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Suppose we have a structure $M$, that is, a set $S$ with some designated functions and/or relations on that set. We can define automorphisms for this structure. What is the term in the standard logic literature for functions, constants, or relations that are invariant with respect to this structure under all automorphisms of this structure?

Gregory Grant
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user107952
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2 Answers2

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Edited after the comment of Levon Haykazyan whom I thank.

Assume $M$ is $\omega$-homogeneous, otherwise there is little I can say. Let $\equiv_n$ be the relation of elementary equivalence $n$-tuples of $M$. Countable homogeneity says that the orbits under automorphisms of $n$-tuples are the equivalence classes of $\equiv_n$. Invariant sets are exactly union of orbits of tuples, hence union of equivalence classes of $\equiv_n$.

The situation is exquisitely simple when $M$ is a countable model of an $\omega$-categorical theory. Then, by the Engler, Ryll-Nardzewski, Svenonius theorem $\equiv_n$ has finitely many equivalence classes. It follows that invariant = type-definable = definable.

In general the equivalence relation $\equiv_n$ has $2^{|L|}$ many classes. Hence there at most $2^{2^{|L|}}$ invariant sets (the bound may be attained). This is a lot, consider that there are at most $|L|$ definable sets and at most $2^{|L|}$ type-definable sets. So, the situation is complicated.

Even if you assume that $M$ saturated invariant sets may be still difficult to describe. It is only possible to say something about invariant sets that are compact in the topology on $M^n$ generated by the definable sets. Below a few examples.

If $D\subseteq M$ is definable over some sets and it is invariant, then $D$ is definable over the empty set. The same holds with type-definable for definable (if the type has $<|M|$ parameters). It is also true that if $D\subseteq M$ is invariant and $\langle M,D\rangle$ is saturated, then $D$ is definable. Where $\langle M,D\rangle$ the expansion of $M$ with a predicate for $D$.

Primo Petri
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    Your claim that invariant subsets in a saturated model are definable is not correct. Say any type definable set is invariant under automorphisms fixing its base. – Levon Haykazyan May 25 '15 at 19:33
  • @LevonHaykazyan Your are right, this is the $\omega$-categorcal case. I'll correct the error a.s.a.p. Thank you very much. – Primo Petri May 26 '15 at 01:12
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Primo Petri has written a nice and informative answer. I'd just like to add a direct answer your question "What is the term in the standard logic literature for functions, constants, or relations that are invariant with respect to this structure under all automorphisms of this structure?" These are usually just called "automorphism invariant".

In model theory, the term "invariant" is usually used for "automorphism invariant", but in the context of of a large saturated structure (or "monster model") $\mathbb{M}$. Saturation implies that for any small set $A$, there is an automorphism moving the tuple $\overline{a}$ to the tuple $\overline{b}$ and fixing $A$ pointwise if and only if $\text{tp}(\overline{a}/A) = \text{tp}(\overline{b}/A)$. Hence a relation $R$ on $\mathbb{M}$ is $A$-invariant (stable under all automorphisms fixing $A$ pointwise, so "invariant" = "$\emptyset$-invariant") if and only if the truth value of $R(\overline{a})$ only depends on $\text{tp}(\overline{a}/A)$. And analogous statements hold about invariant functions, constants, and even more general things. Easily googlable instances of this terminology are "invariant equivalence relation", "invariant type", and "invariant Keisler measure".

Edit: One more comment that you might be interested in. If your structure $M$ is countable, then one consequence of Scott's isomorphism theorem is that a relation on $M$ is isomorphism invariant if and only if it is definable in the infinitary logic $\mathcal{L}_{\omega_1,\omega}$.

Alex Kruckman
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