Analytic Theory of Modular Forms
Spring 2012, taught by Jacob Tsimerman.
Contents
1 Elliptic Functions
2
2 Hecke Operators
4
3 L-Functions Associated to Modular Forms
8
4 The Functional Equation for the Zeta Function
10
5 Smooth Sums and Sharp Sums
12
6 The Approximate Functional Equation
15
7 Poincaré Series
17
7.1
The Fourier Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
7.2
Properties of Jk−1 and K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
8 Congruence Subgroups
22
9 The Theta Function
25
10 Quadratic Forms
27
11 Harmonic Polynomials
34
12 Bounds for Fourier Coefficients
37
13 Metaplectic Groups and Half-Integral Weight Hecke Operators
38
14 Bounds in the Half-Integral Weight Case
41
14.1 Trivial Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
14.2 A Toy Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
14.3 A Return to the Original Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
14.3.1 Estimate I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
14.3.2 Estimate II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
14.3.3 The Final Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
1
15 Heegner Points and Geodesics
52
16 Maass Forms
53
17 Subconvexity of L-Functions
64
18 The Burgess Estimate
69
19 Amplification
70
20 Applications of Subconvexity
78
1
Elliptic Functions
w1
Given w1 , w2 ∈ C with w
∈
/ R, let Λ be Zw1 ⊕ Zw2 . f is elliptic if f is meromorphic on C and
2
Λ-periodic. Define EΛ = C/Λ; then elliptic functions are meromorphic on EΛ .
An example of an elliptic function is
1
℘(z; Λ) = 2 +
z
X
w∈Λ\{0}
1
1
− 2
2
(z − w)
w
.
(1.1)
The poles of ℘ consist precisely of a double pole at 0 (mod Λ).
P
Define Gk (Λ) = w∈Λ w1k . Then for |z| < minw∈Λ\{0} |w|, we have
℘(z) = z −2 +
X
(k + 1)Gk+2 (Λ)z k .
(1.2)
k≥1
After differentiating and comparing principal parts, let g2 = 60G4 (Λ) and g3 = 140G6 (Λ). Then
= 4℘(z)3 − g2 ℘(z) − g3 . We also find that for k > 6, Gk is a polynomial in G4 and G6 with
integer coefficients, by equating higher terms.
Observe that Gk (λw1 , λw2 ) = λ−k Gk (w1 , w2 ), and if ac db ∈ SL2 (Z), then Gk (aw1 + bw2 , cw1 +
dw2 ) = Gk (w1 , w2 ). Let Gk (z) = Gk (1, z); then
℘0 (z)2
Gk (
az + b
) = (cz + d)k Gk (z).
cz + d
(1.3)
For H the upper half-plane, Gk can be viewed as a function H → C. f : H → C is a modular
function of weight k if f satisfies (1.3) for every ac db ∈ SL2 (Z). If f is also holomorphic and
“bounded at ∞”, we say f is a modular form.
Define F = {z ∈ H : |z| > 1, <(z) < 21 }. Then F is a fundamental domain for SL2 (Z) on H.
That is, for every z ∈ H, there exists γ ∈ SL2 (Z) such that γ · z ∈ F ∪ ∂F, and if z1 , z2 ∈ F such
that there exists γ ∈ SL2 (Z) with γ · z1 = z2 , then z1 = z2 .
If f is a modular function, then f (z + 1) = f (z), so f is a function of q = e2πiz defined for
q ∈ D \ {0}. “Bounded at ∞” means this function can be extended to q ∈ D.
2
k
zeros modulo SL2 (Z). What this means
Lemma 1.1. f (z) modular of weight k implies f (z) has 12
1
is that for p ∈ C, define m(p) = 2 |stabSL2 (Z) (p)|, so m(i) = 2 and m(ρ) = 3. Then define mf (p) =
ordp f and mf (∞) = ordq=0 f (q). Then we have
X
p
(mod SL2 (Z))
mf (p)
k
= .
m(p)
12
(1.4)
f0
f dz
over ∂F, with appropriate tweaking.
k
−2
a b
Alternatively, d( az+b
∈ SL2 (Z), so f (z)(dz) 2 is invariant under
c d
cz+d ) = (cz + d) dz for
SL2 (Z), so descends to SL2 (Z)\H.
Sketch of proof. Integrate
Denote the space SL2 (Z)\H by Y0 (1), and let X0 (1) = Y0 (1) ∪ {∞}, so X0 (1) is a compact
Riemann surface of genus 0. Let Ω be the line bundle of complex differential 1-forms on X0 (1).
k
k
Then f (z)(dz) 2 induces a global meromorphic section of Ω 2 .
dq
; near i, w = (z − i)2 so dw = 2(z − i)dz;
In local coordinates: near ∞, q = e2πiz so dz = 2πiq
near ρ, w = (z − ρ)3 so dw = 3(z − ρ)2 dz. We obtain

deg
k
Ω2
k
2
(f (z)(dz) ) = 

X
p6=i,ρ,∞
=
X mf (p)
p
m(p)
As X0 (1) has genus 0, deg
Corollary 1.2.
k
mf (p) + mf (∞) −
+
2
k
Ω2
−
mf (i) −
2
k
2
!
13k
.
12
+
mf (ρ) − k
3
(1.5)
(1.6)
= −k, implying the desired result.
• G4 (z) is modular of weight 4, so G4 has ρ as its only zero modulo SL2 (Z).
• G6 has i as its only zero modulo SL2 (Z).
Define ∆(z) = g2 (z)3 − 27g3 (z)2 , so that ∆ is modular of weight 12 and not identically zero (as
∆(i) 6= 0). Furthermore,
∆(∞) = (60 · 2ζ(4))3 − 27(140 · 2ζ(6))2 = 0.
(1.7)
So ∆ is nonvanishing in C.
3
2 (z))
so j is a modular function of weight 0, having a single pole at ∞,
Now define j(z) = (12g
∆(z)
so j descends to a map X0 (1) → P1 of degree 1, therefore an isomorphism.
Theorem 1.3. For k even,
 k


 12
dim Mk (SL2 (Z)) = 
k


+1
12
k≡2
(mod 12)
(1.8)
otherwise
Moreover, Mk (SL2 (Z)) is spanned by the Ga4 Gb6 , where 4a + 6b = k.
3
To do There’s a gap here. Fill in? (1)
2
Hecke Operators
Recall that for L the space of lattices in C, we have an equivalence between modular functions of
weight k under SL2 (Z) and G : L → C such that for λ ∈ C× , G(λΛ) = λ−k G(Λ). Now for n ∈ N,
define
X
(Tn G)(Λ) = nk−1
G(Λ0 ).
(2.1)
Λ0 ⊆Λ
[Λ:Λ0 ]=n
Lemma 2.1.
• If gcd(n, m) = 1, then Tn ◦ Tm = Tnm .
• For p prime, Tp ◦ Tpa = Tpa+1 + pk−1 Tpa−1 .
Proof. The first part is easy to show. For the second, consider lattices up to scale, L/R× , and draw
an edge between Λ1 and Λ2 if there exists λ ∈ R× with λΛ2 ⊆ Λ1 and [Λ1 : λΛ2 ] = p. (Then
[Λ2 : λp Λ1 ] = p, so the graph is symmetric.) Then Tpa consists of averaging over points a away.
Alternatively, we could split up the sum for Tpa+1 , keeping track of homogeneity.
Remark. The graph above can be thought of as P GL2 (Qp )/P GL2 (Zp ).
Corollary 2.2. Tpa ∈ Z[Tp ].
Corollary 2.3. Tn ◦ Tm = Tm ◦ Tn for every m, n. In fact,
Tn ◦ Tm =
X
dk−1 T mn
.
2
(2.2)
d
d| gcd(m,n)
T = Z[Tn ]n∈N is called the Hecke algebra.
Let M2 (n) ⊆ M2 (Z) be the matrices of determinant n. Then the space of index n sublattices is
parameterized by
[ a b SL2 (Z)\M2 (n) =
.
0 d
(2.3)
ad=n
b mod d
Now if f is a modular function of weight k, then
(Tn f )(z) := (Tn F )h1, zi = n
k−1
X
F hd, az + bi = n
k−1
X
ad=n
ad=n
b mod d
b mod d
−k
d
f
az + b
d
.
(2.4)
Corollary 2.4. T preserves the spaces Mk of modular forms and Sk of cusp forms of weight k.
4
Now if f (z) =
P∞
m=1 af (m)e(mz),
then
∞
X nk−1 X
maz + mb
(Tn f )(z) =
,
af (m)e
d
dk
(2.5)
m=1
ad=n
b mod d
which, after summing over b first, gives
∞
X
af (m)
m=0
X
ak−1 e
maz d
ad=n
.
(2.6)
d|m
So for fn = Tn f , we have
∞
X
X
af (m) =
m=0
dk−1 af
mn d|(m,n)
d2
.
(2.7)
In particular, afn (1) = af (n).
Remark. These operators exist since for α ∈ M (n), αSL2 (Z)α−1 ∩ SL2 (Z) has finite index in
SL2 (Z). The set of such α is called COM (SL2 (Z)) and equals SL2 (Q). But for “most” discrete
Γ ∈ SL2 (R), COM (Γ) = Γ.
Recall that ∆ generates S12 , so is an eigenfunction of T. We get (Tn ∆)(z) = τ (n)∆(z). It follows
11
that τ is multiplicative, and if αp satisfies τ (p) = p 2 (αp + αp−1 ), then
τ (pa ) = p
11a
2
a
X
αpa−2b .
(2.8)
b=0
As a result, the Ramanujan conjecture for ∆ is equivalent to |αp | = 1.
Claim. For each δ > 0, if |τ (n)| = O (nδ+ ), then |τ (n)| ≤ σ0 (n)nδ .
This follows from the above structure: assume |αp | ≤ 1. Then |τ (pa )| ≤ (a + 1)|αp |a p
11a
2
. Now
−(a+2)
τ (pa ) = p
If |αp | > 1, then this is Ωp (p
11a
2
11a
2
·
αpa+2 − αp
αp2 − αp−2
.
(2.9)
|αp |a ).
Now we want a basis of Sk or Mk consisting of eigenfunctions of T. To get existence, it is enough
to show that each Tn is semisimple.
2
2
Consider H with the metric dx y+dy
, invariant under SL2 (R), and the invariant measure
2
For f (z), g(z) ∈ Mk , with f or g in Sk , define the Petersson inner product
Z
f (z)g(z)y k
hf, gik =
SL2 (Z)\H
Lemma 2.5. If g ∈ Sk , then hg, Ek i = 0.
5
dx dy
.
y2
dx dy
.
y2
(2.10)
Proof.
Z
hg, Ek i =
g(z)
SL2 (Z)\H
Z
= ζ(k)
g(z)
F
yk
dx dy
k
(cz + d) y 2
(c,d)6=(0,0)
X
(2.11)
yk
dx dy
.
k
y2
(cz
+
d)
(c,d)=1
X
(2.12)
Define h(z) = g(z)y k . Then
h
az + b
cz + d
= (cz + d)k g(z)
yk
yk
=
g(z)
.
|cz + d|2k
(cz + d)k
∼
Define Γ∞ = {( 01 n1 ) , n ∈ Z}; then Γ∞ \SL2 (Z) −
→ {(c, d) = 1} via
a b
c d
(2.13)
7→ (c, d).
Now our integral is
Z
X
ζ(k)
h(γz)
F γ∈Γ \SL (Z)
∞
2
dx dy
= ζ(k)
y2
Z
X
h(z)
γ∈Γ∞ \SL2 (Z) γF
Z
= 2ζ(k)
Z
h(z)
∞Z
=
0
0
Γ∞ \H
1
g(z)y k
dx dy
y2
dx dy
y2
dx dy
y2
=0
(2.14)
(2.15)
(2.16)
(2.17)
since g is a cusp form.
Lemma 2.6. For f, g ∈ Sk , hTn f, gi = hf, Tn gi.
Proof. Define Lp = {(Λ, Λ0 ) : Λ0 ⊆ Λ, [Λ : Λ0 ] = p}, and let wp : Lp /R× → Lp /R× be the
involution (Λ, Λ0 ) 7→ (Λ0 , pΛ). We also have a map H → Lp /C× by z 7→ (h1, zi, h1, pzi). We have
Lp ,→L × L via the natural embedding. We obtain a map H → L/C× × L/C× . Defining Γ0 (p) =
a b ∈ SL (Z), p|c , then Γ (p) preserves h1, zi and h1, pzi. Now Γ (p)\H (L → L/C× ×
2
0
0
p
c d
L/C× ) with wp =
have
0 −p−1/2
p1/2
0
hf, Tp gi = p
k−1
1
. (Note that wp (z) = − pz
, and that wp Γ0 (p)wp−1 = Γ0 (p).) Now we
Z
(2.18)
f (z)y k
(2.19)
Γ0 (p)\H
= pk−1
Z
Γ0 (1)\H
dx dy
y2
!
p−1 X
z+i
dx dy
g
+ pk−1 g(pz)
.
p
y2
f (z)g(pz)y k
i=0
1
Now setting w = − pz
, we get
6
pk−1
Z
f
Γ0 (p)\H
=p
k−1
1
1
v k du dv
−
g −
pw
w pk |w|2k v 2
Z
f (pw)(p|w|k )g(w)(|w|k )
Γ0 (p)\H
= pk−1
Z
f (pw)g(w)v k
Γ0 (p)\H
v k du dv
pk |w|2k v 2
du dv
v2
(2.20)
(2.21)
(2.22)
= hTp f, gi.
(2.23)
As the Tp generate T, we get the desired result.
The lemma implies that there exist {f1 , . . . , fk } an orthogonal basis of Sk such that fi is an
eigenfunction of every Tn , called a Hecke cusp form.
Theorem 2.7 (Multiplicity One). If f, g ∈ Sk are Hecke cusp forms with the same eigenvalues
under all of the Tn , then f and g differ by a constant.
This is true since Tn determines the Fourier coefficients.
For f a Hecke cusp form, let λn (f ) =
Tn f
f .
Define Mk (Z) ⊆ Mk to be the set of modular forms having integral Fourier coefficients. Since
E4a E6b are in Mk (Z) after an appropriate scalar multiplication, Mk (Z) is a lattice in Mk . And T
preserves Mk (Z), so the λn (f ) are algebraic integers.
Corollary 2.8. For (λn (f ))n∈N the Hecke eigenvalues for f ∈ Sk , and γ ∈ Gal(Q/Q), then there
exists f γ ∈ Sk such that λn (f γ ) = (λn (f ))γ .
The square-root cancellation heuristic: for a1 , . . . , aN complex numbers which are “random” and
|ai | ∼ 1, then
N
X
√
ai ∼ N .
(2.24)
i=1
Theorem 2.9. Fix f ∈ Sk . Then
N
X
|af (n)| f n
k+1
2
.
(2.25)
f (x + iy)e(−nx) dx,
(2.26)
n=1
That is, the sum is Of (n
k+1
2
).
Proof. |f (z)|2 y k = O(1). Then bound af (n) by
e
−2πiny
Z
af (n) =
1
0
trivially bounding, and optimizing with respect to y.
7
A better bound can be obtained using Parseval’s identity:
1
Z
|f (θ)|2 dθ =
0
X
|an |2 .
(2.27)
n∈Z
Applying to the above, we get
∞
X
|af (n)|2 e−4πy =
1
N,
|f (x + iy)|2 dx y −k .
(2.28)
0
n=1
For y =
1
Z
we get
N
X
|af (n)|2 e−
4πn
N
N k =⇒
n=1
N
X
|af (n)|2 N k .
(2.29)
n=1
Then Cauchy implies the theorem.
3
L-Functions Associated to Modular Forms
Suppose f is a Hecke cusp form of weight k. Define the (analytically normalized) L-function
L(f, s) =
∞
X
λf (n)
n=1
ns+
k−1
2
.
(3.1)
This converges absolutelt for <(s) > 1 by the above. Because λf (n) is multiplicative,
∞
Y X
λf (pn )
L(f, s) =
p
Letting λf (p) = p
k−1
2
n=0
pn(s+
!
k−1
)
2
.
(3.2)
−1
(αp,f + αp,f
),
!
n k−1
2
n
X
λf (p ) = p
−b
a
αp,f
αp,f
.
(3.3)
a+b=n
We finally end up with
L(f, s) =
1
Y
p
(1 − αp,f
p−s )(1
−1 −s
− αp,f
p )
.
Here is how to obtain a meromorphic continuation and functional equation:
holomorphic for every s. It equals
8
(3.4)
R∞
0
f (iy)y s dy
y is
Z
∞
∞X
0
af (n)e
−2πny s dy
y
y
n=1
=
∞
X
n=1
∞
X
Z
∞
af (n)
e−2πny y s
0
dy
y
Z ∞
af (n)
dy
−s
e−y y s
=
(2π)
s
n
y
0
n=1
k−1
(2π)−s Γ(s).
= L f, s −
2
(3.5)
(3.6)
(3.7)
But f ( yi ) = f (iy)(iy)k , so
∞
Z
0
∞
i
dy
f (iy)y
f
y s−k
=i
y
y
y
Z0 ∞
dy
= ik
f (iy)y k−s
y
0
k+1
= L f,
− s (2π)s−k Γ(k − s).
2
s dy
k
Z
(3.8)
(3.9)
(3.10)
We conclude that
L(f, s)(2π)
k−1
−s
2
k−1
k+1
k
s− k+1
2 Γ
= i L(f, 1 − s)(2π)
−s .
Γ s+
2
2
(3.11)
L is also holomorphic everywhere since the integral and Γ1 are. We also find that L(f, s) has
k+1
“trivial zeros” at − k−1
2 , − 2 , . . .. The generalized Riemann Hypothesis asks if all other zeros of
1
L(f, s) lie on <(s) = 2 .
Theorem 3.1 (Strong Multiplicity One). If f, g are Hecke cusp forms of weight k succh that for
all but finitely many primes p (∀0 p) λf (p) = λg (p), then f and g differ by a constant factor.
Proof. Let R(s) =
L(f,s)
L(g,s)
=
Qr
i=1 Ri (s),
Ri (s) =
for
−1 −s
(1 − αpi ,f p−s
i )(1 − αpi ,f pi )
−1 −s
(1 − αpi ,g p−s
i )(1 − αpi ,g pi )
.
(3.12)
We have R(s) = R(1 − s). As a result, it turns out that for every i, Ri (s) = Ri (1 − s). (We can
expand the Dirichlet series, and need a finite product to expand for s < 12 .) Letting X = p−s
i , the
rational function
(1 − αpi ,f X)(1 − αp−1
X)
i ,f
(1 − αpi ,g X)(1 − αp−1
i ,g X)
(3.13)
pi
is invariant under X 7→ X
. Either we get {αpi ,f , αp−1
} = {αpi ,g , αp−1
} (we’re done if this is true
i ,g
i ,f
pi
for every i), or the poles of Ri (s) are invariant under X 7→ X
, which is impossible.
9
The subconvexity problem: for F a family of L-functions L(ϕi , s), we want to understand
|L(ϕi , 12 )|. This is typically done for ϕi the Hecke cusp forms of weight k, or consider |L(f, 12 + it)|
as t → ∞ for f a Hecke cusp form.
For σ > 1, |L(f, σ + it)| is bounded as a function of t. By the functional equation, for σ < 0,
|L(f, σ + it)| ∼ |t|ψ(σ) for some explicit linear function ψ. We can then interpolate to get |L(f, σ +
it)| ∼ |t|ψ0 (σ) .
4
The Functional Equation for the Zeta Function
For f ∈ L2 (R), define fb by fb(y) =
Fact. If for every j ≤ A,
dj f
dxj
R∞
Then fb ∈ L2 (R), and f 7→ fb is an isometry.
−∞ e(xy)f (x) dx.
∈ L1 ∩ L∞ , then fb(y) = O((1 + |y|)−A ).
For integration by parts gives fb0 (y) = −2πifb(y).
P
P
Let g(x) = n∈Z f (x + n), so g ∈ C 0 (R/Z). We have g(x) = n∈Z gb(n)e(nx) for
Z
gb(n) =
Z
g(x)e(−nx) dx =
f (x)e(−nx) dx = fb(−n).
R/Z
(4.1)
R
By taking x = 0, we recover
Theorem 4.1 (Poisson Summation Formula).
X
X
f (n) =
fb(n).
n∈Z
(4.2)
n∈Z
2
Apply this to ht (x) = e−πtx . We have
Z
b
ht (y) =
e
R
= e−
πy 2
t
= e−
πy 2
t
= e−
πy 2
t
ix2 t
+ xy dx
2
Z 2
ix t
iy 2
e
+ xy −
dx
2
2t
R
!
Z
it
iy 2
x−
dx
e
2
4t
R
Z
2
e−πtx dx
=(x)==( iy
)
4t
πy 2
e− t
= √
t
By a contour argument, the integral is
Z
2
=(x
R
Re
√
t)==( iy
)
4t
−πx2
e−πx dx.
(4.3)
(4.4)
(4.5)
(4.6)
(4.7)
dx = 1, so we have
2
πy
h 1 (y)
e− t
t
b
.
ht (y) = √ = √
t
t
10
(4.8)
Now for =(z) > 0, define
θ0 (z) =
X
2z
eπin
n∈Z
=
X
h−iz (n).
(4.9)
n∈Z
By Poisson, we get
1 X
1 1
i 2
i 2
h i (n) =
θ0 −
.
z
2
2
z
(4.10)
n∈Z
θ0 is a modular form of half-integral weight.
R∞
s
Now for <(s) > 1, 0 θ0 (iy)−1
y 2 dy
2
y converges. It equals
∞ Z
X
∞
2
s
e−πn y y 2
n=1 0
s
s
dy
= π 2 ζ(s)Γ
.
y
2
But we can also express the left hand integral by splitting into
(4.11)
R∞
1
and
1
Z ∞
θ0 (iy) − 1 s dy
θ0 (iy)y 2 − 1 − s dy
2
y
+
y 2
2
y
2
y
1
1
Z ∞
dy
1−s
1
1
θ0 (iy) − 1 s
y2 + y 2
+
− .
=
2
y
s−1 s
1
Z
R1
0
. We obtain
∞
(4.12)
(4.13)
This is meromorphic everywhere with poles only at 0 and 1 and symmetric under s 7→ 1 − s. We
get
s
2
π ζ(s)Γ
s
An exercise is to show that ζ(s) =
√
2
1
s−1
=π
1−s
2
ζ(1 − s)Γ
1−s
2
.
(4.14)
+ γ + · · · as s → 1.
For σ ∈ R, define ψ(σ) = inf{r : |ζ(σi t)| |t|r }. For σ > 1, ψ(σ) = 0. Also |Γ(σ + it)| ≈
1
π|t|
2π|t|σ− 2 e− 2 for each σ and t → ∞. This and the functional equation gives us, for σ < 0,
ψ(σ) = 12 − σ. In fact, for every σ, ψ(σ) = ψ(1 − σ) + 12 − σ.
Theorem 4.2. ψ(σ) is convex in σ.
Lemma 4.3. Suppose f (s) is holomorphic for a ≤ σ ≤ b, |f (s)| ≤ M on σ = a and σ = b, and
α
|f (σ + it)| = O(e|t| ). Then |f (s)| ≤ M for a ≤ σ ≤ b.
m
Proof of lemma. Pick m ≡ 2 mod 4 with m > |α| and look at f (s)e−s . This decays at ∞, so we
m
can apply the maximum modulus principle. We get |f (s)| ≤ maxσ=a,b M |es | for every , and then
take → 0.
Proof of theorem. Let L(s) =
lemma.
b−s
b−a ψ(a)
+
a−s
a−b ψ(b)
and consider ζ(s)(−is)−L(s) , and then apply the
11
The theorem implies that for 0 ≤ σ ≤ 1, ψ(σ) ≤
Hypothesis is that ψ( 21 ) = 0.
1
2
− σ2 . In particular, ψ( 12 ) ≤ 14 . The Lindelof
Theorem 4.4. The Riemann Hypothesis implies the Lindelof Hypothesis.
Proof. Assuming Riemann, log(ζ(s)(s − 1)) is holomorphic for <(s) > 12 . And for any σ >
<(log(ζ(s)(s − 1))) ≤ C log |t|. By complex analysis, we obtain | log(ζ(s)(s − 1))| log |t|.
1
2,
(Borel-Catheodory: if f (z) is holomorphic on |z| < R with |f (z)| ≤ M on |z| < r, and sup <(f ) ≤
N , then |f (z)| ≤ ϕ(M, N, r, R) for some ϕ.)
Then convexity gives | log(ζ(s)(s−1))| (log |t|)γ for some γ < 1. Exponentiating gives Lindelof.
5
Smooth Sums and Sharp Sums
1
For <(s) > 1, ζ(s)
=
Riemann Hypothesis.
P∞
µ(n)
n=1 ns .
We can then understand partial sums of µ(n), assuming the
Mellin Transform: For ϕ(x) ∈ C 0 (R+ ),
∞
Z
ϕ(s)
e =
ϕ(x)xs
0
dx
x
(5.1)
converges for a < σ < b. For every c with a < c < b,
1
2πi
Z
σ=c
−s
ϕ(s)x
e
dx = ϕ(x).
(5.2)
(After substituting x = ey , this becomes Fourier inversion.)
j
Suppose that for j ≤ A, ddxϕj ∈ L1 ∩ L∞ (R+ ). Then |ϕ(σ
e + it)| = O((1 + |t|)−A ). In particular,
if ϕ ∈ Cc∞ (R+ ), then |ϕ(σ
e + it)| = O((1 + |t|)−A ) for every A.
Theorem 5.1. The Riemann Hypothesis is true if and only if for every ϕ ∈ Cc∞ (R+ ),
∞
X
µ(n)ϕ
n
x
n=1
1
= O (x 2 + ).
(5.3)
Proof. For the “only if” implication, consider
1
2πi
Z
σ=a
∞
X
1
1
s
ϕ(s)x
e
dx =
µ(n) ·
ζ(s)
2πi
=
n=1
∞
X
n=1
for a where
and |t| large.
1
ζ(s)
µ(n)ϕ
Z
σ=a
ϕ(s)
e
n
12
(5.4)
(5.5)
x
converges. Assuming Riemann, we can take a =
xs
ds
ns
1
2
+ and bound for t near 0
For the “if” implication, given the bound, then
∞X
Z
µ(n)ϕ
n
0
x
x−s
dx
x
(5.6)
converges for σ > 12 , so we get ζ −1 (s)ϕ(s)
e
holomorphic for σ > 21 . But for every s, there exists
ϕ ∈ Cc∞ (R+ ) with ϕ(s)
e 6= 0. So ζ −1 (s) is holomorphic for σ > 12 .
P Givenn a sequence (an )n∈N , we can either consider
n an ϕ( x ). Here are some examples:
• For an = 1,
P
n≤x 1
P
n≤x an ,
or choose ϕ ∈ Cc∞ and consider
= bxc = x + O(1), while
X
X n
= xϕ(0)
b +x
ϕ(nx)
b
= xϕ(0)
b + OA (x−A ).
ϕ
x
n
(5.7)
n6=0
P
1
• Let r2 (n) = #{a, b ∈ Z : a2 + b2 = n}. Then n r2 (n) = πx + O(x 2 ). The Gauss circle
problem is to lower the exponent as much as possible (it is believed that 14 + is possible).
But
X
r2 (n)ϕ
n
x
n
=
X
ϕ
a,b∈Z
a2 + b2
x
.
(5.8)
For ψ(a, b) = ϕ(a2 + b2 ), we get
b 21 , dx 12 ) = xψ(0,
b 0) + OA (x−A ).
xψ(cx
X
b 0) +
xψ(0,
(5.9)
(c,d)6=(0,0)
• Suppose f ∈ Sk . Then
Z
X an (f ) n 1
ϕ
=
L(f, s)x−s ϕ(s)
e ds
k−1
x
2πi
σ=2
n n 2
Z
1
=
L(f, s)x−s ϕ(s)
e ds
2πi σ=−A
= OA (x−A ).
(5.10)
(5.11)
(5.12)
Computing sharp sums is in fact more difficult than computing smooth sums, in that sharp sum
estimates imply smooth sum estimates. For if ϕ ∈ Cc∞ (R+ ), then
(5.13)
N
X
n X
N +1
N δ
an ϕ
N ϕ
−ϕ
xδ .
n
x x
x (5.14)
n
So if
PN
n=1 an
N
X X
! N +1
N
an
ϕ
−ϕ
.
x
x
X
an ϕ
n
x
=
N
n=1
= O(xδ ), then
n=1
13
Theorem 5.2. The Riemann Hypothesis holds if and only if
X
1
µ(n) = O (x 2 + ).
(5.15)
n≤x
We only need to prove the “only if” implication.
Perron’s Formula: for δ > 0 and y > 0,
1
2πi


