Simple Abelian Topological Groups. Luke Dominic Bush Hipwood. Mathematics Institute

M A E NS G I T A T MOLEM UNIVERSITAS WARWICENSIS Simple Abelian Topological Groups by Luke Dominic Bush Hipwood supervised by Dr Dmitriy Rumynin 4th Year Project Submitted to The University of Warwick Mathematics Institute 1 st 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, topological 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 {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 C p with the discrete topology, where C p denotes the cyclic group of order p, and p is a prime number. We write G = C p in this case. Conversely if a complete, metrizable, topologically simple abelian group, is necessarily C p 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, topologically simple, abelian group, then G is topologically isomorphic with C p 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 compact 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 example. 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 component 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 = R n 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 = (a i ), with a i 2 i. Let a-adic integers be denoted by a. We define the topological group a as follows. As a set a = {0, 1,..., a i 1}. i=1 2

For each n N, put Λ n = {x = (x i ) a : x 1 = x 2 = = x n 1 = 0}. {Λ n } 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 0. Let n and m, be the least integers such that x n, y m 0. Without loss of generality, n m. For i < n, set z i = 0. For i = n, write x n + y n = t n a n + z n, with t n Z, and z n {0, 1,..., a n 1}. For i > n, write x i + y i + t i 1 = t i a i + z i. Set z = (z i ). 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 = (a i ) be sequence of integers with each a i 2. Let a-adic solenoid be denoted by Σ a. The topological group Σ a 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 Σ a = (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 = Σ m a p P np p, where P is the set of all prime numbers, m and every n p 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, nonmetrizable 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 V U = {(x, y) G G : x 1 y U}. 3

Define U (G) = {V U : 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]. satisfying A filter on G is a set F of subsets of G 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 {G α } α A be a collection of topological groups. Put G = G α. Then G is Hausdorff if and only if every G α is Hausdorff. α A 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, {G α } α A a collection of Hausdorff groups and {f βα : G β G α 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 = f γα = f βα f γβ. Then, the inverse limit is a closed subgroup of the product group G α. Corollary α A 2.15. [B, II 3.5]. Let (G α, f βα ) be an inverse mapping system. If each G α is Hausdorff and complete then so is the inverse limit. Proposition 2.16. [mc, 3.4]. Let G be an abelian topological group (written additively), with norm. H G a subgroup. Then the quotient space G/H has norm x + H = inf { x + y }. y H 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) 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 = C p where p is a prime number. Proof. Let C denote the connected component to the identity e G, it is a closed subgroup 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 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 = C p 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 = R n 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 = R or 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 = (Σ a ) m p P np p, where P is the set of all primes, m and the n p s are arbitrary cardinal numbers, and 6

a = (2, 3, 4,... ). Σ a 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 Λ 2 = {x p : x 1 = 0}. So p is not topologically simple. Since a product of non-trivial Hausdorff groups is not topologically simple, n p = 0 p P. (R Λ 2 )/B is an open and closed subgroup of Σ a, thus Σ a 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 n (1) ) : 1 1 1, 2 1 1,..., n 1 1, (n + 1)1 1,... (y n (2) ) : 1 2 1, 2 2 1,..., n 2 1, (n + 1)2 1,........ (y n (m) ) : 1m 1, 2m 1,..., nm 1, (n + 1)m 1,... (y n (m+1) ) : 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 n (m) ) as n. We define the sequence (x n ) as follows. Set x 1 = 1. Choose x 2 in (y n (1) ) to be such that x 2 > 2 2 x 1. (This is possible since (y n (1) ) ). Now pick x 3 in (y n (2) ) such that x 3 > 2 3 x 2. (Again possible as (y n (2) ) ). Once x k(k+1) 2 is defined (we write K = k(k+1) 2 for ease of notation) choose: x K+1 in (y n (1) ) such that x K+1 > 2 K+1 x K ; x K+2 in (y n (2) ) such that x K+2 > 2 K+2 x K+1 ;... ; x K+k+1 in (y n (k+1) ) such that x K+k+1 > 2 K+k+1 x K+k. As K + k + 1 = (k+1)(k+2) 2, this defines (x n ) inductively. It is such that: 7..

