Reflexive group topologies on Abelian groups

REFLEXIVE GR OUP TOPOLOGIES ON ABELIAN GR OUPS S.S. GABRIYEL Y AN Abstract. It is pro v ed that an y infinite Ab elian group of infinite exponen t admits a non-discrete reflexive group topology . Introduction F or a top ological g roup G , the group G ∧ of c o n tin uous homomorphisms (char- acters) int o the tor us T = { z ∈ C : | z | = 1 } e ndowed with the compact-op en top ology is calle d the char acter gr oup of G and G is named Pontryagin r efl ex ive or r eflexive if the canonical homomorphism α G : G → G ∧∧ , g 7→ ( χ 7→ ( χ, g )) is a top ological isomorphism. In the article w e consider the f ollowing question. Problem 1. Is any infinite A b elian gr oup a dmits a n on- discr ete r eflexive gr oup top olo gy ? A group G with the discrete topology is denoted b y G d . The exp onen t of G (=the least co mmo n m ultiple of the or ders of the elements of G ) is denoted by exp G . The subgroup of G which gener ated by an element g is denoted by h g i . F ollowing E.G.Zelenyuk and I.V.Protasov [5], we say that a se quence u = { u n } in a gro up G is a T - se quenc e if there is a Hausdorff group top ology o n G fo r which u n conv erges to zer o. The gro up G e q uipped with the finest gr o up top ology with this prop erty is denoted by ( G, u ). Using the metho d o f T -sequences, they pr o ved that every infinite Abelian gr o up admits a complete group topo logy for whic h c haracter s do not sepa rate points. Using this metho d, we give the p ositive answer to Problem 1 for gr oups of infinite exp onent. W e prove the following theo rem. Theorem 1. Any infinite Ab elian gr oup G such that exp G = ∞ admits a non- discr ete r eflexive gr oup t op olo gy. Let G be a Bor el subg roup of a Polish group X . G is called p olishable if there exists a Polish group top ology τ on G such that the identit y map i : ( G, τ ) → X , i ( g ) = g , is contin uous. A δ -neighbor ho o d of z ero in a Polish gro up is denoted by U δ . Let X be a compa c t metrizable group and u = { u n } a sequence of elements of X ∧ . W e denote by s u ( X ) the set o f all x ∈ X such that ( u n , x ) → 1. Let G be a s ubgroup of X . If G = s u ( X ) w e sa y that u char acterizes G and that G is Date : may 2009. 2000 Mathematics Subje ct Classific ation. Primary 22A10, 43A40; Secondary 54H11. Key wor ds and phr ases. Characterized group, T -sequence, dual group, Polish gr oup, reflexive group. The author was partially supported by Israel Ministry of Immigrant Absorption. 1 2 S.S. GABRIYEL Y AN char acterize d (by u ). By Theor em 1 [3], if G is characterized, then it is p olishable by the following metric (0.1) ρ ( x, y ) = d ( x, y ) + sup {| ( u n , x ) − ( u n , y ) | , n ∈ N } , where d is the initial metric o n X . G w ith the metric ρ is denoted by G u . The integral part of a real num ber x is denoted by [ x ]. By k x k we denote the distance o f a rea l num ber x to the near est integer. W e also use the follo wing inequality π | ϕ | 6 | 1 − e 2 π iϕ | 6 2 π | ϕ | , ϕ ∈ [ − 1 2 ; 1 2 ). 