An orthogonality relation for a thin family of $GL(3)$ Maass forms

0.18 MB·English

by  João Guerreiro

AN ORTHOGONALITY RELATION FOR A THIN FAMILY OF GL(3) MAASS
FORMS
JOA˜O GUERREIRO
5
1 Abstract. We prove an orthogonality relation for the Fourier-Whittaker coefficients of a thin
0
family of GL(3) Maass forms containing all self-dual forms. This is obtained by analysing the
2
Kuznetsov trace formula on GL(3) for a certain family of test functions. The method also yields

n Weyl’s law for thesame family of Maass forms.
a
J

8

1. Introduction
]
T
A well known fact about Dirichlet characters is the following orthogonality relation
N
. φ(q), if n m modq
h
χ(n)χ(m) = ≡ (1.1)
at (0, otherwise.
χmodq
m X
for integers m,n coprime to q, where the sum on the left is over all characters (mod q). Since
[
Dirichlet characters can be viewed as automorphic representations of GL(1,A ), this result can be
Q
1
interpreted as the simplest case of the orthogonality relation conjectured by Zhou [19] concerning
v
Fourier-Whittaker coefficients of Maass forms on the space SL (Z) SL (R)/SO (R), n 2. This
8 n n n
\ ≥
9 orthogonality relation conjectured by Zhou was proved by Bruggeman [2] in the case n = 2 and by
9 Goldfeld-Kontorovich [6] and Blomer [1] in the case n = 3. Versions of this result have applications
1
totheSato-TateproblemforHeckeoperators,bothintheholomorphic[3],[17]andnon-holomorphic
0
. setting [15], [19], as well as to theproblem of determiningsymmetry types of families of L-functions
1
[6] as introduced in the work of Katz-Sarnak [9]. In this paper, we prove an orthogonality relation
0
5 for the Fourier-Whittaker coefficients of a thin family of Maass forms on SL (Z) which contains
3
1 all self-dual forms. The main tool used is the Kuznetsov trace formula for GL(3) developed by
:
v Blomer [1] and Goldfeld-Kontorovich [6]. The same methods yield a weighted Weyl’s law for the
i same family of Maass forms.
X
Weyl’s law was first proved by Selberg [16] for Maass forms on SL (Z) using the Selberg trace
r 2
a formula. Miller[13]latershowedWeyl’s lawforMaassformsonthespaceSL (Z) SL (R)/SO (R).
3 3 3
\
Thisresult hassince been obtained in moregeneral settings in [11], [12], [14]. Theversion of Weyl’s
law presented here tells us that the family of Maass forms being studied has zero density in the set
of all Maass forms on SL (Z). We also note that a lower bound Weyl’s law for self-dual forms on
3
SL (Z) follows from the Gelbart-Jacquet lift [4] and Weyl’s law for Maass forms on SL (Z). This
3 2
work is asteptowards aproofof theGelbart-Jacquet lift byisolating thecontribution of symmetric
square lifts from GL(2) in the GL(3) Kuznetsov trace formula, in the spirit of Langlands’ “Beyond
Endoscopy” [10].
Letφ (j = 0,1,2,...),whereφ isaconstantfunction,beasetoforthogonalGL(3)Hecke-Maass
j 0
forms with spectral parameters ν(j) = ν(j),ν(j) , and define ν(j) = ν(j) ν(j). The parameters
1 2 3 − 1 − 2
are normalized such that for tempered(cid:16) forms th(cid:17)ey are purely imaginary. For a particular non-
constant Maass form φ with spectral parameters ν = (ν ,ν ) define ν = ν ν , and define
1 2 3 1 2
− −
Langlands parameters
α = ν ν , α = ν ν , α = ν ν .
1 1 3 2 2 1 3 3 2
− − −
Date: January 12, 2015.
1
2 JOA˜OGUERREIRO
The Laplace eigenvalue of φ is given by
λ = 1 3(ν2+ν2+ν2)= 1 (α2+α2+α2).
