Infinitesimal generators and quasi non-archimedean topological groups
aa r X i v : . [ m a t h . G R ] M a r Infinitesimal topological generators and quasinon-archimedean topological groups
Tsachik Gelander ∗ and François Le Maître † Abstract
We show that connected separable locally compact groups are infinitesimallyfinitely generated, meaning that there is an integer n such that every neighborhoodof the identity contains n elements generating a dense subgroup. We generalize atheorem of Schreier and Ulam by showing that any separable connected compactgroup is infinitesimally -generated.Inspired by a result of Kechris, we introduce the notion of a quasi non-archimedeangroup. We observe that full groups are quasi non-archimedean, and that every con-tinuous homomorphism from an infinitesimally finitely generated group into a quasinon-archimedean group is trivial. We prove that a locally compact group is quasinon-archimedean if and only if it is totally disconnected, and provide various exam-ples which show that the picture is much richer for Polish groups. In particular, weget an example of a Polish group which is infinitesimally -generated but totally dis-connected, strengthening Stevens’ negative answer to Problem 160 from the Scottishbook. One of the simplest invariants one can come up with for a topological group G is itstopological rank t ( G ) , that is, the minimum number of elements needed to generate adense subgroup of G . For this invariant to have a chance to be finite, one needs to assumethat G is separable since every finitely generated group is countable.The oldest result in this topic is probably due to Kronecker in 1884 [Kro31], and saysthat an n -tuple ( a , ..., a n ) of real numbers projects down to a topological generator of T n = R n / Z n if and only if (1 , a , ..., a n ) is Q -linearly independent. Since such n -tuplesalways exist, the topological rank of the compact connected abelian group T n is equal toone.Using this result, it is then an amusing exercise to show that t ( R n ) = n +1 . This meansthat as a topological group, R n remembers its vector space dimension. In particular, wesee that the topological rank sometimes contains useful information (another somehowsimilar instance of this phenomenon was recently discovered by the second-named authorfor full groups, see [LM14]).In order to deal with general connected Lie groups, it is useful to introduce the fol-lowing definition. ∗ Research supported by the ISF, Moked grant 2095/15. † Research supported by the Interuniversity Attraction Pole DYGEST and Projet ANR-14-CE25-0004GAMME. efinition 1.1. The infinitesimal rank of a topological group G is the minimum n ∈ N ∪ { + ∞} such that every neighborhood of the identity in G contains n elements g , ..., g n which generate a dense subgroup of G . We denote it by t I ( G ) .One can easily show that t I ( R n ) = t ( R n ) = n + 1 . This is relevant for the study ofthe infinitesimal rank of real Lie groups because if we know that the Lie algebra of aconnected Lie group G is generated (as a Lie algebra) by n elements, then using the factfact that t I ( R ) = 2 we can deduce that t I ( G ) n (see the well-known Lemma 4.2).For various classes of connected Lie groups one can say more. In particular this is thesituation for compact connected Lie groups, in which case a uniform result is useful in viewof the important role of compact groups in the structure theory of locally compact groups.Auerbach showed that for every compact connected Lie group G , one has t I ( G ) [Aue34]. Moreover, he could show that the set of pairs of topological generators of G hasfull measure, which then led Schreier and Ulam to the following general result, for whichwe will include a short proof. Theorem 1.2 (Schreier-Ulam, [SU35]) . Let G be a connected compact metrisable group.Then almost every pair of elements of G generates a dense subgroup of G . In particular, t I ( G ) = t ( G ) = 2 for any non-abelian such G . Note that there is a vast area between metrizable compact groups and separable ones;for instance ( T ) R is compact separable, but not metrizable. Conversely, being separableis a minimal assumption for a group to have finite topological rank. Going back tothe n -torus, Kronecker’s result admits a far-reaching generalization due to Halmos andSamelson, which settles the abelian case. Theorem 1.3 (Halmos-Samelson, [HS42, Corollary]) . The topological rank of every con-nected separable compact abelian group is equal to one.
Our first result is a generalization of Schreier and Ulam’s Theorem to the separablecompact case. One of the difficulties is that we cannot use their Haar measure argumentanymore.
Theorem 1.4 (see Theorem 3.1) . Let G be a separable compact connected group. If G isnon-abelian, then t I ( G ) = 2 , and if G is abelian, then t I ( G ) = 1 . The fact that the topological rank t ( G ) = 2 was proved by Hoffmann and Morris[HM90, Theorem. 4.13]. Note that it follows that t I ( G ) = t ( G ) for separable compactconnected group.It is natural to ask if the following stronger statement is true: Question 1.5.
Let G be a separable compact connected group. Is the set of pairs in G which topologically generate G necessarily of full measure in G × G ?Say that a topological group is infinitesimally finitely generated if it has finiteinfinitesimal rank. Using the previous result, and the solution to Hilbert’s fifth problem,we establish the following result, which was noted by Schreier and Ulam in the abeliancase (see the last paragraph of [SU35]). Theorem 1.6 (see Theorem. 4.1) . Let G be a separable locally compact group. Then G is infinitesimally finitely generated if and only if G is connected. Polish groups , i.e. separable topological groups whose topology admits a compatiblemetric. These groups abound in analysis, for instance the unitary group of a separableHilbert space or the group of measure-preserving transformations of a standard probabilityspace are Polish groups. Moreover, they form a robust class of groups, e.g. every countableproduct of Polish groups is Polish (see [Gao09] for other properties of this flavour). Recallthat a locally compact group is Polish if and only if it is second-countable.It is not hard to show that Theorem 1.6 fails for Polish groups: for instance R N isconnected but not topologically finitely generated, in particular it is not infinitesimallyfinitely generated. The question of the converse is more interesting, even for the followingweaker property. Definition 1.7.