1. The Proof Let G b e a n infinite Ab elian g roup and H be its infinite subgroup. If H admits a non-trivial reflexive g roup topolog y τ , then w e ca n extend τ to G such that H will b e an op en subgroup. Then, b y [1], G a lso will be reflexive. It is w ell k no wn that any Ab elian gr oup G of infinite exp onen t contains at least one o f the following groups: (1) Z ; (2) Z ( p ∞ ) fo r so me prime num b er p ; (3) G = L n Z ( b n ), whe r e b 1 6 b 2 6 . . . , b n → ∞ . Thu s it is enough to prov e Theorem 1 for these three case s only . 1.1. The case Z . A non- tr ivial reflexiv e group top ology on Z is constructed in Theorem 2 [2]. 1.2. The case Z ( p ∞ ) . Pr o of. Set u = { u k } , u k = 1 p n k +1 , where n 1 < n 2 < . . . and n k +1 − n k → ∞ . By (25 .2) [4], if ω = ( a n ) ∈ ∆ p = Z ( p ∞ ) ∧ , where 0 6 a n < p , then (1.1) ( u k , ω ) = exp  2 π i p n k +1 ( a 0 + pa 1 + · · · + p n k a n k )  . By (1 0.4) [4], if ω 1 6 = ω 2 ∈ ∆ p , then d ( ω 1 , ω 2 ) = 2 − n , where n is the minima l index such that a 1 n 6 = a 2 n . F or a n y ω = ( a n ) ∈ ∆ p and k > 1, we put m k = m k ( ω ) = max { d k , n k − 1 } , where d k = n k if 0 < a n k < p − 1 , or d k = min { j : either a s = 0 for every j < s 6 n k , or a s = p − 1 for every j < s 6 n k } . Then n k − 1 6 m k 6 n k and the equality m k = n k is p ossible only if p 6 = 2 . Set ω 0 = (1 , 0 , 0 , . . . ) ∈ ∆ p . It is clea r, b y (1.1), that ω 0 ∈ s u (∆ p ). Since h ω 0 i is dense in ∆ p , then, b y Theore m 3 [3], u is a T -sequence and ( Z ( p ∞ ) , u ) ∧ = G u , where the Polish gro up G u is s u (∆ p ) with the Polish group metric ρ [see (0.1)]. W e need the following three lemmas. Lemma 2. ω ∈ s u (∆ p ) ⇔ n k − m k → ∞ . REFLEXIVE GR OUP TOPOLOGIES ON ABELIAN GROUPS 3 Pr o of. W e can r ewrite (1.1) as follows: (a) if a s = 0 for m k < s 6 n k , then (mo d 1) (1.2) 1 2 π i Arg( u k , ω ) = 1 p n k − m k +1 m k X l =0 a l p m k − l ; (b) if a s = p − 1 for m k < s 6 n k , then (mo d 1) 1 2 π i Arg( u k , ω ) = 1 p n k − m k +1 m k X l =0 a l p m k − l + n k X l = m k +1 p − 1 p n k +1 − l = (mo d 1 ) (1.3) = 1 p n k − m k +1 m k X l =0 a l p m k − l − 1 p n k − m k ; (c) if m k = n k (and, hence, p 6 = 2 ), then (mo d 1) (1.4) 1 2 π i Arg( u k , ω ) = n k − 1 X l =0 a l p n k +1 − l + a n k p . Assume that ω ∈ s u (∆ p ) and n k − m k 6→ ∞ . Let ca se (a) b e fulfilled for k 1 < k 2 < . . . and n k t − m k t = r > 0 . Then, b y (1.2), we hav e (mo d 1) 1 p r +1 6 a m k t p r +1 6 1 2 π i Arg( u k t , ω ) < 1 p r , and, hence, ( u k t , ω ) 6→ 1. It is a contradiction. Let ca se (b) b e fulfilled for k 1 < k 2 < . . . and n k t − m k t = r > 0. Since a m k t 6 p − 2 , then m k t X l =0 a l p m k t − l < p − 2 + 1 p ∞ X l =0 p − 1 p l = p − 2 + p − 1 p · p p − 1 = p − 1 . Then, by (1.3), we hav e (mo d 1) − 1 p r 6 1 2 π i