Lecture notes for Topology MMA100

Lecture notes for Topology MMA100 J A S, S-11 1 Simplicial Complexes 1.1 Affine independence A collection of points v 0, v 1,..., v n in some Euclidean space R N are affinely independent if the (affine hyper-) plane of least dimension containing them has dimension n. Synonymously, the points are in general or generic position. This is equivalent to v 1 v 0, v 2 v 0,..., v n v 0 being linearly independent vectors (or similarly with any v i replacing v 0 ). Thus e.g. four points in R 3 can be in generic position (but not five). Another way to express genericity is to say that t i v i = 0, where t i are scalars with t i = 0, only if all t i = 0. Yet another criterion is to use the embedding R N R N+1 given by adding the scalar 1 as a first coordinate: x (1, x). Then v 0, v 1,..., v n are in generic position iff (1, v 0 ), (1, v 1 ),..., (1, v n ) are linearly independent vectors. v 3 v 0 v 1 v 2 Figure 1: Four points in R 3 in general position Obviously, if v 0, v 1,..., v n are linearly independent as vectors they are also affinely independent as points (considering vectors as position vectors for points). An point p is an affine combination of the points v 0, v 1,..., v n if p = t i v i, where t i = 1. If the points are in generic position, the affine coordinates t i are uniquely determined by p. The (n-dimensional) plane spanned by points v 0, v 1,..., v n consists of all affine combinations of them. E.g. the (two-dimensional) plane spanned by three points v 0, v 1, v 2 in R 3 consist of all points x = t 0 v 0 + t 1 v 1 + t 2 v 2, with t 0 + t 1 + t 2 = 1, or differently put: x = v 0 + t 1 (v 1 v 0 ) + t 2 (v 2 v 0 ), with no condition on the scalars. This is a standard parametric equation for the plane. 1.2 Convex closure As you know a subset of R N is convex if whenever it contains two points p and q it contains the whole line segment between them. This is parametrized by (1 s)p + sq, where s [0, 1]. It follows directly that the intersection of any number of convex sets is again convex. Consequently, for any subset A R N there is a smallest convex set containing it: the intersection of all convex sets containing it. This is the convex closure, co(a) of A. An affine combination t i p i of points p 0, p 1,..., p n in R N, is a convex combination if all t i [0, 1] (in addition to t i = 1). An easy induction on n shows that if a 0, a 1,..., a n are points in A, and C is a convex set containing A, then any convex combination of the points is again in C. In particular the applies to C = co(a). It s equally easy to see that if p and q are convex combinations of points in A, then so is any (1 s)p + sq, where s [0, 1]. Thus, co(a) can be described intrinsically as the set of all convex combinations of points in A. 1

1.3 Simplexes An n-simplex in R N, denoted σ n = σ, is the convex closure of a set of n + 1 points v 0, v 1,..., v n R N in generic position. It will be denoted by v 0, v 1,..., v n, a notation that will (hopefully) only be used under the proviso that the points are affinely independent. It s natural to ask if v 0, v 1,..., v n = w 0, w 1,..., w m is possible without n = m and w 0, w 1,... w n being some reordering of v 0, v 1,..., v n. The answer is no, as will be demonstrated later on. To wit, a simplex is the convex closure of exactly one set of affinely independent points. This observation makes it clear the vertexes can be used to define properties and derive concepts of simplexes, as is done bellow. Thus σ consist of all points p that can be (uniquely) expressed as p = t i v i, where t i = 1 and all t 1 [0, 1]. In this connection the scalars t i are called the barycentric coordinates of p and the v i :s are vertexes of σ. The dimension of σ is one less than the number of vertexes, and is thus the same as the dimension of the smallest (affine hyper) plane containing them. A simplex of dimension 0 is hence the same as a single vertex (or point). We allow