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Arithmetic of generalized Dedekind sums and their modularity

  • Dohoon Choi , Byungheup Jun , Jungyun Lee EMAIL logo and Subong Lim
Published/Copyright: September 1, 2018

Abstract

Dedekind sums were introduced by Dedekind to study the transformation properties of Dedekind η function under the action of SL2(ℤ). In this paper, we study properties of generalized Dedekind sums si,j(p, q). We prove an asymptotic expansion of a function on ℚ defined in terms of generalized Dedekind sums by using its modular property. We also prove an equidistribution property of generalized Dedekind sums.

MSC 2010: 11F20

1 Introduction

Dedekind sums are defined by

s(b,c):=h(modc)B¯1(hc)B¯1(bhc)(1)

for coprime integers b and c, where

B¯1(x):=x[x]12ifxRZ,0ifxZ.

These sums were introduced by Dedekind [1], and have been studied with applications in diverse areas of mathematics (for example, see [2,3,4]).

The Dedekind sum has been generalized and studied by many other authors from diverse standing (for example, see [5,6,7]). In this paper, we consider generalized Dedekind sums defined as follows. The Bernoulli polynomial Bi(x) is defined by the exponential generating function

i=0Bi(x)tii!=tetxet1(2)

and Bi(x) denotes the periodic Bernoulli polynomial

B¯i(x):=Bi(x)fori1orxZ,0fori=1andxZ,(3)

where 〈x〉 := x – [x] and [x] denotes the greatest integer not exceeding x. Let i and j be nonnegative integers. Suppose that p is an integer and q is a positive integer with gcd(p, q) = 1. Then, generalized Dedekind sums are defined as

si,j(p,q):=k=1qB¯i(kq)B¯j(pkq)(4)

and the number i + j is called the weight of si,j(p, q). For convenience, we let

hi,j(p,q):=(1)i+j1i!j!(si,j(p,q)di,jBiBj),(5)

where Bn is the nth Bernoulli number and di,j is given by

di,j:=1ifi=1orj=1,0otherwise.

Let {e1, …, em} be the standard basis of ℚm. We define a vector FN(p, q) ∈ ℚN for an even positive integer N by

FN(p,q):=i=1Nfi(p,q)ei,(6)

where fi(p, q) = qN21pN2+i1hi1,N+1i(p,q) for i = 1, …, N. Then we define a function GN : ℚ → ℚN as

GN(x)=GN(pq):=FN(p,q).(7)

We remark that any rational number x can be uniquely written as pq for a positive integer q and an integer p which are relatively prime. Thus, the above mapping GN is well-defined, and has the following asymptotic expansion expressed in terms of Bernoulli numbers.

Theorem 1.1

We have an asymptotic expansion of the form

GN1n((1)N2+1CN)GN(n)+AN1n1EN1n+2B12δN,2e2

as n → ∞, where

δi,j=0ifij,1ifi=j,
AN(x)=ANpq:=qN2i+1piN2(1)i+jj1Ni1i,jN,CN:=(1)j+1NjNi1i,jN,(8)

andEN(x) = i=1Nei(x)eiis a vector-valued function defined by

ei(x)=eipq:=q1ipi(1)iNNiBNN!+piNqNi+1(1)NiNi1BNN!+BNiBi(Ni)!i!q(1)i1+BNi+1Bi1(Ni+1)!(i1)!p(1)i1.(9)

Here, ab is defined by

ab=1ifb=0,a(a1)(ab+1)b!ifb>0,(10)

for nonnegative integers a and b. Theorem 1.1 is proved by using the fact that GN can be understood as a modular object, which was recently introduced by Zagier [8].

Example 1.2

For a givenN, we can compute the asymptotic expansion ofGNexplicitly and it is written in terms of Bernoulli numbersBn. For example, ifN = 2, then

GN1n1011GN(n)+012nB2+B12+12nB2

asn → ∞. IfN = 4, then

GN1n1000310032101111GN(n)+0124n2B416nB3B1+14B2216nB3B1+124n2B4124n2B4+14B2213nB3B1+18n2B4124n2B416nB3B1+18n2B4

asn → ∞.

Dedekind sums also have a special property of distribution. For an even positive integer N and positive integers p, q which are coprime, we consider another vector-valued function

HN(p,q):=qN2h1,N1(p,q),,hN1,1(p,q)QN1.(11)

Let

X:=x1,x2,,xN1[0,1)N1

be the fractional part of the vector X = (x1, x2, … xN–1). The following theorem states that HN(p, q) satisfies the property of equidistribution.

Theorem 1.3

For an even positive integerN, there exists an integerRNsuch that the set of rational numbers

RNHN(p,q)|(p,q)=1,0<p<q(12)

is equidistributed in [0, 1)N–1.

