ij is a function of r i,r j. In fact, in hyperbolic geometry, l ij is determined by cosh l ij =coshr i cosh r j + I ij sinh r i sinh r j. Similarly, we obtain l jk,l ki. Lemma 9. The set R ijk H := {(r i,r j,r k ) R 3 > l ij,l jk,l ki satisfy triangle inequality} is a connected and simply connected open subset in R 3 >. Proof. Let ρ H : R 3 > R 3 > be the map sending (r i,r j,r k )to(l ij,l jk,l ki ). We claim that ρ H is a smooth embedding. First the determinant of the Jacobian matrix of ρ H is nonzero. Next, we will show ρ H is one-to-one. That means, for fixed numbers I ij,i jk,i ki [, ) and a fixed hyperbolic triangle, there is at most one configuration of three circles centered at the vertices of the triangle and having the prescribed inversive distance I ij,i jk,i ki. The statement is true in Euclidean geometry as proved in the proof of Lemma 3. We use the result in Euclidean geometry to prove a similar result in hyperbolic geometry. Let s assume there are two configurations of three circles in a hyperbolic plane with the fixed triangle and inversive distance. As in Figure, the stereographic projection sends the two configurations to Euclidean plane C. Since the

14 477 REN GUO stereographic projection preserves the inversive distance, there will be two different configurations of three circles with the same Euclidean triangle and the same inversive distance. This is impossible. To determine the image ρ H (R 3 >), we consider the image of the boundary of R 3. If r i =, then r k = l ki,r j = l ij. Thus cosh l jk =coshr j cosh r k + I jk sinh r j sinh r k =coshl ij cosh l ki + I jk sinh l ij sinh l ki. On the other hand, if r i >, then r k <l ki,r j <l ij. We have cosh l jk < cosh l ij cosh l ki + I jk sinh l ij sinh l ki. Therefore the image ρ H (R 3 > ) is the subset of vectors (l ij,l jk,l ki ) satisfying cosh l jk < cosh l ij cosh l ki + I jk sinh l ij sinh l ki, cosh l ki < cosh l jk cosh l ij + I ki sinh l jk sinh l ij, cosh l ij < cosh l ki cosh l jk + I ij sinh l ki sinh l jk. In fact, ρ H (R 3 >) is a cone bounded by three surfaces. In R 3, the intersection of any of the two surfaces is a straight line. For example, if cosh l jk =coshl ij cosh l ki + I jk sinh l ij sinh l ki and cosh l ki =coshl jk cosh l ij + I ki sinh l jk sinh l ij, then the sum of the two equation gives =(coshl ki +coshl jk )(cosh l ij ) + (I jk sinh l ki + I ki sinh l jk ) sinh l ij =(coshl ki +coshl jk ) sinh l ij +(I jk sinh l ki + I ki sinh l jk ) cosh l ij sinh l ij. The only possibility is sinh l ij =. Therefore l ij =andl jk = l ki. Hence the three lines of intersection are l ij =andl jk = l ki, l jk =andl ki = l ij, l ki =andl ij = l jk. Recall that L is the subset of R 3 > formed by vectors (l ij,l jk,l ki ) satisfying the triangle inequality. Therefore L is a cone bounded by three planes. The three lines of intersection of the three planes are the same as that of the boundary surfaces of ρ H (R 3 >). We claim that ρ H (R 3 > ) Lis a cone. In fact, if I ij [, ], then in R 3 > we have {(l ij,l jk,l ki ) cosh l ij < cosh l ki cosh l jk + I ij sinh l ki sinh l jk } {(l ij,l jk,l ki ) l ij <l ki + l jk }. If I ij (, ), then in R 3 > we have {(l ij,l jk,l ki ) cosh l ij < cosh l ki cosh l jk + I ij sinh l ki sinh l jk } {(l ij,l jk,l ki ) l ij <l ki + l jk }. Applying the same argument in Euclidean geometry, we see ρ H (R 3 >) L is a cone. Therefore R ijk H = ρ H (ρ H(R 3 >) L) is connected and simply connected.

