Signs on Magmas in Abstract Algebra

A previous discussion on a more intuitive variant of a UFD led naturally to discussing signs on commutative monoids, where instead of thinking of signs as additive inverses in a ring, we generalized to defining a more abstract concept of sign. Upon further thought, we can probably generalize this definition even further, from a commutative monoid to a most general kind of algebra — namely, a magma. In this post, we take this approach, which has a “reverse mathematics” benefit of seeing the most general settings in which certain theorems hold and certain questions can be formulated.

Our fundamental definition, which was first given for commutative monoids in this post but can nevertheless be generalized almost word-for-word to magmas, is as follows:

Definition. Let $M$ be a magma. A sign on $M$ is a unary operation $s:M \rightarrow M$ satisfying the following:

$s(a) \neq a$ for all $a$
$s\left( s(a) \right) = a$ for all $a$
$s(ab) = as(b) = s(a)b$ for all $a,b$

As implied by the discussion in that post, an equivalent definition is given by:

Theorem. A unary operation $s:M \rightarrow M$ is a sign if and only if it satisfies the following:

$s$ is a permutation that can be decomposed into disjoint 2-cycles
$s(ab) = as(b) = s(a)b$ for all $a,b$

Note how strong the $s(a) \neq a$ requirement is; in particular, for this definition to apply say to rings, we must exclude 0. (So $a \rightarrow - a$ wouldn’t be a sign on a ring; instead, it would be a sign on the ring excluding 0.)

Also note that if $M$ is commutative, then (3) can be equivalently shortened either to $s(ab) = as(b)$ or $s(ab) = s(a)b$ .

We now attempt to generalize some of the results as well as conjectures in that post to this setting.

—

Are the three axioms independent? We showed that (1) and (2) don’t imply (3) for commutative monoids, so they certainly don’t in general for magmas, and similarly (2) and (3) don’t imply (1). Do (1) and (3) imply (2)?

We can restrict to commutative monoids, or indeed as rich a structure as we’d like — that’d only make the counterexample stronger.

Actually, on second thought, this is obvious — just take a commutative ring with $a \rightarrow - a$ as the sign. Clearly, this doesn’t satisfy (1) since the ring contains 0, but it always satisfies (2) and (3). Hence, (2) and (3) don’t imply (1).

It follows that the three axioms are independent.

—

Can we develop some of “polynomial algebra” in a very general setting here? Specifically, can we show:

Conjecture. Let $s$ be a sign on $M$ . Then, $a^{2} = b^{2}$ if and only if $a = b$ or $a = s(b)$ .

This would yield that, if we have an equation $x^{2} = m$ on $M$ with unknown $x$ , then if there is a solution, there must be exactly two solutions, which are both signs of each other.

Before we work on this, let’s note some basic facts that hold in this setting. We have $a = s(b)$ if and only if $b = s(a)$ , and $s(a)s(b) = ab$ , thus $s(a)^{2} = a^{2}$ . (For a general magma, we have \hfill\break $s(a)s(b) = s\left( as(b) \right) = as\left( s(b) \right) = ab$ .) In particular, to prove the conjecture we only need one direction: if $a^{2} = b^{2}$ and $a \neq b$ then $a = s(b)$ . If we have a signed semigroup $S$ (where multiplication is associative), then we can write $x^{n}$ for $x \in S$ , and then we have that $s(a)^{2n} = a^{2n}$ and $s(a)^{2n + 1} = s(a)a^{2n}$ ; this then inspires similar questions about “roots” of elements in a signed semigroup. We can extrapolate consequences and generalizations of the conjecture:

Lemma. Let $S$ be a signed semigroup. Consider the equation $x^{2n} = s$ in unknown $x$ . Then, the solution set can be partitioned into two subsets where each subset is the signs of the elements in the other. Consequently, if it has a finite number of solutions, then this number is even.

