MA

EGNS

IT A T

MOLEM

UN

IVERSITAS WARWICENSISSimple Abelian Topological Groups

by

Luke Dominic Bush Hipwood

supervised by Dr Dmitriy Rumynin

4th Year Project

Submitted to The University of Warwick

Mathematics Institute

1st April, 2014

Table of contents

1 Introduction 1

2 Preliminaries 2

3 Locally Compact Case 6

4 Non-complete Example 7

5 Non-metrizable example 17

6 The general Conjecture 24

7 Appendix 26

ii

1 Introduction

In the following we take the convention 0 /∈ N. All sums are over N unless explicitly

stated. We write T to be the subgroup {z ∈ C : |z| = 1} of C×, equipped with the

subspace topology. For a group G, and an element x ∈ G we denote 〈x〉 to be the cyclic

subgroup of G generated by x. For a subset Y of a topological space X, denote the

closure of Y in X by Y (usually X will be a group and Y a subgroup.) If there exists a

map that is both an isomorphism and a homeomorphism between two topological groups

G and H, we say that they are topologically isomorphic. In this case we write G ∼= H.

If they are isomorphic as groups only, we still write G ∼= H but explicitly say ’as groups.’

In 1976 R.C.Hooper gave an example of an infinite, complete, metrizable, topologi-

cal group whose only locally compact subgroup is the trivial one [Ho]. After which he

asked the question ’whether, if H is an infinite, complete, metrizable, topological abelian

group, H has a non-trivial, proper closed subgroup.’ [Ho]

Definition 1.1. An abelian topological group G, is said to be topologically simple if

G 6= {e} and contains no closed subgroup, other than the trivial one and itself.

Let G be a complete, metrizable, topologically simple, abelian group. Suppose every

infinite, complete, metrizable topological abelian group has a non-trivial, proper closed

subgroup. It follows that G cannot be infinite. As G is metrizable and finite, it’s topology

must be the discrete topology. Hence the condition that G contains no closed subgroup,

other than the trivial one and itself becomes G contains no non-trivial proper subgroup.

That is, G is a simple group. Hence G is topologically isomorphic to Cp with the discrete

topology, where Cp denotes the cyclic group of order p, and p is a prime number. We

write G ∼= Cp in this case.

Conversely if a complete, metrizable, topologically simple abelian group, is neces-

sarily Cp with the discrete topology (up to isomorphism), then, certainly, there can not

be an infinite, complete, metrizable, topological abelian group that has a no non-trivial,

proper closed subgroups.

Thus we can reformulate Hooper’s conjecture as ’If G is a complete, metrizable, topo-

logically simple, abelian group, then G is topologically isomorphic with Cp equipped with

the discrete topology, where p is prime.’

This project investigates Hooper’s conjecture. There are three main chapters. The

first proves the conjecture to be true when G is assumed to be locally compact. We in

1

fact do this with Hausdorff in place of metrizable, and without the complete assumption

altogether. We make use of the heavy machinery available with regards to locally com-

pact abelian groups. In the second we constructed an example of an infinite, metrizable,

topologically simple abelian group. This is done by placing an appropriate norm on the

integers. We owe thanks to Dr Dmitriy Rumynin, Supervisor, for a sketch to this exam-

ple. In our final main chapter we give an example of an infinite, complete, topologically

simple, Hausdorff abelian group. To do this we construct a net of complete, metrizable

abelian groups, each a completion of Z, and such that the subgroup Z is topologically

simple. Then taking the inverse limit, we recover Z, with the required topology. We owe

thanks to Dr Stanislav Shkarin [S] for a sketch of this example. Our final chapter looks

at the general conjecture. We look at the properties that a counter example must have.

2 Preliminaries

Here we state a few results that will be used in the proceeding. They are listed in the

approximate order in which they are used.

Proposition 2.1. [HR, 5.3]. Let G be a topological group, and H a subgroup. Then

H is also a subgroup.

Proposition 2.2. [HR, 7.1]. Let G be a topological group and C the connected com-

ponent of the identity. Then C is a closed normal subgroup of G.

Proposition 2.3. [HR, 7.7]. Let G be a totally disconnected, locally compact group.

Then every neighbourhood of the identity contains a compact open subgroup.

Proposition 2.4. [HR, 9.14]. Let G be a locally compact abelian group, and let C be

the connected component of the identity. Then

C ∼= Rn × E

where n is a non-negative integer and E is a compact connected abelian group.

Definition/Proposition 2.5. [HR, 10.1-10.15]. Fix a sequence of integers a = (ai),

with ai > 2 ∀i. Let a-adic integers be denoted by ∆a. We define the topological group

∆a as follows. As a set

∆a =

∞∏i=1

{0, 1, . . . , ai − 1}.

2

For each n ∈ N, put

Λn = {x = (xi) ∈ ∆a : x1 = x2 = · · · = xn−1 = 0}.

{Λn}n∈N, forms an open basis of the topology at 0 = (0, 0, . . . ) (which will be the

identity), on ∆a. Given x,y ∈ ∆a, x + y is defined as follows. If y = 0 = (0, 0, . . . ), we

set x + 0 = 0 + x = x, and similarly if x = 0. Suppose x,y 6= 0. Let n and m, be the

least integers such that xn, ym 6= 0. Without loss of generality, n 6 m. For i < n, set

zi = 0. For i = n, write xn + yn = tnan + zn, with tn ∈ Z, and zn ∈ {0, 1, . . . , an − 1}.For i > n, write xi + yi + ti−1 = tiai + zi. Set z = (zi). We define x + y = z. The

above make ∆a into a Hausdorff, locally compact, compact, abelian group, with a dense

subgroup isomorphic to Z.

For prime p (or indeed any integer greater or equal to 2), ∆p denotes ∆a, where

a = (p, p, . . . ).

Let a = (ai) be sequence of integers with each ai > 2. Let a-adic solenoid be

denoted by Σa. The topological group Σa is defined as follows. Set u to be the element

(1, 0, 0, . . . ) ∈ ∆a. Let B be the subgroup {(n, nu)}n∈Z ⊂ R×∆a. Then as a topological

group

Σa = (R×∆a)/B.

It is a compact connected abelain group.

Proposition 2.6. [HR, 25.8]. A compact abelian group G is torsion-free if and only if

G ∼= Σma ×

∏p∈P

∆npp ,

where P is the set of all prime numbers, m and every np are arbitrary cardinal numbers,

and a = (2, 3, 4, . . . ).

Theorem 2.7 (Baire’s Category Theorem). A complete metric space is not a countable

union of nowhere dense closed sets.

Our final example is of an infinite complete non-metrizable topologically simple

abelian group. To do this we need to first know what is meant by a complete, non-

metrizable abelian group.

Definition 2.8. [HR, 4.11]. Let G be an abelian group, with identity e. For each open

neighbourhood of U of e, set

VU = {(x, y) ∈ G×G : x−1y ∈ U}.

3

Define

U (G) = {VU : U is a neighbourhood of e},

it is called the uniform structure on G.

Note that if G is non-abelian then defined above is the left uniform structure on G,

with the right uniform structure on G defined analogously.

Proposition 2.9. [B, III 3.1]. U (G) is such that:

• If V ∈ U (G) and W ⊃ V then W ∈ U (G);

• U (G) is closed under finite intersection;

• Put ∆ = {(x, x) ∈ G×G}. Then ∆ ⊂ V for all V ∈ U (G);

• If V ∈ U (G) then V −1 ∈ U (G);

• For every V ∈ U (G) the exists W ∈ U (G) such that W 2 ∈ U (G).

