$\Phi_{\mathcal{U}}^{\boxtimes n}:D^{b}_{{\mathfrak{S}}_{n}}(M^{n})\rightarrow
D^{b}_{{\mathfrak{S}}_{n}}(S^{n})$

and its conjugate

(7.1) $\Phi_{\mathcal{U}}^{[n]}:=BKR\circ\Phi_{\mathcal{U}}^{\boxtimes n}\circ
BKR^{-1}:D^{b}(M^{[n]})\rightarrow D^{b}(S^{[n]}).$

Let ${\mathfrak{S}}_{n,\Delta}$ be the diagonal subgroup of
${\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}$. Let ${\mathcal{U}}^{\boxtimes
n}:={\mathcal{U}}\boxtimes\cdots\boxtimes{\mathcal{U}}$ be the exterior tensor
power of ${\mathcal{U}}$ over $M^{n}\times S^{n}$. Then
${\mathcal{U}}^{\boxtimes n}$ admits the permutation
${\mathfrak{S}}_{n,\Delta}$-linearization $\rho_{\boxtimes}$. Denote by $\chi$
the sign character of ${\mathfrak{S}}_{n,\Delta}$. We get the object
$({\mathcal{U}}^{\boxtimes n},\rho_{\boxtimes}\otimes\chi)$ in
$D^{b}_{{\mathfrak{S}}_{n,\Delta}}(M^{n}\times S^{n})$.

We use the same notation for the morphisms $M^{n}\stackrel{{\scriptstyle
b}}{{\leftarrow}}\Gamma_{M}\stackrel{{\scriptstyle q}}{{\rightarrow}}M^{[n]}$
from the isospectral Hilbert scheme of $M$. We get the product morphisms

$M^{n}\times S^{n}\stackrel{{\scriptstyle b\times
b}}{{\longleftarrow}}\Gamma_{M}\times\Gamma_{S}\stackrel{{\scriptstyle q\times
q}}{{\longrightarrow}}M^{[n]}\times S^{[n]}.$

######  Lemma 7.1.

The equivalence $\Phi_{\mathcal{U}}^{[n]}$ has the rank $n!r^{n}$ locally free
Fourier-Mukai kernel

(7.2) $E:=\pi_{M^{[n]}}^{*}({\mathcal{O}}_{M^{[n]}}(\delta))\otimes(q\times
q)_{*}^{{\mathfrak{S}}_{n,\Delta}}((b\times b)^{*}({\mathcal{U}}^{\boxtimes
n},\rho_{\boxtimes}\otimes\chi)).$

###### Proof.

The statement is proved in [Ma6, Sec. 11.1.2, Eq. (11.6)] ∎

###  7.2. The functor
$\widetilde{\Theta}_{n}:{\mathcal{G}}^{[1]}\rightarrow{\mathcal{G}}^{[n]}$

In the remainder of Section 7 we will keep the assumption that the rank of the
object ${\mathcal{U}}\in D^{b}(S\times M)$ is non-zero, but drop the
assumption that ${\mathcal{U}}$ is represented by a coherent sheaf. Let
${\mathcal{G}}^{[n]}$ be the subgroupoid of the groupoid ${\mathcal{G}}$ given
in (1.1) whose objects are of $K3^{[n]}$-type. We define next the functor
$\widetilde{\Theta}_{n}:{\mathcal{G}}^{[1]}\rightarrow{\mathcal{G}}^{[n]}$.
Define $\widetilde{\Theta}_{n}(X,\epsilon)=(X^{[n]},\epsilon^{[n]})$, where
the orientation $\epsilon^{[n]}$ of $H^{2}(X^{[n]},{\mathbb{Q}})$ is such that
the class $\delta$ extends a basis compatible with the orientation $\epsilon$
to one compatible with $\epsilon^{[n]}$.

(1) $\widetilde{\Theta}_{n}$ maps a parallel-transport operator, associated to
two fibers over two points $b_{0}$ and $b_{1}$ in the base of a family
${\mathcal{X}}\rightarrow B$ of $K3$ surfaces and a path $\gamma$ from $b_{0}$
to $b_{1}$, to the parallel-transport operator associated to the same path and
the relative Hilbert scheme ${\mathcal{X}}^{[n]}\rightarrow B$. Denote by

(7.3) $\theta:H^{2}(S,{\mathbb{Z}})\rightarrow H^{2}(S^{[n]},{\mathbb{Z}})$

the composition of the isomorphism $H^{2}(S,{\mathbb{Z}})\cong
H^{2}(S^{(n)},{\mathbb{Z}})$ with the pullback
$H^{2}(S^{(n)},{\mathbb{Z}})\rightarrow H^{2}(S^{[n]},{\mathbb{Z}})$ via the
Hilbert-Chow morphism.

(2) $\widetilde{\Theta}_{n}$ sends cup product with $\exp(\lambda)$,
$\lambda\in H^{2}(X,{\mathbb{Q}})$, to cup product with
$\exp(\theta(\lambda))$.

(3) $\widetilde{\Theta}_{n}$ sends the isometry
$\phi_{\mathcal{P}}:H^{*}(X,{\mathbb{Q}})\rightarrow H^{*}(Y,{\mathbb{Q}})$
induced by the correspondence $\sqrt{td_{X\times Y}}ch({\mathcal{P}})$, where
${\mathcal{P}}\in D^{b}(X\times Y)$ is the Fourier-Mukai kernel of an
equivalence $\Phi_{\mathcal{P}}:D^{b}(X)\rightarrow D^{b}(Y)$, to the isometry
induced by the correspondence $\sqrt{td_{X^{[n]}\times
Y^{[n]}}}ch({\mathcal{P}}^{[n]})$, where ${\mathcal{P}}^{[n]}$ is the Fourier-
Mukai kernel in (7.2) of the BKR-conjugate of $\Phi_{\mathcal{P}}^{\boxtimes
n}$ as in (7.1).

The image of morphisms under the functor $\widetilde{\Theta}_{n}$ was defined
above separately for the three types of morphisms. Let
$[{\mathcal{O}}_{\Gamma_{S}}]$ be the class of ${\mathcal{O}}_{\Gamma_{S}}$,
with its natural linearization, in the topological
${\mathfrak{S}}_{n}$-equivariant K-ring $K_{{\mathfrak{S}}_{n}}(S^{n}\times
S^{[n]})$. We get the correspondence
$[{\mathcal{O}}_{\Gamma_{S}}]_{*}:K_{{\mathfrak{S}}_{n}}(S^{n})\rightarrow
K(S^{[n]})$ on the level of topological K-rings. The inverse is induced by a
class in $K_{{\mathfrak{S}}_{n}}(S^{n}\times S^{[n]})$ as well. The value of
$\widetilde{\Theta}_{n}$ on a morphism in ${\mathcal{G}}^{[1]}$ of type (3)
associated to an equivalence $\Phi_{\mathcal{U}}$ with Fourier-Mukai kernel
${\mathcal{U}}$ involves first composing the correspondence
$[{\mathcal{U}}^{\boxtimes
n},\rho_{\boxtimes}]_{*}:K_{{\mathfrak{S}}_{n}}(M^{n})\rightarrow
K_{{\mathfrak{S}}_{n}}(S^{n})$ with $[{\mathcal{O}}_{\Gamma_{S}}]_{*}$ and the
inverse of $[{\mathcal{O}}_{\Gamma_{M}}]_{*}$, to obtain the correspondence
from $K(M^{[n]})$ to $K(S^{[n]})$, and then using the Chern character
isomorphism and its inverse to obtain the cohomological morphism in ${\rm
Hom}_{{\mathcal{G}}^{[n]}}(M^{[n]},S^{[n]})$ (see [BKR, Sec. 10.1]). The value
of $\widetilde{\Theta}_{n}$ on morphisms of types (1) and (2) may be obtained
by first conjugating via the Chern character their cartesian powers to lift
these morphisms to correspondences from $K_{{\mathfrak{S}}_{n}}(M^{n})$ to
$K_{{\mathfrak{S}}_{n}}(S^{n})$. The check that $\widetilde{\Theta}_{n}$ maps
compositions of morphisms to compositions reduces to showing that its value on
(1) morphisms that are parallel-transport operators or (2) morphisms that are
multiplication by $\exp(\lambda)$, $\lambda\in H^{2}(X,{\mathbb{Q}})$, is
induced by composing the correspondences from $K_{{\mathfrak{S}}_{n}}(M^{n})$
to $K_{{\mathfrak{S}}_{n}}(S^{n})$ with the same correspondences,
$[{\mathcal{O}}_{\Gamma_{S}}]_{*}$ and the inverse of
$[{\mathcal{O}}_{\Gamma_{M}}]_{*}$, used for morphisms of type (3). For
parallel-transport operators this is the case, since the class
$[{\mathcal{O}}_{\Gamma_{S}}]$ in $K_{{\mathfrak{S}}_{n}}(S^{n}\times
S^{[n]})$ is a global class in the corresponding local system of K-rings over
the base of every family of $K3$ surfaces, hence the correspondences
$[{\mathcal{O}}_{\Gamma_{S}}]$ and $[{\mathcal{O}}_{\Gamma_{M}}]$ intertwine
parallel-transport operators associated to paths in the base of such families.
The check for multiplication by $\exp(\lambda)$, $\lambda\in
H^{2}(X,{\mathbb{Q}})$, reduces to checking for integral classes, as the
correspondences are all linear. In this case the action corresponds to
tensorization by topological line bundles on the level of K-rings. The
correspondence $[{\mathcal{O}}_{\Gamma_{S}}]_{*}$ intertwines tensorization in
$K_{{\mathfrak{S}}_{n}}(S^{n})$ by the cartesian power $L^{\boxtimes n}$ of a
line bundle on $S$ and by the line bundle $\tilde{L}$ on $S^{[n]}$ with
$c_{1}(\tilde{L})=\theta(c_{1}(L))$, since $\theta(c_{1}(L))$ is defined as
the pullback of the class associated to $c_{1}(L)$ in the symmetric product
$S^{(n)}$, and the isospectral Hilbert scheme is a reduced fiber product of
$S^{n}$ and $S^{[n]}$ over $S^{(n)}$. Hence, the projection formula yields
that $[{\mathcal{O}}_{\Gamma_{S}}]_{*}$ indeed intertwines the two
tensorizations.