Proof. If $a$ is a solution, then so is $s(a)$ , for $a^{2} = s(a)^{2} \rightarrow a^{2n} = s(a)^{2n}$ . Thus, if the solution set is $T$ , then it must consist of elements that can be separated into pairs $\left( a,s(a) \right)$ (the elements in these pairs are distinct by axiom (1) of signs.) Thus, $T$ can be partitioned into two subsets in the desired manner, and hence if it is finite then it must consist of an even number of elements. $\blacksquare$

Lemma. Let $S$ be a signed semigroup, and assume $S$ is cancellative. Consider the equation $x^{2n + 1} = s$ in unknown $x$ (with odd exponent this time.) If $x$ is a solution then $s(x)$ isn’t.

Proof. Assume for the sake of contradiction there existed $x$ such that $x^{2n + 1} = s = s(x)^{2n + 1}$ . We have $s(x)^{2n + 1} = s(x)x^{2n}$ , thus $s(x)x^{2n} = xx^{2n}$ . But since $S$ is cancellative we can yield $s(x) = x$ , contradicting the sign axioms. $\blacksquare$

(It seems like it would be hard / maybe not possible to derive this result if $S$ isn’t cancellative, although we could try something maybe with the disjoint 2-cycle structure of the sign and Pigeonhole Principle.)

Conjecture. Let $S$ be a signed semigroup. Consider the equation $x^{2n + 1} = s$ in unknown $x$ . If the equation has a finite number of solutions, then this number is odd.

Attempt. Assume for the sake of contradiction that the number of solutions was even.

Actually, this seems a bit trickier on second thought … it remains to investigate this further. $\blacksquare$

Restatement of the Conjecture. In a general magma, the solution set of $x^{2} = m$ in unknown $x$ must have size less than or equal to 2. (From the lemmas, equivalently, the size must be 0 or 2.)

Proof. Clearly, the conjecture implies this, and if the size is less than or equal to 2, then if $a^{2} = b^{2}$ , there must be at most two possible values for $a$ in terms of $b$ ; both $a = b$ and $a = s(b)$ work, and these are different, so these must be the only solutions. $\blacksquare$

Conjectured Consequence of the Conjecture. In a semigroup, the solution set of $x^{n} = s$ in unknown $x$ must have size less than or equal to $n$ .

Attempt. We proceed by strong induction on $n$ .

For the base case, $n = 1$ , we trivially have that there is exactly one solution to $x = s$ .

Now, for the induction step, we split into cases depending on parity of $n$ .

First, assume $n$ is even, $n = 2k$ , and assume the result holds for all $n < 2k$ ; we prove it for $n$ . The equation in consideration is $x^{2k} = s$ , which implies $\left( x^{k} \right)^{2} = s$ . By the conjecture, there are at most two possible values of $x^{k}$ , say $a$ and $s(a)$ . But by the induction hypothesis, there are at most $k$ values of $x$ for $x^{k} = a$ and similarly for $x^{k} = s(a)$ , thus there are at most $2k = n$ values of $x$ , as desired.

Otherwise, assume $n$ is odd, $n = 2k + 1$ , and assume the result holds for all $n < 2k + 1$ ; we prove it for $n$ . The equation in consideration is $x^{2k + 1} = s$ , or $\left( x^{k} \right)^{2}x = s$ . Actually, it seems trickier to yield something now analogously to the case of $n$ being even … we could try Pigeonhole Principle maybe with the disjoint 2-cycle structure of the sign, assuming $S$ is finite. Even if $S$ is cancellative or a division semigroup, if we produced $\left( x^{k} \right)^{2} = \frac{s}{x}$ , we’d still have a variable on both sides, which would invalidate applying the conjecture assuming the RHS is a constant. Thus, odd seems trickier. $\blacksquare$

At the very least, this is what we’ve shown:

Consequence of the Conjecture. In a signed semigroup $S$ , the number of solutions of $x^{n} = s$ for unknown $x$ is finite and at most $n$ , if $n$ is a power of 2.

It remains to investigate the conjecture.

—

We can generalize many basic concepts about counting numbers and integers in a straightforward manner. From the disjoint-cycle permutation structure of a sign, it is clear that a sign partitions the magma into two subsets that are signs of each other’s elements. Furthermore, if we know the multiplication table for one of these halves, then that determines the multiplication table for the whole magma.

