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Proceedings of the American Mathematical Society

Published by the American Mathematical Society, the Proceedings of the American Mathematical Society (PROC) is devoted to research articles of the highest quality in all areas of pure and applied mathematics.

ISSN 1088-6826 (online) ISSN 0002-9939 (print)

The 2020 MCQ for Proceedings of the American Mathematical Society is 0.85.

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Lifting invertibles in von Neumann algebras
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by Robert R. Rogers PDF
Proc. Amer. Math. Soc. 113 (1991), 381-388 Request permission

Abstract:

Given $\mathcal {B}(\mathcal {H})$, the algebra of bounded operators on a separable Hilbert space $\mathcal {H}$, and $\mathcal {K}$, the ideal of compact operators, it is a well-known fact that $T$ in $\mathcal {B}(\mathcal {H})$ is a Fredholm operator if and only if $\pi (T)$ is invertible in $\mathcal {B}(\mathcal {H})$ where $\pi$ is the canonical quotient map. A natural question arises: When can a Fredholm operator be perturbed by a compact operator to obtain an invertible operator? Equivalently, when does the invertible element $\pi (T)$ lift to an invertible operator? The answer is well known: $T$ may be perturbed by a compact operator to obtain an invertible operator if and only if the Fredholm Index of $T$ is 0. In this case, the perturbation may be made as small in norm as we wish. Using the generalized Fredholm index for a von Neumann algebra developed by C. L. Olsen [3], the following generalization is obtained: let $\mathfrak {A}$ be a von Neumann algebra with norm closed ideal $\mathfrak {F}$ and canonical quotient map $\pi$. Let $T \in \mathfrak {A}$ be such that $\pi (T)$ is invertible. Then there exists $K \in \mathfrak {F}$ such that $T + K$ is invertible if and only if $\operatorname {ind} (T) = 0$.
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Additional Information
  • © Copyright 1991 American Mathematical Society
  • Journal: Proc. Amer. Math. Soc. 113 (1991), 381-388
  • MSC: Primary 46L10
  • DOI: https://doi.org/10.1090/S0002-9939-1991-1087469-1
  • MathSciNet review: 1087469