Introduction to the Lascar Group

Martin Ziegler^{∗}

June 13, 2001

1 Introduction

The aim of this article is to give a short introduction to the Lascar Galois group Gal_{L}(T) of a complete first order theory T. We prove that Gal_{L}(T) is a quasicompact topological group in section 5. Gal_{L}(T) has two canonical normal closed subgroups: Γ_{1}(T), the topological closure of the identity, and Gal_{L}^{0}(T), the connected component. In section 6 we characterize these two groups by the way they act on bounded hyperimaginaries. In the last section we give examples which show that every compact group occurs as a Lascar Galois group and an example in which Γ_{1}(T) is non–trivial.

None of the results, except possibly Corollary 26, are new, but some technical lemmas and proofs are. In particular, the treatment of the topology of Gal_{L}(T) in sections 4 and 5 avoids ultraproducts, by which the topology was originally defined in [6]. Most of the theory expounded here was taken from that article, and the more recent [7], [4] and [2].

I thank Katrin Tent for reading the manuscript carefully, Markus Tressl, who found a serious mistake in an earlier version, and Anand Pillay, who helped me with the proof of Theorem 23.

2 The group

We fix a complete theory T. Let ℂ be a saturated^{1} model of T, of cardinality larger than as 2^{|T |}, and let Aut(ℂ) its automorphism group. The subgroup Autf_{L}(ℂ) generated by all point–wise stabilizers Aut_{M}(ℂ) of elementary^{2} submodels M is called the group of Lascar strong automorphisms. Autf_{L}(ℂ) is a normal subgroup of Aut(ℂ). The quotient is the Lascar (Galois) group of ℂ:

∗Partially supported by the Mittag–Leffler Institute, Stockholm

1T may not have saturated models. In this case we take for ℂ a special model (see [3] Chapter 10.4) of T and use the cf|ℂ| instead of |ℂ|. Especially we assume that cf|ℂ| > 2^{|T |}.

2In the sequel submodel will always mean elementary submodel.

Gal_{L}(ℂ) = Aut(ℂ)/Autf_{L}(ℂ).

We will show that Gal_{L}(ℂ) does not depend on the choice of ℂ.

Lemma 1 Let M and N be two small ^{3} submodels of ℂ and f an automorphism. Then the class of f in Gal_{L}(ℂ) is determined by the type of f(M) over N.

Proof: Let (m_{i})_{i∈I} be an enumeration of M. By the type of f(M) over N we mean the type of the infinite tuple (f(m_{i}))_{i∈I} over N. This is a type in variables (x_{i})_{i∈I}. We denote by S_{I}(N) the set of all such types over N.

Let g(M) have the same type over N as f(M). Choose an automorphism s which fixes N and maps f(M) to g(M). Then s is a Lascar strong automorphism, as is t = (sf)^{−1}g, which fixes M. Now we see that g = sft and f have the same class in Gal_{L}(ℂ). □

Two possibly infinite tuples a and b from ℂ are said to have the same Lascar strong type iff f(a) = b for a Lascar strong automorphism f.

Lemma 2 a and b have the same Lascar strong type iff there is a sequence of tuples a = a_{0}, …, a_{n} = b and a sequence of small submodels N_{1}, …, N_{n} such that, for each i, a_{i−1} and a_{i} have the same type over N_{i}.

Proof: Clear □

Corollary 3 a and b have the same Lascar strong type in ℂ if they have the same Lascar strong type in an elementary extension of ℂ.

Proof: If a_{0}, …, N_{n} exist in an elementary extension of ℂ, we find by saturation in ℂ a sequence a′_{0}, …, N′_{n} which has the same type over ab as a_{0}, …, N_{n}. This sequence shows that a and b have the same Lascar strong type in ℂ. □

Theorem 4 ([6]) Gal_{L}(ℂ) depends only on T and not on the choice of ℂ.

Proof: If ℂ′ is another big saturated model of T we can assume that ℂ′ is an elementary extension of ℂ and of larger cardinality. We can extend every automorphism f of ℂ to an automorphism f′ of ℂ′. Since all such f′ differ only by elements of Aut_{ℂ}(ℂ′), this defines a homomorphism Aut(ℂ) → Gal_{L}(ℂ′). If f Lascar strong, f′ is Lascar strong as well. Whence we have a well defined natural map

Gal_{L}(ℂ) → Gal_{L}(ℂ′),

3of smaller cardinality than ℂ

which will turn out to be an isomorphism.

To prove surjectivity, fix an automorphism g of ℂ′. Choose two small submodels M and N of ℂ. By saturation we find a submodel M′ of ℂ which has the same type over N as g(M). There is an automorphism f of ℂ which maps M to M′. Extend f to an automorphism f′ of ℂ′. Then f′(M) and g(M) have the same type over N. Whence, by the last lemma f′ and g represent the same element of Gal_{L}(ℂ′).

Now assume that f ∈ Aut(ℂ) extends to a Lascar strong automorphism f′ of ℂ′. Fix a small submodel M of ℂ. Then M and f(M) have the same Lascar strong type in ℂ′, whence also in ℂ by Corollary 3. So M can be mapped to f(M) by a Lascar strong automorphism of ℂ. Such an automorphism agrees with f on M, whence f is also strong. This shows that Gal_{L}(ℂ) → Gal_{L}(ℂ′) is injective. □

Definition The Lascar group of T is the quotient

Gal_{L}(T) = Aut(ℂ)/Autf_{L}(ℂ),

where ℂ is any big saturated model of T.

