14 From modular forms to automorphic representations

Transcription

1 14 From modular forms to automorphic representations We fix an even integer k and N > 0 as before. Let f M k (N) be a modular form. We would like to product a function on GL 2 (A Q ) out of it. Recall that we have the action (γ k f)(τ) := j(γ 1, τ) k f(γ 1 τ) for γ GL 2 (R) + for which f is invariant by Γ = Γ 0 (N) under this action. We now define φ : GL 2 (R) + C by φ (g) = f(gi)j(g, i) k for any g GL 2 (R) +. We have φ (γg) = φ (g) for any γ Γ and any g GL 2 (R) +. The stabilizer of i H under GL 2 (R) + is R + SO 2 ; here R + GL 2 (R) + is in the center and we write K = SO 2 be the group of orthogonal matrices with determinant 1 on the ( standard real ) inner product space R 2. Every element in K may be expressed as cos θ sin θ k θ =. We have sin θ cos θ φ (zk θ ) = e iθk φ (e) where e GL 2 (R) + is the identity. For any z R + and k θ as above. This implies for any g GL 2 (R) + φ (gzk θ ) = e iθk φ (g). (1) In particular, (1) allows us to easily recover the weight of f. Lastly, we would like to interpret the holomorphy of f in terms of φ. As holomorphy of functions is described by a first order partial equation (i.e. Cauchy-Riemann equation), it is natural to consider the Lie algebra of GL 2 (R) acting φ. More precisely, for any 2 2 matrix X in R and any g GL 2 (R) +, we put φ (ge tx ) φ (g) R X (φ )(g) := lim. t 0 t This is evidently R-linear in X, and we shall extend it to X M 2 (C) by C-linearity. We now put H = 0 i, X i 0 + = 1 1 i, X 2 i 1 = i. i 1 They satisfy 1 X + X X X + = H, HX + X + H = 2X + and HX X H = 2X. They are also a basis for sl 2 (C), the space of 2 2 matrices in C with trace zero. Proposition The holomorphy of f is equivalent to R X (φ ) = 0. 1 That is to say, they form a so-called sl 2 -triple. 1

2 Proof. We will only show that R X (φ ) = 0 when f is holomorphic. The differential of the map g τi is an R-linear map δ : M 2 (R) T i (H) = R 2. By extending the map C-linearly we have δ C : M 2 (C) C 2. But we may instead use the complex structure on T i (H) = C to have δ : M 2 (C) C. One computes directly that δ C (X ) = 0. This means that δ C (X ) lies in the anti-holomorphic tangent space of T i (H). Write δ 0 (g) := f(gi). Then the above implies R X (δ 0 )(Id) = 0. Moreover, as GL 2 (R) + acts on H by conformal maps, we have in fact R X (δ 0 )(g) 0 for any g GL 2 (R). This leaves the j(g, i)-term to worry. In other words, with j (g) := j(g, i) we would like to have also R X (j )(g) 0. One notes that j(ge tx, i) = j(g, e tx i)j(e tx, i). The derivative of the first term is also zero as X is in the anti-holomorphic tangent space. For ( the) second term a b we would like R X (j )(Id) = 0. This is a property of the function g = (ci + d) c d that one may check. Let us collect a bit about what we have Proposition The association f φ is an isomorphism between M k (N) to the space of smooth functions on GL 2 (R) + satisfying (i) φ (γgzk θ ) = e iθk φ (g) for any γ Γ, g GL 2 (R) + arbitrary, and z, k θ as in (1). (ii) φ has polynomial growth in the entries of g and g 1. (iii) R X (φ ) 0. It suffices to explain (ii). As j(g, i) 1 has polynomial growth in the coefficients of g and g 1, if φ comes from a modular form it has polynomial growth. Conversely, a Laurent series in q that has logarithmic growth in q is automatically holomorphic, thus (ii) is enough to ensure holomorphy at cusps. We will write A = A Q, and A := p< Q p be the restrict product of the finite completions. To translate functions on GL 2 (R) + to GL 2 (A) we rely on the following results Theorem (Strong approximation for the additive group) Let v be any place of Q and Q v be the corresponding completion. Then Q v is dense in Q\A. Theorem (Strong approximation for SL 2 ) Let v be any place of Q and Q v be the corresponding completion. Then SL 2 (Q v ) is dense in SL 2 (Q)\SL 2 (A). Proof. We want to prove that SL 2 (Q)SL 2 (Q v ) is dense( in SL) 2 (A). This ( follows ) from Theorem 14.3 and the observation that elements of the form and with any x A 1 0 x 1 generate SL 2 (A). 2

