1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Jan 28 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. $\endgroup$ – Joseph Van Name Jan 28 at 21:23
3
$\begingroup$

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 $H$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.