Bounds for the rank of the finite part of operator K-Theory
aa r X i v : . [ m a t h . K T ] M a y BOUNDS FOR THE RANK OF THE FINITE PART OFOPERATOR K -THEORY SÜLEYMAN KAĞAN SAMURKAŞ
Abstract.
We derive a lower and an upper bound for the rank of the finite part ofoperator K -theory groups of maximal and reduced C ∗ -algebras of finitely generatedgroups. The lower bound is based on the amount of polynomially growing conjugacyclasses of finite order elements in the group. The upper bound is based on theamount of torsion elements in the group. We use the lower bound to give lowerbounds for the structure group S ( M ) and the group of positive scalar curvaturemetrics P ( M ) for an oriented manifold M .We define a class of groups called “polynomially full groups” for which the upperbound and the lower bound we derive are the same. We show that the class ofpolynomially full groups contains all virtually nilpotent groups. As example, wegive explicit formulas for the ranks of the finite parts of operator K -theory groupsfor the finitely generated abelian groups, the symmetric groups and the dihedralgroups. Introduction
The purpose of this paper is to derive a lower bound for the rank of the finite partof the operator K -theory of maximal and reduced C ∗ -algebras, and use that lowerbound to study the non-rigidity of the manifolds and the space of positive scalarcurvature metrics on a manifold. Moreover, we derive an upper bound for the rankof the finite part, and introduce a class of groups called “polynomially full groups”for which the upper bound and the lower bound we derive are the same.Given a manifold M, we can ask the following question: How many “distinct”manifolds exist which are homotopy equivalent to M ? Two manifolds are consideredto be “distinct” if they are not homeomorphic. The answer to the question above isobviously related to the non-rigidity of the manifold M. The more “distinct” manifoldshomotopy equivalent to M exists, the less “rigid” M is.Given a compact oriented manifold M, the structure group S ( M ) of M is definedto be the abelian group generated by the equivalence classes of elements of the form ( f, M ′ ) , where M ′ is a compact oriented manifold and f : M ′ → M is an orientationpreserving homotopy equivalence.If a compact smooth spin manifold M has a positive scalar curvature metric andthe dimension of M is greater than or equal to , then P ( M ) is defined to be theabelian group of equivalent classes of all positive scalar curvature metrics on M. Fora more precise definition we refer to [10, Section 4].Weinberger, Xie and Yu [11] use the higher rho invariant to study the structuregroup S ( M ) . Mathematics Subject Classification.
Primary: 46L80. Secondary: 19M05.
Key words and phrases. finite part of operator K-theory, structure group, positive scalar curvaturemetric, polynomially full groups.
Weinberger and Yu [10] use linearly independent elements in K ( C ∗ G ) whose linearspan intersects trivially with the image of the assembly map µ : K G ( EG ) → K ( C ∗ G ) , to give lower bounds for the ranks of the groups S ( M ) and P ( M ) , where G is thefundamental group of M and EG is the universal cover of the classifying space BG.
So in order to give lower bounds for the ranks of the groups S ( M ) and P ( M ) , a general strategy would be to construct linearly independent elements in K ( C ∗ G ) whose linear span intersects trivially with the image of the assembly map µ : K G ( EG ) → K ( C ∗ G ) . Given a finite order element g ∈ G with order( g ) = d, define p g = 1 + g + g + · · · + g d − d ∈ C G ⊆ C ∗ G. It is not hard to show that p g is a projection (i.e. p g = p ∗ g = p g ). So p g gives anelement in K ( C ∗ G ) . For distinct finite order elements g , g , . . . , g n , in order to prove that p g , p g , . . . , p g n are linearly independent in K ( C ∗ G ) , we can use homomorphisms ρ i : K ( C ∗ G ) → C , by mapping the p g j ’s to C n and showing that their images are linearly independentin C n . So the problem reduces to find homomorphisms ρ i : K ( C ∗ G ) → C . One way of finding such homomorphisms is to use trace maps on the algebra C ∗ G. However, it is difficult (if possible) to construct different trace maps τ : C ∗ G → C . Fortunately, we don’t have to construct such traces.We say a subalgebra A of C ∗ r G is smooth if it is stable under holomorphic functionalcalculus. For a smooth dense subalgebra A of C ∗ r G, we have K ( A ) ∼ = K ( C ∗ r G ) , where the isomorphism is induced by the inclusion map. Hence, if we find trace maps τ i : A → C , then they induce homomorphisms τ i : K ( C ∗ r G ) ∼ = K ( A ) → C . Thus, composing with the homomorphism K ( C ∗ G ) → K ( C ∗ r G ) we get homomor-phisms ρ i : K ( C ∗ G ) → C . For all h ∈ G, let τ h : C G → C be defined as: τ h (cid:16) X g ∈ G a g · g (cid:17) := X g ∈ C ( h ) a g , where C ( h ) is the conjugacy class of h in G. It is easy to see that τ h is a trace map on C G [9]. So the problem reduces to lifting the τ h ’s to trace maps on a suitable smoothand dense subalgebra A of C ∗ r G. OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 3 For this we can define seminorms on C G and take the completion of C G withrespect to those seminorms. The completion should be as small as possible so thatwe can lift the τ h ’s.Now we describe our results in more detail.For a group G, let G fin be the subset of G of finite order elements. Now define arelation ∼ fin on G fin as follows: g ∼ fin h if and only if there exists γ ∈ G such that p g = γp h γ − . It is easy to see that ∼ fin is an equivalence relation on G fin . Now wedefine F G := | G fin / ∼ fin | . In the following, we give an equivalent definition for F G that we use in the proofs.For all d ∈ N define G find := { g ∈ G fin | order( g ) = d } . In the following, we define an equivalence relation on G find . Definition 1.1.
For a group G we define the relation ∼ d on G find as follows: g ∼ d h if and only if ∃ a ∈ N such that g a ∈ C ( h ) , where C ( h ) = { f hf − | f ∈ G } . It is easy to verify that ∼ d is an equivalence relation on G find . Next, we give an equivalent definition for F G . Definition 1.2.
Let e G find := G find / ∼ d . Define F G := P ∞ d =1 | e G find | . Remark 1.3.
It is not hard to see that given g, h ∈ G fin we have g ∼ fin h ⇔ ∃ d ∈ N , order( g ) = order( h ) = d and g ∼ d h. So the two definitions of F G are the same.Following Weinberger and Yu [10], we define K fin ( C ∗ G ) to be the subgroup of K ( C ∗ G ) generated by the set { [ p g ] : g ∈ G fin } , where [ p g ] denotes the class of the projection p g in K ( C ∗ G ) . We define K fin ( C ∗ r G ) similarly. Given g, h ∈ G fin with g ∼ fin h, we have [ p g ] = [ p h ] ∈ K ( C ∗ G ) . Hence, K fin ( C ∗ G ) has rank at most F G . Using the natural surjection K fin ( C ∗ G ) ։ K fin ( C ∗ r G ) induced by the identity map id : C G → C G, we can conclude that K fin ( C ∗ r G ) alsohas rank at most F G . The following result is proved in Section 2.
Theorem 1.4.
