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Arithmetic of generalized Dedekind sums and their modularity

Dohoon Choi
/ Byungheup Jun
/ Jungyun Lee
• Corresponding author
• Institute of Mathematical science, Ewha womans university, 11-1 Daehyun-Dong, Seodaemun-gu, Seoul, 120-750, Republic of Korea
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• Other articles by this author:
/ Subong Lim
• Department of Mathematics Education, Sungkyunkwan University, 25-2, Sungkyunkwan-ro, Jongno-gu, Seoul 03063, Republic of Korea
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Published Online: 2018-09-01 | DOI: https://doi.org/10.1515/math-2018-0082

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.

Keywords: Dedekind sum; Quantum modular form

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

$B¯1(x):=x−[x]−12if x∈R∖Z,0if x∈Z.$

These sums were introduced by Dedekind , 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

$∑i=0∞Bi(x)tii!=tetxet−1$(2)

and Bi(x) denotes the periodic Bernoulli polynomial

$B¯i(x):=Bi(〈x〉)for i≠1 or x∉Z,0for i=1 and x∈Z,$(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

$di,j:=1if i=1 or j=1,0otherwise.$

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) = $\begin{array}{}{q}^{\frac{N}{2}-1}{p}^{-\frac{N}{2}+i-1}{h}_{i-1,N+1-i}\left(p,q\right)\end{array}$ 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 $\begin{array}{}\frac{p}{q}\end{array}$ 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

$GN1n∼((−1)N2+1CN)GN(−n)+AN1n−1EN1n+2B12δN,2e2$

as n → ∞, where

$δi,j=0if i≠j,1if i=j,$

$AN(x) =ANpq:=qN2−i+1pi−N2(−1)i+jj−1N−i1≤i,j≤N, CN:=(−1)j+1N−jN−i1≤i,j≤N,$(8)

and EN(x) = $\begin{array}{}\sum _{i=1}^{N}{e}_{i}\left(x\right){\mathbf{e}}_{i}\end{array}$ is a vector-valued function defined by

$ei(x)=eipq:=q1−ipi(−1)iNN−iBNN!+pi−NqN−i+1(−1)N−iNi−1BNN!+BN−iBi(N−i)!i!q(−1)i−1+BN−i+1Bi−1(N−i+1)!(i−1)!p(−1)i−1.$(9)

Here, $\begin{array}{}\left(\begin{array}{c}a\\ b\end{array}\right)\end{array}$ is defined by

$ab=1if b=0,a(a−1)⋯(a−b+1)b!if b>0,$(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 .

Example 1.2

For a given N, we can compute the asymptotic expansion of GN explicitly and it is written in terms of Bernoulli numbers Bn. For example, if N = 2, then

$GN1n∼101−1GN(−n)+012nB2+B12+12nB2$

as n → ∞. If N = 4, then

$GN1n∼−1000−3100−32−10−11−11GN(−n)+0124n2B4−16nB3B1+14B22−16nB3B1+124n2B4−124n2B4+14B22−13nB3B1+18n2B4124n2B4−16nB3B1+18n2B4$

as n → ∞.

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

$HN(p,q):=qN−2h1,N−1(p,q),…,hN−1,1(p,q)∈QN−1.$(11)

Let

$〈X〉:=〈x1〉,〈x2〉,…,〈xN−1〉∈[0,1)N−1$

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 integer N, there exists an integer RN such that the set of rational numbers

$〈RNHN(p,q)〉|(p,q)=1,0(12)

is equidistributed in [0, 1)N–1.

Remark 1.4

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

$RN=N!βNrN,$

where βk is a positive integer such that $\begin{array}{}\frac{{\alpha }_{k}}{{\beta }_{k}}\end{array}$ is the reduced fraction of Bk ≠ 0 and

$rN:=lcmDenominator ofβNNi+1Bi+1BN−i−1|0≤i≤N−2.$(13)

The following is the table for values of RN for N = 2, 4, 6, 8, 10.

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  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.