Remark 1.4

Once for an integerRNthe set (12) is equidistributed in [0, 1)N–1, the same holds for any integer multiple ofRN. An example ofRNis given by

RN=N!βNrN,

whereβkis a positive integer such thatαkβkis the reduced fraction ofBk ≠ 0 and

rN:=lcmDenominatorofβNNi+1Bi+1BNi1|0iN2.(13)

The following is the table for values ofRN forN = 2, 4, 6, 8, 10.

N246810
RN12720604803628800479001600.

The rest of the paper is organized as follows. Section 2 summarizes the properties of generalized Dedekind sums. In Section 3, we show that generalized Dedekind sums satisfy a modular property introduced by Zagier [8] and prove Theorem 1.1. In Section 4, we describe the distribution of generalized Dedekind sums and prove Theorem 1.3.

2 Reciprocity formulas

In this section, we prove the reciprocity formulas of generalized Dedekind sums, which can be induced from those of Dedekind-Rademacher sums.

2.1 Dedekind-Rademacher sums

In this subsection, we briefly review the definition and the reciprocity formulas of Dedekind-Rademacher sums based on the paper [6]. The following is the definition of Dedekind-Rademacher sums.

Definition 2.1

For a, b, c ∈ ℕ andx, y, z ∈ ℝ/ℤ, the Dedekind-Rademacher sum is defined by

Sm,nabcxyz:=h(modc)B¯ma(h+z)cxB¯nb(h+z)cy.

Since reciprocity relations mix various pairs of indices (m, n), Hall, Wilson and Zagier [6] stated them in terms of the generating function

δabcxyzXYZ:=m,n01m!n!Sm,nabcxyzXam1Ybn1,

where X and Y are nonzero variables and the variable Z is defined as –XY. The following is the precise statement of the reciprocity formula of Dedekind-Rademacher sums.

Theorem 2.2

([6, [Section 4]). Leta, b, cbe three positive integers with no common factor, x, y, zthree real numbers, andX, Y, Zthree variables with sum zero. Then

δabcxyzXYZ+δbcayzxYZX+δcabzxyZXY=14if(x,y,z)(a,b,c)R+Z3,0otherwise.

2.2 Reciprocity formulas of generalized Dedekind sums

The following are the reciprocity formulas of generalized Dedekind sums, which are induced from Theorem 2.2.

Theorem 2.3

Letpandqbe positive integers with gcd(p, q) = 1. The generalized Dedekind sums satisfy the following reciprocity formulas.

  1. Fora, b ≥ 1 with odda + b,

    qa1i=0bha1+i,bi(p,q)(p)ia1+ia1+pb1j=0ahb1+j,aj(q,p)(q)jb1+jb1=Ba1Bb(a1)!b!q(1)b1+BaBb1a!(b1)!p(1)b1.(14)
  2. Fora = 0 and odd b ≥ 1,

    hb1,0(q,p)=h0,b1(q,p)=p2bBb1(b1)!.

Proof

  1. By the definition of Dedekind-Rademacher sums, we have the following equations

    Sm,n1pq000=h(modq)B¯mhqB¯nphq=sm,n(p,q),Sm,nq1p000=h(modq)B¯nhpB¯mqhp=sn,m(q,p),

    and

    Sm,npq1000=B¯m(0)B¯n(0).

    Then Theorem 2.2 implies that the sum

    m,n01m!n!(sm,n(p,q)Xm1Ypn1+B¯m(0)B¯n(0)Ypm1Zqn1+sn,m(q,p)Zqm1Xn1)(15)

    is equal to 14.

    Suppose that m + n is even. Then, we have

    hm,n(p,q)=1m!n!(sm,n(p,q)+dm,nBmBn).

    Since Bk = 0 for odd integer k > 1 and B1 = 12, we see that

    hm,n(p,q)=1m!n!sm,n(p,q)14em,n,

    where

    em,n:=1ifn=m=1,0otherwise.

    Moreover, we have

    B¯m(0)B¯n(0)=BmBn14em,n.

    Therefore, the following can be induced from the sum (15)

    m,n0m+neven(hm,n(p,q)Xm1Ypn1+1m!n!BmBnYpm1Zqn1+hn,m(q,p)Zqm1Xn1)=0.(16)

    If we multiply 1pqXYZ on both sides of equation (16), then we obtain

    m,n0m+neven(hm,n(p,q)XmYpnZq+1m!n!BmBnXYpmZqn+hn,m(q,p)XnZqmYp)=0.(17)

    From the relation X + Y + Z = 0, the equation (17) is rewritten as

    m,n0m+nevenhm,n(p,q)(1)m(Y+Z)mYpnZq+hn,m(q,p)(1)nZqm(Y+Z)nYp=m,n0m+neven1m!n!BmBnYpmZqn(Y+Z).(18)

    By the binomial expansion of (Y + Z)n one can see that

    m,n0m+nevenk1=0mhm,n(p,q)(1)mq1pnmk1Yk1+nZmk1+1+m,n0m+nevenk2=0nhn,m(q,p)(1)nqmp1nk2Ynk2+1Zk2+m=m,n0m+neven1m!n!BmBnYpmZqn(Y+Z).(19)

    Now, we compare the coefficients of YbZa on both sides of equation (19). If a, b ≥ 1, then we have

    i=0bq1p(bi)(1)a+i1ha+i1,bi(p,q)a+i1i+j=0aq(aj)p1(1)b+j1hb+j1,aj(q,p)b+j1j=1(b1)!a!Bb1Bap(b1)qa+1b!(a1)!BbBa1pbq(a1).(20)

    This gives the first result after multiplying by qapb (–1)b–1 in both sides of equation (20).

