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Say that structures $\mathfrak{A},\mathfrak{B}$ with the same underlying set are parametrically equivalent iff every primitive relation/function in one is definable (with parameters) in the other. For example, every group $\mathfrak{G}=(G;*,{}^{-1})$ is parametrically equivalent to its "torsor reduct" $\mathfrak{T}_\mathfrak{G}=(G; (x,y,z)\mapsto x*y^{-1}*z)$.

Parametrically equivalent structures can still have very different combinatorial properties. In analogy with the notion of vertex transitive graphs, say that a structure $\mathfrak{A}$ is point-transitive iff the natural action of its automorphism group is $1$-transitive - that is, if for every $a,b\in\mathfrak{A}$ there is some $f\in Aut(\mathfrak{A})$ with $f(a)=b$. If $\mathfrak{G}$ is a nontrivial group then $\mathfrak{G}$ is not point-transitive (no automorphism can move the identity) but $\mathfrak{T_G}$ is point-transitive (for each $g\in G$ the map $a\mapsto g*a$ is in $Aut(\mathfrak{T_G}$)).

I'm broadly interested in understanding the information captured by the family of automorphism groups of parametric equivalents of a given structure (see also here). Motivated by the group/torsor example, one question which seems like it should be easy to answer is: which structures have a parametric equivalent whose automorphism group acts $1$-transitively? But even for simple structures things aren't very clear to me. So I'd like to start with a simple example:

Is the field of rationals $\mathfrak{Q}=(\mathbb{Q};+,\times)$ parametrically equivalent to a point-transitive structure?

Per the torsor example above, the answer is affirmative for $(\mathbb{Q};+)$. It's also affirmative for $(\mathbb{Q};+,<)$ and various similar expansions, by the same construction, so the existence of such a parametric equivalent isn't connected to any obvious model-theoretic tameness property. However, multiplication seems to complicate things.

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  • $\begingroup$ How about the rationals with the 8-ary relation (a-b)(c-d)=(e-f)(g-h)? Then -y is the unique e such that (y-0)(0-1)=(e-0)(1-0); x+y is the unique e such that (x-(-y))(1-0)=(e-0)(1-0); and xy is the unique e such that (x-0)(y-0)=(e-0)(1-0). $\endgroup$
    – user44143
    Aug 9, 2021 at 6:43
  • $\begingroup$ @MattF. Oh yes that's enough (it's late, I'm tired :P). Looks like that works! Now I wonder if there's a snappy characterization of "parametrically equivalent to a point-transitive structure" after all ... $\endgroup$ Aug 9, 2021 at 6:48
  • $\begingroup$ The 8-ary relation above is longer than the 6-ary one that I used in the answer, but has the advantage of being doubly transitive. $\endgroup$
    – user44143
    Aug 9, 2021 at 14:32

1 Answer 1

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Consider $(\mathbb{Q},R)$, where $R$ is the 6-ary relation defined by $$R(a,b,c,d,e,f) \iff (a-b)(c-d)=(e-f)$$

The automorphism group of this structure includes at least the translations $x\to x+h$. Since those translations include $x \to x+(b-a)$, there is always a translation taking $a$ to $b$, so the group of automorphisms is point-transitive.

We can define $x+y$ and $xy$ in terms of $R$ as the unique $s$ and $t$ for which $R(1,0,s,x,y,0)$ and $R(x,0,y,0,t,0)$ respectively -- i.e. using the equations \begin{align} (1-0)(s-x)&=(y-0)\\ (x-0)(y-0)&=(t-0) \end{align}

Finally, we can define $R$ in terms of $+$ and $\times$ directly by $ac+bd+f=ad+bc+e$.

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  • $\begingroup$ Looks right, thanks! $\endgroup$ Aug 9, 2021 at 7:07

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