BAD GROUPS IN THE SENSE OF CHERLIN

OLIVIER FRÉCON

Abstract. There exists no bad group (in the sense of Gregory Cherlin), namely any simple group of Morley rank 3 is isomorphic to PSL_{2}(K) for an algebraically closed field K.

1. Introduction

Model theory is a branch of mathematical logic concerned with the study of classes of mathematical structures. There are numerous interactions between model theory and the other areas of mathematics, sometimes spectacular as Wilkie’s article on the field of real numbers with the exponential function, Hrushovski’s proofs of the geometric Mordell-Lang conjecture and of the Manin-Mumford conjecture, Sela’s work connecting logic with geometric group theory, Pila’s paper on the André-Oort conjecture, the manuscript on Berkovich spaces by Hrushovski and Loeser, or the theory of motivic integration of Cluckers, Loeser and others.

Model theoreticians study mathematical structures by considering first-order sentences and formulas. The Morley rank is a model-theoretical notion of abstract dimension. It generalizes the dimension of an algebraic variety (when the ground field is algebraically closed). There are other notions of abstract dimension, the importance of the Morley rank lies on Morley’s Categoricity Theorem below, which “can be thought of as the beginning of modern model theory” (David Marker [11, p. 2]) and the following Baldwin and Zilber Theorems.

We remember that a theory is a set of first-order ℒ -sentences for a language ℒ, it is complete if for any sentence ϕ, either ϕ or ¬ϕ belongs to T, and a theory is κ-categorical for some cardinal κ if, up to isomorphism, it has exactly one model of cardinality κ (cf. [11, Chapters 1 and 2] for more details).

Fact 1.1. – Let T be a complete theory in a countable language.

• (Morley’s Categoricity Theorem, [12]) If T is κ-categorical for some uncountable κ, then T is κ-categorical for every uncountable κ.

• (Baldwin, [2]) If T is uncountably categorical, then it is of finite Morley rank.

• (Zilber, [18]) The theory of an infinite simple group of finite Morley rank is uncountably categorical.

In this paper, we are concerned with groups of finite Morley rank. The main example of such a group is an algebraic group defined over an algebraically closed field in the field language (Zilber, [18]). In the late seventies, Gregory Cherlin [6, §6] and Boris Zilber [18] formulated independently the following algebraicity conjecture.

Date: July 1, 2016.

2000 Mathematics Subject Classification. 20F11, 03C45, 20A15.

Key words and phrases. Groups of finite Morley rank, Bad groups, Projective space.

Conjecture 1.2. – (Cherlin-Zilber Conjecture or Algebraicity Conjecture) An infinite simple group of finite Morley rank is algebraic over an algebraically closed field.

This is the main conjecture on groups of finite Morley rank, and it is still open. Most of studies on groups of finite Morley rank focus on this conjecture. Actually, the original Cherlin Conjecture concerned simple ω-stable groups, but the substantial litterature on the Algebraicity Conjecture treats only the finite Morley rank case.

The Algebraicity Conjecture has been proved for several important classes of groups including locally finite groups [17]. The main theorem on groups of finite Morley rank ensures that any simple group of finite Morley rank with an infinite abelian subgroup of exponent 2 satisfies the Cherlin-Zilber Conjecture [1].

However, in despite of numerous papers on the subject, the Cherlin-Zilber Conjecture is still open, even for groups of Morley rank 3. As a matter of fact, in [6], the Algebraicity Conjecture was formulated as a result from an analysis of simple groups of Morley rank 3. The main result of [6] can be summarized as follows, where a bad group is a nonsolvable group of Morley rank 3 containing no definable subgroup of Morley rank 2.

Fact 1.3. – (Cherlin, [6]) Let G be an infinite simple group of Morley rank at most 3. Then G has Morley rank 3, and one of the following two assertions is satisfied:

• there is an algebraically closed field K such that G ≃ PSL_{2}(K),

• G is a bad group.

Thus bad groups became a major obstacle to the Cherlin-Zilber Conjecture. These groups have been studied in [6, 14] and [15], whose results are summarized in Facts 2.3 and 2.4 respectively. Later, it was shown that no bad group is existentially closed [10] or linear [13]. However, these groups appeared very resistant, and very sparse other information was known on bad groups.

Furthermore, Nesin has shown in [15] that a bad group acts on a natural geometry, which is not very far from being a non-Desarguesian projective plane of Morley rank 2. However, Baldwin discovered non-Desarguesian projective planes of Morley rank 2 [3]. Thus, the question of the existence, or not, of a bad group was still fully open. In this paper, we show that bad groups do not exist.

Main Theorem 1.4. – There is no bad group.

Note other more general notions of bad groups have been introduced independently by Corredor [7] and by Borovik and Poizat [4], where a bad group is defined to be a nonsolvable connected group of finite Morley rank all of whose proper connected definable subgroups are nilpotent. Such a bad group has similar properties to original bad groups. Moreover, later Jaligot will introduce a more general notion of bad groups [9], and he will obtain similar results. However, we recall that, in this paper, a bad group is defined to be nonsolvable, of Morley rank 3, and containing no definable subgroup of Morley rank 2.

Our proof of Main Theorem goes as follows. First we note that it is sufficying to study simple bad groups since for any bad group G, the quotient group G/Z(G) is a simple bad group by [14, §4, Introduction].

Then we fix a simple bad group G, and we introduce a notion of lines as cosets of Borel subgroups of G (Definition 3.1). In §3, we study their behavior, mainly in regards with conjugacy classes of elements of G.

In §4, we propose a definition of a plane (Definition 4.1). This section is dedicated to prove that G contains a plane (Theorem 4.14). This result is the key point of our demonstration. Roughly speaking, we show that for each nontrivial element g of G such that g = [u, v] for (u, v) ∈ G × G, the union of the preimages of g, by maps of the form ad_{v} : G → G defined by ad_{v}(x) = [x, v], is almost a plane, and from this, we obtain a plane.

In last section §5, we try to show that our notions of lines and planes provide a structure of projective space over the group G. Indeed, such a structure would provide a division ring (see [8, p. 124, Theorem 7.15]), and probably it would be easy to conclude. However, a contradiction occurs along the way, and achieves our proof.

2. Background material

A thorough analysis of groups of finite Morley rank can be found in [5] and [1]. In this section we recall some definitions and known results.

2.1. Axioms. We present the rank axioms exactly as in [1, p. 23-24].

“We consider a group G equipped with additional structure. We suppose that G carries a rank function in the sense of Borovik and Poizat, namely a function “rk” which assigns to each set S definable over G, its “dimension” rk (S), and which satisfies the following four axioms. Monotonicity For any integer n, rk (S) > n if and only if S contains an infinite

family of disjoint definable subsets S_{i} of rank n. Additivity If f : A → B is definable and surjective, and if the fibers f^{−1}(b) have

constant rank r for b ∈ B, then rk (A) = r + rk (B). Definability For any uniformly definable family {S_{b} : b ∈ B} of definable sets,

and for any n ∈ ℕ, the set {b ∈ B : rk (S_{b}) = n} is also definable. Finite Bounds For any uniformly definable family 𝔉 of finite subsets, the sizes of

the sets in 𝔉 are bounded. The definable subsets over G are quotients by definable equivalence relations of definable subsets of G^{n} for some n. A family {S_{b} : b ∈ B} is uniformly definable if, first, B is itself definable, and second, the relation “x ∈ S_{y}” is definable.”

It is shown in [16] that the groups (G, · , ⋯) as above are precisely the groups of finite Morley rank, and that the function rk assigns to each definable set its Morley rank. In particular, in this paper, the Morley rank of a definable set S will be denoted by rk S, as in [5] and [1].

2.2. Morley degree. A nonempty definable set A is said to have Morley degree 1 if for any definable subset B of A, either rk B < rk A or rk (A \ B) < rk A. The set A is said to have Morley degree d if A is the disjoint union of d definable sets of Morley degree 1 and Morley rank rk A.

Fact 2.1. –

• [5, Lemmas 4.12 and 4.14] Every nonempty definable set has a unique degree.

• [5, Proposition 4.2] Let X and Y be definable subsets of Morley degree d and d′ respectively. Then X × Y has Morley degree dd′.

• [6, §2.2] A group of finite Morley rank has Morley degree 1 if and only if it is connected, namely it has no proper definable subgroup of finite index.

Moreover, the following elementary result will be useful for us.

Fact 2.2. – Let f : E → F be a definable map. If the set E has Morley degree 1 and r = rk f^{−1}(y) is constant for y ∈ F, then the Morley degree of F is 1.