(I) x 1 = 1; (II) for every m N the sequence (y n (m) ) contains a subsequence, that itself is a subsequence of (x n ); (III) x n+1 > 2 n+1 x n n N. Now note that for every a Z we can always write a = a i x i for some a i Z all but finitely many 0. This is because x 1 = 1, and therefore ax 1 is a possibility. For a Z we define a = inf a i 2 i where the infimum ranges over all possible ways of writing a = a i x i as above. The sequence (x n ) 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 = e i x i, with e i {0, 1} all but finitely many zero. Then e = e i 2 i Proof. Let e = x i1 + x i2 + + x in. We proceed by induction on n. Suppose n = 1, then e = x k for some k N. Clearly x k 2 k. Assume x k < 2 k. So we can write x k = a i x i with all but finitely many a i = 0 and a i 2 i < 2 k. We see that a i = 0 i k and that a i < 2 i k i k + 1. Now also for every l N, l a i x i < x l+1. Since otherwise, because x i+1 > 2 i+1 x i i, we would have i=1 l a i x i x l+1 i=1 > (2 l+1 1)x l + x l > (2 l+1 1)x l + (2 l 1)x l 1 + x l 1. > (2 l+1 1)x l + + (2 3 1)x 2 + x 2 > (2 l+1 1)x l + + (2 3 1)x 2 + 2 2 x 1. 8

Therefore would require at least one i with a i > 2 i+1 1, contradicting a i < 2 i k i. Let m be maximal with a m 0. By the above m k + 1. So x k = a k+1 x k+1 + a k+2 x k+2 + + a m x m. ( ) If a m < 0 then as x k > 0, we would require a 1 x 1 + + a m 1 x m 1 > x m. This l contradicts a i x i < x l+1 l. So a m > 0 (a m 0 by definition). i=1 Now for each i, x i+1 > 2 i+1 x i. This gives a m x m x k x m x k > x m x k+1 > (2 m 1)x m 1 + x m 1 x k+1 > (2 m 1)x m 1 + (2 m 1 1)x m 2 + x m 2 x k+1. > (2 m 1)x m 1 + + (2 k+3 1)x k+2 + x k+2 x k+1 > (2 m 1)x m 1 + + (2 k+3 1)x k+2 + 2 k+2 x k+1 x k+1 > (2 m 1)x m 1 + + (2 k+3 1)x k+2 + (2 k+2 1)x k+1 Therefore, for ( ) to hold, we require at least one i {k + 1,..., m} such that a i > 2 i+1 1. Contradicting a i < 2 i k i. Thus x k 2 k, and hence x k = 2 k. Now let n > 1. Clearly e e i 2 i. Assume e < e i 2 i. Write e = a i x i, with all but finitely many a i = 0 and a i 2 i < e i 2 i. As e i 2 i < 1, we see l a i < 2 i i. As in n = 1 case, we also have a i x i < x l+1 l (same proof works here too). Let m be maximal with a m 0. We have e = x i1 + x i2 + + x in = a 1 x 1 + a 2 x 2 + + a m x m. ( ) Similarly to the n = 1 case, a m > 0. Suppose m > i n. Then n a m x m (x i1 + x i2 + + x in ) > 2 2 x 1 + (2 ij+1 2)x ij + i=1 j=1 m i 1,i 1,...,i n (2 i+1 1)x i Hence for ( ) to hold we need at least one j {1,..., m 1} with a j > 2 j+1 2. This contradicts a i < 2 i i. So m i n. 9

