Notation 2.4. To simplify notations, if A is a small set of parameters, we write $|A|$ instead of $|A|+|\mathcal {L}|+\aleph _0$ .

Fact 2.6 (See [Reference Pillay13, Proposition 1.9] and [Reference Simon16, Section 2.2, discussion before Lemma 2.18])

Let A be a small subset of a model M. Let $p \in S(M)$ be A-invariant. Then, for any model $N \supseteq M$ , the type p has a unique extension $p|_N \in S(N)$ which is A-invariant.

Moreover, if p is A-definable, then $p|_N$ is A-definable, using the same defining scheme as p, and the whole conclusion holds even if A is not small with respect to M.

Proof. The first point is a consequence of the definition of forking, and the fact that B-indiscernible sequences are A-indiscernible. For the second point, see [Reference Tent and Ziegler17, Lemma 7.1.11].

For the third point, see [Reference Conant and Kruckman5, Proposition 2.12]. Let us prove the fourth point. By the third point, we deduce that . This implies that the formula $x=a$ does not fork over A. In particular, it does not divide over A. Then, one can check that no infinite A-indiscernible sequence containing a is injective. This implies that $a \in acl(A)$ , as required. For the fifth point, see [Reference Tent and Ziegler17, Exercise 7.1.4]. For the sixth point, see [Reference Simon16, Lemma 5.18].

Proof. Let us prove the first point, the second one being easier. If $b \in acl(Ca)$ , let $k < \omega $ , and $\phi (x,y)$ be a formula over C such that $\models \phi (a, b)$ and $\models \forall x \, \exists ^{\leq k} \, y \, \phi (x,y)$ . By Definition 2.5(5), let $M \supset D$ be a sufficiently saturated model and a C-definable extension $p \in S(M)$ of $tp(a/D)$ . Note that p is not necessarily unique, but we only want to find some $acl(C)$ -definable extension of $tp(b/D)$ . To simplify notations, let us assume that a realizes p. Let $\psi (y, z)$ be a formula without parameters. Let q denote $tp(b/M)$ . Let us consider the following C-definable binary relation: $d_1 E d_2$ if and only if $\models d_p x [\forall y \, \phi (x,y) \rightarrow [\psi (y,d_1)\leftrightarrow \psi (y,d_2)]]$ . Then, for $d_1, d_2 \in M$ , if $d_1 E d_2$ , then $\models \psi (b, d_1)\leftrightarrow \psi (b, d_2)$ . Indeed, in this context, the element a realizes $p|_{C d_1 d_2}$ .

Then, by elimination of imaginaries, let $c_1/E, \dots , c_r/E$ be the codes of the classes modulo E, where the $c_i$ are in M. These codes are in $acl(C)$ . To construct the definition $d_q y \, \psi (y,z)$ , let $I \subseteq \lbrace 1, \dots , r \rbrace $ be the set of indices i such that $\models \psi (b, c_i)$ . Using the definition of the relation E, one can check that the formula $\bigvee _{i \in I} z E c_i$ is an appropriate definition of q for the formula $\psi (y,z)$ . Moreover, since the $c_i / E$ are in $acl(C)$ , this formula is equivalent to a formula defined over $acl(C)$ .

Proof. For points 2, 3, 4, and 6, see Proposition 2.1 in [Reference Pillay and Tanović15]. For point 1, see the proof of [Reference Pillay and Tanović15, Proposition 2.1.i]. Let us now prove point 5. Let $B \subset M$ be a small set such that p is B-invariant. Then, by point 3, p is A-definable, so B-definable as well. By elimination if imaginaries, the type p is thus $dcl(A)\cap dcl(B)$ -definable. Let $C = dcl(A)\cap dcl(B)$ .

Then, unfolding the definition of generic stability, we deduce that p is generically stable over $dcl(B) \supseteq C$ , so over B as well.

Fact 2.18. If $q \in S(M)$ does not fork over A, where $M \supset A$ is a (not necessarily sufficiently saturated) model, then $q|_A$ is generically stable, in the sense of Definition 2.17(1), if and only if q is generically stable over A in the sense of Definition 2.17(2). If M is sufficiently saturated, this is also equivalent to q being generically stable over A, in the sense of Definition 2.13(2).

Proof. Let $M \supset C$ be a sufficiently saturated model, and let $p' \in S(M)$ be the nonforking extension of the type $p \in S(A)$ . We know that q does not fork over A. So, by Proposition 2.10(2), q has an extension $q' \in S(M)$ which does not fork over A. Then, $q'|_A =q|_A = p$ . So, by stationarity of p, we have $q' = p'$ . So $q'$ is generically stable over A, so a fortiori over B. Then, by Proposition 2.15, $q'|_B = q$ is stationary. Therefore, $q'$ is indeed the unique nonforking extension of q, and $q'$ is generically stable over B. Since we have also proved the equality $q' = p'$ , we are done.