the set of vertexes to be empty, so that the empty set,, is simplex of dimension 1. The barycenter of σ, or the center of gravity, is the point 1 n + 1 (v 0 + v 1 + + v n ) = 1 n + 1 v 0 + 1 n + 1 v 1 + + 1 n + 1 v n σ. Any (proper) subset of the vertexes spans a simplex τ which is a (proper) face of σ, written τ σ (or τ < σ). The boundary of a simplex σ, denoted by σ, is the union of all its proper faces. In terms of convex combinations it consists of all such t i v i, where at least one coordinate is 0. Beware that this is generally not the topological boundary of σ as a subset of some R N. The interior σ of a simplex σ is σ σ. This is the set of all t i v i, where all t i > 0. This is generally not the same as the interior of σ in a containing Euclidean space. Lemma 1.1 A simplex of dimension > 0 is (homeomorphic to) the cone on its boundary. Proof Let σ = v 0, v 1,..., v n denote the simplex and b its barycenter. Define f : σ I σ by f(p, t) = (1 t)p+tb. The target is correct by the convexity of σ and f is continuous since it is the restriction of the continuous map R N I R N, given by the same formula. Obviously, f maps the top of the cylinder on σ to b. The usual compact-hausdorff argument gives that, in order to prove that f induces a homeomorphism C( σ) = σ, we are left with checking that f restricts to a bijection σ [0, 1) σ {b}. v 0 t σ p b v 3 q v 1 v 2 Figure 2: Polar coordinates for q First note that if q b is a point in σ and q = tb + (1 t)p, where 0 t < 1 and p σ, then t/(n + 1) is the minimal barycentric coordinate of q, as some coordinate of p is 0. This shows the uniqueness of t and hence of p, so the restriction of f is injective. On the other hand with q as above, let t < 1/(n + 1) be the minimal barycentric coordinate of q. Write ˆt = (n + 1) t < 1. Then q = (n + 1) tb + i (t i t )v i = ˆtb + (1 ˆt ) i t i t 1 ˆt v i = ˆtb + (1 ˆt )p. Notice that the sum of the barycentric coordinates of p is 1, and that at least one of them is 0, so p is a point in σ. Hence the restriction of f is surjective. If q σ is q = tp + (1 t)b, with p σ and 0 t 1, then (t, p) are the polar coordinates for q. They are unique, with the exception that (0, p) corresponds to b for any p σ The standard n-simplex n in R n+1 is the convex closure of the standard basis of R n+1, which in this connection is enumerated e 0, e 1,..., e n. Explicitly, this e 2 2 2 e 1 e 0 Figure 3: The standard 2- simplex sits in R 3

means that n = {(t 0, t 1,..., t n ) t i = 1 and t i [0, 1], for all i}. Lemma 1.2 An n-simplex is homeomorphic to the standard n-simplex hence compact. Proof Define f : n v 0, v 1,..., v n by f(t) = t i v i. Then f is continuous. In fact, if v 0, v 1,... v n R m the same formula gives a continuous function R n+1 R m. The map f is obviously a bijection, n compact and v 0, v 1,..., v n Hausdorff, so f is a homeomorphism. Lemma 1.3 An n-simplex (n 1) is homeomorphic to the cone on any of its n 1-dimensional faces. Proof Let σ be an n+1-simplex with vertexes v 0, v 1,..., v n and τ the face with the same vertexes except v n. Define a map f : τ I σ by f(p, t) = (1 t)p + tv n. It s injective, by uniqueness of barycentric coordinates, with the exception that f(p, 1) = e n, for all p. All we need to check to see that f induces a homeomorphism on from the cone on τ to σ is the surjectivity of f : τ [0, 1) σ {v n }. If q = t i v i σ and p v n, i.e. t n 1, then q = t n v n + (1 t n ) i n t i 1 t n v i = t n v n + (1 t n )p. Notice that the (positive!) barycentric coordinates of p has sum 1 and at least one is 0. Thus p τ and f(p, t n ) = q. Lemma 1.4 An n-simplex is homeomorphic to an n-disk by a homeomorphism respecting the boundaries of the two spaces. Proof It s enough to show this for the standard n-simplex n. This simplex is the graph of the map f : T n R, f(t 0,..., t n 1 ) = 1 Σ i t i over the set T n R n consisting of those t R n with each coordinate t i [0, 1] and Σ i t i 1. Thus n is homeomorphic to T