AUTOMORPHIC FORMS FOR
GL(1,AQ ) - TATE'S THESIS
NAVA CHITRIK
Referenced heavily from Tate's thesis and from Goldfeld and Hundley (2011), Automorphic Representations and L-Functions for the General Linear Group, Cambridge University Press
Contents
Automorphic Forms for GL(1,AQ )
1
Dirichlet L-Functions
4
Integration and Poisson Summation on
4
A
The Functional Equation of the Riemann
ζ -Function
5
The Zeta-Integral and its Functional Equation
7
The Local Zeta Integral
9
The L-Function and its
-Factors
14
Automorphic Forms for GL(1,AQ )
Automorphic forms for GL(1,AQ ) are a special type of function on
Q\A.
They
will turn out to be no more than the classical objects, the Dirichlet characters.
First we will make some denitions for automorphic forms which are capable of
generalization to
n>1
then we will show that the one dimensional automorphic
forms are associated to a unique Dirichlet character. In this way, we present Tate's
thesis as the bottom rung in the theory of Automorphic forms. Many of the techniques encountered here will be used again in the theory of Automorphic forms for
GL(2, A).
Denition 1. (Unitary Hecke Character) A unitary Hecke character of the ideles
is a homomorphism,
q ∈ Q.
×
ω : A×
Q → C such
that
|ω(x)| = 1
and
ω(q) = 1
whenever
It should also be continuous. This is sometimes more concisely dened as a
homomorphism
1
ω : Q× \A×
Q →S .
1
AUTOMORPHIC FORMS FOR
GL(1,AQ )
Denition 2. (Moderate Growth) A function
moderate growth if for each adele
- TATE'S THESIS
φ
g = {g∞ , g2 , ...},
2
on the adeles is said to have
there are constants such that
φ({tg∞ , g2 , ...}) < C(1 + |t|∞ )M
for all
t ∈ R.
Denition 3. (Automorphic form) Let
ω
be a xed unitary Hecke character. An
automorphic form is a function
φ : Q× \AQ × → C×
satisfying:
(1)
φ(zg) = ω(z)φ(g) ∀z, g ∈ AQ ×
(2)
φ
has moderate growth (this is automatic)
While the above denition seems a bit silly- it implies that automorphic forms
are simply scaled unitary Hecke characters - this denition is only to mimic the
denitions for higher dimension when these functions can be more complicated.
Denition 4. Dirichlet Character (χ
C×
we can lift
χ
to all of
Z
mod q ).
χ(n) = χ(n̄)
by
χ : (Z/qZ)× →
For any character
where
n ≡ n̄( mod q)
and
χ(n) = 0
if
(q, n) 6= 1.
Denition 5. (Idelic lift of a Dirichlet character). Let
We will associate a unitary Hecke character
the idelic lift of
χ,
χidelic : Q \AQ
as follows:
χidelic (g) =
Y
p≤∞
where
and
χ∞(g∞ ) =
×
χ be a character mod pf


