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Circle patterns with obtuse exterior intersection angles

Thurston's Circle Pattern Theorem studies existence and rigidity of circle patterns of a given combinatorial type and the given non-obtuse exterior intersection angles. Using topological degree theory, variational principle, Teichmuller theory, and Sard's Theorem, this paper generalizes Circle Pattern Theorem to the case of obtuse exterior intersection angles.


Introduction
The patterns of circles were introduced as useful tools to study hyperbolic 3-manifolds by Thurston [27]. He also conjectured that, under a procedure of refinement, the hexagonal circle packings converge to the classical Riemann mapping [28]. In 1987 this conjecture was resolved by Rodin-Sullivan [23]. Over the past decades, circle patterns (packings) have been bridging combinatorics [25,18], discrete and computational geometry [8,26], minimal surfaces [4] and others.
Let S be an oriented closed surface and T be a triangulation of S with the sets of vertices, edges and triangles V, E, F. Suppose that S is equipped with a constant curvature metric µ. A circle pattern P on (S , µ) is a collection of oriented circles. Say P is T -type if there exists a geodesic triangulation T (µ) of (S , µ) with the following properties: (i) T (µ) is isotopic to T ; (ii) the vertices of T (µ) coincide with the centers of circles in P. In this paper we focus on these circle patterns P = {C v : v ∈ V} such that C u , C w intersect with each other whenever there exists an edge between u and w. Then we have the exterior intersection angle Θ(e) ∈ [0, π) for any e ∈ E. One refers to Stephenson's monograph [26] for more background.
Given a function Θ : E → [0, π) defined on the edge set of T , let us consider the following question: Does there exist a T -type circle pattern whose exterior intersection angle function is given by Θ? If it does, to what extent is the circle pattern unique? A celebrated answer to this question is the following Circle Pattern Theorem due to Thurston [27,Chap. 13]. Theorem 1.1 (Thurston). Let T be a triangulation of an oriented closed surface S of genus g > 0. Suppose that Θ : E → [0, π/2] is a function satisfying the following conditions: (i) If the edges e 1 , e 2 , e 3 form a null-homotopic closed curve in S , and if 3 i=1 Θ(e i ) ≥ π, then these edges form the boundary of a triangle of T .
(ii) If the edges e 1 , e 2 , e 3 , e 4 form a null-homotopic closed curve in S and if 4 i=1 Θ(e i ) = 2π, then these edges form the boundary of the union of two adjacent triangles.
Then there exists a constant curvature (equal to 0 for g = 1 and equal to −1 for g > 1) metric µ on S such that (S , µ) supports a T -type circle pattern P with the exterior intersection angles given by Θ. Moreover, the pair (µ, P) is unique up to isometry if g > 1, and up to similarity if g = 1.
There are many results relating to Circle Pattern Theorem. See, for example, the works of Andreev [2], Marden-Rodin [19], Colin de Verdière [9], Chow-Luo [7] and others. Meanwhile, a natural problem arises: can we relax the requirement of non-obtuse angles in the above theorem? So far few progress has been made, except for some cases considered by Rivin [21,22], Bao-Bonahon [3], Bobenko-Springborn [5] and Schlenker [24], respectively. The purpose of this paper is to establish some general results.
First let us explain some of our terminologies. A closed (not necessarily simple) curve γ in S is called a pseudo-Jordan curve, if the complement S \γ contains a simply connected component whose boundary is equal to γ. For a pseudo-Jordan curve γ in S , an enclosing vertex set of γ consists of all vertices covered by Ω, where Ω is any simply connected component of S \ γ such that ∂Ω = γ. A pseudo-Jordan curve is said to be non-vacant if one of its enclosing vertex sets is non-empty. Theorem 1.2. Let T be a triangulation of an oriented closed surface S of genus g > 1. Suppose that Θ : E → [0, π) is a function satisfying the following conditions: (C1) If the edges e 1 , e 2 , e 3 form the boundary of a triangle of T , and if 3 i=1 Θ(e i ) > π, then Θ(e 1 ) + Θ(e 2 ) < π + Θ(e 3 ), Θ(e 2 ) + Θ(e 3 ) < π + Θ(e 1 ), Θ(e 3 ) + Θ(e 1 ) < π + Θ(e 2 ).
(C2) If the edges e 1 , e 2 , · · · , e s form a non-vacant pseudo-Jordan curve in S , Then there exists a hyperbolic metric µ on S such that (S , µ) supports a T -type circle pattern P with the exterior intersection angles given by Θ.