Arg( u k t , ω ) < 1 p r +1 · ( p − 1) − 1 p r = − 1 p r +1 , and, hence, ( u k t , ω ) 6→ 1. It is a contradiction. Let case (c) be fulfilled for k 1 < k 2 < . . . and n k t = m k t . Then p > 2 a nd 0 < a n k t < p − 1 . Th us, by (1.4), we hav e (mo d 1) 1 p 6 a n k t p 6 1 2 π i Arg( u k t , ω ) < a n k t p + 1 p 6 p − 1 p , and, hence, ( u k t , ω ) 6→ 1. It is a contradiction. The conv erse a ssertion evidently follows from (1.2) and (1.3).  Lemma 3. h ω 0 i is dense in G u and, henc e, G u is monothetic. Pr o of. Let ε > 0 a nd ω = ( a n ) ∈ G u . Cho ose r such that 1 p r < ε 10 . By Lemma 2, we c a n c hoo se k 0 such tha t 1 2 n k 0 − 1 < ε 10 and n k − m k > r + 1 for ev ery k > k 0 . Cho ose q suc h that ω − ω q 0 = (0 , . . . , 0 m k 0 , a m k 0 +1 , . . . ) . 4 S.S. GABRIYEL Y AN Thu s, if k < k 0 , then, ( u k , ω − ω q 0 ) = 1. If k > k 0 , then, by (1.2) and (1.3), | 1 − ( u k , ω − ω q 0 ) | 6 2 π · 1 p n k − m k < 0 . 7 ε. So ρ ( ω , ω q 0 ) < ε and h ω 0 i is dense in G u .  Lemma 4. G ∧ u = Z ( p ∞ ) algebr aic al ly. Pr o of. Since h ω 0 i is dense in G u , any contin uous c haracter χ of G u is defined b y its v alue o n ω 0 . Let ( χ, ω 0 ) = exp { 2 π iα } for so me α ∈ [0 , 1). It is eno ugh to pr o ve that α ∈ Z ( p ∞ ). Let α = ∞ X i =1 b i p i , where 0 6 b i < p. Let 0 < ε < 1 p 2 . Since χ is contin uous, then there ex is ts δ > 0 such that (1.5) | 1 − ( χ, ω ) | < ε, ∀ ω ∈ U δ . Cho ose r 0 and k 0 such that 1 p r 0 < δ 10 and 1 2 n k 0 < δ 10 . Let ω = ( a n ) ∈ s u (∆ p ) hav e the fo llo wing form (i) There exist k 0 < k 1 < · · · < k s , s ∈ N , suc h that a n ∈ [0 , p − 1] if n ∈ [ n k i + 1 , n k i +1 − r 0 − 1] , i = 0 , . . . , s − 1 , and a n = 0 o therwise. Then, by (0.1), (1.1) and (1.2), w e hav e ρ (0 , ω ) 6 δ 10 + sup {| 1 − ( u n , ω ) | , n ∈ N } 6 δ 10 + 2 π p r 0 < δ, i.e. ω ∈ U δ . 1) L et p > 2. Set R ( k ) = { i : n k − 1 + 1 < i 6 n k + 1 and 0 < b i < p − 1 } . If R ( k ) is not empty , we set r k = min { i : i ∈ R ( k ) } . L et us pr ove that ther e exists C 1 > 0 su ch that for every k > k 0 if i ∈ R ( k ) , then 0 6 n k + 1 − i < C 1 . Henc e ther e exists k ′ 0 > k 0 such that for any k > k ′ 0 if n k − 1 + 1 < i 6 n k + 1 − C 1 , then either b i = p − 1 or b i = 0 . Assume the co n v erse a nd there exists a subsequence r k q such that n k q − r k q → ∞ . W e can assume that n k q − r k q > r 0 . Since ω p r k q − 1 0 = (0 , . . . , 0 n k q − 1 , . . . , 0 , 1 r k q , 0 , . . . ) satisfies conditio n (i), then it is contained in U δ . On the o ther hand,  χ, ω p r k q − 1 0  = exp ( 2 π i b r k q p + ∞ X i =1 b r k q + i p i +1 !) and 1 p 6 b r k q p 6 b r k q p + ∞ X i =1 b r k q + i p i +1 6 b r k q + 1 p < p − 1 p . So     1 −  χ, ω p r k q − 1 0      > π p > ε . This inequality contradicts to (1 .5). Now we ca n choose k ′ 0 > k 0 such that n k − 1 < n k − C 1 , ∀ k > k ′ 0 . 