Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
Surely you mean "trisecAnts", not "trisecEnts", right? I don't know a hell of a lot of algebraic geometry, so maybe that's really a word, but I make spelling errors all the time, so generally assume that a word I don't know is also a spelling error :) In any case, I'd recommend rewriting the title to include the question --- see mathoverflow.net/howtoask#yourtitle . Something like "Given a smooth algebraic curve in $\mathbb{CP}^n$, why is the variety of trisecants at most two-dimensional?" –  Theo Johnson-Freyd Apr 28 '10 at 3:17
    
I cleaned up the title for the OP. –  Andy Putman Apr 28 '10 at 4:24

1 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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