Denote by $e_{\sigma(a)}$ the endomorphism of $H^*(X,\CC)$ of multiplication by $\sigma(a)$, for $a\in HT^2(X)$.
Equation (<ref>) yields

\begin{equation}
\label{eq-eta-conjugates-m-sigma-a-to-m-a}
\eta\circ e_{\sigma(a)}\circ \eta^{-1}=m_{a}
\end{equation}

and $(e_{\sigma(a)},h,f)$ is a $\LieAlg{sl}_2$ triple, if and only if $(m_a,h',\eta\circ f\circ \eta^{-1})$
is. Thus, $\eta$ conjugates  $\LieAlg{g}_\CC$ to $\LieAlg{g}'$.

\begin{equation}
\label{eq-eta-conjugates-Lie-algebra-g-to-g-prime}
\eta\LieAlg{g}_\CC\eta^{-1}=\LieAlg{g}'.
\end{equation}

The group $\Spin(\tilde{H}(X,\CC))$ acts on $H^*(X,\CC)$ integrating the infinitesimal action of the LLV Lie algebra $\LieAlg{g}_\CC:=\LieAlg{so}(\tilde{H}(X,\CC)).$
Denote the resulting representation by

\begin{equation}
\label{eq-rho}
\rho: \Spin(\tilde{H}(X,\CC)) \rightarrow GL(H^*(X,\CC)).
\end{equation}

It is known that  $\rho$ is injective, if the odd cohomology of $X$ is non trivial. Furthermore, the restriction of $\rho$ to the even cohomology factors through an injective homomorphism

\begin{equation}
\label{eq-bar-rho}
\bar{\rho}:SO(\tilde{H}(X,\CC))\rightarrow GL(H^{ev}(X,\CC)).
\end{equation}

The following is a fundamental result of Verbitsky.

The operator $\eta$ is an involution and it belongs to the image of $\rho$.

Consequently, we get the equality

\begin{equation}
\label{eq-of-two-Lie-algebras}
\LieAlg{g}_\CC=\LieAlg{g}',
\end{equation}

since the left hand side of (<ref>) is equal to $\LieAlg{g}_\CC$. An alternative proof of Equality (<ref>) is provided in <cit.>.

The image $m(HT^2(X)))$ of $HT^2(X)$ in $\End(H\Omega_*(X))$, via the homomorphism $m$ given in (<ref>), is contained in $\LieAlg{g}_\CC$, yielding

\begin{equation}
\label{eq-m-from-HT-2-into-LLV-algebra}
m: HT^2(X)\rightarrow \LieAlg{g}_\CC.
\end{equation}

The operator $h'$ belongs to $\LieAlg{g}_\CC$.

The operator $h'$ belongs to $\LieAlg{g}'$, by definition, hence also to $\LieAlg{g}_\CC$.
Similarly, $m_a$ belongs to $\LieAlg{g}'$, for a Zariski dense open subset of $HT^2(X)$, by definition, hence also to $\LieAlg{g}_\CC$.

The $\LieAlg{g}$-equivariance of $\Psi$, given in (<ref>), and the fact that $\Psi$ maps the class $1\in SH^0(X,\CC)$
to $\alpha^n/n!\in \Sym^n(\tilde{H}(X,\CC))$ and the class $[pt]\in SH^{4n}(X,\CC)$ to $\beta^n/c_X$, induces a Hodge structure on the image of $\Psi$ such that $\alpha^n$ is of type $(0,0)$, $\beta^n$ is of type $(2n,2n)$, $\Psi$ is a morphism of Hodge structures, and the Hodge direct summand of weight $(p,q)$ in $Im(\Psi)$
is the common eigenspace of $h$ and  $h'$ of weights $p+q-2n$ and $q-p$ respectively. The Lie algebra $\LieAlg{g}$ is a sub Hodge structure of $\End(H^*(X,\QQ))$ and, given a class $\lambda\in H^{p,q}(X,\CC)$ with $p+q=2$, the operator
$e_\lambda:SH^*(X,\CC)\rightarrow SH^*(X,\CC)$ has Hodge type $(p,q)$. Hence, the same holds for the operator $e_\lambda$ acting on $Im(\Psi)$ with respect to the induced Hodge structure. Any $\LieAlg{g}$-module gets endowed with a Hodge structure and in particular so does the rational Hodge lattice $\tilde{H}(X,\QQ)$ and the latter induces one on $\Sym^n(\tilde{H}(X,\CC))$ of which the image of $\Psi$ is a sub Hodge structure with respect to the same Hodge structure induced above.
We will denote by $\tilde{H}^{p,q}_{\fine}(X)$ the Hodge direct summand of weight $(p,q)$, in order to distinguish it from the Hodge decomposition of the pure Hodge structure of weight $0$ on $\tilde{H}(X,\QQ)$ introduced in Section <ref>.
The class $\alpha$ has weight $(0,0)$ and the class $\beta$ has weight $(2,2)$ and the direct summand $H^2(X,\QQ)$ in $\tilde{H}(X,\QQ)$ is a sub Hodge structure with its original weight $2$ Hodge structure.

Theorem <ref> determines an action of $\eta$ on $\tilde{H}(X,\CC)$,
by the inverse of $\bar{\rho}$, given in (<ref>), of the restriction of $\eta$ to the even cohomology of $X$.
Conjugation by the operator $\eta$ interchanges $h$ and $h'$. Hence, $\eta(\tilde{H}^{p,q}_\fine(X))=\tilde{H}^{2-p,q}_\fine(X)$.
Denote by $\sigma$ the class in $\tilde{H}^{2,0}_\fine(X,\CC)$ corresponding to the chosen symplectic form under the identification
Given $\lambda\in H^i(\wedge^jTX)$, $i+j=2$, the operator $m_\lambda$, given in (<ref>), has Hodge type $(-j,i).$
Hence, $m_\lambda(\sigma)$ has Hodge type $(i,i)$. We get the homomorphism

\begin{eqnarray}
\label{eq-embedding-of-HT-2-in-Mukai-lattice}
\mu:HT^2(X)&\rightarrow & \tilde{H}(X,\CC)
\\
\nonumber
\lambda & \mapsto & m_\lambda(\sigma).
\end{eqnarray}

We have $(\eta\circ e_{\sigma(\lambda)}\circ\eta^{-1})(\sigma)=m_\lambda(\sigma)$, by
(<ref>). Now $\eta^{-1}(\sigma)=c\alpha$, for a non-zero scalar $c$, and the homomorphism
$\lambda\mapsto e_{\sigma(\lambda)}(\alpha)$ is injective. Hence, the homomorphism

$\mu$ is injective as well. We have the commutative diagram

\[
\xymatrix{
H^2(X,\CC) \ar[r]^e \ar@/^2pc/[rr]^{\iota} & \LieAlg{g}_\CC \ar[r]^{ev_\alpha} & \tilde{H}(X,\CC)
\\
HT^2(X) \ar[u]^\sigma \ar[r]_m \ar@/_2pc/[rr]_{\mu}&  \LieAlg{g}_\CC \ar[u]^{Ad_\eta} \ar[r]_{ev_\sigma} &
\tilde{H}(X,\CC), \ar[u]_{\frac{1}{c}\eta}
\]

where $\iota$ is the natural inclusion and $ev_\alpha$ and $ev_\sigma$ are the evaluations of an endomorphism of $\tilde{H}(X,\CC)$ on the elements $\alpha$ and $\sigma$. The top dome commutes by Equation (<ref>) and the bottom by definition of $\mu$.
The left square commutes by Equation
(<ref>) and the right by the equality $\eta^{-1}(\sigma)=c\alpha$ and the fact that $\eta=\eta^{-1}$
(Theorem <ref>). Note the equalities

\begin{equation}
\label{eq-mu-takes-Poisson-tensor-to-alpha}
\mu(H^0(\wedge^2TX))=\CC\alpha, \ \ \ \mu(H^1(TX))=H^{1,1}(X), \ \ \ \mbox{and} \ \ \ \mu(H^2(\StructureSheaf{X}))=\CC\beta.
\end{equation}

The Hodge structure of Section <ref> is related to the above by setting
$\tilde{H}^{1,-1}(X)=\tilde{H}^{2,0}_\fine(X)$, $\tilde{H}^{-1,1}(X)=\tilde{H}^{0,2}_\fine(X)$, and

\[
\tilde{H}^{0,0}(X)=\tilde{H}^{0,0}_\fine(X)\oplus \tilde{H}^{1,1}_\fine(X)\oplus \tilde{H}^{2,2}_\fine(X).
\]

In particular, the homomorphism $\mu$, given in (<ref>), is an isomorphism
between $HT^2(X)$ and $\tilde{H}^{0,0}(X)$. The homomorphism $\mu$ thus pulls back the Mukai pairing to a non-degenerate pairing on $HT^2(X)$ satisfying

\begin{equation}
\label{eq-Mukai-pairing-on-HT-2}
\end{equation}

for all $\lambda_1,\lambda_2\in HT^2(X)$.
The homomorphism $\mu$

depends only on the choice of $\sigma$, and is thus canonical up to a scalar multiple. Hence, so is the pairing (<ref>). In particular, a hyperplane $\Sigma$ in
$HT^2(X)$ determines the orthogonal line $\Sigma^\perp$ in  $HT^2(X)$, and hence the line $\ell(\Sigma)$ in
$\tilde{H}^{0,0}(X)$, which is the image of $\Sigma^\perp$ via (<ref>).

\begin{equation}
\label{eq-ell-Sigma}
\ell(\Sigma) := \mu(\Sigma^\perp):=\{m_\lambda(\sigma) \ : \ \lambda\in\Sigma^\perp\}.
\end{equation}

Assume $\sigma$ is normalized so that $(\sigma,\bar{\sigma})=1$.

The following equality holds for all $\lambda_1,\lambda_2\in HT^2(X)$.

\begin{equation}
\label{eq-formula-for-m-lambda}
\end{equation}

We have

\[
\]

where the first equality is due to the fact that $m_{\lambda_1}$ belongs to $\LieAlg{g}_\CC$, by Corollary <ref>, and the second equality follows by the definition of $\mu$.
Now $m_\lambda$ has degree $2$ with respect to $h'$, for all $\lambda\in HT^2(X)$,
and so $m_{\lambda_1}(m_{\lambda_2}(\sigma))$ belongs to $\tilde{H}^{0,2}_\fine(X)$ and is thus of the form $t\bar{\sigma}$, for a scalar $t$. We get that $t=(\sigma,t\bar{\sigma})=(\sigma,m_{\lambda_1}(m_{\lambda_2}(\sigma)))=-(\mu(\lambda_1),\mu(\lambda_2))$.

§.§ Subalgebras of the  Hochschild Lie algebra generated by a hyperplane in $HT^2(X)$

Let $\Sigma$ be a hyperplane in $HT^2(X)$. Let $\ell_0$ be a line in $\tilde{H}^{0,0}(X)=\CC\alpha\oplus H^{1,1}(X)\oplus \CC\beta$. The image  $m(\Sigma)$ of $\Sigma$ in $\LieAlg{g}_\CC$ via (<ref>) is contained in the subalgebra $\LieAlg{g}_{\CC,\ell_0}$ of $\LieAlg{g}_\CC$ annihilating $\ell_0$, if and only if  $\ell_0=\ell(\Sigma)$.

Let $\lambda_0$ be a non-zero class in $\Sigma^\perp$. Then $m_{\lambda_0}(\sigma)$ spans $\ell(\Sigma)$. Given a class $\lambda\in HT^2(X)$ we have
by Equation (<ref>) and the definition of the pairing on $HH^2(T)$.

We conclude that $m_\lambda$ annihilates $\ell(\Sigma)$, if and only if $\lambda$ belongs to $\Sigma$.

Let $g\in O(\tilde{H}(X,\CC))$ be an isometry which leaves invariant every element of $\span\{\sigma,\bar{\sigma}\}$.

\begin{equation}
\label{eq-conjugating-m-lambda-by-g}
gm_\lambda g^{-1}=m_{\mu^{-1}(g(\mu(\lambda)))},
\end{equation}

where $\mu^{-1}:\tilde{H}^{0,0}(X)\rightarrow HT^2(X)$ is the inverse of $\mu$.

We have
$(gm_\lambda g^{-1})(\sigma)=g(m_\lambda(\sigma))=g(\mu(\lambda))=m_{\mu^{-1}(g(\mu(\lambda)))}(\sigma).$
Every element of $\tilde{H}^{0,0}(X)$ is of the form $\mu(\lambda')$, for some $\lambda'\in HT^2(X)$, and

\[
\]

\[
(gm_\lambda g^{-1})(\mu(\lambda'))=-g[(\mu(\lambda),g^{-1}(\mu(\lambda'))) \bar{\sigma}]=
\]

Finally, both sides of (<ref>) annihilate $\bar{\sigma}$.

Let $U_0\subset HT^2(X)$ be a $2$-dimensional subspace such that the subspace $\mu(U_0)$ of $\tilde{H}(X,\CC)$ is defined over $\RR$ and the restriction of the pairing to $\mu(U_0)$ is non-degenerate.

Let $U_0^\perp$ be the subspace of $HT^2(X)$ which is the orthogonal complement of $U_0$ with respect to the pairing (<ref>). Then the subalgebra of
$\LieAlg{g}_\CC:=\LieAlg{so}(\tilde{H}(X,\CC))$ generated by $m(U_0^\perp)$ and its complex conjugate
$\overline{m(U_0^\perp)}$ is equal to the subalgebra
$\LieAlg{so}[\tilde{H}^{1,-1}(X)\oplus \mu(U_0^\perp)\oplus \tilde{H}^{-1,1}(X)]$ annihilating $\mu(U_0)$.

The proof is identical to Step 2 in the proof of Lemma <ref> replacing $\iota:H^1(TX)\rightarrow \LieAlg{g}_\CC$ by $m:U_0^\perp\rightarrow \LieAlg{g}_\CC$.

Let $g\in O(\tilde{H}(X,\RR))$ be an isometry which leaves invariant every element of
$\tilde{H}(X,\RR)\cap \span\{\sigma,\bar{\sigma}\}$ (so that $\tilde{H}^{0,0}(X)$ is $g$ invariant) and maps $\mu(U_0)$ to $U_\CC=\span_\CC\{\alpha,\beta\}$. Then $g(\mu(U_0^\perp))=\mu(H^1(TX))$.
Given $\lambda\in U_0^\perp$, we have
$gm_\lambda g^{-1}=m_{\mu^{-1}(g(\mu(\lambda))}$, by Lemma <ref>,
and $\mu^{-1}(g(\mu(\lambda)))$ belongs to $H^1(TX)$.
The Lie algebra generated by $m(H^1(TX))$ and its complex conjugate
$\overline{m(H^1(TX))}$ is $\bar{\LieAlg{g}}=\LieAlg{so}[\tilde{H}^{1,-1}(X)\oplus \mu(H^1(TX))\oplus \tilde{H}^{-1,1}(X)]$, by the proof of Lemma <ref>.
Hence, $m(U_0^\perp)$ and its complex conjugate
$\overline{m(U_0^\perp)}$ generate

\[
g^{-1}\bar{\LieAlg{g}}g=\LieAlg{so}[\tilde{H}^{1,-1}(X)\oplus \mu(U_0^\perp)\oplus \tilde{H}^{-1,1}(X)]
\]

as claimed.

Let $\ell\subset HT^2(X)$ be a $1$-dimensional subspace such that the line $\mu(\ell)$ in $\tilde{H}(X,\CC)$ is defined over $\RR$. Let $\Sigma$ be the subspace of $HT^2(X)$ which is the orthogonal complement of $\ell$ with respect to the pairing (<ref>). Then the subalgebra of
$\LieAlg{g}_\CC$ generated by $m(\Sigma)$ and its complex conjugate
$\overline{m(\Sigma)}$ is equal to the subalgebra
$\LieAlg{g}_{\CC,\ell}$ annihilating $\mu(\ell)$.

The subspace $m(\Sigma)$ of $\LieAlg{g}_\CC$ is contained in $\LieAlg{g}_{\CC,\ell}$,
by Lemma <ref>, and  its complex conjugate $\overline{m(\Sigma)}$
is contained in  $\LieAlg{g}_{\CC,\ell}$, since $\mu(\ell)$ is defined over $\RR$. Hence, the subalgebra $\LieAlg{a}$  of $\LieAlg{g}_\CC$ generated by $m(\Sigma)$ and
$\overline{m(\Sigma)}$ is contained in $\LieAlg{g}_{\CC,\ell}$.
Furthermore, $\LieAlg{a}$  contains the subalgebra $\LieAlg{g}_{\CC,U_0}$ annihilating $U_0$, for every two dimensional subspace $U_0$ of $\tilde{H}^{0,0}(X)$, which is defined over $\RR$, contains $\mu(\ell)$,
and such that the induced pairing on $U_0$ is non-degenerate,

by Lemma <ref>.

Choose such a $U_0$. We get the flag $\LieAlg{g}_{\CC,U_0}\subset \LieAlg{a}\subseteq \LieAlg{g}_{\CC,\ell}$, where the first inclusion is strict.

The quotient
$\LieAlg{g}_\CC/\LieAlg{g}_{\CC,U_0}$ is isomorphic to the direct sum of $\LieAlg{so}(U_0)$ and $U_0\otimes U_0^\perp$.

If $\LieAlg{a}$ contains  $U_0\otimes U_0^\perp$ then it is the whole of $\LieAlg{g}_\CC$, since
$\LieAlg{so}(U_0)$ is contained[
Choose and orthogonal basis $\{e_i\}$ of $\tilde{H}(X,\CC)$ with $e_1, e_2\in U_0$. Let
$\{e_i^*\}$ be the dual basis. Set $A=(e_1,e_1)e_1^*\otimes e_3-(e_3,e_3)e_3^*\otimes e_1$,
$B=(e_2,e_2)e_2^*\otimes e_3-(e_3,e_3)e_3^*\otimes e_2$, and
$C=(e_1,e_1)e_1^*\otimes e_2-(e_2,e_2)e_2^*\otimes e_1$. Then $A$ and $B$ belong to the submodule of $\LieAlg{g}_\CC$ isomorphic to
$U_0\otimes U_0^\perp$, $C$ spans $\LieAlg{so}(U_0)$, and
in the Lie bracket of $U_0\otimes U_0^\perp$ with itself.
Hence, $\LieAlg{a}$ intersects  $U_0\otimes U_0^\perp$ either trivially, or along
$\ell_0\otimes U_0^\perp$, for some line $\ell_0\subset U_0$.
If the intersection is trivial, then $\LieAlg{a}=\LieAlg{so}(U_0)\times \LieAlg{g}_{\CC,U_0}$, which is impossible, since
$\LieAlg{so}(U_0)$ is not contained in $\LieAlg{g}_{\CC,\ell}$.  Hence, the intersection is
$\ell_0\otimes U_0^\perp$, for some line $\ell_0\subset U_0$.
Finally, $\ell_0$ must be orthogonal to $\mu(\ell)$, since $\LieAlg{a}$ is contained in $\LieAlg{g}_{\CC,\ell}$.

If we replace in Lemma <ref> the assumption that $\mu(\ell)$ is defined over $\RR$ by the assumption that $U_0:=\mu(\ell)+\overline{\mu(\ell)}$ is two dimensional and the induced pairing on $U_0$ is non-degenerate, then the subalgebra $\LieAlg{a}$ of
$\LieAlg{g}_\CC$ generated by $m(\Sigma)$ and its complex conjugate
$\overline{m(\Sigma)}$ is the whole of $\LieAlg{g}_\CC$. Indeed, $\LieAlg{a}$ contains $\LieAlg{g}_{\CC,U_0}$,
by Lemma <ref>,
the proof of Lemma <ref>
shows that the subalgebra $\LieAlg{b}$ generated by $m(\Sigma)$ and $\LieAlg{g}_{\CC,U_0}$ is
the subalgebra $\LieAlg{c}$ generated by $\overline{m(\Sigma)}$ and $\LieAlg{g}_{\CC,U_0}$ is
$\LieAlg{g}_{\CC,\overline{\mu(\ell)}}$. Hence, $\LieAlg{a}$ contains both  $\LieAlg{g}_{\CC,\mu(\ell)}$ and $\LieAlg{g}_{\CC,\overline{\mu(\ell)}}$ and so $\LieAlg{a}/\LieAlg{g}_{\CC,U_0}$ contains $U_0\otimes U_0^\perp$. It follows that $\LieAlg{a}$
is the whole of $\LieAlg{g}_\CC$, as we saw in the proof of Lemma <ref>.

Let $K$ be a subfield of $\CC$,
let $\ell$ be a line in $\widetilde{H}(X,\CC)$, and let $j$ a positive integer. If the line $\ell^j$ in $\Sym^j\widetilde{H}(X,\CC)$
is defined over $K$, then so does $\ell$.

The Veronese embedding $\varphi:\PP(\widetilde{H}(X,\CC))\rightarrow \PP(\Sym^{j}\widetilde{H}(X,\CC))$, $\ell\mapsto \ell^j$, is defined over $\QQ$ and is thus equivariant with respect to every automorphism $g$ of $\CC$.
Assume that $\ell^j$ is defined over $K$.
The injectivity of  $\varphi$ implies that $\ell$ is invariant with respect to every automorphism of $\CC$ leaving $K$ invariant.
Hence, $\ell$ is defined over $K$.

Let $f$ be a non-zero element of the kernel of $\Delta:\Sym^n(\tilde{H}(X,\QQ))\rightarrow \Sym^{n-2}(\tilde{H}(X,\QQ))$ of Hodge type $(0,0)$.
Let $\Sigma$ be a subspace of $HT^2(X)$ of co-dimension $1$.
Assume that $m(\Sigma)$ is contained in the subalgebra $\LieAlg{g}_{\CC,f}$ annihilating $f$. Then the line $\ell(\Sigma)\subset \tilde{H}(X,\CC)$, given in (<ref>), is defined over $\QQ$, is contained in $\tilde{H}^{0,0}(X)$,  and
$f$ spans the image of the projection of $\ell(\Sigma)^n$ to $\ker(\Delta)$. Furthermore, $\LieAlg{g}_{\CC,f}=\LieAlg{g}_{\CC,\ell(\Sigma)}$.

The element $f$ has the form
\sum_{i=0}^{\lfloor n/2\rfloor}(\sigma\bar{\sigma})^i f_i,
where $f_i$ belongs to $\Sym^{n-2i}(\tilde{H}^{0,0}(X))$, since $f$ has Hodge type $(0,0)$.