If a in = 0 (i.e. m < i n ), then, a 1 x 1 + + a m x m > x in x m+1 > (2 m+1 1)x m + + (2 3 1)x 2 + 2 2 x 1. Meaning we would need an i with a i > 2 i+1 1; a contradiction. So m = i n. We know from above a in = a m > 0. Consider f = e x i1 = x i1 + + x in 1. By the induction hypothesis, f = 2 i 1 + + 2 i n 1. But also f = a 1 x 1 + a 2 x 2 + + (a in 1)x in. Since a 1 2 1 + a 2 2 2 + + a in 2 in < 2 i 1 + + 2 in and a in > 0, a 1 2 1 + a 2 2 2 + + a in 1 2 in < 2 i 1 + + 2 i n 1. Thus f < 2 i 1 + + 2 in 1 ; a contradiction. So e = e i 2 i. The result now follows by induction. Now to begin proving we have constructed an infinite, metrizable, topologically simple, 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 a + b a, b Z. Proof. (i) is obvious. For (ii) we can note that, if a = a i x i then a = a i x i, and hence a = a. (iv): Let a, b Z. Write a = a i x i and b = b i x i, with a i, b i Z all but finitely many zero. Then clearly, a + b = (a i + b i )x i. We then see, a + b a i + b i 2 i a i 2 i + b i 2 i = a i 2 i + b i 2 i. 10

Since a i x i and b i x i 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 a + b. (iii): Clearly 0 = 0. Suppose a 0. We need to show a 0. Since a = a, we may assume a > 0. Now as (x i ) and is strictly increasing, there is a (unique) N such that x N a < x N+1. We claim a 2 N. In particular a 0. If a = x N, then, by Proposition 4.2, a = 2 N. So suppose x N < a < x N+1. Assume, for a contradiction, that a < 2 N. Write a = a i x i with a i 2 i < 2 N. We see that a i = 0 i N, and a i < 2 i N i > N. n Now, as in the proof Proposition 4.2, a i x i < x n+1 n (same proof holds here). i=1 We will show inductively that the above in fact implies either a i x i < x N or ai x i x N+1, in both cases, a contradiction. First we note that the first possible non-zero a i is a N+1. Since x N+1 > x N it is clear that either a N+1 x N+1 < x N or a N+1 x N+1 x N+1. Now let k N + 1, and suppose either a N+1 x N+1 + + a k x k < x N or a N+1 x N+1 + + a k x k x N+1. Assume first that a N+1 x N+1 + + a k x k < x N. If a k+1 0, it is obvious that a N+1 x N+1 + + a k+1 x k+1 < x N. So assume a k+1 > 0. Since each a i < 2 i N, we have a N+1 x N+1 + + a k+1 x k+1 a N+1 x N+1 + + a k x k + x k+1 (2 1 1)x N+1 (2 k N 1)x k + x k+1. Now for each i, x i+1 > 2 i+1 x i. This gives x k+1 > (2 k+1 1)x k + + (2 N+3 1)x N+2 + 2 N+2 x N+1 > (2 k N 1)x k + + (2 2 1)x N+2 + 2x N+1. Putting these two inequalities together we get, a N+1 x N+1 + + a k+1 x k+1 > x N+1. Now assume a N+1 x N+1 + + a k x k x N+1. If a k+1 0, it s obvious that a N+1 x N+1 + + a k+1 x k+1 x N+1. So suppose a k+1 < 0. Because a N+1 x N+1 + + a k x k < x k+1, it s clear that a N+1 x N+1 + + a k+1 x k+1 < 0 < x N. Since the sum a i x i must be a finite sum we indeed have either a i x i < x N or ai x i x N+1. That is either a < x N or a x N+1, contradicting x N < a < x N+1. 11