Definition 2.10. [B, I 6.1, I 7.1, II 3.1]. A filter on G is a set F of subsets of G

satisfying

• If U ∈ F and V ⊃ U then V ∈ F ;

• F is closed under finite intersections;

• ∅ /∈ F .

A filter F is said to be convergent to x ∈ G, if for every neighbourhood of U of x,

there is a V ∈ F such that V ⊂ U .

A filter F is said to be cauchy if for every V ∈ U (G) there is a U ∈ F such that

U × U ∈ V .

A convergent filter is Cauchy, however in general the converse is not true. [B] II 3.1

Definition 2.11. [B, II 3.3]. G is complete if every Cauchy filter converges.

This coincides with the notion of completeness when G is assumed to be metrizable.

Also in our non-metrizable we use inverse limits.

Proposition 2.12. [HR, 6.4]. Let {Gα}α∈A be a collection of topological groups. Put

G =∏α∈A

Gα. Then G is Hausdorff if and only if every Gα is Hausdorff.

4

Proposition 2.13. [B, II 3.4-3.5].

• Every closed subgroup of a complete group is complete..

• Every product of complete groups is complete.

Proposition 2.14. [HR, 6.14]. Let (A,≤) be a partially ordered set, {Gα}α∈A a col-

lection of Hausdorff groups and {fβα : Gβ → Gα |α, β ∈ A,α < β} a collection of

continuous homomorphisms. Suppose the above forms an inverse mapping system, that

is, such that α < β < γ =⇒ fγα = fβα ◦ fγβ. Then, the inverse limit is a closed

subgroup of the product group∏α∈A

Gα.

Corollary 2.15. [B, II 3.5]. Let (Gα, fβα) be an inverse mapping system. If each Gα

is Hausdorff and complete then so is the inverse limit.

Proposition 2.16. [mC, 3.4]. Let G be an abelian topological group (written addi-

tively), with norm ‖ · ‖. H ⊂ G a subgroup. Then the quotient space G/H has norm

‖x+H‖′ = infy∈H{‖x+ y‖}.

Proposition 2.17. [mC, 3.25]. Let G be a complete abelian group with norm ‖ · ‖.Suppose H ⊂ G is a closed subgroup. Then the quotient group G/H is complete.

Proposition 2.18. [HR, 8.3]. Let G be a Hausdorff group. Then G is metrizable if and

only if there is a countable open basis at the identity. In this case the metric can be

taken to be left invariant.

In the final chapter ’The General Conjecture’ we briefly look at dual groups.

Definition/Proposition 2.19. [HR, 23.1, 23.2, 23.15]. A character of a topological

group G, is a continuous homomorphism G→ T.

The set of characters is made into an abelian group via point-wise multiplication.

This group is called the dual group of G and is denoted G∗.

For every compact set F ⊂ G and every ε > 0, put

U(F, ε) = {χ ∈ G∗ : |χ(x)− 1| < ε ∀x ∈ F}.

With all the sets U(F, ε) as an open basis at 1 ∈ G∗, G∗ is a topological group.

Proposition 2.20. [AT, 9.5.7]. The dual group of the discrete group Z is topologically

isomorphic with T.

5

Proposition 2.21. [HR, 24.2]. Let G be abelian topological group, and such that for

every x ∈ G \ {e}, there exists a character χ ∈ G∗ such that χ(x) 6= 1, then G embeds

as a topological subgroup into G∗∗ in the conical way.

3 Locally Compact Case

Theorem 3.1. Let G be a topologically simple, locally compact, Hausdorff, abelian

group. Then

G ∼= Cp where p is a prime number.

Proof. Let C denote the connected component to the identity e ∈ G, it is a closed sub-

group of G, hence either C = {e} or C = G.

Case 1: C = {e}In this case G is totally disconnected. Proposition 2.3 then says every neighbourhood

of e contains a compact open subgroup. Choose a neighbourhood of e, U 6= G (which

exists as we are assuming G is Hausdorff and non-trivial), we find a open, hence closed,

subgroup H s.t.

e ∈ H ⊆ U ( G,

forcing H = {e}. In particular {e} is open, hence G is discrete. It follows that G must

be simple as a group, and hence G ∼= Cp for some prime p.

Case 2: G = C, i.e. G is a connected abelian group.

Since G is locally compact we can apply Proposition 2.4, and deduce G ∼= Rn×E with E

a compact connected abelian group. Because G is topologically simple and a product of

non-trivial Hausdorff groups is not topologically simple (see appendix 7.1), either G ∼= Ror G ∼= E. Since clearly R is clearly not topologically simple we can rule out the first,

and deduce G ∼= E. i.e. G is also compact.

Let x be a torsion element. Consider the subgroup 〈x〉. It has finite order and hence,

as G is Hausdorff, it is discrete, and therefore closed. So we must have either 〈x〉 = G or

〈x〉 = {e}. But the first is impossible, as the G would be discrete and so not connected.

Thus we must have x = e. i.e. G is torsion-free. Proposition 2.6 then says

G ∼= (Σa)m ×∏p∈P

∆npp ,

where P is the set of all primes, m and the np’s are arbitrary cardinal numbers, and

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a = (2, 3, 4, . . . ). Σa denotes the a-adic solenoid and ∆p denotes the p-adic integers (as

described in the Preliminaries 2.5).

∆p has open, and therefore closed, proper subgroup Λ2 = {x ∈ ∆p : x1 = 0}. So

∆p is not topologically simple. Since a product of non-trivial Hausdorff groups is not

topologically simple, np = 0 ∀p ∈ P.

(R× Λ2)/B is an open and closed subgroup of Σa, thus Σa is not topologically simple.

Hence m = 0 also. But then G is trivial. Since we excluded G = {e} in the definition

of topologically simple, G is not simple; a contradiction. Meaning case 1 is the only

possibility.

4 Non-complete Example

In this chapter we aim to prove the following theorem:

Theorem 4.1. There exists an infinite metrizable, topologically simple, abelian group.

To do this we construct a norm on Z. This norm ensures that 1 is in the closure of

every subgroup, that is, ensures topological simplicity.

Construction. Consider the following sequences:

(y(1)n ) : 1 · 1− 1 , 2 · 1− 1 , . . . , n · 1− 1 , (n+ 1)1− 1 , . . .

(y(2)n ) : 1 · 2− 1 , 2 · 2− 1 , . . . , n · 2− 1 , (n+ 1)2− 1 , . . ....

......

......

(y(m)n ) : 1m− 1 , 2m− 1 , . . . , nm− 1 , (n+ 1)m− 1 , . . .

(y(m+1)n ) : 1(m+ 1)− 1 , 2(m+ 1)− 1 , . . . , n(m+ 1)− 1 , (n+ 1)(m+ 1)− 1 , . . .

......

......

...

Notice for every m ∈ N, the sequence (y(m)n ) −→∞ as n −→∞.

We define the sequence (xn) as follows. Set x1 = 1. Choose x2 in (y(1)n ) to be such

that x2 > 22x1. (This is possible since (y(1)n ) → ∞). Now pick x3 in (y

(2)n ) such that

x3 > 23x2. (Again possible as (y(2)n )→∞).

Once x k(k+1)2

is defined (we write K = k(k+1)2 for ease of notation) choose: xK+1 in

(y(1)n ) such that xK+1 > 2K+1xK ; xK+2 in (y

(2)n ) such that xK+2 > 2K+2xK+1; . . . ;

xK+k+1 in (y(k+1)n ) such that xK+k+1 > 2K+k+1xK+k. As K + k + 1 = (k+1)(k+2)

2 , this

defines (xn) inductively.

