Skip to main content
added 35 characters in body
Source Link
Henri Cohen
  • 13.1k
  • 1
  • 34
  • 62

Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?

More generally, in even weight and nontrivial quadratic character the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?

Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?

More generally, in even weight the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?

Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?

More generally, in even weight and nontrivial quadratic character the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?

Source Link
Henri Cohen
  • 13.1k
  • 1
  • 34
  • 62

Rational eigenforms

Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?

More generally, in even weight the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?