sign(g∞ )
if χ(−1) = −1

1
if χ(−1) = 1
χp (gp )
×
→C
×
.
which is called
AUTOMORPHIC FORMS FOR


χ(v)m
χv (gv ) =

χ(j)−1
GL(1,AQ )
- TATE'S THESIS
3
if v 6= p where gv ∈ v m Z×
v
if v = p where gv = pk (j + pf Zp )
This is a unitary Hecke character: unitary since
χ
which a character of a nite
group must lie on the unit circle. The other properties can be checked easily.
More generally, any Dirichlet character factors into characters mod
case we dene
χidelic as
pf ,
and in this
the products of these lifts. Thus we can lift an arbitrary
Dirichlet character to a unitary Hecke character.
Every automorphic form for GL(1,AQ ) is uniquely of the form
Theorem 6.
φ(g) = c · χidelic (g) · |g|it
where c ∈ C, t ∈ R and χidelic is the idelic lift of a Dirichlet character as dened
above.
Proof.
φ(g) = c · ω(g)where c = φ(1, 1, ...)
As we noted in the denition,
unitary Hecke character. By continuity,
ω
and
ω
is a
factors into a product of a characters of
Q×
v so we need to classify the individual multiplicative characters
ωv
of each local
eld. Also from continuity, for all but nitely many primes, the factors must satisfy
ωv (v) = 1.
R×
Characters of
we know are just
|r|it
|r|it · sign(r)
or
(if you like, this follows
from Pontryagin duality).
Characters of
that
Q×
p
ω(u) = 1
form on
Q×
p:
n
come in two avors. Suppose rst that
whenever
n
, so that we can fully describe
ω
ω
a local character and
takes on a very simple
u · pn ,
this implies that
by the single piece of data,
case is called the unramied case.
The other possibility is that
n
u∈
is a unit, i.e.
ω is
since any element of the eld is of the form
ω(u · p ) = ω(p)
ω(p).This
u
Z×
p . Then
such that
ω(Z×
p ) 6≡ 1.
ω(1 + pk Zp )≡ 1
In this case we still have a minimal integer
since by continuity, the kernel of
ω
must contain an
open subgroup of this form. We call such a character ramied with conductor
Now, for an arbitrary eld element
addition to specifying
simpler, on
u · pn ,
we have that
pk .
ω(u · pn ) = ω(u)ω(p)n ,so
in
ω(p) we must also specify how ω acts on the group Z×
p , or even
k
k
Z×
p /(1 + p Zp ) ' Z/p Z.
So
ω
is actually given locally by a Dirichlet
AUTOMORPHIC FORMS FOR
character mod
pk .
GL(1,AQ )
- TATE'S THESIS
A ramied local character is simply given by an unramied local
character (that is we have to specify
ω(p)
) and a Dirichlet character!
Note: for the archimedian places, we say a character is ramied if
ω(u) = −ω(−u)
4
e.g.
ω(u)
the sign function is present in the formula.
denition consistent, we say
ω
is odd, i.e.
To make the
is ramied if it doesn't factor through the absolute
value.
We have left out some tedious calculation but it follows that
You may wonder where the sign(g∞ ) has gone.
other components, indeed we have that
ω(g) = χidelic (g)|g|it .
This parity disappears into the
ω(−1) = ω∞ (−1)χ(−1).
Dirichlet L-Functions
Denition 7. (The L-function of an Automorphic form) Since any automorphic
form
φ
has an associated Dirichlet character
L(s, φ) =
χ,
−1
Y
χ(p)
1− s
p
p
Example 8. If we take the automorphic form
L(s, φ) = ζ(s) =
we can dene the complex function
Y
φ(g) = 1,
1 − p−s
p
−1
=
we get that
∞
X
n−s
n=1
The L-function of an automorphic form, as dened above converges absolutely for
<(s) > 1
(easily seen from the summation).
The so-called analytic continuation
via functional equation was solved in the 1800's by means of theta functions and
other complicated mechanisms. In his thesis, Tate presented an elegant way to get
these continuations using adelic integrals.
This process will be described in the
remainder of these notes.
Integration and Poisson Summation on A
We need a couple of denitions. The adelic Bruhat-Schwartz space is the space of
linear combinations of factorizable functions, where each factor is locally constant
and compactly supported (nite places), or Schwartz (innite places), and which is
the characteristic function
1Zp
at all but nitely many places. We can dene the
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
5
integral of a function in such a space by using linearity and taking the product of
the integral, for purely factorizable functions, i.e. if
ˆ
×
φ(x)d x =
A×
Q
Yˆ
Q×
v
p∈S
φ(x) =
Q
p
φp (xp )
then
φv (xv )d× xv
Let's say explicitly what these Haar measures are:
d× x =

