multiple dirichlet series associated to prehomogeneous
TRANSCRIPT
Multiple Dirichlet Series
Associated to Prehomogeneous
Vector Spaces and the Relation
with GL3(Z) Eisenstein Series
byLi-Mei Lim
B.S., Massachusetts Institute of Technology, 2008M.Sc., Brown University, 2010
A dissertation submitted in partial ful�llment of therequirements for the degree of Doctor of Philosophyin Department of Mathematics at Brown University
Providence, Rhode Island
May 2013
c© Copyright 2013 by Li-Mei Lim
This dissertation by Li-Mei Lim is accepted in its present form
by Department of Mathematics as satisfying the
dissertation requirement for the degree of Doctor of Philosophy.
Date
Je�rey Ho�stein, Advisor
Recommended to the Graduate Council
Date
Gautam Chinta, Reader
Date
Michael Rosen, Reader
Approved by the Graduate Council
Date
Peter M. Weber, Dean of the Graduate School
Vitae
Li-Mei Lim was born at Emerson Hospital in Concord, MA on November 13,
1985. She grew up in Lexington, MA and is a graduate of Lexington High School.
During high school, she attended PROMYS (the Program in Mathematics for Young
Scientists) at Boston University, �rst as a student, then as a junior counselor. Li-Mei
was an undergraduate at MIT, where she majored in mathematics and continued
her involvement with PROMYS as a counselor. She graduated from MIT with her
Bachelor of Science in 2008.
Li-Mei began her graduate studies in the fall of 2008 at Brown University. She
received her Master of Science in 2010 and started her dissertation under the direction
of Je�rey Ho�stein. With her collaborators, Thomas Hulse, Eren Mehmet K�ral and
Chan Ieong Kuan, Li-Mei published a paper, �The Sign of Fourier Coe�cients of
Half-integral Weight Cusp Forms� in the International Journal of Number Theory.
Also as a graduate student, Li-Mei was �rst a teaching assistant, then a teaching
fellow. She participated in programs run by the Sheridan Center for Teaching and
Learning and served as the graduate student liaison to the Sheridan Center for
the mathematics department for two years. In the fall of 2012, Li-Mei received a
departmental award for excellent teaching. She graduates from Brown University
with her Ph.D. in mathematics in May 2013.
iv
Acknowledgements
I would not have made it this far without the generous help of many people. Though
I will certainly miss many, I would like to list some of them here.
First of all, I would like to thank my advisor, Je� Ho�stein, for his constant
guidance and good humor. The value of his mentorship and support cannot be
overstated. To Gautam Chinta, thank you for introducing me to this problem and
suggesting this beautiful proof. Gautam's careful reading and suggestions have cer-
tainly improved this document. I would also like to thank Michael Rosen for reading
my thesis and serving on my thesis committee.
I am very grateful to my mathematical siblings, Thomas Hulse, Eren Mehmet
K�ral, Chan Ieong Kuan and David Lowry. Thank you for the many delightful
conversations, mathematical and otherwise. To Miles Wheeler, thank you for your
LATEX expertise and for helping to make this document more beautiful.
I would also like to thank the wonderful Brown math department sta� for pa-
tiently guiding me through graduate school. Thanks also to Jill Pipher for generously
giving me funding for semesters o� teaching.
Finally, I am grateful to have a fantastic family. Thanks to my father, Beoleong,
my mother, SweeCheng, my brother, Shenkiat, and my sister, Li-Ann for their un-
conditional love and support. And last but not certainly not least, thanks to my
husband, Ben Harris, for always being there for me.
v
Contents
Vitae iv
Acknowledgments v
1 Introduction 1
2 Background 4
2.1 The Hurwitz Zeta Function . . . . . . . . . . . . . . . . . . . . . . . 42.2 Binary Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 The Case of the GL2 Eisenstein Series 7
3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 The Multiple Dirichlet Series as a Sum of Eisenstein Series Values 17
4.1 The SL3 (Z) Eisenstein Series . . . . . . . . . . . . . . . . . . . . . . 174.2 The Multiple Dirichlet Series for the space of Ternary Quadratic Forms 184.3 Relating the Two Eisenstein Series . . . . . . . . . . . . . . . . . . . 20
5 The Multiple Dirichlet Series as a Product of Double Dirichlet
Series 23
5.1 Explicit Coset Representatives . . . . . . . . . . . . . . . . . . . . . . 235.2 Rewriting the Sum with Genus Theory . . . . . . . . . . . . . . . . . 255.3 The Sum of ζQ(s) Over a Genus Class . . . . . . . . . . . . . . . . . 26
6 Genus Theory 29
6.1 Cosetting to Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.2 Computing Ak(Q) for Principal Forms . . . . . . . . . . . . . . . . . 36
6.2.1 The Case d ≡ 0 mod 4 . . . . . . . . . . . . . . . . . . . . . . 366.2.2 The Case d ≡ 1 mod 4 . . . . . . . . . . . . . . . . . . . . . . 47
7 The Main Result 49
7.1 Tying Up the Loose Ends into Beautiful Bows . . . . . . . . . . . . . 49
8 Conclusion 54
8.1 Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
vi
Chapter 1
Introduction
In [9], Katok and Sarnak show that for a GL2 Maass form ϕ,
∑z∈Λd
ϕ(z) =∑
Shim(Fj)=ϕ
ρj(1)ρj(d),
where the sum on the left side is over Heegner points of discriminant d, the sum on
the right side is over all half-integral weight forms that lift to ϕ under the Shimura
correspondence described in [14], and ρj(n) denotes the n-th Fourier coe�cient of
Fj.
Katok and Sarnak's result is a generalization of a well-known result for GL2
Eisenstein series. Setting E(z, s) to be the real analytic GL2 Eisenstein series, the
following is true.
Proposition 1.1. If d a fundamental discriminant,
2s
ds/2
∑z∈Λd
E(z, s) = ζ(s)L(s, χd) = ζQ(√d)(s),
where χd is the quadratic character corresponding to the �eld extension Q(√d)/Q.
This result for GL2 Eisenstein series is extended by Chinta and O�en in [3]. In
this paper, they compute a particular orthogonal period of the GL3 Eisenstein series
1
2
and show that a special value of the Eisenstein series is equal to a linear combination
of products of double Dirichlet series.
In fact, all these results are special cases of Jacquet's conjecture, stated in [7],
which relates integrals of G = GLr cusp forms over orthogonal subgroups of G to
Whittaker coe�cients on the double cover of G via a relative trace formula. In this
dissertation, I generalize the results of Chinta and O�en for GL3 Eisenstein series,
providing further evidence for Jacquet's conjecture. In particular, in [3] Chinta and
O�en compute the case for the discriminant d = 1, and in this thesis, I obtain the
result for all negative d.
The method I use involves looking at multiple Dirichlet series associated to pre-
homogeneous vector spaces. A prehomogeneous vector space is pair (G, V ) of a
complex algebraic group G and a complex vector space V such that G acts on V
with an open dense orbit. A key fact is that the algebra of relative invariants of
such an action is �nitely generated. For example, one could take the space of binary
quadratic forms with the action of GL2(C). The algebra of relative invariants is then
just generated by the usual discriminant b2 − 4ac.
M. Sato and Shintani, in [13], de�ne zeta functions associated to prehomoge-
neous vector spaces with a single relative invariant. They prove some basic results,
for example showing that these zeta functions have an analytic continuation and
satisfy a functional equation. The study of multiple Dirichlet series coming from
prehomogeneous vector spaces was originated by F. Sato in [10], [12], [11].
The multiple Dirichlet series I construct come from the action of a non-reductive
group and have multiple relative invariants rather than a single invariant. This
construction gives a function of several, rather than just one, complex variables.
3
This thesis has three parts:
• Background: I record some facts about the Hurwitz zeta function and binary
quadratic forms. None of these results are new.
• The GL2 case: Although the result in this part is well-known, the proof given
is (as far as I know) new. This part is included, in part, to give the idea for
the GL3 case, which is similar, but slightly more complicated.
• The GL3 case: This section contains my original results, using methods similar
to the previous section and is broken into multiple chapters. The �rst chapter
in this section constructs the multiple Dirichlet series and relates it to the
GL3 Eisenstein series. The second rewrites the series using genus theory and
recovers products of double Dirichlet series. The third chapter investigates the
genus theory in more depth, explicitly computing the necessary coe�cients.
Finally, in the last chapter, the main result is achieved by combining all these
results.
Chapter 2
Background
� 2.1 The Hurwitz Zeta Function
It will be necessary to use some basic facts about Hurwitz zeta functions when
manipulating the multiple Dirichlet series attached to the space of ternary quadratic
forms. Therefore, we give a brief introduction to the theory of Hurwitz zeta functions.
For 0 < a ≤ 1, the Hurwitz zeta function is de�ned to be
ζ(s, a) =∞∑n=0
1
(n+ a)s.
