Petersson's Trace Formula and the Hecke

PETERSSON’S TRACE FORMULA AND THE HECKE EIGENVALUES OF HILBERT MODULAR FORMS

ANDREW KNIGHTLY AND CHARLES LI

Abstract. Using an explicit relative trace formula, we obtain a Petersson trace formula for holomorphic Hilbert modular forms. Our main result ex- presses a sum (over a Hecke eigenbasis) of products of Fourier coeﬃcients and Hecke eigenvalues in terms of generalized Kloosterman sums and Bessel functions. As an application we show that the normalized Hecke eigenvalues for a ﬁxed prime p have an asymptotic weighted equidistribution relative to a polynomial times the Sato-Tate measure, as the norm of the level goes to ∞.

1. Introduction

Let h ∈ Sk(Γ0(N)) be a Hecke eigenform (k even), and for a prime p - N define h the normalized Hecke eigenvalue νp by k−1 − 2 h p Tph = νp h. h The Ramanujan-Petersson conjecture asserts that |νp | ≤ 2. This is a theorem of Deligne. Because we have assumed trivial central character, the operator Tp is self-adjoint, so its eigenvalues are real numbers, and thus h νp ∈ [−2, 2]. For a fixed non-CM newform h, the Sato-Tate conjecture predicts that the set h {νp : p - N} is equidistributed in [−2, 2] relative to the Sato-Tate measure ( q 1 1 − x2 dx when x ∈ [−2, 2], (1) dµ∞(x) = π 4 0 otherwise. Taylor has recently proven this in many cases when k = 2 [Ta]. When the prime p is fixed, the normalized eigenvalues of Tp on the space Sk(Γ0(N)) are asymptotically equidistributed relative to the measure p + 1 dµ (x) = dµ (x) p (p1/2 + p−1/2)2 − x2 ∞ as k + N → ∞. This was proven in the late 1990’s by Serre [Se], and independently (for N = 1) by Conrey, Duke and Farmer [CDF]. An extension of the result (in the level aspect only) to Hilbert modular forms is given in [Li2]. These results have an antecedent in work of Bruggeman in 1978, who proved that the eigenvalues of Tp on Maass forms of level 1 exhibit the same distribution ([Br], Sect. 4; see also Sarnak [Sa]). Even earlier, in 1968 Birch proved a similar result for sizes of elliptic curves over Fp [Bi].

The ﬁrst author was supported by NSA grant H98230-06-1-0039. Document date: Jan. 26, 2008. 2 ANDREW KNIGHTLY AND CHARLES LI

In this paper we prove a different result on the asymptotic distribution of Hecke eigenvalues, valid over a totally real number field F . The following is a special case of a more general result (cf. Theorem 6.6). Notation and terminology will be defined precisely later on. Theorem 1.1. Let F be a totally real number field, and let m be a totally positive −1 element of the inverse different d ⊂ F . For a cusp form ϕ on GL2(F )\ GL2(AF ) ϕ th with trivial central character, let Wm denote its m Fourier coefficient (see (10)). Define a weight |W ϕ (1)|2 w = m . ϕ kϕk2 Fix a prime ideal p - md. Let {ϕ} be an orthogonal basis of eigenforms of prime-to-p level N and weight (k1,..., kr), with all kj > 2 and even. For such ϕ, let ϕ νp ∈ [−2, 2] denote the associated normalized eigenvalue of the Hecke operator Tp. Then the ϕ wϕ-weighted distribution of the eigenvalues νp is asymptotically uniform relative to the Sato-Tate measure as the norm of N goes to ∞. This means that for any continuous function f : R −→ C, P ϕ Z ϕ f(νp )wϕ lim P = f(x)dµ∞(x). N(N)→∞ ϕ wϕ R Noteworthy is the fact that here the measure and the weights are independent of p, in contrast to Serre’s (unweighted) result. The Sato-Tate measure dµ∞ is nonzero on any subinterval of [−2, 2], so in particular the above illustrates the density of the Hecke eigenvalues in [−2, 2]. The proof of the above theorem involves a trace formula which may be of independent interest. In a previous paper [KL1], we detailed the way in which the classical 1932 Petersson trace formula can be realized as an explicit relative trace formula, as a conceptual alternative to the usual method using Poincaréseries. Here we extend the technique and result to cusp forms on GL2(AF ), where F is an arbitrary totally real number field. We work with spaces Ak(N, ω) of holomorphic Hilbert cusp forms of weight k = (k1,..., kr), all strictly greater than 2, on the Hecke congruence subgroups Γ0(N). Our main result is Theorem 5.11 in which a sum (over a Hecke eigenbasis) of terms involving the associated Fourier coefficients, Petersson norms and eigenvalues of a Hecke operator Tn, is expressed in terms of generalized Kloosterman sums and Bessel functions. The incorporation of Hecke eigenvalues is a novel feature of this generalized sum formula. In Section 6 we prove a more general version of the above weighted distribution theorem by an argument based on bounding the terms of the sum formula in terms of N(N), following the technique for the F = Q case from [Li1]. This is analogous to the way in which Serre’s result follows from bounding the terms of the Eichler-Selberg formula for tr(Tpm ). In Section 7 we give some variants of the generalized Petersson formula, including a simplified version for the case where F has narrow class number 1. In Corollary 7.3, setting n = 1 for the trivial Hecke operator, we recover the classical extension of Petersson’s formula to Hilbert modular forms, as follows from a 1954 paper of Gundlach, [Gu]. His results are valid for cusp forms of uniform weight k ≥ 2. Refer to §2 of [Lu] for an overview of the classical derivation. PETERSSON’S FORMULA AND HECKE EIGENVALUES 3

In 1980 Kuznetsov gave an analog of Petersson’s formula involving Maass forms and Eisenstein series on the spectral side, [Ku]. As an application, he used his formula to give estimates for sums of Kloosterman sums. His work was reformulated in a representation-theoretic setting by various authors, see [CPS] and its references, notably [MW] where the general rank-1 case is treated. More recently Bruggeman, Miatello and Pacharoni gave a general Kuznetsov trace formula for automorphic forms on SL2(F∞) of uniform even weight ([BMP], Theorem 2.7.1). An important application of their formula is an estimate for sums of Kloosterman sums. In contrast to the above results, here we are concerned only with holomorphic cusp forms, i.e. those whose infinity types are discrete series. For our test function, we take discrete series matrix coefficients at the infinite components. This serves to isolate the holomorphic part of the cuspidal spectrum. At the finite places we take Hecke operators, which introduces Hecke eigenvalues into the final formula. Thanks: We would like to thank Don Blasius for a helpful discussion about the Ramanujan conjecture. We also thank the referee for many insightful comments.

2. General setting We recall the general setting of Jacquet’s relative trace formula [Ja]. Full details for the discussion below are given in §2 of [KL1]. Let F be a number field, with adele ring A. Let G be a reductive algebraic group defined over F . Let H be an F -subgroup of G × G, with H(A) unimodular. We assume that H(F )\H(A) is −1 compact. Define a right action of H on G by g(x, y) = x gy. For g ∈ G, let Hg be the stabilizer of g, i.e. −1 Hg = {(x, y) ∈ H| x gy = g}. For δ ∈ G(F ), let [δ] be the H(F )-orbit of δ in G(F ), i.e. [δ] = {x−1δy| (x, y) ∈ H(F )}. Each element of [δ] can be expressed uniquely in the form u−1δv for some (u, v) ∈ Hδ(F )\H(F ). Let f be a continuous function on G(A), and let X (2) K(x, y) = f(x−1γy)(x, y ∈ G(A)) γ∈G(F ) be the associated kernel function. We assume that the above sum is uniformly absolutely convergent on compact subsets of H(A). In particular, K(x, y) is a continuous function on the compact set H(F )\H(A). Let χ(x, y) be a character of H(A), invariant under H(F ). Consider the expression Z (3) K(x, y)χ(x, y)d(x, y), H(F )\H(A) where d(x, y) is a Haar measure on H(A). A relative trace formula results from computing this integral using spectral and geometric expressions for K(x, y). Using the geometric expression (2), it is straightforward to show that the integral (3) is P equal to [δ] Iδ(f), where Z −1 Iδ(f) = f(x δy)χ(x, y)d(x, y). Hδ (F )\H(A) 4 ANDREW KNIGHTLY AND CHARLES LI

We see that Z "Z # −1 Iδ(f) = f(x δy) χ(rx, sy)d(r, s) d(x, y). Hδ (A)\H(A) Hδ (F )\Hδ (A)

An orbit [δ] is relevant if χ is trivial on Hδ(A). From the above, we see that Iδ = 0 whenever [δ] is not relevant. Indeed the expression in the brackets is equal R to χ(x0, y0) χ(r, s)d(r, s), where (x0, y0) ∈ H(A) is any representative Hδ (F )\Hδ (A) for (x, y), and this integral vanishes unless δ is relevant.

3. Preliminaries 3.1. Notation. Throughout this paper we work over a totally real number field F 6= Q. Let r = [F : Q], and let σ1, . . . , σr be the distinct embeddings F,→ R. Let ∞1,..., ∞r denote the corresponding archimedean valuations. Let O be the ring of integers of F . We will generally use Gothic letters a, b etc. to denote fractional ideals of F . We reserve p for prime ideals. Let v = vp be the discrete valuation corresponding to p. Let Ov be the ring of integers in the local field Fv, and let $v ∈ Ov be a generator of the maximal ideal pv = pOv. Let N : F → Q denote the norm map. For a nonzero ideal a ⊂ O, let N(a) = |O/a| denote the absolute norm. This extends by multiplicativity to the group of nonzero fractional ideals of F . For α ∈ F ∗, we define N(α) = N(αO) = | N(α)|. We also use the above norms in the local setting with the analogous meanings. We −1 normalize the absolute value on Fv by |$v|v = N(p) . By Dirichlet’s unit theorem, the unit group of F is O∗ =∼ (Z/2Z) × Zr−1. Letting O∗2 denote the subgroup consisting of squares of units, we see that O∗/O∗2 =∼ (Z/2Z)r is a finite group. Let ∗ (4) U = {u1, . . . , u2r } ⊂ O be a fixed set of representatives for O∗/O∗2. For a fractional ideal a ⊂ F , let ordv a (or ordp a) denote the order. Let av (or ordv (a) ap) denote its localization, so av = $v Ov. Let A denote the adele ring of F , with finite adeles Afin, so that

