Attraction in Laver tables

Question

If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x*x^{[n]}$. The motivation for this question comes from the following fact about self-distributivity on one generator.

Theorem: Suppose that $X$ is a self-distributive algebra generated by one element. Then for all $x,y\in X$, there are $m,n$ with $x^{[m]}=y^{[n]}$.

Define the Fibonacci terms by letting $t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$.

An algebra $(X,*,1)$ is said to be a reduced permutative algebra if $x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$ for all $x,y,z\in X$ and where for all $x,y\in X$, there is some $n$ where $t_{n}(x,y)=1$. We say that a reduced permutative algebra $(X,*,1)$ is critically simple if whenever $\simeq$ is a congruence, either $\simeq$ is the trivial congruence or $x\simeq 1$ for some $x\neq 1$.

Suppose that $(X,*,1)$ is a finite reduced permutative algebra and $x,y\in X$ and suppose that $p\geq 0$. Then does there a reduced permutative algebra $(Y,*,1)$, natural numbers $m,n$ and a homomorphism $\phi:Y\rightarrow X$ and $x',y'\in Y$ where $\phi(y')=y,\phi(x')=x$ and $x'^{[m]}=y'^{[n]}$ and where $x'^{[m+p]}\neq 1$? What if $(X,*,1)$ is critically simple?

Answer 1

I have a method of generating counterexamples even if $(X,*,1)$ is critically simple. In this post, all algebras will be assumed to be finite reduced permutative self-distributive algebras. To do this, we provide an example of an algebra algebra $X$ generated by $x,y$ where there does not exist a algebra $Y$ generated by $x_1,y_1$ and a surjective homomorphism $\phi:Y\rightarrow X$ with $\phi(x_1)=x,\phi(y_1)=y$ and where $Y$ has more critical points than $X$.

The examples of the algebras $X$ are quite rare, and I currently only know of one technique for producing such an algebra. Let $X_1$ be the algebra with only one element which we shall denote by $x_1$ and also by $y_1$. We then by induction build a sequence of critically simple algebra $(X_1,\dots,X_n)$ where $|\text{crit}[X_j]|=j$ for $1\leq j\leq n$ and where $X_j$ is generated by 2 elements $x_j,y_j$ where $y_j*y_j=1$ and where there is a surjective homomorphism $\phi_j:X_j\rightarrow X_{j-1}$ with $\phi(x_j)=x_{j-1},\phi(y_j)=y_{j-1}$, and we quite the induction process when we cannot extend the sequence $(X_1,\dots,X_n)$ any further in the sense that there does not exist an algebra $X_{n+1}$ with $\text{crit}[X_{n+1}]=n+1$ and a surjective homomorphism $\phi_{n+1}:X_{n+1}\rightarrow X_n$ along with generators $x_{n+1},y_{n+1}$ such that $\phi_{n+1}(x_{n+1})=x_n,\phi_{n+1}(y_{n+1})=y_n$ and $y_{n+1}*y_{n+1}=1$. After we recursively build our sequence of critically simple algebras $(X_1,\dots,X_n)$, we extend the sequence some more to obtain more critically simple algebras $(X_n,\dots,X_m)$ and homomorphisms $\phi_j:X_j\rightarrow X_{j-1}$ where $|\text{crit}[X_j]|=j$ and $\phi_j(x_j)=x_{j-1},\phi_j(y_j)=y_j$ and where $X_j$ is generated by $x_j,y_j$ until we cannot extend $X_m$ any more to an algebra $X_{m+1}$ generated by $x_{m+1},y_{m+1}$ with $|\text{crit}[X_{m+1}]|=m+1$ and $\phi_{m+1}(x_{m+1})=x_m,\phi_{m+1}(y_{m+1})=y_m$.

In this case, $Y_m$ with the generators $x_m,y_m$ is our counterexample.

To go from $X_j$ to $X_{j+1}$, we use my (possibly inefficient) backtracking algorithm that takes a critically simple algebra $X$ and generating set $(x_a)_{a\in A}$ and returns all possible (up-to generator preserving isomorphism) critically simple algebras $Y$ with generating sets $(y_a)_{a\in A}$ and homomorphisms $\phi:Y\rightarrow X$ with $\phi(y_a)=x_a$ for $a\in A$ and where $|\text{crit}[Y]|=|\text{crit}[X]|+1$.

For critically simple algebras $X$ with multiple generators $(x_a)_{a\in A}$, I have no algorithm or theory other than my exhaustive search that finds all critically simple algebras $Y$ generated by $(y_a)_{a\in A}$ along with a homomorphism $\phi:Y\rightarrow X$ with $\phi(y_a)=x_a$ for $a\in A$, and I have no way of knowing what critically simple algebras $Y$ might look like, so I have no way of obtaining a different kind of counterexample.