0
$\begingroup$

Let $X\subset\mathbb{P}^n$ be a smooth nondegenerate (i.e. not contained in any hyperplane) curve over $\mathbb{C}$. Is it possible that every collection of $n-3$ points on $X$ lies on a $n-1$-secant (i.e. a linear space spanned by $n-1$ points on $X$) that has dimension at most $n-3$? If yes, can $X$ even be chosen to be projectively normal?

For example, for $n=4$, is there a curve $X\subset\mathbb{P}^4$ such that every $p\in X$ lies on a trisecant line?

My hope and intuition is that this cannot happen but I don't have a proof.

$\endgroup$
3
$\begingroup$

I will give an example for $n = 4$. Let $$ S = \mathbb{P}_{\mathbb{P}^1}(\mathcal{O}(-1) \oplus \mathcal{O}(-2)) \subset \mathbb{P}^4 $$ be a cubic scroll and let $$ X \subset S $$ be a curve which is a 3-section of the morphism $S \to \mathbb{P}^1$. Then every point of $X$ lies on a trisecant line (the ruling of the scroll).

$\endgroup$

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.