n. Write t s = Σ i t i and t m = max(t 0,..., t n 1 ) Define g : T n I n by g(t) = ( t s / t m )t. It s obviously continuous except possibly at t = 0. But g(t) m = t s 0, as t tends to 0. The map g is easily checked to be a homeomorphism (with inverse g 1 (p) = ( p m / p s )p). Now, I n is homeomorphic to [ 1, 1] n which in turn is homeomorphic to D n by the usual normalization. Notice that the homeomorphism n D n restricts to a homeomorphism δ n = δd n = S n 1. Suppose you have a set A in R N which you some how know to be a simplex. How do you determine it s vertexes? Are they indeed uniquely 1determined by the set A itself? Lemma 1.5 A point p in a simplex is a vertex exactly when every line through it intersects the simplex in a subset of the line having p as a boundary point. Proof Let p be a point of the simplex. The intersection of all faces (with respect to some choice of vertexes) containing p is the minimal face containing p. Suppose this is v 0, v 1,..., v n and p = t i v i where all t i > 0 by minimality. Suppose t 0 < 1, then p = t 0 v 0 + (1 t 0 ) i>0 t i 1 t 0 v i = t 0 v 0 + (1 t 0 )q, where the sum, call it q, is a convex combination of vertexes and hence in the simplex. Thus the line trough p and q intersects the simplex in a line segment containing v 0 and q with p in between. Thus if p satisfies the criterion of the lemma t 0 has to be 1 and p = v 0. 3

On the other hand let v 0 be one of he chosen vertexes and suppose a line intersects the simplex in another point p = t i v i. The line through v 0 and p is parametrized by (1 s)v 0 + s t i v i = (1 s(1 t 0 ))v 0 + st i v i, i>0 which is an affine combinations of the vertexes for any real s. In order for this to be a convex combination (and hence a point of the simplex) we need 0 1 s(1 t 0 ) 1 Figure 4: for vertex Testing and 0 st i 1, for all 1 i. This implies s 0, so v 0 is a boundary point for the intersecting line segment. 1.4 Simplicial Complexes A (geometric) simplicial complex in R N is a collection K of simplexes (in R N ) such that 1. if σ is in K, so is any of its faces, 2. if σ and τ are in K, then σ τ is a face of both σ and τ (which might be the empty simplex). In this course all simplicial complexes will be finite (i.e. K is a finite collection). A subcomplex L of K, L K, is a subset closed under taking faces. For example let n be a positive integer and let K (n) be the sub-collection of K consisting of all simplexes in K of dimension n. Then K (n) is a subcomplex of K, called the n-skeleton of K. The dimension, dimk, of K is the largest dimension of the simplexes in K. If σ is a simplex of K, let K σ be the collection of all faces of σ. Then K σ is a subcomplex of K. It might seem a bit odd to require 1), but it makes it easy to define the notion of subcomplex. A simplex in K is a facet if it is not a proper face of any other simplex of K. The facet obviously determines the rest of K by taking faces. One way to specify a simplicial complex is to pinpoint some simplexes (facets) that pairwise intersect in common faces. Taking all the faces of the facets, to satisfy 1), then defines a simplicial complex. Indeed, 2) follows from the proper intersection of facets, since faces of a fixed simplex always intersect in a common face. The set of zero dimensional simplexes in K is the set of vertexes of K and we denote it by V (K). The polyhedron K defined by K, or underlying K, is the union of all simplexes in K : K = σ R N. σ K Notice that K is a subset of R N, as opposed to K which is a collection of subsets K. Obviously the polyhedron can equally be described as the union of all facets of K. The polyhedron is compact since it is a finite union of (compact) simplexes. In general a subset of R N is a polyhedron if it is K for some simplicial complex K. In this case K is a triangulation of the polyhedron. Different simplicial complexes can very well have the same underlying polyhedron. In this case we say that they are different triangulations of the same space. Intersecting all simplexes containing a point x in the polyhedron of K, we see that there is a unique smallest simplex containing it. It s called the carrier of x, and we denote it by carr(x) K. Obviously, x carr(x), so x is an interior point of its carrier. An alternative characterization of carr(x) is that it is the unique simplex with x in its interior. We have a function carr : K K. 1.5 Subdivision In this section we will denote the barycenter of a simplex σ by ˆσ = i 1 n + 1 v i σ. 4