Like the Riemann zeta function, which the Hurwitz zeta function specializes to when
a = 1, this converges absolutely for <s = σ > 1 and has an analytic continuation to
the complex plane with simple pole at s = 1, as shown in [1].
The Hurwitz zeta function is important because we can express it as a combina-
tion of Dirichlet L-series.
Proposition 2.1. If k and m are relatively prime integers, then
ζ(w, k
m
)=
mw
ϕ(m)
∑χmod m
χ̄(k)L(w, χ).
Here, the sum is over all characters χ, primitive and imprimitive, mod m.
4
5
Proof. This proposition follows from the orthogonality of characters, that is, the fact
that ∑χmod m
χ(n)χ̄(k) =
ϕ(m) if n ≡ k mod m
0 otherwise
.
Then we have
mw
ϕ(m)
∑χmod m
χ̄(k)L(w, χ) =mw
ϕ(m)
∑χmod m
∑n>0
χ(n)χ̄(k)
nw
= mw∑
n≡kmod mn>0
1
nw
=∑
n≡kmod mn>0
1
(n/m)w
=∞∑`=0
1
(`+ k/m)w
= ζ(w, k
m
),
completing the proof.
� 2.2 Binary Quadratic Forms
To show that the multiple Dirichlet series associated to the space of ternary
quadratic forms is equal to a product of double Dirichlet series, we will need some
standard facts about binary quadratic forms.
Let X2 be the space of binary quadratic forms. The group SL2(Z) acts on X2 by
linear substitution. Namely,
γ ◦Q(x, y) = Q(γ(x, y)).
6
For a �xed discriminant D, there are �nitely many SL2(Z)-orbits. When choosing a
representative from each orbit we have the following proposition.
Proposition 2.2. A representative Q(x, y) = ax2 + bxy + cy2 for an SL2(Z) orbit
in X2 can be chosen such that (a, b) = 1.
This statement can be found in the exercises of [4] with some hints at the proof.
Chapter 3
The Case of the GL2 Eisenstein Series
Although the following result is well-known, we present a proof of it here to give
a simple example of the methods we will use in the case of GL3.
� 3.1 Preliminaries
Let E(z, s) denote the real analytic Eisenstein series for GL2. Namely, setting
Γ = SL2(Z) and Γ∞ =
{(1 n0 1
)∣∣∣∣n ∈ Z}
we have
E(z, s) =∑
γ∈Γ∞\Γ
=(γz)s.
Next, let E∗(z, s) be the half-integral weight Eisenstein series for GL2 studied by
Goldfeld and Ho�stein in [6]. To de�ne E∗(z, s), let
Γ0(4) =
{(a bc d
)∈ SL2(Z)
∣∣∣∣c ≡ 0 mod 4
}.
For γ ∈ Γ0(4), de�ne jγ(z) to be the factor of automorphy of the theta series:
jγ(z) =( cd
)ε−1d (cz + d)
12
7
8
where(cd
)is the extension of the Jacobi symbol described by Shimura in [14] and
εd =
1 if d ≡ 1 mod 4
i if d ≡ 3 mod 4
.
Then we recall the well-known fact mentioned in the introduction, Proposition
1.1, which says that a sum of special values of the Eisenstein series is equal to the
product of a zeta function and a quadratic Dirichlet L-function.
In [6], it is shown that the d-th Fourier coe�cient of E∗(z, s) is essentially L(s, χd),
so the proposition can be rewritten as
Proposition 3.1. Let d < 0. Then
∑z∈Λd
E(z, s) = ρ(1)ρ(d)
where ρ(n) is the n-th Fourier coe�cient of E∗(z, s).
� 3.2 Proof
The idea of the proof in this case is to look at the space X+ of positive de�nite
integral binary quadratic forms. To a binary quadratic formQ(x, y) = ax2+bxy+cy2,
we associate the matrix
Q =
(2a bb 2c
)so that (abusing notation), Q(x, y) = 1
2(x, y)Q(x, y)t.
Usually, we think of Γ = SL2(Z) acting on the space of binary quadratic forms.
In this case, the space of invariants of the action is one-dimensional. Namely, it is
9
generated by the discriminant, b2 − 4ac. The key for this proof, though, is to only
consider the action of Γ∞ on X+. As usual, the action is by linear substitution of
variables, so that
γ ◦Q(x, y) = Q(γ(x, y)).
In terms of matrices, this means that
γ ◦Q = γtQγ.
With this action, the space of invariants is two-dimensional. The discriminant is
still invariant, of course, but so is a = Q(1, 0). For clarity, set
r1(Q) = discQ = b2 − 4ac
r2(Q) = Q(1, 0) = a
to be the two generating invariants.
We form the double Dirichlet series associated to the space X+ with the action
of Γ∞ by
Z(s, w) =∑
Q∈Γ∞\X+
1
|r1(Q)|w|r2(Q)|s.
By rewriting this sum in two di�erent ways, we will obtain, on the one hand,
a sum of values of the Eisenstein series, and on the other, the quadratic Dirichlet
L-series. The �rst way involves writing the quotient Γ∞\X+ explicitly and using
Hurwitz zeta functions. The second way splits the sum up further.
To get the quotient explicitly, we have the following lemma.
10
Lemma 3.1. A set of representatives for the quotient Γ∞\X+ is
{ (2a bb 2c
)∣∣∣∣a > 0 and 0 ≤ b ≤ 2a− 1 and c >b2
4a, c ∈ Z
}
Proof. We compute
(1 01 1
)(2a bb 2c
)(1 10 1
)=
(2a 2a+ b
2a+ b b+ 2c
).
Therefore, we see that Q1 ∼ Q2 modulo the action of Γ∞ if b1 ≡ b2 mod 2a, and
a condition on the set of representatives of Γ∞\X+ is that 0 ≤ b ≤ 2a − 1. Since
the form is positive de�nite, a > 0 and also 4ac− b2 > 0. Therefore, c ranges freely
over the integers greater than b2
4a. So a set of representatives for Γ∞\X+ has a > 0,
0 ≤ b ≤ 2a− 1 and c > b2
4a, c ∈ Z as desired.
Therefore, we can rewrite our double Dirichlet series as
Z(s, w) =∑a>0
∑0≤b≤2a−1c>b2/4a
1
as1
|b2 − 4ac|w
=∑a>0
1
as
∑0≤b≤2a−1c>b2/4a
1
|b2 − 4ac|w
Focusing on the inner sum, we see that
∑0≤b≤2a−1c>b2/4a
1
|b2 − 4ac|w=
1
(4a)w
∑0≤b≤2a−1c>b2/4a
1∣∣ b24a− c∣∣w
=1
(4a)w
∑0≤b≤2a−1
ζ(w, α)
where α is −b2
4amod 1.
11
Using Proposition 2.1, we can rewrite this sum of Hurwitz zeta functions in terms
of L-series, but we need to be careful to ensure that α is a fraction expressed in lowest
terms.
Let d = (b2, 4a). Then d can be expressed as d = d21d0, where d0 is square-free.
Since d | b2, in fact the values of b with (b2, 4a) = d = d21d0 and b mod 2a will be
the kd1d0 with k between 0 and b2a/d1d0c inclusive such that (k, 4a/d) = 1. With
this notation, and with the understanding that ζ(w, a) might mean ζ(w, a mod 1),
our sum becomes
1
(4a)w
∑0≤b≤2a−1(b2,4a)=d
ζ(w, −b2/d
4a/d) =
1
(4a)w
∑d|4a
d=d21d0
b2a/d1d0c∑k=0
(k,4a/d)=1
ζ(w, −k2d0
4a/d21d0)
=1
(4a)w
∑d|4a
d=d21d0
b2a/d1d0c∑k=0
(k,4a/d)=1
(4a/d)w
ϕ(4a/d)
∑χ mod 4a/d
χ̄(−k2d0)L(w, χ)
=∑d|4a
d=d21d0
1
dwϕ(4a/d)
∑χ mod 4a/d
χ̄(−d0)
b2a/d1d0c∑k=0
(k,4a/d)=1
χ̄(k2)
L(w, χ).
But since the character sum in parentheses is just
b2a/d1d0c∑k=0
(k,4a/d)=1
χ̄(k2) =
d12ϕ(4a/d) if χ trivial or quadratic
0 otherwise
,
12
this sum is in fact
∑d|4a
d=d21d0
d1
2dw
∑χ mod 4a/d
trivial or quadratic
χ̄(−d0)L(w, χ).
With this, we can rewrite the original double Dirichlet series Z(s, w) as
Z(s, w) =∑a>0
1
as
∑d|4a
d=d21d0
d1
2dw
∑χ mod 4a/d
trivial or quadratic
χ̄(−d0)L(w, χ).
Carefully exchanging the order of the �rst two sums, we get
Z(s, w) =∑d odd
d1
2dw
∞∑k=1
1
(dk)s
∑χ mod 4k
triv. or quad.
χ̄(−d0)L(w, χ)
+∑2‖d
d1
2dw
∞∑k=1
1
(dk/2)s
∑χ mod 2k
triv. or quad.
