$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}$. This is the complex of
graded $\mathcal{S}_{B}$-modules with terms

$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}^{i}=\bigoplus_{p+q=i}\pi_{\ast}\mathcal{A}_{M}^{n+p,n+q}\otimes_{\mathscr{O}_{M}}\operatorname{Sym}^{\bullet+p}(\mathscr{T}_{B}),$

and with differential (in local coordinates)

$d_{\pi}\bigl{(}\alpha\otimes P\bigr{)}=\bar{\partial}\alpha\otimes
P+\sum_{j=1}^{n}\pi^{\ast}(\mathit{dt}_{j})\wedge\alpha\otimes\partial_{j}P.$

From Saito’s theory, we get an isomorphism

$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}\cong\bigoplus_{i=-n}^{n}\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}[-i]$

in the derived category of graded $\mathcal{S}_{B}$-modules; here the
decomposition is induced by the one in the decomposition theorem. The
calculation in 2 shows that
$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}\cong\mathbf{L}_{B}(M,d)$,
where $(M,d)$ is the complex of graded $\Omega_{B}$-modules with

$M_{k}^{i}=\pi_{\ast}\mathcal{A}_{M}^{n+k,n+i}\quad\text{and}\quad
d=(-1)^{k}\bar{\partial}.$

The $\Omega_{B}$-module structure is the obvious one: a holomorphic form
$\beta\in\Omega_{B}^{j}$ acts via wedge product with the pullback
$\pi^{\ast}\beta$. Since the BGG correspondence is an equivalence of
categories, this means that we have an isomorphism

$G\cong(M,d)$

in the derived category of graded $\Omega_{B}$-modules. This says in
particular that the complex $(M,d)$ splits in the derived category (because of
the decomposition theorem), which is an extremely deep fact about proper
morphisms between Kähler manifolds.

####  94 .

Note that $G$ is also related to the direct sum of all the sheaves of
holomorphic forms on $M$, because Saito’s formula in (39.1) gives us an
isomorphism

$G\cong\bigoplus_{k=-n}^{n}\mathbf{R}\pi_{\ast}\Omega_{M}^{n+k}[n]$

in the derived category of graded $\mathscr{O}_{B}$-modules.

####  95 .

Now we introduce the second object. This is the complex

$G_{v}=\bigoplus_{i,k}G_{i,k}[k],$

in which the summand $G_{i,k}[k]$ sits in graded degree $-k$. As a complex of
$\mathscr{O}_{B}$-modules, this is of course isomorphic to $G$, but the
grading and the $\Omega_{B}$-module structure are different. We turn $G_{v}$
into a complex of graded $\Omega_{B}$-modules in the following way. Recall
that we have the Weil element $\mathsf{w}_{\sigma}$, associated to the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation on
$\bigoplus_{k}\Omega_{M}^{n+k}$. It determines an automorphism of the complex
$(M,d)$ that maps $M_{k}^{i}$ isomorphically to $M_{-k}^{i}$. According to 7,
for any local section $\beta$ of $\Omega_{B}^{1}$, it also exchanges the
action by the holomorphic form $\pi^{\ast}\beta$ agains the action by the
holomorphic vector field $v(\beta)$. In other words,

$\mathsf{w}_{\sigma}\colon G\to G_{v}$

is an isomorphism between $G$ and $G_{v}$, as complexes of graded
$\mathscr{O}_{B}$-modules. Note that $\mathsf{w}_{\sigma}$ respects the
grading, but there is no reason why it should preserve the individual summands
$G_{i,k}[k]$ in the decompositions of $G$ and $G_{v}$.

####  96 .

We use this isomorphism to give $G_{v}$ the structure of a complex of graded
$\Omega_{B}$-modules. 7 tells us that we have, for every local section
$\beta\in\Omega_{B}^{1}$, a commutative diagram

${G}$${G_{v}}$${G}$${G_{v}.}$$\scriptstyle{\mathsf{w}_{\sigma}}$$\scriptstyle{\beta}$$\scriptstyle{v(\beta)}$$\scriptstyle{\mathsf{w}_{\sigma}}$

The $\Omega_{B}$-module structure on $G_{v}$ is therefore the unique one for
which a locally defined holomorphic form $\beta\in\Omega_{B}^{1}$ acts via the
collection of graded morphisms

$v(\beta)\colon G_{i,k}[k]\to G_{i-1,k-1}[k-1].$

Let me stress that this step of the construction works on the level of smooth
forms, meaning on the complex $(M,d)$, because the symplectic form is a
holomorphic form of type $(2,0)$. Neither Hodge theory nor the decomposition
theorem are needed here.

####  97 .

Let us now describe the third object. This is the complex of graded
$\mathscr{O}_{B}$-modules

$G_{f}=\bigoplus_{i,k}G_{i,k}[i+k]$

in which the term $G_{i,k}[k]$ sits in graded degree $i-k$. The reason for
this choice of grading is the following. The topmost component of the Kähler
form gives us a morphism of complexes $\omega_{2}\colon G_{i,k}\to
G_{i+2,k+1}[2]$; from the resulting representation of the Lie algebra
$\mathfrak{sl}_{2}(\mathbb{C})$, we get a second Weil element
$\mathsf{w}_{\omega_{2}}$. For every $i,k\in\mathbb{Z}$, it induces an
isomorphism of complexes

$\mathsf{w}_{\omega_{2}}\colon G_{i,k}[k]\to G_{-i,k-i}[k-2i].$

The resulting isomorphism of complexes of $\mathscr{O}_{B}$-modules

$\mathsf{w}_{\omega_{2}}\colon G_{v}\to G_{f}$

therefore respects the grading exactly when we put the summand $G_{i,k}[i+k]$
in the complex $G_{f}$ in graded degree $i-k$. Unlike the other isomorphism,
this one cannot be defined on the level of smooth forms, because
$\mathsf{w}_{\omega_{2}}$ does not commute with the operator $\bar{\partial}$;
instead, we have to rely on difficult results from Hodge theory (such as the
relative Hard Lefschetz theorem) to do the work for us.

####  98 .

As in the previous step, we use the isomorphism $\mathsf{w}_{\omega_{2}}\colon
G_{v}\to G_{f}$ to turn the object $G_{f}$ into a complex of graded
$\Omega_{B}$-modules. We can again describe the $\Omega_{B}$-module structure
on $G_{f}$ very concretely with the help of the results from Section 6. Recall
from 12 that the operator $v(\beta)$ is primitive of weight $-1$ with respect
to the representation of $\mathfrak{sl}_{2}(\mathbb{C})$ determined by
$\omega_{2}$. Consequently, we have

$\operatorname{Ad}\mathsf{w}_{\omega_{2}}\bigl{(}v(\beta)\bigr{)}=\mathsf{w}_{\omega_{2}}v(\beta)\,\mathsf{w}_{\omega_{2}}^{-1}=[\omega_{2},v(\beta)]=f(\beta)_{1}.$

For every local section $\beta$ of $\Omega_{B}^{1}$, we therefore get another
commutative diagram

${G_{v}}$${G_{f}}$${G_{v}}$${G_{f}.}$$\scriptstyle{\mathsf{w}_{\omega_{2}}}$$\scriptstyle{v(\beta)}$$\scriptstyle{f(\beta)_{1}}$$\scriptstyle{\mathsf{w}_{\omega_{2}}}$

It follows that the $\Omega_{B}$-module structure on the complex $G_{f}$ is
the unique one for which a local section $\beta$ of $\Omega_{B}^{1}$ acts via
the collection of graded morphisms

$f(\beta)_{1}\colon G_{i,k}[i+k]\to G_{i+1,k}[i+k+1].$

This is compatible with the grading (because the left-hand side sits in graded
degree $i-k$ and the right-hand side in graded degree $i-k+1$.)

