The deformation Space of Geodesic Triangulations and Generalized Tutte’s Embedding Theorem
Abstract.
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.
Key words and phrases:
geodesic triangulations, Tutte’s embedding1. 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 [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 is a connected closed orientable smooth surface with a smooth Riemannian metric of non-positive Gaussian curvature. A topological triangulation of can be identified as a homeomorphism from to , where is the carrier of a 2-dimensional simplicial complex with the vertex set , the edge set , and the face set . For convenience, we label the vertices as where is the number of vertices. The edge in determined by vertices and is denoted as . Each edge is identified with the closed unit interval .
Let be the 1-skeleton of , and denote as the space of geodesic triangulations homotopic to . More specifically, contains all the embeddings satisfying that
-
(1)
The restriction of on the edge is a geodesic parameterized with constant speed, and
-
(2)
is homotopic to .
It has been proved by Colin de Verdière [8] that such is always non-empty. Further, is naturally a metric space, with the distance function
Then our main theorem is formally stated as follows.
Theorem 1.4.
If has strictly negative Gaussian curvature, then is contractible. In particular, it is connected.
1.2. Generalized Tutte’s Embedding
Let be the super space of , containing all the continuous maps satisfying that
-
(1)
The restriction of on the edge is geodesic parameterized with constant speed, and
-
(2)
is homotopic to .
Notice that elements in may not be embeddings of to . The space is also naturally a metric space, with the same distance function
We call an element in a geodesic mapping. A geodesic mapping is determined by the positions of the vertices and the homotopy classes of relative to the endpoints and . In particular, this holds for geodesic triangulations. Since we can perturb the vertices of a geodesic triangulation to generate another, is a dimensional manifold.
Let be the directed edge starting from the vertex and ending at the vertex . Denote as the set of directed edges of . A positive vector is called a weight of . For any weight and geodesic mapping , we say is -balanced if for any ,
Here is defined with the exponential map such that for .
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 has strictly negative Gaussian curvature. For any weight , there exists a unique geodesic mapping that is -balanced. Such induced map is continuous from to .
Theorem 1.6.
If is -balanced for some weight , then .
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 in the homotopy class of is a geodesic triangulation. Here is the geodesic length of in . The minimizer is a -balanced geodesic mapping with for . 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 from to . For the oppositie direction, we can construct a weight for a geodesic embedding , using mean value coordinates which was firstly introduced by Floater [11]. Given , the mean value coordinates are defined to be
where equals to the geodesic length of , and and are the two inner angles in at the vertex sharing the edge . The construction of mean value coordinates gives a continuous map from to . Further, by Floater’s mean value theorem (see Proposition 1 in [11]), any is -balanced. Namely, . Then Theorem 1.4 is a direct consequence of Theorem 1.5 and 1.6.
Proof of Theorem 1.4.
Since is contractible, is homotopic to the identity map. Since , is homotopy equivalent to the contractible space . ∎
2. Proof of Theorem 1.5
Theorem 1.5 consists of three parts: the existence of -balanced geodesic mapping, the uniqueness of -balanced geodesic mapping and the continuity of the map . In this section, we will first parametrize by , where is the universal covering of , and then prove the three parts in Subsection 2.1 and 2.2 and 2.3 respectively.
Assume that is the covering map from to , and is the corresponding group of deck transformations of the covering so that . For any , fix a lifting of . For any edge , denote as the lifting of such that . Then , and there exists a unique deck transformation such that . It is easy to see that for any edge .
Equip with the natural pullback Riemannian metric of with negative Gaussian curvature. This metric is equivariant with respect to . For any , there exists a unique geodesic with constant speed parameterization such that and . We can naturally parametrize as follows.
Theorem 2.1.
For any , define as
for any and . Then such is a well-defined geodesic mapping in , and the map is a homeomorphism from to .
Here we omit the proof of Theorem 2.1 which is routine but lengthy. In the remaining of this section, for any and , we denote
-
(1)
as the intrinsic distance between in , and
-
(2)
, and
-
(3)
as the geodesic triangle in with vertices , which could possibly be degenerate, and
-
(4)
as the inner angle of at if and , and
-
(5)
as the norm of under the metric , and
-
(6)
as the inner product of and under the metric .
