Our goal in this paper is to give a new estimate for the number of integer lattice points lying in a sphere of radius R centred at the origin. Thus we define S(R)  = #{x ∈ ZZ3: ||x| |  ≤ R}

Heath-Brown, D. R.

English

D. R. Heath-brown

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Lattice points in the sphere

Heath-Brown, DR

Oxford University Research Archive (ORA)

Oxford University Research Archive

Lattice Points in the SphereD.R. Heath-BrownMagdalen College, OxfordFor Professor Andrzej SchinzelOn the Occasion of his Sixtieth Birthday1 IntroductionOur goal in this paper is to give a new estimate for the number of integer latticepoints lying in a sphere of radius R centred at the origin. Thus we defineS(R) = #{x ∈ ZZ3 : ||x|| ≤ R},and seek an asymptotic formulaS(R) =43πR3 + O(Rθ)with as small an exponent θ as possible. It is known that the error term isΩ(R(log R)1/2), and it is conjectured that it should be Oε(R1+ε) for any ε > 0.However the best result established to date isS(R) =43πR3 + Oε(R29/22+ε), (1)due to Chamizo and Iwaniec [1]. This sharpened a long standing result, dueindependently to Chen [3] and Vinogradov [5], in which the exponent was 4/3.Here we shall build on the method of Chamizo and Iwaniec to obtain thefollowing bound.Theorem For any ε > 0 we haveS(R) =43πR3 + Oε(R21/16+ε).The extraneous part in the exponent has been reduced from 29/22−1 = 7/22to 21/16− 1 = 5/16. Since722 − 516722=1561the new result may be regarded as being roughly a 2% improvement on (1).Luckily we are able to use much of Chamizo and Iwaniec’s work withoutmajor changes. They divide the problem into two parts, one involving the esti-mation of a three-dimensional exponential sum, and the other involving boundsfor an average of real character sums. The key to their advance was the discoverythat character sums could be brought into play. Indeed they treat the exponen-tial sum in essentially the same way as Chen and Vinogradov. Following theirlead we shall concentrate on the character sums, and provide better bounds,using a mean-value estimate (Lemma 4) obtained recently by the author [4].We remark that it seems likely that the methods of this paper might beapplied to Gauss’s problem of the average of the class number, in much thesame way as has been done by Chamizo and Iwaniec [2]. One would then hopeto show, in the usual notation, that∑n≤Nh(−n) = π18ζ(3)N3/2 − 32π2N + Oε(N21/32+ε),for any ε > 0.2 The Method of Chamizo and IwaniecFollowing the work of Chamizo and Iwaniec, we may view the estimates ofChen and Vinogradov as arising from a Tauberian argument, in which oneapproximates S(R) by a sum of the formS(H)(R) =∑x∈Z 3fH(||x||),with fH(t) = 1 or 0 for t ≤ R and t ≥ R + H respectively, andfH(t) =R(R + H − t)tH, for R ≤ t ≤ R + H.Here the parameter H will be taken to lie in the range R−1 ≤ H < 1. Thenumber of solutions of ||x||2 = n with x ∈ ZZ3 is r3(n), where r3(n) countsrepresentations as a sum of 3 squares. Since r3(n) ¿ε n1/2+ε, it follows thatS(R) = S(H)(R) + Oε(HR2+ε), (2)on re-defining ε.One may now apply the Poisson summation formula to evaluate S(H)(R)via an infinite series involving terms of the form r3(n) exp(2πiR√n). The effectof the weight function fH(t) is to damp the above terms so that the main2contribution is from those with n ¿ H−2. One then has to consider a three-dimensional exponential sum of the form∑a,b,cexp(2πiR√a2 + b2 + c2).A suitable bound for this