φ − 1 2 3 − 1 2 3
Let
=Res L(s,φ φ˜).
j s=1 j j
L ×
For T 1, ν = (ν ,ν ) C2 (with ν = ν ν ) and R > 0, to be fixed later, we define
1 2 3 1 2
≫ ∈ − −
2
3
Γ 2+R+3νj Γ 2+R−3νj
4 4
hT,R(ν) = Tα21/R2 +Tα22/R2 +Tα23/R2 2e2(α21+α22+α23)/T2 jQ=1 (cid:16) (cid:17) (cid:16) (cid:17)! . (1.2)
3
(cid:16) (cid:17) Γ 1+3νj Γ 1−3νj
2 2
j=1
(cid:16) (cid:17) (cid:16) (cid:17)
Q
This function is essentially supported on the region where ν , ν , ν T. Let φ be tempered
1 2 3
| | | | | | ≪
and ν , ν , ν < T. Note that in this situation, h (ν) is real valued and positive. We estimate
1 2 3 T,R
| | | | | |
this function in three regions. If one of the α is equal to 0, then
i
c(1)[(1+ ν )(1+ ν )(1+ ν )]R h (ν) c(2)[(1+ ν )(1+ ν )(1+ ν )]R, (1.3)
R | 1| | 2| | 3| ≤ T,R ≤ R | 1| | 2| | 3|
(2) (1)
for some c c > 0. If one of the α is smaller than R, then
R ≥ R | i|
[(1+ ν )(1+ ν )(1+ ν )]R
1 2 3
h (ν) | | | | | | . (1.4)
T,R ≫R T2
If α , α , α > 2R3/2, then
1 2 3
| | | | | |
1
h (ν) . (1.5)
T,R ≪ TR
Theorem 1.1. For j = 0,1,2,..., let φ be a set of orthogonal GL(3) Hecke-Maass forms with
j
spectral parameters ν(j) = ν(j),ν(j) . Here φ is a constant function. Let h be given as in
1 2 0 T,R
(1.2). We have that for fixe(cid:16)d R > 50(cid:17)and some cR > 0,
h (ν(j)) T4+3R
T,R = c +O(T3+3R), (T ). (1.6)
R
√logT → ∞
(cid:12) j (cid:12)
j≥1 L
X(cid:12) (cid:12)
Moreover, for fixed ǫ > 0 and positive integers m ,m ,n ,n , we have
1 2 1 2
A (m ,m )A (n ,n ) hT,R(ν(j)) =  j |hT,RL(jν(j))| +OR,ǫ(cid:18)(m1m2n1n2)6T3R+3+ǫ(cid:19), if mm12==nn12
j 1 2 j 1 2
P
Xj (cid:12)(cid:12) Lj (cid:12)(cid:12) OR,ǫ (m1m2n1n2)6T3R+3+ǫ , otherwise.
(cid:18) (cid:19)
(1.7)



Remark 1.2. Note that an analogous result in [6] for all GL(3) Hecke-Maass forms with spectral
parameters ν , ν , ν T yields a main term of size T5+3R. This implies that the family of
1 2 3
| | | | | | ≪
GL(3) Maass forms picked out by our test function h is indeed significantly thinner (by a factor
T,R
of T√logT) than the family of all GL(3) Hecke-Maass forms.
AN ORTHOGONALITY RELATION FOR A THIN FAMILY OF GL(3) MAASS FORMS 3
Remark 1.3. The test functions h appearing in this theorem are a product of three terms chosen
T,R
with the following objectives: the first exponential term contributes with polynomial decay when all
α , α , α 0 and exponential decay when all α , α , α Tǫ; the second exponential
1 2 3 1 2 3
| | | | | | ≫ | | | | | | ≫
term contributes with exponential decay when one of ν > T1+ǫ; the product of Gamma factors is
i
| |
already partially present in the Kuznetsov trace formula and its particular form is such that it has
polynomial growth in ν. We note that the methods presented here have also been carried out for a
different family of test functions in [6]. It would be interesting to generalize 1.1 for a broader class
of test functions.
