Counting solutions to a_p = λ

We know that the eigenvalue of T_2 on \Delta is 24. Are there any other level one cusp forms with the same Hecke eigenvalue? Maeda’s conjecture in its strongest form certainly implies that there does not. But what can one prove along these lines? Conjecturally, one would certainly predict the following:

Conjecture: Fix a tame level N prime to p. If \lambda \ne 0, there are finitely many eigenforms of level N an arbitrary weight such that a_p = \lambda. If \lambda = 0, there are finitely many eigenforms with the additional condition that they do not have CM by a quadratic field in which p is inert.

I have no idea how to prove this conjecture. If one counts the number of such forms of weight \le X, then the trivial bound for eigenforms with a_p = \lambda is O(X^2). When I visited Princeton a few weeks ago, Naser Sardari, a student of Sarnak, showed me a short preprint he is writing which improves this bound by a power saving (additionally, it gives a power saving for each individual weight as well). The most interesting case of this result is when \lambda = 0, but today I want to talk about the much easier case when \lambda \ne 0, where, via some p-adic tricks, one can obtain a substantial improvement on the trivial bound. Let’s start from the following:

Proposition I: Let S_{\lambda}(X) denote the number of cuspforms of level N and weights \le X such that a_p = \lambda. Assume that \lambda \ne 0. Then

S_{\lambda}(X) = O(X).

Proof: Since \lambda \ne 0, the p-adic valuation of \lambda is finite. However, all forms with bounded slope belong to one of finitely many Coleman families, so the number of such forms in any weight is bounded. Using Wan’s explicit results, one can even give an explicit bound here that depends only on N, p, and the valuation of \lambda.

The point of this post, however, is to give an improvement on this bound.

Proposition II: Let S_{\lambda}(X) denote the number of cuspforms of level N and weight \le X such that a_p = \lambda. Assume that \lambda \ne 0. Then, as X \rightarrow \infty,

S_{\lambda}(X) \ll_{\lambda} \log \log \log \log \log \log \log X.

The argument will (obviously) allow for an arbitrary number of logs. But then the statement would become more cumbersome.

Proof: As in the proof of the previous result, we may reduce to the case where we are considering a single Coleman family \mathcal{F}. Over this family, the function U_p is continuous, and hence so is U_p(U_p - \lambda). More importantly, over a small enough disc, it is an Iwasawa function. Let \Sigma denote an infinite set of integral weight such that, for the relevant points of \mathcal{F}, we have T_p = \lambda, or

U_p(U_p - \lambda) = - p^{k-1}.

If s is a limit point of \Sigma, then certainly U_p(U_p - \lambda) will vanish at s. Since this function is a non-zero bounded function on a disc, it has only finitely many zeros, and so the set of weights \Sigma will have only finitely many limit points. Thus, we may reduce to the case where the set of weights has a single limit point. In particular, if S_{\lambda}(X) is not bounded, we may imagine that the set \Sigma consists of a sequence of integers (which we may assume to be increasing in the Archimedean norm): k_0, k_1, k_2, \ldots which converge p-adically to s, and, at the relevant point of \mathcal{F}, correspond to an eigenform which satisfies the equation

U_p(U_p - \lambda)(k_i) = - p^{k_i - 1}.

Around a zero s, any Iwasawa function has an asymptotic expansion of the form

F(s + \epsilon) \simeq A \cdot \epsilon^m + \ldots

where the LHS has the same valuation as the leading term of the RHS for sufficiently small \epsilon. If F = U_p(U_p - \lambda), we deduce that, for sufficiently large integers k_i,

v(s - k_n) = r k_n + c

for some r = 1/m > 0, which implies that v(k_{n+1} - k_n) = v(s - k_n), and hence also that

k_{n+1} - k_n > C p^{r k_n}

for some r  > 0. This iterated exponential growth proves the result. QED.

The argument also shows that if the set \Sigma is infinite, the limit roots of U_p - \lambda = 0 will be transcendental Liouville numbers, which seems unlikely. The result also applies if one replaces \lambda by a sufficiently continuous function without zeros, say a_2 = 24(1 + 2(k -12)^2). On the other hand, I don’t think these analytic methods will ever be enough to prove the conjectural bound, which is O(1).

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2 Responses to Counting solutions to a_p = λ

  1. Nicolas Templier says:

    There is a standard notation with which the bound reads \ll_{N,\lambda} \mathrm{slog} X. The Sloane sequence is A001069, see also iterated logarithm.

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