###  7.3. The functor
$\Theta_{n}:{\mathcal{G}}^{[1]}\rightarrow{\mathcal{G}}^{[n]}$

Let
$\phi:\widetilde{H}(M,{\mathbb{Z}})\rightarrow\widetilde{H}(S,{\mathbb{Z}})$
be the isometry induced by the correspondence
$ch({\mathcal{U}})\sqrt{td_{M\times S}}$. The morphism
$\widetilde{\Theta}_{n}(\phi):H^{*}(M^{[n]},{\mathbb{Q}})\rightarrow
H^{*}(S^{[n]},{\mathbb{Q}})$ is induced by the correspondence

$ch(E)\sqrt{td_{M^{[n]}\times S^{[n]}}},$

where $E$ is given in terms of ${\mathcal{U}}$ in Equation (7.2). We recall
next the relationship between $\phi$ and
$\widetilde{H}(\widetilde{\Theta}_{n}(\phi)):\widetilde{H}(M^{[n]},{\mathbb{Q}})\rightarrow\widetilde{H}(S^{[n]},{\mathbb{Q}})$.
Let

(7.4)
$\tilde{\theta}:\widetilde{H}(S,{\mathbb{Q}})\rightarrow\widetilde{H}(S^{[n]},{\mathbb{Q}})$

be the extension of $\theta:H^{2}(S,{\mathbb{Z}})\rightarrow
H^{2}(S^{[n]},{\mathbb{Z}})$, given in (7.3), mapping $\alpha$ to $\alpha$ and
$\beta$ to $\beta$. Let

(7.5) $\tilde{\iota}:O(\widetilde{H}(S,{\mathbb{Q}}))\rightarrow
O(\widetilde{H}(S^{[n]},{\mathbb{Q}}))$

be the embedding given by
$\tilde{\iota}_{g}(\tilde{\theta}(x))=\tilde{\theta}(g(x))$ and
$\tilde{\iota}_{g}$ acts as the identity on the $1$-dimensional subspace
orthogonal to the image of $\tilde{\theta}$. Given an isometry
$g:\widetilde{H}(M,{\mathbb{Q}})\rightarrow\widetilde{H}(S,{\mathbb{Q}})$ we
denote by
$\tilde{\iota}_{g}:\widetilde{H}(M^{[n]},{\mathbb{Q}})\rightarrow\widetilde{H}(S^{[n]},{\mathbb{Q}})$
the isometry satisfying
$\tilde{\iota}_{g}(\tilde{\theta}(x))=\tilde{\theta}(g(x))$ and mapping
$\delta\in H^{2}(M^{[n]},{\mathbb{Q}})$ to $\delta\in
H^{2}(S^{[n]},{\mathbb{Q}})$.

Choose markings $\eta_{S}:H^{2}(S,{\mathbb{Z}})\rightarrow\Lambda_{K3}$ and
$\eta_{M}:H^{2}(M,{\mathbb{Z}})\rightarrow\Lambda_{K3}$, where $\Lambda_{K3}$
is the $K3$ lattice, and let $\Lambda$ be the orthogonal direct sum

(7.6) $\Lambda_{K3}\oplus{\mathbb{Z}}\delta_{0},$

where $(\delta_{0},\delta_{0})=2-2n$, $n\geq 2$. Let

(7.7) $\iota:O(\Lambda_{K3})\rightarrow O(\Lambda)$

be the embedding, where $\iota_{g}$ is the extension of $g$ satisfying
$\iota_{g}(\delta_{0})=\delta_{0}$. Choose an orientation of $\Lambda_{K3}$
and an orientation of $\Lambda$, so that $\delta_{0}$ extends a basis
compatible with the orientation of the former to one compatible with that of
the latter. Set $\det(\phi):=\det(\eta_{S}\circ\phi\circ\eta_{M}^{-1})$.
Extend the marking $\eta_{S}$ to a marking

(7.8) $\eta_{S^{[n]}}:H^{2}(S^{[n]},{\mathbb{Z}})\rightarrow\Lambda$

sending $\delta$ to $\delta_{0}$, where $\delta$ is half the class of the
divisor of non-reduced subschemes. Define $\eta_{M^{[n]}}$ similarly. Define
$\widetilde{H}(\widetilde{\Theta}_{n}(\phi))$ using Convention 5.4.

######  Lemma 7.2.

$\widetilde{H}(\widetilde{\Theta}_{n}(\phi))=\det(\phi)^{n+1}(B_{-\delta/2}\circ\tilde{\iota}_{\phi}\circ
B_{\delta/2}).$

###### Proof.

Let $\zeta:H^{*}(M,{\mathbb{Q}})\rightarrow H^{*}(S,{\mathbb{Q}})$ be a
parallel-transport operator such that
$\det(\eta_{S}\circ\bar{\zeta}\circ\eta_{M}^{-1})=1,$ where $\bar{\zeta}$ is
the restriction of $\zeta$ to $H^{2}(M,{\mathbb{Q}})$. Then $\zeta$ lifts to a
parallel-transport operator
$\zeta^{[n]}:H^{*}(M^{[n]},{\mathbb{Q}})\rightarrow
H^{*}(S^{[n]},{\mathbb{Q}})$, since associated to a family of $K3$ surfaces is
a canonical family of relative Douady spaces. Let $\bar{\zeta}^{[n]}$ be the
restriction of $\zeta^{[n]}$ to $H^{2}(M^{[n]},{\mathbb{Q}})$. Then
$\det(\eta_{S^{[n]}}\circ\bar{\zeta}^{[n]}\circ\eta_{M^{[n]}}^{-1})=1$. We get
$\widetilde{H}(\zeta^{[n]})=\tilde{\iota}_{\zeta}=B_{-\delta/2}\circ\tilde{\iota}_{\zeta}\circ
B_{\delta/2}$, where the first equality is by Remark 5.6. The composition
$\phi_{1}:=\zeta\circ\phi$ is an element of ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S)$. We conclude that the equality in the statement
of the lemma is equivalent to the equality

$\widetilde{H}(\widetilde{\Theta}_{n}(\phi_{1}))=\det(\phi_{1})^{n+1}(B_{-\delta/2}\circ\tilde{\iota}_{\phi_{1}}\circ
B_{\delta/2}).$

The latter equality holds for all element in ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S)$ by [Ma6, Theorem 12.2] (see also [Bec, Theorem
7.4] and in case $n=2$ Taelman’s earlier result [Ta, Theorem 9.4]). ∎

Let
$\varphi:\widetilde{H}(M,{\mathbb{Q}})\rightarrow\widetilde{H}(S,{\mathbb{Q}})$
be the isometry induced by $\kappa({\mathcal{U}})\sqrt{td_{M\times S}}$. Let

$\varphi^{[n]}:H^{*}(M^{[n]},{\mathbb{Q}})\rightarrow
H^{*}(S^{[n]},{\mathbb{Q}})$

be the isometry induced by the correspondence
$\kappa(E)\sqrt{td_{M^{[n]}\times S^{[n]}}}.$ Again, $\varphi^{[n]}$ is a
morphism in ${\rm Hom}_{{\mathcal{G}}^{[n]}}(M^{[n]},S^{[n]})$. As above, we
denote by $\widetilde{H}(\varphi)_{0}$ the restriction of $\varphi$ to
$H^{2}(M,{\mathbb{Q}})$ and set
$\psi_{\mathcal{U}}:=\widetilde{H}_{0}(\varphi)$, where
$\widetilde{H}_{0}(\varphi):=\nu(\widetilde{H}(\varphi)_{0})\widetilde{H}(\varphi)_{0}$
as in (5.5). We define $\widetilde{H}(\varphi^{[n]})_{0}$ as the restriction
of $\widetilde{H}(\varphi^{[n]})$ to $H^{2}(M^{[n]},{\mathbb{Q}})$ and set
$\psi_{E}:=\widetilde{H}_{0}(\varphi^{[n]})$, where
$\widetilde{H}_{0}(\varphi^{[n]}):=\nu(\widetilde{H}(\varphi^{[n]})_{0})\widetilde{H}(\varphi^{[n]})_{0}$.

######  Corollary 7.3.

(7.9)
$\widetilde{H}(\varphi^{[n]})=\det(\varphi)^{n+1}\tilde{\iota}_{\varphi}.$

Consequently, $\psi_{E}\circ\theta=\theta\circ\psi_{\mathcal{U}}$ and
$\psi_{E}(\delta)=\delta$.

###### Proof.

Let $a_{1}\in H^{2}(M,{\mathbb{Z}})$ and $a_{2}\in H^{2}(S,{\mathbb{Z}})$ be
the classes satisfying
$c_{1}({\mathcal{U}})=\pi_{M}^{*}(a_{1})+\pi_{S}^{*}(a_{2})$. Let $e_{1}\in
H^{2}(M^{[n]},{\mathbb{Z}})$ and $e_{2}\in H^{2}(S^{[n]},{\mathbb{Z}})$ be the
classes satisfying
$c_{1}(E)=\pi_{M^{[n]}}^{*}(e_{1})+\pi_{S^{[n]}}^{*}(e_{2})$. Set
$R:=n!r^{n}$. We have

$\displaystyle\widetilde{H}(\varphi^{[n]})$ $\displaystyle=$ $\displaystyle
B_{-e_{2}/R}\circ\widetilde{H}(\widetilde{\Theta}_{n}(\phi))\circ
B_{-e_{1}/R}$ $\displaystyle=$
$\displaystyle\det(\phi)^{n+1}B_{-(\delta/2+e_{2}/R)}\circ\iota_{\phi}\circ
B_{\delta/2-e_{1}/R}$ $\displaystyle=$
$\displaystyle\det(\phi)^{n+1}B_{(\theta(a_{2})/r-\delta/2-e_{2}/R)}\circ\iota_{\varphi}\circ
B_{(\theta(a_{1})/r+\delta/2-e_{1}/R)}$ $\displaystyle=$
$\displaystyle\det(\phi)^{n+1}B_{\lambda_{2}}\circ\iota_{\varphi}\circ
B_{\lambda_{1}},$

where $\lambda_{1}=\theta(a_{1})/r+\delta/2-e_{1}/R$ and
$\lambda_{2}=\theta(a_{2})/r-\delta/2-e_{2}/R$. The first equality follows
from the definition of $\kappa(E)$, the second from Lemma 7.2, the third from
the definition of $\kappa({\mathcal{U}})$, and the fourth is obvious. Note
that $\det(\phi)=\det(\varphi)$. In order to establish Equation (7.9) it
remains to prove that $\lambda_{i}=0$, $i=1,2$, or equivalently,

(7.10) $\displaystyle e_{1}$ $\displaystyle=$ $\displaystyle
n!r^{n}\left(\frac{\theta(a_{1})}{r}+\frac{\delta}{2}\right),$ (7.11)
$\displaystyle e_{2}$ $\displaystyle=$ $\displaystyle
n!r^{n}\left(\frac{\theta(a_{2})}{r}-\frac{\delta}{2}\right).$

Both $\varphi$ and $\varphi^{[n]}$ are degree reversing, by Lemma 4.1. Hence,
both $\widetilde{H}(\varphi^{[n]})$ and $\iota_{\varphi}$ are degree
reversing, the first by Lemma 3.4. If $\lambda_{1}\neq 0$, then
$(\lambda_{1},\lambda^{\prime}_{1})\neq 0$, for some $\lambda_{1}^{\prime}\in
H^{2}(M^{[n]},{\mathbb{Z}})$ and the coefficient of $\beta$ in
$B_{\lambda_{1}}(\lambda^{\prime}_{1})$ is non-zero, which implies that the
coefficient of $\alpha$ in
$\iota_{\varphi}(B_{\lambda_{1}}(\lambda^{\prime}_{1}))$ is non-zero, and
hence so is the coefficient of $\alpha$ in
$\widetilde{H}(\varphi^{[n]})(\lambda_{1}^{\prime})$. This contradicts the
fact that $\widetilde{H}(\varphi^{[n]})$ maps $H^{2}(M^{[n]},{\mathbb{Q}})$ to
$H^{2}(S^{[n]},{\mathbb{Q}})$. Hence, $\lambda_{1}=0$.