1
ys
ds = 0

s
1
Z
σ=1+δ
y>1
0<y<1
y=1
(5.16)
ds X
=
µ(n).
s
(5.17)
2
Given this,
1
2πi
Z
ζ −1 (s)xs
σ=1+δ
n≤x
Proposition 5.3 (Quantitative Perron). For δ > 0 and y > 0,
Z
1+δ+iT
1+δ−iT

1+δ y


1 + O

s
T log y
y
ds =
1+δ 
s
y


O
T log y
y>1
(5.18)
0<y<1
Proof. For y > 1, consider
To do Rectangular contour. (2)
Integrating over the rectangle is 1. The top is bounded by
1+δ
Z
−A
yσ
1+δ
dσ ≤
.
T
T log y
(5.19)
The bottom is similar. For the left, we get at most
Z
T
−T
y −A
dt → 0
|A + it|
(5.20)
as A → ∞.
Proof of theorem. Pick y to be a half-integer. Then
Z
1+δ+iT
ζ
1+δ−iT
We now estimate
−1
(s)y
s ds
s
=
X
n≤y
∞
X
(y/n)1+δ
µ(n) + O
T | log(y/n)|
.
(5.21)
n=1
(y/n)1+δ
y 1+δ
n
n=1 T | log(y/n)| . If 1 ≤ y ≤ 2 , then | log(y/n)| ≥ log 2, so we get O( T ζ(1 +
1+δ
get O( y T ζ(1 + δ)). Finally, for y2 ≤ n ≤ 2y, | log ny | = | log(1 + n−y
y )| ∼
P∞
δ)). If n ≥ 2y, we also
1
| n−y
y | ≥ 2y , so
!
14
2y
X
x= y2
We could get
y 1+δ log y
T
Z
2y
X y 2+δ
(y/n)1+δ
y2
≤2
≤
.
1+δ T
T | log(y/n)|
T
y n
(5.22)
n= 2
by a more careful approximation. So
1+δ+iT
ζ
−1
(s)y
s ds
s
1+δ−iT
=
X
µ(n) + Oδ
n≤y
y 2 y 1+δ
+
T
T
.
(5.23)
Now consider the contour
To do Rectangle (3)
Assuming Riemann, the integral over the rectangle is 0. Now
Z
top, bottom
Z
1+δ
|ζ −1 (σ + it)|y σ
σ= 12 +δ
dσ
,
T
(5.24)
1+δ
which by Riemann again is O ( Ty 1− ). And
Z
left
Z 1 +δ+iT
2
1
1
ds
|ζ −1 (s)|y 2 +δ
= O (y 2 +δ T ).
1
|s|
2
(5.25)
+δ−iT
We obtain
X
1
µ(n) = O (y 2 +δ T ) + Oδ
n≤y
y 2 y 1+δ
+
T
T
.
(5.26)
1
For T = y 2 , we get O (y 2 + ). (Had we instead only known that |ζ −1 (σ + iT )| O (T a+ ), we
2
1+δ
1
would get O (y 2 +δ T a ) + Oδ ( yT + Ty 1−a ) and would optimize accordingly.)
6
The Approximate Functional Equation
For 0 < <(s0 ) < 1, choose ϕ(x) ∈ Cc∞ (R+ ). The idea is to consider
1
2πi
Z
σ=2
ϕ(s0 + s)xs ϕ
e
ds X −s0 n =
n ϕ
.
s
x
n
(6.1)
This also equals
1
2πi
ds x−s0 ϕ(−s0 )
ζ(s0 + s)xs ϕ(s)
e
−
+ ϕ(0)ζ(s
e
0 ).
s
s0
σ=−2
|
{z
}
Z
(?)
Now (?) equals
15
(6.2)
Z
ζ(s0 − s)x
−
−s
σ=−2
ds (?)
ϕ(−s)
e
=
s
Z
π
1+s−s0
)
2
ζ(1
s0 −s
Γ( 2 )
−1−2s+2s0 Γ(
σ=2
+ s − s0 )x−s ϕ(−s)
e
ds
.
s
(6.3)
Fix Me (4)
We need to estimate
1
2πi
Z
y −s
σ=2
0
Γ( 1+s−s
)
ds
2
ϕ(−s)
e
.
s0 −s
s
Γ( 2 )
(6.4)
s
Let Λ(s) = ζ(s)π 2 Γ( 2s ), and choose ϕ such that ϕ(x) = ϕ( x1 ), so that ϕ(−s)
e
= ϕ(s).
e
Then
1
2πi
Z
Λ(s + s0 ) s
ds
x−s0 ϕ(s
e 0)
Λ(s0 )
1
x
ϕ(s)
e
=
−
+ ϕ(0)
e
s0
s0
s0 +
Γ( 2 )
s
s0 Γ( 2 )
Γ( 2 ) 2πi
Z
σ=−2
|
Λ(s + s0 ) s
ds
x ϕ(s)
e
.
s0
Γ( 2 )
s
{z
}
(6.5)
(?)
(?) equals
Z
−
σ=2
ds
Λ(s0 − s) −s
x ϕ(s)
e
=−
s0
Γ( 2 )
s
Z
σ=2
Λ(1 + s − s0 ) −s
ds
x ϕ(s)
e
.
s0
Γ( 2 )
s
(6.6)
We need to understand
1
Vs0 (y) =
2πi
Lemma 6.1.
Z
σ=2
Γ( s02+s ) −s
ds
y ϕ(s)
e
.
s0
Γ( 2 )
s


!σ0 −δ 

y




1 + Oδ  p


1 + |s0 |


Vs0 (y) =
!−A 




 p y



OA
1 + |s |
(6.7)
0≤y≤1
(6.8)
y≥1
0
and Vs00 (y) = Oδ
√ y
1+|s0 |
σ0 −δ !
for 0 ≤ y ≤ 1.
Proof.
Γ( s0 +s ) σ0 +σ− 12
s +s
s
− π2 (|=( 02 )|−|=( 20 )|) |t0 + t|
2
.
≈e
1
Γ( s20 ) |t0 |σ− 2
(6.9)
We have | t02+t | − | t20 | ≤ |t|. Now
1
Vs0 (y) = 1 +
2πi
Z
σ=−σ0 +δ
16
Γ( s02+s ) −s
ds
y ϕ(s)
e
s0
Γ( 2 )
s
(6.10)
1
and for t small, the second integral is
Vs0 (y) =
|t0 +t|δ− 2
|t0 |
1
2πi
σ0 −δ
−σ0 +δ− 1
2
Z
σ=A
Oδ ( yσ0 −δ ). Also
t0
Γ( s02+s ) −A
ds
y ϕ(s)
e
.
Γ( s20 )
s
(6.11)
For Vs00 (y), differentiating eliminates the pole at s = 0.
For x > 0, we have
ζ(s0 )π
s0
2
X
=
n
−s0
V s0
n
x
n
1−s0
+
X
n
s0 −1
n
x1−s0 π 2
V1−s0 (nx) +
.
Γ( s20 )
√
Pick s0 = 21 + iT . Then the two sums have length x T and
(roughly) that
ζ
1
+ iT
2
π
=
Γ(
iT
2
1
+iT
2
2
+
1
X
)
n
=
X
n
1
+iT
2
Vs0 (n) +
√
T
x ,
1
X
n
n
so both
1
−iT
2
√
(6.12)
T if x = 1. We get
Vs0 (n).
(6.13)
Morally,
ζ
1
+ iT
2
1
n− 2 −iT +
√
n≤ T
X
1
n− 2 +iT ,
(6.14)
√
n≤ T
so estimating,
X
1
1
ζ 1 + iT ≤ 2
n− 2 ∼ T 4 .
2
√
(6.15)
n≤ T
In order to improve this, we need to exploit some cancellation in the above sums.
7
Poincaré Series
Fix a positive even integer k and write jk ac db , z = (cz +d)−k . For γ ∈ SL2 (Z), write (f |k γ)(z) =
f (γz)jk (γ, z) so that f ∈ Mk ⇐⇒ f |k γ = γ for every γ ∈ SL2 (Z).
Also, for γ1 , γ2 ∈ SL2 (Z), we have the cocycle relation jk (γ1 γ2 , z) = jk (γ1 , γ2 z)jk (γ2 , z).
Corollary 7.1. SL2 (Z) acts on C ∞ (H) by f 7→ f |k γ.
Recall that
Ek (z) =
1
2
X
jk (γ, z) =
γ∈Γ∞ \Γ
1
2
X
(1|k γ)(z),
(7.1)
γ∈Γ∞ \Γ
so we see easily that Ek is a modular form. More generally, for m ≥ 0, let em (z) = e(mz), and
define the Poincaré series
17
pm,k (z) =
1
2
X
(em |k γ)(z).
(7.2)
γ∈Γ∞ \Γ
For m ≥ 0 and k > 2, pm,k ∈ Mk .
Claim. If m > 0, then pm,k ∈ Sk .
Proof.
pm,k (z) = e(mz) +
X
−k
(cz + d)
c>0
az + b
e m
cz + d
.
(7.3)
d∈Z
As =(z) → ∞, e(mz) → 0 and
P
c>0
d∈Z
Lemma 7.2. For f ∈ Sk , hf, pm,k ik =
Proof. For dµ(z) =
dx dy
,
y2
|cz + d|−k → 0, so pm,k (z) → 0.
af (m)
Γ(k
(4πm)k−1
we have
1
hf, pm,k ik =
2
Z
1
=
2
Z
=
Z
Γ\H
0
apm,k (n)
nk−1
=
(7.4)
f (γz)em (γz)=(γz)k dµ(z)
(7.5)
Γ\H γ∈Γ \Γ
∞
f (z)e(mz)y k
dx dy
y2
f (z)e(mz)y k
0
dx dy
y2
af (m)
Γ(k − 1).
(4πm)k−1
apn,k (m)
mk−1
em (γz) jk (γ, z)y k dµ(z)
γ∈Γ∞ \Γ
X
=
=
X
f (z)
Γ∞ \H
Z ∞Z 1
Corollary 7.3.
− 1).
(7.6)
(7.7)
(7.8)
.
This follows from hpm,k , pn,k ik = hpn,k , pm,k ik .
Corollary 7.4. The pm,k over m > 0 span Sk .
As an exercise, it can be shown that for d = dim Sk , (p1,k , . . . , pd,k ) is a basis for Sk .
Let hf1 , . . . , fd i be the orthonormal Hecke basis for Sk . Then we have
pm,k (z) =
d
X
hpm,k , fi ifi (z),
i=1
so
18
(7.9)
d
X
afi (m)afi (n)
apm,k (n) = Γ(k − 1)
.
(4πm)k−1
(7.10)
i=1
In particular, the apm,k (n) are real. Also,
d
Γ(k − 1) X
apm,k (m) =
afi (m)2 .
(4πm)k−1
(7.11)
i=1
7.1
The Fourier Expansion
We have
1+iy
Z
apm,k (n) =
pm,k (z)e(−nz) dz

iy
(7.12)