A topological group G is infinitesimally generated if every neigh-bourhood of the identity generates G .Clearly every connected group is infinitesimally generated, and every infinitesimallyfinitely generated group is infinitesimally generated. Moreover it follows from van Dantzig’stheorem that every infinitesimally generated locally compact group is connected. Question 1.8 (Mazur’s Problem 160 [Mau81]) . Must an infinitesimally generated Polishgroup be connected?In [Ste86], Stevens exhibited the first examples of infinitesimally generated Polishgroup which are totally disconnected. We show that her examples actually have infinites-imal rank and then provide the following stronger negative answer to Question 1.8. Theorem 1.9 (see Theorem. 5.15) . There exists a Polish group of infinitesimal rank which is totally disconnected. Let us now introduce the quasi non-archimedean property which is a strong negationof being infinitesimally finitely generated.
Definition 1.10.
A topological group is quasi non-archimedean if for every neighbor-hood of the identity U in G and every n ∈ N , there exists a neighborhood of the identity V such that for every g , ..., g n ∈ V , the group generated by g , ..., g n is contained in U . Remark.
If we switch the quantifiers and ask for a V which works for every n ∈ N , it isnot hard to see that the definition then becomes that of a non-archimedean topologicalgroup (i.e. admitting a basis of neighborhoods of the identity made of open subgroups).Our inspiration for the above definition comes from the following result of Kechris:every continuous homomorphism from an infinitesimally finitely generated group into afull group is trivial (see the paragraph just before Section (E) of Chapter 4 in [Kec10]).We upgrade this by showing that every full group is quasi non-archimedean, and that anycontinuous homomorphism from an infinitesimally finitely generated group into a quasinon-archimedean group is trivial (see Proposition 5.5). For locally compact groups, weobtain the following characterisation. Theorem 1.11 (see Theorem. 5.8) . Let G be a separable locally compact group. Then G is quasi non-archimedean if and only if G is totally disconnected. -generated. Section 4 is devoted to the proof that every connected sepa-rable locally compact group is infinitesimally finitely generated. In Section 5 we introducequasi non-archimedean groups and study their basic properties. We also give numerousexamples, and show that a separable locally compact group is totally disconnected if andonly if it is quasi non-archimedean. Finally, a Polish group into which no non-discretelocally compact group can embed is built in Section 6, where we also ask three questionsraised by this work. Acknowledgements.
We warmly thank Pierre-Emmanuel Caprace for coming up withthe terminology “infinitesimally generated”, as well as for useful conversations around thistopic. We also thank Yves de Cornulier for his helpful remarks on a first version of thepaper.
We collect some results which will serve us in the preceding sections.
Proposition 2.1.
Let G be a connected locally compact group. Suppose that K is aprofinite normal subgroup of G such that G/K is infinitesimally finitely generated. Then t I ( G ) = t I ( G/K ) . Moreover, if ¯ g , . . . , ¯ g k topologically generate the group G/K , and g , . . . , g k are arbitrary respective lifts in G , then g , . . . , g k topologically generate G . The proof of Proposition 2.1 will rely on the following two lemmas:
Lemma 2.2.
Let H be a connected locally compact group and f : H ։ L a finite coveringmap. Let { l , . . . l k } be a topological generating set for L and pick h i ∈ f − ( l i ) , i = 1 , . . . , k arbitrarily. Then h , . . . , h k topologically generate H .Proof. Set F = h h , . . . , h k i . Note that f , being a finite cover, is a closed map, and hence f ( F ) is closed in L . Since l , . . . , l k ∈ f ( F ) we have f ( F ) = L . Thus Ker f · F = H .Hence by the Baire category theorem F has a non-empty interior. Since H is connected,this implies that F = H . Lemma 2.3.
Let G be a (connected) locally compact group admitting a pro-finite normalsubgroup K ⊳ G such that L = G/K is a Lie group. Then G is an inverse limit G = lim ←− L α of finite (central) extensions L α of L . Although the Lemma holds without the assumption that G is connected, since itsignificantly simplify the proof while being sufficient for our needs, we will prove it onlyunder the connectedness assumption. 4 roof. Given k ∈ K , the image of the orbit map G → K, g gkg − is at the same timeconnected, since G is connected and the map is continuous, and totally disconnected asthe image lies in K . It follows that it is constant and k is central. Since k is arbitrary wededuce that K is central in G . In particular every subgroup of K is normal in G .Let K α be a net of open subgroups in K with trivial intersection. Then K = lim ←− K/K α and G = lim ←− G/K α . Set L α = G/K α and note that as K α is open in K , it is of finiteindex there. Thus the map L α → L is a finite covering. Proof of Proposition 2.1.
Let G and K be as in Proposition 2.1. By Lemma 2.3, G =lim ←− L α is an inverse limit of finite covers L α of L = G/K . Let ¯ g , . . . , ¯ g k be topologicalgenerators of L , let g , . . . , g k be arbitrary lifts in G and denote by g αi the projection of g i in L α , for every i, α . By Lemma 2.2, h g αi : i = 1 , . . . , k i is dense in L α . That is, the group h g i : i = 1 , . . . , k i projects densely to all L α . This implies that it is dense in G . -generated Our aim in this section is to prove the following result. Recall that t ( G ) is the minimalnumber of topological generators of G , while t I ( G ) is the minimal n ∈ N such that everyneighborhood of the identity contains n elements which topologically generate G . Theorem 3.1.
Let G be a separable compact connected group. Then t I ( G ) = t ( G ) = 2 if G is nonabelian and t I ( G ) = t ( G ) = 1 if G is abelian. The case where G is metrizable was proved by Schreier and Ulam [SU35]. Recall that acompact group is metrizable if and only if it is first countable. In that case one can showthat almost every pair of elements in G (or a single element if G is abelian) topologicallygenerates G . Let us give a short argument for Theorem 1.2 in that case. First recallthat by the Peter–Weyl theorem, G is an inverse limit of compact Lie groups and, beingfirst countable, the limit is over a countable net. Since a countable intersection of fullmeasured sets is of full measure, it is enough to prove the analog statement for compactconnected Lie groups.Note also that, in complete generality, a subgroup H ≤ G is dense if and only if • H ∩ G ′ is dense in G ′ , where G ′ is the commutator subgroup in G , and • HG ′ is dense in G .A connected compact Lie group G is reductive, hence an almost direct product of itscommutator G ′ with its centre Z . Moreover G ′ is connected and a finite cover of G/Z which is a semisimple group of adjoint type. In view of Lemma 2.2, we deduce:
Lemma 3.2.
