THE DISTRIBUTION OF THE EIGENVALUES

OF HECKE OPERATORS

J.B. Conrey W. Duke D.W. Farmer

Abstract. For each prime p, we determine the distribution of the p^{th} Fourier coefficients of the Hecke eigenforms of large weight for the full modular group. As p → ∞, this distribution tends to the Sato–Tate distribution.

1. Introduction and statement of results

Let

f(z) = ∑_{n=1}^{∞} a_{f,n}e(nz)

be a cusp form of weight k for the full modular group. We assume that f(z) is an eigenfunction of all the Hecke operators T_{n}, and we set a_{f}(n) = n^{\frac{1−k}{2}} a_{f,n}. The assumption that f(z) is a Hecke eigenform gives:

a_{f}(mn) = a_{f}(n)a_{f}(m)  if (m, n) = 1

a_{f}(p^{j}) = a_{f}(p)a_{f}(p^{j−1}) − a_{f}(p^{j−2})  for p prime

|a_{f}(n)| ≤ d(n),

where d(n) is the divisor function. The above relations were first conjectured by Ramanujan based on a computation of the coefficients of the weight 12 cusp form Δ(z) = ∑ τ(n)e(nz). The two equations were proven for τ(n) by Mordell, using what are now known as the Hecke operators. The inequality was proven by Deligne as a consequence of his proof of the Weil conjectures. Those results determine everything about a_{f}(n) except for the distribution of the a_{f}(p) ∈ [−2, 2].

Define θ_{f}(p) ∈ [0, π] by a_{f}(p) = 2 cos θ_{f}(p). It is conjectured that for each f the θ_{f}(p) are uniformly distributed with respect to the Sato–Tate measure

\frac{2}{π} sin^{2} θ dθ.

Typeset by AMS-TEX

Research of the first and second author supported in part by a grant from the NSF.

This conjecture was first made for cusp forms associated to non–CM elliptic curves, and was extended to Hecke eigenforms for the full modular group by Serre [Ser]. This conjecture appears to be quite deep, for it has been shown [Ogg] [Mur] to be equivalent to the nonvanishing on the line Re(s) = 1 of all the m^{th} symmetric power L–functions L_{m}(s) associated to f(z).

In this paper we change perspective slightly by fixing p and looking at the distribution of a_{f}(p) as f(z) varies. That is, we are looking at the distribution of the eigenvalues of T_{p}.

We write S_{k}(Γ(1)) for the space of cuspforms of weight k for the full modular group, and we write f ∈ S_{k} to mean f ∈ S_{k}(Γ(1)) and f is a Hecke eigenform. Following Selberg, we write σ_{k}(T_{n}) for the trace of the Hecke operator T_{n} acting on S_{k}(Γ(1)). Because of our normalization, σ_{k}(T_{n}) = n^{(k−1)/2} ∑_{f∈S_{k}} a_{f}(n).

Theorem 1. Suppose p is prime. As k → ∞, the set {θ_{f}(p) : f ∈ S_{k}} becomes uniformly distributed with respect to the measure

\frac{2}{π} (1 + \frac{1}{p}) \frac{sin^{2} θ}{(1 − \frac{1}{p})^{2} + \frac{4}{p} sin^{2} θ} dθ.

The same distribution was found by Sarnak [Sar] for the p^{th} coefficient of Maass forms averaged over the Laplacian eigenvalues. Sarnak has pointed out to us that the above measure is the p–adic Plancherel measure, and it is also the spectral measure of the nearest–neighbor Laplacian on a p + 1 regular tree (see [LPS]).

Theorem 2. As p → ∞, with p prime, and k → ∞ with k > e^{p}, the set {θ_{f}(p) : f ∈ S_{k}} becomes uniformly distributed with respect to the Sato–Tate measure.

A result similar to this was obtained by Birch [B], who considered the distribution of θ(p) over the elliptic curves over F_{p}. Similar results have also been found for Kloosterman sums [K] [A] [Mic] and cubic exponential sums [L]. In each case, results have been obtained by fixing the prime p and varying the other parameters in the problem. The original problem of varying the prime, keeping the other parameters fixed, remains inaccessible.

The idea behind the proofs of the Theorems is as follows. Using the Selberg trace formula we obtain a simple expression for ∑_{f∈S_{k}} a_{f}(p^{j}), valid for large k. Using the recurrence relation satisfied by a_{f}(p^{j}) leads to an expression for ∑_{f∈S_{k}} a_{f}(p)^{j}. Such an expression for all j is sufficient to determine the distribution of {a_{f}(p) : f ∈ S_{k}}.

2. Initial Lemmas

Our starting point is the Selberg trace formula applied to the Hecke operators. The following is formula (4.5) in [Sel].

Lemma 1. We have

\begin{aligned}σ_{k}(T_{n}) = & − \frac{1}{2} ∑_{−2\sqrt{n}<m<2\sqrt{n}} H(4n − m^{2})\frac{η_{m}^{k−1} − \overline{η}_{m}^{k−1}}{η_{m} − \overline{η}_{m}} \\ & − ∑_{{d|n \atop d≤\sqrt{n}}}′d^{k−1} + δ(\sqrt{n})\frac{k − 1}{12}n^{\frac{k}{2} −1}, \end{aligned}

where H(d) is the class number, δ(x) = 1 if x is an integer and is 0 otherwise, ∑′ means that a term with d = \sqrt{n} is counted with weight 1/2, and

η_{m} = \frac{m + i(4n − m^{2})^{\frac{1}{2}}}{2} .