χ̄(−d0)L(w, χ)
+∑4|d
d1
2dw
∞∑k=1
1
(dk/4)s
∑χ mod k
triv. or quad.
χ̄(−d0)L(w, χ).
Note that the sums over characters include all characters, primitive and imprim-
itive. We can change these sums to include only primitive quadratic characters by
summing over square-free, odd divisors of k (or 2k or 4k). Modulo a square-free odd
number m0, there is a unique primitive quadratic character, which we will call χm0 .
Of course, there are also primitive quadratic characters mod 4 and 8 to be added in
if 4 or 8 divide k, 2k or 4k. Modulo 4, there is a unique primitive quadratic character
χ4 and modulo 8 there are two primitive quadratic characters, χ8 and χ′8. With this
notation, we rewrite the series as
13
Z(s, w) =∑d odd
d1
2dw+s
∞∑k=1
1
ks
(χ̄4(−d0)P (k, χ4, w)L(w, χ4)+gd(1, k, w)
+∑m0|4k
odd, sq. free
χ̄m0(−d0)P ( 4km0, χm0 , w)L(w, χm0)
)
+∑2‖d
d1
2dw+s
∞∑k=1
2s
ks
(fd(1, k, w) + gd(2, k, w)
+∑m0|2k
odd, sq. free
χ̄m0(−d0)P ( 2km0, χm0 , w)L(w, χm0)
)
+∑4|d
d1
2dw+s
∞∑k=1
4s
ks
(fd(2, k, w) + gd(3, k, w)
+∑m0|k
odd, sq. free
χ̄m0(−d0)P ( km0, χm0 , w)L(w, χm0)
)
where we de�ne the functions fd(`, k, w) and gd(`, k, w) to be
fd(`, k, w) =
χ̄4(−d0)P ( k
2`, χ4, w)L(w, χ4) if 2` | k
0 otherwise
(3.1)
gd(`, k, w)=
P ( k
2`, χ8, w)χ̄8(−d0)L(w,χ8)+P ( k
2`, χ′8, w)χ̄′8(−d0)L(w,χ′8) if 2` | k
0 otherwise
(3.2)
and the functions P (n,w) are correction polynomials
P (n, χ, w) =∏p|n
(1− χ(p)
pw
).
14
On the other hand, we can rewrite Z(s, w) in terms of the Eisenstein series. We
�rst rewrite Z(s, w) by splitting the sum further, as follows.
∑Q∈Γ∞\X
1
|r1(Q)|w|r2(Q)|s=∑
Q∈Γ\X
1
| discQ|w∑
γ∈Γ∞\Γ
1
|γ ◦Q(1, 0)|s
We can compute γ ◦Q(1, 0) to �nd that if
γ =
(A BC D
)
then
γ ◦Q(1, 0) = A2a+ ACb+ C2c.
Note, however, that this is equal to
γ ◦Q(1, 0) = a|Cz + A|2
where
z =b+√b2 − 4ac
2a.
Then, setting z as above, the inner sum becomes
∑γ∈Γ∞\Γ
1
|γ ◦Q(1, 0)|s=
1
as
∑γ∈Γ∞\Γγ=(A B
C D )
1
|Bz + A|2s
=1
asysE(z, s)
=2s
ds/2E(z, s).
15
Therefore, we see that
Z(s, w) =∑
Q∈Γ\X+
2s
|b2 − 4ac|w+s/2E(z, s).
But �xing the discriminant of Q, we can rewrite the sum as
Z(s, w) = 2s∑d<0
1
|d|w+s/2
∑z∈Λd
E(z, s)
where Λd is the set of Heegner points of discriminant d.
With these two representations of our original double Dirichlet series Z(s, w),
we recover the original result. Restricting to the case of d an odd fundamental
discriminant, that is, d is square-free and 1 mod 4, we can compare d−w coe�cients.
On the one had, we have
2s
ds/2
∑z∈Λd
E(z, s),
namely the sum of special values of the Eisenstein series.
On the other hand, if we ignore the problems at 2, we can compute the d−w
coe�cient from the �rst representation of Z(s, w). The �rst thing we have to do
is group all the terms with w's in them. Ignoring the problems at the prime 2
and assuming all numbers are �nice� (that is, quadratic reciprocity works perfectly),
16
Z(s, w) is
∑` oddsq. free
1
2`s+w
∞∑k=1
1
ks
∑m0|k
odd, sq. free
χm0(−d)L(w, χm0) =∑`
1
2`w+s
∑k
1
ks
∑m0|k
χm0(−d)∞∑n=1
χm0(n)
nw
=∑`
∑n
1
2(`n)w`s
∑k
1
ks
∑m0|k
χm0(−`n)
=∑`,n
1
2(`n)w`s
∑m0
χm0(`n)
ms0
∑k
1
ks
=∑d
1
2dw
∑`|d
τ(d)
`sζ(s)
∑m0
χd(m0)
ms0
=∑d
1
2dw
∑`|d
τ(d)
`sζ(s)L(s, χd).
So restricting to a particular d, we get, up to some factors, that the d−w coe�cient
is ζ(s)L(s, χd), as desired.
Putting this all together, we recover the known result, that a sum of special
values of the Eisenstein series is (up to some factors) equal to the product of the �rst
and d-th Fourier coe�cients of the half-integral weight Eisenstein series.
Chapter 4
The Multiple Dirichlet Series as a Sum
of Eisenstein Series Values
In this chapter, we will construct the multiple Dirichlet series associated to the
space of ternary quadratic forms and prove its relationship with the SL3(Z) Eisenstein
series.
� 4.1 The SL3 (Z) Eisenstein Series
The SL3 Z Eisenstein Series is described in detail in [2]. For completeness, we
record the pertinent de�nitions and theorems here.
Let τ ∈ H = G/ZK, where G = GL3(R), K = O(3) and Z is the set of scalar
matrices. Then τ will be given in coordinates by
τ =
(y1y2 y1x2 x3
0 y1 x1
0 0 1
). (4.1)
Also, let Γ denote SL3(Z) and let P be the minimal parabolic subgroup in SL3(Z),
namely
P =
{(1 ∗ ∗0 1 ∗0 0 1
)}.
17
18
Then the SL3(Z) Eisenstein series is de�ned to be
E(ν1, ν2, τ) =∑g∈P\Γ
I(ν1,ν2)(g · τ),
where
I(ν1,ν2)(τ) = y2ν1+ν21 yν1+2ν2
2 .
Setting x1x2 = x3 + x4, the Eisenstein series can be written explicitly as
E(ν1, ν2, τ) = I(ν1,ν2)(τ)
×∑
(A1,B1,C1)=1(A2,B2,C2)=1
A1C2+B1B2+C1A2=0
((A1x3 +B1x1 + C1)2 + (A1x2 +B1)2y2
1 + A21y
21y
22
)−3ν1/2
(4.2)
·((A2x4 −B2x2 + C2)2 + (A2x1 −B2)2y2
2 + A22y
21y
22
)−3ν2/2 .
The A's, B's and C's are the invariants of an orbit of G∞\G. For g = (aij) ∈ G,
the invariants are:
A1 = −a31 A2 = a22a31 − a21a32
B1 = −a32 B2 = a21a33 − a23a31 (4.3)
C1 = −a33 C2 = a23a32 − a22a33.
� 4.2 The Multiple Dirichlet Series for the space of Ternary
Quadratic Forms
We want to construct a multiple Dirichlet from the space of ternary quadratic
forms which can be interpreted as both a sum of values of the SL3(Z) Eisenstein
series and as a product of double Dirichlet series. To do this, let X3 be the space
19
of ternary quadratic forms, and let X+3 be the space of positive de�nite ternary
quadratic forms. The minimal parabolic subgroup in SL3(Z) P acts on X+3 by linear
substitution, and the action has three relative invariants. Namely the invariants are
r1(T ) = r(s2 − 4pq) + (pu2 − stu+ qt2) = disc(T )
r2(T ) = s2 − 4pq = disc(T1)
r3(T ) = p
when T =
(2p s ts 2q ut u 2r
)and T1 =
(2p ss 2q
).
Then we form the multiple Dirichlet series by
Z(s1, s2, s3) =∑
T∈P\X+3 (Z)
1
|r1(T )|s1|r2(T )|s2|r3(T )|s3. (4.4)
In this chapter, we will split up the sum further to show that this series can be
interpreted as a sum of special values of the minimal parabolic Eisenstein series. The
�rst step is as follows.
Z(s1, s2, s3) =∑
T∈P\X+3 (Z)
1
|r1(T )|s1|r2(T )|s2|r3(T )|s3
=∑
T∈Γ\X+3 (Z)
∑γ∈P\Γ
1
|r1(γ ◦ T )|s1|r2(γ ◦ T )|s2|r3(γ ◦ T )|s3
=∑
T∈Γ\X+3 (Z)
1
| disc(T )|s1∑γ∈P\Γ
1
|r2(γ ◦ T )|s2 |r3(γ ◦ T )|s3.
