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AUTOMORPHIC L-FUNCTIONS NOTES

HAO (BILLY) LEE

Abstract. These are notes I took in class, taught by Professor Ngo Bao Chau. I claim no credit to the originality of the

contents of these notes. Nor do I claim that they are without errors, nor readable.

Full course notes available at math.uchicago/~ngo/GJ.pdf

1. Introduction and Overview

Denition 1.1. Dene the Riemann zeta function to be

ζ(s) =∑n≥1

1

ns=∏p

1

1− 1ps

.

Both sides of these converge absolutely on Re(s) > 1.

Riemann found that ζ(s) can be extended to a meromorphic function of s ∈ C with a single simple pole at s = 1. It

satises a functional equation of the follwing form. Let ξ(s) = πs/2Γ( s2 )ζ(s) then

ξ(s) = ξ(1− s).

Tate's Thesis uses Fourier analysis on adèles to explain more about this Euler product. For local elds R and Qp, theyare locally compact groups. Therefore, they have invariant measures on them, which gives us Fourier analysis.

Denition 1.2. Dene AQ = R ×∏′pQp, where the restricted product means that x = (xv) ∈ AQ is such that xv ∈ Ov

for almost all v. We know that A is locally compact, so also have a Haar measure.

Denition 1.3. For a locally compact abelian group A, dene the Pontryagin dual to be

A =x : A→ C1 unitary characters : continuous homomorphism

.

We will call χ : A→ C× a character.

The dual A will be equipped with the compact open topology, making it a locally compact abelian group.

Theorem 1.4. (Pontryagin Theorem) A ∼= ˆA.

Example 1.5. For A = Z, we know that A = C1. Then the theorem says that all χ : C1 → C1 is of the form χ(x) = xn

for some n ∈ Z.

Example 1.6. Suppose A = F is a local eld. Then A = F (not canonical). This is because for x ∈ F , we can dene

χx : F → C1 given by y 7→ ψ1(xy) where ψ1 : F → C1 is some preferred character. When F = R, let ψ1(x) = exp(2πix).

Example 1.7. When A = AQ, then A = A as well.

Denition 1.8. For ϕ ∈ L1(A), dene ϕ : A→ C be

ϕ(χ) =

∫A

ϕ(x)χ(x)−1dx

for some chosen Haar measure dx. Then ϕ is continuous and 0 at innity.1

AUTOMORPHIC L-FUNCTIONS NOTES 2

Denition 1.9. When A = F , dene the Schwartz space to be

S (A) =

F = R smooth, rapid decay, like p(x)e−x2

F = Qp locally unipotent compact support, like 1Zp

.

The functions in S (F ) are stable under Fourier transform.

Note, when we have a Haar measure on A, we have it on A as well. By using Fourier inversion,∫A

ϕ(χ)dχ = ϕ(0).

For A = F , there exists a unique dx self-dual so that∫F

ϕ(x)dx = ϕ(0).

For b ∈ S (F ), b = b self-dual. When F = R, let b∞(x) = e−πx2

. When F = Qp, let bp = 1Zp . These are called basic

functions.

We have

S (AQ) =

nite linear combination of functions ϕ = ⊗vϕwhere ϕv ∈ S (Fv) and ϕv = bv for almost all v

.

Theorem 1.10. (Poisson Summation formula). For all ϕ ∈ S (A),∑α∈k

ϕ(α) =∑α∈k

ϕ(α).

Tate applies Mellin transform to this whole package. Dene

Ω(F×) =χ : F× → C× : continuous, homomorphic

.

We have module character F× → R+ (absolute value). For α ∈ F×, d(αx) = |α| dx. We have a homomorphism

C→ Ω(F×) given by

s 7→ (x 7→ |x|s) .

The image Ω0 (F×) is neutral connected component of Ω(F×). The map

C→ Ω0(F×)

is an isomorphism when F = R, and kernel 2πi log p if F = Qp (second one may be a bit o).

Denition 1.11. Mellin transform. Start with ϕ ∈ S (F ) 7→Mϕ : Ω(F×)→ C where

Mϕ(χ) =

∫F×

ϕ(x)χ(x)−1d×x

where d×x = dx|x| is a Haar measure on F×.

Notice that this map factors through the restriction to F×. However, note that ϕ |F× no longer has rapid decay, but

Mϕ is a meromorphic function on Ω(F×). When F× = Q×p , Mϕ has a simple pole at s = 0.

Example 1.12. For F = Qp,

M`p(χ) =

0 if χ /∈ Ω0(F×)

(1− p−s)−1 if χ = |·|s.

For F = R,M`∞(χ) = π−s/2Γ(

s

2).

Question: what is the eect of Fourier transform on the Mellin transform? How to relate Mϕ with Mϕ?


Consider ϕ ∈ S (F ), maps to ϕa(x) = ϕ(ax) for some a ∈ F×. Then ϕa = |a|−1ϕ1/a. This is the analog of s↔ 1− s

(the 1 is the |a|−1).

We can compute that

Mϕ(χ) = γ(χ)Mϕ

(|·|χ−1

),

where γ is a factor depending solely on χ and not on ϕ. This is called the Tate local functional equation.

ϕ(y) =

∫ϕ(x−1)e(xy)dx

where ϕ(x) = ϕ(x−1). We write it this way, because it looks like a convolution. This will behave well with the Mellin

transform.

Mellin transform for A. Suppose ϕ ∈ S (A), we want to map it toMϕ : Ω (K×\A×)→ C. First, we have |·| : A× → R+.

Let A1 be the kernel of |·|, then K×\A1 is compact. Similar to the above, we have

C→ Ω(K×\A×

)by

s 7→ (x 7→ |x|s)

and C Ω0 (K×\A×).

First, suppose ϕ = b∞ ⊗ bp.Mϕ(s) = π−s/2Γ(

s

2)∏p

(1− p−s

)−1

for Re(s) > 1. For every ϕ ∈ S (A), dene

Mϕ : Ω(K×\A

)→ C

holomorphic for Re(χ) > 1. Can show that Mϕ can be meromorphically continued to all of Ω (K×\A) satisfying

Mϕ(χ) = Mϕ

(|·|χ−1

).

This is proved with Poisson summation formula, which comes together with the meromorphic continuation. The functional

equation is the same as saying the product of the γ-factors∏γv(χ) = 1.

However, this never converges, so does not really make sense.

Note, Ω (K×\A×) is automorphic forms of GL1. We will also talk about generalizations of this due to Langlands, to

automorphic forms on any reduction group G/K . These are irreducible representation of G(A), which is

π ⊆ A (G(Q)\G(A))

which breaks down into π = ⊕′vπv where πv is unramied for almost all v and πG(Ov)v 6= 0.

Langlands dene a dual group G/C whose root system is dual to that of G, for G split. These unramied representations

are the same as semisimple conjugacy class of αp ∈ G. With these datum, we can dene L-functions.

Let ϕ be a representation

ϕ : G→ GL(n).

Dene

L(s, π, ρ) =∏v

L (s, πv, ρ)

where

L (s, πv, ρ) =

det (1− ρ(αp)p−s)−1

πv unramied

hard... else.


Langlands showed that L(s, π, b) converges for Re(s) > N . It is conjectured that it can be meromorphically continued

and has a functional equation.

This functional equation is very important. When we are given a general representation, its associated L-function has

these functional equation properties i the representation is automorphic.

These are known for G = GLn and ρ is standard representation (by Godement-Jacquet). There is another conjecture

of Langlands that all other cases can be reduced to this one (functoriality).

2. Adèles and Idèles

2.1. For Q.

Theorem 2.1. (Ostrowski) There are no other norms on Q other than |·|∞ and |·|p (up to equivalence).

Proposition 2.2. (Product formula) For x ∈ Q, |x|∞∏p |x|p = 1.

Proof. For x ∈ N, we can write x = pr11 ...prnn . Then |x|pi = p−rii and |x|∞ = x.

Denition 2.3. Adèles over Q is dened to be

AQ =

′∏v

Qv = (x∞, xp, ..., ) : x∞ ∈ R, xp ∈ Qp, xp ∈ Zp for a.e. p .

Notice that R and Qp are locally compact, with [−1, 1] and Zp compact neighbourhood of 0. Notice that

[0, 1] =

[0,

1

n

]∪ ... ∪

[n− 1

n, 1

], and Zp = ∪

p−1p

r=0 (r + pZp).

Similar, on AQ, we have

AQ = ∪N∈NR× (xp)p : Nxp ∈ Zp = ∪N∈NR×N−1Z.

We give AQ the nest topology such that R×N−1Z → A is continuous, where R×N−1Z is given the product topology.

Then

[−1, 1]× Z

is a compact neighbourhood of 0 (by Tychonov's theorem, Z is compact). There is a canonical embedding Q → Adiagonally.

Theorem 2.4. Q is a discrete co-compact subgroup of A. Moreover, there is an exact sequence of topological group

0→ Z→ A/Q→ R/Z→ 0.