Remark 1.3.
Under the assumption that every exterior intersection angle is non-obtuse, when s > 4, it is easy to see s i=1 Θ(e i ) ≤ sπ/2 < (s − 2)π. Therefore, Thurston's conditions imply (C2). Remark 1.4. The contact graph of a circle pattern is a graph which has a vertex for each circle and an edge between two vertices for each intersection component of the corresponding closed disks. For a T -type circle pattern P with acute exterior intersection angles, the contact graph G(P) is isomorphic to the 1-skeleton of T . However in obtuse angle cases the statement may not hold. There is a discussion on relations of G(P) and T in an early arXiv version of this paper. To relieve the burden of involved details, we do not include it here. i=1 Θ(e i ) ≤ π or each side of the corresponding spherical triangle has length less than or equal to π/2. In this way one finds that (C1) is strictly wider than (R1). See Proposition 2.7 for details.
As a consequence of Theorem 1.2 and Theorem 1.5, we obtain the following result which is related to the Hyperideal Circle Pattern Theorem due to Schlenker [24]. Theorem 1.7. Let T be a triangulation of an oriented closed surface S of genus g > 1. Suppose whenever e 1 , e 2 , · · · , e s form a pseudo-Jordan curve in S . Then there exists a hyperbolic metric µ on S such that (S , µ) supports a T -type circle pattern P with the exterior intersection angles given by Θ. Moreover, the pair (µ, P) is unique up to isometry.
One may ask the following question: what can be said regarding rigidity when (R1) is not satisfied? Let W denote the set of functions Θ : E → [0, π) satisfying (C1), (C2). Then W is a convex subset of [0, π) |E| . Below is a part answer. Theorem 1.8. For almost every Θ ∈ W, there are at most finitely many T -type circle pattern pairs (µ, P), up to isometry, with the exterior intersection angles given by Θ.
The paper is organized as follows: In next section we introduce some properties of three-circle configurations. In Section 3 we prove Theorem 1.2 by using topological degree theory. In Section 4, applying variational principle, we derive Theorem 1.5. As a corollary, Theorem 1.7 is established. In Section 5 we deduce Theorem 1.8 through a combination of Teichmüller theory and Sard's Theorem. The last section contains an appendix concerning some results from manifold theory.
Throughout this paper, we denote by | · | the cardinality of a set, and denote by χ(·) the Euler characteristic of a manifold.

Three-circle configurations
The following three lemmas played crucial roles in the proof of Circle Patten Theorem. Please refer to [27,7] for more information.
Lemma 2.1. For any three positive numbers r i , r j , r k and three non-obtuse angles Θ i , Θ j , Θ k , there exists a configuration of three mutually intersecting circles in hyperbolic geometry, unique up to isometry, having radii r i , r j , r k and meeting in exterior intersection angles Θ i , Θ j , Θ k .  Figure 1, let ϑ i , ϑ j , ϑ k denote the corresponding inner angles of the triangle of centers.

Lemma 2.2.
Under the above condition, we have lim (r i ,r j ,r k )→(0,0,0) where a, b, c are positive constants.

Some new observations
In search of generalizing Circle Pattern Theorem, one needs to go a step further to study three-circle configurations with obtuse exterior intersection angles. Below are some new observations.
For any three positive numbers r i , r j , r k , there exists a configuration of three mutually intersecting circles in hyperbolic geometry, unique up to isometry, having radii r i , r j , r k and meeting in exterior intersection angles Θ i , Θ j , Θ k .
Proof. Set l i = cosh −1 (cosh r j cosh r k + cos Θ i sinh r j sinh r k ) and l j , l k similarly. It suffices to check that l i , l j , l k satisfy the triangle inequalities. Namely, Equivalently, one needs to show To simplify notations, for m = i, j, k, set Then Substituting (6) into (5), we need to prove where Now we divide the proof into the following two cases: Then Thus we deduce (7).
Then there exists a spherical triangle with angles Θ i , Θ j , Θ k . Denote by φ i , φ j , φ k the lengths of sides opposite to Θ i , Θ j , Θ k , respectively. By the second cosine law of spherical triangles, Note that (7) is equivalent to where Set y i = sin Θ i a i x j x k , y j = sin Θ j a j x k x i and y k = sin Θ k a k x i x j . Substituting (9) into (10), it remains to check Completing the square gives By the cosine law of spherical triangles, one obtains It follows that y 2 i + y 2 j + y 2 k + 2 cos φ i y j y k + 2 cos φ j y k y i + 2 cos φ k y i y j ≥ sin 2 φ j y 2 k + sin 2 φ k y 2 j + 2 cos Θ i sin φ j sin φ k y j y k = (sin φ j y k + cos Θ i sin φ k y j ) 2 + sin 2 Θ i sin 2 φ k y 2 Thus the lemma is proved.

