Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $f$ is a weight $k$ newform for $\Gamma_1(N)$ with attached $p$-adic Galois representation $\rho_f$. Denote by $\rho_{f,p}$ the restriction of $\rho_f$ to a decomposition group at $p$. When is $\rho_{f,p}$ semistable (as a representation of $\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$?

To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$.

Certainly if $N$ is prime to $p$ then $\rho_{f,p}$ is in fact crystalline, while if $p$ divides $N$ exactly once then $\rho_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X_1(N)$ over $\mathbf{Q}$ at $p$). For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable. Can we say more? For example, are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific extension? Can we bound the degree and ramification of the minimal extension over which $\rho_{f,p}$ becomes semistable in terms of numerical data attached to $f$? Can it happen that $N$ is highly divisible by $p$ and yet $\rho_{f,p}$ is semistable over $\mathbf{Q}_p$?

I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use...

As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho_{f,p}$ becomes semistable over an abelian extension of $\mathbf{Q}p$ and even becomes crystalline over such an extension provided that the Hecke eigenvalues of $f$ for the action of $\mu_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators are not all 1.

share|improve this question
    
can you give some references of the known results (for a learner) such as the pst of the attached Galois reps? Thank you! –  natura Feb 10 '10 at 18:17
add comment

2 Answers

up vote 11 down vote accepted

The right way to do this sort of question is to apply Saito's local-global theorem, which says that the (semisimplification of the) Weil-Deligne representation built from $D_{pst}(\rho_{f,p})$ by forgetting the filtration is precisely the one attached to $\pi_p$, the representation of $GL_2(\mathbf{Q}_p)$ attached to the form via local Langlands. Your suggestions about the $p$-adic valuation of $N$ and so on are rather "coarse" invariants---$\pi_p$ tells you everything and is the invariant you really need to study.

So now you can just list everything that's going on. If $\pi_p$ is principal series, then $\rho$ will become crystalline after an abelian extension---the one killing the ramification of the characters involved in the principal series. If $\pi_p$ is a twist of Steinberg by a character, $\rho_{f,p}$ will become semistable non-crystalline after you've made an abelian extension making the character unramified. And if $\pi_p$ is supercuspidal, $\rho_{f,p}$ will become crystalline after a finite non-trivial extension that could be either abelian or non-abelian, and figuring out which is a question about $\pi_p$ (it will be a base change from a quadratic extension if $p>2$ and you have to bash out the possibilities).

Seems to me then that semistable $\rho$s will show up precisely when $\pi_p$ is either unramified principal series or Steinberg, so the answer to your question is (if I've got everything right) that $\rho_{f,p}$ will be semistable iff either $N$ (the level of the newform) is prime to $p$, or $p$ divides $N$ exactly once and the component at $p$ of the character of $f$ is trivial. Any other observations you need should also be readable from this sort of data in the same way.

One consequence of this I guess is that $\rho_{f,p}$ is semi-stable iff the $\ell$-adic representation attached to $f$ is semistable at $p$.

share|improve this answer
add comment

Since f is potentially semi-stable, you can look at its attached filtered (φ, N, Gal(L/Qp))-representation (where ρf,p becomes semi-stable when restricted to GL). If its N is zero, then it is potentially cristalline, otherwise it is not.

As for the ordinary case, I'm not sure what definition you're using. Under Greenberg's definition, an ordinary p-adic Galois representation is semi-stable (see Perrin-Riou's article in the Bures). Also, the Tate curve is ordinary at p, but not potentially cristalline (once something is semi-stable and non-cristalline, it can't be potentially cristalline).

share|improve this answer
    
Thanks for your thoughts, Rob. I was thinking of ordinary as $p$ not dividing $a_p$ and was concentrating on weight 2. My statement about crystalline/ semistable over an abelian extension comes from analyzing the ordinary parts of the p-divisible groups of modular abelian varieties, as in the work of Mazur-Wiles and Tilouine. I agree that the filtered module contains all the information I'm asking about, but this seems like more of a rephrasing of the question, as knowledge of the filtered module requires understanding $L$ first... –  B. Cais Feb 7 '10 at 18:11
    
If $p$ doesn't divided $a_p$, then $\rho=\rho_{f,p}$ is ordinary (in the sense of Greenberg) by a theorem of Wiles and hence is semistable. If, in addition, $p$ divides $N$ and $f$ is new, then $\rho$ is not cristalline and hence can't become cristalline over any extension since it is already semistable. Am I missing something? –  Rob Harron Feb 7 '10 at 19:42
    
Rob: What Wiles shows is that $p$ not dividing $a_p$ implies potentially ordinary, as he only works with his Galois representations up to $\overline{\mathbf{Q}}_p$-equivalence. The point is that the ordinary filtration on the Galois side may not be defined over $\mathbf{Q}_p$. Any weight 2 neworm of level $p^r$ primitive nebentypus has associated Galois representation that is potentially crystalline but non-crystalline over $\mathbf{Q}_p$ if $r>0$. Wiles + Perrin-Riou as you indicate only gives $p$ not dividing $a_p$ implies potentially semistable, which one has anyway... –  B. Cais Feb 8 '10 at 0:49
    
I see, your nebentype is ramified, so Wiles' result only gives nearly ordinary (I guess this is where your condition on the action of the Diamond operators swoops in basically killing the Tate curve case; this is what I was missing). Sorry bout that. –  Rob Harron Feb 8 '10 at 5:31
    
On the automorphic side, what's happening here is that one can check via a representation-theoretic calculation that if f has level p^r and character of conductor p^r too (r>=1) then pi_p must be principal series associated to one ramified (conductor p^r) and one unramified character. Note also that whether or not p divides a_p has consequences on the shape of the Galois representation, but doesn't seem to me to have any consequences for D_{pst}(rho_{f,p}) (in some vague sense). –  Kevin Buzzard Feb 8 '10 at 7:53
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.