Branch locus of a 6:1 cover of the grassmannian G(1,3) - MathOverflow

Branch locus of a 6:1 cover of the grassmannian G(1,3)

2012-12-10T11:45:17Z

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?

Answer by IMeasy for Branch locus of a 6:1 cover of the grassmannian G(1,3)

2012-12-10T13:20:55Z

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).

Answer by Sasha for Branch locus of a 6:1 cover of the grassmannian G(1,3)

2012-12-10T16:52:55Z

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).