If $\lambda_{2}\neq 0$, then $(\lambda_{2},\lambda^{\prime}_{2})\neq 0$, for
some $\lambda_{2}^{\prime}\in H^{2}(S^{[n]},{\mathbb{Z}})$ and the coefficient
of $\beta$ in $B_{\lambda_{2}}(\lambda^{\prime}_{2})$ is non-zero. There
exists a class $\lambda_{1}^{\prime}\in H^{2}(M^{[n]},{\mathbb{Q}})$, such
that $\iota_{\varphi}(\lambda_{1}^{\prime})=\lambda_{2}^{\prime}$, as
$\iota_{\varphi}$ is an isometry. Hence, the coefficient of $\beta$ in
$\widetilde{H}(\varphi^{[n]})(\lambda_{1}^{\prime})$ does not vanish. A
contradiction. We conclude that $\lambda_{2}=0$.

Equality (7.9) implies that
$\nu(\widetilde{H}(\varphi^{[n]})_{0})=\det(\varphi)^{n+1}\nu(\widetilde{H}(\varphi)_{0})$.
Hence $\psi_{E}$ is the extension of $\psi_{\mathcal{U}}$ mapping $\delta\in
H^{2}(M^{[n]},{\mathbb{Q}})$ to $\delta\in H^{2}(S^{[n]},{\mathbb{Q}})$. ∎

Let ${\mathcal{G}}^{[n]}$ be the subgroupoid of the groupoid ${\mathcal{G}}$
given in (1.1) whose objects are of $K3^{[n]}$-type. Let

(7.12) $\Theta_{n}:{\mathcal{G}}^{[1]}\rightarrow{\mathcal{G}}^{[n]}$

be the following normalization of the functor $\widetilde{\Theta}_{n}$ defined
in Section 7.2. The functor $\Theta_{n}$ sends an object of
${\mathcal{G}}^{[1]}$ to the same object that $\widetilde{\Theta}_{n}$ does.
Given a morphism $\phi\in{\rm
Hom}_{{\mathcal{G}}^{[1]}}((X,\epsilon),(Y,\epsilon^{\prime}))$, we define

$\Theta_{n}(\phi)=B_{\delta/2}\circ\widetilde{\Theta}_{n}(\phi)\circ
B_{-\delta/2},$

where we denote by $\delta$ half the class of the divisor of non-reduced
subschemes in both $X^{[n]}$ and $Y^{[n]}$ and $B_{\lambda}$ is cup product
with $\exp(\lambda)$. Note that if $\phi$ is $B_{\lambda}\in{\rm
Aut}(X,\epsilon)$, $\lambda\in H^{2}(X,{\mathbb{Q}})$, or if $\phi$ is a
parallel-transport operator, then
$\Theta_{n}(\phi)=\widetilde{\Theta}_{n}(\phi)$.

######  Corollary 7.4.

The functor $\Theta_{n}$ maps ${\mathcal{G}}^{[1]}_{an}$ to
${\mathcal{G}}^{[n]}_{an}$. Furthermore, $\Theta_{n}(\varphi)=\varphi^{[n]},$
where $\varphi$ and $\varphi^{[n]}$ are the morphisms in Corollary 7.3.

###### Proof.

The statement is clear for parallel-transport operators. Hence, it sufices to
prove the equality $\Theta_{n}(\varphi)=\varphi^{[n]}$. Keep the notation of
the proof of Corollary 7.3. The second equality below follows from Equations
(7.10) and (7.11).

$\displaystyle\varphi^{[n]}$ $\displaystyle=$ $\displaystyle
B_{-(e_{2}/R)}\circ\widetilde{\Theta}_{n}(\phi)\circ
B_{-(e_{1}/R)}=B_{-(\theta(a_{2}/r))}\circ
B_{\delta/2}\circ\widetilde{\Theta}_{n}(\phi)\circ B_{-\delta/2}\circ
B_{-(\theta(a_{1})/r)}$ $\displaystyle=$
$\displaystyle\Theta_{n}(B_{-a_{2}/r})\circ\Theta_{n}(\phi)\circ\Theta_{n}(B_{-a_{1}/r})=\Theta_{n}(B_{-a_{2}/r}\circ\phi\circ
B_{-a_{1}/r})=\Theta_{n}(\varphi).$

The equality $\Theta_{n}(\varphi)=\varphi^{[n]}$ follows. ∎

##  8\. Deforming the universal vector bundle over $M^{[n]}\times S^{[n]}$

We consider in this section a sequence of examples of $2$-dimensional moduli
spaces $M$ of stable vector bundles over a $K3$ surface $S$, such that the
vector bundle $E$ over $M^{[n]}\times S^{[n]}$, given in (7.2) as the Fourier-
Mukai kernel of an equivalence of derived categories, can be deformed with
$M^{[n]}\times S^{[n]}$ over the whole connected component
${\mathfrak{M}}_{\psi}^{0}$ of a moduli space of rational Hodge isometries
introduced in Section 5.2.

Keep the notation of Section 7. Assume that ${\rm Pic}(S)$ is cyclic,
generated by the class $h:=c_{1}(H)$ of an ample line bundle $H$ with
$(h,h)=2rs$, where $r$ and $s$ are positive integers and $\gcd(r,s)=1$. The
Mukai vector $v:=(r,h,s)$ is then isotropic. The following properties of the
moduli space $M$ of $H$-stable sheaves with Mukai vector $v$ are proved in
[Mu3]. $M$ is a $K3$ surface with a cyclic Picard group generated by an ample
line bundle $\hat{H}$ with class $\hat{h}:=c_{1}(\hat{H})$ satisfying
$(\hat{h},\hat{h})=(h,h)$. $M$ parametrizes locally free $H$-slope-stable
sheaves. For each integer $k$, such that $ks\equiv 1$ (mod $r$) there exists
over $M\times S$ a universal vector bundle ${\mathcal{U}}$ with
$c_{1}({\mathcal{U}})=\pi_{M}^{*}(k\hat{h})+\pi_{S}^{*}(h)$. The universal
vector bundle ${\mathcal{U}}$ restricts to an $\hat{H}$-slope-stable vector
bundle on $M\times\\{x\\}$ of Mukai vector $\hat{v}:=(r,k\hat{h},k^{2}s)$, for
each point $x\in S$, and the classification morphism induces an isomorphism
between $S$ and the moduli space $M_{\hat{H}}(\hat{v})$ of $\hat{H}$-stable
sheaves on $M$, by [Mu3, Theorem 1.2].

######  Proposition 8.1.

([Bu, Prop. 4.19]131313Buskin states it for $\kappa({\mathcal{U}}^{*})$, but
the latter is equal to $\kappa({\mathcal{U}})$, by Corollary 4.2. ) The class
$-\kappa({\mathcal{U}})\sqrt{td_{M\times S}}$ induces a degree reversing
rational Hodge isometry
$\phi_{\mathcal{U}}:\widetilde{H}(M,{\mathbb{Q}})\rightarrow\widetilde{H}(S,{\mathbb{Q}})$,
which restricts to a rational Hodge isometry
$\psi_{\mathcal{U}}:H^{2}(M,{\mathbb{Q}})\rightarrow H^{2}(S,{\mathbb{Q}})$
satisfying $\psi_{\mathcal{U}}(\hat{h})=h$, such that
$\eta_{S}\circ\psi_{\mathcal{U}}\circ\eta_{M}^{-1}$ belongs to the double
orbit $O(\Lambda_{K3})\rho_{u}O(\Lambda_{K3})$ of a reflection $\rho_{u}$ in
the co-rank $1$ lattice $u^{\perp}\subset\Lambda_{K3}$ of a primitive class
$u$ with $(u,u)=2r$.

We claim that ${\mathcal{U}}$ is slope-stable with respect to every Kähler
class on $M\times S$. It suffice to show it for every ample class, i.e., for
$\omega:=a\pi_{S}^{*}(h)+b\pi_{M}^{*}(\hat{h})$, for all $a>0$ and $b>0$. Let
${\mathcal{W}}$ be a saturated non-zero proper subsheaf of ${\mathcal{U}}$.
Then ${\mathcal{W}}$ is reflexive, hence it is locally free away from a closed
subvariety of dimension at most $1$. It follows that the restriction of
${\mathcal{W}}$ to the generic fiber of $\pi_{S}$ and of $\pi_{M}$ is locally
free. Let $W_{x}$ (resp. $W_{m}$) be such a restriction to the fiber over
$x\in S$ (resp. $m\in M$). Define $U_{x}$ and $U_{m}$ similarly in terms of
${\mathcal{U}}$. Then
$c_{1}({\mathcal{W}})=\pi_{M}^{*}c_{1}(W_{x})+\pi_{S}^{*}c_{1}(W_{m})$ and
$\omega^{3}=a^{2}bh^{2}\hat{h}+ab^{2}h\hat{h}^{2}$. Set $\rho:={\rm
rank}({\mathcal{W}})$. We have

$\displaystyle\mu_{\omega}({\mathcal{W}})$ $\displaystyle=$
$\displaystyle\frac{1}{\rho}c_{1}({\mathcal{W}})\omega^{3}=a^{2}b(h,h)\mu_{\hat{h}}(W_{m})+ab^{2}(\hat{h},\hat{h})\mu_{h}(W_{x})$
$\displaystyle<$ $\displaystyle
a^{2}b(h,h)\mu_{\hat{h}}(U_{m})+ab^{2}(\hat{h},\hat{h})\mu_{h}(U_{x})=\mu_{\omega}({\mathcal{U}}).$

Hence, ${\mathcal{U}}$ is slope-stable with respect to every Kähler class on
$M\times S$. The universal vector bundle ${\mathcal{U}}$ thus satisfies the
assumptions of Proposition 5.21 (stated for $E$). Set
$\psi:=\eta_{S}\circ\psi_{\mathcal{U}}\circ\eta_{M}^{-1}$. Let
${\mathfrak{M}}_{\psi}^{0}$ be the connected component of
${\mathfrak{M}}_{\psi}$ through $q_{0}:=(M,\eta_{M},S,\eta_{S})$.

Choose a generic twistor line $Q$ in ${\mathfrak{M}}_{\psi}^{0}$ through
$q_{0}$ and let $(S^{\prime},\eta_{S^{\prime}},M^{\prime},\eta_{M^{\prime}})$
be a quadruple in this line with ${\rm Pic}(S^{\prime})=0$. There exists
classes $\zeta_{S^{\prime}}\in
H^{2}(S^{\prime},{\mathcal{O}}_{S^{\prime}}^{*})$ and $\zeta_{M^{\prime}}\in
H^{2}(M^{\prime},{\mathcal{O}}_{M^{\prime}}^{*})$ and a $\zeta$-twisted
locally free sheaf ${\mathcal{U}}^{\prime}$ over $M^{\prime}\times
S^{\prime}$, with
$\zeta:=\pi_{S^{\prime}}^{*}(\zeta_{S^{\prime}})\pi_{M^{\prime}}^{*}(\zeta_{M^{\prime}})$,
such that $(M^{\prime}\times
S^{\prime},{\mathcal{E}}nd({\mathcal{U}}^{\prime}))$ is deformation equivalent
to $(M\times S,{\mathcal{E}}nd({\mathcal{U}}))$, by Proposition 5.21.
Moreover, ${\mathcal{U}}^{\prime}$ is $\omega$-slope-stable with respect to
some Kähler class on $S^{\prime}\times M^{\prime}$, again by Proposition 5.21.
We note that the existence of such ${\mathcal{U}}^{\prime}$ follows already
from [Bu, Theorem 5.1] in this case where $n=1$. It follows that
${\mathcal{U}}^{\prime}$ does not have any non-zero proper saturated subsheaf,
since ${\rm Pic}(M^{\prime}\times S^{\prime})$ is trivial.

