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Let $\pi$ be an automorphic representation (say, trivial central character) of $GL(2)$. Let $\alpha(p)$ and $\beta(p)$ denote its spectral parameters at the place $p$, that is to say the associated local L-function takes the form $$L_p(s, \pi) = (1-\alpha(p)p^{-s})^{-1}(1-\beta(p)p^{-s})^{-1}.$$

Using the expression of $L(s, \pi)$ as a Dirichlet series $\sum_n c_n n^{-s}$ and the Hecke relations, I can deduce the usual relations

$$\alpha(p) + \beta(p) = c_p \qquad \text{and} \qquad \alpha(p)^2 + \beta(p)^2 = c_{p^2} - 1.$$

Is there a relation in a similar fashion for spectral parameters and coefficients in the case of "twisted", like $$\alpha(p_1)\alpha(p_2) + \beta(p_1)\beta(p_2) \qquad \text{or} \qquad \alpha(p_1)^{v_1}\beta(p_2)^{v_2}+\alpha(p_2)^{v_1}\beta(p_1)^{v_2}$$

for some powers $v_i$, even $1$ or $2$? I have no idea how to get it from L-functions or other arguments (symmetric square or Rankin-Selberg?).

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  • $\begingroup$ Do you mean $\alpha_1(p)$ instead of $\alpha(p_1)$ ? $\endgroup$
    – reuns
    Feb 15, 2020 at 17:14
  • $\begingroup$ If $p_1$ and $p_2$ are distinct primes, then there is no way to write these expressions in terms of polynomials of the form $\sum_{i,j} a_{ij} c_{p_1^i} c_{p_2^j}$ with coefficients $a_{ij} \in \mathbb{C}$. You can only do this if the expressions are symmetric in both $(\alpha(p_1),\beta(p_1)) \mapsto (\beta(p_1),\alpha(p_1))$ and $(\alpha(p_2),\beta(p_2)) \mapsto (\beta(p_2),\alpha(p_2))$ $\endgroup$ Feb 15, 2020 at 17:36
  • $\begingroup$ @PeterHumphries This could be already very interesting, if they have such symmetries, how could I get suitable relations as polynomials in the coefficients? $\endgroup$
    – Gory
    Feb 16, 2020 at 1:20
  • $\begingroup$ Not sure - it's more of a problem of algebraic geometry. Basically, Hecke eigenvalues are Schur polynomials in $\alpha,\beta$, and Schur polynomials form a basis for the vector space of symmetric polynomials, and so every symmetric polynomial in $\alpha,\beta$ can be written as a linear combination of Hecke eigenvalues. $\endgroup$ Feb 16, 2020 at 16:32

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