# Which varieties are compatible with the classical Laver tables?

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.”

• How do you manage to ask so many long questions so often? – Monroe Eskew Feb 4 at 11:16
• @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? – Joseph Van Name Feb 4 at 14:00