analysis / Operator algebras; uniform tracial completions

Schafhauser-Tikuisis-White Problem XXI: tracial real rank zero implies uniform property Gamma

Theorem 1.1: for AA unital, separable, simple, infinite-dimensional, nuclear and stably finite with T(A)≠∅T(A)\neq\emptyset, if BωB^\omega has real rank zero then there is a projection p∈Bω∩B′p\in B^\omega\cap B' with λ(px)=λ(x)/2\lambda(px)=\lambda(x)/2 for all x∈Bx\in B and all limit traces λ\lambda, so AA has uniform property Γ\Gamma; it also identifies all traces on BB. It does not show that real rank zero of BωB^\omega holds in general, and Toms (2026) has shown nuclearity alone does not give uniform property Γ\Gamma.

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analysisSep 23, 2026Significance 15/100Registry: unreviewed

Schafhauser-Tikuisis-White Problem XXI: tracial real rank zero implies uniform property Gamma

Prior state unknown→proved

Theorem 1.1: for AA unital, separable, simple, infinite-dimensional, nuclear and stably finite with T(A)≠∅T(A)\neq\emptyset, if BωB^\omega has real rank zero then there is a projection p∈Bω∩B′p\in B^\omega\cap B' with λ(px)=λ(x)/2\lambda(px)=\lambda(x)/2 for all x∈Bx\in B and all limit traces λ\lambda, so AA has uniform property Γ\Gamma; it also identifies all traces on BB. It does not show that real rank zero of BωB^\omega holds…

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Theorem 1.1: for AA unital, separable, simple, infinite-dimensional, nuclear and stably finite with T(A)≠∅T(A)\neq\emptyset, if BωB^\omega has real rank zero then there is a projection p∈Bω∩B′p\in B^\omega\cap B' with λ(px)=λ(x)/2\lambda(px)=\lambda(x)/2 for all x∈Bx\in B and all limit traces λ\lambda, so AA has uniform property Γ\Gamma; it also identifies all traces on BB. It does not show that real rank zero of BωB^\omega holds in general, and Toms (2026) has shown nuclearity alone does not give uniform property Γ\Gamma.

Theorem 1.1: for AA unital, separable, simple, infinite-dimensional, nuclear and stably finite with T(A)≠∅T(A)\neq\emptyset, if BωB^\omega has real rank zero then there is a projection p∈Bω∩B′p\in B^\omega\cap B' with λ(px)=λ(x)/2\lambda(px)=\lambda(x)/2 for all x∈Bx\in B and all limit traces λ\lambda, so AA has uniform property Γ\Gamma; it also identifies all traces on BB. It does not show that real rank zero of BωB^\omega holds in general, and Toms (2026) has shown nuclearity alone does not give uniform property Γ\Gamma.

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Schafhauser-Tikuisis-White Problem XXI: tracial real rank zero implies uniform property Gamma — Mathematical Frontier Network