# Example of idempotent left quasigroups which are right-distributive but not left-distributive

I am looking for examples of the following algebraic structure: a set (X,.) which satisfy the axioms

(idempotent) x.x = x

(left quasigroup) the equation a.x = b has a unique solution denoted by x = a*b

(right distributive) (x ? y) ! z = (x ! z) ? (y ! z) , where ? and ! are any of the operations . or *

but not the axiom

(left distributive) x ! (y ? z) = (x ! y) ? (x ! z) for any choice of the operations ? and ! among . and *

Remark that an idempotent left quasigroup which is left distributive is called a quandle. The question can be rephrased as: give examples of idempotent left quasigroups which are right distributive but not quandles.

With the help of the (idempotent) axiom, the (right distributive) and (left distributive) may be rewritten as (right medial) and (left medial) axioms, so the question may be rephrased as: give (as many as possible) examples of idempotent left quasigroups which are right medial but not left medial.

The motivation for the question is described in the notes A problem concerning emergent algebras.

• Since you want to satisfy the identity x ! (y ? z) = (x ! y) ? (x ! z) for any choice of ? and ! among . and *, I would call this axiom 'bi-left distributivity' instead of 'left distributivity' since this is a version of left distributivity involving two operations that distribute over each other and themselves. This terminology is consistent with the terminology by Patrick Dehornoy. Apr 27, 2021 at 21:47
• Agree with your "bi-left", "bi-right" terminology. Would you point to relevant references to Dehornoy? thank you. Apr 29, 2021 at 7:00
• The main textbook I am referring to is the book Braids and Self-Distributivity. This book has plenty of examples of self-distributive algebraic structures. In Chapter 10, he talks about multi-left distributive algebraic structures. I highly recommend the entire book. Apr 29, 2021 at 12:09
• I proved that the original question (from the linked notes) has a negative answer: there is no emergent algebra which satisfies (COLIN) but not (LIN). The proof is described here: chorasimilarity.wordpress.com/2021/06/02/colin-implies-lin Jun 2, 2021 at 15:18

We can use artificial intelligence to generate some bi-right-distributive idempotent left quasigroups $$(X,*,+)$$.

To do this, we first of all need to represent the algebraic structure $$(X,*,+)$$ in a way that is easy for the artificial intelligence to work with.

We shall represent $$(X,*,+)$$ as a system $$(g_{x})_{x\in X}$$ where $$g_{x}:X\rightarrow X$$ is the permutation for each $$x\in X$$ such that $$x*y=g_{x}(y),x+y=g_{x}^{-1}(y)$$ whenever $$x,y\in X$$. Here, $$(X,*,+)$$ satisfies the identities $$x+(x*y)=y=x*(x+y)$$ for any system of permutations $$(g_{x})_{x\in X}$$.

Now we define a loss function $$L$$ where the domain of $$L$$ is the set of all tuples $$(g_{x})_{x\in X}$$ such that $$g_{x}$$ is a bijection from $$X$$ to $$X$$ for each $$x\in X$$ as follows:

$$L((g_{x})_{x\in X})=|\{x\in X\mid g_{x}(x)\neq x\}|$$ $$+\{(x,y,z)\mid(x*y)+z\neq(x+z)*(y+z)\}+\{(x,y,z)\mid(x+y)+z\neq(x+z)+(y+z)\}$$ $$+\{(x,y,z)\mid(x+y)*z\neq(x*z)+(y*z)\}+\{(x,y,z)\mid (x*y)*z\neq(x*z)*(y*z)\}.$$

The motivation for the loss function $$L$$ is that $$L((g_{x})_{x\in X})$$ is a measure of how close the algebra $$(X,+,*)$$ is to being a bi-right-distributive idempotent left quasigroups, and $$L((g_{x})_{x\in X})=0$$ if and only if $$(X,+,*)$$ is a bi-right-distributive idempotent left quasigroup.

Now, one can construct bi-right-distributive idempotent left quasigroups by starting off with a random system of permutations $$(g_{x})_{x\in X}$$ and then repeatedly modifying this system of permutations in such a way that $$L((g_{x})_{x\in X})$$ gets lower and lower. Eventually you will get $$L((g_{x})_{x\in X})=0$$.

