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Let $C$ be a smooth, complex, irreducible, nondegenerate curve of degree $d$ and genus $g$ in $\mathbb{P}^3$. What is the class of the variety of lines that are secant to $C$?

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  • $\begingroup$ What do you mean by "class" here? $\endgroup$ – Hoot Aug 17 '15 at 0:02
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I guess "class" means homology (or cohomology) class in $H_4(\mathbf G, \mathbf Z)$ where here $\mathbf G = \mathbf G(1,3)$ is the Grassmannian of lines in $\mathbf P^3$.

The homology group $H_4(\mathbf G, \mathbf Z)$ has a basis consisting of the Schubert cells $\sigma_{2,0}$ and $\sigma_{1,1}$, where $\sigma_{2,0}$ is the subvariety of all lines passing through a given point, and $\sigma_{1,1}$ is the subvariety of all lines contained in a given plane $\mathbf P^2 \subset \mathbf P^3$.

There are rules (called Pieri's formula and Giambelli's formula) for calculating the intersections of arbitrary Schubert cells in any Grassmannian. Here we see the answer by more elementary means (exercise): we get

$$ \sigma_{2,0}^2= \sigma_{1,1}^2=1\\ \sigma_{2,0} \cdot \sigma_{1,1}=0.$$

So to figure out the coefficients of the expression

$$[S] = a \, \sigma_{2,0} + b \, \sigma{1,1}$$

for the class of your subvariety $S$, we just need to calculate $[S] \cdot \sigma_{2,0}$ and $S \cdot \sigma_{1,1}$ and solve some linear equations.

Now $[S] \cdot \sigma_{1,1}$ is the number of secant lines in a general plane, which (exercise) equals ${n \choose 2}$; $[S] \cdot \sigma_{2,0}$ is the number of secant lines passing through a general point. You can find that by projecting away from the point: the image of your curve is now a plane curve of degree $d$ with some number of nodes that you can calculate (exercise) in terms of $g$. Nodes on the plane curve correspond to secant lines through the projection point, and so you have all the information you need.

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