Semigroups of trivial charts

The elements of an inverse semigroup are called charts. Amongst the set of charts of an inverse semigroup, there are the charts containing no more then one element, which we call trivial. The non-empty semigroups constructed from these trivial charts are equivalent to the semigroups with zero of thin semigroupoids.

The complete Brandt semigroup of trivial charts:

Let $PS_n$ be the symmetric inverse semigroup on $n$ elements. Then the semigroup with zero of the complete thin groupoid $K_n + 0$ is a Brandt semigroup consisting of all trivial charts on $n$ elements. Chart representation makes $K_n + 0$ a subsemigroup of $PS_n$.

	$\emptyset$	(0,0)	(1,1)	(0,1)	(1,0)
$\emptyset$	$\emptyset$	$\emptyset$	$\emptyset$	$\emptyset$	$\emptyset$
(0,0)	$\emptyset$	(0,0)	$\emptyset$	(0,1)	$\emptyset$
(1,1)	$\emptyset$	$\emptyset$	(1,1)	$\emptyset$	(1,0)
(0,1)	$\emptyset$	$\emptyset$	(0,1)	$\emptyset$	(0,0)
(1,0)	$\emptyset$	(1,0)	$\emptyset$	(0,0)	$\emptyset$

The Brandt semigroup on five elements is the smallest non-commutative inverse semigroup, and it is the special case of the semigroup with zero of a category. The non-zero elements a complete Brandt semigroup of trivial charts are ordered pairs, and their inverses are the ordered pairs with elements reversed.

Definition. a semigroup of trivial charts is a subsemigroup of $K_n + 0$.

Theorem 1. subsemigroups of $K_n + 0$ are either trivial or they contain zero.

Proof. (1) in the trivial case we have the empty semigroup and the semigroups formed by any of the idempotents of $K_n + 0$. (2) if $S$ contains nilpotent charts then it must contain zero, so suppose that $S$ has no nilpotent charts and at least two non-zero elements. Then those non-trivial elements must be idempotents, but any two idempotents in $K_n + 0$ combine to produce zero, so a non-trivial subsemigroup must contain zero. $\square$

Theorem 2. let $S$ be a semigroupoid, then subsemigroupoids $Sub(S)$ are equivalent to zero-preserving subsemigroups of $Sub(S + 0)$.

Proof. (1) let $T$ be a subsemigroupoid of $S$. Then for all $a,b \in T$ either $ab \in T$ or $ab = 0$, it follows that $ab \in T + 0$. Suppose $X$ is a zero preserving subsemigroup of $S + 0$. Then $ab \in X-0$ or $ab = 0$. Let $T = X - 0$, then $T \subseteq S$ and forall $a,b$ either $ab \in T$ or $ab = 0$, so $ab$ is a subsemigroupoid of $S$. $\square$

Theorem 3. every non-empty semigroup of trivial charts is isomorphic to the semigroup with zero of a thin semigroupoid.

By theorem 1 a non-empty semigroup of trivial charts is equal to the trivial monoid with a single element. This is simply the semigroup with zero of the empty category. Then in the case that $S$ is non-trivial, by theorem one it is a zero-preserving subsemigroup of $K_n + 0$, which by theorem 2 is the semigroup with zero of a thin semigroupoid. $\square$

This allows us to create a direct correspondonce between a concept in semigroup theory and a concept in semigroupoid theory, so that thin semigroupoids can be studied by semigroups of trivial charts. Thin categories are simply a special case, and so they produce a special class of semigroups.

Proposition. a non-empty semigroup of trivial charts is the semigroup with zero of a thin category provided that every nilpotent chart absorbs two idempotents.

Proof. by theorem 3 every non-empty semigroup of trivial charts comes from a thin semigroupoid. There are two elements in a thin semigroupoid: nilpotents and idempotents corresponding to non-endofunctions and endofunctions respectively. In a thin category there are no non-identity idempotents to preserve, so all that is required is that each nilpotent morphism preserves two idempotents which are necessarily identities.

If $(a,b)$ is a morphism in a thin category, then the idempotents preserved by it are precisely the identities that it absorbs: $(a,a) \circ (a,b) = (a,b)$ and $(a,b) \circ (b,b) = (a,b)$ of which there can only be two. So if a nilpotent charts absorbs two different idempotents it preserves identities. $\square$

This is a purely semigroup theoretic characterization of thin categories. These are a very restricted class of semigroups, because concepts of semigroup theory most readily translates to semigroupoid theory rather then category theory. This correspondence forms the basis of the subject at hand.

We now need to characterize the Green's relations in the semigroup with zero of the complete thin groupoid $K_n + 0$ and prove that it is indeed in a Brandt semigroup. We have mentioned that $K_n + 0$ is a Brandt semigroup, and we can see this in the case of five elements by simple inspection, but we have not proved this result in general.

Lemma 1. let $K_n$ be a complete thin groupoid. Then the $L$ preorder on $K_n + 0$ has any ordered pair $(a,b)$ less then $0$ and any pair of ordered pairs are related if they have the same right elements. Dually, the right preorder $R$ has everything less then zero and it preserves left elements.

Proof. let $(a,b)$ be an ordered pair, then by left action $0(a,b) = 0$ so that $(a,b) \subseteq 0$. Then by left action with another ordered pair $(c,a)(a,b) = (c,b)$ two elements are left action related provided they have the same right elements. In other direction, $(a,b)0 = 0$ by right action and $(a,b)(b,c) = (a,c)$ so that ordered pairs are right action related provided they preserve the same left elements.

It follows that $K_n + 0$ has two $J$ classes: zero and all ordered pairs. It follows that $K_n + 0$ is a 0-simple semigroup. In the minimal $D$ class, $L$ and $R$ permute with one another (as is necessary in any $D$ class) and furthermore they form direct products of one another in the lattice of partitions (which is the quotient lattice of the topos of sets). Their direct product is the set of all ordered pairs. $\square$

Theorem 4. $K_n + 0$ is an H-trivial Brandt semigroup.

Proof. (1) let $a$, $b$ be idempotents in $K_n + 0$ then $a$ and $b$ can be represented as charts $(a,a)$ or $(b,b)$ so that $(a,a)(b,b) = 0$ which implies that $ab = 0$. Alternatively, $a = 0$ or $b = 0$ implies that $ab = 0$. As any two idempotents compose to zero, they commute. So that $K_n + 0$ is idempotent commutative.

(2) let $a$ be any element in $K_n = 0$. Then suppose that $a$ is equal to zero, then $a$ is idempotent which implies that it is a regular element. Suppose that $a$ is non-zero, then it is an ordered pair $(a,b)$ so it has an inverse $(b,a)$ then $(a,b)(b,a)(a,b) = (a,b)$ so that $a$ is a regular element.

(3) by lemma 1, $K_n + 0$ is 0-simple. Every trivial chart is periodic, so that every element in $K_n + 0$ is group bound. It follows that $K_n + 0$ is completely 0-simple. Every idempotent commutative regular semigroup is inverse, so by parts (1) and (2) it follows that $K_n + 0$ is an inverse semigroup. As it is a completely 0-simple inverse semigroup, it is a Brandt-semigroup. By lemma 1, the intersection of its L and R relations is trivial, so it is H-trivial. $\square$.

Proposition. $K_n + 0$ is a two-sided inverse subsemigroup ideal of the symetric inverse semigroup $PS(n)$.

Proof. $K_n + 0$ is closed under inverses, with $(a,b)^{-1} = (b,a)$ so it is an inverse subsemigroup of $PS(n)$. It is also a two-sided semigroup ideal because with respect to function composition for any $fg$ then $|fg| \leq |f|,|g|$, so function composition can only reduce the cardinality of partial transformations. It follows that composition can only make trivial charts smaller to become empty, which is included as a trivial chart. So that $K_n+0$ is a two sided semigroup ideal. $\square$

This demonstrates that the semigroup with zero of a thin category is a subsemigroup of an inverse semigroup, so among other things it is idempotent commutative. In general, every thin category embdes in a groupoid whose semigroup with zero is an inverse semigroup.

Zero divisor digraphs:

The zero divisor digraph $Z(S)$ is an important component in a number of constructions related to semigroups of trivial charts including the commuting graph $Com(S)$ and the construction of the underlying thin semigroupoid or category of the semigroup.

Definition. let $S$ be a semigroup, then the zero divisor digraph $Z(S)$ is the fiber of $0$. Then $(a,b) \in Z(S) \Leftrightarrow ab = 0$.

The zero divisor graph $Z(S)$ for a semigroup of trivial charts, can be described by the composition of trivial charts.

Proposition. let $S$ be a semigroup of trivial charts then $a,b \in S$ and $(a,b) \in Z(S)$ provided that $a = 0$, $b = 0$, or $a,b \not = 0$ and $a = (a_1,a_2), b = (b_1,b_2)$ with $a_2 \not= b_1$.

With this, we can show that the commuting graph of a semigroup of trivial charts is a subgraph of the zero divisor graph. As it is symmetric, it is naturally embedded in the symmetric component of the zero divisor graph. In fact it is equal to it, as we are about to show.

Theorem 5. let $S$ be a semigroup of trivial charts, then $Com(S) \subseteq Z(S)$.

Proof. the composition any elements with zero is zero, therefore in order for two elements to not be zero divisors of one another they must both be non-zero. Let $a,b$ be non-zero then by trivial chart representation they are $(a_1,a_2)$ and $(b_1,b_2)$ then if $(a_1,a_2)(b_1,b_2) = (a_1,b_2) = (b_1,a_2) = (b_1,b_2)(a_1,a_2)$ we have that $a_1 = a_2$ and $b_1 = b_2$ which implies that $(a_1,a_2) = (b_1,b_2)$. It follows that $a = b$ so the only non-zero elements that commute are equal. $\square$

We define a semigroupoid such as a category from a semigroup of trivial charts, not from the zero divisor graph but rather from its complement: the non-zero divisor graph. The domain of a semigroup is a complete binary relation, but the domain of a semigroupoid or a category is a non-zero divisor graph of a semigroup with zero.

Definition. let $S$ be the composition function of a non-empty semigroup of trivial charts. Then the underlying thin semigroupoid of $S$ is the subobject (in the topos of functions) of $S$ is the partial semigroup with domain equal to the non-zero divisor graph of $S$.