Let G be a connected compact Lie group and H ≤ G a subgroup. Then H is dense in G if and only if both HG ′ and HZ ( G ) are dense in G . Therefore, for a pair ( x, y ) ∈ G to generate a dense subgroup, it is sufficient if • the projections of x, y to G/Z generate a dense subgroup in
G/Z , and5 the projection of x to G/G ′ generates a dense subgroup in G/G ′ .It is easy to check that both conditions are satisfied with Haar probability (cf. [Gel08,Lem. 1.4 and Lem. 1.10]). Let us first deal with abelian groups. By the Halmos–Samelson theorem, whenever G isconnected compact separable abelian, one has t ( G ) = 1 . From their result, we deduce thefollowing consequence. Lemma 3.3.
Let G be a compact connected separable abelian group. Then t I ( G ) = 1 .Proof. By the Halmos–Samelson theorem, we may and do pick g ∈ G such that h g i isdense in G . Let U be a neighborhood of the identity in G , and fix n ∈ Z \ { } suchthat g n ∈ U . Then since h g n i has finite index in h g i , its closure h g n i has finite index in h g i = G . Since G is connected, we must have h g n i = G .Now suppose that G is any connected compact group. By the Peter–Weil theorem G = lim ←− G α is an inverse limit of compact connected Lie groups G α .Observe that a surjective map f : G ։ G between groups always satisfies f ( G ′ ) = G ′ and f ( Z ( G )) ⊂ Z ( G ) , while for reductive Lie groups we also have f ( Z ( G )) = Z ( G ) . Since every connectedcompact Lie group is reductive, hence the product of its centre and its commutator, wededuce that the same hold for general compact connected groups, i.e. Z ( G ) = lim ←− Z ( G α ) , G ′ = lim ←− G ′ α and G = Z ( G ) G ′ . Moreover, since the G ′ α are semisimple, and in particular perfect, G ′ is also perfect,i.e. G ′ = G ′′ . It follows that if Γ ≤ G ′ is dense, then Γ ′ is dense in G ′ . Hence we have: Claim.
Suppose that a, b ∈ G ′ topologically generate G ′ , and h ∈ Z ( G ) is an elementwhose image mod G ′ topologically generates G/G ′ . Then ah and b topologically generate G . Suppose from now on that G is separable. Then every quotient of G is also separable.Now G/G ′ is connected and abelian, so we have t I ( G/G ′ ) = 1 by Lemma 3.3.Since Z ( G ) surjects onto G/G ′ , every identity neighbourhood in G contains a centralelement h whose image in G/G ′ generate a dense subgroup. Thus we are left to showthat t I ( G ′ ) = 2 . The centre of G ′ is totally disconnected since it can be written as Z ( G ′ ) = lim ←− Z ( G ′ α ) , and every G ′ α has finite center. In view of Proposition 2.1 we maythus suppose that G ′ is center-free. In order to simplify notations, let us suppose belowthat G itself is center-free. Note that: Lemma 3.4.
A center-free connected compact group is a direct product of simple Liegroups.
Thus, G is of the form G = Q α ∈ I S α with S α being connected adjoint simple Lie group. Lemma 3.5.
The group G = Q α ∈ I S α is separable (if and) only if Card ( I ) ≤ ℵ . ( I ) ≤ ℵ implies that G has a two generated dense subgroup. Suppose by way ofcontradiction that Card ( I ) > ℵ . Since there are only countably many isomorphism typesof (compact adjoint) simple Lie groups, we deduce that there is some compact simple Liegroup S and a cardinality κ > ℵ such that G admits a factor isomorphic to S κ . Howeverby the cardinals version of the pigeon hole principal, if D ⊂ S κ is a countable subset,there must be two factors S , S of S κ such that the projection of D to S × S lies in thediagonal. In particular, D cannot be dense, confirming the desired contradiction.Thus, we may suppose below that Card ( I ) ≤ ℵ . Definition 3.6.
Let S , S be two groups, F a set and f i : F → S i , i = 1 , two maps.We shall say that the maps f and f are isomorphically related if there is an isomorphism φ : S → S such that the following diagram is commutative F f −→ S ց f ↓ φ S Lemma 3.7.
Let Q ki =1 S i ( k ≥ be a product of simple groups and let R < Q ki =1 S i be aproper subgroup that projects onto every S i . Then there are two factors S i , S j , i = j suchthat the restrictions of the quotient maps R → S i and R → S j are isomorphically related.Proof. For a subset J ⊂ { , . . . , k } let us denote S J = Q i ∈ J S i and R J = Proj S J ( R ) . Let J ⊂ { , . . . , k } be a minimal subset such that R J is a proper subgroup of S J . By ourassumption J exists and satisfies | J | > . We claim that | J | = 2 . To see this, we mayreorder the indices so that J = { , . . . , j } and suppose by way of contradiction that j ≥ .This, together with the minimality of J , implies that for every g ∈ S and every < i ≤ j there is an element in R J whose first coordinate is g and whose i ’th coordinate is .However, multiplying commutators of elements as above, forcing that at each coordinate < i ≤ j at least one of the elements we use is trivial, and using the fact that S isperfect, we deduce that S ≤ R J . In the same way we get that S i ≤ R J for all i ∈ J contradicting the assumption that R J is proper in S J . Thus J = { , } . Moreover, as S i is simple and normal, and R J projects onto S i , while R J is proper in S J , it follows that R J ∩ S i is trivial, for i = 1 , . Therefore, the restriction of the projection from R J to S i is an isomorphism, for i = 1 , , hence the maps R → S and R → S are isomorphicallyrelated.Assembling together isomorphic factors of G , we may decompose G as a direct product G = Q n ∈ N G n where different n ’s correspond to non-isomorphic simple Lie factors S n , andeach G n is of the form G n = S I n n with | I n | ≤ ℵ . For x ∈ G we shall denote by x αn , α ∈ I n its corresponding coordinates.Two elements x, y ∈ G generate a dense subgroup if and only if for any finite set ofindices F = { ( n i , α i ) } the projections of x and y generate a dense subgroup in Q F S α i n i .In view of Lemma 3.7 we have: Corollary 3.8.
