Geometric angle structures on triangulated surfaces

In this paper we characterize a function defined on the set of edges of a triangulated surface such that there is a spherical angle structure having the function as the edge invariant (or Delaunay invariant). We also characterize a function such that there is a hyperbolic angle structure having the function as the edge invariant.


§1. Introduction
Suppose S is a closed surface and T is a triangulation of S. Here by a triangulation we mean the following: take a finite collection of triangles and identify their edges in pairs by homeomorphism. Let V, E, F be the sets of all vertices, edges and triangles in T respectively. If a, b are two simplices in triangulation T , we use a < b to denote that a is a face of b. Let C(S, T ) = {(e, f )|e ∈ E, f ∈ F, such that e < f } be set of all corners of the triangulation. An angle structure on a triangulated surface (S, T ) assigns each corner of (S, T ) a number in (0, π). A Euclidean (or hyperbolic, or spherical) angles structure is an angle structure so that each triangle with the angle assignment is Euclidean (or hyperbolic, or spherical). More precisely, a Euclidean angle structure is a map x : C(S, T ) → (0, π) assigning every corner i (for simplicity of notation, we use one letter to denote a corner) a positive number x i such that x i +x j +x k = π whenever i, j, k are three corners of a triangle. A hyperbolic angle structure is a map x : C(S, T ) → (0, π) such that x i +x j +x k < π. A spherical angle structure is a map x : C(S, T ) → (0, π) such that Actually it is proved in [B] that positive numbers x i , x j , x k are three inner angles of a spherical triangle if and only if they satisfy conditions (1).
Given an angle structure x : C(S, T ) → (0, π), we define its edge invariant which is a function D x : E → (0, 2π) such that D x (e) = x i + x i ′ where i = (e, f ), i ′ = (e, f ′ ) are two opposite corners facing the edge e. And we define its Delaunay invariant which is a function D opposite corners facing the edge e and j, k(or j ′ , k ′ ) are the other two corners of the triangle f (or f ′ ).
For the simplicity of natation, we use G to denote a fixed geometry, where G = E, H or S means the Euclidean, hyperbolic or spherical geometry respectively. Now given a function D : E → (0, 2π) (or D : E → (−2π, 2π)), we use AG(S, T ; D) (or AG(S, T ; D)) to denote the set fo all G angle structures having D (or D) as the edge (or Delaunay) invariant.
The motivation of considering these sets is the study of geometric cone metrics with prescribed edge invariant or Delaunay invariant on triangulated surfaces from the variational point of view. A Euclidean (or hyperbolic, or spherical) cone metric assigns each edge in T a positive number such that the numbers on any three edges of a triangle in T form three edge length of a Euclidean (or hyperbolic, or spherical) triangle. The variational method contains a variational problem and a linear programming problem. The variational problem is to show that the unique maximal point of a convex "capacity" defined on the set AG(S, T ; D) (or AG(S, T ; D)) gives the unique geometric cone metric. The linear programming problem is to characterize the function D (or D) such that the set AG(S, T ; D) (or AG(S, T ; D)) is nonempty.
For Euclidean angle structures, the Delaunay invariant and the edge invariant are related by 2D x (e) + D x (e) = 2π for any e. Thus given two functions D and D satisfying 2D(e)+D(e) = 2π for any e, we have AE(S, T ; D) = AE(S, T ; D). Therefore the problem of Euclidean cone metric with given edge invariant is eqivalent to the problem of Euclidean cone metric with given Delaunay invariant. Rivin [Ri1] [Ri2] worked out the variational problem and the linear programming problem about AE(S, T ; D). Leibon [Le] worked out the variational problem and the linear programming problem about AH(S, T ; D). Luo [Lu] worked out the variational problem about AS(S, T ; D) the linear programming problem about which will be solved in this paper (theorem 1). Although the variational problems about AH(S, T ; D) and AS(S, T ; D) are still open, we will solve the linear programming problem about them in this paper (theorem 2 and 3).
The main results are the following. For a triangulated surface (S, T ), a subset X ⊆ F, we use |X| to denote the number of triangles in X and we use E(X) to denote the set of all edges of triangles in X.
The paper is organized as follows. In section 2, we prove theorem 1 by using Leibon's result. In section 3, we recall the duality theorem in linear programming. In section 4, following Rivin's method, we prove theorem 2 and 3 by using the duality theorem.
