A graph-theoretic approach to an exercise concerning the simplicial homology

   Part a of Exercise 16.7 in Topology - Huynh Quang Vu asks students to consider a simplicial complex X and prove that if the underlying space |X| is connected, then the homology group H_0(X) is isomorphic to the additive group of integers. In particular, this homology group can be viewed as the free abelian group generated by any vertex v of X. 

   To solve this problem, first, let v(1), v(2), ..., v(n) be all vertices of X. Consider a graph G with vertex set V = {v(1), v(2), ..., v(n)} such that two vertices are adjacent if and only if they are connected by an edge (or equivalently a one-dimensional simplex) in X. Since the underlying space |X| is connected, the graph G must also be connected (and finite). Therefore, it has a spanning tree T. We will work with this spanning tree in order to determine the image of the mapping considered in the homology. 

   Now, back to the homology group H_0(X) = Ker(\partial_0)/Im(\partial_1), since the "next" group in the chain after S_0 is obviously the trivial group, it can be seen that Ker(\partial_0) must be S_0 itself, which is isomorphic to Z^n. The difficulties lie on Im(\partial_1). Since the value on each (directed) edge (v(i), v(j)) is v(j) - v(i), which means that the sum of the coefficients corresponding to all the n vertices is zero, we conclude that Im(\partial_1) must be a subgroup of the subgroup of Z^n consisting of all points whose sum of coordinates is zero. More precisely, we are consider the subgroup of Z^n consisting of all n-tuples of integers (k(1), k(2), ..., k(n)) satisfying k(1) + k(2) + ... + k(n) = 0. But is every such n-tuple an element of Im(\partial_1)? This is where the spanning tree works. 

   Consider an arbitrary n-tuple (k(1), k(2),  ..., k(n)) satisfying k(1) + k(2) + ... + k(n) = 0. Set S = 0. Pick an index i such that the degree of v(i) in T is exactly one. Assume that v(i) is only adjacent to v(j) in T, we set S := S + k(i)(v(i) - v(j)) and k(j) := k(j) + k(i). Remove v(i) from T, we again obtain a tree with exactly n - 1 vertices.    

   Moreover, after such algorithms, we again obtain a new (n-1)-tuple (k(1), k(2), ..., k(i - 1), k(i + 1), k(i + 2), ..., k(n)). The sum of these n - 1 numbers is again zero, thus the previously. mentioned steps can be repeated. again in a new setting. with a smaller number of vertices. After running this algorithm n times, S becomes k(1)v(1) + k(2)v(2) + ... + k(n)v(n), which is the desired element. 

   Therefore, Im(\partial_1) is the subgroup of Z^n consisting of all n-tuples of integers (k(1), k(2), ..., k(n)) satisfying k(1) + k(2) + ... + k(n) = 0. As a result, H_0(X) = Ker(\partial_0)/Im(\partial_1) is isomorphic to the additive group of integers. 

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