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I think this may be a silly question, but here goes. Let $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\overline{\mathbf{F}_p})$ be a representation; say $\rho$ is of S-type if it is continuous, unramified almost everywhere, and the determinant of complex conjugation is $-1$. Serre's conjecture, now a theorem of Khare-Wintenberger, states that every $\rho$ of S-type arises from some modular form $f=\sum a_n e(nz)$ in the sense that $\mathrm{tr}\rho(\mathrm{Frob}_l)=a_l\;( \mathrm{mod}\;p)$ for (almost all) primes $l$.

Question: Are S-type representations $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\mathbf{Z}/p^n\mathbf{Z})$ for $n\geq 2$ also expected/known to be modular?

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2 Answers 2

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In what sense? If you mean "come from the reduction of $\rho_f$ for some Hecke eigenform $f$'', no, they are not.

If you mean "come from the reduction of $\rho$ where $\rho:G_{\mathbb Q} \to GL_2(\mathbb T)$ is the Galois rep'n attached to the Hecke algebra $\mathbb T$ acting on modular forms of some sufficiently large level, then the answer is known to be yes in most cases (i.e. with comparitively minor technical restrictions on $\rho$). This is the content of so-called big $R = $ big $\mathbb T$ theorems, due to Gouvea--Mazur, Boeckle, and others (combined with Serre's conjecture to know that $\overline{\rho}$ is modular).

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  • $\begingroup$ By "modular" I just meant whether the traces on Frobenii agree mod $p^n$ with modular forms coefficients. $\endgroup$ Nov 30, 2010 at 14:26
  • $\begingroup$ Then, as I wrote, the answer to your question is "no, this is not expected". $\endgroup$
    – Emerton
    Nov 30, 2010 at 16:44
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    $\begingroup$ @David: Emerton's answer is correct but your clarification, where you define "modular", is unclear. The traces of Frobenii will agree mod $p^n$ with coefficients of modular forms, but not with coefficients of eigenforms. The point is that the forms you have to use will not be eigenforms, but they will be eigenforms mod $p^n$ (i.e. $Tf=\lambda f+p^ng$ with $g$ having integral coefficients). $\endgroup$ Nov 30, 2010 at 18:08
  • $\begingroup$ Dear Kevin, Thanks for that. I had implicitly understood "modular forms" in David's initial comment to mean "eigenforms", hence the "no" in my reply. $\endgroup$
    – Emerton
    Dec 1, 2010 at 3:47
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To elaborate on Emerton's answer, an arbitrary (finitely ramified and odd) representation $\rho_n:G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{Z}/p^n\mathbb{Z})$ can't lift to one coming from an eigenform because of some reasons which may be explained locally. Let $\bar{\rho}$ denote the mod $p$ representation and $Ad^0\bar{\rho}$ the $\mathbb{F}_p[G_{\mathbb{Q}}]$-module of trace zero $2\times 2$ matrices over $\mathbb{F}_p$ equipped with the adjoint action, i.e.$g\in G_{\mathbb{Q}}$ acts on a matrix $A$ by conjugation $g\cdot A:=\bar{\rho}(g) A\bar{\rho}(g)^{-1}$. Let $l$ be a prime at which $H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})\neq 0$ (here $G_{\mathbb{Q}_l}$ is a decomposition subgroup at $l$). The obstruction-class associated with ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$ is a cohomology class $O_l({\rho_n}_{\restriction G_{\mathbb{Q}_l}})\in H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})$ which is non-zero if there is no lift of the local representation ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$ to $GL_2(\mathbb{Z}/p^{n+1}\mathbb{Z})$. This local obstructedness phenomenon is an issue when $n\geq 2$ though not so for $n=1$. This can be further explained as follows, at each prime $l$ there is a smooth subscheme in the scheme parametrizing local deformations of $\bar{\rho}_{\restriction G_{\mathbb{Q}_l}}$, if for $n\geq 2$ at some prime $l$ where $H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})\neq 0$ it is possible that ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$ does not lie on this smooth subscheme in the space of all deformations of ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$. In this case, ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$ cannot lift one more step (and thus the global representation $\rho_n$ will not lift one more step either). The construction of these smooth schemes in the spaces of local deformations (dubbed smooth/liftable local deformation conditions) is carried out in the paper of Ramakrishna: "Deforming Galois representations and the Conjectures of Serre and Fontaine-Mazur" and at the prime $p$ this was previously carried out in his paper "On a Variation of Mazur's Deformation Functor".

There is now another issue, one does require that the lift to characteristic zero must satisfy a $p$ adic Hodge theoretic condition at $p$, this can be done when $n=1$ but there can be issues when $n\geq 2$.

It would be of interest to know if in case the local representations ${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$ are all unobstructed (i.e. their obstruction classes are trivial, i.e. they lift to $GL_2(\mathbb{Z}/p^{n+1}\mathbb{Z})$)) at the finitely many primes at which it is unramified, and if $\rho_{\restriction G_{\mathbb{Q}_p}}$ satisfies a further condition, then if indeed it lifts to the representation associated to a cuspidal eigenform/ big Hecke algebra?

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  • $\begingroup$ Why should $\rho : G_{\mathbf{Q}} \rightarrow \text{GL}_2(\mathbf{Z}/p^n\mathbf{Z})$ coming from an eigenform necessarily lifts to $\text{GL}_2(\mathbf{Z}/p^{n+1}\mathbf{Z})$ (even locally)? $\endgroup$ Nov 8, 2018 at 15:21
  • $\begingroup$ Let $\rho_f:G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{Z}_p)$ be the representation attached to an eigenform such that $\rho_f$ mod $p^n$ is $\rho$. Then $\rho$ must lift to $\rho_f$ mod $p^{n+1}$, perhaps I don't understand what you have in mind exactly. $\endgroup$
    – user130124
    Nov 8, 2018 at 15:26
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    $\begingroup$ The point is: the coefficient ring of $f$ is not necessarily $\mathbf{Z}_p$. So it could be some finite extension $\mathcal{O}$ of $\mathbf{Z}_p$ admitting a surjection to $\mathbf{Z}/p^n\mathbf{Z}$ but not to $\mathbf{Z}/p^{n+1}\mathbf{Z}$. $\endgroup$ Nov 8, 2018 at 15:37
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    $\begingroup$ I can't think of an example where $\mathcal{O}/\pi^n\simeq Z/p^n$ for $n\geq 2$ and $\mathcal{O}/\pi^{n+1}$ not isomorphic to $Z/p^{n+1}$. Can you construct examples for large values of $n$? $\endgroup$
    – user130124
    Nov 8, 2018 at 16:09
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    $\begingroup$ Sorry, I was thinking about the case $n=1$. But indeed for $n\geq 2$, $\mathcal{O}/\pi^n\mathcal{O} \simeq \mathbf{Z}/p^n\mathbf{Z}$ implies $\mathcal{O} \simeq \mathbf{Z}_p$. I think that this fact is not true anymore if $\mathcal{O}$ is a more complicated ring (not a DVR, like the Hecke algebra $\mathbb{T}$ in many cases). So if $\rho$ is simply assumed to come from the Hecke algebra (and not from a modular form), then the lifiting does necessarily exist. $\endgroup$ Nov 8, 2018 at 23:52

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