aa r X i v : . [ m a t h . F A ] A ug AN EXAMPLE OF A NON-COMMUTATIVE UNIFORMBANACH GROUP
MICHAL DOUCHA
Abstract.
Benyamini and Lindenstrauss mention in their mono-graph
Geometric nonlinear functional analysis Vol. 1., Ameri-can Mathematical Society Colloquium Publications, 48. AmericanMathematical Society, Providence, RI, 2000 that there is no knownexample of a non-commutative uniform Banach group. Prassidisand Weston also asked whether there is a non-commutative ex-ample. We answer this problem affirmatively. We construct anon-commutative uniform Banach group which has the free groupof countably many generators as a dense subgroup.Moreover, we show that our example is a free one-generateduniform Banach group whose metric induced by the norm is bi-invariant.
Introduction
A uniform Banach group is a Banach space equipped with an addi-tional group structure so that the group unit coincides with the Banachspace zero and the group operations are uniformly continuous with re-spect to norm. Uniform Banach groups were introduced and studiedby Enflo in [4], [5] with connection to the infinite-dimensional versionof the Hilbert’s fifth problem. Typical example comes when we aregiven two Banach spaces X and Y and a uniform homeomorphism φ : X → Y between them such that φ (0) = 0. Then we can define a(commutative) group operation · on X as follows: for x, y ∈ X we set x · y = φ − ( φ ( x ) + φ ( y )). Note that unless φ is linear there is no a prioriconnection between the two group operations + (resp. + X ) and · .A comprehensive source of information about uniform Banach groupsis Chapter 17 in [2]. As mentioned there, the following problem wasleft open. Does there exist a non-commutative uniform Banach group?
This question was also asked by Prassidis and Weston in [9] and [10].
Mathematics Subject Classification.
Key words and phrases. uniform Banach group, free group, Hilbert’s fifth prob-lem, Lipschitz-free space.The author was supported by IMPAN’s international fellowship programme par-tially sponsored by PCOFUND-GA-2012-600415.
Here we give a positive answer to this question. The following is themain result.
Theorem 0.1.
There exists an infinite dimensional separable Banachspace ( X , + , , k· k ) equipped with an additional group structure ( · , − , whose unit coincides with the Banach space zero, the group multiplica-tion · is invariant with respect to the norm k · k , and F ∞ , the free groupof countably many generators, is a dense subgroup of ( X , · , − , .In particular, there exists a non-commutative uniform Banach group. Preliminaries
We assume the reader to know basic facts about uniform spaces. Werefer to Chapter 8 in [6] for more information.Recall that for any topological group G there are two distinguishedcompatible uniformities: the left uniformity U L generated by basic en-tourages of the form { ( g, h ) : g − h ∈ U } , where U is a basic neigh-borhood of the identity in G ; and the right uniformity U R which isgenerated by basic entourages of the form { ( g, h ) : hg − ∈ U } , where U is again a basic neighborhood of the identity in G .We start with the following definition given by Enflo in [4]. Definition 1.1.
Let G be a topological group. G is called uniform ifthere exists a compatible uniformity on G such that the group multi-plication is uniformly continuous with respect to that uniformity.Below, we collect some basic facts about uniform groups. Fact 1.2. (1) (see Proposition 1.1.3. in [4] ) A topological group G is uniformif and only if the left and right uniformities coincide.(2) (folklore) That is in turn equivalent with the fact that there ex-ists a neighborhood basis of the unit of G consisting of open setsclosed under conjugation. Such groups are more often called SIN (small invariant neighborhood) groups, or also balanced groups.(3) (folklore) In case that G is metrizable, i.e. the neighborhood ba-sis can be taken countable, G is uniform, resp. SIN, if and onlyif it admits a compatible bi-invariant metric; i.e. metric d suchthat for any x, y, a, b ∈ G it holds that d ( x, y ) = d ( axb, ayb ) (the same reasoning gives that if G is not metrizable then itstopology is given by a family of bi-invariant pseudometrics). ON-COMMUTATIVE UNIFORM BANACH GROUP 3
