Sara Kališnik January 10, 2018 Sara Kališnik January 10, 2018 1 / 34
Homology Homology groups provide a mathematical language for the holes in a topological space. Perhaps surprisingly, they capture holes indirectly, by focusing on what surrounds them. Their main ingredients are group operations and maps that relate topologically meaningful subsets of a space with each other. b B e a A F T C d E D c Simplicial complex for which we will carry out a computation as an example. Sara Kališnik January 10, 2018 2 / 34
Homology Fix a dimension i and a field F. i-chains An i-chain is a formal sum of i-simplices c i σ i, where c i F and the sum is taken over all possible i-simplices σ i K. The set of all such i-chains is denoted C i (K). We can add two i-chains: given c = c i σ i and d = s i σ i, we define multiply i-chains by scalars c + d = (c i + s i )σ i. Sara Kališnik January 10, 2018 3 / 34
Homology Fix a dimension i and a field F. the vector space of i-chains Hence C i (K) forms a vector space over F, and is called the vector space of i-chains in K. Note: The set of i-simplices forms an example of a basis for C i (K). Hence the rank of C i (K) is simply n i, the number of i-simplices in K. For i < 0 and i > dim(k), we have C i (K) = 0 since there are no simplices in those dimensions. Sara Kališnik January 10, 2018 4 / 34
Homology From now on, we will make the simplifying assumption that our field F is simply the binary field Z/2Z; in this case, an i-chain can be thought of as just a collection of i-simplices and adding two i-chains corresponds to taking the symmetric difference of the collections. Sara Kališnik January 10, 2018 5 / 34
Homology The Boundary of a Simplex To relate the vector spaces of chains, we define the boundary of an i-simplex as the sum of its (i 1)-dimensional faces. Writing σ = [u 0, u 1,..., u i ] for the simplex spanned by the listed vertices, its boundary is i i σ = [u 0, u 1,..., û j,..., u i ] j=0 where the hat indicates that u j is omitted. Sara Kališnik January 10, 2018 6 / 34
Homology The Boundary Map For an i-chain, c = i c iσ i, the boundary is the sum of the boundaries of its simplices, i c = i c i σ i. Hence, taking the boundary maps an i-chain to a (i 1)-chain, and we write i : C i (K) C i 1 (K). This map is linear and we will refer to it as the boundary homomorphism or, shorter, the boundary map for chains. Sara Kališnik January 10, 2018 7 / 34
Homology Example b B e a A F T C d E D c Then 2 (T ) = B + C + F C 1 (K), and 1 (F ) = b + d. Note also that 1 ( 2 (T )) = 1 (B+C+F ) = 1 (B)+ 1 (C)+ 1 (F ) = b+e+e+d+b+d = 0. Sara Kališnik January 10, 2018 8 / 34
Homology The Chain Complex The chain complex is the sequence of chain vector spaces connected by boundary maps,... i+2 C i+1 (K) i+1 C i (K) i C i 1 (K) i 1... C 0 (K) 0 Sara Kališnik January 10, 2018 9 / 34
Computing Homology b B e a A F T C d E D c C 0 (K) = a, b, c, d, e, C 1 (K) = A, B, C, D, E, C 2 (K) = T, C i (K) = 0 for i 3. Sara Kališnik January 10, 2018 10 / 34
Computing Homology b B e a A F T C d E D c The boundary maps are 2 = 0 1 1 0 0 1, 1 = A B C D E F a 1 0 0 0 1 0 b 1 1 0 0 0 1 c 0 0 0 1 1 0, 0 = (0). d 0 0 1 1 0 1 e 0 1 1 0 0 0 Sara Kališnik January 10, 2018 11 / 34
Homology i-cycles An i-cycle is an i-chain with empty boundary, that is, i C i (K) is an i-cycle iff i (c) = 0. The set of all such i-cycles forms a subspace of C i (K), which we denote by Z i (K) = ker i ; we set its rank to z i. Sara Kališnik January 10, 2018 12 / 34
Homology Example b B e a A F T C d E D c The group of 1-cycles Z 1 (K) contains the chains B + C + D, A + D + F + G, as well as many more. On the other hand, Z 2 (K) = 0 as can be seen by direct computation. Sara Kališnik January 10, 2018 13 / 34
Homology i-boundaries Given an i-chain c C i (K), we say that c is an i-boundary if there exists d C i+1 (K) such that c = i+1 (d). The set of i-boundaries also forms a subspace of C i (K), and we denote it by B i (K) = Im i+1 ; we set its rank to b i. Sara Kališnik January 10, 2018 14 / 34