The construction of the sheaf $E$ over $M^{[n]}\times S^{[n]}$, given in
(7.2), has a natural relative extension over the chosen twistor line $Q$ in
${\mathfrak{M}}^{0}_{\psi}$ through $q_{0}$ resulting in the sheaf
$E^{\prime}$ over $M^{\prime[n]}\times S^{\prime[n]}$ given by substituting
${\mathcal{U}}^{\prime}$ for ${\mathcal{U}}$ in equation (7.2). The check that
equation (7.2) is still meaningful in the twisted setting reduces to checking
that $(q\times q)_{*}^{{\mathfrak{S}}_{n,\Delta}}$ is well defined. This is
the case, since ${\mathcal{U}}^{\prime\boxtimes n}$ is $\zeta^{\boxtimes
n}$-twisted, and $(b\times b)^{*}(\zeta^{\boxtimes n})$ is the pullback via
$q\times q$ of a Brauer class on $M^{\prime[n]}\times S^{\prime[n]}$ (which is
itself the pullback of a Brauer class on $M^{\prime(n)}\times S^{\prime(n)}$),
as explained in [Ma6, Sec. 10, Proof of Theorem 1.4].

Let $\psi_{{\mathcal{U}}^{\prime}}:H^{2}(M^{\prime},{\mathbb{Q}})\rightarrow
H^{2}(S^{\prime},{\mathbb{Q}})$ and
$\psi_{E^{\prime}}:H^{2}(M^{\prime[n]},{\mathbb{Q}})\rightarrow
H^{2}(S^{\prime[n]},{\mathbb{Q}})$ be the parallel-transports of
$\psi_{\mathcal{U}}$ and $\psi_{\mathcal{E}}$ in the relevant local systems
over the chosen twistor line $Q$.

######  Lemma 8.2.

The isometry $\psi_{E^{\prime}}$ maps some Kähler class $\omega_{1}$ on
$M^{\prime[n]}$ to a Kähler class on $S^{\prime[n]}$.

###### Proof.

The isometry $\psi_{{\mathcal{U}}^{\prime}}$ maps some Kähler class
$\omega_{M^{\prime}}$ to a Kähler class $\omega_{S^{\prime}}$, as
$(S^{\prime},\eta_{S^{\prime}},M^{\prime},\eta_{M^{\prime}})$ is a point of
${\mathfrak{M}}_{\psi}$. The equalities
$\psi_{E^{\prime}}\circ\theta=\theta\circ\psi_{{\mathcal{U}}^{\prime}}$ and
$\psi_{E^{\prime}}(\delta)=\delta$ follow from Corollary 7.3. The Picard
groups of both Douady spaces $M^{\prime[n]}$ and $S^{\prime[n]}$ are cyclic
generated by the classes $\delta$ and their cones of effective curves are thus
rays spanned by the class $\delta^{\vee}:=\frac{2(\delta,\bullet)}{n-1}$ in
$H^{2}(M^{\prime[n]},{\mathbb{Q}})^{*}$ (were we identify
$H_{2}(M^{\prime[n]},{\mathbb{Q}})$ with
$H^{2}(M^{\prime[n]},{\mathbb{Q}})^{*}$), by [Ma3, Cor. 3.6]. Choose a
sufficiently large real number $t$, such that
$(t\theta(\omega_{M^{\prime}})-\delta,t\theta(\omega_{M^{\prime}})-\delta)>0$.
Then $\omega_{1}:=t\theta(\omega_{M^{\prime}})-\delta$ is a class in the
connected component of the positive cone in
$H^{1,1}(M^{\prime[n]},{\mathbb{R}})$, which contains the Kähler cone, as it
pairs positively with $\theta(\omega_{M^{\prime}})$, and $\omega_{1}$ pairs
positively with the class of every effective curve. Hence, $\omega_{1}$ is a
Kähler class, by [Hu6, Cor. 3.4]. Now,
$\psi_{E^{\prime}}(\omega_{1})=t\theta(\omega_{S^{\prime}})-\delta$ is a
Kähler class, by the same reason. ∎

Let $\eta_{S^{\prime[n]}}$ be the marking of $S^{\prime[n]}$ induced by
extending the marking $\eta_{S^{[n]}}$ given in (7.8) to a trivialization of
the local system over the chosen twistor path $Q$. Define
$\eta_{M^{\prime[n]}}$ similarly in terms of $\eta_{M^{[n]}}$. Set
$\psi^{[n]}:=\eta_{S^{\prime[n]}}^{-1}\psi_{E^{\prime}}\eta_{M^{\prime[n]}}.$
Let

(8.1) ${\mathfrak{M}}^{0}_{\psi^{[n]}}$

be the connected component of ${\mathfrak{M}}_{\psi^{[n]}}$ containing the
quadruple
$(M^{\prime[n]},\eta_{M^{\prime[n]}},S^{\prime[n]},\eta_{S^{\prime[n]}})$.

######  Lemma 8.3.

The vector bundle $E^{\prime}$ is slope-stable with respect to every Kähler
class on $M^{\prime[n]}\times S^{\prime[n]}$.

###### Proof.

We have already noted that ${\mathcal{U}}^{\prime}$ does not have any non-zero
proper saturated subsheaf. Hence, ${\mathcal{U}}^{\prime\boxtimes n}$ does not
have any non-zero proper saturated subsheaves, by [Ma6, Prop. 10.1]. Given
$\sigma\in{\mathfrak{S}}_{n}$, let $(1\times\sigma):M^{n}\times
S^{n}\rightarrow M^{n}\times S^{n}$ act as the identity on $M^{n}$ and via
$\sigma$ on $S^{n}$. The vector bundles ${\mathcal{U}}^{\boxtimes n}$ and
$(1\times\sigma)^{*}{\mathcal{U}}^{\boxtimes n}$ are non-isomorphic, for every
$\sigma\in{\mathfrak{S}}_{n}$, $\sigma\neq 1$. Furthermore, ${\rm
Hom}({\mathcal{U}}^{\boxtimes n},(1\times\sigma)^{*}{\mathcal{U}}^{\boxtimes
n})=0,$ for such $\sigma$. Now the dimensions of ${\rm
Hom}({\mathcal{U}}^{\boxtimes n},(1\times\sigma)^{*}{\mathcal{U}}^{\boxtimes
n})$ and ${\rm Hom}({\mathcal{U}}^{\prime\boxtimes
n},(1\times\sigma)^{*}{\mathcal{U}}^{\prime\boxtimes n})$ are equal, by the
invariance of the dimension of sheaf cohomologies of hyperholomorphic sheaves
under twistor deformations [Ve2, Sec. 3.5]. Hence

$\oplus_{\tau\in{\mathfrak{S}}_{n}}(1\times\tau)^{*}({\mathcal{U}}^{\prime\boxtimes
n})$

does not have any non-zero proper saturated ${\mathfrak{S}}_{n}$-invariant
subsheaf, by [Ma6, Lemma 11.5]. The same is true for $(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes n}$, as $b\times b$ is a birational
morphism.

Consider the commutative diagram

$\textstyle{(q\times q)^{*}(q\times q)_{*}^{{\mathfrak{S}}_{n,\Delta}}(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p\circ
ev\circ
e}$$\scriptstyle{e}$$\textstyle{\oplus_{\tau\in{\mathfrak{S}}_{n}}(1\times\tau)^{*}(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes n}}$$\textstyle{(q\times q)^{*}(q\times
q)_{*}(b\times b)^{*}{\mathcal{U}}^{\prime\boxtimes
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{ev}$$\textstyle{\oplus_{(\sigma,\tau)\in{\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}}(\sigma\times\tau)^{*}(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes
n}.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$

Above, $e$ is the natural injective homomorphism from the
${\mathfrak{S}}_{n,\Delta}$-invariant subsheaf with respect to the
linearization $\chi\otimes\rho_{\boxtimes}$ of ${\mathcal{U}}^{\prime\boxtimes
n}$. The right arrow $p$ is the projection on the subset of direct summands.
For each $(\sigma,\tau)\in{\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}$ we have
the evaluation homomorphism from the bottom left vector bundle to the
corresponding direct summand of the bottom right vector bundle due to the
equality $(q\times q)\circ(\sigma\times\tau)=(q\times q)$. The homomorphism
$ev$ is the resulting diagonal homomorphism. The homomorphism $ev$ restricts
to an isomorphism over the complement of the ramification divisor of $q\times
q$.

The homomorphism $e$ is ${\mathfrak{S}}_{n,\Delta}$-equivariant. The
homomorphism $ev$ is ${\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}$-equivariant.
The homomorphism $p$ is
${\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}$-equivariant, where the latter
group acts on the top right vector bundle via the projection
${\mathfrak{S}}_{n}\times{\mathfrak{S}}_{n}\rightarrow{\mathfrak{S}}_{n}$ onto
the second factor. It follows that the homomorphism $p\circ ev\circ e$ is
${\mathfrak{S}}_{n,\Delta}$-equivariant, where the latter group acts via the
isomorphism ${\mathfrak{S}}_{n,\Delta}\rightarrow{\mathfrak{S}}_{n}$, given by
$(\sigma,\sigma)\mapsto\sigma$ on the top right vector bundle.

The homomorphism $p\circ ev\circ e$ restricts to an isomorphism over the
complement of the ramification divisor of $q\times q$. Hence, the pullback via
$q\times q$ of any saturated non-zero proper subsheaf of $(q\times
q)_{*}^{{\mathfrak{S}}_{n,\Delta}}(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes n}$ maps via $p\circ ev\circ e$ to an
${\mathfrak{S}}_{n}$-invariant non-zero subsheaf of the top right vector
bundle, whose saturation is a proper subsheaf. We have seen that such a
subsheaf does not exist. It follows that $(q\times
q)_{*}^{{\mathfrak{S}}_{n,\Delta}}(b\times
b)^{*}{\mathcal{U}}^{\prime\boxtimes n}$ does not have any saturated non-zero
proper subsheaf. Hence neither does $E^{\prime}$, as the latter is the tensor
product of the former with a line bundle. ∎

######  Theorem 8.4.

Every quadruple $(X_{1},\eta_{1},X_{2},\eta_{2})$ in the component
${\mathfrak{M}}_{\psi^{[n]}}^{0}$, given in (8.1), supports a compatible
vector bundle (Definition 5.20).

###### Proof.

Lemma 8.2 verifies Assumption 5.9(2) for the locally free sheaf $E^{\prime}$
supported by the quadruple
$(M^{\prime[n]},\eta_{M^{\prime[n]}},S^{\prime[n]},\eta_{S^{\prime[n]}})$ in
the component ${\mathfrak{M}}^{0}_{\psi^{[n]}}$. Lemma 8.3 verifies the
stability condition in Assumption 5.9(3) for $E^{\prime}$. The statement thus
follows from Proposition 5.21. ∎

Let $\Lambda_{K3}$ be the $K3$ lattice, $\Lambda$ the lattice given in (7.6),
and $\iota:O(\Lambda_{K3})\rightarrow O(\Lambda)$ the inclusion (7.7). Let
$u\in\Lambda_{K3}$ be a primitive element with $(u,u)>0$. Let $\rho_{u}$ be
the reflection in the co-rank $1$ sublattice $u^{\perp}\subset\Lambda_{K3}$.
Let ${\rm Mon}(\Lambda)\subset O^{+}(\Lambda)$ be the subgroup which acts on
the discriminant group $\Lambda^{*}/\Lambda$ via $\pm 1$.