Proof of theorem 1. To show the conditions are necessary, for any X ⊆ F , we have e∈E(X) D(e) = e∈E(X) (x i +x i ′ ), where i, i ′ are two opposite corners facing the edge e. It turns out that the right hand side of the equation is equal To show the conditions are sufficient, let us define a function D : E → (0, 2π) by setting D(e) = 2π−2D(e). Thus the conditions π|X| < e∈E(X) D(e) are equivalent to π|X| < e∈E(X) (π− 1 2 D(e)) which guarantee AH(S, T ; D) is nonempty by theorem 4. It follows that there is a solution for the inequalities Let us define new variables y i for all i ∈ C(S, T ) by setting provided i, j, k are three corners of a triangle. And since D(e) = 2π − 2D(e), the inequalities above are equivalent to      y i + y j + y k > π i, j, k are three corners of a triangle y i + y i ′ = D(e) i, i ′ are two opposite corners facing an edge e y j + y k < π j, k are two corners of a triangle This solution obviously satisfies y i + y j + y k > π i, j, k are three corners of a triangle y i + y i ′ = D(e) i, i ′ are two opposite corners facing an edge e y j + y k − y i < π i, j, k are three corners of a triangle y i > 0 Thus we obtain an angle structure in AS(S, T ; D). QED §3. Duality Theorem We fix the notations as follows: x = (x 1 , ..., x n ) t is a column vector in R n . The standard inner product in R n is denoted by a t x. If A : R n → R m is a linear transformation, we denote its transpose by A t : R m → R n . Given two vectors x, a in R n , we say x ≥ a if x i ≥ a i for all indices i. Also x > a means x i > a i for all indices i.
A linear programming problem (P ) is to minimize an objective function z = a t x subject to the restrain conditions We call a point x satisfying the restrain conditions a feasible solution and denote the set of all the feasible solutions by D(P ) = {x ∈ R n |Ax = b, x ≥ 0}. An optimal solution x for (P ) is a feasible solution so that the objective function z realizes the minimal value. The dual problem (P * ) of (P ) is to maximize z = b t y subject to A t y ≤ a, y ∈ R m . Let us recall the duality theorem in linear programming. The proof of the theorem can be found in the book [KB].
Theorem 5. The following statements are equivalent. (a) Problem (P) has an optimal solution. (b) D(P ) = ∅ and D(P * ) = ∅. (c) Both problem (P ) and problem (P * ) have optimal solutions so that the minimal value of (P ) is equal to the maximal value of (P * ).
In applications that we are interested, there is a special case that the objective function z = 0 for (P ). Thus the optimal solution exists if and only if D(P ) = ∅. Thus we obtain the following corollary.
Corollary 6. For A : R n → R m and b ∈ R m , the set {x ∈ R n |Ax = b, x ≥ 0} = ∅ if and only if the maximal value of z = b t y on {y ∈ R m |A t y ≤ 0} is non-positive. §4. Proof of theorem 2 and 3 By following Rivin's method in [Ri2], we will prove a lemma about the closure of AH(S, T ; D) in R 3|F | = {(x i ) t , i ∈ C(S, T )}. The closure of AH(S, T ; D) consists of all the points satisfying Proof. The linear programming problem (P ) with variables x = (..., x i , ..., t f , ...) indexed by C(S, T ) ∪ F is to minimize the objective function z = 0 subject to the restrain conditions i, i ′ are two opposite corners facing an edge e The dual problem (P * ) with variable y = (..., y f , ..., y e , ...) indexed by E ∪ F is to maximize the objective function z = f ∈F πy f + e∈E D(e)y e subject to the restrain conditions Since the closure of AH(S, T ; D) is nonempty is equivalent to that the set D(P ) is nonempty, by corollary 6, the latter one is equivalent to that the maximal value of the objective function of (P * ) is non-positive.
To show the conditions π(|F | − |X|) ≥ e / ∈E(X) D(e) for any X ⊂ F are necessary, for any X ⊂ F, let We claim that (y f , y e ) is a feasible solution. In fact, given a pair e < f, if f ∈ X, we must have e ∈ E(X), then y f + y e = 0. If f / ∈ X, then y f + y e = −1 + y e ≤ 0.
