The Trace and Integrable Commutators of the Measurable Operators Affiliated to a Semifinite von Neumann Algebra

Pub Date : 2024-05-29 DOI:10.1134/s0037446624030030
A. M. Bikchentaev
{"title":"The Trace and Integrable Commutators of the Measurable Operators Affiliated to a Semifinite von Neumann Algebra","authors":"A. M. Bikchentaev","doi":"10.1134/s0037446624030030","DOIUrl":null,"url":null,"abstract":"<p>Assume that <span>\\( \\tau \\)</span> is a faithful normal semifinite trace\non a von Neumann algebra <span>\\( {\\mathcal{M}} \\)</span>, <span>\\( I \\)</span> is the unit of <span>\\( \\mathcal{M} \\)</span>,\n<span>\\( S({\\mathcal{M}},\\tau) \\)</span> is the <span>\\( * \\)</span>-algebra of <span>\\( \\tau \\)</span>-measurable operators,\nand <span>\\( L_{1}({\\mathcal{M}},\\tau) \\)</span> is the Banach space of <span>\\( \\tau \\)</span>-integrable operators.\nWe present a new proof of the following generalization\nof Putnam’s theorem (1951):\nNo positive self-commutator\n<span>\\( [A^{*},A] \\)</span>\nwith\n<span>\\( A\\in S({\\mathcal{M}},\\tau) \\)</span>\nis invertible in <span>\\( {\\mathcal{M}} \\)</span>.\nIf <span>\\( \\tau \\)</span>\nis infinite\nthen no positive self-commutator\n<span>\\( [A^{*},A] \\)</span>\nwith\n<span>\\( A\\in S({\\mathcal{M}},\\tau) \\)</span>\ncan be of the form\n<span>\\( \\lambda I+K \\)</span>,\nwhere <span>\\( \\lambda \\)</span>\nis a nonzero complex number and <span>\\( K \\)</span>\nis a <span>\\( \\tau \\)</span>-compact operator.\nGiven\n<span>\\( A,B\\in S({\\mathcal{M}},\\tau) \\)</span>\nwith\n<span>\\( [A,B]\\in L_{1}({\\mathcal{M}},\\tau) \\)</span>\nwe seek for the conditions that\n<span>\\( \\tau([A,B])=0 \\)</span>.\nIf\n<span>\\( X\\in S({\\mathcal{M}},\\tau) \\)</span>\nand\n<span>\\( Y=Y^{3}\\in{\\mathcal{M}} \\)</span>\nwith\n<span>\\( [X,Y]\\in L_{1}({\\mathcal{M}},\\tau) \\)</span>\nthen\n<span>\\( \\tau([X,Y])=0 \\)</span>.\nIf\n<span>\\( A^{2}=A\\in S({\\mathcal{M}},\\tau) \\)</span>\nand\n<span>\\( [A^{*},A]\\in L_{1}({\\mathcal{M}},\\tau) \\)</span>\nthen\n<span>\\( \\tau([A^{*},A])=0 \\)</span>.\nIf a partial isometry <span>\\( U \\)</span>\nlies in <span>\\( {\\mathcal{M}} \\)</span>\nand\n<span>\\( U^{n}=0 \\)</span>\nfor some integer\n<span>\\( n\\geq 2 \\)</span>\nthen <span>\\( U^{n-1} \\)</span>\nis a commutator\nand\n<span>\\( U^{n-1}\\in L_{1}({\\mathcal{M}},\\tau) \\)</span>\nimplies that\n<span>\\( \\tau(U^{n-1})=0 \\)</span>.</p>","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2024-05-29","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.1134/s0037446624030030","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 0

Abstract

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 \( \tau(U^{n-1})=0 \).

分享
查看原文
半inite von Neumann 代数所属可测算子的迹和积分换元子
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).
本文章由计算机程序翻译,如有差异,请以英文原文为准。
求助全文
约1分钟内获得全文 求助全文
×
引用
GB/T 7714-2015
复制
MLA
复制
APA
复制
导出至
BibTeX EndNote RefMan NoteFirst NoteExpress
×
提示
您的信息不完整,为了账户安全,请先补充。
现在去补充
×
提示
您因"违规操作"
具体请查看互助需知
我知道了
×
提示
确定
请完成安全验证×
copy
已复制链接
快去分享给好友吧!
我知道了
右上角分享
点击右上角分享
0
联系我们:info@booksci.cn Book学术提供免费学术资源搜索服务,方便国内外学者检索中英文文献。致力于提供最便捷和优质的服务体验。 Copyright © 2023 布克学术 All rights reserved.
京ICP备2023020795号-1
ghs 京公网安备 11010802042870号
Book学术文献互助
Book学术文献互助群
群 号:481959085
Book学术官方微信