Let ${\mathfrak{M}}_{\Lambda}^{0}$ be the connected component of the moduli
space ${\mathfrak{M}}_{\Lambda}$ of marked pairs $(X,\eta)$, with $X$ of
$K3^{[n]}$-type, containing the marked pair $(S^{[n]},\eta_{S^{[n]}})$, given
in (7.8).

######  Corollary 8.5.

Let $\tilde{\psi}\in O^{+}(\Lambda_{\mathbb{Q}})$ be an isometry in the double
orbit ${\rm Mon}(\Lambda)(-\iota(\rho_{u})){\rm Mon}(\Lambda)$. There exists a
non-empty connected component ${\mathfrak{M}}_{\tilde{\psi}}^{0}$ of
${\mathfrak{M}}_{\tilde{\psi}}$ parametrizing quadruples
$(X_{1},\eta_{1},X_{2},\eta_{2})$, with $(X_{1},\eta_{1})$ and
$(X_{2},\eta_{2})$ in ${\mathfrak{M}}_{\Lambda}^{0}$, such that each quadruple
in ${\mathfrak{M}}_{\tilde{\psi}}^{0}$ supports a compatible vector bundle
(Definition 5.20).

###### Proof.

The double orbit $O^{+}(L)(-\rho_{u})O^{+}(L)$ depends only on $(u,u)$, by
[Bu, Prop. 3.3]. Hence, the double orbit ${\rm
Mon}(\Lambda)(-\iota(\rho_{u})){\rm Mon}(\Lambda)$ depends only on $(u,u)$,
and so we may assume that
$\tilde{\psi}=\gamma_{1}\iota(\eta_{S}\psi_{\mathcal{U}}\eta_{M}^{-1})\gamma_{0}$,
for some $\gamma_{1},\gamma_{2}\in{\rm Mon}(\Lambda)$, where
$\psi_{\mathcal{U}}$ is the isometry in Proposition 8.1. Set
$\psi:=\eta_{S}\psi_{\mathcal{U}}\eta_{M}^{-1}$. The isometry $\psi^{[n]}$,
given in (8.1), is equal to $\iota(\psi)$, by Corollary 7.3, and so
$\tilde{\psi}=\gamma_{2}\psi^{[n]}\gamma_{1}$. We get the isomorphism
$f:{\mathfrak{M}}_{\psi^{[n]}}\rightarrow{\mathfrak{M}}_{\tilde{\psi}}$ given
by

$f(X_{1},\eta_{1},X_{2},\eta_{2})=(X_{1},\gamma_{1}^{-1}\eta_{1},X_{2},\gamma_{2}\eta_{2}),$

which maps connected components to connected components, since
$\eta_{i}^{-1}{\rm Mon}(\Lambda)\eta_{i}$ is the monodromy group ${\rm
Mon}^{2}(X_{i})$ of $X_{i}$, $i=1,2$, by [Ma2, Theorem 1.6 and Lemma 4.10].
Furthermore, the marked pair $(X_{1},\gamma_{1}^{-1}\eta_{1})$ belongs to the
connected component ${\mathfrak{M}}_{\Lambda}^{0}$ of $(X_{1},\eta_{1})$,
since $\gamma_{1}^{-1}$ belongs to ${\rm Mon}(\Lambda)$ and ${\rm
Mon}(\Lambda)=\eta_{1}{\rm Mon}^{2}(X_{1})\eta_{1}^{-1}$. Similarly,
$(X_{2},\gamma_{2}\eta_{2})$ belongs to the same connected component
${\mathfrak{M}}_{\Lambda}^{0}$ of $(X_{2},\eta_{2})$. Clearly, if
$q\in{\mathfrak{M}}_{\psi^{[n]}}$ supports a compatible vector bundle, then
$f(q)$ supports (the same) compatible vector bundle. The statement now follows
from Theorem 8.4. ∎

##  9\. Algebraicity of a rational Hodge isometry

We prove Theorems 1.1 and 1.4 in this section. Let ${\mathcal{G}}_{an}$ be the
subgroupoid of ${\mathcal{G}}$ given in (1.2). Note that morphisms in ${\rm
Hom}_{{\mathcal{G}}_{an}}((X,\epsilon),(Y,\epsilon^{\prime}))$ are induced by
analytic correspondences in $X\times Y$ and are degree preserving up to sign.

######  Theorem 9.1.

Let $X$ and $Y$ be irreducible holomorphic symplectic manifolds of
$K3^{[n]}$-type and let $f:H^{2}(X,{\mathbb{Q}})\rightarrow
H^{2}(Y,{\mathbb{Q}})$ be a rational Hodge isometry. For every orientations
$\epsilon$ and $\epsilon^{\prime}$ there exists a morphism $\phi$ in ${\rm
Hom}_{{\mathcal{G}}_{an}}((X,\epsilon),(Y,\epsilon^{\prime}))$, such that
$\widetilde{H}_{0}(\phi)$ is equal to $f$ or $-f$

###### Proof.

The set ${\rm Hom}_{{\mathcal{G}}_{an}}((X,\epsilon),(Y,\epsilon^{\prime}))$
is independent of the orientations as is the functor $\widetilde{H}_{0}$, so
the existence of $\phi$ is independent of the orientations. We may assume that
$\nu(f)=1$, possibly after replacing $f$ by $-f$. Choose markings $\eta_{X}$
and $\eta_{Y}$, such that $(X,\eta_{X})$ and $(Y,\eta_{Y})$ belong to the same
connected component ${\mathfrak{M}}^{0}_{\Lambda}$ of
${\mathfrak{M}}_{\Lambda}$. Choose an orientation of $\Lambda$ and endow
$H^{2}(X,{\mathbb{Z}})$ and $H^{2}(Y,{\mathbb{Z}})$ with the orientations
$\epsilon$ and $\epsilon^{\prime}$, so that $\eta_{X}$ and $\eta_{Y}$ are
orientation preserving. We write ${\rm Hom}_{{\mathcal{G}}_{an}}(X,Y)$ instead
of ${\rm Hom}_{{\mathcal{G}}_{an}}((X,\epsilon),(Y,\epsilon^{\prime}))$, as in
Convention 5.4, and we will continue to use the convention below. Set
$\psi:=\eta_{Y}\circ f\circ\eta_{X}^{-1}$. Then $\psi$ belongs to
$O^{+}(\Lambda_{\mathbb{Q}})$ and so it decomposes as

$\psi=\psi_{k}\circ\psi_{k-1}\circ\cdots\circ\psi_{1},$

where each $\psi_{i}$ belongs to the double orbit ${\rm
Mon}(\Lambda)\iota(-\rho_{u}){\rm Mon}(\Lambda)$, for some element
$u\in\Lambda_{K3}$ with $(u,u)>0$, by Corollary 2.5. The proof proceeds by
induction on $k$.

Case $k=0$. In this case $\psi=id$ and $f=\eta_{Y}^{-1}\eta_{X}$ is a Hodge
isometry. There exists a parallel-transport operator
$\phi:H^{*}(X,{\mathbb{Q}})\rightarrow H^{*}(Y,{\mathbb{Q}})$, which is a
morphism in ${\rm Hom}_{{\mathcal{G}}_{an}}(X,Y)$, and such that $\phi$
restricts to $H^{2}(X,{\mathbb{Q}})$ as $f$, by Theorem 5.2. In this case
$\widetilde{H}_{0}(\phi)$ is also equal to the restriction of $\phi$ to
$H^{2}(X,{\mathbb{Q}})$, by Remark 5.6.

Assume that $k\geq 1$ and that the statement holds for $k-1$. Set
$\psi^{\prime}=\psi_{k-1}\circ\cdots\circ\psi_{1}$, so that
$\psi=\psi_{k}\psi^{\prime}$. Set $\ell_{0}:=P(X,\eta_{X})$,
$\ell_{k-1}:=\psi^{\prime}(\ell_{0})$, and $\ell_{k}=\psi(\ell_{0})$. Note
that $\ell_{k}=P(Y,\eta_{Y})$. Let $(X_{k-1},\eta_{k-1})$ be a marked pair in
${\mathfrak{M}}_{\Lambda}^{0}$, such that $P(X_{k-1},\eta_{k-1})=\ell_{k-1}$.
Such a marked pair exists, by the surjectivity of the period map [Hu1]. Set
$f^{\prime}:=\eta_{k-1}^{-1}\circ\psi^{\prime}\circ\eta_{X}$. There exists a
morphism $\phi^{\prime}\in{\rm Hom}_{{\mathcal{G}}_{an}}(X,X_{k-1})$, such
that $\widetilde{H}_{0}(\phi^{\prime})=f^{\prime}$, by the induction
hypothesis. The quadruple $(X_{k-1},\eta_{k-1},Y,\eta_{Y})$ satisfies the
condition that $\eta_{Y}^{-1}\psi_{k}\eta_{k-1}$ is a Hodge isometry, since

$\eta_{Y}^{-1}\psi_{k}\eta_{k-1}(H^{2,0}(X_{k-1}))=\eta_{Y}^{-1}(\psi_{k}(\ell_{k-1}))=\eta_{Y}^{-1}(\psi(\ell_{0}))=\eta_{Y}^{-1}(P(Y,\eta_{Y}))=H^{2,0}(Y),$

but it may not belong to ${\mathfrak{M}}_{\psi_{k}}$, since
$\eta_{Y}^{-1}\psi_{k}\eta_{k-1}$ need not map a Kähler class to a Kähler
class.

