THE DEFORMATION SPACE OF GEODESIC TRIANGULATIONS AND GENERALIZED TUTTE’S EMBEDDING THEOREM

A BSTRACT . We proved the contractibility of the deformation space of the geodesic triangulations on a closed surface of negative curvature. This solves an open problem proposed by Connelly et al. in 1983 [7], in the case of hyperbolic surfaces. The main part of the proof is a generalization of Tutte’s embedding theorem for closed surfaces of negative curvature.


INTRODUCTION
In this paper, we study the deformation space of geodesic triangulations of a surface within a fixed homotopy type.Such space can be viewed as a discrete analogue of the space of surface diffeomorphisms homotopic to the identity.Our main theorem is the following.
Theorem 1.1.For a closed orientable surface of negative curvature, the space of geodesic triangulations in a homotopy class is contractible.In particular, it is connected.
The group of diffeomorphisms of a smooth surface is one of the fundamental objects in the study of low dimensional topology.Determining the homotopy types of diffeomorphism groups has profound implications to a wide range of problems in Teichmuller spaces, mapping class groups, and geometry and topology of 3-manifolds.Smale [16] proved that the group of diffeomorphisms of a closed 2-disk which fix the boundary pointwisely is contractible.This enables him to show that the group of orientation-preserving diffeomorphisms of the 2-sphere is homotopic equivalent to SO(3) [16].Earle-Eells [9] identified the homotopy type of the group of the diffeomorphisms homotopic to the identity for any closed surface.In particular, such topological group is contractible for a closed orientable surface with genus greater than one, consisting with our Theorem 1.1 for the discrete analogue.
Cairns [5] initiated the investigation of the topology of the space of geodesic triangulations, and proved that if the surface is a geometric triangle in the Euclidean plane, the space of geodesic triangulations with fixed boundary edges is connected.A series of further developments culminated in a discrete version of Smale's theorem proved by Bloch-Connelly-Henderson [2] as follows.
Theorem 1.2.The space of geodesic triangulations of a convex polygon with fixed boundary edges is homeomorphic to some Euclidean space.In particular, it is contractible.
A simple proof of the contractibility of the space above is provided in [15] using Tutte's embedding theorem [17].It also provides examples showing that the homotopy type of this space can be complicated if the boundary of the polygon is not convex.For closed surfaces, it is conjectured in [7] that Conjecture 1.3.The space of geodesic triangulations of a closed orientable surface with constant curvature deformation retracts to the group of isometries of the surface homotopic to the identity.
The connectivity of these spaces has been explored in [5,6,14].Awartani-Henderson [1] identified a contractible subspace in the space of geodesic triangulations of the 2-sphere.Hass-Scott [14] showed that the space of geodesic triangulation of a surface with a hyperbolic metric is contractible if the triangulation contains only one vertex.The main result of this paper affirms conjecture 1.3 in the case of hyperbolic surfaces.
1.1.Set Up and the Main Theorem.Assume M is a connected closed orientable smooth surface with a smooth Riemannian metric g of non-positive Gaussian curvature.A topological triangulation of M can be identified as a homeomorphism ψ from |T | to M, where |T | is the carrier of a 2-dimensional simplicial complex T = (V, E, F ) with the vertex set V , the edge set E, and the face set F .For convenience, we label the vertices as 1, 2, ..., n where n = |V | is the number of vertices.The edge in E determined by vertices i and j is denoted as ij.Each edge is identified with the closed unit interval [0, 1].
Let T (1) be the 1-skeleton of T , and denote X = X(M, T, ψ) as the space of geodesic triangulations homotopic to ψ| T (1) .More specifically, X contains all the embeddings ϕ : The restriction ϕ ij of ϕ on the edge ij is a geodesic parameterized with constant speed, and (2) ϕ is homotopic to ψ| T (1) .It has been proved by Colin de Verdière [8] that such X(M, T, ψ) is always non-empty.Further, X is naturally a metric space, with the distance function Then our main theorem is formally stated as follows.
