# Splitting quasireductive supergroups and volumes of supergrassmannians

Vera Serganova, Alexander Sherman

###### Abstract.

We introduce the notion of splitting subgroups of quasireducitve supergroups,
and explain their significance. For $GL(m|n)$, $Q(n)$, and defect one basic
classical supergroups, we give explicit splitting subgroups and prove they are
minimal up to conjugacy, except in the $GL(m|n)$ case where it remains a
conjecture. A key tool in the proof is the computation of the volumes of
complex supergrassmannians, which is of interest in its own right.

##  1\. Introduction

In this paper we deal with representations of quasireductive supergroups $G$,
meaning that $G_{0}$, the underlying group of $G$, is reductive. Many
important supergroups (including $GL(m|n)$, $SOSp(m|2n)$, $Q(n)$,..) are
quasireductive, and although their representation theory is similar to that of
reductive groups, semisimplicity of $\operatorname{Rep}G$ is very rare. On the
other hand, the property of being quasireductive is equivalent to
$\operatorname{Rep}G$ being Frobenius, a property shared by representations of
finite groups in positive characteristic.

We say a quasireductive subgroup $K\subseteq G$ is _splitting_ if the trivial
$G$-submodule of $\mathbb{C}[G/K]$ spanned by the constant function 1 splits
off. Here $\mathbb{C}[G/K]$ denotes the superalgebra of polynomial functions
on $G/K$. In modular representation theory, the analogous notion is very
natural. Indeed, if $G$ is a finite group and $\Bbbk$ is a field of
characteristic $p>0$, then the $G$-submodule of constant functions in
$\Bbbk[G/K]$ splits off if and only if $|G|/|K|$ is prime to $p$. In
particular, the minimal ‘splitting’ subgroups in the modular setting would be
the Sylow $p$-subgroups of $G$, which are all conjugate to one another. In
modular representation theory in characteristic $p$, one may often reduce
questions about $G$-modules to analogous questions about modules over a
(normalizer of a) Sylow $p$-subgroup (see [B] for more on modular
representation theory).

Heuristically speaking, a splitting subgroup $K$ of $G$ has representation
theory that is at least as complicated as that of $G$, despite being a smaller
group. For example, if the trivial subgroup of $G$ is splitting, then $G$ has
semisimple representation theory. Further, projectivity of a $G$-module can
always be checked by restricting to a splitting subgroup.

###  1.1. Main theorem

In this paper we initiate a study of minimal splitting subgroups $K$ of
distinguished quasireductive supergroups $G$; or if not minimal, then at least
conjecturally so. The supergroups we consider are $GL(m|n),Q(n),$ and all
other supergroups $G$ such that $\operatorname{Lie}G$ is either
$\mathfrak{p}\mathfrak{s}\mathfrak{l}(n|n)$ or simple basic classical of
defect 1. In particular we show that following:

######  Theorem 1.1.

The following subgroups are splitting.

  1. (1)

$SL(1|1)^{\min(m,n)}\subseteq GL(m|n),\ \ \ P\left(SL(1|1)^{n}\right)\subseteq
PSL(n|n);$

  2. (2)

$Q(2)^{n}\subseteq Q(2n),\ \ \ \ Q(2)^{n}\times Q(1)\subseteq Q(2n+1);$

  3. (3)

for any $G$ with $\operatorname{Lie}G$ a defect 1 basic classical Lie
superalgebra:

$SL(1|1)\subseteq G.$

Further, in cases (2) and (3), these subgroups are minimal splitting
subgroups, and are unique up to conjugacy with this property.

In the above, the subgroups $SL(1|1)^{\operatorname{def}\mathfrak{g}}$, where
$\operatorname{def}\mathfrak{g}$ is the defect of $\mathfrak{g}$ (see Section
3.2), are those that arise from the root subgroup of
$(\operatorname{def}\mathfrak{g})$-many isotropic, mutually orthogonal odd
roots of $\mathfrak{g}$. In the case of $Q(n)$ we take the natural root
subgroup of the specified type (in fact it is a Levi subgroup).

The proof of (3) can be done using straightforward techniques in the algebraic
category, in particular by studying the space of functions on supersymmetric
spaces for the defect one supergroups. However for the infinite series
$GL(m|n)$ and $Q(n)$ we need techniques from differential geometry.

###  1.2. Connections to support varieties

In [BoKN1], [BoKN2], and [GGNW], cohomological support varieties are studied
for quasireductive Lie superalgebras $\mathfrak{g}$. There, they prove that
for basic Lie superalgebras and $\mathfrak{q}(n)$, there are detecting
subalgebras $\mathfrak{e},\mathfrak{f}\subseteq\mathfrak{g}$ on which one can
compute the support variety of a given $\mathfrak{g}$-module $M$ after
restricting to $\mathfrak{e}$ or $\mathfrak{f}$. In the basic classical cases,
their subalgebras $\mathfrak{f}$ agree with the Lie superalgebras of our
splitting subgroups above. For $\mathfrak{q}(n)$, they have
$\mathfrak{e}=\mathfrak{f}=\mathfrak{q}(1)^{n}$, so their detecting
subalgebras are smaller.

In a forthcoming work, we will study rank varieties determined by the Duflo-
Serganova functor for homological elements, i.e.
$\mathfrak{g}_{\overline{1}}^{hom}$ (see Section 2.6). Our splitting subgroups
will play an important role in these rank varieties; it will be interesting to
understand the connections of these rank varieties to the ones defined in
[BoKN1].

###  1.3. Volumes of supergrassmannians

In order to prove parts (1) and (2) of Theorem 1.1, we use a unitary trick.
First of all, by Lemma 2.11, we may reduce (1) to showing that
$GL(1|1)^{\min(m,n)}\subseteq GL(m|n)$ is splitting. Also, the case of
$PSL(n|n)$ may be reduced to that of $GL(n|n)$, so we put this case aside in
this introductory discussion.

Now both $GL(m|n)$ and $Q(n)$ admit real compact forms $U(m|n)$ and $UQ(n|n)$.
If we consider the natural embeddings

$U(1|1)^{\min(m,n)}\subseteq U(m|n),\ \ \ \ UQ(2)^{n}(\times UQ(1))\subseteq
UQ(2n(+1)),$

then their complexifications are the subgroups we are interested in showing
are splitting. On the other hand, their quotients are compact supermanifolds
admitting invariant volume forms; thus if we can show their volumes are
nonzero, we may obtain our desired splitting via the integration map.

Instead of computing these volumes directly, we answer a separate question,
which is sufficient for our purposes via an inductive argument for splitting
subgroups. Consider the supergrassmannian $Gr(r|s,m|n)$ or $(r|s)$-planes in
$\mathbb{C}^{m|n}$. This is a homogeneous space for $U(m|n)$ with stabilizer
of a point given by $U(r|s)\times U(m-r|n-s)$. The study of supergrassmannians
was initiated in the early 80s by Manin and his students, see [M] and [PS].

We prove that:

######  Theorem 1.2.

Let $\omega$ be an invariant volume from on $\mathcal{M}:=Gr(r|s,m|n)$; then

$\int_{\mathcal{M}}\omega\neq 0$

if and only if $\operatorname{sdim}\mathcal{M}\geq 0$.

An explicit formula for this volume is given in Corollary 5.18. We note that
while our formula does not agree with conjecture given in [V], the two
formulas are consistent on when the volume is nonzero.

Now using the unitary trick, Theorem 1.2 implies that $GL(r|s)\times
GL(m-r|n-s)\subseteq GL(m|n)$ is splitting if and only if
$(r-s)((m-r)-(n-s))\geq 0$. With the help of the transitivity of splitting
subgroups (see Corollary 2.7), one can obtain part (1) of Theorem 1.1.

For the $Q(n)$ case, we take $\mathbb{C}^{n|n}$ with a fixed odd linear
automorphism $F:\mathbb{C}^{n|n}\to\mathbb{C}^{n|n}$, and consider the space
$QGr(r,n)$, which is the supergrassmannian of $(r|r)$-dimensional, $F$-stable
subspaces of $\mathbb{C}^{n|n}$. This is a homogeneous space for $UQ(n)$ with
stabilizer $UQ(r)\times UQ(n-r)$.

######  Theorem 1.3.

Let $\omega$ be an invariant volume form on $\mathcal{M}:=QGr(r,n)$; then

$\int_{\mathcal{M}}\omega\neq 0$

if and only if $r(n-r)$ is even.

From this we can again use the unitary trick and transitivity to prove part
(2) of Theorem 1.1.

###  1.4. Localization of integrals

In order to prove Theorems 1.2 and 1.3, we use the results of [SZ], where a
localization theorem was proven for integrals on supermanifolds. We use the
following version of it.

Let $\mathcal{M}=Gr(r|r,n|n)$ or $QGr(r,n)$, and $\mathcal{U}=U(n|n)$ or
$UQ(n)$ in the respective cases; then there exists an element
$Q\in(\operatorname{Lie}\mathcal{U})_{\overline{1}}$ such that the zeroes of
$Q$ are isolated and nondegenerate; in other words if $p$ is an zero of $Q$,
then the natural endomorphism $Q:T_{p}\mathcal{M}\to T_{p}\mathcal{M}$ is an
isomorphism.

In this case, the result of [SZ] implies that the integral
$\int_{\mathcal{M}}\omega$ localizes to $Z(Q)$, the zero set of $Q$. In
particular we have that

$\int_{\mathcal{M}}\omega=(2\pi)^{\operatorname{dim}\mathcal{M}_{0}}\sum\limits_{p\in
Z(Q)}\alpha(T_{p}\mathcal{M},\omega_{p}),$

where $\alpha(T_{p}\mathcal{M},\omega_{p})\in\mathbb{C}$ can be computed in
terms of the form $\omega_{p}$ and the action of $Q$ on $T_{p}\mathcal{M}$
(see Section 5.1 for the precise definition). In our situation we are able to
explicitly compute $\alpha(T_{p}\mathcal{M},\omega_{p})$ to obtain the values
of our integrals and show they are nonzero. To deal with the general case of
$Gr(r|s,m|n)$, either the localization theorem applies similarly to the above,
or we can apply a trick of realizing the two step flag variety
$Fl(r_{1}|s_{1},r_{2}|s_{2},m|n)$ as a fiber bundle in two differ ways,
allowing us to obtain relationships between volumes of different
Grassmannians.

###  1.5. Summary of sections

Section 2 introduces splitting subgroups and proves general properties they
satisfy. Section 3 defines contragredient supergroups and their defect
subgroups, which are the natural candidates for minimal splitting subgroups.
We prove they are minimal in the defect one case, and also discuss what
happens for $G=Q(n)$. In Section 4 we prove defect subgroups are splitting for
all defect one contragredient supergroups via transitivity and the use of
symmetric spaces. Section 5 computes the volume of the supergrassmannians for
$GL(m|n)$ and $Q(n)$ via the localization technique of [SZ]. Finally Section 6
applies the volume computations to proving Theorem 1.1.

