The theory of J-trivial semigroups

Starting from partial orders, we can form semilattices from partial orders satisfying certain conditions. Semilattices are restricted to semigroups that are commutative and idempotent. Relaxing the restriction of idempotence, we arrived at finite commutative aperiodic semigroups. The next step, of relaxing commutativity, produces J-trivial semigroups. It can be proven that finite commutative aperiodic semigroups are J-trivial. We already described the simplest J-trivial semigroup that is not commutative, the T3 special combiner on a totally ordered set of three elements.

These non-commutative J-trivial semigroups produce upper bounds of elements in a partial order in an argument-dependent manner. This opens up a whole wide range of possibilities that were not available previously and it gives an approach to considering different upper bound producing functions on partial orders. Actually, J-trivial semigroups constitute all associative upper bound producing functions on partial orders. As we discussed previously, there are two types of non-monotonic behavior that can occur on J-trivial semigroups: chain non-monotonic behavior between comparable elements and antichain non-monotonic behavior on incomparable elements.

• Chain non-monotonic: non-monotonic products of comparable elements
• Antichain non-monotonic: non-monotonic products of comparable elements

We already demonstrated examples of chain non-monotonic and antichain non-monotonic behavior. These can occur argument-order dependently, essentially what this means is that given two elements which are either related to one another or not related to one another by the partial ordering comparability relation, these elements can produce a larger result in one argument order then in another argument order. This makes it so that one argument order is greater then another argument order.

In larger semigroups, smaller semigroups like the totally ordered T3 combiner can be combined in different directions, so they are not isomorphic even though they are antiisomorphic. In such semigroups for certain elements one argument order can be greater then another, well in other elements a greater result is produced between them by arguments in a different order. In other cases they can be combined in the same argument order. So these non-commutative J-trivial semigroups can either favor one argument order or another, it it can maximize them both equally. This is one thing that adds to the complexity of J-trivial semigroups.

Non-commutative version of the exceptional commutative semigroup

The exceptional small commutative semigroup of order four was defined primarily by the different product of its incomparable minimal elements. Rather then then producing the least upper bound it produced the maximal element of the partial order. Amongst the non-commutative semigroups there is a similar semigroup that produces either the maximum or the join depending upon the argument order.

[ [ 1, 1, 1, 1 ], 
  [ 1, 1, 1, 1 ], 
  [ 1, 1, 2, 1 ], 
  [ 1, 1, 2, 2 ] ]

So in total for the same indices and the same factorisation partial order there are three different types of behavior: the semilattice like behavior, the maximizing behavior, and the argument dependent half way case. All of these semigroups, just like the T3 special semigroup on three elements discussed previously are J-trivial. So we will be considering J-trivial semigroups which are associative bound-producing functions on partial orders.

Smallest non-commutative totally ordered semigroup

We considered commutative aperiodic semigroups such as semilattices upper bound functions on partial orders. We even considered how these upper bound functions can themselves be partially ordered by "how upper" their bounds are. Now we will move to the non-commutative case. Towards that end, we should consider the simplest case of a non-commutative upper bound function:

[[1 1 1]
 [1 1 1]
 [1 2 3]]

The first thing we notice is that this semigroup is actually totally ordered [3,2,1]. The second thing that we notice about this semigroup is that the middle element is index two. Therefore, the corresponding monotonic semigroup is M2,1 + identity or the index two aperiodic monogenic semigroup with an identity element adjoined to it.

We can learn about this semigroup by comparing it to its monotonic commutative aperiodic counterpart. The only difference is that 2*3=1 rather then 2 this means that the middle and minimal elements can either produce their least upper bound which is the middle element or the maximal element. Since this produces a greater upper bound then it would otherwise would among comparable elements I call this chain non-monotonic. This special T3 semigroup often appears in larger partially ordered semigroups as a subsemigroup.

Algebraic description of ordered fields

An ordered field can be described relationally $(S,<,+,*)$ or it can be described purely algebraically $(S,\wedge,\vee,+,*)$. In this way, there are four different operations associated with the ordered field.

• Min: the meet of the total order
• Max: the join of the total order
• Addition: the addition of the field
• Multiplication: the multiplication of the field

The axioms for the four algebraic operations are the field axioms for addition and multiplication, the lattice axioms for the minimum and maximum operations and the totality of the two operations which means that for any two elements $a$ and $b$ the minimum and maximum both produce one of the two elements. For non-total orders the lattice operations can produce values that are not contained in the total order. In addition to this, there is the additive monotonicity and positivity preservation which can easily be translated to the algebraic formulation. A result of this description is that all the operations are commutative.