1.2.Generalized Tutte's Embedding.Let X = X(M, T, ψ) be the super space of X, containing all the continuous maps ϕ : T (1) → M satisfying that (1) The restriction ϕ ij of ϕ on the edge ij is geodesic parameterized with constant speed, and (2) ϕ is homotopic to ψ| T (1) .Notice that elements in X may not be embeddings of T (1) to M. The space X is also naturally a metric space, with the same distance function We call an element in X a geodesic mapping.A geodesic mapping is determined by the positions q i = ϕ(i) of the vertices and the homotopy classes of ϕ ij relative to the endpoints q i and q j .In particular, this holds for geodesic triangulations.Since we can perturb the vertices of a geodesic triangulation to generate another, X is a 2n dimensional manifold.
Let (i, j) be the directed edge starting from the vertex i and ending at the vertex j.Denote E = {(i, j) : ij ∈ E} as the set of directed edges of T .A positive vector w ∈ R E >0 is called a weight of T .For any weight w and geodesic mapping ϕ ∈ X, we say ϕ is w-balanced if for any i ∈ V , j:ij∈E Here v ij ∈ T q i M is defined with the exponential map exp : The main part of the proof of Theorem 1.4 is to generalize Tutte's embedding theorem (see Theorem 9.2 in [17] or Theorem 6.1 in [10]) to closed surfaces of negative curvature.Specifically, we will prove the following two theorems.
Theorem 1.5.Assume (M, g) has strictly negative Gaussian curvature.For any weight w, there exists a unique geodesic mapping ϕ ∈ X(M, T, ψ) that is w-balanced.Such induced map Φ(w) = ϕ is continuous from R E >0 to X.
Theorem 1.6.If ϕ ∈ X is w-balanced for some weight w, then ϕ ∈ X.
Theorem 1.6 can be regarded as a generalization of the embedding theorems by Colin de Verdière (see Theorem 2 in [8]) and Hass-Scott (see Lemma 10.12 in [14]), which imply that the minimizer of the following discrete Dirichlet energy among the maps ϕ in the homotopy class of ψ| T (1) is a geodesic triangulation.Here l ij is the geodesic length of ϕ ij in M. The minimizer is a w-balanced geodesic mapping with w ij = w ji for ij ∈ E. Hence, Theorem 1.6 extends the previous results from the cases of symmetric weights to non-symmetric weights.We believe that, the proofs in Colin de Verdière [8] and Hass-Scott [14] could be easily modified to work with our non-symmetric case.Nevertheless, we will give a new proof in Section 3 to make the paper self-contained.
1.3.Mean Value Coordinates and the Proof of Theorem 1.4.Theorem 1.5 and 1.6 give a continuous map Φ from R E >0 to X.For the oppositie direction, we can construct a weight w for a geodesic embedding ϕ ∈ X, using mean value coordinates which was firstly introduced by Floater [11].Given ϕ ∈ X, the mean value coordinates are defined to be where |v ij | equals to the geodesic length of ϕ ij ([0, 1]), and α ij and β ij are the two inner angles in ϕ(T (1) ) at the vertex ϕ(i) sharing the edge ϕ ij ([0, 1]).The construction of mean value coordinates gives a continuous map Ψ from X to R E >0 .Further, by Floater's mean value theorem (see Proposition 1 in [11]), any ϕ ∈ X is Ψ(ϕ)-balanced.Namely, Φ • Ψ = id X .Then Theorem 1.4 is a direct consequence of Theorem 1.5 and 1.6.
In the remaining of the paper, we will prove Theorem 1.5 in Section 2 and Theorem 1.6 in section 3.
2. PROOF OF THEOREM 1.5 Theorem 1.5 consists of three parts: the existence of w-balanced geodesic mapping, the uniqueness of w-balanced geodesic mapping and the continuity of the map Φ.In this section, we will first parametrize X by Mn , where M is the universal covering of M, and then prove the three parts in Subsection 2.1 and 2.2 and 2.3 respectively.
