Timeline for simple Lie groups over C [closed]
Current License: CC BY-SA 4.0
10 events
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| Mar 28, 2020 at 17:31 | history | closed |
LSpice Bugs Bunny abx user44191 Ben McKay |
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| Mar 27, 2020 at 12:57 | comment | added | Ben McKay | There are some very good books of Tony Springer, of J. Humphreys, and of A. Borel, all with the title Linear Algebraic Groups, which cover this material in detail. | |
| Mar 27, 2020 at 10:20 | review | Close votes | |||
| Mar 28, 2020 at 17:31 | |||||
| Mar 27, 2020 at 10:12 | comment | added | YCor | Yes: if $G$ is a (finite-dimensional) Lie group over a complete normed field of characteristic zero (a) if its Lie algebra is not semisimple, then $G$ has a nontrivial abelian normal subgroup (b) if its Lie algebra is semisimple and not simple, then $G$ has a positive-dimensional proper normal subgroup, obtained as centralizer of some simple factor. | |
| Mar 27, 2020 at 10:09 | comment | added | Rupert | Thanks, seems as though that would generalise easily enough to non-archimedean local fields of characteristic zero? | |
| Mar 27, 2020 at 10:02 | comment | added | YCor | The sketch would be: if $\mathfrak{g}$ has a nontrivial abelian ideal, then $G(\mathbf{C})$ has a nontrivial abelian normal subgroup (taking the exponential), hence its Zariski closure yields a contradiction. So $\mathfrak{g}$ is semisimple (and nontrivial: I assume the group has positive dimension since otherwise the claim is false). If $\mathfrak{g}$ were not simple, it would have a simple factor: then the centralizer of this factor is not trivial, and hence the centralizer of this factor in the group is a nontrivial proper Zariski-closed subgroup. | |
| Mar 27, 2020 at 10:00 | comment | added | Rupert | Thank you for reminding me of that point, so I made a few modifications to my question which I hope now make it a more sensible question. | |
| Mar 27, 2020 at 10:00 | history | edited | Rupert | CC BY-SA 4.0 |
added 48 characters in body
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| Mar 27, 2020 at 9:56 | comment | added | YCor | For a Lie group, "Zariski closed" makes no sense in general. It turns out to make sense in a unique way in a simple complex Lie group, but this takes some energy to prove. Maybe you just want to start with $G(\mathbf{C})$ for some complex algebraic group. | |
| Mar 27, 2020 at 9:54 | history | asked | Rupert | CC BY-SA 4.0 |