###  1.6. Acknowledgements

The authors are grateful to A. Schwarz and T. Voronov for numerous helpful
discussions. The first author was supported in part by NSF grant 2001191. The
second author was supported in part by ISF grant 711/18 and NSF-BSF grant
2019694.

##  2\. Splitting subgroups

###  2.1. Preliminaries and notation

We work throughout over the field $\mathbb{C}$ of complex numbers. For a super
vector space $V$, we write $V=V_{\overline{0}}\oplus V_{\overline{1}}$ for its
$\mathbb{Z}_{2}$-gradation. We will write $\mathcal{M}$ for a supervariety,
and let $\mathbb{C}[\mathcal{M}]$ denote the algebra of global functions on
$\mathcal{M}$. If $\mathcal{M}$ is smooth, we set
$\operatorname{dim}\mathcal{M}_{1}$ to be the odd dimension of $\mathcal{M}$.

###  2.2. Quasireductive supergroups

Recall that a linear algebraic supergroup $G$ is called quasireductive if the
underlying algebraic group $G_{0}$ is reductive. Let
$\mathfrak{g}:=\operatorname{Lie}G$ denote the Lie superalgebra of $G$. Then
by definition, a representation of $G$ (or a $G$-module) is a
$\mathfrak{g}$-module $V$ such that the action of
$\mathfrak{g}_{\overline{0}}$ on $V$ integrates to an action of $G_{0}$. We
write $\operatorname{Rep}G$ for the category of all representations of $G$,
and $\mathcal{F}(G)$ for the subcategory of finite-dimensional $G$-modules.

For a quasireductive supergroup $G$, the category $\mathcal{F}(G)$ is
Frobenius i.e., there are enough projective objects and every projective
object is injective, see [S1]. Since $\operatorname{Rep}G$ is the coinductive
completion of $\mathcal{F}(G)$, one may check that the class of projectives
and the class of injectives coincide in $\operatorname{Rep}G$, and that they
consist exactly of arbitrary direct sums of finite-dimensional indecomposable
projectives (or equivalently injectives).

###  2.3. Homogeneous spaces and induction

Let $K$ be a quasireductive subgroup of $G$. We may form the homogeneous
supervariety $G/K$, and we write $\mathbb{C}[G/K]$ for its space of functions.
Because $K$ and $G$ are quasireductive, $G/K$ in fact will be affine. See
[MaT] for more on the constructions of quotients, and proof of the affinity of
$G/K$. Recall the Frobenius reciprocity tells us that
$\operatorname{Res}_{K}^{G}$ is left adjoint to $\operatorname{Ind}_{K}^{G}$.

######  Lemma 2.1.

If $K\subset G$ are quasireductive supergroups then the induction functor
$\operatorname{Ind}^{G}_{K}:\operatorname{Rep}K\to\operatorname{Rep}G$ is
exact and takes injectives to injectives.

###### Proof.

By the geometric definition of induction, $\operatorname{Ind}^{G}_{K}M$ is
isomorphic to the functor of global sections of the $G$-equivariant vector
bundle $G\times_{K}M$. Exactness now follows from the affinity of $G/K$. By
Frobenius reciprocity, injectivity of a $K$-module $M$ implies injectivity of
$\operatorname{Ind}^{G}_{K}M$. ∎

######  Corollary 2.2.

If $K\subset G$ are quasireductive supergroups, then the restriction functor
$\operatorname{Res}_{K}:\operatorname{Rep}G\to\operatorname{Rep}K$ takes
projectives to projectives.

###### Proof.

By Frobenius reciprocity, $\operatorname{Res}_{K}^{G}$ is left adjoint to
$\operatorname{Ind}_{K}^{G}$. ∎

We need the following analogue of Shapiro’s lemma for quasireductive
supergroups.

######  Lemma 2.3 (Frobenius Reciprocity).

Let $K\subset G$ be quasireductive supergroups and $N$ be a $K$-module. For a
$G$-module $M$, and for all $i$, we have natural isomorphisms

$\operatorname{Ext}^{i}_{G}(M,\operatorname{Ind}^{G}_{K}N)\cong\operatorname{Ext}^{i}_{K}(\operatorname{Res}_{K}M,N).$

###### Proof.

If $I^{\cdot}$ is an injective resolution of $N$ then
$\operatorname{Ind}^{G}_{K}I^{\cdot}$ is an injective resolution of
$\operatorname{Ind}^{G}_{K}N$. The statement now follows from Frobenius
reciprocity:

$\operatorname{Hom}_{G}(M,\operatorname{Ind}^{G}_{K}I^{s})\simeq\operatorname{Hom}_{K}(\operatorname{Res}_{K}M,I^{s}).$

∎

###  2.4. Splitting subgroups

The $G$-module $\mathbb{C}[G/K]$ contains a trivial submodule
$\mathbb{C}\subseteq\mathbb{C}[G/K]$ spanned by the constant function $1$.

######  Proposition 2.4.

Assume that $G$ is quasireductive and $K$ is its quasireductive closed
subsupergroup. The following are equivalent:

  1. (1)

The trivial summand $\mathbb{C}$ splits off $\mathbb{C}[G/K]$ as a $G$-module.

  2. (2)

For any $G$-module $M^{\prime}$ the natural injective morphism
$\varphi:M^{\prime}\to\operatorname{Ind}^{G}_{K}M^{\prime}$ splits.

  3. (3)

For any pair of $G$-modules $M,M^{\prime}$, the restriction morphism

$\operatorname{Ext}^{i}_{G}(M,M^{\prime})\to\operatorname{Ext}^{i}_{K}(M,M^{\prime})$

is injective for all $i$.

######  Definition 2.5.

We say that subgroup $K\subseteq G$ is a splitting subgroup in $G$ if it is
quasireductive and satisfies the above equivalent conditions.

###### Proof.

(1)$\Rightarrow$(2) follows from the isomorphism

$\tau:\operatorname{Ind}^{G}_{K}M^{\prime}\to(\operatorname{Ind}^{G}_{K}\mathbb{C})\otimes
M^{\prime}.$

Indeed, if $\pi:\operatorname{Ind}^{G}_{K}\mathbb{C}\to\mathbb{C}$ is a
splitting homomorphism then $(\pi\otimes
1_{M^{\prime}})\tau\varphi=1_{M^{\prime}}$.

Let us show that (2)$\Rightarrow$(3). Note that by (2) the natural
homomorphism

$\varphi:\text{Ext}_{G}^{i}(M,M^{\prime})\to\text{Ext}^{i}_{G}(M,\operatorname{Ind}^{G}_{K}M^{\prime})$

is an embedding. Using Lemma 2.3 we have an isomorphism

$\psi:\text{Ext}^{i}_{G}(M,\operatorname{Ind}^{G}_{K}M^{\prime})\to\text{Ext}^{i}_{K}(M,M^{\prime}).$

The composition $\psi\varphi$ is the restriction homomorphism.

For $(3)\Rightarrow(1)$, we have a short exact sequence of $G$-modules

$0\to\mathbb{C}\to\mathbb{C}[G/K]\to\mathbb{C}[G/K]/\mathbb{C}\to 0.$

The evaluation morphism at $eK$ provides a splitting as $K$-modules, and thus
by (3) the sequence must also split over $G$. ∎

######  Corollary 2.6.

If $K$ is a splitting subgroup of $G$, then a $G$-module $M$ is projective if
and only if its restriction to $K$ is projective.

###  2.5. Various sufficient criteria for splitting

In what follows we always assume that $G$ and $K$ are quasireductive. Our
first result follows immediately from the definition of splitting.

######  Corollary 2.7.

Let $K\subseteq H\subseteq G$.

  1. (1)

If $K$ is a splitting subgroup in $G$, then $H$ is also a splitting subgroup
in $G$.

  2. (2)

If $K$ is a splitting subgroup of $H$, and $H$ is a splitting subgroup of $G$,
then $K$ is a splitting subgroup of $G$.

######  Remark 2.8.

By Corollary 2.7, we may work inductively to prove that a given subgroup of
$G$ is splitting, in the following sense: if $K\subseteq G$ is a subgroup then
it is splitting if and only there exists a chain of inclusions
$K=H_{0}\subseteq H_{1}\subseteq\cdots\subseteq H_{k}=G$ such that $H_{i}$ is
a splitting subgroup of $H_{i+1}$ for all $i$.

######  Lemma 2.9.

Suppose that $R$ is a normal subgroup of $G$ such that $R^{\circ}$ is
reductive, and let $H=G/R$. Write $\pi:G\to H$ for the projection map. Let
$K\subseteq G$ be a subgroup, and let

$\overline{K}:=\pi(K)=K/K\cap R\subseteq H.$

Then if $\overline{K}$ is a splitting subgroup of $H$, $K$ is a splitting
subgroup of $G$.

###### Proof.

Write $\varphi:\mathbb{C}[H/\overline{K}]\to\mathbb{C}$ for the
$H$-equivariant, and thus $G$-equivariant splitting ($H$ being a quotient of
$G$). Because $R$ has semisimple representation theory, we have a
$G$-equivariant projection
$\mathbb{C}[G/K]\to\mathbb{C}[G/K]^{R}=\mathbb{C}[H/\overline{K}]$. Taking a
composition gives our desired splitting. ∎

######  Lemma 2.10.

Suppose that $N\subseteq G$ is a normal subgroup of $G$, an write $H=G/N$.
Suppose that $K\subseteq G$ is a splitting subgroup. Then $\overline{K}$ is a
splitting subgroup of $H$, where $\overline{K}=K/K\cap N$.

###### Proof.

We have a natural $G$-equivariant map $G/K\to H/\overline{K}$. Thus we have an
inclusion $\mathbb{C}[H/\overline{K}]\to\mathbb{C}[G/K]$, so composing with
the splitting $\mathbb{C}[G/K]\to\mathbb{C}$ gives the result. ∎

######  Lemma 2.11.

Let $K\subset G$ be a subgroup such that
$(\operatorname{Lie}G)_{\overline{1}}=(\operatorname{Lie}K)_{\overline{1}}$.
Then $K$ is splitting in $G$.

###### Proof.

We have $G/K=G_{0}/K_{0}$. Hence the statement. ∎

For the proof of the following, see Section 6 of [Sh2].

######  Lemma 2.12.

Suppose that $G$ is quasireductive, $K$ is a splitting subgroup of $G$, and
$\mathfrak{h}\subseteq\mathfrak{g}_{\overline{1}}$ is an odd abelian ideal.
Then $\mathfrak{h}\subseteq\mathfrak{k}$.

