6
$\begingroup$

I'm learning about rational normal curves of degree n. And the book says that rational normal curves of degree 3 cannot be written by intersection of two quadrics. I can visualize the situation in my head, but cannot formulate a rigorous argument...Could anyone help me?

$\endgroup$
1
  • 1
    $\begingroup$ Do you know about degrees of (embedded) projective varieties? Two quadrics will intersect in a curve of degree 4. $\endgroup$ Oct 20 '14 at 4:03
0
$\begingroup$

If one of the quadrics has maximal rank $4$ (or better one of the quadrics in the pencil, which is in fact true), then it is projectively equivalent to the Segre variety. In this case, the other quadric cuts out a curve of bidegree $(2, 2)$ on $\mathbb{P}^1 \times \mathbb{P}^1$. As the twisted cubic has bidegree $(1, 2)$, it follows that we get not only the twisted cubic, but a line, something of bidegree $(1, 0)$, so that the union has bidegree $(2, 2)$. Now, the line is a fiber of one of the rulings, and a general fiber meets the cubic in two points, since a quadratic polynomial has two roots in general.

$\endgroup$

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.