A = F∞ × Afin, ∼ r Q where F∞ = F ⊗ R = R . Let Ob = v<∞ Ov. Generally if a is a fractional ideal Q of F , we write ba = aOb = v<∞ av ⊂ Afin. Let Cl(F ) be the class group of F , of cardinality h(F ). For a fractional ideal a, let [a] represent its image in the class group. + + Let F denote the set of totally positive elements of F . We let F∞ denote the subset of F∞ of vectors whose entries are all positive. Let −1 F d = {x ∈ F | trQ(xO) ⊂ Z} −1 −1 + denote the inverse different. We set d+ = d ∩ F . PETERSSON’S FORMULA AND HECKE EIGENVALUES 5

We use the mathtt font to represent finite ideles. Thus we write ∗ (5) a ∈ Afin, ba = aOb, a = O ∩ ba. ∗ In the other direction, given a fractional ideal a ⊂ F , there is an element a ∈ Afin ordv a such that (5) holds. Explicitly, we can take av = $v . The element a is unique up to Ob∗. We define norms of ideles by taking the products of the local norms. For example, in the situation of (5), we have Y Y −1 −1 −1 |a|fin = |av|v = N(av) = N(a) = N(a) . v<∞ v<∞ We use Roman or Greek letters for rational elements a, α ∈ F . 3.2. Haar measure. We use Lebesgue measure on R, and take the product mea- ∼ r sure on F∞ = R . We normalize Haar measure on each non-archimedean comple- tion Fv by taking meas(Ov) = 1. This choice induces a Haar measure on Afin with meas(Ob) = 1, and because A = F + F∞ × Ob,

meas(F \A) = meas((F ∩ Ob)\(F∞ × Ob)) = meas((O\F∞) × Ob) 1/2 (6) = meas(O\F∞) = dF , where dF is the discriminant of F . The resulting measure on A is not self-dual. Let G = GL2. We normalize Haar measure on G(R) as follows. Use Lebesgue measure dx to define a measure dn on the unipotent subgroup N(R) =∼ R. On R∗ we use the measure dx/|x|. We take the product measure dm on the diagonal ∼ ∗ ∗ subgroup M(R) = R × R , and normalize dk on SO2(R) to have total measure 1. On G(R) = M(R)N(R) SO2(R) we take dg = dm dn dk. Let Y Y Kfin = Kv = G(Ov) = G(Ob) v<∞ v<∞ be the standard maximal compact subgroup of G(Afin). We normalize Haar measure on G(Afin) by taking meas(Kv) = 1 for all v < ∞. We let Z denote the center of G, and write G = G/Z.

We normalize Haar measure on G(Fv) by taking meas(Kv) = 1. 3.3. Hilbert modular forms. Let N ⊂ O be an integral ideal of F . Let k = (k1,..., kr) be an r-tuple of positive integers, each greater than or equal to 2. Let ω : F ∗\A∗ → C∗ Q be a unitary Hecke character. Write ω = v ωv. We assume that: (1) the conductor of ω divides N kj (2) ω∞j (x) = sgn(x) for all 1 ≤ j ≤ r. The first condition means that ωv is trivial on 1 + Nv for all v|N, and unramified for all v - N. In other words, ωfin is trivial on the open set ∗ def ∗ Y Y ∗ ObN = Ob ∩ (1 + NOb) = (1 + Nv) Ov. v|N v-N Thus ω can be viewed as a character of the ray class group mod N ∼ ∗ ∗ + ∗ Cl(N) = A /[F (F∞ × ObN)]. 6 ANDREW KNIGHTLY AND CHARLES LI

We freely identify Z(A) with A∗ throughout this paper. For example if z ∈ Z(A) we write ω(z). Let G = GL2 and let L2(ω) = L2(G(F )\G(A), ω) be the space of left G(F )-invariant functions on G(A) which transform by ω under the center and which are square integrable over G(F )\G(A). Let R denote the right 2 2 regular representation of G(A) on L (ω). Let L0(ω) be the subspace of cuspidal 2 functions. We know that the restriction of R to L0(ω) decomposes as a discrete sum of irreducible representations. These are the cuspidal representations π. Every such π factorizes as a restricted tensor product of admissible local representations ⊗πv. Deﬁne groups a b K (N) = { ∈ K | c ∈ NO} 0 c d ﬁn b and a b K (N) = { ∈ K (N)| d ∈ 1 + NO}. 1 c d 0 b

These are open compact subgroups of G(Afin). Let M (7) Hk(N, ω) = π, 2 where π runs through all cuspidal representations in L0(ω) for which: (1) πfin = ⊗v<∞πv contains a nonzero K1(N)-fixed vector

(2) π∞j = πkj is the discrete series representation of G(R) of weight kj, for j = 1, . . . , r.

The central character of πkj is given by χ (x) = sgn(x)kj = ω (x). πkj ∞j For a discrete series representation π , let v be a lowest weight vector, ∞j π∞j unique up to nonzero multiples. Deﬁne the subspace M K (N) (8) A (N, ω) = Cv ⊗ · · · ⊗ v ⊗ π 1 , k π∞1 π∞r ﬁn π

K1(N) where π runs through all irreducible summands of Hk(N, ω). Here πfin is the subspace of K1(N)-fixed vectors in the space of πfin. 2 Proposition 3.1. The space Ak(N, ω) defined above is equal to the set of ϕ ∈ L0(ω) satisfying:

(1) ϕ(gkﬁn) = ϕ(g) for all kﬁn ∈ K1(N) r Y ikj θj Q r (2) ϕ(gk∞) = e ϕ(g) for all k∞ = j kθj ∈ K∞ = SO2(R) j=1

(3) For any ﬁxed x ∈ G(A) and 1 ≤ j ≤ r, the function g∞j 7→ ϕ(xg∞j ) is 1 −i annihilated by R(E−), where E− = ∈ gl (C) and R denotes −i −1 2

the right regular action of G(F∞j ). The elements of Ak(N, ω) are continuous functions on G(A). Proof. The proof is the same as for the case F = Q given in Theorem 12.6 of [KL2]. For the continuity, see Lemma 3.3 of [Li2]. PETERSSON’S FORMULA AND HECKE EIGENVALUES 7

By a general theorem of Harish-Chandra, the space Ak(N, ω) is finite-dimensional (cf. [HC], [BJ]). In particular the set of π in (7) is finite. 3.4. Fourier coefficients. Let θ : A −→ C∗ be the standard character of A. Explicitly,

−2πi(x1+···+xr ) (1) θ∞(x) = e for x = (x1, . . . , xr) ∈ F∞ (2) For v < ∞, θv is the composition

Fv tr (2πi·) Qp e ∗ θv : Fv −−−→ Qp → Qp/Zp ,→ Q/Z −−−−→ C . −1 −1 Note that θv is trivial precisely on the local inverse different dv = d Ov = {x ∈ F | trFv (x) ∈ Z }. Recall that θ is trivial on F , and that every character on F \A v Qp p is of the form θm(x) = θ(−mx) for some m ∈ F . This identifies the discrete group F with the dual group F[\A. 1 ∗ Consider the unipotent subgroup N = { 1 } of G. As topological groups, ∼ N(A) = A, so characters of the two can be identified. For m ∈ F , we identify θm with a character on N(F )\N(A) in the obvious way. For any smooth cusp form ϕ on G(A) and g ∈ G(A), the map n 7→ ϕ(ng) is a continuous function on N(F )\N(A), with a Fourier expansion 1 x 1 X (9) ϕ( g) = W ϕ (g)θ (x). 1 1/2 m m dF m∈F The coefficients are Whittaker functions defined by Z ϕ 1 x (10) Wm(g) = ϕ( g)θ(mx)dx. F \A 1 −1/2 The purpose of the factor of dF in (9) is to balance the fact that by our choice, 1/2 meas(F \A) = dF . This non-selfdual measure is convenient for the calculations later on. ∗ ∗ ∗ ∗ Let y1,..., yh ∈ Afin be representatives for Afin/F Ob , the class group of F . Then h [ y G(A) = G(F ) B(F )+K × i K (N) ∞ ∞ 1 1 i=1 ([Hi], §9.1). Here B denotes the subgroup of invertible upper triangular matrices. It follows that an element ϕ ∈ Ak(N, ω) is determined by the values 1 x y yi ϕ( 1 ∞ 1 ∞ 1 fin) + ∗ for y ∈ F∞ and x ∈ F∞. We define for y ∈ A the notation ϕ ϕ y (11) Wm(y) = Wm( 1 ). ϕ ∗ By the above remarks, ϕ is determined by the coefficients Wm(y) for y ∈ A .

Proposition 3.2. For ϕ ∈ Ak(N, ω), ϕ ϕ m ∗ • Wm(g) = W1 ( 1 g) for all m ∈ F and g ∈ G(A) ∗ ∗ ϕ ϕ ∗ • For any u ∈ Ob and y ∈ A , Wm(y) = Wm(uy) for all m ∈ F ∗ ϕ −1 • For y ∈ Afin, W1 (y) 6= 0 only if y ∈ bd . ∗ ∗ ∗ (Here and throughout, we identify Afin with {1∞} × Afin ⊂ A .) 8 ANDREW KNIGHTLY AND CHARLES LI

Proof. These are standard facts. The ﬁrst follows by a change of variables in (10) plus the left G(F )-invariance of ϕ. The K1(N)-invariance of ϕ gives the second, and also implies that for u ∈ Ob 1 x y 1 x y 1 u 1 x+uy y ϕ( 1 1 ) = ϕ( 1 1 1 ) = ϕ( 1 1 ). ϕ ϕ This means that Wm(y) = Wm(y)θ(−muy) for all u ∈ Ob, which gives the third when m = 1. (See also §9.1 of [Hi].) + ϕ Proposition 3.3. For any ϕ ∈ Ak(N, ω) and any y ∈ F∞, Wm(y) = 0 unless −1 −1 + m ∈ d+ = d ∩ F .