Let S ⊆ G fin . If there exists a smooth subalgebra A of C ∗ r G containing C G and if ∀ h ∈ S there exists a trace function e τ h : A → C extending the trace function τ h : C G → C , then we have rank( K fin ( C ∗ r G )) ≥ | S/ ∼ fin | and for the assembly map µ : K G ( EG ) → K ( C ∗ G ) , we have Im µ ∩ K S ( C ∗ G ) = { } , where K S ( C ∗ G ) is the subgroup of K ( C ∗ G ) generated by the set { [ p g ] : g ∈ S } and EG is the universal cover of the classifying space BG.
S.K.SAMURKAŞ
Given a finitely generated group G with word length norm k . k w and given h ∈ G, wedefine C ( h ) := { ghg − : g ∈ G } , C l ( h ) := { g ∈ C ( h ) : k g k w = l } and n h,l := | C l ( h ) | . Definition 1.5.
We say that C ( h ) has polynomial growth, if ∃ c ∈ R > ∃ d ∈ N suchthat n h,l ≤ c · l d for all l ∈ N and define G pol := { g ∈ G | C ( g ) has polynomial growth } . In the following, we define the maximum number of non-equivalent finite orderelements we can choose from G pol . Definition 1.6.
Let G be a finitely generated group. We define F polG := | ( G pol ∩ G fin ) / ∼ fin | . In Section 4, we prove that the hypotheses of Theorem 1.4 are satisfied for S = G pol ∩ G fin and A = C polS G defined in the proof.Hence, we get the following main result about polynomial growth in conjugacyclasses: Theorem 1.7.
Let G be a finitely generated group. We have F polG ≤ rank( K fin ( C ∗ r G )) ≤ rank( K fin ( C ∗ G )) ≤ F G and for the assembly map µ : K G ( EG ) → K ( C ∗ G ) , we have Im( µ ) ∩ K fin,pol ( C ∗ G ) = { } , where K fin,pol ( C ∗ G ) is the subgroup of K ( C ∗ G ) generated by the set { [ p g ] : g ∈ G fin ∩ G pol } . The proof of this result is in Section 4 of this paper. Note that this result followsfrom the injectivity part of the Baum-Connes conjecture [1].Gong [4] finds a lower bound for the rank of K fin ( C ∗ r G ) for the groups with property(RD) and conjugacy classes having poynomial growth. In our results, we don’t requireproperty (RD) and also improve the lower bound.The importance of these results lies in the following: Theorem 4.1. [10]
Let M be a compact oriented manifold with dimension k − k > . Suppose π ( M ) = G and g , · · · , g n be finite order elements in G such that g i = e for all i and { [ p g ] , · · · , [ p g n ] } generates an abelian group of K ( C ∗ G ) with rankn. Suppose that any nonzero element in the abelian subgroup of K ( C ∗ G ) generatedby { [ p g ] , · · · , [ p g n ] } is not in the image of the map µ : K G ( EG ) → K ( C ∗ G ) , thenthe rank of the structure group S ( M ) is greater than or equal to n. Combining this result with our results we get:
Corollary 4.2.
For a compact oriented manifold M with dimension k − k > , the rank of the structure group S ( M ) is greater than or equal to F polG − , where G = π ( M ) . Another application of our results is the following:Let r fin ( G ) be the rank of the abelian group K fin ( C ∗ G ) generated by [ p g ] for allfinite order elements g ∈ G. Here g is allowed to be the identity element e. So wehave F G ≥ r fin ( G ) = rank( K fin ( C ∗ G )) ≥ F polG . Theorem 4.3. [10]
OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 5 (1) Let M be a compact smooth spin manifold with a positive scalar curvaturemetric and dimension k − k > . The rank of the abelian group P ( M ) isgreater than or equal to r fin ( G ) − . (2) Let M be a compact smooth spin manifold with a positive scalar curvaturemetric and dimension k − k > . The rank of the abelian group P ( M ) isgreater than or equal to r fin ( G ) . Combining this result with our results, we get:
Corollary 4.4.
Let M be a compact smooth spin manifold with a positive scalarcurvature metric. Let G = π ( M ) . (1) If M has dimension k − k > , then the rank of the abelian group P ( M ) is greater than or equal to F polG − . (2) If M has dimension k − k > , then the rank of the abelian group P ( M ) is greater than or equal to F polG . This paper consists of 5 sections (including the introduction): • In Section 2, we prove our framework theorem (Theorem 1.4). • In Section 3, we recall dominating functions from [7]. As Engel did in [2],using the dominating functions, we define a seminorm k . k µ,h on C G for each h ∈ G pol . Using the seminorm and the operator norm, we complete C G andget a smooth dense subalgebra of C ∗ r G. We call that algebra C polh G. We alsorecall the trace functions on C G corresponding to an element in G. Using theproperties of the seminorms, we lift τ h to a trace function e τ h on C polh G. • In Section 4, we prove Theorem 1.7. As applications, we derive lower boundsfor the ranks of the structure group and the group of positive scalar curvaturemetrics of manifolds. • In Section 5, we define the class of polynomially full groups. We show thatsubgroups, products, and finite extensions of polynomially full groups are alsopolynomially full. For polynomially full group G, we show that K fin ( C ∗ r G ) ∼ = K fin ( C ∗ G ) ∼ = F G M i =1 Z . The class of polynomially full groups includes trivially all finite groups andfinitely generated torsion-free groups. We show that it also includes all finitelygenerated virtually nilpotent groups. At the end of the section, we deriveformulas for the number F G , where G is finitely generated abelian group,dihedral group, or symmetric group. Acknowledgements
The author would like to acknowledge Guoliang Yu for his invaluable guidance.The author would also like to acknowledge Alexander Engel and Bogdan Nica fortheir valuable suggestions. 2.
Proof of Theorem 1.4
Proof.
Since A is a smooth and dense subalgebra of C ∗ r G, we have K ( A ) ∼ = K ( C ∗ r G ) , S.K.SAMURKAŞ where the isomorphism is induced by the inclusion map i : A → C ∗ r G .