It is such that:

7

(I) x1 = 1;

(II) for every m ∈ N the sequence (y(m)n ) contains a subsequence, that itself is a

subsequence of (xn);

(III) xn+1 > 2n+1xn ∀n ∈ N.

Now note that for every a ∈ Z we can always write a =∑aixi for some ai ∈ Z all

but finitely many 0. This is because x1 = 1, and therefore ax1 is a possibility. For a ∈ Zwe define

‖a‖ = inf∑|ai|2−i

where the infimum ranges over all possible ways of writing a =∑aixi as above.

The sequence (xn) is the crucial ingredient in proving Theorem 4.1 (and in fact plays a

vital role in proving Theorem 5.1 later). Property (I) allows us to define the norm, and

will prove useful throughout this and the next chapter. (II) is the key for the simplicity.

The purpose of (III) is to ensure we have indeed defined a norm, and will be our main

tool in approximating values of ‖ · ‖.We start by calculating the value of ‖ · ‖ on a certain set of integers. The techniques

used in the proof of Proposition 4.2 will be crop up throughout this chapter.

Proposition 4.2. Let e =∑eixi, with ei ∈ {0, 1} all but finitely many zero. Then

‖e‖ =∑ei2−i

Proof. Let e = xi1 + xi2 + · · ·+ xin . We proceed by induction on n.

Suppose n = 1, then e = xk for some k ∈ N. Clearly ‖xk‖ > 2−k. Assume ‖xk‖ <2−k. So we can write xk =

∑aixi with all but finitely many ai = 0 and

∑|ai|2−i < 2−k.

We see that ai = 0 ∀i 6 k and that |ai| < 2i−k ∀i > k + 1. Now also for every l ∈ N,

|l∑

i=1aixi| < xl+1. Since otherwise, because xi+1 > 2i+1xi ∀i, we would have

|l∑

i=1aixi| > xl+1

> (2l+1 − 1)xl + xl

> (2l+1 − 1)xl + (2l − 1)xl−1 + xl−1...

> (2l+1 − 1)xl + · · ·+ (23 − 1)x2 + x2

> (2l+1 − 1)xl + · · ·+ (23 − 1)x2 + 22x1.

8

Therefore would require at least one i with |ai| > 2i+1 − 1, contradicting |ai| < 2i−k ∀i.Let m be maximal with am 6= 0. By the above m > k + 1. So

xk = ak+1xk+1 + ak+2xk+2 + · · ·+ amxm. (∗)

If am < 0 then as xk > 0, we would require |a1x1 + · · · + am−1xm−1| > xm. This

contradicts |l∑

i=1aixi| < xl+1 ∀l. So am > 0 (am 6= 0 by definition).

Now for each i, xi+1 > 2i+1xi. This gives

amxm − xk > xm − xk> xm − xk+1

> (2m − 1)xm−1 + xm−1 − xk+1

> (2m − 1)xm−1 + (2m−1 − 1)xm−2 + xm−2 − xk+1

...

> (2m − 1)xm−1 + · · ·+ (2k+3 − 1)xk+2 + xk+2 − xk+1

> (2m − 1)xm−1 + · · ·+ (2k+3 − 1)xk+2 + 2k+2xk+1 − xk+1

> (2m − 1)xm−1 + · · ·+ (2k+3 − 1)xk+2 + (2k+2 − 1)xk+1

Therefore, for (∗) to hold, we require at least one i ∈ {k + 1, . . . ,m} such that |ai| >2i+1 − 1. Contradicting |ai| < 2i−k ∀i. Thus ‖xk‖ > 2k, and hence ‖xk‖ = 2−k.

Now let n > 1. Clearly ‖e‖ 6∑ei2−i. Assume ‖e‖ <

∑ei2−i. Write e =

∑aixi,

with all but finitely many ai = 0 and∑|ai|2−i <

∑ei2−i. As

∑ei2−i < 1, we see

|ai| < 2i ∀i. As in n = 1 case, we also have |l∑

i=1aixi| < xl+1 ∀l (same proof works here

too).

Let m be maximal with am 6= 0. We have

e = xi1 + xi2 + · · ·+ xin = a1x1 + a2x2 + · · ·+ amxm. (†)

Similarly to the n = 1 case, am > 0. Suppose m > in. Then

amxm − (xi1 + xi2 + · · ·+ xin) > 22x1 +

n∑j=1

(2ij+1 − 2)xij +

m∑i 6=1,i1,...,in

(2i+1 − 1)xi

Hence for (†) to hold we need at least one j ∈ {1, . . . ,m− 1} with |aj | > 2j+1 − 2. This

contradicts |ai| < 2i ∀i. So m 6 in.

9

If ain = 0 (i.e. m < in), then,

a1x1 + · · ·+ amxm > xin

> xm+1

> (2m+1 − 1)xm + · · ·+ (23 − 1)x2 + 22x1.

Meaning we would need an i with ai > 2i+1 − 1; a contradiction. So m = in. We know

from above ain = am > 0.

Consider f = e − xi1 = xi1 + · · · + xin−1 . By the induction hypothesis, ‖f‖ =

2i1 + · · ·+ 2in−1 . But also f = a1x1 + a2x2 + · · ·+ (ain − 1)xin . Since

|a1|2−1 + |a2|2−2 + · · ·+ |ain |2−in < 2−i1 + · · ·+ 2−in

and ain > 0,

|a1|2−1 + |a2|2−2 + · · ·+ |ain − 1|2−in < 2−i1 + · · ·+ 2−in−1 .

Thus ‖f‖ < 2−i1 + · · ·+ 2−in1 ; a contradiction. So ‖e‖ =∑ei2

i. The result now follows

by induction.

Now to begin proving we have constructed an infinite, metrizable, topologically sim-

ple, abelian group.

Lemma 4.3. ‖ · ‖ defines a norm on Z. That is, satisfies

(i) ‖a‖ > 0 ∀a ∈ Z;

(ii) ‖a‖ = ‖ − a‖ ∀a ∈ Z;

(iii) ‖a‖ = 0⇔ a = 0;

(iv) ‖a+ b‖ 6 ‖a‖+ ‖b‖ ∀a, b ∈ Z.

Proof. (i) is obvious. For (ii) we can note that, if a =∑aixi then −a =

∑−aixi, and

hence ‖a‖ = ‖ − a‖.(iv): Let a, b ∈ Z. Write a =

∑aixi and b =

∑bixi, with ai, bi ∈ Z all but finitely

many zero. Then clearly, a+ b =∑

(ai + bi)xi. We then see,

‖a+ b‖ 6∑|ai + bi|2−i 6

∑|ai|2−i + |bi|2−i =

∑|ai|2−i +

∑|bi|2−i.

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Since∑aixi and

∑bixi are arbitrary ways of writing a and b in this form, this inequality

holds for every possible way for writing a and b in this form. Hence we indeed have,

‖a+ b‖ 6 ‖a‖+ ‖b‖.(iii): Clearly ‖0‖ = 0. Suppose a 6= 0. We need to show ‖a‖ 6= 0. Since ‖a‖ = ‖−a‖,

we may assume a > 0. Now as (xi) −→∞ and is strictly increasing, there is a (unique)

N such that xN 6 a < xN+1. We claim ‖a‖ > 2−N . In particular ‖a‖ 6= 0.

If a = xN , then, by Proposition 4.2, ‖a‖ = 2−N . So suppose xN < a < xN+1.

