Timeline for Why is the radical of a reductive group equal to the connected component of the center?
Current License: CC BY-SA 3.0
9 events
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| Jun 5, 2017 at 17:44 | comment | added | Jim Humphreys | A comment on your last sentence: Relative to an arbitrary field of definition $k$ (as treated by Borel and Tits), the immediate problem is that $G$ need not possess any Borel subgroups defined over $k$. Indeed, in the extreme case where $G$ is $k$-anisotropic, little can be said about $G$ relative to $k$ from the viewpoint of the Borel-Tits structure theory. (A smaller comment is that "the connected component of the center" doesn't make sense; I'd prefer to say "the identity component of the center", meaning that component which contains the identity element.) | |
| Jun 5, 2017 at 10:10 | review | Close votes | |||
| Jun 5, 2017 at 10:57 | |||||
| Jun 5, 2017 at 9:56 | comment | added | Not a grad student | @Arkandias Ah, it's Theorem 5.1(b) on page 261 in those notes. Thanks! | |
| Jun 5, 2017 at 9:35 | comment | added | Arkandias | In fact $k$ needs not be perfect. This is the proof from Milne's AGS course. The radical $RG_{\bar k}$ of $G_{\bar k}$ is a connected smooth solvable group with trivial unipotent radical (since $G$ is reductive), thus it is a torus. Since $(RG)_{\bar k}$ is a smooth connected subgroup of $RG_{\bar k}$, it is a subtorus. Since $G$ is connected, its action by conjugation on $RG$ is trivial (by rigidity). Hence $RG \subseteq (ZG)_{\mathrm{red}}^\circ$. The other inclusion is clear (since $(ZG)_{\mathrm{red}}^\circ$ is a smooth connected commutative group) . | |
| Jun 5, 2017 at 9:33 | comment | added | Not a grad student | I just added some clarifying comments. | |
| Jun 5, 2017 at 9:27 | history | edited | Not a grad student | CC BY-SA 3.0 |
added 500 characters in body
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| Jun 5, 2017 at 9:13 | history | edited | Not a grad student | CC BY-SA 3.0 |
changed "field" to "perfect field"
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| Jun 5, 2017 at 8:51 | comment | added | Jason Starr | What do you mean by "radical"? I know the unipotent radical (which will be trivial for a reductive group) and the solvable radical (which can be nontrivial, but is typically not equal to the connected component of the center). | |
| Jun 5, 2017 at 8:29 | history | asked | Not a grad student | CC BY-SA 3.0 |