# Hodge classes of type $(2,2)$ on Hilbert squares of projective K3 surfaces

Simone Novario  Università degli Studi di Milano, Dipartimento di Matematica
“F. Enriques”, Via Cesare Saldini 50, 20133 Milano, Italy.
Université de Poitiers, Laboratoire de Mathématiques et Applications, Téléport
2, Boulevard Marie et Pierre Curie, 86962 Futuroscope Chasseneuil, France.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We give a basis for the vector space generated by rational Hodge classes of
type $(2,2)$ on the Hilbert square of a projective K3 surface general in its
rank, which is a subspace of the singular cohomology ring with rational
coefficients: we use Nakajima operators and an algebraic model developed by
Lehn and Sorger as main tools. We then obtain a basis of the lattice generated
by integral Hodge classes of type $(2,2)$ on the Hilbert square of a
projective K3 surface general in its rank: we exploit lattice theory, a
theorem by Qin and Wang and a result by Ellingsrud, Göttsche and Lehn.

##  1\. Introduction

Irreducible holomorphic symplectic (IHS) manifolds are a generalization to
higher dimensions of K3 surfaces. The interest in these varieties has been
increasing thanks to the Beauville–Bogomolov decomposition theorem: up to a
finite étale cover, any compact Kähler manifold with trivial first Chern class
is the product of a complex torus, Calabi–Yau manifolds and IHS manifolds, see
[Bea83, Théorème 2]. In this paper we deal with Hilbert schemes of $2$ points
on a K3 surface, also known as _Hilbert squares of K3 surfaces_ , the first
example of IHS manifold other than K3 surfaces to be found, see [Fuj87]: if
$S$ is a K3 surface, we denote by $S^{[2]}$ its Hilbert square. There exists a
quadratic form on the second cohomology group $H^{2}(S^{[2]},\mathbb{Z})$,
called _Beauville–Bogomolov–Fujiki_ (BBF) form and denoted by $q_{S^{[2]}}$,
so that $(H^{2}(S^{[2]},\mathbb{Z}),q_{S^{[2]}})$ is a lattice.

If $S$ is a K3 surface, we denote by $T(S)=\textnormal{NS}(S)^{\perp}$ the
_transcendental lattice_ of $S$, where $\textnormal{NS}(S)$ is the
Néron–Severi group of $S$ and the orthogonal is taken with respect to the
intersection product. Moreover, $T(S)_{\mathbb{Q}}=T(S)\otimes\mathbb{Q}$ is a
rational Hodge structure of weight $2$, and we denote by

$E_{S}:=\text{Hom}_{0}(T(S)_{\mathbb{Q}},T(S)_{\mathbb{Q}})$

the algebra of endomorphisms on $T(S)_{\mathbb{Q}}$ of weight $0$. See [Huy16,
$\S 3$] for details on Hodge structures. Zahrin showed in [Zar83] that $E_{S}$
is either $\mathbb{Q}$, or a totally real field or a CM field. We say that $S$
is _general in its rank_ , or simply _general_ , if $E_{S}\cong\mathbb{Q}$. As
remarked in [vG08], see also [EJ16, $\S 1.1$], K3 surfaces which are not
general gives a locus of positive codimension in the analytic moduli space of
K3 surfaces of fixed Picard rank, unless the Picard rank is $20$, in which
case $E_{S}$ is always a CM field.

In this paper we study the $\mathbb{Q}$-vector space of _rational Hodge
classes_ of type $(2,2)$ and the lattice of _integral Hodge classes_ of type
$(2,2)$ on the Hilbert square of a general projective K3 surface $S$, defined
respectively as

$H^{2,2}(S^{[2]},\mathbb{Q}):=H^{4}(S^{[2]},\mathbb{Q})\cap
H^{2,2}(S^{[2]}),\qquad
H^{2,2}(S^{[2]},\mathbb{Z}):=H^{4}(S^{[2]},\mathbb{Z})\cap H^{2,2}(S^{[2]}).$

The integral bilinear form considered on $H^{2,2}(S^{[2]},\mathbb{Z})$ is the
cup product. Hodge classes are usually studied in the context of the so-called
_Hodge conjecture_ : this states that given a smooth complex projective
variety $Y$, the subspace of $H^{2k}(Y,\mathbb{Q})$ generated by algebraic
cycles, i.e., classes which are obtained as fundamental cohomological classes
$[Z]$ of subvarieties $Z\subset Y$, coincides with $H^{k,k}(Y,\mathbb{Q})$,
which is by definition the set $H^{2k}(Y,\mathbb{Q})\cap H^{k,k}(Y)$ of
_rational Hodge classes_ of type $(k,k)$. Similarly one defines _integral
Hodge classes_ of type $(k,k)$ as the elements which belong to the set
$H^{k,k}(Y,\mathbb{Z}):=H^{2k}(Y,\mathbb{Z})\cap H^{k,k}(Y)$. We now present
the main result of this paper, which gives a basis of the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$, where $S$ is a general projective K3 surface
whose Picard group is known, cf. Theorem 8.3.

######  Theorem.

Let $S$ be a general projective K3 surface and let $\\{b_{1},\dots,b_{r}\\}$
be a basis of $\textnormal{Pic}(S)$. Then the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$ is odd. Moreover,
$\textnormal{rk}(H^{2,2}(S^{[2]},\mathbb{Z}))=\frac{(r+1)r}{2}+r+2$, and a
basis of $H^{2,2}(S^{[2]},\mathbb{Z})$ is the following:

$\left\\{b_{i}b_{j},\frac{b_{i}^{2}-b_{i}\delta}{2},\frac{1}{8}\left(\delta^{2}+\frac{1}{3}c_{2}(S^{[2]})\right),\delta^{2}\right\\}_{1\leq
i\leq j\leq r,}$ (1.1)

where $c_{2}(S^{[2]})\in H^{2,2}(S^{[2]},\mathbb{Z})$ is the second Chern
class of $S^{[2]}$.

Clearly the elements in (1.1) give a basis also for the $\mathbb{Q}$-vector
space $H^{2,2}(S^{[2]},\mathbb{Q})$: a basis in terms of Nakajima operators is
described in Theorem 6.2. Topics discussed in this paper are related to
[She16], where Shen studies Hyperkähler manifolds of Jacobian type.

The paper is organised as follows. In Section 2 we recall some notions of
lattice theory. In Section 3 we give the definition of IHS manifold, together
with general properties. In Section 4 we present some useful results which
hold for an IHS manifold $X$ of $K3^{[2]}$-type, in particular the relation
between the BBF bilinear form and the cup product between elements in
$H^{4}(X,\mathbb{Q})$, and the definition of the dual $q_{X}^{\vee}\in
H^{2,2}(X,\mathbb{Q})$ of the BBF quadratic form. In Section 5 we introduce
Nakajima operators, following [Nak97] and [Leh99], and the algebraic model
developed by Lehn and Sorger in [LS03]. In Section 6, using Nakajima operators
and the model of Lehn–Sorger, we prove Theorem 6.2, which gives a basis of the
$\mathbb{Q}$-vector space of rational Hodge classes of type $(2,2)$ on the
Hilbert square of a general projective K3 surface whose Picard group is known:
the basis is presented in terms of Nakajima operators. In Section 7, using a
result by Ellingsrud, Göttsche and Lehn in [EGL01], we obtain an explicit
description of the second Chern class of the Hilbert square of a projective K3
surface depending on Nakajima operators, cf. Proposition 7.4. We use this
description in Section 8, where we pass to study integral Hodge classes of
type $(2,2)$ on the Hilbert square of a general projective K3 surface. The
main result of this section is Theorem 8.3, which describes a basis for
$H^{2,2}(S^{[2]},\mathbb{Z})$ for a general projective K3 surface $S$: the
strategy of the proof is to combine Proposition 7.4 with a result by Qin and
Wang in [QW05], together with the property that the lattice
$H^{2}(S,\mathbb{Z})$ is unimodular.

A _generic K3 surface of degree_ $2t$ is a general projective K3 surface
$S_{2t}$ whose Picard group is generated by the class of an ample divisor $H$
with $H^{2}=2t$ with respect to the intersection form. This paper is based on
Chapter 3 of the author’s PhD thesis, where Theorem 8.3 is used to show that
Hilbert squares of general K3 surfaces of degree $2t$, with $t\neq 2$,
admitting an ample divisor $D$ with $q_{X}(D)=2$ are double EPW sextics, see
[O’G06] for the definition, where $q_{X}$ represents the BBF quadratic form on
$X$. Proofs and details that will be omitted in this paper can be found in the
author’s PhD thesis, see [Nov].

Acknowledgements: I thank Samuel Boissière for his constant support and for
useful clarifications on the theory of Nakajima operators. I thank Bert van
Geemen for many useful discussions. I thank Simon Brandhorst for precious
clarifications about lattice theory and the software Sage. I deeply thank
Olivier Benoist and Daniel Huybrechts for useful comments, summarised in
Remark 6.3, on a first draft of this paper, which improved the statement of
the main Theorem. I also thank Pietro Beri and Ángel David Ríos Ortiz for many
useful discussions.

##  2\. Lattices

In this section we recall the most important definitions and results of
lattice theory that we need. General references are [Nik80] and [CS13].

######  Definition 2.1.

A _lattice_ $L$ is a free $\mathbb{Z}$-module of finite rank together with a
symmetric bilinear form $b:L\times L\rightarrow\mathbb{Z}$. We denote by
$q:L\rightarrow\mathbb{Z}$ the quadratic form defined by $q(x):=b(x,x)$ for
every $x\in L$.

If $L$ is a lattice of rank $n$ and $\mathcal{B}:=\\{e_{1},\dots,e_{n}\\}$ is
a $\mathbb{Z}$-basis of $L$, we call _Gram matrix_ of $L$ associated to
$\mathcal{B}$ the following symmetric matrix:

$\begin{pmatrix}b(e_{1},e_{1})&\cdots&b(e_{1},e_{n})\\\
\vdots&\ddots&\vdots\\\ b(e_{n},e_{1})&\cdots&b(e_{n},e_{n})\end{pmatrix}.$

A _non-degenerate_ lattice is a lattice $L$ of rank $n$ such that for any non-
zero $l\in L$ there exists $l^{\prime}\in L$ such that $b(l,l^{\prime})\neq
0$, equivalently, $\text{det}(G)\neq 0$ if $G$ is a Gram matrix of $L$. A
lattice $L$ is _even_ if $b(l,l)\in 2\mathbb{Z}$ for every $l\in L$, and _odd_
if it is not even. A _sublattice_ of a lattice $L$ is a free submodule
$L^{\prime}\subseteq L$ with symmetric bilinear form
$b^{\prime}:=b|_{L^{\prime}\times L^{\prime}}$. A sublattice
$L^{\prime}\subseteq L$ is _primitive_ if $L/L^{\prime}$ is a free module. The
_direct sum_ of two lattices $L_{1}$ and $L_{2}$ is by definition the lattice
$L_{1}\oplus L_{2}$ whose bilinear form is
$b(v_{1}+v_{2},w_{1}+w_{2}):=b_{1}(v_{1},w_{1})+b_{2}(v_{2},w_{2})$ for every
$v_{1},w_{1}\in L_{1}$ and $v_{2},w_{2}\in L_{2}$, where $b_{1}$ and $b_{2}$
are the bilinear forms of $L_{1}$ and $L_{2}$ respectively.

For a lattice $L$ of rank $n$ we write
$L_{\mathbb{R}}:=L\otimes_{\mathbb{Z}}\mathbb{R}$ and we extend
$\mathbb{R}$-bilinearly the bilinear form $b$ to $L_{\mathbb{R}}$, similarly
we extend $q$ to $L_{\mathbb{R}}$. If the lattice is non-degenerate, the
_signature_ of $L$ is the signature $(l_{(+)},l_{(-)})$ of the quadratic form
on $L_{\mathbb{R}}$. A non-degenerate lattice is _positive definite_ if
$l_{(-)}=0$, similarly it is _negative definite_ if $l_{(+)}=0$, while it is
_indefinite_ if $l_{(+)},l_{(-)}\neq 0$.

The _determinant_ of a lattice $L$ is the determinant of a Gram matrix $G$ of
the lattice, and the _discriminant_ of $L$ is
$\text{disc}(L):=|\text{det}(G)|$. A _unimodular_ lattice is a lattice $L$
such that $\text{disc}(L)=1$: if $L$ is a unimodular lattice, for every $x\in
L$ there exists $y\in L$ such that $b(x,y)=1$. Let $k$ be an integer: the
easiest example of lattice is $\langle k\rangle$, which is the rank one
lattice $L=\mathbb{Z}e$ with bilinear form $b(e,e)=k$. Other two important
examples of lattices are the _hyperbolic lattice_ , which is denoted by $U$,
and $E_{8}(-1)$: the former is the unique unimodular lattice of rank $2$ and
signature $(1,1)$, the latter is an even unimodular lattice of signature
$(0,8)$. The following two matrices are Gram matrices respectively for $U$ and
$E_{8}(-1)$:

$\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\qquad\begin{pmatrix}-2&1&&&&&&\\\
1&-2&1&&&&&\\\ &1&-2&1&&&&1\\\ &&1&-2&1&&&\\\ &&&1&-2&1&&\\\ &&&&1&-2&1&\\\
&&&&&1&-2&\\\ &&1&&&&&-2\\\ \end{pmatrix}.$ (2.1)

A _morphism of lattices_ $\varphi:L\rightarrow L^{\prime}$ between two
lattices $L$ and $L^{\prime}$ whose bilinear forms are respectively $b$ and
$b^{\prime}$ is a morphism of $\mathbb{Z}$-modules such that for every
$l_{1},l_{2}\in L$ we have
$b(l_{1},l_{2})=b^{\prime}(\varphi(l_{1}),\varphi(l_{2}))$. Morphisms between
two non-degenerate lattices are necessarily injective. We call _isometry_ a
bijective morphism of lattices. A lattice $L$ _embeds primitively_ in a
lattice $L^{\prime}$ if there is a morphism $\varphi:L\rightarrow L^{\prime}$
such that $\varphi(L)$ is a primitive sublattice of $L^{\prime}$.

##  3\. Generalities on IHS manifolds

In this section we recall basics on irreducible holomorphic symplectic
manifolds.

######  Definition 3.1.

An _irreducible holomorphic symplectic_ (IHS) _manifold_ is a simply connected
compact complex Kähler manifold $X$ such that $H^{0}(X,\Omega^{2}_{X})$ is
generated by a non-degenerate holomorphic $2$-form, called _symplectic form_.

The dimension of an IHS manifold is necessarily even as a consequence of the
existence of a symplectic form. The Enriques–Kodaira classification of compact
complex surfaces shows that the only IHS manifolds of dimension $2$ are K3
surfaces. If $X$ is an IHS manifold, then the $\mathbb{C}$-vector space
$H^{0}(X,\Omega^{p}_{X})$ is zero if $p$ is odd, and it is generated by
$\sigma^{\frac{p}{2}}$ if $0\leq p\leq\textnormal{dim}(X)$ is even, where
$\sigma$ is a symplectic form, see [Bea83, Proposition 3]. The Picard group
$\textnormal{Pic}(X)$ is isomorphic to the Néron–Severi group
$\textnormal{NS}(X)=H^{1,1}(X)_{\mathbb{R}}\cap H^{2}(X,\mathbb{Z})$, and this
embeds in the second cohomology group $H^{2}(X,\mathbb{Z})$. By the universal
coefficient theorem, the second singular cohomology group
$H^{2}(X,\mathbb{Z})$ is torsion free. Moreover, by a result due to Beauville,
Bogomolov and Fujiki, $H^{2}(X,\mathbb{Z})$ can be equipped with a non-
degenerate integral quadratic form, denoted by $q_{X}$ and called
_Beauville–Bogomolov–Fujiki_ (BBF) form, see [Bea83] and [Fuj87] for details:
in particular $(H^{2}(X,\mathbb{Z}),q_{X})$ has a structure of even lattice of
signature $(3,b_{2}(X)-3)$, where $b_{2}(X)$ is the second Betti number of
$X$. For K3 surfaces the BBF form coincides with the intersection form. The
_transcendental lattice_ of an IHS manifold $X$ is defined as
$T(X):=(\textnormal{NS}(X))^{\perp}$: the orthogonal is taken with respect to
the BBF form.

