I would like to know why every nondegenerate irreducible projective curve has a threedimensional secant variety. It is clear to me that the dimension can't be larger.
Thanks for your help!
I would like to know why every nondegenerate irreducible projective curve has a threedimensional secant variety. It is clear to me that the dimension can't be larger. Thanks for your help! 


Suppose that the secant variety of your curve is a surface $S$. This implies that $S$ is covered by lines in such a way that there is a $1$dimensional family of such lines through every general point of $S$. Now let $P$ be a general smooth point on $S$. The lines contained in $S$ through $P$ are necessarily included in the plane $T_P S$. Since there is a $1$dimensional family of such lines, these are exactly the lines in $T_P S$ through $P$. These lines cover $T_P S$ so that $T_P S\subset S$. Since the dimensions agree, and by irreducibility of $S$, $T_P S=S$. This shows that the curve is included in the plane $T_P S$, hence degenerate. 


Let $C\subset\mathbb{P}^{3}$ be a nondegenerate curve. Let us assume that $S=Sec_{2}(C)$ is a surface. Let $p\in S$ be a general point. Since $expdim(Sec_{2}(C))dim(Sec_{2}(C)) = 1$ through $p$ there is a $1$dimensional family of lines contained in $S$. Therefore $S\cong\mathbb{P}^{2}$ and $C$ is a plane curve. A contradiction. 