######  Lemma 2.13.

Suppose that $K\subseteq H\subseteq G$ are quasireductive supergroups such
that $\mathfrak{g}=\mathfrak{h}\rtimes\mathfrak{l}$ for a quasireductive Lie
superalgebra $\mathfrak{l}$. Suppose further that $\mathfrak{l}$ stabilizes
$\mathfrak{k}$. If $K\subseteq G$ is splitting, then $K\subseteq H$ is
splitting as well.

###### Proof.

Applying Lemma 2.10, we may pass to a finite cover of $G$ so that $G_{0}\cong
H_{0}\times L_{0}$. Then by the theory of Harish-Chandra pairs for
supergroups, we obtain a decomposition $G=H\rtimes L$. Since $\mathfrak{l}$
stabilizes $\mathfrak{k}$, $L$ must similarly stabilize $K$, and thus
$K\rtimes L\subseteq G$ is a subgroup, and since it contains $K$ it is
splitting. Consider the natural $H$-equivariant map

$H/K\to(H\rtimes L)/(K\rtimes L).$

It is a surjective closed embedding, and since the supervarieties are smooth
of the same dimension, it must be an isomorphism. Since the trivial summand
splits off functions on the RHS as an $H$-module, it must do the same on the
LHS, and we are done. ∎

###  2.6. Geometric criterion

Let $\mathfrak{g}=\operatorname{Lie}G$. An element
$x\in\mathfrak{g}_{\overline{1}}$ is called homological if
$x^{2}:=\frac{1}{2}[x,x]$ is a semisimple element in
$\mathfrak{g}_{\overline{0}}$. For example, if $[x,x]=0$ then $x$ is
homological. If $M$ is a $G$-module then $[x,x]$ acts semisimply on $M$. One
can define a tensor functor
$DS_{x}:\operatorname{Rep}G\to\operatorname{SVect}$ defined by

$DS_{x}M:=\operatorname{ker}x/(xM\cap\operatorname{ker}x)=H(x,M^{[x,x]}).$

Assume that $\mathcal{M}$ is a affine algebraic supervariety with a vector
field $Q$ such that $Q^{2}$ acts semisimply on $\mathbb{C}[\mathcal{M}]$. We
say that a point $p\in\mathcal{M}_{0}$ is a zero of $Q$ if
$Q(\mathbb{C}[\mathcal{M}])\subset I_{p}$, where $I_{p}$ is the maximal ideal
determined by $p$. Let $Z(Q)$ denote zero locus of $Q$.

######  Lemma 2.14.

If $Z(Q)=\emptyset$ then $DS_{Q}\mathbb{C}[\mathcal{M}]=0$.

###### Proof.

Let $\mathcal{J}(\mathcal{M})$ denote the ideal generated by odd functions on
$\mathcal{M}$ and consider the $\mathcal{J}(\mathcal{M})$-adic filtration on
$\mathbb{C}[\mathcal{M}]$. Consider the corresponding graded
$\mathbb{C}[\mathcal{M}_{0}]$-module
$\operatorname{gr}\mathbb{C}(\mathcal{M})$. Note that it has a compatible
structure of $Q^{2}$-module since $\mathcal{J}(\mathcal{M})$ is
$Q^{2}$-stable. The operator
$Q:\mathbb{C}[\mathcal{M}]\to\mathbb{C}[\mathcal{M}]$ induces a vector field
$\operatorname{gr}Q$ in $\operatorname{gr}\mathbb{C}[\mathcal{M}]$ of degree
$-1$. In particular, $\operatorname{gr}Q$ defines a morphism
$\mathcal{J}(\mathcal{M})/\mathcal{J}^{2}(\mathcal{M})\to\mathbb{C}[\mathcal{M}]/\mathcal{J}(\mathcal{M})$,
which defines a section $\gamma$ of the normal bundle
$\mathcal{N}_{\mathcal{M}_{0}}$.

The condition $Z(Q)=\emptyset$ implies that $\gamma$ has no zeros on
$\mathcal{M}_{0}$. Therefore $\gamma$ defines an injective morphism
$\gamma:\mathcal{L}\to\mathcal{N}_{\mathcal{M}_{0}}$ of a trivial line bundle
$\mathcal{L}$. The dual morphism
$\mathcal{N}^{*}_{\mathcal{M}_{0}}\to\mathcal{L}^{*}$ splits by affinity of
$\mathcal{M}_{0}$. This implies the existence of $Q^{2}$-invariant
$f\in\mathcal{J}(\mathcal{M})$ such that
$\operatorname{gr}Q(\operatorname{gr}f)=1$.

We have thus shown that

$Q(f)=1+\eta$

where $\eta\in\mathcal{J}(\mathcal{M})$ is even, nilpotent, and $Q$ invariant.
Thus if we set $g=f/(1+\eta)$ we see that $Q(g)=1$, and we are done. ∎

######  Theorem 2.15.

Let $G$ be a quasireducitve supergroup and $K\subset G$ be a splitting
subgroup. Then for any homological $x\in\mathfrak{g}_{\overline{1}}$ the orbit
$G_{0}x$ has non-trivial intersection with $\mathfrak{k}_{\overline{1}}$;
equivalently,
$G_{0}\cdot\mathfrak{k}_{\overline{1}}^{hom}=\mathfrak{g}_{\overline{1}}^{hom}$.

###### Proof.

Assume that there exists $x\in\mathfrak{g}_{\overline{1}}$ such that
$G_{0}x\cap\mathfrak{k}=\emptyset$. Let $Q$ be the odd vector field on $G/K$
corresponding to $x$. Then the Lie algebra of the stabilizer of any
$p\in(G/K)_{0}$ does not contain $x$ and hence $Z(Q)=\emptyset$. By Lemma 2.14
we have $DS_{x}\mathbb{C}[G/K]=0$; but if $\mathbb{C}$ splits as a direct
summand we have $\mathbb{C}\subset DS_{x}\mathbb{C}[G/K]$. A contradiction. ∎

######  Corollary 2.16.

Suppose that $\mathfrak{g}_{\overline{1}}^{hom}$ and
$\mathfrak{k}_{\overline{1}}^{hom}$ contain dense open subsets of
$\mathfrak{g}_{\overline{1}}$ and $\mathfrak{k}_{\overline{1}}$, respectively.
Then if $K$ is splitting in $G$ we must have $\operatorname{sdim}G/K\geq 0$.

###### Proof.

Since $K$ is splitting in $G$, by Theorem 2.15,
$G_{0}\cdot\mathfrak{k}_{\overline{1}}^{hom}=\mathfrak{g}_{\overline{1}}^{hom}$.
By our assumptions, this implies that the image of the map
$G_{0}\times\mathfrak{k}_{\overline{1}}\to\mathfrak{g}_{\overline{1}}$
contains an open subset of $\mathfrak{g}_{\overline{1}}$. Since we work over
$\mathbb{C}$, it follows that for some $(g,x)\in
G_{0}\times\mathfrak{k}_{\overline{1}}$, the map of tangent spaces

$T_{(g,x)}(G_{0}\times\mathfrak{k}_{\overline{1}})\to T_{g\cdot
w}\mathfrak{g}_{\overline{1}}$

is surjective. We may act by $g^{-1}$ so that WLOG $g=e$. Then we have

$T_{(e,x)}(G_{0}\times\mathfrak{k}_{\overline{1}})=\mathfrak{g}_{0}\times\mathfrak{k}_{\overline{1}}\to\mathfrak{g}_{\overline{1}},$

given by

$(u,x)\mapsto[u,x]+x.$

Since $\mathfrak{k}_{\overline{0}}$ preserves $W$, if we choose a splitting of
vector spaces
$\mathfrak{g}_{\overline{0}}=\mathfrak{k}_{\overline{0}}\oplus\mathfrak{m}$,
we obtain that the restriction of the above map to:

$\mathfrak{m}\times\mathfrak{k}_{\overline{1}}\to\mathfrak{g}_{\overline{1}}$

remains surjective. This implies that

$\operatorname{dim}\mathfrak{g}_{\overline{0}}-\operatorname{dim}\mathfrak{k}_{\overline{0}}+\operatorname{dim}\mathfrak{k}_{\overline{1}}\geq\operatorname{dim}\mathfrak{g}_{\overline{1}}.$

From this we obtain $\operatorname{sdim}G/K\geq 0$. ∎

##  3\. Defect subgroups

###  3.1. Contragredient superalgebras

We call a finite-dimensional Lie superalgebra $\mathfrak{g}$ contragredient if
$\mathfrak{g}\simeq\mathfrak{g}(A)$ is a finite-dimensional Kac-Moody
superalgebra with symmetrizable Cartan matrix. Then
$\mathfrak{g}=\bigoplus_{i=1}^{s}\mathfrak{g}_{i}$ is a direct sum of Kac-
Moody superalgebras with indecomposable Cartan matrix. Each $\mathfrak{g}_{i}$
is either simple (basic classical or exceptional) or isomorphic to
$\mathfrak{g}\mathfrak{l}(n|n)$ for $n\geq 1$. We call $\mathfrak{g}$
indecomposable if $s=1$. Furthermore, $\mathfrak{g}$ satisfies the following
properties:

  1. (1)

$\mathfrak{g}_{\overline{0}}$ is reductive and $\mathfrak{g}_{\overline{1}}$
is a semisimple $\mathfrak{g}_{\overline{0}}$-module;

  2. (2)

The centralizer of the Cartan subalgebra $\mathfrak{h}$ in $\mathfrak{g}$
coincides with $\mathfrak{h}$;

  3. (3)

In the root decomposition

$\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Delta}\mathfrak{g}_{\alpha},$

all roots spaces $\mathfrak{g}_{\alpha}$ have dimension $1|0$ or $0|1$;

  4. (4)

$\mathfrak{g}$ admits a nondegenerate, invariant supersymmetric form $(-,-)$,
which we fix.

We say that $\alpha\in\Delta$ is either even or odd if the corresponding root
space is of that parity. In this section we denote by $G$ a connected
quasireductive algebraic supergroup with Lie superalgebra $\mathfrak{g}$.

######  Lemma 3.1.

For any $h\in\mathfrak{h}$ the centralizer $\mathfrak{g}^{h}$ is a direct sum
of a contragredient Lie superalgebras and a subalgebra of $\mathfrak{h}$.

