A Symplectic Test of the L-Functions Ratios Conjecture
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE
STEVEN J. MILLER
ABSTRACT. Recently Conrey, Farmer and Zirnbauer [CFZ1, CFZ2] conjectured for-
mulas for the averages over a family of ratios of products of shifted L-functions. Their
L-functions Ratios Conjecture predicts both the main and lower order terms for many
problems, ranging from n-level correlations and densities to mollifiers and moments
to vanishing at the central point. There are now many results showing agreement be-
tween the main terms of number theory and random matrix theory; however, there are
very few families where the lower order terms are known. These terms often depend
on subtle arithmetic properties of the family, and provide a way to break the universal-
ity of behavior. The L-functions Ratios Conjecture provides a powerful and tractable
way to predict these terms. We test a specific case here, that of the 1-level density for
the symplectic family of quadratic Dirichlet characters arising from even fundamental
discriminants d ≤ X . For test functions supported in (−1/3, 1/3) we calculate all
the lower order terms up to size O(X−1/2+ǫ) and observe perfect agreement with the
conjecture (for test functions supported in (−1, 1) we show agreement up to errors of
size O(X−ǫ) for any ǫ). Thus for this family and suitably restricted test functions, we
completely verify the Ratios Conjecture’s prediction for the 1-level density.
1. INTRODUCTION
Montgomery’s [Mon] analysis of the pair correlation of zeros of ζ(s) revealed a strik-
ing similarity to the behavior of eigenvalues of ensembles of random matrices. Since
then, this connection has been a tremendous predictive aid to researchers in number
theory in modeling the behavior of zeros and values of L-functions, ranging from spac-
ings between adjacent zeros [Mon, Hej, Od1, Od2, RS] to moments of L-functions
[CF, CFKRS]. Katz and Sarnak [KaSa1, KaSa2] conjectured that, in the limit as the
conductors tend to infinity, the behavior of the normalized zeros near the central point
agree with theN → ∞ scaling limit of the normalized eigenvalues near 1 of a subgroup
of U(N). One way to test this correspondence is through the n-level density of a family
F of L-functions L(s, f); we concentrate on this statistic in this paper. The n-level
density is
Dn,F(φ) :=
ℓ1,...,ℓn
ℓi 6=±ℓk
γf,ℓ1
logQf
· · ·φn
γf,ℓn
logQf
, (1.1)
Date: October 25, 2018.
2000 Mathematics Subject Classification. 11M26 (primary), 11M41, 15A52 (secondary).
Key words and phrases. 1-Level Density, Dirichlet L-functions, Low Lying Zeros, Ratios Conjecture.
The author would like to thank Eduardo Dueñez, Chris Hughes, Duc Khiem Huynh, Jon Keating,
Nina Snaith and Sergei Treil for many enlightening conversations, Jeffrey Stopple for finding a typo in
the proof of Lemma 3.2, and the University of Bristol for its hospitality (where much of this work was
done). This work was partly supported by NSF grant DMS0600848.
http://arxiv.org/abs/0704.0927v5
2 STEVEN J. MILLER
where the φi are even Schwartz test functions whose Fourier transforms have compact
support, 1
+ iγf,ℓ runs through the non-trivial zeros of L(s, f), and Qf is the analytic
conductor of f . As the φi are even Schwartz functions, most of the contribution to
Dn,F(φ) arises from the zeros near the central point; thus this statistic is well-suited to
investigating the low-lying zeros.
There are now many examples where the main term in number theory agrees with
the Katz-Sarnak conjectures (at least for suitably restricted test functions), such as all
Dirichlet characters, quadratic Dirichlet characters, L(s, ψ) with ψ a character of the
ideal class group of the imaginary quadratic field Q(
−D), families of elliptic curves,
weight k level N cuspidal newforms, symmetric powers of GL(2) L-functions, and
certain families of GL(4) and GL(6) L-functions (see [DM1, FI, Gü, HR, HM, ILS,
KaSa2, Mil1, OS2, RR, Ro, Rub1, Yo2]).
For families of L-functions over function fields, the corresponding classical compact
group can be identified through the monodromy. While the situation is less clear for L-
functions over number fields, there has been some recent progress. Dueñez and Miller
[DM2] show that for sufficiently nice families and sufficiently small support, the main
term in the 1-level density is determined by the first and second moments of the Satake
parameters, and a symmetry constant (which identifies the corresponding classical com-
pact group) may be associated to any nice family such that the symmetry constant of the
Rankin-Selberg convolution of two families is the product of the symmetry constants.
There are two avenues for further research. The first is to increase the support of
the test functions, which often leads to questions of arithmetic interest (see for example
Hypothesis S in [ILS]). Another is to identify lower order terms in the 1-level density,
which is the subject of this paper. The main term in the 1-level density is independent of
the arithmetic of the family, which surfaces in the lower order terms. This is very similar
to the Central Limit Theorem. For nice densities the distribution of the normalized
sample mean converges to the standard normal. The main term is controlled by the
first two moments (the mean and the variance of the density) and the higher moments
surface in the rate of convergence. This is similar to our situation, where the universal
main terms arise from the first and second moments of the Satake parameters.
There are now several families where lower order terms have been isolated in the
1-level density [FI, Mil2, Mil3, Yo1]; see also [BoKe], where the Hardy-Littlewood
conjectures are related to lower order terms in the pair correlation of zeros of ζ(s) (see
for example [Be, BeKe, CS2, Ke] for more on lower terms of correlations of Riemann
zeros). Recently Conrey, Farmer and Zirnbauer [CFZ1, CFZ2] formulated conjectures
for the averages over families ofL-functions of ratios of products of shiftedL-functions,
such as
+ α, χd
+ γ, χd
ζ(1 + 2α)
ζ(1 + α + γ)
AD(α; γ)
) ζ(1− 2α)
ζ(1− α + γ)
AD(−α; γ)
+ O(X1/2+ǫ) (1.2)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 3
(here d ranges over even fundamental discriminants, −1/4 < ℜ(α) < 1/4, 1/ logX ≪
ℜ(γ) < 1/4, and AD (we only give the definition for α = γ, as that is the only in-
stance that occurs in our applications) is defined in (1.4)). Their L-functions Ratios
Conjecture arises from using the approximate functional equation, integrating term by
term, and retaining only the diagonal pieces (which they then ‘complete’); they also
assume uniformity in the parameters so that the resulting expressions may be differen-
tiated (this is an essential ingredient for 1-level density calculations). It is worth noting
the incredible detail of the conjecture, predicting all terms down to O(X1/2+ǫ).
There are many difficult computations whose answers can easily be predicted through
applications of theL-functions Ratios Conjecture, ranging from n-level correlations and
densities to mollifiers and moments to vanishing at the central point (see [CS1]). While
these are not proofs, it is extremely useful for researchers to have a sense of what the
answer should be. One common difficulty in the subject is that often the number theory
and random matrix theory answers appear different at first, and much effort must be
spent on combinatorics to prove agreement (see for example [Gao, HM, Rub1, RS]); the
analysis is significantly easier if one knows what the final answer should be. Further,
the Ratios Conjecture often suggest a more enlightening way to group terms (see for
instance Remark 1.4).
Our goal in this paper is to test the predictions of the Ratios Conjecture for a specific
family, that of quadratic Dirichlet characters. We let d be a fundamental discriminant.