3 In particular, SL 2 (Q)SL 2 (R) is dense in SL 2 (A). Hence for any open subgroup K SL 2 (A ) we have SL 2 (A) = SL 2 (Q)SL 2 (R)K. Now consider in GL 2 : a b K = K 0 (N) := {g GL 2 (Ẑ) g = with N c} GL c d 2 (A ). In particular K := K SL 2 (A ) is open. As A = Q R +Ẑ and determinants of elements in K can take all values in Ẑ, we obtain GL 2 (A) = GL 2 (Q)GL 2 (R) + K. One also easily see that GL 2 (Q) GL 2 (R) + K = Γ = Γ 0 (N). Thus we may define φ : GL 2 (A) C by φ(γg k) := φ (g ), for any γ GL 2 (Q), g GL 2 (R) +, k K. As a summary, we now have Theorem The association f φ is an isomorphism between M k (N) and the space of smooth functions on GL 2 (A) satisfying (i) For any γ GL 2 (Q), and k K, we have φ(γgk) = φ(g) for any g GL 2 (A). cos θ sin θ (ii) For any k θ = GL sin θ cos θ 2 (R) + as in (1), we have φ(gk θ ) = e iθk φ(g) for any g GL 2 (A). (iii) φ is translation invariant by the center A GL 2 (A). (iv) φ(g) is of polynomial growth in the entries of g and g 1. (v) R X (φ) 0. Moreover the subspace S k (N) is mapped to cuspidal function on GL 2 (A), namely those with (vi) For any g GL 2 (A), Q\A φ( g)dx = 0 Proof. It remains to prove the assertion with (vi). Write g = γg k as any decomposition with γ GL 2 (Q), g GL 2 (R) + and k( K. ) By continuity, there exists some open + u compact subgroup U g A such that φ( g) = φ( g) for any u U g. By Theorem 14.3 we have Q\A is covered by R U g, and thus integration over Q\A becomes (up to a constant) integration over Ūg\R where Ūg = Q (R U g ). In fact we may assume 3

4 Ū g is of the form MZ, and by shrinking U g ( we may ) assume M is as divisible as we like. In any case we arrive at the integral φ( g)dx. MZ\R Choose any decomposition g = γg k with γ GL 2 (Q), g GL 2 (R) + and k K as before. By condition (i) in th theorem we then have, for x R, φ( g) = φ( γg k) = φ(γ 1 γg ) = φ (γ 1 γ g ) where γ is the GL 2 (R)-component of γ GL 2 (Q) GL 2 (A), i.e. the image of γ in GL 2 (R). Now by Euclid s algorithm one may find γ SL 2 (Z) GL 2 (R) such that γ = bγ where b GL 2 (R) is upper triangular. This gives φ( g) = φ (γ 1 ( ) γ g ) = φ (γ 1 b 1 bγ g ). But conjugation by b simply scales x by some non-zero rational number! We thus get, up to a constant, φ( g)dx = φ Q\A (γ 1 γ g M Z\A )dx. If γ Γ, then φ (γ 1 γ g ) = φ ( γ g ) and the vanishing of the integral says f has a zero at q = 0. For general γ SL 2 (Z), this says f is a cusp form. Nevertheless, this function on GL 2 (A) should live inside a particular representation of GL 2 (A)! On a first thought, we would like to have GL 2 (A) acting on a subspace (containing φ) of the space of functions on GL 2 (A) by right translation. Our experience at the real place however suggest something less. We have seen the action of gl 2 (C) := M 2 (C) by differentiation (of the right translation), and the action of K = SO 2 by right translation. Let us also recall that gl 2 (C) can be generated by 0 i H =, X i 0 + = 1 1 i, X 2 i 1 = 1 1 i, and Z = 2 i Let us still denote by R X the action of X gl 2 (C). From basic differential geometry we have R [XY Y X] = [R X, R Y ] for X, Y gl 2 (C), where [X, Y ] := XY Y X and [R X, R Y ] := R X R Y R Y R X. This suggests to consider the universal enveloping algebra U(gl 2 ) of gl 2 (C), defined to be the quotient, of the free associative algebra generated by gl 2 (C) (as a C-vector space, i.e. generated by any basis of gl 2 (C), by the ideal generated by XY Y X [X, Y ]. The algebra U(gl 2 ) act on smooth functions on GL 2 (R) and thus those on GL 2 (A). By direct computation one may prove 4

