Timeline for Why are we defining character groups differently for additive and multiplicative group in Tate's thesis?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2023 at 4:50 | comment | added | Rits | @paulgarrett Ohh, right. I missed that. Thanks a lot! | |
| Jul 18, 2023 at 18:26 | comment | added | paul garrett | @Rits, well, it's not quite right to say that non-archimedean $k$ has no unitary characters of $k^\times$: for real $t$, $x\to |x|_p^{it}$ on $\mathbb Q_p^\times$ is such, and these play a large role in treating Hecke characters... | |
| Jul 18, 2023 at 2:39 | comment | added | Rits | @LSpice haha yes, I completely failed to realize that. Even to answer my second question, there aren't any characters from $k^{*} \to \mathbb{S}^{1}$, except the trivial character, which is clear from Matt's response. | |
| Jul 17, 2023 at 22:00 | comment | added | LSpice | @Rits, re, as @MattYoung's later answer indicates, we don't look at non-unitary, continuous characters of the additive group of $k$ because there aren't any! | |
| Jul 17, 2023 at 19:29 | comment | added | Rits | Thanks for the response. So why not even look at such quasi characters for $k^{+}$? Is it because they are just not interesting? Or maybe there is no proper structure to them, as we have for the $S^{1}$ case? Also, these quasi characters dont have a proper structure to them it seems. If we look at only the characters from $k^{*} \to $S^{1}$, can we expect them to have a nice structure? Have people tried doing it? | |
| Jul 17, 2023 at 18:25 | history | answered | Stopple | CC BY-SA 4.0 |