A simplicial complex L is a subdivision of K if L = K and every simplex of L is contained in a simplex of K. This is (sometimes) written L K. In this situation, let p be a point in L = K. Then its carrier with respect to L, carr L (p) is contained in a smallest simplex µ of K. If carr L (p) = u 0,..., u m then p = i s iu i, for some 0 < s i < 1. Each u i is in µ = v 0,... v n, so u i = j t ijv j, some 0 t ij 1, where for each j at least one t ij 0, since otherwise carr L (P ) would be cotained in a proper face of µ. This gives p = j ( i s it ij )v j where each coefficient of the vertexes is > 0. Thus carr L (p) µ = carr K (p). A consequence of this is that any simplex in K is a union of simplexes in L. Indeed, take a point p in σ K. Then carr L (p) carr K (p) σ. Note that if L K and K is a subcomplex of K, then the set of simplexes in L contained in simplexes of K is a subcomplex L of L with L K. Lemma 1.6 Suppose K is a subcomplex of K and L a subdivision of K. Then there is a subdivision L of K containing L as a subcomplex. Put differently: A subdivision of a subcomplex can be extended to a subdivision of the total complex. Proof By induction on the number of simplexes in K K. Let σ be a simplex of lowest dimension in K K. Then the faces of σ is a subcomplex K σ of K corresponding to a certain subcomplex L σ of L which is a subdivision of K σ. It suffices to show that there is a subdivision of K σ extending L σ. For each simplex ν of L σ we let νˆσ which is the simplex spanned by the vertexes of ν and the barycenter ˆσ of σ. The case when ν = gives the 0-simplex ˆσ. Obviously νˆσ σ and by polar coordinates σ = ν νˆσ. Also, as a consequence of the essential uniqueness of these coordinates ν 1ˆσ ν 2ˆσ = (ν 1 ν 2 )ˆσ and ν 1 (ν 2ˆσ) = ν 1 ν 2. This shows that the collection L σ = {ν ν L σ } {νˆσ ν L σ } is a simplicial complex which is a subdivision of K σ extending L σ. Figure 5: Extending a subdivision Then L L σ is a subdivision of K {σ} extending L. Proceeding by induction we get a subdivision L of K extending L. Now, let K be a simplicial complex. If σ 0 < σ 1 <... < σ n is a sequence of faces of a simples σ n in K, then the barycenters ˆσ 0, ˆσ 1,... ˆσ n are affinely independent and hence spans a simplex ˆσ 0, ˆσ 1,... ˆσ n σ. Let sdk be the collection of all these for all such sequences of faces of simplexes in K. Using the construction in the poof above it follows (by induction) on that sdk is a simplicial complex which is a subdivision of K. It s called the barycentric subdivision of K. This process can be repeated and we let sd r K denote the r:th barycentric subdivision of K, which we get by subdividing successively r times. ˆσ 3 σ 1 ˆσ 1 σ 3 ˆσ 0 σ 0 Figure 6: The sequence σ 0 < σ 1 < σ 2 determines a 2-simplex in sdk We will see that the diameter of the simplexes in sdk is smaller than the diameter of the simplexes in K in a uniform way. To do this we first need a few observations: Lemma 1.7 The diameter of a simplex is the maximal distance between its vertexes. Proof. Let σ = v 0, v 1,..., v n be an n-simplex and suppose x = t i v i σ. Denote max{d(v i, v j ) 0 i, j n} by d. Then, for any vertex v i0 x v i0 = t i (v i v i0) t i d = d. Now, suppose y = s i v i σ. Then, by the above, x y = s i (x v i ) s i d = d. 5

