$\eta\in\mathfrak{gl}(\mathbf{T}(M)^{*})$ and $v\in\mathfrak{g}$}.$

If $\operatorname{ad}_{-1}i(\eta)=0$, then $\eta(v)=0$ for any
$v\in\mathfrak{g}$. Since $G$ acts on $M$ transitively, we have

$\operatorname{ev}_{x}(\mathfrak{g})=T_{x}(M)$ for all $x\in M$,

and hence $\eta=0$. Thus, $\operatorname{ad}_{-1}$ is faithful on
$i(\mathfrak{gl}(\mathbf{T}(M)^{*}))$. Let $\mathfrak{a}$ denote the radical
of $i(\mathfrak{gl}(\mathbf{T}(M)^{*}))$. It is non-trivial: it contains
$\langle\varepsilon\rangle=\langle i(\operatorname{id})\rangle$. Since
$\operatorname{ad}_{-1}$ is irreducible, its image is a reductive Lie algebra
with radical
$\langle\operatorname{id}\rangle=\langle\operatorname{ad}_{-1}\varepsilon\rangle$.
By (5.60), $\mathfrak{a}$ coincides with the radical of
$\mathfrak{v}(M,\Omega)_{0}$, and hence
$\mathfrak{a}~{}=\langle\varepsilon\rangle$. It follows that
$i(\mathfrak{gl}(\mathbf{T}(M)^{*}))$ is reductive, and

$i(\mathfrak{gl}(\mathbf{T}(M)^{*}))=\langle\varepsilon\rangle\oplus\mathfrak{s},$

where $\mathfrak{s}$ is a semisimple Lie algebra. We have to prove that
$\mathfrak{s}=0$. Clearly, $\mathfrak{s}$ is invariant under
$\operatorname{ad}_{0}(l(\mathfrak{g}))$, and hence we get the homomorphism

$\operatorname{ad}_{0}(l):\mathfrak{g}\longrightarrow\mathfrak{der}\,\mathfrak{s}=\operatorname{ad}\mathfrak{s}$

which is injective if $[l(\mathfrak{g}),\mathfrak{s}]\neq 0$. In this latter
case, we obtain therefore an injective homomorphism
$h:\mathfrak{g}\longrightarrow\mathfrak{s}$ satisfying

$[h(u),z]=[l(u),z]\ \text{~{}for any $u\in\mathfrak{g},\;z\in\mathfrak{s}$}.$

In particular,

$h([u,v])=[h(u),h(v)]=[l(u),h(v)]\ \text{~{}for any $u,v\in\mathfrak{g}$}.$
(5.61)

Now, we note that
$\mathfrak{gl}(\mathbf{T}(M)^{*})=\Gamma(\mathbf{T}(M)^{*}\otimes\mathbf{T}(M))$
is the vector space of holomorphic sections of the homogeneous vector bundle
$\mathbf{T}(M)^{*}\otimes\mathbf{T}(M)=\mathbf{E}_{\varphi}$, where
$\varphi=\tau^{*}\tau$. From (5.59) we deduce that the induced representation
$\Phi$ of $G$ in $\Gamma(\mathbf{T}(M)^{*}\otimes\mathbf{T}(M))$ satisfies the
following condition

$i(d\Phi(u)\eta)=[l(u),i(\eta)]\ \ \text{~{}for any
$u\in\mathfrak{g},\;\eta\in\mathfrak{gl}(\mathbf{T}(M)^{*})$}.$ (5.62)

Suppose that $[l(\mathfrak{g}),\mathfrak{s}]\neq 0$. Then, eqs. (5.61) and
(5.62) imply that $\operatorname{Im}h$ determines a $G$-submodule of
$\mathfrak{gl}(\mathbf{T}(M)^{*})$, where the adjoint representation of $G$ is
realized. Thus, the highest root $\delta$ is a highest weight of $\Phi$. By
Corollary 5.59, $\delta$ is a highest weight of $\varphi$. But the weights of
$\varphi$ have the form $\alpha-\beta$, where $\alpha,\beta\in\Delta_{+}$.
This yields a contradiction. Thus, we have proved that
$[l(\mathfrak{g}),\mathfrak{s}]=0$. It follows that
$\operatorname{ad}_{-1}(\mathfrak{s})$ commutes with the irreducible linear
Lie algebra $\operatorname{ad}_{-1}l(\mathfrak{g})$. By the Schur lemma,
$\operatorname{ad}_{-1}(\mathfrak{s})=0$, and hence $\mathfrak{s}=0$. ∎

A graded Lie superalgebra of the form

$\mathfrak{v}=\bigoplus_{p\geq 1}\mathfrak{v}_{p}$ (5.63)

is called transitive if for any $p\geq 1$ it satisfies

$\\{x\in\mathfrak{v}_{p}\mid[x,\mathfrak{v}_{-1}]=0\\}=0.$ (5.64)

A graded Lie superalgebra of the form (5.63) is called irreducible if the
representation $\operatorname{ad}_{-1}$ of $\mathfrak{v}_{0}$ is irreducible.
All irreducible transitive complex graded Lie superalgebras of finite
dimension were classified in [24] (see also [48]). [$\mathfrak{v}(G/P,\Omega)$
for $G$ simple]

  1. $1)$

For any flag manifold $M=G/P$ of a simple complex Lie group $G$, the graded
Lie superalgebra $\mathfrak{v}(M,\Omega)$ is transitive and irreducible.

  2. $2)$

Under the above assumptions, suppose that
$\mathfrak{v}(M,\mathcal{O})=\mathfrak{g}$. Then,

$\displaystyle\mathfrak{v}(M,\Omega)_{-1}$ $\displaystyle=i(\mathfrak{g}),$
$\displaystyle\mathfrak{v}(M,\Omega)_{0}$
$\displaystyle=\langle\varepsilon\rangle\oplus l(\mathfrak{g}),$
$\displaystyle\mathfrak{v}(M,\Omega)_{1}$ $\displaystyle=\langle d\rangle,$
$\displaystyle\mathfrak{v}(M,\Omega)_{p}$ $\displaystyle=0\text{ for any
}p\geq 2.$

###### Proof.

By Lemma 5.6, $\mathfrak{v}(M,\Omega)$ is irreducible and satisfies condition
(5.64) for $p=0$. Thus, we need to prove (5.64) for any $p>0$. We will use the
following fact: if $\varphi\in\Omega^{p}$, where $p>0$, and if $i(v)\varphi=0$
for all $v\in\mathfrak{g}$, then $\varphi=0$. To prove this, we note that

$(i(v)\varphi)_{x}(v_{1},\ldots,v_{p})=\varphi_{x}(\operatorname{ev}_{x}(v),v_{1},\ldots,v_{p}),\
\ v_{i}\in T_{x}(M)\text{ for }x\in M.$

Since $\mathfrak{g}$ acts transitively, the condition $i(v)\varphi=0$ yields
$\varphi_{x}=0$ for any $x\in M$. Now, suppose that a vector field
$u\in\mathfrak{v}(M,\Omega)_{p}$ for $p>0$ satisfies $[u,i(v)]=0$ for all
$v\in\mathfrak{g}$. Then, for any $f\in\mathcal{F}$, we have

$[i(v),u](f)=i(v)u(f)=0.$

By the above, we have $u(f)=0$. Therefore, for any $\varphi\in\Omega^{1}$, we
get

$[i(v),u](\varphi)=i(v)u(\varphi)+(-1)^{p+1}u(i(v)(\varphi))=i(v)u(\varphi)=0.$

Since $u(\varphi)\in\Omega^{p+1}$, this implies $u(\varphi)=0$. Thus, $u=0$,
and item (1) is proved. The item (2) for $p=-1,0$ follows from (5.60) and
Lemma 5.6. It is also clear that $\langle
d\rangle\subset\mathfrak{v}(M,\Omega)_{1}$. In particular, we see that the
representation $\operatorname{ad}_{-1}$ of
$\mathfrak{v}(M,\Omega)_{0}\simeq\mathfrak{g}$ is the adjoint one, while the
representation $\operatorname{ad}_{1}$ of this Lie algebra contains a trivial
component of dimension 1. The classification of transitive irreducible graded
Lie superalgebras $\mathfrak{v}$ given in [24, Theorem 4], shows that if
$\mathfrak{v}$ satisfies the above conditions, then $\dim\mathfrak{v}_{1}=1$
and $\mathfrak{v}_{p}=0$ for $p\geq 2$. Thus, item (2) follows from item (1).
∎

###  5.8 A family of non-split supermanifolds

Here, we apply the construction of Subsection 5.2 to the case where $M=G/P$ is
a flag manifold of a simple complex Lie group $G$. We show that in this
situation one always obtains a non-empty family of non-split supermanifolds
having $(M,\Omega)$ as their retract. We also study holomorphic vector fields
on these supermanifolds. Let $M=G/P$ be a flag manifold, where $G$ is simple
and $\dim M>1$, and denote $r:=|\Pi\setminus S|$. Then, there exists a family
of distinct non-split supermanifolds that have $(M,\Omega)$ as their retract,
parametrized by $\mathbb{CP}^{r-1}/\Sigma$. Here $\Sigma$ is the finite group
from (5.57). If $P$ is maximal, then this family consists of a unique
supermanifold, which is isomorphic to the canonical one.

###### Proof.

The group $\operatorname{Aut}\mathbf{T}(M)^{*}$ naturally acts on
$H^{1,1}(M,\mathbb{C})$, and the mapping

$\mu^{*}\hat{\delta}:H^{1,1}(M,\mathbb{C})\longrightarrow
H^{1}(M,\mathcal{A}ut_{(2)}\Omega)$

is equivariant. By Theorems 4.1 and 4.14, this mapping determines a family of
distinct non-split supermanifolds having $(M,\Omega)$ as their retract which
is parametrized by the set
$\left(H^{1,1}(M,\mathbb{C})\setminus\\{0\\}\right)/\operatorname{Aut}\mathbf{T}(M)^{*}$.
On the other hand, $\operatorname{GL}(\mathbf{T}(M)^{*})=\mathbb{C}^{\times}$
(see Lemma 5.6), and (3.31) yields

$\operatorname{Aut}\mathbf{T}(M)^{*}=\mathbb{C}^{\times}\times\operatorname{Bih}M.$

Thanks to (5.57), we see that

$\operatorname{Aut}\mathbf{T}(M)^{*}=\mathbb{C}^{\times}\times(G\rtimes\Sigma).$

Clearly, the action of $G$ on $H^{1,1}(M,\mathbb{C})$ is trivial. Using Lemma
4.40, we deduce that

$(H^{1,1}(M,\mathbb{C})\setminus\\{0\\})/\operatorname{Aut}\mathbf{T}(M)^{*}=(H^{1,1}(M,\mathbb{C})\setminus\\{0\\})/(\mathbb{C}^{\times}\times\Sigma)=\mathbb{P}(H^{1,1}(M,\mathbb{C}))/\Sigma.$

Due to (5.56), this implies our first assertion. To prove the second claim, we
note that the canonical supermanifold corresponding to $M$ is non-split, since
in our case the canonical form $\omega$ is positive-definite [28] and hence
$[\omega]\neq 0$. Thus, it enters the family just constructed. But if $P$ is
maximal, then $r=1$, and $\mathbb{CP}^{r-1}$ contains only one point. ∎

Now we will study holomorphic vector fields on supermanifolds
$(M,\mathcal{O})$ of the family constructed above, applying Proposition 3.1 to
$\mathcal{O}_{\operatorname{gr}}=\Omega$. We have to settle what derivations
of $\Omega$ described in Theorem 5.64 can be lifted to $(M,\mathcal{O})$. We
need the following lemma. Let $M=G/P$ be a flag manifold. Then, both
homomorphisms of the sequence

$H^{1,1}(M,\mathbb{C})\overset{\hat{\delta}}{\longrightarrow}H^{1}(M,\mathcal{Z}\Omega^{1})\overset{\beta^{*}}{\longrightarrow}H^{1}(M,\Omega^{1})$

(see Theorem 4.14) are isomorphisms, and $H^{1}(M,\mathcal{Z}\Omega^{1})$ is a
trivial $G$-module.

###### Proof.

Clearly, we have the exact sequence of sheaves

$0\longrightarrow\mathcal{Z}\Omega^{1}\overset{\beta}{\longrightarrow}\Omega^{1}\overset{d}{\longrightarrow}\mathcal{Z}\Omega^{2}\longrightarrow
0,$

where $\mathcal{Z}\Omega^{2}$ is the sheaf of closed forms from $\Omega^{2}$.
Consider the corresponding cohomology exact sequence:

$H^{0}(M,\mathcal{Z}\Omega^{2})\longrightarrow
H^{1}(M,\mathcal{Z}\Omega^{1})\overset{\beta^{*}}{\longrightarrow}H^{1}(M,\Omega^{1}).$

Since $M$ is Kähler, all holomorphic forms on it are closed, and hence

$H^{0}(M,\mathcal{Z}\Omega^{2})=H^{0}(M,\Omega^{2})\simeq
H^{2,0}(M,\mathbb{C}).$

By (5.55), this group is trivial, and therefore $\beta^{*}$ is injective. It
is also surjective, since $\beta^{*}\hat{\delta}$ is the Dolbeault
isomorphism. It follows that $\beta^{*}$ and $\hat{\delta}$ are isomorphisms.
The natural $G$-action on $H^{1}(M,\mathcal{Z}\Omega^{1})$ is trivial, since
this is true for $H^{1,1}(M,\mathbb{C})$. ∎

[Technical] Let $M=G/P$ be as in Theorem 5.8, $\mathfrak{g}=\mathfrak{v}(M)$,
and let $(M,\mathcal{O})$ be any non-split supermanifold of the family
described in Theorem 5.8. Then, $l(v)$ for $v\in\mathfrak{g}$, and $d$ can be
lifted to $(M,\mathcal{O})$, and we have

$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(0)}$
$\displaystyle=\mathfrak{v}(M,\mathcal{O})_{\bar{0}}\oplus\langle\hat{d}\rangle,$
$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(1)}$
$\displaystyle=\langle\hat{d}\rangle,$
$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(p)}$ $\displaystyle=0\text{ for
}p\geq 2.$

Here
$\sigma_{0}:\mathfrak{v}(M,\mathcal{O})_{\bar{0}}\longrightarrow\mathfrak{g}$
is an isomorphism, $\hat{d}\neq 0$, $\sigma_{1}(\hat{d})=d$ and
$[\hat{d},\hat{d}]=[\hat{d},v]=0$ for all
$v\in\mathfrak{v}(M,\mathcal{O})_{\bar{0}}$.

###### Proof.

Consider the exact sequence (4.45) for
$\mathcal{O}_{\operatorname{gr}}=\Omega$. From Theorem 5.64 we deduce that
$\mathfrak{v}(M,\mathcal{O})_{(p)}=0$ for $p\geq 2$ and that
$\sigma_{1}:\mathfrak{v}(M,\mathcal{O})_{(1)}\longrightarrow\mathfrak{v}(M,\Omega)_{1}=\langle
d\rangle$ is injective. By Proposition 4.19, we see that
$\mathfrak{v}(M,\mathcal{O})_{(1)}=\langle\hat{d}\rangle$, where $\hat{d}$ is
odd and $\sigma_{1}(\hat{d})=d$. For $p=0$, the exact sequence has the form

$0\longrightarrow\mathfrak{v}(M,\mathcal{O})_{(1)}\longrightarrow\mathfrak{v}(M,\mathcal{O})_{(0)}\overset{\sigma_{0}}{\longrightarrow}\mathfrak{v}(M,\Omega)_{0}=\langle\varepsilon\rangle\oplus
l(\mathfrak{g}).$

By Lemma 5.8 and Corollary 4.19, any $v\in l(\mathfrak{g})$ lifts to
$(M,\mathcal{O})$. On the other hand, $\varepsilon$ does not lift by
Proposition 4.45, since
$[\varepsilon,\lambda_{2}(\gamma)]=2\lambda_{2}(\gamma)\neq 0$ by eq. (2.2).
Hence, $\operatorname{Im}\sigma_{0}=l(\mathfrak{g})$. This implies our
assertion concerning $\mathfrak{v}(M,\mathcal{O})_{(0)}$. Since
$\mathfrak{v}(M,\mathcal{O})_{(0)}$ is a subalgebra of
$\mathfrak{v}(M,\mathcal{O})$, it follows that $\hat{d}$ is a weight vector of
the representation $\operatorname{ad}_{\bar{1}}$ of
$\mathfrak{v}(M,\mathcal{O})_{\bar{0}}$, but the corresponding weight is 0,
since $\mathfrak{g}$ is simple. Thus, $[\hat{d},v]=0$ for all
$v\in\mathfrak{v}(M,\mathcal{O})_{\bar{0}}$. It follows that
$[\hat{d},\hat{d}]$ lies in the center of
$\mathfrak{v}(M,\mathcal{O})_{\bar{0}}$, whence $[\hat{d},\hat{d}]=0$. ∎

One can ask whether $\mathfrak{v}(M,\mathcal{O})$ coincides with it subalgebra
$\mathfrak{v}(M,\mathcal{O})_{(0)}$ calculated in Proposition 5.8. This is not
true in general, and in Theorem 5.31 I give the complete answer for the case
where $M$ is an irreducible Hermitian symmetric space.