####  99 .

The rest of the proof consists mostly in applying the BGG correspondence for
the two graded algebras $\mathcal{S}_{B}=\operatorname{Sym}(\mathscr{T}_{B})$
and $\Omega_{B}=\bigoplus_{j}\Omega_{B}^{j}$. Recall that we have

$G=\mathbf{R}_{B}\left(\bigoplus_{i=-n}^{n}\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}[-i]\right).$

By construction, $\mathsf{w}_{\omega_{2}}\mathsf{w}_{\sigma}\colon G\to G_{f}$
is an isomorphism in the derived category of graded $\Omega_{B}$-modules; the
$\Omega_{B}$-module structure on $G_{f}$ has the property that a local section
$\beta$ of $\Omega_{B}^{1}$ acts via the operator $f(\beta)_{1}$. Since
$f(\beta)_{1}\colon G_{i,k}[i]\to G_{i+1,k}[i+1]$ only changes the index $i$,
it is obvious that $G_{f}$ decomposes, as a complex of graded
$\Omega_{B}$-modules, into a direct sum

$G_{f}=\bigoplus_{k=-n}^{n}\left(\bigoplus_{i=-n}^{n}G_{i,k}[i]\right)[k].$

Let $F_{-k}$ denote the complex of graded $\mathcal{S}_{B}$-modules that the
BGG correspondence associates to the $k$-th summand in this decomposition; in
symbols,

$F_{-k}=\mathbf{L}_{B}\left(\bigoplus_{i=-n}^{n}G_{i,k}[i]\right).$

Since the BGG correspondence is an equivalence of categories (by 10), we
conclude from the isomorphism $G\cong G_{f}$ in the derived category of graded
$\Omega_{B}$-modules that

$\bigoplus_{i=-n}^{n}\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}[-i]\cong\bigoplus_{k=-n}^{n}F_{k}[-k],$
(99.1)

in the derived category of graded $\mathcal{S}_{B}$-modules.

####  100 .

What we need to do now is to prove that
$F_{k}\cong\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{k}$ for all
$k=-n,\dotsc,n$. Since we know next to nothing about the complexes $F_{k}$,
this may seem impossible – but in fact, we have just enough information to
make it work.

####  101 .

The left-hand side of (99.1) is a direct sum of graded
$\mathcal{S}_{B}$-modules, each of which is an $n$-dimensional Cohen-Macaulay
module. It follows that each complex $F_{k}$ also splits into a direct sum of
$n$-dimensional Cohen-Macaulay modules; consequently, we get a decomposition

$F_{k}\cong\bigoplus_{\ell\in\mathbb{Z}}F_{k,\ell}[-\ell]$

in the derived category of graded $\mathcal{S}_{B}$-modules. Each $F_{k,\ell}$
is a graded $\mathcal{S}_{B}$-module that is $n$-dimensional and Cohen-
Macaulay. Our task now becomes showing that $F_{k,\ell}=0$ for $\ell\neq 0$.
It turns out that this is a purely formal consequence of what we know about
the complexes $G_{i,k}$. On the one hand, the bound on the amplitude of the
complex $G_{i,k}$ (in 3) implies that $F_{k}$ is concentrated in nonpositive
degrees.

######  Lemma 13.

We have $F_{k,\ell}=0$ for $\ell>0$.

###### Proof.

Recall that we defined $F_{k}$ with the help of the BGG correspondence as

$F_{k}=\mathbf{L}_{B}\left(\bigoplus_{i=-n}^{n}G_{i,-k}[i]\right).$

According to 3, we have $\mathcal{H}^{j}G_{i,k}=0$ for $j>k$, and therefore
$\mathcal{H}^{j}\bigl{(}G_{i,-k}[i]\bigr{)}=\mathcal{H}^{i+j}G_{i,-k}=0$ for
$i+j+k>0$. Since the summand $G_{i,-k}[i]$ has degree $i+k$ with respect to
the grading on $G_{f}$ (and therefore on the object in parentheses), this is
exactly what we need in order to apply 11. The conclusion is that
$\mathcal{H}^{j}F_{k}=0$ for $j>0$. ∎

####  102 .

We can use duality to prove the vanishing of the graded
$\mathcal{S}_{B}$-modules $F_{k,\ell}$ for $\ell<0$. For the sake of clarity,
let us temporarily write

$G_{k}=\bigoplus_{i=-n}^{n}G_{i,-k}[i]$

for the complex of graded $\Omega_{B}$-modules on the right-hand side; then
$F_{k}=\mathbf{L}_{B}(G_{k})$. Recall from (43.1) that
$G_{-i,-k}\cong\mathbf{R}\mathcal{H}\mathit{om}_{\mathscr{O}_{B}}\bigl{(}G_{i,k},\omega_{B}[n]\bigr{)}$.
Therefore

$\widehat{G}_{k}=\mathbf{R}\mathcal{H}\mathit{om}_{\mathscr{O}_{B}}\Bigl{(}G_{k},\omega_{B}[n]\Bigr{)}\cong\bigoplus_{i=-n}^{n}G_{i,k}[i],$

where the notation $\widehat{G}_{k}$ comes from the proof of 12. Note that the
summand $G_{i,k}[i]$ again ends up having degree $i-k$ with respect to the
induced grading on $\widehat{G}_{k}$.777In fact, it should be the case that
$\widehat{G}_{k}\cong G_{-k}$, but this would be tedious to check, and
fortunately it turns out to be irrelevant for the proof. By the same argument
as in 13, the complex $\mathbf{L}_{B}\bigl{(}\widehat{G}_{k}\bigr{)}$ is
concentrated in degrees $\leq 0$. Now 12 gives

$\displaystyle\mathbf{L}_{B}(\widehat{G}_{k})$
$\displaystyle\cong\mathbf{R}\mathcal{H}\mathit{om}_{\mathcal{S}_{B}}\Bigl{(}\mathbf{L}_{B}(G_{k}),\mathcal{S}_{B}\otimes\omega_{B}[n]\Bigr{)}\cong\omega_{B}\otimes\mathbf{R}\mathcal{H}\mathit{om}_{\mathcal{S}_{B}}\Bigl{(}F_{k},\mathcal{S}_{B}[n]\Bigr{)}$
$\displaystyle\cong\bigoplus_{\ell\leq
0}\omega_{B}\otimes\mathbf{R}\mathcal{H}\mathit{om}_{\mathcal{S}_{B}}\Bigl{(}F_{k,\ell},\mathcal{S}_{B}[n]\Bigr{)}[\ell].$

Since each $F_{k,\ell}$ is an $n$-dimensional Cohen-Macaulay module, the
complex on the right-hand side is concentrated in degrees $\geq 0$. But the
complex on the left-hand side is concentrated in degrees $\leq 0$, and so it
must be that $F_{k,\ell}=0$ for $\ell<0$.

####  103 .

The conclusion is that each complex $F_{k}$ is actually a single graded
$\mathcal{S}_{B}$-module (in cohomological degree $0$). Because of the
isomorphism in (99.1), we then get

$F_{k}\cong\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{k}.$

If we now use the BGG correspondence in the other direction, we find that

$\bigoplus_{i=-n}^{n}G_{i,-k}[i]\cong\bigoplus_{i=-n}^{n}G_{k,i}[i+k]$ (103.1)

are isomorphic in the derived category of graded $\Omega_{B}$-modules. We will
now take a short break from proving 3 and turn to the symplectic relative Hard
Lefschetz theorem (in 1) and the symmetry conjecture of Shen and Yin (in 1).

####  104 .

If we forget the grading and the $\Omega_{B}$-module structure, (103.1)
becomes an isomorphism in the derived category of $\mathscr{O}_{B}$-modules.
After replacing $k$ by $-k$ and using the relative Hard Lefschetz isomorphism
in (48.1), we can put it into the form

$\bigoplus_{i=-n}^{n}G_{i,k}[i+k]\cong\bigoplus_{i=-n}^{n}G_{-k,i}[i]\cong\bigoplus_{i=-n}^{n}G_{k,i+k}[i+2k]\cong\bigoplus_{i=-n}^{n}G_{k,i}[i+k].$

This is true for every $k=-n,\dotsc,n$, and so we are in a position where we
can apply 8. The conclusion is that the morphism

$\sigma_{1}^{k}\colon G_{i,-k}\to G_{i,k}[2k]$

is an isomorphism for every $k\geq 1$; this establishes the “symplectic
relative Hard Lefschetz theorem”. It also follows that the complexes $G_{i,k}$
and $G_{k,i}$ are isomorphic in the derived category, as predicted by the
symmetry conjecture of Shen and Yin.

