Return to Question

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two distributive laws $$(a+x+b)\wedge(a+y+b) = a + (x\wedge y) + b$$ and $$(a+x+b)\vee(a+y+b) = a + (x\vee y) + b$$ both hold. A lattice-ordered group is defined similarly. It is well known that the lattice reduct $(A,{\vee},{\wedge})$ of a lattice-ordered group is always distributive, but this result does not extend to lattice-ordered monoids.

Are there quasi-equational conditions on the monoid $(A,0,{+})$ that guarantee that the lattice reduct $(A,{\vee},{\wedge})$ is distributive? I am particularly interested in the case where the monoid is commutative.

I'm not looking for an exact characterization (which is probably impossible in terms of quasi-identities) but for sufficient conditions.

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two distributive laws $$(a+x+b)\wedge(a+y+b) = a + (x\wedge y) + b$$ and $$(a+x+b)\vee(a+y+b) = a + (x\vee y) + b$$ both hold. A lattice-ordered group is defined similarly. 

It is well known that the lattice reduct $(A,{\vee},{\wedge})$ of a lattice-ordered group is always distributive, but this result does not extend to lattice-ordered monoids.

Are there quasi-equational conditions on the monoid $(A,0,{+})$ that guarantee that the lattice reduct $(A,{\vee},{\wedge})$ is distributive? I am particularly interested in the case where the monoid is commutative.

I'm not looking for an exact characterization (which is probably impossible in terms of quasi-identities) but for sufficient conditions.

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two distributive laws $$(a+x+b)\wedge(a+y+b) = a + (x\wedge y) + b$$ and $$(a+x+b)\vee(a+y+b) = a + (x\vee y) + b$$ both hold. A lattice-ordered group is defined similarly. It is well known that the lattice reduct $(A,{\vee},{\wedge})$ of a lattice-ordered group is always distributive, but this result does not extend to lattice-ordered monoids.

Are there useful quasi-equational conditions on the monoid $(A,0,{+})$ that guarantee that the lattice reduct $(A,{\vee},{\wedge})$ is distributive? I am particularly interested in the case where the monoid is commutative.

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two distributive laws $$(a+x+b)\wedge(a+y+b) = a + (x\wedge y) + b$$ and $$(a+x+b)\vee(a+y+b) = a + (x\vee y) + b$$ both hold. A lattice-ordered group is defined similarly. It is well known that the lattice reduct $(A,{\vee},{\wedge})$ of a lattice-ordered group is always distributive, but this result does not extend to lattice-ordered monoids.

Are there quasi-equational conditions on the monoid $(A,0,{+})$ that guarantee that the lattice reduct $(A,{\vee},{\wedge})$ is distributive? I am particularly interested in the case where the monoid is commutative.

I'm not looking for an exact characterization (which is probably impossible in terms of quasi-identities) but for sufficient conditions.