Given a smooth algebraic curve in CP^n, why is the variety of trisecants at most two-dimensional?

Question

I'm trying to understand the following statement which I read somewhere without proof. Let $C$ be a smooth algebraic curve in $\mathbb{CP}^n$. Define $X_k$ to be the subvariety of $\mathbb{CP}^n$ consisting of points that lie on lines that intersect $C$ at least $k$ times. Then $X_3$ is at most $2$ dimensional.

Here's how I've been trying to understand this result. It is easy to see that $X_2$ is at most $3$-dimensional. Indeed, let $Y \subset C \times C \times \mathbb{CP}^n$ be the closure of the set of points $(v,v',w)$, where $v \neq v'$ and $w$ lies on the line through $v$ and $v'$. Then $Y$ is clearly at most $3$ dimensional and $X_2$ is the projection of $Y$ onto its third factor.

We have $X_3 \subset X_2$, and the desired statement would follow if we could prove that $X_3 \neq X_2$. However, this need not hold -- for instance, $C$ could be a high degree curve in some $\mathbb{CP}^2$ in $\mathbb{CP}^n$ (in that case, of course, the desired result is trivial!). How do I complete this proof?

Answer 1

Dear Undergraduate Student, first of all, congratulations on the beautiful geometry you chose to study so early in your studies.

To prove $dim X_3 \leq 2$ it is indeed enough to prove that there exists a secant to $C$ which is not a trisecant, since the space of secants is irreducible (Harris, Algebraic Geometry, p.144).

This existence is proved on page 110 of the book Geometry of Algebraic Curves by Arbarello, Cornalba, Griffiths, Harris (Springer, Grundlehren 267).They assume the curve is nondegenerate i.e. not contained in a hyperplane of $\mathbb P^n, n \geq 3$, and you can reduce to this for a non plane curve. For a plane curve the result $dim X_3 \leq 2$ is evident.