Groups in Simple Theories

Frank Wagner

Institut Girard Desargues Université Lyon 1

21, avenue Claude Bernard 69622 Villeurbanne cedex France

wagner@desargues.univ-lyon1.fr
www.desargues.univ-lyon1.fr/home/wagner

Generic types and elements. Let G be a group ∅-interpretable in a simple theory, and S_{G}(A) the set of types over A implying x ∈ G. A type p(x) ∈ S_{G}(A) is generic for G if gq does not fork over ∅ for all g ∈ G and any non-forking extension q of p to A, g. We call g generic over A if tp(g/A) is generic.

• If g is generic over A and B | ⌣_{A} g then g is generic over AB.

• If g is generic over A, then g^{−1} is generic over A.

• If g is generic over A and h | ⌣_{A} g, then hg and gh are generic.

• If p is generic over A and q a non-forking extension of p to A, g, then pg does not fork over ∅.

• Generic types exist.

• Generic types are precisely the types of maximal local stratified rank.

• If SU(G) < ∞, generic types are those of maximal SU-rank.

Proof. Step 1: There is a family {D_{i}(.) : i < α} of ω-valued local ranks on partial types, such that for all i < α, all n < ω and all g ∈ G

• D_{i}(π(x, A)) ≥ n is a type-definable condition on A.

• D_{i}(φ ∨ ψ) = max{D_{i}(φ), D_{i}(ψ)}.

• D_{i}(π) = min{D_{i}(φ) : π ⊢ φ}.

• g | ⌣_{A} B iff D_{i}(g/A) = D_{i}(g/AB) for all i < α.

• D_{i}(hg/A) = D_{i}(g/A) for all h ∈ G ∩ bdd(A).

Step 2: Put ⋀_{j<0} Γ_{j} := G, and define inductively for i < α

n_{i} = D_{i}(⋀_{j<i} Γ_{j}) and Γ_{i} = ⋀_{j<i} Γ_{j} ∪ {¬φ(x) ∈ L(A) : D_{i}(φ) < n_{i}}.

Let p(x) be a completion of ⋀_{i<α} Γ_{i} over A. Then D_{i}(p) = n_{i} for all i < α. Consider g ⊧ p and h ∈ G with h | ⌣_{A} g. Then for all i < α

n_{i} = D_{i}(g/A) = D_{i}(g/A, h) = D_{i}(hg/A, h) ≤ D_{i}(hg/∅) ; as n_{i} was chosen maximal possible, we get equality, whence hg | ⌣ A, h.

Remark. The family D_{i} will depend on the definability properties of G.

If G is a type-definable group, let {(φ_{i}(x, ȳ), k_{i}) : i < |T |} be an enumeration of all parameter-free formulas φ_{i} and positive integers k_{i}. Choose a definable superset X of G such that multiplication is defined and associative on X^{3}, and let D_{i}(π) be the local D(π(x), φ_{i}(y · x, ȳ), k_{i})-rank restricted to those instances of yȳ such that y ∈ G and φ_{i}(x, ȳ) defines a subset of X. This will work.

Coset spaces. Consider a subgroup H ≤ G. A type p(x) ∈ S_{G/H}(A) is generic for G/H if gq does not fork over ∅ for all g ∈ G and any non-forking extension q of p to A, g. Then the generic types for G/H are precisely the images of the generic types for G.

If g ∈ G, then the generic types for the coset gH (over any A ∋ g) are the translates by g of the generic types for H, and thus the types p(x) ∈ S_{gH} satisfying D_{i}(p) = D_{i}(gH) = D_{i}(H) for all i. If g is generic, and ḡ the class of g under the equivalence relation x^{−1}y ∈ H, then gH is ḡ-invariant, and g is generic for gH over ḡ.

A subgroup H ≤ G has bounded index if |G : H| is bounded independently of the ambient model.

Proposition. Let H ≤ G and X ⊆ G. Then the index of H in XH is bounded iff D_{i}(XH) = D_{i}(H) for all i.

Corollary. |G : H| is bounded iff D_{i}(G) = D_{i}(H) for all i.

Corollary. Any descending chain of subgroups (H_{i} : i < α) with |H_{i} : H_{i+1}| = ∞ for all i + 1 < α has length less than |T |^{+}.

The A-connected component G_{A}^{0} of G is the smallest A-interpretable subgroup of G of bounded index.

• G_{A}^{0} is normal.

• |G : G_{A}^{0}| ≤ 2^{|T (A)|}.

The generic types of G_{A}^{0} are called principal generic types for G over A.

Stabilizers. For p(x) ∈ S_{G}(A) put

S(p) = {g ∈ G : p ∪ gp does not fork over A},

and let stab_{G}(p) be S(p) · S(p), the stabilizer of p. Proposition. If A = bdd(A), then stab(p) is a subgroup, and S(p) contains its generic types. Proof. Consider g, g′ ∈ S(p) with g | ⌣_{A} g′. By the Independence Theorem there is x ⊧ p ∪ gp ∪ g′p with x | ⌣_{A} g, g′. So g^{−1}x ⊧ p ∪ g^{−1}g′p, and g^{−1}x | ⌣_{A,g} g^{−1}g′. As

D_{i}(g^{−1}x/A, g) = D_{i}(x/A, g) = D_{i}(p) = D_{i}(g^{−1}x/A) for all i, we get g^{−1}x | ⌣_{A} g,whence g^{−1}x | ⌣_{A} g^{−1}g′ and g^{−1}g′ ∈ S(p). We now finish by the following Lemma.

