Kuznietsov Trace Formula and Weighted Distribution of Hecke Eigenvalues

Charles C.C. Li

December 20, 2005

Abstract Let (X, μ) be a measurable topological space. Let S_{1}, S_{2}, ⋯ be a family of finite subsets of X. Suppose each x ∈ S_{i} has a weight w_{ix} ∈ R^{+} assigned to it. We say {S_{i}} is {w_{i}}-distributed with respect to the measure μ if for any continuous function f on X, we have lim_{i→∞} \frac{∑_{x∈S_{i}} w_{ix}f(x)}{∑_{x∈S_{i}} w_{ix}} =

∫_{X} f(x)dμ(x).

Let S(N, k) be the space of modular cusp forms over Γ_{0}(N) of weight k and let E(N, k) ⊂ S(N, k) be a basis consists of Hecke eigenforms. Let a_{r}(h) be the r^{th} Fourier coefficient of h. Let x_{p}^{h} be the eigenvalue of h relative to the normalized Hecke operator T′_{p}. Let ∥ · ∥ be the Petersson norm on S(N, k). In this paper we will show that for any even integer k ≥ 3, \{x_{p}^{h} : h ∈ E(N, k)\}, p ∤ N is \{ \frac{|a_{r}(h)|^{2}e^{−4πr}}{∥h∥^{2}} \}-distributed with respect to a polynomial times the Sato-Tate measure when N → ∞.

1 Introduction

Let S(N, k) be the space of modular cusp forms on Γ_{0}(N) of weight k and for p ∤ N, let T_{p} be the Hecke operator on S(N, k) as defined in [Se2]. We shall consider T′_{p} = p^{−(k−1)/2}T_{p}. Denote the eigenvalue of a Hecke eigenform h relative to T′_{p} by x_{p}^{h}. The Ramanujan-Petersson conjecture (Deligne’s Theorem) asserts that |x_{p}^{h}| ≤ 2 for (p, N) = 1. Furthermore, it is conjectured that the set {x_{p}^{h} : (p, N) = 1} is equidistributed with respect to the Sato-Tate measure(refer to [Se1] Chapter 1)

dμ_{∞}(x) = \{ \begin{matrix}\frac{1}{π} \sqrt{1 − \frac{x^{2}}{4}} dx & when x ∈ [−2, 2], \\ 0 & otherwise. \end{matrix}

Let E(N, k) be a basis of S(N, k) consisting of Hecke eigenforms. Serre [Se2] considered the distribution of {x_{p}^{h} : h ∈ E(N, k)} for fixed p. He used the Selberg trace formula for the Hecke operators and showed that when N → ∞, the set {x_{p}^{h}} is equidistributed as

dμ_{p}(x) = \frac{p + 1}{(p^{1/2} + p^{−1/2})^{2} − x^{2}} dμ_{∞}(x).  (1)

In this paper, we established a weighted distribution for {x_{p}^{h} : h ∈ E(N, k)} for fixed p.

Let (X, μ) be a measurable topological space. Let S_{1}, S_{2}, ⋯, S_{i}, ⋯ be a family of finite subsets of X. Suppose each x ∈ S_{i} has a weight w_{ix} ∈ R^{+} assigned to it. Let δ_{x} be the Dirac measure at x. Define

dμ_{i} = \frac{∑_{x∈S_{i}} w_{ix}δ_{x}}{∑_{x∈S_{i}} w_{ix}} .

We say {S_{i}} is {w_{i}}-distributed with respect to measure dμ if

lim_{i→∞} dμ_{i} = lim_{i→∞} \frac{∑_{x∈S_{i}} w_{ix}δ_{x}}{∑_{x∈S_{i}} w_{ix}} = dμ.

This means for any continuous function f on X, we have

lim_{i→∞} ∫_{X} f(x)dμ_{i}(x) = lim_{i→∞} \frac{∑_{x∈S_{i}} w_{ix}f(x)}{∑_{x∈S_{i}} w_{ix}} = ∫_{X} f(x)dμ(x).

When w_{i,x} = 1, the definition is the same as the definition of equidistribution given in Serre [Se2] Section 1.

In this paper, we will use Kuznietsov trace formula to obtain a certain weighted distributions.

Suppose h ∈ S(N, k) is a Hecke eigenform with Fourier expansion

h(z) = ∑_{r=1}^{∞} a_{r}e^{2rπiz}, Re z > 0.

Write a_{r}(h) = a_{r}. Let ∥ · ∥ be the Petersson norm on S(N, k) ([Ge] p.24 (2.6)). We can assume k is even because S(N, k) is empty when k is odd. Define polynomials X_{n} by

X_{n}(2 cos ϕ) = \frac{sin(n + 1)ϕ}{sin ϕ}.

Let r be a positive integer. Let p be a fixed prime. Let r_{p} = ord_{p} r. Then we have

Theorem 1.1 Let k be an even number ≥ 3. Consider the family of sets S_{N} = \{(x_{p}^{h}) : h ∈ E(N, k)\}, p ∤ N with weight w_{h}^{r} = \frac{|a_{r}(h)|^{2}e^{−4πr}}{∥h∥^{2}} assigned to x_{p}^{h}. Then the family of sets {S_{N} : p ∤ N} is {w_{h}^{r}}-distributed with respect to

∑_{0≤i≤r_{p}} X_{2i}(x)dμ_{∞}(x)

when N → ∞.