Examples: (a) All abelian and compact topological groups are uniform groups.It is obvious for the former. For the latter, consider a compacttopological group G . Notice that by the continuity of the groupoperations, for any open neighborhood U of the identity and forany group element g there are open neighborhoods V g of g and W g ⊆ U of the identity so that V g · W g · V − g ⊆ U . Then bycompactness one can find finitely many elements g , . . . , g n ∈ G sothat V g , . . . , V g n cover G . Take W = T i ≤ n W g i and notice that forany g ∈ G we have g · W · g − ⊆ U . Clearly, S g ∈ G g · W · g − is thena conjugacy-invariant open neighborhood of the identity containedin U .(b) The Heisenberg group U T ( R ) consisting of the upper triangular3 × U T ( R ) is metrizableit is sufficient to show that it does not admit a compatible bi-invariant metric. Suppose for contradicition that d is a compatiblebi-invariant metric on U T ( R ). Given matrices A ′ = a ′ c ′ b ′ , A = a c b , a computation gives that A − A ′ A = a ′ c ′ − ab ′ + a ′ b b ′ . Since d is compatible there exist r > R > R with respect to d centred at I , the identitymatrix, is contained in the open set of matrices x z y with | z | < r . Since d is compatible, we can choose A ′ with a ′ = 0and d ( A ′ , I ) < R . Then by choosing b sufficiently large we canguarantee that | c ′ − ab ′ + a ′ b | > r . It follows that d ( A − A ′ A, I ) ≥ R .So d ( A ′ , I ) = d ( A − A ′ A, I ), and thus d is not bi-invariant.(c) However, an example of a uniform group that is finite-dimensionalEuclidean is the group of unitary n × n -matrices. That follows eitherfrom the fact that this group is compact, or one can check thatthe Hilbert-Schmidt norm there induces a compatible bi-invariantmetric.(d) Let G be a group that acts faithfully by isometries on a boundedmetric space ( M, d M ). The following is then a bi-invariant metric M. DOUCHA (that makes G a topological uniform group): for g, h ∈ G set d ( g, h ) = sup x ∈ M d M ( gx, hx ) . Every group with bi-invariant metric is of this form. Indeed, if G is a group with bi-invariant metric d G , then G has a faithfullaction by isometries on itself induced by left translations. Theformula above defining a metric gives the original metric d G . Definition 1.3 (Enflo, [5]) . A uniform group G is Banach if it isuniformly homeomorphic to some Banach space.The preceding definition is readily checked to be equivalent with thatone mentioned in the introduction; i.e. a Banach space with additionalgroup structure, where the additional group operations are uniformlycontinuous with respect to norm, and the additional group unit coin-cides with the Banach space 0.Recall the canonical example of a uniform Banach group given bysome uniform homeomorphism between Banach spaces that was alsomentioned in the introduction. Let us mention here the result of Enflowhich says that under some conditions the converse is true. We refer to[4] and [5] for unexplained notions from the statement of the theorem.
Theorem 1.4 (Enflo, [5]) . Let G be a commutative uniform Banachgroup such that the corresponding Banach space has roundness p > and such that moreover: • G is uniformly dissipative, • for every x , x , y ∈ G we have k x · y − x · y k = o ( k x − x k /p ) uniformly in x , x , y as k x − x k → .Then G is isomorphic to the additive group of some Banach space. Preliminary discussion of the proof.
Before proceding to theproof of Theorem 0.1, let us roughly explain the ideas behind it. Weshall construct a countable set X equipped with two group structures.Under the first group structure, X is isomorphic to the free group ofcountably many generators. Under the second one, X is isomorphic tothe minimal subgroup of the real vector space with countable Hamelbasis that contains the elements of this Hamel basis and is closed undermultiplication by scalars that are dyadic rationals. Moreover, X willget equipped with a metric which is bi-invariant with respect to thenon-commutative group operation and which behaves like a norm withrespect to the latter group operation. In particular, the completion of X with respect to this metric will become a real Banach space.Let us mention two ‘peculiarities’ of the construction. First, X isconstructed so that there is no connection between these two algebraic ON-COMMUTATIVE UNIFORM BANACH GROUP 5 structures (i.e. non-commutative group structure and commutativegroup structure with dyadic scalar multiplication) in the sense that forany x, y ∈ X , the elements x, y, x · y, x − are linearly independent (inthe commutative structure), and similarly, for any x, y ∈ X and dyadicrationals α, β we have that αx + βy is a new free group generator (inthe non-commutative structure).The second thing to mention is that the metric on X is constructedby induction on finite fragments X n , n ∈ N , of X . This will give us abetter control of the metric we are defining.The construction is inspired and in some sense analogous to theconstruction of Lipschitz-free Banach spaces (we refer the reader tothe book [11] as the main reference on this subject) and also to theconstruction of free groups with the Graev metric (see [7]). Indeed, theuniform Banach group we construct can be viewed as a free object in anappropriate category and we refer the reader to the last section of thepaper where this is discussed. We also discuss there in more detail thesimilarity between our Banach group and Lipschitz-free Banach spacesand groups with the Graev metric.1.2. Some notation regarding free groups and vector spaces.