This means (see §5 of [Da]) that either d is a square-free number congruent to 1 mod-
ulo 4, or d/4 is square-free and congruent to 2 or 3 modulo 4. If χd is the quadratic
character associated to the fundamental discriminant d, then if χd(−1) = 1 (resp., −1)
we say d is even (resp., odd). If d is a fundamental discriminant then it is even (resp.,
odd) if d > 0 (resp., d < 0). We concentrate on even fundamental discriminants below,
though with very few changes our arguments hold for odd discriminants (for example,
if d is odd there is an extra 1/2 in certain Gamma factors in the explicit formula).
For notational convenience we adopt the following conventions throughout the paper:
• Let X∗ denote the number of even fundamental discriminants at most X; thus
X∗ = 3X/π2 +O(X1/2), and X/π2 +O(X1/2) of these have 4|d (see Lemma
B.1 for a proof).
• In any sum over d, d will range over even fundamental discriminants unless
otherwise specified.
The goal of these notes is to calculate the lower order terms (on the number theory
side) as much as possible, as unconditionally as possible, and then compare our answer
to the prediction from the L-functions Ratios Conjecture, given in the theorem below.
Theorem 1.1 (One-level density from the Ratios Conjecture [CS1]). Let g be an even
Schwartz test function such that ĝ has finite support. Let X∗ denote the number of
even fundamental discriminants at most X , and let d denote a typical even fundamental
discriminant. Assuming the Ratios Conjecture for
d≤X L(
+ α, χd)/L(
+ γ, χd),
4 STEVEN J. MILLER
we have
X∗ logX
X∗ logX
+ A′D
− e−2πiτ log(d/π)/ logX
− πiτ
+ πiτ
1− 4πiτ
− 2πiτ
+ O(X−
+ǫ), (1.3)
AD(−r, r) =
(p+ 1)p1−2r
A′D(r; r) =
log p
(p+ 1)(p1+2r − 1)
. (1.4)
The above is
1− sin(2πx)
, (1.5)
which is the 1-level density for the scaling limit of USp(2N). If supp(ĝ) ⊂ (−1, 1),
then the integral of g(x) against − sin(2πx)/2πx is −g(0)/2.
If we assume the Riemann Hypothesis, for supp(ĝ) ⊂ (−σ, σ) ⊂ (−1, 1) we have
X∗ logX
g(τ) e
−2πiτ
log(d/π)
− πiτ
+ πiτ
1− 4πiτ
− 2πiτ
= −g(0)
+O(X−
(1−σ)+ǫ); (1.6)
the error term may be absorbed into the O(X−1/2+ǫ) error in (1.3) if σ < 1/3.
The conclusions of the above theorem are phenomenal, and demonstrate the power of
the Ratios Conjecture. Not only does its main term agree with the Katz-Sarnak conjec-
tures for arbitrary support, but it calculates the lower order terms up to sizeO(X−1/2+ǫ).
While Theorem 1.1 is conditional on the Ratios Conjecture, the following theorem is
not, and provides highly non-trivial support for the Ratios Conjecture.
Theorem 1.2 (One-level density for quadratic Dirichlet characters). Let the notation be
as in Theorem 1.1, with supp(ĝ) ⊂ (−σ, σ).
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 5
(1) Up to terms of size O(X−(1−σ)/2+ǫ), the 1-level density for the family of qua-
dratic Dirichlet characters with even fundamental discriminants at most X
agrees with (1.3) (the prediction from the Ratios Conjecture).
(2) If we instead consider the family {8d : 0 < d ≤ X, d an odd, positive square-
free fundamental discriminant}, then the 1-level density agrees with the predic-
tion from the Ratios Conjecture up to terms of size O(X−1/2 + X−(1−
σ)+ǫ +
(1−σ)+ǫ). In particular, if σ < 1/3 then the number theory calculation
agrees with the Ratios Conjecture up to errors at most O(X−1/2+ǫ).
Remark 1.3. The above theorem indicates that, at least for the family of quadratic
Dirichlet characters and suitably restricted test functions, the Ratios Conjecture is pre-
dicting all lower order terms up to size O(X−1/2+ǫ). This is phenomenal agreement
between theory and conjecture. Previous investigations of lower order terms in 1-level
densities went as far as O(logN X) for some N ; here we are getting square-root agree-
ment, and strong evidence in favor of the Ratios Conjecture.
Remark 1.4 (Influence of zeros of ζ(s) on lower order terms). From the expansion
in (1.3) we see that one of the lower order terms (arising from the integral of g(τ)
against ζ ′(1 + 4πiτ/ logX)/ζ(1 + 4πiτ/ logX)) in the 1-level density for the family
of quadratic Dirichlet characters is controlled by the non-trivial zeros of ζ(s). This
phenomenon has been noted by other researchers (Bogomolny, Conrey, Keating, Ru-
binstein, Snaith); see [CS1, BoKe, HKS, Rub2] for more details, especially [Rub2] for
a plot of the influence of zeros of ζ(s) on zeros of L-functions of quadratic Dirichlet
characters.
The proof of Theorem 1.2 starts with the Explicit Formula, which relates sums over
zeros to sums over primes (for completeness a proof is given in Appendix A). For
convenience to researchers interested in odd fundamental discriminants, we state it in
more generality than we need.
Theorem 1.5 (Explicit Formula for a family of Quadratic Dirichlet Characters). Let g
be an even Schwartz test function such that ĝ has finite support. For d a fundamental
discriminant let a(χd) = 0 if d is even (χd(−1) = 1) and 1 otherwise. Consider a
family F(X) of fundamental discriminants at most X in absolute value. We have
|F(X)|
d∈F(X)
|F(X)| logX
d∈F(X)
a(χd)
a(χd)
|F(X)|
d∈F(X)
χd(p)
k log p
pk/2 logX
log pk
(1.7)
As our family has only even fundamental discriminants, all a(χd) = 0. The terms
arising from the conductors (the log(|d|/π) and the Γ′/Γ terms) agree with the Ratios
Conjecture. We are reduced to analyzing the sums of χd(p)
k and showing they agree
6 STEVEN J. MILLER
with the remaining terms in the Ratios Conjecture. As our characters are quadratic, this
reduces to understanding sums of χd(p) and χd(p)
2. We first analyze the terms from
the Ratios Conjecture in §2 and then we analyze the character sums in §3. We proceed
in this order as one of the main uses of the Ratios Conjecture is in predicting simple
forms of the answer; in particular, it suggests non-obvious simplifications of the number
theory sums.
2. ANALYSIS OF THE TERMS FROM THE RATIOS CONJECTURE.
We analyze the terms in the 1-level density from the Ratios Conjecture (Theorem
1.1). The first piece (involving log(d/π) and Γ′/Γ factors) is already matched with the
terms in the Explicit Formula arising from the conductors and Γ-factors in the functional
equation. In §3 we match the next two terms (the integral of g(τ) against ζ ′/ζ and A′D)
to the contributions from the sum over χd(p)
k for k even; we do this for test functions
with arbitrary support. The number theory is almost equal to this; the difference is the
presence of a factor −g(0)/2 from the even k terms, which we match to the remaining
piece from the Ratios Conjecture.
This remaining piece is the hardest to analyze. We denote it by
R(g;X) = −
X∗ logX
g(τ)e
−2πiτ
log(d/π)
− πiτ
+ πiτ
dτ, (2.1)
with (see (1.4))
AD(−r, r) =
(p+ 1)p1−2r
. (2.2)
There is a contribution to R(g;X) from the pole of ζ(s). The other terms are at
most O(1/ logX); however, if the support of ĝ is sufficiently small then these terms
contribute significantly less.