Proof. By stationarity and Proposition 2.20, one can check that $tp(a/C) = p|_C$ .

Proof. The first point is actually a consequence of Proposition 2.10(1) and (6), and holds in general. Let us prove the second point. We know that , in particular $tp(bc / A)$ does not fork over A. Since $tp(d / A)$ is generically stable, we can apply symmetry, to deduce . Also by symmetry, we have . So, by the first point applied to $(d, c, b)$ , we have . In particular, . So, by symmetry again, we get , as required.

Proof. Let $\kappa $ be an infinite cardinal. Let $N \supseteq A$ be $\kappa $ -saturated and $\kappa $ -strongly homogeneous. Let $\alpha $ realize $p|_N$ . Then, we have $\alpha \equiv _A a$ , so there exists an automorphism $\sigma \in Aut(\mathbb {U} / A)$ such that $\sigma (a) = \alpha $ . Let $M= \sigma (N)$ . Then, the model M is $\kappa $ -saturated and $\kappa $ -strongly homogeneous, since it is isomorphic to N. Also, we have $\alpha \models p|_N$ , and the type $p|_{\mathbb {U}}$ is A-invariant. Since $\sigma \in Aut(\mathbb {U} / A)$ , this implies that the element $\sigma (\alpha ) = a$ realizes the type $p|_{\sigma (N)} = p|_M$ , as required.

Proof. For the first point, see [Reference Dobrowolski and Krupinski6, Proposition 1.2]. Although the statement there only deals with the case $b \in dcl(a)$ , the proof can easily be adapted to the hypothesis $b \in dcl(Aa)$ .

Let us then prove the second point. Note that both the hypotheses and the conclusion only depend on $tp(ab / M)$ . By Lemma 2.24, let $N \succeq M$ be a veryFootnote ² saturated and very strongly homogeneous model such that a realizes $p|_N$ . By Lemma 2.11, we know that $q=tp(b/N)$ is definable over $acl(A)$ , so is a fortiori $acl(A)$ -invariant. Let $\phi (y,x,n)$ be a formula with parameters in N such that $\phi (y,a,n)$ isolates the type of b over $Na$ . Let $M_1 \preceq N$ be a $|M|^+$ saturated model containing $M n$ , and which is small with respect to N.

We will show that q is generically stable over $M_1$ . Let $r = tp(ab / M_1)$ . Then, by construction, we have $p(x) \cup r(x,y) \models q(y)$ . So, by Theorem 3.5(3) in [Reference Mennui11], the type q is generically stable over $M_1$ . Since it is $acl(A)$ -invariant, we conclude by point 5 of Proposition 2.15 that it is generically stable over $acl(A)$ , as required.

Notation 2.28. If a type p is definable and admits a unique definable extension to a model, we let $Cb(p)$ denote its canonical basis, i.e., the definable closure of the codes of the formulas in its defining scheme. Similarly, if $tp(a/B)$ is definable and admits a unique definable extension to a model, we write $Cb(a/B)$ for the canonical basis of $tp(a/B)$ .

Proof. To simplify notations, let $C = Cb(a f_c(a) / B)$ . First note that the result does not depend on the choice of the realization a. So, we may assume that a realizes $p|_M$ , for some model M containing B.

Proof. By definition of C, the type q defined by

$$\begin{align*}q(x,y)=tp(a f_c(a) / M)(x,y),\end{align*}$$

is C-definable. We use compactness. Let d be such that $d \equiv _{AC} c$ . It suffices to show that $[f_d]=[f_c]$ . We know, by choice of the type q, that $q(x,y) \models f_c(x)=y$ . In other words, $\models d_q xy \, (f_c(x) = y)$ . Note that the formula $\phi (z) = d_q xy \, (f_z(x) = y)$ is over $AC$ , since q is C-definable and f is A-definable. By choice of d, we have $d \equiv _{AC} c$ . Thus $\models d_q xy \, (f_d(x) = y)$ . So, if $q'$ is the C-definable extension of q to $Md$ , we have

$$\begin{align*}q'(x,y) \models f_d(x) = y \, \wedge \,f_c(x) = y.\end{align*}$$

Finally, $q'(x,y) \models f_d(x) = f_c(x)$ , so $p(x)|_{Md} \models f_d(x) = f_c(x)$ , i.e., $[f_c] = [f_d]$ , as desired.

So, we have indeed proved that $[f_c]$ and C are interdefinable over A.

Proof. Let $c \in X$ , and $a \models p|_{Ac}$ . Then, by [Reference Adler, Casanovas and Pillay1, Theorem 2.2], we have $f_c(a) \in dcl(A a [f_c])$ . We now wish to make this fact uniform in c, i.e., we look for a suitable A-definable family F of definable maps. By compactness, there is a finite collection of A-definable families of definable maps $F_1, \dots , F_k$ such that, for all $c \in X$ , we have $\models d_p x \, f_c(x) = {(F_i)}_{[f_c]}(x)$ for some $i = 1, \dots , k$ . It then suffices to glue these families of maps together into a single F to conclude the proof.

