MATH G9905 RESEARCH SEMINAR IN NUMBER THEORY
(FALL 2014)
LECTURE 1 (SEPTEMBER 9, 2014)
DORIAN GOLDFELD
MAASS–SELBERG RELATIONS FOR EISENSTEIN SERIES ON GL(2)
NOTES TAKEN BY PAK-HIN LEE
1. Introduction
This is a research seminar that will go on for the entire year. This is aimed for graduate
students who are working on a project to present their work, but postdocs and faculty are
welcome too. There will occasionally be outside speakers.
I will give two lectures on the Maass–Selberg relations — the first on GL(2) and the second
on GL(n). This is joint work with Xiaoqing Li, who will continue my lectures. In 1899, de
la Vallée-Poussin proved that
ζ(s) 6= 0
c
for σ > 1 − log(|t|+2) , where s = σ + it and c > 0 is a constant. This is equivalent to
c
ζ(1 + it) >
.
log(|t| + 2)
The constant c is effective. This result was improved by his joint work with Hadamard.
In a paper by Jacquet–Shalika, they showed that if π is an automorphic cuspidal representation for GL(n, A), then
L(1 + it, π) 6= 0
for t ∈ R. In the last ten years, people have been trying to make this effective. Now we can
prove that
c
L(1 + it, π) >
.
log(|t| + 2)
This is due to Sarnak–Gelbart.
P a(n)
What about the Rankin–Selberg L-functions L(s, π × π)? If L(s, π) =
, then on
ns
GL(2) we have
X a(n)2
L(s, π × π) =
.
ns
(The formula is more complicated for groups other than GL(2), but can be made explicit
using Euler products.) We don’t know that π × π is automorphic unless π is on GL(2)
(Ramakrishnan). Thus we can’t use the theorem of Jacquet–Shalika. Brumley proved that
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L(1 + it, π × π) .
(2 + |t|)N
Last updated: October 9, 2014. Please send corrections and comments to [email protected].
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Recently, Gelbart–Lapid proved that
1
(2 + |t|)N
where the πi ’s are on any reductive groups. This result is computable but the bounds are
very bad. This is the best of what’s known. The proof of this result uses the Maass–Selberg
relations.
L(1 + it, π1 × π2 · · · × πr ) Theorem 1 (G.–Li). Let π be a cuspidal automorphic representation on GL(n, AQ ). Assume
π is tempered at all places. Then
c
L(1 + it, π × π) >
.
(log(2 + |t|))3+
The proof has not been written up yet. This is the first result on a logarithmic lower
bound. One of the main ingredients is the Maass–Selberg relations, which first appeared in
Gelbart–Sarnak.
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In the relative trace formula, we need to integrate terms of the form L(1+it)
, so it is
important to know that L(1 + it) is non-zero. This is why we are interested in the lower
bounds for L(1 + it).
2. Maass–Selberg Relation (Simplest Case)
∗ ∗
Let G = GL(2, R), K = O(2, R), Γ = SL(2, Z), P =
⊂ G be the parabolic
0 ∗
1 ∗
∗ 0
subgroup, N =
⊂ G be the nilpotent radical, and M =
⊂ G be the Levi
0 1
0 ∗
subgroup. Let F = Γ\G/(K · R× ).
Given a function φ : G/(K · R× ) → C satisfying
φ(ng) = φ(g)
for all n ∈ N and g ∈ G, we can construct the Eisenstein series
X
E(g, φ) :=
φ(γg).
γ∈(P ∩Γ)\Γ
s
The classical Eisenstein series is when we take φs (g) = y for g =
y x
, y > 0, x ∈ R.
0 1
Then
E(g, s) = E(g, φs ) =
X
c,d∈Z
(c,d)=1
ys
|cz + d|2s
= y s + cs y 1−s + higher terms in (∗)e2πilx (l 6= 0)
where
ζ ∗ (2s − 1)
ζ ∗ (2s)
s
s
ζ ∗ (s) = π − 2 Γ
ζ(s).
2
cs =
and
2
Definition 2 (Constant term along
a parabolic P ). Let φ : G/(K · R× ) → C. Then the
∗ ∗
constant term along P =
is cP φ where
0 ∗
Z 1 1 u
φ
g du.
cP φ(g) :=
0 1
0
On GL(n), there are more parabolics and the constant terms get more complicated.
For example, we have
cP E(g, s) = y s + cs y 1−s .
It is well-known that the Eisenstein series is not square-integrable:
2
ZZ
dxdy
y x
E
,
s
y 2 = ∞.
