A measure for perfect PAC fields with pro-cyclic Galois group

Immanuel Halupczok

Institut Camille Jordan Université Claude Bernard Lyon 1 43 boulevard du 11 novembre 1918 69622 Villeurbanne cedex France

Abstract In “Pseudo-finite fields and related structures” Hrushovski asks (in a more general context) whether the notion of measure on definable sets in pseudo-finite fields can be extended to perfect PAC fields whose Galois group is bounded but not ℤ̂. We define a suitable generalization of measure and give an answer to this question: yes if the Galois group is pro-cyclic, no otherwise. Key words: pseudo-algebraically closed fields, measure, definable sets, bounded Galois group 1991 MSC: 03C60, 12L12, 12E30, 12F10

Contents 1 Introduction 2 1.1 Overview over the article 3 1.2 Thanks 4 2 Preliminaries 4 2.1 Setting and notation 4 2.2 Galois covers 5 2.3 The Artin symbol 6 2.4 Galois stratifications 7 2.5 Definable sets and Galois stratifications 8 Email address: math@karimmi.de (Immanuel Halupczok).

Preprint submitted to Elsevier 1 September 2006 3 The main theorem 9 3.1 The statement 9 3.2 Sketch of the proof of existence 10 3.3 Sketch of the proof of uniqueness 11 4 Some useful lemmas 11 4.1 Some fiber sizes 11 4.2 A closer look on Galois covers 12 4.3 Which of the X(V ↠^{G} W, C) are empty? 13 5 The measure on the groups 18 5.1 The theorem for groups 18 5.2 The proof for groups 19 6 Proof of the main theorem 22 6.1 Existence 22 6.2 Uniqueness 26 7 Generalizations 28 7.1 If the Galois group is not pro-cyclic 28 7.2 Some open questions 29 References 30

1 Introduction

To understand definable sets up to definable bijections, it is helpful to have invariants of these sets. In the case of pseudo-finite fields, one well-known such invariant is the measure defined in [1]. By the “measure” of a definable set X, one should think of the size of the part of highest dimension of X. In [2], Hrushovski asks (in a more general context) whether this measure can be generalized to cases where the Galois group is not ℤ̂. More precisely, of the three conditions on pseudo-finite fields—pseudo algebraically closed (PAC), perfect, and the Galois group is ℤ̂—we only keep the first two, and the third one is weakened to the requirement that the Galois group is bounded, i.e. it has only a finite number of quotients of each cardinality. In such fields, one still has almost-quantifier-elimination as in pseudo-finite fields, so one can hope to control the definable sets. The question is whether one can find an appropriate definition of a “measure” on the definable sets of such fields.

Of course, such a measure can not have all the properties of the measure on pseudo-finite fields. (This was proven in [3].) In particular, one can not hope to have a “Fubini theorem” as in the case of pseudo-finite fields. Fubini states, more or less, that for a surjective map X ↠ Y with constant fiber size, the measure of X is equal to the measure of Y times the measure of a fiber. This is already false, for example, for the map K^{×} → K^{×}, x ↦ x^{2} if K is algebraically closed.

In this article, instead of requiring the measure to satisfy Fubini, we will just require that it is invariant under definable bijections. The main theorem (Theorem 15) states that if the Galois group is pro-cyclic, then such a measure indeed exists, and is—under one additional hypothesis—even unique. In the case of pseudo-finite fields our measure is just the usual measure of [1]. In the case of algebraically closed fields, it counts the irreducible components of the sets. Our definition of measure can be seen as an attempt to find a common generalization of these two extreme cases.

Note that uniqueness of the measure is a new result even in the case of pseudofinite fields: up to now, it was known only if one requires the measure to satisfy Fubini.

The main theorem only gives a measure if the Galois group is pro-cyclic. Indeed, if the Galois group is not pro-cyclic but still bounded, then (in general) no measure exists. In Subsection 7.1, we will give a counter-example which works for a lot of different non-pro-cyclic Galois groups.

So for fields, the question of Hrushovski is answered: a measure does exist if and only if the Galois group is cyclic—at least for measures in our sense. See Subsection 7.2 for further weakenings of the notion of measure.

1.1 Overview over the article

In Section 2 we will define all our notation and state some basic facts. In Section 3 we state the main theorem and give an idea of the proof. The proof itself is done in Section 6, after proving some lemmas in the previous section. In particular, we prove a variant of Chebotarevs density theorem on bounded Galois groups in Subsection 4.3. In the last section (Section 7) we give the counter-example when the Galois group is not pro-cyclic and we mention some open problems.

1.2 Thanks

This article emerged from the work group on definable measures that took place in Lyon. I want to thank the other participants, in particular Thomas Blossier, Amador Martin Pizarro, and Juan Pons Llopis for their contribution to that workgroup and for the plenty of helpful discussions.

I also want to thank the CNRS for financial support during the year 2005/2006.

2 Preliminaries

2.1 Setting and notation

In the whole article, K will be a perfect PAC field with bounded absolute Galois group. In some sections, we will additionally require that the Galois group is pro-cyclic.

We will denote the algebraic closure of K by K̃ and its absolute Galois group by Gal_{K} := Gal(K̃/K).

As the language of schemes is not really necessary in the present work, we will avoid it as much as possible. The reader can think of an algebraic set as a set defined by polynomials. Morphisms between algebraic sets will always be considered to be morphisms between the K̃-valued points of the sets. In particular, if we have a morphism f: V → W and a point w ∈ W (K) over K, then f^{−1}(w) denotes the set {v ∈ V (K̃) | f(v) = w} of all points over K̃ in the fiber, not only the ones over K. (In the language of schemes, this means: we always work with geometric points and we write W (K) for those geometric points which factor through K.)

All our definable sets will be definable in the language of rings and with parameters in K. Also, all our varieties, algebraic sets, and morphisms between such will be defined over K, with one exception: absolutely irreducible components of varieties will (of course) only be defined over K̃.

Sometimes, we will have to consider homomorphisms between groups, in particular from Gal_{K} to some finite group G. We will always want these homomorphisms to be continuous. Nevertheless, we will only write Hom(Gal_{K}, G) to keep the notation light. Think of it as working in the category of topological groups, where finite groups have the discrete topology.

We will need a notion of dimension. One possible definition of the dimension of a definable set X ⊂ K^{n} is to take for dim X the (usual algebraic) dimension of the Zariski closure of X in K̃^{n} (see [4]). As K is supersimple of rank 1, another equivalent definition is to take the SU-Rank of X.

As customary (see e.g. [5]), we will describe definable sets in terms of “Galois stratifications”. In the remainder of this section, we repeat the necessary definitions and basic facts. However, we have to work in a slightly more general context than usual: our Galois group need not be ℤ̂, so we will need a generalized version of the Artin symbol.

2.2 Galois covers

We use the following definition of Galois cover (note the slightly unusual requirement that W is absolutely irreducible).

Definition 1 A Galois cover is a morphism f: V ↠ W such that V is irreducible, W is absolutely irreducible, f is finite and étale and Aut_{W} (V) acts regularly on the fibers of f (i.e. for any two elements v_{1}, v_{2} in one fiber, there exists exactly one morphism in Aut_{W} (V) which maps v_{1} to v_{2}).

The group G := Aut_{W} (V)^{opp}, which acts from the right on V, is called the group of the Galois cover. (We will avoid to call G “Galois group” to avoid confusion with the Galois group of K.) We will write f: V ↠^{G} W to say that f is a Galois cover with group G.

As right actions are a bit difficult to read, we will always write them with a lower dot, i.e. for example “v.g” if g acts on v.

If V ↠^{G} W is a Galois cover, then we have canonically W ≅ V/G.

The requirement that Aut_{W} (V) acts freely follows from the irreducibility of V anyway. If V is not irreducible (and Aut_{W} (V) maybe does not act freely), then one can get a Galois cover by replacing V by one of its irreducible components.

Definition 2 For two Galois covers f: V ↠^{G} W and f′: V′ ↠^{G′} W of the same set W, we say that V′ ↠^{G′} W is a refinement of V ↠^{G} W if there is a finite étale map g: V′ ↠ V such that f′ = f ∘ g.

If V′ ↠^{G′} W is a refinement of V ↠^{G} W, then we also have a natural surjective map G′ ↠ G.

Definition 3 If f: V ↠^{G} W is a Galois cover, W′ ⊂ W an algebraic subset, and V′ ⊂ V an irreducible component of f^{−1}(W′), then f↾_{V′}: V′ ↠ W′ is again a Galois cover. We call it a restriction of f to W′.

Remark 4 Denote the group of the restricted cover V′ ↠ W′ by G′. If W′ is dense in W, then f^{−1}(W′) is irreducible and G′ = G. In general, we have G′ ↪ Aut_{W′}(f^{−1}(W′)) ↩ G.

Here are some more well-known facts about Galois covers which we will be using without further mentioning:

Fact 5 (1) If V ↠^{G} W is a Galois cover and H ⊂ G is a subgroup, then V ↠^{H} V/H is also a Galois cover. If H is normal, then V/H ↠^{G/H} W is a Galois cover, too.