Commutative structures

What algebraic structures can we form if we put the limitation that all operations have to be commutative? This is purely interesting as a thought experiment and to gather information on commutative semigroups. We can then form a restricted outline of algebraic structures. We can exclude vector spaces for now because scalar multiplication is typically not defined to be commutative.

Single binary operation:

These structures are all of the form $(S,*)$ where $*$ is some commutative semigroup. The most important particular cases are the commutative groups and commutative group-free (aperiodic) semigroups. That is to say, the most important property is rather then semigroup is a group or it has any groups embedded in it. So we already split up our ontology into two special cases. Semilattices are a special case of aperiodic semigroups because they are group-free.

Two commutative binary operations:

When we are given two commutative binary operations like $+$ and $*$ generally speaking they are going to be related by the distributive law, which connects them so that we don't need to view them separately. This leads to commutative semigroup pairs, commutative hemirings, commutative rings, integral domains, fields, etc. All the structures of commutative algebra are in this category. The other case is that the operations are related by the absorportion law which produces lattices, modular lattices, distributive lattices, etc.

Four commutative binary operations:

Finally we get at ordered fields which as we described at the start can actually be described entirely using commutative operations. Ordered fields and related structures are the most advanced structures I know of that can be formed entirely with commutative operations as they combine both lattice and field structures.

Necessity of formal realness

We can define an ordered field axiomatically as a totally ordered field that has two conditions (1) additive monotonicity which is that if $a < b$ then for all $x$ it is the case that $a+x < b+x$ and (2) multiplicative positivity preservation which means that for all $0 < a$ and $0 < b$ then $0 < ab$. We will prove several theorems that describe other properties of ordered fields.

Theorem. the sum of positive elements is positive

Let $a$ and $b$ be two positive elements. By positivity we have that $0 < a$ and $0 < b$. By the fact that $0 < a$ we can add $b$ to both sides to get $b < a+b$. Then connecting that with the previous inequality $0 < b$ we get $0 < b < a+b$ which by transitivity means that $0 < a+b$.

Theorem. the additive inverse of a negative number is positive

Let $a$ be a negative number. By negativity we have that $a < 0$. By additive monotonicity we know that $-a$ is an additive monotone element which can be added to both sides to get $a+(-a) < 0+(-a)$ which equals $0 < (-a)$ which means that the additive inverse $(-a)$ is greater then zero and therefore positive.

Theorem. addition must be torsion-free

In order for a group to be torsion-free it must not have any elements that are of finite order. Suppose that we have a field that has a non-identity additive torsion element, then there are two cases (1) the element is positive in which case the fact that the element can be added to itself to get zero defies additive positivity preservation or (2) the element is negative in which case its additive inverse is a positive torsion element which can be added to itself to get zero which also defies additive positivity preservation. So by contradiction we know that the additive group of the ordered field must be torsion-free.

Previously we touched on the two essentially types of properties of algebraic operations (1) an element being non-periodic and (2) an element being torsion-free. Being torsion-free is the weaker condition. These two properties are related to rather or not an algebraic structure can be ordered. It is actually intuitive that a non torsion-free group cannot be ordered, because it has some element that exhibits unordered cyclical behavior. So we can get a special case of this theorem, which is that any ordered group must be torsion-free.

Corollary. the characteristic of the field is zero

Theorem. the square of any non-zero number is positive.

By the total ordering of the field we can split this up into two cases (1) the element is positive and (2) the element is negative. In the first case that the element is positive we know that its square is positive by positivity preservation which is axiomatic. In the second case that the element is negative, by a previous theorem we know that its additive inverse is a positive number $p$. By the fact that the additive inverse is an involution this means that this number can be expressed as $-p$. Then consider the square as $(-p)^2$. This is equal to $(-1)^2*p^2$. We can cancel out the $(-1)^2$ to get $p^2$. By positivity preservation this again a positive number. This means that the product of any non-zero number with itself is positive.

Theorem. there are no proper zero sums of squares in an ordered field.

By the fact that the square of a non-zero number is positive and that the sum of positive numbers is again a positive number, we know that the sum of the squares of non-zero numbers is positive and therefore not equal to zero. This means that there are no proper zero sums of squares.

Conclusion. in order for a field to be orderable it must be formally real

On formally real fields

We can define formally real fields entirely algebraically.

