I'm trying to show that this dimension is three, but I'm stuck. Could anyone give me a hint?
Dimension of Secant variety of an irreducible projective curve not contained in a plane
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algebraic-geometry
algebraic-curves
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0@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 $r$. – 2015-03-15