1+iy 
X

az + b
e(mz) +
=
(cz + d)−k 
e m
 e(−nz) dz

cz + d
iy
c>0
Z
(7.13)
(c,d)=1
X Z
= δm,n +
1+iy
az + b
e m
− nz (cz + d)−k dz.
cz + d
iy
c>0
(7.14)
(c,d)=1
Now
az+b
cz+d
=
a
c
−
ad−bc
c(cz+d)
δm,n +
=
X Z
a
c
1
c(cz+d) ,
−
1+iy
e
iy
c>0
so we get
ma
m
−
− nz (cz + d)−k dz
c
c(cz + d)
(7.15)
(c,d)=1
= δm,n +
Z
X
e
=(z)=y
c>0
ma
m
−
− nz (cz + d)−k dz
c
c(cz + d)
(7.16)
d∈(Z/cZ)×
z7→z− dc
=
δm,n +
X
c>0
e
ma nd
+
c
c
Z
m
(cz)−k e − 2 − nz dz.
c z
=(z)=y
(7.17)
d∈(Z/cZ)×
Define the Kloosterman sums
X
K(m, n; c) =
x∈(Z/cZ)×
e
mx + nx
c
.
(7.18)
xx=1
Also define the Bessel functions
k
Jk−1 (4πx) = (2πi)
Z
x
z −k e −xz −
dz.
z
=(z)=y
19
(7.19)
So we have the formula
√
k−1
X
4π mn K(m, n; c)
2
k n
(2πi)
apm,k (n) = δm,n +
Jk−1
.
m
c
c
(7.20)
c>0
7.2
Properties of Jk−1 and K
• Jk−1 maps R to R.
• Jk−1 (x) k min{xk−1 , √1x }.
• |K(m, n; c)| ≤ c.
k
Corollary 7.5. apm,k (n) = O (n 2 + ).
This is almost Hecke’s bound.
More properties of K:
1. K(m, n; c) = K(n, m; c).
2. If (a, c) = 1, then K(am, n; c) = K(m, an; c).
3. If d|(m, n, c), then K(m, n; c) =
ϕ(c)
m n c
ϕ( dc ) K( d , d ; d ).
4. If c = c1 c2 where (c1 , c2 ) = 1, then K(m, n; c) = K(mc1 , nc1 ; c2 )K(mc2 , nc2 ; c1 ) where, as
before, the bar denotes multiplicative inverse.
5. An exercise (due to Selberg):
K(m, n; c) =
X
dK
mn
d|(m,n,c)
d2
, 1;
c
.
d
(7.21)
As an example, consider k = 12. For m > 0, pm,12 (z) = cm ∆(z) where
cm =
hpm,12 , ∆i12
Γ(11)
τ (m)
=
,
h∆, ∆i12
(2πm)11 h∆, ∆i12
(7.22)
(n)
so apm,12 (n) = v τ (m)τ
for some constant v. Now (5) implies
m11
apm,k (n) = m1−k
X
dk−1 ap1,k
d|(m,n)
mn d2
,
(7.23)
so we find that
τ (m)τ (n) = v −1 m11 am,12 (n) =
X
d|(m,n)
recovering the Hecke identity.
Now consider K(1, m; pa ) for p a prime.
20
d11 τ
mn d2
,
(7.24)
Lemma 7.6. For p a prime, |K(1, m; p2a )| ≤ 2pa .
Proof. x0 + pa y ≡ x0 − x0 2 pa y (mod p2a ), so
X
X
x0 mod pa
y mod pa
2a
K(1, m; p ) =
e
x0 + mx0
p2a
1 − mx0 2
e y
p2a
(7.25)
p-x0
=p
X
a
e
p-x0
x0 + mx0
p2a
(7.26)
x20 ≡m mod pa
with the sum having at most two terms.
1
Lemma 7.7. |K(1, m; p2a+1 )| ≤ 2pa+ 2 for a > 0.
Proof. x0 + pa+1 y = x0 − x0 2 pa+1 y, so by the same method as before,
K(1, m, p
2a+1
)=p
X
a
e
x0 mod pa+1 , p-x0
x20 ≡m mod pa
x0 + mx0
p2a+1
.
(7.27)
Now x0 + pa z = x0 − x0 2 pa z + x0 3 p2a z 2 , so we get
p
a
X
x0 mod pa
e
x0 + mx0
p2a+1
a
X
!
p
p z(1 − mx0 2 ) + x0 3 p2a z 2
e
.
p2a+1
(7.28)
z=1
x20 ≡m mod pa
To get the desired bound, we use:
P
√
p
nx2 ×
Lemma 7.8. For n ∈ (Z/pZ) , x=1 e( p ) = p.
Proof. We have
p 2
2
p
X
X
nx2 nx − ny 2
e
e
.
=
p p
x=1
(7.29)
x,y=1
For p > 2, take a = x − y and b = x + y to get
p
X
a,b=1
e
nab
p
21
= p.
(7.30)
It remains to deal with K(1, m; p). The Weil Conjectures imply K(1, m; p) = αp + βp where
√
√
|αp | = |βp | = p, so |K(1, m; p)| ≤ 2 p. We’ll prove this result soon.
k
1
The result is that apm,k (n) = O (n 2 − 4 + ).
1
3
Theorem 7.9. |K(1, a; p)| < 2 4 p 4 .
Proof. Let Vp =
Pp−1
a=1 |K(1, a; p)|
4.
Note that
(
−1
a 6≡ 0 (mod p)
K(0, a; p) =
p − 1 a ≡ 0 (mod p)
(7.31)
Now by property (2),
(p − 1)Vp =
p−1
X
|K(a, b; p)|4
(7.32)
|K(b, a; p)|4 − 2(p − 1) − (p − 1)4 .
(7.33)
a,b=1
=
p−1
X
a,b=0
The above sum equals
p−1
X
p−1
X
e
a,b=0 x,y,z,w=1
2
b(x + y − z − w) + a(x + y − z − w)
p
= p · #{x, y, z, w ∈ (Z/pZ)× : x + y = z + w, x + y = z + w}.
(7.34)
(7.35)
This occurs if and only if either x = −y and z = −w, or x + y = z + w and xy = zw (implying
{x, y} = {z, w}). We end up with #{ } = (p − 1)(2p − 5), so
Vp = p2 (2p − 5) − 2 − (p − 1)3 < 2p3 .
8
(7.36)
Congruence Subgroups
ϕ
∗∗
Fix N ∈ N, and let Γ(N ) = ker(SL2 (Z) −−N
→ SL2 (Z/N Z)), Γ0 (n) = ϕ−1
N ({( 0 ∗ )}), and Γ1 (N ) =
−1
ϕN ({( 10 ∗1 )}). Define Y (N ) = Γ(N )\H and define Y0 (N ) and Y1 (N ) similarly.
22
Fact.
Y
1
1− 2
p
p|N
Y
1
1+
|Γ : Γ0 (N )| = N
p
p|N
Y
1
2
1− 2
|Γ : Γ1 (N )| = N
p
|Γ : Γ(N )| = N 3
(8.1)
(8.2)
(8.3)
p|N
Lemma 8.1. Let F be a fundamental domainS
for Γ, and γ1 , . . . , γd left representatives of Γ0 \Γ, for
Γ0 a subgroup of Γ of finite index. Then F 0 = di=1 γi F is a fundamental domain for Γ0 .
As an example, for Γ0 = Γ0 (N ), SL2 (Z/N Z) acts on P1 (Z/N Z) = {[x, y] : x, y ∈ Z/N Z, (x, y) =
a b
1}/(scaling by (Z/N Z)× ). Then Γ0 (N
) = stab([0 : 1]). So Γ → Γ1 00(N )\Γ is c d = [c : d]. For the
0 −1
case N = p, we can take γi = 1 i for 1 ≤ i ≤ p and γp+1 = ( 0 1 ).
To do Fundamental domain picture (5)
Now look at the cusps. For H ⊆ P1 (C), ∂(H) = P1 (R), and SL2 (Z) acts transitively on P1 (Q).
For Γ0 ⊆ SL2 (Z), Γ0 \P1 (Q) are the cusps of Γ0 \H.
Suppose that c ∈ Γ0 \P1 (Q). Then there exists σc ∈ SL2 (Z) such that σc (∞) = c, so σc−1 F is a
fundamental domain taking c to ∞ (and two arcs meeting at c to vertical lines).
For Γc = Γ0 ∩ σc Γ∞ σc−1 , Γc is precisely the set of matrices in Γ0 preserving c. Define X0 (N ) =
Y0 (N ) ∪ Γ0 (N )\P1 (Q).
S
Fact.
• P1 (Q) =
Γ0 (N ) · uv where the union is disjoint, taken over v|N , (u, v) = 1, and
N
u mod (v, v ).
• Determine the genus of X0 (N ).
To do (6)
Now for k ∈ N, possibly odd, and Γ0 ⊆ Γ, define Mk (Γ0 ) to be the set of holomorphic f : H → C
such that for every γ ∈ Γ0 , f |k γ 0 = f , and f is holomorphic at the cusps. That is, for any c ∈
Γ0 \P1 (Q),
(f |k σc )(z) =
X
af,c
n≥0
n nz e
m
m
(8.4)
for m = [Γ∞ : Γ∞ ∩ σc−1 Γ∞ σc ].
We say that f ∈ Sk (Γ0 ) if for every c ∈ Γ0 \P1 (Q), af,c (0) = 0.
We know that Γ1 (N ) Γ0 (N ) with Γ0 (N )/Γ1 (N ) ∼
= (Z/N Z)× . So for χ : (Z/N Z)× → C× , let
Mk (Γ0 (N ), χ) = f ∈ Mk (Γ1 (N )) : f |k
L
Then Mk (Γ1 (N )) = χ Mk (Γ0 (N ), χ).
a b
c d
= χ(a)f ∀
a b
c d
∈ Γ0 (N ) .
(8.5)
We can extend the construction of th Poincaré series as follows: for k > 2, define
pm,N,k =
1
2
X
(em |k γ)(z),
pcm,N,k =
γ∈Γ∞ \Γ0 (N )
23
1
2
X
(em |k σc−1 γ)(z).
γ∈Γc \Γ0 (N )
(8.6)
c .
Denote pc0,N,k by EN,k
Theorem 8.2.
•
c (c0 )
EN,k
• If m > 0, then pcm,N,k ∈ Sk (Γ0 (N )).
(
1
=
0
c = c0
.
otherwise
Proof. First suppose c 6= c0 . Then lim=(z)→∞ (pcm,N,k |k σc0 )(z) = pm,N,k (c0 ). But we have
(pcm,N,k |k σc0 )(z) =
1
2
X
(em |k σc−1 γσc0 )(z).
(8.7)
γ∈Γc \Γ0 (N )
For c 6= c0 , 1 6∈ σc−1 Γ0 (N )σc0 , so
1
c
(pm,N,k |k σc0 )(z) ≤
2
X
|jk (γ, z)|
(8.8)
γ∈Γ∞ \SL2 (Z)
γ6=1
which tends to 0 as =(z) → ∞.
Now if c = c0 , then


(pcm,N,k |k σc )(z) = e(mz) + O 


X
γ∈Γ∞ \SL2 (Z)

|jk (γ, z)|
.
(8.9)
γ6=1
The second sum tends to 0 as =(z) → ∞, while e(mz) tends to 1 if m = 0 and 0 if m > 0.
L c
· C.
Corollary 8.3. For k > 2, Mk (Γ0 (N )) = Sk (Γ0 (N )) ⊕ c EN,k
If f ∈ Sk (Γ0 (N )) and g ∈ Mk (Γ0 (N )), define
Z
N hf, gik
f (z)g(z)y k
=
Γ0 (N )\H
We have
L
c
c EN,k
dx dy
.
y2
(8.10)
· C = Sk (Γ0 (N ))⊥ by unfolding. And we have the identity
N hf, pm,N,k ik
=
Γ(k − 1)
af (m)
(4πm)k−1
(8.11)
for every f ∈ Sk (Γ0 (N )).
The Fourier expansion of pm,N,k is given by
apm,N,k (n) = δm,n +
P
n>0 apm,N,k (n)e(nz),
n k−1 X
2
m
c>0
N |c
24
Jk−1
where
√
4π mn K(m, n; c)
.
c
c
(8.12)
1
Iwaniec’s trick: Sk ,→ Sk (Γ0 (N )) by inclusion. We have N kf kk = [Γ : Γ0 (N )] 2 kf kk . Let
f
1 . Take {f1 , . . . , fd } an orthonormal basis for Sk . Then for some gi ,
Nf
=
[Γ:Γ0 (N )] 2
{N f1 , . . . ,N fd , g1 , . . . , gr }
(8.13)
is an orthonormal basis for Sk . Then
pm,N,k
Γ(k − 1)
=
(4πm)k−1
d
X
afi (m)
i=1
[Γ : Γ0 (N )] 2
1
fi +
r
X
!
.
agi (m)gi
(8.14)
i=1
Taking mth Fourier coefficients,
1+
X
Jk−1
c>0
4πm
c
d
r
X
X
|afi (m)|2
+
|agi (m)|2
[Γ : Γ0 (N )]
K(m, m; c)
Γ(k − 1)
=
c
(4πm)k−1
i=1
N |c
(8.15)
m δ−1
Nm
|afi (m)|2
,
N 1+o(1)
or 1 +
mδ
N
1
2
so (N + mδ ) |afi (m)|.
For the left hand side, the length of the sum is about
−1
2 m
m
would expect the sum to behave as √ mN =
N
9
.
i=1
Theorem 8.4. If |K(m, m; c)| ≤ cδ for some δ < 1, then 1 +
|afi (m)|2
,
N 1+o(1)
!
1
and each term is about m− 2 , so we
m
N,
√1 .
N
The Theta Function
√
P
πim2 z , we had θ (z + 2) = θ (z) and θ (− 1 ) =
Recall that for θ0 (z) =
−izθ0 (z). Fix
0
0
0
m∈Z e
z
γ = ac db ∈ SL2 (Z) with a ≡ b ≡ 0 (mod 2). Then θ0 ( az+b
)
is
given
by
cz+d
θ0
a
1
−
c c(cz + d)
=
2
X
e
2
iπm
iπ mc a − c(cz+d)
=
X
eπi
m2 a
c
m
2
c
cz+d
X
m
2
e−iπ( c +t)
c
cz+d
.
(9.2)
t∈Z
m mod c
Now let f (t) = e−iπ( c +t)
(9.1)
m∈Z
2
. Then by scaling and shifting by e−iπt ,
1
1
ms
d
2
fb(s) = (ic)− 2 (cz + d) 2 e2πi c +iπs (z+ c ) .
(9.3)
Apply Poisson summation to get
θ0
az + b
cz + d
1
1
= (ic)− 2 (cz + d) 2
X
m mod c
Now for a = 2α,
25
eiπ
m2 a
c
X
s∈Z
e2πi
ms
+iπs2 (z+ dc )
c
.
(9.4)
X
eiπ
m2 a
+2πi ms
c
c
m mod c
αm2 + ms
c
m mod c
2
X
m + ms
m=αn
e α
=
c
m mod c
X
α(m + s · 2)2
4s2 α
=
e
e −
c
c
m mod c
X
αm2 −iπ s2 d
ad≡1 mod c
c
=
e
e
c
m mod c
X
s2 d
αm2
=
e
e−iπ c .
c
X
=
e
(9.5)
(9.6)
(9.7)
(9.8)
(9.9)
m mod c
Fix Me Maybe deal with the overset alignments? (7)
Let β(α, c) =
z 7→ − z1 , we get
αm2
m mod c e( c ).
P
θ0
1
1
−2
Then θ0 ( az+b
(cz + d) 2 β(α, c)θ0 (z). After mapping
cz+d ) = (−ic)
bz − a
dz − c
=
β(α, c)
c
1
2
1
(dz − c) 2 θ0 (z).
(9.10)
We will now compute the Gauss sum β(α, c). We have the following properties:
• If c = c1 c2 with gcd(c1 , c2 ) = 1, then β(α, c) = β(αc1 , c2 )β(αc2 , c1 ).
• If c = pn with p prime and n > 1, then β(α, pn ) = pβ(α, pn−2 ).
• If gcd(x, p) = 1, then β(αx2 , c) = β(α, c).
• If x is such that xp = −1, then β(1, p) + β(x, p) = 0.
So it suffices to determine β(1, p).
Theorem 9.1.
Sketch.
( 1
p2
β(1, p) =
1
ip 2
p ≡ 1 (mod 4)
p ≡ 3 (mod 4)
(9.11)
Fix Me We may have confused notation. (8)
2
This proof works even for composite p. Let f (x) = e( xp ) for 0 ≤ x ≤ p and 0 otherwise. Then
we have
p
X
i=0
#
p−1
X
X Z p
1
1
= f (0) +
f (i) + f (p) = lim
f (x)e(−mx) dx
N →∞
2
2
0
!"
0
f (i)
i=1
|m|≤N
so that
26
(9.12)
p
X Z
β(1, p) = lim
N →∞
= lim
N →∞
0
|m|≤N
X
e
x2
− mx
p
dx
(9.13)
1
Z
e(p(m2 − mx)) dx
p
(9.14)
0
|m|≤N
Z 1 m2 p
m 2
pe −
= lim
e p x−
dx
N →∞
4
2
0
|m|≤N
Z 1− m
X
2
m2 p
pe −
= lim
e(px2 ) dx
m
N →∞
4
−
2
|m|≤N
Z ∞
e(px2 ) dx
= p(1 + i−p )
−∞
Z ∞
√
= p(1 + i−p )
e(x2 ) dx.
X
(9.15)
(9.16)
(9.17)
(9.18)
−∞
By plugging in p = 1, we get
R∞
2
−∞ e(x ) dx
i
1+i ,
=
−p
√
so β(1, p) = i p 1+i
1+i .
√
So forodd c, let c be 1 for c ≡ 1 (mod 4) and i for c ≡ 3 (mod 4). Then β(α, c) = c c
where αc is the Jacobi symbol. Therefore if b ≡ c ≡ 0 (mod 2),
θ0
az + b
cz + d
=
2c
d
−1
1
d (cz + d) 2 θ0 (z).
Now let θ(z) = θ0 (2z). Then ∀, γ ∈ Γ0 (4), if γ =
1
2
where j 1 (γ, z) = dc −1
d (cz + d) .
a b
c d
α
c
(9.19)
, then we have θ(γz) = j 1 (γ, z)θ(z),
2
2
For k ∈ Z, 4|N , we define M k (Γ0 (N ), χ) to be the set of f : H → C such that f (γz) =
2
χ(γ)j 1 (γ, z)k f (z), and f is holomorphic on H and at the cusps.
2
10
Quadratic Forms
Now suppose Q is a positive definite quadratic form on Zg such that ∀ v ∈ Zg , Q[v] ≡ 0 (mod 2).
Theorem 10.1. For N ∈ N, A a positive definite symmetric matrix with even diagonal entries,
such that N A−1 has integral entries (so (Zg )∗ ⊆ N −1 Zg , with the dual being with respect to A),
then define
θQ (z) =
X
e
m∈Zg
1
Q[m]z
2
where Q[v] = v T QV . Then θQ ∈ M g (Γ0 (2N ), χQ ) for χQ
2
Let rQ (n) = #{m ∈ Zg : Q[m] = 2n}. Then
27
(10.1)
a b
c d
=
2g ·|A|
d
.
θQ (z) =
X
rQ (n)e(nz).
(10.2)
n≥0
We have that
g
1 X
1
rQ (n) =
#{m ∈ ZG : Q[m] ≤ 2M } ∼ cQ M 2 −1 .
M
M
(10.3)
n≤M
But if f is a cusp form of weight g2 , then
g
1
1 X
|af (n)| ∼ M 4 − 4 .
M
(10.4)
n≤M
We’ll later find Siegel’s mass formula, which determines the Eisenstein series part of θQ (z), so
it would remain to deal with the cusp form part.
P
2
As an exercise, n≡a (b) q n is a modular form of weight 21 for some congruence group Γ. The
P
2
same holds for n≡a (b) q rn for r ∈ Q, r > 0. So a product of k such sums will be a modular form
of weight k2 .
We can treat quadratic forms either as a positive definite matrix Q in Zk , or a lattice L in Rk .
(Here L = Q[Zk ].) If L is a lattice, then
θL =
X
1
q 2 hv,vi =
v∈L
X
exp(−πthv, vi)
(10.5)
v∈L
for q = e2πiz and z = it. To be invariant under some integer translation, we want our quadratic
form to have rational values.
θL also satisfies a Poisson summation formula: in general,
X
1
Fb(v ∗ )
vol(Rk /L) ∗ ∗
v ∈L
v∈L
Z
Fb(x∗ ) =
e(hx∗ , xi)F (x) dx.
X
F (v) =
(10.6)
(10.7)
Rk
We obtain
θL =
X
1
− k2
t
exp(−(πt−1 hv ∗ , v ∗ i)).
vol(Rk /L)
∗
∗
(10.8)
v ∈L
We can also consider the restricted sum over a “shifted lattice” v ≡ v0 (mod L0 ) where [L :
L0 ] < ∞.
Also, we have θL1 ⊕L2 = θL1 θL2 . The exercise will show that shifted rectangular lattices are
modular forms. But then any lattice L has a rectangular sublattice L0 of finite index. Then choose
appropriate L0 .
28
The same will hold for
lattices). As an example,
P
L ω(v)q
1
hv,vi
2
1
η = q 24
for ω(v) L0 -periodic (this is a linear combination of shifted
∞
Y
(1 − q n ) =
n=1
1 2
1X
χ(n)q 24 n
2
(10.9)
n∈Z
for some χ by Euler’s pentagonal number formula.
We have
X
v
ω(v)F (v) =
X
1
ω
b (v ∗ )Fb(v ∗ )
|Rk /L0 | ∗ ∗
(10.10)
v ∈L0
for ω
b the discrete Fourier transform. Also,
1
η 3 = q 8 (1 − 3q + 5q 3 − 7q 6 + 9q 10 − · · · ) =
for some character χ. Also, this shows that in weight
expected weight.
3
2
1 2
1X
χ(n)nq 8 n
2 n
(10.11)
or 12 , coefficients can have larger than
Fix Me This part appears in the notes after equidistribution. May want to organize differently? (9)
Let Q be a positive definite quadratic form over Z in d variables. For each m ∈ N, let Sm = {v ∈
d−1
given by v 7→ √ v . This can either be
Zd : Q(v) = m}. We have a measure ϕ : Rd \ {0} → SQ
Q(v)
d−1
induced from the metric given on Q, or the µ-unique invariant
thought of as the µ-measure on SQ
measure under
OQ (R) = {A ∈ GLn (R) : ∀ v, Q(Av) = Q(v)}.
(10.12)
Then:
d
Theorem 10.2. Given {mi } ∈ N, assume rQ (mi ) mi2
equidistributed with respect to µ.
−1−
, and d ≥ 4. Then ϕ(Smi ) become
For R any ring, a quadratic form over R is a symmetric matrix Q ∈ Md (R), with Q(v) = v T Qv.
We say that Q1 ∼R Q2 if there exists γ ∈ GLn (R) such that γQ1 γ T = Q2 .
Theorem 10.3 (Hasse-Minkowski). If Q1 , Q2 are quadratic forms over Q in d variables, then
Q1 ∼Q Q2 if and only if Q1 ∼R Q2 and Q1 ∼Qp Q2 for every prime p.
For Q1 , Q2 quadratic forms over Z in d variables, we say that Q1 ∈ genus(Q2 ) if Q1 ∼R Q2 and
Q1 ∼Zp Q2 for every prime p.
Remark. This is weaker than equivalence. For example, consider Q1 = ( 1 14 ) and Q2 = ( 2 7 ).
Theorem 10.4. For any quadratic form Q over Z, genus(Q) (as a set of equivalence classes) is
finite.
29
e ∈ genus(Q), then det Q
e = det Q. Then use Minkowski’s
Proof in case Q is positive definite. If Q
d
e
e
induction theory: Pick v1 ∈ Z such that Q(v1 ) = minv∈Zd \{0} Q(v), and inductively choose vi+1 =
e
minv∈Zd \{0} Q(v)
such that hv1 , . . . , vi+1 i is a primitive sublattice of Zd (can be extended to a basis).
Fact.
e d
det Q
d
Y
e i ).
Q(v
(10.13)
i=1
e j | are all bounded, and determine Q.
e As they are integers, there are finitely
As a result, |viT Qv
many options.
Now let L be a lattice over Z and Q a quadratic form on L. For each p, let Lp be LT
⊗Z Z(p) , a
lattice over Z(p) , and all contained in the Q-space LQ = L ⊗Z Q. We can recover L as p Lp . For
L0 ⊆ LQ another lattice, we have L0p = Lp ∀0 p (for all but finitely many p).
Now (L, Q) determines some integral quadratic form, and so does (L0 , Q) if Q(L0 ) ⊆ Z. We
have (L0p , Q) ∼Zp (Lp , Q) if and only if there exists γp ∈ OQ (Qp ) such that γp Lp = L0p . Likewise,
(L0 , Q) ∼Z (L, Q) if and only if there exists γ ∈ OQ (Q) such that γL = L0 .
Q
0
Let OQ (Af ) = 0p OQ (Qp ) (consisting
T of (γ2 , γ3 , γ5 , . . .) such that γp ∈ OQ (Qp ), and ∀ p, γp ∈
OQ (Zp )). For γ ∈ OQ (Af ), let γ · L = p (γp · Lp ). We have that
-
,
genus(L) = OQ (Q)
OQ (Af )
!
Y
OQ (Zp )
(10.14)
p
via γ · L →7 γ. Given (L, Q), define
1
#Aut(L, Q)
(L)
w(L) = P
o(L) =
L0 ∈genus(L) o(L
Let rgenus(L,Q) (n) =
(10.15)
0)
.
(10.16)
0
L0 ∈genus(Q) w(L )rL0 (n).
P
Theorem 10.5. Let (L, Q) be a Z-lattice with a positive definite quadratic form, and n > 0 such
that for every p, there exists vp ∈ Lp such that Q(vp ) = n. Then rgenus(L,Q) 6= 0.
Proof. By Hasse-Minkowski, there exists v ∈ LQ such that Q(v) = n. ∀0 p, v ∈ Lp . Consider
wp = v − vp , and let
γp (w) = w − 2
wT Qwp
wp .
Q(wp )
Then γp ∈ OQ (Zp ) with γp (vp ) = v. So v ∈ γp · Lp , implying v ∈
We have γp = 1 if v ∈ Lp , so L0 ∈ genus(L).
Define
30
(10.17)
T
p γp
· Lp ; define this to be L0 .
θgen(L) (z) =
X
w(L0 )θL0 (z) =
L0 ∈gen(L)
X
rgen(L) (n)e(nz).
(10.18)
n≥0
1. If L0 ∈ gen(L), then θL (z) − θL0 (z) is a cusp form.
Theorem 10.6 (Siegel-Weil).
2. θgen(L) (z) lies in the Eisenstein spectrum. That is, for all cusp forms f , hθgen(L) (z), f (z)i = 0.
Q
3. rgen(L) (m) =
r
(m)
· r∞ (m), where:
p p
• For prime p and d = dim L,
#{v ∈ L/pr L : Q(v) ≡ m
r→∞
pr(d−1)
rp (m) = lim
(mod pr )}
= volZp (Q−1 (m)).
(10.19)
(The limit actually stabilizes.)
d
• r∞ (m) = volR (Q−1 (m)) = m 2 −1 volR (Q−1 (1)).
The product converges absolutely for d ≥ 4, and conditionally for d = 2, 3.
P
Remark. (2) =⇒ (3), theoretically speaking. For we then have θgen(L) (z) = c ac Ec (z) for some
ac , and we know the Fourier coefficients of Ec .
We can also show that (3) =⇒ (2), by the theory of Hecke operators. Pick p - 2 det Q, and m
such that 2 det(Q) - m. Then
rgen(L) (p2 m)
#{Fp : Q(v) = m}
= pd−2 ·
.
rgen(L) (m)
#{Fp : Q(v) = 0}
(10.20)
This shows that θgen(L) is, ∀0 p, an eigenfunction for the p2 Hecke operator with a very large
eigenvalue.
P
As an example, take Q = Id , so Q(x1 , . . . , xd ) = di=1 x2i . Pick p ≡ 3 (mod 4). The number of
Fp solutions to Q(v) = m is given by:
m
d\
1
2
3
4
5
2
(F×
p)
0
1
2
1
p+1
p2
p2 − p
p3 + p2 − p p3 − p
···
p4 − p2
× 2
F×
p \ (Fp )
0
p−1
p2 + p
p3 − p
p4 + p2
(10.21)
e = Q + x2 + y 2 , we have
Observe that for Q
rQe (m) =
X
rx2 +y2 (n) · rQ (m − n)
(10.22)
m∈Fp
= (p + 1)
X
n
d
rQ (n) − prQ (m)
= pd+1 + p − prQ (m).
31
(10.23)
(10.24)
It then follows that ∀0 p,
p
d−1
rp (m) =