  2. For the second result, suppose that a = 0 and b is an odd integer with b ≥ 1. Note that

    sb1,0(p,q)=k=1qB¯b1kqB¯0pkq=k=1qB¯b1pkqB¯0p(pk)q=k=1qB¯b1pkqB¯0kq=s0,b1(p,q).

    Here, we used the fact that B0(x) = 1 and gcd(p, q) = 1. If we compare the coefficients of Yb in both sides of equation (19), then we have

    hb1,0(q,p)(1)b1p1=1(b1)!Bb1p(b1),

    which gives the second result.□

The reciprocity formulas of generalized Dedekind sums can be expressed in terms of the vectors FN(p, q).

Corollary 2.4

Letpandqbe positive integers with gcd(p, q) = 1, andNan even positive integer. Then

ANpqFN(p,q)+BN(p,q)FN(p,q)=ENpq,

whereANpq, FN(p, q) andENpqare given as in (8), (6), and (9), respectively. Here, we defineBN(p, q) byBN(p, q) = (βi,j(p, q))1≤i,jNand

βi,j(p,q)=qN2i+1piN2(1)i+jj1i1.(21)

Proof

Let N = a + b – 1 for a, b ≥ 1. By Theorem 2.3 (2), we have

qa1i=0bha1+i,bi(p,q)(p)ia1+ia1=j=0b1qN2+apN2a+1fa+j(p,q)(1)j+1a1+ja1qa1pb(1)bNa1hN,0(p,q)=j=0b1qN2+apN2a+1fa+j(p,q)(1)j+1a1+ja1+qaNpb(1)b+1Na1BNN!(22)

and

pb1j=0ahb1+j,aj(q,p)(q)jb1+jb1=k=0a1pN2+bqN2b+1fb+k(q,p)(1)kb1+kb1+pb1qa(1)aNb1hN,0(q,p)=k=0a1pN2+bqN2b+1fb+k(q,p)(1)kb1+kb1+pbNqa(1)aNb1BNN!.(23)

If we change variables aNi + 1, jja, bi, and kjb in (22) and (23), then (14) can be written as

j=0NqN2i+1piN2(1)i+jj1Nifj(p,q)+j=0NpiN2qN2i+1(1)i+jj1i1fj(q,p)=q1ipi(1)iNNiBNN!+piNqNi+1(1)NiNi1BNN!+BNiBi(Ni)!i!q(1)i1+BNi+1Bi1(Ni+1)!(i1)!p(1)i1.

This gives the desired result.□

Besides the reciprocity formulas, generalized Dedekind sums satisfy the following properties.

Theorem 2.5

LetNbe an even positive integer, andi, jbe nonnegative integers with i + j = N. Then we have the following.

  1. hi,j(p,q)=(1)jhi,j(p,q)2B12,ifi=j=1;(1)jhi,j(p,q),otherwise.

  2. hi,j(p + q, q) = hi,j(p, q).

Proof

  1. By the definition of si,j(p, q), we see that

    si,j(p,q)=k=0q1B¯ikqB¯jpkq.

    It is known that Bj(1 – x) = (–1)jBj(x) for j ≥ 0. Therefore, we have

    B¯jpkq=B¯j1pkq=(1)jB¯jpkq

    for j ≥ 0. From this, we obtain

    si,j(p,q)=(1)jsi,j(p,q).

    If i = j = 1, then we have

    hi,j(p,q)=si,j(p,q)B12=si,j(p,q)B12=hi,j(p,q)2B12.

    Otherwise, we see that

    hi,j(p,q)=(1)i+j1i!j!si,j(p,q)=(1)j(1)i+j1i!j!si,j(p,q)=(1)jhi,j(p,q).
  2. Note that si,j(p + q, q) is equal to

    k=0q1B¯ikqB¯j(p+q)kq=k=0q1B¯ikqB¯jpkq+k=k=0q1B¯ikqB¯jpkq,

    which is si,j(p, q) by its definition. From this, we obtain the desired result that hi,j(p + q, q) = hi,j(p, q).□

3 Modular properties of generalized Dedekind sums

In this section, we show that the vector-valued function GN defined in (7) satisfies the modular property, which was introduced by Zagier [8].

