Is the reciprocal of a Satake parameter a Satake parameter?

Question

Let $F$ be an automorphic L-function for $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$. Is it known whether for all but finitely many primes $p$ the set of Satake parameters of $F$ at $p$ is invariant under the map $x\mapsto x^{-1}$?

Answer 1

No, this surely is false. First off, you shouldn't write $F$ an automorphic $L$-function; this is saying that $F(s) = L(s,\pi)$, which is terrible notation. Just write "let $\pi$ be an automorphic representation of $\mathrm{GL}_n(\mathbb{A_Q})$".

Anyway, your question is almost correct in one specific case: $n = 2$ and $\omega_{\pi} = 1$, where $\omega_{\pi}$ is the central character of $\pi$. This corresponds to the case of the $L$-function $L(s,f)$ of a newform $f$ on $\Gamma_0(q) \backslash \mathbb{H}$ of conductor $q$ with principal nebentypus $\chi = \chi_{0(q)}$. Then the $L$-function is $L(s,f) = \prod_p L_p(s,f)$ with \[L_p(s,f) = \frac{1}{1 - \lambda_f(p) p^{-s} + \chi_{0(q)}(p) p^{-2s}} = \frac{1}{(1 - \alpha_{f,1}(p) p^{-s}) (1 - \alpha_{f,2}(p) p^{-s})},\] where $\lambda_f(p) = \alpha_{f,1}(p) + \alpha_{f,2}(p)$ is the $p$-th Hecke eigenvalue of $f$ and $\alpha_{f,1}(p), \alpha_{f,2}(p)$ are the two Satake parameters at $p$. Then one sees that \[\alpha_{f,1}(p) \alpha_{f,2}(p) = \chi_{0(p)}(p),\] which is $1$ if $p \nmid q$ and $0$ if $p \mid q$, so indeed the Satake parameters do satisfy the invariance map for all but the primes dividing the conductor $q$.

In general, however, this is not the case. Let $\pi$ be an automorphic representation of $\mathrm{GL}_n(\mathbb{A_Q})$ with central character $\omega_{\pi}$, and let $\alpha_{\pi,1}(p),\ldots,\alpha_{\pi,n}(p)$ denote the Satake parameters at a prime $p$. Then it is the case that \[\alpha_{\pi,1}(p) \cdots \alpha_{\pi,n}(p) = \omega_{\pi}(p).\] But in general, one shouldn't expect that $\alpha_{\pi,j}(p)^{-1}$ is equal to $\alpha_{\pi,k}(p)$ for some $k$.