Corollary 5 The cardinality of Gal_{L}(T) is bounded by 2^{|T |}.

Proof: The class of f in Gal_{L}(T) is determined by the type of f(M) over N. If M and N are chosen to be of cardinality T, there are at most 2^{|T |} possible types. □

3 Digression: Lascar strong types and thick formulas

Definition Let θ(x, y) be a formula in two tuples of variables x and y having the same length. θ(x, y) is thick, if it has no infinite antichain, that is a sequence of tuples a_{0}, a_{1}, … such that ℂ ⊧ ¬ θ(a_{i}, a_{j}) for all i < j.

Clearly θ(x, y) is thick iff there is no indiscernible sequence a_{0}, a_{1}, … such that ℂ ⊧ ¬ θ(a_{0}, a_{1}). With this description it is easy to see that the intersection of two thick formulas is thick again and that a formulas remains thick if one interchanges the role of x and y.

Lemma 6 Let Θ(x, y) be the set of all thick formulas in x and y and let a and b two tuples of the same length. Then the following are equivalent:

a) ℂ ⊧ Θ(a, b) b) a and b belong to an infinite indiscernible sequence.

Proof: Assume ℂ ⊧ Θ(a, b). Then, if ψ(x, y) is satisfied by ab, ¬ψ is not thick, so there is an infinite sequence of indiscernibles a_{0}, a_{1}, … such that ψ(a_{0}, a_{1}) is true. Whence, by compactness, there is one infinite sequence of indiscernibles such that a_{0}a_{1} has the same type as ab. If conversely a, b are the first two elements of an infinite indiscernible sequence they have to satisfy all thick formulas □

{Lemma 7 \atop 1.} If ℂ ⊧ Θ(a, b), there is a model over which a and b have the same type.

2. If a and b have the same type over some model, the pair ab satisfies the relational product Θ ∘ Θ. I.e. there is a tuple a′ such that ℂ ⊧ Θ(a, a′) and ℂ ⊧ Θ(a′, b).

Proof:

1. Let I be an infinite sequence of indiscernibles and M any small model. Then there are indiscernibles I′ over M of the same type as I. Whence there is a model M′ of the same type as M over which I is indiscernible. Therefore, if a, b are the first elements of some I, they have the same type over some model M′. Now apply Lemma 6.

A more direct proof, which avoids Lemma 6, uses the observation that two sequences a and b of the same length have the same type over a model iff ab satisfies all formulas of the form

∃z φ(z) → ∃z (φ(z) ∧ ⋀_{i=1}^{n} ψ_{i}(x, z) ↔ ψ_{i}(y, z))  (1)

for all finite variable tuples z and formulas φ(z), ψ_{1}(x, z), …, ψ_{n}(x, z). All formulas (1) are thick, antichains have length at most 2^{n}.

2. Assume that a and b have the same type over M. If θ is a thick formula, consider a maximal antichain a_{1}, …, a_{n} for θ in M. Then, since M is an elementary substructure, a_{1}, …, a_{n} is also a maximal antichain in ℂ. Whence ℂ ⊧ θ(a_{i}, a) for some i. Since b has the same type over M, we have ℂ ⊧ θ(a_{i}, b). This proves that for every finite subset Θ_{0} of Θ there is an a′ such that ℂ ⊧ Θ_{0}(a′, a) and ℂ ⊧ Θ_{0}(a′, b). This proves the claim using compactness and the observation that Θ defines a symmetric relation. □

Corollary 8 The relation of having the same Lascar strong type is the transitive closure of the relation defined by Θ. □

Let π be a type defined over the empty set. A formula θ(x, y) is thick on π if θ has no infinite antichain in π(ℂ). Let Θ_{π} be the set of all formulas which are thick over π.

Corollary 9 Two realizations of π, a and b, have the same Lascar strong type if the pair (a, b) is in the transitive closure of the relation defined by Θ_{π}.

Proof: Assume that a and b have the same type over a model M. The proof of Lemma 7 (1) shows that we can assume that M is ω–saturated. If θ is thick on π, let a_{1}, …, a_{n} be a maximal antichain for θ in π(M). Then, since is ω– saturated, a_{1}, …, a_{n} is also maximal in π(ℂ). Now proceed as in Lemma 7 (2). □

4 The topology

Let M and N be two small submodels of ℂ. Assign to every automorphism f of ℂ the type of f(M) over N. This defines a surjective map μ from Aut(ℂ) to S_{M}(N), the set all types over N of conjugates of M. By Lemma 1 the projection Aut(ℂ) → Gal_{L}(T) factors through μ:

Aut(ℂ) ⟶^{μ} S_{M}(N) ⟶^{ν} Gal_{L}(T).

S_{M}(N), as a closed subspace of S_{I}(N), is a boolean space. We give Gal_{L}(T) the quotient topology with respect to ν.

To show that this does not depend on the choice of M and N we consider another pair M′ and N′. We may assume that M ⊂ M′ and N ⊂ N′. The map S_{M′}(N′) ⟶ Gal_{L}(T) then factors as

S_{M′}(N′) ⟶ S_{M}(N) ⟶^{ν} Gal_{L}(T),

where the first map is restriction of types. Since restriction is continuous and the spaces are compact, S_{M}(N) carries the quotient topology of S_{M′}(N′), which implies that on Gal_{L}(T) the two topologies, coming from S_{M′}(N′) and S_{M}(N), are the same.