Proof. By the deﬁnition of cuspidality, the constant term ϕN (g) vanishes for a.e. g ∈ G(A). Because ϕ is actually continuous, it follows that ϕN (g) = 0 for all g. Therefore when m = 0, ϕ (12) W0 (g) = ϕN (g) = 0. To ϕ we can attach a holomorphic function on Hr by r y x Y −kj /2 (13) h(x + iy) = ϕ( 1 ∞ × 1ﬁn) yj . j=1

Then h is a Hilbert modular form for the group Γ1(N) = SL2(F ) ∩ K1(N), so it has a Fourier expansion of the form X −2π tr(my) h(x + iy) = a0(h) + am(h)e θ∞,m(x), −1 m∈d+ Pr + where tr(my) = j=1 σj(m)yj. For any y ∈ F∞, the Fourier coefficients am(h) ϕ and Wm(y) are related by r ϕ 1/2 Y kj /2 −2π tr(my) (14) Wm(y) = dF ( yj )e am(h). j=1 This follows immediately by equating the classical and adelic Fourier expansions of y∞ x∞ ϕ −1 ϕ( 0 1 ). Together with (12), this implies that Wm(y) = 0 unless m ∈ d+ . 4. Construction of the test function The right regular action of an element f ∈ L1(G(A), ω−1) on L2(ω) is given by Z (15) R(f)φ(x) = f(g)φ(xg)dg. G(A) In this section we will construct a continuous integrable function f such that ⊥ R(f) has finite rank (vanishing on Ak(N, ω) ), and acts like a Hecke operator on Ak(N, ω). The function will be defined locally: r Y Y f = f∞ffin = f∞j fv. j=1 v<∞ PETERSSON’S FORMULA AND HECKE EIGENVALUES 9

4.1. Archimedean test functions. For j = 1, . . . , r let vkj be a lowest weight unit vector for πkj and let dkj be the formal degree of πkj relative to the measure on G(R) ﬁxed in Sect. 3.2. We take f∞j (g) to be the normalized matrix coeﬃcient a b dkj πkj (g)vkj , vkj . Explicitly, if g = c d , then (cf. [KL2], Theorem 14.5)

 kj /2 kj (kj − 1) det(g) (2i)  if det(g) > 0  4π (−b + c + (a + d)i)kj (16) f∞j (g) =   0 otherwise.

This function is integrable over G(R) if and only if kj > 2. Therefore in order for (15) to converge we must assume henceforth that

kj > 2 (j = 1, . . . , r). Q Proposition 4.1. Let f∞ = j f∞j be as above, and suppose ffin is a bi-K1(N)- −1 invariant function on G(Afin) satisfying ffin(zg) = ωfin(z) ffin(g) and whose sup- ⊥ port is compact mod Z(Afin). Then R(f) vanishes on Ak(N, ω) (the orthogonal 2 complement in L (ω)) and its image is a subspace of Ak(N, ω). Proof. The case F = Q is proven in Corollary 13.13 of [KL2], and the general case R is no different. The main point is that N(R) f∞j (g1ng2)dn = 0 for each j. Using 2 this, one shows that R(f)φ is cuspidal for each φ ∈ L (ω). By the left K1(N)- invariance of ffin, it is easy to see that R(f)φ is K1(N)-invariant, while the matrix coefficients project onto the span of the lowest weight vectors of the discrete series of the appropriate weight. Thus R(f)φ ∈ Ak(N, ω). (See also [Li2], Prop. 2.2). 4.2. Non-Archimedean test functions. We now specify the local factors of f more precisely. Fix a discrete valuation v of F , and let p be the corresponding prime ideal of O. Let N ⊂ O be the ideal fixed earlier, and let Nv = NOv be its localization. Let Gv = G(Fv), and similarly for its subgroups Zv = Z(Fv), etc. Suppose fv : Gv → C is a locally constant function whose support is compact modulo Zv. Then the Hecke operator R(fv) is defined by Z R(fv)φ(x) = fv(g)φ(xg)dg Gv for any continuous function φ on Gv. Note that the integrand is not always well- defined modulo Zv. In our situation, φ will be a function satisfying φ(zg) = ωv(z)φ(g) for all z ∈ Zv and g ∈ Gv. Therefore we must require fv to transform −1 under the center by ωv . The Hecke algebra of bi-K1(N)v-invariant functions is generated by functions supported on sets of the form

ZvK1(N)v xv K1(N)v for xv ∈ Gv. Fix an integral ideal nv in Ov. We assume that nv is coprime to Nv, i.e., that either nv or Nv is equal to Ov. Deﬁne a set a b M(n , N ) = { ∈ M (O )| c ∈ N , (ad − bc)O = n }. v v c d 2 v v v v 10 ANDREW KNIGHTLY AND CHARLES LI

ordv (nv ) ∗ The determinant condition is equivalent to det g ∈ $v Ov. By the Cartan decomposition of Gv we have  i  S $v  Kv j Kv if v - N  i+j=ordv (nv ) $v M(nv, Nv) = i≥j≥0    K0(N)v if v|N.

Clearly this is a compact set. We need to deﬁne a bi-K1(N)v-invariant function fnv , supported on ZvM(nv, Nv), −1 a b and with central character ωv . If v|N, for k = c d ∈ K0(N)v deﬁne

ωv(k) = ωv(d).

Because c ∈ Nv, one easily sees that this is a character of K0(N)v. Now for z ∈ Zv and m ∈ M(nv, Nv), deﬁne  −1  ωv(z) if v - N (17) fnv (zm) = −1 −1  ψ(Nv)ωv(z) ωv(m) if v|N. Here, when p|N,

−1 ordv (N) −1 ψ(Nv) = meas(K1(N)v) = [Kv : K0(N)v] = N(p) (1 + N(p) ).

It is straightforward to show that fnv is well-deﬁned. a b Lemma 4.2. Suppose g = c d ∈ G(Fv) and that (det g) = nv. Then fnv (g) 6= 0 if and only if g ∈ M2(Ov) and c ∈ Nv.

Proof. Note that fnv (g) 6= 0 if and only if g = zm, with z ∈ Z(Fv), m ∈ M(nv, Nv). ∗ Taking determinants we see that z is a unit in Ov (identifying Zv with Fv ). Thus z can be absorbed into m, so in fact g ∈ M(nv, Nv) as required. 2 Proposition 4.3. The adjoint of the operator R(f ` ) on L (G , ω ) is given by pv v v ∗ ` −1 R(f ` ) = ω ($ ) R(f ` ). pv v v pv ∗ ∗ ∗ −1 Proof. We have R(fp` ) = R(f ` ) where f ` (g) = fp` (g ). If g = zm, then v pv pv v −1 −1 −1 −1 −` ` −1 g = z m = (z $v )($vm ) ∈ ZvM(nv, Nv). The proposition follows easily from this. (Note that ` = 0 if v|N.)

When v - N, the functions fnv deﬁned above linearly span the spherical Hecke algebra of bi-Kv-invariant complex-valued functions on Gv with central character −1 ωv . ` a b Now suppose v - N and write nv = pv for ` > 0. Let χ( c d ) = χ1(a)χ2(d) be an unramiﬁed character of the Borel subgroup B(Fv), and let (π, Vχ) be the representation of Gv obtained from χ by normalized induction. We assume that

χ1(z)χ2(z) = ωv(z) for all z ∈ Zv. Deﬁne a function φ0 ∈ Vχ by a b 1/2 (18) φ0( 0 d k) = |a/d|v χ1(a)χ2(d).

Then φ0 spans the 1-dimensional space of Kv-ﬁxed vectors in Vχ. PETERSSON’S FORMULA AND HECKE EIGENVALUES 11

Proposition 4.4. Let q = (p ). Then π(f ` )φ = λ ` φ , where v N v pv 0 pv 0 ` `/2 X j `−j λ ` = q χ ($ ) χ ($ ) . pv v 1 v 2 v j=0

Proof. Because f ` is left K -invariant, R(f ` )φ is again ﬁxed by K , and hence pv v pv 0 v R(f ` )φ = λφ for some λ ∈ C. The action of π is the same as the action of pv 0 0 R, so it suﬃces to compute λ = R(f ` )φ (1). Using the well-known left coset pv 0 decomposition ([KL2], Lemma 13.4)

` j ` [ [ $v a M(pv, Nv) = `−j Kv, $v j=0 j a∈Ov /pv we see that Z Z λ = f ` (g)φ (g)dg = φ (g)dg pv 0 0 ` Gv M(pv ,Nv ) ` j X X $v a = φ0( `−j ) (meas(Kv) = 1) $v j=0 j a∈Ov /pv ` ` X j j `−j 1/2 j `−j `/2 X j `−j = qv $v/$v v χ1($v) χ2($v) = qv χ1($v) χ2($v) , j=0 j=0 as claimed. Proposition 4.5. With notation as above, −`/2 −`/2 −1/2 −1/2 (19) q ω ($ ) λ ` = X (q ω ($ ) λ ) v v v pv ` v v v pv where sin(` + 1)θ X (2 cos θ) = = ei`θ + ei(`−2)θ + ··· + e−i`θ ` sin θ is the Chebyshev polynomial of degree `. −1/2 −1/2 Proof. Let αp = ωv($v) χ1($v) and βp = ωv($v) χ2($v). Note that iθ −iθ αpβp = 1. Hence we may write αp = e , βp = e , and αp + βp = 2 cos θ for some θ ∈ C. By the previous proposition, the left-hand side of (19) is ` ` ` X j `−j X ijθ −i(`−j)θ X i(2j−`)θ αpβp = e e = e = X`(2 cos θ). j=0 j=0 j=0

−1/2 −1/2 This proves the result since qv ω($v) λpv = αp + βp = 2 cos θ. 4.3. Global Hecke operators. Finally we define the global Hecke operator. Fix an ideal n in O, relatively prime to N. Define a function on Afin by Y fn = fnv , v where fnv is defined as in the previous subsection. Then fn is bi-K1(N)-invariant, and supported on Z(Afin)M(n, N), where Y a b M(n, N) = M(n , N ) = ∈ M (O)| c ∈ N, (ad − bc)O = n . v v c d 2 b b b b v<∞ 12 ANDREW KNIGHTLY AND CHARLES LI

It is also clear that −1 fn(zg) = ωfin(z) fn(g)(z ∈ Z(Afin), g ∈ G(Afin)). We define the operator Tn = R(f∞ × fn) 2 on L0(ω), which we can view as an operator on Ak(N, ω) by Proposition 4.1. The family of operators Tn for (n, N) = 1 is simultaneously diagonalizable (see Lemma 6.3 below). The following proposition and its corollaries spell out the connection between Hecke eigenvalues and Fourier coefficients.