(2) If f ∘ g is a Galois cover and both f and g are finite and étale, then g is a Galois cover, too.

(3) Given two Galois covers V_{1} ↠^{G_{1}} W and V_{2} ↠^{G_{2}} W of the same set, there exists a common refinement of both.

(4) If W is absolutely irreducible and f: V → W is a finite étale map, then there exists an algebraic set Ṽ and a finite étale map g: Ṽ → V such that f ∘ g is a Galois cover. (In particular, g is also a Galois cover.) If f is only finite and dominant, then this is still true after replacing W by some suitable dense open subset W′ of W and V by f^{−1}(W′).

2.3 The Artin symbol

Given a Galois cover V ↠^{G} W and an element w ∈ W (K), the usual Artin symbol of w is a conjugacy class of elements of G, which one gets as images of a generator of Gal_{K} under certain maps from Gal_{K} to G. When Gal_{K} is not ℤ̂, it is better to consider directly conjugacy classes of maps from Gal_{K} to G. Note that Hom(Gal_{K}, G) is a group (under pointwise multiplication), but by “conjugacy class in Hom(Gal_{K}, G)”, we will always mean classes under conjugation by elements of G (i.e. ρ and ρ′ are conjugate if ρ′ = ρ^{g} := Int(g)∘ρ for some g ∈ G).

Note also that in our setting, the group Hom(Gal_{K}, G) is always finite: by the boundedness of Gal_{K}, there are only finitely many quotients of Gal_{K} whose cardinality is less or equal to the cardinality of G and each homomorphism ρ ∈ Hom(Gal_{K}, G) factors through such a quotient. (And of course, there are only finitely many homomorphisms from that quotient to G.)

Another remark: When G_{1} is a subgroup of G_{2}, we will often identify Hom(Gal_{K}, G_{1}) with the corresponding subgroup of Hom(Gal_{K}, G_{2}).

Definition 6 Suppose f: V ↠^{G} W is a Galois cover and w ∈ W (K). Then we have a (left) action of Gal_{K} on the fiber f^{−1}(w) ⊂ V (K̃). Suppose v ∈ f^{−1}(w) lies in that fiber. By the regularity of the action of G on f^{−1}(w), there is a unique map ρ: Gal_{K} → G such that σv = v.ρ(σ) for all σ ∈ Gal_{K}. We call this map the Frobenius symbol of v and denote it by Fr(v).

For w ∈ W (K), we call the set {Fr(v) | v ∈ f^{−1}(w)} the Artin symbol of w and denote it by Ar(w).

Remark 7 It is easy to check that Fr(v) is a continuous group homomorphism from Gal_{K} to G and that Ar(w) is exactly one conjugacy class in Hom(Gal_{K}, G) (under conjugation by G).

Definition 8 Given a Galois cover f: V ↠^{G} W and a union C of conjugacy classes in Hom(Gal_{K}, G), we define the subset of W (K) consisting of those elements whose Artin symbol lies in C:

X(V ↠^{G} W, C) := {w ∈ W (K) | Ar(w) ⊂ C} .

Remark 9 Suppose that f: V ↠^{G} W and f′: V′ ↠^{G′} W′ are two Galois covers and that C ⊂ Hom(Gal_{K}, G) is closed under conjugation.

• If f′ is a refinement of f, then X(V ↠^{G} W, C) = X(V′ ↠^{G′} W′, C′), where C′ = {ρ ∈ Hom(Gal_{K}, G′) | π ∘ ρ ∈ C} and π is the natural map G′ ↠ G.

• If f′ is a restriction of f to W′, then X(V ↠^{G} W, C)∩W′ = X(V′ ↠^{G′} W′, C′), where C′ = C ∩G′ (here, both G and G′ are identified with the corresponding subgroup of Aut_{W′}(f^{−1}(W′))).

2.4 Galois stratifications

The main idea to get hold of definable sets is that given a definable set X ⊂ K^{n}, we can cut K^{n} into locally closed subsets such that on each of these subsets, X has the form X(V ↠^{G} W, C). To make this precise, we define Galois stratifications.

Definition 10 A Galois stratification A of a variety W consists of:

• a partition of W into finitely many absolutely irreducible locally closed subsets W_{i} ⊂ W (i ∈ I),

• for each i ∈ I, a Galois cover f_{i}: V_{i} ↠^{G_{i}} W_{i}, • for each i ∈ I, a union C_{i} of conjugacy classes of Hom(Gal_{K}, G_{i}).

We shall say that such a Galois stratification defines the following subset of W (K):

⋃_{i∈I} X(V_{i} ↠^{G_{i}} W_{i}, C_{i})

The data of a Galois stratification denoted by A will always be denoted by V_{i}, W_{i}, G_{i}, C_{i}, and analogously with primes for A′, A′′, etc. This will not always be explicitely mentioned.

Definition 11 Suppose A and A′ are two Galois stratifications. We say that A′ is a refinement of A, if:

• Each W_{i} is a union ⋃_{j∈J_{i}} W′_{j} for some J_{i} ⊂ I.

• For each i ∈ I and each j ∈ J_{i}, the Galois cover V′_{j} ↠^{G′_{j}} W′_{j} is a refinement of a restriction of the Galois cover V_{i} ↠^{G_{i}} W_{i} to the set W′_{j}.

• C′_{j} is constructed out of C_{i} as described in Remark 9, such that X(V′_{j} ↠^{G′_{j}}

W′_{j}, C′_{j}) = X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) ∩ W′_{j}.

By the third condition, A and A′ define the same set.

One important reason for Galois stratifications being handy to use is the following well-known lemma:

Lemma 12 If A and A′ are two Galois stratifications, then there exist refinements Ã and Ã′ of A resp. A′ which differ only in the sets C̃_{i} (resp. C̃′_{i}).

2.5 Definable sets and Galois stratifications

Lemma 13 For each definable set X ⊂ K^{n}, there exists a Galois stratification A of 𝔸^{n} which defines X.

This is well known if K is a pseudo-finite field. In the case of a perfect PAC field with bounded Galois group, the same proof works. We give a short sketch of it.

The following lemma is proven in [2] in a very general context (Corollary 1.16).

Lemma 14 Each definable subset X of K^{n} can be defined by a formula of the form ∃\overline{y} ϕ(\overline{x}, \overline{y}), where ϕ is quantifier-free and such that for each \overline{a} ∈ K^{n}, there are only finitely many \overline{b} ∈ K^{m} such that ϕ(\overline{a}, \overline{b}) holds.

Sketch of a proof of Lemma 13. One can easily eliminate negations in ϕ by replacing p(\overline{x}, \overline{y}) ≠ 0 by p(\overline{x}, \overline{y}) · z = 1. So we can suppose that ϕ defines an algebraic subset of 𝔸^{n+m}, which we will denote by V.

Let W be the closure of the image of V in 𝔸^{n} and write f for the map V → W. To prove the lemma, we proceed by induction on the dimension of W.

Suppose first that V is absolutely irreducible. Then W is also absolutely irreducible. After restricting to a dense open subset W′ of W and to V′ := f^{−1}(W′) ⊂ V we get a Galois cover f̃ = f ∘ g: Ṽ ↠^{G} W′ where g: Ṽ ↠^{H} V′ is also a Galois cover, with group H ⊂ G. We have

\begin{aligned}X ∩ W′ & = f(V′(K)) = f̃(\{ṽ ∈ Ṽ (K̃) | σ(g(ṽ)) = g(ṽ) ∀σ ∈ Gal_{K}\}) \\ & = f̃(\{ṽ ∈ Ṽ (K̃) | im Fr(g) ⊂ H\}) = X(Ṽ ↠^{G} W′, C) , \end{aligned}

where C := Hom(Gal_{K}, H)^{G}. By induction, we already have a Galois stratification defining X ∩ (W ∖ W′) = f(V ∖ V′). Together with X(Ṽ ↠^{G} W′, C), this gives a Galois stratification defining X.

Now suppose V is not absolutely irreducible. By the “decomposition-intersection procedure” described in [1], we can suppose that each irreducible component of V is already absolutely irreducible. Apply the above argument to each irreducible component of V. We get that X is the union of sets defined by Galois stratifications. By refining these Galois stratifications, we can suppose that they only differ in the sets C_{i}. Define one new Galois stratification by taking the union of the sets C_{i}. This Galois stratification defines X. □

Note that the converse of Lemma 13 is also true: Any set defined by a Galois stratification is definable in the usual sense. Indeed, to speak about the Artin symbol of an element w ∈ W (K) with respect to a given Galois cover V ↠^{G} W, it is enough to work in a finite extension of K (this uses the boundedness of the Galois group), and finite extensions of K are interpretable in K.

3 The main theorem

3.1 The statement

The main result of this article is the following:

Theorem 15 Let K be a perfect PAC field, with pro-cyclic Galois group. Then there is exactly one function μ (a “measure”) from the definable sets over K to ℚ satisfying:

(1) μ(K) = 1.

(2) If there is a definable bijection between X_{1} and X_{2}, then μ(X_{1}) = μ(X_{2}).

