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Dear MO,

This is a follow up to a previous question here in MO, but I will make this question self-contained for convenience. Those already familiar with the following paper [G] by Gross can safely skip down to the questions below.

In [G],

  • Benedict H. Gross, "A tameness criterion for Galois representations associated to modular forms (mod $p$)", Duke Math. Journal, Vol. 61, No. 2, 1990,

Gross shows criteria (A) and (B) below. Let $p$ be a prime, and let $f=\sum a_nq^n$ be a normalized cuspidal eigenform of weight $k$ and character $\varepsilon$ for $\Gamma_1(N)$, with coefficients in a finite field $\mathbb{F}$ of characteristic $p$. Let $$\rho_f:\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \operatorname{GL}_2(\mathbb{F})$$ be a continuous semi-simple Galois representation such that $\rho_f$ is unramified for all primes $\ell$ such that $\gcd(\ell,Np)=1$, and the matrix $\rho_f(Frob_\ell)$ has characteristic polynomial $x^2-a_\ell x+\varepsilon(\ell)\ell^{k-1}$.

Assume $2\leq k\leq p$ and $a_p\neq 0$ (if $k=p$, assume $a_p^2\neq \varepsilon(p)$. The criterion says as follows:

  • (A) The representation $\rho_f$ is completely reducible when restricted to a decomposition group at $p$ if and only if there is a companion form for $f$, i.e., there is $g=\sum b_nq^n$ of weight $k'=p+1-k$ and character $\varepsilon$ for $\Gamma_1(N)$ over $\mathbb{F}$, such that $n^kb_n=na_n$, for all $n\geq 1$.

The statements with all the details are in [G], Proposition 13.8 and Theorem 13.10.

Now jump to p. 514 of [G]. Suppose that $f$ is the reduction mod $p$ of a newform $F$ of weight $2$, where $F$ has trivial character on $\Gamma_1(N)$, and has coefficients in $\mathbb{Z}$. Then, $F$ corresponds to an elliptic curve $E/\mathbb{Q}$ of conductor $N$. Let $$\rho_{E,p} : \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \operatorname{GL}_2(\mathbb{Z}_p)$$ be the representation associated to the action of Galois on the Tate module $T_p(E)$. Let us denote $\rho_{E,p} \bmod p$ by ${\overline{\rho}}$. Then Gross deduces the following criterion, when $E$ has good ordinary reduction at $p$, with $j(E)\neq 0, 1728$, and $\overline{\rho}$ is irreducible:

  • (B) The restriction of $\rho_{E,p} \bmod p^n$ to a decomposition group at $p$ is diagonalizable if and only if $j(E)\equiv j_0 \bmod 2p^{n+1}$, where $j_0$ is the $j$-invariant of the "canonical lifting" to $\mathbb{Q}_p$ of $E \bmod p$.

Now my questions:

  • Question 1: I take it that Gross assumes that $\overline{\rho}$ be irreducible in (B) so it coincides with its semi-simplification, and therefore $\overline{\rho}$ is equivalent to $\rho_f$ and (A) applies. But, does criterion (B) still hold if $\overline{\rho}$ is reducible? I.e., what if $E$ has a $\mathbb{Q}$-rational $p$-isogeny? For an example, see my previous question here in MO, where I specify my interest in the curve $E$ with Cremona label 1225h1, which has a rational $37$-isogeny.

  • Question 2: If (B) does not hold when $\overline{\rho}$ is reducible (or if it is not known whether it holds), do any of the two directions of the if-and-only-if hold when $\overline{\rho}$ is reducible?

  • Question 3: If $E/\mathbb{Q}$ has potential good ordinary reduction, does (B) still hold?

Thank you!

share|cite|improve this question
I don't think Gross ever says anything non-trivial about reducible Galois representations in that paper. In fact it even surprises me that he proves (A) in the paper. Is the first line of the proof of (A) "if the mod $p$ rep is reducible then just let $g$ be an appropriate Eisenstein series?" [i.e. not even a cusp form]? Back in those days, the assumption that the mod $p$ representation was irreducible was commonplace. I certainly don't know the answer to your questions by the way. It would surprise me a little if Gross' methods could say anything about the reducible case. – Kevin Buzzard Mar 17 '12 at 9:03
@Kevin: Thanks. Indeed, in the first paragraph of the proof of Cor. 13.11, Gross says "If $\rho_f$ is reducible [...] one can show, using the theory of Eisenstein series, that a form $h$ satisfying (2) also exists" where (2) is the condition that describes a companion form, and then briefly describes how to this in a particular case. – Álvaro Lozano-Robledo Mar 17 '12 at 14:52

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