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Lectures on Maass Forms Postech, March 25-27, 2007

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Lectures on Maass Forms Postech, March 25-27, 2007 LECTURES ON MAASS FORMS POSTECH, MARCH 25-27, 2007 Jianya Liu Shandong University 1. Introduction 1.1. The aim of these lecture series. The aim of these lecture series is to explain the basic concepts and ideas of ² Maass forms; ² their automorphic L-functions; ² the Kuznetsov trace formula; ² subconvexity of Rankin-Selberg L-functions with applications. For simplicity, only the theory for ½µ ¶ ¾ a b ¡ = SL (Z) = : a; b; c; d 2 Z; ad ¡ bc = 1 (1.1) 2 c d will be treated; the general theory for congruence subgroups ¡0(N) is similar in philosophy but more involved in details. For detailed treatment of Maass forms in book form, the reader is referred to e.g. Borel [2], Bump [3], Iwaniec [4], and Ye [20]. Even for the simplest case (1.1), there are already big amount of materials in the literature. To explain these materials in such a short time, I have to omit most of the proofs, and content myself just with illustrations and explanations. The contents are as follows: x2. Maass forms for SL2(Z) x3. Fourier expansion for Maass forms x4. Spectral decomposition of non-Euclidean Laplacian ¢ x5. Hecke theory for Maass forms x6. The Kuznetsov trace formula x7. Automorphic L-functions 1 x8. Number theoretical background x9. Subconvexity bounds for Rankin-Selberg L-functions x10. An application of subconvexity bounds x11. A proof of subconvexity 1.2. Notation. Throughout the lecture series, ¡ is always SL2(Z), as in (1.1). We use s = σ + it and z = x + iy to denote complex variables. The Vinogradov symbol A ¿ B means that A = O(B), and as usual e(x) = e2¼ix. 2. Maass forms for SL2(Z) The standard Laplace operator on the complex plane C is de¯ned by @2 @2 ¢e = + ; @x2 @y2 but on the upper half-plane H, the non-Euclidean Laplace operator is given by µ ¶ @2 @2 ¢ = ¡y2 + : @x2 @y2 The operator ¢ is invariant under the action of SL (R); that is, for any smooth function f ³ ´ 2 a b on H, any element g = c d 2 SL2(R), (¢f) ± g = ¢(f ± g): (2.1) ³ ´ a b The last equation means that, for any smooth function f on H, any element g = c d 2 SL2(R), and any z 2 H, (¢f)(gz) = ¢(f(gz)); (2.2) where we recall that az + b gz = : cz + d ¡ 0 ¡1 ¢ ¡ 1 1 ¢ It is well known that SL2(R) is generated by 1 0 and 0 1 . Thus in order to show (2.1), it is su±cient to check it for the generators. De¯nition 2.1. A smooth function f 6= 0 on H is called a Maass form for the group ¡ if it satis¯es the following properties: ² For all g 2 ¡ and all z 2 H; f(gz) = f(z); 2 ² f is an eigenfunction of the non-Euclidean Laplacian, ¢f = ¸f; where ¸ is an eigenvalue of ¢; ² There exists a positive integer N such that f(z) ¿ yN ; y ! +1: De¯nition 2.2. A Maass form f is said to be a cusp form if Z µµ ¶ ¶ Z 1 1 b 1 f z db = f(z + b)db = 0 0 0 1 0 holds for all z 2 H. As an example of Maass forms, we consider non-analytic Eisenstein series ¡(s) X ys E¤(z; s) = : (2.3) 2¼s jmz + nj2s m;n2Z (m;n)6=(0;0) Proposition 2.3. Let s = σ + it. Then ² The Eisenstein series E¤(z; s) is absolutely convergent for σ > 1; ² Although E¤(s; z) is not analytic as a function of z 2 H, it is strictly automorphic, i.e. for σ > 1, E¤(gz; s) = E¤(z; s); g 2 ¡: (2.4) ² For σ > 1, the Eisenstein series E¤(s; z) is an eigenfunction of ¢, ¢E¤(z; s) = s(1 ¡ s)E¤(z; s); (2.5) Proof. We have ys jyjσ ¿ : jmz + nj2s min(jmyj; jnj)2σ Therefore, for σ > 1, X ys X X 1 X X 1 ¿ jyjσ + jyjσ + jyj¡σ jmz + nj2s n2σ m2σ m;n2Z m¸1 n·mjyj n¸1 m·n=jyj (m;n)6=(0;0) jyj1¡σ jyjσ¡1 ¿ jyjσ + + : 2σ ¡ 2 2σ ¡ 2 3 And hence the series in (2.3) is absolutely convergenet in σ > 1. It is not analytic as a function of z 2 H. To see E¤(s; z) is strictly automorphic, we introduce E(z; s) by the following formula E¤(z; s) = E(z; s)»(s); (2.6) where »(s) = ¼¡s¡(s)³(2s) and ³(s) is