Proof – Let B be a definable subset of F of Morley rank rk F. We show that rk (F \ B) < rk F. By the additivity axiom, we have rk E = r + rk F and

rk f^{−1}(B) = r + rk B = r + rk F = rk E

Since E has Morley degree 1, we obtain rk f^{−1}(F \ B) = rk (E \ f^{−1}(B)) < rk E, and by the additivity axiom again,

rk (F \ B) = rk f^{−1}(F \ B) − r < rk E − r = rk F

so F has Morley degree 1. □

2.3. Bad groups. Main properties of bad groups are summarized in the following facts, where a Borel subgroup of a bad group G is defined to be an infinite definable proper subgroup of G.

Fact 2.3. – ([6, §5.2] and [14]) Let G be a simple bad group, and B be a Borel subgroup of G.

(1) B = C_{G}(b) for b ∈ G \ {1},

(2) B is connected, abelian, self-normalizing and of Morley rank 1,

(3) C_{G}(x) is a Borel subgroup for each nontrivial element x of G,

(4) if A is another Borel subgroup of G, then A is conjugate with B, and either A = B or A ∩ B = {1},

(5) G = ⋃_{g∈G} B^{g},

(6) G has no involution.

Fact 2.4. – [15, Lemma 18] Let A and B be two distinct Borel subgroups of a simple bad group G. Then rk (ABA) = 3, rk (AB) = 2, and AB has Morley degree 1.

3. Lines

In this paper, G denotes a fixed simple bad group. We fix a Borel subgroup B of G and we denote by ℬ the set of Borel subgroups of G.

In this section, we define a line of G, and we provide their basic properties. We note that, by conjugation of Borel subgroups (Fact 2.3 (4)), any Borel subgroup is a line in the following sense.

Definition 3.1. – A line of G is a subset of the form uBv for two elements u and v of G.

We denote by Λ the set of lines of G.

We note that, by Fact 2.3 (2), each line has Morley rank 1 and Morley degree 1.

3.1. Basic properties. The elementary properties below of lines are used throughout this paper. In particular, they show that any two distinct elements of G belong to a unique line (Lemma 3.4).

Lemma 3.2. – Let uBv and rBs be two lines. Then uBv = rBs if and only if uB = rB and Bv = Bs.

Proof – We may assume that uBv = rBs. Then we have

B = u^{−1}rBsv^{−1} = u^{−1}rsv^{−1}B^{sv^{−1}}

so u^{−1}rsv^{−1} ∈ B^{sv^{−1}} and B = B^{sv^{−1}}. Now sv^{−1} belongs to B since B is selfnormalizing by Fact 2.3. Hence we obtain Bv = Bs, and the equality uB = rB follows from uBv = rBs. □

By the above lemma, the set Λ identifies with (G/B)_{l} × (G/B)_{r} where (G/B)_{l} (resp. (G/B)_{r}) denotes the set of left cosets (resp. right cosets) of B in G. Then Λ is a definable set. Moreover, since G is connected of Morley rank 3 and B has Morley rank 1, the Morley rank of Λ is 4 and its Morley degree is 1.

Lemma 3.3. – The set Λ is a uniformly definable family.

Proof – We consider the set A = {(x, u, v) ∈ G × G × G | x = uv} and the map f : A → G × Λ defined by f(x, u, v) = (x, (uB, Bv)). Then f is definable, so its image f(A) is definable too. But if (x, (uB, Bv)) belongs to f(A), then there are u′ ∈ uB and v′ ∈ Bv such that x = u′v′, so x belongs to uBv (which identifies with (uB, Bv)). Moreover, for any (x, (uB, Bv)) ∈ G × Λ such that x belongs to uBv, there is b ∈ B such that x = ubv, so (x, ub, v) belongs to A and (x, (uB, Bv)) = f(x, ub, v) ∈ f(A). Hence f(A) is the graph of the membership relation ∈ of an element of G to a line of G, so ∈ is definable, and Λ is a uniformly definable family. □

Lemma 3.4. – Two distinct elements x and y of G lie in one and only one line l(x, y).

Proof – By Fact 2.3 (5), there exists v ∈ G such that y^{−1}x belongs to B^{v}. Then x and y lie in uBv for u = yv^{−1}.

Now, if rBs is a line containing x and y, then we find two elements b_{1} and b_{2} of B such that x = rb_{1}s and y = rb_{2}s. Thus y^{−1}x = s^{−1}b_{2}^{−1}b_{1}s is a nontrivial element of B^{s}. But y^{−1}x belongs to B^{v} by the choice of v, hence we have B^{s} = B^{v} (Fact 2.3 (4)). Since B is self-normalizing, sv^{−1} belongs to B and we obtain Bs = Bv, so there exists b ∈ B such that s = bv. This implies that u = yv^{−1} = (rb_{2}s)(s^{−1}b) = rb_{2}b belongs to rB, and rBs = uBv. □

Corollary 3.5. – The map l : {(x, y) ∈ G × G | x ≠ y} → Λ is definable.

Proof – We consider the set

Γ = {((x, y), uBv) ∈ (G × G) × Λ | x ≠ y, x ∈ uBv, y ∈ uBv}

By Lemma 3.3, it is a definable subset of {(x, y) ∈ G × G | x ≠ y} × Λ. But by Lemma 3.4, it is the graph of the map l, hence l is definable. □

Lemma 3.6. – If uBv = (uBv)^{g} for uBv ∈ Λ \ ℬ and g ∈ G, then g = 1.

Proof – We have uBv = g^{−1}uBvg, so uB = g^{−1}uB and Bv = Bvg by Lemma 3.2, and g belongs to the Borel subgroups B^{u^{−1}} and B^{v}. If g is nontrivial, then B^{u^{−1}} = B^{v} (Fact 2.3 (4)), and vu belongs to N_{G}(B) = B. Consequently u belongs to v^{−1}B, and we obtain uBv = B^{v}, contradicting uBv ∉ ℬ. Thus g = 1. □

3.2. Sets of lines. By Lemma 3.3, the set Λ is uniformly definable. In this section, we are interested by its definable subsets.

Definition 3.7. – For each g ∈ G and each definable subset X of G, we consider the following subsets of Λ:

ℒ (g, X) = {l(g, x) ∈ Λ | x ∈ X \ {g}}
Λ_{X} = {λ ∈ Λ | λ ∩ X is infinite}

Lemma 3.8. – Let X be a definable subset of G. Then the sets Λ_{X} and ℒ (g, X) are definable subsets of Λ for each g ∈ G.

Proof – The set ℒ (g, X) is definable since it is the image of the definable set {g} × (X \ {g}) by the map l, which is definable by Corollary 3.5.

Now we consider the definable map l_{X} from {(x, y) ∈ X × X | x ≠ y} to Λ defined by l_{X}(x, y) = l(x, y). We show that Λ_{X} = {λ ∈ Λ | rk (l_{X}^{−1}(λ)) = 2}. We note that l_{X}^{−1}(λ) is contained in (λ ∩ X) × (λ ∩ X). Thus, if λ does not belong to Λ_{X}, then λ ∩ X is finite, and l_{X}^{−1}(λ) has not Morley rank 2. If λ belongs to Λ_{X}, then λ ∩ X is infinite, and for each pair (x, y) of elements of λ ∩ X with x ≠ y, we have l_{X}(x, y) = λ. This implies that l_{X}^{−1}(λ) is a generic subset of (λ ∩ X) × (λ ∩ X), with Morley rank 2. Consequently Λ_{X} = {λ ∈ Λ | rk (l_{X}^{−1}(λ)) = 2}, and Λ_{X} is a definable subset of Λ. □

Lemma 3.9. – Let λ_{1}, …, λ_{n} be n lines. Then λ_{1} ∪ ⋯ ∪ λ_{n} is a definable set of Morley rank 1 and Morley degree n.

Proof – For each i, the set A_{i} = λ_{i} ∩ (⋃_{j≠i} λ_{j}) has at most n − 1 elements by Lemma 3.4, and λ_{1} ∪⋯∪λ_{n} is the disjoint union of λ_{1} \A_{1}, …, λ_{n} \A_{n}, A_{1}, …, A_{n}. Since each line λ_{i} has Morley rank 1 and Morley degree 1 (Fact 2.3 (2)), the result follows. □

Lemma 3.10. – If Λ_{0} is a definable subset of Λ, then ⋃ Λ_{0} is a definable subset of G. Moreover, if Λ_{0} is infinite, then ⋃ Λ_{0} has Morley rank at least 2.

Proof – Since Λ_{0} is a definable subset of Λ, the set ⋃ Λ_{0} = {x ∈ G | ∃λ ∈ Λ_{0}, x ∈ λ} is definable by Lemma 3.3. Moreover, if Λ_{0} is infinite, then ⋃ Λ_{0} has Morley rank at least 2 by Lemma 3.9. □

Corollary 3.11. – The subset ⋃ Λ_{X} of G is definable for each definable subset X of G.

