Math 679: Automorphic Forms

MATH 679: AUTOMORPHIC FORMS

LECTURES BY PROF. TASHO KALETHA; NOTES BY ALEKSANDER HORAWA

These are notes from Math 679 taught by Professor Tasho Kaletha in Winter 2019, LATEX’ed by Aleksander Horawa (who is the only person responsible for any mistakes that may be found in them). This version is from September 26, 2019. Check for the latest version of these notes at

http://www-personal.umich.edu/~ahorawa/index.html If you ﬁnd any typos or mistakes, please let me know at [email protected].

Contents

1. Introduction: Review of modular forms1 2. Overview of harmonic analysis on LCA groups 13 3. Harmonic analysis on local ﬁelds and adeles 19 4. Tate’s thesis 25 5. Artin L-functions 42 6. Non-abelian class ﬁeld theory? 59

7. Automorphic representations of SL2(R) 61

8. Automorphic representations of SL2(Qp) 81 9. Automorphic representations of GL(2, A) 106

1. Introduction: Review of modular forms

1.1. Automorphy and functional equations of L-functions. Let H = {z ∈ C | Im(z) > 0} be the upper half plane. It has an action of the Lie group SL2(R) of 2 × 2 matrices with R-coeﬃcients and determinant 1 by a b az + b z = . c d cz + d Deﬁnition 1.1. A holomorphic modular form of weifght k ∈ Z is a holomorphic function f : H → C Date: September 26, 2019. 1 2 TASHO KALETHA satisfying a b • f(γz) = (cz + d)kf(z) for all γ = ∈ SL ( ), c d 2 Z • f is bounded on domains of the form {z ∈ H | Im(z) > C > 0}. The second condition is equivalent to f being holomorphic at i∞.

Remark 1.2. The group SL2(Z) is generated by 1 1 1 0 and , 0 1 1 1 and by 1 1 0 1 and . 0 1 −1 0

1 1 In particular, taking γ = , we get 0 1 f(z + 1) = f(z), so f has a Fourier expansion X n 2πiz anq for q = e . n∈Z Then the second condition is also equivalent to saying

an = 0 for n < 0. By Remark 1.2, the ﬁrst condition is equivalent to 1 f(z + 1) = f(z) and f − = (−z)kf(z). z

How can we generalize this?

(1) We can replace SL2(Z) by a subgroup (usually of ﬁnite index). Deﬁnition 1.3. For N ∈ N, let Γ(N) = ker(SL2(Z) → SL2(Z/NZ)). A subgroup Γ ⊆ SL2(Z) is congruence if it contains Γ(N) for some N. Example 1.4. The subgroup Γ(2) is generated by 1 2 1 0 −1 0 , , . 0 1 2 1 0 −1 Let ∆ ⊆ Γ(2) be the subgroup generated by these. Notice that ∆ is normal in SL2(Z). Consider Γ(2)/∆. An element of Γ(2)/∆ can be represented as an alternating 1 1 1 0 word in , . Compute to check that the smallest length of a word 0 1 −1 1 contained in Γ(2) has length 6 and it is equal to −1 0 ∈ ∆. 0 −1 MATH 679: AUTOMORPHIC FORMS 3

1 2 0 1 Example 1.5. The subgroup Γ generated by , called the theta group. 0 1 −1 0

(2) We can allow half-integral√ weight. On the√ slices complex place C \ R<0. We have an analytic map z 7→ z determined by Re( z) √> 0. √ This can be extended non-continuously by −r = i r for r ∈ R>0. k We can try to deﬁne, for k ∈ Z, a modular form of weight 2 by requiring that f(γz) = (cz + d)k/2f(z). We will write ν(γ, z) = (cz + d). Remark 1.6. We have that ν(γδ, z) = ν(γ, δz)ν(δ, z). This is no longer true for 1 ν(γ, z) 2 . It is only true up to a ± sign. Therefore, we introduce a multiplicator system:

µ:Γ → µ∞(C) such that k µ(γ) · ν(γ, z) 2 has the desired behavior. Example 1.7. The theta multiplicator system is deﬁned by 1 2 0 1 µ = 1, µ = eπi/4. 0 1 −1 0 Example 1.8. The Jacobi theta function X 2 θ(z) = eπin z n∈Z 1 is a modular forms of weight 2 for Γ the theta group. This converges locally uniformly. Take a domain {Im(z) > C > 0}. If z = x + iy is in this domain, then

2 2 2 2 πin z πin x −πn y −πn C e = e · e < e is bounded uniformly. 1 2 0 1 To check automorphy, note that γ = is obvious. The case γ = is 0 1 −1 0 interesting. Z 2πixy Fourier analysis. Let f : R → C and fˆ = f(x)e dx. We have the inversion

R formula ˆ fˆ = f(−x) and the Poisson summation formula X X f(n) = fˆ(n). n∈Z n∈Z 4 TASHO KALETHA

The Gaussian f(x) = e−πx2 is its own Fourier transform:

Z fˆ(y) = e−πx2+2πixydx

R Z = e−π(x2−2ixy+(iy)2)e−πy2 dx

R Z = e−πy2 e−π(x−iy)2 dx.

We claim that the last integral is equal to 1. (a) The contour integral of e−πz2 over the region R which is a rectangle with corner −T, T, iy − T, iy + T is 0 because the function is holomorphic. Therefore, letting T → ∞, Z Z e−π(x−iy)2 dx = e−πx2 dx. R R (b) Rescale to see that

Z 2 1 Z 2 e−πx dx = √ e−x dx. R π R

Z 2 Z e−x2 dx = e−x2−y2 dxdy

Z 2π Z ∞ = e−r2 rdrdϕ changing to polar coordinates 0 0 " 2 #∞ e−r = 2π −2 0 = π.

1/2 Let ft(x) = f(xt ) for t ∈ R>0. Note that

1 1 fˆ (y) = fˆ(y/t) = f(y/t) t t1/2 t1/2

We want to show that

1 z 1/2 θ − = eπi/y · (−z)1/2θ(z) = θ(z). z i MATH 679: AUTOMORPHIC FORMS 5

z Since i is in the right half plane where the square root is analytic, both sides are analytic functions, so it is enough to take z = it, t ∈ R>0. Then X θ(it) = ft(n) n∈Z X ˆ = ft(n) Poisson summation n∈Z 1 X = f (n/t) t1/2 t n∈Z 1 i = θ . t1/2 t Recall that the Riemann zeta function is deﬁned as ∞ X 1 ζ(s) = for Re(s) > 1. ns n=1 We want to show

(1) ζ has an analytic continuation to C with a pole at s = 1, (2) ζ obeys a functional equation.

Note that ζ is incomplete. We have the Euler product: Y ζ(s) = (1 − p−s)−1. p In hindsight, we are missing an Euler factor for p = ∞. Deﬁne Λ(s) = π−s/2Γ(s/2)ζ(s), where Z ∞ dt Γ(s) = tse−t 0 t is the Γ-function, deﬁned for Re(s) > 0.

Remark 1.9. This is a “Fourier transform” on (R×, ·) in place of (R, +). Indeed, normally we integrate the function against the character x 7→ e2πixy of (R, +) with respect to the addition-invariant measure dt. Here, we integrate against the character t 7→ ts of (R×, ·) dt with respect to the multiplication-invariant measure t . This is called a Mellin transform.

We have X 2 θ(z) = eπin z n∈Z ∞ X 2 = 1 + 2 eπin z . n=1 | {z } w(z) 6 TASHO KALETHA

Let us apply the Mellin transform to t 7→ w(it): Z ∞ Z ∞ ∞ dt X 2 dt w(it)ts = e−πn tts t t 0 0 n=1 ∞ Z ∞ X 2 dt = e−πn tts t n=1 0 ∞ s X Z ∞ t dt = e−t πn2 t n=1 0 ∞ X 1 Z ∞ dt = e−tts (πn2)s t n=1 0 = π−sζ(2s)Γ(s) = Λ(2s).

From the automorphy of θ, we get 1 1 1 w − = θ − − 1 z 2 z 1 z 1/2 = θ(z) − 1 2 i 1 1 = (−iz)1/2w(z) + (−iz)1/2 − . 2 2 Therefore, i 1 1 w = t1/2w(it) + t1/2 − . t z 2 Using this, we get that Z ∞ dt Λ(2s) = w(it)ts 0 t Z ∞ dt Z 1 dt = w(it)ts + w(it)ts 1 t 0 t Z ∞ dt Z ∞ i dt = w(it)ts + w t−s 1 t 1 t t Z ∞ Z ∞ s −s+ 1 dt 1 −s+1/2 −s dt = w(it)(t + t 2 ) + (t − t ) . 1 t 2 1 t We consider these two summands separately. The second summand is 1 Z ∞ 1 t1/2−s ∞ 1 t−s ∞ 1 1 1 t−1/2−s − t−1−sdt = − = + = . 2 2 1 2 −s 1 − 2s 2s 2s(1 − 2s) 1 2 − s 1 1 Let us now consider the ﬁrst summand. We claim that |w(it)| decays exponentially as t → ∞. Recall that writing ∞ X n θ(z) = anq , n=0 MATH 679: AUTOMORPHIC FORMS 7 we get that ∞ ∞ 1 X 1 X w(z) = a qn = q a qn . 2 n 2 n+1 n=1 n=0 | {z } bounded around 0 Therefore, the ﬁrst integral converges for any s, and gives an analytic function f(s), invariant 1 under s 7→ 2 − s Altogether, we conclude that

1 1 (1) Λ(2s) = 1−2s + 2s + f(s) |{z} analytic (2) Λ(1 − s) = Λ(s). Morals. (1) The automorphy of θ implies the analytic continuation and functional equation of ζ. This generalizes. (2) Fourier transforms and Poisson summation are important for automorphy, hence for L-functions. The theory of L-functions for automorphic forms generalizes to connected reductive groups. In this course, we will not emphasize the group theory, but the methods in appropriate examples:

(1) G = GL1 (Tate’s thesis); this will be a warm up, (2) G = GL2 (Jacquet–Langlands theory); this is the true goal of the course — this is the simplest non-abelian reductive group. We will study representations of the Lie group GL2(R), the p-adic Lie group GL2(Qp), and ﬁnally we will study representations of the group GL2(A). The argument we gave proving the analytic continuation and functional equation for ζ using the automorphy of θ was generalized by Hecke.

Theorem 1.10 (Hecke). Let a0, a1, a2,... be a sequence of complex numbers such that an = O(nc) for some c > 0. Fix a positive even integer k. Consider ∞ X n 2πiz f(z) = anq , q = e , n=0 ∞ X an φ(s) = , ns n=1 Φ(s) = (2π)−sΓ(s)φ(s). Then the following are equivalent:

k/2 a0 (−1) a0 (1)Φ( s) + s + k−s is entire, bounded in vertical strips, and satisﬁes the functional equation Φ(s) = (−1)k/2Φ(k − s), (2) f is a modular form of weight k and level 1 (i.e. for the full modular group). 8 TASHO KALETHA

The proof of this theorem is entirely analogous to the argument above, so we do not reproduce it here. Remark 1.11. The implication “(2) implies (1)” generalizes fairly directly to higher level, but one has to pay attention to the normalizing factors.

The implication “(1) implies (2)” also generalizes, but since Γ(N), or Γ0(N), has many more generators, it is not enough to consider L(f, s) (i.e. φ(s)). We need all χ-twists for Dirichlet characters χ. This leads to Weil’s converse theorem.

1.2. Translation to automorphic forms. Today, we will set up the language for the rest of the course. In other words, we will translate the notion of modular forms to a form which we will deal with for the rest of the class.

Fact 1.12. The action of SL2(R) on H is transitive and

Stab(i, SL2(R)) = SO(2).

1 −x y−1/2 0 Proof. If z = x+iy, then z = iy and iy = i. This proves transitivity 0 1 0 −y1/2 and calculating the stabilizer is immediate. Corollary 1.13. We have the (canonical) isomorphism

SL2(R)/ SO(2) → H, g 7→ g · i.

In particular, from a modular form f, we obtain a function

SL2(R) → C g 7→ f(gi) which is invariant on the right by SO(2). a b Recall that we deﬁned ν(g, z) = (cz + d) if g = . Deﬁne c d

−k φf (g) = f(g · i)ν(g, i) , where k is the weight of f.

Now, φf : SL2(R) → C has the following properties: (1) it is invariant under Γ on the left, (2) under the action of SO(2) on the right, we have the following: writing e(ϕ) = cos ϕ sin ϕ , − sin ϕ cos ϕ φ(ge(ϕ)) = eikϕφ(g), i.e. φ transforms by a character under the action of SO(2). (3) growth condition, (4) φ satisﬁes a diﬀerential equation. MATH 679: AUTOMORPHIC FORMS 9

We will only discuss conditions (3) and (4) later, when we have the structure theory of SL2(R). Since we will ﬁrst discuss the theory for GL(1), we will not need this until we reach the theory for GL(2).

Remark 1.14. This is essentially the deﬁnition of and automorphic form on SL2(R). In fact, we will slightly weaken condition (2) to allow for other objects such a Maass forms (which are not holomorphic).

We now translate things to the adelic setting. We brieﬂy recall what adeles are. Consider the ﬁeld Q of rational numbers. • We have the archimedean absolute value

| · |∞ : Q → Q≥0 and the completion Q is R, a locally compact, connected, topological ﬁeld. • For each prime number p, we have the p-adic absolute value

| · |p : Q → Q≥0 −r r a given by |x|p = p where x = p b for a, b ∈ Z such that p does not divide a or b. We write r = ordp(x). Completing Q with respect to | · |p gives Qp, a locally compact, totally d isconnected, topological field. It has a compact open subring Zp ⊆ Qp. The adeles are defined as fin A = R × A where the finite adeles are ( ) 0 fin Y Y A = Qp = (xp) ∈ Qp | | xp ∈ Zp for almost all p . p p In other words: Y Y fin = lim × A −→ Qp Zp S p∈S p6∈S where S runs over finite sets of places. Then is a locally compact topological ring. The topology is the lim topology of the product A −→ topology. Explicitly, it is generated by sets of the form Y Up p≥∞ where Up ⊆ Qp is open and Up = Zp for almost all p. Here, we are using the notation Q∞ = R. Remark 1.15. The diagonal embedding Q ,→ A is discrete. Fact 1.16. We have that Y0 SL2(A) = SL2(Qp) p≤∞ where the prime means restricted direct product again, and the restriction is with respect to SL2(Zp) 10 TASHO KALETHA

Proposition 1.17.

(1) The embedding SL2(Q) ,→ SL2(A) has discrete image. (2) We have that Y ∼ SL2(Q) SL2(A) SL2(Zp) = SL2(Z) SL2(R), p<∞ i.e. we can thing of automorphic forms as functions on the left hand side. Q rp (3) Let N = p p . Deﬁne r Kp,r = ker(SL2(Zp) → SL2(Z/p Z)). Then Y ∼ SL2(Q) SL2(A) Kp,rp = Γ(N) SL2(R), p<∞ Remark 1.18. For a general simply connected algebraic group, the analogous theorem is called the strong approximation theorem. This is harder to prove.

Remark 1.19. We will write G for SL2.

Proof of Proposition 1.17. For (1), note that Q ,→ A is discrete so Qn ,→ An is discrete, and hence X(Q) ,→ X(A) is discrete for any aﬃne variety X.

For (2), ﬁx a prime p. Given g ∈ SL2(Qp), there exists γp ∈ SL2(Q) such that

(1) γ · g ∈ SL2(Zp), (2) γ ∈ SL2(Z`) for all ` 6= p. This is left as an exercise.

Given g = (gp) ∈ G(A), there is a ﬁnite set S of primes such that gp ∈ G(Zp) for p 6∈ S. Choose γp as above for each p ∈ S. Then Y Y γpg ∈ G(R) G(Zp). p∈S p<∞

Thus G(R) meets each coset in Y ∼ SL2(Q) SL2(A) SL2(Zp) = SL2(Z) SL2(R). p<∞

Two elements of G(R) represent the same coset if and only if their diﬀerent lies in Y G(Q) ∪ G(Zp) = G(Z), p<∞ since Q ∩ Zˆ = Z. Part (3) is similar and hence left as an exercise. MATH 679: AUTOMORPHIC FORMS 11

Remark 1.20. The group Kp,r is compact and open in G(Qp). For r = 0, it is maximal such. Y Hence K = K∞ × Kp ⊆ G(A) is maximal compact, where K∞ = SO(2) and Kp = Kp,0. p<∞ Y Moreover, KN = K∞ × Kp,rp ⊆ G(A) is compact and has ﬁnite index in K. p<∞

Corollary 1.21. Extending φ to a function φ: SL2(A) → C via Proposition 1.17 gives a function with the following properties:

(1) left-invariant under SL2(Q), (2) K-finite on the right (i.e. the span of the orbit of φ under K is finite-dimensional), (3) growth condition, (4) differential equation.

Proof. To prove (2), note that K∞ acts on φ by a character. If f is full level, then the span of Kf is 1-dimensional. If f has higher level, the span will have higher dimension because φ Q is only invariant under a ﬁnite index subgroup of Kp. p<∞

Deﬁnition 1.22. An automorphic form on SL2(A) is the automorphic form is a function SL( A) → C satisfying conditions (1)–(4) of Corollary 1.21.

Remark 1.23. We may switch from SL2 to GL2 using the identiﬁcation Y Y SL2(Q) SL2(A) SL2(Zp) = Z(A) GL2(Q) GL2(A) GL2(Zp), p<∞ p<∞ a 0 where Z = Z(GL ) = . For modular forms, the automorphy factor for GL(2) is 2 0 a

ν(g, z) = (cz + d) · det(g)1/2.

For now on, G = GL2.

1.3. Hecke operators. Recall that for any prime p, we have the operator X az + b T (p)f(z) = pk−1 d−kf . d a≥1, ad=p 0≤b

These play an important role:

• they are self-adjoint with respect to the Petersson scalar product for (p, N) = 1, so we can expect eigenfunctions for them, ∞ X n • if f = anq is a cusp form, normalized such that a1 = 1, which is an eigenform, n=1 then T (p)f = apf. 12 TASHO KALETHA

Thus, the Fourier coefficients may be thought of as eigenvalues of the Hecke operators. In the adelic language, the Hecke operators has a nicer description. Definition 1.24. For a prime p, define for φ: G(A) → C, g ∈ G(A), h ∈ G(Qp) ,→ G(A) 3 hp = (1, 1,..., 1, h, 1,...), the Hecke operator as the integral: Z Te(p)φ(g) = φ(g · hp)dh. p 0 K K p0 1 p

Remark 1.25. Here, dh is the Haar measure on G(Qp), normalized so that the compact open subgroup Kp has measure 1. We will discuss this later in the class. Fact 1.26. For (p, N) = 1, we have that k/2−1 p · Te(p)φf = φT (p)f .

Here, f is a modular form of weight k for the Hecke group Γ0(N).

Proof. First, express the double coset as a union of single cosets:

ξt ξp p−1 z }| { z }| { p 0 [ p −t 1 0 K K = K ∪ K . p 0 1 p 0 1 p 0 p p t=0 Checking this is left as an exercise. Then p X Te(p)φf (g) = φf (g · ξt,p), t=0

because φf is right Kp-invariant. Write

g = γ · g∞ · k + for γ ∈ G(Q), g∞ ∈ G(R) , k ∈ K. Consider γg∞k · ξt,p. Note that p 0 k · ξ ∈ K K p t,p p 0 1 p 0 0 so there exist t and kp such that 0 kp · ξt,p = ξt0,p · kp. 0 0 0 We avoid computing t based on t by the observation: if t1 6= t2, then t1 6= t2. Therefore, we get p X 0 Te(p)φf (g) = φf (γ · g∞ · ξt,p · k ) t=0 and we may write 0 −1 0 Y −1 γ · g∞ · ξt,p · k = (γξt) · (ξt,∞ · g∞) · kp ξt,` k` |{z} | {z } `6=p ∈G(Q) + ∈G∞ | {z } ∈Kﬁn MATH 679: AUTOMORPHIC FORMS 13

and hence p X −1 −1 −k Te(p)φf (g) = f(ξt,∞g∞ · i) · ν(ξt,∞g∞, i) . t=0 | {z } | {z } −1 −1 −k −k f(ξt,∞·z) ν(ξt,∞,z) ν(g∞,i) For 0 ≤ t < p, we get z + t f(ξ−1 z) = f t,∞ p −1 −1/2 ν(ξt,∞, z) = p . For t = p, we get −1 f(ξt,∞z) = f (pz) −1 1/2 ν(ξt,∞, z) = p . Putting all of this together gives the result. Fact 1.27. If f is cuspidal, then 2 φf ∈ L (Z(A)G(Q)\G(A)). The right hand side has a unitary action of G(A). If f is a cuspidal, new eigenform, then the subrepresentation it spans is an irreducible subrepresentation. Conversely, one can recover from any irreducible subrepresentation of G(A) in this Hilbert space such a vector φ.

Therefore, this reduces the question of studying automorphic forms to studying the irreducible constituents of this representation. Theorem 1.28. If π is an irreducible representation of G(A), then O0 π = πp

where πp is an irreducible representation of G(Qp).

Therefore, we need to study irreducible representations of the groups G(Qp).

2. Overview of harmonic analysis on LCA groups

We will soon develop the theory for GL1, following Tate’s thesis, but we start by discussing harmonic analysis on LCA groups. The reference for this is Hewitt–Ross.

2.1. LCA groups. Deﬁnition 2.1. An LCA group is an abelian topological group G, whose topology is locally compact (i.e. Hausdorﬀ and every points has a compact neighborhood).

× × × Example 2.2. The groups (R, +), (R , ·), (Qp, +), (Qp , ·), (A, +), (A , ·) are all LCA. Deﬁnition 2.3. We write LCA for the category of locally compact abelian groups with continuous group homomorphism.

Recall that a morphism f in any category is 14 TASHO KALETHA

• monic if and only if f · g1 = g ◦ g2 implies g1 = g2, • epic if and only if g1 ◦ f = g2 ◦ f implies g1 = g2. In LCA:

• monic is equivalent to injective, • epic is equivalent to having dense image.

Example 2.4. Let r ∈ R be irrational. Consider 2 2 R → R /Z t 7→ (t, rt) This is monic and epic, but it is not an isomorphism.

Therefore, LCA is not an abelian category. It is only additive. We can make it into an exact category, i.e. we will declare which monics and epics are admissible. Lemma 2.5. Let G be LCA and H ⊆ G be a closed subgroup. Then H is LCA and G/H is also LCA. Conversely, if H ⊆ G is a topological subgroup, which is LCA, then H is closed.

Recall that a continuous map f : X → Y of topological space is an immersion if it is injective and a homeomorphism onto its image. Moreover, f is an open/closed immersion if f is an immersion and it is open/closed; equivalently, if f is an immersion and f(X) is open/closed. A map f : X → Y is a quotient map if f is surjective and open. Deﬁnition 2.6. A admissible monic is a closed immersion. An admissible epic is a quotient.

f An sequence 1 A B C 1 is exact in LCA if

• it is exact in Ab, • f is an admissible monic, • g is an admissible epic. Theorem 2.7 (Open mapping). Let f : G → H be a surjective morphism of LCA. Assume G is a countable union of compact sets. Then f is open.

2.2. Pontryagin duality. Let G be LCA.

Definition 2.8. A character of G is an LCA morphism χ: G → C×. We say that χ is unitary if χ(G) ⊆ S1. The dual group is Gˆ = {χ: G → S1} with compact open topology U(K,V ) = {χ | χ(K) ⊆ V } where K is a compact in G and V is an open in S1. Proposition 2.9. The group Gˆ is LCA. Definition 2.10. If f : G → H is a morphism, define fˆ: Hˆ → Gˆ by fˆ(χ) = χ ◦ f. MATH 679: AUTOMORPHIC FORMS 15

Remark 2.11. If V is a ﬁnite-dimensional vector space, then V ∼= V ∗ non-canonically, but there is a canonical isomorphism ev: V → V ∗∗ such that hev(v), ξi = hv, ξi. Theorem 2.12 (Pontryagin duality). The map G → Gˆ is an exact self-equivalence of LCA. ˆ ˆ Moreover, ev: G → G is an isomorphism from the functor idLCA to the functor G 7→ G. In particular, if f : H → G is a closed immersion, then fˆ: Gˆ → Hˆ is surjective, so every unitary character of H extends to a unitary character of G. Remark 2.13. An LCA group G is compact if and only if its dual Gˆ is discrete. Remark 2.14. An isomorphism G → Gˆ is the same as a non-degenerate bi-character h·, ·i: G × G → S1. Examples 2.15.

• The dual of Z is S1 by definition. • The dual of S1 is Z by duality. • For G = R, the bi-character (x, y) 7→ eixy is non-degenerate, so the dual of R is isomorphic to R. I.e. every unitary character of R is of the form x 7→ eixy for y ∈ R. Moreover, every character R → C× is of the form x 7→ esx for s ∈ C. s • Every character of R>0 is of the form x 7→ x for s ∈ C. It is unitary if and only if s ∈ iR. • Every character of R× is of the form x 7→ |x|s · sgn(x)m for s ∈ C and m ∈ Z/2Z. • Every character of C× is of the form z 7→ zazb for a, b ∈ C, a − b ∈ Z. Exercise. Prove this without using Potryagin duality. Definition 2.16. Given a closed subgroup H ⊆ G, define H⊥ = {χ ∈ Gˆ | hh, xi = 1 for all h ∈ H}. Given a closed subgroup H0 ⊆ Gˆ, define ⊥H0 = {g ∈ G | hg, χi for all x ∈ H0}. ˆ Proposition 2.17. The isomorphism G → Gˆ identifies H with ⊥(H⊥). Proposition 2.18. If H ⊆ G is a closed subgroup, then

G/H[ ∼= H⊥ canonically. In other words, the short exact sequences

1 H G G/H 1

1 H⊥ Gˆ G/Hˆ ⊥ 1

are dual to each other. 16 TASHO KALETHA

2.3. Haar measure. Let G be a locally compact group. Deﬁnition 2.19. A Haar measure on G is a Borel measure µ such that it is

• left-invariant, • countably additive, • ﬁnite on compact sets, • inner regular: if S ⊆ G is measurable, then µ(S) = lim{µ(K) | K ⊆ S compact}, • outer regular: if S ⊆ G is measurable, then µ(S) = lim{µ(O) | S ⊆ O open}.

Theorem 2.20 (Haar). A Haar measure exists and is unique up to R>0. Remark 2.21. In general, no canonical normalization. In some special cases, there are standard normalizations:

• if G is compact, we can take µ(G) = 1, • if G is discrete, we take the counting measure µ({1}) = 1.

In practice, there will always be a natural way to choose the normalization. Proposition 2.22. Let 1 → A → B → C → 1 be an exact sequence and da, db be Haar db measures on A and B. There is a unique Haar measure dc = da on C such that Z Z Z f(b)fb = f(ac)dadc. B C A

2.4. Fourier transforms. Let G be LCA and dg be a Haar measure on G. We have the spaces Lp(G), independent of the choice of dg.

1 ˆ ˆ Definition 2.23. For f ∈ L (G), define the Fourier transform fdg : G → C by Z ˆ fdg(ξ) = f(g)hf, ξidg. G We will often omit the dg from the notation. Proposition 2.24 (Riemann-Lebesgue). The function fˆ is continuous and vanishes towards infinity. Theorem 2.25 (Inversion theorem). There exists a unique Haar measure dξ on Gˆ such that if f ∈ L1(G) and fˆ ∈ L1(Gˆ), then ˆ ∧ (fdg)dξ(x) = f(−x). Definition 2.26. The Haar measure dξ is the dual measure to dg. Exercise. If G is compact and vol(G, dg) = 1, then the dual measure dξ on Gˆ is the counting measure. MATH 679: AUTOMORPHIC FORMS 17

Theorem 2.27 (Plancharet theorem). If f ∈ L2(G) ∩ L1(G) and fˆ ∈ L2(Gˆ), and Z Z |f(g)|2dg = |fˆ(ξ)|2dξ. G Gˆ The Fourier transform extends to an isometric isomorphism L2(G) → L2(Gˆ). Theorem 2.28 (Poisson summation). Let 1 → A → B → C → 1 be an exact sequence in db ˆ LCA. Fix Haar measures da, db on A and B. Let dc = da , db be dual to db and daˆ be dual to da. Let f ∈ L1(B) be such that

ˆ 1 ⊥ (1) f|A⊥ ∈ L (A ), (2) a 7→ f(a + b) is in L1(A) for any b ∈ B, R (3) b 7→ A f(a + b)da is continuous. Then Z Z f(a)da = fˆ(ˆc)dc,ˆ A Cˆ where Cˆ = A⊥ and the measure dcˆ is the dual of dc (or, equivalently, the unique measure dˆb such that daˆ = dcˆ).

