On a class of lattice ordered inverse semigroups
It was shown by \textit{T. Saito} [Proc. Japan Acad. 50, 837-838 (1974; Zbl 0338.06004)] that the free inverse semigroup \(S\) on a nonempty set cannot be totally ordered (i.e., there is no total order \(\leq\) on \(S\) such that \(a\leq b\) implies \(ac\leq bc\) and \(ca\leq cb\) for every \(c\in S\)). In the paper under review it is proved that every free inverse monoid can be lattice-ordered (i.e., there exists a lattice order such that multiplication distributes over join and meet from the left and the right). McAlister's representation theorem for \(E\)-unitary inverse semigroups is used to construct all lattice-ordered \(E\)-unitary inverse semigroups which are lexicographically ordered (i.e., there exists an order-preserving homomorphism \(\theta\) onto a partially ordered group such that \(a\theta< b\theta\) imlies \(a< b\)). The particular case where the lattice order \(\leq\) on \(S\) is natural (in the sense that \(e= ef= fe\) and \(e,f\in E(S)\) implies \(e\leq f\)) is also considered. Furthermore, the compatible lattice orderings on the free monogenic inverse monoid are specified: there are, up to conjugacy, precisely two such orderings. Also, it is shown that an inverse semigroup is an idempotent separating homomorphic image of a naturally lattice ordered inverse semigroup if and only if its idempotents form a distributive lattice under the usual ordering. Finally, it is proved that a lattice-ordered \(E\)-unitary inverse semigroup which is lexicographically ordered, is a distributive lattice under the imposed partial order if and only if its set of idempotents forms a distributive lattice.
scientific article; zbMATH DE number 1512977