Étale groupoids arising from products of shifts of finite type

Abstract

Two conjectures about homology groups, K-groups and topological full groups of minimalétale groupoids on Cantor sets are formulated. We verify these conjectures for many examples ofétale groupoids including products ofétale groupoids arising from one-sided shifts of finite type. Furthermore, we completely determine when these product groupoids are mutually isomorphic. Also, the abelianization of their topological full groups are computed. They are viewed as generalizations of the higher dimensional Thompson groups.

Topological full groups of ample groupoids with applications to graph algebras

“…For finite, strongly connected graphs, this was proved directly, using socalled zipper actions, by Matui in [Mat15b]. Later, in [Mat16a], Matui proved that for any finite, strongly connected graph E, G E embeds into G E 2 . In fact, he proved even more, namely that G E 2 could be replaced by any groupoid with similar properties (see [Mat16a,Proposition 5.14] for the details).…”

Homology of odometers

“…Let G be a second countable, ample, essentially principal and minimal groupoid with unit space homeomorphic to the Cantor set. Matui conjectured in [15] (AH conjecture) that there is an exact sequence…”

“…The HK conjecture has been verified for several classes of groupoids (see [19] and [10] for recent developments). The second conjecture (AH conjecture) relates the abelianization of the topological full group of G with the two first homology groups of G. It has been verified for principal, almost finite groupoids, and groupoids arising from products of one-sided shifts of finite type ( [15]).…”

Tensor Products of Steinberg Algebras

“…Since every unit in O n1 × · · · × O n k has isotropy group Z k and every unit in O m1 × · · · × O m l has isotropy group Z l , deduce that k = l. The adjacency matrix of the directed graph with one vertex and n edges is the 1 × 1 matrix [n]. The Bowen-Franks group associated to [n] is defined (see [25,28]) as BF([n]) = Z/(n − 1)Z. Therefore, BF([n k ]) ∼ = BF([m k ]) if and only if n k = m k .…”

“…Therefore, BF([n k ]) ∼ = BF([m k ]) if and only if n k = m k . By [28,Theorem 5.12], O n1 × · · · × O n k ∼ = O m1 × · · · × O m l implies (n 1 , . .…”