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math.RA — Rings and Algebras

Maximal sublattices of finite semidistributive lattices: the complement need not be an interval

Contributed by Deep Bhattacharjee

Adams, Freese, Nation and Schmid proved that the complement of a maximal sublattice of a finite bounded lattice is an interval, and asked whether the same holds for finite semidistributive lattices; Adaricheva, Mata, Silberger and Zamojska-Dzienio conjectured that it does. It does not. In a 26-element semidistributive sublattice of the square of a 14-element semidistributive lattice, the diagonal is a maximal sublattice whose complement has two minimal elements.

A mix of human-written and AI-generated textHuman understanding: some partsbounded latticeformal verificationintervalmaximal sublatticesemidistributive lattice

Oct 23, 2026PDF ↗View work →

hexagon:2610.00495v1

math.RA — Rings and Algebras

Maximal sublattices of convex geometries: four counterexamples

Contributed by Deep Bhattacharjee

Adaricheva, Mata, Silberger and Zamojska-Dzienio conjectured that in a finite convex geometry, and more generally in a finite join-semidistributive lattice, the complement of a maximal sublattice is a union of intervals with a common least element, and proved this for convex dimension two. We give convex geometries of convex dimension four and three in which such a complement has two and three minimal elements, so their theorem does not extend. The same examples refute their Hypothesis 4 and one half of their Question 2; an eleven-element convex geometry refutes the other half, and a ten-element lattice their Hypothesis 3.

A mix of human-written and AI-generated textHuman understanding: some partsconvex geometryformal verificationjoin-semidistributive latticemaximal sublatticemeet-irreducible element

Oct 23, 2026PDF ↗View work →

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Maximal sublattices of finite semidistributive lattices: the complement need not be an interval

Maximal sublattices of convex geometries: four counterexamples

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