leads, in Chamizo and Iwaniec’s version (see [1; (5.3)]),to the estimateS(H)(R) =4π3R3 + 2πHR2 + Oε(H−ε{R21/16 + RH−1/2 + R9/8H−1/8}), (3)providing that H ≤ R−1/2.If one now choses H = R−2/3 it follows from (2) and (3) thatS(R) =43πR3 + Oε(R4/3+ε),which is the result of Chen and Vinogradov.The new idea introduced by Chamizo and Iwaniec, was to examine the dif-ference S(H)(R)−S(R), which was previously estimated in (2) by a crude upperbound. Since we may rewrite S(R) as1 +∑n≤R2r3(n),this leads one to consider sums of the form∑N<n≤N+Kr3(n),where, roughly speaking, one has N = R2 and K = RH.One can now use Gauss’s formula for the number of primitive representationsas a sum of three squares, in terms of class numbers. One then obtains anexpression for S(H)(R) − S(R) in terms of averages of L-functions L(1, χn),whereχn(m) = (−4nm),by applying Dirichlet’s class number formula . To be specific, the analysis ofChamizo and Iwaniec [1; pp. 418 & 426], may be used to establish the followingrelationship.Lemma 1 Suppose that E(N,K) is a non-negative function, increasing withrespect to K. Suppose further that the estimate∑N<n≤N+Kn≡ν(mod 8)L(1, χn) =3ζ(2)28ζ(3)K + O(E(N,K))3holds for 0 < K ≤ N1/2 and for each of ν = 1, 2, 3, 5 and 6. ThenS(H)(R)− S(R)− 2πHR2 ¿ε R1+ε +∑d≤2√RRdE((Rd)2 ,3RHd2), (4)for 0 < H ≤ 1/3 and R ≥ 1.Chamizo and Iwaniec show thatE(N, K) = K7/8Nε + K2/3N1/32+εis admissable, for any ε > 0. The estimate (4) then producesS(H)(R)− S(R)− 2πHR2 ¿ε R1+ε + H7/8R15/8+ε + H2/3R83/48+ε.When this is combined with (3) the choice H = R−7/11 leads to the bound (1).The remainder of this paper will be devoted to the proof of the followingresult.Lemma 2 One may takeE(N, K) = Nε(K5/6 + N2/15 + N1/6 min{1 , K−1/4})in Lemma 1, for any ε > 0.As above this producesS(H)(R)− S(R)− 2πHR2¿ε R1+ε + R11/6+2εH5/6 + R19/15+2ε + R7/6+2εH−1/6. (5)If we combine this with (3) we now obtainS(R) =4π3R3 + Oε(H−ε{R21/16 + RH−1/2 + R9/8H−1/8})+Oε(R2ε{R + R11/6H5/6 + R19/15 + R7/6H−1/6}.The choice H = R−5/8 then leads to the estimate claimed in our theorem. Weshould observe at this point that the error terms in (3) can be improved a little,although the term RH−1/2 remains. It is therefore the terms RH−1/2 in (3)and R11/6H5/6 in (5) that produce the exponent 21/16 in our theorem, ratherthan the term R21/16 in (3).Before turning to the proof of Lemma 2 we remark that the GeneralizedRiemann Hypothesis (or indeed the Generalized Lindelöf Hypothesis) wouldpermit one to take E(N,K) = NεK1/2. As remarked above, improvements on(3) are possible, but it is difficult to see how the term RH−1/2 can be avoided.It therefore appears that an error term ε(R5/4+ε) in our theorem is the verybest that one can hope for by these methods.43 Averages of Real Character Sums: IBy the Pólya-Vinogradov inequality we haveL(1, χn) =∑m≤Nm−1(−4nm) + Oε(N−1/2+ε),for N < n ≤ N + K, whence∑N<n≤N+Kn≡ν(mod 8)L(1, χn) =∑N<n≤N+Kn≡ν(mod 8)∑m≤Nm−1(−4nm) + Oε(Nε).We begin by examining the contribution of square values of m. Followingthe argument of Chamizo and Iwaniec [1; page 425], we see that this is∑N<n≤N+Kn≡ν(mod 8)∑k≤√N, (k,4n)=1k−2 ==3ζ(2)28ζ(3)K + O(1).We proceed to investigate sums of the formS(K, M) =∑N<n≤N+Kn≡ν(mod 8)∑M<m≤2M∗ (−4nm),where Σ∗ indicates that only non-square values of m are to be included. Weobserve at the outset that Lemma 2 is trivial unless K ≥ 1, as we henceforthassume. We shall show that, for any positive integer r and any