MATH G9905 RESEARCH SEMINAR IN NUMBER THEORY
(FALL 2014)
LECTURE 5 (OCTOBER 7, 2014)
XIAOQING LI
LOWER BOUNDS FOR RANKIN–SELBERG TYPE INTEGRALS
NOTES TAKEN BY PAK-HIN LEE
Abstract. In this talk, we will derive an almost sharp lower bound for a Rankin–Selberg
type integral involving truncated Eisenstein series on GL(2n). Some orthogonality relations
in the Fourier expansion of automorphic functions on GL(n) will be discovered and results
from sieve methods will be applied. This is a joint work with Goldfeld.
1. The modified truncated Eisenstein series
We will start by explaining the relationship between Authur’s truncated Eisenstein series
ˆ
EA and the modified truncated Eisenstein series EˆA∗ .
Let f be a Maass form for SL(n, Z), and
φ(nmk) := | det m1 |ns | det m2 |−ns f (m1 )f (m2 ).
Here
nmk is the Iwasawa decomposition: n is in the unipotent radical of Pn,n , m =
m1
is in the Levi, and k ∈ O(2n, R). The cuspidal Eisenstein series is
m2
X
E(z, s) :=
φ(γz).
γ∈Pn,n \Γ
Arthur’s truncated Eisenstein series is
EˆA (z, s) = E(z, s) −
X
cP E|γ
γ∈Pn,n \ SL2n
h(γ·z)≥A
det m1 is the height function and cP E is the constant term of E along P .
where h(z) = det
m2 We don’t know how to compute the Fourier series of Arthur’s truncation, so we introduce
a modified version.
Definition 1.
n
Λ(2ns − 2n, f × f )
1
∗
ˆ
2
EA (z, s) := E(z, s) − A E z, s −
+
E(z, 2 − s)
2
Λ(1 + 2ns − 2n, f × f )
Last updated: October 9, 2014. Please send corrections and comments to [email protected].
1
−
X
n
A2
1−
h(z)
f (m1 )f (m2 )
n
h(z) 2
γ
ns
γ∈Pn,n \ SL2n
h(γ·z)≥A
Λ(2ns − 2n, f × f )
−
Λ(1 + 2ns − 2n, f × f )
X
n(2−s)
h(z)
γ∈Pn,n \ SL2n
h(γ·z)≥A
n
A2
1−
f (m1 )f (m2 ) .
n
h(z) 2
γ
We can compute the Fourier expansion of EˆA∗ because
(
Z
1
x−w dw
if x > 1,
= 1−x
0
if x ≤ 1.
(2) w(w + 1)
See Lemma 4 below.
√
Proposition 2. h(γ · z) ≤ h(z) for yi ≥ 23 , 1 ≤ i ≤ 2n − 1 and γ ∈ SL2n (Z).
Proof. Let ν = 0, · · · , 0, n1 , 0, · · · , 0 be the (2n − 1)-tuple with n1 at the n-th place. Then
Iν (z) :=
2n−1
Y 2n−1
Y
i=1
b
yi i,j
νj
= h(z).
j=1
Here
bi,j
(
ij
if i + j ≤ 2n,
=
(2n − i)(2n − j) if i + j ≥ 2n.
We have
h(γ · z) = Iν (γ · z) = ken+1 γ · z ∧ · · · ∧ e2n−1 γ · z ∧ e2n γ · zk−2 ken+1 z ∧ · · · ∧ e2n−1 z ∧ e2n zk2 h(z).
As we mentioned last time, ken+1 M ∧ · · · ∧ e2n M k2 is the sum of squares of all the n × n
minors of the last n rows of M . Using the Cauchy–Binet formula, we can see that this is
≤ h(z).
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Corollary 3. For A tn ,
√
3
2
≤ yi ≤ (t1+ )n(2n−1) , 1 ≤ i ≤ 2n − 1, we have
n
1
∗
EˆA (z, s) = EˆA (z, s) − A 2 EˆA z, s −
+ cs− 1 EˆA (z, 2 − s).
2
2
In other words, if we choose A to be very big, the truncated terms in the modified truncated
Eisenstein series are all 0.
Lemma 4.
n
1
Λ(2ns − 2n, f × f )
∗
ˆ
2
EA (z, s) = E(z, s) − A E z, s −
+
E(z, 2 − s)
2
Λ(1 + 2ns − 2n, f × f )
Z
n
1
A− 2 w
w
−
E z, s +
dw
2πi (2) w(w + 1)
2
Z
n
Λ(2ns − 2n, f × f )
1
A− 2 w
w
−
·
E z, 2 − s +
dw.
Λ(1 + 2ns − 2n, f × f ) 2πi (2) w(w + 1)
2
2
2. Orthogonality of Fourier coefficients
Recall the integral
Z
2
I = |L(1 + 2int, f × f )| ·
P2n−1,1 (Z)\η
Z
2n
∞
EˆA∗ (z, 1 + it)g
0
n
Here η 2n = GL(2n, R)/ O(2n, R)R× , g, ψ ∈ C∞
c ([1, 2]), β = t
Last time we derived an almost sharp upper bound
2 A dA det z
ψ
d× z.
β A
δ
10
2
and δ = β −1 (log log t)n .
I β n δ −1 log2 t|L(1 + it, f × f )|
using the Fourier expansion of EˆA∗ and the Maass–Selberg relations. Today we will prove a
sharp lower bound, which will only use the Fourier expanion.
Proposition 5. I β n δ −1
.
log t
Recall the Fourier expansion of a general automorphic function on GL(n).
Lemma 6. Suppose F is an automorphic function for SL(k, Z), k ≥ 4. Then
 