d d
d−1

− (−1) 2 p 2 −1
p
d−1 d−1

pd−1 + (−1) 2 p 2
2|d
m
p
(10.25)
2 - d.
d
For d ≥ 4, rp (m) = 1 + O(p−2 ). To show that rgen(Q) (m) m 2 −1− , we need to consider
p| det Q. Given x0 , y0 , z0 with Q(x0 , y0 , z0 ) ≡ m (mod pr ), we want to lift
Q(x0 +pr x, y0 +pr y, z0 +pr z) ≡ Q(x0 , y0 , z0 )+pr Qx (x0 , y0 , z0 )x+pr Qy (x0 , y0 , z0 )y+pr Qz (x0 , y0 , z0 )z.
(10.26)
If (x0 , y0 , z0 ) 6= 0, we have exactly p2 lifts.
For p| det(Q), m, rp (m) p−m` for ` such that p` km. For n ≥ 5, rp (m) p−n .
n
Corollary 10.7. For n ≥ 5, rgen(Q) (m) m 2 −1− if m is locally represented. The same holds for
n = 4 and squarefree m.
As a consequence, |rgen(Q) (m)| → ∞ implies equidistribution for n ≥ 4.
Let A ∈ Mn (Z) such that AT = A, and v T Av > 0 for v 6= 0. Fix N ∈ N such that N A−1 ∈
Mn (Z), and assume diag(A) ≡ 0 (mod 2). Then A determines a quadratic form. Let A[v] = v T Av,
and define
θA (z) =
X
e
v∈Zn
1
A[v]z ,
2
(10.27)
and H ⊆ Zn /N Zn consisting of h such that Ah ≡ 0 (mod N ). (So then (Zn )∗ =
define the theta function
X
θA (z; h) =
v≡h (N )
e
A[v]z
2N 2
HZn
N .)
For h ∈ H,
.
(10.28)
Then θA (z; 0) = θA (z).
Theorem 10.8. For h ∈ H,
θA
T n X
1
h A`
− 21
e
− ; h = det(A) (−iz) 2
θA (z; `).
z
2N 2
(10.29)
`∈H
Observe that h`, hi 7→ e
Proof. Let f (x) = e
1
2 A[x]z
`T Ah
2N 2
is well-defined and nondegenerate on H.
. Then
Z
fb(y) =
e(−xT y)f (x) dx
Rn
− 12
= det(A)
n T i 2
h y
e
.
z
N
32
(10.30)
(10.31)
Poisson summation for f h yields
N
X
v∈Zn
e
n X i 2
1
h
A−1 [v] hT v
− 21
z = det(A)
.
A v+
e −
+
2
N
z
2z
N
n
(10.32)
v∈Z
So
− 12
θA (z; h) = det(A)
n X X
A−1 [v] hT Av
i 2
e −
+
.
z
2zN 2
N2
(10.33)
`∈H v≡` (N )
Theorem 10.9. For B also a nondegenerate symmetric positive definite matrix, B ∈ gen(A) ⇐⇒
∀ d ∈ Z, ∃ γ ∈ GLn (Z/dZ) such that γAγ T ≡ B (mod d).
Proof. We have N B −1 ∈ Mn (Z) by assumption. We can assume that A ≡ B (mod N ). Then:
Claim. ∀ c ∈ P1 (Q) and ∀ ` ∈ H,
((θA (; `) − θB (; `))| n2 σc )(∞) = 0,
(10.34)
where σc ∈ SL2 (Z) such that σc (∞) = c.
1
Proof of claim. For c = ∞, θA (∞; `) = δ` =
`). Now
T go from c to − c by the transformation
θTB (∞;
formula and the fact that for all ` ∈ H, e `2NAh
= e `2NBh
. Finally, we can go from c to c + 1
2
2
T and similarly for B.
since θA (z + 1; `) = θA (c; `)e `2NA`
2
In fact, we just needed the case d = N . This is a reflection of the following:
Theorem 10.10. If A, B are positive definite quadratic forms with det(A) = det(B) and A ≡ B
(mod Z/ det(A) det(B)Z), if diag(A) is even, then A ∈ gen(B).
Proof of modularity of θA (z). Take V = hθA (z; `)i`∈H ∼
= L2 (H). We have a right action of B by
SL2 (Z), by θA (z; `)|γ = θA (z; `)| n2 γ. We have
f |( 1 1 ) (`) = f (`)h`, `i
01
1 X
f (h)h`, −hi
f| 0 1
(`) =
1
−1 0
|H| 2 h∈H
(10.35)
= fb(`).
(10.37)
(Here h`, hi is as before.) Then we have
33
(10.36)
(H, h , i) ∼
=
n
M
i=1
for ha, bi = e
ab
ni
(Z/ni Z, h , i)
|
{z
}
(10.38)
ϕni
.
a b
c d
Claim. Γ(2ni ) ⊆ ker ϕni , and if
∈ Γ0 (2ni ), then γ · δ0 =
d
ni
· δ0 .
Proof. Consider θ(ni z).
Claim. For ac db ∈ Γ0 (4N ),
az + b
θ N·
cz + d
= j1
2
d
a b
,z ·
θ(N z).
c d
N
(10.39)
Proof.
az + b
θ N·
cz + d
=θ
a(N z) + bN
(c/N )(N z) + d
.
(10.40)
By the modularity of θ(z), we get
1
2
θ(N z) = d (cz + d) ·
(This trick works since
11
N
1
a b
c d
c/N
d
N −1
= θ(N z) · j 1
2
1
=
a b/N
c/N d
d
a b
,z
.
c d
N
(10.41)
.)
Harmonic Polynomials
Let Σ be the unit sphere in Rk , and let Sn = { √vn : v ∈ L, hv, vi = n, possible congruence conditions}.
In general, take Sn finite nonempty subsets of Σ. We say that the Sn are equidistributed as n → ∞
n ∩R)
if the Sn are nonempty and for every open ball R ⊆ Σ, #(S
→ µ(R) as n → ∞. Equivalently,
#Sn
for every f ∈ C(Σ, R),
Z
1 X
f (v) →
f dµ.
#Sn
Σ
v∈Sn
It actually suffices to check for polynomial f .
Recall the operators
34
(11.1)
X ∂2
∂x2i
i
X
∂
E=
xi
∂xi
i
X
F =
x2i .
∆=
(11.2)
(11.3)
(11.4)
i
Let Pd denote the homogeneous polynomials of degree d, and
Pd0 = ker(∆ : Pd → Pd−2 ) = ker(F ∆ : Pd → Pd ).
(11.5)
Theorem 11.1. Pd decomposes as
Pd =
M
0
F k Pd−2k
(11.6)
k
0
where each F k Pd−2k
is an eigenspace of F ∆, with different eigenvalues.
Proof. We will show that F ∆|F k P 0
d−2k
= λk (d) where the λk (d) are distinct. But also
0
0
≥ dim Pd−2k − dim Pd−2k−2 ,
dim F k Pd−2k
= dim Pd−2k
(11.7)
so a dimension count forces the sum to be all of Pd . Also, equality holds everywhere. In particular,
∆ : Pd → Pd−2 is surjective.
We can check easily that [∆, F ] = 4E + 2n. So if f is harmonic of degree d, then F ∆(F f ) =
(4d + 2n)F f . More generally, if (F ∆)f = λf , then (F ∆)(F f ) = (4d + 2n + λ)F f . We end up with
λk (d) = k(4(d − k − 1) + 2n).
(We also have [E, ∆] = −2∆ and [E, F ] = 2F .)
Because of Theorem 11.1, it suffices to check equidistribution for the Pd0 , because F is the identity
map on
We now claim that if P is harmonic of degree d and L is a rational lattice,
P the unit sphere.
1
then v∈L P (v)q 2 hv,vi is modular (with congruence group depending on L) of weight d + 21 rank(L),
and is a cusp form if d > 0.
Proposition 11.2. Define
Gt (P ) = eπthx,xi P . (Then Gt F = F Gt .) If f = Gt P for P ∈ Pd , then
P
n
0
0
fb = G 1 Pb where Pb = d0 ≤d Pbd0 with Pbd0 = id t−( 2 +d ) P .
t
Fix Me Are they both d0 , or does d appear? (10)
∂
b
\
\
Proof. We have G
t xj P = xj Gt P = c ∂xj (G 1 P ) for appropriate c. Then use induction.
t
Now suppose P ∈ Pd0 . Then ∆P = 0 and EP = dP . We have 0 = (Gt ∆G−1
t )Gt P . We can show
−1
−1
that Gt F G−1
=
F
,
G
EG
=
E
+
2πtF
,
and
G
∆G
=
∆
+
πt(4E
+
2n)
+
(2πt)2 F . We can then
t
t
t
t
t
n
−(
+d)
d
0
check that Pb = i t 2
P ⇐⇒ P ∈ Pd .
35
R
Now Rn e−πthx,xi P (x) dx = cPb(0) = 0 if P is harmonic of degree at least 1, so the Sj are
equidistributed if and only if for every nonconstant harmonic polynomial P ,
1 X
P (v) → 0.
|Sj |
(11.8)
j
d
1
1 X
P (v) =
· j − 2 · (coefficient of q 2 ).
|Sj |
|Sj |
(11.9)
v∈Sj
But we have that
v∈Sj
d
n
So we need this coefficient to be o(|Sj | · j 2 ). If we have that |Sj | = j 2 −1− and that the q j
n
d
coefficient of a cusp form is o(j 2 −1−+ 2 ), then we are done.
n
d
The trivial bound o(j 4 + 2 ) suffices if n ≥ 5. For n = 4, any power estimate better than the
n
d
1
trivial bound will suffice. (The Kloosterman bound is o(j 4 + 2 − 4 + ).) For n = 3, any power estimate
better than the Kloosterman bound would suffice.
Claim. Let p be an odd prime. Then
γ=
p−1
X
j=0
2πij 2
e
p
 (
√
X a 2πia
p p≡1
p
=

e
= √
p
i p p≡3
a (p)

(mod 4)
(mod 4).
(11.10)
2πijk
Proof. Let V = (e p )p−1
j,k=0 , the discrete Fourier transform modulo p. Our desired expression is
√
tr(V ). We can show that the spectrum is p · (1, i, −1, −i, 1, i, . . . , ip−1 ). To get the spectrum, we
first know that


1
1

1 


1


1



2
V = p



(11.11)
1
which is p times a matrix with square the identity and trace 1, so V 2 has spectrum p with
√
p−1
multiplicity p+1
2 and −p with multiplicity 2 . This implies V has spectrum {± p} with multiplicity
p−1
√
p+1
p−1
2
2 p shows the spectrum is as close to
2 and {±i p} with multiplicity 2 . Then γ = (−1)
balanced as possible. Finally,
det V =
Y k
j
−e
.
e
p
p
j<k
Now e(α) − e(β) has argument
π
2
+ 12 (α + β).
36
(11.12)
12
Bounds for Fourier Coefficients
Recall that K(m, n; c) =
formula,
mx+nx
c
P
x (c) e
, and that for f ∈ Sk (SL2 (Z)), the Petersson trace
X
|af (m)|2
Γ(k − 1)
positive terms +
Jk−1
≤
1
+
(4πm)k−1 [Γ : Γ0 (N )] · kf k
c>0
4πm
c
K(m, m; c)
.
c
(12.1)
N |c
Pick large Q and sum over Q ≤ N ≤ 2Q. We have [Γ : Γ0 (N )] = N 1+o(1) , so
2Q
X
N =Q
1
= Qo(1) .
[Γ : Γ0 (N )]
(12.2)
So we find that
1−k
m
2
o(1)
af (m) Q
2Q X
∞
X
1
4πm
≤Q+
Jk−1
K(m, m; qr).
qr
qr
(12.3)
q=Q r=1
Lemma 12.1.
1 X
|K(am, am; r)|2 ≤ 4 gcd(m, r)σ0 (r) · r.
r
(12.4)
a (r)
Proof.
LHS =
1
r
X
e
a,x,y (r)
amx + amx + amy + amy
r
(12.5)
xx=yy=1
= #{x, y (r) : mx + mx + my + my ≡ 0 (r)}
(12.6)
= #{x, y (r) : m(x + y)(xy + 1) ≡ 0 (r), gcd(xy, r) = 1}.
(12.7)
Sum over prime powers and count.
Now if gcd(q, r) = 1, K(m, m; qr) = K(mq, mq; r)K(mr, mr; q). Let
A(Q, R) =
X X
|K(mq, mq; r)|2
(12.8)
q≤Q r≤R
(q,r)=1
≤
X Q
r≤R
≤
X
r
+1
X
|K(am, am; r)|2
(12.9)
q (r)
4(Q + r) gcd(m, r)rσ0 (r)
(12.10)
r≤R
≤ (Q + R)m R2+ .
37
(12.11)
Next, let
B(Q, R) =
X X
|K(m, m; qr)|
(12.12)
|K(mq, mq; r)||K(mr, mr; q)|
(12.13)
q∈Q r≤R
(q,r)=1
=
X X
q≤Q r≤R
(q,r)=1
1
1
≤ A(Q, R) 2 B(R, Q) 2
(12.14)
≤ (Q + R)m Q1+ R1+
(12.15)
where the inequality (12.14) follows from Cauchy-Schwarz. Finally, let
C(Q, R) =
XX
|K(m, m; qr)|
(12.16)
q≤Q r≤R
≤
X X
X
a≤Q q0 ≤ Q
r0 ≤ R
a
a
|K(m, m, q0 r0 a2 )|
(12.17)
|K(m, m, q0 r0 )|a1+
(12.18)
(q0 ,r0 )=1
≤
≤
X
X X
a≤Q
q0 ≤ Q
a
X
B
a≤Q
1+
r0 ≤ R
a
(q0 ,r0 )=1
Q R
,
a a
a1+
(12.19)
R1+ (Q + R)m .
(12.20)
X X |K(m, m; qr)|
≤ (Q + R)1+ m .
qr
(12.21)
≤Q
By dyadically summing, we find that
q≤Q r≤R
1
We also have Jk−1 (x) min{xk−1 , x− 2 }, and
1−k
m
∞
XX
4πm m 1+ 1 Jk−1
|K(m, m; qr)| ≤ Q + Q +
|af (m)| Q +
m + O(1).
qr qr Q
2
q≤Q r=1
(12.22)
1
2
Taking Q = m , m1−k |af (m)|2 m
13
1
+
2
, so |af (m)| m
k
− 41 +
2
.
Metaplectic Groups and Half-Integral Weight Hecke Operators
(Reference: Shimura, “On Modular Forms of Half-Integral Weight”.)
Define the group
38
e=
G
cz + d
a b
+
2
(α, ϕ) : α =
∈ GL2 (Q), ϕ holomorphic on H with ϕ(z) = ± √
.
c d
det α
(13.1)
Then (α, ϕ(z))(β, ψ(z)) = (αβ, ϕ(βz)ψ(z)) defines a group multiplication. We have an exact
sequence
e → GL+ (Q) → 1.
1 → µ4 → G
2
(13.2)
e does not arise from an algebraic group.
But G
e by i(γ) = (γ, j 1 (γ, z)). If 4|N and k ∈ 1 Z, define (f |k ge)(z) = f (g(z))ϕg (z)−2k ,
Define i : Γ0 (4) ,→ G
2
2
for ge = (g, ϕg ). Then f ∈ Mk (Γ0 (N )) iff f |k i(γ) = f for every γ ∈ Γ0 (N ).
0
e let Γ^
Now if α
e ∈ G,
e−1 Γ0 (N )e
α ∩ Γ^
0 (N ) = i(Γ0 (N )) and Γ = α
0 (N ).
0
Claim. [Γ^
0 (N ) : Γ ] < ∞.
(We omit the proof of the claim.)
Sr
Pr
0 e . Let f | [e
So we can write Γ^
eγ
ei .
0 (N ) =
i
k α] =
i=1 Γ γ
i=1 f |k α
Claim. f 7→ f |k [e
α] sends Mk (Γ0 (N )) → Mk (Γ0 (N )). (Also, this is well-defined on Mk (Γ0 (N )).)
For k ∈ Z, Tp = [e
α] where
α
e=
p 0
− 12
,p
.
0 1
(13.3)
1
0
Theorem 13.1. Let αm = ( m
em = (α, m− 4 ). Then unless k ∈ Z or m is a perfect square,
0 1 ) and α
f |k [e
αm ] = 0.
Proof.
−1 Γ (N )α , and let γ ∈
• Take γ ∈ Γ0 (N ) ∩ αm
0
m
−1
i(αm γαm
)
a b
c d
. Then we have
1 c/m 2
=
+d
d
m
d
1 c m z
2
−1
= αm γαm , c + d d
m
d
d
m −1
=α
em i(γ)e
αm
· 1,
.
d
−1
αm γαm
,
cz
(13.4)
(13.5)
(13.6)
−1 Γ (N )α such that i(γ) 6∈ Γ0 .
So if m is not a perfect square, then there exists γ ∈ Γ0 (N )∩αm
0
m
−1 Γ
^
• i(γ)e
αm
αm i(γ)−1 is equal to
0 (N )e
m
m −1
−1
−1 −1
−1 ^
) ·α
em = α
em
Γ0 (N )e
αm ,
α
em
i(αm γαm
) 1,
· Γ^
· i(αm γαm
0 (N ) · 1,
d
d
−1 Γ
^
so i(γ) stabilizes α
em
αm , and therefore Γ0 , under conjugation.
0 (N )e
39
(13.7)
• We have
f |k [e
αm ] =
r
X
f |k α
em γ
ei
(13.8)
f |k α
em · i(γ) · γ
ei
(13.9)
i=1
=
r
X
i=1
=
r
X
−1
f |k i(αm γαm
)e
αm · (1, −1) · γ
ei ,
(13.10)
f |k α
em · γi = −f |k [e
αm ].
(13.11)
i=1
which for k 6∈ Z, is
−
r
X
i=1
Now define
2 n 0
− 12
,n
Tn2 f = f |k
.
0 1
(13.12)
^ e−1
As an exercise, we have that Γ^
Γ^
(N )e
αp2 ) decomposes as
0 (N )/(Γ0 (N ) ∩ α
p2 0
2 −1 p[
b=0
If f (z) =
P∞
n=0 af (n)e(nz),
p−1
[ p h
1
−h
1 b
−1
, p2 ∪
, p
0 p2
0 p
p
h=1
2 1
p 0
∪
, p− 2 .
0 1
then we have (Tp2 f )(z) =
2
bf (n) = af (np ) +
−1
p
P∞
n=0 bf (n)e(nz),
(13.13)
(13.14)
where
k− 1 2
n
n
k− 32
k−2
·p
· af (n) + p af
.
p
p2
(13.15)
(At least this is true if gcd(p, N ) = 1.)
For f, g ∈ Sk (Γ0 (N )), define
Z
f (z)g(z)y k
hf, gik =
Γ0 (N )\H
dx dy
.
y2
(13.16)
Then the Tp2 are self-adjoint with respect to the h , ik . So the Tp2 are simultaneously diagonalizable.
Suppose f ∈ Sk (Γ0 (N )) is such that for every p, Tp2 f = up f . Then for t ∈ N squarefree,
40
∞
X
af (tn2 )
n=1
ns
= af (t) ·
1−
Y
1 − up ·
p
t
p
p−s
3
pk− 2 −s
+ p2k−2−2s
.
(13.17)
Theorem 13.2 (Waldspurger,Shimura,Shintani,Konen-Zagier).
• Let k ∈ 21 + Z, k ≥ 32 , and f ∈ Sk (Γ0 (N )) such that for all p, Tp2 f = up f . Then there exists
g ∈ M2k−1 (Γ0 (N/2)) such that for every p, Tp g = up g. So then
∞
X
ag (n)
i=1
ns
Y
=
1 − up
p
p−s
1
.
+ p2k−1−2s
(13.18)
• Let D be squarefree and gcd(D, N ) = 1. Define
L(g ⊗ χD , s) =
∞
X
ag (n)
n=1
n
D
ns+k−1
.
(13.19)
Then
k − 32 ! |D|k−1
|af (D)|2
1
=
·
L g ⊗ χD ,
.
hf, f ik
π
hg, gik
2
(13.20)
So Ramanujan for half-integral weight is equvialent to Ramanujan for integral weight along with
Lindelof in the D-aspect.
We have g ∈ S2k−1 (Γ0 (N/2)) if k ≥ 52 , or if k =
functions.
3
2
and f is orthogonal to one-variable theta
For a, b ∈ N and χ : (Z/bZ)× → C× , define
θa,χ (z) =
∞
X
nχ(n)e(an2 z).
(13.21)
n=1
Then θa,χ (z) ∈ S 3 (Γ0 (4ab2 ), ψ) for ψ(d) =
1
2
2
3/2
− 14
2
−a
d
χ(d). The coefficients, at their largest, have size
roughly n = n
. Under the Shimura correspondence, they correspond to Eisenstein series, and
Ramanujan fails for these.
14
Bounds in the Half-Integral Weight Case
Fix Me Disgtinguish between p and P . (11)
Recall that
pm
k,N (z) =
1
2
X
(em |γ )(z)
γ∈Γ∞ \Γ0 (N )
for em (z) = e(mz). If f ∈ Sk (Γ0 (N )), hf, pm
k,N i is a constant times af (m), specifically
41
(14.1)
Γ(k − 1)
af (m).
(4πm)k−1
hf, pm
k,N i =
(14.2)
So it suffices to bound coefficients for Poincaré series. We have
Z
i+1
pm
k,N (z)e(−nz) dz