Proof. Define Z F (c) = f(a + b)da A if b 7→ c. Then, by Fourier inversion, Z Z f(a)da = F (0) = Fˆ(ˆc)dc.ˆ A Cˆ Now, Z Fˆ(ˆc) = F (c)hc, cîdc C Z Z = f(a + c)hc, cîdadc C A Z Z = f(a + c)ha + c, cîdadc C A Z = f(b)hb, cîdb B = fˆ(ˆc). This completes the proof.

2.5. Self-duality. Let B be LCA, j : B → Bˆ be an isomorphism, given by the non- degenerate bi-character h·, ·i: B × B → S1. We have ˆ ˆj : Bˆ ∼= B → B.ˆ Exercise. Show that 18 TASHO KALETHA

(1) ˆj = j, (2) ˆj = j if and only if h·, ·i is symmetric.

For any f ∈ L1(G), we get a function ˆ ˆ fj = f ◦ j : G → C. Again, we will omit the j from the notation occasionally. Explicitly, Z ˆ 0 0 0 fj(b) = f(b )hb , bidb . B ˆ Deﬁnition 2.29. The measure db is called self-dual if j∗(db) is the measure on B dual to db.

Note that this deﬁnition depends on a choice of bi-character. Exercise. A self-dual measure exists an is unique. Exercise. Assume ˆj = j and db is self-dual. Then ˆ ∧ (fj)j (b) = f(−b). Lemma 2.30. Let A ⊆ B be closed. Assume that j identiﬁes A with A⊥. For any Haar ⊥ ˆ db measure da on A, j∗(da) is the Haar measure on A = C dual to dc = da , where db is self-dual.

Proof. Both j∗(da) and dcˆ are Haar measures, so j∗(da) = kdcˆ for some k ∈ R>0. By Poisson summation 2.28, Z Z Z ˆ ˆ f(a)da = f(ˆc)dcˆ = k fj(a)da. A Cˆ A Applying this equation twice, we get that Z Z Z Z 2 ˆ ∧ 2 2 f(a)da = k (fj)j (a)da = k f(−a)da = k f(a)da. A a a A 2 Hence k = 1, so k = 1. Corollary 2.31 (Possion summation). For any suﬃciently nice function f : B → C (see Theorem 2.28), we have Z Z f(a)da = fˆ(a)da, A A where fˆ: B → C is formed with respect to the self-dual measure on B. Fact 2.32. Let G be a compact group and χ: G → C× is non-trivial. Then Z χ(g)dg = 0. A

This is analogous to the statement for ﬁnite groups that: X χ(g) = 0. g∈G The proof is also analogous. MATH 679: AUTOMORPHIC FORMS 19

3. Harmonic analysis on local fields and adeles

Recall that = lim /pk , = ⊗ = p−1 . Zp ←− Z Z Qp Zp Z Q Zp k Then x ∈ Zp can be written as ∞ X k x = akp k=0 and x ∈ Qp can be written as X k x = anp k>−N

for ak ∈ {0, . . . , p − 1}. Fact 3.1.

1 1 (1) We have that Qp = Zp + p∞ Z and p∞ Z ∩ Zp = Z. (2) The map

Qp/Zp → Q/Z x = y + z 7→ z

is an injective group homomorphism, inverse to Q/Z → Qp/Zp induced by Q ,→ Qp. These identify Qp/Zp with the p-power torsion gorup of Q/Z. 1 Deﬁnition 3.2. Deﬁne the character ψp : Qp/Zp → S as the composition of Qp/Zp ,→ Q/Z with 1 Q/Z → S x 7→ e2πix.

1 k Fact 3.3. Let ψ : Qp → S be any character. There exists k such that ψ(p Z) = 1.

Proof. Take an open neighborhood U ⊆ S1 not contained in a subgroup of S1. Then ψ−1(U) −1 is an open subset of Qp, containing the subgroup ψ {1}. This open subset contains an open k subgroup p Zp for large enough k, whose image in contained in U. Therefore, the image of k p Zp is {1}. Deﬁnition 3.4. The level of ψ is the smallest k such that ψ(pkZ) = 1. Lemma 3.5. The bi-character 1 Qp × Qp → S

(x, y) 7→ ψp(x · y) ˆ ⊥ is non-degenerate. Under the isomorphism Qp → Qp, we have that Zp = Zp.

ˆ Proof. Injectivity and continuity of Qp → Qp are clear. Once surjectivity is proved, openness S k will follow from the open mapping theorem (Qp = (1/p )Zp). 20 TASHO KALETHA

1 We show surjectivity. Given ψ : Qp → S , want y ∈ Qp such that ψ(x) = ψp(xy). Let k be k−1 the level of ψ. Then on p Zp, we have that −k ψ(x) = ψp(u0p x) × for some u1 ∈ Zp , well-defined up to 1 + pZp. k−1 × Now, look at p Zp. Get u2 ∈ Zp such that u2 ≡ u1 mod 1 + pZp and −k ψ(x) = ψp(u2p x) k−2 × n for all x ∈ p Zp. Inductively, get a sequence (un), un ∈ Zp , un ≡ un−1 mod 1 + p Zp. × −k This sequence is Cauchy, so it converges to u ∈ Zp such that ψ(x) = ψp(p ux) for all x ∈ Qp. Recall that 0 fin Y A = Qp p<∞ contains ˆ Y Z = Zp. p<∞ Moreover, A = R × Afin. The topology is generated by Y Y Up × Zp p∈S p6∈S

for finite sets S of places and Up ⊆ Qp open. Remark 3.6. The sequence x + n = (1, 1, 1,..., 1, 0, 0,...) with n ones coverges to 1 ∈ A. Fact 3.7. The map Q → A has discrete image and A/Q is compact. Q Proof. Consider the following open set U = (−0.5, 0.5) × Zp. Take q ∈ Q ∩ U. Then p qp ∈ Zp for all < ∞ implies that q ∈ Z. Then q∞ ∈ (−0.5, 0.5) implies that q = 0. This proves discreteness. Q Now, consider the compact set [−0.5, 0.5] × Zp. Then p Q + W = A. Thus restriction of A → A/Q to W is surjective and hence A/Q is compact. Remark 3.8. The embedding Q ,→ A is a closed immersion, if Q is given the discrete topology. From now on, we will always give Q the discrete topology. Corollary 3.9. The embedding Q ,→ Afin induces an isomorphism =∼ fin Q/Z → A /Zˆ, which translates the p-primary decomposition of Q/Z into the canonical decomposition fin ˆ Y A /Z = Qp/Zp. p<∞

This gives MATH 679: AUTOMORPHIC FORMS 21

ﬁn ˆ 1 • a character ψﬁn : A /Z → S via the diagram

ψﬁn Aﬁn/Zˆ S1 e2πix

Q/Z

• ψﬁn|Qp = ψ deﬁned before.

Deﬁne

1 ψ∞ : R/Z → S x 7→ e−2πix.

Note the minus sign.

1 Then deﬁne ψA = (ψ∞, ψﬁn): A → S . Lemma 3.10.

(1) The character ψA is trivial on Q. (2) The bi-character

1 A × A → S

(x, y) 7→ ψA(x, y) is non-degenerate. (3) Under this character, Q⊥ = Q.

Proof. Part (1) is immediate from construction and the sign convention. As in the local case, it is enough to prove surjectivity: given ψ : A → S1, there is a y ∈ A such that ψ(x) = ψA(xy).

For each p, let yp ∈ Qp be such that ψ| p (xp) = ψp(xpyp). Note that ψ is trivial on a Q Q subgroup of the form Zp. Therefore, yp ∈ Zp for all p 6∈ S. Therefore, y ∈ A and this p6∈S completes the proof of (2). Consider Q⊥ ⊆ A.

• Since A/Q is compact, Q⊥ is discrete. • By (1), Q ⊆ Q⊥. Therefore, Q⊥/Q is a discrete subgroup of the compact A/Q, hence ﬁnite. • By deﬁnition, Q⊥ is invariant under multiplication by Q, and hence it is a Q-vector space.

⊥ Hence Q /Q = {1}, proving (3). Exercise. We have the following self-dual exact sequences 22 TASHO KALETHA

0 Zp Qp Qp/Zp 0,

0 Z R R/Z 0,

0 Q A A/Q 0.

Additive Haar measures.

Fact 3.11. The Lebesgue measure on R is Haar and self-dual with respect to ψ∞.

Fact 3.12. Let dxp be the Haar measure on Qp, self dual with respect to ψp. Then 1 ˆ (1) if f = Zp , then f = f, (2) vol(Zp, dxp) = 1.

1 Proof. We compute for f = Zp and any Haar measure dy: Z ˆ 1 f(x) = Zp (y)ψp(xy)dy

Qp Z = ψp(xy)dy

Zp ( vol( , dx) for x ∈ = Zp Zp 0 for x 6∈ Zp In the second case, we note that Z Z −1 ψp(xy)dy = ψp(y)d(x y) = 0

Zp x·Zp Thus: ˆ f = vol(Zp, dx) · f and ˆ 2 f = vol(Zp, dx) · f. ˆ If dx is self dual, then f = f, proving both statements. −πx2 ˆ 1 Remark 3.13. Recall that for the Gaussian f = e on R satisﬁes f = f. Therefore, Zp is a Qp-analog of the Gaussian. Remark 3.14. Note that we can by-pass Haar’s existence theorem. Indeed, if we can reverse −k k engineer the measure from what we know: if vol(Zp, dx) = 1, vol(p Zp, dx) = p . Now, −k any open subset of Qp may be written as a union of translates of p Zp, and we may have deﬁned the measure this way.

× Exercise. Suppose p ≤ ∞. For any a ∈ Qp , d(ax) = |a|pdx. MATH 679: AUTOMORPHIC FORMS 23

Fact 3.15. Let dxp be a Haar measure on Qp. Assume vol(Zp, dxp) = 1 for almost all p. Then O dx = dxp is a measure on A. If each dxp is ψp-self-dual, then dx is ψA-self-dual. Corollary 3.16. We have that vol(A/Q, da/dq) = 1 where da is self-dual and dq is the counting measure.

Proof. Let δ0 : Q → C be the delta function. Then Z ˆ δ0(ˆq) = δ0(q)hq, qˆidq = 1. Q By Fourier inversion, this shows that Z ˆ ˆ vol(Q) = δ0(ˆq)dqˆ = δ0(0) = 1. Qˆ

Consider the short exact sequence 0 Q A A/Q 0. Then Lemma 2.30 says that dqˆ is the quotient of da by j∗(dq), which is the counting measure. This completes the proof.

Recall that there is an adelic norm | · |A : A → R.

Fact 3.17. We have that d(ax) = |a|Adx.

Schwartz functions. Deﬁnition 3.18.

(1) The space S(R) of Schwartz functions consists of smooth functions f : R → C such that p (q) x · f (x) → 0 as x → ±∞. ∞ (2) For p < ∞, we set S(Qp) = Cc (Qp), the space of locally constant compactly supported functions Qp → C (Schwartz–Bruhat functions). O0 N (3) S(A) = S(Qp), i.e. the vector space generated by f = fp for fp ∈ S(Qp) such p≤∞ p that fp = 1 for almost all p. We call these generators factorizable. Zp Q Explicitly, f(a) = fp(ap) is well-deﬁned, since almost all factors are 1. p

Fact 3.19. For p ≤ ∞, Fourier transforms gives a bijection S(Qp) → S(Qp).

1ˆ 1 ∞ Proof. For p < ∞, use Zp = Zp and the fact that f ∈ Cc (Qp) is a ﬁnite linear combination 1 of additive and multiplicative transforms of Zp .

0 ˆ ˆ 0 For p = ∞, use fb (x) = xf(x) and xf\(x) = (f) (x). N ˆ N ˆ Fact 3.20. If f = fp ∈ S(A) then f = fp. p p Corollary 3.21. Fourier transform is a bijection S(A) → S(A). 24 TASHO KALETHA

We can apply the Poisson summation formula 2.28 to the case of adeles. Corollary 3.22 (Adelic Poisson summation). For f ∈ S(A), we have that X X f(q) = fˆ(q). q∈Q q∈Q Corollary 3.23. For f ∈ S(A) and a ∈ A×, X X f(aq) = |a|−1 fˆ(a−1q). A q∈Q q∈Q

Proof. This is immediate: if f a(x) = f(ax), then fˆa(x) = |a|−1fˆ(a−1x). A ˆ Remark 3.24. Note that f is formed with respect to the ﬁxed character ψA. Any other 1 character ψ : A/Q → S is of the form ψ(x) = ψA(a · x) for some a ∈ Q.

So the ψA-self-dual measure is also the ψ-self-dual measure (by the product formula). Also, the Poisson summation formula still holds.

Multiplicative Haar measures. −1 × Fact 3.25. Let dx be a Haar measure on Qp. Then |x|p dx is a Haar measure on Qp . (Here, p ≤ ∞.)

× Proof. Note that the measure dx on Qp restricted to Qp is still a measure. However, it is −1 not invariant. The factor of |x|p makes it invariant.

The case of A× is more subtle. This is because the topology of A× is not the subspace topology:

• A× is not closed in A, • inversion is not continuous. This is what we do instead. We recognize that: 0 × × Y × × A = R [Qp , Zp ] p<∞ × × (the restricted direct product of Qp with respect to Zp ). Therefore, we take the topology to be generated by Y Y × × Up × Zp for Up ⊆ Qp open, and S ﬁnite. p∈S p∈S × × × × Fact 3.26. If d xp is a Haar measure on Qp such that vol(Zp , d xp) = 1 for almost all p, then × O × d x = d xp is a Haar measure on A×. × −1 Fact 3.27. If d xp = |x|p dxp for dxp the ψp-self-dual measure, then 1 vol( ×, |x|−1dx ) = 1 − . Zp p p p MATH 679: AUTOMORPHIC FORMS 25

Remark 3.28. Note that Y O Y 1 vol([−0.5, 0.5] × ×, |x|−1dx ) = 1 − = ζ(1)−1 = 0. Zp p p p p p<∞ × p −1 Deﬁnition 3.29. Deﬁne d xp = p−1 |x|p dxp.

Multiplicative characters. × × Deﬁnition 3.30. For p < ∞, a character χ: → is called unramiﬁed if χ| × = 1. Qp C Zp

We have a short exact sequence

× × val 1 Zp Qp Z 0.

Therefore, unramified characters are characters on Z. More precisely, unramified characters s s·log |x|p χ are of the form χ(x) = |x|p = e for some s ∈ C. Note that this s is unique up to 2πi . log(p)Z × Therefore, the set of unramified characters of Qp has the structure of a Riemann surface, namely 2πi . C log(p)Z Since the group of unramified characters acts freely on the group of all characters, we get a structure of a Riemann surface on the latter.

× Recall: | · |A : A → R>0. Deﬁne 1 A = ker(| · |). By the product formula, Q× ⊆ A1. Fact 3.31. The group Q× is discrete in A1 and A1/Q× is compact.

The proof is left as an exercise. Deﬁnition 3.32. A Hecke character is a character A×/Q× → C×. Example 3.33. For any s ∈ , x 7→ |x|s is a Hecke character. C A

We again get a Riemann surfact structure on all Hecke characters (by letting this group of characters act on the set of all characters).

4. Tate’s thesis

Local and global zeta integrals. × × Deﬁnition 4.1. Let p ≤ ∞, χ: Qp → C , f ∈ S(Qp). Deﬁne Z × Zp(f, χ) = f(x)χ(x)d xp. × Qp 26 TASHO KALETHA

Definition 4.2. Let χ be a Hecke character, f ∈ S(A). Define Z Z(f, χ) = f(x)χ(x)d×x. A× Remark 4.3. We can study meromorphy on each component of the Riemann surfact of characters. Definition 4.4. Assume χ is unitary. Then define

s Zp(s, f, χ) = Zp(f, χ · | · |p), s Z(s, f, χ) = Z(f, χ · | · |A). Fact 4.5. In both the local and the global setting, Z(f, χ) is linear in f and Z(f a, χ) = χ(a)−1Z(f, χ), where f a(x) = f(ax). Lemma 4.6.

(1) The integral defining Zp(s, f, χ) converges for Re(s) > 0. (2) If there is a neighborhood U of 0 such that f|U = 0, then the integral defining Zp(s, f, χ) converges for all values of s, and defines an entire function. (3) For p < ∞ and for Re(s) > 0, ( (1 − χ(p)p−s)−1 if χ is unramified, Z (s, 1 , χ) = p Zp 0 if χ is ramified.

Proof. Split the integration domain into |x|p > 1 and |x|p ≤ 1. On |x|p > 1, the Schwartzness s of f dominates the behavior of |x|p so the integral converges absolutely to a holomorphic function.

Now, consider the domain |x|p ≤ 1. If f|U = 0 for some open set U around 0, then the × integration is over {x | |x|p ≤ 1}\ U, compact in Qp ,so the result converges absolutely to a holomorphic function. This proves (2). If there is no such U, we have |f(x)| is continuous on |x| ≤ 1, and hence is it bounded. Therefore, it is enough to study Z s × |x|pd x. |x|p≤1 For p < ∞, this integral is Z Z X s × X k s × |x|pd x = |p x|pd x k × × k≥0 p Zp k≥0 Zp X Z = p−ks · 1 · d×x . × k≥0 Zp | {z } =1 This series converges when Re(s) > 0. MATH 679: AUTOMORPHIC FORMS 27

For p = ∞, we have Z 1 s dt |t|∞ . −1 |t|∞ Consider Z 1 ts−1dt, 0 which converges for Re(s) > 0. This completes the proof of (2). For (3), as above, we compute Z 1 X s × Zp(s, Zp , χ) = χ(x)|x|pd x k × k≥0 p Zp Z X k k s × = χ(p x)|p x|pd x × k≥0 Zp X Z = χ(p)kp−ks χ(x)d×x × k≥0 Zp ! Z X k = χ(p)p−s · χ(x)d×x × k≥0 Zp Z = (1 − χ(p)p−s)−1 · χ(x)d×x for Re(s) > 0. × Zp Finally, ( Z 1 if χ is unramified, χ(x)d×x = × Zp 0 if χ is ramified. This completes the proof. Corollary 4.7. The integral defining Z(s, f, χ) converges for Re(s) > 1 and satisfies for a N factorizable function f = fp: p Y Z(s, f, χ) = Zp(s, fp, χp). p≤∞

N Proof. For convergence, we may also assume that f = fp by linearity. By deﬁnition of p × N × d x = d xp, the integral is a product of local integrals. When does this product converge? By the previous lemma, all local integrals converge for Re(s) > 0, and almost all equal −s −1 (1 − χ(p)p ) , so the convergence of the Euler product for ζ implies the claim.

Analytic continuation of Zp. Lemma 4.8. Let p < ∞.

(1) The zeta integral Zp(s, f, χ) has a meromorphic continuation to C as a rational function in p−s. 28 TASHO KALETHA

(2) If χ is unramiﬁed, Zp(f, χ) has a simple pole at χ = 1 and

Zp(f, χ)(1 − χ(p)) is entire. (3) If χ is ramiﬁed, Zp(s, f, χ) is entire.

1 ∞ × Proof. Let g = f − f(0) Zp . Then g ∈ Cc (Qp ). By Lemma 4.6, Zp(s, g, χ) is entire for 1 any χ and we have explicit formulas for Zp . When χ is ramiﬁed, we have

Zp(f, χ) = Zp(g, χ). When χ is unramiﬁed, we have −s −1 Zp(s, g, χ) = Zp(s, f, χ) − f(0)(1 − χ(p)p ) . This proves the result.

Now, consider the case p = ∞. Recall that Landau notation: for f, g : R → R, we have

f • f = Ox→c(g) if and only if g is bounded as x → c, f • f = ox→c(g) if and only if g → 0 as x → c. Fact 4.9. For f ∈ C∞(R), we have n X f (k)(0) f(x) = xk + R(x) k! k=0 with R(x) = O(xn+1).

Proof. By Taylor’s theorem: n X f (k)(0) f(x) = xk + o(xn). k! k=0 Therefore, n+1 X f (k)(0) f(x) = xk + o(xn+1) k! k=0 proves the result.

Lemma 4.10. The zeta integral Z∞(s, f, χ) has meromorphic continuation to C with (at most) simple poles at even non-positive integers when χ(x) = 1 and odd non-positive integers f (n)(0) when χ(x) = sgn(x). The residue at s = −n is n! .

Proof. As last time, we split the integral into domain |x| ≤ 1 and |x| > 1. The integral over |x| > 1 is absolutely convergent to an analytic function in s, so we just need to consider the region |x| ≤ 1.

Decompose f = fe+f0 into a sum of even and odd function, and consider fr and f0 separately. We may hence assume f is even or odd. MATH 679: AUTOMORPHIC FORMS 29

The integral is 0 unless f is odd when χ = sgn and f is even when χ = triv. Assume this is the case (by decomposing f into a sum ). We are now studying the integral: Z 1 dx f(x)xs . 0 x We use Fact 4.9 to get n X f (k)(0) Z 1 dx Z 1 dx xk+s + R(x)xs . k! 0 x 0 x k=0 | {z } | {z } = 1 absolutely convergent k+s and analytic for Re(s)>−n This completes the proof.

Invariant distributions.

∞ Definition 4.11. Let p < ∞.A distribution on Qp is any linear functional on Cc (Qp). ∞ Definition 4.12. A distribution on R is a continuous linear functional on Cc (R). ∞ Remark 4.13. Note that we did not define a topology on Cc (R) and we will avoid this for now. Instead, we will refine this definition slightly. Definition 4.14. A tempered distribution on R is a continuous linear functional on S(R), where S(R) is topologized with respect to seminorms: a (b) kϕka,b = sup |x ϕ (x)|. x∈R Fact 4.15. If V is a C-vector space, topologized with respect to a family A of semi-norms and λ: V → C is a continuous linear functional, then there exist B ⊆ A finite and C > 0 such that |λ(v)| ≤ C · max p(v). p∈B

Proof. Recall that a basis of neighborhoods of x ∈ V is given by

UB,δ = {y ∈ V | p(x − y) < δ for all p ∈ B} for B ⊆ A ﬁnite and δ > 0.

Let now λ: V → C be continuous. Choose ε > 0. There exist B, δ such that λ(UU,δ(0)) ⊆ (−ε, ). Let v ∈ V be arbitrary, deﬁne kvkB = max p(v). p∈B

× −1 × If kvkB = 0, then zv ∈ UB,δ(0) for all z ∈ C . Therefore, |λ(v)| ≤ |z| ε for all z ∈ C . This shows that λ(v) = 0.

δ Otherwise, kvk · v ∈ UB,δ(0), so B ε λ(v) ≤ kvk . δ B ε This prove the fact with C = δ . 30 TASHO KALETHA

Let D(Qp) be the space of (tempered) distributions. On this space, we have • a C-vector space structure, • for p = ∞, we have differentiation given by hλ, fi = −hλ, ∂fi, • Fourier transform given by hλ,ˆ fi = hλ, fî, × a a−1 a • an action of Qp by hλ , fi = hλ, f i where f (x) = f(ax). × × Definition 4.16. Let χ: Qp → C , The space of χ-eigendistributions is χ a D(Qp) = {λ ∈ D(Qp) | λ = χ(a)λ}. χ ˆ Fact 4.17. If λ = D(Qp) , then λ ∈ D(Qp)χ−1·|·|.

Proof. Check that f rˆ−1 (x) = |r| · fˆ(rx). Then trace through the deﬁnitions this to get the result. χ Example 4.18. The distribution Zp(χ): f 7→ Zp(f, χ) is an element of D(Qp) , provided that it is deﬁned at χ.

χ Theorem 4.19. The dimension of D(Qp) over C is 1.

∞ × Proof. Step 1. Consider Cc (Qp ) ⊆ S(Qp). This dualizes to

∞ × 0 0 D(Qp)0 D(Qp) C0 (Qp ) 0.

0 Here, the denotes the dual space. The space D(Qp)0 is deﬁned to be the kernel and its elements are called distributions supported at 0. Taking χ-eigenspaces, we get

χ χ ∞ × 0χ 0 D(Qp)0 D(Qp) C0 (Qp ) .

∞ × 0χ Step 2. We claim that dim Cc (Qp ) = 1. × × Note that for χ1, χ2 : Qp → C , we have an isomorphism

∞ × 0χ1 ∞ × 0χ2 Cc (Qp ) → Cc (Qp ) λ1 7→ λ2 −1 ∞ × 01 where λ2(f) = λ1(f · χ1 · χ2 ). It is hence enough to consider Cc (Qp ) . By the existence and uniqueness of the Haar measure, this proves the claim.

Step 3. Suppose p < ∞. We claim that D(Qp)0 = Cδ0. In particular, ( 1 χ C · δ0 χ = , D(Qp)0 = {0} χ 6= 1.

∞ To prove this claim, let λ ∈ D(Qp)0. For f ∈ Cc (Qp), we have that 1 ∞ × f − f(0) Zp ∈ Cc (Qp ), 1 1 so λ(f − f(0) Zp ) = 0, showing that λ(f) = f(0) · λ( Zp ). Then note that f(0) = δ0(f) and 1 r C = λ( Zp ) is a constant. Moreover, δ0(f ) = δ0(f), showing that C = 1. MATH 679: AUTOMORPHIC FORMS 31

Step 4. We are done for p < ∞, χ 6= 1.

χ χ Indeed, steps 1, 2, and 3 show that D(Qp) ≤ 1. But 0 6= z(χ) ∈ D(Qp) . Step 5. Consider p < ∞ and χ = 1.

χ Steps 1,2, and 3 show that dim D(Qp) ≤ 2. We need to show that Z µ = f(x)d×x × Qp 1 does not lift to D(Qp) . Observe that Z 1 × λ(f) = f(x) − f(0) Zp d x × Qp 1 lifts µ. More formally, f 7→ f − f(0) Zp is a section of ∞ ∞ × Cc (Qp) → Cc (Qp ) and we are using its dual.

Any lift of µ is of the form λ + aδ0, and will be unramiﬁed if and only if λ is invariant. Now, 1 λ( Zp ) = 0. Moreover, −1 λp(1 ) = λ(1p ) = λ(1 ). Zp Zp pZp Now, −µ(1 × ) = −1, showing that the measure is not invariant. Zp

M k Step 6. Suppose p = ∞. We claim that D(R)0 = C∂ δ0. In particular, k≥0

( k −k χ C · ∂ δ0 χ(x) = x , D(R)0 = {0} χ(x) 6= x−k.

Let λ ∈ D(R)0. By Fact 4.15, there are C > 0 and a ∈ Z≥0 such that |λ(f)| ≤ C max sup |xbf (c)(x)|. b+c≤a x∈R By Fact 4.9, a X f (k)(0) f(x) = xk + R(x) k! k=0 where R(x) = O(xa+1). Then a X λ(xk) λ(f) = f (k)(0) · + λ(R). k! k=0 (k) k Note that f (0) = ∂ δ0f by deﬁnition. We just need to show λ(R) = 0. First, k k r k d −1 h(∂ δ0) , fi = (−1) k f(r x). dx x=0 Therefore, k k k −k d −1 h∂ δ0, fi = (−1) r k f (r x). dx =0 32 TASHO KALETHA

∞ Let ξ ∈ Cc (R) be a bump function such that

ξ|(−0.5,0.5) = 1, ξ|(−1,1)c = 0. ε−1 ∞ × Then R − ξ · R ∈ Cc (R ), so λ(R) = λ(ξε−1 R) ≤ C max sup |xb∂c(ξε−1 R)(x)|. b+c≤a x∈R Using the Leibnitz rule, we can show that this is bounded by C · εa+1−c < C · ε.

−k χ Step 7. We are done unless χ(x) = x . Indeed, then D(Qp) ≤ 1, but Z(s, χ) is a non-trivial element. Step 8. Consider the case χ(x) = x−k. We need to show that Z dx f 7→ f(x)x−k R× |x| ∞ × χ on Cc (R ) does not lift to D(R) . See Remark 4.20 for a general strategy. We treat the case k = 0, the other cases are similar. We have Z ∞ dx f 7→ 2 fe(x) , 0 x where fe(x) is the even part of f(x). We have the lift λ to D(R) given by Z 1 dx Z ∞ f 7→ 2 (fe(x) − fe(0)) + 2 fe(x)dx. 0 x 1 Compute Z f Z ∞ r−1 dx dx λ (f) = (fe(x) − fe(0)) + fe(x) . 0 x r x Therefore, r −1 Z dx λr (f) − λ(f) = −f(0) . 1 x | {z } 6=0 Therefore, r−1 λ = λ + Cδ0, where C 6= 0. Remark 4.20. General strategy employed in Step 8 of the proof. Suppose we have some integral Z(f, s), defined at some s but potentially at some poles. We want to define a distribution associated to this integral. We can then take the finite part at a pole s = s0,

f.p.s=s0 Z(s, f), which is the constant term of the Laurent expansion of Z(s, f) at s = s0. Remark 4.21. We introduced Z s × Zp(s, χ, f) = f(x)χ(x)|x|pd x × Qp s for f ∈ S(Qp). This looks like a Fourier transform. First, note that χ(x)|x|p is not unitary, × but we could consider the Fourier transform of f(x)|x|p evaluated at the character χ(x). MATH 679: AUTOMORPHIC FORMS 33

× However, this is the Fourier transform on Qp , so this only makes sense if f(x) vanishes at x = 0. One could hence think about it as extending the Fourier transform around x = 0. However, s we really want to think of |x|p as part of the character, not part of the function. Therefore, it is useful to call these zeta integrals, not just an extension of Fourier transforms.

