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Chapter 3 – Cohomology

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Chapter 3 – Cohomology Chapter 3 { Cohomology 3.0 { Cohomology Introduction Let X be a 1-dimensional ∆-complex, so X is an oriented graph (or pseudo-graph if edges that begin and end at the same vertex are allowed). If v0 and v1 are two vertices in X and e is an oriented edge from v0 to v1, then notationally we'll denote e by [v0; v1]. Recall that, formally, 0 1 a \vertex" is a map σα : ∆ ! X, and an \edge" is a map τβ : ∆ ! X that, when restricted 1 to each endpoint (or \face") of ∆ , becomes one of the maps σα when pre-composed with the appropriate canonical linear homeomorphism ∆0 7! (face of ∆1). So our edge e is in fact a 1 1 map σe : ∆ ! X such that, if we denote ∆ by [u0; u1] (a line segment), then σe(u0) = v0 1 and σe(u1) = v1; thus, if indeed v0 6= v1, it follows that σe(∆ )|truly our edge e in X in the graphical sense|is homeomorphic to the standard 1-simplex ∆1, and so it makes sense to represent e using the simplex notation [v0; v1] (especially since it conveys information about the orientation of e). If v0 = v1 it still makes sense to represent e by [v0; v1] = [v0; v0] to maintain consistent notation even though the corresponding edge is not homeomorphic to any kind of simplex.1 Let G be an abelian group (not necessarily free), V the set of vertices of X, and E the set of edges of X. Define ∆0(X; G) = f' : V ! G j ' is a functiong and ∆1(X; G) = f : E ! G j is a functiong 0 0 Note ∆ (X; G) forms an abelian group: if '1;'2 2 ∆ (X; G), then '1 + '2 given by ('1 + '2)(v) = '1(v) + '2(v) 0 shows that '1 + '2 2 ∆ (X; G) since '1(v) 2 G and '2(v) 2 G implies that '1(v) + '2(v) 2 G. In similar fashion ∆1(X; G) is also an abelian group. 0 1 0 Now define a homomorphism δ1 : ∆ (X; G) ! ∆ (X; G) as follows: for ' 2 ∆ (X; G) let 1 δ1' 2 ∆ (X; G) be such that, for [v0; v1] 2 E, δ1'([v0; v1]) = '(v1) − '(v0). Set up a chain complex · · · −! 0 −−−!δ0 ∆0(X; G) −−−!δ1 ∆1(X; G) −−−!δ2 0 −! · · · : By definition the homology groups associated with this chain complex are the simplicial n cohomology groups H∆(X; G) of X. In particular we have 0 1 0 Ker[δ1 : ∆ (X; G) ! ∆ (X; G)] ∼ H∆(X; G) = 0 = Ker δ1 (1) Im[δ0 : 0 ! ∆ (X; G)] since Im δ0 = 0, and 1 1 1 Ker[δ2 : ∆ (X; G) ! 0] ∼ ∆ (X; G) H∆(X; G) = 0 1 = 0 1 : (2) Im[δ1 : ∆ (X; G) ! ∆ (X; G)] Im[δ1 : ∆ (X; G) ! ∆ (X; G)] 1 Recall that the map σe is also called a 1-simplex. 2 0 ∼ 0 So H∆(X; G) = f' 2 ∆ (X; G): δ1' = 0g, where 0(e) := 0 2 G for every e 2 E. Thus 0 ' 2 H∆(X; G) implies that δ1'([v1; v0]) = '(v1) − '(v0) = 0; or '(v1) = '(v0) for every vertex v0 and v1 that is connected by an edge in X. This in turn implies that ' must be constant on each component of X. Let fXαgα2I be the components of X, and let Vα be the set of vertices for the subgraph Xα. Then 0 Ker δ1 = f' 2 ∆ (X; G): 8 α 2 I' is constant on Vαg 0 = f' 2 ∆ (X; G): 8 α 2 I 9 gα 2 G s.t. ' ≡ gα on Vαg: Q 2 It follows that each element of the group Ker δ1 corresponds to some fgαgα2I 2 α2I G; where it is necessary to consider the direct product of copies of G as opposed to the direct sum since it may well be that gα 6= 0 for an infinite number of index values α. The one-to-one Q correspondence Ker δ1 7! α2I G can easily be shown to be an isomorphism, so therefore from (1) we obtain 0 ∼ ∼ Y H∆(X; G) = Ker δ1 = G: α2I 1 Next, from (2) is can be seen that H∆(X; G) = 0 iff δ1 is surjective, which is to say that for 1 0 each 2 ∆ (X; G) there exists some ' 2 ∆ (X; G) such that δ1' = . This will be the case whenever the components fXαgα2I of X are trees since the path between any two vertices in a 1 tree is unique: for a given α 2 I,[v0; v1] = e 2 Eα (the set of edges in Xα) and 2 ∆ (X; G), 0 we need only define ' 2 ∆ (X; G) such that '(v1) − '(v0) = (e), with the choice of definition being unique up to a constant; then the values of ' at