tracial topological rank
Recently Published Documents


TOTAL DOCUMENTS

14
(FIVE YEARS 1)

H-INDEX

6
(FIVE YEARS 0)

2011 ◽  
Vol 54 (11) ◽  
pp. 2295-2307 ◽  
Author(s):  
XiaoChun Fang ◽  
YiLe Zhao

2005 ◽  
Vol 48 (3) ◽  
pp. 673-690
Author(s):  
Huaxin Lin

AbstractWe introduce the notion of tracial equivalence for $C^*$-algebras. Let $A$ and $B$ be two unital separable $C^*$-algebras. If they are tracially equivalent, then there are two sequences of asymptotically multiplicative contractive completely positive linear maps $\phi_n:A\to B$ and $\psi_n:B\to A$ with a tracial condition such that $\{\phi_n\circ\psi_n\}$ and $\{\psi_n\circ\phi_n\}$ are tracially approximately inner. Let $A$ and $B$ be two unital separable simple $C^*$-algebras with tracial topological rank zero. It is proved that $A$ and $B$ are tracially equivalent if and only if $A$ and $B$ have order isomorphic ranges of tracial states. For the Cantor minimal systems $(X_1,\sigma_1)$ and $(X_2,\sigma_2)$, using a result of Giordano, Putnam and Skau, we show that two such dynamical systems are (topological) orbit equivalent if and only if the associated crossed products $C(X_1)\times_{\sigma_1}\mathbb{Z}$ and $C(X_2)\times_{\sigma_2}\mathbb{Z}$ are tracially equivalent.


2005 ◽  
Vol 218 (2) ◽  
pp. 475-494 ◽  
Author(s):  
Huaxin Lin ◽  
Hiroyuki Osaka

2004 ◽  
Vol 94 (1) ◽  
pp. 125 ◽  
Author(s):  
Shanwen Hu ◽  
Huaxin Lin ◽  
Yifeng Xue

Let $0\to \mathcal J\to \mathcal A\to \mathcal A / \mathcal J\to 0$ be a short exact sequence of separable $C^*$-algebras. We introduce the notion of tracially quasidiagonal extension. Suppose that $\mathcal J$ and $\mathcal A/J$ have tracial topological rank zero. We prove that if $(\mathcal A, \mathcal J)$ is tracially quasidiagonal, then $\mathcal A$ has tracial topological rank zero.


Sign in / Sign up

Export Citation Format

Share Document