χ Canonical basis for D(Qp) . First, consider the case p < ∞. • If χ is ramified, define 0 χ·|·|s Zp (s, χ) = Zp(s, χ) ∈ D(Qp) . • If χ is unramified, define 0 p−1 Zp (s, χ)(f) = Zp(s, χ)(f − f ). Since f − f p−1 is 0 at x = 0, this function is entire in s. 0 Fact 4.22. For p < ∞ and χ unramified, Zp (s, χ) is “natural” in the sense that 0 1 Zp (s, χ)( Zp ) = 1.

Proof. We have that 1 1 1 −s 1 Zp(s, χ)( Zp ) − Zp(s, χ)( Zp ) = Zp(s, χ)( Zp ) − χ(p)p Zp(s, χ)( Zp ) 1 −s 1 = Zp(s, χ)( Zp ) − χ(p)p Zp(s, χ)( Zp ) = (1 − χ(p)p−s)−1 + χ(p)p−s(1 − χ(p)p−s)−1 by Lemma 4.6 = 1 This completes the proof.

Now, consider p = ∞. In this case, a character is unramiﬁed if it is independent of the sign 2 of x ∈ R×. Take f 0(x) = e−πx ∈ S(R) and demand that 0 0 Z∞(s, 1)(f ) = 1. Then 0 0 Z∞(s, 1/x)(xf ) = 1.

Local L-factors. × 1 Deﬁnition 4.23. Let p ≤ ∞ and χ: Qp → S . The L-factor Lp(s, χ) is deﬁned as

Zp(s, χ) 0 . Zp (s, χ)

0 In other words, Lp(s, χ) is the change of basis between the basis Zp(s, χ) and Zp (s, χ) of χ D(Qp) . Lemma 4.24.

(1) For p < ∞, χ ramiﬁed, Lp(s, χ) = 1. 1 −s −1 (2) For p < ∞, χ unramiﬁed, Lp(s, χ) = Zp(s, χ)( Zp ) = (1 − χ(p)p ) . 34 TASHO KALETHA

1 (3) For p = ∞, χ(x) = 1 or χ(x) = x , then 0 −s/2 Lp(s, χ) = Z∞(s, χ)(fχ) = π Γ(s/2) 0 0 0 0 1 where f1(x) = f , fχ = xf for χ(x) = x . s −s s (4) If p < ∞, then Lp(s, χ) is a generator for the C[p , p ]-submodule of C(p ) given by

{Zp(s, χ)(f) | f ∈ S(Qp)}. (5) For any p,

0 Zp(s, χ) Zp (s, χ) = Lp(s, χ) is entire.

Proof. Parts (1) and (2) are clear. For part (3), we compute ∞ Z 2 dx Z 2 dx e−πx |x|s = 2 e−πx xs R× |x| 0 x Z ∞ dx = e−πxxs/2 0 x Z ∞ dx = π−s/2 e−xxs/2 . 0 x

0 Parts (4) and (5) amount to proving that Zp (s, χ) is entire. For p < ∞, this is Lemma 4.6. For p = ∞, this follows from (3) and Lemma 4.10.

χ ˆ Local -factors and local functional equation. Recall that for λ ∈ D(Qp) , λ ∈ χ−1·|·| D(Qp) . 0 χ Note that Zp(s, χ),Zp (s, χ) ∈ D(Qp) . Deﬁnition 4.25. Deﬁne

ˆ0 −1 Zp (1 − s, χ ) (1) the -factor as p(s, χ, ψp) = 0 Zp (s, χ), ˆ −1 Zp(1 − s, χ ) (2) the γ-factor as γp(s, χ, ψp) = . Zp(s, χ)

ˆ0 −1 0 In other words, p(s, χ, ψp) is the change of basis between the basis Zp (1−s, χ ) and Zp (s, χ) and similarly for γp(s, χ, ψp). × Remark 4.26. Recall that ψp : Qp → C is a non-trivial additive character. It is used to ˆ deﬁne f. We had a natural such character, that we denoted by ψp. From now on, we will 0 call it ψp, and ψp will be any such. So far, we speciﬁed dx to be the Lebesgue measure when p = ∞ and normalized so that 0 vol(Zp) = 1 when p < ∞. We then showed that dx is ψp-self-dual. Form now on, we take dx = dxψ, the ψ-self-dual measure. Fact 4.27. MATH 679: AUTOMORPHIC FORMS 35

(1) The -factor p(s, χ, ψp) is entire. (2) We have −1 Lp(1 − s, χ ) γp(s, χ, ψp) = p(s, χ, ψp). Lp(s, χ) (3) Local functional equation: ˆ −1 Zp(1 − s, f, χ ) = γp(s, χ, ψp)Zp(s, f, χ), 0 ˆ −1 0 Zp (1 − s, f, χ ) = p(s, χ, ψp)Zp (s, f, χ).

Proof. This is immediate from the deﬁnitions. Lemma 4.28. We have that

t (1) p(s, χ| · | , ψp) = p(s + t, χ, ψp), 1 a s− 2 (2) p(s, χ, ψp ) = |a|p χ(a)p(s, χ, ψp).

ˆ Proof. Part (1) follows from the deﬁnition. For part (2), write fψ,dx for the Fourier transform formed with respect to the character ψ and measure dx. Then ˆ ˆ a fψa,dx = (fψ,dx) , ˆ ˆ fψ,|a|dx = |a|fψ,dx. ˆ ˆ Write fψ = fψ,dxψ . Then 1/2 dxψa = |a|p dxψ, ˆ 1/2 ˆ a fψa = |a|p (fχ) . Compute 0 −1 0 −1 1/2 a a ˆ a ˆ Zp (1 − s, χ , dxψ )(fψ ) Zp (1 − s, χ , dxψ)(|a|p fψ) 1/2 s−1 0 = 0 = |a|p χ(a)|a|p p(s, χ, ψ). Zp (s, χ, dxψa )(f) Zp (s, χ, dxψ)(f) This completes the proof of (2).

Lemma 4.29. The factor p(s, χ, ψp) can be computed explicitly as follows: (1) when p = ∞, 1 0 ∞(s, , ψ∞) = 1 1 (s, , ψ0 ) = i, ∞ x ∞ 0 (2) when p < ∞, χ is unramified, p(s, χ, ψp) = 1, c c−1 (3) when p < ∞, χ is ramified and specifically trivial on Up , non-trivial on Up where c c 0 × Up = 1 + p Zp when c > 0 and Up = Zp , then 1 0 c c(s− 2 ) 0 p(s, χ, ψp) = χ(p )p g(χ, ψp) where the Gauss sum

1 Z 0 − 2 c −1 0 −c g(χ, ψp) = p χ(y) ψp(p y)dy × Zp 36 TASHO KALETHA

is a square root of χ(−1). (It is a subtle thing to determine which square root it is.)

ˆ0 −1 0 Proof. We can plug in any test function into Zp (1−s, χ )/Zp (s, χ) (since this is a constant). 0 ˆ0 01 For (1), we plug in fχ. We check f1 = f1 In the other case,

0 0 d 0 d 0 0 0 fˆ = xfd1 = −2πi fˆ1 = −2πi f1 = ixf1 = if . χ dx dx χ 0 −1 0 This proves the result, given that Z∞(s, χ ) = Z∞(s + 2, χ). 0 1 ˆ0 0 In (2), take f = Zp and use f = f . Finally, in (3), take ( χ(x)−1, x ∈ ×, f(x) = Zp 0, otherwise.

0 Then Zp (s, χ)(f) = Zp(s, χ)(f) = 1. Recall that Z fˆ(y) = χ(y)−1ψ(xy)dy. × Zp × Note that dy is the additive Haar measure but on Zp gives a multiplicative Haar measure. ˆ −c −k × We claim that f is supported on p Zp. Write x = p x0 for x0 ∈ Zp . If k > c, then Z Z fˆ(x) = χ(y)−1 ψ(xyz)dzdy.

× c z∈1+pc y∈Zp /(1+p Zp) Zp c Write z = 1 + p z0 for z0 ∈ Zp. Then c c−k ψ(xyz) = ψ(xz(1 + p z0))ψ(xy) · ψ(x0yz0p ). Therefore: Z Z ˆ −1 c−k f(x) = χ(y) ψ(xy) ψ(x0yp z0)dz0 dy

× c z ∈ y∈Zp /(1+p Zp) 0 Zp | {z } =0 = 0,

+ c−k because x0y ∈ Up , p 6∈ Zp, so the character c−k z0 7→ ψ(x0yp z0)

is non-trivial on Zp. If k < c, then Z Z fˆ(x) = ψ(xyz)χ(yz)−1dzdy,

× c−1 c−1 y∈Zp /Up z∈Up MATH 679: AUTOMORPHIC FORMS 37

c−1−k c−1−k and ψ(xyz) = ψ(xy)ψ(x0yp z0) where x0yp z0 ∈ Zp. Therefore, Z Z fˆ(x) = ψ(xy)χ(y)−1 χ(z)−1dz dy.

× c−1 c−1 y∈Zp /Up z∈Up | {z } =0

Finally, if k = c, then Z ˆ −1 −c f(x) = χ(y) ψ(p x0y)dy

× Zp Z −1 −c = χ(x0) χ(y) ψ(p y)dy . × | {z } Zp f(p−cc) | {z } call this G

Therefore, for all x ∈ Qp, (1) fˆ(x) = f(p−cx) · G (Here, G is the unnormalized Gauss sum.) We can use Fourier inversion on f, we get that G2 = pcχ(−1). Indeed, ( χ(x), x ∈ ×, f(x) = Zp 0, otherwise. and χ−1(x) also has conductor pc if χ(x) does. Therefore, the formula (1) applies to f and yields:

(2) fˆ(x) = f(p−cx) · G.

× Finally, we get that for x ∈ Zp ˆ f(−x) = fˆ(x) Fourier inversion

= f\(p−cx) · G equation (1)

p−c = fd (x) · G

−c ˆ c −1 a−1 = p f(p x) · G fca = |a| (fˆ) = p−c · f(x) · G2 equation2 −c 2 −1 × = p · G · χ(−1) f(−x) as f(x) = χ(x) for x ∈ Zp . This shows that G2 = pcχ(−1) and so g = p−c/2G satisﬁes g2 = χ(−1). 38 TASHO KALETHA

Moreover,

0 −1 ˆ −1 p−c c c(s−1) −1 ¯ (s, χ, ψp) = Zp(1 − s, χ )(f) = Zp(1 − s, χ )(G · f ) = G · χ(p )p Zp(1 − s, χ , f) . | {z } =1 This completes the proof.

Analytic continuation and functional equation of the global zeta integral. We deﬁne the global zeta integral to be Z Z(s, χ, f) = f(x)χ(x)|x|sd×x. A× Theorem 4.30.

(1) The function Z(s, χ, f) is meromorphic on all s ∈ C. (2) If χ(x) = |x|t, then Z(s, χ, f) has a simple pole at s = −t with residue −f(0) · 1 vol(A1/Q×, d×x), and at s = 1 − t with residue fˆ(0) · vol(A1/Q×, d×x). (3) For all other s, Z(s, χ, f) is holomorphic. (4) The functional equation holds: Z(s, χ, f) = Z(1 − s, χ−1, fˆ).

Proof. We integrate in stages, according to

1 × |·| 1 A A R>0 1

and obtain Z Z |r|sχ(r) χ(a)f(ra)d×ad×r.

r∈R>0 a∈A1 Depending on s, |r|s explodes as r → ∞ or r → 0. Since f ∈ S(A), the explosion for r → ∞ is controlled. We split the integral over (0, ∞) into (0, 1) ∪ (1, ∞), and worry about (0, 1). To treat (0, 1), we split further according to

1 Q× A1 A1/Q× 1

and get r=1 Z Z X |r|sχ(r) χ(a) f(rqa)d×ad×r. q∈ × r=0 a∈A1/Q× Q

1 We normalized the Haar measure d×x so that the volumes are 1. However, if you replace Q with another number ﬁeld, these volumes will give a non-trivial contribution. MATH 679: AUTOMORPHIC FORMS 39

By (the Corollary to) Poisson summation formula 3.23, this is equal to

r=1    Z Z s −1 X −1 −1 × × |r| χ(r) χ(a) f(0) + r fˆ(0) + fˆ(r a q) d ad r. q∈ × r=0 a∈A1/Q× Q This is equal to

r=1   Z Z s−1 X −1 −1 × × |r| χ(r) χ(a)  fˆ(r a q) d ad r q∈ × r=0 a∈A1/Q× Q r=1 Z Z −f(0) rsχ(r) χ(a)d×ad×r

r=0 a∈A1\Q× r=1 Z Z +fˆ(0) rs−1χ(r) χ(a)d×ad×r.

r=0 a∈A1\Q× In the ﬁrst time, substitute r 7→ r−1 and a 7→ a−1 to get

∞ Z Z X r1−sχ(r)−1 χ(a)−1 fˆ(raq)d×ad×r. q∈ × r=1 a∈A1/Q× Q

t The inner integral in the second line is equal to 0 unless χ|A1 = 1, i.e. χ(x) = |x| , in which case it is equal to vol(A1/Q×), and we get

1 Z 1 × s+t × − vol(A /Q ) · f(0) · r d r . r=0

| {z1 } s+t Similarly, the third line is equal to 1 fˆ(0) · vol( 1/ ×) · . A Q s − 1 + t

We collect everything together to get that Z(s, f, χ) is equal to

∞     Z Z Z s × 1−s −1 × × r χ(ra)f(ra)d a + r χ (ra)fˆ(ra)d a d r

r=1 a∈A1 a∈A1 1 1 − vol( 1/ ×) · f(0) + vol( 1/ ×) · fˆ(0) . A Q s + t A Q s − 1 + t The ﬁrst line is holomorphic, and the entire expression is invariant under s 7→ 1−s, χ 7→ χ−1, ˆ f 7→ f. 40 TASHO KALETHA

Adelic eigendistributions. Recall that:

• a tempered distribution on R is a continuous linear functional S(R) → C. ∞ • a (tempered) distribution on Qp is any linear function S(Qp) = Cc (Qp) → C.

There is a canonical topology on S(Qp) for p < ∞. Note that C∞( ) = lim C(H /H0 ) c Qp −→ p p 0 Hp⊆Hp 0 0 0 where Hp ⊆ Hp run over lattices on Qp. Since Hp/Hp is finite, so C(Hp/Hp) is a finite- dimensional C-vector space, so it has a canonical topology. We get a colimit topology on ∞ Cc (Qp). With respect to this topology, every linear functional is continuous. Recall that O0 S(A) = S(Qp). p≤∞ Equivalently, S( ) = lim S( × (H/H0)). A −→ R 0 H ⊆H⊆Afin Zˆ-lattices We have isomorphisms

=∼ Y =∼ S(R × H/H0) S(R) S(R) ⊗ C(H/H0) x∈H/H0

f (f(−, x))x∈H/H0 P (fx)x∈H/H0 x∈H/H0 fx ⊗ δx

Y Put the product topology of S(R) on S(R × (H/H0)), and take the colimit topology x∈H/H0 on S(A). Deﬁnition 4.31. A tempered distribution on A is a continuous linear function on S(A). The set of tempered distributions is denote D(A). 0 Lemma 4.32. Let (λp)p≤∞ be a collection of λp ∈ D(Qp) such that λp(fp ) = 1 for almost all p. Deﬁne λ: S(A) → C by Y λ(⊗fp) = λp(fp). p≤∞ Then λ ∈ D(A) and any element of λ ∈ D(A) is of this form.

Proof. To show that λ = ⊗λp is continuous, restrict to 0 0 S(R × (H/H )) = S(R) ⊗ C(H/H ).

Then λ = λ∞ ⊗ λfin, and λfin must be a linear combination of δ-distributions. So in terms of 0 Y S(R × (H/H )) = S(R) MATH 679: AUTOMORPHIC FORMS 41 we see that λ is X cxλ∞ x∈H/H0 for cx ∈ C, so it is continuous. N Let λ ∈ D(A) be arbitrary non-zero. Let f = fp such that λ(f) = 1. For each p, define p N f = fq. Define q6=p p λp(gp) = λ(gp ⊗ f ). Then λp ∈ D(Qp) for p < ∞, and for p = ∞, we just need to check continuity. There are 0 lattices H ⊆ H ⊆ Afin such that f ∞ ∈ C(H/H0). The map 0 S(R) → S(R) ⊗ C(H/H ) ∞ g∞ 7→ g∞ ⊗ f is continuous, and hence so is λ∞. 0 Moreover, λp(fp) = λ(f) = 1. For almost all p, fp = fp , and thus (λp) satisfies the assumption. N Exercise. Check that λ = λp. Corollary 4.33. Let χ: A×/Q× → C× be a Hecke character. Then

χ O χp D(A) = D(Qp) p≤∞ is one-dimensional.

In particular, we have Z(s, χ),Z0(s, χ) ∈ D(A)χ, and O Z(s, χ) = Zp(s, χp), p 0 O 0 Z (s, χ) = Zp (s, χp). p Deﬁnition 4.34. We deﬁne the completed L-function and the global -factor as: Λ(s, χ) = Z(s, χ)/Z0(s, χ), (s, χ) = Zˆ0(1 − s, χ−1)/Z0(s, χ). Fact 4.35. We have that: (1) Y Λ(s, χ) = Lp(s, χp), p Y (s, χ) = p(s, χp, ψp). p (2) (s, χ) does not depend on ψ, 42 TASHO KALETHA

(3) (s, χ) is entire, (4)Λ( s, χ) has simple poles at s = −t and s = 1 − t when χ = | · |t . A

Proof. For (2), note that ψp is deﬁned up to a rational unit and scaling ψp by an element of Q× multiplies by a factor. For (3), note that Z0(s, χ) is entire and has no zeros.

Part (4) follows simply from Theorem 4.30. Theorem 4.36 (Functional equation). We have that (s, χ)Λ(1 − s, χ−1) = Λ(s, χ).

Proof. Immediate from Theorem 4.30.

This completes Tate’s thesis.

5. Artin L-functions

Review of algebraic number theory. Let K be a number field (finite extension of Q). Let OK ⊆ K be the ring of integers (integral closure of Z in K). It is a Dedekind domain. Any prime ideal p ⊆ OK is maximal and we define

k(p) = OK /p,N(p) = |k(p)|.

Being an Dedekind domain then implies that any ideal I ⊆ OK can be written (uniquely) as Y I = pnp . p

If L/K is any ﬁnite extension and pK ⊆ OK , then

Y npL pK ·OL = pL . pL

We say that pL|pK if npL > 0, i.e. pK ⊆ pL.

If L/K is Galois with Galois group G, then G acts transitively on {pL|pK }. Hence:

e e (1) In OL, pK = pL,1 ····· pL,k and we say that pK ramified if e > 1; we call e the ramification degree. (2) The finite fields k(pL) for pL|pK are all isomorphic, of degree f, called the inertial degree. (3) With the above notation, [L : K] = e · f · k.

(4) Let GpL be the stabilizer of pL|pK . These subgroups of G are all conjugate. The

action of GpL on k(pL) gives a short exact sequence

1 I(pL) GpL Gal(k(pL)/k(pK )) 1.

Here, I(pL) is the inertia group, which it has degree e. Moreover, Gal(k(pL)/k(pK ))

is canonically generated by the Frobenius element FrobpL . MATH 679: AUTOMORPHIC FORMS 43

We can form the completion LpL (either topologically, with respect to the valuation given by p , or algebraically, as the quotient ring of O where O = lim O /pn ). The action L L,pL L,pL ←− L L GpL on L extends to a continuous action of GpL on LpL , and gives ∼ GpL = Gal(LpL /KpK ).

Any prime pK of K is called a ﬁnite place of K. An inﬁnite place of K is either jK : K,→ R (a real place) or a pair (jK , jk) with jK : K,→ C (a complex place).

If jL is an inﬁnite place of L, we say jL|jK if and only if jL|K = jK . Note that G acts on {jL|jK } transitively and ∼ GjL = Gal(LjL /KjK ). Notation. If v is a place,  p KpK v ↔ K , ∼  Kv = R v ↔ jK real,  C v ↔ (jK , jK ) complex. Then ∼ GvL = Gal(LvL /KvK ).

Artin L-functions. Let K be a number field. Let GK = Gal(K/K) and consider a representation ρ: Gk → GL(V ) for a finite-dimensional Q-vector-space V . Since GL(V ) has no small subgroups, there is a finite Galois extension E/K such that ρ factors as ρ: GK /GE → GL(V ).

For an unramified prime p ⊆ OK , define −s −1 Lp(s, ρ) = det(I − N(p) ρ(FrobP)) where P ⊆ OE and P|p. Note that this is independent of the choice of P, because FrobP is well-defined up to conjugation and we are taking the determinant. If ρ = χ is a character and K = Q, then this gives (1 − p−sχ(p))−1 which agrees with our previous definition of a local L-factor. When p is (possibly) ramified, we define −s I(P) −1 Lp(s, ρ) = det(I − N(p) ρ(FrobP) | V ) . This agrees with the previous definition when I(P) = {1}, i.e. p is unramified. The global Artin L-function is defined to be Y L(s, ρ) = Lp(s, ρ).

p⊆OK Lemma 5.1. Each local factor is analytic for Re(s) > 0 and the product converges when Re(s) < 1. 44 TASHO KALETHA

−1 Proof. Note that Lp(s, ρ) = 0 if and only if there is an eigenvalue of ρ(Frobp) equal to N(p)s. But ρ is unitary with respect to a suitable scalar product on V , so all eigenvalues have absolute value 1.

Let z1(p), . . . , zn(p) be the eigenvalues. Then

n Y Y −s L(s, ρ) = (1 − N(p) zi). p i=1

Fact. If 0 ≤ xn < 1, then Y X (1 − xn) converges if and only if xn converges.

Therefore, we want to know if n X −s X N(p) zi(p) p i=1 | {z } ≤n converges. Note that #{p|p} ≤ [K : Q] and N(p)−s ≤ p−s. This proves convergence by the convergence of the Riemann zeta function.

Fact 5.2. The local L-factor Lp(s, ρ) depends only on ρ|Gp for one P|p.

Therefore, we can simply deﬁne for K local ﬁeld and ρ: GK → GL(V ):

−s I(P) −1 L(s, ρ) := det(I − N(p) ρ(FrobP) | V ) . Then the global L-function is Y L(s, ρ) = L(s, ρ|Gp ). p Fact 5.3. The function L(s, ρ) is independent of the choice of E.

Proof. If E0/E/K and P0|P|p, then

0 0 I(P ) Gal(E /K) FrobP0

I(P) Gal(E/K) FrobP .

This completes the proof.

Fact 5.4. We have that L(s, ρ1 ⊕ ρ2) = L(s, ρ1) · L(s, ρ2). Corollary 5.5. The L-fucntion L(s, ρ) makes sense for virtual representation ρ, i.e. a formal Z-linear combination of irreducible representations of GK . MATH 679: AUTOMORPHIC FORMS 45

Completed Artin L-functions. Given a finite Galois extension L/K of archimedean local fields, GL/K is {1} or Z/2Z. Since the Artin L-function is supposed to be additive, we just need to define the local L-factor at infinite for the trivial representations and the non-trivial representation of Z/2Z. Define

−s/2 ΓR(s) = π Γ(s/2), −s ΓC(s) = 2(2π) Γ(s).

For an irreducible representation ρ of Gal(L/K), we deﬁne  Γ (s) K = , ρ = 1,  R R L(s, ρ) = ΓR(s + 1) K = R, ρ = sgn,  ΓC(s) K = C.

By deﬁnition, L(s, ρ) is additive and independent of L.

Deﬁnition 5.6. The completed Artin L-function is Y Λ(s, ρ) = L(s, ρ|Gv ). v Corollary 5.7. The completed Artin L-function Λ is additive and independent of E.

Review of representation theory. If H ⊆ G are ﬁnite groups and ρH : H → GL(V ) is a ﬁnite-dimensional C-representation, then

G IndH ρH = {f : G → V | f(hg) = ρH (h)f(g)}. This is a representation of G via the action by right translation.

Fact 5.8 (Transitivity). If H1 ⊆ H2 ⊆ G, then

IndG ρ = IndG IndH2 ρ H1,H2 H1 H2 H1 H1

G We also have ResH π for π a representation of G. We have a H-representation homomorphism:

G G ev1 : ResH IndH V → V f 7→ f(1)

Fact 5.9 (Frobenius reciprocity). We have natural isomorphism

G =∼ G HomG(π, IndH ρ) → HomH (ResH π, ρ)

F 7→ ev1 ◦F. 46 TASHO KALETHA

Fact 5.10 (Mackey formula). Consider two subgroups H1,H2 ⊆ G and a ﬁnite-dimensional C-representation ρ: H1 → GL(V ). Then −1 G G M H1 gH1g g Res Ind ρ = Ind −1 Res −1 (ρ ) H2 H1 gH1g ∩H2 gH1g ∩H2 g∈H2\G/H1 −1 −1 g M gH2g H1 = Ind −1 Res −1 ρ , H1∩gH2g H1∩gH2g g∈H1\G/H2

g −1 −1 where ρ is the representation of gH1g given by ρ ◦ Ad(g ) and −1 Ad(g): H1 → gH1g , x 7→ gxg−1.

Proof. Exercise. Theorem 5.11 (Brauer induction). The Grothendieck group of virtual representations of G G is generated by the elements IndH χ, where H ⊆ G is a subgroup and χ is a 1-dimensional representation of H.

Inductivity of Artin L-functions. Recall that Y L(s, ρ) = L(s, ResGK ρ) GKv˙ v finite Y Λ(s, ρ) = L(s, ResGK ρ), GKv˙ v all places wherev ˙ is any place of E above v and the representation ρ factors through the finite extension E. Proposition 5.12. Let L/K be a finite Galois extension of local fields or number fields and let ρL : GL → GL(V ) be a finite-dimensional C-representation. Then L(s, IndGK ρ ) = L(s, ρ ). GL L L In the global case, also Λ(s, IndGK ρ ) = Λ(s, ρ ). GL L L

Proof. Step 1: Reduce the global statement to the local statement. Fix a large Galois extension E/L and write GK , GL for Gal(E/K) and Gal(E/L). Moreover, write Gal(E/K) ρK = IndGal(E/L) ρL.

It is enough to show that for a given v, a place of K, Y L(s, ResGK ρ ) = L(s, ResGL ρ ). GK,v˙ K GL,w˙ L w|v Indeed, one can compare the Euler factors on both sides. As abovev ˙ andw ˙ are any places of E above v and w. MATH 679: AUTOMORPHIC FORMS 47

By Mackey formula 5.10, −1 M G g ResGK IndGK ρ = Ind K,gv˙ ResGL ρ . GK,v˙ GL L GK,gv˙ ∩GL GK,gv˙ ∩GL L g∈GL\GK /GK,v˙

Moreover, GK,gv˙ ∩ GL = GL,gv˙ and

GL\GK /GK,v˙ → {w | v} g 7→ p(gv˙) where p is the restriction map p: {places of E} → {places of L}. We are now done by additivity and inductivity in the local case. Step 2: Local archimedean case. The only possibility is

K = R,L = C, ρL = 1, and

ρK = Ind 1GL = 1GK ⊕ sgn . Then L(s, ρ) = Γ (s)Γ (s + 1) = π−s/2Γ(s/2)π−s/2−1/2Γ(s/2 + 1/2). R R √ By the duplication formula (Γ(z)Γ(z + 1/2) = 21−2z πΓ(2z)) for the Gamma function, we get that

L(s, ρ) = ΓC(s). Step 3: Local non-archimedean case. Let L/K be a finite extension. Let K0 ⊆ L be the maximal unramified subextension. The L/K0/K is a tower where L/K0 is totally ramified and K0/K is unramified. By transitivity of induction 5.8, it is enough to deal with each case separately.

Step 3a: L/K is totally ramiﬁed. Let IK ⊆ GK be the inertia subgroup. Then

IK \GK /GL = {1}, so the product is trivial. We hence have an isomorphism

IL\GL → IK \GK ,

FrobL 7→ FrobK . Then ResGK IndGK ρ = IndIK ResGL ρ and hence by Frobenius reciprocity 5.9, ρIK = ρIL . IK GL L IL IL L K L Since the Frobenii match, this shows that the L-factors match.

