Relation between Fourier coefficients and Satake parameters
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Let $L(s)$ be an automorphic L-function (attached to a self contragredient automorphic representation on $GL(3)$), according to the following notations for $s$ of sufficiently large real part:
$$L(s) = sum_k=0^infty fraca_nn^s = prod_p left( 1 - alpha(p)p^-s right)^-1 left( 1 - beta(p)p^-s right)^-1 left( 1 - gamma(p)p^-s right)^-1$$
Straightforwardly developing the Euler product provides expressions of the Fourier coefficients $a_n$'s in terms of the Satake parameters $alpha(p)$, $beta(p)$ and $gamma(p)$. I am not particularly aware of others standard useful relations between them. I bumped into the following one:
$$a_p^k = frac
left
left$$
I guess this can be verified, but even the case $k=1$ seems obscure to me. I do not want to believe such a formula to be a (verifiable) accident. Despite it works computationally, am I missing something lying behind? How strongly is the self-contragredience assumption necessary?
Any insight is welcome, as well as other ways to embrace the relations between spectral parameters and coefficients.
nt.number-theory automorphic-forms l-functions
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up vote
4
down vote
favorite
Let $L(s)$ be an automorphic L-function (attached to a self contragredient automorphic representation on $GL(3)$), according to the following notations for $s$ of sufficiently large real part:
$$L(s) = sum_k=0^infty fraca_nn^s = prod_p left( 1 - alpha(p)p^-s right)^-1 left( 1 - beta(p)p^-s right)^-1 left( 1 - gamma(p)p^-s right)^-1$$
Straightforwardly developing the Euler product provides expressions of the Fourier coefficients $a_n$'s in terms of the Satake parameters $alpha(p)$, $beta(p)$ and $gamma(p)$. I am not particularly aware of others standard useful relations between them. I bumped into the following one:
$$a_p^k = frac
left
left$$
I guess this can be verified, but even the case $k=1$ seems obscure to me. I do not want to believe such a formula to be a (verifiable) accident. Despite it works computationally, am I missing something lying behind? How strongly is the self-contragredience assumption necessary?
Any insight is welcome, as well as other ways to embrace the relations between spectral parameters and coefficients.
nt.number-theory automorphic-forms l-functions
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $L(s)$ be an automorphic L-function (attached to a self contragredient automorphic representation on $GL(3)$), according to the following notations for $s$ of sufficiently large real part:
$$L(s) = sum_k=0^infty fraca_nn^s = prod_p left( 1 - alpha(p)p^-s right)^-1 left( 1 - beta(p)p^-s right)^-1 left( 1 - gamma(p)p^-s right)^-1$$
Straightforwardly developing the Euler product provides expressions of the Fourier coefficients $a_n$'s in terms of the Satake parameters $alpha(p)$, $beta(p)$ and $gamma(p)$. I am not particularly aware of others standard useful relations between them. I bumped into the following one:
$$a_p^k = frac
left
left$$
I guess this can be verified, but even the case $k=1$ seems obscure to me. I do not want to believe such a formula to be a (verifiable) accident. Despite it works computationally, am I missing something lying behind? How strongly is the self-contragredience assumption necessary?
Any insight is welcome, as well as other ways to embrace the relations between spectral parameters and coefficients.
nt.number-theory automorphic-forms l-functions
Let $L(s)$ be an automorphic L-function (attached to a self contragredient automorphic representation on $GL(3)$), according to the following notations for $s$ of sufficiently large real part:
$$L(s) = sum_k=0^infty fraca_nn^s = prod_p left( 1 - alpha(p)p^-s right)^-1 left( 1 - beta(p)p^-s right)^-1 left( 1 - gamma(p)p^-s right)^-1$$
Straightforwardly developing the Euler product provides expressions of the Fourier coefficients $a_n$'s in terms of the Satake parameters $alpha(p)$, $beta(p)$ and $gamma(p)$. I am not particularly aware of others standard useful relations between them. I bumped into the following one:
$$a_p^k = frac
left
left$$
I guess this can be verified, but even the case $k=1$ seems obscure to me. I do not want to believe such a formula to be a (verifiable) accident. Despite it works computationally, am I missing something lying behind? How strongly is the self-contragredience assumption necessary?
Any insight is welcome, as well as other ways to embrace the relations between spectral parameters and coefficients.
nt.number-theory automorphic-forms l-functions
nt.number-theory automorphic-forms l-functions
edited 2 hours ago
asked 2 hours ago
Desiderius Severus
2,15121847
2,15121847
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1 Answer
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There is no coincidence, this is the Weyl character formula for the representation $operatornameSym^k$ of $GL_3$. The reason that the Langlands dual group comes up is, unsurprisingly, the Satake isomorphism.
The general statement is: For an automorphic representation associated to a group $G$ with dual group $hatG$, the coefficient of $p^k$ in the $L$-function associated to a representation $rho$ of $hatG$ is equal to the trace of the Satake parameter (a conjugacy class on $hatG$) acting on $Sym^k rho$.
No assumption beyond unramifiedness should be necessary.
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
There is no coincidence, this is the Weyl character formula for the representation $operatornameSym^k$ of $GL_3$. The reason that the Langlands dual group comes up is, unsurprisingly, the Satake isomorphism.
The general statement is: For an automorphic representation associated to a group $G$ with dual group $hatG$, the coefficient of $p^k$ in the $L$-function associated to a representation $rho$ of $hatG$ is equal to the trace of the Satake parameter (a conjugacy class on $hatG$) acting on $Sym^k rho$.
No assumption beyond unramifiedness should be necessary.
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
add a comment |Â
up vote
2
down vote
There is no coincidence, this is the Weyl character formula for the representation $operatornameSym^k$ of $GL_3$. The reason that the Langlands dual group comes up is, unsurprisingly, the Satake isomorphism.
The general statement is: For an automorphic representation associated to a group $G$ with dual group $hatG$, the coefficient of $p^k$ in the $L$-function associated to a representation $rho$ of $hatG$ is equal to the trace of the Satake parameter (a conjugacy class on $hatG$) acting on $Sym^k rho$.
No assumption beyond unramifiedness should be necessary.
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There is no coincidence, this is the Weyl character formula for the representation $operatornameSym^k$ of $GL_3$. The reason that the Langlands dual group comes up is, unsurprisingly, the Satake isomorphism.
The general statement is: For an automorphic representation associated to a group $G$ with dual group $hatG$, the coefficient of $p^k$ in the $L$-function associated to a representation $rho$ of $hatG$ is equal to the trace of the Satake parameter (a conjugacy class on $hatG$) acting on $Sym^k rho$.
No assumption beyond unramifiedness should be necessary.
There is no coincidence, this is the Weyl character formula for the representation $operatornameSym^k$ of $GL_3$. The reason that the Langlands dual group comes up is, unsurprisingly, the Satake isomorphism.
The general statement is: For an automorphic representation associated to a group $G$ with dual group $hatG$, the coefficient of $p^k$ in the $L$-function associated to a representation $rho$ of $hatG$ is equal to the trace of the Satake parameter (a conjugacy class on $hatG$) acting on $Sym^k rho$.
No assumption beyond unramifiedness should be necessary.
edited 58 mins ago
Desiderius Severus
2,15121847
2,15121847
answered 1 hour ago
Will Sawin
65.2k6131273
65.2k6131273
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
add a comment |Â
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
â Desiderius Severus
58 mins ago
add a comment |Â
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