###### Proof.

Without loss of generality we may assume that $\alpha(h)\in\mathbb{Q}$ for all
roots $\alpha$. We have the decomposition

$\mathfrak{g}^{h}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Delta(\mathfrak{g}^{h})}\mathfrak{g}_{\alpha},\quad\Delta(\mathfrak{g}^{h})=\\{\alpha\in\Delta\mid\alpha(h)=0\\}.$

Let $\varepsilon$ denote the minimum of $\alpha(h)$ for all roots $\alpha$
such that $\alpha(h)>0$. Let $h_{0}\in\mathfrak{h}$ be some regular element
such that $|\alpha(h_{0})|<1$ for $\alpha\in\Delta$. Set
$h^{\prime}=h+\frac{\varepsilon}{2}h_{0}$ and let $\Delta^{+}$ be the set of
positive roots defined by $h^{\prime}$. If $B$ is the corresponding set of
simple roots, then $B^{\prime}=B\cap\Delta(\mathfrak{g}^{h})$ is the base of
of $\mathfrak{g}^{h}$. By construction $B^{\prime}$ is linearly independent.
That implies
$\mathfrak{g}^{\mathfrak{h}}=\mathfrak{a}\oplus\mathfrak{g}^{\prime}$ where
$\mathfrak{a}\subset Z(\mathfrak{g}^{h})$ is the subspace complementary to
$[\mathfrak{g}^{h},\mathfrak{g}^{h}]\cap Z(\mathfrak{g}^{h})$. Then
$\mathfrak{g}^{\prime}$ is a quotient of the Kac-Moody superalgebra by some
central ideal $\mathfrak{c}$. Since the restriction of the invariant form on
$\mathfrak{g}^{\prime}$ is nondegenerate $\mathfrak{c}=0$.

∎

###  3.2. Defect subgroups

For an odd root $\alpha\in\Delta$, we say that $\alpha$ is isotropic if
$(\alpha,\alpha)=0$, and otherwise we say $\alpha$ is non-isotropic.

The defect of $\mathfrak{g}$ is the maximal number $d$ of mutually orthogonal
linearly independent isotropic roots of $\mathfrak{g}$.

For an odd isotropic root $\alpha$, choose non-zero elements
$x\in\mathfrak{g}_{\alpha}$ and $y\in\mathfrak{g}_{-\alpha}$ so that
$h:=[x,y]\neq 0$. Then we obtain a subalgebra

$\mathfrak{s}=\mathbb{C}\langle
h,x,y\rangle\cong\mathfrak{s}\mathfrak{l}(1|1).$

We write $S$ for the corresponding subgroup of $G$, so that $S\cong SL(1|1)$.

######  Definition 3.2.

A quasireductive subgroup $D\subset G$ is called a defect subgroup if its Lie
superalgebra $\mathfrak{d}$ satisfies the following:

  1. (1)

the $G_{0}$-orbit of any homological element of $\mathfrak{g}_{\overline{1}}$
has a non-trivial intersection with $\mathfrak{d}_{\overline{1}}$;

  2. (2)

$\operatorname{dim}\mathfrak{d}_{\overline{1}}=2d$;

  3. (3)

$[\mathfrak{d}_{\overline{1}},\mathfrak{d}_{\overline{1}}]=\mathfrak{d}_{\overline{0}}$,
$[\mathfrak{d}_{\overline{1}},\mathfrak{d}_{\overline{0}}]=0$.

######  Proposition 3.3.

Let $\\{\alpha_{1},\dots,\alpha_{d}\\}$ be the maximal set of mutually
orthogonal linearly independent isotropic roots in $\mathfrak{g}$. Then the
supergoup $D$ with Lie superalgebra $\mathfrak{d}$ generated by
$\\{\mathfrak{g}_{\pm\alpha_{i}}\mid i=1,\dots d\\}$ is a defect subgroup, and
any defect subgroup is conjugate to $D$.

###### Proof.

Let us prove the first assertion. Let $x$ be a homological odd element such
that $[x,x]=h$. Without loss of generality we may assume that $h$ lies in the
fixed Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. Note that
$x\in\mathfrak{g}^{h}$ and that $\mathfrak{g}^{h}$ is a direct sum of a
contragredient superalgebra and a purely even center by Lemma 3.1. Thus, if
$h$ is not central, then we can reduce the statement to the contragredient
superalgebra of smaller dimension and central $h$. Thus we can assume that $h$
is central and $\mathfrak{g}$ is indecomposable. If the center of
$\mathfrak{g}$ is trivial, then $[x,x]=0$ and the statement follows from
description of self-commuting cone in [GHSS]. If
$\mathfrak{g}=\mathfrak{g}\mathfrak{l}(n|n)$ one can check by direct
computation that $x$ is conjugate to the matrix proportional to
$\left(\begin{matrix}0&1_{n}\\\ 1_{n}&0\end{matrix}\right)$ which immediately
implies the statement.

Now we deal with the second assertion. It is possible to choose
$x_{i}\in\mathfrak{g}_{\alpha_{i}},y_{i}\in\mathfrak{g}_{-\alpha_{i}}$ such
that if we set $x=\sum_{i=1}^{d}(x_{i}+y_{i})$ then the Lie algebra of the
minimal torus containing $h=[x,x]$ coincides with $\mathfrak{d}_{0}$. Since
$x$ is homological, using the action of $G_{0}$ we can assume without loss of
generality that Lie superalgebra $\mathfrak{d}^{\prime}$ of a defect subgroup
$D^{\prime}$ contains $x$ and hence $\mathfrak{d}_{\overline{0}}$.
Furthermore, by direct inspection of root systems of all exceptional and basic
classical superalgebras we learn that the centralizer of any generic element
of $\mathfrak{d}_{\overline{0}}$ is isomorphic to
$\mathfrak{d}\oplus\mathfrak{l}$ where $\mathfrak{l}$ is a reductive Lie
superalgebra. Therefore $\mathfrak{l}_{\overline{1}}$ does not contain non-
zero homological elements. Hence
$\mathfrak{d}^{\prime}_{\overline{1}}\subset\mathfrak{d}_{\overline{1}}$ and
by dimension count $\mathfrak{d}=\mathfrak{d}^{\prime}$. ∎

The main theorem of this text is:

######  Theorem 3.4.

If $G$ be a quasireductive supergroup with contragredient Lie superalgebra
$\mathfrak{g}$ such that every direct summand of $\mathfrak{g}$ has defect
$\leq 1$ or is isomorphic $\mathfrak{sl}(m|n)$ with $m\neq n$,
$\mathfrak{gl}(n|n)$. Then a defect subgroup is a splitting subgroup.

###### Proof.

If $G=G_{1}\times G_{2}$ and $K=K_{1}\times K_{2}$. If $K_{1}$ and $K_{2}$ are
splitting in $G_{1}$ and $G_{2}$ respectively then $K_{1}\times K_{2}$ is
splitting in $G_{1}\times G_{2}$. Therefore it suffices to prove the statement
for all simple Lie algebras of defect $1$ and for $GL(m|n)$. The former is
done in Section 4 and the latter in Section 6. ∎

######  Corollary 3.5.

Let $S\subseteq GL(n|n)$ be a defect subgroup. Then

  1. (1)

$S$ lies in $SL(n|n)$ and is splitting;

  2. (2)

the image of $S$ in $PGL(n|n)$ and $PSL(n|n)$ is splitting.

###### Proof.

The first statement follows from Lemma 2.13, and the second from Lemma 2.10. ∎

For $Q(n)$ we have:

######  Theorem 3.6.

The subgroups

$Q(2)^{n}\subseteq Q(2n),\ \ \ \ Q(2)^{n}\times Q(1)\subseteq Q(2n+1),$

are splitting.

######  Corollary 3.7.

Let $K\subseteq Q(n)$ be the splitting subgroup given in Theorem 3.6. Then we
have:

  1. (1)

$K\cap SQ(n)$ is splitting in $SQ(n)$;

  2. (2)

the image of $K$ in $PQ(n)$ and $K\cap SQ(n)$ in $PSQ(n)$ are splitting.

###### Proof.

For (1), consider the natural $SQ(n)$-equivariant map

$SQ(n)/(K\cap SQ(n))\to Q(n)/K.$

It is a surjective closed embedding between smooth supervarieties of the same
dimension, and thus is an isomorphism. Therefore since $\mathbb{C}$ is splits
off of $\mathbb{C}[Q(n)/K]$ over $SQ(n)$, the same must be true in
$\mathbb{C}[SQ(n)/(K\cap SQ(n))]$.

Now (2) follows from Lemma 2.10. ∎

###  3.3. Density of $\mathfrak{g}_{\overline{1}}^{hom}$

######  Lemma 3.8.

Let $\mathfrak{g}$ be a contragredient Lie superalgebra.

  1. (1)

For any $x\in\mathfrak{g}_{\overline{1}}$ we have
$\operatorname{dim}G_{0}\cdot
x=\operatorname{dim}\mathfrak{g}_{\overline{1}}-\operatorname{dim}(\mathfrak{g}^{x})_{\overline{1}}$
where $\mathfrak{g}^{x}$ denotes the centralizer of $x$ in $\mathfrak{g}$.

  2. (2)

There exists a Zariski open set set $U\subset\mathfrak{d}_{\overline{1}}$ such
that for all $x\in U$

$\operatorname{codim}G_{0}\cdot
x=\operatorname{dim}(DS_{x}{\mathfrak{g}})_{\overline{1}}+\operatorname{dim}(\mathfrak{d}^{x})_{\overline{1}}.$

###### Proof.

(1) follows from the fact that $\operatorname{sdim}G\cdot x=0$ see [GHSS]
since $G_{0}\cdot x$ has an odd symplectic form. (2) For
$x\in\mathfrak{d}_{\overline{1}}$ set $y=[x,x]$. There exists a Zariski open
dense set $U$ such that
$\mathfrak{g}^{y}=\mathfrak{h}+\mathfrak{d}+DS_{x}(\mathfrak{g})$ for any
$x\in U$. Then
$DS_{x}(\mathfrak{g})\subset\mathfrak{g}^{x}\subset\mathfrak{g}^{y}$ implies
$(\mathfrak{g}^{x})_{\overline{1}}=(DS_{x}{\mathfrak{g}})_{\overline{1}}\oplus(\mathfrak{d}^{x})_{\overline{1}}$.
∎

######  Proposition 3.9.

Let $\mathfrak{g}$ be a contragredient Lie superalgebra. Then
$\mathfrak{g}^{hom}_{\overline{1}}$ contains a Zariski open subset in
$\mathfrak{g}_{\overline{1}}$ if and only if
$DS_{x}(\mathfrak{g})_{\overline{1}}=0$ for some
$x\in\mathfrak{g}^{hom}_{\overline{1}}$.

