To appear in the Proc. of Orsatty Conference, Padova, Italy, 1997, Abelian groups, Modules and Topology, Lecture Notes in Pure and Appl. Algebra, Marcel Dekker.

G-Minimal Topological Groups

Michael Megrelishvili Department of Mathematics and Computer Science Bar-Ilan University, 52 900 Ramat-Gan, Israel

Abstract. We study sufficient conditions for G-minimality of topological G-groups X and minimality of semidirect products X ⋋G arising under some natural constructions.

Introduction.

A Hausdorff topological group X is called minimal [S] if it does not admit a strictly coarser Hausdorff group topology. The development of the theory of minimal topological groups is described in the monograph [DPS] and in several survey articles [CHR], [D2]. Here we discuss only facts that directly concern the present paper. It was shown by Stephenson [S] that every locally compact abelian minimal group must be compact. Nevertheless, every locally compact abelian Hausdorff group is a group retract of a locally compact minimal group (see [M]). The first example of a locally compact (noncompact) minimal group was found by Dierolf and Schwanengel [DS]. They proved that the semidirect product ℝ⋋ℝ_{+} of the multiplicative group ℝ_{+} of all positive real numbers with ℝ is minimal. Later Remus and Stoyanov [RS] established the minimality of ℝ^{n} ⋋ H for every closed subgroup H of GL(n, ℝ) which contains all diagonal matrices with positive entries.

In the present article we generalize these results in several directions. In section 1 we show that for every normed space X there exists an abelian subgroup H of GL(X) such that X ⋋ H is minimal. This result helps us to establish that there exists one and only one nontrivial group topology on X which preserves the continuity of the action of GL(X) on X. At the same time, the question about minimality of X ⋋ GL(X) is still open even for Hilbert space X = ℓ_{2}.

In section 2 we prove that if a locally compact abelian Hausdorff group X contains a copy of the circle group 𝕋 = ℝ/ℤ, then X ⋋ Aut(X) is minimal.

Typeset by AMS-TEX

Section 3 is devoted to the following general question:

Question. For which topological fields K and subgroups H of GL(n, K) are the groups K^{n} ⋋ H minimal? In particular, when is K ⋋ K^{×} minimal (here K^{×} = K \ {0})?

This question leads us to the important class of locally retrobounded division rings introduced by Nachbin [N].

In the last section, resolving a question of Dikranjan [D2], we construct a totally disconnected minimal group which is not zero-dimensional. This answers negatively a question posed earlier by Archangel’skij: does a totally disconnected group always admit a coarser zero-dimensional Hausdorff group topology (see [D1], [D2])? Our construction uses Erdös’ classical example of all rational points in ℓ_{2}.

The author thanks D. Dikranjan and V. Pestov for stimulating discussions.

§1. Preliminaries.

Let (X, τ) and (G, σ) be topological groups and

α : G × X → X, α(g, x) = gx

be a fixed action. We say that X is a G-group if α is continuous and every g− transition α^{g} : X → X, α^{g}(x) = gx is a group automorphism of X. For every G-group X denote by X ⋋_{α} G the standard semidirect product [Bo]. As usual, its normal subgorup X ×{e_{G}} will be identified with X and, analogously, the subgroup {e_{X}} × G will be identified with G.

Definition 1.1. Let (G, σ) be a Hausdorff topological group and let (X, τ) be a Hausdorff (G, σ)-group with respect to an action α.

(a) [RS] X is a G-minimal group if there is no strictly coarser Hausdorff group topology τ′ ⊆ τ on X which preserves the continuity (actually, the (σ, τ′, τ′)- continuity) of α.

(b) X is strongly G-minimal if X ⋋_{α} G is a minimal group.

(c) (see [M, Def. 1.8]) The action α is topologically exact (t-exact, in short) if there is no strictly coarser group (not necessarily Hausdorff) topology σ′ ⊆ σ such that α is (σ′, τ, τ)-continuous. α is hereditarily t-exact (ht -exact) if for every topological subgroup P of G, the corresponding action of P on X is t-exact.

Note that, since in the definition (G, σ) is Hausdorff, every t-exact action α is necessarily algebraically exact, that is, the normal subgroup Ker(α) = {g ∈ G|gx = x ∀x ∈ X} is trivial.

Lemma 1.2. Let (X ⋋_{α} G, γ) be a semidirect product of Hausdorff topological groups (X, τ) and (G, σ). Suppose that a minimal group Y is a topological subgroup of X ⋋_{α} G.

