拟对角迹不构成面
Quasidiagonal traces need not form a face
- University of Toronto（多伦多大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
本文构造了一个可分单位剩余有限维C*-代数，其拟对角迹不构成迹态空间的面，通过加权表示阶梯连接两个副本并证明拟对角迹对两副本等权。
AI中文摘要：
设 \\(G\\) 为具有Kazhdan性质\\((T)\\)的无限剩余有限可数离散群。假设其有限维不可约酉表示具有穷举排序，且维数非降趋于无穷，相邻维数之比有界。基于此数据，我们构造一个可分的单位剩余有限维\\(C^*\\)-代数，其具有忠实的拟对角迹态 \\(\tau=\frac14\mu_1+\frac34\mu_2\\)，其中 \\(\mu_1\\) 不是拟对角的。对于所得代数，拟对角迹不构成迹态空间的面。该构造包含两个加权表示阶梯的副本，在其顶层连接。我们证明每个拟对角迹对两个副本赋予相等的权重。证明结合了均匀Kazhdan谱投影、舍入有限秩压缩的精确秩，以及控制所有表示块的加权Hilbert--Schmidt比较。
英文摘要：
Let \(G\) be an infinite residually finite countable discrete group with Kazhdan's property~\((T)\). Assume that its finite-dimensional irreducible unitary representations admit an exhaustive ordering with nondecreasing dimensions tending to infinity and bounded consecutive dimension ratios. From this data we construct a separable unital residually finite-dimensional \(C^*\)-algebra with a faithful quasidiagonal tracial state \(τ=\frac14μ_1+\frac34μ_2\), where \(μ_1\) is not quasidiagonal. For the resulting algebra, quasidiagonal traces do not form a face of the tracial state space. The construction has two copies of a weighted representation ladder joined at their top levels. We prove that every quasidiagonal trace gives equal weight to the two copies. The proof combines uniform Kazhdan spectral projections, exact ranks of rounded finite-rank compressions, and a weighted Hilbert--Schmidt comparison that controls all representation blocks.
