Do almost-point-transitive algebras generate almost-point-transitive varieties?

Question

Say that an algebra $\mathfrak{A}$ (in the sense of universal algebra) is point-transitive iff for every $a,b\in\mathfrak{A}$ there is a $\pi\in Aut(\mathfrak{A})$ with $\pi(a)=b$. While genuinely point-transitive algebras are somewhat rare, many naturally-occurring algebras yield the same clone as a point-transitive algebra. Specifically, given an algebra $\mathfrak{A}$ let $\mathsf{Cl}(\mathfrak{A})$ be the smallest set of functions from some finite Cartesian power of $\mathfrak{A}$ to $\mathfrak{A}$ which contains all the constant functions and projection functions, all the primitive functions of $\mathfrak{A}$ itself, and is closed under composition. (EDIT: as Keith Kearnes states below, this is actually the polynomial clone of $\mathfrak{A}$.) Then:

• Each group $(G;*,{}^{-1},e)$ yields the same clone as its "torsor reduct" $(G;(a,b,c)\mapsto a*(b^{-1}*c))$, which is clearly point-transitive since each $x\mapsto y*x$ is an automorphism.

• A similar trick works with rings: each ring $(R;0,1,+,-,\times)$ yields the same clone as $(R; (a,b,c,d,e)\mapsto (a-b)(c-d)+e)$, and the latter is again point-transitive via maps of the form $x\mapsto x+y$. (This is basically due to Matt F.)

Say that an algebra $\mathfrak{A}$ is almost-point-transitive iff we have $\mathsf{Cl}(\mathfrak{A})=\mathsf{Cl}(\mathfrak{B})$ for some point-transitive algebra $\mathfrak{B}$ with the same underlying set. I'd like to just ask "Which algebras are almost-point-transitive?," but I don't see how to make that precise in the right way (e.g. to avoid "the almost-point-transitive ones" as an answer). Instead, the following seems like it might be more immediately approachable:

If $\mathfrak{A}$ is almost-point-transitive and $\mathfrak{B}\in\mathsf{HSP}(\mathfrak{A})$, must $\mathfrak{B}$ be almost-point-transitive as well?

I strongly suspect that the answer is negative but I don't see how to construct a counterexample.

Answer 1

Let me distinguish between clone and polynomial clone. The former is the smallest composition-closed collection of operations on $A$ containing the primitive operations of $\mathbb A$ and the projections, while the latter is the smallest composition-closed collection of operations on $A$ containing the primitive operations of $\mathbb A$, the projections, and the constant operations. In fact, let me write $\mathbb A_A$ for the constant expansion of $\mathbb A$, and then refer to the clone of $\mathbb A_A$ when I want to talk about the polynomial clone of $\mathbb A$.

The $1$-unary algebra $\mathbb A =\langle \mathbb Z; f(x)\rangle$, where $f^{\mathbb A}(x)=x+1$, provides a negative answer to the question. This algebra has transitive automorphism group, since all powers of $f^{\mathbb A}(x)$ are automorphisms.

The terms in this language in the variable $x$ are $x, f(x), f^2(x), \ldots$, so each proper subvariety of the variety of all $1$-unary algebras is axiomatizable by some set of identities of the form $f^m(x)\approx f^n(x)$, $m\neq n$, or of the form $f^m(x)\approx f^n(y)$. None of these hold in $\mathbb A$, so $\mathbb A$ generates the variety of all $1$-unary algebras.

Let $\mathbb B = \langle \mathbb Z; f(x)\rangle$ where $f^{\mathbb B}(x)=|x|$. $\mathbb B$ is a $1$-unary algebra whose basic operation has proper range. The algebra $\mathbb B_B$ has the property that its clone consists of essentially unary operations only, it contains all the constant operations on the domain, and it contains a nonsurjective unary operation $f^{\mathbb B}(x)$.

Let $\mathbb C$ be any algebra defined on the same set as $\mathbb B$ whose polynomial clone agrees with that of $\mathbb B$. Then $\mathbb C_C$ has the property that its clone consists of essentially unary operations only, it contains all the constant operations on the domain, and it contains a nonsurjective unary operation $f^{\mathbb B}(x)$. Since the clone of the reduct $\mathbb C$ contains all the nonconstant operations of the clone of $\mathbb C_C$, the clone of $\mathbb C$ must contain $f^{\mathbb B}(x)$. The range of $f^{\mathbb B}(x)$ is closed under any automorphism of $\mathbb C$, so the automorphism group of $\mathbb C$ is not transitive.

Altogether this shows that (i) $\mathbb A$ has transitive automorphism group (so $\mathbb A$ is point-transitive), but (ii) the variety generated by $\mathbb A$ contains an algebra $\mathbb B$ that is not almost-point-transitive.