Hilbert squares of degeneracy loci

Abstract

Let S be the first degeneracy locus of a morphism of vector bundles corresponding to a general matrix of linear forms in ${\mathbb {P}}^s$. We prove that, under certain positivity conditions, its Hilbert square ${{\mathrm{Hilb}}}^2(S)$ is isomorphic to the zero locus of a global section of an irreducible homogeneous vector bundle on a product of Grassmannians. Our construction involves a naturally associated Fano variety, and an explicit description of the isomorphism.

Appendices

Appendix 1: Euler characteristic of Hilbert squares

The goal of this appendix is to give a detailed proof of Proposition 2.4. We shall exploit a nontrivial Chern class calculation on (smooth) degeneracy loci following Pragacz [27].

Fix $m = 1$ throughout this section. Let $s \in \{3,4\}$, and consider, as ever, a general map $\varphi :{\mathcal {F}} \rightarrow {\mathcal {E}}$ between vector bundles ${\mathcal {F}}={\mathcal {O}}_{{\mathbb {P}}^s}^{\oplus n+1}$ and ${\mathcal {E}}={\mathcal {O}}_{{\mathbb {P}}^s}(1)^{\oplus n}$. The k-th degeneracy locus of $\varphi$ is the closed subscheme $D_k(\varphi ) \subset {\mathbb {P}}^s$ defined by the condition ${{\,\mathrm{rank}\,}}(\varphi ) \le k$, which is (locally) equivalent to the vanishing of the $(k+1)$-minors of $\varphi$. We are interested in the case $k=n-1$, which leads to $D_{n-2}(\varphi )$ of expected codimension 6, and $D_{n-1}(\varphi )$ of expected codimension 2. Since $\varphi$ is general, we have $D_{n-2}(\varphi )=\emptyset$, so that $D_{n-1}(\varphi ) \subset {\mathbb {P}}^s$ is a smooth subvariety of codimension 2. In the case $s=4$, we shall denote it by $S_n \subset {\mathbb {P}}^4$, whereas in the case $s=3$ we shall denote it by $C_n \subset {\mathbb {P}}^3$.

We start assuming $s=4$, the case $s=3$ being essentially a truncation of the case $s=4$. Let $H \in A^1({\mathbb {P}}^4)$ denote the first Chern class of ${\mathcal {O}}_{{\mathbb {P}}^4}(1)$. The ordinary Segre class of ${\mathcal {E}}$ is the class

$$\begin{aligned} {{\widetilde{s}}}({\mathcal {E}}) = \sum _{0\le i\le 4}\widetilde{s}_i({\mathcal {E}}) = (1+H)^{-n}, \end{aligned}$$

with ${{\widetilde{s}}}_i({\mathcal {E}}) \in A^i({\mathbb {P}}^4) = {\mathbb {Z}}[H^i]$ sitting in codimension i. Inverting the Chern class

$$\begin{aligned} c({\mathcal {E}}) = 1+nH+\left( {\begin{array}{c}n\\ 2\end{array}}\right) H^2+\left( {\begin{array}{c}n\\ 3\end{array}}\right) H^3+\left( {\begin{array}{c}n\\ 4\end{array}}\right) H^4 \end{aligned}$$