###### Proof.

Consider the action map
$\varphi:G_{0}\times\mathfrak{d}_{\overline{1}}\to\mathfrak{g}_{\overline{1}}$
defined by $(g,x)\mapsto gx$. We have to check when its image contains a dense
open subset. It suffices to check that

$D\varphi_{(e,x)}:\mathfrak{g}_{\overline{0}}\oplus\mathfrak{d}_{\overline{1}}\to\mathfrak{g}_{\overline{1}}$

is surjective for some $x\in\mathfrak{d}_{\overline{1}}$. Moreover, we my
assume that $x\in U$ as defined in Lemma 3.8(2).

Let $\tilde{\mathfrak{d}}:=\mathfrak{h}+\mathfrak{d}$. Then the restriction of
the form to $\tilde{\mathfrak{d}}$ is nondegenerate. Let us choose
$x\in\mathfrak{d}_{\overline{1}}$ as in the proof of Proposition 3.3. Then

$[\mathfrak{h},x]=[\mathfrak{g}_{\overline{0}},x]\cap\mathfrak{d}_{\overline{1}}.$

Furthermore, the restriction of the form on $[\mathfrak{h},x]$ is non-
degenerate and the orthogonal complement
$[\mathfrak{h},x]^{\perp}\cap\mathfrak{d}_{\overline{1}}$ coincides with
$(\mathfrak{d}^{x})_{\overline{1}}$. Therefore we obtain

$\mathfrak{d}_{\overline{1}}=([\mathfrak{g}_{\overline{0}},x]\cap{\mathfrak{d}}_{\overline{0}})\oplus(\mathfrak{d}^{x})_{\overline{1}}.$

By Lemma 3.8 (2)

$\operatorname{dim}[\mathfrak{g}_{\overline{0}},x]=\operatorname{dim}G_{0}\cdot
x=\operatorname{dim}\mathfrak{g}_{\overline{1}}-\operatorname{dim}(\mathfrak{d}^{x})_{\overline{1}}-\operatorname{dim}(DS_{x}\mathfrak{g})_{\overline{1}}.$

The differential $D\varphi_{(e,x)}$ is given by the formula
$D\varphi_{(e,x)}(u,y)=y+[u,x]$ for all $u\in\mathfrak{g}_{\overline{0}}$,
$y\in\mathfrak{d}_{\overline{1}}$. Therefore

$\operatorname{Im}D\varphi_{(e,x)}=[\mathfrak{g}_{\overline{0}},x]+\mathfrak{d}_{\overline{1}}$

and

$\operatorname{dim}\operatorname{Im}D\varphi_{(e,x)}=\operatorname{dim}[\mathfrak{g}_{\overline{0}},x]+\operatorname{dim}\mathfrak{d}_{\overline{1}}-\operatorname{dim}([\mathfrak{g}_{\overline{0}},x]\cap\mathfrak{d}_{\overline{1}})=\operatorname{dim}[\mathfrak{g}_{\overline{0}},x]+\operatorname{dim}(\mathfrak{d}^{x})_{\overline{1}}.$

We have

$\operatorname{dim}\operatorname{Im}D\varphi_{(e,x)}=\operatorname{dim}\mathfrak{g}_{\overline{1}}-\operatorname{dim}(DS_{x}{\mathfrak{g}})_{\overline{1}}.$

Thus, we obtain that $D\varphi_{(e,x)}$ is surjective if and only if
$(DS_{x}\mathfrak{g})$ is an even Lie algebra. ∎

######  Corollary 3.10.

If all roots of $\mathfrak{g}$ are isotropic then
$\mathfrak{g}^{hom}_{\overline{1}}$ contains a Zariski open subset in
$\mathfrak{g}_{\overline{1}}$.

Note that Corollary 3.10 is not ‘tight’; indeed,
$\mathfrak{o}\mathfrak{s}\mathfrak{p}(2m+1|2n)$ satisfies the hypotheses of
Proposition 3.9 whenever $m\geq n$, as does $\mathfrak{g}(2|1)$.

###  3.4. Minimal splitting subgroups

By part (1) of Corollary 2.7, the condition of being a splitting subgroup is
closed under taking supersets (i.e. sets containing a given one) in the poset
of quasireductive subgroups of $G$. Clearly this poset is stable under
conjugation.

######  Conjecture 3.11.

If $G$ is a quasireductive supergroup with $\operatorname{Lie}G$
contragredient, then up to conjugacy a defect subgroup $D$ of $G$ is the
unique minimal splitting subgroup.

######  Proposition 3.12.

Conjecture 3.11 holds for all $G$ of defect 1.

###### Proof.

Let $K\subseteq G$ be a splitting subgroup; then necessarily
$\mathfrak{k}_{\overline{1}}$ contains elements
$x,x^{\prime}\in\mathfrak{g}_{\overline{1}}^{hom}$ such that $h:=x^{2}$ and
$h^{\prime}:=(x^{\prime})^{2}$ are semisimple, nonzero, and non-conjugate. We
may conjugate so that there exists an odd isotropic root $\alpha$ such that
$x\in\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{-\alpha}$; this is by
Proposition 3.3 and the fact that $G$ is defect one.

Choose a maximal torus $\mathfrak{t}$ of $\mathfrak{g}$ containing $h$, and
conjugate within $K$ (using reductivity of $K_{0}$) so that $h^{\prime}$ also
lies in $\mathfrak{t}$. Now the centralizer of $\mathbb{C}\langle x,h\rangle$
in $\mathfrak{g}$ is known to be $\mathbb{C}\langle
h,y\rangle\oplus\mathfrak{g}_{x}$, where $\mathfrak{g}_{x}$ is a reductive Lie
superalgebra (i.e. has $(\mathfrak{g}_{x})_{\overline{1}}^{hom}=0$) and
$y\in\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{-\alpha}$ has $[y,x]=0$ (see
[GHSS]). If $\alpha(h^{\prime})\neq 0$, then we obtain $\mathfrak{g}_{\alpha}$
and $\mathfrak{g}_{-\alpha}$, giving the corresponding defect subgroup in $K$,
and we would be done. Thus suppose instead that $\alpha(h^{\prime})=0$, which
implies that $h^{\prime}\in\mathfrak{g}_{x}\oplus\mathbb{C}\langle h\rangle$.

By the Jacobi identity we have $[x^{\prime},[x,x^{\prime}]]=0$, which implies
that $[x,x^{\prime}]\in\mathbb{C}\langle
h^{\prime}\rangle\oplus\mathfrak{g}_{x^{\prime}}$. However since
$\mathfrak{g}_{x^{\prime}}$ splits off $\mathfrak{g}$ as a trivial module
under $\langle x^{\prime},h^{\prime}\rangle,$ we must in fact have
$[x,x^{\prime}]=rh^{\prime}$ for some $r\in\mathbb{C}$. If $r=0$, then
$[x,x^{\prime}]=0$ which implies that $x^{\prime}\in(\mathbb{C}\langle
h,y\rangle\times\mathfrak{g}_{x})^{hom}=\mathbb{C}\langle y\rangle$. It
follows that $x$ is a nonzero multiple of $y$, and since $\langle
x,y^{\prime}\rangle=\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{-\alpha}$ we are
done.

Therefore suppose instead that $r\neq 0$; then since
$h^{\prime}\in\mathfrak{g}_{x}\oplus\mathbb{C}\langle h\rangle$ and
$\mathfrak{g}_{x}$ splits off $\mathfrak{g}$ as a trivial module under
$\operatorname{ad}(x)$, we must have $h^{\prime}=sh$ for some
$s\in\mathbb{C}^{\times}$.

From here, if we decompose $x^{\prime}=\sum\limits x_{\beta}^{\prime}$ into a
sum of root vectors, for all $\beta\notin\\{\pm\alpha\\}$ we must have
$\beta(h^{\prime})\neq 0$; on the other hand $[h^{\prime},x^{\prime}]=0$, so
it follows that $x^{\prime}=x_{\alpha}^{\prime}+x_{-\alpha}^{\prime}$. From
here it is easy to see that $x,x^{\prime}$ generate the defect subgroup
determined by $\alpha$, and since this will lie in $K$ we are done. ∎

For $Q(n)$ we have the following clean result:

######  Proposition 3.13.

Let $G=Q(n)$ and $K$ be a splitting subgroup of $G$, then $K$ contains a
subgroup conjugate to $Q(2)^{d}$ for $n=2d$ and to $Q(2)^{d}\times Q(1)$ for
$n=2d+1$.

###### Proof.

In the Lie superalgebra $\mathfrak{g}=\mathfrak{q}(n)$, the odd component is
an adjoint $G_{0}=GL(n)$-module. Theorem 2.15 implies that without loss of
generality we may assume that $\mathfrak{k}_{\overline{1}}$ contains a
diagonal matrix $u$ with distinct eigenvalues $\lambda_{1},\dots,\lambda_{n}$
such that $\lambda^{2}_{1},\dots,\lambda_{n}^{2}$ are linearly independent
over $\mathbb{Q}$. Since $K_{0}\subset G_{0}$ is a reductive subgroup and
$[u,u]\in\mathfrak{k}_{0}$, this imply that $K_{0}$ contains a maximal torus
of $G_{0}$. Hence $\mathfrak{k}$ contains an even part $\mathfrak{h}_{0}$ of
the Cartan subalgebra $\mathfrak{h}\subset\mathfrak{g}$. Theorem 2.15 implies
that $\mathfrak{k}_{\overline{1}}$ contains any $v$ of full rank with
eigenvalues $\mu_{1},\dots,\mu_{n}$. Then $[v,v]$ is conjugate to some element
$t\in\mathfrak{h}_{0}$ which is also of full rank. Therefore there exists an
odd element $t^{\prime}\in\mathfrak{k}$ of full rank such that
$[t^{\prime},t^{\prime}]=t$. If eigenvalues of $t$ are distinct then the
condition $[t^{\prime},t]=0$ implies
$t^{\prime}\in\mathfrak{h}_{\overline{1}}$. That means
$\mathfrak{k}_{\overline{1}}\cap\mathfrak{h}_{\overline{1}}$ contains a
Zariski dense subset of $\mathfrak{h}_{\overline{1}}$, hence
$\mathfrak{k}\cap\mathfrak{h}=\mathfrak{h}$. Now we have that

$\mathfrak{k}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Delta_{\mathfrak{k}}}\mathfrak{g}_{\alpha}$

for some subset $\Delta_{\mathfrak{k}}$ of roots. Since $\mathfrak{k}_{0}$ is
a reductive root subalgebra of $\mathfrak{g}_{0}=\mathfrak{gl}(n)$, we get
$\mathfrak{k}_{0}=\mathfrak{gl}(n_{1})\oplus\dots\mathfrak{gl}(n_{p})$ for
some $n_{1}+\dots+n_{p}=n$. Hence
$\mathfrak{k}=\mathfrak{q}(n_{1})\oplus\dots\oplus\mathfrak{q}(n_{p})$. The
maximal rank of any $u\in k_{\overline{1}}$ such that $[u,u]=0$ equals
$2\sum\lfloor n_{i}/2\rfloor$. But the maximal rank of a self-commuting odd
element in $\mathfrak{q}(n)$ is $2\lfloor n/2\rfloor$. Since
$\mathfrak{k}_{\overline{1}}$ must contain a self-commuting element of maximal
rank we obtain that for even $n$ all $n_{i}$ are even and for odd $n$ at most
one $n_{i}$ is odd. Therefore $K$ contains $Q(2)^{d}$ for $n=2d$ and
$Q(2)^{d}\times Q(1)$ for $n=2d+1$. ∎

##  4\. Defect one cases: splitting for symmetric pairs

In this section we prove Theorem 3.4 for all defect one supergroups. We do
this via the application of Corollary 2.7 to the following symmetric
subgroups, which we will show are splitting:

  1. (1)

$GL(m|n-1)\times GL(1)\subseteq GL(m|n),\ \ n>m$;

  2. (2)

$SOSp(m-1|2n)\subseteq SOSp(m|2n),\ \ m\text{ odd, or }\ m>2n$;

  3. (3)

$SOSp(2m|2n-2)\times Sp(2)\subseteq SOSp(2m|2n),\ \ n>m$;

  4. (4)

$SOSp(2|2)\times SO(2)\subseteq D(2,1;\alpha)$;

  5. (5)

$D(2,1;3)\subseteq G(1|2)$;

  6. (6)

$D(1,2;2)\times SL(2)\subseteq F(1|3)$.