####  105 .

We can now go back and finish the proof of 3. The relative Hard Lefschetz
theorem gives us a representation of the Lie algebra
$\mathfrak{sl}_{2}(\mathbb{C})$ on the direct sum of all the $G_{i,k}$ (with
appropriate shifts). To simplify the notation, let us denote the three
operators by the symbols $\mathsf{X}_{1},\mathsf{Y}_{1},\mathsf{H}_{1}$, and
the Weil element by the symbol $\mathsf{w}_{1}$. Concretely,

$\mathsf{X}_{1}=\omega_{2}\colon G_{i,k}\to
G_{i+2,k+1}[2]\quad\text{and}\quad\mathsf{Y}_{1}=\mathsf{Y}_{\omega_{2}}\colon
G_{i,k}\to G_{i-2,k-1}[-2],$

$\mathsf{H}_{1}$ acts on $G_{i,k}$ as multiplication by the integer $i$, and
the Weil element is the isomorphism

$\mathsf{w}_{1}=\mathsf{w}_{\omega_{2}}\colon G_{i,k}\to G_{-i,k-i}[-2i].$

Likewise, the symplectic relative Hard Lefschetz theorem gives us a second
representation of $\mathfrak{sl}_{2}(\mathbb{C})$, for which we use the
symbols $\mathsf{X}_{2},\mathsf{Y}_{2},\mathsf{H}_{2}$ and $\mathsf{w}_{2}$.
Once again,

$\mathsf{X}_{2}=\sigma_{1}\colon G_{i,k}\to
G_{i+1,k+2}[2]\quad\text{and}\quad\mathsf{Y}_{2}=\mathsf{Y}_{\sigma_{1}}\colon
G_{i,k}\to G_{i-1,k-2}[-2],$

$\mathsf{H}_{2}$ acts on $G_{i,k}$ as multiplication by the integer $k$, and
the Weil element is the isomorphism

$\mathsf{w}_{2}=\mathsf{w}_{\sigma_{1}}\colon G_{i,k}\to G_{i-k,-k}[-2k].$

We know from 2 that $[\omega_{2},\sigma_{1}]=0$, which translates into the
relation $[X_{1},X_{2}]=0$.

####  106 .

We can now establish a direct relationship between
$\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}$ and
$\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]$ with the help of the two
$\mathfrak{sl}_{2}(\mathbb{C})$-representations. Recall from 4 that

$\mathbf{R}_{B}(\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i})=\bigoplus_{k=-n}^{n}G_{i,k}[i+k].$

Consider the following chain of isomorphisms

${G_{i,k}}$${G_{-i,k-i}[-2i]}$${G_{-k,i-k}[-2k]}$${G_{k,i}}$$\scriptstyle{\mathsf{w}_{1}}$$\scriptstyle{\mathsf{w}_{2}}$$\scriptstyle{\mathsf{w}_{1}}$

that, very concretely, realizes the symmetry between $G_{i,k}$ and $G_{k,i}$.
(In the drawing in 16, the composition
$\mathsf{w}_{1}\mathsf{w}_{2}\mathsf{w}_{1}$ is exactly reflection in the
third diagonal of the hexagon.)

####  107 .

Let $\beta\in\Omega_{B}^{1}$ be a locally defined holomorphic $1$-form. From
the construction in Section 6, we get three operators (that are defined on the
same open set as $\beta$, of course):

$\displaystyle\pi^{\ast}\beta=(\pi^{\ast}\beta)_{0}$ $\displaystyle\colon
G_{i,k}\to G_{i,k+1}[1]$ $\displaystyle v(\beta)=v(\beta)_{-1}$
$\displaystyle\colon G_{i,k}\to G_{i-1,k-1}[-1]$ $\displaystyle
f(\beta)_{1}=[X_{1},v(\beta)]$ $\displaystyle\colon G_{i,k}\to G_{i,k+1}[1]$

Let us see how these three operators interact with the two Weil elements
$\mathsf{w}_{1}$ and $\mathsf{w}_{2}$. We know that
$\operatorname{ad}X_{1}(\pi^{\ast}\beta)=0$ (from 11), which implies that

$\operatorname{Ad}\mathsf{w}_{1}\bigl{(}\pi^{\ast}\beta\bigr{)}=\mathsf{w}_{1}\bigl{(}\pi^{\ast}\beta\bigr{)}\mathsf{w}_{1}^{-1}=\pi^{\ast}\beta.$

We also showed (in 12) that $(\operatorname{ad}X_{1})^{2}v(\beta)=0$; because
$v(\beta)$ has weight $-1$ with respect to $\mathsf{H}_{1}$, it follows that

$\operatorname{Ad}\mathsf{w}_{1}\bigl{(}v(\beta)\bigr{)}=[X_{1},v(\beta)]=f(\beta)_{1}.$

Lastly, we know from 7 that $[\sigma,v(\beta)]=\pi^{\ast}\beta$, and because
$v(\beta)$ and $\pi^{\ast}\beta$ have weight $-1$ and $0$ with respect to
$\mathsf{H}_{1}$, we get
$[X_{2},v(\beta)]=[\sigma_{1},v(\beta)]=\pi^{\ast}\beta$. At the same time,
$\pi^{\ast}\beta$ has weight $1$ with respect to $\mathsf{H}_{2}$ and commutes
with $X_{2}$, and therefore

$\operatorname{Ad}\mathsf{w}_{2}\bigl{(}\pi^{\ast}\beta\bigr{)}=-[\mathsf{Y}_{2},\pi^{\ast}\beta]=v(\beta).$

####  108 .

The conclusion of all these computations is that we get a commutative diagram

${G_{i,k}}$${G_{-i,k-i}[-2i]}$${G_{-k,i-k}[-2k]}$${G_{k,i}}$${G_{i,k+1}[1]}$${G_{-i,k-i+1}[-2i+1]}$${G_{-k-1,i-k-1}[-2k-1]}$${G_{k+1,i}[1]}$$\scriptstyle{\mathsf{w}_{1}}$$\scriptstyle{\pi^{\ast}\beta}$$\scriptstyle{\mathsf{w}_{2}}$$\scriptstyle{\pi^{\ast}\beta}$$\scriptstyle{\mathsf{w}_{1}}$$\scriptstyle{v(\beta)}$$\scriptstyle{f(\beta)_{1}}$$\scriptstyle{\mathsf{w}_{1}}$$\scriptstyle{\mathsf{w}_{2}}$$\scriptstyle{\mathsf{w}_{1}}$

in which all the horizontal morphisms are isomorphisms. After shifting
everything by $i+k$ and taking the direct sum over all $k\in\mathbb{Z}$, this
gives us an isomorphism between

$\mathbf{R}_{B}(\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i})=\bigoplus_{k=-n}^{n}G_{i,k}[i+k]$

and the complex of graded $\Omega_{B}$-modules

$\bigoplus_{k=-n}^{n}G_{k,i}[i+k],$ (108.1)

in the derived category of graded $\Omega_{B}$-modules. Here the
$\Omega_{B}$-module structure on the second complex is the unique one where
$\beta\in\Omega_{B}^{1}$ acts through the collection of graded morphisms

$f(\beta)_{1}\colon G_{k,i}[i+k]\to G_{k+1,i}[i+k+1],$

and the grading has the summand $G_{k,i}[i+k]$ in graded degree $k$. The
argument above shows that this object is isomorphic, via
$\mathsf{w}_{1}\mathsf{w}_{2}\mathsf{w}_{1}$, to the object

$\mathbf{R}_{B}(\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i})=\bigoplus_{k=-n}^{n}G_{i,k}[i+k],$

which is the image of $\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}$ under
the BGG correspondence.

####  109 .

To conclude the proof, we only need to describe how the object in (108.1) is
related to $\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]$. In the derived category
of $\mathscr{O}_{B}$-modules, we do have an isomorphism

$\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]\cong\bigoplus_{k=-n}^{n}G_{k,i}[i].$
(109.1)

The object on the left-hand side is naturally a module over the algebra

$\mathbf{R}\pi_{\ast}\mathscr{O}_{M}\cong\bigoplus_{j=0}^{n}\Omega_{B}^{j}[-j],$

where the isomorphism comes from Matsushita’s theorem (in 7). If we take the
associated graded of this action with respect to the perverse filtration,
which is just the filtration by increasing values of $k$ in the above
decomposition, (109.1) becomes an isomorphism of graded modules

$\operatorname{gr}_{\bullet}^{P}\Bigl{(}\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]\Bigr{)}\cong\bigoplus_{k=-n}^{n}G_{k,i}[i].$

Moreover, $\Omega_{B}^{1}[-1]$ now acts on the object on the right-hand side
exactly through the collection of morphisms $f(\beta)_{1}\colon G_{i,k}\to
G_{i+1,k}[1]$. We can turn this into an honest action by $\Omega_{B}$ by
adding a shift by $k$ to the $k$-th term in the sum; in this way, we arrive at
the object

$\bigoplus_{k=-n}^{n}G_{k,i}[i+k]$

with the $\Omega_{B}$-module structure and the grading that appeared in the
proof above.