Homology The fundamental property that makes homology work is that the boundary of a boundary is necessarily zero. Fundamental Lemma of Homology i i+1 d = 0 for every integer i and every (i + 1)-chain d. We just need to show that i i+1 σ = 0 for an (i + 1)-simplex σ. The boundary, i+1 σ, consists of all i-faces of σ. Every (i 1)-face of σ belongs to exactly two i-faces, so i ( i+1 σ) = 0. Sara Kališnik January 10, 2018 15 / 34
Homology i-th homology The i-th homology group of K is H i (K) = Z i (K)/B i (K). The rank of this group is called the i-th Betti number of K, β i = rank(h i (K)). The elements of H i are called i-dimensional homology classes, and they correspond to cosets c + B i, where c is an i- cycle. Any two cycles in the same class are called homologous; note that c, c Z i (K) are homologous iff there is a (i + 1)-chain d such that c + c = (d). Sara Kališnik January 10, 2018 16 / 34
Computing Homology To find homology, we need to study pairs of matrices (A, B) with A B = 0. Let B : U V and A: V W be such that A B = 0. Let F : U U, G: V V and H : W W be invertible linear transformations. Then HAG 1 GBF = 0 and there is an isomorphism of vector spaces Ker(A)/Im(B) = Ker(HAG 1 )/Im(GBF ). Sara Kališnik January 10, 2018 17 / 34
Computing Homology This implies that we can use the following operations on a pair (A, B): 1 An arbitrary row operation on A. 2 An arbitrary column operation on B. 3 Perform a column operation on A and a row operation on B simultaneously, with the operations related as follows: If the column operation on A is multiplication of the i-th column by x, then the row operation on B is multiplication of the i-th row by x 1. If the column operation on A is the transposition of two columns, then the row operation on B is the transposition of the corresponding rows of B. If the column operation on A is the addition of x times the i-th column to the j-th column, then the row operation on B is the subtraction of x times the j-th row from the i-th row. Sara Kališnik January 10, 2018 18 / 34
Computing Homology Given a pair (A, B) with A B = 0, we can perform operations of the type just described to obtain a pair (A, B ), with I n 0 0 0 0 0 0 0 0, 0 0 0. 0 0 0 0 0 I m Here k, l and m are columns in the leftmost, middle and rightmost blocks of column of A respectively. The dimension of homology is l. The pair (A, B ) is uniquely determined by (A, B). Sara Kališnik January 10, 2018 19 / 34
Computing Homology Let us compute H 1 of A B C D a b e d c T E F The pair we observe is A B C D E F a 1 0 0 0 1 0 b 1 1 0 0 0 1 c 0 0 0 1 1 0 d 0 0 1 1 0 1 e 0 1 1 0 0 0, 0 1 1 0 0 1 Sara Kališnik January 10, 2018 20 / 34
Computing Homology We first do row operations on 1 : A B C D E F a 1 0 0 0 1 0 a+b 0 1 0 0 1 1 c 0 0 0 1 1 0, d 0 0 1 1 0 1 e 0 1 1 0 0 0 T A 0 B 1 C 1 D 0 E 0 F 1 Sara Kališnik January 10, 2018 21 / 34
Computing Homology We first do row operations on 1 : A B C D E F a 1 0 0 0 1 0 a+b 0 1 0 0 1 1 c 0 0 0 1 1 0, d 0 0 1 1 0 1 a+b+e 0 0 1 0 1 1 T A 0 B 1 C 1 D 0 E 0 F 1 Sara Kališnik January 10, 2018 22 / 34
Computing Homology We first do row operations on 1 : A B C D E F a 1 0 0 0 1 0 a+b 0 1 0 0 1 1 a+b+e 0 0 1 0 1 1, c 0 0 0 1 1 0 d 0 0 1 1 0 1 T A 0 B 1 C 1 D 0 E 0 F 1 Sara Kališnik January 10, 2018 23 / 34
Computing Homology We first do row operations on 1 : A B C D E F a 1 0 0 0 1 0 a+b 0 1 0 0 1 1 a+b+e 0 0 1 0 1 1, c 0 0 0 1 1 0 a+b+e+d 0 0 0 1 1 0 T A 0 B 1 C 1 D 0 E 0 F 1 Sara Kališnik January 10, 2018 24 / 34
Computing Homology We first do row operations on 1 : A B C D E F a 1 0 0 0 1 0 a+b 0 1 0 0 1 1 a+b+e 0 0 1 0 1 1, c 0 0 0 1 1 0 a+b+c+e+d 0 0 0 0 0 0 T A 0 B 1 C 1 D 0 E 0 F 1 Sara Kališnik January 10, 2018 25 / 34
