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In Universal Algebra, it is possible to say that two presentations denote the "same" kind of algebraic structures, if the two corresponding varieties are "rationally equivalent" (Mal'cev 1958). In Model Theory, such "synonymity" is defined by various non-equivalent equivalence relations, such that definitional equivalence, mutual interpretability or bi-interpretability.

If I consider the universal closures of two equational presentations, what is the model-theoretical counterpart of the rational equivalence?

If there is a standard reference for a proof of this correspondance, many thanks, in advance!

PS: given a universal algebra $ \mathcal{A} = <A, \sigma>$ of signature $\sigma$, denote by $Ter(\sigma)$ the collection of all terms of signature $\sigma$ and by $Ter(\mathcal{A})$ the collection of all termal functions of $\mathcal{A}$. Two algebras $\mathcal{A} = <A, \sigma_{1}>$ and $\mathcal{B} = <B, \sigma_{2}>$ are rationally equivalent whenever there exists a bijection $\phi$ from the set $A$ to the set $B$ such that:

  • $\phi f ( \phi^{-1}(x_{1},...,\phi^{-1}(x_{n})) \in Ter(\mathcal{B})$ for every signature function $f$ of $\mathcal{A}$

  • and $\phi^{-1} g (\phi(x_{1},...,\phi(x_{n})) \in Ter(\mathcal{A})$ for every signature function $g$ of $\mathcal{B}$

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    $\begingroup$ You would have to provide more of the definition of rational equivalence. It sounds though like bi-interpretability might be closest. I'm guessing that you essentially want terms in each language so that each basic operation in the other language is given by such terms. Sort of like isomorphism of clones. My example is Boolean rings versus Boolean algebras. Gerhard "Doesn't Know About Rational Equivalence" Paseman, 2015.03.23 $\endgroup$ – Gerhard Paseman Mar 23 '15 at 19:40
  • $\begingroup$ Here is a definition of "rationally equivalent". Many thanks for these indications! $\endgroup$ – Sylvain Mar 23 '15 at 20:33
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    $\begingroup$ This is a special case of definitional equivalence (up to isomorphism). $\endgroup$ – Emil Jeřábek Mar 23 '15 at 21:04
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    $\begingroup$ Well, I’m not in that much of a touch with universal algebra, but I’m in touch with model theory. The definition of rational equivalence given in the question is even stricter than definitional equivalence (the strictest of the three model-theoretic notions mentioned). The only difference between the two is that for rational equivalence, functions of one structure are required to be defined by terms in the other structure, whereas definitional equivalence allows functions (and relations) defined by arbitrary first-order formulas. $\endgroup$ – Emil Jeřábek Mar 23 '15 at 21:57
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    $\begingroup$ Thanks a lot for all these informations! A reference about Mal'cev rational equivalence: "Rational equivalence of algebras, its clone generalizations and clone categoricity", by A.G. Pinus, Siberian Mathematical Journal, 54.3, 2013 (p.533). $\endgroup$ – Sylvain Mar 23 '15 at 22:07
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If you assume that $A=B$ and $\phi$ is the identity, the all term functions of one algebra are also term functions in the other algebra, so in today's language (of universal algebra) I would just say that the two algebras have the same clones.

If $\phi$ is not the identity, you would say the clones are the same up to conjugation.

This is not the same as having isomorphic clones. For example, if you take any linearly ordered set $(X,<)$, then the clone generated by the binary $\min$ function is the essentially the set of all $\min$-functions of any arity, and this is isomorphic (as an abstract clone) to the clone generated by the $\max$ function.

I am not a model theorist, but I don't think that model theorists would distinguish between the two structures at all. Formulas, types, embeddings, automorphisms etc would be essentially the same for $A$ and $B$.

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  • $\begingroup$ Thanks a lot! So if I understand you well, two rationally equivalent varieties of algebras do not necessarily have isomorphic clones? Concerning the model-theoretical point of view, it seems to me that this 'essentially sameness' is precisely the issue : if the signatures are different, the structures cannot satisfy the same formulas. $\endgroup$ – Sylvain Mar 24 '15 at 10:56
  • $\begingroup$ Contrariwise. "rational equivalence" of two algebras implies that their clones are isomorphic. It is the other direction that is not necessarily true. $\endgroup$ – Goldstern Mar 24 '15 at 12:02
  • $\begingroup$ Concerning the model-theoretical point of view: The two structures use different languages, so formally the formulas satisfied in one structure are not the same as those satisfied in the other. But the assumption of "rational equivalence" implies that there are natural translations in both directions. $\endgroup$ – Goldstern Mar 24 '15 at 12:04

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