Timeline for What is the subgroup generated by involutions?
Current License: CC BY-SA 2.5
11 events
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| Dec 1, 2010 at 9:36 | comment | added | BCnrd | Dear Pete: Yes, it was a small goof. But just move on: when rank-2 is too small, a standard device is to use the rank-3 cases to handle everything else. That is why (in my various comments) I keep suggesting to look into those basic rank-3 cases, which would then handle everything else split and simply connected by using Bruhat decomposition, torus centralizers, and the like (as Jim says). | |
| Nov 30, 2010 at 23:08 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 30, 2010 at 23:03 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 30, 2010 at 19:52 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 30, 2010 at 17:04 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 30, 2010 at 13:11 | comment | added | Pete L. Clark | @BCnrd: not yet, no. My point is that I find the sentence "On the other hand, $SL_2$ is almost simple, while its subgroup generated by involutions is obviously normal and too big to be the center" to be confusing. (Or, less politely but more plainly, it seems not to be true.) | |
| Nov 30, 2010 at 4:08 | comment | added | BCnrd | Dear Pete: sure, but one can get at all split simply connected examples other than ${\rm{SL}}_2$ by using the rank-3 cases (SL$_3$, G$_2$, and Sp$_4$), so have you looked into those basic building blocks instead? | |
| Nov 30, 2010 at 0:09 | comment | added | Pete L. Clark | I'm confused: as in the above comments, isn't $-1$ the only involution in $\operatorname{SL}_2$? | |
| Nov 29, 2010 at 21:46 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 29, 2010 at 19:03 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
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| Nov 29, 2010 at 18:58 | history | answered | Jim Humphreys | CC BY-SA 2.5 |