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Given a general quartic surface $S$ in $\mathbf{P}^3$, there is a natural 6:1 surjective map $\phi: Hilb^2(S) \to G(1,3)$ sending $\{P,Q\}$ to the line through them in $\mathbf{P}^3$.

Can you describe the branch locus of $\phi$ in terms of Schubert classes?

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up vote 11 down vote accepted

Since you are interested in a divisor, you only need to know its degree, that is its intersection with a line. A generic line on $Gr(1,3)$ is given by the set of all lines contained in a plane $P$ and passing through a point $Q$. So, you want to know how many tangents to $S$ pass through $Q$ and lie in $P$.

Consider the intersection $S_P = S \cap P$. Since $P$ is generic $S_P$ is a quartic curve. The number of tangents passing through generic point is nothing but the degree of the projectively dual curve which is known to be $d(d-1) = 4\cdot 3 = 12$.

So, the answer is that the branch locus is given by $12\sigma_1$ (honestly, I don't remember whether the standard notation for the Schubert class of codimension 1 is $\sigma_1$ or not).

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I assume the surface smooth since you take it general. The branch locus is given by the lines that cut out on $S$ divisors of type $2p+q+t$, for any $p,q,t \in S$. Computing the exact shubert classes requires a little more time (and work!) but it should work using the standard exact sequences on G(1,3).

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As you point out, the branch locus $B$ is the set of tangent lines to $S$, so it is the image of the projectivized tangent bundle of $S$ via a birational maps. $B$ has a 2-dimensional double locus, corresponding to the image of the bitangent lines. Also, the map factors through the involution of $Hilb$ that maps $p,q\in S$ to the remainining 2 points on the line joining $p$ and $q$. Maybe this can help with the computation. – rita Dec 10 '12 at 13:54

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