In other words, A/Q is the pro-universal covering of the circle.

lim←

R/Z

where R/Z→ R/Z is given by x 7→ nx.

Proof. Let B = (−1, 1)× Z. Then

B ∩Q = x ∈ Q : x ∈ (−1, 1), x ∈ Zp for all p = (−1, 1) ∩ Z = 0.

This gives the commutative diagram

0 R× Z A ⊕pQp/Zp 0

0 0 Q Q 0

Z


By the Snake lemma,

0→ Z diagonal→ Z× R→ A/Q→ 0

and so A/Q ∼=(Z× R

)/Z and so

0→ Z→ A/Q→ R/Z→ 0.

There is another proof for compactness of A/Q. It is enough to construct a compact subset B ⊆ A mapping surjectively

in the map A→ A/Q. TakeB = (x∞, xp) : |x|∞ ≤ 1, |xp|p ≤ 1 ,

which is compact. By the same exact sequence,

B + Q→ A

is surjective, and so A/Q is compact. It is important to note that here, B is just a set, not a group.

Denition 2.5. Dene the idèles A× to be

A×Q =

(x∞, xp) : x∞ ∈ R×, xp ∈ Q×p , xp ∈ Z×p for almost all p.

This is also a locally compact abelian group, by giving the topology where A× → A given by x 7→ x and x 7→ x−1 are

both continuous.

Consider the map Q× → A×|·|A→ R×+ where |x|A =

∏v |xv|v. Dene A1 to be ker |·|A1 ⊆ A, which contains Q× by the

product formula.

Theorem 2.6. Q× is a discrete cocompact subgroup of A1.

Proof. Similar to the additive case. The claim is the

Q× × Z× × R×+ → A×

is an isomorphism (the map is given by multiplication since each is a subgroup of A×). Then

A1/Q× ∼= Z×

is compact.

Remark 2.7. For general number elds, everything extends the same way, except for this theorem. It becomes quite hard

and fundamental.

2.2. General number elds. Let K be a number eld over Q.

Theorem 2.8. (Ostrowski). For all rational primes p,

K ⊗Q Qp =∏v|p

Kv.

Similarly,

k∞ = k ⊗Q R = Rr ⊗ Cs.

Let x ∈ K, dene |x|v =∣∣NKv/Qp(x)

∣∣v. Then these are all the absolute values on K.

Proposition 2.9. (Product formula) For all x ∈ K×,∏v

|x|v = 1.


Denition 2.10. Dene

AK =

′∏v

Kv = (xv) : xv ∈ Kv, xv ∈ Ov for almost all nite v .

Similarly, dene

A×K =

(xv) : xv ∈ K×v , |xv| = 1 for almost all nite v.

Theorem 2.11. The embedding K → AK diagonally is a discrete cocompact group. Since AK = AQ ⊗Q K, we have

AK/K ∼= (AQ/Q)r,

where r = [K : Q].

Similarly, K× is a discrete cocompact subgroup of A1K .

Remark 2.12. This will imply the niteness of class number, and Dirichlet's unit theorem.

Proof. Discreteness is similar to the additive case. Use the box

Bc = (xv) : |xv|v ≤ cv

for some c = (cv) and cv ∈ R×+, cv = 1 for almost all v. For all c as above, k× ∩Bc is nite.

Lemma 2.13. (Minkowski) There exists a constant C > 0 depending only on the discriminant of K/Q, such that for all

c = (cv), cv ∈ R×+, cv = 1 for almost all v, such that∏cv > C, then K× ∩Bc 6= ∅.

Lemma 2.14. For all sequences c = (cv) as above, Bc ∩ A1K is compact.

Proof. For (xv) ∈ Bc, |xv| ≤ cv. For x ∈ Bc ∩ A1K , we hav e

|c|−1cv ≤ |xv| ≤ cv

where |c| =∏cv. Then

Bc ∩ A1K ⊆

∏Uv

where Uv ⊆ K×v is

Uv =xv ∈ K×v : |c|−1

cv ≤ |xv| ≤ cv.

Since for almost all c, cv = 1, we have

|c|−1 ≤ |xv| ≤ 1.

When qv > cv (the size of the residue eld), the inequality implies that |xv| = 1. That is, Uv = O×v .

For |c| > C from Minkowski,

A1K ⊆ ∪α∈K×αBc

and αBc ∩ A1K compact. Hence, A1

K/K× is compact.

From the theorem, we get an exact sequence

0→(O×K ×K

1R

)/O×K → A1

K/K× → ClK → 0

where K1R are the norm one elements are the innite places. Since ClK is discrete, and A1

K/K× is compact, this implies

that K1R/O

×K is compact. That is, K1

R and O×K have the same rank.

Let F be a local eld. Given a, b ∈ R×+, the sets

F×a,b =x ∈ F× : a ≤ |x| ≤ b

is compact. For K a global eld, v places, av, bv ∈ R×+ the sets∏

v

F×a,b


is compact by Tychono's theorem. It is however unclear whether or not∏v

F×a,b ∩ A×

is compact in A×. Unless F×a,b = Z×v for almost all v, there's no way that this set is compact. This is because it can be

written as innite disjoint union of open subsets. This is because a system of open neighbourhoods of A× is∏v∈S

Uv ×∏v/∈S

Z×v

for some nite set S, where Uv ⊆ F×v is some open subset. However,∏v F×a,b ∩A× is compact is F×a,b = Z×p for almost all

v.

2.3. Haar Meausre. Let Fv be a local eld. Pick dx a Haar measure on Fv self-dual with the choice of a preferred

character ψF : F → C1. We have a linear form Cc(Fv)→ C given by

ϕ 7→∫F

ϕ(x)dx.

For adèles it also has a Haar measure dx =∏dxv. We also have a linear form Cc(A)→

∫A ϕ(x)dx. We map

ϕ = ⊗vϕv ∈ Cc(A) 7→∏v

∫Fv

ϕv(x)dx

where ϕv = 1Ov for almost all v.

Locally, for dx a Haar measure on Fv,dx|x| is our choice of Haar measure on F×v . Then

d(xy)

|xy|=|y| dx|xy|

=dx

|x|

so it is invariant under multiplication. However, ∏v

dxv|xv|

is not a Haar measure on A×. Suppose ϕ = ⊗ϕv with ϕv = 1O×v for almost all v. When ϕv = 1O×v ,∫F×v

ϕv(x)dx

|x|=

∫F×v

ϕv(x)dx =p− 1

p

so the innite product goes to 0, which is bad.

2.3.1. Review on Pontryagin duality. Generalization of Fourier series and Fourier transform to all locally compact abelian

group.

Let G be a locally compact abelian group.

Example 2.15. Take G = Z,C1, F, Fv, ...

Let Cc(G) be the continuous functions with compact support. We say fn → f if for every compact K ⊆ G, fn |K→ f |Kuniformly. Then

Cc(G) = ∪K⊆GCK(G).

A random measure is G is a continuous linear form µ : Cc(G)→ C. Dene `x(ϕ)(y) = ϕ(x−1y).

Denition 2.16. The measure µ is said to be left invariant, if for all x ∈ G,

µ (`xϕ) = µ(ϕ)

for all ϕ ∈ Cc(G).

Theorem 2.17. (Haar, Von Neuman) Every locally compact abelian group has a left invariant measure, unique up to a

non-zero scalar.


Denition 2.18. Let

Ω(G) =χ : G→ C× continuous

be the set of cocharacters of G and

Λ(G) =χ : G→ C1 continuous

be the set of unitary characters.

These groups are topological groups with topology the compact-open topology. Λ(G) as an topological abelian group

is the Pontryagin dual of C.

Example 2.19. Λ(Z) = C1 and Λ(C1) = Z.Λ(R) = R (identication by a preferred character), Λ(Qp) = Qp and Λ(A) = A.Meanwhile, Λ(R×) = Λ

(± × R×+

)= ±1 × Λ(R×+) = ±1 × R.

Denition 2.20. Fourier transform Let ϕ ∈ L1 (G, dx). Dene ϕ : Λ(G)→ C by

ϕ(χ) =

∫G

ϕ(x)χ(x)−1dx.

Theorem 2.21. (Riemann-Lebesgue) For ϕ ∈ L1 (G, dx) then ϕ is a continuous function tending to 0 as χ→∞.

Remark 2.22. This means the following. For all a > 0, let

Ka = χ ∈ Λ(G) : |ϕ(χ)| ≥ a .

We mean that for all a > 0, Ka is compact.

For any X locally compact topological space, we can do the one point compactication by X ∪ ∞. A system of

neighbourhood of ∞ is declared to be the complement of compact subsets of X.

We say ϕ : Λ(G) → C tends to 0 as χ → ∞ by ϕ extended to Λ(G) ∪ ∞ by ϕ(∞) = 0 is continuous. Here, we are

assuming the following theorem.

Theorem 2.23. If G is a locally compact abelian group, then Λ(G) is abelian locally compact. This is a consequence of

the Riemann-Lebesgue lemma.

