Timeline for Unirational implies rationally connected
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2013 at 13:40 | answer | added | Puzzled | timeline score: 2 | |
| Dec 9, 2012 at 5:02 | vote | accept | HNuer | ||
| Dec 8, 2012 at 0:16 | answer | added | Sándor Kovács | timeline score: 1 | |
| Dec 6, 2011 at 22:00 | comment | added | HNuer | I have edited my question to ask why any two points can be joined by a chain of rational curves. I agree if we're using your defition then it's obvious. | |
| Dec 6, 2011 at 21:58 | history | edited | HNuer | CC BY-SA 3.0 |
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| Dec 6, 2011 at 21:54 | comment | added | Jack Huizenga | You need only join a general pair of points, not every pair. The general pair of points lies in the image of the map from projective space. It's not obvious why the various definitions of rationally connected are equivalent in characteristic zero, however; this typically requires deformation theory. See Debarre, "Higher Dimensional Algebraic Geometry," chapter 4, for this. | |
| Dec 6, 2011 at 21:50 | comment | added | HNuer | @Jack Why is it completely obvious as you say? For example if my points are outside of the image of rational map from projective space, I don't see why it should be clear I can get a rational curve connecting these points, or even a chain of curves. | |
| Dec 6, 2011 at 21:47 | comment | added | Jack Huizenga | You should emphasize your hypotheses more. The reason this would be merely a remark in many birational geometry papers is that it is completely trivial in case X is smooth and the characteristic is zero: there rationally connected is equivalent to being able to join a general pair of points by a rational curve, which is obvious for unirational varieties. | |
| Dec 6, 2011 at 21:21 | history | asked | HNuer | CC BY-SA 3.0 |