∗ Proposition 4.6. Given n, choose n, d ∈ Aﬁn such that nOb = bn and dOb = bd. Then for any ϕ ∈ Ak(N, ω),

Tnϕ ϕ W1 (1/d) = N(n)W1 (n/d). Proof. We use the left coset decomposition [ [ r t M(n, N) = K (N), 0 s 0 ∗ r,s∈Ob/Ob t∈Ob/rOb rsOb=nb which is proven as in [KL2], Lemma 13.5. We see that y x Z y x T ϕ( ) = f (g)ϕ( g)dg n 1 n 1 G(Afin) X X y x r t = ϕ( ). 1 s r,s t Note that y x r t yr yt + xs yr yt + x ϕ( ) = ϕ( ) = ω (s)ϕ( s s ) 1 s s fin 1 " # ωfin(s) yr X myr yt = W ϕ( ) + W ϕ( )θ ( + x) . 1/2 0 s 1 s m s dF m∈F ∗ Therefore 1/2 X yr X yt W Tnϕ(y) = d · [coeff. of θ (x)] = ω (s)W ϕ( ) θ(− ). 1 F 1 fin 1 s s r,s t

Now suppose y∞ = 1 and identify y with yﬁn. Then we can assume that yr/s ∈ −1 ϕ yr bd since otherwise W1 ( s ) = 0. Then θ(−yt/s) is a well-deﬁned character of t ∈ Ob/rOb, and ( X (r) if − y/s ∈ d−1 θ(−yt/s) = N b 0 otherwise. t∈Ob/rOb Thus

Tn(ϕ) X ϕ yr ∗ (20) W1 (y) = ωﬁn(s)N(r)W1 ( )(y ∈ Aﬁn). r,s s −1 y/s∈db PETERSSON’S FORMULA AND HECKE EIGENVALUES 13

Now take y = 1/d. Then y/s ∈ bd−1 only if s = 1 ∈ Ob/Ob∗. We can therefore take r = n and s = 1, so Tnϕ ϕ W1 (1/d) = N(n)W1 (n/d), as claimed. Using the above, we can express Hecke eigenvalues in terms of Fourier coeﬃcients and vice versa.

Corollary 4.7. Suppose ϕ is an eigenvector of Tn with eigenvalue λn. Then if ϕ W1 (1/d) 6= 0, ϕ N(n)W1 (n/d) λn = ϕ . W1 (1/d)

Corollary 4.8. If (md, N) = 1, then for any Tmd-eigenfunction ϕ ∈ Ak(N, ω) with ϕ W1 (1/d) = 1 and Tmdϕ = λmdϕ, we have

2πr Qr kj /2−1 e σj(m) W ϕ (1) = j=1 λ . m 2π tr(m) md dF e Proof. Apply Cor. 4.7 with n = md and n = md. We get ϕ ϕ −1 (21) λmd = N(md)W1 (1∞ × mﬁn) = N(md)Wm(m∞ × 1ﬁn). + Here m∞ = (σ1(m), . . . , σr(m)) ∈ F∞. Using (14), it is straightforward to show that r ϕ −1 Y −kj /2 2π tr(m) −2π tr(1) ϕ Wm(m∞ ) = ( σj(m) )e e Wm(1). j=1 Substituting this into (21) and using N(md) = dF N(m) gives the result. 5. A Hilbert modular Petersson trace formula

Let f = f∞ × fn for an ideal (n, N) = 1, and recall that Tn = R(f) is the associated Hecke operator. Let F be any orthogonal basis for Ak(N, ω). (Later we will require F to consist of eigenvectors of Tn.) Then the kernel of R(f) is given by X R(f)ϕ(x)ϕ(y) X K(x, y) = = f(x−1γy), kϕk2 ϕ∈F γ∈G(F ) The ﬁrst expression is the spectral expansion of the kernel, and the second is the geometric expansion. The equality of the two hinges on the continuity (in x, y) of the geometric expansion. The proof of this continuity given in Prop 3.2. of [Li2] carries over easily to the case of nontrivial central character we consider here. In this section we apply the technique of Section 2 to the kernel function given above, taking H = N×N. We need to ﬁx a character on N(F )\N(A)×N(F )\N(A), which amounts to choosing two characters on F \A. As discussed earlier, every character on F \A is of the form

θm(x) = θ(−mx) for some m ∈ F . Fix m1, m2 ∈ F . Our goal is to obtain a trace formula by computing Z Z

(22) K(n1, n2)θm1 (n1)θm2 (n2)dn1dn2 N(F )\N(A) N(F )\N(A) with the two expressions for the kernel. 14 ANDREW KNIGHTLY AND CHARLES LI

5.1. The spectral side. Using the spectral expansion of the kernel, expression (22) is easily computed in terms of Hecke eigenvalues and Fourier coeﬃcients of cusp forms. Suppose the basis F consists of eigenfunctions of Tn. Then for ϕ ∈ F ϕ ϕ we have R(f)ϕ = λn ϕ for some scalar λn ∈ C. Hence (22) is ϕ Z Z X λn = ϕ(n )θ (n )dn ϕ(n )θ (n )dn kϕk2 1 m1 1 1 2 m2 2 2 ϕ∈F N(F )\N(A) N(F )\N(A)

ϕ ϕ ϕ X λn Wm (1)Wm2 (1) (23) = 1 , kϕk2 ϕ∈F as in (10). By Proposition 3.3, the above expression is nonzero only if −1 m1, m2 ∈ d+ . We may assume that this holds from now on. 5.2. The geometric side. Here we use the method of Section 2 to compute (22) P using the geometric expansion of the kernel. This gives a sum [δ] Iδ(f), where Z −1 Iδ(f) = f(n1 δn2)θm1 (n1)θm2 (n2)dn1dn2. Hδ (F )\H(A) The orbits [δ] are in one-to-one correspondence with the double cosets N(F )\G(F )/N(F ). Let M be the group of invertible diagonal matrices. The Bruhat decomposition is the following partition of G(F ) into two cells: 0 1 G(F ) = N(F )M(F ) ∪ N(F )M(F ) N(F ). 1 0 We call these the ﬁrst and second Bruhat cells respectively. This gives γ 0 ∗ [ 0 µ ∗ N(F )\G(F )/N(F ) = {[ 0 1 ] γ ∈ F } {[ 1 0 ] µ ∈ F }. We need to determine which of these orbits are relevant in the sense of §2. γ 0 1 t1 1 t2 First let δ = 0 1 ∈ G(F ). If ( 1 , 1 ) ∈ Hδ(A), then 1 −t γ 0 1 t γ 0 1 2 = z , 1 0 1 1 0 1 for some z ∈ Z(F ). A simple calculation shows that z = 1 and t1 = γt2, so 1 γt 1 t (24) H (A) = ( , )| t ∈ A . δ 0 1 0 1 Thus δ is relevant if and only if

θ((m1γ − m2)t) = 1 −1 for all t ∈ A, or equivalently, γ = m2/m1 (since m1 ∈ d+ is nonzero). 0 µ On the other hand, if δ = ∈ G(F ), one sees easily that 1 0

Hδ(A) = {(e, e)}, so all of these matrices are relevant. PETERSSON’S FORMULA AND HECKE EIGENVALUES 15

−1 5.2.1. Computation of the first type of Iδ. Here we take m1, m2 ∈ d+ , and δ = γ 1 where γ = m2/m1. By (24), Z 1 −t1 γ 0 1 t2 Iδ(f) = f( )θ(m1t1 − m2t2)dt1dt2 {(γt,t)∈F 2}\A×A 1 0 1 1 Z γ γt2 − t1 = f( )θ(m1t1 − m2t2)dt1dt2. {(γt,t)∈F 2}\A×A 0 1 0 0 Let t1 = γt2 − t1 and t2 = t2. Then because m1γ = m2, we have m1t1 − m2t2 = 0 −m1t1, so Z 0 γ t1 0 0 0 Iδ = f( )θ(−m1t1)dt1dt2 {0}×F \(A×A) 0 1 Z m2/m1 t = meas(F \A) f( )θ(−m1t)dt A 0 1 Z 1/2 m2 m1t = dF f( )θ(−m1t)dt A m1 Z 1/2 m2 t = dF f( )θ(−t)dt. A 0 m1 Here we used (6) and the fact that f(zg) = f(g) for z ∈ Z(F ). We factorize the above integral into (Iδ)∞(Iδ)fin, and incorporate the coefficient 1/2 dF into (Iδ)∞. First we consider Z m t (I ) = f ( 2 )θ (−t)dt. δ fin n m fin Afin 1

m2 t We shall compute this locally, as we may since fnv ( )θv(−t) ≡ 1 for m1 t∈Ov almost all v. For any ﬁnite place v we have Z m t (I ) = f ( 2 )θ (−t)dt. δ v nv m v Fv 1 The integrand is nonzero only if m1m2 ordv( 2 ) = ordv(nv) sv ∗ for some sv ∈ Fv . Supposing this is the case, Z m2 t sv sv sv (Iδ)v = fnv ( m1 )θv(−t)dt sv Fv sv

m2 m1 t is nonzero only if , , ∈ Ov by Lemma 4.2. Assuming this, we have sv sv sv Z m2 t −1 sv sv (Iδ)v = ωv(sv) fnv ( m1 )θv(−t)dt. sv Ov sv

Now suppose v - N. Then Z −1 (Iδ)v = ωv(sv) θv(−t)dt, sv Ov −1 which as before vanishes unless sv ∈ dv . So in this case, −1 (Iδ)v = |sv|v ωv(sv) , v - N. Define ( ωv if v|N (25) ωN,v = 1 if v - N, Q Q ∗ and let ωN = ωN,v = v|N ωv. Notice that ωN is a character on Afin. Multiplying the above local results together, we obtain the following. m2 0 Proposition 5.1. Suppose δ = m1 ∈ G(F ). The integral (I ) is nonzero 0 1 δ fin ∗ if and only if there exists a finite idele s ∈ Afin such that m1 m2 • s , s ∈ Ob m1m2 • ordv( 2 ) = ordv(n) for all v < ∞ sv • s ∈ bd−1. Under these conditions, −1 −1 (Iδ)fin = |s|finωfin(s) ψ(N)ωN(m1/s) where ψ(N) = [Kfin : K0(N)]. Remark: The expression is independent of the choice of s. Indeed if s0 = us for u ∈ Ob∗, then 0 −1 0 Y 0 Y −1 ωfin(s ) ωN(s ) = ωv(sv) = ωv(sv) = ωfin(s) ωN(s) v-N v-N since ωv is unramified for v - N.

For purposes of computation, the following lemma is helpful and easily veriﬁed.