2) Set T ( k ) = { i : n k − 1 + 1 < i < n k + 1 such that b i = p − 1 and b i +1 = 0 } . If T ( k ) is not empt y , we set t k = min { i : i ∈ T ( k ) } . REFLEXIVE GR OUP TOPOLOGIES ON ABELIAN GROUPS 5 L et u s pr ove that t her e exist s C 2 > ( C 1 ) > 0 such that for every k > k ′ 0 if i ∈ T ( k ) , t hen 0 6 n k − i < C 2 . Henc e ther e exists k ′′ 0 > k ′ 0 such that for every k > k ′′ 0 and i ∈ T ( k ) if b i > 0 , then b i = b i +1 = · · · = b n k − C 2 = p − 1 . Assume the conv erse and there exists a subsequence t k q such that n k q − t k q → ∞ . W e can assume that k 1 > k 0 and n k q − r k q > r 0 . Since ω p t k q − 1 0 = (0 , . . . , 0 n k q − 1 , . . . , 0 , 1 t k q , 0 , . . . ) satisfies conditio n (i), then it is contained in U δ . On the o ther hand,  χ, ω p t k q − 1 0  = exp ( 2 π i p − 1 p + ∞ X i =1 b t k q + i p i +2 !) and 1 − 1 p 6 p − 1 p + ∞ X i =1 b t k q + i p i +2 6 p − 1 p + p − 1 p 3 1 1 − 1 /p = 1 − p − 1 p 2 . So     1 −  χ, ω p t k q − 1 0      > π p 2 > ε . This ineq ua lit y co n tradicts to (1.5). Cho ose k ′′ 0 > k ′ 0 such that n k − 1 < n k − C 2 , ∀ k > k ′′ 0 . 3) L et us pr ove that ther e exist C 3 > C 2 and k ′′′ 0 > k ′′ 0 such that for every k > k ′′′ 0 either b n k − 1 +2 = b n k − 1 +3 = · · · = b n k − C 3 = p − 1 or b n k − 1 +2 = b n k − 1 +3 = · · · = b n k − C 3 = 0 . Set S ( k ) = { i : n k − 1 + 1 < i < n k − C 2 and b i = p − 1 } . Denote by s k = min { i ∈ S ( k ) } if S ( k ) 6 = ∅ and s k = n k − C 2 otherwise. By item 2), it is enough to prove that the sequence { n k − s k , wher e k is chosen such that s k > n k − 1 + 2 } is b ounded (then we can put C 3 is the maximum of this sequence and choose k ′′′ 0 > k ′′ 0 such that n k − 1 + 2 < n k − C 3 , ∀ k > k ′′′ 0 ). Assume the conv erse a nd there ex ists a subsequenc e s k q such that s k q > n k q − 1 + 2 and n k q − s k q → ∞ . Then b i = 0 for n k q − 1 + 1 < i < s k q and ω p s k q − 2 0 = (0 , . . . , 0 n k − 1 , . . . , 0 , 1 s k q − 1 , 0 , . . . ) ∈ U δ for eno ugh big q . On the other hand,  χ, ω p s k q − 2 0  = exp ( 2 π i p − 1 p 2 + ∞ X i =1 b s k q + i p i +2 !) and p − 1 p 2 6 p − 1 p 2 + ∞ X i =1 b s k q + i p i +2 6 p − 1 p 2 + p − 1 p 3 1 1 − 1 /p = 1 p . So     1 −  χ, ω p s k q − 2 0      > π ( p − 1) p 2 > ε . This inequality contradicts to (1.5). 4) Set A = { k : k > k ′′′ 0 and b n k − 1 +2 = b n k − 1 +3 = · · · = b n k − C 3 = 0 } . W e can assume that A is infinite. Indeed, if A is finite, then ( − χ, ω 0 ) = exp ( 2 π i ∞ X i =1 p − 1 − b i p i !) and we can consider the character − χ instead o f χ . Denote b y l ( k ) = min { i : whe r e n k − C 3 < i a nd b i > 0 } , k > k ′′′ 0 . 