20
For now, we will de�ne the Eisenstein series on quadratic forms to be exactly
E(s2s3, Q) =∑
γ∈Γ∞\Γ
|r2(γ ◦Q)|−s2 |r3(γ ◦Q)|−s3 ,
so
Z(s1, s2, s3) =∑
T∈Γ\X+3 (Z)
1
| disc(T )|s1E(s2, s3, T ). (4.5)
The next section is devoted to showing how the Eisenstein series de�ned in Section
4.1 is related to this new Eisenstein series. The two are, in fact, very closely related.
� 4.3 Relating the Two Eisenstein Series
In this section, we will prove the following proposition, giving the relationship
between the two Eisenstein series de�ned previously in this chapter.
Proposition 4.1. Let T be the binary quadratic form given by 2ττT , where τ is as
shown in (4.1) and Iν1,ν2, r2 and r3 are as de�ned above. Then
E(ν1, ν2, τ) = disc(τ)(ν1+2ν2)/2∑P\Γ
r2(γ ◦ T )−3ν2/2r3(γ ◦ T )−3ν1/2.
Proof. First, recall from equation 4.2 that the Eisenstein series is a sum over P\Γ
of terms involving the invariants A1, B1, C1, A2, B2, and C2. These terms are the
product of powers of
f1(τ) := (A1x3 +B1x1 + C1)2 + (A1x2 +B1)2y21 + A2
1y21y
22
and
f2(τ) := (A2x4 −B2x2 + C2)2 + (A2x1 −B2)2y22 + A2
2y21y
22.
21
Focusing on f1(τ), we can expand and rewrite it to get
f1(τ) = (A1x3 +B1x1 + C1)2 + (A1x2 +B1)2y21 + A2
1y21y
22
= A21(x2
3 + x22y
21 + y2
1y22) +B2
1(x21 + y2
1) + C21 + 2A1B1(x1x3 + x2y
21)
+ 2A1C1x3 + 2B1C1x1
= T (A1, B1, C1)
where T is given by
2ττT = 2 ·
(y2
1y22 + x2
2y21 + x2
3 x1x3 + x2y21 x3
x1x3 + x2y21 x2
1 + y21 x1
x3 x1 1
).
Note that this seemingly extraneous factor of 2 comes from the fact that we are
normalizing our forms so that (again abusing notation),
Q(a, b, c) =1
2(a, b, c)Q(a, b, c)T .
Now looking at f2(τ) and rewriting it, we get
f2(τ) = (A2x4 −B2x2 + C2)2 + (A2x1 −B2)2y22 + A2
2y21y
22
= A22(y2
1y22 + x2
4 + x21y
22) +B2
2(x22 + y2
2) + C22 − 2A2B2(x2x4 + x1y
22)
+ 2A2C2x4 − 2B2C2x2
=1
y21
T ′(C2, B2, A2)
where T ′ is the adjoint matrix of T .
This means that, if γ has top row (A1, B1, C1) and 2-by-2 minors C2, B2 and A2,
22
then we can express f1(τ) and f2(τ) as
f1(τ) = T ((1, 0, 0)γ) f2(τ) =1
y21
T ′((0, 0, 1)γ′)
= γ ◦ T (1, 0, 0) =1
y21
(γ ◦ T )′(0, 0, 1)
= r3(γ ◦ T ) =1
y21
r2(γ ◦ T )
where γ′ denotes the adjoint matrix for γ. Therefore, we see that in fact the functions
f1 and f2 are closely related to the invariants r2 and r3, and that
E(ν1, ν2, τ) = y3ν21 Iν1,ν2(τ)
∑P\Γ
r2(γ ◦ T )−3ν2/2r3(γ ◦ T )−3ν1/2
= y2ν1+4ν21 yν1+2ν2
2
∑P\Γ
r2(γ ◦ T )−3ν2/2r3(γ ◦ T )−3ν1/2
= disc(τ)(ν1+2ν2)/2∑P\Γ
r2(γ ◦ T )−3ν2/2r3(γ ◦ T )−3ν1/2
as desired.
Chapter 5
The Multiple Dirichlet Series as a
Product of Double Dirichlet Series
The goal of this chapter is to show that the multiple Dirichlet series Z(s1, s2, s3)
is a Dirichlet series in the variable s3 whose coe�cients are (as functions of s1 and
s2) linear combinations of products of double Dirichlet series. To do this, we will
make use of genus theory and the results of [3].
� 5.1 Explicit Coset Representatives
We begin by writing down explicit coset representatives for the quotient P\X+3 .
Given a ternary quadratic form, we can represent it by a symmetric matrix
T =
(2p s ts 2q ut u 2r
).
Then we can describe the coset representatives with restrictions on p, q, r, s, t and
u as follows:
Proposition 5.1. A set of coset representatives for B(3)\X+3 is
{T =
(2p s ts 2q ut u 2r
)∣∣∣∣∣ p > 0, 0 ≤ s ≤ 2p− 1, q ∈ Z, q > s2
4p
r ∈ Z, r > pu2−stu+qt2
s2−4pq, (t, u) ∈ Z2/QZ2
}
where Q is the matrix
(2p ss 2p
).
23
24
Proof. Considering how B(3) acts on X+3 , we compute
(1 0 0a 1 0b c 1
)(2p s ts 2q ut u 2r
)(1 a b0 1 c0 0 1
)
=
(2p 2pa+s 2pb+sc+t
2pa+s 2pa2+2as+2q 2abp+bs+acs+2cq+at+u2pb+sc+t 2abp+acs+at+bs+2cq+u 2pb2+2bcs+bt+2qc2+2cu+b+2r
)
So we see that T1 ∼ T2 if s1 ≡ s2 mod 2p. But also, note that if T1 ∼ T2, then
(t1u1
)=
(t2u2
)+
(2p ss 2q
)(bc
)+
(0at1
).
Now, if T1 ∼ T2 and s1 = s2, then a is zero since s1 = 2p1a + s2. So we see
that T1 ∼ T2 implies that (t1, u1) and (t2, u2) lie in the same coset of Z2/QZ2, where
Q =
(2p ss 2p
).
Therefore, we see that
{T =
(2p s ts 2q ut u 2r
)∣∣∣∣∣ p > 0, 0 ≤ s ≤ 2p− 1, q ∈ Z, q > s2
4p
r ∈ Z, r > pu2−stu+qt2
s2−4pq, (t, u) ∈ Z2/QZ2
}
is a set of coset representatives for the quotient B(3)\X+3 .
25
� 5.2 Rewriting the Sum with Genus Theory
Using the result of the previous section, we can rewrite the multiple Dirichlet
series as
Z(s1, s2, s3) =∑p>0
smod 2pq>s2/4p
r>(pu2−stu+qt2)/(s2−4pq)(t,u)∈Z2/QZ2
1
|r(s2 − 4pq) + (pu2 − stu+ qt2)|s1|s2 − 4pq|s2 |p|s3
=∑p>0
1
|p|s3∑
q>s2/4p0≤s≤2p−1
1
|s2−4pq|s2∑
r> pu2−stu+qt2
s2−4pq
(t,u)∈Z2/QZ2
1
|r(s2−4pq)+(pu2−stu+qt2)|s1.
Notice, however, that the conditions on the �rst two sums are exactly the same as
the bounds we had in the GL2 case. Therefore, we can express Z(s1, s2, s3) as
Z(s1, s2, s3) =∑d<0
1
|d|s2∑
Q∈Γ∞\X+2
1
|Q(1, 0)|s3∑
rd+Q′(t,u)<0(t,u)∈Z2/QZ2
1
|rd+Q′(t, u)|s1.
De�ning
ζQ(s) =∑
γ∈Γ∞\Γ
1
|γ ◦Q(1, 0)|s,
and noting that a translation by γ ∈ Γ will not change the innermost sum, we can
rewrite the series as
Z(s1, s2, s3) =∑d<0
1
|d|s2∑
Q∈Γ\X+2
discQ=d
ζQ(s3)∑
rd+Q′(t,u)<0(t,u)∈Z2/QZ2
1
|rd+Q′(t, u)|s1.
26
Now the innermost sum is just some Dirichlet series, and we can express it as
∑rd+Q′(t,u)<0(t,u)∈Z2/QZ2
1
|rd+Q′(t, u)|s1=∑k<0
Ak(Q)
|k|s1,
where
Ak(Q) = #{
(t, u) ∈ Z2/QZ2∣∣Q′(t, u) ≡ k mod d
}. (5.1)
Now, Ak(Q) depends only on the genus class of Q (modulo d) since forms in the same
genus class are precisely those which represent the same congruence classes modulo
d. Therefore, we can break the sum over SL2(Z) classes of binary quadratic forms
into a sum over the genus classes. Denote that two forms are in the same genus class
by Q1 ∼ Q2. This gives
Z(s1, s2, s3) =∑d<0
1
|d|s2∑
[Qi] genusclasses mod d
∑k<0
Ak(Qi)
|k|s1∑Q∼Qi
ζQ(s3). (5.2)
� 5.3 The Sum of ζQ(s) Over a Genus Class
The innermost sum of (5.2) is computed in [3] in Proposition 4.1. Namely, Chinta
and O�en show that
∑Q∼Qi
ζQ(s) =#O×d
2ω(d)−1
∑χ∈X (d)
χ(Qi)LOd(s, χ) (5.3)
where X (d) denotes the group of genus class characters of the group of SL2(Z)
equivalence classes of primitive integral binary quadratic forms Cl(d), Od is the
27
order Z[d+√d/2], and LOd
(s, χ) is de�ned to be
LOd(s, χ) =
∑a
χ(a)
N(a)s.