Lemma 2.1. Assume the Riemann Hypothesis. If supp(ĝ) ⊂ (−σ, σ) then
R(g;X) = −
+O(X−
(1−σ)+ǫ). (2.3)
In particular, if σ < 1/3 then R(g;X) = −1
g(0) +O(X−
Remark 2.2. If we do not assume the Riemann Hypothesis we may prove a similar re-
sult. The error term is replaced with O(X−(1−
)(1−σ)+ǫ), where θ is the supremum
of the real parts of zeros of ζ(s). As θ ≤ 1, we may always bound the error by
O(X−(1−σ)/2+ǫ). Interestingly, this is the error we get in analyzing the number the-
ory terms χ(p)k with k odd by applying Jutila’s bound (see §3.2.1); we obtain a better
bound of O(X−(1−
σ)) by using Poisson summation to convert long character sums to
shorter ones (see §3.2.2).
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 7
Remark 2.3. The proof of Lemma 2.1 follows from shifting contours and keeping
track of poles of ratios of Gamma and zeta functions. We can prove a related result with
significantly less work. Specifically, if for supp(ĝ) ⊂ (−1, 1) we are willing to accept
error terms of size O(log−N X) for any N then we may proceed as follows: (1) modify
Lemma B.2 to replace the d-sum with X∗e−2πi(1−
log π
logX )τ
1− 2πiτ
+O(X1/2); (2)
use the decay properties of g to restrict the τ sum to |τ | ≤ logX and then Taylor expand
everything but g, which gives a small error term and
|τ |≤logX
lognX
(2πiτ)ne
−2πi(1− logπlogX )τdτ
lognX
|τ |≤logX
(2πiτ)ng(τ)e
−2πi(1− log πlogX )τdτ ; (2.4)
(3) use the decay properties of g to extend the τ -integral to all of R (it is essential here
that N is fixed and finite!) and note that for n ≥ 0 the above is the Fourier transform of
g(n) (the nth derivative of g) at 1− π
, and this is zero if supp(ĝ) ⊂ (−1, 1).
We prove Lemma 2.1 in §2.1; this completes our analysis of the terms from the
Ratios Conjecture. We analyze the lower order term of size 1/ logX (present only if
supp(ĝ) 6⊂ (−1, 1)) in Lemma 2.6 of §2.2. We explicitly calculate this contribution
because in many applications all that is required are the main and first lower order
terms. One example of this is that zeros at height T are modeled not by the N → ∞
scaling limits of a classical compact group but by matrices of size N ∼ log(T/2π)
[KeSn1, KeSn2]. In fact, even better agreement is obtained by changing N slightly due
to the first lower order term (see [BBLM, DHKMS]).
2.1. Analysis of R(g;X). Before proving Lemma 2.1 we collect several useful facts.
Lemma 2.4. In all statements below r = 2πiτ/ logX and supp(ĝ) ⊂ (−σ, σ) ⊂
(−1, 1).
(1) AD(−r, r) = ζ(2)/ζ(2− 2r).
(2) If |r| ≥ ǫ then |ζ(−3− 2r)/ζ(−2− 2r)| ≪ǫ (1 + |r|).
(3) For w ≥ 0, g
τ − iw logX
≪ Xσw
τ 2 + (w logX
for any B ≥ 0.
(4) For 0 < a < b we have |Γ(a± iy)/Γ(b± iy)| = Oa,b(1).
Proof. (1): From simple algebra, as we may rewrite each factor as
p2−2r
p2−2r
. (2.5)
(2): By the functional equations of the Gamma and zeta functions Γ(s/2)π−s/2ζ(s)
= Γ((1− s)/2)π−(1−s)/2ζ(1− s) and Γ(1 + x) = xΓ(x) gives
ζ(−3− 2r)
ζ(−2− 2r)
Γ(1− (−1 − r))π−2−rΓ(−1− r)π1+rζ(4 + 2r)
− r)π 32+rΓ(1− (−3
− r))(3
+ r)−1π−
+rζ(3 + 2r)
. (2.6)
Using
Γ(x)Γ(1− x) = π/ sin πx = 2πi/(eiπx − e−iπx), (2.7)
8 STEVEN J. MILLER
we see the ratio of the Gamma factors have the same growth as |r| → ∞ (if r = 0 then
there is a pole from the zero of ζ(s) at s = −2), and the two zeta functions are bounded
away from 0 and infinity.
(3): As g(τ) =
ĝ(ξ)e2πiξτdξ, we have
g(τ − iy) =
ĝ(ξ)e2πi(τ−iy)ξdξ
ĝ(2n)(ξ)(2πi(τ − iy))−ne2πi(τ−iy)ξdξ
≪ e2πyσ(τ − iy))−2n; (2.8)
the claim follows by taking y = (w logX)/2π.
(4): As |Γ(x− iy)| = |Γ(x+ iy)|, we may assume all signs are positive. The claim
follows from the definition of the Beta function:
Γ(a + iy)Γ(b− a)
Γ(b+ iy)
ta+iy−1(1− t)b−a−1 = Oa,b(1); (2.9)
see [ET] for additional estimates of the size of ratios of Gamma functions. �
Proof of Lemma 2.1. By Lemma 2.4 we may replace AD(−2πiτ/ logX, 2πiτ/ logX)
with ζ(2)/ζ(2−4πiτ/ logX). We replace τ with τ − iw logX
with w = 0 (we will shift
the contour in a moment). Thus
R(g;X) = − 2
X∗ logX
τ − iw logX
−2πi(τ−iw logX2π )
log(d/π)
− πiτ
+ πiτ
ζ(2)ζ
1− w − 4πiτ
2− 2w − 4πiτ
) dτ. (2.10)
We now shift the contour to w = 2. There are two different residue contributions as
we shift (remember we are assuming the Riemann Hypothesis, so that if ζ(ρ) = 0 then
either ρ = 1
+ iγ for some γ ∈ R or ρ is a negative even integer), arising from
• the pole of ζ
1− w − 4πiτ
at w = τ = 0;
• the zeros of ζ
2− 2w − 4πiτ
when w = 3/4 and τ = γ logX
(while potentially there is a residue from the pole of Γ
− πiτ
when w = 1/2
and τ = 0, this is canceled by the pole of ζ
2− 2w − 4πiτ
in the denominator).
We claim the contribution from the pole of ζ
1− w − 4πiτ
at w = τ = 0 is
−g(0)/2. As w = τ = 0, the d-sum is just X∗. As the pole of ζ(s) is 1/(s− 1), since
s = 1 − 4πiτ
the 1/τ term from the zeta function has coefficient − logX
. We lose the
factor of 1/2πi when we apply the residue theorem, there is a minus sign outside the
integral and another from the direction we integrate (we replace the integral from −ǫ to
ǫ with a semi-circle oriented clockwise; this gives us a minus sign as well as a factor of
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 9
1/2 since we only have half the contour), and everything else evaluated at τ = 0 is g(0).
We now analyze the contribution from the zeros of ζ(s) as we shift w to 2. Thus
w = 3/2 and we sum over τ = γ logX
with ζ(1
+ iγ) = 0. We use Lemma B.2 (with
z = τ − iw logX
) to replace the d-sum with
−2πi(1− log πlogX )τ
− 2πiτ
2 log π
logX +O(logX). (2.11)
The contribution from the O(logX) term is dwarfed by the main term (which is of size
X1/4+ǫ). From (3) of Lemma 2.4 we have
≪ X3σ/4(τ 2 + 1)−B (2.12)
for any B > 0. From (4) of Lemma 2.4, we see that the ratio of the Gamma factors
is bounded by a power of |τ | (the reason it is a power is that we may need to shift a
few times so that the conditions are met; none of these factors will every vanish as we
are not evaluating at integral arguments). Finally, the zeta function in the numerator is
bounded by |τ |2. Thus the contribution from the critical zeros of ζ(s) is bounded by
+iγ)=0
X∗ logX
·X1/4 ·
X3σ/4
(γ2 + 1)B
· (|γ logX|+ 1)n. (2.13)
For sufficiently largeB the sum over γ will converge. This term is of sizeO(X−
(1−σ)+ǫ).