Let S be some non-empty set and let w be a word over the alphabet S ∪ S − ∪ { e } , where S − = { s − : s ∈ S } is a disjoint copy of S interpreted as a set of inverses of elements from S and e is an elementnot belonging to either S or S − which shall be interpreted as a groupunit. We say that w = w . . . w n is irreducible if either n = 1 and w = w = e , or n > i ≤ n , w i = e and there is no i < n such that w i = w − i +1 . If w is a word that is not irreducible,then by w ′ we shall denote the unique irreducible word obtained from w by deleting each occurence of the letter e and each occurence ofneighbouring letters a and a − (if this procedure leads to an emptyword, then we set w ′ to be e ).It is well-known and easy to observe that elements of F ( S ), the freegroup of free generators coming from the set S with e as a unit, arein one-to-one correspondence with irreducible words over the alphabet S ∪ S − ∪ { e } .Let n ∈ N . By W n ( S ) we shall denote the set of all irreducible wordsof length at most n over the alphabet S ∪ S − ∪ { e } .Let now similarly B be some non-empty set not containing the dis-tinguished element e . The vector space over some field F with B asthe maximal linearly independent set and e representing zero can beviewed as a set of all functions from B to K that have finite support, M. DOUCHA where f has finite support if for all but finitely many b ∈ B we have f ( b ) = 0.In our case, we shall work with F = R (resp. Q ), however since weshall need to work with only finitely many vectors at any given timewe restrict to functions whose range is some specified finite subset of R (also, the set B will be always finite).Let K ⊆ fin R some finite subset of reals. Then by V K ( B ) we shalldenote the set of all functions from B to K with finite support. Therequirement on finite support will be in our construction superfluoussince B will be at any given time finite as already mentioned.2. The proof of Theorem 0.1
We start by describing the underlying countable set X mentionedabove in the preliminaries.2.1. The underlying dense set.
We now describe a countably infi-nite set X , constructed as an increasing union of finite sets X ⊆ X ⊆ . . . , which will also carry a multiplicative group operation and the cor-responding group inverse operation so that X will be isomorphic to afree group of countably many generators, and it will also carry an addi-tive (abelian) group operation together with multiplication by scalarsthat are dyadic rationals so that it is a proper subgroup of an infinitedimensional vector space over the rationals. Moreover, the unit foraddition and multiplication will be the same.Later, we shall define a metric on X that will be invariant underboth addition and multiplication and will preserve scalar multiplica-tion. The completion then will be a Banach space over the reals whichis also equipped with multiplication with free group as a dense part.Let n ≥ D n we shall denote the set of dyadicrationals a n , where a ∈ [ − n , n ]. Clearly, D = S n D n is the set ofall dyadic rational numbers.We set X = { e } . Let S = { x } be some singleton. We set X = W ( S ) = { e, x, x − } . Let B = X \ X = { x, x − } . We set X = V D ( B ) = { αx + βx − : α, β ∈ D } .Suppose now that X n has been constructed. We need to construct X n +1 and X n +2 . Set S n +1 to be S n − ∪ ( X n \ X n − ). Then we set X n +1 to be W n +1 ( S n +1 ).Next we set B n +2 to be B n ∪ ( X n +1 \ X n ). Then we set X n +2 tobe V D n +1 ( B n +2 ). ON-COMMUTATIVE UNIFORM BANACH GROUP 7
This finishes the inductive construction. Note that X = S n X n = S n W n +1 ( S n +1 ) = S n V D n +1 ( B n +2 ). It follows that if we set S = S n S n +1 and B = S n B n +2 , then X is also naturally isomorphic to F ( S ), the free group of countably many generators coming from the set S , and it is a (additive) proper subgroup of the rational (or real) vectorspace with B as the maximal linearly independent set - the minimalsubgroup that contains free abelian group with B as a set of generatorsthat is closed under multiplication by scalars from D .We define inductively a rank function r : X → ω . For any x ∈ X weset r ( x ) = 0 if neither x nor x − is possible to write as α y + . . . + α m y m ,where m > α i >
0, for every i , and α + . . . + α m = 1.If x or x − is possible to write as α y + . . . + α m y m , where m > α i >
0, for every i , and α + . . . + α m = 1, then we set r ( x ) = max { r ( y i ) : i ≤ m } + 1.2.2. Construction of the metric.