Example 2.24. For G = C1, we have the Fourier coecients

ϕ(n) = an =

∫C1

ϕ(x)xndx

and an → 0 as n→∞.

Example 2.25. Let G = R. Thenϕ(ξ) =

∫Rϕ(x) exp(−2πiξx)dx

is continuous, and ϕ(ξ)→ 0 as ξ →∞.

2.3.2. Fourier inversion theorem. Let ϕ : C1 → C, then when does ϕ(x) =∑n∈Z anx

−n converge. The goal is to

generalize this.

Theorem 2.26. Fourier adjunction formula. Let f ∈ L1(G) and g ∈ L1(Λ(G)). Then∫G

f(x)g(x)dx =

∫Λ(G)

f(χ)g(χ)dχ =

∫G

∫G

f(x)g(χ)χ(x)−1dxdχ.

Remark 2.27. At this point, dχ can be anything. However, we do have a canonical choice. That is, choose it so that we

have Fourier inversion formula (to come).


Consider the space of functions f : G→ C and f are continuous and in L1. Then we want dχ so that∫Λ(G)

f(χ)dχ = f(0)

for all f . This will follow from the adjunction formula, and the existance of a Dirac sequence. When can just apply g → δ0

and g → 1 to the adjunction formula.

Example 2.28. For G = R, let b∞(x) = e−πx2

then b∞(ξ) = e−πξ2

. We can scale x 7→ b∞(tx). Then

limn→∞

∫G

f(x)gn(x)dx = f(0).

Example 2.29. For F = Qp, we can let bp(x) = 1Zp then bp(ξ) = 1Zp . The Dirac sequence is obtained by scaling the

basis functions.

Theorem 2.30. (Pontryagin) G→ Λ (Λ(G)) by x 7→ (χ 7→ χ(x)) is an isomorphism of locally compact groups.

More properties:

(1) Let (`xf)(y) = f(x−1y) then xf(χ) = χ(x)−1f(x)

(2) There is a multiplication on L1 (G, dx) given by convolution. For f, g ∈ L1 (G, dx) then

(f ? g) (x) =

∫G

f(y)g(y−1x)dy.

Then f ? g ∈ L1 (G, dx). Note, the map y 7→ f(y)g(y−1x) is only integrable for almost all x (not necessarily all).

Additionally, ∫G

(f ? g)(x) =

∫G

f(x)dx

∫G

g(x)dx.

(3) Fubini's theorem implies that f ? g(χ) = f(χ)g(χ).

Remark 2.31. That is, Fourier transform converts convolution product into pointwise multiplication.

We can rewrite `xf = δx ? f .

2.3.3. Topology on Λ(G). We gave it the compact open topology (so a sequence of function converges, if they uniformly

converge on every compact subset).

Theorem 2.32. Λ(G) with compact open topology is locally compact.

Lemma 2.33. (Riemann-Lebesgue) Suppose f ∈ L1(G), then f is continuous on Λ(G) (it tends to 0 as χ → ∞). f

extends as a continuous function to the one point compactication Λ(G)∪∞, by setting f(∞) = 0. The topology on the

compactication is that the neighbourhood of ∞ is the complement of compact subsets of Λ(G).

Proof. For all ε > 0, consider

Kε =χ ∈ Λ(G) :

∣∣∣f(χ)∣∣∣ ≥ ε

is compact. To prove this, use Ascoli theorem and the fact that translations `x is continuous on L1(G).

Consider the map ψ : G→ Λ (Λ(G)) by x 7→ (χ 7→ χ(x)).

Proposition 2.34. Fourier adjunction formula. For f ∈ L1 (G, dx) and g ∈ L1 (Λ(G), dχ) then∫G

f(χ)g(χ)dχ =

∫G

f(x)g(ψ(x))dx.

Proof. Application of Fubini.

Theorem 2.35. Fourier inversion theorem. Let G be a locally compact abelian group and Λ(G) its Pontryagin dual. For

all Haar measure dx on G, there exists a unique Haar measure dχ on Λ(G) satisfying:


• For all f ∈ L1 (G, dx) such that f ∈ L1 (Λ(G), dχ) then

f(0) =

∫G

f(χ)dχ.

In this case, both f and f are continuous and vanishing at ∞.

Proof. Use Dirac sequence.

Theorem 2.36. (Pontraygin) The map G→ Λ (Λ(G)) is an isomorphism.

Proof. Fourier adjunction and inversion.

Proposition 2.37. Absract Poisson Summation Formula. Let G be a locally compact abelian group. Let H ⊆ G be a

closed subgroup.

0→ H → G→ G/H → 0.

Let

H⊥ =χ ∈ G : χ |H= 1

which is a closed subgroup of G giving us

0→ H⊥ → G→ G/H⊥ → 0.

Then Λ(H) = G/H⊥ and Λ(H⊥) = G/H.

Proof. We compare the Fourier transforms. Let f ∈ L1(G) and ϕ(x) =∫Hf(xh)dh which is a function of G/H. This is

well-dened almost everywhere by Fubini, and

ϕ ∈ L1 (G/H, dg/dh) .

Then ϕ ∈ C(H⊥

)which is the restriction of f to H⊥ almost everywhere.

Example 2.38. Let G = R and H = Z. Let f ∈ L1(G), then

ϕ(x) =∑n∈Z

f(x+ n)

a function on R/Z. Assume that everything converges, f, f ∈ L1, ϕ, φ absolutely converges. Apply Fourier inversion to ϕ

then

ϕ(0) =∑n∈Z

f(n)

and ∫Zϕ =

∑n∈Z

f(n).

The usual Poisson summation formula says ∑n∈Z

f(n) =∑n∈Z

f(n).

2.4. Fourier theory on additive groups. Let F be a local eld (just R,C,Qp or F/Qp nite extension). What is

Λ(F )? It is just F .

We have a multiplication action of F× on F , and so on Λ(F ). Therefore, Λ(F ) must be a F -vector space. If Λ(F ) = 0

then Λ (Λ(F )) = 0 which is a contradiction. If dimF Λ(F ) ≥ 2, then Λ(F ) = F ⊕H. Then Λ (Λ(F )) = Λ(F ) ⊕ Λ(H) so

the dimension is too great. Hence, Λ(F ) = F . Note that if F = Q, then F is not closed in Λ(F ), so we can't write this

Λ(F ) = F ⊕H.

The Pontryagin biduality theorem implies that dimF Λ(F ) = 1. We will construct a preferred element. If F = R,ψ1(x) = exp (2πix) . If F = C, ψC

1 (x + iy) = ψR1 (2x). When F = Qp, for x ∈ Qp of the form x = a + b where a ∈ Zp,

b = mpn with 0 ≤ m < pn. Then ψ

Qp1 (x) = exp (−2πib) . Extend this to all of Qp.


Then for all x ∈ Q, ∏v

ψFv1 (x) = 1.

For F/Qp nite extension, letψF1 (x) = ψ

Qp1

(Trf/Qp(x)

).

Then for all number elds K/Q, and x ∈ K, ∏v

ψFv1 (x) = 1.

With the choice of a square root of −1, we have constructed in a consistent way, additive characters ψF1 for all local elds

F .

The characters (non-unitary) of R and Qp are very dierent. The character R→ R+ given by x 7→ exp(x) is non-unitary.

On the other hand, all characters of Qp are unitary. This is because Qp is a union of compact subgroups and they must

be mapped to C1.

Let µ be a Haar measure of F , and v dual Haar measure on Λ(F ) = F . For all α ∈ R×+, the dual Haar measure of αµ

is α−1µ. We will now compute self-dual Haar measure on local elds.

For F = R, this is just the usual Lebesgue measure. For F = C, it's the measure 2dxdy. If F = Qp, it's the one wherevol(Zp) = 1 (so 1Zp is self-dual). For F/Qp,

1O⊥F = x ∈ F : for all a ∈ OF , Tr(ax) ∈ Zp .

Therefore, the dual of 1OF is the volume of OF . The self-dual Haar measure is then where vol(OF ) =∣∣Nm(O⊥F )

∣∣−1.

ψF1 : F → C1 identication F ∼= Λ(F ) and so gives a canonical self-dual measure. Fix dFx on F , giving dAx on A.Since ∫

1Ovdx =∣∣Nm (O⊥v )∣∣−1/2

and ϕ = ⊗ϕv with ϕv = 1Ov for almost all v, ∫Aϕ =

∏∫ϕv

and the second part is 1 for almost all v.

For bv = 1Ov , bv = vol(Ov) · 1O⊥v and so∫bvdx = vol (Ov) vol

(O⊥v)

= |volOv|2∣∣Nm(O⊥v )

∣∣ = 1.

Remark 2.39. Since A/K is compact, for a ∈ K×and∏v |a|v = 1, for a ∈ F×, d(ax) = |a| dx. Let

Bc = x ∈ A : |xv| ≤ cv

where c = (cv) is such that cv = 1 for almost all v. Bc is a fundamental domain of K ⊆ A.Let f = 1Bc then ϕ(x) =

∑a∈k f(x+ a) ∈ L1 (A/K) and ϕ(x) = 1 for almost all x.

vol (Bc, dx) = vol (A/K, dy)

and ∫Af(x)dx =

∫A/K

ϕ(y)dy.