Remark 1.4. Blomer [1] shows that the residues are bounded by
j
L
−1 ǫ
(j) (j) (j) (j) (j) (j)
(1+ ν )(1+ ν )(1+ ν ) (1+ ν )(1+ ν )(1+ ν ) (1.8)
| 1 | | 2 | | 3 | ≪ Lj ≪ǫ | 1 | | 2 | | 3 |
and, c(cid:16)onjecturally, the lower bound is e(cid:17)xpected to be (cid:16) (cid:17)
−ǫ
(j) (j) (j)
1+ ν )(1+ ν )(1+ ν ) . (1.9)
| 1 | | 2 | | 3 | ≪ǫ Lj
(cid:16) (cid:17)
4 JOA˜OGUERREIRO
2. GL(3) Kuznetsov trace formula
Following ([1], Proposition 4) and [6], the GL(3) Kuznetsov trace formula takes the form
+ + = Σ +Σ +Σ +Σ , (2.1)
min max 1 2a 2b 3
C E E
where all the quantities are explained below.
Definition 2.1. Let ν C2 and z h with Iwasawa coordinates
∈ ∈
1 x x y y 0 0
2 3 1 2
z = xy = 0 1 x 0 y 0 .
1 1
  
0 0 1 0 0 1
Let   
I (z) = yν1+2ν2y2ν1+ν2, (z = xy h).
ν 1 2 ∈
Define the Whittaker function
1+3ν 1+3ν 1+3ν +3ν
W (z) = π−ν1−ν2Γ 1 Γ 2 Γ 1 2
ν
2 2 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1 1 u u
2 3
I 1 1 u e( u u )du du du , (2.2)
ν 1 1 2 1 2 3
×    − −
ZZZ 1 1
R3
  
where e(x) = e2πix.
Let F :R2 C be a bounded function with the following decay
+ →
F(y ,y ) (y y )2+ǫ, (2.3)
1 2 1 2
| |≪
asy 0ory 0. ForsuchafunctiondefineitsLebedev-KontorovichtransformF# :D D C
1 2
→ → × →
as
dy dy
F#(ν) = F(y)W (y) 1 2, (2.4)
ν
y y
1 2
ZZ
R2
+
where W (y) is the Whittaker function and D C is of the form ( δ,δ) iR for some δ > 0.
ν
⊂ − ×
For m,n Z, m,n > 0, let A (m,n)betheFourier-Whittaker coefficient of aMaass formφas in
φ
∈
[6]. We normalize the GL(3) Hecke-Maass forms by choosing the first Fourier-Whittaker coefficient
to be A (1,1) = 1. Fix positive integers m ,m ,n ,n . The cuspidal contribution is given by
φ 1 2 1 2
C
2
F# ν(j),ν(j)
1 2
= A (m ,m )A (n ,n ) , (2.5)
C j 1 2 j 1 2 3 (cid:12)(cid:12) 1(cid:16)+3ν(j) (cid:17)(cid:12)(cid:12)1−3ν(j)
j 6 Γ k Γ k
X Lj (cid:12) 2 (cid:12) 2
k=1 (cid:18) (cid:19) (cid:18) (cid:19)
with j ranging over cuspidal Hecke-Maass forms on SQL(3,Z). The minimal Eisenstein series is
1 i∞ i∞ F#(ν ,ν ) 2
1 2
= A (m ,m )A (n ,n ) dν dν (2.6)
Emin (4πi)2 ν 1 2 ν 1 2 3 1 2
Z Z ζ((cid:12)1+3ν )Γ (cid:12)1+3νk 2
−i∞−i∞ (cid:12) k (cid:12) 2
k=1
where the Fourier coefficients satisfy Aν(n1,n2) ǫ (nQ1n2(cid:12)(cid:12))ǫ. The maxim(cid:0)al Eis(cid:1)e(cid:12)(cid:12)nstein series is
| | ≪
AN ORTHOGONALITY RELATION FOR A THIN FAMILY OF GL(3) MAASS FORMS 5
2
i∞ B (m ,m )B (n ,n ) F# ν irj,2irj
c ν,j 1 2 ν,j 1 2 − 3 3
= dν, (2.7)
Emax 2πi Xj −Zi∞ L(1,Ad uj)|L(1+3ν,uj)|2 Γ 1+3ν2(cid:12)(cid:12)(cid:12)−irj (cid:16)Γ 1+22irj Γ(cid:17)(cid:12)(cid:12)(cid:12)1+3ν2+irj 2
(cid:12) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)(cid:12)
where c is an absolute constant and u is an or(cid:12)thogonal basis of Hecke-Maass forms (cid:12)on SL(2,Z)
{ j} (cid:12) (cid:12)
with eigenvalues 1/4+r2. The Fourier coefficients satisfy B (n ,n ) (n n )1/2+ǫ. Set
j | ν,j 1 2 | ≪ǫ 1 2
1, if AB
1 = |
A|B
(0, otherwise.
