Let $p_1,...,p_8\in\mathbb{P}^{3}$ be points in linear general position. Then there exists a unique elliptic curve $C$ of degree $4$ passing through $p_1,...,p_8$. I am interested in what happens for nodal rational curves of degree $4$.

Now, let us suppose we have seven points $p_1,...,p_7\in\mathbb{P}^{3}$ in linear general position.

How many irreducible, rational curves of degree $4$ pass through $p_2,...,p_7$ and have a singular point of multiplicity $2$ at $p_1$ ?

Such a curve can be constructed as a complete intersection of two quadric surfaces which are tangent at $p_1$, or as a projection of a rational normal quartic curve $C$ in $\mathbb{P}^{4}$ from a point lying on a secant line or an a tangent line of $C$.


Zero. Indeed, if the intersection $Q_1 \cap Q_2$ of two quadrics is singular at $p_1$, then there is a quadric $Q$ in the pencil generated by $Q_1$ and $Q_2$ which is singular at $p_1$. On the other hand, if a quadric cone with vertex at $p_1$ passes through $p_2,\dots,p_7$, then the images of $p_2,\dots,p_7$ in $\mathbb P^2$, obtained by the projection from $p_1$, all lie on a conic. But a general 6-tuple of points in $\mathbb P^2$ does not lie on a conic.

  • $\begingroup$ 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? $\endgroup$
    – F_L
    Mar 8 '14 at 16:52
  • $\begingroup$ $p_1,\cdots,p_7$ 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. $\endgroup$
    – Will Sawin
    Mar 8 '14 at 20:05
  • $\begingroup$ 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. $\endgroup$ Mar 8 '14 at 22:06
  • $\begingroup$ @Jeremy Blanc: you are right of course, it is much simpler to project the curve and do not care about the quadrics. $\endgroup$
    – Sasha
    Mar 9 '14 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.