###  5.10 Irreducible Hermitian symmetric spaces

A Hermitian symmetric space is, by definition, a connected complex manifold
$M$, endowed with a Hermitian structure and satisfying the following
condition: for any $x\in M$, there exists a holomorphic isometry $s_{x}$ of
$M$ such that $d_{x}s_{x}=-\operatorname{id}$. Let $M$ be a compact Hermitian
symmetric space. Let $K$ be the identity component of the group of all
holomorphic isometries of $M$; this is a compact Lie group. It is known (see
[21]) that $M$ is a homogeneous space of $K$, and hence can be regarded as the
coset space $K/L$, where $L$ is the stabilizer $K_{o}$ of a point $o\in M$. In
what follows, we suppose that $M$ is simply connected and irreducible (as a
Hermitian space). It is known (see [21]) that if $M$ is simply connected, then
$L$ is the centralizer of a torus in $K$, containing the symmetry $s_{o}$.
Now, $G=(\operatorname{Bih}\,M)^{\circ}$ is the complexification
$G=K(\mathbb{C})$, and $M=G/P$, where $P=G_{o}$ is a parabolic subgroup of
$G$. Thus, $M$ is a flag manifold of a special type. Now, a simply connected
compact Hermitian symmetric space $M$ is irreducible if and only if $K$ and
$G$ are simple. In this case, $P$ is maximal. Let $G$ be a connected simple
complex Lie group. We retain the notation of Subsection 5.1 and suppose that a
maximal torus $T$ and a Borel subgroup $B\supset T$ of $G$ are chosen.
Consider the decomposition (5.53) of the highest root $\delta$. A simple root
$\alpha\in\Pi$ will be called special if $n_{\alpha}=1$. Let $P$ be a
parabolic subgroup of $G$ containing the Borel subgroup $B_{-}$. In the above
notation, the flag manifold $M=G/P$ is Hermitian symmetric if and only if the
subset $S\subset\Pi$ defining $P$ has the form
$S=\Pi\setminus\\{\alpha_{0}\\}$, where $\alpha_{0}$ is a special simple root.
Thus, in this case,

$\Delta(P)=\Delta_{-}\cup[\Pi\setminus\\{\alpha_{0}\\}].$

It follows that

$\Delta(N_{-})=\Delta_{-}\setminus[\Pi\setminus\\{\alpha_{0}\\}],$

i.e., this is the set of those negative roots $-\beta$ of $G$, whose
expression through simple roots contains $\alpha_{0}$ (necessarily with
coefficient 1). The subgroups $N_{+}$ and $N_{-}$ are commutative. The
isotropy representation
$\tau:P\longrightarrow\operatorname{GL}(\mathfrak{n}_{+})$ is irreducible; in
particular, $\tau|N_{-}$ is trivial, and $\tau$ is completely determined by
its restriction onto $R$. For all simple Lie groups $G$ of rank $>1$ that have
special simple roots, [36, Table 6] shows the Dynkin diagrams of extended
systems of simple roots $\widetilde{\Pi}=\Pi\cup\\{-\delta\\}$, where for any
$\alpha\in\Pi$ the coefficient $n_{\alpha}$ is indicated. Note that a simple
root is special if and only if it lies in the same orbit as $-\delta$ under
the symmetry group of $\widetilde{\Pi}$. On the other hand, the nontrivial
symmetries of $\Pi$, existing for the types $A_{l},\;D_{l},\;E_{6}$, transform
special roots into special roots, and we have to consider special roots up to
these symmetries. In all cases, we have $(\alpha_{0},\alpha_{0})=2$ for a
special root $\alpha_{0}$.

###  5.11 The three cases

We also see that any irreducible symmetric Hermitian space of dimension $\geq
2$ satisfies to one of the following three conditions, depending on the choice
of $G$ and of a special simple root $\alpha_{0}$: I. $(\delta,\alpha_{0})=0$,
and there exists a unique $\alpha_{1}\in\Pi$ such that
$(\alpha_{0},\alpha_{1})\neq 0$. This case occurs for the groups $G$ of types
$B_{l},\;C_{l},\;D_{l},\;E_{6},\;E_{7}$ and for any special root$\alpha_{0}$;
we have $n_{\alpha_{1}}=2$. II. $(\delta,\alpha_{0})=0$, and there exist two
different simple roots $\alpha_{1},\;\alpha_{2}$, such that
$(\alpha_{0},\alpha_{i})\neq 0$ for $i=1,2$. This case occurs for the groups
$G$ of type $A_{l},\;l\geq 3$, for any $\alpha_{0}$ corresponding to the
interior vertices of the (non-extended) Dynkin diagram. Here, we have
$n_{\alpha_{1}}=n_{\alpha_{2}}=1$. The manifolds $M$ are the Grassmannians
$\operatorname{Gr}^{l+1}_{s}$ for $1<s<l$. III. $(\delta,\alpha_{0})\neq 0$.
This case occurs for the groups $G$ of type $A_{l}$ for $l\geq 2$, for any of
the two roots $\alpha_{0}$ corresponding to the end vertices of the (non-
extended) Dynkin diagram. There exists a unique $\alpha_{1}\in\Pi$ such that
$(\alpha_{0},\alpha_{1})\neq 0$, and we have $n_{\alpha_{1}}=1$. The manifolds
$M$ are the projective spaces $\mathbb{CP}^{l}$, where $l\geq 2$. The roots
$\alpha_{1},\;\alpha_{2}\in\Pi$, described above, will be called the neighbors
of $\alpha_{0}$. We admit a numbering of simple roots
$\Pi=\\{\alpha_{0},\alpha_{1},\ldots,\alpha_{l-1}\\}$ using this notation. For
a weight $\lambda$ of $G$, we will denote by $m(\lambda)$ the coefficient at
$\alpha_{1}$ (or the sum of the coefficients of $\alpha_{1},\;\alpha_{2}$) in
the expression of $\lambda$ in terms of $\Pi$. In particular, we see that

$m(\delta)=\begin{cases}2\text{ in the cases I, II}\\\ 1\text{ in the case
III}.\end{cases}$

Clearly, the weight system of the irreducible representation $\tau$ coincides
with $\Delta(N_{+})$, the highest weight being $\delta$ and the lowest one
$\alpha_{0}$. Similarly, the weight system of $\tau^{*}$ is $\Delta(N_{-})$,
the highest weight being $-\alpha_{0}$ and the lowest one $-\delta$.

###  5.12 Invariant vector-valued forms

In this Subsection, we discuss invariant vector-valued forms on flag manifolds
and the invariant cohomology $H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{G}$ of irreducible Hermitian
symmetric spaces. Retaining the notation of Subsection 5.1, consider a flag
manifold $M=G/P=K/L$. Clearly, $K$ naturally acts on the vector space
$\Gamma(M,\Phi\otimes\Theta)$ of all smooth vector-valued forms on $M$. The
well-known É. Cartan principle of reducing invariants of a transitive action
to invariants of the isotropy group (see, e.g., [38, Theorem 4.2]) gives The
evaluation mapping of $\Gamma(M,\Phi\otimes\Theta)$ onto

$\bigwedge(T^{1,0}_{o}(M)\oplus T^{0,1}_{o}(M))^{*}\otimes T^{1,0}_{o}(M)$

given by $\varphi\mapsto\varphi_{o}$ determines an isomorphism of the bigraded
vector spaces

$\displaystyle\Gamma(M,\Phi\otimes\Theta)^{K}$
$\displaystyle\to(\bigwedge(T^{1,0}_{o}(M)\oplus T^{0,1}_{o}(M))^{*}\otimes
T^{1,0}_{o}(M))^{L}$
$\displaystyle=(\bigwedge(\mathfrak{n}_{+}\oplus\mathfrak{n}_{-}))^{*}\otimes\mathfrak{n}_{+})^{L}$
$\displaystyle=(\bigwedge(\mathfrak{n}_{+}\oplus\mathfrak{n}_{-}))^{*}\otimes\mathfrak{n}_{+})^{R}$

preserving the operations $\barwedge$ and $\\{-,-\\}$. This isomorphism maps
$\Gamma(M,\Phi^{p,q}\otimes\Theta)^{K}$ onto

$((\bigwedge^{p}\mathfrak{n}_{+}\otimes\bigwedge^{q}\mathfrak{n}_{-})^{*}\otimes\mathfrak{n}_{+})^{R}=(\bigwedge^{p}\mathfrak{n}_{+}^{*}\otimes\bigwedge^{q}\mathfrak{n}_{-}^{*}\otimes\mathfrak{n}_{+})^{R}=(\bigwedge^{p}\mathfrak{n}_{-}\otimes\bigwedge^{q}\mathfrak{n}_{+}\otimes\mathfrak{n}_{+})^{R}.$

Now we give examples of invariant vector-valued forms.
[$\omega\in\Gamma(M,\Phi^{1,1})$ on a complex manifold $M$] Let $M$ be a
complex manifold and $\omega\in\Gamma(M,\Phi^{1,1})$. Consider the form

$\theta_{2}=\widetilde{\jmath}(\omega)\in\Gamma(M,\Phi^{2,1}\otimes\Theta)$

given by the formula (4.43). Thus,

$\theta_{2}(u_{1},u_{2},v)=\omega(u_{1},v)u_{2}-\omega(u_{2},v)u_{1},\ \
u_{1},u_{2}\in\Theta,\;v\in\bar{\Theta}.$ (5.65)

More generally, we can construct the following vector-valued $(p,p-1)$-form
$\theta_{p}$ for $p\geq 1$:

$\theta_{p}(u_{1},\ldots,u_{p},v_{1},\ldots,v_{p-1})=(p-1)!\begin{vmatrix}\omega(u_{1},v_{1})&\ldots&\omega(u_{1},v_{p-1})&u_{1}\\\
\omega(u_{2},v_{1})&\ldots&\omega(u_{2},v_{p-1})&u_{2}\\\
\vdots&&\vdots&\vdots\\\
\omega(u_{p},v_{1})&\ldots&\omega(u_{p},v_{p-1})&u_{p}\end{vmatrix},$ (5.66)

where $u_{i}\in\Theta$ and $v_{j}\in\bar{\Theta}$. In particular,
$\theta_{1}=\operatorname{id}$ and $\theta_{2}$ is as in (5.65). Clearly,
$\theta_{p}\neq 0$ for $p\leq n=\dim M$. We note that (see [40])

$\theta_{p}\barwedge\theta_{q}=p\theta_{p+q-1}.$ (5.67)

By Proposition 5.12, the form $\theta_{p}$ is completely determined by its
value at $o\in M$ which is expressed by the same formula (5.66) through the
value $\omega_{o}$ at $o$. As $\omega_{o}$, we can choose any $L$-invariant
$(1,1)$-form at $o$. For example, the Killing form on $\mathfrak{g}$
determines an invariant form $\omega$ satisfying

$\omega_{o}(u,v)=(u,v),\ \ u\in\mathfrak{n}_{+},\;v\in\mathfrak{n}_{-}.$
(5.68)

In what follows, we consider the case where $M$ is an irreducible compact
Hermitian symmetric space. Then, the isotropy representation is irreducible,
and hence the form (5.68) is the only (up to a constant factor) $L$-invariant
$(1,1)$-form on $T_{o}(M)$. [$M=\operatorname{Gr}^{n}_{s}$] Consider the
complex Grassmannian $M=\operatorname{Gr}^{n}_{s}$, where $s$ is an integer in
$\\{1,\dotsc,n-1\\}$. This is an irreducible compact Hermitian symmetric space
with $G=\operatorname{SL}_{n}(\mathbb{C})$ and $K=\operatorname{SU}_{n}$. It
is convenient to regard $M$ as a homogeneous space of the group
$G_{0}=\operatorname{GL}_{n}(\mathbb{C})$ with a natural action on $M$ (this
action is not effective). As usual, we choose in $G_{0}$ the maximal torus $T$
consisting of all diagonal matrices and the Borel subgroup $B$ consisting of
all upper triangular matrices. Then, $B_{-}$ is the subgroup of all lower
triangular matrices. Denote $r:=n-s$. For $o$ we take the point $\langle
e_{r+1},\ldots,e_{n}\rangle\in\operatorname{Gr}^{n}_{s}$. Then, the isotropy
subgroup $P$ of $G$ at $o$ is parabolic and contains $B_{-}$; it consists of
all matrices of the form

$\begin{pmatrix}A_{1}&0\\\ V&A_{2}\end{pmatrix},$ (5.69)

where $A_{1}\in\operatorname{GL}_{r}(\mathbb{C})$ and
$A_{2}\in\operatorname{GL}_{s}(\mathbb{C})$. Its maximal reductive subgroup
$R$ consists of matrices of the form (5.69) with $V=0$ and can be identified
with
$\operatorname{GL}_{r}(\mathbb{C})\times\operatorname{GL}_{s}(\mathbb{C})$,
while the unipotent radical $N_{-}$ is abelian and consists of matrices of the
form (5.69) with $A_{1}=I_{r}$ and $A_{2}=I_{s}$. The subalgebras
$\mathfrak{n}_{-}$ and $\mathfrak{n}_{+}$ consist of matrices of the form

$\begin{pmatrix}0&0\\\ v&0\end{pmatrix},\ \ \begin{pmatrix}0&u\\\
0&0\end{pmatrix},$

respectively, $v$ being an $(s\times r)$-matrix and $u$ an $(r\times
s)$-matrix. We will identify $\mathfrak{n}_{-}$ and $\mathfrak{n}_{+}$ with
the vector spaces of matrices $\operatorname{M}_{s,r}(\mathbb{C})$ and
$\operatorname{M}_{r,s}(\mathbb{C})$, respectively. The isotropy
representation $\tau$ of $P$ on $\mathfrak{n}_{+}=T_{o}(M)$ is as follows:

$\tau\left(\begin{pmatrix}A_{1}&0\\\
V&A_{2}\end{pmatrix}\right)(u)=A_{1}uA_{2}^{-1}.$ (5.70)

Let us replace the Killing form by the following invariant bilinear form on
$\mathfrak{gl}_{n}(\mathbb{C})$:

$(X,Y)=\operatorname{tr}\,XY.$

Then, using (5.68), we can define the $K$-invariant vector-valued forms
$\theta_{p}$ on $M$ by (5.66). Now, we construct new examples of $K$-invariant
vector-valued (2,1)- and (3,2)-forms. Note that the same method permits to
construct certain invariant vector-valued $(p,p-1)$-forms for any $p\geq 1$.
Define the $K$-invariant vector-valued (2,1)-form $\eta$ by its $L$-invariant
value

$\eta_{o}(u_{1},u_{2},v)=u_{1}vu_{2}-u_{2}vu_{1},\ \
u_{1},u_{2}\in\mathfrak{n}_{+},v\in\mathfrak{n}_{-}.$ (5.71)

The forms $\theta_{2}$ and $\eta$ are linearly independent whenever $1<s<n-1$,
and they coincide for $s=1$ or $s=n-1$. Similarly, we define the $K$-invariant
vector-valued (3,2)-forms $\eta_{1},\eta_{2},\eta_{3}$, whose $L$-invariant
values at $o$ are as follows:

$\displaystyle(\eta_{1})_{o}$
$\displaystyle=\operatorname{Alt}\,\operatorname{tr}(u_{1}v_{1}u_{2}v_{2})u_{3}=\operatorname{Alt}\,(u_{1}v_{1},u_{2}v_{2})u_{3}$
(5.72)
$\displaystyle=2((u_{1}v_{1},u_{2}v_{2})u_{3}+(u_{2}v_{1},u_{3}v_{2})u_{1}+(u_{3}v_{1},u_{1}v_{2})u_{2}$
$\displaystyle-(u_{2}v_{1},u_{1}v_{2})u_{3}-(u_{3}v_{1},u_{2}v_{2})u_{1}-(u_{1}v_{1},u_{3}v_{2})u_{2}),$
$\displaystyle(\eta_{2})_{o}$
$\displaystyle=\operatorname{Alt}\,\operatorname{tr}(u_{1}v_{1})u_{2}v_{2}u_{3}=\operatorname{Alt}\,(u_{1},v_{1})u_{2}v_{2}u_{3}$
(5.73)
$\displaystyle=(u_{1},v_{1})u_{2}v_{2}u_{3}+(u_{2},v_{1})u_{3}v_{2}u_{1}+(u_{3},v_{1})u_{1}v_{2}u_{2}$
$\displaystyle-(u_{2},v_{1})u_{1}v_{2}u_{3}-(u_{3},v_{1})u_{2}v_{2}u_{1}-(u_{1},v_{1})u_{3}v_{2}u_{2}$
$\displaystyle-(u_{1},v_{2})u_{2}v_{1}u_{3}-(u_{2},v_{2})u_{3}v_{1}u_{1}-(u_{3},v_{2})u_{1}v_{1}u_{2}$
$\displaystyle+(u_{2},v_{2})u_{1}v_{1}u_{3}+(u_{3},v_{2})u_{2}v_{1}u_{1}+(u_{1},v_{2})u_{3}v_{1}u_{2},$
$\displaystyle(\eta_{3})_{o}=\operatorname{Alt}\,u_{1}v_{1}u_{2}v_{2}u_{3}$
$\displaystyle=u_{1}v_{1}u_{2}v_{2}u_{3}+u_{2}v_{1}u_{3}v_{2}u_{1}+u_{3}v_{1}u_{1}v_{2}u_{2}$
(5.74) $\displaystyle-
u_{2}v_{1}u_{1}v_{2}u_{3}-u_{3}v_{1}u_{2}v_{2}u_{1}-u_{1}v_{1}u_{3}v_{2}u_{2}$
$\displaystyle-
u_{1}v_{2}u_{2}v_{1}u_{3}-u_{2}v_{2}u_{3}v_{1}u_{1}-u_{3}v_{2}u_{1}v_{1}u_{2}$
$\displaystyle+u_{2}v_{2}u_{1}v_{1}u_{3}+u_{3}v_{2}u_{2}v_{1}u_{1}+u_{1}v_{2}u_{3}v_{1}u_{2}.$

Here, $u_{1},u_{2},u_{3}\in\mathfrak{n}_{+}$ for
$v_{1},v_{2}\in\mathfrak{n}_{-}$. We will need the following properties of the
forms introduced in Examples 5.12 and 5.68. Suppose that
$M=\operatorname{Gr}^{n}_{s}$ and (recall, $r:=n-s$)

  1. $(1)$

The forms $\theta_{3},\,\eta_{1},\,\eta_{2},\,\eta_{3}$ are linearly
independent whenever $s,r\geq 3$.