nth term =
i
Z
=
i+1 
i
(14.3)

2k az + b 
X
−k −2k c
 e(−nz) dz (14.4)
(cz + d) d
e m
d
cz + d 
X
e(mz) +

(d,c)=1
c>0
c≡0 (N )
X
= δm,n +
X
c>0
−2k
d
(d,c)=1
c 2k Z
i+∞
d
i−∞
X
−k
(cz + d)−k e
c≡0 (N ) d (c)
= δm,n +
m k−1
2
n
−k
(2πi)
c
Jk−1
c>0
c≡0 (N )
ma
m
−
− nz
c
c(cz + d)
dz (14.5)
√
4π mn X d
md + nd
e
· d−2k .
c
c
c
d (c)
|
{z
}
denoted Sk (m,n;c)
(14.6)
Lemma 14.1. If c = (q, r) and 4|r, then Sk (m, n; c) = Sk−q+1 (mq, nq; r)S(mr, nr; q) for
X d md + nd e
.
c
c
S(m, n; c) =
(14.7)
d (c)
(d,c)=1
This lemma is an exercise.
S(m, n; c) is called the Salié sum.
Lemma 14.2. If gcd(m, q) = gcd(n, q) = 1, then (in addition to S(m, n; q) = S(n, m; q)):
1. S(m, n; q) =
m
q
2. S(1, m; q) = 0 if
S(1, mn; q).
m
q
= −1.
√ P
3. S(1, n2 ; q) = q q x2 ≡1
Proof.
(q) e
2xn
q
.
1. Use the change of variables d → dm.
2. S(1, m; q) = S(m, 1; q) =
m
q S(1, m; q).
3. Define
X x x + n2 x h(n) =
e
.
q
q
x (q)
42
(14.8)
Then we have
mn
h(n)e −
q
n (q)
X x x X x(n2 − xmn) =
e
e
q
q
q
n
x (q)
X x
x
xm2
√ x
e
e −
q q
=
q
q
4
q
x
m2
x
√ X
= q q
1−
.
e
q
4
b
h(m) =
X
(14.9)
(14.10)
(14.11)
(14.12)
(x,q)=1
So
√
b
h(m) = q q
X
q
µ
d
d|(q,m2 −4)
d.
(14.13)
Fourier inversion implies
1 X b
mn
h(n) =
h(m)e
q
q
(14.14)
m (q)
q X
=√
n
X
m (q) d|(q,m2 −4)
q mn dµ
e
.
d
q
(14.15)
Switch summations. Then as gcd(n, q) = 1, the sum vanishes if d 6= q. So we get
q
h(n) = √ q
q
X
e
m2 ≡4 (q)
mn
q
.
(14.16)
Now suppose c is odd and relatively prime to n. We have a bijection between {x : x2 ≡ 1
(mod c)} and {ab = c, (a, b) = 1} by x 7→ ((x − 1, c), (x + 1, c)). Then
x
a b
aa − bb
= − =
.
c
b a
c
(14.17)
We then get
2
√
S(1, n ; c) = c c
X
ab=c
(a,b)=1
so that
43
e 2n
a b
−
b a
(14.18)
S(n, n; c) =
√ n X
a b
e 2n
.
S(1, n ; c) = c c
−
c
c
b a
n
2
(14.19)
ab=c
(a,b)=1
Theorem 14.3. For k ∈
1
2
+ Z with k ≥ 52 , and f ∈ Sk (Γ0 (N )), if n is squarefree, then
k
3
|af (n)| n 2 − 7 + .
(14.20)
For every q, Sk (Γ0 (N )) ,→ Sk (Γ0 (N q)). The Petersson trace formula implies
X
|af
4πn 1 + 2π c−1 Kk (n, n; c)Jk−1
.
[Γ0 (N ) : Γ0 (N q)]
c c≡0 (qN )
(n)|2
14.1
(14.21)
Trivial Bounds
4πn
c
If q is prime, then [Γ0 (N ) : Γ0 (N q)] = q + 1, Jk−1
1
min
X
n
c
n
1
2
,
n
c
3 o
2
, and |Kk (n, n; c)| ≤
1
σ0 (c)(n, c) 2 c 2 . By convention, write L = O (n ).
If we fix C ≤ N ≤ D, then
X
c−1 Kk (n, n; c)Jk−1
c≤C
4πn
c
1
1
n− 2 (c, n) 2 σ0 (c)
(14.22)
c≤C
c≡0 (qN )
c≡0 (qN )
L·C
1
,
(14.23)
qn 2
and
X
−1
c
Kk (n, n; c)Jk−1
c≥D
4πn
c
c≡0 (qN )
X
c≥D
3
1
n 2 (c, n) 2
σ0 (c)
c2
(14.24)
c≡0 (qN )
3
n2
.
L·
qD
14.2
(14.25)
A Toy Problem
We will try to bound
KN (x) =
X S(n, n; c)
√
c
c≤x
N |c
for x ∼ n. The trivial bound is xL.
44
(14.26)
Let
X n a b e 2n
−
Fy (A, B; N ) =
ab
b a
(14.27)
where the sum ranges over a, b such that A ≤ a ≤ 2A, B ≤ b ≤ 2B, y ≤ ab ≤ 2y, and N |ab.
We have the reciprocity formula
1
a b
+ ≡
b a
ab
(mod 1).
(14.28)
So
X n 4na 2n Fy (A, B; N ) =
e −
.
e
ab
b
ab
Summation by parts: if f (x) ∈ C 1 (R) and {an }n∈N are given, let Sx =
2A
X
an f (n) = f (2A)S2A −
n=A
2A
X
P
A≤n≤x an .
Then
Sn (f (n + 1) − f (n))
n=A
Z 2A
= f (2A)S2A −
(14.29)
(14.30)
Sx f 0 (x) dx.
(14.31)
A
In particular,
2A
X
an f (n) ≤ A max |f 0 (x)| max |Sx | + |f (2A)||S2A |.
[A,2A]
[A,2A]
(14.32)
n=A
0
Take f (x) = e − 2n
bx so that f (x) =
Hence
2n
e
bx2
0
− 2n
bx , giving A max[A,2A] |f (x)| X
X
N n
4na e
Fy (A, B; N ) 1+
AB a
b N |ab
(n,b)=1
A≤a≤m
B≤b≤2B
n
A
BA2
=
n
BA .
(14.33)
b
with mb ≤ 2A. Notice that
X n 4na =0
e
a
b
(14.34)
a (nb)
for n squarefree (or just n not a square).
Completing the sum: let F : Z → R be m-periodic. Let
Fb(r) =
X
s (m)
rs F (s)e −
.
m
45
(14.35)
Then
x
X
F (u) =
u=0
x
ur X
1 X b
F (r)e
m
m
u=0
(14.36)
r (m)
x
ur xFb(0)
1 X X b
F (r)e
+
m
m
m
u=0 r6≡0 (m)
X
b
x|F (0)|
1
1 b
F (r)
+ r
m
m
1−e m r6≡0 (m)
=
X
xFb(0)
1
+
max |Fb(r)|
m
m r6≡0 (m)
r6≡0 (m)
(14.37)
(14.38)
m
r
(14.39)
x|Fb(0)|
+ max |Fb(r)| log m.
m
r6≡0 (m)
4nx
b
We can apply this to m = nb and F (x) = e
Fb(r) =
(14.40)
. Then Fb(0) = 0. If r 6= 0, then
X n 4nnx − rx e
.
x
nb
(14.41)
x (nb)
After using partial fractions, we can factor
this into a Salié sum and a Kloosterman sum. If we
√
assume the Weil bound, we get |Fb(r)| nbL. So
X
|Fy (A, B; N )| 1+
(n,b)=1
1
n · L · (nb) 2
AB
(14.42)
B≤b≤2B
n 1 3
n2 B 2 .
1+
AB
(14.43)
1
If AB = n, the trivial bound is AB, or n, but the above is better if B n 3 .
If A, B are both large, then we can “average over the level”. Let
X
FP,y (A, B; N ) =
|Fy (A, B; pN )|.
(14.44)
P ≤p≤2P
p prime,p-n
We will choose P to be a small power of n. We have
FP,y (A, B; N ) =
X
P ≤p≤2P
X n a b λp
e 2n
−
ab
b a
(14.45)
where λp = sgn(Fy (A, B; pN )) and the inner sum runs over a, b such that A ≤ a ≤ 2A, B ≤ b ≤
2B, y ≤ ab ≤ 2y, (a, b) = 1, and pN |ab. Now we can write
46
FP,y (A, B; N ) = Fy0 (A/P, B; N ) +Fy0 (A, B/P ; N ).
{z
}
|
(14.46)
X X
n
ap
b
e 2n
−
|Fy0 (A/P, B; N )| ≤
λp
p
b
ap
p1 ≤p≤p2
(14.47)
terms with p|a
We have
where the outer sum runsnover PA ≤ oa ≤ 2A
P , B ≤ b ≤ 2B, (a, b) = 1, and N |ab. Here p1 =
y A
2y 2A
max P, ab , a and p2 = min 2P, ab , a . By Cauchy-Schwarz,