Also, given a non-signed magma $M$ , we can construct a new magma $M^{'}$ from $M$ that is signed. We deem this construction to be canonical, and it preserves many algebraic properties of $M$ (like commutativity or associativity.) Specifically, we define

$\displaystyle M^{'} = \left\{ (b,m):m \in M,b \in \left\{ 0,1 \right\} \right\},$

with multiplication and sign on $M^{'}$ defined in the natural manner.

—

That statement can be interpreted in some sense as yielding the multiplication structure given the sign. Can we go the other way around? Specifically, given the multiplication structure on all of $M$ , can we constrain what a sign could be? Can we determine conditions for whether or not there exists a sign? Can we for example do this if we are given a sign on a subset of $M$ (much like how we can be given a sign on $M$ and multiplication on a suitable half of $M$ to yield multiplication on all of $M$ )?

Let’s ask this question as follows. Let $M^{'}$ be a magma whose full multiplication table is known, and say we are given $M \subseteq M^{'}$ and a sign on $M$ . Are there conditions on $M,M^{'}$ , and $s$ that allow $s$ to be extended to a sign $s^{'}$ on all of $M^{'}$ ?

Well, it seems natural to require that $M$ be closed under multiplication, aka $M$ is a submagma of $M^{'}$ . So we know the multiplication tables of $M,M^{'}$ , and have a sign $s$ on $M$ . Given that we are dealing with a general magma, it is hard to require more conditions beyond this that would seem to help. So let’s say this is our setup. Then, can we extend $s$ to $s^{'}$ ? Furthermore, is the extension unique?

Actually, we of course need to require $M$ to be closed under sign too. So we’d need $M$ to be closed under multiplication and sign.

Let’s try a toy example: look at $\mathbb{R -}\left\{ 0 \right\}$ and $\left\{ 1, - 1 \right\}$ , under multiplication. So we have $M = \left\{ 1, - 1 \right\}$ and we have a sign on $M$ . Given that we know how nonzero real numbers multiply, does that determine the sign on all of $\mathbb{R -}\left\{ 0 \right\}$ ?

Let $r\mathbb{\in R}$ ; we are trying to find $s(r)$ . We are given $s( \pm 1) = \mp 1$ . If say $r\mathbb{\in Z}$ , then we could generate $r$ from $1$ or $- 1$ , and then that would yield $s(r) = - r$ . But wait … that generation requires additivity, which we are not considering in the signed magma structure. And actually, it depends on $s(a + b) = s(a) + s(b)$ , which is not a general assumption for magmas. Thus, that logic wouldn’t work for magmas.

It seems we can say this clearly: if an element is generated from $M$ , then its sign is determined by the sign on $M$ and repeated application of (3). This generation could be done without associativity too, if words in $M$ can contain parentheses; for example, if $a,b,c \in M$ , then $(ab)c \in M$ , and $s\left( (ab)c \right) = s(ab)c = \left( s(a)b \right)c$ .

But actually, if $M$ is closed under multiplication, then all elements generated from $M$ would just stay within $M$ . So that wouldn’t be helpful. Instead, we can say the following: if $M$ is closed under sign and multiplicatively generates $M^{'}$ , then the sign on $M$ uniquely determines the sign on $M^{'}$ .

Hmm … but the fact that (3) gave two expressions for $s(ab)$ , and furthermore asserted their equivalence, seems to me to show that there are additional “redundancies” that we are not considering, that could allow us to derive more information here. In fact, it seems like the existence of a sign could lead to implications for the multiplicative structure. Certainly, many different “expression trees” in a non-associative signed magma must evaluate to the same thing, and the fact that $s$ is bijective means that this could presumably yield some version of associativity as a consequence, or maybe a weaker form of associativity.