Corollary. the partial semigroup of a thin semigroupoid is anticommutative.

We have primarily studied thin semigroupoids in terms of their semigroups of trivial charts, which naturally emerge from their completions. But in the case of thin semigroupoids they are also anticommutative, which produces a special relationship with rectangular bands.

Theorem 6. let $\circ : R \to M$ with $R \subseteq M^2$ be a thin semigroupoid with object set $O$. Then $\circ$ is a subobject in the topos of functions of the composition function of a rectangular band.

Proof. the composition function of a thin semigoupoid has $(a,b)(c,d) = (a,d)$ when $b = c$. The composition function in a rectangular band has $(a,b)(c,d) = (a,d)$ no matter what. Let $O^2$ be the rectangular band of ordered pairs on $O$, this leads to a function $\cdot : (O^2)^2 \to O^2$. Then we can embed $\circ$ in $\cdot$ by an ordered pair of inclusion monomorphisms $(R \hookrightarrow (O^2)^2, M \hookrightarrow M)$. $\square$

This is of course a different kind of embedding of the composition function of a semigroup or category into a semigroup then we are used to, but this demonstrates a special relationship that exists between rectangular bands and thin categories. This is a consequence of the fact that thin categories are anticommutative.

Theorem 7. let $S$ be a semigroup of trivial charts and $a,b \in S$. If $ab \not= 0$ and $ba \not= 0$ then $a$ and $b$ are non-zero nilpotents and inverses of one another.

Proof. suppose that $a$ or $b$ are idempotent. Then $ab = 0$ and $ba = 0$ because idempotents always compose to zero. So that means $a$ and $b$ must both be non-idempotents and therefore non-zero nilpotent. Let $(a,b)$ and $(c,d)$ be their values then if they are composable then $b = c$ and $a = d$ which means that $(c,d) = (b,a)$ so that they are equal to $(a,b)$ and $(b,a)$ which are inverses of one another. $\square$

With this, we can get an important property of the zero divisors graphs of thin skeletal categories and semigroupoids arising from posets and strict orders.

Theorem 8. let $S$ be an antisymemtric thin semigroupoid. Then the zero divisor digraph of $S$ is total.

Proof. by theorem 7 in order for two elements to not compose to zero with one another they must be inverses of one another. An antisymmetric thin semigroupoid is inverse-free, so that for each $a,b \in S$ we have $ab = 0$ or $ba = 0$ which implies that the zero divisor digraph is total.

In particular, the semigroups with zero of thin categories always have at least one pair of elements compose to zero. Then same is true for the semigroups of trivial charts of strict orders.

Theorem 9. let $S$ be a commutative semigroup of trivial charts. Then $S$ is a maximum chain length two partially ordered commutative J-trivial semigroup, and all such commutative semigroups emerge in this way.

Proof. by the fact that the commuting graph of a semigroup of trivial charts is the symmetric component of the zero divisor graph, if $S$ is commutative this means that $ab = 0$ and $ba = 0$ for all $a,b$ with $a \not= b$. It follows that $S$ is a maximum chain length two J-trivial semigroup, with maximum chains equal to elements together with zero. In the other direction, every height two J-trivial commutative semigroup is classified by its set of idempontents and nilpotents. Idempotents can be represented by permutation charts and nilpotents by nilpotent charts, to get a trivial chart representation of the commutative semigroup. $\square$

Let $S$ be a semigroup with automorphism group $Aut(S)$ and suppose that $p \in Aut(S)$. Then if $ab = ba$ we have $p(ab) = p(ba)$ which implies that $p(a)p(b) = p(b)p(a)$ so that if $(a,b) \in Com(G)$ then $(p(a),p(b)) \in Com(G))$ so that automorphisms of a semigroup are automorphisms of its commuting graph.

Lemma. let $K_n + 0$ be a complete Brandt semigroup of trivial charts. Then let $f \in S_n$ be a permutation on the underlying set $n$. Then define $f' : K_n + 0 \to K_n + 0$ with $f'(0) = 0$ and $f'((a,b) = (f(a),f(b))$ then $f'$ is an automorphism.

Proof. let $ab \in K_n + 0$. Then suppose that $a = 0$ then $ab = 0$ and $f'(ab) = 0f'(b) = f'(0) = 0$ or if $b = 0$ then $f'(ab) = f'(a)0 = f'(0) = 0$. Suppose that $a \not = 0$ and $b \not= 0$ then $$f'((a_1,a_2)(b_1,b_2)) = f'((a_1,b_2)) = (f(a_1),f(b_2))$$ $$f'(a_1,a_2)f'(b_1,b_2)) = (f(a_1),f(a_2))(f(b_1),f(b_2)) = (f(a_1),f(b_2))$$ Then $f'((a_1,a_2)(b_1,b_2)) = (f(a_1),f(b_2)) = f'((a_1,a_2))f'((b_1,b_2))$. In the special case in which $a_2 \not = b_1$ then this implies that $f(a_2) \not= f(b_1)$ because $f$ reflects equality since its a permutation. So $f'$ preserves zeros. It follows that $f'$ is a semigroup automorphism.

Corollary. the Brand semigroup $K_n + 0$ has an automorphism group with there orbits: zero, non-zero idempotents, and non-zero nilpotents

We can use this result as an organizing principle in the theory of the centralizers of $K_n + 0$. By this result, we know that the centralizers belong into three classes. In the following theorem we will characterize all of them.

Theorem 10. let $K_n + 0$ be the complete semigroup of trivial charts, then the centralizers of $K_n + 0$ come in three forms:

1. The entire semigroup $K_n + 0$
2. $K_{n-1} + 0$ plus a side idempotent which has $(n-1)^2 + 1$ elements
3. A special case which has $(n-1)^2$ elements and a side nilpotent.

Proof. (1) let $0$ be the zero element of $K_n + 0$, then $0$ is a central element so its centralizer $C(a)$ is the entire semigroup $K_n + 0$.

(2) let $a$ be an idempotent non-zero element. Then it is equal to an element $(a,a)$ and so its centralizer is all elements disjoint from $(a,a)$ so besides $(a,a)$ it consits of the $D$ class of all $(x_1,x_2)$ with $x_1 \not = a$ and $x_2 \not = a$. The commuting graph is a subgraph of the zero divisor graph, so $ac = 0$ for any $c \in C(a)$ which implies that $a$ is a side idempotent.

(3) let $a$ be a non-zero nilpotent, then as before for any $c \in C(a)$ we have $ac = 0$ and $ca = 0$ so that $a$ is a side nilpotent element. Then let $(x_1,x_2)$ be the chart of $a$. Any idempotent element has the form $(y_1,y_2)$ with $x_2 \not= y_1$ and $x_1 \not= y_2$. These come in three forms $x_1 = y_1$, $x_2 = y_2$ and $x_1 \not= y_1$ and $x_2 \not= y_2$.

Those with $x_1 \not= y_1$ and $x_2 \not= y_2$ form a single D class with $L$ and $R$ classified by their components. Then those with $x_1 = y_1$ form a R-total D class and those with $x_2 = y_2$ form an L-total D class. Together, these constitute a complement set of J classes of $K_n + 0$. This results in a subsemigroup whose J class ordering has the form $[\{[1,\{1,1\}],1\},1]$. $\square$

The commuting graph of the Brandt semigroup on five elements is the cricket graph. In general, by theorem 10 we have that any commuting graph of a complete Brandt semigroup of trivial charts is a nearly-regular graph (in the sense that degrees can only differ by at most one) with a zero element adjoined.

Special cases:

We have defined semigroups of trivial charts by their embeddings in the partial inverse semigroup $PS(n)$ and its intermediary ideal $K_n + 0$, but we have not created a theory of recognising which semigroups of trivial charts without embeddings. We will now do that.

Theorem 11. let $S$ be a non-empty subsemigroup of $K_n+0$. Then $S$ has the following properties

1. $S$ is idempotent commutative, with a max height two semilattice of idempotents, and the idempotent action poset is 0-trivial, in the sense that non-zero elements form an antichain.
2. $S$ is group-free
3. $S$ is a semigroup with zero
4. The commuting graph of $S$ is a subgraph of its zero divisor graph
5. $S$ is max order two aperiodic.
6. Only elements that are inverses of one another are non-zero dividing as pairs

Proof. (1) as a subsemigroup of an inverse semigroup, $S$ is idempotent commutative. By theorem 9, the semilattice of idempotents of $S$ is max height two. Then for any idempotent $e$ and any element $x$ we have $ex = x$ or $ex = 0$, so that the idempotent action poset is 0-trivial. Furthermore, as a semigroup of trivial charts an inverse semigroup of trivial charts can only have a 0-trivial natural partial ordering.

(2) By theorem 4 $K_n + 0$ is H-trivial which by Green's theorem means it is group-free. So its subsemigroups are group-free as well.

(3) By theorem 1, $S$ is a semigroup with zero.

(4) By theorem 5, $Com(S) \subseteq Z(S)$ so that the only elements that commute are ones that both compose to zero.

(5) The charts in $K_n+0$ take two forms: they are idempotent or they are non-zero nilpotent. A non-zero nilpotent $(a,b)$ composed with itself is zero, so every non-zero nilpotent has index two. So $S$ is max index two as an aperiodic semigroup.

(6) By theorem 7, only inverses can be non-zero dividing as pairs. $\square$

The morphism preordering of a thin category $C$ is antisymmetric. This is translated into semigroup theoretic terms by the statement that $C+0$ is a J-trivial semigroup. This is encoded in the following theorem.

Theorem 12. a non-empty semigroup of trivial charts is J-trivial iff it comes from an antisymmetric thin semigroupoid.

Proof. (1) if $S$ is a thin semigroupoid with symmetric pair $(a,b)$ and $(b,a)$ then $(b,a)(a,a)(a,b) = (b,b)$ and $(a,b)(b,b)(b,a) = (a,a)$ so that $a$ and $b$ are in the same $J$ class. So if its semigroup $S + 0$ is J-trivial it must be antisymmetric.

(2) if it is antisymmetric then for $(a,b) \subseteq (c,d)$ then $c \subseteq a$ and $b \subseteq d$ and $(c,d) \subseteq (a,b)$ means $a \subseteq c$ and $d \subseteq b$. By antisymmetry if $a \subseteq b$ and $b \subseteq a$ then $a = b$ and if $c \subseteq d$ and $d \subseteq c$ then $c = d$ so that $(a,b) = (c,d)$ which implies that $S + 0$ is J-trivial. $\square$

Theorem 13. $S$ is a nilpotent semigroup iff it comes from a strict order.