For all a ∈ K×, if BC is a fundamental domain, so is B|av|cv . Therefore,

vol(Bc0) = vol(B|av|cv

)= vol (A/K) .

Proposition 2.40. If dx is the self-dual Haar measure on A with respect to ψA1 , then vol (A/K) = 1.

Proof. Explicitly show that vol (Ov) = |discriminant|−1/2.


Another proof is using the Poisson summation formula.

0→ K → A→ A/K → 0.

Here, K is discrete and A/K is compact. Using ψ1, we get

0→ K⊥ = K → Λ(A) = A→ A/k → 0.

Let f ∈ L1(A) assume that f ∈ L1(A) (which implies they are continuous). Then ϕ ∈ L1 (A/K) maps to ϕ = f |K by

Fubini.

Assume ϕ ∈ L1(K) = L1 (Λ(A/K)) (that is,∑a∈K

∣∣∣f(a)∣∣∣ converges). Fourier inversion says that

ϕ(0) =

∫K

ϕ(a)da.

Here, da is the Haar measure on K, dual to dy on A/K. This is

= vol (A/K, dy)∑a∈K

ϕ(a).

Since ϕ and ϕ are continuous, and ϕ(x) =∑a∈K f(x + a), we see that ϕ(0) =

∑a∈K f(a). Exchange f and f then

vol (A/K, dy) .

Corollary 2.41. If f ∈ L1(A) and f ∈ L1(A) then∑a∈K f(a) =

∑a∈K f(a) as long as both sums converge absolutely.

2.5. Schwartz Functions. Let V be a nite dimensional R-vector space. We sayϕ : V → C is a rapid decaying function

if for all polynomial functions p : V → C, we have |pϕ| is bounded. Then it must be that ϕ ∈ L1(V ). Let

S (V ) = ϕ measurable function on V such that ϕ and ϕ are of rapid decay .

Proposition 2.42. ϕ ∈ S (V ) i ϕ is smooth and for all polynomials p and n,∣∣pϕ(n)

∣∣is bounded.Proof. Let V = R. Then

ϕ(x) =

∫ϕ(y)e(xy)dy.

Since ϕ is of rapid decay, we can dierentiate under the integral sign. This implies that ϕ is smooth, and similarly for

ϕ.

Recall that for R, bα = e−πx2

is of rapid decay, and is its own Fourier transform. It is a Schwartz function.

Now, for non-archimedean elds F . Then V is totally disconnected locally compact abelian groups with V = Λ(V ).

Let ϕ be a continuous function on V with compact support.

S (V ) = ϕ continuous function on V with compact support, and same with ϕ .

Proposition 2.43. ϕ ∈ S (V ) i ϕ is a linear combination of characteristic functions of compact open subsets.

Proof. Since V is totally disconnected, we have a system of neighbourhoods of 0 consisting of compact open subgroups

Kα with ∩Kα = 0. Then ∪K⊥α = V .

ϕ is invariant under Kα for some α, and

ϕ(x) =

∫V

ϕ(y)e(xy)dy.

If supp(ϕ) ⊆ K⊥α then this is equal to ∫K⊥α

ϕ(y)e(xy)dy.

For y ∈ K⊥α , x 7→ e(xy) is Kα-invariant.


We have

S (AK) = S (Afin)⊗S (KR) = nite linear combinations of ϕfin ⊗ ϕα

where ϕfin ∈ S (Afin) and ϕα ∈ S (KR).

For ϕ ∈ S (A), ∑a∈K

ϕ(a)

is absolutely convergent and so ∑a∈K

ϕ(a) =∑a∈K

ϕ(a).

To prove absolute convergence, we reduce to the usual case. Since supp(ϕfin) ⊆ K for some compact open subgroup of

Afin, ∑a∈K|ϕ(a)| ≤ c

∑a∈K∩K

ϕα′(a).

The K ∩ K is a lattice in KR (think Q ∩ Z = Z ⊆ R). Therefore, this converges.

2.6. Tempered Distribution on F . Dene

S ?(F ) = continuous linear form on S (F ) .

When F is p-adic, we use the discrete topology. Otherwise, we use the norm topology.

Example 2.44. For x ∈ F , let δx(ϕ) = ϕ(x). This is the Dirac function.

Example 2.45. Let f be a continuous function on K. Then f ∈ S ?(F ) as the map ϕ 7→∫ϕ(x)f(x)dx.

Example 2.46. For a ∈ S ?(F ) get a ∈ S ?(F ) dented by a(ϕ) = a(ϕ).

Example 2.47. δx = 1F and δψ = ψ−1x (mapping χ 7→ χ(x)−1).

Suppose χ : F× → C×is a character. What is χ? If χ(x) = |x|s with Re(s) > 0, then x can be extended contiuously

to 0 and χ ∈ S ? (F ) (really, just need Re(s) > −1).

We want to extend almost all χ to a tempered distribution. We will get a formula

χ |·|−1= γ(χ)χ−1

where γ(χ) is some constant.

3. Module in Local Fields

Let F be a local eld, with invariant measure µ. For f ∈ Cc(F ), we let µ(f) =∫Xf(x)µ(x) (here, X is the support of

f). We have an action of F× on F , on functions, and on measures. For t ∈ F×, x ∈ F , ϕ ∈ Cc(F ) and µ a measure, we

act by

(t, x) 7→ tx, ϕt(x) = ϕ(t−1x) and µt(ϕ) = µ(ϕt−1).

Then

〈µt, ϕt〉 = 〈µ, ϕ〉 .

Let µ be the additive invariant measure on F (for instance, the self-dual measure depending on ψ : F → C×). µt is thenalso invariant measure, with

µt = |t|−1µ

and here, |t|−1is called the module (modulus?). This is because∫

ϕ(x)µt(x) =

∫ϕt−1(x)µ(x) =

∫ϕ(tx)µ(x)


then use chage of variable with y = tx.

Use the usual absolute value over R, and the absolute value |x+ iy| = x2 + y2 on C . For F a local non-archimedean

eld, let |a|F = q−val(a). Dene the absolute value on A×F by

(xv) 7→∏|xv|Kv .

Now, how do we normalize the multiplicative measure. Let ω1 : F× → C× be given by x 7→ |x|. Consider the measure

µ on Cc (F×) be the same as the measure on Cc(F ) via extension by 0.

Claim 3.1. ω−1µ is an invariant measure on F× (the measure µ(x)|x| ).

For F = Q×p , vol(Z×p , ω−1µ

)= p−1

p = 1− 1p and vol (Zp, µ) = 1.

For F non-archimedean local eld,

vol(O×F , µ

)= vol

(O×F , ω

−1µ)

=(1− q−1

)disc

−1/2F/Qp =

(1− q−1

)vol (OF , µ) .

We know A× = A×fin × K×R . The rst is totally disconnected, and the latter is (R×)

r1 × (C×)r2 . Take the invariant

measure to be µ× = µ×fin × µ×kR. The invariant measure on A×f is determined by

vol(∏

Ov, µ)> 0,

because it is a compact open subgroup of A×f . If we choose µ =∏|ω|−1

v µv and when K = Q, we have

vol =∏(

1− p−1)

= 0.

This means that we need to normalize so that vol (O×v , µ×) = 1. That is, let µ×v =(1− q−1

)−1disc−1/2ω−1µv.

Suppose F is archimdean, and take µ× = ω−1µ. We have the exact sequence

0→ A1 → A× |·|→ R×+ → 0

which has a splitting, by choice of archimedean places. This means A× = A1 × R×+, so let µ×A = µA1 × µ×R+to give a

measure on A1. Recall that A1/K× is compact.

Proposition 3.2. vol(A1/K×, µA1

)= 2r1 (2π)r2hR√

|d|wwhere h is the class number, R regulator, d is the discriminant and w

is the number of unit roots.

Proof. Hint:

0→ K×R /lattices→ A1/K×∏O×v → CL(K)→ 0.

Recall

Ω(F×) =χ : F× → C× not necessarily unitary

and it has a distinct character ω1 (the module character). For all s ∈ C, let ωs(x) = |x|s. This denes a homomorphism

C→ Ω (F×), and this is injective if F = R or C.For F non-archimediean, ω1 : F× → R×+ factors through qZ. This means that we have

0→ 2πi

log qZ→ C→ Ω(F×)→ Ω

(kerω1

)→ 0

The kernel ker(ω1)is a compact subgroup of F×. When F = R, this is ±1, when F = C, this is C1 and for non-archimdean,

it is O×F .Then, Ω(F×) is an extension of the discrete group Λ

(kerω1

)= Ω

(kerω1

)by C over the quotient C by Z.