In the de�nition of LOd(s, χ), the sum runs over all invertible ideals of Od.
Following [3], we can de�ne the genus class characters as follows. First, we de�ne
χ(p) on Cl(d) by
χ(p)(Q) =
χp′(a) if (p, a) = 1
χp′(c) if (p, c) = 1
where p′ = (−1)(p−1)/2p. Of course, if Q is primitive, then one of the two conditions
will be satis�ed. Now, writing d = d1d2 where d1 is an even fundamental discriminant
and d2 is an odd discriminant, we can set d0 to be d1 times each distinct prime factor
of d2. Then we can de�ne the genus class character χe′1,e′2 , where e′1e′2 = d0, as follows:
χe′1,e′2 =∏p|e1
χ(p).
As noted by Chinta and O�en, as e1 ranges over positive odd squarefree divisors of d,
χe′1,e′2 ranges over all the genus characters exactly once if d is even and exactly twice
if d is odd. Therefore, we can rewrite the innermost sum of (5.2) (that is, (5.3)) as
∑Q∼Qi
ζQ(s) =#O×d
2ω(d)−1
∑e1|d
odd, squarefree
χe′1,e′2(Qi)LOd(s, χe′1,e′2).
Now we can use the results of Kaneko in [8] and Chinta and O�en in [3] to express
this L-function in terms of Dirichlet L-functions. Namely, in Proposition 4.2 of [3],
28
it is shown that for fundamental discriminants e1 and e2 with d = e1e2f2, that
LOd(s, χe1,e2) = L(s, χe1)L(s, χe2)
∏pk‖f
Pk(p−s, χe1(p), χe2(p)) (5.4)
where Pk(p−s, χe1(p), χe2(p)) is the correction polynomial de�ned by the generating
series
F (u,X;α, β) =∑k≥0
Pk(u, α, β)Xk =(1− αuX)(1− βuX)
(1−X)(1− pu2X).
We see, then, that we get products of Dirichlet L-functions coming into the series.
In particular,
Z(s1, s2, s3) =∑d<0
1
|d|s2∑
[Qi] genusclasses mod d
∑k<0
Ak(Qi)
|k|s1· #O×d
2ω(d)−1
×∑e1|d
odd, squarefree
χe′1,e′2(Qi)L(s3, χe1)L(s3, χe2)∏pi‖f
Pi(p−s3 , χe1(p), χe2(p)).
(5.5)
In the next chapter, we will carefully compute the values of the coe�cients Ak(Qi)
so that this expression for Z(s1, s2, s3) becomes explicit.
Chapter 6
Genus Theory
The goal of this chapter is to compute the coe�cients Ak(Q). That is, we need
to count
Ak(Q) = #{
(t, u) ∈ Z2/QZ2∣∣Q′(t, u) ≡ k mod d
}where d = discQ.
We �rst show that it is enough to consider just the principal forms of discriminant
d, then compute the coe�cients for principal forms.
� 6.1 Cosetting to Freedom
First de�ne SQ to be the multi-set
SQ ={Q′(t, u)
∣∣(t, u) ∈ Z2/QZ2}.
That is, consider the values of Q′(t, u) and retain the multiplicities. Then computing
Ak(Q) is the same as counting the multiplicity of k in SQ. Now we prove the following
lemma, showing that we need only compute Ak(Q) in the case where Q is a principal
form.
Lemma 6.1. Let Q0 be the principal form with discriminant d, and let Q be a
29
30
primitive binary quadratic form with the same discriminant. Then there exists α in
Z/|d|Z× such that
SQ = αSQ0 ={αQ′0(t, u)
∣∣(t, u) ∈ Z2/Q0Z2},
retaining multiplicites as before.
Proof. We consider two cases. The �rst is when d ≡ 0 mod 4 and the second is
when d ≡ 1 mod 4.
Case 1: d ≡ 0 mod 4. In this case, let n = −d4. Then we have
Q0 =
(2 00 2n
)Q′0 =
(2n 00 2
).
Now suppose we have another binary quadratic form Q with discriminant d, and Q
and Q′ are given by
Q =
(2a bb 2c
)Q′ =
(2c −b−b 2a
).
Since Q and Q′ are primitive, either a or c is relatively prime to d. Assume that
(a, d) = 1. The case of (c, d) = 1 follows from the same argument.
Then note that Q′(t, u) = ct2 − btu+ au2, so
aQ′(t, u) = act2 − abtu+ a2u2
= nt2 +
(au− b
2t
)2
= Q′0
(t, au− b
2t
).
31
Of course, since d ≡ 0 mod 4, b must be even, and b2is an integer. So we see that
SQ ={a−1Q′0(t, au− bt/2)
∣∣(t, u) ∈ Z2/QZ2}.
Now, if we could show that
{Q′0(t, au− bt/2)
∣∣(t, u) ∈ Z2/QZ2}
={Q′0(t, u)
∣∣(t, u) ∈ Z2/Q0Z2}
we'd be done.
Consider
T =
{(t, au− bt/2)
∣∣∣∣(t, u) ∈ Z2/
(2a bb 2c
)Z2
}.
Then T can be rewritten as
T =
{(t, v) ∈
(1 0−b/2 a
)Z2/
(2a b0 2n
)Z2
}.
But now, we will see that a set of representatives for T is also a set of represen-
tatives for Z2/Q0Z2, so
{Q′0(t, au− bt/2)
∣∣(t, u) ∈ Z2/QZ2}
={Q′0(t, u)
∣∣(t, u) ∈ Z2/Q0Z2}.
If (t1, v1) ∼ (t2, v2) mod
(2a b0 2n
), where (ti, vi) ∈
(1 0−b/2 a
)Z2, then
(t1 − t2, v1 − v2) =
(2a b0 2n
)(x, y)t
32
for some (x, y) ∈ Z2. Then notice that
(2a b0 2n
)(x, y)t =
(2 00 2n
)(a b/20 1
)(x, y)t,
so in fact, (t1, v1) ∼ (t2, v2) mod Q0.
On the other hand, if (t1, v1) ∼ (t2, v2) mod Q0 for (ti, vi) ∈(
1 0−b/2 a
)Z2,
then, letting t1 − t2 = t and v1 − v2 = v,
(t, v) =
(2 00 2n
)(x, y)t
for some (x, y) ∈ Z2. But then
(2 00 2n
)(x, y)t =
(2 b0 2n
)(1/a −b/2a0 1
)(x, y)t.
But also, since (ti, vi) ∈(
1 0−b/2 a
)Z2,
(t, v) =
(1 0−b/2 a
)(x′, y′)t
for some (x′, y′) ∈ Z2, and in particular, t = x′ = 2x and v = −bx′/2 + ay′ =
−bx+ ay′ = 2ny. So
2ny + bx = ay′ = (2ac− b2/2)y + bx
and bx− b2
2y ≡ 0 mod a. But since (b, a) = 1, we have
a|(x− by/2),
33
implying (1/a −b/2a0 1
)(x, y)t ∈ Z2.
Thus, (t1, v1) ∼ (t2, v2) mod
(2a b0 2n
). This tells us that a set of represen-
tatives for T is a set of representatives for Z2/Q0Z2, and vice versa, giving us the
desired result.
Case 2: d = 1 mod 4. In this case, we set n = 1−d4. Then
Q0 =
(2 11 2n
)Q′0 =
(2n −1−1 2
).
Now suppose we have another binary quadratic form Q of discriminant d, and Q
and Q′ are given by
Q =
(2a bb 2c
)Q′ =
(2c −b−b 2a
).
Again, since Q and Q′ are primitive, either a or c is relatively prime to d. Assume
that (a, d) = 1. The case of (c, d) = 1 follows from the same argument.
Then note that, since Q′(t, u) = ct2 − btu+ au2,
aQ′(t, u) = act2 − abtu+ a2u2
= nt2 − t(au+
1− b2
t
)+
(au+
1− b2
t
)2
= Q′0
(t, au+
1− b2
t
).
34
Of course, since d = b2 − 4ac ≡ 1 mod 4, 1−b2∈ Z. Therefore, we see that
SQ ={a−1Q0(t, au+ (1− b)t/2)
∣∣(t, u) ∈ Z2/QZ2}.
