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Are Lascar strong types (definition below) in models of fragments of arithmetic always type definable? (They trivially are, in models of full induction.)

Definition Given a saturated model ${\cal M}$ and a set $A\subseteq{\cal M}$ the Lascar graph over $A$ has an arc between $a,b\in{\cal M}$ if $a\equiv_Mb$ for some $A\subseteq M\preceq{\cal M}$. The Lascar strong type of $a$ is the set of the $c\in{\cal M}$ that are in the same connected component of $a$.

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  • $\begingroup$ The usual definition of the Lascar graph is slightly different (and not equivalent, afaik): an edge connects two points if they are two elements of an infinite indiscernible sequence. The connected components are the same, though. $\endgroup$
    – tomasz
    Commented Jun 21, 2015 at 19:28
  • $\begingroup$ It should also be pointed out that the Lascar strong types are type-definable iff components of the Lascar graph have uniformly bounded diameter. What do you mean by full induction? I'm not very familiar with theories of arithmetic, but it seems like it would be odd for them to be G-compact, considering they are rather wild. $\endgroup$
    – tomasz
    Commented Jun 21, 2015 at 19:35
  • $\begingroup$ @tomasz: The point is that full induction implies the existence of definable Skolem functions, in which case $M$ can be fixed as the Skolem hull/definable closure of $A$. $\endgroup$
    – Emil Jeřábek
    Commented Jun 21, 2015 at 20:34
  • $\begingroup$ @tomasz 1.Type definability of the Lascar strong type is only mildly related to tameness/wildness (cfr. the comment of Emil Jeřábek). 2. Full induction means induction for all first-order formulas. Fragments have induction only for formulas in some fixed complexity class. $\endgroup$
    – Domenico Zambella
    Commented Jun 21, 2015 at 21:03
  • $\begingroup$ @EmilJeřábek: Ah. Indeed, that makes it trivial. Thanks for the clarification. $\endgroup$
    – tomasz
    Commented Jun 21, 2015 at 21:36

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