(3) Suppose X_{1} and X_{2} are disjoint definable sets. If dim X_{1} = dim X_{2}, then μ(X_{1} ∪ X_{2}) = μ(X_{1}) + μ(X_{2}); if dim X_{1} > dim X_{2}, then μ(X_{1} ∪ X_{2}) = μ(X_{1}).

(4) Suppose V ↠^{G} W is a Galois cover where V is absolutely irreducible and suppose C ⊂ Hom(Gal_{K}, G) is a conjugacy class. Then the measure of X(V ↠^{G} W, C) only depends on C and on G as an abstract group (and not on V, W, the map V → W and the action of G on V).

This measure additionally satisfies:

(6) μ(X) ≥ 0 for any X, and μ(X) = 0 if and only if X is empty.

(7) μ(X_{1} × X_{2}) = μ(X_{1}) · μ(X_{2})

Remark 16 Condition (4) is the “additional hypothesis” needed to get uniqueness mentioned in the introduction. It can be seen in the following way:

To get uniqueness, one has to fix the measure of absolutely irreducible varieties. This is done by Condition (4) applied to X(id: W ↠^{1} W, 1) for any absolutely irreducible W (together with Condition (1)). In this sense, Condition (4) is a generalization of fixing the measure of absolutely irreducible varieties.

We now give the main steps of the proofs. The details will be done later.

3.2 Sketch of the proof of existence

Condition (4) suggests to define a “measure” which associates a value μ_{G}(C) to each finite group G and each each conjugacy class C ⊂ Hom(Gal_{K}, G), and then to define μ(V ↠^{G} W, C) := μ_{G}(C). Indeed, to prove the existence, we will state a theorem similar to Theorem 15 about such a measure μ_{G}(C) on the groups (Theorem 21). In this manner, the whole proof is divided into one part in which the real work of finding a measure is done, but which is only group theoretical (Section 5), and one part which consists in transfering the result to definable sets (Subsection 6.1).

There is one technical complication to this: Condition (4) only treats the case when V is absolutely irreducible. If V is not absolutely irreducible, then one has to take into account the action of the Galois group on the absolutely irreducible components of V. This makes the measure on the groups somewhat uglier.

The definition of the measure we will give can be interpreted as follows: In the case of pseudo-finite fields, a measure exists. If the Galois group of K is a subgroup of ℤ̂, then in a certain sense the lanugage is a “simplification” of the language of pseudo-finite fields: when defining sets using Galois covers, then there are fewer possible Artin symbols. Using an embedding Gal_{K} ↪ ℤ̂, we will therefore be able to “pull back” the well-known measure on pseudo-finite fields to K.

3.3 Sketch of the proof of uniqueness

One idea would be to use the same approach as for the existence: Prove the uniqueness of a measure on groups and then transfer this result to definable sets. However, there are a some technical reasons which would make this proof unnecessarily complicated. In particular we would need a version of Condition (4) which also treats non-irreducible V. To avoid this, we will prove the uniqueness directly. However, while doing this, we will still have the above idea in mind.

4 Some useful lemmas

Before we really start with the proof of the main theorem, we need some lemmas. In this section, we require the Galois group of K only to be bounded (not necesarrily pro-cyclic).

4.1 Some fiber sizes

Lemma 17 Suppose f: V ↠^{G} W is a Galois cover and w ∈ W (K) has Artin symbol C := Ar(w) ⊂ Hom(Gal_{K}, G). Choose a ρ ∈ C. Then the number

# {v ∈ V (K̃) | f(v) = w ∧ Fr(v) = ρ}

of elements in the fiber of w with Frobenius symbol ρ is \frac{|G|}{|C|}.

Proof. For ρ ∈ C, write F (ρ) := {v ∈ V (K̃) | f(v) = w ∧ Fr(v) = ρ} for the part of the fiber with ρ as Frobenius symbol. We will show that for any two ρ, ρ′ ∈ C, F (ρ) and F (ρ′) have the same cardinality. As the whole preimage of w consists of |G| elements, the lemma follows.

Given ρ, ρ′ ∈ C, there exists a g ∈ G such that ρ′ = ρ^{g}. Suppose v ∈ F (ρ), i.e. σv = v.ρ(σ) for all σ ∈ Gal_{K}. Then σv.g^{−1} = v.ρ(σ)g^{−1} = v.g^{−1}ρ(σ)^{g} = v.g^{−1}ρ′(σ), i.e. v.g^{−1} ∈ F (ρ′). So g^{−1} maps F (ρ) to F (ρ′). By the same argument, g maps F (ρ′) to F (ρ), so we have a bijection. □

Lemma 18 Suppose we have the following diagram, where the maps f_{1} : V → W_{1} and f_{2} : V → W_{2} are Galois covers with groups G_{1} resp. G_{2}. Note that we have naturally G_{1} ⊂ G_{2} and Hom(Gal_{K}, G_{1}) ⊂ Hom(Gal_{K}, G_{2}).

\begin{matrix}V \\ W_{1} & W_{2} \end{matrix}f_{1} f_{2} ϕ

Suppose additionally that C_{1} ⊂ Hom(Gal_{K}, G_{1}) is a conjugacy class and set C_{2} := C_{1}^{G_{2}} ⊂ Hom(Gal_{K}, G_{2}). Then the image under ϕ of X_{1} := X(V ↠^{G_{1}} W_{1}, C_{1}) is X_{2} := X(V ↠^{G_{2}} W_{2}, C_{2}). In addition, ϕ restricts to a bijection X_{1} → X_{2} if and only if \frac{|G_{1}|}{|C_{1}|} = \frac{|G_{2}|}{|C_{2}|}.

Proof. “ϕ(X_{1}) ⊂ X_{2}”: Suppose w_{1} ∈ X_{1} and suppose v ∈ f_{1}^{−1}(w_{1}) is a preimage. Then Fr(v) ∈ C_{1}. But v is also a preimage of ϕ(w_{1}), so Ar(ϕ(w_{1})) contains Fr(v) and is therefore equal to C_{1}^{G_{2}}.

“ϕ(X_{1}) ⊃ X_{2}”: Suppose w_{2} ∈ X_{2}. As Ar(w_{2}) contains C_{1}, we may choose some preimage v ∈ V of w_{2} with Fr(v) ∈ C_{1}. In particular, im Fr(v) ⊂ G_{1}, which means that Gal_{K} fixes v.G_{1} ∈ V/G_{1} ≅ W_{1}. So the image f_{1}(v) lies in W_{1}(K). As Ar(f_{1}(v)) contains Fr(v), we have Ar(f_{1}(v)) = C_{1}, so f_{1}(v) is a preimage of w_{2}.

It remains to check that the fibers of ϕ↾_{X_{1}} have size one if and only if \frac{|G_{1}|}{|C_{1}|} = \frac{|G_{2}|}{|C_{2}|}. Indeed, we prove that the size of the fibers is \frac{|G_{2}|·|C_{1}|}{|C_{2}|·|G_{1}|}: Fix any element ρ ∈ C_{1}. For any w_{2} ∈ X_{2}, consider the set F_{2} := {v ∈ f_{2}^{−1}(w_{2}) | Fr(v) = ρ}. By Lemma 17, this set has \frac{|G_{2}|}{|C_{2}|} elements.

For any v ∈ F_{2}, f_{1}(v) lies in X_{1} ∩ ϕ^{−1}(w_{2}) (lying in X_{1} holds by the same argument as in “ϕ(X_{1}) ⊃ X_{2}”). So the sets F_{1}(w_{1}) := {v ∈ f_{1}^{−1}(w_{1}) | Fr(v) = ρ\}, w_{1} ∈ X_{1} ∩ ϕ^{−1}(w_{2}), form a partition of F_{2}. By Lemma 17, F_{1}(w_{1}) has \frac{|G_{1}|}{|C_{1}|} elements, so w_{2} has \frac{|G_{2}|·|C_{1}|}{|C_{2}|·|G_{1}|} preimages in X_{1}. □

4.2 A closer look on Galois covers

A Galois cover comes with some additional data which we will need several times. We now fix notation for that data once and for all. So suppose V ↠^{G} W is a Galois cover.

Suppose g ∈ G fixes some absolutely irreducible component V_{0} of V. Any other component V′_{0} can be written as V′_{0} = σV_{0} for some σ ∈ Gal_{K}, so V′_{0}.g = σV_{0}.g = σV_{0} = V′_{0}. That is, if g fixes any component, then it fixes all components. One also deduces that the set of g ∈ G fixing the components form a normal subgroup. We will write S for this subgroup, T := G/S for the quotient, and τ: G ↠ T for the canonical homomorphism. Note that T acts regularly on the set of components of V.

Now choose one absolutely irreducible component V_{0} of V. This yields a map η: Gal_{K} → T defined by V_{0}.η(σ) = σV_{0}. Using the fact that the actions of T and of Gal_{K} on the set of absolutely irreducible componenents of V commute, we get that η is a (continuous) group homomorphism. As Gal_{K} acts transitively on the set of components, η is surjective.

\begin{matrix}Gal & K \\ S & G & T \end{matrix}ητ

Note that another choice of V_{0} would yield a map Gal_{K} → T which is conjugate to η by some element of T.