5 Lemma The center Z(gl 2 ) of the algebra U(gl 2 ) is generated by H 2 +2X + X +2X X + and Z. We can now state what kind of representations we are studying. ambient space. Let χ : A Q C be a (continuous) unitary character. First we state the Definition An automorphic form of GL 2 over Q with central character χ is a function φ : GL 2 (A) C satisfying (i) For any γ GL 2 (Q), we have φ(γg) = φ(g). a 0 (ii) For any a A Q, φ(g ) = χ(a)φ(g). 0 a (iii) There exists an open subgroup K GL 2 (A ) such that φ is right K -invariant. (iv) φ is right K -finite, i.e. vector space. right K -translates of φ generate a finite-dimensional (v) φ is smooth at the real place, and right Z(gl 2 )-finite. (vi) φ has polynomial growth, and φ(g) 2 dg < where A GL 2 (A) GL 2 (Q)\GL 2 (A)/A is the center. Moreover, the automorphic form is cuspidal if it satisfies (vii) φ( g)dx = 0 for any g GL 2 (A). Q\A We denote by A(GL 2, χ) the space of automorphic forms and A o (GL 2, χ) the subspace of cuspidal automorphic forms. They carry actions of K, U(gl 2 ) and GL 2 (A ) by right translation. We remark that the action of K and U(gl 2 ) have to satisfy certain compatibility property. Such a representation of K and U(gl 2 ) is typically called a (gl 2 (C), K )-module (or just (g, K)-module). Definition An automorphic representation (with central character χ) is an irreducible subquotient representation of A(GL 2, χ). A cuspidal automorphic representation is a irreducible subrepresentation of A o (GL 2, χ). A fundamental feature about automorphic representations is that they factor into local components. Let (π, V ) be an automorphic representation. To avoid difficulties at the real place, we focus on those vectors for which R X = 0 and k θ K acts by e iθk for some fixed positive integer k. In other words, let V V be the space of elements satisfying the above archimedean condition, and we assume V 0. The following is a result of representation theory of GL 2 (R) that we will not prove. Lemma V is an irreducible representation of GL 2 (A ). 5

