Triangulations of surfaces

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1 Triangulations of surfaces Yiwang Chen May 7, 2014 Abstract Triangulation is a convenient tool to consider when we are determining some properties of a topological space. Specifically, we can compute the homology and cohomology groups using the simplicial homology and cohomology of the triangulated space which is much easier than involving more complicated theories. In this paper, we will review the standard proof by Radó of the existence of triangulation of the surfaces, which Ahlfors and Sario introduced in the book Riemann Surfaces [3]. 1 Introduction In the part we will begin to introduce some concepts we will use in the theorem. The basic setting is based on the setting by Ahlfors and Sario [3]. To begin with, we define surface to be the connected 2-dimensional manifold which is Hausdorff and second-countable. We begin by the definition of the abstract simplicial complex. Definition 1.1. An abstract simplicial complex is defined as a family K of non-empty finite subsets of a universal set K with the property that for every set X K, and every non-empty subset Y X, Y K. We then can define the simplices to be the subsets in the family, and the dimension of the simplices defined by one less than the number of the elements and dimension of the complex K to be the maximum dimension of its simplices. Then we can construct a corresponding geometric complex K g such that a point of it is a real-valued function t defined by: 1

2 t(a) 0 for all a K. Simplex is formed by such a with t(a) > 0. a K t(a) = 1. We can then define the triangulation on a surface F as following. Definition 1.2. A triangulation on a surface F or bordered surface F is an abstract 2- dimensional complex K, with a map σ : K F (OrF ) by s σ(s) (s is a simplex in K), satisfying: σ(s 1 s2 ) = σ(s 1 ) σ(s 2 ). There exists a homeomorphism of s g onto σ(s) which maps each s g induced by s s, onto σ(s ). σ(s) = F (OrF ). Every point on F (OrF ) has a neighbourhood which meets only on finite number of σ(s). We also need to first introduce a concept before getting to the theorem. The Jordan region is defined as an open set with the closure mapped topologically onto a closed disk and the definition of Jordan arc and Jordan curve can be induced by it. We then know that every surface F possess an open cover of Jordan regions since every point on the surface can be homeomorphically mapped to a center of a closed disk. Denote the Jordan regions as J, and the boundaries to be γ, we can thus define the following: Definition 1.3. The covering of closed Jordan regions of a space is of finite character if it satisfies: Each J meets at most a finite number of the others. The intersection of any two boundaries γ consists of at most a finite number of points or arcs. To prove the main theorem, we need to assume the Jordan-Schoenflies theorem without proving it. Theorem 1.4. If γ is a Jordan curve in the plane X, then γ can be homeomorphically mapped to a circle by a map φ : X X. To have another usable version of the theorem, we need to first define the concept of a 2

3 cross-cut. Definition 1.5. A cross-cut of J is defined as the interior of an arc in J with only the end points lie on the boundary of J. And actually, under finite character property, if γ n J m (n, m I where I is the index, n m) then it intersects J m along finite cross-cuts. Thus by the theorem we have a corollary as following. Corollary. A cross-cut of a Jordan region divide it into two Jordan regions. And boundary of the new Jordan regions are consists of the cross-cut and the boundary arcs. Proof. Given the Jordan region J, the cross-cut γ and the boundary arcs γ 1, γ 2 between the end points p 1, p 2 of cross cut. Thenwe denote the Jordan regions bounded by γ γ 1, γ γ 2 as J 1 and J 2. Since J is connected, we then have J 1 J and J 2 J and J 1 is distinct from J 2. Then by identifying all the boundary points of J we have that γ with the end point becomes a Jordan curve dividing a sphere J, and thus only divides J into two Jordan regions, J 1 and J 2 in this case. 2 Main Theorem Theorem 2.1. Every surface permits a triangulation. Proof. The proof is consisted of two parts. So we first prove that the surface has a triangulation if it possesses an open covering by Jordan regions which is of finite character and thus able to be triangulated. And then we will construct an open covering for the surface. We now proceed the proof of the first part. For the Jordan regions J n we can assume it has no such t with that J t J s for some s t since if it is the case we can always throw out terms with this property and it will still cover the space since otherwise the space is not of finite character. Therefore, by the property of finite character, we have that γ n J m (n, m I where I is the index) or γ n J m. We will do the simpler one first. Take γ n J m (n m). Then since J n J m for n m, so the complement of J n is contained in J m and thus makes the surface to be a sphere, and a triangulation can be found. Then we check the case γ n J m (n, m I where I is the index, n m). Then we 3

4 have that each γ n have a finite number of cross-cuts, and thus combined with that it meets also finitely many γ n s, we have that J m can be divided into finitely many Jordan regions by the finite cross-cuts by using Jordan-Schoenflies theorem. Now denote that J mi to be the closed subregion of J m obtained in this way. Then given any regions J mi, J nj we have that they have no boundary points in the interior of another and thus they are either identical or have distinct interior. Similarly, we consider the boundary arcs γ m. Divided by the points on γ n and end points common to γ m and γ n, we have arcs γ mi defined in this way. Then by construction of the Jordan regions J mi, we have that γ mi and γ nj are either identical or have at most end points coincide. We then proceed by constructing a triangulation on the surface F. Define an abstract 2-dimentional complex K as following. Define the vertices of K to be the end points of all the arcs γ mi and also an interior point of each γ mi and each region J nj. And define the edges of K to be induced by the arcs γ mi and also the edges by adjoining the interior point of J nj to vertices defined on the boundary of it, as the graph below indicates. The we have a natural definition of the 1- and 2-simplices on K. By check the triangulation condition we have that K is the triangulation on F. Figure 1: Triangulation on one Jordan region We then left to show that a surface possesses an open covering by Jordan regions of finite character. The construction of the finite character covering is proceeded by several important claims. 4

