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Jan 26 at 5:29 comment added Peter Humphries No, you need the representations to be unitary and $\psi$ to be unramified. As to your second question, the functional equation should be $\epsilon(s,\pi,\psi) \epsilon(1-s,\widetilde{\pi},\psi^{-1}) = 1$.
Jan 26 at 4:55 comment added L-JS But, in the Godement-Jacquet case the epsilon factor satisfy the FE $$\epsilon(s, \pi, \psi)\epsilon(1-s, \widetilde \pi, \psi) = \omega_\pi(-1)$$ whereas the Rankin-Selberg case, the epsilon factor satisfy the FE $$\epsilon(s, \pi \times \pi', \psi) \epsilon(1-s, \widetilde \pi \times \widetilde \pi', \psi^{-1}) = 1.$$
Jan 26 at 4:55 comment added L-JS @PeterHumphries thank you for the references. Also, in the answer provided you have provided in MO "On the consistency of the definition of the conductor for automorphic forms", do we also implicitly have $|\epsilon(1/2, \pi \times \pi', \psi)| = 1$? where $\pi, \pi'$ are irreducible admissible representations, and no conditions imposed on $\psi$?
Jan 25 at 22:59 comment added Peter Humphries @L-JS It is stated in Theorem 3.3 (4) of "Zeta functions of simple algebras" by Godement and Jacquet that $\epsilon(s,\pi,\psi)$ is equal to $q^{-ms}$ times a nonzero constant. From here, see Section 5 of "Conducteur des représentations du groupe linéaire" to see that $m = c(\pi)$. The fact that $|\epsilon(1/2,\pi,\psi)| = 1$ when $\psi$ is unramified and $\pi$ is a generic irreducible admissible unitary representation follows by applying the local functional equation twice for the Godement-Jacquet zeta integral.
Jan 25 at 17:46 comment added L-JS @PeterHumphries Can I clarify for the expression $$\epsilon(s, \pi, \psi) = \epsilon( \frac{1}{2}, \pi, \psi) q^{-c(\pi) (s - \frac{1}{2})}$$ it is found in Jacquet's "Conducteur des représentations du groupe linéaire" paper? As the paper only indicates that epsilon factor can be written as a monomial, which follows from the local FE. Furthermore, do you happen to have the reference to show that $|\epsilon(s, \pi, \psi)| =1$ for any irreducible admissible representation $\pi$, with $\psi$ being unramified? This is from Cogdell et. al's "Lectures on Automorphic L-functions". Thank you!
Apr 19, 2023 at 15:49 vote accept user15243
Dec 11, 2019 at 10:05 comment added Subhajit Jana I was originally thinking about the analytic conductor, but if you like you may include either of them.
Dec 11, 2019 at 9:51 comment added Peter Humphries @SubhajitJana depends on whether you want me to include the archimedean case for the algebraic conductor exponent or the analytic conductor
Dec 11, 2019 at 8:47 comment added Subhajit Jana Would you like to include the archimedean case? Otherwise, I might spell my comment out with a bit details.
Dec 10, 2019 at 15:23 history answered Peter Humphries CC BY-SA 4.0