Step 3b: L/K is unramiﬁed. Then IL = IK ⊆ GL ⊆ GK . Thus GL is normal in GK and GK /GL is cyclic of order f = [L : K] = [kL : kK ], generated by the image of FrobK ∈ GK /IK . f We have that FrobL = FrobK . By Mackey formula 5.10: −1 −1 g GK GK M gIK g GL Res Ind ρL = Ind −1 Res −1 ρL IK GL gIK g ∩GL gIK g ∩G g∈GL\GK /IK M g−1 = IndIK ResGL ρ . IL IL L g∈GK /GL 48 TASHO KALETHA

Let A ∈ GL(ρL) be ρL(FrobL). Then ρK (FrobK ) be given by 0 1   0 1    IL B =   ∈ Mf×f (End(V ))   L  1 A 0

−s IL We want to compute det(I −t·B) where t = qK . Let R ⊆ End(VL ) be generated by scalars and A.

Formula. If R ⊆ Mn×n(F ) is a commutative subgring, and B ∈ Matm×m(R) ⊆ Matnm×nm(F ), then

detF (B) = detF (detR(B)). We apply this to I − t · B:  I −tI   I −tI      .    −tI −tA I We have that f−1 f f detR(I − tB) = I + (−1) (−t) A = I − t A. By the formula above f detF (I − tB) = detF (I − t A). f We use the fact that FrobL = FrobK to conclude that L(s, IndGK ρ ) = L(s, ρ ). GL L L This completes the proof.

Review of class ﬁeld theory. Let K be local non-archimedean with Galois group GK . We have the Weil group WK ⊆ GK given by the diagram

WK GK

Z Zˆ

but not with the subspace topology but rather with IK ⊆ WK open. Theorem 5.13 (Local Class Field theory).

(1) There is a canonical isomorphism

ab =∼ × WK → K × mapping inertia to OK and any Frobenius element a uniformizer. (2) For a ﬁnite Galois extension L/K, we have two diagrams: MATH 679: AUTOMORPHIC FORMS 49

ab × ab × WK K WK K

inclab NL/K tr

ab × ab × WL L WL L (3) If L/K is abelian, the ﬁrst diagram gives

× × K /NL/K (L ) → GL/K .

For any ﬁnite extnesion L/K, we have the the relative Weil group WL/K = WK /[WL,WL]. Then

× 1 L WL/K GL/K 1

and conversely W = lim W . K ←− L/K L In the archimedean case, there is also a relative Weil group which ﬁts in the above short exact sequence.

Let K be a number ﬁeld. There exists a Weil group WK equipped with a map WK → GK satisfying the following properties. Theorem 5.14.

(1) There exists an isomorphism

ab × × WK → A /K = CK . (2) For any ﬁnite Galois extension L/K, we have two diagrams

ab ab WK CK WK CK

inclab NL/K tr

ab × WL CL CL L (3) When L/K is abelian, we have an isomorphism ∼ CK /NL/K (CL) = GL/K .

Construction of the Weil groups in the global case:

2 (1) Identify a canonical class in H (GL/K ,CL). (2) Obtain the short exact sequence

1 CL WL/K GL/K 1. (3) Deﬁne W = lim W . K ←− L/K × Notation. When K is local, we set CK = K . Theorem 5.15. If v˙ is a place of K, then we have a diagram 50 TASHO KALETHA

ab WK CK

W ab C . Kv˙ Kv˙

Fact 5.16. Every finite-dimensional representation of WK factors through WL/K for some L/K finite. It extends to GK if and only if its image is finite.

Remark 5.17. If ρ is an irreducible representation of WK (primitive if K is a number ﬁeld — not induced from a subgroup), then ρ = ρ1 ⊗ χ where ρ1 has ﬁnite image and χ is a character.

Remark 5.18. We can deﬁne L-functions for representations of WK . Everything goes through, with convergence in some right half-plane.

Abelian case.

× Proposition 5.19. Let K be local or global and ρ: GK → C be a 1-dimensional represen- t tation, χ: CK → C imes the corresponding character. Then L(s, ρ) = L(s, χ).

Proof. Due to compatibility of local and global Artin reciprocity maps, we are reduce to × the local case. Since the local map IK → OK sends Frobenius to the uniformizer, we are done. × Corollary 5.20. Let K be a number ﬁeld and ρ: GK → C . Then L(s, ρ) has meromorphic continuation and Λ(s, ρ) satisﬁes the functional equation.

Meromorphic continuation of non-abelian Artin L-functions and the Artin conjecture.

Theorem 5.21. Let K be a number ﬁeld and ρ: GK → GL(V ) be a representation. Then L(s, ρ) has meromorphic continuation and Λ(s, ρ) satisﬁes a funtional equation.

Proof. By Brauer Induction Theorem 5.11, X ρ = uiρi for u ∈ , ρ = IndGK χ where χ : H → × and H ⊆ G . By inductivity 5.12 and i Z i Hi i i i C i K additivity 5.7, we have that

Y ni L(s, ρ) = L(s, χi) and similarly for Λ. Conjecture (Artin). If ρ is irreducible and not the trivial representation, then L(s, ρ) is entire.

Remark 5.22. In the proof, the ni can be negative, so a zero of L(s, χi) will contribute to a pole of L(s, ρ). The Artin conjecture says that these poles cancel out to give a holomorphic function. MATH 679: AUTOMORPHIC FORMS 51

Note that we have not deﬁned an -factor which is to appear in the functional equation. We may simply deﬁne it as Λ(s, ρ) (s, ρ) = . Λ(1 − s, ρ∨) This makes it meromorphic, inductive, and additive. By inductivity and additivity, it is entire and has no zeros, since this is true in the 1-dimensional case. Let (s, ρ) = |(s, ρ)| · W (ρ) where W (ρ) is the root number. Each of these pieces is inductive and additive, and we want to describe them separately. The absolute value is described using the theory of Artin conductors. The root number is harder to understand but we will still describe it as best we can.

Filtrations of the Galois group of a local field. Let L/K be a finite extension of non-archimedean local fields, G = Gal(L/K). Definition 5.23. The lower numbering higher ramification groups are

Gi = {σ ∈ G | σ| i+1 = 1} OL/pL where G−1 = G by deﬁnition.

Note that G0 is the inertia group. We can write

Gi = {σ ∈ G | ιG(σ) > i} where ιG(σ) = max{vL(σ(x) − x) | x ∈ OL}. Fact 5.24.

−1 (1) For σ, τ ∈ G, ιG(στσ ) = ιG(τ). (2) If H ⊆ G, ιH = ιG|H .

In particular, Hi = Gi ∩ H. Deﬁnition 5.25.

(1) For r ∈ R, r ≥ −1. define Gr = Gdre. Z u −1 −1 (2) Define ϕ(u) = [G0 : Gt] dt, ψ = ϕ . 0 u (3) Define the upper numbering higher ramification groups to be G = Gψ(u). m P gi Remark 5.26. Note that ϕ(−1) = −1, ϕ(0) = 0, ϕ(m) = for gi = #Gi. g0 i=1 Fact 5.27. Let N ⊆ G be normal.

−1 X (1) We have that ιG/N (σ) = eL/LN ιG(σ). σ7→σ (2) Therefore, (G/N)t = GtN/N. 52 TASHO KALETHA

Theorem 5.28 (Hasse-Arf). If L/K is abelian, then the jumps in the upper numbering are integers. Theorem 5.29. The local Artin reciprocity maps are ab,t ∼ t × GK = UK ⊆ OK .

Review of representation theorem. Let G be a ﬁnite group. (1) Let (ρ, V ) be a representation of G. Then the character of ρ is deﬁne to be

ρρ(g) = trρ(g)

(a class function). The map (ρ, V ) 7→ χρ is injective. (2) Each representation is a direct sum of irreducibles. (3) The characters of irreducible representations form an orthonormal basis of the space of class functions with respect to 1 X (f , f ) = f (g)f (g). 1 1 G #G 1 2 g∈G We have the monoid of isomorphism classes of rerpesentations of G with ⊕. Take the abelian group generated by this monoid; this is called the (Grothendieck) group of virtual representations. This group is isomorphism to #(conjugacy classes) Z . This group embeds into the group of complex-valued class functions. If π, σ are representations and σ is irreducible, then

(χπ, χσ) = dim HomG(σ, π).

If H ⊆ G is a subgroup and π is a representation of G, σ a representation of H, then

(χπ, χ G G)G = (χ G , χσ)H . IndH σ ResH π

G In particular, IndH 1 contains every irreducible representation ρ of G with multiplicity dim ρH .

Review of discriminants. Let L/K be a ﬁnite extension of local number ﬁelds. We have a symmetric bilinear form L ⊗ L → K,

x ⊗ y 7→ trL/K (xy). Deﬁnition 5.30.

(1) The complementary module of OL is ∗ OL = {x ∈ L | (x, y) ⊆ OK for all y ∈ OL}. ∗ −1 (2) Then DL/K = (OL) is the diﬀerent (as ideal of OL). (3) Finally, the discriminant is dL/K = NL/K (DL/K ). Theorem 5.31. MATH 679: AUTOMORPHIC FORMS 53

(1) For a tower M/L/K,

DM/K = DM/LDL/K

[M:L] dM/K = dL/K NL/K dM/L. (2) For L/K an extension of number fields, Y DL/K = DLP/Kp p and similarly for d. (3) If OL/OK is generated by α ∈ OL and g be the minimal polynomial ofα. Then DL/K is generated by g0(α). (4) The discriminant dL/K is generated by the discriminants of all bases of L/K contained in OL. (5) A prime P ⊆ OL ramifies if and only if P|DL/K . A prime p ⊆ OK ramifies if and only if p|dL/K .

× Fact 5.32. Let ψ : /K → . Let u be such that ψ| uv = 1 but ψ uv−1 6= 1. Then AK C v pv pv

Y −nv qv v is independent of ψ and equals dK/Q.

Proof. Since the dual of AK /K is K, the independence is immediate from the product formula. We can use 0 ψ = ψ ◦ trK/Q . −nv Then pv = DKv/Qp . Then

Y Y Y −nv dK/Q = dKv/Qp = #(OKv /dKv/Qp ) = qv . v v v This completes the proof.

The Artin conductor. Let L/K be a Galois extension of local non-archimedean ﬁelds. Deﬁne  −fιG(σ) σ 6= 1 aG(σ) = P f ιG(τ) σ = 1  τ6=1

−1 Fact 5.33. The function aG is a class function and aG(σ ) = aG(σ).

Therefore, X aG = f(ρ) · χρ ρ∈Irr(G)

for f(ρ) = (aG, χρ)G. The main goal is to show that f(ρ) ∈ Z≥0.

Lemma 5.34. We have that aG(1) = f · ord(DL/K ). 54 TASHO KALETHA

Proof. Choose a generator α of OL over OK . Let g be the minimal polynomial of α. Then DL/K is generated by Y g0(α) = (τα − α). τ∈G τ6=1 Then X ordL(DL/K ) = ordL(τα − α) τ∈G τ6=1 X = ιG(τ). τ6=1

This completes the proof.

It is immediate from the deﬁnition of aG that (aG, 1G)G = 0.

Gi Lemma 5.35. For i, let ui = Ind1 1 − 1 be the augmentation representation. Then ∞ X a = [G : G ]−1 IndG u . G 0 i Gi i i=0

Proof. Case 1: σ ∈ G, σ 6= 1.

Let k be unique such that σ ∈ Gk \ Gk+1. Then

aG(σ) = −f(k + 1). On the other hand, IndG u = IndG 1 − IndG 1. Gi i 1 Gi Then the character of IndG u is g/g if i ≤ k and 0 otherwise. Gi i i Case 2: σ = 1. P We have aG = 0. This is equivalent to (aG, 1G)G = 0. It is hence enough to show that σ∈G

(RHS, 1G)G = 0. This is true because (IndG u , 1 ) = (u , 1 ) = 0 Gi i G G i Gi Gi by Frobenius reciprocity.

Proposition 5.36. If ρ has dimension 1, then let j be maximal such that ρ|Gj 6= 1. Then f(ρ) = ϕ(j) + 1, and this is an integer. MATH 679: AUTOMORPHIC FORMS 55

Proof. Using Lemma 5.35,

f(ρ) = (aG, χρ)G ∞ X −1 = [Gi : G0] (ui, χρ|Gi )Gi i=0 | {z } 1 i ≤ j

0 i > j

j X −1 = (gi/gi) i=0 = ϕ(j) + 1. We just have to check that ϕ(j) is an integer. Let K0 = Lker(ρ) and r = ϕ(j). Then Gal(K0/K)t = im(Gt). We have that Gal(K0/K)t 6= 0 but Gal(K0/K)t+ = 0 for > 0. Therefore, t is a jump in the upper numbering for the abelian extension K0/K, so the Hasse–Arf Theorem 5.28, t is an integer. H 0 H Lemma 5.37. Let H ⊆ G be a subgroup and let rH = Ind1 1. Let K = L . Then

aG|H = ordK (dK0/K ) · rH + fK0/K aH .

Proof. Let σ ∈ H. Recall that ιG|H = ιH . Case 1: σ 6= 1. Then

aG(σ) = −fL/K ιG(σ)

= −fL/K0 fK0/K ιH (σ)

= fK0/K aH (σ) by deﬁnition, while rH (σ) = 0. Case 2: σ = 1. Then

rH (σ) = [H : 1]

= eL/K0 · fL/L0 . We then have that: 0 RHS(1) = ordK (∂K0/K )eL/K fL/K0 + fK0/K fL/K0 ordL(D(L/K )) by Lemma 5.34

= ordK (NK0/LDK0/K )eL/K0 fL/K0 + fL/K ordL(DL/K0 )

= ordK0 (DK0/K )fK0/K eL/K fL/K0 + fL/K ordL(DL/K0 )

= fL/K ordL(DK0/K ) + fL/K ordL(DL/K0 )

= fL/K ordL(DL/K )

= aG(1) by Lemma 5.34 This gives the result.

Theorem 5.38. For any irreducible representation ρ of G, f(ρ) ∈ Z≥0. 56 TASHO KALETHA

Proof. Using Brauer induction 5.11, we know that X ρ = u IndG ρ i Hi i i for ui ∈ Z and 1-dimensional ρi. Then

f(ρ) = (aG, χρ)G X = ui(aG|Hi , ρi)Hi i X = ui(ord(dKi/K )(rHi , ρi)Hi + fKi/K (aHi , ρi)Hi ) Lemma 5.37 i

Finally, (rHi , ρi) ∈ Z and (aHi , ρi)Hi ∈ Z by Proposition 5.36.

On the other hand, by Lemma 5.35, g0 ·aG is a representation, so f(ρ) ∈ Q≥0. This completes the proof.

Remark 5.39. This theorem is equivalent to showing that aG is a representation of G, called the Artin representation. There is no explicit construction of this representation. Definition 5.40. The Artin conductor f(ρ) of a (virtual) representation ρ is defined as (aG, χρ)G. Fact 5.41. If ρ is unramified, then f(ρ) = 0.

Proof. Use Proposition 5.36. Proposition 5.42. The Artin conductor f(ρ) is compatible with inﬂation: if M/L/K is a tower then ρ: GL/K → GL(V ) satisﬁes fM/K (ρ) = fL/K (ρ).

Proof. We have that 0 −1 −1 X aGL/K (σ ) = fL/K eM/LdM/K aGM/K (σ) σ7→σ0 −1 X = [M : L] aGM/K (σ). σ7→σ0 Then

(aGM/K , χρ)GM/L = (aGL/K , ρρ)GL/K , completing the proof.

Let L/K be a ﬁnite Galois extension of number ﬁelds and ρ: GL/K → GL(V ) be a complex representation.

Deﬁnition 5.43. The Artin conductor f(ρ) = fL/K (ρ) of ρ is Y f(ρ| ) p GL/K,P p for any choice of P|p. MATH 679: AUTOMORPHIC FORMS 57

Remark 5.44. The conductor is independent on the choice of P|p, because the resulting representations are isomorphic for different choices. Also, note that f(ρ|GL/K ,P) = 0 for almost all p so this product is well-defined. Fact 5.45. The Artin conductor f is additive and compatible with inflation. Proposition 5.46. The Artin conductor f is compatible with induction as follows: for a tower M/L/K, ρL : GM/L → GL(V ), we have that

dim(ρL) fM/K (Ind GM/K ρL) = dL/K · NL/K (fM/L(ρL)). GM/L

Proof. The proof is left as an exercise. Hint: use Mackey formula 5.10 and established results.

Definition 5.47. Let L/K be a finite Galois extension of number fields and ρ: GL/K → GL(V ) be a representation. Define c (ρ) = ddim(ρ)N (f (ρ)) L/K K/Q K/Q L/K as an ideal of Z; equivalently, as a positive integer.

Corollary 5.48. The function cL/K (ρ) is compatible with addition, inﬂation, and induction.

Proof. This is immediate from the above results.

From now on, we may refer to cL/K (ρ) as simply cK (ρ), because it is independent of the choice of L. 1 −s Theorem 5.49. The absolute value of the epsilon factor |(s, ρ)| = cK (ρ) 2 .

Proof. Both sides are compatible with induction and additive. By Brauer induction 5.11, this reduces to the case dim ρ = 1. Both sides are products of local factors. For the left side, × we need to choose a non-trivial additive character ψK : AK /K → C to write Y (s, ρ) = (s, ρv, ψKv ). v We take ψ0 = ψ0 ◦ tr . Now it is enough to compare |(s, ρ , ψ0 )| with K Q K/Q v Kv 1 −s f(ρv) 2 pKv/Qp · NKv/Qp pv . Note that when v|∞, both sides are 1.

To lighten the notation, we write K = Kv for any v, a local non-archimedean ﬁeld. Recall that we have computed the local -factors in Lemma 4.29 for a special choice of character (in the case K = Qp, but this generalizes easily). 0 n n−1 0 0 n Let n be such that ψK is trivial on pK , not on pK . Then ψK (x) = ψK (π x) is of level 0, and 0 n(s−1/2) 0 |(s, ρ, ψK )| = q |(s, ρ, ψK )| −n by Lemma 4.28. By Fact 5.32, dK/Qp = q . Thus we are left with showing that: 1 0 f(p) 2 −s |(s, ρ, ψK )| = (NK/Qp (p )) . We can ﬁnally use Lemma 4.29. There are two cases: 58 TASHO KALETHA

0 (1) When ρ is unramiﬁed, (s, ρ, ψK ) = 1, f(ρ) = 0. (2) When ρ is ramiﬁed, let c be smallest so that ρ| c is trivial. Then UK

0 c(1/2−s) c 0 (s, ρ, ψK ) = qK ρ(π )g(ρ, ψK ). At the same time, Proposition 5.36 shows that f(ρ) = c.

ab,c ∼ c Finally, we use the local Artin map GK = UK to get the result.

This completes the proof.

The root number. The root number W (ρ) is a root of unity, independent of s. We will deﬁne it soon.

Aside. Let G be a ﬁnite group and K0(G) be the Grothendieck group of virtual representations. We have a map

dim: K0(G) → Z. If H ⊆ G,

Ind K0(H) K0(G). Res Theorem 5.50 (Deligne/Langlands).

× (1) Let K be a local ﬁeld, ψK be a non-trivial character K → C . For any ﬁnite extension L/K, ψL = ψK ◦ trL/K . There is exactly one assignment

(L/K, ρL : GL → GL(V )) 7→ (ρL, ψL) ∈ C such that 0 0 (a) (ρL ⊕ ρL, ψL) = (ρL, ψL) · (ρL, ψL), (b) if ρL has dimension 1, then

(ρL, ψL) = (1/2, χ, ψL), × × where χ: L → C corresponds to ρL under the local Artin map, (c) is additive in degree 0. × (2) Let K be a global field, ψK : AK /K → C be a non-trivial character. Then 1 Y , ρ = W (ρ) = (ρ , ψ ). 2 v Kv v Definition 5.51. Let L/K be an extension of local fields. Define

GK (Ind 1L, ψK ) λ(L/K, ψ ) = (IndGK 1 , ψ ) = GL . K GL L K (1L, ψL)

Fact 5.52. For any ρL : GL → GL(V ),

(IndGK ρ , ψ ) = λ(L/K, ψ )dim ρL (ρ , ψ ). GL L K K L L MATH 679: AUTOMORPHIC FORMS 59

Remark 5.53. The proof of the Theorem is hard. Langlands takes around 250 pages to do it purely locally with some very diﬃcult computations. Deligne’s idea is to use the existence of the global epsilon factor and specialize it to deduce something about the local epsilon factors. This turns out to be much easier.

6. Non-abelian class field theory?

Class ﬁeld theory was the crucial input for describing the 1-dimensional Artin representations in terms of Hecke characters. What about representations ρ of the Galois group that do not factor through an abelian quotient? Conjecture (Langlands). Let K be a number ﬁeld. Let ρ be an irreducible n-dimensional complex representation of GK . Then there exists a cuspidal automorphic representation π of GLn(AK ) such that L(s, ρ) = L(s, π).

What is a cuspidal automorphic representation? We have

GLn(AK ) — a locally compact group with Haar measure which contains

GLn(K) — a discrete subspace with the counting measure. We can hence form GLn(K)/ GLn(AK ), which is a coset space with a measure. This is the non-abelian (n-dimensional) version of × × AK /K . Consider 2 L (GLn(K)/ GLn(AK )), which is a Hilbert space with GLn(AK )-action. Deﬁnition 6.1. An automorphic representation is a “constituent” of 2 L (GLn(K)/ GLn(AK )).

Caution. This L2-space is not the direct sum of irreducibles. Therefore, it is not clear what “constituent” means. Instead of deﬁning constituents in general, we will reduce to a simpler case.

× × 2 Note that Z(GLn) = GL1 ,→ GLn acts on GLn. This gives an action of AK /K on L . Using Fourier theory, we can decompose the space Z ⊕ 2 2 L (GLn(K)/ GLn(AK )) = L (GLn(K)/ GLn(AK ), ψ)dψ ψ as a direct integral. 60 TASHO KALETHA

2 We deal with L (GLn(K)/ GLn(AK ), ψ). This still does not decompose discretely, but there is a subspace 2 2 L0(GLn(K)/ GLn(AK ), ψ) ⊆ L (GLn(K)/ GLn(AK ), ψ) of cusp forms.

2 Theorem 6.2. The space L0(GLn(K)/ GLn(AK ), ψ) decomposes as a direct sum of irreducible representations with ﬁnite multiplicities.

Remark 6.3. It is another big theorem that for GLn the multiplicities must in fact be one. This fails for other reductive groups (while the stated theorem is still true). Deﬁnition 6.4. A cuspidal automorphic representation is an irreducible constituent of 2 L0(−, ψ). Remark 6.5. More generally, we have a decomposition

2 M 2 L (−, ψ) = LP,ψ. P parabolic

2 2 When P = G, L0 ⊆ LG,ψ is the discrete spectrum, which decomposes discretely. When P 6= G, 2 ∼ G 2 LP,ψ = IP (LM,ψ,dx) G where IP is parabolic induction and M is a Levi subgroup. This was one of the ﬁrst contribu- tions of Langlands to this subject. The isomorphism is proved using the theory of Eisenstein series.

What is the L-function of a π, L(s, π)? O0 We can decompose π = πv if πv is an irreducible representation of GLn(Kv). Then v deﬁne Y L(s, π) = L(s, πv). v

Defining L(s, πv) uses representation theory and harmonic analysis of GLn(Kv). We will see this definition only make sense when π is an automorphic representation, and we will see how the decomposition with respect to ψ is relevant. There is a local version of the above theorem. Theorem 6.6 (Harris–Taylor, Henniart). Let K be a non-archimedean local field. There is a bijection

r : Irrn(WK ) → Cusp(GLn(K)) preserving L and factors (and satisfying extra properties).

Theorem 6.7. Let LK = WK × SL2(C) be the Langlands group. There is a bijection

r : Repn(LK ) → Irr(GLn(K)) preserving L and factors (and satisfying extra properties). MATH 679: AUTOMORPHIC FORMS 61

ˆ Sometimes, a representation ρ: Gk → GL(V ) factors through another linear algebraic group G. For example, ρ might preserve a symplectic or orthogonal pairing. Then Langlands associates to Gˆ a reductive group G/K. Conjecture (Local Langlands Conjecture). Therre exists a ﬁnite-to-one map: ˆ Irr(G(K)) → Rep(LK , G) with a description of the ﬁbers (L-packets), satisfying some extra properties. 0 O 2 Conjecture (Global Langlands Conjecture). Given π = πv, the multiplicity of π in L is X (∗)

ρ: GK →Gˆ ρv↔πv where (∗) depends on the L-packets in the Local Langlands Conjecture. Remark 6.8. There is also a Functoriality conjecture which says that any L-homomorphism between L-groups induces a transfer map between automorphic forms.

7. Automorphic representations of SL2(R)

We are interested in representations of GL2(A). We noted above (without proof) that such representations decompose O0 π = πp p≤∞ (if admissible) where πp is a representation of GL2(Qp). We therefore start by discussing representation theory of GL2(Qp) for all p. In this chapter, we deal with p = ∞.

The Lie group SL2(R).

Remark 7.1. Most representations of SL2(R) are inﬁnite-dimensional. The next example describes all the ﬁnite-dimensional ones.

Example 7.2. For each N ∈ N, let MN be the set of homogeneous degree N polynomials in 2 variables, i.e. n o X i j ai,jX Y ai,j ∈ C, i + j = N . Then SL2(R) acts on MN . Moreover, MN is a ﬁnite-dimensional irreducible representation of SL2(R) of dimension N + 1. It is unique such up to isomorphism. Deﬁnition 7.3. A representation of G is a complex Hilbert space V , together with a group homomorphism π : G → Autcts(V ) such that the map G × V → V is continuous. A representation (π, V ) is unitary if π takes values in the unitary operators (i.e. it preserves the scalar product). 62 TASHO KALETHA

We have the following useful subgroups: ∗ 1 ∗ ∗ ∗ cos ϕ sin ϕ T = ,U = ,B = = U T,K = ∼= S1. ∗ 0 1 0 ∗ o − sin ϕ cos ϕ

Definition 7.4 (Parabolic induction). Let χ: T → C× be continuous character of T . Define G 1/2 IB χ = f : G → C | f(tug) = δB(t) χ(t)f(g) . This is a representation of G by right translations. The functions f satisfy extra properties, soon to be defined.

Questions.

(1) What is δB and why is it there? (2) Qhat kind of functions f are we actually taking?

Aside on Haar measures. Let H be locally compact topological group. Let H be a Haar measure dh (left-invariant2). For g ∈ H, consider the functionals Z Z f 7→ f(h)dh, f 7→ f(hg)dh. H H

These are two Haar functionals, so by uniqueness, there is δH (g) ∈ R>0 such that Z Z f(hg)dh = δH (g) f(h)dh H H for all f.

Exercise. The function δH : H → R>0 is a continuous group homomorphism.

Deﬁnition 7.5. The function δH is the modulus character. The group H is called unimodular if δH = 1. Examples 7.6. Unimodular groups include: abelian groups, compact groups, and reductive groups. In particular, T, U, K, G are all unimodular. However, B is not unimodular.

When H is a real Lie group,

δH (h) = | det(Ad(h) | Lie(H))|. Here, Ad(h): H → H and Ad(h) = dAd(h): Lie(H) → Lie(H). In the case of H = B, we have that t1 u Ad = |t1/t2|. 0 t2 This is the character which appeared in Deﬁnition 7.4. This (partially) answers question (1). Consider H ⊆ G locally compact. We have the G-homogeneous space G/H. Question. Is there a G-invariant measure on G/H?

2In the ﬁrst part of the course, we dealt with abelian groups, so we never had to specify if we’re using right or left-invariant measures. MATH 679: AUTOMORPHIC FORMS 63

If G, H are real Lie groups, such a measure amounts to an H-invariant top form on T1(G/H) = Lie(G)/ Lie(H). Note that H acts on G/H by conjugation. Check that

(Ad(h) | T1(G/H)) = (Ad(h) | Lie(G)/ Lie(H)). Therefore det(Ad(h) | Lie(G)) δ (h) = G . det(Ad(h) | Lie(H)) δH (h)

Answer. A G-invariant measure on G/H exists if and only if δG|H = δH .

But for B ⊆ G, this is false, since δG = 1, δB 6= 1. In particular, there will never be a G-invariant measure on G/H. Lemma 7.7. There exists a G-invariant Haar functional on the space of functions

δG(h) f : G → C such that f(gh) = f(g), δH (h) which are continuous and compactly supported mod H.

Proof. Let ω be a top form on Lie(G)/ Lie(H). Consider the diﬀerential form of degree dim(G) − dim(H) on G given by ∗ ω(g) = Lgω

where LG denotes left translation by g. It satisﬁes δ (h) ω(gh) = H · ω(g). δG(h) Therefore, for f as above, f · ω is H-invariant, and hence a diﬀerential form of top degree on G/H. Integrating against it is the desired Haar functional.