####  110 .

In other words, we need to take the associated graded of
$\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]$ with respect to the perverse
filtration, in order to extract from the action by $f(\beta)$ its topmost
component $f(\beta)_{1}$. After adding appropriate shifts (determined by the
degree with respect to the grading), we then obtain a complex of graded
$\Omega_{B}$-modules, and under the BGG correspondence, this goes to the
graded $\mathcal{S}_{B}$-module
$\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}$. With this, we have proved
all the claims that were made in the introduction.

####  111 .

The interesting point is that the process we have described does relate
$\mathbf{R}\pi_{\ast}\Omega_{M}^{n+i}[n]$ to the filtered $\mathscr{D}$-module
$(\mathcal{P}_{i},F_{\bullet}\mathcal{P}_{i})$, but only after we take the
associated graded on both sides: with respect to the perverse filtration on
one side, and with respect to the Hodge filtration on the other. I do not know
whether one can expect to relate the two sides without going to the associated
graded objects.

##  9 Symmetries and the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$

####  112 .

In this chapter, we explain why the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ acts on the direct sum of the
complexes $G_{i,k}$ (with suitable shifts). As in Section 6, this relies on
computations with differential forms on $M$; the main point is to understand
how the Weil element $\mathsf{w}_{\sigma}$ for the holomorphic symplectic form
$\sigma$ interacts with the Kähler form $\omega$.

####  113 .

Recall that the holomorphic symplectic form $\sigma$ determines an isomorphism
$\mathscr{T}_{M}\cong\Omega_{M}^{1}$. If we think of the Kähler form $\omega$
as a $\bar{\partial}$-closed $(0,1)$-form with coefficients in
$\Omega_{M}^{1}$, hence as an element $\omega\in A^{0,1}(M,\Omega_{M}^{1})$,
the isomorphism provides us with another element

$i(\omega)\in A^{0,1}(M,\mathscr{T}_{M}),$

which is again $\bar{\partial}$-closed. We can also express the relation
between $\omega$ and $i(\omega)$ as

$i(\omega)\operatorname{\lrcorner}\sigma=\omega.$

More generally, contraction with $i(\omega)$ is an operator

$i(\omega)\operatorname{\lrcorner}\colon A^{p,q}(M)\to A^{p-1,q+1}(M),$

that acts as follows. Let $z_{1},\dotsc,z_{2n}$ be local holomorphic
coordinates on $M$. Then

$i(\omega)=\sum_{j,k}f_{j,k}\frac{\partial}{\partial z_{j}}\otimes
d\bar{z}_{k},$

and we define the contraction against $(p,q)$-forms as

$i(\omega)\operatorname{\lrcorner}\mathit{dz}_{J}\wedge
d\bar{z}_{K}=(-1)^{p}\sum_{j,k}f_{j,k}\operatorname{sgn}(J,j)\mathit{dz}_{J\setminus\\{j\\}}\otimes
d\bar{z}_{k}\wedge d\bar{z}_{K},$

where $\lvert J\rvert=p$ and $\lvert K\rvert=q$. The sign $(-1)^{p}$ is caused
by swapping the order of $d\bar{z}_{k}$ and $\mathit{dz}_{J}$.

####  114 .

The following lemma is proved in exactly the same way as 7.

######  Lemma 14.

For every $k,q\in\mathbb{Z}$, the following diagram commutes:

${A^{n-k,q}(M)}$${A^{n+k,q}(M)}$${A^{n-k+1,q+1}(M)}$${A^{n+k-1,q+1}(M)}$$\scriptstyle{\omega\wedge}$$\scriptstyle{\mathsf{w}_{\sigma}}$$\scriptstyle{i(\omega)\operatorname{\lrcorner}}$$\scriptstyle{\mathsf{w}_{\sigma}}$

####  115 .

Now let us try to understand the commutator of the two operators
$\alpha\mapsto\omega\wedge\alpha$ and $\alpha\mapsto
i(\omega)\operatorname{\lrcorner}\alpha$. We notice that the contraction

$\Theta=-i(\omega)\operatorname{\lrcorner}\omega\in A^{0,2}(M)$

is a $\bar{\partial}$-closed $(0,2)$-form. The following lemma shows that the
commutator in question is nothing but wedge product with $\Theta$.

######  Lemma 15.

We have
$\bigl{[}\omega\wedge,i(\omega)\operatorname{\lrcorner}\bigr{]}=\Theta\wedge$.

###### Proof.

This is due to the identity

$i(\omega)\operatorname{\lrcorner}\bigl{(}\omega\wedge\alpha\bigr{)}=\bigl{(}i(\omega)\operatorname{\lrcorner}\omega\bigr{)}\wedge\alpha+\omega\wedge\bigl{(}i(\omega)\operatorname{\lrcorner}\alpha\bigr{)}=-\Theta\wedge\alpha+\omega\wedge\bigl{(}i(\omega)\operatorname{\lrcorner}\alpha\bigr{)},$

which is easily proved by a computation in local coordinates. ∎

####  116 .

Because $\Theta=-i(\omega)\operatorname{\lrcorner}\omega$ is a
$\bar{\partial}$-closed $(0,2)$-form, it acts on the complex
$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}$. Using the decomposition
theorem, it therefore determines a morphism (in the derived category)

$\Theta\colon\bigoplus_{i=-n}^{n}\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}[-i]\to\bigoplus_{i=-n}^{n}\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}[2-i].$

As in earlier chapters, this morphism breaks up into a finite sum
$\Theta=\Theta_{2}+\Theta_{1}+\Theta_{0}+\dotsb$; each component $\Theta_{j}$
is a morphism

$\Theta_{j}\colon\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i}\to\operatorname{gr}_{\bullet}^{F}\mathcal{P}_{i+j}[2-j]$

in the derived category of graded $\operatorname{Sym}\mathscr{T}_{B}$-modules.
We get induced morphisms

$\Theta_{j}\colon G_{i,k}\to G_{i+j,k}[2].$

What matters for our computation is that $\Theta_{j}=0$ for $j\geq 3$. This
holds because $\Theta$ is a $\bar{\partial}$-closed $(0,2)$-form and therefore
commutes with the differential in the complex
$\operatorname{gr}_{\bullet}^{F}\mathcal{C}_{\pi}$.

####  117 .

We now consider how $i(\omega)$ and $\Theta$ act on the complex $(M,d)$, where
$M_{k}^{i}=\pi_{\ast}\mathcal{A}_{M}^{n+k,n+i}$ and
$d=(-1)^{k}\bar{\partial}$. Recall that this complex is isomorphic (in the
derived category) to

$G=\bigoplus_{i,k=-n}^{n}G_{i,k}[k]\cong\bigoplus_{k=-n}^{n}\mathbf{R}\pi_{\ast}\Omega_{M}^{n+k}[n],$

but now we purposely forget the action by $\Omega_{B}$, because contraction
with $i(\omega)$ does not preserve it. Contraction against $i(\omega)$ and
wedge product with $\Theta$ define two morphisms

$i(\omega)\colon G\to G(-1)[1]\quad\text{and}\quad\Theta\colon G\to G[2]$

in the derived category; the first morphism changes the grading, but the
second one preserves it. The analysis in the previous two paragraphs shows
that

$\Theta=-\bigl{[}i(\omega),\omega\bigr{]}=-\bigl{[}\mathsf{w}_{\sigma}\cdot\omega\cdot\mathsf{w}_{\sigma}^{-1},\omega\bigr{]},$

where $\mathsf{w}_{\sigma}$ is the Weil element for the representation of
$\mathfrak{sl}_{2}(\mathbb{C})$ on

$\bigoplus_{k=-n}^{n}\Omega_{M}^{n+k}$

coming from the action of the symplectic form $\sigma$. For the sake of
clarity, let us denote the third generator of this representation by the
symbol $\mathsf{Y}_{\sigma}$.

####  118 .

As $[\sigma,\omega]=0$, the operator $\omega$ is primitive with respect to
this representation, and so

$\mathsf{w}_{\sigma}\cdot\omega\cdot\mathsf{w}_{\sigma}^{-1}=-[\mathsf{Y}_{\sigma},\omega].$

After combining this with the formula for $\Theta$, we get

$\Theta=-\bigl{[}\mathsf{w}_{\sigma}\cdot\omega\cdot\mathsf{w}_{\sigma}^{-1},\omega\bigr{]}=\bigl{[}[\mathsf{Y}_{\sigma},\omega],\omega\bigr{]}=\bigl{[}\omega,[\omega,\mathsf{Y}_{\sigma}]\bigr{]}.$

Now we look at the topmost component on each side, with respect to the
filtration by the first index $i$. On the right-hand side, by 17, this is the
iterated commutator

$\Bigl{[}\omega_{2},\bigl{[}\omega_{2},\mathsf{Y}_{\sigma_{1}}\bigr{]}\Bigr{]},$

which is a morphism $G_{i,k}\to G_{i+3,k}[2]$. The corresponding term on the
left-hand side is $\Theta_{3}$, and we already know that $\Theta_{3}=0$. This
observation proves the following lemma.