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

Definition 2.1

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

$Sm,nabcxyz:=∑h (mod c)B¯ma(h+z)c−xB¯nb(h+z)c−y.$

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

$δabcxyzXYZ:=∑m,n≥01m!n!Sm,nabcxyzXam−1Ybn−1,$

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]). Let a, b, c be three positive integers with no common factor, x, y, z three real numbers, and X, Y, Z three 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

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

1. For a, b ≥ 1 with odd a + b,

$−qa−1∑i=0bha−1+i,b−i(p,q)(−p)ia−1+ia−1+pb−1∑j=0ahb−1+j,a−j(q,p)(−q)jb−1+jb−1=Ba−1Bb(a−1)!b!q(−1)b−1+BaBb−1a!(b−1)!p(−1)b−1.$(14)

2. For a = 0 and odd b ≥ 1,

$hb−1,0(q,p)=h0,b−1(q,p)=p2−bBb−1(b−1)!.$

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

$∑m,n≥01m!n!(sm,n(p,q)Xm−1Ypn−1+B¯m(0)B¯n(0)Ypm−1Zqn−1+sn,m(q,p)Zqm−1Xn−1)$(15)

is equal to $\begin{array}{}\frac{1}{4}\end{array}$.

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 = $\begin{array}{}-\frac{1}{2}\end{array}$, we see that

$hm,n(p,q)=1m!n!sm,n(p,q)−14em,n,$

where

$em,n:=1if n=m=1,0otherwise.$

Moreover, we have

$B¯m(0)B¯n(0)=BmBn−14em,n.$

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

$∑m,n≥0m+neven(hm,n(p,q)Xm−1Ypn−1+1m!n!BmBnYpm−1Zqn−1+hn,m(q,p)Zqm−1Xn−1)=0.$(16)

If we multiply $\begin{array}{}\frac{1}{pq}XYZ\end{array}$ on both sides of equation (16), then we obtain

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

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

$∑m,n≥0m+nevenhm,n(p,q)(−1)m(Y+Z)mYpnZq+hn,m(q,p)(−1)nZqm(Y+Z)nYp=∑m,n≥0m+neven1m!n!BmBnYpmZqn(Y+Z).$(18)

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

$∑m,n≥0m+neven∑k1=0mhm,n(p,q)(−1)mq−1p−nmk1Yk1+nZm−k1+1+∑m,n≥0m+neven∑k2=0nhn,m(q,p)(−1)nq−mp−1nk2Yn−k2+1Zk2+m=∑m,n≥0m+neven1m!n!BmBnYpmZqn(Y+Z).$(19)

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

$∑i=0bq−1p−(b−i)(−1)a+i−1ha+i−1,b−i(p,q)a+i−1i+∑j=0aq−(a−j)p−1(−1)b+j−1hb+j−1,a−j(q,p)b+j−1j=1(b−1)!a!Bb−1Bap−(b−1)q−a+1b!(a−1)!BbBa−1p−bq−(a−1).$(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

$sb−1,0(p,q)=∑k=1qB¯b−1kqB¯0pkq=∑k=1qB¯b−1pkqB¯0p(pk)q=∑k=1qB¯b−1pkqB¯0kq =s0,b−1(p,q).$

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

$hb−1,0(q,p)(−1)b−1p−1=1(b−1)!Bb−1p−(b−1),$

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

Let p and q be positive integers with gcd(p, q) = 1, and N an even positive integer. Then

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

where AN $\begin{array}{}\left(\frac{p}{q}\right)\end{array}$, FN(p, q) and EN $\begin{array}{}\left(\frac{p}{q}\right)\end{array}$ are given as in (8), (6), and (9), respectively. Here, we define BN(p, q) by BN(p, q) = (βi,j(p, q))1≤i,jN and

$βi,j(p,q)=qN2−i+1pi−N2(−1)i+jj−1i−1.$(21)