 dx∞

where the
´
Zp
dxv
dxp = 1
1
1−p−1
dxp
|xp |p
´
Z×
p
µ(a + pn Zp ) = p−n .
This gives that
d× x = 1
(Adelic Poisson Summation)
X
h(αx) =
α∈Q
Proof.
at a nite place
are derived from the metric, e.g.
and
Theorem 9.
at the innite place
x∞
1 X α
ĥ
|x|
x
α∈Q
omitted
(The Fourier transform is taken with respect to a self-dual Haar measure i.e. formula
ˆ
fˆ(x) = f (−x)
holds)
The Functional Equation of the Riemann ζ -Function
Example 10. In this example, we will compute the functional equation of the
Riemann
ζ -function
using poisson summation of the adeles.
Let us consider the adelic function
2
Φ(g) = e−πg∞
1Zp
Y
p<∞
and notice that
Φ(g) = Φ̂(g),
and
Φ(0) = Φ̂(0)=
1.
We construct the adelic integral
ˆ
ˆ
∞
Φ(g)|g|s d× x =
−∞
A×
Q
Now, since
Zp \{0} =
∞
[
n=0
2
e−πx |x|s−1 dx ·
pn Z×
p
ˆ
Y
p<∞
|gp |s d× gp = (∗)
Zp \{0}
AUTOMORPHIC FORMS FOR
ˆ
|gp |s d× gp =
ˆ
∞
2
e−πx xs
(∗) = 2 ·
0
- TATE'S THESIS
ˆ
p−ns
d× x =
pn Z×
p
n=0
Zp \{0}
So that
∞
X
GL(1,AQ )
6
1
1 − p−s
s
dx Y
1
−s/2
=
π
Γ
ζ(s)
·
x p<∞ 1 − p−s
2
On the other hand, using the strong approximation
1
we can rewrite (*) as
ˆ
X
(∗) =
Φ(αx)|x|s d× x
α∈Q×
Q× \A×
Q
ˆ
ˆ
X
=
×
Q× \A
Q
X
s ×
Φ(αx)|x| d x +
α∈Q×
×
Q× \A
Q
|x|≤1
Φ(αx)|x|s d× x
α∈Q×
|x|≥1
Now we'll apply Poisson summation to the rst of these two integrals
ˆ
X
Q× \A
×
Q
Φ(αx)|x|s d× x
´
=
P
( α∈Q× Φ̂
|x|−1 − Φ(0) + Φ̂(0)|x|−1 )|x|s d× x
α
x
|x|s−1 d× x +
×
Q× \A
Q
α∈Q×
|x|≤1
|x|≤1
=
´
P
α∈Q×
α
x
Φ̂
×
Q× \A
Q
=
´
P
α∈Q×
1−s ×
Φ̂ (αx) |x|
d x+
´1
0
×
Q× \A
Q
=
(|x|s − |x|s−1 )d× x
×
Q× \A
Q
|x|≤1
|x|≥1
´
´
P
α∈Q×
|x|≤1
(xs − xs−1 ) dx
x
Φ̂ (αx) |x|s−1 d× x +
×
Q× \A
Q
|x|≥1
In total, we have that
ˆ
X (∗) =
×
Q× \A
Q
α∈Q×
1
1
Φ̂ (αx) |x|s−1 + Φ(αx)|x|s d× x + −
s s−1
|x|≥1
which is symmetric under the simultaneous changes
1
∼
Q× \A×
Q = (0, ∞) ·
Y
Z×
p
p
Φ ↔ Φ̂, s ↔ 1 − s.
1
s
−
Q ´
1
s−1
p Z×
p
d× x
AUTOMORPHIC FORMS FOR
Therefore, since (*) was also equal to
π
−s/2
Γ
s
2
GL(1,AQ )
π −s/2 Γ
ζ(s) = π
− 1−s
2
s
2
Γ
- TATE'S THESIS
ζ(s)
1−s
2
7
we have that
ζ(1 − s)
You may have noticed a couple of thing about the above calculations. One major
observation is that to get the functional equation we didn't even need to use the
fact that
on a
Φ = Φ̂ so that there is some reason to suspect that we will not need to rely
particular
function
Φ
and the theory will have to explain the independence.
The remainder of these notes will be devoted to generalizing this process to obtain
functional equations for more general Dirichlet L-functions.
The Zeta-Integral and its Functional Equation
Denition 11. Let
Φ=
Q
p
φp
be an adelic Bruhat-Schwartz function, and let
ω
be a unitary Hecke character. We dene the global zeta-integral
ˆ
Φ(x)ω(x)|x|s d× x
Z(s, Φ, ω) =
A×
Q
This denition is by analogy with the Mellin Transform of real functions which is
useful in classical analytic number theory.
Theorem 12.
(Functional equation of the global Zeta integral)
Z(s, Φ, ω) = Z(1 − s, Φ̂, ω̄)
Proof.
We compute:
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
8
ˆ
Z(s, Φ, ω)
Φ(x)ω(x)|x|s d× x
=
A×
Q
ˆ
X
=
Q× \A
×
Q
Φ(αx)ω(x)|x|s d× x
by strong approximation
α∈Q×
and the fact that
 ˆ
X