######  Lemma 16.

We have $\bigl{[}\omega_{2},[\omega_{2},\mathsf{Y}_{\sigma_{1}}]\bigr{]}=0$,
and therefore $[\mathsf{Y}_{\omega_{2}},\mathsf{Y}_{\sigma_{1}}]=0$.

###### Proof.

The first assertion is clear. Since $\mathsf{Y}_{\sigma_{1}}\colon G_{i,k}\to
G_{i-1,k-2}[-2]$, this is saying that $\mathsf{Y}_{\sigma_{1}}$ is primitive
of weight $-1$ (with respect to the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation determined by $\omega_{2}$),
which gives the second assertion. ∎

####  119 .

We can now show that the two $\mathfrak{sl}_{2}(\mathbb{C})$-representations
determined by $\omega_{2}$ and $\sigma_{1}$ can be combined into a single
representation of the larger Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$. Let us briefly review its
structure. The Lie algebra $\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ is
associated to the Dynkin diagram of type $\mathsf{A}_{2}$, and so it has two
simple roots, and the resulting Cartan matrix is

$\begin{pmatrix}a_{1,1}&a_{1,2}\\\
a_{2,1}&a_{2,2}\end{pmatrix}=\begin{pmatrix}2&-1\\\ -1&2\end{pmatrix}.$

According to a theorem by Serre,
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ is therefore generated as a Lie
algebra by six elements $\mathsf{e}_{1},\mathsf{f}_{1},\mathsf{h}_{1}$,
$\mathsf{e}_{2},\mathsf{f}_{2},\mathsf{h}_{2}$, subject to the following
relations:

  1. 1.

$[\mathsf{h}_{i},\mathsf{h}_{j}]=0$,
$[\mathsf{h}_{i},\mathsf{e}_{j}]=a_{i,j}\mathsf{e}_{j}$, and
$[\mathsf{h}_{i},\mathsf{f}_{j}]=-a_{i,j}\mathsf{f}_{j}$

  2. 2.

$[\mathsf{e}_{i},\mathsf{f}_{j}]=\delta_{i,j}\mathsf{h}_{j}$, where
$\delta_{i,j}=1$ if $i=j$, and $0$ otherwise

  3. 3.

$(\operatorname{ad}\mathsf{e}_{i})^{1-a_{i,j}}\mathsf{e}_{j}=0$ and
$(\operatorname{ad}\mathsf{f}_{i})^{1-a_{i,j}}\mathsf{f}_{j}=0$.

####  120 .

The point is that the operators $\omega_{2}$ and $\sigma_{1}$ satisfy the
Serre relations for $\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$.

######  Proposition 3.

If we define $\mathsf{e}_{1}=\omega_{2}$,
$\mathsf{f}_{1}=\mathsf{Y}_{\omega_{2}}$,
$\mathsf{e}_{2}=\mathsf{Y}_{\sigma_{1}}$, and $\mathsf{f}_{2}=\sigma_{1}$, and
let $\mathsf{h}_{1}$ and $\mathsf{h}_{2}$ act on the complex $G_{i,k}$ as
multiplication by $i$ respectively $-k$, then these six operators satisfy the
Serre relations for the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$.

###### Proof.

The relations on the first line hold because $\omega_{2}$ and $\sigma_{1}$
each determine a representation of the Lie algebra
$\mathfrak{sl}_{2}(\mathbb{C})$, and because $\omega_{2}\colon G_{i,k}\to
G_{i+2,k+1}[2]$ and $\sigma_{1}\colon G_{i,k}\to G_{i+1,k+2}[2]$. The
relations on the second line hold because
$[\mathsf{e}_{1},\mathsf{f}_{2}]=[\omega_{2},\sigma_{1}]=0$, and because
$[\mathsf{f}_{1},\mathsf{e}_{2}]=[\mathsf{Y}_{\omega_{2}},\mathsf{Y}_{\sigma_{1}}]=0$
by 16. The relations on the third line now follow from the finite-dimensional
representation theory of $\mathfrak{sl}_{2}(\mathbb{C})$: for example,
$[\mathsf{h}_{1},\mathsf{e}_{2}]=-\mathsf{e}_{2}$ and
$[\mathsf{f}_{1},\mathsf{e}_{2}]=0$ are saying that $\mathsf{e}_{2}$ is
primitive of weight $-1$ with respect to the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation
$\mathsf{e}_{1},\mathsf{f}_{1},\mathsf{h}_{1}$, and consequently
$(\operatorname{ad}\mathsf{e}_{1})^{2}\mathsf{e}_{2}=0$. ∎

####  121 .

Because the operators $\omega_{2}$ and $\sigma_{1}$ each act with a shift, we
need to be a little bit careful if we want to say exactly which object in the
derived category the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ acts on. There are two possible
answers. One possibility is to take the direct sum

$\bigoplus_{i,k,\ell\in\mathbb{Z}}G_{i,k}[\ell].$

All six operators $\mathsf{e}_{1},\mathsf{f}_{1},\mathsf{h}_{1}$ and
$\mathsf{e}_{2},\mathsf{f}_{2},\mathsf{h}_{2}$ clearly act on this object; the
disadvantage is that there are infinitely many nonzero terms. The other
possibility is to look at the direct sum

$\bigoplus_{i,k=-n}^{n}G_{i,k}\Bigl{[}\lfloor\tfrac{2}{3}(i+k)\rfloor\Bigr{]}.$

This only has finitely many nonzero terms, and all six operators act on it;
the only disadvantage is that the extra shift – by the integer part of
$\tfrac{2}{3}(i+k)$ – seems somewhat artificial.

####  122 .

The proof above depends on knowing the topmost component of the operator
$\mathsf{Y}_{\sigma}$. The following lemma, whose proof uses the symplectic
relative Hard Lefschetz theorem (in 1), shows that this is exactly
$\mathsf{Y}_{\sigma_{1}}$.

######  Lemma 17.

The difference $\mathsf{Y}_{\sigma}-\mathsf{Y}_{\sigma_{1}}$ maps $G_{i,k}$
into the sum of the $G_{i+j,k-2}[-2]$ with $j\leq-2$.

###### Proof.

To simplify the notation, let us again denote the operators in the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation determined by $\sigma_{1}$ by
the letters $\mathsf{X}_{2}=\sigma_{1}$, $\mathsf{Y}_{2}$, and
$\mathsf{H}_{2}$. Let us also set $\tau=\mathsf{Y}_{\sigma}$, and expand this
according to its degree in $i$ as $\tau=\sum_{j}\tau_{j}$; the individual
components are operators

$\tau_{j}\colon G_{i,k}\to G_{i+j,k-2}[-2].$

In these terms, the lemma is asserting that $\tau_{j}=0$ for $j\geq 0$, and
that $\tau_{-1}=\mathsf{Y}_{2}$. Let $j$ be the largest integer such that
$\tau_{j}\neq 0$. If $j\geq 0$, then after expanding the relation

$\mathsf{H}_{2}=[\sigma,\tau]=[\mathsf{X}_{2}+\sigma_{0}+\dotsb,\tau_{j}+\tau_{j-1}+\dotsb]$

by degree, we would get $[\mathsf{X}_{2},\tau_{j}]=0$. But this is impossible
because $\tau_{j}$ has weight $-2$ with respect to
$\operatorname{ad}\mathsf{H}_{2}$. The same reasoning shows that
$\mathsf{H}_{2}=[\mathsf{X}_{2},\tau_{-1}]$, and as $\mathsf{Y}_{2}$ is
uniquely determined by $\mathsf{H}_{2}$ and $\mathsf{X}_{2}$, it follows that
$\tau_{-1}=\mathsf{Y}_{2}$. Consequently,
$\tau=\mathsf{Y}_{2}+\tau_{-2}+\dotsb$. ∎

####  123 .

The relations for the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ can also be interpreted nicely
in terms of the reflections given by the two Weil elements $\mathsf{w}_{1}$
and $\mathsf{w}_{2}$.