Proof. every non-zero idempotent is of the form $(a,a)$. It follows that if $S$ is nilpotent, it must avoid every element of the form $(a,a)$ which means it is irreflexive. If $(a,b)$ and $(b,a)$ are in $S$ then $(a,b)(b,a) = (a,a)$ and $(b,a)(a,b) = (b,b)$ so that $(a,a)$ and $(b,b)$ are in $S$ it follows that irreflexive transitive are antisymmetric, in which case they are called strict orders. So only strict orders have nilpotent semigroups, and strict orders are nilpotent because they are irreflexive. $\square$

In theorem 9, we characterized from a semigroup perspective the commutative semigroups of charts. In the other direction, we can characterize the thin semigroupoids with commutative semigroup completions.

Theorem 14. let $S$ be a thin semigroupoid, then $S+0$ is commutative iff $S$ is a loop isolated maximum chain length two antisymmetric thin semigroupoid.

Proof. there are cases whereby two morphisms can be composable (1) if we have a loop and a non-loop edge $(a,a)(a,b)$ or $(a,b)(b,b)$ so to forbid this $S$ must be loop isolated (2) if we have two edges in a symmetric pair $(a,b)(b,a)$ which must be forbidden so that $S$ is antisymmetric (3) we have a chain of length three $(a,b)(b,c)$ which means that $S$ must be maximum chain length two. $\square$

Corollary. let $S$ be a thin semigroupoid, then $S+0$ is a null semigroup iff $S$ is a maximum chain length two strict order.

Proof. (1) by theorem 9 $S$ must have maximum chain length to be commutative and by theorem 13 it must be nilpotent, so to be a commutative nilpotent semigroup like a null semigroup it must be a maximu mchain length two strict order (2) by theorem 14 the fact that $S+0$ is commutative implies that $S$ is a maximum chain length two antisymmetric thin semigroupoid, and by theorem 13 we know it is irreflexive. So by combining the two $S$ is a maximum chain length two strict order. $\square$

Semilattices are an important special case in semigroup theory. By theorem 9, we know that every such semilattice is a maximum chain length two semilattice. So every semilattice associated to a thin category is isomorphic, it follows that in order to classify the thin semigroupoids associated with semilattices we need a class of semigroupoids classified by their cardinalities. These are precisely the discrete categories.

Theorem 15. let $S$ be a thin semigroupoid, then $S+0$ is a semilattice if $S$ is a discrete category.

Proof. if $S+0$ is a semilattice then every element of $S$ is an idempotent, which means it is a loop. It follows that $S$ is coreflexive, so that every element is a loop. The only thin semigroupoids that are coreflexive are the discrete categories, so $S$ is a discrete category. Then if $S+0$ is a semilattice, then every element of $S$ must still be idempotent, so that it must be a discrete category. $\square$

We started this discussion by considering the symmetric inverse semigroup $PS_n$ whose elements consist of charts on at most $n$ elements. These charts all have permutation and nilpotent parts, and they can be represented as sets of ordered pairs. If they have at most one ordered pair, they are trivial. So inverse semigroups have played an important role in this entire theory.

We return to the question of inverse semigroups. It remains to characterize which semigroups of trivial charts are indeed inverse semigroups. These are then inverse subsemigroups of the symmetric inverse semigroup $PS_n$. This is a fundamental relationship between groupoids and inverse semigroups.

Theorem 16. let $G$ be a thin groupoid, then $G+0$ is an inverse semigroup. Every inverse semigroup with zero of trivial charts emerges in this way.

Proof. let $(x,y)$ be an element of the thin groupoid, then $(y,x) \in G$ so that $(x,y)(y,x)(x,y) = (x,y)$. It follows that $G+0$ is a regular semigroup, and by theorem 11 it is idempotent commutative so it is an inverse semigroup. Then let $S+0$ be a semigroup with zero then every element $(x,y)$ has an inverse $(y,x)$ so that the underlying semigroupoid $S$ is a groupoid. $\square$

The semigroups of trivial charts are part of the basic relationship between category theory and semigroup theory, because the semigroup completion of any thin category is a semigroup of trivial charts. Further, if we generalize to the semigroup completion of any arbitrary category $C$, then hom class equivalence forms a congruence on $C+0$ whose quotient is a semigroup of trivial charts.

It follows that the theory of semigroups of trivial charts, like those dealt with in this post are part of the basic semigroup theory of categories. Properties of the semigroup completions of categories can be inferred from their quotient semigroups of trivial charts. This suggests a new direction to take the semigroup theory of categories in.

Completion of partial semigroups

A partial semigroup $f: R \to X$ with $R \subseteq X^2$ can be completed to form a total semigroup by adding a zero element, provided that it satisfies a number of conditions. These conditions are summarized in the following diagram presented below. The forbidden existence conditions are highlighted. Theorem 1. let $f: R \to X$ be a partial semigroup then adding a zero to $f$ completes it provided that:

1. If $(xy)z$ and $x(yz)$ both exist then $(xy)z = x(yz)$
2. $(xy)z$ exists is logically equivalent to $x(yz)$ existing (equivalently the four highlighted cases in the diagram above are forbidden)

Proof. there are three cases (1) they both exist in which case they both coincide by condition 1, (2) neither of them exist in which case they both produce zero and so they coincide, (3) one output exists and the other doesn't in which case they wouldn't coincide but condition 2 means this never happens. $\square$

Categories as partial semigroups:

A natural question is where do categories fit into this mathematical universe of semigroups and semirings? This is answered in two parts (1) categories are partial semigroups embedded in semigroups with zero (2) categories are certain types of idempotent semirings.

As this perspective on categories is not commonly dealt with (one mention the semigroup perspective on categories exists in stackoverflow), I want to make the exposition of this post as clear as possible. Most introductions to category theory don't mention that they are partial semigroups.

With so few mentions of the semigroup-theoretic perspective on categories, one might wonder if this perspective is even valid. I intend this post to be so simple and clear that there is no doubt that categories are indeed partial semigroups. If there is any doubt, you can always recheck the diagram at the start.

Theorem 2. Let $C$ be a category. The composition function $\circ$ of $C$ is a partial semigroup satisfying the conditions of theorem 1.

Proof. (1) by associativity $\circ$ satisfies condition 1 of theorem 1, (2) let $x,y,z$ be morphisms and suppose that $(xy)z$ exists. Then the output object of $xy$ and the output object of $y$ coincide, so that $(xy)z$ exists implies that $yz$ exists. Then the input object of $yz$ and $y$ coincide, so that $xy$ existing implies that $x(yz)$ exists, and vice versa. So $\circ$ satisfies condition 2 of theorem 1. $\square$

Then by the use of theorem 1 in combination with theorem 2, we have that we can complete the composition function of any category by adjoining a zero element.

Corollary. let $C$ be a category with composition $\circ$. Then $\circ + 0$ is a semigroup.

We can now use this to get something of the semigroup theoretic perspective on categories.

The issue of partiality

The most desirable properties in abstract algebra are associativity and distributivity. Semigroups and semirings are the most general structures containing these desirable properties. Although those are most desirable properties in abstract algebra, this doesn't entail anything about partiality.

A partial semigroup such as a category doesn't need to lose any of the most desirable properties, like associativity that make the structure convenient to work with. Indeed, a category may as well be a semigroup, as we have seen by adjoining a zero element. So the only real issue is partiality.

Question. what should the return value of composition of incompatible morphisms in a category be?

If the return value is nil, then the composition function of a category is just a semigroup: the semigroup with zero. If instead we decide to throw an error whenever incompatible morphisms are composed, that might cause a program to throw too many errors.

The adjoining of a zero/nil/null element is a possible solution, and that makes the composition function of the category a semigroup. This might just be an implementation detail, but it gets to what distinguishes categories from semigroups.

Polynomials in the noncommutative case

Let $R$ be a non-commutative ring. Then we can define the semigroup ring of $R$ over the free non-commutative semigroup $F^{\rightarrow}(A)$. Then this semigroup ring $RF^{\rightarrow}(A)$ consists of ordered words over the alphabet $A$ with coefficients in $R$. This mirrors the construction of the polynomial ring $R[x_1,x_2,...]$ in commutative algebra as a free commutative semigroup ring.

Elements of the free non-commutative semigroup ring $RF^{\rightarrow}(x,y,z)$ consist of ordered words with coefficients in $R$. For example, we might get a polynomial like $5xyx + 6xz + 7zx$. A term like $xyx$ is not the same as $x^2y$ and $xz$ and $zx$ are not the same as one another. This non-commutativity would naturally add a great deal of complexity to compuations over non-commutative rings.

The issue with this free semigroup ring, is that it is not truly the ring of polynomial functions over $R$. Consider $c_1,c_2,...$ to be constants and $x,y,z,...$ to be indeterminates. Then a coefficient may not commute with its indeterminate, and $c_1 x \not= xc_1$. So for polynomial functions over a non-commutative ring you ultimately get potentially nasty terms like $c_1 x c_2 c_3 y c_4 x c_5 z c_6$.

A term of a polynomial function in a non-commutative ring is an alternating sequence of coefficients and indeterminates, where the coefficients can be identities. So for example $c_1xyc_2yxc_3$ is a term except the coefficients between $x$ and $y$ and then between $y$ and $x$ are simply identities, but there is always the possibility of having coefficients between indeterminates.

The terms of the free semigroup ring are already complicated enough, but terms in the ring of polynomial functions with scattered coefficients are even more messy. This is a serious impediment to doing noncommutative algebraic geometry, so what is the big idea? It seems that noncommutative geometry has more to do with functional analysis and operator theory then polynomials.

Noncommutative spaces can instead be studied using concepts of functional analysis and operator theory like C*-algebras. In particular, the Gelfand representation related C* algebras to locally compact Hausdorff spaces. The study of noncommutative geometry based upon techniques from functional analysis is an interesting direction, albiet one that is very different from you might expect.

Rings of multivariable Laurent polynomials

The commutative group ring of the free commutative group $F^{\circ}(X)$ over a field $k$ is a ring extension of the commutative semigroup ring of the free commutative monoid $F(X)$ over $k$. Although, $F^{\circ}(X)$ is a natural semigroup extension of $F(X)$, rings of multivariable Laurent polynomials don't have the same level of importance as polynomial rings in commutative algebra.