(i) If X ⊆ Y then X is G-minimal.

(ii) If X is strongly G-minimal then X is G-minimal.

(iii) If G ⊆ Y and α is algebraically exact then α is t-exact.

Proof. (i) Let τ′ ⊆ τ be a coarser Hausdorff group topology on X such that the action α is (σ, τ′, τ′)-continuous. Then the semidirect product (X, τ′) ⋋ (G, σ) is well-defined. Denote the corresponding product Hausdorff group topology by γ′. Since γ′ ⊆ γ and Y is minimal then γ′|_{Y} = γ|_{Y} . Therefore, γ′|_{X} = γ|_{X}. On the other hand, γ′|_{X} = τ′ and τ = γ|_{X}. Thus, we obtain τ′ = τ.

(ii) Directly follows from (i).

(iii) Let σ′ ⊆ σ be a strictly coarser group topology on G such that α is (σ′, τ, τ)-continuous. Since α is algebraically exact then necessarily σ′ is Hausdorff. The rest of the proof, via the semidirect product (G, α′) ⋋_{α} (X, τ), is similar to the case of (i). □

Fact 1.3. [M, Corollary 2.8] Let (X ⋋_{α} G, γ) be a topological semidirect product and let α be t-exact. Suppose that X is abelian and γ′ ⊆ γ is a group topology which agrees with γ on X. Then γ′ = γ.

Theorem 1.4. Let (G, σ) be a Hausdorff group and let (X, τ) be a Hausdorff abelian G-group. If X is G-minimal and the action α is t-exact then X ⋋ G is minimal (i.e., X is strongly G-minimal).

Proof. Denote by γ the original product topology of (X, τ) ⋋_{α} (G, σ). Assume that γ′ ⊆ γ is a coarser Hausdorff group topology on X ⋋_{α} G. Then the map

α : (G, γ′|_{G}) × (X, γ′|_{X}) → (X, γ′|_{X})

is continuous (see [Bo, Ch.III, §2]). Clearly, γ′|_{G} ⊆ γ|_{G} = σ and γ′|_{X} = γ|_{X}. By Fact 1.3 we get γ′ = γ. Therefore X ⋋_{α} G is minimal, as required. □

The last result is not true in general for non-abelian X. It was shown in [EDS, Example 10] that there exists a totally minimal precompact torsion group X such that a certain semidirect product X ⋋ G is not minimal, where G = ℤ_{2} is the discrete cyclic group of order 2.

A topological group G is called perfectly minimal [St, p.107] if G×H is minimal for every minimal group H. Lemma 1.5. Let G be a minimal group. Then G will be perfectly minimal in each of the following cases:

(a) [EDS] G is complete in its two-sided uniformity (e.g., compact);

(b) [M, Th.1.14] The center Z(G) is perfectly minimal (e.g., compact);

(c) Algebraically, G is a semidirect product X ⋋_{α} H such that the subgroups Ker(α) and Coker(α) = {x ∈ X | gx = x ∀g ∈ H} are compact (e.g., trivial).

Proof. Z(X ⋋_{α} H) = Coker(α) × Ker(α) generally holds. Therefore, (c) directly follows from (b). □ Fact 1.6. [M, Th.1.15] The arbitrary product ∏ X_{i} of minimal groups X_{i} with trivial center is perfectly minimal.

§2. GL-minimality of normed spaces.

Let E be a real normed space and let α : GL(E) × E → E be the usual continuous action of GL(E) on E, where GL(E) denotes the topological group of all topological automorphisms of E endowed with the uniform operator topology. The standard dual normed space of E will be denoted by E^{∗}. We need a method from [M] which enables us to construct a minimal group which is denoted by M_{+}(E) and called induced group of the canonical bilinear form E ×E^{∗} → ℝ, (x, f) ↦ f(x). Summarizing for convenience some definitions of [M], recall that M_{+}(E) is the group

M_{+}(R) = ((ℝ × E) ⋋_{ν} E^{∗}) ⋋_{π} ℝ_{+}

where the actions ν and π are defined as follows:

ν : E^{∗} × (ℝ × E) → ℝ × E, ν(f, (a, x)) = (a + f(x), x)
π : ℝ_{+} × ((ℝ × E) ⋋_{ν} E^{∗}) → (ℝ × E) ⋋_{ν} E^{∗}, π(t, (a, x, f) = (ta, tx, f).