There exists a connected component ${\mathfrak{M}}^{0}_{\psi_{k}}$ of
${\mathfrak{M}}_{\psi_{k}}$ in which every quadruple supports a compatible
vector bundle, and such that
$\Pi_{i}:{\mathfrak{M}}^{0}_{\psi_{k}}\rightarrow{\mathfrak{M}}_{\Lambda}$ has
image in the connected component ${\mathfrak{M}}^{0}_{\Lambda}$, for $i=1,2$,
by Corollary 8.5. Extend the pair $(X_{k-1},\eta_{k-1})$ to a quadruple
$q:=(X_{k-1},\eta_{k-1},Y^{\prime},\eta_{Y^{\prime}})$ in
${\mathfrak{M}}^{0}_{\psi_{k}}$. Such a point $q$ exists, by the surjectivity
Lemma 5.14. Let $F_{q}$ be the compatible vector bundle supported by $q$. Set
$\phi_{k}^{\prime}:=\left[\kappa(F_{q})\sqrt{td_{X_{k-1}\times
Y^{\prime}}}\right]_{*}:H^{*}(X_{k-1},{\mathbb{Q}})\rightarrow
H^{*}(Y^{\prime},{\mathbb{Q}})$. Then
$\widetilde{H}_{0}\left(\phi_{k}^{\prime}\right)=\eta_{Y^{\prime}}^{-1}\circ\psi_{k}\circ\eta_{k-1},$
by Definition 5.20 of the compatibility of $F_{q}$. We have
$P(Y^{\prime},\eta_{Y^{\prime}})=\psi_{k}(P(X_{k-1},\eta_{k-1}))=P(Y,\eta_{Y})$,
where the first equality is by Lemma 5.11. Hence, there exists a morphism
$\phi_{k}\in{\rm Hom}_{{\mathcal{G}}_{an}}(Y^{\prime},Y)$, such that
$\widetilde{H}_{0}(\phi_{k})=\eta_{Y}^{-1}\eta_{Y^{\prime}}$, by the case
$k=0$ of the statement. We conclude that
$\phi:=\phi_{k}\circ\phi_{k}^{\prime}\circ\phi^{\prime}$ is a morphism in
${\rm Hom}_{{\mathcal{G}}_{an}}(X,Y)$, and $\widetilde{H}_{0}(\phi)=f$. ∎

###### Proof of Theorem 1.4.

The Theorem is a rephrasing of Theorem 9.1. ∎

###### Proof of Theorem 1.1.

Let $f:H^{2}(X,{\mathbb{Q}})\rightarrow H^{2}(Y,{\mathbb{Q}})$ be a Hodge
isometry. We may assume that $\nu(f)=1$, possibly after replacing $f$ by $-f$
(and $\tilde{f}$ by $-\tilde{f}$ in the statement of the theorem). There exist
orientations $\epsilon$ and $\epsilon^{\prime}$ and a morphism $\phi\in{\rm
Hom}_{{\mathcal{G}}_{an}}((X,\epsilon),(Y,\epsilon^{\prime}))$, such that
$\widetilde{H}_{0}(\phi)=f$, by Theorem 9.1. If $\phi$ is degree preserving,
then $\widetilde{H}_{0}(\phi)$ is a rational scalar multiple of the
restriction of $\phi$ to $H^{2}(X,{\mathbb{Q}})$, by Lemma 3.6, and so we
choose $\tilde{f}$ to be the appropriate scalar multiple of $\phi$. If $\phi$
is degree reversing, then setting $\tilde{f}=\phi$ we get an analytic
correspondence with the desired properties, by Lemma 3.7. ∎

##  10\. The Pontryagin product on the cohomology of IHSMs of $K3^{[n]}$-type

In Section 10.1 we introduce the Pontryagin product on the cohomology of an
IHSM with vanishing odd cohomology. The results still apply to the even
cohomology if the odd cohomology does not vanish. In Section 10.2 we prove
Conjecture 1.8 for morphisms in the image of the functor
$\Theta_{n}:{\mathcal{G}}^{[1]}_{an}\rightarrow{\mathcal{G}}^{[n]}_{an}$,
given in (7.12). In particular, morphisms in the image of $\Theta_{n}$, once
normalized as in Conjecture 1.8, conjugate the cup product to itself, if they
are degree preserving, and to the Pontryagin product if they are degree
reversing.

###  10.1. The Pontryagin product for an IHSM with vanishing odd cohomology

Let $X$ be an IHSM, let
${\mathfrak{g}}\subset{\mathfrak{s}l}(H^{*}(X,{\mathbb{Q}}))$ be its LLV
algebra, and let $h\in{\mathfrak{g}}$ be the grading operator given in (3.1).
The commutator of $h$ in ${\mathfrak{g}}$ is
$\bar{{\mathfrak{g}}}\oplus{\mathbb{Q}}\cdot h$ and composing the isomorphism
$\dot{\rho}^{-1}:{\mathfrak{g}}\cong{\mathfrak{s}o}(\widetilde{H}(X,{\mathbb{Q}}))$,
given in (3.2), with restriction to $H^{2}(X,{\mathbb{Q}})$ yields an
isomorphism $\bar{{\mathfrak{g}}}\cong{\mathfrak{s}o}(H^{2}(X,{\mathbb{Q}}))$,
by Lemma 3.5. Assume that the odd cohomology of $X$ vanishes. If the second
Betti number $b_{2}(X)$ of $X$ is even, assume that the monordomy group of $X$
contains reflections and $dim(X)=2n$ for odd $n$. The latter assumption holds
true for all currently known examples of irreducible holomorphic symplectic
manifolds with even $b_{2}(X)$ ($K3$ surfaces and IHSMs of O’Grady $6$
deformation type). The ${\mathfrak{g}}$ action on $H^{*}(X,{\mathbb{C}})$
integrates to an action of $SO(\widetilde{H}(X,{\mathbb{C}}))$ and the
commutator of $h$ is ${\mathbb{C}}^{\times}\times SO(H^{2}(X,{\mathbb{C}}))$,
where an element $t$ of the first factor ${\mathbb{C}}^{\times}$ maps $\alpha$
to $t\alpha$ and $\beta$ to $\frac{1}{t}\beta$. The image of the second factor
$SO(H^{2}(X,{\mathbb{C}}))$ in $GL(H^{*}(X,{\mathbb{C}}))$ acts on
$H^{*}(X,{\mathbb{C}})$ via ring automorphisms, by [LL, Prop. 4.5(ii)]. The
coset $\\{g\in SO(\widetilde{H}(X,{\mathbb{C}}))\ :\ ghg^{-1}=-h\\}$ of the
commutator is the product of the component of
$O({\mathbb{C}}\alpha\oplus{\mathbb{C}}\beta)$ with determinant $-1$ with the
component of $O(H^{2}(X,{\mathbb{C}}))$ with determinant $-1$. If the
monodromy group contains an element $R$ of order $2$ acting as a reflection on
$H^{2}(X,{\mathbb{C}})$, then the factor $O(H^{2}(X,{\mathbb{C}}))$ acts on
$H^{*}(X,{\mathbb{C}})$ via ring automorphisms. Such a monodromy operator $R$
acts on $\widetilde{H}(X,{\mathbb{Q}})$ via the operator $\widetilde{H}(R)$
stabilizing both $\alpha$ and $\beta$ and restricting to
$H^{2}(X,{\mathbb{Q}})$ as the monodromy reflection. A choice of such an
element $R\in{\rm Mon}(X)$ yields an action of
$O(\widetilde{H}(X,{\mathbb{C}}))$ on $H^{*}(X,{\mathbb{C}})$

(10.1) $\rho:O(\widetilde{H}(X,{\mathbb{C}}))\rightarrow
GL(H^{*}(X,{\mathbb{C}}))$

extending the above homomorphism
$\rho:SO(\widetilde{H}(X,{\mathbb{C}}))\rightarrow GL(H^{*}(X,{\mathbb{C}}))$
and satisfying $\rho(\widetilde{H}(R))=R$.

Let $\tau$ be the element of $O(\widetilde{H}(X,{\mathbb{Q}}))$ interchanging
$\alpha$ and $\beta$ and acting on $H^{2}(X,{\mathbb{C}})$ via multiplication
by $-1$. If $b_{2}(X)$ is odd, then $\tau$ belongs to
$SO(\widetilde{H}(X,{\mathbb{Q}}))$ and $\rho_{\tau}$ is well defined, and if
$b_{2}(X)$ is even we define $\rho_{\tau}$ via (10.1). Define the Pontryagin
product $\star$ on $H^{*}(X,{\mathbb{Q}})$ by

(10.2)
$\gamma\star\delta:=\rho_{\tau}(\rho_{\tau}(\gamma)\cup\rho_{\tau}(\delta)).$

######  Remark 10.1.

The subring $SH^{*}(X,{\mathbb{C}})$ of $H^{*}(X,{\mathbb{C}})$ generated by
$H^{2}(X,{\mathbb{C}})$ is invariant under monodromy operators and it is an
LLV-subrepresentation, hence it is invariant under elements in the image of
$\rho$. The defining Equation (10.2) implies that $SH^{*}(X,{\mathbb{C}})$ is
a subring with respect to the Pontryagin product.

The values of the representation $\rho$ on the connected component with
determinant $-1$ depends on our choice of the monodromy reflection $R$ (see
[Ma2, Lemma 4.13]). Thus, if $b_{2}(X)$ is even, then $\rho_{\tau}$ depends on
the choice of $R$. Nevertheless, the Pontryagin product does not depend on
this choice as we show next (and generalize in Remark 10.5 below).

######  Lemma 10.2.

  1. (1)

The Pontryagin product $\star$ is independent of the choice of the monodromy
reflection $R$.

  2. (2)

Parallel-transport operators are isomorphisms of the Pontryagin rings.

  3. (3)

The unit with respect to $\star$ is $\rho_{\tau}(1)$ and is independent of the
choice of $R$.

###### Proof.

(1) The statement is clear if $b_{2}(X)$ is odd. Assume that $b_{2}(X)$ is
even. Let $R^{\prime}\in{\rm Mon}(X)$ be another monodromy reflection and let
$\rho^{\prime}:O(\widetilde{H}(X,{\mathbb{C}}))\rightarrow
GL(H^{*}(X,{\mathbb{C}}))$ be the resulting representation. Let
$f,f^{\prime}\in SO(\widetilde{H}(X,{\mathbb{C}}))$ be the elements satisfying
$\tau=f\widetilde{H}(R)=f^{\prime}\widetilde{H}(R^{\prime})$. Then
$\rho_{f^{\prime}}=\rho^{\prime}_{f^{\prime}}$, as the restrictions of $\rho$
and $\rho^{\prime}$ to $SO(\widetilde{H}(X,{\mathbb{C}}))$ are equal. Now,
$f^{\prime}=f\widetilde{H}(R)\widetilde{H}(R^{\prime})=f\widetilde{H}(RR^{\prime})$.
So,

$\rho^{\prime}_{\tau}=\rho_{f^{\prime}}R^{\prime}=\rho_{f\widetilde{H}(RR^{\prime})}R^{\prime}=\rho_{f}\rho_{\widetilde{H}(RR^{\prime})}R^{\prime}=\rho_{f}RR\rho_{\widetilde{H}(RR^{\prime})}R^{\prime}=\rho_{\tau}(R\rho_{\widetilde{H}(RR^{\prime})}R^{\prime}).$

Set $g:=R\rho_{\widetilde{H}(RR^{\prime})}R^{\prime}$. Then
$\rho^{\prime}_{\tau}=\rho_{\tau}g$, and $g$ is an automorphism of
$H^{*}(X,{\mathbb{C}})$ with respect to cup product, since each of $R$,
$\rho_{\widetilde{H}(RR^{\prime})}$, and $R^{\prime}$ is. Furthermore,
$\rho_{\tau}g\rho_{\tau}g=(\rho^{\prime}_{\tau})^{2}=id$, so
$g\rho_{\tau}g=\rho_{\tau}$. Now compute:

$\displaystyle\rho^{\prime}_{\tau}(\rho^{\prime}_{\tau}(\gamma)\cup\rho^{\prime}_{\tau}(\delta))$
$\displaystyle=$
$\displaystyle(\rho_{\tau}g)(\rho_{\tau}(g(\gamma))\cup\rho_{\tau}(g(\delta)))=\rho_{\tau}((g\rho_{\tau}g)(\gamma)\cup(g\rho_{\tau}g)(\delta))$
$\displaystyle=$
$\displaystyle\rho_{\tau}(\rho_{\tau}(\gamma)\cup\rho_{\tau}(\delta))=\gamma\star\delta.$

(2) Let $f:H^{*}(X,{\mathbb{C}})\rightarrow H^{*}(Y,{\mathbb{C}})$ be a
parallel-transport operator. We get the commutative diagram

$\textstyle{SO(\widetilde{H}(X,{\mathbb{C}}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Ad_{\widetilde{H}(f)}}$$\textstyle{GL(H^{*}(X,{\mathbb{C}}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Ad_{f}}$$\textstyle{SO(\widetilde{H}(Y,{\mathbb{C}}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{GL(H^{*}(Y,{\mathbb{C}})),}$

where the horizontal arrows are the integration of the infinitesimal LLV Lie
algebras actions. Clearly, $\widetilde{H}(f)$ conjugates $\tau$ to $\tau$.
Parallel-transport operators conjugate a monodromy reflection to such, and so
the representation $\rho$, given in (10.1), get conjugated via the pair $f$
and $\widetilde{H}(f)$ to
$\rho^{\prime}:O(\widetilde{H}(Y,{\mathbb{C}}))\rightarrow
GL(H^{*}(Y,{\mathbb{C}}))$, as in part (1), and so $f$ conjugates
$\rho_{\tau}$ to $\rho^{\prime}_{\tau}$. The statement of part (2) now follows
from part (1).

