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On the homology groups of clique complexes of strongly re...
[Submitted on 25 Jun 2026] · 2026-06-26 · via math.CO updates on arXiv.org

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Abstract:In this paper, we study the first homology groups of clique complexes of strongly regular graphs over arbitrary fields and prove that most of these graphs have trivial first clique homology groups. Using Neumaier's classification of strongly regular graphs with smallest integral eigenvalue, we show that a non-vanishing first homology group may occur only in a short collection of cases: the Petersen graph, the Shrikhande graph, the complete bipartite graphs, the conference graphs on at most $255$ vertices, the lattice graphs, and the exceptional families $E_m$ in Neumaier's classification of strongly regular graphs with smallest adjacency eigenvalue $-m$, for some integer $m \geq 3$. Let $\text{Cl}(G)$ denote the clique complex of a graph $G$, $H_i(\text{Cl}(G),\mathbb{F})$ be the $i$-th homology group of $\text{Cl}(G)$ over the field $\mathbb{F}$, for some $i\geq 1$, and $\lambda_{min}(G)$ denote the smallest eigenvalue of the adjacency matrix of $G$. We prove that if $(G_n)_{n\geq 1}$ is an infinite family of pairwise distinct strongly regular graphs and $(\mathbb{F}_n)_{n\geq 1}$ is a sequence of fields such that $H_1(\text{Cl}(G_n), \mathbb{F}_n)\not=0$ for every $n$, then either $G_n$ is a lattice graph for infinitely many $n$, or $\lim_{n\rightarrow +\infty} \lambda_{\min}(G_n)=-\infty$.
For Latin square graphs, we determine the clique homologies over arbitrary fields and show that if $G$ is the strongly regular graph associated with a Latin square $M$ of order $n \geq 5$ and $\mathbb{F}$ is any field, then $H_i(\text{Cl}(G),\mathbb{F})=0$ for $i=1$ or $i \geq 3$, and $\dim H_2(\text{Cl}(G),\mathbb{F})=(n-1)^3-I(M),$ where $I(M)$ is the number of $2 \times 2$ Latin subsquares or intercalates in $M$.

Submission history

From: Mutasim Mim [view email]
[v1] Thu, 25 Jun 2026 17:41:10 UTC (40 KB)