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Proceedings of the American Mathematical Society

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



The selfadjoint operators of a von Neumann algebra form a conditionally complete lattice

Author: Milton Philip Olson
Journal: Proc. Amer. Math. Soc. 28 (1971), 537-544
MSC: Primary 46.65
MathSciNet review: 0276788
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Abstract: The bounded resolutions of the identity in a von Neumann algebra can be ordered by $ \{ {E_S}(u)\} \preccurlyeq \{ {E_T}(u)\} $ if $ {E_S}(u) \geqq {E_T}(u),u \in R$. The selfadjoint operators in the algebra are partially ordered by this relation and are shown to form a conditionally complete lattice. The lattice operations are (essentially) defined by $ {E_{V{S_\alpha }}}(u) = \wedge {E_{{S_\alpha }}}(u)$ for all $ u$ contained in $ R$. This order is called spectral order and agrees with the usual order on commutative subalgebras. For positive operators, $ S$ is greater than or equal to $ T$ in spectral order if and only if $ {S^n}$ is greater than or equal to $ {T^n}$ in the usual order for all $ n \geqq 1$. Kadison's well-known counterexample is shown to fail. The operator lattice defined by spectral order differs from a vector lattice in the fact that $ S \succcurlyeq T$ does not imply that $ S + C \succcurlyeq T + C$.

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Keywords: von Neumann algebra, resolution of the identity, projection lattice, conditionally complete lattice, spectral integral
Article copyright: © Copyright 1971 American Mathematical Society