Timeline for Subgroup containing a given torus as a maximal torus
Current License: CC BY-SA 3.0
11 events
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| Oct 28, 2013 at 16:15 | comment | added | Marguax | @michipili: maybe you also need to assume (absolutely) simple, which is to say an irreducible root system (as otherwise one could take a direct product of my example with whatever you have in mind)? | |
| Oct 28, 2013 at 15:44 | comment | added | Michaël Le Barbier | Degenerate means contained in the kernel of a root. | |
| Oct 28, 2013 at 15:42 | history | edited | Michaël Le Barbier | CC BY-SA 3.0 |
Describe the maximal torus o the result
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| Oct 28, 2013 at 15:30 | comment | added | user41972 |
Does degenerate' mean anisotropic'. It seems to me that Margaux's objection does not hold up in the case T is anisotropic?
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| Oct 28, 2013 at 15:27 | comment | added | Michaël Le Barbier | @Marguax You are right, the construction I proposed only works in the case of where $T$ is degenerated. | |
| Oct 28, 2013 at 15:27 | comment | added | user41972 | Maybe I should add that I am interested in the case where T is anisotropic over Q... | |
| Oct 28, 2013 at 15:18 | comment | added | Marguax | This construction does not seem to work. Let $S$ be a maximal torus in $G$ and let $\lambda:{\rm{GL}}_1 \rightarrow S$ be a "generic" cocharacter. Then $Z_G(\lambda) = S$ and its derived group is 1. So if one begins with $T = \lambda({\rm{GL}}_1)$ and chooses $T'$ an isogeny complement to $T$ in $S$ then $T' \cdot T = S$, etc. Am I misunderstanding the proposed answer? | |
| Oct 28, 2013 at 15:13 | history | edited | Michaël Le Barbier | CC BY-SA 3.0 |
Derive to kill teh centre
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| Oct 28, 2013 at 15:10 | comment | added | Michaël Le Barbier | My bad, you need to take the derived group! | |
| Oct 28, 2013 at 15:08 | comment | added | user41972 | Many thanks for this reply! I just do not quite see why T is a maximal torus of this H; isn't T X T' in H? | |
| Oct 28, 2013 at 14:50 | history | answered | Michaël Le Barbier | CC BY-SA 3.0 |