4. Planes

Our aim is to find a definable structure of projective space on our bad group G. In this section, we introduce a notion of planes, and we show that G has such a plane (Theorem 4.14).

Definition 4.1. – A definable subset X of G is said to be a plane if it has Morley rank 2 and satisfies \overline{X} = X where

\overline{X} = \{g ∈ G | rk (ℒ (g, X)) = 1\}

Lemma 4.2. – Let X be a definable set of Morley rank 2. For each g ∈ G, we have rk (ℒ (g, X)) ∈ \{1, 2\}. In particular, \overline{X} = \{g ∈ G | rk (ℒ (g, X)) ≤ 1\} and G \ \overline{X} = \{g ∈ G | rk (ℒ (g, X)) = 2\}.

Proof – Let g ∈ G. We remember that, by Lemma 3.8, the set ℒ (g, X) is definable. We consider the map l_{g} : X \ {g} → ℒ (g, X) defined by l_{g}(x) = l(g, x). This map is definable and surjective, and since rk X = 2, we have rk ℒ (g, X) ≤ 2.

Moreover, the set ∪ℒ (g, X) is definable (Lemma 3.10), and it contains X. Since rk X = 2, we obtain rk (∪ℒ (g, X)) ≥ 2, and the set ℒ (g, X) is infinite (Lemma 3.9), so rk ℒ (g, X) ≥ 1. □

Lemma 4.3. – Let X be a definable subset of G of Morley rank 2. Then the set \overline{X} is a definable subset of ⋃ Λ_{X}.

Proof – If g ∈ G does not belong to ⋃ Λ_{X}, then l(g, x) ∩ X is finite for each x ∈ X, and since X has Morley rank 2, the set ℒ (g, X) has Morley rank 2, so g ∉ \overline{X}. Thus \overline{X} is contained in ⋃ Λ_{X}.

We show that \overline{X} is definable. We consider the set

A = {(g, λ) ∈ G × Λ | g ∈ λ, ∃x ∈ X \ {g}, x ∈ λ}

and the map f : A → G defined by f(g, λ) = g. We note that A is definable by Lemma 3.3, so f is definable too. Moreover, the preimage by f of each g ∈ G is f^{−1}(g) = {g} × ℒ (g, X), and we have rk (f^{−1}(g)) = rk ℒ (g, X). Consequently, we obtain \overline{X} = \{g ∈ G | rk (f^{−1}(g)) = 1\}, and \overline{X} is definable. □

Lemma 4.4. – Let X be a definable subset of G of Morley rank 2. Then rk Λ_{X} and rk ⋃ Λ_{X} are at most 2. Moreover, Λ_{X} is infinite if and only if rk ⋃ Λ_{X} = 2.

Proof – We consider the surjective definable map

l_{0} : (X × X) ∩ l^{−1}(Λ_{X}) → Λ_{X}

defined by l_{0}(x, y) = l(x, y). For each λ ∈ Λ_{X}, we have l_{0}^{−1}(λ) = {(x, y) ∈ (λ∩X)× (λ ∩ X) | x ≠ y}, and since rk λ = 1, we obtain rk (λ ∩ X) = 1 and rk l_{0}^{−1}(λ) = 2. But we have

rk ((X × X) ∩ l^{−1}(Λ_{X})) ≤ rk (X × X) = 2rk X = 4

hence rk Λ_{X} is at most 4 − 2 = 2.

We show that rk ⋃ Λ_{X} ≤ 2. We consider the definable set

A = {(x, λ) ∈ G × Λ_{X} | x ∈ λ \ X}

and the definable map l_{1} : A → Λ_{X} defined by l_{1}(x, λ) = λ. For each λ ∈ Λ_{X}, we have rk λ = 1 = rk (λ ∩ X), so l_{1}^{−1}(λ) is finite. Consequently we obtain rk A ≤ rk Λ_{X} ≤ 2. But the definable map l_{2} : A → (⋃ Λ_{X}) \ X, defined by l_{2}(x, λ) = x, is surjective, hence the Morley rank of (⋃ Λ_{X}) \ X is at most rk A ≤ 2. Since X has Morley rank 2, we obtain rk ⋃ Λ_{X} ≤ 2.

Now it follows from Lemmas 3.9 and 3.10 that Λ_{X} is infinite if and only if rk ⋃ Λ_{X} = 2. □

For each a ∈ G, let ℒ (a) = ℒ (a, G) be the set of lines containing a. It is definable by Lemma 3.8. Moreover, we note that ℒ (1) = ℬ.

Lemma 4.5. – Let Λ_{0} be a definable subset of Λ. If rk ∪ Λ_{0} = 2, then we have rk (ℒ (g) ∩ Λ_{0}) ≤ 1 for each g ∈ G.

Moreover, if further rk Λ_{0} = 2, then the set {g ∈ G | rk (ℒ (g) ∩ Λ_{0}) = 1} has Morley rank 2.

Proof – We recall that ℒ (g) = ℒ (g, G) is definable by Lemma 3.8. We show that rk (ℒ (g) ∩ Λ_{0}) ≤ 1 for each g ∈ G. Let g ∈ G and l_{g} : ∪(ℒ (g) ∩ Λ_{0}) \ {g} → ℒ (g) ∩ Λ_{0} be the map defined by l_{g}(x) = l(g, x). Since each line has Morley rank 1, the preimage of each element of ℒ (g) ∩ Λ_{0} has Morley rank 1. Consequently, we have

rk (ℒ (g) ∩ Λ_{0}) = rk ∪ (ℒ (g) ∩ Λ_{0}) − 1 ≤ rk ∪ Λ_{0} − 1 = 1

as desired.

We suppose further that rk Λ_{0} = 2, and we show that {g ∈ G | rk (ℒ (g) ∩ Λ_{0}) = 1} has Morley rank 2. Let U = ∪Λ_{0}, A = {(u, λ) ∈ U ×Λ_{0} | u ∈ λ} and f : A → Λ_{0} be the map defined by f(u, λ) = λ. Then A and f are definable, and the preimage f^{−1}(λ) of each λ ∈ Λ_{0} has Morley rank rk λ = 1, so rk A = 1 + rk Λ_{0} = 3. Now let h : A → U be the map defined by g(u, λ) = u. It is a definable map, and the preimage h^{−1}(u) of each u ∈ U has Morley rank either 0, or 1 by the previous paragraph.

But the preimage of U_{0} = {u ∈ U | rk h^{−1}(u) = 0} has Morley rank

rk h^{−1}(U_{0}) = rk U_{0} ≤ rk U = 2 < rk A

so the preimage of U_{1} = {u ∈ U | rk h^{−1}(u) = 1} has Morley rank 3. Hence we obtain rk U_{1} = 3 − 1 = 2. Moreover, we note that

U_{1} = {u ∈ U | rk (ℒ (u) ∩ Λ_{0}) = 1} = {g ∈ G | rk (ℒ (g) ∩ Λ_{0}) = 1}

so {g ∈ G | rk (ℒ (g) ∩ Λ_{0}) = 1} has Morley rank 2. □

Proposition 4.6. – Let X be a definable subset of G of Morley rank 2. For each g ∈ \overline{X}, we have rk (ℒ (g) ∩ Λ_{X}) = 1.

Moreover, if X has Morley degree 1, then \overline{X} = \{g ∈ G | rk (ℒ (g) ∩ Λ_{X}) = 1\} and G \ \overline{X} = \{g ∈ G | ℒ (g) ∩ Λ_{X} is finite\}.

Proof – First we note that ℒ (g) ∩ Λ_{X} = ℒ (g, X) ∩ Λ_{X} for any g ∈ G. For each g ∈ G, we consider the definable map l_{g} : X \ {g} → ℒ (g, X) defined by l_{g}(x) = l(g, x). In particular, the preimage l_{g}^{−1}(λ) of each λ ∈ ℒ (g, X) is (λ ∩ X) \ {g}.

We show that rk (ℒ (g, X) ∩ Λ_{X}) ≤ 1 for each g ∈ G. We may assume that Λ_{X} is infinite. Then, by Lemma 4.4, the set ∪Λ_{X} has Morley rank 2, and by Lemma 4.5, we obtain rk (ℒ (g, X) ∩ Λ_{X}) ≤ 1 for each g ∈ G.