Two elements x, y ∈ G generate a dense subgroup in G if and only if • h x αn , y αn i is dense in S αn for every n ∈ N and every α ∈ I n , and • For every n, m ∈ N and α ∈ I n , β ∈ I m the (projection) maps { x, y } → S αn , { x, y } → S βm are not isomorphically related. ( x αn , y αn ) arbitrarilyclose to the identity in S αn , for every n ∈ N , α ∈ I n , which generate a dense subgroup of S n , such that for every pair ( n, α ) = ( m, β ) there is no isomorphism S n → S m taking ( x αn , y αn ) to ( x βm , y βm ) . Since for n = m there is no isomorphism between S n and S m weshould only consider the case n = m .Let S = S n . Recall that there is an identity neighbourhood (a Zassenhaus neighbour-hood) Ω ⊂ S on which the logarithm is well defined, and for x, y ∈ Ω the group h x, y i isdense in S if log( x ) and log( y ) generate the Lie algebra Lie ( S ) , (see [Kur49]). Thus wemay pick some open set U × U ⊂ Ω such that for all ( x, y ) ∈ U × U the group h x, y i isdense in S (cf. [GŻuk02]). Furthermore, we may pick U , U arbitrarily close to the iden-tity. Since Out ( S ) is finite and S is compact, we may also suppose, by taking U and U sufficiently small, that for ( x, y ) ∈ U × U , if f ∈ Aut ( S ) satisfies ( f ( x ) , f ( y )) ∈ U × U ,then f is inner. Finally, since the orbit of each ( x, y ) ∈ U × U under the action of S on S × S by conjugation, is dim( S ) dimensional, while dim( U × U ) = 2 dim( S ) , we canpick a section S transversal to the orbits foliation, inside U × U . Clearly the cardinal-ity of that section is ℵ , hence we may imbed I n inside this section. This imbeddingyields a choice of ( x α , y α ) ∈ S ⊂ U × U for every α ∈ I n , and we have that each pair ( x α , y α ) generates a dense subgroup in S , and no two pairs are isomorphically related.This completes the proof of Theorem 3.1. Recall that a topological group G is infinitesimally finitely generated if t I ( G ) < + ∞ ,that is, if there exists n ∈ N such that every neighborhood of the identity contains n topological generators for G .Similarly, we say that G is topologically finitely generated if t ( G ) < ∞ , that is,if G admits a dense finitely generated subgroup.Last but not least, a topological group is infinitesimally generated if it is generatedby every neighborhood of the identity.Note that if a group is infinitesimally finitely generated, then it is also topologicallyfinitely generated and infinitesimally generated. Our main goal in this section is to provethe following result. Theorem 4.1.
Let G be a separable connected locally compact group. Then G is infinites-imally finitely generated. As already noted, the separability condition is necessary for a group to be topologicallyfinitely generated. The connectedness assumption is also necessary (for local generation)since otherwise
G/G ◦ , the group of connected components, is non-trivial and by the vanDantzig’s theorem admits a base of identity neighbourhood consisting of open compactsubgroups. In particular, if O ≤ G/G ◦ is a proper open subgroup, its pre-image in G cannot contain a topological generating set.Let us start by dealing with the nicest connected locally compact groups: Lie groups.For these, one can use the Lie algebra to produce topological generators. The followinglemma is well known, but we include a proof for the reader’s convenience. Lemma 4.2 (Folklore) . Let G be a connected Lie group, and let g be its Lie algebra.Suppose that g is generated as a Lie algebra by X , ..., X n . Then every neighborhood of he identity in G contains n elements g , ..., g n which generate a dense subgroup in G .In particular, G is infinitesimally finitely generated.Proof. Let V be a neighborhood of the identity in G . Let U be a small enough neigh-borhood of 0 in g such that exp : U → G is a homeomorphism onto its image, and exp( U ) ⊆ V . Fix n elements Y , ..., Y n of U such that for every i ∈ { , ..., n } , { Y i , Y i +1 } generates a dense subgroup of R X i .For all i ∈ { , ..., n } , let g i = exp( Y i ) , we will show that these elements topologicallygenerate G . Let H be the closed subgroup generated by the set { g i } ni =1 . Note that forall i ∈ { , ..., n } , the group H contains exp( R X i ) since the restriction of the exponentialmap to R X i is a continuous group homomorphism and R X i is topologically generated by Y i and Y i +1 which are mapped to g i ∈ H and g i +1 ∈ H .Furthermore, H is a Lie group by Cartan’s theorem, and since H contains every exp( R X i ) the Lie algebra h of H contains every X i , so h = g . Because G is connected,we get that G = H .We deduce from the above lemma that for any connected Lie group, t I ( G ) G ) .Better bounds on t I ( G ) can be deduced from the analysis in [BG03, BGSS06, Gel08].Let now G be a general connected locally compact group. Recall the celebratedGleason-Yamabe theorem (cf. [Kap71, Page 137]): Theorem 4.3. (Gleason–Yamabe) Let G be a connected locally compact group. Thenthere is a compact normal subgroup K ⊳ G such that G/K is a Lie group.