Fact 2.1. [M, Theorem 3.10] For every normed space E the induced group M_{+}(E) is perfectly minimal.

Theorem 2.2. For every normed space E there exists an abelian subgroup H of GL(E) such that the corresponding semidirect product E ⋋ H is perfectly minimal. Hence E is GL(E)-minimal.

Proof. The conclusion part follows from Lemma 1.2(i). The proof of the main part of our assertion is divided into two subcases.

Case I. dim E = 1. In this case use minimality of ℝ ⋋ ℝ_{+} [DS] and Lemma 1.5.

Case II. dim E > 1. Represent E as a product ℝ × X, where X is a closed linear subspace of codimension 1. Consider the topological group product X^{∗} × ℝ_{+} and the following action

α : (X^{∗} × ℝ_{+}) × E → E, α((f, t), (a, x)) = (ta + tf(x), tx).

It is straightforward to show that:

(a) α is a continuous linear action;

(b) The corresponding homomorphism i_{α} : X^{∗} × ℝ_{+} → GL(E) is a topological group embedding;

(c) The semidirect product E⋋_{α}(X^{∗}×ℝ_{+}) is naturally topologically isomorphic to the induced group M_{+}(X).

By Fact 2.1, the group M_{+}(X) is minimal. Therefore we can complete the proof by defining H := X^{∗} × ℝ_{+}. □

Note that, as it was pointed out us by the referee, the abelian group H, as well as the resulting semidirect product E ⋋ H, are both connected Banach-Lie groups rather than merely topological groups.

Proposition 2.3. Let z be a fixed non-zero element of a normed space (E, ∥ ∥). Then the orbit map

z̃: GL(E) → E \ {0}, z̃(g) = gz

is surjective and open.

Proof. Assign to every pair (f, y) ∈ E^{∗} × E a continuous linear operator

A_{f,y} : E → E, A_{f,y}(x) = x + f(x)y.

It is easy to show that:

(a) ∥A_{f,y} − I∥ = ∥f∥ · ∥y∥ (where I denotes the neutral element of GL(E));

(b) A_{f,y} is invertible if and only if f(y) ≠ −1. The inverse is A_{f,y}^{−1} = A_{tf,y}, where t = −(1 + f(y))^{−1}.

In order to prove that z̃ is surjective, consider an arbitrary non-zero element x ∈ E\{0}. There exists a continuous functional f such that f(z) = 1 and f(x) ≠ 0. Define y = x − z and observe that f(y) ≠ −1. Therefore, A_{f,y} ∈ GL(E). At the same time, A_{f,y} moves z into x.

Now we prove that z̃ is open. It suffices to show (see, for example [RD, Proposition 5.7]) that for every neighborhood U of I in GL(E) the image z̃(U) = Uz is a neighborhood of z in X \ {0}. There exists an ε > 0 such that ε < 1 and

V_{ε} := {g ∈ GL(E) | ∥g − I∥ < ε} ⊆ U.

The properties (a), (b) imply that the set

P_{ε} := {A_{f,y} | ∥f∥ ≤ 1, ∥y∥ < ε}

is contained in V_{ε}. Thus, P_{ε}z ⊆ V_{ε}z ⊆ Uz. On the other hand, P_{ε}z contains the Ball B(z, r) with center z and radius r := ∥z∥ε. Therefore, Uz is a neighborhood of z, as required. □

Note that the use of the automorphisms A_{f,y} in the proof of Proposition 2.3 was inspired by [P].

Theorem 2.4. Let τ_{∥ ∥} be the norm topology of a normed space (E, ∥ ∥) and let σ_{∥ ∥} be the uniform operator topology on GL(E). Suppose that τ′ is a non-trivial group topology on E such that the action

α : (GL(E), σ_{∥ ∥}) × (E, τ′) → (E, τ′)

remains continuous. Then τ′ = τ_{∥ ∥}.

Proof. Case I. τ′ is not Hausdorff. Then the τ′-closure cl_{τ′}{0} of {0} contains a non-zero element. Moreover, since α is (σ_{∥ ∥}, τ′, τ′)-continuous and {0} is GL(E)- invariant then the subgroup cl_{τ′}{0} is GL(E)-invariant too. By the transitivity of the action on E \ {0} we obtain that cl_{τ′}{0} = E. Therefore τ′ is a trivial topology contradicting our assumption.

