Topologically independent sets in precompact groups

TL;DR

This paper proves that in precompact abelian groups, a subset generates a subgroup as a Tychonoff direct sum of cyclic groups if and only if it is topologically independent, removing the need for absolute Cauchy summability.

Contribution

It establishes that absolute Cauchy summability is unnecessary for the main result in precompact groups, extending previous work on topological independence.

Findings

In precompact abelian groups, topological independence characterizes subgroup generation as a Tychonoff direct sum.

Absolute Cauchy summability can be omitted in the precompact setting.

The result does not extend to locally compact groups, only precompact ones.

Abstract

It is a simple fact that a subgroup generated by a subset $A$ of an abelian group is the direct sum of the cyclic groups $\langle a\rangle$, $a\in A$ if and only if the set $A$ is independent. In [5] the concept of an $independent$ set in an abelian group was generalized to a $topologically$ $independent$ $set$ in a topological abelian group (these two notions coincide in discrete abelian groups). It was proved that a topological subgroup generated by a subset $A$ of an abelian topological group is the Tychonoff direct sum of the cyclic topological groups $\langle a\rangle$, $a\in A$ if and only if the set $A$ is topologically independent and absolutely Cauchy summable. Further, it was shown, that the assumption of absolute Cauchy summability of $A$ can not be removed in general in this result. In our paper we show that it can be removed in precompact groups. In other words, we prove that if $A$ is a subset of a {\em precompact} abelian group, then the topological subgroup generated by $A$ is the Tychonoff direct sum of the topological cyclic subgroups $\langle a\rangle$, $a\in A$ if and only if $A$ is topologically independent. We show that precompactness can not be replaced by local compactness in this result.

Equations

$$S_A=\underset{a\in A}{⨁}⟨a⟩,$$

$${\mathcal{K}}_A:S_A\to G$$

$$\mbox{there exists finite $F\subset A$ such that }\langle{A\setminus F}\rangle\subset U.$$

$$V_{\delta }=\{e^{it}:t\in (-\delta ,\delta )\},$$

$$\mathcal{U}({\chi }_1,\dots ,{\chi }_n;\delta )=\{g\in G:{\chi }_i(g)\in V_{\delta }\text{mbox}foralli=1,\dots ,n\},$$

$$sup{p}_A(\chi )=\{a\in A:\chi (a)\ne 1\}.$$

$$Ua\cap Ub\ne \varnothing .$$

$$z_{a}a\notin W\text{mbox}foralla\in sup{p}_A(\chi ).$$

$$z_{a}a-z_{b}b\in \mathcal{U}({\chi }_1,\dots ,{\chi }_n;\delta ).$$

$$s_a=({\chi }_1(z_{a}a),\dots ,{\chi }_n(z_{a}a)),$$

$${\chi }_i(z_{a}a-z_{b}b)\in V_{\delta }\text{mbox}foralli=1,\dots ,n.$$

$${\mathcal{K}}_A:S_A\to (G,\tau )\text{mbox}isanisomorphismandahomeomorphism.$$

$$\chi (a)={\chi }_a(a)\ne 1$$

$${\mathcal{K}}_A:S_A\to (G,T_H)$$

Full text

In other words, we prove that if $A$ is a subset of a precompact abelian group, then the topological subgroup generated by $A$ is the Tychonoff direct sum of the topological cyclic subgroups $\langle{a}\rangle$, $a\in A$ if and only if $A$ is topologically independent. We show that precompactness can not be replaced by local compactness in this result.

All groups in this paper are assumed to be abelian and all topological groups are assumed to be Hausdorff. A topological group is precompact if it is a topological subgroup of a compact group. As usually, the symbols ${\mathbb{N}}$ and ${\mathbb{Z}}$ stay for the sets of natural numbers and integers respectively.

Given an abelian group $G$, by $0_{G}$ we denote the zero element of $G$, and the subscript is omitted when there is no danger of confusion. Given a subset $A$ of $G$, the symbol $\langle{A}\rangle$ stays for the subgroup of $G$ generated by $A$. For ${a}\in G$, we use the symbol $\langle{a}\rangle$ to denote $\langle{\{a\}}\rangle$. Following [2], the symbol $S_{A}$ stays for the direct sum

[TABLE]

and by $\mathcal{K}_{A}$ we denote the unique group homomorphism

[TABLE]

which extends each natural inclusion map $\langle{a}\rangle\to G$ for $a\in A$. As in [2], we call the map $\mathcal{K}_{A}$ the Kalton map associated with $A$.