Lemma 5.2. With notation as above, (Iδ)fin is nonzero if and only if for every finite valuation v of F , def 1 −1 • ξv = 2 ordv(m1m2n ) ∈ Z • ordv(mi) ≥ ξv ≥ − ordv(dv) for i = 1, 2. ξv If these conditions hold, then we can define s by taking sv = $v for all v. For the infinite part, we have the following. m2 0 Proposition 5.3. Let δ = m1 with m , m ∈ F ∗. Then (I ) is nonzero if 0 1 1 2 δ ∞ + and only if m1, m2 ∈ F . Under this condition, r 1/2 kj −1 kj /2 dF Y (4π) σj(m1m2) (Iδ)∞ = F . 2π trQ(m1+m2) (k − 2)! e j=1 j PETERSSON’S FORMULA AND HECKE EIGENVALUES 17

Proof. We have r Z ∞ 1/2 Y σj(m2) t (Iδ)∞ = dF f∞j ( )θ∞j (−t)dt. σj(m1) j=1 −∞

The integrand is nonzero only if σj(m1m2) > 0. By the formula for f∞j , Z ∞ σj(m2) t f∞j ( )θ∞j (−t)dt −∞ σj(m1) − 1 Z ∞ e2πit kj kj /2 kj = σj(m1m2) (2i) dt. kj 4π −∞ (−t + iσj(m1 + m2)) Using a complex contour integral, this is nonzero only if σj(m1 + m2) > 0, in which case it equals (4π)kj −1 kj /2 −2πσj (m1+m2) σj(m1m2) e . (kj − 2)! Full details are given in [KL1], Proposition 3.4. Because σj(m1) and σj(m2) have the same sign, the condition σj(m1 + m2) > 0 is equivalent to σj(m1), σj(m2) > 0. m2 0 −1 Proposition 5.4. Let δ = m1 for m , m ∈ d . Then I (f) is nonzero if 0 1 1 2 + δ

m1m2 m1 m2 −1 and only if s2 Ob = bn and s , s ∈ Ob for some s ∈ bd . Under this condition,  kj −1  r p 4π σj(m1m2) d1/2 (n)1/2ψ(N) Y  F N Iδ = F .  (k − 2)!  2π trQ(m1+m2) j=1 j e ωN(m1/s)ωﬁn(s)

(ordv (m1m2)−ordv (n))/2 ∗ Proof. By Lemma 5.2, we can take sv = $v . Choose n ∈ Afin such that nOb = bn. Then 1/2 |m1m2|fin |s|fin = . |n|fin + Note that by the product formula and the fact that m1, m2 ∈ F , r Y −1 |m1m2|fin = σj(m1m2) . j=1 Hence r 1/2 Y −1/2 |s|fin = N(n) σj(m1m2) . j=1

The proposition now follows immediately upon multiplying (Iδ)ﬁn and (Iδ)∞ together. 18 ANDREW KNIGHTLY AND CHARLES LI

5.2.2. Computation of the second type of Iδ. Let v < ∞ be a ﬁnite valuation of F . −1 Let nv ∈ Ov − {0} and let m1v, m2v ∈ dv . Recall that the conductor of ωv divides Nv. For cv ∈ Nv − {0}, deﬁne a generalized (local) Kloosterman sum by

X m1vs1 + m2vs2 −1 Sωv (m1v, m2v; nv; cv) = θv( )ωv(s2) . cv s1,s2∈Ov /cv Ov s1s2≡nv mod cv Ov ∗ The value of the sum is 1 if cv ∈ Ov. (The above notation is not quite consistent −1 with [KL1], which has ωv(s1) .) ∗ ∗ −1 For n ∈ Ob ∩ Afin, c ∈ Nb ∩ Afin, m1, m2 ∈ bd , and ωN as in (25), we define X m1s1 + m2s2 S (m , m ; n; c) = θ ( )ω (s )−1. ωN 1 2 fin c N 2 s1,s2∈Ob/cOb s1s2≡n mod cOb Then Y SωN (m1, m2; n; c) = SωN,v (m1v, m2v; nv; cv). v<∞ 0 µ When δ = 1 0 , we have seen that Hδ = {(e, e)}, so ZZ −1 0 µ Iδ(f) = f(n1 n2) θm1 (n1) θm2 (n2)dn1dn2. N(A)×N(A) 1 0

1 ti We will compute this locally. Write ni = 1 , i = 1, 2. Then −1 0 µ −t1 µ − t1t2 (26) n1 n2 = . 1 0 1 t2 µ Proposition 5.5. Let m , m ∈ d−1. If δ = , then (I ) is nonzero only 1 2 1 δ fin 2 2 if c µOb = bn for some c ∈ Nb . Under this condition, let n = −c µ be a generator of bn (the negative sign is for convenience). Then  r  Y kj (Iδ)fin =  (−1)  ψ(N)ωfin(c)SωN (m1, m2; n; c). j=1 Proof. For a discrete place v, ZZ −t µ − t t (I ) = f ( 1 1 2 )θ (m t − m t )dt dt . δ v nv 1 t v 1 1 2 2 1 2 Fv ×Fv 2 ∗ The integrand is nonzero only if there exists cv ∈ Fv such that cv −t1 µ − t1t2 ∈ M(nv, Nv). cv 1 t2 This means

(1) cv ∈ Nv (2) cvt1, cvt2 ∈ Ov 2 (3) ordv(cvµ) = ordv(nv). PETERSSON’S FORMULA AND HECKE EIGENVALUES 19

Note that the ﬁrst and third conditions determine cv up to units. By the third 2 condition, cvµ = −nv for some generator nv ∈ nv. Now make substitutions in the 1 1 integral by replacing t1 and t2 by t1 and t2 respectively. Then cv cv ZZ −nv −t1t2 −2 −1 −t1 m1t1 − m2t2 (I ) = |c | f (c cv )θ ( )dt dt . δ v v v nv v c t v 1 2 Ov ×Ov v 2 cv The integrand is nonzero if and only if

(1) cv ∈ Nv (2) t1t2 ≡ −nv mod cvOv. Assuming these hold, the value  −1 −nv −t1t2 ωv(cv)ψ(Nv)ωv(t2) if v|N −t1  −1 cv fnv (cv ) = cv t2  ωv(cv) otherwise depends only on the residue class of t2 modulo Nv. Furthermore, because θv is −1 −1 m1t1−m2t2 trivial on d and m1, m2 ∈ d , the value θv( ) depends only on the v cv cosets t1 + cvOv and t2 + cvOv. Thus the entire integrand is constant on cosets of cvOv. Each of these cosets has measure |cv|v, so these measures for t1 and t2 will −2 cancel the coeﬃcient |cv|v in the integral. Therefore  X −1 m1s1 − m2s2  ψ(Nv)ωv(cv) ωv(s2) θv( ) if v|N  cv  s1,s2∈Ov /cv Ov  s1s2≡−nv mod cv Ov (Iδ)v =  X m1s1 − m2s2  ωv(cv) θv( ) if v - N.  cv  s1,s2∈Ov /cv Ov s1s2≡−nv mod cv Ov

Replacing s2 by −s2, in either case we see that

(Iδ)v = ψ(Nv)ωN,v(−1)ωv(cv)SωN,v (m1, m2; nv; cv). Multiplying the local results together, we obtain

(Iδ)ﬁn = ψ(N)ωN(−1)ωﬁn(c)SωN (m1, m2; n; c).

The ﬁnal point is that because ω(−1) = 1 and ωv is unramiﬁed for v - N, r Y Y −1 Y kj ωN(−1) = ωv(−1) = ωv(−1) = ω∞(−1) = (−1) . v|N v<∞ j=1 µ Proposition 5.6. Let δ = ∈ G(F ). Then 1 ZZ −t µ − t t (27) (I ) = f ( 1 1 2 )θ (m t − m t )dt dt δ ∞ ∞ 1 t ∞ 1 1 2 2 1 2 F∞×F∞ 2 is nonzero only if m1, m2, −µ are all totally positive. Under these conditions,

kj −1 1/2 r kj p p N(−µ) Y (4πi) σj(m1m2) Jkj −1(4π −σj(µm1m2)) (Iδ)∞ = F 2π trQ(m1+m2) 2(k − 2)! e j=1 j where Jk is the Bessel J-function. 20 ANDREW KNIGHTLY AND CHARLES LI

Proof. The computation of (Iδ)∞j is given in [KL1] Proposition 3.6, and the above is the product of these local computations. Note that since −µ ∈ F +, N(−µ) = Qr N(−µ) = j=1 σj(−µ). Multiplying the above results together, we obtain the following. µ Proposition 5.7. Let δ = ∈ G(F ), and m , m ∈ d−1. Then I is nonzero 1 1 2 + δ only if (1) −µ is totally positive 2 ∗ (2) c µOb = bn for some c ∈ Nb ∩ Afin. Under these conditions, letting n = −c2µ, we have (−µ)1/2 I = ψ(N)ω (c)S (m , m ; n; c) N δ fin ωN 1 2 2π trF (m +m ) 2re Q 1 2 kj −1 r kj p Y (−4πi) σj(m1m2) q × J (4π −σ (µm m )). (k − 2)! kj −1 j 1 2 j=1 j 5.2.3. Total contribution of the second Bruhat cell. The total contribution of the ∗ second type of δ is a summation of Iδ over all µ ∈ F satisfying the conditions in Proposition 5.7. For this we need to give a more systematic description of the set of such µ. ∗ Let c ∈ Afin, and let c = cOb ∩ F be the associated fractional ideal of F . Then condition (2) of Proposition 5.7 is equivalent to (28) c2(µ) = n and (29) N|c. We need to determine the elements µ which satisfy (28) for some c as in (29). Consider the following equation in the ideal class group (30) 1 = [b]2[n]. If no solution b exists, then the contribution of this Bruhat cell is 0. Some examples are given at the end of this section. Otherwise, let

[b1], [b2],..., [bt] 2 be the distinct solutions of (30). We take the bi to be integral ideals. Because bi n is principal, there exists a nonzero element ηi ∈ O such that 2 (31) (ηi) = bi n.