Step 1: We derive a system (<ref>) and (<ref>)
of partial differential equations that the $f_i$ satisfy.
Let $\{x_1, \dots, x_N\}$ be an orthonormal basis of $\tilde{H}^{0,0}(X)$.
We have

\[
\Delta\left((\sigma\bar{\sigma})^i f_i\right)=
\]

\begin{equation}
\label{eq-first-recursive-equation}
\Delta(f_{i-1})+i^2f_{i}=0,
\end{equation}

for $1\leq i\leq\lfloor n/2\rfloor$,
since $f$ belongs to the kernel of $\Delta$.

Given $\lambda\in HT^2(X)$, we have $m_\lambda(x_i)=-(\mu(\lambda),x_i)\bar{\sigma}$, by Equation (<ref>). Thus,

\[
(\sigma\bar{\sigma})^i x_{i_1}\cdots x_{i_{n-2i}}
\right)=
i\mu(\lambda)\sigma^{i-1}\bar{\sigma}^i x_{i_1}\cdots x_{i_{n-2i}}-
\sigma^i \bar{\sigma}^{i+1} \sum_{j=1}^{n-2i}(\mu(\lambda),x_{i_j})x_{i_1}\cdots \hat{x}_{i_j} \cdots x_{i_{n-2i}}.
\]

Let $\mu(\lambda)^*$ be the element of $\tilde{H}^{0,0}(X)^*$ satisfying $\mu(\lambda)^*(\lambda')=(\mu(\lambda),\lambda')$, for all $\lambda'\in \tilde{H}^{0,0}(X)$. Then

\[
m_\lambda((\sigma\bar{\sigma})^i f_i)=i\mu(\lambda)\sigma^{i-1}\bar{\sigma}^i f_i
- \sigma^i \bar{\sigma}^{i+1} [f_i\lrcorner \mu(\lambda)^*].
\]

\[
\sum_{i=1}^{\lfloor n/2\rfloor+1} \left(
i\mu(\lambda)f_i-[f_{i-1}\lrcorner \mu(\lambda)^*]
\right)\sigma^{i-1}\bar{\sigma}^i=0,
%-[f_0\lrcorner \mu(\lambda)^*]\bar{\sigma}+\mu(\lambda)f_1-\sigma\bar{\sigma}^2[f_1\lrcorner \mu(\lambda)^*]=0,
\]

for all $\lambda\in \Sigma$, where we set $f_{\lfloor n/2\rfloor+1}=0$.
Hence, for all $\lambda\in \Sigma$ and $0<i\leq \lfloor n/2\rfloor+1$ we have

\begin{equation}
\label{eq-second-recursive-equation}
i\mu(\lambda)f_i-[f_{i-1}\lrcorner \mu(\lambda)^*]=0.
\end{equation}

ix_jf_i=\frac{\partial f_{i-1}}{\partial x_j},

if $x_j$ belongs to $\mu(\Sigma)$.

Step 2: We prove that $\ell(\Sigma)$ is defined over $\RR$.
If $n$ is odd, then $f_{\lfloor n/2\rfloor}$ is linear and Equation (<ref>) for $i=1+\lfloor n/2\rfloor$ implies that $f_{\lfloor n/2\rfloor}$ is orthogonal to $\mu(\Sigma)$ and so belongs to $\ell(\Sigma)$. Note that $f_i$ is real, for all $i$, since $f$ is rational.
If $f_{\lfloor n/2\rfloor}$ does not vanish, then  $\ell(\Sigma)$ is defined over $\RR$.
If $f_{\lfloor n/2\rfloor}$ does vanish, let $j$ be the maximal index, such that $f_j\neq 0$, but $f_{j+1}=0$.
Equation  (<ref>) for $i=j+1$ implies that $f_j$ belongs to $\ell(\Sigma)^{n-2j}$, which is thus defined over $\RR$, and again we get that
$\ell(\Sigma)$ is defined over $\RR$, by Lemma <ref>.

If $n$ is even, then $f_{n/2}$ is constant. If it vanishes, then the above argument proves that $\ell(\Sigma)$ is defined over $\RR$. Assume that $f_{n/2}=t$, where $t\neq 0$. Set $g:=f_{n/2-1}$. Assume first that $\ell(\Sigma)$ is not isotropic. We can then assume that $x_1$ belongs to $\ell(\Sigma)$.
Equation (<ref>) for $i={n/2}$ implies that
$\frac{\partial g}{\partial x_j}=\frac{nt}{2}x_j$, for $j\geq 2$. So $g=\frac{nt}{4}(x_2^2+ \ \cdots \ +x_N^2)+cx_1^2$.
Equation (<ref>) yields $-\frac{tn^2}{4}=\Delta(g)=\frac{nt(N-1)}{4}+c$, so
We get

\[
f_{n/2-1}=\frac{nt}{4}(x_1^2+ \ \cdots \ +x_N^2)-\left(\frac{tn(N+n)}{4}\right)x_1^2.
\]

Now, $t$ is real, as is $f_{n/2-1}$. The element $x_1^2+ \ \cdots \ +x_N^2$ is independent of the choice of an orthonormal basis and is thus real. The coefficient $\frac{tn(N+n)}{4}$ is a non-zero real number. Hence, $x_1^2$ is real and the line
$\ell(\Sigma)$ is defined over $\RR$.

Assume that $n$ is even and $\ell(\Sigma)$ is isotropic. We claim that $f_{n/2}=0.$
Set $t:=f_{n/2}$ and $g:=f_{n/2-1}$. We may assume that $\ell(\Sigma)$ is spanned by $x_1+\sqrt{-1}x_2$. Then
$x_1+\sqrt{-1}x_2$ belongs to $\mu(\Sigma)$.
Equation (<ref>) for $i={n/2}$ implies that
$\frac{\partial g}{\partial x_j}=\frac{nt}{2}x_j$, for $j\geq 3$ and

\[
(n/2)(x_1+\sqrt{-1}x_2)t=g\lrcorner (x_1+\sqrt{-1}x_2)^*
\]

\[
f_{n/2-1}=\frac{nt}{4}(x_3^2+ \ \cdots \ +x_N^2)+\frac{nt}{4}(x_1+\sqrt{-1}x_2)(x_1-\sqrt{-1}x_2)
\]

for some scalar $a$. Equation (<ref>) yields

\[
-\frac{tn^2}{4}= \Delta(f_{n/2-1})=\frac{nt}{4}(N-2)+\frac{nt}{2}=\frac{nNt}{4}.
\]

Hence, $t=0$ as claimed. We conclude once again that $\ell(\Sigma)$ is defined over $\RR$.

Step 3: The subalgebra $\LieAlg{a}$ of $\LieAlg{g}_\CC$ generated by $m(\Sigma)$ and its complex conjugate
$\overline{m(\Sigma)}$ is equal to the subalgebra $\LieAlg{g}_{\CC,\ell(\Sigma)}$  of $\LieAlg{g}_\CC$ annihilating $\ell(\Sigma)$,
by Lemma <ref>. The subalgebra $\overline{m(\Sigma)}$
is contained in $\LieAlg{g}_{\CC,f}$, since $f$ is rational and $m(\Sigma)$ is contained in $\LieAlg{g}_{\CC,f}$. Hence,
$\LieAlg{g}_{\CC,\ell(\Sigma)}$ is contained in $\LieAlg{g}_{\CC,f}$.

Let $\lambda$ be a non-zero element of $\ell(\Sigma)$.
The trivial $\LieAlg{g}_{\CC,\ell(\Sigma)}$-submodule in $\Sym^n(\tilde{H}(X,\CC))$ has the basis
$\{\lambda^{n-2i}\tilde{q}^i\}_{i=0}^{\lfloor n/2\rfloor}$, where $\tilde{q}$ is the $\LieAlg{g}_\CC$-invariant degree $2$
element given in Example <ref>.
$\Delta:\Sym^n(\tilde{H}(X,\CC))\rightarrow \Sym^{n-2}(\tilde{H}(X,\CC))$ is a surjective $\LieAlg{g}_\CC$-modules homomorphism, hence it induces a surjective homomorphism of the respective trivial $\LieAlg{g}_{\CC,\ell(\Sigma)}$-submodules.
We conclude that the trivial $\LieAlg{g}_{\CC,\ell(\Sigma)}$-submodule $\ker(\Delta)^{\LieAlg{g}_{\CC,\ell(\Sigma)}}$
of $\ker(\Delta)$ is one-dimensional.
If $\ell(\Sigma)$ is isotropic, then $\lambda^n$ spans $\ker(\Delta)^{\LieAlg{g}_{\CC,\ell(\Sigma)}}$,
and so $f$ is a scalar multiple of $\lambda^n$. We conclude that $\ell(\Sigma)^n$ is defined over $\QQ$, hence so is $\ell(\Sigma)$, by Lemma <ref>.

If $\ell(\Sigma)$ is non-isotropic, then the projection of $\lambda^n$ to $\ker(\Delta)$ spans
$\ker(\Delta)^{\LieAlg{g}_{\CC,\ell(\Sigma)}}$, and so $f$ belongs to the projection of $\ell(\Sigma)^n.$
Hence, $\LieAlg{g}_{\CC,f}$ is contained in $\LieAlg{g}_{\CC,\ell(\Sigma)}$.
The equality $\LieAlg{g}_{\CC,\ell(\Sigma)}=\LieAlg{g}_{\CC,f}$ follows, regardless of whether $\ell(\Sigma)$ is isotropic.

It remains to prove that $\ell(\Sigma)$ is defined over $\QQ$ when $\ell(\Sigma)$ is non-isotropic.
The quadric $Q$ in $\PP(\widetilde{H}(X,\CC))$ of isotropic lines is defined over $\QQ$ and the morphism

\[
\PP(\widetilde{H}(X,\CC))\setminus Q\rightarrow Gr\left(\Choose{N+1}{2},\LieAlg{g}_\CC\right),
\]

sending $\ell$ to $\LieAlg{g}_{\CC,\ell}$, is defined over $\QQ$ (and is thus equivariant with respect to the group of field automorphisms of $\CC$, and is injective. Hence, if $\LieAlg{g}_{\CC,\ell}$ is defined over $\QQ$, then so is $\ell$, as both are invariant under all field automorphisms of $\CC$. We already know that
$\LieAlg{g}_{\CC,\ell(\Sigma)}=\LieAlg{g}_{\CC,f}$ is defined over $\QQ$. Hence, so it $\ell(\Sigma)$.

§.§ The LLV subspace of an object in $D^b(X)$

The Mukai vector  of an object $F\in D^b(X)$ is the class $v(F):=ch(F)\sqrt{td_{X}}$ in $H^*(X,\RationalNumbers)$.
Let $\LieAlg{g}_{v(F)}$ be the subalgebra of $\LieAlg{g}$ annihilating $v(F)$.
Let $\obs^{HT}_F$ be the obstruction map (<ref>) and set $\Sigma(F):=\ker\left(\obs^{HT}_F\right)$.
Let $\tilde{\Sigma}(F)$ be the image of the subspace $\Sigma(F)$  of $HT^2(X)$ via the graded Duflo isomorphism
$D:HT^*(X)\rightarrow HT^*(X)$, given by contraction with $\sqrt{td_X}$.

\[
\tilde{\Sigma}(F):=D(\Sigma(F))=D\left(\ker\left(\obs^{HT}_F\right)\right).
\]

Let $X$ be a projective irreducible holomorphic symplectic manifold and $F\in D^b(X)$ an object.
The image $m(\tilde{\Sigma}(F))$ of the subspace $\tilde{\Sigma}(F)$  of $HT^2(X)$ via (<ref>) is contained in the complexification of $\LieAlg{g}_{v(F)}$.

The image $m(\tilde{\Sigma}(F))$ is contained in $\LieAlg{g}_\CC$, by Corollary <ref>. We prove that it annihilates $v(F)$ in two steps.

Step 1: Let $\Delta:X\rightarrow X\times X$ be the diagonal embedding.
Then $HH^i(X):=\Hom(\Delta_*\StructureSheaf{X},\Delta_*\StructureSheaf{X}[i])$ and
$HH_i(X):=\Hom(\Delta_!\StructureSheaf{X}[i],\Delta_*\StructureSheaf{X})$, where $\Delta_!$ is the left adjoint of $\Delta^*$.
We have $\Delta_!\StructureSheaf{X}\cong \Delta_*\omega_X^{-1}[-2n]$, where $\dim(X)=2n$.
Hence, $HH_0(X)=\Hom(\Delta_*\omega_X^{-1}[-2n],\Delta_*\StructureSheaf{X})$.
Elements of $HH^i(X)$ yield natural transformations from the identity functor $id$ of $D^b(X)$ to $id[i]$. We get the homomorphism
\obs_F:HH^2(X)\rightarrow \Hom(F,F[2])
sending $\alpha$ to the value $\alpha_F$ of the natural transformation on the object $F$.
The Chern character $\widetilde{ch}(F)$ is defined as an element of $HH_0(X)$ <cit.>.
We prove first that the kernel $\Sigma'(F)$ of $\obs_F$ is contained in the ideal of $HH^*(X)$ annihilating $\widetilde{ch}(F)$.

Given a smooth complex projective variety $Y$ and an object $E$ of $D^b(Y)$, let
Let $Tr_Y:\Hom(E,E\otimes\omega_Y[\dim Y])\rightarrow \CC$ be the Serre's Duality trace (see <cit.>).
The Chern character $\widetilde{ch}(F)$ satisfies

\[
Tr_{X\times X}(\widetilde{ch}(F)\circ \nu)=Tr_X(\nu_F),
\]

for all $\nu\in \Hom(\Delta_*\StructureSheaf{X},\Delta_*\omega_X[2n])$, where $\nu_F\in \Hom(F,F\otimes\omega_X[2n])$
is the value on $F$ of the natural transformation $\nu$ from $id$ to the Serre functor of tensorization by $\omega_X[2n]$
<cit.>. In particular, for every $\alpha\in HH^2(X)$ and $\beta\in \Hom(\Delta_*\StructureSheaf{X},\Delta_*\omega_X[2n-2])$, setting $\nu=\alpha\circ\beta$ above we get

\[
Tr_{X\times X}([\widetilde{ch}(F)\circ \alpha]\circ\beta)=Tr_X(\alpha_F\circ \beta_F).
\]

If $\alpha_F=0$, then the element $\widetilde{ch}(F)\circ \alpha$ of $HH_{-2}(X)=\Hom(\Delta_*\omega_X^{-1}[-2-2n],\Delta_*\StructureSheaf{X})$ pairs to zero with every class $\beta$ as above. Hence, $\widetilde{ch}(F)\circ \alpha=0,$
because the trace pairing is non-degenerate.

Step 2:
Consider the  Hochschild-Kostant-Rosenberg isomorphisms

\begin{eqnarray*}
I^{HKR} : HT^*(X) & \rightarrow & HH^*(X),
\\
I_{HKR} : HH_*(X) & \rightarrow & H\Omega_*(X).
\end{eqnarray*}

Set $I^K:=I^{HKR}\circ D^{-1}$ and let $I_K$ be given by $I_K(\alpha)=I_{HKR}(\alpha)\sqrt{td_X}$.
We have $I_K(\widetilde{ch}(F))=ch(F)\sqrt{td_X}=v(F)$, by <cit.>,
$I_K(I^K(\alpha)\circ \widetilde{ch}(F))=\alpha\cdot I_K(\widetilde{ch}(F))$,  for all $\alpha\in HT^*(X)$, by <cit.>.
Hence, the image via $(I^K)^{-1}$ of
the kernel $\Sigma'(F)$ of $\obs_F$ annihilates $v(F)$, by Step 1 of the proof.

\begin{equation}
\label{I^K-inverse-maps-Sigma-prime-to-stabilizer}
m\left((I^K)^{-1}(\Sigma'(F))\right)\subset \LieAlg{g}_{\CC,v(F)}.
\end{equation}

We have $\obs_F(I^{HKR}(\alpha))=(\exp a_F)(\alpha)$, for all $\alpha\in HT^2(X)$, by <cit.>.
Hence,  $I^{HKR}$ maps
the kernel $\Sigma(F)$ of the obstruction map $\obs^{HT}_F$, given in (<ref>), to
the kernel $\Sigma'(F)$ of $\obs_F$. So $(I^K)^{-1}(\Sigma'(F))=D(\Sigma(F))=\tilde{\Sigma}(F)$.
We conclude that  $m(\tilde{\Sigma}(F))$ annihilates $v(F)$, by Equation (<ref>).

Let $X$ be a $2n$-dimensional irreducible holomorphic symplectic manifold. Composing the homomorphism $m:HT^2(X)\rightarrow \LieAlg{g}_\CC$, given in Equation (<ref>), with the inclusion
$\LieAlg{g}_\CC\subset \End(H^*(X,\CC))$ followed by evaluation at $v(F)$ we get the cohomological obstruction map

\[
\obs_F^{Coh}:HT^2(X)\rightarrow H^*(X,\CC).
\]

* We say that an object $F$ in $D^b(X)$
deforms in co-dimension $c$, if the kernel $\Sigma(F)$ of the obstruction map (<ref>) has co-dimension $c$ in $HT^2(X)$.

We say that $F$ has a rank $c$ cohomological obstruction map, if $\rank(\obs_F^{Coh})=c$, i.e., if
the annihilator $\LieAlg{g}_{v(F)}$ of the Mukai vector $v(F)$ intersects $m(HT^2(X))$ in a co-dimension $c$ subspace of the latter.

* If $F$ is $\theta$-twisted of non-zero rank, we say that
$F$ has a rank $c$ cohomological obstruction map

if the assumption in Part (<ref>) holds with $v(F)$ replaced by

If $F$ deforms in co-dimension $c$, then $F$ has a rank $\leq c$ cohomological obstruction map,

since $m(\tilde{\Sigma}(F))$ is contained in $\LieAlg{g}_{v(F)}$
by Lemma <ref>.

Important examples of modular sheaves, which deform in co-dimension $2$, are constructed in [51] in the $K3^{[n]}$ deformation type and in [56] in the generalized Kummer deformation type.

Let $k$ be a positive integer. If $F_i$, $1\leq i\leq k$, each deforms in co-dimension one and each has positive rank, then $\otimes_{i=1}^k F_i$ deforms in co-dimension $\leq 2$. Indeed, in this case $\Sigma(F_i)$ does not contain $H^2(\StructureSheaf{X})$, as was explained in the proof of
Proposition <ref>,
and so the intersection $\Sigma(F_i)\cap [H^1(TX)\oplus H^2(\StructureSheaf{X})]$
is the graph of a linear homomorphism $\theta_i:H^1(TX)\rightarrow H^2(\StructureSheaf{X})$
and $F_i$ deforms in the direction $\xi+\theta_i(\xi)$ for all $\xi\in H^1(TX)$. Then $\otimes_{i=1}^k F_i$ deforms in the direction
$\xi+(\sum_{i=1}^k\theta_i)(\xi)$, for each $\xi\in H^1(TX)$. So $\Sigma(\otimes_{i=1}^k F_i)\cap [H^1(TX)\oplus H^2(\StructureSheaf{X})]$ contains the graph of $\sum_{i=1}^k\theta_i$.
The linear homomorphism $\theta_i$ is computed in Lemma <ref> below.

The LLV subspace of an object $F$ in $D^b(X)$, is the subspace of $\tilde{H}(X,\QQ)$ given by

\[
\]

generalizing Equation (<ref>).

Note that $F$ deforms in co-dimension $\dim(W(F))$.
If one-dimensional, the LLV subspace $W(\otimes_{i=1}^k F_i)$ of the tensor product discussed above is described by
Theorem <ref> (<ref>), while if it is two dimensional it is computed in Lemma <ref>.
Given a subspace $W\subset \tilde{H}(X,\RationalNumbers)$, let $\LieAlg{g}_W$ be the subalgebra of $\LieAlg{g}$ annihilating $W$.
If $F$  has a rank $1$ cohomological obstruction map,

and the projection $\hat{v}(F)$ of $v(F)$ to $SH^*(X)$ does not vanish,
then $\LieAlg{g}_{v(F)}=\LieAlg{g}_{\ell(F)}$, for a rational line $\ell(F)$ in
$\tilde{H}(X,\QQ)$, by Lemma <ref> applied with $f=\Psi(\hat{v}(F))$. The equality $\ell(F)=W(F)$ holds, if furthermore $F$ deforms in co-dimension $1$.

§.§ The LLV line

Let $X$ be a projective irreducible holomorphic symplectic manifold and $F\in D^b(X)$ an object which deforms in co-dimension $1$, so that $\tilde{\Sigma}(F)$ has co-dimension one in $HT^2(X)$.
Denote by $\ell(F)$ the line $\ell(\tilde{\Sigma}(F))$ in  $\tilde{H}(X,\CC)$ given in (<ref>).
Assume[The assumption holds if $\rank(F)\neq 0$, or $c_1(F)\neq 0$. The assumption depends only on the orbit of $v(F)$ under the derived monodromy group.]
that the projection $\hat{v}(F)$ of $v(F)$ to $SH^*(X)$  via (<ref>) does not vanish.

The line $\ell(F)$ is defined over $\QQ$ and
$\LieAlg{g}_{v(F)}$ is equal to the Lie subalgebra $\LieAlg{g}_{\ell(F)}$ annihilating $\ell(F)$.

The embedding
$\Psi$, given in (<ref>), maps $\hat{v}(F)$

into the  image of $\ell(F)^n$ via the projection (<ref>).

If the rank $r$ of $F$ does not vanish, then
the line $\ell(F)$ is spanned by an element of the form $r\alpha+c_1(F)+s(F)\beta$,

for some $s(F)\in\RationalNumbers$.

Given an equivalence $\Phi:D^b(X)\rightarrow D^b(Y)$, let $\tilde{H}(\phi):\tilde{H}(X,\QQ)\rightarrow
\tilde{H}(Y,\QQ)$ be its associated isometry via the functor (<ref>).

We have

The inclusion
$\LieAlg{g}_{\ell(F)} \subset \LieAlg{g}_{v(F)}$
follows from the assumption that $\Sigma(F)$ has co-dimension one in $HT^2(X)$, Lemma <ref>, and Lemma <ref>.
The equality $\LieAlg{g}_{\ell(F)}=\LieAlg{g}_{\hat{v}(F)}$ and the fact that $\ell(F)$ is defined over $\QQ$ both follow from
Lemma <ref>, by letting $f$ be $\Psi(\hat{v}(F))$.
We conclude that $\LieAlg{g}_{\ell(F)}=\LieAlg{g}_{v(F)}$ from the evident inclusion $\LieAlg{g}_{v(F)}\subset \LieAlg{g}_{\hat{v}(F)}$.