This error is O(X−ǫ) whenever σ < 1, and if σ < 1/3 then the error is at most
O(X−1/2+ǫ).
The proof is completed by showing that the integral over w = 2 is negligible. We
use Lemma B.2 (with z = τ − i2 logX
) to show the d-sum is O(X∗X−2+ǫ). Arguing
as above shows the integral is bounded by O(X−2+2σ+ǫ). (Note: some care is required,
as there is a pole when w = 2 coming from the trivial zero of ζ(s) at s = −2. The
contribution from the residue here is negligible; we could also adjust the contour to
include a semi-circle around w = 2 and use the residue theorem.) �
Remark 2.5. We sketch an alternate start of the proof of Lemma 2.1. One difficulty is
that R(g;X) is defined as an integral and there is a pole on the line of integration. We
may write
ζ(s) = (s− 1)−1 +
ζ(s)− (s− 1)−1
. (2.14)
For us s = 1 − 4πiτ
, so the first factor is just − logX
. As g(τ) is an even function, the
main term of the integral of this piece is
e−2πiτ
e−2πiτ
e2πiτ
sin(2πτ)
dτ = −g(0)
, (2.15)
10 STEVEN J. MILLER
where the last equality is a consequence of supp(ĝ) ⊂ (−1, 1). The other terms from
the (s − 1)−1 factor and the terms from the ζ(s) − (s − 1)−1 piece are analyzed in a
similar manner as the terms in the proof of Lemma 2.1.
2.2. Secondary term (of size 1/ logX) of R(g;X).
Lemma 2.6. Let supp(ĝ) ⊂ (−σ, σ); we do not assume σ < 1. Then the 1/ logX term
in the expansion of R(g;X) is
ζ′(2)
− 2γ + 2 log π
ĝ(1). (2.16)
It is important to note that this piece is only present if the support of ĝ exceeds (−1, 1)
(i.e., if σ > 1).
Proof. We sketch the determination of the main and secondary terms of R(g;X). We
may restrict the integrals to |τ | ≤ log1/4X with negligible error; this will allow us to
Taylor expand certain expressions and maintain good control over the errors. As g is
a Schwartz function, for any B > 0 we have g(τ) ≪ (1 + τ 2)−4B . The ratio of the
Gamma factors is of absolute value 1, and AD(−r; r) = ζ(2)/ζ(2− 2r) = O(1). Thus
the contribution from |τ | ≥ log1/4X is bounded by
|τ |≥log1/4 X
(1 + τ 2)−4B ·max
logC τ
dτ ≪ (logX)−B (2.17)
for B sufficiently large.
We use Lemma B.2 to evaluate the d-sum in (2.1) for |τ | ≤ log1/4X; the error term
is negligible and may be absorbed into the O(log−BX) error. We now Taylor expand
the three factors in (2.1). The main contribution comes from the pole of ζ ; the other
pieces contribute at the 1/ logX level.
We first expand the Gamma factors. We have
− πiτ
+ πiτ
) = 1−
log2X
. (2.18)
As AD(−r; r) = ζ(2)/ζ(2− 2r),
= 1 + 2
ζ ′(2)
log2X
. (2.19)
Finally we expand the ζ-piece. We have (see [Da]) that
ζ(1 + iy) =
+ γ +O(y), (2.20)
where γ is Euler’s constant. Thus
1− 4πiτ
= − logX
+ γ +O
. (2.21)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 11
We combine the Taylor expansions for the three pieces (the ratio of the Gamma fac-
tors, the ζ-function and AD), and keep only the first two terms:
− logX
. (2.22)
Finally, we Taylor expand the d-sum, which was evaluated in Lemma B.2. We may
ignore the error term there because it is O(X1/2). The main term is
−2πi(1− log πlogX )τ
1− 2πiτ
= X∗e
−2πi(1− logπlogX )τ
log2X
(2.23)
R(g;X) =
X∗ logX
∫ logX
− log1/4 X
g(τ) ·X∗e
log π
log1/4 X
log2X
− logX
dτ +O
logBX
∫ log1/4 X
− log1/4 X
g(τ) · e−2πi(1−
log π
logX )τ ·
ζ ′(2)
log5/4X
. (2.24)
We may write
−2πi(1− logπlogX )τ = e−2πiτ ·
2πiτ log π
log2X
. (2.25)
The effect of this expansion is to change the 1/ logX term above by adding log π
Because g is a Schwartz function, we may extend the integration to all τ and absorb
the error into our error term. The main term is from (logX)/4πiτ ; it equals −g(0)/2
(see the analysis in §2.1). The secondary term is easily evaluated, as it is just the Fourier
transform of g at 1. Thus
R(g;X) = −g(0)
ζ′(2)
− 2γ + 2 log π
ĝ(1) +O
log5/4X
(2.26)
3. ANALYSIS OF THE TERMS FROM NUMBER THEORY
We now prove Theorem 1.2. The starting point is the Explicit Formula (Theorem
1.5, with each d an even fundamental discriminant). As the log(d/π) and the Γ′/Γ
terms already appear in the expansion from the Ratios Conjecture (Theorem 1.1), we
12 STEVEN J. MILLER
need only study the sums of χd(p)
k. The analysis splits depending on whether or not k
is even. Set
Seven = −
χd(p)
2 log p
pℓ logX
log pℓ
Sodd = −
χd(p) log p
p(2ℓ+1)/2 logX
log p2ℓ+1
. (3.1)
Based on our analysis of the terms from the Ratios Conjecture, the proof of Theorem
1.2 is completed by the following lemma.
Lemma 3.1. Let supp(ĝ) ⊂ (−σ, σ) ⊂ (−1, 1). Then
Seven = −
g(τ)A′D
+O(X−
Sodd = O(X
− 1−σ
2 log6X). (3.2)
If instead we consider the family of characters χ8d for odd, positive square-free d ∈
(0, X) (d a fundamental discriminant), then
Sodd = O(X
−1/2+ǫ +X−(1−
σ)+ǫ). (3.3)
We prove Lemma 3.1 by analyzing Seven in §3.1 (in Lemmas 3.2 and 3.3) and Sodd
in §3.2 (in Lemmas 3.4, 3.5 and 3.6).
3.1. Contribution from k even. The contribution from k even from the Explicit For-
mula is
Seven = −
χd(p)
2 log p
pℓ logX
log pℓ
, (3.4)
where
d≤X 1 = X
∗, the cardinality of our family. Each χd(p)
2 = 1 except when p|d.
We replace χd(p)
2 with 1, and subtract off the contribution from when p|d. We find
Seven = −2
log p
pℓ logX
log pℓ
log p
pℓ logX
log pℓ
= Seven;1 + Seven;2. (3.5)
In the next subsections we prove the following lemmas, which completes the analysis
of the even k terms.