By scaling the metric if necessary, we may assume that the Gaussian curvatures of and are bounded above by .
2.1. Proof of the Uniqueness
We first prove the following Lemma 2.2 using CAT() geometry. See Theorem 4.3.5 in [4] and Theorem 1A.6 in [3] for the well-known comparison theorems.
Lemma 2.2.
Assume , then
-
(1)
, and
-
(2)
,
and the equality holds if and only if is degenerate.
Proof.
If is degenerate, then there exists a geodesic in such that , and then the proof is straightforward. So we assume that is non-degenerate.
(1) Three points , and in determine a Euclidean triangle, where , and and the angle between and is equal to . Then by the CAT(0) comparison theorem,
(2) Let be such that
Then by the CAT comparison theorem, , and . Hence,
∎
Proof of the uniqueness part in Theorem 1.5 .
If not, assume and are two different geodesic mappings that are both -balanced for some weight . We are going to prove a discrete maximum principle for the function . Assume is such that . By lifting the -balanced assumption to , we have that
(1) |
and
(2) |
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 of , , and is on the geodesic determined by and . Then the one-ring neighborhood of in degenerates to a geodesic arc. By the connectedness of the surface, we can repeat the above argument and deduce that for any . Further, for any triangle , degenerates to a geodesic arc.
It is not difficult to extend to a continuous map from to , such that for any triangle , being a geodesic arc.
It is also not difficult to prove that is homotopic to . Therefore, is degree-one and surjective. This is contradictory to that 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 of , there exists a compact subset of such that for any , there exists a -balanced geodesic mapping .
Lemma 2.4.
Suppose is the unit ball in , and is a continuous map such that for any with . Then has a zero in .
Proof.
If not, is a continuous map from to . Since is contractible, is null-homotopic, and thus is also null-homotopic. Since , it is easy to verify that
is a homotopy between and . This contradicts to that is not null-homotopic. ∎
Lemma 2.5.
We fix an arbitrary point . If and satisfies that
(3) |
for any , then
for some constant which depends only on and
The vector in Figure 1
is defined as the residue vector at of with respect to the weight . Notice that a geodesic mapping is -balanced if and only if all its residue vectors vanish with respect to . Lemma 2.5 means that if all the residue vectors are dragging ’s away from , then all the ’s must stay not far away from .
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 , and then by Lemma 2.5 we can pick a sufficiently large constant such that if
there exists such that
We will prove that the compact set
is satisfactory.
For any , let be the parallel transport along the geodesic . Set
as a Euclidean -dimensional unit ball, and construct a map in the following three steps. Firstly, we construct points as Secondly, we compute the residue vector at each as
Lastly, we pull back the residues to as
Notice that the map is a homeomorphism from to , and if and only if the corresponding in is -balanced map. Hence, it suffices to prove that has a zero in . By Lemma 2.4 it suffices to prove that for any ,
Suppose is an arbitrary point on , and then it suffices to prove that there exists such that .
Notice that satisfy that , so by our assumption on , there exists such that
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 is very large, then by a standard compactness argument, there exists a long edge in the geodesic mapping . Assume , then the corresponding long edge in is pulling towards . This implies that there exists another long edge dragging away from , otherwise the residue vector would not drag away from . It can be shown that . Repeating the above steps, we can find an arbitrary long sequence of vertices such that the distance from each of these vertices to is increasing. This is impossible as we only have finitely many vertices.

Here is a listing of properties serving as the building blocks of the proof of Lemma 2.5.
Lemma 2.6.
(a) For any constant , there exists a constant such that if
then
(b) There exists a constant such that if
then
(c) There exists a constant such that if satisfy that
then
(d) There exists a constant such that if satisfy that
then
(e) For any constant , there exists a constant such that if
then there exists such that
(f) For any constant , there exists a constant such that if
for some edge , then there exists such that
(g) For any constant , there exists a constant such that if
then
(h) For any constant , there exists a constant such that if
then

Proof of Lemma 2.5 assuming Lemma 2.6.
