Let $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ denote the $n$-th classical Laver table. The operation $*_{n}$ is the unique binary operation on $\{1,\dots,2^{n}\}$ such that $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and $$x*_{n}1=x+1\mod 2^{n}$$ for all $x,y,z\in A_{n}$.

Let $t:\{1,\dots,2^{n}\}^{r}\rightarrow\{1,\dots,2^{n}\}$ be an operation. Then we say that $t$ is compatible with $*_{n}$ if $$t(x*_{n}x_{1},\dots,x*_{n}x_{r})=x*_{n}t(x_{1},\dots,x_{r})$$ for all $$x,x_{1},\dots,x_{r}\in\{1,\dots,2^{n}\}.$$

We say that an algebra $(\{1,\dots,2^{n}\},(t_{i})_{i\in I})$ is compatible with the operation $*_{n}$ if each individual operation $t_{i}$ is compatible with $*_{n}$.

For which varieties $\mathcal{V}$ does there exist an algebra in $\mathcal{V}$ which is compatible with the operation $*_{n}$?

I want to know examples and non-examples of varieties $\mathcal{V}$ where there exists an algebra in $\mathcal{V}$ compatible with $*_{n}$. I am especially interested in varieties $\mathcal{V}$ such that the algebras in $\mathcal{V}$ compatible with the operations of the form $*_{n}$ generate the variety $\mathcal{V}$ (in this case, we say $\mathcal{V}$ is compatible with the classical Laver tables).

Let me start off with a few examples that I have found.

Example: LD-monoids: An LD-monoid is an algebra $(X,*,\circ,1)$ that satisfies the following identities:

  1. $(X,\circ,1)$ is a monoid.

  2. $x*1=1,1*x=x$

  3. $x*(y*z)=(x*y)*(x*z)$ (self-distributivity)

  4. $x*(y\circ z)=(x*y)\circ(x*z)$ (distributivity)

  5. $x\circ y=(x*y)\circ x$ (braid law)

  6. $(x\circ y)*z=x*(y*z)$ (Composition of functions)

There exists a unique operation $\circ_{n}$ on $(A_{n},*_{n})$ such that $(A_{n},*_{n},\circ_{n},2^{n})$ is an LD-monoid, and this LD-monoid is compatible with $*_{n}$. Under strong large cardinal assumptions, the variety generated by $(A_{n},*_{n},\circ_{n},2^{n})$ contains the free LD-monoid on one generator.

Example: Distributive lattices: For all $n$, there exists a linear ordering $\leq_{n}$ on $A_{n}$ such that if $y\leq_{n}z$, then $x*_{n}y\leq_{n}x*_{n}z$ for all $x,y,z\in A_{n}$. The join and meet operations obtained from $\leq_{n}$ are compatible with $*_{n}$.

Non-Example: Groups: Suppose that $n\geq 4$. Suppose that $\cdot$ is a group operation on $A_{4}$ such that $x*_{4}(y\cdot z)=(x*_{4}y)\cdot(x*_{4}z)$. Then $$\{x*_{4}y|y\in A_{4}\}$$ is a subgroup of $(A_{4},\cdot)$ for all $x\in A_{4}$. Therefore $$\{2,12,14,16\}=\{1*_{4}y|y\in A_{4}\}$$ and $$\{9,10,11,12,13,14,15,16\}=\{9*_{4}y|y\in A_{4}\}$$ are subgroups of $(A_{4},\cdot)$ which implies that $$\{12,14,16\}=\{2,12,14,16\}\cap\{9,10,11,12,13,14,15,16\}$$ is a subgroup of $(A_{4},\cdot)$ which contradicts Lagrange's theorem since $|\{12,14,16\}|=3$.

Non-Example: Define $u_{a}(x)=(x-1)*_{n}a$ for $x>1$ and $u_{a}(1)=a$. Then I claim that the unary operations $u_{a}$ are precisely unary the operations on $A_{n}$ which are compatible with $*_{n}$. First of all, we have $u_{1}(x)=x$ and $u_{a+1}(x)=u_{a}(x)*x$. By induction, one can show that $u_{a}(x)$ is compatible with $*_{n}$. Suppose that $u$ is a unary operation on $A_{n}$ compatible with $*_{n}$. Let $a=u(1)$. Then $u(x)=u((x-1)*_{n}1)=(x-1)*_{n}u(1)=(x-1)*_{n}a=u_{a}(x)$. Therefore, $u=u_{a}$.

In particular, if $u,v$ are inverse unary operations on $A_{n}$ compatible with $*_{n}$, then $u,v$ must be the identity function. Furthermore, if $c$ is a constant compatible with $*_{n}$, then $c=2^{n}$.

From these examples, we observe that the varieties compatible with the classical Laver tables must have some sort of “acyclicity.”

  • $\begingroup$ How do you manage to ask so many long questions so often? $\endgroup$ – Monroe Eskew Feb 4 at 11:16
  • $\begingroup$ @MonroeEskew. There are many natural and interesting problems about the algebras of elementary embeddings, so I naturally have quite a few questions about these structures. Should I slow down to give people some time to ponder these questions that are already up here a little more? $\endgroup$ – Joseph Van Name Feb 4 at 14:00

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