6
$\begingroup$

Let $f$ be a primitive weight one Hilbert modular form for the totally real field $F$ (the weight of $f$ is parallel). Assume that $f$ is $N$-new, $p$-stable and cuspidal. Let $\rho:G_F \rightarrow \mathrm{GL}_2(\mathbb{C})$ be the galois representation attached to $f$.

I have the following question : Let $q$ be a prime ideal of the ring of integers of $F$ such that $q$ is prime to $p$ and $ q \mid N$:

  • If $U_q (f)=a_q.f$ with $a_q \ne 0$. Is it equivalent to the fact that there exits a line of $\mathbb{C}^2$ fixed by the inertia group $I_q$ and on which $\mathrm{Frob}_q$ acts via $a_q$ ?

  • If $U_q(f)=0$. Is it equivalent to the fact that $\rho^{I_q}=0$ ?

Since $f$ is $p$-stable and of weight one, $f$ is $p$-ordinary and we can put $f$ is a Hida family $F$ of level $N$, and we know that $U_q^s (U_q^2 - Tr \rho_F(\sigma) U_q + \det \rho_F(\sigma))=0$, where $\sigma$ is a lift of the Frobenius at $q$, $s$ is an integer such that $q^s \mid N $ and $q^{s+1} \nmid N$ and $\rho_F$ is the galois representation attached to the hida family $F$, and by density we can proof what I want. But is there a simple argument without using Hida family.

$\endgroup$

1 Answer 1

2
$\begingroup$

These are all simple instances of a much more powerful statement, which is local-global compatibility in the Langlands program: the Weil-Deligne representation obtained from $\rho |_{D_q}$ should correspond under the Langlands program to $\pi_{f, q}$. When the weights are all $\ge 2$, this goes back to Carayol. For parallel weight 1 forms I believe this is now entirely known, due to Jarvis and Newton. (There are a few cases that are not yet fully resolved for partial weight 1 forms, but these involve Steinberg primes and those cannot occur for parallel weight 1.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.