Parameter space for complete intersections and their discriminant

Question

Consider globally complete intersections in $\mathbb{P}^n$, of codimension $k$, of some fixed multi-degree $(d_1,\dots,d_k)$.

1. Is there some nice (i.e. "explicit") parameter space for them? (even if all the $d_i$'s are distinct one cannot take just the product of linear systems :) Or, does the corresponding locus in the Hilbert scheme admit some nice/explicit description? (a relevant question on this locus in Hilbert scheme)

2. Consider the discriminant in the parameter space of complete intersections, i.e. the locus parametrizing singular c.i.'s. I need standard results: it is an irreducible, reduced hypersurface, singular in codimension one etc. (I guess many of these properties do not depend on the particular choice of parameter space.) Where can I read about this? Somehow I did not find this in Gel'fand-Kapranov-Zelevinsky. And the book of Looijenga treats the local case only.

a somewhat related question

Answer 1

The description of the Hilbert scheme of complete intersections (obtained by taking in an iterative way open subsets of grassmannian bundles, as explained in the answer above) may be found in part 2.2 of my thesis http://www.math.ens.fr/~obenoist/articles/Thesefinale.pdf. If the distinct degrees are $\delta_1<\dots<\delta_r$, at the $i$-th step you construct a grassmannian bundle corresponding to choosing the equations of degree $\delta_i$ up to multiples of the equations of lower degree, and you remove the closed locus where your equations do not define a complete intersection.

Note that this space is smooth. It is not complete unless $k=1$. However, the construction above provides an explicit compactification if $d_1\leq d_2=\dots=d_k$. This compactification is studied in part 2 of arXiv:1111.1589.

The Picard group of this Hilbert scheme is easy to describe (its rank increases by one with each grassmannian bundle and it is not changed when removing the small closed loci).The class of the discriminant in this Picard group (i.e. the degrees of the discriminant) is calculated in arXiv:1009.0704 Théorème 1.3, building on Gel'fand, Kapranov and Zelevinsky's work.

Answer 2

(As Will Sawin has answered the question, in a comment above, I record here the detailed answer. Still, I'd like to know some references where all this is written.)

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(upd: there are several mistakes below, see the answer of O.Benoist. Let's keep this entry, for historical reasons.)

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1. As some of the multi-degrees can coincide, arrange them as follows: $d_1^{(1)}=d_1^{(2)}=\cdots=d_1^{(r_1)}< d_2^{1}=\cdots=d_2^{(r_2)}<\cdots<d_k^{(r_k)}$. Let $GCI_{d^{r_1}_1\cdots d^{r_k}_k}$ denote the parameter space of globally complete intersections of the multidegree as above. This space is the chain of fibrations as follows.

$GCI_{d^{(r_1)}_1}=Gr(\mathbb{P}^{r_1-1},|\mathcal{O}_{\mathbb{P}^n}(d_1)|)$. And the forgetful projection $GCI_{d^{(r_1)}_1\cdots d^{(r_{j+1})}_{j+1}}\to GCI_{d^{(r_1)}_1\cdots d^{(r_{j})}_{j}}$ is the projectivization of the vector bundle. Its fibre over a point $I=(f_{1,d_1},\dots,f_{r_1,d_1},f_{1,d_2},\dots,f_{r_2,d_2},\dots,f_{r_j,d_j})\subset k[x_0,\dots,x_n]$ of $GCI_{d^{(r_1)}_1\cdots d^{(r_{j})}_{j}}$ is: $Gr(\mathbb{P}^{r_{j+1}-1},\mathbb{P}(H^0(\mathcal{O}_{\mathbb{P}^n}(d_{j+1}))/I[d_{j+1}])$ . Here $I[d_{j+1}]$ is the homogeneous part of $I$ consisting of polynomials of degree $d_{j+1}$.

This is a smooth and not too complicated parameter space.

2. Many properties of the discriminant, $\Delta\subset GCI_{d^{r_1}_1\cdots d^{r_k}_k}$ can be obtained from the local consideration (Looijenga's book). It is a reduced divisor. To prove the irreducibility one considers the standard Nash modification, i.e. the incidence variety, $\tilde{\Delta}\subset \mathbb{P}^n\times GCI_{d^{r_1}_1\cdots d^{r_k}_k}$, consisting of pairs: the complete intersection and (one of) its singular point. The projection $\tilde{\Delta}\to\Delta$ is birational, while all the fibres of the projection onto $\mathbb{P}^n$ are linear spaces. Thus $\tilde{\Delta}$ is smooth, in particular $\Delta$ is irreducible.