For R× = R×+ × ±1 and Ω(R×+) = C and Ω(±1) = ±1. This means that Ω (R×) = Cq C.For C× = R×+ × C, Ω(C×) = Ω

(R×+)× Ω(C1) = C× Z.

For F× = O×F × Z, Ω(F×) = Ω(O×F )× Ω(Z) = Ω(O×F )× C/Z (which are 2πilog q 's).


Let χ : F× → C× then we can dene |χ| : F× → R×+. This will satisfy |χ| (x) = |x|s for some s ∈ C, because|χ| |O×F = 1. We dene Re(χ) = Re(s). Then dene an exponent map Ω(F×)→ R by χ 7→ Re(χ).

This can be applied to A×/K× ωA→ R×+. ker(ωA) = A1/K× some gigantic compact group.

3.1. Mellin transform over R×+. Let ϕ : R×+ → C. We know Ω(R×+)

= C. Dene

Mϕ(s) =

∫R×+

ϕ(x)xsµ×(x)

for all s ∈ C.Recall

S (R×+) =ϕ : R×+ → C of rapid decay at 0 and ∞

That is,

∣∣xσϕ(n)(x)∣∣ bounded for all σ, and so if σ > 0, bounded as x→∞; and if σ < 0, as x→ 0.

Theorem 3.3. For ϕ ∈ S (R×+), Mϕ(s) is absolutely converge for all s ∈ C and denes a holomorphic function g = Mϕ

of variable s. They lie in

Z(C) = g : C→ C hol., and ∀P (s) polynomial, p(s)g(s) is bounded in every vertical strip .

If g ∈ Z(C) then

ϕ = M−1g (x) =

1

2πi

∫σ+iR

g(x)x−sds.

Proof. We have ∫ α

σ

ϕ(x)xsµ×(x) =

∫x<1

+

∫x>1

.

For the x < 1 portion, we see that |ϕ(x)xs| < cx2 and for x > 1, |ϕ(x)xs| < cx−2. Therefore, we have absolute convergence.

We know that ∂sxs = log(x)xs. ∫

x<1

ϕ log(x)xs +

∫x>1

ϕ(x) log(x)xs

absolutely converges. Since Mϕ has complex derivatives, it is holomorphic.

Write g = Mϕ. Consider R→ R×+ by y 7→ x = ey.

M(ϕ)(σ + it) =

∫Rϕ(ey)eyσeiytµ(y).

The eiyt is an additive character, and so Mϕ(σ + it) viewed as function of t is the Fourier transform of ϕ(ey)eyσ. The

function y 7→ ϕ(ey)eyσ is in L1 and so Mϕ(σ + it) as function of t is continuous and bounded (by the L1-norm). This

means it is bounded on vertical strips.

For boundedness, use invariant dierential operator xdx (with respect to multiplication).

M (xδxϕ) =

∫ α

a

δxϕ(x)xs+1dx = −sMϕ(s)

which shows that |sMϕ(s)| is bounded.Now,

M−1g(x) =1

2πi

∫ α

−αg(σ + it)x−(σ+it)µ(t)

is again a Fourier transform. Since P (s)g(s) is bounded on vertical strips, g(σ + it) in L1 (for xed σ). Apply Fourier

inverse to get M−1g = ϕ.

Need to check M−1g is smooth, rapid decay and things. Smooth follows from dierentiating inside the integral. For

rapid decay, it's from the x−σ factor.

ϕ(x)xσ =1

2πi

∫ α

−αg(σ + it)x−itµ(t)

is bounded, because g is L1. Identify R→ R×+ by y 7→ eiy. Then ϕ(x)xσ = ϕ(ey)eyσ gives Fourier transform g(σ+ it).


Let S+

(R×+)be the set of ϕ : R×+ → C smooth, such that

• for all σ > 0,∣∣xσϕ(n)(x)

∣∣ is bounded (rapid decary as x→∞)

• ϕ has a Taylor expansion at 0,∑n≥0 anx

n formal series (not analytic). That is, for all n, ϕ(x) =∑ni=0 aix

i+O(xn).

Theorem 3.4. Let Z+(C) be meromorphic functions g of variable s ∈ C, with at most simple poles at 0,−1,−2, ..., such

that for all polynomial P (s), P (s)g(s) is bounded on every vertical strip often deleting a small disk centered at the poles.

If ϕ ∈ S+(R×+) then g = Mϕ ∈ Z+(C). If g ∈ Z+(C), then ϕ = Ng ∈ S+(R×+) with σ > 0. The M and N are inverses

of each other.

The Taylor coecients a0, a1, ... of ϕ at 0 are exactly the residues of g at 0,−1,−2, ....

Proof. The fact that∣∣ϕ(n)(x)xσ

∣∣ is bounded for σ > 0, implies that

Mϕ(s) =

∫R×+

ϕ(x)xsµ×(x)

is integrable, for σ = Re(s) > 0.

Meromorphic continuation of Mϕ, to get for Re(s) > 1

M(∂xϕ)(s) =

∫ ∞0

∂xϕ(x)xs−1dx = −(s− 1)

∫ ∞0

ϕ(x)xs−2dx = −(s− 1)Mϕ(s− 1).

We take this to be the formula,

Mϕ(s− 1) = −(s− 1)−1M(∂xϕ)(s)

giving Mϕ dened for Re(s) > −1 with a pole at 0. We can repeat this, to get pole at 0,−1,−2, ....

For p(s)g(s) bounded on vertical strips, use invariant dierential operator x∂x to get

M(x∂xϕ)(s) = −sMϕ(s).

Now,

s−1 (s− 1)−1...(s− n)−1

is bounded on C− (D0 ∪D1 ∪ ... ∪Dn) (the vertical strips). This proves the boundedness of p(s)g(s).

Suppose g : C→ C meromorphic with poles at 0,−1, ...

1

2πi

∫σ+iR

g(s)x−sds.

If −(n− 1) < σ < −n, this intergral is just the sum of the residues

ϕ(x) =

−n∑i=0

resi(g(s)x−sds

)+O(x−σ) =

n∑i=0

xires−ig(s)ds+O(x−σ).

They become the Taylor coecients as required.

Proposition 3.5. S (R×+) is an algebra with respect to convolution, that is

(ϕ1 ? ϕ2)(x) =

∫R×+

ϕ1(y)ϕ2(y−1x)µ×(y).

Additionally,

Mϕ1?ϕ2(s) = Mϕ1

(s)Mϕ2(s).

Remark 3.6. It won't hold of S+(R×+), because when we take Mellin transform, we pick up simple poles. The product

will then get double poles.

A good reference is Iguasa higher form...

However, S+(R×+) is a module over S (R×+) with generator e−x.

M(e−x)(s) =

∫ ∞0

e−xxsdx

x= Γ(s)


for Re(s) > 0. Recall that we prove meromorphic continuation of Γ(s) exactly by applying the dierential operator, and

get Γ(s+ 1) = sΓ(s).

Γ has simple pole at 0,−1,−2, ... and is a generator of Z+(C). Γ is non-vanishing away from 0,−1, ...

Γ(s)Γ(1− s) =2πi

eπis − e−πis

has simple pole in Z and non-vanishing on C− Z.For all ϕ ∈ S+(R), ϕ = e−x ? ϕ0 for some ϕ0 ∈ S (R×+).

3.2. Mellin Transform on R×. We can write R× = R×+ × ±1. We know Ω(R×) = CqC, where the rst s ∈ C maps

to (x 7→ |x|s) the other maps x 7→ sgn(x) |x|s.For ϕ ∈ S (R), χ ∈ Ω(R×), let

Mϕ(χ) =

∫R×

ϕ(x)χ(x)µ×(x)

which converges for all σ = Re(s) > 0. Have meromorphic continuation of Ω(R×) with simple poles at 0,−2,−4, ... in

Ω+(R×) and −1,−3, .... in Ω−(R×).

We write ϕ = ϕ+ + ϕ−, the even and odd Schwartz functions. Then

Mϕ+|Ω−= 0 = Mϕ− |Ω+

.

ϕ+ only has even Taylor coecients, and ϕ− only has odd ones.

Let L(Ω(R×)) be the set of meromorphic functions on Ω(R×) with simple poles 0,−2, ... on Ω+ and −1,−3, ... on

Ω−(R). Also, p(s)g(s) bounded on vertical strips, as before.

Our L-function is a generator of L (Ω(R×)) (with some choice), as a module on holomorphic functions with boundedness

condition on vertical strips.

Recall we have a basic function b∞(x) = e−πx2

which is self-dual under Fourier transform, and is an even function.

Then

Mϕ∞ |Ω−= 0 and Mb(ωs) = π−s2 Γ(

s

2)

This latter part is what you multiply with ζ(s) to get functional equation!

Exercise 3.7. Work out Mellin transform on C×.

3.3. Mellin Transform on non-archimedean local elds. Let F be a non-archimedean local eld.

S (F ) = C∞c (F ) ⊇ C∞c (F×).

This is similar for S (R) ⊇ S (R×), but this time, we just have codimension 1 (value at 0).