A quotient of a quasicompact space remains quasicompact. So we have

Lemma 10 Gal_{L}(T) is quasicompact. □

Let p and q be types in S_{M}(N). Two realizations M′ and M′′ of p and q have the same Lascar strong type iff ν(p) = ν(q). Whence, by Corollary 8, the equivalence relation

p ≈ q ⇔ ν(p) = ν(q)

is the transitive closure of the relation D, where D(p, q) holds if p and q have realizations M′ and M′′ with ℂ ⊧ Θ(M′, M′′).

{Lemma 11 \atop 1.} D is a closed subset of S_{M}(N) × S_{M}(N)

2. ≈ is a F_{σ}–set, i.e. a countable union of closed sets.

Proof:

1. This is clear, because

D(p, q) ⇔ p(x) ∪ q(y) ∪ Θ(x, y) consistent.

2. ≈ is the union of all powers

D^{n} = \underbrace{D ∘ ⋯ ∘ D}_{n times} .

So, is suffices to show that all D^{i} are closed. This follows from the fact that, in compact spaces, the product of two closed relations is closed again. To see this, note that, for binary relations R and S, R ∘ S is the projection of {(p, q, r)|R(p, q) ∧ S(q, r)} onto the first and third variable. □

In general the map S_{M}(N) ⟶^{ν} Gal_{L}(T) is not open.^{4} But it has a property that comes close to openness. Define for p ∈ S_{M}(N)

D[p] = {q ∈ S_{M}(N) | D(p, q)}.

Lemma 12 If D[p] is contained in the interior of some subset O ⊂ S_{M}(N), then ν(p) is an inner point of ν(O).

Proof: D[p] is the intersection of all

D_{δ}[p] = {q ∈ S_{M}(N) | p(x) ∪ q(y) ∪ {δ(x, y)} consistent}, (δ ∈ Θ).

By compactness some D_{δ}[p] is contained in (the interior of) O.

\underline{Claim 1}: p is an inner point of D_{δ}[p].

Proof: Since δ is thick, there is a finite set {H_{1}, …, H_{n}} of realizations of p such that for every other realization H we have ℂ ⊧ δ(H_{i}, H) for some i. By compactness this is true for every realization H of any p′ contained in a small enough neighborhood C of p, which implies that C is contained in D_{δ}[p].

After replacing O by ν^{−1}(νO) we can assume that O is closed under ≈ (i.e. is a union of ≈–classes.) We set

U = {q ∈ S_{M}(N) | D_{δ}[q] ⊂ O for some δ ∈ Θ}.

4If Aut(ℂ) is endowed with the topology of point–wise convergence, μ becomes continuous (see Lemma 29). If ν were always open, Aut(ℂ) → Gal_{L}(T) would be open too: If a, b are two (finite) tuples, choose N, M in such a way that a, b ∈ M = N. Then the basic open set {f ∈ Aut(ℂ)|f(a) = b} will be mapped onto an open subset of S_{M} (N) and whence, by assumption, onto an open subset of Gal_{L}(T). Whence, the closedness of Autf_{L}(ℂ) would imply that Gal_{L}(T) is hausdorff. That this is not true shows one of the examples in [2] (T h(M^{∗}) in Proposition 4.5).

U contains p.

\underline{Claim 2}: U is closed under ≈.

Proof: Let q be in U, witnessed by D_{δ}[q] ⊂ O, and q ≈ r. Then a realization H of q is mapped by a Lascar strong automorphism f to a realization f(H) = K of r. In order to show that r belongs to U we fix an element r′ of D_{δ}[r]. We have then a realization K′ of r′ such that ℂ ⊧ δ(K, K′). Let q′ be the type of H′ = f^{−1}(K′) over M. Since ℂ ⊧ δ(H, H′), q′ belongs to D_{δ}[q] and therefore to O. Since q ≈ q′ and O is closed under ≈, we have q′ ∈ O. It follows D_{δ}[r] ⊂ O.

\underline{Claim 3}: U is open.

Proof: U is a subset of the interior of O by Claim 1. Since U is closed under ≈, it is contained in the open set

U′ = {q ∈ S_{M}(N) | D[q] ⊂ interior(O)},

which, by compactness, equals

U′′ = {q ∈ S_{M}(N) | D_{δ}[q] ⊂ interior(O) for some δ ∈ Θ}.

But U′′ is contained in U, which shows that U = U′.

By Claims 2 and 3 the projection of U is an open subset of ν(O) and contains ν(p). This completes the proof of Lemma. □

Corollary 13 If L is countable, Gal_{L}(T) has a countable basis.

Proof: If L is countable we can choose countable M and N. S_{M}(N) has then a countable base, B. We can assume that B is closed under finite unions. Let us show that the set of all ν(B)^{∘}, (B ∈ B), is a basis of Gal_{L}(T). Let Ω be open and α ∈ Ω. Choose a preimage p of α and a basic open set B, such that D[p] ⊂ B ⊂ ν^{−1}(Ω). This is possible, since B is compact and B closed under finite unions. Then ν(B)^{∘} ⊂ Ω is an open neighborhood of p. □

The following corollary is a reformulation of Corollary 3.5 in [2].

Corollary 14 Let X be a subset of Gal_{L}(T). Then

\overline{X} = ν(\overline{ν^{−1}(X)}).