Proof

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

$−qa−1∑i=0bha−1+i,b−i(p,q)(−p)ia−1+ia−1=∑j=0b−1q−N2+apN2−a+1fa+j(p,q)(−1)j+1a−1+ja−1−qa−1pb(−1)bNa−1hN,0(p,q)=∑j=0b−1q−N2+apN2−a+1fa+j(p,q)(−1)j+1a−1+ja−1+qa−Npb(−1)b+1Na−1BNN!$(22)

and

$pb−1∑j=0ahb−1+j,a−j(q,p)(−q)jb−1+jb−1=∑k=0a−1p−N2+bqN2−b+1fb+k(q,p)(−1)kb−1+kb−1+pb−1qa(−1)aNb−1hN,0(q,p)=∑k=0a−1p−N2+bqN2−b+1fb+k(q,p)(−1)kb−1+kb−1+pb−Nqa(−1)aNb−1BNN!.$(23)

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

$∑j=0NqN2−i+1pi−N2(−1)i+jj−1N−ifj(p,q)+∑j=0Npi−N2qN2−i+1(−1)i+jj−1i−1fj(q,p)=q1−ipi(−1)iNN−iBNN!+pi−NqN−i+1(−1)N−iNi−1BNN!+BN−iBi(N−i)!i!q(−1)i−1+BN−i+1Bi−1(N−i+1)!(i−1)!p(−1)i−1.$

This gives the desired result.□

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

Theorem 2.5

Let N be an even positive integer, and i, j be nonnegative integers with i + j = N. Then we have the following.

1. $\begin{array}{}{h}_{i,j}\left(-p,q\right)=\left\{\begin{array}{ll}\left(-1{\right)}^{j}{h}_{i,j}\left(p,q\right)-2{B}_{1}^{2},& if\text{\hspace{0.17em}}i=j=1;\\ \left(-1{\right)}^{j}{h}_{i,j}\left(p,q\right),& otherwise.\end{array}\right\\end{array}$

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

Proof

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

$si,j(−p,q)=∑k=0q−1B¯ikqB¯j−pkq.$

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

$B¯j−pkq=B¯j1−pkq=(−1)jB¯jpkq$

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

$∑k=0q−1B¯ikqB¯j(p+q)kq=∑k=0q−1B¯ikqB¯jpkq+k=∑k=0q−1B¯ikqB¯jpkq,$

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 .

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 rank m is a function ρ: SL2(ℤ)×ℍ → GLm(ℂ) satisfying the following conditions.

1. The function ρ satisfies the cocycle relation

$ρ(γ1γ2,x)=ρ(γ1,γ2x)ρ(γ2,x)$(24)

for γ1, γ2 ∈ SL2(ℤ) and x ∈ ℍ, where $\begin{array}{}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)x=\frac{ax+b}{cx+d}.\end{array}$

2. For a fixed γ ∈ SL2(ℤ), entries of ρ(γ, x) are rational functions of x with 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 $\begin{array}{}S:=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),\phantom{\rule{thinmathspace}{0ex}}T:=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\in {\mathrm{S}\mathrm{L}}_{2}\left(\mathbb{Z}\right).\end{array}$ We define

$ρN(S,x):=((−1)N2+1CN)−1$(25)

and

$ρN(T,x):=DN(x)−1,$(26)

where

$CN:=(−1)jN−jN−i1≤i,j≤N$

and

$DN(x):=δi,jxx+1−N2+i−11≤i,j≤N.$

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

Let N be a positive integer. Let i and j be integers with 1 ≤ i, jN.

1. If N is even and we let

$Λi,j:=∑k=1N(−1)k+j−1N−ki−1j−1N−k,$

then Λi,j = δi,j.

2. $\begin{array}{}\sum _{k=1}^{N}\left(\begin{array}{c}N-k\\ i-1\end{array}\right)\left(\begin{array}{c}j-1\\ k-1\end{array}\right)\left(-1{\right)}^{k-1}=\left(\begin{array}{c}N-j\\ N-i\end{array}\right).\end{array}$

Proof

1. Recall that

$ab=1if b=0,a(a−1)⋯(a−b+1)b!if b>0,$(27)

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

$(1−(1−T))j−1=∑k=0N−1j−1k(−1)k(1−T)k.$

In the above equality, we used the fact that $\begin{array}{}\left(\begin{array}{c}a\\ b\end{array}\right)\end{array}$ = 0 if a < b. Then we have

