Timeline for Metaplectic groups over non-archimedean local fields of characteristic>2
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2016 at 13:02 | comment | added | m07kl | As far as I know, metaplectic groups over reals are not linear algebraic groups, because linear algebraic groups have finitely many topological components (proved by Whitney '57) and in this case metaplectic groups contain $\mathbb{Z}$ as a central subgroup (center has infinitely many components ). Metaplectic groups over complex number are linear, in fact, metaplectic Groups over complex number are equal to symplectic groups. However, I am interested in function fields case. Wee Teck Gan is interested in $Q_p$ case. | |
| Jan 3, 2016 at 2:37 | comment | added | Venkataramana | @jim: the symplectic group over ${\mathbb R}$ is not simply connected (has $\mathbb Z$ as fundamental group) but over $\mathbb C$, it is simply connected. | |
| Jan 3, 2016 at 1:30 | comment | added | Jim Humphreys | @Venkataramana: Yes, it gets more complicated over the reals, but your example just has rank 1. What is true for symplectic groups of higher rank? Anyway, the complex case is the crucial one here. | |
| Jan 3, 2016 at 0:55 | comment | added | Venkataramana | @jim: not over the reals ($SL_2({\mathbb R})$ is not simply connected) but over the complex numbers | |
| Jan 2, 2016 at 21:24 | comment | added | Jim Humphreys | Concerning your Q1, over any sort of reasonable field a symplectic group of rank at least 2 is "simply connected" in the sense of algebraic groups (which coincides with the usual notion over the real or complex fields). So a non-split extension should typically not be a linear algebraic group. | |
| Jan 1, 2016 at 17:02 | history | asked | m07kl | CC BY-SA 3.0 |