1 0 · · · 0 u1k
Z 1  1 · · · 0 u2k  
Z 1
 ×

.. 
. . ..


F
···
F (z) =
. .
. 
 z d u

0
0
 

1 u
k−1,k
1
+
X
X
1≤l≤k−2 mk−1 6=0
···
X
Z
X
0 ···
0 ···
...
Z
1
···
mk−l 6=0 γ∈P̃k−1,l \ SLk−1

1 0 ···
 1 · · ·


F


1
0
0


u1,k−1 u1k

u2,k−1 u2k  
γ
..
.. 
k−1

z
.
. 


1

1
uk−1,k 
1
· e(−mk−1 uk−1,k − · · · − mk−l uk−l,k−l+1 )d× u
Z 1
Z 1
X
X
X
+
···
···
mk−1 6=0
m1 6=0 γ∈Uk−1 \ SLk−1
0
0


1 u12 · · · u1k


1 · · · u2k  γ



F
z
.
.

. . .. 
1 
1

· e(−mk−1 uk−1,k − · · · − m1 u12 )d× u.
The last term is called the nondegenerate term ND(F ). The other terms are the degenereate term D(F ). We need the orthogonality of these Fourier coefficients.
3
Proposition 7.
Z
D(F ) · ND(F )d× z = 0
Pk−1,1
\η k
Proof. Since
[
γ
1
γ∈Uk−1 \ SLk−1
· Pk−1,1 \η k ∼
= Uk (z)\η k ,
the integral unfolds as
Z
D(F ) · ND(F )d× z
Pk−1,1 \η k
=
X
···
m1 6=0
Z
X Z
mk−1 6=0
Z
1
···
D(F )(z)
Uk (Z)\η k
1
0
0
 
1 u12 · · · u1k

1 · · · u2k  

z
e(−m1 u12 − · · · − mk−1 uk−1,k )d× ud× z
F
.. 
..



.
.
1
Z ∞
Z 1
Z 1
X
XZ ∞
···
···
=
···
D(F )(z)e(m1 x12 + · · · + mk−1 xk−1,k )d× x

m1
mk−1
y1 =0
y2n−1 =0
x12 =0
xk−1,k =0
 

1 u12 · · · u1k
Z 1
Z 1

1 · · · u2k  
×
×


·
···
F
.. 
..
 y  e(−m1 u12 − · · · − mk−1 uk−1,k )d ud y.