Case II. τ is Hausdorff. First denote τ_{0} := τ_{∥ ∥} |_{E\{0}} , τ′_{0} := τ′ |_{E\{0}} and fix a non-zero element z of E. Then by Proposition 2.3, the GL(E)-space (E \ {0}, τ_{0}) can be identified with the topological coset GL(E)-space GL(E)/St_{z}, where St_{z} is the stabilizer {g ∈ GL(E) | gz = z}. On the other hand, the (σ_{∥ ∥}, τ′, τ′)-continuity of α implies the continuity of the orbit map:

z̃: (GL(E), σ_{∥ ∥}) → (X \ {0}, τ′_{0}).

By Proposition 2.3 and the characterization property of the quotient topology τ_{0} on GL(X)/St_{z} = X \ {0} we obtain that τ′_{0} ⊆ τ_{0}. Since X \ {0} is open under group topologies τ′ and τ_{∥ ∥}, then τ′ ⊆ τ_{∥ ∥}. Now, by Theorem 2.2, we can conclude τ′ = τ_{∥ ∥}. □

Question 2.5. For what infinite-dimensional normed spaces E is the group E ⋋ GL(E) minimal?

By Theorem 2.2, Theorem 1.4 and Lemma 1.2, it is equivalent to ask:

Question 2.6. Is the action of GL(E) on a normed space E t-exact?

This question seems to be unclear even for E = ℓ_{2}.

§3. Aut-minimality of locally compact groups.

Let (X, τ) be a locally compact group and let Aut(X) be the group of all topological automorphisms of X endowed with the Birkhoff topology τ_{B} (see for example [HR, §26]). Recall that τ_{B} has a local base at the identity formed by the sets:

B(C, U) := {φ ∈ Aut(X) | φ(c) ∈ Uc and φ^{−1}(c) ∈ Uc ∀c ∈ C},

where C runs over the compact subsets of X and U runs over the neighborhoods of e in X.

Lemma 3.1. For every locally compact Hausdorff group X the action

α : (Aut(X), τ_{B}) × (X, τ) → (X, τ)

is hereditarily t-exact.

Proof. See, for example [M, Remark 1.9]. □

Lemma 3.2. For every locally compact abelian Hausdorff group X TFAE: (a) X is Aut(X)-minimal;

(b) X ⋋ Aut(X) is minimal;

(c) X ⋋ H is minimal for a certain topological subgroup H of Aut(X).

Proof. The first implication (a) ⇒ (b) follows from Theorem 1.4 and Lemma 3.1. The implication (b) ⇒ (c) is trivial. If X ⋋ H is minimal then by Lemma 1.2(ii), X is H-minimal. Then, clearly, X is Aut(X)-minimal too. Hence, (c) ⇒ (a). □

In the sequel we say that a locally compact group X is Aut-minimal if X is Aut(X)-minimal. Theorem 3.3. Let a locally compact abelian Hausdorff group X contain as a subgroup the circle group 𝕋 = ℝ/ℤ. Then X is Aut-minimal. Proof. The group 𝕋 splits in X by [HR, §25.31]. Thus, X can be represented as a topological group product 𝕋 × G. Denote by G^{∗} the dual group of all continuous characters of G endowed with the compact open topology. Following [M] consider the continuous action:

α : G^{∗} × (𝕋 × G) → 𝕋 × G, α(χ, (t, g)) = (t + χ(g), g).

The dual form of [M, Theorem 2.11] shows that the semidirect product M(G) := (𝕋 × G) ⋋_{α} G^{∗} is a minimal group. Moreover, it is straightforward to show that the homomorphism π_{α} : G^{∗} → Aut(𝕋 × G) associated to the action α is a topological group embedding. Therefore, there exists a topological subgroup H := π_{α}(G^{∗}) of Aut(𝕋 × G) such that the group X ⋋ H (being topologically isomorphic to M(G)) is minimal. Now, we can apply the implication (c) ⇒ (a) from Lemma 3.2. □

§4. Minimal groups generated by division rings.

Let (K, τ) be a Hausdorff topological division ring. Denote by (K^{×}, τ^{×}) the topological multiplicative group consisting of all non-zero elements endowed with the subspace topology τ^{×} := τ|_{K^{×}}. The natural group action

α : (K^{×}, τ^{×}) × (K, τ) → (K, τ), α(a, b) = ab

is t-exact. Indeed, if η ⊆ τ^{×} is a coarser group topology such that the action is (η, τ, τ)-continuous, then the map

(K^{×}, η) → (K^{×}, τ^{×}), g ↦ g · 1 is continuous. Therefore, η ⊇ τ^{×}. Thus, η = τ^{×}, as required.