Assume that p is the covering map from M to M, and Γ is the corresponding group of deck transformations of the covering so that M /Γ = M.For any i ∈ V , fix a lifting qi ∈ M of q i ∈ M. For any edge ij, denote φij (t) as the lifting of ϕ ij (t) such that φij (0) = qi .Then p( φij (1)) = ϕ ij (1) = q j = p(q j ), and there exists a unique deck transformation ji for any edge ij.Equip M with the natural pullback Riemannian metric g of g with negative Gaussian curvature.This metric is equivariant with respect to Γ.For any x, y ∈ M , there exists a unique geodesic with constant speed parameterization γ x,y : [0, 1] → M such that γ x,y (0) = x and γ x,y (1) = y.We can naturally parametrize X as follows.
for any ij ∈ E and t ∈ [0, 1].Then such ϕ is a well-defined geodesic mapping in X, and the map Here we omit the proof of Theorem 2.1 which is routine but lengthy.In the remaining of this section, for any x, y, z ∈ M and u, v ∈ T x M , we denote (1) d(x, y) as the intrinsic distance between x, y in ( M, g), and (2) v(x, y) = exp −1  x y ∈ T x M , and (3) △xyz as the geodesic triangle in M with vertices x, y, z, which could possibly be degenerate, and (4) ∠yxz as the inner angle of △xyz at x if d(x, y) > 0 and d(x, z) > 0, and (5) |v| as the norm of v under the metric gx , and (6) u • v as the inner product of u and v under the metric gx .By scaling the metric if necessary, we may assume that the Gaussian curvatures of (M, g) and ( M , g) are bounded above by −1.

2.1.
Proof of the Uniqueness.We first prove the following Lemma 2.2 using CAT(0) geometry.See Theorem 4.3.5 in [4] and Theorem 1A.6 in [3] for the well-known comparison theorems.
, and the equality holds if and only if △xyz is degenerate.
Proof.If △xyz is degenerate, then there exists a geodesic γ in M such that x, y, z ∈ γ, and then the proof is straightforward.So we assume that △xyz is non-degenerate.
(1) Three points v(z, x), v(z, y), and 0 in T z M determine a Euclidean triangle, where |v(z, x)| = d(x, z), and |v(z, y)| = d(z, y) and the angle between v(z, x) and v(z, y) is equal to ∠xzy.
Then by the CAT(0) comparison theorem, |v(z, x) − v(z, y)| < d(x, y). ( Then by the CAT(0) comparison theorem, ∠yxz < ∠y ′ x ′ z ′ , and ∠xyz < ∠x ′ y ′ z ′ .Hence, Proof of the uniqueness part in Theorem 1.5 .If not, assume ϕ[x 1 , ..., x n ] and ϕ[x ′ 1 , ..., x ′ n ] are two different geodesic mappings that are both w-balanced for some weight w.We are going to prove a discrete maximum principle for the function j By lifting the w-balanced assumption to M , we have that (1) Then by part (1) of Lemma 2.2 and equation (1), By part (2) of Lemma 2.2, equation ( 2), and the Cauchy-Schwartz inequality, Therefore, the equalities hold in both inequalities above.Then for any neighbor , and A ij x j is on the geodesic determined by x i and Then the one-ring neighborhood of p(x i ) in ϕ[x 1 , ..., x n ](T (1) ) degenerates to a geodesic arc.By the connectedness of the surface, we can repeat the above argument and deduce that d(x j , x ′ j ) = d(x i , x ′ i ) for any j ∈ V .Further, for any triangle σ ∈ F , ϕ[x 1 , ..., x n ](∂σ) degenerates to a geodesic arc.
It is also not difficult to prove that φ is homotopic to ψ.Therefore, φ is degree-one and surjective.This is contradictory to that φ(|T |) is a finite union of geodesic arcs.
2.2.Proof of the Existence.Here we prove a stronger existence result.
Theorem 2.3.Given a compact subset K of R E >0 , there exists a compact subset K ′ = K ′ (M, T, ψ, K) of X such that for any w ∈ K, there exists a w-balanced geodesic mapping ϕ ∈ K ′ .
