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Mar 12, 2020 at 18:31 comment added LSpice Although you wrote that clearly, I skipped over it both in the parenthesis and in the statement of the theorem. Sorry! (Still, is it true algebraically, in characteristic 0, that automorphisms of a semisimple Lie algebra exponentiate to automorphisms of the simply connected group with that Lie algebra? I believe it, but don't know a reference.)
Mar 12, 2020 at 18:18 comment added Mikhail Borovoi @LSpice: In my answer ${\rm char}\,R=0$ !
Mar 12, 2020 at 17:52 comment added LSpice Finally, why does $\operatorname{Aut} \mathfrak g^{\text{ss}}$ equal $\operatorname{Aut} G^{\text{sc}}$? For example, why can't there be some exotic Lie-algebra automorphisms in positive characteristic?
Mar 12, 2020 at 17:48 comment added LSpice The decomposition $\mathfrak g = \mathfrak z \oplus \mathfrak g^{\text{ss}}$ can fail, for example, if $R = \mathbb F_p$ and $G = \operatorname{SL}_p$. (Also the equality $\operatorname{Lie} G^{\text{sc}} = \mathfrak g^{\text{ss}}$.)
Mar 12, 2020 at 16:52 history answered Mikhail Borovoi CC BY-SA 4.0