Weirdos but algebraic

Question

Weirdos generalize Abelian groups as well as an algebra of arithmetic mean of reals (or geometric mean of positive reals). But first, I'll define eccentrics. (I will not ask about eccentrics here since we should ask only one question per post).

An eccentric is a universal algebra $\ (X\ \sigma\ \lambda\ \rho)\ $ such that operations $\ \sigma\ \lambda\ \rho\,:\,X\times X\to X\ $ satisfy:

1. $\quad \forall_{x\ y\,\in X}\quad \lambda(\sigma(x\ y)\ y)\ =\ x; $
2. $\quad \forall_{x\ y\,\in X}\quad \rho\,(x\ \sigma(x\ y))\ =\ y; $

A weirdo is eccentric $\ (X\ \sigma\ \lambda\ \rho)\ $ such that

3. $\quad \forall_{u\ w\ x\ y\,\in\,X}\quad \sigma(\sigma(u\ w)\ \sigma(x\ y)) \ =\ \sigma(\sigma(u\ x)\ \sigma(w\ y)) $

When $\ |X|=1\ $ then such a weirdo is trivial. Also, weirdos (like all universal algebras) admit the direct product operation (simply the Cartesian product with induced operations).

Open Challenge   Classify the indecomposable weirdos (i.e. non-trivial weirdos which are not direct products of two non-trivial weirdos).

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Two classes of examples of weirdos:

• Abelian groups, where $\ \sigma\ $ is the group operation, and $\ \lambda\ \rho\ $ are the left and right inverse operations as described by above conditions 1. and 2.; if a weirdo is an arbitrary group as described in the abelian case then it follows form 1. 2. 3. that this group must be abelian;

• Weirdos based on certain modules:

let $\ L\ R\ $ be rings with element $\ 1.\ $ Let $\ a\in L\ $ and $\ b\in R\ $ be invertible. Let $\ X\ $ be an $L$-$R$-module. Then we define:

$$ \forall_{x\ y\,\in X}\quad \sigma(x\ y)\ :=\ a\cdot x + y\cdot b $$ and $$ \forall_{x\ y\,\in X}\quad \lambda(x\ y)\ :=\ \frac 1a\cdot x - \frac 1a\cdot y\cdot b; $$ $$ \forall_{x\ y\,\in X}\quad \rho(x\ y)\ :=\ y\cdot\frac 1b - a\cdot x\cdot\frac 1b. $$

When $\ a=b\ $ then such a weirdo is commutative (i.e. $\ \sigma\ $ is commutative).

A specific example: let $\ L=R=X=\Bbb Z[\frac 16];\ $ Then we define

$$\forall_{x\ y\,\in\,X}\quad \sigma(x\ y)\ :=\ \frac 23\cdot x+\frac 13\cdot y; $$ $$\forall_{x\ y\,\in\,X}\quad \lambda(x\ y)\ :=\ \frac 32\cdot x-\frac 12\cdot y; $$ $$\forall_{x\ y\,\in\,X}\quad \rho(x\ y)\ :=\ 3\cdot y - 2\cdot x. $$

Obviously, this weirdo is not commutative.

When $\ a+b=1,\ $ as in the above specific example, then our weirdo describes an averaging operation, i.e. it satisfies the property:

$$\forall_{x\,\in\,X}\quad \sigma(x\ x)\ =\ x $$

If an arbitrary weirdo $\ X\ $satisfies the above averaging property and is commutative $\ (\sigma(x\ y)\,=\,\sigma(y\ x)),\ $ then (as I proved in 1961/62) weirdo $\ X\ $ is a module over $\ \Bbb Z[\frac 12],\ $ with

$$ \forall_{x\,\in\,X}\quad \sigma(x\ y):=\frac{x+y}2.$$

E.g., with respect to the above averaging operation, every Abelian torsion group that has all elements of odd order, carries a weirdo structure.

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An observation: always with respect to $\ \sigma,\ $ some weirdos are associative and not self-distributive -- for instance the abelian groups while some other weirdos are not associative but self-distributive -- for instance, the last two examples based on modules $\ \Bbb Z[\frac 16]\ $ and $\ \Bbb Z[\frac 12].$

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EDIT:

I didn't concentrate enough (I took this topic for granted, unfortunately). Formally, everything is fine, a definition is a definition -- but this is hardly a consolation. In fact, I meant also to have properties:

• $\ \forall_{a\ b\,\in\,X}\quad \sigma(\lambda(b\ a)\ a)\ =\ b;$
• $\ \forall_{a\ b\,\in\,X}\quad \sigma(a\ \rho(a\ b))\ =\ b.$

Big thanks to @KeithKearnes for instantly opening my eyes.

Answer 1

I think one can clarify this in a clear-cut way. Partially following OP's terminology, I'll call E-structure a set $X$ endowed with three binary laws $\sigma=\cdot$, $\lambda$, $\rho$ satisfying the given axioms 1,2. I'll write $\sigma(x,y)=xy$. It is a W-structure if moreover it satisfies the 3rd axiom (depending only on $\sigma$), namely $(xy)(zt)=(xz)(yt)$.