Let $S$ be a K3 surface: we denote by $S^{[n]}$ the _Hilbert scheme of_ $n$
_points on_ $S$, which is the scheme which parametrises zero-dimensional
closed subschemes of length $n$ on $S$. By [Bea83, Théorème 3] the variety
$S^{[n]}$ is an IHS manifold of dimension $2n$. If $S^{(n)}$ is the quotient
of $S^{n}=S\times\dots\times S$ by the symmetric group of $n$ elements, the
morphism $\rho:S^{[n]}\rightarrow S^{(n)}$ defined by
$\rho([\xi])=\sum_{x}l(\mathcal{O}_{\xi,x})\cdot x$ is called _Hilbert–Chow
morphism_ , see for instance [Ive06]. In particular $\rho$ is a
desingularisation of $S^{(n)}$, whose singular locus is the so-called
diagonal, i.e., the set of cycles $p_{1}+\dots+p_{n}$ such that there exist
$i$ and $j$ with $i\neq j$ and $p_{i}=p_{j}$. The pre-image of the diagonal in
$S^{(n)}$ is an irreducible divisor $E$ on $S^{[n]}$, and there exists a
primitive class $\delta\in\textnormal{Pic}(S^{[n]})$ such that $2\delta=[E]$.
An important fact is the following: there exists a primitive embedding of
lattices

$i:H^{2}(S,\mathbb{Z})\hookrightarrow H^{2}(S^{[n]},\mathbb{Z})$

such that
$H^{2}(S^{[n]},\mathbb{Z})=i(H^{2}(S,\mathbb{Z}))\oplus\mathbb{Z}\delta$, and
$q_{S^{[n]}}(\delta)=-2(n-1)$. For a K3 surface $S$ there exists an isometry
of lattices $H^{2}(S,\mathbb{Z})\cong U^{\oplus 3}\oplus E_{8}(-1)^{\oplus
2}$, see for instance [BHPVdV15, $\S\text{VII.3}$], in particular
$H^{2}(S,\mathbb{Z})$ is an even unimodular lattice of signature $(3,19)$.
Hence there is an isometry of lattices

$H^{2}(S^{[n]},\mathbb{Z})\cong U^{\oplus 3}\oplus E_{8}(-1)^{\oplus
2}\oplus\langle-2(n-1)\rangle,$

similarly
$\textnormal{Pic}(S^{[n]})=i(\textnormal{Pic}(S))\oplus\mathbb{Z}\delta$: see
[Bea83, $\S 6$] for details. Moreover, the singular cohomology ring
$H^{*}(S^{[n]},\mathbb{Z})$ for a K3 surface $S$ and $n\geq 1$ is torsion free
by [Mar07, Theorem 1]. When $n=2$, we call $S^{[2]}$ the _Hilbert square of_
$S$. An IHS manifold which is deformation equivalent to the Hilbert square of
a K3 surface is said to be of $K3^{[2]}$-_type_.

The other known examples of IHS manifolds up to deformation equivalence are
_generalised Kummer varieties_ , see [Bea83, $\S 7$], an isolated example of
dimension $10$ and second Betti number $b_{2}=24$, see [O’G99], and an
isolated example of dimension $6$ and second Betti number $b_{2}=8$, see
[O’G03]. We do not discuss details on these examples since in this paper we
deal only with Hilbert squares of K3 surfaces.

##  4\. Generalities on IHS manifolds of $K3^{[2]}$-type

Let $X$ be an IHS manifold of dimension $4$ of $K3^{[2]}$-type. In this
section we recall some useful properties from [O’G08, $\S 2$], in particular
the link between the intersection pairing on $H^{4}(X,\mathbb{Q})$ and the BBF
form on $H^{2}(X,\mathbb{Q})$.

First of all, we state the following corollary of Verbitsky’s results in
[Ver96], obtained by Guan in [Gua01], see also [O’G10, Corollary 2.5]. We
denote by $b_{i}(X)$ the $i$-th Betti number of $X$.

######  Proposition 4.1.

Let $X$ be an IHS manifold of dimension $4$. Then $b_{2}(X)\leq 23$. If
equality holds then $b_{3}(X)=0$ and the map

$\textnormal{Sym}^{2}H^{2}(X,\mathbb{Q})\rightarrow H^{4}(X,\mathbb{Q})$

induced by the cup product is an isomorphism. In particular this happens when
$X$ is an IHS fourfold of $K3^{[2]}$-type.

We denote by $\langle\,\cdot\,,\,\cdot\,\rangle$ the intersection pairing on
the singular cohomology group $H^{4}(X,\mathbb{Z})$ induced by the cup product
and defined by $\langle\alpha,\beta\rangle:=\int_{X}\alpha\beta$. We use the
same notation $\langle\,\cdot\,,\,\cdot\,\rangle$ for the
$\mathbb{Q}$-bilinear extension of the intersection pairing above to
$H^{4}(X,\mathbb{Q})$, obtaining a $\mathbb{Q}$-valued intersection pairing on
$\textnormal{Sym}^{2}H^{2}(X,\mathbb{Q})$. Let $X$ be an IHS manifold of
$K3^{[2]}$-type: then the intersection pairing
$\langle\,\cdot\,,\,\cdot\,\rangle$ can be described in terms of the
$\mathbb{Q}$-extension of the BBF bilinear form on $H^{2}(X,\mathbb{Q})$
thanks to the following Proposition, see [O’G08, Remark 2.1].

######  Proposition 4.2 (O’Grady).

Let $X$ be an IHS fourfold of $K3^{[2]}$-type. The intersection pairing
$\langle\,\cdot\,,\,\cdot\,\rangle$ defined above coincides with the bilinear
form on $\textnormal{Sym}^{2}H^{2}(X,\mathbb{Q})$ given by

$\langle\alpha_{1}\alpha_{2},\alpha_{3}\alpha_{4}\rangle=(\alpha_{1},\alpha_{2})(\alpha_{3},\alpha_{4})+(\alpha_{1},\alpha_{3})(\alpha_{2},\alpha_{4})+(\alpha_{1},\alpha_{4})(\alpha_{2},\alpha_{3})$

for every $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in
H^{2}(X,\mathbb{Q})$, where $(\,\cdot\,,\,\cdot\,)$ denotes the BBF bilinear
form on $H^{2}(X,\mathbb{Q})$.

Let $q_{X}$ be the BBF quadratic form on $X$ and $\\{e_{1},\dots,e_{23}\\}$ be
a basis of $H^{2}(X,\mathbb{Q})$. Let $\\{e_{1}^{\vee},\dots,e_{23}^{\vee}\\}$
be the dual basis in $H^{2}(X,\mathbb{Q})^{\vee}$, i.e.,
$e_{i}^{\vee}(e_{j})=\delta_{i,j}$. Then we have

$q_{X}=\displaystyle\sum_{i,j}g_{i,j}e_{i}^{\vee}\otimes e_{j}^{\vee},\qquad
q_{X}^{\vee}=\displaystyle\sum_{i,j}m_{i,j}e_{i}e_{j},$

where $g_{i,j}:=(e_{i},e_{j})$, the matrix $(g_{i,j})$ is symmetric and
$(m_{i,j})=(g_{i,j})^{-1}$. As shown in [O’G08, Proposition 2.2], the products
$\langle q_{X}^{\vee},\alpha\rangle$ for every $\alpha\in H^{4}(X,\mathbb{Q})$
can be computed exploiting the BBF bilinear form on $H^{2}(X,\mathbb{Q})$.

######  Proposition 4.3 (O’Grady).

Let $X$ be an IHS fourfold of $K3^{[2]}$-type. Let
$\langle\,\cdot\,,\,\cdot\,\rangle$ be the bilinear form described in
Proposition 4.2. Then $\langle\,\cdot\,,\,\cdot\,\rangle$ is non-degenerate
and

$\begin{array}[]{l}\langle
q_{X}^{\vee},\alpha\beta\rangle=25(\alpha,\beta)\qquad\text{for
all}\,\,\alpha,\beta\in H^{2}(X,\mathbb{Q}),\\\\[4.30554pt] \langle
q_{X}^{\vee},q_{X}^{\vee}\rangle=23\cdot 25.\end{array}$

O’Grady has shown in [O’G08, $\S 3$] that $q_{X}^{\vee}$ is a rational
multiple of $c_{2}(X)$, the second Chern class of the tangent bundle of $X$,
in particular it is an element of $H^{2,2}(X,\mathbb{Q})$, i.e., it is a
rational Hodge class of type $(2,2)$.

######  Proposition 4.4 (O’Grady).

Let $X$ be an IHS fourfold of $K3^{[2]}$-type. Then $q_{X}^{\vee}\in
H^{2,2}(X,\mathbb{Q})$, i.e., $q_{X}^{\vee}$ is a rational Hodge class of $X$
of type $(2,2)$, and

$\frac{6}{5}q_{X}^{\vee}=c_{2}(X)\in H^{2,2}(X,\mathbb{Z}).$

Moreover, $\frac{2}{5}q_{X}^{\vee}\in H^{2,2}(X,\mathbb{Z})$ is an integral
Hodge class of $X$ of type $(2,2)$.

##  5\. Nakajima operators and the Lehn–Sorger model

In this section we introduce Nakajima operators following [Nak97] and [Leh99],
and the algebraic model developed by Lehn and Sorger in [LS03].

Let $S$ be a smooth complex projective surface and let $S^{[n]}$ be the
Hilbert scheme of $n$ points on $S$ for any integer $n>0$. We define

$\mathbb{H}_{n}^{S}:=\displaystyle\oplus_{i=0}^{4n}H^{i}(S^{[n]},\mathbb{Q}),\qquad\mathbb{H}^{S}:=\displaystyle\oplus_{n\geq
0}\mathbb{H}^{S}_{n}.$

The unit of the $\mathbb{Q}$-algebra, with cup product,
$\mathbb{H}^{S}_{0}\cong\mathbb{Q}$ is called _vacuum vector_ and it is
denoted by $|0\rangle$. If $1_{S^{[n]}}$ denotes the unit for the cup product
in $\mathbb{H}_{n}^{S}$ for every $n$, the unit in $\mathbb{H}^{S}$ for the
cup product is given by $|1\rangle:=\sum_{n\geq 0}1_{S^{[n]}}$. The space
$\mathbb{H}^{S}$ is double graded by $(n,i)$: we say that $n$ is the
_conformal weight_ and $i$ is the _cohomological degree_ , denoted by
$|\,\cdot\,|$. Let $\mathfrak{f}\in\text{End}(\mathbb{H}^{S})$ be a linear
operator. We say that $\mathfrak{f}$ is _homogeneous of bidegree_
$(\nu,\iota)$ if for any $n$ and $i$ we have
$\mathfrak{f}(H^{i}(S^{[n]},\mathbb{Q}))\subset
H^{i+\iota}(S^{[n+\nu]},\mathbb{Q})$. The _commutator_ of two homogeneous
operators $\mathfrak{f},\mathfrak{g}\in\text{End}(\mathbb{H}^{S})$ is defined
by

$[\mathfrak{f},\mathfrak{g}]:=\mathfrak{f}\circ\mathfrak{g}-(-1)^{|\mathfrak{f}|\cdot|\mathfrak{g}|}\mathfrak{g}\circ\mathfrak{f}.$

We now define an intersection pairing $\langle\,\cdot\,,\,\cdot\,\rangle$ on
$\mathbb{H}^{S}$. First of all, fix an integer $n\geq 0$ and let
$\alpha,\beta\in\mathbb{H}^{S}_{n}$: we set

$\langle\alpha,\beta\rangle:=\displaystyle\int_{S^{[n]}}\alpha\beta$ (5.1)

Note that $\langle\alpha,\beta\rangle=0$ if $|\alpha|+|\beta|\neq 4n$. Then
$\langle\,\cdot\,,\,\cdot\,\rangle$ extends naturally to a non-degenerate
graded symmetric bilinear form on $\mathbb{H}^{S}$, which we denote again by
$\langle\,\cdot\,,\,\cdot\,\rangle$. If
$\mathfrak{f}\in\text{End}(\mathbb{H}^{S})$ is a homogeneous operator, we
define the _adjoint operator_ $\mathfrak{f}^{\dagger}$ as the homogeneous
operator characterised by the relation

$\langle\mathfrak{f}(\alpha),\beta\rangle=(-1)^{|\mathfrak{f}|\cdot|\alpha|}\langle\alpha,\mathfrak{f}^{\dagger}(\beta)\rangle.$

Following [Leh99, $\S 1.2$], we define an irreducible subvariety $S^{[n,n+k]}$
of $S^{[n]}\times S\times S^{[n+k]}$ for any integers $n\geq 0,k>0$. Let
$X^{[n,n+k]}\subset S^{[n]}\times S^{[n+k]}$ be the uniquely determined closed
subscheme with the property that any morphism $f=(f_{1},f_{2}):T\rightarrow
S^{[n]}\times S^{[n+k]}$ from an arbitrary variety $T$ factors through
$X^{[n,n+k]}$ if and only if the following holds:

$(f_{1}\times\text{id}_{S})^{-1}(\Xi_{n}^{S})\subset(f_{2}\times\text{id}_{S})^{-1}(\Xi^{S}_{n+k}),$

where $\Xi_{n}^{S}\subset S^{[n]}\times S$ is the universal family of
subschemes parametrised by $S^{[n]}$. Closed points in $X^{[n,n+k]}$
corresponds to pairs $(\xi,\xi^{\prime})$ of subschemes with
$\xi\subseteq\xi^{\prime}$. Then one obtains a morphism similar to the
Hilbert–Chow morphism:

$\rho:X^{[n,n+k]}\rightarrow\text{Sym}^{k}S.$

We set $X_{0}^{[n,n+k]}:=\rho^{-1}(\Delta)$, where
$\Delta\subset\textnormal{Sym}^{k}S$ is the small diagonal, and we consider on
$X_{0}^{[n,n+k]}$ the reduced induced subscheme structure. Set-theoretically
we can identify $X_{0}^{[n,n+k]}$ with the following subset of $S^{[n]}\times
S\times S^{[n+k]}$:

$X_{0}^{[n,n+k]}:=\\{(\xi,x,\xi^{\prime})\,|\,\xi\subseteq\xi^{\prime}\,\,\text{and}\,\,\text{Supp}(\mathcal{I}_{\xi}/\mathcal{I}_{\xi^{\prime}})=x\\},$

where $\mathcal{I}_{\xi}$ is the ideal sheaf of $\xi$. We define $S^{[n,n+k]}$
as

$S^{[n,n+k]}:=\overline{\\{(\xi,x,\xi^{\prime})\in
X_{0}^{[n,n+k]}\,|\,l(\xi_{x})=0\\}},$

where the closure is the Zariski closure. Then $S^{[n,n+k]}$ is an irreducible
subvariety of $S^{[n]}\times S\times S^{[n+k]}$ of complex dimension $2n+k+1$
by [Leh99, Lemma 1.1]. Consider the following diagram

${S^{[n]}\times S\times
S^{[n+k]}}$${S^{[n]}}$${S}$${S^{[n+k]},}$$\scriptstyle{\varphi}$$\scriptstyle{\rho}$$\scriptstyle{\psi}$
(5.2)

where $\varphi$, $\rho$ and $\psi$ are the projections respectively on
$S^{[n]}$, $S$ and $S^{[n+k]}$.

######  Definition 5.1.

Let $S$ be a smooth complex projective surface. The _Nakajima creation
operators_ , known also as _Heisenberg operators_ , are defined as follows:
for $\alpha\in H^{*}(S,\mathbb{Q})$ and $x\in H^{*}(S^{[n]},\mathbb{Q})$ we
set for $k>0$

$\mathfrak{q}_{k}:H^{*}(S,\mathbb{Q})\rightarrow\text{End}(\mathbb{H}^{S}),\qquad\mathfrak{q}_{k}(\alpha)(x):=\psi_{*}\left(PD^{-1}\left[S^{[n,n+k]}\right]\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),$
(5.3)

where $[S^{[n,n+k]}]\in H_{4n+2k+2}(S^{[n]}\times S\times
S^{[n+k]},\mathbb{Q})$ is the fundamental class of $S^{[n,n+k]}$, we denote by
$PD$ the Poincaré duality, the dot is the cup product, and $\psi_{*}$ is the
Gysin homomorphism, i.e., $\psi_{*}=PD^{-1}\psi_{*}PD$, where $\psi_{*}$ in
the right-hand side is the pushforward in homology. By convention
$\mathfrak{q}_{0}=0$, and the _Nakajima annihilation operators_ are defined as

$\mathfrak{q}_{k}(\alpha):=(-1)^{-k}\mathfrak{q}_{-k}(\alpha)^{\dagger}\qquad\text{for
all}\,\,k<0.$

Note that the unit $|1\rangle\in\mathbb{H}^{S}$ can be represented as
$|1\rangle=\sum_{n\geq
0}1_{S^{[n]}}=\text{exp}(\mathfrak{q}_{1}(1_{S}))|0\rangle$, which gives

$1_{S^{[n]}}=\frac{1}{n!}\mathfrak{q}_{1}(1)^{n}|0\rangle.$ (5.4)

The following commutation formula was obtained by Nakajima in [Nak97].

