Bi-interpretability implies isomorphism of the automorphism groups?

Question

The theorem of Ahlbrandt and Ziegler says that two countable $\omega$-categorical structures $M$ and $N$ are bi-interpretable if and only if $Aut(M)\cong Aut(N)$ as topological groups.

Dropping the hypothesis of $\omega$-categoricity, the implication from left to right it is still true? I have read the original paper "Quasi finitely axiomatizable totally categorical theories" but there are a lot of details left to the reader.

So,in another word, my question is :

Let $M,N$ two countable structure. If they are bi-interpretable, then $Aut(M)\cong Aut(N)$ as topological groups?

Is there some references about it?

Answer 1

Yes, this is true even without the countability hypothesis. See Hodges Model Theory Section 5.4 Exercise 8(b) on p. 226.

This is a fairly straightforward exercise (though there are a lot of details to check). Here's an outline.

• Let $M$ and $N$ be arbitrary structures. Let $\Gamma$ be an interpretation of $N$ in $M$. Then we have a definable set $D\subseteq M^n$ for some $n$ and a surjective map $f\colon D\to N$ such that the equivalence relation $a{E_f}b\leftrightarrow f(a) = f(b)$ is definable on $D$. Every automorphism $\sigma\in \mathrm{Aut}(M)$ extends coordinatewise to a permutation of $M^n$ which fixes $D$ setwise, so it restricts to a permutation of $D$. This permutation preserves the definable equivalence relation $E_f$, so it pushes forward along $f$ to a permutation of $N$. You need to check that this gives an automorphism of $N$. This is all a special case of Theorem 5.3.4 in Hodges.
• The above construction establishes a function $\mathrm{Aut}(M)\to \mathrm{Aut}(N)$. In fact, this is a continuous group homomorphism. Continuity is Theorem 5.3.5 in Hodges.
• Finally, if $M$ and $N$ are mutually interpretable, we get continuous group homomorphisms in each direction between $\mathrm{Aut}(M)$ and $\mathrm{Aut}(N)$. You need to establish that if $M$ and $N$ are bi-interpretable, then these homomorphisms are inverses. This is the content of Exercise 5.4.8 in Hodges.