(3) The unit of a ring is unique, and so $\rho_{\tau}(1)$ is independent of
$R$, by Part 1. ∎

We calculate the unit explicitly next.

######  Lemma 10.3.

The unit of $H^{*}(X,{\mathbb{Q}})$ with respect to the $\star$ product is
$c_{X}[pt]/n!$, where $\dim(X)=2n$ and $c_{X}$ is the Fujiki constant.

###### Proof.

Consider the $SO(\widetilde{H}(X,{\mathbb{Q}}))$-equivariant embedding

$\Psi:SH^{*}(X,{\mathbb{Q}})\rightarrow{\rm
Sym}^{n}(\widetilde{H}(X,{\mathbb{Q}})),$

given in (3.3), where $SO(\widetilde{H}(X,{\mathbb{Q}}))$ acts on
$SH^{*}(X,{\mathbb{Q}})$ via the representation $\rho$ integrating the
infinitesimal LLV action and via the natural representation on ${\rm
Sym}^{n}(\widetilde{H}(X,{\mathbb{Q}}))$. If $b_{2}(X)$ is odd, then $\tau$
belongs to $SO(\widetilde{H}(X,{\mathbb{C}}))$. If $b_{2}(X)$ is even, then
$X$ admits monodromy reflections, by assumptions, and $\Psi$ is equivariant
with respect to those as well, and so $\Psi$ is equivariant with respect to
the $\rho_{\tau}$ action on its domain and the $\tau$ action of its codomain.
The element $1$ belongs to $SH^{*}(X,{\mathbb{C}})$ and
$\Psi(1)=\alpha^{n}/n!$, by [Ta, Proposition 3.5]. Now, $\tau(\alpha)=\beta$
and so
$\Psi(\rho_{\tau}(1))=\rho_{\tau}(\Psi(1))=\beta^{n}/n!=\Psi(c_{X}[pt]/n!)$,
where the last equality follows from [Ta, Lemma 3.6]. Hence,
$\rho_{\tau}(1)=c_{X}[pt]/n!$. ∎

Let $\mu_{t}$ be the element of
$SO({\mathbb{C}}\alpha\oplus{\mathbb{C}}\beta)$, given by
$\mu_{t}(\alpha)=t^{-1}\alpha$ and $\mu_{t}(\beta)=t\beta$, and $\mu_{t}$
restricts to $H^{2}(X,{\mathbb{C}})$ as the identity. Then $\rho_{\mu_{t}}$
acts on $H^{2k}(X,{\mathbb{C}})[2n]$ via multiplication by $t^{k}$.

######  Lemma 10.4.

Let $g$ be an element of $O(\widetilde{H}(X,{\mathbb{C}}))$.

  1. (1)

If $ghg^{-1}=h$, then $g(\alpha)=t\alpha$, for some
$t\in{\mathbb{C}}^{\times}$, and $\rho_{\mu_{t}g}$ is an automorphism of the
cohomology ring $H^{*}(X,{\mathbb{C}})$ with respect to the usual cup product.

  2. (2)

If $ghg^{-1}=-h$, then $g(\alpha)=t^{-1}\beta$, for some
$t\in{\mathbb{C}}^{\times}$, and

$\rho_{\mu_{t}g}(\gamma\cup\delta)=\rho_{\mu_{t}g}(\gamma)\star\rho_{\mu_{t}g}(\delta),$

for all $\gamma,\delta\in H^{*}(X,{\mathbb{C}})$.

###### Proof.

(1) If $ghg^{-1}=h$ and $g(\alpha)=t\alpha$, then $\mu_{t}g$ belongs to the
factor $OH^{2}(X,{\mathbb{C}})$ of the commutator of $h$, and so
$\rho_{\mu_{t}g}$ acts by ring automorphisms.

(2) If $ghg^{-1}=-h$ and $g(\alpha)=t^{-1}\beta$, then $f:=\tau\mu_{t}g$
commutes with $h$ and the second equality below follows from part (1).

$\displaystyle\rho_{\mu_{t}g}(\gamma\cup\delta)$ $\displaystyle=$
$\displaystyle\rho_{\tau
f}(\gamma\cup\delta)=\rho_{\tau}(\rho_{f}(\gamma)\cup\rho_{f}(\delta))=\rho_{\tau}(\rho_{\tau}(\rho_{\tau
f}(\gamma))\cup\rho_{\tau}(\rho_{\tau f}(\delta)))$ $\displaystyle=$
$\displaystyle\rho_{\tau f}(\gamma)\star\rho_{\tau
f}(\delta)=\rho_{\mu_{t}g}(\gamma)\star\rho_{\mu_{t}g}(\delta).$

∎

######  Remark 10.5.

Lemma 10.4(2) equivalently states that the same Pontryagin product can be
defined in terms of any degree reversing isometry $g\in
O(\widetilde{H}(X,{\mathbb{Q}}))$, by

$\gamma\star\delta=\rho_{g^{-1}\mu_{t^{-1}}}(\rho_{\mu_{t}g}(\gamma)\cup\rho_{\mu_{t}g}(\delta)).$

###  10.2. The Pontryagin product for an IHSM of $K3^{[n]}$-type

Assume from now on that $X$ is of $K3^{[n]}$-type. The following lemma states
that the functor $\Theta_{n}$, given in (7.12), integrates the infinitesimal
LLV Lie algebra action when restricted to automorphisms of determinant $1$.

######  Lemma 10.6.

The restriction of $\Theta_{n}$ to the subgroup $S{\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$, of elements of ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$ of determinant $1$, is equal to the
restriction of the composition of
$\tilde{\iota}:SO(\widetilde{H}(S,{\mathbb{C}}))\rightarrow
SO(\widetilde{H}(S^{[n]},{\mathbb{C}}))$, given in (7.5), with the
representation $\rho$ given in (10.1).

###### Proof.

The infinitesimal $LLV$-action $\dot{\rho}$, given in (3.2), sends
$e_{\theta(\lambda)}\in{\mathfrak{s}o}(\widetilde{H}(S^{[n]},{\mathbb{Q}}))$
to cup product with $\theta(\lambda)$, by the
${\mathfrak{s}o}(\widetilde{H}(S^{[n]},{\mathbb{Q}}))$-equivariance of $\Psi$
in (3.3). Thus, the equality $\Theta_{n}(\phi)=\rho_{\tilde{\iota}(\phi)}$
holds, when $\phi$ is cup product with $\exp(e_{\lambda})$, $\lambda\in
H^{2}(S,{\mathbb{Q}})$, by definition of $\Theta_{n}$. The map
$\Theta_{n}:{\rm Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)\rightarrow{\rm
Aut}_{{\mathcal{G}}^{[1]}}(S^{[n]},\epsilon^{[n]})$ is induced by an algebraic
map, due to its topological nature via Grothendieck-Riemann-Roch, and so
extends to the Zariski closure $O(\widetilde{H}(S,{\mathbb{C}}))$ of ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$. The statement thus reduces to the
equality of the differentials of $\Theta_{n}$ and $\rho\circ\tilde{\iota}$. We
already established the equality
$d\Theta_{n}(e_{\lambda})=\dot{\rho}(d\tilde{\iota}(e_{\lambda}))=\dot{\rho}(e_{\theta(\lambda)}).$
We will denote $\dot{\rho}(e_{\theta(\lambda)})$ by $e_{\theta(\lambda)}$ as
well, so that the latter equality becomes

(10.3) $d\Theta_{n}(e_{\lambda})=e_{\theta(\lambda)}.$

Choose $(M,\epsilon^{\prime})$ and $\phi\in{\rm
Hom}_{{\mathcal{G}}^{[1]}}((S,\epsilon),(M,\epsilon^{\prime}))$, so that
$\phi$ is degree reversing. We can choose, for example,
$(M,\epsilon^{\prime})=(S,\epsilon)$ and
$\phi:=[ch({\mathcal{I}}_{\Delta})\sqrt{td_{S\times S}}]_{*}$, where
${\mathcal{I}}_{\Delta}$ is the ideal sheaf of the diagonal. Note that
$\phi=\widetilde{H}(\phi)$, as
$H^{*}(S,{\mathbb{Q}})=\widetilde{H}(S,{\mathbb{Q}})$ and similarly for $M$.
Let $t\in{\mathbb{Q}}$, so that $\phi(\beta)=t\alpha$. Let $\lambda\in
H^{2}(S,{\mathbb{Q}})$ be an element with $(\lambda,\lambda)\neq 0$. Then

(10.4)
$\phi^{-1}e_{\phi(\lambda)}\phi=\frac{t(\lambda,\lambda)}{2}e_{\lambda}^{\vee},$

by Lemma 1.6. Above $e_{\lambda}^{\vee}$ is the dual Lefschetz operator with
respect to the grading operator of the LLV algebra ${\mathfrak{g}}_{S}$ of
$S$. The morphism $\widetilde{H}(\Theta_{n}(\phi))$ is degree reversing as
well, by Corollaries 7.3 and 7.4. Hence, so is $\Theta_{n}(\phi)$, by Lemma
3.4. Say $(\widetilde{H}(\Theta_{n}(\phi)))(\beta)=t^{\prime}\alpha$. Then

$\Theta_{n}(\phi)^{-1}e_{(\widetilde{H}(\Theta_{n}(\phi)))(\theta(\lambda))}\Theta_{n}(\phi)=\frac{t^{\prime}(\lambda,\lambda)}{2}e_{\theta(\lambda)}^{\vee}.$

by Lemma 1.6. Above $e_{\theta(\lambda)}^{\vee}$ is the dual Lefschetz
operator with respect to the grading operator of the LLV algebra
${\mathfrak{g}}_{S^{[n]}}$ of $S^{[n]}$ given in (3.1). Note that
$t^{\prime}=\det(\phi)^{n+1}t$ and
$\widetilde{H}(\Theta_{n}(\phi))=\det(\phi)^{n+1}\tilde{\iota}_{\phi},$ by
Corollary 7.3. The second equality below follows (using also that
$e_{\tilde{\iota}_{\phi}(\theta(\lambda))}=e_{\theta(\phi(\lambda))})$.

