R-trivial semigroups

The monoid of increasing actions on a poset is R-trivial. This leads to the following question: can all R-trivial semigroups be embedded within a monoid of increasing actions on a poset. We will prove the affirmative, and demonstrate a couple of ways to go about producing an embedding.

Lemma 1. let $S$ be an R-faithful R-trivial semigroup, then its representation by left actions produces an embedding into the monoid of increasing actions on $\subseteq_R$.

Proof. $S$ induces a monoid action on itself by left actions, this monoid action in turn induces a preorder on $S$. If $R$ is trivial, then left actions on $S$ are antisymmetric ($R$ follows from defining right principal ideals as $aS^1$). Then this produces a semigroup homomorphism from $S$ into a monoid of increasing actions on $\subseteq_R$. This is injective provided that the representation of $S$ by left actions is faithful. $\square$

Lemma 2. let $S$ be a monoid then it is left and right faithful.

Proof. $S$ has an identity element $e$, so that for all $x,y$ we have $ex \not= ey$ and $xe \not= ye$ so that $x$ and $y$ cannot have equal left or right actions. $\square$

Theorem 1. every $R$ trivial monoid is a monoid of increasing actions on itself ordered by $\subseteq_R$.

Proof. by lemma 2 $R$ has a faithful representation by left actions and by lemma 1, it therefore has an embedding into the monoid of increasing actions on itself ordered by $\subseteq_R$. $\square$

As an example, a right zero band is a faithless extension of the unique increasing identity action on an antichain. In such a R-trivial band, each element is also a left identity. Although its actions are all increasing, its faithlessness means it cannot be represented by left actions. We can fix this by adjoining an identity, to get an R-trivial monoid.

Theorem 2. let $S$ be an R-trivial semigroup, then it can be embedded in a monoid of increasing actions.

Proof. $S$ is an R-trivial monoid, then let $S^1$ be the monoid constructed from $S$ by adjoining an identity $e$. Then $e$ is a J-trivial element, so that $S^1$ is R-trivial. $S$ is also an ideal in $S^1$. Therefore, by theorem 1 this produces an embedding of $S$ into $S^1$ which is a monoid of increasing actions on itself. $\square$

We saw that a rectangular band cannot faithfully be represented by its increasing actions on itself. By embedding it in a monoid we have resolved that issue. Consider the height two tree ordered set $[1,n]$, then its monoid of increasing actions is precisely a rectangular band plus the identity. So the rectangular band is a subsemigroup of increasing actions produced by removing the identity action.

Every category is associated to its dual category, so a R-trivial monoid of increasing actions has an order dual L-trivial monoid of increasing actions. By duality, this produces a classification of both L-trivial and R-trivial monoids in terms of increasing actions on partial orders.

Corollary. every $L$ trivial monoid is dual to a monoid of increasing actions.

The difference between $L$ triviality and $R$ triviality isn't so important, and it is only a matter of representation. In either case, L-trivial and R-trivial monoids can be studied by increasing actions on certain posets. A J-trivial monoid can also be considered to be a system of increasing actions on a poset, in either direction.

Example 1. the commutative J-trivial monoid $(\mathbb{N},+)$ can be seen as a monoid of increasing actions: each addition operation by a non-negative integer produces a larger number. Dually with respect to $(\mathbb{N},*)$ over the divisibility ordering.

Example 2. the three element non-commutative totally ordered semigroup $T_3^*$ can be embedded in the monoid of increasing actions on a total order on three elements $T_3$ by $[1,2,3],[2,3,3],[3,3,3]$. In this case, $T_3^*$ is also embeddable in the submonoid of increasing monotone actions. Its dual semigroup is not faithful.

Example 3. in a rectangular band $S$ both $L$ and $R$ form congruences with $L$ trivial and $R$ trivial quotients, and $S$ is the direct product of $\frac{S}{L}$ and $\frac{S}{R}$. As a direct product, $L$ and $R$ also form direct products in the semilattice of set partitions.