The proof will be given at the end of this paper. A more general result is given in Theorem 5.7.

Corollary 1.2 Let k, S_{N}, w_{h}^{r} be as above. If p ∤ r, then the family of sets {S_{N} : p ∤ N} , N → ∞ is {w_{h}^{r}}-distributed with respect to the Sato-Tate measure.

Proof. If p ∤ r, r_{p} = ord_{p} r = 0 and X_{0}(x) = 1. The corollary follows easily. □

The technique used here can be generalized to other groups (example G = GSp(2k)), refer to [Li] for the generalizations.

2 Construction of test functions

Let G = GL_{2}. The unipotent group is N = \binom{1 ∗}{1} ⊂ G. Write e =

(\begin{matrix}1 \\ & 1 \end{matrix}). Let Z be the center of G, let M be the diagonal subgroup of G.

Denote \overline{U} = U/Z for any subset U of GL_{2}. Define K_{∞} = \{(\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix})\}.

When p < ∞, define K_{p} = GL_{2}(Z_{p}). Define K_{0}(N)_{p} = \{(\begin{matrix}a & b \\ c & d \end{matrix}) ∈ K_{p} : c ≡ 0

(mod N)}, K_{0}(N) = ∏_{p<∞} K_{0}(N)_{p}.

Let A be the adeles of Q. Let A_{fin} be the finite component of A.

Let L^{2} be the Hilbert space of continuous functions φ on Z(A)G(Q)\G(A) such that ∫_{Z(A)G(Q)\G(A)} |f(g)|^{2}dg < ∞. The subset of cuspidal functions in L^{2} is denoted by L_{0}^{2}. Let R be the right regular representation of G(A) on L^{2}.

There is an embedding S(N, k) → L_{0}^{2} ([Ge], p. 42). The map is denoted by f ↦ φ_{f}. Denote the image of the map by A(N, k).

Suppose (π, V) is a representation of a measurable topological group G and f is a continuous function on G. Define π(f)v = ∫_{G} f(g)π(g)vdg.

We are going to construct a function f = f_{∞}f_{fin} on \overline{G(A)} = \overline{G(R)}×\overline{G(A_{fin})}. The main property of this function is given in Proposition 2.1. The results quoted below are well-known.

The function f_{∞} = f_{k} is a function defined on \overline{GL_{2}(R)}. It is the conjugate of a normalized matrix coefficient. Explicitly it is defined by f_{k}(g) = d_{π_{k}} \overline{⟨π(g)v_{0}, v_{0}⟩} where π_{k} is the discrete series representation of lowest weight k, v_{0} is the lowest weight unit vector and d_{π_{k}} is the formal degree. Explicitly we can take

f_{k}(g) = \{ \begin{matrix}\frac{k−1}{4π} \frac{2^{k}(det g)^{k/2}}{((a+d)+i(b−c))^{k}} & if g = (\begin{matrix}a & b \\ c & d \end{matrix}) and det g > 0, \\ 0 & if det g < 0. \end{matrix}  (2)

Refer to [Va] p192 or [KL] for the details.

Define ψ(N) = [Γ_{0}(1) : Γ_{0}(N)]. We take meas(K_{p}) = 1. One can easily show that meas(\overline{K_{0}(N)}) = \frac{1}{[K_{0}(1):K_{0}(N)]} = \frac{1}{ψ(N)}.

If R be a ring, let M_{2}(R) be all the 2 × 2 matrix over R. Now we define

M(n, N) = \{g = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ M_{2}(∏ Z_{p}) : det(g) ∈ n ∏ Z_{p}^{∗} and c ≡ 0 (mod N)\},

Define

f^{n}(g) = \{ \begin{matrix}\frac{1}{meas \overline{K_{0}(N)}} = ψ(N) & if g = zm, z ∈ Z(A_{fin}), m ∈ M(n, N), \\ 0 & otherwise. \end{matrix}

Proposition 2.1 Suppose f = f_{k}f^{n}, R(f) vanishes on A(N, k)^{⊥}. On A(N, k) it acts by

R(f)φ_{h} = φ_{n^{−( \frac{k}{2} −1)}T_{n}h}

Proof. The idea of the proof can be found in [Ge] Lemma 3.7 and also [Ro] Lemma 2.12. A complete proof can be found in [KL]. □

Corollary 2.2 Let h ∈ E(N, k), f = f_{k}f^{n} then

R(f)φ_{h} = n^{1/2} (∏_{p|n} X_{n_{p}}(x_{p}^{h})) φ_{h}

Proof. Recall that n = ∏ p^{n_{p}}. We have T′_{n} = ∏_{p} T′_{p^{np}}. By [Se2] Section 2 and Section 3 Lemma 1, T′_{p^{np}} = X_{n_{p}}(T′_{p}). The corollary follows easily from the previous proposition. □

3 Kuznietsov trace formula

Let f be a continuous function on \overline{G(A)}. The kernel of R(f) is defined as

K(x, y) = ∑_{γ∈\overline{G(Q)}} f(x^{−1}γy).  (3)