Part (<ref>)
follows from Lemma <ref>, by letting $f$ be $\Psi(\hat{v}(F))$.

The class $\kappa(F)$ is $\bar{\LieAlg{g}}$-invariant, by Proposition <ref>.
Hence, so is $\kappa(F)\sqrt{td_X}$.
Set $g:=\exp(e_{-c_1(F)/r})\in SO(\tilde{H}(X,\QQ))$. Then $g$ acts on $H^{ev}(X,\CC)$, via (<ref>), and  $\kappa(F)\sqrt{td_X}=g(v(G)),$ by Theorem <ref>.
The map $F\mapsto \ell(F)$ factors through $v$ and the map $v(F)\mapsto \ell(F)$ extends to an $SO(\tilde{H}(X,\CC))$-equivariant map from the $SO(\tilde{H}(X,\CC))$-orbit of $v(F)$ to the $SO(\tilde{H}(X,\CC))$-orbit of $\ell(F)$,

since the  embedding $\Psi$, the two projections
(<ref>) and
(<ref>), and the $n$-power map $\ell(F)\mapsto \ell(F)^n$, are all
Hence, $g(\ell(F))$ is $\bar{\LieAlg{g}}$-invariant. So
$g(\ell(F))$ is contained in $U_\QQ$.

Let $w$ be a non-zero vector in $\ell(F)$.
Write $w=r_0\alpha+\lambda+s\beta\in \tilde{H}(X,\RationalNumbers)$, with $r_0,s\in\RationalNumbers$,  and $\lambda\in H^2(X,\RationalNumbers)$.
Note that $r_0\neq 0$. Indeed, $\exp(e_{\lambda'})(v(F))$
is not $\bar{\LieAlg{g}}$-invariant, for $\lambda'\neq -c_1(F)/r$, $\lambda'\in H^2(X,\QQ)$, and so
neither is $\exp(e_{\lambda'})(\ell(F))$,
by the equivariance of the map $v(F)\mapsto \ell(F)$. But if $r_0=0$, then $\ell(F)$ is $\bar{\LieAlg{g}}$-invariant, if and only if
$\exp(e_{\lambda'})(\ell(F))$ is, for all $\lambda'\in H^2(X,\QQ)$.
\left(s-\frac{(\lambda,\lambda')}{r'}+\frac{r_0(\lambda,\lambda')}{2(r')^2}\right)\beta$

belongs to $U_\RationalNumbers$ if and only if $\lambda'/r'=\lambda/r_0$. We know that
$\exp(e_{-c_1(F)/r})w$ belongs to $U_\QQ$. Hence, $\lambda/r_0=c_1(F)/r$ and  $(r/r_0)w$ has the stated form.

(<ref>) Let $\phi:H^*(X,\QQ)\rightarrow H^*(Y,\QQ)$ be the isomorphism associated to $\Phi$.
Then $\phi(v(F))=v(\Phi(F))$ and so $\phi(\hat{v}(F))=\hat{v}(\Phi(F))$.

Theorem <ref> yields the commutative diagram

\[
\xymatrix{
SH^*(X,\QQ) \ar[d]_{\Psi_X} \ar[r]^{\epsilon(\tilde{H}(\phi))\phi} & SH^*(Y,\QQ) \ar[d]^{\Psi_Y}
\\
\ker(\Delta_X) \ar[r]_{\Sym^n(\tilde{H}(\phi))} & \ker(\Delta_Y),
\]

where $\Delta_X:\Sym^n(\tilde{H}(X,\QQ))\rightarrow \Sym^{n-2}(\tilde{H}(X,\QQ))$ is given in (<ref>) and $\Delta_Y$ is its analogue.

Hence, $\Sym^n(\tilde{H}(\phi))$ maps $\Psi(\hat{v}(F))$ to $\Psi(\hat{v}(\Phi(F)))$.
This means that $\Sym^n(\tilde{H}(\phi))$ maps the projection of $\ell(F)^n$ in $\ker(\Delta_X)$ to the projection of $\ell(F)^n$
in $\ker(\Delta_Y)$. The map

sending $\ell(F)$ to the projection of $\ell(F)^n$, is injective, by the equality $\LieAlg{g}_{\CC,\hat{v}(F)}=\LieAlg{g}_{\CC,\ell(F)}$ in Lemma

<ref>. Hence, $\tilde{H}(\phi)(\ell(F))=\ell(\Phi(F))$.

The class $\kappa(F)$ is $\bar{\LieAlg{g}}$-invariant, by Proposition <ref>.
Hence, so is $\kappa(F)\sqrt{td_X}$. Set $n=\dim_\CC(X)/2$ and let $h$ be the grading operator acting on
$H^k(X,\RationalNumbers)$ by $k-2n$. The operator $h$ belongs to $\LieAlg{g}$  and commutes with $\bar{\LieAlg{g}}$ and hence belongs to $\LieAlg{so}(U_\RationalNumbers)$, yet it does not annihilate $\kappa(F)\sqrt{td_X}$, since the graded summand in $H^0(X,\RationalNumbers)$ of the latter is $r$ and so it does not vanish,  by assumption. Hence,
the  class $\kappa(F)\sqrt{td_X}$ is not invariant under the Lie algebra $\LieAlg{so}(U_\RationalNumbers)$.

The Lie algebra $\LieAlg{g}_{\kappa(F)\sqrt{td_X}}$ is strictly bigger than $\bar{\LieAlg{g}}$, since $\Sigma(F)$ has co-dimension one in $HT^2(X)$ and its image
n $\End(H^*(X,\RationalNumbers))$ is contained in the complexification of $\LieAlg{g}_{v(F)}$, by
Lemma <ref>.
(??? need the following to relate $\Sigma(F)$, which is contained in the stabilizer of $v(F)=ch(F)\sqrt{td_X}$, to its translate by $\exp(-c_1(F)/r)$ which is contained in the stabilizer of $\kappa(F)\sqrt{td_X}$: ???) Note that given an object $F\in D^b(X)$, we have $H^1(TX)\subset\Sigma(F)$, if and only if the subalgebra
is contained in $\LieAlg{g}_{v(F)}$, by Lemma <ref>.
$\Pic(X)$ embeds in $\Aut(D^b(X))$ and hence in
$DMon(X)$. $\Aut(D^b(X))$ acts on $HT^2(X)$, hence so does $\Pic(X)$, as well as $\Pic(X)_\RationalNumbers$.
The map $Ob(D^b(X)) \ni F\mapsto \Sigma(F)\subset HT^2(X)$ is $\Aut(D^b(X))$ equivariant. Hence, if the rank $r$ of $F$ is non-zero, then
$H^1(X,TX)\subset \det(F)^{-1/r}\otimes \Sigma(F)$

if and only if  $\bar{\LieAlg{g}}$ is contained in

There are two types of Lie subalgebras $\LieAlg{a}$ satisfying $\bar{\LieAlg{g}} \subset \LieAlg{a}\subset \LieAlg{g}$, with both inclusions strict, one is $\LieAlg{so}(U_\RationalNumbers)\times \bar{\LieAlg{g}}$, and the other is the stabilizer
$\LieAlg{g}_\ell$ of a line $\ell$ in $U_\RationalNumbers$. Indeed, the quotient
$\LieAlg{g}/\bar{\LieAlg{g}}$ is isomorphic to the direct sum of $\LieAlg{so}(U_\RationalNumbers)$ and $U_\RationalNumbers\otimes H^2(X,\RationalNumbers)$ as a $\bar{\LieAlg{g}}\times \LieAlg{so}(U_\RationalNumbers)$-module, and both direct summands are irreducible.
If $\LieAlg{a}$ contains $\LieAlg{so}(U_\RationalNumbers)$, then $\LieAlg{a}$
must be $\bar{\LieAlg{g}}\times \LieAlg{so}(U_\RationalNumbers)$.
If $\LieAlg{a}$ contains the $\bar{\LieAlg{g}}\times \LieAlg{so}(U_\RationalNumbers)$-submodule
$U_\RationalNumbers\otimes H^2(X,\RationalNumbers)$ of $\LieAlg{g}$, then it contains $\LieAlg{so}(U_\RationalNumbers)$, as the latter is contained[
Choose and orthogonal basis $\{e_i\}$ of $\tilde{H}(X,\RationalNumbers)$ with $e_1, e_2\in U_\QQ$. Let
$\{e_i^*\}$ be the dual basis. Set $A=(e_1,e_1)e_1^*\otimes e_3-(e_3,e_3)e_3^*\otimes e_1$,
$B=(e_2,e_2)e_2^*\otimes e_3-(e_3,e_3)e_3^*\otimes e_2$, and
$C=(e_1,e_1)e_1^*\otimes e_2-(e_2,e_2)e_2^*\otimes e_1$. Then $A$ and $B$ belong to the submodule of $\LieAlg{g}$ isomorphic to
$U_\RationalNumbers\otimes H^2(X,\RationalNumbers)$, $C$ spans $\LieAlg{so}(U_\RationalNumbers)$, and
in the Lie bracket of the former with itself. Hence, if  $\LieAlg{a}$ does not contain $\LieAlg{so}(U_\RationalNumbers)$, then
the quotient $\LieAlg{a}/\LieAlg{g}$ is a $\bar{\LieAlg{g}}$-module, which is not a $\LieAlg{so}(U_\RationalNumbers)$-module, hence equal to $\ell'\otimes H^2(X,\RationalNumbers)$, for a line $\ell'$ in $U_\RationalNumbers$. We have seen that $\LieAlg{g}_{\kappa(F)\sqrt{td_X}}$ does not contain $\LieAlg{so}(U_\RationalNumbers)$.
We conclude that $\LieAlg{g}_{\kappa(F)\sqrt{td_X}}$ is equal to $\LieAlg{g}_{\ell'}$, for some line $\ell'$ in $U_\RationalNumbers$. Hence, $\LieAlg{g}_{v(F)}=\LieAlg{g}_{\ell(F)}$, where $\ell(F)=\exp(c_1(F)/r)\ell'$.

The equality $\ell(F)\otimes_\QQ\CC=\ell(\Sigma(F))$ follows from
Lemma <ref>
and the fact that $m(\Sigma(F))$ is contained in the complexification of $\LieAlg{g}_{v(F)}$ and hence of $\LieAlg{g}_{\ell(F)}$.

The orthogonal projection of $v(F)$ to
$SH^*(X,\RationalNumbers)$ is non-zero, since $r\neq 0$.
If $v(F)$ is $\LieAlg{g}_\ell$-invariant, then so is
its projection to $SH^*(X,\RationalNumbers)$.
It remains to show that the projection of $\ell^n$ into the kernel of
$\Delta:\Sym^n(\tilde{H}(X,\RationalNumbers))\rightarrow \Sym^{n-2}(\tilde{H}(X,\RationalNumbers))$
surjects onto
the $\LieAlg{g}_\ell$-invariant subspace of $\ker(\Delta)$.

Assume first that $\ell$ is non-isotropic.
We have the direct sum decomposition

\[
\Sym^{n}(\tilde{H}(X,\RationalNumbers))\cong
\oplus_{i=0}^n[\ell^{n-i}\otimes\Sym^{i}(\ell^\perp)].
\]

Choose an element $\lambda\in \ell$.
Set $\tilde{q}_{\ell^\perp}:=\tilde{q}-\frac{1}{(\lambda,\lambda)\dim\tilde{H}(X,\RationalNumbers)}\lambda^2$.
Then the trivial $\LieAlg{g}_\ell$-submodule $\Sym^{i}(\ell^\perp)^{\LieAlg{g}_\ell}$ in $\Sym^{i}(\ell^\perp)$
vanishes, for odd $i$, and it is spanned by $\tilde{q}_{\ell^\perp}^{i/2}$ for even $i$. Hence, the trivial $\LieAlg{g}_\ell$-submodule $\Sym^{n}(\tilde{H}(X,\RationalNumbers))^{\LieAlg{g}_\ell}$ in
$\Sym^{n}(\tilde{H}(X,\RationalNumbers))$ is spanned by $\{\lambda^{n-2i}\tilde{q}_{\ell^\perp}^i\}_{i=0}^{\lfloor\frac{n}{2}\rfloor}$.
Now $\Delta$ is a surjective homomorphism of $\LieAlg{g}$-modules, hence also of $\LieAlg{g}_\ell$-modules.
We see that
$\Delta:\Sym^{n}(\tilde{H}(X,\RationalNumbers))^{\LieAlg{g}_\ell}\rightarrow \Sym^{n-2}(\tilde{H}(X,\RationalNumbers))^{\LieAlg{g}_\ell}$ is surjective and so its kernel is one-dimensional.
The  direct sum decomposition $\Sym^{n}(\tilde{H}(X,\RationalNumbers))=\ker(\Delta)\oplus \tilde{q}\Sym^{n-2}(\tilde{H}(X,\RationalNumbers))$ is $\LieAlg{g}$-equivariant and $\ell^n$ is not contained in the second direct summand, hence it projects to the first non-trivially and so surjectively onto $\ker(\Delta)^{\LieAlg{g}_\ell}.$

If $\ell$ is isotropic then $\ell^n$ is contained in $\ker(\Delta)$ is thus  equal to its projection into the latter. The two projections, of $v(F)$ and of $\ell^n$ do not vanish and the equality of the latter to the image of the former then follows by continuity from the non-isotropic case.

Assume that the rank of $F$ is non-zero and $X$ is of $K3^{[2]}$ or $K3^{[3]}$-type.
Then $v(F)$ is determined by its rank and the line $\ell(F)$, by  parts
(<ref>) and (<ref>) of
Theorem <ref>, since $\hat{v}(F)=v(F)$ in this case.
If $X$ is of $K3^{[2]}$-type the equality $\hat{v}(F)=v(F)$ follows from the equality $H^*(X,\QQ)=SH^*(X,\QQ)$.
When $X$ is of $K3^{[3]}$-type, $H^*(X,\QQ)$ decomposes as the direct sum of two irreducible $\LieAlg{g}$-module
$H^*(X,\QQ)\cong SH^*(X,\QQ)\oplus \wedge^2\tilde{H}(X,\QQ)$, by <cit.>.
The projection of $v(F)$ to $\wedge^2\tilde{H}(X,\QQ)$ must vanish, since the $\LieAlg{g}_\ell$-invariant subspace
of the latter vanishes. Hence, $v(F)$ is contained in $SH^*(X,\QQ)$ and again we get the equality $\hat{v}(F)=v(F)$.

Let $\tilde{D}:\tilde{H}(X,\QQ)\rightarrow \tilde{H}(X,\QQ)$ be given by $\tilde{D}(r\alpha+\lambda+s\beta)=r\alpha-\lambda+s\beta$.
Let $D:H^{ev}(X,\QQ)\rightarrow H^{ev}(X,\QQ)$ be the automorphism of the even cohomology ring acting on
$H^{2i}(X,\QQ)$ by $(-1)^i$.
If $\Sigma(F)$ has codimension one in $HT^2(X)$, then so does the subspace $\Sigma(F^\vee)$ of the dual object $F^\vee$.

Let $\Pi:\Sym^n(\tilde{H}(X,\QQ))\rightarrow \ker(\Delta)$ be the projection (<ref>).
Then $\Pi$ commutes with $\Sym^n(\tilde{D})$. The projection (<ref>) commutes with $D$.
The isometry $\Psi:SH^*(X,\QQ)\rightarrow \ker(\Delta)$, given in (<ref>),
satisfies $\Psi(D(\theta))=\Sym^n(\tilde{D})(\Psi(\theta))$, for all $\theta\in SH^*(X,\QQ)$.
The $n$-th power map satisfies $\Sym^n(\tilde{D})(\lambda^n)=(\tilde{D}(\lambda))^n$, for all
$\lambda\in \tilde{H}(X,\QQ)$. Let $\lambda$ be a non-zero vector in $\ell(F)$ satisfying
$\Psi(\hat{v}(F))=\Pi(\lambda^n)$. Then

\[
\Psi(\hat{v}(F^\vee))=\Psi(D(\hat{v}(F)))=\Sym^n(\tilde{D})(\Psi(\hat{v}(F)))=\Sym^n(\tilde{D})(\Pi(\lambda^n))=
\Pi(\tilde{D}(\lambda)^n).
\]

Hence, $\tilde{D}(\lambda)$ spans $\ell(F^\vee)$.

Let $X$ be a $2n$-dimensional irreducible holomorphic symplectic manifold.
Given a closed point $z\in X$ we have $\ell(\StructureSheaf{z})=\span_\QQ\{\beta\}.$
Indeed, $v(\StructureSheaf{z})=[pt]$ belongs to $SH^*(X,\QQ)$ and $\Psi(1)=\beta^n/c_X$, by (<ref>).

We will need the following
multiplication table in the cohomology of $X$ of $K3^{[2]}$type, where $\alpha$ is a class in $H^2(X)$ and $c_2$ is $c_2(TX)$.

\begin{equation}
\label{eq-multiplication-table}
\left[
\begin{array}{cccccc}
1 & \alpha & \alpha^2 & c_2 & c_2\alpha & [\mbox{pt}]
\\
\alpha & \alpha^2 & \frac{(\alpha,\alpha)}{10}c_2\alpha &
c_2\alpha & 30(\alpha,\alpha)[\mbox{pt}] & 0
\\
\alpha^2 & \frac{(\alpha,\alpha)}{10}c_2\alpha & 3(\alpha,\alpha)^2[\mbox{pt}]&
30(\alpha,\alpha)[\mbox{pt}] & 0 & 0
\\
c_2 & c_2\alpha & 30(\alpha,\alpha)[\mbox{pt}] & 828[\mbox{pt}] & 0 & 0
\\
c_2\alpha & 30(\alpha,\alpha)[\mbox{pt}] & 0 & 0 & 0 & 0
\\
\mbox{[pt]} & 0 & 0 & 0 & 0 & 0
\end{array}
\right]
\end{equation}

Given classes $\alpha$, $\beta$, $\gamma$, $\delta$ in $H^2(X)$, we have

\[
\alpha\beta\gamma\delta=[(\alpha,\beta)(\gamma,\delta)+(\alpha,\gamma)(\beta,\delta)+(\alpha,\delta)(\beta,\gamma)][pt].
\]

Assume $X$ is of $K3^{[2]}$-type. We claim that
\ell(\StructureSheaf{X})=\span_\QQ\{4\alpha+5\beta\}.
We have $v(\StructureSheaf{X})=\sqrt{td_X}=1+\frac{1}{24}c_2(X)+\frac{25}{32}[pt].$

\begin{equation}
\label{eq-b-X}
\end{equation}

where $\{y_i\}$ is an orthogonal basis of $H^2(X,\QQ)$.
We have $\int_Xy_i^4=3(y_i,y_i)^2$ and for  $i\neq j$ we have $\int_X y_i^2y_j^2=(y_i,y_i)(y_j,y_j)$, due to the vanishing of $(y_i,y_j)$. Hence,

\begin{eqnarray*}
\int_Xb_X^2=\frac{1}{23^2}\left(\sum_{i<j}\frac{2}{(y_i,y_i)(y_j,y_j)}\int_Xy_i^2y_j^2+
\sum_{i=1}^{23} \frac{1}{(y_i,y_i)^2}\int_X y_i^4
\right)=\frac{25}{23},
\\
\int_Xb_Xy_i^2=\frac{1}{23(y_i,y_i)}\int_Xy_i^4+\frac{1}{23}\sum_{j=1,j\neq i}^{23}\frac{1}{(y_j,y_j)}\int_Xy_i^2y_j^2=
\frac{25}{23}(y_i,y_i)=\frac{5}{23\cdot 6}\int_Xc_2(X)y_i^2.
\end{eqnarray*}

It follows that

\begin{equation}
\label{eq-b-X-in-terms-of-c-2}
c_2(X)=\frac{23\cdot 6}{5}b_X,
\end{equation}

as both span the one-dimensional space of $O(H^2(X,\QQ))$-invariant classes.
We have

\begin{equation}
\label{eq-values-of-Psi}
\Psi(1)=\frac{\alpha^2}{2}, \
\Psi([pt])=\beta^2, \ \mbox{and} \
\Psi(b_X)=\tilde{b}_X+\alpha\beta,
\end{equation}

where $\tilde{b}_X$ denotes the element of $\Sym^2(\tilde{H}(X,\QQ))$ given by the right hand side of Equation (<ref>).

\begin{eqnarray}
\label{eq-Psi-of-c-2}
\Psi(c_2(TX)) & = & 30(\tilde{q}+\alpha\beta),
\\
\label{eq-Psi-of-sqrt-td-class}
\Psi(\sqrt{td_X})&=&\frac{1}{2}\left[\alpha^2+\frac{5}{2}(\tilde{q}+\alpha\beta)+\frac{25}{16}\beta^2\right]=
\frac{1}{2}\left[(\alpha+\frac{5}{4}\beta)^2+\frac{5}{2}\tilde{q}\right],
\end{eqnarray}

where $\tilde{q}$ is defined in Example <ref>  and we used the equality

\begin{equation}
\label{eq-tilde-q-in-terms-of-tilde-b-X}
\tilde{q}=\frac{1}{25}[23\tilde{b}_X-2\alpha\beta].
\end{equation}

The right hand side of the latter displayed equation  is the projection of $\frac{1}{2}(\alpha+\frac{5}{4}\beta)^2$ from $\Sym^2(\tilde{H}(X,\QQ))$
to the image of $\Psi$, by Example <ref>. So
\ell(\StructureSheaf{X})=\span_\QQ\{4\alpha+5\beta\}
$ as claimed.

More generally, we have the following.

Let $X$ be a $2n$-dimensional irreducible holomorphic symplectic manifold. There exists a rational number $t$, such that
$t^n=\frac{n!}{c_X}\int_X\sqrt{td_X}$ and
\ell(\StructureSheaf{X})=\span_\QQ\{\alpha+t\beta\}.
If  $X$ is of $K3^{[n]}$-type, $n\geq 2$, then $\ell(\StructureSheaf{X})=\span_\QQ\{4\alpha+(n+3)\beta\}$.
If  $X$ is deformation equivalent to a $2n$-dimensional generalized Kummer, $n\geq 2$, then $\ell(\StructureSheaf{X})=\span_\QQ\{4\alpha+(n+1)\beta\}$.

