Timeline for Does viewing multiplicative functions as morphisms from $\mathbb{Q}^*\to\mathbb{C}^*$ have any consequnces?
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 11, 2020 at 15:45 | vote | accept | Milo Moses | ||
| Nov 11, 2020 at 11:27 | comment | added | reuns | For $f$ completely multiplicative $\Bbb{Z}\to \Bbb{C}$ then $f$ extends continuously to $\Bbb{A_Q}^\times$ in the obvious way: $F (a_\infty\prod_p (a_pp^{e_p})_p)=f(sign(a_\infty))\prod_{p,f(p)\ne 0} f(p)^{e_p}$. The point is that for $f=\chi$ a Dirichlet character (modulo a prime power $q^k$) we have another way to extend continuously to the ideles : $G(a_\infty\prod_p (a_pp^{e_p})_p)=\chi(a_q \bmod q^k)$. Then $F(x)/G(x)=1$ for each $x\in \Bbb{Q}^\times$ embedded diagonally in the ideles. $H=F/G$ is an automorphic form on $ GL_1(\Bbb{Q})\setminus GL_1(\Bbb{A_Q})$ | |
| Nov 11, 2020 at 9:21 | history | answered | Daniel Loughran | CC BY-SA 4.0 |