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 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 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) 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, (z n ) say, in a converging to 1, or equivalently, such that (z n 1) converges to 0. Now notice that the sequence (y n (a) ) is the sequence (na 1) n N, and by construction contains a subsequence that is itself a subsequence of (x n ), say (x kn ). We can write this sequence in the form (z n 1) for some sequence (z n ) in a. By construction, for each n, z n 1 = x kn = 2 kn. Hence indeed (z n 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 (s n ) defined by s n = n x i. It is Cauchy, but has no limit in Z. Proof. It is Cauchy since: for m > n > 0, by Proposition 4.2, s m s n = x n+1 + + x m = 2 (n+1) + + 2 m < 2 n. This clearly converges to zero as n, m. We will show that (s n ) does not converge in Z. Let a Z be a candidate limit. By 4.2, lim s n = 2 n = 1, so a 0. Let N N be such that x N 1 a < x N. Let s first suppose a < 0. Fix n > N. We claim that s n a 2 N. Assume for a contradiction that s n a < 2 N. So we can write s n a = a i x i with ai 2 i < 2 N. We immediately see that a i = 0 i N and that a i < 2 i N i > N. k As previously (Proposition 4.2 and Lemma 4.3(iii)), we have that a i x i < x k+1 k (the same proof holds here too). Let m the largest number such that a m 0, so s n a = x 1 + + x n a = a N+1 x N+1 + + a m x m. Since a < 0, and, x N is larger then both a and x 1 + + x N 1 clearly i=1 i=1 x N + + x n < a N+1 x N+1 + + a m x m < 3x N + x N+1 + + x n. ( ) Suppose m > n. If a m < 0 then, as s n a > 0, a N+1 x N+1 + + a m 1 x m 1 > x m, 13

k which contradicts a i x i < x k+1 k. So a m > 0 (a m 0 by definition). We have So i=1 a m x m x m > (2 m 1)x m 1 + + (2 N+2 1)x N+1 + 2 N+1 x N. a m x m (3x N +x N+1 + +x n ) > (2 N+1 3)x N + n i=n+1 (2 i+1 2)x i + m 1 i=n+1 (2 i+1 1)x i. Hence for ( ) to be true we would need at least one i {N + 1,..., m 1} with a i > 2 i > 2 i N, which is a contradiction. So a i = 0 i > n. Now if a n 0, because ( ) implies x n < a N+1 x N+1 + + a n x n, we would need a N+1 x N+1 + + a n 1 x n 1 > x n ; a contradiction. Suppose a n > 1. Then a n x n 2x n > 2[(2 n 1)x n 1 + + (2 N+2 1)x N+1 + 2 N+1 x N ] > (2 n 1)x n 1 + + (2 N+2 1)x N+1 + 2 N+1 x N + 3x N + x N+1 + + x n. Hence we have the inequality a n x n (3x N + x N+1 + + x n ) > (2 n 1)x n 1 + + (2 N+2 1)x N+1 + 2 N+1 x N. So, for ( ) to hold, we require at least one a i 2 i ; a contradiction. Thus a n = 1. ( ) now reads x N + + x n < a N+1 x N+1 + + a n 1 x n 1 + x n < 3x N + x N+1 + + x n. Subtracting x n from this we get x N + + x n 1 < a N+1 x N+1 + + a n 1 x n 1 < 3x N + x N+1 + + x n 1. The same argument now applies to show a n 1 = 1. After this we repeat over and yield a i = 1 i : N + 2 i n. We eventually get x N + x N+1 < a N+1 x N+1 < 3x N + x N+1. 14