Lemma 3.2. Notation as above,
Seven;1 = −
dτ. (3.6)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 13
Lemma 3.3. Notation as above,
Seven;2 =
g(τ)A′D
+O(X−
+ǫ). (3.7)
3.1.1. Analysis of Seven;1.
Proof of Lemma 3.2. We have
Seven;1 =
. (3.8)
We use Perron’s formula to re-write Seven;1 as a contour integral. For any ǫ > 0 set
ℜ(z)=1+ǫ
(2z − 2) logA
dz; (3.9)
we will later take A = X1/2. We write z = 1 + ǫ + iy and use (A.6) (replacing φ with
g) to write g(x+ iy) in terms of the integral of ĝ(u). We have
y logA
− iǫ logA
e−iy lognidy
ĝ(u)eǫu logA
e−2πi
−y logA
e−iy logndy. (3.10)
We let hǫ(u) = ĝ(u)e
ǫu logA. Note that hǫ is a smooth, compactly supported function
hǫ(w) = hǫ(−w). Thus
−y logA
e−iy logndy
ĥǫ(y)e
−y log n
− log n
log n
eǫ logn
log n
. (3.11)
By taking A = X1/2 we find
Seven;1 =
= −I1. (3.12)
We now re-write I1 by shifting contours; we will not pass any poles as we shift. For
each δ > 0 we consider the contour made up of three pieces: (1 − i∞, 1 − iδ], Cδ,
14 STEVEN J. MILLER
and [1 − iδ, 1 + i∞), where Cδ = {z : z − 1 = δeiθ, θ ∈ [−π/2, π/2]} is the semi-
circle going counter-clockwise from 1− iδ to 1+ iδ. By Cauchy’s residue theorem, we
may shift the contour in I1 from ℜ(z) = 1 + ǫ to the three curves above. Noting that∑
nΛ(n)n
−z = −ζ ′(z)/ζ(z), we find that
[∫ 1−iδ
∫ 1+i∞
(2z − 2) logA
−ζ ′(z)
. (3.13)
The integral over Cδ is easily evaluated. As ζ(s) has a pole at s = 1, it is just half the
residue of g
(2z−2) logA
(the minus sign in front of ζ ′(z)/ζ(z) cancels the minus sign
from the pole). Thus the Cδ piece is g(0)/2. We now take the limit as δ → 0:
− lim
[∫ −δ
y logA
ζ ′(1 + iy)
ζ(1 + iy)
. (3.14)
As g is an even Schwartz function, the limit of the integral above is well-defined (for
large y this follows from the decay of g, while for small y it follows from the fact that
ζ ′(1+ iy)/ζ(1+ iy) has a simple pole at y = 0 and g is even). We again takeA = X1/2,
and change variables to τ = y logA
= y logX
. Thus
dτ, (3.15)
which completes the proof of Lemma 3.2. �
3.1.2. Analysis of Seven;2.
Proof of Lemma 3.3. Recall
Seven;2 =
log p
pℓ logX
log pℓ
. (3.16)
We may restrict the prime sum to p ≤ X1/2 at a cost of O(log logX/X). We sketch
the proof of this claim. Since ĝ has finite support, p ≤ Xσ and thus the p-sum is finite.
Since d ≤ X and p ≥ X1/2, there are at most 2 primes which divide a given d. Thus
p=X1/2
log p
pℓ logX
log pℓ
p=X1/2
p>X1/2
≪ log logX
.(3.17)
In Lemma B.1 we show that
X +O(X1/2) (3.18)
and that for p ≤ X1/2 we have
+O(X1/2). (3.19)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 15
Using these facts we may complete the analysis of Seven;2:
Seven;2 =
p≤X1/2
log p
pℓ logX
log pℓ
log logX
p≤X1/2
log p
pℓ logX
log pℓ
d≤X, p|d
log logX
p≤X1/2
log p
pℓ logX
log pℓ
p≤X1/2
log logX
p≤X1/2
log p
pℓ logX
log pℓ
+O(X−
+ǫ). (3.20)
We re-write ĝ(2 log pℓ/ logX) by expanding the Fourier transform.
Seven;2 = 2
p≤X1/2
log p
(p+ 1)pℓ logX
g(τ)e−2πiτ ·2 log p
ℓ/ logXdτ +O(X−
p≤X1/2
log p
(p+ 1) logX
p−ℓ · p−2πiτ ·2ℓ/ logXdτ +O(X−
p≤X1/2
log p
(p+ 1) logX
g(τ)p
−(1+2· 2πiτ
1− p−(1+2·
dτ +O(X−
(3.21)
We may extend the p-sum to be over all primes at a cost of O(X−1/2+ǫ); this is because
the summands are O(log p/p2) and g is Schwartz. Recalling the definition of A′D(r; r)
in (1.4), we see that the resulting p-sum is just A′D(2πiτ/ logX ; 2πiτ/ logX); this
completes the proof of Lemma 3.3. �
3.2. Contribution from k odd. As k is odd, χd(p)
k = χd(p). Thus we must analyze
the sum
Sodd = −
χd(p) log p
p(2ℓ+1)/2 logX
log p2ℓ+1
. (3.22)
If supp(ĝ) ⊂ (−1, 1), Rubinstein [Rub1] showed (by applying Jutila’s bound [Ju1,
Ju2, Ju3] for quadratic character sums) that if our family is all discriminants then
Sodd = O(X
−ǫ/2). In his dissertation Gao [Gao] extended these results to show that
the odd terms do not contribute to the main term provided that supp(ĝ) ⊂ (−2, 2). His
analysis proceeds by using Poisson summation to convert long character sums to shorter
ones. We shall analyze Sodd using both methods: Jutila’s bound gives a self-contained
presentation, but a much weaker result; the Poisson summation approach gives a better
16 STEVEN J. MILLER
bound but requires a careful book-keeping of many of Gao’s lemmas (as well as an
improvement of one of his estimates).
3.2.1. Analyzing Sodd with Jutila’s bound.
Lemma 3.4. Let supp(ĝ) ⊂ (−σ, σ). Then Sodd = O(X−
2 log6X).
Proof. Jutila’s bound (see (3.4) of [Ju3]) is
1<n≤N
n non−square
∣∣∣∣∣∣
0<d≤X
d fund. disc.
χd(n)
∣∣∣∣∣∣
≪ NX log10N (3.23)
(note the d-sum is over even fundamental discriminants at most X). As 2ℓ + 1 is odd,
p2ℓ+1 is never a square. Thus Jutila’s bound gives
p(2ℓ+1)/2≤Xσ
∣∣∣∣∣
χd(p)
∣∣∣∣∣
2 log5X. (3.24)
Recall
Sodd = −
log p
p(2ℓ+1)/2 logX
log p2ℓ+1
χd(p). (3.25)
We apply Cauchy-Schwartz, and find
|Sodd| ≤
p2ℓ+1≤Xσ
log p
p(2ℓ+1)/2 logX
log p2ℓ+1
)∣∣∣∣
p2ℓ+1≤Xσ
∣∣∣∣∣
χd(p)
∣∣∣∣∣
2 log5X
2 log6X ; (3.26)
thus there is a power savings if σ < 1. �
3.2.2. Analyzing Sodd with Poisson Summation.
Gao analyzes the contribution from Sodd by applying Poisson summation to the char-
acter sums. The computations are simplified if the character χ2(n) =
is not present.
He therefore studies the family of odd, positive square-free d (where d is a fundamental
discriminant). His family is
{8d : X < d ≤ 2X, d an odd square− free fundamental discriminant}; (3.27)
we discuss in Lemma 3.6 how to easily modify the arguments to handle the related
family with 0 < d ≤ X . The calculation of the terms from the Ratios Conjecture
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 17
proceeds similarly (the only modification is to X∗, which also leads to a trivial mod-
ification of Lemma B.2 which does not change any terms larger than O(X−1/2+ǫ) if
supp(ĝ) ⊂ (−1/3, 1/3)), as does the contribution from χ(p)k with k even. We are left
with bounding the contribution from Sodd. The following lemma shows that we can
improve on the estimate obtained by applying Jutila’s bound.
