eu THE HECKE ALGEBRA H ( P Q , P Z ) ARISING IN NUMBER THEORY

Abstract: The algebra H(P Q , PZ) arises in number theory and has been studied in [1] of Bost and Connes. Laca and Raeburn continued the study in [3] and gave an improvement of the theorem of Bost and Connes. This leads us to consider a closely related algebra H(PQ, PZ) because of its interesting connections with C∗-algebras and group algebras. In this paper we give a detailed proof of Laca and Raeburn’s theorem. Then we define a new Hecke pair (PQ, PZ) and show that the Hecke algebra H(PQ, PZ) is a universal ∗algebra generated by the elements {μn : n ∈ N ∗}, {e(r) : r ∈ Q/Z} and a new element


Introduction
A Hecke pair (G, S) consists of a discrete group G and a subgroup S such that every double coset consists of finitely many left cosets.
The Hecke algebra H(P + Q , P Z ) first arose in Bost and Connes study [2] and they have proved that it is a universal * -algebra generated by elements {µ n : n ∈ N * } and {e(r) : r ∈ Q/Z} subject to six relations.M. Laca and I. Raeburn studied also the Hecke algebra of Bost and Connes in [1] and they gave an improvement of Bost and Connes Theorem by showing that there are two redundant relations among the six in Bost and Connes Theorem.
In this paper, we first introduce Hecke algebras, and prove some results about them.We then studied Bost and Connes Hecke algebra H(P + Q , P Z ), where and then we gave a proof for the improvement of their Theorem, given by Laca and Raeburn in [1], after we stated and proved some lemmas that lead to the proof.
In the last part of this work we define a new Hecke pair (P Q , P Z ), where and we show that this Hecke algebra H(P Q , P Z ) is a universal * -algebra generated by elements {µ n : n ∈ N * }, {e(r) : r ∈ Q/Z} and a new element which we called u.

Hecke Algebras
Definition 2.1.Let G be a discrete group and S a subgroup of G.The pair (G, S) ia called a Hecke pair if each double coset StS can be written as a finite union of left cosets.This * -algebra is called a Hecke algebra.
Lemma 2.3.The sum in (2.1) is finite and depends only on StS.
Proof.The sum is finite because f (SrS) = 0 for only finitely many SrS ∈ S\G/S and these finitely many double cosets contain finitely many left cosets.For the second part of this lemma replace t in the right-hand side by s 1 ts 2 for some s 1 , s 2 ∈ S.
Proof of proposition 2.2.Since H(G, S) is obviously a vector space, and the distributive laws are pretty clear, we need to check that (f * g) * h = f * (g * h) and the properties of the involution.On the one hand, we have and on the other hand, By writing r = qd, and noting that for fixed qS, rS = dS , we find For λ, µ ∈ C, we have λf Finally, we compute and by writing p = tr we have that r = t −1 p, so Remark 2.5.Let S be any subgroup of a group G. Then the map Sp → p −1 S is a one to one correspondence between G/S and S\G.
Remark 2.6.If S is a normal subgroup of G, then each double coset StS is equal to tS, so S\G/S = G/S, and the Hecke algebra H(G, S) reduces to the group algebra C(G/S).
Lemma 2.7.Let (G, S) be a Hecke pair and r, t ∈ G. Then StS = SrS if and only if t ∈ SrS.
Proof.First suppose that e is the identity element of the group G and that StS = SrS.Since t = ete ∈ StS then t ∈ SrS.Secondly, suppose that t ∈ SrS.Then t = s 1 rs 2 for some s 1 , s 2 ∈ S, so StS = S(s 1 rs 2 )S = SrS.Lemma 2.8.Suppose that G is a subgroup of the group K and S is a subgroup of G such that (K, S) is a Hecke pair.Then there is an injective * -homomorphism ι from H(G, S) into H(K, S) such that Proof.That ι is linear follows directly from the definition, so we need to check that ι is * -preserving, multiplicative and injective.Now, let f ∈ H(G, S).Then Thus ι is multiplicative.Finally, let f, h ∈ H(G, S) and suppose that ι(f )(SkS) = ι(h)(SkS).Then ι(f − h)(SkS) = 0, since ι is linear and this equivalent to which is equivalent to say that Thus ι is one to one.

Representations of H(G,S)
Let S be any subgroup of a group G, and define Then ℓ 2 (G/S) is a Hilbert space with inner product (f |h) = pS∈G/S f (pS)h(pS).
Lemma 3.1.Let (G,S) be a Hecke pair.Then the formula λ Proof.To prove this lemma we need to show that the operator λ G/S t is isometric, onto and that λ Thus, λ G/S t defines a unitary representation of G on ℓ 2 (G/S).Proposition 3.2.Let (G, S) be a Hecke pair, and define ).On the one hand we have and on the other, By writing pb = v, and noting that b = p −1 v, we find Next, we compute R(h) * .Since R(h) * is a bounded linear operator on ℓ 2 (G/S) then for all f, g ∈ ℓ 2 (G/S) we have

The Hecke Pair of Bost and Connes
Bost and Connes defined and and showed that (P + Q , P Z ) is a Hecke pair.To see this, let G = P + Q and S = P Z .Then for arbitrary t = 1 a 0 r ∈ G, the left coset associated to t is and the double coset associated to t is By expressing r = p/q, where p and q are relatively prime in N (we can do this since r ∈ Q and r > 0), each double coset StS may be written as the disjoint union of q single cosets t i S by where t i = 1 a+ri 0 r .

Presentation of the Hecke Algebra
Let us first consider the Hecke pair (G, S) = (P + Q , P Z ).We shall use the following notations: µ n (SpS)f (Sp −1 tS).

Now µ n (SpS) vanishes unless
so there is exactly one non zero term in the sum, occuring when pS = 1 0 0 n S, and then µ n (SpS) = n −1/2 .Thus Now e(r)(Sp −1 tS) vanishes unless and by taking the inverse of both sides we have that so there is exactly one non zero term in the sum, occuring when t −1 pS = 1 r 0 1 −1 S, and then e(r)(Sp −1 tS) = 1.Thus Lemma 5.5.Consider the Hecke pair (G, S) where G = P + Q and S = P Z , then for f ∈ H(G, S) and n ∈ N * we have and by taking the inverse of both sides we get that so there are n non zero terms in the sum, occuring when pS = Proof of theorem 5.1 .We have to show, first, that H(P + Q , P Z ) is generated by elements µ n , e(r) satisfying (a)-(d).
For (a), let t ∈ G. Then Lemma 5.5 says that and For (b), let t ∈ G. Then Lemma 5.2 says that Now µ n S 1 0 0 1/m tS vanishes unless and this is equivalent to saying that So the non zero value of µ n S 1 0 0 1/m tS occurs when and then µ n S 1 0 0 1/m tS = n −1/2 .Thus Thus e(r) * = e(−r).For the second part of (c) Lemma 5.3 says that for t ∈ G we have and this equivalent to saying that so the non zero value of e(r 2 ) S 1 −r 1 0 1 tS occurs when While e(r 1 + r 2 )(StS) vanishes unless and this is equivalent to saying that Thus by comparing the two results we get that e(r 1 ) * e(r 2 ) (StS) = e(r 1 + r 2 )(StS).
For part (d), let t ∈ G.By applying first Lemma 5.2 and then Lemma 5.4 we get and since (P + Q , P Z ) is a Hecke pair we have Thus for some 1 Hence The right-hand side of (d) gives Our next step is to show that the elements {µ n : n ∈ N * } and {e(r Proof.On one hand, we have which means that there exist α ∈ Z and 1 ≤ j ≤ m such that On the other hand, by applying Lemma 5.5 and Lemma 5.3 we have which is equivalent to and hence to Thus (5.1) holds if and only if there exists i ∈ Z such that and thus if and only if , and if so 0 n/m S ∈ S\G/S is a basis for the vector space H(P + Q , P Z ).For the last part of the proof suppose that {μ n : n ∈ N * }, and {ê(r) : r ∈ Q/Z} are elements in a * -algebra A which also satisfy the relations (a)-(d).
This shows that the set uµ * m e(r)µ n spans H(P Q , P Z ), and that the set B = n m (uµ * m e(a/n)µ n ) : S 1 a 0 −n/m S ∈ S\G/S is a basis for the vector space H(P Q , P Z ) by the proof of Theorem 5.1.
For the last part of the proof suppose that μn : n ∈ N * , ê(r) : r ∈ Q/Z and û are elements in a * -algebra A which also satisfy the relations (a)-(g).We need to show that there is a * -homomorphism φ : H(P Q , P Z ) −→ A such that φ(µ n ) = μn , φ(e(r)) = ê(r) and φ(u) = û.But since B is a basis, there is a linear transformation φ satisfying these things.Proof.We still need to prove that φ is * -preserving and that it is multiplicative.
For * -preserving we will use relation (g), relation (f) and its adjoint So φ is * -preserving.
For the last part let q = [n, k] (the least common multiple), and let ν 1 , ν 2 ∈ H(P Q , P Z ).Then we will have four cases here Since φ is a homomorphism on H(P + Q , P Z ) we conclude that Thus φ is multiplicative.
Hence, the proof of our theorem is finished.

. 2 )
This shows that the set {µ * m * e(r) * µ n } spans H(P + Q , P Z ), and that any µ * m e(r)µ n and µ * k e(s)µ l are either of the same characteristic function or have orthogonal supports.So the elements n 1/2 m −1/2 µ * m * e(a/n) * µ n are linearly independent.Hence the set B = n 1/2 m −1/2 µ * m * e(a/n) * µ n : S 1 a For the last part we shall use the relations (a)-(d).Now n m µ * m e(r 1 )µ n l k µ * k e(r 2 )µ l = nl mk µ * m e(r 1 )µ n µ * k e(r 2 )µ l .