Lemma. Let X ⊆ G be such that x^{−1}y ∈ X for any independent x, y ∈ X. Then H := X · X is a subgroup of G, and X contains all its generic types.

Proof. We may assume X = X^{−1}. Choose p(x) ∈ S_{X} such that D_{i}(p) = n_{i} is maximal, subject to D_{j}(p) = n_{j} for all j < i.

If x, x′, x′′ ∈ X, choose y ⊧ p with y | ⌣ x, x′, x′′. Then x′y ∈ X and x′y | ⌣_{x′} x, x′′. Since

n_{i} = D_{i}(y/x′) = D_{i}(x′y/x′) ≤ D_{i}(x′y) ≤ n_{i}

for all i, we get equality for all i, and x′y | ⌣ x′, whence x′y | ⌣ x, x′, x′′. Thus xx′y ∈ X. As y^{−1}x′′ ∈ X, we get xx′x′′ ∈ H, so H is a subgroup.

If g is generic for H, choose x, x′ ∈ X with g = xx′, and y ⊧ p with y | ⌣ x, x′. Then x′y ∈ X and

n_{i} = D_{i}(y/x′) = D_{i}(x′y/x′) ≤ D_{i}(x′y) ≤ n_{i}

for all i, whence x′y | ⌣ x′, and x′y | ⌣ x, x′. So gy = xx′y ∈ X; as y^{−1} ∈ X with y^{−1} | ⌣ gy by genericity of g, we get g ∈ X.

Properties. Let p ∈ S_{G}(A), where A = bdd(A).

• D_{i}(stab(p)^{op}) = D_{i}(S(p)^{op}) ≤ D_{i}(p) for all i.

• stab(p) ≤ G_{A}^{0}; equality holds iff p is generic.

• If q is a non-forking extension of p, then |stab(p) : stab(q)| is bounded.

• Every type in S_{G}(A) specifies its coset modulo G_{A}^{0}.

Proposition Let p ∈ S_{G}(A) be such that a^{−1}b | ⌣_{A} a for independent a, b ⊧ p, where A = bdd(A). Then a left translate of some non-forking extension of p is a generic type of stab(p)^{op}. Proof. a^{−1}b ∈ S(p)^{op}, so for all i

\begin{aligned}D_{i}(p) & ≥ D_{i}(stab(p)^{op}) ≥ D_{i}(a^{−1}b/A) \\ & = D_{i}(a^{−1}b/A, a) = D_{i}(b/A, a) = D_{i}(b/A) = D_{i}(p). \end{aligned}

Hence a^{−1}b is generic for stab(p)^{op} over A, a. Thus b is generic for a · stab(p)^{op} over A, a.

Lemma. Suppose p, q, r are generic for G with p·q·G_{A}^{0} = r·G_{A}^{0}. Then there are pairwise independent realizations a, b, c of p, q, r with ab = c.

Proof. Let b, c realize q, r independently, and a′ = cb^{−1}. So a′ ∈ pG_{A}^{0}, and a′, b, c are pairwise independent. Let a ⊧ p with a′ | ⌣_{A} a, and put d = a^{−1}a′. Then d is principal generic, d | ⌣_{A} a′, and d ⊧ S(q). By the Independence Theorem there is d′ ⊧ lstp(d/A) with d′ | ⌣_{A} a′, b such that d′b ⊧ q and a′d^{′−1} ⊧ p. So a′d^{′−1}, d′b, c will do.

Corollary. Let K be a simple field, T ≤ K^{×} of bounded index, e ∈ K^{×}, and p, q generic. Then there are independent a, b realizing p, q with a + b ∈ eT.

Two subgroups H_{1}, H_{2} of G are commensurable if H_{1} ∩ H_{2} has finite index in H_{1} and in H_{2}; they are commensurate if the index is bounded.

Commensurativity is type-definable if D_{i}(H_{1}) = D_{i}(H_{2}) is fixed for all i.

Schlichting’s Theorem. Let G be a group, and ℌ a family of uniformly commensurable subgroups. Then there is N ≤ G commensurable with ℌ, invariant under aut_{{ℌ}}(G), and with ⋂ ℌ ≤ N ≤ ⟨ℌ⟩.

Theorem. Let ℌ be an A-hyperdefinable family of commensurate subgroups of G (in a simple theory). Then there is A-hyperdefinable N ≤ G commensurable with ℌ with ⋂ ℌ ≤ N ≤ ⟨ℌ⟩.

Proof. We may assume ℌ = {H(a) : a ⊧ p}, where p is a complete Lascar strong type over A. Put

X = { g ∈ G : ∃ a ⊧ p [ a | ⌣_{A} g ∧ g ∈ H(a) ] }.