Let me now consider this as a magma $(X,\sigma)$ and then discuss existence and uniqueness of $\lambda,\rho$.

A first remark is that in the presence of a unit, Axiom 3 implies commutativity. Also (commutativity + associativity) implies Axiom 3, so Axiom 3 for a monoid means commutative.

The existence of $\lambda$ means that $(xy,y)=(x'y',y')$ implies $x=x'$. That is, $xy=x'y$ implies $x=x'$. This just means that the magma $(X,\cdot)$ is right-cancelative. Similarly the existence of $\rho$ means left-cancelative, and

For a magma $(X,\cdot)$, there exists an E-structure with $\sigma$-law is the magma product $\cdot$ iff $(X,\cdot)$ is cancelative. (And it is an W-structure iff it satisfies identically $(xy)(zt)=(xz)(yt)$.

About uniqueness: clearly $\lambda$ is uniquely determined on the image of the map $X^2\to X^2$, $(x,y)\mapsto (xy,y)$, and can be arbitrarily modified on the complement of this image $I=\bigcup_{y}Xy\times\{y\}$. A similar thing happens for $\rho$, with $J=\bigcup_x\{x\}\times xX$. So an E-structure (resp. W-structure) is a cancelative magma, along with some choice of maps on these complements (which seem not to be the point of interest, as OP focusses on the law $\cdot=\sigma$). (Nevertheless, modifying $\lambda$ and $\rho$ might affect direct indecomposability.)

Actually $I=X^2$ iff $Xy=X$ for all $y$, etc. Hence, the pair $(\lambda,\rho)$ is at most unique if and only if all left and right multiplications in the magma $X$ are surjective. As seen above, its existence means they're injective. So the existence and uniqueness of $(\lambda,\rho)$ means that left and right multiplications are bijective. (For a semigroup, this means being a group.)

OP's theorem, as stated in the post can be restated as: if $(X,\cdot)$ is a cancelative commutative idempotent magma, then it admits a structure of $\mathbf{Z}[1/2]$-module such that $xy=\frac12(x+y)$ for all $x,y$. [This is precisely equivalent to the formulation in the post. OP claims in a comment that the result is weaker, namely that every cancelative commutative idempotent magma embeds into another one that admits such a structure of $\mathbf{Z}[1/2]$-module inducing the magma law as average map.]

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Update: OP has added axioms $\lambda(yx)x=y$, $x\rho(xy)=y$. These additional axioms just mean that right and left multiplications are not only injective, but bijective. For finite magmas this doesn't change anything, we get cancelative ones. For infinite ones of course this changes: for instance in the associative case, we get the groups instead of the semigroups. It also discards Jónsson algebras (that is magmas for which the law $X\times X\to X$ is bijective), which satisfied the original axioms 1,2 among E-structures. Of course it also changes the notions of free E-structure and free W-structures. Last and not least, it ensures that $(\lambda,\rho)$ is determined by the law (i.e., by $\sigma$).

Eventually, with these axioms the OP's E-structures are known as quasigroups and the W-structures are known as medial quasigroups (thanks to user zeb for the terminology). The Bruck–Murdoch-Toyoda theorem (same link, also mentioned by user zeb) says that a medial quasigroup is always of the form $(A,\ast)$ defined from an abelian group $(A,+)$, for some commuting pair of automorphisms $(\varphi,\psi)$ of $A$ and constant $c\in A$, with $x\ast y=\varphi(x)+\psi(y)+c$.

(This is quite restrictive: for instance, it implies that for any two $x,x'$, writing $L_x(y)=xy$, we have $L_x^{-1}L_{x'}(y)=y+\psi^{-1}(\varphi(x)-\varphi(x'))$, so that $L_{x}^{-1}L_{x'}$ is a translation. In particular, if $A$ is finite with $|A|\notin 4\mathbf{Z}+2$, all permutations $L_x$ have the same signature.) Similarly for right $\ast$-multiplications.

Answer 2

Edit 2020.05.20: with new conditions added, it appears sigma is onto, so all weirdos have empty heads (using the terminology below). So a generator of a free algebra in this variety (with lambda and rho also) is also a term, so now the free algebras may also be decomposable as a direct product (they are already isomorphic to a sub algebra of a direct product). End Edit 2020.05.20.

Edit: easy, but incorrect. It needs to be modified to : if there is a finite head, then there must be a finite (possibly trivial) factor with empty head.

An easy observation: such a structure with a finite nonempty head is not directly decomposable.

Let sigma not be onto. That part of the base set outside the range of sigma I call the head. Then invertibility implies the base set is infinite. If one has two structures with one having a nonempty head, their product will have a nonempty head that is infinite. Therefore any such structure with a finite nonempty head is not directly decomposable. The free finitely generated structures in this variety gives a class of such examples.

Gerhard "Nothing Up My Sleeve... Presto!" Paseman, 2020.05.19.