$Ad_{\Theta_{n}(\phi)^{-1}}(d\Theta_{n}(e_{\phi(\lambda)}))\stackrel{{\scriptstyle(\ref{eq-
differential-of-Theta-n-sends-e-lambda-
to})}}{{=}}\Theta_{n}(\phi)^{-1}e_{\theta(\phi(\lambda))}\Theta_{n}(\phi)=\frac{t(\lambda,\lambda)}{2}e_{\theta(\lambda)}^{\vee}.$

Given a morphism $\phi\in{\rm
Hom}_{{\mathcal{G}}^{[1]}}((S,\epsilon),(M,\epsilon^{\prime}))$, we have

$d\Theta_{n}\circ Ad_{\phi}=Ad_{\Theta_{n}(\phi)}\circ d\Theta_{n}$

since
$\Theta_{n}(\phi\psi\phi^{-1})=\Theta_{n}(\phi)\Theta_{n}(\psi)\Theta_{n}(\phi)^{-1}$.
We conclude that

$d\Theta_{n}(Ad_{\phi^{-1}}(e_{\phi(\lambda)}))=\frac{t(\lambda,\lambda)}{2}e_{\theta(\lambda)}^{\vee}.$

The left hand side is
$d\Theta_{n}\left(\frac{t(\lambda,\lambda)}{2}e_{\lambda}^{\vee}\right)$, by
Equation (10.4). Hence,
$d\Theta_{n}(e_{\lambda}^{\vee})=e_{\theta(\lambda)}^{\vee}$. Now
$\dot{\rho}:{\mathfrak{s}o}(\widetilde{H}(S^{[n]},{\mathbb{Q}}))\rightarrow{\mathfrak{g}}_{S^{[n]}}$
maps the grading operator of
${\mathfrak{s}o}(\widetilde{H}(S^{[n]},{\mathbb{Q}}))$ to that of
${\mathfrak{g}}_{S^{[n]}}$ and the differential of $\tilde{\iota}$ sends the
grading operator of ${\mathfrak{g}}_{S}$ to that of
${\mathfrak{s}o}(\widetilde{H}(S^{[n]},{\mathbb{Q}}))$. Hence, the
differential of $\rho\circ\tilde{\iota}$ sends $e_{\lambda}^{\vee}$ to
$e_{\theta(\lambda)}^{\vee}$ and we get the equality

$d\Theta_{n}(e_{\lambda}^{\vee})=d(\rho\circ\tilde{\iota})(e_{\lambda}^{\vee}).$

The statement follows, since ${\mathfrak{s}o}(H^{*}(S,{\mathbb{Q}}))$ is
generated by the pairs $(e_{\lambda},e^{\vee}_{\lambda})$, as $\lambda$ varies
over elements of $H^{2}(S,{\mathbb{Q}})$ with $(\lambda,\lambda)\neq 0$. ∎

Let ${\mathcal{R}}^{[n]}_{an}$ be the subgroupoid of
${\mathcal{G}}_{an}^{[n]}$ given in (1.6).

######  Proposition 10.7.

The restriction of the functor $\widetilde{H}_{0}$ to the subgroupoid
${\mathcal{R}}^{[n]}_{an}$ remains full.

###### Proof.

It suffices to prove that all morphisms of ${\mathcal{G}}^{[n]}_{an}$, which
were used in the proof of Theorem 1.4 to prove the fullness of
$\widetilde{H}_{0}$, are already morphisms of ${\mathcal{R}}^{[n]}_{an}$.
Parallel-transport operators, which are Hodge isometries, are morphisms in
${\mathcal{R}}^{[n]}_{an}$. The morphisms other than parallel-transport
operators where all deformations of those that appear in Corollary 7.4, hence
conjugates of those that appear in Corollary 7.4 via parallel-transport
operators. It suffices to prove that the morphism $\Theta_{n}(\varphi)$ of
${\mathcal{G}}^{[n]}_{an}$ that appears in Corollary 7.4 belongs to
${\mathcal{R}}^{[n]}_{an}$, in other words, that
$\mu_{\chi(\Theta_{n}(\varphi))}\Theta_{n}(\varphi)$ conjugates the cup
product of $H^{*}(S^{[n]},{\mathbb{Q}})$ to the Pontryagin product.

Let $f:H^{*}(S,{\mathbb{Z}})\rightarrow H^{*}(M,{\mathbb{Z}})$ be a parallel-
transport operator. Then $f\circ\varphi$ belongs to ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(M)$,
$\mu_{\chi(\Theta_{n}(f\circ\varphi))}=\mu_{\chi(\Theta_{n}(\varphi))}$ and
$\mu_{\chi(\Theta_{n}(\varphi))}\Theta_{n}(\varphi)$ conjugates the cup
product of $H^{*}(M^{[n]},{\mathbb{Q}})$ to the Pontryagin product, if and
only if $\mu_{\chi(\Theta_{n}(f\circ\varphi))}\Theta_{n}(f\circ\varphi)$ does,
by Lemma 10.2(2). Hence, it suffices to prove the statement of the Proposition
for automorphisms in the image of $\Theta_{n}$.

The restriction of $\Theta_{n}$ to the subgroup $S{\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$ of elements of ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$ of determinant $1$ is equal to the
restriction of the composition of
$\tilde{\iota}:SO(\widetilde{H}(S,{\mathbb{C}}))\rightarrow
SO(\widetilde{H}(S^{[n]},{\mathbb{C}}))$ with the representation $\rho$ given
in (10.1), by Lemma 10.6. The two agree on the whole of ${\rm
Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$ if we choose the reflection $R$ in the
definition of $\rho$ to be the image of a monodromy reflection of $S$. Hence,
if $\varphi\in{\rm Aut}_{{\mathcal{G}}^{[1]}}(S,\epsilon)$ conjugates $h$ to
$-h$, then
$\mu_{\chi(\Theta_{n}(\varphi))}\Theta_{n}(\varphi)=\rho(\mu_{\chi(\varphi)}\varphi)$
conjugates the cup product to the Pontryagin product and if $\varphi$ commutes
with $h$, then $\mu_{\chi(\Theta_{n}(\varphi))}\Theta_{n}(\varphi)$ is a ring
automorphism with respect to cup product, by Lemmas 10.2 and 10.4. ∎

###### Proof of Corollary 1.10.

Choose an IHSM $X$ and a twisted vector bundle $E$ over $X\times Y$ as in
Theorem 1.5. There exists a rational number $t$, such the the isomorphism
$\phi:=\mu_{t}\circ[\kappa(E)\sqrt{td_{X\times
Y}}]_{*}:H^{*}(X,{\mathbb{Q}})\rightarrow H^{*}(Y,{\mathbb{Q}})$ satisfies
Equation (1.7), by the proof of Proposition 10.7. Given classes
$\gamma_{1},\gamma_{2}\in H^{*}(Y,{\mathbb{Q}})$, we thus have

(10.5)
$\phi(\phi^{-1}(\gamma_{1})\cup\phi^{-1}(\gamma_{2}))=\gamma_{1}\star\gamma_{2}.$

The isomorphism $\mu_{t}$ is algebraic, since the Lefschetz standard
conjecture holds for $Y$, by [CM], and so the Künneth factors of the diagonal
in $H^{i}(Y,{\mathbb{Q}})\otimes H^{4n-i}(Y,{\mathbb{Q}})$, $0\leq i\leq 4n$,
are algebraic, by [K, Prop. 1.4.4]. Hence, the isomorphism $\phi$ is induced
by an algebraic correspondence. Now
$\phi^{-1}:H^{*}(Y,{\mathbb{Q}})\rightarrow H^{*}(X,{\mathbb{Q}})$ is equal to
$[\kappa(E^{*})\sqrt{td_{Y\times X}}]_{*}\circ\mu_{t^{-1}}$ and is algebraic
as well. Denote by $(\phi^{-1})^{t}:H^{*}(X,{\mathbb{Q}})\rightarrow
H^{*}(Y,{\mathbb{Q}})$ the isomorphism induce by the class in $H^{*}(X\times
Y,{\mathbb{Q}})$, which is the transpose of the class in $H^{*}(Y\times
X,{\mathbb{Q}})$ inducing $\phi^{-1}$. Then
$(\phi^{-1})^{t}\times(\phi^{-1})^{t}\times\phi:H^{*}(X^{3},{\mathbb{Q}})\rightarrow
H^{*}(Y^{3},{\mathbb{Q}})$ is an algebraic correspondence taking the the class
$[\Delta]$ of the diagonal to an algebraic class ${\rm Pont}$ inducing the
Pontryagin product. The latter statement is equivalent to the equality

$\phi\left(\pi_{3,*}\left(\pi_{1}^{*}(\phi^{-1}(\gamma_{1}))\cup\pi_{2}^{*}(\phi^{-1}(\gamma_{2}))\cup[\Delta]\right)\right)=\pi_{3,*}\left(\pi_{1}^{*}(\gamma_{1})\cup\pi_{2}^{*}(\gamma_{2})\cup{\rm
Pont}\right),$

by Equation (10.5). The above equation is verified as follows. Let $a,b,c$ be
classes in $H^{*}(X,{\mathbb{Q}})$.

$\displaystyle\phi\left(\pi_{3,*}\left(\pi_{1}^{*}[\phi^{-1}(\gamma_{1})\cup
a]\cup\pi_{2}^{*}[\phi^{-1}(\gamma_{2})\cup b]\cup\pi_{3}^{*}c\right)\right)$
$\displaystyle=$ $\displaystyle\left(\int_{X}\phi^{-1}(\gamma_{1})\cup
a\right)\left(\int_{X}\phi^{-1}(\gamma_{2})\cup b\right)\phi(c)$
$\displaystyle=$
$\displaystyle\left(\int_{X}\gamma_{1}\cup(\phi^{-1})^{t}(a)\right)\left(\int_{X}\gamma_{2}\cup(\phi^{-1})^{t}(b)\right)\phi(c)$
$\displaystyle=$
$\displaystyle\pi_{3,*}\left(\pi_{1}^{*}(\gamma_{1})\cup\pi_{2}^{*}(\gamma_{2})\cup\left((\phi^{-1})^{t}\otimes(\phi^{-1})^{t}\otimes\phi\right)(\pi_{1}^{*}(a)\cup\pi_{2}^{*}(b)\cup\pi_{3}^{*}(c))\right).$

Finally, we observe that $[\kappa(E^{*})\sqrt{td_{Y\times
X}}]_{*}\circ\mu_{t^{-1}}=\mu_{t}\circ[\kappa(E^{*})\sqrt{td_{Y\times
X}}]_{*},$ as $[\kappa(E^{*})\sqrt{td_{Y\times X}}]_{*}$ is degree reversing,
and $\kappa(E^{*})=\kappa(E)$, by Corollary 4.2. Hence,
$\phi=(\phi^{-1})^{t}$. ∎

##  11\. Acknowledgements

This work was partially supported by a grant from the Simons Foundation
(#427110). Part of this work was carried out while the author was visiting the
Mathematics Institute at Bonn University. I thank Daniel Huybrechts for the
invitation, his hospitality, and for stimulating conversations. I thank Georg
Oberdieck for his comments on an earlier draft of this paper. I thank Chenyu
Bai for pointing out the need for Lemma 5.3.

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