  2. $(2)$

If $r=2$ for $s\geq 3$, then $\theta_{3},\,\eta_{1},\,\eta_{2}$ are linearly
independent, while

$\eta_{3}=\eta_{2}+\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\eta_{1}-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\theta_{3}.$
(5.75)

  3. $(3)$

If $s=2$ for $r\geq 3$, then $\theta_{3},\,\eta_{1},\,\eta_{2}$ are linearly
independent, while

$\eta_{3}=-\eta_{2}-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\eta_{1}-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\theta_{3}.$
(5.76)

  4. $(4)$

If $s=r=2$, then $\theta_{3},\,\eta_{1}$ are linearly independent, while

$\eta_{2}=-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\eta_{1},\
\
\eta_{3}=-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}\theta_{3}.$
(5.77)

###### Proof.

To check the relations (5.75), (5.76), (5.77), we use the following simple
fact: for any two $2\times 2$-matrices $A,B$ we have

$AB+BA=(\operatorname{tr}A)B+(\operatorname{tr}B)A+(\operatorname{tr}AB-(\operatorname{tr}A)(\operatorname{tr}B))I.$
(5.78)

In the case (2), we obtain (5.75) by applying (5.78) to $A=u_{1}v_{1}$ and
$B=u_{2}v_{2}$, and alternating the resulting expression of

$u_{1}v_{1}u_{2}v_{2}u_{3}+u_{2}v_{2}u_{1}v_{1}u_{3}=(u_{1}v_{1}u_{2}v_{2}+u_{2}v_{2}u_{1}v_{1})u_{3}.$

Similarly, in the case (3) we apply (5.78) to $A=v_{1}u_{2}$ and
$B=v_{2}u_{3}$ and alternate the resulting expression of

$u_{1}v_{1}u_{2}v_{2}u_{3}+u_{1}v_{2}u_{3}v_{1}u_{2}=u_{1}(v_{1}u_{2}v_{2}u_{3}+v_{2}u_{3}v_{1}u_{2}).$

Now, (5.75) and (5.76) imply (5.77) in the case (4). I skip the proof of
linear independence. ∎

It is well known (thanks to É. Cartan) that the real cohomology algebra of a
compact Riemannian symmetric space $M=K/L$ is naturally isomorphic to the
algebra of $K$-invariant differential forms on $M$ (see, e.g., [38, Corollary
of Theorem 9.7]). We want to prove a similar assertion concerning the
invariant cohomology $H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{K}$ of a simply connected
compact Hermitian symmetric space $M$. We use the fine resolution
$(\Phi\otimes\Theta,\bar{\partial})$ of the sheaf $\Omega\otimes\Theta$. By
the Dolbeault-Serre theorem, the sheaf cohomology $H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)$ and the cohomology of the
complex $(\Gamma(M,\Phi\otimes\Theta),\bar{\partial})$ are isomorphic.
Actually, we have

$H^{q}(M,\Omega^{p}\otimes\Theta)\simeq
H^{p,q}(\Gamma(M,\Phi\otimes\Theta),\bar{\partial}).$

Under this isomorphism, the algebraic and the FN-bracket in
$H^{\mathbin{\hbox{\raise 1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)$ are
induced by the same operations in $\Gamma(M,\Phi\otimes\Theta)$. Denote the
operator in $\Gamma(M,\Phi\otimes\Theta)$ conjugate to $\bar{\partial}$ with
respect to the given $K$-invariant Hermitian metric on $M$ by
$\bar{\partial}^{*}$ and the Beltrami–Laplace operator by
$\square=\bar{\partial}\bar{\partial}^{*}+\bar{\partial}^{*}\bar{\partial}$.
As usual, a form $\varphi\in\Gamma(M,\Phi\otimes\Theta)$ is called harmonic if
$\square\varphi=0$. For a harmonic $\varphi$, we have
$\bar{\partial}\varphi=0$, and any cohomology class contains precisely one
harmonic form. Let $M$ be a simply connected compact Hermitian symmetric
space, $K$ the identity component of the group of all holomorphic isometries
of $M$. Then,

$\Gamma(M,\Phi^{r}\otimes\Theta)^{K}=0\text{~{}~{}whenever $r$ is even.}$

Moreover, any $\varphi\in\Gamma(M,\Phi\otimes\Theta)^{K}$ is harmonic, and
hence $\bar{\partial}$-closed. Assigning to a form
$\varphi\in\Gamma(M,\Phi\otimes\Theta)^{K}$ its cohomology class, we get an
isomorphism of bigraded algebras

$\lambda:\Gamma(M,\Phi\otimes\Theta)^{K}\longrightarrow
H^{\mathbin{\hbox{\raise 1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{G}$

both for the algebraic and the FN-brackets. The FN-bracket in
$H^{\mathbin{\hbox{\raise 1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{G}$
is identically 0.

###### Proof.

For any form $\varphi\in\Gamma(M,\Phi^{r}\otimes\Theta)^{K}$ we have
$s^{*}\varphi=\varphi$. Since $ds_{o}=-\operatorname{id}$, we see that
${(s^{*}\varphi)_{o}=(-1)^{r+1}\varphi_{o}}$. If $r$ is even, then
$\varphi_{o}=0$, and hence $\varphi=0$. This proves the first assertion.
Moreover, in the same situation we have
$\bar{\partial}\varphi\in\Gamma(M,\Phi^{r+1}\otimes\Theta)^{K}$. If $r$ is
odd, then $\bar{\partial}\varphi=0$. Similarly, $\bar{\partial}^{*}\varphi=0$,
and hence $\varphi$ is harmonic. It follows that

$\lambda:\Gamma(M,\Phi\otimes\Theta)^{K}\longrightarrow
H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{K}=H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\Omega\otimes\Theta)^{G}$

is defined and injective. To prove that $\lambda$ is surjective, suppose that
$\varphi\in\Gamma(M,\Phi\otimes\Theta)$ is a harmonic form representing a
$G$-invariant cohomology class. Then, for any $k\in K$, the form
$k^{*}\varphi$ is harmonic and lies in the same cohomology class as $\varphi$.
Therefore, $k^{*}\varphi=\varphi$ for $k\in K$, so
${\varphi\in\Gamma(M,\Phi\otimes\Theta)^{K}}$. Clearly,
$\Gamma(M,\Phi\otimes\Theta)^{K}$ is a subalgebra under both brackets and
$\lambda$ is an isomorphism of algebras. The FN-bracket is 0, since
$H^{q}(M,\Omega^{p}\otimes\Theta)^{G}=0$ whenever $p+q$ is even. ∎

Under assumptions of Proposition 5.12, we have (recall definition of $R$ under
(5.69))

$H^{q}(M,\Omega^{p}\otimes\Theta)^{G}\simeq(\bigwedge^{p}\mathfrak{n}_{-}\otimes\bigwedge^{q}\mathfrak{n}_{+}\otimes\mathfrak{n}_{+})^{R}.$

Now we are going to calculate certain invariant cohomology groups assuming
that $M$ is irreducible. First of all, we will find the degrees for which they
are non-zero. For any simply connected irreducible compact Hermitian symmetric
space $M$ of dimension $n\geq 2$ we have
$H^{q}(M,\Omega^{p}\otimes\Theta)^{G}\neq 0$ if and only if $q=p-1$ for any
$p=1,\ldots,n$.

###### Proof.

Let $\omega$ be the Kähler form on $M$ corresponding to the given Kähler
metric. Consider the invariant forms $\theta_{p}$ for $p=1,\ldots,n$, given by
the formula (4.14). By Proposition 5.12, they determine non-zero cohomology
classes in $H^{p-1}(M,\Omega^{p}\otimes\Theta)^{G}$. By Corollary 5.12, it is
sufficient to show that the representation of $R$ induced in
$\bigwedge^{p}\mathfrak{n}_{-}\otimes\bigwedge^{q}\mathfrak{n}_{+}\otimes\mathfrak{n}_{+}$
has no zero weights whenever $q\neq p-1$. But each weight of this
representation has the form

$\lambda=(-p+q+1)\alpha_{0}+\sum_{j=1}^{l-1}k_{j}\alpha_{j}.$

If $\lambda=0$, then $q=p-1$. ∎

It was proved in [40] that

$H^{p-1}(M,\Omega^{p}\otimes\Theta)^{G}\simeq\mathbb{C},\
\text{where~{}~{}}p=1,\ldots,n,$

for $M=\mathbb{CP}^{n}$ (case III). We investigate now the degrees $p=1,2,3$
in the general case. By Lemma 5.6, we have

$H^{0}(M,\Omega^{1}\otimes\Theta)^{G}\simeq\mathbb{C}.$

For the case $p=2$ we need the following fact, implied by a result of Kostant
(see [26])]. We will denote by $\sigma_{j}$ the reflection
$\sigma_{\alpha_{j}}\in W$ corresponding to the simple root $\alpha_{j}$. The
irreducible components of the $R$-module $\bigwedge^{2}\mathfrak{n}_{+}$
correspond one-to-one to those simple roots $\alpha_{k}$ of $G$ that are
neighbors of $\alpha_{0}$. The component that corresponds to $\alpha_{k}$ has
the lowest weight $2\alpha_{0}+\alpha_{k}$ and the lowest weight vector
$e_{\alpha_{0}}\wedge e_{\alpha_{0}+\alpha_{k}}$. Thus,
$\bigwedge^{2}\mathfrak{n}_{+}$ is irreducible in the cases I, III and has two
irreducible components in the case II.

###### Proof.

By [26, Section 8], the irreducible components of
$\bigwedge^{2}\mathfrak{n}_{+}$ correspond to those elements $\sigma\in W$
satisfying

  1. 1.

$\sigma=\sigma_{j}\sigma_{k}$ for $j\neq k$,

  2. 2.

$\Phi_{\sigma}=\sigma\Delta_{-}\cap\Delta_{+}\subset\Delta(N_{+})$.

The set $\Phi_{\sigma}=\\{\alpha,\beta\\}$ can be determined from the relation

$\sigma\gamma=\gamma-\alpha-\beta.$

The lowest weight of the component corresponding to $\sigma$ is
$\alpha+\beta$, and the lowest weight vector is $e_{\alpha}\wedge e_{\beta}$.
Clearly,

$\sigma\gamma=\sigma_{j}\sigma_{k}\gamma=\sigma_{j}(\gamma-\alpha_{k})=\gamma-\alpha_{j}-\sigma_{j}\alpha_{k}.$

Hence, $\alpha+\beta=\alpha_{j}+\sigma_{j}\alpha_{k}$, where
$\sigma_{j}\alpha_{k}=\alpha_{k}-\langle\alpha_{k},\alpha_{j}\rangle\alpha_{j}$.
Since $\alpha$ and $\beta$ contain $\alpha_{0}$ with coefficient 1, the same
property must have the roots $\alpha_{j}$ and $\sigma_{j}\alpha_{k}$. It
follows that $j=0$ and $\langle\alpha_{k},\alpha_{0}\rangle\neq 0$, i.e.,
$\alpha_{k}$ is a neighbor of $\alpha_{0}$. Since
$\langle\alpha_{k},\alpha_{0}\rangle=-1$, we have

$\alpha+\beta=\alpha_{0}+(\alpha_{0}+\alpha_{k}).$

This easily implies that $\alpha=\alpha_{0}$ and $\beta=\alpha_{0}+\alpha_{k}$
(or vice versa). ∎

[$H^{1}(M,\Omega^{2}\otimes\Theta)^{G}$] We have

$H^{1}(M,\Omega^{2}\otimes\Theta)^{G}\simeq\begin{cases}\mathbb{C}&\text{in
the cases I, III}\\\ \mathbb{C}^{2}&\text{in the case II. }\end{cases}$

###### Proof.

By Corollary 5.2,

$H^{1}(M,\Omega^{2}\otimes\Theta)^{G}\simeq(\bigwedge^{2}\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}\otimes\mathfrak{n}_{+})^{R}.$

Now,

$\mathfrak{n}_{+}\otimes\mathfrak{n}_{+}=\bigwedge^{2}\mathfrak{n}_{+}\oplus\operatorname{S}^{2}\mathfrak{n}_{+}.$

By Lemma 5.12, $\bigwedge^{2}\mathfrak{n}_{+}$ is the irreducible $R$-module
with lowest weight $2\alpha_{0}+\alpha_{1}$ in the cases I, III. It is easy to
prove that it is not isomorphic to any submodule of
$\operatorname{S}^{2}\mathfrak{n}_{+}$. Indeed, the lowest weight vector of
such a submodule must be $e_{\alpha_{0}}e_{\alpha_{0}+\alpha_{1}}$, which is
impossible. Thus, $\mathfrak{n}_{+}\otimes\mathfrak{n}_{+}$ contains precisely
one component dual to $\bigwedge^{2}\mathfrak{n}_{-}$, implying the result.
The case II is considered similarly. ∎

Clearly, in the cases I, III, a basic element of
$H^{1}(M,\Omega^{2}\otimes\Theta)^{G}$ is determined by the invariant form
$\theta_{2}$ given by the formula (5.65), $\omega$ being determined by the
formula (5.68). In the case II, a basis of
$H^{1}(M,\Omega^{2}\otimes\Theta)^{G}$ is formed by the cohomology classes of
$\theta_{2}$ and $\eta$, where $\eta$ is given by the formula (5.71), see
Example 5.68. The following proposition can be proved by case-by-case
verification using the decompositions into irreducible components. We omit the
proof, since we will not use the result. [$\dim
H^{2}(M,\Omega^{3}\otimes\Theta)^{G}$] The dimension $k=\dim
H^{2}(M,\Omega^{3}\otimes\Theta)^{G}$ is as follows:

  1. $(1)$

$k=0$ in the case III for $l=2$ ($M=\mathbb{CP}^{2}$);

  2. $(2)$

$k=1$ in the case I for types $B_{l},E_{l}$ and $D_{l}$ for $l>4$, if $M$ is a
quadric, and in the case III for $l>2$;

  3. $(3)$

$k=2$ in the case I for types $C_{l}$ and $D_{l}$, if $M$ is the isotropic
Grassmannian of maximal type, and in the case II, if $l=s+1=3$
($M=\operatorname{Gr}^{4}_{2}$);

  4. $(4)$

$k=3$ in the case II whenever $2<s=l-1$ or $2=s<l-1$;

  5. $(5)$

$k=4$ in the case II whenever $2<s<l-1$.

In the case II, a basis of $H^{2}(M,\Omega^{3}\otimes\Theta)^{G}$ is given by
the cohomology classes of the following forms:

$\displaystyle\theta_{2},\eta_{1}\text{ for }s=t=2,$
$\displaystyle\theta_{2},\eta_{1},\eta_{2}\text{ for }s=2\text{ for }t\geq
3\text{ or }s\geq 3,\;t=2,$
$\displaystyle\theta_{2},\eta_{1},\eta_{2},\eta_{3}\text{ for }s,t\geq 3$

(see Example 5.68 and Lemma 5.12).