the Riemann zeta-function. Since µ ¶ y ¤ ¤ =gz = ; for g = 2 ¡; jcz + dj2 c d we have 1 X ys 1 X E(z; s) = ys + = (=gz)s; (2.7) 2 jcz + dj2s 2 (c;d)=1 g2¡1n¡ c6=0 ¡ 1 1 ¢ where ¡1 is the subgroup of ¡ generated by 0 1 . From (2.7), it is easy to see that E(z; s) (and also E¤(z; s)) is invariant under ¡, hence (2.4). That E(z; s) is an eigenfunction of ¢ in σ > 1 can be checked term by term in the expansion (2.7). In fact, we have µ ¶ @2 @2 ¢ys = ¡y2 + (ys) = s(1 ¡ s)ys; @x2 @y2 so that the ¯rst term ys in (2.7) is an eigenfunction of ¢ with eigenvalue ¸ = s(1 ¡ s). The ys general term jcz+dj2s in (2.7) is also an eigenfunction of ¢ with eigenvalue ¸ = s(1 ¡ s), since ¢ is invariant under ¡, and ys ³¤ ¤´ (=gz)s = for g = 2 ¡: jcz + dj2s c d Therefore, ¢E(z; s) = s(1 ¡ s)E(z; s); (2.8) and hence (2.7). ¤ We remark that Proposition 2.3 does not prove that E¤(z; s) is a Maass form, since Propo- sition 2.3 holds only for σ > 1, and the growth condition in De¯nition 2.1 is to be checked. These will be done via the analytic continuation of E¤(z; s). 4 3. Fourier expansion for Maass forms 3.1. Fourier expansion of Maass forms. We will ¯rstly give the Fourier expansion of a Maass form f. Theorem 3.1. Let f(z) be a Maass form for ¡. ² Then it has the Fourier expansion X p f(z) = an yKiº(2¼jnjy)e(nx): (3.1) n2Z where Kiº is the modi¯ed Bessel function of the third kind. ² For ¸ ¸ 1=4, a square-integrable Maass form f(z) must be a cusp form, and its Fourier expansion takes the form X p f(z) = an yKiº(2¼jnjy)e(nx): n6=0 Proof. Since µ ¶ 1 1 f(gz) = f(z); for g = 2 ¡; 0 1 we have f(z + 1) = f(x + 1 + iy) = f(z) = f(x + iy); and therefore f(z) must have the Fourier expansion X X 2¼inx f(z) = cn(y)e = cn(y)e(nx): (3.2) n2Z n2Z It remains to determine the coe±cients cn(y). By de¯nition, we have ¢f = ¸f, and hence cn(y) satis¯es 2 00 2 2 2 ¡y cn + 4¼ n y cn = ¸cn: (3.3) If n 6= 0, this is a modi¯ed Bessel equation, which has the general solution p p cn(y) = an yKiº(2¼jnjy) + bn yIiº(2¼jnjy): (3.4) p Here º = ¸ ¡ 1=4, Kiº is the modi¯ed Bessel function of the third kind which is rapidly decreasing when y ! 1, and Iiº(y) is the modi¯ed Bessel function of the ¯rst kind which is rapidly increasing. More precisely, we have µ µ ¶¶ ey 1 + jºj2 I = p 1 + O ; iº 2¼y y 5 and r µ µ ¶¶ ¼ 1 + jºj2 K = e¡y 1 + O : iº 2y y From (iii) in De¯nition 2.1, we see that bn must equal zero, and hence (3.1) follows from (3.2) and (3.4). p 1=2+iº If n = 0, the equation (3.3) has the solution c0(y) = y , where º = § ¸ ¡ 1=4. If ¸ ¸ 1=4, then º is real, and consequently, Z Z 1=2 Z 1 1+2iº Z 1 2 jy j dy jc0(y)j dz ¸ dx 2 dy ¸ = 1: ¡nH ¡1=2 1 y 1 y For a square-integrable Maass form f, we then must have c0(y) = 0. Therefore for ¸ ¸ 1=4, a square-integrable Maass form must be a cusp form, and its Fourier expansion takes the form X p f(z) = an yKiº(2¼jnjy)e(nx): n6=0 This proves the theorem. ¤ Let ¶ : H ! H be the antiholomorphic involution ¶(x + iy) = ¡x + iy: If f is an eigenfunction of ¢, then f ± ¶ is an eigenfunction with the same eigenvalue. Since ¶2 = 1, its eigenvalue are §1. We may therefore diagonalize the Maass cusp forms with respect to ¶. If f ± ¶ = f, we call f even. In this case an = a¡n. If f ± ¶ = ¡f, then we call f odd. Then an = ¡an. 3.2. Analytic continuation of Eisenstein series. We will establish the analytic continu- ation of E¤(z; s) via its Fourier expansion. Theorem 3.2. Let s 2 C. ² The Eisenstein series E¤(z; s), originally de¯ned for σ > 1, has meromorphic con- tinuation to all s; it is analytic except at s = 1 and s = 0, where it has simple poles. The residue at s = 1 is the constant function z = 1=2. ² The Eisenstein series satis¯es the functional equation E¤(z; s) = E¤(z; 1 ¡ s): (3.5) ² We have E¤(x + iy; s) ¿ ymax(σ;1¡σ); y ! 1: (3.6) 6 It follows that the Eisenstein series E¤(z; s) is a Maass form with Laplace eigenvalue s(1¡s).
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