G We now return to the discussion of parabolic induction (Deﬁnition 7.4), IB χ. Consider the G case when χ is unitary. Then f ∈ IB χ satisﬁes −1 (f · f)(tug) = δB(t) ff(g). This belongs to the space of functions in Lemma 7.7, where we can integrate.

• We demand that f ∈ L2(B\G). G • The representation IB χ is unitary. This completes the answer to questions (1) and (2) above in the case when χ is unitary. When χ is not unitary, we need the following lemma. Lemma 7.8 (Iwasawa decomposition). There is a decomposition G = B · K, i.e. g ∈ G is the product g = b · k for b ∈ B, k ∈ K. Moreover, B ∩ K = {±1} = ZG.

Proof. Exericse. 64 TASHO KALETHA

Caution. Here, neither of the two groups B and K normalize each other. Therefore, this is not just the usal product of groups from group theory. By this Lemma, we have that G ∼ IB χ = {f : K → C | f(k) = χ()f(k) for ∈ ZG}.

Which functions do we take? We simply require that f ∈ L2(K). The two previous deﬁnitions agree by the integral formula: Z Z Z f(g)dg = f(bk)dbdk. G K B

Eventually, we will see that any irreducible representation is obtained by parabolic induction. For now, we need to do some more work.

Smooth and K-ﬁnite vectors. Remark 7.9. Let V be a ﬁnite-dimensional representaion of G, π : G → GL(V ). Then dπ : Lie(G) → End(V ) is a representation of Lie(G).

Why does this work?

(1) Continuity of G × V → V implies continuity of π : G → GL(V ). (2) The group GL(V ) is a Lie group, and any continuous homomorphism of Lie groups is diﬀerentiable; in fact, real analytic.

However, this clear does not work for ﬁnite-dimensional representations. Deﬁnition 7.10. A vector v ∈ V is called

(1) diﬀerentiable if for each X ∈ g = Lie(G), the limit

d π(exp(tX))v − v XV = π(exp(tX))v = lim t→0 dt t=0 t exists. (2) k-times differentiable if it is differenitable and for all X ∈ g, Xv is (k−1)-differentiable, (3) smooth if it is k-differentiable for all k.

We write V ∞ ⊆ V for the subspace of smooth vector. Fact 7.11. The subspace V ∞ is G-invariant, but not closed, and is a representation of g.

From now on, the real vector space g will be replaced by its complexiﬁcation g⊗R C. This still has a representation V ∞ by extending scalars. We do this to use the tools from representation theory of complex Lie algebras. MATH 679: AUTOMORPHIC FORMS 65

Universal enveloping algebra. Let L be a complex Lie algebra. Deﬁne M T (L) = L⊗n n≥0 and let I be the inhomogeneous ideal generated by X ⊗ Y − Y ⊗ X − [X,Y ]. The universal enveloping algebra is T (L) U(L) = . I

Theorem 7.12 (Poincaré–Birkhoff–Witt(PBW)). If (x1, . . . , xn) is an ordered basis of the C-vector space L. Then k1 kn (x1 . . . xn ) is a basis of the C-vector space U(L). In particular, the map L,→ U(L) is injective. Fact 7.13 (Universal property). Let A be an associative algebra. Define the Lie bracket on A by [a, a0] = aa0 − a0a. Then any Lie algebra homomorphism L → A has a unique extension to an algebra homomorphism U(L) → A. In particular, V ∞ is a U(G)-representation.

The intuition is that the Lie algebra contains “level 1” differential operators, while the universal enveloping algebra contains all level operators. Definition 7.14. A vector v ∈ V is K-finite if span{π(k)v | k ∈ K} is finite-dimensional. Let V fin be the space of K-finite vectors.

ﬁn G Then V depends only on ResK V . This is a unitary representation (because we assumed V is a Hilbert space). Theorem 7.15 (Peter–Weyl). Let K be a compact group.

(0) Any irreducible unitary representation of a compact group is finite-dimensional. (1) The K-finite vectors in the space of continuous functions on K are dense with respect to the sup norm, and (if K is a Lie group) they are automatically smooth, because they are precisely the matrix coefficients of irreducible representations (which are finite-dimensional by (0)). (2) If V is a unitary representation of K, then V is the Hilbert direct sum of irreducible representations. The space of K-finite vectors is the algebraic direct sum of these, hence dense.

In particular, V ﬁn is dense in V . How to produce a smooth K-ﬁnite vectors? 66 TASHO KALETHA

Given a continuous compactly supported function f : G → C, deﬁne π(f): V → V Z v 7→ f(g)π(g)vdg. G Fact 7.16.

(1) If f is smooth, so is π(f)v and Xπ(f)v = π(Xf)v. (2) If f is K-finite on the left, then π(f)v is K-finite. Lemma 7.17. The space V fin ∩ V ∞ is dense in V . In particular, V ∞ is dense.

Proof. Recall the Iwasawa decomposition G = BK = KB, Lemma 7.8. Let f1 : K → C be K-ﬁnite (and hence smooth), and f2 : B → C smooth and compactly-supported. Then f(kb) = f1(k)f2(b) is smooth and left K-ﬁnite. Given v ∈ V , ε > 0, consider the open subgroup {g ∈ G | |π(g)v − v| < ε}.

Choose f1, f2 such that f is supported inside it. Then |π(f)v − v| < and π(f)v is smooth and K-ﬁnite by Fact 7.16.

Admissible representations. Theorem 7.18. Let (π, V ) be an irreducible unitary representation of G. The multiplicity of τ ∈ Irr(K) in V fin is ≤ dim(τ). Definition 7.19. A representation (π, V ) of G is admissible if V fin is a direct sum of irreducible representations of K with finite multiplicities. Examples 7.20.

(1) Unitary representations are admissible by Theorem 7.18. (2) Parabolically induced representations are admissible. In fact, Casselman proved that any irreducible admissible representation is a subrepresentation of a parabolically induced one. We will work towards proving this theorem, see Theorem 7.41. Lemma 7.21. If (π, V ) is an admissible representation, then V ﬁn ⊆ V ∞ and V ﬁn is g- invariant.

Proof. For each τ ∈ Irr(K), (V fin∩V ∞)[τ] is dense in V [τ], but the latter is finite-dimensional, so (V fin ∩ V ∞)[τ] = V [τ]. Therefore, V [τ] consists of smooth vectors, and since M V fin = V [τ], we are done. We now show that V fin is stable under g. Taking v ∈ V fin, let W ⊆ V fin be finite-dimensional and K-stable, with v ∈ W . Observe that gW is finite-dimensional (but not necessarily a MATH 679: AUTOMORPHIC FORMS 67

representation of g, since g is not associative). For w ∈ W , X ∈ k = Lie(K)C, Y ∈ g, we have that XY w = Y Xw + [X,Y ]w ∈ gW. | {z } | {z } ∈gW ∈gW In particular, this shows that the vector Y w is K-ﬁnite.

By this lemma, we obtain a representation of g and of K on V ﬁn. However, note that V ﬁn is not invariant under G.

Deﬁnition 7.22. A(g,K)-module is a C-vector space which is a representation of g and K, subject to the conditions:

(1) differentiating the K-action gives the same action as the restriction of the g-action to k, (2) `Xv = (Ad(`)X)`v for ` ∈ K, X ∈ g. Definition 7.23. A(g,K)-module W is admissible if W is the direct sum of irreducible representations of K with finite multiplicity. These are also known as Harish-Chandra modules. Remark 7.24. If V is an admissible representation, then V fin is an admissible (g,K)-module. Definition 7.25.

(1)A G-representation V is irreducible if it has no non-trivial proper closed invariant subspaces. (2)A( g,K)-module is irreducible if it has no non-trivial proper subspaces.

The natural question to ask is: if a representation of G is irreducible, is the associated (g,K)-module irreducible, and vice versa. Definition 7.26. Two representations of G are called infinitesimally equivalent if their (g,K)-modules are isomorphic. Remark 7.27. If two representations are isomorphic, then they are clearly infinitesimally equivalent. The converse is actually not true in general. However, a deep theorem due to Harish-Chandra states that for irreducible unitary representations, these two notions are equivalent.

Matrix coefficients. Let (π, V ) be an admissible representation. Definition 7.28. For v ∈ V , η ∈ V ∨, define the matrix coefficient as

fv,η : G → C

fv,η(g) = hπ(g)v, ηi.

If v, η are K-finite, we call fv,η K-finite. Theorem 7.29. A K-finite matrix coefficient is a real analytic function.

Observe that for D ∈ U(g),

(3) Dfv,η = fπ(D)v,η. 68 TASHO KALETHA

ﬁn Corollary 7.30. If U0 ⊆ V is g-invariant, then U = U0 (the closure in V ) is G-invariant.

Proof. Since G acts by continuous automorphisms, it is enough to check that GU0 ⊆ U. Fix ⊥ ∨ v ∈ U0 and η ∈ U ⊆ V . For small X ∈ g, ∞ X n hπ(exp(X))v, ηi = X hπ(x)v, ηi n=0 x=1 ∞ X 1 = hXnv, ηi by equation (3) n! n=0 n = 0 as X v ∈ U0.

The matrix coeﬃcient fv,η is zero on exp(g) ⊆ G, an open neighborhood. By Theorem 7.29, this shows that fv,η = 0, i.e. hgv, ηi = 0. This shows that gv ∈ U. Corollary 7.31. Taking closures gives a bijection g-invariant subspaces G-invariant subspaces ↔ of V ﬁn of V

In particular, V ﬁn is irreducible if and only if V is irreducible.

Corollary 7.32. Let (π1,V1) and (π2,V2) be an admissible representation of G.

(1) If π1 and π2 are infinitesimally equivalent, then they have the same K-finite matrix coefficients. (2) If π1 and π2 are irreducible and share a K-finite matrix coefficient, then π1 is infinitesimally equivalent to π2.

∨ Proof. We ﬁrst prove (1). Let v ∈ V1 and η ∈ V1 be K-ﬁnite. By Theorem 7.29, fv,η is uniquely characterized by its derivatives at 1, which by equation (3) are given by

hπ1(D)v, ηi. Inﬁnitesimal equivalence preserves these.

For (2), assume πi are irreducible and

hπ1(g)v, ηi = hπ2(g)v2, η2i. ∞ Let V3 ⊆ C (G) be the subspace spanned by U(g)fv,η. Any v ∈ V1 is of the form π1(D)v1. Deﬁne ϕ: V1 → V3 by

ϕ(π1(D)v1) = D(fv1,η). This is bijective by irreducibility and one checks equivariance to complete the proof. This allows us to prove a version of Schur’s Lemma. Note that Schur’s Lemma normally only holds for ﬁnite-dimensional representations. Corollary 7.33 (Schur’s Lemma). Assume V is irreducible. Then any endomorphism of the g-module V ﬁn is a scalar. In particular, the center of the universal enveloping algebra Z(U(g)) acts by a character. MATH 679: AUTOMORPHIC FORMS 69

Remark 7.34. In general, the algebra Z(U(g)) is a polynomial algebra over C. This is a theorem due to Harish-Chandra, but we omit it here. In our case, it will be a 1-dimensional polynomial algebra and we will explicitly write the generator.

Proof. If ϕ commutes with g, it does so with k, so preserves K-types, which are ﬁnite- dimensional, so has an eigenvalue, and the eigenspace is g-invariant and non-trivial, so all of ﬁn V .

The Casimir element. Recall the Killing form on a Lie algebra L over C is the map κ: L × L → C (X,Y ) 7→ tr(ad(X), ad(Y )), where ad(X): L → L is the map Z 7→ [X,Z]. Then κ is L-invariant and symmetric. Theorem 7.35. The Lie algebra L is semisimple if and only if κ is non-degenerate.

Therefore:

(1) κ: L → L∗ is an isomorphism, ∼ (2) κ ∈ L∗ ⊗ L∗ →= L ⊗ L → U(L) 3 C.

The element C is called the Casimir operator. Fact 7.36. The Casimir operator is in the center, C ∈ Z(U(g)).

The differential equation satisfied by the matrix coefficients. Let (π, V ) be an irreducible admissible representation of G. Then C acts on V f in by a scalar λ ∈ C by Schur’s Lemma 7.33. By equation (3), we have that

(4) Cfv,η = λ · fv,η for K-ﬁnite v and η. Let us introduce the following coordinates: 1 0 0 1 0 0 H = ,E = ,F = a basis for g 0 −1 0 0 1 0

0 1 0 1 Y = ,V = . −1 0 1 0

Then 1 C = (H2 + V 2 − Y 2). 4 This computation is left as an exercise. Fact 7.37 (Cartan decomposition). Let t A− = h(t) = −1 0 < t ≤ 1 t 70 TASHO KALETHA

and <1 t A = h(t) = −1 0 < t < 1 . − t The multiplication map

K × A− × K → G is surjective, and <1 K × A− × K → G \ K is a double cover (local diﬀeomorphism of degree 2).

∼ 1 <1 In the coordinates on K = S given by angles φ1, φ2 and on A− given by t ∈ (0, 1), we have that ∂ 2 t2 + t−2 ∂ C = t − 2 · t ∂t t2 − t−2 ∂t ! 4 ∂ 2 ∂ 2 + 2 −2 2 φ1 + φ2 (t − t ) ∂φ1 ∂φ2 4(t2 + t−2) ∂ ∂ − 2 −2 2 φ1 φ2 . (t − t ) ∂φ1 ∂φ2

Assume v, η have pure K-type n (i.e. associated to the representation z 7→ zn of S1). Substitute z = t2. Then equation (4) gives the diﬀerential equation ! ∂ 2 (1 + z2)z 1 ∂ n2 λ (5) − − − − f(h(t)) = 0. ∂z 1 − z2 z ∂z z(1 + z)2 z2

<1 Here u(t) is the function on A− . In general,

iφ1 iφ2 in(φ1+φ2) fv,η(e h(t)e ) = e · fv,η(h(t)).

The diﬀerential equation (5) makes sense for z ∈ D ⊆ C, the unit disc. It is an ODE with regular singularities and there is a theory for solving such equations. √ 1 i Lemma 7.38. Let µi = 2 (1 + (−1) 4λ + 1) for i = 1, 2. There are holomorphic functions F1,F2 on D with F1(0) = F2(0) = 1 such that the space of solutions of (5) on D \ (−1, 0) is given by

µ1 µ2 (1) az F1(z) + bz F2(z) if µ1 − µ2 6∈ Z, µ1 µ2 (2) az F1(z) + bz log(z)F2(z) if µ1 − µ2 ∈ Z.

Thus

2µ1 2 4 2µ2 2 fv,η(h(t)) = t (a0 + a1t + a2t + ··· ) + t (b0 + b1t + ··· )

where a0 and b0 are not both 0. For v, η which are not of pure type, we still have this asymptotic expansion, but it may be true that a0 = b0 = 0. MATH 679: AUTOMORPHIC FORMS 71

Casselman submodule theorem. Recall that for a character χ: T → C×, G G 1/2 IB χ = IndB(χ ⊗ δB ), where we inﬂate the representation χ of T to a representation of B = U o T on the right hand side. Fact 7.39 (Frobenius reciprocity). For any (g,K)-module V, we have G ﬁn Hom(g,K)(V, (IB χ) ) = Hom(t,KT )(V/u ·V, C 1/2 ), χ+δB

where t = Lie(T ), u = Lie(U), and KT = K ∩ T = {±} = ZG.

The proof of this is essentially the same as Frobenius reciprocity. Theorem 7.40. If V = V fin is a finitely-generated admissible non-zero (g,K)-module, then V/uV is a finitely-generated non-zero (t,K)-module.

The assumption that V comes from an admissible representation V is unnecessary. We make it only because we did not introduce matrix coeﬃcients for general (g,K)-modules (which can be done).

Proof. Let v1, . . . , vn generate V. By admissibility, these are contained in a ﬁnite-dimensional K-invariant subspace. Add some more vectors to span this K-invariant C-vector space. Recall that 1 0 0 1 0 1 H = ,E = ,Y = 0 −1 + 0 0 −1 0

is a basis of g. Recall that H spans t, E+ spans u, Y spans k. α β γ Then PBW 7.12 gives E+H Y as a basis of U(g). Therefore, α β γ {E+H Y vi} α β γ are generators for the C-vector space V. If α > 0, the E+H Y vi = 0 in V/uV. On the other γ β hand, Y vi is a linear combination of v1, . . . , vn. Thus H vi span the C-vector space V/uV, so vi span the (t,KT )- module V/uV. To show that uV ( V, let us assume that V is irreducible. Let v ∈ V, η ∈ V∨. We have the asymptotic expansion of fv,η. We have 2 Ad(h(t))(E+)v = t E+v. Therefore,

fE+v,η(h(t)) = hh(t)E + v, ηi 2 = t hE+h(t)v, ηi 2 = −t hh(t)v, E+ηi 2 = −t fv,E+η(h(t))

So any matrix coeﬃcient for vector in uV goes to zero by t2 faster than any matrix coeﬃcient for vectors in V. Therefore, uV ( V. 72 TASHO KALETHA

The proof in fact shows that E+v is not of pure K-type even if v is. Otherwise, the matrix coeﬃcient fE+v,η would have the same asymptotic expansion as before. Theorem 7.41 (Casselman submodule theorem). Any irreducible G-representation V is G inﬁnitesimally equivalent to a subrepresentation of IB χ.

Proof. By Theorem 7.40, V/uV is a non-zero and finitely-generated (t,KT )-module. Sicne KT = Z/2Z, this is a direct sum of two t-modules. Any t-module is a U(t)-module and U(t) = C[X]. This has an irreducible quotient, which is C with X acting by a complex number s, i.e. V/uV → C(s,m) for s ∈ C, m ∈ Z/2Z, where C[X] acts as Xv = sv and Z/2Z acts as sgn(·)m. Then Frobenius reciprocity5.9 implies that fin G fin V → (IB χ) , s m where χ(t) = |t| sgn(t) .

G The structure of IB χ. Recall that G 1/2 IB χ = {f : G → C | f(tug) = χ(t)δB (t)f(g)} Suppose f is of pure K-type n. Recall the Iwasawa decomposition 7.8: G = B · K. Thus 1/2 inϕ f(g) = f(tuk(ϕ)) = χ(t)δB (t)e f(1).

Thus f ∈ C · ψn, where 1/2 inϕ ψn(tuk(ϕ)) = χ(t)δB (t)e . Therefore, G ﬁn M (IB χ) = C · ψn, n≡m (2) where χ(t) = |t|s sgn(t)m. 1 0 0 1 Recall that H = ,V = . Deﬁne P = H +iV , P = H −iV . Then P ,P ,Y 0 −1 1 0 + − + − are a basis of g. Lemma 7.42. We have that

P+ψn = (1 + s + n)ψn+2

P−ψn = (1 + s − n)ψn−2.

Proof. Check that 2iϕ Ad(k(ϕ))P+ = e P+ −2iϕ Ad(k(ϕ))P− = e P−.

Therefore, we know the K-type of P+ψn is n + 2, and hence

P+ψn = αψn+2, MATH 679: AUTOMORPHIC FORMS 73 and we can evaluate it at 1 to get

(P+ψn)(1) = αψn+2(1) = α.

Now, P+ = H + 2iE+ − iY . Compute:

Hψn(1) = s + 1,

E+ψn(1) = 0,

Y ψn(1) = in.

The same argument shows the statement for P−. G ﬁn s2−1 Corollary 7.43. Casimir acts on (IB χ) by the scalar 4 .

Proof. We have that Y 2 P P + P P C = − + + − − + 4 8 and a computation gives the result.

We have the following picture:

P+ P+ P+ P+ P+ P+ ··· • • • • • ···

P− P− P− P− P− P− Cψ−4 Cψ−2 Cψ0 Cψ2 Cψ4 and Lemma 7.42 tells us when P+ or P− can be zero. Corollary 7.44.

G (1) If s 6∈ m + 1 + 2Z, then IB χ is irreducible. (2) If s ∈ m + 1 + 2Z, then G + − (a) s = 0, then m = 1 and IB = I0 ⊕ I0 with the diagram 0 ··· • • • • ··· 0 −3 −1 1 3 and ± M I0 = ψ±k. k∈1+2Z≥0 (b) If s > 0, write n = s + 1 ∈ Z and n ≡ m (2). We have the diagram 0 6=0 ··· • • • ··· • • • ··· 6=0 0 − + where we have marked in red, a submodule Dn , and in green, a submodule Dn . We hence have a non-split exact sequence − + G 0 Dn ⊕ Dn IB χ Mn−2 0. where the quotient Mn−2 is marked above in blue. (c) If s < 0, write n = −s + 1 ∈ Z. we have the non-split exact sequence 74 TASHO KALETHA

G − + 0 Mn−2 IB χ Dn ⊕ Dn 0.

Now, we have the irreducible modules

G • IB χ, χ ↔ (s, m) and s 6∈ m + 1 + 2Z, + − • I0 ,I0 , + − •D n ,Dn for n = 2, 3, 4,..., •M n−2 for n = 2, 3,....

By consider K-types and Casimir eigenvalues, we see that the only possible isomorphisms G G −1 among these are IB χ → IB χ when these are irreducible. These are provided by the theory of intertwining operators. We have the following diagram showing the representations parameterized by s:

s ∈ C

+ − I0 ⊕ I0

+ − Mk−2, Dk ⊕ Dk

All the representations we marked above ((−1, 1)∪Z∪iR) are unitary. The representations on the imaginary axis are principal series, the representations on (0, 1) are called complementary ± series, and the representations Dk are the discrete series.

Intertwining operators. Consider χ(t) = |t|s sgn(t)m. Let

G L∞ : IB χ → C be given by Z L∞(f) = f(w · u)dw U for 0 1 w = . −1 0

G We write IB χ = V.

Lemma 7.45. The map L∞ is a (t,KT )-module homomorphism

V/uV → C −1 1/2 . χ ·δB MATH 679: AUTOMORPHIC FORMS 75

Proof. The map L∞ is U-invariant by construction, so it factors through V/uV. For t ∈ T , Z L∞(t · f) = f(w · u · t)du U Z = δB(t) f(w · t · u)du U Z −1 = δB(t) f(t wu)du U Z −1 −1/2 = χ(t) δB(t) δB(t) f(wu)du. U This gives the result.

By Frobenius reciprocity 7.39, we get a map G G −1 T : IB χ → IB χ .

By Iwasawa decomposition 7.8, restriction gives an isomorphism G 2 m IB χ → {f ∈ L (K) | f(k) = f(k), ∈ {±1} = ZG}. On the left hand side, the action is independent of s but the vector space depends on s. On the right hand side, one can check that it is the opposite: the vector space visible does not depend on s, but the action of G depends on s. Consider 2 L∞(s): L (K) → C as a function of s ∈ C. Proposition 7.46. There exists a meromorphic function α: C → C such that

Lf∞(s) = α(s)L∞(s) is always deﬁned and never 0.

Proof. Take ψn and apply L∞ to it. Note that 0 1 1 u √ 1 √−u −u 1 = 1+u2 √ 1+u2 · # # , 2 −1 −u −1 0 0 1 0 1 + u # # and hence √ n √ 0 1 1 u 2 −s−1 −u + i 2 −s−1−|n| |n| ψn = ( 1 + u ) √ = ( 1 + u ) (−u ± i) . −1 0 0 1 1 + u2

We integrate this function over u ∈ (−∞, ∞). Applying binomial theorem to (−u ± i)|n|, we get  Z ∞ √ 0 k odd, ( 1 + u2)−s−1−|n| · ukdu = −1 s+|n|+k k+1 s+|n|+1 −∞ Γ 2 Γ 2 Γ 2 otherwise.

This completes the proof. 76 TASHO KALETHA

Unitary and discrete series. Let (π, V ) be an irreducible representation. We have the asymptotic expansion of its matrix coeﬃcient, based on the eigenvalue λ ∈ C of the Casimir operator: 2µ1 2µ2 fv,η(h(t)) ∼ t , t √ 1 where µi = 2 (1 + ± 4λ + 1). G s m For IB (χ) where χ(t) = |t| sgn(t) , we checked in Corollary 7.43 that s2 − 1 λ = . 4 If π is unitary, then

•| fv,η(g)| ≤ |v| · |η| (so the matrix coeﬃcients are uniformly bounded), • λ ∈ R.

Thus s ∈ R or s ∈ iR.

1 1 (i) If λ ≤ − 4 (equivalently, s ∈ iR), then µ1, µ2 ∈ 2 + iR, so fv,η go to zero as fast as t. G At the same time IB χ is unitary, and the scalar product is:

G G −1 G 1/2 Haar ((·, ·)): IB χ × IB χ IB δB C. When χ is unitary, we have χ−1 = χ. 1 2µ1 2µ2 (ii) If − 4 ≤ λ ≤ 0, then s ∈ [−1, 1], and both t , t → 0, but one of them is fast. (iii) If λ > 0, then one of t2µi goes to 0 fast, but the other does not go to 0 (actually, usually explodes).

In the previous section, we constructed an intertwining operator G G −1 T : IB χ → IB χ .

The approach is as follows: for s ∈ R, define G G IB χ × IB χ → C, (f, g) 7→ ((f, T g)). This is a G-invariant Hermitian pairing, so it is a scalar product if and only if it is positive definite. For (ii), one can check that for s = 0, the map G G T : IB χ → IB χ is the identity. (Note that χ = χ−1 in this case). Thus ((f, g)) = ((f, T g)) G is a scalar product. Since IB χ is irreducible for s ∈ (0, 1), by continuity, we see that ((f, T g)) is still a scalar product. Definition 7.47. The representations in case (i) are called principal series and the ones in case (ii) are called complementary series. MATH 679: AUTOMORPHIC FORMS 77

G In case (iii), we see that IB χ cannot be unitary, unless the oﬀending exponent does not contribute. For s = k − 1 ∈ Z, the representation is reducible, and we have

G T − + IB χ/Mk−2 → Dk ⊕ Dk , =∼ or G − + T IB χ/Dk ⊕ Dk → Mk−2. =∼

+ − This ((f, T g)) gives a scalar product on Dk and Dk . One checks that

((ψn, T ψn)) = ((ψn, Lf∞(ψn)ψn)) = Lf∞(ψn) · ((ψn, ψn)) . | {z } =1

A computation of Lf∞(ψn) using Proposition 7.46 shows that these numbers are positive. ± Corollary 7.48. The matrix coeﬃcients of Dk are square-integrable. When k > 2, the matrix coeﬃcients are even integrable.

2µ1 2µ2 1 1 Proof. The dominant terms are t and t where µ1 = 2 k, µ2 = 1 − 2 k. Then, by boundedness, the term for µ2 must vanish. (Some more work is required for k = 2, but we skip it here for the same of time.)

k Thus fv,η(h(t)) ∼ t . We need the Haar measure in terms of the Cartan decomposition G = KAK: Z Z 2π Z 1 Z 2π 2 −2 dt f(g)dg = f(k(φ1)h(t)k(φ2))dφ1dφ2(t − t ) . G 0 0 0 t From this, we see that if f(h(t)) ∼ ta, then f is integrable if and only if a > 2. This completes the proof. G Fact 7.49. The matrix coeﬃcients for IB χ, χ unitary, lie in L2+(G) for any > 0.

1 Proof. We know that s ∈ iR. Then fv,η(h(t)) ∼ t , and we use the above formula to prove the statement. Deﬁnition 7.50. A representation of G is called

• discrete series if its matrix coeﬃcients lie in L2(G), • tempered if its matrix ceoﬃcients lie in L2+(G) for all > 0.

− + G In particular, Dk and Dk are the discrete series representations, and IB χ for χ unitary are + − tempered. The representations I0 , I0 are sometimes called limits of discrete series, even tough they are not actually discrete series 78 TASHO KALETHA

s ∈ C complementary series unitary principal series

limits of discrete series discrete series

± This is one reason to call Dk discrete series. There is another reason to call them this. They are the only irreducible representations that occurs are discrete constituents of L2(G). There is a verson of Plancharet’s Theorem that says that Z ⊕ L2(G) = πdπ where the measure dπ is supported on tempered representations and is called Plancharet measure. This is an analog of the decomposition Z ⊕ 2 ixy L (R) = Ce dy given by Fourier transforms and Fourier series. The discrete series are the only representations which are given non-zero measure with respect to this measure, in fact: ± vol(Dk , dπ) = k − 1 for ±k = 2, 3,....

Langlands correspondence. Let ΓC/R = Gal(C/R). We have the Weil group WC/R deﬁned as an extension:

× 1 C WC/R ΓC/R 1

generated by C× and j (lifting complex conjugation) subject to the relations jzj−1 = z, j2 = −1. Deﬁnition 7.51.

(1)A Langlands parameter is a continuous semisimple group homomorphism

WC/R → PGL2(C).