By the assumption that the maximal value of the objective function of (P * ) is non-positive, since (y f , y e ) is feasible, we have 0 ≥ z(y f , y e ) = f / ∈X πy f + e / ∈E(X) D(e)y e = π(|X| − |F |) + e / ∈E(X) D(e). To show the conditions are sufficient, take an arbitrary feasible solution (y f , y e ). If y f = 0 for all f , from y f + y e ≤ 0, we know y e ≤ 0. Hence z(y f , y e ) = e / ∈E D(e)y e ≤ 0, since D(e) ∈ [0, 2π]. Otherwise, define X = {f ∈ F |y f = 0} ⊂ F , and let a = max{y f , f / ∈ X}. We have a < 0. Define We claim that (y (1) f , y (1) e ) is a feasible solution. In fact, y (1) f ≤ 0. Given a pair e < f, if f ∈ X, we must have e ∈ E(X), then y (1) f + y (1) e = y f + y e ≤ 0. If f / ∈ X and e / ∈ E(X), then y (1) f + y (1) e = y f − a + y e + a ≤ 0. If f / ∈ X but e ∈ E(X), there exists another triangle f ′ ∈ X so that e < f ′ , then y e = y e + y f ′ ≤ 0. Therefore y (1) (1) f } is more than that in {y f }. By the same procedure, after finite steps, it ends at a feasible solution (y Since the value of the objective function does not increase, therefore 0 ≥ z(y a i + a j + a k + 3ε ≤ π i, j, k are three corners of a triangle a i + a j + 2ε = D(e) i, j are two opposite corners facing an edge e a i ≥ 0 ε ≥ 0 The dual problem (P * ) with variable y = (..., y f , ..., y e , ...) indexed by E ∪ F is to maximize the objective function z = f ∈F πy f + e∈E D(e)y e subject to the restrain conditions By the theorem 5(c), the maximal value of the objective function of (P * ) is negative is equivalent to that the minimal value of the objective function of (P ) is negative. The latter one is equivalent to that there exists a feasible solution a i ≥ 0, ε > 0. Therefore the set AH(S, T ; D) is nonempty.
We only need to show that the maximal value of the objective function of (P * ) is negative is equivalent to the conditions π(|F | − |X|) > e / ∈E(X) D(e) for any X ⊂ F.
To show the conditions are necessary, for any X ⊂ F, we have 2|E(X)| > 3|X| or 2|E(X)| ≥ 3|X| + 1. Let We claim that (y f , y e ) is a feasible solution. If fact, as in lemma 7, we can check y f + y e ≤ 0 for any pair e < f. Furthermore is feasible implies that z(y f , y e ) < 0 which is equivalent to π(|F | − |X|) < e / ∈E(X) D(e). To show the conditions are sufficient, by the proof of lemma 7 we know the maximal value of the objective function of (P * ) is ≤ 0 under the conditions. We try to show it can not be 0. Assume that (y f , y e ) is a feasible solution satisfying z(y f , y e ) = 0. We claim that y f = 0 for all f . Otherwise, as in the proof of lemma 7, we can find another feasible solution (y (1) f , y (1) e ) and we can check that z(y (1) f , y (1) e ) = z(y f , y e ) + a(π(|X| − |F |) + e / ∈E(X) D(e)) > z(y f , y e ) = 0, according to the conditions. It is contradiction since the maximal value of the objective function of (P * ) is ≤ 0.
Now from y f = 0 for all f we see y e ≤ 0. Since 0 = z(y f , y e ) = e∈E D(e)y e and D(e) > 0, we get y e = 0 for all e and therefore (y f , y e ) = (0, 0). But (y f , y e ) = (0, 0) does not satisfy 3 f ∈F y f + 2 e∈E y e ≤ −1. It is a contradiction since we assume that (y f , y e ) is a feasible solution. This proves that the maximal value of the objective function of (P * ) is negative. QED Proof of theorem 3. Given two functions D : E → (0, 2π) and D : E → (−2π, 2π) satisfying 2D(e) + D(e) = 2π for any e, we claim that AH(S, T ; D) = ∅ is eqivalent to AS(S, T ; D) = ∅. By this claim, theorem 3 is true as a corollary of theorem 2.
In fact, AS(S, T ; D) is the set of solutions for the inequalities i, j, k are three corners of a triangle x j + x k − x i < π i, j, k are three corners of a triangle x i > 0 Let us define new variables y i for all i ∈ C(S, T ) by setting provided i, j, k are three corners of a triangle. Since 2D(e) + D(e) = 2π, we see that the inequalities above are equivalent to          y i + y j + y k < π i, j, k are three corners of a triangle y i > 0 y i + y i ′ = D(e) i, i ′ are two opposite corners facing an edge e y j + y k < π j, k are two corners of a triangle Since y i + y j + y k < π implies y j + y k < π, we can omit the latter one. Equivalently, we get      y i + y j + y k < π i, j, k are three corners of a triangle y i > 0 y i + y i ′ = D(e) i, i ′ are two opposite corners facing an edge e Now the set of solutions of the inequalities above is exactly AH(S, T ; D). Thus we see AH(S, T ; D) = ∅ is eqivalent to AS(S, T ; D) = ∅. QED