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I found an old MOF post about representations attached to p-adic modular forms: Representations attached to p-adic modular forms and I have some follow up questions on the same topic.

If we have a classical form of weight k, it is know that the p-adic representation attached to the form is crystalline if p does not divide the level. It is de Rham in general and it has Hodge-Tate weights 0 and k-1.

My question is that if the form is not classical, then the weight is not an integer. Is this representation crystalline or de Rham? Can we say something about the Hodge-Tate weights?

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There are two subtleties regarding how to formulate this question.

Firstly, there are several notions of "p-adic modular form". There's Hida's ordinary p-adic modular forms (a very small space); there's Coleman's overconvergent p-adic modular forms (a much bigger space); and there's Serre and Katz's space of p-adic modular forms, which is a vastly bigger space (too big to have a reasonable theory of eigenforms).

The second subtlety is that for the latter two definitions (Coleman or Serre-Katz) it is not true that a non-classical form has non-integer weight; there are non-classical forms of integer weight (lots of them).

With that out of the way: if f is in any of these spaces and its weight is non-integral, then the associated Galois representation isn't even Hodge--Tate, so it's certainly not de Rham or crystalline. The eigenvalues of its Sen operator are what they should be, i.e. 0 and $k - 1$ where $k$ is the weight (but you have to be careful what this means!).

If $f$ is of integer weight and it's ordinary, then it's classical and the local representation is de Rham. If it's overconvergent and has finite slope (i.e. defines a point on the eigencurve), then it might be non-classical, in which case then the local Galois rep is Hodge--Tate (with weights 0 and $k-1$), but it is never de Rham; this is a theorem of Kisin, from his big paper "Overconvergent modular forms and the Fontaine-Mazur conjecture". Kisin's work also shows that if $f$ is an overconvergent finite-slope non-classical eigenform, then $\mathbf{D}_{\mathrm{cris}}V_p(f)$ is 1-dimensional (and this crystalline period varies analytically in families).

As for the case of infinite-slope non-classical eigenforms, or non-overconvergent eigenforms, I don't know.

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  • $\begingroup$ Kisin proved that $\mathbb{D}_{cris} \mathrm{V}_p(f)^{\phi=a_p}$ is one dimensional, where $U_p(f)=a_p.f$. Why $\mathbb{D}_{cris} \mathrm{V}_p(f)$ is one dimensional? $\endgroup$
    – Adel BETINA
    Commented Apr 26, 2016 at 9:21
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    $\begingroup$ If it were two-dimensional $\mathbf{D}_{\mathrm{cris}}(V_p(f)$ would be crystalline, hence de Rham, but Kisin proved that $V_p(f)$ is only de Rham when $f$ is classical. $\endgroup$
    – David Loeffler
    Commented Apr 26, 2016 at 14:30

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