3.1 Automorphic factor

To introduce a modular property for a vector-valued function, we need an automorphic factor for a vector-valued function.

Definition 3.1

An automorphic factor of rankmis a functionρ: SL2(ℤ)×ℍ → GLm(ℂ) satisfying the following conditions.

  1. The functionρsatisfies the cocycle relation

    ρ(γ1γ2,x)=ρ(γ1,γ2x)ρ(γ2,x)(24)

    forγ1, γ2 ∈ SL2(ℤ) and x ∈ ℍ, whereabcdx=ax+bcx+d.

  2. For a fixedγ ∈ SL2(ℤ), entries ofρ(γ, x) are rational functions ofxwith coefficients in ℚ.

For a fixed γ ∈ SL2(ℤ), the action of ρ(γ, x) on ℂm is defined by

ρ(γ,x)ej=i=1mρij(γ,x)ei,

where ρi,j(γ, x) is the (i, j)th entry of ρ(γ, x).

Let N be an even positive integer. Now, we define an automorphic factor ρN of rank N as follows. Let S:=0110,T:=1101SL2(Z). We define

ρN(S,x):=((1)N2+1CN)1(25)

and

ρN(T,x):=DN(x)1,(26)

where

CN:=(1)jNjNi1i,jN

and

DN(x):=δi,jxx+1N2+i11i,jN.

Here, δi,j is the Kronecker delta. Then, it induces an automorphic factor of rank N.

Note that S and T generate SL2(ℤ). Therefore, from (25) and (26), we can compute ρN(γ) for any γ ∈ SL2(ℤ) by using the cocycle condition as in (24). To prove that ρN is an automorphic factor of rank N, it suffices to check that

ρN(S,Sx)ρN(S,x)=ρN(U,U2x)ρN(U,Ux)ρN(U,x)

and

(ρN(S,Sx)ρN(S,x))2=I,

where U := TS, ρN(U, x) = ρN(T, Sx)ρN(S, x) and I denotes the N × N identity matrix. To prove this, we need the following lemmas.

Lemma 3.2

LetNbe a positive integer. Letiandjbe integers with 1 ≤ i, jN.

  1. IfNis even and we let

    Λi,j:=k=1N(1)k+j1Nki1j1Nk,

    then Λi,j = δi,j.

  2. k=1NNki1j1k1(1)k1=NjNi.

Proof

  1. Recall that

    ab=1ifb=0,a(a1)(ab+1)b!ifb>0,(27)

    for nonnegative integers a and b. If we consider the binomial expansion of (1 –(1 – T))j–1, then we see that

    (1(1T))j1=k=0N1j1k(1)k(1T)k.

    In the above equality, we used the fact that ab = 0 if a < b. Then we have

    (1(1T))j1=k=1Nj1Nk(1)Nk(1T)Nk=k=1Nj1Nk(1)Nki=0N1Nki(1)iTi=k=1Nj1Nk(1)Nki=1NNki1(1)i1Ti1=k=1Ni=1Nj1NkNki1(1)Nk+i1Ti1=i=1NΛi,j(1)N+i+jTi1.

    Since (1 –(1 – T))j–1 = Tj–1, we see that

    Λi,j(1)N+i+j=0ifij,1ifi=j.

    If i = j, then (–1)N+i+j = 1 since N is even. Therefore, we obtain the desired result that Λi,j = δi,j for 1 ≤ i, jN.

  2. We will use the induction on N. If N = 1, it is easy to see that (2) is true. Suppose that (2) is true for N ≥ 1. Let 1 < i < N + 1 and 1 ≤ j < N + 1. By the recursive formula, we have

    N+1jN+1i=NjN(i1)+NjNi.

    Then, the induction hypothesis implies that

    N+1jN+1i=k=1NNki2j1k1(1)k1+k=1NNki1j1k1(1)k1.

    If we use the recursive formula again, then we have

    N+1jN+1i=k=1NN+1ki1j1k1(1)k1=k=1N+1N+1ki1j1k1(1)k1.

    The last equality follows from that N+1ki1=0i1=0ifk=N+1 since i – 1 > 0.

    Now, we will check the remaining three cases: i = 1, i = N + 1, and j = N + 1. If i = 1 and 1 ≤ jN + 1, then we have

    k=1N+1N+1k0j1k1(1)k1=k=1N+1j1k1(1)k1=k=0Nj1k(1)k=δj,1=N+1jN.

    If i = N + 1 and 1 ≤ jN + 1, then we obtain

    k=1N+1N+1kNj1k1(1)k1=NN=1=N+1j0.

    Suppose that j = N + 1 and 2 ≤ iN. Then, we have

    k=1N+1N+1ki1Nk1(1)k1=k=1N+1Ni1N+1ik1(1)k1=Ni1k=1N+1N+1ik1(1)k1=Ni1k=0NN+1ik(1)k=Ni1(11)N+1i=0=0N+1i.