On the geometric side we have
dy dy
Σ1 = 1{mm12==nn12}ZZ |F(y1,y2)|2 (y11y2)23 = 1{mm12==nn12}hF,Fi, (2.8)
R+
2
S˜(δm ,n ,n ,D ,D ) m n n
Σ = 1 1 2 1 2 ˜ 1 1 2 , (2.9)
2a δ
D D J D D
δX=±1 DX1|D2 1 2 (cid:18)r 1 2 (cid:19)
m2D12=n1D2
S˜(δm ,n ,n ,D ,D ) m n n
Σ = 2 2 1 2 1 ˜ 2 1 2 , (2.10)
2b δ
D D J D D
δX=±1 DX2|D1 1 2 (cid:18)r 1 2 (cid:19)
m1D22=n2D1
S(δ m ,δ m ,n ,n ,D ,D ) √m n D √m n D
1 1 2 2 1 2 1 2 1 2 1 2 1 2
Σ = , , (2.11)
3 D D Jδ1,δ2 D D
δ1,Xδ2=±1DX1,D2 1 2 (cid:18) 2 1 (cid:19)
where
1+x2+x2 A 1+x2
˜(A) = A−2 F(Ay ,y )e( δAx y )F y 1 2, 1
Jδ 1 2 − 1 1 2 1+x2 y y 1+x2+x2
ZZ ZZ p 1 1 2 p 1 2!
R2 R2
+
x x A x dx dx dy dy
1 2 2 1 2 1 2
e y + , (2.12)
× 21+x2 y y 1+x2+x2 y y2
(cid:18) 1 1 2 1 2(cid:19) 1 2
(A ,A ) = (A A )−2 F(A y ,A y )e( δ A x y δ A x y )
Jδ1,δ2 1 2 1 2 1 1 2 2 − 1 1 1 1− 2 2 2 2
ZZ ZZZ
R2 R3
+
A (x x x )2+x2+1 A 1+x2+x2
F 2 1 2− 3 1 , 1 2 3
× y 1+x2+x2 y (x x x )2+x2+1
2 p 2 3 1 1 p2− 3 1 !
A x x +x A x (x x x )+x dx dx dx dy dy
2 1 3 2 1 2 1 2 3 1 1 2 3 1 2
e − , (2.13)
× −y 1+x2+x2 − y (x x x )2+x2+1 y y
(cid:18) 2 2 3 1 1 2− 3 1 (cid:19) 1 2
m C +n C C n C
S˜(m ,n ,n ,D ,D ) = 1 e 1 1 1 1 2 e 2 2 , (2.14)
1 1 2 1 2 D1|D2 D D /D
C1(modDX1),XC2(modD2) (cid:18) 1 (cid:19) (cid:18) 2 1(cid:19)
(C1,D1)=1=(C2,D2/D1)
6 JOA˜OGUERREIRO
m B +n (Y D Z B )
S˜(m ,m ,n ,n ,D ,D ) = e 1 1 1 1 2− 1 2
1 2 1 2 1 2
D
XB1,CX1(mXodDX1) (cid:18) 1 (cid:19)
B2,C2(modD2)
(B1,C1,D1)=1=(B2,C2,D2)
B1B2+C1D2+C2D1≡0(modD1D2)
m B +n (Y D Z B )
2 2 2 2 1 2 1
e − , (2.15)
× D
(cid:18) 2 (cid:19)
and Y ,Y ,Z ,Z are such that
1 2 1 2
Y B +Z C 1 (mod D ) and Y B +Z C 1 (mod D ). (2.16)
1 1 1 1 1 2 2 2 2 2
≡ ≡
AN ORTHOGONALITY RELATION FOR A THIN FAMILY OF GL(3) MAASS FORMS 7
3. Bounding the inverse transform
Let
3
2+R+3ν 2+R 3ν
FT#,R(ν1,ν2) = Tα21/R2 +Tα22/R2 +Tα23/R2 e(α21+α22+α23)/T2 Γ 4 j Γ 4− j .
