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:
$$a_p^k = frac
left
$$



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
    2












    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:
    $$a_p^k = frac
    left
    $$



    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







    share|cite|improve this question

























      up vote
      4
      down vote

      favorite
      2









      up vote
      4
      down vote

      favorite
      2






      2





      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:
      $$a_p^k = frac
      left
      $$



      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







      share|cite|improve this question















      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:
      $$a_p^k = frac
      left
      $$



      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






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      edited 2 hours ago

























      asked 2 hours ago









      Desiderius Severus

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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.






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          • Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
            – Desiderius Severus
            52 mins ago










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          1 Answer
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          active

          oldest

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          oldest

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          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.






          share|cite|improve this answer






















          • Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
            – Desiderius Severus
            52 mins ago














          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.






          share|cite|improve this answer






















          • Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
            – Desiderius Severus
            52 mins ago












          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 52 mins ago









          Desiderius Severus

          2,15121847




          2,15121847










          answered 58 mins 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?
            – Desiderius Severus
            52 mins ago
















          • Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
            – Desiderius Severus
            52 mins ago















          Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
          – Desiderius Severus
          52 mins ago




          Thanks for your great and enlightening answer. Do you have any reference with some more details on why this statement is true?
          – Desiderius Severus
          52 mins ago

















           

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