ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
RIZWANUR KHAN
Abstract. Fix g a self-dual Hecke-Maass form for SL3 (Z). Let f be a holomorphic newform of
prime level q and fixed weight. Conditional on a lower bound for a short sum of squares of Fourier
coefficients of f , we prove a subconvexity bound in the q aspect for L(s, g × f ) at the central point.
1. Introduction
An outstanding problem in analytic number theory is to understand the size of an L-function
at its central point. For an L-function L(s) from the Selberg class with analytic conductor C and
functional equation relating values at s and 1 − s, the Lindelöf hypothesis L( 21 ) C is expected for
any > 0, where the implied constant depends on and the degree of L(s). Given an average version
of the Ramanujan conjecture (which in many cases is available by the works of Iwaniec [8], Molteni
[18], and Xiannan Li [15]), it only requires the functional equation to prove the so called convexity
1
bound L( 21 ) C 4 + . This (or a refinement of this due to Heath-Brown [7]) is considered to be the
trivial bound and was the best known in general until Soundararajan [20] recently proved, on a weak
1
form of the Ramanujan conjecture, that L( 12 ) C 4 (log C)−1+ . The subconvexity problem is to
1
save a power of C, that is to prove that L( 21 ) C 4 −δ for some δ > 0. For automorphic L-functions
of degree one or two, the problem is completely solved. This involves the work of many authors,
but the contribution of Friedlander and Iwaniec is particularly noteworthy for their invention of
the amplifier method [3, 8] to attack the subconvexity problem. For higher degree L-functions,
a subconvexity bound is known only in a limited number of cases and remains a challenging and
important goal.
In this paper we study certain degree six L-functions, the Rankin-Selberg GL(3) × GL(2) Lfunctions. In a recent breakthrough, Xiaoqing Li [16] proved a subconvexity bound for the L-function
of a fixed self-dual Hecke-Maass form for SL3 (Z) twisted by a Hecke-Maass form for SL2 (Z), or by
a holomorphic Hecke cusp form for SL2 (Z), in the eigenvalue aspect, or respectively in the weight
aspect, of the GL(2) form. Blomer [2] considered this problem in the level aspect and proved
subconvexity for GL(3) × GL(2) L-functions where the twist is by special Hecke-Maass forms of
prime square level. For prime level however, subconvexity is still unknown. This appears to be a
very deep problem, perhaps out of reach of current technology in full generality, but we are able to
make partial progress.
Let Hk? (q) denote the set of holomorphic cusp forms of weight k which are newforms of level q
with trivial nebentypus in the sense of Atkin-Lehner Theory [1]. Fix g a self-dual Hecke-Maass form
for SL3 (Z) which is unramified at infinity. Let L(s, g × f ) denote the Rankin-Selberg convolution
of g with f ∈ Hk? (q). Kim and Shahidi [13] have shown that this is in fact an GL(6) automorphic
L-function. We normalize to have the central point at s = 12 . The analytic conductor in the q aspect
3
equals q 3 , so that the convexity bound is q 4 + .
In the works of Xiaoqing Li and Blomer, a study of the first moment of the L-function at s = 21 is
enough to yield subconvexity. For example, in the weight aspect, the analytic conductor of L(s, g×f )
3
equals k 6 so that the convexity bound is k 2 + . We know by a result of Lapid [14] that L(g×f, 12 ) ≥ 0.
2010 Mathematics Subject Classification: 11M99
The author was supported by a grant from the European Research Council (grant agreement number 258713).
1
2
RIZWANUR KHAN
Hence if we had the expected (by the Lindelöf hypothesis) upper bound
X
L( 12 , g × f ) q k 1+ ,
(1.1)
f ∈Hk? (q)
then dropping all but one term of this sum would immediately yield subconvexity. Although Xiaoqing
Li does not establish (1.1), she studies a similar first moment with an extra averaging over k. The
situation in the level aspect is more challenging. From the expected estimate
X
(1.2)
L( 12 , g × f ) k q 1+ ,
f ∈Hk? (q)
dropping all but one term does not yield any useful bound in the q aspect. One could try to estimate
the second moment, but this seems to be difficult. Thus a new ingredient is needed. We make use
of an amplifier to prove
Theorem 1.1. Fix k > 106 an even number. Let q be a prime number. Suppose that for some
f0 ∈ Hk? (q) we have
X
(1.3)
af0 (n)2 L1−
n<L
for all L > q
(1.4)
1/4+1/2001
, where af0 (n) is the n-th Fourier coefficient of f0 as defined in (1.5). Then
L( 12 , g × f0 ) q 3/4−1/2001 .
