Timeline for Counting curves of degree 4 in $\mathbb{P}^{3}$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2014 at 16:12 | vote | accept | Puzzled | ||
| Mar 9, 2014 at 15:57 | comment | added | Sasha | @Jeremy Blanc: you are right of course, it is much simpler to project the curve and do not care about the quadrics. | |
| Mar 8, 2014 at 22:06 | comment | added | Jérémy Blanc | Just two comments: 1) your argument is nice and but you don't need in fact to consider any cones or quadrics, just project directly from $p_1$ and observe that the image of the quartic curve is a conic. 2) In the question it was written "in general linear position", but what you probably mean is "in general position", because there are indeed $7$ pts in general linear position where the projection by one of the pts is a plane an the image of the $6$ others are on a conic. | |
| Mar 8, 2014 at 20:05 | comment | added | Will Sawin | $p_1,\cdots,p_7$ are points in $\mathbb P^3$. If you want to lift them to $\mathbb P^4$, you don't get to change the point $p$. Otherwise by varying $p$, you could get different arrangements of $7$ points in $\mathbb P^3$. So you probably want to fix $p,p_1,\cdots,p_7$. Then this is generically impossible for the same reason - projection to $\mathbb P^2$. Alternately, you can view this as a dimension-counting argument, I believe. | |
| Mar 8, 2014 at 16:52 | comment | added | Puzzled | This would imply that the following situation is impossible: There exist a point $p\in\mathbb{P}^{4}$ and a rational normal curve $C\subset\mathbb{P}^{4}$ such that $\left\langle p,p_1\right\rangle$ is secant to $C$ and $\left\langle p,p_i\right\rangle$ intersects $C$ for any $i = 2,...,7$. Is this clear to you? | |
| Mar 8, 2014 at 15:40 | history | answered | Sasha | CC BY-SA 3.0 |