We say that $\langle{A}\rangle$ is the direct sum of cyclic groups $\langle{a}\rangle$, $a\in A$ provided that the Kalton map $\mathcal{K}_{A}$ is an isomorphic embedding. When $G$ is a topological group, we always consider $\langle{a}\rangle$ with the subgroup topology inherited from $G$ and $S_{A}$ with the subgroup topology inherited from the Tychonoff product $\prod_{a\in A}\langle{a}\rangle$. Finally, we say that $\langle{A}\rangle$ is a Tychonoff direct sum of cyclic groups $\langle{a}\rangle$, $a\in A$ if the Kalton map $\mathcal{K}_{A}$ is at the same time an isomorphic embedding and a homeomorphic embedding.

1. Introduction

The concept of compactness allows to transfer some purely non-topological issues into the realm of topology. A nice example of this phenomenon is the paper of Nagao and Shakhmatov (see [3]), where the classical, purely combinatorial result of Landau on the existence of kings in finite tournaments, where finite tournament means a finite directed complete graph, is generalized by means of continuous weak selections to continuous tournaments for which the set of players is a compact Hausdorff space. In our paper we provide another example of this phenomenon which non-trivially transfers a result from the area of abelian groups to the realm of precompact abelian groups.

Recall that a subset $A$ of nonzero elements of a group $G$ is independent provided that for every finite set $B\subset A$ and every family $(z_{a})_{a\in B}$ of integers the equality $\sum_{a\in B}z_{a}a=0$ implies $z_{a}a=0$ for all $a\in B$.

Similarly, a subset $A$ of nonzero elements of a topological group $G$ is topologically independent (see [2, Definition 4.1]) provided that for every neighborhood $W$ of $0_{G}$ there exists neighborhood $U$ of $0_{G}$ such that for every finite set $B\subset A$ and every family $(z_{a})_{a\in B}$ of integers the inclusion $\sum_{a\in B}z_{a}a\in U$ implies $z_{a}a\in W$ for all $a\in B$. This neighborhood $U$ is called a $W$-witness of the topological independence of $A$.

One can readily verify that in (Hausdorff) topological groups every topologically independent set is independent (see [2, Lemma 4.2]) and that these two notions coincide in discrete groups. Thus topological independence can be viewed as a natural generalization of independence.

Let us recall a basic and simple fact about independent sets.

Fact**.**

A set $A$ of nonzero elements of a group is independent if and only if the subgroup generated by $A$ is a direct sum of the cyclic groups $\langle{a}\rangle$, $a\in A$.

The aim of this paper is to prove the following counterpart of the above fact. Its proof is postponed to the end of the next section.

Theorem 1.1**.**

A set $A$ of nonzero elements of a precompact group is topologically independent if and only if the topological subgroup generated by $A$ is a Tychonoff direct sum of the cyclic topological groups $\langle{a}\rangle$, $a\in A$.

Example 3.1 demonstrates, that precompactness can not be replaced by local compactness in Theorem 1.1.

In [2] a result closely related to Theorem 1.1 was obtained. In order to state it, recall that by [2, Definition 3.1] a subset $A$ of a topological group is absolutely Cauchy summable provided that for every neighborhood $U$ of $0_{G}$

[TABLE]

Let us state the promised result (see [2, Theorem 5.1]).

Fact 1.2**.**

A subset $A$ of nonzero elements of a topological group $G$ is at the same time topologically independent and absolutely Cauchy summable if and only if the topological subgroup generated by $A$ is a Tychonoff direct sum of the cyclic topological groups $\langle{a}\rangle$, $a\in A$.