Since the finite parts of K ( A ) and K ( C ∗ r G ) are coming from C G, we have K fin ( A ) ∼ = K fin ( C ∗ r G ) . Hence, for the first part of the theorem, it suffices to show that rank( K fin ( A )) ≥ | S/ ∼ fin | . Let K S ( A ) be the subgroup of K fin ( A ) generated by the set { [ p g ] : g ∈ S } . Thus, itsuffices to show that rank( K S ( A )) ≥ | S/ ∼ fin | . Let { s , ..., s n } be an arbitrary subset of S such that, we have s i ≁ fin s j for i = j. We are going to show that, the subgroup of K S ( A ) generated by the set { [ p s ] , · · · [ p s n ] } has rank n. Therefore, we are going to conclude that rank( K S ( A )) ≥ | S/ ∼ fin | . For all i ∈ { , , · · · , n } let d i = order( s i ) and assume d ≤ d ≤ · · · ≤ d n . We haveall the traces e τ s i : A → C defined. This gives us the homomorphisms (with abuse ofnotation) e τ s i : K S ( A ) → C . Now define M n = e τ s ([ p s ]) e τ s ([ p s ]) · · · e τ s ([ p s n ]) e τ s ([ p s ]) e τ s ([ p s ]) · · · e τ s ([ p s n ]) ... ... . . . ... e τ s n ([ p s ]) e τ s n ([ p s ]) · · · e τ s n ([ p s n ]) , where p s j = s j + ... + s dj − j d j ∈ C G ⊆ A and [ p s j ] shows the class in K S ( A ) representedby the projection p s j . Now ∀ i, j ∈ { , , ..., n } with i > j , there are 2 cases: Case 1 ( d i > d j ). In this case, we have e τ s i ([ p s j ]) = τ s i ( p s j ) = τ s i s j + ... + s d j − j d j ! and since ∀ a ∈ N order( s aj ) ≤ order( s j ) = d j < d i = order( s i ) , we have ∀ a ∈ N s aj / ∈ C ( s i ) (all elements from C ( s i ) have order d i ). Thus, e τ s i ([ p s j ]) = 0 . Case 2 ( d i = d j ) and s j ≁ d s i . In this case, we have ∀ a ∈ N s aj / ∈ C ( s i ) by definitionof ∼ d . So e τ s i ([ p s j ]) = 0 . Hence, M n is an upper triangular matrix.Now ∀ i ∈ { , , ..., n } , we have s i ∈ C ( s i ) so, e τ s i ([ p s i ]) = τ s i ( p s i )= τ s i s i + ... + s d i − i d i ! ≥ d i . OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 7 Thus, the elements in the diagonal of M n are non-zero. Hence, det ( M n ) = 0 . So M n has full rank. Thus, in K S ( A ) , the elements [ p s ] , ..., [ p s n ] are Z -linearlyindependent. Therefore, rank( K S ( A )) = n. Thus, we get rank( K fin ( A )) ≥ | S/ ∼ fin | . Now, let’s make some preliminary definitions for the proof of the second part ofthe Theorem 1.4:Let H be an infinite dimensional separable Hilbert space. Let B ( H ) be the set ofbounded linear operators on H . We define S p := { T ∈ B ( H ) | tr (( T ∗ T ) p ) < ∞} , where tr ( P ) := P n ∈ N h P e n , e n i for an orthonormal basis { e n } ∞ n =1 and for a boundedlinear operator P ∈ B ( H ) . We remark that, trace does not depend on the particularchoice of an orthonormal basis. We call S p the ring of Schatten p -class operators onan infinite dimensional and separable Hilbert space. Now define S := ∪ ∞ p =1 S p . Thering S is called the ring of Schatten class operators. Let S G be the group algebraover the ring S [13]. Let j : C G → S G be the inclusion homomorphism defined by: j ( a ) = p a for all a ∈ C G, where p is a rank one projection in S . In the following, we show that nonzero elements in the finite part of K ( C ∗ G ) generated by the set { [ p g ] : g ∈ S } are not in the image of the assembly map µ : K G ( EG ) → K ( C ∗ G ) , where EG is the universal space for proper and free G -action. In the proof, we use the n -cocycle τ ( n ) g on S m G introduced in [10], whichgives in some sense the extension of the classical trace τ g . So we have a commutativediagram K ( S m G ) ψ / / ( τ ( n ) g ) ∗ (cid:15) (cid:15) K ( C ∗ G ) e τ g w w ♣♣♣♣♣♣♣♣♣♣♣♣ C where (with abuse of notation) e τ g : K ( C ∗ G ) → C is the pullback of the homomor-phism e τ g : K ( C ∗ r G ) → C . Recall that K ( A ) ∼ = K ( C ∗ r G ) . Assume there exists a non-zero z ∈ Im( µ ) ∩ K S ( C ∗ G ) . Then z = P si =1 c i · [ p g i ] for some pairwise non-equivalent g , · · · , g s ∈ S and c , · · · , c s ∈ Z \ { } . For all i ∈{ , · · · , s } let d i = order( g i ) . Without loss of generality, we can assume d ≤ · · · ≤ d s . Now let g = g s . Now let z ′ = P si =1 c i · [ j ( p g i )] . We have z ′ ∈ K ( S m G ) for some m ∈ N . Then weget z ′ ∈ Im( A ) , where A : H OrG ( EG, K ( S m ) −∞ ) → K ( S m G ) is the assembly map.Let n = 2 k be the smallest even number greater than or equal to m. Define an n -cocycle τ ( n ) g on S m G by: τ ( n ) g ( a , a , · · · , a n ) := X γ ∈ C ( g ) tr ( γ − a a · · · a n ) for all a i ∈ S m G, where tr : S G → C , is the trace defined by: tr ( X γ ∈ G b γ γ ) := trace ( b e ) . S.K.SAMURKAŞ
Since τ ( n ) g is an n -cocycle, it induces a homomorphism ( τ ( n ) g ) ∗ : K ( S m G ) → C . It is shown in [10] that ( τ ( n ) g ) ∗ ([ j ( p )]) = τ g ( p ) for all projections p ∈ C G and ( τ ( n ) g ) ∗ ( z ′ ) = 0 . So we have the commutative diagram K ( S m G ) ψ / / ( τ ( n ) g ) ∗ (cid:15) (cid:15) K ( C ∗ G ) e τ g w w ♣♣♣♣♣♣♣♣♣♣♣♣ C where (with abuse of notation) e τ g : K ( C ∗ G ) → C is the pullback of the homomor-phism e τ g : K ( C ∗ r G ) → C . We have ψ ( z ′ ) = z. Hence, we get e τ g ( z ) = e τ g ( ψ ( z ′ )) = ( τ ( n ) g ) ∗ ( z ′ ) = 0 . However, we have e τ g ( z ) = e τ g ( P si =1 c i · [ p g i ]) = P si =1 c i · e τ g ([ p g i ]) = k · c s d s for some k ∈ N . Thus, e τ g ( z ) = 0 . Contradiction shows that
Im( µ ) ∩ K S ( C ∗ G ) = { } . (cid:3) Remark 2.1.
For any group G, we can build up a matrix similar to the matrix in theproof of Theorem 1.4, and show that p ′ g s corresponding to pairwise non-equivalent g ′ s in G fin are linearly independent in K ( C G ) , since all the traces τ g : C G → C are already defined. Hence, we can conclude that K fin ( C G ) ∼ = F G M i =1 Z , as soon as F G ≤ ℵ , where ℵ is the cardinality of the set of the natural numbers N , and K fin ( C G ) is the subgroup of K ( C G ) generated by the idempotents { [ p g ] : g ∈ G fin } . Dominating Functions, Seminorms, and Trace Functions
In the first part of this section, we recall dominating functions from [7]. As Engeldid in [2], using the dominating functions, we define a seminorm k . k µ,h on C G foreach h ∈ G pol . Using the seminorms and the operator norm, we complete C G and geta smooth dense subalgebra of C ∗ r G. We call that algebra C polh G. In the second partof this section, we recall the trace functions on C G corresponding to an element in G. Using the properties of the seminorms, we lift τ h to a trace function e τ h on C polh G for each h ∈ G pol . Dominating Functions.
In the first part of this section, we recall the domi-nating functions, prove some properties about them, and using those functions, wedefine seminorms on C G. Completing C G with respect to those seminorms and theoperator norm, we construct smooth dense subalgebras C polh G of C ∗ r G for h ∈ G pol . In the following, we recall preliminary notions for the definition of the dominatingfunctions. We use R > and R ≥ for the sets of positive and non-negative real numbers,respectively. OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 9 Definition 3.1.