For any , there exists a sufficiently large constant determined from (a), (e), (f) and (g) in Lemma 2.6 such that if
then there exist three vertices , , and shown in Figure 3 with
Moreover, by (g), (e) and (f) of Lemma 2.6, we can find another vertex such that
if the constant is sufficiently large.
Inductively, we can find a sequence such that
This contradicts to the fact that only has different elements. ∎
Proof of Lemma 2.6.
(a) By a standard compactness argument, the set
is a compact subset of . Notice that is a homeomorphism from to and
Therefore,
is compact and the conclusion follows.
(b) We claim that the constant , which is determined by
is satisfactory. Let be the hyperbolic triangle with the corresponding edge lengths
Since is a CAT() space, it suffices to show that . By the hyperbolic law of sine,
(c) We claim that the constant determined by
is satisfactory. Let be the hyperbolic triangle with the corresponding edge lengths
Since is a CAT() space, it suffices to show that . By the hyperbolic law of sine
(d) We claim that the constant determined by
is satisfactory. Let be the hyperbolic triangle with the corresponding edge lengths
Since is a CAT() space, it suffices to show that . By the hyperbolic law of cosine,
Then,
Thus, .

(e) We claim that the constant determined by
is satisfactory. Assume and , and we have two cases shown in Figure 4.
If , then by part (d)
and
If , then . By part (b) and part (d),
and . Therefore,
(f) We claim that the constant determined by
is satisfactory. If not, for any , we have
Then
and it is a contradiction.
(h) We claim that the constant determined by
is satisfactory. Notice that
Then by part (c), , and by the triangle inequality,
Therefore,
∎
2.3. Proof of the Continuity
Proof of the continuity part of Theorem 1.5.
If not, there exists and a weight and a sequence of weights such that
-
(1)
converge to , and
-
(2)
for any .
By the stronger existence result Theorem 2.3, the sequence are in some fixed compact subset of . By picking a subsequence, we may assume that converge to some . Since is -balanced, then by the continuity of the residue vectors , is -balanced, and thus , which is contradictory to that does not converge to . ∎
3. Proof of Theorem 1.6
3.1. Set up and preparations
Assume is -balanced for some weight , and we will prove that is an embedding. Recall that for each , and denote as the length of for any . It is not difficult to show that has a continuous extension defined on , such that for any triangle a continuous lifting map of from to will
-
(1)
map to a geodesic triangle in homeomorphically if does not degenerate to a geodesic, and
-
(2)
map to if 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 , even for the degenerate triangles. A convenient way is to assign a “direction” to each edge, even for the degenerate edges with zero length.
Definition 3.1.
A direction field is a map satisfying that
-
(1)
for any , and
-
(2)
for any .
Given a direction field , define the inner angle of the triangle at the vertex as
where is the origin and is the angle between and in .
A direction field assigns a unit tangent vector in to each directed edge starting from , and determines the inner angles in .
Definition 3.2.
A direction field is admissible if
-
(1)
if , and
-
(2)
in if , and
-
(3)
for a fixed vertex , if for any neighbor of , then there exist neighbors and of such that , and
-
(4)
if and , then .
An admissible direction field encodes the directions of the non-degenerate edges in , and the induced angle sum of a degenerate triangle is always . Then for any admissible and triangle , by the Gauss-Bonnet formula
(4) |
Here is the area form on or .
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.
Lemma 3.3.
If is admissible and , then for any ,
and for any , has area .
Based on Lemma 3.3, if admissible direction fields exist, the image of the star of each vertex determined by does not contain any flipped triangles overlapping with each other. If does not degenerate to a geodesic arc for any triangle , then 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 as follows. Two vertices are equivalent if there exists a sequence of vertices such that . This equivalence relation introduces a partition Denote as the only point in . For any and , denote as and are parallel, i.e., there exists such that .
The following Lemma 3.4 shows that there are plenty of choices of admissible direction fields.