$(1−(1−T))j−1=∑k=1Nj−1N−k(−1)N−k(1−T)N−k=∑k=1Nj−1N−k(−1)N−k∑i=0N−1N−ki(−1)iTi=∑k=1Nj−1N−k(−1)N−k∑i=1NN−ki−1(−1)i−1Ti−1=∑k=1N∑i=1Nj−1N−kN−ki−1(−1)N−k+i−1Ti−1=∑i=1NΛi,j(−1)N+i+jTi−1.$

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

$Λi,j(−1)N+i+j=0if i≠j,1if i=j.$

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

$N+1−jN+1−i=N−jN−(i−1)+N−jN−i.$

Then, the induction hypothesis implies that

$N+1−jN+1−i=∑k=1NN−ki−2j−1k−1(−1)k−1+∑k=1NN−ki−1j−1k−1(−1)k−1.$

If we use the recursive formula again, then we have

$N+1−jN+1−i=∑k=1NN+1−ki−1j−1k−1(−1)k−1=∑k=1N+1N+1−ki−1j−1k−1(−1)k−1.$

The last equality follows from that $\begin{array}{}\left(\begin{array}{c}N+1-k\\ i-1\end{array}\right)=\left(\begin{array}{c}0\\ i-1\end{array}\right)=0\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}k=N+1\end{array}$ 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

$∑k=1N+1N+1−k0j−1k−1(−1)k−1=∑k=1N+1j−1k−1(−1)k−1=∑k=0Nj−1k(−1)k=δj,1=N+1−jN.$

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

$∑k=1N+1N+1−kNj−1k−1(−1)k−1=NN=1=N+1−j0.$

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

$∑k=1N+1N+1−ki−1Nk−1(−1)k−1=∑k=1N+1Ni−1N+1−ik−1(−1)k−1=Ni−1∑k=1N+1N+1−ik−1(−1)k−1=Ni−1∑k=0NN+1−ik(−1)k=Ni−1(1−1)N+1−i=0=0N+1−i.$

Here, we used the identity

$N−(k−1)i−1Nk−1=Ni−1N−(i−1)k−1.$

Lemma 3.3

Let i and j be integers with Nj > i ≥1. Then

$xjyi∑k,l=1Nzxkyzljkklli=(x+y+z)j−ij!i!(j−i)!.$

Proof

If we apply the binomial expansion twice, then we obtain

$(x+y+z)j−i=∑k=0j−i∑l=0kj−ikklxj−i−kylzk−l.$

Then, one can see that

$(x+y+z)j−i=∑k=ij∑l=ikj−ik−ik−il−ixj−kyl−izk−l=i!(j−i)!j!xjyi∑k=ij∑l=ik(zx)k(yz)ljkklli.$

By the definition of $\begin{array}{}\left(\begin{array}{c}a\\ b\end{array}\right)\end{array}$ as in (27), we obtain the desired result. □

Lemma 3.4

Suppose that N is an even positive integer. Let i and j be integers with 1 ≤ i, jN.

1. $\begin{array}{}\sum _{k=1}^{N}\left(\begin{array}{c}N-k\\ N-i\end{array}\right)\left(\begin{array}{c}N-j\\ N-k\end{array}\right)\left(-1{\right)}^{j+k}={\delta }_{i,j}\end{array}$.

2. Let

$Δi,j:=∑k,l=1Nx−1xN2+1−ixN2+1−l11−xN2+1−k(−1)l+k+jN−lN−iN−kN−lN−jN−k.$

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 $\begin{array}{}\left(\begin{array}{c}a\\ b\end{array}\right)\end{array}$ as in (27), we have

$N−lN−iN−kN−lN−jN−k≠0$

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 $\begin{array}{}{\mathit{\Delta }}_{i,j}=\left(-1{\right)}^{\frac{N}{2}+1}\end{array}$ . In the case of i > j, one can see that