Remark 2.5.
On the other hand, if the triangle inequalities hold for all triples of positive numbers r i , r j , r k , then the condition of Lemma 2.4 must be satisfied. Namely, the above lemma is in the optimal form.
the above proof can be simplified because (7) automatically holds. In fact the following Proposition 2.7 manifests that this condition is stronger than the condition of Lemma 2.4.
if and only if one of the following properties holds: (ii) Θ i , Θ j , Θ k are the angles of a spherical triangle with each side less than or equal to π/2.
Proof. The "if" part is an immediate consequence of (8) and (9).
To show the "only if" part, assume Θ i + Θ j + Θ k > π. It suffices to verify property (ii). First a routine calculation gives If Θ k = 0, similar arguments lead to a contradiction. Thus Θ j , Θ k ∈ (0, π), which implies As a result, Similarly, Hence there exists a spherical triangle with angles Θ i , Θ j , Θ k . Using (9), it is easy to see each side of this spherical triangle has length less than or equal to π/2. Proof. The proof of (1) is similar to the related results in Ge-Jiang [11] and Ge-Xu [12]. Due to the cosine law of hyperbolic triangles, we have .
As a result, Hence A → 0 as r i → +∞. Meanwhile, cosh l i = cosh r j cosh r k + cos Θ i sinh r j sinh r k ≤ 2 cosh r j cosh r k .
It follows that 0 < B ≤ cosh l i cosh l j cosh l k ≤ 2 (1 + c k )(1 + c j ) cosh 2 r i .
Then B → 0 as r i → +∞, which concludes (1). The formula (2) is derived from a direct computation. Let us consider (3). By the cosine law of hyperbolic triangles, we have As (r i , r j , r k ) → (0, 0, c), one obtains l i → c, l j → c, l k → 0.
Consequently, cos ϑ i + cos ϑ j → 0, which yields It remains to prove (4). As (r i , r j , r k ) → (0, 0, 0), the area of the triangle tends to zero. Due to the Gauss-Bonnet formula, one obtains the desired result. Remark 2.9. From the above proof, one finds that (1) holds uniformly, no matter how the other two radii behave, even if either or both of them tend to infinity or zero. In addition, regarding ϑ i as the function of r i , r j , r k , Θ i , Θ j , Θ k , the formula is true as is any compact set such that the condition of Lemma 2.4 is satisfied.

Existence
To prove Theorem 1.2, a natural strategy is to follow Thurston [27]. Unfortunately, for three-circle configurations with obtuse exterior intersection angles, the result similar to Lemma 2.2 may not hold (see Remark 4.2). Part of his method does not work. In order to overcome the difficulty, we will employ the topological degree theory. As a comparison, this approach has the advantage of being independent on rigidity.

Thurston's construction
Recall that S is an oriented closed surface of genus g > 1 and T is a triangulation of S with the sets of vertices, edges and triangles V, E, F. Let r ∈ R |V| + be a radius vector, which assigns each vertex v ∈ V a positive number r(v). Then r together with the function Θ : E → [0, π) in Theorem 1.2 determines a hyperbolic cone metric structure on S as follows.
For each triangle (v i v j v k ) of T , one associates it with the hyperbolic triangle determined by the centers of three mutually intersecting circles with radii and exterior intersection angles Because Θ satisfies (C1), Lemma 2.4 implies the above procedure works well.
Gluing these hyperbolic triangles produces a metric surface S , µ(r) which is locally hyperbolic with possible cone type singularities at the vertices. For each v ∈ V, the vertex curvature k(v) is defined by where σ(v) denotes the cone angle at v. More precisely, σ(v) is equal to the sum of all inner angles having vertex v.
Write k(v) as k(v)(Θ, r). If there exists a radius vector r * such that k(v)(Θ, r * ) = 0 for all v ∈ V, then it produces a smooth hyperbolic metric on S . For every v ∈ V, on S , µ(r * ) drawing the circle centered at v with radius r * (v), one then obtains the demanded circle pattern. Consider the following curvature map The major purpose of this section is to show that the origin O = (0, 0, · · · , 0) ∈ R |V| belongs to the image of T h(Θ, ·).