It s immediate that a N+1 2. But then a N+1 x N+1 2x N+1 > x N+1 + 2 N+1 x N > x N+1 + 3x N ; a contradiction. Now we assume a > 0. Fix n N + 2. If a = x N 1 then, by Proposition 4.2, n s n x N = 2 N + 2 i 2 (N+1), and (s j ) cannot converge to a. So assume a > x N 1. Now because a < x N < x 1 + +x N, and x 1 + + x N a < x N+1 < 2x N+1, 1=1 x N+1 + + x n < s n a < 3x N+1 + x N+2 + + x n. Assume that s n a < 2 (N+1) and write a i x i = s n a with a i 2 i < 2 (N+1). We can conclude similarly to the a < 0 case, that a i = 0 for all i N + 1 and i n + 1, and a i = 1 i: N + 3 i n. Hence reducing to the inequality x N+1 + x N+2 < a N+2 x N+2 < 3x N+1 + x N+2. Similar to before, this is a contradiction. We have proved that for every non-zero a Z. Let N be such that x N 1 a < x N. Then if n N + 2, s n a 2 (N+1). In particular (s k ) cannot converge to a. We have already ruled out s k 0 so (s k ) 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 (s n ) in Proposition 4.7. We prove that the subgroup s is closed in A. Being countable ensures s A. To show s is closed we prove that ε > 0 such that ts ε t Z \ {0}. Then for elements ts t s in s, ts t s = (t t )s ε, proving s is discrete, and hence closed. Since addition and subtraction are continuous, ts n ts, and therefore ts n ts. Thus it is sufficient to prove for n large enough ts n ε. Let t be a non-zero integer. As ts n = ts n n, we may assume t > 0. Choose l N such that 2 l 1 t < 2 l. We claim for n l + 2, ts n 1 8. 15

Suppose ts n < 1 8 for some n l + 2. Write ts n = a i x i, with a i Z all but finitely many a i = 0 and a i 2 i < 1 8. This immediately forces a i < 2 i 3 i. As before k (Proposition 4.2, Lemma 4.3(iii) and Proposition 4.7), we have that a i x i < x k+1 k. Let m N be maximal such that a m 0, so Now ts l < 2 l (x 1 + + x l ) < 2 l 2x l < x l+1, ts n = tx 1 + + tx n = a 1 x 1 + + a m x m. so ts n < x l+1 + tx l+1 + + tx n < (1 + 2 l )x l+1 + 2 l x l+2 + + 2 l x n, i.e. a 1 x 1 + + a m x m < (1 + 2 l )x l+1 + 2 l x l+2 + + 2 l x n. Since, a 1 x 1 + + a l x l < t(x 1 + + x l ) (as a i < 2 i 3 < 2 l 1 < t i l), and in turn t(x 1 + + x l ) < 2 l (x 1 + + x l ) < x l+1, it follows that a l+1 x l+1 + + a m x m < 2x l+1 + n i=l+1 i=1 2 l x i. (1) Now suppose m > n. If a m < 0, then as ts n > 0, we would need m 1 a i x i > x m ; a contradiction. So a m > 0 (a m 0 by definition). Now therefore a m x m [2x l+1 + a m x m x m > (2 m 1)x m 1 + + (2 l+3 1)x l+2 + 2 l+2 x l+1, n i=l+1 i=1 2 l x i ] > (2 l+2 2 l 4)x l+1 + n (2 i+1 2 l 1)x i + i=l+2 m 1 i=n+1 (2 i+1 1)x i. Thus, for (1) to hold, i {1,..., n 1} with at least a i > 2 i+2 2 l 2. But for all i N, 2 i+2 2 l 2 > 2 i 3. Hence we have a contradiction. So we conclude m n, i.e. a k = 0 k > n. Hence we have We now aim to show a n = t. a 1 x 1 + + a n x n = tx 1 + + tx n. (2) If a n < t, then a 1 x 1 + + a n 1 x n 1 > (t a n )x n x n ; a contradiction. So a n t. 16

Assume a n > t, then, because ts n 1 < 2 l (x 1 + + x n 1 ) < 2 l+1 x n 1 2 (n 1) x n 1 = 1 2 2n x n 1 < 1 2 x n (recall n l + 2), we have a n x n ts n x n ts n 1 > 1 2 x n > 1 2 [(2n 1)x n 1 + + (2 3 1)x 2 + 2 2 x 1 ]. Thus for (2) to be true we need at least one i {1,..., n 1} with a i > 1 2 (2i+1 1) > 2 i 3 ; a contradiction. Hence indeed a n = t. Equation (2) now reduces to a 1 x 1 + + a n 1 x n 1 = tx 1 + + tx n 1. We can now repeat the above over, yielding a l+2 = = a n = t. In particular, a l+2 = t, and thus ai 2 i t2 (l+2) 2 l 1 2 (l+2) = 1 8. Contradicting our initial assumption. So this assumption must be false, that is, ts n 1 8. 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 A. Any non-zero a A with a < 1 8 will not be in s, a = x 4 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 = (a n ) satisfying: 17