Given u ∈ ℓ G define Supp u := { g ∈ G : u ( g ) = 0 } . Now for all S ⊆ G and R ∈ R ≥ , define B R ( S ) := { g ∈ G : d w ( g, S ) ≤ R } , where d w is themetric induced by k . k w . Define k u k S := ( P g ∈ S | u ( g ) | ) . In the following, we recall the dominating function µ A for an operator A ∈ B ( ℓ G ) . We use these dominating functions to define some seminorms on C G. Definition 3.2. [2] For all A ∈ B ( ℓ G ) define µ A : R > → R ≥ as µ A ( R ) := inf { C ∈ R > : k Au k G \ B R (Supp u ) ≤ C · k u k , for all u ∈ ℓ G } . The following is a triangular inequality result we use at several places in our paper.
Lemma 3.3.
For all
A, B ∈ B ( ℓ G ) and R ∈ R > we have µ A + B ( R ) ≤ µ A ( R ) + µ B ( R ) . Proof.
For all R ∈ R > and u ∈ ℓ G, we have k ( A + B ) u k G \ B R ( Supp u ) ≤ k Au k G \ B R ( Supp u ) + k Bu k G \ B R ( Supp u ) ≤ µ A ( R ) · k u k + µ B ( R ) · k u k = ( µ A ( R ) + µ B ( R )) · k u k . Thus, we get µ A + B ( R ) ≤ µ A ( R ) + µ B ( R ) for all R ∈ R > . (cid:3) In the following, we estimate the dominating function with the operator norm.
Lemma 3.4.
For all A ∈ B ( ℓ G ) , we have µ A ( R ) ≤ k A k op for all R ∈ R > . Proof.
For all A ∈ B ( ℓ G ) , R ∈ R > , u ∈ ℓ G, we have k Au k G \ B R ( Supp u ) ≤ k Au k ≤ k A k op · k u k . Thus, we get µ A ( R ) ≤ k A k op . (cid:3) In the following, we use the previous estimate to show that convergence in theoperator norm implies point-wise convergence in the dominating functions. We usethis result in the proof of the smoothness of the subalgebras C polh G of C ∗ r G for h ∈ G pol . Lemma 3.5.
Let { A n } ∞ n =1 be a sequence of operators in B ( ℓ G ) converging (in k . k op norm) to A ∈ B ( ℓ G ) . Then { µ A n } ∞ n =1 converges to µ A point-wise.Proof. Given R ∈ R > , we have µ A n ( R ) = µ A +( A n − A ) ( R ) ≤ µ A ( R ) + µ A n − A ( R ) ≤ µ A ( R ) + k A n − A k op . Similarly, we get µ A ( R ) ≤ µ A n ( R ) + k A − A n k op . Thus, we have µ A ( R ) − k A − A n k op ≤ µ A n ( R ) ≤ µ A ( R ) + k A n − A k op . Hence, we get lim n →∞ µ A n ( R ) = µ A ( R ) , ∀ R ∈ R > . (cid:3) Remark 3.6.
Actually we have uniform convergence of { µ A n } ∞ n =1 to µ A . However,point-wise convergence is enough for our purposes.In the following, we are defining the seminorms we use to build the smooth densesubalgebras C polh G of C ∗ r G for h ∈ G pol . Definition 3.7.
Recall that given h ∈ G pol ∃ C h ∈ R > and d h ∈ N such that ∀ l ∈ N we have n h,l ≤ C h · l d h . Let b h be a natural number greater than or equal to d h + 2 . Now define k A k µ,h := inf { D ∈ R > | µ A ( R ) ≤ D · R − b h ∀ R ∈ R > } . Lemma 3.8. k . k µ,h is a seminorm on C G. Proof.
For all A ∈ C G, we have obviously k A k µ,h ≥ . Now for all
A, B ∈ C G and R > we have µ A + B ( R ) ≤ µ A ( R ) + µ B ( R ) ≤ k A k µ,h R − b h + k B k µ,h R − b h = ( k A k µ,h + k B k µ,h ) R − b h . Therefore k A + B k µ,h ≤ k A k µ,h + k B k µ,h . Hence, k . k µ,h is a seminorm on C G. (cid:3) In the following, we define our main gadget (a smooth dense subalgebra of C ∗ r G ).We use the properties of the seminorm to lift the trace function τ h (originally on C G )to this subalgebra of C ∗ r G. Definition 3.9.
For each h ∈ G pol , we define C polh G as the completion of C G withrespect to the norm k . k op and the seminorm k . k µ,h . Since C polh G contains C G , it is dense (in the operator norm) in C ∗ r G. In the following, we show that C polh G is an algebra over the complex numbers. Theonly nontrivial part is to show that it is closed under multiplication. Lemma 3.10. C polh G is an algebra over C .Proof. Given
A, B ∈ C polh G, there exist sequences { A n } ∞ n =1 and { B n } ∞ n =1 in C G con-verging (in both norms) to A and B respectively.The only nontrivial part is to show that lim n →∞ k AB − A n B n k µ,h = 0 . We have k AB − A n B n k µ,h = k ( AB − A n B ) + ( A n B − A n B n ) k µ,h = k ( A − A n ) B + A n ( B − B n ) k µ,h ≤ k ( A − A n ) B k µ,h + k A n ( B − B n ) k µ,h . We only show lim n →∞ k ( A − A n ) B k µ,h = 0 : Let C n = A − A n then, for all R ∈ R > , we have (the first inequality below is from[7, Prop. 5.2]) µ C n B ( R ) ≤ k C n k op µ B ( R/
2) + k B k op µ C n ( R/
2) + 2 µ C n ( R/ µ B ( R/ ≤ k C n k op k B k µ,h ( R/ − b h + k B k op k C n k µ,h ( R/ − b h + 2 k C n k µ,h k B k µ,h ( R/ − b h = { b h +1 k C n k op k B k µ,h + 2 b h k C n k µ,h k B k op + 2 b h +1 k C n k µ,h k B k µ,h R − b h } R − b h . Now if R ≥ , then b h +1 k C n k op k B k µ,h + 2 b h k C n k µ,h k B k op + 2 b h +1 k C n k µ,h k B k µ,h R − b h ≤ b h +1 k C n k op k B k µ,h + 2 b h k C n k µ,h k B k op + 2 b h +1 k C n k µ,h k B k µ,h . Let D n = 2 b h +1 k C n k op k B k µ,h + 2 b h k C n k µ,h k B k op + 2 b h +1 k C n k µ,h k B k µ,h . We use a notation different than in [2].
OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 11 If < R < , then µ C n B ( R ) ≤ k C n B k op ≤ k C n k op k B k op ≤ k C n k op k B k op · R − b h . So for all R ∈ R > , we have µ C n B ( R ) ≤ max { D n , k C n k op k B k op } · R − b h . Hence, weget k C n B k µ,h ≤ max { D n , k C n k op k B k op } . Since we have lim n →∞ D n = lim n →∞ k C n k op k B k op = 0 , we get lim n →∞ k C n B k µ,h = 0 . Similarly, we can show that lim n →∞ k A n ( B − B n ) k µ,h = 0 . Thus, we get lim n →∞ k AB − A n B n k µ,h = 0 . Hence, we have lim n →∞ A n B n = AB. So AB ∈ C polh G. Thus, C polh G is an algebra. (cid:3) In the following, we are giving an estimate that is used in the proof of smoothnessof C polh G. It can be proven by induction on n. Lemma 3.11. [2]
Given A ∈ C polh G and n ∈ N , we have µ (Id − A ) n ( R ) ≤ n − X k =1 k k Id − A k n − op µ A (cid:16) R k (cid:17) . In the following, we show that C polh G is a smooth subalgebra of C ∗ r G. Lemma 3.12. [2] C polh G is closed under holomorphic functional calculus.Proof. Given A ∈ C polh G with k Id − A k op < ǫ , where ǫ =
12 15 · bh . We have µ A − − P Nn =0 (Id − A ) n ( R ) ≤ ∞ X n = N +1 µ (Id − A ) n ( R ) ≤ ∞ X n = N +1 n − X k =1 k ǫ n − µ A (cid:16) R k (cid:17) = N X k =1 n k µ A (cid:16) R k (cid:17) · ∞ X n = N +1 ǫ n − o + ∞ X k = N +1 n k µ A (cid:16) R k (cid:17) · ∞ X n = k +1 ǫ n − o = ǫ N − ǫ N X k =1 k µ A (cid:16) R k (cid:17) + 11 − ǫ ∞ X k = N +1 (5 ǫ ) k µ A (cid:16) R k (cid:17) ≤ ǫ N · k A k µ,h − ǫ · R − b h · N X k =1 (5 · b h ) k + k A k µ,h − ǫ · R − b h · ∞ X k = N +1 (5 · ǫ · b h ) k = k A k µ,h − ǫ · (cid:26) ǫ N N X k =1 (5 · b h ) k + ∞ X k = N +1 ( 12 ) k (cid:27) · R − b h . Thus, we have k A − − P Nn =0 (Id − A ) n k µ,h ≤ k A k µ,h − ǫ · { ǫ N P Nk =1 (5 · b h ) k + P ∞ k = N +1 ( ) k } . Now since lim N →∞ { ǫ N P Nk =1 (5 · b h ) k + P ∞ k = N +1 ( ) k } = 0 , we get lim N →∞ k A − − N X n =0 (Id − A ) n k µ,h = 0 . Hence we have A − ∈ C polh G. So C polh G is closed under holomorphic functionalcalculus by [8, Lemma 1.2] and [3, Lemma 3.38]. (cid:3) Trace Functions.
In this section, we recall the trace function τ h on C G corre-sponding to an element h ∈ G. If h ∈ G pol , then we extend this trace to a trace e τ h on C polh G. In the following, we are recalling the classical trace on C G corresponding to anelement h ∈ G. Definition 3.13.
For all h ∈ G, let τ h : C G → C be defined as: τ h ( X g ∈ G a g .g ) := X g ∈ C ( h ) a g , where C ( h ) is the conjugacy class of h. It is clear that τ h is a trace function on C G [9].In the following, we introduce a notation so that, we can use operators as matrices. Definition 3.14.
Given A ∈ B ( ℓ G ) , define A ( g, f ) := ( Aδ f )( g ) for all g, f ∈ G, where δ f ( k ) = ( if k = f otherwise . The following equivariance property is used in the proof that liftings e τ h : C polh G → C are trace functions. It can be shown with a direct calculation. Lemma 3.15.
Given A ∈ C ∗ r G and g, f, h ∈ G , we have A ( g, f ) = A ( gh, f h ) . In the following, we define a lifting of the classical trace function τ h : C G → C . Definition 3.16.
For each h ∈ G pol , define e τ h : C polh G → C as, e τ h ( A ) = X g ∈ C ( h ) A ( g, e ) . In the following, we prove an inequality that we use in the proof of the Theo-rem 3.18.
Lemma 3.17.
For all A ∈ B ( ℓ G ) and R ∈ R > we have (cid:18) X k g k w >R | A ( g, e ) | (cid:19) ≤ µ A ( R ) . Proof.
For all A ∈ B ( ℓ G ) and R ∈ R > we have (cid:18) X k g k w >R | A ( g, e ) | (cid:19) = (cid:18) X g ∈ G \ B R ( { e } ) | ( Aδ e )( g ) | (cid:19) = k Aδ e k G \ B R (Supp δ e ) ≤ µ A ( R ) k δ e k = µ A ( R ) . (cid:3) OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 13 Since we defined e τ h to be a sum over the (possibly infinite) set C ( h ) , we need toprove that the sum converges. In the following, we show that the sum absolutelyconverges. Theorem 3.18. e τ h : C polh G → C is well defined and continuous.Proof. Given A ∈ C polh G, we have 2 cases:If h = e , then | e τ h ( A ) | = | A ( e, e ) | < ∞ . If h = e , then we have | e τ h ( A ) | ≤ ∞ X l =1 X g ∈ C l ( h ) | A ( g, e ) | (since h = e, l starts from ) ≤ ∞ X l =1 √ n h,l · (cid:18) X g ∈ C l ( h ) | A ( g, e ) | (cid:19) (Cauchy-Schwarz inequality) ≤ ∞ X l =1 √ n h,l · (cid:18) X k g k w > ( l − ) | A ( g, e ) | (cid:19) ≤ ∞ X l =1 √ n h,l · µ A (cid:18) l − (cid:19) (Lemma 3.17) ≤ ∞ X l =1 p C h · l dh · k A k µ,h · (cid:18) l − (cid:19) − b h ≤ C · p C h · k A k µ,h · ∞ X l =1 l − , for some C ∈ R > . ( b h ≥ d h < ∞ . Hence e τ h : C polh G → C is well-defined and continuous. (cid:3) In the following, we show indeed e τ h : C polh G → C is a trace function extending theclassical trace function τ h : C G → C . Theorem 3.19.
For all h ∈ G pol , e τ h is a trace function on C polh G extending τ h . Proof.
The only nontrivial part is to show that ∀ A, B ∈ C polh G we have e τ h ( AB ) = e τ h ( BA ) : Given
A, B ∈ C polh G , we have e τ h ( AB ) = X f ∈ G X g ∈ C ( h ) A ( gf − , e ) B ( f, e )= X f ∈ G X k ∈ C ( h ) A ( f − k, e ) B ( f, e ) ( k = f gf − )= X k ∈ C ( h ) X f ∈ G A ( f − k, e ) B ( f, e )= X k ∈ C ( h ) X l ∈ G A ( l, e ) B ( kl − , e ) ( l = f − k )= e τ h ( BA ) . (cid:3) We have absolute convergence in the sums (by the proof of Lemma . ). So we can changethe order of summation as we want. proof of Theorem 1.7 and its applications In this section, we present the proof of Theorem 1.7 and apply the result to derivelower bounds for the groups S ( M ) and P ( M ) . Proof of Theorem 1.7.