######  Theorem 5.2 (Nakajima).

The operators $\mathfrak{q}_{i}$ satisfy the following commutation formula:

$[\mathfrak{q}_{i}(\alpha),\mathfrak{q}_{j}(\beta)]=i\cdot\delta_{i+j,0}\cdot\displaystyle\int_{S}\alpha\beta\cdot\textnormal{id}_{\mathbb{H}^{S}},$

where $\delta$ is the Kronecker delta.

We now introduce the _boundary operator_. Fix a positive integer $n>0$ and let
$\lambda=\\{\lambda_{1}\geq\lambda_{2}\geq\dots\geq\lambda_{s}>0\\}$ be a
partition of $n$, i.e., an $s$-uple of ordered positive integers such that
$\lambda_{1}+\lambda_{2}+\dots+\lambda_{s}=n$. Consider the Hilbert–Chow
morphism $\rho:S^{[n]}\rightarrow\text{Sym}^{n}S$. We set

$\text{Sym}^{n}_{\lambda}S:=\\{\alpha\in\text{Sym}^{n}S\,|\,\alpha=\displaystyle\sum_{1\leq
i\leq s}\lambda_{i}x_{i},\,\,x_{i}\in S\,\,\text{pairwise distinct}\\}$

for a fixed partition $\lambda$. By a theorem of Briançon
$S^{[n]}_{\lambda}:=\rho^{-1}(\text{Sym}^{n}_{\lambda}S)$ is irreducible of
dimension $n+s$, see [Bri77]. For example, $S^{[n]}_{(1,1,\dots 1)}$ is the
open subset of $S^{[n]}$ which corresponds to the configuration space of
unordered $n$-tuples of pairwise distinct points: it is the only stratum which
is open. The only stratum of codimension $1$ is $S^{[n]}_{(2,1,\dots,1)}$.

######  Definition 5.3.

The _boundary_ of $S^{[n]}$ for $n\geq 2$ is the irreducible divisor

$\partial
S^{[n]}:=\displaystyle\bigcup_{\lambda\neq(1,1,\dots,1)}S^{[n]}_{\lambda}=\overline{S^{[n]}_{(2,1,\dots,1)}}.$

######  Remark 5.4.

For $n=1$ the boundary $\partial S^{[n]}$ is the empty set.

There is a description of the divisor $\partial S^{[n]}$ in sheaf theoretic
terms. Consider the universal family of subschemes $\Xi_{n}\subset
S^{[n]}\times S$. Let $p:\Xi_{n}\rightarrow S^{[n]}$ be the projection on the
first factor. Since $p$ is a flat morphism of finite degree $n$, we have that
$\mathcal{O}_{S}^{[n]}:=p_{*}(\mathcal{O}_{\Xi_{n}})$ is a locally free sheaf
of rank $n$. Then $[\partial S^{[n]}]=-2c_{1}(\mathcal{O}_{S}^{[n]})$ by
[Leh99, Lemma 3.7].

We can now define the boundary operator and the _derivative_ of a linear
operator $\mathfrak{f}\in\text{End}(\mathbb{H}^{S})$.

######  Definition 5.5.

The _boundary operator_ $\mathfrak{d}:\mathbb{H^{S}}\rightarrow\mathbb{H^{S}}$
is the homogeneous linear map of bidegree $(0,2)$ given by

$\mathfrak{d}(x):=c_{1}(\mathcal{O}_{S}^{[n]})\cdot x=-\frac{1}{2}[\partial
S^{[n]}]\cdot x\qquad\text{for all}\,\,x\in H^{*}(S^{[n]},\mathbb{Q}).$

For any endomorphism $\mathfrak{f}\in\text{End}(\mathbb{H}^{S})$ we define the
_derivative_ of $\mathfrak{f}$ as

$\mathfrak{f}^{\prime}:=[\mathfrak{d},\mathfrak{f}]=\mathfrak{d}\circ\mathfrak{f}-\mathfrak{f}\circ\mathfrak{d}.$

We denote by $\mathfrak{f}^{(n)}$ the higher derivatives.

The following relation was obtained by Lehn, see [Leh99, Theorem 3.10,
Proposition 3.12].

######  Theorem 5.6 (Lehn).

Let $S$ be a smooth complex projective surface. Then for any integers $n,m$
such that $n+m\neq 0$ and cohomology classes $\alpha,\beta\in
H^{*}(S,\mathbb{Q})$ we have

$[\mathfrak{q}_{n}^{\prime}(\alpha),\mathfrak{q}_{m}(\beta)]=-nm\cdot\mathfrak{q}_{n+m}(\alpha\beta).$

We now state the _Ellingsrud–Göttsche–Lehn_ formula. Let $\Xi_{n}\subset
S^{[n]}\times S$ be the universal family of subschemes parametrised by
$S^{[n]}$. By [Leh99, Theorem 1.9] we have that $S^{[n,n+1]}$ is a smooth
irreducible variety isomorphic to the blow-up of $S^{[n]}\times S$ along the
universal family, i.e., $S^{[n,n+1]}\cong\text{Bl}_{\Xi_{n}^{S}}(S^{[n]}\times
S)$. We get the following diagram

${S^{[n,n+1]}}$${S^{[n+1]}}$${S^{[n]}\times
S}$${S^{[n]}}$${\Xi_{n}}$${S,}$$\scriptstyle{\varphi}$$\scriptstyle{\rho}$$\scriptstyle{\sigma}$$\scriptstyle{\psi}$$\scriptstyle{p}$$\scriptstyle{q}$$\scriptstyle{\iota}$
(5.5)

where $\sigma$ is the blow-up of $S^{[n]}\times S$ in $\Xi_{n}^{S}$, the
morphism $\iota$ is the inclusion and the other maps are the projections. If
$N$ is the exceptional divisor of $\sigma$, let
$\mathcal{L}:=\mathcal{O}_{S^{[n,n+1]}}(-N)$. Let $\mathcal{T}_{n}$ be the
tangent bundle $\mathcal{T}S^{[n]}$ and $\omega_{S}$ be the canonical bundle
of $S$. Given a smooth projective variety $X$ and $F\in\text{Coh}(X)$,
following [HL10, $\S 1.1$] we define the _dual_ $F^{\vee}$ as

$F^{\vee}:=\displaystyle\sum_{i}(-1)^{i}\mathcal{E}\textit{xt}^{\,i}(F,\mathcal{O}_{X}).$
(5.6)

Let $K(X)$ be the Grothendieck group of vector bundles on $X$. We denote by
$f^{!}$ the pullback between the Grothendieck groups of a morphism $f$ of
smooth projective varieties, see [Ful13, $\S 15.1$] for more details. We can
now state the _Ellingsrud–Göttsche–Lehn formula_ , see [EGL01, Proposition
2.3].

######  Proposition 5.7 (EGL formula).

Keep notation as above. The following relation holds in $K(S^{[n,n+1]})$:

$\begin{array}[]{lll}\psi^{!}\mathcal{T}_{n+1}&=&\varphi^{!}\mathcal{T}_{n}+\mathcal{L}-\mathcal{L}\cdot\sigma^{!}(\mathcal{O}^{\vee}_{\Xi_{n}})+\mathcal{L}^{\vee}\cdot\rho^{!}\omega_{S}^{\vee}\\\\[5.16663pt]
&&-\mathcal{L}^{\vee}\cdot\sigma^{!}(\mathcal{O}_{\Xi_{n}})\cdot\rho^{!}\omega^{\vee}_{S}-\rho^{!}(\mathcal{O}_{S}-\mathcal{T}_{S}+\omega_{S}^{\vee}).\end{array}$

Let now $S$ be a projective K3 surface. The following theorem by Qin and Wang
gives integral basis of $H^{2}(S^{[2]},\mathbb{Z})$ and
$H^{4}(S^{[2]},\mathbb{Z})$ in terms of Nakajima operators, see [BNWS13,
p.17]. See [QW05, Theorem 5.4, Remark 5.6] for a more general statement.

######  Theorem 5.8 (Qin–Wang).

Let $S$ be a projective K3 surface and $X=S^{[2]}$ be its Hilbert square. Let
$\\{\alpha_{i}\\}_{i=1,\dots,22}$ be an integral basis of
$H^{2}(S,\mathbb{Z})$. Denote by $1\in H^{0}(S,\mathbb{Z})$ the unit and by
$x\in H^{4}(S,\mathbb{Z})$ the class of a point.

  1. (i)

An integral basis of $H^{2}(X,\mathbb{Z})$ in terms of Nakajima operators is
given by

$\left\\{\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle,\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\alpha_{i})|0\rangle\right\\}_{i=1,\dots,22.}$

  2. (ii)

An integral basis of $H^{4}(X,\mathbb{Z})$ in terms of Nakajima operators is
given by the following elements:

$\begin{array}[]{c}\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\qquad\mathfrak{q}_{2}(\alpha_{i})|0\rangle,\qquad\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\,\,\text{with}\,\,i<j,\\\\[4.30554pt]
\mathfrak{m}_{1,1}(\alpha_{i})|0\rangle:=\frac{1}{2}\left(\mathfrak{q}_{1}(\alpha_{i})^{2}-\mathfrak{q}_{2}(\alpha_{i})\right)|0\rangle,\end{array}$

with $i,j=1,\dots,22$.

######  Remark 5.9.

By Definition 5.1 with $k=2$, $n=0$, we have
$\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle=\delta$, where $\delta\in
H^{2}(S^{[2]},\mathbb{Z})$ is such that $2\delta$ is the class of the
exceptional divisor of the Hilbert–Chow morphism. Moreover, if $\alpha\in
H^{2}(S,\mathbb{Z})$, then by Definition 5.1 with $n=k=1$ we obtain that
$\alpha$ seen as an element of $H^{2}(S^{[2]},\mathbb{Z})$ is represented by
$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\alpha)|0\rangle$.

Consider the diagonal embedding $\tau_{2}:S\rightarrow S\times S$ of the K3
surface $S$. We denote the Gysin homomorphism followed by the Künneth
isomorphism by $\tau_{2*}:H^{*}(S,\mathbb{Z})\rightarrow
H^{*}(S,\mathbb{Z})\otimes H^{*}(S,\mathbb{Z})$. We take the following basis
$\\{\alpha_{1},\dots,\alpha_{22}\\}$ for the lattice $H^{2}(S,\mathbb{Z})$:
let $\\{\alpha_{1},\dots,\alpha_{8}\\}$ and
$\\{\alpha_{9},\dots,\alpha_{16}\\}$ be the basis of the two copies of
$E_{8}(-1)$ whose Gram matrix is the second matrix of Example 2.1 and
$\\{\alpha_{17},\alpha_{18}\\}$, $\\{\alpha_{19},\alpha_{20}\\}$,
$\\{\alpha_{21},\alpha_{22}\\}$ be the basis of the three copies of $U$ whose
Gram matrix is the first matrix in Example 2.1. The following Lemma will be
useful, see [BNWS13, p.18].

######  Lemma 5.10.

Let $X=S^{[2]}$ be the Hilbert square of a projective K3 surface $S$. Assume
that $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of the lattice
$H^{2}(S,\mathbb{Z})$ constructed above. Then

$\tau_{2*}1=\sum_{i,j}\mu_{i,j}\alpha_{i}\otimes\alpha_{j}+1\otimes x+x\otimes
1,$

where the $\mu_{i,j}$’s are represented in Table 1 (we write down only the
$\mu_{i,j}$’s which are non zero and such that $i\leq j$):

$\mu_{1,1}=-4$ | $\mu_{1,2}=-7$ | $\mu_{1,3}=-10$ | $\mu_{1,4}=-8$ | $\mu_{1,5}=-6$
---|---|---|---|---
$\mu_{1,6}=-4$ | $\mu_{1,7}=-2$ | $\mu_{1,8}=-5$ | $\mu_{2,2}=-14$ | $\mu_{2,3}=-20$
$\mu_{2,4}=-16$ | $\mu_{2,5}=-12$ | $\mu_{2,6}=-8$ | $\mu_{2,7}=-4$ | $\mu_{2,8}=-10$
$\mu_{3,3}=-30$ | $\mu_{3,4}=-24$ | $\mu_{3,5}=-18$ | $\mu_{3,6}=-12$ | $\mu_{3,7}=-6$
$\mu_{3,8}=-15$ | $\mu_{4,4}=-20$ | $\mu_{4,5}=-15$ | $\mu_{4,6}=-10$ | $\mu_{4,7}=-5$
$\mu_{4,8}=-12$ | $\mu_{5,5}=-12$ | $\mu_{5,6}=-8$ | $\mu_{5,7}=-4$ | $\mu_{5,8}=-9$
$\mu_{6,6}=-6$ | $\mu_{6,7}=-3$ | $\mu_{6,8}=-6$ | $\mu_{7,7}=-2$ | $\mu_{7,8}=-3$
$\mu_{8,8}=-8$ | $\mu_{9,9}=-4$ | $\mu_{9,10}=-7$ | $\mu_{9,11}=-10$ | $\mu_{9,12}=-8$
$\mu_{9,13}=-6$ | $\mu_{9,14}=-4$ | $\mu_{9,15}=-2$ | $\mu_{9,16}=-5$ | $\mu_{10,10}=-14$
$\mu_{10,11}=-20$ | $\mu_{10,12}=-16$ | $\mu_{10,13}=-12$ | $\mu_{10,14}=-8$ | $\mu_{10,15}=-4$
$\mu_{10,16}=-10$ | $\mu_{11,11}=-30$ | $\mu_{11,12}=-24$ | $\mu_{11,13}=-18$ | $\mu_{11,14}=-12$
$\mu_{11,15}=-6$ | $\mu_{11,16}=-15$ | $\mu_{12,12}=-20$ | $\mu_{12,13}=-15$ | $\mu_{12,14}=-10$
$\mu_{12,15}=-5$ | $\mu_{12,16}=-12$ | $\mu_{13,13}=-12$ | $\mu_{13,14}=-8$ | $\mu_{13,15}=-4$
$\mu_{13,16}=-9$ | $\mu_{14,14}=-6$ | $\mu_{14,15}=-3$ | $\mu_{14,16}=-6$ | $\mu_{15,15}=-2$
$\mu_{15,16}=-3$ | $\mu_{16,16}=-8$ | $\mu_{17,18}=1$ | $\mu_{19,20}=1$ | $\mu_{21,22}=1$
Table 1.  The $\mu_{i,j}$’s.

