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Let $\bar\rho$ be an odd, absolutely irreducible, 2-dimensional mod $p$ representation of $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$ (with coefficients in some finite extension $k / \mathbb{F}_p$). Then one can form the universal deformation ring $R_\bar\rho$ of $\bar \rho$, parametrizing equivalence classes of lifts of $\bar \rho$ to characteristic 0 with some fixed Artin conductor away from $p$ (but no local hypotheses at $p$). Under some mild hypotheses this ring is known to be isomorphic to a power series ring in three variables over $W(k)$.

There are some results ("big R = big T" theorems), due to Boeckle, Emerton and others, which show that (under suitable hypotheses on $\bar\rho$) this ring $R_{\bar \rho}$ is canonically isomorphic to a localization of a Hecke algebra, acting on a suitable inverse limit of the cohomology of modular curves of $p$-power levels. See e.g. section 3.3.1 of Emerton's ICM survey.

In this setting, can one also exhibit the free rank 2 $R_{\bar\rho}$-module $M$ that realizes the universal deformation $$\rho^{\mathrm{univ}} : \operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to \operatorname{GL}(M)$$ as a quotient of some module built from cohomology of modular curves?

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This is not always possible.

Suppose we could always find such an $M$. Take $\bar{\rho}$ reducible at $p$ with scalar image of the Frobenius. Then $M\otimes_{R_{\bar{\rho}}}R^{\operatorname{ord}}_{\bar{\rho}}$ is the universal deformation with the ordinary condition at $p$ and also occurs as a quotient in the cohomology of a (tower of) modular curves (because the ordinary Hecke algebra is just the image of the Hecke algebra under the ordinary projection) and furthermore one could descend to a fixed weight and level in the same way. So there exists a lattice in the cohomology of modular curves with comparable properties for forms of given weight and level. But this is known not to be true, as explained in the answer to this old question of mine, which is also the one considered by MO to be the most related to yours (an indication that MO does a good job).

Above is a specific counterexample which you might want to discard (for instance by taking $p\neq 2$) but proving a positive result in the direction you wish (assuming suitable supplementary hypotheses) seems to me to be very hard.

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  • $\begingroup$ OK, so the non-$p$-distinguished case is bad; but I'm happy to rule this out -- I'd be content if there were any non-empty set of $\bar\rho$ for which such a construction were possible. $\endgroup$
    – David Loeffler
    Commented May 13, 2014 at 14:55
  • $\begingroup$ I thought you might say this. This is why I added the final paragraph. You are in much better shape if you are fine with inverting $p$ and with free sheaves on the corresponding rigid analytic space. $\endgroup$
    – Olivier
    Commented May 13, 2014 at 15:25
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    $\begingroup$ If your "why" was more of a philosophical nature, to me it all boils down to the fact that the Hecke algebra is "very nice" after inverting $p$ (étale over $\mathbb Q_{p}$, formally smooth in a classical neighborhood...). $\endgroup$
    – Olivier
    Commented May 14, 2014 at 9:23
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    $\begingroup$ No, I am trying to do mathematics here, not philosophy. I am fully aware of the existence of sheaves of Gal.reps. on the eigencurve coming from completed cohomology, but that is the answer to a different question: this is not the right rigid space -- it is 2-dimensional (if you include twists by characters) and parametrizes deformations of $\bar\rho$ with a choice of $p$-refinement, but I want the 3-dimensional space parametrizing arbitrary deformations which are not necessarily finite-slope at $p$. $\endgroup$
    – David Loeffler
    Commented May 14, 2014 at 9:27
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    $\begingroup$ Look, all I'm pointing out is that a positive answer to your question implies a positive answer for the "different" question, which is already hard (read: false in general) but at least admits a positive answer if you invert $p$. That's why I said you have better chances in that case. But these are just comments: my official answer remains that I think your problem is a hard one. $\endgroup$
    – Olivier
    Commented May 14, 2014 at 14:20

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