Topological degree
Let us make use of the topological degree theory. Specifically, one finds a relatively compact open set Λ ⊂ R |V| + and determines the degree it follows from Theorem 6.8 that O is in the image of T h(Θ, ·). We shall compute deg(T h(Θ, ·), Λ, O) via homotopy method. Namely, deform T h(Θ, ·) to another map which is relatively easier to manipulate. For each t ∈ [0, 1], note that the function Θ t = tΘ satisfies (C1). Applying Thurston's construction, T h t (·) = T h(Θ t , ·) is well-defined. Moreover, T h t forms a homotopy from T h(Θ, ·) to T h(0, ·).

Lemma 3.1. There exists a relatively compact open set
Proof. Let us exhaust R |V| + by an increasing sequence of relatively compact open sets {Λ n }. Assume on the contrary that the lemma is not true. For each n, one obtains t n ∈ [0, 1] and r n ∈ R |V| Because {Λ n } exhausts R |V| + , there exist v 0 ∈ V and {r n k } such that r n k (v 0 ) → +∞ or r n k (v 0 ) → 0.
In the first case, by Lemma 2.8 and Remark 2.9, it follows from formula (1) that k(v 0 )(t n k Θ, r n k ) → 2π, which contradicts to (11).
In the second case, let V 0 ⊂ V be the set of vertices v ∈ V for which r n k (v) → 0. Then V 0 is a non-empty subset of V. We denote by S (V 0 ) the union of these q-cells (q = 0, 1, 2) of T that have at least one vertex in V 0 , and denote by Lk(V 0 ) the set of pairs (e, v) an edge e and a vertex v with the following properties: Evidently, S (V 0 ) is an open set of S and thus is a surface. Suppose that t n k converges to a number t * ∈ [0, 1]. Otherwise, one picks up a convergent subsequence. By the following Proposition 3.2, we have (11) gives

Combining with formula
Without loss of generality, assume that S (V 0 ) is connected. Otherwise, one considers the connected component of S (V 0 ). Note that where g 0 , h 0 are the genus and the number of boundary components of S (V 0 ). Hence If h 0 = 0, then S (V 0 ) has no boundary component. It follows that Substituting these into (12), one derives which contradicts to the condition g > 1.
Let h 0 > 0. If g 0 > 0 or h 0 > 1, it is easy to see (12) leads to a contradiction. Let g 0 = 0 and h 0 = 1. Then S (V 0 ) is a simply connected domain in S . Suppose that (12) gives Meanwhile, one finds that e 1 , · · · , e s form a non-vacant pseudo-Jordan curve. According to (C2), we have which leads to a contradiction.

Proposition 3.2.
Let S (V 0 ) and Lk(V 0 ) be as above. Then Proof. For m = 1, 2, 3, let F m (V 0 ) be the set of triangles having exactly m vertices in V 0 . Using Lemma 2.8, it follows from formulas (2), (3) and (4) that Let E(V 0 ), F(V 0 ) denote the sets of edges and triangles having at least one vertex in V 0 . It is easy to see Combining the above relations gives Proof. It suffices to compute deg(T h 0 , Λ, O). By Theorem 1.1, Λ ∩ T h −1 0 (O) consists of a unique point. Using Lemma 2.2, one shows that the Jacobian matrix of T h 0 (·) has positive diagonal entries and strictly diagonally dominant columns. As a result, the determinant of this matrix is positive and the tangent map preserves the orientation. Therefore,

As a result, we have
Owing to Lemma 3.1 and Theorem 6.7, one deduces the conclusion.
Proof of Theorem 1.2. Because of Theorem 3.3 and Theorem 6.8, O is in the image of the map T h(Θ, ·). Consequently, there exists a hyperbolic metric µ on S such that (S , µ) supports a T -type circle pattern P with the exterior intersection angles given by Θ.