In the remainder of this article, whenever we have a Galois cover denoted by V ↠^{G} W, then S, T, τ, η will always denote the objects described here, and analogously for V′ ↠^{G′} W′, V_{i} ↠^{G_{i}} W_{i}, etc. with primes resp. indices. This will not always be explicitely mentioned. (We will of course have to pay attention to the fact that η depends on the choice of V_{0}.)

4.3 Which of the X(V ↠^{G} W, C) are empty?

Let f: V ↠^{G} W be a Galois cover. To construct the measure, we need to know precisely for which conjugacy classes C ⊂ Hom(Gal_{K}, G) the sets X(V ↠^{G} W, C) are empty and for which they are not. This is what the following lemma states. It can be seen as a variant of Chebotarev’s density theorem. In fact, the idea of the proof was taken from the proof of Chebotarev’s density theorem in [5].

The second assertion of the lemma will be needed in the proof of the uniqueness of the measure. It is part of the same lemma because the proof is the same.

Lemma 19 Suppose V ↠^{G} W is a Galois cover and C ⊂ Hom(Gal_{K}, G) is a conjugacy class. Define S, T, τ, η as in section 4.2. Then X(V ↠^{G} W, C) is non-empty if and only if there is a ρ ∈ C such that τ ∘ ρ = η. In that case,

dim X(V ↠^{G} W, C) = dim W.

If the conjugacy class C consists of a single element ρ and τ ∘ ρ = η, then there exists another Galois cover V̂ ↠^{S} W where V̂ is absolutely irreducible and S is as above, and such that X(V ↠^{G} W, C) = X(V̂ ↠^{S} W, {1}).

Remark 20 Concerning the choice involved in η, note that the condition that C contains an element ρ satisfying τ ∘ ρ = η does not change if η is replaced by an element conjugate to it.

Proof. We start with the easy direction of the first statement: Suppose X(f: V ↠^{G} W, C) is not empty. Choose w ∈ X(V ↠^{G} W, C) and choose a preimage v ∈ f^{−1}(w) ∩ V_{0}, where V_{0} is the absolutely irreducible component of V which gave rise to η as in Subsection 4.2. The corresponding Frobenius symbol ρ := Fr(v) is an element of C. This ρ satisfies τ ∘ ρ = η: For any σ ∈ Gal_{K}, look at the component of V containing the left hand side resp. the right hand side of v.ρ(σ) = σv. The one on the left is V_{0}.ρ(σ) = V_{0}.τ(ρ(σ)), the one on the right is σV_{0} = V_{0}.η(σ). As T acts freely on the set of components, it follows that τ(ρ(σ)) = η(σ).

For the remainder of the lemma, the goal is to construct a surjective map f̂: V̂ ↠ W such that V̂ is absolutely irreducible and f̂(V̂ (K)) ⊂ X(V ↠^{G} W, C). As K is PAC, it follows that X(V ↠^{G} W, C) is not empty and has the right dimension. If |C| = 1, then this f̂ will be a Galois cover and we will have an equality X(V̂ ↠^{S} W, {1}) = f̂(V̂ (K)) = X(V ↠^{G} W, C).

Here is a sketch of the construction of V̂ and f̂:

\begin{aligned}& V′ & ⊂ V × U \\ V′/H =: V̂ & & V \\ & W \end{aligned}H G f̂ G′

(1) Define some Galois cover U ↠^{Q} {Pt} and consider the product V × U.

(2) Restrict the map V × U ↠ W to an irreducible component V′ of V × U. This yields a Galois cover with group G′ ⊂ G × Q.

(3) Find a set C′ ⊂ Hom(Gal_{K}, G′) such that X(V ↠^{G′} W, C′) = X(V ↠^{G} W, C).

(4) Choose a subgroup H ⊂ G′ such that V̂ := V′/H is absolutely irreducible. V′ ↠^{H} V̂ is a Galois cover. Choose a subset C′′ ⊂ Hom(Gal_{K}, H) such that

f̂(X(V′ ↠^{H} V̂ , C′′)) ⊂ X(V′ ↠^{G′} W, C′), with equality if |C| = 1.

(5) Check that X(V′ ↠^{H} V̂ , C′′) = V̂ (K).

(6) Check that, if |C| = 1, then f̂: V̂ → W is a Galois cover with group S.

Now let’s get to work:

Step 1: First choose ρ_{0} ∈ C such that τ ∘ ρ_{0} = η. (This exists by the hypothesis.)

Denote the image im ρ_{0} ⊂ G by Q and write π_{0} for the induced map Gal_{K} ↠ Q.

\begin{aligned}Gal(L/K) & & Gal_{K} \\ & Q \\ S & & G & T \end{aligned}ητ ρ_{0} π_{0} ⊂_{∼}_{\overline{π}_{0}}

Let L be the extension of K corresponding to the quotient Q of Gal_{K}. In particular, π_{0} factors over Gal(L/K) and the induced map \overline{π}_{0} : Gal(L/K) → Q is an isomorphism.

As K is perfect, L can be generated by a single element: L = K(u_{0}). Let f be the minimal polynomial of u_{0} over K and define U := V (f) to be the set of its zeros (as an algebraic set defined over K). One easily checks that the group of automorphisms of U is isomorphic to Q^{opp} and that the isomorphism can be choosen such that for any σ ∈ Gal_{K}, we have σu_{0} = u_{0}.π_{0}(σ), where the right action of π_{0}(σ) ∈ Q on U is the one induced by the isomorphism Q → Aut(U)^{opp}. It is also easy to check that this yields a Galois cover U ↠^{Q} {Pt}.

In other words, U := spec L as a scheme over K. As K is perfect, Aut(U)^{opp} = Gal(L/K) = Q and U ↠^{Q} {Pt} is a Galois cover. Let u_{0} be the point of U defined by the inclusion L ↪ K̃. Then we have σu_{0} = u_{0}.π_{0}(σ) for any σ ∈ Gal_{K}.

Step 2: Let V_{0} be the absolutely irreducible component of V as in Subsection 4.2. Then for any σ ∈ Gal_{K} we have

σV_{0} = V_{0}.η(σ) = V_{0}.ρ_{0}(σ) = V_{0}.π_{0}(σ) .  (1)

Let V′ be the irreducible component of V × U containing the absolutely irreducible component V_{0} × {u_{0}}. We have:

V′ = ⋃_{σ∈Gal_{K}} σ(V_{0} × {u_{0}}) = ⋃_{σ∈Gal_{K}} σV_{0} × σ{u_{0}}

=^{(1)} ⋃_{σ∈Gal_{K}} V_{0}.π_{0}(σ) × {u_{0}}.π_{0}(σ) = ⋃_{q∈Q} V_{0}.q × {u_{0}}.q .

(2)

In particular, the absolutely irreducible components of V′ are in bijection with U(K̃) and the diagonal action of Q on these components is regular.

The action of G on V and the action of Q on U induce an action of G × Q on V × U. This gives rise to a Galois cover V′ ↠^{G′} W, where G′ is the subgroup of G × Q fixing V′. We want to understand a bit more precisely how G′ looks like.

An element (g, q) ∈ G×Q fixes V′ if and only if it maps V_{0}×{u_{0}} to some other absolutely irreducible component of V′. By (2) these absolutely irreducible components are of the form V_{0}.q′ × {u_{0}}.q′ for some q′ ∈ Q. In other words, (g, q) ∈ G′ is equivalent to: there exists a q′ ∈ Q such that V_{0}.g = V_{0}.q′ and u_{0}.q = u_{0}.q′. As Q acts regularly on U(K̃), it follows that q = q′, so we finally get:

\begin{aligned}G′ & = \{(g, q) ∈ G × Q | V_{0}.g = V_{0}.q\} \\ & = \{(g, q) ∈ G × Q | τ(g) = τ(q)\} \end{aligned}  (3)

Step 3: We now want to find a union of conjugacy classes C′ ⊂ Hom(Gal_{K}, G′)

such that X(V ↠^{G} W, C) = X(V′ ↠^{G′} W, C′).

We have canonically

Hom(Gal_{K}, G′) ⊂ Hom(Gal_{K}, G × Q) ≅ Hom(Gal_{K}, G) × Hom(Gal_{K}, Q) .

Using these identifications, define C′ := (C × π_{0}^{Q}) ∩ Hom(Gal_{K}, G′). (C′ might consist of several conjugacy classes.)

Claim: X(V ↠^{G} W, C) = X(V′ ↠^{G′} W, C′).

Proof of the claim: For any w ∈ W (K), we can choose a preimage v ∈ V_{0}(K̃). This also yields a preimage (v, u_{0}) ∈ V′(K̃).

w is an element of the left hand side of the claim if and only if there exists a ρ ∈ C such that v.ρ(σ) = σv for all σ ∈ Gal_{K}. w is an element of the right hand side of the claim if and only if there exists a (ρ, π) ∈ C′ such that v.ρ(σ) = σv and u_{0}.π(σ) = σu_{0} for all σ ∈ Gal_{K}.

Now “⊃” is clear (using the same ρ).

