Skip to main content
+ top level tag (nt.)
Source Link
Myshkin
  • 17.6k
  • 5
  • 71
  • 137

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.

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.

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.

added 491 characters in body
Source Link
Adel BETINA
  • 1.1k
  • 8
  • 12

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.

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 $\rho^{I_q}=0$ ?

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.

Source Link
Adel BETINA
  • 1.1k
  • 8
  • 12

Galois representation and weight one Hilbert modular form

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 $\rho^{I_q}=0$ ?