Let $\tau_{2i}$ be the graded summand of $\sqrt{td_X}$ in $H^{4i}(X,\QQ)$. So $\tau_0=1$ and $\tau_{2n}=s[pt]$ for $s:=\int_X\sqrt{td_X}$.
It is known that $s>0$, by <cit.>.
If $X$ is of $K3^{[n]}$-type, then $s=\frac{(n+3)^n}{4^nn!}$, by <cit.>.
If  $X$ is deformation equivalent to a $2n$-dimensional generalized Kummer, then
by <cit.>.

Choose a basis of $\widetilde{H}(X,\QQ)$ consisting of $\alpha$, $\beta$, and a basis of $H^2(X,\QQ)$.
We get a well defined notion of the degree $\deg_\alpha(f)$ of an element of $\Sym^n\widetilde{H}(X,\QQ)$ in $\alpha$.  Define
$\deg_\beta(f)$ similarly.
Let $\bar{\tau}_{2i}$ be the projection of $\tau_{2i}$ to $SH^*(X)$. Note that $\bar{\tau}_{0}=\tau_0=1$ and
Then $\Psi(\bar{\tau}_0)=\alpha^n/n!$,
$\Psi(\bar{\tau}_{2n})=s\beta^n/c_X$, and for $1\leq i\leq n-1$ we have $\deg_{\alpha}(\Psi(\bar{\tau}_{2i}))<n$
and $\deg_{\beta}(\Psi(\bar{\tau}_{2i}))<n$, by definition of $\Psi$. We know that
$\ell(\StructureSheaf{X})=\span\{\alpha+t\beta\}$, for some $t\in \QQ$, by Theorem <ref>.
Furthermore, the projection of $(\alpha+t\beta)^n$ to $\ker(\Delta)$ is a rational scalar multiple of $\Psi(\sum_{i=0}^n\bar{\tau}_{2i})$,
by Theorem <ref>. Let $V(d)$ be the kernel of $\Delta:\Sym^d(\widetilde{H}(X,\QQ))\rightarrow \Sym^{d-2}(\widetilde{H}(X,\QQ))$.
We have the decomposition into irreducible $\LieAlg{so}(\widetilde{H}(X,\QQ))$-sub-representations

\[
\Sym^n(\widetilde{H}(X,\QQ))=\oplus_{j=0}^{\lfloor n/2\rfloor}\tilde{q}^jV(n-2j),
\]

where $\tilde{q}$ is defined in Example <ref> (see the proof of <cit.>).

Now $\deg_\alpha(\tilde{q})=\deg_\beta(\tilde{q})=1$, and so
only the direct summand $V(n)$ in the above decomposition of $\Sym^n(\widetilde{H}(X,\QQ))$ contains elements $f$ with
$\deg_\alpha(f)=n$ and similarly for $\deg_\beta$.
Thus the coefficient of $\alpha^n$
in  the projection of $\frac{1}{n!}(\alpha+t\beta)^n$ to $\ker(\Delta)=V(n)$ is $\frac{1}{n!}$ and the coefficient of $\beta^n$ is
Hence, $t^n=\frac{s(n!)}{c_X}$. We see that $t^n=\left(\frac{n+3}{4}\right)^n$ in the $K3^{[n]}$-type and $t^n=\left(\frac{n+1}{4}\right)^n$ in the generalized Kummer case.
If $n$ is odd, we are done. If $n$ is even it remains to prove that $t>0$.
The inequality $t>0$  is
equivalent to $(\alpha+t\beta,\alpha+t\beta)<0$.
The inequality $(\alpha+t\beta,\alpha+t\beta)<0$ follows from Remark <ref> for $X$ of $K3^{[n]}$-type. Below we will prove the inequality $t>0$ directly in the $K3^{[n]}$ and generalized Kummer deformation types.

The coefficient of $\alpha^{n-1}\beta$
in  the projection of $(\alpha+t\beta)^n$ to $\ker(\Delta)=V(n)$ has the same sign as $t$.
Indeed, we have $\Delta(\tilde{b}_X)=\Delta(\tilde{q})=1$,

\begin{eqnarray*}
\Delta\left(\alpha^{n-1}\beta+(n-1)\alpha^{n-2}\tilde{b}_X
\right)&=& (n-1)(\alpha,\beta)\alpha^{n-2}+(n-1)\alpha^{n-2}\Delta(\tilde{b}_X)=0,
\\
\tilde{b}_X&=& \left(\frac{b_2(X)+2}{b_2(X)}\right)\tilde{q}+\left(\frac{2}{b_2(X)}\right)\alpha\beta,
\\
\alpha^{n-1}\beta+(n-1)\alpha^{n-2}\tilde{b}_X&=&
\left( \frac{(n-1)(b_2(X)+2)}{b_2(X)}\right)\alpha^{n-2}\tilde{q}+\left(\frac{b_2(X)+2n-2}{b_2(X)}\right)\alpha^{n-1}\beta.
\end{eqnarray*}

Multiplying both sides of the last equation by
$\frac{b_2(X)}{b_2(X)+2n-2}$ we see that the projection of $\alpha^{n-1}\beta$ to $\ker(\Delta)$ is

\[
\left(\frac{b_2(X)}{b_2(X)+2n-2} \right)
\left[
\alpha^{n-1}\beta+(n-1)\alpha^{n-2}\tilde{b}_X
\right],
\]

and the coefficient of $\alpha^{n-1}\beta$ is indeed positive.

The only term in $\sum_{i=0}^n\Psi(\bar{\tau}_{2i})$ that contributes to the coefficient of $\alpha^{n-1}\beta$ is $\Psi(\bar{\tau}_2)$, by definition of $\Psi$. It suffices to show that the coefficient of $\alpha^{n-1}\beta$ in $\Psi(\bar{\tau}_2)$ is positive.
Now, $\tau_2=c_2(TX)/24$. The projection of $c_2(TX)$ to $SH^4(X)$ is monodromy invariant, hence a  multiple of $b_X$, and we claim that it is a positive multiple.
Indeed, given a line bundle $L$ on $X$, the coefficient of $(c_1(L),c_1(L))^{n-1}$ in the formula for
$\int_X ch(L)td_X$ depends only on $\int_X \frac{c_1(L)^{2n-2}}{(2n-2)!} \frac{c_2(TX)}{12}$ and it is
positive, by <cit.>, as is the coefficient of $(c_1(L),c_1(L))^{n-1}$ in the formula for
$\int_X b_Xc_1(L)^{2n-2}$, by <cit.>.
We conclude that $\Psi(\bar{\tau}_2)$ is a positive scalar multiple of $\Psi(b_X)$.
Finally, the coefficient of $\alpha^{n-1}\beta$ in $\Psi(b_X)=\frac{1}{(n-2)!}\alpha^{n-2}[\tilde{b}_X+\alpha\beta]$ is positive.

The proof of Lemma <ref> shows that the positive $n$-th root $\sqrt[n]{\frac{s(n!)}{c_X}}$ of $\frac{s(n!)}{c_X}$ is rational
and $\ell(\StructureSheaf{X})=\span\left\{\alpha+\epsilon\sqrt[n]{\frac{s(n!)}{c_X}}\beta\right\}$, with $\epsilon=1$, if $n$ is odd, and $\epsilon=\pm 1$, if $n$ is even.

In [73] the BBF form $b(\bullet,\bullet)$ on a $2d$-dimensional irreducible holomorphic symplectic manifold $X$ is normalized as

\[
\int_X\lambda^{2d}=\frac{(2d)!}{2^dd!}b(\lambda,\lambda)^d,
\]

for $\lambda\in H^2(X,\QQ)$.
If $X$ is a $2d$-dimensional irreducible holomorphic symplectic manifold deformation equivalent to a generalized Kummer, then
the integral and primitive BBF form $b_0(\bullet,\bullet)$ satisfies

\[
\int_X\lambda^{2d}=(d+1)(2d-1)!!b_0(\lambda,\lambda)^d,
\]

for $\lambda\in H^2(X,\QQ)$, by O'Grady's arxiv:1805.12075.v1 formula (2.2.5), where the double factorial is defined in formula
(1.4.1) of that paper.  When $d=2$, then
$(d+1)(2d-1)!!=9$, while $\frac{(2d)!}{2^dd!}=3$. Hence
$b(\lambda,\lambda)^2=3b_0(\lambda,\lambda)^2$. Thus, pairing $(\bullet,\bullet)$ has values in $\sqrt{3}\QQ$.
The Chern numbers of a fourfold $X$ of generalized Kummer type
are $\int_Xc_2(X)^2=756$, and $\int_Xc_4(X)=108$, by Appendix A in [63]. Hence, $td_X=1+\frac{1}{12}c_2(X)+3[pt]$ and

\[
\sqrt{td_X}=1+\frac{1}{24}c_2+\frac{27}{32}[pt].
\]

If $X$ is $2n$-dimensional of  generalized Kummer deformation type, then
$c_X=n+1$ and

\[
\int_X\sqrt{td_X}=\frac{(n+1)^{n+1}}{4^n(n!)},
\]

by <cit.>.
We see that $\frac{s(n!)}{c_X}=\left(\frac{n+1}{4}\right)^n$ and
$\ell(\StructureSheaf{X})=\span\{\alpha\pm \left(\frac{n+1}{4}\right)\beta\}$.

The computations of  Example <ref> generalize to yield  the following statement.

Let $X$ be of $K3^{[2]}$-type and $F$ an object of $D^b(X)$ of non-zero rank.

If $F$ deforms in co-dimension $1$, then $\kappa_3(F)=0$ and
\kappa(F)=x+yc_2(TX)+z[pt],
where $x, y, z\in \RationalNumbers$ and

\begin{equation}
\label{eq-necessary-condition--on-kappa-class-for-F-to-deform-in-co-dimension-1}
\end{equation}

In this case,

\begin{equation}
\label{eq-LLV-line-of-an-objects-which-deforms-in-co-dimension-1}
\ell(F)=\span\left\{x^2\alpha+xc_1(F)+\left(\frac{5x^2}{4}+30xy+\frac{(c_1(F),c_1(F))}{2}\right)\beta\right\}.
\end{equation}

Conversely, if $\kappa_3(F)=0$,
\kappa(F)=x+yc_2(TX)+z[pt],
and $450y^2+3xy-xz=0$, then
$F$  has a rank $1$ cohomological obstruction map.

If $F_1, F_2\in D^b(X)$ are of non-zero rank and $F_1$, $F_2$, and $F_1\otimes F_2$, all deform in co-dimension $1$,
then $\kappa(F_i)$ is a scalar multiple of $\kappa(\StructureSheaf{X})$, for $i=1$ or $i=2$.

The class $\kappa(F)$ remains of Hodge type under every Kähler deformation of $X$, by Proposition <ref>. Hence, $\kappa_3(F)=0$ and $\kappa_2(F)$ is a scalar multiple of $c_2(TX)$.
We have

\begin{eqnarray*}
\kappa(F)\sqrt{td_X}&=&(x+yc_2(X)+z[pt])\left(1+\frac{1}{24}c_2(X)+\frac{25}{32}[pt]\right)
\\&=& x+(x/24+y)c_2(X)+(z+25x/32+69y/2)[pt].
\end{eqnarray*}

Set $t=x/24+y$ and $s=z+25x/32+69y/2$.
We have

\begin{eqnarray*}
\Psi(x+tc_2(X)+s[pt])&=&
\\
&=& 30t\tilde{q}+\frac{x}{2}\left(\alpha^2+\frac{60t}{x}\alpha\beta+\frac{2s}{x}\beta^2\right).
\end{eqnarray*}

The latter belongs to the projection via (<ref>) of the  square of a line, if and only if $450t^2=sx$, in which case it belongs to the projection of the square of
$\span\{\alpha+\frac{30t}{x}\beta\}=\span\{\alpha+\left(\frac{5}{4}+\frac{30y}{x}\right)\beta\}$, by Example <ref>. This is the case, if and only if
$450y^2+3xy-xz=0$. The equality $v(F)=\exp(e_{c_1(F)/x})\kappa(F)\sqrt{td_X}$ implies that $\ell(F)$ is spanned by
$\exp(e_{c_1(F)/x})\left(\alpha+\left(\frac{5}{4}+\frac{30y}{x}\right)\beta\right)$, by the $SO(\widetilde{H}(X,\QQ))$-equivariance of $\Psi$.
Finally observe that

\[
\exp(e_{c_1(F)/x})\left(\alpha+\left(\frac{5}{4}+\frac{30y}{x}\right)\beta\right)=
\alpha+c_1(F)/x+\left(\frac{5}{4}+30y/x+\frac{(c_1(F),c_1(F))}{2x^2}\right)\beta.
\]

The assumptions imply that $\Psi(\kappa(F)\sqrt{td_X})$ is invariant under the stabilizer of the line in Equation (<ref>) in the LLV algebra $\LieAlg{g}:=\LieAlg{so}(\tilde{H}(X,\QQ))$. Hence,
$\kappa(F)\sqrt{td_X}$ is invariant under this subalgebra, by the $\LieAlg{g}$-equivariance of $\Psi$.

Set $\kappa(F_i)=x_i+y_ic_2(X)+z_i[pt]$. Then

\[
\kappa(F_1\otimes F_2)=
\]

Set $\bar{y}_i:=y_i/x_i$ and $\bar{z_i}=z_i/x_i$.
Equation (<ref>) for $F_1\otimes F_2$ translates to the vanishing of

\begin{eqnarray*}
\\
\end{eqnarray*}

Hence, $y_i=0$, for $i=1$ or $i=2$. Equation (<ref>) and the vanishing of $y_i$ imply that $z_i=0$ as well.

Theorem <ref> imposes the following cohomological constraint on the Todd class of a $2n$ dimensional irreducible holomorphic symplectic manifold $X$.

Denote by $\hat{c}_{2k}$
the projection of $c_{2k}(X)$ to the primitive subspace of $H^{4k}(X,\QQ)$ defined in <cit.>.
Then $\hat{c}_{2k}$ generates an irreducible subrepresentation $\LieAlg{g}(\hat{c}_{2k})$ of the LLV algebra $\LieAlg{g}$  isomorphic to the subspace $V(n\!-\!2k)$ of
$\Sym^{n-2k}\tilde{H}(X,\QQ)$ spanned by powers of isotropic classes, by <cit.>.

The direct sum  $\oplus_{k=0}^{\lfloor n/2\rfloor}\LieAlg{g}(\hat{c}_{2k})$ is the minimal $\LieAlg{g}$ invariant subspace containing $\sqrt{td_X}$. Theorem <ref> implies that
the projection of $\sqrt{td_X}$ to $\LieAlg{g}(\hat{c}_{2k})$ spans the image of the power $\ell(\StructureSheaf{X})^{n-2k}$ of the same line $\ell(\StructureSheaf{X})$ under the unique, up to scalar, isomorphism $\LieAlg{g}(\hat{c}_{2k})\cong V(n\!-\!2k)$. Note that when $X$ is of $K3^{[n]}$-type and $n\geq 4$, then $\hat{c}_{2k}$ is non-trivial
for $1\leq k\leq n/2$, by  <cit.>.

Theorem <ref> lists topological consequences about the Mukai vector of an object $F$
deforming in co-dimension $1$ (Definition <ref>).

We emphasize next the evident fact that these topological consequences remain valid under deformations of the pair $(X,F)$. For simplicity we restrict to the case of locally free sheaves.
Let $\pi:\X\rightarrow B$ be a smooth and proper morphism over a connected analytic space $B$, all of which fibers are irreducible holomorphic symplectic manifolds. Let  $\theta$ be a class in $H^2(\X,\StructureSheaf{\X}^*)$, in the analytic topology,
and let $\F$ be a locally free $\theta$-twisted sheaf of rank $r$. Denote by $X_b$ the fiber of $\pi$ over a point $b$ of $B$ and by $\theta_b\in H^2(X_b,\StructureSheaf{X_b}^*)$ and $F_b$ the restrictions of $\theta$ and $\F$ to $X_b$. The rational LLV lattices $\tilde{H}(X_b,\QQ)$ then form a local system $\underline{\QQ\alpha}\oplus R^2\pi_*\QQ\oplus \underline{\QQ\beta}$, where
$\underline{\QQ\alpha}\oplus\underline{\QQ\beta}$ is a trivial local subsystem.

Denote by $\LieAlg{g}(X_b)$ the LLV Lie algebra of $X_b$.
Set $\bar{v}(F_b):=\kappa(F_b)\sqrt{td_{X_b}}$ and let $\hat{\bar{v}}(F_b)$ be its projection to $SH^*(X_b,\QQ)$.

Assume that for some point $b_0$ of $B$ the class $\theta_{b_0}$ is trivial[Strictly speaking, $\F$ is a twisted sheaf with respect to a co-cycle $\tilde{\theta}$ representing the class $\theta$ and $F_{b_0}$ is a twisted sheaf with respect to the restriction $\tilde{\theta}_{b_0}$ of $\tilde{\theta}$. The co-cycle $\tilde{\theta}_{b_0}$ is a co-boundary. We then abuse notation and denote by $F_{b_0}$ also an untwisted coherent sheaf under an equivalence of the category of coherent sheaves on $X_{b_0}$ with the category of $\tilde{\theta}_{b_0}$-twisted sheaves. The isomorphism depends on a choice of a $1$-co-chain whose co-boundary is $\tilde{\theta}_{b_0}$, and so is unique up to tensorization by a line-bundle. In particular, the condition that
$F_{b_0}$ deforms in co-dimension $1$

is independent of the choice of the equivalence.]
and that $F_{b_0}$ deforms in co-dimension $1$.

\[
\ell_0:=\exp\left(e_{-c_1(F_{b_0})/r}\right)(\ell(F_{b_0}))
\]

is a line in $\QQ\alpha\oplus\QQ\beta$ and the following statements hold for every $b\in B$.

* The subalgebra $\LieAlg{g}(X_b)_{\ell_0}$ of $\LieAlg{g}(X_b)$ annihilating the line $\ell_0$ is equal to the subalgebra
$\LieAlg{g}(X_b)_{\bar{v}(F_b)}$ annihilating the class $\bar{v}(F_b)$. In particular, $\bar{v}(F_b)$ is annihilated by

* The  embedding
$\Psi:SH^*(X_b,\QQ)\rightarrow \Sym^n\tilde{H}(X_b,\QQ)$, given in (<ref>), maps $\hat{\bar{v}}(F_b)$

into the  image of the $n$-th power $\ell_0^n$ of $\ell_0$ via the projection (<ref>).

The line $\ell_0$ is in $\QQ\alpha\oplus\QQ\beta$, by Theorem <ref>(<ref>).
The statements hold for $X_{b_0}$, by the equivariance of $\Psi$ with respect to the isometry
$\exp\left(e_{-c_1(F_{b_0})/r}\right)$ and by Theorem <ref> parts (<ref>) and (<ref>). It follows for all $b\in B$, by the flatness of the section associated to $\kappa(\F)\sqrt{td_\pi}$
in $R\pi_*\QQ$ and the assumption that the base $B$ is connected.

Assume that $F_i$, $1\leq i\leq k$, each deforms in co-dimension one and its rank $r_i$ is positive.
Then the following statements hold.

$\Sigma(F_i)\cap [H^1(TX)\oplus H^2(\StructureSheaf{X})]$ is the graph of the homomorphism
$\theta_i:H^1(TX)\rightarrow H^2(\StructureSheaf{X})$ given by
$\lambda\mapsto -m_\lambda\left(c_1(F_i)/r_i\right)$.

The LLV subspace  $W(\otimes_{i=1}^k F_i)$ is contained in
\exp\left(e_{\sum_{i=1}^k c_1(F_i)/r_i}\right)(\span_\QQ\{\alpha,\beta\}).

The isomorphism $\mu: HT^2(X)\rightarrow \tilde{H}^{0,0}(X,\CC)$, given by $\mu(\lambda)=m_\lambda(\sigma)$
in (<ref>),
maps $H^1(TX)\oplus H^2(\StructureSheaf{X})$ to $H^{1,1}(X)\oplus \ComplexNumbers\beta$
(see Equation (<ref>)).
The homomorphism $\mu$ maps $\Sigma(F_i)\cap [H^1(TX)\oplus H^2(\StructureSheaf{X})]$ to
$\ell(F_i)^\perp\cap[H^{1,1}(X)\oplus \CC\beta]$, hence to
the graph of $\tilde{\theta}_i:H^{1,1}(X)\rightarrow \CC\beta$ given by
$\tilde{\theta}_i(x)=(x,c_1(F_i)/r_i)\beta$, by Theorem <ref>(<ref>).
We have $m_{\theta_i(\lambda)}(\sigma)=\mu(\theta_i(\lambda))=\tilde{\theta}_i(\mu(\lambda))=(m_\lambda(\sigma),c_1(F_i)/r_i)\beta$, for all $\lambda\in H^1(TX)$. Pairing with $\alpha$, we get:

\begin{eqnarray*}
(\alpha,\mu(\theta_i(\lambda)))&=& -(m_{\theta_i(\lambda)}(\alpha),\sigma),
\\
(\alpha,\tilde{\theta}_i(\mu(\lambda)))&=& -(m_\lambda(\sigma),c_1(F_i)/r_i)=(\sigma,m_\lambda(c_1(F_i)/r_i)).
\end{eqnarray*}

It remains to prove that $m_{\theta_i(\lambda)}(\alpha)=\theta_i(\lambda)$.
Now, $m_{\theta_i(\lambda)}=\eta\circ e_{\theta_i(\lambda)}\circ \eta^{-1}$,
by Equation (<ref>), and $\eta\circ e_{\theta_i(\lambda)}\circ \eta^{-1}=e_{\theta_i(\lambda)},$
since $\theta_i(\lambda)$ belongs to $H^2(\wedge^{0}TX)$, and $e_{\theta_i(\lambda)}(\alpha)=\theta_i(\lambda)$, by definition
(Eq. (<ref>)).

The intersection $\Sigma(\otimes_{i=1}^k F_i)\cap [H^1(TX)\oplus H^2(\StructureSheaf{X})]$
is the graph of $\sum_{i=1}^k\theta_i:H^1(TX)\rightarrow H^2(\StructureSheaf{X})$,
as observed in Section <ref>. Hence,
$\mu$ maps the intersection to the graph of the homomorphism from $H^{1,1}(X)$ to $\CC\beta$ given by

\[
x\mapsto \left(x,\sum_{i=1}^k c_1(F_i)/r_i\right)\beta,
\]

by the proof of Part (<ref>).
The latter graph is precisely
$\exp\left(e_{\sum_{i=1}^k c_1(F_i)/r_i}\right)(H^{1,1}(X))$. The subspace
$W(\otimes_{i=1}^k F_i)$ is thus contained in the subspace ${\displaystyle
\exp\left(e_{\sum_{i=1}^k c_1(F_i)/r_i}\right)(\span_\QQ\{\alpha,\beta\})
orthogonal to $\exp\left(e_{\sum_{i=1}^k c_1(F_i)/r_i}\right)(H^{1,1}(X))$.