Computing Homology Next we perform column operations on 1 : A B C D A+B+C+D+E F a 1 0 0 0 0 0 a+b 0 1 0 0 0 1 a+b+e 0 0 1 0 0 1, c 0 0 0 1 0 0 a+b+c+e+d 0 0 0 0 0 0 T A 0 B 1 C 1 D 0 A+B+C+D+E 0 F 1 Sara Kališnik January 10, 2018 26 / 34
Computing Homology Next we perform column operations on 1 : A B C D A+B+C+D+E B+C+F a 1 0 0 0 0 0 a+b 0 1 0 0 0 0 a+b+e 0 0 1 0 0 0 c 0 0 0 1 0 0 a+b+c+e+d 0 0 0 0 0 0, T A 0 B 0 C 0 D 0 A+B+C+D+E 0 B+C+F 1 After we are done, the pair is in the desired form. From here we can read off that β 1 = 1. Also H 1 (K) = A + B + C + D + E = Z/2Z. Sara Kališnik January 10, 2018 27 / 34
Making homology more sensitive It is useful to ask how sensitive a measure homology is of the shape of a simplicial complex, but considering a simple shape recognition task, namely the recognition of printed letters. We begin with the first three letters of the alphabet, and find that H 1 succeeds at distinguishing between them. Sara Kališnik January 10, 2018 28 / 34
Making homology more sensitive However, after this initial success, we see that every other letter has the same first Betti number as one of these three. Homology can be refined to discriminate more finely between the letters. To understand how this works, we digress a bit to discuss how an analogous problem in manifold topology is approached. Sara Kališnik January 10, 2018 29 / 34
Making homology more sensitive The homology groups H i (R n ) vanish for all i > 0. What this means is that homology is unable to distinguish between R m and R n when m = n. From the point of view of a topologist who is interested in distinguishing different manifolds from each other, this means that homology is in some ways a relatively weak invariant. This failure can be addressed by computing homology on auxiliary or derived spaces, constructed using various geometric constructions. Sara Kališnik January 10, 2018 30 / 34
Making homology more sensitive Removing a point While the homology groups of R n vanish, the homology groups of R n \ {0} do not. To see this, we observe that we have the inclusion i: S n 1 R n \ {0} as well as the map r : R n \ {0} S n 1 defined by r(v) = f v. r i is equal to the identity map for Sn 1, and the other composite i r is homotopic to the identity map of R n \ {0} via the straight line homotopy H(v, t) = (1 t)i r + tid R n \{0}. With this we can detect the difference between R n and R m, with m = n, by recognizing that the homology of the result of removing a single point from the two spaces are different, since homology detects the difference between spheres of different dimensions. Sara Kališnik January 10, 2018 31 / 34
Making homology more sensitive Removing singular points Consider the space given by the crossing of two lines. This space is also contractible, as one can easily see by retracting each line segment onto the crossing point A. On the other hand, we recognize that its shape has features which distinguish it from an interval or a circle, and might want to detect that homologically. If we remove the singular point A, we will find that the space remaining breaks up into four distinct components, which can be detected by H 0. Sara Kališnik January 10, 2018 32 / 34
Making homology more sensitive If one uses homology on suitably constructed auxiliary spaces, one can obtain classification criteria for many interesting problems in shape discrimination. Try 2 To think through how we might apply these ideas to the problem of distinguishing between letters, let us define an end of a space X to be a point x X so that there is a neighborhood N of x which is homeomorphic to [0, 1), and so that the homeomorphism carries x to 0. Sara Kališnik January 10, 2018 33 / 34
Making homology more sensitive Try 2 In this case, the auxiliary or derived space is the set of ends of the space, e(x), and we can compute its zero-dimensional homology H 0, to get the Betti number β 0. We now obtain the following partition of the set of letters. Sara Kališnik January 10, 2018 34 / 34