Proof. Let us prove the first point. One direction is straightforward. Assume that $dcl(Aa) = dcl(Ab)$ . Let f be an A-definable map sending a to b, and g an A-definable map sending b to a. So, we have $g \circ f (a) = a$ (and $f\circ g (b) = b$ ). Then, let X be the A-definable set $\lbrace \alpha \, | \, g \circ f (\alpha ) = \alpha \rbrace $ . By definition, we have $a \in X$ . Hence, the A-definable map $f|_X$ is injective, and sends a to b. One may thus pick $\sigma = f|_X$ .

Let us now prove the second point. Let a realize p. So, by hypothesis, we have $f(a) = g(a) =:b$ . By construction, the element b realizes the definable type $f_*(p) = g_*(p) \in S(M)$ , and we have $f^{-1}(b) = g^{-1}(b) = a$ . Thus, b is a witness for the equality of the germs $[f^{-1}]$ and $[g^{-1}]$ .

Fact 2.39 (Commutativity see [Reference Conant, Gannon and Hanson4, Remark 5.18])

Let $p, q \in S(M)$ be A-invariant types, where A is a small set contained in M. Assume that p is generically stable over A. Then, $p(x) \otimes q(y) = q(y) \otimes p(x)$ , this equality being between A-invariant types (see Remark 2.7 for an explanation of this idea).

Proof. Let $q(y)$ be an element of $\mathcal {F}$ . By hypothesis on $\mathcal {F}$ , there exists a small set $C \supseteq B$ such that $q(y)$ is C-invariant. Let us show that $h_{*} p(x)\otimes q(y) = q(y)\otimes h_{*} p(x)$ . Let $D \supseteq C$ , and let $(k,b)$ realize $h_{*} p(x)\otimes q(y) |_{D}$ .

Then, k realizes $h_{*} p|_{Db}$ . So, by Remark 2.31 applied to $Db$ , there exists a realizing $p|_{Db}$ such that $h(a)=k$ . So $(a,b)$ realizes $p \otimes q|_{D}$ . Since p commutes with $\mathcal {F}$ , the pair $(b,a)$ realizes $q\otimes p|_{D}$ , so $b \models q|_{Da}$ , a fortiori $b \models q|_{Dh(a)}=q|_{Dk}$ . Therefore, $(b,k)$ realizes $q \otimes h_{*} p |_{D}$ . We have proved that $h_{*} p$ commutes with $\mathcal {F}$ .

Proof. We may assume that M is sufficiently saturated. Let $c \in Stab(p)(M)$ . Then, we compute $c \cdot f_*(p) = f_*(c \cdot p)= f_*(p)$ , so $c \in Stab(f_*(p))$ , as required.

Proof. The implication $2. \implies 1.$ is straightforward: by definition, the stabilizer of any generic type is of bounded index. Let us prove $1. \implies 2.$ Let p be a generic type for G, whose stabilizer is G itself. Let $H \leq G$ be a type-definable subgroup of bounded index. Then, if M is a sufficiently saturated model containing all the parameters involved, it represents every coset of H. Then, $p|_M$ concentrates on a coset of H. So, the stabilizer $Stab(p)$ is contained in a conjugate of H. As $Stab(p) = G$ , we deduce that $H=G$ , as desired.

Proof. Let $\alpha , \beta $ realize $(p\otimes p)|_B$ . Then, since $\beta ^{-1} \in Stab(p) = G$ and $\alpha \models p|_{B\beta }$ , we know that $\beta ^{-1} \cdot \alpha $ realizes $p|_B$ . Then, by Fact 2.39, we know that $\alpha \beta \equiv _B \beta \alpha $ . So $\alpha ^{-1} \cdot \beta $ realizes $p|_B$ . Then, since $p|_B$ is a complete type, we have shown that, for all elements c realizing $p|_B$ , the element $c^{-1}$ realizes $p|_B$ , as desired.

Let us then prove the second point. If g is in $ G$ and a realizes $p|_{Ag}$ , then, by the first point, we know that $a^{-1}$ realizes $p|_{Ag}$ . So, by hypothesis on the stabilizer of p, the element $g^{-1} \cdot a^{-1}$ realizes $p|_{Ag}$ . Then, again by the first point, the element $a \cdot g = (g^{-1} \cdot a^{-1})^{-1}$ realizes $p|_{Ag}$ , as stated.

For the third point, see [Reference Pillay and Tanović15, Lemma 2.1]. Finally, let us prove the fourth point. Let $g \in G$ , and $a \models p|_{Ag}$ . Then, we have $g= (g \cdot a) \cdot a^{-1}$ ,where $g \cdot a$ realizes p because $Stab(p) = G$ , and $a^{-1}$ realizes p by the first point.