“⊂”: We use the same ρ and we set π := π_{0}. Then we have v.ρ(σ) = σv and u_{0}.π(σ) = σu_{0}, and it remains to check that (ρ, π_{0}) is indeed an element of C′. The only thing which is not clear here is that it is an element of Hom(Gal_{K}, G′), i.e. that the image of (ρ, π_{0}) lies in G′. But (v, u_{0}).(ρ, π_{0})(σ) = σ(v, u_{0}), so (ρ, π_{0})(σ) maps the irreducible component V′ of V × U to itself, and therefore (ρ, π_{0})(σ) ∈ G′.

So the claim is proven.

Step 4: Define H := {(q, q) ∈ G × Q | q ∈ Q} to be the diagonal embedding of Q in G×Q. By (2), H is a subgroup of G′ (as it fixes V′) and acts regularly on the absolutely irreducible components of V′. It follows that V̂ := V′/H is absolutely irreducible. Denote the map from V̂ to W by f̂.

Obviously, im(ρ_{0}, π_{0}) ⊂ H, so we can set C′′ := (ρ_{0}, π_{0})^{H} ⊂ Hom(Gal_{K}, H). By Lemma 18, the image of X(V′ ↠^{H} V̂ , C′′) under f̂ is X(V′ ↠^{G′} W, (ρ_{0}, π_{0})^{G′}), which is included in X(V′ ↠^{G′} W, C′).

If |C| = 1, then the conjugation action of G on ρ_{0} is trivial, i.e. the image of ρ_{0} lies in the center of G. In particular, Q is abelian. It follows that C′ = (C ×π_{0}^{Q})∩Hom(Gal_{K}, G′) = {(ρ_{0}, π_{0})}, so in particular, f̂(X(V′ ↠^{H} V̂ , C′′)) =

X(V′ ↠^{G′} W, (ρ_{0}, π_{0})^{G′}) = X(V′ ↠^{G′} W, C′).

Step 5: We want to check that X(V′ ↠^{H} V̂ , C′′) is the whole of V̂ (K). Suppose w ∈ V̂ (K). The preimage of w in V′(K̃) has exactly one element (v, u) for each second coordinate u ∈ U(K̃), as the composition H ↪ G′ ↠ Q is an isomorphism and Q acts regularly on U. Suppose (v, u_{0}) is the preimage with second coordinate u_{0}. We have to check that σ(v, u_{0}) = (v, u_{0}).(ρ_{0}, π_{0})(σ). Both the left and the right hand side are preimages of w, so it is enough to compare their second coordinates. But indeed, σ(u_{0}) = u_{0}.π_{0}(σ).

Step 6: Now suppose |C| = 1. Then Q is abelian and H ⊂ Z(G) × Q, so H is a normal subgroup of G × Q and in particular of G′. Therefore, V̂ → W is a Galois cover with group G′/H. It remains to check that this group is isomorphic to S.

\begin{matrix}(g, q) & & gq^{−1} \\ G × Q & & G \\ ∪ & & ∪ \\ H ⊂ G′ & G′/H & S \end{matrix}ϕ

Consider the map ϕ: G′ → G, (g, q) ↦ gq^{−1}. The kernel of ϕ is exactly H (by definition of H), so we get an injective map from G′/H to G. By (3), τ(gq^{−1}) = 1, so the image of ϕ lies in the kernel of τ, which is S. On the other hand, given an element g ∈ S, we get (g, 1) ∈ G′ (again by (3)) and ϕ(g, 1) = g; so the image of ϕ is the whole S. Therefore, the map G′/H → S is an isomorphism. □

5 The measure on the groups

As mentioned in the sketch of the proof of the main theorem, to get the existence of the measure, we first define a measure on the groups. In order to do that, we first have to clarify what exactly we want to associate a measure to. Then we will state a theorem about the existence of such a measure, and finally, we will prove it.

In this section, Gal_{K} will just be an abstract profinite group for us. However, we still write Gal_{K} to avoid unnecessary confusion.

5.1 The theorem for groups

Fix a subgroup Gal_{K} of ℤ̂. Suppose we are given the following data:

• A finite group G.

• A quotient T of G. We will write τ for the canonical map G ↠ T and S for the kernel of τ.

• A continuous surjective map η: Gal_{K} ↠ T.

• A subset C ⊂ Hom(Gal_{K}, G) which is closed under conjugation by G.

It is to these data that we want to associate a measure. We will denote that measure by μ_{G,η}(C).

Theorem 21 Suppose Gal_{K} is a subgroup of ℤ̂. Then, there exists a map μ which associates to each G, T, η, C as above a rational number μ_{G,η}(C) ∈ ℚ satisfying the following conditions:

(1) If G, T and η are trivial and C = Hom(Gal_{K}, G) is the only existing morphism, then μ_{G,η}(C) = 1.

(2) If C_{1}, C_{2} ⊂ Hom(Gal_{K}, G) are disjoint, then μ_{G,η}(C_{1} ∪ C_{2}) = μ_{G,η}(C_{1}) + μ_{G,η}(C_{2}).

(3) Suppose we have the following commutative diagram of groups, where S resp. S′ is the kernel of τ resp. τ′:

\begin{aligned}S & ⊂ & G & T \\ Gal & K \\ S′ & ⊂ & G′ & T′ \end{aligned}τ τ′ η η′ π

Suppose further that C′ ⊂ Hom(Gal_{K}, G′) is a conjugacy class and that C := {ρ ∈ Hom(Gal_{K}, G) | π ∘ ρ ∈ C′} is its preimage. Then μ_{G,η}(C) = μ_{G′,η′}(C′).

(4) Suppose we have the following commutative diagram of groups, where S_{i} is the kernel of τ_{i}:

\begin{aligned}S_{1} & ⊂ & G_{1} \\ ∩ & & ∩ & T & Gal_{K} \\ S_{2} & ⊂ & G_{2} \end{aligned}τ_{1} τ_{2} η

Suppose further that we have a single conjugacy class C_{1} ⊂ Hom(Gal_{K}, G_{1}), that C_{2} := C_{1}^{G_{2}} is the induced conjugacy class in Hom(Gal_{K}, G_{2}), and that these classes satisfy \frac{|G_{1}|}{|C_{1}|} = \frac{|G_{2}|}{|C_{2}|}. Then μ_{G_{1},η}(C_{1}) = μ_{G_{2},η}(C_{2}).

(5) μ_{G,η}(C) ≥ 0 for any G, η and C, and μ_{G,η}(C) > 0 if and only if C contains an element ρ which is compatible with η, i.e. such that τ ∘ρ = η.

(6) μ_{G_{1}×G_{2},(η_{1},η_{2})}(C_{1} × C_{2}) = μ_{G_{1},η_{1}}(C_{1}) · μ_{G_{2},η_{2}}(C_{2}).

5.2 The proof for groups

Proof of Theorem 21. Fix any injection ι: Gal_{K} ↪ ℤ̂. For any finite group G, composition with ι defines a map from Hom(ℤ̂, G) to Hom(Gal_{K}, G). Denote this map by ϕ.

Suppose G, S, T, τ, η are given as in Subsection 5.1. Then first note that there is exactly one continuous map η̂ ∈ Hom(ℤ̂, T) such that η̂ ∘ ι = η. Indeed, ℤ̂ is the product of all ℤ_{p} and Gal_{K} = ∏_{p∈P} ℤ_{p} is a product of some of them. As T is a quotient of Gal_{K}, the prime factors of |T | lie in P. Therefore, the only map from ℤ_{p′} to T for any p′ ∉ P is the trivial one.

It follows that for any ρ̂ ∈ Hom(ℤ̂, G), we have τ ∘ ρ̂ = η̂ if and only if τ ∘ ϕ(ρ̂) = η.

Gal_{K}  ℤ̂_{G  T}

ι τ η η̂

Hom(Gal_{K}, G)  Hom(ℤ̂, G)
∪  ∪
{ρ | τ ∘ ρ = η} {ρ̂ | τ ∘ ρ̂ = η̂}

ϕ ϕ

Here is a definition of the measure:

μ_{G,η}(C) := \frac{#\{ρ̂ ∈ ϕ^{−1}(C) | τ ∘ ρ̂ = η̂\}}{|S|} = \frac{|ϕ^{−1}(\{ρ ∈ C | τ ∘ ρ = η\})|}{|S|} .

We check all the conditions required by the theorem.

Conditions (1) and (2) are clear.

Condition (3): We have the following commutative diagram:

\begin{aligned}S & ⊂ & G & T \\ & & & & Gal_{K} & ℤ̂ \\ S′ & ⊂ & G′ & T′ \end{aligned}τ τ′ η η′ η̂ η̂′ π

We also have a conjugacy class C′ ⊂ Hom(Gal_{K}, G′) and its preimage C := {ρ ∈ Hom(Gal_{K}, G) | π ∘ ρ ∈ C′}. Fix a generator σ_{0} of ℤ̂. Then maps from ℤ̂ to a finite group are in bijection to the elements of that finite group by taking the image of σ_{0}.