15 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 477 In the construction of the energy function, we will use a natural variable u = (u,u,..., u V )whereu i =lntanh r i. Let τ H : R 3 R 3 > be the map sending (u i,u j,u k )to(r i,r j,r k ). Since τ H is a diffeomorphism, we have Corollary. The set U ijk H open subset in R 3 >. := τ H (Rijk H ) is a connected and simply connected To prove Theorem, we construct a strictly concave energy function on U ijk H for each triangle ijk F. As in the case of Euclidean geometry, the space U ijk H is not always convex (for example, when I ij = I jk = I ki = ); we cannot obtain global rigidity. Therefore we are not able to determine the space of cone angles in the case of inversive distance circle packing. 3.. Energy function. To prove Theorem, we will construct an energy function on the space of radius vectors. First, we construct an energy function on the space for one triangle. To simplify notation, consider the triangle 3 F and the map U 3 H R 3 H L R 3 >, (u,u,u 3 ) (r,r,r 3 ) (l,l,l 3 ) (α,α,α 3 ), where r i =ln +eu i e u i for i =,, 3. For three fixed positive numbers I,I 3,I 3, we have cosh l i = coshr j cosh r k + I ij sinh r j sinh r k for {i, j, k} = {,, 3}. For simplicity of notation, we set (l,l,l 3 )=(l 3,l 3,l ). Also, α,α,α 3 are inner angles of a hyperbolic triangle with edge lengths l,l,l 3. By the cosine law, cos α i = cosh l i +coshl j cosh l k sinh l j sinh l k for {i, j, k} = {,, 3}. Therefore α,α,α 3 are functions in terms of u,u,u 3. Lemma. The Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 is symmetric. Proof. Due to the cosine law of a hyperbolic triangles, α,α,α 3 are functions of l,l,l 3. The following formula can be proved by direct calculation. For {i, j, k} = {,, 3}, dα dα = sinh l sinh l sin α dα i sinh l j sinh l k 3 sinhl 3 cosα 3 cos α cos α 3 cosα dl dl. cos α cos α dl 3 Next, by the construction cosh l i =coshr j cosh r k +I jk sinh r j sinh r k, for {i, j, k} = {,, 3}, where I ij is a constant. Differentiating the two sides of the equality, we obtain dl i = cosh r k +coshl i cosh r j dr j + cosh r j +coshl i cosh r k dr k. sinh l i sinh r j sinh l i sinh r k Finally, r i =ln +eu i e implies sinh r u i i du i = dr i,i=,, 3.

16 477 REN GUO Combining the three relations, we have, for {i, j, k} = {,, 3}, dα dα = dα sin α i sinh l j sinh l k 3 sinh l sinh l cosα 3 cos α cos α 3 cosα sinhl 3 cos α cos α cosh r 3 +coshl cosh r cosh r +coshl cosh r 3 sinh l sinh r sinh l sinh r 3 cosh r 3 +coshl cosh r cosh r +coshl cosh r 3 sinh l sinh r sinh l sinh r 3 cosh r +coshl 3 cosh r cosh r +coshl 3 cosh r sinh l 3 sinh r sinh l 3 sinh r sinh r sinh r du du. sinhr 3 du 3 We write the above formula as dα dα dα 3 = sin α i sinh l j sinh l k M du du du 3 where M is a product of four matrixes. To show that the Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 is symmetric is equivalent to showing that the matrix M is symmetric. To simplify notation, let a := cosh l,b := cosh l,c := cosh l 3,x := cosh r,y := cosh r,z := cosh r 3. Bythecosinelaw,wehave M = a ab c ca b ab c b bc a ca b bc a c a b c, ay z az y bx z bz x cx y cy x We check that M is symmetric, for example, by showing M = M = z ac b c x bc a c y. Lemma. The Jacobian matrix of functions α,α,α 3 in terms of u,u,u 3 is negative definite. Proof. Using the notation of Lemma, we need to verify that the matrix M is positive definite. The determinant of M is positive since the determinant of each factor is positive. Note that the second matrix is a Gram matrix up to the negative sign. Since, by Corollary, the set UH 3 is connected, to show that M is positive definite, it is enough to check that M is positive definite at one vector in UH 3. It is equivalent to picking up a vector (r,r,r 3 ) such that the resulting l,l,l 3 satisfy the triangle inequality. We pick up (r,r,r 3 )=(s, s, s). In the following, we show that such kind of number s exists. For fixed I,I 3,I 3 [, ), we have cosh l i =coshr j cosh r k +I jk sinh r j sinh r k = cosh s+i jk sinh s cosh s, for {i, j, k}={,, 3}. Therefore sinh l i cosh 4 s for i=,, 3..