###  5.22 An application of a theorem of Bott

Let again $M$ be an irreducible simply connected compact Hermitian symmetric
space. In this subsection, we apply Theorem 5.58 to calculation of the
cohomology $H^{q}(M,\Omega^{p}\otimes\Theta)$ for $q=1,2$. We regard
$\Omega^{p}\otimes\Theta$ as the sheaf of holomorphic sections of the
homogeneous vector bundle $\bigwedge^{p}\mathbf{T}(M)^{*}\otimes\mathbf{T}(M)$
corresponding to the completely reducible representation
$\tau\bigwedge^{p}\tau^{*}$ of $P$. The following well-known property of
dominant weights will be used (see [22, $\operatorname{S}$ 13, Exercise 8]).
If $\lambda$ is a non-zero dominant weight of a simple group $G$, then in the
expression

$\lambda=\sum_{i=0}^{l-1}k_{i}\alpha_{i}$

we have $k_{i}>0$ for all $i=0,\ldots,l-1$. A weight $\lambda$ of $G$ will be
called $R$-dominant if $(\lambda,\alpha_{i})\geq 0$ for all $i=1,\ldots,l-1$.
Any highest weight of a representation of $R$ is, evidently, $R$-dominant.
Recall that in the theory of Bott the operation $\xi\mapsto\xi^{*}$ given by
the formula (5.58) is essential. Note that if $\sigma=\sigma_{i}$ is the
reflection corresponding to the simple root $\alpha_{i}$, then

$\xi^{*}=\sigma_{i}\xi-\alpha_{i}=\xi-(1+\langle\xi,\alpha_{i}\rangle)\alpha_{i}.$
(5.79)

We also need the following lemmas. Let $\lambda$ be an $R$-dominant weight of
$G$. The weight $\lambda+\gamma$ has index 1 if and only if
$\lambda^{*}=\sigma_{0}(\lambda+\gamma)-\gamma$ is dominant.

###### Proof.

Clearly, the condition is sufficient. Now suppose that $\lambda+\gamma$ has
index 1. Then, $\alpha_{0}$ is the only positive root of $G$ such that
$(\lambda+\gamma,\alpha_{0})<0$. For any $i>0$, we have

$(\lambda^{*},\alpha_{i})=(\sigma_{0}(\lambda+\gamma),\alpha_{i})-1=(\lambda+\gamma,\sigma_{0}\alpha_{i})-1.$

Since
$\sigma_{0}\alpha_{i}=\alpha_{i}-\langle\alpha_{i},\alpha_{0}\rangle\alpha_{0}$
is a positive root not equal to $\alpha_{0}$, this number is non-negative.
Also

$(\lambda^{*},\alpha_{0})=(\sigma_{0}(\lambda+\gamma),\alpha_{0})-1=-(\lambda+\gamma,\alpha_{0})-1\geq
0.$

Thus, $\lambda^{*}$ is dominant. ∎

  1. $(1)$

A root $\alpha\in\Delta(N_{+})$ satisfies $m(\alpha)=0$ if and only if
$\alpha=\alpha_{0}$.

  2. $(2)$

Let $\lambda$ be a weight of the representation $\tau\bigwedge^{p}\tau^{*}$
for $p\geq 1$, i.e.,

$\lambda=\alpha-\beta_{1}-\ldots-\beta_{p},$ (5.80)

where $\alpha,\beta_{i}\in\Delta(N_{+}),\;\beta_{i}$ are all distinct. Then,

$m(\lambda)\leq\begin{cases}3-p\text{ in the cases }I,II\\\ 2-p\text{ in the
case }III.\end{cases}$

If the equality takes place here, then $m(\alpha)=m(\delta)$, one of
$\beta_{i}$ coincides with $\alpha_{0}$, and we have $m(\beta_{i})=1$ for all
$\beta_{i}\neq\alpha_{0}$.

###### Proof.

(1) If $\alpha$ does not coincide with the lowest root $\alpha_{0}$ of $\tau$,
then there exists a sequence of simple roots
$\alpha_{j_{1}},\ldots,\alpha_{j_{k}}$ such that

$\alpha=(\ldots((\alpha_{0}+\alpha_{j_{1}})+\alpha_{j_{2}})+\ldots)+\alpha_{j_{k}},$

where any sum in parentheses is a root. In particular, we have that
$\alpha_{0}+\alpha_{j_{1}}\in\Delta(N_{+})$; whence
$(\alpha_{0},\alpha_{j_{1}})<0$, and $\alpha_{j_{1}}$ is a neighbor of
$\alpha_{0}$. (2) The number $m(\lambda)$ attains its maximum whenever
$m(\alpha)$ is maximal (that is, whenever $m(\alpha)=m(\delta)$) and
$m(\beta_{i})$ are minimal (that is, $=0,1$). Due to item (1),
$m(\beta_{i})=0$ for only one root $\beta_{i}=\alpha_{0}$. Therefore,

$m(\lambda)\leq\begin{cases}2-(p-1)=3-p\text{ in the cases I, II}\\\
1-(p-1)=2-p\text{ in the case III},\end{cases}$

and the equality takes place in the situation described above. ∎

[$H^{p}(M,\Omega^{1}\otimes\Theta)$] We have

$\displaystyle H^{1}(M,\Omega^{1}\otimes\Theta)$
$\displaystyle\simeq\begin{cases}\mathfrak{g}\text{ in the cases I, II,}\\\
0\text{ in the case III,}\end{cases}$ $\displaystyle
H^{p}(M,\Omega^{1}\otimes\Theta)$ $\displaystyle=0\text{ for }p\geq 2.$

###### Proof.

The representation $\tau^{*}\tau$ contains a unique irreducible component with
highest weight $\lambda_{0}=\delta-\alpha_{0}$. By (5.79),

$\lambda_{0}^{*}=\sigma_{0}(\lambda_{0}+\gamma)-\gamma=\sigma_{0}\delta=\delta-\langle\delta,\alpha_{0}\rangle\alpha_{0}.$

In the cases I and II, $\lambda_{0}^{*}=\delta$ is dominant. By Lemma 5.22,
$\lambda_{0}+\gamma$ has index 1, and, by Bott’s theorem, we get a unique
$G$-submodule of $H^{1}(M,\Omega^{1}\otimes\Theta)$ isomorphic to
$\mathfrak{g}$. In the case III, we have
$\langle\lambda_{0},\alpha_{0}\rangle=-1$. Therefore, $\lambda_{0}+\gamma$ is
singular, and, by Bott’s theorem, our component gives nothing to the
cohomology. Now, it suffices to prove that any non-dominant highest weight
$\lambda$ of $\tau^{*}\tau$, such that $\lambda+\gamma$ is regular, coincides
with $\lambda_{0}$. Clearly, $\lambda=\alpha-\beta$, where
$\alpha,\beta\in\Delta(N_{+})$ for $\alpha\neq\beta$. Since $\lambda$ does not
contain $\alpha_{0}$, we have

$\lambda=\sum_{j=1}^{l-1}k_{j}\alpha_{j},$

where $k_{j}\in\mathbb{Z}$. Since $\lambda\prec\alpha\preceq\delta$, we have
$k_{j}\leq n_{\alpha_{j}}$ for $j=1,\ldots,l-1$. In particular,
$m(\lambda)\leq m(\delta)$. Since $\lambda$ is $R$-dominant, but not dominant,
and $\lambda+\gamma$ is regular, it follows that
$\langle\lambda,\alpha_{0}\rangle\leq-2$. On the other hand,
$\langle\lambda,\alpha_{0}\rangle=-m(\lambda)$, whence $m(\lambda)\geq 2$. We
see that the case III is impossible and that in the cases I and II we have
$m(\lambda)=m(\delta)=2$. Then, Lemma 5.22(2) implies that $\beta=\alpha_{0}$.
Thus, $\lambda=\alpha-\alpha_{0}$ is the only expression of the weight
$\lambda$ as a difference of two roots from $\Delta(N_{+})$. It follows that
the corresponding highest vector $v\in\mathfrak{n}_{+}\otimes\mathfrak{n}_{-}$
has the form

$v=e_{\alpha}\otimes e_{-\alpha_{0}}.$

But this vector cannot be a highest one if $\alpha\neq\delta$. Thus,
$\lambda=\lambda_{0}$. ∎

The next proposition reduces calculation of $H^{1}(M,\Omega^{p}\otimes\Theta)$
for $p\geq 2$ to the results of Subsection 5.6, where its invariant part has
been calculated.
[$H^{1}(M,\Omega^{p}\otimes\Theta)=H^{1}(M,\Omega^{p}\otimes\Theta)^{G}$] For
$p\geq 2$, we have

$H^{1}(M,\Omega^{p}\otimes\Theta)=H^{1}(M,\Omega^{p}\otimes\Theta)^{G}.$

###### Proof.

Let $\lambda$ be a highest weight of $\tau\bigwedge^{p}\tau^{*}$ for $p\geq
2$. Then, $\lambda$ has the form (5.80). Hence,

$\lambda=(1-p)\alpha_{0}+\mu,$ (5.81)

where

$\mu=\sum_{j=1}^{l-1}k_{j}\alpha_{j}\text{ for $k_{j}\in\mathbb{Z},\;k_{j}\leq
n_{\alpha_{j}},\;j=1,\ldots,l-1$}.$ (5.82)

Since $1-p<0$, it follows that $\lambda$ is not dominant due to Lemma 5.22.
Hence $(\lambda,\alpha_{0})<0$. But it is $R$-dominant, and hence
$\lambda+\gamma$ has index 1 if and only if
$\lambda^{*}=\sigma_{0}(\lambda+\gamma)-\gamma$ is dominant (see Lemma 5.22).
Clearly,

$\lambda^{*}=\sigma_{0}\lambda-\alpha_{0}=(p-2)\alpha_{0}+\sigma_{0}\mu=(p-2-\langle\mu,\alpha_{0}\rangle)\alpha_{0}+\mu=(p-2+m(\lambda))\alpha_{0}+\mu.$

By Bott’s theorem we have to show that $\lambda^{*}$ cannot be dominant and
non-zero. Suppose that the weight $\lambda^{*}$ is dominant and non-zero.
Then, by Lemma 5.22,

$k_{j}>0\text{~{}~{}for $j=1,\ldots,l-1;\ \ m(\lambda)>2-p$}.$

Applying Lemma 5.22(2), we see that the case III is impossible and that in the
cases I and II we have $m(\lambda)=3-p$. Since $m(\lambda)>0$, it follows that
$p=2$ and $m(\lambda)=1$. Thus,

$\lambda^{*}=\alpha_{0}+\mu,\ \ \lambda=-\alpha_{0}+\mu.$

In the case II, we have $m(\lambda)=k_{1}+k_{2}=2$ which gives a
contradiction. Now me must consider the case I only. Clearly,
$\alpha_{0}+\alpha_{1}=\sigma_{0}\alpha_{1}$ is a positive root of $G$. Since
$\lambda+\gamma$ has index 1 and $(\lambda,\alpha_{0})<0$, we get
$(\lambda,\alpha_{0}+\alpha_{1})\geq 0$. On the other hand,

$(\lambda,\alpha_{0}+\alpha_{1})=(-\alpha_{0}+\alpha_{1}+\sum_{j=2}^{l-1}k_{j}\alpha_{j},\alpha_{0}+\alpha_{1})=-2+(\alpha_{1},\alpha_{1})+\sum_{j=2}^{l-1}k_{j}(\alpha_{j},\alpha_{1}).$

If $l\geq 3$, we get $(\lambda,\alpha_{0}+\alpha_{1})<0$ which is a
contradiction. If $l=2$, then $G$ is of type $B_{2}$, and
$(\lambda,\alpha_{0}+\alpha_{1})=-1<0$, too. ∎

[$H^{2}(M,\Omega^{p}\otimes\Theta)$] For $p=2$ or $p\geq 4$, we have

$H^{2}(M,\Omega^{p}\otimes\Theta)=0.$

Also,

$H^{2}(M,\Omega^{3}\otimes\Theta)=H^{2}(M,\Omega^{3}\otimes\Theta)^{G}.$

###### Proof.

Let $\lambda$ be a highest weight of $\tau\bigwedge^{p}\tau^{*}$ for $p\geq
2$. Then, as in Proposition 5.22, statements (5.80), (5.81) and (5.82) hold.
Similarly, $\lambda$ is $R$-dominant, but not dominant, and hence
${(\lambda,\alpha_{0})<0}$. Suppose that the index of $\lambda+\gamma$ is 2.
As in the proof of Proposition 5.22,

$\sigma_{0}(\lambda+\gamma)=(p-2+m(\lambda))\alpha_{0}+\mu+\gamma.$

We have

$\displaystyle(\sigma_{0}(\lambda+\gamma),\alpha_{0})$
$\displaystyle=-(\lambda+\gamma,\alpha_{0})>0,$
$\displaystyle(\sigma_{0}(\lambda+\gamma),\alpha_{j})$
$\displaystyle=(\lambda+\gamma,\sigma_{0}\alpha_{j})=(\lambda+\gamma,\alpha_{j})>0,$

if $\alpha_{j}$ is not a neighbor of $\alpha_{0}$. Since the index is equal to
2, $\sigma_{0}(\lambda+\gamma)$ is regular and non-dominant, and hence

$(\sigma_{0}(\lambda+\gamma),\alpha_{1})<0$

for a neighbor $\alpha_{1}$ of $\alpha_{0}$. Then, the weight

$\lambda^{*}=\sigma_{1}\sigma_{0}(\lambda+\gamma)-\gamma$

must be dominant. Using (5.79), we get

$\lambda^{*}=(p-2+m(\lambda))\alpha_{0}+((-p+2-m(\lambda))\langle\alpha_{0},\alpha_{1}\rangle-\langle\mu,\alpha_{1}\rangle+k_{1}-1)\alpha_{1}+\mu^{\prime},$
(5.83)

where

$\mu^{\prime}=\sum_{j=2}^{l-1}k_{j}\alpha_{j}.$

By Proposition 5.12, $\lambda^{*}\neq 0$, if $p\neq 3$. Suppose that
$\lambda^{*}\neq 0$ for $p=3$, too. Then, by Lemma 5.22, all the coefficients
in (5.83) are positive. In particular,

$k_{j}>0\text{ for $j=2,\ldots,l-1$ and $m(\lambda)>2-p$}.$

Applying Lemma 5.22(2), we see that the case III is impossible and that in the
cases I and II we have $m(\lambda)=3-p$. Now consider the weight

$\widetilde{\lambda}=\sigma_{0}(\lambda+\gamma)-\gamma.$

Clearly, $\widetilde{\lambda}+\gamma$ is of index 1. As we saw,
$(\widetilde{\lambda}+\gamma,\alpha_{1})<0$, and hence $\alpha_{1}$ is the
only positive root with this property. It follows from formula (5.79) that

$\widetilde{\lambda}=\alpha_{0}+\mu.$

Therefore,

$(\widetilde{\lambda},\alpha_{0})=2-m(\lambda)=p-1.$

To get a contradiction, we consider separately three cases. 1) Case II. We
have, evidently, $k_{2}=\ldots=k_{l-1}=1$, and hence

$\widetilde{\lambda}=\alpha_{0}+(2-p)\alpha_{1}+\alpha_{2}+\ldots+\alpha_{l-1}.$

Therefore,

$(\widetilde{\lambda},\alpha_{1})=\begin{cases}3-2p\text{ if $\alpha_{1}$
corresponds to an end vertex of the Dynkin diagram,}\\\ 2-2p\text{
otherwise}.\end{cases}$

Hence,

$(\widetilde{\lambda},\alpha_{1})<0$

for all $p\geq 2$. If $p=2$, then the first case is impossible, because
$\widetilde{\lambda}+\gamma$ is singular. Now,

$(\widetilde{\lambda},\alpha_{1}+\alpha_{0})=(p-1)+(2-2p)=1-p<0$

for $p\geq 2$. This gives a contradiction. 2) Case I, the type of $G$ is not
$C_{l}$. We have

$\widetilde{\lambda}=\alpha_{0}+(3-p)\alpha_{1}+\mu^{\prime},$ (5.84)

where $\mu^{\prime}\neq 0$ (since $l\geq 3$) and $\alpha_{1}$ is long. Hence,

$(\widetilde{\lambda},\alpha_{1})=5-2p+\sum_{j=2}^{l-1}k_{j}(\alpha_{j},\alpha_{1})<5-2p,$

and

$(\widetilde{\lambda},\alpha_{0}+\alpha_{1})<4-p.$

This gives a contradiction whenever $p\geq 4$. If $p=2$, then

$(\widetilde{\lambda},\alpha_{0}+\alpha_{1})=1+(\widetilde{\lambda},\alpha_{1})\leq-1,$

since $(\widetilde{\lambda},\alpha_{1})\leq-2$, and we get a contradiction as
well. For $p=3$, the same argument shows that
$(\widetilde{\lambda},\alpha_{1})=-2$. Then, we see from (5.84) that there
exists precisely one root $\alpha_{j}$ (say, for $j=2$) such that
$(\alpha_{j},\alpha_{1})\neq 0$, and we have $k_{2}=1$ for
$(\alpha_{2},\alpha_{1})=-1$. Then, $\alpha_{1}+\alpha_{2}\in\Delta_{+}$ and
hence

$0\leq(\widetilde{\lambda},\alpha_{1}+\alpha_{2})=-2+(\widetilde{\lambda},\alpha_{2}).$

Thus, $(\widetilde{\lambda},\alpha_{2})\geq 2$. But

$\widetilde{\lambda}=\alpha_{0}+\alpha_{1}+\alpha_{2}+\sum_{j=3}^{l-1}k_{j}\alpha_{j},$

whence

$(\widetilde{\lambda},\alpha_{2})=1+\sum_{j=3}^{l-1}k_{j}(\alpha_{j},\alpha_{2})\leq
1.$

3) Case I, the type of $G$ is $C_{l}$. Here, equality (5.84) holds as well,
but $\alpha_{1}$ is short. Hence,

$(\widetilde{\lambda},\alpha_{1})=2-p+\sum_{j=2}^{l-1}k_{j}(\alpha_{j},\alpha_{1})\leq
2-p.$

But in this case, $\alpha_{0}+2\alpha_{1}\in\Delta(N_{+})$, and

$(\widetilde{\lambda},\alpha_{0}+2\alpha_{1})\leq 3-p.$

This gives a contradiction whenever $p\geq 4$. On the other hand,
$\langle\widetilde{\lambda},\alpha_{1}\rangle\leq-2$, whence
$(\widetilde{\lambda},\alpha_{1})\leq-1$, and
$(\widetilde{\lambda},\alpha_{0}+2\alpha_{1})\leq p-3$, which gives a
contradiction for $p=2$. For $p=3$, we see that the equality
$(\widetilde{\lambda},\alpha_{1})\leq-1$ is compatible with (5.84) only for
$\widetilde{\lambda}=\alpha_{0}$ for $l=2$. But then $\lambda=-2\alpha_{0}$,
and it is easy to see that in this case

$-2\alpha_{0}\neq\alpha-\alpha_{0}-\beta_{1}-\beta_{2}$ for any
$\alpha,\beta_{1},\beta_{2}\in\Delta(N_{+})$.