Since G is connected, G/K is a connected Lie group and hence t I ( G/K ) is finite.However, K may not be connected. Example 4.4. (1) (The solenoid) For every n , let T n be a copy of the circle group { z ∈ C : | z | = 1 } , and whenever m divides n let f n,m : T n → T m be the n/m sheeted cover f n,m ( z ) = z n/m . Let T = lim ←− T n be the inverse limit group. Then T is connected, abelianand locally compact, but admits no connected co-Lie subgroups. (2) Similarly, as SL ( R ) is homotopic to a circle, we can define, for every n ∈ N , G n asthe n sheeted cover of SL ( R ) . Then whenever m divides n there is a canonical coveringmorphisms ψ n,m : G n → G m , and we may let G be the inverse limit G = lim ←− G n . Then G is a connected locally compact group which admits no nontrivial connected compactnormal subgroups.In order to prove Theorem 4.1, we need one last elementary lemma. Lemma 4.5.
Let G be a topological group and N a normal subgroup. Then t I ( G ) ≤ t I ( N ) + t I ( G/N ) . In particular if G/N and N are infinitesimally finitely generated thenso is G .Proof of Theorem 4.1. Let G be a connected separable locally compact group. Let K ⊳ G be a compact normal subgroup such that G/K is a Lie group (see Theorem 4.3), and let K ◦ be its identity connected component. Then K ◦ is characteristic in K and therefore normalin G . Being a closed subgroup of a locally compact separable group, K ◦ is separable[CI77]. Let H = G/K ◦ and K t = K/K ◦ . By the isomorphism theorem, G/K ∼ = H/K t .Note that K t is a pro-finite group, hence by Proposition 2.1 H is infinitesimally finitelygenerated. By Theorem 3.1, K ◦ is infinitesimally finitely generated, hence, by Lemma4.5, G is infinitesimally finitely generated. 9 Quasi non-archimedean groups
A topological group is non-archimedean if it has a basis of neighborhoods of the identitymade of open subgroups. Equivalently, every neighborhood of the identity V contains asmaller neighborhood of the identity U such that the group generated by U is contained in V , which is the same as requiring that for every n ∈ N and g , ..., g n ∈ U , the group gen-erated by g , ..., g n is contained in V . The definition that follows is obtained by switchingtwo quantifiers in the above condition. Definition 5.1.
Say a topological group G is quasi non-archimedean if for all n ∈ N and all neighborhood of the identity V ⊆ G , there exists a neighborhood of the identity U ⊆ V such that for all g , ..., g n ∈ U , the group generated by g , ..., g n is contained in V .Clearly every non-archimedean group is also quasi non-archimedean. Let us give rightaway the motivating example for this definition. We fix a standard probability space ( X, µ ) , that is, a probability space which is isomorphic to the interval [0 , with its Borel σ -algebra and the Lebesgue-measure.A Borel bijection T of X is called a non-singular automorphism if for all mea-surable A ⊆ X , one has µ ( A ) = 0 if and only if µ ( T − ( A )) = 0 . The group of all theseautomorphisms is denoted by Aut ∗ ( X, µ ) , two such automorphisms being identified if theycoincide on a full measure set. We then define the uniform metric d u on Aut ∗ ( X, µ ) by:for all T, U ∈ Aut ∗ ( X, µ ) , d u ( T, U ) = µ ( { x ∈ X : T ( x ) = U ( x ) } ) . This is a complete metric, though far from being separable (e.g. the group S acts freelyon itself, yielding an uncountable discrete subgroup of ( Aut ∗ ( X, µ ) , d u ) ). But amongclosed subgroups of ( Aut ∗ ( X, µ ) , d u ) , full groups are separable. Full groups are invariantsof orbit equivalence attached to nonsingular actions of countable groups on ( X, µ ) : givena non-singular action of a countable group Γ on ( X, µ ) , its full group [ R Γ ] is the group ofall T ∈ Aut ∗ ( X, µ ) such that for every x ∈ X , T ( x ) ∈ Γ · x .Since every subgroup of a quasi non-archimedean group is quasi non-archimedeanfor the induced topology, the following result implies that full groups are quasi non-archimedean. Theorem 5.2 (Kechris) . Aut ∗ ( X, µ ) is quasi non-archimedean for the uniform metric.Proof. Define the support of T ∈ Aut ∗ ( X, µ ) to be the set of all x ∈ X such that T ( x ) = x . Note that d u (id X , T ) is precisely the measure of the support of T .Let ǫ > and n ∈ N , and consider the open ball U := B d u (id X , ǫ ) . Suppose that g , ..., g n belong to V := B d u (id X , ǫ/n ) , and let A be the reunion of their supports. Byassumption, A has measure less than ǫ . Then the group generated by g , ..., g n is containedin the group of elements supported in A , which is itself a subset of U = B d u (id X , ǫ ) .The following proposition shows that the class of quasi non-archimedean groups sat-isfies basically the same closure properties as the class of non-archimedean groups. Proposition 5.3.
The class of quasi non-archimedean groups is closed under taking sub-groups (with the induced topology), products and quotients. ∗ ( X, µ ) endowed with the weak topology, or the unitary group of a separable Hilbertspace endowed with the strong operator topology. Let us point out that a locally compactgroup is Polish if and only if it is second-countable (see [Kec95, Thm. 5.3]).Recall that if G is a Polish group and ( X, µ ) is a standard (non-atomic) probabilityspace, then the group L ( X, µ, G ) of measurable maps from X to G is a Polish group forthe topology of convergence in measure, two such maps being identified if they coincideon a full measure set. A basis of neighborhoods of the identity for this topology is givenby the sets ˜ U ǫ = { f ∈ L ( X, µ, G ) : µ ( { x ∈ X : f ( x ) U } ) < ǫ } , where U is an open neighborhood of the identity in G and ǫ > . The Polish group L ( X, µ, G ) enjoys the two following nice properties (see e.g. [Kec10, Chap. 19]) . • G embeds into L ( X, µ, G ) via constant maps. • L ( X, µ, G ) is connected, in fact contractible. Proposition 5.4.
