Lifting to GL(2) over a division quaternion algebra and an explicit construction of CAP representations

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1 Lifting to GL) over a division quaternion algebra and an exlicit construction of CAP reresentations Masanori Muto, Hiro-aki Narita and Ameya Pitale Abstract The aim of this aer is to carry out an exlicit construction of CAP reresentations of GL) over a division quaternion algebra with discriminant two. We first construct cus forms on such grou exlicitly by lifting from Maass cus forms for the congruence subgrou Γ 0 ). We show that this lifting is non-zero and Hecke-equivariant. This allows us to determine each local comonent of a cusidal reresentation generated by such a lifting. We then know that our cusidal reresentations rovide examles of CAP reresentations, and in fact, counterexamles of the Generalized Ramanujan conjecture. Introduction One of the fundamental roblems in the theory of automorhic forms or reresentations is to study the generalized) Ramanujan conjecture. To review it let G be a reductive algebraic grou over a number field F and let A := v F v be the ring of adeles for F, where F v denotes the local field at a lace v. Conjecture. Generalized Ramanujan Conjecture) Let π = v π v be an irreducible cusidal reresentation of GA), where π v denotes the local comonent of π at a lace v. Then π v is temered for every v. Nowadays it is widely known that counterexamles of this conjecture are found. A well-known examle is given by the Saito-Kurokawa lifting to holomorhic Siegel cus forms of degree two cf. [7]). As another well-known examle there is the work by Howe and Piatetski-Shairo [], which gives a counterexamle by a theta lifting from O) to S4). In fact, it is exected that liftings from automorhic forms on a smaller grou rovide such counterexamles. On the other hand, let us recall that there is the notion of CAP reresentation Cusidal reresentation Associated to a Parabolic subgrou), which has been originally introduced by Piatetski-Shairo [5]. This is a reresentation theoretic aroach to find counterexamles of the Ramanujan conjecture. We now note that the Ramanujan conjecture for the general linear grou GLn) is strongly believed. In fact, by Jacquet and Shalika [4], it can be shown that the CAP henomenon never Mathematical Subject Classification 00. Primary F55, F70. Partly suorted by Grant-in-Aid for Scientific Research C) , Jaan Society for the Promotion of Science. Partly suorted by National Science Foundation grant DMS-0054.

2 M.Muto, H.Narita and A.Pitale occurs for GLn). More generally, it is exected that the conjecture would hold for generic cusidal reresentations of quasi-slit reductive grous. In view of the Langlands functoriality rincile for quasi-slit grous and their inner forms, the Ramanujan conjecture for the inner forms are quite natural and interesting to study. We can thereby say that CAP reresentations of the cases of the inner forms are significant to discuss. To define the notion of CAP reresentations for the inner forms we follow Gan [8] and Pitale [] cf. Definition 6.6). Let us note that the Saito-Kurokawa lifting deals with the case of the slit symlectic grou GS4) of degree two. The case of the non-slit inner form GS, ) GSin, 4) of GS4) is considered by [8] and [] etc. In this aer we take u the case of GL B) over the division quaternion algebra B with discriminant two, where note that GL B) is an inner form of the slit grou GL4). Our results rovide an exlicit construction of cus forms on GL B) with lifts from Maass cus forms for the congruence subgrou Γ 0 ), and show that the cusidal automorhic reresentations generated by such lifts are CAP reresentations of GL B). The method of our construction of the lifting is what follows [], which deals with an exlicit construction of lifting to GS, ). In fact, note that the cus forms constructed by our lifting are viewed as Maass cus forms on the 5- dimensional real hyerbolic sace, while the lifting considered in [] rovides Maass cus forms on the 4-dimensional real hyerbolic sace. As in [], to rove the automorhy of our lifts, we use the converse theorem [9] by Maass, which is useful for real hyerbolic saces of arbitrary dimension. We exlain the exlicit construction of our lifting. Let f be a Maass cus form for Γ 0 ) which is an eigenfunction of the Atkin-Lehner involution. Let {cn)} n Z\{0} be Fourier coefficients of f. From cn) s we define numbers Aβ) s cf. 4.)) for β B \ {0} in order to construct our lifting to a cus form F f on GL H) in the non-adelic setting. Actually Aβ) s are nothing but Fourier coefficients of F f. The statement of our first result is as follows: cf. Theorem 4.3) Theorem. Let f be a non-zero Mass cus form which is an eigenfunction of the Atkin- Lehner involution. Then F f is a non-zero cus form on GL H). Another result is that the cusidal reresentations generated by F f s are CAP reresentations of GL B) and rovide counterexamles of the Generalized Ramanujan conjecture. To be more recise, assume that f is a Hecke eigenform. We can regard F f as a cus form on the adele grou GA) with G = GL B). We can show that F f is a Hecke eigenform cf. Section 5). Then the strong multilicity one theorem roved by Badulescu and Renard [], [3] imlies that F f generates an irreducible cusidal reresentation π := π of GA). By our detailed study on Hecke eigenvalues of F f we can determine local reresentations π for every <. We can also determine π exlicitly at = by the calculation of the eigenvalue for the Casimir oerator. For this we note that every π is unramified at < ) or sherical at = ). We can show that π resectively π ) is non-temered at every odd rime resectively temered at = ). If we further assume that f is a new form, we can also show the non-temeredness of π at =. These lead to our another result as follows cf. Theorem 6.7, Theorem 6.8): Theorem.3 ) Let f be a non-zero Hecke eigen cus form and F f be the lift. Let σ f and π F be irreducible cusidal reresentations generated by f and F = F f resectively. Then π F is nearly equivalent to an irreducible comonent of Ind GL 4A) P A) det / σ det / σ). Here P is

3 Lifting to GL) over a division quaternion algebra the standard arabolic subgrou of GL 4 with Levi subgrou GL GL. Namely π F is a CAP reresentation. ) The cusidal reresentations π F s are counterexamles of the Ramanujan conjecture. Let π be the unique irreducible quotient of Ind GL 4A) P A) det / σ det / σ). This is denoted by MWσ, ) in Section 8 of [3], which is a non-cusidal, discrete series reresentation of GL 4 A). Since σ is not the image of a cusidal reresentation of B A under the Jacquet- Langlands corresondence, π is B-comatible according to Proosition 8., art a) of [3]. Hence there exists a discrete series reresentation π of GL B A ) which mas to π under the Jacquet-Langlands corresondence. Also, from Proosition 8., art b) of [3], the reresentation π has to be cusidal. By the strong multilicity one theorem for GL B), the reresentation π has to be exactly the same as π F obtained from the classical construction. The novelty of our method is that we obtain exlicit formula for the lift in terms of Fourier exansions which are valid for non-hecke eigenforms as well. In addition, the classical method immediately shows that the lifting is a linear non-zero ma. Let us remark that Grobner [9] has also obtained examles of CAP reresentations for GL B) using the results of [3]. The examle by Grobner [9] has a non-temered local comonent at the archimedean lace, while our cusidal reresentation π has a temered local comonent at the lace as is remarked above. The outline of this aer is as follows. In Section we first introduce basic notation of algebraic grous and Lie grous. Then we next introduce automorhic forms in our concern. In Section 3 we study a zeta-integral attached to a Maass cus form f, which is necessary to use the converse theorem by Maass later. Then the exlicit construction of cus forms on GL H) is given by lifts F f s in Section 4. In Section 5 we view F f as a cus form on the adele grou GA) and rove that F f is a Hecke eigenform at every finite lace. Then, in Section 6, we determine local comonents π of the cusidal reresentation π generated by F f for all laces. We thus see that π is a CAP reresentation and rovides a counterexamle of the Ramanujan conjecture. Basic notations. Algebraic grou, real Lie grous and the 5-dimensional hyerbolic sace Let B be the definite quaternion algebra over Q with discriminant d B =. The algebra B is given by B = Q + Qi + Qj + Qk with a basis {, i, j, k} characterized by the conditions i = j = k =, ij = ji = k. Let G be the Q-algebraic grou defined by its grou of Q-rational oints GQ) = GL B). Here GL B) is the general linear grou over B, which consists of elements in M B) whose reduced norms are non-zero. Let H = B Q R, which is nothing but the Hamilton quaternion algebra R+Ri+Rj +Rk. Let H x x H denote the main involution of H, and trx) = x+ x and νx) := x x be the reduced trace and the reduced norm of x H resectively. In what follows, 3

4 M.Muto, H.Narita and A.Pitale we often use the notation β := νβ) for β H. We ut H := {x H trx) = 0} to be the set of ure quaternions, and H := {x H νx) = }. Denote by G := GL H) the general linear grou of degree two with coefficients in the Hamilton quaternion algebra H. The Lie grou G admits an Iwasawa decomosition G = Z + NAK, where {[ ] } { [ ] Z + c 0 := c R x 0 c +, N := nx) = x H},.) 0 { [ ] } y 0 A := a y := 0 y y R +, K := {k G t kk = }. The subgrou Z + is contained in the center of G and K is a maximal comact subgrou of G, which is isomorhic to the definite symlectic grou S ). Let us consider the quotient G/Z + K, which is realized as {[ y x 0 ] y R +, x H }. This gives a realization of the 5-dimensional hyerbolic sace.. Lie algebras The Lie algebra g of G is nothing but M H), and has an Iwasawa decomosition g = z n a k, where {[ ] {[ ] c 0 0 x z := c R}, n := x H},.) 0 c 0 0 {[ ] t 0 a := t R}, k := {X M 0 t H) t X + X = 0 }, where z, n, a and k are the Lie algebras of Z +, N, A and K resectively. [ ] 0 We next consider the root sace decomosition of g with resect to a. Let H := 0 and α be the linear form of a such that αh) =. Then {±α} is the set of roots for g, a). For z H we ut [ ] [ ] E z) 0 z α :=, E z) α :=. z 0 The set {E ) α, Ei) α, Ej) α, Ek) α } n resectively a basis of n := resectively {E) α, Ei) α, Ej) α, Ek) α }) forms a basis of ] x H} ). Let z a k) := {X k [X, A] = 0 A a}, {[ 0 0 x 0 which coincides with {[ a 0 0 d ] } a, d H. 4

5 Lifting to GL) over a division quaternion algebra Then z z a k) a is the eigen-sace with the eigenvalue zero. We then see from the root sace decomosition of g with resect to a that g decomoses into g = z z a k) a) n n. We also introduce the simle Lie grou SL H) consisting of elements in GL H) with their reduced norms. The Lie algebra g 0 = sl H) of SL H) is the Lie algebra consisting of elements in M H) with their reduced traces zero. For this we note that GL H)/Z + SL H), g/z g 0. We introduce the differential oerator Ω defined by the infinitesimal action of Ω := 6 H H + 4 z {,i,j,k} E z) α..3) This differential oerator Ω coincides with the infinitesimal action of the Casimir element of g 0 see [5,.93]) on the sace of right K-invariant smooth functions of G/Z +. To check this ) we note [E z) α, E z) α ] = H for z H and Iwasawa decomositions E z) 0 z α = Ez) α + for z 0 z H. In what follows, we call Ω the Casimir oerator..3 Automorhic forms For λ C and a discrete subgrou Γ SL R) we denote by SΓ, λ) the sace of Maass cus forms of weight 0 on the comlex uer half lane h whose eigenvalue with resect to the hyerbolic Lalacian is λ. For a discrete subgrou Γ GL H) and r C we denote by MΓ, r) the sace of smooth functions F on GL H) satisfying the following conditions:. Ω F = r 4 + )F, where Ω is the Casimir oerator defined in.3),. for any z, γ, g, k) Z + Γ G K, we have F zγgk) = F g), 3. F is of moderate growth. Let K α, with α C, denote the modified Bessel function see [, Section 4.]), which satisfies the differential equation y d K α dy + y dk α dy y + α )K α = 0. Proosition. Let Γ be an arithmetic subgrou of GL H), and let L Γ := {x H nx) N Γ} and L ˆ Γ be the dual lattice of L Γ with resect to tr. Then F MΓ, r) admits a Fourier exansion F nx)a y ) = uy) + Cβ)y K r 4π β y)eπ trβx), with a smooth function u on R >0. β ˆ L Γ \{0} 5

6 M.Muto, H.Narita and A.Pitale Proof. A Maass form F MΓ; r) is left-invariant with resect to {nβ) β L Γ }. This imlies that F nx + α)g) = F nx)g) holds for α L Γ and g G. Therefore F has a exansion F nx)a y ) = W β y) ex π trβx) β L ˆ Γ with a smooth function W β on R +. For ξ H \ {0} we ut Ŵξy) := y 3 W ξ y). From the condition Ω F = r 4 + )F we deduce that Ŵβ satisfies the differential equation d dy )) ) r Y Y Ŵ β = 0 8π β for β L ˆ Γ \{0}, where Y := 8π β y. This is recisely the differential equation for the Whittaker function see [, Section 4.3]). With the Whittaker function W 0, r arametrized by 0, r) we thereby see that F nx)a y ) = uy) + C β)y 3 W0, r 8π β y) ex π trβx), β ˆ L Γ \{0} with constants C β) deending only on β.. We now note the relation W y 0, r y) = π K r y) see [, Section 3, 3.8 iii)]). statement of the roosition. This means that F has the Fourier exansion as in the We shall consider the automorhic forms above with secified discrete subgrous of SL R) and GL H). As a discrete subgrou of SL R) we take the congruence subgrou Γ 0 ) of level. For a choice of a discrete subgrou of GL H) we recall that B denotes the definite quaternion algebra over Q with discriminant d B =. This has a unique maximal order O given by O = Z + Zi + Zj + Z + i + j + ij, which is called the Hurwitz order. As a discrete subgrou of GL H) we mainly take GL O). Proosition. The grou GL O) is generated by {[ ] [ ] [ ] 0 u 0 v,, u O, v O} Proof. Any element of the form [ ] α β GL 0 δ O) can be exressed as [ α 0 0 ] [ 0 0 δ ] [ α β 0 6 ].

7 Lifting to GL) over a division quaternion algebra We see that α, δ O and thus α β O. We note that [ ] [ ] [ ] [ ] [ ] 3 [ ] 0 u =, = u 0 0 [ ] α β These imly the assertion for GL 0 δ O). Next, we have the following claim. For a, b O with b 0 there exists c, d O such that a = cb + d, νd) < νb). [ This follows ] from [6, Chater I, Section, Corollary.8]. This reduces the general case of α β GL γ δ O), γ 0, to the revious case. This comletes the roof of the roosition. Let GA) = GL B A ), where B A denotes the adelization of B, and let U be the comact subgrou of GA) given by < GL O ), where O denotes the -adic comletion of O at a finite rime. Then the class number of G with resect to U is defined as the number of cosets in UGR)\GA)/GQ). We next{[ ut P to ] be a standard } Q-arabolic subgrou of G whose grou of Q-rational oints α β is PQ) = GQ). We now recall that, for an arithmetic subgrou Γ GQ), the 0 δ cosets Γ\GQ)/PQ) are called the set of Γ-cuss. Lemma.3. The class number of G with resect to U is one, namely we have GA) = GQ)GR)U.. The number of cuss with resect to Γ := GL O) is one. Proof. According to [4, Theorem 8.] the class number of a reductive grou over a number field F is not greater than that of a arabolic F -subgrou of it. Furthermore the class number of a arabolic F -subgrou is not greater than that of its Levi subgrou see [4, Proosition 5.4]). Hence, the class number of G with resect to U turns out to be not greater than that of the Levi subgrou L defined by the Q-rational oints B B. Since the class number of B with resect to < O is one, the class number of G is also one, which means GA) = GQ)GR)U = UGR)GQ). This comletes the roof of. The first assertion imlies that there is a bijection Γ\GQ)/PQ) GR)U\GA)/PQ). Furthermore note an Iwasawa decomosition { GL O ) P Q ) v = < ) GQ v ) = K P R) v = ) at every lace v. We can then reduce the counting of the number of Γ-cuss to that of the class number of the Levi subgrou L. This imlies that the number of cuss with resect to Γ is one and comletes the roof of the lemma. We define Γ T as a subgrou GL O) generated by [ ] [ ] 0 β, 0 0 β O)..4) 7

8 M.Muto, H.Narita and A.Pitale In what follows, we will deal mainly with SΓ 0 ); 4 + r ) )), MΓ T ; r) and MGL O), r). For this we should note that the Selberg conjecture for Γ 0 ) is verified cf. [, Corollary.5]). This means the following: Proosition.4 If SΓ 0 ); 4 + r ) )) {0}, we have 4 + r ) 4, namely we can assume that r R. By Proosition., the Fourier exansion of F MGL O), r) is then written as F nx)a y ) = uy) + Cβ)y K r 4πνβ)y)eπ β S\{0} = uy) + Aβ)y K r π β y)eπ Reβx) β S\{0} with a smooth function u on R >0. Here trβx) S := Z ij) + Z i ij) + Z j ij) + Z ij.5) is the dual lattice of O with resect to the bilinear form on H H defined by Re = tr. Now we introduce ϖ = + i, which is a uniformizer of B Q Q. We can verify the following lemma by a direct comutation. Lemma.5 We have S = ϖ O. 3 Some zeta integral of convolution tye In this section, we study certain zeta integrals which will lay a crucial role in the roof of automorhy in Section 4. Theta functions Consider the sace of harmonic olynomials of degree l on H. These are homogeneous olynomials of degree l and are annihilated by the Lalace oerator in 4 variables. We can act on this sace by the cyclic grou of order 8 generated by +i H. Let {P l,ν } ν denote a basis for this sace consisting of eigenvectors under the above action. Hence, P l,ν + i x) = ϵ l,ν P l,ν x), 3.) for some eighth root of unity ϵ l,v. Define the following theta function Θ l,ν z) := β S P l,ν β)e π β z = m=0 bm)e π mz 3.) on h, where bm) := β S, β =m P l,νβ). Since, S is invariant under β β and P l,ν x) = ) l P l,ν x), we see that Θ l,ν z) is the zero function if l is odd. 8

9 Lifting to GL) over a division quaternion algebra Lemma 3. Let l be an even non-negative integer. Let Θ l,ν be as defined in 3.), with P l,ν satisfying 3.). Then Θ l,ν is a holomorhic modular form of weight l + with resect to Γ 0 ), and is a cus form if l. Moreover, we have the following transformation formula Proof. Set B = Θ l,ν z ) = ϵ l,ν l + z l+ Θ l,ν z). 3.3) and A = t BB. By.5), we see that the ma x Bx is a bijection from Z 4 to S. Here, we consider x as a column vector. For P l,ν in the statement of the lemma, set P x) := P l,ν Bx), x Z 4. Then P is a homogeneous olynomial in 4 variables of degree l annihilated by the oerator A = 4 i,j= One can then see that Θ l,ν z) = Θz; A, P ), where b i,j x i x j, where A = b i,j ). Θz; A, P ) = m Z 4 P m)e π t mam z is as defined in [0, Corollary 4.9.5]. Since, all diagonal entries of A and A are even, art 3) of Corollary of [0] imlies that Θ l,ν is a holomorhic modular form of weight l + with resect to Γ 0 ), and is a cus form if l. Once again, by art 3) of Corollary of [0], we have Θ z ; A, P ) = l+ z l+ Θz; A, P ), where A = A and P x) = P A x) = P l,ν t B x). Note, that the ma x t B x gives a bijection between Z 4 and O. Hence, we see that Θz; A, P ) = β O P l,ν β)e π β z = ϵ l,ν l m=0 bm)e π mz = ϵ l,ν l Θl,ν z). Here, we have used Lemma.5 and 3.). This comletes the roof of the lemma. Eisenstein series with resect to Γ 0 ) We introduce an Eisenstein series Ẽ z, s) := 4π) l Γs + + l) π s Γs)ζs)) Γs) on h with a comlex arameter s, where Γ := satisfies the following functional equation. 9 γ Γ \Γ 0 ) {[ m 0 cz + d cz + d )l+ Imz) cz + d )s 3.4) ] m Z}. The Eisenstein series

10 M.Muto, H.Narita and A.Pitale Lemma 3. Let Ẽ0z, s) := z z )l+ Ẽ z, s). Then the functional equation holds. Ẽ z, s) = s s Ẽ z, s) + s s ) s Ẽ 0 z, s) Proof. This is settled by the argument in [7, Lemma 7] with the hel of the formula for the scattering matrices in [, Section.]. Let f SΓ 0 ); 4 + r ) )) be an eigenfunction with resect to the Atkin-Lehner involution, i.e. f z ) = ϵfz) with some ϵ {±}. Suose the form f has the Fourier exansion fx + y) = cn)w 0, r 4π n y)e π nx, n Z\{0} where W 0, r denotes the Whittaker function with the arameter 0, r ). Let l be an even non-negative integer. Let Θ l,ν be as defined in 3.), with P l,ν satisfying 3.). Let Ẽ z, s) be the Eisenstein series defined in 3.4). Let the zeta integral Is) be defined by Is) := Γ 0 )\h fz)θ l,ν z)ẽ z, s)y l+ dxdy y. 3.5) By Lemma 3., the above integral is well-defined. Let us now state the theorem of this section. Theorem 3.3 The zeta integral Is) is entire and is bounded on vertical stris. When ϵϵ l,ν = we have s )Is) = s )I s). When ϵϵ l,ν = we have s + )Is) = s + )I s). Proof. The entireness and boundedness on vertical stris of Is) is verified by the same argument as [, Section 3.]. We ut I 0 s) := fz)θ l,ν z)ẽ0z, s)y l+ dxdy y. Γ 0 )\h [ / ] Since, Γ 0 ) is stable under conjugation by SL R), we can make a change of variable z /z). Now, using the assumtion f z ) = ϵfz), 3.3) and the definition of Ẽ 0 z, s), we have I 0 s) = f Γ 0 )\h z )Θ l,ν z )Ẽ0 z, s)im ) l+ dxdy z y [ ][ ][ = ϵfz) ϵ l,ν l + z l+ z ) l+ẽ y ] l+ dxdy Θ l,ν z) z, s)][ z z y Γ 0 )\h 0

11 Lifting to GL) over a division quaternion algebra = ϵ l,v ϵis). From Lemma 3., we deduce I s) = s s Is) + s s ) s I 0 s). Observe that s s ) ϵ s l,ν ϵs s = s s s + s + if ϵ l,ν ϵ =, if ϵ l,ν ϵ =. We have therefore roved the theorem. 4 Construction of the lifting We now construct a lifting ma from SΓ 0 ); 4 + r 4 )) to MGL O); r), which is an analogue of Pitale []. The fundamental tool of our study is the converse theorem by Maass [9]. Theorem 4. Maass) Let {Aβ)} β S\{0} be a sequence of comlex numbers such that and ut F nx)a y ) := Aβ) = O β κ ) κ > 0) β S\{0} Aβ)y K r π β y)eπ Reβx). For a harmonic olynomial P on H of degree l we introduce r r ξs, P ) := π s Γs + )Γs ) β S\{0} Aβ) P β) β s, which converges for Res) > l+4+κ. Let {P l,ν } ν be a basis of harmonic olynomials on H of degree l. Then F MΓ T ; r) is equivalent to the condition that, for any l, ν, the ξs, P l,ν ) satisfies the following three conditions.. it has analytic continuation to the whole comlex lane.. it is bounded on any vertical stri of the comlex lane. 3. the functional equation ξ + l s, P l,ν ) = ) l ξs, ˆP l,ν ) holds, where ˆP x) := P x) for x H.

12 M.Muto, H.Narita and A.Pitale Recall that Γ T is defined in.4). For this theorem we remark that the infinitesimal action of the Casimir oerator Ω on the sace of smooth right K-invariant {[ functions ] on G/Z can be } identified y x with a constant multile of the hyerbolic Lalacian on x H, y R 0 + for the hyerbolic Lalacian see [9, 3)]). We can therefore follow the argument in [9] to see that this theorem is useful also for our situation. We wish to define {Aβ)} β S\{0} from Fourier coefficients cn) of f SΓ 0 ); 4 + r 4 )). Let ϖ = + i, as before. An easy comutation shows that ϖ O = Oϖ. This allows us to write any β O uniquely as β = ϖ udβ, where u 0, an odd integer d and β O is neither of the form ϖ β 0 with some non-zero β 0 O nor a multile of an element of O by an odd integer. Hence, we can define ϖ m β by m u with u as above. Recall that, by Lemma.5, we have S = ϖ O. It thus makes sense to define the set S rim of rimitive elements in S by S rim := {β S \ {0} ϖ β, ϖ β, d β for all odd integer d}, 4.) where d β for d Z means that β is not a multile of an element in S by d. Proosition 4. Let β S \ {0} be exressed as β = ϖ u dβ 0, where u is a non-negative integer, d an odd integer and β 0 S rim. Given f SΓ 0 ); 4 + r 4 )) with Fourier coefficients cn) and eigenvalue ϵ {±} of the Atkin-Lehner involution, we set Aβ) := β u ϵ) t c β t+ ). 4.) n t=0 n d Let {P l,ν } ν be a basis of harmonic olynomials on H of degree l satisfying 3.). Then we have ξs + l +, P l,ν) = { l π l+) s ϵϵ l,ν )Is) if ϵ l,ν {±}, 0 if ϵ l,ν {±}) and the ξs, P l,ν ) satisfies the three analytic conditions in Theorem ) Proof. Note that, if l is odd, then ξs, P l,ν ) 0, since Aβ) is invariant under β β and P l,ν is homogeneous of degree l. For this we remark that, from Section 3, Is) 0 also holds when l is odd. From now on we will assume that l is even. Now suose that the condition ϵ l,ν {±} is satisfied, which is equivalent to ϵ l,ν. We see that P l,νiβ) = ϵ l,ν P l,νβ) P l,ν β) for β H. In addition we note that Aiβ) iβ s = Aβ) β s for β S. In the definition of ξs, P l,ν ) we can relace S by is. Hence, we obtain ξs, P l,ν ) 0, if ϵ l,ν {±}. We now assume that ϵ l,ν {±}. By a formal calculation similar to [, Proosition 3.5] we get Is) = π s s Γs + l + + r )Γs + l + r )ζs) c m)bm). m= m s+ l

13 Lifting to GL) over a division quaternion algebra Put Γ l,r s) := π s+l+) Γs + l + + r )Γs + l + r ). We have ξs + l +, P Aβ)P l,ν β) l,ν) = Γ l,r s) β s+ l + ) = Γ l,r s) = Γ l,r s) = Γ l,r s) β S\{0} β S\{0} u β t=0 n d c ) ϵ) t P t+ n l,ν β) u β S\{0} t=0 n d c β u u=0 d β S rim t=0 n d d:odd u = Γ l,r s) = Γ l,r s) = l u=0 d d:odd u=0 t=0 β ) s+ l t+ n ) ϵ) t P l,ν t n ) s β ) s+ l t n n d β S rim s ϵ l,ν ϵ) u d d:odd d s c ϖu t d β t/ n ) n β 0 ) ϵϵ l,ν ) t P l,ν ϖ u t d n β 0) t n ) s ϖ u t d n β 0 s+ l ) c ϖt nβ )P l,ν ϖ t nβ) s ϵ l,ν ϵ) u t d n )s ϖ t nβ s+ l ) c m) m= s s + ϵ l,ν ϵ s Γ l,r s)ζs) = l π l+) s ϵ l,ν ϵ)is). β S, β =m m) s+ l c m)bm) m= m s+ l P l,ν β) We note that ξs, ˆP l,ν ) = ξs, P l,ν ) since S = S. The formula just roved and Theorem 3.3 then imly that ξs, P l,ν ) satisfies the desired three analytic roerties in Theorem 4.. Theorem 4.3 Let f SΓ 0 ); 4 + r 4 )) with Fourier coefficients cn) and with eigenvalue ϵ of the Atkin Lehner involution. Define F f nx)a y ) := Aβ)y K r π β y)eπ Reβx) β S\{0} with {Aβ)} β S\{0} defined by 4.). Then we have F f MGL O); r) and F f is a cus form. Furthermore, F f 0. [ ] u 0 Proof. We can verify the left invariance of F f with resect to { u O 0 } in a straightforward way. Proosition., Theorem 4. and Proosition 4. thus imly F f MGL O); r). Since GL O) has only one cus see Lemma.3), the Fourier exansion of F f means that F f is cusidal. To show the non-vanishing we need the following lemma: Lemma 4.4 Let f SΓ 0 ); 4 + r 4 )) with Fourier coefficients cn) and with eigenvalue ϵ of the Atkin Lehner involution. Then, there exist N > 0, N Z, such that c N) 0. 3

14 M.Muto, H.Narita and A.Pitale Proof. Assume that cn) = 0 for all n < 0. Set f z) = fz) + f z))/ and f z) = fz) f z))/. Then, f, f are elements of SΓ 0 ); 4 + r 4 )) with the same eigenvalue ϵ of the Atkin Lehner involution as f. In addition, f is an even Maass form and f is an odd Maass form, with the roerty that they have the exact same Fourier coefficients corresonding to ositive indices. This imlies that the L-functions for f and f satisfy Ls, f ) = Ls, f ). On the other hand, Ls, f ) and Ls, f ) satisfy functional equations with the gamma factors shifted by. Here, we use that both f and f have the same Atkin Lehner eigenvalue. If Ls, f ) 0, we obtain an identity of gamma factors, which can be checked to be imossible. This gives us that f has to be zero, a contradiction. Let N 0 be the smallest ositive integer such that c N 0 ) 0. Let β 0 O be such that β 0 = N 0. Choose, β = ϖ β 0. Then, by the choice of N 0 and definition of Aβ), we see that Aβ) = N 0 c N 0 ) 0, as required. Remark 4.5 Weyl s law for congruence subgrous of SL Z) by Selberg cf. [, Section.]) imlies that there exist Maass cus forms for Γ 0 ). This and the theorem above imly the existence of non-zero lifts F f. 5 Actions of Hecke oerators on the lifting 5. Adelization of automorhic forms To study the actions of Hecke oerators on our cus forms constructed by the lifting we need both adelic and non-adelic treatments of automorhic forms. For a comlex number r C we introduce another sace MGA), r) of automorhic forms for G. Definition 5. Let MGA), r) be the sace of smooth functions Φ on GA) satisfying the following conditions:. Φzγgu f u ) = Φg) for any z, γ, g, u f, u ) Z A GQ) GA) U K, where Z A denotes the center of GA),. Ω Φg ) = r 4 + )Φg ) for any g GR) = GL H), 3. Φ is of moderate growth. According to art ) of Lemma.3, the class number of G with resect to U is one, which means that GA) = GQ)GR)U. We can thus view F MGL O), r) as a smooth function Φ F on GA) by Φ F γg u f ) = Φ F g ) γ, g, u f ) GQ) GR) U. We therefore see the following: Lemma 5. We have an isomorhism MGL O), r) MGA), r). 4

15 Lifting to GL) over a division quaternion algebra 5. Hecke oerators For each lace let G := GL B ) with B = B Q Q. For a finite rime, we have GL B ) GL 4 Q ). Let O be the -adic comletion of O for <. For a finite rime, O M Z ) and GL O ) GL 4 Z ). Set K = GL O ) for <. We denote by H the Hecke algebra for GL B ) with resect to GL O ) for <. According to [6, Section 8, Theorem 6], H has the following generators: { {φ ±, φ } if =, {ϕ ±, ϕ, ϕ 3, ϕ 4 } if. Here φ, φ denote the characteristic functions for [ ] [ ] ϖ 0 ϖ 0 K K 0 ϖ, K K 0 5.) resectively, and ϕ, ϕ, ϕ 3, ϕ 4 denote the characteristic functions for K K, K K, K K, K K 5.) resectively when. Recall that ϖ denotes a rime element of B. We want to obtain the single coset decomosition for the above double cosets. For that, we next review the Bruhat decomosition of K given by K = T wt, w W where W denotes the Weyl grou of GL B ), and T the subgrou of elements in K which are uer triangular modulo. Let N be the standard maximal uniotent subgrou of GL B ) defined over Q. For this we note that N for an odd is not isomorhic to that for =. We ut N 0 Z ) := T t N Q ). We furthermore introduce NZ ) = N Q ) K and DZ ) = D Q ) K, where D denotes the subgrou of diagonal matrices in GL B ). Then we have T = NZ )DZ )N 0 Z ) see [3, Theorem.5]). Let h be one of 4 or ),,,, [ ] or ϖ. Lemma 5.3 K hk = NZ )whk, w W /W h) where W h) := {w W whw = h}. 5

16 M.Muto, H.Narita and A.Pitale To describe this coset decomosition of K hk exlicitly we need a set of reresentatives for W /W h). [ ] ϖ 0 Lemma 5.4. Let =. For h = we can take 0 { [ ]} 0, 0 as a set of reresentatives for W /W h).. Let. a) When h = 4, we can take , , as a set of reresentatives for W /W h). b) When h = we can take , 0 0 0, 0 0 0, as a set of reresentatives for W /W h). c) When h = we can take { 4, , , } as a set of reresentatives for W /W h) , , 0 0 0

17 Lifting to GL) over a division quaternion algebra By Lemma 5.3 and Lemma 5.4 we are now able to write down the coset decomosition of K hk exlicitly. [ ] ϖ 0 Lemma 5.5. Let = and h =. We have 0 [ ] 0 [ ] [ ϖ 0 ϖ K hk = K 0 ϖ x ] K Let. a) For h = K hk = b) For h = we have x 4,x 4,x 34 Z /Z x Z /Z x 3,x 3 Z /Z K hk = we have x,x 3,x 4 Z /Z x 34 Z /Z x 3,x 4 Z /Z x O /ϖ O x 4 x 4 x 34 K x K x 3 x 3 K 7 K. x x 3 x 4 K x 34 K x 3 x 4 K

18 M.Muto, H.Narita and A.Pitale K. c) For h = we have K hk = x 3,x 4,x 3,x 4 Z /Z x 3 x 4 x 3 x 4 K x 4,x 34 Z /Z x 4 x 34 K x 3 Z /Z x 3 K x,x 4,x 34 Z /Z x x 4 x 34 K x,x 3 Z /Z x x 3 K K. We can now describe the actions of Hecke oerators defined by K hk s above. With the invariant measure dx of G normalized so that K dx =, we define K hk Φ by K hk Φ)g) := G char K hk x)φgx)dx for Φ MGA), r), where char K hk denotes the characteristic function for K hk. We rovide the non-adelic descrition of K hk Φ, which enables us to describe exlicitly the influence of the K hk -action on Fourier coefficients of the lifting F f. To this end we need the two following lemmas. 8

19 Lifting to GL) over a division quaternion algebra Lemma 5.6 Let Φ MGA), r) be a cus form. For all finite rime, Φ satisfies Φn g) = Φg) for any n, g) N GA), where we view N as a subgrou of GA). Proof. The GA)-module generated by Φ is a finite sum of irreducible cusidal reresentations of GA). As is well-known, an irreducible automorhic reresentation of GA) decomoses into a restricted tensor roduct of irreducible admissible reresentations of G v = GQ v ) over all laces v. Since Φ is right K -invariant for any finite rime, every non-archimedean local comonent of an irreducible summand for the cusidal reresentation generated by Φ is an irreducible subreresentation of an unramified rincial series reresentation see [4, 4.4]). The statement is a consequence of the left N -invariance of such irreducible admissible reresentations of G for <. We now introduce the set C := {α O να) = }/O, C := {x M Z ) detx) = }/GL Z ). 5.3) Lemma 5.7. There is a bijection {[ ] C 0, 0 [ ] b b Z/Z}. 0. For an odd rime the isomorhism O M Z ) induces the bijection C C. Proof. The first assertion is verified by a direct calculation. We rove the second assertion. As is remarked in the roof of [, Proosition 5.] we have #C = +. Under the isomorhism O M Z ), we can regard any elements in O as those in M Z ). Any two inequivalent reresentatives of C are not equivalent to each other in {x M Z ) detx) = }/GL Z ). Otherwise there are two inequivalent reresentatives α and α of C which are equivalent under O l -action for all rime l, since α /α O l for all rime l. Here note that O l GL Z l ) for any odd rime l. This however imlies that such two reresentatives are equivalent to each other in C. We therefore know that there is a injection from C into {x M Z ) detx) = }/GL Z ). Since the latter set also has + reresentatives as in the statement the injection is actually a bijection. Let F MGL O), r) corresond to Φ. By K hk F we denote the cus form in MGL O), r) corresonding to K hk Φ. Due to Lemma.3, GQ)\GA)/U has a comlete set of reresentatives in GR), where GR) is viewed as a subgrou of GA) in the usual manner. The cus form K hk Φ is thereby determined by its restriction to GR), which is nothing but K hk F. Moreover, we remark that an element in MGL O), r) is determined by its restriction to NA = {nx)a y x H, y R + } see.)). 9

20 M.Muto, H.Narita and A.Pitale Proosition 5.8 Let K hk F be as above and let nx) and a y be as defined in.). For odd, let C be as defined in 5.3).. Let =. We have. Let. a) When h = b) When h = c) When h = K hk F )nx)a y ) = F nϖ x)a y ) + F nϖ x)a y ). we have K hk F )nx)a y ) = we have K hk F )nx)a y ) = we have α C F nαx)a y ) + α C F nxα)a y ) + α C F nxα )a y ). α C F nα x)a y ). K hk F )nx)a y ) = F nx)a y ) + 4 F n x)a y) + xα )a y ). α,α ) C C F nα Proof. We rove only c). The other cases are settled similarly. The left coset decomosition of K hk in art ) iii) of Lemma 5.5 can be rewritten as x 3,x 4,x 3,x 4 Z /Z x 4,x 34 Z /Z x 3 x 4 x 3 x 4 x 4 0 x 34 K K

21 Lifting to GL) over a division quaternion algebra x 3 K x 4 x x 34 K x 3 x K x 3 Z /Z x,x 4,x 34 Z /Z x,x 3 Z /Z K. We regard the identity resectively the zero 0) of B as the identity v resectively 0 v ) of B v for v. Taking Lemma 5.6 into consideration, the contribution of the first and the last left cosets to K hk Φg) can be written as 4 Φ v, v [ v 0 v 0 v v ] [ ] 0 )g) + Φ 0 v, v [ v 0 v 0 v v and that of the remaining four cosets to K hk Φg) as [ ] [ ] v 0 Φ v α 0 0 v v 0 α )g) for g GR) GA). In GQ)\GA)/U, v, v v, v α,α ) C C [ v 0 v 0 v v [ v 0 v 0 v v ] ] v, v [ ] 0 = [ v 0 v 0 0 v v v< [ ] 0 = [ v 0 v 0 0 v v v< ] ] ] [ ] 0 )g) 0 [ ] 0, 0 [ ] 0 0. By Lemma 5.7 the cosets reresented by [ ] [ ] v 0 v α v, v 0 0 v v 0 α with α, α C are [ ] α in bijection with the cosets in GL O)\GR) reresented by { 0 0 α α, α ) C C }. Now let us ut g := [ ] v 0 v v< nx)a 0 v y GR) GA). We therefore have v [ ] [ ] 4 v 0 Φ v 0 [ ] [ ] v 0 )g) + Φ v 0 )g) 0 v v 0 0 v v 0 v, v v, v [ ] [ ] = F nx)a 0 y ) + F 0 nx)a y )

22 M.Muto, H.Narita and A.Pitale and α,α ) C C Φ v, v [ v 0 v 0 v v ] [ ] α 0 0 α )g) = F α,α ) C C [ α 0 0 α ] nx)a y )). Noting the invariance of F with resect to K and Z + see.) for Z + ), we deduce the assertion from this by a direct comutation. For this roosition we remark that the formulas above do not deend [ on ] the choices of u reresentatives of C since F is left and right invariant with resect to { u, u O }. Let the Fourier decomosition of K hk F ) be given by K hk F )nx)a y ) = K hk F ) β y K r π β y)eπ Reβx). β S\{0} The next roosition rovides a formula for K hk F ) β in terms of the Fourier coefficients Aβ) of F. u Proosition 5.9. Let =. We obtain. Let be an odd rime and β S \ {0}. a) When h =, b) When h = K hk F ) β = Aβϖ ) + Aβϖ )). K hk F ) β =, K hk F ) β = α C Aβᾱ ) + α C Aα β) + α C Aᾱβ)). α C Aβα)). c) When h =, K hk F ) β = A β) + Aβ) + Aα βα )). α,α ) C C

23 Lifting to GL) over a division quaternion algebra [ u For this roosition we note that the automorhy of F with resect to { u ] u, u O } imlies Au βu ) = Aβ) for β S \ {0} and u, u O. From this we see that the formulas in )b) and )c) do not deend on the choices of reresentatives for C. As for )a) we furthermore see that, given any comlete set {α i i + } of reresentatives for C, {ᾱ i i + } also forms such a set. As a result we see that the formula in )a) is also not deendent on the choices of reresentatives for C. 5.3 Hecke equivariance for = Let f SΓ 0 ); 4 + r 4 )) be a new form for the definition see [, Section 8.5]) with Hecke eigenvalue λ for =. By the Hecke eigenvalue λ we [ mean ] the eigenvalue of f for the U) oerator defined by the action of the double coset Γ 0 ) Γ 0 ). Let us also assume that f is an eigenfunction of the Atkin Lehner involution with eigenvalue ϵ. It can be checked that λ and ϵ are related by λ = ϵ. 5.4) Using the single coset decomosition [ ] [ ] [ ] Γ 0 ) Γ 0 ) = Γ 0 ) Γ 0 ), we get f z + ) + fz ) = λ fz). In terms of Fourier coefficients of f, using 5.4), we get cm) = λ cm) = ϵ cm), for all m Z. 5.5) Proosition 5.0 Let f SΓ 0 ); 4 + r 4 )) be a new form with Hecke eigenvalue λ for = and an eigenfunction of the Atkin Lehner involution with eigenvalue ϵ. Let F = F f be as defined in Theorem 4.3. Then [ ] ϖ K K )F = 3 ϵf. 5.6) Proof. Let β = ϖ u dβ 0 be a decomosition according to Proosition 4.. Hence, u 0, d is odd and β 0 S rim. Using 5.4) and 5.5) we see that Aβ) = u+ ) β β ) c n, n d Aβϖ ) = u+ ) ϵ β β ) c n, n d Aβϖ ) = u ) ϵ ) β β ) c n. n d 3

24 M.Muto, H.Narita and A.Pitale Note that, if u = 0, then Aβϖ ) = 0 and so is the right hand side of the third equality above. We have u+ + u ) = 3 u+ ). Hence, we have ) Aβϖ ) + Aβϖ ) = 3 ϵaβ). The roosition now follows from art. of Proosition Hecke equivariance for odd rimes We assume that f SΓ 0 ); 4 + r 4 )) is a Hecke eigenform with Hecke eigenvalue λ for every odd rime but do not assume that f is a new form. In terms of Fourier coefficients of f, the Hecke relation is given by cn) + cn/) = λ cn), 5.7) where cn/) is assumed to be zero if does not divide n. The following lemma will lay a key role in the comutation of the Hecke oerator. Lemma 5. Let β S rim. Then #{α C : βα} = #{α C : αβ} = { if β, 0 if β. 5.8) In addition, does not divide αβ or βα for any α C. Proof. Note that, by taking conjugates, it is enough to rove the statement of the lemma for {α C : βα} for all β. Taking norms, it is clear that does not divide βα if does not divide β. Hence, assume that divides β. Let β = β + β i + β 3 j + β 4 ij. The conditions β and β, imly that there is a air amongst the set {β, β, β 3, β 4 } which does not satisfy x +y 0 mod ). From the roof it will be clear that we can take, without loss of generality, β3 + β 4 0 mod ). Let α = α + α i + α 3 j + α 4 ij. The condition βα is equivalent to the following matrix equation modulo. β β β 3 β 4 α 0 0 β β β 4 β 3 β 3 β 4 β β β 4 } β 3 β {{ β } P β α α 3 α 4 =. The matrix P β considered over Z has rank. By our assumtion β3 + β 4 0 mod ), we see that the kernel of P β is sanned by the first two rows of P β. Hence, α is given by α = aβ + bβ, α = aβ + bβ, α 3 = aβ 3 bβ 4, α 4 = aβ 4 + bβ 3, for some a, b Z. This gives us α3 +α 4 = a +b )β3 +β 4 ) 0. This is because, by assumtion β3 + β 4 0, and α = a + b ) β and does not divide α

25 Lifting to GL) over a division quaternion algebra On Pg 69 of [3], it has been shown that the set S = {α C : α3 + α 4 0 mod )} is in bijection with the set S = {x, y) Z Z : x + y + 0 mod )}. The ma from S to S is given as follows. For α S, we obtain x α, y α ) as the solution to the matrix equation modulo given by [ ] [ ] [ ] α3 α 4 xα α =. α 4 α 3 y α α One can check that x, y) S for the following choice of x and y. x = β β 3 β β 4 β 3 + β 4, y = β β 3 β 4 β β ) β 4 If one takes α S to be the re-image of the above x, y), then we can check that α KerP β ). On the other hand, if α C belongs to KerP β ), then it can also be checked that the corresonding x α, y α ) are equivalent modulo to those in 5.9). This comletes the roof of 5.8). If divides βα or αβ for some α C then it is clear that β cannot be rimitive. This comletes the roof of the lemma. Proosition 5. Let f SΓ 0 ); 4 + r 4 )) be a Hecke eigenform with Hecke eigenvalue λ for every odd rime. Let F = F f be as defined in Theorem 4.3. For an odd rime we then have K K )F = K K )F = + )λ F. 5.0) Proof. Using Proosition 5.9 we can show that, if the Fourier coefficients satisfy Aβ) = A β) for all β S and the second equality in 5.0) holds, then so does the first equality. Since, the Fourier coefficients of F = F f satisfy the above condition, we are reduced to showing the second equality in 5.0). We will comute the action of the Hecke oerator on the Fourier coefficients Aβ) of F. Since, all the comutations only involve the rime, it will be enough to consider the case β = s β 0 with s 0 and β 0 S rim. For such a β, we have Aβ) = s β 0 s β0 s k ) c. Note that α β 0 = ᾱβ 0 for α B with να) =. Hence, for such α, α β 0 S if and only if divides ᾱβ 0. Let us first consider the case where does not divide β 0. Hence, by Lemma 5., we see that α β 0 S and does not divide β 0 α for any α C. Hence, for any α C, we have Aβα) = s β 0 s β0 s+ k ) c 5.) 5

26 M.Muto, H.Narita and A.Pitale and Aα β) = s s β0 s k ) β 0 c. 5.) Note that, if s = 0, both the left and right hand side of the last equation are zero. Now, using 5.7), we get for any α C, s Aα β) + Aβα) = s β 0 / β0 s k ) c + / β0 s+ k )) c + / β0 )) c s = s β0 s k ) β0 )) β 0 λ c + λ c = λ Aβ). Now, using the fact that the number of elements in C is + and art b) of Proosition 5.9, we get the result. Next, let us assume that divides β 0. By Lemma 5., there is a unique α C such that divides β 0 α and a unique α C such that α β 0 S. If α C but α α then the formula for Aβα) is the same as in 5.). For α = α, we have Aβα ) = s+ s β0 s+ k ) β 0 c. For α C but α α, the formula for Aα β) is the same as in 5.). For α = α, we have Hence, we get the following, α C Aα β) + Aβα)) Aα β) = s β 0 s β0 s k ) c. [ s = s β 0 / β0 s+ k ) s c + / β0 s k )] c + s β 0 = s β 0 [ s+ [ s / β0 s+ k c β0 s k λ c ) + ) + / c s / β0 s k )] c β0 )] [ s + s β0 s k ) β 0 λ c + / β0 )] c 6

27 Lifting to GL) over a division quaternion algebra [ s = λ Aβ) + s β0 s k ) β 0 λ c + / β0 ) c + / β0 )] c = λ Aβ) + λ Aβ) = + )λ Aβ). This comletes the roof of the roosition. Proosition 5.3 Let f SΓ 0 ); 4 + r 4 )) be a Hecke eigenform with Hecke eigenvalue λ for every odd rime. Let F = F f be as defined in Theorem 4.3. For an odd rime we then have K Proof. First observe that, using 5.7), one can show that, for all n, K )F = λ ) F. 5.3) cn ) = λ )cn) / λ cn/). 5.4) If we assume that n, then we can get another identity given by cn ) + cn/ ) = λ )cn). 5.5) As in the roof of Proosition 5., we can assume that β = s β 0, where s 0 and β 0 S rim. Let us abbreviate ν β) = s. For such a β we have Hence, and Aβ) = s β 0 s β0 s k ) c. s+ Aβ) = s+ β0 s k+ ) β 0 c s A β) = s β0 s k ) β 0 c. Next, we need to comute Aα βα ) where the sum is over all α, α in C. We consider three cases deending on whether β 0 or β 0 but β 0 or β 0. Case : Let us assume that β 0. Alying Lemma 5. to β 0, we see that β 0 α S rim for all α C. Again alying Lemma 5. to β 0 α for a fixed α, we see there is a unique α, C such that ν α, βα ) = s and, for all α α,, we have ν α βα ) = s. Hence, Aα βα ) = α,α C α C Aα, βα ) + 7 α C α α, ) Aα βα )

28 M.Muto, H.Narita and A.Pitale s = + )Aβ) + + ) s β0 s k ) β 0 c. Putting this all together, we see that Aβ) + A β)) + α,α Aα βα ) is equal to s+ s+ β0 s k+ ) β 0 c s + s β 0 s + + )Aβ) + + ) s β0 s k ) β 0 c β0 s k )) c s = s β 0 λ β0 s k ) β0 ) β0 )) ) c + c + c s + + )Aβ) + + ) s β0 s k ) β 0 c s = s β 0 λ β0 s k ) ) c λ β0 )c β0 + c = λ ) + + ) + + ) ) Aβ) = λ ) Aβ). ) + λ )c )) + + )Aβ) + + ) Aβ) + ) s β 0 c β0 ) β0 ) Here, we have used both 5.4) and 5.5). Case : Let β 0 but β 0. Alying Lemma 5. to β 0, we see that there is a unique ˆα C such that β 0 ˆα. For α ˆα, we have β 0 α S rim. Let β 0 ˆα = β 0. Then β 0 Srim and β 0 since we have assumed that β 0 ). Hence, by Lemma 5., we see that, for all α C, we have α β 0 ˆα = ᾱ β 0 Srim. This imlies ν α β ˆα ) = s for all α C. If α ˆα, then Lemma 5. imlies that there is a unique α, C such that ν α, βα ) = s. For all α α,, we have ν α βα ) = s. This gives us Aα βα ) = Aα β ˆα ) + α,α C α C α C α ˆα Aα, βα ) + α C α α, s = + )Aβ) + Aβ) + s β0 s k ) β 0 c. ) Aα βα ) Putting this all together, we see that Aβ) + A β)) + α,α Aα βα ) is equal to s s β 0 λ β0 s k ) β0 ) β0 )) ) c + c + c s + + )Aβ) + Aβ) + 3 s β0 s k ) β 0 c 8

29 Lifting to GL) over a division quaternion algebra = λ )Aβ) + s β 0 λ β0 ) )c + λ β0 ) β0 )) )c + c + + )Aβ) + Aβ) + 3 Aβ) 3 s β0 ) β 0 c = λ ) + + ) + + 3) Aβ) = λ ) Aβ). Here, we have used 5.5) and β 0. Case 3: Let β 0. As in Case above, Lemma 5. alied to β 0 imlies that there is a unique ˆα C such that β 0 ˆα. For α ˆα, we have β 0 α S rim. Let β 0 ˆα = β 0. Then β 0 Srim and β 0 since we have assumed β 0 ). Hence by Lemma 5., there is a unique ˆα, C such that ν ˆα, β ˆα ) = s +, and for all α ˆα,, we have ν α β ˆα ) = s. If α ˆα, then Lemma 5. imlies that there is a unique α, C such that ν α, βα ) = s. For all α α,, we have ν α βα ) = s. This gives us that Aα βα ) is equal to Aˆα, β ˆα ) + Aα β ˆα ) + Aα, βα ) + ) Aα βα ) α ˆα, α ˆα α α, s+ = s β0 s k ) s β 0 c + Aβ) + Aβ) + s β0 s k ) β 0 c =Aβ) + s β0 ) β 0 c + Aβ) + Aβ) + Aβ) s β0 ) β 0 c. Putting this all together, we see that Aβ) + A β)) + α,α Aα βα ) is equal to λ )Aβ) + s β 0 λ )c ) β0 ) + c + c β0 ) β0 )) c β )Aβ) + s β 0 c = λ ) ) ) Aβ) = λ ) Aβ). β0 )) Here, we have used 5.5) and β 0. This comletes the roof of the roosition. 6 The automorhic reresentation corresonding to the lifting In this section, we will use the Hecke equivariance from the revious section to determine the local comonents of the automorhic reresentation corresonding to the lifting. This will lead us to the conclusion that we have obtained a CAP reresentation and have found a couterexamle of the Ramanujan conjecture. 9

30 M.Muto, H.Narita and A.Pitale 6. The local comonents of the automorhic reresentation Let f SΓ 0 ); 4 + r 4 )) be a Hecke eigenform with Hecke eigenvalue λ for every odd rime. Let F = F f be as defined in Theorem 4.3. Let Φ F : GA) C be defined by Φ F γg u f ) = F g ) γ, g, u f ) GQ) GR) U. See Section 5. for details. Let π F be the irreducible cusidal automorhic reresentation of GA) generated by the right translates of Φ F. Note that the irreducibility follows from the strong multilicity one result for GA) see [], [3]). The reresentation π F is cusidal since F is a cus form. Let π F = π, where π is an irreducible admissible reresentation of GQ ) for < and π is an irreducible admissible reresentation of GR). Recall U = < K where K is the maximal comact subgrou of G cf. Section.3). Hence, for <, the reresentation π is a sherical reresentation and can be realized as a subreresentation of an unramified rincial series reresentation, i.e. a reresentation induced from an unramified character of the Borel subgrou. The reresentation π is comletely determined by the action of the Hecke algebra HG, K ) on the sherical vector in π, which in turn, is comletely determined by the Hecke eigenvalues of F obtained in the revious section. See [4] for details. For = we need to assume that f is a new form for the determination of Hecke eigenvalue of F f cf. Section 5.3). Descrition of π for odd If is an odd rime, then we have G = GL 4 Q ) and K = GL 4 Z ). Given 4 unramified characters χ, χ, χ 3, χ 4 of Q, we obtain a character χ of the Borel subgrou P of uer triangular matrices in G, by The modulus character δ P is given by a χ a a 3 ) = χ a )χ a )χ 3 a 3 )χ 4 a 4 ). 6.) a 4 a δ P a a 3 ) = a3 a a a 4 3 a 3 4, 6.) where denotes the -adic absolute value. The unramified rincial reresentation corresonding to χ is given by Iχ) which consists of locally constant functions f : GL 4 Q ) C, satisfying fbg) = δ P b) / χb)fg), for all b P, g GL 4 Z ). The action of the Hecke algebra is as follows. If ϕ HGL 4 Q ), GL 4 Z )) and f Iχ), define ϕ f ) g) = GL 4 Q ) ϕh)fgh)dh. 6.3) 30

31 Lifting to GL) over a division quaternion algebra Recall that we have normalized the measure dh on GL 4 Q ) so that the volume of GL 4 Z ) is. Let f = f 0, the unique vector in Iχ) that is right invariant under GL 4 Z ) and f 0 ) =, and ϕ = ϕ h a characteristic function of GL 4 Z )hgl 4 Z ) = i h i GL 4 Z ). It follows from 6.3) that ) ϕh f 0 ) = f 0 h i ) = µ h, 6.4) where µ h is determined by the reresentation π. The Hecke algebra HGL 4 Q ), GL 4 Z )) is generated by {ϕ ±, ϕ, ϕ 3, ϕ 4 }, defined in 5.). Lemma 6. Let µ, µ, µ 3, µ 4 be the constants obtained by the action of ϕ i, i =,, 3, 4 on the sherical vector f 0 in π according to 6.4). Then µ =, µ = µ 4 = + )λ and µ 3 = λ Proof. The lemma follows from the fact that the action of the -adic Hecke algebra on the sherical vector in π is exactly the same as the action of the -art of the classical Hecke algebra on F. First note that ϕ acts as the identity oerator, which imlies µ =. The other Hecke eigenvalues follow from Proositions 5. and 5.3. Recall that Proosition 5.5 gives the double coset decomositions which can be used to determine the action of ϕ on f 0. Let us abbreviate α i = χ i ) for i =,, 3, 4. Working in the induced model Iχ) of π, we see that i ϕ f 0 )) = α α α 3 α 4, 6.5) ϕ f 0 )) = 3 3/ α α α 3 + / α α 3 α 4 + / α α α 4 + 3/ α α 3 α 4 ) = 3/ α α α 3 α 4 α + α + α3 + α4, ϕ4 f 0 )) = 3 3/ α + / α 3 + / α + 3/ α 4 = 3/ α + α + α 3 + α 4 ), ϕ3 f 0 )) = 4 α α + α α 3 + α α α α 3 + α α 4 + α 3 α 4 = α α + α α 3 + α α 4 + α α 3 + α α 4 + α 3 α 4 ). Proosition 6. Let f SΓ 0 ); 4 + r 4 )) be a Hecke eigenform with Hecke eigenvalue λ for every odd rime. Let F = F f be as defined in Theorem 4.3. Let π F = π be the corresonding irreducible cusidal automorhic reresentation of GA). For an odd rime, the reresentation π is the unique sherical constituent of the unramified rincial series reresentation Iχ) where, u to the action of the Weyl grou of GL 4, the character χ is given by λ χ ) = / + λ 4 λ, χ ) = / λ 4, 6.6) λ χ 3 ) = / + λ 4 λ, χ 4 ) = / λ 4. 3