######  Lemma 18.

We have
$(\operatorname{Ad}\mathsf{w}_{1}\circ\operatorname{Ad}\mathsf{w}_{2}\circ\operatorname{Ad}\mathsf{w}_{1})(\omega_{2})=\sigma_{1}$.

###### Proof.

Recall from 105 that $\mathsf{X}_{1}=\omega_{2}$ and that
$\mathsf{X}_{2}=\sigma_{1}$. The definition of the Weil element shows that
$\operatorname{Ad}\mathsf{w}_{1}(\mathsf{X}_{1})=\mathsf{w}_{1}\mathsf{X}_{1}\mathsf{w}_{1}^{-1}=-\mathsf{Y}_{1}$.
The relations $[\mathsf{Y}_{2},\mathsf{Y}_{1}]=0$ and
$[\mathsf{H}_{2},\mathsf{Y}_{1}]=-\mathsf{Y}_{1}$ are saying that
$\mathsf{Y}_{1}$ is primitive of weight $-1$ with respect to the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation
$\mathsf{X}_{2},\mathsf{Y}_{2},\mathsf{H}_{2}$, and therefore
$\operatorname{Ad}\mathsf{w}_{2}(-\mathsf{Y}_{1})=[\mathsf{X}_{2},-\mathsf{Y}_{1}]=[\mathsf{Y}_{1},\mathsf{X}_{2}]$.
Because $(\operatorname{ad}\mathsf{Y}_{1})^{2}\mathsf{X}_{2}=0$, this is now
primitive of weight $-1$ with respect to the
$\mathfrak{sl}_{2}(\mathbb{C})$-representation
$\mathsf{X}_{1},\mathsf{Y}_{1},\mathsf{H}_{1}$, and so finally

$(\operatorname{Ad}\mathsf{w}_{1}\circ\operatorname{Ad}\mathsf{w}_{2}\circ\operatorname{Ad}\mathsf{w}_{1})(\mathsf{X}_{1})=\operatorname{Ad}\mathsf{w}_{1}\bigl{(}[\mathsf{Y}_{1},\mathsf{X}_{2}]\bigr{)}=\bigl{[}\mathsf{X}_{1},[\mathsf{Y}_{1},\mathsf{X}_{2}]\bigr{]}=\mathsf{X}_{2}.$

This proves the assertion. ∎

##  10 The compact case

####  124 .

We end this paper with a short analysis of the compact case. Most notably, we
prove that the action by the Lie algebra
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$ can be upgraded, in the case
where $M$ and $B$ are compact, to an action by the Lie algebra
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$. Since
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})\cong\operatorname{\mathfrak{so}}_{6}(\mathbb{C})$,
this gives an alternative proof (not relying on the existence of a hyperkähler
metric) for a result by Looijenga-Lunts [LL, §4] and Verbitsky [Verbitsky]:
the cohomology of an irreducible compact hyperkähler manifold with a
Lagrangian fibration carries an action by
$\operatorname{\mathfrak{so}}_{6}(\mathbb{C})$.

####  125 .

We assume from now on that $M$ and $B$ are both compact; in other words, $M$
is a holomorphic symplectic compact Kähler manifold of dimension $2n$, and the
base $B$ of our Lagrangian fibration $\pi\colon M\to B$ is a compact Kähler
manifold of dimension $n$. In this case, the holomorphic symplectic form
$\sigma\in H^{0}(M,\Omega_{M}^{2})$ is automatically closed. If $M$ is simply
connected, then it has a hyperkähler metric (by Yau’s theorem) and $B$ is a
product of projective spaces (by Hwang’s theorem), but we are not assuming
that this is the case.

####  126 .

Let $\lambda\in A^{1,1}(B)$ be a Kähler class on $B$. Recall that we have

$\mathbf{R}\pi_{\ast}\mathbb{Q}_{M}(n)[2n]\cong\bigoplus_{i=-n}^{n}P_{i}[-i],$

and that each $P_{i}$ is a polarizable Hodge module of weight $i$ on $B$.
Since $B$ is a compact Kähler manifold, the cohomology groups $H^{j}(B,P_{i})$
therefore have Hodge structures of weight $i+j$, and we have an isomorphism of
Hodge structures

$H^{2n+j}(M,\mathbb{Q})(n)\cong\bigoplus_{i=-n}^{n}H^{j-i}(B,P_{i}).$

By the Hard Lefschetz theorem (for polarizable Hodge modules on compact Kähler
manifolds), cup product with $\lambda$ determines, for each $j\geq 1$, an
isomorphism

$\lambda^{j}\colon H^{-j}(B,P_{i})\to H^{j}(B,P_{i})(j)$

of Hodge structures of weight $i-j$.

####  127 .

Since $\lambda$ is closed, its pullback $\pi^{\ast}\lambda\in A^{1,1}(M)$ acts
on the complex $\mathcal{C}_{\pi}$ from 34. Let us check that, with respect to
the isomorphism

$\mathcal{C}_{\pi}\cong\bigoplus_{i=-n}^{n}(\mathcal{P}_{i},F_{\bullet}\mathcal{P}_{i})[-i]$

from (34.1), the action by $\pi^{\ast}\lambda$ is diagonal. With our usual
notation, this amounts to saying that
$\pi^{\ast}\lambda=(\pi^{\ast}\lambda)_{0}$, and that the morphism

$(\pi^{\ast}\lambda)_{0}\colon(\mathcal{P}_{i},F_{\bullet}\mathcal{P}_{i})\to(\mathcal{P}_{i},F_{\bullet-1}\mathcal{P}_{i})[2]$

is the one induced by the Kähler form $\lambda\in A^{1,1}(B)$. This is a
consequence of the projection formula: if we view $\lambda$ as a morphism
$\lambda\colon\mathbb{C}_{B}\to\mathbb{C}_{B}[2]$ in the derived category of
constructible sheaves, then the claim is that

$\mathbf{R}\pi_{\ast}\left(\mathbb{C}_{M}\xrightarrow{\pi^{\ast}\lambda}\mathbb{C}_{M}[2]\right)\cong\left(\mathbb{C}_{B}\xrightarrow{\lambda}\mathbb{C}_{B}[2]\right)\otimes\mathbf{R}\pi_{\ast}\mathbb{C}_{M}.$

Using the de Rham resolutions of $\mathbb{C}_{M}$ and $\mathbb{C}_{B}$ by
differential forms, the projection formula in this case amounts to the
following statement.

######  Lemma 19.

The morphism of complexes

$(\mathcal{A}_{B}^{\bullet},d)\otimes(\pi_{\ast}\mathcal{A}_{M}^{\bullet},d)\to(\pi_{\ast}\mathcal{A}_{M}^{\bullet},d),\quad\alpha\otimes\beta\mapsto\pi^{\ast}\alpha\wedge\beta,$

is a quasi-isomorphism.

###### Proof.

Since we are using Deligne’s sign conventions, the tensor product has terms

$\bigoplus_{i+j=k}\mathcal{A}_{B}^{i}\otimes\pi_{\ast}\mathcal{A}_{M}^{j}$

and differential
$d(\alpha\otimes\beta)=d\alpha\otimes\beta+(-1)^{i}\alpha\otimes d\beta$. The
morphism of complexes is

$\bigoplus_{i+j=k}\mathcal{A}_{B}^{i}\otimes\pi_{\ast}\mathcal{A}_{M}^{j}\to\pi_{\ast}\mathcal{A}_{M}^{k},\quad\sum_{i+j=k}\alpha_{i}\otimes\beta_{j}\mapsto\sum_{i+j=k}\pi^{\ast}\alpha_{i}\wedge\beta_{j}.$

This is clearly surjective: if $U\subseteq B$ is open, then the element
$1\otimes\beta\in A^{0}(U)\otimes A^{k}\bigl{(}\pi^{-1}(U)\bigr{)}$ is a
preimage for $\beta\in A^{k}\bigl{(}\pi^{-1}(U)\bigr{)}$. To show that the
morphism is a quasi-isomorphism, one then argues locally with the help of the
Poincaré lemma, which says that $\mathcal{A}_{B}^{\bullet}$ is a resolution of
the constant sheaf $\mathbb{C}_{B}$. ∎

####  128 .

The Hard Lefschetz theorem for the Kähler form $\lambda$ gives us a third
action by the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$. In the rest of this
chapter, we shall argue that all three actions can be combined into an action
of the Lie algebra $\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$. Consider
the vector spaces