X
AB

|Fy0 (A/P, B; N )|2 P
X
λp1 λp2
P1 ≤p1 ,p2 ≤P2
a,b

ap1 p2
n
b

−
e 2n(p2 − p1 )
p1 p2
b
ap1 p2
(14.48)
so then
X X
ap
p
AB
b
1 2
.
|Fy (A/P ; B; N )|2 −
e(2n(p2 − p1 )
P p ,p b
ap1 p2 1 2 a,b
(14.49)
Summing over a first, the summand is
ap1 p2
e 4n(p2 − p1 )
b
2 −p1 )
so |xf 0 (x)| Let f (x) = e 2n(p
xbp1 p2
F (x) = e 4n(p2 −pb1 )xp1 p2 . Then
Fb(0) =
n
AB
2n(p2 − p1 )
e
.
abp1 p2
(14.50)
for x ∈ [ PA , 2A
P ]. Completing the sum, let m = b and
X
e
x (b)
(p2 − p1 )x
b
(14.51)
(x,b)=1
=
X
dµ
n
d|(b,p2 −p1 )
d
≤ (p2 − p1 , b)
(14.52)
(14.53)
for p1 6= p2 . For r 6= 0,
Fb(r) =
X 4n(p2 − p1 )xp1 p2 − rx e
b
(14.54)
x (b)
1
1
b 2 (p2 − p1 , b) 2 L.
47
(14.55)
Therefore for p1 6= p2 ,
X
1
1
ap1 p2
b
n A(p2 − p1 , b)
L 1+
e 2n(p2 − p1 )
−
+ b 2 (p2 − p1 , b) 2 .
b
ap1 p2
AB
Pb
2A
A
≤a≤ P
P
(14.56)
Now we have
X
p1 ,p2 ,b
X 1 1
3
n n AB
A(p1 − p2 , b)
A
L 1+
L 1+
(p2 − p1 , b) 2 b 2 +
B2 +
+P ·
AB
Pb
AB
P
P
p1 6=p2
(14.57)
n 3 2
L 1+
(B 2 P + AP ) + AB.
AB
(14.58)
Hence
|F (A/P, B)| 1
5
1
1
n 12
+ 1+
· L(A 2 B 4 P 2 + AB 2 ).
AB
(14.59)
5
1
1
1
1
n 21
+ 1+
· L(A 4 B 2 P 4 + A 2 BP − 2 ).
AB
(14.60)
AB
1
P2
Doing the b-sum first,
|F (A/P, B)| AB
1
P2
14.3
A Return to the Original Problem
Again take k ∈
1
2
+ Z, 4|N , and n squarefree. For 4|c,
c md + nd e
.
d
c
(14.61)
Kk (m, n; qr) = Kk (mq, nq; r)S(mr, nr; q).
(14.62)
X
Kk (m, n; c) =
d
2k
d (c)
If (q, 2r) = 1, then
Now for N |Q, define
KQ (x) =
X
1
c− 2 Kk (n, n; c)
(14.63)
c≤x
Q|c
X Kk (nq, nq; r) S(nr, nr; q)
√
=
.
√
q
r
48
(14.64)
where the sum runs over q, r with qr ≤ x, (q, r) = 1, (q, nN ) = 1, N |r, r|(N n)∞ , and Q|qr. Let
w(c) = #{P < p < 2P : p|c, p - n}. Then w(c) < log c. Then we have
X
KN P (x) =
P <p<2P
X X X Kk ( ) S( )
√ √
q
r
p
N |r
(14.65)
q≤x
r|(N n)∞ p|q
=
X X w(q)
√ Kk ( )S( ).
qr
x
r
(14.66)
q≤ r
We have
Kk (nq, nq; r) =
X
fr (2s)e
s (r)
2nqs
r
X r
fr (2s) =
d 2k .
d
(14.67)
(14.68)
d (r)
d+d=2s
Therefore
X w(q)
X 1 X
2nqs
√
fr (2s)
.
KN P (x) =
√ S(nr, nr; q)e
q
r
r
r
P <p<2P
q≤ xr
s (r)
|
{z
}
X
(14.69)
call this Tr,s (x)
We then have
Tr,s (x) = fr (2s)
X
nr ra rb sab w(ab)
e 2n
−
+
ab
b
a
r
(14.70)
with the sum running over a, b with ab ≤ xr , (a, b) = 1, and N |ab. Let fy (A, B; N ) be the above
summand over y ≤ a, b ≤ 2y, A ≤ a ≤ 2A, B ≤ b ≤ 2B, and (a, b) = 1.
14.3.1
Estimate I
Modulo 1,
ra rb
ra rbrb
−
≡
−
b
a
b
arb
ra 1 − aa
≡
−
b
abr
ra
a
1
≡
+
−
.
b
br abr
So
49
(14.71)
(14.72)
(14.73)
a b
2nra 2na
2n
e 2nr
=e
e −
−
+
b a
b
br
abr
2n(1 + rr)a
2n
=e
e −
.
br
abr
(14.74)
(14.75)
Partial summation with respect to a then gives
To do Can the (?) be referenced? (12)
n X X
2n(1 + rr + sbb)q nr Fy (A, B; N ) 1 +
e
w(ab)
abr
br
a
(n,b)=1 |
{z
}
B≤b≤2B (?)
(14.76)
with the sum in (?) running over A ≤ a ≤ 2A, y ≤ ab ≤ 2y, (a, b) = 1, and N |ab. Then (a, b) = 1
implies w(ab) = w(a) + w(b). (?) becomes
w(b)
X
e( )
a
nr a
X
+
P ≤p≤2P
X
···
(14.77)
p|a
A≤a≤2A
is nontrivial in arithmetic progressions
Now complete the sum over A. It’s important that nr
a
with modulus P N with respect to a. (This is where n being squarefree is used.)
1
We get (?) L(nbr) 2 P , so
3
1
n
Fy (A, B; N ) 1 +
LB 2 (nr) 2 P.
y
(14.78)
Summing over b first would replace B with A.
14.3.2
Estimate II
Fy (A, B; N ) = Fy0 (A/P, B; N ) + Fy0 (A, B/P ; N ) for
Fy0 (A/P, B; N )
X
=
P ≤p≤2P
X rap
rb
sabp
nr
λp
e 2n
−
+
.
b
ap
r
pab
(14.79)
To do What is the set the sum runs over? (13)
By Cauchy-Schwarz,
|Fy0 (A/P, B; N )|2 ap1 p2 r
br
sabp1 p2
AB X X
−
.
e 2n(p2 − p1 )
+
P p ,p
b
ap1 p2
r
1
2
a,b
Partial summation and completing the sum (the a sum first),
50
(14.80)
Fy0 (A/P, B; N )
1
1
1
n 2 1 1 5 1
+L 1+
(r 4 A 2 B 4 P 2 + r− 4 AB 2 ).
y
(14.81)
1
1
1
1
n 2 1 5 1 1
(r 4 A 4 B 2 P 4 + r− 4 A 2 BP − 2 ).
+L 1+
y
(14.82)
AB
1
P2
If the b sum is done first,
Fy0 (A/P, B; N )
AB
1
P2
Optimizing,
5
1
1
5
1
1
7
3
min{A 2 B 4 P 2 , A 4 B 2 P 4 } ≤ (AB) 8 P 8 .
(14.83)
Therefore
Fy (A, B; N ) yr
14.3.3
−1
n
p +L 1+
y
1
2
1
2
5
7
1
3
1
5
[y 8 r− 8 p 8 + (A− 2 + B − 2 )yr− 4 ].
(14.84)
The Final Result
Recall that Estimate I gives
n
Fy (A, B; N ) L min{A, B} (nr)
1+
.
y
(14.85)
1
n −4 −1 −7 1 −1
n 2r 8y2p 2.
min{A, B} 1 +
y
(14.86)
3
2
1
2
We’ll use Estimate I if
We then get
|Tr,s (2y) − Tr,s (y)| yr
−1 − 21
p
5
7
5 3
1 3
13 1
n 8
+ 1+
· [y 8 r− 8 p 8 + n 8 y 4 r− 16 p 4 ].
y
(14.87)
We get a similar bound for |Tr,s (y)| as a result. So
X
1 X
√
|Tr,s (x)|
r
X
KN P (x) L
p-n
N |r
P <p<2P
r|(nN )∞
L
s (r)
5
1 3
1
1 1
x
8
4
8
8
8
4
+
(x
+
p)
[x
p
+
n
x
p
]
.
√
p
Now if f ∈ Sk (Γ0 (N )), we have
51
(14.88)
(14.89)
|af (n)|2
1 + 2πi−2k
nk−1 p
X
−1
c
Kk (n, n; c)Jk−1
c≡0 (pN )
4πn
c
.
(14.90)
Fix Me Should all of 2πi be raised to power? (14)
Summing over P < p < 2P , p prime, and p - n,
X
|af (n)|2
4πn − 1
P
− 12
c w(c)Kk (n, n; c)Jk−1
c 2.
+
log P
c
nk−1 log P
(14.91)
c≡0 (N )
1
5
Partial summing with respect to c, using (x− 2 Jk−1 (x))0 min{x− 2 , 1}, we get
|af (n)|2
L
nk−1 log P
k
1
1
1
3
3
1
1
P
+ n2 P−2 + n8 P 8 + n8 P 4
log P
2
For P = n 7 , we get |af (n)| n 2 − 7 + . (Note that
half-integral weight.)
2
7
1
4,
>
.
(14.92)
so we improved our bound for
Recall that there exists g ∈ S2k−1 (Γ0 (N )) such that for t squarefree,
af (tn2 ) = af (t)
X
3
µ(d)dk− 2 ag
n
d
d|n
1
k
1
.
(14.93)
δ
So ag (m) mk− 2 −δ+o(1) ⇐⇒ ∀ t, af (tn2 ) (n2 ) 2 − 4 − 2 . The trivial bound δ = 0 gives
k
k
1
3
af (tn2 ) (n2 ) 2 − 4 +o(1) , while δ = 14 gives af (tn2 ) (n2 ) 2 − 8 +o(1) . So this technology would have
imporoved our bounds even more.
15
Heegner Points and Geodesics
For quadratic forms with negative discriminant, let Q(x, y, z) = x2 − 4yz, then #{Q(a, b, c) = n}
equals either 0 or ∞, bu this happens because SOQ (Z) is infinite.
Let Vn = {(a, b, c) : b2 − 4ac = n}. For n < 0, we have
SOQ (Z)\Vn (Z) ,→ SOQ (Z)\V−1 (R)
a
b
c
(a, b, c) 7→ √ , √ , √
.
−n −n −n
(15.1)
(15.2)
√
We also have V−1 (R) ∼
= H by (x, y, z) 7→
SL2 (Z) action on H. So we get
−x±
ϕ
x2 −4yz
.
2y
The SOQ (Z) action translates to the
SOQ (Z)\Vn (Z) −
→ SL2 (Z)\H.
(15.3)
Let Hn = im ϕ, called the Heegner points of discriminant n. These will be shown to be equidistributed with respect to dxy2dy for squarefree n → ∞.
52
√
−b− b2 −4ac
2a
For n > 0, let γa,b,c be the geodesic in H connecting
b
centered at − 2a
). Then we have
/
SL2 (Z)
and
√
−b+ b2 −4ac
2a
(a semicircle
ψ
[
γa,b,c −
→ SL2 (Z)\H.
(15.4)
b2 −4ac=n
Let Sn = im ψ.
Fix Me More stuff on this topic appears later. Reorganize? (15)
16
Maass Forms
Let ∆ be the Laplacian with respect to the hyperbolic metric; that is,
SL2 (R)-invariant.
−y 2
∂2
∂x2
+
∂2
∂y 2
. ∆ is
We say that ϕ(z) is a maass form of level N if:
• ∆ϕ = λϕ · ϕ for some λϕ ∈ R.
• ϕ is invariant under Γ0 (N ).
• ϕ(x + iy) = O(y A ) as y → ∞, for some A.
As an example, for s ∈ R, ∆y s = s(1 − s)y s . If <(s) > 1, take Γ = SL2 (Z) for simplicity and
take
E(z.s) =
1 X
im(γz)s
2
(16.1)
ys
1 X
.
2
|cz + d|2s
(16.2)
Γ∞ \Γ
=
(c,d)=1
Then ∆E(z, s) = s(1 − s)E(z, s) and E(γz, s) = E(z, s). We have that
1
ζ(2s)E(z, s) = y s
2
We can write E(z, s) =
y
−s
P
r∈Z ar (y, s)e(rx)
X
|cz + d|−2s .
(16.3)
(c,d)6=(0,0)
since E is 1-periodic. We have
ζ(2s)ar (y, s) = δ0,r ζ(2s) +
XXZ
c>0 d∈Z
= δ0,r ζ(2s) +
1
|cz + d|−2s e(−rx) dx
X Z
c (c2 x2 + c2 y 2 )−s e(−rx) dx.
c|r
Now let
53
R
(16.4)
0
(16.5)
Ks (y) =
1
2
Z
∞
−1 )/2
e−y(t+t
0
ts
dt
.
t
(16.6)
Ks (y) converges for y > 0, decays exponentially as y → ∞, and Ks (y) = K−s (y). We end up
with
(
π −s Γ(s)ζ(2s)y s + π s−1 Γ(1 − s)ζ(2 − 2s)y 1−s
Γ(s)π −s ζ(2s)ar (y, s) =
1
√
2|r|s− 2 σ1−2s (|r|) yKs− 1 (2π|r|y)
r=0
r 6= 0
(16.7)
2
We learn that ζ(2s)E(z, s) continues meromorphically in s to C, with a unique pole at s = 1,
π
and Ress=1 E(z, s) = ζ(2)
= π3 .
Theorem 16.1. vol(Γ\H) = π3 .
Idea of proof. Use
Z
Z
dx dy
y2
(16.8)
=(γz)s δ(=(γz) ≤ T ).
(16.9)
π
dx dy
1 2 = Ress=1
y
3
Γ\H
E(z, s)
Γ\H
and unfold.
Proof. Take T > 1 and let
ET (z, s) =
1
2
X
γ∈Γ∞ \Γ
This is still invariant under Γ, and if z ∈ F, then ET (z, s) = E(z, s) − y s δ(y > T ). Therefore
Ress=1 ET (z, s) = π3 . Now
Z
π
dx dy
Ress=1 ET (z, s) 2
3 Γ\H
y
Z
π
dx dy
y s δ(y ≤ T ) 2
= Ress=1
3
y
Γ∞ \H
vol(Γ\H) =
T s−1
π
Ress=1
3
s−1
π
= .
3
=
(16.10)
(16.11)
(16.12)
(16.13)
We can compute vol(G(Z)\G(R)) for many G similarly (such as G split of rank one).
P
Suppose ϕ(z) is a maass form. Then ϕ(z) = r∈Z fr (y)e(rx), and
∆ϕ(z) =
X
− y 2 fr00 (y) + 4πr2 y 2 fr (y) e(rx)
r∈Z
54
(16.14)
implying fr (y)(4πr2 y 2 − λϕ ) − y 2 fr00 (y) = 0. This and fr (y) = O(y A ) for some A implies
√
fr (y) = aϕ (r) yKv (2π|r|y)
for r 6= 0, where λϕ =
1
4
(16.15)
− v2.
We say that ϕ is even if ϕ(x + iy) =Rϕ(−x + iy) and odd if ϕ(x + iy) = −ϕ(−x + iy). We say
1
that ϕ is a maass cusp form if for all y, 0 ϕ(x + iy) dx = 0. Let L20 (Γ\H) be the space of cuspidal
functions. We have
Z
hf, gi =
f (z)g(z)
Γ\H
dx dy
.
y2
(16.16)
Lemma 16.2. For f, ∈ L2 (Γ\H), h∆f, yi = h∆f, gi.
R
Proof. We need Γ\H ω = 0 for ω = (fxx + fyy )g − (gxx + gyy )f dx dy. But ω = d (fx g − gx f )dy +
(fy g − f gy )dx . By Stokes’ theorem, we’re done.
We also have that for f > 0,
Z
hf, ∆f i =
−f (fxx + fyy ) dx dy
(16.17)
(fx2 + fy2 ) dx dy
(16.18)
Γ\H
Z
=
Γ\H
>0
(16.19)
because the integrands differ by d(f fx dy + f fy dx). As a result (and some spectral theory), we
get that
L20 (Γ\H) =
M
(16.20)
Cϕi
i∈N
∆ϕi = λi ϕi .
As y → ∞, E(z, s) ∼ Ay s + By 1−s so
ϕ(r) ∈ Cc∞ (R), define
2 dx dy
Γ\H |E(z, s)| y 2
R
(16.21)
doesn’t quite converge. But for
1
Eϕ (z) =
E z, + ir ϕ(r) dr.
2
R
(16.22)
Eis = hEϕi : ϕi ∈ Cc∞ (R)i.
(16.23)
Z
Then Eϕ ∈ L2 (Γ\H). Define
Then we have:
55
Theorem 16.3. L2 (Γ\H) = L20 (Γ\H) ⊕ Eis ⊕ C · 1.
Theorem 16.4. For y ∈ R+ , {x + iy : 0 ≤ x ≤ 1} = Iy → Γ\H; then Iy equidistribute to
Equivalently,
Z
1
Z
f (x + iy) dx →
f
Γ\H
0
dx dy
y2
dx dy
.
y2
(16.24)
as y → 0 for each f ∈ Cc∞ (Γ\H).
Proof. Use the decomposition. For f = 1, this is clear. For f ∈ L20 , the left hand side is identically
zero, and
Z
f
Γ\H
dx dy
= hf, 1i = 0.
y2
(16.25)
Finally,
Z
0
1
1
1
Eϕ (x + iy) dx =
ϕ(r)
E x + iy, + ir dx
2
R
0
Z
1
1
=
ϕ(r) f1 (r)y 2 +ir + f2 (r)y 2 −ir dr
Z
Z
(16.26)
(16.27)
R
1
= O(y 2 )
(16.28)
as y → 0.
Fix Me We have more quadratic stuff now. Reorganize? (16)
√
Suppose √D is squarefree, negative, and for simplicity D ≡ 1 (mod 4). Let K = Q( D) and
OK = Z[ 1+2 D ]. Let IK be the group of fractional ideals, so K × \IK = Cl(K). Consider triples
(I, α, β) for I ⊆ K a fractional ideal, α and β generating I, and =(αβ) > 0. We can also mod
this out by K × , acting by scaling all of I, α, and β. Call the quotient RD . Now we have a map
ψD : RD → H by (I, α, β) 7→ αβ . Then ψD is SL2 (Z)-equivariant, with SL2 (Z) acting on RD by
change of basis.
Lemma 16.5. z ∈ im ψD iff z =
√
−b+ D
2a
for a, b, c ∈ Z with b2 − 4ac = D.
Proof. For the “if” implication, take β = 2a and α = −b +
√
D. Check that α and β span an ideal.
√
1+ D
∈ span(α, β),
2
b2 −D
that 4a ∈ Z.
For the “only if” implication, after scaling, we can assume that β = 1. Then
so α =
√
−b+ D
2a
for some a ∈ N and b ∈ Q. Then
√
1+ D
2 α
∈ span(α, 1) shows
Corollary 16.6. |HD | = |Cl(K)|.
Fix I1 , . . . , Ir representatives in Cl(K). Then for <(s) > 1,
X
z∈HD
r X s
X
α
E(z, s) =
=
β
i=1
56
(16.29)
αβ
with the inner sum ranging over α, β ∈ Ir , α (mod β), and =(αβ) > 0. Now =( αβ ) = =( |β|
2 ).
We get that
r X
X
i=1
√ !s
r
s X X
N (Ii ) D
2
=D
N (βIi−1 )−s
|β|2
i=1 β∈Ii
s X
= D2
N (I)−s
(16.30)
(16.31)
IOK
s
2
= D ζK (s).
(16.32)
∼
We have π : P GL+
→ H by g 7→ g · i. Let
2 (R)/P SO(2) −
T ⊆
P GL+
2 (R)
x 0
×
=
: x, y ∈ R , xy > 0 .
0 y
(16.33)
Lemma 16.7. For g ∈ SL2 (R), π(gT ) is a geodesic, and all geodesics arise this way.
Proof. For g = e, π(T ) = {ix : x > 0} which is a geodesic. In general, π(gT ) = gπ(T ) and z 7→ gz
is an isometry. If g = ac db and tx = ( x0 10 ), then
π(g · tx ) =
We get a geodesic joining
a
c
axi + b
ax b
.
·i=
cx d
cxi + d
(16.34)
and db .
For D > 0 and a, b, c ∈ Z with b2 − 4ac = D, let γ[a,b,c] be the geodesic joining
√
√
√ −b− D
=b+ D −b− D T . Now gT = (gT g −1 )g; we have
.
This
is
given
by
π
2a
2a
2a
√
−b+ D
2a
and
√ √
√ 1
x − y −b 2c
x+y 1 0
x 0
2a
b + √D
−b + D −b − D
√ = √
−
.
0 y
0 1
2a
2a
−2a −b + D 4a D
2
2 D 2a b
(16.35)
√
−2c . Then M 2
Let M[a,b,c] = −b
[a,b,c] = D · I. For K = Q( D), we get i[a,b,c] : K ,→ M2 (Q) by
2a b
√
u + v D 7→ u + vM[a,b,c] . Observe:
√
D+
• i−1
2
[a,b,c] (M2 (Z)) = Z[
this is OK .
D
], the unique quadratic ring of discriminant D. If D is squarefree,
• For g ∈ SL2 (Z), gM[a,b,c] g −1 = Mg[a,b,c] .
• If M ∈ M2 (Z) with M 2 = D, then M =
−b −2c
2a b
∼
for some a, b, c ∈ Z with b2 − 4ac = D.
−1
Given [a, b, c], consider θ[a,b,c] : K −
→ Q2 by λ 7→ i[a,b,c] (λ) ( 10 ). Let I[a,b,c] = θ[a,b,c]
(Z2 ). Then I
is a fractional ideal of K. For g ∈ SL2 (Z),
57
1
−1 1
ig[a,b,c] (λ)
= gi[a,b,c] (λ)g
.
0
0
(16.36)
We know that there exists λ ∈ K × such that λg −1 ( 10 ) = ( 10 ). Then I[a,b,c] λ = Ig[a,b,c] . So the
ideal class of I[a,b,c] in Cl(K) is independent of G-orbit. Conversely, if I OK is an ideal, and α, β
is a basis of I, we get i : K ,→ M2 (Q) by
a
α β i(λ)
= λ(aα + bβ).
b
(16.37)
As I is an ideal, i−1 (M2 (Z)) = OK and θi−1 (Z2 ) = α−1 I. This provides a bijection
SL2 (Z)\{(a, b, c) : b2 − 4ac = D} ↔ Cl(K).
(16.38)
Let
[
SD =
T[a,b,c] Z[a,b,c]
(16.39)
SL2 (Z)\[a,b,c]
Z[a,b,c] =
√ √
−b + D −b − D
.
2a
2a
(16.40)
Q
Q
Let A be the adele ring 0p Qp × R, where 0 consists of sequences (ap ) such that ap ∈ Zp for
all but finitely many p. Then Q ,→ A and A/Q is compact. We have
ϕ
+
+
→ P GL+
P GL+
2 (Q)\P GL2 (A)
2 (Z)\P GL2 (R)/P SO(2) −
.Y
P GL2 (Zp ) × P SO(2)
(16.41)
p
by ϕ(g)∞ = g, ϕ(g)p = 1.
Theorem 16.8 (Strong approximation). ϕ is an isomorphism.
+
+
Proof. For
(Q)
Q injectivity, suppose g ∈ P GL2 (R) such that ϕ(g) = 1. Then there exists hQ ∈ P GL2−1
and k ∈ p P GL2 (Zp ) × P SO(2) such that ϕ(g) = hQ k. The for every p, hQ · kp = 1 so hQ = kp ∈
P GL2 (Zp ) implying hQ ∈ P GL+
2 (Z). Now considering the infinite place, g = hQ k∞ is trivial in
+
P GL+
(Z)\P
GL
(R)/P
SO(2).
2
2
For surjectivity, it is enough to show that for all primes p,
1
P GL2 (Zp ) · P GL2 Z
= P GL2 (Qp ).
p
(16.42)
Suppose g = ac db ∈ P GL2 (Qp ). We have ϕm : Qp → Qp /pm Zp ≈ Z[ p1 ]/pm Zp . We can pick
A B ∈ P GL (Z[ 1 ]) such that h + pm Z = ϕ (g) for m 0. So g h−1 ∈ P GL (Z ).
hp = C
2
p
p
m
p p
2
p
D
p
58
Now define TD = {(x, y) : x2 − Dy 2 is a unit} with
(x1 , y1 ) · (x2 , y2 ) = (x1 y1 + Dx2 y2 , x1 y2 + x2 y1 ).
(16.43)
For example, TD (Q) ≈ K × and
TD (A) ≈ A×
K =
0
Y
Kp× × (K ⊗Q R)×
(16.44)
p
√
where K = Q( D). We have
.Y
∼
→ Cl(K).
TD (Z)p −
TD (Q)\TD (Af )
(16.45)
p
We have i[a,b,c] : TD (A) → GL2 (A) → P GL+
2 (A). Examine TD (A) · Z[a,b,c] :
P GL+
2 (Q)
(t∞ , t2 , . . . , tp , . . .)
√
√ −b − D −b + D
, 12 , . . . , 1p , . . .
P GL2 (Zp ) (16.46)
2a
2a
∞
Claim. TD (A) · Z[a,b,c] = SD .
√
D
. For z ∈ H,
We can also do the case D < 0 adelically. For b2 − 4ac = D, let Z[a,b,c] = −b+
2a
+
−1
Kz = stabz ⊆ P GL2 (R). So stabi = P SO(2) and stabg(i) = gP SO(2)g . Therefore
√
KZ[a,b,c]
x
D −b
=
−y
0
2a
y
b
= xI + √
D −2a
|
{z
1
2a √b
√
0
D 2a D
2c
.
−b
}
y
x
(16.47)
(16.48)
M[a,b,c]
√
√
We get i[a,b,c] : K ,→ M2 (Q) for K = Q( D) by u + v D 7→ uI + vM[a,b,c] . We have TD (A) ·
Z[a,b,c] = HD .
For points on a sphere: Let H be the quaternions (over R). We have a quadratic form hz, wi =
+ wz). Let V ⊆ H be the subset of purely imaginary quaternions; that is, those α such that
α = −α. Then S 2 ⊆ V as (a, b, c) = ai + bj + ck, as the elements of norm 1.
1
2 (zw
∼
Let P1 (H× ) = H× /Z(H× ) −
→ SO(3) by
ϕ
ϕ(α)β = αβα−1 =
αβα
.
|α|2
(16.49)
For β ∈ V \ {0}. Z(β) = R ⊕ Rβ ≈ C by u + vβ 7→ u + v|β|i. Now consider
{(a, b, c) : a2 + b2 + c2 = D, a, b, c, D ∈ Z}.
59
(16.50)
√
√
We have stab ai+bj+ck
−D)
=
P(R
⊕
R(ai
+
bj
+
ck)).
We
obtain
i
:
K
,→
H
for
K
=
Q(
[a,b,c]
D
√
√
D+ −D
−1
by u + v −D 7→ u + v(ai + bj + ck), and i[a,b,c] (H(Z)) = Z[
].
2
Our analogies are:
P GL2
H
Heegner points/geodesics
hyperbolic Laplacian
maass forms
P(H× )
S2
elements of fixed norm
Laplacian
harmonic polynomials
√
) are coefficients of theta functions
Recall that for harmonic polynomials P , a2 +b2 +c2 =D P ( (a,b,c)
D
(modular forms of half-integral weight). So we might expect that for ϕ a maass form, we would get
a “weight 12 ” maass form ϕ
e with
P
Z
X
ϕ(z) ∼
ϕ(z) dz.
(16.51)
SD
z∈HD
Fix Me One of these should have a tilde. (17)
Fix Me Now we’re doing L-functions of two modular forms. (18)
Suppose f (z) ∈ Sk (Γ) and g(z) ∈ Sk (Γ) are Hecke cusp forms. Then
Z
f (z)g(z)y k E(z, s)
As =
Γ\H
dx dy
y2
(16.52)
is meromorphic in C, and is holomorphic for <(s) > 1. We have that
Z
f (z)g(z)y k
Ress=1 As =
Γ\H
=
(
0
3 dx dy
π y2
f=
6 c·g
f = g.
3
π hf, f i
(16.53)
(16.54)
But also
Z
f (z)g(z)y k ·
As =
Γ\H
Z
X
=(γz)s
γ∈Γ∞ \Γ
dx dy
y2
f (z)g(z)y k+s−2 dx dy
=
Γ∞ \H
Z ∞Z 1
=
0
Z
1
2
X
0 n>0
∞
=
0
X
af (n)e(nz)
X
ag (m)e(−mz)y k+s−2 dx dy
(16.55)
(16.56)
(16.57)
m>0
af (n)ag (n)e−4πny y k+s−2 dy
(16.58)
n>0
X af (n)ag (n)
=
Γ(k + s − 1)
(4πn)k+s−1
(16.59)
= L(f × g, s)(4π)−(k+s−1) Γ(k + s − 1)
(16.60)
n>0
60
where
X af (n)ag (n)
L(f × g, s) =
n>0
nk+s−1
.
(16.61)
In the case f = g, L(f × f, s) is meromorphic in C is a unique pole at s = 1. Also L(f × f, s)
has polynomial growth in vertical strips. We also have a functional equation relating L(f × g, s) to
L(f × g, 1 − s).
Theorem 16.9. There exists c such that
X af (n)2
∼ cX.
nk−1
(16.62)
n≤X
In fact c = Ress=1 L(f × f, s).
P
a (n)N
Remark. If n>0 f ns were reasonably behaved for every N , we would know Ramanujan.
Recall that
L(f, s) =
1
Y
p
(1 − αf,p
p−s )(1
−1 −s
− αf,p
p )
.
(16.63)
We can do something similar for L(f × g, s):
!
X af (pm )ag (pm )
L(f × g, s) =
pm(k+s−1)
p
m>0
!
P
r1 −r2 P
s1 −s2
Y X r1 +r2 =m αf,p
α
g,p
s1 +s2 =m
=
m(s−1)
p
p
m>0
Y
(16.64)
(16.65)
!−1
=
Y
−s
(1 − αf,p αg,p p
)(1 −
−1 −s
αf,p αg,p
p )(1
−
−1
αf,p
αg,p p−s )(1
−
−1 −1 −s
αf,p
αg,p p )
.
p
(16.66)
The Langlands philosophy: think of
α
f,p
0
0
α−1
f,p
as a (semisimple) conjugacy class in GL2 , or
SL2 . We have a block diagonal action GL2 × GL2 → GL4 by (v, w) 7→ v ⊗ w, preserving conjugacy
classes, and
αf,p
−1
αf,p
×
αg,p
*
−1
αg,p
7→
!+
αf,p αg,p
..
.
.
(16.67)
For the case f = g,
!−1
L(f × f, s) = ζ(s)
Y
−2 −s
2
(1 − αf,p
p−s )(1 − αf,p
p )(1 − p−s )
p
61
.
(16.68)
We have a symmetric square map Sym2 : GL2 → GL3 with
α
α−1
*α2