.
.
0
0
1
The first integral
Z
1
Z
···
0
1
D(F )(z)e(m1 x12 + · · · + mk−1 xk−1,k )d× x
0
can be shown to be 0 by induction.
Applying this to the automorphic function
Z ∞ A ˆ∗
dA
g
EA (z, s) ,
β
A
0
we get
Z ∞
Z ∞
X
2
I |L(1 + 2nit, f × f )|
···
y1 =0
y2n−1 =0 N ≤p≤2N



Z 1
1 · · · u1,2n
Z 1Z ∞
A dA
.
.
∗ 
ˆ


..
..
···
EA
y, 1 + it g
β A
0
0
0
1
2
det z
× · e(−pu2n−1,2n − u2n−2,2n−1 − · · · − u12 )d u ψ
d× y.
δ
4
The u-integral picks up the nondegenerate (1, · · · , 1, p) coefficients of EˆA∗ , which are related
to Jacquet’s Whittaker function, so we need Stade’s formula.
Lemma 8 (Stade).
Z ∞
Z ∞
|W (y, β)|2 | det y|w d× y
···
0
0
−2
m−1
1
1
π aw 2b Y Y
1
+
β
−
β
m−k
m−k+j
= mw π − 2 − 2 (βm−k −βm−k+j ) Γ
2
Γ 2 j=1 j≤k≤m−1
m Y
m
Y
w + βj − βk
·
.
Γ
2
j=1 k=1
Here (β1 , · · · , βm ) are the Langlands parameters of W (y, β).
Fact 9. The (1, · · · , 1, p) Fourier coefficient of E z, 12 + it is
λ(p)σnit (p)
2n−1
2
where σs (n) :=
P
W (my, α0 )
1
L(1 + 2int, f × f )
p
d|n d and λ(p) is the p-th Hecke eigenvalue of f .
s
Now we want to get a lower bound for
X
|λ(p)σnit (p)|2 .
N ≤p≤2N
3. Lower bounds for Hecke eigenvalues
Lemma 10. For full density of primes N ≤ p ≤ 2N ,
1
σnit (p) .
log log t
Here N ≥ t2 .
In his GL(2) paper, Sarnak used σnit (p) 1 for a positive proportion of primes.
Proof. We have
|σnit (p)| = |p2int + 1| ≥ | cos 2nt log p + 1|.
If 2nt log p is close to (2m + 1)π, then the prime is bad. The condition
1
|2nt log p − (2m + 1)π| ≤ √
log log t
holds for p in short intervals only. Counting primes in short intervals by sieve theory, we see
that the bad primes only contribute to lower order terms.
What about λ(p)? If it gets too small, we have no hope.
Lemma 11. For tempered Hecke–Maass f for SLn (Z), there exist at least
in [N, 2N ] such that the p-th Hecke eigenvalue satisfies
1
|λ(p)| .
log log t
5
1
N
100n2 log N
primes
This is the only place where we need f to be tempered, i.e. to satisfy the Ramanujan
conjecture.
Proof. By Liu–Wang–Ye,
X
Λ(n)|λ(n)|2 ∼ N
N ≤n≤2N
(
log p if n = pk ,
where Λ(n) =
Here |λ(p)| ≤ n by the Ramanujan conjecture.
0
otherwise.
Suppose the conclusion is wrong. Then we split the sum into
X
X
Λ(p)|λ(p)|2 +
Λ(p)|λ(p)|2
good p
≤ n2
≤
bad p
1
N
+
Λ(p) 2
log N N ≤P ≤2N
log log t
X
50n2
N
N
+
2
50 log N
log log t
which is a contradiction.
Hence
X
|λ(p)σnit (p)|2 f
N ≤p≤2N
Corollary 12. I N
.
log N
β n δ −1
.
log t
The p-th Hecke eigenvalue of E is λ(p)σnit (p). Recall that
φ(z) := h(z)ns f (m1 )f (m2 )
and
E(z, s) =
X
φ(z)|γ .
For the Rankin–Selberg L-function L(s, f × f ), the eigenvalue is λ(p)ps + λ(p)p−s , which
we can bound by taking out λ(p). For L(s, f1 × f2 ) with two different f1 and f2 , the upper
bound still works but we haven’t found a trick to obtain a lower bound yet.
4. Reference
Some background materials on the Maass–Selberg relations are available on Paul Garrett’s
website, in the following two papers:
• Simplest Example of Truncation and Maass–Selberg Relations 1,
• Truncation and Maass–Selberg Relations 2.
1Available
2Available
at http://www-users.math.umn.edu/~garrett/m/v/maass_selberg_trivial.pdf.
at http://www.math.umn.edu/~garrett/m/v/maass_selberg.pdf.
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