Lemma 4.1. For every Hausdorff topological division ring K the group K ⋋_{α} K^{×} is minimal if and only if K is a K^{×}-minimal group.

Proof. As we have seen the action α is t-exact. Hence we may apply Lemma 1.2 and Theorem 1.4. □

Let (E, σ) be a Hausdorff topological K-vector space. A Hausdorff topology σ′ is called admissible (with respect to σ) if σ′ ⊆ σ and (E, σ′) is also a K-vector space. As usual, (E, σ) is called a minimal K-vector space if σ is the unique admissible topology on E. In the particular case, when E = K is a one-dimensional K-vector space, we obtain the definition of strictly minimal division ring in the sense of Nachbin [N] (see also straight division ring [W]).

Fact 4.2. ([N, Theorem 2], [W, Theorem 24.2]) Let K be a non-discrete locally retrobounded division ring. Then K is strictly minimal.

Recall some important resources of locally retrobounded division rings.

Fact 4.3. ([W, §19], [Wi, §5]) A Hausdorff topological division ring K is locally retrobounded in each of the following cases:

(a) K is locally compact;

(b) K is topologized by an absolute value or a valuation;

(c) K is a linearly ordered field.

Lemma 4.4. Let (K, σ) be a Hausdorff topological division ring and let (E, τ) be a K-vector space with respect to the action α : K × E → E.

(i) α is continuous if and only if the restriction α^{×} : K^{×}×E → E is continuous. (ii) If K is a linearly ordered field then α is continuous if and only if the re-

striction α_{+} : K_{+} × E → E is continuous, where K_{+} is the multiplicative group of all positive elements in K.

Proof. (i) If α is continuous then clearly α^{×} is continuous too. Now suppose that α^{×} is continuous. We have to show that α is continuous at each point (t, x) ∈ K ×X. If t is not the zero element 0_{K} of K then the proof becomes trivial because K^{×} is open in K. Thus, it is enough to establish the continuity of α at (0_{K}, x) for arbitrarily fixed x ∈ E. Let U be a neighborhood of the zero element 0_{E} = 0_{K} · x of E. Choose a neighborhood U_{1} of 0_{E} such that U_{1} − U_{1} ⊆ U. Let 1 be the multiplicative unit of K^{×}. Since α^{×} is continuous at (1, x) we can pick a σ^{×}-neighborhood W of 1 in K^{×} and a τ-neighborhood V of x in E such that W V ⊆ U_{1}. Since K^{×} is σ-open in K then W is a σ-neighborhood of 1 in K. Whence W_{1} := W − 1 is a σ-neighborhood of 0_{K}. Then W_{1}V ⊆ U. Indeed, by our choice W V = (W_{1} + 1)V ⊆ U_{1}. Thus W_{1}V + V ⊆ U_{1} and hence W_{1}V ⊆ U_{1} − V ⊆ U_{1} − U_{1} ⊆ U. Therefore the proof of (i) is completed.

(ii) The proof is similar to (i) taking into account that K_{+} is open in K. □ Corollary 4.5. Let K be a Hausdorff topological division ring and let E be a topological K-vector space. TFAE:

(i) E is a minimal K-vector space.

(ii) E is a K^{×}-minimal group.

Theorem 4.6. For every Hausdorff topological division ring K TFAE: (i) K ⋋ K^{×} is minimal. (ii) K is strictly minimal.

Proof. By Lemma 4.1, K ⋋ K^{×} is minimal iff K is a K^{×}-minimal group. On the other hand, applying Corollary 4.5 for E := K we obtain that K is K^{×}-minimal iff K is strictly minimal. Thus, (i) ⇔ (ii) is proved. □

For every integer n ≥ 1 denote by K^{n} the standard n-dimensional K-vector space endowed with the product topology. Then the scalar action K × K^{n} → K^{n} leads to the well-defined semidirect product K^{n} ⋋ K^{×}. Theorem 4.7. Let K be a non-discrete locally retrobounded division ring and let ν be an arbitrary cardinal.

(a) The group (K ⋋ K^{×})^{ν} is perfectly minimal.

