Mixing
Why does the Frobenius-semisimplicity of a Weil representation not depend on the choice of the Frobenius...
0
$begingroup$
Definition: Let $K$ be a (non-Archimedean) local field and $k$ its residue field.
- A Frobenius element of the absolute Galois group $G_K$ is any element of $G_K$ which is a lift of the Frobenius automorphism $x mapsto x^{|k|}$ in $G_k simeq G_K/I_K$.
- A Weil representation $rho$ is called Frobenius-semisimple if $rho(Phi)$ is diagonalizable for some Frobenius element $Phi$ in $G_K$.
Question: If $rho$ is a Frobenius-semisimple Weil representation why is $rho(Phi')$ diagonalizable for any Frobenius element $Phi'$ in $G_K$?
Attempts:
- I know that the Frobenius element in $G_K$ is unique up to an element in $I_K$, i.e. if $Phi, Phi'$ are two Frobenius elements in $G_K$, there exists an element $sigma in I_K$ with $Phi' = Phi circ sigma$.
- Let us say that $rho(Phi')$ is diagonalizable and $(v_1,dots,v_n)$ is a basis of the representation space consisting of eigenvectors of $rho(Phi')$. The only thing I was able to show with that $(v_1,dots,v_n)$ is also a basis of eigenvectors of $rho(sigma^{-1}circ Phi circ sigma)$. But this is not helpful since I want to find a basis of eigenvectors for $Phi$.
Could you please help me with this problem? Thank you!
abstract-algebra number-theory representation-theory algebraic-number-theory galois-representations
asked Jan 10 at 10:47
DiglettDiglett
9571521
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0
$begingroup$
Definition: Let $K$ be a (non-Archimedean) local field and $k$ its residue field.
- A Frobenius element of the absolute Galois group $G_K$ is any element of $G_K$ which is a lift of the Frobenius automorphism $x mapsto x^{|k|}$ in $G_k simeq G_K/I_K$.
- A Weil representation $rho$ is called Frobenius-semisimple if $rho(Phi)$ is diagonalizable for some Frobenius element $Phi$ in $G_K$.
Question: If $rho$ is a Frobenius-semisimple Weil representation why is $rho(Phi')$ diagonalizable for any Frobenius element $Phi'$ in $G_K$?
Attempts:
- I know that the Frobenius element in $G_K$ is unique up to an element in $I_K$, i.e. if $Phi, Phi'$ are two Frobenius elements in $G_K$, there exists an element $sigma in I_K$ with $Phi' = Phi circ sigma$.
- Let us say that $rho(Phi')$ is diagonalizable and $(v_1,dots,v_n)$ is a basis of the representation space consisting of eigenvectors of $rho(Phi')$. The only thing I was able to show with that $(v_1,dots,v_n)$ is also a basis of eigenvectors of $rho(sigma^{-1}circ Phi circ sigma)$. But this is not helpful since I want to find a basis of eigenvectors for $Phi$.
Could you please help me with this problem? Thank you!
abstract-algebra number-theory representation-theory algebraic-number-theory galois-representations
asked Jan 10 at 10:47
DiglettDiglett
9571521
$endgroup$
$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14
add a comment |
0
0
0
$begingroup$
Definition: Let $K$ be a (non-Archimedean) local field and $k$ its residue field.
- A Frobenius element of the absolute Galois group $G_K$ is any element of $G_K$ which is a lift of the Frobenius automorphism $x mapsto x^{|k|}$ in $G_k simeq G_K/I_K$.
- A Weil representation $rho$ is called Frobenius-semisimple if $rho(Phi)$ is diagonalizable for some Frobenius element $Phi$ in $G_K$.
Question: If $rho$ is a Frobenius-semisimple Weil representation why is $rho(Phi')$ diagonalizable for any Frobenius element $Phi'$ in $G_K$?
Attempts:
- I know that the Frobenius element in $G_K$ is unique up to an element in $I_K$, i.e. if $Phi, Phi'$ are two Frobenius elements in $G_K$, there exists an element $sigma in I_K$ with $Phi' = Phi circ sigma$.
- Let us say that $rho(Phi')$ is diagonalizable and $(v_1,dots,v_n)$ is a basis of the representation space consisting of eigenvectors of $rho(Phi')$. The only thing I was able to show with that $(v_1,dots,v_n)$ is also a basis of eigenvectors of $rho(sigma^{-1}circ Phi circ sigma)$. But this is not helpful since I want to find a basis of eigenvectors for $Phi$.
Could you please help me with this problem? Thank you!
abstract-algebra number-theory representation-theory algebraic-number-theory galois-representations
asked Jan 10 at 10:47
DiglettDiglett
9571521
$endgroup$
Definition: Let $K$ be a (non-Archimedean) local field and $k$ its residue field.
- A Frobenius element of the absolute Galois group $G_K$ is any element of $G_K$ which is a lift of the Frobenius automorphism $x mapsto x^{|k|}$ in $G_k simeq G_K/I_K$.
- A Weil representation $rho$ is called Frobenius-semisimple if $rho(Phi)$ is diagonalizable for some Frobenius element $Phi$ in $G_K$.
Question: If $rho$ is a Frobenius-semisimple Weil representation why is $rho(Phi')$ diagonalizable for any Frobenius element $Phi'$ in $G_K$?
Attempts:
- I know that the Frobenius element in $G_K$ is unique up to an element in $I_K$, i.e. if $Phi, Phi'$ are two Frobenius elements in $G_K$, there exists an element $sigma in I_K$ with $Phi' = Phi circ sigma$.
- Let us say that $rho(Phi')$ is diagonalizable and $(v_1,dots,v_n)$ is a basis of the representation space consisting of eigenvectors of $rho(Phi')$. The only thing I was able to show with that $(v_1,dots,v_n)$ is also a basis of eigenvectors of $rho(sigma^{-1}circ Phi circ sigma)$. But this is not helpful since I want to find a basis of eigenvectors for $Phi$.
Could you please help me with this problem? Thank you!
abstract-algebra number-theory representation-theory algebraic-number-theory galois-representations
abstract-algebra number-theory representation-theory algebraic-number-theory galois-representations
asked Jan 10 at 10:47
DiglettDiglett
9571521
asked Jan 10 at 10:47
DiglettDiglett
9571521
asked Jan 10 at 10:47
DiglettDiglett
9571521
asked Jan 10 at 10:47
DiglettDiglett
9571521
asked Jan 10 at 10:47
DiglettDiglett
9571521
9571521
$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14
add a comment |
$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14
$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14
add a comment |
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$begingroup$
You probably want to use that fact that $rho$ is continuous, so that $rho(I_K)$ is finite. It turns out for a Weil representation, Frobenius-semisimple is the same as just being semisimple. The notions become different when you consider Weil-Deligne representations.
$endgroup$
– Mathmo123
Jan 10 at 13:01
$begingroup$
@Mathmo123: Thanks for your remark! It is$rho(I_K)$ is finite, so let $rho(sigma_1), dots rho(sigma_m)$ for $sigma_i in I_K$ be these elements. Then the image of a Frobenius element must be one of the values $rho(Phi circ sigma_i)$ for $i=1,dots,m$. Could this be helpful? I have problems to see how this can lead to the semisimplicity.
$endgroup$
– Diglett
Jan 10 at 14:14