§.§ Torsion sheaves which deform in co-dimension $1$ are supported on lagrangian subvarieties or points

Let $X$ be a $2n$-dimensional irreducible holomorphic symplectic manifold.
Let $\sigma$ be a non-zero class in $H^0(X,\Omega^2_X)$.

Let $F$ be an object of $D^b(X)$ such that
$F$  has a rank $1$ cohomological obstruction map (Def. <ref>).

Assume that the line $\ell(F)$ in
$\tilde{H}(X,\QQ)$ is spanned by a class of the form $0\alpha+\lambda+s\beta$, where $\lambda$ is a class in $H^2(X,\QQ)$.
Then $ch_i(F)=0$, for $i<n$, and $ch_n(F)\cup \sigma=0$.

Let $\bar{\sigma}\in H^2(X,\StructureSheaf{X})$ be the class conjugate to that of $\sigma$ and let $m_{\bar{\sigma}}$ be the image of $\bar{\sigma}$ in $\LieAlg{g}_\CC$ via (<ref>).
The element $m_{\bar{\sigma}}$ has weight $(0,2)$ and thus it maps the direct summand $\tilde{H}_\fine^{p,q}(X,\CC)$ to
$\tilde{H}_\fine^{p,q+2}(X,\CC)$. It follows that $m_{\bar{\sigma}}$ annihilates $\tilde{H}_\fine^{p,q}(X,\CC)$ if $(p,q)$ is $(1,1)$ or $(2,2)$.

Consequently, $m_{\bar{\sigma}}$
annihilates $\ell(F)$, as it is contained in $\tilde{H}_\fine^{1,1}(X,\CC)\oplus \tilde{H}_\fine^{2,2}(X,\CC)$.

The subalgebra $\LieAlg{g}_{v(F)}$ annihilating $v(F)$ is equal to $\LieAlg{g}_{\ell(F)}$
(use the proof of Theorem <ref> part (<ref>)

replacing the use of Lemma <ref> by the assumption that
$F$  has a rank $1$ cohomological obstruction map).

Hence, $m_{\bar{\sigma}}$ belongs to $\LieAlg{g}_{v(F)}$.

Wedge product with $\sigma^k$ induces an isomorphism $\Omega_X^{n-k}\rightarrow\Omega_X^{n+k}$, for $0\leq k\leq n$, and thus an isomorphism $H^{n-k,q}(X)\cong H^{n+k,q}(X)$. It follows that
cup product with the class of $\sigma$ induces an injective homomorphism from $H^{p,p}(X,\QQ)$ into $H^{p+2,p}(X,\QQ)$, for $p<n$. Thus, cup product with $\bar{\sigma}$
induces an injective homomorphism from $H^{p,p}(X,\QQ)$ into $H^{p,p+2}(X,\QQ)$, for $p<n$. We conclude that the direct summand $v_p(F)$ of $v(F)$ in $H^{p,p}(X,\QQ)$ vanishes, for $p<n$, as the operator $m_{\bar{\sigma}}$ acts on $H^*(X,\QQ)$ via cup product with $\bar{\sigma}$. The equality $v(F)=ch(F)\sqrt{td_X}$ implies that the first non zero graded summand of $ch(F)$ and $v(F)$ are equal. Thus,  $ch_p(F)=0$, for $p<n$, and
$ch_n(F)=v_n(F)$. Thus $ch_n(F)\cup \sigma$ vanishes, being the complex conjugate of $v_n(F)\cup \bar{\sigma}$ which vanishes.

Let $\iota:Z\hookrightarrow X$ be an embedding of an irreducible variety $Z$ of dimension $< 2n$ and let $F$ be a torsion free sheaf over $Z$. Assume that $v(\iota_*F)$ deforms in co-dimension $1$ and the projection $\hat{v}(\iota_*F)$ does not vanish. Then $\iota(Z)$ is either a lagrangian subvariety or a point.

Let $p$ be the complex co-dimension of $\iota(Z)$ in $X$.
Let $[Z]\in H^{p,p}(X,\ZZ)$ be the class Poincaré dual to $\iota(Z)$.
Then $ch_i(\iota_*F)=0$, for $i<p$, and $v_p(\iota_*F))=ch_p(\iota_*F)=\rank(F)[Z]$. The inequality $p\geq n$ and the vanishing of $ch_n(\iota_*F)\cup\sigma$ follow, by
Lemma <ref>.

Let $r\alpha+\lambda+s\beta$ be a non-zero element of $\ell(\iota_*F)$. Then $r=0$, since $\iota(Z)$ is a proper subvariety of $X$. If $\lambda=0$, then $v(F)$ is annihilated by $\LieAlg{g}_\beta$.  In particular, $v(F)$ is annihilated by
$L_\omega$, for every Kähler class $\omega$ on $X$. Hence, $v_p(\iota_*F)=\rank(F)[Z]$
is annihilated by
$L_\omega$, for every Kähler class $\omega$ on $X$.
Hence, $\iota^*\omega=0$, for every Kähler class $\omega$ on $X$, and so $Z$ is a point.

Assume that $\lambda\neq 0$. Let $\omega$ be a Kähler class on $X$ such that $(\lambda,\omega)\neq 0$.
Then $L_\omega^n((\lambda+s\beta)^n)=n!(\omega,\lambda)^n\beta^n$ does not vanish.
Now $L_\omega$ belongs to $\LieAlg{g}_\CC$, $\Psi$ is $\LieAlg{g}_\CC$-equivariant, and
$\Psi(\hat{v}(\iota_*F))$ spans $\ell(\iota_*F)^n$.
Hence, $L_\omega^n$ does not annihilate $\hat{v}(\iota_*F)$ and thus neither $v(\iota_*F)$.
Now $L_\omega^n$ annihilates
the graded summand $v_i(\iota_*F)$ in $H^{2i}(X,\QQ)$, for $i>n$. Thus $L_\omega^n(v_n(\iota_*F))\neq 0$.
We conclude that  $Z$ is $n$-dimensional.

The subvariety $Z$ is lagrangian by the vanishing of

Let $\iota:Z\rightarrow X$ be an embedding of a compact Kähler manifold as a holomorphic lagrangian submanifold.

Assume that the obstruction map (<ref>) from $HT^2(X)$ to $\Ext^2(\iota_*\StructureSheaf{Z},\iota_*\StructureSheaf{Z})$ has rank $1$. Then the homomorphism $H^1(X,TX)\rightarrow H^1(Z,N_{Z/X})$,
induced by the sheaf homomorphism $TX\rightarrow \iota_*N_{Z/X}$, has rank $1$. Conjugating via the isomorphisms
$TX\cong \Omega^1_X$ and $N_{Z/X}\cong \Omega^1_Z$, induced by the symplectic structure, we get that
$H^1(X,\Omega^1_X)\rightarrow H^1(Z,\Omega^1_Z)$ has rank $1$.
The restriction homomorphism $H^2(X,\CC)\rightarrow H^2(Z,\CC)$ factors through the latter, since $Z$ is lagrangian, and hence
has rank $1$.
Let $L\in\Pic(X)$ be a line bundle, such that the sublattice $c_1(L)^\perp$ orthogonal to $c_1(L)$ with respect to the BBF-pairing is equal to the kernel of the restriction homomorphism $\iota^*:H^2(X,\ZZ)\rightarrow H^2(Z,\ZZ)$.

Assume that there exists
a class $\tau\in H^2(X,\QQ)$, such that
$\iota^*(\tau)=c_1(\omega_Z).$ This is the case if $h^{1,1}(Z)=1$, as in
Example <ref>, if $Z$ is a complex torus, or if $Z$ is the zero subscheme of a global section of a rank $n$ vector bundle on $X$.

$\ell(\iota_*\StructureSheaf{Z})$ is spanned by ${\displaystyle

Let $\gamma:=r\alpha+\lambda+s\beta$ be a non-zero element of $\ell(\iota_*\StructureSheaf{Z})$.
The restriction homomorphism $\iota^*:H^2(X,\StructureSheaf{X})\rightarrow H^2(Z,\StructureSheaf{Z})$ vanishes, so $\iota_*\StructureSheaf{Z}$ deforms along gerby deformations of $X$. Hence,
$H^2(X,\StructureSheaf{X})$ is contained in $\Sigma(\iota_*\StructureSheaf{Z})$. The Duflo automorphism of $HT^*(X)$ restricts to $H^2(X,\StructureSheaf{X})$ as the identity, and so $H^2(X,\StructureSheaf{X})$ is contained in $\tilde{\Sigma}(\iota_*\StructureSheaf{Z})$.
$\mu(H^2(X,\StructureSheaf{X}))$ is orthogonal to $\ell(\iota_*\StructureSheaf{Z})$, by Theorem <ref>.
Now, $\mu(H^2(X,\StructureSheaf{X}))=\CC\beta$ (see Equation (<ref>)).
Hence, $\beta$ is orthogonal to $\gamma$ and thus the coefficient $r$ of $\alpha$  is $0$.

The sheaf $\iota_*\StructureSheaf{Z}$ deforms with $X$ in a direction $\xi\in H^1(TX)$, if and only if the line bundle $L$ does, by
[83]. Hence, $\mu(\tilde{\Sigma}(\iota_*\StructureSheaf{Z})\cap H^1(TX))=c_1(L)^\perp\cap H^{1,1}(X)$ and
$\lambda$ is a non-zero scalar multiple of $c_1(L)$. We may and  thus do choose $\lambda$ to be $c_1(L)$, so that $\gamma=c_1(L)+s\beta$.

We have $\ell((\iota_*\StructureSheaf{Z})^\vee)=\tilde{D}(\gamma)=-c_1(L)+s\beta$, by Lemma <ref>.
Local duality yields $(\iota_*\StructureSheaf{Z})^\vee\cong\iota_*(\omega_Z)[-n]$, so that $v((\iota_*\StructureSheaf{Z})^\vee)=(-1)^n\exp(\tau)\cdot v(\iota_*\StructureSheaf{Z})$.   Hence,
$\ell((\iota_*\StructureSheaf{Z})^\vee)=-\exp(e_{\tau})(\ell(\iota_*\StructureSheaf{Z}))$, which is spanned by
$-\exp(e_{\tau})(\gamma)=-c_1(L)-[s+(\tau,c_1(L))]\beta$. Comparing coefficients of $\beta$ we get
$s=-s-(\tau,c_1(L))$, so $s=-\frac{(\tau,c_1(L))}{2}$.

Let $S$ be a $K3$ surface and let $\iota:C\rightarrow S$ be an embedding of a smooth rational curve $C$.
Set $X:=S^{[n]}$, $n\geq 2$. We get the embedding
$\iota^{[n]}:C^{(n)}\rightarrow X$ of the $n$-th symmetric product of $C$, hence a lagrangian $\PP^n$ in $X$.
Example <ref> exhibits an auto-equivalence $\Phi:D^b(S)\rightarrow D^b(S)$ which maps $\StructureSheaf{S}$ to $\iota_*\omega_C$. We get the auto-equivalence $\Phi^{[n]}:D^b(X)\rightarrow D^b(X)$ which maps $\StructureSheaf{X}$ to the object in $D^b(X)$ corresponding to the $\fS_n$-equivariant sheaf
$\boxtimes_{i=1}^n \iota_*\omega_C$ over $S^n$ via the BKR-correspondence.
The latter object in $D^b(X)$ is isomorphic to $\iota^{[n]}_*\omega_{C^{(n)}}$. Hence, the isometry $\phi^{[n]}$ of
$\tilde{H}(X,\QQ)$ associated to $\Phi^{[n]}$  maps $\ell(\StructureSheaf{X})$ to $\ell(\iota^{[n]}_*\omega_{C^{(n)}})$.
The latter is spanned by a class $\gamma$ of the form $0\alpha+\lambda+s\beta$, where $s\in\QQ$ and $\lambda$ is a primitive class in $H^2(X,\ZZ)$, such that $\lambda^\perp$ is the kernel of the restriction homomorphism $H^2(X,\ZZ)\rightarrow H^2(C^{(n)},\ZZ)$. The class $\lambda$ has BBF self-intersection $-2(n+3)$

This agrees with the fact that $\ell(\StructureSheaf{X})$
is spanned by the class $2\alpha+\frac{n+3}{2}\beta$ of self-intersection $-2(n+3)$ (see Lemma <ref>).

§ LAGRANGIAN SURFACES IN IHSMS OF $K3^{[2]}$-TYPE

In Section <ref>
we use the determination of $\ell(\iota_*\StructureSheaf{Z})$ in Lemma <ref>
to calculate the Chern character of $\iota_*\StructureSheaf{Z}$ for the embedding
$\iota:Z\rightarrow X$ of a smooth lagrangian surface $Z$ in $X$ of $K3^{[2]}$-type (Lemma <ref>).
In Section <ref>
we show that $\iota_*\StructureSheaf{Z}$ has a rank $1$ cohomological obstruction map if $Z$ is a
smooth lagrangian surface in $X$ of $K3^{[2]}$-type with a rank $1$ restriction homomorphism $H^2(X)\rightarrow H^2(Z)$ with an image containing $c_1(\omega_Z)$  (Lemma <ref>).
Integral constraints on the topological invariants of such surfaces are presented in Section <ref> (Lemma <ref>).
In Section <ref> we explain why examples of such smooth lagrangian subvarieties with a non-vanishing first Betti number may yield new examples of holomorphic symplectic varieties.

§.§ The Chern character of $\StructureSheaf{Z}$ for a maximally deformable lagrangian surface $Z$

Keep the notation and assumptions of Lemma <ref>.
Set $\lambda:=c_1(L)$ and assume that $(\lambda,\lambda)\neq 0$ and $c_1(\omega_Z)=\iota^*\tau$,
for some $\tau\in H^2(X,\QQ)$. Then

\[
\]

for some[In Section <ref> we will assume that $\lambda$ is a primitive class and derive arithmetic constraints on $\chi(Z)$ and $(\lambda,\lambda)$. This is the reason we introduce the variable $t$, rather than allow $\lambda$ to be a rational class.
$t\in\QQ$, since $\lambda$ does not belong to $\lambda^\perp$. Set

\begin{equation}
\label{eq-c}
\end{equation}

where $\chi(Z)$ is the topological Euler characteristic of $Z$. The following Lemma expresses $ch(\iota_*\StructureSheaf{Z})$
in terms of $\chi(Z)$ and $\lambda$.

Assume, furthermore, that $X$ is of $K3^{[2]}$-type.

Then $\chi(Z)>0$ and the following equalities hold.

\begin{eqnarray}
\label{eq-t}
t^2 & = & \frac{4\sqrt{3\chi(Z)}}{5|(\lambda,\lambda)|}-\frac{6}{5(\lambda,\lambda)}.
\\
\label{eq-Euler-characteristic-of-structure-sheaf-of-lagrangian-surface}
\chi(\StructureSheaf{Z})&=& c\left(\frac{(\lambda,\lambda)}{10}+\frac{t^2(\lambda,\lambda)^2}{4}
\right).
\\
\label{eq-ch-2-of-lagrangian-structure-sheaf}
ch_2(\iota_*\StructureSheaf{Z}) & = & [Z]=c\left(\lambda^2-\frac{(\lambda,\lambda)}{30}c_2(X)\right).
\\
\label{eq-ch-3-of-lagrangian-structure-sheaf}
ch_3(\iota_*\StructureSheaf{Z}) & = & -\frac{ct}{3}\lambda^3.
\\
\label{eq-ch-4-of-lagrangian-structure-sheaf}
ch_4(\iota_*\StructureSheaf{Z}) & = & c\left(
\frac{t^2(\lambda,\lambda)^2}{4}-\frac{(\lambda,\lambda)}{10}
\right)[pt].
\end{eqnarray}

Lemma <ref>
yields the equality

\begin{equation}
\label{eq-Psi-of-Mukai-vector-of-lagrangian-structure-sheaf}
\Psi(ch(\iota_*\StructureSheaf{Z})\sqrt{td_X})
\right),
\end{equation}

for some $a\in\QQ$, where $\Pi$ is the projection from $\Sym^2(\tilde{H}(X,\QQ))$ to $\ker(\Delta)$.
We get

\begin{eqnarray*}
& &
\Psi\left([Z]+ch_3(\iota_*\StructureSheaf{Z})+ch_4(\iota_*\StructureSheaf{Z})+\frac{1}{24}[Z]c_2(X)
\right) \ \ \ =
\\
& &
\lambda^2-(\lambda,\lambda)\tilde{q}-t(\lambda,\lambda)\lambda\beta+\frac{t^2(\lambda,\lambda)^2}{4}\beta^2
\right)
\end{eqnarray*}

Comparing the weight $(2,2)$ summands we get Equation (<ref>), with $a$ instead of $c$, using also Example <ref>, Equation (<ref>), and the equality $\Psi(\lambda^2)=\lambda^2+(\lambda,\lambda)\alpha\beta$. We have

\[
\chi(Z)= \int_X[Z]^2=a^2\int_X\left(\lambda^4-\frac{(\lambda,\lambda)}{15}c_2(X)\lambda^2+\frac{(\lambda,\lambda)^2}{900}c_2(X)^2
\right)
= a^2\frac{48}{25}(\lambda,\lambda)^2.
\]

Hence, $\chi(Z)>0$ and $a=c$.
Equation (<ref>) follows from the equality $\Psi(\lambda^3)=3(\lambda,\lambda)\lambda\beta$, by comparing the weight $(3,3)$ summands. Equation (<ref>)
follows by equating the weight $(4,4)$ summands using Equation
(<ref>). Equation (<ref>) follows from the equality

\[
\chi(\StructureSheaf{Z})=\chi(\iota_*\StructureSheaf{Z})=\int_Xch(\iota_*\StructureSheaf{Z})td_X=
\int_X\left(ch_4(\iota_*\StructureSheaf{Z})+\frac{1}{12}[Z]c_2(X)\right)
\]

and  (<ref>).

It remains to prove Equation (<ref>).
We have the short exact sequence

\[
0\rightarrow TZ\rightarrow \iota^*TX \rightarrow N_{Z/X}\rightarrow 0.
\]

The isomorphism $TZ\cong N^*_{Z/X}$ yields

\[
\]

Hence, $c_2(TZ)=\frac{1}{2}\left[\iota^*c_2(TX)+c_1(\omega_Z)^2\right].$
The equality $\chi(Z)=\int_Zc_2(TZ)$ yields

\[
\int_Zc_1(\omega_Z)^2=2\chi(Z)-\int_X[Z]c_2(TX).
\]

Hirzebruch-Riemann-Roch over $Z$ yields the first equality below.

\[
\chi(\StructureSheaf{Z})=\frac{1}{12} \int_Z\left(c_1(\omega_Z)^2+c_2(TZ)\right)=
\frac{\chi(Z)}{4}-\frac{1}{12}\int_X[Z]c_2(TX).
\]

Equation (<ref>) and the definition of $c$ yield

\begin{equation}
\label{eq-Euler-characteristic-of-structure-sheaf-depends-only-on-topological-Euler-characteristic}
\chi(\StructureSheaf{Z}) = \frac{\chi(Z)}{4}-\frac{c(\lambda,\lambda)}{5}=\frac{1}{4}\left(
\chi(Z)-\frac{(\lambda,\lambda)}{|(\lambda,\lambda)|}\sqrt{\frac{\chi(Z)}{3}}
\right).
\end{equation}

Comparing the right hand sides of the above equation with that of Equation (<ref>) and solving for $t^2$ we get Equation (<ref>).

Note that Equation (<ref>) and the definition of $c$ yield

\begin{equation}
\label{eq-t-squared}
\end{equation}

The latter, combined with
Equation (<ref>), yields

\begin{equation}
\label{eq-two-Euler-characteristics}
\chi(\StructureSheaf{Z}) = \frac{1}{4}\left(
\chi(Z)-\frac{(\lambda,\lambda)}{|(\lambda,\lambda)|}\sqrt{\frac{\chi(Z)}{3}}
\right).
\end{equation}

The above and the definition of $\chi(Z)$ yield the two linear equations in three variables.

\begin{eqnarray*}
2\pm \sqrt{\frac{\chi(Z)}{3}}&=&h^{1,1}(Z)-2h^{2,0}(Z),
\\
\chi(Z)-2&=&h^{1,1}(Z)+2h^{2,0}(Z)-4h^{1,0}(Z),
\end{eqnarray*}

where the sign in the first equation is that of $(\lambda,\lambda)$.
If $(\lambda,\lambda)>0$, then

\begin{eqnarray}
\label{eq-h-2-0-when-h-1-0-vanishes}
\\
\nonumber
h^{1,1}(Z) & = & \frac{1}{2}\left[\chi(Z)+\sqrt{\frac{\chi(Z)}{3}}+4h^{1,0}(Z)\right].
\end{eqnarray}

Equation (<ref>) yields $\chi(Z)=3\left[h^{1,1}(Z)-2-2h^{2,0}(Z)\right]^2$ (the Euler characteristic is $3$ times the signature minus $2$ squared), or equivalently,

\[
\right).
\]

If $h^{1,1}(Z)=1$, then $h^{1,0}(Z)=-\frac{1}{2}(5h^{2,0}(Z)+6h^{2,0}(Z)^2)$, so $h^{1,0}(Z)=h^{2,0}(Z)=0$.
If, for example, $h^{2,0}(Z)=0$, then the pair $(h^{1,0}(Z),h^{1,1}(Z))$ is either $(0,1)$, or $(1,2)$. The former is realized by $\PP^2$, while the latter implies $\chi(Z)=0$ which is impossible if $(\lambda,\lambda)\neq 0$, by Lemma <ref>. If $h^{2,0}=1$, then the pair $(h^{1,0}(Z),h^{1,1}(Z))$ is either
$(1,3)$, or $(2,4)$. The latter is realized by a lagrangian abelian surface, while the former can not occur if $(\lambda,\lambda)>0$,
by Equation (<ref>). More generally, the last displayed equality implies that $h^{1,1}(Z)$ and $2+2h^{2,0}(Z)$ are “close”, in the sense that if $2+2h^{2,0}(Z)=ah^{1,1}(Z)$, then
$h^{1,1}(Z)\leq \frac{a+1}{3(a-1)^2}$, since $h^{1,0}(Z)\geq 0$, so $a$ is close to $1$ for large values of $h^{1,1}(Z)$.

