Timeline for Product of two algebraic subgroups of a (solvable) group = another algebraic subgroup?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 16, 2011 at 13:30 | answer | added | Paul Ziegler | timeline score: 1 | |
| May 14, 2011 at 13:34 | comment | added | H A Helfgott | I know, I know, I was just being handwavy. | |
| May 14, 2011 at 8:28 | comment | added | Kevin Buzzard | Harald: I think it's a bit dangerous to think of a constructible set as "a variety with perhaps a few varieties of lower dimension deleted from it". Take for example the affine plane and then remove the x-axis and then re-insert the origin. That's constructible, but in some sense near the origin it is very far from being a variety: there is no sensible notion of an algebraic function in a neighbourhood of the origin as far as I know. In particular it seems to me to be "worse" then a variety with some bits missing---it's a variety with a bit added in quite a strange way. | |
| May 14, 2011 at 8:03 | answer | added | zroslav | timeline score: 0 | |
| May 13, 2011 at 22:14 | history | edited | Jim Humphreys |
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| May 13, 2011 at 21:24 | answer | added | David E Speyer | timeline score: 6 | |
| May 13, 2011 at 21:00 | answer | added | Jim Humphreys | timeline score: 2 | |
| May 13, 2011 at 16:56 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| May 13, 2011 at 16:55 | comment | added | H A Helfgott | Yes, I should really change the phrasing of my question to ask what I wanted to ask (and thereby break Paul Ziegler's answer below...) | |
| May 13, 2011 at 14:17 | comment | added | Felipe Voloch | A pedantic remark. If $K$ is a finite field then, as finite groups are algebraic groups for trivial reasons, when your set is a group, it is automatically an algebraic group. It's probably not the group you want though. | |
| May 13, 2011 at 13:10 | answer | added | Paul Ziegler | timeline score: 8 | |
| May 13, 2011 at 13:03 | comment | added | H A Helfgott | You are right of course - I mistyped. Still, it would be interesting to know whether one can say anything more if one knows H_1 H_2 to be a group. | |
| May 13, 2011 at 13:02 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| May 13, 2011 at 11:34 | comment | added | James Cranch | Apologies for the trivial comment, but... There's another group-theoretical issue, surely: in general you wouldn't expect the image to be a subgroup unless $H_1$ and $H_2$ are commuting subgroups. | |
| May 13, 2011 at 11:05 | history | asked | H A Helfgott | CC BY-SA 3.0 |