The hypothesis (1.3), that Fourier coefficients are not very small on average, is natural and appears
to be unrelated to our subconvexity problem. One can show that it holds for almost all cusp forms
and it is certainly believed to be true individually (in fact it would follow from expected bounds
for the symmetric square L-function associated to f0 ). Historically, the first proof of subconvexity
for Hecke-Maass cusp forms in the eigenvalue aspect, due to Iwaniec [8], was conditional on an
assumption similar to (1.3). Iwaniec was later able to get around the assumption by using an
ingenious amplifier based on the observation that the p-th and p2 -th Hecke eigenvalues cannot be
simultaneously small for a prime p. Unfortunately we cannot use his idea here because in contrast
to (1.2), Iwaniec had an average of L-function values which already yielded the convexity bound and
only needed an amplifier of arbitrarily small length to do better.
The exponent in our subconvexity bound and the lower bound for k are not optimal. In an
effort to present the method as clearly and transparently as possible, we have only concentrated on
obtaining a convexity breaking exponent.
1.2. L-functions. Every f ∈ Hk? (q) has a Fourier expansion of the type
(1.5)
f (z) =
∞
X
af (n)n
k−1
2
e(nz)
n=1
for =z > 0, where e(z) = e2πiz , af (n) ∈ R and af (1) = 1. The coefficients af (n) satisfy the
multiplicative relation
nm X
(1.6)
af (n)af (m) =
af
d2
d|(n,m)
(d,q)=1
and Deligne’s bound af (n) ≤ d(n) n . Here and throughout the paper, denotes an arbitrarily
small positive constant, but not necessarily the same one from one occurrence to the next, and
any implied constant may depend on . Also, q will always be a prime number. The L-function
associated to f is the entire function given for <(s) > 1 by the absolutely convergent series
∞
X
af (n)
(1.7)
L(s, f ) =
.
ns
n=1
ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
3
This satisfies the functional equation
(1.8)
s
q 2 π −s Γ
s +
k−1
2
2
s + k+1 1 − s +
1−s
2
Γ
L(s, f ) = f q 2 π 1−s Γ
2
2
k−1
2
1 − s +
Γ
2
k+1
2
L(1−s, f ),
where
1
f = −ik af (q)q 2 = ±1.
(1.9)
The facts above can be found in [9].
We fix a self-dual Hecke-Maass form g of type (ν, ν) for SL3 (Z). We refer to [4], especially
Chapter 6, and follow its notation. We write A(n, m) = A(m, n) for the Fourier coefficients of g in
the Fourier expansion (6.2.1) of [4], normalized so that A(1, 1) = 1. The L-function associated to g
is the entire function given for <(s) > 1 by
(1.10)
L(s, g) =
∞
X
A(n, 1)
.
ns
n=1
This satisfies the functional equation
s + 3ν − 1 s s + 1 − 3ν (1.11) π −3s/2 Γ
Γ
Γ
L(s, g)
2
2
2
1 − s + 3ν − 1 1 − s 1 − s + 1 − 3ν = π 3(1−s)/2 Γ
Γ
Γ
L(1 − s, g).
2
2
2
The coefficients A(n, 1) are real. L(s, g) is actually the symmetric-square L-function of a HeckeMaass form for SL2 (Z), by the work of Soudry [19]. This implies, by the work of Kim and Sarnak
[12], that
|A(n, 1)| n7/32+
(1.12)
and, by the work of Selberg, that
<(3ν − 1) = 0.
(1.13)
We have the Hecke relation
(1.14)
A(n, m) =
X
µ(d)A
d|(n,m)
n
m
, 1 A 1,
,
d
d
and if (n1 m1 , n2 m2 ) = 1, we have
(1.15)
A(n1 n2 , m1 m2 ) = A(n1 , m1 )A(n2 , m2 ).
By (1.12) and (1.14) we have
A(n, m) (nm)7/32+ .
