Relation between Fourier coefficients and Satake parameters

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

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.

Answer 1

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.

Answer 2

Let $s_{k}(\alpha_1(p),\ldots,\alpha_n(p))$ be the complete homogeneous symmetric polynomial of degree $k$ in variables $\{\alpha_1(p),\ldots,\alpha_n(p)\}$. If $\mathrm{Re}(s)$ is sufficiently large, then

$(*)~\sum_{k=0}^{\infty}\cfrac{a_{p^k}}{p^{ks}}=\prod_{j=1}^n (1-\alpha_j(p) p^{-s})^{-1} = \sum_{k=0}^{\infty}s_{k}(\alpha_1(p),\ldots,\alpha_n(p))p^{-ks}$.

This can be easily verified since the LHS is a product of geometric sums. To recover your second identity, we realize that if we let $\lambda(1)=k$ and $\lambda(j)=0$ for all $j\geq 2$, then

$s_k(\alpha_1(p),\ldots,\alpha_n(p))=\cfrac{\det[(\alpha_i(p)^{\lambda(j)+n-j})_{ij}]}{\det[(\alpha_i(p)^{n-j})_{ij}]}$,

where $i,j\in\{1,\ldots,n\}$. Note that the numerator is an alternating symmetric polynomial in $\{\alpha_1(p),\ldots,\alpha_n(p)\}$ (this follows from standard determinant properties), and hence is divisible by the Vandermonde determinant in the denominator. None of this relies on whether the underlying representation is self-contragredient.

The polynomial $s_k(\alpha_1,\ldots,\alpha_n)$ is a special case of the more general Schur polynomials. These are the characters of polynomial irreducible representations of $\mathrm{GL}(n)$. Moreover, $(*)$ is a special case of Cauchy's identity, which serves as a sort of orthogonality statement for Schur polynomials.