§.§ Lagrangian surfaces $Z$ with $\StructureSheaf{Z}$ having a rank $1$ cohomological obstruction map

Assume that $X$ is of $K3^{[2]}$-type and that $\iota^*:H^2(X,\ZZ)\rightarrow H^2(Z,\ZZ)$ has rank $1$,
Let $L$ be the line bundle over $X$ such that $\ker(\iota^*)=c_1(L)^\perp$. Assume that $(c_1(L),c_1(L))\neq 0$
and that $c_1(\omega_Z)=t\iota^*c_1(L)$, for some $t\in \QQ$.

Then $\LieAlg{g}_{v(\iota_*\StructureSheaf{Z})}=\LieAlg{g}_\ell$, where $\ell$ is spanned by

Set $\lambda:=c_1(L)$.
It suffices to verify Equation (<ref>) with $a=c$.
Equation (<ref>) would follow once we verify Equations
(<ref>), (<ref>), and (<ref>).
The class $[Z]\in H^4(X,\QQ)$ is invariant under the stabilizer of $\lambda$ in the monodromy group of $X$.
Hence, $[Z]=a\lambda^2+bc_2(TX)$, for some $a, b\in\QQ$. Let $\sigma$ be a non-zero element of $H^0(X,\Omega^2_X)$. The pull back $\iota^*\sigma$ vanishes.
Hence, the cohomology class $[Z]\cup \sigma$ vanishes, since $\int_X[Z]\cup\sigma \cup (\bullet)=\int_Z\iota^*(\sigma) \cup (\bullet)$.
We drop the symbol $\cup$ below.
We have

\[
0=\int_X[Z]\sigma\bar{\sigma}=\int_X [a\lambda^2+bc_2(TX)]\sigma\bar{\sigma}=
\]

and $(\lambda,\sigma)=0$.
Hence, $b=-a(\lambda,\lambda)/30$ and Equation (<ref>) holds with $c$ replaced by $a$. The equality $a=c$ is proven using the equality $\chi(Z)=\int_X[Z]^2$ as in the proof of
Lemma <ref>.

Equation (<ref>) is equivalent to Equation (<ref>), by Hirzebruch-Riemann-Roch (and the fact that Equation (<ref>) is verified and the vanishing of the direct summand of $td_X$ in $H^2(X,\QQ)$). Now Equation (<ref>) follows from Equation (<ref>) and the fact that $Z$ is lagrangian, see the derivation of Equation (<ref>) which exhibits $\chi(\StructureSheaf{Z})$ as a function of only  $\chi(Z)$ and the sign of $(\lambda,\lambda)$.

It remains to prove Equation (<ref>). The class $ch_3(\iota_*\StructureSheaf{Z})$
must be equal to $a\lambda^3$, for some $a\in\QQ$, as it is invariant under the stabilizer of $\lambda$ in the monodromy group of $X$. We need to prove the equality $a=-ct/3$. We calculate $\chi(\iota^*(L^{\otimes nt}))$ in two ways.
Using Grothendieck-Riemann-Roch on $X$ we get the polynomial in $n$

\[
\chi(\iota^*(L^{\otimes nt}))=
\int_X\left(1+nt\lambda+\dots +\frac{n^4t^4}{4!}\lambda^4\right)ch(\iota_*\StructureSheaf{Z})td_X.
\]

The coefficient of $n^1$ is $3at(\lambda,\lambda)^2$.
On the other hand, Hirzebruch-Riemann-Roch on $Z$ yields the second equality below:

\[
\chi(\iota^*(L^{\otimes nt}))=\chi(\omega_Z^{\otimes n})=
\chi(\StructureSheaf{Z})+\frac{n(n-1)}{2}\int_Z c_1(\omega_Z)^2.
\]

The coefficient of $n^1$ is

\[
\int_Z c_1(\omega_Z)^2
= -\frac{1}{2}\int_X c(\lambda^2-\frac{(\lambda,\lambda)}{30}c_2(TX)) t^2\lambda^2 = -ct^2(\lambda,\lambda)^2.
\]

Comparing the two coefficients of $n^1$ we get the equality $a=-ct/3$.

If we let $Z=\PP^2$ in Lemma <ref>, then $(\lambda,\lambda)=-10$, the divisibility (Definition <ref>) of $\lambda$ is $2$, $c=\frac{1}{8}$, and $t=\pm\frac{3}{5}$. If $Z$ is the Fano variety of lines on a hyperplane section of a cubic $4$-fold, as in Example <ref>(<ref>), then $(\lambda,\lambda)=6$, the divisibility of $\lambda$ is $2$, $c=\frac{5}{8}$, and $t=1$.
If $Z$ is the fixed locus of an anti-symplectic involution, as in Example <ref>, then $(\lambda,\lambda)=2$, the divisibility of $\lambda$ is $1$, $c=5$ (since $\chi(Z)=192$), and $t=3$, by <cit.>.

§.§ Integral constraints for the existence of smooth lagrangian surfaces

Let  $X$ be of $K3^{[2]}$-type and let $\lambda$ be a primitive class in $H^2(X,\ZZ)$.

* If $\div(\lambda)=1$, then $\frac{1}{\gcd(5,(\lambda,\lambda))}\left(5\lambda^2-\frac{(\lambda,\lambda)}{6}c_2(TX)\right)$
is a primitive class in $H^4(X,\ZZ)$.

* If $\div(\lambda)=2$, then $\frac{1}{\gcd\left(40,5+\frac{(\lambda,\lambda)}{2}\right)}\left(5\lambda^2-\frac{(\lambda,\lambda)}{6}c_2(TX)\right)$
is a primitive class in $H^4(X,\ZZ)$.

The proof of Lemma <ref> uses the integral generators for the cohomology ring $H^*(X,\ZZ)$ found in [47]. The proof is omitted.

Smooth lagrangian surfaces satisfying the hypotheses of Lemma <ref>
must satisfy the following arithmetic constraints. Let $c$ be the rational number given in Equation (<ref>).

Let $\iota:Z\rightarrow X$ be an embedding of a smooth projective lagrangian surface in $X$ of $K3^{[2]}$-type. Assume that
the kernel of the restriction homomorphism $\iota^*:H^2(X,\ZZ)\rightarrow H^2(Z,\ZZ)$ is equal to $c_1(L)^\perp$ for a line bundle $L$ with primitive $\lambda:=c_1(L)$, such that $(\lambda,\lambda)>0$.
Assume furthermore that $c_1(\omega_Z)=t\iota^*\tau$, for some class $\tau\in H^2(X,\ZZ)$ and $t\in\QQ$. Then

$5$ does not divide $(\lambda,\lambda)$.

If $3$ divides $x:=\frac{(\lambda,\lambda)}{2}$,

$\div(\lambda)=2$, $x\equiv 3$ modulo $8$,  $8c/5$ is an integer, and both
$3x$ and $16cx-5$ are perfect squares.

$3$ does not divide

$x:=\frac{(\lambda,\lambda)}{2}$, then $x$ and $3(16cx-5)$ are perfect squares.
If $\div(\lambda)=1$, then $c/5$ is an integer. If $\div(\lambda)=2$, then $\gcd\left(8,5+\frac{(\lambda,\lambda)}{2}\right)\frac{c}{5}$ is an integer.

Assume first that $\gcd(5,(\lambda,\lambda))=1$.
If $\div(\lambda)=1$, set $n:=0$. If $\div(\lambda)=2$, let $n$ the integer, such that  $2^n:=\gcd\left(8,5+\frac{(\lambda,\lambda)}{2}\right)$.
Then $[Z]=\frac{k}{2^n}\left(5\lambda^2-\frac{(\lambda,\lambda)}{6}c_2(TX)\right)$, for an integer $k=\left\{
\begin{array}{ccl}
c/5 & \mbox{if} & \div(\lambda)=1,
\\
2^nc/5 & \mbox{if} & \div(\lambda)=2,
\end{array}
\right.$
by Lemmas <ref> and <ref>.

\[
\frac{48k}{5\cdot 2^n}-\frac{6}{5(\lambda,\lambda)}
\ \ = \ \
\frac{6(2^{3-n}k(\lambda,\lambda)-1)}{5(\lambda,\lambda)}
\]

must be the square of the rational number $t$, by Equation (<ref>).

\[
\]

is a perfect square.

Set $x:=\frac{(\lambda,\lambda)}{2}$.
Then $15x(2^{4-n}kx-1)$ is a perfect square.
Let $p$ be a prime integer. If $p$ divides $x$, then
$p$ does not divide $(2^{4-n}kx-1)$.
Thus, either both $3x$  and $5(2^{4-n}kx-1)$ are perfect squares, or both
$x$  and $15(2^{4-n}kx-1)$ are perfect squares. Part (<ref>) follow from the second case.
The first case implies that $n=3$, since otherwise $5(2^{4-n}kx-1)\equiv -1$ modulo $4$ and it can not be a square.
Hence, if $3$ divides $x$, then
$\div(\lambda)=2$, $x=(\lambda,\lambda)/2\equiv 3$ modulo $8$, $k=\frac{8c}{5}$ is an integer, and both $3x$  and $5(2kx-1)$ are perfect squares. Part (<ref>) follows.

Assume next that $5$ divides $(\lambda,\lambda)$.
Then the primitive lagrangian class is
$\frac{1}{2^n}\left(\lambda^2-\frac{(\lambda,\lambda)}{30}c_2(TX)\right)$, where $n=0$, if
$\div(\lambda)=1$, or
$2^n=\gcd\left(8,\frac{(\lambda,\lambda)}{10}+1\right)$, if
$\div(\lambda)=2$, by Lemma
Thus, $[Z]=\frac{k}{2^n}\left(\lambda^2-\frac{(\lambda,\lambda)}{30}c_2(TX)\right)$, for the integer $k=2^nc$.
Set $x:=\frac{(\lambda,\lambda)}{10}$. Then

\[
\sqrt{\frac{48k}{25\cdot 2^n}-\frac{6}{5(\lambda,\lambda)}}
\frac{1}{5}\sqrt{\frac{3(2^{4-n}kx-1)}{x}}
\]

should be a rational number. Hence, $3x(2^{4-n}kx-1)$ is a perfect square.
If a prime $p$ divides $x$, then it does not divide $(2^{4-n}kx-1)$.
Hence, either both $x$ and  $3(2^{4-n}kx-1)$ are perfect squares,
or both $3x$ and $2^{4-n}kx-1$ are perfect squares.
If both $x$ and  $3(2^{4-n}kx-1)$ are perfect squares, then $n\geq 2$, since $-3$ is not a square modulo $8$. Hence $\div(\lambda)=2$. It not possible for
both $3x$ and $2^{4-n}kx-1$ to be perfect squares, since in that case

$3$ divides $x$ and so $2^{4-n}kx-1\equiv -1$ modulo $3$, which is not a square.
We conclude that $x$ is a perfect square. Now $10x\equiv 2x$ modulo $8$ and $10x=(\lambda,\lambda)\equiv 6$ modulo $8$, since $\div(\lambda)=2$. Hence $x\equiv 3$ or $-1$ modulo $8$. The latter contradicts the fact that $x$ is a perfect square. Thus $5$ does not divide $(\lambda,\lambda)$ and
Part (<ref>) follows.

Following are two cases satisfying the arithmetic constraints above.
The case $(\lambda,\lambda)=8$, $\div(\lambda)$ either $1$ or $2$, $c=620$ (so that $\chi(Z)=3\cdot 2^{14}\cdot 31^2$), and $t=69/2$.
The case $(\lambda,\lambda)=54$, $\div(\lambda)=2$, $c=245/8$ (so $\chi(Z)=3^77^4$), and $t=23/3$.
A statement similar to Lemma <ref> can be formulated for $(\lambda,\lambda)<0$ and proven essentially with the same argument.

The moduli space of sheaves on $X$, which are deformations of $\iota_*\StructureSheaf{Z}$,
with $Z$ as in Example <ref>(<ref>),
is related to O'Grady's $10$-dimensional example of an irreducible holomorphic symplectic manifold with second Betti number equal to $24$ [46, 84]. Though rather constrained, additional examples
of lagrangian surfaces satisfying the assumptions of Lemma <ref>
would be potentially interesting (see Remark <ref>).

Singular lagrangian ruled surfaces, which deform with a polarized holomorphic symplectic manifold $(X,L)$ of $K3^{[2]}$ deformation type, are more common and their existence is proven[The existence of a uniruled divisor in $X$ in the linear system of $L$ is proven  in [18] and a choice of a curve in the base of the ruling determines a lagrangian surface.
in [18] for all but finitely many deformation types of the polarization $L$.

§.§ Lagrangian subvarieties with a positive LLV line may lead to new examples of IHSMs

Though rather constrained, additional examples
of lagrangian surfaces satisfying the assumptions of Lemma <ref>
would be potentially interesting.

Smooth lagrangian subvarieties $Z$ of projective irreducible holomorphic symplectic manifolds
$X$ satisfying the conditions

* $H^{1,0}(Z)$ does not vanish,

* the restriction homomorphism $\rho:H^2(X,\ZZ)\rightarrow H^2(Z,\ZZ)$ has rank $1$,

* the lattice orthogonal to $\ker(\rho)$ is spanned by an ample class $\lambda$,

are interesting from the point of view of the classification of holomorphic symplectic varieties for the following reason. The ampleness of $\lambda$ implies that $(\lambda,\lambda)>0$
and so $\lambda^\perp$ is a non-degenerate sublattice of co-rank $1$ in $H^2(X,\ZZ)$.
The pair $(X,Z)$ deforms over an open subset of the period domain $\Omega_{\lambda^\perp}$ of weight $2$ Hodge structures  associated to the lattice $\lambda^\perp$ [83]. The class $\lambda$ remains ample over an open dense subset in moduli.
Note that $\dim(\Omega_{\lambda^\perp})=h^{1,1}(X)-1$. Let $M$ be the connected component of the moduli space of $\lambda$-stable sheaves on $X$ containing
$\iota_*L$, where $\iota:Z\rightarrow X$ is the inclusion and $L$ is a line bundle on $Z$ with
$c_1(L)\in \span_\QQ\{\iota^*\lambda,c_1(\omega_Z)\}$. Let $B$ be the irreducible component of the
Hilbert scheme of $X$ containing $Z$.
Then $Z$ is a smooth point of $B$ and $\dim(B)=h^{1,0}(Z)$, since the normal bundle of $Z$ in $X$ is isomorphic to
$T^*Z$ [83].
A Zariski open subset $M^0$ of $M$ parametrizes pairs $(Z',\iota_*L')$, where $Z'$ is a smooth lagrangian subvariety of $X$ and
$L'$ is a line bundle over $Z'$. $M^0$ is holomorphic symplectic and the support morphism $\pi:M^0\rightarrow B$ is projective and a lagrangian fibration, by
<cit.>. Being quasi-projective, $M^0$ admits an ample divisor class, which restricts non-trivially to the projective fibers of $\pi$, as well as the pull back of an ample class from $B$.
In particular, the rank of  the Neron-Severi group  of any compactification $\bar{M}$ of $M^0$
is at least $2$.

Assume that $M^0$ admits a compactification $\bar{M}$, which is an irreducible symplectic variety with at worst $\QQ$-factorial terminal singularities.
Then the moduli space of complex deformations of $\bar{M}$ admits a surjective and generically injective period map onto the period domain $\Omega_\Lambda$ of weight $2$ pure Hodge structures on the lattice $\Lambda:=H^2(\bar{M},\ZZ)$, by [7]. In particular, the moduli space of $\bar{M}$ has dimension $h^{1,1}(\bar{M})$ and
the locus $\LB$ in moduli of $\bar{M}$, arising from varying $(X,Z)$ via the above construction, has co-dimension $\geq 2$,
since such symplectic varieties with Picard rank $r$ vary in co-dimension $r$ in moduli. Hence, the dimension of $\LB$ is at most
The locus $\LB$ is expected

to have the same dimension $h^{1,1}(X)-1$ as $\Omega_{\lambda^\perp}$, which would imply the inequality $h^{1,1}(\bar{M})\geq h^{1,1}(X)+1$. Only one deformation type of irreducible symplectic varieties with second Betti number $\geq 24$ is known, namely O'Grady's desingularization of a $10$ dimensional moduli space of sheaves on a $K3$ surface
[66, 69].
The latter is related to the above strategy in the papers [46, 84].

§ EFFECTIVE LLV LINES

Let $X$ be a projective irreducible holomorphic symplectic manifold.
We say that a line $\ell$ in $\tilde{H}(X,\QQ)$ is effective, if $\ell=\ell(F)$ for an object $F$ in $D^b(X)$,

such that
$F$  has a rank $1$ cohomological obstruction map

(Definition <ref>).
We say that a line $\ell$ in $\tilde{H}(X,\QQ)$ is potentially effective,

if there exists a projective irreducible holomorphic symplectic manifold $Y$ and a derived parallel transport operator
$\tilde{\phi}:\tilde{H}(X,\QQ)\rightarrow \tilde{H}(Y,\QQ)$, such that $\tilde{\phi}(\ell)$ is effective.

Clearly, if $\ell$ is potentially effective, then so is every line in the orbit of $\ell$ under the derived monodromy group. Theorem <ref> makes the following plausible.

A line $\ell$  in $\tilde{H}(X,\QQ)$ is effective, if it is potentially effective and is contained in $\tilde{H}^{0,0}(X,\QQ)$.

Assume that $X$ is of $K3^{[2]}$-type.
Choose a class $\delta\in H^2(X,\Integers)$ of divisibility $2$ satisfying $(\delta,\delta)=-2$.
Set $B_\lambda:=\exp(e_\lambda)$, for $\lambda\in H^2(X,\QQ)$.

\begin{equation}
\label{eq-Mukai-lattice-K3-2-case}
\Lambda_X:=B_{-\delta/2}(\ZZ\alpha+H^2(X,\ZZ)+\Integers\beta)\subset \tilde{H}(X,\QQ).
\end{equation}

The subgroup $\Lambda_X$ of $\tilde{H}(X,\QQ)$ is independent of the choice of $\delta$. We refer to $\Lambda_X$ as the integral LLV lattice, since it is invariant under the derived monodromy group of $X$, by <cit.>.

Assume that $X$ is of $K3^{[n]}$-type.
Choose a primitive class $\delta\in H^2(X,\Integers)$ of divisibility $2n-2$ satisfying $(\delta,\delta)=2-2n$.
Define $\Lambda_X$ as in Equation (<ref>).

Let us show that the lattice $\Lambda_X$ depends only on the
$\Mon(X)$-orbit of $\delta$. Notice that $B_{-\delta/2}(H^2(X,\ZZ)+\Integers\beta)=H^2(X,\ZZ)+\Integers\beta$ and
$B_{-\delta/2}(\alpha)=\alpha-\delta/2+\frac{1-n}{4}\beta$. Given $g\in O(H^2(X,\Integers))$ we get that
$B_{-g(\delta)/2}(\alpha)-B_{-\delta/2}(\alpha)=(\delta-g(\delta))/2$. It suffices to show that the latter difference belongs to $H^2(X,\ZZ)$.
Indeed, the discriminant group $H^2(X,\ZZ)^*/H^2(X,\ZZ)$ is cyclic generated by $\delta/(2n-2)$ and the coset $\delta/2+H^2(X,\ZZ)$ is
invariant under $\Mon(X)$, as the monodromy action on $H^2(X,\ZZ)$ factors through a (surjective) homomorphism  onto
the subgroup of
$O^+(H^2(X,\ZZ))$ of elements acting on the discriminant group by $\pm1$, by <cit.>.

* The lattice $\Lambda_X$ is invariant under $\Mon(X)$, by part (<ref>).
Indeed, we have the equalities $g\circ B_{-\delta/2}=(g\circ B_{-\delta/2}\circ g^{-1})\circ g=B_{-g(\delta)/2}\circ g$, which takes $\Integers\alpha\oplus H^2(X,\Integers)\oplus \Integers\beta$ to $\Lambda_X$
if $g\in \Mon(X)$.

When $X=S^{[n]}$, for some $K3$-surface $S$, then
$\Lambda_X$ is $DMon(S)$-invariant, by Theorem <ref>, as well as with respect to
the auto-equivalence $\Phi_\chi^{[n]}$ which is the $BKR$-conjugate of tensorization with the sign character of $\fS_n$ (Lemma
<ref>). Hence, $\Lambda_X$ is invariant by the subgroup of $DMon(S^{[n]})$ generated by these three types of elements.

* Following is an intrinsic definition of a $DMon(X)$-invariant lattice in $\widetilde{H}(X,\QQ)$.
Let $v:K_{top}(X)\rightarrow H^*(X,\QQ)$ send a class $x$ to $ch(x)\sqrt{td_X}$. Let $\hat{v}:K_{top}(X)\rightarrow SH^*(X,\QQ)$
be the composition of $v$ with the projection to $SH^*(X,\QQ)$. We get the lattice $\Lambda_{X,n}:=\Psi(\hat{v}(K_{top}(X)))$
in the kernel of $\Delta:\Sym^n(\widetilde{H}(X,\QQ))\rightarrow \Sym^{n-2}(\widetilde{H}(X,\QQ))$.
A class $\gamma$ in $\widetilde{H}(X,\QQ)$ is an integral  class, if
the projection of $\gamma^n$ to $\ker(\Delta)$ belongs to $\Lambda_{X,n}$.

Let $\Lambda'_X\subset \widetilde{H}(X,\QQ)$
be the abelian subgroup generated by integral classes.

$\Lambda'_X$ is $DMon(X)$-invariant, since equivalences of derived categories map objects, which remain of Hodge type in co-dimension $1$, to such objects.
Set $t_X:=\min\{t\in \QQ \ : \ t>0 \ \mbox{and} \ t^nc_X\in \ZZ\}$, where $c_X$ is the Fujiki constant.
The class $t_X\beta$ belongs to $\Lambda_X'$, by Equation (<ref>).
If $X$ is of $K3^{[n]}$-type, then $t_X=c_X=1$ and so
$\Lambda'_X$ contains $\beta$ and $\alpha+\frac{n+3}{4}\beta$,
by Lemma <ref>. If $X$ is of generalized Kummer deformation type, then
$t_X=1$, as $c_X=n+1$ is not divisible by an $n$-th power of an integer larger than $1$. Thus
the classes $\beta$ and $\alpha+\frac{n+1}{4}\beta$ belongs to $\Lambda'_X$, by Lemma <ref>.

Assume that $X$ is of $K3^{[2]}$-type.
A rank $1$ sublattice of $\Lambda_X$ determines a unique orbit of primitive elements in
$\Lambda_X$ for the action of the
derived monodromy group $DMon(X)$.

