The algebras of elementary embeddings have been studied from a set theoretic perspective and an algebraic perspective. I wonder is there is a purely model theoretic approach to the self-distributive algebras of elementary embeddings that does not mention any set theory.

Let $L$ be the language with a single binary function symbol $*$. Let $T$ be the theory over the language generated by the sentences of the form $\forall x,x_{1},...,x_{n},\phi(x_{1},...,x_{n})\leftrightarrow\phi(x*x_{1},...,x*x_{n})$ where $\phi$ is any first order formula.

Observe that $T$ implies the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$ and that $*$ is left-cancellative. All racks and quandles satisfy the theory $T$. The motivation behind the theory $T$ is that the theory $T$ axiomatizes the notion of an inner-elementary embedding.

Let $\theta$ denote the sentence $\forall x\exists y\forall z,y*z=x$ and let $T^{+}=T\cup\{\theta\}$.


  1. Every algebra that satisfies $T$ also satisfies $\forall x\exists y,t(y)*y=x$ and also generalized versions of this sentence.

  2. The algebras that satisfy $T$ are closed under reduced products (this result follows from the Feferman-Vaught theorem). The algebras from $T$ are also closed under reduced products.

  3. Let $X$ be a set and let $f:X\rightarrow X$ be an injective function. Then define $*$ by $x*y=f(y)$. Then $(X,*)$ satisfies $T$.


I am looking for constructions in ZFC that do not rely upon large cardinals.

  1. Could you give an example of an algebra that satisfies the theory $T^{+}$ or which do not satisfy the identity $(x*x)*y=x*y$ (algebras that satisfy this identity embed into racks)? I prefer computable structures rather than structures obtained by the compactness theorem. I also prefer complicated structures (like braid groups) as opposed to basic structures (like the integers).

  2. Is there an algebra $X$ that satisfies $T$ (or $T^{+}$) where each $x\in X$ freely generates a subalgebra of $X$?

  3. Does there exist an algebra $X$ that satisfies $T$ (or $T^{+}$) where each for all $x\in X$ there exists some $y\in X$ with $y\neq x$ and where $x,y$ freely generate a subalgebra of $X$?

More background.

We take note that the free self-distributive algebra on one generator does not contain a free self-distributive subalgebra on two generators. However, the free self-distributive algebra on two generators has a free self-distributive algebra on infinitely many generators. The word problem for the free self-distributive algebras is solvable. The free self-distributive algebras on multiple generators occur naturally as subalgebras of self-distributive operations (known as shifted conjugacy) on the charged braid groups and the Larue groups. The self-distributive operations on braid groups do not satisfy $T$.


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