A distinguishing property of ordinary polynomials is that they can be cast into functions, so that over affine space $\mathbb{A}^n$ they are functions $f: \mathbb{A}^n \to k$ and this works for any commutative ring. On the other hand, for Laurent polynomials to be cast into functions $f : \mathbb{A}^n \to k$ we need to have some notion of division to deal with negative degree exponents, so we need to work over a field.

In those cases when we take the commutative group ring of a commutative ring $R$ which is not a field, with respect to $F^{\circ}(X)$ then what we get is certainly still a commutative ring, it is just not a commutative ring of functions. Consider the commutative group ring $\mathbb{Z}(\mathbb{Z}^n,+)$ consisting of polynomials with integer exponents and integer coefficients. Then this is a valid commutative ring, but its elements are not functions because $\mathbb{Z}$ is not a field.

So in some sense we ought to have our free commutative group ring be over a field $k$. Then we get terms like $5\frac{x}{yz} + \frac{6y}{xz}$ consisting of fractional monomials, but then given any term like this we can add to get $\frac{5x^2 + 6y^2}{xyz}$ which is always a polynomial with a monomial in the denominator. So we see that multivariable Laurent polynomials are simply rational functions with monomials in the denominator, so we have an embedding. $$kF^{\circ}(x,y,z) \subseteq k(x,y,z)$$ The ring of Laurent polynomials is simply the localisation of the polynomial ring by the multiplicative set of monomials, and so they are merely a special case of rational functions. Now it is clear why we don't see Laurent polynomials so much in commutative alebra. They are simply part of the far more important and general concept of fields of rational functions $k(x,y,z)$.

Just as we don't tend to restricted localisations of the integers like the dyadic rationals that much, we won't see rings of multivariable Laurent polynomials showing up as much. In general, we always want to deal with the largest localisation of a domain, which is its field of fractions.

So although the group completion of the free commutative semigroup $F(X)$ is an important and natural concept in commutative semigroup theory, it doesn't have the same role and level of importance in commutative algebra, as determined by commutative semigroup rings. This demonstrates that not everything in commutative algebra is a consequence of commutative semigroup rings, but the use of semigroup rings is still an infinitely powerful technique in the construction of commutative rings.

Centralizers of division rings

We have that centralizers of commuting graphs determine subsemigroups and subrings, and corresponding to this there is the fact that centralizers of groups are subgroups and centralizers of division algebras are not simply subrings but division subalgebras. This can be used for example to determine the maximal subfields of a division algebra.

Subalgebras of division rings:
Let $R$ be a division ring, then $R$ as a ring is associated to a lattice of subrings. A meet subsemilattice of this lattice of subrings is the lattice of division subrings $Sub(R)$ of $R$, which consists of subrings that are also inverse closed with respect to non-zero elements.

The special case of centralizers:
By the theory of commutativity necessary subrings, we know that the centralizers of $R$ are subrings. The centralizers of $R$ together form a lattice: the centralizer lattice of the commuting graph of $R$. It remains to show that they are not only subrings but also division subrings.

Lemma 1. let $G$ be a group and $x$ an element of $G$ then the centralizer $C(x)$ is a subgroup of $G$.

Proof. the centralizer $C(x)$ is a submonoid, so let $y \in C(x)$ then $yx = xy$. Let $y^{-1}\in G$ then we want to show that $y^{-1}x = xy^{-1}$. By the fact that $G$ is a group, we can multiply both sides by $yx^{-1}$ to get $y^{-1}xyx^{-1} = 1$. Then by the fact that $x$ and $y$ commute this is equivalent to $y^{-1}yxx^{-1} = 1$ which is true, so that $y^{-1} \in C(x)$. $\square$

Lemma 2. let $S$ be a group with zero (respectively a Clifford 0-simple semigroup). Then centralizers of elements in $S$ are inverse closed.

Proof. let $x \in S$ and suppose that $x = 0$. Then the centralizer of $0$ is the entire semigroup, which is inverse closed. Suppose $x \not= 0$ then $x \in S-\{0\}$ which is a group, so the centralizer of $x$ in $S-\{0\}$ is a subgroup by lemma 1. It is therefore inverse closed. $\square$

Theorem. let $R$ be a division ring, then centralizers $C(x)$ are division subalgebras.

Proof. by commutativity necessary subrings, $C(x)$ is a subring and by lemma 2 it is inverse closed, so it is a division subalgebra. $\square$

The maximal cliques of the commuting graph of a division ring are precisely the maximal fields of the division ring. All the commutativity necessary subfields of a division ring are determined by maximal cliques and their intersections, which form a semilattice.

Centers of division rings:
The center of a division ring is precisely the intersection of all the maximal fields of the division ring. As it is the intersection of fields, it is a field. The division ring can then be seen to be a vector space over its central field. The measure of the commutativity of a division ring is the dimension of its extension over a central field: finite dimensional division rings (like the quaternions) are the most commutative.

Maximal subfields of a division ring:
An immediate corollary of this is that the maximal subfields of a division ring can be determined by the same general mechanism by which we determine maximal commutative subalgebras of semigroups, groups, rings, and semirings: they are maximal cliques of the commuting graph.

Proposition. let $D$ be a division ring then $F$ is a maximal subfield of $D$ iff $D$ is equal to its own centralizer: $C(D) = D$

It is not hard too see by the same reasoning that applies to any other algebraic structure like a group or ring, that every element of a division ring belongs to some maximal field.

Proposition. every element of a division ring belongs to some maximal field.

Maximal subfields of a division ring are a vital tool in the study of division rings. Their further properties can be studied using tensor products of division rings described as algebras over fields [1].

References:
[1] A first course in non-commutative rings
T.Y Lam

Applications of commutative semigroup rings

Let $R$ be a commutative ring. Then every commutative semigroup $S$ is naturally associated to a commutative ring extension of $R$, the commutative semigroup ring of $R$ by $S$. It is not hard to see that this construction is full of applications in commutative algebra and algebraic geometry. We will utilize commutative semigroup rings as an organizing principle in the theory of polynomial rings, which is an important part of algebraic geometry.

Polynomial rings:

The free $\mathbb{N}$-semimodule $F(X)$ is a very familiar object of commutative semigroup theory. It is not hard to see that the polynomial ring $R[x_1,x_2,...]$ is merely the commutative semigroup ring of $R$ by $F(x_1,x_2,...)$ : $RF(x_1,x_2,..)$. As a consequence, the polynomial rings that are so fundamental in algebraic geometry, can be considered to be a special case of a commutative semigroup ring.

Subalgebras of polynomial rings

Let $S$ be a finitely generated torsion-free cancellative J-trivial commutative semigroup. Then $S$ embeds into the free commutative semigroup $F(X)$ on a finite set of generators $X$. As a consequence, we can embed the commutative semigroup ring $RS$ into the polynomial ring $R[x_1,x_2,...]$.

As an example, any numerical semigroup can be embedded in the polynomial ring on a single generator. The polynomial subring $R[x^2,x^3]$ for example is merely the commutative semigroup ring of the numerical semigroup $\{2,3\}$. If we had $R[x^2y,yz^3]$ for example it would be generated by the commutative semigroup $(x^2y,yz^3) \in F(x,y)$, and so on.

Extensions of polynomial rings:

It is a basic fact of commutative algebra that $F(X)$ is a cancellative semigroup. Therefore, the free $\mathbb{N}$ semimodule $F(X)$ can be embedded in the free $\mathbb{Z}$-module $F^{\circ}(X)$. As a consequence, the commutative semigroup ring of multivariable polynomials $RF(X)$ can be embedded in the ring of multivariable Laurent polynomials $RF^{\circ}(X)$.

This can be further extended by considering rings of Puiseux polynomials $R\mathbb{Q}^n$ consisting of polynomials that have rational exponents, or this could even be embedded in $R\mathbb{R}^n$ to have arbitrary real exponents, so that we can have a complete extension of the ordinary polynomial ring $R[x_1,x_2,...]$.

Coordinate rings of varieties

Let $Y$ be an algebraic variety defined by a system of polynomial equations in $R[x_1,x_2,...]$. Then by now means is it the case that the coordinate ring $A(Y)$ can always be defined by a commutative semigroup ring. However, there is an important case in which they can be: algebraic varieties defined by differences of monomials. These correspond to relations in the presentation of a commutative semigroup.

Therefore, we can use commutative semigroup rings in algebraic geometry in order to deal with the important special case of varieties determined by differences of monomials. For example, consider the hyperbola $\frac{R[x,y]}{xy=1}$. Then this clearly produces a presentation of the commutative group $\mathbb{Z}$ so this is a ring of Laurent polynomials. As you can see, this is a very useful concept of commutative algebra.

References:
[1] Commutative semigroup rings by Gilmer

Multiset addition semigroups

The class of all multisets on a set forms a semigroup $F(S)$ with multiset addition as its operation. The additive property of $F(S)$ is formalied by a semigroup homomorphism $f : F(S) \to \mathbb{N}$ which maps each multiset to its cardinality. Given a free commutative semigroup $F(S)$ then there are two ways to form semigroups from it by taking a quotient to get a commutative semigroup presentation or taking a subalgebra to get a multiset addition semigroup.

Properties of free $\mathbb{N}$-semimodules

The free $\mathbb{N}$ semimodule $F(S)$ on a set $S$ has a number of properties that are inherited by its subalgebras. Although every commutative semigroup is a quotient of some $F(S)$, only a small number of commutative semigroups can be embedded in the free commutative semigroup $F(S)$.

Theorem. the free $\mathbb{N}$ semimodule $F(S)$ is:

1. $\mathbb{N}^S$ distributive lattice ordered
2. Cancellative
3. Torsion-free
4. Commutative and a monoid

Proof. (1) the semiring $\mathbb{N}$ is totally ordered. Therefore, $F(S)$ is ordered by the product ordering $\mathbb{N}^S$ having terms in $S$ with multiplicities in $S$. Distributive lattices are a variety of lattices that include total orders, so the product ordering on $F(S)$ is distributive.

(2) the semiring $\mathbb{N}$ is additively cancellative so that $a + b = a + c$ implies that $b = c$. It follows from additive cancellativity that we have $a + b = a + c$ for addition in the free $\mathbb{N}$ semimodule.

