Here is a question I raised at the Puerto-Rico conference during one of the “problem sessions.” Toby Gee seems to remember that I had some half-baked heuristics that predicted both **A** and **B** below, but perhaps one of my readers has a more sophisticated suggestion, or even a similarly wild guess (or even a similarly contradictory collection of guesses).

Fix a pair of distinct odd primes and . Now consider a random normalized Hecke eigenform of weight two and level , where is squarefree and prime to both and . Now take the Hecke eigenvalue and reduce it modulo a random prime above .

**Question**: As one ranges over all newforms of conductor , what is the resulting distribution — if it even exists — of ?

Let me not be too precise about what random means — for example, there's a question about whether one wants to normalize in some way for Galois conjugates of eigenforms, but none of this will really matter for the very weak questions I have in mind. For example, consider the following two possibilities:

**A**: The element lies in at least 100% of the time.**B**: The element lies in at least 100% of the time.

Here by 100% I mean as a proportion of all forms as , although I confess that I can’t even rule out the extreme version of **A** where 100% really means every single form.

The specifications on the level are designed to rule out some “trivial” examples. At level with squarefree there will be no CM-forms, which is one cheap way to generate large coefficient fields. The level also prevents twisting by characters (an even cheaper trick). Finally, in general, it is possible to generate large coefficient fields for Galois representations by imposing a local condition at some auxiliary prime . For example, one can impose some supercuspidal condition so that the *local* residual representation at does not land in for any not divisible by an arbitrary fixed integer chosen in advance. However, this too is not possible if is forced to be either unramified or special (up to unramified quadratic twist).

Note that there do exist infinitely many semi-stable modular elliptic curves, so will lie in at least infinitely often. This disproves the “extreme” version of **B**, but doesn’t go very far towards disproving the asymptotic version of **B**. As for **A**, every single time you write down a normalized eigenform with coefficients in some field , you disprove the extreme version of **A** for a positive density of pairs . But no finite collection of such forms can disprove **A** even for a single and varying , because there will always be (many) primes which split completely in any finite collection of number fields.

Here are three questions:

- Can you disprove the extreme version of
**A**for all and ? - Can you disprove the super-extreme version of
**A**, namely, show that for all primes , there exists a newform of squarefree level prime to such that the residual representation is not defined over ? (equivalently, replace by the collection of all with prime to .) - Can you give any heuristic that suggests that either
**A**or**B**(in the weaker form) is either strong or true? - Do you have any guesses as to the distribution of the ?

Right now, as you read this, KB’s computer is churning away in sage generating some data, which will be the topic of Part II. But until then, I would like to hear your opinions/guesses. For me, I think that **A** is probably false, but I honestly have no feeling for **B**.

I’m not sure why you’re saying “at least 100%”. My emails do suggest that you originally conjectured (during the problem session?) that most of the time they were in F_p, and possibly 100% of the time asymptotically, but that you then flipflopped on that. I think the second conversation took place in the sea, though, so unfortunately I have no notes to back this claim up.

Wow, what heuristic supports A??

I would have thought that, most of the time the eigenvalues live in bigger and bigger extensions, so to speak, and therefore no limiting distribution? I’m sure you’ve thought of both of these, but both thinking about charpoly(T_l) like a random polynomial, and thinking on the Galois side

seems to point against (A). On the Galois side it is a bit unclear, perhaps, how much a fixed residual representation deforms, but without thinking carefully I’d imagine that Cohen-Lenstra predicts “not too much”.

To be fair, my flirtation with

Awas relatively short. (In other words, I was for it before I was against it, or something like that.) Suppose you just count . Then is the expected number of level forms with image containing something like a constant that decreases rapidly with ? I remember you told me the heuristics here, but I can never quite remember the numbers. If you are correct, though, then surely it’s embarrassing that one can’t disproveA?I think actually doesn’t decrease with …

You are right, it is embarrassing. I didn’t think this through, but one thing we could try is this: if the strong form of (

A) holds, then the trace of will be zero for any other Hecke operator , and we can try to show this doesn’t happen (at least for many ) via trace formula. At the least, the class numbers that show up here don’t depend on , and thus we could at the least hope to show this way that a fixed doesn’t satisfy extreme-Afor most with a finite amount of computation. There are also terms in the trace formula like the genus of , which we could arrange to be indivisible by even if we know nothing about class numbers, so one might optimistically hope to get more this way.The class number terms are really hard to understand modulo , in some sense. For example, if you go to high weight (and level one, say), then you know that the Newton Polygon of is bounded below by some quadratic, and so the coefficients of the trace formula (for higher coefficients of the char poly, which are related to the trace of powers of ) become more and more divisible by . But I don’t think you can even prove just by looking at the formula that they are even divisible by . It’s one of those things in which hard work will lead to only modest rewards. (That reminds of the result, [maybe by Silverman?] that if you assume the ABC conjecture, you can prove that there are primes with for .)

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