$Δi,j=x−1xN2+1−ix−N2+111−x−N2+1(−1)j ×∑k,l=1N−11−xN−k(x−1)N−lN−lN−iN−kN−lN−jN−k.$

By Lemma 3.3, Δi,j is equal to

$(N−j)!(N−i)!(i−j)!x−1xN2+1−ixN2+1−i11−xN2+1−j(−1)j×((1−x)+x+(−1))i−j=0.$

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

Let ρN be 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 $\begin{array}{}{C}_{N}^{2}\end{array}$ = I, and hence we have

$ρN(S,Sx)ρN(S,x)=I.$

2. For U = TS, we see that

$ρN(U,x)=(−1)N2+111−xN2+1−iN−jN−i(−1)j1≤i,j≤N, ρN(U,Ux)=(−1)N2+1xN2+1−iN−jN−i(−1)j1≤i,j≤N,$

and

$ρN(U,U2x)=(−1)N2+1x−1xN2+1−iN−jN−i(−1)j1≤i,j≤N.$

By Lemma 3.4 (2), we have

$ρN(U,U2x)ρN(U,Ux)ρN(U,x)=(−1)N2+1(Δi,j)1≤i,j≤N=I.$

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) := z6 Δ(z), where Δ(z) is defined by

$Δ(z)=q∏n=1∞(1−qn)24$

and q = e2πiz for z ∈ ℍ. Then, it is known that Δ(z) is a modular form of weight 12 on SL2(ℤ). Note that

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

and

$F−1z=F(z).$

Hence, we see that F(z) is a modular form associated with ρ, 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 $\begin{array}{}f=\sum _{i=1}^{m}{f}_{i}{\mathbf{e}}_{i}\end{array}$ , where fi is a function on ℚ for i = 1, …, m. Then we define

$(f|ρNγ)(x):=∑i=1mfi(γx)ρN−1(γ,x)ei$

for x ∈ ℚ and γ ∈ SL2(ℤ).

Lemma 3.7

The vector-valued function GN : ℚ → ℂN satisfies the functional equations

$GN−GN|ρNS(x)=AN(x)−1EN(x)+2B12δN,2e2$

for all positive rational numbers x and

$GN−GN|ρNT(x)=0$

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 AN $\begin{array}{}\left(\frac{p}{q}\right)\end{array}$ , BN(p, q), FN(p, q), and EN $\begin{array}{}\left(\frac{p}{q}\right)\end{array}$ are defined as in (8), (21), (6), and (9), respectively.

Lemma 3.2 (1) implies that

$ANpq−1=(vi,j(p,q))1≤i,j≤N,vi,j=−q−N2+j−1pN2−jN−ji−1$

since (vi,j(p, q))1≤i,jN × AN $\begin{array}{}\left(\frac{p}{q}\right)\end{array}$ = (Λi,j)1≤i,jN, which is the identity matrix. Lemma 3.2 (2) implies that

$ANpq−1BN(p,q)=CN.$

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 = $\begin{array}{}\frac{p}{q}\end{array}$ . By Theorem 2.5 (1), we see that

$GN(−x)=(−1)N2GN(x)−2B12e2if N=2,(−1)N2GN(x)if N≥4.$

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 = $\begin{array}{}\frac{p}{q}\end{array}$ ∈ 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 . 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 = $\begin{array}{}\frac{1}{n}\end{array}$ for n ∈ ℕ, then

$GN1n∼((−1)N2+1CN)GN(−n)+AN1n−1EN1n+2B12δN,2CNe2$

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( $\begin{array}{}\frac{p}{q}\end{array}$ ) in terms of the negative continued fraction of $\begin{array}{}\frac{p}{q}\end{array}$ . For this, we recall the followings:

1. $\begin{array}{}{G}_{2}\left(\frac{p}{q}\right)=\left(\begin{array}{c}{p}^{-1}{h}_{0,2}\left(p,q\right)\\ {h}_{1,1}\left(p,q\right)\end{array}\right)=\left(\begin{array}{c}\frac{1}{12}{p}^{-1}{q}^{-1}\\ {s}_{1,1}\left(p,q\right)+\frac{1}{4}\end{array}\right)\end{array}$,

