# Secant variety of an irreducible non-degenerate projective curve

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.

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What about plane curves? –  M P Aug 10 '12 at 9:54
This isn't a research level question. Problems like this are studied in Harris' "first course" book. Ask at math.stackexchange.com if you're still stuck. –  Jack Huizenga Aug 10 '12 at 9:58
For a joke answer: if it isn't 3-dimensional, then projection from a general codimension 3 subspace gives an isomorphism from the curve to a smooth plane curve of the same degree and genus. But the Castelnuovo bound forbids this. More realistically, try differentiating a map like $X\times X \times \mathbb{A}^1\to Sec X$ in suitable local coordinates. –  Jack Huizenga Aug 10 '12 at 10:10
I'm sorry for this (I just didn't get any answers on the Stack exchange) –  Miklos Aug 10 '12 at 13:26

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.