The above formula directly leads to an expression for ∑_{f∈S_{k}} a_{f}(p^{j}). Using the estimate H(d) ≪ d^{\frac{1}{2} +ε} gives

Lemma 2. Suppose p is prime. As k → ∞,

∑_{f∈S_{k}} a_{f}(p^{j}) = \frac{k}{12} \frac{1}{p^{\frac{j}{2}}} + O(p^{\frac{j}{2} +ε})  j even

= O(p^{\frac{j}{2} +ε})  j odd

To make use of the above Lemma, it is necessary to relate a_{f}(p)^{n} to a_{f}(p^{j}). This involves the Chebyshev polynomials U_{n}(x). We have

Lemma 3. Suppose p is prime and write a_{f}(p) = 2 cos θ_{f}(p). Then

a_{f}(p^{j}) = U_{j}(cos θ_{f}(p))

and

a_{f}(p)^{n} = ∑_{j=0}^{n} h_{n}(j)a_{f}(p^{j}),

where

h_{n}(j) = \frac{2^{n+1}}{π} ∫_{0}^{π} cos^{n} θ sin(j + 1)θ sin θ dθ.

Proof. The first expression follows directly from the recurrence relation a_{f}(p^{j}) = a_{f}(p)a_{f}(p^{j−1}) − a_{f}(p^{j−2}) and the fact that the U_{j}(x) satisfy essentially the same recurrence. To prove the second, note that the h_{m}(j) satisfy

x^{n} = ∑_{j=0}^{n} h_{n}(j)U_{n}(x/2).

Since the U_{n}(x) are orthonormal on [−1, 1] with respect to the measure \frac{2}{π} \sqrt{1 − x^{2}} dx, we have

h_{n}(j) = \frac{2^{n+1}}{π} ∫_{−1}^{1} x^{n}U_{n}(x)\sqrt{1 − x^{2}} dx.

Changing variables x ↦ cos θ gives the stated formula.

3. Proof of the Theorems

Both Theorems follow immediately from this Proposition.

Proposition 1. Suppose p is prime. As k → ∞,

∑_{f∈S_{k}} a_{f}(p)^{n} = \frac{k}{6π} (1 + \frac{1}{p}) ∫_{0}^{π} 2^{n} cos^{n} θ \frac{sin^{2} θ}{1 + \frac{1}{p^{2}} − \frac{2}{p} cos 2θ} dθ + O (p^{\frac{n}{2} +ε})

Proof. We assume n is even, the case of odd n being identical. Starting with the second expression in Lemma 3, and then applying Lemma 2 gives,

∑_{f∈S_{k}} a_{f}(p)^{n} = ∑_{f∈S_{k}} ∑_{j=0}^{n} h_{n}(j)a_{f}(p^{j})

= \frac{k}{12} ∑_{j=0}^{n/2} \frac{h_{n}(2j)}{p^{j}} + O (p^{\frac{n}{2} +ε})

= \frac{k}{12} ∑_{j=0}^{∞} \frac{h_{n}(2j)}{p^{j}} + O (p^{\frac{n}{2} +ε}) ,

the completion of the sum on j being valid because h_{n}(j) = 0 for j > n. Now inserting the expression for h_{n}(j) from Lemma 3, switching the sum and integral, and then evaluating the sum on j gives,

∑_{f∈S_{k}} a_{f}(p)^{n} = \frac{k}{6π} ∑_{j=0}^{∞} \frac{1}{p^{j}} ∫_{0}^{π} 2^{n} cos^{n} θ sin(2j + 1)θ sin θ dθ + O (p^{\frac{n}{2} +ε})

= \frac{k}{6π} ∫_{0}^{π} 2^{n} cos^{n} θ sin θ ∑_{j=0}^{∞} \frac{sin(2j + 1)θ}{p^{j}} dθ + O (p^{\frac{n}{2} +ε})

= \frac{k}{6π} (1 + \frac{1}{p}) ∫_{0}^{π} 2^{n} cos^{n} θ \frac{sin^{2} θ}{1 + \frac{1}{p^{2}} − \frac{2}{p} cos 2θ} dθ + O (p^{\frac{n}{2} +ε}) ,

as claimed.

The deduction of the Theorems from Proposition 1 is by Weyl’s criterion: the sequence (b_{n}) is uniformly distributed on [a, b] with respect to the measure dμ provided

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} g(b_{n}) = ∫_{a}^{b} g dμ,

for all g ∈ C([a, b]). Proposition 1 gives the necessary equality for g(x) = x^{n}, uniformly in k, which is sufficient because the polynomials are dense in C([a, b]). This completes the proofs of the Theorems.

References

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Department of Mathematics, Oklahoma State University, Stillwater, OK 74078

Department of Mathematics, Rutgers University, New Brunswick, NJ 08903