Assume, for a contradiction, that ‖a‖ < 2−N . Write a =∑aixi with

∑|ai|2−i < 2−N .

We see that ai = 0 ∀i 6 N , and |ai| < 2i−N ∀i > N .

Now, as in the proof Proposition 4.2, |n∑i=1

aixi| < xn+1 ∀n (same proof holds here).

We will show inductively that the above in fact implies either∑aixi < xN or∑

aixi > xN+1, in both cases, a contradiction.

First we note that the first possible non-zero ai is aN+1. Since xN+1 > xN it is clear

that either aN+1xN+1 < xN or aN+1xN+1 > xN+1. Now let k > N + 1, and suppose

either

aN+1xN+1 + · · ·+ akxk < xN or aN+1xN+1 + · · ·+ akxk > xN+1.

Assume first that aN+1xN+1 + · · ·+ akxk < xN . If ak+1 6 0, it is obvious that

aN+1xN+1 + · · ·+ak+1xk+1 < xN . So assume ak+1 > 0. Since each |ai| < 2i−N , we have

aN+1xN+1 + · · ·+ ak+1xk+1 > aN+1xN+1 + · · ·+ akxk + xk+1

> −(21 − 1)xN+1 − · · · − (2k−N − 1)xk + xk+1.

Now for each i, xi+1 > 2i+1xi. This gives

xk+1 > (2k+1 − 1)xk + · · ·+ (2N+3 − 1)xN+2 + 2N+2xN+1

> (2k−N − 1)xk + · · ·+ (22 − 1)xN+2 + 2xN+1.

Putting these two inequalities together we get,

aN+1xN+1 + · · ·+ ak+1xk+1 > xN+1.

Now assume aN+1xN+1 + · · · + akxk > xN+1. If ak+1 > 0, it’s obvious that

aN+1xN+1 + · · · + ak+1xk+1 > xN+1. So suppose ak+1 < 0. Because |aN+1xN+1 +

· · ·+ akxk| < xk+1, it’s clear that aN+1xN+1 + · · ·+ ak+1xk+1 < 0 < xN .

Since the sum∑|ai|xi must be a finite sum we indeed have either

∑aixi < xN or∑

aixi > xN+1. That is either a < xN or a > xN+1, contradicting xN < a < xN+1.

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Thus ‖a‖ > 2−N , verifying (iii).

Before we start talking about topological simplicity we need to make sure we indeed

have a topological group. We in fact prove a more general result.

Lemma 4.4. Let G be an abelian group (written additively). Let ‖ · ‖ : G −→ R be a

norm on G, that is, satisfies

(i) ‖a‖ > 0 ∀a ∈ G;

(ii) ‖a‖ = ‖ − a‖ ∀a ∈ G;

(iii) ‖a‖ = 0⇔ a = 0;

(iv) ‖a+ b‖ 6 ‖a‖+ ‖b‖ ∀a, b ∈ G.

Then G with ‖ · ‖, is a topological group.

Proof. Fix c ∈ G, and ε > 0, as ‖a‖ = ‖ − a‖ ∀a ∈ G, if ‖a − c‖ < ε, then clearly

‖(−a)− (−c)‖ < ε. So a 7→ −a is continuous.

Now ‖(a, b)‖′ = ‖a‖+ ‖b‖ defines a norm on G×G that coincides with the product

topology (see appendix 7.2). Fix (c, d) ∈ G×G and ε > 0. Then if ‖(a, b)− (c, d)‖′ < ε,

then, by the triangle inequality,

‖a+ b− (c+ d)‖ 6 ‖a− c‖+ ‖b− d‖ = ‖(a, b)− (c, d)‖ < ε.

Thus addition is continuous.

Finally we check the simplicity of our example.

Lemma 4.5. Z with this norm is topologically simple.

Proof. Let a ∈ Z non-zero. We need to prove 〈a〉 = Z. As 1 generates Z, and the closure

of a subgroup is again a subgroup, it suffices to prove 1 ∈ 〈a〉. Since 〈a〉 = 〈−a〉 we

may assume a > 0. We need to find a sequence, (zn) say, in 〈a〉 converging to 1, or

equivalently, such that (zn − 1) converges to 0.

Now notice that the sequence (y(a)n ) is the sequence (na−1)n∈N, and by construction

contains a subsequence that is itself a subsequence of (xn), say (xkn). We can write this

sequence in the form (zn − 1) for some sequence (zn) in 〈a〉. By construction, for each

n, ‖zn − 1‖ = ‖xkn‖ = 2−kn . Hence indeed (zn − 1) −→ 0.

12

With these Lemmas above we have now proven Theorem 4.1.

We now ask if we have stumbled upon a counter example to Hooper’s conjuncture,

i.e. whether Z with is norm is complete. The answer (unsurprisingly) is no.

Proposition 4.6. Z, with the norm ‖ · ‖, is not complete.

Proof. This follow form Baire’s Category Theorem. Since single points are certainly

closed and nowhere dense, and Z is countable, Z, with any metric, can not be complete.

We can realise the incompleteness more explicitly.

Proposition 4.7. Consider the sequence (sn) defined by sn =n∑i=1

xi. It is Cauchy, but

has no limit in Z.

Proof. It is Cauchy since: for m > n > 0, by Proposition 4.2,

‖sm − sn‖ = ‖xn+1 + · · ·+ xm‖ = 2−(n+1) + · · ·+ 2−m < 2−n.

This clearly converges to zero as n,m −→∞. We will show that (sn) does not converge

in Z.

Let a ∈ Z be a candidate limit. By 4.2, lim ‖sn‖ =∑

2−n = 1, so a 6= 0. Let N ∈ Nbe such that xN−1 6 |a| < xN . Let’s first suppose a < 0. Fix n > N . We claim that

‖sn − a‖ > 2−N .

Assume for a contradiction that ‖sn−a‖ < 2−N . So we can write sn−a =∑aixi with∑

|ai|2−i < 2−N . We immediately see that ai = 0 ∀i 6 N and that |ai| < 2i−N ∀i > N .

As previously (Proposition 4.2 and Lemma 4.3(iii)), we have that |k∑i=1

aixi| < xk+1 ∀k

(the same proof holds here too). Let m the largest number such that am 6= 0, so

sn − a = x1 + · · · + xn − a = aN+1xN+1 + · · · + amxm. Since a < 0, and, xN is larger

then both −a and x1 + · · ·+ xN−1 clearly

xN + · · ·+ xn < aN+1xN+1 + · · ·+ amxm < 3xN + xN+1 + · · ·+ xn. (∗)

Suppose m > n. If am < 0 then, as sn − a > 0,

|aN+1xN+1 + · · ·+ am−1xm−1| > xm,

13

which contradicts |k∑i=1

aixi| < xk+1 ∀k . So am > 0 (am 6= 0 by definition). We have

amxm > xm > (2m − 1)xm−1 + · · ·+ (2N+2 − 1)xN+1 + 2N+1xN .

So

amxm− (3xN +xN+1 + · · ·+xn) > (2N+1−3)xN +n∑

i=N+1

(2i+1−2)xi+m−1∑i=n+1

(2i+1−1)xi.

Hence for (∗) to be true we would need at least one i ∈ {N + 1, . . . ,m − 1} with

|ai| > 2i > 2i−N , which is a contradiction. So ai = 0 ∀i > n.

Now if an 6 0, because (∗) implies xn < aN+1xN+1 + · · · + anxn, we would need

|aN+1xN+1 + · · ·+ an−1xn−1| > xn; a contradiction.

