Timeline for Certain central extensions of simply connected simple algebraic groups
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23, 2013 at 2:09 | comment | added | David Stewart | @Yves Of course... | |
| Jul 22, 2013 at 23:00 | comment | added | YCor | @David: $H$ is not a perfect group. | |
| Jul 22, 2013 at 20:52 | comment | added | David Stewart | So I think this won't work for positive characteristic. If we let $G=SL_p(F)$ with char $F=p$ then as abstract groups $G$ is also isomorphic to $PGL_p(F)$. So there is a (non-algebraic) map from $H=GL_p(F)$ to $G$ with non-trivial kernel being the centre of $H$. Does that make sense? | |
| Jul 22, 2013 at 7:51 | comment | added | YCor | I mean: (unlike in the OP's question): let $H$ be a perfect central extension of $G$ ($G$ being as in the question), is every linear rep of $H$ trivial on the kernel $Z$ of $H\to G$? The OP's question seems equivalent to whether every linear rep of $G$ is unipotent in restriction to $Z$ for every such $H$. | |
| Jul 22, 2013 at 0:00 | comment | added | Venkataramana | Yes, it is too optimistic: $H$ has by definition a faithful linear representation. If you wish to prove that the kernel is trivial, it is equivalent to your statement. | |
| Jul 21, 2013 at 23:41 | comment | added | YCor | Is it too optimistic to expect that every linear representation of a perfect central extension $H$ of $G$ is trivial on the kernel of $H\to G$? | |
| Jul 21, 2013 at 21:52 | comment | added | David Stewart | Would it be stupid to hope all this is controlled by $H^2(G,F)$ and $H^1(G,F)$? Which are trivial, even in positive characteristic ... That book of Serre's where he thinks about extensions of algebraic groups comes to mind. | |
| Jul 21, 2013 at 19:59 | comment | added | Jim Humphreys | @Aakumadula: Quite right. I left out one crucial part of the original question, that $H$ should be perfect (equal to its derived group). (This assumption is the abstract version of "connected" in the setting of central extensions.) | |
| Jul 21, 2013 at 18:54 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
added 149 characters in body
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| Jul 21, 2013 at 17:48 | comment | added | Venkataramana | What is to prevent us from taking $H= Z \times G$ where $Z$ is an Abelian linear algebraic group over $F$? | |
| Jul 21, 2013 at 16:14 | history | asked | Jim Humphreys | CC BY-SA 3.0 |