###### Proof.

We can write

$\tau_{2*}1=\sum_{i,j}\mu_{i,j}\alpha_{i}\otimes\alpha_{j}+a(1\otimes
x)+b(x\otimes 1)$

for some $\mu_{i,j}\in\mathbb{Z}$ such that $\mu_{i,j}=\mu_{j,i}$ with
$i,j\in\\{1,\dots,22\\}$, and some $a,b\in\mathbb{Z}$. Let
$\langle\,\cdot\,,\,\cdot\,\rangle$ be the intersection pairing of
$\mathbb{H}^{S}$ in (5.1). Since $\tau_{2*}$ is the adjoint of the cup
product, we have the relation
$\langle\tau_{2*}1,\alpha_{k}\otimes\alpha_{l}\rangle=\int_{S}\alpha_{k}\alpha_{l}$,
which gives, together with
$\tau_{2*}1=\sum_{i,j}\mu_{i,j}\alpha_{i}\otimes\alpha_{j}+a(1\otimes
x)+b(x\otimes 1)$, the following system:

$\displaystyle\sum_{i,j}\mu_{i,j}\int_{S}\alpha_{i}\alpha_{k}\int_{S}\alpha_{j}\alpha_{l}=\int_{S}\alpha_{k}\alpha_{l}.$
(5.7)

From (5.7), with the help of a computer, we can compute the coefficients
$\mu_{i,j}$. The solution of the system is given in Table 1 (we have written
down only the $\mu_{i,j}$’s which are non zero and such that $i\leq j$, since
$\mu_{i,j}=\mu_{j,i}$). Similarly, from the relations

$\langle\tau_{2*}1,1\otimes
x\rangle=\displaystyle\int_{S}x=1,\qquad\langle\tau_{2*}1,x\otimes
1\rangle=\displaystyle\int_{S}x=1,$

we obtain $a=b=1$. ∎

We now introduce the algebraic model developed by Lehn and Sorger in [LS03].
Recall that a _graded Frobenius algebra_ of degree $d$ is a finite dimensional
graded vector space $A=\bigoplus_{i=-d}^{d}A^{i}$ endowed with a graded
commutative and associative multiplication $A\otimes A\rightarrow A$ of degree
$d$, i.e., $\textnormal{deg}(ab)=\textnormal{deg}(a)+\textnormal{deg}(b)+d$,
and unit element $1$, necessarily of degree $-d$, together with a linear form
$T:A\rightarrow\mathbb{Q}$ of degree $-d$ such that the induced symmetric
bilinear form $\langle a,b\rangle:=T(ab)$ is non-degenerate and of degree $0$.
Note that $d$ must be an even number since $1\cdot 1=1$. An example of graded
Frobenius algebra is the shifted cohomology ring $H^{*}(X,\mathbb{Q})[d]$ of a
compact complex manifold $X$ of complex dimension $d$. From now on we consider
$A:=H^{*}(S,\mathbb{Q})[2]$, where $S$ is a projective K3 surface. The linear
form $T:A\rightarrow\mathbb{Q}$ is $T(\alpha):=-\int_{S}\alpha$, and the
induced bilinear form is
$\langle\alpha,\beta\rangle=T(\alpha\beta)=-\int_{S}\alpha\beta$, which is the
intersection pairing changed of sign. The graded Frobenius algebra structure
of $A$ induces a structure of graded Frobenius algebra on $A^{\otimes n}$: as
remarked in [HHT12], since $S$ is a K3 surface, $A$ has only graded pieces of
even degree, so the general construction of Lehn-Sorger simplifies. In this
case, the multiplication induced on $A^{\otimes n}$ is

$(a_{1}\otimes\dots\otimes a_{n})\cdot(b_{1}\otimes\dots\otimes
b_{n})=(a_{1}b_{1})\otimes\dots\otimes(a_{n}b_{n}),$

and the linear form is

$T:A^{\otimes n}\rightarrow\mathbb{Q},\qquad a_{1}\otimes\dots\otimes
a_{n}\mapsto T(a_{1})T(a_{2})\dots T(a_{n}).$

The symmetric group $S_{n}$ of order $n$ acts on $A^{\otimes n}$ as

$\pi(a_{1}\otimes\dots\otimes a_{n}):=a_{\pi^{-1}(1)}\otimes\dots\otimes
a_{\pi^{-1}(n)}.$

Given a partition $n=n_{1}+\dots+n_{k}$ with
$n_{1},\dots,n_{k}\in\mathbb{Z}_{>0}$, there is a generalised multiplication
map $A^{\otimes n}\rightarrow A^{\otimes k}$ defined by

$a_{1}\otimes\dots\otimes a_{n}\mapsto(a_{1}\dots
a_{n_{1}})\otimes\dots\otimes(a_{n_{1}+\dots+n_{k-1}+1}\dots
a_{n_{1}+\dots+n_{k}}).$

Given a finite set $I\subset\\{1,\dots,n\\}$, let $A^{\otimes I}$ denote the
tensor power with factors indexed by elements of $I$. Let $n$ be a fixed
positive integer, $\pi\in S_{n}$ be a permutation and
$\langle\pi\rangle\subset S_{n}$ be the subgroup generated by $\pi$. If
$[n]:=\\{1,\dots,n\\}$, denote by $\langle\pi\rangle\setminus[n]$ the set of
orbits. Then we set

$A\\{S_{n}\\}:=\displaystyle\bigoplus_{\pi\in
S_{n}}A^{\otimes\langle\pi\rangle\setminus[n]}\cdot\pi.$

######  Example 5.11.

If $n=2$, we have $S_{2}=\\{\text{id},(1\,2)\\}$. Moreover,
$\langle\text{id}\rangle\setminus[2]=\\{\\{1\\},\\{2\\}\\}$ and
$\langle(1\,2)\rangle\setminus[2]=\\{\\{1,2\\}\\}$, hence we obtain
$A\\{S_{2}\\}=A^{\otimes 2}\text{id}\oplus A(1\,2)$. Similarly for $n=3$ we
have

$A\\{S_{3}\\}=A^{\otimes 3}\text{id}\oplus A^{\otimes 2}(1\,2)\oplus
A^{\otimes 2}(1\,3)\oplus A^{\otimes 2}(2\,3)\oplus A(1\,2\,3)\oplus
A(1\,3\,2).$

Let $\sigma\in S_{n}$. There is a bijection

$\sigma:\langle\pi\rangle\setminus[n]\rightarrow\langle\sigma\pi\sigma^{-1}\rangle\setminus[n],\qquad
x\mapsto\sigma x$

and an isomorphism

$\tilde{\sigma}:A\\{S_{n}\\}\rightarrow A\\{S_{n}\\},\qquad
a\pi\mapsto\sigma(a)\sigma\pi\sigma^{-1}.$

Thus we obtain an action of the symmetric group $S_{n}$ on $A\\{S_{n}\\}$. We
denote by

$A^{[n]}:=(A\\{S_{n}\\})^{S_{n}}$

the subspace of invariants. We can now state the main theorem of [LS03].

######  Theorem 5.12 (Theorem 3.2 in [LS03]).

Let $S$ be a projective K3 surface. Then there is a canonical isomorphism of
graded rings

$(H^{*}(S,\mathbb{Q})[2])^{[n]}\xrightarrow{\sim}H^{*}(S^{[n]},\mathbb{Q})[2n].$

The structure of graded Frobenius algebra of $H^{*}(S^{[n]},\mathbb{Q})[2n]$
is obtained by setting

$T(a):=(-1)^{n}\displaystyle\int_{S^{[n]}}a\qquad\text{for all}\,\,a\in
H^{*}(S^{[n]},\mathbb{Q}).$

We sketch the description of the isomorphism of Theorem 5.12. Let
$\mathcal{V}(A):=\text{Sym}^{*}(A\otimes t^{-1}\mathbb{Q}[t^{-1}])$, which is
bigraded by degree and weight: an element $a\otimes t^{-m}\in A\otimes t^{-m}$
has degree $|a|$ and weight $m$. The component of $\mathcal{V}(A)$ of constant
weight $n$ is the graded vector space

$\mathcal{V}(A)_{n}\cong\displaystyle\bigoplus_{||\alpha||=n}\displaystyle\bigotimes_{i}\text{Sym}^{\alpha_{i}}A,$

where the direct sum is taken over all the possible partitions
$\alpha=(1^{\alpha_{1}},2^{\alpha_{2}},\dots)$ of $n$, and
$||\alpha||:=\alpha_{1}\cdot 1+\alpha_{2}\cdot 2+\dots$. We now fix $\pi\in
S_{n}$. Let $f:\\{1,2,\dots,N\\}\rightarrow\langle\pi\rangle\setminus[n]$ be
an enumeration of the orbits of $\pi\in S_{n}$. We denote by $l_{i}$ the
length of the $i$-th orbit, i.e., $l_{i}:=|f(i)|$. We define
$\Phi^{\prime}:A^{\otimes N}\rightarrow\mathcal{V}(A)$ as

$a_{1}\otimes\dots\otimes
a_{N}\xmapsto{\Phi^{\prime}}\frac{1}{n!}(a_{1}\otimes
t^{-l_{1}})\dots(a_{N}\otimes t^{-l_{N}}).$ (5.8)

Let $\Phi:\bigoplus_{n\geq 0}A\\{S_{n}\\}\rightarrow\mathcal{V}(A)$ be defined
on each summand $A^{\otimes\langle\pi\rangle\setminus[n]}\cdot\pi$ by the
composition

$A^{\otimes\langle\pi\rangle\setminus[n]}\cdot\pi\xrightarrow{\tilde{f}^{-1}}A^{\otimes
N}\xrightarrow{\Phi^{\prime}}\mathcal{V}(A),$

where $\tilde{f}^{-1}$ denotes the identification of
$A^{\otimes\langle\pi\rangle\setminus[n]}\cdot\pi$ with $A^{\otimes N}$
through the enumeration $f$. Then $\Phi$ induces an isomorphism of graded
vector spaces $A^{[n]}\xrightarrow{\sim}\mathcal{V}(A)_{n}$ by [LS03,
Proposition 2.11]. Moreover, there is an isomorphism of graded vector spaces
$\Psi:\mathcal{V}(A)\rightarrow\bigoplus_{n\geq
0}H^{*}(S^{[n]},\mathbb{Q})[2n]$ given by

$(a_{1}t^{-n_{1}})\dots(a_{s}t^{-n_{s}})\xmapsto{\Psi}\mathfrak{q}_{n_{1}}(a_{1})\dots\mathfrak{q}_{n_{s}}(a_{s})|0\rangle,$
(5.9)

see [LS03, Theorem 3.6]. Combining (5.8) and (5.9) we can obtain a basis of
$H^{*}(S^{[n]},\mathbb{Q})[2n]$ from a basis of $H^{*}(S,\mathbb{Q})[2]$. In
order to use correctly the isomorphism of Theorem 5.12, if
$A:=H^{*}(S,\mathbb{Q})[2]$, we have to work with elements of $A\\{S_{n}\\}$
which are invariant for the action of $S_{n}$. The basis of
$H^{*}(S^{[n]},\mathbb{Q})[2n]$ obtained is clearly also a basis of
$H^{*}(S^{[n]},\mathbb{Q})$ with the standard grading. For instance, let
$\\{\alpha_{1},\dots,\alpha_{22}\\}$ be a basis of $H^{2}(S,\mathbb{Q})$,
denote by $1\in H^{0}(S,\mathbb{Z})$ the unit and by $x\in
H^{4}(S,\mathbb{Z})$ the class of a point: one can show that
$\\{\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle,\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\alpha_{i})|0\rangle\\}_{i=1,\dots,22}$
is a basis of $H^{2}(S^{[2]},\mathbb{Q})$ and
$\\{\frac{1}{2}\mathfrak{q}_{2}(\alpha_{i})|0\rangle,\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\\}_{1\leq
i\leq j\leq 22}$ is a basis of $H^{4}(S^{[2]},\mathbb{Q})$.

We omit the definition of the product on $(H^{*}(S,\mathbb{Q})[2])^{[n]}$
defined in [LS03] to give the structure of ring. We state the following result
which describes the cup product between elements in
$H^{2}(S^{[2]},\mathbb{Z})$, see [BNWS13, p.13].

######  Lemma 5.13.

Let $X=S^{[2]}$ be the Hilbert square of a projective K3 surface $S$. Let
$\\{\alpha_{1},\dots,\alpha_{22}\\}$ be the basis of the lattice
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10. Then the following equalities hold
in $H^{4}(X,\mathbb{Z})$.

  1. (i)

For every $\alpha\in H^{2}(S,\mathbb{Z})$ we have

$\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cup\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\alpha)|0\rangle=\mathfrak{q}_{2}(\alpha)|0\rangle.$

  2. (ii)

For every $\alpha,\beta\in H^{2}(S,\mathbb{Z})$ we have

$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\alpha)|0\rangle\cup\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(\beta)|0\rangle=\left(\int_{S}\alpha\beta\right)\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle+\mathfrak{q}_{1}(\alpha)\mathfrak{q}_{1}(\beta)|0\rangle.$

  3. (iii)

If $\mu_{i,j}$, with $i,j=1,\dots,22$, are the coefficients computed in Lemma
5.10, then

$\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cup\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle=-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.$

######  Remark 5.14.

The result for the product
$\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cup\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle$
given in [BNWS13, p.18] is not correct: a change of sign is needed in the
right-hand side. The map $\Delta_{*}$ in [BNWS13, p.18] corresponds to
$\tau_{2*}$ in our notation: in that article the cohomology ring taken is
$H^{*}(S,\mathbb{Q})$, without the shifting used in the Lehn–Sorger model,
which gives the change of sign of the intersection pairing on
$H^{*}(S,\mathbb{Q})[2]$, as seen above. See also [HHT12, Remark 3.1].

##  6\. Rational Hodge classes of type $(2,2)$ on Hilbert squares of K3
surfaces

Let $S$ be a projective K3 surface. In this section we use Theorem 5.12 to
compute a basis of the $\mathbb{Q}$-vector space $H^{2,2}(S^{[2]},\mathbb{Q})$
of rational Hodge classes of type $(2,2)$ on $S^{[2]}$.

Recall that the rational cohomology groups $H^{2i}(S,\mathbb{Q})$ and
$H^{2j}(S^{[n]},\mathbb{Q})$, where $i,j\in\mathbb{Z}_{\geq 0}$, are Hodge
structures of weight $2i$ and $2j$ respectively, and the shifted cohomology
groups $H^{2i}(S,\mathbb{Q})[2]$ and $H^{2j}(S^{[n]},\mathbb{Q})[2n]$ are
Hodge structures of weight $2i-2$ and $2j-2n$ respectively, with the following
Hodge decompositions:

$H^{2i}(S,\mathbb{C})[2]=\displaystyle\bigoplus_{p+q=2i-2}H^{p,q}(S)[2],\qquad
H^{2j}(S^{[n]},\mathbb{Q})[2n]=\displaystyle\bigoplus_{r+s=2j-2n}H^{r,s}(S^{[n]})[2n],$

where $p,q\in\\{-1,0,\dots,i-1\\}$ and $r,s\in\\{-n,1-n,\dots,j-n\\}$, and

$H^{p,q}(S)[2]=H^{p+1,q+1}(S),\qquad
H^{r,s}(S^{[n]})[2n]=H^{r+n,s+n}(S^{[n]}).$

For details on Hodge structures, see [Huy16, $\S 3$]. If
$A:=H^{*}(S,\mathbb{Q})[2]$, recall that
$A^{[n]}\cong\bigoplus_{||\alpha||=n}\bigotimes_{i}\text{Sym}^{\alpha_{i}}A$.
We denote by $(A^{[n]})^{2i}$ the component of degree $2i-2n$ of $A^{[n]}$.
Then the Hodge structures $H^{2j}(S,\mathbb{Q})[2]$ give rise to a Hodge
structure on $(A^{[n]})^{2i}$. Since the weights of the Hodge structures
considered depend only on the (shifted) cohomological degrees, we have that
$(A^{[n]})^{2i}$ is a Hodge structure of weight $2i-2n$. Note that
$H^{2i}(S^{[n]},\mathbb{Q})[2n]$ is a Hodge structure of weight $2i-2n$ and
$(A^{[n]})^{2i}\cong H^{2i}(S^{[n]},\mathbb{Q})[2n]$ by Theorem 5.12. Recall
that, given two Hodge structures $V$ and $W$, a morphism of weight $k$ is a
linear map $f:V\rightarrow W$ such that its $\mathbb{C}$-linear extension
satisfy $f(V^{p,q})\subseteq W^{p+k,q+k}$. The pullback and the Gysin
homomorphism induced by a morphism $f:X\rightarrow Y$ between two complex
projective manifolds are morphisms of Hodge structures of weight respectively
$0$ and $r=\textnormal{dim}_{\mathbb{C}}(Y)-\textnormal{dim}_{\mathbb{C}}(X)$,
see [Voi02, $\S 7.3.2$]. Using this property, Definition 5.1 and the
isomorphisms (5.8) and (5.9) one can show the following result, which is well
known to experts and implicitly given in [LS03].

######  Theorem 6.1.

Let $S$ be a projective K3 surface. Let $A:=H^{*}(S,\mathbb{Q})[2]$ and
consider $(A^{[n]})^{2i}\cong H^{2i}(S^{[n]},\mathbb{Q})[2n]$, the isomorphism
induced by Theorem 5.12 on the components of (shifted) cohomological degree
$2i-2n$. Take on $(A^{[n]})^{2i}$ the Hodge structure of weight $2i-2n$
described above and on $H^{2i}(S^{[n]},\mathbb{Q})[2n]$ the Hodge structure of
weight $2i-2n$ induced by shifted cohomology. Then

$(A^{[n]})^{2i}\xrightarrow{\sim}H^{2i}(S^{[n]},\mathbb{Q})[2n]$

is an isomorphism of Hodge structures of weight $0$.