Suppose an > 1. Then

anxn > 2xn

> 2[(2n − 1)xn−1 + · · ·+ (2N+2 − 1)xN+1 + 2N+1xN ]

> (2n − 1)xn−1 + · · ·+ (2N+2 − 1)xN+1 + 2N+1xN + 3xN + xN+1 + · · ·+ xn.

Hence we have the inequality

anxn − (3xN + xN+1 + · · ·+ xn) > (2n − 1)xn−1 + · · ·+ (2N+2 − 1)xN+1 + 2N+1xN .

So, for (∗) to hold, we require at least one ai > 2i; a contradiction. Thus an = 1.

(∗) now reads

xN + · · ·+ xn < aN+1xN+1 + · · ·+ an−1xn−1 + xn < 3xN + xN+1 + · · ·+ xn.

Subtracting xn from this we get

xN + · · ·+ xn−1 < aN+1xN+1 + · · ·+ an−1xn−1 < 3xN + xN+1 + · · ·+ xn−1.

The same argument now applies to show an−1 = 1. After this we repeat over and yield

ai = 1 ∀i : N + 2 6 i 6 n. We eventually get

xN + xN+1 < aN+1xN+1 < 3xN + xN+1.

14

It’s immediate that aN+1 > 2. But then

aN+1xN+1 > 2xN+1 > xN+1 + 2N+1xN > xN+1 + 3xN ;

a contradiction.

Now we assume a > 0. Fix n > N + 2. If a = xN−1 then, by Proposition 4.2,

‖sn − xN‖ = −2−N +

n∑1=1

2−i > 2−(N+1),

and (sj) cannot converge to a. So assume a > xN−1. Now because a < xN < x1+· · ·+xN ,

and x1 + · · ·+ xN − a < xN+1 < 2xN+1,

xN+1 + · · ·+ xn < sn − a < 3xN+1 + xN+2 + · · ·+ xn.

Assume that ‖sn − a‖ < 2−(N+1) and write∑aixi = sn − a with

∑|ai|2−i < 2−(N+1).

We can conclude similarly to the a < 0 case, that ai = 0 for all i 6 N + 1 and i > n+ 1,

and ai = 1 ∀i: N + 3 6 i 6 n. Hence reducing to the inequality

xN+1 + xN+2 < aN+2xN+2 < 3xN+1 + xN+2.

Similar to before, this is a contradiction.

We have proved that for every non-zero a ∈ Z. Let N be such that xN−1 6 |a| < xN .

Then if n > N + 2, ‖sn − a‖ > 2−(N+1). In particular (sk) cannot converge to a. We

have already ruled out sk −→ 0 so (sk) cannot converge to any a ∈ Z.

Consider the completion of Z with respect to this norm. Denote it by A. A natural

question to ask is whether A remains topologically simple. The answer is no.

Proposition 4.8. A has a non-trivial proper closed subgroup.

Proof. Let s ∈ A be the limit of the Cauchy sequence (sn) in Proposition 4.7. We prove

that the subgroup 〈s〉 is closed in A. Being countable ensures 〈s〉 6= A.

To show 〈s〉 is closed we prove that ∃ε > 0 such that ‖ts‖ > ε ∀t ∈ Z \ {0}. Then

for elements ts 6= t′s in 〈s〉, ‖ts − t′s‖ = ‖(t − t′)s‖ > ε, proving 〈s〉 is discrete, and

hence closed. Since addition and subtraction are continuous, tsn −→ ts, and therefore

‖tsn‖ −→ ‖ts‖. Thus it is sufficient to prove for n large enough ‖tsn‖ > ε.

Let t be a non-zero integer. As ‖tsn‖ = ‖ − tsn‖ ∀n, we may assume t > 0. Choose

l ∈ N such that 2l−1 6 t < 2l. We claim for n > l + 2, ‖tsn‖ > 18 .

15

Suppose ‖tsn‖ < 18 for some n > l + 2. Write tsn =

∑aixi, with ai ∈ Z all but

finitely many ai = 0 and∑|ai|2−i < 1

8 . This immediately forces |ai| < 2i−3 ∀i. As before

(Proposition 4.2, Lemma 4.3(iii) and Proposition 4.7), we have that |k∑i=1

aixi| < xk+1 ∀k.

Let m ∈ N be maximal such that am 6= 0, so

tsn = tx1 + · · ·+ txn = a1x1 + · · ·+ amxm.

Now tsl < 2l(x1 + · · ·+ xl) < 2l2xl < xl+1,

so tsn < xl+1 + txl+1 + · · ·+ txn < (1 + 2l)xl+1 + 2lxl+2 + · · ·+ 2lxn,

i.e. a1x1 + · · ·+ amxm < (1 + 2l)xl+1 + 2lxl+2 + · · ·+ 2lxn.

Since, |a1x1 + · · ·+ alxl| < t(x1 + · · ·+xl) (as |ai| < 2i−3 < 2l−1 < t ∀i 6 l), and in turn

t(x1 + · · ·+ xl) < 2l(x1 + · · ·+ xl) < xl+1, it follows that

al+1xl+1 + · · ·+ amxm < 2xl+1 +

n∑i=l+1

2lxi. (1)

Now suppose m > n.

If am < 0, then as tsn > 0, we would need |m−1∑i=1

aixi| > xm; a contradiction. So am > 0

(am 6= 0 by definition).

Now

amxm > xm > (2m − 1)xm−1 + · · ·+ (2l+3 − 1)xl+2 + 2l+2xl+1,

therefore

amxm−[2xl+1+n∑

i=l+1

2lxi] > (2l+2−2l−4)xl+1+

n∑i=l+2

(2i+1−2l−1)xi+m−1∑i=n+1

(2i+1−1)xi.

Thus, for (1) to hold, ∃i ∈ {1, . . . , n − 1} with at least |ai| > 2i+2 − 2l − 2. But for all

i ∈ N, 2i+2 − 2l − 2 > 2i−3. Hence we have a contradiction. So we conclude m 6 n, i.e.

ak = 0 ∀k > n. Hence we have

a1x1 + · · ·+ anxn = tx1 + · · ·+ txn. (2)

We now aim to show an = t.

If an < t, then |a1x1 + · · ·+an−1xn−1| > (t−an)xn > xn; a contradiction. So an > t.

16

Assume an > t, then, because

tsn−1 < 2l(x1 + · · ·+ xn−1) < 2l+1xn−1 6 2(n−1)xn−1 =1

22nxn−1 <

1

2xn

(recall n > l + 2), we have

anxn − tsn > xn − tsn−1 >1

2xn >

1

2[(2n − 1)xn−1 + · · ·+ (23 − 1)x2 + 22x1].

Thus for (2) to be true we need at least one i ∈ {1, . . . , n− 1} with |ai| > 12(2i+1 − 1) >

2i−3; a contradiction. Hence indeed an = t.

Equation (2) now reduces to

a1x1 + · · ·+ an−1xn−1 = tx1 + · · ·+ txn−1.

We can now repeat the above over, yielding al+2 = · · · = an = t. In particular, al+2 = t,

and thus ∑|ai|2−i > t2−(l+2) > 2l−12−(l+2) =

1

8.

Contradicting our initial assumption. So this assumption must be false, that is, ‖tsn‖ >18 .

Hence we conclude 〈s〉 is a proper non-trival closed subgroup, that is, A is not

topologically simple.

Notice that this proof allows us to see more explicitly that 〈s〉 6= A. Any non-zero

a ∈ A with ‖a‖ < 18 will not be in 〈s〉, a = x4 for example.