Proof. Let us first prove the existence of the type-connected component $G^{00}$ . Let $p \in S(N)$ be a generically stable generic type for G, where N is sufficiently saturated. Then, as $H=Stab(p)$ is of bounded index, every left coset and every right coset of $Stab(p)$ is represented in N. So p concentrates on some left coset of H, and on some right coset of H as well. Let $g \in G(N)$ be such that p concentrates on $H \cdot g$ . Now, let q be the translate $p \cdot g^{-1}$ . Then, the type q concentrates on H. Also, by Proposition 2.25, it is generically stable.

Now, by Lemma 3.5, we have $Stab(q) \geq Stab(p) = H$ . Then, since q concentrates on H, a fortiori it concentrates on $H_1:= Stab(q)$ . So, by Proposition 3.6, the type-definable group $H_1$ has no proper type-definable subgroup of bounded index. Moreover, as $H \leq H_1 \leq G$ , we know that $H_1$ is of bounded index in G. Hence, $H_1$ is indeed the smallest type-definable subgroup of bounded index of G. This implies that $H_1 = G^{00}$ is normal, and even invariant under definable automorphisms, in G, and the inclusion $H \leq H_1$ implies that $H_1 = H = G^{00}$ .

Since q is generically stable, concentrates on $G^{00}$ , and is stabilized by $G^{00}$ , we may apply Lemma 3.7, to deduce that the group $G^{00}$ has a unique generic type, which is q.

Now, let $p_1$ be some generic of G. By definition, $Stab(p_1)$ is of bounded index, so it contains $G^{00}$ . On the other hand, the complete type $p_1$ concentrates on some coset of $G^{00}$ , so $Stab(p_1)$ is contained in some conjugate of $G^{00}$ . Since the latter is normal, we have in fact $Stab(p_1) = G^{00}$ , as desired.

Proof. Let us show that the second point follows from the first. We know that the type-connected component $G^{00}$ of G exists: it is the stabilizer of any generic of G. Then, we consider the action of $G^{00}$ on X. By the first point, each $G^{00}$ -orbit contains a unique generic type, whose stabilizer is $G^{00}$ . Note that, since $G^{00}$ is of bounded index in G and the action is transitive, there are only boundedly many $G^{00}$ -orbits. Now, let M be a sufficiently saturated model over which everything is defined. So M contains a point in each $G^{00}$ -orbit. Let $q_1, q_2 \in S(M)$ be two generic types of X. Let $x_1, x_2 \in X(M)$ be in the $G^{00}$ -orbits of (the realizations of) $q_1$ and $q_2$ respectively, and let $g \in G(M)$ be such that $g \cdot x_1 = x_2$ . We shall prove that g sends the type $q_1$ to $q_2$ . Since $G^{00}$ is normal, we have $g \cdot G^{00} = G^{00} \cdot g$ . Thus, we can compute $G^{00}(M) \cdot g \cdot q_1 = g \cdot G^{00}(M) \cdot q_1 = g \cdot q_1$ , so the type $g \cdot q_1$ is generic. Also, since g sends $x_1$ to $x_2$ , the type $g \cdot q_1$ concentrates on the same $G^{00}$ -orbit as $q_2$ . Hence, by the first point, we have $g \cdot q_1 = q_2$ , as desired.

Let us now prove the first point. Assume that G is type-connected. By Lemma 3.7, let p be the unique generic type of G. We know that p is generically stable. Let M be a big enough model containing all the parameters involved, and let $x_0 \in X(M)$ . Let $f: G \rightarrow X$ be the definable map $g \mapsto g \cdot x_0$ . By transitivity of the action, it is onto. Let q be the type $f_*(p)$ . By Proposition 2.25, it is generically stable. We know that $Stab(q)(M) = G(M)$ , because $Stab(p) = G$ . So, since M is sufficiently saturated, we have $Stab(q) = G$ , so q is generic in X.

Now, let $q_1$ be another generic type in X. Without loss of generality, we may assume that $q_1$ is M-definable. We want to show that $q_1 = q$ . Let $x_1 \models q_1|_M$ , and $g_1 \in G$ such that $f(g_1) = x_1$ . Let $g \models p|_{Mg_1}$ . Then, by Lemma 3.7, we have $g \cdot g_1 \models p|_{M g_1}$ , so $f(g \cdot g_1) \models f_*(p)|_{M g_1} = q|_{M g_1}$ . In particular, we have $f(g \cdot g_1) = g \cdot x_1 \models q|_M$ . On the other hand, since $q_1$ is generic in X, and G is type-connected, we have $Stab(q_1) = G$ . Moreover, by Fact 2.39, we have $(x_1, g) \models (q_1 \otimes p)|_M$ . So $g \cdot x_1 \models q_1|_{M g}$ . Therefore, $g\cdot x_1$ realizes both $q|_M$ and $q_1|_M$ . So $q=q_1$ , as desired.

