Congruences of finite semidistributive lattices

J. B. Nation

doi:10.56754/0719-0646.2603.443

Authors

J. B. Nation Department of Mathematics, University of Hawaii, Honolulu, HI 96822, USA. https://orcid.org/0000-0001-5959-3304

DOI:

https://doi.org/10.56754/0719-0646.2603.443

Keywords:

Distributive lattice, semidistributive lattice, congruence lattice

Abstract

We show that there are finite distributive lattices that are not the congruence lattice of any finite semidistributive lattice. For \(0 \leq k \leq 2\), the distributive lattice \((\mathbf{B}_k)_{++} = \mathbf{2} + \mathbf{B}_k\), where \(\mathbf{B}_k\) denotes the boolean lattice with \(k\) atoms, is not the congruence lattice of any finite semidistributive lattice. Neither can these lattices be a filter in the congruence lattice of a finite semidistributive lattice. However, each \((\mathbf{B}_k)_{++}\) with \(k \geq 3\) is the congruence lattice of a finite semidistributive lattice, say \(\mathbf{L}_k\). These lattices \(\mathbf{L}_k\) cannot be bounded (in the sense of McKenzie), as no \((\mathbf{B}_k)_{++}\) \((k \geq 0)\) is the congruence lattice of a finite bounded lattice. A companion paper shows that every \((\mathbf{B}_k)_{++}\) \((k \geq 0)\) can be represented as the congruence lattice of an infinite semidistributive lattice. We also find sufficient conditions for a finite distributive lattice to be representable as the congruence lattice of a finite bounded (and hence semidistributive) lattice.

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