Φ(αx)ω(x)|x|s d× x



×

Q× \A× α∈Q
ˆQ X
=



Φ(αx)ω(x)|x|s d× x



 × × α∈Q×
Q
for
|x| ≥ 1
for
|x| ≤ 1
|α|s ω(α) = 1
+
\A
Q
= I + II
We we will now use Poisson summation on the rst integral. We rewrite the Poisson
summation formula as
X
Φ(αx) =
α∈Q×
Then integral
II
X 1 α
Φ̂(0)
Φ̂
− Φ(0) +
|x|
x
|x|
×
α∈Q
turns into:
ˆ
X
II =
|x|≤1
×
Q
Q× \A
ˆ
=
|x|≤1
×
Q
Φ(αx)ω(x)|x|s d× x
α∈Q×
ˆ
X 1 α
s ×
Φ̂
ω(x)|x| d x − Φ(0)
|x|
x
×
α∈Q
|x|≤1
×
Q
ˆ
|x|≥1
×
Q
|x|≤1
×
Q× \A
Q
ˆ
X
Φ̂ (αx) ω̄(x)|x|
α∈Q×
Q× \A
Now we have that
1−s ×
ω(x)|x|s−1 d× x
ω(x)|x| d x + Φ̂(0)
Q× \A
Q× \A
=
ˆ
s ×
ˆ
s ×
d x − Φ(0)
ω(x)|x| d x + Φ̂(0)
|x|≤1
×
Q
Q× \A
|x|≤1
×
Q× \A
Q
ω(x)|x|s−1 d× x
AUTOMORPHIC FORMS FOR
(0.1)
GL(1,AQ )
- TATE'S THESIS
Z(s, Φ, ω) = I + II
ˆ
X =
Φ̂ (αx) ω̄(x)|x|1−s − Φ (αx) ω(x)|x|s d× x
|x|≥1
×
Q
α∈Q×
Q× \A
ˆ
ˆ
s ×
− Φ(0)
ω
has ramication, since
trivial on one of the
If
ω
Zp
ω(x)|x|s−1 d× x
ω(x)|x| d x + Φ̂(0)
|x|≤1
×
Q× \A
Q
If
9
|x|≤1
×
Q× \A
Q
Q ×
∼
Q× \A×
Zp
Q = (0, ∞) ·
and
ω
on this domain isn't
then by standard tricks, the last two integrals will be zero.
has no ramication then
ω
is trivial on the
become, by integrating just the real place from
Q
0
Z×
p
to
part and the last two pieces
1:
1
1
−Φ(0) + Φ̂(0)
s
s−1
In summary, by inspection of 0.1 and the above discussion, we have that
Z(s, Φ, ω) =
Z(1 − s, Φ̂, ω̄)
Summary of above proof:
(1) Use strong approximation
(2) Break into
|x| ≤ 1
(3) Use P.S. on
(4) let
x → 1/x
and
|x| ≥ 1
|x| ≤ 1
in the big piece of the formula
(5) determine if ramication occurs
Although the global zeta integral satises such a neat functional equation, the local
pieces (dened below) require a slight modication for their functional equations.
The Local Zeta Integral
Denition 13. (Local Zeta Integral) Let
s ∈ C
with
<(s) > 0, φ(x)
a Bruhat-
Schwartz function on the local eld (i.e. Locally constant and compactly supported
on
Q×
p , or a Schwartz-function on R ).
Let
ω be a local unitary character.
ˆ
φ(x)ω(x)|x|s d× x
Zv (s, φ, ω) =
Q×
v
We dene
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
10
Does the above converge? Well we can use the standard trick of splitting it into
two pieces.
If we consider the integral when
|x| > 1
then it certainly converges
there since we are dealing with a (Bruhat-) Schwartz function.
hand, for
∞
X
ˆ
|x| ≤ 1
|x|σ d× x =
0 pn Z
p
On the other
the function is bounded by some constant times
∞
X
p−nσ