Let $S$ be a K3 surface and $T(S)$ be its transcendental lattice, which is a
Hodge structure of weight $2$. Denote by
$E_{S}:=\text{Hom}_{0}(T(S)_{\mathbb{Q}},T(S)_{\mathbb{Q}})$ the algebra of
endomorphisms on $T(S)_{\mathbb{Q}}$ of weight $0$. Recall that a K3 surface
$S$ is _general_ if $E_{S}\cong\mathbb{Q}$. We can now give a basis of the
$\mathbb{Q}$-vector space $H^{2,2}(S^{[2]},\mathbb{Q})$ of rational Hodge
classes of type $(2,2)$ on the Hilbert square of a general projective K3
surface.

######  Theorem 6.2.

Let $S$ be a general projective K3 surface and let $\\{b_{1},\dots,b_{r}\\}$
be a basis of $\textnormal{Pic}(S)$. Then:

  1. (i)

$\textnormal{dim}(H^{2,2}(S^{[2]},\mathbb{Q}))=\frac{(r+1)r}{2}+r+2$.

  2. (ii)

A basis of $H^{2,2}(S^{[2]},\mathbb{Q})$ is given by the following elements:

     * •

$\frac{1}{2}\mathfrak{q}_{2}(b_{i})|0\rangle$ for $i=1,\dots,r$.

     * •

$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$, where $1\in
H^{0}(S,\mathbb{Q})$ is the unit and $x\in H^{4}(S,\mathbb{Q})$ is the class
of a point.

     * •

$\frac{1}{2}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle$ for $i=1,\dots,r$.

     * •

$\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle$ for $1\leq i<j\leq
r$.

     * •

$-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$,
where $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are given in
Table 1.

###### Proof.

By Theorem 5.12 and Theorem 6.1 the following is an isomorphism of Hodge
structures of weight $0$:

$H^{4}(S^{[2]},\mathbb{Q})\cong
H^{2}(S,\mathbb{Q})\oplus\left(H^{0}(S,\mathbb{Q})\otimes
H^{4}(S,\mathbb{Q})\right)\oplus\textnormal{Sym}^{2}(H^{2}(S,\mathbb{Q})),$
(6.1)

we omit the shiftings of the cohomology groups. The Hodge classes of $S^{[2]}$
of bidegree $(2,2)$ have bidegree $(0,0)$ in the shifted cohomology, so we
look for the components of (shifted) bidegree $(0,0)$ in the right-hand side
of $(\ref{isoojojm})$. If $\textnormal{NS}(S)$ is the Néron–Severi group of
$S$ and $T(S)=(\textnormal{NS}(S))^{\perp}$ is the transcendental lattice,
$H^{2}(S,\mathbb{Q})$ can be decomposed as

$H^{2}(S,\mathbb{Q})\cong\textnormal{NS}(S)_{\mathbb{Q}}\oplus
T(S)_{\mathbb{Q}},$ (6.2)

where $\textnormal{NS}(S)_{\mathbb{Q}}=\textnormal{NS}(S)\otimes\mathbb{Q}$
and $T(S)_{\mathbb{Q}}=T(S)\otimes\mathbb{Q}$.
The first summand of $(\ref{isoojojm})$ has the $\mathbb{Q}$-vector space
$\textnormal{NS}(S)_{\mathbb{Q}}[2]$ as component of bidegree $(0,0)$. Since
$\textnormal{NS}(S)\cong\textnormal{Pic}(S)$ and
$\textnormal{rk}(\textnormal{Pic}(S))=r$ by assumption, the component of
bidegree $(0,0)$ of $H^{2}(S,\mathbb{Q})[2]$ has dimension $r$. By (5.8) and
(5.9) we have
$b_{i}\xmapsto{\Psi\circ\Phi^{\prime}}\frac{1}{2}\mathfrak{q}_{2}(b_{i})|0\rangle$,
so $\frac{1}{2}\mathfrak{q}_{2}(b_{i})|0\rangle$ is in a basis of
$H^{2,2}(S^{[2]},\mathbb{Q})$ for $i=1,\dots,r$.
The second summand of $(\ref{isoojojm})$ is $H^{0}(S,\mathbb{Q})[2]\otimes
H^{4}(S,\mathbb{Q})[2]$, which is a vector space over $\mathbb{Q}$ of
dimension $1$. This is generated by $1\otimes x$, which is an element of
bidegree $(0,0)$. Since $1\otimes x+x\otimes
1\xmapsto{\Psi\circ\Phi^{\prime}}\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$,
the element $\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$ is in a basis of
$H^{2,2}(S^{[2]},\mathbb{Q})$.
Consider $\textnormal{Sym}^{2}(H^{2}(S,\mathbb{Q})[2])$, the third summand of
$(\ref{isoojojm})$. Using $(\ref{eq dec H2(S) con NS e T(S)})$, we can
decompose it as

$\textnormal{Sym}^{2}(H^{2}(S,\mathbb{Q})[2])\cong\textnormal{Sym}^{2}(\textnormal{Pic}(S)_{\mathbb{Q}}[2])\oplus\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2])\oplus\left(\textnormal{Pic}(S)_{\mathbb{Q}}[2]\otimes
T(S)_{\mathbb{Q}}[2]\right).$ (6.3)

By assumption $\textnormal{rk}(\textnormal{Pic}(S))=r$, so
$\textnormal{Sym}^{2}(\textnormal{Pic}(S)_{\mathbb{Q}}[2])$, whose elements
have all bidegree $(0,0)$, has dimension $\frac{(r+1)r}{2}$ as
$\mathbb{Q}$-vector space. By (5.8) and (5.9) we have:

$b_{i}\otimes
b_{i}\xmapsto{\Psi\circ\Phi^{\prime}}\frac{1}{2}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle,\qquad
b_{i}\otimes b_{j}+b_{j}\otimes
b_{i}\xmapsto{\Psi\circ\Phi^{\prime}}\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle$

for $i,j\in\\{1,\dots,r\\}$ and $i<j$. Then the elements
$\frac{1}{2}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle$, for $i=1,\dots,r$, and
$\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle$, for $1\leq i<j\leq
r$, are in a basis of $H^{2,2}(S^{[2]},\mathbb{Q})$. Note that
$\textnormal{Pic}(S)_{\mathbb{Q}}[2]\otimes T(S)_{\mathbb{Q}}[2]$ does not
contain any element of bidegree $(0,0)$. It remains to determine the elements
of bidegree $(0,0)$ in $\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2])$, i.e.,

$(\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2]))^{0,0}\cap(\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2])).$

Consider $T(S)_{\mathbb{Q}}\otimes T(S)_{\mathbb{Q}}$ with the standard
grading. Then $T(S)_{\mathbb{Q}}$ is the minimal sub-Hodge structure of
$H^{2}(S,\mathbb{Q})$ with $H^{2,0}(S)=T(S)_{\mathbb{Q}}^{2,0}$, see [Huy16,
Definition 3.2.5, Lemma 3.3.1], in particular $T(S)_{\mathbb{Q}}$ is a Hodge
structure of weight $2$. By [Huy16, Example 3.1.3, (iii)] the dual
$T(S)_{\mathbb{Q}}^{*}=\text{Hom}(T(S)_{\mathbb{Q}},\mathbb{Q})$ is a Hodge
structure of weight $-2$, and there is an isomorphism of Hodge structures of
weight $-2$ from $T(S)_{\mathbb{Q}}$ to $T(S)_{\mathbb{Q}}^{*}$. This implies
that

$(T(S)_{\mathbb{Q}}\otimes
T(S)_{\mathbb{Q}})^{2,2}\xrightarrow{\sim}(T(S)_{\mathbb{Q}}^{*}\otimes
T(S)_{\mathbb{Q}})^{0,0},$

and by [Huy16, Example 3.1.4, (iv)] we have

$E_{S}=\text{Hom}_{0}(T(S)_{\mathbb{Q}},T(S)_{\mathbb{Q}})\cong(T(S)_{\mathbb{Q}}^{*}\otimes
T(S)_{\mathbb{Q}})\cap(T(S)_{\mathbb{Q}}^{*}\otimes T(S)_{\mathbb{Q}})^{0,0},$

where $\text{Hom}_{0}(T(S)_{\mathbb{Q}},T(S)_{\mathbb{Q}})$ denotes the space
of Hodge endomorphisms on $T(S)_{\mathbb{Q}}$ of weight $0$. Since by
assumption $S$ is a general K3 surface we have
$\text{Hom}_{0}(T(S)_{\mathbb{Q}},T(S)_{\mathbb{Q}})\cong\mathbb{Q}\cdot\text{id}$.
Passing to the shifted cohomology groups, this implies that the
$\mathbb{Q}$-vector space
$(\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2]))^{0,0}\cap(\textnormal{Sym}^{2}(T(S)_{\mathbb{Q}}[2]))$
has dimension $1$. We now describe the element induced by its generator on
$H^{4}(S^{[2]},\mathbb{Q})$. Let $\\{\beta_{r+1},\dots,\beta_{22}\\}$ be an
orthogonal basis of $T(S)_{\mathbb{Q}}$ with respect to the intersection form,
and let $\\{\beta_{r+1}^{\vee},\dots,\beta_{22}^{\vee}\\}$ be the basis of
$T(S)^{*}_{\mathbb{Q}}$ given by
$\beta_{i}^{\vee}:=(\beta_{i},\,\cdot\,)\in\text{Hom}(T(S)_{\mathbb{Q}},\mathbb{Q})\cong
T(S)_{\mathbb{Q}}^{*}$ for $i\in\\{r+1,\dots,22\\}$. Then

$\textnormal{id}=\displaystyle\sum_{i=r+1}^{22}\frac{1}{(\beta_{i},\beta_{i})}\beta_{i}^{\vee}\otimes\beta_{i}\in
T(S)_{\mathbb{Q}}^{*}\otimes T(S)_{\mathbb{Q}}$

since for every $k\in\\{r+1,\dots,22\\}$ we have

$\left(\displaystyle\sum_{i=r+1}^{22}\frac{1}{(\beta_{i},\beta_{i})}\beta_{i}^{\vee}\otimes\beta_{i}\right)(\beta_{k})=\displaystyle\sum_{i=r+1}^{22}\frac{1}{(\beta_{i},\beta_{i})}(\beta_{i},\beta_{k})\cdot\beta_{i}=\beta_{k}.$

Note that $(\beta_{i},\beta_{i})\neq 0$ since
$\\{\beta_{r+1},\dots,\beta_{22}\\}$ is an orthogonal basis and the
intersection form on $H^{2}(S,\mathbb{Q})$ is non-degenerate. Since
$T(S)_{\mathbb{Q}}\cong T(S)_{\mathbb{Q}}^{*}$ by the map
$\beta_{i}\mapsto\beta_{i}^{\vee}$, the identity, seen as element in
$T(S)_{\mathbb{Q}}\otimes T(S)_{\mathbb{Q}}$, is

$\textnormal{id}=\displaystyle\sum_{i=r+1}^{22}\frac{1}{(\beta_{i},\beta_{i})}\beta_{i}\otimes\beta_{i}\in
T(S)_{\mathbb{Q}}\otimes T(S)_{\mathbb{Q}}.$ (6.4)

We see that (6.4) is invariant for the action of the symmetric group $S_{2}$
on $A\\{S_{2}\\}$, where $A:=H^{*}(S,\mathbb{Q})[2]$, so we obtain the
following element of $H^{4}(S^{[2]},\mathbb{Q})[4]$:

$\textnormal{id}\xmapsto{\Psi\circ\Phi^{\prime}}\frac{1}{2}\displaystyle\sum_{i=r+1}^{22}\frac{1}{(\beta_{i},\beta_{i})}\mathfrak{q}_{1}(\beta_{i})^{2}|0\rangle.$
(6.5)

Hence the last element of the basis of $H^{2,2}(S^{[2]},\mathbb{Q})$ is
$(\ref{smoooch})$. After some tedious computations, one can show that the
element in (6.5) can be substituted in the basis of
$H^{2,2}(S^{[2]},\mathbb{Q})$ obtained by the following element:

$\delta^{2}=-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.$

We omit this part of the proof. We conclude that
$\textnormal{dim}(H^{2,2}(S^{[2]},\mathbb{Q}))=\frac{(r+1)r}{2}+r+2$ and a
basis of the $\mathbb{Q}$-vector space $H^{2,2}(S^{[2]},\mathbb{Q})$ is the
one presented in the statement of the theorem. ∎

######  Remark 6.3.

Note that the assumption that $S$ is a _general_ K3 surface, i.e.,
$E_{S}\cong\mathbb{Q}$, is necessary in the statement of Theorem 6.2. If $S$
is general, as seen in the proof, the component
$\text{Sym}^{2}(T(S)_{\mathbb{Q}}[2])$ in $(\ref{eq decomposition Sym2(H2)})$
gives only the element (6.4) in a basis of $H^{2,2}(S^{[2]},\mathbb{Q})$ (we
have substituted it with $\delta^{2}$ in the end of the proof). If $S$ is not
general, $E_{S}$ is bigger than $\mathbb{Q}$, hence the component
$\text{Sym}^{2}(T(S)_{\mathbb{Q}}[2])$ in $(\ref{eq decomposition Sym2(H2)})$
gives not only (6.4), but also other elements in a basis of
$H^{2,2}(S^{[2]},\mathbb{Q})$.

One can combine Theorem 6.2 with Theorem 5.8 to find a basis, in terms of
Nakajima operators, of the lattice $H^{2,2}(S^{[2]},\mathbb{Z})$ of integral
Hodge classes of type $(2,2)$ of a _generic_ K3 surface $S$, which is by
definition a general projective K3 surface with Picard group of rank $1$. The
proof of this result is very similar to the one of Theorem 8.3, which
generalises it to the Hilbert square of _any_ general projective K3 surface.
For this reason, we only sketch the proof: details can be found in the
author’s PhD thesis, see [Nov, Theorem 3.3.17]. While Theorem 8.3, as we will
see, needs Theorem 8.1 to be proven, Theorem 6.4 can be shown without using
it.

######  Theorem 6.4.

Let $S$ be a generic K3 surface and $h\in\textnormal{Pic}(S)$ be the ample
generator of $\textnormal{Pic}(S)$. Then a basis of the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$ of integral Hodge classes of type $(2,2)$ is
given by the following elements:

$\begin{array}[]{c}\mathfrak{q}_{2}(h)|0\rangle,\qquad\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\qquad\frac{1}{2}\left(\mathfrak{q}_{1}(h)^{2}-\mathfrak{q}_{2}(h)\right)|0\rangle,\\\\[4.30554pt]
-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\end{array}$
(6.6)

where $x\in H^{4}(S,\mathbb{Z})$ is the class of a point, $1\in
H^{0}(S,\mathbb{Z})$ is the unit, $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the
basis of $H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the
$\mu_{i,j}^{\prime}s$ are the integers given by Table 1.

###### Sketch of the proof.

After a slight modification of the basis given in Theorem 6.2, we see that
(6.6) is a basis of $H^{2,2}(S^{[2]},\mathbb{Q})$. In order to prove the
theorem, we give a basis of the lattice $H^{4}(S^{[2]},\mathbb{Z})$ which
contains the elements in (6.6). The Picard group
$\textnormal{Pic}(S)\cong\mathbb{Z}h$ can be primitively embedded in a unique
way up to isometries in $H^{2}(S,\mathbb{Z})$, see [Huy16, Theorem 14.1.12]:
we can identify $h$ with $\alpha_{17}+t\alpha_{18}$. After some computations,
using Theorem 5.8 and the fact that $\mu_{21,22}=1$ by Table 1 (compare with
the proof of Theorem 8.3), it is possible to show that the following is a
basis of $H^{4}(S^{[2]},\mathbb{Z})$:

$\begin{array}[]{l}\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\\\\[4.30554pt]
\mathfrak{q}_{2}(\beta_{i})|0\rangle\,\,\text{for}\,\,i=1,\dots,22,\\\\[4.30554pt]
\mathfrak{q}_{1}(\beta_{i})\mathfrak{q}_{1}(\beta_{j})|0\rangle\,\,\text{for}\,\,1\leq
i<j\leq 22\,\,\,\text{and}\,\,(i,j)\neq(21,22),\\\\[4.30554pt]
\frac{1}{2}\left(\mathfrak{q}_{1}(\beta_{i})^{2}-\mathfrak{q}_{2}(\beta_{i})\right)|0\rangle\,\,\text{for}\,\,i=1,\dots,22,\\\\[4.30554pt]
-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\end{array}$
(6.7)

where $\beta_{i}:=\alpha_{i}$ for $i\neq 17$ and
$\beta_{17}:=\alpha_{17}+t\alpha_{18}$. Since the elements in (6.6) are
contained in (6.7) and $\beta_{17}=h$, we conclude that (6.6) is a basis of
$H^{2,2}(S^{[2]},\mathbb{Z})$. ∎

##  7\. Second Chern class of the Hilbert square of a K3 surface

Let $S$ be a projective K3 surface and $X:=S^{[2]}$ be its Hilbert square. In
this section we look for a representation of the second Chern class $c_{2}(X)$
of $X$ in terms of Nakajima operators: this will be fundamental to find a
basis of the lattice $H^{2,2}(X,\mathbb{Z})$ of integral Hodge classes of type
$(2,2)$ on the Hilbert square of a general projective K3 surface. Let
$q_{X}^{\vee}\in H^{2,2}(X,\mathbb{Q})$ be the dual of the BBF-form. As always
we denote by $\delta\in\textnormal{Pic}(X)$ the class such that $2\delta$ is
the class of the exceptional divisor of the Hilbert–Chow morphism. First of
all, the following proposition gives a basis of $H^{2,2}(X,\mathbb{Q})$ which
does not depend on Nakajima operators, where $X$ is the Hilbert square of a
generic K3 surface: we omit the proof.