(1.16)
By (1.12) and Rankin-Selberg theory we have (cf. [2] for a proof):
X
(1.17)
|A(na, b)|2 x(ab)7/16+ .
n≤x
This together with (1.14) and the Cauchy-Schwarz inequality yields
X
(1.18)
|A(na, mb)| (xy)1+ (ab)7/32+ .
n<x
m<y
The Rankin-Selberg L-function L(s, g × f ) is defined as the entire function given for <(s) > 1 by
X A(r, n)af (n)
(1.19)
.
L(s, g × f ) =
(r2 n)s
n,r≥1
4
RIZWANUR KHAN
It satisfies the functional equation
(1.20)
q
3s
2
G(s)L(s, g × f ) = g×f q
3(1−s)
2
G(1 − s)L(1 − s, g × f ),
where g×f = (f )3 = f and
s + k+1 + 1 − 3ν + 3ν − 1 s + k+1
2
2
Γ
Γ
2
2
2
s + k−1 + 3ν − 1 s + k−1 s +
2
2
×Γ
Γ
Γ
2
2
To study L(1/2, g × f ), we first express it as a weighted Dirichlet series.
s +
(1.21) G(s) = π −3s Γ
k+1
2
k−1
2
+ 1 − 3ν .
2
Lemma 1.3. Approximate functional equation
Let
Z
G( 1 + s) ds
1
V (x) =
(1.22)
x−s 2 1
2πi (σ)
G( 2 ) s
for x, σ > 0. We have
(1.23) L( 12 , f × g) =
X af (n)A(r, n) X af (n)A(r, n) r2 n
r2 n
√
√
V 3/2+1/300 + g×f
V 3/2−1/300 .
r n
r n
q
q
n,r≥1
n,r≥1
For any A > 0 and integer B ≥ 0 we have that
V (B) (x) B x−B (1 + x)−A ,
(1.24)
so that the first sum in (1.23) is essentially supported on r2 n < q 3/2+1/300+ and the second sum is
essentially supported on r2 n < q 3/2−1/300+ .
Proof. The proof of this standard result may be found in Theorem 5.3 of [10].
Note that the sums in (1.23) are of different lengths. This will result in less work with the second
sum, which contains the root number g×f .
1.4. Trace formula. We have Weil’s estimate for the Kloosterman sum:
X? nh + mh 1 1
(1.25)
e
|S(n, m, c)| = ≤ (n, m, c) 2 c 2 d(c).
c
h mod c
Here ? denotes that the summation is restricted to (h, c) = 1, and hh ≡ 1 mod c. We will also need
the following estimates for the J-Bessel function (see [6] and [21]). For x > 0 we have
Jk−1 (x) min(xk−1 , x−1/2 ).
(1.26)
For x > 0 and integers B > 0 we have
(B)
Jk−1 (x) B x−B + x−1/2 .
(1.27)
For any complex numbers αf , define the weighted sum
XP
X
αf
αf =
(1.28)
,
−1 L(1, sym2 f )
ζ(2)
?
?
f ∈Hk (q)
f ∈Hk (q)
2
where L(s, sym f ) denotes the symmetric-square L-function associated to f . The arithmetic weights
above occur naturally in the Petersson trace formula (1.30) and the following trace formula for
newforms. Define
XP
12
(1.29)
∆?k,q (n, m) =
af (n)af (m).
q(k − 1)
?
f ∈Hk (q)
ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
5
Lemma 1.5. Trace formula.
(i) We have
∆?k,1 (n, m) = δ(n, m) + 2πi−k
(1.30)
X S(n, m, c)
c
c≥1
Jk−1
4π √nm c
,
where δ(n, m) equals 1 if n = m and 0 otherwise.
(ii) Let q be a prime. If (m, q) = 1 and q 2 - n then
(1.31)
∆?k,q (n, m)
−k
= δ(n, m) + 2πi
X S(n, m, cq)
c≥1
cq
Jk−1
4π √nm cq
∞
−
X
1
q −i ∆?k,1 (n, mq 2i ).
q[Γ0 (1) : Γ0 ((n, q))] i=0
Proof. (1.30) can be found in [9]. See Proposition 2.8 of [11] for (1.31).
Note that if q|n then the second line of (1.31) is q −2+ (nm) since [Γ0 (1) : Γ0 (q)] > q.
1.6. Voronoi summation. The GL(3) Voronoi summation formula (1.32) was found by Miller and
Schmid [17]. Goldfeld and Li [5] later gave another proof.