Here is the code that produces such an algebra in the language GAP (this code is not optimized for speed but is instead designed to be simple; occasionally the algorithm gets stuck in a local minimum).

 loss:=function(l) local c,i,j,k,f,g; f:=function(x,y) return y^l[x]; end; g:=function(x,y) return y^(l[x]^(-1)); end; c:=0; for i in [1..n] do if not f(i,i)=i then c:=c+1; fi; for j in [1..n] do for k in [1..n] do if not f(f(i,j),k)=f(f(i,k),f(j,k)) then c:=c+1; fi; if not g(g(i,j),k)=g(g(i,k),g(j,k)) then c:=c+1; fi; if not g(f(i,j),k)=f(g(i,k),g(j,k)) then c:=c+1; fi; if not f(g(i,j),k)=g(f(i,k),f(j,k)) then c:=c+1; fi; od; od; od; return c; end; n:=10; l:=[]; for i in [1..n] do l[i]:=Random(SymmetricGroup(n)); od; p:=loss(l); while true do ll:=StructuralCopy(l); a:=Random([1..n]); r:=1; s:=1; while r=s do r:=Random([1..n]); s:=Random([1..n]); od; if Random([1..30])=1 then ll[a]:=ll[a]*(r,s); else ll[a]:=(r,s)*ll[a]*(r,s); fi; q:=loss(ll); if q<=p then p:=q; l:=ll; fi; Display(p); if p=0 then break; fi; od; l; 

You can also minimize the loss using the following evolutionary algorithm.

 list:=[]; for i in [1..10] do list[i]:=[]; for j in [1..n] do list[i][j]:=Random(SymmetricGroup(n)); od; od; while true do for i in [11..100] do if Random([true,false]) then qq:=StructuralCopy(list[Random([1..10])]); else qq:=[]; for j in [1..10] do qq[j]:=list[Random([1..10])][j]; od; fi; a:=Random([1..n]);b:=1;c:=1; while b=c do b:=Random([1..10]); c:=Random([1..10]); od; if Random([true,false]) then qq[a]:=qq[a]*(b,c); else qq[a]:=(b,c)*qq[a]*(b,c); fi; list[i]:=qq; od; SortBy(list,loss); while Length(list)>10 do Remove(list); od; qar:=List(list,loss); Display(qar); if Sum(qar)=0 then break; fi; od; list; 

An algebra produced using artificial intelligence

Here is a right-distributive idempotent left quasigroup produced using our algorithms.

$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|} * &\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}& \mathbf{5}&\mathbf{6}&\mathbf{7}&\mathbf{8}&\mathbf{9}&\mathbf{10}\\ \hline \mathbf{1}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{1}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{1}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{1}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{1}& 8& 9& 2& 1& 5& 6& 7& 4& 3& 10 \\ \hline \mathbf{1}& 4& 2& 3& 8& 10& 6& 7& 1& 9& 5 \\ \hline \mathbf{1}& 4& 2& 3& 8& 10& 6& 7& 1& 9& 5 \\ \hline \mathbf{1}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{1}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{1}& 8& 9& 2& 1& 5& 6& 7& 4& 3& 10 \end{array}$$

$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|} + &\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}& \mathbf{5}&\mathbf{6}&\mathbf{7}&\mathbf{8}&\mathbf{9}&\mathbf{10}\\ \hline \mathbf{1}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{2}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{3}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{4}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{5}& 4& 3& 9& 8& 5& 6& 7& 1& 2& 10 \\ \hline \mathbf{6}& 8& 2& 3& 1& 10& 6& 7& 4& 9& 5 \\ \hline \mathbf{7}& 8& 2& 3& 1& 10& 6& 7& 4& 9& 5 \\ \hline \mathbf{8}& 1& 9& 3& 4& 10& 6& 7& 8& 2& 5 \\ \hline \mathbf{9}& 1& 2& 3& 8& 10& 7& 6& 4& 9& 5 \\ \hline \mathbf{10}& 4& 3& 9& 8& 5& 6& 7& 1& 2& 10 \\ \hline \end{array}$$

Generalization

Since this question was motivated by emergent algebras, let me generalize this case to a broader class of structures that more closely resembles emergent algebras.

Suppose that $$G$$ is a group. Then a $$G$$-multi-antiquandle is an algebraic structure $$(X,(*_{g})_{g\in G})$$ that satisfies the following identities:

1. $$x*_{e}y=y$$

2. $$x*_{f}(x*_{g}y)=x*_{fg}y$$

3. $$(x*_{f}y)*_{g}z=(x*_{g}z)*_{f}(y*_{g}z)$$

4. $$x*_{f}x=x$$.

whenever $$x,y,z\in X,f,g\in G$$.

Our artificial intelligence techniques easily generalize to generating $$G$$-multi-antiquandles even when $$G$$ is a non-abelian group. The way to do this is to first choose a finitely generated free group $$F$$ and use our artificial intelligence algorithm to produce an $$F$$-multi-antiquandle $$(X,(*_{f})_{f\in F})$$ where $$X$$ is finite. Then there is a normal subgroup $$N$$ such that $$F/N$$ is finite and where we can define an $$F/N$$-multi-anti-quandle $$(X,(*_{fN})_{f\in F/N})$$ simply by setting $$*_{fN}=*_{f}$$ for each $$f\in F$$.