Let G be a quasi non-archimedean Polish group. Then L ( X, µ, G ) isquasi non-archimedean. In particular any quasi non-archimedean Polish group embeds ina connected quasi non-archimedean Polish group.Proof. Let ˜ U ǫ = { f : µ ( { x ∈ X : f ( x ) U } ) < ǫ } be a basic neighborhood of theidentity in L ( X, µ, G ) . Let n ∈ N and V be a corresponding neighborhood of the identitywitnessing that G is quasi non-archimedean. Consider the following open neighborhoodof the identity in L ( X, µ, G ) : ˜ V ǫ/n = { f : µ ( { x ∈ X : f ( x ) V } ) < ǫ/n } . Then if we let f , ..., f n ∈ ˜ V ǫ,n , the reunion of the sets { x ∈ X : f i ( x ) V } has measureless than ǫ . By the definition of V and U the group generated by f , ..., f n is a subset of ˜ U ǫ . Remark.
Since non-archimedean groups are totally disconnected, the above propositionimplies there are a lot more quasi non-archimedean groups than the non-archimedeanones.The following proposition is inspired by Section (D) of Chapter 4 in [Kec10], whereit is shown that any continuous homomorphism from an infinitesimally finitely generatedgroup into a full group is trivial.
Proposition 5.5.
Any continuous homomorphism from an infinitesimally finitely gener-ated group into a quasi non-archimedean group is trivial.Proof.
Let ϕ : G → H be such a morphism, let V be any neighborhood of the identityin H , and let n = t I ( G ) . Then there is a neighborhood of the identity U in H suchthat any subgroup of H generated by n elements of U is contained in V . Since ϕ iscontinuous, ϕ − ( U ) is a neighborhood of the identity in G , and because t I ( G ) = n wemay find g , ..., g n ∈ ϕ − ( U ) which generate a dense subgroup in G . Then the closure of ϕ ( G ) coincides with the closure of the group generated by ϕ ( g ) , ..., ϕ ( g n ) ∈ U , which byassumption is contained in V . So ϕ ( G ) is contained in the closure of any neighborhoodof the identity in H , and since H is Hausdorff this means that ϕ is trivial.11 orollary 5.6. The only topological group which is both infinitesimally finitely generatedand quasi non-archimedean is the trivial group.
Corollary 5.7.
Every continuous homomorphism from a connected separable locally com-pact group into ( Aut ∗ ( X, µ ) , d u ) is trivial.Proof. Since every connected separable locally compact group is infinitesimally finitelygenerated by Theorem 4.1 and Aut ∗ ( X, µ ) is quasi non-archimedean by Theorem 5.2, theprevious proposition readily applies.As a consequence of the previous proposition and Theorem 1.6, we have the follow-ing interesting characterizations of connectedness and total disconnectedness for locallycompact separable groups. Theorem 5.8.
Let G be a locally compact separable group. Then the following hold:(1) G is connected if and only if G is infinitesimally finitely generated.(2) G is totally disconnected if and only if G is quasi non-archimedean.Proof. If G is not connected but infinitesimally finitely generated, then G/G must alsobe infinitesimally finitely generated, which is impossible by van Dantzig’s theorem. Theconverse is provided by Theorem 1.6.If G is totally disconnected, then G is non-archimedean by van Dantzig’s theorem. Butthis implies that G is quasi non-archimedean. For the converse, suppose G is quasi non-archimedean. Then G also is, but then by (1) and Proposition 5.5 it must be trivial. Remark.
As was pointed out by Caprace and Cornulier, one can prove (2) more directly.Indeed, if G is non-trivial then it admits a non-trivial one-parameter subgroup whichis in particular infinitesimally -generated, contradicting Proposition 5.5. This actuallygives a proof that a locally compact group is quasi non-archimedean if and only if it istotally disconnected, regardless of its separability.Let us now give an example of a totally disconnected Polish group which is quasinon-archimedean, but not non-archimedean. This class of examples was introduced byTsankov [Tsa06, Sec. 5], using work of Solecki [Sol99]. We denote by S ∞ the group of allpermutations of the integers, equipped with its Polish topology of pointwise convergence.Recall that every non-archimedean Polish group arises as a closed subgroup of S ∞ (see[BK96, Thm. 1.5.1]). Here, the groups that we will consider are subgroups of S ∞ , butequipped with a Polish topology which refines the topology of pointwise convergence. Definition 5.9. A lower semi-continuous submeasure on N is a function λ : P ( N ) → [0 , + ∞ ] such that the following hold: • λ ( ∅ ) = 0 ; • for all n ∈ N , we have < λ ( { n } ) < + ∞ ; • for all A ⊆ B ⊆ N , we have λ ( A ) λ ( B ) ; • ( subadditivity ) for all A, B ⊆ N , we have λ ( A ∪ B ) λ ( A ) + λ ( B ) ; • ( lower semi-continuity ) for every increasing sequence ( A k ) k ∈ N of subsets of N , wehave λ ( S k ∈ N A k ) = lim k ∈ N λ ( A k ) . 12e associate to every lower semi-continuous submeasure λ on N a subgroup of S ∞ ,denoted by S λ , defined by S λ = { σ ∈ S ∞ : λ (supp σ \ { , ..., n } ) → n → + ∞ ] } , where supp σ = { n ∈ N : σ ( n ) = n } . Note that since µ ( { , ..., n } ) < + ∞ for every n ∈ N , the support of every σ ∈ S λ has finite measure. Also, if λ is actually a measure,then S λ = { σ ∈ S ∞ : λ (supp σ ) < + ∞} ; furthermore if λ is a probability measure then S λ = S ∞ .The group S λ is equipped with a natural left-invariant metric d λ analogous to theuniform metric on Aut ∗ ( X, µ ) defined by d λ ( σ, σ ′ ) = λ ( { n ∈ N : σ ( n ) = σ ′ ( n ) } ) . Note that the condition λ (supp σ \ { , ..., n } ) → ensures that the countable group ofpermutations of finite support is dense in S λ which is thus separable. It is a theoremof Tsankov that S λ is actually a Polish group. The following result is a straightforwardadaptation of Theorem 5.2, replacing d u by d λ . Proposition 5.10.