(3) the semiring $\mathbb{N}$ is multiplicatively cancellative for $n \not= 0$. It follows that $na = nb$ implies that $a = b$ for $n \not= 0$. Therefore, $F(S)$ is torsion-free.

(4) every semimodule is an additive commutative monoid, therefore so too is $F(S)$. $\square$

Although $F(S)$ is a distributive lattice ordered torsion-free commutative cancellative semigroup, not all of these propreties are inherited by its subsemigroups. In particular, the distributive lattice ordering is not preserved.

Definition. a commutative semigroup is subfinite provided that each element is contained in a finite number of principal ideals.

Corollary. every subsemigroup of $F(S)$ is a subfinite J-trivial commutative cancellative torsion-free semigroup

Proof. let $A \subseteq B$ be semigroups, then the algebraic preorder on $A$ is a suborder of that of $B$. Therefore, given $C \subseteq F(S)$ then the algebraic preorder on $C$ is a subpreorder of a subfinite partial order, so it is a subfinite partial order making $C$ subfinite J-trivial. It is also commutative, cancellative, and torsion-free as these properties are hereditary. $\square$

This demonstrates that not all J-trivial commutative cancellative torsion-free semigroups can be embedded in a free commutative semigroup $F(S)$. The Puiseux monoid $(\mathbb{Q}_{\ge 0}, +)$ is a J-trivial commutative cancellative semigroup but it is not subfinite so there is no way of achieving an embedding.

A notable property of the Puiseux monoid $(\mathbb{Q}_{\ge 0},+)$ is that is infinitely generated. This suggests perhaps we can produce an embedding in the finitely generated case. This is an easy corollary of Grillet's theorem.

Grillet's theorem. a monoid is finitely generated commutative cancellative reduced monoid iff it is embeddable in $\mathbb{N}^n$.

Corollary. a finitely generated commutative cancellative J-trivial monoid is embeddable in $\mathbb{N}^n$ iff it is torsion-free

This demonstrates that the only properties necessary to demonstrate that a finitely generated commutative J-trivial semigroup is embeddable in $\mathbb{N}^n$ is that it is cancellative and torsion-free. These semigroups can therefore be expressed as multiset addition semigroups.

Factorisation in $F(S)$ subsemimodules

The free commutative semigroup $F(S)$ is a $\mathbb{N}$ semimodule, and so most important operations over it can be solved by linear algebra over the natural numbers. A finitely generated subsemirgoup of $F(S)$ can be described by the span of a multiset system $\{M_1,M_2,...\}$ which is the set of all linear combinations of the system of multisets. Each solution is a different factorisation.

Example 1. consider the subsemigroup $xy,x^2,y^2$. Then every factorisation of $x^n,y^m$ is a solution of the following system of linear equations: $$\begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 2 \end{bmatrix} * v = \begin{bmatrix} n \\ m \end{bmatrix}$$ Each solution to the system of linear equations produces a different factorisation of the multiset. For example, $x^4y^4$ has three factorisations: $(x^2)^2(y^2)^2$,$(xy)^2x^2y^2$,$(xy)^4$.

Example 2. consider the subsemigroup $x^3,x^2y,xy^2,y^3$. Then a factorisation of $x^n,y^m$ is a solution to the following system of linear equations: $$\begin{bmatrix} 3 & 2 & 1 & 0 \\ 0 & 1 & 2 & 3 \end{bmatrix} * v = \begin{bmatrix} n \\ m \end{bmatrix}$$ Now $x^6 y^6$ has five factorisations: $(x^3)^2(y^3)^2$, $(xy^2)^2 (x^2y)^2$, $x^3 y^3 x^2y xy^2$, $x^3 (xy^2)^3$, $y^3 (yx^2)^3$.

This demonstrates by linear factorisation, that not all commutative J-trivial cancellative torsion-free finitely generated semigroups have unique factorisations. Although $F(S)$ does have unique factorisations, so that each element is uniquely expressed as a multiset.

Definition. a commutative J-trivial semigroup is called factorial provided that every element has a unique factorisation.

Example. the condensation $\frac{*}{H}$ of the multiplicative semigroup $*$ of a UFD is a factorial commutative cancellative J-trivial semigroup.

Notice that $x^2,y^2,xy$ determines a commutative subsemigroup but $x^2,x^2,xy,x^2y^2$ determines the same semigroup. We therefore need one more concept in order to enable computations on multiset systems related to the subsemigroups they generate:

Definition. a multiset system $S$ is sum minimal provided that $\forall x : x \not\in (S-x)$ so that no element $x$ is generated by the other elements in the multiset system.

For example, we can describe a numerical semigroup by a minimal set of generators, which is a simple combinatorial data structure we can work with. With this definition, it is a fairly simple procedure to create an algorithm to check if a given multiset system is sum minimal by solving a system of linear equations to check for factorisations of each element.

Proposition. the category of free commutative monoids is equivalent to the category of $\mathbb{N}$ semimodules with natural matrices between them.

The linear algebraic approach to free $\mathbb{N}$ semimodules allows us to describe any homomorphism of $\mathbb{N}$ semimodules by natural matrices. In particular, the endomorphism semiring $End(F(S))$ of a free commutative semimodule is equivalent to a matrix ring $Mat_S(\mathbb{N})$ over the semiring of natural numbers.

The factorisation of multisets can be determined by solving systems of linear equations over the natural numbers, or by determining the inverse image of a natural-valued matrix. This leads to the linear algebraic approach to $\mathbb{N}$ semimodules.

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.

Directed preorders and semigroups

It should be readily apparent that not all posets can appear as the orderings of J-trivial semigroups. By the same token, not all preorders can arise from semigroups in general. A first thought is that perhaps all orderings of J-trivial semigroups are semilattices, but of course this isn't the case for semigroups of order greater then four: because there is a unique [2,2,1] weak ordered commutative J-trivial semigroup. What we can get is a generalisation of semilattices: the class of directed posets.

Theorem. let $S$ be a semigroup then its algebraic $J$ preorder $\subseteq$ is a directed preorder, which means that every two elements $a$ and $b$ have an upper bound.

Proof. let $a,b \in S$ then we can construct an upper bound by $ab$ $\square$

Corollary. every J-trivial semigroup is a directed poset.

Directed posets can be seen to be the posets that have some kind of compositional process associated to them, even if that compositional process is not a semilattice. Upper bounds correspond to the composition of elements.

Example. the free non-commutative semigroup $S$ on a set of generators $X$ is a J-trivial semigroup, whose partial ordering is the consecutive subsequence ordering. This is not a semilattice, but it is directed because any two words have an upper bound provided by their composition.

The example of the free non-commutative semigroup should demonstrate that $L$ and $R$ preorders are not directed in general, even though $J$ is. The prefix and postfix orderings on the free non-commutative semigroups are trees, so elements don't have an upper bound unless they are comparable. There is also the example of a pure rectangular band, which has one of its $L$ or $R$ preorders an antichain.

The reason every pair of elements in a semigroup has an upper bound, is every element has a composition that exists. It follows that in a partial semigroup such as a category, its preorder need not be directed. Morphisms in different connected components don't need to have an upper bound.

• The $L$ and $R$ preorders of a semigroup need not be directed.
• The morphism preorder of a category need not be directed.

A directed preorder describes a set of elements for which a compositional process exists, but those don't need to exist in a partial semigroup such as a category. Nonetheless, we can use directed posets in the study of J-trivial semigroups. Directed posets still have suprema they just don't need to be unique, so we have to adjust the theory a little bit.

Rectangular band-free semigroups

The least commutative semigroups are rectangular bands, so the condition that a semigroup is rectangular band free is a natural generalization of commutativity. It means that there are no non-trivial anticommutative subsemigroups. Idempotent commutative semigroups such as inverse semigroups are rectangular band free.

It can be seen to be dual to E-semigroups, because all rectangular-band free E-semigroups are idempotent commutative. This follows from Clifford's theorem which characterizes all rectangular band free semigroups as semilattices. As a result, the intersection of the class of rectangular band free semigroups and E-semigroups is the class of idempotent commutative semigroups.

Definition. a semigroup is rectangular band free if it contains no subsemigroups of order two or greater that are rectangular bands

The relationship between rectangular band freeness and idempotent commutativity are dealt with in the following ontology of classes of semigroups. Idempotent commutative semigroups, as well as commutative semigroups themselves are rectangular-band free. While the relationship between idempotent commutativity and rectangular band freeness is interesting. It is also noticeable that rectangular band free semigroups tend to be divisibility commutative (L=R) for the smallest semigroups. Rectangular bands are responsible for different L and R relations in the smallest semigroups.

The weakest generalisation of commutativity is that semigroups be anticommutative semigroup free, because the anticommutative semigroups clearly have the least commutative behaviour (the most different Green's L and R relations, the least commutativity, etc). As a result, it is worth highlighting this as a possible generalisation of commutativity.

Proposition. a semigroup of order four or less is divisibility commutative iff it is rectangular band free

We see that semigroups that are of order four or less are rectangular band free iff they are divisibility commutative, but by no means is this the case in general. If it were, then every inverse semigroup would be Clifford.

Example 1. the inverse semigroup on five elements containing all non-permutation charts on two elements is the smallest non-Clifford inverse semigroup. It is idempotent commutative but not divisibility commutative. In general, non-trivial symmetric inverse semigroups are rectangular band free but not divisibility commutative.

Example 2. the monoid of increasing monotone increasings on a three element total order, is a four element tree ordered J-trivial semiband. As a non-commutative J-trivial semiband it is rectangular band free and not idempotent commutativity. In general, non-commutative J-trivial semibands are rectangular band free and not idempotent commutative.

The existence of non-commutative J-trivial semibands as well as non-Clifford inverse semigroups demonstrates that while rectangular band freeness is bascially the weakest generalisation of commutativity (by forbidding anticommutative components) it doesn't imply any of the other nice conditions like divisibility commutative and idempotent commutativity.

There are a couple of things that can be said about the commuting graphs of rectangular band free semigroups. A graph is idempotent commutative provided that all singleton centralizers form a clique. It is rectangular band free only if no non-trivial centralizers are independent sets. The cyclic graph $C_4$ is an ordered pair of rectangular bands in two different ways, so it cannot be rectangular band free.

