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Stream: learning: questions

Topic: biinterpretations and automorphisms

Oscar Cunningham (Sep 01 2021 at 08:19):

I'm trying to understand the effect that a biinterpretation between two first order theories has on the automorphisms of their models.

My expectation was that if models of each theory correspond to each other under the biinterpretation, then they have isomophic automorphism groups. This seems to agree with the nLab here saying that 'bi-interpretable theories have equivalent categories of models'.

However, consider $T_R$ the theory of real closed fields, and $T_C$ the corresponding theory for complex numbers, where importantly we include conjugation in the language of $T_C$. I believe $T_C$ can be stated as the theory of algebraically closed fields equipped with a self-inverse automorphism $z\mapsto\bar{z}$ such that $z\bar{z}=-1$ has no solutions. Then $T_R$ is interpretable in $T_C$ by sending a model of $T_C$ to the fixed points of $z\mapsto\bar{z}$, and $T_C$ is interpretable in $T_R$ by taking a model of $T_R$ and equipping pairs $(r_0,r_1)$ with operations to make them act like $r_0 + r_1i$. And it seems clear to me that this is a biinterpretation. Indeed Joel David Hamkins and Noah Schweber say so here.

But $\mathbb R$ has only the trivial automorphism, whereas $\mathbb C$ also has complex conjugation. What gives?

Jens Hemelaer (Sep 01 2021 at 08:56):

I'm not sure about the definitions, but maybe the automorphisms have to preserve the complex conjugation? Then complex conjugation isn't an automorphism anymore, because $\overline{\overline{z}} \neq \overline{z}$.

Oscar Cunningham (Sep 01 2021 at 09:01):

I think the relevant equation for an automorphism $f$ respecting conjugation isn't $f(\bar{z}) = f(z)$ but rather $f(\bar{z}) = \overline{f(z)}$. So this holds for conjugation itself since $\overline{\bar{z}} = \overline{\bar{z}}$.

Jens Hemelaer (Sep 01 2021 at 09:09):

Yes, that makes sense!

Jens Hemelaer (Sep 07 2021 at 13:51):

Hi @Oscar Cunningham, did you find already what is going on?

Maybe the situation is that $T_R$ is interpretable in $T_C$ and $T_C$ is interpretable in $T_R$, but the two interpretations are not inverse to each other.

Oscar Cunningham (Sep 07 2021 at 14:07):

Yes, I suspect you're right.

In particular I think the composite in the $T_R\to T_C\to T_R$ direction is the identity, because the fixed points of $z\mapsto \bar{z}$ in $\mathbb{R}^2$ is definitely $\mathbb{R}$.
But in the $T_C\to T_R\to T_C$ direction I think we don't get the identity because the isomorphism between models isn't first order definable. If I say 'the' isomorphism $\mathbb{R}[X]/\langle x^2+1\rangle\to\mathbb{C}$, how do you know whether to send $X$ to $i$ or $-i$?

Perhaps this is similar to this example of functors $D$ and $A$ such that $D(A(V))\simeq V$ for all $V$ and $A(D(U))\simeq U$ for all $U$, but there's no equivalence because the second isomorphism isn't natural in $U$.

I would still like to see someone with more experience spell out the answer clearly though, if anyone else wants to help!

Oscar Cunningham (Sep 07 2021 at 14:09):

(This doesn't necessarily contradict what people were saying in the thread I linked, because they might have been saying that there was a biinterpretation of models, rather than theories.)

Mike Shulman (Sep 07 2021 at 14:31):

Unfortunately, the word "biinterpretable" seems to be used by many people to mean that the two interpretations are inverse to each other. This constantly trips me up. So under that definition, your two theories are not "biinterpretable" even though they are "mutually interpretable".

Oscar Cunningham (Sep 07 2021 at 14:43):

That would make sense. But since it's Joel Hamkins making the distinction between those two terms at the link you gave, what's he talking about here where he says these two theories are biinterpretable?

Oscar Cunningham (Sep 07 2021 at 15:57):

I've asked a MathOverflow question about this here.