ε > 0 we haveS(K, M) ¿r,ε K1−1/2r(M + M1/2N1/2r)(MN)ε, (6)andS(K, M) ¿ε (M7/5 + M3/2K−1/4 + M5/4K1/4)(MK)ε. (7)We shall use (6) with r = 3 for M ≥ N1/3, and (7) for M < N1/3. Thus (7)yieldsS(K,M) ¿ε M(N2/15 + N1/6K−1/4 + N1/12K1/4)(NK)ε,whenceS(K,M) ¿ε M(K5/6 + N2/15 + N1/6K−1/4 + N1/12K1/4)N2ε¿ε M(K5/6 + N2/15 + N1/6K−1/4)N2ε,5in either case. This clearly suffices for Lemma 2.For the proof of (6) we apply Hölder’s inequality to obtainS(K, M)2r ¿r K2r−1∑N<n≤N+K|Sn|2,whereSn = {∑M<m≤2M∗ χn(m)}r =∑v≤(2M)ravχn(v),say, so that av ¿r,ε Mε. We now observe that no two of the characters χn canbe equivalent. For if αβ2 6= αγ2 both lie in (N, N + K], with β < γ, say, thenK > αγ2 − αβ2 ≥ α(β + 1)2 − αβ2 > 2αβ ≥ 2α1/2β ≥ 2√N,which contradicts our initial assumption.We may therefore call on the following result.Lemma 3 Let χ run over a set of inequivalent real characters of modulus atmost Q. Then∑χ|∑v≤Vavχ(v)|2 ¿ε QεV 1+ε(Q + V )maxv≤V|av|2,for any ε > 0.This shows thatS(K, M)2r ¿r,ε K2r−1NεMr+rε(N + Mr)M2ε,whenceS(K, M) ¿r,ε K1−1/2r(M + M1/2N1/2r)(MN)ε,on re-defining ε if necessary. This proves (6).Lemma 3 is a simple extension of the following bound of the author [4;Corollary 3].Lemma 4 Let χ run over the set of primitive real characters of modulus atmost Q. Then∑χ|∑v≤Vavχ(v)|2 ¿ε QεV 1+ε(Q + V )maxv≤V|av|2,for any ε > 0.For each character χ in Lemma 3 we write χ0 for the corresponding primitivecharacter, and we let the moduli of χ and χ0 be qj and q respectively. Then∑χ|∑v≤Vavχ(v))|2 =∑χ0|∑v≤V, (v,j)=1avχ0(v)|2.6However|∑v≤V, (v,j)=1avχ0(v)|2 = |∑d|jµ(d)∑w≤V/dadwχ0(d)χ0(w)|2≤ (∑d|j1)(∑d|j|∑w≤V/dadwχ0(d)χ0(w)|2)Since j ≤ Q we deduce that∑χ|∑v≤Vavχ(v))|2 ¿ε Qε∑d≤Q∑χ0|∑w≤V/dadwχ0(d)χ0(w)|2.We may now apply Lemma 4 for each value of d to obtain∑χ|∑v≤Vavχ(v))|2 ¿ε Qε∑d≤QQε(Vd)1+ε(Q +Vd)maxv≤V|av|2¿ε Q2εV 1+ε(Q + V )maxv≤V|av|2.Lemma 3 now follows, on re-defining ε.4 Averages of Real Character Sums: IIOur second bound for S(K, M) comes from the following result.Lemma 5 For any ε > 0 we have∑M<m≤2M∗∗ |∑N<n≤N+Kn≡ν(mod 8)(−4nm)| ¿ε (MK)ε{M7/5 + M3/2K−1/4 + M5/4K1/4},for M ≥ 1, uniformly in N , where Σ∗∗ indicates that the integers m are re-stricted to be square-free.Since we may remove a factor (−4/m) from the inner sum it suffices toconsider ∑M<m≤2M∗∗ |∑N<n≤N+Kn≡ν(mod 8)(nm)|,where m must be odd as well as square-free. We may regard ν as fixed, and wetake N ′ = [(N − ν)/8] and N ′ + K ′ = [(N + K − ν)/8], so that the conditionon n becomes n = 8n′ + ν with N ′ < n′ ≤ N ′ + K ′.7We can now transform the inner sum as∑a(mod m)(am)∑n1m∑b(mod m)e(b(n− a)m),where e(x) = exp(2πix) as usual. We proceed to perform the summations overa and n explicitly. On writing τm for the Gauss sum, the above then becomesτm2im(−1m)∑b(mod m)(bm){e(Ab/m)− e(Bb/m)}cosec(8πb/m),whereA = 8N ′ + 8K ′ + 4 + ν, and B = 8N ′ − 4 + ν.Since m cannot be not equal to 1, the term corresponding to b = 0 vanishes.We may therefore think of the sum over b as being for 1 ≤ |b| ≤ m/2. We shallconfine our attention to the interval 1 ≤ b ≤ m/2, since the alternative rangecan be handled analogously. When b ≤ M/K we remove a factor{e(8K ′b/m)− 1}cosec(8πb/m)by partial summation, to show that∑b≤M/K(bm){e(Ab/m)− e(Bb/m)}cosec(8πb/m)¿ K|Sm(M/K)|+ K2M−1∫ M/K0|Sm(t)|dt,withSm(t) =∑b≤t(bm)e(Bb/m).For the remaining range M/K < b ≤ m/2 we consider only terms involvinge(Bb/m), since the treatment of those containing e(Ab/m) is identical. Here