In general, a complete bipartite graph $K_{n,m}$ with $n,m \ge 2$ is an ordered pair of rectangular bands, and so it cannot be rectangular band free, for example. The path graph $P_4$ on four elements is not idempotent commutative, but it doesn't have any commutativity necessary rectangular band subsemigroups, so it is an example of the weaker condition.

See also:
[1] Divisibility commutativity

[2] Idempotent commutativity

Finite commutative totally preordered semigroups

The subject of this examination is the location of the non-trivial non-group H classes in a commutative semigroup. This can be used to solve the inverse condensation problem of which semigroups have a given finite total order commutative condensation type $\frac{S}{J}$. This naturally follows from semilattice decompositions, which reduce such semigroups into their Archimedean components.

A consequence of this is that we can completely characterize all commutative totally preorder semigroups by semilattice decompositions. This naturally generalizes to condensible totally preordered semigroups with commutative condensation, by replacing the J-classes with other simple semigroups. Generalizations of the structure theorem will be briefly considered.

Lemma 1. non-trivial group $H$ classes in commutative semigroups are enclosed

Proof. (1) suppose that a given non-group $H$ class is maximal then $H^2 \cap H = \emptyset$ means that $H$ iterates to produce something greater then it, but it is maximal so there is nothing for it to produce which is a contradiction. (2) suppose that a given non-group $H$ class is minimal then since $H^2 \cap H = \emptyset$ the action representatives moving elements around in $H$ must come from its predecessors, but no such predecessors because it is minimal. $\square$

This is why commutative semigroups of order four or less are all group symmetric. The existence of non-trivial non-group $H$ classes necessitates the existence of extra elements to inclose the $H$ classes.

Lemma 2. let $S$ be a commutative semigroup with finite monogenic condensation $\frac{S}{H}$, then all non-maximal $H$ classes in $S$ are singletons.

Proof. let $C$ be a non-maximal $H$ class in $S$, then since $S$ has monogenic condensation all non-maximal $H$ classes are non-groups. Suppose that $C$ is non-trivial and has at least two elements, then there is some action representative predecessor of it that moves any given element $x \in C$ to any other element $y \in C$. This means that there is some other $H$ class $D$ such that $CD = C$.

Naturally, absorpotion $CD = C$ means that $CD^n = C$. Every finite monogenic semigroup generates an idempotent, so the condition that $CD^n = C$ means that there must be some idempotent predecessor of $C$, but there are no idempotent predecessors of an element in a finite monogenic J-trivial semigroup. So $C$ must be a singleton. $\square$

We can now apply this lemma to totally preordered commutative semigroups to produce the structure theorem. This makes use of semilattice decompositions, condensation, and the characterization theorem for finite totally ordered commutative semigroups.

Theorem 1. a finite commutative totally preordered semigroup is a semilattice of Archimedean commutative semigroups with total preorder types [1 1 1 ... n]

Proof. $S$ is a commutative totally preordered semigroup so can be condensed $\frac{S}{H}$ into a finite commutative totally ordered semigroup. This is then a semilattice of totally ordered monogenic semigroups. Each of the Archimedean components in this semilattice is a semigroup with monogenic condensation as in lemma 2, and so all non-maximal $H$ classes are singletons. $\square$

A finite Archimedean commutative totally preordered semigroup can be identified with the data of a finite commutative group $G$ and an element of a given index that generates some cyclic subgroup of $G$. Finite monogenic semigroups are precisely the commutative totally preordered semigroups with an element that generates the entire group $G$.

Lemma. let $S$ be a monogenic semigroup, then $S$ is a totally preordered semigroup

Proof. let $x$ be the generator for $S$ then every element is of the form $x^n$ for some $x \in \mathbb{N}$. Then we have a total order $x \subseteq x^2 ...$ containing all the elements of $S$, so the algebraic preorder of $S$ must be a total preorder. $\square$

We can now construct an ontology: there are a number of classes of commutative semigroups dealt with in this classification determined by the layers of the decomposition provided by condensation, semilattice decomposition, and the fundamental theorem of finite commutative groups. The class of Clifford finite commutative totally preordered semigroups, is the class of linear semilattices of groups. In that case, the only relevant detail not covered by the structure theorem is the action of predecessor groups on their successors, which can actually be reduced to the action of a predecessor group to its parent's identity element.

This can be generalized first to divisibility commutative semigroups with commutative total order condensation, by replacing the finite commutative groups with any finite groups in general. Then secondly, it could be generalized to all totally preordered semigroups by replacing groups with other J-total semigroups like rectangular bands.

Chain conditions on commutative semigroups

The chain conditions on commutative semigroups can occur on either subalgebras, ideals, or principal ideals. The stongest of these are the chain conditions on subalgebras.

Theorem. let $S$ be a commutative semigroup for which $Sub(S)$ satisfies both the ACC and the DCC then $S$ is finite.

Proof. (1) the ACC on subalgbras means that $S$ is finitely generated, for if it were not then an infinite gneerating set for it would create an infinite ascending chain (2) the DCC on subalgebras means that $S$ is monofinite, meaning that for each $x$ in $S$ the principal subalgebra $(x)$ is finite. If it were not finite, then $(x)$ would generate a $(\mathbb{Z}_+,+)$ semigroup which has an infinite descending chain of subalgebras (3) $S$ is finitely generated, each element generates a finite number of elements, and $S$ is commutative so $S$ is finite. $\square$

The situation with respect to $\mathbb{Z}$-modules is a bit different. In that case, there is no distinction between subalgebras, ideals, and principal ideals: there are only submodules. An analogous result shows that $\mathbb{Z}$ modules satisfying both the ACC and DCC on subalgebras are finite. This theorem is in the same vein of the following familiar theorem from commutative algebra:

Proposition. let $M$ be a module satisfying both the ACC and the DCC then $M$ has finite length. [1]

This is one case in which commutativity is necessary. We have that finitely generated commutative semigroups in which each element has a finite principal subalgebra are finite, but the converse does not need to be the case for non-commutative semigroups or groups. It has been shown that there is a finitely generated torsion group that is not finite for example [2].

References:
[1] Commutative algebra volume one Zariski and Samuel

[2] Burnside problem

Rees factors of J-trivial semigroups

The arithmetical properties of finite boolean algebras and partition lattices are Rees factor semigroups of addition and multiplication. This naturally lends us to a further study of the Rees factor semigroups of J-trivial semigroups. J-trivial semigroups are subalgebra closed, but not quotient closed so a first step is to check if Rees factor semigroups always preserve J-triviality.

A necessary condition in that direction is to determine if all the congruence classes of a Rees semigroup congruence in a J-trivial semigroup are convex. It is common when dealing with ordered algebraic structures to require congruence classes are convex, for example all congruence classes of a lattice congruence are convex. This is an easy verification.

Proposition. let $S$ be a J-trivial semigroup and $I$ an ideal of $S$ then the congruence classes of the Rees semigroup congruence $C$ associated to $I$ are all convex.

Proof. (1) singleton sets are convex (2) ideals are convex because if $a,b \in I$ such that $a \subseteq c \subseteq b$ then $a \subseteq c$ implies $c$ is in $I$. So congruence classes are convex. $\square$

It is an easy verification that congruence classes of the Rees semigroup congruence are convex. It is a bit harder to show that their quotient is J-trivial.

Theorem. let $S$ be a J-trivial semigroup and $I$ an ideal then $\frac{S}{I}$ is J-trivial with partial order $I^C + 0$.

Proof. (1) suppose that $a,b \in I^C$ then $a \subseteq b$ in $\frac{S}{I}$ provided that $\exists x,y \in (\frac{S}{I})^1 : xay = b$. Then suppose either $x$ or $y$ is $I$ then $xay = b$ implies that $I \subseteq b$ which means that because $b$ is a filter than $b \in I$ but that is a contradiction because we had that $b \in I^C$ so $x,y \in I^C$ which means that the order of $I^C$ is a suborder of $S$.

Then to prove that the ideal $I$ is an adjoined zero element in $\frac{S}{I}$ we need to prove two conditions (1) $I$ is not less then any element in $I^C$. This follows because $I$ is an ideal, so it includes all its successors. (2) every element is less then $I$ in $\frac{S}{I}$. This one follows from the fact that J-trivial semigroups are directed posets, so every element has a successor in an ideal. $\square$

As this proof involves the characterization of the algebraic preorders of Rees factor semigroups, we can translate this into order theoretic terms to get the Rees factors of a preorder directly. In general, its good to have our terminology from order theory and semigroup theory coincide.

Definition. the Rees factor of a preorder with respect to an upper set (respectively a lower set) is the quotient order with respect to ordinary precedence produce by collapsing the upper set (respectively a lower set) into a single element and preserving every other element.

The partial orders of finite J-trivial semigroups are not always semilattices. Therefore, it is not an entirely trivial result that the Rees factors of $(\mathbb{Z}_+,*)$ by total order upper sets are semilattice ordered.

Proposition. The Rees factor semigroups of $(\mathbb{Z}_+,*)$ by total order upper sets $\{x : x \ge n\}$ are semilattice ordered.

The partition lattices are important because they provide the logical foundations for the semantics of multiplication. We can now characterize some of the properties of the arithmetic properties of finite partition lattices by Rees factors.

Proposition. let $\frac{(\mathbb{Z}_+,*)}{\ge n}$ be the Rees factor semigroup of positive integer multiplication. Then it has at most two idempotents (1 and $n$) and every other element has index equal to the number of powers strictly less then $n$ plus one.

Proof. (1) $n$ and the $\ge n$ class are preserved as idempotents (2) let $x$ be any other element then $x^m$ is equal to $n$ provided that $x^m \ge n$ so the index of $x$ is the number of $x^m \not= n$ plus one for $x^m = n$. $\square$

Additional properties of these Rees factor semigroups could be determined analytically, such as by applying the prime number theorem, so a variety of methods can be applied to study Rees factors of the positive integers multiplication semigroup.

Condensation of commutative cancellative semigroups

The class of commutative cancellative semigroups is subalgebra closed, but it is not quotient closed. The free commutative monoid $F(x)$ is commutative and cancellative but $\frac{F(x)}{x^2=x}$ is not. If it is in the case that the condensation $\frac{S}{H}$ is cancellative, it is therefore not an inherent property of the class of commutative cancellative semigroups.