2. $\begin{array}{}{\rho }_{2}\left(S,x{\right)}^{-1}=\left(\begin{array}{cc}-1& 0\\ -1& 1\end{array}\right)\end{array}$,

3. $\begin{array}{}{\rho }_{2}\left(T,x{\right)}^{-1}=\left(\begin{array}{cc}\frac{x+1}{x}& 0\\ 0& 1\end{array}\right)\end{array}$,

4. $\begin{array}{}{G}_{2}\left(x\right)-\left({G}_{2}{|}_{{\rho }_{2}}S\right)\left(x\right)=\left(\begin{array}{c}0\\ \frac{{B}_{2}}{2}\frac{1}{x}-3{B}_{1}^{2}+\frac{{B}_{2}}{2}x\end{array}\right)\end{array}$,

5. G2(x) − (G2|ρ2T)(x) = 0.

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

$G2(x)−(G2|ρ2TaS)(x)=G2(x)−ρ2−1(S,x)ρ2(Ta,Sx)−1G2(TaSx)=G2(x)−ρ2−1(S,x)G2(Sx)=0B221x−3B12+B22x$(28)

for a positive integer a.

Now, we express $\begin{array}{}\frac{p}{q}\end{array}$ by using the negative continued fraction as follows

$pq=[a1,a2,…,an]:=a1−1a2−1⋯−1an$

and define

$piqi:=[ai,ai+1,…,an]$

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

$ai−110pi+1qi+1=piqi.$(29)

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

$G2pi+1qi+1−ρ2−1ai−110,pi+1qi+1G2piqi=E2pi+1qi+1.$(30)

For simplicity, we define

$Fi:=ρ2−1ai−110,pi+1qi+1=piqi+10piqi+11.$

By combining equations in (30), we have

$G2pq=F1−1F2−1⋯Fn−1−1G2pnqn−E2p2q2−E2p3q3−⋯−E2pnqn.$

Note that

$F1−1F2−1⋯Fn−1−1=anpq0−anq′q1,$

where

$q′p′=[a2,a3,…,an−1].$

Moreover, we have

$G2pnqn=112an−13,G2pq=112p−1q−112s1,1(p,q)+3,$

and

$E2p2q2+E2p3q3+⋯+E2pnqn=0112−pq+a1+a2+⋯+an−n−14.$

Finally, we obtain

$12s1,1(p,q)=p−q′q−(a1+a2+⋯+an)+3(n−1).$

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 by $\begin{array}{}S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\end{array}$ . 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

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

together with the continued fraction of $\begin{array}{}\frac{p}{q}\end{array}$ (cf. ).

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

$14!s0,4(p,q)13!s1,3(p,q)12!2!s2,2(p,q)13!s3,1(p,q)+−p3q3000−3p2q3p2q200−3pq32pq2pq0−1q31q2−1q114!s0,4(q,p)13!s1,3(q,p)12!2!s2,2(q,p)13!s3,1(q,p)=F(p,q)$

for a column vector F(p, q) = (f1(p, q), …, f4(p, q))T consisting of Laurent polynomials in p and q. Thus, we are able to compute algorithmically the (vector of) generalized Dedekind sums of (p, q) of fixed weight using the continued fraction of $\begin{array}{}\frac{p}{q}\end{array}$ , 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 .

Theorem 4.1

(Weyl’s equidistribution criterion). A sequence

${sk=(s1(k),s2(k),…,sn(k))∈[0,1)n}k∈N$

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

$limT→∞1T∑k=1Te(m⋅sk)=0,$

where $\begin{array}{}m\cdot {s}_{k}=\sum _{i=1}^{n}{m}_{i}{s}_{i}^{\left(k\right)}\end{array}$ and e(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(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(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 integer q and F(x) ∈ ℤ[x, x−1], we define the exponential sum of modulus q of F(x) as

$K(F,q):=∑x∈(Z/qZ)∗eqF(x),$

where $\begin{array}{}{\mathbf{e}}_{q}\left(x\right):=\mathrm{exp}\left(2\pi i\frac{x}{q}\right)\end{array}$.