In view of Fact 1.2 we can see that Theorem 1.1 reads as follows: Every topologically independent set in a precompact group is absolutely Cauchy summable. Let us note, that absolutely Cauchy summable sets can be far away from topologically independent sets even in the realm of compact groups. Indeed, by [2, Remark 5.3], every null sequence in the compact metric group ${\mathbb{Z}}_{p}$ of $p$-adic integers is absolutely Cauchy summable while every topologically independent subset of ${\mathbb{Z}}_{p}$ is a singleton.

The “only if” part of Fact 1.2 is based on two straightforward results. First one states that absolute Cauchy summability of a set $A$ is equivalent to the continuity of the Kalton map $\mathcal{K}_{A}$ ([2, Theorem 3.5]), while the second says that topological independence of a set $A$ implies that the Kalton map $\mathcal{K}_{A}:S_{A}\to\langle{A}\rangle$ is an open isomorphism ([2, Lemma 4.2 and Proposition 4.7 (i)]). One may ask, whether the implication in the latter statement can be reversed:

Question 1.3**.**

Let $A$ be a subset of nonzero elements of a topological group $G$ such that the Kalton map $\mathcal{K}_{A}:S_{A}\to\langle{A}\rangle$ is an open isomorphism. Must $A$ be topologically independent?

Example 3.3 provides a negative answer to this question even in the case, when $G$ is precompact.

2. Precompact group topologies on direct sums

By ${\mathbb{S}}$ we denote the unit circle in the complex plain which is a compact group with respect to multiplication of complex numbers.

Given a group $G$ we denote by $G^{\prime}$ its character group (the group of all homomorphisms from $G$ to ${\mathbb{S}}$). Elements of $G^{\prime}$ are called characters. If $G$ is a topological group, then the symbol $G^{*}$ stays for the subgroup of $G^{\prime}$ consisting of all continuous characters.

Let $H$ be a group of characters on $G$. We denote by $T_{H}$ the coarsest group topology on $G$ making all characters of $H$ continuous. Since the collection $\{V_{\delta}:\delta\in(0;\pi)\}$, where

[TABLE]

is a local base of the topology of ${\mathbb{S}}$ at the identity, the topology $T_{H}$ has a local base at $0_{G}$ consisting of all the sets of the form

[TABLE]

where $\delta\in(0;\pi)$, $n\in{\mathbb{N}}$ and $\chi_{1},\ldots,\chi_{n}\in H$. The topology is Hausdorff whenever the characters of $H$ separate points of $G$.

Let us recall a basic fact about the topology $T_{H}$ for precompact groups (see [1, Theorem 2.3.2]).

Fact 2.1**.**

Let $G$ be a topological group. Then the topology of $G$ is precompact if and only if it is equal to $T_{G^{*}}$ and the characters of $G^{*}$ separate points of $G$.

Definition 2.2**.**

Given a subset $A$ of a group $G$ and a character $\chi\in G^{\prime}$ we define the $A$-support of $\chi$ by the formula

[TABLE]

We say that the character $\chi$ is finitely $A$-supported provided that $supp_{A}(\chi)$ is finite.

Lemma 2.3**.**

Let $A$ be an infinite subset of a compact group $G$, and $V$ an open neighborhood of the identity element of $G$. Then there are distinct $a,b\in A$ such that $ab^{-1}\in V$.

Proof.

Let $U$ be an open neighborhood of the identity element of $G$ such that $U^{-1}U\subset V$. Since $G$ is compact, there are $a,b\in A$ such that

[TABLE]

Otherwise $A$ would be an infinite closed discrete subset of a compact group $G$ - a contradiction. Now, (2) yields $ab^{-1}\in U^{-1}U\subset V$. ∎

The following lemma is obvious.

Lemma 2.4**.**

Let $A$ be a subset of a topological group $G$ and $\mathcal{V}$ a fixed local base at $0_{G}$ of the topology of $G$. Assume that $W$ is a neighborhood of $0_{G}$ such that no element of $\mathcal{V}$ is a $W$-witness of the topological independence of $A$. Then $A$ is not topologically independent.

Our next proposition plays the key role in the proof of Theorem 1.1.

Proposition 2.5**.**

Let $A$ be a topologically independent subset of a precompact group $G$. Then each character of $G^{*}$ is finitely $A$-supported.

Proof.

