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!
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Sign up to join this communityI 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!
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 non-degenerate 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.