Is the Lascar group over $A$ trivial when $T=T^{eq}$ and $A = acl(A)$? Let $T$ be a first-order theory which eliminates imaginaries, and let $A$ be an algebraically closed set in a model of $T$. Let $Gal_L(T[A])$ be the Lascar group of the theory $T[A]$, which is $T$ with constants added for $A$. Is $Gal_L(T[A])$ trivial? I would think not, but I don't know a counterexample.
 A: There are several notions of strong type in model theory. $a$ and $b$ have the same:


*

*Lascar strong type over $A$, if they are equivalent under every bounded $A$-invariant equivalence relation.

*Kim-Pillay strong type (aka compact strong type) over $A$, if they are equivalent under every bounded $A$-type-definable equivalence relation.

*Shelah strong type (aka just strong type) over $A$, if they are equivalent under every finite $A$-definable equivalence relation.


Same Lascar strong type $\implies$ same Kim-Pillay strong type $\implies$ same Shelah strong type $\implies$ same type.
To say that $Gal_L(T[A])$ is trivial is to say that Lascar strong types over $A$ coincide with types over $A$. This is always true if $A$ is a model.
If $A$ is just $acl^{eq}$-closed, Shelah strong types over $A$ coincide with types over $A$. So you're looking for examples of theories in which Lascar strong types and Shelah strong types differ.
The easiest example is the circle of diameter $1$, equipped with binary relations $(R_n)_{n\geq 1}$ such that $R_n(a,b)$ if and only if $d(a,b)<1/n$. The type-definable equivalence relation $\{R_n(x,y)\mid n\geq 1\}$ is bounded but not finite (it has $2^{\aleph_0}$ classes).
In that example, Shelah strong type $\neq$ Kim-Pillay strong type. There are also lots of examples of theories in which Kim-Pillay strong type $\neq$ Lascar strong type, see this paper for example. On the other hand, a theory is called $G$-compact if Kim-Pillay strong type $=$ Lascar strong type.
By the way, in every stable theory, every supersimple theory, every o-minimal theory, and every C-minimal theory, Lascar strong type $=$ Shelah strong type. Simple theories are always $G$-compact, but the question of whether Kim-Pillay strong type = Shelah strong type in general simple theories is open.