(2) Two such are equivalent if they are conjugate under PGL2(C). (3) A parameter ϕ is called discrete if any of the following equivalent conditions are satisﬁed: • the image is not contained in B up to conjugation, • it is an irreducible projective representation, • Centr(ϕ, PGL2(C)) is ﬁnite. (4) A parameter ϕ is tempered if its image is bounded. MATH 679: AUTOMORPHIC FORMS 79

Recall: the characters of C× are × × C → C z 7→ zazb for a, b ∈ C, a − b ∈ Z. If z = reiφ, zazb = ra+beiφ(a−b). Lemma 7.52. Any parameter is equivalent to one of the following: (zz)s 0 (−1)m 0 (1) z 7→ , j 7→ for s ∈ , m ∈ /2 , 0 1 0 1 C Z Z (z/z)a 0 0 1 (2) z 7→ , j 7→ for a ∈ 1 . 0 1 1 0 2 Z

Proof. Restrict ϕ to C×. The image is a connected Lie subgroup consisting of commuting diagonalizable matrices. (a) Suppose the image is trivial. Then j maps to an elements of order 2, and this is case 1 with s = 0. ∗ 0 (b) Suppose the image is non-trivial. After conjugating, the image is in ∼= × 0 ∗ C where the isomorphism is given by renormalizing to make the right bottom entry 1. Thus × × ϕ|C× : C → C zazb 0 z 7→ . 0 1 ∗ 0 Now, the image of j normalizes and hence it is one of the following 0 ∗ d 0 0 d (b1): , (b2): . 0 1 1 0 (b1) Then from jzj−1 = z, we see that a = b, and hence (zz)a 0 z 7→ , 0 1 and from j2 = −1, we get d2 = 1. Thus this is case (1). (b2) From jzj−1 = z, we get a = −b. Therefore, (z/z)d 0 z 7→ 0 1 0 d ∗ 0 0 1 and now is conjugate under to . We are hence in case (2). 1 0 0 ∗ 1 0

This completes the proof. Remark 7.53. In case (1), the only two equivalent representations correspond to s and −s. In case (2), a and −a are the only equivalent representations. Between cases (1) and (2), the only equivalences are between s = 0, m = 1 and a = 0. 80 TASHO KALETHA

We compute the centralizers. In case (1) but not s = 0, m = 1, the centralizer is connected. ±1 0 0 ∗ In case (2), a 6= 0, centralizer is . In case (2) when a = 0, the centralizer is , 0 1 ∗ 0 ∗ 0 . 0 ∗ Then discrete parameters are: case (2), a 6= 0. The tempered parameters are: discrete and case (1), s ∈ iR. We now describe the correspondence between irreducible admissible representations of G and representations of the Weil group.

Local Langlands correspondence for SL2(R). • In case (2), a 6= 0, + − ϕ ↔ {Dk , Dk } + − where k = 2d + 1. Note that {Dk , Dk } is in bijection with Irr(π0(Centr(ϕ))). • In case (2), a = 0, + − ϕ ↔ {I0 ,I0 } + − and again {I0 ,I0 } is in bijection with Irr(π0(Centr(ϕ))). • In case (1), not (s = 0, m = 1), we have × ϕ: WC/R → C ,→ PGL2(C). Moreover, W ab ∼= × via the local Artin map C/R R z 7→ zz, j 7→ −1. Therefore, ϕ corresponds to χ: R× → C× with χ(t) = |t|s sgn(t)m. In this case, G ϕ ↔ IB χ when the right hand side is irreducible, and

ϕ ↔ Mk−2 otherwise.

Formal degree. Let G = SL2(R) or SL2(Qp). Let (π, V ) be a discrete series representation of G. We have not discussed what this means in the latter case, but once we do the following discussion will apply.

∨ ∨ Lemma 7.54. There is a unique real number deg(π, dg) such that for all v1, v2 ∈ V , v1 , v2 ∈ V ∨, Z ∨ −1 ∨ ∨ ∨ −1 hgv1, v1 ihg v2, v2 idg = hv1, v2 ihv2, v1 i deg(π, dg) . G Deﬁnition 7.55. The constant deg(π, dg) in Lemma 7.54 is called the formal degree.

± Exercise. Show that deg(Dk , dg) = cdg(k − 1), where cdg is a constant dependent only on the measure dg. MATH 679: AUTOMORPHIC FORMS 81

Proof of Lemma 7.54. The representation V ⊗ V ∨ of G × G is irreducible. We deﬁne two G × G-equivariant pairings on V ⊗ V ∨:

(−, −)1 = LHS,

(−, −)2 = RHS without deg . By irreducibility, these are scalar multiples of each other. We show that this scalar is nonzero and, in fact, belongs to R>0. 2 For this, recall that π is unitary (embed into L (G) via a matrix coeﬃcient). Let (−, −)h be ∨ ∨ an invariant scalar product on V . Choose 0 6= v ∈ V and let v = (−, v)h ∈ V . Then Z ∨ ∨ 2 (v ⊗ v , v ⊗ v )1 = |(gv, v)h| dg > 0, G ∨ ∨ 2 (v ⊗ v , v ⊗ v )2 = |(v, v)h| > 0, showing that the scalar is a positive real number.

8. Automorphic representations of SL2(Qp)

∼ × ∼ Basic structure of SL2(Qp). Let G = SL2(Qp). We have groups T = Qp , U = Qp and B = TU, as before. However, in this case, a maximal compact is K0 = SL2(Zp), which is non-abelian. S Lemma 8.1 (Cartan decomposition). The group G decomposes as G = K0λn(p)K0, a n≥0 disjoint union, where pn 0 λ (p) = . n 0 p−n Furthermore, we have the integral formula Z Z Z X 2n f(g)dg = p f(k1λn(p)k2)dk1dk2 G n≥0 K K

Proof. Elementary divisors. t 0 Note that this is the analog of G = KAK where A− = −1 0 < t ≤ 1 , but we may 0 t get rid of the units here and the group A− is now discrete.

Remark 8.2. The proof by elementary divisors works for G = SL2(Qp). For other Gs, one needs to develop a general theory, Bruhat–Tits theory.

Lemma 8.3 (Iwasawa decomposition). We have a decomposition G = TUK0 and we have the integral formula Z Z Z Z f(g)dg = f(tuk)dtdudk. G T U K0

2 Proof. For g ∈ G, look where ghe1i ⊆ Qp goes. We can return this line to he1i via k ∈ K0. But B is the stabilizer of he1i. 82 TASHO KALETHA

Lemma 8.4 (Bruhat decomposition). We have a decomposition G = B ∪ BwB for w = 0 1 . −1 0

Smooth and admissible representations. Deﬁnition 8.5. Let (π, V ) be a representation of G on a complex vector space. (1) A vector v ∈ V is called smooth if Stab(v) ⊆ G contains a compact open subgroup. (2) The representation (π, V ) is called smooth if all of its vectors are smooth. (3) The representation (π, V ) is called admissible if it is smooth and for each compact K open subgroup (c.o.s.) K ⊆ G, dimC V < ∞. Remark 8.6. If (π, V ) is any representation of G, then V ∞ = S V K is the space of smooth K⊆G c.o.s. vectors. Unlike over R where V ∞ was only a representation of the Lie algebra g, V ∞ is still a G-representation. The Lie algebra in this case plays a diﬀerent role. Theorem 8.7 (Harish-Chandra). If (π, V ) is an irreducible unitary representation of G, then V ∞ is dense in V and an admissible representation.

This result is diﬃcult and we will not attempt to prove it. Remark 8.8. In fact, if π is smooth and irreducible, then π is admissible. We will almost prove this, but skip enough results that we will not present a complete prove. Lemma 8.9. The only irreducible smooth ﬁnite-dimensional representation of G is the trivial representation.

Proof. The intersection of the stabilizers of a basis is the kernel of the representation and contains a compact open subgroup. Since the kernel is normal and bigger than the center, it equals all of G. Lemma 8.10. (1) Any ﬁnitely generated smooth representation has an irreducible quotient. (2) Any smooth representation has an irreducible subquotient.

Proof. Note that (2) follows from (1). For (1), consider the set of all proper subrepresenta- tions , ordered by inclusion. By ﬁnite generation assumption, this set has the tower property. By Zorn’s Lemma, this set has a maximal element. Quotienting by this maximal element gives an irreducible quotient. × Example 8.11. A smooth ﬁnite-dimensional representation of Qp need not be reducible (i.e. a direct sum of irreducible representations).

2 × Consider V = C with Qp acting via × ord Qp → Z → GL2(C), 1 1 1 7→ . 0 1 This is clearly smooth, but is clearly not reducible. MATH 679: AUTOMORPHIC FORMS 83

× Fact 8.12. Let V be a smooth n-dimensional representation of Qp and let (χ1, . . . , χn) be the irreducible subquotients of a Jordan-H¨olderseries. If χi are pairwise distinct, then M V = C(χi).

Proof. Exercise.

Fact 8.13. A smooth ﬁnite-dimensional representation of Qp is the direct sum of irreducibles.

−n −n Proof. The group Qp is exhausted by p Zp, compact opens. The restriction V |p Zp de- −n composes as a sum of characters of p Zp, which are compatible with n. This completes the proof.

Induced representations. From now on, (π, V ) is always assumed to be admissible. For χ: T → C× smooth character, the parabolic induction of χ is G 1/2 IB χ = {f : G → C | f(tng) = χ(t)δB(t) f(g)},

where f is right-invariant under some compact open subgroup Kf . One can show that in 2 this case δB(t) = |t| . G Clearly, IB χ is a smooth representation by the restriction we made on f. G Lemma 8.14. The parabolic induction IB χ is ﬁnitely-generated and admissible.

We skip the proof of this lemma for the sake of time.

G Just like over R, if χ is unitary, so is IB χ. The same argument as we presented for R works G in this case. However, note that IB χ is not complete. G Deﬁnition 8.15. Given a representation (π, V ) of G, we deﬁne its Jacquet module RBπ as the representation of T on VU = V/V (U), the coinvariance, where V (U) = hv = π(u)v | v ∈ V, u ∈ Ui, −1/2 and the natural T -action is multiplied by δB .

−1/2 We twist the action by δB for convenience of notation. We did not do this for SL2(R) and −1/2 so a lot of the formulas involved δB . Fact 8.16 (Frobenius reciprocity). For any representation (π, V ) of G, we have G ∼ G HomG(π, IB χ) = HomT (RBπ, χ). Lemma 8.17. We have Z

V (U) = v ∈ V there exists a c.o.s. U0 ⊆ Usuch that π(u)vdu = 0 . U0

Corollary 8.18. The functor V 7→ VU on the category of smooth B-representations is exact.

Proof. Right-exactness is immediate (taking coinvariance is always right exact). Left exactness follows from Lemma 8.17. G Corollary 8.19. The functor RB is exact. 84 TASHO KALETHA

We ﬁrst prove a lemma. We will use the notation: K eK : V → V 1 Z v 7→ π(k)vdk. vol(K) K Lemma 8.20. For a representation (π, V ) of G, the following are equivalent:

G (1) RBπ = 0, (2) for any compact open subgroup K ⊆ G and any vector v ∈ V , the function g 7→ eK π(g)v is compactly supported, (3) any matrix coeﬃcient of π is compactly supported.

Proof. We ﬁrst show that (2) is equivalent to (3). Assume (2) and show (3). Given v ∈ V , η ∈ V ∨, choose K ⊆ G c.o.s. such that v ∈ V K , η ∈ (V ∨)K (one exists by smoothness). Then

fv,η(g) = hgv, ηi

= hgv, eK ηi

= heK gv, ηi which is compactly-supported since eK gv is. Conversely, given v ∈ V , choose a basis ∨ K η1, . . . , ηn for (V ) . Then eK gv = 0 if and only if heK gv, ηii = 0 for all i, and hence [ supp(eK gv) ⊆ supp(fv,ηi ) i and the latter is compact. This shows that (2) is equivalent to (3).

Let us now show that (2) implies (1). Let v ∈ V and K be a congruence subgroup of SL2(Zp) p 0 such that v ∈ V K . Let t = λ (p) = . Then, by (2), e tmv = 0 for suﬃciently large 1 0 p−1 K m. Then m m eK t v = t et−mKtm v = 0 for m 0. Write U for the lower-triangular matrices with 1 on the diagonal. Then K = (K ∩ U) · (K ∩ T ) · (K ∩ U), | {z } | {z } | {z } Ku KT KU see Remark 8.21. Now,

0 = e −m m v = e −m m e −m m e −m m v = e −m m v. t Kt t KU t t KT t t KU t t KU t By Lemma 8.17, v ∈ V (U). We just have to prove (1) implies (3), which is similar to the argument for (2) implies ∨ (1). Suppose v ∈ V , η ∈ V are such that fv,η is not compactly supported. By Cartan decopmosition, the support of fv,η meets infinitely many K0 double cosets. Choose K ⊆ K0, K ∨ K compact open, such that v ∈ V , η ∈ (V ) . Since K0/K is finite, there exists k1, k2 ∈ K0 such that the support of fv,η meets m kiKt Kk2 −1 for infinitely many m. Replace v by k2v and η by k1 η to remove k1, k2 from the notation. MATH 679: AUTOMORPHIC FORMS 85

Now, m 0 6= fv,η(t ) m = fv,eK∩U η(t ) m = ht v, eK∩U ηi m = heK∩U t v, ηi m = ht et−mK∩Utm v, ηi. −m Therefore, et−mK∩Utm v 6= 0 for infinitely many m > 0. Since t acts expanding on K ∩ U, we see that eKU v 6= 0 for any compact open KU ⊆ U. By Lemma 8.17, this shows that v 6∈ V (U), so the image of v in VU is non-zero, so VU 6= 0. Remark 8.21. We skipped a few details. First, we did not define the dual of an admissible representations: it is defined as (V ∨)∞ (first, take the algebraic dual, and then take the smooth vectors). Second, we did not prove that the dual of an admissible representation is admissible. This is true and elementary, but we skipped it for the sake of time. Remark 8.22. An compact open subgroup K ⊆ G has a Iwahori decomposition if and only if the multiplication map (K ∩ U) × (K ∩ T ) × (K ∩ U) → K is bijective. The fact we used in the proof above is that congruence subgroups have Iwahori decomposition. Moreover, the element t = λ1(p), acting by conjugating, fixes K ∩ T , is contracting on K ∩ U, and is expanding on K ∩ U. Definition 8.23. A representation (π, V ) that satisfies the equivalent conditions of Lemma 8.20 is called supercuspidal.

Remark 8.24. Note that there are no supercuspidal representations of GL2(R). Indeed, the matrix coeﬃcients satisfy a diﬀerential equation and cannot be compactly supported unless they are 0. Lemma 8.25. An irreducible supercuspidal representation is both injective and projective in the category of smooth representations.

Proof. The two statements are equivalent under duality.3 So we prove projectivity. Let (π, V ) be an irreducible supercuspidal representation. Then, in particular, it is discrete 4 ∨ ∨ series. Choose v0 ∈ V , v0 ∈ V such that ∨ hv0, v0 i = deg(π, dg). Let (σ, W ) be a representation with a surjective map f :(σ, W ) → (π, V ).

Choose a preimage w0 ∈ W of v0. Deﬁne h:(π, V ) → (σ, W )

3The dual of a supercuspidal representation is supercuspidal, because their matrix coefficients are the same. 4Recall that discrete series are representations whose matrix coefficients are square integrable. This is definitely true if they are compactly supported. 86 TASHO KALETHA

by Z −1 ∨ h(v) = hπ(g) v, v0 iσ(g)w0dg. G This integral clearly converges. For any v∨ ∈ V ∨, Z ∨ −1 ∨ ∨ hf(h(v)), v i = hg v, v0 ihgv0, v idg G ∨ ∨ −1 = hv0, v0 ihv, v i deg(π, dg) by Lemma 7.54 = hv, v∨i.

Therefore, f(h(v)) = v, and hence h is a splitting of f. G Corollary 8.26. No irreducible subquotient of IB χ is supercuspidal.

G Proof. Suppose there is V1 ( V2 ⊆ IB χ such that V1/V1 is irreducible and supercuspidal. By Lemma 8.25, the map V2 → V2/V1 splits as G V2/V1 ,→ V@ ,→ IB χ. By Frobenius reciprocity, we get a non-zero map G RB(V2/V1) → C(χ), G and hence RB(V2/V1) 6= 0, which is a contradiction. Lemma 8.27. There is a short exact sequence of T -modules

−1 G G 0 χ RBIB χ χ 0

and it splits if and only if χ 6= 1.

G G G Proof. The representation RBIB χ only depends on the restriction of IB χ to B. So we are looking at B-double costs of G. Recall that Bruhat decomposition . G = B ∪ BwB G where B is closed and BwB is open and dense. Let V0 ⊆ IB χ be the subspace of functions supported on BwB. It is a B-submodule. The functions in V0 are precisely those f such that f(1) = 0. Thus we have a short exact sequence of B-modules:

G f7→f(1) 1/2 0 V0 IB χ C(χ · δB ) 0.

Since the Jacquet module is exact, we get a short exact sequence

G G 0 V0/V0(U) RBIB χ C(χ) 0.

We claim that for any f ∈ V0, there exists a compact open subgroup U1 ⊆ U such that f is supported on BwU1. Since f is locally constant, there exists a compact open subgroup U ⊆ U such that f| ≡ 0. 1 U1 MATH 679: AUTOMORPHIC FORMS 87

Now, a matrix computation shows that: 1 0 1 x−1 −x−1 0 1 x−1 (6) = w . x 1 0 1 0 −x 0 1 1 0 This shows that if U is described by such that ord(x) ≥ k, then we may take U to 1 x 1 1 1 x be for ord(x) > −k. This proves the claim. 0 1 Therefore, the integral Z f(w · u)du U

converges for f ∈ V0. Deﬁne Z L∞ : V0 → C,L∞(f) = f(w · u)du. U Then we get a map of T -modules:

−1 1/2 L∞ : V0/V0(U) → C(χ δB ). It is clearly surjective (for a good choice of f, we can show that the integral is non-zero). By Lemma 8.17, this map is an isomorphism. Therefore, the exact sequence above gives

−1 G G 0 C(χ ) RBIB χ C(χ) 0

∼ −1 1/2 via the isomorphism V0/V0(U) = C(χ δB ). Now, we come to the splitting. By Fact 8.12, if χ 6= χ−1, the sequence splits.

−1 G Now suppose χ 6= χ , χ 6= 1. We try to extend L∞ to all of IB χ. There is a convergence s/2 issue, but we can use analytic continuation to overcome it. Let χs = χ · δB for s ∈ C. Consider G −1 1/2 L∞(s): IB χs → C(χs δB ). G Note that IB χs is independent of s via restriction ot K. We are looking at Z 1 u f w du. 0 1 Qp

We split this as two integrals according to Qp = Zp ∪ Qp \ Zp. The integral over Zp converges so we only deal with the other part. Now,6 shows that 1 u 1 −u−1 −u−1 0 1 0 w = . 0 1 0 1 0 −u u−1 1

−1 The integral over Qp \ Zp via the substitution u 7→ u becomes Z 1 0 χ (−u)δ1/2(−u)|u|−2f du. s B u 1 u∈pZp 88 TASHO KALETHA

The convergence problem is not with f but with the characters in the integrand. There is N such that

f   1 0  N  p Zp 1 is constant. The problematic part is (after substituting u 7→ −u) Z Z −1 X n × χs(u) |u| du = χs(p ) χ(u)d u. × u∈Zp N | {z× } n≥N p Zp d u If χ is ramiﬁed, this integral is 0. In this case, the analytic continuation comes for free.5 When χ is unramiﬁed, we get that the sum is equal to 1 vol( ×, d×u) = . Zp 1 − χ(p)p−s This has a pole at s = 0 if χ is trivial and no pole at s = 0 if χ 6= 1. This shows the splitting when χ 6= 1. Dealing with the case χ = 1 would take a whole other lecture, so we will skip it. We already 1 ∗ know that T acts on RGχ as and one has to compute ∗ explicitly. B 0 1

G G Lemma 8.28. Let σ be a subquotient of IB χ. Then σ is a submodule of either IB χ or G −1 IB χ .

G G G Proof. Take V1 ⊆ V2 ⊆ IB χ such that σ1 is V2/V1. Applying RB, we get RBσ as a subquotient G G G G −1 G G of RBIB χ. By Corollary 8.26, RBχ 6= 0, so by Lemma 8.27, RBσ is one of χ, χ or RBIB χ. Applying Frobenius reciprocity completes the proof.

G Recall that (π, V ) is supercuspidal if and only if RBπ = 0. Therefore, π is not supercusp- G idal, then RBπ 6= 0 and is ﬁnitely generated (as a representation of T ), and hence has an G irreducible quotient which is a character. Frobenius reciprocity then shows that π ,→ IB χ. G To describe all non-supercuspidal representations, we need to study the structure of IB χ. G Lemma 8.29. The representation IB χ has length 1 or 2.

G G G Proof. We claim that if V1 ( V2 ⊆ IB χ, then RBV1 ( RBV2.

The representation V2/V1 has an irreducible subquotient by Lemma 8.10 (2), i.e. we have

V1 ⊆ V3 ( V4 ⊆ V2 and V4/V3 is irreducible. By Corollary 8.26, is not supercuspidal. Thus, by exactness of Jacquet functor (Corollary 8.19), G G G G RBV1 ⊆ RBV3 ( RBV4 ⊆ RBV2.

Assume by contradiction that G V1 ( V2 ( IB χ.

5Note that this was also the case in Tate’s thesis. MATH 679: AUTOMORPHIC FORMS 89

By the claim we proved above, G G G RBV1 ( RBV2 ( RBIB χ, contradicting Lemma 8.27. G −1 Corollary 8.30. If χ is unitary, then IB χ is irreducible unless χ = χ and χ 6= 1, in which G case IB χ is the direct sum of two inequivalent irreducible representations.

G Proof. Since χ is unitary, so is IB χ, so composition series splits, and hence it is enough to compute G dim EndG(IB χ). G G G −1 But EndG(IB χ) = HomT (RBIB χ, χ) had dimension 1 unless χ = χ and χ 6= 1 (by Lemma 8.27). We now have to deal with the case where χ is not unitary. Lemma 8.31. The following are equivalent:

G (1) the G-module IB χ has a one-dimensional submodule, G (2) the U-module IB χ has a one-dimensional submodule, −1/2 (3) χ = δB .

−1/2 Proof. It is clear that (1) implies (2). To prove that (3) implies (1), note that when χ = δB , G then IB χ is the set of smooth functions of B\G, and we have a 1-dimensional submodule of constant functions. We need to prove that (2) implies (3). Assume there is a 1-dimensional U-submodule spanned G × by f ∈ IB χ. Thus there is a character ψ : U → C such that 0 1/2 0 (7) f(tugu ) = δB (t)χ(t)f(g)ψ(u ). In particular, the support of f is a union of (B,U)-double cosets. By Bruhat decomposition 8.4, the options are: (i) B, (ii) BwU, (iii) G. Clearly, (i) is not possible, because f is locally constant. Also, (ii) is no possible, because otherwise f(1) = 0, and by the proof of Lemma 8.27 we know that f is supported on BwU1 for some compact open U1 ⊆ U, and this contradicts7. Therefore, (iii) is true, and in particular f(1) 6= 0. Now, f(1) = f(u) = f(1) · ψ(u) and hence ψ ≡ 1. By the matrix computation6 in the proof of Lemma 8.27, we see that 1 0 f(1) = f = χ(−x−1)δ1/2(−x−1)f(w) x 1 B

−1 1/2 −1 for suﬃciently small x. So the value χ(−x )δB (−x ) is constant for small x, hence −1/2 χ = δB . 90 TASHO KALETHA

Corollary 8.32. Assume χ is not unitary.

±1/2 G (1) If χ 6= δB , then IB χ is irreducible. −1/2 (2) If χ = δB , then we have the non-split exact sequence G 0 C IB χ St 0. 1/2 (3) If χ = δB , we have the non-split exact sequence G 0 St IB χ C 0. Deﬁnition 8.33. The representation St in Corollary 8.32 is the Steinberg representation.

G G Proof. Assume IB χ is irreducible. Let V0 ⊆ IB χ be the C-submodule {f | f(1) = 0}. 6 Fact. The module V0(U) is irreducible as a B-module. G The composition series of IB χ as a B-module has length 3 with two 1-dimensional subquo- G tients and one infinite-dimensional. Assume IB χ has a finite-dimensional B-submodule, so a −1/2 finite-dimensional U-submodule, and hence a 1-dimensional one. By Lemma 8.31, χ = δB . G Since the length of IB χ is 2 by Lemma 8.29, we are done. G G ∨ If IB χ does not have a finite-dimensional B-submodule, then (IB χ) does, so we are almost done by dualizing. We just need to show that the two representations St are isomorphic. This follows from Lemma 8.34 below. Lemma 8.34. Suppose χ 6= χ−1. Then there is a non-trivial intertwining operator G G −1 IB χ → IB χ . G In particular, if IB χ is irreducible, then this is an isomorphism.

G G −1 Proof. Apply Lemma 8.27 to get RBIB χ → χ and Frobenius reciprocity to get the result. Remark 8.35. So far, we have all the non-supercuspidal irreducible representations:

(1) the trivial representation C, (2) the Steinberg St, G 2 (3) the two distinct constituents of IB χ when χ = 1, χ 6= 1, G ∼ G −1 (4) the irreducible IB χ = IB χ . There are no isomorphisms between them by looking at the Jacquet modules.

Unitarity, temperedness, and discreteness. We list the properties of the representations listed above. Most of these things followed from what we discussed above, but we point out when something needs a proof.

G • We know that IB χ is unitary if χ is. 6This fact is proved using the theory of Kirillov models, which we will not discuss. It takes a while to set up and it is something that is only useful for GL(n), but does not generalize to other reductive groups. It a b uses the mirabolic subgroup and the reason for the name is that such subgroups only exist for GL(n). 0 1 MATH 679: AUTOMORPHIC FORMS 91

s/2 • We have complementary series for χ = δB for −1 < s < 1, unitary. The proof is the same as in the real case. Theorem 8.36 (Casselman’s pairing). For (π, V ) irreducible, admissible representation, there is a non-degenerate pairing G G ∨ h | i: RBπ ⊗ RBπ → C satisfying the following condition: given v ∈ V , v∨ ∈ V ∨, there is > 0 such that for × t ∈ T = Qp , |t| < , ∨ 1/2 G ∨ hπ(t)v, v i = δB (t)hRBπ(t)vu, vu iB ∨ ∨ where vu ∈ Vu, vu ∈ VU are the images.

The proof of this theorem would take a few lectures, so we will omit it here. This theorem reduces the study of matrix coeﬃcients to the Jacquet module, which has a simple description. We apply this together with Lemma 8.27 to see that:

G • The matrix coeﬃcient of IB χ has the form 1/2 1/2 −1 a · δB χ(t) + bδB (t)χ (t). These are bounded as t → 0 if and only if |χ(t)| = |t|s with −1 < Re(s) < 1. G • The representation IB χ is tempered if and only if s ∈ iR, i.e. χ is unitary. G 1/2 2 • For Steinberg, RB St = CδB , so the matrix coeﬃcients are δB(t) = |t| which are not integrable, but square-integrable. This is hence a discrete series representation. Fact 8.37. Steinberg is the unique discrete series representation of G that is not supercuspidal.

Here is the analogy between the real and the p-adic case: real case p-adic case non-unitary: G IB χ and χ not unitary (same) unitary non-tempered: composition series (same) tempered but not discrete G 2 IB χ with χ unitary or its two pieces when χ = 1, χ 6= 1 (same) discrete: discrete: + − Dk ⊕ Dk Steinberg St one for each ﬁnite-dimensional representation (same) supercupidal: supercuspidal: none they exist!

We will see that supercuspidal representations exist in the p-adic case next time. One can think of the discrete series together with supercuspidal case as one case, in which case the analogy does not break down. This can be seen using Harish–Chandra parameters, which we deﬁne below (Deﬁnition 8.41). 92 TASHO KALETHA

Hecke algebra. The Hecke algebra is the locally constant functions on G: ∞ H = H(G) = Cc (G), which is an algebra under convolutions: Z (f ∗ g)(x) = f(xy−1)g(y)dy G where we normalize the measure so that vol(K0, dy) = 1. Then H is an associative, non-commutative algebra with no unit. For a compact open subgroup K ⊆ G, consider −1 eK = 1K · vol(K, dy) .

Then eK ∗ eK = eK and we deﬁne

HK = eK ∗ H ∗ eK , which is an associative algebra, with unit eK . In fact, it is the subspace of K-biinvariant functions in H and [ H = HK . K

Given a smooth G-representation (π, V ), we promote it to an H-module by deﬁning the action as Z π(f)v = f(g)π(g)vdg. G K Moreover, V becomes a HK -module this way. Proposition 8.38. The functor V 7→ V K gives a bijection between isomorphism classes of K irreducible representations with V 6= 0 and isomorphism classes of irreducible HK -module.

Deﬁnition 8.39. The distribution Θπ(f) = tr(π(f)) is the Harish–Chandra distribution character of π.

