Universal valued Abelian groups

UNIVERSA L V ALUED ABELIAN GR OUPS PIOTR NIEMIEC Abstract. The counterparts of the Uryso hn universal spa ce in category of metric spaces a nd the Gurari ˇ ı space in categ ory of Ba- nach spaces are constructed for separable v alued Ab e lian gr oups of ﬁxed (ﬁnite) exp onents (and for v alued g roups of s imilar type) and their uniquene s s is established. Geometry of these groups, denoted by G r ( N ), is inv estigated a nd it is shown that each o f G r ( N )’s is homeomorphic to the Hilb ert spa ce l 2 . Those of G r ( N )’s which are Urysohn as metric spaces are recognize d. ‘Linear-like’ s truc- tures on G r ( N ) are studied and it is pr ov ed that every separable metrizable top olog ical vector space may be enlar ged to G r (0) with a ‘linear-like’ structur e w hich extends the linear structure o f the given space. 2010 MSC: Primary 54H11 ; Se c ondary 22K45, 22A05, 46A99. Key words: v a lued Abelia n gr oup; Ab elian gr oup of ﬁnite exp o- nent ; Polish group; universal P olish Abelian g roup; Ur y sohn uni- versal metric space; extending con tinuous homomo r phisms; uni- versal disp osition prop erty; top olo gical pse udovector g roup. 1. Introd uction In sev eral branc hes of mathematics, suc h as metric space theory or functional analysis, there are known examples of the so- called univer- sal spaces. Probably the most famous example in this sub ject is the Banac h space C ([0 , 1]) of all real-v alued contin uous functions on [0 , 1] whic h is univ ersal for separable normed vec tor spaces (o v er R ) as w ell as for separable metric spaces. Here unive rsalit y means that eve ry separable normed v ector space (resp ectiv ely separable metric space) is isometrically-linearly isomorphic (repsectiv ely isometric) to a linear subspace (to a subse t) of C ([0 , 1]) . This is kno wn as the Ba nac h-Mazur theorem. Undoubtedly , the ‘concreteness’ of the univers al space in this theorem merits its fame. Ho w ev er, it is no so easy to c haracter- ize C ([0 , 1]) among Banach ( or metric) spaces. There is also a general idea, in the spirit of F ra ¨ ıss ´ e limits, of constructing univ ersal spaces whic h are sumiltaneously uniquely determined b y certain conditions related to the so-called unive rs a l disp osition pr op erty (see b elow ; with this terminology we fo llo w e.g. Wies la w Kubi ´ s). Surely , the ﬁrst ex- ample of a univ ersal space of this kind w as giv en b y Urysohn [2 4, 25]. The Urysohn univ ersal metric space, usually denoted b y U , is a unique (up to isometry) separable complete metric space whic h has univers al disp osition prop ert y for ﬁnite metric spaces. That is: 1 2 P . NI EMIEC Every is o metric map of a subset of a ﬁn i te metric sp ac e X i n to U is e x tendable to an is o metric map of the whole sp ac e X into U . There is also a b ounded v ersion of U . It ma y be deﬁned in the follo wing w ay . A metric space X is said to b e Urysohn iﬀ (U0) X is separable and complete, (U1) (unive rsalit y) ev ery separable metric space of diameter no greater than the diameter of X is isometrically em b eddable in to X , (U2) ( ω -homogeneity ) e v ery isometry b et wee n t wo ar bitrary ﬁn ite s ub- sets o f X is extendable to an isometry of X onto X . A fundamen t al res ult on Ury sohn spaces states that for eve ry r ∈ [0 , ∞ ] there is a unique (up to isometry ) Urysohn metric space ( denoted by U r ) of diameter r . It is also easily show n that U r is uniquely determined (among separable complete metric spaces of diameter no great er than r ) b y univ ersal disposition prop ert y for ﬁnite metric spaces o f diameter no greater than r . If we pass from the category of metric spaces to Banac h spaces, univ ersal disp osition prop erty ma y b e deﬁned as follows (cf. [8]). A Banac h space E is said to ha v e unive rs a l disp osition pr op erty (for ﬁnite dimensional Banach sp ac es) iﬀ ev ery isometric linear map of a linear subspace of a ﬁnite dimensional Banac h space F in to E is e xtendable to an isometric linear map of F in to E . G urari ˇ ı [8] has pro v ed that there is no separable Banac h space with univ ersal disp osition prop ert y . In the same pap er he ha s constructed a separable Ba nac h space G , which is now ada ys called the Gur ari ˇ ı space, with almost universal disp osition pr op erty deﬁned in the following w a y (the same space was also built by Lazar and Lindenstrauss [11] in a diﬀerent contex t): Every iso metric line ar map ψ : E → G o f a line ar sub- sp ac e E of a ﬁnite di m ensional Banach sp ac e F admits a li n e ar e xtension b ψ : F → G such that (1 − ε ) k x k 6 k b ψ ( x ) k 6 (1 + ε ) k x k for al l x ∈ F wher e ε is an arbi- tr a ril y give n r e al numb er fr om (0 , 1) . Gurari ˇ ı has also sho wn that G is unive rsal for separable Banac h spaces (i.e. tha t ev ery separable Banac h space admits an isometric linear em b edding in to G ). It was Lusky [12] who ﬁrst prov ed the uniquenes s o f G up to isometric linear isomorphism (other pro of of the uniquenes s is con tained in [13 ]; the pro of whic h inv olv es the bac k-a nd-forth tec hnique w as giv en b y Solec ki [20]). Thus , the Gurari ˇ ı space is a unique separable Banac h space with almost univ ersal disp osition prop ert y . There is a striking resem blance b et w een (almost) unive rsal disp osi- tion prop erties of U and G . In a sense, b oth these sp aces corresp ond to eac h other in categories of metric spaces and Banach spaces. The aim of this pap er is to prov e the existence (together with uniqueness) UNIVERS AL V ALUED ABELIA N GR OUPS 3 of a unive rsal (in the sense of em b edding by isometric gr oup homomor- phisms) group with univers al disp osition prop ert y (for ﬁnite g roups) in the class (and some of their sub classes) of all separable v alued Ab elian groups o f class O 0 , deﬁned b y the follow ing condition: A v alued Abelian group ( G, + , p ) is said to b e of class O 0 iﬀ lim n →∞ p ( na ) n = 0 for ev ery a ∈ G . (In particular, if p is a v a lue on an Ab elian gr oup ( G, +), then eac h of the v alued groups ( G, + , min( p, 1)), ( G, + , p p +1 ) and ( G, + , p α ) for 0 < α < 1 is of class O 0 .) What w e exactly mean is explained b elo w. Denote b y G the class of all separable v alued Ab elian groups. Let G ∞ (0) and G 1 (0) stand for the classes of all g roups ( G, + , p ) ∈ G of class O 0 and, resp ectiv ely , for whic h p 6 1. Note that G 1 (0) ⊂ G ∞ (0). Additionally , for natural N > 1 let G ∞ ( N ) ( ⊂ G ∞ (0)) consist of all groups ( G, + , p ) ∈ G of exp onen t N . Finally , put G 1 ( N ) = G 1 (0) ∩ G ∞ ( N ). W e sa y a function ω : [0 , ∞ ) → [0 , ∞ ) is a mo dulus o f c ontinuity iﬀ ( ω 1) ω is monotone increasing, that is, ω ( x ) 6 ω ( y ) provided 0 6 x 6 y , ( ω 2) ω ( x + y ) 6 ω ( x ) + ω ( y ) for an y x, y > 0, ( ω 3) lim t → 0 + ω ( t ) = ω ( 0) = 0. (Observ e that we allo w the zero function to be a mo dulus of con tinuit y .) The main t wo results of the pap er are: 1.1. Theorem. L et r ∈ { 1 , ∞} and N ∈ { 0 , 2 , 3 , 4 , . . . } . Ther e is a unique (up to isometric gr oup isomorp hism) value d Ab elia n gr o up, denote d by G r ( N ) , with the fol lowin g thr e e pr op erties: (G1) G r ( N ) is c omplete an d G r ( N ) ∈ G r ( N ) , (G2) whenever ( H , + , q ) is a ﬁnite Ab el i a n gr oup (of exp onent N , pr o- vide d N 6 = 0 ) with q 6 r , K is a sub gr oup of H and ϕ : K → G r ( N ) is an isometric gr oup homomo rphism, then for every ε ∈ (0 , 1) ther e is a gr oup homo m orphism ϕ ε : H → G r ( N ) such that max x ∈ K p ( ϕ ( x ) − ϕ ε ( x )) 6 ε and (1-1) (1 − ε ) q ( y ) 6 p ( ϕ ε ( y )) 6 (1 + ε ) q ( y ) for y ∈ H , wher e p i s the value of G r ( N ) , (G3) if N = 0 , ﬁnite r ank elements of G r ( N ) form a d e nse subset of G r ( N ) . 1.2. Theorem. L et r ∈ { 1 , ∞} and N ∈ { 0 , 2 , 3 , 4 , . . . } and let p stand for the value of G r ( N ) . (A) L et ( H , + , q ) ∈ G r ( N ) , K b e a close d sub gr oup of H and ϕ : K → G r ( N ) b e a c ontinuous gr oup homomorphis m whose r ange has c omp act closur e in G r ( N ) . 4 P . NI EMIEC (i) Ther e ar e mo duli of c ontinuity ω 6≡ 0 ,  6≡ 0 and τ such that for e ach x ∈ K , (1-2) p ( ϕ ( x )) 6 ( ω ◦ q )( x ) , (1-3) τ (dist q ( x, k er ϕ )) 6 (  ◦ p )( ϕ ( x )) (wher e dist q ( x, k er ϕ ) = inf { q ( x − y ) : y ∈ ker ϕ } ) and ( ω -  - τ ) ( τ ( t ) +  ( s ) 6  ( ω ( t ) + s ) for any t, s > 0 , τ (1) 6  (1) pro vided r = 1 . If ϕ is o p en as a map of K onto ϕ ( K ) , the ab o ve τ may b e chosen nonzer o. (ii) F or every mo dulus of c ontinuity ω 6≡ 0 such that (1 -2) is fulﬁl le d for al l x ∈ K ther e is a c on tinuous gr oup homomor- phism ϕ ω : H → G r ( N ) e x tend ing ϕ and s a tisfying (1-2) fo r e ach x ∈ H with ϕ r eplac e d by ϕ ω , and such that k er ϕ ω = k er ϕ and whenever (1-3) and ( ω -  - τ ) ar e fulﬁl le d for τ and  6≡ 0 (and x ∈ K ), then (1 -3) is satisﬁe d for any x ∈ H with ϕ r eplac e d by ϕ ω . In p articular, if ϕ is isometric, it admits an extens i o n b eing an isometric gr oup hom omorphism. (B) Every (top olo gic al) isomorphism b etwe en two c om p ac t sub g r oups of G r ( N ) is extendable to an automorphism of the top olo gic al gr oup G r ( N ) . What is mor e, if K is a c omp act sub gr oup of G r ( N ) , ϕ : K → G r ( N ) is a gr oup homomorphism, ω 6≡ 0 and τ 6≡ 0 ar e mo duli of c ontinuity such that ( ω - τ -id) and ( τ - ω -id) ar e fulﬁl le d (wher e id denotes the id entity map on [0 , ∞ ) ) and for e ac h x ∈ K , (1-4) p ( ϕ ( x )) 6 ( ω ◦ p )( x ) and p ( x ) 6 ( τ ◦ p )( ϕ ( x )) , then ther e is a gr oup automorphism ψ : G r ( N ) → G r ( N ) e xtending ϕ such that (1-4) is satisﬁe d for e ach x ∈ G r ( N ) with ϕ r eplac e d by ψ . In p articular, if ϕ i s isometric, it ad m its an extensio n b eing an isometric automorphism. Note that the p o in t (G2) (in the statemen t of Theorem 1.1) is a coun- terpart of the condition, prop o sed b y Solec ki [20 ], whic h c ha racterizes the ab ov e men tio ned Gura ri ˇ ı space up to isometric linear isomorphism. Observ e also that (G3) is quite natural, since (G 2) refers o nly to ﬁnite rank elemen ts of t he gro up. Theorem 1.2 implies that ev ery member of G r ( N ) admits an em- b edding to G r ( N ) b y an isometric gro up homomo rphism and that in condition (G2) o ne ma y substitute ε = 0, that is, G r ( N ) ha s univ ersal disp osition prop ert y for ﬁnite gr oups of class G r ( N ). W e shall also show that the g roup G r ( N ) a s a metric space is uni- v ersal fo r separable metric spaces of diameter no greater than r , that UNIVERS AL V ALUED ABELIA N GR OUPS 5 is, that ev ery suc h space is isometrically em b eddable in to the metric space G r ( N ). It turns out that G r ( N ) (as a metric space) is Urysohn iﬀ N ∈ { 0 , 2 } . In particular, the groups G r (2) are Bo olean Urysohn groups in tro duced b y us in [17]. What is m ore, for diﬀeren t pairs ( N , r ) and ( M , s ) the metric spaces G r ( N ) a nd G s ( M ) are isometric iﬀ r = s and { N , M } = { 0 , 2 } . Also all the groups G r ( N ) are pairwise noniso- morphic as top olo gical groups. In Section 8 w e pro v e that each of the topolo gical spaces G r ( N ) is homeomorphic to the Hilb ert space l 2 . T o establish this r esult w e de- v elop our earlier study of the so-called top olo gic al pseudove ctor gr oups , in tro duced in [18]. Namely , a pseudove ctor Ab elian gr oup is a triple ( G, + , ∗ ) suc h that ( G, +) is an Ab elian g roup, ‘ ∗ ’ is an action of [0 , ∞ ) on G satisfying the f ollo wing axioms: 0 ∗ x = 0 G , 1 ∗ x = x , ( st ) ∗ x = s ∗ ( t ∗ x ) for all x ∈ G and s, t > 0 , and for ev ery t > 0 the function G ∋ x 7→ t ∗ x ∈ G is a gr oup homomorphism. If, in addition, G is a top olog ical group, it is said to be a top olo gic al pseu- dove ctor gr oup provided the action ‘ ∗ ’, as a function of [0 , ∞ ) × G in to G , is con tin uo us. A v alue p o n the pseudo ve ctor Ab elian group G is called a norm if p ( t ∗ x ) = tp ( x ) for an y t > 0 and x ∈ G , and it is called a subnorm iﬀ p ( x ) 6 p ( s ∗ x ) 6 sp ( x ) for eac h s > 1 and x ∈ G . If a (sub)norm p induces the to p ology on G with resp ect to whic h the action ‘ ∗ ’ is con tin uous, then the quadruple ( G, + , ∗ , p ) is called a (sub)norme d top ological pseudo vec tor group. In Section 7 w e show that ev ery metrizable top ological pseudov ector group admits a subnorm inducing its top olo gy . The main result on pseudo v ector structures of G r ( N ) is the follow ing 1.3. Theorem. L et r ∈ { 1 , ∞} and N ∈ { 0 , 2 , 3 , 4 , . . . } . Every no n - trivial (sub)norme d top olo gic al pseudove ctor gr oup ( G, + , ∗ , k · k G ) such that ( G, + , k · k G ) ∈ G r ( N ) ma y b e enlar ge d to a (sub)norme d pseu- dove ctor top olo gic al gr oup ( ˜ G, + , ∗ , k · k ) such that the value d gr oups ˜ G and G r ( N ) ar e isometric al ly gr oup isomorphi c . In the abov e result one m a y erase the w ord ‘nontrivial’ pro vided that r = ∞ or t he ﬁnal v a lue k·k ha s to b e a subnorm rather tha n a norm. As a main application of the ab ov e result w e obtain the ab o v e men tio ned theorem whic h sa ys tha t G r ( N ) is homeomorphic to l 2 . (This result ma y b e immediately deduced for N = 0 , 2 from the fact that the me tric spaces G r ( N ) with N = 0 , 2 are Urysohn and from Uspenskij’s theorem [28] on the top ology o f the Urysohn space.) The pro of of the existence of the gro ups G r ( N ) is based on the gen- eral tec hnique of F ra ¨ ıss ´ e limits. First w e construct a coun ta ble v alued Ab elian group QG r ( N ), a counte rpart of the so-called ratio nal Urysohn metric space, and then w e prov e tha t its completion is the group w e searc h fo r. Although this approac h is just an ada ptation of the or iginal construction of the Urysohn space U describ ed in [24, 25], the details 6 P . NI EMIEC are more complicated, mainly b ecause ﬁnite groups admit no one-p oin t extensions, in the opp osite to ﬁnite metric spaces. The unus ual prop- ert y of compact subgroups of G r ( N ), formulated in Theorem 1.2, cor- resp onds to Huh unai ˇ svili’s theorem [9] for compact subsets of U (whic h sa ys that ev ery isometry b etw een tw o compact subsets of U is extend- able to an isometry of U on to itself ). It is not so strong for N 6 = 0, since eve ry compact metric Ab elian group of ﬁnite exp onen t is totally disconnected. In contrast, G r (0) con tains a cop y o f eve ry metrizable compact Ab elian group, among whic h one ma y ﬁnd groups whic h are univ ersal (in the sense of top ological embedding), as top ological sp a- ces, for separable metrizable top olog ical spaces, suc h a s the coun table inﬁnite Cartesian p ow er of R / Z . There are kno wn examples o f the so-called univ ersal P olish groups, that is, of completely metrizable separable (non-Ab elian) top ological groups G suc h tha t every P olish gr oup is isomorphic (as a top ological group) to a closed subgroup o f G . F or example, Usp enskij ha ve shown that the homeomorphism group of the Hilb ert cub e [26] as w ell as the isometry group of t he Urysohn unive rsal metric space [27] are (noniso- morphic) univ ersal P olish groups (cf. Remark just after Theorem 5 .2 of [15]). In the oppo site to this, the author kno ws no example of a univ ersal Polish Ab elian group (i.e. a P olish Ab elian gro up whic h is univ ersal for P o lish Ab elian groups). In this pa p er w e giv e tw o suc h examples: G 1 (0) and G ∞ (0). (Bo th these v alued g roups are of course ‘metrically’ univ ersal fo r separable v alued Ab elian gro ups with v alues b ounded b y 1.) The pa rt on pseudo v ector structures exte nds our earlier (intro duc- tory) w ork [18] in this sub ject. The pro ofs presen ted here are quite new and m uc h more general. Theorem 1.3 generalizes and strengthens Theorem 4.3 of [18]. Since ev ery no rm o n a nontrivial pseudo v ector group is unbounded, to equip the groups G 1 ( N ) with ‘normed-lik e’ pseudo v ector structures w e ha ve to extend the no tion of a norm to a subnorm, whic h is done in the recen t pap er. It seems to b e inter- esting whether one may distinguish a sp ecial subnormed top ological pseudo v ector structure on G r ( N ) whic h will mak e this gro up univ er- sal for subnormed top ological ps eudo ve ctor groups of ‘ G r ( N )-like ’ class (this cannot b e infered from Theorem 1.3). The one idea is to deﬁn e a coun terpart of the Gurari ˇ ı space for pseudo ve ctor groups of suitable class (this is discussed in details in Section 9). W e do not kno w y et whether suc h a pseudo v ector group, denoted b y PVG r ( N ), exists and w e leav e this as an o p en problem. How ev er, w e pro ve t hat if it only exists, it has to b e isometrically g roup isomorphic to G r ( N ) and that PVG r ( N ) as a PV gro up is unique up to isometric linear isomorphism. What is more, ev ery subnormed top ological PV group of suitable class (to whic h PVG r ( N ) b elongs) is embeddable in to PVG r ( N ) b y means of a isometric linear homomorphism. UNIVERS AL V ALUED ABELIA N GR OUPS 7 F or well understanding of this exp osition it is enough to kno w basic facts on metrizable and v alued Ab elian groups, e.g. the material of Chapter 1 of [2 ]. The reader in terested in Urysohn univ ersal space is referred to a surv ey article on the sub ject [15] or to [5], [10], [24, 25]. Notation and ter minology . In this pap er R , Q and Z stand for the sets of all real, all rat ional and, respectiv ely , all integer n um b ers. The sym b ol ‘id’ is r eserv ed to denote the iden tit y map. All g roups are Ab elian and w e use the additiv e nota tion. F or simplicit y , the action of an y group is alw a ys denoted by (the same sign) ‘+’ and its neutral elemen t is denoted by 0. If G is a group, a ∈ G and k ∈ Z , the v a lue at k of a unique group homomorphism of Z in t o G whic h sends 1 to a is denoted by k a or k · a . The gr oup G is said to b e of exp onent N (where N ∈ Z , N > 2) iﬀ N · x = 0 for an y x ∈ G . The subgroup of G consisting of all ﬁnite rank elemen ts of G is denoted b y G f in . F or s, t ∈ [0 , ∞ ] let s ∧ t and s ∨ t stand for the minimum and t he maxim um of s and t (resp ectiv ely). Similar ly , if f is a real-v alued function a nd t ∈ [0 , ∞ ], f ∧ t is a function with the same domain a s f and ( f ∧ t )( x ) = f ( x ) ∧ t . In a similar manner we deﬁne f ∨ t , and f ∧ g and f ∨ g for t wo real-v a lued functions f and g with common domain. A value on a group G is a function p : G → [0 , ∞ ) suc h that for an y x, y ∈ G , (V1) p ( x ) = 0 ⇐ ⇒ x = 0, (V2) p ( − x ) = p ( x ), (V3) p ( x + y ) 6 p ( x ) + p ( y ). If in the ab o v e the condition (V1) is replaced b y ‘ p (0 ) = 0’, p is called a semiv a lue . Ev ery v alue p on G induces an in v ariant metric G × G ∋ ( x, y ) 7→ p ( x − y ) ∈ [0 , ∞ ). The top olo gy induced b y the v alue p is the top ology induc ed b y the latter metric. So, w e ma y speak of a separable, complete, compact (etc.) v alued group. Tw o v alues on a group are e quivalent iﬀ they induce the same top olog y . A v alue on a top ological group is said to be c omp atible if it induces the given top ology of the group. By δ G w e denote the discr ete v alue on G deﬁned b y δ G ( g ) = 1 for g ∈ G \ { 0 } . F or tw o v alued groups ( G, + , p ) and ( G ′ , + , p ′ ) w e shall write ( G ′ , + , p ′ ) ⊃ ( G, + , p ) iﬀ G ⊂ G ′ , p ′ extends p and the addition of G ′ extends the addition of G . If this happ ens, w e sa y that ( G, + , p ) is enla r ge d to ( G ′ , + , p ′ ). Subgroups need not b e closed. A subgroup generated by a sub- set A of a g roup G is denoted b y h A i . W e write h a 1 , . . . , a n i instead of h{ a 1 , . . . , a n }i . If ψ : G → H is a homomorphism b et w een v alued groups ( G, + , p ) and ( H, + , q ), w e use the term a gr oup homomor- phism to underline that w e make no additional assumptions on the top ological b eha vior of ψ . The group homomorphism is isometric if q ( ψ ( x )) = p ( x ) for eac h x ∈ G . Adapting terminology of func- tional analysis, w e say ψ is ε -alm o s t isometric with ε ∈ (0 , 1) pro vided 8 P . NI EMIEC (1 − ε ) p ( x ) 6 q ( ψ ( x )) 6 (1 + ε ) p ( x ) for ev ery x ∈ G (compare with (1-1)). The classes G and G r ( N ) (with r ∈ { 1 , ∞} and N ∈ Z + \ { 1 } ) ha ve the same meaning as in In tro duction. A metric space ( X , d ) is said to b e top olo gic al ly (resp ectiv ely metric al ly ) universal for a class of metric spaces prov ided ev ery mem b er of the class is homeomorphic (resp ectiv ely isometric) to a subset of X . Whenev er f and g are tw o real- v a lued f unctions deﬁned on a common nonempt y domain, the suprem um distance ( ∈ [0 , ∞ ]) of f and g is denoted b y k f − g k ∞ . Similar ly , if f and g tak e v alues in a v alued group ( G, + , p ) and there is no danger of confusion, w e shall also write k f − g k ∞ for the suprem um distance o f f and g induced by p . The diameter of a metric space ( X , d ) is denoted by diam X or diam( X, d ). When ( G, + , p ) is a v a lued group, w e shall also write diam( G, p ). F or any subset A of R let A + stand for the set A ∩ [0 , ∞ ). In partic- ular, R + = [0 , ∞ ) a nd Z + = { 0 , 1 , 2 , 3 , . . . } . F or t w o in tegers k and l w e write k | l if k divides l (i.e. if l = mk for some m ∈ Z ). Notice tha t k | 0 for each k ∈ Z and 0 | l iﬀ l = 0. 