[Paper Review] Homology of subgroups of right-angled Artin groups
This paper computes the integral homology of coordinate subgroups $N_{\rho}$ in right-angled Artin groups as modules over $\mathbb{Z}[\mathbb{Z}^m]$, using exterior Stanley-Reisner rings and combinatorial commutative algebra. The key result establishes a duality isomorphism between homology and Ext modules over the polynomial ring $S$, showing that for Cohen-Macaulay clique complexes, $H_q(\mathcal{Z}_K(S^1), \mathbb{k}[\mathbb{Z}^m]) \cong \mathrm{Ext}^{d+1-q}_S(F_K, R)$, with cohomology vanishing outside degree $d+1$.
We describe the (co)homology of a certain family of normal subgroups of right-angled Artin groups that contain the commutator subgroup, as modules over the quotient group. We do so in terms of (skew) commutative algebra of squarefree monomial ideals.
Motivation & Objective
- To describe the homology of normal subgroups $N_{\rho}$ of right-angled Artin groups as modules over $\mathbb{Z}[\mathbb{Z}^m]$.
- To relate the cohomology of these subgroups to the simplicial topology of the clique complex $K_\Gamma$ via combinatorial commutative algebra.
- To establish a duality isomorphism between homology and Ext modules when $K_\Gamma$ is Cohen-Macaulay.
- To compute the Krull dimension of each homology module $H_p(N_{\rho}, \mathbb{Z})$ using the exterior Stanley-Reisner ring.
Proposed method
- Use the partial product complex $\mathcal{Z}_K(S^1)$ to model the classifying space of the right-angled Artin group $G_\Gamma$, with $K$ the clique complex of $\Gamma$.
- Represent the homology of the coordinate subgroup $N_{\rho}$ as the homology of the universal abelian cover $\mathcal{Z}_K(S^1) \times_{\mathbb{Z}^n} \mathbb{Z}^m$.
- Model the group ring $\mathbb{Z}[\mathbb{Z}^m]$ as a quotient of the exterior algebra $E$ and use the Koszul complex to compute cohomology.
- Apply Bernstein-Gelfand-Gelfand duality to relate $H_q(\mathcal{Z}_K(S^1), \mathbb{k}[\mathbb{Z}^m])$ to $\mathrm{Ext}^{d+1-q}_S(F_K, R)$, where $F_K$ is the Cartan complex module.
- Use the Eagon–Reiner theorem to link the Cohen-Macaulay property of $K$ to the linear resolution of the dual Stanley-Reisner ideal $I_{K^\star}$.
- Apply the Künneth formula and induction on $m$ to generalize results from the Bestvina–Brady case to arbitrary coordinate homomorphisms.
Experimental results
Research questions
- RQ1How can the homology of coordinate subgroups $N_{\rho}$ in right-angled Artin groups be described as modules over $\mathbb{Z}[\mathbb{Z}^m]$?
- RQ2What is the relationship between the cohomology of $N_{\rho}$ and the simplicial topology of the clique complex $K_\Gamma$?
- RQ3Under what conditions does the cohomology $H^q(G_\Gamma, \mathbb{k}[G_\Gamma^{ab}])$ vanish outside a single degree?
- RQ4How does the Cohen-Macaulay property of $K_\Gamma$ influence the structure of the homology of $N_{\rho}$?
- RQ5Can the homology of $N_{\rho}$ be computed via a duality isomorphism involving Ext and Tor functors over the polynomial ring $S$?
Key findings
- For a Cohen-Macaulay complex $K$ of dimension $d$, the cohomology $H^q(G_\Gamma, \mathbb{k}[G_\Gamma^{ab}])$ vanishes for $q \leq m - n + d$ and $q > d + 1$, with non-zero cohomology only in degree $d+1$.
- The homology $H_q(\mathcal{Z}_K(S^1), \mathbb{k}[\mathbb{Z}^m])$ is isomorphic to $\mathrm{Ext}^{d+1-q}_S(F_K, R)$ as $S$-modules when $K$ is Cohen-Macaulay of dimension $d$.
- The cohomology with compact support $H^q_c(\mathcal{Z}_K(S^1), \mathbb{k}[\mathbb{Z}^m])$ is isomorphic to $\mathrm{Tor}^{S}_{d+1-q}(F_K, R)$, establishing a dual description.
- The Krull dimension of each module $H_p(N_{\rho}, \mathbb{Z})$ can be computed via the exterior Stanley-Reisner ring of the clique complex $K_\Gamma$.
- When $K$ is a homology sphere (Gorenstein*), $F_K \cong I_{K^\star}$ as $S$-modules, simplifying the duality isomorphism.
- The result recovers and extends Leary and Saadetoğlu’s finite-dimensionality criterion for the Bestvina–Brady group, showing that $H^q(\mathcal{Z}_K(S^1), \mathbb{k}[\mathbb{Z}])$ is finite-dimensional iff $\widetilde{H}^{q-1}(K, \mathbb{k}) = 0$.
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This review was created by AI and reviewed by human editors.