Timeline for Regular embeddings of reductive groups
Current License: CC BY-SA 3.0
7 events
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| Mar 11, 2015 at 1:26 | comment | added | Matthias Klupsch | Yes, this was known to me. But from that it seems to me that I can only deduce that I have a bijective morphism $(S \times H) / Z \rightarrow G$ of algebraic groups (which is thus an isomorphism of abstract groups). What I do not know is why this is automatically an isomorphism of algebraic groups? | |
| Mar 10, 2015 at 20:32 | comment | added | Jim Humphreys | The basic structure of any connected reductive $G$ is laid out by Borel-Tits (1965), with treatments in books on linear algebraic groups as well as a nice summary in Chap. 0 of the text by Digne-Michel (emphasizing finite fields). To be precise, $G$ is the almost direct product of its semisimple derived group $H$ and its radical $S$, the latter being a central torus; here $Z:=Z(H) \supset H∩S$ is finite. If $Z(G)$ is connected, then it equals $S$ with $Z \subset S$; so the surjective map $S \times H \rightarrow G$ has kernel consisting of the pairs $(z,z^{−1})$ with $z \in Z$. | |
| Mar 10, 2015 at 7:23 | comment | added | Matthias Klupsch | If this were true, all my problems would be solved indeed. I was not aware of this. Some time ago I ran into a problem (math.stackexchange.com/questions/1136933/…) where in a similar situation I got convinced that I do not necessarily have this isomorphism between e.g. $G'$ and the quotient of $[G,G] \times Z(G')$. Is this easy to see or do you have some reference for this? | |
| Mar 9, 2015 at 18:24 | comment | added | Jim Humphreys | I hope I'm not oversimplifying, but for a connected reductive group with connected center such as your $G', G''$, the group is just isomorphic to the direct product of its center (a torus) and the derived group $H \cong [G,G]$ modulo the diagonal action of $Z(H)$ (which lies in the big center). So both $G', G''$ embed naturally into $G''':=(S \times H)/Z(H)$ for a torus $S$ containing both of their centers. | |
| Mar 9, 2015 at 7:43 | comment | added | Matthias Klupsch | Thank you for your answer. Note that my change in definition is not really one. If we assume $[G',G'] \subseteq \varphi(G)$, then we have $G' = Z(G')\varphi(G)$ and so the commutator subgroups coincide. The idea you formulated is essentially what I wanted to do and what I actually did when only considering the special regular embeddings from the standard construction. However, I do not understand how this should work in the general case where the regular embeddings given are not of the form $(G \times T)/Z$. How would you get the desired maps into $S \times H/Z(H)$? | |
| Mar 8, 2015 at 1:31 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
added 512 characters in body
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| Mar 7, 2015 at 23:32 | history | answered | Jim Humphreys | CC BY-SA 3.0 |