Definition 1.1 The mesh of a simplicial complex is the maximum of the diameters of its simplexes: mesh(k) = max{diam(σ) σ K}. Lemma 1.8 For any n-simplex σ the distance from its barycenter ˆσ to any other point in σ is nd/(n + 1), where d is the diameter of σ Proof Let v 0, v 1,..., v n be the vertexes of σ. then ˆσ v i = 1 n n + 1 (v j v i ) nd n + 1 as v j v i = 0, when j = i. Now, suppose y = s i v i σ. Then, j=0 ˆσ y = s i (ˆσ v i ) nd s i n + 1 = nd n + 1. Lemma 1.9 Suppose σ 0 < σ 1 <... < σ n and d is the diameter of σ n, which has dimension m, then the diameter of ˆσ 0, ˆσ 1,..., ˆσ n is md/(m + 1) Proof By induction ˆσ 0, ˆσ 1,..., ˆσ n 1 has diameter m d /(m + 1), where m is the dimension of σ n 1 and d is the diameter. Note that m /(m + 1) = 1 1/(m + 1) m/(m + 1), since m > m, and d d, so m d /(m + 1) md/(m + 1). Taking the previous lemma in to account the distance between any two vertexes of ˆσ 0, ˆσ 1,..., ˆσ n is md/(m + 1), so the simplex has diameter md/(m + 1). It now follows that mesh(sdk) N N + 1 mesh(k), where N is the maximal dimension of a simplex of K. This is the same as the maximal dimension of a simplex of sdk, so which approaches 0 as r tends to infinity. mesh(sd r K) ( N ) rmesh(k) N + 1 2 Simplicial Maps Let K and L be two simplicial complexes. A map f : K L is simplicial if 1. it maps vertexes to vertexes, 2. if v 0, v 1,..., v n K the (the not necessarily different) vertexes f(v 0 ), f(v 1 ),..., f(v n ) are vertexes of a simplex of L 3. when restricted to a simplex f is affine, i.e., f( i t i v i ) = i t i f(v i ). It s usual to denote simplicial maps by f : K L (i.e. without the decorations for the polyhedrons). Thus a simplicial map is completely determined by its restriction to the set of vertexes. Conversely any map f : V (K) V (L) satisfying (2) gives a simplicial map f : K L, by turning (3) into a definition. 6

Definition 2.1 A simplicial approximation to a map f : K L is a simplicial map g : K L, such that g(x) carr(f(x)), for each x K The last condition is equivalent to: any simplex containing f(x) also contains g(x). It s not true that any map has a simplicial approximation. We will prove, though, that it has after an iterated subdivision of K. If f restricts to a map f : K L, where K and L are subcomplexes of K and L respectively, then for each point x K the carrier of f(x) will be in L. Thus g(x) is in L and g : K L is a simplicial approximation to f. Theorem 2.1 If g is a simplicial approximation to f, then f g. Proof Define a map F : K I R N, where R N is a euclidean space containing L, by F (x, t) = (1 t)f(x) + tg(x). We note that the image of this map is in L since for any x there is a simplex of L containing both f(x) and g(x). Definition 2.2 For a vertex v of a simplicial complex K define the open star of v as st(v) = {x K v carr(x)} Alternatively, it s the union of the interiors of all simplexes containing v as a vertex. The complement of st(v) is the union of the simplexes not having v as a vertex, and so is a closed subset of K (being the union of a finite number of compact simplexes). Obviously, the open stars of all vertexes of K forms an open cover of K. Lemma 2.1 Suppose that v 0, v 1,..., v n are vertexes of a simplicial complex K, then n v 0, v 1,..., v n K st(v i ). Proof Assume that σ = v 0, v 1,..., v n K then ˆσ st(v i ) for i = 0, 1,..., n. Thus the intersection is non-empty. Assume that x n i=0 st(v i), then each v i is a vertex of carr(x), so v 0, v 1,..., v n is a face of carr(x) and thus a simplex of K. i=0 Figure 7: Open star of two vertexes Theorem 2.2 Suppose f : K L is a map where K and L are simplicial complexes. Then there is an integer r 0, such that f has a simplicial approximation g : sd r K L. Proof The open cover f 1 (st(v)), v V (L) of the compact metric space K has a Lebesgue s number δ > 0. We can choose r so that mesh(sd r K) δ/2. Then each simplex of K = sd r K has diameter less than δ/2. A consequence of this is that for any vertex w of K the open star st(w) has diameter less than δ. This, in turns, tells us that there is a vertex v of L such that f(st(w)) st(v). Let