Theorem 8.40. There is a (unique) locally integrable function θπ : G → C such that Z Θπ(f) = f(g)θπ(g)dg. G

Deﬁnition 8.41. The Harish-Chandra character is θπ.

These can be computed explicitly for all the representations we discussed.

Guest lecture by Stephen DeBacker (4/3/19): supercuspidal representations. We start with some motivation. ˆ (1) We know that for ﬁnite groups, the set of characters of representations {θπ | π ∈ G} is supposed to separate conjugacy classes. reg For p-adic groups, the Harish–Chandra character θπ is a function on G for any ˆ admissible π, so we kind of expect that in the p-adic setting {θπ : π ∈ G} should separate regular semisimple orbits. MATH 679: AUTOMORPHIC FORMS 93

(2) What is Greg? In general,

reg 0 G = {γ ∈ G | CG(γ) is a torus} = {γ ∈ G | DG(γ) 6= 0},

where DG(γ) is given by

··· ` detg(t − (Ad(γ) − 1)) = t + DG(γ)t where ` is the semisimple rank of G. reg For GL2 or SL2, γ ∈ G if and only if γ has distinct eigenvalues. In this case, 0 2 0 DG(γ) = (λ − λ ) where λ and λ are the distinct eigenvalues of γ. Explicitly: a b D = (a + d)2 − 4. G c d (3) Let

T = {S ⊆ G | S maximal Qp-torus}.

We write k = Qp. What do elements of T look like? They could be a 0 • Split — tori which are G-conjugate to A = a, d ∈ k, ad = 1 . 0 d −1 a bη • Elliptic — tori of the form T θ,η = a2 − b2θ = 1 for θ ∈ k× \ bθη a × 2 × × 2 (k ) , θ ∈ {, ω, ω} where ∈ (Zp ) \ ((Zp) ) and ω is the uniformizer of k. The element η ∈ k× can be arbitrary for now, but we will next write down all the conjugacy classes. We have that

|NG(A)/A| = 2 and ( √ 1 −1 6∈ N(k( θ)), |N (T θ,η)/T θ,η| = √ G 2 −1 ∈ N(k( θ)).

Elliptic tori in SL2(k) up tp G-conjugation: for f = Fp, the residue ﬁeld of k, – if −1 ∈ (f×)2: T ,1,T ,ω,T ω,1,T ω,, T ω,1, T ω,, – if −1 6∈ (f×)2: T ,1, T ,ω, T ω,1, T ω,1. Fact 8.42. Fix a maximal k-torus T and H ≤ G open subgroup. The map H × (T ∩ Greg) → Greg (h, t) 7→ ht = hth−1 is a submersion. In particular, H (T ∩ Greg) is open. Corollary 8.43. We have that [ Greg = H (T ∩ Greg) T ∈T /G-conj. is a disjoint union of opens. 94 TASHO KALETHA

Thus, in analogy with the situation for ﬁnite groups of Lie type7, we expect irreducible representations of G to be, roughly, parameterized by pairs (T, ψ), where T is a maximal k-torus and ψ ∈ Tˆ.

The goal will be to make this formal in order to write down supercupsidal repreentations. We already saw that if χ ∈ Aˆ, then we can extend it toχ ˜ ∈ Bˆ where a 0 1 x a 0 χ = χ . e 0 d 0 1 0 d

G 2 2 We then have IB (χ) and if χ 6= 1, this is irreducible, and if χ = 1, we need to do some work and pin down the irreducible components. Basically, we have that a 0 a 0 a−1 0 θIGχ −1 = (constants) · χ −1 + χ . B 0 a 0 a 0 a What about the Harish–Chandra parameters for non-split groups? G Recall Mackey theory which says that for σ ﬁnite-dimensional, IndH σ is irreducible if and only if

(1) σ ∈ Hˆ ,

(2) if g ∈ G intertwines σ with itself (i.e. hχσ, χg∗σig−1 H∩H 6= 0), then g ∈ H. The idea is that for an elliptic maximal torus T , we have a map B(T ) → B(G). Here, B(−) is the building of a group. Since T is elliptic B(T ) is a point. Since B(G) is an inﬁnite tree, each vertex has valence p + 1. Although we assume p 6= 2, we can draw the tree of B(SL2(Q2)) to illustrate what is looks like. It is an inﬁnite tree with valency 3 so it looks as follows:

x y C

7For ﬁnite groups of Lie type Deligne–Lusztig theory is the correct approach. The relevant representations G are ±RT,ψ, classiﬁed by a maximal torus T and a character χ. MATH 679: AUTOMORPHIC FORMS 95

The group G acts on the tree. For any part of the tree P ⊆ B(G), StabG(P ) is a compact open subgroup of G. It is not too much of a lie to say that interesting compact open subgroups of G arise as StabG(P ). Pick any vertices x, y on the tree and C be the edge connecting them (as in the picture above). Then C is a fundamental domain for the action of G on B(G). We set things up so that

StabSL2(k)(x) = SL2(R) = K0, where R = Zp is the ring of integers of k, and π StabSL2(k)(y) = K0 = Q0, 0 1 where π = , and ω 0

StabSL2(k)(C) = K0 ∩ Q0 = I, the Iwahori subgroup. Then B(T ,1) = x, B(T ,ω) = y and all the other elliptic maximal tori give points in C. We have the Moy-Prasad ﬁltration

Gx,0 = K0 = SL2(R) a b Gx,0+ = Gx,1 = K1 = a, b, c, d ∈ 1 + p c d a b 2 Gx,1+ = Gx,2 = a, b, c, d ∈ 1 + p c d . .

In general, Gx,r = Gx,dre for r > 0. Roughly, this ﬁltration is obtained by looking at

StabSL2(k)(Cr) where the region Cr of the graph is a circle of radius r centered at x. To get the ﬁltration for y, we conjugate by the matrix π. For any maximal elliptic torus

z ∈ C, we have a similar ﬁltration Gz,0 ⊇ Gz,0+ = Gz, 1 ⊇ G 1 + and so on. One can write 2 z,( 2 ) it down explicitly, but we omit this here.

Construction of supercuspidal representations. Let T be a maximal elliptic torus in SL2(k). This gives a point x ∈ B(G). Let T0 be the connected component of the identity, which lives between T0+ and T . Deﬁne

Tr = T0 ∩ Gx,r. Then √ ∼ 1 0 T0 = Norm (k( θ)) (the connected component√ of identity of norm 1 elements). The ﬁltration Tr corresponds to the obvious ﬁltration of k( θ)×. Take a character ψ ∈ Tˆ. Suppose Res ψ 6= 1 but Res ψ1. Then %(ψ) = r be the depth Tr Tr+ of ψ. 96 TASHO KALETHA

See Rabinoﬀ’s senior thesis for an exposition of the case r = 0. In this lecture, suppose r > 0 and ﬁx Λ: k+ → C of conductor p.

By the magic of SL2, there exists Xψ ∈ t−r \ t−r+ such that −1 ψ(γ) = Λ(Tr(c (γ)Xψ))

for all γ ∈ Tr, where

c: g0+ → G0+ X 1 + 2 X 7→ X . 1 − 2

∼ ∼ Fact. T\r/Tr+ = t\r/tr+ = t−r/t−r+ .

We fatten ψ using the character Λ and the element XΨ. Note that Xψ ∈ t−r ⊆ gx,−r. As above, ∼ gx,−r/gx,−r = Gx,r\/Gx,r+ .

In fact, G r + /Gx,r+ is abelian and x, 2 ∼ gx,−r Gx,\r + /Gx,r+ = . 2 g r + x,− 2 ˆ Take ψe ∈ TG r + given by x, 2 −1 ψe(tk) = ψ(t)Λ(Tr(c (k))XΨ). Then Stab(ψ,e T G r + ) = TGx, r . x, 2 2 G If G r = G r + , then Ind ψ is irreducible and supercuspidal. x, 2 x, TG r e 2 x, 2

If G r + 6= Gx, r , then let x, 2 2

r + N = ker(ResG r + ψe) ≤ Gx, . x, 2 2 It turns out that

TG r + /N x, 2 is the center; the derived group of TG r /N. This is hence a Heisenberg group, i.e. it ﬁts in x, 2 a short exact sequence of the form:

1 A H A/H 1.

By the Stone–von-Neumann Theorem, there is a unique irreducible representation σ of ψe TGx, r , whose restriction to TG r + contains ψe. Then 2 x, 2 G IndTG r σψ x, 2 e is an irreducible supercuspidal representation. One can classify all irreducible supercuspidal representations this way by tracking through this argument. MATH 679: AUTOMORPHIC FORMS 97

The guest lecture ended here and Tasho Kaletha continues from next time.

A review of supercuspidal representations. Let F be any ﬁeld. Deﬁnition 8.44.

× (1)A maximal torus of GL2(F ) is an embedded copy fo A where A/F is a quadratic ´etalealgebra. (2)A maximal torus of SL2(F ) is the intersection of one for GL2(F ).

There are two possibilities for A: • A = F ⊕ F , • A = E/F , a separatble quadratic ﬁeld extension. To embed A into Mat(2, 2,F ), choose a basis of A/F and let A act on A by multiplication. Then A → Mat(2, 2,F ) →det F 1 1 is the norm N : A → F . Thus for SL2, we get a map A → SL2(F ), where A denotes the norm 1 elements. Any A has a non-trivial Galois automorphism τ and N(a) = a · τ(a). Note that for A = F ⊕ F , τ permutes the two factors and hence: A1 = {(x, x−1) | x ∈ F ×}. In the non-split case, choose η ∈ E/F , then a + ηb ∈ E acts on E in term of the basis a b {1, η−1} by the matrix . These are the non-split tori deﬁned last time. η2b a

× Fact 8.45. All embeddings A → GL2(F ) are conjugate.

This is not true for SL2. 1 × × Fact 8.46. The embeddings of A into SL2(F ) are a torsor under F /N(A ). Examples 8.47.

• Let F = R. The only non-split extension E = C, i.e. the only non-split maximal 1 torus in SL2(R) is isomorphic to S . We get two embeddings this way, which have the same image but differ by inversion. • Now let F = Qp for p 6= 2. There are 3 isomorphism classes of non-split maximal tori. When E/F is unramified, the two embeddings have different images and are dis- tinguished by the two vertices in the Bruhat–Tits building.8 For E/F ramified, two embeddings have the same image if and only if −1 ∈ F ×,2 and then differ by inversion. The point in the Bruhat–Tits building associated to this torus is the center of an edge. • When F = Q, have infinitely many isomorphism classes of maximal toris, and in each infinitely many conjugacy classes. 8 SL2 does not act transitively on the vertices of the building, but it acts transitively on the edges. Two two embeddings correspond to the two endpoints, labeled x and y in the picture above. 98 TASHO KALETHA

Theorem 8.48. Let F = Qp, p 6= 2. The irreducible supercupsidal representations are classiﬁed as follows.

× (1) For each pair (S, θ) where S ⊆ SL2 is a maximal torus and θ : S → C is a smooth character with θ2 6= 1, get one. Two such representations are isomorphism if and only if the pairs are conjugate. 1 2 (2) Each of the two embeddings of (Qp) together with an unramiﬁed sign character gives 2 supercuspidal irreducible representations, for a totall of 4.

Representations in the ﬁrst class are called ordinary supercuspidals (depth is equal to the depth of the character θ, discussed in the previous lecture), and in the second class are called exceptional supercuspidals (corresponding to depth 0, which has not discussed in the previous lecture).

1 Remark 8.49 (Discrete series of SL2(R).). The non-trivial characters of S are of the form z 7→ zk ∈ Z \{0}. It is enough to ﬁx one embedding and consider all k, or look at both and + − only k > 0. For k > 0, have Dk = Dk and D−k = Dk . Therefore, the discrete series are classiﬁed by (S, θ) as in the ordinary supercupsidal case. The exceptional supercuspidals have no analog in SL2(R).

The Satake transform and unramiﬁed representations.

Deﬁnition 8.50. A smooth irreducible representation π is called unramiﬁed (K0-spherical) if πK0 6= 0.

Note that

{Unramiﬁed irreducible representations} ↔ {unramiﬁed HK0 -modules},

where we recall that HK0 is the Hecke algebra.

Given t ∈ T , f ∈ HK0 , deﬁne Z B 1/2 f (t) = δB(t) f(tu) du. U B × × Then f is a compactly supported function on T = Qp and T0 = Zp -invariant. In other B words, f ∈ HT0 (T ). Theorem 8.51 (Satake isomorphism). The map W S : HK0 (G) → HT0 (T ) , f 7→ f B is an isomorphism of algebras, where W = N(T )/T .

Before we prove this theorem, we discuss some consequences.

Since HT0 (T ) is commutative, so is HK0 (G). Moreover, unramiﬁed representations of HK0 (G) W correspond to characters η : HT0 (T ) → C. We have a short exact sequence

1 T0 T Z 0, MATH 679: AUTOMORPHIC FORMS 99

and so HT0 (T ) = C[X]. Therefore, unramiﬁed representations correspond to complex numbers.

s/2 Deﬁnition 8.52. A character of T is unramiﬁed if it is trivial on T0, i.e. it is t 7→ δB(t) for s ∈ C.

G K0 Lemma 8.53. Let χ be any character of T . Then (IB χ) is 1-dimensional if χ is unramiﬁed and 0-dimensional otherwise.

Proof. Iwasawa decomposition.

Lemma 8.54. For any unramiﬁed character χ and f ∈ HK0 (G), B Θ G (f) = Θχ(f ). IB χ

This is the adjunction formula between induction and taking the constant term f 7→ f B.

Proof. We have that f = eK0 ∗ f ∗ eK0 , so π(f): Vπ → Vπ factors as

π(f) K0 K0 Vπ Vπ Vπ Vπ .

K0 To compute the trace of this operator, take any 0 6= φ ∈ Vπ and note that tr(π(f)) = π(f) · φ. Normalize φ such that φ(1) = 1. Therefore: tr(π(f)) = (π(f)φ)(1) Z = f(g)φ(g) dg G Z Z Z = f(tuk) φ(tuk) dt du dk T U K0 | {z } 1/2 δB χ(t) Z Z 1/2 = χ(t) δB (t) f(tu) du dt. T U | {z } f B (t) This gives the result.

Corollary 8.55. The character ηπ : C[X] → C corresponding to the unramiﬁed subquotient G s/2 s of IB χ sneds X to χ(p). In particular, χ = δB and χ(p) = p .

Proof. Note that X ∈ C[X] corresponds to λ1(p) ∈ HT0 (T ). Let f be its preimage in HK0 (G). K0 Then ηπ(x) be the action of π(f) on V , which is equal to Θ G (f) = θχ(λ1(p)) = χ(p). π IB χ

We now turn to the proof of the Satake isomorphism 8.51. We start with a lemma.

G K0 G −1 K0 Lemma 8.56. For χ unramiﬁed, the HK0 -modules (IB χ) and (IB χ ) are isomorphic. 100 TASHO KALETHA

±1/2 G G −1 Proof. If χ 6= δB , we know that IB χ is irreducible and isomorphic to IB χ . Let now −1/2 χ = δB . Then we have a short exact sequence

G 0 C IB (χ) St 0.

∼ G K0 K0 Thus, C = (IB χ) and St = 0. The intertwining operator now gives

= G −1 K0 C (IB χ )

G G −1 IB χ IB χ

G K0 = (IB χ) C

and the diagonal dotted arrow is an isomorphism.

Proof of the Satake isomorphism 8.51. We will use Lemma 8.54 throughout without speciﬁc references. We start by proving the W -invariance of the image:

w Θχ((Sf) ) = Θwχ(Sf)

= Θχ−1 (Sf)

= Θ G −1 (f) IB χ

= Θ G (f) IB χ = Θχ(Sf). We then check that it is an algebra homomorphism. Linearity is obvious, so we just check multiplication:

Θχ((Sf1) ∗ (Sf2)) = Θχ(Sf1) ◦ Θχ(Sf2)

= Θ G (f1) ◦ Θ G (f2) IB χ IB χ = Θ G (f1 ∗ f2) IB χ = Θχ(S(f1 ∗ f2)). We now check bijectivity. Note that

W •H T0 (T ) has basis (λn(p) + λ−n(p))T0 for n ∈ Z/ ± 1,

•H K0 (G) has basis K0λn(p)K0 for n ∈ Z/ ± 1 (by the Cartan decomposition). Therefore, X S(1K0λn(p)K0 ) = an,m1λm(p)+λ−m(p), m∈Z/±1 MATH 679: AUTOMORPHIC FORMS 101

where

an,m = S(1K0λn(p)K0 )(λm(p)) Z 1/2 = δB(λn(p)) 1K0λn(p)K0 (λm(p)u) du U u = p vol(K0λn(p)K0 ∩ K0λm(p)U).

We claim (and leave the proof as an exercise) that a matrix belongs to K0λn(p)K0 if and only if the minimal valuation of its entries is −n.

By Iwasawa decomposition, this implies that an,n 6= 0 and an,m = 0 if m > n. Therefore, the matrix is upper-triangular with non-zero entries on the diagonal, and hence invertible.

Local Langlands correspondence for tori. Next time, we will discuss the local Lang- lands correspondence for SL2(Qp), but first, we need to discuss this correspondence for tori. We will consider the cases S = F × and S = E1 where E/F is a quadratic eztension. Here, F = R or F = Qp. Define Sˆ = C× with Γ-action. This is trivial if S = F × and the union non-trivial represen- 1 tation of ΓE/F if S = E . Define L ˆ S = S o WF , the Langlands dual of the torus S. Theorem 8.57. There is a group isomorphism between the group of continuous isomorphisms

L WF S id

ˆ × up to conjugation by S and Homcts(S, C ).

1 ˆ ˆ Write ϕ(w) = (ϕ0(w), w). Then ϕ0 ∈ Z (Wf , S) and the S-conjugacy class of ϕ corresponds 1 ˆ to [ϕ0] ∈ H (WF , S). Another way to state the theorem is that 1 ˆ ∼ × H (WF , S) = Homcts(S, C ).

Proof. In the split case, 1 ˆ × × × × H (WF , S) = Hom(WF , C ) = Hom(F , C ) = Hom(S, C ).

In the non-split case, we claim that corestriction gives an isomorphism

1 ˆ Cor 1 × H (WE, S)ΓE/F → H (WF , C ). =∼ 102 TASHO KALETHA

E This follows from the following formula for ϕ0 = Corϕ0 . Fix w ∈ WF \ WE. For x ∈ WE, E w−1 E −1 ϕ0(x) = ϕ0 (x) · ϕ0 (wxw ), E w−1 E −1 2 ϕ0(xw) = ϕ0 (x) · ϕ0 (wxw · w ). Now, H1(W , Sˆ) ∼ H1(W , IndWF ×) = Hom(W , ×) = Hom(E×, ×). F = E WE C E C C This completes the proof.

Note that: E1 ,→ E× × × C× ← C × C a ← (a, b). b [ On the other hand, E1 → E1 x x 7→ τx × × × C × C ← C (a, a−1) ← a. [ This is a manifestation of a general phenomenon, called functoriality. If S,T are tori over F and S → T is a homomorphism, there are dual homomorphism Tˆ → Sˆ LT 7→ LS and the local Langlands for tori is functorial.

Embedding the L-group of a maximal torus. Let S = E×, S1 = E1. We want to ﬁnd embeddings: L S → GL2(C) L 1 S → PGL2(C). Turns out we need a choice: we can send L × × j (C × C ) × WE/F → GL2(C) a 0 (a, b) 7→ 0 b but it is not clear where to sends elements of the Weil group. Recall that

× 1 E WE.F ΓE/F 1

τ˙ τ MATH 679: AUTOMORPHIC FORMS 103

whereτ ˙ is any lift. Take e ∈ E×. Then Lj(1 o e) commutes with Lj((a, b) o 1), so L χ1(e) 0 j(1 o e) = 0 χ2(e) 0 a for χ , χ : E× → ×. Now, what remain is Lj(τ ˙). It must be where the − is 1 2 C −a−1 0 there to make determinant 1. There are more general reasons for putting the − there, but we do not go into them here. We know thatτe ˙ τ˙ −1 = τ e. Thus

χ2(e) = χ1(¯e). Thus, χ(e) 0 Lj(e) = 0 χ(¯e) × × 2 × × 2 τ × for one χ: E → C . We know thatτ ˙ ∈ F \NE.F (E ). Otherwise,τ ˙ = x· x for x ∈ E and (x−1τ˙)2 = 1, contradicting that the Weil group is non-split. But we also know that −1 0 Lj(τ ˙)2 = . 0 −1

Therefore, χ|F × is the local class ﬁeld theory sign character of E/F . × Fact 8.58. Any χ: E → C such that χ|F × is the sign character for E/F given an L- L embedding S → GL2(C), well-deﬁned up to GL2(C)-conjugation.

Warning. Diﬀerent χ will in general lead to diﬀerent conjugacy classes of embeddings. Take χ and consider

L L jX S GL2(C)

L L 0 jX S PGL2(C),

where we note that χ(e/e) 0 Lj (e) = . X 0 1

What are the choices of χ. In general, they are a torsor under (E×/F ×)∗.

• When F = R, E = C, there is a preferred choice for χ, the complex argument. • When F = Qp, E unramified, the preferred choice of χ is the unramified quadratic character. • When F = Qp, E ramified (assuming again that p 6= 2 here), there are two tamely ramified χ and their quotient is the unramified quadratic character. For PGL2, it turns out that this choice is irrelevant.

Altogether, we have shown the following fact. 104 TASHO KALETHA

Fact 8.59. When p 6= 2, there is always a preferred choice of a map L 1 S → PGL2(C). Remark 8.60. Here is another embedding which may seem much more natural: a 0 Lj((a, b) 1) = , o 0 b 0 1 Lj(1 τ) = , o 1 0 1 0 Lj(1 e) = . o 0 1 It leads to the na¨ıveLanglands correspondence, which turns out not to be true. This is why we need to go through the more complicated process above.

The Local Langlands correspondence for SL2(Qp). Let ϕ: WF × SL2(C) → PGL2(C) be a Langlands parameter. We describe below the L-packet associated to it.

(0) If ϕ|SL2 6= 1, then ϕ: SL2(C) → PGL2(C) is the natural projection. Then ϕ|WF = 1. We associate to it the Steinberg rerpresentation, Πϕ = {St}.

(1) If ϕ|SL2(C) = 1, note that ϕ(PF ), where PF is the wild inertia group, is a ﬁnite subgroup of p-power order—it has to be cyclic, generated by a semisimple element, ∗ 0 ∗ 0 so a subgroup of . Therefore, ϕ(W ) is a subgroup of the normalizer of , 0 ∗ F 0 ∗ i.e. ∗ 0 0 ∗ ∪ . 0 ∗ ∗ 0 ∗ 0 (a) Suppose ϕ(W ) ⊆ ∼= ×. We get ϕ: W → × and by local Langlands F 0 ∗ C F C × × G correspondence for tori, this gives χ: F → C . We then get IB χ, which is ±1/2 semisimple unless χ = δB , in which case the composition factors are C and St. ±1/2 If χ = δB , let Πϕ = {C}. χ(w) 0 If χ 6= δ±1/2, let Π be the constituents of IGχ. We have that ϕ(w) = . B ϕ B 0 1 ˆ We compute the centralizers Cent(ϕ(WK ), G): (i) ϕ ≡ 1, then it is PGL2(C), ∗ 0 ∗ 0 0 ∗ (ii) ϕ 6≡ 1, the centralizer is between and ∪ and it is 0 ∗ 0 ∗ ∗ 0 equal to the latter if and only if χ2 = 1. In general, this veriﬁes that

Πϕ ↔ Irr(π0(Sϕ)). ∗ 0 (b) Suppose ϕ(W ) 6⊆ . Then we have that F 0 ∗ ϕ ∗ 0 W → Norm → /2 . F 0 ∗ Z Z MATH 679: AUTOMORPHIC FORMS 105

We get a quadratic extension E/F , so we get the torus S = E1. We have a preferred map L S PGL2(C)

ϕS ϕ

WF × so we get θ : S → C . We have 2 embeddings S,→ SL2(Qp), together with θ, so we get 2 supercuspidal representations (unless S = E1 for E unramiﬁed and θ = sgn where we get 4). Take them as Πϕ. Let us now compute the centralizers. We have θ(e/e¯)χ(e)2 0 0 1 ϕ(e) = , ϕ(τ ˙) = . 0 1 −1 0 Assume θ2 6= 1. The the matrix is regular, so ± 0 S = . ϕ 0 1 The remaining case, up to conjugation, has image ±1 0 0 1 ∪ = ( /2 )2 0 1 ±1 0 Z Z

and in self-centralizing in PGL2(C).

Notice that in (1)(a), χ is unramified if and only if ϕ is unramified (trivial in IF ). Then ϕ s 0 s−1 0 is determined by ϕ(Fr), a semisimple conjugacy class in PGL ( ), i.e. ≡ . 2 C 1 0 1 0 Similarly, an unramified χ satisfies χ(p) = ps, χ(x) = |x|t = et log |x|, where t corresponds to s via et = s. Moreover, s is also the Satake parameter (see Theorem 8.51).

The local Langlands correspondence for GL2(F ), F = R, F = Qp.

Deﬁnition 8.61. An irreducible representation π of GL2(F ) is called

(1) unitary (same as before), (2) tempered if unitary and matrix coefficients are in L2+(G/Z), (3) discrete series if unitary and matrix coefficients are in L2(G/Z), (4) essentially tempered/discrete series if for some n ∈ Z, π⊗| det |n is tempered/discrete series, (5) supercuspidal if the matrix coefficients are compactly supported modulo the center.

The following diagram describes the relation between all these notions:

supercupsidal essentially discrete series essentially tempered

discrete series temperedbook unitary. 106 TASHO KALETHA

Remark 8.62. The representation theory for GL2 becomes much simpler than that of SL2. However, we presented the theory for SL2 because a lot of the methods presented generalize to other reductive groups.

G × × × For F = R, we consider IB χ for χ: F × F → C . This is irreducible, unless χ1/χ2 = n x · sgn(x) for n ∈ Z \{0}, un which case we get a ﬁnite-dimensional representation Mn−2 and one discrete series Dn. G ±1 For F = Qp, IB χ is irreducible unless χ1/χ2 = |x| . This is equivalent to ±1/2 χ = (φ ◦ det) ⊗ δB for some φ: F × → C×. Then we get two constituents: φ ◦ det and φ ◦ det ⊗ St . × × × Moreover, for each maximal torus E ⊆ GL2(F ) and non-trivial character θ : E → C , we get one supercuspidal. On the Galois side, we consider

ϕ: WF × SL2(C) → GL2(C).

× (0) If ϕ: SL2(C) → GL2(C) is the natural map, then ϕ: WF → C ,→ GL2(C) corresponds to φ: F × → C×, and we associate to it St ⊗(φ ◦ det). ∗ 0 (1) (a) ϕ: W → ⊆ GL ( ) corresponds to χ: F × × F × → × and we either F 0 ∗ 2 C C G take IB χ if irreducible, or (φ ◦ det) if not. L L (b) We must be careful about which j : S → GL2(C) to take. We do not go to these detials. All centralizers are connected in this case.

In the unramified case, χ unramified if and only if ϕ is unramified. If χ is unramified it cor- × responds to an unordered pair s1, s2 ∈ C . Similarly, when ϕ is unramified, it corresponds to × × semisimple conjugacy classes in GL2(C), i.e. unordered pairs s1, s2 ∈ C . Again, s1, s2 inC are the Satake parameters.

9. Automorphic representations of GL(2, A)

The blueprint for this theory is Tate’s thesis.

Local factors. Let M2(F ) be 2 × 2 matrices with F -coeﬃcients. Let S(M2(F )) be the set of Schwartz–Bruhat functions on M2(F ), deﬁned in the same way as in Tate’s thesis. There is a Fourier transform

S(M2(F )) → S(M2(F )) with respect to the bicharacter 0 ψp(tr(X · Y )), 0 where ψp is the chosen character from Tate’s thesis. MATH 679: AUTOMORPHIC FORMS 107

Deﬁnition 9.1. Let π be an irreducible admissible representation of G(F ), β a matrix coeﬃcient of π and φ ∈ S(M2(F )). Then the zeta integral is Z s+ 1 Z(s, φ, β) = φ(g)β(g)| det(g)| 2 dg G where dg is the Haar measure on G, normalized so that vol(K,dg) = 1 if F is p-adic and K0 = GL2(Zp). Fact 9.2. The zeta integral converges in a right half-plane.

+ −2 Proof. The additive Haar measure on M2(F ) is given by dg = | det(g)| dg. Therefore: Z Z(s, φ, β) = φ(g)β(g)| det(g)|s−3/2d+g. M2(F )

Since φ is Schwartz, going to inﬁnity in M2(F ) (as a vector space) is not an issue. The other issue is the matrix coeﬃcient β has a singular set, but making Re(s) 0, we can dominate the asymptotic expansion of β as g goes to the singular set. Remark 9.3. Since the leading exponents of β do not depend on the particular choice of β (but only on the representation), there is a right half plane such that Z(s, φ, β) converges there for all β. Theorem 9.4.