Three-circle configurations revisited
For a triple of indices i, j, k, recall that λ i jk = cos Θ i + cos Θ j cos Θ k . Under the condition that λ i jk ≥ 0, λ jki ≥ 0, λ ki j ≥ 0, it follows from Remark 2.6 that the inner angles ϑ i , ϑ j , ϑ k are well-defined functions of (r i , r j , r k ). Our key observation in this section is the following generalization of Lemma 2.2.
Proof. We mention that part of computations here is parallel to these in Colin de Verdière [9], Chow-Luo [7], Guo-Luo [14], Guo [15], Xu [29] and others. For simplicity, we use the same notations in the proof of Lemma 2.4. By the cosine law of hyperbolic triangles, Taking the partial derivative with respect to l i gives where K i jk = sin ϑ i sinh l j sinh l k .
By the sine law of hyperbolic triangles, it is easy to see Hence Similarly, Meanwhile, Combining (13), (14) and (15), one obtains It follows that where the equality holds if and only if Θ k = 0 and Θ i + Θ j = π. Fix r j , r k and let r i vary. Then l i stays constant. Because ∂ϑ j /∂r i ≥ 0 and ∂ϑ k /∂r i ≥ 0, that means the other two sides move outwards or remain unchanged as r i increases. Note that these two sides can not stay unchanged simultaneously. Thus the area Area( ) of the triangle is a strictly increasing function of r i . By the Gauss-Bonnet formula, one derives

Remark 4.2.
Assume that λ i jk < 0. For any t > 0, set Using (16), one finds that ∂ϑ i /∂r j < 0 when t is sufficiently small. Thus the condition of Lemma 4.1 can not be relaxed further.

Variational principle
Let us establish rigidity via the variational approach poineered by Colin de Verdière [9].
Remind that similar methods have been used by Rivin [21,22], Bobenko-Springborn [5], Guo-Luo [14], Guo [15] and others in many situations. Suppose that V = {v 1 , · · · , v |V| }. To simplify notations, for i = 1, 2, · · · , |V|, set For a vertex v i ∈ V and a triangle ∈ F incident to v i , denote by ϑ i the inner angle of the triangle at v i . Then Consider the change of variables u i = ln tanh(r i /2). It is easy to see Proof. In case that v i , v j is a pair of non-adjacent vertices, then

Regarding vertex curvatures as functions of u induces the map
In case that [v i , v j ] = e ∈ E, there exist triangles 1 , 2 adjacent to e. Lemma 4.1 gives Hence the Jacobian matrix of T h(Θ, ·) in terms of u is symmetric.
Meanwhile, a simple computation shows It follows that The Jacobian matrix is strictly diagonally dominant and hence is positive definite. Now it is ready to prove Theorem 1.5.
Proof of Theorem 1.5. Consider the following 1-form Because ∂k i /∂u j = ∂k j /∂u i , it is easy to see ω is closed. Thus the function ω is well-defined and does not depend on the particular choice of a piecewise smooth arc in R |V| − from an initial point u(0) to u. Note that the Hessian of Φ(·) is equal to the Jacobian of T h(Θ, ·), which is positive definite by Lemma 4.3. That means Φ is a strictly convex function of u.
From Thurston's construction, the demanded circle pattern is related to the critical point of Φ. Because Φ is strictly convex, the critical point must be unique. We thus finish the proof.

Remark 4.4.
Under the condition (R1), the above analysis shows that the Jacobian matrix of T h(Θ, ·) is non-singular. It follows from the Implicit Function Theorem that the demanded radius vector r * depends on Θ smoothly.

Remark 4.5.
Based on observations in above parts, Xu [29] recently studied rigidity of circle patterns with inversive distances. His work generalizes the results of Guo-Luo [14] and Theorem 1.5. It follows from Theorem 1.5 that the rigidity holds.

Finiteness
To obtain the proof Theorem 1.8, some knowledge on Teichmüller theory is needed. One refers to [1,17] for basic background. See also [6] from the geometric and topological viewpoint.