Another way to express the kernel is

K(x, y) = ∑_{ϕ} R(f)ϕ(x)\overline{ϕ(y)},

here ϕ runs through an orthonormal basis of L^{2}. When f = f_{k}f^{n}, R(f) annihilates A(N, k)^{⊥}. We can sum over an orthonormal basis of A(N, k). An orthonormal basis can be taken as \{\frac{φ_{h}(x)}{∥φ_{h}∥} : h ∈ E(N, k)\}. It is easy to show that

K(x, y) = ∑_{h∈E(N,k)} \frac{R(f)φ_{h}(x)}{∥φ_{h}∥} \frac{\overline{φ_{h}(y)}}{∥φ_{h}∥}.  (4)

Use θ to denote a character on Q\A. We can decompose θ into θ_{∞}θ_{fin}. Here θ_{∞}(resp. θ_{fin}) is the infinite(resp. finite) component of θ. There exists r ∈ Q, such that θ_{∞}(x) = e^{2πirx}. We assume r ∈ Z^{+} throughout the whole paper.

Under this assumption θ_{fin} is trivial on ∏_{p} Z_{p}. The character θ can also be regarded as a character on N(Q)\N(A).

We factorize r into ∏ p^{r_{p}}.

Normalize measure on Q_{p} by taking meas(Z_{p}) = 1. Define measure on A by using the product measure. We can show that meas(Q\A) = 1. Measure on N(A) is defined by identifying A with N(A).

From (3), the kernel K(x, y) is invariant under left multiplication of elements in N(Q) × N(Q).

Kuznietsov trace formula is the equality obtained by expanding the following integral using the two formulas (3) and (4)

KTF(f) = ∫_{N(Q)\N(A)} ∫_{N(Q)\N(A)} K(n_{1}, n_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.  (5)

The integral is convergent because Q\A is compact. The expression obtained using formula (3) is called the geometric side. Using formula (4) we obtain the spectral side.

Proposition 3.1 When f = f_{k}f^{n}, then KTF(f) is equal to

n^{1/2} ∑_{h∈E(N,k)} (∏_{p|n} X_{n_{p}}(x_{p}^{h})) w_{h}^{r},

recall that

w_{h}^{r} = \frac{|a_{r}(h)|^{2}e^{−4πr}}{∥h∥^{2}} .

Proof. Using (4) and Corollary 2.2

K(x, y) = n^{1/2} ∑_{h∈E(N,k)} (∏_{p|n} X_{n_{p}}(x_{p}^{h})) \frac{φ_{h}(x)\overline{φ_{h}(y)}}{∥h∥^{2}}.

Thus the spectral side of (5) is

n^{1/2} ∑_{h∈E(N,k)} (∏_{p|n} X_{n_{p}}(x_{p}^{h})) \frac{1}{∥h∥^{2}} |∫_{N(Q)\N(A)} φ_{h}(n)θ(n^{−1})dn|^{2} .

From [Ge] Chapter 3 lemma 3.6,

∫_{N(Q)\N(A)} φ_{h}(n)θ(n^{−1})dn = \{ \begin{matrix}a_{r}e^{−2πr} & if r ∈ Z^{+} \\ 0 & otherwise. \end{matrix}

The proposition follows easily. □

Let δ ∈ G. We define

N_{δ} = {(n_{1}, n_{2}) ∈ N × N : n_{1}^{−1}δn_{2} ∼ δ};

here g_{1} ∼ g_{2} if g_{1} = zg_{2} for some z in the center. Denote the image of δ in N(Q)\\overline{G(Q)}/N(Q) by [δ] Proposition 3.2

KTF(f) = ∑_{[δ]∈N(Q)\\overline{G(Q)}/N(Q)} ∫_{N_{δ}(Q)\N(A)×N(A)} f(n_{1}^{−1}δn_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

Proof. The geometric side of KTF(f)

= ∫_{N(Q)\N(A)} ∫_{N(Q)\N(A)} ∑_{γ∈\overline{G(Q)}} f(n_{1}^{−1}γn_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}

= ∑_{[δ]∈N(Q)\\overline{G(Q)}/N(Q)} ∫_{N(Q)\N(A)} ∫_{N(Q)\N(A)} ∑_{γ∈\overline{N(Q)δN(Q)}} f(n_{1}^{−1}γn_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

= ∑_{[δ]} ∫ ∫ ∑_{(m_{1},m_{2})∈N_{δ}(Q)\N(Q)×N(Q)} f(n_{1}^{−1}m_{1}^{−1}δm_{2}n_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

Replace n_{1} by m_{1}^{−1}n_{1}, n_{2} by m_{2}^{−1}n_{2}. Since θ is trivial on m_{1}, m_{2} ∈ N(Q), KTF(f) is equal to

∑_{[δ]∈N(Q)\\overline{G(Q)}/N(Q)} ∫_{N_{δ}(Q)\N(A)×N(A)} f(n_{1}^{−1}δn_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

□

Denote

I_{δ}(f) = ∫_{N_{δ}(Q)\N(A)×N(A)} f(n_{1}^{−1}δn_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

An element δ ∈ \overline{G(Q)} is said to be admissible if the map

N_{δ}(A) → C : (n_{1}, n_{2}) ↦ θ(n_{1}^{−1}n_{2})

is trivial.

Lemma 3.3 If δ is not admissible, then I_{δ}(f) = 0.