we find

$$\begin{aligned} {{\widetilde{s}}}_1({\mathcal {E}})&= -c_1({\mathcal {E}})=-nH \\ {{\widetilde{s}}}_2({\mathcal {E}})&= s_1({\mathcal {E}})^2-c_2({\mathcal {E}}) = \left[ n^2-\left( {\begin{array}{c}n\\ 2\end{array}}\right) \right] H^2\\ {{\widetilde{s}}}_3({\mathcal {E}})&= -s_1({\mathcal {E}})c_2({\mathcal {E}})-s_2({\mathcal {E}})c_1({\mathcal {E}})-c_3({\mathcal {E}}) = \left[ -n^3-\left( {\begin{array}{c}n\\ 3\end{array}}\right) +2n\left( {\begin{array}{c}n\\ 2\end{array}}\right) \right] H^3\\ {{\widetilde{s}}}_4({\mathcal {E}})&= -s_1({\mathcal {E}})c_3({\mathcal {E}})-s_2({\mathcal {E}})c_2({\mathcal {E}})-s_3({\mathcal {E}})c_1({\mathcal {E}})-c_4({\mathcal {E}}) \\&= \left[ n^4+2n\left( {\begin{array}{c}n\\ 3\end{array}}\right) -3n^2\left( {\begin{array}{c}n\\ 2\end{array}}\right) +\left( {\begin{array}{c}n\\ 2\end{array}}\right) ^2-\left( {\begin{array}{c}n\\ 4\end{array}}\right) \right] H^4. \end{aligned}$$

We set $s_i = (-1)^i \widetilde{s_i}({\mathcal {E}})$ for $0\le i\le 4$. Then, unraveling [27, Example 5.8 (ii)], we have, for the smooth surface $S_n \subset {\mathbb {P}}^4$, an identity

$$\begin{aligned} e_{\mathrm {top}}(S_n) = s_2c_2({\mathbb {P}}^4)-\left[ s_{(2,1)}+2s_3\right] c_1({\mathbb {P}}^4)+s_{(2,1,1)}+3s_{(3,1)}+3s_4, \end{aligned}$$

(7.1)

given the Schur polynomials

$$\begin{aligned} s_{(2,1)}&= \begin{vmatrix} s_2&s_3 \\ s_0&s_1 \end{vmatrix} = \begin{vmatrix} s_2&s_3 \\ 1&s_1 \end{vmatrix} = s_2s_1-s_3 \\ s_{(3,1)}&= \begin{vmatrix} s_3&s_4 \\ s_0&s_1 \end{vmatrix} = \begin{vmatrix} s_3&s_4 \\ 1&s_1 \end{vmatrix} = s_3s_1-s_4\\ s_{(2,1,1)}&= \begin{vmatrix} s_2&s_3&s_4 \\ s_0&s_1&s_2 \\ 0&s_0&s_1 \end{vmatrix} = \begin{vmatrix} s_2&s_3&s_4 \\ 1&s_1&s_2 \\ 0&1&s_1 \end{vmatrix}=s_2(s_1^2-s_2)-(s_1s_3-s_4). \end{aligned}$$

Expanding, we obtain

$$\begin{aligned} s_2c_2({\mathbb {P}}^4)&= 10n^2-10\left( {\begin{array}{c}n\\ 2\end{array}}\right) \\ \left[ s_{(2,1)} + 2s_3\right] c_1({\mathbb {P}}^4)&= 5(s_2s_1+s_3)H = 10n^3 - 15n\left( {\begin{array}{c}n\\ 2\end{array}}\right) +5\left( {\begin{array}{c}n\\ 3\end{array}}\right) \\ s_{(2,1,1)}&= n\left( {\begin{array}{c}n\\ 3\end{array}}\right) -\left( {\begin{array}{c}n\\ 4\end{array}}\right) \\ 3s_{(3,1)}&= \left( {\begin{array}{c}n\\ 2\end{array}}\right) \left[ 3n^2-3\left( {\begin{array}{c}n\\ 2\end{array}}\right) \right] -3n\left( {\begin{array}{c}n\\ 3\end{array}}\right) +3\left( {\begin{array}{c}n\\ 4\end{array}}\right) \\ 3s_4&= 3n^4+6n\left( {\begin{array}{c}n\\ 3\end{array}}\right) -9n^2\left( {\begin{array}{c}n\\ 2\end{array}}\right) +3\left( {\begin{array}{c}n\\ 2\end{array}}\right) ^2-3\left( {\begin{array}{c}n\\ 4\end{array}}\right) . \end{aligned}$$