2. Preliminaries The results of this section will b e used la ter. T o make t he pap er b etter orga nized, t he section is divided in to parts. In some of results w e will deal with v alues on groups whic h t ak es v alues in a subset Q of R suc h that : (2-1) Q + is dense in R + , 0 ∈ Q and Q + + Q + ⊂ Q. Suc h considerations will ﬁnd applications in next sections. 2A. Boundedness. Denote by Ω the set of all mo duli of contin uit y , that is, all functions (including the zero one) satisfying conditions ( ω 1)– ( ω 3). Additio nally , let Ω ∗ = Ω \ { 0 } . Note that whenev er ω ∈ Ω ∗ and p is a (semi-)v alue on a g roup G , so is ω ◦ p and if q is a semiv alue on G suc h that (2-2) q 6 ω ◦ p, then q is contin uous with resp ect to p . Ho w eve r, it ma y turn out that for some contin uous (with resp ect to p ) semiv alue q there is no ω ∈ Ω witnessing (2-2) (for example, this happ ens if p is b ounded and q is not, e.g. p = q ∧ 1). So, the condition (2 -2) is stronger than con tinuit y . Whenev er it is fulﬁlled for some ω ∈ Ω ∗ , w e sa y that q is p -b ounde d . With eve ry gr oup homomorphism ψ : ( G, + , p ) → ( G ′ , + , p ′ ) w e may asso ciate a semiv alue p ψ on G : p ψ = p ′ ◦ ψ . It is easily seen that ψ is con tinuous iﬀ p ψ is contin uous (with resp ect to p ). W e sa y that ψ is b ounde d iﬀ p ψ is p - b ounded. That is: ψ is b ounded ⇐ ⇒ ∃ ω ∈ Ω : p ′ ◦ ψ 6 ω ◦ p. UNIVERS AL V ALUED ABELIA N GR OUPS 9 The following are left as exercises. Some of them are quite easy , other are well-kno wn results of real analysis: (MC1) F or eve ry ω ∈ Ω there is a ﬁnite limit lim x →∞ ω ( x ) x and it is equal to inf x> 0 ω ( x ) x . (MC2) If ω , τ ∈ Ω, then ω ◦ τ ∈ Ω. (MC3) Eve ry ω ∈ Ω is unifo rmly contin uous, and ω − 1 ( { 0 } ) = { 0 } pro vided ω 6≡ 0. (MC4) If ω ∈ Ω ∗ and ( x n ) ∞ n =1 ⊂ R + , then lim n →∞ ω ( x n ) = 0 ⇐ ⇒ lim n →∞ x n = 0. F or each r ∈ [0 , ∞ ) denote by Ω r the set of all ω ∈ Ω suc h that lim x →∞ ω ( x ) x 6 r (cf. (MC1)). F or us, the set Ω 0 is of great imp ortance. 2.1. Example. Ev ery b ounded mo dulus of con tin uity b elongs to Ω 0 . If ω ∈ Ω and τ ∈ Ω 0 , then ω ◦ τ , τ ◦ ω ∈ Ω 0 . The following functions are members of Ω 0 : x 7→ x ∧ 1, x 7→ x x +1 and x 7→ x α (0 < α < 1). The next result is w ell-kno wn. It is a v ariation of [1 , § 1, Theorem 1]. W e shall use it to c haracterize b ounded group homomorphisms. 2.2. Lemma. F o r a function f : R + → R + the fol lowing c onditions ar e e quivalent: (i) ther e is ω ∈ Ω such that f 6 ω , (ii) lim sup x → 0 + f ( x ) = f (0) = 0 and r := lim sup n →∞ sup f ([0 , 2 n ]) 2 n < ∞ . What is mor e, if (ii) is fulﬁl le d [and f is b ounde d by M ], ther e is ω ∈ Ω 2 r [b ounde d by M ] such that f 6 ω . As an immediate consequence of Lemma 2.2 w e obtain 2.3. Prop osition. L et ψ : ( G, + , p ) → ( G ′ , + , p ′ ) b e a gr oup h o momor- phism and let f : R + ∋ ξ 7→ sup { p ′ ( ψ ( x )) : x ∈ G, p ( x ) 6 ξ } ∈ [0 , ∞ ] . Then: (a) ψ is c ontinuous iﬀ lim t → 0 + f ( t ) = 0 , (b) ψ is b ounde d iﬀ lim t → 0 + f ( t ) = 0 and lim sup t →∞ f ( t ) t < ∞ . 2.4. Corollary . If ψ : ( G, + , p ) → ( G ′ , + , p ′ ) is a c ontinuous gr o up ho - momorphism such that the se t ψ ( G ) is b ounde d in the metric sp a c e ( G ′ , p ′ ) , then ψ is b ounde d. 2.5. Corollary . A gr oup hom omorphism ψ : ( G, + , p ) → ( G ′ , + , p ′ ) is c ontinuous iﬀ ther e a r e ω , τ ∈ Ω ∗ such that τ ◦ p ′ ◦ ψ 6 ω ◦ p . Pr o o f. Suﬃciency is clear (thanks to (MC4)). T o prov e the necessit y , put τ ( t ) = t ∧ 1 and apply Lemma 2.2 for suitable f : R + → [0 , 1].  10 P . NI EMIEC 2B. Extending a v alue. 2.6. Lemma. L et Q b e as in (2-1) . L et ( D , + , λ ) b e a ﬁnite value d gr oup and D 0 its sub gr oup such that λ ( D 0 ) ⊂ Q . Then fo r every ε > 0 ther e is a value ¯ λ on D extending λ   D 0 such that ¯ λ ( D ) ⊂ Q and k λ − ¯ λ k ∞ 6 ε . What is mor e, if λ 6 1 , then ¯ λ 6 1 . Pr o o f. W e may assume that ε ∈ (0 , 1). F or h ∈ D 0 let λ ′ ( h ) = 0 and for h ∈ D \ D 0 let λ ′ ( h ) ∈ [0 , ε ] b e suc h that λ ′ ( − h ) = λ ′ ( h ) and λ ( h ) + λ ′ ( h ) ∈ Q . F or x ∈ D put ¯ λ ( x ) = inf { n X j =1 ( λ ( h j ) + λ ′ ( h j )) : n > 1 , h 1 , . . . , h n ∈ D , x = n X j =1 h j } . It is easily seen that ¯ λ is a v alue on D suc h that λ 6 ¯ λ 6 λ + λ ′ . This yields that k λ − ¯ λ k ∞ 6 ε a nd ¯ λ = λ on D 0 . F urther, since D is ﬁnite, the inﬁm um in the formula fo r ¯ λ ( x ) is reache d and therefore ¯ λ ( D ) ⊂ Q . Finally , if λ 6 1 , tak e M ∈ Q ∩ [1 − ε, 1] suc h that λ ( D 0 ) ⊂ [0 , M ] and replace ¯ λ by ¯ λ ∧ M .  2.7. Lemma. L et r ∈ { 1 , ∞} . L et ( D, + , λ ) b e a value d gr oup with λ 6 r , D 0 its close d sub gr oup, λ 0 a semivalue on D 0 and let ω ∈ Ω ∗ b e such that λ 0 6 ( ω ◦ λ )   D 0 . Then ther e is a semivalue ¯ λ on D which extends λ 0 and satisﬁe s the fol lowing c onditions: (a) ¯ λ 6 ω ◦ λ , (b) ¯ λ − 1 ( { 0 } ) = λ − 1 0 ( { 0 } ) ; in p articular: ¯ λ is a val ue p r ovid e d so is λ 0 , (c) if λ 0 6 r , then ¯ λ 6 r , (d) if λ 0 is a value e quivalent to λ   D 0 , then ¯ λ is e quivalent to λ , (e) if for some τ ,  ∈ Ω with  6≡ 0 a nd ( ω -  - τ ) one has τ (dist λ ( h, λ − 1 0 ( { 0 } ))) 6 (  ◦ λ 0 )( h ) for e v e ry h ∈ D 0 , then τ (dist λ ( x, λ − 1 0 ( { 0 } ))) 6 (  ◦ ¯ λ )( x ) for x ∈ D . Pr o o f. F or x ∈ D put ¯ λ ( x ) = inf { ( ω ◦ λ )( x − h ) + λ 0 ( h ) : h ∈ D 0 } . F urther, if λ 0 6 r , replace ¯ λ by ¯ λ ∧ r . One easily v eriﬁes that ¯ λ is a semiv alue extending λ 0 whic h satisﬁes (a) and (d). The p oint (b) follo ws from the closedness of D 0 . W e shall only show (e). F or simplicit y , put q ( x ) = dist λ ( x, λ − 1 0 ( { 0 } )) ( x ∈ D ). Thanks to the con tinuit y of  , it suﬃces to ch ec k that ( τ ◦ q )( x ) 6  (( ω ◦ λ )( x − h ) + λ 0 ( h )) for x ∈ D and h ∈ D 0 (the second condition in ( ω -  - τ ) a llo ws us to replace ¯ λ by ¯ λ ∧ r ). But ( τ ◦ q )( x ) 6 ( τ ◦ λ )( x − h ) + (  ◦ λ 0 )( h ) and hence it is enough to sho w that ( τ ◦ λ )( x − h ) + (  ◦ λ 0 )( h ) 6  (( ω ◦ λ )( x − h ) + λ 0 ( h )) , whic h is the ﬁrst condition in ( ω -  - τ ) with t = λ ( x − h ) and s = λ 0 ( h ).  UNIVERS AL V ALUED ABELIA N GR OUPS 11 2.8. Example. Let ω 0 ,  0 ∈ Ω ∗ , τ 0 ∈ Ω and r ∈ { 1 , ∞} . Put ω = ω 0 ∨ τ 0 ,  =  0 + id and τ = τ 0 ∧  ( r ) (  ( ∞ ) := lim t →∞  ( t )). Then ω ,  and τ are mo duli of contin uit y suc h that τ 6 τ 0 ,  0 6  , ω 0 6 ω and ( ω -  - τ ) is fulﬁlled. The example sho ws that w e ma y alw ays replace mo duli of contin uit y app earing in the inequalities in p oints (a) and (e) of Lemma 2.7 by other mo duli in such a w ay that these inequalities are still satisﬁed and the strange condition ( ω -  - τ ) is fulﬁlled. 2C. T he class O 0 . W e call a gro up of class O f in iﬀ ev ery its elemen t has ﬁnite rank. That is, G is of class O f in if G = G f in . Let ( G, + , p ) b e an arbitrary v alued group and x its elemen t. Since the sequence a n := p ( nx ) ( n > 1) satisﬁes the condition a n + m 6 a n + a m ( n, m > 1) , the sequence ( a n n ) ∞ n =1 has ﬁnite limit and lim n →∞ a n n = inf n > 1 a n n . Put p 0 ∗ : G → R + , p 0 ∗ ( x ) = lim n →∞ p ( nx ) n . Observ e that p 0 ∗ is a semiv alue on G s uc h that p 0 ∗ 6 p . Th us p 0 ∗ is p -b ounded and hence the set G 0 ∗ = p − 1 0 ∗ ( { 0 } ) is a closed subgroup of ( G, + , p ). L et G ′ b e the quotien t g roup G/G 0 ∗ equipped with the v alue p ′ (naturally) induced by p 0 ∗ . Observ e t hat (2-3) p ′ ( k x ) = | k | p ′ ( x ) for ev ery x ∈ G ′ and k ∈ Z and hence p ′ 0 ∗ = p ′ . W e say the v alued group ( G, + , p ) is o f class O 0 iﬀ G 0 ∗ = G or, equiv alently , if p 0 ∗ ≡ 0. It is of class O ∞ iﬀ p 0 ∗ = p (that is, if (2-3) is fulﬁlled with p ′ replaced b y p ). Finally , ( G, + , p ) is of class O 1 iﬀ p 0 ∗ is a v alue on G or, equiv alently , if G 0 ∗ is trivial. Note tha t O ∞ ⊂ O 1 . W e b egin with an in teresting c haracterization of mem b ers of the ab ov e men t ioned classes. Making use of (2- 3) and repeating the pro of of the Hahn-Banac h theorem one easily gets 2.9. Lemma. L et ( G, + , p ) b e a v a lue d gr oup. (A) If ψ : G → R is a gr oup homomorphism such that for e ach x ∈ G , (2-4) | ψ ( x ) | 6 p ( x ) , then | ψ ( x ) | 6 p 0 ∗ ( x ) for every x ∈ G . (B) F or e ach a ∈ G ther e is a gr oup ho m omorphism ψ : G → R s a tis- fying (2-4 ) for al l x ∈ G such that ψ ( a ) = p 0 ∗ ( a ) . Let us call a group homomorphism ψ : G → R satisfying (2 -4) non- exp ansive . The a b o v e result yields 2.10. Theorem. L et ( G, + , p ) b e a value d gr oup. 12 P . NI EMIEC (I) G is of class O 0 iﬀ G adm i ts no nonzer o b ounde d g r oup homo- morphisms into B a nach sp ac es. (I I) G is of class O ∞ iﬀ G admits an isome tric gr oup homomorphism into a Ban ach sp ac e. (I I I) G is of class O 1 iﬀ G admits a no nexp ansive gr oup homomor- phism with trivial kernel into a Banach sp ac e, iﬀ G admits a b ounde d gr oup homomorp hism with trivia l k e rnel into a Ba n ach sp ac e. Pr o o f. (I): Suppo se ( G, + , p ) is of class O 0 , ( E , k · k ) is a Banac h space, ω ∈ Ω and ψ : G → E is a group homomorphism suc h that k ψ ( x ) k 6 ( ω ◦ p )( x ) fo r x ∈ G . Then w e hav e (2-5) k ψ ( x ) k 6 ( ω ◦ p )( nx ) n = ω ( n · p ( nx ) n ) n 6 p ( nx ) n → 0 . The inv erse implication f ollo ws from the p oint (B) of Lemma 2.9. (I I): Assume ( G, + , p ) is of class O ∞ . Let Z be the set of all nonex- pansiv e group homomorphisms of G into R a nd let E b e the Banac h space of all real-v alued bo unded f unctions on Z , equipped with the suprem um norm. F or g ∈ G let e g ∈ E b e given b y e g ( ξ ) = ξ ( g ). Then the function G ∋ g 7→ e g ∈ E is an isometric group homomor phism (again by Lemma 2.9). The in v erse implication is immediate. (I I I): If ( G, + , p ) is of class O 1 , then ( G, + , p 0 ∗ ) is a v alued group of class O ∞ and thus the assertion follows from (I I). Conv ersely , if ψ : G → E is a b ounded g roup homomorphism of G into a Ba nac h space E , (2-5) shows that p 0 ∗ ( x ) > k ψ ( x ) k .  2.11. R emark. Theorem 2 .10 expre sses ho w far is b oundedness from con tinuit y: o nly v alued groups of class O 1 admit b ounded group ho- momorphisms with trivial k ernels into Banach spaces. In con trast, there are v alued groups of class O 0 whic h are group isomorphic to Ba- nac h spaces: if ( E , k · k ) is a Banach space, then ( E , k · k ∧ 1) is of class O 0 . By the w ay , t he latter example shows that mem b ership to any of the classes O 0 , O 1 , O ∞ is not a t op ological gr oup inv ariant. Theorem 2 .10 also implies that ev ery v alued group G has a unique closed subgroup H (namely , G 0 ∗ ) suc h that H is of class O 0 and G/H admits a no nexpansiv e gro up homomo rphism with trivial k ernel in to a Banac h space. It also f ollo ws f rom Theorem 2.10 and the w ell-kno wn theorem of Banac h and Mazur that the Banac h space C ([0 , 1]) of all contin uous real-v alued functions o n [0 , 1], as a v alued group, is univ ersal for sepa- rable v alued Ab elian groups of class O ∞ . F ro m now on, w e are only in terested in v alued gro ups of class O 0 . The reader will easily c hec k that ev ery group with b ounded v alue is of this class and that O f in ⊂ O 0 . It turns out that, in a sense, O 0 is the ‘closure’ of O f in . F ormally this is formulated in the follow ing UNIVERS AL V ALU ED ABELIA N GR OUPS 13 2.12. Theorem. A value d gr oup ( G, + , p ) is of class O 0 iﬀ it may b e enlar ge d to a value d gr oup ( ˜ G, + , ˜ p ) such that ˜ G f in is dense in ˜ G . Pr o o f. The suﬃciency is clear, since ˜ G 0 ∗ is close d and con tains ˜ G f in . T o pro v e the necessit y , thanks to transﬁnite induction, it is enough to sho w that if p 0 ∗ ( a ) = 0, then for ev ery ε > 0 there is a v alued group ( ˜ G, + , ˜ p ) ⊃ ( G, + , p ) and b ∈ ˜ G f in suc h that ˜ p ( a − b ) 6 ε . Let N > 2 b e suc h that (2-6) p ( na ) 6 εn for n > N . T ak e M > 0 suc h tha t M > p ( j a ) fo r j = 1 , . . . , N . Let H = h b i b e a cyclic group of rank N . W e iden tify g ∈ G with ( g , 0) ∈ G × H =: ˜ G . Deﬁne ˜ p : ˜ G → R + b y ˜ p ( g , h ) = inf { p ( g − k a ) + | k | ε + M δ H ( k b + h ) : k ∈ Z } . It is clear that ˜ p is a v alue on ˜ G suc h that ˜ p ( g , 0 ) 6 p ( g ) fo r g ∈ G and ˜ p (( a, 0) − (0 , b )) 6 ε . So, it suﬃces to sho w that p ( g ) 6 ˜ p ( g , 0), which is equiv alen t to (2-7) p ( k a ) 6 | k | ε + M δ H ( k b ) for k ∈ Z . W e may assume k 6 = 0. If | k | > N , (2-7) is cov ered b y (2-6). Finally , if 0 < | k | < N , then p ( k a ) 6 M = M · δ H ( k b ) and we are do ne.  Note that if in the ab o v e t heorem G is separable, ˜ G ma y b e con- structed to b e separable as w ell and tha t in that case the pro of is constructiv e. It is also easily seen that w e ma y force ˜ G to hav e the same diameter as G . 2.13. Example. Let ( G, + , p ) b e a v alued group and ω ∈ Ω ∗ . If G is of class O 0 or ω ∈ Ω 0 , t hen ( G, + , ω ◦ p ) is of class O 0 . In particular, for ev ery separable v alued group ( G, + , p ), ( G, + , p α ) ∈ G ∞ (0) for 0 < α < 1 and ( G, + , p ∧ 1 ) , ( G, + , p p +1 ) ∈ G 1 (0). The v a lue p ma y b e reconstruct from each of the v alues p α (0 < α < 1) and p p +1 , a nd since p ( x ) = lim α → 1 − p α ( x ) for ev ery x ∈ G , one ma y say that eac h v alued group can b e ‘appro ximated’ b y v alued gro ups of class O 0 . It is also clear that ev ery metrizable top o logical group admits a compatible v a lue under whic h it b ecomes a v alued group of class O 0 . T aking in to accoun t R emark 2.11, w e see that the image of a v alued group of class O 0 under a contin uous gro up homomorphism nee d no t b e of class O 0 . How ev er, the reader will easily c hec k that 2.14. Prop osition. A sub gr oup (e quipp e d with the inherite d v a lue) and the c om pletion of a value d gr oup of class O 0 is of class O 0 as wel l. Th e image of a value d gr oup of clas s O 0 under a b ounde d gr oup homom or- phism is of class O 0 . 14 P . NI EMIEC 2D. Enla rging a ﬁnite v alued group. A v ery useful metho d of constructing the Urysohn univ ersal space, in tro duced b y Kat ˇ etov [1 0], in volv es the tec hnique of the so-called Katˇ etov maps which corresp o nd to one-p oin t extensions of metric spaces. In this part w e shall describe ho w to enlarge v alued gro ups b y means of these maps. As a corolla ry of the results presen ted b elow w e shall obta in in the sequel a rather surprising result that G r ( N ) is Urysohn iﬀ N ∈ { 0 , 2 } . Recall t hat a Kat ˇ eto v map on a metric space ( X , d ) is a function f : X → R + suc h that | f ( x ) − f ( y ) | 6 d ( x, y ) 6 f ( x ) + f ( y ) fo r any x, y ∈ X . The set of all Katˇ eto v maps on X is denoted b y E ( X ). Additionally , for r ∈ [0 , ∞ ] let E r ( X ) b e the set of all f ∈ E ( X ) suc h that f ( x ) 6 r f or eac h x ∈ X , and let E i ( X ) = E diam X ( X ). Kat ˇ etov maps b elonging to E i ( X ) are called inner . If A ⊂ B ⊂ X , a Katˇ etov map f : B → R + is said to b e trivial on A in X iﬀ there is b ∈ X suc h that f ( a ) = d ( a, b ) for eac h a ∈ A . The map f is trivial in X if f is trivial on its do main in X . A fundamen tal result on the Urysohn space sa ys that a nonempty separable complete metric space X is Urysohn iﬀ ev ery inner Kat ˇ etov map is trivial in X o n ev ery ﬁnite subset of the space (see e.g. [15]). W e b egin with 2.15. Prop osition. L et ( G, + , p ) b e a va lue d gr oup of exp onent N > 3 . If A ⊂ G and f ∈ E ( A ) is trivial in G , then for any a 1 , . . . , a N ∈ A , (2-8) | p ( N X k =1 a k ) − f ( a N ) | 6 N − 1 X k =1 f ( a k ) . Pr o o f. Let x ∈ G b e suc h that f ( a ) = p ( x − a ) for a ∈ A . Then, since N · x = 0, | p ( N X k =1 a k ) − f ( a N ) | 6 p ( N X k =1 a k + ( x − a N )) = p ( N − 1 X k =1 ( a k − x )) 6 6 N − 1 X k =1 p ( a k − x ) = N − 1 X k =1 f ( a k ) .  