weremove the factor cosec(8πb/m) by partial summation, producing∑M/K<b≤m/2(bm)e(Bb/m)cosec(8πb/m)¿ K|Sm(M/K)|+ |Sm(m/2)|+ M∫ MM/Kt−2|Sm(t)|dt,with Sm(t) as before.The standard procedure for estimating incomplete Gauss sums shows thatSm(t) ¿ε m1/2+ε. We use this bound both for Sm(m/2) and for that part of8the integral in which t ≥ M3/5. In view of the above results we then obtain thebound∑N<n≤N+Kn≡ν(mod 8)(nm) ¿ε M2/5+ε + K√M|Sm(MK)|+ K2M3/2∫ M/K0|Sm(t)|dt+M1/2∫M/K≤t≤M3/5t−2|Sm(t)|dt. (8)We now proceed to show that∑M<m≤2M∗∗ |Sm(t)| ¿ε Mε(Mt3/4 + M3/4t5/4) (9)for 1 ≤ t ≤ M . Taken in conjunction with (8), one readily checks that thissuffices to establish Lemma 5.In order to establish (9) we begin by applying Hölder’s inequality, whence∑M<m≤2M∗∗ |Sm(t)| ¿ M3/4{∑M<m≤2M∗∗ |Sm(t)|4}1/4.Now comes the key idea. For each value of m we set Bm = B − m[B/m], sothatSm(t) =∑b≤t(bm)e(Bmb/m).We then define intervalsIm,h = [(h− 1)mt,hmt), h = 1, . . . , [t + 1],and divide the possible m into subsets J(h), according to the value of h forwhich Bm ∈ Im,h. It follows that there is some h for which∑M<m≤2M∗∗ |Sm(t)| ¿ M3/4t1/4{∑m∈J(h)∗∗ |∑b≤t(bm)e(Bmb/m)|4}1/4.We can now estimate the inner sum by partial summation, knowing that Bm =hm/t−mηm/t, where ηm lies in [0, 1) and is independent of b. Thus, if we setS(0)m (v) =∑b≤v(bm)e(hb/t),we have ∑b≤t(bm)e(Bmb/m) ¿ |S(0)m (t)|+ t−1∫ t0|S(0)m (v)|dv,9whence∑m∈J(h)∗∗ |∑b≤t(bm)e(Bmb/m)|4 ¿∑m∈J(h)∗∗ {|S(0)m (t)|4 + t−1∫ t0|S(0)m (v)|4dv},by a further application of Hölder’s inequality. We therefore conclude that itsuffices, for the proof of (9), to show that∑M<m≤2M∗∗ |S(0)m (v)|4 ¿ε {Mt2 + t4}M4ε, (10)uniformly for 1 ≤ v ≤ t ≤ M . The important point here is that Sm(t) containsan exponential factor which varies with m, while S(0)m (v) does not. This hasbeen achieved at a cost of a factor t however.In order to prove (10) we writeS(0)m (v)2 =∑w≤v2bw(wm),where the coefficents bw are independent of m and satisfy bw ¿ε vε. Lemma 4then shows that ∑M<m≤2M∗∗ |∑w≤v2bw(wm)|2 ¿ε Mεv2+4ε(M + v2),and (10) follows, on re-defining ε.To complete the proof of our theorem it remains to demonstrate how (7)may be derived from Lemma 5. We have therefore to show how the assumptionthat m should be square-free may be replaced by the condition that m is not asquare. To this end we observe thatS(K, M) ¿∑M<m≤2M∗ |∑N<n≤N+Kn≡ν(mod 8)(−4nm)|.We write m = qs2, where q > 1 is square-free. ThusS(K,M) =∑s≤√M2 |/ s∑Ms−2<q≤2Ms−2∗∗ |∑N<n≤N+K, (n,s)=1n≡ν(mod 8)(nq)|=∑s≤√M2 |/ s∑Ms−2<q≤2Ms−2∗∗ |∑N<n≤N+Kn≡ν(mod 8)∑d|s, d|nµ(d)(nq)|≤∑s≤√M2 |/ s∑d|s∑Ms−2<q≤2Ms−2∗∗ |∑N/d<n′≤(N+K)/dn′≡ν′(mod 8)(n′q)|,10where n = dn′ and dν′ ≡ ν (mod 8). We may apply Lemma 5 to the sums overq and n′ to obtainS(K, M)¿ε∑s≤√M2 |/ s∑d|s(MK)ε{(Ms2)7/5 + (Ms2)3/2(Kd)−1/4 + (Ms2)5/4(Kd)1/4}¿ε (MK)ε{M7/5 + M3/2K−1/4 + M5/4K1/4},as required.References[1] F. Chamizo and H. Iwaniec, On the Sphere Problem, Rev. Mat. Iberoamer-icana, 11 (1995), 417-429.[2] F. Chamizo and H. Iwaniec, On the Gauss mean-value formula for the classnumber, Nagoya Math. J., (to appear).[3] J.-R. Chen, Improvement on the asymptotic formulas for the lattice pointsin a region of the three dimensions (II), Sci. Sinica, 12 (1963), 751-764.[4] D.R. Heath-Brown, A mean value estimate for real character sums, ActaArithmetica, 72 (1995), 235-275.[5] I.M. Vinogradov, On the number of integer points in a sphere, Izv. Akad.Nauk SSSR Ser. Mat., 27 (1963), 957-968.11