Lemma 3.4.
For any , there exists an admissible such that if and .
The following Lemma 3.5 shows that for any with at least two vertices, the image of its “neighborhood” lies in a geodesic.
Lemma 3.5.
If , then there exists such that if and .
Now let be as Lemma 3.5 if , and arbitrary if . Then construct an admissible direction field as in Lemma 3.4, with induced inner angles . If the image of a triangle under degenerates to a geodesic, then its inner angles are or . Let be the set of degenerate triangles under .
Lemma 3.6.
If , , and , then for any in the star neighborhood of the vertex .
Let be a connected component of the interior of , and be the completion of under the natural path metric on . Notice that could be different from the closure of in .
Since is surjective, and and has non-empty boundary. Then is a connected surface with a natural triangulation , and
Assume is the set of interior vertices, and is the set of boundary vertices, and is the set of interior edges, and is the set of boundary edges of . Then , and by Lemma 3.6, if and , then . Therefore,
Thus,
Therefore, . Since has non-empty boundary, or . In either case, it contradicts with the fact that is a simplicial complex.
3.3. Proof of Lemma 3.3
We claim that for any ,
If for any neighbor of , this is a consequence of condition (3) in the definition of an admissible direction field. If , by the -balanced condition, should not be contained in any open half unit circle, and the angle sum around should be at least .
By the fact that is surjective and equation (4), we have
and
Hence, the inequalities above are equalities. This fact implies that
Since each term in this summation is non-positive, then . The statement on the area follows similarly.
3.4. Proof of Lemma 3.4
We claim that for any , there exists a map such that
-
(1)
if , and
-
(2)
for a fixed , if for any neighbor of , then there exist neighbors and of in such that .
Given such , set as
where sgn is the sign function. It is easy to verify that such is satisfactory.
To construct such function , we will prove the following lemma, which is more general than our claim.
Lemma 3.7.
Assume is a subgraph of the -skeleton , and . Denote
and . Then there exists such that
-
(1)
if , and
-
(2)
for any there exist neighbors and of in such that .
Proof.
We prove by the induction on the size of . The case is trivial. For the case , first notice that for any proper subgraph of . Assign distinct values to each , then solve the discrete harmonic equation
with the given Dirichlet boundary condition on .
Let be the all distinct values that appear in . Then consider the subgraphs defined as
and
Notice that , so and for any . By the induction hypothesis, there exists a function such that
-
(1)
if , and
-
(2)
for any there exist neighbors of in such that .
Define if , then for sufficiently small positive ,
is a desirable function. ∎
3.5. Proof of Lemma 3.5
It amounts to prove that if and and and , then . Let
which is a closed path-connected set in . For any , if and only if there exists with . Therefore, it suffices to prove that
-
(1)
for any and edges with and , , and
-
(2)
for any satisfying that , there exists such that , and thus , and
-
(3)
is connected.
For part (1), if it is not true, then there exists and and such that and and is not parallel to . Assuming this claim, by the -balanced condition, there exists with , and the three vectors are not contained in any closed half-space in . Assume and , and without loss of generality, are ordered counter-clockwise in the one-ring neighborhood of in . By Lemma 3.4, there exists an admissible such that . By possibly changing a sign, we may assume that .

Part (2) is straightforward and we will prove part (3). By our assumption on the extension , contains only one point, then the embedding map from to is homotopic to the constant map , meaning that is contractible in . If contains at least two boundary components, then it is not difficult to show that has a connected component homeomorphic to an open disk. Let be a lifting of . Then contains only a single point . Then by the -balanced condition, it is not difficult to derive a maximum principle and show that equals to constant . Then by the definition of it is easy to see that should be a subset of . It is contradictory.
3.6. Proof of Lemma 3.6
Assume and are two edges in . If the conclusion is not true, then there exists such that and is not parallel to . Notice that , and we have
Then the equality holds in the above inequality, and for any , should be on the half circle that contains . Let be the middle point of this half circle, then
This contradicts to the fact that is -balanced.
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