Let g ∈ \overline{X}. We show that rk (ℒ (g) ∩ Λ_{X}) = 1. For each λ ∈ ℒ (g, X) \ Λ_{X}, the set l_{g}^{−1}(λ) = (λ ∩ X) \ \{g\} is finite, and since g ∈ \overline{X}, we have rk ℒ (g, X) = 1. Consequently, l_{g}^{−1}(ℒ (g, X)\Λ_{X}) has Morley rank at most 1, and l_{g}^{−1}(ℒ (g, X)∩Λ_{X}) has Morley rank rk X = 2. But the set l_{g}^{−1}(λ) = (λ ∩ X) \ {g} is infinite of Morley rank 1 for each λ ∈ ℒ (g, X)∩Λ_{X}. Hence we obtain rk (ℒ (g, X)∩Λ_{X}) = 2−1 = 1.

Now we assume that X has Morley degree 1. Let g ∈ G such that rk (ℒ (g, X) ∩ Λ_{X}) = 1. We show that g ∈ \overline{X}. Since the set l_{g}^{−1}(λ) = (λ ∩ X) \ \{g\} is infinite of Morley rank 1 for each λ ∈ ℒ (g, X) ∩ Λ_{X}, the set l_{g}^{−1}(ℒ (g, X) ∩ Λ_{X}) has Morley rank

1 + rk (ℒ (g, X) ∩ Λ_{X}) = 2 = rk X

Then, since X has Morley degree 1, the preimage of ℒ (g, X) \ Λ_{X} has Morley rank at most 1. Moreover, for each λ ∈ ℒ (g, X)\Λ_{X}, the preimage l_{g}^{−1}(λ) = (λ∩X)\{g} is finite and non-empty, so we obtain

rk (ℒ (g, X) \ Λ_{X}) = rk l_{g}^{−1}(ℒ (g, X) \ Λ_{X}) ≤ 1

This shows that rk ℒ (g, X) = 1 and g ∈ \overline{X}.

Furthermore, since rk (ℒ (g, X) ∩ Λ_{X}) ≤ 1 for each g ∈ G, we obtain G \ \overline{X} = {g ∈ G | ℒ (g) ∩ Λ_{X} is finite}, as desired. □

Corollary 4.7. – If X is a definable subset of G of Morley rank 2, then rk (\overline{X}\X) ≤ 1.

Proof – We remember that \overline{X} is definable by Lemma 4.3, so the sets Y = \overline{X} \X and A = {(y, λ) ∈ Y × Λ_{X} | y ∈ λ} are definable too. Let l_{Y} : A → Y and l_{D} : A → Λ_{X} be the definable maps defined by l_{Y} (y, λ) = y and l_{D}(y, λ) = λ respectively. On the one hand, for each λ ∈ Λ_{X}, the set λ ∩ X is infinite, and since λ has Morley rank 1 and Morley degree 1 (Fact 2.3 (2)), the set λ ∩ Y is finite and l_{D}^{−1}(λ) has Morley rank at most 0. This implies rk A ≤ rk Λ_{X} ≤ 2 (Lemma 4.4). On the other hand, for each y ∈ Y, we have rk (ℒ (y) ∩ Λ_{X}) = 1 by Proposition 4.6, so l_{Y}^{−1}(y) has Morley rank 1, and we obtain rk A = 1 + rk Y. Consequently, the Morley rank of Y is at most 1. □

Corollary 4.8. – Let X be a definable subset of G of Morley rank 2. If the Morley degree of X is not 1, then rk \overline{X} < 2. In particular, any plane has Morley degree 1.

Proof – Let n be the Morley degree of X, and X_{1}, …, X_{n} be n definable subsets of X of Morley rank 2 and Morley degree 1 such that X is the disjoint union of X_{1}, …, X_{n}. For each g ∈ \overline{X}, we have rk ℒ (g, X) ≤ 1 (Lemma 4.2), so we obtain rk ℒ (g, X_{i}) ≤ 1 for each i, and g ∈ \overline{X_{i}} for each i by Lemma 4.2 again. Thus \overline{X} is contained in \overline{X_{1}} ∩ \overline{X_{2}}. Since X_{1} ∩ X_{2} = ∅, the set \overline{X} is contained in (X_{1} ∩ Y_{2}) ∪ (Y_{1} ∩ X_{2}) ∪ (Y_{1} ∩ Y_{2}) where Y_{1} = \overline{X_{1}} \ X_{1} and Y_{2} = \overline{X_{2}} \ X_{2}. Since Y_{1} and Y_{2} have Morley rank at most 1 by Corollary 4.7, we obtain rk \overline{X} < 2. □

Proposition 4.9. – Let X be a definable subset of G of Morley rank 2 and Morley degree 1. Then rk \overline{X} = 2 if and only if Λ_{X} has Morley rank 2.

In this case, Λ_{X} and \overline{X} have Morley degree 1, and \overline{X} contains a generic definable subset of X.

Proof – We consider the definable set A = \{(x, λ) ∈ \overline{X} × Λ_{X} | x ∈ λ\} and the definable maps l_{1} : A → \overline{X} and l_{2} : A → Λ_{X} defined by l_{1}(x, λ) = x and l_{2}(x, λ) = λ respectively. By Proposition 4.6, the preimage l_{1}^{−1}(g) of each element g of \overline{X} has Morley rank 1, so rk A = 1 + rk \overline{X}. Moreover, the preimage l_{2}^{−1}(λ) of each λ ∈ Λ_{X} has Morley rank at most 1, so rk A ≤ 1 + rk Λ_{X}. Then we obtain rk \overline{X} ≤ rk Λ_{X}. In particular, it follows from Lemma 4.4 that if rk \overline{X} = 2, then rk Λ_{X} = 2. Hence we may assume that rk Λ_{X} = 2.

At this stage, Lemma 4.4 gives rk Λ_{X} = 2, and by Lemma 4.5 and Proposition 4.6, we obtain rk \overline{X} = 2. Moreover, it follows from Corollary 4.7 that \overline{X} has Morley degree 1 and that X ∩ \overline{X} is a generic definable subset of \overline{X}.

We show that the Morley degree of Λ_{X} is 1. Let l_{0} : {(x, y) ∈ X×X | x ≠ y} → Λ be the definable map defined by l_{0}(x, y) = l(x, y). Since the Morley degree of X is 1, the one of {(x, y) ∈ X × X | x ≠ y} is 1 too. For each λ ∈ Λ_{X}, we have rk l_{0}^{−1}(λ) = rk ((λ ∩ X) × (λ ∩ X)) = 2. Since rk Λ_{X} = 2, we obtain

rk l_{0}^{−1}(Λ_{X}) = 2 + rk Λ_{X} = 4 = rk {(x, y) ∈ X × X | x ≠ y}

and since the Morley degree of {(x, y) ∈ X × X | x ≠ y} is 1, the Morley degree of l_{0}^{−1}(Λ_{X}) is 1 too. Now the Morley degree of Λ_{X} is 1 by Fact 2.2. □

Lemma 4.10. – Let X and Y be two definable subsets of G of Morley rank 2 and Morley degree 1. If X ∩ Y has Morley rank 2, then \overline{X} = \overline{Y}.

Proof – Let g ∈ G. If g belongs to \overline{X ∩ Y}, then we have rk ℒ (g, X ∩ Y) ≤ 1 (Lemma 4.2). Since X has Morley degree 1 and X ∩ Y has Morley rank 2, the set X \Y has Morley rank at most 1, and the set ℒ (g, X \Y) has Morley rank at most 1. Thus ℒ (g, X) has Morley rank at most 1, and g belongs to \overline{X} by Lemma 4.2 again.

Conversely, if g ∈ \overline{X}, then ℒ (g, X) has Morley rank 1, so ℒ (g, X∩Y) ⊆ ℒ (g, X) has Morley rank at most 1. Then Lemma 4.2 gives g ∈ \overline{X ∩ Y}. This shows that \overline{X ∩ Y} = \overline{X}. By the same way, we obtain \overline{X ∩ Y} = \overline{Y}, so \overline{X} = \overline{Y}. □

Lemma 4.11. – Let g be a nontrivial element such that g = [u, v] for (u, v) ∈ G×G. Then we have {x ∈ G | [x, v] = g} = C_{G}(v)u and {y ∈ G | [u, y] = g} = C_{G}(u)v. In particular, they are two lines and have Morley rank 1 and Morley degree 1.

Proof – The equalities are obvious. Moreover, by Fact 2.3, the sets C_{G}(v)u and C_{G}(u)v are two lines, and they have Morley rank 1 and Morley degree 1. □

Lemma 4.12. – For each a ∈ G, the set a^{G} ∩ B has exactly one element.

