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Let $\bar{\rho}_p$ be a two-dimensional irreducible local Galois representation of $Gal(\bar{\mathbb{Q}}_p / \mathbb{Q}_p)$ on a $k$-vector space, where $k$ is a finite extension of $\mathbb{F}_p$. Let $\bar{\rho}$ be a modular irreducible representation of $Gal(\bar{\mathbb{Q}} / \mathbb{Q})$ unramified outside $\Sigma$ such that restriction of $\bar{\rho}$ to $Gal(\bar{\mathbb{Q}}_p / \mathbb{Q}_p)$ is isomorphic to $\bar{\rho} _p$. Let $\mathfrak{m}$ be the maximal ideal in the Hecke algebra $\mathbb{T} _{ \Sigma}$ corresponding to $\bar{\rho}$ (where by $\mathbb{T} _{ \Sigma}$ I mean a completed Hecke algebra unramified outside $\Sigma$, i.e. a projective limit of Hecke algebras of all finite levels which are unramified outside $\Sigma$).

Question: Let $R(\bar{\rho}_p)^{det}$ be a local deformation ring which classifies deformations $\bar{\rho _p}$ with a given fixed central character (so this ring is of the form $\mathcal{O}[[x_1,x_2,x_3]]$). Does there exist a surjection:

$$R(\bar{\rho}_p)^{det} \twoheadrightarrow \mathbb{T} _{\mathfrak{m}, \Sigma}$$

where $\mathbb{T} _{\mathfrak{m}}$ denotes localisation at $\mathfrak{m}$? Observe that the morphism itself is provided for example by a result of Carayol on Galois representations over local rings.

Remark: The answer is yes if instead of $R(\bar{\rho}_p)^{det}$ we would take a global deformation ring (i.e. ring which classifies deformations of $\bar{\rho}$).

If this is not a surjection, can we say anything meaningful about the morphism above?

It might not be a surjection for every $\mathfrak{m}$ and $\Sigma$ but can we at least always find one pair ($\mathfrak{m}, \Sigma$) for a fixed $\bar{\rho}_p$ such that it is a surjection?

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Dear Przemyslaw, This map needn't be surjective in general (e.g. you could have two non-isomorphic elliptic curves with isomorphic $p$-torsion unramified outside $\Sigma$; then they would give two points in the Spec of the global Hecke algebra which have the same local behaviour at $p$ (say if $p$ is large enough, so that knowing $a_p$ modulo $p$ actually determines $a_p$)). However (under some mild assumptions, at least) it is finite. This will appear in a forthcoming paper of mine with Paskunas. Regards, Matthew – Emerton Jun 23 '12 at 5:21
Dear Matthew, thank you for your answer! Could you possibly send me a preprint of your paper with Paskunas? Best wishes, Przemyslaw – Przemyslaw Chojecki Jun 23 '12 at 9:43
The audio of Matt's talk in Toronto on this topic is available at – vytas Jun 23 '12 at 10:36
Dear Przemyslaw, Once the preprint exists in a form suitable for public consumption, I'll let you know! Best wishes, Matthew – Emerton Jun 25 '12 at 2:55
I thank you both, Matthew and Vytas! – Przemyslaw Chojecki Jun 25 '12 at 12:54

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