Theorem. let $S$ be a commutative cancellative semigroup then $\frac{S}{H}$ is cancellative.

Proof. suppose $a =H a'$ and $ax =_H a'y$ then $a = ba'$ and $a' = ca$ for some $b,c$. Furthermore, $ax \lambda = a'y$ ad $ax = a'y \mu$. By plugging in we get $$(ba')\lambda x = a'y \Rightarrow a'(b\lambda x) = a'y$$ $$ax = (ca)\mu y \Rightarrow a(x) = a(c \mu y)$$ By the cancellative property this yields $$ax = a(c\mu y) \Rightarrow x = (c\mu)y$$ $$a'(b\lambda x) = a'y \Rightarrow y=(b \lambda)x$$ Thusly, $x =_H y$. We can conclude that $a =H a'$ and $ax =_H a'y$ implies that $x =H y$. This is our first step. Let $H_1,H_2,H_3$ be three $H$ classes and suppose that $H_1 H_2 = H_1 H_3$ then we have: $$\exists h_1,h_1' \in H_1, h_2 \in H_2,h_3 \in H_3 : h_1 h_2 =_H h_1' h_3$$ We have that $h_1 =_H h_1'$ as well ast that $h_1 h_2 =_H h_1' h_3$ but as we previously concluded that this means that $h_2 =_H h_3$. If $h_2 \in H_2$ and $h_3 \in H_3$ then $H_2 = H_3$ because $H$ classes are disjoint.

Then $H_1 H_2 = H_1 H_3 \Rightarrow H_2 = H_3$ implies that the condensation semigroup $\frac{S}{H}$ is a commutative cancellative J-trivial semigroup. $\square$

This is an infinite source of commutative cancellative J-trivial semigroups. For example, it follows from this theorem that the ideal multiplication semigroup of a principal ideal domain is a commutative cancellative J-trivial semigroup with zero.

Subgroups form a sublattice

Every group $G$ is associated with a lattice of subgroups $Sub(G,*,1,/)$ but it also associated with a lattice of submonoids $Sub(G,*,1)$. By this, we will see that the subgroups lattice of $G$ is a sublattice of the submonoid lattice.

Lemma 1. let $G$ be a group and $S$ an inverse-closed subset then composition closure $cl(S)$ is inverse closed.

Proof. let $a,b \in S$ then by composition closure $ab \in cl(S)$. By inverse closure $a^{-1} \in S$ and $b^{-1} \in S$. So by composition closure, $b^{-1}a^{-1} \in cl(S)$, but this is equal to $(ab)^{-1}$ so $(ab)^{-1} \in cl(S)$. It follows that $cl(S)$ is inverse closed. $\square$

Theorem. let $G$ be a group then the lattice of subgroups $Sub(G,*,1,/)$ is a sublattice of the lattice of submonoids $Sub(G,*,1)$.

Proof. let $A,B$ be subgroups of $G$. Then $A,B$ are inverse closed, so $A \cup B$ is inverse closed. By lemma 1, $cl(A \cup B)$ is a subgroup. With respect to the lattice of submonoids $A \vee B$ coincides with $cl(A \cup B)$ so the monoid join of subgroups is a group. It follows that subgroups are a sublattice of submonoids. $\square$

By extending previous results we get the following chain of sublattices: normal subgroups $\subseteq$ groups $\subseteq$ monoids $\subseteq$ subsemigroups. Previously we also saw that the lattice of partitions $Part(A)$ is a sublattice of the lattice of preorders. With similar results from both order theory and semigroup theory, it is reasonable to assume that this can be extended to categories.

This is indeed possible using lattice of subcategories, but for now I am interested in applications of this approad to semigroup theory. A significant result to that effect is that if $S$ is a finitely generated cancellative submonoid of a group $G$ then its subgroup closure is finitely generated as well.

Lemma 2. let $G$ be a group and $S$ a submonoid then if $S$ is finitely generated its group closure $H$ is a finitely generated as well.

Proof. let $X$ be the finite generating set of $S$. Then since $X$ is finite $X \cup -X$ is finite as well. By lemma 1, $X \cup -X$ is a finite generating set for $H$, so $H$ is finitely generated as well. $\square$

Theorem. finitely generated commutative cancellative torsion-free J-trivial monoids are affine

Proof. let $S$ be a finitely generated commutative cancellative J-trivial monoid. Then let $G$ be its Grothendeick group, then since $S$ is J-trivial $G$ is torsion-free. Since $S$ is finitely generated, $G$ is as well. By the classification of torsion-free finitely generated commutative groups, $G$ is isomorphic to $\mathbb{Z}^d$ for some $d$. $\square$

It is very easy to construct affine semigroups, by taking subsets of $\mathbb{Z}^d$. However, it is harder to determine when a given commutative cancellative semigroup is affine. This is a partial result to that effect.

Arithmetical properties of semilattices

Classical abstract algebra deals with homomorphisms between algebraic structures defined by certain equalities, by generalizing to inequalities we will treat the arithmetic properties of semilattices. A motivating application of this subject is that addition and multiplication of positive integers are properties of semilattices. Generalizing from this, we provide a general framework for associating commutative magmas to semilattices.

Order theoretic foundations of arithmetic

Every finite set is associated with a cardinal number. This associates the union semilattice of sets to addition in the following way: $$|A \cap B| = 0 \Rightarrow |A| \cup |B| = |A| + |B|$$ This forms a partial homomorphism from the union semilattice of finite sets to the addition commmutative semigroup of the non-negative integers. More then this, we have the following inequality: $$|A \cup B| \leq |A| + |B|$$ Thusly, addition is the maximal cardinality provided by the union of two sets with respect to their cardinalities. We can think of addition as a process by which we can combine two disjoint sets together. $$|\{0,1\} \cup \{2,3\}| = |\{0,1\}| + |\{2,3\}|$$ The multiplication of positive integers $(\mathbb{Z}_+,*)$ is associated to the intersection semilattice of set partitions. Two set partitions form an intersection matrix. The entries of this intersection matrix are the intersections of pairs of elements from each set partition.

	0,1	2,3	4,5
0,2	0	2	$\emptyset$
1,4	1	$\emptyset$	4
3,5	$\emptyset$	3	5

In the preceding example we see that not all entries in the intersection matrix of the two set partitions $\{\{0,1\},\{2,3\},\{4,5\}\}$ and $\{\{0,2\},\{1,4\},\{3,5\}$ are filled with non-empty elements. That means that they don't multiply, instead we get that $|P| = 3$, $|Q| = 3$, and $|P \cap Q| = 6$. Two set partitions multiply provided that all entries in their intersection matrix are non-empty as in the following example.

	0,1	2,3,4
0,2,4	0	2,4
1,3	1	3

These two set partitions form an indirect product of one another, because not every element in their intersection matrix is a singleton. The special case of a direct product occurs when each pairwise intersection of the two set partitions has cardinality one such as in the example below.

	0,1,2	3,4,5	6,7,8
0,3,6	0	3	6
1,4,7	1	4	7
6,7,8	6	7	8

With these intersection matrices, we see that there is a special case in which the interesction of set partitions is multiplicative much like how the union of sets is additive. This is described in the formula below: $$\forall s \in P, t \in Q : |s \cap t| \not= 0 \Rightarrow |P \cap Q| = |P| + |Q|$$ Multiplication can be thought of as determining the area of a rectangle where the two numbers are the width or height of the rectangle. In practice this rectangular area is the area enclosed by the intersection matrix of the two set partitions. As all entries of the intersection are enclosed in this rectangle, we have the following inequality: $$|P \cap Q| \leq |P|*|Q|$$ Addition is associated to the union of sets and multiplication is associated to the intersection of partitions. This is encodeded in the following table:

Table of commutative semigroups

	Semilattice	Arithmetic
Additive theory	Set union	Addition
Multiplicative theory	Partition intersection	Multiplication

The columns of this table can be combined to form a topos or a commutative semiring. The rows of this table form order morphisms, which describe certain inequalities inherent to semilattices. These order morphisms are our object of study today.

Order morphisms:

One of the central concepts of category theory is that of an arrow category. Using arrow categories, homomorphisms can be described by certain morphisms of morphisms: $(i,o): f \to g$. This commutative diagram can be described in symbolic form as: $$o(f(x)) = g(i(x))$$ The idea of an order morphism is to replace the equality of a commutative diagram with an inequality: $$o(f(x)) \leq g(i(x))$$ $$o(f(x)) \geq g(i(x))$$ An order-theoretic morphism of functions has two forms: $f \ge g$ and $g \ge f$, determined by the two ways we change the equality in the commutative diagram into an inequality. An example from commutative algebra is the valuation map from a field to a totally ordered group, which has the following inequality: $$v(a+b) \geq min(v(a),v(b))$$ If we look back at the inequalities that relate addition and multiplication to their respective semilattices, they take the same form. This puts these inequalities into a broad class, which we can be apply to a number of different algebraic structures. $$|A \cup B| \leq |A| + |B|$$ $$|P \cap Q| \leq |P|*|Q|$$ To simplify things, we tend to focus on an order morphism of form $(v^2,v)$ which can be defined by a single function $v$. The function $v : A \to B$ can have anything from a semilattice to a field as its input, but its output must always be always be a poset of some kind.

If $B$ is a poset, then the hom class $Hom(B^2,B)$ is partially ordered so that $g_1 : B^2 \to B$ is less then or equal to $g_2 : B^2 \to B$ provided that $\forall b : g_1(b) \leq g_2(b)$. Then given a function $f : A^2 \to A$ and a function $v: A \to B$ there is a subset $S$ of the hom class $Hom(B^2,B)$ with $g \in S$ provided that: $$v(f(x,y)) \leq g(v(x),v(y))$$ Suppose that we have two functions $g_1, g_2 : B^2 \to B$ such that $g_1 \leq g_2$ then if $v(f(x,y)) \leq g_1(v(x),v(y))$ then $v(f(x,y)) \leq g_2(v(x),v(y))$. So that set of all $g \in S$ is an upper set in $Hom(B^2,B)$. The function $g$ is an upper bound of $f$ with respect to $v$.

Definition. let $v : A \to B$ be a function to a poset $B$ and let $f: A^2 \to A$. Then $g : B^2 \to B$ is a functional upper bound provided that $v(f(x,y)) \leq g(v(x),v(y))$. Dually, it is a functional lower bound provided that $g(v(x),v(y)) \leq v(f(x,y))$.

