I would like to know why every non-degenerate irreducible projective curve has a three-dimensional secant variety. It is clear to me that the dimension can't be larger.


Thanks for your help!



Secant variety of an irreducible non-degenerate projective curve
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.