We shall now define a metric ρ and a norm k · k on X . Actually, the metric and the norm will beone and the same in the sense that for any x, y ∈ X we shall have ρ ( x, y ) = k x − y k and k x k = ρ ( x, e ). The distinguishing is done onlyfor practical notational reasons since we understand X as both a freegroup and a subgroup of a vector space. In the former case, it is morenatural to consider a metric there, while in the latter to consider anorm there. By induction, we shall define functions ρ n : X n → R (therange will actually be a subset of non-negative rationals) for odd n and functions k · k n : X n → R for even n that satisfy the followingproperties:(1) for every odd n we have that ρ n is a symmetric function thatis equal to zero only on diagonal, i.e. ρ n ( x, y ) = 0 iff x = y for x, y ∈ X n ; similarly, for every even n we have that k x k n = 0 iff x = e for x ∈ X n ,(2) for every even n , k · k n extends ρ n − , i.e. for every a, b ∈ X n − we have k a − b k n = ρ n − ( a, b ) , similarly for every odd n , ρ n extends k · k n − , i.e. for every a, b ∈ X n − such that a − b ∈ X n − we have ρ n ( a, b ) = k a − b k n − , (3) for every odd n , for any words (not necessarily irreducible) w , w , v , v over the alphabet S n ∪ S − n ∪ { e } such that M. DOUCHA w ′ , w ′ , v ′ , v ′ , ( w w ) ′ , ( v v ) ′ ∈ X n , we have ρ n (( w w ) ′ , ( v v ) ′ ) ≤ ρ n ( w ′ , v ′ ) + ρ n ( w ′ , v ′ ) , and for every a, b ∈ X n we have ρ n ( a, b ) = ρ n ( a − , b − ) , (4) for every odd n and for every a ∈ X n and for every b ∈ X n suchthat b = α c + . . . + α m c m , where m > α i ≥ c i ∈ B n − ,for every i , and α + . . . + α m = 1, we have ρ n ( a, b ) ≤ α · ρ n ( a, c ) + . . . + α m · ρ n ( a, c m ) , (5) for every even n and for every a ∈ X n and any scalar α suchthat αa ∈ X n we have k αa k n = | α | · k a k n , and for every a, b ∈ X n such that also a + b ∈ X n we have k a + b k n ≤ k a k n + k b k n , (6) for every odd n , for every a ∈ X n and for every b ∈ X n suchthat b = ( α c + . . . + α m c m ) − , where, m > α i ≥ c i ∈ B n − , for every i , and α + . . . + α m = 1, we have ρ n ( a, b ) ≤ α · ρ n ( a, c − ) + . . . + α m · ρ n ( a, c − m ) , and similarly, for every even n , for every a ∈ X n and for every b ∈ X n − such that b = ( α c + . . . + α m c m ) − , where m > α i ≥ c i ∈ B n − , for every i , and α + . . . + α m = 1, wehave k a − b k n ≤ α · k a − c − k n + . . . + α m · k a − c − m k n . We define ρ on X by setting ρ ( x, e ) = ρ ( x − , e ) = 1 and ρ ( x, x − ) =2. Obviously, this satisfies all the requirements.Let us now define k · k on X , i.e. we have to define k αx + βx − k for every α, β ∈ D . We set k αx + βx − k = min {| γ | · ρ ( x, e ) + | γ | · ρ ( x − , e ) + | γ | · ρ ( x, x − ) : αx + βx − = γ x + γ x − + γ ( x − x − ) , γ , γ , γ ∈ R } . Condition (6) is automatically satisfied as there is no b ∈ X of theform ( α c + . . . + α m c m ) − for appropriate α ’s and c ’s. Condition (1)is also obvious. Thus we need to check the conditions (2) and (5).Let us do the former. We need to check that k · k extends ρ .We shall check that k x k = ρ ( x, e ). The other cases are similar.Clearly, k x k ≤ ρ ( x, e ). Suppose that k x k = | γ | · ρ ( x, e ) + | γ | · ON-COMMUTATIVE UNIFORM BANACH GROUP 9 ρ ( x − , e ) + | γ | · ρ ( x, x − ), where x = γ x + γ x − + γ ( x − x − ). Itfollows that necessarily γ = γ and γ + γ = 1. By triangle inequalitywe have | γ | · ρ ( x − , e ) + | γ = γ | · ρ ( x, x − ) ≥ | γ | · ρ ( x, e ), thus | γ | · ρ ( x, e ) + | γ | · ρ ( x − , e ) + | γ | · ρ ( x, x − ) ≥ | γ | · ρ ( x, e ) + | γ | · ρ ( x, e ) ≥ | γ + γ | · ρ ( x, e ) = ρ ( x, e ).We now check (5). Fix some a ∈ X and α such that αa ∈ X .Suppose that k a k = | γ | · ρ ( x, e ) + | γ | · ρ ( x − , e ) + | γ | · ρ ( x, x − ),where we have a = γ x + γ x − + γ ( x − x − ). Then since αa = α · γ x + α · γ x − + α · γ ( x − x − ) we have that k αa k ≤ | αγ | · ρ ( x, e ) + | αγ | · ρ ( x − , e ) + | αγ | · ρ ( x, x − ) = | α | · k a k . The other inequalityis analogous. By a similar argument one can also show that for any a, b ∈ X such that also a + b ∈ X we have k a + b k ≤ k a k + k b k andwe leave this to the reader. In fact, we note here that this definitionof k · k is equivalent with that one that says that k · k is the great-est function that satisfies condition (4) and such that k x k ≤ ρ ( x, e ), k x − k ≤ ρ ( x − , e ) and k x − x − k ≤ ρ ( x, x − ). Extending the metric.