Proof. Assume that δ is not admissible. Let (ν_{1}, ν_{2}) ∈ N_{δ}(A) such that θ(ν_{1}^{−1}ν_{2}) ≠ 1. Replace n_{1}, n_{2} by ν_{1}n_{1}, ν_{2}n_{2} respectively.

I_{δ}(f) = ∫_{N_{δ}(Q)\N(A)×N(A)} f(n_{1}^{−1}ν_{1}^{−1}δν_{2}n_{2})θ(n_{1}^{−1}ν_{1}^{−1}ν_{2}n_{2})dn_{1}dn_{2}.

Thus I_{δ}(f) = θ(ν_{1}^{−1}ν_{2})I_{δ} ⇒ I_{δ} = 0. □

Theorem 3.4 KTF(f) =

∫_{N(A)} f(n)θ(n)dn + ∑_{μ∈Q^{∗}} ∫_{N(A)} ∫_{N(A)} f(n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

Proof. By the Bruhat decomposition G = NM ∪ NM (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) N. Thus a

representative set of N(Q)\\overline{G(Q)}/N(Q) is \{(\begin{matrix}γ & 0 \\ 0 & 1 \end{matrix}) : γ ∈ Q^{∗}\} ⋃\{(\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) : μ ∈ Q^{∗}}. From Theorem 3.2, the geometric side = ∑_{[δ]} I_{δ}. By Lemma 3.3, I_{δ}(f) = 0 unless δ is admissible.

When δ = (\begin{matrix}γ & 0 \\ 0 & 1 \end{matrix}), simple calculation shows that N_{δ} = \{(\binom{1 γt}{1} , \binom{1 t}{1})\}. If δ is admissible, we have θ((γ−1)t) = 1 for any t ∈ A. Because θ is non-trivial, this cannot happen unless γ = 1. Thus

I_{e}(f) = ∫_{N_{δ}(Q)\N(A)×N(A)} f(n_{1}^{−1} (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) n_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}

= ∫_{diagonal \N(A)×N(A)} f(n_{1}^{−1}n_{2})θ(n_{1}^{−1}n_{2})dn_{2}dn_{1}.

Letting m_{1} = n_{1}, m_{2} = n_{1}^{−1}n_{2}, the diagonal becomes (m_{1}, e). The integral becomes

∫_{N(Q)×e\N(A)×N(A)} f(m_{2})θ(m_{2})dm_{2}dm_{1} = ∫_{N(A)} f(m_{2})θ(m_{2})dm_{2}.

When δ = (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}), simple calculation shows that N_{δ}(Q) = \{(e, e)\}. Thus δ is admissible. We have

I_{δ}(f) = ∫_{N(A)×N(A)} f(n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2})θ(n_{1}^{−1}n_{2})dn_{1}dn_{2}.

We can prove the theorem by summing up all the terms. □

4 Evaluation of Integrals

Lemma 4.1 For u ∈ Q, θ_{fin}(u) = θ_{∞}(−u) = e^{−2πru} Proof. We have 1 = θ(u) = θ_{∞}(u)θ_{fin}(u). The lemma follows easily. □

The following lemmas show us how to evaluate f^{n}. Suppose R is a ring, we denote R^{∗2} = {x^{2} : x ∈ R^{∗}}.

Lemma 4.2 Suppose g = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ G(A_{fin}) and det(g) ∈ n ∏ Z_{p}^{∗}. Then g ∈ supp f^{n} if and only if g ∈ M_{2}(∏ Z_{p}) and c ≡ 0 mod N.

Proof. Write g = zm, z = (\begin{matrix}ζ \\ & ζ \end{matrix}) ∈ Z(A_{fin}), m ∈ M(n, N). Taking the

determinant on both sides, we see that ζ is in ∏ Z_{p}^{∗}. Thus z can be absorbed into m, so g ∈ M(n, N). It is easy to see that g ∈ M(n, N) if and only if g satisfies the conditions in the lemma. □

Lemma 4.3 Suppose g = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ G(A_{fin}), then g is in supp f^{n} only if

det g ∈ nA_{fin}^{∗2} ∏ Z_{p}^{∗}.

Under this assumption, say det g = nζ^{2}u for ζ ∈ A_{fin}^{∗}, u ∈ ∏ Z_{p}^{∗}. Let

z = (\begin{matrix}ζ \\ & ζ \end{matrix}). Let m = z^{−1}g = (\begin{matrix}m_{11} & m_{12} \\ m_{21} & m_{22} \end{matrix}). Then g ∈ supp f^{n} if and only if

m ∈ M_{2}(∏ Z_{p}) and m_{21} ≡ 0 mod N.

Proof. If g is in supp f^{n}, then g = zm with z = (\begin{matrix}ζ \\ & ζ \end{matrix}) ∈ Z(A_{fin}) and

m ∈ M(n, N). Thus det g = ζ^{2} det m ∈ nA_{fin}^{∗2} ∏ Z_{p}^{∗}. This proves the first part.

Suppose det g = nζ^{2}u as stated in the lemma. Let z = (\begin{matrix}ζ \\ & ζ \end{matrix}). Obvi-

ously g ∈ supp f^{n} if and only if z^{−1}g ∈ supp f^{n}. One can easily show that det z^{−1}g = nu ∈ n ∏ Z_{p}^{∗}. The lemma follows easily from the above lemma. □

From now on we call the z = (\begin{matrix}ζ \\ & ζ \end{matrix}) appearing in the previous lemma a

z-part of g.