Suppose that $$X$$ and $$I$$ are finite sets. Suppose that $$g_{x,i}:X\rightarrow X$$ is a permutation for $$x\in X,i\in I$$ and that $$x*_{i}y=g_{x,i}(y)$$ whenever $$x,y\in X,i\in I$$. Then define a loss function $$L$$ by letting $$L((X,*_{i}))=|\{(i,j,x,y,z)\mid i,j\in I,x,y,z\in X,(x*_{i}y)*_{j}z\neq(x*_{j}z)*_{i}(y*_{j}z)\}|$$ $$+|\{(i,x)\mid i\in I,x\in X,x*_{i}x\neq x\}.$$

By minimizing the loss function $$L$$, one should obtain an algebra $$(X,(*_{i})_{i\in I})$$ with $$(X,(*_{i})_{i\in I})=0$$. Now, let $$F$$ be the free group generated by $$(z_{i})_{i\in I}$$. Then if $$v=z_{i_{1}}^{\alpha_{1}}\dots z_{i_{n}}^{\alpha_{n}}$$, then define $$x*_{v}y=g_{x,i_{1}}^{\alpha_{1}}\cdots g_{x,i_{n}}^{\alpha_{n}}(y)$$. Then $$(X,(*_{v})_{v\in F})$$ should hopefully be an $$F$$-multi-anti-quandle.

I am sure there is an interesting theory behind the $$G$$-multi-anti-quandles, but I do not think anyone has studied these kinds of structures before.

• Wow I have to think and to have means to check your results. Thank you for this, can you contact me at mbuliga@pm.me ? Why don't you make a github repo with this? Apr 30, 2021 at 15:51
• Two more: 1) Your G-multi-anti-quandle is interesting! I thought about this a long time ago, it was my first article imar.ro/~mbuliga/Top_deriv_I.pdf . Because I am more interested in the possibility to pass to the limit with g \in G (because it gives a notion of intrinsic differential calculus), I think that with your notation g_{x,f} is derivable implies that f commutes with all elements of G. But there is a big place to play with this notion. 2) how do you know that you find examples which satisfy one condition, say bi-left, but not the other (bi-right). Is your example like this? Apr 30, 2021 at 16:25

We shall call an algebraic structure $$(X,*_{0},*_{1})$$ bi-right distributive if $$(x*_{i}y)*_{j}z=(x*_{j}z)*_{i}(y*_{j}z)$$ whenever $$x,y,z\in X,i,j\in\{0,1\}$$, and we call $$(X,*_{0},*_{1})$$ bi-left distributive if $$x*_{i}(y*_{j}z)=(x*_{i}y)*_{j}(x*_{i}z)$$ whenever $$x,y,z\in X,i,j\in\{0,1\}$$.

Suppose that $$R$$ is a commutative ring with unity and $$M$$ is a left $$R$$-module. Then for each $$\lambda\in R$$, define an operation $$*_{\lambda}$$ on $$M$$ by letting $$x*_{\lambda}y=(1-\lambda)x+\lambda y$$. Then we satisfy the following identities:

1. $$x*_{\lambda}x=x$$.

2. $$(x*_{\lambda}y)*_{\mu}z=(x*_{\mu}z)*_{\lambda}(y*_{\mu}z)$$.

3. $$x*_{\lambda}(y*_{\mu}z)=(x*_{\lambda}y)*_{\mu}(x*_{\lambda}z)$$.

4. $$x*_{\lambda}y=y*_{1-\lambda}x$$.

5. $$x*_{\lambda\mu}y=x*_{\lambda}(x*_{\mu}y)$$.

6. $$x*_{1}y=y$$

Therefore, if $$\lambda\in R$$ and $$\lambda$$ is invertible, then $$(M,*_{\lambda},*_{\lambda^{-1}})$$ is a bi-left distributive, bi-right distributive idempotent left-quasigroup (and also a quandle). If $$2\lambda=1$$, then $$*_{\lambda}$$ is commutative, and if $$\lambda^{2}=\lambda$$, then $$*_{\lambda}$$ is associative. Therefore, if $$(X,*,+)$$ is a bi-right distributive idempotent left-quasigroup that is not left-distributive, then the direct product $$(X,*,+)\times (M,*_{\lambda},*_{\lambda^{-1}})$$ is also a bi-right distributive idempotent left-quasigroup that is not left-distributive. Therefore, $$(M,*_{\lambda},*_{\lambda^{-1}})$$ is a tool to produce bi-right distributive idempotent left-quasigroups that are not left-distributive.

• I am working on a second answer where I use completely different techniques to produce bi-right distributive algebraic structures. Apr 27, 2021 at 21:48
• The first part of your answer is "bi-right" and "bi-left", or I am looking for examples which are "bi-right" but not "bi-left". Apr 29, 2021 at 7:01