Suppose we have defined k · k n for some even n ≥
2. We now define ρ n +1 on X n +1 . First we inductively define an auxiliary function δ on X n +1 . Fix a pair x, y ∈ X n +1 . If r ( x ) = r ( y ) = 0, then we set δ ( x, y ) = min {k a − b k n + . . . + k a m − b m k n : x = ( a . . . a m ) ′ , y = ( b . . . b m ) ′ , ∀ i ≤ m ( a i , b i , a i − b i ∈ X n ) } . Note that the minimum is indeed attained as X n is finite. Note againthat for any z ∈ X n +1 if r ( z ) > z or z − belongsto X n . So if r ( x ) > r ( y ) >
0, then we set δ ( x, y ) = k x − y k n if x, y ∈ X n k x − − y − k n if x − , y − ∈ X n min {k x − z k + k z − − y − k : z, z − ∈ X n } if x, y − ∈ X n min {k x − − z − k + k z − y k : z, z − ∈ X n } if x − , y ∈ X n . Now we suppose that for one of the elements, say x , we have r ( x ) = 0,and for the other one we have r ( y ) >
0. The following is done byinduction on r ( y ). First suppose that y = α z + . . . + α m z m , where m > α i >
0, for all i , and α + . . . + α m = 1. Then we set δ ( x, y ) = α · δ ( x, z ) + . . . + α m · δ ( x, z m ) . Similarly, if y = ( α z + . . . + α m z m ) − , where m > α i >
0, for all i ,and α + . . . + α m = 1, then we set δ ( x, y ) = α · δ ( x, z − ) + . . . + α m · δ ( x, z − m ) . We are now ready to define ρ n +1 . Thus fix now again a pair x, y ∈ X n +1 and set ρ n +1 ( x, y ) = min { δ ( a , b ) + . . . + δ ( a m , b m ) : x = ( a . . . a m ) ′ , y = ( b . . . b m ) ′ , a , . . . , a m , b , . . . , b m ∈ X n +1 } . First notice that since X n +1 is finite the minimum is attained, thusin particular we have for x = y that ρ n +1 ( x, y ) >
0. Since ρ n +1 isclearly symmetric we get it satisfies the condition (1).We claim that ρ n +1 is the greatest function satisfying:(a) ρ n +1 ( x, y ) ≤ k x − y k n for every x, y ∈ X n such that x − y ∈ X n ,(b) ρ n +1 ( x, y ) = ρ n +1 ( x − , y − ) for every x, y ∈ X n +1 ,(c) ρ n +1 ( ab, cd ) ≤ ρ n +1 ( a, c ) + ρ n +1 ( b, d ) for every a, b, c, d ∈ X n +1 suchthat ab, cd ∈ X n +1 ,(d) ρ n +1 ( x, ( α z + . . . + α m z m )) ≤ α · ρ n +1 ( x, z )+ . . . + α m · ρ n +1 ( x, z m ),where m > α i >
0, for all i , α + . . . + α m = 1 and α z + . . . + α m z m ∈ X n +1 ,(e) ρ n +1 ( x, ( α z + . . . + α m z m ) − ) ≤ α · ρ n +1 ( x, z − ) + . . . + α m · ρ n +1 ( x, z − m ), where m > α i >
0, for all i , α + . . . + α m = 1 and( α z + . . . + α m z m ) − ∈ X n +1 .First of all, it is clear from the definitions of ρ n +1 (and of δ ) that ρ n +1 satisfies all these conditions. Thus in particular, we get that ρ n +1 satisfies the conditions (3),(4) and (6). Next, if ξ is any other functionsatisfying conditions (a)-(e), then it is readily checked that ξ ≤ δ andbecause of (c) also ξ ≤ ρ n +1 .We shall now conclude from that that ρ n +1 also satisfies (2). Indeed,let X ′ n ⊆ X n be such that for every x, y ∈ X ′ n we have x − y ∈ X n .Then consider the metric ξ on X ′ n defined as ξ ( x, y ) = k x − y k n . Weclaim it satisfies the conditions (a)-(e) above. Condition (a) is satisfiedsince ξ ( x, y ) = k x − y k n for appropriate x, y . Take some x, y ∈ X ′ n such that x − , y − ∈ X ′ n . Necessarily x, y, x − , y − ∈ X n − and since k · k n extends ρ n − we get ξ ( x, y ) = ρ n − ( x, y ) = ρ n − ( x − , y − ) = ξ ( x − , y − ). Take now some a, b, c, d ∈ X ′ n such that ab, cd ∈ X ′ n .We again necessarily have that a, b, c, d, ab, cd ∈ X n − and since k · k n extends ρ n − we again obtain ξ ( ab, cd ) = ρ n − ( ab, cd ) ≤ ρ n − ( a, c ) + ρ n − ( b, d ) = ξ ( a, c ) + ξ ( b, d ). We have verified conditions (b) and (c).Condition (d) follows since k·k n satisfies the condition (5) further above.Finally, take some x, ( α z + . . . + α m z m ) − ∈ X ′ n , where m > α i > i , α + . . . + α m = 1. We have ξ ( x, ( α z + . . . + α m z m ) − ) = k x − ( α z + . . . + α m z m ) − k n ≤ α · k x − z − k n + . . . + α m · k x − z m k n = ON-COMMUTATIVE UNIFORM BANACH GROUP 11 α · ξ ( x, z ) + . . . + α m · ξ ( x, z m ), where the middle inequality followsfrom the property (6) above.We thus get k x − y k n = ξ ( x, y ) ≤ ρ n +1 ( x, y ) for every x, y ∈ X n suchthat x − y ∈ X n . Since by assumption ρ n +1 ( x, y ) ≤ k x − y k n we get ρ n +1 ( x, y ) = k x − y k n and we are done. Extending the norm.