$H^{i,j,k}=H^{i+j}(B,G_{i,k})=H^{j}\bigl{(}B,\operatorname{gr}_{-k}^{F}\operatorname{DR}(\mathcal{P}_{i})\bigr{)}$
(128.1)

for $i,j,k\in\mathbb{Z}$. These vector spaces refine the Hodge decomposition
on $M$, taking into account the decomposition theorem for $\pi\colon M\to B$.
Indeed, according to the direct image theorem for the cohomology of the
polarizable Hodge module $P_{i}$ on the compact Kähler manifold $B$, the
$\mathbb{Q}$-vector space $H^{j}(B,P_{i})$ carries a Hodge structure of weight
$i+j$, and

$\operatorname{gr}_{-k}^{F}\bigl{(}H^{j}(B,P_{i})\otimes_{\mathbb{Q}}\mathbb{C}\bigr{)}\cong\operatorname{gr}_{-k}^{F}H^{j}\bigl{(}B,\operatorname{DR}(\mathcal{P}_{i})\bigr{)}\cong
H^{j}\bigl{(}B,\operatorname{gr}_{-k}^{F}\operatorname{DR}(\mathcal{P}_{i})\bigr{)}.$

This leads to the following isomorphism with the Hodge decomposition on
$H^{j}(B,P_{i})$:

$H^{i,j,k}\cong\bigl{(}H^{j}(B,P_{i})\otimes_{\mathbb{Q}}\mathbb{C}\bigr{)}^{k,i+j-k}$
(128.2)

In order to relate this to the Hodge decomposition on $M$, we use the
isomorphism

$\mathbf{R}\pi_{\ast}\Omega_{M}^{n+k}[n-k]\cong\bigoplus_{i=-n}^{n}G_{i,k}$

from (39.1). Substituting in the definition of $H^{i,j,k}$, we see immediately
that

$H^{n+k,n+j}(M)\cong
H^{n+j}(M,\Omega_{M}^{n+k})\cong\bigoplus_{i=-n}^{n}H^{j+k}(B,G_{i,k})=\bigoplus_{i=-n}^{n}H^{i,j+k-i,k}.$
(128.3)

The vector spaces $H^{i,j,k}$ also interact with duality in the expected way.
Indeed, because of (43.1), the Grothendieck dual of $G_{i,k}$ is isomorphic to
$G_{-i,-k}$, and so

$\operatorname{Hom}_{\mathbb{C}}\bigl{(}H^{i,j,k},\mathbb{C}\bigr{)}\cong
H^{-i,-j,-k}.$ (128.4)

####  129 .

The discussion in the preceding paragraph leads to the following concrete
interpretation for the three indices $i,j,k$, similar to what happens for the
complexes $G_{i,k}$:

  1. 1.

The first index $i$ records the cohomological degree along the fibers of
$\pi$, in the sense that $H^{i,j,k}$ is associated with the $(n+i)$-th
cohomology groups of the fibers.

  2. 2.

The second index $j$ records the cohomological degree along the base $B$, in
the sense that $H^{i,j,k}$ is associated with the $(n+j)$-th cohomology groups
on $B$.

  3. 3.

The third index $k$ records the holomorphic degree, in the sense that
$H^{i,j,k}$ is associated with the sheaf $\Omega_{M}^{n+k}$ of holomorphic
forms of degree $(n+k)$.

####  130 .

The Kähler form $\omega\in A^{1,1}(M)$ induces a morphism

$\omega_{2}\colon H^{i,j,k}=H^{i+j}(B,G_{i,k})\to
H^{i+j+2}(B,G_{i+2,k+1})=H^{i+2,j,k+1},$

and the relative Hard Lefschetz theorem for $\omega$ becomes an isomorphism

$\omega_{2}^{i}\colon H^{-i,j,k}\to H^{i,j,k+i}\quad\text{for $i\geq 1$.}$
(130.1)

Similarly, the holomorphic symplectic form $\sigma\in H^{0}(M,\Omega_{M}^{2})$
induces a morphism

$\sigma_{1}\colon H^{i,j,k}=H^{i+j}(B,G_{i,k})\to
H^{i+j+2}(B,G_{i+1,k+2})=H^{i+1,j+1,k+2},$

and the symplectic relative Hard Lefschetz theorem (1) becomes an isomorphism

$\sigma_{1}^{k}\colon H^{i,j,-k}\to H^{i+k,j+k,k}\quad\text{for $k\geq 1$.}$
(130.2)

Finally, by (128.2), the Kähler form $\lambda\in A^{1,1}(B)$ induces a
morphism

$\lambda\colon
H^{i,j,k}\cong\bigl{(}H^{j}(B,P_{i})\otimes_{\mathbb{Q}}\mathbb{C}\bigr{)}^{k,i+j-k}\to\bigl{(}H^{j+2}(B,P_{i})\otimes_{\mathbb{Q}}\mathbb{C}\bigr{)}^{k+1,i+j-k+1}\cong
H^{i,j+2,k+1},$

and the Hard Lefschetz theorem for $\lambda$ becomes an isomorphism

$\lambda^{j}\colon H^{i,-j,k}\to H^{i,j,k+j}\quad\text{for $j\geq 1$.}$
(130.3)

####  131 .

From these isomorphisms and 3, one can deduce the following result.

######  Lemma 20.

We have $H^{i,j,k}=0$ unless $\max\bigl{(}\lvert i\rvert,\lvert j\rvert,\lvert
k\rvert,\lvert i-k\rvert,\lvert j-k\rvert,\lvert i+j-k\rvert\bigr{)}\leq n$.

We may arrange the $H^{i,j,k}$ on a three-dimensional grid, by putting the
vector space for the multi-index $(i,j,k)$ at the point with coordinates
$\frac{1}{2}\bigl{(}2k-i-j,i\sqrt{3},j\sqrt{3}\bigr{)}$. In geometric terms,
the conditions in the lemma are describing a _rhombic dodecahedron_ :

The lemma is asserting that all the points corresponding to nonzero
$H^{i,j,k}$ lie inside this rhombic dodecahedron. The example of a projection
$\pi\colon A\times B\to B$ from the product of two $n$-dimensional abelian
varieties shows that this is the best one can expect in general. For
irreducible compact hyperkähler manifolds, Nagai’s conjecture would imply that
the convex hull of the points corresponding to nonzero $H^{i,j,k}$ is actually
an octahedron [LF, §3.8].

####  132 .

The three isomorphisms in (130.1), (130.2), and (130.3) give us three
reflections

$(i,j,k)\to(-i,j,k-i),\quad(i,j,k)\to(i,-j,k-j),\quad(i,j,k)\to(i-k,j-k,-k),$

and a little bit of calculation shows that these three reflections together
generate a total of $24$ symmetries, with a group structure isomorphic to the
symmetric group $S_{4}$. This is not surprising, because $S_{4}$ is the Weyl
group of $\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$. One of the symmetries
is

$(i,j,k)\to(j,i,i+j-k),$

and because of (128.3), it induces an isomorphism $H^{p,q}(M)\cong
H^{q,p}(M)$. It would be interesting if one could explain this particular
symmetry in a geometric way.

####  133 .

As in the previous chapter, we need to establish one additional identity in
order to have an action by the Lie algebra
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$. Once again, the isomorphism
$\mathscr{T}_{M}\cong\Omega_{M}^{1}$ associates to the cohomology class of
$\pi^{\ast}\lambda\in H^{1}(M,\Omega_{M}^{1})$ a new element

$i(\pi^{\ast}\lambda)\in A^{0,1}(M,\mathscr{T}_{M}),$

that is $\bar{\partial}$-closed and satisfies
$i(\pi^{\ast}\lambda)\operatorname{\lrcorner}\sigma=\pi^{\ast}\lambda$. The
following lemma is also proved in the same way as 7.

######  Lemma 21.

For every $k,q\in\mathbb{Z}$, the following diagram commutes:

${A^{n-k,q}(M)}$${A^{n+k,q}(M)}$${A^{n-k+1,q+1}(M)}$${A^{n+k-1,q+1}(M)}$$\scriptstyle{(\pi^{\ast}\lambda)\wedge}$$\scriptstyle{\mathsf{w}_{\sigma}}$$\scriptstyle{i(\pi^{\ast}\lambda)\operatorname{\lrcorner}}$$\scriptstyle{\mathsf{w}_{\sigma}}$

####  134 .

The crucial observation is that the commutator of the two operators
$\alpha\mapsto(\pi^{\ast}\lambda)\wedge\alpha$ and $\alpha\mapsto
i(\omega)\operatorname{\lrcorner}\alpha$ vanishes.

