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Lifting invertibles in von Neumann algebras


Author: Robert R. Rogers
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
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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

DOI: https://doi.org/10.1090/S0002-9939-1991-1087469-1
Keywords: von Neumann Algebra, Fredholm, generalized dimension function, generalized index
Article copyright: © Copyright 1991 American Mathematical Society

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