# Dimension of Secant variety of an irreducible projective curve not contained in a plane

I'm trying to show that this dimension is three, but I'm stuck. Could anyone give me a hint?

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It's great to know that your curve is not contained in a plane, but where exactly is it? And what does curve mean, by the way? – Georges Elencwajg Aug 8 '12 at 20:21
@georges It is exercise $11.25$ in Harris's algebraic geometry book:"Show that if $X\subset\mathbb{P}^{n}$ is an irreducible curve then its chordal variety $S(X)$ is three-dimensional unless $X$ is contained in a plane." – Srinivasa Granujan Mar 8 '15 at 17:18
@Gauloises Doesn't it follow from Proposition 11.24, which says $\operatorname{dim} S(X) \le 3$? You only have to show that if $X$ is contained in a plane, then for a general point $p \in \overline{qr}$ lying on a secant line to $X$ lies on an infinite finite number of secant lines to $X$. If $X$ is linear, then $\operatorname{dim} S(X) = 1$ trivially. If not, then should be infinitely many points on $X \setminus \overline{qr}$ that determine distinct secant lines through $r$. – Takumi Murayama Mar 15 '15 at 16:35