Lemma 3.5. Let supp(ĝ) ⊂ (−σ, σ) ⊂ (−1, 1). Then for the family given in (3.27),
Sodd = O(X
+ǫ +X−(1−
σ)+ǫ). In particular, if σ < 1/3 then Sodd = O(X
−1/2+ǫ).
Proof. Gao is only concerned with main terms for the n-level density (for any n) for
all sums. As we only care about Sodd for the 1-level density, many of his terms are not
present. We highlight the arguments. We concentrate on the ℓ = 0 term in (3.22) (the
other ℓ ≪ logX terms are handled similarly, and the finite support of ĝ implies that
Sodd(ℓ) = 0 for ℓ≫ logX):
Sodd = −
χd(p) log p
p(2ℓ+1)/2 logX
log p2ℓ+1
Sodd(ℓ). (3.28)
Let Y = Xσ, where supp(ĝ) ⊂ (−σ, σ). Our sum Sodd(0) is S(X, Y, ĝ) in Gao’s
thesis:
S(X, Y, ĝ) =
X<d<2X
(2,d)=1
µ(d)2
log p
χ8d(p)ĝ
log p
. (3.29)
Let Φ be a smooth function supported on (1, 2) such that Φ(t) = 1 for t ∈ (1 +
U−1, 2 − U−1) and Φ(j)(t) ≪j U j for all j ≥ 0. We show that S(X, Y, ĝ) is well
approximated by the smoothed sum S(X, Y, ĝ,Φ), where
S(X, Y, ĝ,Φ) =
(d,2)=1
µ(d)2
log p
χ8d(p)ĝ
log p
. (3.30)
To see this, note the difference between the two involves summing d ∈ (X,X +X/U)
and d ∈ (2X−X/U, 2X). We trivially bound the prime sum for each fixed d by log7X
(see Proposition III.1 of [Gao]). As there are O(X/U) choices of d and Φ(d/X) ≪ 1,
we have
S(X, Y, ĝ)− S(X, Y, ĝ,Φ) ≪ X log
. (3.31)
We will take U =
X . Thus upon dividing by X∗ ≫ X (the cardinality of the family),
this difference is O(X−1/2+ǫ). The proof is completed by bounding S(X, Y, ĝ,Φ).
To analyze S(X, Y, ĝ,Φ), we write it as SM(X, Y, ĝ,Φ) + SR(X, Y, ĝ,Φ), with
SM(X, Y, ĝ,Φ) =
(d,2)=1
MZ(d)
log p
χ8d(p)ĝ
log p
SR(X, Y, ĝ,Φ) =
(d,2)=1
RZ(d)
log p
χ8d(p)ĝ
log p
, (3.32)
18 STEVEN J. MILLER
where
µ(d)2 = MZ(d) +RZ(d)
MZ(d) =
µ(ℓ), RZ(d) =
µ(ℓ); (3.33)
here Z is a parameter to be chosen later, and SM(X, Y, ĝ,Φ) will be the main term (for
a general n-level density sum) and SR(X, Y, ĝ,Φ) the error term. In our situation, both
will be small.
In Lemma III.2 of [Gao], Gao proves that SR(X, Y, ĝ,Φ) ≪ (X log3X)/Z. We
haven’t divided any of our sums by the cardinality of the family (which is of size X).
Thus for this term to yield contributions of size X−1/2+ǫ, we need Z ≥ X1/2.
We now analyze SM(X, Y, ĝ,Φ). Applying Poisson summation we convert long char-
acter sums to short ones. We need certain Gauss-type sums:
1 + i
Gm(k) =
a mod k
e2πiam/k . (3.34)
For a Schwartz function F let
F̃ (ξ) =
1 + i
F̂ (ξ) +
F̂ (−ξ). (3.35)
Using Lemma 2.6 of [So], we have (see page 32 of [Gao])
SM(X, Y, ĝ,Φ) =
2<p<Y
log p
log p
(α,2p)=1
(−1)mGm(p)Φ̃
. (3.36)
We follow the arguments in Chapter 3 of [Gao]. The m = 0 term is analyzed in §3.3
for the general n-level density calculations. It is zero if n is odd, and we do not need
to worry about this error term (thus we do not see the error terms of size X logn−1X
or (X lognX)/Z which appear in his later estimates). In §3.4 he analyzes the contri-
butions from the non-square m in (3.36). In his notation, we have k = 1, k2 = 0,
k1 = 0, α1 = 1 and α0 = 0, and these terms’ contribution is ≪ (U2Z
Y log7X)/X
(remember we haven’t divided by the cardinality of the family, which is of order X).
This is too large for our purposes (we have seen that we must take U = Z =
and Y = Xσ). We perform a more careful analysis of these terms in Appendix C, and
bound these terms’ contribution by
Y log7X
UZY 3/2 log4X
Z3U2Y 7/2 log4X
X4018−2ǫ
. (3.37)
Lastly, we must analyze the contribution from m a square in (3.36). From Lemma
III.3 of [Gao] we have that Gm(p) = 0 if p|m. If p |r m and m is a square, then
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 19
Gm(p) =
p. Arguing as in [Gao], we are left with
(p,2)=1
log p
log p
(α,2p)=1
(−1)mΦ̃
(−1) emΦ̃
p2m̃2X
(3.38)
If we assume supp(ĝ) ⊂ (−1, 1), then arguing as on page 41 of [Gao] we find the
m-sum above is ≪ α
p/X , which leads to a contribution ≪
Y/X logX logZ; the
m̃-sum is ≪ α/
pX and is thus dominated by the contribution from the m-sum.
Collecting all our bounds, we see a careful book-keeping leads to smaller errors than
in §3.6 of [Gao] (this is because (1) many of the error terms only arise from n-level
density sums with n even, where there are main terms and (2) we did a more careful
analysis of some of the errors). We find that
S(X, Y, ĝ,Φ) ≪ X log
Y log7X
UZY 3/2 log4X
Y logX logZ√
(3.39)
We divide this by X∗ ≫ X (the cardinality of the family). By choosing Z = X1/2,
Y = Xσ with σ < 1, and U =
X (remember we need such a large U to handle
the error from smoothing the d-sum, i.e., showing |S(X, Y, ĝ) − S(X, Y, ĝ,Φ)|/X ≪
X−1/2+ǫ), we find
S(X, Y, ĝ,Φ)/X ≪ X−1/2+ǫ +X−(1−
σ)+ǫ, (3.40)
which yields
Sodd ≪ X−1/2+ǫ +X−(1−
σ)+ǫ. (3.41)
Note that if σ < 1/3 then Sodd ≪ X−1/2+ǫ. �
Lemma 3.6. Let supp(ĝ) ⊂ (−σ, σ) ⊂ (−1, 1). Then for the family
{8d : 0 < d ≤ X, d an odd square− free fundamental discriminant} (3.42)
we have Sodd = O(X
−1/2+ǫ + X−(1−
σ)+ǫ). In particular, if σ < 1/3 then Sodd =
O(X−1/2+ǫ).