######  Proposition 7.1.

Let $X=S^{[2]}$ be the Hilbert square of a generic K3 surface and let
$h\in\textnormal{Pic}(X)$ be the class induced by the ample generator of
$\textnormal{Pic}(S)$. Then
$\\{h^{2},h\delta,\delta^{2},\frac{2}{5}q_{X}^{\vee}\\}$ is a basis of the
$\mathbb{Q}$-vector space $H^{2,2}(X,\mathbb{Q})$.

Theorem 6.4 gives us a basis of the lattice $H^{2,2}(S^{[2]},\mathbb{Z})$,
where $S$ is a generic K3 surface, in terms of Nakajima operators. It is
natural to wonder how to describe this basis without Nakajima operators, in
particular we want to obtain a basis of $H^{2,2}(S^{[2]},\mathbb{Z})$ in terms
of some rational linear combination of
$h^{2},h\delta,\delta^{2},\frac{2}{5}q_{X}^{\vee}\in
H^{2,2}(S^{[2]},\mathbb{Q})$ of Proposition 7.1. Note that using Lemma 5.13 we
can represent $h^{2}$, $h\delta$ and $\delta^{2}$ in terms of Nakajima
operators, while we do not know how to write
$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$ of Theorem 6.4 in terms of
$h^{2},h\delta,\delta^{2},\frac{2}{5}q_{X}^{\vee}$. Hence we need to express
$\frac{2}{5}q_{X}^{\vee}\in H^{2,2}(X,\mathbb{Z})$ in terms of Nakajima
operators. Recall that $\frac{6}{5}q_{X}^{\vee}=c_{2}(X)$ by Proposition 4.4,
so we look for a representation of $c_{2}(X)\in H^{2,2}(X,\mathbb{Z})$ in
terms of Nakajima operators. The tool that we use is the EGL formula of
Proposition 5.7

Let $S$ be a projective K3 surface. We denote by
$\mathcal{T}_{2}:=\mathcal{T}_{S^{[2]}}$ the tangent bundle of $S^{[2]}$. From
now on, we will denote by $\,\cdot\,$ the cup product. We define the following
operator on the cohomology ring $H^{*}(S^{[2]},\mathbb{Q})$:

$\textnormal{ch}(\mathcal{T}_{2}):H^{*}(S^{[2]},\mathbb{Q})\rightarrow
H^{*}(S^{[2]},\mathbb{Q}),\qquad x\mapsto\textnormal{ch}(\mathcal{T}_{2})\cdot
x.$

By the general construction of $S^{[n,n+k]}$ seen in Section 5, if
$\Delta\subset S\times S$ is the diagonal, we have
$S^{[1,2]}\cong\text{Bl}_{\Delta}(S^{2})$. Diagram $(\ref{diagram EGL})$ for
$n=1$ gives

${S^{[1,2]}}$${S^{[2]}}$${S\times
S}$${S}$${\Delta}$${S.}$$\scriptstyle{\varphi}$$\scriptstyle{\rho}$$\scriptstyle{\sigma}$$\scriptstyle{\psi}$$\scriptstyle{p}$$\scriptstyle{q}$$\scriptstyle{\iota}$
(7.1)

Note that the morphisms $\varphi,\rho$ and $\psi$ appearing in diagram
$(\ref{diagram EGL n=1})$ correspond to the morphisms $\varphi,\rho$ and
$\psi$ of diagram $(\ref{diagram Nakajima projections})$, with $n=k=1$,
precomposed with the inclusion of $S^{[1,2]}$ in $S\times S\times S^{[2]}$.
With this notation, the definition of the Nakajima operator $\mathfrak{q}_{1}$
is the same of $(\ref{eq defn Nakajima op})$ without the component
$PD^{-1}[S^{[1,2]}]$, i.e., for $\alpha,x\in H^{*}(S)$ we have

$\mathfrak{q}_{1}(\alpha)(x)=\psi_{*}\left(\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right).$

By properties of cup and cap product, the latter denoted by $\cap$, we have
$\varphi^{*}(x)\cdot\rho^{*}(\alpha)=PD^{-1}\left([S^{[1,2]}]\cap\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)$
in $H^{*}(S^{[1,2]})$ for every $\alpha,x\in H^{*}(S)$. Then we get

$\begin{array}[]{lll}\textnormal{ch}(\mathcal{T}_{2})\cdot\mathfrak{q}_{1}(\alpha)(x)&=&\textnormal{ch}(\mathcal{T}_{2})\cdot\psi_{*}\left(\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&=&\textnormal{ch}(\mathcal{T}_{2})\cdot
PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&=&PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\psi^{*}(\textnormal{ch}(\mathcal{T}_{2}))\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&=&PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\psi^{!}\mathcal{T}_{2})\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),\end{array}$

where $\psi_{*}$ in the first equality is the Gysin homomorphism, while in the
other equalities is the pushforward in homology, and the third equality comes
from the projection formula. Applying Proposition 5.7 with $n=1$ and
$\omega_{S}$ trivial we get:

$\begin{array}[]{lll}\textnormal{ch}(\mathcal{T}_{2})\cdot\mathfrak{q}_{1}(\alpha)(x)&=&PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\varphi^{*}(\textnormal{ch}(\mathcal{T}_{S})\cdot
x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&&+PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L})\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&&-PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L})\cdot\sigma^{*}(\textnormal{ch}(\mathcal{O}_{\Delta}^{\vee}))\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&&+PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L^{\vee}})\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&&-PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L}^{\vee})\cdot\sigma^{*}(\textnormal{ch}(\mathcal{O}_{\Delta}))\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right)\\\\[4.30554pt]
&&-PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\varphi^{*}(x)\cdot\rho^{*}(\textnormal{ch}(2\mathcal{O}_{S}-\mathcal{T}_{S}))\right),\end{array}$
(7.2)

where $\mathcal{L}:=\mathcal{O}_{S^{[1,2]}}(-N)$ and $N$ is the exceptional
divisor of the blowing up $\sigma:\text{Bl}_{\Delta}(S^{2})\rightarrow S^{2}$.
If we set $x:=\mathfrak{q}_{1}(1)|0\rangle$ and $\alpha:=1$, we can use
formula (7.2) to compute $c_{2}(S^{[2]})$ in terms of Nakajima operators.
Recall that here the dual is defined by (5.6), while we denote by
$F^{*}:=\text{Hom}(F,\mathcal{O}_{X})$ the classical dual of a coherent sheaf
$F$. We now study the duals appearing in the right-hand side of $(\ref{eq
Chern character T2 Nakajima R1 R6})$.

######  Lemma 7.2.

Keep notation as above. Then:

  1. (i)

The dual $\mathcal{L}^{\vee}$ is isomorphic to the dual
$\mathcal{L}^{*}=\textnormal{Hom}(\mathcal{L},\mathcal{O}_{S^{[1,2]}})$.

  2. (ii)

$\mathcal{O}_{\Delta}^{\vee}=\mathcal{O}_{\Delta}$.

###### Proof.

  1. (i)

By [Har13, Proposition III.6.7] we have
$\mathcal{E}\textit{xt}^{\,i}(\mathcal{L},\mathcal{O}_{S^{[1,2]}})\cong\mathcal{E}\textit{xt}^{\,i}(\mathcal{O}_{S^{[1,2]}},\mathcal{O}_{S^{[1,2]}})\otimes\mathcal{L}^{*}$.
Moreover, by [Har13, Proposition III.6.3] we have that
$\mathcal{E}\textit{xt}^{\,i}(\mathcal{O}_{S^{[1,2]}},\mathcal{O}_{S^{[1,2]}})$
is $0$ for $i>0$ and it is equal to $\mathcal{O}_{S^{[1,2]}}$ if $i=0$. We
conclude that
$\sum_{i}(-1)^{i}\mathcal{E}\textit{xt}^{\,i}(\mathcal{L},\mathcal{O}_{S^{[1,2]}})\cong\mathcal{L}^{*}$.

  2. (ii)

We apply [Sch, Lemma 1] with $X=S\times S$, $Z=\Delta$ and
$\mathcal{L}=\mathcal{O}_{S\times S}$. We have
$\mathcal{E}\textit{xt}^{\,i}(\mathcal{O}_{\Delta},\mathcal{O}_{S\times S})=0$
if $i\neq 2$ and
$\mathcal{E}\textit{xt}^{\,2}(\mathcal{O}_{\Delta},\mathcal{O}_{S\times
S})=\text{det}\,N_{\Delta|S\times S}$. Moreover, $N_{\Delta|S\times
S}=\mathcal{T}_{S}$ and $\text{det}\,\mathcal{T}_{S}\cong\mathcal{O}_{S}$
since $S$ is a K3 surface. We conclude that
$\sum_{i}(-1)^{1}\mathcal{E}\textit{xt}^{\,i}(\mathcal{O}_{\Delta},\mathcal{O}_{S\times
S})=\mathcal{O}_{\Delta}$, as we wanted.

∎

We recall the computation of the Chern character of
$\mathcal{O}_{\Delta}:=i_{*}\mathcal{O}_{\Delta}$, where
$i:\Delta\hookrightarrow S\times S$ is the inclusion.

######  Lemma 7.3.

Let $S$ be a K3 surface and $\Delta\subset S\times S$ be the diagonal. Denote
by $[\Delta]\in H^{4}(S\times S,\mathbb{Z})$ the fundamental cohomological
class of $\Delta$ in $S\times S$. Then
$\textnormal{ch}(\mathcal{O}_{\Delta})=[\Delta]-2y$, where $y\in H^{8}(S\times
S,\mathbb{Q})$ is the class of a point in $S\times S$.

###### Proof.

Let $i:\Delta\hookrightarrow S\times S$ be the inclusion. By
Grothendieck–Riemann–Roch Theorem, see [AH59, Theorem 1] and [HA62], we have

$\begin{array}[]{lll}\textnormal{ch}(\mathcal{O}_{\Delta})&=&i_{*}\left(\textnormal{ch}(\mathcal{O}_{\Delta})\cdot\textnormal{td}(S)\right)\cdot\textnormal{td}(S\times
S)^{-1}\\\ &=&i_{*}\left(\textnormal{td}(S)\cdot i^{*}\textnormal{td}(S\times
S)^{-1}\right)\\\ &=&i_{*}\left(\textnormal{td}(S)^{-1}\right),\end{array}$

where $i_{*}$ is the Gysin homomorphism and $\mathcal{O}_{\Delta}$ in the
left-hand side is $i_{*}\mathcal{O}_{\Delta}$. Let $x\in H^{4}(S,\mathbb{Q})$
be the class of a point of $S$. The Todd class of a K3 surface is
$\text{td}(S)=1+2x$, hence we obtain
$\textnormal{ch}(\mathcal{O}_{\Delta})=[\Delta]-2y$. ∎

Consider formula $(\ref{eq Chern character T2 Nakajima R1 R6})$. We introduce
the following notation:

$\begin{array}[]{l}L1:=\textnormal{ch}(\mathcal{T}_{2})\cdot\mathfrak{q}_{1}(\alpha)(x),\\\\[3.44444pt]
R1:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\varphi^{*}(\textnormal{ch}(\mathcal{T}_{S})\cdot
x)\cdot\rho^{*}(\alpha)\right),\\\\[3.44444pt]
R2:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L})\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),\\\\[3.44444pt]
R3:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L})\cdot\sigma^{*}(\textnormal{ch}(\mathcal{O}^{\vee}_{\Delta}))\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),\\\\[3.44444pt]
R4:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L}^{\vee})\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),\\\\[3.44444pt]
R5:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\textnormal{ch}(\mathcal{L}^{\vee})\cdot\sigma^{*}(\textnormal{ch}(\mathcal{O}_{\Delta}))\cdot\varphi^{*}(x)\cdot\rho^{*}(\alpha)\right),\\\\[3.44444pt]
R6:=PD^{-1}\psi_{*}\left([S^{[1,2]}]\cap\varphi^{*}(x)\cdot\rho^{*}(\textnormal{ch}(2\mathcal{O}_{S}-\mathcal{T}_{S})\cdot
x)\right).\end{array}$

Recall that we have taken $x=\mathfrak{q}_{1}(1)|0\rangle$ and $\alpha=1$. We
can now compute $c_{2}(S^{[2]})$ in terms of Nakajima operators.

######  Proposition 7.4.

Let $S$ be a projective K3 surface. Then the second Chern class
$c_{2}(S^{[2]})\in H^{2,2}(S^{[2]},\mathbb{Z})$ of $S^{[2]}$ in terms of
Nakajima operators is

$c_{2}(S^{[2]})=27\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle+3\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+\frac{3}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle,$
(7.3)

where $1\in H^{0}(S,\mathbb{Z})$ is the unit and $x\in H^{4}(S,\mathbb{Z})$ is
the class of a point, $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of the
lattice $H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are
given in Table 1.

Note that by Table 1 the integers $\mu_{i,i}$ are all even, so (7.3) is an
element of $H^{2,2}(S^{[2]},\mathbb{Z})$.

###### Proof.

We make some computations on $L1,R1,\dots,R6$ introduced above.

  * •

We have
$L1=\textnormal{ch}(\mathcal{T}_{2})\cdot\mathfrak{q}_{1}(\alpha)(x)=\textnormal{ch}(\mathcal{T}_{2})\cdot\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(1)|0\rangle$.
By the definition of exponential Chern character and by $(\ref{formula 1sn
Nakajima})$, which gives
$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(1)|0\rangle=2\cdot 1_{S^{[2]}}$, we
obtain

$L1=8\cdot
1_{S^{[2]}}-2c_{2}(S^{[2]})+\frac{1}{6}c_{2}(S^{[2]})^{2}-\frac{1}{3}c_{4}(S^{[2]}).$

  * •

Since $S$ is a K3 surface we have $c_{1}(S)=0$ and $c_{2}(S)=24x$, where $x\in
H^{4}(S,\mathbb{Z})$ is the class of a point on $S$. Hence
$\textnormal{ch}(\mathcal{T}_{S})=2-24x$ and we obtain

$\begin{array}[]{lll}R1&=&2\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(1)|0\rangle-24\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\\\\[4.30554pt]
&=&4\cdot
1_{S^{[2]}}-24\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\end{array}$

where the second equality comes from $(\ref{formula 1sn Nakajima})$.