ORA - Oxford University Research Archive

Heath-Brown, David

Mathematical Institute Eprints Archive

Lattice Points in the Sphere
D.R. Heath-Brown
Magdalen College, Oxford
For Professor Andrzej Schinzel
On the Occasion of his Sixtieth Birthday
1 Introduction
Our goal in this paper is to give a new estimate for the number of integer lattice
points lying in a sphere of radius R centred at the origin. Thus we define
S(R) = #{x ∈ ZZ3 : ||x|| ≤ R},
and seek an asymptotic formula
S(R) =
4
3
piR3 +O(Rθ)
with as small an exponent θ as possible. It is known that the error term is
Ω(R(logR)1/2), and it is conjectured that it should be Oε(R1+ε) for any ε > 0.
However the best result established to date is
S(R) =
4
3
piR3 +Oε(R29/22+ε), (1)
due to Chamizo and Iwaniec [1]. This sharpened a long standing result, due
independently to Chen [3] and Vinogradov [5], in which the exponent was 4/3.
Here we shall build on the method of Chamizo and Iwaniec to obtain the
following bound.
Theorem For any ε > 0 we have
S(R) =
4
3
piR3 +Oε(R21/16+ε).
The extraneous part in the exponent has been reduced from 29/22−1 = 7/22
to 21/16− 1 = 5/16. Since
7
22 − 516
7
22
=
1
56
1
the new result may be regarded as being roughly a 2% improvement on (1).
Luckily we are able to use much of Chamizo and Iwaniec’s work without
major changes. They divide the problem into two parts, one involving the esti-
mation of a three-dimensional exponential sum, and the other involving bounds
for an average of real character sums. The key to their advance was the discovery
that character sums could be brought into play. Indeed they treat the exponen-
tial sum in essentially the same way as Chen and Vinogradov. Following their
lead we shall concentrate on the character sums, and provide better bounds,
using a mean-value estimate (Lemma 4) obtained recently by the author [4].
We remark that it seems likely that the methods of this paper might be
applied to Gauss’s problem of the average of the class number, in much the
same way as has been done by Chamizo and Iwaniec [2]. One would then hope
to show, in the usual notation, that∑
n≤N
h(−n) = pi
18ζ(3)
N3/2 − 3
2pi2
N +Oε(N21/32+ε),
for any ε > 0.
2 The Method of Chamizo and Iwaniec
Following the work of Chamizo and Iwaniec, we may view the estimates of
Chen and Vinogradov as arising from a Tauberian argument, in which one
approximates S(R) by a sum of the form
S(H)(R) =
∑
x∈Z 3
fH(||x||),
with fH(t) = 1 or 0 for t ≤ R and t ≥ R+H respectively, and
fH(t) =
R(R+H − t)
tH
, for R ≤ t ≤ R+H.
Here the parameter H will be taken to lie in the range R−1 ≤ H < 1. The
number of solutions of ||x||2 = n with x ∈ ZZ3 is r3(n), where r3(n) counts
representations as a sum of 3 squares. Since r3(n)¿ε n1/2+ε, it follows that
S(R) = S(H)(R) +Oε(HR2+ε), (2)
on re-defining ε.
One may now apply the Poisson summation formula to evaluate S(H)(R)
via an infinite series involving terms of the form r3(n) exp(2piiR
√
n). The effect
of the weight function fH(t) is to damp the above terms so that the main
2
contribution is from those with n ¿ H−2. One then has to consider a three-
dimensional exponential sum of the form∑
a,b,c
exp(2piiR
√
a2 + b2 + c2).
A suitable bound for this leads, in Chamizo and Iwaniec’s version (see [1; (5.3)]),
to the estimate
S(H)(R) =
4pi
3
R3 + 2piHR2 +Oε(H−ε{R21/16 +RH−1/2 +R9/8H−1/8}), (3)
providing that H ≤ R−1/2.
If one now choses H = R−2/3 it follows from (2) and (3) that
S(R) =
4
3
piR3 +Oε(R4/3+ε),
which is the result of Chen and Vinogradov.
The new idea introduced by Chamizo and Iwaniec, was to examine the dif-