(b) If K is complete then (K^{n} ⋋ K^{×})^{ν} is perfectly minimal.

(c) If K is an ordered field then K^{×} can be replaced by K_{+} in both cases (a) and (b).

Proof. (a) By Fact 4.1 and Theorem 4.6, the group K ⋋ K^{×} is minimal. Since Z(K ⋋ K^{×}) is trivial we may apply Fact 1.6.

(b) By Nachbin’s uniqueness theorem [N, Theorem 7] K^{n} has only one admissible topology. Therefore, K^{n} is a minimal K-vector space. By Corollary 4.5, K^{n} is a K^{×}-minimal group. Theorem 1.4 implies that K^{n} ⋋K^{×} is minimal. Then Fact 1.6 is again applicable.

(c) The given proofs may be easily modified taking into account Lemma 4.4(ii). □

Let K be a Hausdorff topological field. Denote by GL(n, K) the topological group of all invertible n × n matrices over K endowed with the topology inherited from K^{n^{2}}.

Theorem 4.8. Let K be a non-discrete locally retrobounded topological field and let H be a topological subgroup of GL(n, K). Then (K^{n} ⋋ H)^{ν} is perfectly minimal in each of the following cases:

(a) H contains all diagonal matrices with non-zero entries.

(b) K is complete and H contains all scalar matrices with non-zero entries.

(c) K is a (complete) ordered field and H contains all (resp., scalar) diagonal matrices with positive entries.

Proof. In each case Z(K^{n} ⋋H) is trivial (see the proof of Lemma 1.5(c)). Therefore by Lemma 1.5(b) and fact 1.6 it suffices to show the minimality of K^{n}⋋H. Moreover, since H carries the topology of pointwise convergence then the action is t-exact. By Theorem 1.4 we have only to establish that K^{n} is H-minimal. Consider the group (K ⋋ K^{×})^{n}. As we know from Theorem 4.7, this group is minimal. On the other hand, (K ⋋ K^{×})^{n} is topologically isomorphic to Y := K^{n} ⋋ (K^{×})^{n}. If the assertion (a) is satisfied then K^{n} ⊆ Y ⊆ K^{n} ⋋ H. Now Lemma 1.2(i) implies that K^{n} is H-minimal. Therefore, (a) is proved.

The proofs of (b) and (c) are similar to the case (a). For (b) we need only to replace Y by the group K^{n} ⋋ K^{×}. In the case (c) we consider (K ⋋ K_{+})^{n} (or, respectively K^{n} ⋋ K_{+}). □

Using completely different methods, Remus and Stoyanov [RS] have shown that (ℝ^{n} ⋋ H)^{ν} is minimal for every cardinal ν and a closed subgroup H of GL(n, ℝ) which contains all diagonal matrices with positive entries. Our Theorems 4.7 and 4.8 improve this result. Even in the case K = ℝ we may consider any (not necesarily closed) subgroup H which contains all scalar (not necessarily all diagonal) matrices with positive entries. In particular, this implies that ℝ^{n} ⋋ ℝ_{+} is minimal. This seems to be new even for n = 2.

§5. A counterexample.

Recently Dikranjan asked^{1} [D2, Question 7.9]: is a minimal totally disconnected group always zero-dimensional? This was inspired by the more general question of Arhangel’skij [D1, Question 1.2]: does a totally disconnected group always admit a coarser zero-dimensional group topology?

In this section we negatively answer these questions by constructing the following

Example 5.1. There exists a totally disconnected (separable metrizable) perfectly minimal group G which is not zero-dimensional.

Construction. Let ω : ℓ_{2} × ℓ_{2} → ℝ be the standard scalar product, where ω(a, b) = ∑_{i=1}^{∞} a_{i}b_{i} for a = ⟨a_{i}⟩, b = ⟨b_{i}⟩ ∈ ℓ_{2}. As was mentioned above in section 2, the induced group

M_{+}(ω) = ((ℝ × ℓ_{2}) ⋋ ℓ_{2}) ⋋ ℝ_{+}

is minimal. It will be convenient to present here explicitly the operations of this group.

Lemma 5.2. Let u_{1} = (r_{1}, x_{1}, y_{1}, t_{1}) and u_{2} = (r_{2}, x_{2}, y_{2}, t_{2}) be two elements of M_{+}(w). Then

(a) u_{1} · u_{2} = (r_{1} + t_{1}r_{2} + t_{1}w(y_{1}, x_{2}), x_{1} + t_{1}x_{2}, y_{1} + y_{2}, t_{1}t_{2}).