Configuration spaces
Recall that S is an oriented closed surface of genus g > 1. Denote by T (S ) the Teichmüller space of S , which parameterizes the equivalence classes of marked hyperbolic metrics on S . Here a marking is an isotopy class of orientation-preserving homeomorphism from S to itself. And two marked hyperbolic metrics µ, µ ∈ T (S ) are equivalent if there exists an isometry φ : (S , µ) → (S , µ ) isotopic to the identity map. T (S ) admits the structure of a smooth manifold of dimension 6g − 6.
Let p i , p j ∈ S be two points and let γ be a simple curve with endpoints p i , p j . For any µ ∈ T (S ), on (S , µ) there exists a unique geodesic in the isotopy class [γ] with the endpoints p i , p j . Let d [γ] (µ, p i , p j ) denote the length of this geodesic. Under the above manifold structure, d [γ] (µ, p i , p j ) depends on µ smoothly. See Buser's monograph [6,Chap. 6] for details.
We endow S with a smooth structure and define a space Z = T (S ) × (S × R + ) |V| . Then Z is a smooth manifold of dimension dim (Z) = 6g − 6 + 3|V|.
A point z = (µ, p 1 , r 1 , · · · , p |V| , r |V| ) ∈ Z is called a configuration, since it gives a choice of a marked hyperbolic metric µ on S and assigns each vertex v i an oriented circle C i on (S , µ), where C i denotes the circle centered at p i with radius r i . To summarise, a configuration z gives a marked hyperbolic metric µ on S together with a circle pattern P on (S , µ). Briefly, we say z gives a circle pattern pair (µ, P). Let Z T ⊂ Z denote the subspace of configurations that give T -type circle pattern pairs. More precisely, z ∈ Z T if and only if there exists a geodesic triangulation T (z) of (S , µ) such that T (z) is isotopic to T and the vertices of T (z) are p 1 , p 2 , · · · , p |V| . Clearly, Z T is open in Z and thus is a smooth manifolds of dimension |E|.
For each e = [v i , v j ] ∈ E, the inversive distance I(e, z) is defined by where [γ e ] represents the isotopy class of the edge e. Let Z P ⊂ Z T be the subspace of configurations z for which −1 < I(e, z) < 1, ∀ e ∈ E.
Because I(e, z) is continuous, Z P is open in Z T , which implies Z P is a smooth manifold of dimension |E|. Obviously, a configuration z ∈ Z P gives a T -type circle pattern with the exterior intersection angle Θ(e, z) ∈ (0, π) for each e ∈ E, where Θ(e, z) = arccos I(e, z).

One establishes the following smooth map
Ev :

Rigidity from the viewpoint of Sard's Theorem
Note that a regular point of the above map gives a circle pattern pair which is locally determined up to isometry by its exterior intersection angles. Sard's Theorem indicates that local rigidity is a generic property.
Proof of Theorem 1.8. Set W 0 = Ev(Z 0 ), where Z 0 ⊂ Z P denotes the set of critical points of Ev. Owing to Sard's Theorem (Theorem 6.2), W 0 has zero measure. Meanwhile, the boundary set ∂W of W also has zero measure. Thus W \ (W 0 ∪ ∂W) is a subset of W with full measure. Note that dim (Z P ) = dim (Y) = |E|.
For each Θ ∈ W \ (W 0 ∪ ∂W) ⊂ Y, the Regular Value Theorem (Theorem 6.1) implies that Ev −1 (Θ) is a discrete set. Combining with the following Proposition 5.1, this set must be finite. Thus the theorem is proved.
Proof. It suffices to show any sequence {z n } ⊂ Ev −1 (Θ) contains a convergent subsequence in Z P . Let (µ n , P n ) be the circle pattern pair give by z n . First we claim that there exists a subsequence {z n k } converging to a point z * ∈ Z. In view of the topology of Z = T (S ) × (S × R + ) |V| , one needs to check the following two properties: Now assume that M, N are oriented manifolds of the same dimensions. Let Λ ⊂ M be a relatively compact open subset. Namely, Λ has compact closure in M. For a point y ∈ N, we use f Λ y to denote that y is a regular value of the restriction map f : Λ → N.
Suppose f ∈ C 0 (Λ, N) ∩ C ∞ (Λ, N), f Λ y, and f (∂Λ) ∈ N \ {y}. If Λ ∩ f −1 (y) is empty, the topological degree deg( f, Λ, y) is defined to be zero. If Λ∩ f −1 (y) is non-empty, the Regular Value Theorem implies that it consists of finite points. The following lemma is a consequence of Sard's Theorem.