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J-subspace lattices and subspace M-bases

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The class of J-lattices was defined in the second author's thesis. A subspace lattice on a Banach space X which is also a J-lattice is called a J- subspace lattice}, abbreviated JSL. Every atomic Boolean subspace lattice, abbreviated ABSL, is a JSL. Any commutative JSL on Hilbert space, as well as any JSL on finite-dimensional space, is an ABSL. For any JSL ℒ both LatAlg ℒ and $ℒ^⊥$ (on reflexive space) are JSL's. Those families of subspaces which arise as the set of atoms of some JSL on X are characterised in a way similar to that previously found for ABSL's. This leads to a definition of a subspace M-basis of X which extends that of a vector M-basis. New subspace M-bases arise from old ones in several ways. In particular, if ${M_γ}_{γ∈Γ}$ is a subspace M-basis of X, then (i) ${(M_γ')^⊥}_{γ∈Γ}$ is a subspace M-basis of $V_{γ∈Γ}^(M_γ')^⊥$, (ii) ${K_γ}_{γ∈Γ}$ is a subspace M-basis of $V_{γ ∈Γ}^K_γ$ for every family {K_γ}_{γ∈Γ}$ of subspaces satisfying $(0)≠ K_γ⊆M_γ(γ ∈Γ)$ and (iii) if X is reflexive, then ${⋂_{β ≠ γ}^M_β'}_{γ∈Γ}$ is a subspace M-basis of X. (Here $M_γ'$ is given by $M_γ' = V_{β ≠ γ}^M_β$.)
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The decomposability of operators relative to two subspaces

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Let M and N be nonzero subspaces of a Hilbert space H satisfying M ∩ N = {0} and M ∨ N = H and let T ∈ ℬ(H). Consider the question: If T leaves each of M and N invariant, respectively, intertwines M and N, does T decompose as a sum of two operators with the same property and each of which, in addition, annihilates one of the subspaces? If the angle between M and N is positive the answer is affirmative. If the angle is zero, the answer is still affirmative for finite rank operators but there are even trace class operators for which it is negative. An application gives an alternative proof that no distance estimate holds for the algebra of operators leaving M and N invariant if the angle is zero, and an analogous result is obtained for the set of operators intertwining M and N.
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