Define D := {ρ̂(σ_{0}) ∈ G | ρ̂ ∈ ϕ^{−1}(C)} and D′ := {ρ̂′(σ_{0}) ∈ G′ | ρ̂′ ∈ ϕ^{−1}(C′)}. Using this, what we have to check translates to

\frac{#\{g ∈ D | τ(g) = η̂(σ_{0})\}}{|S|} = \frac{#\{g′ ∈ D′ | τ(g′) = η̂′(σ_{0})\}}{|S′|} .  (4)

Choose g_{0} ∈ G such that τ(g_{0}) = η̂(σ_{0}). Then the sets in the numerators of (4) can be rewritten as D ∩g_{0}S resp. D′ ∩π(g_{0})S′. The homomorphism π restricts to a map g_{0}S ↠ π(g_{0})S′ with fiber size \frac{|S|}{|S′|}. So we are finished as soon as we have checked that D is the preimage of D′ under π. But this follows from the hypothesis that C is the preimage of C′: Applying ϕ^{−1} to C yields

ϕ^{−1}(C) = {ρ̂ ∈ Hom(ℤ̂, G) | π∘ρ̂∘ι ∈ C′} = {ρ̂ ∈ Hom(ℤ̂, G) | π∘ρ̂ ∈ ϕ^{−1}(C′)} .

From this, we get D = {g ∈ G | π(g) ∈ D′}.

Condition (4): We have the following commutative diagram of groups:

\begin{aligned}S_{1} & ⊂ & G_{1} & & Gal_{K} \\ ∩ & & ∩ & T \\ S_{1} & ⊂ & G_{2} & & & ℤ̂ \end{aligned}η̂ηι

We also have a single conjugacy class C_{1} ⊂ Hom(Gal_{K}, G_{1}) and the induced class C_{2} := C_{1}^{G_{2}} ⊂ Hom(Gal_{K}, G_{2}), and these conjugacy classes satisfy \frac{|G_{1}|}{|C_{1}|} = \frac{|G_{2}|}{|C_{2}|}. We want to check that μ_{G_{1},η}(C_{1}) = μ_{G_{2},η}(C_{2}).

To avoid confusion, we will write ϕ_{i} (instead of just ϕ) for the composition with ι from Hom(ℤ̂, G_{i}) to Hom(Gal_{K}, G_{i}).

C_{2} ⊂ Hom(Gal_{K}, G_{2}) Hom(ℤ̂, G_{2})
∪  ∪  ∪
C_{1} ⊂ Hom(Gal_{K}, G_{1}) Hom(ℤ̂, G_{1})

ϕ_{2} ϕ_{1}

We will check

\frac{#\{ρ ∈ C_{1} | τ_{1} ∘ ρ = η\}}{|C_{1}|} = \frac{#\{ρ ∈ C_{2} | τ_{2} ∘ ρ = η\}}{|C_{2}|}  (5)

and

\frac{|ϕ_{1}^{−1}(\{ρ ∈ C_{1} | τ_{1} ∘ ρ = η\})|}{#\{ρ ∈ C_{1} | τ_{1} ∘ ρ = η\}} = \frac{|ϕ_{2}^{−1}(\{ρ ∈ C_{2} | τ_{2} ∘ ρ = η\})|}{#\{ρ ∈ C_{2} | τ_{2} ∘ ρ = η\}} .  (6)

Together with \frac{|S_{2}|}{|S_{1}|} = \frac{|G_{2}|}{|G_{1}|} = \frac{|C_{2}|}{|C_{1}|}, this implies

\frac{|ϕ_{1}^{−1}(\{ρ ∈ C_{1} | τ_{1} ∘ ρ = η\})|}{|S_{1}|} = \frac{|ϕ_{2}^{−1}(\{ρ ∈ C_{2} | τ_{2} ∘ ρ = η\})|}{|S_{2}|} ,  (7)

which is what we have to show. (In (6), it is possible that a denominator is zero. However by (5) then both of them have to be zero and it is clear that in that case both numerators are zero, too, so (7) still holds.)

For the proof of (5), consider the image of C_{i} in Hom(Gal_{K}, T) under composition with τ_{i}. It is exactly one conjugacy class, and it does not depend on i, as (τ_{1}∘)(C_{1}) ⊂ (τ_{2}∘)(C_{2}). If this conjugacy class does not contain η, then both sides of (5) are zero, and there is nothing more to prove. So suppose now that (τ_{1}∘)(C_{1}) = (τ_{2}∘)(C_{2}) = η^{T}.

The fibers of the restricted composition map (τ_{i}∘)↾_{C_{i}} all have the same size, as they are conjugate. The set {ρ ∈ C_{i} | τ_{i} ∘ ρ = η} is one such fiber, so |C_{i}| = #{ρ ∈ C_{i} | τ_{i} ∘ ρ = η} · |η^{T} |. As this is true for both i, (5) follows.

For the proof of (6), consider the map ϕ_{2} restricted to ϕ_{2}^{−1}(C_{2}) → C_{2}. The fibers are conjugate by elements of G_{2} and therefore all have the same size, which we denote by k. So the right hand side of (6) is k.

Now consider ϕ_{2}^{−1}(C_{1}). A priori, this is a set of maps from ℤ̂ to G_{2}. However, we will check that the image of these maps lies in G_{1}. This means that ϕ_{2}^{−1}(C_{1}) = ϕ_{1}^{−1}(C_{1}) and it follows that the left hand side of (6) is k, too.

So suppose ρ̂ ∈ Hom(ℤ̂, G_{2}) is such that ρ := ϕ_{2}(ρ̂) ∈ C_{1}. As | C_{G_{1}}(im ρ)| = \frac{|G_{1}|}{|C_{1}|} = \frac{|G_{2}|}{|C_{2}|} = | C_{G_{2}}(im ρ)| and C_{G_{1}}(im ρ) ⊂ C_{G_{2}}(im ρ), we have C_{G_{1}}(im ρ) = C_{G_{2}}(im ρ), so in particular C_{G_{2}}(im ρ) ⊂ G_{1}. Now im ρ̂ is abelian (as ℤ̂ is abelian) and contains im ρ, so im ρ̂ ⊂ C_{G_{2}}(im ρ) ⊂ G_{1}.

Condition (5) is clear.

Condition (6) is straight forward to check. □

6 Proof of the main theorem

6.1 Existence

We know that there exists a measure on the groups as described by Theorem 21. Now we have to define a measure on the definable sets. We do this by defining the measure of a Galois stratification and then showing that this measure only depends on the set the Galois stratification defines.

Let A = (f_{i}: V_{i} ↠^{G_{i}} W_{i}, C_{i})_{i∈I} be a Galois stratification of K^{n} for some n, and let T_{i}, τ_{i}, η_{i} be as usual. Let d be the maximal dimension of those W_{i} where μ_{G_{i},η_{i}}(C_{i}) ≠ 0, or d := 0 if all μ_{G_{i},η_{i}}(C_{i}) = 0. Then we define the measure of A to be

μ(A) := ∑_{{i∈I|dim W_{i}=d}}μ_{G_{i},η_{i}}(C_{i}) .

Remember that η_{i} is only well defined up to conjugacy (Subsection 4.2), so for this definition to make sense, we have to check that μ_{G_{i},η_{i}}(C_{i}) does only depend on the conjugacy class of η_{i}. Indeed, if η′_{i} = η_{i}^{t} for some t ∈ T, then choose a preimage g ∈ τ_{i}^{−1}(t) and apply Condition (3) of Theorem 21 with G = G′ and π := Int g the corresponding internal automorphism.

Remark 22 By Condition (5) of Theorem 21, μ_{G_{i},η_{i}}(C_{i}) ≠ 0 if and only if there is a ρ ∈ C_{i} such that τ_{i} ∘ ρ = η_{i}. By Lemma 19, this is equivalent to X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) being non-empty. As in the latter case dim X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) = dim W_{i} (also by Lemma 19), it follows that d is the dimension of the set defined by A.

Lemma 23 If two Galois stratifications A and A′ define the same set, then their measures (as defined above) are equal.

Proof. By Lemma 12, there exist refinements Ã resp. Ã′ of A resp. A′ which only differ in the sets C̃_{i} resp. C̃′_{i}, and the refinements define the same sets as the originals. So it is enough to check that the measure of two stratifications A and A′ is equal in two cases: if A′ is a refinement of A and if A and A′ differ only in the C_{i}.

Suppose first that A = (f_{i}: V_{i} ↠^{G_{i}} W_{i}, C_{i})_{i∈I} and A′ = (f_{i}: V_{i} ↠^{G_{i}} W_{i}, C′_{i})_{i∈I}

differ only in the sets C_{i} and that both define the same set. We show that for all i ∈ I, μ_{G_{i},η_{i}}(C_{i}) = μ_{G_{i},η_{i}}(C′_{i}). For this, it is enough to check that for any conjugacy class C in Hom(Gal_{K}, G_{i}) with μ_{G_{i},η_{i}}(C) ≠ 0, either both or none of C_{i} and C′_{i} contain C.