1
7
→
+
α−2
 .
(16.69)
More generally, we have SymN : GL2 → GLN +1 , and an L-function
!−1
N
L(Sym f, s) =
Y
N −2 −s
−N −s
N −s
(1 − αf,p
p )(1 − αf,p
p ) · · · (1 − αf,p
p )
.
(16.70)
p
It’s conjectured that this extends holomorphically to C.
Fix Me Back to maass forms, maybe? (19)
θ sin θ
1x
Let G= SL2
(R), K =SO2 (R) = kθ = −cos
sin θ cos θ : θ ∈ R/2πZ , N = {nx = ( 0 1 ) : x ∈ R},
1
2
and A = ay = y −0 1 : y ∈ R+ a maximal torus. Then we have the Iwasawa decomposition:
0 y 2
1
1
y 2 xy − 2
= x + iy.
G = N AK, G/K ≈ N A ≈ H by g 7→ g · i, and nx ay =
−1
0
y
2
Recall the Laplacian on G,
∆G = −y
2
∂2
∂2
+
∂x2 ∂y 2
+y
∂2
.
∂x∂θ
(16.71)
Any f ∈ C ∞ (G) can be written
f (nx ay kθ) =
X
fk (x + iy)e(kθ).
(16.72)
k∈Z
Applying ∆G , we have


X
∂2
∂ 
 2 ∂2

(∆G f )(nx ay kθ ) =
+
+
iky
−y
 fk (x + iy)e(kθ).
2
2
∂x
∂y
∂x 

k∈Z |
{z
}
(16.73)
∆k
For Γ ⊆ G discrete, we need to study L2 (Γ\G) under the right action of G, fh (g) = f (gh).
Under ∆G , if f ∈ C ∞ (Γ\G) with ∆G f = λf , then ∆k fk = λfk . Now for γ ∈ Γ given by γ = ac db ,
If for z = x + iy, we let
a b
c d
az+b
cz+d
1
y2
0
1
xy − 2
1
y− 2
!
=y
− 12
ay ax + b
.
cy cx + d
(16.74)
= x0 + iy 0 , then get
1
√ 0
y
cx+d
|cz+d|
cy
|cz+d|
0
y x0
0 1
|
cy
− |cz+d|
.
cx+d
|cz+d|
{z
kz
62
!
}
(16.75)
Now for g = hx ay kθ , we have
f (g) = f (γg)
(16.76)
= f (γax hy kθ )
(16.77)
= f (hx0 ay0 kz kθ )
X az + b cz + d k
e(kθ).
=
fk
cz + d
|cz + d|
(16.78)
(16.79)
k∈Z
For f ∈ C(H) and γ =
a b
c d
, define f |k γ to now be
f |k γ = f
az + b
cz + d
cz + d
|cz + d|
k
.
(16.80)
We say that f is a maass form of weight k and level N if for every γ ∈ Γ0 (N ), f |k γ = f ,
∆k f = λf for some λ, and f (x + iy) = O(y A ) for some A.
k
Theorem 16.10. For f (z) ∈ Mk (Γ0 (N )), let F (z) = y 2 f (z). Then F is a maass form of weight k
and level N .
k
Proof. We have F (x + iy) = O(y 2 ), ∆k f = k2 (1 − k2 )f , and F |k γ = F for γ ∈ Γ0 (N ).
1
Let θ1 (z) = y 4 θ(z). Then for γ ∈ Γ0 (4), define
θ1 (γz)
θ1 (z)
1
cz + d 2 c =
.
d
|cz + d|
d
J(γ, z) =
(16.81)
(16.82)
For k ∈ 21 Z, f is a maass form of weight k and level N if f (x + iy) = O(y A ) for some A,
∂
f (γz) = J(γ, z)2k f (z), and f is an eigenfunction of ∆ 1 = ∆ + iy
2 ∂x .
2
For any s > 0, ∆k
ys
= s(1 − s)ys, so for s > 1, letting =s (z) = y s , we can define
Ek (s, z) =
=
1
2
X
(=s |k γ)(z)
(16.83)
γ∈Γ∞ \Γ0 (4)
2k
X y s (cz + d)k
2k c
.
|cz + d|k+2s d d
(16.84)
(c,d)=1
4|c
For f ∈ Sk (Γ) and g ∈ S` (Γ), consider
Z
f (z)g(z)Ek−` (s, z)y
Γ\H
63
k+`
2
dx dy
.
y2
(16.85)
Take k =
Z
1
2
and N = 4. Then for n 6= 0, we get
iy+1
E 1 (s)e(−nx) dx = y
s
X Z
2
iy
(c,d)=1
iy+1
d
iy
c
d
k
k
(cz + d) 2 −s (cz + d)− 2 −s e(−nx) dx
(16.86)
4|c
= ys
X
c−2s
c6=0
X nd c Z i+∞ k
k
d
z 2 +s z 2 −s e(−nx) dx
e −
c
d i−∞
d (c)
|
{z
}
(16.87)
call this W 1 ,s (4π|n|y)
4
2
= y W 1 ,s (4π|n|y)
X
−2s
c
4
K 1 (−n, 0; c).
(16.88)
2
c6=0
For c = 4p, K 1 (−n, 0; c) = K 1 (np−1 , 0; 4)S(n · 4−1 , 0; p). We want to determine
2
2
X x ax S(a, 0; p) =
e
.
p
p
(16.89)
x (p)
Then S(ab, 0; p) = S(a, 0; p)
b
p
for (b, p) = 1. And, taking α to be any nonsquare modulo p,
X x x
e
S(1, 0; p) =
p
p
x (p)
2
2
1 X
1 X
x
αx
=
−
e
e
2
p
2
p
x (p)
x (p)
1 √
1 √
= p p − − p p
2
2
√
= p p.
(16.90)
(16.91)
(16.92)
(16.93)
We end up with
Z
iy+1
iy
E 1 (z, s)e(−nx) dx = W 1 ,s (4π|n|y)
2
X
4
c6=0
c
n
c2s
.
(16.94)
| {z }
L(χn ,2s)
17
Subconvexity of L-Functions
Recall that
X
s
E(z, s) = |D| 2 L(s, χD )ζ(s)
z∈HD
Also, we have
64
(?).
(17.1)
L2 (SL2 (Z)\H) = 1 ⊕ L20 (SL2 (Z)\H) ⊕ Eis,
(17.2)
Eis = hEϕ : ϕ ∈ Cc∞ (R)i.
(17.3)
1
E z, + it ϕ(t) dt.
Eϕ (z) =
2
R
(17.4)
where
Here
Z
Corollary 17.1. To conclude equidistribution, we need that as |D| → ∞,
1 X
1
E z, + it → 0.
|HD |
2
(17.5)
z∈HD
Observe that the residue at s = 1 of (?) gives
3
π |HD |
1
= |D| 2 L(1, χD ).
Theorem 17.2. L(1, χD ) = |D|o(1) .
Hence to prove the corollary, we need to show that there exists δ > 0 such that for every t,
1
L( 12 + it, χD ) = O(|D| 4 −δ ). This is a subconvexity problem.
For n ∈ N, χ : (Z/mZ)× → C× is called a Dirichlet character to modulus m. We say that χ is
primitive if there does not exist a proper divisor n of m such that χ factors through (Z/nZ)× . In
this case, we say that m is the conductor of χ.
There are exactly m characters to modulus m. The number of primitive characters is
X
ϕ(d)µ
n
d
d|m
If m is squarefree, the above equals
m1− .
Q
=
Y
(ϕ(pr ) − ϕ(pr−1 )).
(17.6)
pr kn
p|m (p
− 2). If 2km, this number is zero, otherwise it is
Observe that if D is odd and squarefree, then χD (n) =
P
n
Define τ (x) = n (m) χ(n)e m
, the Gauss sum for χ.
√
Lemma 17.3. If χ is primitive, then |τ (χ)| = m.
n
D
is primitive.
Proof. Consider χ : Z/mZ → C by extension by zero. Then χ
b(r) =
nr
n (m) χ(n)e m .
(mod m
c ) such that
P
If (r, m) = c 6= 1, then since χ is primitive, there exists s ≡ 1
Then n 7→ ns gives χ
b(r) = χ(s)b
χ(r) so χ
b(r) = 0.
χ(s) 6= 1.
On the other hand, if (r, m) = 1, n 7→ nr gives χ
b(r) = χ(r)τ (χ).
Now Plancherel implies
X
n (m)
|χ(n)|2 =
1 X
|b
χ(r)|2
m
r (m)
65
(17.7)
so ϕ(m) =
ϕ(m)|τ (χ)|2
,
m
giving |τ (χ)| =
√
m.
We can now write
χ(r) =
nr 1 X
χ(n)e
τ (χ)
m
(17.8)
nr τ (χ)χ(−1) X
χ(n)e
m
m
(17.9)
n (m)
=
n (m)
because τ (χ) = τ (χ)χ(−1).
Twisted Poisson summation formula: For f ∈ C 2 (R), with f (x) = O((1+|x|)−2 ), and χ primitive
of conductor m, then
X
χ(n)f (n) =
n∈Z
X
χ(n)
X
f (km + n)
(17.10)
k∈Z
n (m)
X χ(n) X k kn fb
=
e −
m
m
m
k∈Z
n (m)
τ (χ)χ(−1) X
k
.
=
χ(k)fb
m
m
(17.11)
(17.12)
k∈Z
Suppose χ is even (meaning χ(−1) = 1). Then let
θχ (t) =
X
2
χ(n)e−πn t .
(17.13)
n∈Z
Twisted Poisson summation yields
τ (χ)
θχ (t) = √ θχ
m t
For L(χ, s) =
is nontrivial.
χ(n)
n∈N ns
P
1
m2 t
.
(17.14)
converges absolutely for <(s) > 1, and conditionally for <(s) > 0 if χ
−1
Assume that χ is nontrivial. Then θχ (t) = O(e−t ) as t → ∞, and θχ (t) = O( e√tt ) as t → 0.
R∞
s
Then 0 θχ (t)t 2 dtt is entire in s, and if <(s) > 1, is equal to
XZ
n∈Z 0
Applying t 7→
1
,
m2 t
∞
2
s
χ(n)e−πn t t 2
s
s
dt
= 2π − 2 L(χ, s)Γ
= Λ(χ, s).
t
2
we obtain Λ(χ, s) = τ (χ)m−s Λ(χ, 1 − s).
Corollary 17.4. L(χ, s) is entire, with trivial zeros at the nonpositive even integers.
66
(17.15)
Now suppose that χ is odd, meaning χ(−1) = −1. Then define
X
θχ (t) =
2
χ(n)ne−πn t .
(17.16)
n∈Z
We end up with
s+1
Λ(χ, s) = π
Γ
L(χ, s)
2
Λ(χ, s) = (−i)τ (χ)m−s Λ(χ, 1 − s).
− s+1
2
L(χ, s) has an Euler product
Q
p (1
(17.17)
(17.18)
− χ(p)p−s )−1 in the domain of convergence.
Convexity bounds: let s = σ + it. For σ > 1, L(χ, σ + it) = O(1). For σ < 0, the functional
1
1
equation gives L(χ, σ + it) = O((1 + |t|) 2 −σ , m 2 −σ ). The Phragmen-Lindelof principle shows that
1
L χ, + it
2
Define Snχ =
Pm
i=1 χ(i).
1
1
= O ((1 + |t|) 4 + m 4 + ).
(17.19)
Then by summation by parts,
X
∞
1
1
1
χ
L χ,
=
Sn √ − √
2
n
n−1
n=1
∞
X
1
− 52
χ
=
Sn
) .
3 + O(n
2n 2
n=1
(17.20)
(17.21)
We have |Snχ | ≤ ϕ(m) ≤ m by the orthogonality relations (so Snχ = Snχ (m) ). Also, we have that
√
√
for every r, |b
χ(r)| ≤ m, as well as χ
b(0) = 0, so |Snχ | m log m. Finally, we have the trivial
bound |Snχ | ≤ n. Now
√
m log m
∞
X
X
1
n
L χ,
3 +
2
2n 2 √
n=1
1
m log m
√
m log m
(17.22)
3
n2
1
m 4 (log m) 2 .
(17.23)
1
1
To prove subconvexity, we need a bound of the form |Snχ | m 2 −δ for n ∼ m 2 .
1
1
Return to the Riemann
O (|t| 4 + ).
P∞ zeta1function. The convexity bound for ζ( 2 + it) as t → ∞ isP
Observe that if t 6= 0, n=1 1 +it converges conditionally. For subconvexity, we need nr=1 rit n2
√
1
−δ
2
t
for n ∼ t. Let
67
M
X
M
RN
=
nit
(17.24)
n=N
M
−N
X
it
=N
= N it
n=0
M
−N
X
1+
n it
N
1
n
(17.25)
n 2
n 3
+O(( N
) ))
eit( N − 2 ( N )
.
(17.26)
n=0
P −N it(( n )− 1 ( n )2 )
N
2 N
= o(1). Then we need M
to be small.
n=0 e
PN
Model problem: for α, β ∈ (0, 1], we want to bound n=0 e(αn + βn2 ). We have
We can ignore the third tem if
tn3
N3
2
N
N
X
X
e(αn + βn2 ) =
e(α(n − m))e(β(n − m)(n + m))
n=0
(17.27)
m,n=0
=
2N −|c|
N
X
X
e(αc)
c=−N
e(βcd)
(17.28)
d=|c|
d≡c (2)
N
X
min
c=−N
1
,N
|1 − e(2βc)|
N log N.
(17.29)
(17.30)
As an exercise, one can show that for k ∈ N,
N
X
e(nk α) N
1−
1
2k
+
.
(17.31)
n=1
Returning to Snχ for n ∼
√
m, consider
X
χ (m)
This implies |Snχ | √
χ 2
χ (m) |Sn | ,
P
|Snχ |2 =
n
X X
We get
χ(i)χ(j)
(17.32)
χ (m) i,j=1
= m · #{i ≤ n : (i, m) = 1}
(17.33)
mn.
(17.34)
mn, which is not useful. Now try
68
X
|Snχ |4 =
X
X
χ(i1 i2 )χ(j1 j2 )
(17.35)
χ (m) i1 ,i2 ,j1 ,j2
χ (m)
m · #{i1 , i2 , j1 , j2 ∈ [1, m] : i1 i2 ≡ j1 j2
(mod m)}
(17.36)
2
n
X
m
τr (c1 )τr (c2 )
(17.37)
c1 ,c2 =1
c1 ≡c2 (m)
1+
m
1
1
So |Snχ | m 4 + n 2 +
18
n
1
n4
n +
m
2
.
(17.38)
, and we only obtain the trivial bound.
m4
The Burgess Estimate
P
1
2
We want to bound SχN = 2N
n=N χ(n) for χ a primitive character modulo q and N ∼ q . Here is the
idea: if ab is small compared to N ,
SχN =
2N
X
χ(n + ab) + O(ab)
(18.1)
n=N
= χ(a)
2N
X
χ(an + b) + O(ab).
(18.2)
n=N
Let A, B ∈ N with AB small compared to N . Then
ABSχN
B X
2N
A X
X
χ(n + ab) + O(A2 B 2 )
a=1 b=1 n=N
A
B
X
X
=
χ(a)
a=1
(18.3)
χ(an + b) + O(A2 B 2 )
b=1
B
2N
X X
χ(an
≤
+
b)
+ O(A2 B 2 )
a=1 n=N b=1
B
X
X
χ(y + b) + O(A2 B 2 )
≤
v(y) A
X
y (q)
(18.4)
(18.5)
(18.6)
b=1
for v(y) = #{1 ≤ a ≤ A, N ≤ n ≤ 2N : an = y}. We have
X
v(y) = AN
(18.7)
y (q)
X
v(y)2 = #{(a1 , n1 ), (a2 , n2 ) : a1 n2 ≡ a2 n1
y (q)
69
(mod q)}.
(18.8)
If AN < q, congruence implies equality, so
X
2
v(y) ≤
AN
X
σ0 (r)
(18.9)
r=1
y (q)
AN q .
(18.10)
Hölder: pick r ∈ N. Then
1− 1 
1
B
2r  2r1 
r
2r
X
X
X X
2
2 2

v(y)
v(q)  .
A B +
χ(y + b)  

ABSχN
y (q) b=1
y (q)
(18.11)
y (q)
Therefore
1
ABSχN q (AN )1− 2r

 1
2r
 r

X
X
Y

y + bi 
 .
χ


y (q) 1≤b1 ,...,br ≤B i=1 y + ci 
| {z } 1≤c1 ,...,cr ≤B
(18.12)
Q(y)
P
1
The Weil bound gives y (q) χ(Q(y)) q 2 unless χ(Q(y)) ≡ 1. This occurs if ord χ = k ord Q(y)
is a kth power. The number of bi , ci for which this happens is B r , since bi , ci take on at most
2r
k ≤ r values. So now
1
1
1
ABSχN A2 B 2 + q (AN )1− 2r (B r q + B 2r q 2 ) 2r
2
2
1
1− 2r
A B + q (AN )
1
2r−1
1
2
(B q
1
2r
(18.13)
1
4r
+ Bq ).
(18.14)
1
For B = q 2r and A = N 2r+1 q 2(2r+1) ,
−1
2
− r+1
SχN q N N 2r+1 q 2r(2r+1)
.
r+1
(18.15)
1
1
This is useful if N > q 4r . We can choose r large enough if N > q 4 +δ . If N ∼ q 2 , we can pick
1
1
r = 2, lowering our bound by q 24 . (We could get q 16 savings if we do more work to improve the
initial O(ab) bound.)
19
Amplification
Recall that we have
X
|SχN |4 N 2 q 1+ .
χ (q)
Here are some ideas to improve this:
70
(19.1)
• Use higher moments (such as replacing 4 with 6), but this usually doesn’t help much.
• Shorten the character sum; this works in certain cases, like q highly composite, but not if q is
prime.
• Use amplification.
We’ll explain this last part. Let L be a small power of N . Then look at
X
χ (q)
L
2
X
|SχN |4 c` χ(`)
(19.2)
`=1
for (c` ) general in C. We hope for a bound of the form N 2 q 1+
PL
`=1 |c` |
2,
which we call (?).
Expanding our sum,
X
`1 ,`2
2N
X
c`1 c`2
X
χ(m1 n2 n3 n4 `1 `2 ) =
n1 ,n2 ,n3 ,n4 =N χ (q)
N ≤n ,n ,n ,n ≤2N 1 2 3 4
c`1 c`2 ϕ(q)# n1 n2 `1 ≡ n3 n4 `2 .
X
(19.3)
`1 ,`2
If we assume N 2 L < q, we get
2
N
X
#{n1 n2 `1 = n3 n4 `2 } σ0 (r)
(19.4)
.
(19.5)
r=1
`1
|r
(`1 ,`2 )
N 2+
`1
(`1 ,`2 )
So our sum is
ϕ(a)N 2+
L
X
|c` |2 .
(19.6)
`=1
1
But we want a similar bound for N 2 L > q. Suppose that (?) holds for N ∼ q 2 and L = q δ for
2 q 1+
some δ > 0. Then take c` = χ(`). Then |SχN |4 N L
. We now deal with
X
L
2
X
χ 6= 1|SχN |4 c` χ(`) .
(19.7)
`=1
Fact.