• The field is characteristic zero
• For all sums of squares equal to zero $x^2+y^2=0$ it is the case that $x,y$=0

A proper zero sum of squares is a sum of squares $x^2+y^2=0$ such that $x,y\ne 0$. These are used to define formally real fields. This is reminiscent of the idea of a zero divisor $ab=0$ such that $a$=$b$ used to define integral domains. All formally real fields are of course integral domains as well, so they also have no proper zero divisors.

The simplest counter example is the set of Gaussian rationals $\mathbb{Q}(i)$ which is not a formally real field because $i^2 + 1^2 = 0$ That is to say that {1,i} together form a proper zero sum of their squares. The complex numbers $\mathbb{C}$ of course also do not form a formally real field for the same reason. It is easy to see that if a given field has a subfield that is not formally real, then it is not formally real itself. This means that being formally real is subfield closed.

Theorem. no algebraically closed field is formally real

This follows from that fact that $\mathbb{Q}(i)$ is a non-real subfield of any characteristic zero algebraically closed field. Since the property of being formally real is subfield closed, this demonstrates that the field is not closed.

The purpose of spending so much time on this concept, is that it is fundamental to the definition of an ordered field. A given field can only be ordered only if it is formally real. As a result, we have a concept of order theory that emerges purely from commutative algebra.

The possibility of different orderings: if a given field is an algebraic extension of its prime subfield then it can ordered in a unique manner. The ordering of the formally real field in this case is determined by its embedding within the totally ordered field of real numbers. The resulting ordered field is archimedean by this embedding. The rational numbers $\mathbb{Q}$ as well as real algebaric number fields like $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{2},\sqrt{3})$, and so on are all standard ordered in this way.

Therefore in order for a given formally real field to have multiple orderings it must have at least one transcendental element in it with respect to a subfield. The simplest example is $\mathbb{Q}(x)$ the field of rational functions over the rational numbers. This is a transcendental extension because the variable $x$ is transcendental. Two different orderings are $\mathbb{Q}(e)$ and $\mathbb{Q}(\infty)$. These are a isomorphic as fields, but different as ordered fields as one is archimedean and the other is not. Everyone knows that $e$ is a transcendental number, but not everyone knows that $\infty$ and related numbers are also transcendental in an ordered field. This similarity is seen here, as an infinite transcendental and a finite transcendental number are indistinguishable algebraically and only distinguished through the ordering.

As a result we see three different cases: (1) fields that cannot be ordered like $\mathbb{Q}(i)$, $\mathbb{C}$, etc which are not formally real, (2) fields that can be uniquely ordered which are algebraic extensions of the rational numbers, and (3) other fields that can be ordered in multiple different ways like the field of rational functions over the rational numbers. This addresses the issue of determining how algebraic properties determine orderings.

Torsion free and aperiodic semigroups

I described previously how $(\mathbb{N},+,*)$ has aperiodic addition and multiplication. Addition is completely free and multiplication is free on non-zero elements, so that addition and multiplication are essentially multiset combination. As aperiodic commutative operations these operations produce non-minimal upper bounds well the least upper bounds are produced by the max and lcm operations instead. This keeps the natural operations in the category of aperiodic commutative operations.

The operations of the integers $(\mathbb{Z},+,*)$ are certainly not aperiodic. Neither of them are, but I noticed there is something special about addition different from multiplication. Well every element has an inverse, there are no elements that invert themselves under iteration. Elements exist separately from their inverses. But since the inverses exist they are not aperiodic. To avoid confusion, we can call the first type of semigroup aperiodic and the other type torsion free.

So well neither $(\mathbb{Z},+)$ nor $(\mathbb{Z},*)$ are aperiodic we have that $(\mathbb{Z},+)$ is torsion-free well $(\mathbb{Z},*)$ is neither torsion-free nor aperiodic. The importance of this comes in the construction of the complex numbers $\mathbb{C}$ from the integers $\mathbb{Z}$ because the property of having torsion-free addition is preserved from the integers all the way to the complex numbers well the fact that multiplication is not torsion free is key to the construction of complex multiplication which includes elements of all orders as the unit circle group extends the torsion group of integer multiplication.

We have a certain intuitive sense of "groupness" of semigroups if you allow me a figure of expression. Groups are operations that always have inverses for all of their elements. But since infinite torsion free groups allow you to separate elements from their inverses they have subsemigroups that are not subgroups. Indeed we can get an infinite width distributive lattice ordered subsemigroup of $\mathbb{Q}_{\ne 0}$ as seen by the natural numbers. We can get ordered subsets out of unordered structures when they are not torsion. So in a sense torsion groups are the most strongly group-like of operations among the groups.