∎

###  5.26 Cohomology of $\mathcal{T}$

Summarizing the results of Subsections 5.6 and 5.31, we now describe the
structure of the cohomology $H^{q}(M,\Omega^{p}\otimes\Theta)$ for $q=0,1,2$
under our assumptions about $M$. Suppose that $M$ is a simply connected
irreducible compact Hermitian symmetric space of dimension $\geq 2$. The
$G$-modules $H^{q}(M,\Omega^{p}\otimes\Theta)$ for $q=0,1,2$, are listed in
the following tables: Case I:

$\begin{matrix}&p&\vrule&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\---&\---\\\ 0&&\vrule&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&\mathfrak{g}&\mathbb{C}&0&0\\\ 2&&\vrule&0&0&0&\mathbb{C}^{k}&0\\\
\end{matrix}$

Case II:

$\begin{matrix}&p&\vrule&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\---&\---\\\ 0&&\vrule&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&\mathfrak{g}&\mathbb{C}^{2}&0&0\\\
2&&\vrule&0&0&0&\mathbb{C}^{k}&0\\\ \end{matrix}$

Case III:

$\begin{matrix}&p&\vrule&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\---&\---\\\ 0&&\vrule&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&0&\mathbb{C}&0&0\\\ 2&&\vrule&0&0&0&\mathbb{C}^{k}&0\\\
\end{matrix}$

where we denote by $\mathbb{C}$ the trivial $G$-module and by $\mathfrak{g}$
the adjoint one, and the number $k$ is to be found in Proposition 5.12. Due to
Proposition 3.32, this result permits us to describe
$H^{q}(M,\mathcal{T}_{p})$ for $q=0,1,2$. [The $G$-modules
$H^{q}(M,\mathcal{T}_{p})$] Suppose that $M$ is a simply connected irreducible
compact Hermitian symmetric space of dimension $\geq 2$. The $G$-modules
$H^{q}(M,\mathcal{T}_{p})$ for $q=0,1,2$, where
$\mathcal{T}=\mathcal{D}er\,\Omega$, are listed in the following tables: Case
I:

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\----&\----&\----&\---&\---&\---\\\
0&&\vrule&i^{*}(\mathfrak{g})&l^{*}(\mathfrak{g})\oplus
i^{*}(\mathbb{C})&l^{*}(\mathbb{C})&0&0&0\\\
1&&\vrule&0&i^{*}(\mathfrak{g})&l^{*}(\mathfrak{g})\oplus
i^{*}(\mathbb{C})&l^{*}(\mathbb{C})&0&0\\\
2&&\vrule&0&0&0&i^{*}(\mathbb{C}^{k})&l^{*}(\mathbb{C}^{k})&0\\\ \end{matrix}$

Case II:

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\----&\----&\----&\---&\---&\---\\\
0&&\vrule&i^{*}(\mathfrak{g})&l^{*}(\mathfrak{g})\oplus
i^{*}(\mathbb{C})&l^{*}(\mathbb{C})&0&0&0\\\
1&&\vrule&0&i^{*}(\mathfrak{g})&l^{*}(\mathfrak{g})\oplus
i^{*}(\mathbb{C}^{2})&l^{*}(\mathbb{C}^{2})&0&0\\\
2&&\vrule&0&0&0&i^{*}(\mathbb{C}^{k})&l^{*}(\mathbb{C}^{k})&0\\\ \end{matrix}$

Case III:

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\----&\----&\----&\---&\---&\---\\\
0&&\vrule&i^{*}(\mathfrak{g})&l^{*}(\mathfrak{g})\oplus
i^{*}(\mathbb{C})&l^{*}(\mathbb{C})&0&0&0\\\
1&&\vrule&0&0&i^{*}(\mathbb{C})&l^{*}(\mathbb{C})&0&0\\\
2&&\vrule&0&0&0&i^{*}(\mathbb{C})&l^{*}(\mathbb{C}^{k})&0\\\ \end{matrix}$

where we denote by $\mathbb{C}$ the trivial $G$-module and by $\mathfrak{g}$
the adjoint one, and the number $k$ is to be found in Proposition 5.12. Using
Proposition 3.32, it is also possible to calculate the Lie bracket $[-,-]$ for
the part of the algebra $H^{\mathbin{\hbox{\raise
1.20552pt\hbox{\bf.}}}}(M,\mathcal{T})$ that is described in Theorem 5.26.
Here we calculate only the adjoint operator $\operatorname{ad}\zeta$, where
$\zeta\in H^{1}(M,\mathcal{T}_{2})$. Recall a result of Bott (see [4, Theorem
I and Corollary 2 of Theorem W], and also [26]) that describes the cohomology
of a flag manifold $M=G/P$ with values in the sheaf of holomorphic sections of
a homogeneous vector bundle $\mathbf{E}\longrightarrow M$ in terms of the
cohomology of the Lie algebra $\mathfrak{n}_{-}$. Suppose that
$\mathbf{E}=\mathbf{E}_{\varphi}$, where $\varphi$ is a holomorphic
representation of $P$. In contrast to Theorem 5.58, this description is valid
for arbitrary $\varphi$. Let a holomorphic representation of $G$ in a finite-
dimensional vector space $V$ be given. Then,

$\operatorname{Hom}_{G}(V,H^{q}(M,\mathcal{E}))\simeq
H^{q}(\mathfrak{n}_{-},\operatorname{Hom}(V,E_{o}))^{R},$ (5.85)

where the representation of $\mathfrak{n}_{-}$ in $V$ is the restriction of
the differential of the given representation of $G$, and that in $E_{o}$ is
the restriction of $\varphi$.

###### Proof.

oof (a sketch of) By the Dolbeault–Serre theorem, $H^{q}(M,\mathcal{E})$ can
be identified with the $q$-th cohomology of the complex
$(\Gamma(M,\Phi^{0,*}\otimes\mathcal{E}),\bar{\partial})$ of
$\mathbf{E}$-valued forms of type $(0,*)$. The vector space
$\Gamma(M,\Phi^{0,q}\otimes\mathcal{E})$ is the space of smooth sections of
the homogeneous vector bundle
$\bigwedge^{q}\mathbf{T}^{0,1}(M)^{*}\otimes\mathbf{E}$, whose fiber at $o$
can be identified with $\bigwedge^{q}\mathfrak{n}_{-}^{*}\otimes E_{o}$. By
the Frobenius reciprocity law,

$\displaystyle\operatorname{Hom}_{K}(V,\Gamma(M,\Phi^{0,q}\otimes\mathcal{E})$
$\displaystyle\simeq\operatorname{Hom}_{L}(V,\bigwedge^{q}\mathfrak{n}_{-}^{*}\otimes
E_{o})$
$\displaystyle=\operatorname{Hom}_{R}(V,\bigwedge^{q}\mathfrak{n}_{-}^{*}\otimes
E_{o})=C^{q}(\mathfrak{n}_{-},\operatorname{Hom}(V,E_{o}))^{R}.$

The isomorphism here is defined by the formula

$h\mapsto\widetilde{h},\text{ where $\widetilde{h}(v)=h(v)(o)$ for any $v\in
V$},$ (5.86)

and we denote by $C^{q}(\mathfrak{n}_{-},\operatorname{Hom}(V,E_{o}))$ the
vector space of $q$-cochains of the Lie algebra $\mathfrak{n}_{-}$ with values
in $\operatorname{Hom}(V,E_{o})$. Passing to the cohomology, we get the
isomorphism (5.85).∎

Let $\zeta=l^{*}([\theta])\in H^{1}(M,\mathcal{T}_{2})$, where $[\theta]\in
H^{1}(M,\Omega^{2}\otimes\Theta)$ is the cohomology class of the form
$\theta\in\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$. The map
$\operatorname{ad}_{\zeta}:H^{0}(M,\mathcal{T}_{-1})\to
H^{1}(M,\mathcal{T}_{1})$ is as follows:

  1. $(1)$

an isomorphism of the $G$-modules

$H^{0}(M,\mathcal{T}_{-1})=i^{*}(H^{0}(M,\Theta))\longrightarrow
l^{*}(H^{1}(M,\Omega^{1}\otimes\Theta))$

for any $\theta\neq 0$ in the case I and for any $\theta=a\theta_{2}+b\eta$,
where $a\neq 0$, in the case II;

  2. $(2)$

0 for $\theta=b\eta$ in the cases II and III.

###### Proof.

For any $w\in\mathfrak{g}$ we have, by (3.28),

$[l(\theta),i(w)]=[i(w),l(\theta)]=l(\theta\barwedge w)-i([w,\theta]).$

Since $\theta$ is $K$-invariant, we see that $[w,\theta]=0$. By Proposition
3.32, $[l^{*}([\theta]),i^{*}(w)]$ is determined by the cocycle
$l(\theta\barwedge w)$. Thus, our problem is reduced to the study of the
mapping

$H^{0}(M,\Theta)\to H^{1}(M,\Omega^{1}\otimes\Theta)$

defined on the cochain level by $w\mapsto\theta\barwedge w$. Recall that the
form $\theta\barwedge w\in\Gamma(M,\Phi^{1,1}\otimes\Theta)$ is given by the
formula

$(\theta\barwedge w)(u,v)=\theta(w,u,v)\text{ for
}u\in\Theta,v\in\bar{\Theta}.$

We will use the isomorphism

$\operatorname{Hom}_{G}(\mathfrak{g},H^{1}(M,\Omega^{1}\otimes\Theta))\simeq
H^{1}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{+}))^{R}$
(5.87)

that follows from (5.85) if we identify the fiber $E_{o}$ of the bundle
$\mathbf{E}=\mathbf{T}(M)^{*}\otimes\mathbf{T}(M)$ with
$\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{+}$. As it was noticed above, this
isomorphism on the cochain level is determined by (5.86). Let

$h:\mathfrak{g}\longrightarrow\Gamma(M,\Phi^{1,1}\otimes\Theta)\;(=\Gamma(M,\Phi^{0,1}\otimes\Omega^{1}\otimes\Theta))$

be given by the formula $h(w)=\theta\barwedge w$. Then, $h$ determines the
mapping

$\widetilde{h}:\mathfrak{g}\to(\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{-}^{*})\otimes\mathfrak{n}_{+}=\operatorname{Hom}(\mathfrak{n}_{+}\otimes\mathfrak{n}_{-},\mathfrak{n}_{+})$

given by the formula

$\widetilde{h}(w)(u,v)=\theta_{o}(\pi(w),u,v),\ \ u\in\mathfrak{n}_{+}\text{
for }v\in\mathfrak{n}_{-},$

where we identify the value $w(o)$ of the vector field $w$ at $o$ with
$\pi(w)$, where $\pi:\mathfrak{g}\longrightarrow\mathfrak{n}_{+}$ is the
projection along $\mathfrak{p}$ in the decomposition (5.54). In order to
interprete $\widetilde{h}(w)$ as an element of

$\mathfrak{n}_{-}^{*}\otimes(\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{+})=\operatorname{Hom}(\mathfrak{n}_{-},\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{+}),$

we choose a basis $e_{1},\ldots,e_{n}$ of $\mathfrak{n}_{+}$ and denote by
$e_{1}^{*},\ldots,e_{n}^{*}$ the dual basis of $\mathfrak{n}_{+}^{*}$. Then,

$\widetilde{h}(w)(v)=\sum_{i=1}^{n}e_{i}^{*}\otimes\widetilde{h}(w)(u,v)=\sum_{i=1}^{n}e_{i}^{*}\otimes\theta_{o}(\pi(w),e_{i},v),\
\ v\in\mathfrak{n}_{-}.$

Now, this form is viewed as the following cochain $c_{\gamma}\in
C^{1}(\mathfrak{n}_{+}^{*},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}))$:

$c_{\theta}(v)(w)=\sum_{i=1}^{n}e_{i}^{*}\otimes\theta_{o}(\pi(w),e_{i},v),\ \
v\in\mathfrak{n}_{-},\;w\in\mathfrak{g}.$ (5.88)

This cochain is an $R$-invariant cocycle of $\mathfrak{n}_{-}$, and we have to
understand what is its cohomology class. By Proposition 5.26, we have

$H^{1}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{+}^{*}\otimes\mathfrak{n}_{+}))^{R}\simeq\begin{cases}\mathbb{C}&\text{in
the cases I, II, }\\\ 0&\text{in the case III. }\end{cases}$

It is convenient to identify $\mathfrak{n}_{+}^{*}$ with $\mathfrak{n}_{-}$
using the Killing form. Then, we have to consider the cochain complex
$C^{*}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}))^{R},\delta)$.
Let us describe the space of 1-coboundaries $\delta
C^{0}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}))^{R}$.
Clearly,

$C^{0}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}))^{R}=\operatorname{Hom}_{R}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}).$

For any
$c\in\operatorname{Hom}_{R}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+})$,
we have $\delta c(y)=yc$ for any $y\in\mathfrak{n}_{-}$, i.e.,

$\delta c(y)(z)=c([y,z]),\text{~{}~{}for any
$y\in\mathfrak{n}_{-},\;z\in\mathfrak{g}$},$

since $d\tau(\mathfrak{n}_{-})=0$. Clearly,
$[\mathfrak{n}_{-},\mathfrak{g}]=\mathfrak{n}_{-}\oplus\mathfrak{r}$. Since
$c$ is a homomorphism of $R$-modules, it follows that $(\delta
c)(\mathfrak{n}_{-})(\mathfrak{g})$ is contained in the vector subspace of
$\mathfrak{n}_{-}\oplus\mathfrak{n}_{+}$ spanned by all $e_{-\alpha}\otimes
e_{\beta}$, where $\alpha,\beta\in\Delta(N_{+})$ and
$\beta-\alpha\in\Delta(R)$ or $\alpha=\beta$. In the cases I and II this
subspace does not coincide with $\mathfrak{n}_{-}\oplus\mathfrak{n}_{+}$,
i.e., there exist $\alpha,\beta\in\Delta(N_{+})$ such that
$\beta-\alpha\notin\Delta(R)$ and $\alpha\neq\beta$. Indeed, we can take
$\beta=\delta$ and $\alpha=\alpha_{0}$. Suppose that $\theta=\theta_{2}$. By
(5.88),

$\displaystyle c_{\theta_{2}}(v)(w)$
$\displaystyle=\sum_{i=1}^{n}e_{i}^{*}\otimes((e_{i},v)\pi(w)-(\pi(w),v)e_{i})$
$\displaystyle=\sum_{i=1}^{n}(e_{i},v)e_{i}^{*}\otimes\pi(w)-(\pi(w),v)\sum_{i=1}^{n}e_{i}^{*}\otimes
e_{i}$ $\displaystyle=v\otimes\pi(w)-(\pi(w),v)\sum_{i=1}^{n}e_{i}^{*}\otimes
e_{i}.$

In particular,

$c_{\theta_{2}}(e_{-\alpha_{0}})(e_{\delta})=e_{-\alpha_{0}}\otimes
e_{\delta}.$

It follows that $c_{\theta_{2}}\notin\delta
C^{0}(\mathfrak{n}_{-},\operatorname{Hom}(\mathfrak{g},\mathfrak{n}_{-}\otimes\mathfrak{n}_{+}))^{R}$.
Thus, $\theta=\theta_{2}$ defines a non-zero homomorphism in the cases I and
II. Now consider the case II, i.e., suppose that
$M=\operatorname{Gr}^{n}_{s}(\mathbb{C})$, where $1<s<n-1$. We will use the
notation of Example 5.68. Then, $\mathfrak{n}_{+}$ and $\mathfrak{n}_{-}$ are
the following subspaces of $\mathfrak{gl}_{n}(\mathbb{C})$:

$\displaystyle\mathfrak{n}_{+}$ $\displaystyle=\langle E_{i\beta}\mid 1\leq
i\leq s,\;s+1\leq\beta\leq m\rangle,$ (5.89) $\displaystyle\mathfrak{n}_{-}$
$\displaystyle=\langle E_{\alpha j}\mid s+1\leq\alpha\leq m,\;1\leq j\leq
s\rangle.$

Here, $E_{\alpha j}=E_{j\alpha}^{*}$ form the basis dual to $E_{j\alpha}$. If
$\theta=\eta$, then the cochain (5.88) has the form

$c_{\eta}(v)(w)=\sum_{i,\alpha}E_{\alpha
i}\otimes(E_{i\alpha}v\pi(w)-\pi(w)vE_{i\alpha}).$

We write $v=\sum_{\beta j}v_{\beta j}E_{\beta
j},\;\pi(w)=\sum_{j\beta}w_{j\beta}E_{j\beta}$. Then,

$\displaystyle E_{i\alpha}v\pi(w)$ $\displaystyle=E_{i\alpha}(\sum_{\beta
j\rho}v_{\beta j}w_{j\rho})E_{\beta\rho}=\sum_{j\rho}v_{\alpha
j}w_{j\rho}E_{i\rho},$ $\displaystyle\pi(w)vE_{i\alpha})$
$\displaystyle=(\sum_{jk\rho}w_{j\rho}v_{\rho
k})E_{jk}E_{i\alpha}=(\sum_{j\rho}w_{j\rho}v_{\rho i})E_{j\alpha}.$

Hence,

$c_{\eta}(v)(w)=\sum_{ij\alpha\rho}E_{\alpha i}\otimes\sum_{j\rho}(v_{\alpha
j}w_{j\rho})E_{i\rho}-\sum_{ij\alpha\rho}E_{\alpha
i}\otimes(\sum_{j\rho}w_{j\rho}v_{\rho i})E_{j\alpha}.$

Consider the 0-cochain
$c\in\operatorname{Hom}_{R}(\mathfrak{gl}_{n}(\mathbb{C}),\mathfrak{n}_{-}\otimes\mathfrak{n}_{+})$
given by the formula

$\displaystyle c(\mathfrak{n}_{+})$ $\displaystyle=c(\mathfrak{n}_{-})=0,$
$\displaystyle c(E_{ij})$ $\displaystyle=\sum_{\rho}E_{\rho j}\otimes
E_{i\rho},$ $\displaystyle c(E_{\alpha\beta})$
$\displaystyle=\sum_{k}E_{\alpha k}\otimes E_{k\rho}$

and restrict it to $\mathfrak{g}=\mathfrak{sl}_{n}(\mathbb{C})$. Then, for any
$v\in\mathfrak{n}_{-}$ and $w\in\mathfrak{g}$, we have

$\displaystyle\delta c(v)(w)$ $\displaystyle=c([v,w])=c([v,\pi(w)])$
$\displaystyle=c(\sum_{j\alpha\rho}v_{\alpha
j}w_{j\rho}E_{\alpha\rho}-\sum_{ij\rho}v_{\rho i}w_{j\rho}E_{ji})$
$\displaystyle=\sum_{jk\alpha\rho}v_{\alpha j}w_{j\rho}E_{\alpha k}\otimes
E_{k\rho}-\sum_{ij\alpha\rho}v_{\rho i}w_{j\rho}E_{\alpha i}\otimes
E_{j\alpha}$ $\displaystyle=c_{\eta}(v)(w).$

Thus, $c_{\eta}=\delta c$, and $\eta$ defines the zero homomorphism.
Evidently, this implies the statement. ∎

Let $M=\operatorname{Gr}^{n}_{s}$ for $2\leq s\leq n-2$, and let
$\theta,\varphi\in\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$. If $n\geq 5$, then
$\theta\barwedge\varphi=0$ implies $\theta=0$ or $\varphi=0$. For
$M=\operatorname{Gr}^{4}_{2}$, the only solutions of
$\theta\barwedge\varphi=0$, up to a constant factor, are
$\theta=\sqrt{2}\theta_{2}\pm\eta,\;\varphi=\theta_{2}\pm\sqrt{2}\eta$.

###### Proof.

By (5.67),

$\theta_{2}\barwedge\theta_{2}=2\theta_{3}.$

From (5.72), (5.73), and (5.74) we easily deduce the following relations:

$\displaystyle\theta_{2}\barwedge\eta$ $\displaystyle=2(\eta_{1}+\eta_{2}),$
$\displaystyle\eta\barwedge\theta_{2}$ $\displaystyle=4\eta_{2},$
$\displaystyle\eta\barwedge\eta$ $\displaystyle=4\eta_{3}.$

Write $\theta=a\theta_{2}+b\eta,\;\varphi=c\theta_{2}+d\eta$ with
$a,b,c,d\in\mathbb{C}$. It follows that

$\theta\barwedge\varphi=2ac\theta_{3}+2ad\eta_{1}+2(ad+2bc)\eta_{2}+4bd\eta_{2}.$

Suppose that $3\leq s\leq n-3$. By Lemma 5.12, $\theta\barwedge\varphi=0$
yields

$ac=ad=ad+bc=bd=0.$

Clearly, this implies $(a,b)=0$ or $(c,d)=0$. If $n-s=2$ and $s\geq 3$, then,
by (5.75),

$\theta\barwedge\varphi=2(ac-bd)\theta_{3}+2(ad-
bd)\eta_{1}+2(ad+2bc+2bd)\eta_{2}.$

By Lemma 5.12, $\theta\barwedge\varphi=0$ yields $ac-bd=ad+bd=ad+2bc+2bd=0$.
Clearly, this implies $(a,b)=0$ or $(c,d)=0$. The case $s=2,\;n-s\geq 3$ is
considered similarly. Suppose now that $n=4$ and $k=2$. It follows from (5.77)
that

$\theta\barwedge\varphi=2(ac-bd)\theta_{3}+(ad-2bc)\eta_{1}.$

If $\theta\barwedge\varphi=0$, then $ac-bd=ad-2bc=0$. If $(a,b)\neq 0$, then
this implies

$\begin{vmatrix}c&-d\\\ d&-2c\end{vmatrix}=-2c^{2}+d^{2}=0,$

whence $d=\pm\sqrt{2}c$. If $(c,d)\neq 0$, then $a=\pm\sqrt{2}b$. ∎

###  5.31 Non-split supermanifolds

In this subsection, we apply our results to the problem of classification of
non-split supermanifolds. Theorem 5.26 implies that the split supermanifold
$(M,\Omega)$ satisfies the conditions of Theorem 4.4. Thus, in this case the
mapping

$\lambda_{2}^{*}:H^{1}(M,\mathcal{A}ut_{(2)}\Omega)\longrightarrow
H^{1}(M,\mathcal{T}_{2})$

is bijective. By Theorem 4.1, we can parametrize non-split supermanifolds with
retract $(M,\Omega)$ (up to isomorphism) by orbits of the group
$\operatorname{Aut}\,\mathbf{T}(M)^{*}$ in
$H^{1}(M,\mathcal{T}_{2})\setminus\\{0\\}$. By Propositions 5.22 and 5.12, one
can identify
$H^{1}(M,\mathcal{T}_{2})=l^{*}(H^{1}(M,\Omega^{2}\otimes\Theta))$ with the
vector space of $K$-invariant vector-valued (2,1)-forms
$\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$ using the Dolbeault–Serre isomorphism.
We use this parametrization in the statement of the following classification
theorem. Suppose that $M$ is a simply connected irreducible compact Hermitian
symmetric space of dimension $\geq 2$.

  1. $1)$

If $M$ is of type I or III, then there exists (up to an isomorphism) precisely
one non-split supermanifold with retract $(M,\Omega)$, namely, the canonical
one. The corresponding invariant vector-valued $(2,1)$-form is the form
$\theta_{2}$ given by the formula $\eqref{(4.13)}$, $\omega$ being determined
by $\eqref{(4.16)}$.

  2. $2)$

If $M=\operatorname{Gr}^{n}_{s},\;1<s<n-1$ is of type II, then non-split
supermanifolds with retract $(M,\Omega)$ are parametrized by
$\mathbb{CP}^{1}/\Sigma$, where

$\Sigma=\begin{cases}\mathbb{Z}_{2}&\text{if }n=2s\\\
\\{e\\}&\text{otherwise}.\end{cases}$

The corresponding invariant vector-valued $(2,1)$-forms are
$a\theta_{2}+b\eta$, where $\eta_{o}$ is given by the formula $\eqref{(4.19)}$
and $a,b\in\mathbb{C}$ serve as homogeneous coordinates in $\mathbb{CP}^{1}$.
For $n=2s$, the action of the generator $\sigma$ of $\Sigma$ is expressed in
these coordinates as follows: $\sigma(a\,:\,b)=(a\,:\,-b)$.

###### Proof.

Similarly to the proof of Theorem 5.8, we have

$(H^{1}(M,\mathcal{T}_{2})\setminus\\{0\\})/\operatorname{Aut}\,\mathbf{T}(M)^{*}=\mathbb{P}(H^{1}(M,\mathcal{T}_{2}))/\Sigma=\mathbb{P}(\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K})/\Sigma.$

Then, one applies Proposition 5.12. Suppose that
$M=\operatorname{Gr}^{2s}_{s}$ for $s\geq 2$. It is known (one deduces this
from [38, $\operatorname{S}$ 15, Theorem 3]) that the generator $\sigma$ of
$\Sigma$, being regarded as a biholomorphic transformation of $M$, acts as
follows:

$\sigma(gP)=A(g)P,\ \ g\in G=\operatorname{SL}_{2s}(\mathbb{C}),$

where $A$ is the automorphism of $G$ given by the formula

$A(g)=\begin{pmatrix}0&I_{s}\\\
I_{s}&0\end{pmatrix}(g^{\top})^{-1}\begin{pmatrix}0&I_{s}\\\
I_{s}&0\end{pmatrix}.$

We easily check that the automorphism $d_{e}A$ acts on $\mathfrak{n}_{\pm}$ by

$d_{e}A(u)=-u^{\top},\ \ u\in\mathfrak{n}_{\pm}.$

By formula (5.71),

$\eta_{o}(-u_{1}^{\top},-u_{2}^{\top},-v^{\top})=\eta(u_{1},u_{2},v)^{\top}.$

Therefore, $\sigma^{*}\eta=-\eta$. Clearly, $\sigma^{*}\theta_{2}=\theta_{2}$.
Thus, $\sigma^{*}(a\theta_{2}+b\eta)=a\theta_{2}+(-b)\eta$. ∎

Comparing Theorem 5.31 with Theorem 5.8, we see that the construction of
Subsection 5.8 gives all non-split supermanifolds with retract $\Omega$ in the
cases I and III, while this is not true in the case II (one uses Proposition
4.43, see Example 5.12). Let us now fix a non-split supermanifold
$(M,\mathcal{O})$ with retract $(M,\Omega)$, where $M$ is a compact
irreducible Hermitian symmetric space. Changing the notation, we will denote
by $\mathcal{T}$ the tangent sheaf $\mathcal{D}er\,\mathcal{O}$ of
$(M,\mathcal{O})$, setting
$\mathcal{T}_{\operatorname{gr}}=\mathcal{D}er\,\Omega$. Our goal is to
calculate the cohomology groups $H^{q}(M,\mathcal{T})$ for $q=0,1$. These
groups depend on the non-zero form
$\theta\in\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$ which parametrizes the
supermanifolds $(M,\mathcal{O})$, as it has been described above. Let $M$ be a
simply connected irreducible compact Hermitian symmetric space of dimension
$\geq 2$. Let $(M,\mathcal{O})$ be a non-split supermanifold with retract
$(M,\Omega)$, the tangent sheaf $\mathcal{T}$, and the corresponding vector-
valued form $\theta\in\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$.

  1. $(1)$

Let $M$ be of type I and $\theta=\theta_{2}$, where $\omega$ is determined by
$\eqref{(4.16)}$. Then,

$H^{0}(M,\mathcal{T}_{\bar{0}})=\mathfrak{v}(M,\mathcal{O})_{\bar{0}}\simeq\mathfrak{g}\textup{~{}~{}(as
Lie algebras),}$

while (as $\mathfrak{g}$-modules)

$\displaystyle H^{0}(M,\mathcal{T}_{\bar{1}})$
$\displaystyle=\mathfrak{v}(M,\mathcal{O})_{\bar{1}}\simeq\mathbb{C}$
$\displaystyle H^{1}(M,\mathcal{T})$
$\displaystyle=H^{1}(M,\mathcal{T}_{\bar{0}})\simeq\mathfrak{g}.$

The basic element $\hat{d}\in\mathfrak{v}(M,\mathcal{O})_{\bar{1}}$ satisfies
$[\hat{d},\hat{d}]=0$.

  2. $(2)$

Let $M$ be of type II, i.e., $M=\operatorname{Gr}^{n}_{s}$ for $2\leq s\leq
n-2$, and $\theta=a\theta_{2}+b\eta$, where $a\neq 0$. If $n\geq 5$, or $n=4$
and $(a,b)$ is not proportional to $(\sqrt{2},\pm 1)$, then
$H^{0}(M,\mathcal{T})$ is as in $(1)$, while

$H^{1}(M,\mathcal{T})=H^{1}(M,\mathcal{T}_{\bar{0}})\simeq\mathfrak{g}\oplus\mathbb{C}\textup{~{}~{}(as
$\mathfrak{g}$-modules).}$

  3. $(3)$

Let $M=\operatorname{Gr}^{4}_{2},\;\theta=\sqrt{2}\theta_{2}+\eta$. Then,
$H^{0}(M,\mathcal{T})$ is as in $(1)$, while (as $\mathfrak{g}$-modules)

$\displaystyle H^{1}(M,\mathcal{T}_{\bar{0}})$
$\displaystyle\simeq\mathfrak{g}\oplus\mathbb{C},$ $\displaystyle
H^{1}(M,\mathcal{T}_{\bar{1}})$ $\displaystyle\simeq\mathbb{C}$

  4. $(4)$

Let $M$ be of type II and $\theta=\eta$. Then,

$H^{0}(M,\mathcal{T}_{\bar{0}})=\mathfrak{v}(M,\mathcal{O})_{\bar{0}}\simeq\mathfrak{g}\textup{~{}~{}(as
Lie algebras),}$

while (as $\mathfrak{g}$-modules)

$\displaystyle H^{0}(M,\mathcal{T}_{\bar{1}})$
$\displaystyle=\mathfrak{v}(M,\mathcal{O})_{\bar{1}}\simeq\mathfrak{g}\oplus\mathbb{C},$
$\displaystyle H^{1}(M,\mathcal{T}_{\bar{0}})$
$\displaystyle\simeq\mathfrak{g}\oplus\mathbb{C},$ $\displaystyle
H^{1}(M,\mathcal{T}_{\bar{1}})$ $\displaystyle\simeq\mathfrak{g}.$

  5. $(5)$

Let $M$ be of type III, i.e., $M=\mathbb{CP}^{n-1}$ for $n\geq 3$, and
$\theta=\theta_{2}=\eta$. Then, $H^{0}(M,\mathcal{T})$ is as in $(4)$, while
(as $\mathfrak{g}$-modules)

$H^{1}(M,\mathcal{T})=H^{1}(M,\mathcal{T}_{\bar{1}})\simeq\begin{cases}0&\text{for
}n\geq 4,\\\ \mathbb{C}&\text{for }n=3.\end{cases}$

####  5.32.1 Comment and Open problem.

Due to the isomorphism between $gr_{p}H^{q}(M,\mathcal{T})$ and
$E_{\infty}^{p,q-p}$, in the proof we need $H^{q}(M,\mathcal{T})$ only for
$q=0$ and 1, so for our purposes it is not necessary to compute $E^{p,q-p}$
for $q=2$ for any $p$. Therefore, some terms (denoted by ”?” in the tables
below) remain unknown and should be calculated for completeness.

