Lifting invertibles in von Neumann algebras
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- by Robert R. Rogers
- Proc. Amer. Math. Soc. 113 (1991), 381-388
- DOI: https://doi.org/10.1090/S0002-9939-1991-1087469-1
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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 $\operatorname {ind} (T) = 0$.References
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Bibliographic 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