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Certain Homology Cycles of the Independence Complex of Grids
Authors:Jakob Jonsson
Institution:1. Department of Mathematics, KTH, 10044, Stockholm, Sweden
Abstract:Let G be an infinite graph such that the automorphism group of G contains a subgroup K ?? d with the property that G/K is finite. We examine the homology of the independence complex Σ(G/I) of G/I for subgroups I of K of full rank, focusing on the case that G is the square, triangular, or hexagonal grid. Specifically, we look for a certain kind of homology cycles that we refer to as “cross-cycles,” the rationale for the terminology being that they are fundamental cycles of the boundary complex of some cross-polytope. For the special cases just mentioned, we determine the set Q(G,K) of rational numbers r such that there is a group I with the property that Σ(G/I) contains cross-cycles of degree exactly r?|G/I|?1; |G/I| denotes the size of the vertex set of G/I. In each of the three cases, Q(G,K) turns out to be an interval of the form a,b]∩?={r∈?:arb}. For example, for the square grid, we obtain the interval $\frac{1}{5},\frac{1}{4}]\cap \mathbb{Q}Let G be an infinite graph such that the automorphism group of G contains a subgroup K d with the property that G/K is finite. We examine the homology of the independence complex Σ(G/I) of G/I for subgroups I of K of full rank, focusing on the case that G is the square, triangular, or hexagonal grid. Specifically, we look for a certain kind of homology cycles that we refer to as “cross-cycles,” the rationale for the terminology being that they are fundamental cycles of the boundary complex of some cross-polytope. For the special cases just mentioned, we determine the set Q(G,K) of rational numbers r such that there is a group I with the property that Σ(G/I) contains cross-cycles of degree exactly r⋅|G/I|−1; |G/I| denotes the size of the vertex set of G/I. In each of the three cases, Q(G,K) turns out to be an interval of the form a,b]∩ℚ={r∈ℚ:arb}. For example, for the square grid, we obtain the interval \frac15,\frac14]?\mathbbQ\frac{1}{5},\frac{1}{4}]\cap \mathbb{Q}.
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