###### Proof.

Consider the spectral sequence $(E_{r})$ associated with $(M,\mathcal{O})$ due
to Theorem 4.52. By this theorem,
$E_{2}^{p,q-p}=H^{q}(M,(\mathcal{T}_{\operatorname{gr}})_{p})$ and
$d_{2}=\operatorname{ad}_{l}^{*}([\theta])$, where $[\theta]\in
H^{1}(M,\Omega^{2}\otimes\Theta)^{G}$ is the cohomology class of $\theta$.
Clearly, $d_{2}$ is $G$-equivariant. We are going to calculate $d_{2}$ on
$E_{2}^{p,q-p}$ for $q=0,1$. The case $q=0,\;p=-1$ is settled by Proposition
5.26. In the case where $q=p=0$, we see that

$\displaystyle{}[l^{*}([\theta]),l(v)]$
$\displaystyle=l^{*}([\theta,v])=l^{*}([[\theta,v]])=0,$
$\displaystyle{}[l^{*}([\theta]),\varepsilon]$
$\displaystyle=-2l^{*}([\theta]).$

Clearly, $d_{2}(E_{2}^{0,1})=d_{2}(E_{2}^{2,-1})=0$. The mapping
$d_{2}:E_{2}^{0,1}\longrightarrow E_{2}^{2,0}$ is 0, too, since $E_{2}^{2,0}$
is a trivial $G$-module. Similarly, $d_{2}=0$ on
$l^{*}(H^{1}(M,\Omega^{1}\otimes\Theta)\subset E_{2}^{1,0}$. Now, for any
$\varphi\in\Gamma(M,\Phi^{2,1}\otimes\Theta)^{K}$ we have

$[l(\theta),i(\varphi)]=[i(\varphi),l(\theta)]=l(\theta\barwedge\varphi),$

due to (3.28), since $[\varphi,\theta]=0$ by Proposition 5.12. By Theorem
4.14, $d_{2}i^{*}(\varphi)=l^{*}([\theta\barwedge\varphi])$. This class can be
calculated with the help of Proposition 5.26 (note that, by Theorem 5.26, the
forms $\sqrt{2}\theta_{2}\pm\eta$ determine isomorphic non-split
supermanifolds). This settles the case $p=q=1$. Summarizing, we see that the
terms $E_{3}^{p,q-p}=E_{4}^{p,q-p}$ for $q=0,1,2$, are as follows (for the
definition of $s$, see (5.89)): Case I, $\theta=\theta_{2}$:

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&-&\--&\--&\--&\---&\---\\\ 0&&\vrule&0&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&\mathfrak{g}&0&0&0&0\\\ 2&&\vrule&0&0&0&?&\mathbb{C}^{s-1}&0\\\
\end{matrix}$

Case II, $\theta=a\theta_{2}+b\eta,\;a\neq 0$ :

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\--&\---&\---\\\
0&&\vrule&0&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&\mathfrak{g}&0&\mathbb{C}&0&0\\\
2&&\vrule&0&0&0&?&\mathbb{C}^{s-2}&0\\\ \end{matrix}$

Case II, $\theta=\eta$ :

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\--&\---&\---\\\
0&&\vrule&\mathfrak{g}&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&\mathfrak{g}&\mathfrak{g}&\mathbb{C}&0&0\\\
2&&\vrule&0&0&0&?&\mathbb{C}^{s-2}&0\\\ \end{matrix}$

Case III, $n\geq 4,\;\theta=\theta_{2}=\eta$ :

$\begin{matrix}&p&\vrule&-1&0&1&2&3&4\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\--&\--&\---\\\
0&&\vrule&\mathfrak{g}&\mathfrak{g}&\mathbb{C}&0&0&0\\\
1&&\vrule&0&0&0&0&0&0\\\ 2&&\vrule&0&0&0&?&0&0\\\ \end{matrix}$

Case III, $n=3,\;\theta=\theta_{2}=\eta$ :

$\begin{matrix}&p&\vrule&-1&0&1&2\ldots\\\ q&&\vrule\\\
-&-&-&\--&\--&\--&\---\\\ 0&&\vrule&\mathfrak{g}&\mathfrak{g}&\mathbb{C}&0\\\
1&&\vrule&0&0&\mathbb{C}&0\\\ 2&&\vrule&0&0&0&0\\\ \end{matrix}$

Clearly, for $q=0,1$ we have $d_{4}=d_{6}=\ldots=0$, and hence
$E_{3}^{p,q-p}=E_{\infty}^{p,q-p}$ for all $p\geq 0$. This implies our
theorem. ∎

Under assumptions of Theorem 5.31, we have

$\displaystyle\mathfrak{v}(M,\mathcal{O})_{\bar{0}}$
$\displaystyle\simeq\mathfrak{g},$
$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(0)}$
$\displaystyle\simeq\mathfrak{g}\oplus\mathbb{C},$
$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(1)}$
$\displaystyle\simeq\mathbb{C},$
$\displaystyle\mathfrak{v}(M,\mathcal{O})_{(p)}$ $\displaystyle=0\text{ for
}p\geq 2.$

In the cases $(1),\,(2),\,(3)$,
$\mathfrak{v}(M,\mathcal{O})=\mathfrak{v}(M,\mathcal{O})_{(0)}$, and the
supermanifold $(M,\mathcal{O})$ is not homogeneous. In the remaining cases,
$\mathfrak{v}(M,\mathcal{O})\neq\mathfrak{v}(M,\mathcal{O})_{(0)}$.

###### Proof.

The claims about $\mathfrak{v}(M,\mathcal{O})_{(p)}$ are implied by the
calculation of the spectral sequence $(E_{r})$. It follows that
$\mathfrak{v}(M,\mathcal{O})_{\bar{0}}\simeq\mathfrak{g}$. In the cases
$(1),\,(2),\,(3)$, we see that
$\mathfrak{v}(M,\mathcal{O})=\mathfrak{v}(M,\mathcal{O})_{(0)}$. Therefore,
$\operatorname{ev}_{x}(v)=0$ for all
$v\in\mathfrak{v}(M,\mathcal{O})_{\bar{1}},\;x\in M$, and hence
$(M,\mathcal{O})$ is not homogeneous. ∎

##  6 The $\Pi$-symmetric super-Grassmannian

Consider the supermanifold $\Pi\operatorname{Gr}^{n|n}_{s|s}$ defined in
Example 2.12. Its reduction is the submanifold $M$ of
$\operatorname{Gr}^{n}_{s}\times\operatorname{Gr}^{n}_{s}$ consisting of the
vector subsuperspaces $L\subset\mathbb{C}^{n|n}$ of dimension $s|s$ satisfying
$L_{\bar{1}}=\Pi(L_{\bar{0}})$. Projecting $M$ onto the first factor, we
identify this manifold with $\operatorname{Gr}^{n}_{s}$. Denoting $r=n-s$, we
suppose that $r,s\geq 1$. Assume that $\Pi$ is given in the standard basis by
the matrix

$\Pi=\begin{pmatrix}0&I_{n}\\\ I_{n}&0\end{pmatrix},$

and define local coordinates in a neighborhood of the point $o=\langle
e_{r+1},\ldots,e_{n},f_{1},\ldots,f_{s}\rangle$ in $M$ identified with
$\langle e_{r+1},\ldots,e_{n}\rangle\in\operatorname{Gr}^{n}_{s}$. Clearly,
the subsupermanifold $\Pi\operatorname{Gr}^{n|n}_{s|s}$ of
$\operatorname{Gr}^{n|n}_{s|s}$ is defined in terms of the coordinate matrix
(2.14) by the equations

$Y=X,\ \ \operatorname{H}=\Xi.$

Thus, the coordinate matrix has the form

$Z=\begin{pmatrix}X&\Xi\\\ I_{s}&0\\\ \Xi&X\\\ 0&I_{s}\end{pmatrix},$ (6.90)

where $X$ and $\Xi$ are $(r\times s)$-matrices. Denoting $X:=(x_{i\alpha})$
and $\Xi:=(\xi_{i\alpha})$, we get the even local coordinates $x_{i\alpha}$
and the odd ones $\xi_{i\alpha},\;1\leq i\leq r,\;1\leq\alpha\leq s$, in a
neighborhood of the point $o$. Denote by $Q_{n}(\mathbb{C})$ the subsupergroup
of $\operatorname{GL}_{n|n}(\mathbb{C})$ that preserves $\Pi$. Its coordinate
matrix has the form

$\begin{pmatrix}A&B\\\ B&A\end{pmatrix},$ (6.91)

where $A$ and $B$ are $(n\times n)$-matrices of even and odd coordinates,
respectively, $\det A\neq 0$. The reduction $G_{0}$ of $Q_{n}(\mathbb{C})$ can
be identified, in an obvious way, with $\operatorname{GL}_{n}(\mathbb{C})$.
The Lie superalgebra $\mathfrak{q}_{n}(\mathbb{C})$ of $Q_{n}(\mathbb{C})$
consists of all complex matrices of the form (6.91) with arbitrary $(n\times
n)$-matrices $A$ and $B$, and its even part $\mathfrak{g}_{0}$ can be
identified with $\mathfrak{gl}_{n}(\mathbb{C})$. The supermanifold
$\Pi\operatorname{Gr}^{n|n}_{s|s}$ admits the standard action of
$Q_{n}(\mathbb{C})$, which is expressed in coordinates as the multiplication
of $Z$ from the left by the coordinate matrix (6.91). This action induces,
clearly, the standard transitive action of the Lie group
$G_{0}=\operatorname{GL}_{n}(\mathbb{C})$ on $M=\operatorname{Gr}^{n}_{s}$.
Let $P$ denote the isotropy subgroup of $G_{0}$ at the point $o\in M$; it
consists of all matrices of the form (4.7). We will use the notation
introduced in Example 5.68. Let us denote by $a\mapsto a^{*}$ the differential
of the standard action of $Q_{n}(\mathbb{C})$ on
$\Pi\operatorname{Gr}^{n|n}_{s|s}$. This is a homomorphism of the Lie
superalgebra $\mathfrak{q}_{n}(\mathbb{C})$ into the Lie superalgebra
$\mathfrak{v}(\Pi\operatorname{Gr}^{n|n}_{s|s})$ of holomorphic vector fields
on $\Pi\operatorname{Gr}^{n|n}_{s|s}$. In what follows, we need the expression
of this homomorphism restricted to $\mathfrak{p}$. The holomorphic vector
fields on $\Pi\operatorname{Gr}^{n|n}_{s|s}$ will be written in terms of the
local coordinates in a neighborhood of $o$ given by the matrix (6.90). Denote
the elements of $\mathfrak{p}$ by

$a_{1}=\begin{pmatrix}(a_{ij})&0\\\
0&0\end{pmatrix},\;a_{2}=\begin{pmatrix}0&0\\\
0&(b_{\alpha\beta})\end{pmatrix},\;v=\begin{pmatrix}0&0\\\ (v_{\alpha
j})&0\end{pmatrix},$

where
$(a_{ij})\in\mathfrak{gl}_{r}(\mathbb{C}),\;(b_{\alpha\beta})\in\mathfrak{gl}_{s}(\mathbb{C})$,
and $(v_{\alpha j})$ is an $(s\times r)$-matrix. We want to calculate the
corresponding fundamental vector fields. Clearly,

$\begin{pmatrix}A_{1}&0&0&0\\\ 0&I_{s}&0&0\\\ 0&0&A_{1}&0\\\
0&0&0&I_{s}\end{pmatrix}\begin{pmatrix}X&\Xi\\\ I_{s}&0\\\ \Xi&X\\\
0&I_{s}\end{pmatrix}=\begin{pmatrix}A_{1}X&A_{1}\Xi\\\ I_{s}&0\\\
A_{1}\Xi&A_{1}X\\\ 0&I_{s}\end{pmatrix}.$

By substituting $A_{1}=\exp ta_{1}$ with $t\in\mathbb{C}$, by differentiating
at $t=0$ and changing the signs, we get

$a_{1}^{*}(x_{i\alpha})=-(a_{1}X)_{i\alpha},\ \
a_{1}^{*}(\xi_{i\alpha})=-(a_{1}\Xi)_{i\alpha},$ (6.92)

where we identify $a_{1}$ with $(a_{ij})$. Similarly, we find that

$\begin{pmatrix}I_{r}&0&0&0\\\ 0&A_{2}&0&0\\\ 0&0&I_{r}&0\\\
0&0&0&A_{2}\end{pmatrix}\begin{pmatrix}X&\Xi\\\ I_{s}&0\\\ \Xi&X\\\
0&I_{s}\end{pmatrix}\sim\begin{pmatrix}XA_{2}^{-1}&\Xi A_{2}^{-1}\\\
I_{s}&0\\\ \Xi A_{2}^{-1}&XA_{2}^{-1}\\\ 0&I_{r}\end{pmatrix},$

whence

$a_{2}^{*}(x_{i\alpha})=(Xa_{2})_{i\alpha},\ \ a_{2}^{*}(\xi_{i\alpha})=(\Xi
a_{2})_{i\alpha},$ (6.93)

where we identify $a_{2}$ with $(b_{\alpha\beta})$. Further, for any
$t\in\mathbb{C}$, we get

$\begin{pmatrix}I_{r}&0&0&0\\\ tv&I_{s}&0&0\\\ 0&0&I_{r}&0\\\
0&0&tv&I_{s}\end{pmatrix}\begin{pmatrix}X&\Xi\\\ I_{s}&0\\\ \Xi&X\\\
0&I_{s}\end{pmatrix}=\begin{pmatrix}X&\Xi\\\ I_{s}+tvX&tv\Xi\\\ \Xi&X\\\
tv\Xi&I_{r}+tvX\end{pmatrix}.$

Multiplying the result from the right by

$\begin{pmatrix}I_{s}+tvX&tv\Xi\\\
tv\Xi&I_{r}+tvX\end{pmatrix}^{-1}=\begin{pmatrix}I_{s}-tvX+\ldots&-tv\Xi+\ldots\\\
-tv\Xi+\ldots&I_{r}-tvX+\ldots\end{pmatrix},$

where the omitted terms are of order $>1$ in $t$, we get the matrix

$\begin{pmatrix}X-t(XvX+\Xi v\Xi)+\ldots&\Xi-t(\Xi vX+Xv\Xi)+\ldots\\\
I_{s}&0\\\ \Xi-t(\Xi vX+Xv\Xi)+\ldots&X-t(XvX+\Xi v\Xi)+\ldots\\\
0&I_{r}\end{pmatrix}.$

Therefore,

$\displaystyle v^{*}(x_{i\alpha})$ $\displaystyle=(XvX+\Xi v\Xi)_{i\alpha},$
(6.94) $\displaystyle v^{*}(\xi_{i\alpha})$ $\displaystyle=(\Xi
vX+Xv\Xi)_{i\alpha},$

where we identify $v$ with $(v_{\alpha j})$. From (6.92), (6.93), and(6.94) we
get [Explicit formulas of vector fields] We have

$\displaystyle a_{1}^{*}$
$\displaystyle=-\sum_{i,k=1}^{r}\sum_{\alpha=1}^{s}a_{ik}x_{k\alpha}\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}-\sum_{i,k=1}^{r}\sum_{\alpha=1}^{s}a_{ik}\xi_{k\alpha}\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}},$
$\displaystyle a_{2}^{*}$
$\displaystyle=\sum_{\alpha,\beta=1}^{s}\sum_{i=1}^{r}b_{\beta\alpha}x_{i\beta}\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}+\sum_{\alpha\beta=1}^{s}\sum_{i=1}^{r}b_{\beta\alpha}\xi_{i\beta}\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}},$
$\displaystyle v^{*}$
$\displaystyle=\sum_{i,j=1}^{r}\sum_{\alpha,\beta=1}^{s}v_{\beta
j}(x_{i\beta}x_{j\alpha}+\xi_{i\beta}\xi_{j\alpha})\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{x_{i\alpha}}}$}}$
$\displaystyle+\sum_{i,j=1}^{r}\sum_{\alpha,\beta=1}^{s}v_{\beta
j}(\xi_{i\beta}x_{j\alpha}+x_{i\beta}\xi_{j\alpha})\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}.$

Let $\mathcal{O}$ denote the structure sheaf of the supermanifold
$\Pi\operatorname{Gr}^{n|n}_{s|s}$. Clearly, the action of $G$ on
$(M,\mathcal{O})$ determines a linear representation of the group $P$ by
automorphisms of the superalgebra $\mathcal{O}_{o}$, which gives a linear
representation $\chi=\chi_{\bar{0}}+\chi_{\bar{1}}$ of this group in
$T_{o}(M,\mathcal{O})$, called the isotropy representation. Proposition 6.94
easily implies its explicit expression. Indeed, denote the tautological
representations of $\operatorname{GL}_{r}(\mathbb{C})$ and
$\operatorname{GL}_{s}(\mathbb{C})$ by $\rho_{1}$ and $\rho_{2}$,
respectively. Let $\widetilde{m}_{o}$ be the linear span of germs at $o$ of
all coordinate functions $x_{i\alpha},\;\xi_{i\alpha}$ in $m_{o}$. Then,
$m_{o}=\widetilde{m}_{o}\oplus m_{o}^{2}$. As Proposition 6.94 shows,
$v^{*}(\widetilde{m}_{o})\subset m_{o}^{2}$ for all $v\in\mathfrak{n}_{-}$,
and hence $\mathfrak{n}_{-}$ trivially acts on $m_{o}/m_{o}^{2}$. The same
proposition implies that $\widetilde{m}_{o}$ is invariant under $\mathfrak{r}$
(or $R$), inducing in both components $(\widetilde{m}_{o})_{\bar{0}}$ and
$(\widetilde{m}_{o})_{\bar{1}}$ the representation
$\rho_{1}^{*}\otimes\rho_{2}$ of $R$. As in Example 5.68, we consider the
maximal algebraic torus $T$ of $R$ and $G_{0}$ consisting of all diagonal
matrices. We will write the matrices of the corresponding Cartan subalgebra
$\mathfrak{t}$ in the form

$H=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{t},\lambda_{t+1},\ldots,\lambda_{n}),\;\lambda_{i}\in\mathbb{C}.$

Proposition 6.94 also implies that the germs of $x_{i\alpha},\;\xi_{\alpha
i},\;\eta_{i\alpha}$ form a weight basis for the representation $\chi^{*}$ in
$\widetilde{m}_{o}\simeq m_{o}/m_{o}^{2}$ with respect to $T$, the
corresponding weights being $-\lambda_{i}+\lambda_{t+\alpha}$, where $1\leq
i\leq r,\;1\leq\alpha\leq s$ (with multiplicity 2). Thus, we got

  1. $(1)$

The isotropy representation $\chi$ is completely reducible, and the
restrictions of its even and odd components onto $R$ are as follows:

$\chi_{\bar{0}}|_{R}\simeq\chi_{\bar{1}}|_{R}\simeq\rho_{1}\otimes\rho_{2}^{*}.$

  2. $(2)$

The germs of $x_{i\alpha},\;\xi_{i\alpha}$ form a weight basis with respect to
$T$ in their linear span $\widetilde{m}_{o}$, the corresponding weights being
in both cases $-\lambda_{i}+\lambda_{t+\alpha}$, where $1\leq i\leq r$ and
$1\leq\alpha\leq s$.