As in the previous case, we'll consider the quotient we're summing over, and show
that a set of representatives for this quotient is a set of representatives for Z2/Q0Z2.
We let
T =
{(t, au+
1− b2
)∣∣∣∣(t, u) ∈ Z2/QZ2
},
and see that
T =
{(t, v) ∈
(1 0
1−b2
a
)Z2/
(2a ba b−1
2+ 2n
)Z2
}.
So we now want (t1, v1) ∼ (t2, v2) mod
(2a ba b−1
2+ 2n
)if and only if (t1, v1) ∼
(t2, v2) mod
(2 11 2n
)for (ti, vi) ∈
(1 0
1−b2
a
)Z2.
First suppose (t1, v1) ∼ (t2, v2) mod
(2a ba b−1
2+ 2n
). Then, letting t = t1 − t2
and v = v1 − v2, we know that for some (x, y) in Z2,
(t, v)t =
(2a ba b−1
2+ 2n
)(x, y)t
=
(2 11 2n
)(a b−1
20 1
)(x, y)t,
showing that (t1, v1) ∼ (t2, v2) mod
(2 11 2n
).
On the other hand, if (t1, v1) ∼ (t2, v2) mod
(2 11 2n
), with (ti, vi) ∈
(1 0
1−b2
a
)Z2,
35
we have, for some (x, y) ∈ Z2,
(t, v)t =
(2 11 2n
)(x, y)t
=
(2a ba b−1
2+ 2n
)(1a
1−b2a
0 1
)(x, y)t.
We need only determine, now, that
(1a
1−b2a
0 1
)(x, y)t is in Z2, or in other words,
that a divides x+ 1−b2y. Since (ti, vi) ∈
(1 0
1−b2
a
)Z2,
(t, v) = (2x+ y, x+ 2ny) = (x′, (1− b)x′/2 + ay′)
for some x′ and y′ in Z. Using the system of equations obtained from this relation,
we �nd that
x+1− b
2y =
1
4n− 1
((2n+
b− 1
2
)x′ − ay′ +
(−b+ b2
2
)x′ + (a− ab)y′
)=
1
4n− 1
((2n+
b2 − 1
2
)x′ − aby′
)=
a
4n− 1(2cx′ − by′).
However, since (a, b) = 1 and 4n − 1 = −d = 4ac − b2, this implies that
(a, 4n− 1) = 1. Therefore, a divides x+ 1−b2y as desired.
Hence, we arrive at the desired result. Namely, in both cases,
SQ = αSQ0 ,
36
where in fact,
α =
a−1 if (a, d) = 1
c−1 if (c, d) = 1
.
� 6.2 Computing Ak(Q) for Principal Forms
As a consequence of the previous section, we only need to compute Ak(Q) for Q
a principal form of discriminant d. As before, we split into two cases: d ≡ 0 mod 4
and d ≡ 1 mod 4.
6.2.1 The Case d ≡ 0 mod 4
This is the more complicated of the two cases because of the factors of 2. In this
case, we have the following proposition.
Proposition 6.1. Let Q be the principal binary quadratic form with discriminant
d = −4n, and let ω(a) denote the number of distinct odd prime factors of a. Suppose
(d, k) = m where m = m0m21 with m0 squarefree, and let m0 = 2δp1 . . . p` (where
δ = 0 or 1) and m1 = 22α0q2α11 . . . q2αr
r . Let d′ = d/m and k′ = k/m. Assume that
37
m−10 k′ is a square modulo every odd prime power of d′. Then
Ak(Q) =
m12ω(d/m)+γ if d′m0 ≡ 3 mod 4
or 4 | d′ and m0 · d′
4≡ m0k
′ ≡ 1 mod 4
or 2 - m0 and 32 | d′
m12ω(d/m)+γ−1 if 2‖d′ and 2 - m0
or 2 - d′ and 2 | m0
or m0d′ ≡ 1 mod 4
or 4 | d′ and 2 | m0 and (m0/2)−1(k′ − d′
4) ≡ 2 mod 8
or 4 | d′ and m0 · d′
4≡ 3 mod 4
or 8‖d′ and m−10 k′ ≡ 1 mod 8
or 8‖d′ and m−10 k′ ≡ 3 mod 8 and 2m−1
0 · d′
8≡ 2 mod 8
or 8‖d′ and m−10 k′ ≡ 7 mod 8 and 2m−1
0 · d′
8≡ 6 mod 8
or 16‖d′ and m−10 k′ ≡ 1 mod 8
or 16‖d′ and m−10 k′ ≡ 5 mod 8
0 otherwise,
where
γ =
0 if 2 - d′ or 2‖d′
1 if 4‖d′
2 if 8 | d′
.
In the course of the proof, it will also become clear that if m−10 k′ is not a square
modulo every prime power divisor of d′, there are zero solutions and Ak(Q) = 0.
38
Proof. When Q is the principal form of discriminant d = −4n, then
Q(x, y) = x2 + ny2 and Q′(x, y) = nx2 + y2.
Also, Z2/QZ2 has representatives
Z2/QZ2 = {(t, u)|0 ≤ t ≤ 1 and u ≤ 0 ≤ 2n− 1} .
Therefore, the set SQ is
SQ ={u2∣∣0 ≤ u ≤ 2n− 1
}∪{u2 + n
∣∣0 ≤ u ≤ 2n− 1},
and we have to count the solutions to u2 = k and u2 + n = k.
First consider the equation
u2 ≡ k mod d.
Since (d, k) = m = m0m21, then u
2 must also be a multiple of m and in particular,
u = m0m1x where x ranges from 0 to (d′/2)m1 where d′ = d/m. Our equation then
becomes
m20m
21x
2 ≡ m0m21k′ mod m0m
21d′
which we can reduce to
m0x2 ≡ k′ mod d′
Notice that (d′, k′) = 1 since we factored out the greatest common divisor of d and
39
k. Also, for the equation to have solutions, we must have (m0, d′) = 1. If not, that
is, if p divides m0 and d′, then p would also divide k′. Therefore, we can invert m0
modulo d′ and solve
x2 ≡ m−10 k′ mod d′.
Now we use the Chinese Remainder Theorem to say that if this equation has solu-
tions, there are 2ω(d′)+γ of them modulo d′ where ω(d′) is the number of distinct odd
prime factors of d′ and
γ =
0 if 2 - d′ or 2‖d′
1 if 4‖d′
2 if 8 | d′
.
However, remember that x is ranging from 0 to m1bd′/2c, so we actually have
m12ω(d′)+γ−1 solutions to this equation. Also, note that there are solutions when
(m0, d′) = 1 and m−1
0 k′ is a square modulo every prime power dividing d′.
With this in mind, we turn out attention to the number of solutions to u2 +n ≡ k
mod d. Recall the notation (d, k) = m0m21 = 2δp1 . . . p`2
2α0q2α11 . . . q2αr
r . Then a
solution to our equation must have u = 2δ+α0−εp1 . . . p`qα11 . . . qαr
r x where ε = 0 or 1.
In particular,
ε =
1 if δ = 0 and d′ odd
0 otherwise
.
40
Using that a solution u must have this form, we rewrite the equation in question as
22δ+2α0−2εp21 . . . p
2`q
2α11 . . . q2αr
r x2 − 2δp1 . . . p`q2α11 . . . q2αr
r d′
4≡ mk′ mod md′
and further reduce to
2δ−2ε+2p1 . . . p`x2 + d′ ≡ 4k′ mod 4d′.
At this point, there are six cases, coming from the two possibilities for δ (that is,
whether m0 is even or odd), and from three possibilities for the two-divisibility of d′
(namely, d′ odd, 2‖d′ and 4 | d′). However, only three of these cases have solutions.
First, if 2‖d′, then ε = 0 and we have
2δ+2p1 . . . p`x2 + d′ ≡ 4k′ mod 4d′.
But if 2‖d′, then this cannot have solutions. However, we need to look at the number
of solutions to this equation in combination with the solutions to u2 ≡ k mod d.
For this particular con�guration of k and d, if m0 is even, then there are still no
solutions since (m0, d′) = 1 is a necessary condition for solutions to exist. However,
if m0 is odd, there are solutions to u2 ≡ k mod d assuming m−10 k′ = (p1 . . . p`)
−1k′
is a square modulo each prime power divisor of d′.
Next, if d′ is odd and m0 is even (i.e. δ = 1), then ε = 0 and we have
8p1 . . . p`x2 + d′ ≡ 4k′ mod 4d′.
But d′ odd implies that this has no solutions either. Again looking at the cor-
responding equation u2 ≡ k mod d in this case, we see that there solutions if
41
m−10 k′ = (2p1 . . . p`)
−1k′ is a square modulo each prime power divisor of d′.
In the remaining cases, we reduce to three di�erent equations, each which must
be considered separately.
Case 1: m0 odd, 2 - d′. In this case, ε = 1 and our equation looks like
p1 . . . p`x2 + d′ ≡ 4k′ mod 4d′,
and x ranges from 0 to b4d′ · m1
4c.