If X(V_{i} ↠^{G_{i}} W_{i}, C) is not empty, then indeed both or none of C_{i} and C′_{i} contain C, as by assumtion, we have X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) = X(V_{i} ↠^{G_{i}} W_{i}, C′_{i}). If on the other hand X(V_{i} ↠^{G_{i}} W_{i}, C) is empty, then by Lemma 19, C does not contain any element ρ such that τ_{i} ∘ ρ = η_{i}. By Condition (5) of Theorem 21 it follows that μ_{G_{i},η_{i}}(C) = 0.

Now suppose A′ is a refinement of A. We have to check that

∑_{{i∈I|dim W_{i}=d}}μ_{G_{i},η_{i}}(C_{i}) = ∑_{{i∈I|dim W′_{i}=d′}}μ_{G′_{i},η′_{i}}(C′_{i}) ,

where d resp. d′ is the dimension of the set defined by A resp. A′. As A and A′ define the same set, we have d = d′.

For i ∈ I, write J_{i} of the set of j ∈ I′ such that W′_{j} ⊂ W_{i} and set

μ_{i} := ∑_{{j∈J_{i}|dim W′_{j}=d}}μ_{G′_{j},η′_{j}}(C′_{j}) .

The assertion follows from the following claim: If dim W_{i} < d, then μ_{i} = 0. Otherwise, μ_{i} = μ_{G_{i},η_{i}}(C_{i}).

If dim W_{i} < d, then dim W′_{j} < d for all j ∈ J_{i}, so the sum is empty and the claim is true. If dim W_{i} = d, then (as W_{i} is irreducible), there exists exactly one W′_{j} (j ∈ J_{i}) such that dim W′_{j} = d. So it remains to check that μ_{G′_{j},η′_{j}}(C′_{j}) = μ_{G_{i},η_{i}}(C_{i}) for this specific j.

By definition of refinement of a Galois stratification, the Galois cover V′_{j} ↠^{G′_{j}} W′_{j} is a refinement of a restriction of V_{i} ↠^{G_{i}} W_{i} to W′_{j}. As W′_{j} is dense in W_{i}, this restriction is f_{i}↾_{V} : V ↠^{G_{i}} W′_{j} where V := f_{i}^{−1}(W′_{j}) ⊂ V_{i}. This means we have the following diagram:

V′_{j}

V

W′_{j}

G′_{j} G_{i}

On the level of groups, we get:

\begin{aligned}S′_{j} & ⊂ & G′_{j} & T′_{j} \\ Gal & K \\ S_{i} & ⊂ & G_{i} & T_{i} \end{aligned}τ_{i} τ′_{j} η_{i} η′_{j} π

Using C′_{j} = {ρ ∈ Hom(Gal_{K}, G′_{j}) | π ∘ρ ∈ C_{i}} (also by definition of refinement of a Galois stratification) we get μ_{G′_{j},η′_{j}}(C′_{j}) = μ_{G_{i},η_{i}}(C_{i}) by Condition (3) of Theorem 21. □

We proved that the measure of a Galois stratification only depends on the set it defines, so we can define: For any definable set X, choose a Galois stratification A defining X and set μ(X) := μ(A).

Now we have to check that this measures satisfies all the conditions stated in Theorem 15. Before we start, note that if a definable set X has can be defined by a single Galois cover, i.e. X = X(V ↠^{G} W, C), then its measure is μ_{G,η}(C) (as expected). Indeed, this is clear if we take a stratification which has W as one of its W_{i}.

Conditions (1), (3) and (7) are straight forward. (For Condition (3), one chooses Galois stratifications defining the sets X, X′ and X′′ := X ˙∪ X′ which only differ in the sets C_{i} and such that C′′_{i} = C_{i} ˙∪ C′_{i}.)

Condition (2): Suppose we have two definable sets X_{1} ⊂ K^{n_{1}} and X_{2} ⊂ K^{n_{2}} and a definable bijection ϕ between them. We have to check that X_{1} and X_{2} have the same measure. We will do this by simplifying the situation more and more until we can apply Lemma 18 and Condition (4) of Theorem 21.

The projections K^{n_{1}+n_{2}} ↠ K^{n_{i}} induce definable bijections between the graph of ϕ and X_{i}. So it is enough to check that the measure is invariant under definable bijections X → X′ which are induced by the projection π: K^{n} ↠ K^{n′}, where X ⊂ K^{n}, X′ ⊂ K^{n′} and n ≥ n′. (In fact, we are not so much interested in the fact that π is a projection; we only need that it is an algebraic map.)

Now choose a Galois stratification A defining X. Then π↾_{X} decomposes into definable bijections X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) → π(X(V_{i} ↠^{G_{i}} W_{i}, C_{i})) (i ∈ I). So it is enough to check the condition when X is of the form X(f: V ↠^{G} W, C). (Note that we already checked that the measure is compatible with disjoint union.)

Now choose a Galois stratification A′ defining X′. Again, we can decompose π↾_{X}, this time into X ∩ π^{−1}(W′_{i}) → X(V′_{i} ↠^{G′_{i}} W′_{i}, C′_{i}). Indeed, an element of

X∩π^{−1}(W′_{i}) is mapped to X′∩W′_{i}, which is precisely X(V′_{i} ↠^{G′_{i}} W′_{i}, C′_{i}). For any i ∈ I′, the set X ∩π^{−1}(W′_{i}) is defined by a Galois cover which is a restriction of X(f: V ↠^{G} W, C) to W ∩π^{−1}(W′_{i}). So it is enough to check the condition when X and X′ both are defined by single Galois covers: X = X(f: V ↠^{G} W, C) and X′ = X(f′: V′ ↠^{G′} W′, C′). Write ϕ for the map π restricted to W → W′.

We claim that we can suppose V = V′. Indeed: X′ is contained in im ϕ and is Zariski dense in W′, so ϕ is dominant. As dim W = dim X = dim X′ = dim W′, ϕ is finite on a dense open subset of W′. By replacing W′ by that dense open subset (and V′, W and V by the corresponding preimages), we can suppose that ϕ is finite everywhere. (We are allowed to do that because removing parts of lower dimension doesn’t change the measure.) So ϕ∘f is finite and dominant, too, and after shrinking W′ a second time, we can extend this composition to a Galois cover V̂ ↠ W′.

Ṽ ; V̂ ; V  V′ ; W  W′f f′ ϕ

Now we take a common refinement Ṽ ↠ W′ of the Galois covers V̂ ↠ W′ and V′ ↠^{G′} W′. As Ṽ ↠ W′ is a Galois cover, so is Ṽ ↠ W. Replace V and V′ by Ṽ. (X and X′ can be defined using the Galois covers Ṽ ↠ W and Ṽ ↠ W′, as these are refinements of V ↠^{G} W resp. V′ ↠^{G′} W′.) So we can

suppose X = X(V ↠^{G} W, C) and X′ = X(V ↠^{G′} W′, C′) and G ⊂ G′.

We decompose ϕ↾_{X} one last time: into ϕ↾_{X(Ṽ ↠^{G̃}W,D)} where D ⊂ C ranges over the conjugacy classes in C. By Lemma 18, the image ϕ(X(V ↠^{G} W, D)) is X(V ↠^{G′} W′, D^{G′}). So we can suppose that C consists of a single conjugacy class and that C′ = C^{G′} is the induced class.

As ϕ↾_{X} is a bijection, we can apply the second statement of Lemma 18 and get \frac{|G|}{|C|} = \frac{|G′|}{|C′|}. One esily verifies that on the level of groups, we have the diagram of Condition (4) of Theorem 21. (T = T′ follows from the fact that W is absolutely irreducible.) Therefore, we can apply that condition and get μ(X) = μ_{G,η}(C) = μ_{G′,η′}(C′) = μ(X′).

Condition (4): As already noted, a set of the form X(V ↠^{G} W, C) has measure μ_{G,η}(C). If V is absolutely irreducible, then T and η are trivial, so μ(X(V ↠^{G} W, C)) = μ_{G,1}(C) only depends on G and C.

Condition (6): μ(X) ≥ 0 is clear and μ(∅) = 0, too. Now suppose μ(X) = 0. Choose any Galois stratification A defining X. All μ_{G_{i},η_{i}}(C_{i}) have to be zero. By Condition (5) of Theorem 21, this implies that C_{i} contains no ρ such that τ_{i} ∘ ρ = η_{i}. By Lemma 19, it follows that X(V_{i} ↠^{G_{i}} W_{i}, C_{i}) is empty; so X is empty.

6.2 Uniqueness

Suppose μ is any function satifying the conditions of the theorem.

First note that by Lemma 19, given a fixed Galois cover V ↠^{G} W, all sets X(V ↠^{G} W, C) (for some C ⊂ Hom(Gal_{K}, G) closed under conjugation) which are not empty have the same dimension. It follows that if C_{1} and C_{2} are

disjoint, then μ(X(V ↠^{G} W, C_{1} ∪ C_{2})) = μ(X(V ↠^{G} W, C_{1})) + μ(X(V ↠^{G}

W, C_{2})).

Each definable set can be written as disjoint union of sets of the form X(f: V ↠^{G} W, C). We can further decompose these sets so that each C is a single conjugacy class. Using Condition (3) of the theorem, it is therefore enough to prove the uniqueness for such sets. We do this by induction on the size of the group G.