2.16. R emark. If ( G, + , p ) is a v alued group of exp onen t N > 3 and f is a Katˇ etov map whose domain H is a subgroup of G , then (2-8) is equiv alent to (2-9) f ( − N − 1 X k =1 a k ) 6 N − 1 X k =1 f ( a k ) for any a 1 , . . . , a N − 1 ∈ H. Indeed, if a 1 , . . . , a N − 1 ∈ H are ﬁxed, then, since − P N − 1 k =1 a k ∈ H , sup a N ∈ H | p ( P N k =1 a k ) − f ( a N ) | = f ( − P N − 1 k =1 a k ). UNIVERS AL V ALU ED ABELIA N GR OUPS 15 Our aim is to pr o ve, in a sense, the con ve rse of Prop osition 2.15. Namely , 2.17. T heorem. L et r ∈ { 0 , ∞} , N ∈ Z + \ { 1 } and let ( G, + , p ) ∈ G r ( N ) b e a b ounde d value d gr oup. L et A b e a nonempty subset of G and f ∈ E r ( A ) . If N > 2 , we assume that f satisﬁes (2-8) for al l a 1 , . . . , a N ∈ A . The n ther e is a b ounde d value d gr oup ( ˜ G, + , ˜ p ) ∈ G r ( N ) such that ( ˜ G, + , ˜ p ) ⊃ ( G, + , p ) and f is trivial in ˜ G . If, in addition, Q is as in (2-1) , G is ﬁni te and p ( G ) ∪ f ( A ) ⊂ Q , the gr oup ˜ G may b e taken so that it is ﬁnite and ˜ p ( ˜ G ) ⊂ Q . Pr o o f. First of all observ e that (2-8) is fulﬁlled for an y a 1 , . . . , a N ∈ A also when N = 2, b ecause then a 1 + a 2 = a 1 − a 2 . Note that if inf f ( A ) = 0, then f is trivial in the completion of G , so w e may and do assume that c := inf f ( A ) > 0 . F urther, let M = sup p ( G ) ∨ sup f ( A ) ( M ∈ Q provided all additional conditions of the theorem are fulﬁlled; f is b ounded b ecause G is so). If N 6 = 0, let m = N . Otherwise, take m > 2 suc h that m − 1 > M c . Let H = h b i b e a cyclic group of rank m . Put ˜ G = G × H . W e identify each g ∈ G with ( g , 0) ∈ ˜ G . Let us agree that P b j = a = 0 pro vided a > b . Deﬁne ˜ p : ˜ G → R + b y the form ula ˜ p ( g , h ) = inf { p ( g − n X j =1 ε j a j ) + n X j =1 f ( a j ) + M δ H ( n X j =1 ε j b − h ) : n > 0 , a 1 , . . . , a n ∈ A, ε 1 , . . . , ε n ∈ {− 1 , 1 }} . It is easy to v erify that ˜ p is a v alue. What is more, if all additional conditions of t he theorem a re satisﬁed, the inﬁm um in the form ula for ˜ p ( g , h ) is reache d a nd th us ˜ p ( ˜ G ) ⊂ Q . Observ e t hat ˜ p ( g , 0) 6 p ( g ) fo r eac h g ∈ G a nd ˜ p ( a, b ) 6 f ( a ) for a ∈ A . If w e sho w that in b oth these inequalities one ma y put the equalit y sign, the pro of will be completed. (Indeed, to make then the v alue ˜ p b ounded b y r , it suﬃces to replace ˜ p b y ˜ p ∧ M .) First w e will chec k that ˜ p ( g , 0) = p ( g ) for g ∈ G . Equiv alently , w e ha ve to sho w that (2-10) n X j =1 f ( a j ) + M δ H ( n X j =1 ε j b ) > p ( n X j =1 ε j a j ) for any n > 1 (f or n = 0 it is clear), a 1 , . . . , a n ∈ A and ε 1 , . . . , ε n ∈ {− 1 , 1 } . If P n j =1 ε j b 6 = 0, then δ H ( P n j =1 ε j b ) = 1 and hence (2-10) is fulﬁlled, b y the deﬁnition o f M . Th us we ma y a ssume that P n j =1 ε j b = 0, that is, m | P n j =1 ε j . If P n j =1 ε j 6 = 0 and N = 0, then n > | P n j =1 ε j | > m , so P n j =1 f ( a j ) > nc > mc > M > p ( P n j =1 ε j a j ) and consequen tly (2-10) is fulﬁlled. If P n j =1 ε j 6 = 0 and N 6 = 0, then m = N and P n j =1 ε j = l N for some l ∈ Z \ { 0 } . This means that, af ter ren umeration of ε j ’s (a nd 16 P . NI EMIEC a j ’s), w e ma y assume ε j = l / | l | for j = 1 , . . . , | l | N and P n j = | l | N +1 ε j = 0. No w making use of (2- 8) we infer that P | l | k =1 p ( P k N j =( k − 1) N +1 a j ) 6 P | l | N j =1 f ( a j ). So, if only p ( P n j = | l | N +1 ε j a j ) 6 P n j = | l | N +1 f ( a j ), (2-1 0) will b e satisﬁed. This reduc es the problem to the case when P n j =1 ε j = 0 (and N is arbitra ry). Finally , if P n j =1 ε j = 0, then n is ev en, sa y n = 2 k , and—after ren umeration—we may assume ε j = ( − 1) j − 1 for j = 1 , . . . , n . But then p ( n X j =1 ε j a j ) = p ( k X j =1 ( a 2 j − 1 − a 2 j )) 6 6 k X j =1 ( f ( a 2 j − 1 ) + f ( a 2 j )) = n X j =1 f ( a j ) whic h ﬁnishes the pro of of (2 -10). No w w e pa ss to pro ving that ˜ p ( a, b ) = f ( a ) for a ∈ A . This is equiv alen t to (2-11) p ( a − n X j =1 ε j a j ) + n X j =1 f ( a j ) + M δ H ( n X j =1 ε j b − b ) > f ( a ) for each n > 0 a nd arbitrary a 1 , . . . , a n ∈ A and ε 1 , . . . , ε n ∈ {− 1 , 1 } . As b efore, if P n j =1 ε j b 6 = b , then, tha nks t o the deﬁnition of M , (2-11 ) is fulﬁlled. Th us w e may assume m | P n j =1 ε j − 1 (so, n > 0). If P n j =1 ε j 6 = 1 and N = 0, n > | P n j =1 ε j | > m − 1 > M c and hence P n j =1 f ( a j ) > nc > M > f ( a ). If P n j =1 ε j 6 = 1 and N 6 = 0, then P n j =1 ε j = l N + 1 for some l ∈ Z \ { 0 } . Then either ε 1 = . . . = ε n = − 1 or at least | l | N of ε j ’s are equal to l/ | l | . In t he ﬁrst case w e get n = | l | N − 1 and, | l | times making use of (2-8): p ( a + n X j =1 a j ) + n X j =1 f ( a j ) > p ( a + N − 1 X j =1 a | l | N − j ) + N − 1 X j =1 f ( a | l | N − j ) + | l |− 1 X k =1 [ N − 1 X j =0 f ( a k N − j ) − p ( N − 1 X j =0 a k N − j )] > f ( a ) , whic h giv es ( 2-11). In the second case w e may assume, after r en umer- ation, that ε j = l/ | l | for j = 1 , . . . , | l | N and P n j = | l | N +1 ε j = 1. Then, again b y (2-8), P | l | k =1 [ P N − 1 j =0 f ( a k N − j ) − p ( P N − 1 j =0 a k N − j ] > 0. So , if only p ( a − P n j = | l | N +1 ε j a j ) + P n j = | l | N +1 f ( a j ) > f ( a ), (2- 11) will b e satisﬁed. This reduces the problem to the case when P n j =1 ε j = 1 (and N is arbitrary). UNIVERS AL V ALU ED ABELIA N GR OUPS 17 Finally , when P n j =1 ε j = 1, then n is o dd, say n = 2 k + 1, and (after ren umeration) ε j = ( − 1) j − 1 . But then p ( a − n X j =1 ε j a j ) + n X j =1 f ( a j ) > p ( a − a n ) + f ( a n ) + k X j =1 [ f ( a 2 j − 1 ) + f ( a 2 j ) − p ( a 2 j − a 2 j − 1 )] > f ( a ) .  In [1 7] we hav e sho wn that t he Urysohn unive rsal metric space U admits a (unique) structure of a metric gr oup of expo nen t 2 (whic h, in fact, will turn out t o b e isometrically gro up isomorphic to G ∞ (2)). A t the end of [1 7] w e ask ed (follo wing the referee) whe ther U admits group structures of other exp onents. Applying Remark 2.16, b elow we giv e a partial (negative ) answ er to this problem. 2.18. Prop osit ion. Ther e is no nontrivial value d Ab elian gr oup of ex- p onent 3 whic h is Urysohn as a m etric sp ac e. Pr o o f. Let ( H , + , q ) be a non trivial v alued Ab elian group. T ak e h ∈ H suc h that 0 < q ( h ) < 1 2 sup q ( H ) (if there is no suc h h , then H is automatically non-Urysohn). Notice that q ( h ) = q ( − h ) = q ( h − ( − h )). Let f : { 0 , h, − h } → R + b e given b y f ( h ) = f ( − h ) = 1 2 q ( h ) and f (0 ) = 3 2 q ( h ). It is easily seen that f is a Katˇ etov map b ounded b y the diameter of H . If f w as trivial in H , w e w ould hav e (b y (2-9)) f (0 ) 6 f ( h ) + f ( − h ), whic h fails to b e true.  W e do not kno w whether it is p ossible to show, using only (2-9), the coun terpart of Prop osition 2.18 for expo nen ts gr eater than 3. 2E. Amalgamation lemmas. In constructions of F ra ¨ ıs s ´ e limits, p os- sibilit y o f amalg amating ( i.e. ‘gluing’) spaces is the crucial axiom. Ho wev er, only t he ﬁrst o f the results stated b elow will b e used for this purp o se. F rom no w to the end of the section, r ∈ { 1 , ∞} and N ∈ Z + \ { 1 } are ﬁxed. T o a void rep etitions, let us mak e the following preliminary annotation (whic h will b e used in the proo fs of the nex t three lemmas): (2-12)        ˜ D = ( D 1 × D 2 ) / ˜ D 0 π : D 1 × D 2 → ˜ D t he quotient group homomorphism ˜ ψ 1 : D 1 ∋ x 7→ π ( x, 0) ∈ ˜ D ˜ ψ 2 : D 2 ∋ y 7→ π (0 , y ) ∈ ˜ D Note t hat b elo w Prop osition 2.14 is applied sev eral times without re- ferring to it. 18 P . NI EMIEC 2.19. Lemma. L et ( D j , + , λ j ) ∈ G r ( N ) ( j = 0 , 1 , 2 ) an d fo r j = 1 , 2 let ϕ j : D 0 → D j b e a n isometric gr oup homomo rphism. Then ther e exist a va lue d gr oup ( D , + , λ ) ∈ G r ( N ) a nd isometric gr oup hom omorphisms ψ j : D j → D ( j = 1 , 2 ) such that ψ 2 ◦ ϕ 2 = ψ 1 ◦ ϕ 1 . If, in addition , Q is as in (2-1) , D 1 and D 2 ar e ﬁnite and λ j ( D j ) ⊂ Q for j = 1 , 2 , then D is ﬁnite as we l l an d λ ( D ) ⊂ Q . Pr o o f. Let ˜ D 0 = { ( ϕ 1 ( x ) , − ϕ 2 ( x )) : x ∈ D 0 } ⊂ D 1 × D 2 and ˜ D , π , ˜ ψ 1 and ˜ ψ 2 b e as in (2-1 2). Then ˜ ψ 2 ◦ ϕ 2 = ˜ ψ 1 ◦ ϕ 1 . Deﬁne ˜ λ : ˜ D → R + b y the fo rm ula ˜ λ ( z ) = inf { λ 1 ( x 1 ) + λ 2 ( x 2 ) : ( x 1 , x 2 ) ∈ π − 1 ( { z } ) } . It is clear that ˜ λ is a w ell deﬁned semiv alue on ˜ D . No w put D = ˜ D / ˜ λ − 1 ( { 0 } ) and deﬁne λ , ψ 1 and ψ 2 b y the rules λ ◦ ˜ π = ˜ λ and ψ j = ˜ π ◦ ˜ ψ j ( j = 1 , 2) where ˜ π : ˜ D → D is the quotien t group homomorphism. Finally , replace λ b y λ ∧ 1 if r = 1. W e leav e this as a simple exercise that all assertions are satisﬁed.  2.20. Lemma . L et ( D 1 , + , λ 1 ) , ( D 2 , + , λ 2 ) ∈ G r ( N ) b e ﬁnite value d gr oups and D 0 b e a sub gr oup of D 1 . L et u : D 0 → D 2 and v : D 1 → D 2 b e an isometric and , r esp e ctively, an ε -almost isometric gr oup homo- morphism (wher e ε ∈ (0 , 1) ) such that (2-13) k u − v   D 0 k ∞ 6 ε. Then ther e ar e a ﬁnite va l ue d gr oup ( D, + , λ ) ∈ G r ( N ) a n d isometric gr oup homomo rp h isms w j : D j → D ( j = 1 , 2 ) such that w 1   D 0 = w 2 ◦ u and (2-14) k w 1 − w 2 ◦ v k ∞ 6 Aε wher e A = 1 + diam ( D 1 , λ 1 ) . Pr o o f. Let ˜ D 0 = { ( x, − u ( x )) : x ∈ D 0 } ⊂ D 1 × D 2 and ˜ D , π , ˜ ψ 1 and ˜ ψ 2 b e as in (2-12). Put D = ˜ D and w j = ˜ ψ j ( j = 1 , 2). It is easily c hec ked that w 2 ◦ u = w 1   D 0 and D = w 1 ( D 1 ) + w 2 ( D 2 ). Deﬁne λ : D → R + b y λ ( z ) = inf { λ 1 ( x 1 − x 0 ) + Aεδ D ( w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 )) + λ 2 ( x 2 + v ( x 0 )) : x 0 , x 1 ∈ D 1 , x 2 ∈ D 2 , z = w 1 ( x 1 ) + w 2 ( x 2 ) } . W e see that λ is a semiv alue on D and, since D 1 and D 2 are ﬁnite, the inﬁm um in the for m ula for λ ( z ) is reac hed. Therefore, if λ ( z ) = 0, then for some x 0 , x 1 ∈ D 1 and x 2 ∈ D 2 one has z = w 1 ( x 1 ) + w 2 ( x 2 ) and x 1 − x 0 = 0, w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 ) = 0 and x 2 + v ( x 0 ) = 0. Thes e yield x 1 = x 0 , x 2 = − v ( x 1 ) and w 1 ( x 1 ) = − w 2 ( x 2 ) and consequen tly z = 0. So, λ is a v alue. Moreo ve r, it follo ws from the deﬁnition of λ that (2-14) is satisﬁed a nd that λ ( w j ( x )) 6 λ j ( x ) for x ∈ D j ( j = 1 , 2). W e UNIVERS AL V ALU ED ABELIA N GR OUPS 19 shall now pro v e t hat in the la tter inequalities one ma y put the equalit y sign. Fix j ∈ { 1 , 2 } . F or x ∈ D j w e ha ve to sho w that (2-15) λ 1 ( x 1 − x 0 )+ Aεδ D ( w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 ))+ λ 2 ( x 2 + v ( x 0 )) > λ j ( x ) pro vided x 0 , x 1 ∈ D 1 , x 2 ∈ D 2 and (2-16) w 1 ( x 1 ) + w 2 ( x 2 ) = w j ( x ) . First assume j = 1. In that case (2- 16) means that ( x 1 − x, x 2 ) ∈ ˜ D 0 and th us h := x − x 1 ∈ D 0 and x 2 = u ( h ). So, (2 -15) has the form λ 1 ( x 1 − x 0 ) + Aεδ D ( w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 )) + λ 2 ( u ( h ) + v ( x 0 )) > λ 1 ( x 1 + h ) whic h, after substitution x 1 := x 0 , is equiv alen t to (2-17) Aεδ D ( w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 )) + λ 2 ( u ( h ) + v ( x 0 )) > λ 1 ( x 0 + h ) . If w 1 ( x 0 ) = w 2 ( v ( x 0 )), then x 0 ∈ D 0 and v ( x 0 ) = u ( x 0 ) and conse- quen tly (2 -17) c hanges in to λ 2 ( u ( h ) + u ( x 0 )) > λ 1 ( x 0 + h ) whic h is fulﬁlled b ecause u is isometric. So, w e may assume that w 1 ( x 0 ) 6 = w 2 ( v ( x 0 )) a nd then we hav e to sho w that Aε + λ 2 ( u ( h ) + v ( x 0 )) > λ 1 ( x 0 + h ). But v is ε -almost isometric, hence λ 2 ( v ( x 0 + h )) > (1 − ε ) λ 1 ( x 0 + h ). So, thanks to ( 2-13), we obtain Aε + λ 2 ( u ( h ) + v ( x 0 )) > ελ 1 ( x 0 + h ) + ε + λ 2 ( v ( x 0 ) + v ( h )) − λ 2 ( u ( h ) − v ( h )) > λ 1 ( x 0 + h ) . When j = 2, w e argue in a similar wa y . In that case (2-16) giv es h := x 1 ∈ D 0 and x − x 2 = u ( h ). After substitution x 2 := − v ( x 0 ) the inequalit y (2-15 ) changes in to λ 1 ( h − x 0 ) + Aεδ D ( w 1 ( x 0 ) − ( w 2 ◦ v )( x 0 )) > λ 2 ( u ( h ) − v ( x 0 )). As b efore, the situation when w 1 ( x 0 ) = w 2 ( v ( x 0 )) is simple. Otherwise w e ha v e to prov e that Aε + λ 1 ( h − x 0 ) > λ 2 ( u ( h ) − v ( x 0 )). W e ha ve λ 2 ( u ( h ) − v ( x 0 )) 6 λ 2 ( u ( h ) − v ( h )) + λ 2 ( v ( h ) − v ( x 0 )) 6 ε + (1 + ε ) λ 1 ( h − x 0 ) 6 Aε + λ 1 ( h − x 0 ) . T o end the pro of, replace λ b y λ ∧ r to g uaran tee that ( D, + , λ ) ∈ G r ( N ).  2.21. Lemma. L et ( D j , + , λ j ) ∈ G r ( N ) ( j = 1 , 2 ), E 1 and E 2 b e sub- gr oups of D 1 and ϕ j : E j → D 2 ( j = 1 , 2 ) b e isometric gr oup ho m o- morphisms such that for al l ( x 1 , x 2 ) ∈ E 1 × E 2 , (2-18) | λ 2 ( ϕ 1 ( x 1 ) − ϕ 2 ( x 2 )) − λ 1 ( x 1 − x 2 ) | 6 ε (wher e ε > 0 ). Then ther e exist a v a lue d gr oup ( D , + , λ ) ∈ G r ( N ) and isometric gr oup homomorphism s ψ j : D j → D ( j = 1 , 2 ) such that ψ 2 ◦ ϕ 1 = ψ 1   E 1 and (2-19) k ψ 1   E 2 − ψ 2 ◦ ϕ 2 k ∞ 6 ε. 20 P . NI EMIEC Pr o o f. Let ˜ D 0 = { ( x, − ϕ 1 ( x )) : x ∈ E 1 } ⊂ D 1 × D 2 and ˜ D , π , ˜ ψ 1 and ˜ ψ 2 b e as in (2-12). Deﬁne ˜ λ : ˜ D → R + b y ˜ λ ( z ) = inf { λ 1 ( x 1 − x 2 ) + εδ ˜ D ( ˜ ψ 1 ( x 2 ) − ( ˜ ψ 2 ◦ ϕ 2 )( x 2 )) + λ 2 ( y + ϕ 2 ( x 2 )) : x 1 ∈ D 1 , x 2 ∈ E 2 , y ∈ D 2 , z = ˜ ψ 1 ( x 1 ) + ˜ ψ 2 ( y ) } . As usual, ˜ λ is a semiv alue on ˜ D . W e see that ˜ λ ( ˜ ψ 1 ( x 2 ) − ( ˜ ψ 2 ◦ ϕ 2 )( x 2 )) 6 ε for x 2 ∈ E 2 (this corresp onds to (2 -19)) and ˜ λ ( ˜ ψ j ( x )) 6 λ j ( x ) for x ∈ D j ( j = 1 , 2). W e wan t to sho w that in fact ˜ λ ( ˜ ψ j ( z )) = λ j ( z ) for z ∈ D j , which is equiv alen t to (2-20) λ 1 ( x − h ) + εδ ˜ D ( ˜ ψ 1 ( h ) − ( ˜ ψ 2 ◦ ϕ 2 )( h )) + λ 2 ( y + ϕ 2 ( h )) > λ j ( z ) pro vided x ∈ D 1 , h ∈ E 2 , y ∈ D 2 and (2-21) ˜ ψ j ( z ) = ˜ ψ 1 ( x ) + ˜ ψ 2 ( y ) . First a ssume j = 1. W e infer from ( 2-21) that k := x − z ∈ E 1 and y = − ϕ 1 ( k ). Then (2-20 ) is equiv alen t to (2-22) εδ ˜ D ( ˜ ψ 1 ( h ) − ( ˜ ψ 2 ◦ ϕ 2 )( h )) + λ 2 ( ϕ 2 ( h ) − ϕ 1 ( k )) > λ 1 ( h − k ) . When ˜ ψ 1 ( h ) = ( ˜ ψ 2 ◦ ϕ 2 )( h ), ϕ 2 ( h ) = ϕ 1 ( h ) and (2-2 2) is satisﬁed, thanks to the isometricit y of ϕ 1 . Otherwise, (2- 22) follows from (2- 18). The case of j = 2 is similar and is left for the reader. T o ﬁnish the pro of, deﬁne D , λ , ψ 1 and ψ 2 in exactly the same wa y as at the end of the pro of of Lemma 2.19.  3. Proof of Theorem 1.1 F or the dura tion of this section r ∈ { 1 , ∞} , N ∈ Z + \ { 1 } and a set Q ⊂ R suc h as in (2-1) are ﬁxed. Note that it is not assumed that Q is coun table. Ho we v er, in some results its coun t abilit y is necessary and then w e add this assumption to their statements . F or simplicit y , w e put the followin g 3.1. Deﬁnition. Supp ose Q is countable. A v alued gr oup ( G, + , p ) is said to b e a Q -gr oup iﬀ G is ﬁnite or a coun ta ble gro up o f class O f in and p ( G ) ⊂ Q . As a sp ecial case of t he general tec hnique of F ra ¨ ıss ´ e limits, w e get 3.2. Theorem. Supp ose Q is c ountable. Ther e is a unique (up to iso- metric gr oup isomorphi s m ) Q -gr oup Q G r ( N ) ∈ G r ( N ) with the f o l low- ing pr op erty. Whenever ( H, + , q ) ∈ G r ( N ) is a ﬁnite Q -g r oup and K is a sub gr oup of H , ev e ry isometric gr oup homo m orphism of K into Q G r ( N ) is extendable to an isometric gr oup homomorphi s m of H in to Q G r ( N ) . UNIVERS AL V ALU ED ABELIA N GR OUPS 21 Pr o o f. The uniqueness follows from the bac k- and-forth metho d. The existence ma y b e provide in a standard w ay . There are only coun tably man y (up to isometric g roup isomorphism) ﬁnite Q -g roups b elonging to G r ( N ) and th us there is a se quence ( H n , + , q n ) ∞ n =1 ⊂ G r ( N ) of ﬁnite Q - groups in whic h ev ery suc h group app ears inﬁnitely many times . Using rep eatedly Lemma 2 .19, inductiv ely deﬁne a sequence of ﬁnite Q - groups ( G n , + , p n ) ∈ G r ( N ), starting with G 0 = { 0 } , such that ( G n , + , p n ) ⊂ ( G n +1 , + , p n +1 ) and ev ery isometric gr oup homomorphism of a sub- group of H n is extendable to an isometric group homomorphism of H n in to G n +1 . Finally , put ( Q G r ( N ) , + , p ) = S ∞ n =1 ( G n , + , p n ). The details are left for the reader.  It is clear that the completion o f a mem b er of G r ( N ) is of t he same class. In what follows, we ﬁx the follo wing situation. W e a ssume that ( G, + , p ) ∈ G r ( N ) is complete and fo r some dense subgroup G 0 of G the fo llo wing conditions are fulﬁlled: (QG1) whenev er ( H , + , q ) ∈ G r ( N ) is a ﬁnite group with Q -v alued v a lue, K is its subgroup and ϕ : K → G 0 is an isometric g roup homomorphism, then for every ε ∈ (0 , 1) there is an ε -almost isometric group homomorphism ψ : H → G 0 suc h that k ψ   K − ϕ k ∞ 6 ε , (QG2) p ( G 0 ) ⊂ Q and the set of ﬁnite rank elemen ts of G 0 is dense in G 0 . (W e underline that it is not assumed here t hat Q is coun ta ble.) O ur aim is to show that (UEP) whenev er ( H , + , q ) ∈ G r ( N ) is a ﬁnite v alued group, K is its subgroup and ϕ : K → G is a n isometric group homomorphism, there is an isometric gr oup homomorphism ψ : H → G whic h extends ϕ . The pro of of (UEP) is preceded by a f ew lemmas. 3.3. Lemma. L et a ∈ G b e of ﬁnite r ank k > 2 . F or every ε > 0 ther e is b ∈ G 0 such that ra nk( b ) = k and p ( j a − j b ) 6 ε for e ach j ∈ Z . Pr o o f. Let δ ∈ (0 , 1) b e suc h that δ 6 ε 6 k . By (QG2), t here is a ﬁnite rank elemen t c ∈ G 0 for whic h p ( a − c ) 6 δ . Let H = h a, c i and K = h c i ⊂ H . Notice that p ( K ) ⊂ Q . W e conclude from Lemma 2.6 that there is a Q -v alued v alue q on H which extends p   K , is b ounded b y r and satisﬁes k p   H − q k ∞ 6 δ . W e see that ( H, + , q ) ∈ G r ( N ). So, b y (QG1) applied to id : K → G 0 , there is a δ -almost isometric (with resp ect to q ) group h omomorphism ϕ : H → G 0 suc h that p ( x − ϕ ( x )) 6 δ f or x ∈ K . Put b = ϕ ( a ). Then rank( b ) = ra nk( a ) and p ( a − b ) 6 p ( a − c ) + p ( c − ϕ ( c )) + p ( ϕ ( c ) − ϕ ( a )) 6 2 δ + (1 + δ ) q ( c − a ) 6 2 δ + 2( p ( c − a ) + δ ) 6 6 δ 6 ε k , 22 P . NI EMIEC so p ( j a − j b ) 6 ε for j = 0 , 1 , . . . , k − 1 and w e are done.  3.4. Lemma. L et H b e a ﬁnite sub gr oup of G . F or e ach ε > 0 ther e is a gr oup homomorphi s m ϕ : H → G 0 with trivial kernel such that p ( ϕ ( x ) − x ) 6 ε f or e ach x ∈ H . Pr o o f. W e assume H is non trivial. Since ev ery ﬁnite Ab elian gro up is group isomorphic to a direct product of cyclic gr oups, there is s > 1 and a 1 , . . . , a s suc h that the function Φ : h a 1 i × . . . × h a s i ∋ ( x 1 , . . . , x s ) 7→ P s j =1 x j ∈ H is a gro up isomorphism. Let µ = min { p ( x ) : x ∈ H \ { 0 }} > 0. W e assume ε < µ . By Lemma 3.3, there ar e b 1 , . . . , b s ∈ G 0 with rank( b j ) = rank( a j ) and p ( l b j − la j ) 6 ε/ s for j = 1 , . . . , s and l ∈ Z . W e infer from this that there are gro up isomorphisms Λ j : h a j i → h b j i suc h that Λ j ( a j ) = b j . Let Λ = Λ 1 × . . . × Λ s : h a 1 i × . . . × h a s i → h b 1 i × . . . × h b s i (Λ( x 1 , . . . , x s ) = (Λ 1 ( x 1 ) , . . . , Λ s ( x s ))). F urther, if l 1 , . . . , l s ∈ Z , then p ( P s j =1 l j a j − P s j =1 l j b j ) 6 P s j =1 p ( l j a j − l j b j ) 6 ε < µ and th us P s j =1 l j a j = 0 iﬀ P s j =1 l j b j = 0. This yields that the group homo- morphism Ψ : h b 1 i × . . . × h b s i ∋ ( y 1 , . . . , y s ) 7→ s X j =1 y j ∈ G 0 has trivial k ernel. Now it suﬃces to put ϕ = Ψ ◦ Λ ◦ Φ − 1 .  