Now suppose we have defined ρ n on X n for some odd n >
2. We define k · k n +1 on X n +1 . We again at first define, inductively, an auxiliaryfunction γ : X n +1 → R . First, for every x, y ∈ X n we set γ ( x − y ) = ρ n ( x, y ) . Next, for every x, y ∈ X n +1 such that x − y ∈ X n +1 and x ∈ X n +1 \ X n we define γ ( x − y ) by induction on r ( y ). If r ( y ) = 0 then we set γ ( x − y ) = min {| β | · γ ( v ) + . . . + | β i | · γ ( v i ) : x − y = β v + . . . + β i v i , ∀ j ≤ i ∃ a j , b j ∈ X n ( v j = a j − b j ) } . If r ( y ) > y = ( α z + . . . + α m z m ) − , where m > α i >
0, forevery i , and α + . . . + α m = 1, then we set γ ( x − y ) = α · γ ( x − z − ) + . . . + α m · γ ( x − z − m ) . Finally, for any x ∈ X n +1 we set k x k n +1 = min { γ ( y ) + . . . + γ ( y i ) : x = y + . . . + y i , y , . . . , y i ∈ X n +1 } . First thing to observe is again that for any x ∈ X n +1 we have k x k n +1 = 0 iff x = e . It follows that condition (1) is satisfied for k · k n +1 .Next we claim that k · k n +1 is the greatest function satisfying:(a) k x − y k n +1 ≤ ρ n ( x, y ) for every x, y ∈ X n ,(b) k αx + βy k n +1 ≤ k αx k n +1 + k βy k n +1 = | α | · k x k n +1 + | β | · k y k n +1 for x, y ∈ X n +1 such that also αx + βy ∈ X n +1 ,(c) k x − ( α z + . . . + α m z m ) − k n +1 ≤ α · k x − z − k n +1 + . . . + α m · k x − z − m k n +1 , where m > α i >
0, for every i , and α + . . . + α m = 1.First, it follows from the definitions of γ and k · k n +1 that k · k n +1 satisfies all these conditions, thus we have that it satisfies the condi-tions (5) and (6) further above. If ξ is any other functions satisfyingconditions (a)-(c) then it again follows that necessarily ξ ≤ γ and thusalso ξ ≤ k · k n +1 .We are ready to verify the remaining condition (2) for k · k n +1 that itextends ρ n . Let X ′ n +1 ⊆ X n +1 be the set { x − y : x, y ∈ X n } . Define ξ on X ′ n +1 as follows: ξ ( x − y ) = ρ n ( x, y ) for every x − y ∈ X ′ n +1 . Then ξ satisfies (a) since ξ ( x − y ) = ρ n ( x, y ) for appropriate x, y . Next we check (b). Take some αx + βy . If α = 1 and β = − ξ ( x − y ) = ρ n ( x, y ) ≤ ρ n ( x, e ) + ρ n ( e, y ) = ξ ( x ) + ξ ( y ), where themiddle inequality follows from the condition (3) that ρ n satisfies (whichimplies the triangle inequality). Finally, for ( α z + . . . + α m z m ) − ,where m > α i >
0, for every i , and α + . . . + α m = 1, wehave ξ ( x − ( α z + . . . + α m z m ) − ) = ρ n ( x, ( α z + . . . + α m z m ) − ) ≤ α · ρ n ( x, z − )+ . . . + α m · ρ n ( x, z − m ) = α · ξ ( x − z − )+ . . . + α m · ξ ( x − z − m ),since ρ n satisfies the condition (6), and we are done.Now we can define k · k on X by putting k · k = [ i k · k i and analogously we can define ρ on X by putting ρ = [ i ρ i . By the condition (2) we have that for any x, y ∈ X we have k x − y k = ρ ( x, y ). By (1) we have that for x = y ∈ X we have ρ ( x, y ) > x = e ∈ X we have k x k > ρ is a bi-invariant metric on X when considered asa (free) group. By (1) is symmetric. We use the following simple fact. Fact 2.1.