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

$FN(x):=f1x,f2x,…,fN−1x,$(33)

where $\begin{array}{}{f}_{i}\left(x\right)={\alpha }_{N}{r}_{N}\left(\left(\genfrac{}{}{0em}{}{N-1}{i}\right){x}^{-i}+\left(\genfrac{}{}{0em}{}{N-1}{N-i}\right){x}^{N-i}\right)\end{array}$ . 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

$m⋅FN(x)=αNrN∑i=1N−1miN−1ix−i+miN−1N−ixN−i.$(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] Let N be an even positive integer. For positive integers i and j with i + j = N,

$RNqN−2hij(p,q)−αNrN(p′)iN−1i+pjN−1jq∈Z,$

where p′ is an integer such that p′p ≡ 1 (mod q), and RN = N! βN rN with βN being the denominator of BN.

By Theorem 4.3, one can see that

$∑0

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

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

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 integers i and j, let $\begin{array}{}F\left(x\right)=\sum _{k=-j}^{i}{a}_{k}{x}^{k}\end{array}$ be a Laurent polynomial with integer coefficients such that ai ≠ 0 and aj ≠ 0. Let p be any prime with pak for some k ≠ 0. Then

$K(F,p)≤(i+j)p.$

Proof

See Theorem 1.3 in . □

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

Proposition 4.5

Let m be a nonzero integer vector and mFN(x) be the Laurent polynomial as give in (33). Put d = gcd(m1, m2, …, mN−1). Then for any positive integer pαN rN d, we have

$K(m⋅FN,p)≤2(N−1)p.$(35)

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

Proposition 4.6

Suppose that q is a positive integer, and that N is an even positive integer. Let DFN be a positive integer such that DFN||FN. Let ω(q) be the number of prime factors of q. Then

$K(m⋅FN,q)≤DFN(12N−12)ω(q)q(1−13N−2).$

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. Let p be a fixed prime and pβ|| Then for α > β, we have

$K(F,pα)=pβK(F~,pα−β),$

where $\begin{array}{}\stackrel{~}{F}\left(x\right)=\frac{1}{{p}^{\beta }}F\left(x\right)\end{array}$.

Proof

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

$K(F,pα)=∑z∈(Z/pαZ)∗eFpα=∑x∈Z/pβZ∑y∈(Z/pα−βZ)∗eF~pα−β=pβK(F~,pα−β).$

After the previous lemma, we can pull out p-factors out of the coefficients of mFN(x). For positive integers i and j, let $\begin{array}{}F\left(x\right)=\sum _{k=-j}^{i}{a}_{k}{x}^{k}\end{array}$ 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  with the trivial character.

Lemma 4.8

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

$K(F,pα)≤4(i+2j)pα(1−1i+2j+1).$

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

Proposition 4.9

Let p be a prime. Suppose that m is any fixed nonzero vector inN−1, and that pβ||mFN(x) for some integer β. Suppose that α ≥ 2 is an integer such that α > β. Then for any p, we have

$K(m⋅FN,pα)≤pβ12(N−1)pα(1−13N−2)≤DFN12(N−1)pα(1−13N−2),$

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 mod qi Chinese remainder of F(x) for i = 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<x ϕ(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

$ω(q)≤clog⁡qloglog⁡q$(36)

for some constant c. For sufficiently large q,

$(12N−12)ω(q)≤(12N−12)clog⁡qloglog⁡q≤(q)clog⁡(12N−12)loglog⁡q.$

Thus, we obtain that for any ϵ > 0,

$(12N−12)ω(q)≪qϵ.$

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

Proposition 4.11

Let N be an even positive integer, and let m be a nonzero integer vector of rank N − 1. Then

$K(m⋅FN,q)≪q(1+ϵ−13N−2)$

for all ϵ > 0.

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

$∑0(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(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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Accepted: 2018-06-01

Published Online: 2018-09-01

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 967–985, ISSN (Online) 2391-5455,

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