We fix such generators η1, . . . , ηt once and for all. −1 −1 ∗ Suppose (28) holds. Then [c] = [bi] for some i, so c = sbi for some s ∈ F . −1 −2 −ηiu Substituting c = sbi into (28), we see that (µ) = (s ηi), i.e. µ = s2 for ∗ −ηiu ∗ some u ∈ O . Conversely if µ = s2 for some s ∈ F , then it satisfies (28) with −1 c = sbi . For each i = 1, . . . , t fix a generator bi ∈ bi. We obtain the following lemma: PETERSSON’S FORMULA AND HECKE EIGENVALUES 21

Lemma 5.8. An element µ ∈ F ∗ satisﬁes condition (2) of Proposition 5.7 if and only if −η u (32) µ = i s2 ∗ for some i ∈ {1, . . . , t}, u ∈ O and s ∈ biN. If this condition holds, then we can −1 −2 take c = sbi and n = ηiubi in Proposition 5.7. Proof. The above discussion shows that µ satisﬁes (28) if and only if it is given by −1 (32) where c = sbi . Then it is easy to check that N|c if and only if s ∈ biN. The last part of the lemma follows by a simple calculation. Lemma 5.9. In the notation above, if 0 −η u −η 0 u i = i , s2 s02 then i = i0 and uO∗2 = u0O∗2. Furthermore if we assume u, u0 ∈ U as in (4), then u = u0 and s = ±s0. 2 −2 2 0−2 −1 2 0−1 2 Proof. By (31), we have bi ns = bi0 ns . Therefore (bibi0 ) = (ss ) . By −1 0−1 unique factorization of ideals, we have bibi0 = (ss ). Therefore [bi] = [bi0 ], and hence i = i0. With i = i0, we have uu0−1 = (ss0−1)2. This implies that uu0−1 ∈ O∗ ∩ F ∗2 = O∗2. Then if u, u0 ∈ U, it is immediate that u = u0, and hence 0 s = ±s . Proposition 5.10. The total contribution of the second Bruhat cell is

kj −1 r kj p ψ(N) Y (−4πi) σj(m1m2) (33) F r 2π trQ(m1+m2) (k − 2)! 2 e j=1 j

t 1/2 X X X (ηiu) × ω (sb−1)S (m , m ; η ub−2; sb−1)N ﬁn i ωN 1 2 i i i (s) i=1 u∈U s∈biN/± N + ηiu∈F s6=0  r p Y σj(ηium1m2)  × J (4π ) . kj −1 |σ (s)| j=1 j 

−1 ϕ 5.3. Main result. For ϕ ∈ Ak(N, ω) and m ∈ d+ , let Wm denote the Fourier coeﬃcient of ϕ as in §3.4. If ϕ is an eigenfunction of the Hecke operator Tn = R(f), we write ϕ Tnϕ = λn ϕ. Equating the geometric and spectral computations of the previous sections, we obtain the following upon multiplying both sides by F 2π tr (m1+m2) r e Q Y (kj − 2)! . ψ(N) p kj −1 j=1 (4π σj(m1m2))

Theorem 5.11. Let n and N be integral ideals with (n, N) = 1. Let k = (k1,..., kr) with all kj > 2. Let F be an orthogonal basis for Ak(N, ω) consisting of eigenfunc- −1 tions for the Hecke operator Tn. Then for any m1, m2 ∈ d+ ,   2π trF (m +m ) r ϕ ϕ ϕ Q 1 2 λ W (1)W (1) e Y (kj − 2)! X n m1 m2   2 ψ(N) p kj −1 kϕk j=1 (4π σj(m1m2)) ϕ∈F 22 ANDREW KNIGHTLY AND CHARLES LI

1/2 1/2 dF N(n) = Tb(m1, m2, n) ωN(m1/s)ωfin(s) t 1/2 X X X (ηiu) + ω (sb−1)S (m , m ; η ub−2; sb−1)N fin i ωN 1 2 i i i (s) i=1 u∈U s∈biN/± N + ηiu∈F s6=0  r p Y 2π σj(ηium1m2)  × √ Jkj −1(4π ) , ( −1)kj |σ (s)| j=1 j  where: −1 • Tb(m1, m2, n) ∈ {0, 1} is nonzero if and only if there exists s ∈ bd such 2 that m1, m2 ∈ sOb and m1m2Ob = s bn (see also Lemma 5.2) • U is a set of representatives for O∗/O∗2 • biOb = bi for bi as in (30) (for i = 1, . . . , t) 2 • ηi ∈ F generates the principal ideal bi n Y Y • ωN = ωv × 1. v|N v-N Remarks: (1) The above represents only part of a larger picture because we have treated just those Fourier coefficients coming from the identity component of G(F )\G(A)/K∞K0(N). The general case would result from integrating yj yi K( 1 n1, 1 n2), ∗ where y1,..., yh ∈ Afin are representatives for the class group of F . (2) We can choose the basis so that each ϕ ∈ Vπ for some cuspidal representation ϕ π. Then the eigenvalue λn depends only on π, and not on ϕ. This can be seen from (38) and (39) below. √ Example 5.12. Suppose F = Q[ d] is a real quadratic field with narrow class number 1 (e.g. d = 2, 5). For such fields, the fundamental unit ε has negative norm. We can take U = {±1, ±ε}, and n = (η), with η ∈ F +. See Theorem 7.2 below. √ Example 5.13. Let F = Q[ d] be a real quadratic field with class number 1 and narrow class number 2 (e.g. d = 3). In this case, the fundamental unit ε has positive norm. If n does not have a totally positive generator (i.e. if n = (η) is nontrivial in the narrow class group), then there is no u ∈ U for which ηu ∈ F +. Thus for such n, the entire Kloosterman term in the above theorem vanishes.

For general F , ηi can only make a nontrivial contribution to the Kloosterman term if it satisﬁes the sign condition

σ1 σr σ1 σr (sgn ηi ,..., sgn ηi ) = ±(sgn ε ,..., sgn ε ) for some unit ε in a fundamental unit system. We now give some examples to illustrate various possibilities for the ideals bi. Example 5.14. F has class number 2. PETERSSON’S FORMULA AND HECKE EIGENVALUES 23

In the class group Cl(F ), the equation [b]2[n] = 1 has solutions if and only if [n] is trivial (since x2 = 1 for all x ∈ Cl(F )). Hence the Kloosterman term is nonzero only if n = (η) is a principal ideal. In this case, we can take b1 to be any non-principal ideal and b2 = O. Example 5.15. F has class number 3. Now we have x2 = x−1 for all x ∈ Cl(F ). Therefore the equation [b]2[n] = 1 has a unique solution in Cl(F ), and we can take b = n. Example 5.16. F has odd class number. Suppose h = | Cl(F )| is odd. Then the equation [b]2[n] = 1 has a unique solution. For example we can take b = n(h−1)/2.

6. Weighted distribution of Hecke eigenvalues As an application of Theorem 5.11, we will show that relative to a certain measure, the (normalized) eigenvalues of the Hecke operator Tp have a weighted equidistribution in the interval [−2, 2] as N(N) → ∞. 6.1. Estimates. As a function of N, the contribution of the first Bruhat cell (given in Proposition 5.4) has order Y 1 ψ(N) = [K : K (N)] = (N) (1 + ). fin 0 N p p|N N Here we will show that this is the dominant term in the Petersson trace formula as N(N) → ∞. For this, we need to show that the contribution (33) of the second Bruhat cell is small in comparison. We start with the following naive estimate (essentially the triangle inequality) for the Kloosterman sums. −1 ∗ Lemma 6.1. For any m1, m2 ∈ d , nonzero n ∈ Ob and c ∈ Nb ∩ Afin,

|SωN (m1, m2; n; c)| ≤ N(n)N(c). Proof. It suﬃces to prove the lemma locally. Note that Y Y N(c) = N(cv) = |Ov/cvOv|. v<∞ v<∞ We have

|SωN,v (m1, m2; nv; cv)|

X m1s1 + m2s2 −1 X ≤ |θv( )ωN,v(s2) | = 1. cv s1,s2∈Ov /cv Ov s1,s2∈Ov /cv Ov s1s2≡nv mod cv Ov s1s2≡nv mod cv Ov

In the case where ordv(nv) > ordv(cv), we can replace nv by cv without loss of generality. Thus assuming 0 ≤ ordv(nv) ≤ ordv(cv), we need to count the elements of the following set: 2 (34) {(s1, s2) ∈ (Ov/cvOv) | s1s2 ≡ nv mod cvOv}. 24 ANDREW KNIGHTLY AND CHARLES LI

For a ﬁxed s1 ∈ Ov/cvOv (we can assume that 0 ≤ ordv(s1) ≤ ordv(cv)), there exists a solution s2 to the congruence if and only if ordv(s1) ≤ ordv(nv). If this condition holds, then the number of solutions s2 is cv | Ov/cvOv| = |Ov/s1Ov| = N(s1). s1 Thus the cardinality of (34) is

ordv nv X X cv (s) = ($` ) · |(O / O )∗| N N v v $` v s∈Ov /cv Ov `=0 v ordv (s)≤ordv (nv )

ordv nv X ` ` (35) ≤ N($v)N(cv/$v) = N(cv)(ordv(nv) + 1) ≤ N(cv)N(nv) `=0 as desired. Proposition 6.2. As a function of N, the contribution (33) of the second Bruhat ψ(N) cell is 2−ε for any 0 < ε < 1 as (N) → ∞. Here the implied constant N(N) N depends on n, k and ε. ε0 0 Remark: The middle expression of (35) is ε0 N(cv)N(nv) for any ε > 0. Using 0 this, the dependence of (33) on n can easily be shown to be N(n)3/2+ε −ε/2 for 0 < ε < 1. Proof. By Lemma 6.1, (33) is

t 1/2 r p X X X (ηiu) (s) (ηiu) Y 4π σj(ηium1m2) N N N N ψ( ) 3 Jkj −1( ) (bi) (s) |σj(s)| i=1 u∈U s∈biN/± N N j=1 + ηiu∈F s6=0 t r p X X X X Y 4π σj(ηium1m2) ψ(N) Jkj −1( ). ∗ |σj(s)||σj(a)| i=1 u∈U s∈biN/O a∈O∗/{±1} j=1 + ηiu∈F s6=0 ∗ We remark that |σj(s)| is not well-deﬁned for s ∈ biN/O , so the summands depend on a choice of representatives s, which we regard as ﬁxed. By the units theorem, the set ∗ {(log |σ1(a)|,..., log |σr(a)|): a ∈ O } r is a full lattice Λ in the (r−1)-dimensional hyperplane x1 +···+xr = 0 in R . Thus ∗ the summation over a ∈ O /{±1} can be replaced by a sum over λ = (λ1, . . . , λr) ∈ Λ. Recall that 1 2 J (x) min(|x|−1/2, |x|kj −1) ≤ ≤ . kj −1 max(|x|1/2, |x|−2) |x|1/2 + |x|−2

σj (s) As a result, if we let τj = log √ and λj = log |σj(a)|, we can bound 4π σj (ηium1m2) −(λj +τj ) Jkj −1(e ) to give r p X Y 4π σj(ηium1m2) X 1 J ( ) . kj −1 Qr −(λ +τ )/2 2(λ +τ ) |σj(s)||σj(a)| (e j j + e j j ) a∈O∗/{±1} j=1 λ∈Λ j=1 PETERSSON’S FORMULA AND HECKE EIGENVALUES 25