(cid:16) (cid:17) jY=1 (cid:18) (cid:19) (cid:18) (cid:19)
 (3.1) 
#
As F has enough exponential decay on a strip (ν ), (ν ) < ǫ then by Lebedev-Whittaker
T,R |ℜ 1 | |ℜ 2 |
inversion as in section 2.2 of [6],
i∞ i∞
1 dν dν
# 1 2
F (y) = F (ν)W (y) . (3.2)
T,R (πi)2 T,R ν 3
Z Z Γ 3νj Γ 3νj
−i∞−i∞ 2 − 2
j=1
(cid:16) (cid:17) (cid:16) (cid:17)
We also have the Parseval-type identity Q
2
i∞ i∞ #
F (ν) dν dν
F ,F = F (y ,y )2 dy1dy2 = 1 T,R 1 2 = F# ,F# .
h T,R T,Ri | T,R 1 2 | (y y )3 (πi)2 3(cid:12) (cid:12) T,R T,R
ZR+Z 1 2 −Zi∞−Zi∞ (cid:12)(cid:12) Γ 32νj (cid:12)(cid:12) Γ −32νj D E
2 j=1
(cid:16) (cid:17) (cid:16) (cid:17)
Q (3.3)
For proofs of these results refer to [5].
Proposition 3.1. Fix C ,C > 0, R > 3max(C ,C )+6 and ǫ > 0. For any y ,y > 0, T 1,
1 2 1 2 1 2
≫
we have
F (y) y y T3R/2+11/2+C1/2+C2/2+ǫ y1 C1 y2 C2. (3.4)
| T,R | ≪C1,C2,R,ǫ 1 2 T T
(cid:16) (cid:17) (cid:16) (cid:17)
Proof. We start by writing out the representation of W as an inverse Mellin transform [18],
ν
3
Γ s1+αj Γ s2−αj
W (y) = y1y2π3/2 j=1 (cid:16) 2 (cid:17) (cid:16) 2 (cid:17)y−s1y−s2ds ds
ν (2πi)2 Q 4πs1+s2Γ s1+s2 1 2 1 2
Z Z 2
(C2)(C1)
for C ,C > 0. Combining this with the Lebedev-Whitt(cid:0)aker in(cid:1)verse transform of F , we observe
1 2 T,R
that
3
Γ s1−αj Γ s2+αj
F# (ν ,ν ) 2 2
F (y) = T,R 1 2 j=1 (cid:16) (cid:17) (cid:16) (cid:17)y1−s1y1−s2ds ds dν dν .
T,R Z Z Z Z 3 Γ 3νj Γ 3νj Q16πs1+s2+5/2Γ s1+2s2 1 2 1 2 1 2
(0)(0)(C2)(C1) 2 − 2
j=1 (cid:0) (cid:1)
(cid:16) (cid:17) (cid:16) (cid:17)
Q
Assume 2k < C < 2k +2 for some non-negative interger k and pull s from the contour (C ) to
1 1 1
the contour ( C ). Then
1
−
k 3
F (y) = + , (3.5)
T,R m,l
M R
m=0 l=1
X X
where is the main term and is given by the same expression as F with the contour (C )
T,R 1
M
replaced by ( C ). Here are the residues corresponding to the poles of the integrand at
1 m,l
− R
s = α 2m for m = 0, ,k and l = 1,2,3. These residues are given by
1 l
− ···
8 JOA˜OGUERREIRO
3
Γ αl−2m−αj Γ s2+αj
F# (ν ,ν ) 2 2
= C T,R 1 2 j6=l (cid:16) (cid:17)j=1 (cid:16) (cid:17)y1−αl+2my1−s2ds dν dν
Rm,l m,lZ Z Z 3 Γ 3νj Γ 3νj Q παl−2m+s2Γ αQl−22m+s2 1 2 2 1 2
(0)(0)(C2) 2 − 2
j=1 (cid:0) (cid:1)
(cid:16) (cid:17) (cid:16) (cid:17)
Q
for some constant C . Note that there will be no contribution of residues of higher order poles,
m,l
whichcouldpossiblyshowupwhentwo oftheα differbyaneven integer. Inthissituation, atleast
i
one of 3ν , 3ν , 3ν is equal to a non-positive even integer, making the integrand identically
1 2 3
± ± ±
zero in that region.