6 S.S. GABRIYEL Y AN Assume that α 6∈ Z ( p ∞ ). Then, by item 3), ther e ex is t a > 0 and a subsequence k q such that n k q + 2 − l ( k q ) = a . Choo se e k > k ′′′ 0 such that n k − n k − 1 > r 0 + a + 3 for every k > e k . Set h ( q ) = n k q − ( r 0 + a + 3). Then l ( k q ) − h ( q ) = r 0 + 5 and ω p h ( q ) − 1 0 = (0 , . . . , 0 n k q − 1 , . . . , 0 , 1 h ( q ) , 0 , . . . ) satisfies conditio n (i). Put w ( j ) = ( h (1) − 1) + ( h (2) − 1) + · · · + ( h ( j ) − 1). Then ω p w ( j ) 0 also s a tisfies condition (i) for every j and, hence, it is co n tained in U δ . Since  χ, ω p h ( q ) − 1 0  = exp ( 2 π i b l k q p r 0 +6 + ∞ X i =1 b i p r 0 +6+ i !) and b l k q p r 0 +6 < b l k q p r 0 +6 + ∞ X i =1 b i p r 0 +6+ i < b l k q + 1 p r 0 +6 , then (mo d 1) 1 p r 0 +6 j X q =1 b l k q 6 1 2 π i Arg  χ, ω p w ( j ) 0  6 1 p r 0 +6 j X q =1 ( b l k q + 1) . It is clear that ther e exists j s uc h that    1 −  χ, ω p m ( j ) 0     > ε . This inequa lit y contradicts to (1.5). Thus α ∈ Z ( p ∞ ).  L et us pr ove The or em 1 for the c ase Z ( p ∞ ). By Lemma 4, G ∧ u = Z ( p ∞ ) alge- braically . By Prop osition 1 [3], G u is re fle x iv e. So Z ( p ∞ ) with the top ology of G ∧ u is als o reflexive.  1.3. The case G = L n Z ( b n ) . Pr o of. Assume that G = L n Z ( b n ), whe r e b 1 6 b 2 6 . . . , b n → ∞ . The metric d on G ∧ d = Q n Z ( b n ) is defined as follo ws: if ω 1 6 = ω 2 ∈ G ∧ d , then d ( ω 1 , ω 2 ) = 2 − n , where n is the minimal index such that a 1 n 6 = a 2 n . Set u = { u n } , wher e u n = e n is a g enerator of Z ( b n ). Then s u ( G ∧ d ) =  ω = ( a n ) ∈ G ∧ d : ( u n , ω ) = exp  2 π i a n b n  → 1  . So: ω ∈ s u ( G ∧ d ) if and only if    a n b n    → 0. Evidently , G is dense in G ∧ d and G ⊂ s u ( G ∧ d ). By Theorem 3 [3], u is a T - sequence and ( G, u ) ∧ = G u , where the Polish group G u is s u ( G ∧ d ) with the Polish group metric ρ [see (0.1)]. 1. L et us pr ove tha t G is dense in G u . Let ω = ( a n ) ∈ G u and ε > 0 . Cho ose n 0 such that | 1 − ( u n , ω ) | < ε/ 10 for all n > n 0 . Cho ose m > n 0 such tha t d ( ω , ω m ) < ε/ 10, where ω m = ( a 1 , . . . , a m , 0 , . . . ). Then | ( u n , ω ) − ( u n , ω m ) | = 1 for n < n 0 and ρ ( ω , ω m ) < ε/ 10 + ε/ 1 0 < ε . Thus G is dense in G u . 2. L et us pr ove tha t G ∧ u = G algebr aic al ly . By item 1, the c onjugate homomorphis m G ∧ u → G ∧ d is a monomor phis m. So any χ ∈ G ∧ u we can repr esen t in the for m χ = ( c n ) ∈ G ∧ d , 0 6 c n < b n . W e need to prov e only tha t c n = 0 for all enough big n . REFLEXIVE GR OUP TOPOLOGIES ON ABELIAN GROUPS 7 Let ε > 0. Since χ is contin uous, ther e exists an integer M > 10 such that | 1 − ( χ, ω ) | < ε, ∀ ω ∈ U 1 / M . By the de finitio n of the metric ρ on G u (0.1), there exists k 0 such that if ω ∈ G u has the form ω = (0 , . . . , 0 n k 0 − 1 , a n k 0 , a n k 0 +1 , . . . ) and (1.6)     1 − exp  2 π i a n b n      < 1 M for a ll n > n k 0 , then ω ∈ U 1 / M , and, in par ticular, | 1 − ( χ, ω ) | < ε . Now assume the converse and there ex ists a sequence n 1 < n 2 < . . . such tha t c n k > 0. W e can assume that c n k b n k conv erges to λ ∈ [0 , 1]. Ther e exist three po ssibilities. 