6 By (iii) of Definition 14.7, every vector in V is invariant under some open subgroup of GL 2 (A ). We may assume this open subgroup is of the form K = p< K p with K p GL 2 (Q p ) and K p = GL 2 (Z p ) for almost all p. We may consider the space V := V K of K -fixed vectors in V. If K = K 0 (N), then we have seen in Theorem 14.5 that V is a subspace of S k (N) and hence finite-dimensional by complex function theory. It s possible to show this finite-dimensional property in general 2, and we now assume dim C V <. We have the Hecke algebra H := H(GL 2 (Q p ), K p ) acting on V. As V is irreducible, we may prove as in local theory (Lemma 12.3) that V is an irreducible H -module. Let S be a finite set of primes so that K p = GL 2 (Z p ) for p S. For any p S, as V is finite-dimensional, the algebra H p := H(GL 2 (Q p ), GL 2 (Z p )) has to act through a maximal ideal I p H p for if any element in H p doesn t act by a constant, its eigenspace will give a proper subrepresentation. We write V p := H p /I p as a H p -module. For p S, as an H p := H(GL 2 (Q p ), GL 2 (Z p ))-module V has to decompose into a direct sum of isomorphic irreducible submodule for the same reason. We call any such irreducible submodule V p. We then have V = Vp as an H -module. Note that the RHS is finite-dimensional exactly because almost all V p are 1-dimensional. For every p S, we know that V p arises as V p = (V p ) Kp for some GL 2 (Q p )- representation V p. In general, V p may be explicitly found within V as follows: with a fixed p, consider a basis of open compact subgroups K p,1 K p,2... of GL 2 (Q p ). Let Ki be as K with K p replaced by K p,i, and V i := V K i. Then V 1 V 2... and for each we have V i = Vp,i V p p p, where V p,i is an irreducible H p,i := H(GL 2 Q p, K p,i )-module. By choosing embeddings V p,i V p,i+1 (unique up to constant) we have V p := Vp,i is a representation of GL 2 (Q p ) such that V p,i = (V p ) K p,i. In particular V p = (V p ) Kp. Moreover, we may enlarge shrink finitely many K p (and there enlarge V p and H p ) at the same time. Since all vectors in V are invariant under some open subgroup of GL 2 (A ), we then have V = V p where the restricted tensor product means the following: a vector in V p is a finite linear combination of v p for v p V p where almost all v p V p. On the other hand, suppose we begin with a sequence of irreducible smooth representations W p of GL 2 (Q p ) and form their restricted tensor product W = W p consisting of linear combinations of v p with v p W GL 2(Z p) p for almost all p, then W is also an irreducible smooth representation of GL 2 (A ). Here smoothness means every vector in W is invariant by an open subgroup of GL 2 (A ). In fact, with a bit of the archimedean theory, we can also take V to be any irreducible (gl 2 (C), K )-submodule of V and have 2 Without the holomorphic condition, it will be a result of elliptic partial differential equations. 6

7 Theorem The automorphic representation V is a restricted tensor product of all V v for all places v of Q. Note that we have only used property (iii), (iv) and (v) in Definition 14.7, but not the most important (i). In other words, there is no need in the theorem that V is automorphic. It is then natural to wonder that what condition (i) in Definition 14.7 means. That is, what it means to be automorphic? Let us look at instead the GL 1 -example. An automorphic representation of GL 1 can be readily checked to be a Hecke character χ : Q \A Q C. Instead, a character of A Q is a tensor product of characters of Q v that are unramified for almost all v. Let us furthermore suppose that χ R 1. Then class field theory tells us that + this character is trivial on Q if and only if the Galois characters rec(χ Q p ) : G Qp C obtained via class field theory combines the be a character of G Q! If we are not willing to restrict to the case χ R 1, then we still need χ + R (a ) = a n + for some integer n. For example, if χ is the character unramified everywhere given by χ(a ) = a n for any a R and χ(p) = p n for p Q p. Then we sees that χ is automorphic, i.e. trivial on Q and rec χ Q p agrees with χ n cyc,l. This suggests that for general χ with χ R (a ) = a n +, it is trivial on Q if and only if rec(χ Q p ) combines to a 1-dimensional Galois representation from ρ : G Q Q l such that ρ GQl is χ n cyc,l twisted by a finite order character. We are now ready to state the global Langlands conjecture. Conjecture There is a bijection π rec(π) between the set of isomorphic classes of automorphic representations (π, V ) whose real part V is algebraic, and the set of isomorphic classes of semisimple Galois representations ρ : G Q GL n ( Q l ) that are unramified almost everywhere and de Rham at l. This bijection satisfies: (i) For any p l, rec(v p ) is isomorphic to the Frobenius-semisimplification of ρ GQp. (ii) The real part V is given by certain (p-adic Hodge theoretic) invariant of ρ GQl, called Hodge-Tate weights. (iii) Cuspidal automorphic representations correspond to irreducible Galois representations. If ρ = ρ i and ρ i = rec π i with π being cuspidal automorphic representations of GL ni (A Q ) (where n i = dim ρ i and n = n i ), then π can be formed from π i by a formation generalizing Eisenstein series. We need some more precise statement about how cusp forms enter the picture. To see this, we need the following global analytic result that we will not prove: Theorem (Multiplicity one) Let V, V A o (GL 2, χ) be two cuspidal automorphic representations. Suppose V p = V p for almost all p. Then V = V. 7