Heir extension of an invariant type. I was asked to prove that given a definable set $G$ in a model $\mathcal{M}$ and $\mathcal{M}\preccurlyeq \mathcal{N}$. If a type $p\in S_G(M)$ is $G(M)$-invariant, then any of its heir extension to $\mathcal{N}$ is $G(N)$-invariant.
I tried assuming the extension $q$ of $p$ is not $G(N)$-invariant. Then, there are $b,b'\in N$ with $tp(b/G(N))=tp(b'/G(N))$ and $\varphi(x,y)$ a $L(M)$-formula such that $\varphi(x,b)\wedge \neg \varphi(x,b')\in q$. As $q$ is a heir to $p$, there are tuples $m,m'\in M$ such that $\varphi(x,m)\wedge \neg\varphi(x,m')\in p$. Here I don't know how to continue. I would like to have $tp(m/G(M))=tp(m'/G(M))$ but I don't see how to deduce it.
 A: Unfortunately this is not true as written. For instance, let $M=(\mathbb{Q},<)$, and let $N$ be any elementary extension of $M$ containing elements $a,c$ such that $0<a<c<q$ for every $q\in\mathbb{Q}$ such that $q>0$. Now, define the formula $G(x)\equiv x\leqslant 0$ and let $p(x)\in S_1(M)$ be the unique type containing the formulas $$\{x>q:q\in\mathbb{Q},q\leqslant 0\}\cup\{x<q:q\in\mathbb{Q},q>0\}.$$ Now, $p(x)$ is $G(M)$-invariant. Indeed, suppose two elements $q_1,q_2\in\mathbb{Q}$ have the same type over $G(M)$. If $q_1\neq q_2$, then we must have $q_1,q_2>0$, and so we have $0<q_1\in p$ and $0<q_2\in p$. By quantifier elimination for $\text{DLO}$, every formula is equivalent to a Boolean combination of formulas of the form $x=y$ and $x<y$, so this suffices to show that $p(x)$ is $G(M)$-invariant. (As an aside, note too that, again by quantifier elimination, $p$ is a definable type, since $x<q$ lies in $p$ for an element $q\in\mathbb{Q}$ if and only if $q>0$.)
However, consider the unique type $q(x)\in S_1(N)$ containing the formulas $$\{x>n:n\in N,n\leqslant a\}\cup\{x<n:n\in N,n>a\}.$$ Certainly $q(x)\supseteq p(x)$, and it is also an heir of $p$, again by quantifier elimination for $\text{DLO}$. But $q(x)$ is not $G(N)$-invariant, since it contains $x>a$ and $x<c$ even though $\operatorname{tp}(a/G(N))=\operatorname{tp}(c/G(N))$. So this gives the desired counterexample.

The desired result does hold if $M$ is $\aleph_0$-saturated and if $p$ is a definable type. To see this, suppose $q\supseteq p$ is an heir of $p$ in $S(N)$ that is not $G(N)$-invariant. Then there is a formula $\varphi(x,y)$ and elements $n,n'\in N$ such that $\varphi(x,n)\wedge\neg\varphi(x,n')\in q$ but $\operatorname{tp}(n/G(N))=\operatorname{tp}(n'/G(N))$. Now, if $p$ is definable, then there is a formula $d_p x\varphi(x,y)$ such that $$\varphi(x,m)\in p\iff M\models d_px\varphi(x,m),$$ for any $m\in M$. Now, for any formula $\psi(y,z_1,\dots,z_n)$ without parameters, define a corresponding formula $$\widehat{\psi}(y,y')\equiv\forall z_1\dots\forall z_n\left[\bigwedge_{i=1}^n G(z_i)\to \left(\psi(y,\overline{z})\leftrightarrow\psi(y',\overline{z})\right)\right],$$ and consider the family of formulas $$r(y,y')=\left\{d_px\varphi(x,y),\neg d_px\varphi(x,y')\right\}\cup\left\{\widehat{\psi}(y,y'):\psi(y,\overline{z})\in\text{Form}\right\},$$ where $\text{Form}$ is the set of formulas without parameters. We claim that $r$ is finitely satisfiable in $M$. To see this, note that by taking conjunctions we may replace the set on the right with a single formula $\widehat{\psi}(y,y')$ for some $\psi(y,\overline{z})\in\text{Form}$. Now, we know that $\operatorname{tp}(n/G(N))=\operatorname{tp}(n'/G(N))$, so we have $N\models \widehat\psi(n,n')$. In particular, the formula $\varphi(x,n)\wedge\neg\varphi(x,n')\wedge\widehat\psi(n,n')$ lies in $q$. But $q$ is an heir of $p$, so there are $m,m'\in M$ such that the same formula but with $n,n'$ replaced by $m,m'$ lies in $p$. In particular, we have $\varphi(x,m)\in p$ and $\varphi(x,m')\notin p$, so $$M\models d_px\varphi(x,m)\wedge \neg d_px\varphi(x,m'),$$ and $M\models\widehat\psi(m,m')$. Thus $m$ and $m'$ give the desired witnesses, and hence $r(y,y')$ is consistent.
Now, $r(y,y')$ contains only finitely many parameters, all of which lie in $M$, so by $\aleph_0$-saturation of $M$ there exist $m,m'\in M$ with $M\models r(m,m')$. By construction, we have $M\models\widehat{\psi}(m,m')$ for every formula $\psi(y,\overline{z})$, whence $\operatorname{tp}(m/G(M))=\operatorname{tp}(m'/G(M))$. But also $$M\models d_px\varphi(x,m)\wedge \neg d_px\varphi(x,m'),$$ so $\varphi(x,m)\wedge\neg\varphi(x,m')\in p$, and hence $p$ is not $G(M)$-invariant, as desired.
A: Your comment to Atticus's answer suggests a different interpretation, under which the question has an easy positive answer.
Let $G$ be a definable group in a theory $T$. For any model $M\models T$, let $S_G(M)$ be the space of types over $M$ which contain the formula defining $G$. Then there is a natural action of $G(M)$ on $S_G(M)$ by translation: for $g\in G(M)$ and a type $p\in S_G(M)$, define $g\cdot p = \text{tp}((g\cdot a) / M)$ for any realization $a$ of $p$ in an elementary extension $M\preceq N$ (here $(g\cdot a)$ refers to the product of $g$ and $a$ in the group $G(N)$). More concretely, $g\cdot p = \{\varphi(x,m)\mid \varphi(g\cdot x,m)\in p\}$.
Now let $M\preceq N$ be models of $T$, and let $p\in S_G(M)$ be $G(M)$-invariant, meaning that it is fixed by the action of $G(M)$ on $S_G(M)$ defined above. Let $q\in S_G(N)$ be an heir of $p$. Then the claim is that $q$ is $G(N)$-invariant.
Suppose $q$ is not $G(N)$-invariant. Then there is some $g\in G(N)$ and some formula $\varphi(x,n)$ with $n\in N$ such that $\varphi(x,n)\in q$ but $\varphi(g\cdot x,n)\notin q$. Then the conjunction $G(g)\land \varphi(x,n)\land \lnot \varphi(g\cdot x,n)$ is in $q$, and since $q$ is an heir of $p$, there are some $h,m\in M$ such that $G(h)\land \varphi(x,m)\land \lnot \varphi(h\cdot x,m)$ is in $p$. But this contradicts $G(M)$-invariance of $p$.
