# Is this a pre-ordered commutative semigroup?

### Motivation

I'm studying an approach to axiomatic thermodynamics based on the notion of commutative semigroup $(S,+)$ with a preorder relation $\to$ on $S$. In other words, $S$ is non-empty set, the operation $+: S \times S \to S$ is commutative and associative, the relation $\to$ on $S$ is reflexive and transitive, and in addition there is a compatibility condition: $$(\forall a,b,c \in S) \qquad a \to b \iff a + c \to b + c$$

### Question

Is this a standard (or well-known) structure? And if so, is there an accepted term for it?

Naively I thought that it would be a pre-ordered commutative semigroup, but I googled and found very little with this name and the little I found suggests that the compatibility condition obeyed by a pre-ordered commutative semigroup is the weaker $$(\forall a,b,c \in S) \qquad a \to b \implies a + c \to b + c$$

• I don't know an answer to your question. But as a side note: There is an interesting paper by Lieb and Ynvason concerning your motivation: arxiv.org/abs/cond-mat/9708200. Perhaps the references therein could be helpful to answer your question. Oct 2, 2010 at 15:09
• Thanks -- I know the papers of Lieb and Yngvason and in fact theirs are among the papers I'm studying. My question, though, derives from earlier work: a book by Giles and the PhD thesis of Hans Duistermaat. Oct 2, 2010 at 15:49

Your condition certainly was not considered by algebraists studying commutative semigroups. It implies, for example, that if the semigroup has a $0$ (i.e. $0+x=0$), then for every two $a,b$ $a\to b$ and $b\to a$. Your condition makes more sense for semigroups satisfying cancelation law: $a+c=b+c\to a=b$. Then you can embed your semigroup into a group where the two conditions are equivalent.

• Thanks. The semigroup $S$ does not have a $0$. In the context I've encountered this, the compatibility condition is used in order to induce a preorder in the Grothendieck group $K(S)$ of $S$. In this context elements of $S$ are called states and elements of $K(S)$ are processes. The relation $\to$ is akin to "accessibility", in case this is of any use. Oct 2, 2010 at 15:51
• Even if your semigroup does not contain 0, but does not satisfy the cancelation law, then there are very many pairs $a,b$ with $a\to b$ and $b\to a$, and so it is perhaps not very interesting. If it does, the Grothendieck group is just a group of fractions (consisting of $a-b$, since you use the additive notation).
– user6976
Oct 2, 2010 at 16:14
• In the thermodynamic context, pairs which obey $a\to b$ and $b \to a$ correspond to reversible processes, so they do have a physical significance. Oct 2, 2010 at 20:02
• @José: I see, but I do not think anybody used pre-ordered commutative semigroups in your sense except groups before. If $G$ is a pre-ordered commutative semigroup, then the relation $a\sim b$ iff $a\to b$ and $b\to a$ is a congruence. The semigroup $G/\sim$ is partially ordered ($[a]\to [b]$ if $a\to b$). Moreover, this partial order satisfies a very strong condition: if $a+c\to b+c$, then $a\to b$. In particular $G/\sim$ must satisfy the cancelation law. So it must be a subgroup of a partially ordered commutative group with induced order.
– user6976
Oct 2, 2010 at 21:06
• Continued: Conversely, if $G$ is any commutative semigroup, $\phi$ is a homomorphism of $G$ into a partially ordered commutative group, then one can pull the partial order from $\phi(G)$ to a pre-order of $G$ satisfying your condition. I guess this is the best algebraic description you can get. At least I do not see a better one.
– user6976
Oct 2, 2010 at 21:07

In addition to transitivity and reflexivity, to obtain the definition of a pre ordered (sometimes quasi ordered) semigroup, you need only have that the binary operation is monotone in each coordinate. That is, if $a\rightarrow b$ then for every $c$ you have that $a+c \rightarrow b+c$ and $c+a \rightarrow c+b$. There is no requirement for the converse implication to be true.

Without the existence of a nullary operation you cannot necessarily obtain the other implication.

In general, quasi ordered algebras only need each component of their operations to be either order preserving or order reversing, they don't need to be surjective on the order structure.

• Actually, I'm only looking at commutative semigroups; although I suppose that the question arises for any semigroup. Oct 18, 2010 at 17:45
• It doesn't change the order structure at all, just the underlying algebra structure. Oct 19, 2010 at 19:48