Lemma 1.7. GL(3) Voronoi Summation. Let ψ be a smooth, compactly supported function on
the positive real numbers and (b, d) = 1. We have
nb n X π 32 X A(n, l) X
dr ± n
A(r, n)e
d
S rb, ±n,
(1.32)
ψ
=
Ψ
,
d
N
2
nl
l
d3 N −1 rl−2
±
n≥1
n≥1
l|dr
where we define
(1.33)
Z
1
e − s)ds,
(π 3 X)−s H ± (s)ψ(1
2πi (σ)
1+s+3ν−1
s
s+1−3ν
1+s
1+s+1−3ν
Γ s+3ν−1
Γ
Γ
Γ
Γ
Γ
2
2
2
2
2
2
∓i H ± (s) = 1−s
1−s+3ν−1
2−s
2−s+3ν−1
1−s−1+3ν
2−s+1−3ν
Γ
Γ
Γ
Γ
Γ
Γ
2
2
2
2
2
2
Ψ± (X) = X
for σ > 0, where ψe denotes the Mellin transform of ψ.
Writing s = σ + it, by Stirling’s approximation of the gamma function we have
H ± (s) σ (1 + |t|)3σ .
(1.34)
1.8. Amplifier method. Let f0 ∈ Hk? (q) satisfy (1.3). Define the amplifier
X
af0 (n)af (n)
√
(1.35)
A(f ) =
.
n
1/4+1/2000
n<q
The assumption (1.3) implies that A(f ) is ‘amplified’ at f = f0 ; that is,
X
af0 (n)2
√
(1.36)
A(f0 ) =
q 1/8+1/4000− .
n
1/4+1/2000
n<q
This can be seen by partial summation together with the following upper bound given in [18]:
X
(1.37)
af0 (n)2 L1+
n<L
for all L > q . Theorem 1.1 will be deduced from the following.
6
RIZWANUR KHAN
Proposition 1.9. We have
XP
(1.38)
L( 12 , g × f )A(f )2 q 1+ .
f ∈Hk? (q)
By Lapid’s work, we have that L( 12 , g × f ) ≥ 0. Now if we drop all but the term corresponding to
f0 then we have
1
L( 1 , g × f0 )A(f0 )2 q 1+ .
L(1, sym2 f0 ) 2
(1.39)
Using the trivial bound L(1, sym2 f0 ) q and (1.36), the subconvexity bound (1.4) follows.
By (1.6) we may write
X
xm af (m)
√
A(f )2 =
(1.40)
,
m
1/2+1/1000
m<q
for some numbers xm q . By (1.18), Proposition 1.9 follows from
Proposition 1.10. Let m < q 1/2+1/1000 be a natural number. We have
X |A(r, m)| q 1 XP
(1.41)
L( 21 , g × f )af (m) √
1+
.
q
r
m
?
2
f ∈Hk (q)
r<q
2. Proof of Proposition 1.10
In this section we reduce the proof of Proposition 1.10 to two claims. By Lemma 1.3 and (1.6),
we have that the left hand side of (1.41) is bounded by
|L| + |S|,
(2.1)
where
L=
(2.2)
X A(r, n) r2 n
√ V 3/2+1/300 ∆?k,q (n, m)
r n
q
n,r≥1
is the longer sum and
S = q 1/2
(2.3)
X A(r, n) r2 n
√ V 3/2−1/300 ∆?k,q (nq, m)
r n
q
n,r≥1
is the shorter sum. By Lemma 1.5 we have that
X S(n, m, cq)
4π √nm X |A(r, m)| X A(r, n) r2 n
√
√
+
V 3/2+1/300
Jk−1
(2.4) L cq
cq
r
m
r
n
q
2
r<q
c≥1
n,r≥1
+ q −1+
X
X A(r, n)a 0 (n) r2 n
√f
V 3/2+1/300 .
r n
q
f 0 ∈Hk? (1) n,r≥1
Using (1.22), the last sum over n and r above can be written as an integral involving L(s, g × f 0 ),
an L-function independent of q. The line of integration can be moved to −∞, picking up a residue
L(1/2, g × f 0 ) 1 at s = 0, to see that the second line of (2.4) is q −1+ . We will prove
Lemma 2.1. Let m < q 1/2+1/1000 . We have
X S(n, m, cq)
4π √nm X A(r, n) r2 n
q
√ V 3/2+1/300
(2.5)
Jk−1
√ .
cq
cq
r n
m
q
n,r≥1
c≥1
ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
7
For S, we first consider the contribution of the terms with (n, q) = 1. By Lemma 1.5 and the
remark immediately following, the contribution of such terms is
X S(nq, m, cq)
4π √nqm X A(r, n) r2 n
√ V 3/2−1/300
(2.6)
q 1/2
+ q −1/2 .
