Math 679: Automorphic Forms
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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 = fz 2 C j Im(z) > 0g 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 2 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 = 2 SL ( ), c d 2 Z • f is bounded on domains of the form fz 2 H j Im(z) > C > 0g. The second condition is equivalent to f being holomorphic at i1. 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 : n2Z 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 2 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 2 ∆: 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-integralp weight. On thep slices complex place C n R<0. We have an analytic map z 7! z determined by Re( z) p> 0. p This can be extended non-continuously by −r = i r for r 2 R>0. k We can try to define, for k 2 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 ! µ1(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 n2Z 1 is a modular forms of weight 2 for Γ the theta group. This converges locally uniformly. Take a domain fIm(z) > C > 0g. 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 f^ = f(x)e dx. We have the inversion R formula ^ f^ = f(−x) and the Poisson summation formula X X f(n) = f^(n): n2Z n2Z 4 TASHO KALETHA The Gaussian f(x) = e−πx2 is its own Fourier transform: Z f^(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: R 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 ! 1, 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 = p e−x dx: R π R (c) Take square to see that Z 2 Z e−x2 dx = e−x2−y2 dxdy Z 2π Z 1 = e−r2 rdrd' changing to polar coordinates 0 0 " 2 #1 e−r = 2π −2 0 = π: 1=2 Let ft(x) = f(xt ) for t 2 R>0. Note that 1 1 f^ (y) = f^(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 2 R>0. Then X θ(it) = ft(n) n2Z X ^ = ft(n) Poisson summation n2Z 1 X = f (n=t) t1=2 t n2Z 1 i = θ : t1=2 t Recall that the Riemann zeta function is defined as 1 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 = 1. Define Λ(s) = π−s=2Γ(s=2)ζ(s); where Z 1 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 n2Z 1 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 1 Z 1 1 dt X 2 dt w(it)ts = e−πn tts t t 0 0 n=1 1 Z 1 X 2 dt = e−πn tts t n=1 0 1 s X Z 1 t dt = e−t πn2 t n=1 0 1 X 1 Z 1 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 1 dt Λ(2s) = w(it)ts 0 t Z 1 dt Z 1 dt = w(it)ts + w(it)ts 1 t 0 t Z 1 dt Z 1 i dt = w(it)ts + w t−s 1 t 1 t t Z 1 Z 1 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 1 t1=2−s 1 1 t−s 1 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 jw(it)j decays exponentially as t ! 1. Recall that writing 1 X n θ(z) = anq ; n=0 MATH 679: AUTOMORPHIC FORMS 7 we get that 1 1 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).