Publications Meetings The Profession Membership Programs Math Samplings Policy & Advocacy In the News About the AMS

   
Mobile Device Pairing
Green Open Access
Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

 

$ K\sb{1}$ of the compact operators is zero


Authors: L. G. Brown and Claude Schochet
Journal: Proc. Amer. Math. Soc. 59 (1976), 119-122
MSC: Primary 47B05; Secondary 58G15
MathSciNet review: 0412863
Full-text PDF Free Access

Abstract | References | Similar Articles | Additional Information

Abstract: We prove that $ {K_1}$ of the compact operators is zero. This theorem has the following operator-theoretic formulation: any invertible operator of the form (identity) $ + $ (compact) is the product of (at most eight) multiplicative commutators $ {({A_j}{B_j}A_j^{ - 1}B_j^{ - 1})^{ \pm 1}}$, where each $ {B_j}$ is of the form (identity) $ + $ (compact). The proof uses results of L. G. Brown, R. G. Douglas, and P. A. Fillmore on essentially normal operators and a theorem of A. Brown and C. Pearcy on multiplicative commutators.


References [Enhancements On Off] (What's this?)


Similar Articles

Retrieve articles in Proceedings of the American Mathematical Society with MSC: 47B05, 58G15

Retrieve articles in all journals with MSC: 47B05, 58G15


Additional Information

DOI: http://dx.doi.org/10.1090/S0002-9939-1976-0412863-0
PII: S 0002-9939(1976)0412863-0
Keywords: Compact operator, essentially normal operator, algebraic $ K$-theory, extensions of $ {C^{\ast}}$-algebras
Article copyright: © Copyright 1976 American Mathematical Society