Jk−1
cq
cq
r n
q
n,r≥1
(n,q)=1
c≥1
In the sum above, the contribution of the terms with c > q 1/2 is q −100 by (1.26) and having
k > 106 . Thus we may assume that (c, q) = 1, so that we have S(nq, m, cq) = −S(nq, m, c). We
may extend the sum to all natural numbers n, with an error of
1 X S(n, m, c)
4π √nm X A(r, nq) r2 n
√
q −1/2 ,
(2.7)
Jk−1
V 1/2−1/300
q
c
c
r n
q
1/2
n,r≥1
1≤c≤q
on observing that the c-sum is 1 +
∆?k,1 (n, m)
q and using (1.18). We will prove
Lemma 2.2. Let m < q 1/2+1/1000 and A > 0. We have
X S(nq, m, c)
4π √nmq X A(r, n) r2 n
√ V 3/2−1/300
(2.8)
Jk−1
A q −A .
q 1/2
cq
cq
r n
q
c≥1
n,r≥1
Finally we must consider the terms of S with q|n. The contribution of these terms is
X A(r, nq) r2 n
√
V 1/2−1/300 ∆?k,q (nq 2 , m) q −1/2 ,
(2.9)
r n
q
n,r≥1
on using (1.6) and (1.9) to see that ∆?k,q (nq 2 , m) = q −1 ∆?k,q (n, m) and then using (1.18) to bound
the sum absolutely. This completes the proof of Proposition 1.10.
3. Proof of Lemma 2.2
P?
Write S(nq, m, c) =
h mod c e(nqh/c)e(mh, c). As noted above, by (1.26) we may assume that
1/2
c < q . By (1.24), we may also assume that r2 < q 3/2 . Thus to prove Lemma 2.2, it is enough to
show that
4π √nmq X
r2 n
(3.1)
A(r, n)e(nqh/c)n−1/2 Jk−1
V 3/2−1/300 A q −A
cq
q
n
for any A > 0. We consider this sum in dyadic intervals. For N > 0, let ω(x) be a smooth function,
compactly supported on [1, 2] and satisfying ω (B) (x) B 1 and let
4π pxN mq −1 xr2 N −1/2
(3.2)
W (x) = x
Jk−1
V 3/2−1/300 ω(x).
c
q
By (1.27) and (3.5) we have for B > 0,
(3.3)
W
(B)
(x) B
pN mq −1 B
c
5
q B/10 .
It is enough to show that
X
(3.4)
A(r, n)e(nqh/c)W
n
n
N
A q −A
for
p
5
N mq −1
> q −1/10 .
c
We enforce the condition (3.5) since otherwise (3.4) follows easily by (1.24) and (1.26).
(3.5)
2
r N <q
3/2−1/300+
,
8
RIZWANUR KHAN
Applying Lemma 1.7 to (3.4), it is enough to show that
nN l2 X A(n, l)
A q −A ,
(3.6)
S(rhq, ±n, rc/l)W
nl
c3 r
n≥1
l|rc
where
Z
f (1 − s)ds,
X 1−s H ± (s)W
W(X) =
(3.7)
(σ)
for σ > 0 and
Z
f (1 − s) =
W
(3.8)
2
x−s W (x)dx.
1
Writing s = σ + it, we have by (3.3) and integration by parts B times,
pN mq −1 B
5
−B
f (1 − s) B |t|
(3.9)
q B/10 .
W
c
Using this bound with B = b3σ + 5c and (1.34), we have
nN l2 2
2 −σ pN mq −1 3σ
3 σ/2
5
3σ/105
10
2 −σ r N m
10 nl N
(3.10) W
q
q
(nl
)
q 3σ/10 .
q
σ
c3 r
c3 r
c
q3
4
By (3.5) and the assumptions of the lemma, we have r2 N m3 q 3−1/10 . Thus taking σ large
enough proves (3.6).