(r, q) 6= 1
0
X
χ(r) = −1
(r, q) = 1, r 6≡ 1 (mod q)


χ6=1
ϕ(q) − 1 r ≡ 1 (mod q).
71
(19.8)
So we get
L
X
c`1 c`2 (ϕ(q) − 1)#{N ≤ n1 , n2 , n3 , n4 ≤ 2N : n1 n3 `2 = n2 n4 `1 }
(19.9)
`1 ,`2 =1
− #{N ≤ n1 , n2 , n3 , n4 ≤ 2N : n1 n3 `2 6= n2 n4 `1 } .
(19.10)
Remark. For n1 , n2 , n3 , n4 , n1 n2 n3 n4 has one of ϕ(q) values.If `1 = `2 , we have at least 2N 2 solutions.
Now n1 n3 `2 = n2 n4 `1 =⇒ ∃ h : −L ≤ h ≤ L such that n1 n3 `2 − n2 n4 `1 = hq. We want to
understand the asymptotics of the shifted divisor sum
2
4N
X
σ0 (a)σ0 (b).
(19.11)
a,b=N 2
`2 a−`1 b<qh
A model problem is given by
X
X
σ0 (a)σ0 (a + 1) =
X
X
σ0 (a2 + a).
(19.12)
a=1
a=1
Non-shifted sums are easier. For example, fixing σ > 1 and σ 0 < 1,
X
Z
ζ(s)2 xs
σ0 (n) =
<(s)=σ
n≤x
Z
ds
s
(19.13)
ds
ζ(s)2 xs
+ Ress=1
s
s
(19.14)
ζ(s)2 xs
=
<(s)=σ 0
= x log x + x + O(x1−δ ).
(19.15)
1
Alternatively, count lattice points under a hyperbola. This gives a bound of O(x 2 ), and we can
improve this further with more work.
The shifted convolution problem: let f, g be (cuspidal) maass forms, and we want to understand
the sum
X
X
λf (n)λg (n).
(19.16)
m,n=1
`1 m−`2 n=h
Morally speaking,
X
σ0 (m)σ0 (n) ≈
P
|n|≤X
σ0 (an2 + bn + c).
(19.17)
|n|≤X
m,n≤X
`1 m−`2 n=k
Theorem 19.1 (Bykovskii).
X
2
σ0 (n2 + D) = c1 (D)X log X + c0 (D)X + O,D (X 3 + ).
72
Proof. σ0 (b2 + D) = #{a, c ∈ N : b2 + D = ac}. Let K(D) = {(a, b, c) ∈ Z : b2 − ac =√ −D}.
−1
b −a
Γ = SL2 (Z) acts on K(D) by γ · cb −a
γ . Also K(D) → H by (a, b, c) 7→ b+ a−D =
−b = γ c −b
√
pc
D
Z(a,b,c) . We have |Za,b,c | S=
a and =(Za,b,c ) = a . By finiteness of the class group, there exist
Z1 , . . . , Zλ(D) such that i Γ/ΓZi = im(K(D)).
q
P
2 +D
2
Choose h ∈ C(R+ ) such that n∈Z σ0 (n + D)h( n D
) converges absolutely. We can rewrite
this sum as
∞
X X
√
ac
D
h
b∈Z a,c=1
λ(D)
X
=
1 X
h
|ΓZk |
σ∈Γ
k=1
|σZk |
=(σzk )
(19.18)
λ(D)
1
ϕ(h, Zk );
|ΓZk |
k=1
X |σz| ϕ(h, z) =
.
h
=(σz)
X
=
(19.19)
(19.20)
σ∈Γ
Observe that ϕ is Γ-invariant.
Idea: L2 (Γ\H) = C · 1 ⊕ Eis ⊕ L20 (Γ\H) with
L20 (Γ\H) =
∞
M
(19.21)
Cuj
i=1
∆uj = λj uj
X
√
uj (x + iy) =
ρj (n) yKirj (2π|n|y)e(nx)
(19.22)
(19.23)
n∈Z
1
+ rj2 = λj .
4
(19.24)
Now we have
ϕ(h, z) = c0 +
X
hϕ, uj iuj (z) + (Eis part).
(19.25)
j
But ϕ 6∈ L2 .
Define
u (z, s, s0 ) =
X =(σz) s
|σz|
σ∈Γ
(=(σz))s0 .
(19.26)
If <(s0 ) > 0, then u(z, s, s0 ) ∈ L2 (Γ\H). Now
u (z, s, s0 ) =
X
σ∈Γ∞ \Γ
=(σz)s+s0
X
n∈Z
73
s
((=(σz))2 + (<(σz) + n)2 )− 2 .
(19.27)
Applying Poisson summation with respect to x to
√
X
s+1
(=(σz))
σ∈Γ∞ \Γ
1
n∈Z (y 2 +(x+n)2 ) 2s
P
, we get
πΓ s−1
2
(=(σz))1−s
Γ 2s
(19.28)
!
s
s−1
1−s
2π 2 X
+
|m| 2 (=(σz)) 2 K s−1 (2π|m|=(σz))e(m=(σz))
2
Γ 2s m6=0
(19.29)
√
s+1
πΓ s−1
2(2π)−s0 i− 2 X
2
E(1 + s0 , z) +
|m|−1−s ρm (z, s, s0 )
=
√
Γ 2s
πΓ 2s
m6=0
(19.30)
where
ρm,s,s0 (z) =
X
(2π|m|=(σz))
σ∈Γ∞ \Γ |
s+1
+s0
2
K s−1 (2π|m|=(σz)) e(m<(σz)).
{z 2
}
(19.31)
Fm (s,s0 ,=(σz))
Now
Z
hρm,s,s0 , 1i =
Fm (s, s0 , y)e(mx) dx
(19.32)
Γ∞ \H
=0
(19.33)
√
hρm,s,s0 , uj i = pj (m) 2πm
Z
Γ∞ \H
ρm,s,s0 , E
√
1
+ it, •
= 2πm
2
Γ
Fm (s, s0 , y)Kirj (2π|m|y)
1
1
2
2π 2 −it
d 1 (m)
− it ζ(1 − 2it) 2 +it
dy
3
(19.34)
y2
Z
··· .
(19.35)
We end up with
1
u(z, s, s0 ) = · · · +
L uj , + s0 b(uj ) + Eis
(19.36)
2
j
Z ∞
∞
X
1
1
12 log u
2
ϕ(h, z) =
h(u)
+ c1 + c2 (n)s η(z) du +
b0,h (rij )L uj ,
uj (z) + Eis.
π
2
1
X
j=1
(19.37)
Kronecker’s limit formula:
E(z, s) =
1
3
6
(s − 1) + (γ − log(z) − log |y 2 η(z)|) + O((s − 1)2 ).
π
π
Fi Me Where does this proof actually end, if ever? (20)
Now for f, g ∈ Sk (Γ), consider L(f × g, 12 ). We need to understand
74
(19.38)
X
λf (n)λg (n)
(19.39)
m,n≤X
`1 m−`2 n=h
where f (z) =
P∞
m=1 m
k−1
2
λf (m)e(mz). Now for N = `1 `2 , define
X
ph,s =
=(γz)s e(−h<(γ1 γ2 ))
(19.40)
γ∈Γ∞ \Γ0 (N )
V (z) = f (`1 z)g(`2 z)y k .
(19.41)
Then V (z) is Γ0 (N )-equivariant. We have
Z
hph,s , V (z)iN =
Γh,s V (z)
Γ0 (N )\H
Z
dx dy
y2
(19.42)
y k+s f (`1 z)g(`2 z)e(−hx)
=
Γ∞ \H
Z
y k+s
=
Γ∞ \H
X
dx dy
y2
λf (m)λg (n)(mn)
k−1
2
(19.43)
e((m`1 − n`2 − h)x)e2π(m`1 +n`2 )y
m,n
dx dy
y2
(19.44)
=
Γ(k + s − 1)
(2π)k+s−1
X
λf (m)λg (n)
m,n
`1 m−`2 n=h
√
mn
`1 m + `2 n
k−1 ,
(`1 m + `2 n)s .
{z
|
(19.45)
}
D(f,g,s)
We have
Z
X
s
e
D(f, g, s)G(s)x
ds =
<(s)=2
λf (m)λg (n)
m,n
`1 m+`2 n=h
√
mn
`1 m + `2 n
k−1
G
`1 m + `2 n
x
. (19.46)
Now
hph,s , 1i = 0
Z
hph,s , uj i =
(19.47)
Γ\H
z s−1 ρj (h)
=
Γ
|h|s−1
So for
1
4
dx dy
y2
!
s − 21 + itj
Γ
2
e(−hx)y s uj (z)
+ t2j = λj , we obtain
75
(19.48)
s−
1
2
− itj
2
!
.
(19.49)
ph,s (z) =
∞
X
j=1
z s−1 ρj (h) Y
Γ
uj (z)
|h|s−1 ±
s−
!
1
2
± itj
1
+ Eis.
(19.50)
To continue this beyond s = 1, we need <(s − 12 + itj ) > 0. Since the uj are self-adjoint under
∆, + t2j = λj > 0. We need that for some δ > 0, λj > δ for every uj and for every N .
1
4
Selberg’s eigenvalue conjecture: λj ≥ 41 ; equivalently tj ∈ R.
We have
D(f, g, s) =
X
hy k f (`1 z)g(`2 z), uj (z)i.
(19.51)
j
Understanding
Z
s
e
D(f, g, s)G(s)x
ds
(19.52)
<(s)=1−δ
requires knowing the growth of D(f, g, s) on vertical strips to be controlled. We have D(f, g, s) =
hph,s , V i. By Plancherel, this is
X
hph, suj ihV, uj i + Eis =
X 2s−1 ρj (h) Y
j
Suppose s =
1
2
j
|h|s−1
Γ
s−
1
2
±
± itj
2
!
hV, uj i.
(19.53)
+ iR. Then
Γ
iR + itj
2
π|t|
iR − itj
Γ
: Γ(it) ≈ e− 2 .
2
(19.54)
Consider the following:
• |R| > |tj | : e−
• |R| < |tj | : e−
π|R|
2
.
π|tj |
2
.
What now needs to be shown is:
Proposition 19.2 (Sarnak’s triple product estimate).
k
k
hV, uj i = hy 2 f (z)y 2 g(z), uj (z)i = O(e−
π|tj |
2
|t|A ).
(19.55)
Recall that
Dh (f, g, s) =
(2π)s+k−1 (`
1 `2
s− 12
|h|
1
)k− 2 2s
·
∞
X
Γ
ρj (−h)
j=1
76
s− 12 +itj
2
1
s− 2 −itj
Γ
2
Γ(s + k − 1)
hV, uj i + Eis.
(19.56)
Suppose s = σ + iR. Then Γ(s) ∼ e−
|R+tj |+|R−tj |−2R
−π
2
RA e
πR
2
Rσ−1 so the absolute value of the Γ factors is ∼
. In order to control the growth of the terms with small tj , we need
|ρj (−h)|hV, uj i |tj |A
(19.57)
for some A > 0.
As an aside, applying ∆ A times, ∆(A) V =
∞
X
A
j cj λj uj
P
+ Eis for cj = hV, uj i, so
(A)
c2j t2A
V, ∆(A) V i.
j ≤ h∆
(19.58)
j=1
So we already know polynomial decay of the cj . But this is not strong enough!
πt
Lemma 19.3. e 2 (tj N )−1− |ρj (1)| e
πtj
2
(tj N ) .
Proof.
1 = huj , uj i
Z
dx dy
π
uj (z)2 E(z, s) 2
= Ress=1
3
y
Γ\H
Z ∞Z 1
π
dx dy
= Ress=1
uj (x + iy)2 y s 2
3
y
0
0
Z
∞
∞
X
dy
|ρj (n)|2 Kiy (2π|n|y)y s
= Ress=1
y
n=1 0
∞
X
|ρj (n)|2
1
1
= Ress=1
Γ
+
it
−
it
Γ
j
j .
ns
2
2
n=1
{z
}
|
(19.59)
(19.60)
(19.61)
(19.62)
(19.63)
∼e−πtj
Let L(s) =
P∞
n=1
|ρj (n)|2
ns
X
n
and c = Ress=1 L(s). Let h ∈ Cc∞ ( 21 , 1). Then
2
|ρj (n)| h
n
x
Z
L(s)e
h(s)xs s
=
(19.64)
<(s)=2
Z
= ce
h(1)x +
<(s)= 12
1
L(s)e
h(s)xs dx.
1
(19.65)
By the convexity bound, the second integral is x 2 P
O ((tj N ) 2 + ). Also
ρj (n) = ρj (1)aj (n) for
mn
aj (n) the Hecke eigenvalues of uj . Then aj (n)aj (m) = d|(m,n) aj d2 . Now we have
77
X
X
!2
aj (n)2
X
X
=
n=1
(aj (n)aj (m))2
(19.66)
n,m=1
X
≤
mn 2
aj
d
m,n=1
· d(m, n)
(19.67)
d|(m,n)
2
≤
X
X
|aj (n)|2 σ0 (n)
n=1
X σ0 (m)
.
m
m≤X
|
{z
}
(19.68)
(log 3X)2
PX
2
2 2
2
2
Pick X c12 (tj N )1+ . Then
n=1 aj (n) ∼ cX, so c X cX (log 3X) implying c P
1
2
(log 3X)2 so that c (tj N ) . For the lower bound on c, X
n=1 |aj (n)| ≥ 1 so c ≥ X .
In fact (we won’t prove this): |ρj (1)| ∼ e
20
πtj
2
|tj N |o(1) .
Applications of Subconvexity
• Let f, g be distinct Hecke cusp forms. We want to know the smallest n = N (f, g) such that
λf (n) 6= λg (n). Suppose f ∈ S2 (Γ0 (Nf )) and g ∈ S2 (Γ0 (Ng )). Then for q = e(z),
f (z)dz − g(z)dz =
X
(λf (n) − λg (n))q n−1 dq
(20.1)
n≥1
so N (f, g) is the order of vanishing of (f (z) − g(z)dz at ∞, implying
N (f, g) ≤ deg Ωx0 [Nf ,Ng ] ≤ 2gen(· · · ) + 1 = [Nf , Ng ]1+o(1) .
(20.2)
Now suppose V ∈ Cc∞ ( 21 , 1). Consider
X
(f, g, N ) =
X
λf (n)λg (n)V
n
n
N
.
P
P
P
Note that if N = N (f, g), then (f, g, N ) = (f, f, N ). We have L(f ×g, s) = n
so that
X
1
(f, g, N ) =
2πi
Z
L(f × g, s)Ve (s)N s d
(20.3)
λf (n)λg (n)
,
ns
(20.4)
(2)
Z
= δf =g Ve (1)N Ress=1 L(f × f, s) +
( 21 )
L(f × g, s)Ve (s)N s ds.
1
(20.5)
The convexity bound is L(f × g, 21 + it) (|t| + [Nf , Ng ]) 2 + . Also Ress=1 L(f × f, s) =
[Nf , Ng ]o(1) . We end up with N (f, g) [Nf , Ng ]1+ . A subconvexity bound L(f ×g, 21 +it) 1
(|t| + [Nf , Ng ]) 2 +−δ would give N (f, g) [Nf , Ng ]1+−2δ .
78
1
.
Fact (Kowalki-Michel-Vanderkam). This holds for δ = 12 − 82
√
• Let K = Q( −D) for D positive and squarefree. Recall
√ the Minkowski bound: for every
√
2
a ∈ Cl(K), there exists I OK with [I] = a and N (I) ≤ π D. As an aside, I = (2a, −b+ D)
√
√
√
D
∈ H with h(zI ) = 2aD . So constant D is optimal.
has N (I) = 2a, and have zI = −b+
2a
Fix a character ψ : Cl(K) → C× . We want to know N (ψ); that is, we want to determine the
existence of I with N (I) = N (ψ) and ψ(I) 6= 1. Let
θψ (z) =
X
IOK
• ψ(I)e(N (I)z) ∈ M1 Γ0 (D),
.
D
(20.6)
θψ is cuspidal unless ψ is real. We have
L(ψ, s) =
ζK (s) =
X ψ(I)
N (I)s
IOK
X
1
IOK
N (I)s
(20.7)
.
(20.8)
Taking X = N (ψ),
X
N (I)≤X
V
N (I)
X
=
X
ψ(I)V
N (I)≤X
N (I)
X
(20.9)
so that
Z
L(ψ, s)Ve (s)X s ds =
(2)
Z
ζK (s)Ve (s)X s ds.
(20.10)
(2)
1
1
A subconvexity bound of the form L(ψ, 21 ) D 4 +−δ implies N (ψ) D 2 +−2δ . It turns
1
out that this is possible for δ = 24001
.
Suppose Cl(K) is cyclic. We want to know the smallest N (I) among I generating Cl(K). Let
G = Cl(K) and δG (a) be 1 if a generates G and 0 otherwise. Then
δG (a) =
X µ(d) X
ψ(a).
d
d
d||G|
(20.11)
ψ =1
• Quantum Chaos: Let X be a (compact) hyperbolic manifold, ∆x a (the?) Fix Me (21) LaplaceBeltrami operator, and u1 , u2 , . . . , uj , . . . with ∆x uj = λj uj . We have λj → ∞ and that the
uj span L2 . Consider the measure d|uj |2 on X. It is conjectured that d|uj |2 converges to the
hyperbolic measure on X.
Arithmetic Q.U.E.: In L2 (Γ\H), u1 , . . . , uj , . . . Hecke maass forms. Then:
Theorem 20.1. For ϕ ∈ Cc∞ (Γ\H),
Z
Γ\H
dx dy j→∞
ϕ(z)|uj (z)|2 2 −−−→
y
79
Z
ϕ(z)
Γ\H
dx dy
.
y2
(20.12)
An example is given by |E( 12 + it, z)|2 dxy2dy as t → ∞.
Proof. We have
Z
u1 (z)E
Γ\H
1
+ it, z
2
1
1
dx dy
u1 (z)E
+ it, z y 2 −it 2
2
y
Γ∞ \H
1
1
L u1 , + it Gt
= L u1 ,
2
2
dx dy
=
y2
Z
(20.13)
(20.14)
1
for Gt consisting of gamma functions. So Gt ∼ |t|− 2 . So decay to 0 of the integral amounts to
a subconvex bound on L(u1 , 12 + it).
For the case of cusp forms, a subconvex bound on L(f ⊗ g ⊗ h, 12 + it) would imply the result,
but we don’t know this.
80
To do. . .
2 1 (p. 4): There’s a gap here. Fill in?
2 2 (p. 14): Rectangular contour.
2 3 (p. 15): Rectangle
2 4 (p. 16): Fix Me Fix the (?).
2 5 (p. 23): Fundamental domain picture
2 6 (p. 23): Actually determine the genus.
2 7 (p. 26): Fix Me Maybe deal with the overset alignments?
2 8 (p. 26): Fix Me We may have confused notation.
2 9 (p. 29): Fix Me This part appears in the notes after equidistribution. May want to
organize differently?
2 10 (p. 35): Fix Me Are they both d0 , or does d appear?
2 11 (p. 41): Fix Me Disgtinguish between p and P .
2 12 (p. 50): Can the (?) be referenced?
2 13 (p. 50): What is the set the sum runs over?
2 14 (p. 52): Fix Me Should all of 2πi be raised to power?
2 15 (p. 53): Fix Me More stuff on this topic appears later. Reorganize?
2 16 (p. 56): Fix Me We have more quadratic stuff now. Reorganize?
2 17 (p. 60): Fix Me One of these should have a tilde.
2 18 (p. 60): Fix Me Now we’re doing L-functions of two modular forms.
2 19 (p. 62): Fix Me Back to maass forms, maybe?
2 20 (p. 74): Fi Me Where does this proof actually end, if ever?
2 21 (p. 79): Fix Me Which word should go here?
81