Skip to main content
13 events
when toggle format what by license comment
Dec 18, 2022 at 12:02 vote accept Alireza Shavali
Dec 15, 2022 at 18:58 answer added Yuri Zarhin timeline score: 4
Dec 15, 2022 at 10:42 comment added YCor @AlirezaShavali Yes, because the normalizer of the Lie algebra in $\mathrm{SL}_2$ is Zariski-closed (since the action of the group on the Lie algebra is algebraic). So Zariski-density of the group, and the fact that the group normalizes the Lie algebra, implies that the Lie algebra is normalized by $\mathrm{SL}_2$, and hence is $\{0\}$ or $\mathfrak{sl}_2$.
Dec 15, 2022 at 10:20 comment added Alireza Shavali @YCor Is this clear that Zariski density forces the Lie subalgebra to actually be a Lie ideal?
Dec 12, 2022 at 11:22 comment added YCor Since this is Zariski-dense, the Lie algebra is an ideal in $\mathfrak{sl}_2$, hence $\{0\}$ or equal to $\mathfrak{sl}_2$. So if it's not $\mathfrak{sl}_2$, the image has to be finite (since it's compact), hence not Zariski-dense, contradiction.
Dec 12, 2022 at 9:07 history edited Alireza Shavali CC BY-SA 4.0
corrected spelling
Dec 11, 2022 at 16:12 comment added LSpice Re, ah, sorry, I had not considered that gap. If it is easy to fix the gap, then I do not know how. I have retracted my close-as-duplicate vote.
Dec 11, 2022 at 11:38 comment added Alireza Shavali @LSpice Thanks for the link. From what I understand, this only implies that the Zariski closure of the image contains $SL_2$. How does this imply that the $\ell$-adic Lie algebra (which might not be algebraic) contains $\mathfrak{sl}_2$?
Dec 8, 2022 at 16:08 review Close votes
Dec 11, 2022 at 16:19
Dec 8, 2022 at 15:42 comment added LSpice I think this is a duplicate of Reductive subgroups of $\operatorname{GL}_2$ over an algebraically closed field of characteristic zero.
Dec 8, 2022 at 15:30 history edited YCor CC BY-SA 4.0
formatting
S Dec 8, 2022 at 13:24 review First questions
Dec 8, 2022 at 20:11
S Dec 8, 2022 at 13:24 history asked Alireza Shavali CC BY-SA 4.0