Timeline for Why is the semisimple quotient of a reductive group with semisimple rank 1 equal to PGL2?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2014 at 19:54 | vote | accept | Daniel Barter | ||
| Feb 26, 2014 at 3:37 | comment | added | user76758 | @KeerthiMadapusiPera: Is $I$ defined as the identity component of the intersection? I think one has to control the entire intersection, and its centrality in the entirety of $G$ has to be proved. The "problem" is that the good behavior of centrality under quotients, a familiar feature of connected reductive groups, is not true for connected linear algebraic groups more generally, so at this early stage in the theory some care is needed for handling centrality claims. | |
| Feb 26, 2014 at 3:34 | answer | added | user76758 | timeline score: 5 | |
| Feb 26, 2014 at 3:30 | comment | added | Keerthi Madapusi | @OP: Now that I've read your question more carefully, I see that what you're really asking is: Why is the identity component of the intersection of the Borels the radical? This is because all Borels are conjugate, which means that their intersection is a normal solvable sub-group of $G$. But then its connected component must be contained in the radical! | |
| Feb 26, 2014 at 3:15 | comment | added | Keerthi Madapusi | That's certainly possible! I'm no expert on the classification. | |
| Feb 26, 2014 at 3:11 | comment | added | user76758 | @KeerthiMadapusiPera: I think your suggestion is circular, since the classification with Dynkin diagrams (which, oddly enough, is not discussed in Borel's textbook) requires a huge amount of structure theory, the entire foundation of which rests on knowing the classification in semisimple-rank 1 to get a root system... | |
| Feb 25, 2014 at 23:45 | comment | added | Keerthi Madapusi | I don't know what Humphreys's proof is, but this is certainly a consequence of the classification of adjoint semsimple groups (over algebraically closed fields) by Dynkin diagrams: There is only one Dynkin diagram with a single vertex. | |
| Feb 25, 2014 at 23:42 | history | edited | Daniel Barter | CC BY-SA 3.0 |
added 233 characters in body
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| Feb 25, 2014 at 23:14 | history | asked | Daniel Barter | CC BY-SA 3.0 |