which converges for
σ = <(s) > 0.
´
Zp
|x|σ d× x =
So the integral
0
Zv (s, φ, ω)
converges for
<(s) > 0.
We will next determine that it has an meromor-
phic continuation via a functional equation.
Theorem 14.
There exists a meromorphic function γ(s, ω), which is independent
of φ such that for 0 < <(s) < 1
Zv (s, φ, ω) = γ(s, ω) · Zv (1 − s, φ̂, ω̄)
Proof.
We only need to show that the ratio
which is inependent of the function
functions
φ
and
ψ
on
Qv .
φ.
Zv (s,φ,ω)
is a meromorphic function
Zv (1−s,φ̂,ω̄)
Consider two arbitrary Bruhat-Schwartz
We compute:
¯ =
Zv (s, φ, ω) · Zv (1 − s, ψ̂, ω)
ˆ ˆ
φ(x)ψ̂(y)|x|s |y|1−s ω(x)ω̄(y)d× xd× y
×
Q×
p Qp
By letting
ˆ ˆ
y → xy
φ(x)ψ̂(xy)|x| · |y|1−s ω̄(y)d× xd× y
=
×
Q×
p Qp
Using the denition of the fourier transform
ˆ ˆ ˆ
φ(x)ψ(z)ev (−xyz)|x| · |y|1−s ω̄(y)d× x · d× y · dz
=
× Qp
Q×
p Qp
dx
d× x = c
and the point 0 is measureless
|x|
ˆ ˆ ˆ
=c
φ(x)ψ(z)ev (−xyz)|y|1−s ω̄(y)dx · d× y · dz
Since
Qp Q×
Qp
p
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
The last expression is independent over
φ ↔ ψ
, therefore
¯ = Zv (1 − s, φ̂, ω) · Zv (1 − s, ψ̂, ω)
¯
s, ψ̂, ω)
Zv (s, φ, ω)
Zv (1 − s, φ̂, ω)
which shows that
γ(s, ω)
11
Zv (s, φ, ω) · Zv (1 −
and so
=
Zv (s, φψ, ω)
Zv (1 − s, ψ̂, ω)
is indeed independent of the test-function
It turns out that it is not very hard to compute
γ(s, ω)
in each case.
All that
Zv (s,φ,ω)
is required is a particularly easy test function and the comparison of
Zv (1−s,φ̂,ω)
for that easy function
whether
one on
ω
φ.
is an unramied unitary character (recall this means that it is identically
Z×
p)
or ramied.
Example 15. Suppose
ω(x) = 1
The excplicit computation will basically depend only on
for
x ∈ R.
v = ∞ and ω(x) is unramied.
Then we choose the test function
ˆ
2
e−πx |x|s
Zv (s, φ, ω) =
Recall that this means that
2
φ(x) = e−πx
. Then,
dx
|x|
R×
− 2s
ˆ
s
Γ( )
2
=π
ˆ
2
dx
¯
Zv (1 − s, φ̂ω) = e−πx |x|1−s
|x|
some
u − substitution
and recall
∞
e−x xs
Γ(s) =
0
R×
= π−
1−s
2
Γ(
1−s
)
2
just take the ratio of the two, to obtain
Example 16. Suppose
v=∞
Then we use the test funtion
and
γ(s, ω)
ω(x)
is ramied, that is,
2
φ(x) = xe−πx
ω(x) = sign(x) =
and notice that
φ̂(x) = iφ(x)
|x|
x .
so we
dx
x
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
12
have that
ˆ
2
e−πx x2 |x|s+1
Zv (s, φ, ω) =
dx
|x|
R×
− s+1
2
ˆ
s+1
Γ(
)
2
=π
ˆ
¯ = ie−πx2 x2 |x|(1−s)+1 dx
Zv (1 − s, φ̂ω)
|x|
some
u − substitution