3.5. Lemma. L et ( H , + , q ) ∈ G r ( N ) b e a ﬁnite gr oup, K its sub gr oup and let ϕ : K → G b e an isom e tric g r oup homomorphism. Then for every ε ∈ (0 , 1) ther e is an ε -almos t isome tric g r oup homo morphism ψ : H → G such that k ψ   K − ϕ k ∞ 6 ε . Pr o o f. Again, w e assume H is nontrivial. Let µ = min { q ( h ) : h ∈ H \ { 0 }} . T ak e δ suc h tha t (3-1) δ ∈ (0 , 1 2 ) , δ < c, (1 + 2 δ ) 2 6 1 + ε, (1 − 2 δ ) 2 > 1 − ε. By Lemma 3 .4, there is a g roup homomorphism κ : ϕ ( K ) → G 0 with trivial kernel suc h that p ( κ ( x ) − x ) 6 δ 2 . Put ψ 0 = κ ◦ ϕ : K → G 0 . Then ψ 0 has trivial k ernel and (3-2) k ψ 0 − ϕ k ∞ 6 δ 2 . Let λ 0 : K ∋ x 7→ p ( ψ 0 ( x )) ∈ R + . Observ e that λ 0 is a v a lue b ounded b y r and λ 0 ( K ) ⊂ Q (see (QG2)). Moreo ve r, for x ∈ K \ { 0 } w e hav e (thanks to (3-1 ) and ( 3-2)): q ( x ) = p ( ϕ ( x )) 6 p ( ψ 0 ( x )) + δ 2 6 λ 0 ( x ) + δ q ( x ) and λ 0 ( x ) 6 p ( ϕ ( x )) + δ 2 6 q ( x ) + δ q ( x ) = (1 + δ ) q ( x ) . UNIVERS AL V ALU ED ABELIA N GR OUPS 23 The ab ov e estimations giv es q   K 6 1 1 − δ λ 0 and λ 0 6 (1 + δ ) q   K . Now b y Lemma 2.7 (with ω ( t ) = (1 + δ ) t ,  ( t ) = t 1 − δ and τ ( t ) = t ), there is a v alue λ 1 on H whic h extends λ 0 , is b ounded b y r and satisﬁes (3-3) (1 − δ ) q 6 λ 1 6 (1 + δ ) q . F urther, since λ 1 ( K ) = λ 0 ( K ) ⊂ Q , w e infer from Lemma 2.6 that there is a v alue λ o n H b ounded b y r , exte nding λ 1   K and satisfying λ ( H ) ⊂ Q and (3-4) k λ 1 − λ k ∞ 6 δ 2 . Note that ψ 0 : K → G 0 is isometric with respect to λ and there- fore, b y (Q G1), there is a δ -almost isometric group homomorphism ψ : ( H , + , λ ) → ( G, + , p ) suc h that k ψ   K − ψ 0 k ∞ 6 δ . The la tter in- equalit y com bined with (3 -1) a nd (3-2) giv es k ψ   K − ϕ k ∞ 6 ε . So, w e only need to c hec k that ψ , as a gro up homomorphism of ( H , + , q ) into ( G, + , p ), is ε -almost isometric. F or h ∈ H \ { 0 } w e ha v e, thanks to (3-1), (3- 3) and (3-4): p ( ψ ( h )) 6 (1 + δ ) λ ( h ) 6 (1 + δ )( λ 1 ( h ) + δ 2 ) 6 (1 + δ )[(1 + δ ) q ( h ) + δ q ( h )] 6 (1 + 2 δ ) 2 q ( h ) 6 (1 + ε ) q ( h ) and p ( ψ ( h )) > (1 − δ ) λ ( h ) > (1 − δ )( λ 1 ( h ) − δ 2 ) > (1 − δ )[(1 − δ ) q ( h ) − δ q ( h )) > (1 − 2 δ ) 2 q ( h ) > (1 − ε ) q ( h ) whic h ﬁnishes the pro of.  3.6. R emark. F rom the b eginning of the section to this momen t w e ha ve nev er used the assumption that G is complete. So, the assertion of Lemma 3.5 is fulﬁlled for ev ery group G whic h is ‘b et w een’ G 0 and its completion. In particular, Lemma 3.5 ho lds true for G = G 0 . 3.7. Lemma. L et ( H , + , q ) ∈ G r ( N ) b e a ﬁnite gr oup and K its sub- gr oup. F urther, let ϕ : K → G and ψ : H → G b e, r esp e ctively, an isometric and an ε -almost isometric gr o up homomorphism (whe r e ε ∈ (0 , 1) ) such that k ψ   K − ϕ k ∞ 6 ε . F or every δ ∈ (0 , ε ) ther e is a δ -almost isometric gr oup homomorphism ψ ′ : H → G such that k ψ ′   K − ϕ k ∞ 6 δ and k ψ − ψ ′ k ∞ 6 C ε wh er e C = 3 + 2 diam( H , q ) . Pr o o f. Let L = ϕ ( K ) + ψ ( H ) and p ′ = p   L . T hen ( L, + , p ′ ) ∈ G r ( N ) and L is ﬁnite. By Lemma 2.20, there are a ﬁnite v alued gr oup ( D , + , λ ) ∈ G r ( N ) and isometric group homomo rphisms w H : H → D and w L : L → D such that (3-5) w L ◦ ϕ = w H   K and k w H − w L ◦ ψ k ∞ 6 Aε 24 P . NI EMIEC where A = 1 + diam( H , q ). Put D 0 = w L ( L ). O bserv e that w − 1 L : D 0 → L ⊂ G is isome tric. Hence, b y Lemma 3.5, there is a δ -almost isome tric group homomorphism v : D → G suc h that (3-6) k v   D 0 − w − 1 L k ∞ 6 δ . Then ψ ′ = v ◦ w H : H → G is δ - almost isometric. Now if x ∈ K , then b y (3-5), w H ( x ) = ( w L ◦ ϕ )( x ) and th us p ( ϕ ( x ) − ψ ′ ( x )) = p ( w − 1 L (( w L ◦ ϕ )( x )) − v (( w L ◦ ϕ )( x ))) 6 δ, thanks to ( 3-6). Finally , if h ∈ H , then ψ ( h ) ∈ L and (by (3- 5) and (3-6)): p ( ψ ( h ) − ψ ′ ( h )) 6 p ( w − 1 L (( w L ◦ ψ )( h )) − v (( w L ◦ ψ )( h ))) + p ( v (( w L ◦ ψ )( h )) − v ( w H ( h ))) 6 k w − 1 L − v   D 0 k ∞ + (1 + δ ) p (( w L ◦ ψ )( h ) − w H ( h )) 6 δ + 2 k w L ◦ ψ − w H k ∞ 6 ε + 2 Aε = C ε.  No w w e are ready to pro v e 3.8. Theorem. Th e gr oup G satisﬁes (UEP) . Pr o o f. Let ( H, + , q ), K and ϕ b e as in (UEP ). Pu t C = 3 + 2 diam( H , q ) and ε n = 2 − n ( n > 1). By Lemma 3.5, there is an ε 1 -almost isometric group homomorphism ψ 1 : H → G suc h that k ψ 1   K − ϕ k ∞ 6 ε 1 . Sup- p ose w e ha v e ψ n − 1 for some n > 2. Applying Lemma 3 .7, w e obtain ψ n : H → G suc h that (1 n ) ψ n is an ε n -almost isometric group homomorphism, (2 n ) k ψ n   K − ϕ k ∞ 6 ε n , (3 n ) k ψ n − ψ n − 1 k ∞ 6 C ε n − 1 . The condition (3 n ) ensures that the sequence ( ψ n ( h )) ∞ n =1 con ve rges fo r ev ery h ∈ H and thus w e ma y deﬁne a group homomorphism ψ : H → G by ψ ( h ) = lim n →∞ ψ n ( h ). Then (1 n ) yields that ψ is isometric and (2 n ) give s ψ   K = ϕ .  Pr o o f of The or em 1.1. Let G b e the completion of QG r ( N ) (cf. Theorem 3.2). By Theorem 3.8, G satisﬁes (UEP). This giv es the existence of G r ( N ). The uniqueness follo ws aga in f rom Theorem 3 .8, applied to the situation when Q = R , (G 3) and the back -and-forth metho d.  Note that w e hav e pro v ed that G r ( N ) fulﬁlls (UEP). This fa ct will b e used in the next section. The results o f this section giv es more than just the assertion o f Theorem 1.1. Namely , 3.9. Prop osition. In e ach of the fol lowing c ases the c omple tion of a value d gr oup ( G, + , p ) is isometric al ly gr oup isomorph ic to G r ( N ) . UNIVERS AL V ALU ED ABELIA N GR OUPS 25 (A) G = Q G r ( N ) wher e Q is a c ountable set as in (2-1) (cf. The o- r em 3.2 ). (B) ( G, + , p ) ∈ G r ( N ) and G satisﬁes c o nditions (G2) and (G3) of The or em 1.1 (with G r ( N ) r eplac e d by G ). (C) ( G, + , p ) satisﬁes c o nditions (Q G1) and (QG2) (with G 0 r eplac e d by G ) for some set Q as in (2-1) . 4. Proof of Theorem 1.2 As in the previous section, w e ﬁx r ∈ { 1 , ∞} and N ∈ Z + \ { 1 } . Our ﬁrst aim of this part is to sho w that (CEP) Whenev er ( L, + , q ) ∈ G r ( N ), K and H are, respectiv ely , a compact and a ﬁnite subgroup of L , and ϕ : K → G r ( N ) is an isometric group homomorphism, then there is an isometric group homomorphism ψ : K + H → G r ( N ) whic h extends ϕ . Similarly as in the previous section, the pro of of (CEP) is preceded b y a few auxiliary lemmas. In some of them we use the Hausdorﬀ distance, whic h is denoted by us b y dist q ( A, B ) if only A and B are tw o compact nonempt y subsets of a v a lued group with v alue q . 4.1. Lemma. L e t a ∈ G r (0) b e such that the c losur e K of h a i is c om- p act. Then for every ε > 0 ther e is a ﬁnite r ank element b of G r (0) such that dist p ( h b i , K ) 6 ε wher e p is the value of G r (0) . Pr o o f. W e asssume rank( a ) = ∞ . Sinc e the ele men ts of the sequence ( na ) ∞ n =1 form a dense subset of K , there is m > 2 suc h that the set { 0 , a, . . . , ( m − 1) a } is an ( ε / 4)-net for K and (4-1) p ( ma ) 6 ε 8 . By (G 3), there is a ﬁnite rank elemen t c ∈ G r (0) for whic h (4-2) p ( a − c ) 6 ε 4 m . Put H = h c i and let H ′ = h c ′ i b e a cyclic group of rank m . F or eac h k ∈ Z let s ( k ) ∈ { 0 , 1 , . . . , m − 1 } b e suc h that m | k − s ( k ). Deﬁne q 0 : H × H ′ → R + b y q 0 ( x, l c ′ ) = p ( x + s ( l ) a ) + δ H ′ ( l c ′ ) ε/ 8 ( l ∈ Z ). Observ e that q 0 is w ell deﬁne d; q 0 ( x, y ) = 0 iﬀ x = 0 and y = 0; and q 0 ( x, 0) = p ( x ). W e claim that q 0 satisﬁes the triang le inequalit y , that is, q 0 ( x + x ′ , ( l + l ′ ) c ′ ) 6 q 0 ( x, l c ′ ) + q 0 ( x ′ , l ′ c ′ ). Indeed, if s ( l ) + s ( l ′ ) < m , then s ( l + l ′ ) = s ( l ) + s ( l ′ ) and then the la tter inequalit y is immediate. And when s ( l ) + s ( l ′ ) > m , we ha v e s ( l ) > 0, s ( l ′ ) > 0 and s ( l + l ′ ) = s ( l ) + s ( l ′ ) − m and hence, b y ( 4-1), q 0 ( x, l c ′ ) + q 0 ( x ′ , l ′ c ′ ) > p ( x + x ′ + ( s ( l ) + s ( l ′ )) a ) + ε 8 + ε 8 > p ( x + x ′ + s ( l + l ′ ) a − ma ) + p ( ma ) + ε 8 δ H ′ (( l + l ′ ) c ′ ) > q 0 ( x + x ′ , ( l + l ′ ) c ′ ) . 26 P . NI EMIEC No w let q : H × H ′ → R + b e giv en b y q ( x, y ) = [ 1 2 ( q 0 ( x, y ) + q 0 ( − x, − y ))] ∧ r . It follo ws from the prop erties of q 0 that q is a v alue and q ( x, 0) = p ( x ) for x ∈ H . Moreo v er, for j ∈ { 0 , 1 , . . . , m − 1 } one has (4-3) q ( j c, − j c ′ ) 6 ε 2 . Indeed, w e may assume that j 6 = 0 and then s ( j ) = j and s ( − j ) = m − j . So, b y (4-1) and (4-2), q 0 ( j c, − j c ′ ) 6 p ( j ( c − a )) + p ( ma ) + ε 8 6 ε 2 and q 0 ( − j c, j c ′ ) = p ( j ( c − a )) + ε 8 6 ε 2 . F urther, since the function H × { 0 } ∋ ( h, 0) 7→ h ∈ G r (0), w e conclude from (UEP) that there exists an isometric group homomorphism ψ : H × H ′ → G r (0) with ψ ( h, 0) = h for an y h ∈ H . Put b = ψ (0 , c ′ ). Then h b i = { 0 , b, . . . , ( m − 1) b } . No w if j ∈ { 0 , 1 , . . . , m − 1 } , w e get (see (4-2) and (4-3)): p ( j b − j a ) 6 p ( j c − j b ) + p ( j c − j a ) 6 p ( ψ ( j c, 0) − j ψ (0 , c ′ )) + j p ( c − a ) 6 p ( ψ ( j c, − j c ′ )) + ε 4 = q ( j c, − j c ′ ) + ε 4 6 3 4 ε. Con ve rsely , if x ∈ K , there is j ∈ { 0 , 1 , . . . , m − 1 } suc h that p ( x − j a ) 6 ε/ 4 a nd then p ( x − j b ) 6 p ( x − j a ) + p ( j a − j b ) 6 ε whic h ﬁnally giv es dist p ( h b i , K ) 6 ε .  F ro m no w to the end of the next section p denotes the v alue of G r ( N ). 4.2. Lemma. L et K b e a c omp ac t sub g r oup of G r ( N ) . F or e ach ε > 0 ther e is a ﬁnite sub gr oup H of G r ( N ) such that dist p ( H , K ) 6 ε . Pr o o f. There is s > 2 and a 1 , . . . , a s ∈ K suc h that { a 1 , . . . , a s } is an ( ε/ 2)-net for K . If N 6 = 0, put H = h a 1 , . . . , a s i and notice that dist p ( H , K ) 6 ε/ 2 since { a 1 , . . . , a s } ⊂ H ⊂ K . W e no w a ssume N = 0 . By Lemma 4.1, there are ﬁnite rank eleme n ts b 1 , . . . , b s ∈ G r (0) with dist p ( h b j i , K j ) 6 ε/s where K j is the c losure of h a j i ( j = 1 , . . . , s ). Let H = h b 1 , . . . , b s i . F or ev ery x ∈ H there are x j ∈ h b j i ( j = 1 , . . . , s ) for whic h x = P s j =1 x j . Then for ev ery j one can ﬁnd y j ∈ K j suc h that p ( x j − y j ) 6 ε/ s and hence p ( x − y ) 6 ε for y = P s j =1 y j ∈ K . Con ve rsely , if y ∈ K , there is j suc h that p ( x − a j ) 6 ε/ 2, and there is w ∈ h b j i ⊂ H with p ( a j − w ) 6 ε/ s 6 ε/ 2 whic h yields p ( x − w ) 6 ε and we are done.  F or need of the nearest three results let ( L, + , q ), K , H and ϕ b e as in (CEP). 4.3. Lemma. F or e very ε > 0 ther e is an isometric gr oup homomor- phism ψ : H → G r ( N ) such that fo r any h ∈ H and k ∈ K , (4-4) | p ( ψ ( h ) − ϕ ( k )) − q ( h − k ) | 6 ε. UNIVERS AL V ALU ED ABELIA N GR OUPS 27 Pr o o f. Let ˜ K = ϕ ( K ). By Lemma 4.2, there is a ﬁnite subgroup F of G r ( N ) suc h that (4-5) dist p ( F , ˜ K ) 6 ε 2 . It fo llo ws from Lemma 2.19, applied to ϕ 1 = ϕ : K → ˜ K + F and ϕ 2 = id : K → L , that there exist a v alued g roup ( D, + , λ ) ∈ G r ( N ) and isometric group homomorphisms Φ : L → D a nd Ψ : ˜ K + F → D for whic h Ψ ◦ ϕ = Φ   K . F urther, the group Φ( H ) + Ψ( F ) is a ﬁnite subgroup of D and the group homomorphism Ψ − 1   Ψ( F ) : Ψ ( F ) → G r ( N ) is isometric. So, t hanks to (UEP), there is an isometric group homomorphism τ : Φ( H ) + Ψ( F ) → G r ( N ) whic h extends Ψ − 1   Ψ( F ) . Put ψ = τ ◦ Φ   H . Fix h ∈ H and k ∈ K . T ak e f ∈ F suc h that p ( f − ϕ ( k )) 6 ε/ 2 ( see (4 -5)). Then p ( ψ ( h ) − ϕ ( k )) 6 p ( τ (Φ( h )) − f ) + ε/ 2 and p ( τ (Φ( h )) − f ) = p ( τ (Φ( h )) − τ (Ψ( f ))) = λ (Φ( h ) − Ψ( f )) 6 λ (Φ( h ) − Φ( k )) + λ (Φ( k ) − Ψ( f )) = q ( h − k ) + λ (Ψ ( ϕ ( k )) − Ψ( f )) = q ( h − k ) + p ( ϕ ( k ) − f ) 6 q ( h − k ) + ε 2 whic h sho ws that p ( ψ ( h ) − ϕ ( k )) 6 q ( h − k ) + ε . Con v ersely , q ( h − k ) = λ (Φ( h ) − Φ( k )) 6 λ (Φ( h ) − Ψ( f )) + λ (Ψ( f ) − Φ( k )) = p ( τ ( Φ( h )) − τ (Ψ( f ) )) + λ (Ψ( f ) − Ψ( ϕ ( k ))) = p ( ψ ( h ) − f ) + p ( f − ϕ ( k )) 6 p ( ψ ( h ) − ϕ ( k )) + 2 p ( f − ϕ ( k )) whic h ﬁnishes the pro of of (4-4), b ecause 2 p ( f − ϕ ( k )) 6 ε .  4.4. Lemma. L et ψ : H → G r ( N ) b e an isometric g r oup homom o r- phism such that (4-4) is fulﬁl le d for any h ∈ H and k ∈ K . F or e ach δ > 0 ther e ex i sts a n isometric gr oup homomorph i s m ψ ′ : H → G r ( N ) for which k ψ − ψ ′ k ∞ 6 ε + δ and for al l h ∈ H and k ∈ K , (4-6) | p ( ψ ′ ( h ) − ϕ ( k )) − q ( h − k ) | 6 δ . Pr o o f. By Lemma 2.21, there are a v alued group ( D , + , λ ) ∈ G r ( N ) and isometric group homomo rphisms w L : L → D a nd w G : ϕ ( K ) + ψ ( H ) → D suc h that w L   K = w G ◦ ϕ and (4-7) k w L   H − w G ◦ ψ k ∞ 6 ε. Then Z := w G ( ϕ ( K ) + ψ ( H )) and F := w L ( H ) + w G ( ψ ( H )) are, resp ec- tiv ely , a compact and a ﬁnite subgroup of D , and w − 1 G : Z → G r ( N ) is isometric. So, thanks to Lemma 4.3, t here is an isometric group homomorphism ξ : F → G r ( N ) suc h that for any z ∈ Z and f ∈ F , (4-8) | p ( w − 1 G ( z ) − ξ ( f )) − λ ( z − f ) | 6 δ. 28 P . NI EMIEC Put ψ ′ = ξ ◦ w L   H : H → G r ( N ). Now ﬁx h ∈ H and k ∈ K and put z = w G ( ϕ ( k )) ∈ Z a nd f = w L ( h ) ∈ F . Then w e hav e p ( ψ ′ ( h ) − ϕ ( k )) = p ( ξ ( f ) − w − 1 G ( z )) and (since w G ◦ ϕ = w L   K ) q ( h − k ) = λ ( w L ( h ) − w L ( k )) = λ ( w L ( h ) − w G ( ϕ ( k ))) = λ ( f − z ) . Hence (4-6) f ollo ws from (4 -8). It suﬃces t o sho w that k ψ − ψ ′ k ∞ 6 ε + δ . F o r h ∈ H w e get p ( ψ ( h ) − ψ ′ ( h )) 6 p ( ψ ( h ) − ξ (( w G ◦ ψ )( h ))) + p ( ξ ( ( w G ◦ ψ )( h )) − ξ ( w L ( h ))) and, by (4-7), p ( ξ (( w G ◦ ψ )( h )) − ξ ( w L ( h ))) = λ (( w G ◦ ψ )( h ) − w L ( h )) 6 ε . So, it remains to chec k t hat p ( ψ ( h ) − ξ (( w G ◦ ψ )( h ))) 6 δ . But this follo ws from (4-8) f or z = f = ( w G ◦ ψ )( h ) ∈ Z ∩ F .  Finally , w e ha v e 4.5. Theorem. Th e assertion of (CEP) is sa tisﬁ e d. Pr o o f. Put ε n = 1 2 n . By Lemma 4.3, there is an isometric group homomorphism ψ 1 : H → G r ( N ) s uc h that (4 -4) (with ψ replaced b y ψ 1 ) is fulﬁlled for any h ∈ H and k ∈ K . Now supp ose that ψ n − 1 : H → G r ( N ) is constructed. W e infer from Lemma 4 .4 that there is a group homomorphism ψ n : H → G r ( N ) such that (1 n ) ψ n is isometric, (2 n ) k ψ n − 1 − ψ n k ∞ 6 ε n − 1 + ε n 6 2 ε n − 1 , (3 n ) | p ( ψ n ( h ) − ϕ ( k )) − q ( h − k ) | 6 ε n for all h ∈ H and k ∈ K . W e conclude fr om (1 n ) and (2 n ) that a group homomorphism ϕ 0 : H → G r ( N ) giv en by ϕ 0 ( h ) = lim n →∞ ψ n ( h ) is well deﬁned and isometric. What is more, (3 n ) yields (4-9) p ( ϕ 0 ( h ) − ϕ ( k )) = q ( h − k ) for an y h ∈ H and k ∈ K . W e easily deduce from (4-9 ) tha t the form ula ψ ( h + k ) = ϕ 0 ( h ) + ψ ( k ) (where h ∈ H and k ∈ K ) w ell deﬁnes an isometric g roup homomorphism ψ : K + H → G r ( N ) whic h extends ϕ .  Pr o o f of The or em 1.2 . W e b egin with the note that the assertion of part (B) fo llo ws f rom (A), (G3 ) and the ba c k-and-f orth metho d. Thu s, it suﬃce s to pro ve ( A). F irst we shall sho w the easier p o in t, (i). When ϕ is not op en, it s uﬃces to pu t τ ≡ 0,  = id and to apply Corolla ry 2.4 in order to ﬁnd ω ∈ Ω ∗ . Similarly , if ϕ is op en as a map o f K on to ϕ ( K ), the se miv a lue q 0 : K ∋ x 7→ dist q ( x, k er ϕ ) ∈ R + is con tin uous with respect to p ◦ ϕ and hence for τ 0 = id id +1 one may ﬁnd (thanks to Lemma 2.2)  0 ∈ Ω ∗ suc h that τ 0 ◦ q 0 6  0 ◦ p ◦ ϕ . F urther, take suitable ω 0 ∈ Ω ∗ and apply the idea of Example 2.8 t o guarante e ( ω - ρ - τ ). W e no w pass to the main part of the theorem—to p oin t (ii). Put λ 0 : K ∋ x 7→ p ( ϕ ( x )) ∈ R + . Then λ 0 is a semiv alue on K suc h that λ 0 6 ( ω ◦ q )   K . L emma 2.7 ensures us t he existence of a UNIVERS AL V ALU ED ABELIA N GR OUPS 29 semiv alue λ o n H b ounded by r whic h extends λ 0 and satisﬁes suitable conditions. Put H ′ = H / ker ϕ . Let π : H → H ′ b e the quotien t group homomorphism, λ ′ the v alue induced b y λ (recall that λ − 1 ( { 0 } ) = λ − 1 0 ( { 0 } ) = k er ϕ ), K ′ = π ( K ) and let ϕ ′ : K ′ → G r ( N ) b e a gr oup homomorphism suc h that ϕ ′ ◦ π   K = ϕ . Notice that ( H ′ , + , λ ′ ) ∈ G r ( N ) (thanks to Prop osition 2.14) and (4-10) p ( ϕ ′ ( x )) = λ ′ ( x ) for eve ry x ∈ K ′ (b ecause λ extends λ 0 ). No w let ( ¯ H , + , ¯ λ ) ∈ G r ( N ) b e the completion of ( H ′ , + , λ ′ ). The relation (4-10) ensures us that ϕ ′ extends to an isometric group homomorphism ¯ ϕ : ¯ K → G r ( N ) de- ﬁned on a closed subgroup of ¯ H . Since the closure of ϕ ( K ) is compact in G r ( N ), ¯ K is compact. En larging, if needed, the group ¯ H (ma king use of Theorem 2.1 2) w e ma y assume that ¯ H f in is dense in ¯ H . No w thanks to (CEP) and t he induction argumen t w e see t hat there is a n isometric group homomor phism ¯ ψ : ¯ H → G r ( N ) whic h extends ¯ ϕ . De- ﬁne ϕ ω : H → G r ( N ) by ϕ ω = ¯ ψ ◦ π and observ e that k er ϕ = k er ϕ ω and p ( ϕ ω ( h )) = λ ( h ) for eac h h ∈ H . The v eriﬁcation t hat all other assertions a re fulﬁlled is left for the reader.  In the next section we shall show that G r (2) is Urysohn as a metric space. Thus , Theorem 1 .2 extends and strengthens the results of [17]. 4.6. Corollary . (A) The gr oup G r ( N ) is universal for the class G r ( N ) ; that is, every memb er of G r ( N ) admits an isometric gr oup homo- morphism into G r ( N ) . (B) The gr oups G 1 (0) and G ∞ (0) ar e top olo gic al ly universal for the class of sep ar able metrizable top olo gic al Ab elian gr oups. What is mor e, for every ( G, + , q ) ∈ G the va l ue d gr oup ( G, + , q ∧ 1) (r e- sp e ctively ( G, + , q α ) with 0 < α < 1 ) admits an isome tric gr oup homomorphism into G 1 (0) (r esp e ctively into G ∞ (0) ). F or t wo pairs ( r, N ) , ( s, M ) ∈ { 1 , ∞} × ( Z + \{ 1 } ) let us write ( r , N ) 4 ( s, M ) iﬀ r 6 s and N | M . The reader will easily c hec k that 4.7. Pr op osition. L et ( r , N ) ∈ { 1 , ∞} × ( Z + \ { 1 } ) . (A) G s ( M ) is emb e ddable in G r ( N ) by me ans of an isome tric gr oup homomorphism iﬀ ( s, M ) 4 ( r, N ) . (B) L et M | N (and M 6 = 1 ). Then the gr oup G r ( N , M ) := { x ∈ G r ( N ) : M · x = 0 } is isometric al ly gr oup isomorphic to G r ( M ) . The ab ov e simple result has tw o in teresting consequence, form ula ted in the next tw o results. 