Proof: Since ν is continuous the right hand side lies inside \overline{X}. Let ν(p) be an element of Gal_{L}(T) which does not belong to ν(\overline{ν^{−1}(X)}. Then the whole ≈–class of p, which contains D[p], is disjoint from \overline{ν^{−1}(X)}. By Lemma 12 the complement of \overline{ν^{−1}(X)} is mapped to a neighborhood of ν(p), which is disjoint from X. This shows ν(p) ∉ \overline{X}. □

Corollary 15 Gal_{L}(T) is hausdorff iff ≈ is closed.

Proof: “Gal_{L}(T) hausdorff ⇒ ≈ closed” is an easy consequence of the continuity of ν.

Now assume that ≈ is closed. Consider two different elements x, y of Gal_{L}(T). Since ≈ is closed, we can separate each element of ν^{−1}(x) from each element of ν^{−1}(y) by a pair of neighborhoods which projects onto disjoint subsets of Gal_{L}(T). But ν^{−1}(x) and ν^{−1}(y) are compact. This implies that there is one pair of open sets, O and U, which separate ν^{−1}(x) and ν^{−1}(y) and have disjoint projections ν(O) and ν(U), which are, by the lemma, neighborhoods of x and y. □

We will see in section 7 (Theorem 28) that Gal_{L}(T) need not to be hausdorff.

5 The topological group

Theorem 16 (Lascar) Gal_{L}(T) is a topological group.

For the proof we fix again two small submodels M and N and consider the natural mappings

Aut(ℂ) ⟶^{μ} S_{M}(N) ⟶^{ν} Gal_{L}(T).

Lemma 17 The projections of multiplication

M = {(μ(f), μ(g), μ(fg)) | f, g ∈ Aut(ℂ)}

and of inversion

I = {(μ(f), μ(f^{−1})) | f ∈ Aut(ℂ)}

are closed subset of S_{M}(N) × S_{M}(N) × S_{M}(N) and of S_{M}(N) × S_{M}(N), respectively.

Proof: We introduce two unary function symbols F and G and express the fact that F are G automorphisms by the L ∪ {F, G}–theory A(F, G). Then (p, q, r) belongs to M iff there are are functions f, g : ℂ → ℂ which satisfy the theory

B(F, G, p, q, r) = A(F, G) ∪ p(F (M)) ∪ q(G(M)) ∪ r(F (G(M))).

Since ℂ is saturated, B(F, G, p, q, r) can be satisfied in ℂ if it is consistent with the theory of ℂ_{M,N}. This is a closed condition on p, q, r.

The closedness of I is similar. □

The graphs of the multiplication and inversion in Gal_{L}(T) are the projections of M and I. If Gal_{L}(T) is hausdorff, the projections are closed, which, by compactness, implies that multiplication and inversion are continuous in Gal_{L}(T).

For the general case we need the following notation: For two subsets of A and B of S_{M}(N) define

A ∗ B = {r ∈ S_{M}(N) | (p, q, r) ∈ M for a pair (p, q) ∈ A × B}.

Lemma 18 If A and B are closed and A ∗ B is contained in the open set W, there are neighborhoods U and V of A and B such that U ∗ V ⊂ W.

Proof: Let W′ be the complement of W. A × B is disjoint from the projection C of

M ∩ (S_{M}(N) × S_{M}(N) × W′)

on the first two coordinates. Since C is closed (and A and B are compact) there are neighborhoods U and V of A and B such that U × V is disjoint from C. It follows that U ∗ V ⊂ W. □

We can now prove that multiplication in Gal_{L}(T) is continuous. Let α = ν(p) and β = ν(q) be elements of Gal_{L}(T) and Ω an open neighborhood of αβ. Then

D[p] ∗ D[q] ⊂ ν^{−1}(α) ∗ ν^{−1}(β) ⊂ ν^{−1}(αβ) ⊂ ν^{−1}(Ω).

By the last lemma there neighborhoods U and V of D[p] and D[q], respectively, such that U ∗ V ⊂ ν^{−1}(Ω). This implies ν(U)ν(V) ⊂ Ω. Finally, we remark that, by Lemma 12, ν(U) and ν(V) are neighborhoods of α and β.

The continuity of inversion is proved in the same manner, which completes the proof of the theorem.

6 Two subgroups

Gal_{L}(T) has two canonical normal subgroups:

• Γ_{1}(T), the closure of {1}.

• Gal_{L}^{0}(T), the connected component of 1.

Since Gal_{L}(T) is quasicompact, we have

{Lemma 19 \atop 1.} The quotient Gal_{L}^{c}(T) = Gal_{L}(T)/Γ_{1}(T) is a compact group, the closed Galois group of T.

2. Gal_{L}^{0}(T) is the intersection of all closed (normal) subgroups of finite index.

Proof: Gal_{L}^{c}(T) is quasicompact and hausdorff, i.e. compact. For the second part, note that the quotient Gal_{L}(T)/Gal_{L}^{0}(T) is totally disconnected ([12, §2]) and compact, whence a profinite group. In a profinite group the intersection of all normal closed subgroups of finite index is the identity. □

An imaginary element of ℂ is a class of a ∅–definable equivalence relation on a cartesian power ℂ^{n}. Automorphisms of ℂ act in a natural way on imaginaries. An imaginary with only finitely many conjugates under Aut(ℂ) is called algebraic.

Let us prove that algebraic imaginaries are fixed by Lascar strong automorphisms: Let a/E be an algebraic imaginary with k conjugates. This means that E partitions the set of all conjugates of a into k classes. It follows that the type of a contains a formula φ(x) whose realization set meets exactly k equivalence classes. Let f fix the model M. Then φ(M) meets the same classes as φ(ℂ), which implies that a/E contains an element b of M, which must also belong to f(a)/E. It follows that a/E = f(a)/E.

