Eduardo Mart'inez-Pedroza, Luis Jorge S'anchez Saldana
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引用次数: 4
摘要
设$G$和$H$是拟等距有限生成群,设$P\leq G$;是否存在一个$H$的子群$Q$(或子群的集合),其左余集大致反映$G$中$P$的左余集的几何形状?我们探索一个正答案的充分条件。本文考虑$(G,\mathcal{P})$形式的对,其中$G$是有限生成的群,$\mathcal{P}$是有限子群的集合,存在对的拟等距概念,以及子群的拟等距特征集合。如果子群属于一个气特征集合,则子群是气特征的。关于准等距刚性的文献中已经研究了不同种类的子群的气特征集合,本文列举了其中的一些,并给出了其他的例子。本文第一部分证明:如果$G$和$H$是有限生成的拟等距群,$\mathcal{P}$是$G$的子群的一个齐特征集合,那么存在$H$的子群$\mathcal{Q}$的一个集合,使得$ (G, \mathcal{P})$和$(H, \mathcal{Q})$是拟等距对。文章的第二部分研究了Bowditch引入的一对群的过滤端个数$\tilde e (G, P)$,并给出了我们的主要结果的一个应用:如果$G$和$H$是拟等距群,$P\leq G$是齐特征群,则存在$Q\leq H$使得$\tilde e (G, P) = \tilde e (H, Q)$。
Quasi-isometric rigidity of subgroups and filtered ends
Let $G$ and $H$ be quasi-isometric finitely generated groups and let $P\leq G$; is there a subgroup $Q$ (or a collection of subgroups) of $H$ whose left cosets coarsely reflect the geometry of the left cosets of $P$ in $G$? We explore sufficient conditions for a positive answer. The article consider pairs of the form $(G,\mathcal{P})$ where $G$ is a finitely generated group and $\mathcal{P}$ a finite collection of subgroups, there is a notion of quasi-isometry of pairs, and quasi-isometrically characteristic collection of subgroups. A subgroup is qi-characteristic if it belongs to a qi-characteristic collection. Distinct classes of qi-characteristic collections of subgroups have been studied in the literature on quasi-isometric rigidity, we list in the article some of them and provide other examples. The first part of the article proves: if $G$ and $H$ are finitely generated quasi-isometric groups and $\mathcal{P}$ is a qi-characteristic collection of subgroups of $G$, then there is a collection of subgroups $\mathcal{Q}$ of $H$ such that $ (G, \mathcal{P})$ and $(H, \mathcal{Q})$ are quasi-isometric pairs. The second part of the article studies the number of filtered ends $\tilde e (G, P)$ of a pair of groups, a notion introduced by Bowditch, and provides an application of our main result: if $G$ and $H$ are quasi-isometric groups and $P\leq G$ is qi-characterstic, then there is $Q\leq H$ such that $\tilde e (G, P) = \tilde e (H, Q)$.