ference S(H)(R)−S(R), which was previously estimated in (2) by a crude upper
bound. Since we may rewrite S(R) as
1 +
∑
n≤R2
r3(n),
this leads one to consider sums of the form∑
N<n≤N+K
r3(n),
where, roughly speaking, one has N = R2 and K = RH.
One can now use Gauss’s formula for the number of primitive representations
as a sum of three squares, in terms of class numbers. One then obtains an
expression for S(H)(R) − S(R) in terms of averages of L-functions L(1, χn),
where
χn(m) = (
−4n
m
),
by applying Dirichlet’s class number formula . To be specific, the analysis of
Chamizo and Iwaniec [1; pp. 418 & 426], may be used to establish the following
relationship.
Lemma 1 Suppose that E(N,K) is a non-negative function, increasing with
respect to K. Suppose further that the estimate∑
N<n≤N+K
n≡ν(mod 8)
L(1, χn) =
3ζ(2)
28ζ(3)
K +O(E(N,K))
3
holds for 0 < K ≤ N1/2 and for each of ν = 1, 2, 3, 5 and 6. Then
S(H)(R)− S(R)− 2piHR2 ¿ε R1+ε +
∑
d≤2√R
R
d
E
(
(
R
d
)2 ,
3RH
d2
)
, (4)
for 0 < H ≤ 1/3 and R ≥ 1.
Chamizo and Iwaniec show that
E(N,K) = K7/8Nε +K2/3N1/32+ε
is admissable, for any ε > 0. The estimate (4) then produces
S(H)(R)− S(R)− 2piHR2 ¿ε R1+ε +H7/8R15/8+ε +H2/3R83/48+ε.
When this is combined with (3) the choice H = R−7/11 leads to the bound (1).
The remainder of this paper will be devoted to the proof of the following
result.
Lemma 2 One may take
E(N,K) = Nε(K5/6 +N2/15 +N1/6min{1 , K−1/4})
in Lemma 1, for any ε > 0.
As above this produces
S(H)(R)− S(R)− 2piHR2
¿ε R1+ε +R11/6+2εH5/6 +R19/15+2ε +R7/6+2εH−1/6. (5)
If we combine this with (3) we now obtain
S(R) =
4pi
3
R3 +Oε(H−ε{R21/16 +RH−1/2 +R9/8H−1/8})
+Oε(R2ε{R+R11/6H5/6 +R19/15 +R7/6H−1/6}.
The choice H = R−5/8 then leads to the estimate claimed in our theorem. We
should observe at this point that the error terms in (3) can be improved a little,
although the term RH−1/2 remains. It is therefore the terms RH−1/2 in (3)
and R11/6H5/6 in (5) that produce the exponent 21/16 in our theorem, rather
than the term R21/16 in (3).
Before turning to the proof of Lemma 2 we remark that the Generalized
Riemann Hypothesis (or indeed the Generalized Lindelo¨f Hypothesis) would
permit one to take E(N,K) = NεK1/2. As remarked above, improvements on
(3) are possible, but it is difficult to see how the term RH−1/2 can be avoided.
It therefore appears that an error term ε(R5/4+ε) in our theorem is the very
best that one can hope for by these methods.
4
3 Averages of Real Character Sums: I
By the Po´lya-Vinogradov inequality we have
L(1, χn) =
∑
m≤N
m−1(
−4n
m
) +Oε(N−1/2+ε),
for N < n ≤ N +K, whence∑
N<n≤N+K
n≡ν(mod 8)
L(1, χn) =
∑
N<n≤N+K
n≡ν(mod 8)
∑
m≤N
m−1(
−4n
m
) +Oε(Nε).
We begin by examining the contribution of square values of m. Following
the argument of Chamizo and Iwaniec [1; page 425], we see that this is∑
N<n≤N+K
n≡ν(mod 8)
∑
k≤√N, (k,4n)=1
k−2 ==
3ζ(2)
28ζ(3)
K +O(1).
We proceed to investigate sums of the form
S(K,M) =
∑
N<n≤N+K
n≡ν(mod 8)
∑
M<m≤2M
∗ (
−4n
m
),
where Σ∗ indicates that only non-square values of m are to be included. We
observe at the outset that Lemma 2 is trivial unless K ≥ 1, as we henceforth
assume. We shall show that, for any positive integer r and any ε > 0 we have
S(K,M)¿r,ε K1−1/2r(M +M1/2N1/2r)(MN)ε, (6)
and
S(K,M)¿ε (M7/5 +M3/2K−1/4 +M5/4K1/4)(MK)ε. (7)
We shall use (6) with r = 3 for M ≥ N1/3, and (7) for M < N1/3. Thus (7)
yields
S(K,M)¿ε M(N2/15 +N1/6K−1/4 +N1/12K1/4)(NK)ε,
whence
S(K,M) ¿ε M(K5/6 +N2/15 +N1/6K−1/4 +N1/12K1/4)N2ε
¿ε M(K5/6 +N2/15 +N1/6K−1/4)N2ε,
5
in either case. This clearly suffices for Lemma 2.
For the proof of (6) we apply Ho¨lder’s inequality to obtain
S(K,M)2r ¿r K2r−1
∑
N<n≤N+K
|Sn|2,