This result extends easily to hyperimaginaries. Hyperimaginaries are equivalence classes of type–definable equivalence relations E, which are defined by a set of formulas Φ without parameters:

E(a, b) ⇔ ℂ ⊧ Φ(a, b).

a and b are, possible infinite, tuples of elements of ℂ, of length smaller than |ℂ|. A hyperimaginary is bounded if it has less than |ℂ| conjugates.

Lemma 20 Bounded hyperimaginaries are fixed by Lascar strong automorphisms.

Proof: Let a/E be a bounded hyperimaginary and E defined by Φ(x, y). Then Φ ⊂ Θ_{π}, where π = tp(a), since otherwise some θ ∈ Φ would have antichains in π(ℂ) of arbitrary length, contradicting the assumption that a/E is bounded. If f is Lascar strong, a and f(a) have the same Lascar strong type. By Corollary 9, E(a, f(a)). □

If, conversely, a hyperimaginary h is fixed by all Lascar strong automorphisms, f(h) is determined by the class of f in Gal_{L}(T). Whence h has no more than 2^{|T |}–many conjugates and is bounded.

We conclude that Gal_{L}(T) acts on bounded hyperimaginaries in a well defined way.

{Theorem 21 \atop 1.} Γ_{1}(T) is the set of all elements of Gal_{L}(T) which fix all bounded hyperimaginaries.

2. Gal_{L}^{0}(T) is the set of all elements of Gal_{L}(T) which fix all algebraic imaginaries.

Proof:

1. Let a/E be a bounded hyperimaginary and Γ ≤ Gal_{L}(T) the stabilizer of a/E. The preimage of Γ in S_{M}(N) is

ν^{−1}(Γ) = {tp(f(M)/N) | f ∈ Aut(ℂ), E(f(a), a)}.

Choose M containing a, let N = M and E be axiomatized by Φ. Then

ν^{−1}(Γ) = {p(x) ∈ S_{M}(N) | Φ(x′, a) ⊂ p(x)},

where the variables x′ are a subtuple of x, as a is a subtuple of (m_{i}), the enumeration of M. Whence Γ is closed and we conclude Γ_{1}(T) ⊂ Γ. This shows that the elements of Γ_{1}(T) fix all bounded imaginaries.

For the converse consider the inverse image G_{1} of Γ_{1}(T) in Aut(ℂ). For |T |–tuples a, b let E(a, b) denote the equivalence relation of being in the same G_{1}–orbit. Since the index of G_{1} is bounded by 2^{|T |}, E has at most 2^{|T |} classes. Since Γ_{1}(T) is closed, E is type–definable. To see this, write the closed set ν^{−1}(Γ_{1}(T)) as {p(x) ∈ S_{M}(N) | Ψ(x) ⊂ p(x)} for a set Ψ(x) of L(N)–formulas. Then

E(a, b) ⇔ for some f ∈ Aut(ℂ) ℂ ⊧ f(a) = b ∧ Ψ(f(M)).

This shows, by an argument similar to that in the proof of Lemma 17, that E is can be defined by a set of formulas with parameters from M and N. Since Γ_{1}(T) is a normal subgroup, G_{1} is a normal subgroup of Aut(ℂ). This implies that E is invariant under automorphisms, and whence can be defined by a set of formulas without parameters.

Now assume that α ∈ Gal_{L}(T) fixes all bounded hyperimaginaries. Take a model K of cardinality |T | and consider it as a |T |–tuple. Then K/E is a bounded hyperimaginary and fixed by α. This means that α is represented by an automorphism which agrees on K with an automorphism f from G_{1}. Since K is a model, this implies that α is represented by f and belongs to Γ_{1}(T).

2. Let i be an algebraic imaginary and Γ the stabilizer of i in Gal_{L}(T). Γ is closed and has finite index, since the index equals the number of conjugates of i. It follows that Gal_{L}^{0}(T) ⊂ Γ. Thus the elements of Gal_{L}^{0}(T) fix all algebraic imaginaries.

For the converse it suffices to show that every normal closed Γ ≤ Gal_{L}(T) of finite index is the stabilizer of an algebraic imaginary. The first part of the proof shows that Γ, being a normal^{5} closed subgroup, is the stabilizer of a bounded hyperimaginary a/E. Since Γ has finite index, a/E has only a finite number of conjugates. We will show that a/E has the same stabilizer as an algebraic imaginary a/F. (If a is an infinite tuple, we can replace it by the finite subtuple of elements which occur in F.)

5A slight variation of the argument shows that normality is not necessary: Let G be the preimage of Γ, and K a model of size |T |. Define E(a, b) to be true if a = b or, for some f ∈ Aut(ℂ) and g ∈ G, f(K) = a and fg(K) = b. Then K/E is a bounded hyperimaginary with Γ as its stabilizer. See [7, 4.12].