g(w) be a choice of such a v. Now, suppose that w 0, w 1,..., w n are the vertexes of a simplex of K. Then = f( i st(w i )) i st(g(w i )) so g(w 0 ), g(w 1 ),..., g(w n ) are vertexes of a simplex of L. Hence g defines a simplicial map K L. Finally, suppose that w 0, w 1,..., w n are the vertexes of the carrier of x. Then x i st(w i). Applying f we get f(x) f( st(w i )) f(st(w i )) st(g(w i )) i i i Hence each vertex g(w i ) is in the carrier of f(x). As the carrier of g(x) is the simplex spanned by the g(w i ) we get carr(g(x)) carr(f(x)), for each x K and g is a simplicial approximation to f. 7

3 Consequences of simplicial approximation A space X is triangulable if it is homeomorphic to K for some simplicial complex K. A specific choice of such a K and homeomorphism f : X = K gives a triangulation of X. For example the n-sphere s n is triangulable since it is homeomorphic to n = K σ, where K σ is the simplicial complex of all proper faces of the standard simplex n. It s a theorem that all smooth manifolds (not defined in this course), which are the fundamental objects of study in differential geometry are triangulable. Thus triangulable spaces exists in abundance. Theorem 3.1 The set of homotopy classes of maps between two triangulable spaces is countable. Proof It suffices to show that the set of homotopy classes of maps K L is countable for any two simplicial complexes K and L. We know that any map K L is homotopic to a simplicial map K L, where K is some iterated barycentric subdivision of K. Now there is only a finite number of such map for any K, but a countable collection of iterated barycentric subdivisions of K. Theorem 3.2 The inclusion of the 2-skeleton i : K (2) K induces an isomorphism on fundamental groups for any simplicial complex K. Proof The space I is the polyhedron of a simplicial complex with two vertexes and one 1-simplex. Any path-component of K contains a vertex v, so we can consider loops at this point in K. Any such loop ω : I K then maps the subcomplex consisting of 0 and 1 to the subcomplex of K consisting of just the vertex v. Iterated barycentric subdivision of I means introduction of (equidistant) points in I. Thus ω is homotopic to a map ω : I K which is simplicial. As ω maps the subcomplex consisting of 0 and 1 to the subcomplex consisting of v the so will ω. The usual homotopy between ω and ω is therefore stationary on the endpoints of I. Thus ω and ω are in fact path-homotopic. As the image of ω is contained in K (1), this shows that i induces a surjection i : π 1 ( K (2), v) π 1 ( K, v). To show injectivity, consider I 2 as the polyhedron of the simplicial complex having the vertexes v 0 = (0, 0), v 1 = (1, 0), v 2 = (1, 1) and v 3 = (0, 1), the 1-simplexes v 0, v 1, v 1, v 2, v 1, v 2 and v 2, v 3, and the 2-simplexes v 0, v 1, v 2 and v 0, v 2, v 3. It has a subcomplex J consisting of the faces of the 1-simplexes v 0, v 1, v 1, v 2, v 2, v 3, and v 1, v 3. Now suppose that α, β : I K (2) are loops at the vertex v, which are path-homotopic in K. Let F : I 2 K be a path-homotopy between α and β in K. It s restriction to J has image in K (2) and its restriction to J is constant (with value v). Thus if F : I 2 K is a simplicial approximation to F after some subdivision of I 2, the same is true for F. Even more importantly, the image of F has image in K (2), since it s simplicial. From this is follows that α (t) = F (t, 0) and β (t) = F (t, 1) are a simplicial approximation to α and β in K (2). The new paths α and β are in fact path-homotopic (in K (2) ) to α and β, respectively. As F is a path homotopy α β (in K (2) ) we get (in K (2) ) α α β β, so i is injective. Corollary 3.1 The n-sphere S n is simply connected if n 2. Proof S n is homeomorphic to the polyhedron n. When n 2 n and n has the same 2-skeletons. Hence the fundamental group of S n is isomorphic to the fundamental group of n = D n, which is trivial since D n is homotopy equivalent to a point. 8