A variety $V$ is said to be reversible, if for each $n>0$ and fundamental operation $f$ there are $m\geq n$ and $r$ along with terms $T_{2},\dots,T_{r}$ and $S_{1},\dots,S_{m}$ such that if $G,H$ are the following the functions, then $G,H$ are inverses.

  1. $G(x_{1},\dots,x_{m})=(f(x_{1},\dots,x_{n}),T_{2}(x_{1},\dots,x_{m}),\dots,T_{r}(x_{1},\dots,x_{m})$.

  2. $H(x_{1},\dots,x_{r})=(S_{1}(x_{1},\dots,x_{r}),\dots,S_{m}(x_{1},\dots,x_{r})).$

If $V$ is a reversible variety, then for each fundamental operation of arity greater than $0$, we have $|f^{-1}[\{a\}]|=|f^{-1}[\{b\}]|$.

Suppose that $V$ is a variety whose theory is axiomatized by finitely many identities and where $V$ is generated by its finite members.

Furthermore, assume that whenever $X\in V$ and $X$ is finite and $f$ is an $n$-ary fundamental operation, then $$\{(x_{1},\dots,x_{n}):f(x_{1},\dots,x_{n})=a\}=|X|^{n-1}.$$ Then is the variety $V$ reversible?

  • $\begingroup$ Do your assumptions imply that the algebras are congruence regular or Hamiltonian? That might give you a leg up. Gerhard "Not After Hamilton The Rapper" Paseman, 2019.01.28. $\endgroup$ Jan 28, 2019 at 16:41
  • $\begingroup$ If $G$ is a group that contains a non-normal subgroup $H$, then $G$ is not Hamitonian, but $G$ is still reversible and the variety of groups is reversible. Regularity is out of the question since the variety of all quandles is reversible but if $(X,*,*^{-1})$ is a quandle with more than 3 elements and $x*y=x*^{-1}y=y$ for all $x,y\in X$, then every equivalence relation on $(X,*,*^{-1})$ is a congruence on $(X,*,*^{-1})$. In fact, this means reversibility does not imply any non-trivial property characterized by Mal'cev conditions. $\endgroup$ Jan 28, 2019 at 21:23

1 Answer 1


I don't know the answer to this question, but will make an extended remark.

Let $X$ be a finite set and let $f:X^n\to X$ be any $n$-ary operation on $X$, $n>0$.

Claim. The following conditions are equivalent.

(i) $f$ is surjective with uniform kernel.
(Equivalently, for each $a\in X$ the set $f^{-1}(a)=\{(x_{1},\dots,x_{n}):f(x_{1},\dots,x_{n})=a\}$ has size $|X|^{n-1}.$)

(ii) There exist $n$-ary operations on $X$, $T_2,\ldots, T_n$ and $S_1,\ldots, S_n$ such that, if $G,H$ are $$G(\bar{x}) = (f(\bar{x}), T_2(\bar{x}), \ldots, T_n(\bar{x}))$$ and $$H(\bar{x}) = (S_1(\bar{x}), S_2(\bar{x}), \ldots, S_n(\bar{x})),$$ then $G$ and $H$ are inverse bijections between $X^n$ and $X^n$.

The question asks, if $V$ is a variety satisfying:

I. $V$ is finitely axiomatizable.
II. $V$ is generated by its finite members.
III. Item (i) above holds for the interpretation of any fundamental operation of arity at least $1$ on each finite member of $V$,

then must Item (ii) above hold in the strong sense that the $S$'s and $T$'s are term operations, but in the weak sense that we allow other parameters $m$ and $r$ in place of some instances of $n$?

Roughly, this asks if having Item (i) hold throughout the finite part of $V$ implies that Item (ii) is enforced by the equational theory of $V$.

This seems plausible to me, but it also seems that there are some extraneous elements in the question. I don't think that $V$ being finitely axiomatizable is relevant. I don't think the additional flexibility of introducing parameters $m$ and $r$ possibly different from $n$ helps, but I haven't tried to check any examples. (It is clear at least that $m$ must equal $r$ if $V$ is generated by its finite members.) I also think the result, if true, is not a property of varieties; that is, the question can be asked for a single (fundamental) operation of $V$: if $V$ is generated by its finite members and $f$ is a fundamental operation of positive arity satisfying Item (i) above, then must Item (ii) above hold?

Here is a sketch of a proof of the claim.

(ii) implies (i): Let $\pi_1: X^n\to X$ be the first projection map. It is surjective with uniform kernel. Since $G: X^n\to X^n$ is a bijection, $\pi_1\circ G$ ( = $f$) is also surjective with uniform kernel.

(i) implies (ii): For each $a\in X$, choose a bijection $\beta_a: f^{-1}(a)\to X^{n-1}$. (Item (i) is the statement that such a bijection exists.) If $$G: X^n\to X^n: \bar{x}\mapsto (f(\bar{x}),\beta_{f(\bar{x})}(\bar{x})),$$ then $G$ is a bijection, and the appropriate component functions exist for both $G$ and its inverse $H$.


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