Timeline for Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2010 at 10:41 | comment | added | Georg M. | I didn't write this down very clearly, but here is what I meant to say. There are three cases where "two of the three lines" lie in a plane and the remaining line cuts the plane: The remaining line can 1. cut the plane in the point P of intersection of the first two (your first example); 2. cut the plane in one of the lines but not in P (your second example); 3. cut the plane in a point away from the two lines. | |
| Aug 14, 2010 at 5:10 | comment | added | Victor Protsak | I don't understand where does the configuration of the 3 coordinate axes (projectively completed) falls into your classification of triples of lines in $\mathbb{P}^3:$ each pair of lines lies in a plane, yet all lines do not. Also, if $l_1$ and $l_2$ are skew lines and $l_3$ intersects both of them then there are precisely two pairs of coplanar lines. | |
| Aug 13, 2010 at 15:18 | vote | accept | Georg M. | ||
| S Aug 13, 2010 at 15:18 | vote | accept | Georg M. | ||
| Aug 13, 2010 at 15:18 | |||||
| Aug 13, 2010 at 13:39 | answer | added | Torsten Ekedahl | timeline score: 6 | |
| Aug 13, 2010 at 13:19 | comment | added | Georg M. | This is really helpful: Now I know that projective invariance is not the right notion for my problem. I need to reconsider. | |
| Aug 13, 2010 at 13:04 | vote | accept | Georg M. | ||
| S Aug 13, 2010 at 15:18 | |||||
| Aug 13, 2010 at 13:00 | comment | added | Benoît Kloeckner | Are you sure you want to study configurations up to projectivities? Charles Siegel's answer shows that this is not a combinatorial problem. Maybe you want instead to look at configuration of lines up to homotopies that preserve crossing (along the homotopy, lines that initially cross should cross all the time and lines that does not must not cross at any time). | |
| Aug 13, 2010 at 12:54 | comment | added | Charles Siegel | @jvp: You beat me while I was typing! I had started counting configurations by intersection possibilities, and then realized the above, and made it an answer. | |
| Aug 13, 2010 at 12:53 | answer | added | Charles Siegel | timeline score: 5 | |
| Aug 13, 2010 at 12:48 | comment | added | Jorge Vitório Pereira | Four lines through a point in $\mathbb P^2$ is the same as four points on $\mathbb P^1$. These configurations are not all projectively equivalent as the $j$-invariant (symmetrization of the cross-ratio) will distinguish them. Thus there are infinitely many projectively distinct configurations of $n+1$ lines in $\mathbb P^n$ when $n \ge 3$. | |
| Aug 13, 2010 at 11:48 | history | asked | Georg M. | CC BY-SA 2.5 |