P Let β = 2 − ε for 0 < ε < 1. Using j λj = 0, the right-hand sum is X 1 = r r Q eβ(λj +τj ) Q (e−(1/2+β)(λj +τj ) + eε(λj +τj )) λ∈Λ j=1 j=1 X 1 = Pr r ( j=1 τj )β Q −(1/2+β)(λj +τj ) ε(λj +τj ) λ∈Λ e j=1(e + e ) β rp (4π) N(ηium1m2) X 1 = r . (s) Q (e−(1/2+β)(λj +τj ) + eε(λj +τj )) N λ∈Λ j=1 We claim that this last sum over λ is bounded, independently of s. Let L0 be the hyperplane x1 + ··· + xr = 0 containing Λ. Then Z X 1 dx1 ··· dxr−1 r r . Q (e−(1/2+β)(λj +τj ) + eε(λj +τj )) Q (e−(1/2+β)(xj +τj ) + eε(xj +τj )) λ∈Λ j=1 L0 j=1 The rth factor of the integrand is bounded, so the above is Z r−1 Z dx1 ··· dxr−1 Y dx = < ∞ r−1 −(1/2+β)x εx r−1 Q −(1/2+β)(xj +τj ) ε(xj +τj ) e + e R j=1(e + e ) j=1 R as claimed. Therefore (33) is 2−ε t (4π)rp (η um m ) X X X N i 1 2 ψ(N) . ∗ (s) i=1 u∈U s∈biN/O N + ηiu∈F s6=0 ∗ The first two sums are taken over finite sets. The set of s ∈ biN/O is in 1-1 correspondence with the set of principal ideals (s) divisible by biN. Hence we can replace the sum over s by a larger set: all integral ideals a divisible by N. Therefore (33) is X 1 ψ(N) X 1 ψ(N) ψ(N) . (a)2−ε (N)2−ε (a0)2−ε (N)2−ε N|a N N a0 N N P 1 (We use the absolute convergence of the zeta function ζF (s) = 0 0 s for Re s > a N(a ) 1.) 6.2. Weighted equidistribution. Let (X, µ) be a Borel measure space. For each i = 1, 2,... let Fi be a finite nonempty index set, and let Si = {xij}j∈Fi be a + finite sequence of points of X. Suppose each j ∈ Fi is assigned a weight wij ∈ R . Define P wijδx dµ = j∈Fi ij , i P w j∈Fi ij where δxij is the Dirac measure at xij. We say that the sequence {Si} is w- equidistributed with respect to the measure dµ if P w δ j∈Fi ij xij lim dµi = lim = dµ. i→∞ i→∞ P w j∈Fi ij This means that for any continuous function f : X → C, we have Z P w f(x ) Z j∈Fi ij ij lim f(x)dµi(x) = lim = f(x)dµ(x). i→∞ i→∞ P w X j∈Fi ij X 26 ANDREW KNIGHTLY AND CHARLES LI

If wij = 1 for all i, j, then this definition reduces to that of equidistribution given in §1 of [Se]. 6.3. The distribution theorem. Let p be a prime ideal, not dividing the level N. In this section we will apply the main theorem in the case where n = p` (` ≥ 0), −1 and m = m1 = m2 ∈ d+ . For each irreducible summand π of Hk(N, ω) (cf. (7)), let Fπ be an orthogonal basis for the nonzero finite-dimensional subspace π ∩ Ak(N, ω). Let [ (36) F = Fπ π be the resulting orthogonal basis for Ak(N, ω). Lemma 6.3. For any ϕ ∈ F and any prime ideal p - N, the cuspidal representation (π, V ) containing ϕ is unramified at p. Furthermore, ϕ is an eigenfunction of the ϕ global Hecke operator Tp` , and the associated eigenvalue λp` coincides with the local eigenvalue λ ` attached to π in Prop. 4.4. pv p 0 0 Proof. Write πfin = πp⊗π , where π = ⊗ v<∞ πv is a representation of the restricted v6=vp 0 0 Q 0 0 direct product G = Gv. Let K = G ∩ K1(N). Let f = fn with n = (1). Then v<∞ v6=vp K1(N) πfin(f) is the projection operator of Vfin onto πfin , and 0 K1(N) 0 0 0 Kp 0K (37) πfin = πfin(f)Vfin = πp(fvp )Vp ⊗ π (f )V = πp ⊗ π . The middle equality holds e.g. by Prop. 13.17 of [KL2]. ` Now take n = p and f = f∞ × fn. By the definition of Ak(N, ω) (eq. (8)) we can write ϕ = w ⊗ w , where w = ⊗v and w ∈ πK1(N). Because ∞ fin ∞ π∞j fin fin K1(N) 0 6= wfin ∈ πfin , it follows immediately from (37) that πp is unramified. Because πp is also unitary, it follows that πp = πχ is induced from some unramified character a b Kp χ( 0 d ) = χ1(a)χ2(d) of B(Fv). Therefore πp = Cφ0 is one-dimensional (for 0 notation see (18)), and we can write wfin = φ0 ⊗ w as in (37). Thus ϕ = w∞ ⊗ 0 φ0 ⊗ w , and 0 0 0 Tp` ϕ = R(f)ϕ = π∞(f∞)w∞ ⊗ πp(fvp )φ0 ⊗ π (f )w 0 = w ⊗ λ ` φ ⊗ w = λ ` ϕ. ∞ pv 0 pv

The local Langlands class of πp is the GL2(C)-conjugacy class of the matrix g(πp) = χ1($p) . Let q = (p). By the above lemma and Prop. 4.4, the trace of χ2($p) N g(πp) is given by −1/2 ϕ (38) χ1($p) + χ2($p) = q λp . ϕ 1/2 If χ is unitary (i.e. πp is not complementary series), then it is clear that |λp | ≤ 2q . PETERSSON’S FORMULA AND HECKE EIGENVALUES 27

In fact by the Ramanujan conjecture this is always the case.1 ` −`/2 When n = p , the operator ωp($p) R(f) is self-adjoint (cf. Prop. 4.3), so its eigenvalues are real numbers. We let ϕ def −`/2 −`/2 ϕ νp` = ωp($p) q λp` ∈ R π denote this normalized Hecke eigenvalue. We sometimes write νp` to emphasize ϕ that it depends only on the cuspidal representation π. Note that νp ∈ [−2, 2] by the Ramanujan conjecture. By Prop. 4.5, ϕ ϕ (39) νp` = X`(νp ). We will adapt the argument of [Li1] for finding the asymptotic weighted distribution ϕ of the set of νp . −1 Lemma 6.4. In the notation of Theorem 5.11, for any ` ≥ 0 and any m ∈ d+ , Tb(m, m, p`) 6= 0 if and only if both of the following hold: (1) ` = 2`0 is even 0 (2) 0 ≤ ` ≤ ordp(md). Proof. The two conditions of Lemma 5.2 specialize respectively to the two condi- ` tions above in the special case where m1 = m2 = m and n = p . −1 Proposition 6.5. Fix m ∈ d+ and a prime ideal p - N. For each ϕ ∈ F, define a ϕ 2 |Wm(1)| weight wϕ = kϕk2 . Then for any ` ≥ 0 and 0 < ε < 1, ( ψ(N) 0 0 Jψ(N) + O( 2−ε ) if ` = 2` with 0 ≤ ` ≤ ordv md X ϕ N(N) X`(νp )wϕ = ψ(N) O( 2−ε ) otherwise, ϕ∈F N(N) where X` is the Chebyshev polynomial defined in Prop. 4.5, and r 1/2 kj −1 dF Y (4πσj(m)) J = F . 4π trQ(m) (k − 2)! e j=1 j The implied constant depends only on m, p, `, k and ε. Remark: This shows in particular that when N(N) is sufficiently large, (1) Ak(N, ω) is nontrivial ϕ (2) Wm(1) is nonzero for some ϕ ∈ F. Proof. This follows from the generalized Petersson trace formula. We use the form developed in the proof, rather than the final statement in Theorem 5.11. The ` −1 spectral side (23) of the trace formula with n = p and m1 = m2 = m ∈ d+ gives ϕ ϕ 2 X λp` |Wm(1)| X = ω ($ )`/2q`/2X (νϕ)w . kϕk2 p p ` p ϕ ϕ∈F ϕ∈F

1The Ramanujan conjecture was proven at all but a ﬁnite number of unspeciﬁed places for holomorphic cuspidal representations π of GL2(A) with all weights ≥ 2 by Brylinski and Labesse ([BL], Theorem 3.4.6). Recently the full conjecture (at all places, when all weights are ≥ 2) was proven by Blasius, with a parity condition on the weights [Bl]. The parity requirement was removed in the thesis of his student L. Nguyen [Ng]. However, as we remark after Theorem 6.6 below, our results do not actually depend on this deep theorem. 28 ANDREW KNIGHTLY AND CHARLES LI

The above is equal to the geometric side, which by Prop. 5.4 and Prop. 6.2 is ` Tb(m, m, p ) `/2 ψ(N) = q Jψ(N) + O( 2−ε ) ωN(m/s)ωﬁn(s) N(N) for J as above. By the above lemma, the ﬁrst term is nonzero if and only if ` = 2`0 0 with 0 ≤ ` ≤ ordp(md). In this case we can take

pth −`0 s = (. . . , m, m, m$p , m, m, . . .). Q Note that ωN(m/s) = p|N ωp(1) = 1, so −`0 ωN(m/s)ωfin(s) = ωfin(s) = ωfin(m)ωp($p) . −1 + Furthermore ωfin(m) = ω∞(m) = 1 since m ∈ F . Thus in this case the geometric side is `/2 `/2 ψ(N) = ωp($p) q Jψ(N) + O( ). N(N)2−ε The proposition now follows by equating the spectral side with the geometric side `/2 `/2 and dividing by ωp($p) q .

Theorem 6.6. Let p be a prime ideal of F , and let k = (k1,..., kr) be a weight vector with all kj > 2. For each i = 1, 2,...