4. Proof of Lemma 2.1
We consider the n-sum in dyadic intervals. For N1 > 0, let ω1 (x) be a smooth function, compactly
(B)
supported on [1, 2] and satisfying ω1 (x) B 1 and let
4π pxN mq −2 xr2 N 1
1
−1/2
(4.1)
V 3/2+1/300 ω1 (x).
W1 (x) = x
Jk−1
c
q
It is enough to show that
√
n X
q 1+ N1
√
(4.2)
A(r, n)S(n, m, cq)W1
N1
m
n≥1
for
(4.3)
q 3/2−1/990 < N1 < q 3/2+1/300+ , q 1/2−1/290 <m < q 1/2+1/1000 , r < q 1/450 , c < q 1/450 .
We may assume the conditions above, since otherwise (4.2) follows easily by (1.24) and (1.26). Thus
it is enough to prove that
n X
(4.4)
A(r, n)S(n, m, cq)W1
q 3/2−1/990 .
N1
n≥1
P?
We apply Lemma 1.7 to the left hand side of (4.4) after writing S(n, m, cq) =
h mod cq e((nh +
mh)/cq). We need to show that
nN l2 X?
X A(n, l)
1
(4.5)
cq
e(mh/cq)
S(rh, n, qcr/l)W1 3 3
q 3/2−1/990 ,
nl
c q r
h mod cq
n≥1
l|cqr
where
Z
(4.6)
W1 (X) = X
(σ)
f1 (1 − s)ds
(π 3 X)−s H ± (s)W
ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
(B)
for σ > 0. Note that W1
s = σ + it,
(x) B q B/450 so that by integrating by parts B times we have for
Z
f1 (1 − s) =
W
(4.7)
9
2
x−s W1 (x)dx B |t|−B q B/450 .
1
We can use this bound together with (1.34) to estimate W1 (X). If X > q 1/150+ , we can take σ in
(4.6) to be very large and B = b3σ + 5c in (4.7) to see that W1 (X) X −2 q −100 . If X ≤ q 1/150+ ,
we take σ = in (4.6) and B = 2 in (4.7) to see that
W1 (X) q 1/200 X.
(4.8)
So the n-sum in (4.5) is essentially supported on n <
q 3+1/150+ c3 r
N1 l2
< q 3/2+1/60 .
2
1/150+
1l
and
The contribution to (4.5) by the terms with q|l is negligible since if q|l then nN
c3 q 3 r > q
2
nN1 l
−100
1/225
we have just seen that then W1 c3 q3 r q
. Henceforth fix l|cr, so that l < q
.
P?
We open the Kloosterman sum: S(rh, n, qcr/l) =
u mod qcr/l e((rhu + nu)l/qcr). For (4.5), it
is enough to show that
nN l2 X?
X? X A(n, l)e(nul/qcr)
(4.9)
W1 3 3
e(h(ul + m)/cq) q 1/2−1/300 .
n
c q r
u mod qcr/l n≥1
h mod cq
The innermost sum above, a Ramanujan sum, equals
X?
X?
(4.10)
e(h(ul + m)/c) ,
e(h(ul + m)/q)
h mod c
h mod q
P?
since (c, q) = 1. Note that
h mod q e(h(ul + m)/q) equals −1 or q − 1 according as ul 6≡ −m mod q
or ul ≡ −m mod q respectively. So the left hand side of (4.9) equals
nN l2 X?
X A(n, l)
(4.11)
e(hm/c)S(n, qhr, qcr/l)
−
W1 3 3
n
c q r
n≥1
h mod c
nN l2 X?
X A(n, l)
+ q
W1 3 3
e(hm/c)
n
c q r
n≥1
h mod c
X?
u mod qcr/l
u≡−ml mod q
e
uqhr + un qcr/l
.