(1) The function Z(s, φ, β) has a meromorphic continuation to all of C. When p < ∞, it is a rational function of p−s. (2) The family of functions {Z(s, φ, β) | φ, β} has a gcd, that is, there is on number L(s, π) such that Z(s, φ, β) L(s, φ) is entire. When p < ∞, this family is an ideal in C(p−s). (3) The local functional equation holds:

Z(1 − s, φ,ˆ β∨) = γ(s, π)Z(s, φ, β)

with a function γ(s, π), rational in p−s if p < ∞. Here, φˆ is the Fourier transform of φ, and β∨(g) = β(g−1), a matrix coeﬃcient for π∨. Deﬁnition 9.5. The epsilon factor is L(s, π) (s, π) = γ(s, π) . L(1 − s, π∨)

The goal is to prove the above theorem and give a more explicit description of all the functions involved. 108 TASHO KALETHA

G K0 The unramiﬁed case. Let F = Qp, χ = (χ1, χ2) be unramiﬁed, and π = IB χ. Then π is 1-dimensional, spanned by 0 1/2 fχ(tuk) = |a/b| χ1(a)χ2(b) a 0 for t = . 0 b Take 0 0 β(g) = hgfχ, fχ−1 i Z 0 = gfχ(k)fχ−1 (k)dk K0 Z 0 0 = fχ(kg)fχ−1 (k)dk K0 0 = fχ(g).

Take φ = 1M2(Zp). Then

Z 1 0 s+ 2 Z(s, φ, β) = φ(g)fχ(g)| det(g)| dg G Z Z Z × a b 1/2 s+ 1 d a × = φ |a/d| χ1(a)χ2(d)|ad| 2 d ddb × × 0 d |a| Qp Qp Qp Z Z s s × × = χ1(a)χ2(a)|a| |d| d ad d Zp\{0} Zp\{0} = L(s, χ1) · L(s, χ2).

The principal series case. Deﬁnition 9.6. Deﬁne the L-function as: G L(s, IB χ) = L(s, χ1)L(s, χ2) and the -factor as: G (s, IB χ) = (s, χ1)(s, χ2). Lemma 9.7. The function Z(s, φ, β) L(s, π) is entire.

G G −1 Proof. Let fχ ∈ IB χ and fχ−1 ∈ IB χ . Then Z β(g) = fχ(kg)fχ−1 (k). K0 Writing Z Z Z −1 a b 0 0 0 ξ(a, d) = φ k k f (k )f −1 (k)dkdk db, 0 d χ χ Qp K0 K0 MATH 679: AUTOMORPHIC FORMS 109

we get that Z Z s s × × Z(s, φ, β) = ξ(a, d)χ1(a)χ2(d)|a| |d| d ad d. × × Qp Qp One can check that ξ is a Schwartz–Bruhat function on Qp × Qp. Then Z Z s s × × Z(s, φ, β) = ξ(a, b)χ1(a)χ2(a)|a| |d| d ad d × × Qp Qp s −s = Pξ(p , p )L(s, χ1)L(s, χ2) by Tate’s thesis. This completes the proof. ˆ Lemma 9.8. We have that ξψ,βˆ ∨ = ξφ,β.

The supercuspidal case. Lemma 9.9. For any φ, β, Z(s, φ, β) is entire.

a 0 Proof. Let ω : × → × be the central character9 of π, i.e. π = ω(a). Then β(z g) = Qp C 0 a a | {z } za ω(a)β(g). We have that Z Z ! 2s+1 s+1/2 Z(s, φ, β) = φ(zag)ω(a)|a| sa β(g)| det(g)| dg. × G/Z Qp The inner integral is a 1-dimensional ζ integral. The outer integral is entire (i.e. does not give any new poles), since β is compactly supported modulo Z. There are now 2 possibilities. If ω is ramified, the inner integral is entire, so the whole 1 integral is entire. If ω is unramified, then it has a pole at s = − 2 (assuming ω is trivial) with residue φ(0), independent of g. 1 Thus the whole integral has a pole at s = − 2 (ω is trivial) with residue Z φ(0) β(g)| det(g)|s+1/2dg. G/Z But Z 1 x β g dx = 0 0 1 Qp (proving this is an exercise which involves considering the integral description of the Jacquet module).10 Using the Iwasawa decomposition, this shows that the above integral is 0, and hence there is actually no pole. Definition 9.10. Define L(s, π) = 1 for π supercuspidal.

While the L-function of a supercupsidal representation is trivial, the -factor is actually interesting.

9We have not deﬁned this, but this is done in the same way as for ﬁnite groups. 10We call a function β with this property a supercuspidal function. 110 TASHO KALETHA

Deﬁnition 9.11. For any admissible representation (π, V ) and φ ∈ S(M2(F )), deﬁne

Z 1 s+ 2 Z (s, φ, π) = φ(g)π(g)| det(g)| dg ∈ EndC(V ). G Remark 9.12. This does converge in a right half plane and ∨ Z(s, φ, βv,v∨ ) = hZ (s, φ, π)v, v i. ∞ If supp(φ) ⊆ G, then φ ∈ Cc (G) and then Z(s, φ, π) converges for all s and is entire. In fact, we can arrange that φ and φˆ are supported on G. Lemma 9.13. Let π be discrete series. Then every smooth ﬁnite rank operator can be realized as Z (s, φ, π).

1 Proof. We may assume s = − 2 . A smooth ﬁnite rank operator is a linear combination of smooth rank 1 operators, i.e. (v ⊗ v∨)(w) = v · hw, v∨i. Deﬁne ( deg(π, dg)hv, π∨(g)v∨i if | det g| ∈ {1, p}, φ(g) = 0 otherwise. One can check that φˆ is supported on G. Then Z hZ (−1/2, φ, π)w, w∨i = hπ(g)w, w∨ihv, π∨(g)v∨i deg(π, dg)dg G/Z hv, w∨ihw, v∨i Lemma 7.54 hhw, v∨iv, w∨i. This completes the proof. ˆ Lemma 9.14. For φ, ψ ∈ S(M2(F )) such that ψ, ψ are supported on G, we have Z (1 − s, ψ,ˆ π∨)∨ ◦ Z (s, φ, π) = Z (s, ψ, π) ◦ Z (1 − s, φ,ˆ π∨)∨.

Proof. It is enough to check this after applying h(−)v, v∨i. Then the two sides are: Z Z LHS = φ(g)ψˆ(h)hπ(g)v, π(h)v∨| det(g)|s+1/2| det(h)|3/2−sdgdh, G G Z Z RHS = φˆ(g)ψ(h)hπ(g−1)v, π(h−1)v∨| det(g)|3/2−s| det(h)|s+1/2dgdh. G G After a small massage, this is equivalent to the equation Z Z φ(g)ψˆ(g)dg = φˆ(g)ψ(g)dg, G G which is true. Proposition 9.15. Let π be irreducible. There exists a unique scalar valued function γ(s, π) such that Z (1 − s, φ,ˆ π∨)∨ = γ(s, π)Z (s, φ, π) for all φ such that φ and φˆ are supported on G. MATH 679: AUTOMORPHIC FORMS 111

Proof. We ﬁrst show that there is a operator-valued γ(s, π). Assume it existed and take v ∈ V . What is γ(s, π)v? By Lemma 9.13, there exists φv such that Z(s, φv, π) = v. Then ˆ v (8) γ(s, π)v = Z (1 − s, φv, π ee)v by the relation γ should satisfy. This immediately shows uniqueness of γ(s, π). We also see an approach to construct it.

0 We need to show (8) is independent of φv. Assume not an take φv, φv such that ˆ ˆ0 ∨ ∨ Z(1 − s, φv − φv, π ) v = w 6= 0. Choose ψ such that Z(s, ψ, π)w = w 6= 0. Then

ˆ ˆ0 ∨ ∨ 0 6= w = Z (s, ψ, π)Z (1 − s, φv − ψv, π ) v ˆ ∨ ∨ 0 = Z (1 − s, ψ, π ) ◦ Z (s, φv − φv, π)v by Lemma 9.14 = 0 by assumption. This is a contradiction, showign that equation (8) can be used to deﬁne γ(s, π). A variation of this argument proves that γ(s, π) is linear. Looking at how Z transforms under φh(x) = φ(hx), we see that γ(s, π) commutes with π(h) for all h ∈ G, so γ(s, π) is a scalar by Schur’s Lemma.

The computation of local factors for Steinberg is omitted here.

Preservation of local factors. Using the above computation and the description of the local Langlands correspondence, one can show the following theorem. Theorem 9.16. The Local Langlands correspondence preserves the local factors.

Automorphic representations. Let A be the adele ring of Q, G = GL2, K∞ = O(2) ⊆ G(R) maximal compact, Kp = GL2(Zp) ⊆ G(Qp) maximal compact open. Writing A = R × Af , Y KF = Kp ⊆ G(AF ) maximal compact open, p<∞

K = K∞ × Kf ⊆ G(A) maximal compact. Finally, let

Z = Z(U(g∞)) be the center of the universal enveloping algebra.

Deﬁnition 9.17. A function f : G(A) → C is an automorphic form if

(1) f is G(Q)-invariant on the left, (2) for some Hecke character ψ : A×/Q× → C× such that a 0 f(z g) = ψ(a)f(g) for z = a a 0 a

(3) for any y ∈ G(A), the function x∞ 7→ f(x∞y) is smooth and Z -ﬁnite, 112 TASHO KALETHA

(4) f is slowly increasing: for any c > 0 and ω ⊆ G(A) compact, there are C > 0 and N > 0 such that a 0 f g ≤ C|a|N 0 1 for all a ∈ A×, |a| > c, and g ∈ ω. Deﬁnition 9.18. A function f : G(Q)\G(A) → C is called cuspidal if Z 1 x f g dx = 0 0 1 A/Q for almost all g ∈ G(A). Deﬁnition 9.19.

(1) Let A be the space of all automorphic forms. (2) Let A0 be the space of cuspidal automorphic forms. (3) Let L2(G(Q)\G(A), ψ) be the set of all L2-functions on G(Q)\G(A) satisfying (2) from Deﬁnition 9.17. 2 (4) Let L0(G(Q)\G(A), ψ) be the subspace of all cuspidal functions. Remark 9.20.

2 2 • The spaces L and L0 are (unitary) representations of G(A) via right-translation. • The space A is not a G(A)-representation, because G(R) does not act. However, G(Af ) does act, so O0 H(G(Af )) = H(G(Qp)) p<∞ 1 where the restricted tensor product is with respect to Kp . In place of the G(R)-

action, we have a (g∞,K∞)-module structure. In particular, U(g∞) acts and MK∞ 11 which is the convolution algebra of ﬁnite measures on K∞ also acts. We may hence deﬁne

H(G(R)) = U(g∞) ⊗ MK∞ . Overall, A is an H(G(A)) = H(G(R)) ⊗ H(G(Af ))-module.

Fact 9.21. The cuspdial automorphic forms A0 is a dense subspace of K-finite and Z -finite 2 vector in L0. Definition 9.22. An automorphic representation is an H(G(A))-module which is an irreducible subquotient of A. It is cuspidal if it is a subquotient of A0. 2 Theorem 9.23 (Gelfand–Graev–Piatetski-Shapiro). The space L0 is a (completed) Hilbert direct sum of irreducible representations with finite multiplicities.

2 Remark 9.24. The irreducible constituents of L0 are precisely the cuspidal automorphic representations.

Problem. Describe the cuspidal automorphic representations and their multiplicities!

11 R Convolution is deﬁned as(ν ◦ µ)(E) = 1E(xy)dν(x)dµ(y). MATH 679: AUTOMORPHIC FORMS 113

Proposition 9.25 (Flath’s decomposition theorem). An irreducible admissible representation π of G(A) is O0 π = πp p≤∞ where πp is an irreducible admissible representation of G(Qp) and almost all πp are unrami- ﬁed.

We have already classiﬁed all irreducible local representations. One can put any collection like that with almost all πp unramiﬁed into a representation of G(A). Such a representation may not be an automorphic representation and our task is to work out which such representations do occur in the L2-spectrum.

Multiplicity one.

Theorem 9.26 (Multiplicity one). An irreducible admissible representation of G(A) occurs 2 in L0 with multiplicity ≤ 1.

We give a sketch of the proof. There are two components to prove:

• a global component called global genericity of cusp forms, • a local component called uniqueness of Whittaker model. Deﬁnition 9.27.

(1) An irreducible admissible representation πp of G(Qp) is generic if for one (hence any) × character non-trivial ξp : Qp → C ,

HomU (πp, ξp) 6= 0. (2) An automorphic representation π is generic if for one (hence any) non-trivial ξ : A/ Q → C×, HomU (π, ξ) 6= 0. Remark 9.28. We justify the one (hence any) assertion in the deﬁnition. Note that ∼ × × T = Qp × Qp ∼ acts on U = Qp by conjugation, i.e. a (a, b) · u = u. b 1 2 If ξp , ξp are two characters, then 2 1 ξp (x) = ξp (ax) for a ∈ Qp.

∼ × ∼ 2 × ×,2 Note that for SL2, T = Qp acts on U = Qp via (a, u) 7→ a u, so get Qp /Qp -many options for ξp. The global case also follows. Lemma 9.29. If π is a cuspidal automorphic representation, it is generic. 114 TASHO KALETHA

Proof. For any ξ and f ∈ π, we can take the ξ-Fourier coeﬃcient: Z 1 x f (g) = f g ξ(x)dx. ξ 0 1 A/Q Basic Fourier theory says that X f(g) = fξ(g). ξ

Since π is cuspidal, f1 = 0. Therefore, fξ 6= 0 for some ξ 6= 1. Then f 7→ fξ(1) is an element of HomU (π, ξ) 6= 0.

Proposition 9.30. For any irreducible admissible representation πp of G(Qp),

dim HomU (πp, ξp) ≤ 1.

Instead of proving this proposition, we present a toy proof, when F = Fp, where we do not have to worry about the analytic issues and we can just show the main idea.

Proof when F = Fp. We want to prove that G dimC(HomG(π, IndU ξ)) ≤ 1, G i.e. IndU ξ is multiplicity free. This is equivalent to showing that G EndG(IndU ξ) is commutative, G where EndG(IndU ξ) is a convolution algebra 0 0 {f : G → C | f(ugu ) = ξ(u)f(g)ξ(u )}. Bruhat decomposition gives representatives a 0 0 a , for a, b ∈ × 0 b b 0 Qp for U\G/U. We can assign f(g) for each such representative g almost freely. We have 1 x a 0 1 −y a 0 = 0 1 0 b 0 1 0 b if and only if ax = by. Thus, if a 6= b, we can find x 6= y, and get f(g) = ξ(x − y)f(g), so f(g) = 0. The following method of proof is called Gelfand’s trick. We have the anti-involution 0 1 g 7→ Ad gt. 1 1 This fixes all representatives g where f(g) 6= 0, hence gives an anti-involution of the Hecke algebra that is the identity. Remark 9.31. Gelfand’s trick has later been generalized to other contexts. The general setting in which it works is related to so-called Gelfand pairs. Definition 9.32.

(1) A non-zero element of HomU (π, ξ) is called a Whittaker functional. MATH 679: AUTOMORPHIC FORMS 115

G (2) The image of a non-zero element of HomG(π, IndU ξ) is called a Whittaker model for π.

What we proved above is the uniqueness of Whittaker models.

Proof of Multiplicity one Theorem 9.26. The global result tells us O 0 6= HomU (π, ξ) = HomU (πp, ξp). p

The local result says that HomU (π, ξ) ≤ 1. Therefore,

HomU (π, ξ) = C. Therefore, we have the non-zero map

`ξ : A0 → C (taking the ξ-Fourier coeﬃcient): Z 1 x ` (f) = f ξ(x)dx. ξ 0 1 A/Q We obtain 2 HomG(π, L0) ,→ HomU (π, ξ)

α 7→ `ξ ◦ α. Since the latter is 1-dimensional, this completes the proof.

Remark 9.33. This fails for other groups. It works for GLN and SL2 (which was only proved in 2000 by Ramakrishnan), but not for SLN when N > 2 (proved in 1994 by Blasius). In fact, the general Langlands conjectures predict the multiplicity that one should obtain. 1 N 1 2 N 2 Theorem 9.34 (Rigidity). If π = πp and π = πp are cuspidal automorphic repre- 1 1 ∼ 2 sentations and πp for almost all p, then π = π .

The Rigidity Theorem 9.34 together with Multiplicity one Theorem 9.26 gives strong multiplicity one. 1 2 2 Corollary 9.35 (Strong multiplicity one). If π , π ⊆ L0 are irreducible constituents and 1 ∼ 2 1 2 2 πp = πp for almost all p, then π = π (the same subspace of L0).

This theorem allows us to do the following. Consider the set n × o S = {αp, βp | αp, βp ∈ C }p p primes where we identify two collections 0 0 {{αp, βp}p} ∼ {αp, βp}p 0 0 if {αp, βp} = {αp, βp} for almost all p. Strong multiplicity one gives an injection {irreducible cuspidal representations} ,→ S/ ∼ by assigning to π the collection of the Satake paremeters of its local representations. 116 TASHO KALETHA

Problem. Describe the image of this map, i.e. work out what collections complex numbers appear as Satake parameters of representations. This is an extremely diﬃcult thing to do.

Ramanujan conjecture. N Conjecture (Ramanujan). If π = πp is a cuspidal automorphic representation, then each πp is tempered. Remark 9.36. This implies that the complementary series are never local components of cuspidal automorphic representations.

Remark 9.37. Already for Sp4, there are non-tempered cuspidal automorphic representations. There are also non-generic cuspidal automorphic representations. However, all the examples of non-tempered cuspidal automorphic representations are non-generic. This leads to the generalized Ramanujan conjecture.

Conjecture (Generalized Ramanujan). If π is a generic cuspidal automorphic representation of a quasisplit group G, then all πp are tempered.

The Arthur conjecture extends a lot of the Langlands conjecture to non-tempered representation.

Classical modular forms. Recall that if f : H → C is a cusp form, then at the beginning of the class, we associated to it a function

φf : SL2(Z)\ SL2(R) → C given by −k φf (g) = f(g · i)j(g, i) . By strong approximation, we can think of it as a function

φf : GL2(Z)\ GL2(R) → C.

2 Then one can show that φf ∈ L0(G(Q)\G(A)). Moreover, φf spans an irreducible representation if f is a new eigenform. Altogether, we get an injection

new cuspidal eigenforms ,→ cuspidal automorphic representations.

The cuspidal automorphic representation π = π∞ ⊗ πf has ∼ π∞ = Dk where k is the weight of the modular form. MATH 679: AUTOMORPHIC FORMS 117

Cuspidal automorphic L-functions. Recall the space S(M2(Qp)) of Schwartz–Bruhat functions, for p ≤ ∞. We formed the space O0 S(M2(A)) = S(M2(Qp)), p≤∞ 1 restricted with respect to M2(Zp).

Definition 9.38. Let π be a cuspidal automorphic representation. For φ ∈ S(M2(A)) and a matrix coefficient β of π, the global zeta integral is defined as Z Z(s, φ, β) = φ(g)β(g)| det(g)|s+1/2dg. A G(A) Proposition 9.39. This converges in a right half plane and equals to the product over all p of the local zeta integrals.

Proof. Exercise.

1 × × 1 Recall that for A = {x ∈ A | |x|A = 1} we have A = A × R>0. We can similarly let 1 1 G(A) = {g ∈ G(A) | det(g) ∈ A } and 0 ∼ A∞ = Z(G)(R) = R>0. Once again, we have a decomposition

1 G(A) = G(A) × R>0.

In the 1-dimensional case A1\Q× is compact, but in the 2-dimensional case G(Q)\G(A)1 is no longer compact (but it at least has ﬁnite volume). This is related to the classical issue of cusps, which in the automorphic setting correspond to Borel subgroups. If one replaces G with the units in a deﬁnite quaternion algebra, there are no cusps (and no Borel subgroups) and this quotient is compact.

2 On the space L0(G(Q)\G(A), ψ), we have a scalar product Z hf1, f2i = f1(x)f2(x)dx. G(Q)\G(A1) Therefore, Z β(g) = hgf1, f2i = f1(xg)f2(x)dx. G(Q)\G(A)1 Proposition 9.40. The function Z(s, phi, β) has an analytic continuation to s ∈ C to a holomorphic function, and satisﬁes the functional equation

Z(1 − s, φ,ˆ β∨) = Z(s, φ, β). 118 TASHO KALETHA

Proof. We have that: Z Z s+1/2 Z(s, φ, β) = φ(g) f1(xg)f2(x)dx| det(g)| dg G(Q)\G(A)1 G(A) Z Z −1 s+1/2 = φ(x g)f1(g)f2(x)| det(g)| dx dg

G(A) G(Q)\G(A)1 Z Z Z X −1 2s+1 × = φ(x tag)f1(g)f2(x)a dx dg d a

1 1 t∈G(Q) R>0 G(Q)\G(A) G(Q)\G(A) R 1 R ∞ We split the integral over R>0 as a sum 0 + 1 . The second integral is ﬁne: it is absolutely convergent for all s to a holomorphic function. To compute the ﬁrst integral, we want to apply the Poisson summation formula to

X −1 2s+1 φ(x tag)f1(g)f2(x)a . t∈G(Q) As in the GL(1) case, φˆ is the additive Fourier transform, but the sum is over the multiplicative group. We hence add and subtract the terms from M2(Q) \ G(Q), apply Poisson 1 summation formula, and substitute t 7→ t to get Z ∞ Z ∞ Z Z X ˆ −1 3−2s × (··· ) + φ(g tax)f1(g)f2(x)a dx dg d a 1 1 t∈G( ) G(Q)\G(A)1 G(Q)\G(A)1 Q Z ∞ Z Z 2s+1 2s+1 × − φ(0)f1(g)f2(x)a a dgdxd a 1 G(Q)\G(A)1 G(Q)\G(A)1 Z ∞ Z Z X −1 −1 2s+1 × − φ(x ta g)f1(g)f2(x)a dg dx d a 1 t∈M ( ) G(Q)\G(A)1 G(Q)\G(A)1 2 Q 1 Z ∞ Z Z ˆ 2s+1 2s+1 × + φ(0)f1(g)f2(x)a a dgdxd a 1 G(Q)\G(A)1 G(Q)\G(A)1 Z ∞ Z Z X ˆ −1 2s+1 × + φ(g tax)f1(g)f2(x)a dg dx d a. 1 t∈M ( ) G(Q)\G(A)1 G(Q)\G(A)1 2 Q 1 We can use cuspidality to get rid of the boundary terms. The case t = 0. We have Z f1(g)dg = hf1, 1i = 0

G(Q)\G(A)1

because f1 lies in π and 1 lies in the trivial representation 1 and 1 2 1 π, ⊆ Ld(G(Q)\G(A), ) are orthogonal. MATH 679: AUTOMORPHIC FORMS 119

The case t ∈ M2(Q)1. We break M2(Q) into G(Q)-orbits under left multiplication. Each 1 0 orbit has a representative of the form γ for γ ∈ G( ). The stabilizer of such a 0 0 2 2 Q 1 ∗ representative in G( ) is L( ) = . Therefore, the contribution of that orbit to the Q Q 0 ∗ boundary term is: Z ∞ Z Z X 1 0 φ x−1γ−1 γ g f (g)f (x)a2s+1 dg dx d×a 1 0 0 2 1 2 1 γ ∈L( )\G( ) G(Q)\G(A)1 G(Q)\G(A)1 1 Q Q Z ∞ Z Z −1 1 0 2s+1 × = φ y γ2g f1(g)f2(x)a dg dy d a (where y = γ1x). 1 0 0 G(Q)\G(A)1 L(Q)\G(A)1 Notice that, as a function of y, this is invariant not just under L(Q) but under L(A) ⊇ U(A). Therefore we get an integral Z f2(y)dy = 0. U(Q)\U(A) This completes the proof, since we showed vanishing of the boundary terms. Corollary 9.41. When π is a cuspidal automorphic representation, L(s, π) is holomorphic and satisﬁes the functional equation L(s, π) = (s, π)L(1 − s, π∨).

Weil’s converse theorem. Let πp be an irreducible admissible generic representation of × × G(Qp). We have a Whittaker model W (πp) of πp. Given W ∈ W (πp), χ: Qp → C , Z a 0 Z(s, g, χ, W ) = W g χ(a)|a|s−1/2d×a. × 0 1 Qp a 0 This is almost like the 1-dimensional local zeta integral, but the function W g is 0 1 not locally constant, so it is not a Schwartz–Bruhat function.

Deﬁne L(s, πp × χp), (s, πp × χp) as before. When χ = 1. we get the old L and factors. N Theorem 9.42 (Converse theorem). Let π = πp be an irreducible admissible generic representation of G(A). Then π is cuspidal automorphic if and only if L(s, π × χ) is entire, bounded in vertical strips, and satisﬁes a functional equation, for all χ.

2 Sketch of proof. Assume L(s, π × χ) is nice (as above). We want an embedding π ,→ L0 2 or, equivalently, of its Whittaker model W (π) ,→ L0. A Whittaker functionW : G(A) → C satisﬁes for ξ : U(Q)\U(A) → C× and u ∈ U(A) the invariance property ω(ug) = ξ(u)W (g). We want G(Q)-invariance, so we form X a 0 φ (g) = W g . W 0 1 a∈Q× This function is B(Q)-invariant. By Bruhat decomposition G(Q) = B(Q) ∪ B(Q)wB(Q), we just need to check w-invariance:

φW (g) = φW (wg). 120 TASHO KALETHA

For a ﬁxed g, consider the equation a 0 a 0 φ g = φ w g . W 0 1 W 0 1 To show these functions in a ∈ A are equal, we take Mellin transforms (as functions of χ) of both sides. On the left, we get

Y Y Zp(−) Z(s, g, χ, W ) = Z (−) = L(s, π × χ) . p L (−) p p p On the right side, we get −1 −1 Y Zp(1 − s, wg, χp ξp ,Wp) L(1 − s, π∨ × χ−1) . L(1 − s, π∨ × χ−1) p p p The local and global functional equations combined give the equality of these function.

Rigidity. The key input to the Strong Multiplicity One Theorem 9.35 was the Rigidity Theorem 9.34. We discuss the proof brieﬂy. 1 ∼ 2 1 2 Suppose πp = πp for p 6∈ S, S ﬁnite, where π and π are cuspidal automorphic representation. Apply the functional equation of L-functions to get that 1 ∨ −1 1 2 ∨ −1 2 Y L(1 − s, (πp) × χp )(s, πp × χp) Y L(1 − s, (πp) × χp )(s, πp × χp) 1 = 2 . L(s, π × χp) L(s, π × χp) p∈S p p∈S p | {z } 1 ∨ −1 γ(1−s,(πp) ×χp ) One uses this equality as follows.

(1) Stability of γ-factors: L(s, πp × χp) = 1 if χp is very ramiﬁed and (s, πp × χp) =

(s, χp · ωπp )(s, χp), (2) Globalization of characters — find a global character χ which is highly ramified at some primes p and a specific χp at one place. This shows the local factors agree at 1 ∼ 2 each place p ∈ S, and they determine the local representations, so πp = πp for p ∈ S also.

The global Langlands conjecture.

Conjecture (Langlands). Let φ: G(F /F ) → GL2(C). There exists a cuspidal automorphic

representation π of GL2(AF ) such that for all v, φ|G(Fv/Fv) corresponds to φv via the Local Langlands Correspondence.

On the Galois side, we have Ind, Res. Conjecturally, on the automorphic side, induction corresponds to automorphic induction (AI), and restriction corresponds to base change (BC). Since we understand the Local Langlands Correspondence, we can describe the candidates for these representations.

Langlands showed existence of AI and BC in the case G = GL2 and E/F is cyclic. Later Arthur and Clozel generalized this to G = GLn and E/F is cyclic.

Observation. The image of φ (in PGL2) can be MATH 679: AUTOMORPHIC FORMS 121

(1) cyclic: Hecke theory, (2) dihedral, (3) tetrahedral (A4), (4) octahedral (S4), (5) icosahedral (A5) Cases (1)–(3) were handled by Langlands. Case (4) was handled by his student, Tunnel. Case (5) splits into two cases: odd (proved by Khare–Winterburger), even (this is an open problem). We discuss the cases (2) and (3) brieﬂy. Let φ be dihedral or tetrahedral. Then φ = IndWF φ , WE E × G where E/F is cyclic of order 2 or 3. Now, φE is a character on AE, so we may consider IB φE on GL2(AF ). Using the trace formula, one can show that this corresponds to π, a cuspidal automorphic of GL2(AF ), corresponding to φ, as required. One thing we have not discussed in this class is Eisenstein series. They describes fully the automorphic representations which are not cuspidal.