Which primitive elements of $\Lambda_X$ span potentially effective lines in $\tilde{H}(X,\QQ)$?

As is well known, the positive cone $\C^+:=\{x\in \tilde{H}(X,\RR) \ : \ (x,x)>0\}$ is homotopic to the $3$-sphere (see for example <cit.>).
$DMon(X)$ contains the index $2$ subgroup $O^+(\Lambda_X)$ of $O(\Lambda_X)$ acting trivially on $H^3(\C^+,\ZZ)$, by <cit.>.
It follows that the self intersection $(x,x)$ and the divisibility of a class $x\in\Lambda_X$ determine the $DMon(X)$-orbit of $x$,
by <cit.>.

We have seen that the primitive isotropic element $\beta$ of divisibility $1$ is effective, by Example <ref>.
The element $0\alpha+\lambda+0\beta$ is potentially effective, for every primitive isotropic class $\lambda\in H^2(X,\ZZ)$, by Example <ref>(<ref>).
The latter two examples represent the same $DMon(X)$-orbit, since every primitive isotropic class in $\Lambda_X$ has divisibility $1$ (see for Example <cit.>).
The primitive element
$2\alpha+\frac{5}{2}\beta$ of self intersection $-10$ and divisibility $2$ is effective, by Example  <ref> of the trivial line bundle $\StructureSheaf{X}$.
The class $\lambda+3\beta$ is potentially effective, for $\lambda\in H^2(X,\ZZ)$ primitive of divisibility $2$ with $(\lambda,\lambda)=-10$, by Example <ref> of a lagrangian $\PP^2$.
The latter two examples represent the same $DMon(X)$-orbit.
The class $\lambda-3\beta$, for $\lambda\in H^2(X,\ZZ)$  primitive of divisibility $2$ with $(\lambda,\lambda)=6$ is potentially effective,
by Example <ref> of the lagrangian Fano variety of lines on a cubic threefold and Lemma <ref>. So is
the class $\lambda-3\beta$, for $\lambda\in H^2(X,\ZZ)$  primitive of divisibility $1$ with $(\lambda,\lambda)=2$,
by Example <ref> of the lagrangian fixed locus of an anti-symplectic involution and Lemma <ref>. Reference [18] may lead to more effective classes associated to singular lagrangian surfaces (see Remark <ref>).

§ THE CHERN CHARACTER OF A MODULAR VECTOR BUNDLE ISOMORPHIC TO $\PHI(\STRUCTURESHEAF{K3^{[2]}})$

We calculate the Chern character of a modular vector bundle obtained from the structure sheaf of $X$ of $K3^{[2]}$-type via
equivalences of derived categories and deformations. The higher Chern classes of $F$ are determined by its rank and first Chern class and the values of the latter two are constrained.

Let $X$ be of $K3^{[2]}$-type.
Let $\Lambda_X$ be its integral LLV lattice and let $B_{-\delta/2}$ be the isometry of $\tilde{H}(X,\QQ)$ both given in (<ref>).
We have

\begin{eqnarray}
\label{eq-B-minus-delta-over-2-of-alpha}
\\
\nonumber
B_{-\delta/2}(\lambda)&=& \lambda-\frac{(\lambda,\delta)}{2}\beta, \ \ \ \forall \lambda\in H^2(X,\QQ), \ \ \ \mbox{and}
\\
\nonumber
\end{eqnarray}

Note the equality

\begin{equation}
\label{eq-B-takes-H-2-and-beta-to-itself}
\end{equation}

Let $F$ be a vector bundle over $X$ which is obtained via a sequence of equivalences of derived categories and deformations from the structure sheaf $\StructureSheaf{Y}$, where $Y$ is of $K3^{[2]}$-type as well.
Then $\rank(F)=r_0^2$, for some integer $r_0$. Set $h:=c_1(F)$. Then $h=\frac{r_0}{\gcd(r_0,2)}\eta$ for an integral class $\eta\in H^{1,1}(X,\Integers)$. The class $\eta$ satisfies

\begin{eqnarray}
\label{r-0-odd}
r_0\mid(5+2(\eta,\eta)) & \mbox{if} & r_0 \ \mbox{is} \ \mbox{odd},
\\
\label{r-0-congruent-to-2-mod-4}
r_0\mid\left(5+\frac{(\eta,\eta)}{2}\right) & \mbox{if} & r_0
\ \mbox{is} \ \mbox{even}.
%\equiv 2 \ (\mbox{mod} \ 4),
%r_0\mid\left(5+\frac{(\eta,\eta)}{8}\right) & \mbox{if} &
%r_0 \equiv 0 \ (\mbox{mod} \ 4), \ \mbox{in which case} \ (\eta,\eta)\equiv 0 \ (\mbox{mod} \ 8).
\end{eqnarray}

Furthermore, $\Psi(v(F))=\frac{1}{8}(\gamma^2+10\tilde{q})$, where

\begin{equation}
\label{eq-gamma}
\gamma=
\end{equation}

is a primitive class of divisibility $2$ in the lattice $\Lambda_X$
satisfying $(\gamma,\gamma)=-10$. Consequently,

\begin{eqnarray}
\nonumber
ch_2(F) & = & \frac{h^2}{2r_0^2} +\frac{1-r_0^2}{24}c_2(X),
\\
\nonumber
ch_3(F) & = &\frac{h^3}{6r_0^4}+\frac{(1-r_0^2)}{24r_0^2}hc_2(X),
\\
\nonumber
ch_4(F) & = &\frac{4(h,h)^2+20r_0^2(1-r_0^2)(h,h)+25r_0^4-46r_0^6+21r_0^8}{32r_0^6}[pt],
\\
\label{eq-euler-characteristic-of-F}
\chi(F)&=& \frac{4(h,h)^2+20(h,h)r_0^2(r_0^2+1)+25r_0^4(r_0^4+1)+46r_0^6}{32r_0^6}
\\
\nonumber
\end{eqnarray}

Let $\Lambda_Y$ be the integral LLV lattice of $Y$, given in (<ref>). The line $\ell(\StructureSheaf{Y})$ is spanned by the class $4\alpha+5\beta$, by Example <ref>.
We have $(4\alpha+5\beta)=B_{-\delta/2}(4\alpha+2\delta+4\beta)$ and
$2\alpha+\delta+2\beta$ is a primitive class in $\ZZ\alpha+H^2(X,\ZZ)+\Integers\beta$ of divisibility $2$.
Then $\gamma_0:=2\alpha+\frac{5}{2}\beta$ is a primitive class of divisibility $2$ in $\Lambda_Y$
satisfying $(\gamma_0,\gamma_0)=-10$.
Let $\widetilde{H}(\phi):\Lambda_Y\rightarrow \Lambda_X$ be the
integral isometry associated to a
derived parallel transport operator $\phi:H^*(Y,\QQ)\rightarrow H^*(X,\QQ)$, which is induced by

a sequence of equivalences of derived categories and deformations used to obtain $F$ from $\StructureSheaf{Y}$
(see Theorem <ref>).
Set $\gamma:=(\widetilde{H}(\phi))(\gamma_0)$. Then $\gamma$ is a primitive class of divisibility $2$ in $\Lambda_X$
satisfying $(\gamma,\gamma)=-10$, since $\phi$ is an isometry.
Note that

$H^2(X,\ZZ)$ is contained in $\Lambda_X$

and the component in $H^2(X,\QQ)$ of a class in $\Lambda_X$ belongs to $H^2(X,\ZZ)+\Integers\left(\frac{\delta}{2}\right)$,
by Equations (<ref>) and (<ref>).
Write $\gamma=r\alpha+\lambda+s\beta$, with $r,s\in\QQ$ and $\lambda\in H^2(X,\ZZ)+\Integers\left(\frac{\delta}{2}\right)$.
Then $r=-(\gamma,\beta)$ is an even integer by the divisibility of $\gamma$.
Set $r_0:=r/2$.
Furthermore, $s$ belongs to $\frac{1}{2}\ZZ$, since
$(\gamma,B_{-\delta/2}(\alpha))=(\gamma,\alpha-\delta/2-\beta/4)=-s+\frac{r_0}{2}-(\lambda,\delta/2)$ is an even integer and
$(\lambda,\delta/2)$ belongs to $\frac{1}{2}\ZZ$.
We have $v(F)=\phi(v(\StructureSheaf{Y}))$ and the equality

\begin{equation}
\label{eq-Psi-v-F}
\epsilon(\widetilde{H}(\phi))\Psi(v(F))=\frac{1}{8}(\gamma^2+10\tilde{q})=
\frac{1}{8}\left(
r^2\alpha^2+\lambda^2+s^2\beta^2+2r\lambda\alpha+2rs\alpha\beta+2s\lambda\beta +10\tilde{q}
\right),
\end{equation}

by Equation (<ref>) and Theorem <ref>.

The coefficient of $\alpha^2$ in $\Psi(v(F))$ is $\rank(F)/2$ and the coefficient of $\alpha^2$ in the right hand side is
$r^2/8$, so $\epsilon(\widetilde{H}(\phi))=1$ (by Remark <ref>) and $\rank(F)=r_0^2$.

Let $h=c_1(F)$. Note that $\Psi(h)=\Psi(e_h(1))=e_h(\alpha^2/2)=h\alpha$.
The coefficient in $H^2(X,\QQ)$ of $\alpha$ in the left hand side of (<ref>) is $h$ and the coefficient on the right hand side is $(r_0/2)\lambda$. Hence, $h=(r_0/2)\lambda$.

We have
So $s=\frac{5}{2r_0}+\frac{(h,h)}{r_0^3}=\frac{5r_0^2+2(h,h)}{2r_0^3},$ and

\[
\gamma=2r_0\alpha+\frac{2}{r_0}h+\frac{5r_0^2+2(h,h)}{2r_0^3}\beta,
\]

verifying Equation (<ref>).

The above displayed formula yields

\[
\gamma=
\left[\frac{5r_0^2+2(h,h)-2r_0^4}{2r_0^3}+\frac{(h,\delta)}{r_0}\right]\beta
\right).
\]

We see that $\frac{2}{r_0}h$ is an integral class. Hence, $\frac{(h,\delta)}{r_0}$ is an integer, since $\delta$ has divisibility $2$,  and
so $2r_0^3$ divides $5r_0^2+2(h,h)-2r_0^4$.

If $r_0$ is odd, then $h=r_0\eta$, for $\eta\in H^2(X,\ZZ)$, and $r_0^3$ divides $r_0^2(5+2(\eta,\eta))$, so
Equation (<ref>) holds. If

$r_0$ is even,
then $h=(r_0/2)\eta,$ for  $\eta\in H^2(X,\ZZ)$, and $r_0^3$ divides $r_0^2(5+(\eta,\eta)/2)$, so Equation (<ref>) holds.

Let us first calculate $\kappa(F)=\exp(-h/r_0^2)(ch(F))$. Set $\kappa(\gamma):=\exp(-h/r_0^2)(\gamma)=2r_0\alpha+\frac{5}{2r_0}\beta$. Equation (<ref>) yields the first equality below

\[
\]

Hence, $8\kappa(F)\sqrt{td_X}=8r_0^2+\frac{46}{5}b_X+\frac{25}{4r_0^2}[pt]$, by Equation (<ref>),  and using $\frac{46}{5}b_X=\frac{1}{3}c_2(X)$, established in Equation (<ref>), we get

\[
\kappa(F)=(\sqrt{td_X})^{-1}\left(r_0^2+\frac{1}{24}c_2(X)+\frac{25}{32r_0^2}[pt]\right).
\]

Now, $(\sqrt{td_X})^{-1}=1-\frac{1}{24}c_2(X)+\frac{21}{32}[pt]$ and we get

\begin{equation}
\label{eq-kappa-F-depends-only-on-the-rank}
\kappa(F)=r_0^2+\frac{1-r_0^2}{24}c_2(X)+\frac{1}{32}\left(21r_0^2+\frac{25}{r_0^2}-46\right)[pt].
\end{equation}

Multiplying both sides by $1+\frac{h}{r_0^2}+\frac{h^2}{2r_0^4}+\frac{h^3}{6r_0^6}+\frac{h^4}{24r_0^8}$ and using $h^4=3(h,h)^2$ and $\int_Xc_2(X)h^2=30(h,h)$ we get

\begin{eqnarray*}
\left[\frac{h^2}{2r_0^2}+\frac{1-r_0^2}{24}c_2(X)\right]+
\left[\frac{h^3}{6r_0^4}+\frac{(1-r_0^2)}{24r_0^2}hc_2(X)\right]+
\\
\end{eqnarray*}

Formula (<ref>) now follows, by Grothendieck-Riemann-Roch.

The determination of the Chern character of $F$ in
Lemma <ref>
in terms of $\rank(F)$ and $c_1(F)$ agrees with the computation of $ch_2(F)$ in O'Grady's <cit.>.
The formula (<ref>) for $\chi(F)$ agrees with <cit.>, when $r_0=1$.
When $r_0=2$ and $(h,h)=6$ it yields $\chi(F)=6$, as in the case of the Fano variety of lines on a cubic fourfold <cit.>.
When $r_0=2$ and $(h,h)=22$ it yields $\chi(F)=10$, as in the case of Debarre-Voisin varieties <cit.>.
More generally, whenever $F$ is generated by global sections and has vanishing higher cohomology, we get a morphism $\varphi_F:X\rightarrow Gr(r_0^2,\chi(F))$. When $(X,F)=(S^{[2]},G^{[2]})$, as in Theorem <ref>, and the vector bundle $G$ on the $K3$ surface $S$ has vanishing higher cohomology, then
the vector bundle $G^{[2]}$ has vanishing higher cohomology, by the fact that $BKR$ is an equivalence and Equation (<ref>), and formula (<ref>) agrees with the fact that
$\chi(G^{[2]})=\dim H^0(S^{[2]},G^{[2]})=\dim\Sym^2(H^0(S,G)).$
Formula  (<ref>) and the integrality of $\chi(F)$ imply that
$(h,h)+\frac{5r_0^2(r_0^2+1)}{2}$ is an odd multiple of $r_0^3$. In particular,

\[
(h,h)\equiv r_0^3-\frac{5r_0^2(r_0^2+1)}{2} \  (\mbox{mod} \ 2r_0^3).
\]

The latter congruence relation and the equality $h=\frac{r_0}{\gcd(r_0,2)}\eta$ of Lemma <ref> imply the congruence relations involving $r_0$ and $(\eta,\eta)$, which are conjectured in <cit.>.

Let us relate Lemma <ref>
to the lagrangian surfaces in Example <ref>.
The total Segre class of an object in $D^b(X)$ is the inverse of its total Chern class.
The second Segre class of the vector bundle $F$ is Lemma <ref> is

\[
\]

and it is proportional to the lagrangian class associated in Lemma <ref> to the class $\eta=\frac{\gcd(2,r_0)}{r_0}h$, if and only if
$(h,h)=\frac{5r_0^2(r_0^2-1)}{2(r_0^2+1)}$. The only integral solutions of the latter with positive $r_0$ are
$(r_0,(\eta,\eta))\in \{(1,0),(2,6),(3,2)\},$ and $\chi(F)=3, 6, 10$, respectively, by (<ref>).
In all three cases $s_2(F)$ is the primitive lagrangian class in Lemma <ref>.
In the first two cases the generic such $F$ is generated by global sections, and so $s_2(F)=c_2(E)$, where $E$ is the kernel of the evaluation map $ev$ in the short exact sequence

\begin{equation}
\label{eq-E-with-chern-class-equal-segre-class-of-F}
0\rightarrow E\rightarrow H^0(X,F)\otimes_\ComplexNumbers\StructureSheaf{X}\RightArrowOf{ev} F\rightarrow 0,
\end{equation}

$E$ has rank $\chi(F)-r_0^2=2$, and the zero loci of regular sections of $E^*$ are the lagrangian surfaces in Example <ref>(<ref>),(<ref>). In the third case of
the lagrangian class associated to $\eta$ in Lemma <ref> corresponds to the surface $Z$ of fixed points of the anti-symplectic involution in Example <ref>(<ref>), but in this case
the kernel of $ev$ has rank $1$,
and so $F$ is not generated by global sections, since $s_2(F)$ does not vanish.

Finally, let $L$ be the line bundle with $c_1(L)=\eta$ of BBF degree $2$, and let $E_L$ be the kernel of the evaluation homomorphism
$H^0(X,L)\otimes\StructureSheaf{X}\rightarrow L$. The fixed locus $Z$ of the anti-symplectic involution is the set theoretic degeneracy locus
of the differential $d\varphi_L:TX\rightarrow \SheafHom(E_L,L)$ of the morphism from $X$ to $\PP H^0(X,L)^*\cong\PP^5$ exhibiting $X$ as a double cover of an EPW sextic. The class of the scheme theoretic degeneracy locus is
$c_2(TX\rightarrow \SheafHom(E_L,L))$, by the Thom-Porteous formula, and is equal to $3[Z]$.

Lemma  <ref> may also help construct examples of lagrangian surfaces with the two classes in Remark <ref> satisfying the arithmetic constraints of Lemma <ref>. Over the generic IHSM $X$ of $K3^{[2]}$ type with a primitive polarization of class $\eta$ with $(\eta,\eta)=54$ there exists a rank $4$ vector bundle $F$ with $c_1(F)=\eta$ as in
Lemma <ref>, by <cit.>.
when $(\eta,\eta)=8$ there exists over $X$
a rank $9$ vector bundle $F$ with $c_1(F)=3\eta$ as in
Lemma <ref>, by <cit.>.
In both cases the second Segre class $s_2(F)$ minus $c_2(TX)$ is the primitive lagrangian class associated to $\eta$ in Lemma <ref>.

Let $S$ be a $K3$ surface and
$X=S^{[n]}$. Let $F$ be the vector bundle
corresponding via the BKR equivalence to
$G\boxtimes G \boxtimes \cdots \boxtimes G\otimes_\ComplexNumbers \chi^i$, for a rigid slope-stable vector bundle $G$ on $S$.
If $n=2$, Lemma <ref> applies to the vector bundle $F$,
by Remark <ref>. We use Lemma <ref> to show that $\ell(F)$ is spanned by

\[
\left\{
\begin{array}{ccl}
\frac{(c_1(G),c_1(G))+2}{2}+\frac{r_0^2}{4}-\frac{r_0(r_0-1)}{2}
\right]\beta &\mbox{if} & i=0,
\\
\frac{(c_1(G),c_1(G))+2}{2}+\frac{r_0^2}{4}-\frac{r_0(r_0+1)}{2}
\right]\beta &\mbox{if} & i=1.
\end{array}
\right.
\]

We start with comments for all $n$, which will be used in the following Remark <ref> to calculate $\ell(F)$ for all $n$.
The line $\ell(F)$ is invariant under the action of the stabilizer of $\ell(G)$ in the isometry group of the sublattice
$\tilde{H}(S,\RationalNumbers)$ of $\tilde{H}(S^{[n]},\RationalNumbers)$, where the isometry group of the former is embedded as a subgroup of the isometry group of the latter via the homomorphism $h$  making the diagram below commutative.

\[
\xymatrix{
DMon(S) \ar[d]  \ar[r] & DMon(S^{[n]}) \ar[d]
\\
O(\tilde{H}(S,\QQ)) \ar[r]^h & O(\tilde{H}(S^{[n]},\QQ)),
\]

The top horizontal arrow above is $^{[n]}:\Hom_{K3^{[1]}}(S,S)\rightarrow \Hom_{K3^{[n]}}(S^{[n]},S^{[n]})$
induced by the functor (<ref>).
The homomorphism $h$ is given by
$h(g)=\det(g)\cdot (B_{-\delta/2}\circ \iota(g)\circ B_{\delta/2})$, where $\iota(g)$ is the extension acting trivially on $\span\{\delta\}$ (see Example <ref>), and $B_\lambda$ is  $\exp(e_\lambda)$, by
<cit.> when $n=2$ and by Theorem <ref> for general $n$.
It follows that $B_{\delta/2}(\ell(F))$ is a line in the plane spanned by $\ell(G)$ and $\delta$.

\begin{equation}
\label{eq-B-of-ell-F-is-in-the-plane-spanned-by-delta-and-ell-G}
B_{\delta/2}(\ell(F))\subset \ell(G)+\QQ\delta.
\end{equation}

Now, if the rank of $G$ is $r_0$, then the rank $r$ of $F$ is $r_0^n$ and $c_1(F)=r_0^{n-1}c_1(G)+t\delta$, under the inclusion $H^2(S,\Integers)\subset H^2(S^{[n]},\Integers)$ given in (<ref>), where $t$ is an integer
depending on the exponent $i\in \{0, 1\}$ of $\chi$ and $\delta$ is half the class of the divisor of non-reduced subschemes.
The line $\ell(F)$ is spanned by a vector of the form
$r_0^n\alpha+c_1(F)+s(F)\beta$, for some $s(F)\in\QQ$,

by Theorem <ref>(<ref>).

Assume $n=2$. The coefficient $s(F)-t-\frac{r_0^2}{4}$ of $\beta$ in $B_{\delta/2}(r_0^2\alpha+c_1(F)+s(F)\beta)$ must be
equal $r_0 \cdot s(G)$, by (<ref>), yielding $s(F)=r_0\cdot s(G)+\frac{r_0^2}{4}+t$.  Set $\lambda:=c_1(G)$. Then $s(G)=\frac{(\lambda,\lambda)+2}{2r_0}$ and

\[
\]

On the other hand, multiplying the coefficient of $\beta$ in (<ref>) by $r_0/2$ we get that

\[
\]

Comparing the above two expressions for $s(F)$ we get that $t^2+r_0^2t+\frac{r_0^4-r_0^2}{4}=0,$
so that

\[
t\in \left\{
-\frac{r_0(r_0-1)}{2}, \ -\frac{r_0(r_0+1)}{2}
\right\}.
\]

O'Grady calculated in <cit.> that $t=-r_0(r_0-1)/2$ if $i=0$ and $t=-r_0(r_0+1)/2$ if $i=1$ as follows.
Using the notation of (<ref>) we have the short exact sequence over $\Gamma$

\[
0\rightarrow q^*F\rightarrow b^*(G\boxtimes G)\rightarrow \left[(G\boxtimes G\restricted{)}{\hat{D}}\right]^{\chi^{i+1}}\rightarrow 0,
\]

where $\hat{D}\subset \Gamma$ is the exceptional divisor of the blow-up $b:\Gamma\rightarrow S\times S$ centered along the diagonal.
Now $q^*(\delta)=[E]$, $c_1(b^*(G\boxtimes G))=r_0q^*(\lambda)$, and so

\begin{equation}
\label{eq-minus-t}
-t=\rank \left[(G\boxtimes G\restricted{)}{\hat{D}}\right]^{\chi^{i+1}}=
\left\{
\begin{array}{ccl}
r_0(r_0-1)/2 & \mbox{if} & i=0,
\\
r_0(r_0+1)/2 & \mbox{if} & i=1.
\end{array}
\right.
\end{equation}

O'Grady denotes $F$ by $G[2]^+$, if $i=0$, and by $G[2]^-$, if $i=1$.
Note that $G[2]^-$ is isomorphic to $(G^*[2]^+)^*\otimes \StructureSheaf{S^{[2]}}(-\delta)$,
by Lemma <ref>.
Indeed, the above description of $\ell(G[2]^-)$ agrees with that of $\ell((G^*[2]^+)^*\otimes \StructureSheaf{S^{[2]}}(-\delta))$
via Lemma <ref>.