Since, for all g, h ∈ G fin , we have g ∼ fin h = ⇒ [ p g ] = [ p h ] ∈ K fin ( C ∗ G ) , we get rank( K fin ( C ∗ G )) ≤ F G . Using the surjection K fin ( C ∗ G ) ։ K fin ( C ∗ r G ) , wecan conclude that rank( K fin ( C ∗ r G )) ≤ rank( K fin ( C ∗ G )) . For the rest, it suffices to prove that S = G pol ∩ G fin and A = C polS G := T h ∈ S C polh G satisfies the hypotheses of the Theorem 1.4.Since C polh G ′ s are smooth subalgebras of C ∗ r G containing C G, we get C polS G is asmooth subalgebra of C ∗ r G containing C G. Since G pol ∩ G fin ⊆ G pol , for all h ∈ G pol ∩ G fin , τ h : C G → C has a lift e τ h : C polh G → C . Therefore, for all h ∈ S, the trace function τ h : C G → C has a lift e τ h : C polS G → C , which is also a trace function.Hence, we get rank( K fin ( C ∗ r G )) ≥ | S/ ∼ fin | = | ( G pol ∩ G fin ) / ∼ fin | = F polG . For theassembly map µ : K G ( EG ) → K ( C ∗ G ) , we have Im µ ∩ K fin,pol ( C ∗ G ) = { } . (cid:3) Applications.
In this subsection, we combine the results from Weinberger andYu [10] and Theorem 1.7 to derive lower bounds for the ranks of the structure groupand the group of positive scalar curvature metrics of manifolds.Given a compact oriented manifold M, we define the structure group S ( M ) of M to be the abelian group generated by the equivalence classes of elements of the form ( f, M ′ ) , where M ′ is a compact oriented manifold and f : M ′ → M is an orientationpreserving homotopy equivalence. We say ( f , M ) is equivalent to ( f , M ) if thereexists an h-cobordism ( W ; M , M ) and a homotopy equivalence F : W → M suchthat restrictions of F to M and M gives f and f respectively [6, Definition 1.14].We have the following result about the structure group S ( M ) of a compact orientedmanifold M from Weinberger and Yu. Theorem 4.1. [10]
Let M be a compact oriented manifold with dimension k − k > . Suppose π ( M ) = G and g , · · · , g n be finite order elements in G such that g i = e for all i and { [ p g ] , · · · , [ p g n ] } generates an abelian group of K ( C ∗ G ) with rankn. Suppose that any nonzero element in the abelian subgroup of K ( C ∗ G ) generatedby { [ p g ] , · · · , [ p g n ] } is not in the image of the map µ : K G ( EG ) → K ( C ∗ G ) , thenthe rank of the structure group S ( M ) is greater than or equal to n. Now we combine the previous result about S ( M ) with Theorem 1.7, where thelower bound is in terms of F polG . Corollary 4.2.
For a compact oriented manifold M with dimension k − k > , the rank of the structure group S ( M ) is greater than or equal to F polG − , where G = π ( M ) . OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 15 Proof.
We have rank( K fin,pol ( C ∗ G )) ≥ F polG by the proof of Theorem 1.7 and, wehave Im( µ ) ∩ K fin,pol ( C ∗ G ) = { } . Since we have e ∈ G fin ∩ G pol , we get the rank ofthe structure group S ( M ) is greater than or equal to F polG − . (cid:3) Let r fin ( G ) be the rank of the abelian group K fin ( C ∗ G ) generated by [ p g ] for allfinite order elements g ∈ G. Here g is allowed to be the identity element e. So wehave r fin ( G ) = rank( K fin ( C ∗ G )) ≥ F polG . If a compact smooth spin manifold M has a positive scalar curvature metric andthe dimension of M is greater than or equal to , then we define (roughly) P ( M ) tobe the abelian group of equivalent classes of all positive scalar curvature metrics onM. For a more precise definition, we refer to [10, Section 4].We have the following result about the group P ( M ) from Weinberger and Yu. Theorem 4.3. [10](1)
Let M be a compact smooth spin manifold with a positive scalar curvaturemetric and dimension k − k > . The rank of the abelian group P ( M ) isgreater than or equal to r fin ( G ) − . (2) Let M be a compact smooth spin manifold with a positive scalar curvaturemetric and dimension k − k > . The rank of the abelian group P ( M ) isgreater than or equal to r fin ( G ) . In the following, we combine the previous result about P ( M ) with Theorem 1.7.The lower bounds are in terms of F polG . Corollary 4.4.
Let M be a compact smooth spin manifold with a positive scalarcurvature metric and let G = π ( M ) . (1) If M has dimension k − k > , then the rank of the abelian group P ( M ) is greater than or equal to F polG − . (2) If M has dimension k − k > , then the rank of the abelian group P ( M ) is greater than or equal to F polG . Proof.
We have r fin ( G ) ≥ F polG . (cid:3) Polynomially Full Groups
In this section, we define the class of polynomially full groups. We show thatsubgroups, products, and finite extensions of polynomially full groups are also poly-nomially full. For polynomially full group G, we show that K fin ( C ∗ r G ) ∼ = K fin ( C ∗ G ) ∼ = F G M i =1 Z . The class of polynomially full groups includes trivially all finite groups and finitelygenerated torsion-free groups. We show that it also includes all finitely generatedvirtually nilpotent groups. At the end of the section, we derive formulas for thenumber F G , where G is finitely generated abelian group, dihedral group, or symmetricgroup.For a finitely generated group G with a finite generating set S, we denote the word-length norm by k . k , k . k S , or k . k G . For g ∈ G, we denote the conjugacy class of g in G by C G ( g ) . We denote the set of elements in the conjugacy class of g with length(with respect to S ) l by C Gl ( g ) . In the following, we give two equivalent conditions for a group. We use theseconditions to define the class of polynomially full groups.
Proposition 5.1.
For a finitely generated group G the following are equivalent: (1) G fin ⊆ G pol . (2) For all g ∈ G fin there exists h ∈ G fin ∩ G pol such that g ∼ fin h. Proof. (1) = ⇒ (2) : Obvious. (2) = ⇒ (1) : Given g ∈ G fin , there exists h ∈ G fin ∩ G pol such that g ∼ fin h. So there exist a, b ∈ N and f ∈ G such that g a = f hf − and h b ∈ C G ( g ) . Define A l = { α a : α ∈ C G ( g ) , k α k = l } . Define i : C Gl ( g ) → A l by i( α ) = α a and j : A l → C Gl ( g ) by j( β ) = β b . It is easy to see that, j ◦ i = id C Gl ( g ) and i ◦ j = id A l . Hence, we have | C Gl ( g ) | = | A l | . Now, let B l = { β ∈ C G ( h ) : k β k ≤ a · l } . We show that A l ⊆ B l : Given ω ∈ A l , there exists α ∈ C G ( g ) with k α k = l and ω = α a . Since α ∈ C G ( g ) , there exists γ ∈ G such that α = γgγ − . So ω = α a = γg a γ − = γf hf − γ − . Hence, ω ∈ C G ( h ) with k ω k = k α a k ≤ a · k α k = a · l. Thus, ω ∈ B l . Therefore, we have A l ⊆ B l . So we get | C Gl ( g ) | = | A l | ≤ | B l | . Since h ∈ G pol , | B l | is bounded from above by apolynomial of l. Thus, we get g ∈ G pol . (cid:3) Definition 5.2.
Let G be a finitely generated group. We say that G is polynomiallyfull, if it satisfies the conditions from Proposition 5.1.Obviously finite groups and finitely generated torsion-free groups are polynomiallyfull.The following result is the motivation behind the definition of polynomially fullgroups. Theorem 5.3.
For a polynomially full group G we have K fin ( C ∗ r G ) ∼ = K fin ( C ∗ G ) ∼ = F G M i =1 Z . Proof.