5 Lemma 2.2. On surface F we can find two finite or infinite sequences of Jordan regions V n, W n satisfies: V n W n, V n = F No point belongs to infinitely many W n. Proof of Lemma 2.2 By the previous discussion we have that there exist a sequence {U i } of Jordan regions that cover surface F, we then can refine every U i to be countable union of Jordan regions U ij with U ij U i, and thus we can refine one more step further, U ijk with U ijk U ij. By rearrangement, let V n = U ijk, W n = U ij. Then we have V n W n W n O for every open set O and that V n = F. We define n k as following: n 1 = 1 and n k, k > 1 is the least integer satisfying V 1 Vnk 1 V 1 Vnk. We then have n k always exist since V 1 Vnk give a compact map but V n give a open covering. We then need to separate the situation in the case F is compact or non-compact. If F is compact, then there exist n k n k 1 and thus V 1 Vnk 1 V 1 Vnk 1 V 1 Vnk 1 will give that V 1 Vnk 1 is both open and closed and thus V 1 Vnk 1 = F. This gives a finite sequence of {V n } and {W n } and thus satisfying the condition. F is non-compact. Set G k = V 1 Vnk, but then G k 1 G k and we have that Gk = F. Define H k = G k+1 G k G k+2 G k 1. Cover the space G k+2 G k 1 by V n and W n, and there are finitely those sets covering H k since H k is compact, denote as V kl and W kl. We then have a infinite series of V n and W n in this way. Since W ij Wkl = if i k 3 by construction of H n, we have that those sequences V n and W n satisfying the conditions. 5

6 Lemma 2.3. If Γ is a discrete set of Jordan arcs on F, then any two points in a region G F can be joined by a Jordan arc σ G whose intersection with arcs γ Γ consists of finite number of points and subarcs of σ. Proof of Lemma 2.3 We notice that the the joined path relation is transitive. Given p 0 G, and define E to be p 0 and all the points can be joined to p 0 in the needed manner. If p E is does not touch γ Γ, then we have a small open neighbourhood V (p) does not intersect Γ, and thus V (p) E If p E lies on some γ Γ, we can also construct a small open neighbourhood V (p) intersects on exactly the same terms of Γ. Then for q V (p), we connect them by an arc first from q to a intersection with an arc γ but then follow the arc to p. Then it satisfies the manner we desired. So we have V (p) E. Then we have that E is open and similarly G E is open. Since E is nonempty, E = G. Then the lemma is proved. The similar arguments can be extended to p 1, p 2 on the boundary of F is they do not have intersection with any arc γ Γ. Then by using the previous two lemma, claim that one can show that there exist closed Jordan regions J n, such that V n J n W n with boundaries γ n have only a finite number of common points or arcs and thus form a covering of finite character of F by definition. (V n s and W n s are the ones introduced in Lemma 2.2) Proof of the Claim: Take J 1 = V 1 and define the rest of J n in a inductive way. Inductively assume that J 1... J n 1 are well-defined. then construct J n with the desired property, which is that the boundary γ n intersects γ 1... γ n 1 on a finite number of points or arcs. Since W n is a Jordan region, we can then represents W n by a closed disk with the center o V n. We can then find points p 1, p 2 on different radiis of W n, not lying on γ 1... γ n 1 and the boundary circle E n not intersecting V n. Denote the corresponding radial segment s 1 and s 2, extended from the outer circle and to the first intersection with γ n. Then apply the cross-cut Corollary we divide the disk W n into three closed Jordan region, V n, X n1 and X n2, with the boundaries well-defined by the corollary. Then apply Lemma 2.3 and the extended argument of it to E n1 and E n2 adjoining p 1 + p 2, we have that J n defined by the closed Jordan region J n W n with the boundary E n = {p 1 } + {p 2 } + E n1 + E n2 meeting γ 1... γ n 1 on finite number of points and arcs. 6

7 Figure 2: Construction of J n Then only need to show that J n V n. Since E n does not intersects V n, So J n V n or Jn c V n. If the first one occurs, then the proof is done. If the latter occurs, then without lost of generality, X n1 would becomes outside region defined by boundary of J n, and thus X n1 Jn = 0. Since E n1 X n1 Jn, a contradiction. Thus the claim is proved and by the first part proof of the main theorem, we prove that every surface has a triangulation. By the proof above shows, there exist a triangulation for every surface. And the proof can also be modified into the version about the existence of the triangulation of bordered surface F. Remark. The proof in this paper is a standard proof for the result. The same result, however, can also be approached using the smooth model of the surface so that he avoided the using of the Jordan-Schoenflies s theorem which here we use without rigorous proof [1]. 7

8 References [1] Hatcher, Allen (2013), The Kirby torus trick for surfaces, arxiv: [2] Thomassen, Carsten (1992), The Jordan-Schoenflies theorem theorem and the classification of surfaces, American Mathematical Monthly 99 (2): [3] Ahlfors, Lars and Sario, Leo (1946), Riemann Surfaces, Princeton University Press 8

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