Timeline for Is there an almost-direct product decomposition for disconnected reductive algebraic groups?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2013 at 22:42 | vote | accept | Maxime | ||
| Apr 26, 2013 at 3:11 | comment | added | Will Sawin | Actually my idea doesn't quite make sense, so I'll describe a better way. Consider the map from the group to its abelianization - because the connected component of the identity is clearly central, the kernel of this map is clearly finite. The abelianization is an abelian reductive group, hence it's non-canonically the product of a torus and a finite group. Take the kernel of the composed map from the group to to this torus. That's the finite group you want. | |
| Apr 26, 2013 at 2:38 | comment | added | Maxime | @Will Thank you for the answer, it is along the lines of what I was looking for. I can see why the connected component of the identity in a nilpotent reductive group is a central torus. However, it isn't clear to me what you mean by the "universal map from the group to a torus" which yields the finite group in the almost-direct product decomposition. Could you perhaps elaborate a bit on this? | |
| Apr 25, 2013 at 6:03 | history | answered | Will Sawin | CC BY-SA 3.0 |