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

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 $S + C \succcurlyeq T + C$.