Using the above list, we will reduce everything down to the case when
$G=GL(1|1)$, for which the statement is obvious, or $G=SOSP(2|2)$; however
$SOSP(2|2)\cong SL(1|2)$, so in fact this is enough.

We will show the above subgroups are splitting in different ways. For case (1)
we refer to Theorem 6.1. For case (2) we will simply refer to Section 10 of
[Sh1]. Case (4) will be dealt with separately in Section 4.2.

For cases (3), (5) and (6), we will first prove that:

######  Lemma 4.1.

There exists a Borel subgroup $B\subseteq G$ such that if there exists a
$B$-eigenvector $v$ of $\mathbb{C}[G/K]$ of weight $\lambda$, then either
$\lambda=0$ or $\lambda$ does not lie in the principal block.

###### Proof.

See Section 4.1. ∎

######  Proposition 4.2.

If Lemma 4.1 holds, $\mathbb{C}$ splits off $\mathbb{C}[G/K]$.

###### Proof.

Write $W$ for the summand of $\mathbb{C}[G/K]$ in the principal block. Then by
the assumptions of Lemma 4.1, $W$ must be indecomposable with socle
$\mathbb{C}$. If $\mathbb{C}$ does not split off, we may choose a submodule
$V$ of $W$ given by a nonsplit short exact sequence $0\to\mathbb{C}\to V\to
L\to 0$. But then $V$ must be highest weight module because for all groups in
cases (3) (5), and (6), $0$ is a minimal dominant weight. But by assumption
its highest weight will not lie in $\mathfrak{a}^{*}$, a contradiction.
Therefore $\mathbb{C}$ must split off. ∎

###  4.1. Action of the Casimir

For cases (3), (5), and (6) we will show that the action of the Casimir is
enough to obtain the hypotheses of Lemma 4.1.

Let $G/K$ be a supersymmetric space and write $(\mathfrak{g},\mathfrak{k})$
for the corresponding supersymmetric pair from an involution $\theta$ of
$\mathfrak{g}$. We have a decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$, where $\mathfrak{p}$ is the
$(-1)$-eigenspace of $\theta$. Choose a Cartan subspace
$\mathfrak{a}\subseteq\mathfrak{p}$. We may decompose $\mathfrak{g}$ into the
weight spaces of the action of $\mathfrak{a}$, and we write
$\overline{\Delta}$ for the nonzero weights of this action, which are exactly
the restrictions of roots from $\mathfrak{h}^{*}$ to $\mathfrak{a}^{*}$ for a
$\theta$-stable Cartan subalgebra $\mathfrak{h}$ containing $\mathfrak{a}$. We
refer to $\overline{\Delta}$ as the set of restricted roots for this
supersymmetric pair.

If we choose a positive set of restricted roots $\overline{\Delta}^{+}$, we
may define

$\mathfrak{n}:=\bigoplus\limits_{\alpha\in\overline{\Delta}^{+}}\mathfrak{g}_{\alpha}.$

We say $(\mathfrak{g},\mathfrak{k})$ admits an Iwasawa decomposition if
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$ for some
choice of $\mathfrak{n}$ as above. In this case we may choose a Borel
subalgebra $\mathfrak{b}$ containing $\mathfrak{a}\oplus\mathfrak{n}$. We call
such a Borel subalgebra an Iwasawa Borel subalgebra.

######  Lemma 4.3.

Suppose that $(\mathfrak{g},\mathfrak{k})$ admits an Iwasawa decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$, and let
$\mathfrak{b}\supseteq\mathfrak{a}\oplus\mathfrak{n}$ be an Iwasawa Borel
subalgebra.

  1. (1)

The space of $\mathfrak{b}$-eigenfunctions in $\mathbb{C}[G/K]$ of a given
weight is at most one-dimensional, and non-zero $\mathfrak{b}$-eigenfunction
must be non-vanishing at $eK$.

  2. (2)

If there is a $\mathfrak{b}$-eigenfunction of weight $\lambda$ appearing in
$\mathbb{C}[G/K]$, then $\lambda\in\mathfrak{a}^{*}$.

###### Proof.

Because $\mathfrak{b}+\mathfrak{k}=\mathfrak{g}$, $\mathfrak{b}$ acts
spherically on $G/K$ implying the first statement. For the second statement,
if $f\in\mathbb{C}[G/K]$ is a $\mathfrak{b}$-eigenfunction of weight
$\lambda\in\mathfrak{h}^{*}$, then for $t\in\mathfrak{h}\cap\mathfrak{k}$ we
have $0=(tf)(eK)=\lambda(t)f(eK)$. ∎

######  Lemma 4.4.

Assume that $(\mathfrak{g},\mathfrak{k})$ admits an Iwasawa decomposition such
that

  1. (1)

The restricted root system $\overline{\Delta}$ is an indecomposable even root
system;

  2. (2)

Invariant form is definite on the root lattice
$\mathbb{Z}\overline{\Delta}\subset\mathfrak{a}^{*}$;

  3. (3)

If $m_{\alpha}=\operatorname{sdim}(\mathfrak{g}_{\alpha})$,
$\alpha_{1}\dots\alpha_{k}\in\overline{\Delta}^{+}$ are simple roots,
$\rho=\frac{1}{2}\sum_{\alpha\in\overline{\Delta}^{+}}m_{\alpha}\alpha$, then
$\rho=\sum c_{i}\alpha_{i}$ with $c_{i}\geq 0$ for all $i$;

  4. (4)

For every simple root $\alpha_{i}$ the root space $\mathfrak{g}_{\alpha_{i}}$
has a non-trivial even part.

Then the Casimir element $C$ acts with positive eigenvalue on any
$\mathfrak{b}$-eigenfunction $f_{\lambda}$ of weight $\lambda\neq 0$.

###### Proof.

One can compute that $Cf_{\lambda}=(\lambda+2\rho,\lambda)f_{\lambda}$ (I
should write computation). The condition (4) implies the existence of root
$\mathfrak{sl}(2)$ associated with every simple root $\alpha_{i}$. Therefore
the dominance condition on $\lambda$ implies $(\lambda,\alpha_{i})\geq 0$.
Therefore we have

$(\lambda+2\rho,\lambda)=(\lambda,\lambda)+2\sum c_{i}(\lambda,\alpha_{i})>0.$

∎

We now use Lemma 4.4 to prove Lemma 4.1.

  * •

$(\mathfrak{o}\mathfrak{s}\mathfrak{p}(2m|2n),\mathfrak{o}\mathfrak{s}\mathfrak{p}(2m|2n-2)\times\mathfrak{s}\mathfrak{p}(2))$,
$n>m$: here $\mathfrak{a}$ is one-dimensional and our restricted root system
is $BC_{1}$. If we write $\alpha$ for the positive long root, we obtain that

$\rho=(n-m-1)\alpha,$

which is a non-negative multiple of $\alpha$ by assumption.

  * •

$(\mathfrak{g}(1|2),D(1,2;3))$: Here $\mathfrak{a}$ is given by the Cartan
subalgebra of $\mathfrak{g}_{2}$, and thus the form is definite on
$\mathfrak{a}^{*}$. If we write $\alpha_{1},\alpha_{2}$ for the simple roots
of $\mathfrak{g}_{2}$ where

$\frac{(\alpha_{1},\alpha_{1})}{(\alpha_{2},\alpha_{2})}=3,$

we obtain that

$\rho=\alpha_{1}+\alpha_{2}.$

  * •

$(F(3|1),D(1,2;2)\times\mathfrak{s}\mathfrak{l}(2))$: Here the Cartan subspace
is given by the Cartan subalgebra of $F(3|1)$, and thus the form is definite
on $\mathfrak{a}^{*}$. If we write $\alpha_{1},\alpha_{2},\alpha_{3}$ for the
simple roots of the restricted root system where

$\frac{(\alpha_{1},\alpha_{1})}{(\alpha_{2},\alpha_{2})}=3/8,\ \ \
\frac{(\alpha_{2},\alpha_{2})}{(\alpha_{3},\alpha_{3})}=2,$

we obtain that

$\rho=\alpha_{1}+2\alpha_{2}+3\alpha_{3}.$

###  4.2. $SOSp(2|2)\times SO(2)\subseteq D(2,1;\alpha)$

For this subgroup we will need to study the supersymmetric pair
$(D(2,1;\alpha),\mathfrak{o}\mathfrak{s}\mathfrak{p}(2|2)\times\mathfrak{s}\mathfrak{o}(2))$
more closely. Present the root system of $D(2,1;\alpha)$ with basis
$\epsilon_{1},\epsilon_{2},\epsilon_{3}$, even roots $\pm 2\epsilon_{i}$, and
odd roots $\pm\epsilon_{1}\pm\epsilon_{2}\pm\epsilon_{3}$. Then for this
supersymmetric pair we have $\mathfrak{a}^{*}$ spanned by
$\epsilon_{1},\epsilon_{2}$, and we may take as simple root system
$\epsilon_{1}-\epsilon_{2}-\epsilon_{3},2\epsilon_{2},2\epsilon_{3}$.