Proof. As the calculation is standard, we merely sketch the argument. We write
(0, X ] =
log2 X⋃
. (3.43)
Let Xi = X/2
i. For each i, in Lemma 3.5 we replace most of the X’s with Xi, U
with U/
2i, Z with Z/
2i; the X’s we don’t replace are the cardinality of the family
(which we divide by in the end) and the logX which occurs when we evaluate the test
function ĝ at log p/ logX . We do not change Y , which controls the bounds for the
prime sum. As we do not have any main terms, there is no loss in scaling the prime
sums by logX instead of logXi. We do not use much about the test function ĝ in our
estimates. All we use is that the prime sums are restricted to p < Y , and therefore we
will still have bounds of Y (to various powers) for our sums.
20 STEVEN J. MILLER
We now finish the book-keeping. Expressions such as UZ/X in (3.39) are stillO(1),
and expressions such as X/U and X/Z are now smaller. When we divide by the cardi-
nality of the family we still have terms such as Y 3/2/X , and thus the support require-
ments are unchanged (i.e., Sodd ≪ X−1/2+ǫ +X−(1−
σ)+ǫ). �
APPENDIX A. THE EXPLICIT FORMULA
We quickly review some needed facts about Dirichlet characters; see [Da] for details.
Let χd be a primitive quadratic Dirichlet character of modulus |d|. Let c(d, χd) be the
Gauss sum
c(d, χd) =
χd(k)e
2πik/d, (A.1)
which is of modulus
d. Let
L(s, χd) =
1− χd(p)p−s
(A.2)
be the L-function attached to χd; the completed L-function is
Λ(s, χd) = π
−(s+a)/2Γ
d−(s+a)/2L(s, χd) = (−1)a
c(d, χd)√
Λ(1− s, χd),
(A.3)
where
a = a(χd) =
0 if χd(−1) = 1
1 if χd(−1) = −1.
(A.4)
We write the zeros of Λ(s, χd) as
+ iγ; if we assume GRH then γ ∈ R. Let φ be an
even Schwartz function and φ̂ be its Fourier transform (φ̂(ξ) =
φ(x)e−2πixξdx); we
often assume supp(φ̂) ⊂ (−σ, σ) for some σ <∞. We set
H(s) = φ
. (A.5)
While H(s) is initially define only when ℜ(s) = 1/2, because of the compact support
of φ̂ we may extend it to all of C:
φ(x) =
φ̂(ξ)e2πixξdξ
φ(x+ iy) =
φ̂(ξ)e2πi(x+iy)ξdξ
H(x+ iy) =
φ̂(ξ)e2π(x−
· e2πiyξdξ. (A.6)
Note that H(x+ iy) is rapidly decreasing in y (for a fixed x it is the Fourier transform
of a nice function, and thus the claim follows from the Riemann-Lebesgue lemma). We
now derive the Explicit Formula for quadratic characters; note the functional equation
will always be even. We follow the argument given in [RS].
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 21
Proof of the Explicit Formula, Theorem 1.5. We have
Λ(s, χ) = π−(s+a)/2Γ
d(s+a)/2L(s, χd) = Λ(1− s, χd)
Λ′(s, χd)
Λ(s, χd)
log π
log d
L′(s, χd)
L(s, χd)
L′(s, χd)
L(s, χd)
χd(p) log p
1− χd(p)p−s
χd(p)
k log p
. (A.7)
We will not approximate any terms; we are keeping all lower order terms to facilitate
comparison with the L-functions Ratios Conjecture. We set
ℜ(s)=3/2
Λ′(s, χd)
Λ(s, χd)
H(s)ds. (A.8)
We shift the contour to ℜ(s) = −1/2. We pick up contributions from the zeros and
poles of Λ(s, χd). As χd is not the principal character, there is no pole from L(s, χd).
There is also no need to worry about a zero or pole from the Gamma factor Γ
L(1, χd) 6= 0. Thus the only contribution is from the zeros of Λ(s, χd); the residue at a
zero 1
+ iγ is φ(γ). Therefore
φ(γ) +
ℜ(s)=−1/2
Λ′(s, χd)
Λ(s, χd)
H(s)ds. (A.9)
As Λ(1− s, χd) = Λ(s, χd), −Λ′(1− s, χd) = Λ(s, χd) and
φ(γ)−
ℜ(s)=−1/2
Λ′(1− s, χd)
Λ(1− s, χd)
H(s)ds. (A.10)
We change variables (replacing s with 1− s), and then use the functional equation:
φ(γ)−
ℜ(s)=3/2
Λ′(s, χd)
Λ(s, χd)
H(1− s)ds. (A.11)
Recalling the definition of I gives
φ(γ) =
ℜ(s)=3/2
Λ′(s, χd)
Λ(s, χd)
[H(s) +H(1− s)] ds. (A.12)
We expandΛ′(s, χd)/Λ(s, χd) and shift the contours of all terms exceptL
′(s, χd)/L(s, χd)
to ℜ(s) = 1/2 (this is permissible as we do not pass through any zeros or poles of the
other terms); note that if s = 1
+ iy then H(s) = H(1 − s) = φ(y) (φ is even).
22 STEVEN J. MILLER
Expanding the logarithmic derivative of Λ(s, χd) gives
φ(γ) =
φ(y)dy
ℜ(s)=3/2
L′(s, χd)
L(s, χd)
· [H(s) +H(1− s)] ds
φ(y)dy
ℜ(s)=3/2
L′(s, χd)
L(s, χd)
· [H(s) +H(1− s)] ds, (A.13)
where the last line follows from the fact that φ is even.
We use (A.7) to expand L′/L. In the arguments below we shift the contour to ℜs =
1/2; this is permissible because of the compact support of φ̂ (see (A.6)):
ℜ(s)=3/2
(s+ iy) · [H (s) +H (1− s)] dy
χd(p)
k log p
ℜ(s)=3/2
[H (s) +H (1− s)] e−ks log pdy
χd(p)
k log p
φ(y)e−2πiy·
log pk
2π dy
= − 2
χd(p)
k log p
log pk
. (A.14)
We therefore find that
φ(γ) =
φ(y)dy
χd(p)
k log p
log pk
. (A.15)
We replace φ(x) with g(x) = φ
x · logX
. A standard computation gives ĝ(ξ) =
ξ · 2π
. Summing over d ∈ F(X) completes the proof. �
APPENDIX B. SUMS OVER FUNDAMENTAL DISCRIMINANTS
Lemma B.1. Let d denote an even fundamental discriminant at most X , and set X∗ =∑
d≤X 1. Then
X +O(X1/2) (B.1)
and for p ≤ X1/2 we have
+O(X1/2). (B.2)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 23
Proof. We first prove the claim forX∗, and then indicate how to modify the proof when
p|d. We could show this by recognizing certain products as ratios of zeta functions or
by using a Tauberian theorem; instead we shall give a straightforward proof suggested
to us by Tim Browning (see also [OS1]).
We first assume that d ≡ 1 mod 4, so we are considering even fundamental discrim-
inants {d ≤ X : d ≡ 1 mod 4, µ(d)2 = 1}; it is trivial to modify the arguments below
for d such that d/4 ≡ 2 or 3 modulo 4 and µ(d/4)2 = 1. Let χ4(n) be the non-trivial
character modulo 4: χ4(2m) = 0 and
χ4(n) =
1 if n ≡ 1 mod 4
0 if n ≡ 3 mod 4.
(B.3)
We have
S(X) =
µ(d)2=1, d≡1 mod 4
µ(d)2 ·
1 + χ4(d)
µ(d)2 +
µ(d)2χ4(d) = S1(X) + S2(X). (B.4)
By Möbius inversion
µ(m) =
1 if d is square-free
0 otherwise.