Proof – We may assume a ≠ 1. By Fact 2.3 (5), there is g ∈ G such that a^{g} belongs to B. If a^{h} ∈ B for h ∈ G, then a is a nontrivial element of B^{g^{−1}} ∩ B^{h^{−1}}. By Fact 2.3 (4), we obtain B^{g^{−1}} = B^{h^{−1}}, and h^{−1}g belongs to N_{G}(B) = B. But B is abelian (Fact 2.3 (2)), so h^{−1}g centralizes a^{h}, and a^{h} = (a^{h})^{h^{−1}g} = a^{g}. Hence a^{G} ∩ B = {a^{g}}. □

The following result isolates a step of the proof of Theorem 4.14. The proof of second point was initially more complicated, and Bruno Poizat proposed a simplification.

Proposition 4.13. – Let g be a nontrivial element of G. If the set {(x, y) ∈ G × G | [x, y] = g} is non-empty, then it has Morley rank 3 and the definable set {x ∈ G | ∃y ∈ G, [x, y] = g} has Morley rank 2.

Proof – We consider the definable map ad : G × G → G defined by ad(x, y) = [x, y].

1. rk ad^{−1}(g) ≥ 3.

Since ad^{−1}(g) is non-empty, there is (x, y) ∈ G × G such that g = [x, y]. Since g is nontrivial, y is nontrivial too, and C = C_{G}(y) is a Borel subgroup. By the same way, x is nontrivial, and since g is nontrivial, x does not belong to C and 1 ∉ Cx.

We show that the set Y = {dy ∈ G | ∃c ∈ C, d ∈ C_{G}(cx)} has Morley rank at least 2. For each c ∈ C, we have cx ∈ Cx, so cx ≠ 1 and C_{G}(cx) is a Borel subgroup. In particular, there exists u ∈ G such that C_{G}(cx) = B^{u} (Fact 2.3 (4)), and C_{G}(cx)y = u^{−1}Buy is a line of G. If we have C_{G}(c_{1}x)y = C_{G}(c_{2}x)y for c_{1} ∈ C and c_{2} ∈ C such that c_{1} ≠ c_{2}, then C_{G}(c_{1}x) = C_{G}(c_{2}x) is a line containing c_{1}x and c_{2}x, and this line contains 1. But Cx is a line containing c_{1}x and c_{2}x, and Cx does not contain 1. Hence we have Cx ≠ C_{G}(c_{1}x), contradicting Lemma 3.4. This proves that Y = ⋃_{c∈C} C_{G}(cx)y is a union of infinitely many lines, and rk Y is at least 2 by Lemma 3.10.

For each α ∈ Y, there is c ∈ C such that α ∈ C_{G}(cx)y, and we have

ad(cx, α) = [cx, α] = [cx, y] = [x, y] = g

Thus, the infinite set C_{G}(α)cx × {α} is contained in ad^{−1}(g). This shows that the preimage of each element α ∈ Y, by the definable map f : ad^{−1}(g) → Y defined by f(r, s) = s, is infinite. Then we obtain rk ad^{−1}(g) ≥ 1 + rk Y ≥ 3.

2. For each g ∈ G \ {1}, the definable set X(g) = {x ∈ G | ∃y ∈ G, [x, y] = g} has not Morley rank 3.

We assume toward a contradiction that X(g) has Morley rank 3. Let U = {z ∈ G \ {1} | rk X(z) = 3}, and let V = {(x, y) ∈ G × G | [x, y] ∈ U}. For each z ∈ G \ {1} and each x ∈ G, we have x ∈ X(z) if and only if ∃y ∈ G, [x, y] = z, so the family {X(z) | z ∈ G \ {1}} is uniformly definable, and U and V are definable sets.

We recall that, by Fact 2.3, each nontrivial conjugacy class a^{G} of an element a of G has Morley rank rk a^{G} = rk G − rk C_{G}(a) = 2

We consider the definable surjective map f : V → U defined by f(x, y) = [x, y]. For each z ∈ U, we have

f^{−1}(z) = {(x, y) ∈ G × G | x ∈ X(z), [x, y] = z}

and by Lemma 4.11, this set has Morley rank rk f^{−1}(z) = rk X(z) + 1 = 4, so rk U = rk V − 4. But U is a normal set containing g, hence rk U ≥ rk g^{G} = 2, and we obtain rk V = 6 and rk U = 2.

Moreover, since G is connected, G×G is connected too, and since V is a subset of G×G of Morley rank 6, the Morley degree of V is 1. Then, by Fact 2.2, the Morley degree of U is 1 too. Since U is normal in G, and since each nontrivial conjugacy class of G has Morley rank 2, the set U is a union of finitely many conjugacy classes of G, and since the Morley degree of U is 1, there exists a ∈ G such that U = a^{G}.

Now, V is a generic definable subset of G × G, so there is (x, y) ∈ V such that (y, x) belongs to V. Thus [x, y] ∈ a^{G} and its inverse [y, x] ∈ a^{G} are conjugate, and they are equal by Lemma 4.12, contradicting Fact 2.3 (6).

3. Conclusion.

Let p_{X} : ad^{−1}(g) → X(g) be the map defined by p_{X}(x, y) = x. Then p_{X} is a surjective definable map. By Lemma 4.11, we have rk p_{X}^{−1}(x) = 1 for each x ∈ X(g). Consequently, we obtain rk ad^{−1}(g) = 1 + rk X(g), and since rk X(g) ≤ 2 by 2, we find rk ad^{−1}(g) ≤ 3. Thus we have rk ad^{−1}(g) = 3 by 1, so 1 + rk X(g) = rk ad^{−1}(g) = 3 and rk X(g) = 2. This finishes the proof. □

Theorem 4.14. – There is a plane in G.

Proof – It is sufficying to show that there is a definable subset X of G satisfying the following properties:

(1) its Morley rank is 2 and its Morley degree is 1,

(2) Λ_{X} has Morley rank 2.

Indeed, by Proposition 4.9, for such a subset X, the set \overline{X} has Morley rank 2 and Morley degree 1, and it contains a generic definable subset Y of X. At this stage, Lemma 4.10 shows that \overline{X} is a plane.

We fix a nontrivial element g such that g = [u, v] for (u, v) ∈ G × G, and we consider the definable map ad : G × G → G defined by ad(x, y) = [x, y]. Let X(g) = {x ∈ G | ∃y ∈ G, [x, y] = g}. By Proposition 4.13, it is a definable set of Morley rank 2. Let d be its Morley degree. Then X(g) is the disjoint union of definable subsets X_{1}, …, X_{d} of Morley rank 2 and Morley degree 1.

Let Y (g) = {y ∈ G | ∃x ∈ G, [x, y] = g}. Then Y (g) = {y ∈ G | ∃x ∈ G, [y, x] = g^{−1}} has Morley rank 2 by Proposition 4.13. By Lemma 4.11, the set l_{y} = {x ∈ G | [x, y] = g} is a line for each y ∈ Y (g). We consider the following definable subset of Λ:

ℒ_{X} = {λ ∈ Λ | ∃y ∈ G, ∀x ∈ λ, [x, y] = g}

that is ℒ_{X} = {l_{y} ∈ Λ | y ∈ Y (g)}. We note that ∪ℒ_{X} = X(g). Indeed, ∪ℒ_{X} is contained in X(g), and conversely, for each x ∈ X(g) there is y ∈ Y (g) such that [x, y] = g so x ∈ l_{y} ⊆ ∪ℒ_{X}. In particular, ℒ_{X} is contained in Λ_{X(g)}.

We show that l_{a} ≠ l_{b} for any distinct elements a and b of Y (g). Indeed, suppose toward a contradiction that l_{a} = l_{b} for two distinct elements a and b of Y (g). We note that a and b are nontrivial because g is nontrivial. There exist r ∈ G and s ∈ G such that [r, a] = [s, b] = g, and we have l_{a} = C_{G}(a)r and l_{b} = C_{G}(b)s by Lemma 4.11. Since r belongs to l_{a} = l_{b} = C_{G}(b)s, we obtain C_{G}(b)r = C_{G}(b)s = l_{a} = C_{G}(a)r and C_{G}(b) = C_{G}(a). Moreover, since r ∈ l_{b}, we have [r, b] = g = [r, a] and r^{b} = r^{a}, so ba^{−1} is a nontrivial element of C_{G}(r). But [r, a] = g is nontrivial, so r and a are nontrivial too, and C_{G}(r) and C_{G}(a) are Borel subgroups (Fact 2.3). Since they contain ba^{−1}, we find C_{G}(r) = C_{G}(a) and g = [r, a] = 1, contradicting that g is nontrivial. Thus, for each λ ∈ ℒ_{X}, there exists a unique y ∈ Y (g) such that λ = l_{y}.