17 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4773 The fact that l,l,l 3 satisfy the triangle inequality is equivalent to cosh(l i + l j ) > cosh l k cosh l i cosh l j + sinh l i sinh l j > cosh l k =cosh s + I ij sinh s =cosh 4 s +cosh 4 s > cosh s + I ij sinh s. Let x := cosh s. The above inequality is equivalent to (x I ij +)(x ) >. Since x>, we need x> I ij. Therefore l,l,l 3 as functions of (s, s, s) satisfy the triangle inequality if cosh s > max{ I, I 3, I 3 }. Now it is enough to check that M is positive definite at such a vector (s, s, s) when s is sufficiently large. In this case, M = sinh l sinh l cosα 3 cos α cos α 3 cosα sinhl 3 cos α cos α +coshl +coshl sinh l sinh l +coshl +coshl cosh s := M sinh l sinh l cosh s. +coshl 3 +coshl 3 sinh l 3 sinh l 3 The fact that M is positive definite is equivalent to M being positive definite. Since M only involves l,l,l 3,thefactthatM is positive definite is a property of a hyperbolic triangle. In fact, the fact that M is positive definite was proved in the paper of Guo-Luo [3] (Lemma 4.4) by showing that its leading principal and minor is positive. For completeness, we include a proof here. That the leading principal minor is positive is equivalent to +coshl +coshl 3 cos α 3 +cosα >. sinh l sinh l 3 Replacing α s by l s by the cosine law, we obtain (8) cosh l > cosh l +cosh l 3 +coshl +coshl 3. coshl cosh l 3 +coshl +coshl 3 Since l > l l 3, therefore cosh l > cosh(l l 3 ). To show that (8) holds, it is enough to check that (9) cosh l cosh l 3 sinh l sinh l 3 cosh l +cosh l 3 +coshl +coshl 3. coshl cosh l 3 +coshl +coshl 3 We simplify the notation by introducing a := cosh l >,b:= cosh l 3 >. Then (9) is rewritten as () ab (a )(b ) a + b + a + b. ab + a + b () is equivalent to () ab a + b + a + b (a ab + a + b )(b ).

18 4774 REN GUO Since a>,b>, the left hand side of () is positive. To show that () holds, we square the two sides and simplify. We have (ab 4 + a 4 b a 3 b a b 3 )+(a 4 + b 4 a b )+(a 3 + b 3 ab a b) (ab +)(a 3 + b 3 a b ab )+(a b ) (ab +)(a + b)(a b) +(a b ). This shows that the leading principal minor is positive. To show that the leading principal minor of M is positive, for simplifying notation, let t i := +cosh l i sinh l i. Up to a diagonal matrix, M is cosα 3 cos α cos α 3 cosα cos α cos α t t t t t 3 t 3 That the leading principal minor of M is positive is equivalent to (t cos α 3 + t 3 cos α )(t cos α 3 + t 3 cos α ) ( t + t 3 cos α )( t + t 3 cos α ) > (cos α 3 )t t +(cosα cos α 3 +cosα )t t 3 +(cosα cos α 3 +cosα )t t 3 > sin α 3 t t +coshl sin α sin α 3 t t 3 +coshl sin α sin α 3 t t 3 > cosh l sinh l t t 3 +coshl sinh l t t 3 > sinh l 3 t t cosh l sinh l + cosh l sinh l > sinh l 3 t t t 3 cosh l (+coshl )+coshl (+coshl ) > +coshl 3 (cosh l +coshl ) + ( cosh l cosh l cosh l 3 ) >. The cosine law and the sine law of a hyperbolic triangle are used in the above calculation. The last inequality is true since cosh l +coshl > andcoshl cosh l > cosh l cosh l + sinh l sinh l =cosh(l + l ) > cosh l 3. For the special case of intersection angle circle packing, Lemma was proved in Chow-Luo [9] (Lemma 3.) based on Thurston s observation [3] that the angle α i is a decreasing function of r i and an increasing function of r j for j i. But this observation does not hold in the general case of inversive distance circle packing. For example, when l =,l =,l 3 =3,r = r = r 3 =, the matrix M is with eigenvalue 3.5, 5.59, This shows that α r <, while α r 3 > forsome value of l i and r i. Since, by Corollary, the space UH 3 of vectors (u,u,u 3 ) is connected and simply connected, Lemma and Lemma imply Corollary 3. The differential -form 3 i= α idu i is closed. For any c UH 3, the integration w(u,u,u 3 )= (u,u,u 3 ) 3 c i= α idu i is a strictly concave function on UH 3 and satisfying, for i =,, 3, w () = α i. u i.