Formula (7.1) then yields

$$\begin{aligned} e_{\mathrm {top}}(S_n) =n^2(10-10n+3n^2)+\left( {\begin{array}{c}n\\ 2\end{array}}\right) (-10+15n-6n^2)+\left( {\begin{array}{c}n\\ 3\end{array}}\right) (4n-5)-\left( {\begin{array}{c}n\\ 4\end{array}}\right) . \end{aligned}$$

In the case of a smooth determinantal curve $C_n \subset {\mathbb {P}}^3$, i.e. when we set $s=3$, we only need to use

$$\begin{aligned} s_0=1,\quad s_1=nH,\quad s_2 = \left[ n^2-\left( {\begin{array}{c}n\\ 2\end{array}}\right) \right] H^2,\quad s_3= \left[ n^3+\left( {\begin{array}{c}n\\ 3\end{array}}\right) -2n\left( {\begin{array}{c}n\\ 2\end{array}}\right) \right] H^3. \end{aligned}$$

In this case, [27, Example 5.8 (i)] gives

$$\begin{aligned} e_{\mathrm {top}}(C_n)&= s_2c_1({\mathbb {P}}^3)-s_{(2,1)}-2s_3 = 4Hs_2 - (s_2s_1-s_3) -2s_3 = 4Hs_2 - s_2s_1 - s_3 \\&= 4n^2-4\left( {\begin{array}{c}n\\ 2\end{array}}\right) -n^3+n\left( {\begin{array}{c}n\\ 2\end{array}}\right) -n^3-\left( {\begin{array}{c}n\\ 3\end{array}}\right) +2n\left( {\begin{array}{c}n\\ 2\end{array}}\right) \\&= 4n^2-2n^3+(3n-4)\left( {\begin{array}{c}n\\ 2\end{array}}\right) -\left( {\begin{array}{c}n\\ 3\end{array}}\right) . \end{aligned}$$

The formulas for $e_{\mathrm {top}}(S_n)$ and $e_{\mathrm {top}}(C_n)$ prove Proposition 2.4.

Appendix 2: Hodge–Deligne polynomial of Hilbert squares

We again set $m=1$ throughout this section. We shall consider once more smooth (sub-determinantal) degeneracy loci $S = D_{n-1}(\varphi )\subset {\mathbb {P}}^s$ (of dimension 2 or 3), and we shall compute the Hodge–Deligne polynomial

$$\begin{aligned} E({{\,\mathrm{Hilb}\,}}^2(S);u,v) = \sum _{p,q\ge 0}h^{p,q}({{\,\mathrm{Hilb}\,}}^2(S))(-u)^p(-v)^q \,\in \,{\mathbb {Z}}[u,v] \end{aligned}$$

via standard motivic techniques, exploiting the power structure on the Grothendieck ring of varieties $K_0({{\,\mathrm{Var}\,}}_{{\mathbb {C}}})$ [14], as well as our knowledge of the Hodge numbers of S (cf. Sect. 3).

1.1 2.1. Surface case: $(s,n,m)=(4,4,1)$

Let us consider the smooth determinantal surface $S_4 = D_{3}(\varphi ) \subset {\mathbb {P}}^4$. By Göttsche’s formula [13] for the motive of the Hilbert scheme of points on a surface, combined with the main result of [14], there is an identity

$$\begin{aligned} \sum _{n\ge 0}\left[ {{\,\mathrm{Hilb}\,}}^n (S_4)\right] q^n = \prod _{n>0}\left( 1-{\mathbb {L}}^{n-1}q^n \right) ^{-[S_4]} \end{aligned}$$

in $K_0({{\,\mathrm{Var}\,}}_{{\mathbb {C}}})\llbracket q \rrbracket$, where exponentiation is to be thought of in the language of power structures. The Hodge–Deligne polynomial of a smooth projective ${\mathbb {C}}$-variety Y is the polynomial

$$\begin{aligned} E(Y;u,v) = \sum _{p,q \ge 0}h^{p,q}(Y) (-u)^p(-v)^q \,\in \,\mathbb Z[u,v]. \end{aligned}$$