  * •

Let $d:=c_{1}(\mathcal{L})$. Then we have
$\textnormal{ch}(\mathcal{L})=\sum_{\nu\geq 0}\frac{1}{\nu!}d^{\nu}$. Now,
[Leh99, Lemma 3.9] implies that the cycle $[S^{[1,2]}]\cap d^{\nu}$ induces
the operator $\mathfrak{q}_{1}^{(\nu)}$, as observed in the proof of [Leh99,
Lemma 4.2], hence we obtain

$R2=\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(\alpha)\cdot
x=\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(1)\mathfrak{q}_{1}(1)|0\rangle.$

We now compute $\mathfrak{q}_{1}^{(\nu)}(1)\mathfrak{q}_{1}(1)|0\rangle$ for
every $\nu\geq 0$. If $\nu=0$, we have

$\mathfrak{q}_{1}^{(0)}(1)\mathfrak{q}_{1}(1)|0\rangle=\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(1)|0\rangle=2\cdot
1_{S^{[2]}}\in H^{0}(S^{[2]},\mathbb{Z}).$

If $\nu=1$, by Theorem 5.6 we have

$\mathfrak{q}_{1}^{\prime}(1)\mathfrak{q}_{1}(1)|0\rangle=-\mathfrak{q}_{2}(1)|0\rangle\in
H^{2}(S^{[2]},\mathbb{Z}).$ (7.4)

If $\nu=2$, we have

$\begin{array}[]{lll}\mathfrak{q}_{1}^{(2)}(1)\mathfrak{q}_{1}(1)|0\rangle&=&\left(\partial\mathfrak{q}_{1}^{\prime}-\mathfrak{q}_{1}^{\prime}\partial\right)(1)\mathfrak{q}_{1}(1)|0\rangle\\\\[4.30554pt]
&=&\partial\mathfrak{q}_{1}^{\prime}(1)\mathfrak{q}_{1}(1)|0\rangle-\mathfrak{q}_{1}^{\prime}\partial(1)\mathfrak{q}_{1}(1)|0\rangle.\end{array}$

The boundary of $S$ is empty by Remark 5.4, so
$\mathfrak{q}_{1}^{\prime}\partial(1)\mathfrak{q}_{1}(1)|0\rangle=0$.
Moreover, using $(\ref{eq doppio tilda})$, we get
$\partial\mathfrak{q}_{1}^{\prime}(1)\mathfrak{q}_{1}(1)|0\rangle=-\partial\mathfrak{q}_{2}(1)|0\rangle$,
and by Definition 5.5 we obtain

$\mathfrak{q}_{1}^{(2)}(1)\mathfrak{q}_{1}(1)|0\rangle=\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{2}(1)|0\rangle\in
H^{4}(S^{[2]},\mathbb{Z}).$

Similarly for $\nu=3$ and $\nu=4$ we obtain the following:

$\begin{array}[]{lll}\mathfrak{q}_{1}^{(3)}(1)\mathfrak{q}_{1}(1)|0\rangle=-\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{2}(1)|0\rangle\in
H^{6}(S^{[2]},\mathbb{Z}),\\\\[4.30554pt]
\mathfrak{q}_{1}^{(4)}(1)\mathfrak{q}_{1}(1)|0\rangle=\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{2}(1)|0\rangle\in
H^{8}(S^{[2]},\mathbb{Z}).\end{array}$

If $\nu\geq 5$, we obtain an element in $H^{2\nu}(S^{[2]},\mathbb{Z})=0$. We
conclude that

$\begin{array}[]{lll}R2&=&2\cdot
1_{S^{[2]}}-\mathfrak{q}_{2}(1)|0\rangle+\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\\\\[4.30554pt]
&&-\frac{1}{3}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right)\\\\[4.30554pt]
&&+\frac{1}{12}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right).\end{array}$

  * •

As before, we set $d:=c_{1}(\mathcal{L})$. By Lemma 7.2 we have
$\textnormal{ch}(\mathcal{O}_{\Delta})=\textnormal{ch}(\mathcal{O}_{\Delta}^{\vee})$,
and by Lemma 7.3 we have $\textnormal{ch}(\mathcal{O}_{\Delta})=[\Delta]-2y$,
where $y\in H^{8}(S\times S,\mathbb{Z})$ is the class of a point. Moreover,
$[\Delta]=\sum_{i,j}\mu_{i,j}\alpha_{i}\otimes\alpha_{j}+1\otimes x+x\otimes
1$ by Lemma 5.10, where $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are given in
Table 1. Recall the notation of diagram $(\ref{diagram EGL n=1})$. By the
Künneth theorem we have $y=x\otimes x$. Then

$\begin{array}[]{lll}\sigma^{*}(\textnormal{ch}(\mathcal{O}_{\Delta}))&=&\sigma^{*}\left(\displaystyle\sum_{i,j}\mu_{i,j}\alpha_{i}\otimes\alpha_{j}+1\otimes
x+x\otimes 1-2(x\otimes x)\right)\\\\[15.00002pt]
&=&\displaystyle\sum_{i,j}\mu_{i,j}\varphi^{*}(\alpha_{i})\cdot\rho^{*}(\alpha_{j})+\varphi^{*}(1)\cdot\rho^{*}(x)+\varphi^{*}(x)\cdot\rho^{*}(1)-2\left(\varphi^{*}(x)\cdot\rho^{*}(x)\right).\end{array}$

Proceeding as for $R2$ we get

$\begin{array}[]{lll}R3&=&\displaystyle\sum_{i,j}\mu_{i,j}\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(x)\mathfrak{q}_{1}(1)|0\rangle\\\\[15.00002pt]
&&+\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(1)\mathfrak{q}_{1}(x)|0\rangle-2\displaystyle\sum_{\nu\geq
0}\frac{1}{\nu!}\mathfrak{q}_{1}^{(\nu)}(x)\mathfrak{q}_{1}(x)|0\rangle.\end{array}$
(7.5)

We call $R3_{\nu=i}$ the component of $R3$ obtained by putting $\nu=i$ in
$(\ref{eq R3})$ for $i\geq 0$. Using the commutativity rule given by Theorem
5.2 we have

$R3_{\nu=0}=\displaystyle\sum_{i,j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+2\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle-2\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(x)|0\rangle.$

Note that
$\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\in
H^{4}(S^{[2]},\mathbb{Z})$ and
$\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(x)|0\rangle\in
H^{8}(S^{[2]},\mathbb{Z})$. If $\nu\geq 3$ we obtain elements in
$H^{i}(S^{[2]},\mathbb{Q})$ with $i\geq 10$, so these are equal to zero. We do
not compute explicitly $R3_{\nu=1}$: we will see that this is not necessary.
If $\nu=2$, using Definition 5.5 we obtain, after some computations,

$R3_{\nu=2}=\displaystyle\sum_{i,j}\mu_{i,j}\frac{1}{2}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right)\cdot\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,$

which is an element of $H^{8}(S^{[2]},\mathbb{Z})$. We conclude that

$\begin{array}[]{lll}R3&=&\displaystyle\sum_{i,j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+2\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle-2\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(x)|0\rangle+R3_{\nu=1}\\\\[13.00005pt]
&&+\frac{1}{2}\displaystyle\sum_{i,j}\mu_{i,j}\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\\\\[13.00005pt]
&&+\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.\end{array}$

  * •

By Lemma 7.2 we have
$\mathcal{L}^{\vee}\cong\text{Hom}(\mathcal{L},\mathcal{O}_{S^{[1,2]}})$.
Hence if $d:=c_{1}(\mathcal{L})$ we have
$\textnormal{ch}(\mathcal{L}^{\vee})=\sum_{\nu\geq
0}\frac{(-1)^{\nu}}{\nu!}d^{\nu}$, so $R4$ is computed in the same way as
$R2$, with a change of sign for the components obtained when $\nu=1$ and
$\nu=3$. We obtain

$\begin{array}[]{lll}R4&=&2\cdot
1_{S^{[2]}}+\mathfrak{q}_{2}(1)|0\rangle+\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\\\\[4.30554pt]
&&+\frac{1}{3}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right)\\\\[4.30554pt]
&&+\frac{1}{12}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right).\end{array}$

  * •

By Lemma 7.2 we have
$\mathcal{L}^{\vee}\cong\text{Hom}(\mathcal{L},\mathcal{O}_{S^{[1,2]}})$ and
$\mathcal{O}_{\Delta}^{\vee}=\mathcal{O}_{\Delta}$, so $R5$ is computed in the
same way as $R3$, with a change of sign for the component $R3_{\nu=1}$, so we
obtain

$\begin{array}[]{lll}R5&=&\displaystyle\sum_{i,j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+2\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle-2\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(x)|0\rangle-R3_{\nu=1}\\\\[13.00005pt]
&&+\frac{1}{2}\displaystyle\sum_{i,j}\mu_{i,j}\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\\\\[13.00005pt]
&&+\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.\end{array}$

  * •

Since $S$ is a K3 surface, we have
$\textnormal{ch}(2\mathcal{O}_{S}-\mathcal{T}_{S})=24x$, where $x\in
H^{4}(S,\mathbb{Z})$ is the class of a point. Then

$R6=24\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,$

where we have used
$\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(1)|0\rangle=\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$
from Theorem 5.2.

Thus formula $(\ref{eq Chern character T2 Nakajima R1 R6})$ with
$x=\mathfrak{q}_{1}(1)|0\rangle$ and $\alpha=1$ gives

$\begin{array}[]{lll}L1&=&8\cdot 1_{S^{[2]}}\\\\[4.30554pt]
&&-52\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle+2\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right)-2\displaystyle\sum_{i,j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\\\\[4.30554pt]
&&+\frac{1}{6}\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\right)\\\\[4.30554pt]
&&+4\mathfrak{q}_{1}(x)\mathfrak{q}_{1}(x)|0\rangle\\\\[4.30554pt]
&&-\displaystyle\sum_{i,j}\mu_{i,j}\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle\\\\[13.00005pt]
&&-2\left(\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\right),\\\
\end{array}$ (7.6)

where

$L1=8\cdot
1_{S}^{[2]}-2c_{2}(S^{[2]})+\frac{1}{6}c_{2}(S^{[2]})^{2}-\frac{1}{3}c_{4}(S^{[2]}).$
(7.7)

We now impose equalities between elements belonging to
$H^{4}(S^{[2]},\mathbb{Z})$ in the right-hand side of $(\ref{eq L1=})$ and
$(\ref{eq L1})$. We obtain

$c_{2}(S^{[2]})=26\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle-\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle\cdot\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle+\displaystyle\sum_{i,j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle.$

Using the commutativity rule of Theorem 5.2 and Lemma 5.13.(iii), we get

$c_{2}(S^{[2]})=27\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle+3\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+\frac{3}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle,$

and we are done. ∎

##  8\. Integral Hodge classes of type $(2,2)$ on Hilbert squares of any K3
surface

In this section we compute a basis for the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$ of integral Hodge classes of type $(2,2)$ on the
Hilbert square of a general projective K3 surface $S$ whose Picard group is
known. Proposition 7.4 implies the following.

######  Theorem 8.1.

Let $S$ be a projective K3 surface and $X=S^{[2]}$ be its Hilbert square.
Consider $q_{X}^{\vee}\in H^{2,2}(X,\mathbb{Q})$, the dual of the BBF
quadratic form. Then

$\frac{2}{5}q_{X}^{\vee}=9\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle+\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle+\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle\in
H^{2,2}(X,\mathbb{Z}),$ (8.1)

where $1\in H^{0}(S,\mathbb{Z})$ is the unit, $x\in H^{4}(S,\mathbb{Z})$ is
the class of a point, $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are the
integers given in Table 1. Moreover, $\frac{2}{5}q_{X}^{\vee}$ is indivisible
in $H^{2,2}(X,\mathbb{Z})$ and

$\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)=\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\in
H^{2,2}(X,\mathbb{Z}).$

###### Proof.

By Proposition 4.4 we have $\frac{6}{5}q_{X}^{\vee}=c_{2}(X)$, so Proposition
7.4 implies $(\ref{eq 2/5qvee con op Nakajima})$. Moreover, taking the basis
of $H^{4}(X,\mathbb{Z})$ given by Theorem 5.8, we see that
$\frac{2}{5}q_{X}^{\vee}$ is indivisible in $H^{4}(X,\mathbb{Z})$, i.e., there
is no $\alpha\in H^{4}(X,\mathbb{Z})$ such that
$n\alpha=\frac{2}{5}q_{X}^{\vee}$ for some integer $n\in\mathbb{Z}_{>1}$: by
Theorem 5.8 it suffices to find some $\mu_{i,j}$ which are coprime with $9$,
the coefficient of $\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$ in
$(\ref{eq 2/5qvee con op Nakajima})$, in Table 1. This implies that
$\frac{2}{5}q_{X}^{\vee}$ is indivisible also in $H^{2,2}(X,\mathbb{Z})$.
Recall that by Lemma 5.13 we have

$\delta^{2}=-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,$
(8.2)

thus from $(\ref{eq 2/5qvee con op Nakajima})$ and $(\ref{eq delta 2 nella
dim})$ we obtain
$\delta^{2}+\frac{2}{5}q_{X}^{\vee}=8\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\in
H^{2,2}(X,\mathbb{Z})$, which implies

$\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)=\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\in
H^{2,2}(X,\mathbb{Z}).$

∎

Note that the element
$\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)$ appears also in
[She16, Lemma 4.3]. It is possible to verify that Theorem 8.1 is consistent
with the computation of $\textnormal{ch}(S^{[2]})$ given by the Maude program
in [BNW07, $\S 11$], based on results obtained by Boissière in [Boi05], see
[Nov, $\S 3.4.3$] for details.

######  Remark 8.2.

If $S$ is a projective K3 surface, the lattice $H^{2,2}(S^{[2]},\mathbb{Z})$
is always an odd lattice: this follows from the product
$\langle\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right),\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)\rangle=1$.

We can now pass to the main result of this paper. Let $S$ be a general
projective K3 surface. In Theorem 6.2 we have given a basis of the vector
space $H^{2,2}(S^{[2]},\mathbb{Q})$ for a general projective K3 surface $S$.
Then in Theorem 6.4 we have described a basis of the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$ when $S$ is generic, i.e., general with Picard
group of rank $r=1$. We now present a basis of the lattice
$H^{2,2}(S^{[2]},\mathbb{Z})$ for _any general_ projective K3 surface $S$ with
Picard group of rank $r$, where $1\leq r\leq 19$, since $h^{1,1}(S)=20$ for a
K3 surface and, as remarked in Section 1, a K3 surface of Picard rank $20$ is
not general. We will give both a basis in terms of Nakajima operators and a
basis which does not depend on Nakajima operators. In the particular case of
general K3 surfaces, the next theorem will prove the conjecture in Section 7.
We will use results obtained in Section 7 and Theorem 8.1.

######  Theorem 8.3.

Let $S$ be a general projective K3 surface and let $\\{b_{1},\dots,b_{r}\\}$
be a basis of $\textnormal{Pic}(S)$. Then:

  1. (i)

$\textnormal{rk}(H^{2,2}(S^{[2]},\mathbb{Z}))=\frac{(r+1)r}{2}+r+2$.

  2. (ii)

The lattice $H^{2,2}(S^{[2]},\mathbb{Z})$ is odd and a basis is given by the
following elements:

     * •

$\mathfrak{q}_{2}(b_{i})|0\rangle$, for $i=1,\dots,r$,

     * •

$\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$, where $1\in
H^{0}(S,\mathbb{Z})$ is the unit and $x\in H^{4}(S,\mathbb{Z})$ is the class
of a point.

     * •

$\frac{1}{2}\left(\mathfrak{q}_{1}(b_{i})^{2}-\mathfrak{q}_{2}(b_{i})\right)|0\rangle$,
for $i=1,\dots,r$,

     * •

$\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle$, for $1\leq i<j\leq
r$,

     * •

$-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$,
where $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are given in
Table 1.