Since (cr/l, q) = 1, we have S(n, qhr, qcr/l) = S(nq, hr, cr/l)S(0, n, q). This product of a Kloosterman sum and a Ramanujan sum is q 1+1/225 if q|n and q 1/225 otherwise. In any case,
using (1.18), the first line of (4.11) satisfies the bound required in (4.9). Now consider the second
line. By the Chinese Remainder Theorem, for (u, qcr/l) = 1 and u ≡ −ml mod q, we can write
u = −mcr(cr/l) + vq, where crcr ≡ 1 mod q and (v, cr/l) = 1. We have
e
uqhr qcr/l
=e
vhrq cr/l
, e
nu nucr/l nuq −nl2 mcr nvq 2 =e
e
=e
e
.
qcr/l
q
cr/l
q
cr/l
Thus (4.9) is reduced to showing
X A(n, l) −nl2 mcr nN l2 X? hm (4.12)
e
W1 3 3
e
S(nq 2 , hrq, cr/l) q −1/2−1/300 .
n
q
c q r
c
n≥1
h mod c
Now comes a crucial step, a conductor dropping in the size of the modulus of the exponential
factors from q to mcr. By the Chinese Remainder Theorem we have
−nl2 mcr −nl2 nl2 q (4.13)
=e
e
.
e
q
mcrq
mcr
10
RIZWANUR KHAN
−nl2
For n < q 3/2+1/60 , the exponential factor e mcrq
has amplitude at most q 1/30 and will be absorbed
3/2+1/60+
into the weight function. Let N2 < q
, let ω2 (x) be a smooth function, compactly supported
(B)
on [1, 2] and satisfying ω2 (x) B 1 and let
−xN l2 xN N l2 2 1
2
(4.14)
W1
ω2 (x).
W2 (x) = x−1 e
mcrq
c3 q 3 r
It is enough to prove that
nl2 q n X? hm X
W2
S(nq 2 , hrq, cr/l) N2 q −1/2−1/300 .
A(n, l)e
(4.15)
e
mcr
N2
c
n≥1
h mod c
Opening the Kloosterman sum and combining the exponential factors in n, it is enough to prove
that
n X
N2 q −1/2−1/100 .
(4.16)
A(n, l)e(nb/d)W2
N2
n≥1
1/2+1/150
for d < q
and (b, d) = 1. We use the Voronoi summation formula again. Applying Lemma
1.7 to the left hand side of (4.16), it suffices to show that
nN l2 X A(`, n)
2
(4.17)
S(lb, n, dl/`)W2
N2 q −1/2−1/100 .
d
n`
d3 r
n≥1
`|dl
where
Z
W2 (X) = X
(4.18)
f2 (1 − s)ds
(π 3 X)−s H ± (s)W
(σ)
for σ > 0. We need to estimate W2 (X). To this end we first note that for x ∈ [1, 2] and by the same
argument used to derive (4.8), we have
(4.19)
xN N l2 N N l2 Z
2
dB −s π 3 N2 N1 l2 −s ± f
dB
2 1
2 1
1/200 N2 N1 l
H
(s)
W
(1
−
s)ds
q
W
=
(x
)
.
1
B
1
dxB
c3 q 3 r
c3 q 3 r (σ) dxB
c3 q 3 r
c3 q 3 r
Thus
N2 N1 l2
q (B+1)/30 q −3/2 N2 .
c3 q 3 r
Integrating by parts B times, we have for s = σ + it the bound
Z 2
f2 (1 − s) =
W
(4.21)
x−s W2 (x)dx B |t|−B q (B+1)/30 q −3/2 N2 .
(4.20)
(B)
W2
(x) B q B/30 q 1/200
1
Now we can estimate W2 (X). We take σ = 1 in (4.18) and B = 5 above. By (1.34) and (4.21) we
see that
W2 (X) N2 q −3/2+1/5 .
(4.22)
Taking σ = 2, we also observe that the sum in (4.17) can be restricted to n < q 100 , say, with
negligible error.
Using (4.22), to prove (4.17) it is enough to show that
X |A(`, n)|
(4.23)
|S(lb, n, dl/`)| q 1/2−13/60 .
n`
100
n<q
`|dl
By (1.25), we have |S(lb, n, dl/`)| (lb, n, dl/`)1/2 (dl)1/2+ q 1/4+1/100 , since (b, d) = 1. This
establishes (4.23), using (1.18).
ON THE SUBCONVEXITY PROBLEM FOR GL(3) × GL(2) L-FUNCTIONS
11
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Mathematisches Institut, Georg-August Universität Göttingen, Bunsenstraße 3-5, D-37073 Göttingen,
Germany
E-mail address: rrkhan@uni-math.gwdg.de