Let G be a group equipped with a symmetric function d : G → R +0 that is equal to only on diagonal. Then d is a bi-invariantmetric if and only if for every x, y, v, w ∈ G we have d ( x · y, v · w ) ≤ d ( x, v ) + d ( y, w ) .Proof. If d is a bi-invariant metric then the inequality readily followsfrom bi-invariance and using triangle inequality.So suppose that d satisfies such an inequality for every x, y, v, w ∈ G .Fix a, b, c ∈ G . Then d ( a, c ) = d ( a · b − · b, b · b − · c ) ≤ d ( a, b ) + d ( b − , b − )+ d ( b, c ) = d ( a, b )+ d ( b, c ), so d is a metric. Now d ( a · b, a · c ) ≤ d ( a, a )+ d ( b, c ) = d ( b, c ) = d ( a − · a · b, a − · a · c ) ≤ d ( a − , a − )+ d ( a · b, a · c ) = d ( a · b, a · c ) which shows the left-invariance. The right invarianceis done analogously. (cid:3) However, ρ does satisfy the condition from the statement of the factsince it satisfies the condition (3).Similarly, for every x, y and α, β ∈ D (recall that D denotes thedyadic rationals) we have k αx + βy k ≤ k αx k + k βy k = | α |·k x k + | β |·k y k since k · k satisfies the condition (5). ON-COMMUTATIVE UNIFORM BANACH GROUP 13
Denote now by X the completion of X with respect to ρ , or equiva-lently, with respect to k · k . Both the multiplicative and additive groupoperations extend to the completion as well as the scalar multiplicationby dyadic rationals. Moreover, since the dyadic rationals are dense in R , X has well-defined scalar multiplication by all the reals. Thus X isa Banach space.3. Concluding discussion and questions
Let us comment on similarities between the constructions of thepresented Banach group and Lipschitz-free Banach spaces. For anypointed metric space ( X,
0) there is a Banach space LF ( X ), calledLipschitz-free space (or Arens-Eells space) over ( X, X \ { } as the Hamel basis and the point 0 as the Banachspace zero. LF ( X ) contains X isometrically and the norm of LF ( X )is uniquely described as the largest norm on the linear span of X thatextends the metric on X .By similar means, one can define the largest bi-invariant metric on F X \{ } , the free group having 0 as a neutral element and X \ { } as theset of free generators, that extends the metric on X .The following generalization of the Lipschitz-free construction wasgiven in [1]. If Y is a Banach space and Y ⊆ X is a metric extension of Y such that for every x ∈ X the distance function dist( x, · ) : Y → R isconvex, then there is a free space LF Y ( X ) which has Y as a subspace,contains X as a subset so that the elements of X \ Y are linearlyindependent, and the norm on LF Y ( X ) is the largest norm on such avector space that extends the metric on X .An analogous generalization of the Graev metric on groups was givenin [3]; i.e. for any group G with bi-invariant metric and any metricextension G ⊆ X such that G is closed in X there exists the largestbi-invariant metric on the free product G ∗ F X \ G .Roughly speaking, the idea behind our construction from the presentpaper was to alternatively use those two constructions. That means,to start with some pointed metric space ( X, LF ( X ), then the free group F LF ( X ) with the Graevmetric, then LF LF ( X ) ( F LF ( X ) ), etc., and at the end to take the comple-tion. The reason why this approach does not literally work is that for ageneral point x in F LF ( X ) , the distance function dist( x, · ) : LF ( X ) → R is not convex. The remedy was to use ‘finitary’ versions of the con-structions above where we were able to guarantee the convexity of theappropriate distance functions. Freeness of the Banach group.
We shall argue that although X as a Banach space is likely not one of the “well-behaved” spaces itis not completely random either. It can be uniquely characterized asa free one-generated uniform Banach group whose metric induced bythe norm is bi-invariant. That can be described via certain universalproperty. Let us at first define morphisms in the category of uniformBanach groups. The natural definition seems to be a bounded linearoperator that is moreover a group homomorphism in the additionalgroup structure. Note that it is then automatically uniformly continu-ous group homomorphism.We claim that: Theorem 3.1. X is the free uniform Banach group over one generatorwhose metric induced by the norm is bi-invariant. That means, forany uniform Banach group Y whose metric induced by the norm is bi-invariant, and any element y ∈ Y there exists a unique uniform Banachgroup morphism φ : X → Y such that φ ( x ) = y and k φ k = k y k Y .This property characterizes X uniquely up to isomorphism.Remark . Note that every uniform Banach group Z admits a bi-invariant metric that is uniformly continuous with respect to norm.For any y, z ∈ Z set D ( y, z ) = sup v,w ∈ Z k v · y · w − v · z · w k Z . It isalways satisfied that D ( y, z ) ≥ k y − z k and the inequality may be strictin general. Proof. If y = 0 then φ is the zero morphism and there is nothing toprove. So we assume that y = 0.Let us take a closer look on the