Ω(F×) =χ : F× → C×

and

Mϕ(χ) =

∫F×

ϕ(x)χ(x)µ×(x)

where µ×(O×F ) = 1. Mϕ : ϕ ∈ C∞c (F×)

= L (Ω(F )) and Mϕ : ϕ ∈ C∞c (F ) = L+ (Ω(F )) .

We want to study these two spaces, and eventually discovering that L(χ) is a generator of L+ (Ω(F )). Recall that

non-canonically, F× = O×F × ωZ (ω uniformizer). This means that

Ω(F×) = Ω(O×F )× Ω(ωZ) = Λ(O×F )× Ω(ωZ).


Additionally,

Λ(O×F)

= lim→

Λ((OF /ωnOF )×

)and is a union of nite subgroups. Ω(ωZ) = C× by t ∈ C× mapping to χt(ω) = t. χt : F× → C×is trivial on O×F ,χt(x) = |χ|s = ωs and so t = q−s. Hence, Ω(F×) =innite disjoint union of circles of C×.

Denition 3.8. g : Ω(F×)→ C is said to be polynomial of t if

• g vanishies on all but nitely many components

• on each component, g is a polynomial of t

Proposition 3.9. For all ϕ ∈ C∞c (F×), the Mellin integral Mϕ(χ) is convergent for all χ ∈ Ω(F×), and Mϕ is a

polynomial of variable t. Conversely, if g is a polynomial on Ω(F×), then

ϕ(x) = Ng(x) =1

2πi

∫|t|=q−s

g(t)χt(x)−1dt

(the integral being independent of σ) denes ϕ ∈ C∞c (F×).

Proof. Suppose ϕ ∈ C∞c (F×). For x ∈ F×, there is a compact open subgroup Kx of F× such that ϕ is constant on xKx.

Supp(F ) is then a nite disjoint union of xKx as above.

K = ∩nite intersectionKx

is a compact open subgroup of F×. ϕ is then a K-invariant. For all x, ϕ is constant on xK, unless χ |K= 1,∫F×

ϕ(x)χ(x)µ×(x) = 0.

This is because on a coset xK,∫Kχ(x)µ×(x) = 0. Therefore, Mϕ vanishes away from Ω(O×F /K)× C×.

supp(ϕ) is compact, there exists m,n ∈ Z such that

supp(ϕ) ⊆x ∈ F× : m ≤ val(x) ≤ n

.

Then ϕ =∑ni=m ϕi where Supp(ϕi) = tiO×F , and ϕi(tix) =

∑χ∈Ω(O×F /K) ϕ

χi and

ϕχi (tix) = χ(x)ϕχi (ti).

For all χ ∈ Ω(O×F /K

), Mϕ |χ×C× is a linear combination of tm, tm−1, ..., tn is a polynomial of t.

For the inverse Mellin transform, it's just Fourier series.

Proposition 3.10. For all ϕ ∈ C∞c (F ), the Mellin transform is converge for all x ∈ Ω(F×) for |t| < 1, and holomor-

phically continued on all component Ωχ = χ × C×. For χ 6= 1, and meromorphically continued on the component

Ω1 = 1 × C× with a simple complex pole at t = 1. More precisely,

Mϕ ∈ L+(Ω(F×)) =g : Ω(F×)→ C rational function of t such that gL−1(χ) is a polynomial of t

.

Here,

L(χ) =

1− t on the neutral component C×

1 everywhere else.

The inverse Mellin transform is given by

Ng(x) =1

2πi

∫|t|=q−σ

g(t)χt(x)−1dt

for σ > 0.


Proof. Compute the Mellin transform of the basi function bF (x) = 1OF . This is O×F -invariant,MbF vanishes o the netural

component of Ω(F×). On the netural component Ω0(F×) = C×, MbF (t) is absolutely convergent for |t| < 1 because

MbF (t) =

∞∑i=0

ti = (1− t)−1

which we can meromorphically continue.

For ϕ ∈ C∞c (F ), ϕ = ϕ(0)bF + ϕ′ where ϕ′ ∈ C∞c (F×). Mϕ′ is polynomial function on Ω(F×), MbF = (1 − t)−1 on

Ω0(F×) and 0 elsewhere. This means the L function is (1− t)−1 on the neutral component and 1 else.

4. Eigendistribution

F local eld.

• F× acts on F by (t, x) 7→ tx.

• F× acts on S (F ) by ϕt(x) = ϕ(t−1x).

• F× acts on S ′(F ) by 〈at, ϕ〉 = 〈a, ϕt−1〉 i 〈at, ϕt〉 = 〈a, ϕ〉.

S ′(F ) = a : S (F )→ C : continuous linear form is space of tempered distribution. For χ : F× → C×,

S ′(F )χ =a ∈ S ′(F ) : for all t ∈ F×, at = χ(t)a

is the space of eigendistribution.

Theorem 4.1. For all χ ∈ Ω(F×), dim S ′(F×)χ = 1. (By Weil, in a Bourbaki talk on Tate's thesis).

Proof. Recall that ϕ ∈ S (F×) implies that Mϕ ∈ L (Ω(F×)) (all polynomial functions on Ω(F×).

S ′(F )χ = C× evaluation of Mϕ at χ .

ϕ 7→Mϕ(χ) =

∫F×

ϕ(x)χ(x)µ×(x).

Here, χµ× ∈ S ′(F )χ−1

.

Dierence between S ′(F×) and S ′(F ): For F non-archimedean, we have

0→ C∞c (F×)→ C∞c (F )→ C→ 0

and so we get an exact sequence

0→ (C)χ → S ′(F )χ → S ′(F×)χ.

From this, we see that

dim S (F )χ ≤

1 if χ 6= 1

2 if χ = 1.

For χ 6= 1, ϕ 7→Mϕ(χ) denes a non-zero element of S (F )χ. When χ = 1, ϕ 7→ rest=1Mϕ is a generator of S (F )χ.

S (Ω(F×)) is invertible O-module on Ω(F×) and so S (Ω(F×))χ has dimension 1.

dim S ′(F )χ = 1, Re(χ) > 0 (|t| < 1) then ⟨χµ×, f

⟩=

∫f(x)χ(x)µ×(x)

converges and χµ× ∈ S ′(F )χ. Its Fourier transform f 7→⟨χµ×, f

⟩is also a non-zero eigendistribution. χµ× ∈ S ′(F )χw1

(ω1 is |·|).


4.1. Eigendisctribution and γ-function.

Proposition 4.2. Let χ ∈ Ω(F×) then we can extend by zero to χ : F → C. This function is locally integrable if

Re(χ) > −1, and dier by a tempered distribution χ ∈ S ′(F )χ−1ω−1

where ω−1(x) = |x|−1. Its Fourier transform

χ ∈ S ′(F )χ.

Proof. Re(χ) > −1 implies that χ is locally integrable. Check case by case.

For F = R, ∫ 1

0

tσµ(t) = (σ + 1)−1(tσ+1

)|10

and so t 7→ tσ is locally integrable if σ > −1.

For χ : F× → C×, |χ| : F× → R×+ is trivial on O×F . Therefore, |χ| (x) = |x|s for s ∈ C. Dene Re(χ) = Re(s). If

χ is locally integrable, it denes a distribution χ : C∞c (F ) → C. χ is tempered, thus linear functions can be extended

S (F )→ C.Let ϕ ∈ S (F ), ∫

F

ϕ(x)χ(x)µ(x) =

∫|x|≤1

+

∫|x|>1

.

The∫|x|≤1

converges as χ is locally integrable.∫|x|>1

converges because of the rapid decay |ϕ(x)|xσ+1+ε bounded and so

χ is a tempered distribution.

〈χ, ϕ〉 =

∫F

ϕ(x)χ(x)µ(x)

converges.

〈χt, ϕ〉 = 〈χ, ϕt−1〉 =

∫F

ϕ(tx)χ(x)µ(x) = χ(t)−1 |t|−1 〈χ, ϕ〉 .

This means that χ ∈ S ′(F )χ−1ω−1

. For χ ∈ S ′(F ), Fourier transform

〈χ, ϕ〉 = 〈χ, ϕ〉 .

For Re(x) > −1, χ is a tempered distribution and χ ∈ S ′(F )χ.

Lemma 4.3. For t ∈ F×, ϕ ∈ S (F ), (ϕ)t = |t| ϕt−1 . If a ∈ S ′(F ), (a)t = |t|−1at−1 .

(χt) = |t|−1χt−1 = |t|−1

χ−1(t−1)∣∣t−1

∣∣−1χ = χ(t)χ.

Proposition 4.4. (Weil) For every character χ1 : F× → C×, dim S ′(F )χ1 = 1.

Proof. For χ1 a xed character, let χ ∈ Ω(F ) be the variable.

For a ∈ S ′(F )χ1 , at(ϕ) = χ1(t)a(ϕ) for all ϕ ∈ S (F ). This means that a (ϕt−1 − χ1(t)ϕ) = 0. a annihilates all

functions of this form. Apply Mellin transform,

M (ϕt−1 − χ1(t)ϕ) (χ) = χ(t)Mϕ(χ)− χ1(t)Mϕ(χ)

= (χ(t)− χ1(t))Mϕ(χ).