Let E be defined by Φ and let a_{1}/E, …, a_{n}/E be the different conjugates of a/E. By compactness there is a symmetric formula^{6} θ ∈ Φ such that no pair (a_{i}, a_{j}) (i ≠ j) satisfies θ^{2} = θ∘θ.^{7} This means that the sets θ(a_{i}, ℂ) are disjoint. Since they cover the set of conjugates of a, there is a formula φ(x) satisfied by a such that the intersections

D_{i} = φ(ℂ) ∩ θ(a_{i}, ℂ)

form a partition of φ(ℂ). In order to ensure that this partition is invariant under automorphisms, we choose θ ∈ Φ so small that no pair (a_{i}, a_{j}) satisfies θ^{4}. This implies that θ^{2}(c, d) is never true for c ∈ D_{i} and d ∈ D_{j} and, therefore, that

F (x, y) = (¬φ(x) ∧ ¬φ(y)) ∨ (φ(x) ∧ φ(y) ∧ θ^{2}(x, y))

defines an equivalence relation, with classes ¬φ(ℂ), D_{1}, …, D_{n}. Thus a/F is an algebraic imaginary. Since a/E and a/F contain the same conjugates of a, they have the same stabilizer. □

Corollary 22 1. Gal_{L}^{c}(T) is the automorphism group of the set of all bounded hyperimaginaries of length |T |.

2. Gal_{L}(T)/Gal_{L}^{0}(T) is the automorphism group of the set of all algebraic imaginaries.

□

It was shown in [7] that every bounded hyperimaginary has the same (pointwise) stabilizer as a set of bounded hyperimaginaries of finite length. So Gal_{L}^{c}(T) is the automorphism group of the set of all bounded hyperimaginaries of finite length.

The set of algebraic imaginaries is often called acl^{eq}(∅). The group

Gal_{L}(T)/Gal_{L}^{0}(T) = Aut(acl^{eq}(∅))

is the Galois group introduced by Poizat in [9].

For stable T two tuples a and b which have the same strong type (i.e. the same type over acl^{eq}(∅)) have the same type over any model which is independent from ab. It follows that Gal_{L}^{0}(T) = 1. This was extended to supersimple theories in [1]. Whether this is true for all simple^{8} theories is an open problem. All we know is Kim’s result ([5]) that Γ_{1}(T) = 1 for simple T.

6Assume Φ closed under conjunction.

7Recall that θ(x, y) is the formula ∃z θ(x, z) ∧ θ(z, y).

8See [11] for an introduction to simple theories.

7 Two Examples

The first part of this section is concerned with the proof of the following unpublished result of E. Bouscaren, D. Lascar and A. Pillay:

Theorem 23 Any compact Lie group is the Galois group of a countable complete theory.

First we need a lemma on O-minimal structures. Recall that a structure M with a distinguished linear order < is O-minimal if every definable subset of M is a union of finitely many points and intervals with endpoints in M. Note that every structure elementarily equivalent to an O-minimal structure is itself O-minimal.

Lemma 24 Every automorphism of a big saturated O-minimal structure is Lascar strong.

Proof: Let ℂ be a big saturated O-minimal structure. We prove that any two small submodels M, N of the same type have the same type over some model K. This implies, as in the proof of Lemma 1, that every automorphism which maps M to N is the product of an automorphism which fixes M and an automorphism which fixes K.

It is enough (and equivalent, see the proof of Lemma 7 (1)) to show the following : Every consistent formula φ(z) has a realization c over which M and N have the same type.

We prove this by induction on the length of z. Assume that z consists of a tuple z_{1} and a single variable z_{2}. By induction there is a realization c_{1} of ∃z_{2}φ(z_{1}, z_{2}) over which M and N have the same type. Let ψ(m, c_{1}, z_{2}) be any formula over Mc_{1}, and let the tuple n ∈ N correspond to m. By O-minimality, and since m and n have the same type over c_{1}, either both ψ(m, c_{1}, ℂ) and ψ(n, c_{1}, ℂ) contain a non–empty final segment of φ(c_{1}, ℂ) or ¬ψ(m, c_{1}, ℂ) and ¬ψ(n, c_{1}, ℂ) contain a non–empty segment. If we choose c_{2} in the intersection of all these segments, c = c_{1}c_{2} realizes φ(z) and M and N have the same type over c. □

Now fix a compact Lie group G. The group G together with its structure of a real analytic manifold can be defined inside an expansion R of the field ℝ by a finite number of analytic functions which are defined on bounded rectangles. By a result of van den Dries R is O–minimal^{9} (see [10]).

Let R^{∗} a big saturated extension of R and G^{∗} the resulting extension of G. The intersection μ of all ∅–definable neighborhoods of the unit element of G^{∗} is the normal subgroup of infinitesimal elements. The compactness of G implies that every element of G^{∗} differs by an infinitesimal from some element of G. Whence G^{∗} is the semi–direct product of G and μ.

9As A. Pillay has told me, compact Lie groups are semi–algebraic. This means that here (and in the proof of Corollary 26) one can actually assume that R is the field of reals with a finite tuple of named parameters.

Lemma 25 μ is the set of all commutators [φ, h] = h^{−1}φ(h), where h ∈ G^{∗} and φ ∈ Aut(R^{∗}).

Proof: Let φ be an automorphism of R^{∗} and let h differ from h_{0} ∈ G by an infinitesimal ε. Since φ fixes ℝ, it fixes h_{0}. Whence h^{−1}φ(h) = (h_{0}ε)^{−1}φ(h_{0}ε) = ε^{−1}φ(ε) is infinitesimal.

Let conversely ε ∈ μ be given. Consider a generic type p ∈ S(∅) of G (cf. [8]). This means that p can be axiomatized by formulas which define (non–empty) open subsets O(G) of G. Each O(G^{∗}) contains two elements h and hε (pick any h ∈ O(G)). Whence, by saturation, p has two realizations h and hε. Choose an automorphism φ with φ(h) = hε. Then ε = h^{−1}φ(h).^{10} □

Consider the two–sorted structure

M = (R, X, ·)

where · is a regular action of G on the set X. We will show that G is the Galois group of the complete theory of M.