We have, on ${\mathbb {Z}}[u,v]$, the power structure defined by the identity

$$\begin{aligned} \left( 1-q\right) ^{-f(u,v)} = \prod _{i,j}\left( 1-u^iv^jq\right) ^{-p_{ij}} \end{aligned}$$

if $f(u,v) = \sum _{i,j}p_{ij}u^iv^j$. Looking at the Hodge diamond depicted in Sect. 3.2, we deduce

$$\begin{aligned} E(S_4;u,v) = 1+4u^2+45uv+4v^2+u^2v^2, \end{aligned}$$

and since $E(-)$ defines a morphism $K_0({{\,\mathrm{Var}\,}}_{{\mathbb {C}}}) \rightarrow {\mathbb {Z}}[u,v]$ of rings with power structure sending ${\mathbb {L}}\mapsto uv$, we have an identity

$$\begin{aligned} \sum _{n\ge 0} E({{\,\mathrm{Hilb}\,}}^n(S_4);u,v) q^n&= \prod _{n>0} \left( 1-u^{n-1}v^{n-1}q^n \right) ^{-E(S_4;u,v)} \\&= \prod _{n>0} \left( 1-q\right) ^{-E(S_4;u,v)}\big |_{q \mapsto u^{n-1}v^{n-1}q^n} \\&= \prod _{n>0} \left( 1-u^{n-1}v^{n-1}q^n\right) ^{-1} \left( 1-u^{n+1}v^{n-1}q^n \right) ^{-4}\cdot \\&\qquad \qquad \cdot \left( 1-u^{n}v^{n}q^n \right) ^{-45}\left( 1-u^{n-1}v^{n+1}q^n \right) ^{-4}\left( 1-u^{n+1}v^{n+1}q^n \right) ^{-1} \end{aligned}$$

where the substitution $q \mapsto u^{n-1}v^{n-1}q^n$ is possible thanks to the properties of a power structure.

Expanding and isolating the coefficient of $q^2$ gives

$$\begin{aligned} E({{\,\mathrm{Hilb}\,}}^2(S_4);u,v)= & {} 1+46uv+4(u^2+v^2)+1097u^2v^2+184(uv^3+u^3v)+10(u^4+v^4)\\+ & {} 46u^3v^3+4(u^4v^2+u^2v^4) +u^4v^4, \end{aligned}$$

in full agreement with the Hodge diamond depicted in Sect. 3.2.

1.2 2.2. Threefold case: $(s,n,m)=(5,5,1)$

In the case $(s,n,m)=(5,5,1)$, we obtain a smooth threefold $S_{5,5,1} \subset {\mathbb {P}}^5$ outside the ‘good range’ of Theorem A, cf. Example 7.5. There is an identity [14, 29]

$$\begin{aligned} {\mathsf {Z}}_{S_{5,5,1}}(q) = \sum _{n\ge 0}\,\left[ {{\,\mathrm{Hilb}\,}}^n(S_{5,5,1})\right] q^n=\left( \sum _{n\ge 0}\,\bigl [{{\,\mathrm{Hilb}\,}}^n({\mathbb {A}}^3)_0\bigr ]q^n\right) ^{[S_{5,5,1}]} \end{aligned}$$

in $K_0({{\,\mathrm{Var}\,}}_{{\mathbb {C}}})\llbracket q \rrbracket$, where ${{\,\mathrm{Hilb}\,}}^n({\mathbb {A}}^3)_0$ denotes the punctual Hilbert scheme, namely the subscheme of ${{\,\mathrm{Hilb}\,}}^n({\mathbb {A}}^3)$ parametrising subschemes entirely supported at the origin $0 \in {\mathbb {A}}^3$. Let us define classes $\Omega _n \in K_0({{\,\mathrm{Var}\,}}_{{\mathbb {C}}})$ via the relation

$$\begin{aligned} \sum _{n\ge 0}\bigl [{{\,\mathrm{Hilb}\,}}^n({\mathbb {A}}^3)_0\bigr ]q^n = {{\,\mathrm{Exp}\,}}\left( \sum _{n>0}\Omega _nq^n\right) =\prod _{n>0}\,\left( 1-q^n\right) ^{-\Omega _n}. \end{aligned}$$

