Timeline for For tori $S \subseteq T$, every character of $S(k)$ extends to a character of $T(k)$?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2018 at 17:15 | vote | accept | D_S | ||
| Mar 25, 2018 at 16:45 | comment | added | YCor | Can you accept your own answer, so that the robot doesn't regularly bump this (standard) instance of Pontryagin duality to the homepage? | |
| Feb 23, 2018 at 16:52 | comment | added | LSpice | Oh, I see; smoothness. It might make more sense to take for $K_S$ the kernel of $\chi$. | |
| Feb 23, 2018 at 16:51 | comment | added | LSpice | Why do you have your roundabout argument with $K_S$ and $K_T$, rather than simply extending directly from $H = S(k)$ to $G = T(k)$? | |
| Feb 23, 2018 at 15:45 | history | answered | D_S | CC BY-SA 3.0 |