Proof of Proposition 3.14

The algebraicity relations are clear in the case of a free action. In fact, let us deal with the more subtle case of a faithful transitive action. Since any free action is in particular faithful, and using Remark 3.12(1), our proof will also include a proof of the independence relations for the case of a free action.

First, note that the type $tp(g_1, g_2, x / M)$ is definable, since it is the tensor product $p\otimes p \otimes q$ . Then, by Lemma 2.11, the type over M of the sextuple is definable. This is the first point of the definition. The third point is easier to check, since $b_1 = g_1$ , $b_2 = g_2$ and $b_3 = g_2 \cdot g_1$ .

Let us now prove the second point of the definition. First, note that $tp(g_1\cdot x / M) = tp(g_2 \cdot g_1 \cdot x / M) = tp(x / M) = q$ , because $Stab(q) = G$ , and $x \models q|_{M g_1 g_2}$ . Similarly, $tp(g_2\cdot g_1 / M) = p$ . So, to prove the second point, we may use stationarity and commutativity for generically stable types. By saturation of M, let $x_0 \in X(M)$ , and let $f: G \rightarrow X$ be the definable map $g \mapsto g \cdot x_0$ .

Claim 3.18. To prove the second point of the definition, it suffices to do it for the following quadrangle:

Note that, by choice of g, we have $(g_1, g_2, g) \models p^{\otimes 3}|_M$ . In other words, by equivariance of f, we reduced to the case $X=G$ , with the action by left translation. Recall that, by Lemma 3.7, we have $Stab^r(p) = G = Stab(p)$ . Also note that the setting is more symmetric now: we only have elements in the group, they are all generic, and they come from (products of) realizations of a generically stable type, so that we may use commutativity of the tensor product. The proof below relies on this symmetry.

Now, consider the alphabet $\Sigma = \lbrace g_1, g_2, g \rbrace $ . Elements in the diagram above can be represented as words in that alphabet, where the letters only come in increasing order, if we set $g_2 < g_1 < g$ .

To conclude the proof of the second point of the definition of generically stable group configurations, it suffices to combine Claims 3.19, 3.20, and 3.21, keeping in mind Proposition 2.10(3)(b), and the fact that any element on a line is definable over M and the other two.

Finally, let us prove the fourth point of the definition of a definable group configuration. The first part follows from definability over M of the group action. Let us now prove the statements dealing with the canonical bases. We follow [Reference Pillay14, Chapter 5, Remark 4.1]. Since the action is faithful, we may harmlessly identify any element of G with the permutation of X it defines. Then, by Proposition 2.32, to prove that, say $g_1$ is interalgebraic over M with $Cb(x, g_1 \cdot x / acl(M g_1))$ , it suffices to prove that $g_1$ is interalgebraic over M with its germ at q. Let us show that, in fact, for any $h \in G$ , the element h is definable over $M\ \ \widehat{}\ \ [h]$ .

Thus, the type-definable set $\Gamma $ , whose elements are the germs at q of the form $[g]$ , for $g \in G$ , can be equipped with a definable group operation in a natural way, such that the map $g \in G \mapsto [g] \in \Gamma $ is an M-definable surjective group homomorphism. Moreover, by the strong germs property, i.e., Corollary 2.36, the generic action of G on q induces a generic action of $\Gamma $ on q.

Thus, the M-definable group homomorphism $g \in G \mapsto [g] \in \Gamma $ is an isomorphism. This implies in particular that, for any $h \in G$ , we have $dcl(Mh) = dcl(M[h])$ .

Proof. Let $C \subset M$ be a small algebraically closed set of parameters over which everything is defined, and which captures the interalgebraicities. Thus, we have the following configurations, which are equivalent over C:

For $1 \leq i \leq 2$ , let $c_i=(g_i, h_i) \in G \times H$ . Also, let $c_3=c_2 \cdot c_1 = (g_2 \cdot g_1, h_2 \cdot h_1) \in G \times H$ . Let $p_i=tp(c_i / M)$ , for $1 \leq i \leq 3$ .

By assumption and interalgebraicity, we have and . Moreover, by definition, we have $c_2 \cdot c_1 = c_3$ . Thus, we have $c_1=(g_1, h_1) \in Stab^r(p_2, p_3)$ . Let us also define $S=Stab^r(p_2)$ and $Z = Stab^r(p_2, p_3)$ . So S is a type-definable subgroup of $G \times H$ . Let $c'_1 \in M$ be such that $c'_1 \equiv _C c_1$ . So, there exist $g \in G(M)$ and $h \in H(M)$ such that $c'_1 = (g,h) \in G \times H$ .