We consider the definable set

𝒜 = {(λ, y) ∈ ℒ_{X} × Y (g) | ∀x ∈ λ, [x, y] = g}

that is 𝒜 = {(λ, y) ∈ ℒ_{X} × Y (g) | λ = l_{y}}. By the previous paragraph, the projection map p : 𝒜 → ℒ_{X} defined by p(λ, y) = λ is bijective, and since it is definable, we obtain rk 𝒜 = rk ℒ_{X}. Moreover, by Lemma 4.11, the definable map q : 𝒜 → Y (g) defined by q(λ, y) = y is bijective too, so rk 𝒜 = rk Y (g) = 2. Hence the Morley rank of ℒ_{X} is 2. Then, since ℒ_{X} is contained in Λ_{X(g)}, the Morley rank of Λ_{X(g)} is 2 by Lemma 4.4.

Now, for each element λ of Λ_{X(g)}, since λ∩X(g) is infinite and since λ has Morley rank 1 and Morley degree 1, there is a unique i ∈ {1, …, d} such that λ ∩ X_{i} is infinite, that is λ ∈ Λ_{X_{i}}. Thus, each λ ∈ Λ_{X(g)} belongs to a unique definable set Λ_{X_{i}} for i ∈ {1, …, d}. Hence there exists i ∈ {1, …, d} such that rk Λ_{X_{i}} = 2. Now the set X_{i} satisfies the conditions (1) and (2) of the beginning of our proof, so \overline{X_{i}} is a plane. □

5. A projective space ?

In this section, we analyze planes. We remember that, by Theorem 4.14, the group G has a plane, and that by Corollary 4.8, any plane has Morley degree 1. We show that, if X and Y are two distinct planes, then Λ_{X} ∩ Λ_{Y} has a unique element (Proposition 6.1). However, along the way, we will obtain our final contradiction.

Definition 5.1. – For each line λ, we consider the following subset of Λ:

ℒ (λ) = {m ∈ Λ | λ ∩ m is not empty}

Lemma 5.2. – For any line λ, the set ℒ (λ) is definable, it has Morley rank 3 and Morley degree 1.

Proof – We consider the definable map f : λ × (G \ λ) → ℒ (λ) \ {λ} defined by f(x, g) = l(x, g). By Lemma 3.4, for each m ∈ ℒ (λ) \ {λ}, there is a unique element x in λ ∩ m. Moreover, for any g ∈ G \ λ, we have f(x, g) = m if and only if g ∈ m \ {x}. Consequently we have rk f^{−1}(m) = rk m = 1, and

rk ℒ (λ) = rk (λ × (G \ λ)) − 1 = 3

Furthermore, since λ and G have Morley degree 1, the Morley degree of λ × G and λ × (G \ λ) is 1, and the Morley degree of ℒ (λ) \ {λ} and ℒ (λ) is 1 too (Fact 2.2). □

Lemma 5.3. – Let X be a plane, and λ ∈ Λ_{X}. Then ℒ (λ) ∩ Λ_{X} has Morley rank 2.

Proof – Since λ belongs to Λ_{X}, the set λ ∩ X is infinite, and since λ is a line, we have rk (λ ∩ X) = 1. We consider the definable set

𝒜 = {(x, m) ∈ (λ ∩ X) × Λ_{X} | m ≠ λ, x ∈ m}

and the definable maps p : 𝒜 → λ ∩ X and q : 𝒜 → Λ_{X} defined by p(x, m) = x and q(x, m) = m respectively. By Proposition 4.6, the set p^{−1}(x) has Morley rank 1 for each x ∈ λ ∩ X, so rk 𝒜 = 1 + rk (λ ∩ X) = 2.

Moreover, each m ∈ Λ_{X} \ {λ} contains at most one element of λ (Lemma 3.4), so q is an injective map and its image has Morley rank rk 𝒜 = 2. But the image of q is contained in (ℒ (λ) ∩ Λ_{X}) \ {λ}, and we have rk Λ_{X} ≤ 2 (Lemma 4.4), hence ℒ (λ) ∩ Λ_{X} has Morley rank 2. □

Lemma 5.4. – Let λ_{1} and λ_{2} be two distinct lines. Then ℒ (λ_{1}) ∩ ℒ (λ_{2}) has Morley rank 2 and Morley degree 1.

Proof – Let A = {(x, y) ∈ λ_{1} × λ_{2} | x ≠ y}, and let f : λ_{1} × λ_{2} → (ℒ (λ_{1}) ∩ ℒ (λ_{2})) \ {λ_{1}, λ_{2}} be the map defined by f(x, y) = l(x, y). This map is definable by Corollary 3.5 and it is a bijection by Lemma 3.4. Since λ_{1} and λ_{2} are two lines, the set λ_{1} × λ_{2} has Morley rank 2 and Morley degree 1, and since f is a definable bijection, ℒ (λ_{1}) ∩ ℒ (λ_{2}) has Morley rank 2 and Morley degree 1. □

Proposition 5.5. – If X and Y are two distinct planes, then Λ_{X} ∩ Λ_{Y} has at most one element.

Proof – Suppose toward a contradiction that λ_{1} and λ_{2} are two distinct elements of Λ_{X} ∩ Λ_{Y}. By Lemma 5.3, the sets ℒ (λ_{1}) ∩ Λ_{X} and ℒ (λ_{2}) ∩ Λ_{X} have Morley rank 2. But Λ_{X} has Morley rank 2 and Morley degree 1 by Proposition 4.9 and Corollary 4.8, hence ℒ (λ_{1})∩ℒ (λ_{2})∩Λ_{X} has Morley rank 2. By the same way, ℒ (λ_{1}) ∩ ℒ (λ_{2}) ∩ Λ_{Y} has Morley rank 2. Thus, since ℒ (λ_{1}) ∩ ℒ (λ_{2}) has Morley rank 2 and Morley degree 1 (Lemma 5.4), the set Λ_{X} ∩ Λ_{Y} has Morley rank 2.

Since Λ_{X} ∩ Λ_{Y} is infinite, the set U = ∪(Λ_{X} ∩ Λ_{Y}) has Morley rank at least 2 by Lemma 3.10, and since U is contained in ∪Λ_{X}, its Morley rank is exactly 2 (Lemma 4.4). Now the set Z = {g ∈ G | rk (ℒ (g) ∩ Λ_{X} ∩ Λ_{Y}) = 1} has Morley rank 2 by Lemma 4.5. But Proposition 4.6 says that Z is contained in X ∩ Y, hence X ∩ Y has Morley rank 2 and Lemma 4.10 gives X = Y, a contradiction. □

From now on, we try to show that the set Λ_{X} ∩ Λ_{Y} has exactly one element (cf. Proposition 6.1). However, the final contradiction will appear earlier.

Corollary 5.6. – Let X be a plane and (a, b) ∈ G×G. Then the following assertions are equivalent:

• aXb = X

• aΛ_{X}b = Λ_{X}

• aXb ∩ X has Morley rank 2.

Proof – We note that aXb is a plane, and that aΛ_{X}b = Λ_{aXb}. If aXb ∩ X has Morley rank 2, then aXb = X by Lemma 4.10, and if aXb = X, then we have aΛ_{X}b = Λ_{aXb} = Λ_{X}. Moreover, if aΛ_{X}b = Λ_{X}, then we have Λ_{aXb} = Λ_{X} and aXb = X by Proposition 5.5, so aXb ∩ X = X has Morley rank 2. □

By Fact 2.4, if A is a Borel subgroup distinct from B, then rk (ABA) = 3. The following result is slightly more general, and its proof is different.

We recall that, if a group H of finite Morley rank acts definably on a set E, then the stabilizer of any definable subset F of E is defined to be

StabF = {h ∈ H | rk ((h · F)ΔF) < rk (F)}

where Δ stands for the symmetric difference. It is a definable subgroup of H by [5, Lemma 5.11].

Lemma 5.7. – Let A and C be two Borel subgroups distinct from B. Then rk (ABC) = 3.

Proof – We consider the action of G on itself by left multiplication. Then we have b · BC = BC for each b ∈ B, so B is contained in Stab(BC).

We assume toward a contradiction that C is contained in Stab(BC). Since BC has Morley rank 2 and Morley degree 1 (Fact 2.4), we have rk (cBC \ BC) ≤ 1 for each c ∈ C, and since rk C = 1, we obtain rk (CBC \ BC) ≤ 2 and rk (CBC) = 2, contradicting Fact 2.4. Consequently, C is not contained in Stab(BC), and since Stab(BC) contains B, Fact 2.3 implies that Stab(BC) = B.