where
Sn = {
∑
M<m≤2M
∗ χn(m)}r =
∑
v≤(2M)r
avχn(v),
say, so that av ¿r,ε Mε. We now observe that no two of the characters χn can
be equivalent. For if αβ2 6= αγ2 both lie in (N,N +K], with β < γ, say, then
K > αγ2 − αβ2 ≥ α(β + 1)2 − αβ2 > 2αβ ≥ 2α1/2β ≥ 2
√
N,
which contradicts our initial assumption.
We may therefore call on the following result.
Lemma 3 Let χ run over a set of inequivalent real characters of modulus at
most Q. Then ∑
χ
|
∑
v≤V
avχ(v)|2 ¿ε QεV 1+ε(Q+ V )max
v≤V
|av|2,
for any ε > 0.
This shows that
S(K,M)2r ¿r,ε K2r−1NεMr+rε(N +Mr)M2ε,
whence
S(K,M)¿r,ε K1−1/2r(M +M1/2N1/2r)(MN)ε,
on re-defining ε if necessary. This proves (6).
Lemma 3 is a simple extension of the following bound of the author [4;
Corollary 3].
Lemma 4 Let χ run over the set of primitive real characters of modulus at
most Q. Then ∑
χ
|
∑
v≤V
avχ(v)|2 ¿ε QεV 1+ε(Q+ V )max
v≤V
|av|2,
for any ε > 0.
For each character χ in Lemma 3 we write χ0 for the corresponding primitive
character, and we let the moduli of χ and χ0 be qj and q respectively. Then∑
χ
|
∑
v≤V
avχ(v))|2 =
∑
χ0
|
∑
v≤V, (v,j)=1
avχ0(v)|2.
6
However
|
∑
v≤V, (v,j)=1
avχ0(v)|2 = |
∑
d|j
µ(d)
∑
w≤V/d
adwχ0(d)χ0(w)|2
≤ (
∑
d|j
1)(
∑
d|j
|
∑
w≤V/d
adwχ0(d)χ0(w)|2)
Since j ≤ Q we deduce that∑
χ
|
∑
v≤V
avχ(v))|2 ¿ε Qε
∑
d≤Q
∑
χ0
|
∑
w≤V/d
adwχ0(d)χ0(w)|2.
We may now apply Lemma 4 for each value of d to obtain∑
χ
|
∑
v≤V
avχ(v))|2 ¿ε Qε
∑
d≤Q
Qε(
V
d
)1+ε(Q+
V
d
)max
v≤V
|av|2
¿ε Q2εV 1+ε(Q+ V )max
v≤V
|av|2.
Lemma 3 now follows, on re-defining ε.
4 Averages of Real Character Sums: II
Our second bound for S(K,M) comes from the following result.
Lemma 5 For any ε > 0 we have∑
M<m≤2M
∗∗ |
∑
N<n≤N+K
n≡ν(mod 8)
(
−4n
m
)| ¿ε (MK)ε{M7/5 +M3/2K−1/4 +M5/4K1/4},
for M ≥ 1, uniformly in N , where Σ∗∗ indicates that the integers m are re-
stricted to be square-free.
Since we may remove a factor (−4/m) from the inner sum it suffices to
consider ∑
M<m≤2M
∗∗ |
∑
N<n≤N+K
n≡ν(mod 8)
(
n
m
)|,
where m must be odd as well as square-free. We may regard ν as fixed, and we
take N ′ = [(N − ν)/8] and N ′ +K ′ = [(N +K − ν)/8], so that the condition
on n becomes n = 8n′ + ν with N ′ < n′ ≤ N ′ +K ′.
7
We can now transform the inner sum as∑
a(mod m)
(
a
m
)
∑
n
1
m
∑
b(mod m)
e(
b(n− a)
m
),
where e(x) = exp(2piix) as usual. We proceed to perform the summations over
a and n explicitly. On writing τm for the Gauss sum, the above then becomes
τm
2im
(
−1
m
)
∑
b(mod m)
(
b
m
){e(Ab/m)− e(Bb/m)}cosec(8pib/m),
where
A = 8N ′ + 8K ′ + 4 + ν, and B = 8N ′ − 4 + ν.
Since m cannot be not equal to 1, the term corresponding to b = 0 vanishes.
We may therefore think of the sum over b as being for 1 ≤ |b| ≤ m/2. We shall
confine our attention to the interval 1 ≤ b ≤ m/2, since the alternative range
can be handled analogously. When b ≤M/K we remove a factor
{e(8K ′b/m)− 1}cosec(8pib/m)
by partial summation, to show that∑
b≤M/K
(
b
m
){e(Ab/m)− e(Bb/m)}cosec(8pib/m)
¿ K|Sm(M/K)|+K2M−1
∫ M/K
0
|Sm(t)|dt,
with
Sm(t) =
∑
b≤t
(
b
m
)e(Bb/m).
For the remaining range M/K < b ≤ m/2 we consider only terms involving
e(Bb/m), since the treatment of those containing e(Ab/m) is identical. Here we
remove the factor cosec(8pib/m) by partial summation, producing∑
M/K<b≤m/2
(
b
m
)e(Bb/m)cosec(8pib/m)
¿ K|Sm(M/K)|+ |Sm(m/2)|+M
∫ M
M/K
t−2|Sm(t)|dt,
with Sm(t) as before.
The standard procedure for estimating incomplete Gauss sums shows that
Sm(t) ¿ε m1/2+ε. We use this bound both for Sm(m/2) and for that part of
8
the integral in which t ≥M3/5. In view of the above results we then obtain the
bound∑
N<n≤N+K
n≡ν(mod 8)
(
n
m
) ¿ε M2/5+ε + K√
M
|Sm(M
K
)|+ K
2
M3/2