Let λ be a lower semi-continuous submeasure on N . Then S λ is quasinon-archimedean. It is easily checked that the topology of S λ refines the topology induced by S ∞ , sothat S λ is always totally disconnected, and that the open subgroups of S λ separate pointsfrom the identity (in particular, S λ is not locally generated). The following example showsthat it can furthermore fail to be non-archimedean. Note that this is just a particularcase of a more general phenomenon: one can actually characterize when the topology failsto be zero-dimensional (see [Tsa06, Thm. 5.3]; the example below is taken from [Mal15,Cor. 4]). Example 5.11.
Consider the measure λ on N defined by λ ( A ) = X n ∈ A n . Then S λ is not a non-archimedean group. Indeed, if we fix ǫ > and N ∈ N , we canfind a finite family ( A i ) Ni =1 of disjoint subsets of N such that for every i ∈ { , ..., N } , ǫ < λ ( A i ) < ǫ . For all i ∈ { , ..., N } , let σ i be a permutation whose support is equalto A i . Then σ = Q Ni =1 σ i is at distance at least N ǫ/ from the identity, so that the ballof radius ǫ around the identity generates a group which contains elements arbitrarily faraway from the identity. Corollary 5.12.
There exists a totally disconnected Polish group which is quasi non-archimedean, but not non-archimedean.
Let us now give examples of totally disconnected Polish group which are infinitesimallyfinitely generated. To do this, we will upgrade a result of Stevens [Ste86] who showed theexistence of totally disconnected infinitesimally generated Polish groups: we will show thather examples are actually infinitesimally finitely generated. This will be a consequenceof the following general statement, which also implies the well-known fact that R has adense G δ of pairs of topological generators. A topology is zero-dimensional if it has a basis made of clopen sets. heorem 5.13. Let G be an abelian Polish group which contains the group Z [1 / ofdyadic rationals as a dense subgroup, and assume furthermore that in the topology inducedby G on Z [1 / , we have n → n → + ∞ ] . Then for every g ∈ Z [1 / , the set of h ∈ G such that h g, h i generate a dense subgroup of G is dense.In particular there is a dense G δ set of couples of topological generators of G in G and so G is infinitesimally finitely generated with infinitesimal rank at most .Proof. Let g ∈ Z [1 / . In order for a couple ( g , h ) ∈ G to generate a dense subgroupof G , it suffices for the closed subgroup they generate to contain n for every n ∈ N , forthe group Z [1 / is dense in G . So the set T := { h ∈ G : h g , h i = G } may be written asa countable intersection T = T n ∈ N T n , where T n := (cid:26) h ∈ G : 12 n ∈ h g , h i (cid:27) . Since G is Polish the set T n is G δ , so we only need to show that T n is dense. To this end,fix ǫ > , n ∈ N , and let h ∈ Z [1 / . We want to find h ∈ T n such that d ( h, h ) < ǫ ,where d is a fixed compatible metric on G .Write g = k m and h = k m , where k , k ∈ Z and m ∈ N . We will find β ∈ N suchthat for all N ∈ N , if h = h + β m + N , then h g , h i contains N + m , so that in particular itcontains n as soon as N + m > n . The group h g , h i contains N + m if and only if we canfind u, v ∈ Z such that ug + vh = m + N . This condition can be rewritten as N k u + (2 N k + β ) v = 1 . So we want to find β ∈ N such that for all N ∈ N we have that N k and N k + β arerelatively prime. Let us furthermore ask that β is odd, so that we only have to make surethat every odd prime divisor of k does not divide N k + β .Let p , ..., p k list the odd primes which divide both k and k , while p k +1 , ..., p l arethe odd primes which divide k but not k . Then it is easily checked that β = (2 + p · · · p k ) p k +1 · · · p l works: for all i k we have that β is invertible modulo p i and p i divides k so that N k + β is not divisible by p i , while for k < i l , β is null modulo p i while N k is invertible so that N k + β is not divisible by p i .But then, since m + N tends to zero as N tends to + ∞ , we also have β m + N → N → + ∞ ] . Then, as explained before, the group generated by g := g and h := h + β m + N contains m + N , so that ( g, h ) ∈ T n , while d ( g, g ) < ǫ and d ( h, h ) < ǫ if N was chosenlarge enough.So every T n is a dense subset of G , which ends the proof since this furthermore showsthat the set of couples generating a dense subgroup of G is dense in G and this set hasto be a G δ .Let us now apply the previous theorem and describe Steven’s examples of Polish groupswhich are infinitesimally finitely generated but totally disconnected. These groups arise asa Polishable subgroups of the real line, constructed by taking a completion of the dyadicrationals with respect to a well chosen norm which makes − n have much bigger normthan usually.We fix a biinfinite sequence of positive real number ( r i ) i ∈ Z such that r i → i → + ∞ ] and for all i ∈ Z , we have r i +1 r i r i +1 . Then one can define the following group14orm k·k on the ring Z [1 / of dyadic rationals: for every x ∈ Z [1 / , k x k := inf ( n X i = − n | a i | r i : x = n X i = − n a i − i , a i ∈ Z , n ∈ N ) . It is easy to check that this defines a group norm on Z [1 / which refines the usual norm.Using the fact that r i r i +1 , one can easily show that for all x ∈ Z [1 / , k x k = inf ( n X i = − n | a i | r i : x = n X i = − n a i − i , a i ∈ {− , , } , n ∈ N ) . In particular, we see that for all n ∈ N , we have k − n k = r n so that − n → as n → + ∞ .Let Z [1 / k·k denote the completion of Z [1 / with respect to this norm. Since this normrefines the usual norm, Z [1 / k·k is a subgroup of R . Stevens explicitely described theelements of R belonging to Z [1 / k·k and showed that the group Z [1 / k·k is infinitesimallygenerated [Ste86, Thm. 2.1 (ii)], and we see that Theorem 5.13 strengthens this becauseit implies that Z [1 / k·k is infinitesimally -generated.To obtain totally disconnected examples, we need another result of Stevens statingthat the following are equivalent (see [Ste86, Thm. 2.2]):(i) P i ∈ N r i = + ∞ ,(ii) k·k is not equivalent to |·| when restricted to Z [1 / ,(iii) Q ∩ Z [1 / k·k = Z [1 / ,(iv) Z [1 / k·k is totally disconnected, Remark.