(B.5)
S1(X) =
m≤X1/2
µ(m) ·
d ≤ X/m2
m≤X1/2
+O(1)
+O(X1/2)
·X +O(X1/2)
X +O(X1/2) (B.6)
24 STEVEN J. MILLER
(because we are missing the factor corresponding to 2 in 1/ζ(2) above). Arguing in a
similar manner shows S2(X) = O(X
1/2); this is due to the presence of χ4, giving us
S2(X) =
m≤X1/2
2)µ(m)
d≤X/m2
χ4(d) ≪ X1/2 (B.7)
(because we are summing χ4 at consecutive integers, and thus this sum is at most 1). A
similar analysis shows that the number of even fundamental discriminants d ≤ X with
d/4 ≡ 2 or 3 modulo 4 is X/π2 +O(X1/2). Thus
d an even fund. disc.
1 = X∗ =
X +O(X1/2). (B.8)
We may trivially modify the above calculations to determine the number of even
fundamental discriminants d ≤ X with p|d for a fixed prime p. We first assume p ≡
1 mod 4. In (B.4) we replace µ(d)2 with µ(pd)2, d ≤ X with d ≤ X/p, 2 |r d and
(2p, d) = 1. These imply that d ≤ X , p|d and p2 does not divide d. As d and p are
relatively prime, µ(pd) = µ(p)µ(d) and the main term becomes
S1;p(X) =
d≤X/p
(2p,d)=1
m≤(X/p)1/2
(2p,m)=1
µ(m) ·
d ≤ (X/p)/m2
(2p,d)=1
m≤(X/p)1/2
(2p,m)=1
· p− 1
+ O(1)
(p− 1)X
(2p,m)=1
+O(X1/2)
· (p− 1)X
+O(X1/2)
(p+ 1)π2
+O(X1/2), (B.9)
and the cardinality of this piece is reduced by (p + 1)−1 (note above we used #{n ≤
Y : (2p, n) = 1} = p−1
Y + O(1)). A similar analysis holds for S2;p(X), as well as the
even fundamental discriminants d with d/4 ≡ 2 or 3 modulo 4).
We need to trivially modify the above arguments if p ≡ 3 mod 4. If for instance
we require d ≡ 1 mod 4 then instead of replacing µ(d)2 with µ(d)2(1 + χ4(d))/2 we
replace it with µ(pd)2(1− χ4(d))/2, and the rest of the proof proceeds similarly.
It is a completely different story if p = 2. Note if d ≡ 1 mod 4 then 2 never divides
d, while if d/4 ≡ 2 or 3 modulo 4 then 2 always divides d. There are 3X/π2+ o(X1/2)
even fundamental discriminants at most X , and X/π2 + O(x1/2) of these are divisible
by 2. Thus, if our family is all even fundamental discriminants, we do get the factor of
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 25
1/(p+ 1) for p = 2, as one-third (which is 1/(2 + 1) of the fundamental discriminants
in this family are divisible by 2. �
In our analysis of the terms from the L-functions Ratios Conjecture, we shall need a
partial summation consequence of Lemma B.1.
Lemma B.2. Let d denote an even fundamental discriminant at most X and X∗ =∑
d≤X 1 and let z = τ − iw
with w ≥ 1/2. Then
−2πiz
log(d/π)
logX = X∗e
−2πi(1− log πlogX )z
1− 2πiz
+O(logX). (B.10)
Proof. By Lemma B.1 we have
+O(u1/2). (B.11)
Therefore by partial summation we have
−2πiz
log(d/π)
log π
d−2πiz/ logX
log π
3X +O(X1/2)
− 2πiz
logX −
∫ X (3u
+O(u1/2)
−2πiz
(B.12)
As we are assuming w ≥ 1/2, the first error term is of size O(X1/2X−w) = O(1).
The second error term (from the integral) is O(logX) for such w. This is because the
integration begins at 1 and the integrand is bounded by u−
−w. Thus
−2πiz
log(d/π)
log π
e−2πiz +
· 2πiz
u−2πiz/ logXdu
+O(logX)
log π
e−2πiz +
· 2πiz
X1−2πiz/ logX
1− 2πiz/ logX
+O(logX)
= X∗e
logX e−2πiz
+O(logX)
= X∗e
−2πi(1− log πlogX )z
1− 2πiz
+O(logX). (B.13)
26 STEVEN J. MILLER
APPENDIX C. IMPROVED BOUND FOR NON-SQUARE m TERMS IN SM(X, Y, ĝ,Φ)
Gao [Gao] proves that the non-squarem-terms contribute ≪ (U2Z
Y log7X)/X to
SM(X, Y, ĝ,Φ). As this bound is just a little too large for our applications, we perform
a more careful analysis below. Denoting the sum of interest by R,
(α,2)=1
(2α,p)=1
log p
log p
m6=0,✷
(−1)mΦ̃
, (C.14)
Gao shows that
log3X
(R1 +R2 +R3), (C.15)
R1, R2 ≪
Y log4X
Y log7X
. (C.16)
The bounds for R1 and R2 suffice for our purpose, leading to contributions bounded by
Y log4X)/X; however, the R3 bound gives too crude a bound – we need to save
a power of U .
We have (see page 36 of [Gao], with k = 1, k2 = 0, k1 = 0, α1 = 1 and α0 = 0) that
α2V 5/2
log p
(log3m)mΦ̃′
2α2pV
dV. (C.17)
We have (see (3.10) of [Gao]) that
Φ̃′(ξ) ≪ U j−1|ξ|−j for any integer j ≥ 1. (C.18)
Letting M = X2008, we break the m-sum in R3 into m ≤M and m > M . For m ≤M
we use (C.18) with j = 2 while for m > M we use (C.18) with j = 3. (Gao uses j = 3
for all m. While we save a bit for small m by using j = 2, we cannot use this for all m
as the resulting m sum does not converge.)
Thus the small m contribute
α2V 5/2
log p
(log3m)m
U22α4p2V 2
log p
log3m
UY 3/2α2 log4X
(C.19)
A SYMPLECTIC TEST OF THE L-FUNCTIONS RATIOS CONJECTURE 27
(since M = X2008 the m-sum is O(log4X)). The large m contribute
α2V 5/2
log p
(log3m)m
U223α6p3V 3
p log p
log3m
V 1/2dV
α4U2Y 3/2Y 2 logX
X2M2−ǫ
. (C.20)
For our choices of U , Y and Z, the contribution from the large m will be negligible
(due to the M2−ǫ = X4016−2ǫ in the denominator). Thus for these choices
log3X
(R1 +R2 +R3)
Y log7X
UZY 3/2 log4X
Z3U2Y 7/2 log4X
X4018−2ǫ
. (C.21)
The last term is far smaller than the first two. In the first term we save a power of U from
Gao’s bound, and in the second we replace U with Y . As Y = Xσ, for σ sufficiently
small there is a significant savings.
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E-mail address: sjmiller@math.brown.edu
DEPARTMENT OF MATHEMATICS, BROWN UNIVERSITY, PROVIDENCE, RI 02912
	1. Introduction
	2. Analysis of the terms from the Ratios Conjecture.
	2.1. Analysis of R(g;X)
	2.2. Secondary term (of size 1/logX) of R(g;X)
	3. Analysis of the terms from Number Theory
	3.1. Contribution from k even
	3.2. Contribution from k odd
	Appendix A. The Explicit Formula
	Appendix B. Sums over fundamental discriminants
	Appendix C. Improved bound for non-square m terms in SM(X,Y,g"0362g,)
	References