We first introduce a topological Lemma 2.4 and then reduce Theorem 2.3 to Lemma 2.5.
is null-homotopic, and thus g| S n−1 is also null-homotopic.Since g(x) = x, it is easy to verify that is a homotopy between g| S n−1 and −id| S n−1 .This contradicts to that −id| S n−1 is not nullhomotopic.
Lemma 2.5.We fix an arbitrary point q ∈ M .If w ∈ R E >0 and (x 1 , ..., x n ) ∈ Mn satisfies that for some constant R > 0 which depends only on M, T, ψ, q and The vector in Figure 1 r i = j:ij∈E is defined as the residue vector r i at with respect to the weight w.Notice that a geodesic mapping ϕ is w-balanced if and only if all its residue vectors vanish with respect to w. Lemma 2.5 means that if all the residue vectors are dragging x i 's away from q, then all the x i 's must stay not far away from q.The definition of residue vector is similar to the concept of discrete tension field in [12].Proof of Theorem 2.3.Fix an arbitrary base point q ∈ M, and then by Lemma 2.5 we can pick a sufficiently large constant R = R(M, T, ψ, K) > 0 such that if We will prove that the compact set is satisfactory.For any x ∈ M , let P x : T x M → T q M be the parallel transport along the geodesic γ x,q .Set as a Euclidean 2n-dimensional unit ball, and construct a map F : B → (T q M ) n in the following three steps.Firstly, we construct n points x 1 , ..., x n ∈ M as x i (v 1 , ..., v n ) = exp q (Rv i ).
Secondly, we compute the residue vector at each x i as Lastly, we pull back the residues to T q M as F (v 1 , ..., v n ) = (P x 1 (r 1 ), ..., P xn (r n )) .
Notice that the map (v 1 , ..., v n ) → ϕ[x 1 , ..., x n ] is a homeomorphism from B to K ′ , and F (v 1 , ..., v n ) = 0 if and only if the corresponding ϕ[x 1 , ..., x n ] in K ′ is w-balanced map.Hence, it suffices to prove that F has a zero in B. By Lemma 2.4 it suffices to prove that for Suppose (v 1 , ..., v n ) is an arbitrary point on ∂B, and then it suffices to prove that there exists , so by our assumption on R, there exists i ∈ V such that v(x i , q) • j:ij∈E and thus, In the rest of this subsection, we will prove Lemma 2.5 by contradiction.Let us first sketch the idea of the proof.Assume i∈V d(x i , q) 2 is very large, then by a standard compactness argument, there exists a long edge ij in the geodesic mapping ϕ[x 1 , ..., x n ].Assume d(q, x i ) ≥ d(q, x j ), then the corresponding long edge γ x i ,A ij x j in M is pulling x i towards q.This implies that there exists another long edge γ x i ,A ik x k dragging x i away from q, otherwise the residue vector r i would not drag x i away from q.It can be shown that d(q, x k ) > d(q, x i ).Repeating the above steps, we can find an arbitrary long sequence of vertices such that the distance from each of these vertices to q is increasing.This is impossible as we only have finitely many vertices.

FIGURE 2. Triangles in Step (b), (c), and (d).
Here is a listing of properties serving as the building blocks of the proof of Lemma 2.5.
(e) For any constant C > 0, there exists a constant (f) For any constant C > 0, there exists a constant for some edge ij ∈ E, then there exists ik ∈ E such that (g) For any constant C > 0, there exists a constant (h) For any constant C > 0, there exists a constant Proof of Lemma 2.5 assuming Lemma 2.6.For any C > 0, there exists a sufficiently large constant C = C(M, T, ψ, λ w , C) determined from (a), (e), (f) and (g) in Lemma 2.6 such that if then there exist three vertices x i , x j , and x k shown in Figure 3 with Moreover, by (g), (e) and (f) of Lemma 2.6, we can find another vertex x l such that d(x l , q) ≥ d(x k , q) + C ≥ d(x j , q) + 2C, if the constant C(M, T, ψ, λ w , C) is sufficiently large.Inductively, we can find a sequence i 1 , ..., i n+1 ∈ V such that This contradicts to the fact that V only has n different elements.
