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Let $T_R$ be the first-order theory of real closed fields. This is precisely the theory over the language $\{0,1,+,\times\}$ such that the theorems are the formulas that hold in $\Bbb R$. It can be effectively axiomatized by saying that it's a field in which every odd-degree polynomial has a root, and that for all $r$ either $r$ or $-r$ has a square root, but that $-1$ is not a square.

Define $T_C$ to be the analogous theory for $\Bbb C$, but where we include complex conjugation in the language. So it's precisely the first-order theory over the language $\{0,1,+,\times,\bar{}\}$ such that a formula is a theorem if and only if it holds in $\Bbb C$. This can be axiomatized by saying that it's an algebraically closed field in which $z\mapsto\bar z$ is a self-inverse automorphism and $z\bar z=-1$ has no solutions.

Then these two theories can interpret each other. The theory $T_R$ is interpretable in $T_C$ by considering the fixed points of $z\mapsto\bar{z}$. In the other direction, and $T_C$ is interpretable in $T_R$ by equipping pairs $(r_0,r_1)$ with operations to make them act like $r_0 + r_1i$.

I'm interested in whether this pair of interpretations rises to the level of biinterpretation, as defined by Joel David Hamkins here. This definition requires that when we compose the two interpretations above (to get interpretations of $T_R$ and $T_C$ in themselves) these composite interpretations are isomorphic to the identity representations, in a way that is definable and provable within the theories themselves.

At first it seemed to me that this would be true, since applying the two interpretations to a model does indeed yield the original model. And Hamkins himself seemingly calls this a biinterpretation here.

What has me doubting is that the nLab says here that biinterpretations of theories induce equivalences between the categories of models. That can't be true here, since the above interpretations send $\Bbb R$ and $\Bbb C$ to each other, and yet $\Bbb R$ has only the trivial automorphism, whereas $\Bbb C$ also has complex conjugation. An equivalence of categories should preserve automorphism groups.

So is this a biinterpretation or not?


I also asked this question over on the category theory Zulip. The answers weren't conclusive, but they lead me to believe the subtlety lies in the interpretation of $T_C$ in itself. If we adjoin a square root of $-1$ to the fixed points of $z\mapsto\bar{z}$ then we would want to define an isomorphism back to the original field by sending this fixed point to $i$. But how do we define $i$ separately from $-i$?

Is this indeed the problem, and is it unavoidable? Does it also prevent $\Bbb R$ and $\Bbb C$ from being biinterpretable as models?

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    $\begingroup$ $\mathrm{Th}(\mathbb R,0,1,+,\cdot)$ is bi-interpretable with $\mathrm{Th}(\mathbb C,0,1,i,+,\cdot,\bar{})$ (you don’t really need $0$ and $1$ in either structure, they are definable). As you say, bi-interpretations preserve the automorphism group, so the result is not true without the $i$ constant. The thing is, model theorists are generally used to take constants from the model for free. $\endgroup$ Sep 7, 2021 at 16:25
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    $\begingroup$ @Emil: Imagine how much money we can make if we start charging a fee for constants. We can fund whole research centres with minimal bureaucracy! $\endgroup$
    – Asaf Karagila
    Sep 7, 2021 at 19:04
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    $\begingroup$ To axiomatize RCF, in addition to saying that $-1$ is not a sum of squares and every odd-degree polynomial has a root, you need to say that for all $a$, $a$ or $-a$ is a square. $\endgroup$ Sep 7, 2021 at 20:28
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    $\begingroup$ While we are talking axiomatizations, $\mathrm{Th}(\mathbb C,0,1,+,\cdot,\bar{})$ is axiomatized by algebraically closed fields + $\bar z$ is an automorphism of order $2$. $\endgroup$ Sep 8, 2021 at 5:32
  • $\begingroup$ @AlexKruckman Thanks, I've fixed it. $\endgroup$ Sep 8, 2021 at 20:07

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$\DeclareMathOperator\Th{Th}\def\R{\mathbb R}\def\C{\mathbb C}\DeclareMathOperator\Aut{Aut}$Basically, all the statements you mention are correct, we just have to be careful about the details to avoid the apparent contradiction. So let us review how these interpretations work:

  • There is a 1-dimensional interpretation $\rho$ of $\Th(\R,0,1,+,\cdot)$ in $\Th(\C,0,1,+,\cdot,\bar{})$ with domain $\{z:z=\bar z\}$ and all operations absolute.

  • There is a 2-dimensional interpretation $\kappa$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in $\Th(\R,0,1,+,\cdot)$ defined in the usual way so that a pair $(x,y)$ represents $x+iy$.

  • The interpretation $\rho\circ\kappa$ of $\Th(\R,0,1,+,\cdot)$ in itself is definably isomorphic to the identity interpretation, the isomorphism being $(x,0)\mapsto x$.

  • The interpretation $\kappa\circ\rho$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in itself is isomorphic to the identity interpretation, with the isomorphism being $(x,y)\mapsto x+iy$. Now, here comes the trouble: this isomorphism is only definable with a parameter $i$ (which is not definable without parameters, as it is indistinguishable from $-i$). You can read this in two ways:

    • $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,i,+,\cdot,\bar{})$.

    • $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ via a bi-interpretation with parameters. (Actually, the bi-interpretation as such is still parameter-free, only the connecting isomorphism needs a parameter, as explained above.)

    Note that different people have different conventions about what exactly is the default definition of an interpretation, and in particular, whether it allows parameters. I’d guess that for syntactically minded people, it is more natural to consider paarameter-free interpretations, whereas for model-theorists, it is more natural to allow parameters.

  • $\Th(\R,0,1,+,\cdot)$ is not bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ without parameters. An interpretation $\iota$ of $T_1$ in $T_0$ induces a construction of a model $i(M)\models T_1$ from any model $M\models T_0$, and it is easy to check that if $\iota$ is a parameter-free bi-interpretation, then $\Aut(\iota(M))\simeq\Aut(M)$. Now, $\Th(\R,0,1,+,\cdot)$ has rigid models (i.e., with $\Aut(M)=1$) such as $\R$ itself, whereas every model of $\Th(\C,0,1,+,\cdot,\bar{})$ has at least one nontrivial automorphism: $\bar{}$.

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  • $\begingroup$ Perfect! Very clear, thanks. $\endgroup$ Sep 13, 2021 at 17:24

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