and recall
0
R×
= iπ −
(1−s)+1
2
Γ(
1−s+1
)
2
Again, we take their ratio to get
Example 17. Suppose
v = p
γ.
is a nite prime and
ω
is unramied.
This is the
easiest case and the one that comes up in proving the functional equation of the
Riemann
ζ−function.
In this case, we choose
φ = 1Z p
which is its own Fourier
transform.
Computing,
ˆ
1Zp ω(x)|x|s d× x
Zv (s, φ, ω) =
Q×
p
ˆ
ω(x)|x|s d× x
=
Zp \{0}
=
∞
X
ω(p)n p−ns
n=0
1
=
1 − ω(p)p−s
ˆ
¯ =
Zv (1 − s, φ̂ω)
1Zp ω̄(x)|x|1−s d× x
Q×
p
=
1
1 − ω̄(p)p−(1−s)
since
Zp \{0}=
∞
[
n=0
pn Z×
p
∞
e−x xs
Γ(s) =
dx
x
AUTOMORPHIC FORMS FOR
GL(1,AQ )
Example 18. Finally, the last possibility is that
acter
2
ω is ramied with conductor pr .
- TATE'S THESIS
v = p a nite prime and the char-
In this case we choose
Then, computing the Fourier transform of
13
φ(x) = e−2πi{x} 1p−r Zp .
φ
ˆ
φ(y)e−2πi{xy} dy
φ̂(x) =
Qp
ˆ
e−2πi({x(1−y}) dy
=
p−r Zp
This is only nonzero when
x ∈ 1 + pr Zp
ˆ
= 11+p−r Zp ·
dy
p−r Z
p
= pr 11+p−r Zp
Now we'll compute the zeta integrals:
2
Recall that
e−2πi{x}
Transform and where
is the additive character on
{x}
Qp
is the fractional part of
-the one we use to compute the Fourier
x,
that is if
x =
∞
X
−N
−1
X
−N
an pn ∈ Q
an pn
then
{x} =
AUTOMORPHIC FORMS FOR
ˆ
Zv (s, φ, ω) =
Q×
p
GL(1,AQ )
- TATE'S THESIS
14
1p−r Zp e−2πi{x} ω(x)|x|s d× x
ˆ
e−2πi{x} ω(x)|x|s d× x
=
p−r Zp \{0}
=
∞
X
n −ns
ω(p) p
since
n=0
p
r X
X
`=1 j=1
p
r X
X
p
[
`=1
j=1
(j,p)=1
e−2πi{x} ω(x)|x|s d× x
p−` (j+pr Zp )
ˆ
r
=
Zp \{0}=
e
− 2πij
l
p
ω(p)−` ω(j) · p`s
`=1 j=1
d× x
p−` (j+pr Zp )
r
p
r
X
p −r X `s
− 2πij
=
p
p ω(p)−`
e pl ω(j)
p−1
j=1
`=1
r
p
X
−r
p−r+1 rs
=
p ω(p)−r ·
e−2πijp ω(j)
p−1
j=1
ˆ
Zv (1 − s, φ̂, ω̄) =
(j,p)=1
11+p−r Zp pr ω̄(x)|x|1−s d× x
Q×
p
ˆ
pr ω̄(x)|x|1−s d× x
=
1+pr Zp
ˆ
d× x
= pr
1+pr Zp
=
p
pr
p−1
r
r
[
ˆ
r
=
p
−r
ˆ
|x|−1 dx
1+pr Zp
p
=
p−1
The L-Function and its -Factors
Great. So now we have functional equations and meromorphic continuations of the
local Zeta-integrals to the entire complex plane. We now use this fact, together with
p−` (j + pr Zp )
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
15
the global-functional equation to obtain a functional equation and root numbers
for a classical L-function.
Let
ω=
Then
ω
ω
Y
ωv be the unitary Hecke character associated to an automorphic form.
v≤∞
is ramied at only nitely many places. We dene the local L-function of
as follows
Lv (s, ωv )=