0 1
Γ\G/(K·R× )
R
R
This is due to the presence of the constant term: either |y s |2 or |y 1−s |2 is not integrable.
The Petersson inner product is
ZZ
hf1 , f2 i :=
f1 (g)f2 (g)d× g.
F
Maass and Selberg had the idea of cutting off the constant term, so that we can integrate
the Eisenstein series. They proposed
(
E(g, s)
if 0 ≤ y ≤ T,
E T (g, s) =
E(g, s) − cP E(g, s) if y > T.
One problem is that this modified Eisenstein series is not continuous and not automorphic.
But if we simply restrict to the fundamental domain F, hE T , E T i is well-defined. The original
proof is by Green’s theorem, but it isn’t the right way to do it. Arthur proposed another
way to do the truncation.
3. Arthur’s Truncation of Eisenstein series
y x
Let T > 0. As usual we write g =
for g ∈ G/(K · R× ).
0 1
Definition 3 (Truncated constant term). For φ : G/(K · R× ) → C, we put
(
0
if 0 ≤ y ≤ T,
cTP φ(g) :=
cP φ(g) if y > T.
Definition 4 (Truncation operator).
ΛT φ(g) := φ(g) −
X
cTP (γg).
γ∈Γ∩P \Γ
Remark. If φ is automorphic, i.e. φ(γg) = φ(g), then ΛT φ(g) is also automorphic.
Let me prove some properties of this truncation operator.
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Definition 5 (Siegel set Sδ ). Let δ > 0. Then
1
1
Sδ := x + iy : − ≤ x ≤ , y > δ .
2
2
Proposition 6. Let T ≥ 1. Then ΛT E(g, s) is of rapid decay on any Siegel set Sδ with
δ > 0.
Proof. We know that E(g, s) − cP E(g, s) is of rapid decay in Sδ . Let g ∈ F. Then
!


c E(g, s) if γ = 1 0 and y > 1,
P
cTP E(γg) =
0 1


0
otherwise.
This implies that ΛT E(g, s) is of rapid decay on F. Since ΛT E is automorphic, it is of rapid
decay everywhere.
Let s, r ∈ C, and Es = E(g, s). Then hΛT Es , ΛT Er i is well-defined.
Theorem 7 (Maass–Selberg Relation). Let r, s ∈ C with s 6= r and s + r 6= 1. Then
T r−s
T s+r−1
T s−r
T 1−s−r
+ cs
+ cr
+ cs cr
.
s+r−1
r−s
s−r
1−s−r
In the remainder of this lecture I will give Arthur’s proof. Next time we will generalize
this to higher rank Eisenstein series, but the main idea of the proof is exactly the same.
hΛT Es , ΛT Er i =
Proof. The first step is hΛT Es , ΛT Er i = hΛT Es , Er i. We need to prove
*
+
X
I := ΛT Es ,
cTP E(γg, r) = 0
γ∈Γ∩P \Γ
This can be proved by Rankin–Selberg. Unraveling the sum, we get
X
ZZ
dxdy
y x
y x
T
T
I=
Λ E
,s
cP E γ
,r
0 1
0 1
y2
F
γ
Z 1Z ∞
dxdy
y x
y x
T
T
=
Λ E
.
, s cP E
,r
0 1
0 1
y2
0
0
For 0 ≤ y ≤ T , the second term is 0; for y > T , the second term is y r + cr y 1−r but the first
term is a power series in e2πilx with l > 0. So the integral is 0.
The final step is a computation. Unfolding,
Z 1Z ∞
dxdy
T
hΛ Es , Er i =
(y s − cTP E(g, s)) · E(g, r) 2
y
0
0
Z T Z 1
Z ∞ Z 1
dxdy
dxdy
=
y s E(g, r) 2 −
cs y 1−s E(g, r) 2
y
y
y=0 x=0
y=T x=0
Z T
Z ∞
dy
dy
=
y s (y r + cr y 1−r ) 2 − cs
y 1−s (y r + cr y 1−r ) 2
y
y
0
T
which is easy to compute.
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If φ is cusp form on GL(n), it induces a cuspidal automorphic representation on the Levi of
GL(2n). Xiaoqing Li will explain how to use this to get a lower bound for L(1 + it, φ × φ) 1
.
(log t)3+
Reference: Paul Garrett, Simplest Example of Truncation and Maass–Selberg Relations 1.
1Available
on his “Vignettes” webpage at http://www-users.math.umn.edu/~garrett/m/v/maass_
selberg_trivial.pdf.
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