Let $\pi : G \times H \rightarrow G$ be the canonical projection. Since $\pi $ is a group morphism, the claim implies that $\pi (S) \cdot \pi (c'_1) = \pi (Z)$ . In particular, as $c_1=(g_1, h_1) \in Z$ , we get ${g}^{-1} \cdot g_1 \in \pi (S)$ . However, by construction, $g_1$ is, in the group G, generic over M. Since g is in M, this implies that ${g}^{-1} \cdot g_1$ is also generic over M. Therefore, the M-type-definable subgroup $\pi (S) \leq G$ contains an element generic over M. So, the generic of G concentrates on $\pi (S)$ . Since, by Lemma 3.7(4), any element of G is the product of two realizations of this generic type, we deduce that $\pi (S)$ is equal to the whole group G. Symmetrically, the projection of S to the second coordinate is equal to H.

Let $N_1 = \lbrace n_1 \in G \, | \, (n_1,1) \in S \rbrace $ and $N_2 = \lbrace n_2 \in H \, | \, (1, n_2) \in S \rbrace $ . We then have $N_1 \times \lbrace 1 \rbrace \lhd S$ , so $N_1 = \pi (N_1 \times \lbrace 1 \rbrace ) \lhd \pi (S) = G$ . Similarly, we have $N_2 \lhd H$ . Thus, projecting the subgroup $S \leq G \times H$ , we get a subgroup $\Sigma \leq G/N_1 \times H/N_2$ , which is the graph of a definable group isomorphism $G/N_1 \rightarrow H/N_2$ . Indeed, since S projects surjectively onto G and H, the subgroup $\Sigma $ projects surjectively onto $G/N_1$ and $H/N_2$ .

Now, by saturation of M, let $(x_0, y_0) \in X(M) \times Y(M)$ be such that $(x_0, y_0) \equiv _C (g_1 \cdot x, h_1 \cdot y) $ . Let us consider the following type-definable set: $T= S \cdot (x_0, y_0) \subseteq X \times Y$ . Let $q_3 = tp(g_1 \cdot x, h_1 \cdot y / M)$ and $q_1 = tp(g_2 \cdot g_1 \cdot x, h_2 \cdot h_1 \cdot y / M)$ .

Then, by equivalence over C of the configurations, Proposition 2.25 implies that $q_1$ and $q_3$ are generically stable over $C=acl(C)$ . Also, by hypothesis, we have $(g_2, h_2) \in Stab(q_3, q_1)$ .

Thus, we can use the set T to define an S-equivariant finite correspondence between X and Y.

Proof. Let $(a_1, a_2, a_3, b_1, b_2, b_3)$ be a generically stable regular group configuration over A. Let us show that it is a generically stable group configuration over A. The second point of the definition of definable group configuration, i.e., definability of relative types, follows from independence, i.e., the first point of the definition of a regular group configuration, and stationarity of generically stable types. Then, the only properties to check are those regarding the canonical bases $Cb(a_j a_k / acl(A b_i))$ , for $i,j,k$ such that $\lbrace i,j,k \rbrace = \lbrace 1,2,3 \rbrace $ . By symmetry of the context, it suffices to show that $b_1$ and $Cb(a_2 a_3 / acl(A b_1))$ are interalgebraic over A.

First, let N be a sufficiently saturated model containing $A b_1$ , such that , which exists by extensibility of $tp(a_2 / A b_1)$ . Then, by Fact 2.18, the type $tp(a_2 / N)$ is generically stable over $A b_1$ : it is in fact the unique nonforking extension of $tp(a_2 / Ab_1)$ . Then, by Proposition 2.25, the type $tp(a_2 a_3 / N)$ is generically stable over $acl(A b_1)$ , and stationarity yields the equality: $Cb(a_2 a_3 / acl(A b_1)) = Cb(a_2 a_3 / N)$ .

On the one hand, it is clear that $Cb(a_2 a_3 / acl(A b_1)) \subseteq acl(A b_1)$ . Conversely, let us show that $b_1$ is algebraic over $A \cup Cb(a_2 a_3 / acl(A b_1))$ . Let $C=A \cup Cb(a_2 a_3 / acl(A b_1))$ . By assumption, we know that $b_1 \in acl (A a_2 a_3) \subseteq acl( C a_2 a_3)$ . Moreover, since the type $tp(a_2 a_3 / N)$ is generically stable over $acl(Ab_1)$ , and $C \subset N$ contains $Cb(a_2 a_3 / acl(A b_1))$ , the type $tp(a_2 a_3 / N)$ is definable over C. Thus, by Proposition 2.15(5), this type is generically stable over C. In particular, . Thus, since $b_1 \in N$ , we have . Finally, since $b_1 \in acl(C a_2 a_3)$ , we can apply Proposition 2.10(4), which implies that $b_1 \in acl(C)$ , as desired.