4.8. Pr op osition. Ther e is a f a m ily { Φ r,N s,M : G s ( M ) → G r ( N ) } ( s,M ) 4 ( r ,N ) of isometric gr oup homomorphisms such that for e ac h ( r , N ) , ( s, M ) , ( t, L ) w ith ( r, N ) 4 ( s, M ) 4 ( t, L ) : 30 P . NI EMIEC (i) Φ r,N r,N = id G r ( N ) , (ii) Φ t,L s,M ◦ Φ s,M r,N = Φ t,L r,N . Pr o o f. Let Ψ 1 , 0 : G 1 (0) → G ∞ (0) b e an isome tric group homomor- phism. F urther, let Ψ ∞ , 0 = id G ∞ (0) and for N > 2 let Ψ ∞ ,N : G ∞ ( N ) → G ∞ (0 , N ) and Ψ 1 ,N : G 1 ,N → G ∞ (0 , N ) ∩ Ψ 1 , 0 ( G 1 (0)) be isometric group isomorphisms (where G ∞ (0 , N ) is as in the statemen t of Prop o- sition 4 .7). Now it s uﬃces to put Φ r,N s,M = Ψ − 1 r,N ◦ Ψ s,M pro vided ( s, M ) 4 ( r , N ).  4.9. Prop osition. Suup ose N 6 = 0 . If N 1 , . . . , N s > 2 ar e mutual ly c oprime and N = N 1 · . . . · N s , then G r ( N ) and G r ( N 1 ) × . . . × G r ( N s ) ar e isomorph ic as top o lo gic al gr oups. Pr o o f. The assertion f ollo ws from Prop osition 4.7 and the fact that the function G r ( N , N 1 ) × . . . × G r ( N , N s ) ∋ ( x 1 , . . . , x s ) 7→ s X j =1 x j ∈ G r ( N ) is an isomorphism of top ological groups.  In the next section we shall sho w that G r ( N )’s are pairwise noniso- morphic as top olo gical groups. The main assumption of Theorem 1.2 is that the closure of the image of a group homomorphism is compact. One may ask whether one ma y w eak en t his condition. As the next result sho ws, nothing else ma y b e done in this direction for group homomorphims with (metrically) b ounded images. (This result has its na tural w ell-kno wn coun terpart for the Urysohn metric space.) 4.10. Prop osition. L et K b e a c lose d b ounde d sub gr oup of G r ( N ) such that for every ﬁ nite gr oup H and every value q on K × H such that ( K × H, + , q ) ∈ G r ( N ) and q ( x, 0) = p ( x ) for x ∈ K ther e is an isometric gr oup hom o morphism ψ : K × H → G r ( N ) with ψ ( x, 0) = x for x ∈ K . Then K is c om p act. Pr o o f. Supp ose K is noncompact. W e infer f rom the completeness of K that then there exist ε > 0 and a sequence ( x n ) ∞ n =1 ⊂ K suc h that p ( x n − x m ) > ε f or distinct n and m . Put A = { x n : n > 1 } . F or ev ery x ∈ G r ( N ) let e x : A → R + b e giv en b y e x ( a ) = p ( x − a ). It is easily seen that the set E = { e x : x ∈ G r ( N ) } ⊂ E r ( A ) is separable (with resp ect to the suprem um metric), since the map G r ( N ) ∋ x 7→ e x ∈ E is a nonexpansiv e surjection. How ev er, w e shall sho w that E contains an uncoun table discrete subset (whic h will ﬁnish the pro of ). Supp ose f ∈ E r ( A ) is suc h t hat f satisﬁes ( 2-8) for any a 1 , . . . , a N ∈ A provided N > 2. Then, by Theorem 2.17 , f is trivial in some v alued group b elonging to G r ( N ) and of the f orm K × H with H ﬁnite. So , UNIVERS AL V ALU ED ABELIA N GR OUPS 31 thanks to our assumption on K , f is trivial in G r ( N ) whic h means that f ∈ E . Let M = diam ( K ) > 0. T ake δ ∈ (0 , ε ) suc h that δ < M / 2. F or a subset J o f A let f J : A → R + b e give n b y: f J ( a ) = M if a ∈ J and f J ( a ) = M − δ otherwis e. A direct calculation sho ws that f J ∈ E r ( A ) and f J satisﬁes (2-8) provid ed N 6 = 0. So, according to the previous paragraph, f J ∈ E . But k f J − f J ′ k ∞ = δ wh enev er J and J ′ are diﬀeren t subsets of A and hence E cannot b e separable.  4.11. R emark. Mellera y [14, 15] has sho wn that if ev ery isometry b e- t wee n to subsets of the Urysohn metric space U whic h ar e isometric to a giv en separable complete metric space X is extendable to an isome- try of U onto itself, then X is compact. A counterpart o f this result in category o f v alued groups reads as follows: If every is ometric gr oup i somorphism b etwe en two sub- gr oups of G r ( N ) w hich ar e isometric al ly gr oup is o mor- phic to a c omplete gr oup ( H , + , q ) ∈ G r ( N ) is extendab le to an isometric gr oup automorph i s m of G r ( N ) , then H is c omp act. It f ollo ws from Prop osition 4.10 that eve ry g roup H hav ing the ab ov e prop erty and b ounded v alue has to b e compact. Ho w ev er, the problem whether the ab ov e stated result is true in G ∞ ( N ) w e lea v e op en. 5. Geometr y of G r ( N ) ’s The part is mainly dev oted to in ves tigations of the gr oups G r ( N )’s as metric spaces. W e b egin with 5.1. Theorem. The metric sp ac es G r ( N ) w ith N ∈ { 0 , 2 } ar e Urysohn. In p articular, G r (2) is a Bo ole an Urysohn m etric g r oup intr o duc e d in [17] . Pr o o f. This is an almost immediate consequence of (UEP) and The- orem 2 .17. Let G 0 = G r ( N ) f in . L et f ∈ E i ( G 0 ) and B b e a ﬁnite nonempt y subs et o f G 0 . Put H = h B i . By Theorem 2.17, H ma y b e enlarged to a ﬁnite group b elonging to G r ( N ) in whic h f   B is trivial. No w w e infer from (UEP) that f   B is indee d trivial in G 0 . So, it follow s from the w ell- kno wn result on the Ury sohn space (see e.g. [24, 25], [10], [15]) that G r ( N ), as the completion of G 0 , is Urysohn.  5.2. R emark. The same argument as in the pro of of Theorem 5.1 show s that the metric space QG r ( N ) for N ∈ { 0 , 2 } is the so-called ra tional Urysohn space. Theorem 5.1 tells us ‘ev erything’ ab out the metric spaces G r ( N ) with N ∈ { 0 , 2 } . Therefore from now on, we assume N N N > > > 2 2 2 . In Theorem 5 .5 w e shall show that in that case G r ( N ) is not Urysohn. 32 P . NI EMIEC In the same w ay as in the pro of o f Theorem 5.1 one sho ws the next result the pro of of which is left as an exercise (use Theorem 2.17 a nd Remark 2.1 6; recall that all elemen ts a re of ﬁnite ra nk). 5.3. Pr op osition. (A) L et B b e a ﬁn ite nonemp ty subset of G r ( N ) and f ∈ E r ( B ) . Then f is trivial in G r ( N ) iﬀ f fulﬁl ls (2-8) for any a 1 , . . . , a N ∈ B . In p articular, f is trivial in G r ( N ) iﬀ f   A is trivial in G r ( N ) for every subset A of B such that 0 < card( A ) 6 N . (B) L et H b e a ﬁ n ite s ub gr oup of G r ( N ) a n d f ∈ E r ( H ) . T h e map f is trivial in G r ( N ) iﬀ f fulﬁl ls (2- 9) . It turns out that the exp onen t N is determined b y the metric of G r ( N ). This is a consequence of the ab ov e result and the follo wing 5.4. Example. Let Z = h b i b e a cyclic group of rank N and let H = Z N . F or j = 1 , . . . , N let e j = ( e j 1 , . . . , e j N ) ∈ H with e j j = b and e j k = 0 for k 6 = j . F urther, let F = {± e j : j = 1 , . . . , e N } ∪ { e j − e k : j, k = 1 , . . . , N } ⊂ H . of course, F generates H . Let k · k F b e the ‘norm’ on H generated by F (in the terminology of ﬁnitely generated groups, cf. [7]), i.e. k · k F is a v alue suc h that for nonzero h ∈ H , k h k F = min { n > 1 | ∃ f 1 , . . . , f n ∈ F : h = n X s =1 f s } (note t hat F = − F ). W e see that (5-1) k e j − e k k F = 1 for j 6 = k . W e claim that if n 1 , . . . , n N ∈ Z + , (5-2) k N X j =1 n j e j k F 6 N − max( n 1 , . . . , n N ) pro vided N X j =1 n j = N . Indeed, supp ose that e.g. max( n 1 , . . . , n N ) = n N and observ e that P N j =1 n j e j = P N − 1 j =1 n j ( e j − e N ), thanks to the assumption in (5-2). The latter equalit y give s (5-2). F urther, w e ha v e (5-3) k N X j =1 e j k F = N − 1 . T o see this, supp ose (for the con trary) that k P N j =1 e j k F 6 N − 2. This means (since 0 ∈ F ) that there are f 1 , . . . , f N − 2 ∈ F whic h sum up to P N j =1 e j . W rite f s = P N j =1 ε j s e j where ε j s ∈ { 0 , 1 , − 1 } . Since the rank of b is N , we conclude from this that P N − 2 s =1 ε j s = 1 for j = 1 , . . . , N and hence P N j =1 P N − 2 s =1 ε j s = N . Ho we v er, P N j =1 ε j s ∈ {− 1 , 0 , 1 } for s = 1 , . . . , N − 2 (b ecause f s ∈ F ) and therefore P N − 2 s =1 | P N j =1 ε j s | < N − 1 whic h denies earlier conclusion. So, (5- 3) is fulﬁlled (b y (5-2)). UNIVERS AL V ALU ED ABELIA N GR OUPS 33 No w put c = max(1 / 2 , 1 − 2 / N ) and let f : { e 1 , . . . , e N } → R + b e constan tly equal to c . By (5- 1), f is a Katˇ etov map. Observ e that (thanks to (5-3 )) (5-4) k N X j =1 e j k F > N X j =1 f ( e j ) (here is the only momen t where w e need to hav e N > 3). So, f fails to satisfy ( 2-8) and th us H cannot b e enlarged to a v alued group of exp onen t N in whic h f is trivial (b y Proposition 2.15). Ho w ev er, if A is a prop er nonempt y subset of { e 1 , . . . , e N } a nd a 1 , . . . , a N ∈ A , then the inequalit y (2- 8) is fulﬁlled, b y (5-2). Consequen tly , H may b e enlarg ed to a ﬁnite group b elonging to G r ( N ) in which f   A is trivial. If we now replace k · k F b y α k · k F with small enough α > 0 , we shall obtain an analogous example in the class G 1 ( N ). As a corollary of Theorem 5.1, P rop osition 5.3 and the ab o v e example w e obtain 5.5. Theorem. When N > 2 , N is the le as t natur al numb er k with the fol lowing pr op erty. F o r e v ery ﬁn i te nonempty subset A of G r ( N ) and e ach f ∈ E r ( A ) , f is trivial in G r ( N ) iﬀ f   B is trivial in G r ( N ) for any nonem p ty B ⊂ A wi th card( B ) 6 k . In p articular, for two di s tinct p airs ( r , N ) , ( s, M ) ∈ { 1 , ∞} × ( Z + \ { 1 } ) , the metric sp ac es G r ( N ) and G r ( M ) ar e iso m etric iﬀ r = s and { N , M } = { 0 , 2 } . Henc e, G r ( N ) is non-Urysohn f o r N > 2 . F urther geometric prop erties of G r ( N )’s are stated b elow . 5.6. Pr op osition. L et f ∈ E i ( G r ( N )) . (A) F or every two-p o i n t subset B of G r ( N ) the map f   B is trivial in G r ( N ) . (B) If N = 4 , f is trivial (in G r ( N ) ) on every thr e e-p oint subset of the sp ac e. (C) I f N 6 = 4 , ther e is a thr e e-p oint set C ⊂ G r ( N ) and a map g ∈ E r ( C ) which is nontrivial in G r ( N ) . Pr o o f. The p oints ( b) and (c) are left as ex ercises. (T o pro v e (c) for N > 4 , take Z a s in Example 5.4, H = Z 3 , deﬁne e 1 , e 2 , e 3 , F and k · k F in a similar manner and consider f : { e 1 , e 2 , e 3 } → R + constan tly equal to 1 / 2. Sho w that k ( k + r 1 ) e 1 + ( k + r 2 ) e 2 + ( k + r 3 ) e 3 k F > 2 3 ( N − 1) > N 2 for r 1 , r 2 , r 3 ∈ { 0 , 1 , 2 } whic h sum up to r ∈ { 0 , 1 , 2 } suc h that N = 3 k + r .) T o sho w (a), it remains to c hec k that (2-8) is fulﬁlled whenev er a 1 , . . . , a N ∈ { a, b } ⊂ G r ( N ). In that case (2-8) reduces to | p ( j ( a − b )) − f ( a ) | 6 ( j − 1) f ( a ) + ( N − j ) f ( b ) (since ( N − j ) b = − j b ) with j = 2 , . . . , N − 1 (for j = 1 this inequalit y follo ws from the deﬁnition of a Kat ˇ etov map). Of course f ( a ) − p ( j ( a − b )) 6 ( j − 1 ) f ( a ), so w e 34 P . NI EMIEC only need to ch ec k that p ( j ( a − b )) 6 j f ( a ) + ( N − j ) f ( b ). By the symmetry ( j ( a − b ) = − ( N − j ) ( a − b )), we may assume that j 6 N / 2. But t hen p ( j ( a − b )) 6 j ( f ( a ) + f ( b )) 6 j f ( a ) + ( N − j ) f ( b ).  5.7. Corollary . F or any two distinct p oints x and y of G r ( N ) ther e is an isometric ar c of [0 , c ] to G r ( N ) joining x and y , wher e c = p ( x − y ) . Pr o o f. By Pro p osition 5.6, for any x, y ∈ G r ( N ) there is z ∈ G r ( N ) suc h that p ( x − z ) = p ( y − z ) = p ( x − y ) / 2. Since G r ( N ) is complete, the assertion follow s.  The ab o ve result implies that ev ery metric space whic h is the image of G 1 ( N ) under a uniformly con tinuous function has b o unded metric. Since con tin uous group ho momorphisms a re uniformly contin uous, as a consequence of this w e obtain the res ult announced in the prev ious section. 5.8. Corollary . The gr oups G r ( N ) ’s ar e p airwise nonisomorphic as top olo gic al gr oups. The inequalit y (5 -4) in Example 5.4 implies that there are p o in ts x 1 , . . . , x N in G r ( N ) and radii r 1 , . . . , r N > 0 suc h t hat p ( x j − x k ) < r j + r k for an y j, k , but the closed balls ¯ B ( x j , r j ) ( j = 1 , . . . , N ) ha ve empt y in tersection. W e conclude from this that the metric space G r ( N ) fails to ha v e the prop erty of extending isometric maps b et we en ﬁnite subsets of G r ( N ) to Lipsc hitz maps with Lipsc hitz constan t s arbitrarily close to 1. This is wh y it is not so easy (as in case of G r (0) and G r (2) whic h are Urysohn spaces; compare with [28] or [19] where it is sho wn that the Urysohn space is homeomorphic to l 2 ) to prov e that G r ( N ) is an absolute retract ( for metric spaces). W e shall do this in Section 8 . Our last a im of this part is to show that each o f the metric spaces G r ( N )’s is metrically univ ersal for separable metric spaces of diameter no greater than r . In what fo llo ws, N > 2 and Z N = Z / N Z represen ts a cyclic g roup of rank N . Moreo ve r, let e stand for the generator of Z N . Adapting the idea of Lipsc hitz-free Banach spaces generated b y metric spaces (see e.g. [29], [6]), we in tro duce 5.9. Deﬁnition. Let X b e a nonempty set. F or x ∈ X let H x = Z N and let Z N [[ X ]] = L x ∈ X H x b e the direct pro duct of groups H x ’s. That is, Z N [[ X ]] consists of all f unctions f : X → Z N for whic h the set supp f := { x ∈ X : f ( x ) 6 = 0 } is ﬁ nite and Z N [[ X ]] is equipp ed with the p oint wise addition. F urther, let Z N [ X ] = { f ∈ Z N [[ X ]] : P x ∈ X f ( x ) = 0 } . It is clear that Z N [ X ] is a subgroup of Z N [[ X ]]. F or ev ery x ∈ X let b x ∈ Z N [[ X ]] b e suc h that b x ( x ) = e and b x ( y ) = 0 for y 6 = x . The set { b x − b y : x, y ∈ X } generates the group Z N [ X ]. UNIVERS AL V ALU ED ABELIA N GR OUPS 35 Whenev er d is a metric on X , deﬁne p d : Z N [ X ] → R + b y (5-5) p d ( f ) = inf { n X j =1 d ( x j , y j ) : n > 1 , x 1 , y 1 , . . . , x n , y n ∈ X, f = n X j =1 ( b x j − b y j ) } . The triple ( Z N [ X ] , + , p d ) is called the value d A b elian gr oup of exp onent N gener ate d b y the me tric sp ac e ( X , d ). As w e will see in the next result, p d is indeed a v alue. F or need of this, let us in tro duce the follo wing notation. Let ( X , d ) b e a ﬁnite metric space. If card( X ) < 2, let µ ( X ) := 0 . Otherwise let µ ( X ) := max  min { d ( x, y ) : y ∈ X , y 6 = x }  > 0 . Then we ha ve 5.10. P rop osition. L et ( X , d ) b e a non e mpty metric sp ac e. F or every f ∈ Z N [ X ] : (A) p d ( f ) > µ (supp f ) ; in p articular, p d is a value, (B) if N = 2 and f 6 = 0 , (5-6) p d ( f ) = min { k X j =1 d ( x j , y j ) : x 1 , y 1 , . . . , x k , y k are a ll diﬀeren t and { x 1 , y 1 , . . . , x k , y k } = supp( f ) } . Pr o o f. First observ e that—thanks t o the triangle inequalit y—in the form ula (5-5) we ma y consider only such systems x 1 , y 1 , . . . , x n , y n ∈ X that (5-7) { x 1 , . . . , x n } ∩ { y 1 , . . . , y n } = ∅ and—for the same reason—if N = 2, w e ma y a lso restrict to systems x 1 , . . . , x n and y 1 , . . . , y n in whic h elemen ts are diﬀeren t. This pro v es (B). Now assume that N > 3, that (5-8) f = n X j =1 ( b x j − b y j ) and (5-7 ) is fulﬁlled. Let F = { x 1 , y 1 , . . . , x n , y n } . F or a, b ∈ F w e write a ∼ b pro vided there is j ∈ { 1 , . . . , n } with { a, b } = { x j , y j } ; and a ≡ b iﬀ either a = b or there are c 0 , c 1 , . . . , c k ∈ F for whic h c 0 = a , c k = b and c j ∼ c j − 1 for j = 1 , . . . , k . It is clear that ‘ ≡ ’ is an equiv alence on F suc h that (5-9) { j : x j ≡ a } = { j : y j ≡ a } for eac h a ∈ A. 36 P . NI EMIEC W e infer from (5- 7) and (5- 8) that supp( f ) ⊂ F and for eac h a ∈ A , f ( a ) = card( { j ∈ { 1 , . . . , n } : x j = a } ) e o r f ( a ) = − card( { j ∈ { 1 , . . . , n } : y j = a } ) e , and hence (5-10) ( card( { j : a ∈ { x j , y j }} ) 6≡ 0 mo d N if a ∈ supp( f ) , card( { j : a ∈ { x j , y j }} ) ≡ 0 mo d N otherwise . Fix a ∈ A . It follo ws from (5-9) and (5-1 0) that there is b ∈ supp( g ) \ { a } suc h that b ≡ a . So, there are c 0 , . . . , c k ∈ A for whic h c 0 = a , c k = b and c j ∼ c j − 1 for j = 1 , . . . , k . P assing in t o a suitable subset of { 0 , . . . , k } w e may assume that all c j ’s are diﬀeren t. This means tha t there are distinct indices ν 1 , . . . , ν k suc h that { c j , c j − 1 } = { x ν j , y ν j } . But t hen P n j =1 d ( x j , y j ) > P k s =1 d ( c s , c s − 1 ) > d ( a, b ). So, the a ssertion follo ws from the arbitrarit y of a ∈ supp( f ).  5.11. Corollary . L et ( X , d ) b e a no nempty metric sp ac e a n d a ∈ X . The function ( X , d ) ∋ x 7→ b x − b a ∈ ( Z N [ X ] , p d ) is isome tric. If we replace in the ab ov e r esult p d b y p d ∧ r , the a ssertion of Corol- lary 5.11 w ell remain true. It is also clear that Z N [ X ] is separable pro vided so is X . So, an application of Corollary 4.6 yields 5.12. Theorem. Every sep ar able metric sp ac e of diameter no gr e ater than r admits a n isometric e m b e dding into the me tric sp ac e G r ( N ) . 5.13. Example. If X is a nonempt y set, H is a group of exp onen t N , ev ery function u : X → H induces a unique (w ell deﬁned) group homomorphism e u : Z N [ X ] → H such that e u ( b x − b y ) = u ( x ) − u ( y ) for an y x, y ∈ X . What is more, if d is a metric on X , q is a v alue on H and u : ( X , d ) → ( H , q ) is Lipsc hitz, it easily follows from the form ula for p d that e u : ( Z N [ X ] , p d ) → ( H, q ) satisﬁes the Lipsc hitz conditio n with the same constan t as u . Similarly , ev ery function v : X → Y b e- t wee n t w o sets in duces a gr oup homomorphism b v : Z N [ X ] → Z N [ Y ] and if v : ( X , d ) → ( Y ,  ) is Lipsc hitz, then b v : ( Z N [ X ] , p d ) → ( Z N [ Y ] , p  ) satisﬁes the Lipsc hitz condition with the same constan t as v . In particular, if A is a subset of ( X , d ), the inclusion map of A in to X induces a nonexpansiv e gro up homo morphism of Z N [ A ] into Z N [ X ]. In case o f N = 2, the la tter group homomorphism is isometric, whic h easily follo ws from Prop osition 5.10 (similar res ult holds true for the Lipsc hitz-free Banac h spaces generated b y metric spaces—see [29]). It turns out that for N > 2 this homomorphism may not b e isometric, whic h is rather strange. L et us giv e an example based on Example 5.4 . Let X = { 0 , 1 , 2 , . . . , N } a nd A = X \ { 0 } . W e equip X with a metric d suc h that for distinct j, k ∈ A , d ( j, k ) = 1 and d ( 0 , j ) = max(1 / 2 , 1 − 2 / N ). F or clarity , let  = d A × A . Put f = P a ∈ A b a ∈ Z N [ A ]. It f ollo ws UNIVERS AL V ALU ED ABELIA N GR OUPS 37 from the argumen t in Example 5.4 that p  ( f ) = N − 1. How ev er, in Z N [ X ], f = P a ∈ A ( b a − b 0) and th us p d ( f ) 6 N · max (1 / 2 , 1 − 2 / N ) < N − 1. T aking into accoun t The orem 2 .17, Theorem 5.1, Theorem 5.5 and the ab o v e example, w e see that t he case of N > 2 is ‘singular’. 