As before, for there to be solutions, it must be the case that m0 = p1 . . . p` and
4d′ are relatively prime. If not, then d′ and k′ would share a common factor pi.
Therefore, we can invert m0 modulo 4d′ and rewrite the equation as
x2 +m−10 d′ ≡ 4m−1
0 k′ mod 4d′.
Now we can use the Chinese Remainder Theorem and work modulo each prime power
dividing 4d′ to count solutions. Modulo an odd prime power πβ which divides 4d′,
we get the equation
x2 ≡ 4m−10 k′ mod πβ,
which has two solutions modulo πβ if there are any solutions. Then the only thing
to check is what happens modulo 4. Modulo 4, we get
x2 +m−10 d′ ≡ 0 mod 4.
Then there are two solutions (modulo 4) when m−10 d′ ≡ 3 mod 4, and there are no
42
solutions if m−10 d′ ≡ 1 mod 4.
Then, in total, if there are solutions to u2 + n mod d, there are 2ωd′+1 solutions
modulo 4d′. However, since x ranges up to b4d′ · m1
4c, we actually have m12ω(d′)−1
solutions to u2 + n ≡ k mod d.
Thinking about the corresponding equation u2 ≡ k mod d, we see that there are
solutions to this equation exactly when m−10 k′ is a square modulo each prime power
divisor of d′, which is a necessary condition for u2 +n ≡ k mod d to have solutions.
Therefore, in this case, when Ak(Q) is nonzero, it is equal to
Ak(Q) =
m12ω(d′)+γ if m0d
′ ≡ 3 mod 4
m12ω(d′)+γ−1 if m0d′ ≡ 1 mod 4
where γ is de�ned as before (and is zero in this case).
Case 2: m0 odd, 4 | d′. In this case, ε = 0 and our equation looks like
p1 . . . p`x2 + d′′ ≡ 4k′ mod d′,
where d′′ = d′/4 and x ranges from 0 to bd′ · m1
2c.
Again, m0 = p1 . . . p` and d′ must be relatively prime for there to be solutions,
so we may invert m0 to get
x2 +m−10 d′′ ≡ 4m−1
0 k′ mod d′.
Modulo an odd prime power divisor of d′, there are solutions if m−10 k′ is a square.
If there are solutions modulo this odd prime power, there are exactly two solutions.
43
Therefore, we have remaining only the prime 2.
Looking modulo 4 · 2β0 , where 2β0‖d′′, we get the equation
x2 +m−10 2β0d0 ≡ m−1
0 k′ mod 4 · 2β0
where d0 = d′′/2β0 .
Now if β0 = 0, we are just looking modulo 4, and our equation really looks like
x2 +m−10 d0 ≡ m−1
0 k′ mod 4.
This equation has either zero or two solutions, depending on whetherm−10 d′′ = m−1
0 d0
and m−10 k′ are equivalent modulo 4 or not. If both are 1 or 3 mod 4, then there are
two solutions to this equation. However, if they are di�erent, there are no solutions.
Using the Chinese Remainder theorem, we see that if there are solutions to u2 +n
mod d, there are 2ωd′+γ solutions modulo d′. However, since x ranges up to bd′ · m1
2c,
we actually have m12ω(d′)+γ−1 solutions to u2 + n ≡ k mod d.
Looking at the corresponding equation u2 ≡ k mod d, it is a necessary condition
that m−10 k′ be a square modulo each odd prime power of d′ for there to be solutions.
There are solutions modulo 4 if m−10 k′ ≡ 1 mod 4, and there are no solutions if
m−10 k′ ≡ 3 mod 4. Therefore, in this subcase, when Ak(Q) is nonzero, we get
Ak(Q) =
m12ω(d′)+γ if m0d
′′ ≡ m0k′ ≡ 1 mod 4
m12ω(d′)+γ−1 if m0d′′ ≡ m0k
′ ≡ 3 mod 4
m12ω(d′)+γ−1 if m0d′′ ≡ 3 mod 4 and m0k
′ ≡ 1 mod 4
.
44
If β0 = 1, then we look modulo 8, and our equation looks like
x2 + 2m−10 d0 ≡ m−1
0 k′ mod 8.
This equation has either zero or four solutions modulo 8. The possibilities are
2m−10 d0 ≡ 2, 6 mod 8 and m−1
0 k′ ≡ 1, 3, 5, 7 mod 8. The only combinations
that will yield solutions are the case where 2m−10 d0 ≡ 2 mod 8 and m−1
0 k′ ≡ 3
mod 8, and the case where 2m−10 d0 ≡ 6 mod 8 and m−1
0 k′ ≡ 7 mod 8.
Using the Chinese Remainder theorem, we see that if there are solutions to u2 +n
mod d, there are 2ω(d′)+γ solutions modulo d′. However, since x ranges up to bd′ ·m1
2c,
we actually have m12ω(d′)+γ−1 solutions to u2 + n ≡ k mod d.
Considering the corresponding equation u2 ≡ k mod d, we have the same neces-
sary condition for solutions as before, namely that m−10 k′ be a square modulo every
prime power divisor of d′. We also need m−10 k′ ≡ 1 mod 8, which tells us that we
either have solutions to u2 + n ≡ k or u2 ≡ k, but never both. Therefore, in this
subcase, when Ak(Q) is nonzero, we get
Ak(Q) = m12ω(d′)+γ−1.
The case where β0 = 2 is similar. The only di�erence is that, modulo 8, the
equation looks like
x2 + (4 mod 8) ≡ (1, 3, 5, 7) mod 8.
where the expression in the parentheses on the left side is 4m−10 d0 and the expression
on the right is still m−10 k′.
45
Again, to have solutions to u2 ≡ k, we need m−10 k′ ≡ 1 mod 8, but to have
solutions to u2 + n ≡ k, we need m−10 k′ ≡ 5 mod 8. These are again mutually
exclusive, and we see that, as before,
Ak(Q) = m12ω(d′)+γ−1
when Ak(Q) is nonzero.
Finally, if β0 ≥ 3, we have twice as many solutions as in the previous two subcases.
This is because, modulo 8, our equation looks like
x2 ≡ m−10 k′ mod 8
which has 4 solutions ifm−10 k′ ≡ 1 mod 8 and 0 otherwise. The conditions u2+n ≡ k
to have solutions is the same as the condition for u2 ≡ k to have solutions, so in this
subcase, we get that, when Ak(Q) is nonzero, it is
Ak(Q) = m12ω(d′)+γ.
Case 3: m0 even, 4 | d′. In this case, ε = 0 and our equation looks like
2p1 . . . p`x2 + d′′ ≡ k′ mod d′,
where d′′ = d′/4 and x ranges from 0 to bd′ · m1
2c.
Note that, since k′ and d′ must be relatively prime, k′ is odd. This tells us that
d′′ must be odd, or there are no solutions.
46
Modulo any odd prime power πβ dividing d′, we need to solve
m0x2 ≡ k′ mod πβ
but m0 = 2p1 . . . p` is relatively prime to πβ since m0 and d′ can only have powers of
2 as common factors. Therefore, we can invert m0 modulo πβ and solve
x2 ≡ m−10 k′ mod πβ.
Then if m−10 k′ is a square, there are two solutions modulo πβ.
Now we check modulo 8. We know d′′ and k′ are odd, and again, our equation
looks like
2p1 . . . p`x2 + d′′ ≡ k′.
For there to be solutions, we need p−1(k′ − d′′) ≡ 2 mod 8, where p = p1 . . . p`,
so that p−1 · k′−d′′2≡ 1 mod 4. If there are solutions, then there are two solutions
modulo 4 (and 4 is the largest power of 2 dividing d′).
If there are solutions to u2 + n ≡ k in this case, then, there are 2ωd′+γ of them.
Since x ranges up to bd′ · m1
2c, though, we actually have m12ω(d′)+γ−1 solutions.
Notice that if k′ is odd, then there can't be solutions to u2 ≡ k mod d, since our
equation would reduce tom0x2 ≡ k′ mod 4d′′ andm0 is even in this case. Therefore,
in this case, when Ak(Q) is non-zero, it is
Ak(Q) = m12ω(d′)+γ−1.
47
Having considered all the cases, we see that we arrive at the desired result.
6.2.2 The Case d ≡ 1 mod 4
The case where d ≡ 1 mod 4 is considerably simpler. In particular, we have the
following.
Proposition 6.2. Let Q be the principal binary quadratic form of discriminant d =
1− 4n, and let ω(a) denote the number of distinct odd prime factors of a as before.
Suppose (d, k) = m where m = m0m21 with m0 squarefree. Then
Ak(Q) =
m12ω(d/m) if Q represents k
0 otherwise.
Proof. If Q is a principal form with discriminant d where d ≡ 1 mod 4, then, setting
n = (1− d)/4, we have Q(x, y) = x2 + xy + ny2.