Suppose first |G| = 1. Then C = Hom(Gal_{K}, G) is the only existing conjugacy class and f: V → W is an isomorphism. In particular, V is absolutely irreducible (as W is), so (4) applies: μ(X(V ↠^{G} W, C)) is equal to μ(X(id: K ↠^{1} K, 1)). But X(id: K ↠^{1} K, 1) = K, so its measure is 1 by Condition (1).

Now suppose |G| > 1. We first treat the case where |C| > 1. Choose ρ ∈ C and let G′ := C_{G}(im ρ) be the centralizer of ρ in G. Define W′ := V/G′. Then, V ↠^{G′} W′ is a Galois cover and we have the following diagram:

\begin{matrix}V \\ W′ & W \end{matrix}G′ G ϕ

As Gal_{K} is cyclic, so is im ρ. Therefore, C_{G}(im ρ) contains im ρ. So we have ρ ∈ Hom(Gal_{K}, G′), and C′ := {ρ} is a conjugacy class in Hom(Gal_{K}, G′). In addition, \frac{|G|}{|C|} = | C_{G}(im ρ)| = \frac{|G′|}{|C′|}, so we are precisely in the situation of Lemma 18. It follows that ϕ restricts to a bijection X(V ↠^{G′} W′, C′) ⟶^{1:1}

X(V ↠^{G} W, C).

By the assumption |C| > 1, we have |G′| < |G|, so we can apply the induction hypothesis: We know the measure of X(V ↠^{G′} W′, C′). By Condition (2) of Theorem 15, X(V ↠^{G} W, C) has the same measure.

We are left with the case |C| = 1. Here, we can apply the second part of Lemma 19. We get another Galois cover V̂ ↠^{S} W such that X(V ↠^{G} W, C) = X(V̂ ↠^{S} W, {1}). If S ⊊ G, we know the measure of X(V̂ ↠^{S} W, {1}) by induction hypothesis, so we can suppose S = G.

Let Z ⊂ Hom(Gal_{K}, G) be the union of all conjugacy classes of size one (Z = Hom(Gal_{K}, Z(G))). We already know the measure of X(V ↠^{G} W, Hom(Gal_{K}, G)∖ Z). We also know μ(X(V ↠^{G} W, Hom(Gal_{K}, G))) = μ(W (K)) = μ(X(W ↠^{1} W, 1)) = 1. Therefore, we know X(V ↠^{G} W, Z).

Now, for any conjugacy class C ⊂ Z, using Lemma 19 and Condition (4) (S = G implies V absolutely irreducible) we get μ(X(V ↠^{G} W, C)) = μ(X(V̂ ↠^{S} W, {1})) = μ(X(V ↠^{G} W, {1})). In other words, μ(X(V ↠^{G} W, C)) is the same for any conjugacy class C ⊂ Z, so we get μ(X(V ↠^{G} W, C)) = \frac{1}{|Z|}μ(X(V ↠^{G} W, Z)).

7 Generalizations

7.1 If the Galois group is not pro-cyclic

The main theorem proves the existence of a measure if the Galois group is pro-cyclic. In this section, we show that one cannot hope for much more. We give an example which shows that no measure exists for many non-pro-cyclic bounded Galois groups.

Theorem 24 Suppose K is a perfect PAC field and suppose Gal_{K} = P_{1} ∗⋯∗ P_{m} is a free product of finitely many (copies of) subgroups P_{i} of ℤ̂. Suppose further that there exists a prime p such that at least two of the groups P_{i} contain ℤ_{p} as a subgroup. Then there does not exist any function μ from the definable sets over K to ℚ satisfying Conditions (1) to (4) of theorem 15.

Remark 25 The prerequisites are in particular satisfied if Gal_{K} is a free product of more that one copy of ℤ̂.

Proof of Theorem 24. Let the groups P_{i} and the prime p be as in the theorem. Suppose G is any finite p-group and consider the maps from P_{i} (for some fixed i) to G. If P_{i} does contain ℤ_{p} as a subgroup, then these maps are in bijection to the elements of G (by taking the image of some generator of P_{i}); otherwise, there is only the trivial map from P_{i} to G. In the following all our finite groups will be p-groups, so we will fix some generators of the groups P_{i} and identify Hom(Gal_{K}, G) with G^{n}, where n is the number of indices i such that P_{i} contains ℤ_{p}.

Suppose a measure μ does exist. We will construct some definable sets which provide a contradiction.

We will have to deal with Galois covers V ↠^{G} W where V is absolutely irreducible and G ≅ (ℤ/pℤ)^{2}. So suppose we have such a Galois cover and suppose ρ ∈ Hom(Gal_{K}, G) is any homomorphism. As G is abelian, the arguments from the proof of the uniqueness of the main theorem concerning one-element conjugacy classes yield

μ(X(V ↠^{G} W, \{ρ\})) = \frac{1}{| Hom(Gal_{K}, G)|} = p^{−2n} =: μ_{0} ,

which in particular is greater than zero.

We will now construct definable sets X = X_{1} ˙∪ … ˙∪ X_{p^{n−1}} with

μ(X) = μ(X_{1}) = ⋯ = μ(X_{p^{n−1}}) = μ_{0} ,

which (as p^{n−1} > 1) will be a contradiction to Condition (3).

Let G be the group defined by the following generators and relations:

G = ⟨g_{1}, g_{2}, z | g_{1}^{p} = g_{2}^{p} = z^{p} = 1, g_{1}z = zg_{1}, g_{2}z = zg_{2}, g_{1}g_{2} = g_{2}g_{1}z⟩ .

Its center Z := Z(G) is ⟨z⟩ ≅ ℤ/pℤ, and the quotient Ĝ := G/Z is isomorphic to (ℤ/pℤ)^{2}. Write π for the canonical map G ↠ Ĝ. The centralizer of g_{1} is G′ := C_{G}(g_{1}) = ⟨g_{1}, z⟩ ≅ (ℤ/pℤ)^{2}.

Now suppose V ↠^{G} W is a Galois cover with this group and that V is absolutely irreducible. (It is well known how to construct such a Galois cover for any finite group G.) Define W′ := V/G′ and V̂ := V/Z. We get Galois covers V ↠^{G′} W′

and V̂ ↠^{Ĝ} W.

\begin{matrix}V & V̂ \\ W′ & W \end{matrix}G Ĝ ≅ (ℤ/pℤ)^{2} (ℤ/pℤ)^{2} ≅ G′

Choose ρ̂ := (π(g_{1}), 1, 1, …, 1) ∈ Ĝ^{n} ≅ Hom(Gal_{K}, Ĝ) and consider the set X := X(V̂ ↠^{Ĝ} W, {ρ̂}). It has measure μ_{0}.

This set can also be described using the Galois cover V ↠^{G} W: X := X(V ↠^{G} W, C), where C = {ρ ∈ G^{n} | π(ρ) = ρ̂} = (g_{1}Z)×Z×⋯×Z. The G-conjucacy classes in C are of the form (g_{1}Z) × {(z_{2}, … z_{n})} for any z_{2}, …, z_{n} ∈ Z. Denote the elements of the form (g_{1}, z_{2}, … z_{n}) by ρ_{i} (i ∈ I := Z^{n−1}) and the induced conjugacy classes by C_{i} := ρ_{i}^{G}. X is the disjoint union of the sets

X_{i} := X(V ↠^{G} W, C_{i}).

As ρ_{i} ∈ G^{′n}, we may consider the sets X′_{i} := X(V ↠^{G′} W′, {ρ_{i}}). Because of \frac{|G|}{|C_{i}|} = p^{2} = \frac{|G′|}{|\{ρ_{i}\}|}, Lemma 18 applies and we get a bijection X′_{i} → X_{i}. So μ(X_{i}) = μ(X′_{i}) = μ_{0}. This terminates the contradiction. □

Note that this counter-example does not depend very much on the target ring of the measure. In particular, it still works when ℚ is replaced by any ring of characteristic zero.

7.2 Some open questions

• Is Condition (4) of the main theorem really necessary for uniqueness? It would be nice to replace it by a condition which only requires the measure of absolutely irreducible varieties to be 1. Indeed, the only place where this condition is really used is in the proof that for ρ ∈ Hom(Gal_{K}, Z(G)) ⊂ Hom(Gal_{K}, G), the measure of X(V ↠^{G} W, {ρ}) does not depend on ρ.

• Is Condition (4) of the main theorem necessary to construct the counterexample of Subsection 7.1? To provide a counter-example without using that condition, it would be necessary to construct some Galois covers more explicitely.

• We proved that for non-procyclic Galois groups, no measure in the sense of the main theorem can exist. However, one might still hope to be able to define some kind of invariant of definable sets which is not additive.

Note however the following: If one still describes definable sets as unions of sets of the form X(V ↠^{G} W, C), one will need some kind of relation between the measure of a union and the components. However, the counter-example of Subsection 7.1 easily adapts to prove that no sensible invariant exists if there is any function f such that μ(X_{1} ˙∪ X_{2}) = f(μ(X_{1}), μ(X_{2})) for any definable sets X_{1} and X_{2} of the same dimension.

References

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