Let G be a polynomially full group. We have F G = | G fin / ∼ fin | = | ( G fin ∩ G pol ) / ∼ fin | = F polG So we have L F G i =1 Z = L F polG i =1 Z . Let F = ( G fin ∩ G pol ) / ∼ fin . Hence, we have L F polG i =1 Z ∼ = L [ g ] ∈ F Z . Now, define φ : M [ g ] ∈ F Z → K fin ( C ∗ r G ) as φ ( δ [ g ] ) = [ p g ] , OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 17 where δ [ g ] is the canonical basis element corresponding to the [ g ] -component. Since g ∼ fin h implies [ p g ] = [ p h ] , φ is a well defined homomorphism. Recall that, K fin ( C ∗ r G ) is generated by the set { [ p g ] : g ∈ G fin } . Since G is polynomially full, we have { [ p g ] : g ∈ G fin } = { [ p g ] : g ∈ G fin ∩ G pol } . Hence, φ is surjective.On the other hand, by the proof of the Theorem 1.4, φ ( δ [ g ] ) = [ p g ] ′ s are Z -linearlyindependent. Thus, φ is injective. So we have L [ g ] ∈ F Z ∼ = K fin ( C ∗ r G ) . Therefore,we get K fin ( C ∗ r G ) ∼ = L F G i =1 Z . The isomorphism K fin ( C ∗ G ) ∼ = L F G i =1 Z can be shownsimilarly. (cid:3) In the following, we show that finite extensions and images of polynomially fullgroups under homomorphisms with finite kernels are also polynomially full.
Proposition 5.4.
Let F be a finite group and let G, H be finitely generated groups.If we have a short exact sequence F G H , α β then G is polynomially full if and only if H is polynomially full. In this case, we have F H ≤ F G . Proof.
Without loss of generality, assume that the generating set of H is in the imageof the generating set of G under β. Assume G is polynomially full. Given h ∈ H fin , since β is onto, there exists g ∈ G with β ( g ) = h. Since F is finite, we get g ∈ G fin . Now, since G is polynomially full, ∃ g ′ ∈ G fin ∩ G pol such that g ∼ fin g ′ . Now, let h ′ = β ( g ′ ) ∈ H. Since g ′ ∈ G fin , wehave h ′ ∈ H fin . Since g ∼ fin g ′ , we have h ∼ fin h ′ . Now, since F is finite, there exists R ∈ N such that ker β = Im α ⊆ B R ( e G ) , where B R ( e G ) is the closed ball around theidentity element of G with radius R. Since, ∪ Nl =1 C Hl ( h ′ ) ⊆ β ( ∪ N + Rl =1 C Gl ( g ′ )) , we have | ∪ Nl =1 C Hl ( h ′ ) | ≤ | ∪ N + Rl =1 C Gl ( g ′ ) | , andright hand side is bounded by some polynomial of N + R (hence by a polynomial of N ). So we get h ′ ∈ H pol . Hence, h ′ ∈ H fin ∩ H pol with h ′ ∼ fin h. Therefore, H ispolynomially full.For the converse, assume H is polynomially full. Given g ∈ G fin , let h = β ( g ) ∈ H fin . Since H is polynomially full, we have h ∈ H pol . Now we have β ( ∪ li =0 C Gi ( g )) ⊆ ∪ li =0 C Hi ( h ) . Hence, we get | ∪ li =0 C Gi ( g ) | ≤ | F | · | β ( ∪ li =0 C Gi ( g )) | ≤ | F | · | ∪ li =0 C Hi ( h ) | . Since h ∈ H pol and F is finite, right hand side is bounded by a polynomial of l. Thus, we get g ∈ G pol . Therefore, G is polynomially full.Now, given g , g ∈ G fin , g ∼ fin g (in G ) implies β ( g ) ∼ fin β ( g ) (in H ). Thus,we have F H = | H fin / ∼ fin | = | β ( G fin ) / ∼ fin | ≤ | G fin / ∼ fin | = F G . (cid:3) In the following, we prove that the property of being polynomially full is inheritedto finitely generated subgroups.
Proposition 5.5.
Let G be a finitely generated group. Let H be a finitely generatedsubgroup of G. If G is polynomially full, then H is also polynomially full.Proof. Let S and T be finite generating sets of G and H respectively. Without lossof generality, we can assume that T ⊆ S. Now, given h ∈ H fin , we have h ∈ G fin . Since G is polynomially full, we have h ∈ G pol . It is easy to see that C H ( h ) ⊆ C G ( h ) . Now, for all f ∈ H, we have k f k T ≥ k f k S . Hence, C Hl ( h ) ⊆ ∪ li =0 C Gi ( h ) . Thus, we have | C Hl ( h ) | ≤ | ∪ li =0 C Gi ( h ) | . Since h ∈ G pol , right hand side is bounded by a polynomial of l. Therefore, we get h ∈ H pol . Hence, H is polynomially full. (cid:3) In the following, we show that the class of polynomially full groups is closed undertaking direct products.
Proposition 5.6.
Let G and H be finitely generated groups. Then G and H arepolynomially full if and only if G × H is polynomially full.Proof. Let S and T be finite generating sets for G and H respectively. Then, W := S × { e H } ∪ { e G } × T is a finite generating set for G × H. Let k . k G , k . k H , and k . k ( G × H ) be the word-lengthnorms on G, H, and G × H respectively, corresponding to the generating sets S, T, and W respectively.Assume G and H are polynomially full. Given ( g, h ) ∈ ( G × H ) fin , we have g ∈ G fin and h ∈ H fin . Since G and H are polynomially full, we have g ∈ G pol and h ∈ H pol . It is not hard to see that k ( g ′ , h ′ ) k ( G × H ) = k g ′ k G + k h ′ k H for all g ′ ∈ G and h ′ ∈ H. So we have | C G × Hn (( g, h )) | = n X i =0 | C Gi ( g ) | · | C Hn − i ( h ) | . All the terms in the sum are bounded by polynomials of n. Thus, the sum is boundedby a polynomial of n. Hence, ( g, h ) ∈ ( G × H ) pol . Therefore G × H is polynomiallyfull.Converse follows from Proposition 5.5. (cid:3) In the following, we give a sufficient condition for a group to be polynomially full.Recall that a subset of a group is said to grow polynomially if the number of elementsin the intersection of the subset with the closed ball of radius l centered around theidentity element is bounded by a fixed polynomial of l. Lemma 5.7.
Let G be a finitely generated group. If G fin grows polynomially, then G is polynomially full.Proof. For all g ∈ G fin we have C G ( g ) ⊆ G fin . Since G fin grows polynomially, C G ( g ) also grows polynomially. Hence, g ∈ G pol . Therefore, G is polynomially full. (cid:3) Wolf [12, Theorem 3.11] showed that for finitely generated group Σ and a subgroup Γ of finite index, we have that(1) Γ is finitely generated, and(2) if Γ has polynomial growth, then Σ also has polynomial growth.He also showed in [12, Theorem 3.2] that, if Γ is a finitely generated nilpotentgroup, then Γ has polynomial growth.Gromov [5] showed that if a finitely generated group Γ has polynomial growth,then it is virtually nilpotent. Recall that a group is called virtually nilpotent, if itcontains a nilpotent subgroup of finite index.In the following, we show that the class of polynomially full groups includes finitelygenerated virtually nilpotent groups. OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 19 Corollary 5.8.