s


π 2 Γ( 2s )






s+1
π − s+1
2 Γ(
2 )



(1 − ωp (p)p−s )−1





1
if
v=∞
is unramied
if
v=∞
is ramied
if
v=p
is unramied
if
v=p
is ramied
Notice that in the nite places, this looks similar to the
pth
place factor in the
Euler product, so this is a reasonable denition of a local L-function. Also, in the
Lv (s, ωv ) = Zv (s, φv , ω)
unramied places notice that
with
φv
chosen as in the
above examples.
Remark
of
.
19
The local L-function can be thought of as the greatest-common-divisor
Zv (s, φ, ω) as we range over all Bruhat-Schwartz functions φ.
That is, the local L-
Zv (s,φ,ω)
function is the least complicated function so that
Lv (s,ω) is always holomorphic.
For example, if
ω
is ramied, then
Zv (s, φ, ω)
GCD of holomorphic functions is 1.
function
the same
φ
so that
will always be holomorphic, so the
Also true is that we can always nd some
Zv (s, φ, ω) = Lv (s, ω)
but it
won't necessarily be true that for
φ, Zv (1 − s, φ̂, ω̄) = Lv (1 − s, ω̄).
Denition 20. (Local Root Number)
The local root number
v (s, ω)
(0.2)
Zv (1 − s, φ̂, ω̄)
Zv (s, φ, ω)
= v (s, ω)
Lv (1 − s, ω̄)
Lv (s, ω)
which is independent of
φ
is dened by
since the function
Important note: as I said above, whenever
local L-function
ω
was.
is unramied at
was a Z-integral. This implies that
Zv (1 − s, φ̂, ω̄) = Lv (1 − s, ω̄)
Lemma 21.
γ(s, φ)
v
we said that the
Zv (s, φ, ω) = Lv (s, ω) and
Therefore:
For a xed ω , and for all but nitely many places v , v (s, ω) ≡ 1.
AUTOMORPHIC FORMS FOR
GL(1,AQ )
- TATE'S THESIS
16
Let us to back to the global functional equation and choose our test function,
Φ=
Y
φv
v≤∞
where the
and at
that
φv are
chosen so that
∞ we choose φv
11+pr Zp
Zv (s, φv , ω) = Lv (s, ω).
At the unramied places
as in the examples and in the last case, its not hard to show
will do the trick
Denition 22. (The global L-function) Let
the local L-factors
ω be a unitary Hecke character, and let
Lv (s, ω) be dened as above.
The we dene the global L-function
as
L∗ (s, ω) =
Y
Lv (s, ω)
v≤∞
We think of the global L-function as a completed L-function. It looks a lot like
the Dirichlet L-function, but now we also have a component at the innite place,
making it more symmetric and complete. It is this L-function which will have a
natural functional equation coming from the adelic factorization.
Theorem 23.
(Functional equation for the global L-function) There exists a mero-
morphic function (s, ω) such that L∗ (s, ω) = (s, ω)L∗ (1 − s, ω̄)
LetΦ
=
Q
φ as above, where φ is chosen so that at each place Zv (s, φ, ω) = Lv (s, ω)
We can apply the global functional equation:
Z(s, Φ, ω) = Z(1 − s, Φ̂, ω̄)
v
to the product over
of equation 0.2
Y Zv (1 − s, φ̂, ω̄)
Y
Zv (s, φ, ω)
=
(s, ω)
Lv (1 − s, ω̄)
Lv (s, ω)
v≤∞
The
Zv 's
v≤∞
factor out entirely by the global functional equation and we get that
L(s, ω) =
Y
Lv (s, ω) = (s, ω)
v≤∞
where
(s, ω) =
Y
v (s, ω) =
v≤∞
Y
v∈S
Y
Lv (1 − s, ω̄)
v≤∞
v (s, ω),
where S is nite by Lemma 21.
AUTOMORPHIC FORMS FOR
And we're done!
GL(1,AQ )
- TATE'S THESIS
17
We've shown how to construct a completed L-function which
has analytic continuation.