We assume toward a contradiction that rk (ABC) ≠ 3. Since rk (BC) = 2, we have rk (ABC) = 2 and ABC is a disjoint union of finitely many definable subsets E_{1}, …, E_{k} of Morley rank 2 and Morley degree 1. For each a ∈ A, the set aBC has Morley rank rk (BC) = 2 and Morley degree 1, so there exists a unique i ∈ {1, …, k} such that rk (aBC ∩ E_{i}) = 2. Since A is infinite, there are i ∈ {1, …, k} and two distinct elements a and a′ of A such that rk (aBC ∩ E_{i}) = rk (a′BC ∩ E_{i}) = 2. Since E_{i} has Morley degree 1, the Morley rank of aBC ∩ a′BC is 2, and we obtain rk (a^{′−1}aBC ∩ BC) = 2. But BC has Morley degree 1, hence a^{′−1}a belongs to Stab(BC) = B. Thus a^{′−1}a belongs to A∩B = {1} (Fact 2.3 (4)), contradicting that a and a′ are distinct. So we have rk (ABC) = 3, as desired. □

Corollary 5.8. – Let A and C be two distinct Borel subgroups. Then rk (BA ∩ BC) = 1.

Proof – We may assume A ≠ B and C ≠ B. By Fact 2.4, we have

1 = rk B ≤ rk (BA ∩ BC) ≤ rk (BA) = 2

We assume toward a contradiction that rk (BA ∩ BC) = 2. Since BC has Morley rank 2 and Morley degree 1 (Fact 2.4), the set E = BC \ BA has Morley rank at most 1. Consequently, EA has Morley rank at most rk E + rk A = 2, and since (BA ∩ BC)A ⊆ BA has Morley rank 2, we obtain rk (BCA) = rk (EA ∪ (BA ∩ BC)A) = 2, contradicting that BCA has Morley rank 3 (Lemma 5.7). □

Lemma 5.9. – For any plane X, we have BX ≠ X and XB ≠ X.

Proof – We assume toward a contradiction that BX = X for a plane X. Let x ∈ X. Since X is a plane, Proposition 4.6 gives rk (ℒ (x, X) ∩ Λ_{X}) = 1, so ℒ (x, X) ∩ Λ_{X} is infinite. But each line containing x has the form B^{u}x for u ∈ G, hence there exist u ∉ B and v ∉ B such that B^{u} ≠ B^{v}, and such that B^{u}x and B^{v}x belong to ℒ (x, X) ∩ Λ_{X}. In particular, there is a co-finite subset S of B such that S^{u}x and S^{v}x are contained in X.

Now, since BX = X, the sets BS^{u}x and BS^{v}x are contained in X. By Fact 2.4, the set BB^{u^{−1}}, and so Bu^{−1}B, has Morley rank 2, and since Bu^{−1}(B \ S) is a finite union of lines, the set Bu^{−1}(B \S) has Morley rank 1 (Lemma 3.9), so Bu^{−1}S has Morley rank 2. Thus, the sets BS^{u}x = Bu^{−1}Sux and BS^{v}x = Bv^{−1}Svx are subsets of X of Morley rank 2, and since the Morley degree of X is 1 (Corollary 4.8), the set BS^{u}x ∩ BS^{v}x has Morley rank 2. This implies that rk (BB^{u} ∩ BB^{v}) = 2, contradicting Corollary 5.8. Now we have BX ≠ X and by the same way, we show that XB ≠ X. □

Corollary 5.10. – For any plane X, the stabilizer of X for the action of G on itself by left multiplication is finite.

Proof – By Corollary 5.6, we have StabX = {a ∈ G | aX = X}. If StabX is infinite, then it contains a Borel subgroup, that is impossible (Lemma 5.9). □

Proposition 5.11. – Let uBv be a line, and X and Y be two planes such that Λ_{X} and Λ_{Y} contain uBv. Then there exists b ∈ B such that Y = b^{u^{−1}}X.

Proof – We note that the set

⋃_{a∈B} Λ_{a^{u^{−1}} X} = {λ ∈ Λ | (a^{u^{−1}})^{−1}λ ∈ Λ_{X}}

is a definable subset of Λ, and that uBv = a^{u^{−1}}(uBv) ∈ a^{u^{−1}}Λ_{X} = Λ_{a^{u^{−1}} X} for each a ∈ B.

We consider the definable map

f : B × ((ℒ (uBv) ∩ Λ_{X}) \ {uBv}) → (ℒ (uBv) ∩ ( ⋃_{a∈B} Λ_{a^{u^{−1}} X})) \ {uBv}

defined by f(a, λ) = a^{u^{−1}}λ.

We show that f is surjective and that its fibers are finite. For each λ′ ∈ (ℒ (uBv)∩(⋃_{a∈B} Λ_{a^{u^{−1}} X}))\{uBv}, there is a ∈ B such that λ′ belongs to Λ_{a^{u^{−1}} X}, so we have (a^{u^{−1}})^{−1}λ′ ∈ Λ_{X} and f(a, (a^{u^{−1}})^{−1}λ′) = λ′. Thus f is surjective.

Moreover, if f(b, λ) = λ′ for b ∈ B and λ ∈ Λ_{X}, then we have b^{u^{−1}}λ = λ′, so the planes a^{u^{−1}}X and b^{u^{−1}}X contain an infinite subset of λ′ and of uBv, and by Proposition 5.5, they are equal. But by Corollary 5.10, there are finitely many elements a′ of B such that a^{′u^{−1}}X = a^{u^{−1}}X, and for such any element a′, there is a unique line λ′′ = (a^{′−1})^{u^{−1}}λ′ such that f(a′, λ′′) = λ′. Hence the preimage of λ′ is finite.

Consequently, by using Lemma 5.3, we have

rk (ℒ (uBv) ∩ ( ⋃_{a∈B} Λ_{a^{u^{−1}} X})) = rk (B × (ℒ (uBv) ∩ Λ_{X})) = 1 + 2 = 3

By the same way, we have rk (ℒ (uBv) ∩ (⋃_{a∈B} Λ_{a^{u^{−1}} Y})) = 3. Now it follows from Lemma 5.2 that

rk (ℒ (uBv) ∩ ( ⋃_{a∈B} Λ_{a^{u^{−1}} X}) ∩ ( ⋃_{a∈B} Λ_{a^{u^{−1}} Y})) = 3

Thus we find a ∈ B and a′ ∈ B such that Λ_{a^{u^{−1}} X} ∩ Λ_{a^{′u^{−1}} Y} contains a line rBs ≠ uBv, and since we have uBv ∈ Λ_{a^{u^{−1}} X} ∩ Λ_{a^{′u^{−1}} Y}, Proposition 5.5 says that a^{u^{−1}}X = a^{′u^{−1}}Y. Hence we have Y = b^{u^{−1}}X for b = a^{′−1}a. □

Lemma 5.12. – Let X be a plane. Then the set {v ∈ G | ∃u ∈ G, uBv ∈ Λ_{X}} is a generic definable subset of G.

Proof – Let V = {v ∈ G | ∃u ∈ G, uBv ∈ Λ_{X}}. We remember that Λ identifies with (G/B)_{l} × (G/B)_{r} (Lemma 3.2). Let f : Λ_{X} → (G/B)_{r} be the definable map defined by f(uBv) = Bv. Then we have V = h^{−1}(f(Λ_{X})) where h : G → (G/B)_{r} is defined by h(x) = Bx. Thus V is a definable set, and since the preimage h^{−1}(Bx) of each Bx ∈ (G/B)_{r} has Morley rank rk B = 1, the Morley rank of V is 1 + rk (f(Λ_{X})).

We assume toward a contradiction that f is not injective. Then there exist Bv ∈ (G/B)_{r} and two distinct lines in Λ_{X} of the form uBv and rBv. For each b ∈ B, we have (uBv)b^{v} = uBv and (rBv)b^{v} = rBv, so uBv ∈ Λ_{X}b^{v} = Λ_{Xb^{v}} and rBv ∈ Λ_{Xb^{v}}. By Proposition 5.5, we obtain Xb^{v} = X, so XB^{v} = X, contradicting Lemma 5.9. Thus f is injective, and since rk Λ_{X} = 2 (Proposition 4.9), we obtain rk (f(Λ_{X})) = 2 and rk V = 3. □

Proposition 5.13. – Let X be a plane. Then for each plane Y, there exists a unique a ∈ G and a unique b ∈ G such that Y = aX = Xb.

Proof – We show that there exists a ∈ G and b ∈ G such that Y = aX = Xb. By Lemma 5.12, there exists v ∈ G, and u ∈ G and u′ ∈ G, such that uBv ∈ Λ_{X} and u′Bv ∈ Λ_{Y}. Then we have Bv ∈ Λ_{u^{−1}X} and Bv ∈ Λ_{u^{′−1}Y}, and Proposition 5.11 provides b ∈ B such that u^{′−1}Y = bu^{−1}X. Hence we have Y = aX for a = u′bu^{−1}. Thus, each plane has the form aX for a ∈ G, and by the same way, each plane has the form Xb for b ∈ G.