Define

True(statement) = \{ \begin{matrix}1 & if statement is true, \\ 0 & otherwise. \end{matrix}

Proposition 4.4

∫_{N(A_{fin})} f^{n}(n_{fin})θ_{fin}(n_{fin})dn_{fin}  (6)
= ψ(N)n^{1/2} True(n^{1/2} ∈ Z and n^{1/2}|r)

Proof. Write n_{fin} = (\begin{matrix}1 & t \\ 0 & 1 \end{matrix}). Its determinant is 1. From Lemma 4.3, n_{fin} ∈

supp f^{n} only if 1 ∈ nQ_{p}^{∗2}Z_{p}^{∗} for all primes p. Thus n_{p} is even for all p. As a result n ∈ ±Q^{∗2}. We can assume n is positive. Write n = n′^{2}, n′ ∈ Z^{+}.

Now a z-part of n_{fin} can be z = (\begin{matrix}n′ & 0 \\ 0 & n′ \end{matrix})^{−1}. Let m = z^{−1}n_{fin} = (\begin{matrix}n′ & n′t \\ 0 & n′ \end{matrix}).

The lower left entry of m is 0, which is divisible by N. By Lemma 4.3, n_{fin} is

in supp f^{n} if and only if (\begin{matrix}n′ & n′t \\ 0 & n′ \end{matrix}) ∈ M_{2}(∏ Z_{p}). Or equivalently

t′ = n′t ∈ ∏ Z_{p}.

Thus (6) is equal to

ψ(N) ∫_{t∈n′^{−1} ∏ Z_{p}} θ_{fin}(t)dt = ψ(N) ∫_{∏ Z_{p}} θ_{fin}(\frac{t′}{n′})d\frac{t′}{n′} .

Because θ_{fin} is trivial on ∏ Z_{p}, the integral (6) is equal to

ψ(N)n′ ∑_{s∈Z/n′Z} ∫_{s+n′ ∏ Z_{p}} θ_{fin}(\frac{t′}{n′})dt′

= ψ(N)n′ ∑_{s∈Z/n′Z} e(−\frac{sr}{n′}) meas(s + n′ ∏ Z_{p}) = ψ(N) ∑_{s∈Z/n′Z} e(\frac{sr}{n′}).

The result follows easily. □ Next, we evaluate

∫_{A_{fin}×A_{fin}} f^{n}(n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2})θ_{fin}(n_{1}^{−1})θ_{fin}(n_{2})dn_{1}dn_{2}.  (7)

Define

Kl_{u}(n, θ_{∞}) = ∑_{s_{1},s_{2}∈Z/uZ,s_{1}s_{2}≡n (mod u)} θ_{∞}(\frac{s_{1}}{u})θ_{∞}(\frac{s_{2}}{u}).

Proposition 4.5 The integral (7) ≠ 0 only if μ = ±\frac{n}{u^{2}} for some integer u ≡ 0 (mod N). Under this assumption, the integral (7) is equal to ψ(N) Kl_{u}(∓n, θ_{∞}).

Proof. Let n_{i} = \binom{1 t_{i}}{1} , i = 1, 2.

n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2} = (\begin{matrix}−t_{1} & μ − t_{1}t_{2} \\ 1 & t_{2} \end{matrix}) .  (8)

Notice det(n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2}) = −μ. From Lemma 4.3 , (8) ∈ supp f^{n} only if

μ ∈ nQ_{p}^{∗2}Z_{p}^{∗} for all p. Thus we have ord_{p}(μ) ≡ n_{p} (mod 2) for all p. As a

result μ ∈ ±nQ^{2}. Let μ = ±nζ^{2} for some ζ ∈ Q^{+}. We can take z = (\begin{matrix}ζ \\ & ζ \end{matrix})

as the z-part. Write m = z^{−1}n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2} = (\begin{matrix}∗ & ∗ \\ ζ^{−1} & ∗ \end{matrix}). By Lemma 4.3, (8)

is in supp f^{n} only if ζ^{−1} ∈ Z_{p} for all p. Hence ζ = 1/u for some u ∈ Z^{+}. Let μ = ±\frac{n}{u^{2}} , u ∈ Z.

m = (\begin{matrix}−ut_{1} & \frac{±n−(ut_{1})(ut_{2})}{u} \\ u & ut_{2} \end{matrix}) .

Write t′_{1} = ut_{1}, t′_{2} = ut_{2}. By Lemma 4.3 again, (8) is in supp f^{n} if and only if

t′_{1}, t′_{2} ∈ ∏ Z_{p}, u ≡ 0 (mod N), t′_{1}t′_{2} ≡ ±n (mod u).