Let M^{∗} = (R^{∗}, X^{∗}) be a big saturated elementary extension of M. To describe the automorphisms of M^{∗} we fix a base point x_{0} ∈ X^{∗}. Any element of X^{∗} can then uniquely be written as

x = h · x_{0}

for some h ∈ G^{∗}. We extend each automorphism φ of R^{∗} to M^{∗} by

\overline{φ}(x) = φ(h) · x_{0}.

The automorphisms which leave R^{∗} fixed have the form \overline{g}, where

\overline{g}(x) = hg^{−1} · x_{0}.

This implies that every automorphism of M^{∗} is a product

Φ = \overline{g} \overline{φ}.

Note the commutation rule \overline{φ} \overline{g} = \overline{φ(g)} \overline{φ}.

Elementary substructures of M^{∗} have the form (R′, G′ · x), where R′ is an elementary substructure of R^{∗} and x = h · x_{0} is any element of X^{∗}. Therefore an automorphism fixes a submodel iff it can be written as \overline{h}^{−1}\overline{φ} \overline{h}, for some φ which fixes an elementary submodel of R^{∗}. It follows that an automorphism is Lascar strong iff it is a product of conjugates of automorphisms of the form \overline{φ}, for Lascar strong φ.

10A variant of the proof shows that one can find a φ which fixes an elementary submodel of R^{∗}.

By Lemma 24 all φ are Lascar strong. The formula

\overline{h}^{−1}\overline{φ} \overline{h} = \overline{[φ, h]} \overline{φ},

together with the last Lemma, implies that Φ = \overline{g} \overline{φ} is Lascar strong iff g is infinitesimal. We conclude that

g ↦ class of \overline{g}

defines an isomorphism ι : G → Gal_{L}(M^{∗}).^{11}

Finally we have to prove that ι is a homeomorphism. Let U(G) be a ∅– definable neighborhood of 1 ∈ G. Consider the map ν : S_{M}(M) → Gal_{L}(M^{∗}). Then ν^{−1}ι(U(G)) consists of those tp(f(M)/M) for which M^{∗} ⊧ U(f(1)). Whence, if 1 has index 1 in the enumeration of M,

ν^{−1}ι(U(G)) = {p ∈ S_{M}(M) | U(x_{1}) ∈ p}.

This proves that ι(U_{n}) is open. So ι is an open map. Since Gal_{L}(M^{∗}) is quasicompact and G is hausdorff, ε must also be continuous. This completes the proof of Theorem 23.

Corollary 26 Every compact group is the Galois group of a complete theory.

Proof: Let G be a compact group. G is the direct limit of a directed system (G_{i}, f_{i,j})_{i≤j∈I} of compact Lie groups ([12, §25]). Again let R be an expansion of the reals by bounded analytic functions, in which all the G_{i} and the maps f_{ij} can be defined. The elements of G are then given by certain infinite tuples g = (g_{i})_{i∈I} from the direct product of the G_{i}.

G will be the Galois group (of the complete theory) of the many–sorted structure

M = (R, X_{i}, f′_{ij})_{i≤j∈I},

where the directed system of sets (X_{i}, f′_{i,j})_{i≤j∈I} is a copy of (G_{i}, f_{i,j})_{i≤j∈I} and each G_{i} operates (regularly) on X_{i} as it operates on itself by left multiplication.

Let again M^{∗} be a big saturated elementary extension of M and G^{∗} the inverse limit of the G_{i}^{∗}. We call an element ε = (ε_{i}) of G^{∗} infinitesimal if all its components are infinitesimal. Let μ the subgroup of all infinitesimals. It is easy to see that G is isomorphic to the quotient G^{∗}/μ.

Fix a base point x_{0} = (x_{0i})_{i∈I} in the (non–empty) inverse limit of the X_{i}^{∗}. Then every automorphism of M^{∗} has the form Φ = \overline{g} \overline{φ} for φ ∈ Aut(R^{∗}) and g ∈ G^{∗}, where \overline{g} and \overline{φ} are defined as in the proof of the theorem. Thus, it suffices to show that Φ is Lascar strong iff g is infinitesimal.

11The proof shows that two elements of X^{∗} differ by an infinitesimal if they have the same Lascar strong type. It is easy to verify that this happens iff they have the same type over a submodel of M^{∗}.

Assume first that Φ is Lascar strong. Then each g_{i} is infinitesimal, since Φ restricted to (R^{∗}, X_{i}) is Lascar strong. Conversely, if g is infinitesimal, we find for every i an h_{i} ∈ G_{i} such that h_{i} and h_{i}g_{i} have the same type. A compactness argument shows that we can find the sequence (h_{i}) in G^{∗}. Then h and hg have the same type. Let ψ be an automorphism of R^{∗} with ψ(h) = hg. As in the proof of the theorem, it is easy to see that \overline{g}\overline{ψ} = \overline{h}^{−1}\overline{ψ} \overline{h} is Lascar strong. Whence also Φ = (\overline{h}^{−1}\overline{ψ} \overline{h})ψ^{−1}\overline{φ} is Lascar strong. □

We construct our second example from the circle group S, the unit circle in the complex number plane. Let us fix some notation: λ_{s} denotes multiplication by s. R is the cyclic ordering on S, where R(r, s, t) holds if s comes before t in the counter–clockwise ordering of S \ {r}.