5 Non-metrizable example

Our aim in this Chapter is to prove

Theorem 5.1. There exists an infinite, complete, topologically simple, Hausdorff abelian

group.

The example we give will turn out to be Z with an non-metrizable topology. This

will therefore not be a counter-example to Hooper’s conjecture. Our first aim will be to

construct a collection complete of groups, each with Z as a subgroup. Furthermore the

subgroup Z should be topologically simple in each case.

Let A be the set of real sequences a = (an) satisfying:

17

• an ∈ (0, 1] ∀n ∈ N;

• a is non-increasing.

• an −→ 0 as n −→∞

• 2−n 6 an ∀n ∈ N.

For a,b ∈ A write a 6 b whenever an 6 bn ∀n. This makes A into a partially ordered

set. Notice also that A is uncountable. Indeed we have an obvious injection (0, 1] ↪→ A,

r 7→ (r, 0, 0, . . . ).

For each a ∈ A, we will construct a complete group, Ga, with the required proper-

ties. Thus A will be our indexing set for these groups. The reason for partially ordering

A will become clear later, and is crucial in recovering the topologically simple Z.

Fix a ∈ A. Consider the subgroup Ha of Zℵ0 , consisting of the sequences h = (hn)

satisfying∑an|hn| < ∞. It is a subgroup, since, if h,h′ ∈ Ha, with h = (hn) and

h′ = (h′n), then ∑|hn − h′n|an 6

∑|hn|an +

∑|h′n|an <∞

by the triangle inequality. So h− h′ ∈ Ha.

For h ∈ Ha define ‖h‖a =∑an|hn|.

Lemma 5.2. ‖ · ‖a is a norm on Ha and with this norm Ha is a topological group.

Proof. We need to prove

(i) ‖h‖a > 0 ∀h ∈ Ha;

(ii) ‖h‖a = ‖ − h‖a ∀h ∈ Ha;

(iii) ‖h‖a = 0⇔ h = 0 = (0, 0, . . . );

(iv) ‖h + h′‖a 6 ‖h‖a + ‖h′‖a ∀h,h′ ∈ Ha;

(v) Addition and subtraction are continuous.

(i) is obvious. Given h ∈ Ha

‖ − h‖a =∑

an| − hn| =∑

an|hn| = ‖h‖a

18

so (ii) is true. Clearly ‖0‖a = 0. Conversely, if ‖h‖a = 0, i.e.∑an|hn| = 0, then as

an 6= 0 ∀n, hn = 0 ∀n. So h = 0, proving (iii). Given h,h′ ∈ Ha, by the standard

triangle inequality

‖h + h′‖a =∑

an|hn + h′n| 6∑

an|hn|+∑

an|h′n| = ‖h‖a + ‖h′‖a,

so (iv) holds. (v) follows form lemma 4.4.

Lemma 5.3. Ha, with this norm, is complete.

Proof. Let (h(m)) be a cauchy sequence in Ha. This means∑i

ai|h(m)i − h(k)i | −→ 0 as m, k −→∞.

For each i ∈ N, since ai > 0, |h(m)i − h(k)i | −→ 0 as m, k −→ ∞. Therefore every h

(m)i

converges, say to hi. Recalling that h(m)i ∈ Z ∀m, we see hi ∈ Z and that h

(m)i = hi

eventually. We need to show the sequence h = (hn) is in Ha, and that h(m) −→ h.

Let ε > 0, and M ∈ N be such if m, k > M then∑i

ai|h(m)i − h(k)i | < ε.

Indeed take m, k > M . Since each term in the above sum is non-negative, for any n ∈ Nwe certainly have

n∑i=1

ai|h(m)i − h(k)i | < ε.

Letting k →∞ we getn∑i=1

ai|h(m)i − hi| 6 ε.

Since this holds for all n ∈ N, we can send n→∞, yielding

∞∑i=1

ai|h(m)i − hi| 6 ε,

i.e. ‖h(m)−h‖a 6 ε. Hence indeed h(m) −→ h. Note also that the above implies for any

m > M , h(m) − h ∈ Ha. As Ha is a subgroup of Zℵ0 , it follows h ∈ Ha.

We now turn our attention to the simplicity of our example. Let (xn) be the sequence

of positive integers constructed in chapter 4. Recall it satisfies:

19

(I) x1 = 1;

(II) for every m ∈ N the sequence (mn − 1)n∈N contains a subsequence, that itself is

a subsequence of (xn);

(III) xn+1 > 2n+1xn ∀n ∈ N.

Note the condition (III) implies xn+1 > 2 1anxn ∀n. We make use of the latter property

of (xn), and is the reason we imposed the last condition on A.

Let K ⊂ Zℵ0 consist of sequences c = (cn) such that all but finitely many cn = 0,

and∑cnxn = 0. Clearly K is a subgroup and is contained in Ha. Let Ga = Ha/K.

Lemma 5.4. K is closed in Ha, and hence the quotient group Ga is complete

Proof. Claim: ‖c‖a > 1 ∀c ∈ K \ {0}.Let c ∈ K \ {0}. Suppose cn1 , cn2 , . . . , cnk

are exactly all the non-zero terms (note

k > 2). Then, by definition, cn1xn1 + · · ·+ cnkxnk

= 0.

Equivalently cn1xn1 + · · ·+ cnk−1xnk−1

= −cnkxnk

. As xn+1 > 2 1anxn, we therefore have

|cn1xn1 + · · ·+ cnk−1xnk−1

| = |cnkxnk|

> xnk

> 2 1ank−1

xnk−1

> 1ank−1

xnk−1+ 2 1

ank−2xnk−2

...

> 1ank−1

xnk−1+ · · ·+ 1

an2xn2 + 2 1

an1xn1 .

For this to be true we need at least one i with cni >1ani

, say cnl> 1

anl. But then

‖c‖a =∑|ci|ai > |cnl

|anl>

1

anl

anl= 1,

proving the claim.

The claim shows K is a discrete subgroup, hence it is closed. As a quotient of a

complete space by a closed space, Ga is complete.

We claim that the Ga’s are the complete groups we are after.

Lemma 5.5. Z embeds as a cyclic dense subgroup of Ga, via

ι : t 7−→ (t, 0, 0, . . . ) +K.

20

Proof. It is clear that ι is a homomorphism. It is injective since if t ∈ ker(ι), then

(t, 0, 0, . . . ) ∈ K. So by definition tx1 = 0. But x1 = 1, so t = 0.

Put T = {(t, 0, 0, . . . ) : t ∈ Z}. For t ∈ Z, denote t = (t, 0, 0, . . . ). We prove T +K

is dense in Ha, and thus im(ι) is dense in Ga.

Let h ∈ Ha, and ε > 0. Since∑|hi|ai <∞, ∃n ∈ N such that

∞∑i=n+1

|hi|ai < ε.

Define c = (ci) as follows: Put ci = hi for i ∈ {2, 3, . . . , n}, put c1 = −n∑i=2

cixi and

ci = 0 otherwise. Then, by construction,∑cixi = 0 (recall x1 = 1), and hence c ∈ K.

Set t = h1 − c1. Then

‖h− (t + c)‖a = |h1 − (t+ c1)|a1 +n∑i=2

|hi − ci|ai +∞∑

i=n+1

|hi|ai =∞∑

i=n+1

|hi|ai < ε

as required.

Define Z = {(t, 0, 0, . . . ) +K : t ∈ Z} = im(ι). Notice Z ⊂ Gb ∀b ∈ A.