Note that $\chi_{\bar{0}}$ coincides with the isotropy representation $\tau$
of the homogeneous space $\operatorname{Gr}^{n}_{s}(\mathbb{C})$ (see (5.70)).
Clearly, the action of $G_{0}$ on the sheaf $\mathcal{O}$ leaves invariant the
filtration (2.12) and induces an action of this group on the locally free
sheaf $\mathcal{E}=\mathcal{J}/\mathcal{J}^{2}$, and hence on the
corresponding vector bundle $\mathbf{E}$, covering its standard action on $M$.
Thus, $\mathbf{E}$ is a homogeneous vector bundle over $M$. The vector bundle
$\mathbf{E}$ is isomorphic to the cotangent bundle $\mathbf{T}(M)^{*}$. The
retract of the super-Grassmannian $(M,\mathcal{O})$ is isomorphic to the
supermanifold $(M,\Omega)$ from Example 2.7.

###### Proof.

By Proposition 6, the representation of $P$ in
$E_{o}=T_{o}(M,\mathcal{O})_{\bar{1}}^{*}$ is isomorphic to $\tau^{*}$. Hence,
$\mathbf{E}\simeq\mathbf{E}_{\tau^{*}}=(\mathbf{E}_{\tau})^{*}\simeq\mathbf{T}(M)^{*}$.
∎

Next, I want to prove that our super-Grassmannian is, as a rule, non-split.
Note that the canonical action of $G_{0}$ on $(M,\mathcal{O})$ gives rise to a
natural linear action of this groups on the tangent sheaf $\mathcal{T}$
leaving invariant the $\mathbb{Z}_{2}$-grading. As a result, we get a linear
representation of $G_{0}$ in the cohomology groups of $\mathcal{T}$ and, in
particular, in the Lie superalgebra $\mathfrak{v}(M,\mathcal{O})$. The
corresponding linear representation of the Lie algebra $\mathfrak{g}_{0}$ is
given by the formula $u\mapsto\operatorname{ad}_{u^{*}}$. If $r\geq 2$ or
$s\geq 2$, then $\mathfrak{v}(M,\mathcal{O})_{\bar{0}}^{G_{0}}=0$.

###### Proof.

Any $\delta\in\mathfrak{v}(M,\mathcal{O})_{\bar{0}}^{G_{0}}$ determines a
$P$-invariant even derivation of the superalgebra $\mathcal{O}_{o}$ (we denote
it by the same character $\delta$), Clearly, $\delta$ preserves the maximal
ideal $m_{o}$. Consider the vector subspace $\widetilde{m}_{o}\subset m_{o}$,
spanned by the germs of local coordinates at $o$. By Proposition 6, $R$
preserves the even and the odd parts of this subspace, inducing in each part
an irreducible representation, and the germs of local coordinates constitute a
weight basis of $\widetilde{m}_{o}$ with respect to $T$ with the weights
$-\lambda_{i}+\lambda_{r+\alpha}$, where $1\leq i\leq r$ and $1\leq\alpha\leq
s$. Note that the remaining weights of the representation of $R$ in the whole
$m_{o}$ are certain sums of these weights, and hence we see that the weight
subspace of $m_{o}$ corresponding to any of these weights is two-dimensional
(and lies in $\widetilde{m}_{o}$). Since $\delta$ is even and $P$-invariant,
the germs of local coordinates are eigenvectors for $\delta$. Moreover, the
Schur lemma implies that

$\delta(x_{i\alpha})=ax_{i\alpha},\;\delta(\xi_{i\alpha})=b\xi_{i\alpha},$

where $a,b\in\mathbb{C}$. We have $a=0$. Indeed, consider the vector field
$\widetilde{\delta}=\sigma_{0}(\delta)\in\mathfrak{v}(M,\operatorname{gr}\mathcal{O})_{0}$
(see Subsection 3.1). Clearly, $\widetilde{\delta}$ is $G_{0}$-invariant, too,
and hence determines the $G_{0}$-invariant vector field
$\alpha(\widetilde{\delta})$ (see (3.21). But it is well known (see, e.g.,
[38]) that the standard action of $\operatorname{GL}_{n}(\mathbb{C})$ on $M$
is asystatic, i.e., $M$ has no non-zero holomorphic $G_{0}$-invariant vector
fields (for the origin of the term asystatic, see [15*, 16*] and interesting
references therein). This implies that
$\widetilde{\delta}(x_{i\alpha}+\mathcal{J})=0$. Therefore,
$\delta(x_{i\alpha})\in\mathcal{J}^{2}$, whence $a=0$. Now we prove that
$b=0$, using the relation $[\delta,v^{*}]=0$ for all $v\in\mathfrak{n}_{-}$.
Proposition 6.94 implies that

$0=[\delta,E_{r+1,1}^{*}](x_{12})=\delta(E_{r+1,1}^{*}(x_{12}))=\delta(\xi_{11}\xi_{12})=2b\xi_{11}\xi_{12}.$

This implies our assertion whenever $s\geq 2$. To prove the assertion for
$r\geq 2$, one takes $x_{21}$ instead of $x_{12}$. ∎

This result makes it possible to solve the splittness question concerning the
super-Grassmannians studied here. [On splitness of
$\Pi\operatorname{Gr}^{n|n}_{s|s}$] The super-Grassmannian
$\Pi\operatorname{Gr}^{n|n}_{s|s}$ is split if and only if $n=2$ and $s=1$.

###### Proof.

Consider the grading derivation $\varepsilon$ of the $\mathbb{Z}$-graded sheaf
$\operatorname{gr}\mathcal{O}$ defined in Subsection 3.4 and the natural
homomorphism of Lie superalgebras
$\sigma_{0}:H^{0}(M,\mathcal{T}_{\bar{0}})\longrightarrow
H^{0}(M,\widetilde{\mathcal{T}}_{0})$ defined in Subsection 3.1. Proposition 6
implies that $\varepsilon\notin\operatorname{Im}\sigma$ whenever $s\geq 2$ or
$r\geq 2$. Indeed, if $\varepsilon=\sigma_{0}(\delta)$, where $\delta\in
H^{0}(M,\mathcal{T}_{\bar{0}})$, then the complete reducibility of the
representation of $G_{0}$ in $H^{0}(M,\mathcal{T}_{\bar{0}})$ implies that
$\delta$ can be chosen to be $G_{0}$-invariant. But then $\delta=0$, whence
$\varepsilon=0$, which gives a contradiction. If $(M,\mathcal{O})$ is split,
then $\sigma$ is an isomorphism, but this is false whenever $s\geq 2$ or
$r\geq 2$. In the case $n=2,\,s=1$, we can see that the super-Grassmannian is
split, e.g., by calculating its transition functions. ∎

An important property of $\Pi\operatorname{Gr}^{n|n}_{s|s}$ is the
homogeneity, which we are going to prove now.
[$\Pi\operatorname{Gr}^{n|n}_{s|s}$ is homogeneous]

  1. $(1)$

The canonical action of $\mathfrak{q}_{n}(\mathbb{C})$ on the supermanifold
$\Pi\operatorname{Gr}^{n|n}_{s|s}$ is transitive.

  2. $(2)$

The kernel of this action is $\langle I_{n|n}\rangle$.

###### Proof.

To prove (1), we have to calculate the vector fields $y^{*}$ corresponding to
certain odd elements of $\mathfrak{q}_{n}(\mathbb{C})$. More precisely, take
the matrix $y=\begin{pmatrix}0&B\\\
B&0\end{pmatrix}\in\mathfrak{q}_{n}(\mathbb{C})_{\bar{1}}$, where

$B=\begin{pmatrix}0&Y\\\ 0&0\end{pmatrix},$

$Y=(y_{i\alpha})$ being an $(r\times s)$-matrix. Denoting by $\tau$ an odd
parameter, we get

$\begin{pmatrix}I_{n}&\tau B\\\ \tau
B&I_{n}\end{pmatrix}\begin{pmatrix}X&\Xi\\\ I_{s}&0\\\ \Xi&X\\\
0&I_{s}\end{pmatrix}=\begin{pmatrix}X&\Xi+\tau Y\\\ I_{s}&0\\\ \Xi+\tau Y&X\\\
0&I_{s}\end{pmatrix}.$

It follows that

$y^{*}=-\sum_{i,\alpha}y_{i\alpha}\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{0.25pt}{$\dfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial}{\partial{\xi_{i\alpha}}}$}}.$

Clearly, $\operatorname{ev}_{o}(y^{*})$ span the vector space
$T_{o}((M,\mathcal{O}))_{\bar{1}}$. Since our action is $\bar{0}$-transitive,
its transitivity follows from Proposition 3.7(2). Let us denote by
$\mathfrak{q}$ the kernel of our action. We see from Proposition 6.94 that
$I_{n|n}\in\mathfrak{q}$. Since
$\mathfrak{g}_{0}=\mathfrak{gl}_{n}(\mathbb{C})$ acts on $M$ in the standard
way, it follows that $\mathfrak{q}\cap\mathfrak{g}_{0}=\langle
I_{n|n}\rangle$. But it is known (see, e.g., [24]) that the only ideal of
$\mathfrak{q}_{n}(\mathbb{C})$ containing $\langle I_{n|n}\rangle$ is

$\mathfrak{sq}_{n}(\mathbb{C}):=\left\\{\begin{pmatrix}A&B\\\
B&A\end{pmatrix}\,\vrule\,\operatorname{tr}B=0\right\\}.$

As we have seen above, $\mathfrak{q}\neq\mathfrak{sq}_{n}(\mathbb{C})$. Hence,
$\mathfrak{q}=\langle I_{n|n}\rangle$. ∎

Now we are able to prove our main result concerning $\Pi$-symmetric super-
Grassmannians. Let $(M,\mathcal{O})=\Pi\operatorname{Gr}^{n|n}_{s|s}$ and
$n\geq 3$.

  1. $1)$

In the classification of non-split supermanifolds with retract
$(\operatorname{Gr}^{n}_{,s},\Omega)$ given by Theorem 5.26, $(M,\mathcal{O})$
corresponds to the invariant $(2,1)$-form $\eta$.

  2. $2)$

The natural action of the Lie superalgebra $\mathfrak{q}_{n}(\mathbb{C})$ on
$(M,\mathcal{O})$ determines an isomorphism of Lie superalgebras

$\mathfrak{v}(M,\mathcal{O})\simeq{\mathfrak{p}}{\mathfrak{q}}_{n}(\mathbb{C}):=\mathfrak{q}_{n}(\mathbb{C})/\langle
I_{n|n}\rangle.$

  3. $3)$

If $2\geq s\geq n-2$, then

$\displaystyle H^{1}(M,\mathcal{T}_{\bar{0}})$
$\displaystyle\simeq\mathfrak{sl}_{n}(\mathbb{C})\oplus\mathbb{C},$
$\displaystyle H^{1}(M,\mathcal{T}_{\bar{1}})$
$\displaystyle\simeq\mathfrak{sl}_{n}(\mathbb{C}).$

  4. $4)$

If $s=1$ or $n-1$, then

$H^{1}(M,\mathcal{T})=H^{1}(M,\mathcal{T}_{\bar{1}})\simeq\begin{cases}0\text{
for }n\geq 4\\\ \mathbb{C}\text{ for }n=3\end{cases}.$

###### Proof.

By Corollary 4.1, the supermanifold corresponding to the form
$a\theta_{2}+b\eta$ in the case II cannot be homogeneous if $a\neq 0$. By
Proposition 6(1), this implies (1). By Proposition 6(2), the natural action of
$\mathfrak{q}_{n}(\mathbb{C})$ on $(M,\mathcal{O})$ induces an injective
homomorphism $\mathfrak{q}_{n}(\mathbb{C})/\langle
I_{n|n}\rangle\longrightarrow\mathfrak{v}(M,\mathcal{O})$. Comparing this with
Theorem 5.31(4), we see that this homomorphism is surjective. Thus, (2) is
proved. The assertions (2) and (3) follow from (1) and Theorem 5.31. ∎

Theorems 6(1) and 5.26 imply [A family of deformations of
$\Pi\operatorname{Gr}^{n|n}_{s|s}$] The supermanifold
$\Pi\operatorname{Gr}^{n|n}_{s|s}$ for $2\leq s\leq n-2$ is included in a
$1$-parameter family of mutually non-isomorphic supermanifolds with the same
retract. In particular, it is not rigid. To conclude, we note that these
properties of the $\Pi$-symmetric super-Grassmannians contrast with the
rigidity of certain other series of super-Grassmannians (see Examples 2.7,
2.7, 2.12). Let us denote by $(M,\mathcal{O})$ one of these super-
Grassmannians, by $(M,\mathcal{O}_{\operatorname{gr}})$ its retract and by
$\mathcal{T},\;\mathcal{T}_{\operatorname{gr}}$ the corresponding tangent
sheaves. [Rigid super-Grassmannians] Suppose that $(M,\mathcal{O})$ is one of
the following supermanifolds:

$\displaystyle\operatorname{Gr}^{n|m}_{k|l}\text{ with }0<k<m,\;0<l<m,\;$
$\displaystyle(k,l)\neq$
$\displaystyle(1,n-1),\;(m-1,1),\;(1,n-2),\;(m-2,1),\;(2,n-1),\;(m-1,2);$
$\displaystyle\operatorname{I}\operatorname{Gr}^{2r|r}_{2s|s}\text{ with
}r\geq 2,\;(r,s)\neq(2,1);$
$\displaystyle\operatorname{I}_{\operatorname{odd}}\operatorname{Gr}^{n|n}_{s|n-s}\text{
with }4\leq s\leq n-3.$

Then, $(M,\mathcal{O})$ is the only non-split supermanifold with retract
$(M,\mathcal{O}_{\operatorname{gr}})$ and, moreover, $(M,\mathcal{O})$ is
rigid.

###### Proof.

It is known that in all the cases listed above we have

$H^{1}(M,(\mathcal{T}_{\operatorname{gr}})_{p})=\begin{cases}\mathbb{C}&\text{if
}p=2\\\ 0&\text{otherwise}\end{cases}$

(see [37, Theorem 1] for $(M,\mathcal{O})=\operatorname{Gr}^{n|m}_{k|l}$, [43,
Theorem 1] for
$(M,\mathcal{O})=\operatorname{I}\operatorname{Gr}^{2r|r}_{2s|s}$, [44,
Theorem 1] for
$(M,\mathcal{O})=\operatorname{I}_{\operatorname{odd}}\operatorname{Gr}^{n|n}_{s|n-s}$.
Moreover, it was proved in these papers that the supermanifolds
$(M,\mathcal{O})$ are non-split. By Proposition 4.4, $(M,\mathcal{O})$ is the
only non-split supermanifold with retract
$(M,\mathcal{O}_{\operatorname{gr}})$, and the corresponding class
$\lambda_{2}^{*}(\gamma)$ is a basic element of
$H^{1}(M,(\mathcal{T}_{\operatorname{gr}})_{2})$. As in the proof of Theorem
5.31, we have $d_{2}(\varepsilon)=-2\lambda_{2}^{*}(\gamma)$ in the spectral
sequence $(E_{r})$. Theorem 4.52 implies that $H^{1}(M,\mathcal{T})=0$, and
hence $(M,\mathcal{O})$ is rigid. (The vanishing of $H^{1}(M,\mathcal{T})$ was
proved in the cited papers as well.) ∎

Note that the super-Grassmannians listed in Theorem 6, together with the
$\Pi$-symmetric super-Grassmannians, are just the supermanifolds of flags that
can be called symmetric superspaces (see [49]).

### Acknowledgements

The author wishes to thank the Laboratory of Mathematics of the University of
Poitiers (France) and the E. Schrödinger International Institute for
Mathematical Physics (Vienna, Austria) for the hospitality during the spring
semester 1996, when this paper was written. I am grateful to P. Torasso, P.
Michor and M. Eastwood for valuable discussions.

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