Equivalently, the following is a basis of $H^{2,2}(S^{[2]},\mathbb{Z})$:

$\left\\{b_{i}b_{j},\frac{b_{i}^{2}-b_{i}\delta}{2},\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right),\delta^{2}\right\\}_{1\leq
i\leq j\leq r.}$ (8.3)

In particular, if $S=S_{2t}$ is a generic K3 surface of degree $2t$, and
$h\in\textnormal{Pic}(S^{[2]}_{2t})$ is the class induced by the ample
generator of $\textnormal{Pic}(S_{2t})$, then

$H^{2,2}(S^{[2]}_{2t},\mathbb{Z})=\mathbb{Z}h^{2}\oplus\mathbb{Z}\frac{h^{2}-h\delta}{2}\oplus\mathbb{Z}\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)\oplus\mathbb{Z}\delta^{2}.$
(8.4)

Moreover, $\textnormal{disc}(H^{2,2}(S^{[2]}_{2t},\mathbb{Z}))=84t^{3}$ and
the Gram matrix in the basis given above is the following:

$\begin{pmatrix}12t^{2}&6t^{2}&2t&-4t\\\ 6t^{2}&t(3t-1)&t&-2t\\\ 2t&t&1&-1\\\
-4t&-2t&-1&12\end{pmatrix}.$

###### Proof.

By Theorem 6.2 we have
$\textnormal{dim}(H^{2,2}(S^{[2]},\mathbb{Q}))=\frac{(r+1)r}{2}+r+2$. Since by
[Mar07, Theorem 1] the cohomology groups $H^{i}(S^{[2]},\mathbb{Z})$ are
torsion free for $i\geq 0$, we obtain
$\textnormal{rk}(H^{2,2}(S^{[2]},\mathbb{Z}))=\frac{(r+1)r}{2}+r+2$. Remark
8.2 shows that $H^{2,2}(S^{[2]},\mathbb{Z})$ is an odd lattice. After a slight
modification of the basis given in Theorem 6.2, we have that the following is
a basis of $H^{2,2}(S^{[2]},\mathbb{Q})$:

$\begin{array}[]{l}\mathfrak{q}_{2}(b_{i})|0\rangle\,\,\,\text{for}\,\,i=1,\dots,r\\\\[4.30554pt]
\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle\\\\[4.30554pt]
\frac{1}{2}\left(\mathfrak{q}_{1}(b_{i})^{2}-\mathfrak{q}_{2}(b_{i})\right)|0\rangle\,\,\,\text{for}\,\,i=1,\dots,r\\\\[4.30554pt]
\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle\,\,\,\text{for}\,\,1\leq
i<j\leq r\\\\[4.30554pt]
-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.\end{array}$
(8.5)

The strategy of the proof is the following: we look for a sublattice $L$ of
$H^{4}(S^{[2]},\mathbb{Z})$ of maximal rank such that $L\cap
H^{2,2}(S^{[2]},\mathbb{Q})=H^{2,2}(S^{[2]},\mathbb{Z})$ and such that a basis
of $L$ contains the elements in $(\ref{base tesi})$. Since the Picard group
$\textnormal{Pic}(S)$ of $S$ can be primitively embedded in
$H^{2}(S,\mathbb{Z})$, there exists a basis of $H^{2}(S,\mathbb{Z})$ of the
form $\\{b_{1},\dots,b_{r},b_{r+1},\dots,b_{22}\\}$ for some
$b_{r+1},\dots,b_{22}\in H^{2}(S,\mathbb{Z})$. By Theorem 5.8 the following is
a basis of $H^{4}(S^{[2]},\mathbb{Z})$:

$\left\\{\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\mathfrak{q}_{2}(b_{i})|0\rangle,\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle,\frac{1}{2}\left(\mathfrak{q}_{1}(b_{i})^{2}-\mathfrak{q}_{2}(b_{i})\right)|0\rangle\right\\},$
(8.6)

where $i,j\in\\{1,\dots,22\\}$ and $i<j$. Recall that by Lemma 5.13 we have

$\delta^{2}=-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,$
(8.7)

where $\\{\alpha_{1},\dots,\alpha_{22}\\}$ is the basis of the lattice
$H^{2}(S,\mathbb{Z})$ used in Lemma 5.10 and the $\mu_{i,j}$’s are the
integers given in Table 1. Using the same procedure of Lemma 5.10 and Lemma
5.13 with the basis $\\{b_{1},\dots,b_{22}\\}$, we obtain

$\delta^{2}=-\displaystyle\sum_{i<j}\sigma_{i,j}\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\sigma_{i,i}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$
(8.8)

for some integers $\sigma_{i,j}$. Then the $\mu_{i,j}$’s are associated to the
basis $\\{\alpha_{1},\dots,\alpha_{22}\\}$ by the description of $\delta^{2}$
in (8.7), and the $\sigma_{i,j}$’s are associated to the basis
$\\{b_{1},\dots,b_{22}\\}$ by the description of $\delta^{2}$ in (8.8).

We show that there exist positive integers $l$ and $k$ with $l<k$ and $k\geq
r+1$ such that $\sigma_{l,k}\neq 0$. Suppose by contradiction that
$\sigma_{i,j}=0$ for every $(i,j)$ such that $i<j$ and $j\geq r+1$. Then
$(\ref{eq delta^2 triangolo})$ becomes

$\delta^{2}=-\displaystyle\sum_{1\leq i<j\leq
r}\sigma_{i,j}\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i=1}^{22}\sigma_{i,i}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.$
(8.9)

Consider the transcendental lattice $T(S)\cong\textnormal{NS}(S)^{\perp}$. Let
$x\in T(S)$. Since $H^{2}(S,\mathbb{Z})$ is a unimodular lattice, there exists
$y\in H^{2}(S,\mathbb{Z})$ such that $\int_{S}xy=1$. By Proposition 4.2 we
have

$\langle\delta^{2},xy\rangle=-2.$ (8.10)

By $(\ref{delta^2 eq tilde b_i})$, Lemma 5.13 and Theorem 8.1 we have

$\begin{array}[]{lll}\langle\delta^{2},xy\rangle&=&\langle-\displaystyle\sum_{1\leq
i<j\leq
r}\sigma_{i,j}\left[b_{i}b_{j}-\left(\displaystyle\int_{S}b_{i}b_{j}\right)\left(\frac{1}{8}\delta^{2}+\frac{1}{20}q_{X}^{\vee}\right)\right]\\\\[13.00005pt]
&&-\frac{1}{2}\displaystyle\sum_{i=1}^{22}\sigma_{i,i}\left[b_{i}^{2}-\left(\displaystyle\int_{S}b_{i}^{2}\right)\left(\frac{1}{8}\delta^{2}+\frac{1}{20}q_{X}^{\vee}\right)\right]\\\\[11.99998pt]
&&-\left(\frac{1}{8}\delta^{2}+\frac{1}{20}q_{X}^{\vee}\right),xy\rangle.\end{array}$
(8.11)

Since $x\in T(S)$ we have $\int_{S}b_{i}x=0$ for $i=1,\dots,r$, hence the
right-hand side of (8.11) is equal to

$\begin{array}[]{l}-\displaystyle\sum_{1\leq i<j\leq
r}\sigma_{i,j}\left[\displaystyle\int_{S}b_{i}b_{j}+\frac{1}{4}\displaystyle\int_{S}b_{i}b_{j}-\frac{5}{4}\displaystyle\int_{S}b_{i}b_{j}\right]\\\\[13.00005pt]
-\frac{1}{2}\displaystyle\sum_{i=1}^{r}\sigma_{i,i}\left[\displaystyle\int_{S}b_{i}^{2}+\frac{1}{4}\displaystyle\int_{S}b_{i}^{2}-\frac{5}{4}\displaystyle\int_{S}b_{i}^{2}\right]\\\\[13.00005pt]
-\frac{1}{2}\displaystyle\sum_{i=r+1}^{22}\sigma_{i,i}\left[\displaystyle\int_{S}b_{i}^{2}+2\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{i}y+\frac{1}{4}\displaystyle\int_{S}b_{i}^{2}-\frac{5}{4}\displaystyle\int_{S}b_{i}^{2}\right].\end{array}$

Note that
$\int_{S}b_{i}b_{j}+\frac{1}{4}\int_{S}b_{i}b_{j}-\frac{5}{4}\int_{S}b_{i}b_{j}=0$
and
$\int_{S}b_{i}^{2}+\frac{1}{4}\int_{S}b_{i}^{2}-\frac{5}{4}\int_{S}b_{i}^{2}=0$,
hence we finally obtain

$\langle\delta^{2},xy\rangle=-\displaystyle\sum_{i=r+1}^{22}\sigma_{i,i}\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{i}y-1.$
(8.12)

Thus $(\ref{eq 1 proof H^22 Z})$ and $(\ref{eq 2 proof H^22 Z})$ imply

$\displaystyle\sum_{i=r+1}^{22}\sigma_{i,i}\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{i}y=1.$
(8.13)

The $\sigma_{i,i}$’s are all even, otherwise by Theorem 5.8 the element
$\delta^{2}$ in $(\ref{eq delta^2 triangolo})$ would not be integral, since a
basis of $H^{4}(S^{[2]},\mathbb{Z})$ is given by (8.6). Hence the left-hand
side of $(\ref{eq 3 proof H^22 Z})$ is even, so it cannot be equal to $1$. We
get a contradiction, so there exist $l,k$ positive integers with $l<k$ and
$k\geq r+1$ such that $\sigma_{l,k}\neq 0$. Let now $L$ be the sublattice of
$H^{4}(S^{[2]},\mathbb{Z})$ with the following basis:

$\begin{array}[]{ll}(\text{i})&\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\\\\[4.30554pt]
(\text{ii})&\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle\,\,\,\text{with}\,\,1\leq
i<j\leq 22\,\,\,\text{and}\,\,(i,j)\neq(l,k),\\\\[4.30554pt]
(\text{iii})&\delta^{2}=-\displaystyle\sum_{i<j}\sigma_{i,j}\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\sigma_{i,i}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle,\\\\[13.00005pt]
(\text{iv})&\mathfrak{q}_{2}(b_{i})|0\rangle\,\,\,\text{for}\,\,i=1,\dots,22,\\\\[4.30554pt]
(\text{v})&\frac{1}{2}\left(\mathfrak{q}_{1}(b_{i})^{2}-\mathfrak{q}_{2}(b_{i})\right)|0\rangle\,\,\,\text{for}\,\,i=1,\dots,22.\end{array}$
(8.14)

Recall that by (8.8) the element in (iii) is also equal to

$-\displaystyle\sum_{i<j}\mu_{i,j}\mathfrak{q}_{1}(\alpha_{i})\mathfrak{q}_{1}(\alpha_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\mu_{i,i}\mathfrak{q}_{1}(\alpha_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle.$

Since $\sigma_{l,k}\neq 0$, the elements in $(\ref{basis eq proof H^2,2 Z
doppio triangolo})$ give a basis for $H^{4}(S^{[2]},\mathbb{Q})$, thus $L$ is
a sublattice of $H^{4}(S^{[2]},\mathbb{Z})$ of maximal rank. If
$\sigma_{l,k}=\pm 1$, then
$\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle$ can be obtained as
an integral linear combination of elements in $(\ref{basis eq proof H^2,2 Z
doppio triangolo})$, hence every element in the basis $(\ref{base QW b_i})$ of
$H^{4}(S^{[2]},\mathbb{Z})$ is in $L$, so $L=H^{4}(S^{[2]},\mathbb{Z})$ and
(8.5) is a basis of $H^{2,2}(S^{[2]},\mathbb{Z})$: that is what happens in the
proof of Theorem 6.4. If $\sigma_{l,k}\neq\pm 1$, then $L\neq
H^{4}(S^{[2]},\mathbb{Z})$. More precisely, we have

$\frac{H^{4}(S^{[2]},\mathbb{Z})}{L}\cong\frac{\mathbb{Z}}{|\sigma_{l,k}|\mathbb{Z}}$
(8.15)

generated by $\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle$. We
show that

$L\cap H^{2,2}(S^{[2]},\mathbb{Q})=H^{2,2}(S^{[2]},\mathbb{Z}).$

The inclusion $L\cap H^{2,2}(S^{[2]},\mathbb{Q})\subseteq
H^{2,2}(S^{[2]},\mathbb{Z})$ is clear. We now prove the inclusion $L\cap
H^{2,2}(S^{[2]},\mathbb{Q})\supseteq H^{2,2}(S^{[2]},\mathbb{Z})$ by showing
that if $z\not\in L\cap H^{2,2}(S^{[2]},\mathbb{Q})$ then $z\not\in
H^{2,2}(S^{[2]},\mathbb{Z})$. If $z\not\in H^{2,2}(S^{[2]},\mathbb{Q})$ we are
done. Suppose now that $z\not\in L$. Clearly we have
$H^{4}(S^{[2]},\mathbb{Z})\cap
H^{2,2}(S^{[2]},\mathbb{Q})=H^{2,2}(S^{[2]},\mathbb{Z})$. Since the quotient
in (8.15) is generated by
$\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle$, it suffices to show
that $\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle\not\in
H^{2,2}(S^{[2]},\mathbb{Q})$ to get the inclusion. Suppose by contradiction
that $\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle\in
H^{2,2}(S^{[2]},\mathbb{Q})$. Hence
$q_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle$ is a rational linear
combination of elements in (8.5). Recall that the last element in (8.5) can be
written as

$\delta^{2}=-\displaystyle\sum_{i<j}\sigma_{i,j}\mathfrak{q}_{1}(b_{i})\mathfrak{q}_{1}(b_{j})|0\rangle-\frac{1}{2}\displaystyle\sum_{i}\sigma_{i,i}\mathfrak{q}_{1}(b_{i})^{2}|0\rangle-\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(x)|0\rangle$

Since $\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle$ appears only
in $\delta^{2}$ among the elements in (8.5), we see that
$\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle\in
H^{2,2}(S^{[2]},\mathbb{Q})$ only if

$\sigma_{i,j}=0\,\,\text{for}\,\,i\leq j,j\geq
r+1\,\,\text{and}\,\,(i,j)\neq(l,k).$ (8.16)

Let again $x\in T(S)$ and $y\in H^{2}(S,\mathbb{Z})$ such that $\int_{S}xy=1$.
Similarly to $(\ref{eq 2 proof H^22 Z})$ we have

$\langle\delta^{2},xy\rangle=-\displaystyle\sum\limits_{\begin{subarray}{c}i<j\\\
j\geq
r+1\end{subarray}}\sigma_{i,j}\left(\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{j}y+\displaystyle\int_{S}b_{j}x\displaystyle\int_{S}b_{i}y\right)-\frac{1}{2}\displaystyle\sum_{i=r+1}^{22}\sigma_{i,i}\left(2\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{i}y\right)-1,$
(8.17)

which implies by $(\ref{eq 1 proof H^22 Z})$ the following:

$\displaystyle\sum\limits_{\begin{subarray}{c}i<j\\\ j\geq
r+1\end{subarray}}\sigma_{i,j}\left(\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{j}y+\displaystyle\int_{S}b_{j}x\displaystyle\int_{S}b_{i}y\right)+\displaystyle\sum_{i=r+1}^{22}\sigma_{i,i}\left(\displaystyle\int_{S}b_{i}x\displaystyle\int_{S}b_{i}y\right)=1.$

Thus we see that (8.16) is not true, otherwise (8.17) becomes
$\sigma_{l,k}\left(\int_{S}b_{l}x\int_{S}b_{k}y+\int_{S}b_{k}x\int_{S}b_{l}y\right)=1$,
which is false since by assumption $\sigma_{l,k}\neq\pm 1$. Hence there exists
$(\alpha,\beta)\neq(l,k)$ with $\alpha\leq\beta$ and $\beta\geq r+1$ such that
$\sigma_{\alpha,\beta}\neq 0$. As remarked above, this shows that
$\mathfrak{q}_{1}(b_{l})\mathfrak{q}_{1}(b_{k})|0\rangle\not\in
H^{2,2}(S^{[2]},\mathbb{Q})$. Then

$L\cap H^{2,2}(S^{[2]},\mathbb{Q})=H^{2,2}(S^{[2]},\mathbb{Z}).$

We conclude that (8.5) is a basis of $H^{2,2}(S^{[2]},\mathbb{Z})$. Using
Remark 5.9, Lemma 5.13 and Theorem 8.1, the elements in (8.5) can be written
as:

$\begin{array}[]{l}b_{i}\delta\qquad\text{for}\,\,i=1,\dots,r,\\\\[4.30554pt]
\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right),\\\\[4.30554pt]
\frac{1}{2}\left(b_{i}^{2}-\displaystyle\int_{S}b_{i}^{2}\cdot\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)-b_{i}\delta\right)\qquad\text{for}\,\,i=1,\dots,r,\\\\[11.99998pt]
b_{i}b_{j}-\displaystyle\int_{S}b_{i}b_{j}\cdot\frac{1}{8}\left(\delta^{2}+\frac{2}{5}q_{X}^{\vee}\right)\qquad\text{for}\,\,1\leq
i<j\leq r,\\\\[4.30554pt] \delta^{2}.\end{array}$

We write $b_{i}$ both for the element in $\textnormal{Pic}(S)$ and for the
line bundle that this induces on $\textnormal{Pic}(S^{[2]})$. By Remark 5.9 we
have $b_{i}=\mathfrak{q}_{1}(1)\mathfrak{q}_{1}(b_{i})|0\rangle$ and
$\delta=\frac{1}{2}\mathfrak{q}_{2}(1)|0\rangle$ in
$H^{4}(S^{[2]},\mathbb{Z})$. Then, starting from the basis of
$H^{2,2}(S^{[2]},\mathbb{Z})$ found before, using Lemma 5.13 one can show that
(8.3) is a basis of $H^{2,2}(S^{[2]},\mathbb{Z})$. In particular when
$S=S_{2t}$ is a generic K3 surface of degree $2t$ we have (8.4): in this case
the discriminant of $H^{2,2}(S^{[2]}_{2t},\mathbb{Z})$ and the Gram matrix can
be computed using Proposition 4.3. ∎

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