a n (0, 1] n N; a is non-increasing. a n 0 as n 2 n a n n N. For a, b A write a b whenever a n b n 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 (r, 0, 0,... ). For each a A, we will construct a complete group, G a, with the required properties. 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 H a of Z ℵ 0, consisting of the sequences h = (h n ) satisfying a n h n <. It is a subgroup, since, if h, h H a, with h = (h n ) and h = (h n), then hn h n a n h n a n + h n a n < by the triangle inequality. So h h H a. For h H a define h a = a n h n. Lemma 5.2. a is a norm on H a and with this norm H a is a topological group. Proof. We need to prove (i) h a 0 h H a ; (ii) h a = h a h H a ; (iii) h a = 0 h = 0 = (0, 0,... ); (iv) h + h a h a + h a h, h H a ; (v) Addition and subtraction are continuous. (i) is obvious. Given h H a h a = a n h n = a n h n = h a 18

so (ii) is true. Clearly 0 a = 0. Conversely, if h a = 0, i.e. an h n = 0, then as a n 0 n, h n = 0 n. So h = 0, proving (iii). Given h, h H a, by the standard triangle inequality h + h a = a n h n + h n a n h n + a n h n = h a + h a, so (iv) holds. (v) follows form lemma 4.4. Lemma 5.3. H a, with this norm, is complete. Proof. Let (h (m) ) be a cauchy sequence in H a. This means i a i h (m) i For each i N, since a i > 0, h (m) i h (k) i 0 as m, k. h (k) i 0 as m, k. Therefore every h (m) i Z m, we see h i Z and that h (m) i = h i converges, say to h i. Recalling that h (m) i eventually. We need to show the sequence h = (h n ) is in H a, and that h (m) h. Let ε > 0, and M N be such if m, k > M then i a i 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 N we certainly have Letting k we get n i=1 n i=1 a i h (m) i h (k) i < ε. a i h (m) i h i ε. Since this holds for all n N, we can send n, yielding i=1 a i h (m) i h i ε, i.e. h (m) h a ε. Hence indeed h (m) h. Note also that the above implies for any m > M, h (m) h H a. As H a is a subgroup of Z ℵ 0, it follows h H a. We now turn our attention to the simplicity of our example. Let (x n ) be the sequence of positive integers constructed in chapter 4. Recall it satisfies: 19