1) λ ∈ (0 , 1 ). Then w e can as sume that α <    c n k b n k    6 1 2 for so me α > 0 and all k . Let ε < α . Cho ose k > k 0 such that b n k > 10 M and s e t ω = (0 , . . . , 0 , 1 n k , 0 , . . . ) . Since    1 − exp n 2 π i 1 b n k o    < 2 π b n k < 1 M , then, b y (1.6), we hav e ω ∈ U 1 / M . Thus | 1 − ( χ, ω ) | < ε . O n the other hand | 1 − ( χ, ω ) | =     1 − exp  2 π i c n k b n k      > π     c n k b n k     > π α > ε. It is a contradiction. 2) λ = 0. Let ε < 0 . 01 . Choose k > k 0 such that (1.7) c n l b n l < 1 20 π M 3 , for every l > k . Set a n l = h b n l 20 πM c n l i and ε l = b n l 20 πM c n l − a n l < 1. Then, by (1.7), a n l > M > 0. Put ω = (0 , . . . , 0 , a n k , 0 , . . . , 0 , a n k +1 , 0 , . . . , 0 , a n k + M − 1 , 0 , . . . ) . It is clear, by (1.6), that ω ∈ U 1 / M and, hence , | 1 − ( χ, ω ) | < ε . O n the other hand, since k + M − 1 X l = k a n l · c n l b n l < k + M − 1 X l = k b n l 20 π M c n l · c n l b n l = M · 1 20 π M = 1 20 π , and k + M − 1 X l = k a n l · c n l b n l = k + M − 1 X l = k  b n l 20 π M c n l − ε l  · c n l b n l = M · 1 20 π M − k + M − 1 X l = k ε l · c n l b n l > (b y (1.7 )) > 1 20 π − M 20 π M 3 = 1 20 π  1 − 1 M 2  > 0 . 9 20 π , then | 1 − ( χ, ω ) | =      1 − exp ( 2 π i k + M − 1 X l = k a n l · c n l b n l )      > 0 . 04 > ε. It is a contradiction. 3) λ = 1. Let ε < 0 . 01 . Choose k > k 0 such that (1.8) b n l − c n l b n l < 1 20 π M 3 , for every l > k . 8 S.S. GABRIYEL Y AN Set a n l = h b n l 20 πM ( b n l − c n l ) i and ε l = b n l 20 πM ( b n l − c n l ) − a n l < 1. Then, b y (1.8), a n l > M > 0. Put ω = (0 , . . . , 0 , a n k , 0 , . . . , 0 , a n k +1 , 0 , . . . , 0 , a n k + M − 1 , 0 , . . . ) . It is clear, by (1.6), that ω ∈ U 1 / M and, hence , | 1 − ( χ, ω ) | < ε . O n the other hand, since k + M − 1 X l = k a n l · c n l b n l (mo d 1 ) = − k + M − 1 X l = k a n l · b n l − c n l b n l , we ca n rep eat the computations in item 2), and obtain tha t ε < 0 . 04 < | 1 − ( χ, ω ) | . It is a contradiction. So ( G, u ) ∧∧ = G ∧ u = G a lgebraically . 3. L et us pr ove The or em 1 f or the c ase G = L n Z ( b n ). By item 2 , G ∧ u = G algebraic ally . By P ropo sition 1 [3], G u is reflexive. So G with the to polog y of G ∧ u is als o reflexive.  References 1. W. Banaszczyk, M .J. Chasco and E. Martin- P einador. Op en subgroups and Pon try agin dual- ity . Math. Z. 215 (1994), 195-204. 2. S.S. Gabriyely an. Groups of quasi- in v ar iance and t he P on try agin dua lity . arXiv:math.GN/0812.1671. 3. S.S. Gabriy ely an. On T -sequences and cha racterized subgroups. arXi v:math.GN/090 2.0723. 4. E. Hewitt and K.A. Ross. Abstr act Harmonic Analysis , V ol. I, 2nd ed. (Springer-V erlag, Berlin, 1979). 5. E.G. Zelen yuk and I.V. Protaso v. T op ologies on abelian groups. Math. USSR Izv. 37 (1991), 445-460. Russian original: Izv. Akad. Nauk SSSR 54 (1990), 1090–1107 . S.S. Gabriyel y an, Dep ar tment of Ma thema tics, Ben-Gurion Univ ersity of the Negev, Beer-Shev a, P.O. 653 , Israel E-mail addr ess : saa k@math.bg u.ac.il