    Here, we used the identity

    N(k1)i1Nk1=Ni1N(i1)k1.

Lemma 3.3

Letiandjbe integers with Nj > i ≥1. Then

xjyik,l=1Nzxkyzljkklli=(x+y+z)jij!i!(ji)!.

Proof

If we apply the binomial expansion twice, then we obtain

(x+y+z)ji=k=0jil=0kjikklxjikylzkl.

Then, one can see that

(x+y+z)ji=k=ijl=ikjikikilixjkylizkl=i!(ji)!j!xjyik=ijl=ik(zx)k(yz)ljkklli.

By the definition of ab as in (27), we obtain the desired result. □

Lemma 3.4

Suppose thatNis an even positive integer. Letiandjbe integers with 1 ≤ i, jN.

  1. k=1NNkNiNjNk(1)j+k=δi,j.

  2. Let

    Δi,j:=k,l=1Nx1xN2+1ixN2+1l11xN2+1k(1)l+k+jNlNiNkNlNjNk.

    Then

    Δi,j=(1)N2+1δi,j.

Proof

  1. One can obtain (1) from Lemma 3.2 (1) by replacing i (resp. j) with Ni + 1 (resp. Nj + 1).

  2. Note that by the definition of (ab) as in (27), we have

    NlNiNkNlNjNk0

    only when ilkj. Therefore, if i < j, then Δi,j = 0. If i = j, then a nonzero term in the summation appears only when k = l = i = j. So, we have Δi,j=(1)N2+1 . In the case of i > j, one can see that

    Δi,j=x1xN2+1ixN2+111xN2+1(1)j×k,l=1N11xNk(x1)NlNlNiNkNlNjNk.

    By Lemma 3.3, Δi,j is equal to

    (Nj)!(Ni)!(ij)!x1xN2+1ixN2+1i11xN2+1j(1)j×((1x)+x+(1))ij=0.

    This completes the proof. □

Now, we prove that ρN induces an automorphic factor of rank N satisfying the cocycle relation as in (24).

Proposition 3.5

LetρNbe defined by (25) and (26). Then

  1. ρN(S, Sx)ρN(S, x) = I,

  2. ρN(U, U2x)ρN(U, Ux)ρN(U, x) = I.

Proof

  1. Lemma 3.4 (1) implies that CN2 = I, and hence we have

    ρN(S,Sx)ρN(S,x)=I.
  2. For U = TS, we see that

    ρN(U,x)=(1)N2+111xN2+1iNjNi(1)j1i,jN,ρN(U,Ux)=(1)N2+1xN2+1iNjNi(1)j1i,jN,

    and

    ρN(U,U2x)=(1)N2+1x1xN2+1iNjNi(1)j1i,jN.

    By Lemma 3.4 (2), we have

    ρN(U,U2x)ρN(U,Ux)ρN(U,x)=(1)N2+1(Δi,j)1i,jN=I.

    This is the desired result. □

Remark 3.6

These kinds of automorphic factors are closely related with a monomial times a modular form. For example, we consider F(z) := z6Δ(z), whereΔ(z) is defined by

Δ(z)=qn=1(1qn)24

andq = e2πizforz ∈ ℍ. Then, it is known thatΔ(z) is a modular form of weight 12 on SL2(ℤ). Note that

F(z+1)=(z+1)6z6F(z)

and

F1z=F(z).

Hence, we see that F(z) is a modular form associated withρ, which is defined by

ρ(T,z)=z+1z6,ρ(S,z)=1.

3.2 Modular property of GN

With the automorphic factor ρN, we can state the transformation property of the vector-valued function GN. We define a slash operator associated with ρN as follows. Let f be a vector-valued function on ℚ, i.e., f is a sum of functions f=i=1mfiei , where fi is a function on ℚ for i = 1, …, m. Then we define

(f|ρNγ)(x):=i=1mfi(γx)ρN1(γ,x)ei

for x ∈ ℚ and γ ∈ SL2(ℤ).

Lemma 3.7

The vector-valued functionGN : ℚ → ℂNsatisfies the functional equations

GNGN|ρNS(x)=AN(x)1EN(x)+2B12δN,2e2

for all positive rational numbersxand

GNGN|ρNT(x)=0

for all x ∈ ℚ.

Proof

In this proof, we will use the properties of generalized Dedekind sums. For positive integers p and q which are relatively prime, by Corollary 2.4, we have

ANpqFN(p,q)+BN(p,q)FN(q,p)=ENpq,

where ANpq , BN(p, q), FN(p, q), and ENpq are defined as in (8), (21), (6), and (9), respectively.

