Is Ramanujan conjecture equivalent to $\operatorname{Tr}M_{p}(F_{\Pi})^{\dagger}M_{p}(F_{\Pi})=n$?

28 Views Asked by At

Let $\Pi$ an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ and $F_{\Pi}(s):=L(\Pi,s)$ the corresponding L-function. Fix an unramified prime $p$ and consider the diagonal matrix $M_{p}(F_{\Pi})$ formed by the Satake parameters of $\Pi$ at $p$ ordered by increasing arguments taking values in $(-\pi,\pi]$.

Ramanujan conjecture states that all such Satake parameters have absolute value $1$, which entails $\operatorname{Tr} M_{p}(F_{\Pi})^{\dagger}M_{p}(F_{\Pi})=n$.

Is the converse true? In other words, does $\operatorname{Tr} M_{p}(F_{\Pi})^{\dagger}M_{p}(F_{\Pi})=n$ for any unramified prime $p$ imply the Ramanujan conjecture for $\Pi$?