6. Pseudovector groups Pseudo v ector gr oups w ere in tro duced in [18]. Here w e shall general- ize the results of [18] on pseudov ector structures on the (un b ounded) Bo olean Urysohn group, mainly in order to prov e that all the groups G r ( N )’s are homeomorphic to the Hilb ert space. How ev er, some of results of this part ma y b e seen in teresting and can pla y imp ortant role in theory of t op ological pseudo v ector groups. All t erms, b eside the notion of a ∇ -norm, are rep eated from the in tro ductory work [18]. As b efo re, we restrict our considerations only to Ab elian groups. W e b egin with 6.1. Deﬁnition. Let F b e a subﬁeld of R . A triple ( G, + , ∗ ) is said to b e a pseudove ctor (A b elian) gr oup o v er F (brieﬂy , an F -PV group, or a PV group pro vided F = R ) iﬀ ( G, +) is an Ab elian group and ∗ : F + × G → G is an action suc h that for an y s, t ∈ F + and x ∈ G , 0 ∗ x = 0, 1 ∗ x = x , ( st ) ∗ x = s ∗ ( t ∗ x ) and the function G ∋ y 7→ t ∗ y ∈ G is a group homomorphism. A top olo gic al pseudove ctor gr oup ov er F is a n F -PV group ( G, + , ∗ ) equipped with a top olo gy τ such that ( G, + , τ ) is a top olog ical group and t he action ‘ ∗ ’ is con tinuous. A norm on an F -PV gro up ( G, + , ∗ ) is a v a lue k · k : G → R + suc h that k t ∗ x k = t k x k fo r each t ∈ F + and x ∈ G . The norm k · k is top olo gic al iﬀ the action ‘ ∗ ’ is con tin uous with resp ect to k · k . In that case w e speak of a norme d top olo gic al pseudove ctor gr oup o ver F (for short, a NTPV group ov er F ). If G is an F -PV group, K is a subﬁeld of F and A an y subset of G , b y lin K A w e denote the K -PV subgroup of G generared by A , that is, lin K A = { P n j =1 t j ∗ a j : n > 1 , t 1 , . . . , t n ∈ K + , a 1 , . . . , a n ∈ A ∪ ( − A ) ∪ { 0 }} . If K = R , w e write lin A instead of lin R A . A group homomorphism u : G → H betw een t w o F -PV groups is line ar if u ( t ∗ x ) = t ∗ u ( x ) fo r ev ery t ∈ F + and x ∈ G . Our aim is to show that eac h of the gro ups G r ( N )’s may b e endo wed with a ‘normed-like’ top ological pseudo v ector structure. Since ev ery norm o n a non trivial group is un b ounded, none of the groups G 1 ( N ) is isometrically gro up isomorphic to a NTPV group. Th us, w e ha v e to generalize the notion of a norm. 6.2. Deﬁnition. A function κ : R + → R + is said t o b e normi n g iﬀ κ satisﬁes the follow ing conditions: 38 P . NI EMIEC (NF1) κ (1 ) = 1 and κ ( x ) > x for eac h x > 0 , (NF2) κ ( xy ) 6 κ ( x ) κ ( y ) for a n y x, y ∈ R + , (NF3) κ is monotone increasing, i.e. κ ( x ) 6 κ ( y ) provid ed 0 6 x 6 y , (NF4) κ is Lipsc hitz, that is, the re is a constan t L ′ > 0 such that | κ ( x ) − κ ( y ) | 6 L ′ | x − y | for x, y > 0. (Notice that it is not a ssumed that κ (0) = 0.) A (semi)v alue k · k o n an F -PV group ( G, + , ∗ ) is said to b e a κ - (semi)norm iﬀ k t ∗ x k 6 κ ( t ) k x k for eac h t ∈ F + and x ∈ G . When k · k is a κ -norm on an F -PV group ( G, + , ∗ ), the quadruple ( G, + , ∗ , k · k ) is said to b e a κ -n orme d F -PV gr oup . 6.3. R emark. In the whole pap er, we use (NF3) only t w o times: in (P6) (whic h ﬁnds no a pplication in this pap er) and in the pro of of Lemma 6.13 b elo w, whic h ma y b e impro ve d so that (NF3 ) will not b e applied. The r eason for adding (NF3) to the axioms of a norming function is just a matter of our p ersonal ‘taste’. The most imp ortant examples on norming functions ar e id : R + → R + and ∇ = id ∨ 1. W e leav e these as simple exercises that ∇ is indeed norming a nd that a v alue is an id-norm iﬀ it is a norm. W e call a ∇ -norm also a subnorm , and subno rme d means ∇ - normed. In f act we need o nly these t wo functions. How ev er, no additional w ork is needed for arbitrary norming functions and it seems to us instructive to p oint out whic h pro p erties of id and ∇ are relev ant in o ur considerations. The reader should ch ec k with no diﬃculties the fo llo wing prop erties of ev ery norming function κ . Belo w we assume that L ′ is as in (NF4). (P1) If κ (0) = 0, then id 6 κ 6 L ′ id. (P2) If κ (0) 6 = 0, then ∇ 6 κ 6 ( L ′ + 1) ∇ . (P3) Ev ery norm is a κ -norm. (P4) If κ (0) 6 = 0, ev ery ∇ -norm is a κ -norm. (P5) if κ (0) 6 = 0 and k · k is a κ -norm, then k · k ∧ 1, k·k 1+ k·k and k · k α with 0 < α < 1 are κ -norms as w ell. (P6) Comp osition o f tw o norming functions is a norming function. As t he follow ing result shows , ∇ -norms app ear m uc h more often than norms. 6.4. P r op osition. Every top olo gic al ps e udo v e ctor gr oup metrizable a s a top olo gic al sp ac e admits a ∇ -norm in d ucing its top olo gy. Pr o o f. Let ( G, + , ∗ ) be a top ological pseudo v ector group. By a w ell- kno wn t heorem (see e.g. [2, Theorem 8.8]), there is a v alue p on G , b ounded by 1, which induces the top ology of G . Now deﬁne k · k : G → R + b y k x k = sup { p ( α ∗ x ) α ∨ 1 : α > 0 } . W e lea ve this as a n exercise that k · k is a subnorm equiv alen t to p (use the b oundednes s of p and the contin uit y o f ‘ ∗ ’).  UNIVERS AL V ALU ED ABELIA N GR OUPS 39 Note that the ab ov e result may b e applied to an y metriz able top o- logical ve ctor space. F ro m no w to the end of the section κ is a ﬁxed norming function and L > 1 is suc h a constant that for all x, y > 0, (6-1) ( | κ ( x ) − κ ( y ) | 6 L | x − y | , κ ( x ) 6 L ∇ ( x ) (cf. (NF4), (P1) and (P2)). No te also that it follo ws f rom (P1) that κ ( x ) 6 Lx for x > 0 provided κ (0) = 0. Ev erywhere b elo w F is a subﬁeld of R and ( G, + , ∗ , k · k G ) is a κ - normed pseudov ector group. A t this momen t, w e do not a ssume that the action ‘ ∗ ’ is con tin uous. As in [18], w e call an elemen t a ∈ G c ontinuous (resp ectiv ely Lips- chitz ) provided the function F + ∋ t 7→ t ∗ a ∈ G is con tin uous (Lips- c hitz). The s et of all contin uous (Lipsc hitz) ele men ts of G is denoted by C F ( G ) ( L F ( G )) and in case F = R we write C ( G ) ( L ( G ) ). G is said to be Lipschitz if L F ( G ) = G . Similarly , a κ -seminorm k · k 0 on G is Lipschitz iﬀ for ev ery a ∈ G there is C a ∈ R + suc h that k t ∗ a − s ∗ a k 0 6 C a | t − s | for any s, t ∈ R + . The pro ofs of the next t w o results are omitted. 6.5. Pr op osition. (A) F or a ∈ G , a ∈ C F ( G ) iﬀ lim t → 0 t ∈ F + k t ∗ a k G = 0 and lim t → 1 − t ∈ F + k a − t ∗ a k G = 0 . (B) F or a ∈ G , a ∈ L F ( G ) iﬀ the function F ∩ [0 , 1] ∋ t 7→ t ∗ a ∈ G is Lipschitz. (C) T he set C F ( G ) is a clos e d p s e udove ctor sub gr oup (over F ) of G and C F ( G ) is a top olo gic al ps eudove ctor g r oup. The set L F ( G ) is a pseudove ctor sub gr oup (over F ) of C F ( G ) . (D) F or every a ∈ C F ( G ) the map F + ∋ t 7→ t ∗ a ∈ G is unifo rmly c ontinuous (as a map b etwe en metric sp ac es) on every b ounde d subset of F + . 6.6. Prop osition. Supp ose C F ( G ) = G . L et ( ¯ G, + , k · k ) b e the c om- pletion of ( G, + , k · k G ) . Ther e is an action ¯ ∗ : R + × ¯ G → ¯ G such that ( ¯ G, + , ¯ ∗ , k · k ) is a κ -n o rme d top olo gic al PV gr oup and t ∗ x = t ¯ ∗ x for t ∈ F + and x ∈ G . Mor e over, L F ( G ) = L ( ¯ G ) ∩ G . The next result will b e used sev eral times by us. 6.7. Lemma. If ther e is a ∈ G \ { 0 } such that M := sup {k t ∗ a k G : t ∈ F + } < ∞ , then κ (0) 6 = 0 . Pr o o f. F or p ositiv e t, s ∈ F w e hav e k t ∗ x k G 6 κ ( s ) k t s ∗ x k G 6 M κ ( s ) and t h us M 6 M κ ( s ). So, κ ( s ) > 1 and the assertion follows .  40 P . NI EMIEC 6.8. Lemma. F or e very a ∈ C F ( G ) and δ > 0 ther e exists a c onstant M = M ( a, δ ) ∈ R + such that fo r any s, t ∈ R + , (6-2) k s ∗ a − t ∗ a k G 6 M | s − t | + δ κ ( s ∨ t ) . Pr o o f. Let M = L sup  k s ∗ a − t ∗ a k G − δ | s − t | : s, t ∈ F ∩ [0 , 1 ] , s 6 = t  . W e infer from p oin t (D) of Prop osition 6.5 that M < ∞ . W e s hall sho w that M satisﬁes (6-2). First assume that κ (0) = 0 . W e kno w that then κ ( t ) 6 Lt for ev ery t ∈ R + . So, fo r t > s > 0 w e hav e k s ∗ a − t ∗ a k G 6 κ ( t ) k ( s/t ) ∗ a − a k G 6 κ ( t )[( M /L )(1 − s/t ) + δ ] 6 M ( t − s ) + δ κ ( t ). When κ (0) 6 = 0, then κ ( t ) > 1 for ev ery t ∈ R + and hence (6- 2) is fulﬁlled when t ∨ s 6 1, b y the deﬁnition of M . Finally , f or t > s > 0 and t > 1, κ ( t ) 6 Lt (see (6-1)) and therefore we ma y r ep eat the estimations from the previous parag raph to get the assertion.  6.9. Lemma. L et ( D j , + , ∗ , k · k j ) ∈ G r ( N ) b e Lipschitz κ -norme d PV gr oups over F ( j = 0 , 1 , 2 ) and ϕ j : D 0 → D j ( j = 1 , 2 ) b e isometric line ar homomorphi s m s. Then ther e is a Lipschitz PV gr oup ( D , + , ∗ , k · k ) ∈ G r ( N ) over F and is o metric line ar ho m omorphisms ψ j : D j → D ( j = 1 , 2 ) such that ψ 2 ◦ ϕ 2 = ψ 1 ◦ ϕ 1 . Pr o o f. If bot h D 1 and D 2 are trivial, it suﬃces to tak e D trivial a s w ell. If D 1 or D 2 is nontrivial, w e ma y a pply Lemma 6 .7 from whic h it follows that κ (0 ) 6 = 0 if r < ∞ . Th us, in the la tter case w e ma y in volv e (P5). These commen ts ensure us tha t w e may rep eat the pro o f of Lemma 2.19. The details are skipp ed.  F or ev ery v alued g roup ( H , + , q ) let L 0 [ H ] consist of all functions u : R + → H for whic h there are num b ers 0 = t 0 < t 1 < . . . < t n < ∞ suc h that u is constant o n each of the in terv als [ t j − 1 , t j ) ( j = 1 , . . . , n ), sa y u ([ t j − 1 , t j )) = { u j } , and u ( t ) = 0 fo r t > t n . W e s hall abbreviate this by writing u = n X j =1 u j χ [ t j − 1 ,t j ) . L 0 [ H ] is a Lipschitz normed PV group when it is equipp ed with the p oin t wise addition, the action giv en b y ( t ∗ u )( s ) = u ( s/t ) f or t > 0, s > 0 and u ∈ L 0 [ H ] (of course, 0 ∗ u ≡ 0) and the norm k u k q = R ∞ 0 q ( u ( t )) d t (cf. [1 8]). What is mor e, fo r ev ery nonzero u ∈ L 0 [ H ] there are unique n > 1, real num b ers 0 < t 1 < . . . < t n and no nzero elemen ts h 1 , . . . , h n ∈ H suc h tha t u = P n j =1 t j ∗ b h j where f or h ∈ H , b h ( s ) = h for s ∈ [0 , 1) and b h ( s ) = 0 for s > 1. Note that the function H ∋ h 7→ b h ∈ L 0 [ H ] is an isometric group homomorphism and ev ery group homomorphism ψ : H → K o f H into a group K induce s a unique linear ho momorphism b ψ : L 0 [ H ] → K determined b y condition UNIVERS AL V ALU ED ABELIA N GR OUPS 41 b ψ ( b h ) = ψ ( h ) for h ∈ H . Th us, ( L 0 [ H ] , + , ∗ ) ma y b e called the fr e e PV gr oup gener ate d by H . A straightforw ard v eriﬁcation show s that ( L 0 [ H ] , + , k · k q ) is of class O 0 or of exp onen t N provid ed so is ( H , + , q ). F ro m now on, ( G, + , ∗ , k · k G ) is a κ -normed PV group. As in the previous sections, we ﬁx r ∈ { 1 , ∞} and N ∈ Z + \ { 1 } . Additionally , w e assume the following elemen tary prop ert y: (AX) G 6 = { 0 } or κ (0) 6 = 0 or r = ∞ . (This is b ecause of Lemma 6.7. Precisely , w e w an t t o enlarge κ -normed PV groups o f class G r ( N ). When G is trivial and r is ﬁnite, this is imp ossible unless κ (0 ) 6 = 0.) Our aim is to prov e counterparts of sev eral results from Section 1 on enlarging gro ups. As we will see, in case o f PV groups this is m uc h more complicated. 6.10. Theorem. L et ( H , + , q ) b e a ﬁnite value d gr oup, K its sub gr oup and k · k 0 a Lipschitz κ -sem i n orm on L 0 [ K ] s uch that k b h k 0 = q ( h ) for h ∈ K . Then ther e is a Lipschitz κ -seminorm o n L 0 [ H ] such that k f k = k f k 0 for f ∈ L 0 [ K ] and k b h k = q ( h ) for h ∈ H . Pr o o f. W e assume tha t H 6 = K . Since K is ﬁnite a nd k · k 0 is Lipsc hitz, there is a constan t λ > 1 suc h that k s ∗ b h − t ∗ b h k 0 6 λ | t − s | for ev ery h ∈ H and s, t ∈ R + . Let M = max q ( H ), µ = min { q ( h ) : h ∈ H \ { 0 }} and A = ( λ + M L ) /µ . F o r f ∈ L 0 [ H ] put (6-3) k f k = inf { A k f − g − n X j =1 t j ∗ b h j k q + n X j =1 κ ( t j ) q ( h j ) + k g k 0 : n > 1 , t 1 , . . . , t n > 0 , h 1 , . . . , h n ∈ H, g ∈ L 0 [ K ] } . It is clear that k · k is a κ -seminorm on L 0 [ H ] ( b y (NF2) and (P3)). What is more, k · k 6 A k · k q whic h yields that k · k is Lipsc hitz. Th us, w e only need to sho w that k f k = k f k 0 and k b h k = q ( h ) for f ∈ L 0 [ K ] and h ∈ H . The inequalities ‘ 6 ’ are immediate (since κ (1) = 1). T o pro ve the in ve rse inequalities, ﬁrst of all note that (6-4) k f k = inf n k m X k =1 s k ∗ b g k k 0 + A m X j =1 ( s j − s j − 1 ) q ( m X k = j ε k ) + m X k =1 κ ( s k ) q ( f k − f k +1 + g k + ε k ) : m > 1 , 0 = s 0 < . . . < s m , ε 1 , . . . , ε m , f 1 , . . . , f m ∈ H, f m +1 = 0 , g 1 , . . . , g m ∈ K, f = m X k =1 s k ∗ ( b f k − d f k +1 ) o . 42 P . NI EMIEC Indeed, under the no tation of (6-4), it suﬃces to substitute t k = s k , g = − m X k =1 s k ∗ b g k and h k = f k − f k +1 + g k + ε k to obtain the expression as is (6- 3) (observ e tha t f = P m k =1 s k ∗ ( b f k − d f k +1 ) iﬀ f = P m k =1 f k χ [ s k − 1 ,s k ) ; a nd k f − g − P m j =1 s j ∗ b h j k q = P m j =1 ( s j − s j − 1 ) q ( P m k = j ε k )). Con verse ly , in (6-3) w e may ass ume that t 1 , . . . , t n are p ositiv e and diﬀeren t and h j 6 = 0. Consequen tly , w e ma y assume that 0 < t 1 < . . . < t n < ∞ . Then tak e 0 = s 0 < . . . < s m < ∞ suc h that f = P m k =1 f k χ [ s k − 1 ,s k ) , g = P m k =1 v k χ [ s k − 1 ,s k ) ( v k ∈ K ) a nd P n j =1 t j ∗ b h j = P m k =1 u k χ [ s k − 1 ,s k ) . Put f m +1 = g m +1 = u m +1 = 0, g k = v k +1 − v k ∈ K and ε k = u k − u k +1 − ( f k − f k +1 ) − g k and c hec k that P n j =1 κ ( t j ) q ( h j ) = P m k =1 κ ( s k ) q ( u k − u k +1 ), k f − g − P n j =1 t j ∗ b h j k q = P m j =1 ( s j − s j − 1 ) q ( f j − v j − u j ) and f j − v j − u j = − P m k = j ε k . This pro v es the equiv alence of (6-3) and (6-4) (the details are left for the reader). So, if f ∈ L 0 [ K ], the inequalit y ‘ k f k > k f k 0 ’ is equiv alent to m X k =1 κ ( s k ) q ( f k − f k +1 + g k + ε k ) + A m X j =1 ( s j − s j − 1 ) q ( m X k = j ε k ) > > k m X k =1 s k ∗ ( b f k − d f k +1 ) k 0 − k m X k =1 s k ∗ b g k k 0 (under the notation of (6-4); in that case f 1 , . . . , f m ∈ K ). So, sub- stituting h k = f k − f k +1 + g k ∈ K and ∆ s j = s j − s j − 1 , the latter inequalit y follows from (6-5) A m X j =1 (∆ s j ) q ( m X k = j ε k ) + m X k =1 κ ( s k ) q ( h k + ε k ) > k m X k =1 s k ∗ b h k k 0 . W e shall prov e (6-5) by induction on m . When m = 1, w e hav e to show that Asq ( ε ) + κ ( s ) q ( h + ε ) > k s ∗ b h k 0 . If ε = 0, this follow s from the fact that k · k 0 is a κ -seminorm. And if ε 6 = 0, w e hav e Asq ( ε ) + κ ( s ) q ( h + ε ) > Aµs > λ | s − 0 | > k s ∗ b h k 0 . No w assume that (6- 5) is satisﬁed for m − 1. W e distinguish b et w een three cases. Firstly , assume that P m k = j ε k 6 = 0 for eac h j ∈ { 1 , . . . , m } . In that case the left- hand side expres sion of (6-5) is no less than Aµ P m j =1 ∆ s j UNIVERS AL V ALU ED ABELIA N GR OUPS 43 and Aµ m X j =1 ∆ s j > λ m X j =1 | s j − s j − 1 | > > k m X j =1 [ s j ∗ ( m X k = j b h k ) − s j − 1 ∗ ( m X k = j b h k )] k 0 = k m X k =1 s k ∗ b h k k 0 , whic h giv es (6-5). Secondly , if ε m = 0, the a ssertion follows from (6-5) for ‘ m − 1’. Thirdly , assume t hat ε m 6 = 0 and there is z ∈ { 1 , . . . , m } suc h that P m k = z ε k = 0. W e may assume that z is the largest natural n umber with this prop erty . This yields that z < m and (6-6) m X k = z ε k = 0 6 = m X k = z +1 ε k . Let us deﬁne (6-7)      s ′ 0 , . . . , s ′ m − 1 ≡ s 0 , . . . , s z − 1 , s z +1 , . . . , s m , ε ′ 1 , . . . , ε ′ m − 1 ≡ ε 1 , . . . , ε z − 1 , ε z + ε z +1 , ε z +2 , . . . , ε m , h ′ 1 , . . . , h ′ m − 1 ≡ h 1 , . . . , h z − 1 , h z + h z +1 , h z +2 , . . . , h m . W e infer fr om induction h yp othesis that (6-8) A m − 1 X j =1 (∆ s ′ j ) q ( m − 1 X k = j ε ′ k ) + m − 1 X k =1 κ ( s ′ k ) q ( h ′ k + ε ′ k ) > k m − 1 X k =1 s ′ k ∗ b h ′ k k 0 . It is easily ve riﬁed, thanks to (6- 6) and (6-1), that (6-9) A m X j =1 (∆ s j ) q ( m X k = j ε k ) = = A m − 1 X j =1 (∆ s ′ j ) q ( m − 1 X k = j ε ′ k ) + A (∆ s z +1 ) q ( m X k = z +1 ε k ) and (6-10) A (∆ s z +1 ) q ( m X k = z +1 ε k ) > ( λ + M L )(∆ s z +1 ) > > k s z +1 ∗ b h z − s z ∗ b h z k 0 + q ( h z + ε z )[ κ ( s z +1 ) − κ ( s z )] . F urther, (6-11) m X k =1 κ ( s k ) q ( h k + ε k ) = m − 1 X k =1 κ ( s ′ k ) q ( h ′ k + ε ′ k ) + [ κ ( s z ) q ( h z + ε z ) + κ ( s z +1 ) q ( h z +1 + ε z +1 ) − κ ( s ′ z ) q ( h ′ z + ε ′ z )] 44 P . NI EMIEC and (6-12) [ q ( h z + ε z ) + q ( h z +1 + ε z +1 )] κ ( s z +1 ) − κ ( s ′ z ) q ( h ′ z + ε ′ z ) > 0 . So, (6-9), (6-10), (6-11) a nd (6-12 ) yield A m X j =1 (∆ s j ) q ( m X k = j ε k ) + m X k =1 κ ( s k ) q ( h k + ε k ) > A m − 1 X j =1 (∆ s ′ j ) q ( m − 1 X k = j ε ′ k ) + m − 1 X k =1 κ ( s ′ k ) q ( h ′ k + ε ′ k ) + k s z +1 ∗ b h z − s z ∗ b h z k 0 and hence (6-5 ) follo ws from (6-8). W e now pass to the pro o f that k b x k > q ( x ) for x ∈ H \ { 0 } . Under the notation of (6-4), when f = b x , there is l ∈ { 1 , . . . , m } for whic h s l = 1 and then f j = x fo r j = 1 , . . . , l and f j = 0 otherwise. Hence, (6-4) has the fo rm (r ecall that κ (1) = 1) : A m X j =1 (∆ s j ) q ( m X k = j ε k ) + X k 6 = l κ ( s k ) q ( g k + ε k ) + k m X k =1 s k ∗ b g k k 0 + q ( x + g l + ε l ) > q ( x ) , whic h is equiv alen t, after replacing g j b y h j , t o (6-13) A m X j =1 (∆ s j ) q ( m X k = j ε k ) + X k 6 = l κ ( s k ) q ( h k + ε k ) + k m X k =1 s k ∗ b h k k 0 > q ( h l + ε l ) (where h j ∈ K ). As b efore, w e shall prov e (6-13) by induction on m . When m = 1 , also l = 1 and hence we need to sho w that Aq ( ε ) + k b h k 0 > q ( h + ε ) whic h easily follo ws since A > 1 and k b h k 0 = q ( h ) for h ∈ K . Supp ose (6-13) is f ulﬁlled for m − 1. W e distinguish b etw een sev eral cases. When P m k = j ε k 6 = 0 for eac h j , the left-hand side expres sion of (6-13) is no less than Aµs m > M s l > q ( h l + ε l ). If ε m = 0 and m 6 = l , ( 6-13) follow s from induction h yp othesis. No w assume that ε l = 0. Let s ′ j , h ′ j , ε ′ j ( j = 1 , . . . , m − 1) b e systems obtained f rom s j , h j , ε j b y erasing t he l -th mem b ers. Making use of UNIVERS AL V ALU ED ABELIA N GR OUPS 45 (6-5), w e see that A m X j =1 (∆ s j ) q ( m X k = j ε k ) + X k 6 = l κ ( s k ) q ( h k + ε k ) + k m X k =1 s k ∗ b h k k 0 = A m − 1 X j =1 (∆ s ′ j ) q ( m − 1 X k = j ε ′ k ) + m − 1 X k =1 κ ( s ′ k ) q ( h ′ k + ε ′ k ) + k m − 1 X k =1 s ′ k ∗ b h ′ k + s l ∗ b h l k 0 > k m − 1 X k =1 s ′ k ∗ b h ′ k k 0 + k m − 1 X k =1 s ′ k ∗ b h ′ k + b h l k 0 > k b h l k 0 = q ( h l + ε l ) . So, w e ma y no w assume that ε m 6 = 0 and there is z suc h that P m k = z ε k = 0. Again, let z b e the largest natural num b er with this prop erty . W e conclude from this that z < m and (6-6). W e shall separately consider the cases when z / ∈ { l − 1 , l } , z = l − 1 or z = l . When z 6 = l − 1 , l , deﬁne systems s ′ j , ε ′ j and h ′ j as in (6- 7), and let l ′ ∈ { 1 , . . . , m − 1 } corresp onds to l . Note that ( 6-9), (6-10) and (6-12) are f ulﬁlled. Moreov er, instead of (6-1 1) w e hav e X k 6 = l κ ( s k ) q ( h k + ε k ) = X k 6 = z ,z +1 ,l κ ( s k ) q ( h k + ε k ) + [ κ ( s z ) q ( h z + ε z ) + κ ( s z +1 ) q ( h z +1 + ε z +1 )] = X k 1), pro ceed in the same wa y . Here l ′ = l − 1 and instead of ( 6-11) one has X k 6 = l κ ( s k ) q ( h k + ε k ) = X k 6 = l − 1 ,l κ ( s k ) q ( h k + ε k ) + κ ( s l − 1 ) q ( h l − 1 + ε l − 1 ) = X k q ( h l + ε l ) whic h is fulﬁlled b ecause h ′ l ′ = h l − 1 + h l and ε ′ l ′ = ε l − 1 + ε l . 