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Let $C\subset\mathbb{P}^{3}$ be a smooth, non-degenerate curve over an algebraically closed field of characteristic zero. Let $d$ be the degree of $C$ and $g$ be its genus.

Consider the variety $S_{3}\subset\mathbb{G}(1,3)$, in the Grassmannian of lines in $\mathbb{P}^{3}$, parametrizing lines which are $3$-secant to $C$. Then $dim(S_3) = 1$.

Does there exists a formula involving $d$ and $g$ for the geometric genus $g(S_{3})$ ?

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There is a classical formula, due to Cayley, for the degree of $S_3$, as a curve in $\mathbb{G}(1,3)\subset \mathbb{P}^5$: it is equal to $\frac{1}{3} (d-1)(d-2)(d-3)-g(d-2)$. For a modern proof, see for intance P. Le Barz, Formules multisécantes pour les courbes gauches quelconques, in Enumerative Geometry and Classical Geometry, PM 24, Birkhäuser.

I would bet that there is no formula for the genus, i.e. that different components of the Hilbert scheme (for the same $d$ and $g$) give different values for $g(S_3)$. It would be interesting to check it in the lowest example of curves of degree $9$ and genus $10$.

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In this paper:

http://www.ams.org/journals/tran/1986-295-01/S0002-9947-1986-0831191-3/S0002-9947-1986-0831191-3.pdf

you could find some informations on the local structure of $S_3$. In such a generality it could be hard to give a formula for the genus. If $C\subset\mathbb{P}^{3}$ is a general projection of a rational normal curve of degree $4$ in $\mathbb{P}^{4}$, then $S_{3}\cong\mathbb{P}^{1}$.

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    $\begingroup$ Why do you say that $S_3$ equals $\mathbb{P}^1$ for a twisted cubic? I believe that $S_3$ should be empty for the twisted cubic. $\endgroup$ – Jason Starr Feb 25 '14 at 11:26
  • $\begingroup$ Yes, you are right. I meant $C\subset\mathbb{P}^{3}$ a general projection of a rational normal curve of degree $4$ in $\mathbb{P}^{4}$. Therefore degree 4 and genus 0. $\endgroup$ – A. M. Flat_Line Feb 25 '14 at 13:56

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