Let’s actually try to show this. Let’s evaluate $s\left( (ab)c \right)$ and $s\left( a(bc) \right)$ and see if we can show them to be equal. There are actually three different expressions for each, which we see as follows:

$\displaystyle s\left( (ab)c \right) = s(ab)c = \left( s(a)b \right)c$

$\displaystyle s\left( (ab)c \right) = s(ab)c = \left( as(b) \right)c$

$\displaystyle s\left( (ab)c \right) = (ab)s(c)$

$\displaystyle s\left( a(bc) \right) = s(a)(bc)$

$\displaystyle s\left( a(bc) \right) = as(bc) = a\left( bs(c) \right)$

$\displaystyle s\left( a(bc) \right) = as(bc) = a\left( s(b)c \right)$

So what we can yield is this: $(ab)c = a(bc)$ if and only if $\left( as(b) \right)c = a\left( s(b)c \right)$ , and also $(ab)c = a(bc)$ if and only if $(ab)s(c) = a\left( bs(c) \right)$ . Thus, certain associativity triples imply other associativity triples. Actually, these are also equivalent to $\left( s(a)b \right)c = s(a)(bc)$ . In other words, $(a,b,c)$ being an associativity triple is equivalent to the associativity of the triple with either of the three elements replaced by its sign. Let us write a shorthand for $(a,b,c) = \left( a^{'},b^{'},c^{'} \right)$ exactly when they are both associative or both not associative. Then, we have:

$\displaystyle (a,b,c) = \left( s(a),b,c \right)$

$\displaystyle = \left( s(a),s(b),c \right)$

$\displaystyle = \left( s(a),s(b),s(c) \right).$

In fact, any triple is “equivalent” (by satisfaction of associativity) to any triple formed from replacing any number of elements by their signs. Thus, associativity of some triples implies associativity of other triples, and with $s(a) \neq a$ it seems very plausible that many of these triples must often be different. But that works another way too: non-associativity of some triples implies non-associativity of many other different triples.

Can we show that finite signed magmas must be associative? The idea is to use contradiction, and show that if a certain number of triples are not associative, then that implies non-associativity for a much larger set of triples. Then, use Pigeonhole Principle to show a number of triples that must be associative, and construct a contradiction from these two.

So let’s say our finite signed magma has size $2n$ . For any triple $(a,b,c)$ , we can form the expressions $(ab)c$ and $a(bc)$ . Let’s use Pigeonhole Principle where the expressions are the items and the $2n$ elements are the buckets. Actually, this may not work without incorporating sign: just because some of the expressions are equal doesn’t mean that the two expressions for a particular triple are equal. In fact, we can simply order the triples say lexicographically (imposing an order on the magma as well) and then always have $a(bc) = (ab)c + 1$ (mod $2n$ .)

It seems that we can combine this with the associativity equivalency from sign applications to elements in the triple to yield some sort of existence of a certain number of associative triples. I think this proof is doable, but I will come back to it at another time and investigate this further.

—

We can ask some additional questions:

Can we “add 0” to any signed magma and define an addition operation, possibly to turn any signed magma into an additive abelian group with a multiplication that distributes over? In that case, whenever we work with signed magmas, we can immediately make use of the additive structure, which we can take to be canonical (implied whenever we use addition.)
In the case that we can show associativity of finite signed magmas, can we exhibit an example of a non-associative infinite signed magma?
Can we go further and fully classify finite signed magmas? The reason this seems to be possible to me is that the sign axioms seem quite strong: (1) and (2) already imply a disjoint-cycle permutation which lead to further consequences, while including (3) with these seems to constrain the possibilities a lot. My first guess is the following: for any finite signed magma $M$ , $M \cup \left\{ 0 \right\}$ is isomorphic as a “ring without associativity or unit” to $\left\{ - n,\ldots, - 1,0,1,\ldots n \right\}$ , with addition and multiplication defined in some way on $\left\{ 1,\ldots,n \right\}$ , and then extended via sign to the rest of the set.

It remains to study these further.

2 replies on “Signs on Magmas in Abstract Algebra”

You should publish your work.

LikeLiked by 1 person

Thanks Jerry! It’s been a while since I’ve worked on this sort of math, but when I get a chance I’ll revisit it and see how much of it can be published.

LikeLike

Share and like (need WordPress account to like):

2 replies on “Signs on Magmas in Abstract Algebra”

Leave a reply to Nihal Uppugunduri Cancel reply