19 LOCAL RIGIDITY OF INVERSIVE DISTANCE CIRCLE PACKING 4775 Proof of Theorem in hyperbolic geometry. Let s prove the local rigidity of inversive distance circle packing in hyperbolic geometry. For a weighted triangulated closed surface (Σ,T,I) with I ij [, ) foreachij E, let U H be the open subset of R V formed by the vectors u =(u,u,..., u V ) satisfying (u i,u j,u k ) U ijk H whenever ijk F. By Corollary 3, for each triangle ijk F, there is a function w(u i,u j,u k ). Define a function W : U H R by W (u,u,..., u V )= w(u i,u j,u k ), ijk F where the sum is over all triangles in F. By Corollary 3, W is strictly concave on U H so that W u i equals the sum of inner angles having vertex i, i.e., the cone angle at vertex i. That means the gradient of W is exactly the map sending a vector u to its cone angles. The local rigidity of inversive distance circle packing in hyperbolic geometry holds due to Lemma 8. Acknowledgment The author would like to thank Feng Luo and Albert Madern for encouragement and helpful comments. He also thanks the referee for valuable suggestions. References. E. M. Andreev, Convex polyhedra in Lobachevsky spaces. (Russian) Mat. Sb. (N.S.) 8 (3) MR59734 (4:4367). E. M. Andreev, Convex polyhedra of finite volume in Lobachevsky space. (Russian) Mat. Sb. (N.S.) 83 (5) MR735 (4:8388) 3. W. Brägger, Kreispackungen und Triangulierungen. Enseign. Math., 38: 7, 99. MR896 (94b:53) 4. P.L.Bowers,M.K.Hurdal,Planar conformal mappings of piecewise flat surfaces. Visualization and Mathematics III, 3 34, Math. Vis., Springer, Berlin, 3. MR P. L. Bowers, K. Stephenson, Uniformizing dessins and Belyĭ maps via circle packing. Mem. Amer. Math. Soc. 7 (4), no. 85. MR5339 (5a:368) 6. A. I. Bobenko, B. A. Springborn, Variational principles for circle patterns and Koebe s theorem. Trans. Amer. Math. Soc. 356 (4), no., MR75 (5b:554) 7. H. S. M. Coxeter, Inversive distance. Ann. Mat. Pura Appl. (4) MR3568 (34:348) 8. Y. Colin de Verdiére, Un principe variationnel pour les empilements de cercles. Invent. Math., 4: , 99. MR6755 (9h:57) 9. B. Chow, F. Luo, Combinatorial Ricci flows on surfaces. J. Differential Geom. 63 (3), no., MR56 (5a:536). J. Dai, X. D. Gu, F. Luo, Variational principles for discrete surfaces. Advanced Lectures in Mathematics (ALM), 4. International Press, Somerville, MA; Higher Education Press, Beijing, 8. MR D. Glickenstein, Discrete conformal variations and scalar curvature on piecewise flat two and three dimensional manifolds. arxiv:96.56, to appear in J. Differential Geom.. R. Guo, A note on circle patterns on surfaces. Geom. Dedicata 5 (7), MR3547 (8f:57) 3. R. Guo, F. Luo, Rigidity of polyhedral surfaces, II. Geom. Topol. 3 (9), MR Z.-X. He, Rigidity of infinite disk patterns. Ann. of Math. () 49 (999), no., 33. MR6853 (j:368) 5. M. K. Hurdal, K. Stephenson, Discrete conformal methods for cortical brain flattening. Neuroimage, vol. 45, S86-S98, 9.

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