46 P . NI EMIEC It remains to c hec k what happ ens if z = l . F ro m the maximalit y of z w e infer that P k = j ε k 6 = 0 for j > l and t h us (6-14) A m X j = l +1 (∆ s j ) q ( m X k = j ε k ) > Aµ m X j = l +1 ( s j − s j − 1 ) > > λ m X j = l +1 | s j − s j − 1 | > k m X j = l +1 [ s j ∗ ( m X k = j b h k ) − s j − 1 ∗ ( m X k = j b h k )] k 0 = = k m X k = l +1 [ s k ∗ b h k − s l ∗ b h k ] k 0 = k m X k = l +1 s k ∗ b h k − m X k = l +1 b h k k 0 . F urther, it fo llo ws from (6-5) that A l − 1 X j =1 (∆ s j ) q ( l − 1 X k = j ε k ) + l − 1 X k =1 κ ( s k ) q ( h k + ε k ) > k l − 1 X k =1 s k ∗ b h k k 0 . This, combined with (6-14), giv es A m X j =1 (∆ s j ) q ( m X k = j ε k ) + X k 6 = l κ ( s k ) q ( h k + ε k ) + k m X k =1 s k ∗ b h k k 0 = = A l − 1 X j =1 (∆ s j ) q ( l − 1 X k = j ε k ) + l − 1 X k =1 κ ( s k ) q ( h k + ε k ) + m X k = l +1 κ ( s k ) q ( h k + ε k ) + A m X j = l +1 (∆ s j ) q ( m X k = j ε k ) + k m X k =1 s k ∗ b h k k 0 > k l − 1 X k =1 s k ∗ b h k k 0 + m X k = l +1 s k q ( h k + ε k ) + k m X k = l +1 s k ∗ b h k − m X k = l +1 b h k k 0 + k m X k =1 s k ∗ b h k k 0 > > m X k = l +1 q ( h k + ε k ) + k s l ∗ b h l + m X k = l +1 b h k k 0 > > q ( m X k = l +1 h k + m X k = l +1 ε k ) + q ( m X k = l h k ) > q ( h l − m X k = l +1 ε k ) but − P m k = l +1 ε k = ε l (b y (6-6)), whic h ﬁnally ﬁnishes the pro of.  As a corollary o f the ab ov e result, w e obtain 6.11. Theorem. L et ( H , + , q ) ∈ G r ( N ) b e a ﬁ nite value d gr oup, K b e a sub gr oup of H , ( G, + , ∗ , k · k G ) b e a Lipschitz κ -norme d pseudove ctor gr oup such that ( G, + , k · k G ) ∈ G r ( N ) and (AX) is fulﬁl le d. A nd let ϕ : K → G b e an isometric gr oup homomo rp hism. Ther e is a Lipschitz κ -norme d PV gr oup ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) and an isome tric gr oup hom omorphism ψ : H → ¯ G such that ( ¯ G, + , k · k ) ∈ G r ( N ) and ψ   K = ϕ . UNIVERS AL V ALU ED ABELIA N GR OUPS 47 Pr o o f. Let b ϕ : L 0 [ K ] → G b e the linear homomorphism induced b y ϕ . D eﬁne k · k 0 : L 0 [ K ] → R + b y k f k 0 = k b ϕ ( f ) k G . Then k · k 0 is a Lipsc hitz κ -seminorm on L 0 [ K ] suc h t hat k b h k 0 = q ( h ) for h ∈ K (since ϕ is isometric). So, according to Theorem 6.10, there is a Lipsc hitz κ -seminorm k · k H on L 0 [ H ] whic h extends k · k 0 and satisﬁes k b h k H = q ( h ) for h ∈ H . Now let e H = L 0 [ H ] / { f ∈ L 0 [ H ] : k f k H = 0 } b e the PV group equipp ed with the Lipsc hitz κ -norm induced b y k · k H , let π : L 0 [ H ] → e H b e the quotien t linear homomorphism and e K = π ( L 0 [ K ]). Observ e that there is an isometric linear homomorphism e ϕ : e K → G suc h that e ϕ ◦ π   L 0 [ K ] = b ϕ . T o this end, apply Lemma 6.9 to e ϕ and the inclusion map o f e K in to e H in order to obtain ¯ G . Finally , in volv e Lemma 6 .7 and (P5) if needed.  6.12. Lemma. Supp ose a ∈ C ( G ) \ { 0 } is of ﬁnite r ank . F or e ach ε > 0 ther e is a PV gr oup ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) and b ∈ L ( ¯ G ) such that rank( b ) = rank ( a ) and k a − b k 6 ε . Pr o o f. Let H = h a i . The homomorphism Z ∋ k 7→ k a ∈ H ⊂ G in- duces linear homomorphisms ψ G : L 0 [ Z ] → G and ψ H : L 0 [ Z ] → L 0 [ H ] suc h that ψ G ( b 1) = a and ψ H ( b 1) = b a . Let δ be the discrete v alue on H and N > 2 b e the rank of a . W e equip L 0 [ Z ] with the κ - norm k · k κ deﬁned as follo ws. When f ∈ L 0 [ Z ] \ { 0 } , there are unique n > 1, 0 < t 1 < . . . < t n < ∞ and k 1 , . . . , k n ∈ Z \ { 0 } suc h that f = P n j =1 t j ∗ b k j . W e put k f k κ = P n j =1 κ ( t j ). F urt her, let M = M ( a, ε/ N ) b e as in L emma 6.8 . Put A = N M , ¯ G = G × L 0 [ H ] and deﬁne a κ -seminorm k · k on ¯ G by k ( g , f ) k = inf {k g − ψ G ( u ) k G + ε k u k κ + A k ψ H ( u ) − f k δ : u ∈ L 0 [ Z ] } . It is easily seen that k · k is a κ -seminorm on ¯ G suc h that k ( g , f ) k > 0 for f 6 = 0 and k ( g , 0) k 6 k g k G for ev ery g ∈ G ; k ( a, b a ) k 6 ε a nd k (0 , f ) k 6 A k f k δ for f ∈ L 0 [ H ]. The latter will imply that (0 , b a ) ∈ L ( ¯ G ). So, if w e show tha t k ( g , 0) k > k g k , k · k will b e a κ - norm and the pro of will b e completed. The latter is equiv alen t t o (6-15) ε k u k κ + A k ψ H ( u ) k δ > k ψ G ( u ) k G where u ∈ L 0 [ Z ]. W e may assume that u 6 = 0. W riting u = P m k =1 s k ∗ b l k with 0 = s 0 < . . . < s m and l 1 , . . . , l m ∈ Z \ { 0 } , and putting ν j = P m k = j l k w e obtain k u k κ = P m k =1 κ ( s k ), u = P m k =1 ( s k ∗ b ν k − s k − 1 ∗ b ν k ), ψ H ( u ) = P m k =1 ( s k ∗ c ν k a − s k − 1 ∗ c ν k a ) a nd ψ G ( u ) = P m k =1 [ s k ∗ ( ν k a ) − s k − 1 ∗ ( ν k a )]. Consequen tly , k ψ H ( u ) k δ = P m k =1 ( s k − s k − 1 ) δ ( ν k a ) and ε k u k κ + A k ψ H ( u ) k δ − k ψ G ( u ) k G > m X k =1 [ εκ ( s k ) + A ( s k − s k − 1 ) δ ( ν k a ) − k s k ∗ ( ν k a ) − s k − 1 ∗ ( ν k a ) k G ] . 48 P . NI EMIEC So, (6-15) will b e fulﬁlled if only k s ∗ ( ma ) − t ∗ ( ma ) k G 6 A | s − t | δ ( ma ) + εκ ( s ∨ t ) for ev ery s, t ∈ R + and m ∈ Z . Since rank( a ) = N , we may assume that 0 < m < N . But then, by the deﬁnition of M , k s ∗ ( ma ) − t ∗ ( ma ) k G 6 m k s ∗ a − t ∗ a k G 6 N [ M | s − t | + ( ε/ N ) κ ( s ∨ t )] = A | s − t | δ ( ma ) + εκ ( s ∨ t ) and we are done.  6.13. Lemma. Supp ose a ∈ C ( G ) \ { 0 } is such that lim n →∞ k na k G /n = 0 . F or every ε > 0 ther e is a PV gr oup ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) and an e lement b ∈ L ( ¯ G ) f in such that k a − b k 6 ε . Pr o o f. T ak e N > 2 suc h that for eve ry n > N , (6-16) k na k G n 6 ε 2 . F urther, let M = M ( a, ε/ (2 N )) b e as in Lemma 6.8. Put A = N M . Let H = h b i b e a cyclic g roup of rank N and δ denote the discrete v a lue on H . Now put ¯ G = G × L 0 [ H ] and deﬁne k · k : ¯ G → R + b y k ( g , f ) k = inf {k g − m X j =1 s j ∗ ( l j a ) k G + ε m X j =1 κ ( s j )+ A k m X j =1 s j ∗ ( l j b b ) − f k δ : m > 0 , s 1 , . . . , s m > 0 , l 1 , . . . , l m ∈ {− 1 , 1 }} . As in the previous pro of, o bserv e that k · k is a κ -seminorm suc h that k ( g , f ) k > 0 provided f 6 = 0, k ( g , 0) k 6 k g k G , k (0 , f ) k 6 A k f k δ ( g ∈ G , f ∈ L 0 [ H ]) and k ( a, b b ) k 6 ε . So, if only w e sho w that k ( g , 0) k > k g k G for g ∈ G , the pro of will b e completed (b ecause then k · k will b e a κ -norm and b b ∈ L ( ¯ G )). The latter is equiv alen t to (6-17) ε m X k =1 κ ( s k ) + A k m X k =1 s k ∗ ( l k b b ) k δ > k m X k =1 s k ∗ ( l k a ) k G for eve ry m > 0, s 1 , . . . , s m ∈ R + and l 1 , . . . , l m ∈ {− 1 , 1 } . W e ma y assume tha t m > 1 and then, when m and l 1 , . . . , l m are ﬁxed, b oth the left-hand and the right-hand side expressions of (6- 14) are con tinuous with resp ect to s 1 , . . . , s m . Hence w e may assume that s 1 , . . . , s m are p o sitiv e and diﬀeren t. Finally , aft er ren umeration, w e ma y hav e 0 = s 0 < s 1 < . . . < s m . But then k P m k =1 s k ∗ ( l k b b ) k δ = P m j =1 (∆ s j ) δ ( P m k = j l k b ) (where, as usually , ∆ s j = s j − s j − 1 ). So, (6-17) c hanges in to (6-18) ε 2 m X j =1 κ ( s j ) + m X j =1 [ A (∆ s j ) δ ( m X k = j l k b ) + ε 2 κ ( s j )] > > k m X j =1 s j ∗ ( l j a ) k G . UNIVERS AL V ALU ED ABELIA N GR OUPS 49 No w w e shall construct a system of natural num b ers { ν 0 , . . . , ν z } (for some z > 1) for whic h (E0) 0 = ν z < . . . < ν 0 = m + 1, (E1) for ev ery j, k with 1 6 k 6 z and ν k < j < ν k − 1 , | ν k − 1 − 1 X s = j l s | < N , (E2) for ev ery k with 1 6 k < z , | ν k − 1 − 1 X s = ν k l s | = N . Put ν 0 = m + 1 and supp ose ν p is deﬁne d for some p > 0. If either ν p = 1 or | P ν p − 1 s = j l s | < N fo r ev ery j ∈ { 1 , . . . , ν p − 1 } , put ν p +1 = 0 and z = p + 1 and ﬁnish the construction. Otherwise , tak e the greatest natural n um b er ν p +1 ∈ { 1 , . . . , ν p − 1 } suc h that | P ν p − 1 s = ν p +1 l s | > N . Since | l s | = 1 for eve ry s , one has | P ν p − 1 s = ν p +1 l s | = N . The v eriﬁcation that (E0)–(E2) are fulﬁlled is left as an exercis e. F or simplicit y , let l 0 = 0. No w it follows from (E2) that (E3) for ev ery j, k with 0 6 k < z and ν k +1 6 j < ν k , m X s = j l s b = ν k − 1 X s = j l s b (recall tha t rank( b ) = N ). Thanks to (E3), w e ha v e (6-19) m X j =1 [ A (∆ s j ) δ ( m X k = j l k b ) + ε 2 κ ( s j )] > > z − 1 X k =0 ν k − 1 X j = ν k +1 +1 [ A (∆ s j ) δ ( ν k − 1 X q = j l q b ) + ε 2 κ ( s j )] . F urther, when 0 6 k < z , ν k − 1 X j = ν k +1 s j ∗ ( l j a ) = ν k − 1 X j = ν k +1 s j ∗ ( ν k − 1 X q = j l q a − ν k − 1 X q = j +1 l q a ) = = ν k − 1 X j = ν k +1 s j ∗ ( ν k − 1 X q = j l q a ) − ν k X j = ν k +1 +1 s j − 1 ∗ ( ν k − 1 X q = j l q a ) = = ν k − 1 X j = ν k +1 +1 ( ν k − 1 X q = j l q )( s j ∗ a − s j − 1 ∗ a ) + s ν k +1 ∗ ( ν k − 1 X q = ν k +1 l q a ) 50 P . NI EMIEC and t herefore (by (E2), (6-16) and the fa ct that s ν z = 0): k ν k − 1 X j = ν k +1 s j ∗ ( l j a ) k G = k z − 1 X k =0 ν k − 1 X j = ν k +1 s j ∗ ( l j a ) k G 6 6 z − 1 X k =0 ν k − 1 X j = ν k +1 +1    ν k − 1 X q = j l q    · k s j ∗ a − s j − 1 ∗ a k G + z − 1 X k =1 κ ( s ν k ) k N a k G 6 6 z − 1 X k =0 ν k − 1 X j = ν k +1 +1    ν k − 1 X q = j l q    · k s j ∗ a − s j − 1 ∗ a k G + ε 2 N z − 1 X k =1 κ ( s ν k ) . So, taking into accoun t (6 -19), w e see that ( 6-18) will b e satisﬁed pro- vided (6-20)    ν k − 1 X q = j l q    · k s j ∗ a − s j − 1 ∗ a k G 6 A (∆ s j ) δ ( ν k − 1 X q = j l q b ) + ε 2 κ ( s j ) whenev er ν k +1 < j < ν k ( k = 0 , . . . , z − 1) and (6-21) N z − 1 X k =1 κ ( s ν k ) 6 m X j =1 κ ( s j ) . T o prov e (6-20 ), put λ =    P ν k − 1 q = j l q    , t = s j and s = s j − 1 . By (E1), 0 6 λ < N . When λ = 0, (6-20 ) is clear. And if λ 6 = 0, δ ( λb ) = 1 and then, by the deﬁnition of M , λ k t ∗ a − s ∗ a k G 6 N ( M | t − s | + ε 2 N κ ( t ∨ s )) = A | t − s | δ ( λb ) + ε 2 κ ( t ) whic h giv es ( 6-20). No w w e pass to (6-21). By (E2), N 6 ν k − 1 − ν k for k = 1 , . . . , z − 1. So, b y (NF3) w e obtain N z − 1 X k =1 κ ( s ν k ) 6 z − 1 X k =1 ν k − 1 − 1 X j = ν k κ ( s ν k ) 6 z − 1 X k =1 ν k − 1 − 1 X j = ν k κ ( s j ) 6 m X j =1 κ ( s j ) whic h ﬁnishes the pro of.  As an immediate consequence of (P5) and Lemmas 6.7, 6.12 and 6.13 w e obtain 6.14. Theorem. If ( G, + , ∗ , k · k G ) is a κ -norm e d top olo gic al pseudove c- tor gr oup such that ( G, + , k · k G ) ∈ G r ( N ) , ther e is a κ -norme d pse u- dove ctor gr oup ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) such that ( ¯ G, + , k · k ) ∈ G r ( N ) and the set L ( ¯ G ) f in is dense in ¯ G . UNIVERS AL V ALU ED ABELIA N GR OUPS 51 7. Proof of Theorem 1.3 Let r ∈ { 1 , ∞} , N ∈ Z + \ { 1 } a nd κ b e a norming function. As it is easily seen, Theorem 1.3 immediately follow s from 7.1. Theorem. L et ( G, + , ∗ , k · k G ) b e a κ -norme d top olo gic al pseu- dove ctor gr oup such that ( G, + , k · k G ) ∈ G r ( N ) and (AX) i s fulﬁl le d. Ther e is a κ -norme d pseudove ctor gr oup ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) such that the set L ( ¯ G ) f in is dense in ¯ G and the value d gr oups ¯ G and G r ( N ) ar e isometric al ly gr oup iso morphic. The ab ov e result strengthens and generalizes [18, Theorem 4 .3]. The pro of of Theorem 7.1 will b e preceded by 7.2. Lemma. L et ( G ′ , + , ∗ , k · k ′ ) b e a Lipschitz κ -norme d pseudove ctor gr oup such that ( G ′ , + , k · k ′ ) ∈ G r ( N ) a nd (AX) is sa tisﬁ e d. L et D a nd F b e, r esp e ctively, a c ountable subset of G ′ and a c ountable subﬁeld o f R . F or every c ountable family H of ﬁnite value d gr o ups of class G r ( N ) ther e is a Lipschitz κ -norme d pseudove ctor gr oup ( G ′′ , + , ∗ , k · k ′′ ) ⊃ ( G ′ , + , ∗ , k · k ′ ) such that ( G ′′ , + , k · k ′′ ) ∈ G r ( N ) a nd with the fol lowing pr op erty. Wheneve r H ∈ H , K is a sub gr oup of H and ϕ : K → lin F D ⊂ G ′ is a n isome tric gr oup homomorp h ism, ther e is an isometric gr oup homomo rp hism ψ : H → G ′′ which extends ϕ . Pr o o f. Let D ′ = lin F D . Since ev ery mem b er of H is a ﬁnite group and D ′ is coun table, the family { ϕ : K → D ′ } , where K runs o ve r all subgroups of members of H and ϕ is an isometric group homo- morphism, is coun table. So, the assertion fo llo ws fr om induction and Theorem 6 .11.  Pr o o f of The or em 7.1. Thanks to Theorem 6.14, w e ma y and do assume that A = L ( G ) f in is dense in G . Let A 0 b e a countable dense subset of A . Note that lin A 0 is a dense Lipsc hitz PV subgroup of G and, as a group, of class O f in . F or purp ose of this pro of, let us call a κ -normed PV group ( E , + , ∗ , k · k E ) of class L r ( N ) f in pro vided ( E , + , k · k E ) ∈ G r ( N ) and E = L ( E ) f in . W e shall construct, making us e of induction, sequences ( F n ) ∞ n =0 , ( G n , + , ∗ , k · k n ) and ( D n ) ∞ n =0 suc h that f or ev ery n > 0: (1 n ) F n is a coun table subﬁeld of R , (2 n ) ( G n , + , ∗ , k · k n ) ∈ L r ( N ) f in , (3 n ) D n is a coun table subset of G n and G n = lin D n , (4 n ) F 0 = Q and for n > 0, F n ⊃ F n − 1 ∪ {k g k n − 1 : g ∈ lin F n − 1 D n − 1 } , (5 n ) G 0 = lin A 0 and k · k 0 is inherited from k · k G ; a nd for n > 0, ( G n , + , ∗ , k · k n ) ⊃ ( G n − 1 , + , ∗ , k · k n − 1 ) and D n ⊃ D n − 1 , (6 n ) for n > 0: whenev er ( H , + , q ) ∈ G r ( N ) is a ﬁnite v a lued group with q ( H ) ⊂ F n − 1 and ϕ : K → lin F n − 1 D n − 1 is an isometric g roup 52 P . NI EMIEC homomorphism of a subgroup K of H , there is an isometric group homomorphism ψ : H → lin F n D n whic h extends ϕ . Deﬁne F 0 and G 0 as in (4 0 ) and (5 0 ) and put D 0 = A 0 . Supp ose that for some n > 0 w e ha ve deﬁned F n − 1 , G n − 1 and D n − 1 with suitable prop erties. Let H b e a countable fa mily of all (up to isometric group isomorphism) ﬁnite F n − 1 -groups of class G r ( N ) (cf. Deﬁnition 3.1). T ak e a Lipsc hitz κ - normed PV g roup ( G ′′ , + , ∗ , k · k ′′ ) whic h witnesses the assertion of Lemma 7.2 fo r H , G ′ = G n − 1 , F = F n − 1 and D = D n − 1 . F urther, let F b e the family of all pairs ( ϕ, H ) with H ∈ H and ϕ : K → lin F n − 1 D n − 1 an isometric group homomorphisms where K is a subgroup o f H . The collection F is coun ta ble. F or each ( ϕ, H ) ∈ F let b ϕ : H → G ′′ b e a n isometric group homomorphism extending ϕ . No w put D n = D n − 1 ∪ S ( ϕ,H ) ∈ F b ϕ ( H ) and G n = lin D n ⊂ G ′′ . Let k · k n b e the κ - norm on G n inherited from k · k ′′ and F n b e the subﬁeld of R generated b y F n − 1 ∪ {k g k n : g ∈ D n } . It is easy to v erify that conditions (1 n )–(6 n ) are fulﬁlled. Ha ving the sequence s ( F n ) ∞ n =0 , ( D n ) ∞ n =0 and ( G n ) ∞ n =0 , deﬁne the PV group ( ¯ G, + , ∗ , k · k ) as the completion of S ∞ n =0 ( G n , + , ∗ , k · k n ) (cf. Prop osition 6.6), F = S ∞ n =0 F n and D = S ∞ n =0 D n . Note that ( ¯ G, + , ∗ , k · k ) ⊃ ( G, + , ∗ , k · k G ) (since G 0 is dense in G ) and b D = lin F D is dense in ¯ G . F urther, b D is isometrically group isomorphic to FG r ( N ), whic h follo ws from the construction. Thus , to this end it remains to apply P rop osition 3.9.  As sonsequences of Theorem 7.1 w e obtain (see Prop osition 6.4) 7.3. Theorem. L e t r ∈ { 1 , ∞} . Every sep ar able metrizable top olo gic al pseudove ctor (Ab elian) gr oup is isomorphic (as a top olo gic al PV gr oup) to a pseudove ctor sub gr oup of a subnorme d top olo gic al PV gr oup which is isome tric al ly g r oup isomorphic to G r (0) . 8. Topology of G r ( N ) ’s This part is devoted to the pro of of t he following 8.1. Theorem. Each of the top olo gic al sp ac es G r ( N ) ’s is home omor- phic to the Hilb ert sp ac e l 2 . Note t hat Theorem 8.1 fo r N ∈ { 0 , 2 } follows from Theorem 5.1 and the result of Usp enskij [28]. Ho we v er, the pro of presen ted here needs no additional w o rk for suc h N a nd th us we g iv e the pro of of Theorem 8.1 in its full generality . W e shall do this making use of Theorem 1.3. Before w e pass to the pro of of Theorem 8.1, we ha v e to recall some notions a nd results of inﬁnite-dimensional top olo gy . F irst o f all, recall that the space l 2 is the Banach space consisting of a ll square summable real sequences equipp ed with the norm k ( a n ) ∞ n =1 k 2 = ( P ∞ n =1 a 2 n ) 1 / 2 . UNIVERS AL V ALU ED ABELIA N GR OUPS 53 A metrizable space X is said to be an absolute r etr act iﬀ it is a retract of any metrizable space in whic h X is em b edded as a closed set. The space X is homotopic al ly trivial pro vided ev ery map of the b oundary of [0 , 1] n (for ar bitrary n > 1) in to X is extendable to a map of the whole cub e [0 , 1] n in to X . Observ e that under suc h a deﬁnition the empty space is homotopically trivial. A subset Y o f the space X is said to b e c ontr actible in X if there is a map H : Y × [0 , 1] → X such that H ( y , 1) = y f or each y ∈ Y and the map H ( · , 0) is constan t. An elemen tary result concerning contractibilit y and homotopical t rivialit y sa ys that if ev ery compact nonempty subset of a metrizable space M is contractible in M , then M is homotopically trivial. T oru ´ nczyk in his famous works [22, 23] has giv en a c haracterization of metric spaces whic h are homeomorphic to l 2 . Based on his results, Dobrow olski and T oru ´ nczyk [4] has pro v ed the f ollo wing theorem, whic h will b e one of to ols of this section. 8.2. Theorem. A sep ar a ble c ompletely metrizable top olo gic al gr oup is home omorphic to l 2 iﬀ it is a non - l o c al ly c omp act absol ute r etr act. Th us, according to Theorem 8.2, in order to pro v e Theorem 8.1, w e only need to sho w that each of the spaces G r ( N )’s is an absolute retract (that G r ( N ) is non-lo cally compact it easily follo ws from The- orem 5.1 2). W e shall pro v e this in v olving a very conv enien t criterion due to T o ru ´ nczy k [21] ( cf. [16, Theorem ?.??]) a special case of whic h is for m ulated b elo w (compare with the pro of in [28]). 