# How can you use Green's relations to learn about a monoid?

Without having ever formally learned any real monoid theory, I was recently pointed to Green's relations by a friend. It sounds like they're quite useful, and I get the impression they might do what character tables do for representation theory.

However, after looking over the article and playing around with a few interesting monoids, I'm still not totally sure what I can learn about the monoid with Green's relations.

Let's take this monoid, which I find interesting, as the focus of this post:

$$M=\langle c,k \mid c^2=1, k^2=k,kck = kckckck \rangle$$

Here's the egg-box diagram I came up with:

$$\begin{matrix}\{1,c\}\\ &\{(ck)^2,(ck)^3\}&\{(ck)^2c,(ck)^3c\}\\ &\{(kc)^2k,kck\}&\{(kc)^2,(kc)^3\}\\ &&&\{ck\}&\{ckc\}\\ &&&\{k\}&\{kc\}\end{matrix}$$

I'll try to break down the other relations later, but I hope this is enough to begin with for now. How can I learn about this monoid using Green's relations?

Short answer: Green's relations are mostly useful to know about the local structure of a semigroup (see the definition below) although the $\leqslant_\mathcal{J}$ preorder also gives information on the global structure.

Overview. One can probably say that the usefulness of Green's relations in semigroup theory compares with that of character tables in group theory, but they are two very different mathematical objects.

Green's relations are better interpreted geometrically on Cayley graphs. Take for instance the monoid $M$ of your example and its two generators, $c$ and $k$. The right Cayley graph of $M$ has $M$ as set of vertices and edges of the form $(m, a, ma)$ where $m \in M$ and $a$ is one of the generators. The left Cayley graph is defined by acting on the left: edges are of the form $(m, a, am)$. Now the $\mathcal{R}$-classes ($\mathcal{L}$-classes) are the strongly connected components of the right (left) Cayley graph. The $\mathcal{J}$-classes (or $\mathcal{D}$-classes since your monoid is finite) are the strongly connected components of the union of these two graphs.

Back to the example. I have slightly modified the egg-box picture of your monoid. First, I added stars to locate the idempotents, a useful information for a monoid. Secondly, the $\mathcal{J}$-classes are now presented according to the $\leqslant_\mathcal{J}$ order: $D_1 = \{1,c\}$ is the top $\mathcal{J}$-class, $D_2$ is the $\mathcal{J}$-class of $k$ and the remaining $\mathcal{J}$-class $D_3$ is the minimum${}^*$ ideal of $M$

${}^* \scriptsize\text{For some reason, the term *minimal ideal* is used in the literature, although this minimal ideal is unique when it exists.}$ $$\begin{matrix} \{{}^*1,c\}\\ \\ \begin{matrix} \{ck\}&\{{}^*ckc\}\\ \{{}^*k\}&\{kc\} \end{matrix}\\ {}\\ \begin{matrix} &\{{}^*(ck)^2,(ck)^3\}&\{(ck)^2c,{}^*(ck)^3c\}\\ &\{{}^*(kc)^2k,kck\}&\{{}^*(kc)^2,(kc)^3\} \end{matrix}\\ \end{matrix}$$ Now, each $\mathcal{J}$-class $D$ gives rise to a semigroup $(D^0, *)$ defined as follows: $D^0 = D$ if $D$ is a subsemigroup of $M$ and $D = D \cup \{0\}$ otherwise. The product $*$ is defined, for $s, t \in D$ by $$s*t = \begin{cases} st &\text{if st \in D} \\ 0 &\text{otherwise} \end{cases}$$

Local structure of a semigroup. The local structure of a semigroup is precisely the data of these semigroups $D^0$. They are ($0$)-simple semigroups and as such they are isomorphic to a Rees matrix semigroup. I let you study this notion in any book of semigroup theory, but let me describe it on your example.

Clearly $D_1^0 = D^1$ is a cyclic group of order $2$. The semigroup $D_2^0$ is isomorphic the semigroup $B_2 = \{1,2\} \times \{1,2\} \cup \{0\}$ with multiplication defined by $$(i,j)(i',j') = \begin{cases} (i,j') &\text{if j = i'} \\ 0 &\text{otherwise} \end{cases}$$ Finally, the minimum ideal $D_3$ is a semigroup, isomorphic to the semigroup $\{(i,g,j) \mid i \in \{1,2\}, j \in \{1,2\}, g \in \mathbb{Z}/2\mathbb{Z}\}$ with multiplication given by $((i,g,j)(i',g',j') = (i, gg', j')$. On your example, this structure is particularly simple because the product of two idempotents of $D_3$ is an idempotent.

In the general case, the Rees matrix semigroup associated to a regular $\mathcal{D}$-class is given by the set $I$ of $\mathcal{R}$-classes, the set $J$ of $\mathcal{L}$-classes, the structure group $G$ of the $\mathcal{D}$-class and a $J \times I$ matrix $(P_{j,i})$ with entries in $G \cup \{0\}$. The semigroup is defined on $(I \times G \times J) \cup \{0\}$ by the product $$(i,g,j)(i',g',j') = \begin{cases} (i, gp_{j,i'}g', j') &\text{if p_{j,i'} \not= 0} \\ 0 &\text{otherwise} \end{cases}$$ If you compute the torsion matrices $(P_{j,i})$ of each regular $D$-class, you will completely know the structure of each regular $D$-class, the local structure. For instance, you would know that the product $[(kc)^2k][(ck)^2c]$ is equal to $(kc)^3$ just by looking at the egg-box picture. Green's preorders also give you some partial information on the product of elements of two different $\mathcal{D}$-classes: for instance, if $e$ is idempotent, then $s \leqslant_\mathcal{R} e$ if and only if $es = s$.

• Awesome answer! This is exactly the sort of thing I was looking for, and I'm already learning a lot from it. Thank you – rschwieb Nov 19 '15 at 14:39
• What is the $\leqslant_\mathcal{J}$ preorder? At first, I thought it was probably the partial order induced by ideal containment, but then I wondered why "preorder" was used. Are you actually preordering the $\mathcal{H}$ classes, so that upon identification of the $\mathcal{H}$ classes into their respective $\mathcal{J}$(=$\mathcal{D}$) block, you get a partial order? – rschwieb Nov 19 '15 at 14:44
• Your definition is right: $a \leqslant_\mathcal{J} b$ iff $MaM \subseteq MbM$. The associated equivalence relation is the Green's relation $\mathcal{J}$. In other words $a\ \mathcal{J}\ b$ iff $a \leqslant_\mathcal{J} b$ and $b \leqslant_\mathcal{J} a$. Thus $\leqslant_\mathcal{J}$ is not in general antisymmetrical and this is why the term "preorder" is used. – J.-E. Pin Nov 19 '15 at 15:00
• Oh I see, I hadn't thought of the $\mathcal{J}$ relation that way. I had only been thinking of the $\mathcal{J}$ ideals and classes as wholes. Of course you can move it to a preorder on the elements this way. – rschwieb Nov 19 '15 at 15:07
• I had been observing the idempotents of monoids I was working with without much knowledge of what they did in monoids. In ring theory they correspond to module decompositions, but I think without the abelian group structure this intuition breaks down. Can you offer any more observations? – rschwieb Nov 19 '15 at 15:10