Let G be a finitely generated group. If G is virtually nilpotent, then G is polynomially full.Proof. Let H be a nilpotent subgroup of G with finite index. By [12, Theorem 3.11], H is also finitely generated. So by [12, Theorem 3.2], H has polynomial growth. Hence,by [12, Theorem 3.11], G has polynomial growth. Thus, G fin also has polynomialgrowth. Therefore, G is polynomially full by Lemma 5.7. (cid:3) In the following propositions, we derive formulas for the number F G for somepolynomially full groups. Recall that when G is polynomially full, F G is the rank ofthe free abelian groups K fin ( C ∗ r G ) ∼ = K fin ( C ∗ G ) ∼ = L F G i =1 Z . In the following, we give a formula for F G for a finite abelian group G. Proposition 5.9.
For G = Z /n × · · · × Z /n k , we have the following formula F G = X d | n · · · X d k | n k φ ( d ) · · · φ ( d k ) φ (lcm( d , · · · , d k )) , where φ denotes Euler’s totient function, lcm denotes the least common multiplefunction, and the sums run over positive divisors d i ’s of n i ’s.Proof. Let’s define an equivalence relation ∼ on G = Z /n × · · · × Z /n k , which iscoarser than ∼ fin : We say ( x , · · · , x k ) ∼ ( x ′ , · · · , x ′ k ) if and only if, for all i ∈ { , · · · , k } x i ∼ fin x ′ i in Z /n i . It is easy to see that, ∼ is an equivalence relation on G. Since homomorphic images of equivalent elements are equivalent, by looking atthe projections to components, we can conclude that ( x , · · · , x k ) ∼ fin ( y , · · · , y k ) implies ( x , · · · , x k ) ∼ ( y , · · · , y k ) , for all ( x , · · · , x k ) , ( y , · · · , y k ) ∈ G. So ∼ iscoarser than ∼ fin . For all d , · · · , d k ∈ N with d | n , · · · , d k | n k , define G d , ··· ,d k := { ( x , · · · , x k ) ∈ G : gcd( n i , x i ) = n i d i for i ∈ { , · · · , k }} . It is easy to see that G d , ··· ,d k = [( x , · · · , x k )] ∼ for all ( x , · · · , x k ) ∈ G d , ··· ,d k , where [( x , · · · , x k )] ∼ is the equivalence class of the element ( x , · · · , x k ) with respect to ∼ . Now, for all ( x , · · · , x k ) ∈ G d , ··· ,d k , we have | [( x , · · · , x k )] ∼ fin | = φ (order(( x , · · · , x k )))= φ (lcm(order( x ) , · · · , order( x k )))= φ (lcm( d , · · · , d k )) and we have | G d , ··· ,d k | = φ ( d ) · · · φ ( d k ) . Hence, we have | G d , ··· ,d k / ∼ fin | = φ ( d ) ··· φ ( d k ) φ (lcm( d , ··· ,d k )) . Thus, we get F G = | G fin / ∼ fin | = | G/ ∼ fin | = X d | n · · · X d k | n k | G d , ··· ,d k / ∼ fin | = X d | n · · · X d k | n k φ ( d ) · · · φ ( d k ) φ (lcm( d , · · · , d k )) . (cid:3) In the following, we give a formula for F G for a finitely generated abelian group G. Corollary 5.10.
For G = Z /n × · · · × Z /n k × Z m , we have the following formula F G = X d | n · · · X d k | n k φ ( d ) · · · φ ( d k ) φ (lcm( d , · · · , d k )) , where the sums run over positive divisors d i ’s of n i ’s.Proof. Let H = Z /n × · · · × Z /n k . Since G is abelian and G fin ∼ = H, result followsfrom Proposition 5.9. (cid:3) Remark 5.11.
If we take G = Z /n, then the above formula tells that F G is equal tothe number of positive divisors of n. In the following, we give a formula for F G for a dihedral group G. Proposition 5.12.
For G = D n , we have F G = ( F Z /n + 1 if n is odd F Z /n + 2 otherwise ,where D n is the dihedral group of order n. Proof.
Let x, y be the generators of D n with x = y n = ( xy ) = 1 . For all a, b ∈ Z , we have ( xy a ) · y b · ( xy a ) − = ( xy a ) · y b · y − a x = x y − b = y − b , and y a · y b · y − a = y b . Sofor all a ∈ Z , we get [ y a ] ∼ fin ⊆ { , y, · · · , y n − } , where [ y a ] ∼ fin denotes the equivalenceclass of y a in D n . Now let’s show that y a ∼ fin y b if and only if gcd( n, a ) = gcd( n, b ) : For the forward direction, we have y a ∼ fin y b = ⇒ order( y a ) = order( y b )= ⇒ gcd( n, a ) = gcd( n, b ) . For the converse, assume we have d = gcd( n, a ) = gcd( n, b ) for some d ∈ N . So weget order( y a ) = order( y b ) = nd and gcd( nd , ad ) = gcd( nd , bd ) = 1 . Hence, there exists c ∈ N such that c · ad ≡ bd (mod nd ) . Thus, we get ac ≡ b (mod n ) . So ( y a ) c = y ac = y b . Therefore, we get y a ∼ fin y b . Now for all a, b ∈ Z , xy a has order , and ( xy b ) · xy a · ( xy b ) − = ( xy b ) · xy a · y − b x = xy b xy a − b x = xy b x y b − a = xy b y b − a = xy b − a and y b · xy a · y − b = xy − b y a − b = xy a − b . So we get [ xy a ] ∼ fin = { xy a +2 b | b ∈ Z } ∪ { xy − a +2 b | b ∈ Z } = { xy a +2 b | b ∈ Z } . OUNDS FOR THE RANK OF THE FINITE PART OF OPERATOR K -THEORY 21 Hence, xy a ∼ fin xy b if and only if ∃ c ∈ Z such that a + 2 c ≡ b (mod n ) . Thus, we get F G = ( F Z /n + 1 if n is odd F Z /n + 2 otherwise. (cid:3) Remark 5.13.
Let D ∞ be the infinite dihedral group. Let x, y be the elementsgenerating D ∞ with relations x = ( xy ) = 1 . Since D ∞ is virtually nilpotent (itcontains the subgroup h y i ∼ = Z of index 2), it is polynomially full by Corollary 5.8.Straightforward calculation shows that { , x, xy } is a complete set of representativesfor the equivalence classes in D fin ∞ / ∼ fin . Hence, F D ∞ = 3 . In the following, we give a formula of F G for G = S n . Proposition 5.14.
For all n ∈ N , F S n is equal to the number of conjugacy classesin S n , where S n is the symmetric group on a finite set of n symbols.Proof. For all g ∈ S n and for all a ∈ N with gcd( a, order( g )) = 1 , the permutations g and g a have the same cycle structures. So they are conjugates. Now, ∀ g, h ∈ S n , we have g ∼ fin h if and only if ∃ a ∈ N with gcd( a, order( g )) = 1 and g a ∈ C S n ( h ) . Therefore, we get g ∼ fin h if and only if g ∈ C S n ( h ) . (cid:3) References [1] P. Baum, A. Connes, N. Higson.
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Texas A&M University, Department of Mathematics, 77840-College Station, Texas,USA
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