Fix a complete first order theory $T$ and a set of parameters $A$ in the monster model $\mathcal{U}$. Recall that an *$A$-invariant global type* is a type $p(x) \in S\_x(\mathcal{U})$ which is fixed by any automorphism of $\mathcal{U}$ which fixes $A$. An equivalent statement is that for every formula $\varphi(x,\bar{y})$, whether or not $\varphi(x,\bar{b}) \in p(x)$ depends only on $\mathrm{tp}(\bar{b}/A)$ (and $\varphi$).


Given an $A$-invariant type $p(x)$ and some small set of parameters $B \subseteq A$, a *Morley sequence* in $p(x)$ over $B$ is a sequence $\{c\_i\}\_{i \in I}$, where $I$ is some linearly ordered set, such that for every $i \in I$, $\mathrm{tp}(c\_i/Bc\_{<i}) = p \upharpoonright Bc\_{<i}$, where $c\_{<i} = \{c\_j: j < i\}$. (I mention this because there are a couple of slightly incompatible definitions of 'Morley sequence' floating around, and I wanted to be clear about which one I meant.) It's not hard to show that Morley sequences are automatically indiscernible.


Given an $A$-indiscernible sequence $\{b\_i\}\_{i < \omega}$, it is possible to construct an $Ab\_{<\omega}$-invariant type whose Morley sequence somewhat 'resemble' $\{b\_i\}$ by picking an ultrafilter $\mathcal{F}$ on $\omega$ and taking the average type along $\mathcal{F}$, i.e. the type $p(x)$ such that for every formula $\varphi(x,\bar{c})$ with $\bar{c} \in \mathcal{U}$, $\varphi(x,\bar{c})$ is in $p(x)$ if and only if $\{ i < \omega : \mathcal{U} \models \varphi(b\_i,\bar{c})\} \in \mathcal{F}$. This is then a type which is finitely satisfiable over $Ab\_{<\omega}$ (and therefore is $Ab\_{<\omega}$-invariant) and which has the property that the EM type over $A$ of any Morley sequence in $p(x)$ is the (order inverse of the) EM type of $\{b\_i\}$ over $A$.


I'm curious about whether this construction gives Morley sequences that share nice properties of the original indiscernible sequence. Call an invariant type $p(x)$ *self-commutative* if any Morley sequence in $p(x)$ over any small set is an indiscernible set. (Note that this is equivalent to the Morley product of $p(x)$ with itself commuting, hence the name.)



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> **Question 1:** If $A$ is a model and $\{b\_i\}\_{i<\omega}$ is an indiscernible set over $A$, does there always exist a non-principal ultrafilter $\mathcal{F}$ on $\omega$ such that the average type of $\{b\_i\}\_{i<\omega}$ along $\mathcal{F}$ is self-commutative?
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Really I care about the following question, but if it has a positive answer it is possibly because the previous question has a positive answer.



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> **Question 2:** If $T$ admits an infinite indiscernible set over some model, does it necessarily have a self-commutative invariant type?
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I already know that this fails without the 'over some model' part. If we let $M$ be a two sorted structure with sort $S$ given by $\omega$ and sort $O$ given by the collection of all linear orders on $S$ (with a ternary relation encoding this), then $S$ is a $\varnothing$-indiscernible set (by considering automorphisms of $M$), but no indiscernible sequence over any parameter set containing any element of $O$ can be an indiscernible set. (This is a modification of a counterexample given to me by Nick Ramsey.)


As a side note, I would like to know if the concept I've called self-commutativity has an existing name in the literature. It is implied by some common niceness conditions of invariant types (such as generically stable or definable and finitely satisfiable), but is weaker than any of them, as seen in the random graph, which has no non-realized invariant types that are definable and finitely satisfiable but also has the property that any indiscernible sequence of $1$-tuples is an indiscernible set.



Existence of invariant types whose Morley sequences are all indiscernible sets
It's possible to lift the counterexample you described to a counterexample to Question 2 (and hence also to Question 1).


Let $E$ be an equivalence relation with infinitely many infinite classes. Let $\leq$ be preorder linearly ordering the set of $E$-classes. Paint each $E$-class by a copy of the counterexample you described. That is, each class contains an infinite set $S$ and a set $O$ consisting of all linear order on $S$, encoded by a ternary relation (which only holds on triples from the same $E$-class, where the first element is in $O$ and the other two are in $S$). This is our model $M$.


Now for any $E$-class which is not represented in $M$, the set $S$ in that class is an indiscernible set over $M$. But for any global type $p$, invariant over a small model $M'$, $p$ does not self-commute. We may assume $p$ has just one variable. If a Morley sequence in $p$ is contained in a single $E$-class, this class must be represented in $M'$, and any element of $O$ in this class in $M'$ linearly orders the Morley sequence. If a Morley sequence in $p$ is not contained in any $E$-class, then the sequence is linearly ordered by $\leq$.




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On the other hand, in the $\mathrm{NIP}$ setting, Question 1 (and hence also Question 2) has a positive answer, over an arbitrary set $A$.


If $I$ is an indiscernible set in any $\mathrm{NIP}$ theory, then for any formula $\varphi(x;y)$, there is a uniform bound $k\_\varphi$ such that for any $b$, $\varphi(x;b)$ or $\lnot\varphi(x;b)$ holds of at most $k\_\varphi$ elements of $I$.


It follows that for any non-principal ultrafilter $\mathcal{F}$ on an infinite indiscernible set $I$, the average type of $I$ along $\mathcal{F}$ is $$p(x) = \{\varphi(x;b)\mid \varphi(x;b) \text{ holds of at least $k\_\varphi+1$ elements of }I\},$$ which does not depend on $\mathcal{F}$. Further, $p(x)$ is definable (by testing $\varphi(x;b)$ on $(2k\_{\varphi}+1)$ elements of $I$) and finitely satisfiable in $I$, which implies that $p(x)$ self-commutes.