Recall that M : S (F ) ∼= L (Ω(F )). We can apply a on both of these spaces. a annihilates the submodule generated by

(χ(t)− χ1(t)) g.

Therefore,

dimC L(Ω(F×)

)/ 〈(χ(t)− χ1(t))L〉 = 1.

If χ1 /∈ poles then a is a multiple of g 7→ g(χ1). If χ1 ∈ simple poles, a is a multiple of g 7→ resx=χ1g.


When Re(χ) > −1, χ ∈ S ′(F )χ−1ω−1

and χ ∈ S ′(F )χ. If −1 < Re(χ) < 0, then χ′, χ ∈ S ′(F )χ where χ′ = χ−1ω−1.

Since Re(χ) < 0, Re(χ′) > −1. This gives two non-zero elements in the same one dimensional space. Write

χ = γ(χω)χ′

where γ is a function of χ, well dened for 0 < Re(χ) < 1.

The integral representation of γ is given by the following.

γ(χω)χ′(y) =

∫F

χ(x)e(xy)µ(x).

Let y = 1, then

γ(χω) =

∫F

χ(x)e(x)µ(x)

and so

γ(χ) =

∫F

χ(x)e(x) |x|−1µ(x).

In particular, for χ(x) = |x|s,

γ(s) =

∫F

e−2πixxs−1dx

which is very similar to

Γ(s) =

∫F

e−xxs−1dx.

Anyways,

γ(χ) =

∫|x|≤1

+

∫|x|>1

which both converge for 0 < Re(χ) < 1 and both have meromorphic continuation. Therefore, γ has a meromorphic

continuation.

To evaluate γ, evaluate χ = γ(χω)χ′ with a test function ϕ ∈ S (F ). We have

Mϕ(χ) = γ(χ)Mϕ

(χ−1ω

).

Calculate γ by choosing a nice test function.

Let F = R, and let ϕ be the basic function b∞(x) = e−πx2

. Then b∞ = e−πx2

.

Mb∞ : Ω(R×) = Ω+

(R×)q Ω−

(R×)→ C×.

The rst component is generated by |x|s and the latter sgn(x) |x|s.

Mb∞(ωs) = π−s/2Γ(s/2)

and 0 on Ω−(R×).

On Ω+,

γ(ωs) =Mb∞(ωs)

Mb∞(ω1−s)=

π−s/2Γ( s2 )

π−(1−s)/2Γ(

(1−s)2

) = 21−sπs/2 cos(πs

2)Γ(s).

For F = Qp, calculate γ on Ω(F×) = C× the unramied characters. Recall, for t ∈ C×, χt(ωF ) = t and is trivial on

O×F . Let bp = 1Zp . Recall

Mbp =

(1− t)−1 on Ω0

0 else

γ(χt) = 1−p−1t−1

1−t and γ(ωs) = 1−ps−1

1+p−s .


5. Global Picture

Let K be a global eld, A = AK . Let A1 = ker(|·| : A× → R×+

), and K×\A1 is compact. Choose an Archimedean place

of K to get a splitting A×/K× =(A1/K×

)× R×+. Recall, we can normalize our measure so that vol (O×v , µ×v ) = 1 when

v is non-Archimedean and µ×v = µv|·| for Archimedean (self-dual).

vol(A1/K×, µ×

)= class number, regulator stu.

Ω (A×/K×) is a countable union of Ω(R×+)

= C.For ϕ ∈ S (A) maps toMϕ(χ) =

∫A× ϕ(x)χ(x)µ×(x). We will show that this converges forRe(χ) > 1, have meormorphic

continuous, and satisfy a similar functional equation.

We can get an invariant measure µ1 on A1/K× so that µ× = µ1 × µR× . For ϕ functino on A×,∫Aϕ(x)µ×(x) =

∫A×/K×

ϕ+(x)µ×(x)

where ϕ+(x) =∑α∈K× ϕ(αx).

5.1. Zeta Integral. ϕ ∈ S (A) is a linear combination of ⊗vϕv where ϕ ∈ S (Kv) and ϕv = 1Ov for almost all non-

archimedean v.

For χ ∈ ΩK = Ω (A×/K×) = Ω(A1/K×

)× C (the rst is a discrete group).

Mϕ(χ) =

∫A×

ϕ(x)χ(x)µ×(x).

Proposition 5.1. Mϕ(χ) converges absolutely if Re(χ) > 1 and dene a holomorphic function on that region.

Proof. For |χ(x)| = |x|σ for some σ = Re(χ), ∫A×|ϕ(x)| |x|σ µ×

is convergent for σ > 1.

Assume that ϕ(x) =∏v ϕv(xv) where ϕv = 1Ov for all v ∈ |K| − S for some nite set S. Suppose v /∈ S, then∫

K×v

1Ov (x) |x|σ µ×(x) =(1− q−σv

)−1

converges for σ > 0. If σ > 1, then∏v/∈S (1− q−σv )

−1<∞ . This implies absolute convergence, holomorphic etc.

For all ϕ ∈ S (A), Mϕ holomorphic function Ω|χi|>1k then Mϕ has meromorphic continuation and function equation.

Proposition 5.2. For every x ∈ A×, x the image in A×/K×, for all ϕ ∈ S (A), the series

ϕ+(x) =∑α∈K×

ϕ(αx)

is absolutely convergent. Moreover, ϕ+(x) has rapid decay as |x| → ∞. That is, for σ > 0, |ϕ+(x) |x|σ| is bounded.

Assume K = Q. A× = Q× × Z× × R×+. Can assume that x = (xfin, x∞) where xfin ∈ Z× and x∞ ∈ R×+.

Proof. Suppose αx ∈ supp(ϕ). xp ∈ Z×p and so α ∈ Zp (because support of ϕv is Zp) and so α ∈ Z. This means

ϕ+(x) =∑

α∈Z−0

ϕ∞(αx).

ϕ∞ has rapid decary and so the series∑n∈Z−0 ϕ(αx) is convergent and has rapid decay as x→∞.


The analytic problem for ϕ ∈ S (R). We want to understand the Mellin transform of ϕ+. Just take the positive α for

now, we see that ∫R×+

(ϕ(x) + ϕ(2x) + ...)xsd×x =(1 + 2−s + 3−s + ..

) ∫ϕ(x)xsd×x

= ζ(s)

∫ϕ(x)xsd×x.

The problem is to understand the asympototic behaviour of ϕ+(x) as x→ 0. We can solve this by Poisson summation.∑n∈Z

ϕ(n) =∑n∈Z

ϕ(n).

For all x 6= 0, ∑n∈Z

ϕ(nx) = |x|−1∑n∈Z

ϕ(nx−1)

and so ∑n∈Z−0

ϕ(nx) = −ϕ(0) +∣∣x−1

∣∣ϕ(0) +∑

n∈Z−0

ϕ(nx−1)

= −ϕ(0) + |x|−1ϕ(0) + rapid decay.

The result is that as x→ 0,

ϕ+(0) = −ϕ(0) + |x|−1ϕ(0) +O(xσ)

(the rst term contributes a pole at s = 1, the second at s = 1).

Now for the general case. ∫A×

ϕ(x)χ(x)µ×(x) =

∫A×/K×

ϕ+(x)χ(x)µ×(x)

makes sense when the integral converges absolutely for σ = Re(χ) > 1.

=

∫|x|>1

ϕ+(x)χ(x)µ×(x) +

∫|x|≤1

ϕ+(x)χ(x)µ×(x).

Since ϕ+(x) has rapid decay as x→∞,∫|x|>1

is absolutely convergent for all χ. Meanwhile,∫|x|≤1

converges only when

σ > 1. We will write ∫|x|≤1

= (3) + (4) + (5)

where 3 is absolutely convergent for all χ, 4 and 5 are simple functions that can easily be meromorphically continued.

For all x ∈ A×, Poisson summation gives

ϕ+(x) =∑α∈K×

ϕ(αx) = |x|−1∑α∈K×

ϕ(αx−1)

= −ϕ(0) + |x|−1ϕ(0) + |x|−1

ϕ+(x−1)︸︷︷︸rapid decay as x→0

.

Therefore, ∫|x|≤1

= −ϕ(0)

∫|x|≤1

χ(x)µ×(x) + ϕ(0)

∫≤1

|x|−1χ(x)µ×(x)

+

∫|x|>1

ϕ+(x) |x|χ−1(x)d×x.


(This is (4) + (5) + (3)). We can go further to compute this integral. Let χ = χ1 |·|s. Because A1/K× is compact:∫|x|≤1

χ(x)µ×(x) =

∫A1/K×

χ1(x)µ1(x) ·∫ 1

0

xsdx

x

=

0 χ1 6= 1

vol(A1/K×, µ1

)· 1s χ1 = 1

which converges for all s > 0. Similarly,

(5) =

0 if χ1 6= 1

vol(A1/K×

)1s−1 if χ1 = 1

converges i Re(s) > 1 and has meromorphic continuation.

