Math 557 Oct 13

Homogeneous Structures

Recall: We have seen that DLO is \(\omega\)-categorical, meaning up to isomorphism there is only one countable model. Since there are no finite models, by Vaught’s test, DLO is complete.

Question: Are there other examples like this?

Let \(\mathcal{L}\) be a language and \(\mathcal{M}\) be an \(\mathcal{L}\)-structure.

Definition 1
For \(A \subseteq M\), we denote by \(\langle A \rangle^{\mathcal{M}}\) the smallest substructure of \(\mathcal{M}\) whose domain contains \(A\).

We say \(\mathcal{N} \subseteq \mathcal{M}\) is finitely generated if \(\mathcal{N} = \langle A \rangle^{\mathcal{M}}\) for some finite \(A \subseteq M\).

Definition 2
We say \(\mathcal{M}\) is homogeneous if any isomorphism between finitely generated substructures of \(\mathcal{M}\) can be extended to an automorphism of \(\mathcal{M}\).

Let \(\mathcal{L} = \{<\}\) and \(\mathcal{M} = (\mathbb{Q}, <)\). The finitely generated substructures coincide with the finite substructures.

The age of a structure

In the back-and-forth proof of \(\omega\)-categoricity of DLO, we used a homogeneity property, similarly for the proof that \((\mathbb{Q}, <) \cong (\mathbb{R}, <)\).

Definition 3
The age of \(\mathcal{M}\), denoted \(\mathrm{age}(\mathcal{M})\), is the class of all finitely generated \(\mathcal{L}\)-structures isomorphic to a substructure of \(\mathcal{M}\).

\(\mathrm{age}(\mathbb{Q}, <)\) is the class of all finite linear orders.

Lemma 1
Any two countable homogeneous structures with the same age are isomorphic.

Proof (Sketch). Let \(\mathcal{A} \subseteq \mathcal{M}\) and \(\mathcal{B} \subseteq \mathcal{N}\) be finitely generated, and let \(\pi: \mathcal{A} \to \mathcal{B}\) be an isomorphism.

Let \(a \in M \setminus A\). We need to show that \(\pi\) can be extended to an isomorphism \(\mathcal{A} \cup \{a\} \to \mathcal{B}'\). This suffices, since we can then use the back-and-forth argument (remember our structures are countable) to extend everything to an automorphism of \(\mathcal{M}\).

Suppose \(\mathcal{A} = \langle E \rangle^{\mathcal{M}}\) and \(\mathcal{B} = \langle F \rangle^{\mathcal{N}}\).

Let \(\mathcal{A}' = \langle E \cup \{a\} \rangle^{\mathcal{M}} \subseteq \mathcal{M}\). Since \(\mathcal{M}\) and \(\mathcal{N}\) have the same age, there exists \(\mathcal{C} \subseteq \mathcal{N}\) finitely generated such that \(\mathcal{A}' \cong \mathcal{C}\) via some isomorphism \(g\).

The map \(g\) is uniquely determined by its values on \(E \cup \{a\}\). The restriction \(g|_E\) induces an isomorphism \(\langle E \rangle^{\mathcal{M}} \xrightarrow{\cong} \langle g(E) \rangle^{\mathcal{N}}\).

Therefore, \(\pi \circ (g|_E)^{-1}\) is an isomorphism \(\langle g(E) \rangle^{\mathcal{N}} \xrightarrow{\cong} \mathcal{B}\). Call this map \(\tau\). Note that \(\langle g(E) \rangle^{\mathcal{N}}\) is a finitely generated subset of \(\mathcal{N}\). Since \(\mathcal{N}\) is homogeneous, \(\tau\) extends to an automorphism \(\overline{\tau}: \mathcal{N} \to \mathcal{N}\). Let \(\mathcal{B}'\) be the image of \(\mathcal{C}\) under \(\overline{\tau}\). By definition of \(\overline{\tau}\), \(\mathcal{B} \subset \mathcal{B}'\) and \(\mathcal{A}'\) is isomorphic to \(\mathcal{B}'\) via the map \(\overline{\tau} \circ g\).

Question

What do ages of countable homogeneous structures look like?