By the Independence Theorem, x^{−1}y ∈ X for independent x, y ∈ X.

One can check that N := X · X will do.

Corollary. Let ℌ be an A-hyperdefinable family of subgroups of G. Then there is A-hyperdefinable N ≤ ⟨ℌ⟩, containing a small intersection of groups in ℌ.

Proof. Consider the family ℌ_{0} of small intersections of groups in ℌ of minimal D_{i}-ranks.

Definition. A hyperdefinable subgroup of G is locally connected if it is either equal to or not commensurate with its model-theoretic conjugates.

Proposition. For any hyperdefinable subgroup H there is a unique minimal hyperdefinable locally connected commensurate subgroup H^{c}.

Proof. If H = H_{A} is hyperdefinable over A, put A ∼ A′ iff H_{A} and H_{A′} are commensurate, and let Ā = A/∼. The family ℌ of commensurate model-theoretic conjugates of H is Ā-hyperdefinable, so there is a unique minimal Ā-hyperdefinable H^{c} commensurate with H. It is easy to see that H^{c} is locally connected.

Remark. H^{c} has a canonical parameter, Ā.

A type p over A is foreign to a partial type π over B if a | ⌣_{C} b for any C ⊇ AB and a ⊧ p with a | ⌣_{A} C, and b ⊧ π.

Let A = bdd(A), p a type over A, and Σ an A-invariant family of partial types. Then p is Σ-internal if for any a ⊧ p there is B | ⌣_{A} a and realizations c̄ of types in Σ over B, such that a ∈ dcl(ABc̄).

Theorem. If p is generic for G and not foreign to π, then there is a normal hyperdefinable N ≤ G of unbounded index, such that G/N is π-internal.

Proof. Over some model 𝔐 we find a realizing a non-forking extension of p, and π-internal b ∈ dcl(𝔐, a) − 𝔐. Put

X = { g ∈ G : ∃ x, y [ x | ⌣_{𝔐} g ∧ (x, y) ≡_{𝔐} (xg, y) ≡_{𝔐} (a, b) ] }.

By the Independence Theorem x^{−1}y ∈ X for independent x, y ∈ X, so X · X =: H is a subgroup of G, and G/H is tp(b/𝔐)-internal, whence π-internal.

Let ℌ be the family of G-conjugates of H, and obtain N.

A structure is one-based if A | ⌣_{A∩B} B for any two boundedly closed sets.

Proposition. If G is one-based, then a locally connected hyperdefinable subgroup is bdd(∅)-invariant.

Corollary. A one-based group is bounded-by-abelian-by-bounded.

Theorem. A group G is one-based iff every n-type is a generic type for some coset of a hyperdefinable subgroup of G^{n}, for all n < ω.

Let G be an abelian group with regular generic type, and let G_{0} be the subgroup of its non-generic elements. An endogeny of G is a partial hyperdefinable endomorphism G → G/G_{0}, whose domain has bounded index in G.

Theorem. Let G be one-based and ∅-connected. A tuple (g_{i} : i < n) in G is dependent iff there are bdd(∅)-definable endogenies (r_{i} : i < n) with ∑_{i<n} r_{i}(g_{i}) = G_{0}.

Remark. The preceding results can be generalized to locally modular groups and types.

Supersimple groups

The generic types of G are precisely those of maximal SU-rank.

If g is generic for G and H ≤ G, then gH is generic for G/H and g is generic in gH, whence

SU(H) + SU(G/H) ≤ SU(G) ≤ SU(H) ⊕ SU(G/H).

Problem. Does a generic type have maximal D-rank ?

Lemma. Suppose SU(G) = ∑_{i≤k} ω^{α_{i}}n_{i}, and let p ∈ S_{G}(A) with A = bdd(A). If SU(ab^{−1}/A) < SU(p) + ω^{α_{k}} for any two independent realizations a, b ⊧ p, then some non-forking extension of p is a right translate of a generic type for stab(p).

Corollary. G has a normal subgroup of SU-rank ∑_{i≤j} ω^{α_{i}}n_{i} for all j ≤ k ; it is unique up to commensurativity.

Proof. Let H be the stabilizer of a type p of that rank and apply the Lemma. Use the Lascar inequality to see that H is commensurate with its G-conjugates. Finally, replace H by a normal subgroup.

Theorem. Suppose SU(G) < ω^{α+1}, and 𝔛 is a family of hyperdefinable subsets. Then there is a hyperdefinable subgroup H ≤ X_{1}^{±1} ⋯ X_{m}^{±1} for some m < ω and X_{i} ∈ 𝔛, such that for all X ∈ 𝔛 either X is contained in a single coset of H, or SU(XH/H) ≥ ω^{α}. Corollary. A hyperdefinably simple non-abelian group is simple, of SUrank of the form ω^{α}n. Corollary. A definably simple definable non-abelian group is simple. Theorem. A type-definable supersimple group is the intersection of definable groups. Corollary. A type-definable supersimple division ring is definable. Problem. Is a type-definable group in a simple theory the intersection of definable groups ? Theorem. A supersimple division ring is commutative. Problem. Is a supersimple division ring PAC ?