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Let $X$ be a smooth projective variety, let $E$ be a vector bundle of rank $4$ on $X$ and let $L$ be a line budnle on $X$. Consider the projectivization $\mathbb{P}_X(E):=\mathrm{Proj}Sym(E^*)$ of the vector bundle $E$. Denote the projection $\mathbb{P}_X(E)\to X$ by $\pi$ and the Serre sheaf of the projectivization by $\mathcal{O}_{\pi}(1)$. A section of the vector bundle $S^3E^*\otimes L$ defines a family of cubic surfaces $p:\mathcal{S}\to X$ as the zero locus of a section the line bundle $\mathcal{O}_{\pi}(3)\otimes \pi^*L$ on $\mathbb{P}_X(E)$. Let $\Delta\subset X$ be the set of points $x \in X$ such that the cubic surface $\mathcal{S}_x:=p^{-1}(x)$ is singular.

The question: Assume a general cubic surface $\mathcal{S}_x:=p^{-1}(x)$ is smooth. Is it true that $\Delta$ is either a divisor in $X$ or empty? Is there a formula for the divisor $\Delta\subset X$ in terms of $E$ and $L$?

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  • $\begingroup$ To check it is a divisor, we have to locally check it is the vanishing locus of a polynomial. Locally, we may assume $E$ is trivial, so $s$ is given by a cubic polynomial in four variables, and $\Delta$ is the vanishing locus of the discriminant of this cubic polynomial. Is the formula for the discriminant the formula you're looking for, or do you want the formula for the divisor class of $\Delta$? $\endgroup$
    – Will Sawin
    Commented Sep 11, 2021 at 14:00
  • $\begingroup$ Exactly. I need the formula for the divisor $\Delta$. $\endgroup$ Commented Sep 11, 2021 at 14:04

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The discriminant of a degree $d$ polynomial in $n$ variables has degree $n (d-1)^{n-1}$, so the discriminant of a cubic in four variables is $4 \cdot 2^3 = 32$.

The discriminant is, by construction, invariant under $SL_4$. If we look at scalars in $GL_4$, they act on cubic polynomials by multiplication by the inverse cube of the scalar, so they act on polynomials of degree $32$ in the coefficients of a cubic polynomial by multiplication by the inverse $3 \times 32$ power.

Since the discriminant is invariant under $SL_4$ and scalars act by the power $-96$, it must be equivariant under $GL_4$ for the character $\det^{-24}$.

Thus, for a vector bundle $E$, and line bundle $L$, the discriminant of a section of $L \otimes S^3 E^*$ is a section of $L^{32}\otimes \det E^{ - 24}$.

For a consistency check, note that if we tensor $E$ with a line bundle and $L$ with the third power of that line bundle, then both $L \otimes S^3 E^*$ and $L^{32}\otimes \det E^{ - 24}$ are preserved.

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