Proof of Lemma 2.6.(a) By a standard compactness argument, the set is compact and the conclusion follows.
(b) We claim that the constant C 2 , which is determined by is satisfactory.Let △ABC be the hyperbolic triangle with the corresponding edge lengths Since M is a CAT(−1) space, it suffices to show that ∠C ≤ π/8.By the hyperbolic law of sine, (c) We claim that the constant C 3 determined by sinh C 3 = 1 sin π 8 is satisfactory.Let △ABC be the hyperbolic triangle with the corresponding edge lengths Since M is a CAT(−1) space, it suffices to show that ∠C ≤ π/8.By the hyperbolic law of sine (d) We claim that the constant C 4 determined by is satisfactory.Let △ABC be the hyperbolic triangle with the corresponding edge lengths Since M is a CAT(−1) space, it suffices to show that ∠B ≤ π/8.By the hyperbolic law of cosine, cos A = − cos B cos C + sin B sin C cosh a.Then, Thus, ∠B ≤ π/8.
(e) We claim that the constant C 5 determined by is satisfactory.Assume ij ∈ E and d(x i , A ij x j ) ≥ C 5 , and we have two cases shown in Figure 4.
If d(x i , q) ≥ d(A ij x j , q), then by part (d) By part (b) and part (d), and (f) We claim that the constant C 6 determined by and it is a contradiction.(g) We claim that C 7 = C + max ij∈E d(A ij q, q) is satisfactory.Notice that By part (1) of Lemma 2.2, (h) We claim that the constant C 8 determined by Then by part (c), ∠(A ji x i )x j q ≤ π/4, and by the triangle inequality,

Proof of the Continuity.
Proof of the continuity part of Theorem 1.5.If not, there exists ǫ > 0 and a weight w and a sequence of weights w (k) such that (1) w (k) converge to w, and (2) d X (Φ(w (k) ), Φ(w)) ≥ ǫ for any k ≥ 1.
By the stronger existence result Theorem 2.3, the sequence Φ(w (k) ) are in some fixed compact subset K ′ of X.By picking a subsequence, we may assume that Φ(w (k) ) converge to some ϕ ∈ X.Since Φ(w (k) ) is w (k) -balanced, then by the continuity of the residue vectors r i , ϕ is w-balanced, and thus Φ(w) = ϕ, which is contradictory to that Φ(w (k) ) does not converge to Φ(w).
3. PROOF OF THEOREM 1.6 3.1.Set up and preparations.Assume ϕ ∈ X is w-balanced for some weight w, and we will prove that ϕ is an embedding.Recall that q i = ϕ(i) for each i ∈ V , and denote l ij as the length of ϕ ij ([0, 1]) for any ij ∈ E. It is not difficult to show that ϕ has a continuous extension φ defined on |T |, such that for any triangle σ ∈ F a continuous lifting map Φ σ of φ| σ from σ to M will (1) map σ to a geodesic triangle in M homeomorphically if ϕ(∂σ) does not degenerate to a geodesic, and (2) map σ to Φ σ (∂σ) if ϕ(∂σ) degenerates to a geodesic.The main tool to prove Theorem 1.6 is the Gauss-Bonnet formula.We will need to define the inner angles for each triangle in ϕ(T (1) ), even for the degenerate triangles.A convenient way is to assign a "direction" to each edge, even for the degenerate edges with zero length.
Given a direction field v, define the inner angle of the triangle σ = △ijk at the vertex i as , where 0 is the origin and ∠v ij 0v ik is the angle between v ij and v ik in T q i M.
A direction field v assigns a unit tangent vector in T q i M to each directed edge starting from i, and determines the inner angles in T .