Since this decomposition is symmetrical for ϕ← ϕ via χ← ωχ−1, we have

Mϕ(χ) = Mϕ(ω1χ−1)

which is global functional equation. Recall that our local functional equation is

γv(χ)Mϕv (χv) = Mϕv (ω′′χ−1).

the global functional equation may be thought of as the fact that∏γv(χ) = 1. However, Mϕ converges for Re(χ) > 1

and Mϕ(χ) for Re(χ) < 0, so... can't actually write this.

6. Langlands Conjectures

Two main problems in automorphic forms.

• Reciprocity conjectures

• Functoriality conjectures

They are all linked to L-functions.

6.1. Class Field Theory. Understand abelian extensions of number elds. Let K be a number eld. It realtes GabK with

IK = A×K/K× and local-global compatibilities.

Suppose v is a place of K, then we have induced maps

GKv → GK and GabKv → GabK

which are well-dened up to conjugation.

6.1.1. Local Reciprocity. Let F = Qp. If p 6= 2, there are three quadratic extensions, four if we count split extensions

(Qp ×Qp).SupposeK/F is a quadratic extension, then NK/F (K×) ⊆ F× is a subgroup of index 2, so χi : F× → F×/NK/F (K×) =

±1. We can check that F×/F×2 ∼= Z/2Z× Z/2Z.Write F× = O×F × πZ

F where πF is an uniformizer. Then

F×/F×2 = O×F /O×2F × π

ZF /π

2ZF∼= ±1 × ±1 .

The rst ±1 is generated by τ , a non-quadratic residue, and the second by ε, the class of πF .

The unramied extensions are those such that χ : F× → Z/2Z has χ(τ) = 1, and else ramied.


For an extension of number eld K/k, we know that Kv/kv is unramied for almost all v. This means that the collection

χv : k×v → ±1 gives a quadratic character, which is an idele class character

χ : A×k → k×\A×k → ±1 .

The idele class character χ has an associated L-function L (χ |·|s) which is a meromorphic function with complex parameter

s. It is holomoprhic if χ 6= 1 and has a functional equation.

This can be extended to all characters Gk → C×.

6.1.2. Galois representations. Consider Artin representations σ : Gk → GLn(C) irreducible. Then σ must have nite

image. Then σ induces σkv : Gkv → GLn(C).

0→ Iv → Gkv → GFv → 0

where GFv = 〈Frv〉 ∼= Z. σkv (Frv) is a well-dened conjugacy class of GLn(C).

Ls (σ, s) =∏v/∈S

det(1− σv(Frv)q−sv

)−1

Conjecture 6.1. (Artin) Can be extended to an holomorphic function of s (non-trivial irreducible reps).

It has known meromorphic continuation. Langlands program in the early days, were to address this conjecture for

many of theses cases (except icosahedral case).

We can dene using local functions so that the complete L-function has functional equation L(s, σ) = εL (1− s, σ).

6.2. Shimura-Taniyama-Weil Conjecture. There are more representations σ = Gk → Gln(Qp) coming from geometry,

which can have innite image. Just like Artin's L-function, we can write

LS (A, s) =∏v/∈S

det(1− σv(Frv)q−sv

)−1.

This depends on a choice of identication Qp ∼= C.Taniyama explains that such representations coming from elliptic curves must be a Hecke L-function of a modular form

of weight 2. These are automorphic representation of GL2(Q) whose archimedean component is given (a discrete series of

Gl2(R)).

For k = Q,GL2(k)\GL2(A)/SO2(R) ∼= Γ\H.

Discrete series means that ϕ is a holomorphic function of H.

6.3. Langlands conjecture. Langlands: Reciprocity conjectures: Galois representations coming from n-dimensional

representation (`-adic cohomology of algebraic variety) correspond to holomorphic reps of Gln.

Functoriality: can be stated entirely within the automorphic theory

Suppose G/k is a split reductive group. These are classied by Dynkin diagrams. The root data ↔dual root data →G∨/C.π = ⊗v∈|k|πv an automorphic representation of G/k, then πv is unramied for almost all v. π

G(Ov)v 6= 0 ↔ classied by

semi-simple conjugacy class σv ∈ G∨(C).

v /∈ S ↔ σv ∈ G∨(C)/ ∼ .

Langlands introduced another data, ρ : G∨ → GLn(C) representation of G∨, and form the L-function

LS (s, π, ρ) =∏v/∈S

det(1− ρ(σv)q

−sv

)−1.


This has the same analytic properties as Riemann zeta-function (meromorphic continuations and functional equation).

6.4. Langland's L-Group and automorphic L-functions. Let K be a perfect eld, and T a torus. Let Λ =

homk (Gm, T ) ∼= Zr which has a continuous action of Γ = Gal(K/K

).

When r = 1, Aut(Z) = µ2 and Γ → µ2 is a quadratic extension of K. When r > 1, we have a map Γ → GLr(Z) (the

latter being discrete).

There is an equivalence of categories

torus over k ⇐⇒ free nite generated abelian group Λ with nite action of Γ .

Let Sr be the symmetric subgroup of GLr(Z) in which Γ factors through (not necessarily surjects on). It is called the

induced torus, and has an action of Γ. It corresponds to K ′/K nite etale extension of degree r. K ′ = K[x]/P where Γ

acts on the set of zeroes of P .

T = ResK′/KGm the Weil-restriction

and T (K) = (K ′)×.

Let G = GLn. Let B be the upper triangular matrices, the maximal solvable subgroup, the minimal subgroup such

that G/B is projective. It is the Borel subgroup.

Let K = K. Suppose B is a Borel subgroup, and T ⊆ B is a maximal torus. The pair (T,B) is called a Borel pair (all

are conjugate). The normalizer of B is B, and of (T,B) is T .

T acts of Lie(G) = g by characters. Λ = hom (Gm, T ), Λ∨ = hom (T,Gm).

g = t⊕⊕α∈Λ∨gα

where t is the Lie algebra of t. By the theorem of reductive group, dim gα = 1. The action of T on g gives rise to a nite

set Φ ⊆ Λ∨ in which dim gα 6= 0, which is the set of roots.

T, gα, g−α assemble into a subgroup of G.

From each α, can get a coroot α∨ : Gm → T , where 〈α, α∨〉 = 2. Let

Sα(x) = x− 〈x, α∨〉α

acting on Λ∨, and

Sα(x) = x− 〈x, α〉α∨

acting on Λ.

The action genereates a nite group W which stabilizer Φ and Φ∨, called the Weil group.

Reductive groups over algebraically closed eld/isom is equivalent to root datum.

Let Φ+ be so that

Lie (B) = t⊕⊕t∈Φ+gα.

There is a subset ∆ ⊆ Φ+ called simple roots (for GLn, the diagonal, just above the main diagonal). For all α ∈ ∆, x

xα ∈ gα non-zero. The

(T,B, xαα∈∆)

is pinning. The Normalizer of this is ZG (center of G), so no more inner automorphisms.

Proposition 6.2. For G/K reductive, Aut(G) automorphism group of G

0→ Inn(G) = G/ZG → Aut(G)→ Out(G)→ 0

has a splitting. The subgroup of Aut(G) stabilizing a pinning is isomorphic to Out(G).


The K-forms of G is the set of reductive groups GK/K, where GK ⊗K K ∼= G. It corresponds to H1 (K,Aut(G)). We

have a long exact sequence

H1 (K, Inn(G))→ H1 (K,Aut(G))→ H1 (K,Out(G))

that has a lift. Note, H1 (K,Out(G)) = hom (ΓK , Out(G)) .

Reductive groups G/K gives Ψ = (A,Φ,∆) with action of Γ. This gives LG.

Let G/C be such that its root system is (Λ∨,Φ∨). The action of Γ on Ψ gives rise to an action of Γ on G (xing

pinning).

Denition 6.3. LG = Go Γ

Satake: G split reudcitve group over F , non-archimedean local eld. K = G(OF ). Let H(G) = Cc (K\G/K) algebraic

under convolution.

Gelfand proves that this is commutative.

Satake: Cc (T (F )/T (OF ))W

= C[T ]W = C[G]Ad(G) (the last equality is by Langlands).

Unramied representations of G(F ) correspond to irreductible representations with xed values under K correspond

to simple H-modules. It is classied by

Spec(C[G]Ad(G)

)= semi-simple conjugacy class of G.

Now for the non-split case. Let G be an unramied group of F (so Γ action on Ψ is unramied). Then LG = Go〈Frob〉.SimpleH-modules are the same as semi-simple conjugacy classes in G·Frob. This is exactly the equivalent class of splitting

LG → 〈Frob〉 .

Suppose

ρ : LG → GL(Vρ).

Then L(s, π, ρ) =∏v det (1− ρ(σv)q

−sv )−1

like last time. It satises the same properties as Riemann zeta functions.

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