Fix a natural number N, write σ_{N} for λ_{\frac{2πi}{N}} and consider the structure

S_{N} = (S, R, σ_{N}).

Let ℂ_{N} a big saturated elementary extension and f an automorphism of ℂ_{N}. If f is Lascar strong, let |f| be the smallest n such that f is the product of n automorphisms which fix elementary submodels. If f is not Lascar strong, write |f| = ∞.

We will make use of the following lemma, which can be proved from Lemma 24 (see [2] for details).

{Lemma 27 \atop 1.} Every automorphism of ℂ_{N} is the product of some σ_{N}^{n} and some f with |f| ≤ 2.

2. |σ_{N}^{n}| = |n| + 2, whenever 0 < |n| ≤ \frac{N}{2} .

Let S_{∞} be the disjoint union of the S_{1}, S_{2}, … viewed as a many–sorted structure^{12} with saturated extension ℂ_{∞} = (ℂ_{1}, ℂ_{2}, …).

Theorem 28 ([2]) For each N let C_{N} be the N–element cyclic group with generator c_{N}. Let B be the group of all sequences (c_{N}^{e_{N}}) with a bounded sequence (e_{N}) of exponents. Then

Gal_{L}(ℂ_{∞}) ≅ ∏_{N} C_{N}/B.

Gal_{L}(ℂ_{∞}) carries the indiscrete topology.

12We take also the disjoint union of the languages.

Proof: The map (c_{N}^{e_{N}}) ↦ (σ_{N}^{e_{N}}) defines a map from ∏_{N} C_{N} to Aut(ℂ_{∞}), which yields a homomorphism

μ : ∏_{N} C_{N} → Gal_{L}(ℂ_{∞}).

Let (f_{N}) be any automorphism of ℂ_{∞}. If we apply part 1 of the Lemma to each component we see that we can write (f_{N}) as a product of some (σ_{N}^{e_{N}}) and two automorphisms which fix a model. This shows that μ is surjective.

Let (c_{N}^{e_{N}}) be an arbitrary element of ∏_{N} C_{N}. We can assume that |e_{N}| ≤ \frac{N}{2} . Then by part 2 of the lemma it is immediate that (σ_{N}^{e_{N}}) is Lascar strong iff (e_{N}) is bounded, which means that B is the kernel of μ.

It remains to show that the topology of Gal_{L}(ℂ_{∞}) is indiscrete, or

Gal_{L}(ℂ_{∞}) = Γ_{1}(ℂ_{∞}).

The preimage of Γ_{1}(ℂ_{∞}) in Aut(ℂ_{∞}) is, by the next Lemma, a closed subgroup, which contains Autf_{L}(ℂ_{∞}). The automorphisms which fix almost every ℂ_{N} are Lascar strong and form a dense subset of Aut(ℂ_{∞}). Thus the preimage of Γ_{1}(ℂ_{∞}) is the whole Aut(ℂ) group. □

We conclude with a general lemma. Let M be a model of T and consider the topology of point–wise convergence on Aut(M), with basic open sets

U_{a,b} = {f | f(a) = b },

where a, b are finite tuples from M.

Lemma 29 The natural map Aut(M) → Gal_{L}(T) is continuous.

Proof: Let Ω be a neighborhood of the image of f in Gal_{L}(T). The preimage of Ω under^{13} ν : S_{M}(N) → Gal_{L}(T) contains a basic neighborhood

O = {p | φ(x) ∈ p}

of tp(f(M)/N). Let a be the tuple of elements of M which are enumerated by the free variables of φ. Then

O = {tp(g(M)/N) | ℂ ⊧ φ(g(a))} ⊂ {tp(g(M)/N) | g(a) = f(a)}.

Whence U_{a,f(a)} is a neighborhood of f which is mapped into Ω. □

13N can be any small model.

References

[1] Steve Buechler, Anand Pillay, and Frank O. Wagner. Supersimple theories. J. Amer. Math. Soc., 14:109–124, 2001.

[2] E. Casanovas, D. Lascar, A. Pillay, and M. Ziegler. Galois groups of first order theories. 19 pages, submitted, 2000.

[3] Wilfried Hodges. Model Theory. Encyclopedia of Mathematics and its Applications. Cambridge University Press, 1993.

[4] Ehud Hrushovski. Simplicity and the lascar group. Preprint, 1997.

[5] Byunghan Kim. A note on lascar strong types in simple theories. J. Symbolic Logic, 63:926–936, 1998.

[6] Daniel Lascar. The category of models of a complete theory. J. Symbolic Logic, 47:249–266, 1982.

[7] Daniel Lascar and Anand Pillay. Hyperimaginaries and automorphism groups. J. Symbolic Logic, 66(1):127–143, 2001.

[8] Anand Pillay. On groups and fields definable in o–minimal structures. J. Pure Appl. Algebra, 53:239–255, 1988.

[9] Bruno Poizat. Une théorie de galois imaginaire. J. Symbolic Logic, 48(4):1151–1170, December 1983.

[10] Lou van den Dries. A generalization of the Tarski–Seidenberg Theorem and some non–definability results. Bull. Amer. Math. Soc. (N.S.), 15(2):189– 193, October 1986.

[11] Frank O. Wagner. Simple Theories, volume 503 of Mathematics and Its Applications. Kluwer Academic Publishers, Dordrecht, NL, 2000.

[12] André Weil. L’integration dans les groupes topologiques. Actualités scientifiques et industrielles. Hermann, Paris, 2. edition, 1951.