The integral (7) becomes

∫_{t_{1},t_{2}} f^{n}((\begin{matrix}−t_{1} & μ − t_{1}t_{2} \\ 1 & t_{2} \end{matrix}))θ_{fin}(−t_{1})θ_{fin}(t_{2})dt_{1}dt_{2}

= ∫_{t′_{1},t′_{2}∈∏ Z_{p},t′_{1}t′_{2}∈±n+u ∏ Z_{p}} f^{n} (\begin{matrix}−t′_{1} & ∗ \\ u & t′_{2} \end{matrix}) θ_{fin}(−\frac{t′_{1}}{u})θ_{fin}(\frac{t′_{2}}{u})d\frac{t′_{1}}{u} d\frac{t′_{2}}{u}

= ψ(N) ∑_{s_{1},s_{2}∈Z/uZ,s_{1}s_{2}=±n} ∫_{(s_{1}+u ∏ Z_{p})×(s_{2}+u ∏ Z_{p})} θ_{fin}(−\frac{t′_{1}}{u})θ_{fin}(\frac{t′_{2}}{u})d\frac{t′_{1}}{u} d\frac{t′_{2}}{u}

= ψ(N) ∑_{s_{1},s_{2}∈Z/uZ,s_{1}s_{2}=±n} θ_{∞}(\frac{s_{1}}{u})θ_{∞}(−\frac{s_{2}}{u}) meas(∏ Z_{p})^{2}

= ψ(N) ∑_{s_{1},s_{2}∈Z/uZ,s_{1}s_{2}=∓n} θ_{∞}(\frac{s_{1}}{u})θ_{∞}(\frac{s_{2}}{u}) = ψ(N) Kl_{u}(∓n, θ_{∞}).

□

Proposition 4.6 When k ≥ 3,

∫_{N(R)} f_{k}(n)θ_{∞}(n)dn = \{ \begin{matrix}\frac{e^{−4πr}(4πr)^{k−1}}{(k−2)!} & if r > 0, \\ 0 & otherwise. \end{matrix}  (9)

Proof. Write n = \binom{1 t}{1}. By (2), the integral in (9) is equal to

\frac{k − 1}{4π} ∫_{−∞}^{∞} \frac{2^{k}}{(2 + it)^{k}} e^{2πirt}dt.

When r > 0, use the x-axis and the upper semi-circle as the contour. We can get the result easily by evaluating the residue of the integrand at t = 2i.

When r < 0, use the x-axis and the lower semi-circle as the contour. The result follows easily. □

Proposition 4.7 When k ≥ 3,

∫_{N(R)×N(R)} f_{k}(n_{1}^{−1} (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}) n_{2})θ_{∞}(n_{1}^{−1}n_{2})dn_{1}dn_{2}  (10)

is non-zero only if r, −μ are all positive. Under this condition, the integral is equal to

\frac{e^{−4πr}(4πi)^{k}r^{k−1}}{2(k − 2)!} (−μ)^{\frac{1}{2}} J_{k−1} (4πr\sqrt{−μ});

here J_{k} is the Bessel J function.

Proof. When μ > 0, det(n_{1}^{−1} (\begin{matrix}& μ \\ 1 \end{matrix}) n_{2}) = −μ < 0. The value of f_{k} at it is 0.

So we can assume μ < 0. Write n_{i} = \binom{1 t_{i}}{1} , i = 1, 2. By (2), (10) =

\frac{k − 1}{4π} ∫_{−∞}^{∞} ∫_{−∞}^{∞} \frac{2^{k}(−μ)^{k/2}e^{2πir(t_{2}−t_{1})}}{(−t_{1} + t_{2} + i(−1 + μ − t_{1}t_{2}))^{k}} dt_{1}dt_{2}

First assume r > 0, evaluate the residue of the integrand at t_{2} = i + \frac{μ}{t_{1}+i}, (10) becomes

\frac{k − 1}{4π} ∫_{−∞}^{∞} \frac{2πi}{(1 − it_{1})^{k}} \frac{2^{k}(−μ)^{k/2}(2πir)^{k−1}e^{2πir(i+ \frac{μ}{t_{1}+i} −t_{1})}}{(k − 1)!} dt_{1}

Use x-axis and lower semi-circle as the contour, the integral can be calculated by evaluating the residue of the integrand at t_{1} = −i. Notice that the path is counterclockwise.

Refer to [Wa] Chapter 2.1, we have

e^{\frac{1}{2} ξ(τ− \frac{1}{τ})} = ∑_{−∞}^{∞} τ^{n}J_{n}(ξ).

Take ξ = 4πr\sqrt{−μ}, τ = −i\frac{t_{1}+i}{\sqrt{−μ}}, Be the residue theorem, (10) becomes

\frac{k − 1}{4π} \frac{2πi}{(−i)^{k}} (−2πi)\frac{2^{k}(−μ)^{k/2}(2πir)^{k−1}}{(k − 1)!} e^{−4πr}J_{k−1}(4πr\sqrt{−μ})\frac{(−i)^{k−1}}{(−μ)^{\frac{k−1}{2}}} .

We can get the result easily.

When r < 0, use the axis and upper semi-circle as the contour. It is easy to show that (10) is 0. □

Theorem 4.8 Let k be an even number ≥ 3. Let n, N, r be any positive integers. Factorize n into ∏_{p} p^{n_{p}}. Assume GCD(N, n) = 1. Define θ_{∞}(x) = e^{2πirx}, then we have

∑_{h∈E(N,k)} (∏_{p|n} X_{n_{p}}(x_{p}^{h})) \frac{|a_{r}(h)|^{2}e^{−4πr}}{∥h∥^{2}}

= True(n^{1/2} ∈ Z, n^{1/2} | r)\frac{e^{−4πr}(4πr)^{k−1}}{(k − 2)!} ψ(N)  (11)

+\frac{e^{−4πr}(4πi)^{k}r^{k−1}}{2(k − 2)!} ψ(N) ∑_{v=1}^{∞} \frac{1}{Nv} J_{k−1}(\frac{4πn^{1/2}r}{Nv}) Kl_{vN}(n, θ_{∞}).  (12)

Proof. The spectral side is obtained by Proposition 3.1.