Lemma 3.2 (1) implies that

ANpq1=(vi,j(p,q))1i,jN,vi,j=qN2+j1pN2jNji1

since (vi,j(p, q))1≤i,jN × ANpq = (Λi,j)1≤i,jN, which is the identity matrix. Lemma 3.2 (2) implies that

ANpq1BN(p,q)=CN.

Therefore, we see that

GN(x)(CN)GN(1x)=AN(x)1EN(x)

for any positive rational number x since GN(x) = FN(p, q), where x = pq . By Theorem 2.5 (1), we see that

GN(x)=(1)N2GN(x)2B12e2ifN=2,(1)N2GN(x)ifN4.

Therefore, we have the first desired result.

The second result directly comes from Theorem 2.5 (2) and the definition of ρN(T) as in (26). □

Note that AN(x)−1EN(x) is a rational function of p and q, where x = pq ∈ ℚ. But AN(x) and EN(x) are homogeneous of degree 1, and hence AN(x)−1EN(x) is homogeneous of degree 0. Therefore, AN(x)−1EN(x) is actually a rational function of x. This implies that this vector-valued function can be extended to ℝ ∖ {0}. The vector-valued function GN can be understood as a quantum modular form, which is a new modular object on ℚ introduced by Zagier [8]. Quantum modular forms were studied in connection with Maass forms, mock modular forms and Eichler integrals (for example, see [9, 10]).

3.3 Proof of Theorem 1.1

By Lemma 3.7, we have

GN(x)((1)N2+1CN)GN(1x)=AN(x)1EN(x)+2B12δN,2e2

for positive rational numbers x. If we let x = 1n for n ∈ ℕ, then

GN1n((1)N2+1CN)GN(n)+AN1n1EN1n+2B12δN,2CNe2

as n → ∞.

3.4 Application of modular property to the arithmetic of Dedekind sums

With the cocycle property of ρ2, one can obtain an explicit expression of G2( pq ) in terms of the negative continued fraction of pq . For this, we recall the followings:

  1. G2pq=p1h0,2(p,q)h1,1(p,q)=112p1q1s1,1(p,q)+14,

  2. ρ2(S,x)1=1011,

  3. ρ2(T,x)1=x+1x001,

  4. G2(x)(G2|ρ2S)(x)=0B221x3B12+B22x,

  5. G2(x) − (G2|ρ2T)(x) = 0.

By the cocycle condition as in (24), we obtain

G2(x)(G2|ρ2TaS)(x)=G2(x)ρ21(S,x)ρ2(Ta,Sx)1G2(TaSx)=G2(x)ρ21(S,x)G2(Sx)=0B221x3B12+B22x(28)

for a positive integer a.

Now, we express pq by using the negative continued fraction as follows

pq=[a1,a2,,an]:=a11a211an

and define

piqi:=[ai,ai+1,,an]

for 1 ≤ in. Then, for 1 ≤ in − 1, we see that

ai110pi+1qi+1=piqi.(29)

Moreover, for 1 ≤ in − 1, we also have

G2pi+1qi+1ρ21ai110,pi+1qi+1G2piqi=E2pi+1qi+1.(30)

For simplicity, we define

Fi:=ρ21ai110,pi+1qi+1=piqi+10piqi+11.

By combining equations in (30), we have

G2pq=F11F21Fn11G2pnqnE2p2q2E2p3q3E2pnqn.

Note that

F11F21Fn11=anpq0anqq1,

where

qp=[a2,a3,,an1].

Moreover, we have

G2pnqn=112an13,G2pq=112p1q112s1,1(p,q)+3,

and

E2p2q2+E2p3q3++E2pnqn=0112pq+a1+a2++ann14.

Finally, we obtain

12s1,1(p,q)=pqq(a1+a2++an)+3(n1).

Eventually, this formula gives a simple expression in 12 s1,1(p, q) modulo 1 that plays a crucial role of the proof for the equidistribution property of h1,1(x).

Remark 3.8

We note that in above examples, the generalized Dedekind sum s1,1(p, q) is completely determined by the initial values and transformation formulas byS=0110andT=1101 . Along this line, restricting ourselves to the two dimensional case, the Apostol sum s1,n(p, q) is completely determined by a two term relation

(n+1)pqns1,n(p,q)+pnqs1,n(q,p)=i=0n1n+1i(1)iBiBn+1ipiqn+1i+nBn+2

together with the continued fraction ofpq(cf. [5]).

In this paper, we consider a similar property for generalized Dedekind sums si,j(p, q). Usually, generalized Dedekind sums si,j, (i ≠ 1, j ≠ 1) except the outermost case (i.e. Apostol sums) do not have two term reciprocity formula. Namely,

12!2!s2,2(p,q)3p4!q3s0,4(q,p)+p3!q2s1,3(q,p)+p2!2!qs2,2(q,p)=f(p,q)

for a Laurent polynomial f(p, q). In this case, the reciprocity formula for s2,2(p, q) cannot be reduced to a Laurent polynomial, but involves a transcendental term s1,3(p, q).

