半inite von Neumann 代数所属可测算子的迹和积分换元子

Pub Date : 2024-05-29 DOI:10.1134/s0037446624030030
A. M. Bikchentaev
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引用次数: 0

摘要

假设 \( \tau \)是冯-诺依曼代数 \( {\mathcal{M}} \)上的忠实正态半无限迹, \( I \)是 \( \mathcal{M}} \)的单元, \( S({\mathcal{M}}、\是可测算子的代数,并且 L_{1}({\mathcal{M}}, \tau) 是可解算子的巴纳赫空间。我们提出了对普特南定理(1951)的以下概括的新证明:在 S({\mathcal{M}},\tau)中,没有任何与 A ([A^{*},A] )的正自互映器在 ({\mathcal{M}}\)中是可逆的。如果\( \tau \)是无穷的,那么在S({\mathcal{M}},\tau)\([A^{*},A] \)中没有一个与\( A\(在S({\mathcal{M}},\tau)\)的正自互调器可以是形式为\( \lambda I+K \)的,其中\( \lambda \)是一个非零复数,而\( K \)是一个\(\tau \)-紧凑的算子。鉴于( A,B在S({\mathcal{M}},\tau)中)与( [A,B]在L_{1}({\mathcal{M}},\tau)中)我们寻求的条件是( ( (tau([A,B])=0)。If\( X\in S({\mathcal{M}},\tau)\)and\( Y=Y^{3}\in{\mathcal{M}}\)with\( [X,Y]\in L_{1}({\mathcal{M}},\tau)\)then\( \tau([X,Y])=0\).If\( A^{2}=A\in S({\mathcal{M}},\tau)\)and\( [A^{*},A]\in L_{1}({\mathcal{M}},\tau)\)then\( \tau([A^{*},A])=0\).如果一个局部等势\( U \)位于\( {\mathcal{M}} \)并且\( U^{n}=0 \)对于某个整数\( n\geq 2 \),那么\( U^{n-1} \)是一个换元器并且\( U^{n-1}\in L_{1}({\mathcal{M}}、\意味着(U^{n-1} =0).
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The Trace and Integrable Commutators of the Measurable Operators Affiliated to a Semifinite von Neumann Algebra

Assume that \( \tau \) is a faithful normal semifinite trace on a von Neumann algebra \( {\mathcal{M}} \), \( I \) is the unit of \( \mathcal{M} \), \( S({\mathcal{M}},\tau) \) is the \( * \)-algebra of \( \tau \)-measurable operators, and \( L_{1}({\mathcal{M}},\tau) \) is the Banach space of \( \tau \)-integrable operators. We present a new proof of the following generalization of Putnam’s theorem (1951): No positive self-commutator \( [A^{*},A] \) with \( A\in S({\mathcal{M}},\tau) \) is invertible in \( {\mathcal{M}} \). If \( \tau \) is infinite then no positive self-commutator \( [A^{*},A] \) with \( A\in S({\mathcal{M}},\tau) \) can be of the form \( \lambda I+K \), where \( \lambda \) is a nonzero complex number and \( K \) is a \( \tau \)-compact operator. Given \( A,B\in S({\mathcal{M}},\tau) \) with \( [A,B]\in L_{1}({\mathcal{M}},\tau) \) we seek for the conditions that \( \tau([A,B])=0 \). If \( X\in S({\mathcal{M}},\tau) \) and \( Y=Y^{3}\in{\mathcal{M}} \) with \( [X,Y]\in L_{1}({\mathcal{M}},\tau) \) then \( \tau([X,Y])=0 \). If \( A^{2}=A\in S({\mathcal{M}},\tau) \) and \( [A^{*},A]\in L_{1}({\mathcal{M}},\tau) \) then \( \tau([A^{*},A])=0 \). If a partial isometry \( U \) lies in \( {\mathcal{M}} \) and \( U^{n}=0 \) for some integer \( n\geq 2 \) then \( U^{n-1} \) is a commutator and \( U^{n-1}\in L_{1}({\mathcal{M}},\tau) \) implies that \( \tau(U^{n-1})=0 \).

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