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!

  • $\begingroup$ What about plane curves? $\endgroup$
    – M P
    Aug 10 '12 at 9:54
  • $\begingroup$ 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. $\endgroup$ Aug 10 '12 at 9:58
  • $\begingroup$ 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. $\endgroup$ Aug 10 '12 at 10:10
  • $\begingroup$ I'm sorry for this (I just didn't get any answers on the Stack exchange) $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.