Now assume 2k′ < C < 2k′ +2 for some non-negative interger k′. For each term in (3.5) shift
2
variable s from the contour (C ) to ( C ) to get,
2 2 2
−
k 3 k,k′ 3
FT,R(y) = ˜ + m′,l′ + m,l,m′,l′ (3.6)
M M R
m′=0l′=1 m,m′=0l=1 l′6=l
X X X XX
where ˜ is the main term, given by the same expression as with the contour (C ) replaced
2
M M
by ( C2) and m′,l′ are the residues corresponding to the poles of the integrand of at s2 =
− M M
αl′ 2m′ for m′ = 0, ,k′. The terms m,l,m′,l′ are the residues corresponding to the poles of
− − ··· R
the integrand of m,l at s2 = αl′ 2m′ for m′ = 0, ,k′ and l′ = l. The residues m′,l′ are
R − − ··· 6 M
given by
3 Γ s1−αj Γ −αl′−2m′+αj
Mm′,l′ = Cm′ ′,l′(Z0)(Z0)(−ZC1) 3 ΓFT#3,2Rνj(ν1Γ,ν2−)32νj jQ=1 πs(cid:16)1−α2l′−2(cid:17)mj′Q6=Γl′ s(cid:16)1−αl2′−22m′ (cid:17)y11−s1y21+αl′+2m′ds1dν1dν2
j=1 (cid:16) (cid:17)
(cid:16) (cid:17) (cid:16) (cid:17)
Q
and the residues m,l,m′,l′ are given by
R
Γ αl−2m−αj Γ −αl′−2m′+αj
Rm,l,m′,l′ = Cm,l,m′,l′(Z0)(Z0) 3 ΓFT#3,2Rνj(ν1Γ,ν2−)32νj jQ6=4lπα(cid:16)l−2m−2αl′−2m(cid:17)′jΓQ6=l′αl−(cid:16)2m−2αl′2−2m′ (cid:17)y11−αl+2my21+αl′+2m′dν1dν2.
j=1 (cid:16) (cid:17)
(cid:16) (cid:17) (cid:16) (cid:17)
Q
We will now bound each term in (3.6). Let ν = it and s = C +iu . Note that, for (ν ) = 0,
j j j j j j
ℜ
#
the first exponential term in the definition of F is bounded and the second one has exponential
T,R
decay for t > T1+δ. Using Stirling’s formula to estimate the Gamma factors we get
j
| |
˜ y1+C1y1+C2 exp( )du du dt dt ,
|M| ≪ 1 2 P · E 1 2 1 2
ZZ ZZ
|t1|,|t2|<T1+δ R2
where the exponential factor is given by
3 3 3
4
E = 3 t + u +u iu α iu +α ,
j 1 2 1 j 2 j
π | | | |− | − |− | |
j=1 j=1 j=1
X X X
and the polynomial term is given by
AN ORTHOGONALITY RELATION FOR A THIN FAMILY OF GL(3) MAASS FORMS 9
(R+2)/2
3
= (1+ t ) (1+ u +u )(1+C1+C2)/2
j 1 2
P  | |  | |
j=1
Y
  (−C1−1)/2 (−C2−1)/2
3 3
(1+ iu α ) (1+ iu +α ) . (3.7)
1 j 2 j
× | − |   | | 
j=1 j=1
Y Y
   
We now show that the exponential factor is non-positive. As the exponential factor is invariant
under cyclic permutations of (t ,t ,t ), we may assume, without loss of generality, that t and t
1 2 3 1 2
have the same sign. Then α + α = 3t +3t . As u +u iu α + iu +α , we get
1 3 1 2 1 2 1 2 2 2
| | | | | | | | | | ≤ | − | | |
3
4
E 3 t iu α iu α iu +α iu +α
j 1 1 1 3 2 1 2 3
π ≤ | |−| − |−| − |−| |−| |
j=1
X
6t +6t 2(α + α ) = 0.
