We know that the eigenvalue of on is 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 prime to If there are finitely many eigenforms of level an arbitrary weight such that If there are finitely many eigenforms with the additional condition that they do not have CM by a quadratic field in which is inert.

I have no idea how to prove this conjecture. If one counts the number of such forms of weight then the trivial bound for eigenforms with is 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 but today I want to talk about the much easier case when where, via some -adic tricks, one can obtain a substantial improvement on the trivial bound. Let’s start from the following:

**Proposition I:** Let denote the number of cuspforms of level and weights such that . Assume that Then

** Proof:** Since the -adic valuation of 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 and the valuation of

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

**Proposition II:** Let denote the number of cuspforms of level and weight such that . Assume that Then, as ,

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 Over this family, the function is continuous, and hence so is More importantly, over a small enough disc, it is an Iwasawa function. Let denote an infinite set of integral weight such that, for the relevant points of we have or

If is a limit point of then certainly will vanish at Since this function is a non-zero bounded function on a disc, it has only finitely many zeros, and so the set of weights 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 is not bounded, we may imagine that the set consists of a sequence of integers (which we may assume to be increasing in the Archimedean norm): which converge -adically to and, at the relevant point of correspond to an eigenform which satisfies the equation

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

where the LHS has the same valuation as the leading term of the RHS for sufficiently small If we deduce that, for sufficiently large integers

for some which implies that and hence also that

for some . This iterated exponential growth proves the result. **QED.**

The argument also shows that if the set is infinite, the limit roots of will be transcendental Liouville numbers, which seems unlikely. The result also applies if one replaces by a sufficiently continuous function without zeros, say On the other hand, I don’t think these analytic methods will ever be enough to prove the conjectural bound, which is