) for a fixed vertex i ∈ V , if l ij = 0 for any neighbor j of i, then there exist neighbors j and k of i such that v ij = −v ik , and (4) if σ = △ijk ∈ F and l ij = l jk = l ik = 0, then θ i σ (v) + θ j σ (v) + θ k σ (v) = π.An admissible direction field encodes the directions of the non-degenerate edges in ϕ(T (1) ), and the induced angle sum of a degenerate triangle is always π. Then for any admissible v and triangle σ ∈ F , by the Gauss-Bonnet formula KdA.
Here dA is the area form on ( M , g) or (M, g).The concept of the direction field is similar to the discrete one form defined in [13].
3.2.Proof of Theorem 1.6.The proof of Theorem 1.6 uses the four lemmas below.We will postpone their proofs to the the subsequent subsections.Based on Lemma 3.3, if admissible direction fields exist, the image of the star of each vertex determined by φ does not contain any flipped triangles overlapping with each other.If φ(σ) does not degenerate to a geodesic arc for any triangle σ ∈ F , then φ is locally homeomorphic and thus globally homeomorphic as a degree-one map.Therefore, we only need to exclude the existence of degenerate triangles.
Define an equivalence relation on V as follows.Two vertices i, j are equivalent if there exists a sequence of vertices i = i 0 , i 1 , ..., i k = j such that l i For any x ∈ M and u, v ∈ T x M, denote u v as u and v are parallel, i.e., there exists (α, β) = (0, 0) such that αu + βv = 0.
The following Lemma 3.4 shows that there are plenty of choices of admissible direction fields.
Lemma 3.4.For any v 1 ∈ T y 1 M, ..., v m ∈ T ym M, there exists an admissible v such that v ij v k if i ∈ V k and l ij = 0.
The following Lemma 3.5 shows that for any V k with at least two vertices, the image of its "neighborhood" lies in a geodesic.Hence, the inequalities above are equalities.This fact implies that i∈V (2π − σ:i∈σ Since each term in this summation is non-positive, then σ:i∈σ θ i σ = 2π.The statement on the area follows similarly. 3.4.Proof of Lemma 3.4.We claim that for any k, there exists a map h : V k → R such that (1) h(i) = h(j) if i = j, and (2) for a fixed i ∈ V k , if l ij = 0 for any neighbor j of i, then there exist neighbors j and j ′ of i in V k such that h(j) < h(i) < h(j ′ ).Given such h, set v as where sgn is the sign function.It is easy to verify that such v is satisfactory.
To construct such function h, we will prove the following lemma, which is more general than our claim.Lemma 3.7.Assume G = (V ′ , E ′ ) is a subgraph of the 1-skeleton T (1) , and and ∂G = V ′ − int(G).Then there exists h : V ′ → R such that (1) h(i) = h(j) if i = j, and (2) for any i ∈ int(G) there exist neighbors j and j ′ of i in V ′ such that h(j) < h(i) < h(j ′ ).
Proof.We prove by the induction on the size of V ′ .The case |V ′ | = 1 is trivial.For the case |V ′ | ≥ 2, first notice that |∂G| ≥ 2 for any proper subgraph G of T (1) .Assign distinct values h(i) to each i ∈ ∂G, then solve the discrete harmonic equation j:ij∈E ( h(j) − h(i)) = 0, ∀i ∈ int(G), with the given Dirichlet boundary condition on ∂G.Let s 1 < ... < s k be the all distinct values that appear in { h(i) : i ∈ V ′ }.Then consider the subgraphs G i = (V ′ i , E ′ i ) defined as Notice that |∂G| ≥ 2, so k ≥ 2 and |V ′ i | < |V ′ | for any i = 1, ..., k.By the induction hypothesis, there exists a function h i : V ′ i → R such that (1) h i (j) = h i (j ′ ) if j = j ′ , and (2) for any j ∈ int(G i ) there exist neighbors j ′ , j ′′ of j in V ′ i such that h i (j ′ ) < h i (j) < h i (j ′′ ).

FIGURE 3 .
FIGURE 3. Vertices leaving the point q.

Lemma 3 . 5 .
If |V k | ≥ 2, then there exists v k ∈ T y k M such that v k ϕ ′ ij (0) if i ∈ V k and l ij > 0.