If Q(x, y) = x2 + xy + ny2, then it is given by the matrix
Q =
(2 11 2n
),
so Z2/QZ2 has representatives
Z2/QZ2 = {(1, u)|1 ≤ u ≤ 2n− 1} ∪ {(2, u)|2 ≤ u ≤ 2n} ∪ {(0, 0)} .
Also,
Q′(x, y) = nx2 − xy + y2.
48
Now, we will use the facts that
n− u+ u2 ≡ (2n− u)2 mod |d|
and
4n− 2u+ u2 ≡ (u− 1)2 mod |d|.
These congruences come from
n− u+ u2 − (2n− u)2 = n− 4n2 − u+ 4nu ≡ (n− u)(1− 4n) ≡ 0 mod |d|
and
4n− 2u+ u2 − (u− 1)2 = 4n− 1 ≡ 0 mod |d|,
where we are using the fact that 4n ≡ 1 mod |d|.
Therefore, the set SQ is
SQ = {0} ∪{
(2n− u)2∣∣1 ≤ u ≤ 2n− 1
}∪{
(u− 1)2∣∣2 ≤ u ≤ 2n
}= {0} ∪
{u2∣∣1 ≤ u ≤ 4n− 1
},
and computing Ak(Q) amounts to counting the solutions to u2 ≡ k mod |d|, which
was done for the previous case.
Chapter 7
The Main Result
With the results of the previous two chapters, we will now show that special values
of the minimal parabolic GL3 Eisenstein series are equal to linear combinations of
products of double Dirichlet series.
� 7.1 Tying Up the Loose Ends into Beautiful Bows
Recall that on the one hand, from (4.5),
Z(s1, s2, s3) =∑
T∈Γ\X+3 (Z)
1
| disc(T )|s1E(s2, s3, T ),
and on the other hand, from (5.5) that
Z(s1, s2, s3) =∑d<0
1
|d|s2∑
[Qi] genusclasses mod d
∑k<0
Ak(Qi)
|k|s1· #O×d
2ω(d)−1
×∑e1|d
odd, squarefree
χe′1,e′2(Qi)L(s3, χe1)L(s3, χe2)∏pi‖f
Pi(p−s3 , χe1(p), χe2(p)).
By equating |k|−s1 coe�cients, we will get the result.
49
50
From (4.5), we see that the |k|−s1 coe�cient is
∑T∈Γ\X+
3discT=k
E(s2, s3, T ). (7.1)
From (5.5), the |k|−s1 coe�cient is
∑d<0
1
|d|s2∑
[Qi] genusclasses mod d
Ak(Qi)
2ω(d)−1
∑χ∈X (d)
χ(Qi)LOd(s3, χ)
=∑d<0
1
|d|s2∑
[Qi] genusclasses mod d
Ak(Qi)
2ω(d)−1
∑e1|d
odd, sq.free
χe′1,e′2(Qi)LOd(s3, χe′1,e′2). (7.2)
Using the results of the previous chapter, we can replace the Ak(Q) by an explicit
value. Note that if k and d are relatively prime, then exactly one genus class will
represent k. However, if d and k share a factor, then there might be multiple genus
classes representing k. In particular, if d and k are not relatively prime, but the
principal genus class does represent k, then using the coset lemma, we see that
there are m1 other classes that represent k as well, where m = (k, d). Also, if k is
represented by some class and k is not relatively prime to d, we can use the ideas
from the coset lemma to show that the principal class represents k.
In particular, we start by splitting the sum into a sum over d ≡ 1 mod 4 and a
sum over d ≡ 0 mod 4. Set
S1 =∑d<0
d≡1 mod 4
1
|d|s2∑
[Qi] genusclasses mod d
Ak(Qi)
2ω(d)−1
∑e1|d
odd, sq.free
χe′1,e′2(Qi)LOd(s3, χe′1,e′2)
51
and
S2 =∑d<0
d≡0 mod 4
1
|d|s2∑
[Qi] genusclasses mod d
Ak(Qi)
2ω(d)−1
∑e1|d
odd, sq.free
χe′1,e′2(Qi)LOd(s3, χe′1,e′2).
We consider each sum separately. Writing d as e1e2f2 and considering m | k (i.e.
the candidates to be the greatest common divisor of k and d), and adding back in
the integers d which are 3 mod 4, we get that S1 is
S1 =∑m|kodd
∑µ0|m0
µ1|m1
m1
∑e1=µ0ε1ε12
e2=(m0/µ0)ε2ε12
1 + χ−4(e1e2)
(e1e2)s2
∑f=µ1ϕ
2ω(d′)−ω(d)
|µ1ϕ|2s2χe′1,e′2(k
′)LOd(s3, χe′1,e′2)
=∑m|k
m12ω(m1)∑e1,e2
1
(e1e2)s2
∑f>0, odd
1
f 2s2χk′(e1)LOd
(s3, χe′1,e′2)
+∑m|k
m12ω(m1)∑e1,e2
χ−4(e1e2)
(e1e2)s2
∑f>0, odd
1
f 2s2χk′(e1)LOd
(s3, χe′1,e′2),
where the parameters of the sums have the additional conditions that the ei are odd
and squarefree with εi relatively prime to k′ and to each other. As before, d′ denotes
d/m and k′ denotes k/m, and m = m0m21 with m0 squarefree.
As in [3], we can consider the innermost sum (over f) and rewrite it using (5.4).
52
This gives
S1 =∑m|k
m12ω(m1)∑e1,e2
χk′(e1)
(e1e2)s2L(s3, χe1)L(s3, χe2)
∏p-k′p 6=2
∞∑i=0
Pi(p−s3 , χe1(p), χe2(p))p2is2
+∑m|k
m12ω(m1)∑e1,e2
χ−4(e1e2)χk′(e1)
(e1e2)s2L(s3, χe1)L(s3, χe2)
×∏p-k′p 6=2
∞∑i=0
Pi(p−s3 , χe1(p), χe2(p))p2is2
=∑m|k
m12ω(m1)∑e1, e2
χk′(e1)
(e1e2)s2L(s3, χe1)L(s3, χe2)
× ζ2k′(2s2)ζ2k′(2s2 + 2s3 − 1)
L2k′(s3 + 2s2, χe1)L2k′(s3 + 2s2, χe2)
+∑m|k
m12ω(m1)∑e1, e2
χ−4(e1e2)χk′(e1)
(e1e2)s2L(s3, χe1)L(s3, χe2)
× ζ2k′(2s2)ζ2k′(2s2 + 2s3 − 1)
L2k′(s3 + 2s2, χe1)L2k′(s3 + 2s2, χe2),
where ζn and Ln denote zeta functions and L-functions with the Euler factor at primes
dividing n removed. Then we see that S1 can be expressed as a linear combination
of products:
S1 =∑
ψ=1,χ−4
∑m|k
m12ω(m1)ζ2k′(2s2)ζ2k′(2s2 + 2s3 − 1)
×
(∑e1
L(s3, χe1)ψχk′(e1)
es21 L2k′(s3 + 2s2, χe1)
)(∑e2
L(s3, χe2)ψ(e2)
es22 L2k′(s3 + 2s2, χe2)
).
The expressions in parentheses are double Dirichlet series that come up in the Fourier
Whittaker expansion of the GL3 Eisenstein series.
53
The computation of S2 is similar, and we get that S2 is
S2 =∑m|k
m12ω(m1)
4s2ζk′(2s2)ζk′(2s3 + 2s2 − 1)
×
(∑e1
L(s3, χe1)χ−4(e1)
es21 Lk′(s3 + 2s2, χe1)
)(∑e2
L(s3, χ−4e2)
es22 Lk′(s3 + 2s2, χ−4e2)
).
Therefore, we see that the k−s1 coe�cient in the series Z(s1, s2, s3) is both equal
to a sum of special values of the minimal parabolic GL3 Eisenstein series, and a
linear combination of double Dirichlet series that arise as Fourier coe�cients of the
Eisenstein series on the double cover of GL3.
Chapter 8
Conclusion
� 8.1 Future Research
A natural next step is to prove the analogous results for the maximal parabolic
GL3 Eisenstein series described, for example, in [5]. The same methods used in this
dissertation can be used to attack this problem, and in fact, this case is expected to
be easier. The only di�erences will be that instead of acting by the minimal parabolic
subgroup P , we will use the action of the maximal parabolic subgroup, and that in
this case there will be two relative invariants rather than three. In fact, I have some
preliminary results in this direction, and I hope to �nish this problem soon.
Once I have obtained the results for the maximal parabolic GL3 Eisenstein series,
I plan to work on the maximal parabolic GL3 Eisenstein series induced from a GL2
cusp form. This should involve introducing some numerator other than 1 in the
de�nition of the multiple Dirichlet series. Proving results in this case would be very
exciting because they would suggest how to induct up from GL3 to higher GLn.
It seems that similar methods could be applied to the general GLn case. How-
ever, it may also be that the genus theory necessary becomes signi�cantly more
complicated. Still, I remain optimistic.
54
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