The geometric side is obtained by Theorem 3.4. The integral I_{e}(f) is the product of (6) and (9). Using the results in Proposition 4.6 and Corollary 4.4, we can get I_{e}(f).

Let δ = (\begin{matrix}0 & μ \\ 1 & 0 \end{matrix}). Then I_{δ}(f) is product of (7) and (10). By Proposition 4.5

(7) ≠ 0 only if μ = \frac{±n}{u^{2}} and N|u. Write u = Nv. By Proposition 4.7, (10) ≠ 0

only if μ < 0. Thus ∑_{δ} I_{δ}(f) is a sum over \{δ = (\begin{matrix}0 & \frac{−n}{(Nv)^{2}} \\ 1 & 0 \end{matrix} ) : v ∈ Z^{+}\}.

Multiply the results we obtained in Proposition 4.5 and Proposition 4.7. Summing up all the terms in the geometric side. We can obtain the formula by equating the spectral side and geometric side and then dividing both sides by n^{1/2}. □

5 Weighted distribution

Lemma 5.1

| Kl_{u}(n, θ_{∞})| ≤ un.

Proof. Obviously | Kl_{u}(n, θ_{∞})| ≤ |{s_{1}s_{2} ≡ n (mod u)}|. It suffices to prove |{s_{1}s_{2} ≡ n (mod u)}| ≤ un for u = p^{u_{p}} and n = p^{n_{p}}.

If n_{p} ≥ u_{p}, |{s_{1}s_{2} ≡ n (mod u)}| ≤ u^{2} ≤ un.

Assume n_{p} < u_{p}.

|{s_{1}s_{2} ≡ n (mod u)}| = ∑_{s=1}^{u} |{t : st ≡ n (mod u)}|

= ∑_{s=1,gcd(s,u)|n}^{u} gcd(s, u) = ∑_{s_{p}=0}^{n_{p}} p^{s_{p}} · |{s : ord_{p}(s) = s_{p}, 1 ≤ s ≤ u}|

≤ ∑_{s_{p}=0}^{n_{p}} p^{s_{p}} \frac{p^{u_{p}}}{p^{s_{p}}} ≤ (n_{p} + 1)p^{u_{p}} ≤ nu

□

From [Iw] equation (5.16)

J_{k}(x) ≤ min{x^{k}, x^{−1/2}}.

Proposition 5.2 Let k be an even integer ≥ 3, then

\frac{1}{ψ(N)} ∑_{h∈E(N,k)} ∏_{p|n} X_{n_{p}}(x_{p}^{h})w_{h}^{r}

= True(n^{1/2} ∈ Z, n^{1/2}|r)\frac{e^{−4πr}(4πr)^{k−1}}{(k − 2)!} + O(\frac{n^{(k+1)/2}}{N^{k−1}} ).

Here the constant in the O-notation depends on θ only.

Proof. Form the inequalities given above, (12) is

≪ \frac{e^{−4πr}(4π)^{k}r^{k−1}}{2(k − 2)!} ψ(N) ∑_{v=1}^{∞} \frac{1}{Nv} (\frac{4πn^{1/2}r}{Nv} )^{k−1} (Nvn)

≪_{r} \frac{ψ(N)}{N^{k−1}} n^{(k+1)/2} ∑_{v} \frac{1}{v^{k−1}} ≪_{r} \frac{ψ(N)}{N^{k−1}} n^{(k+1)/2}.

The proposition follows easily from Theorem 4.8. □

Corollary 5.3 The average of weight

\frac{1}{ψ(N)} ∑_{h∈E(N,k)} w_{h}^{r} = \frac{e^{−4πr}(4πr)^{k−1}}{(k − 2)!} + O(\frac{1}{N^{k−1}} ).

Proof. Put n = 1 in the above proposition. □

Let p_{1}, p_{2}, ⋯ p_{ℓ} be distinct primes. For any prime p, let I_{p} be an interval in R containing all the possible values of x_{p}^{h} for any Hecke eigenform h. By Deligne’s result, we can take I_{p} = [−2, 2], but we don’t need this strong result. We only need the fact that I_{p} is a finite interval. Refer to [Ro] Proposition 2.9 for the proof of this fact.