(I) x 1 = 1; (II) for every m N the sequence (mn 1) n N contains a subsequence, that itself is a subsequence of (x n ); (III) x n+1 > 2 n+1 x n n N. Note the condition (III) implies x n+1 > 2 1 a n x n n. We make use of the latter property of (x n ), and is the reason we imposed the last condition on A. Let K Z ℵ 0 consist of sequences c = (c n ) such that all but finitely many c n = 0, and c n x n = 0. Clearly K is a subgroup and is contained in H a. Let G a = H a /K. Lemma 5.4. K is closed in H a, and hence the quotient group G a is complete Proof. Claim: c a 1 c K \ {0}. Let c K \ {0}. Suppose c n1, c n2,..., c nk k 2). Then, by definition, c n1 x n1 + + c nk x nk = 0. are exactly all the non-zero terms (note Equivalently c n1 x n1 + + c nk 1 x nk 1 = c nk x nk. As x n+1 > 2 1 a n x n, we therefore have c n1 x n1 + + c nk 1 x nk 1 = c nk x nk x nk > 2 1 a nk 1 x nk 1 > 1 a nk 1 x nk 1 + 2 1 a nk 2 x nk 2. > 1 a nk 1 x nk 1 + + 1 a n2 x n2 + 2 1 a n1 x n1. For this to be true we need at least one i with c ni > 1 a ni, say c nl > 1 a nl. But then proving the claim. c a = c i a i > c nl a nl > 1 a nl a nl = 1, The claim shows K is a discrete subgroup, hence it is closed. As a quotient of a complete space by a closed space, G a is complete. We claim that the G a s are the complete groups we are after. Lemma 5.5. Z embeds as a cyclic dense subgroup of G a, via ι : t (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 tx 1 = 0. But x 1 = 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 H a, and thus im(ι) is dense in G a. Let h H a, and ε > 0. Since h i a i <, n N such that h i a i < ε. i=n+1 Define c = (c i ) as follows: Put c i = h i for i {2, 3,..., n}, put c 1 = n c i x i and c i = 0 otherwise. Then, by construction, c i x i = 0 (recall x 1 = 1), and hence c K. Set t = h 1 c 1. Then h (t + c) a = h 1 (t + c 1 ) a 1 + as required. i=2 n h i c i a i + h i a i = h i a i < ε i=2 i=n+1 i=n+1 Define Z = {(t, 0, 0,... ) + K : t Z} = im(ι). Notice Z G b b A. Remark. Notice that the above proof shows that Z is the quotient of the subgroup {(h n ) H a : all but finitely many h n = 0} by K. Proof. Let S = {(h n ) H a : all but finitely many h n = 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, {G a } a A. Each G a 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 b. If h H b, h = (h n ), then by definition, h i b i <. since a b, it follows that h i a i <, and hence h H a. So H b H a. Let f ba : G b G a, h + K h + K. Lemma 5.6. For every a, b A with a b, f ba is an injective continuous homomorphism. If a, b, c A, with a b c, then f ba f cb = f ca. 21

Proof. Let a, b A with a b. Because K is independent of a and b, and is contained in every Hã, ã A, it is clear that f ba is a well-defined injective homomorphism. It is continuous because h a h b h G b. The last part is obvious. By the above lemma we have an inverse mapping system consisting of A, the G a s and the f ba s. Let Z G a denote the inverse limit. As each G a is metrizable (so a A certainly Hausdroff), Z is a closed subgroup of the complete group G a. Hence Z is complete. We claim Z is the topological group we want. Lemma 5.7. Z = Z as groups. a A Proof. We prove Z = {(t + K) a A a A G a : 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 G a a A, given t + K Z and a, b A with a b, it s clear f ba (t + K) = t + K. Thus R Z. Conversely, assume Z R. Let (h a + K) a A Z \ R. Suppose first that h a Z a A. By choosing different co-set representatives (if necessary) we may assume h a = t a = (t a, 0, 0,... ) a A. Let a, b A (not necessarily comparable). Define c = (c n ), by c n = max{a n, b n }, then c A, and a, b c. f ca (t c + K) = t a + K, i.e. Therefore we must have t c + K = t a + K, and f cb (t c + K) = t b + K, i.e. t c + K = t b + K. Hence t a + K = t b + K. So a, b A, t a + K = t b + K, contradicting (h a + K) a A / R. Hence a A such that h a + K / Z. Write h a = (h i ) i N ; by the remark after Lemma 5.5, there must be infinitely many i such that h i 0. Let (h ni ) be a subsequence of (h i ) consisting entirely of non-zero terms. Define b = (b j ) as follows. For j such that 1 j n 1, put b j = 1; for i 2 and for j such that n i 1 < j n i, put b j = max{a ni 1, 1 i }. Clearly b A, and, by construction, a b (recall ã is non-increasing ã A). So h b + K G b such that f ba (h b + K) = h a + K, i.e. h b + K = h a + K. Hence h a = h b + k, for some k K. In particular h a H b. But a contradiction. Hence Z R. So Z = R, and hence Z = Z. hi b i h ni b ni b ni 1 i = ; 22

The topology on Z is the subspace topology from G a. Thus, considering Z with this topology, a sequence (t n ) in Z converges to a limit, say t, if and only if the sequence ((t n, 0, 0,... ) + K) converges to (t, 0, 0,... ) + K in G a, for every a A. 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 (x n ), it has a subsequence, (x kn ) say, such that, for each n N, x kn = ty n 1, for some positive integers y n. Writing z n = ty n, (z n ) is a sequence in t and is such that (z n 1) = (x kn ). Fix a A and n N. Define c 1 = x kn, c kn = 1, c i = 0 otherwise, and put c = (c i ). Then ci x i = x kn + x kn = 0, so c K. Furthermore (z n 1, 0, 0,... ) + c = (0,..., 0, 1, 0,... ), where the 1 is in the k n th position. Hence (z n 1, 0, 0,... ) + K = (0,..., 0, 1, 0,... ) + K. As (0,..., 0, 1, 0,... ) a = a kn 0 as n, it follows that (z n, 0, 0,... ) + K (1, 0, 0,... ) + K in G a. Since this holds a A, z n 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,... ). i write Let U = {U i } i N, be a countable collection of open sets, each containing 0. For each U i = {(t + K) a A Z : t + K U a (i) }, with all but finitely many U a (i) = Z (recall Z = {(t + K) a A G a a A : t + K Z}.) As A is uncountable and there are only countably many U i, there are uncountably many a A such that U (i) a = Z i N, 23

in particular there is at least one, say b. Let U b be any open subset of G b containing 0 + K, and such that Z U. Then the open set {(t + K) a A Z : t + K U b }, 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 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 Z d denote the group Z equipped with the discrete topology. Proposition 2.20 says Z d = T. A topological isomorphism is given by χ χ(1). Since any continuous map Z T, is certainly a continuous map Z d T, we can embed the group Z into T via χ 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. 24

Proof. It suffices to check the topology is finer at 1 T. Let ε > 0. Write B ε (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 χ(1), we see U(F, ε) is just B ε (1). Thus topology on Z is indeed finer then that of T. Given a continuous χ : Z T, we can define a map ˆχ : G T as follows. For x Z, define ˆχ(x) = χ(x). For x G \ Z, choose a sequence in Z, say (x n ), converging to x (which exists as Z is dense in G). Define ˆχ(x) = lim χ(x n ). Since χ is continuous, and T is complete, ˆχ is a well-defined continuous homomorphism G T, i.e. ˆχ 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 ζ, where ζ is a primitive nth root of unity. Suppose φ is continuous and let ˆφ be it s unique extension to G. Then ker ˆφ is a closed subgroup of G. ker ˆφ {0} since n ker ˆφ, and 1 / ker ˆφ, so ker ˆφ 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 {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}, 25

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 {G α } α A is a collection of non-trivial Hausdorff abelian groups, and G = G α. If G is topologically simple then A = 1. α A Proof. Since we exclude the trivial group in the definition of topologically simple, A 0. Suppose A > 1, possible infinite. Let β A. Set H = {(g α ) G : g α = 0 α β}. Clearly H is a proper subgroup of G. For every α, G α is Hausdorff and so {0} is closed in G α. It follows G α \{0} is open, for every α. For α β put U α = G α \{0}. Set U β =. Then U = {(g α ) : g α U α } 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 B ε (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 Therefore the product topology has basis The topology induced by has basis B 0 = {B ε (a) : a G, ε > 0}. B = {U V : U, V B 0 }. 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. 26

Take U V B, and assume U = B ε (c) and V = B δ (d). With out loss of generality assume ε < δ. Suppose (a, b) B ε(c, d) then a c a c + b d = (a, b) (c, d) < ε, and similarly b d (a, b) (c, d) < ε 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 B ε(c) and b B ε (d), then (a, b) (c, d) = a c + b d < ε + ε = 2ε. So B ε (c) B ε (d) U. 27

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, Göttingen, 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, Atlantis Press, Amsterdam Paris, World Scientific, 2008 [mc] B.MacCluer, Elementary Functional Analysis, Springer, New York, 2009 [S] Dr Stanislav Shkarin, Private Communication 28