Year: 2024
Type: article
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.
Source: Cubo
Author: J. B. Nation
Institutions: University of Hawaiʻi at Mānoa, University of Hawaii System
Language: English
Topic: Advanced Algebra and Logic
Subfield: Computational Theory and Mathematics
Field: Computer Science
Domain: Physical Sciences
Open Access status: diamond
OpenAlex ID: W4403521614