The fact that $Hom(B^2,B)$ is partially ordered means that we can define functional least upper bounds and functional greatest lower bounds: or functional infima and suprema for short. This is the key concept we need to define the arithmetical properties of semilattices.

Definition. let $v : A \to B$ be a function to a poset $B$ and let $f: A^2 \to A$. Then $g : B^2 \to B$ is a functional suprema provided that $v(f(x,y)) \leq g(v(x),v(y))$ and $g$ is minimal amongst all $Hom(B^2,B)$ that have this property. Functional infima are defined dually.

A key realisation is that addition and multiplication are not just upper bounds for set union and partition intersection, they are the functional suprema of these semilattices. So addition and multiplication are inherent properties of their respective semilattices.

Semilattice theory:

In order to define the functional suprema of semilattices, we first need some ranking function to optimize for. A ranking function is naturally provided for any finite semilattice $S$: $$h : S \to \omega$$ Wherein $h(x)$ is the maximal chain length of the principal ideal of $x$. This can naturally be generalized from finite semilattices to any semilattice which forbids the following total order types: $\{\omega^*,\omega+1\}$. Then the maximal chain length of the prinicipal ideal of any element will be finite.

We have a semilattice operation $\vee$ and a ranking function $h$ so now we can define a functional upper bound to be any function $\cdot$ with $h(a \vee b) \leq h(a) \cdot h(b)$. Furthermore, we can define the functional suprema to be the $\omega+1$ join of all heights of joins of elements with a given height pair: $$n \cdot m = \bigvee \{h(a \vee b) : h(a) = n, h(b) = m \}$$ This can be used as a simple algorithm to compute the arithmetical properties of the simplest semilattices. The resulting binary operation is the functional suprema of $\vee$ with respect to $h$. We call this operation the arithmetic operation of the semilattice.

Definition. the arithmetical operation of a well ordered semilattice $\vee$ is the functional suprema of $\vee$ with respect to $h$.

Addition and multiplication of positive integers can be produced in this way, from certain semilattices. As we shall see, there are other commutative operations associated to semilattices.

Finite boolean algebras:
Let $B_n$ be a finite boolean algebra. Then $B_n$ is associated to a commutative aperiodic monogenic monoid $I_n$ .

Proposition. the arithmetic operation of $B_n$ is the commutative aperiodic monogenic monoid $I_n$.

Let $(\mathbb{N},+)$ be the arithmetic operation of finite set union. Then $I_n$ is the Rees factor semigroup of the ideal $(n+1)$. The arithmetical property of a finite boolean algebra can be considered to be a part of the arithmetic of finite set union determined by a Rees semigroup congruence.

Finite partition lattices:
Let $Part(A)$ be the finite partition lattice. Then the arithmetic operation of $Part(A)$ is the Rees factor semigroup of $(\mathbb{Z}_+,*)$ determined by the ideal $(n+1),...$ where $n$ is the cardinality of $A$.

Proposition. the arithmetic of $Part(A)$ for a finite partition lattice is the Rees factor semigroup of $(\mathbb{Z}_+,*)$ determined by the ideal $(n+1),...$.

In this senes, the arithmetical properties of finite partition lattices precisely mirror those of finite boolean algebras. Much as in that case, these commutative semigroups are arrived at by cutting off part of the possibilities of multiplication.

The algebraic preorder of these Rees factor semigroups of multiplication all have meet subsemilattices of the first $n$ integers with an extra number adjoined as their algebraic preorders.

Interval semilattices:
Let $I$ be the set of all non-empty intervals in the range from $1$ to $n$. Then the commutative semigroup associated to $I$ is the null semigroup in which every element goes to zero.

Proposition. the arithmetic operation of the interval semilattice is the null commutative semigroup

This can be clearly seen because in an interval semilattice we can always maximize any pair by choosing elements of a given size at opposite sides of the semilattice. As a result, null semigroups are yet another example of a commutative semigroup associated to a semilattice.

Ordinal sums
Let $S$ and $T$ be semilattices and consider their ordered sum $S + T$ in which every element in $S$ is less then every element in $T$. Then the result is a semilattice, with arithmetic operation equal to the ordered sum of $S$ and $T$.

In particular, if $S$ and $T$ are associated to commutative semigroups $A$ and $B$ then $S+T$ is associated to the commutative semigroup $A+B$ with ${{A},{B}}$ a congruence with ordered pair semilattice quotient. An immediate application of this is that the arithmetic operation of a finite total order is trivial:

Proposition. let $T_n$ be a finite total order semilattice, then its arithmetic operation is $T_n$ itself.

Commutative monogenic aperiodic commutative semigroups can be formed by the finite boolean algebra minus the empty set. Then by the characterization theorem of finite totally ordered commutative semigroups this can be used to construct all totally ordered finite commutative semigroups.

Elementary properties:
Given this formalisation, a number of elementary properties of the arithmetic operations of semilattices can be inferred.

Lemma. $h$ is monotone

Proof. let $a \subseteq b$ then a maximal chain of $a$ can be extended to a maximal chain of $b$. The larger maximal chain of $b$ has a larger cardinality, so the maximal chain length of the principal ideal of $b$ is at least that of $a$. $\square$

The fact that $h$ is monotone means that the functional suprema of $\vee$ with respect to it is biextensive. The fact that semilattices are commutative means that their arithmetical operations are as well. These combine into the characterisation theorem for arithmetical operations of semilattices.

Theorem. let $S$ be a semilattice with finite maximal chain length $n$. Then the arithmetic operation $\cdot$ of $S$ is a bi-extensive commutative magma with $n$ elements.

Proof. The commutative property of $\cdot$ follows immediately from the commutativity of the semilattice $\vee$. By the fact that $h$ is monotone we have $h(x) \subseteq h(x \vee y)$ and $h(y) \subseteq h(x \vee y)$. This implies that $h(x) \vee h(y) \leq h(x \vee y)$. By the fact that $\cdot$ is a functional upper bound $h(x \vee y) \leq h(x)\cdot h(y)$ and so by combining inequalities we have $h(x) \vee h(y) \leq h(x \vee y) \leq h(x) \cdot h(y)$. This can be reduced to $h(x) \vee h(y) \leq h(x) \cdot h(y)$ which implies that $\cdot$ is bi-extensive. $\square$

Commutative semigroups that are biextensive over a partial order are necessarily J-trivial, so the following collary immediately follows:

Corollary. if the arithmetic operation of a semilattice is a semigroup, it is a commutative J-trivial semigroup

It is worth noticing that the addition and multiplication of positive integers are commutative J-trivial semigroups as is any semilattice. The class of commutative J-trivial semigroups generalizes semilattices to allow for elements that are not idempotent.

An exceptional semilattice
The following semilattice is not height associative: This has the following non-associative commutative magma as its arithmetic operation:

[[1 2 3 4 5]
 [2 3 4 4 5]
 [3 4 4 5 5]
 [4 4 5 5 5]
 [5 5 5 5 5]

This demonstrates that semilattices don't necessarily need to have associative arithmetic operations.

Idempotent height classes:
We have seen that a number of commutative J-trivial semigroups arise from the arithmetical properties of semilattices, but there are some restrictions to the set of possible commutative semigroups that arise in this way. The first restriction is a consequence of a property of idempotent height classes.

Theorem. let $n$ be an idempotent height class in a semilattice $S$ then $n$ has at most one element.

Proof. suppose that $h(a) = h(b) = n$ then by the fact that $h$ is monotone $h(a) \vee h(b) \leq h(a \vee b) \leq h(a)\cdot h(b)$ but $h(a) \vee h(b)$ is not either $h(a)$ or $h(b)$ because $a \not= b$ so $h(a) \vee h(b) \le h(a) \cdot h(b)$ which means that $n$ is not idempotent. $\square$

This can be used to characterize the arithmetical properties of finite graded semilattices. If a semilattice is graded, then every element must go through an idempotent height class which reduces the operation to an ordinal sum.

Proposition. every finite graded semilattice has an arithmetic operation which is the ordinal sum of finite commutative unitpotent magmas. In the commutative case it is an ordinal sum of nilpotent semigroups.

In the case of posets that are not graded, this restriction doesn't need to occur. For example, $[{[1,1],1},1]$ is the smallest non-graded tree and it has as an arithmetic operation the unique commutative J-trivial semigroup of order type [2,1] with two idempotents.

Corollary. the only idempotent commutative magmas that arise from finite total orders are the finite total order semilattices.

The smallest commutative J-trivial semigroup that cannot emerge from the arithmetic operation of any semilattice is the semilattice with order type [2,1] which is the smallest semilattice that is not a total order. We can construct commutative semigroups of order type [n,1] with one less idempotent then a semilattice by taking a chain and a singleton and adjoining a common parent to both of them.

Height preserving suborders:
The different arithmetic operations on the height classes of a partial order are partially ordered pointwise. As a consequence, the arithmetic operation is monotone over height preserving order extension.

Proposition. let $P \subseteq E$ be a height preserving suborder then the arithmetic operation of $P$ is less then that of $E$ in the pointwise ordering.

A finite boolean algebra is a case in point, given any maximal chain in $B_n$ we get a total order semilattice which is less then addition: $max(a,b) \leq a+b$. The number of join irreducibles in a finite distributive lattice is one less then the maximal chain length. The maximal chains correspond to linear extenions of the join irreducibles. This creates a height preserving embedding of a finite distributive lattice in a finite boolean algebra, so that the arithmetic operation of such a lattice is less then addition.

Proposition. the arithmetic operations of finite distributive lattices are subadditive

The distributive lattice of sets is inherently linked to addition. The cardinality of the disjoint union of sets is the sum of their cardinalities. This proposition relates finite distributive lattices in general to addition.

References:

[1] Commutative algebra volume one
Zariski and Samuel

[2] Commutative algebra volume two
Zariski and Samuel

[3] Lattice theory: foundation
George Gratzer

[4] Lattice theory: first concepts and distributive lattices
George gratzer

[5] Algebraic theory of lattices
Crawley and Dilworth

[6] Commutative semigroups
Grillet

[7] Finitely generated commutative semigroups
László Rédei

[8] Finitely generated commutative monoids
J.C. Rosales and Pedro A. García-Sánchez