countable dense set X ⊆ X . Onecan see that each element of X is obtained in a unique way from theelement x (the only element of S ) using operations of addition +,multiplication · , additive and multiplicative inverses − , − and by scalarmultiplication by dyadic rationals D . It follows that there is a uniquemap φ ′ : X → Y that preserves those operations and satisfies φ ′ ( x ) = y .We need to show that for every z ∈ X we have(3.1) k φ ′ ( z ) k Y ≤ k y k Y k z k X . Once this is proved we can (uniquely) extend φ ′ to the completion of X to obtain φ ⊇ φ ′ : X → Y . φ is still linear and preserves the additionalgroup structure. By 3.1, we get that k φ k ≤ k y k Y . On the other hand, φ ( x ) = y , thus k φ k = k y k Y .To simplify the notation, we shall without loss of generality assumethat k y k Y = 1. ON-COMMUTATIVE UNIFORM BANACH GROUP 15
We consider the following pseudometric σ , resp. pseudonorm k| · k| ,on X . For any y, z ∈ X we set σ ( y, z ) = k φ ′ ( y ) − φ ′ ( z ) k Y ; similarly, forevery y ∈ X we set k| y k| = k φ ′ ( y ) k Y . For every odd n we denote by σ n the restriction of σ to X n . Similarly, for every even n we denote by k| · k| n the restriction of k| · k| to X n . It suffices to show, by induction,that for every odd n we have σ n ≤ ρ n and for every even n we have k| · k| n ≤ k · k n .Consider the case n = 1. We have (from bi-invariance) σ ( x, e ) = σ ( x − , e ) = ρ ( x, e ) = ρ ( x − , e ) = 1. It follows from the triangleinequality that σ ( x, x − ) ≤ ρ ( x, x − ).Consider now the case n = 2. Take any α, β ∈ D . We have k αx + βx − k = min {| γ | · ρ ( x, e ) + | γ | · ρ ( x − , e ) + | γ | · ρ ( x, x − ) : αx + βx − = γ x + γ x − + γ ( x − x − ) , γ , γ , γ ∈ R } . However, for any γ , γ , γ ∈ R if αx + βx − = γ x + γ x − + γ ( x − x − ), then by thesubadditivity and homogeneity of the norm we must have k| αx + βx − k| = ≤ | γ | · σ ( x, e ) + | γ | · σ ( x − , e ) + | γ | · σ ( x, x − ) ≤| γ | · ρ ( x, e ) + | γ | · ρ ( x − , e ) + | γ | · ρ ( x, x − ) . Thus k| · k| ≤ k · k .Consider now some general odd n . Necessarily, σ n +1 satisfies all thefollowing inequalities (recall the analogous inequalities for ρ n +1 )(a) σ n +1 ( x, y ) ≤ k| x − y k| n for every x, y ∈ X n such that x − y ∈ X n ,(b) σ n +1 ( x, y ) = σ n +1 ( x − , y − ) for every x, y ∈ X n +1 ,(c) σ n +1 ( ab, cd ) ≤ σ n +1 ( a, c ) + σ n +1 ( b, d ) for every a, b, c, d ∈ X n +1 suchthat ab, cd ∈ X n +1 ,(d) σ n +1 ( x, ( α z + . . . + α m z m )) ≤ α · σ n +1 ( x, z )+ . . . + α m · σ n +1 ( x, z m ),where m > α i >
0, for all i , α + . . . + α m = 1 and α z + . . . + α m z m ∈ X n +1 ,(e) σ n +1 ( x, ( α z + . . . + α m z m ) − ) ≤ α · σ n +1 ( x, z − ) + . . . + α m · ρ n +1 ( x, z − m ), where m > α i >
0, for all i , α + . . . + α m = 1 and( α z + . . . + α m z m ) − ∈ X n +1 .Inequalities (a),(b) and (c) are clear; (d) follows since σ n +1 ( x, ( α z + . . . + α m z m )) = k φ ′ ( x − ( α z + . . . + α m z m )) k Y ≤ α k φ ′ ( x − z ) k Y + . . . + α m k φ ′ ( x − z m ) k Y = α · σ n +1 ( x, z ) + . . . + α m · σ n +1 ( x, z m ); (e)follows analogously.Since ρ n +1 has been shown to be the greatest function satisfying theanalogous conditions (a)-(e), it follows from the induction hypothesisthat σ n +1 ≤ ρ n +1 .Finally, for a general even n , a completely analogous argumentation,using the fact that k · k n +1 was the greatest function satisfying certain conditions, gives that k| · k| n +1 ≤ k · k n +1 .The uniqueness is a standard argument using the universality prop-erty. (cid:3) As mentioned above, one can rightfully suspect that X is not one ofthe well-behaved Banach spaces. Actually, in our opinion, it would notbe difficult to enhance the construction above so that X was isometricto the Gurarij space ([8]). It thus seems natural to ask whether onecan get a non-commutative Banach group modeled on a ‘reasonable’space.We mentioned in the preliminary section that the Heisenberg group U T ( R ), modeled on a three-dimensional Banach space, is not a uni-form Banach group, although it is very closed to it (this is also thecontent of Remark 5.2. in [9]). The following question thus arises. Question 3.3.
Does there exist a finite-dimensional non-commutativeuniform Banach group?
We expect the answer to the previous question to be negative. How-ever, one can then ask:
Question 3.4.
Does there exist a non-commutative uniform Banachgroup modeled on a Hilbert space?
As already mentioned in the introduction, uniform Banach groupswere originally introduced with connection to the infinite-dimensionalHilbert’s fifth problem. The following question thus also seems to benatural.
Question 3.5.
Are the (non-commutative) group operations on X (Fr´echet)differentiable? Does there exist a non-commutative uniform Banachgroup that is a Banach-Lie group? References [1] I. Ben-Yaacov,
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Institute of Mathematics, Polish Academy of Sciences, 00-656 Warszawa,Poland
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