We show the uniqueness of a and b. Let S = {g ∈ G | gX = X}. It is a finite subgroup of G by Corollary 5.10. For each α ∈ G, the previous paragraph gives β ∈ G such that αX = Xβ. Then, for each s ∈ S, we have s(αX) = s(Xβ) = Xβ = αX, and we obtain s^{α}X = X and s^{α} ∈ S. Thus any element α ∈ G normalizes the finite subgroup S, and since G is a simple group, S is trivial. This proves the uniqueness of a, and by the same way we obtain the uniqueness of b. □

By the previous result, the set of planes is 𝒫 = {aX | a ∈ G}, and it identifies with G, so it is definable.

Corollary 5.14. – The set 𝒫 of planes is uniformly definable.

Proof – The set Δ = {(u, a) ∈ G × G | a^{−1}u ∈ X} is definable, and it is the graph of the membership relation of an element of G to a plane, where 𝒫 identified with G. Thus, 𝒫 is uniformly definable. □

From now on, we fix a plane X, and we consider the action of G × G on G defined by (u, v) · g = ugv^{−1}. This action induces an action of G × G on 𝒫 defined by (u, v) · aX = uaXv^{−1}. Then for each plane aX, we denote by Stab(aX) the stabilizer of aX for this action of G × G on 𝒫.

Remark 5.15. –

(1) It follows from Lemma 3.2 that the stabilizer of each line uBv ∈ Λ is Stab(uBv) = B^{u^{−1}} × B^{v}.

(2) Furthermore, for each pair (x, y) of nontrivial elements of G, by Fact 2.3 there are a unique pair of Borel subgroups (B^{u^{−1}}, B^{v}) such that x ∈ B^{u^{−1}} and y ∈ B^{v}. Then uBv is the unique line stabilized by (x, y).

Lemma 5.16. – There exists a ∈ G such that a^{−1}Xa ≠ X.

Proof – We assume toward a contradiction that a^{−1}Xa ≠ X for each a ∈ G. Then for each uBv ∈ Λ_{X} and each a ∈ G, we have

(uBv)^{a} ∈ Λ_{X}^{a} = Λ_{X^{a}} = Λ_{X}

Since rk Λ_{X} = 2 (Proposition 4.9), the line uBv is a Borel subgroup (Lemma 3.6), and by conjugacy of Borel subgroups, we obtain Λ_{X} = ℬ. Now we have ∪Λ_{X} = G, so rk ∪ Λ_{X} = 3, contradicting Lemma 4.4. □

Lemma 5.17. – The stabilizer StabX is the graph of a definable automorphism μ of G × G. In particular, StabX and G are definably isomorphic.

Moreover, for each b ∈ G, if ν is the definable automorphism of G whose graph is Stab(Xb), then ν ∘ μ^{−1}(x) = x^{b} for each x ∈ G.

Proof – Since the set 𝒫 has Morley rank rk G = 3, and G × G acts transitively on 𝒫 (Proposition 5.13), the stabilizer StabX has Morley rank 6 − 3 = 3. By Proposition 5.13 that StabX ∩ (G × {1}) and StabX ∩ ({1} × G) are trivial. Hence StabX is the graph of a definable automorphism μ of G.

By the same way, Stab(Xb) is the graph of a definable automorphism ν of G. For each (x, y) ∈ G × G, we have (x, y) ∈ StabX if and only if xXy^{−1} = X, that is xXb(b^{−1}y^{−1}b) = Xb. In other words, (x, y) belongs to StabX if and only if (x, y^{b}) belongs to Stab(Xb). This implies that, for each x ∈ G, we have ν(x) = μ(x)^{b} so (ν ∘ μ^{−1})(x) = x^{b}. □

Lemma 5.18. – The Borel subgroups of StabX are the stabilizers of the lines λ ∈ Λ_{X}, for the action of StabX on X.

Proof – Indeed, StabX has Morley rank 3 (Lemma 5.17), and it acts on the definable set Λ_{X} of Morley rank 2 (Lemma 4.4 and Proposition 4.9). Hence for each line λ ∈ Λ_{X}, the stabilizer StabX ∩ Stab(λ) is either a Borel subgroup of StabX, or StabX. Since Stab(λ) has Morley rank 2 (Remark 5.15 (1)), the stabilizer StabX ∩ Stab(λ) is a Borel subgroup of StabX. Conversely, by conjugacy of the Borel subgroups of StabX ≃ G, each Borel subgroup of StabX stabilizes a line of Λ_{X}. □

Now we are ready for the final contradiction of our paper.

Proof – Let 𝒰 = {Xx^{−1} ∈ 𝒫 | x ∈ X} be the set of the planes containing 1. Since X has Morley rank 2, the set 𝒰 has Morley rank 2.

Let Δ = {(x, y) ∈ G × G | x = y}. We note that Δ is definably isomorphic to G, and it has Morley rank 3. Moreover, the action of Δ on G is equivalent to the action by conjugation of G on itself. In particular, Δ stabilizes 𝒰 , and no element of 𝒰 is stabilized by Δ (Lemma 5.16). Thus, since rk 𝒰 = 2, the stabilizer Δ ∩ StabY of each Y ∈ 𝒰 is a Borel subgroup of Δ.

We fix Y ∈ 𝒰 . Since Δ ∩ StabY is a Borel subgroup of Δ ≃ G, we may assume that Δ ∩ StabY = {(b, b′) ∈ B × B | b′ = b}. Then, since StabY is definably isomorphic to G (Lemma 5.17), the set {(b, b′) ∈ B ×B | b′ = b} is a Borel subgroup of StabY, so {(b, b′) ∈ B × B | b′ = b} is the stabilizer of a line λ ∈ Λ_{Y} for the action of StabY on Y (Lemma 5.18), and this line λ is B (Remark 5.15 (2)). By Proposition 4.6, we have

rk (ℬ ∩ Λ_{Y}) = rk (ℒ (1) ∩ Λ_{Y}) = 1

so there is u ∈ G \ B such that B^{u} belongs to Λ_{Y}. By Lemma 5.18, the stabilizer Stab(B^{u}) ∩ StabY is a Borel subgroup of StabY, and it is contained in B^{u} × B^{u} (Remark 5.15 (1)).

Moreover, since the previous paragraph shows that StabB ∩ StabY = {(b, b′) ∈ B × B | b′ = b}, we have

Stab(B^{u}) ∩ Stab(Y^{u}) = {(c, c′) ∈ B^{u} × B^{u} | c′ = c}

By Lemma 5.17, the sets StabY and Stab(Y^{u}) are the graphs of two automorphisms μ and ν respectively. Then, by Proposition 5.13 and Lemma 5.17 again, there is g ∈ G such that ν ∘ μ^{−1}(x) = x^{g} for each x ∈ G. But, by the previous paragraph, μ(x) belongs to B^{u} for each x ∈ B^{u}, and since Stab(B^{u}) ∩ Stab(Y^{u}) = {(c, c′) ∈ B^{u} × B^{u} | c′ = c}, we have ν(x) = x for each x ∈ B^{u}. Hence, for each x ∈ B^{u}, we have μ(x) = x^{g^{−1}}, and since μ(x) ∈ B^{u}, we obtain μ(x) = x by Lemma 4.12. Thus StabY contains {(c, c′) ∈ B^{u} × B^{u} | c′ = c} ⊆ Δ, contradicting Δ ∩ StabY = {(b, b′) ∈ B × B | b′ = b}. □

6. Annex

Actually, after Lemma 5.18 we were ready for a new step to provide a structure of projective space over G, which was the initial goal of our section.

Proposition 6.1. – If Y ≠ X is a plane, then Λ_{X} ∩ Λ_{Y} has a unique element.

Proof – By Lemma 5.17, the subgroup StabX (resp. StabY) of G × G is the graph of a definable automorphism μ (resp. ν) of G. By Proposition 5.13, there is b ∈ G \ {1} such that Y = Xb. We may assume that b belongs to B. Then Lemma 5.17 shows that (ν ∘ μ^{−1})(x) = x^{b} for each x ∈ G, so μ and ν are equal on B.

Since μ is a definable automorphism of G, the set H = {(b, μ(b)) ∈ G × G | b ∈ B} is a Borel subgroup of StabX ≃ G, and Fact 2.3 provides v ∈ G such that μ(B) = B^{v}. Then Bv is the unique line stabilized by H (Remark 5.15 (2)), and by Lemma 5.18, we have Bv ∈ Λ_{X}, But μ = ν on B, hence Bv belongs to Λ_{Y} too. Now Bv is the unique element of Λ_{X} ∩ Λ_{Y} (Proposition 5.5). □

Acknowledgements

Je remercie Bruno Poizat pour sa lecture attentive de cet article, et sa simplification de la démonstration de la proposition 4.13.

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