Then by [Ger], the dominant weights in the principal block are those of the
form $\lambda_{l}=(l+1)\epsilon_{1}+(l-1)(\epsilon_{2}+\epsilon_{3})$ for
$l\in\mathbb{Z}^{+}$, and $\lambda_{0}=(0,0,0)$. Thus the only $\lambda_{l}$
which lie in $\mathfrak{a}^{*}$ are $\lambda_{0}$ and $\lambda_{1}$; further,
by [Ger], the extension graph of the principal block is $D_{\infty}$, and
looks as follows:

$\textstyle{\lambda_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\lambda_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\lambda_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\cdots}$$\textstyle{\lambda_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

In particular we have $\text{Ext}^{1}(L(\lambda_{1}),L(\lambda_{0}))=0$.

Now let $W$ denote the summand of $\mathbb{C}[G/K]$ with trivial central
character. Suppose that $W$ is not semisimple; then consider its simple
composition factors; from the extension graph we see that necessarily
$L(\lambda_{2})$ must appear. However this means that the highest weight of
$W$ is at least $\lambda_{2}$, which is larger than both $\lambda_{1}$ and
$\lambda_{0}$. Since $\lambda_{i}\notin\mathfrak{a}^{*}$ for $i>1$, we obtain
a contradiction. Thus instead $W$ must be semisimple.

##  5\. Volumes of supergrassmannians

###  5.1. Invariants of representations of $\mathfrak{q}(1)$

In this section the ground field is $\mathbb{R}$. Consider the Lie
superalgebra $\mathfrak{q}$, defined over $\mathbb{R}$, with basis
$\\{Q,Q^{2}\\}$ where $Q$ is an odd element.

######  Lemma 5.1.

Let $V$ be a $\mathfrak{q}$-module such that $Q:V\to V$ is an isomorphism and
$Q^{2}:V\to V$ is a compact linear operator (the closure of the subgroup
generated by $\operatorname{exp}Q^{2}$ is compact in
$GL(V_{\overline{0}})\times GL(V_{\overline{1}})$). Then the following hold:

  1. (1)

$\operatorname{dim}V_{\overline{0}}=\operatorname{dim}V_{\overline{1}}=2n$ for
some positive integer $n$;

  2. (2)

Every $Q^{2}$-invariant nondegenerate symmetric bilinear form $B_{0}$ on
$V_{\overline{0}}$ extends uniquely to a bilinear symmetric
$\mathfrak{q}$-invariant form $B$ on $V$.

###### Proof.

(1) follows from the fact that $Q^{2}$ defines a compact isomorphism on both
$V_{\overline{0}}$ and $V_{\overline{1}}$. For (2), observe that given a
$Q^{2}$-invariant nondenegerate symmetric bilinear form $B_{0}$ on
$V_{\overline{0}}$, for $v,w\in V_{\overline{0}}$ we must have
$B(Qv,Qw)=B(Q^{2}v,w)$. This indeed defines a form with the desired
properties, and uniqueness also follows. ∎

Let us fix a volume form $\omega$ on $V$ and orientation
$\omega_{\overline{0}}$ on $V_{\overline{0}}$. Let $B$ be a
$\mathfrak{q}$-invariant even symmetric bilinear form. Choose bases
$e_{1},\dots,e_{2n}\in V_{\overline{0}}$ and $f_{1},\dots,f_{2n}\in V_{1}$
such that

$\omega(e_{1},\dots,e_{2n},f_{1},\dots,f_{2n})=1,\omega_{\overline{0}}(e_{1},\dots,e_{n})>0.$

Set

$\alpha(V,B,\omega):=(-\mathbf{i})^{n}\exp\left(\frac{\pi\mathbf{i}(\operatorname{dim}V_{\overline{0}}^{+}-\operatorname{dim}V_{\overline{0}}^{-})}{4}\right)\frac{\operatorname{Pf}(B_{1})}{\sqrt{|\det
B_{0}|}},$

where
$\operatorname{dim}V_{\overline{0}}^{+}-\operatorname{dim}V_{\overline{0}}^{-}$
is the signature of $B_{0}$, and $\operatorname{Pf}(B_{1})$ is computed with
respect to our chosen basis.

######  Lemma 5.2.

If $V$ satisfies the conditions of Lemma 5.1 then $\alpha(V,B,\omega)$ does
not depend on a choice of $B$.

###### Proof.

Let $B$ be some invariant form such that $B_{0}$ is positive definite and
$B^{\prime}$ be some other invariant form. Then there is a an
$B_{0}$-orthonormal basis $v_{1},\dots,v_{2n}\in V_{\overline{0}}$ orthogonal
with respect to $B^{\prime}_{0}$. Since $B^{\prime}-tB$ is
$\mathfrak{g}$-invariant for any $t\in\mathbb{R}$ there is a decomposition
$V=V^{1}\oplus\dots\oplus V^{s}$ orthogonal with respect to both forms $B$ and
$B^{\prime}$, where $V^{i}=\operatorname{ker}(B^{\prime}-t_{i}B)$ where
$t_{1},\dots,t_{s}$ are eigenvalues of $B^{\prime}_{0}$. Since each $V^{i}$
satisfies the assumption of Lemma 5.1 we have
$\operatorname{dim}V^{i}=(2n_{i}|2n_{i})$ with $n_{1}+\dots+n_{s}=n$. Let
$B^{i}$ denote the restriction of $B$ to $V^{i}$ and $\omega^{i}$ a volume
form on $V^{i}$ such that $\omega=\prod_{i=1}^{s}\omega^{i}$. Then by direct
inspection
$\alpha(V^{i},B^{i},\omega^{i})=\alpha(V^{i},t_{i}B^{i},\omega^{i})$ and
therefore

$\alpha(V,B,\omega)=\prod_{i=1}^{s}\alpha(V^{i},B^{i},\omega^{i})=\prod_{i=1}^{s}\alpha(V^{i},t_{i}B^{i},\omega^{i}).$

∎

Since the coefficient $\alpha(V,B,\omega)$ does not depend on the choice of
$B$ we just write $\alpha(V,\omega)$. The following lemma gives a practical
way to compute $\alpha(V,\omega)$.

######  Lemma 5.3.

Let $V$ satisfy the conditions of Lemma 5.1. Let $e_{1},\dots,e_{2n}\in
V_{\overline{0}}$ be a positively oriented basis such that the matrix of
$Q_{0}^{2}$ is skew-symmetric in this basis. (Such basis exists by compactness
of $Q^{2}$.) For any basis $f_{1},\dots,f_{2n}$ of $V_{\overline{1}}$ such
that $\omega(e_{1},\dots,e_{2n},f_{1},\dots,f_{2n})=1$, the matrix
$\left(\begin{matrix}0&Q_{01}\\\ Q_{10}&0\end{matrix}\right)$ of $Q$ satisfies
the property $Q^{t}_{01}Q_{10}$ is skew-symmetric and

$\alpha(V,\omega)=\operatorname{Pf}(Q^{t}_{01}Q_{10}^{-1})$

###### Proof.

Choose an invariant form $B$ on $V$ such that $e_{1},\dots,e_{2n}$ is an
orthonormal basis with respect to $B_{0}$; then by skew symmetry of
$Q_{0}^{2}$, $B_{0}$ will be $Q^{2}$-invariant, so may be extended uniquely to
an invariant form $B$ on $V$ using Lemma 5.1. For any $v,w\in
V_{\overline{1}}$ we have

$B(v,w)=B(v,QQ^{-1}w)=B(Qv,Q^{-1}w).$

Therefore $B_{1}=Q^{t}_{01}Q_{10}^{-1}$. That implies that $B_{1}$ is skew-
symmetric. Using the definition of $\alpha(V,B,\omega)$ and noticing that
$B_{0}$ is the identity matrix we get the desired result. ∎

######  Corollary 5.4.

Let $V=\mathbb{C}^{n|n}$ with standard complex basis $u_{1},\dots,u_{n}\in
V_{0},v_{1},\dots,v_{n}\in V_{1}$, and real volume form

$\omega=\prod_{i=1}^{n}\frac{dz_{i}d\bar{z}_{i}}{d\zeta_{i}d\bar{\zeta}_{i}}.$

If the action of $Q$ in the standard basis is given by

$Qu_{i}=(1+\mathbf{i})d_{i}v_{i},\quad Qv_{i}=(1+\mathbf{i})c_{i}v_{i}$

for $c_{i},d_{i}\in\mathbb{R}$, then
$\alpha(V,\omega)=\prod_{i=1}^{n}\frac{c_{i}}{d_{i}}$.

###### Proof.

With respect to our complex basis,

$\overline{Q}^{t}_{01}Q_{10}^{-1}=-\operatorname{diag}(\frac{c_{1}}{d_{1}}\mathbf{i},\dots,\frac{c_{n}}{d_{n}}\mathbf{i}).$

Here we recall that the transpose of a $n\times n$ complex matrix $A$ viewed
as a real $2n\times 2n$-matrix is $\overline{A}^{t}$. From here one easily
checks that the Pfaffian of the corresponding real matrix is as desired. ∎

###  5.2. Schwarz–Zaboronsky formula

This section will deal with Berezinian integration on real supermanifolds. We
refer to [M] for details.

Let $\mathcal{M}$ be a compact real supermanifold with underlying manifold
$\mathcal{M}_{0}$ and a volume form $\omega$. We fix an orientation on the
underlying manifold $\mathcal{M}_{0}$. Let $Q$ be a vector field on
$\mathcal{M}$. A point $p\in\mathcal{M}$ is a zero of $Q$ if for any smooth
function $f\in\mathcal{C}^{\infty}\mathcal{M}$ we have $Q(f)\subset I_{p}$
where $I_{p}$ is the vanishing ideal of $p$. By $Z(Q)$ we denote the set of
all zeros of $Q$. If $p\in Z(Q)$ then $Q$ induces a linear operator on
$T^{*}_{p}\mathcal{M}=I_{p}/I_{p}^{2}$. We say that $p\in Z(Q)$ is isolated if
$Q:T^{*}_{p}\mathcal{M}\to T^{*}_{p}\mathcal{M}$ is an isomorphism. Let us
consider an odd vector field $Q$ on $\mathcal{M}$ satisfying the following
properties:

  1. (1)

$Z(Q)$ is finite and consists of isolated zeros;

  2. (2)

$Q\omega=0$;

  3. (3)

$Q^{2}$ is a compact vector field, i.e.,there exists a compact Lie group
$\mathcal{K}$ acting on $\mathcal{M}$ such that
$Q\in\operatorname{Lie}\mathcal{K}$.

Let us note that by Lemma 5.1 $\operatorname{dim}\mathcal{M}=(2n|2n)$ for some
$n\in\mathbb{N}$. From [SZ] we have:

######  Theorem 5.5.

Let $\mathcal{M}$ be a supermanifold with volume form $\omega$ and odd vector
field $Q$ satisfy the above assumptions. Then there exists an odd function
$\sigma$ such that

  1. (1)

$Q^{2}\sigma=0$;

  2. (2)

$Q\sigma\notin I_{p}$ for any $p\notin Z(Q)$; and

  3. (3)

the Hessian of $Q\sigma$ at each point $p\in Z(Q)$ is nondegenerate.

