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 ) } ≼ { E T ( u ) } \{ {E_S}(u)\} \preccurlyeq \{ {E_T}(u)\} if E S ( u ) ≧ E T ( u ) , u ∈ R {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 α ( u ) = ∧ E S α ( u ) {E_{V{S_\alpha }}}(u) = \wedge {E_{{S_\alpha }}}(u) for all u u contained in R R . This order is called spectral order and agrees with the usual order on commutative subalgebras. For positive operators, S S is greater than or equal to T T in spectral order if and only if S n {S^n} is greater than or equal to T n {T^n} in the usual order for all n ≧ 1 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 ≽ T S \succcurlyeq T does not imply that S + C ≽ T + C S + C \succcurlyeq T + C .