The argument of Example <ref> generalizes to calculate $\ell(F)$ for
the vector bundle $F$
corresponding via the BKR equivalence to
$G\boxtimes G \boxtimes \cdots \boxtimes G\otimes_\ComplexNumbers \chi^i$, for a rigid slope-stable vector bundle $G$ on a $K3$ surface $S$, for all $n$. Again $c_1(F)=r_0^{n-1}c_1(G)+t\delta$ and the coefficient
of $\beta$ in the image
$B_{\delta/2}(r_0^n\alpha+c_1(F)+s(F)\beta)$ must be equal to $r_0^{n-1}s(G)=r_0^{n-2}\left(\frac{(c_1(G),c_1(G))+2}{2}\right)$, by (<ref>).
We get that

\[
\]

It remains to calculate $t$.
The quotient $\Gamma/{\mathfrak A}_n$ of $\Gamma$ by the alternating group is a double cover
of $S^{[n]}$ branched over the divisor of non-reduced subschemes

\[
\Gamma\LongRightArrowOf{q_1} \Gamma/{\mathfrak A}_n \LongRightArrowOf{q_2} S^{[n]}.
\]

We have the short exact sequence over $\Gamma/{\mathfrak A}_n$

\[
0\rightarrow q_2^*F\rightarrow \left(q_{1,*}(b^*(G\boxtimes \ \cdots\ \boxtimes G))\right)^{{\mathfrak A}_n}\rightarrow Q\rightarrow 0.
\]

The quotient sheaf $Q$ is again supported over the inverse image of the divisor of non-reduced subschemes in $S^{[n]}$.
Away from the locus of co-dimension $2$, where more then two points come together, $Q$ is locally free of rank
$r_0^{n-2}$ times the right hand side of (<ref>). We conclude that

\[
\left\{
\begin{array}{ccl}
r_0^{n-1}(r_0-1)/2 & \mbox{if} & i=0,
\\
r_0^{n-1}(r_0+1)/2 & \mbox{if} & i=1.
\end{array}
\right.
\]

Use Taelman's calculation of the derived monodromy group for $X$ of $K3^{[2]}$-type and the integral lattice $\Lambda\subset \tilde{H}(X,\RationalNumbers)$ of <cit.> in order to characterize the LLV vectors in $\Lambda$ admitting a modular vector bundle $F$, which is a lift of the Azumaya algebra deforming $\End(G[2]^{\pm})$ (in O'Grady's notation), for a spherical bundle $G$ on a $K3$ surface. Recover the constraints on $\rank(F)$ and $c_1(F)$ in <cit.>. Relate also those lifts, which correspond to the same LLV line. The rational translate of the LLV line  $\ell(F)$ to the line $\ell'(F)=exp(e_{-c_1(F)/r}\ell(F)$ in $U_\QQ$ should depend only on the rank $r$ of $F$ and whether $F$ is a deformation of $G[2]^+$ or $G[2]^-$ for some spherical stable vector bundle  $G$ on a $K3$ surface.

§ MODULAR VECTOR BUNDLES ISOMORPHIC TO $\PHI(\STRUCTURESHEAF{K3^{[N]}})$

We prove Theorem <ref> in this section.

Let $X_i$, $1\leq i \leq n$, be compact complex manifolds with trivial Picard groups and let $V_i$ be a $\theta_i$-twisted vector bundle over $X_i$ or rank $r_i$. Assume that $V_i$ does not have any non-zero subsheaf of rank less than $r_i$.
Then the vector bundle $V_1\boxtimes V_2\boxtimes \cdots \boxtimes V_n$ does not have any non-zero subsheaf of rank less than $\prod_{i=1}^n r_i$.

The proof of the proposition will use the following  lemma.

Let $X$ be a connected compact complex manifold with a trivial Picard group. Let $V$ be a possibly twisted rank $r$ vector bundle over $X$. Assume that $V$ does not have any non-zero subsheaf of rank less than $r$. Let $U$ be a vector space and $E\subset U\otimes_\CC V$ a saturated non-zero subsheaf of rank $\leq r$. Then there exists a line $\ell\subset U$, such that $E=\ell\otimes_\CC V$.

Note that the statement of the theorem is false if we remove the assumption that $\Pic(X)$ is trivial. Indeed,
take $V=\StructureSheaf{\PP(U)}$ and $E=\StructureSheaf{\PP(U)}(-1)\subset U\otimes_\CC V.$

The proof is by induction on $n:=\dim(U)$. The statement is clear if $n=1$. Assume that $n>1$ and the statement holds for $n-1$.
Choose a non-zero element $\xi\in U^*$. Let $h_\xi:E\rightarrow V$
be the restriction of $\xi\otimes id:U\otimes_\CC V\rightarrow V$ to $E$.
If $h_\xi$ vanishes, then $E$ is a saturated subsheaf of $\ker(\xi)\otimes_\CC V$ and we are done, by the induction hypothesis.
It remains to prove that $h_\xi$ vanishes, for some non-zero $\xi\in U^*$.

The proof is by contradiction.
Assume that $h_\xi$ does not vanish for every non-zero $\xi\in U^*$. Then the image of $h_\xi$ must have rank $r$. Hence, $\wedge^r(h_\xi)$ is a non-zero section of
$\det(E)^*\otimes\det(V)$. The latter is a trivial line-bundle, since $\Pic(X)$ is trivial. Hence, $h_\xi$ is an isomorphism, as $E$ is reflexive, being a saturated subsheaf of a locally free sheaf. The homomorphism $h_\xi$ maps the fiber $E_p$ surjectively onto $V_p$, for all $p\in X$. Hence, $E_p$ maps to an $r$-dimensional subspace of $U\otimes_\CC V_p$, for all $p\in X$.

Over $\PP(U^*)\times X$ we have the composite homomorphism $h:\pi_2^*E\rightarrow \pi_1^*\StructureSheaf{\PP(U^*)}(1)\otimes \pi_2^V$, given by

\[
\pi_2^*E
\rightarrow U\otimes_\CC \pi_2^*V
\RightArrowOf{\pi_1^*(\xi)\otimes id}
\pi_1^*\StructureSheaf{\PP(U^*)}(1)\otimes \pi_2^*V,
\]

where $\xi:U\otimes_\CC\StructureSheaf{\PP(U^*)}\rightarrow \StructureSheaf{\PP(U^*)}(1)$
is the quotient homomorphism. Then $\wedge^r(h)$ is a nowhere-vanishing section of
$\pi_1^*\StructureSheaf{\PP(U^*)}(r)\otimes \pi_2^*(\det(E^*)\otimes\det(V))$, by the above discussion.
Such a section does not exist, as the line bundle $\det(E^*)\otimes\det(V)$ is trivial and any non-zero section would vanish over a degree $r$ hypersurface in $\PP(U^*)$. A contradiction.

Let $X$ be a connected compact Kähler manifold with a trivial Picard group. Let $V$ be a possibly twisted rank $r$ vector bundle over $X$. Assume that $V$ does not have any non-zero subsheaf of rank less than $r$. Let $U$ be a vector space and $E\subset U\otimes_\CC V$ a saturated non-zero subsheaf. Then there exists a subspace $W\subset U$, such that $E=W\otimes_\CC V$.

The vector bundle $V$ is slope-stable with respect to every Kähler class on $X$.
The determinant of $\Hom(E,U\otimes_\CC V)$ is trivial, since $\Pic(X)$ is trivial. The statement is thus a special case of the
the more general fact that saturated subsheaves of polystable sheaves of the same slope are direct summands.

The proof is by induction on $e:=\rank(E)$.
The statement holds if $e\leq r$, by Lemma <ref>.
Assume that $e>r$ and the statement holds for all saturated subsheaves $E'$ with $\rank(E')<e$.
Choose $\xi\in U^*$, such that the restriction $h:E\rightarrow V$ of $\xi\otimes id:U\otimes_\CC V\rightarrow V$ to $E$ does not vanish. Then $h(E)$ is a rank $r$ subsheaf of $V$ and so $E_0:=\ker(h)$ is a saturated subsheaf of $U\otimes_\CC V$ of rank $e-r$. Hence, $E_0=W_0\otimes_\CC V$, for some subspace $W_0\subset U$, by the induction hypothesis.
The quotient $E_1:=E/E_0$ is a saturated subsheaf of $(U/W_0)\otimes_\CC V$  of rank $r$.
Thus, $E_1=\ell\otimes_\CC V$, for some line $\ell\subset U$, by Lemma <ref>. It follows that $E=W\otimes_\CC V$, where $W$ is the unique subspace of $U$ containing $W_0$ and projecting onto $\ell$.

(of Proposition <ref>). The proof is by induction on $n$. The case $n=1$ is clear. Assume that
$n>1$ and the statement holds for $n-1$. Let $E$ be a non-zero saturated subsheaf of $V:=V_1\boxtimes \cdots \boxtimes V_n$.
Denote the rank of $E$ by $e$.
There exists an open subset $U$ of $X_n$, such that the restriction of $E$ to $(\prod_{i=1}^{n-1}X_i)\times\{p\}$ maps to a non-zero subsheaf $E_p$ of rank $e$ of the restriction $V_p$ of $V$, for all $p\in U$. Then there exists a  subspace $W_p$ of the fiber $V_{n,p}$ of $V_n$ at $p$, such that the saturation of $E_p$ in $V_p$ is
$V_1\boxtimes \cdots \boxtimes V_{n-1}\otimes_\CC W_p$ and $e=\dim(W_p)\prod_{i=1}^{n-1}r_i$, by the induction hypothesis and Lemma <ref>. We conclude that there is a
rank $e/\prod_{i=1}^{n-1}r_i$ subsheaf $W$ of $V_p$, such that
$E$ is a subsheaf of $V_1\boxtimes \cdots \boxtimes V_{n-1}\boxtimes W$. But the saturation of $W$ must be equal to $V_n$, as $V_n$ does not have non-zero saturated subsheaves. Hence, $E$ must be equal to $V$.

(of Theorem <ref>).
Being $H$-slope-stable, $G$ is projectively $H$-hyperholomorphic [80] (see also [28]). Hence,
the triple $(S,H,G)$ determines a twistor family $\pi:\S\rightarrow \PP^1$ and a possibly twisted vector bundle $\G$ over $\S$, such that each fiber $S_t$ of $\pi$ over $t\in \PP^1$ is endowed with a Kähler class $\omega_t$ and the restriction $G_t$ of $\G$ to $S_t$ is
$\omega_t$-slope-stable. We may assume, possibly after replacing $H$ by another ample class with respect to which $G$ is slope-stable, that $\Pic(S_t)$ is trivial, for a generic $t\in\PP^1$. Then $G_t$ does not have any non-zero saturated proper subsheaves. Thus, the vector bundle $G_t\boxtimes \cdots \boxtimes G_t$ over $(S_t)^n$ does not have any non-zero saturated proper subsheaves, by Proposition <ref>.

The isospectral Hilbert scheme $\Gamma\subset S^n\times S^{[n]}$ is the reduced fiber product of $S^n$ and $S^{[n]}$ over the symmetric product $S^{(n)}$, by <cit.>. Hence, we have a relative isospectral subscheme over the base of the twistor deformation with fiber $\Gamma_t\subset S_t^n\times S_t^{[n]}$.
Let $b_t:\Gamma_t\rightarrow S^n$ and $q_t:\Gamma_t\rightarrow S_t^{[n]}$ be the restriction of the two projections.
The Brauer class $\alpha_t\in H^2_{an}(S_t,\StructureSheaf{S})$ of $G_t$ is the image of a class $\tilde{\alpha}_t\in H^2(S_t,\mu_r)$,
where $r$ is the rank of $G_t$, as pointed out in Equation (<ref>).
The Brauer class $\alpha_t^{\boxtimes n}$ of $G_t\boxtimes \cdots \boxtimes G_t$ is thus the image of an $\fS_n$-invariant class in
$H^2(S_t^n,\mu_r)$, which is the pullback of a class in $H^2(S_t^{(n)},\mu_r)$.
Hence, the Brauer class $b_t^*(\alpha_t^{\boxtimes n})$ of $b_t^*(G_t\boxtimes \cdots \boxtimes G_t)$
is the pullback via $q_t$ of a class in $H^2_{an}(S_t^{[n]},\StructureSheaf{S_t^{[n]}})$. The morphism $q_t$ is flat, by the proof of
The pushforward $q_{t,*}b_t^*(G_t\boxtimes \cdots \boxtimes G_t)$ is thus a well defined locally free twisted sheaf.

\[
G_t^{[n]}:=(q_{t,*}b_t^*(G_t\boxtimes \cdots \boxtimes G_t))^{\fS_n}.
\]

A non-zero proper saturated subsheaf $E$ of $G_t^{[n]}$ over the Douady space $S_t^{[n]}$ determines an $\fS_n$-invariant non-zero subsheaf $\tilde{E}$ of $G_t\boxtimes \cdots \boxtimes G_t$
of the same rank as $E$, since $q_t$ restricts to a Galois covering over the complement of the divisor in $S_t^{[n]}$ of non-reduced subschemes. We have seen that such a subsheaf $\tilde{E}$ does not exist. Hence,  the vector bundle $G_t^{[n]}$ does not have any non-zero saturated proper subsheaves. Thus, $G_t^{[n]}$ is slope-stable with respect to every Kähler class on $S_t^{[n]}$.

The paragraph before the statement of Theorem <ref> establishes that $\obs_{G^{[n]}}$ has rank $1$. Hence, the class $\kappa(G^{[n]})$ remains of Hodge type under every Kähler deformation of $S^{[n]}$, by Proposition <ref> (applied with $\Phi$ the identity endofunctor). The same thus holds for the class $\kappa(G_t^{[n]})$.
We conclude that $G_t^{[n]}$ deforms with $S_t^{[n]}$ to a pair $(X,F)$ of a vector bundle $F$ over every Kähler deformation $X$ of $S_t^{[n]}$, by <cit.>
(applied via an argument identical to the one used in the proof of Theorem

The vector space $\Ext^j(G_t,G_t)$ is one dimensional, for $j=0$ and $j=2$, and it vanishes otherwise. Hence,
the vector space $\Ext^{j}_{\fS_n}((G_t^{[n]},\rho_{\boxtimes}),(G_t^{[n]},\rho_{\boxtimes}))$ is one dimensional, for $j=2i$, $0\leq i\leq n$, and it vanishes otherwise. We conclude that $\Ext^j(G_t^{[n]},G_t^{[n]})$ is one dimensional, for $j=2i$, $0\leq i\leq n$, and it vanishes otherwise, since $BKR$ is an equivalence of categories. Now
$\Ext^j(G_t^{[n]},G_t^{[n]})$ is isomorphic to $H^j(S_t^{[n]},\SheafEnd(G_t^{[n]}))$ and $\SheafEnd(G_t^{[n]})$
is a poly-stable hyperholomorphic vector bundle.
The dimensions of sheaf cohomologies of poly-stable hyperholomorphic vector bundles are constant along twistor deformations, by
<cit.>. Hence, the deformations of $G_t^{[n]}$ remain $\PP^n$-objects in the sense that $\Ext^{i}(F,F)\cong H^i(\PP^n,\CC)$. In particular, $\Ext^1(F,F)$ vanishes and $F$ is infinitesimally rigid.

§ MODULAR VECTOR BUNDLES WITH ISOTROPIC MUKAI VECTOR OVER HILBERT SCHEMES

In Section <ref> we compute the Chern character of a vector bundle on an irreducible holomorphic symplectic manifold of $K3^{[n]}$-type, which is the Fourier-Mukai image of a sky-scraper sheaf.
The Chern character is determined by the rank and first Chern class of $F$.
Let $S$ be a $K3$ surface.
In Section <ref> we prove Theorem <ref> about the modularity of the BKR images on $S^{[n]}$ of equivariant vector bundles on $S^n$ constructed from $n$ vector bundles $\{G_i\}_{i=1}^n$ on $S$, where all $G_i$ belong to  the same two dimensional  component of the moduli space of stable sheaves on $S$.

§.§ The Chern character of the Fourier-Mukai image of a sky-scraper sheaf

§.§.§ An example of a Fourier-Mukai image of a sky-scraper sheaf

Let $\Phi: D^b(M)\rightarrow D^b(S)$ be an equivalence of derived categories of $K3$ surfaces. Assume that for every point $p\in M$, the image $\Phi(\StructureSheaf{p})$ of the sky-scraper sheaf of $p$ is isomorphic to a

vector bundle of Mukai vector $v$

on $S$. Let $\{p_i\}_{i=1}^n$ be a set of $n$ distinct points on $M$
and denote by $G_i$ the  vector bundle isomorphic to $\Phi(\StructureSheaf{p_i})$.
Let $Z\subset M$ be the length $n$ zero dimensional subscheme of $M$ with support $\{p_i\}_{i=1}^n$.
Denote by $z$ the corresponding point of $M^{[n]}$.
Note that the object $\StructureSheaf{p_{\sigma(1)}}\boxtimes \cdots \boxtimes \StructureSheaf{p_{\sigma(n)}}$ is isomorphic in $D^b(M^n)$ to
the sky scraper sheaf $\StructureSheaf{(p_{\sigma(1)}, \dots, p_{\sigma(n)})}$.
The “usual” BKR equivalence is given by the functor

\begin{equation}
\label{eq-usual-bkr}
bkr:=\Phi_{\StructureSheaf{\Gamma_M},\rho}:D^b(M^{[n]})\rightarrow D^b_{\fS_n}(M^n)
\end{equation}

using the same object as in (<ref>) as Fourier-Mukai kernel, but going in the reverse direction.
We see that the sky-scraper sheaf $\StructureSheaf{z}$ corresponds via $bkr$
to the $\fS_n$-equivariant object

\[
\oplus_{\sigma\in\fS_n}\left[\StructureSheaf{p_{\sigma(1)}}\boxtimes \cdots \boxtimes \StructureSheaf{p_{\sigma(n)}}\right]
\]

over $M^n$. The equivalence $\Phi^{\boxtimes n}:D^b_{\fS_n}(M^n)\rightarrow D^b_{\fS_n}(S^n)$
takes the above displayed object to

\begin{equation}
\label{eq-G-Z}
G_z:=\oplus_{\sigma\in\fS_n}\left[G_{\sigma(1)}\boxtimes \cdots \boxtimes G_{\sigma(n)}\right]
\end{equation}

with its natural $\fS_n$-linearization $\rho$.
Let $\tilde{\Phi}^{[n]}:D^b(M^{[n]})\rightarrow D^b(S^{[n]})$ be the composition

\[
\tilde{\Phi}^{[n]}:= BKR\circ \Phi^{\boxtimes n}\circ bkr,
\]

where $BKR$ is given in (<ref>).
Let $F_z$ be the sheaf over $S^{[n]}$ isomorphic to $BKR(G_z,\rho)$. Then
$F_z$ is isomorphic to $\tilde{\Phi}^{[n]}(\StructureSheaf{z})$ and hence
$\obs_{F_z}$ has rank $1$.

Write $v=(r_0,\lambda,(\lambda,\lambda)/2r_0)$, so that $r_0:=\rank(G_i)$ and $\lambda:=c_1(G_i)$, for $1\leq i\leq n$. Let $S^n\LeftArrowOf{b} \Gamma \RightArrowOf{q} S^{[n]}$ be the morphisms from the  isospectral Hilbert scheme given in (<ref>). Let $\theta:H^2(S,\Integers)\rightarrow H^2(S^{[n]},\Integers)$ be the homomorphism given in (<ref>) and let $\delta$ be half the class of the divisor of non-reduced subschemes.

$F_z$ is a locally free sheaf  of rank $n!r_0^n$, which is isomorphic to $q_*b^*(G_1\boxtimes \cdots \boxtimes G_n),$
and $c_1(F_z)=n!r_0^{n-1}\left(\theta(\lambda)-r_0\delta/2\right).$

Set $G_\sigma:=G_{\sigma(1)}\boxtimes \cdots \boxtimes G_{\sigma(n)}$. Denote the automorphisms of $S^n$ and of $\Gamma$ associated to a permutation $\sigma$ by $\sigma$ as well.
The equality $q=q\circ\sigma$ and the isomorphism $\rho_{\sigma^{-1}}:G_\sigma\rightarrow\sigma_*(G_{id})$ induced by  the linearization give rise to the natural isomorphism

\[
q_*b^*(\rho_{\sigma^{-1}}^{-1}): q_*b^*G_{id}\rightarrow q_*b^*G_\sigma.
\]

The invariant subsheaf $F_z$ of $\oplus_{\sigma\in\fS_n}q_*b^*G_\sigma$ is hence the diagonal embedding of $q_*b^*G_{id}$
into the direct sum.
$F_z$ is locally free, by the flatness of the morphism $q$ (see the proof of <cit.>).
The computation of $c_1(F_z)$ is carried out in Corollary <ref>.

Let $Y\subset S$ be a length $n$ subscheme supported on $n$ distinct points $\{y_1, \dots, y_n\}$. Let $y\in S^{[n]}$ be the point representing $Y$. Let $G_{i,y_{j}}$ be the fiber of $G_i$ over $y_j$.
The fiber $F_{z,y}$ of $F_z$ over $y$ is

\[
F_{z,y}= \left(\bigoplus_{\tau\in\fS_n}\bigoplus_{\sigma\in\fS_n}\left(G_{\sigma(1),y_{\tau(1)}}\otimes \ \cdots \ \otimes G_{\sigma(n),y_{\tau(n)}}\right)\right)^{\fS_n}.
\]

The  $\fS_n$-linearization of $G_z$ identifies the fiber