Timeline for Is every algebraic smooth hypersurface of affine space parallelizable?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2011 at 14:15 | vote | accept | Georges Elencwajg | ||
| Jul 1, 2011 at 17:17 | comment | added | Georges Elencwajg | Dear ulrich, you are absolutely right. I find it amazing that every holomorphic vector bundle on a non compact Riemann surface (and in particular on a complex affine algebraic curve) is trivial, whereas on an affine curve of genus $g\geq 1$ already the Picard group is huge, essentially as big as that of the completed curve. I gave an explicit example just to be specific and because I found it funny to mention a down-to-earth conic in algebraic geometry, a branch of mathematics that has the unfortunate reputation of being very sophisticated and abstract... | |
| Jul 1, 2011 at 15:50 | answer | added | Hailong Dao | timeline score: 3 | |
| Jul 1, 2011 at 15:35 | comment | added | naf | Without the stably trivial assumption on the tangent bundle it is easy to find examples of smooth affine varieties which are not parallelizable but which are holomorphically paralellizable. For example, consider a general curve of genus $g>1$ and remove a point. | |
| Jul 1, 2011 at 14:55 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added remark on non-alg.closed fields.
|
| Jul 1, 2011 at 14:28 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added an "Edit"
|
| Jul 1, 2011 at 9:46 | answer | added | naf | timeline score: 29 | |
| Jul 1, 2011 at 8:18 | history | asked | Georges Elencwajg | CC BY-SA 3.0 |