Since ${{\,\mathrm{Hilb}\,}}^1({\mathbb {A}}^3)_0={{\,\mathrm{Spec}\,}}{\mathbb {C}}$ and ${{\,\mathrm{Hilb}\,}}^2(\mathbb A^3)_0 = {\mathbb {P}}^2$, one can easily compute $\Omega _1=1$ and $\Omega _2={\mathbb {L}}+{\mathbb {L}}^2$. Therefore

$$\begin{aligned} {\mathsf {Z}}_{S_{5,5,1}}(q) = \prod _{n>0}\, \left( 1-q^n\right) ^{-\Omega _n[S_{5,5,1}]}, \end{aligned}$$

which implies

$$\begin{aligned} \sum _{n\ge 0}E({{\,\mathrm{Hilb}\,}}^n(S_{5,5,1});u,v)q^n = \prod _{n>0}\, \left( 1-q^n\right) ^{-E(\Omega _n;u,v)E(S_{5,5,1};u,v)}. \end{aligned}$$

(B.1)

One can compute the Hodge–Deligne polynomial of $S_{5,5,1}$ to be

$$\begin{aligned} E(S_{5,5,1};u,v) = 1+2uv+2u^2v^2+u^3v^3-(5u^3+151u^2v+151uv^2+5v^3), \end{aligned}$$

so that extracting the coefficient of $q^2$ from (B.1), one obtains

$$\begin{aligned} E({{\,\mathrm{Hilb}\,}}^2(S_{5,5,1});u,v)= & {} \left[ \frac{(1-u^3q)^5(1-v^3q)^5(1-uv^2q)^{151}(1-u^2vq)^{151}}{(1-q)(1-uvq)^2(1-u^2v^2q)^2(1-u^3v^3q)}\right] _{q^2}\\+ & {} (uv+u^2v^2)E(S_{5,5,1};u,v). \end{aligned}$$

In particular, the topological Euler characteristic is

$$\begin{aligned} e_{\mathrm {top}}({{\,\mathrm{Hilb}\,}}^2(S_{5,5,1})) = E({{\,\mathrm{Hilb}\,}}^2(S_{5,5,1});1,1) = 46053 = e_{\mathrm {top}}(Z_{5,5,1})-105. \end{aligned}$$

1.3 B.3. Threefold case: $(s,n,m)=(5,6,1)$

In the case $(s,n,m)=(5,6,1)$, we get a smooth threefold $S_{5,6,1} \subset {\mathbb {P}}^5$. Using the Hodge diamond depicted in Sect. 3.3, one has

$$\begin{aligned} E(S_{5,6,1};u,v)=1+2uv+2u^2v^2+u^3v^3-(29u^3+520u^2v+520uv^2+29v^3). \end{aligned}$$

Formula (B.1) applied to this case yields

$$\begin{aligned} E({{\,\mathrm{Hilb}\,}}^2(S_{5,6,1});u,v)= & {} \left[ \frac{(1-u^3q)^{29}(1-v^3q)^{29}(1-uv^2q)^{520}(1-u^2vq)^{520}}{(1-q)(1-uvq)^2(1-u^2v^2q)^2(1-u^3v^3q)}\right] _{q^2}\\+ & {} (uv+u^2v^2)E(S_{5,6,1};u,v). \end{aligned}$$

In particular,

$$\begin{aligned} e_{\mathrm {top}}({{\,\mathrm{Hilb}\,}}^2(S_{5,6,1})) = E({{\,\mathrm{Hilb}\,}}^2(S_{5,6,1});1,1) = 593502, \end{aligned}$$

in complete agreement with what one gets out of the Hodge diamond for Z depicted in Sect. 3.3.