(b) u_{2}^{−1} = (t_{2}^{−1}w(x_{2}, y_{2}) − t_{2}^{−1}r_{2}, −t_{2}^{−1}x_{2}, −y_{2}, t_{2}^{−1}).

(c) [u_{1}, u_{2}] = u_{1}u_{2}u_{1}^{−1}u_{2}^{−1} = = ((1−t_{2})r_{1}−(1−t_{1})r_{2}+t_{1}w(y_{1}, x_{2})−t_{2}w(x_{1}, y_{2}), (1−t_{2})x_{1}−(1−t_{1})x_{2}, 𝕆, 1), where 𝕆 denotes the zero element of ℓ_{2}.

Define two subgroups of ℓ_{2} by letting:

E := {⟨a_{i}⟩ ∈ ℓ_{2} | a_{i} ∈ ℚ for all i ∈ ℕ},
F := {⟨a_{i}⟩ ∈ E | a_{i} = 0 for all but finitely many i’s}

Fact 5.3. [E] E is a totally disconnected but not zero-dimensional group.

Observe that if a ∈ E and b ∈ F then ω(a, b) is a rational number. Therefore, the restriction ω|_{E×F} : E × F → ℚ is a well-defined ℚ-bilinear map. This fact together with Lemma 5.2 easily implies that the subset

1The author thanks Dikranjan for providing him the survey article [D2] before its publication.

G := {(q, x, y, t) ∈ M_{+}(ω) | q ∈ ℚ, x ∈ E, y ∈ F, t ∈ ℚ_{+}}

forms a subgroup of M_{+}(ω). Moreover, G can be represented as G = ((ℚ × E) ⋋ F) ⋋ ℚ_{+}. We show that G is the desired group.

First note that by Fact 5.3, G is a totally disconnected non-zero-dimensional group. By Lemma 1.5(c) it suffices to show that G is a minimal group. Since G is a dense subgroup of a minimal group M_{+}(ω), due to the Banaschewskii criterion [B], we have only to check that G is an essential subgroup of M_{+}(ω). That is, we need to check that G ∩ N is not trivial for every closed non-trivial normal subgroup N of M_{+}(ω). We will prove that G ∩ N always contains the element z = (1, 𝕆, 𝕆, 1) which is different from the neutral element e = (0, 𝕆, 𝕆, 1). Obviously, z ∈ G. It suffices to show that always z ∈ N. Let u_{2} = (r_{2}, x_{2}, y_{2}, t_{2}) ∈ N \ {e}. Since N is normal then the commutator [u_{1}, u_{2}] also belongs to N for arbitrary u_{1} ∈ M_{+}(ω). Our aim is to find u_{1} ∈ M_{+}(ω) such that [u_{1}, u_{2}] = z.

Since u_{2} ≠ e then one of the following possibilities hold:

(1) t_{2} ≠ 1;

(2) t_{2} = 1, y_{2} ≠ 𝕆;

(3) t_{2} = 1, y_{2} = 𝕆, x_{2} ≠ 𝕆;

(4) t_{2} = 1, y_{2} = 𝕆, x_{2} = 𝕆, r_{2} ≠ 0.

We indicate below how we will choose the appropriate u_{1} = (r_{1}, x_{1}, y_{1}, t_{1}) according to each case:

(1) u_{1} = ((1 − t_{2})^{−1}, 𝕆, 𝕆, 1);

(2) u_{1} = (0, x_{1}, 𝕆, 1), where x_{1} ∈ E and ω(x_{1}, y_{2}) = −1;

(3) u_{1} = (0, 𝕆, y_{1}, 1), where y_{1} ∈ F and ω(x_{2}, y_{1}) = 1;

(4) u_{1} = (0, 𝕆, 𝕆, 1 + r_{2}^{−1}).

It should be noted that in the case (4) we may suppose that r_{2} > 0 (otherwise, take u_{2}^{−1} = (−r_{2}, 𝕆, 𝕆, 1) instead of u_{2}). Then 1+r_{2}^{−1} ∈ Q_{+} and hence u_{1} is well-defined. By Lemma 5.2(c) in each case we obtain [u_{1}, u_{2}] = z. □

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E-mail: megereli@macs.biu.ac.il