Wikipedia claims that the Baire space "is [the] automorphism group of [a] countably infinite saturated model $\mathfrak{M}$ of some complete theory $T$", however I see neither an obvious group structure on $\mathcal{N}$ nor an obvious topology on $\operatorname{Aut}(\mathfrak M)$ for a generic $\mathfrak M$.

How is this claim to be interpreted and what can be said about $T$?

  • $\begingroup$ I think, $\mathfrak M$ is (can be regarded) fixed, then take the group of its automorphisms.. $\endgroup$ – Berci Sep 21 '18 at 11:29
  • $\begingroup$ Sure, but who is $\mathfrak M$ then? What's the topology on $\operatorname{Aut}(\mathfrak M)$? $\endgroup$ – Alessandro Codenotti Sep 21 '18 at 11:48
  • $\begingroup$ The basic open subgroups are pointwise stabilizers of finite sets. $\endgroup$ – Asaf Karagila Sep 21 '18 at 17:47

For any structure $\mathfrak{M}$, the automorphism group $\mathrm{Aut}(\mathfrak{M})$ has a standard structure as a topological group, where (as in Asaf's comment) the basic open neighborhoods of the identity are pointwise stabilizers of finite sets, i.e. sets of the form $U_A = \{\sigma\in \mathrm{Aut}(\mathfrak{M})\mid \sigma(a) = a \text{ for all }a\in A\}$, when $A\subseteq \mathfrak{M}$ is finite. When $\mathfrak{M}$ is countably infinite, $\text{Aut}(\mathfrak{M})$ is a Polish group (in fact, it is isomorphic to a closed subgroup of the polish group $S_\infty$, the automorphism group of $\mathbb{N}$ as a pure set).

I agree that the way this claim is written in Wikipedia is confusing, but I think the intention is the following:

If $T$ is any complete theory and $\mathfrak{M}$ is any countable saturated model of $T$ (note that some theories have no countable saturated models!), then the underlying space of $\mathrm{Aut}(\mathfrak{M})$ is homeomorphic to the Baire space.

Of course, different structures $\mathfrak{M}$ usually have non-isomorphic automorphism groups, so they correspond to different group structures on the Baire space.

Here's a proof of the claim. We'll use the characterization of the Baire space as the unique nonempty Polish zero-dimensional space for which all compact subsets have empty interior (Theorem 7.7 in Kechris).

  • Nonempty: $\mathrm{Aut}(\mathfrak{M})$ contains the identity.

  • Polish: We may assume $\mathfrak{M}$ has domain $\omega$, so any automorphism is a function $\omega\to \omega$. This identifies $\mathrm{Aut}(\mathfrak{M})$ with a subspace of the Baire space. The set of all bijections $\omega\to \omega$ is a $G_\delta$ subspace of the Baire space (this is the Polish group $S_\infty$), and $\mathrm{Aut}(\mathfrak{M})$ is a closed subspace of $S_\infty$ (since if a bijection $\sigma\colon \omega\to\omega$ fails to be an automorphism, then there is already some finite tuple $a$ from $\omega$ such that no function $\tau$ with $\tau(a) = \sigma(a)$ is an automorphism), so it is Polish.

  • $\mathrm{Aut}(\mathfrak{M})$ is zero-dimensional: $\mathrm{Aut}(\mathfrak{M})$ inherits this property as a subspace of the Baire space.

  • All compact subsets have empty interior: Suppose for contradiction that $C$ is a compact set with nonempty interior. Then $C$ contains a translation of a basic clopen neighborhood of the identity, which (being a closed subset of a compact set) is compact. So it suffices to show that no basic clopen neighborhood of the identity is compact. Let $U_A$ be such a neighborhood (so $A$ is a finite subset of $\mathfrak{M}$). Since $\mathfrak{M}$ is infinite, there is a consistent type $p(x)\in S(A)$ which is non-algebraic. Since $\mathfrak{M}$ is saturated, the set $p(\mathfrak{M})$ of realizations of $p(x)$ is infinite. Pick some $b\in p(\mathfrak{M})$. For any $b'\in p(\mathfrak{M})$, let $V_{b'}$ be the set of all automorphisms of $\mathfrak{M}$ which fix $A$ and move $b$ to $b'$. Note that $V_{b'}$ is nonempty, since $\text{tp}(b/A) = \text{tp}(b'/A)$ and $\mathfrak{M}$ is saturated, hence homogeneous. Now $\{V_{b'}\mid b'\in p(\mathfrak{M})\}$ is a cover of $U_A$ by infinitely many nonempty disjoint open sets, which shows that $U_A$ is not compact.

  • $\begingroup$ That's very interesting thanks! I think I see why $\operatorname{Aut}(\mathfrak M)$ is Polish when $\mathfrak{M}$ is countable, but do you have a reference for the homeomorphism with $\mathcal{N}$? I don't think this is in either Kechris nor Moschovakis's books and I'm not sure where to look $\endgroup$ – Alessandro Codenotti Sep 21 '18 at 18:37
  • $\begingroup$ I don't have a reference, but I'll add a proof to my answer a bit later. $\endgroup$ – Alex Kruckman Sep 21 '18 at 20:29
  • $\begingroup$ @AlessandroCodenotti Proof added. $\endgroup$ – Alex Kruckman Sep 21 '18 at 21:31
  • $\begingroup$ That's very helpful, thank you so much! $\endgroup$ – Alessandro Codenotti Sep 21 '18 at 22:09

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