Timeline for Simplicity of (complex) orthogonal groups
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2010 at 23:18 | comment | added | BCnrd | Jakub, then just keep reading. Basically, it's a refined group version of the Lie algebra fact that pairs of opposite root spaces generate $\mathfrak{sl}_2$'s. (The "refined" aspect is Bruhat decomposition using the Weyl group to nail down the entire group, not just some open around the identity.) At least over an alg. closed field, the version in Jim's book will give you the group version which I was invoking above. | |
| Jun 21, 2010 at 22:54 | comment | added | JGis | Dear BCnrd, I know some theory of semisimple alg. groups (e.g. from Jim's book). But, I dont see how SO_n "magically" reduces to SL_2. | |
| Jun 21, 2010 at 22:31 | comment | added | BCnrd | Jakub, if the structure theory of semisimple groups over alg. closed fields is unfamiliar, then my initial comment on the commutator aspect will probably not make much sense to you, and then you should follow Jim's advice and try to dig it out of Artin's "Geometric Algebra" or Dieudonne's book on classical groups. Or wait until you learn about connected ss groups, and you'll see how it magically reduces to the known case of SL_2. | |
| Jun 21, 2010 at 22:19 | comment | added | JGis | The book of Grove (chapter 7) provides a proof of simplicity only of "real" orthogonal groups PSO_n(R). I don't see the complex case in that book. | |
| Jun 21, 2010 at 22:06 | comment | added | BCnrd | Jakub, I had in mind the case of split groups (since I usually write ${\rm{SO}}_n$ to denote the special orthogonal group associated to the split quadratic form). Now I see that you specifically defined it relative to the "sum of squares" form, so then there will indeed be problems if the form is anisotropic (for example). Not sure what to suggest in that case, though over the real numbers there must be classical facts from compact groups which clarify the matter. Sorry! | |
| Jun 21, 2010 at 21:04 | comment | added | JGis | I know this simplicity criterion (due to Tit's) with BN pair buissness, but the main point is the perfectness of the group. I don't know how to show that SO_n's are perfect. | |
| Jun 21, 2010 at 20:23 | comment | added | BCnrd | Use Ch. IV, sec. 2.7, Cor. to Thm. 5 in Bourbaki LIE. If $G$ is split conn'd ss gp over field $k$ s.t. $G(k)$ is own commutator subgp and gen'td by $U(k)$'s for unip. radicals $U$ of Borel $k$-subgps $B$ of $G$ then structure theory for ss gps provides Tits system, so Bourbaki ref. implies $G(k)$ mod center is simple abstract gp. If $G$ is abs. simple and s.c. then these hypotheses on $G$ hold, by using "big cell" and structure of Weyl group (gen'td by simple reflections!) to reduce to case of ${\rm{SL}}_2(k)$ (win if $|k| > 3$!). So for $k$ alg. closed, $G(k)$ simple if $G$ is adjoint. | |
| Jun 21, 2010 at 20:05 | comment | added | JGis | How it works in odd dimensions? | |
| Jun 21, 2010 at 19:45 | history | answered | Charles Matthews | CC BY-SA 2.5 |