Remark. Notice that the above proof shows that Z is the quotient of the subgroup

{(hn) ∈ Ha : all but finitely many hn = 0} by K.

Proof. Let S = {(hn) ∈ Ha : all but finitely many hn = 0}, and T be as above. It

suffices to prove T + K = S + K. Clearly T ⊂ S, so T + K ⊂ S + K. Conversely let

s ∈ S. Define c in the same way as above, but with s in place of h. Then s− c ∈ T .

Thus S +K ⊂ T +K.

At this point we indeed have our a collection of complete groups, {Ga}a∈A. Each

Ga has a copy Z sitting inside, namely Z, and 1 is in the closure of every subgroup as

required (see Lemma 5.8 for a proof of this). We look to recover Z form these groups.

To this we take the inverse limit (which is why we made A into a partially ordered set).

Let a,b ∈ A with a 6 b. If h ∈ Hb, h = (hn), then by definition,∑|hi|bi < ∞.

since a 6 b, it follows that∑|hi|ai < ∞, and hence h ∈ Ha. So Hb ⊂ Ha. Let

fba : Gb −→ Ga, h +K 7−→ h +K.

Lemma 5.6. For every a,b ∈ A with a 6 b, fba is an injective continuous homomor-

phism. If a,b, c ∈ A, with a 6 b 6 c, then fba ◦ fcb = fca.

21

Proof. Let a,b ∈ A with a 6 b. Because K is independent of a and b, and is contained

in every Ha, ∀a ∈ A, it is clear that fba is a well-defined injective homomorphism. It is

continuous because ‖h‖a 6 ‖h‖b ∀h ∈ Gb. The last part is obvious.

By the above lemma we have an inverse mapping system consisting of A, the Ga’s

and the fba’s. Let Z ⊂∏a∈A

Ga denote the inverse limit. As each Ga is metrizable (so

certainly Hausdroff), Z is a closed subgroup of the complete group∏a∈A

Ga. Hence Z is

complete.

We claim Z is the topological group we want.

Lemma 5.7. Z ∼= Z as groups.

Proof. We prove

Z = {(t +K)a∈A ∈∏a∈A

Ga : t +K ∈ Z},

which is clearly isomorphic to Z, which in turn is isomorphic to Z by Lemma 5.5.

For now, to ease notation, let R denote the right hand side. As Z ⊂ Ga ∀a ∈ A,

given t +K ∈ Z and a,b ∈ A with a 6 b, it’s clear fba(t +K) = t +K. Thus R ⊂ Z.

Conversely, assume Z * R. Let (ha + K)a∈A ∈ Z \ R. Suppose first that ha ∈Z ∀a ∈ A. By choosing different co-set representatives (if necessary) we may assume

ha = ta = (ta, 0, 0, . . . ) ∀a ∈ A. Let a,b ∈ A (not necessarily comparable). Define

c = (cn), by cn = max{an, bn}, then c ∈ A, and a,b 6 c. Therefore we must have

fca(tc + K) = ta + K , i.e. tc + K = ta + K, and fcb(tc + K) = tb + K, i.e.

tc +K = tb +K. Hence ta +K = tb +K. So ∀a,b ∈ A, ta +K = tb +K, contradicting

(ha +K)a∈A /∈ R. Hence ∃a ∈ A such that ha +K /∈ Z.

Write ha = (hi)i∈N; by the remark after Lemma 5.5, there must be infinitely many

i such that hi 6= 0. Let (hni) be a subsequence of (hi) consisting entirely of non-zero

terms. Define b = (bj) as follows. For j such that 1 6 j 6 n1, put bj = 1; for i > 2

and for j such that ni−1 < j 6 ni, put bj = max{ani−1 ,1i }. Clearly b ∈ A, and, by

construction, a 6 b (recall a is non-increasing ∀a ∈ A). So ∃hb + K ∈ Gb such that

fba(hb +K) = ha +K, i.e. hb +K = ha +K. Hence ha = hb + k, for some k ∈ K. In

particular ha ∈ Hb. But

∑|hi|bi >

∑|hni |bni >

∑bni >

∑ 1

i=∞;

a contradiction. Hence Z ⊂ R.

So Z = R, and hence Z ∼= Z.

22

The topology on Z is the subspace topology from∏a∈A

Ga. Thus, considering Z with

this topology, a sequence (tn) in Z converges to a limit, say t, if and only if the sequence

((tn, 0, 0, . . . ) +K) converges to (t, 0, 0, . . . ) +K in Ga, for every a ∈ A.

Lemma 5.8. Z, with the above topology, is topologically simple.

Proof. Let t ∈ Z, we need to show 1 ∈ 〈t〉. By construction of the sequence (xn), it has

a subsequence, (xkn) say, such that, for each n ∈ N, xkn = tyn − 1, for some positive

integers yn. Writing zn = tyn, (zn) is a sequence in 〈t〉 and is such that (zn−1) = (xkn).

Fix a ∈ A and n ∈ N. Define c1 = −xkn , ckn = 1, ci = 0 otherwise, and put c = (ci).

Then ∑cixi = −xkn + xkn = 0,

so c ∈ K. Furthermore (zn − 1, 0, 0, . . . ) + c = (0, . . . , 0, 1, 0, . . . ), where the 1 is in the

knth position. Hence (zn − 1, 0, 0, . . . ) +K = (0, . . . , 0, 1, 0, . . . ) +K. As

‖(0, . . . , 0, 1, 0, . . . )‖a = akn −→ 0 as n→∞,

it follows that (zn, 0, 0, . . . ) + K −→ (1, 0, 0, . . . ) + K in Ga. Since this holds ∀a ∈ A,

zn −→ 1 in Z. Hence 1 ∈ 〈t〉, so 〈t〉 = Z and Z is topologically simple.

We have now proven Theorem 5.1, that is we have successfully constructed an infinite,

complete, topologically simple, Hausdorff abelian group.

Intuitively, we have no hope of Z, with this topology being metrizable; there are

just too many norms on it. Indeed, it cannot be since we know there are no countable

complete metric spaces by Baire’s Category Theorem. A more direct proof is possible.

Proposition 5.9. Z is not metrizable.

Proof. By proposition 2.18 it suffices to show that there is no countable open basis at

0 = (0 +K, 0 +K, . . . ).

Let U = {Ui}i∈N, be a countable collection of open sets, each containing 0. For each

i write

Ui = {(t +K)a∈A ∈ Z : t +K ∈ U (i)a }, with all but finitely many U

(i)a = Z

(recall Z = {(t + K)a∈A ∈∏a∈A

Ga : t + K ∈ Z}.) As A is uncountable and there are

only countably many Ui, there are uncountably many a ∈ A such that U(i)a = Z ∀i ∈ N,

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in particular there is at least one, say b. Let Ub be any open subset of Gb containing

0 + K, and such that Z * U . Then the open set {(t + K)a∈A ∈ Z : t + K ∈ Ub},contains 0 but cannot be written as a union of sets form U .

Hence there can be no countable open basis a 0, and hence Z is not metrizable.

6 The general Conjecture

Suppose G is a counter-example to Hooper’s conjecture, that is, an infinite, complete,

metrizable, topologically simple abelian group. Let e ∈ G denote the identity. We now

list the obvious properties that G must have.

By Baire’s Category Theorem, a complete metric space cannot be countable. Hence

G is uncountable.

In the locally compact case we proved either G is connected or totally disconnected.

The same proof holds in the general case, and hence the same conclusion holds too.