We will prove the contrapositive. In order to do so, let $supp_{A}(\chi)$ be infinite for some $\chi\in G^{*}$. Put $W=\mathcal{U}(\chi;\frac{\pi}{2})$ and pick $n\in{\mathbb{N}}$, $\chi_{1},\ldots,\chi_{n}\in H$ and $\delta\in(0;\pi)$ arbitrarily. By Lemma 2.4, it suffices to show that $\mathcal{U}(\chi_{1},\ldots,\chi_{n};\delta)$ is not a $W$-witness of the topological independence of $A$.

For every $a\in supp_{A}(\chi)$ we have $\chi(a)\neq 1$. Therefore, we can find $z_{a}\in{\mathbb{Z}}$ such that $\chi(z_{a}a)=\chi(a)^{z_{a}}\not\in V_{\frac{\pi}{2}}$. Consequently,

[TABLE]

To finish the proof, it remains to find distinct $a,b\in supp_{A}(\chi)\subset A$ such that

[TABLE]

Indeed, in this case $\mathcal{U}(\chi_{1},\ldots,\chi_{n};\delta)$ is not a $W$-witness of the topological independence of $A$ by (3). If there are distinct $a,b\in supp_{A}(\chi)$ such that $\chi_{i}(z_{a}a)=\chi_{i}(z_{b}b)$ for all $i=1,\ldots,n$, then (4) holds. Otherwise the set $\{s_{a}:a\in supp_{A}(\chi)\}$, where

[TABLE]

is an infinite subset of the compact group ${\mathbb{S}}^{n}$. The set $V=\underbrace{V_{{\delta}}\times\ldots\times V_{{\delta}}}_{n}$ is an open neighborhood of the identity element $(\underbrace{1,\ldots,1}_{n})$ of ${\mathbb{S}}^{n}$. Thus, by Lemma 2.3, there are distinct $a,b\in supp_{A}(\chi)$ such that $s_{a}s_{b}^{-1}\in V$. Therefore,

[TABLE]

This yields (4) and finishes the proof. ∎

We omit the straightforward proof of the next simple lemma.

Lemma 2.6**.**

Let $A$ be a subset of a topological group $G$ and $\mathcal{V}$ a local subbase at $0_{G}$ of the topology of $G$. Then $A$ is absolutely Cauchy summable if and only if (1) holds for every $U\in\mathcal{V}$.

Proposition 2.7**.**

Let $A$ be a topologically independent subset of a precompact group $G$. Then $A$ is absolutely Cauchy summable.

Proof.

By Fact 2.1, the set $\{\mathcal{U}(\chi;\delta):\chi\in G^{*},\delta\in(0;\pi)\}$ is a local subbase at $0_{G}$ of the topology of $G$. Fix arbitrary $\delta\in(0;\pi)$, $\chi\in G^{*}$, and put $U=\mathcal{U}(\chi;\delta)$. By Lemma 2.6, it suffices to show (1). Put $F=supp_{A}(\chi)$. Then $F$ is finite by Proposition 2.5. Observe that (1) is satisfied. ∎

Now we are in position to prove Theorem 1.1:

Proof of Theorem 1.1.

The proof follows immediately from Proposition 2.7 and Fact 1.2. ∎

3. Examples and remarks

Example 3.1**.**

Let $A$ be an arbitrary infinite independent subset of a discrete (in particular, locally compact) group $G$. Then $A$ is topologically independent but the topological subgroup $\langle{A}\rangle$ is never a Tychonoff direct sum of cyclic topological subgroups $\langle{a}\rangle$, $a\in A$. Indeed, as was noted in the Introduction, independent sets coincide with topologically independent sets in discrete groups. On the other hand a Tychonoff direct sum of infinitely many non-trivial topological groups is always non-discrete while every topological subgroup of a discrete group is discrete.

Remark 3.2**.**

Recall that a topological group is compactly generated if it contains a compact subset $A$ such that $\langle{A}\rangle$ is the whole group. Let us note that the discrete group $G$ from Example 3.1 is not compactly generated as it contains infinite independent subset and consequently each its generating set is infinite and discrete (thus non-compact). We do not know whether there is an infinite topologically independent subset $A$ of a locally compact compactly generated group $G$ such that $\langle{A}\rangle$ is not a Tychonoff direct sum of the cyclic groups $\langle{a}\rangle$ $a\in A$. In other words, we do not know whether the word “precompact” can be replaced by “locally compact compactly generated” in Theorem 1.1.

