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: $$L(s) = \sum_{k=0}^\infty \frac{a_n}{n^s} = \prod_{p} \left( 1 - \alpha(p)p^{-s} \right)^{-1} \left( 1 - \beta(p)p^{-s} \right)^{-1} \left( 1 - \gamma(p)p^{-s} \right)^{-1}$$

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: $$a_{p^k} = \frac{ \left| \begin{array}{ccc} \alpha(p)^{k+2} & \beta(p)^{k+2} & \gamma(p)^{k+2} \\ \alpha(p) & \beta(p) & \gamma(p) \\ 1 & 1 & 1 \end{array} \right| }{ \left| \begin{array}{ccc} \alpha(p)^{2} & \beta(p)^{2} & \gamma(p)^{2} \\ \alpha(p) & \beta(p) & \gamma(p) \\ 1 & 1 & 1 \end{array} \right| }$$

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.

  • 1
    Note that the $a_n$'s are not really Fourier coefficients (well, they are, but only very special ones), but Hecke eigenvalues. On GL(n) the connection between Hecke eigenvalues and Fourier coefficients is less direct than on GL(2). – GH from MO Oct 15 at 4:37

There is no coincidence, this is the Weyl character formula for the representation $\operatorname{Sym}^k$ 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 $\hat{G}$, the coefficient of $p^k$ in the $L$-function associated to a representation $\rho$ of $\hat{G}$ is equal to the trace of the Satake parameter (a conjugacy class on $\hat{G}$) 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? – Desiderius Severus Oct 15 at 3:00
  • @DesideriusSeverus This comes from the definition of the local $L$-factor originally given in Langlands' letter to Weil (publications.ias.edu/sites/default/files/letter-to-Weil-rpl.pdf page 2) plus the generating function identity in the representation ring $\frac{1}{ \sum_{k=0}^{ \dim V} (-x)^k \wedge^k V} = \sum_{k=0}^\infty x^k \operatorname{Sym}^k V$ which follows e.g. from character theory by writing the trace of the left side as a product of eigenvalues. – Will Sawin Oct 15 at 11:17
  • @DesideriusSeverus, I learned all this stuff from Gross's paper On the Satake Isomorphism. math.harvard.edu/~gross/preprints/sat.pdf – Marty Oct 15 at 14:56

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