Let x ∈ G be a torsion element and consider the subgroup 〈x〉. As x has finite order,

〈x〉 is finite. Since G is metrizable, it is closed. Because G is infinite 〈x〉 6= G, and so

〈x〉 = {e}, i.e. x = e. Hence G is torsion-free.

Pick any x ∈ G \ {e}, and consider the subgroup 〈x〉. It is a non-trivial closed

subgroup, hence by assumption, 〈x〉 = G. Now x has infinite order, and therefore as a

group 〈x〉 ∼= Z. Because G is complete, the completion of 〈x〉 is just 〈x〉 = G. Thus we

can view G as a completion of Z.

So G has the following properties:

• G is uncountable;

• Either G is connected or G is totally disconnected;

• G is torsion-free;

• G is the completion of Z with an appropriate metric.

Write G as the completion of Z with respect to some metric. We now consider the dual

groups of G and the topological subgroup Z, and see what they can tell us about G.

Let Zd denote the group Z equipped with the discrete topology. Proposition 2.20

says Z∗d ∼= T. A topological isomorphism is given by χ 7→ χ(1). Since any continuous

map Z −→ T, is certainly a continuous map Zd −→ T, we can embed the group Z∗ into Tvia χ 7→ χ(1). Thus we can view the group Z∗ as a subgroup of T. The question remains

whether the topologies necessarily coincides. It is at least true that the topology on Z∗

is finer then the subgroup to topology inherited form T.

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Proof. It suffices to check the topology is finer at 1 ∈ T. Let ε > 0. Write Bε(1) =

{z ∈ C ∩ Z∗ : |z − 1| < 1}. Choose F = {1} ⊂ Z. Clearly F is compact. By definition

(definition/Proposition 2.19) U(F, ε) = {χ ∈ Z∗ : |χ(1) − 1| < ε} is an open set of Z∗.Recalling the natural embedding Z∗ ↪→ T, is given by χ 7→ χ(1), we see U(F, ε) is just

Bε(1). Thus topology on Z∗ is indeed finer then that of T.

Given a continuous χ : Z→ T, we can define a map χ : G→ T as follows. For x ∈ Z,

define χ(x) = χ(x). For x ∈ G \ Z, choose a sequence in Z, say (xn), converging to x

(which exists as Z is dense in G). Define χ(x) = limχ(xn). Since χ is continuous, and

T is complete, χ is a well-defined continuous homomorphism G→ T, i.e. χ ∈ G∗.Conversely given a character of G, we can restrict to a character of Z. This procedure

clearly gives a topological isomorphism, thus Z∗ ∼= G∗.

Fix a non-zero n ∈ Z, and consider the group homomorphism φ : Z → T; 1 7→ ζ,

where ζ is a primitive nth root of unity. Suppose φ is continuous and let φ be it’s unique

extension to G. Then ker φ is a closed subgroup of G. ker φ 6= {0} since n ∈ ker φ, and

1 /∈ ker φ, so ker φ is a proper non-trivial closed subgroup of G. This contradicts the

simplicity of G. Thus φ is not continuous. In particular the kernel, 〈n〉 is not closed, and

so the subgroup Z is topologically simple as well. Also notice this shows Z∗ is proper

subgroup of T.

We now know the following:

• Z∗ = G∗;

• The subgroup Z is topologically simple also;

• Z∗ is proper subgroup of T;

• The topology on Z∗ is at worst finer then T.

Now either Z∗ = G∗ = {1}, or Z∗ = G∗ 6= {1}. In the latter case, as G is topologically

simple, any character will have trivial kernel, and thus G will satisfy the hypotheses of

proposition 2.20. Hence G conically embeds into G∗∗ which equals Z∗∗. Thus there are

two cases

• Either G∗ = {1},

• or G is a topological subgroup of Z∗∗.

Recalling Z∗ is a proper subgroup of T with some finer topology, call it H ⊂ T, we have

• Either G∗ = {1},

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• or G is a topological subgroup of H∗, where H is a proper subgroup of T with

some finer topology.

We have successfully managed to reduce the group structures we are working with to

the well know Z and T.

7 Appendix

Proposition 7.1. Suppose {Gα}α∈A is a collection of non-trivial Hausdorff abelian

groups, and G =∏α∈A

Gα. If G is topologically simple then |A| = 1.

Proof. Since we exclude the trivial group in the definition of topologically simple, |A| 6= 0.

Suppose |A| > 1, possible infinite. Let β ∈ A. Set H = {(gα) ∈ G : gα = 0 ∀α 6= β}.Clearly H is a proper subgroup of G. For every α, Gα is Hausdorff and so {0} is closed

in Gα. It follows Gα\{0} is open, for every α. For α 6= β put Uα = Gα\{0}. Set Uβ = ∅.Then U = {(gα) : gα ∈ Uα} is open. Hence H = G \U is a proper closed subgroup, i.e.

G is not topologically simple.

Thus |A| = 1.

Proposition 7.2. Let ‖ · ‖ be a norm on an abelian group G, i.e. satisfies (i) - (iv) in

lemma 4.4. Define ‖(a, b)‖′ = ‖a‖ + ‖b‖ on G × G. Then ‖ · ‖′ is a norm on G × G,

compatible with the product topology induced by ‖ · ‖.

Proof. The fact that ‖ · ‖′ is a norm on G × G is well known and follows readily form

‖ · ‖ being a norm on G.

Let Bε(c) = {a ∈ G : ‖a− c‖ < ε}, and

B′δ(c, d) = {(a, b) ∈ G×G : ‖(a, b)− (c, d)‖′ < δ}. A basis for the topology on G is

B0 = {Bε(a) : a ∈ G, ε > 0}.

Therefore the product topology has basis

B = {U × V : U, V ∈ B0}.

The topology induced by ‖ · ‖′ has basis

B′ = {B′δ(a, b) : (a, b) ∈ G×G δ > 0}.

Given an element of B we need to find an element of B′ contained in it, and vice versa.

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Take U × V ∈ B, and assume U = Bε(c) and V = Bδ(d). With out loss of generality

assume ε < δ. Suppose (a, b) ∈ B′ε(c, d) then

‖a− c‖ 6 ‖a− c‖+ ‖b− d‖ = ‖(a, b)− (c, d)‖′ < ε,

and similarly ‖b−d‖ 6 ‖(a, b)−(c, d)‖ < ε 6 δ, so (a, b) ∈ U×V . Hence B′ε(c, d) ⊂ U×V .

Conversely, let U ′ ∈ B′, and assume U ′ = B′2ε(c, d). If a ∈ Bε(c) and b ∈ Bε(d), then

‖(a, b)− (c, d)‖′ = ‖a− c‖+ ‖b− d‖ < ε+ ε = 2ε.

So Bε(c)×Bε(d) ⊂ U ′.

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References

[Ho] R.C.Hooper, Locally Compact Subgroups of Metrizable Topological Abelian Groups,

Proc. Amer. Math. Soc. 57 (1976), no. 1, 159-164

[HR] E.Hewitt, K.A.Ross, Abstract Harmonic Analysis, Volume 1, Springer-Verlag,

Berlin, Gottingen, Heidelberg, 1963

[B] N.Bourbaki, Elements of Mathematics, General Topology, Springer-Verlag Berlin

Heidelberg New York London Paris Tokyo 1989

[AT] A. Arhangel’skii and M.Tkachenko Topological Groups and Related Structures, At-

lantis Press, Amsterdam Paris, World Scientific, 2008

[mC] B.MacCluer, Elementary Functional Analysis, Springer, New York, 2009

[S] Dr Stanislav Shkarin, Private Communication

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