Here's an idea that I've found appealing but have never been able to get anywhere with.

One way to frame the Gouvea-Mazur question (for lack of a better term, since the original conjecture by the same name has been disproven) is as follows: Let $\pi:\mathcal{E}\to \mathcal{W}$ be the eigencurve (of some tame level) with its natural projection to weight space. Given a classical point $x\in \mathcal{E}$, what is the radius of the largest disk around $\pi(x)\in \mathcal{W}$ over which $\pi$ admits a section sending $\pi(x)$ to $x$?

Ramification points of the map $\pi$ provide natural obstructions to the existence of such a section. On the other hand, these ramification points arise naturally as the zeros of a symmetric square $L$-function on $\mathcal{E}$. This is the the main result of Walter Kim's 2006 thesis written under Robert Coleman (which, as far as I know, never appeared anywhere).

I once had a chat with Barry Mazur in which he (very roughly - these are my words from my recollection several years on) said that this symmetric square $L$ function should be thought of as a section of $\Omega^1_{\mathcal{E}}$, namely the pullback of the differential $dt/t$ on $\mathcal{W}$ via $\pi$. (Here $t$ is the coordinate on the disk of radius $1$ about $1 \in \mathbb{C}_p$ and we're using the usual identification of $\mathcal{W}$ with a stack of such disks.)

Clearly $\omega=\pi^*(dt/t)$ has zeros precisely at the ramification points. More generally:

1) Is the "size" of $\omega$ related to the maximal radius of a section to $\pi$?

2) Is this radius then related to the algebraic part of the symmetric square $L$-function?

The first question isn't so well-posed, and may rely on an integral model (formal scheme) giving rise to $\mathcal{E}$.

  • $\begingroup$ "should be thought of as a section" - is some version of this literally true? $\endgroup$ Feb 14, 2011 at 21:52
  • $\begingroup$ @David: I don't know, but the fact that their zeros coincide is somehow compelling. My recollection is that Mazur was musing about "where $L$-values naturally lie" as objects on the eigencurve, and that, in the case of the symmetric square $L$-function, the answer seems to be $\Omega^1$. $\endgroup$
    – Ramsey
    Feb 14, 2011 at 22:07
  • $\begingroup$ This really confuses me. If you consider some $p$-adic symmetric square $L$-function then its value is a $p$-adic number, so how can it be considered as a differential? I had always thought that (special values of) $p$-adic $L$-functions were just going to be meromorphic functions on the eigencurve. $\endgroup$ Feb 14, 2011 at 22:17
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    $\begingroup$ To employ a potentially dangerous analogy - a cusp form $f$ of weight two is just a function and it's values are just complex numbers. Nonetheless, there is good reason for thinking of such a thing as a differential form (in this because of the functional equations satisfied by $f$). It wouldn't surprise me if there was something much deeper behind Mazur's comment, but what I took from the conversation was: since the zeros of $L(Sym^2)$ are exactly the zeros of $\omega$, does this mean that there is some natural sense in which these two objects coincide? $\endgroup$
    – Ramsey
    Feb 14, 2011 at 22:40
  • $\begingroup$ @Ramsey: I don't buy it! A cusp form of weight 2 is a function on the upper half plane, not a function on a modular curve. The sheaf of differentials on the upper half plane is canonically trivial; the sheaf of differentials on a modular curve is not. So from that optic it's not surprising at all that a weight 2 modular form is a function---it's just not a function on the right space :-) $\endgroup$ Feb 15, 2011 at 7:26


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