all other vertices in Xα are set as dictated by the values of on the edges of the unique paths leading to those vertices. The process is repeated for the other components of X to get δ1' = . If a particular component Xα of X is not a tree, then we designate a maximal tree that is a subgraph of Xα which includes all of its vertices but not all edges. It is a fact that, for any choice of maximal tree for a graph, the cardinality of the set of edges omitted from the tree will be the same. For the sake of argument suppose that X is a connected graph that is not a tree, let Y ⊂ X be a maximal tree, let E0 be the set of edges not in Y , and let E00 be the set of edges in Y . The claim will be that 1 ∼ Y H∆(X; G) = G: e2E0 For the construction of the appropriate isomorphism, note that for any 2 ∆1(X; G) there 0 can be found some ' 2 ∆ (X; G) (unique up to a constant) such that δ1'jE00 = jE00 . Now 1 Q define F : H∆(X; G) ! e2E0 G by F ( + Im δ1) = f( − δ1')(e)ge2E0 for some ' such that δ1'jE00 = jE00 (3) 2 Recall that formally fgαgα2I is a function g : I ! G given by g(α) = gα for each α 2 I. 3 The choice for ' is irrelevant since the difference must be a constant: if' ^ = ' + g for some fixed g 2 G (more precisely g : E ! fgg), then δ1'^([v0; v1]) = (' + g)(v1) − (' + g)(v0) = ('(v1) + g(v1)) − ('(v0) + g(v0)) = '(v1) + g − '(v0) − g = δ1'([v0; v1]); and so − δ1'^ = − δ1'. It should first be verified that F is well-defined. We start with a simple case when X is a graph with three vertices and three edges as shown in the figure, so E = fe0; e1; e2g. To 0 00 construct a maximal tree Y we need only omit e0, so that E = fe0g and E = fe1; e2g. 0 Suppose 1 + Im δ1 = 2 + Im δ1. There exist '1;'2 2 ∆ (X; G) such that δ1'1jE00 = 1jE00 and δ1'2jE00 = 2jE00 . To show F ( 1 + Im δ1) = F ( 2 + Im δ1) means to show that ( 1 − δ1'1)(e0) = ( 2 − δ1'2)(e0); or ( 1 − 2)(e0) = ('1(v1) − '1(v0) − ('2(v1) − '2(v0) : (4) Now, 1 + Im δ1 = 2 + Im δ1 implies that 1 − 2 2 Im δ1, and so there exists some ' : V ! G such that (δ1')(ei) = ( 1 − 2)(ei) for all ei 2 E. Hence 1(e0) − 2(e0) = '(v1) − '(v0) (5) 1(e1) − 2(e1) = '(v2) − '(v0) (6) 1(e2) − 2(e2) = '(v2) − '(v1); (7) while from δ1'1jE00 = 1jE00 and δ1'2jE00 = 2jE00 we obtain 1(e1) = '1(v2) − '1(v0) and 2(e1) = '2(v2) − '2(v0) (8) 1(e2) = '1(v2) − '1(v1) and 2(e2) = '2(v2) − '2(v1): (9) Combining (6) and (8) gives '(v2) − '(v0) = ('1(v2) − '1(v0) − ('2(v2) − '2(v0) ; (10) and combining (7) and (9) gives '(v2) − '(v1) = ('1(v2) − '1(v1) − ('2(v2) − '2(v1) : (11) Now, if we subtract (11) from (10) we obtain '(v1) − '(v0) = ('1(v1) − '1(v0) − ('2(v1) − '2(v0) : (12) 4 We now put (12) into (5) and get precisely (4), as desired. A simpler analysis can be employed to show that F is well-defined in the case when a maximal tree is formed by deleting one edge and retaining one edge, which becomes the \base case" for an inductive argument that will establish that F is well-defined when one edge is deleted and n edges are retained, n 2 N arbitrary. This result, once obtained, in turn becomes the base case for another inductive argument that establishes the well-definedness of F in the general case when m edges are deleted and n edges are retained in the forming of a maximal tree, m; n 2 N both arbitrary. All of this can be done under the assumption that X is connected (i.e. has just one component), after which it is easy to extend to an arbitrary number of components. To show that F is a homomorphism of groups, along with addition and integer multiplication of cosets in a quotient group we assume the usual (componentwise) definitions for addition and integer multiplication of elements in a direct product of groups. In what follows X is not assumed to be connected, so E00 is taken to be the set of edges included in the maximal tree for some component of X, and E0 = E − E00. Let m; n 2 Z and ; ^ 2 ∆1(X; G).
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