Let E be an elliptic curve over the rationals with conductor $Mp^2$ with p>5 and M and p coprime, and let $\rho$ be the Galois representation attached to the p-torsion points of E. Is there a way to consider deformations of $\rho$ to representations that still have "additive reduction at p" with Serre weight 2? Actually in that direction, what is the local condition on p that one has to put down? Assuming that's the case, and assuming this is representable by ring R, is there an R=T result in that direction?


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If you fix a prime $\ell$, and consider the Galois action of the decomposition group $D_p$ on the (rational) $\ell$-adic Tate module (here "rational" means tensored with $\mathbb Q_{\ell}$), then (in a standard way, due to Deligne) you can convert this action into a representation of the Weil--Deligne group, and so in particular of the Weil group. Restricting to the inertia group, you get a representation of the inertia group $I_p$, known as the inertial type $\tau$. It is independent of $\ell$. (The only reason to detour through the Weil--Deligne group is to deal with possibly infinite image of tame inertia; if the elliptic curve has potentially good reduction, then we can skip this step and just take the representation of $I_p$ on the $\ell$-adic Tate module, which has finite image and is independent of $\ell$.)

[Added: In the above, one should insist that $\ell \neq p$. If $\ell = p$, then one can also arrive at a Weil-Deligne representation, and hence inertial type, which is the same as the one obtained as above for $\ell \neq p$, but to do this one must use Fontaine's theory: one forms the $D_{pst}$ of the rational $p$-adic Tate module, which then can be converted into a Weil--Deligne representation in a standard way, and hence gives an inertial type.]

Now one can look at the deformation ring $R_{\rho}^{[0,1],\tau}$ parameterizing lifts of $\rho$ of which at $p$ are of inertial type $\tau$ and Hodge--Tate weights $0$ and $1$. (See Kisin's recent JAMS paper about potentially semi-stable deformation rings.)

[Added: Here $\ell = p$, i.e. we are looking at $p$-adic deformations of $\rho$ which are potentially semi-stable at $p$, and whose inertial type, computed via $D_{pst}$ as in the above added remark, is equal to $\tau$. But note that, by the preceding discussion, these deformations do precisely capture the idea of lifts of $\rho$ having the same "reduction type" as the original elliptic curve $E$.]

Let's suppose that $E$ really does have potentially good reduction. Then Kisin's "moduli of finite flat group schemes" paper shows that any lift parameterized by $R^{[0,1],\tau}$ is modular. This shows that $R^{[0,1],\tau} = {\mathbb T},$ for an appropriately chosen ${\mathbb T}$.

One thing to note: unlike in the original Taylor--Wiles setting, from this statement one doesn't get quantitive information about adjoint Selmer groups, and one doesn't get any simple interpretation of what this $R = {\mathbb T}$ theorem means on the integral level. (In other words, Artinian-valued points of $R^{[0,1],\tau}$ have no simple interpretation in terms of a ramification condition at $p$; this is related to the fact that the theory of $D_{pst}$ only applies rationally, i.e. to ${\mathbb Q}_p$-representations, not integrally, i.e. not to representations over $\mathbb Z_p$ or over Artin rings.)

  • $\begingroup$ @Emerton: do you really get an R=T theorem and not an "R=T up to torsion" theorem? $\endgroup$ Commented Apr 10, 2010 at 16:13
  • $\begingroup$ Both $R^{[0,1],\tau}$ and $\mathbb T$ are defined to be torsion-free. (But as I commented, the down-side is that one doesn't have such an explicit interpretation of the Artinian-valued points of the deformation ring.) $\endgroup$
    – Emerton
    Commented Apr 10, 2010 at 18:48
  • $\begingroup$ Aah thanks Emerton, that's the point. $\endgroup$ Commented Apr 10, 2010 at 18:56
  • $\begingroup$ @Emerton: Can $\ell=p$ in above? That is, is the image of $I_p$ on the automorphism group of $T_pE$ the same as $T_\ell E$? $\endgroup$
    – Soroosh
    Commented Apr 11, 2010 at 14:55
  • $\begingroup$ Naively, no, but non-naively, yes! I've edited to explain what I mean. $\endgroup$
    – Emerton
    Commented Apr 12, 2010 at 3:07

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