Timeline for Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2011 at 12:40 | comment | added | Torsten Ekedahl | Finite dimensional representations of the Lie algebra of $SL(2,\mathbb R)$ are easily classified and they are all representations of $SL(2,\mathbb R)$. If the double cover were algebraic it would have a faithful finite dimensional representation. | |
| Jul 20, 2011 at 8:07 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
Fixed typo in title
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| Jul 20, 2011 at 8:00 | comment | added | Marc Palm | Following your suggestion, I edited the title. | |
| Jul 20, 2011 at 7:58 | history | edited | Marc Palm | CC BY-SA 3.0 |
added 10 characters in body; edited title; edited title
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| Jul 8, 2011 at 7:34 | vote | accept | Marc Palm | ||
| Jul 7, 2011 at 21:04 | answer | added | André Henriques | timeline score: 28 | |
| Jul 7, 2011 at 20:41 | comment | added | Qiaochu Yuan | It seems to me that central extensions of an algebraic group should be "non-algebraic until proven algebraic" rather than the other way around. | |
| Jul 7, 2011 at 19:48 | comment | added | Fernando Muro | Discrete group central extensions are classified by the second group cohomology. Find out the cohomology classifying central algebraic group extensions. There will probably be a comparison homomotphism from this cohomology to group cohomology which won't be surjective in general, even in degree $2$. An element which is not in the image of the comparison homomorphism will give you a non-algebraic group extension. | |
| Jul 7, 2011 at 19:30 | history | edited | Marc Palm | CC BY-SA 3.0 |
added 45 characters in body
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| Jul 7, 2011 at 19:25 | history | asked | Marc Palm | CC BY-SA 3.0 |