Denote the set of real valued continuous functions on I = I_{p_{1}} × I_{p_{2}} ⋯ × I_{p_{ℓ}} by C(I_{p_{1}} × I_{p_{2}} ⋯ × I_{p_{ℓ}}). Define a topological structure on it by using the L^{∞} norm ∥f∥_{∞} = max{|f(x)|}. Let k be an even integer ≥ 3, define a functional 𝔉_{N} on C(I_{p_{1}} × I_{p_{2}} ⋯ × I_{p_{ℓ}}) by

𝔉_{N}f = \frac{∑_{h∈E(N,k)} f(x_{p_{1}}^{h}, ⋯, x_{p_{ℓ}}^{h})w_{h}^{r}}{∑_{h∈E(N,k)} w_{h}^{r}}

Proposition 5.4

lim_{N→∞,(N,p_{1},··· ,p_{ℓ})=1} 𝔉_{N}X_{n_{p_{1}}} × ⋯ × X_{n_{p_{ℓ}}} = ∏_{i=1}^{ℓ} True(2|n_{p_{i}}, n_{p_{i}}/2 ≤ r_{p_{i}})

Proof. Let n = p_{1}^{n_{p_{1}}} ⋯ p_{ℓ}^{n_{p_{ℓ}}}. From Corollary 5.2 and 5.3,

𝔉_{N}f = \frac{∏_{i=1}^{ℓ} True(2|n_{p_{i}}, n_{p_{i}}/2 ≤ r_{p_{i}})\frac{e^{−4πr}(4πr)^{k−1}}{(k−2)!} + O(\frac{n^{(k+1)/2}}{N^{k−1}} )}{\frac{e^{−4πr}(4πr)^{k−1}}{(k−2)!} + O(\frac{1}{N^{k−1}} )}.

Letting N → ∞, the proposition follows easily. □

Proposition 5.5

∫_{R} X_{n}(x)X_{m}(x)dμ_{∞}(x) = δ_{nm}.

Proof. A proof can be found in [Se2] Section 2.2.

\frac{π}{2} δ_{nm} = ∫_{0}^{π} \frac{sin nθ}{sin θ} sin^{2} θ \frac{sin mθ}{sin θ} dθ

Recall X_{n}(x) = \frac{sin nθ}{sin θ} , x = 2 cos θ. Make a substitution x = 2 cos θ, we have

\frac{π}{2} δ_{nm} = ∫_{−2}^{2} X_{n}(x)X_{m}(x)\frac{sin θ}{2} dx.

The proposition follows easily. □

Theorem 5.6 Define measure

dμ_{i}(x) = ∑_{0≤n′≤r_{p_{i}}} X_{2n′}(x)dμ_{∞}(x).  (13)

Also define

𝔉(f) = ∫_{I_{p_{1}}} ⋯ ∫_{I_{p_{ℓ}}} f(x_{1}, ⋯ x_{ℓ})dμ_{1}(x_{1}) ⋯ dμ_{h}(x_{ℓ}).

Then for any f ∈ C(I_{p_{1}} × I_{p_{2}} ⋯ × I_{p_{ℓ}}),

lim_{N→∞} 𝔉_{N}(f) = 𝔉(f).

Proof. By the previous proposition

∫_{I_{p_{i}}} X_{n_{p_{i}}} (x)dμ_{i}(x) = True(2|n_{p_{i}}, n_{p_{i}}/2 ≤ r_{p_{i}}).

Take the product over i = 1, ⋯ ℓ and by proposition 5.4, we have

𝔉(X_{n_{p_{1}}} × ⋯ × X_{n_{p_{ℓ}}}) = lim_{N→∞,(N,p_{1},··· ,p_{ℓ})=1} 𝔉_{N}(X_{n_{p_{1}}} × ⋯ × X_{n_{p_{ℓ}}}).

One can easily show that |𝔉_{N}(f)| ≤ ∥f∥_{∞}. Thus 𝔉_{N} is a continuous linear functional. Because deg X_{n} = n, the linear span of {X_{n}} consists of all the one variable polynormals. Thus the linear span of {X_{n_{p_{1}}} × ⋯ × X_{n_{p_{ℓ}}} } consists of all the possible polynomials. The theorem follows by the fact that polynomials are dense in C(I_{p_{1}} × I_{p_{2}} ⋯ × I_{p_{ℓ}}). □

Theorem 5.7 Let k be an even integer ≥ 3. Consider the family of sets S_{N} = {(x_{p_{1}}^{h}, ⋯, x_{p_{ℓ}}^{h}) : h ∈ E(N, k)}, (N, p_{1} ⋯ p_{ℓ}) = 1 with weight {w_{h}^{r}} assigned to (x_{p_{1}}^{h}, ⋯, x_{p_{ℓ}}^{h}). Then the family of sets {S_{N}, p ∤ N} is {w_{h}^{r}}-distributed with respect to the measure dμ_{1} ⋯ dμ_{ℓ} when N → ∞. Here μ_{i} is given by equation (13). The measure μ_{i} has the following properties : (a) it is supported on [−2, 2], (b) it is a polynomial times the Sato-Tate measure on [−2, 2], (c) it depends only on ord_{p}(r).

Proof. Follows easily from the previous theorem. □ Proof of 1.1. Take ℓ = 1 and p_{1} = p. □

References

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[KL] A. Knightly, C.C. Li, Traces of Hecke Operators. In preparation.

[Li] C.C. Li, Kuznietsov trace formula and asymptotic behavior of Hecke eigenvalues UCLA thesis, 2001.

[Ro] J. Rogawski, Modular forms, the Ramanujan conjecture, and the Jacquet-Langlands correspondence. Discrete, Groups, Expanding Graphs and Invariant Measures, by A. Lubotzky, Birhäuser, Basel,1994, pp. 135-176.

[Se1] J-P. Serre, Abelian l-adic representations and elliptic curves. W. A. Benjamin, Inc., New York-Amsterdam 1968

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[Wa] G.N. Watson, Theory of Bessel Functions. Cambridge University Press, 1944.