• let Ni be an ideal coprime to p, with lim (Ni) = ∞, i→∞ N • let ωi be a unitary character as in §3.3 relative to Ni and k, • let Fi be an orthogonal basis for Ak(Ni, ωi) as in (36). −1 ϕ 2 2 Fix any m ∈ d+ , and define weights wϕ = |Wm(1)| /kϕk for ϕ ∈ Fi. For each i, define a sequence ϕ Si = {νp }ϕ∈Fi in the interval [−2, 2]. Then the sequence Si is wϕ-equidistributed relative to the measure ordp(dm) X dµ(x) = X2`0 (x)dµ∞(x), `0=0 where dµ∞(x) is the Sato-Tate measure defined in the Introduction. In other words, for any continuous function h on R, P ϕ h(νp )wϕ Z lim ϕ∈Fi = h(x)dµ(x). i→∞ P w ϕ∈Fi ϕ R

Remarks: (1) When p - md, the measure µ coincides with µ∞ and is independent of p. Further taking all kj even and ωi trivial, we immediately obtain Theorem 1.1. (2) The above result (and its proof) is actually independent of the Ramanujan conjecture. All we need is the existence of a ﬁnite interval Ip which contains all of ϕ the eigenvalues νp . This is elementary ([Ro], Prop. 2.9). π (3) The theorem shows in particular that the Satake traces νp are dense in the interval [−2, 2]. The referee has pointed out that this can also be seen directly by considering CM cusp forms. PETERSSON’S FORMULA AND HECKE EIGENVALUES 29

Proof. Setting ` = 0 in the previous proposition, we have

X ψ(Ni) wϕ = Jψ(Ni) + O( 2−ε ). N(Ni) ϕ∈Fi Therefore for any ` ≥ 0 P ϕ ( 0 0 X`(νp )wϕ 1 when ` = 2` , with 0 ≤ ` ≤ ord md, lim ϕ∈Fi = v i→∞ P w ϕ∈Fi ϕ 0 otherwise Z (40) = X`(x)dµ(x). R

This last equality holds by the orthonormality of the polynomials Xn(x) Z Xi(x)Xj(x)dµ∞(x) = δij R

(cf. [Se]). Because deg X` = `, the set {X`} spans the space of all polynomials. ∞ Because the space of polynomials is dense in L ([−2, 2]), we can replace X` by any continuous function h in (40) and thus obtain the result (see §29.3 of [KL2] for details). 7. Variants and special cases We have four corresponding types of parameters (under various hypotheses): Satake Hecke Whittaker Classical π ←→ ϕ ←→ ϕ ←→ νp λn Wm(y) am(h). For convenience, we give the correspondences explicitly here, so that when possible ϕ ϕ ϕ λn W (1)Wm (1) anyone can rewrite the spectral terms m1 2 in the main formula purely kϕk2 ϕ in terms of their parameter of choice. Let nOb = bn and dOb = bd. If Wm(1/d) = 1, then by Corollaries 4.7 and 4.8,

2πr Qr kj /2−1 e σj(m) λϕ = (n)W ϕ( / ) and W ϕ (1) = j=1 λ , n N 1 n d m 2π tr(m) md dF e either of which can be used if an orthogonal basis of such ϕ is given. Also, using (39) we have ϕ 1/2 1/2 Y π (41) λn = ωﬁn(n) N(n) Xordp(n)(νp ). p|n The classical picture is given explicitly at the end of this section. The case of narrow class number 1. From now on we assume that F has narrow class number 1, i.e. that every fractional ideal in F has a totally positive generator. This implies that every totally positive unit is the square of a unit ([CH], Lemma 11.6). In this case Theorem 5.11 simpliﬁes substantially, and a classical interpretation follows from (14). Fix η, N, d ∈ F + such that (42) (η) = n, (N) = N, (d) = d. 30 ANDREW KNIGHTLY AND CHARLES LI

m2/m1 Proposition 7.1. Suppose F has narrow class number 1. For δ = 1 ,  kj −1  r p T (dm , dm , η)d1/2 (η)1/2ψ(N) 4π σj(m1m2) 1 2 F N Y  Iδ(f) = F , 2π tr (m1+m2) p  q m m  e Q ω ( m η/m ) 1 2 kj N 1 2 j=1 (kj − 2)! sgn(σj( η )) where  1 if a a /a is a square in O for all  i j k T (a1, a2, a3) = distinct i, j, k ∈ {1, 2, 3} 0 otherwise, p p and the square roots are chosen compatibly so that m1m2/η m1η/m2 = m1.

Proof. By Prop. 5.4, we know that Iδ is nonzero only if m1 m2 −1 • s , s ∈ Ob for some s ∈ bd m1m2 • s2 Ob = bn. If these hold, then because the class number is 1, we can actually take s ∈ d−1, so we change fonts to s as a reminder that s ∈ F . Furthermore the second condition is 2 2 + equivalent to m1m2 = s uη for some unit u. Because m1m2, s , η ∈ F , it follows that u ∈ F +, and hence u is the square of a unit. We can absorb this unit into s, so the condition becomes 2 (43) m1m2 = s η. Write s = s0/d for d as in (42). Then the above conditions are equivalent to

dm1 dm2 0 (1) s0 , s0 ∈ O for some s ∈ O (dm1)(dm2) (2) s02 = η. It is easy to check that these conditions hold if and only if T (dm1, dm2, η) = 1. Now suppose the above holds, so that by Prop. 5.4,  kj −1  r p 4π σj(m1m2) d1/2 (n)1/2ψ(N) Y  F N Iδ = F .  (k − 2)!  2π trQ(m1+m2) j=1 j e ωN(m1/s)ωfin(s) p By (43), we can write s = m1m2/η. This element is defined up to ±1, and there√ is no canonical choice since we cannot guarantee that s ∈ F +. (E.g. (1 + 2)2 has no totally positive square root.) In any case, the final result is independent of this choice by the remark on page 16. Using this expression for s, the proposition follows by the fact that r −1 Y kj ωfin(s) = ω∞(s) = sgn(σj(s)) . j=1 µ For the second type of Iδ (with δ = 1 ), the condition on µ becomes µ = −ηu/c2 for some totally positive unit u ∈ U and nonzero c ∈ NO. As in the previous case, u is a square, so it can be absorbed into c. Thus µ = −η/c2 PETERSSON’S FORMULA AND HECKE EIGENVALUES 31 for some c ∈ N. Under this condition, by Prop. 5.7 the contribution is ψ(N) (η)1/2 I = ω (c)S (m , m ; η; c)N δ 2π trF (m +m ) fin ωN 1 2 2re Q 1 2 N(c) kj −1 r kj p p Y (−4πi) σj(m1m2) σj(ηm1m2) × J (4π ). (k − 2)! kj −1 |σ (c)| j=1 j j + Once again, we cannot assume c ∈ F so ωfin(c) may be nontrivial. In fact r Y kj ωfin(c) = sgn(σj(c)) . j=1 Summing over all µ amounts to summing over nonzero c ∈ N/±, and we obtain the following. Theorem 7.2. With notation as above, suppose F has narrow class number 1, k = (k1,..., kr) with all kj > 2, and F is an orthogonal basis for Ak(N, ω) consisting −1 of eigenfunctions for the Hecke operator Tn. Then for any m1, m2 ∈ d+ ,   2π trF (m +m ) r ϕ ϕ ϕ Q 1 2 λ W (1)W (1) e Y (kj − 2)! X n m1 m2 1/2  p  2 1/2 (4π σ (m m ))kj −1 kϕk dF N(η) ψ(N) j=1 j 1 2 ϕ∈F

 r  Y p kj p −1 = T (dm1, dm2, η)  (sgn σj( m1m2/η))  ωN( m1η/m2) j=1 √ r σj (ηm1m2) r Jk −1(4π ) 1 X (2π) Y j |σj (c)| + S (m , m ; η; c) . 1/2 ωN 1 2 k (c) (i sgn(σj(c)) j dF c∈NO/± N j=1 c6=0

Remarks: (1) Because Ob/cOb =∼ O/cO, we see easily that for c ∈ N, X m1t1 + m2t2 S (m , m ; η; c) = ω (t )−1θ ( ). ωN 1 2 N 2 ﬁn c t1,t2∈O/cO t1t2≡η mod cO The formula can be further simpliﬁed if we replace c ∈ N by c = Nuτ for u ∈ ∗ ∗ 0 −1 0 −1 O /{±1} and τ ∈ O/O . Then substituting t1 = u t1, t2 = u t2, we have −1 −2 SωN (m1, m2; η; c) = ωN(u) SωN (m1, m2; u η; Nτ), so we can break the sum over c into sums over u and τ and group together the terms with the same τ. (2) If F = Q, then r = 1. The sum over c ∈ NZ/± is simply a sum over c > 0,N|c, and we recover the generalized Petersson trace formula of [KL1].

ϕ Now take η = 1 so n = O. Then λn = 1 for all ϕ. Because the narrow class 0 number is 1, Ak(N, ω) corresponds (isometrically) to a classical space Sk(N, ω ) of 0 a b 0 −1 a b cusp forms for Γ0(N) as in (13). Here ω ( c d ) = ω (d) = ωN(d) for any c d ∈ 0 Γ0(N) = SL2(F ) ∩ K0(N). Conversely, given a character ω :Γ0(N)/Γ1(N) → ∗ 0 C , the space Sk(N, ω ) is isometric to Ak(N, ω), where ω is the Hecke character 0 ∗ ∗ + ∗ determined by ω using strong approximation A = F (F∞ × Ob ) resulting from ϕ 1/2 −2π tr(m) narrow class number 1. By (14), if ϕ ↔ h, then Wm(1) = dF e am(h), and 32 ANDREW KNIGHTLY AND CHARLES LI we obtain the following direct generalization of Petersson’s original formula and (8) of [Lu].

0 −1 Corollary 7.3. For any orthogonal basis F for Sk(N, ω ) and any m1, m2 ∈ d+ we have 1/2  r  dF Y (kj − 2)! X am1 (h)am2 (h)   2 ψ(N) p kj −1 khk j=1 (4π σj(m1m2)) h∈F √ r σj (m1m2) r Jk −1(4π ) 1 X (2π) Y j |σj (c)| = χ(m , m ) + S 0 (m , m ; c) , 1 2 1/2 ω 1 2 k (c) (i sgn(σj(c)) j dF c∈NO/± N j=1 c6=0 2 ∗ where χ(m1, m2) ∈ {0, 1} is nonzero if and only if m2 = m1u for some u ∈ O P F sm1+sm2 0 and Sω0 (m1, m2; c) = exp(2πi trQ( c ))ω (s). s∈(O/cO)∗, ss=1

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Department of Mathematics & Statistics, University of Maine, Neville Hall, Orono, ME 04469-5752, USA

Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong