无平方右角Artin群之间的代数嵌入与可测嵌入
Algebraic and measurable embeddings between square-free right-angled Artin groups
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中文总结 AI 辅助
本文证明无平方图上的右角Artin群之间的可测嵌入等价于子群嵌入到自由阿贝尔群的图积，并给出算法可检验的图论刻画及二维情形下的简化条件。
中文摘要 AI 辅助
设 $\Gamma,\Delta$ 为有限单纯图，并假设 $\Gamma$ 不含导出正方形。我们证明右角Artin群 $A(\Delta)$ 可测嵌入 $A(\Gamma)$ 当且仅当 $A(\Delta)$ 作为子群嵌入到 $\Gamma$ 上的自由阿贝尔群的图积中。这反过来具有一个可通过算法检验的图论刻画。若此外 $A(\Gamma)$ 和 $A(\Delta)$ 的上同调维数等于二，且 $A(\Gamma)$ 不同构于 $\mathbb{Z}\times F_n$，则我们得到 $A(\Delta)$ 可测嵌入 $A(\Gamma)$ 当且仅当它作为子群嵌入 $A(\Gamma)$。我们的证明依赖于以下代数陈述。设 $K_1,\dots,K_n$ 为无平方图 $\Gamma$（或更一般地，其扩张图 $\Gamma^{\mathrm{ext}}$）中两两不相交的团族，并设 $g_1,\dots,g_n$ 为 $A(\Gamma)$ 中分别具有抛物支撑 $A(K_1),\dots,A(K_n)$ 的元素。则由 $g_1,\dots,g_n$ 生成的 $A(\Gamma)$ 的子群是一个右角Artin群，其中仅有的关系是：若 $K_i\cup K_j$ 包含在 $\Gamma$（或 $\Gamma^{\mathrm{ext}}$）的一个团中，则 $g_i$ 与 $g_j$ 交换。
英文摘要
Let $Γ,Δ$ be finite simplicial graphs, and assume that $Γ$ has no induced squares. We prove that the right-angled Artin group $A(Δ)$ measurably embeds into $A(Γ)$ if and only if $A(Δ)$ embeds as a subgroup in a graph product of free abelian groups over $Γ$. This in turn has a graph-theoretical characterisation which can be checked algorithmically. If additionally $A(Γ)$ and $A(Δ)$ have cohomological dimension equal to two and $A(Γ)$ is not isomorphic to $\mathbb{Z}\times F_n$, we get that $A(Δ)$ measurably embeds into $A(Γ)$ if and only if it embeds as a subgroup in $A(Γ)$. Our proof relies on the following algebraic statement. Let $K_1,\dots,K_n$ be a family of pairwise disjoint cliques in the square-free graph $Γ$ (or more generally in its extension graph $Γ^{\mathrm{ext}}$), and let $g_1,\dots,g_n$ be elements of $A(Γ)$ with respective parabolic supports $A(K_1),\dots,A(K_n)$. Then the subgroup of $A(Γ)$ generated by $g_1,\dots,g_n$ is a right-angled Artin group, where the only relations impose that $g_i$ and $g_j$ commute if $K_i\cup K_j$ is contained in a clique of $Γ$ (or of $Γ^{\mathrm{ext}}$).