######  Lemma 22.

We have
$\bigl{[}(\pi^{\ast}\lambda)\wedge,i(\pi^{\ast}\lambda)\operatorname{\lrcorner}\bigr{]}=0$.

###### Proof.

As before, the commutator is the wedge product with the
$\bar{\partial}$-closed $(0,2)$-form

$-i(\pi^{\ast}\lambda)\operatorname{\lrcorner}(\pi^{\ast}\lambda)\in
A^{0,2}(M).$

In a nutshell, this vanishes because the vector fields in
$i(\pi^{\ast}\lambda)$ are tangent to the fibers of $\pi$. It is enough to
prove this locally, and so we may assume without loss of generality that

$\lambda=\sum_{j=1}^{n}\mathit{dt}_{j}\wedge\theta_{j},$

where $t_{1},\dotsc,t_{n}$ are local coordinates on $B$ and
$\theta_{1},\dotsc,\theta_{n}$ are $\bar{\partial}$-closed $(0,1)$-forms. Let
$\eta_{j}$ be the unique holomorphic vector field such that
$\pi^{\ast}(\mathit{dt}_{j})=\eta_{j}\operatorname{\lrcorner}\sigma$. Then

$i(\pi^{\ast}\lambda)=\sum_{j=1}^{n}\eta_{j}\otimes\pi^{\ast}\theta_{j},$

and consequently

$-i(\pi^{\ast}\lambda)\operatorname{\lrcorner}(\pi^{\ast}\lambda)=\sum_{j,k=1}^{n}\bigl{(}\eta_{j}\operatorname{\lrcorner}(\pi^{\ast}\mathit{dt}_{k})\bigr{)}\cdot\pi^{\ast}(\theta_{j}\wedge\theta_{k})=0,$

due to the fact that the vector fields $\eta_{j}$ are tangent to the fibers of
$\pi$. ∎

####  135 .

We now consider how $i(\pi^{\ast}\omega)$ acts on the complex $(M,d)$, where
$M_{k}^{i}=\pi_{\ast}\mathcal{A}_{M}^{n+k,n+i}$ and
$d=(-1)^{k}\bar{\partial}$. Contraction against $i(\pi^{\ast}\lambda)$ defines
a morphism.

$i(\pi^{\ast}\lambda)\colon G\to G(-1)[1]$

in the derived category; note that this morphism changes the grading. Since
$\pi^{\ast}\lambda$ and $\sigma$ commute (as $2$-forms on $M$), we have
$i(\pi^{\ast}\lambda)=\mathsf{w}_{\sigma}\cdot(\pi^{\ast}\lambda)\cdot\mathsf{w}_{\sigma}^{-1}=-[\mathsf{Y}_{\sigma},\pi^{\ast}\lambda]$,
and therefore

$\bigl{[}\pi^{\ast}\lambda,[\pi^{\ast}\lambda,\mathsf{Y}_{\sigma}]\bigr{]}=-[i(\pi^{\ast}\lambda),\pi^{\ast}\lambda]=0.$

If we now take the topmost component, and remember that the action by
$(\pi^{\ast}\lambda)_{0}$ is diagonal and equal to $\lambda$, we arrive at the
following result.

######  Lemma 23.

We have $\bigl{[}\lambda,[\lambda,\mathsf{Y}_{\sigma_{1}}]\bigr{]}=0$, and
therefore $[\mathsf{Y}_{\lambda},\mathsf{Y}_{\sigma_{1}}]=0$.

####  136 .

We can finally prove that the three
$\mathfrak{sl}_{2}(\mathbb{C})$-representations determined by $\omega_{2}$,
$\sigma_{1}$, and $\lambda$ can be combined into a single representation of
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$. Recall that the Lie algebra
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$ is associated to the Dynkin
diagram of type $\mathsf{A}_{3}$, and so it has three simple roots, and the
resulting Cartan matrix is

$\begin{pmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\\ a_{2,1}&a_{2,2}&a_{2,3}\\\
a_{3,1}&a_{3,2}&a_{3,3}\end{pmatrix}=\begin{pmatrix}2&-1&0\\\ -1&2&-1\\\
0&-1&2\end{pmatrix}.$

By Serre’s theorem, $\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$ is
generated as a Lie algebra by nine elements
$\mathsf{e}_{1},\mathsf{f}_{1},\mathsf{h}_{1}$,
$\mathsf{e}_{2},\mathsf{f}_{2},\mathsf{h}_{2}$,
$\mathsf{e}_{3},\mathsf{f}_{3},\mathsf{h}_{3}$, subject to the following
relations (which have the same shape as before):

  1. 1.

$[\mathsf{h}_{i},\mathsf{h}_{j}]=0$,
$[\mathsf{h}_{i},\mathsf{e}_{j}]=a_{i,j}\mathsf{e}_{j}$, and
$[\mathsf{h}_{i},\mathsf{f}_{j}]=-a_{i,j}\mathsf{f}_{j}$

  2. 2.

$[\mathsf{e}_{i},\mathsf{f}_{j}]=\delta_{i,j}\mathsf{h}_{j}$, where
$\delta_{i,j}=1$ if $i=j$, and $0$ otherwise

  3. 3.

$(\operatorname{ad}\mathsf{e}_{i})^{1-a_{i,j}}\mathsf{e}_{j}=0$ and
$(\operatorname{ad}\mathsf{f}_{i})^{1-a_{i,j}}\mathsf{f}_{j}=0$.

####  137 .

The proof of the Serre relations is very similar to what we did for
$\operatorname{\mathfrak{sl}}_{3}(\mathbb{C})$, and so we will only sketch the
argument.

######  Proposition 4.

If we define $\mathsf{e}_{1}=\omega_{2}$,
$\mathsf{f}_{1}=\mathsf{Y}_{\omega_{2}}$,
$\mathsf{e}_{2}=\mathsf{Y}_{\sigma_{1}}$, $\mathsf{f}_{2}=\sigma_{1}$,
$\mathsf{e}_{3}=\lambda$, and $\mathsf{f}_{3}=\mathsf{Y}_{\lambda}$, and let
$\mathsf{h}_{1}$, $\mathsf{h}_{2}$ and $\mathsf{h}_{3}$ act on the vector
space $H^{i,j,k}$ respectively as multiplication by $i$, $-k$ and $j$, then
these nine operators satisfy the Serre relations for the Lie algebra
$\operatorname{\mathfrak{sl}}_{4}(\mathbb{C})$.

###### Proof.

Since $\omega$, $\sigma$, and $\pi^{\ast}\lambda$ commute (as $2$-forms on
$M$), it is easy to see that
$[\mathsf{e}_{1},\mathsf{e}_{3}]=[\omega_{2},\lambda]=0$ and
$[\mathsf{f}_{2},\mathsf{e}_{3}]=[\sigma_{1},\lambda]=0$. Because
$[\mathsf{h}_{1},\mathsf{e}_{3}]=0$, it follows that
$[\mathsf{f}_{1},\mathsf{e}_{3}]=0$, and so the two Lie algebras generated by
$\mathsf{e}_{1},\mathsf{f}_{1},\mathsf{h}_{1}$ and
$\mathsf{e}_{3},\mathsf{f}_{3},\mathsf{h}_{3}$ commute; this gives about half
of the necessary relations. Among the remaining new relations, the only
nontrivial ones are $(\operatorname{ad}\mathsf{e}_{3})^{2}\mathsf{e}_{2}=0$
and $[\mathsf{f}_{3},\mathsf{e}_{2}]=0$, and these are of course contained in
23. ∎

##  11 Acknowledgements

####  138 .

I got interested in Lagrangian fibrations during a workshop about hyperkähler
manifolds at the Simons Center in January of 2023, and I am grateful to
Ljudmila Kamenova, Giovanni Mongardi, and Alexei Oblomkov for organizing it.
During the workshop, Junliang Shen gave a series of lectures, and I thank him
for explaining the conjectures, and also for some very useful discussions
afterwards. I thank Davesh Maulik for inviting me to MIT to talk about some
details of the proof. I thank Chris Brav for a pleasant afternoon chatting
about the categorical aspects of the BGG correspondence. Finally, I am
grateful to Daniel Huybrechts and Mirko Mauri for their detailed comments
about a first draft of the paper, as well as for their survey paper [LF] about
Lagrangian fibrations, from which I learned a lot of the general theory.

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