However, if we consider a vector consisting of generalized Dedekind sums, then they have a two term reciprocity formula as follows

14!s0,4(p,q)13!s1,3(p,q)12!2!s2,2(p,q)13!s3,1(p,q)+p3q30003p2q3p2q2003pq32pq2pq01q31q21q114!s0,4(q,p)13!s1,3(q,p)12!2!s2,2(q,p)13!s3,1(q,p)=F(p,q)

for a column vector F(p, q) = (f1(p, q), …, f4(p, q))Tconsisting of Laurent polynomials inpandq. Thus, we are able to compute algorithmically the (vector of) generalized Dedekind sums of (p, q) of fixed weight using the continued fraction ofpq , once we obtain the reciprocity formula.

4 Distribution of generalized Dedekind sums

For an even positive integer N and positive integers p and q which are relatively prime, we consider a vector-valued function HN(p, q) defined in (11). In this section, we show that there exists an integer RN such that fractional parts of the vectors 〈RNHN(p, q)〉 are equidistributed. That is the image of these vectors under the projection ℝN onto ℝN/ℤN is equidistributed on the torus. There is a necessary and sufficient condition for this due to Weyl.

4.1 Weyl’s equidistribution criterion

We recall the statement of Weyl’s criterion on torus. For details, we refer to [11].

Theorem 4.1

(Weyl’s equidistribution criterion). A sequence

{sk=(s1(k),s2(k),,sn(k))[0,1)n}kN

is equidistributed in [0, 1)nif and only if for everym = (m1, m2, …, mn) ∈ ℤn − {0},

limT1Tk=1Te(msk)=0,

wheremsk=i=1nmisi(k)ande(x) denotes exp(2πix).

For a nonzero vector m ∈ ℤN−1 and a positive real number x, let E(m, x) be the average of the exponentials of (2πi) mRNHN(p, q) defined by

E(m,x):=1#(p,q)|gcd(p,q)=1,p<qx0<q<x0<p<q(p,q)=1emRNHNp,q.(31)

To apply Theorem 4.1, one needs to show that E(m, x) tends to 0 as x goes to ∞. This is done by relating an exponential sum to

0<p<q(p,q)=1emRNHNp,q.(32)

4.2 Exponential sums of generalized Dedekind sums

We relate (32) to an exponential sum for a Laurent polynomial. Let us first recall the exponential sum for a Laurent polynomial F(x) ∈ ℤ[x, x−1].

Definition 4.2

For a positive integerqand F(x) ∈ ℤ[x, x−1], we define the exponential sum of modulusqof F(x) as

K(F,q):=x(Z/qZ)eqF(x),

whereeq(x):=exp2πixq.

Let FN(x) be the rank (N − 1) vector of Laurent polynomials

FN(x):=f1x,f2x,,fN1x,(33)

where fi(x)=αNrNN1ixi+N1NixNi . Here, αN denotes the numerator of BN, and rN is the integer defined as in (13). For a nonzero integer vector m of rank (N − 1), we have the following Laurent polynomial in x

mFN(x)=αNrNi=1N1miN1ixi+miN1NixNi.(34)

To prove the relation between the two exponential sums (32) and K(mFN, q), we need the following theorem.

Theorem 4.3

[12, Theorem 1.1] LetNbe an even positive integer. For positive integersiandjwith i + j = N,

RNqN2hij(p,q)αNrN(p)iN1i+pjN1jqZ,

where p′ is an integer such that p′p ≡ 1 (mod q), and RN = N! βNrNwithβN being the denominator ofBN.

By Theorem 4.3, one can see that

0<p<q(p,q)=1emRNHNp,q=K(mFN,q).

Therefore, we come to the estimation of the exponential sum of mFN(x).

4.3 Bounds for exponential sums

Let q be a prime. Then, the estimation of K(mFN, q), accompanied with some reductions, will be sufficient in showing Weyl’s criterion for

RNHN(p,q)|(p,q)=1,0<p<q

to be equidistributed. This will complete the proof of Theorem 1.3. To achieve the full estimation, we will follow the steps taken in [12].

The following lemma is necessary to prove that mFN(x) has a Weil type bound for all but finitely many primes p.

Lemma 4.4

For positive integersiandj, letF(x)=k=jiakxkbe a Laurent polynomial with integer coefficients such that ai ≠ 0 and aj ≠ 0. Letpbe any prime withpak for some k ≠ 0. Then

K(F,p)(i+j)p.

Proof

See Theorem 1.3 in [13]. □

Since mFN is written as Laurent polynomial type in Lemma 4.4 (see equation (34)), we have the following.

Proposition 4.5

Letmbe a nonzero integer vector andmFN(x) be the Laurent polynomial as give in (33). Putd = gcd(m1, m2, …, mN−1). Then for any positive integerpαNrNd, we have

K(mFN,p)2(N1)p.(35)

For a general modulus q, we have the following bound.

Proposition 4.6

Suppose thatqis a positive integer, and thatNis an even positive integer. Let DFN be a positive integer such that DFN||FN. Letω(q) be the number of prime factors ofq. Then

K(mFN,q)DFN(12N12)ω(q)q(113N2).

From now on, we justify Proposition 4.6.

4.4 Reduction to prime modulus

If q has many prime factors, the bound is obtained by composing the bound previously obtained for primes dividing q. This is done by the next two reduction steps.

First, we consider the case q being a power of a prime p.

Lemma 4.7

Let F(x) be a Laurent polynomial with integer coefficients. Letpbe a fixed prime and pβ|| Then forα > β, we have

K(F,pα)=pβK(F~,pαβ),

whereF~(x)=1pβF(x).

Proof

We note that an element z ∈ (ℤ/pαℤ) is written uniquely as z = pαβx + y for x ∈ ℤ/pβℤ and y ∈ (ℤ/pαβℤ). Thus, we obtain

K(F,pα)=z(Z/pαZ)eFpα=xZ/pβZy(Z/pαβZ)eF~pαβ=pβK(F~,pαβ).

After the previous lemma, we can pull out p-factors out of the coefficients of mFN(x). For positive integers i and j, let F(x)=k=jiakxk be a Laurent polynomial with integer coefficients such that ai ≠ 0 and aj ≠ 0. Let p be a prime such that F(x) ≢ 0 (mod p) and α be a positive integer.

Now, the following lemma is implied by Corollary 4.1 in [14] with the trivial character.

Lemma 4.8

With the above notation, suppose thatα ≥ 2 is an integer. Then

K(F,pα)4(i+2j)pα(11i+2j+1).

The previous two lemmas imply the following bound for prime powers.

Proposition 4.9

Letpbe a prime. Suppose thatmis any fixed nonzero vector inN−1, and that pβ||mFN(x) for some integerβ. Suppose thatα ≥ 2 is an integer such thatα > β. Then for anyp, we have

K(mFN,pα)pβ12(N1)pα(113N2)DFN12(N1)pα(113N2),

where DFN is the largest positive integer such that DFN|FN.

Let us consider the case when q has several prime factors. We have the following effect of the Chinese remainder theorem for exponential sums.

Lemma 4.10

Let F(x) be a Laurent polynomial with integer coefficients, and let q1 > 1 and q2 > 1 be relatively prime integers. Let Fi(x) be the modqiChinese remainder of F(x) fori = 1, 2 (i.e. F ↦ (F1, F2) under the isomorphism (ℤ/q1q2ℤ)[x, x−1] → (ℤ/q1ℤ)[x, x−1] × (ℤ/q2ℤ)[x, x−1]). Then

K(F,q1q2)=K(F1,q1)K(F2,q2).

Proof

This is a consequence of Fubini theorem. □

4.5 Proof of Theorem 1.3

For x > 1, let ϕ(x) := |(ℤ/[x]ℤ)| be Euler’s phi function. By using ∑q<xϕ(q) ∽ x2 as x → ∞, we obtain the proof of the main theorem from Weyl’s criterion for equidistribution and Proposition 4.6.

Now, we apply Proposition 4.6 to deducing Weyl’s criterion from the bound of exponential sums. Note that ω(q) in Proposition 4.6 has a well-known estimation

ω(q)clogqloglogq(36)

for some constant c. For sufficiently large q,

(12N12)ω(q)(12N12)clogqloglogq(q)clog(12N12)loglogq.

Thus, we obtain that for any ϵ > 0,

(12N12)ω(q)qϵ.

Therefore, by Proposition 4.6, we have the following bound.

Proposition 4.11

LetNbe an even positive integer, and letmbe a nonzero integer vector of rank N − 1. Then

K(mFN,q)q(1+ϵ13N2)

for allϵ > 0.

Weyl’s criterion for HN(p, q) comes from the following estimation

0<q<x0<p<q(p,q)=1emRNHN(p,q)=0<q<xK(mFN,q)x(2+ϵ13N2).(37)

Consequently, Weyl’s criterion is fulfilled for the fractional part of the vector HN(p, q):

E(m,x)=1#(p,q)|gcd(p,q)=1,p<qx0<q<x0<p<q(p,q)=1e(mRNHN(p,q))0(38)

as x → ∞. This completes the proof.

Acknowledgement

The authors would like to thank the referees for the valuable comments and helpful corrections, which improved the paper.

The second named author was supported by (NRF-2015R1D1A1A09059083) and the third named author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2009-0093827) and (NRF-2017R1A6A3A11030486) and the last author was supported by (NRF-2017R1C1B5017409).

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Received: 2018-02-05
Accepted: 2018-06-01
Published Online: 2018-09-01

© 2018 Choi et al., published by De Gruyter.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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