1 2 1 3
≤ | | | |− | | | |
For either u > 5T1+δ or u > 5T1+δ, the exponential factor is bounded above by T1+δ
1 2
| | | | −
giving exponential decay to the integral. Integrating first over u , u we get
1 2
y1+C1y1+C2 du du dt dt y1+C1y1+C2T(3R+11−C1−C2)/2+ǫ
|M| ≪ 1 2 P 1 2 1 2 ≪ 1 2
ZZZZ
|t1|,|t2|<T1+δ
|u1|,|u2|<5T1+δ
by choosing δ appropriately. To bound the residues m′,l′ we start by shifting variables ν1 and ν2
M
to contours (B ) and (B ), respectively, where B , B < R/3 and B′ < C for j = l, defining
1 2 | 1| | 2| j 1 6
B′ = 2B +B , B′ = B B , B′ = B 2B , in a manner similar to the ν . Note that the first
ex1ponent1ial te2rm i2s now2b−oun1ded3by −3T(1m−axB1′,B22′,B3′)2/R2 T. It follows thatj
≤
|Mm′,l′| ≪ y11+C1y21+Bl′′+2m′T P ·exp(E)du1dt1dt2
ZZ Z
|t1|,|t2|<T1+δ R
where the exponential factor is given by
3 3
4
E = 3 tj + u1 (αl′) u1 (αj) (αj αl′),
π | | | −ℑ |− | −ℑ |− |ℑ − |
j=1 j=1 j6=l′
X X X
and the polynomial term is given by
(R+2)/2
3
= (1+ tj ) (1+ u1 (αl′))(1+C1+Bl′′+2m′)/2
P  | |  | −ℑ |
j=1
Y
 3  (−C1−1)/2
(1+ u1 (αj)) (1+ ( αl′ +αj))(Bj′−Bl′′−2m′−1)/2 . (3.8)
× | −ℑ |  |ℑ − |
jY=1 jY6=l′(cid:16) (cid:17)
 
The exponential factor is again non-positive as
3
4
E 3 tj u1 (αj) (αj αl′) 0
π ≤ | |− | −ℑ |− |ℑ − | ≤
j=1 j6=l′ j6=l′
X X X
10 JOA˜OGUERREIRO
by using the triangle inequality on the second sum to get a difference of (α ). We now pick
j
ℑ
B′ = C 2m′ and B′ < 0 for j = l′ to get
l′ 2− j 6
|Mm′,l′| ≪ y11+C1y21+C2T Pdu1dt1dt2 ≪ y11+C1y21+C2T(3R+11−C1−C2)/2+ǫ.
ZZZ
|t1|,|t2|<T1+δ
|u1|<10T1+δ
To bound the residues m,l,m′,l′ we again shift variables ν1 and ν2 to contours (B1) and (B2),
R
respectively, where B , B < R/3. Tosimplifynotation, wewillassumewithoutlossofgenerality
1 2
| | | |
that l = 1 and l′ = 2. We obtain
1−B′+2m 1+B′+2m′
|Rm,1,m′,2|≪ y1 1 y2 2 T P ·exp(E)dt1dt2
ZZ
|t1|,|t2|<T1+δ
where the exponential factor is given by
E
3
4
E = 3 t (α α ) (α α ) + (α α ) = 0
j j 1 j 2 1 2
π | |− |ℑ − |− |ℑ − | |ℑ − |
j=1 j6=1 j6=2
X X X
and the polynomial term is given by
(R+2)/2
3
= (1+ t ) (1+ t )(3B1−1)/2(1+ t )(−3B2−2m′−1)/2(1+ t )(3B3−2m−1)/2,
j 1 2 3
P  | |  | | | | | |
j=1
Y
where B=B +B . Then pick B′ = C +2m and B′ = C 2m′ to get
3 1 2 1 − 1 2 2−
|Rm,1,m′,2|≪ y11+C1y21+C2T Pdt1dt2 ≪ y11+C1y21+C2T(3R+9−2C1−2C2+2m+2m′)/2+ǫ
ZZ
|t1|,|t2|<T1+δ
y1+C1y1+C2T(3R+9−C1−C2)/2+ǫ.
≪ 1 2
Combining all the bounds we get the desired result.
(cid:3)
Remark 3.2. It follows from this proposition that F satisfies the decay condition (2.3).
T,R