Independence algebras, basis algebras and the distributivity condition

TL;DR

This paper explores the relationship between stable basis algebras and independence algebras, especially focusing on the conditions under which endomorphism monoids form certain algebraic orders, and provides an example of an independence algebra not satisfying the distributivity condition.

Contribution

It introduces an example of an independence algebra not satisfying the distributivity condition and characterizes when End(B) is a right or two-sided order in End(A).

Findings

End(B) is a left order in End(A) when B satisfies the distributivity condition.

End(B) is a right order in End(A) under specific algebraic conditions.

An example of an independence algebra not satisfying the distributivity condition is provided.

Abstract

Stable basis algebras were introduced by Fountain and Gould and developed in a series of articles. They form a class of universal algebras, extending that of independence algebras. If a stable basis algebra $\mathbb{B}$ of finite rank satisfies the distributivity condition (a condition satisfied by all the previously known examples), it is a reduct of an independence algebra $\mathbb{A}$. Our first aim is to give an example of an independence algebra not satisfying the distributivity condition. Gould showed that if a stable basis algebra $\mathbb{B}$ with the distributivity condition has finite rank, then so does the independence algebra $\mathbb{A}$ of which it is a reduct, and in this case the endomorphism monoid End$(\mathbb{B})$ of $\mathbb{B}$ is a left order in the endomorphism monoid End$(\mathbb{A})$ of $\mathbb{A}$. We complete the picture by determining when End$(\mathbb{B})$ is a right, and hence a two-sided, order in End$(\mathbb{A})$. In fact (for rank at least 2), this happens precisely when every element of End$(\mathbb{A})$ can be written as $\alpha^\sharp\beta$ where $\alpha,\beta\in$ End$(\mathbb{B})$, $\alpha^\sharp$ is the inverse of $\alpha$ in a subgroup of End$(\mathbb{A})$ and $\alpha$ and $\beta$ have the same kernel. This is equivalent to End$(\mathbb{B})$ being a special kind of left order in End$(\mathbb{A})$ known as straight.