Proof. Indeed, Proposition 2.25 implies that generic stability is preserved. Moreover, the algebraicity relations are preserved, and, in the case of generically stable regular group configurations, so are the independence relations, thanks to Proposition 2.10(3). In the case of generically stable group configurations, the second point of the definition (i.e., definability of the relative types) is preserved thanks to Lemma 2.11: let $\alpha , \beta $ be elements such that $tp(\alpha / A \beta )$ is A-definable, and let $\alpha ', \beta '$ be such that $acl(A \alpha ) = acl(A \alpha ')$ and $acl(A \beta ) = acl(A \beta ')$ . Then, by Definition 2.5(5), there exists a model $M \supset A \beta $ and a type $p \in S(M)$ extending $tp(\alpha / A \beta )$ such that p is A-definable. We may assume that $\alpha $ realizes p. Then, $tp(\alpha / M)$ is $acl(A)$ -definable, so $tp(\alpha ' / M)$ is also definable over $acl(A)$ . Since $M \supset acl(A \beta ) = acl(A \beta ')$ , we have indeed shown that $tp(\alpha ' / acl(A) \beta ')$ is definable over $acl(A)$ .

Lemma 2.11 also implies the required properties for the canonical bases: let $(b_1, a_2, a_3)$ and $(b'_1, a'_2, a'_3)$ be such that $acl(A b_1) = acl(A b'_1)$ , $acl(A a_2) = acl(A a'_2)$ and $acl(A a_3) = acl(A a'_3)$ . Assume that $tp(a_2 a_3 / acl(A b_1))$ is definable, and $b_1$ is interalgebraic over A with the canonical basis of this type. Then, applying Lemma 2.11 once, we deduce that the type $tp(a'_2 a'_3 / acl(A b'_1))$ is definable. Let $C \subseteq acl(A b'_1) = acl(A b_1)$ denote its canonical basis.

This concludes the proof.

Notation 3.34. If $p(x,y)$ is a complete type in several variables, where $x,y$ are tuples of variables, we let $p_{x}$ denote the restriction of p to the tuple of variables x.

Proof. First, let us assume that $\alpha $ is a generically stable (regular) group configuration over A, and prove that it is a generically stable (regular) group configuration over $A_0$ .

The interalgebraicity relations are proved using Proposition 2.35. Let us prove one of the independence relations, say . We know that , that and that $tp(a_1 a_2 / A_0)$ is generically stable. So, by Proposition 2.21, we have , which implies that , as desired. Note that this independence relation, along with stationarity of generically stable types, imply that $tp(a_1 a_2 / A_0 b_1)$ is $A_0$ -definable.

To conclude, let us now prove the statement regarding the canonical bases, in the case of generically stable group configurations. Let c denote $Cb(a_1 a_2 / acl(A_0 b_3))$ . We wish to prove that $b_3 \in acl(A_0 c)$ .

Then, by definition, we have $c \subseteq acl(A_0 b_3)$ . By Lemma 2.24 applied to $b_3$ and $A_0$ , let M be a model containing $A_0$ , such that the type of $b_3$ over M is the canonical extension of $tp(b_3 / A_0)$ , which is also the canonical extension of $tp(b_3 / A)$ . Then, since $c \subseteq acl(A_0 b_3)$ , Proposition 2.25 implies that the type $tp(b_3 c / M)$ is generically stable over $acl(A_0)$ . We also know that $b_3 \in acl(A c)$ , in particular $b_3 \in acl(M c)$ . Thus, by Proposition 2.35, we have $b_3 \in acl(A_0 c)$ , as required.

Let us now prove the converse: assume that $\alpha $ is a generically stable (regular) group configuration over $A_0$ , and prove that it is a generically stable (regular) group configuration over A. The algebraicity relations are clear, because $A \supseteq A_0$ . For the independence relations, we use stationarity: for instance, let us prove . Since $\alpha $ is a generically stable group configuration over $A_0$ , we know that $tp(a_1 a_2 b_1 / A_0)$ is the tensor product $tp(a_1 / A_0) \otimes tp(a_2 / A_0) \otimes tp(b_1 / A_0)$ , and is generically stable. Then, by stationarity, the nonforking extension $tp(a_1 a_2 b_1 / A)$ is the tensor product $tp(a_1 / A) \otimes tp(a_2 / A) \otimes tp(b_1 / A)$ . This implies the independence .

The statement regarding canonical bases follows from equalities of the form

$Cb(a_1 a_2 / acl(A b_3)) = Cb(a_1 a_2 / acl(A_0 b_3))$ , which were proved above.

Proof. First, note that a belongs to the finite set coded by $\alpha $ , so $a \in acl(A \alpha )$ . Let us prove that $\alpha $ is algebraic over $Aa$ . We know that $\alpha $ codes a subset of the finite set of conjugates of a over B, so $\alpha \in acl(B a)$ . Similarly, $\alpha \in acl(B'a)$ . Thus, by Proposition 2.10(4) and the hypothesis , we have $\alpha \in acl(Aa)$ , as required.