Our next example provides a negative answer to Question 1.3.

Example 3.3**.**

There exists a subset $A$ of a precompact group such that the Kalton map $\mathcal{K}_{A}:S_{A}\to\langle{A}\rangle$ is an open isomorphism, but $A$ is not topologically independent. To show this, let $A$ be an arbitrary infinite topologically independent subset of a precompact group $G$ and let $\tau$ denote the topology of $G$. Since subgroups of precompact groups are precompact, we may and will assume that $G=\langle{A}\rangle$. By Theorem 1.1,

[TABLE]

By Fact 2.1, characters of $(G,\tau)^{*}$ separate points of $G$. Therefore, for each $a\in A$ we may pick $\chi_{a}\in(G,\tau)^{*}$ such that $\chi_{a}(a)\neq 1$. The set $A$ is independent as it is topologically independent in $(G,\tau)$. Hence there is (unique) $\chi\in G^{\prime}$ such that

[TABLE]

Consider the subgroup $H$ of $G^{\prime}$ generated by $(G,\tau)^{*}\cup\{\chi\}$. Then $A$ as a subset of $(G,T_{H})$ is as required.

Indeed, the topology $T_{H}$ is precompact by Fact 2.1 (it separates points as $(G,\tau)^{*}$ does). Further, since $A$ is infinite, the character $\chi$ is not finitely $A$-supported by (6). Thus the set $A$ is not topologically independent in $(G,T_{H})$ by Proposition 2.5.

Finally, it follows from (6) that each $\langle{a}\rangle$ has the same subgroup topology in the topology $T_{H}$ as in $\tau$. Therefore, $S_{A}$ as the Tychonoff direct sum has the same topology with respect to $T_{H}$ as well as with respect to $\tau$. Since $T_{H}$ is finer then $\tau$, it follows from (5) that the Kalton map

[TABLE]

is an open isomorphism.

Call a group $G$ monothetic if it has a dense cyclic subgroup $C$. Every generator of $C$ is called a topological generator of $G$.

Lemma 3.4**.**

If $g$ is a topological generator of a topological group $G$, then the singleton $\{g\}$ is a maximal (with respect to inclusion) topologically independent subset of $G$.

Proof.

Let $g,h$ be topologically independent elements of $G$. Since finite sets are absolutely Cauchy summable, Fact 1.2 gives us that $\langle{\{g,h\}}\rangle$ is the Tychonoff direct sum $\langle{g}\rangle\bigoplus\langle{h}\rangle$. Hence $\langle{g}\rangle$ is not dense in $\langle{\{g,h\}}\rangle$ and consequently it is not dense in $G$ as well. Thus $g$ is not a topological generator of $G$. ∎

Our last and simple example provides a warning showing that unlike in Theorem 1.1, some other properties that hold for independent sets may not be transfered to the realm of topologically independent sets in precompact groups.

Example 3.5**.**

Obviously, if $a,b,c$ are distinct non-torsion elements of a group such that the set $\{a,c\}$ as well as the set $\{b,c\}$ is not independent, then also the set $\{a,b\}$ is not independent. On the other hand, there exist a compact topological group $G$ and non-torsion elements $a,b,c\in G$ such that the set $\{a,c\}$ as well as the set $\{b,c\}$ is not topologically independent while the set $\{a,b\}$ is topologically independent.

Put $G={\mathbb{S}}^{2}$, and $a=[g,1],b=[1,g]$, where $g\in{\mathbb{S}}$ is non-torsion. The set $\{a,b\}$ is topologically independent by Fact 1.2. It is a folklore fact, that ${\mathbb{S}}^{2}$ is monothetic. Let $c\in{\mathbb{S}}^{2}$ be its topological generator. Then $c$ is non-torsion and the set $\{a,c\}$ as well as the set $\{b,c\}$ is not topologically independent by Lemma 3.4.