∫ M/K
0
|Sm(t)|dt
+M1/2
∫
M/K≤t≤M3/5
t−2|Sm(t)|dt. (8)
We now proceed to show that∑
M<m≤2M
∗∗ |Sm(t)| ¿ε Mε(Mt3/4 +M3/4t5/4) (9)
for 1 ≤ t ≤ M . Taken in conjunction with (8), one readily checks that this
suffices to establish Lemma 5.
In order to establish (9) we begin by applying Ho¨lder’s inequality, whence∑
M<m≤2M
∗∗ |Sm(t)| ¿M3/4{
∑
M<m≤2M
∗∗ |Sm(t)|4}1/4.
Now comes the key idea. For each value of m we set Bm = B −m[B/m], so
that
Sm(t) =
∑
b≤t
(
b
m
)e(Bmb/m).
We then define intervals
Im,h = [
(h− 1)m
t
,
hm
t
), h = 1, . . . , [t+ 1],
and divide the possible m into subsets J(h), according to the value of h for
which Bm ∈ Im,h. It follows that there is some h for which∑
M<m≤2M
∗∗ |Sm(t)| ¿M3/4t1/4{
∑
m∈J(h)
∗∗ |
∑
b≤t
(
b
m
)e(Bmb/m)|4}1/4.
We can now estimate the inner sum by partial summation, knowing that Bm =
hm/t−mηm/t, where ηm lies in [0, 1) and is independent of b. Thus, if we set
S(0)m (v) =
∑
b≤v
(
b
m
)e(hb/t),
we have ∑
b≤t
(
b
m
)e(Bmb/m)¿ |S(0)m (t)|+ t−1
∫ t
0
|S(0)m (v)|dv,
9
whence∑
m∈J(h)
∗∗ |
∑
b≤t
(
b
m
)e(Bmb/m)|4 ¿
∑
m∈J(h)
∗∗ {|S(0)m (t)|4 + t−1
∫ t
0
|S(0)m (v)|4dv},
by a further application of Ho¨lder’s inequality. We therefore conclude that it
suffices, for the proof of (9), to show that∑
M<m≤2M
∗∗ |S(0)m (v)|4 ¿ε {Mt2 + t4}M4ε, (10)
uniformly for 1 ≤ v ≤ t ≤M . The important point here is that Sm(t) contains
an exponential factor which varies with m, while S(0)m (v) does not. This has
been achieved at a cost of a factor t however.
In order to prove (10) we write
S(0)m (v)
2 =
∑
w≤v2
bw(
w
m
),
where the coefficents bw are independent of m and satisfy bw ¿ε vε. Lemma 4
then shows that ∑
M<m≤2M
∗∗ |
∑
w≤v2
bw(
w
m
)|2 ¿ε Mεv2+4ε(M + v2),
and (10) follows, on re-defining ε.
To complete the proof of our theorem it remains to demonstrate how (7)
may be derived from Lemma 5. We have therefore to show how the assumption
that m should be square-free may be replaced by the condition that m is not a
square. To this end we observe that
S(K,M)¿
∑
M<m≤2M
∗ |
∑
N<n≤N+K
n≡ν(mod 8)
(
−4n
m
)|.
We write m = qs2, where q > 1 is square-free. Thus
S(K,M) =
∑
s≤√M
2 |/ s
∑
Ms−2<q≤2Ms−2
∗∗ |
∑
N<n≤N+K, (n,s)=1
n≡ν(mod 8)
(
n
q
)|
=
∑
s≤√M
2 |/ s
∑
Ms−2<q≤2Ms−2
∗∗ |
∑
N<n≤N+K
n≡ν(mod 8)
∑
d|s, d|n
µ(d)(
n
q
)|
≤
∑
s≤√M
2 |/ s
∑
d|s
∑
Ms−2<q≤2Ms−2
∗∗ |
∑
N/d<n′≤(N+K)/d
n′≡ν′(mod 8)
(
n′
q
)|,
10
where n = dn′ and dν′ ≡ ν (mod 8). We may apply Lemma 5 to the sums over
q and n′ to obtain
S(K,M)
¿ε
∑
s≤√M
2 |/ s
∑
d|s
(MK)ε{(M
s2
)7/5 + (
M
s2
)3/2(
K
d
)−1/4 + (
M
s2
)5/4(
K
d
)1/4}
¿ε (MK)ε{M7/5 +M3/2K−1/4 +M5/4K1/4},
as required.
References
[1] F. Chamizo and H. Iwaniec, On the Sphere Problem, Rev. Mat. Iberoamer-
icana, 11 (1995), 417-429.
[2] F. Chamizo and H. Iwaniec, On the Gauss mean-value formula for the class
number, Nagoya Math. J., (to appear).
[3] J.-R. Chen, Improvement on the asymptotic formulas for the lattice points
in a region of the three dimensions (II), Sci. Sinica, 12 (1963), 751-764.
[4] D.R. Heath-Brown, A mean value estimate for real character sums, Acta
Arithmetica, 72 (1995), 235-275.
[5] I.M. Vinogradov, On the number of integer points in a sphere, Izv. Akad.
Nauk SSSR Ser. Mat., 27 (1963), 957-968.
11


A mean value estimate for real character sums,

Improvement on the asymptotic formulas for the lattice points in a region of the three dimensions (II),

On the Gauss mean-value formula for the class number,

On the number of integer points in a sphere,

On the Sphere Problem,

http://eprints.maths.ox.ac.uk/174/1/sphere.pdf

Lattice points in the sphere

Abstract

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