Proof of Proposition 3.38

By saturation of M, let $b'_1, b"_2 \in M$ be such that $b'_1 b"_2 \equiv _{C_0} b^0_1 b^0_2$ .

Then, let $b'_3, a'_2$ and $b"_3, a"_1$ be such that $a^0_1 b^0_2 a^0_3 b'_1 a'_2 b'_3 \equiv _{C_0} a^0_1 b^0_2 a^0_3 b^0_1 a^0_2 b^0_3$ and $b^0_1 a^0_2 a^0_3 a"_1 b"_2 b"_3 \equiv _{C_0} b^0_1 a^0_2 a^0_3 a^0_1 b^0_2 b^0_3$ . In other words, the following are copies of the original configuration:

Then, let $\widetilde {a_1}$ be the code of the set of conjugates of $a_1^0$ over $C_0 b^0_2 a^0_3 a'_2 b'_3$ . Similarly, let $\widetilde {a_2}$ be the code of the set of conjugates of $a_2^0$ over $C_0 b^0_1 a^0_3 a"_1 b"_3$ .

The motivation for building these copies is that $\widetilde {a_1} \in dcl(C_0 b^0_2 a^0_3 b'_3 a'_2)$ and $\widetilde {a_2} \in dcl(C_0 b^0_1 a^0_3 b"_3 a"_1)$ .

Then, let $(\alpha _1, \alpha _2, \alpha _3, \beta _1, \beta _2, \beta _3)$ denote the following configuration:

Let $C_1 = acl(C_0 b'_1 b"_2) \subset M$ .

Thus, by Proposition 3.32, the tuple $(\alpha _1, \alpha _2, \alpha _3, \beta _1, \beta _2, \beta _3)$ is a generically stable group configuration over $C_1$ , and over M. What we gained is that $\alpha _1 \in dcl(C_1 \alpha _3 \beta _2)$ and $\alpha _2 \in dcl(C_1 \alpha _3 \beta _1)$ . Also note that $\beta _3 = b^0_3$ .

Now, let $\widetilde {\alpha _3}$ be the code of the set of conjugates of $\alpha _3$ over $C_1 \beta _1 \alpha _2 \beta _2 \alpha _1$ . As in Claim 3.42, by Lemma 3.40, we have $acl(C_1 \alpha _3) = acl(C_1 \widetilde {\alpha _3})$ . So, by Proposition 3.32, the quadrangle $(\alpha _1, \alpha _2, \widetilde {\alpha _3}, \beta _1, \beta _2, \beta _3)$ is a generically stable group configuration over $C_1$ , and over M. By construction, we get $\widetilde {\alpha _3} \in dcl(C_1 \beta _1 \alpha _2 \beta _2 \alpha _1)$ .

Now, let $\beta "'_3 \in M$ realize $tp(\beta _3 / C_1) = tp(b^0_3 / C_1)$ . Then, by $C_1$ -invariance of the types $tp(\widetilde {\alpha _3} \beta _2 \alpha _1 / M \beta _3)$ and $tp(\widetilde {\alpha _3} \alpha _2 \beta _1 / M \beta _3)$ , we have $\widetilde {\alpha _3} \beta _2 \alpha _1 \beta _3 \equiv _{C_1} \widetilde {\alpha _3} \beta _2 \alpha _1 \beta "'_3$ and $\widetilde {\alpha _3} \alpha _2 \beta _1 \beta _3 \equiv _{C_1} \widetilde {\alpha _3} \alpha _2 \beta _1 \beta "'_3$ . So, let $\beta "'_1, \alpha "'_2$ and $\beta "'_2, \alpha "'_1$ be such that $\widetilde {\alpha _3}\beta _2 \alpha _1 \beta _3 \beta _1 \alpha _2 \equiv _{C_1} \widetilde {\alpha _3} \beta _2 \alpha _1 \beta "'_3 \beta "'_1 \alpha "'_2 $ and $\widetilde {\alpha _3} \alpha _2 \beta _1 \beta _3 \beta _2 \alpha _1 \equiv _{C_1} \widetilde {\alpha _3} \alpha _2 \beta _1 \beta "'_3 \beta "'_2 \alpha "'_1$ .

In other words, we have two copies of the configuration $(\alpha _1, \alpha _2, \widetilde {\alpha _3}, \beta _1, \beta _2, \beta _3)$ :

Finally, the configuration $(a_1, a_2, a_3, b_1, b_2, b_3)$ we wish to consider is the following: $(\alpha _1 \alpha "'_2, \alpha _2 \alpha "'_1, \widetilde {\alpha _3}, \beta _1 \beta "'_2, \beta _2 \beta "'_1, \beta _3)$ . We let C denote $acl(C_1 \beta "'_3) = acl(C_0 b'_1 b "_2 \beta "'_3)$ .