# Journal of Operator Theory

Volume 46, Issue 2, Fall 2001 pp. 355-380.

Pentagon subspace lattices on Banach spaces**Authors**: A. Katavolos (1) , M.S. Lambrou (2), and W.E. Longstaff (3)

**Author institution:**(1) Department of Mathematics, University of Athens, Panepistimiopolis, 15784 Athens, Greece

(2) Department of Mathematics, University of Crete, 71409 Iraklion, Crete, Greece

(3) Dept. of Mathematics and Statistics, University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australia

**Summary:**If $K, L$ and $M$ are (closed) subspaces of a Banach space $X$ satisfying $K \cap M =$ (0), $K\vee L = X$ and $L \subset M$, then ${\cal P} = \left\{(0), K, L, M, X\right\}$ is a {\it pentagon subspace lattice on $X$}. If ${\cal P}_1$ and ${\cal P}_2$ are pentagons, every (algebraic) isomorphism $\varphi : \Alg{\cal P}_1 \rightarrow \Alg{\cal P}_2$ is quasi-spatial. The SOT-closure of the fin- ite rank subalgebra of $\Alg{\cal P}$ is $\{T \in \Alg{\cal P}: T (M) \subseteq L\}$. On separable Hilbert space $H$ every positive, injective, non-invertible operator $A$ and every non-zero subspace $M$ satisfying $M \cap\Ran(A) = (0)$ give rise to a pentagon ${\cal P}(A;M).$ $\Alg{\cal P}(A;M)$ and $\Alg{\cal P}(B;N)$ are spatially isomorphic if and onl y if $T$ $\Ran(A) = \Ran(B)$ and $T(M) = N$ for an invertible operator $T\in B(H)$. If ${\cal A}(A)$ is the set of operators leaving Ran$(A)$ invariant, every isomorphism $\varphi : {\cal A}(A) \rightarrow {\cal A}(B)$ is implemented by an invertible operator $T$ satisfying $T\Ran(A) = \Ran (B)$.

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