Note that every subgroup G of R which is not equal to R has to be totallydisconnected for the induced topology, since its complement is dense in R so that the setsof the form ] r, + ∞ [ for r ∈ R \ G are clopen in G . In particular, if G is a proper subgroupof R equipped with a topology which refines the usual topology of R then G is totallydisconnected. So conditions (ii) and condition (iii) clearly imply condition (iv).So suppose further that P i ∈ N r i = + ∞ (e.g. take r i = 1 if i and r i = i otherwise).Then we see that Z [1 / k·k is a totally disconnected Polish group which has infinitesimalrank at most . Moreover since this group is a subgroup of the real line endowed with afiner topology (see [Ste86, Thm. 2.1]) it cannot be monothetic, so its infinitesimal rankis actually equal to . Corollary 5.14.
There exists a totally disconnected Polish group which has infinitesimalrank , in particular there is a totally disconnected Polish group which is not quasi non-archimedean. Remark.
Note that one can see directly that Stevens’ groups are not quasi non-archimedean,even for n = 1 . Indeed, if U is a neighborhood of not containing , then if V is anotherneighborhood of there is some N ∈ N such that / N ∈ V , but the group generated by / N contains hence it is not a subset of U . A norm on an abelian group is a function |·| : G → [0 , + ∞ ) such that for any x, y ∈ G , | x + y | | x | + | y | , and | x | = |− x | .
15e know that while the group of the reals has infinitesimal rank , its quotient S = R / Z has infinitesimal rank . The same is true of Stevens’ examples, which is going toyield the following result. Theorem 5.15.
There exists a totally disconnected Polish group which has infinitesimalrank .Proof. Let G be a totally disconnected Polish group obtained by Stevens’ constructionfrom a sequence ; then G is a proper subgroup of R containing Z [1 / as a dense subgroup.Observe that Z is a discrete subgroup of G and we may thus form the Polish group ˜ G := G/ Z .The group ˜ G is a proper dense subgroup of S = R / Z , so S \ ˜ G is thus dense in S .Let A = p ([0 , / where p : R → R / Z is the usual projection, then for all g ∈ S \ ˜ G the set ( g + A ) ∩ ˜ G is clopen in ˜ G . Moreover since S \ ˜ G is dense in S the family of sets (( g + A ) ∩ ˜ G ) g ∈ S \ ˜ G separates points in ˜ G , so ˜ G is totally disconnected.Furthermore, we have by Theorem 5.13 that there is a dense G δ of h ∈ G such thatthe group generated by and h is dense in G . Since ∈ Z and ˜ G = G/ Z we concludethat there is a dense G δ of h ∈ ˜ G which generate a dense subgroup in ˜ G , in particular ˜ G has infinitesimal rank .We don’t know an example of a totally disconnected Polish group which is infinites-imally generated and quasi non-archimedean. Moreover, we want to stress out that allthe examples we know of Polish groups which are quasi non-archimedean actually failthe property even for n = 1 , so it would be very interesting to have examples having a“non-QNA rank” greater than . Let us point out how one can easily build Polish groups into which no non-discrete locallycompact group can embed.
Lemma 6.1.
Let Γ be a countable discrete group without elements of finite order. Thenevery monothetic subgroup of L ( X, µ, Γ) is infinite discrete. In particular, no nontrivialcompact group embeds into L ( X, µ, Γ) .Proof. Given = f ∈ L ( X, µ, Γ) , find A ⊆ X non-null and γ ∈ Γ \ { } such that f ↾ A isconstant equal to γ . By assumption, for all n ∈ Z \ { } , the support of f n contains A ,and so h f i is discrete. Theorem 6.2.
Let Γ be a countable discrete group without elements of finite order, and let G be a separable locally compact group. Then every continuous morphism G → L ( X, µ, Γ) factors through a discrete group.Proof. Let G be the connected component of the identity. Because L ( X, µ, Γ) is quasinon-archimedean, π factors through G/G by Proposition 5.5 and Theorem 1.6. Then byvan Dantzig’s theorem and the previous lemma, the kernel of the later map contains anopen subgroup of G/G , hence it factors through a discrete group. Question 6.3.
Is there a Polish group without any locally compact closed subgroup?16he group L ( X, µ, G ) was originally introduced to show that every Polish groupembeds into a connected group, and we saw that being quasi non-archimedean is somehowopposite to being connected. Because L ( X, µ, G ) can be quasi non-archimedean, onemay ask whether every Polish group embeds into a connected not quasi non-archimedeangroup. The isometry group of the Urysohn space answers this question — it is universalfor Polish groups, connected (see [Mel06, Mel10] for stronger versions of these resultsas well as background on the Urysohn space) and cannot be quasi non-archimedean byuniversality.However, the same question can be asked replacing not quasi non-archimedean byinfinitesimally finitely generated. It seems to be open wether the isometry group of theUrysohn space is infinitesimally finitely generated (it is topologically -generated by aresult of Solecki, see [Sol05]). Question 6.4.
Is the isometry group of the Urysohn space infinitesimally -generated?Let us end this paper by mentioning a question related to ample generics. A Pol-ish group G has ample generics if the diagonal conjugacy action of G onto G n has acomeager orbit for every n ∈ N (see [KR07]). It has been recently discovered that thereexists Polish groups with ample generics which are not non-archimedean (see[KLM15] and[Mal15]). These examples arise either as full groups or as groups of the form S λ , whichare quasi non-archimedean groups by Theorem 5.2 and Proposition 5.10. This motivatesthe following question. Question 6.5.
Is there a Polish group which has ample generics, but which is not quasinon-archimedean?
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