8.3. Theorem. If the interse ction o f every ﬁnite nonempty c o l le ction of op en b al ls in a metric sp ac e ( X , d ) with c entr es in a given dense subset of X is homotopic al ly trivial, then X is an abso l ute r etr a ct. No w we are ready to prov e Theorem 8.1. It turns out that the argumen t is sligh tly more complicated in case of b ounded groups than in the unbounded case (the main reason for this is that there are no non trivial b ounded norms on PV gro ups). Th us, w e divide the pro of in to these t wo cases. Pr o o f of The or em 8.1 for r = ∞ . Let D = G r ( N ) f in and p b e the v alue of G ∞ ( N ). By (G3), D is dense in G ∞ ( N ). T ake a rbitrary p oin ts x 1 , . . . , x n ∈ D and radii r 1 , . . . , r n > 0 . Let B = T n j =1 B p ( x j , r j ) where B p ( x j , r j ) = { z ∈ G ∞ ( N ) : p ( z − x j ) < r j } . W e ma y assume that B is nonempty . W e shall sho w that B is contractible (in itse lf ) . Since translations x 7→ x + a ( a ∈ G ∞ ( N )) are (bijectiv e) isometries on G ∞ ( N ), w e ma y also assume that 0 ∈ B . This means t hat (8-1) p ( x j ) < r j ( j = 1 , . . . , n ) . Let H = h x 1 , . . . , x n i . Observ e that H is a ﬁnite gro up. Denote by q the restriction of p to H . Then ( H , + , q ) ∈ G ∞ ( N ) and consequen tly ( L 0 [ H ] , + , k · k q ) ∈ G ∞ ( N ). Notice that ϕ : H ∋ h 7→ b h ∈ L 0 [ H ] is an 54 P . NI EMIEC isometric group homomorphism. By The orem 1.3, t here is a normed top ological PV gro up ( G, + , ∗ , k · k ) ⊃ ( L 0 [ H ] , + , ∗ , k · k q ) suc h that G and G ∞ ( N ) a re isometrically group isomorphic. So, thanks to (UEP), there is an isometric group isomorphism ψ : G ∞ ( N ) → G whic h extends ϕ . D eﬁne H : B × [0 , 1 ] → G ∞ ( N ) b y H ( y , t ) = ψ − 1 ( t ∗ ψ ( y )). W e see that H is contin uous and H ( y , 0) = 0 and H ( y , 1) = y for y ∈ B . So, it suﬃces to che c k that H ( B × [0 , 1]) ⊂ B . F or y ∈ B , t ∈ [0 , 1] and j ∈ { 1 , . . . , n } w e ha v e, b y (8-1), p ( H ( y , t ) − x j ) = k t ∗ ψ ( y ) − ψ ( x j ) k 6 6 k t ∗ ψ ( y ) − t ∗ ψ ( x j ) k + k t ∗ ψ ( x j ) − ψ ( x j ) k = = t k ψ ( y − x j ) k + k t ∗ b x j − b x j k q = tp ( y − x j ) + (1 − t ) p ( x j ) < r j and we are done.  Pr o o f of The or em 8.1 (for arbitr ary r ) . W e shall impro ve the pre- vious argumen t so that it will w ork also in b ounded case. Let D ; p ; x 1 , . . . , x n ∈ D ; r 1 , . . . , r n > 0 and B b e as in the prev ious pro of. As there, w e ma y and do assume that (8-1) is fulﬁlle d. Let K b e a compact subset of B . W e shall sho w that K is con tractible in B . F or j ∈ { 1 , . . . , n } , let  j = max { p ( z − x j ) : z ∈ K ∪ { 0 }} < r j . Let ε > 0 b e suc h that (8-2)  j + 2 ε < r j ( j = 1 , . . . , n ) . F urther, tak e w 1 , . . . , w l ∈ D for whic h (8-3) B p ( w k , ε ) ∩ K 6 = ∅ ( k = 1 , . . . , l ) and K ⊂ l [ k =1 B p ( w k , ε ) . The ab ov e implies t hat (8-4) p ( w k − x j ) 6  j + ε ( j ∈ { 1 , . . . , n } , k ∈ { 1 , . . . , l } ) . Let H = h x 1 , . . . , x n ; w 1 , . . . , w l i , q denote the restriction o f p to H and let k · k H = k · k q ∧ r . Then ( L 0 [ H ] , + , ∗ , k · k H ) is a Lipsc hitz subnormed PV group suc h that ( L 0 [ H ] , + , k · k H ) ∈ G r ( N ) and the function ϕ : H ∋ h 7→ b h ∈ L 0 [ H ] is an isometric group homomorphism. Again, w e conclude from Theorem 1.3 that there is a subnormed top ological PV group ( G, + , ∗ , k · k ) ⊃ ( L 0 [ H ] , + , ∗ , k · k H ) suc h that G and G r ( N ) are isometrically g roup isomorphic. Therefore, there is an isometric gr oup isomorphism ψ : G r ( N ) → G whic h extends ϕ . D eﬁne H : K × [0 , 1] → G r ( N ) as b efore: H ( y , t ) = ψ − 1 ( t ∗ ψ ( y ) ). W e only need to show that H tak es v alues in B . T o this end, let y ∈ K , t ∈ [0 , 1 ] a nd j ∈ { 1 , . . . , n } . T ak e k ∈ { 1 , . . . , l } suc h that p ( y − w k ) < ε and, using (8-2) and (8-4), UNIVERS AL V ALU ED ABELIA N GR OUPS 55 observ e that p ( H ( y , t ) − x j ) = k t ∗ ψ ( y ) − ψ ( x j ) k 6 6 k t ∗ ψ ( y − w k ) k + k t ∗ ψ ( w k ) − ψ ( x j ) k 6 6 ∇ ( t ) k ψ ( y − w k ) k + k t ∗ c w k − b x j k = = p ( y − w k ) + tp ( w k − x j ) + (1 − t ) p ( x j ) 6 6 ε + t (  j + ε ) + (1 − t )  j 6  j + 2 ε < r j whic h ﬁnishes the pro of.  9. Subnormed topological PV groups of class O 00 In this section all g roups are subnormed top o logical pseudo ve ctor. Observ e that if k · k is a subnorm on G , then f or ev ery g ∈ G the function (0 , ∞ ) ∋ t 7→ k t ∗ g k /t ∈ R + is monotone decreasing. Consequen tly , there is a ﬁnite limit k g k ∗ 0 = lim t →∞ k t ∗ g k t . The function k · k ∗ 0 : G → R + is a seminorm on G suc h that k · k ∗ 0 6 k · k . In particular, G ∗ 0 = { g ∈ G : k g k ∗ 0 = 0 } is a clos ed PV subgroup o f G . W e call G of c lass O ∗ 0 iﬀ k · k ∗ 0 ≡ 0 or , equiv a len tly , if G = G ∗ 0 . 9.1. Deﬁnition. A subnormed PV group G is of class O 00 if G is of b oth classes O 0 and O ∗ 0 . That is, G is of class O 00 if for ev ery g ∈ G , lim t →∞ k t ∗ g k t = lim n →∞ k ng k n = 0 . W e call an elemen t a of G b ounde d provide d lin { a } is a (metrically) b ounded subset of G . Let us denote b y G bd the set of all b ounded elemen ts of G . Let E ( G ) = L ( G ) ∩ G f in ∩ G bd . So, E ( G ) is a PV subgroup of G consisting of all ﬁnite rank elemen ts whic h are b oth Lipsc hitz and b ounded. It is easily seen t hat G is of class O 00 pro vided E ( G ) is dense. W e w ant to pro v e the ‘conv erse’ of this statemen t, namely 9.2. Theorem. A subnorme d top o l o gi c al PV gr oup G is of class O 00 iﬀ it may b e enlar ge d to a subnorm e d PV gr oup e G such that E ( e G ) is dense in e G . Mor e over, if G ∈ G r ( N ) ∩ O 00 , the a b ove e G m ay b e chosen so that e G ∈ G r ( N ) . Theorem 9.2 is a n almost immediate consequ ence of 9.3. Lemma. L et a ∈ C ( G ) \ { 0 } b e such that a ∈ G 0 ∗ ∩ G ∗ 0 . F or e v ery ε > 0 ther e is a subnorme d PV gr oup ( ¯ G, + , ∗ , k · k ) enlar ging G and b ∈ E ( ¯ G ) such that k a − b k 6 ε . What is mor e, if r ank( a ) < ∞ , then rank( b ) = r ank( a ) ; an d k · k is b ounde d by 1 pr o vide d so is the subnorm of G . 56 P . NI EMIEC Pr o o f. W e mimic t he pro ofs of Lemma 6 .12 and Lemma 6.13. Let k · k G denote the subnorm of G . If a has ﬁnite r ank, let N = ra nk( a ). Otherwise let N > 2 be as in (6-16). Put κ = ∇ . T ake T > 1 suc h that for eac h t > T , (9-1) N k t ∗ a k G 6 εt. If rank ( a ) < ∞ , repeat the pro of of L emma 6.1 2 (b elo w w e use the same notatio n a s there) with k · k replaced b y k ( g , f ) k = inf {k g − ψ G ( u ) k G + ε k u k κ + A ( k ψ H ( u ) − f k δ ∧ T ) : u ∈ L 0 [ Z ] } . If rank( a ) = ∞ , rep eat t he pro of of Lemma 6.13 with k ( g , f ) k = inf {k g − m X j =1 s j ∗ ( l j a ) k G + ε m X j =1 κ ( s j ) + A h k m X j =1 s j ∗ ( l j b b ) − f k δ ∧ T i : m > 0 , s 1 , . . . , s m > 0 , l 1 , . . . , l m ∈ {− 1 , 1 }} . In or der to sho w that k ( g , 0) k > k g k G for g ∈ G , distinguish b et w een t wo cases: when k ψ H ( u ) k δ 6 T (respectiv ely k P m j =1 s j ∗ ( l j b b ) k δ 6 T ) and when the latter is false. In t he ﬁrst case just cop y the or iginal pro of of suitable lemma. Here w e only show how to derive the second case. First assume rank( a ) < ∞ . W rite u = P m k =1 s k ∗ b l k with 0 = s 0 < . . . < s m and l 1 , . . . , l m ∈ Z \ { 0 } . Let z ∈ { 0 , . . . , m } b e such that s z 6 T and s z +1 > T . Put u ′ = P z k =0 s k ∗ b l k ∈ L 0 [ Z ]. Notice that k ψ H ( u ′ ) k δ 6 T , k ψ G ( u ) k G 6 k ψ G ( u ′ ) k G + P m k = z +1 k s k ∗ ( l k a ) k G and, by (9-1), k s k ∗ ( l k a ) k G 6 εs k = εκ ( s k ) for k > z . So, k ψ G ( u ) k G 6 k ψ G ( u ′ ) k G + ε m X k = z +1 κ ( s k ) 6 6 ε z X k =1 κ ( s k ) + A k ψ H ( u ′ ) k δ + ε m X k = z +1 κ ( s k ) 6 ε m X k =1 κ ( s k ) + AT = = ε k u k κ + A ( k ψ H ( u ) k δ ∧ T ) whic h giv es t he assertion. When rank( a ) = ∞ , ar gue in a similar wa y . Assuming tha t 0 = s 0 < . . . < s m and t aking z ∈ { 0 , . . . , m } fo r whic h s z 6 T and s z +1 > T , observ e that k P z k =1 s k ∗ ( l k b b ) k δ 6 T and (by Lemma 6 .13) k z X k =1 s k ∗ ( l k a ) k G 6 ε z X k =1 κ ( s k ) + A k z X k =1 s k ∗ ( l k b b ) k δ . UNIVERS AL V ALU ED ABELIA N GR OUPS 57 F urther, it follo ws from (9-1) that k s k ∗ ( l k a ) k G 6 εκ ( s k ) for k > z (recall tha t l k ∈ {− 1 , 1 } ). Hence k m X k =1 s k ∗ ( l k a ) k G 6 k z X k =1 s k ∗ ( l k a ) k G + ε m X k = z +1 κ ( s k ) 6 6 ε m X k =1 κ ( s k ) + AT 6 ε m X k =1 κ ( s k ) + A h k m X k =1 s k ∗ ( l k b b ) k δ ∧ T i whic h is equiv alen t t o k ( g , 0) k > k g k G . Finally , replace k · k b y k · k ∧ 1 if k · k G 6 1. The details are left for the reader.  Theorem 1 .1 and the existence of the G urari ˇ ı Banach space suggests to adapt these ideas in the realm of subnormed t op ological pseudo ve c- tor (Ab elian) g roups. It may b e done in a few wa ys. One of them is prop osed b elo w. W e ﬁx r ∈ { 1 , ∞} , N ∈ Z + \ { 0 } . F or simplic it y , denote b y PV G r ( N ) the class of all subnormed t op ological PV gro ups whic h b elong, as v a lued groups, to G r ( N ) and are of class O 00 . In particular, PV G ∞ (0) consists of all separable subnormed to p ological PV g roups of class O 00 . Let us call a subnormed PV group H ∈ PV G r ( N ) of class E r ( N ) f in if H = E ( H ) and H is ﬁn i tely gener ate d , that is, H = lin F for some ﬁnite subset o f H . Observ e that the subnorm of a PV g roup of class E r ( N ) f in is b ounded. ‘Gurari ˇ ı-lik e’ space in category of subnormed top ological PV groups of class PV G r ( N ) ma y b e deﬁned as follow s. 9.4. Deﬁnition. A subnormed PV group ( G, + , ∗ , k · k G ) is said to b e E r ( N ) -Gur ari ˇ ı iﬀ the f ollo wing tw o conditions are satisﬁed: (GPV1) ( G, + , k · k G ) ∈ G r ( N ), G is complete and the set E ( G ) is dense in G , (GPV2) whenev er ( H, + , ∗ , k · k H ) ∈ E r ( N ) f in , K is a ﬁnitely gen- erated PV subgroup of H and ϕ : K → G is an isometric linear homomorphism, for ev ery ε ∈ (0 , 1) there exists an ε - almost isometric linear homomorphism ψ : H → G suc h tha t k ϕ ( x ) − ψ ( x ) k G 6 ε for ev ery x ∈ K . Notice that, thanks to (G PV1), ev ery E r ( N )-Gur ari ˇ ı PV group is of class P V G r ( N ) (b y Prop osition 6.5 and Theorem 9.2). Belo w we list other pr op erties o f them. 9.5. Pr op osition. Every E r ( N ) -Gur ari ˇ ı p seudove ctor gr oup is isomet- ric al ly gr oup isomorphic to G r ( N ) . Pr o o f. Let G be a subnormed E r ( N )-Gur ari ˇ ı PV gro up. Thanks to Prop osition 3 .9, it suﬃces to chec k that G 0 = E ( G ) satisﬁes conditions 58 P . NI EMIEC (QG1) and (QG2) for Q = R + . But this easily follows from Theo- rem 6.11 and (GPV2). (Indeed, note that the PV group ¯ G o btained from Theorem 6.11 ma y b e constructed so that ¯ G ∈ E r ( N ) f in .)  W e shall no w show that ev ery E r ( N )-Gur ari ˇ ı PV group satisﬁes the coun terpart of (UEP). Precisely , that in (GPV2) one ma y put ε = 0. W e can provide this b y r ep eating the argumen ts of Sections 2 and 3. The crucial p oint here is that the PV groups o f class E r ( N ) f in ha ve b ounded subnorms. W e start with 9.6. Lemma. L et ( D j , + , ∗ , k · k j ) ∈ E r ( N ) f in ( j = 1 , 2 ) and D 0 b e a ﬁnitely gener ate d PV sub g r oup of D 1 . L et u : D 0 → D 2 and v : D 1 → D 2 b e an isometric and, r esp e c tively, an ε -almost isomet- ric line ar homomorph i s m (wher e ε ∈ (0 , 1) ) such that k u − v   D 0 k ∞ 6 ε . Then ther e ar e a subnorme d PV gr oup ( G, + , ∗ , k · k ) ∈ E r ( N ) f in and isometric line ar homomorphisms ψ j : D j → G ( j = 1 , 2 ) such that ψ 1   D 0 = ψ 2 ◦ u and k ψ 1 − ψ 2 ◦ v k ∞ 6 Aε wh er e A = 1 + diam( D 1 , k · k 1 ) . Pr o o f. Deﬁne D , w 1 , w 2 and a semiv alue λ on D in exactly t he same w ay a s in the pro of of Lemma 2.20. O bserv e that D is a PV gr oup, λ is a semisubnorm and w 1 and w 2 are linear homo morphisms. T o this end, let G b e the quotien t subnormed PV gro up D /λ − 1 ( { 0 } ) a nd ψ 1 and ψ 2 b e the homomorphisms naturally induced by w 1 and w 2 .  No w rep eating the pro o fs of L emma 3.7 and Theorem 3 .8 w e o btain the next t wo results. 9.7. Lemma. L e t G b e an E r ( N ) -Gur ari ˇ ı PV gr oup, ( H , + , ∗ , k · k H ) ∈ E r ( N ) f in b e a subnorme d PV gr o up and K its ﬁnitely gene r ate d PV sub gr o up. F urther, let ϕ : K → G and ψ : H → G b e, r esp e ctively, an isometric and a n ε -al m ost is ometric line ar ho momorphism (wher e ε ∈ (0 , 1) ) such that k ψ   K − ϕ k ∞ 6 ε . F or every δ ∈ (0 , ε ) ther e is a δ -almost isome tric line ar homomorphis m ψ ′ : H → G such that k ψ ′   K − ϕ k ∞ 6 δ and k ψ − ψ ′ k ∞ 6 C ε wher e C = 3 + 2 diam( H , k · k H ) . 9.8. Theorem. L et G b e an E r ( N ) -Gur ari ˇ ı PV gr oup, l e t H ∈ E r ( N ) f in and K b e a ﬁnitely ge n er ate d PV sub gr oup of H . Every isometric line ar homo m orphism of K into G is e x tendable to an is ometric line ar homomorphism of H into G . 9.9. Corollary . Each two subnorme d E r ( N ) -Gur ari ˇ ı PV gr oups ar e isometric al ly isomorphic as s ubn o rme d PV gr oups. Thanks to Corolla ry 9 .9, w e may reserv e a special sym b ol for a E r ( N )-Gur ari ˇ ı PV group. W e shall denote it (if it only exists) by PVG r ( N ). It follows from Theorem 9.2 and Theorem 9.8 that 9.10. P rop osition. Every subnorme d PV gr o up of class PV G r ( N ) is emb e ddable i n to PVG r ( N ) by me ans of an is o metric line ar homo mor- phism. UNIVERS AL V ALU ED ABELIA N GR OUPS 59 W e b eliev e PVG r ( N ) exists for ev ery r and N . The existence of it for N = 0 w ould hav e an in teresting application in functional a nalysis (cf. Prop osition 6.4). 9.11. Prop osition. If ther e exists PVG r (0) , ther e is a sep ar able c om - plete subnorme d top olo gic al ve ctor sp ac e V of class P V G r (0) which is universal for al l subn o rme d top olo gic al ve ctor sp ac es of this class . Pr e- cisely, every subnorme d top olo gic al ve ctor sp ac e of class O 00 whose sub- norm is b ounde d by r a dmits an isometric line ar ho momorphism in to V . Mor e over, every sep ar able metrizable top ol o gi c al ve ctor sp ac e admits a home om o rphic line ar emb e dding into V . Pr o o f. Put V = { x ∈ PVG r (0) : ( t + s ) ∗ x = t ∗ x + s ∗ y for ev ery t, s ∈ R + } . Equip V with the subnorm inherited f rom the one of PVG r (0). W e lea ve this as a simple exercise that V is a v ector space (o ver R ) when the multiplication by reals is deﬁned by tv = t ∗ v and ( − t ) v = − ( t ∗ v ) for t ∈ R + and v ∈ V ; and that the assertion of the pro p osition is fulﬁlled (use Prop osition 6.4 and Prop osition 9.10).  Then next result corresp onds to Prop osition 4.7. 9.12. P rop osition. If M | N , then PVG r ( N , M ) := { x ∈ PVG r ( N ) : M · x = 0 } is isome tric al ly i s omorphic as a subn o rme d PV gr oup to PVG r ( M ) . Pr o o f. F or simplicit y put G = PVG r ( N , M ). W e o nly need to c hec k that the set E ( G ) is dense if G . Let a ∈ G , ε > 0 a nd let k · k stand for the subnorm of PVG r ( N ). By (GPV1), there is b ∈ E ( PVG r ( N )) suc h that k a − b k 6 ε . At the same time, a ccording to Theorem 9.2, there is a subnormed PV gro up ( G ′ , + , ∗ , k · k ′ ) of class PV G r ( M ) whic h enlarges lin { a } a nd suc h that E ( G ′ ) is dense in G ′ . In pa rticular, there is a ′ ∈ E ( G ′ ) with k a − a ′ k ′ 6 ε . Now using a standard ar- gumen t of amalgamatio n, w e see that there is a subnormed PV group ( D , + , ∗ , k · k D ) and isometric linear homomorphisms ψ : lin { a, b } → D and ψ ′ : G ′ → D whic h coincide on lin { a } (we omit the details). Put H = lin { ψ ( b ) , ψ ′ ( a ′ ) } ⊂ D and K = lin { ψ ′ ( b ) } ⊂ H . Notice that H ∈ E r ( N ) f in and a pply (GPV2) t o get an ε -almost isometric linear homomorphism ϕ : H → PVG r ( N ) for whic h k b − ϕ ( a ′ ) k 6 ε . T o this end, observ e that ϕ ( a ′ ) ∈ E ( G ) and k a − ϕ ( a ′ ) k 6 2 ε .  9.13. Corollary . PVG r ( N ) exists for e ach N > 2 pr ovide d PVG r (0) exists. 9.14. Example. Let ( H, + , ∗ , k · k ) b e a separable subnormed to p olog- ical PV gro up. Then ( H , + , ∗ , k · k α ) ∈ PV G ∞ (0) for ev ery α ∈ ( 0 , 1). So, as in case of v alued groups, ev ery separable subnormed top ological PV gro up ma y b e ‘approximated’ b y mem b ers of class PV G ∞ (0). 60 P . NI EMIEC 10. Concluding re marks In Proposition 2 .18 w e hav e pro v ed tha t the Urysohn univ ersal space U admits no structure of an Ab elian v a lued group of exp o nen t 3. A t the same time, as it w as shown in [17], U admits a unique structure of a v alued group of exp o nen t 2. T aking in to accoun t these t w o remarks, w e p ose the fo llo wing Conjecture. F or r ∈ { 1 , ∞} , G r (2) is a uniq ue (up to isometric gr oup isomorph ism) value d A b elian gr oup of ﬁnite exp onen t and of diameter r which is Urysohn as a metric sp ac e. If t he ab o v e conjecture w as false, it w ould b e in teresting to know for whic h exp onents suitable groups exist a nd for whic h exp onents they are unique. Cameron and V ershik [3] ha v e sho wn that the Urysohn univ ersal metric space admits a structure of a monothetic v a lued group. T aking this into accoun t, t he following may also b e in teresting. Question 1. Are the top ological groups G 1 (0) and G ∞ (0) mono- thetic? There is a fascinating phenomenom related to univ ersal disposition prop erty . Namely , the Urysohn univ ersal space U is a unique ( sepa- rable complete) metric space with univ ersal disp osition prop ert y for ﬁnite metric spaces; G ∞ (0) is a unique (separable complete with dense set of ﬁnite rank elemen ts) v alued Ab elian g roup with univ ersal dis- p osition prop erty for ﬁnite v alued Ab elian groups and PVG ∞ (0) (if only exists) is a unique ( separable complete with dense set of ﬁnite rank b o unded Lipsc hitz elemen ts) subnormed top ological PV group with unive rsal disp osition prop ert y for ﬁnitely generated b ounded Lip- sc hitz subnormed PV g roups. It follow s from Theorem 5.1 and Prop o- sition 9.5 that all these three metric spaces are isometric and b oth the ab ov e v alued g roups a re isometrically group isomorphic. So, in fact w e just enric h the structure of a single space. It seems t o us imp ortant to answ er the follo wing question with whic h we end the pap er. Question 2. Do there exis t the PV groups PVG 1 (0) and PVG ∞ (0) ? 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[29] N. W eaver, Lipschitz Algebr as , W o r ld Scientiﬁc, 1999. 62 P . NI EMIEC Piotr Niemiec, Jagi ellonian University, Institute of Ma thema tics, ul. Lojasiewicza 6, 30-348 Krak ´ ow, Poland E-mail addr ess : piot r.niem iec@u j.edu.pl

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