某些 C*-代数的最大超高频子代数

Nasser Golestani, Saeid Maleki Oche
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引用次数: 0

摘要

动力学系统中的一个著名结果断言,任何康托最小系统 $(X,T)$ 都有一个最大有理等连续因子 $(Y,S)$,它实际上是一个里程表,并实现了 $(X,T)$ 的 $K_0$ 群的有理子群,即 $\mathbb{Q}(K_0(X,T), 1) \cong K^0(Y,S)$ 。我们将介绍最大超高频子代数的概念,并利用它得到这一结果的 C* 代数对映体。我们说,如果一个单素 C* 代数 $A$ 的超高频子代数 $B$ 包含 $A$ 的单元,那么它就是最大超高频子代数。我们证明,如果$K_0(A)$是不穿孔的并且具有一定的$K_0$提升性质,那么$B$是存在的并且是唯一同构的,特别是,所有三秩为零的简单可分离的单C*代数和所有其$K_0$群是不穿孔的单基希伯格代数都有一个最大超高频子代数。并不是每个单素 C* 代数都有最大的超高频子代数,例如,单素无穷积 $\mathrm{M}_2 \ast_{r}\3$。作为应用,我们给出了阶单位 $u$ 的任意维群 $G$ 的有理子群 $\mathbb{Q}(G,u)$ 的 C* 代数实现,即有一个简单的单素 AF 代数(和一个单素基希贝格代数)$A$ 和一个最大的超高频子代数 $B$,使得 $(G,u)\cong (K_0(A), [1]_0)$ 和 $\mathbb{Q}(G,u)\cong K_0(B)$.
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Maximal UHF subalgebras of certain C*-algebras
A well-known result in dynamical systems asserts that any Cantor minimal system $(X,T)$ has a maximal rational equicontinuous factor $(Y,S)$ which is in fact an odometer, and realizes the rational subgroup of the $K_0$-group of $(X,T)$, that is, $\mathbb{Q}(K_0(X,T), 1) \cong K^0(Y,S)$. We introduce the notion of a maximal UHF subalgebra and use it to obtain the C*-algebraic alonog of this result. We say a UHF subalgebra $B$ of a unital C*-algebra $A$ is a maximal UHF subalgebra if it contains the unit of $A$ any other such C*-subalgebra embeds unitaly into $B$. We prove that if $K_0(A)$ is unperforated and has a certain $K_0$-lifting property, then $B$ exists and is unique up to isomorphism, in particular, all simple separable unital C*-algebras with tracial rank zero and all unital Kirchberg algebras whose $K_0$-groups are unperforated, have a maximal UHF subalgebra. Not every unital C*-algebra has a maximal UHF subalgebra, for instance, the unital universal free product $\mathrm{M}_2 \ast_{r} \mathrm{M}_3$. As an application, we give a C*-algebraic realization of the rational subgroup $\mathbb{Q}(G,u)$ of any dimension group $G$ with order unit $u$, that is, there is a simple unital AF algebra (and a unital Kirchberg algebra) $A$ with a maximal UHF subalgebra $B$ such that $(G,u)\cong (K_0(A), [1]_0)$ and and $\mathbb{Q}(G,u)\cong K_0(B)$.
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