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 $g(S\_{3})$ ?



Curve of 3-secant lines
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 $10$. 