Furthermore,

$\int_{\mathcal{M}}\omega=(2\pi)^{n}\sum_{p\in
Z(Q)}\alpha(T_{p}\mathcal{M},\operatorname{Hess}_{p}(Q\sigma),\omega_{p}),$

where $\operatorname{Hess}_{p}(Q\sigma)$ is the Hessian of $Q\sigma$ at $p$,
which will be an even symmetric $Q$-invariant form on $T_{p}\mathcal{M}$.

It follows from Lemma 5.2 that the function $\sigma$ can be removed from the
Schwarz–Zaboronsky formula:

######  Corollary 5.6.

$\int_{\mathcal{M}}\omega=(2\pi)^{n}\sum_{p\in
Z(Q)}\alpha(T_{p}\mathcal{M},\omega_{p}).$

######  Corollary 5.7.

If $Z(Q)=\emptyset$ then $\int_{\mathcal{M}}\omega=0$.

###  5.3. Supergrassmannian

The supergrassmannian $Gr(r|s,m|n)$ is the superscheme representing the
functor of $(r|s)$-dimensional subspaces in $\mathbb{C}^{m|n}$. We refer to
[M] and [PS] for its construction and properties.

If $G=GL(m|n)$, $K=GL(r|s)\times GL(m-r|n-s)$ and $P\subset G$ a maximal
parabolic subgroup of $G$ containing $K$ then the homogeneous space $G/P$ is
isomorphic to $Gr(r|s,m|n)$. The underlying algebraic variety is a product of
two classical Grassmannians $Gr(r|m)\times Gr(s|n)$. If $p\in Gr(r|s,m|n)$ is
the point with stabilizer $P$ then the tangent space $T_{p}Gr(r|s,m|n)$ can be
identified with $\mathfrak{g}/\mathfrak{p}$. This immediately implies the
formulas for dimension and superdimension of $Gr(r|s,m|n)$:

$\operatorname{dim}Gr(r|s,m|n)=(r(m-r)+s(n-s)|r(n-s)+s(m-r)),$ (5.1)
$\operatorname{sdim}Gr(r|s,m|n)=(r-s)((m-r)-(n-s)).$ (5.2)

Recall that $G$ has a compact real form $\mathcal{U}=U(m|n)$, the unitary
supergroup. The Lie algebra $\mathfrak{u}(m|n)$ is the set of fixed points of
antilinear involution $\theta$ defined by

$\theta\left(\begin{matrix}A&B\\\
C&D\end{matrix}\right)=\left(\begin{matrix}-\bar{A}^{t}&\mathbf{i}\bar{C}^{t}\\\
\mathbf{i}\bar{B}^{t}&-\bar{D}^{t}\end{matrix}\right).$

Let $\mathcal{K}:=\mathcal{U}\cap K=\mathcal{U}\cap P\simeq U(r|s)\times
U(m-r|n-s)$.

######  Proposition 5.8.

The unitary supergroup $\mathcal{U}$ acts transitively on the
supergrassmannian $Gr(r|s,m|n)$ and we have an isomorphism of real
supermanifolds

$Gr(r|s,m|n)\simeq\mathcal{U}/\mathcal{K}.$

Furthermore, $Gr(r|s,m|n)$ has a unique up to normalization
$\mathcal{U}$-invariant volume form.

###### Proof.

The first assertion follows from the corresponding classical fact for the
underlying manifolds

$Gr(r|m)\times Gr(s|n)\simeq(U(m)/(U(r)\times U(m-r))\times(U(n)/(U(s)\times
U(n-s))$

and comparison of odd dimensions of $\mathcal{U}/\mathcal{K}$ and $G/P$.

To prove the second assertion we consider the point $p\in Gr(r|s,m|n)$ which
is stabilized by $\mathcal{K}$. The tangent space $T_{p}Gr(r|s,m|n)$ is a
$\mathcal{K}$-module. One can identify it with the space
$\operatorname{Hom}_{\mathbb{C}}(\mathbb{C}^{r|s},\mathbb{C}^{m-r|n-s})$. In
this way a tangent vector is $(r|s)\times(m-r|n-s)$-matrix $y$. The real part
of the Hermitian form $H(y_{1},y_{2})=\operatorname{str}\theta(y_{1})y_{2}$
defines a non-degenerate symmetric bilinear form $B$ on $T_{p}Gr(r|s,m|n)$
considered as a vector space over $\mathbb{R}$. Since $\mathcal{K}$ is
connected it is clear that $\mathcal{K}\subset SL(T_{p}Gr(r|s,m|n))$ and hence
preserves a volume form $\omega_{p}$ on $T_{p}Gr(r|s,m|n)$. Then using
standard translation argument $\omega_{p}$ induces the form on $Gr(r|s,m|n)$ ∎

We choose an orientation $Gr(r|s,m|n)_{0}$ via its complex structure, i.e. so
that for $z=x+iy$ the basis $dx,dy$ is positively oriented; similarly choose
an orientation on $T_{p}Gr(r|s,m|n)_{\overline{1}}$ according to its complex
structure.

If we let $\operatorname{dim}Gr(r|s,m|n)=(d_{1}|d_{2})$, then choosing local,
positive oriented coordinates
$u_{1},\dots,u_{d_{1}},\xi_{1},\dots,\xi_{d_{2}}$, we normalize $\omega$ so
that in these local coordinates

$\omega=\sqrt{|\operatorname{Ber}B|}\frac{dx_{1}\cdots
dx_{d_{1}}}{d\xi_{1}\cdots d\xi_{d_{2}}},$

where $B$ denotes the matrix form of the metric in our chose local coordinate
system.

######  Definition 5.9.

With the above chosen volume form $\omega$ on $Gr(r|s,m|n)$, we set

$V(r|s,m|n)=\int\limits_{Gr(r|s,m|n)}\omega.$

Observe that we have $V(r|s,m|n)=V(s|r,n|m)$.

###  5.4. Orientation subtlety and volume relationship

Over the real supergroup $\mathcal{U}(m|n)$ there is an obvious identification
of homogeneous supervarieties $Gr(r|s,m|n)\cong Gr(m-r|n-s,m|n)$; however over
$GL(m|n)$ these spaces are _not_ isomorphic. Indeed the corresponding
parabolics will not be conjugate in general. However, the antilinear
involution $\theta$ takes one parabolic to the other, inducing an isomorphism
of real supervarieties

$\Theta:Gr(r|s,m|n)\cong Gr(m-r|n-s,m|n)$

which is $\mathcal{U}(m|n)$-equivariant.

This subtle point is important because the orientation chosen on $Gr(r|s,m|n)$
comes from the complex structure induced as a $GL(m|n)$ homogeneous space.
Thus $\Theta$ will not in general take the chosen orientation on $Gr(r|s,m|n)$
to the chosen orientation on $Gr(m-r|n-s,m|n)$. Since $\theta$ is acting by
complex conjugate on coordinates, the orientation on the even part will be
reversed $\operatorname{dim}_{\mathbb{C}}Gr(r|s,m|n)_{0}$ times. On the other
hand the invariant form $\omega$ will change by a sign of
$(-1)^{\operatorname{dim}_{\mathbb{C}}Gr(r|s,m|n)_{0}+\operatorname{dim}_{\mathbb{C}}Gr(r|s,m|n)_{1}}$.
Therefore the volume will change by
$(-1)^{\operatorname{dim}_{\mathbb{C}}Gr(r|s,m|n)_{1}}$:

######  Lemma 5.10.

$V(r|s,m|n)=(-1)^{r(n-s)+s(m-r)}V(m-r|n-s,m|n).$

###  5.5. Computation of volumes

In this section we seek to show that:

######  Theorem 5.11.

Let $\mathcal{M}=Gr(r|s,m|n)$. Then the integral $\int_{\mathcal{M}}\omega\neq
0$ if and only if $\operatorname{sdim}\mathcal{M}\geq 0$.

######  Remark 5.12.

Note that in the case $\operatorname{sdim}\mathcal{M}=0$, the normalization of
$\omega$ is independent of the choice of invariant form $B$ that we made.

######  Proposition 5.13.

Assume that $\operatorname{sdim}\mathcal{M}<0$. Then
$\int_{\mathcal{M}}\omega=0$.

###### Proof.

By Corollary 5.7 it suffices to construct an odd vector field $Q$ satisfying
conditions (2) and (3) with $Z(Q)=\emptyset$. However here we have
$\mathfrak{u}_{\overline{1}}^{hom}=\mathfrak{u}_{\overline{1}}$ and
$\mathfrak{v}_{\overline{1}}^{hom}=\mathfrak{v}_{\overline{1}}$, where
$\mathfrak{v}=\operatorname{Lie}\mathcal{K}$. The argument of Corollary 2.16
still works in our situation, and thus we find that
$\mathcal{U}_{0}\cdot\mathfrak{v}_{\overline{1}}\neq\mathfrak{u}_{\overline{1}}^{hom}$,
and we are done. ∎

Next we prove

######  Proposition 5.14.

Let $\mathcal{M}=Gr(r|r,n|n)$. Then
$\int_{\mathcal{M}}\omega=(2\pi)^{2(n-r)r}\binom{n}{r}$.

###### Proof.

Consider the standard basis $e_{1},\dots,e_{n},f_{1},\dots,f_{n}$ in
$\mathbb{C}^{n|n}$. Let $p\in\mathcal{M}$ which corresponds to the subspace
spanned by $e_{1},\dots,e_{r},f_{1},\dots,f_{r}$. Choose
$a_{1},\dots,a_{n}\in\mathbb{R}^{*}$ such that $a_{i}\pm a_{j}\neq 0$ for any
$i\neq j$. Let

$Q=\left(\begin{matrix}0&A\\\ A&0\end{matrix}\right),\quad
A=(1+\mathbf{i})\operatorname{diag}(a_{1},\dots,a_{n}).$

Then $Q\in\operatorname{Lie}(\mathcal{K})$. Moreover, it is clear that
$Q:T_{p}\mathcal{M}\to T_{p}\mathcal{M}$ is an isomorphism.

Let us compute $\alpha(T_{p}(\mathcal{M}),\omega_{p})$. We use identification
$T_{p}\mathcal{M}\simeq\operatorname{Hom}_{\mathbb{C}}(\mathbb{C}^{r|r},\mathbb{C}^{n-r|n-r})$
and the $Q$-invariant decomposition

$T_{p}\mathcal{M}=\bigoplus_{1\leq i\leq r,r<j\leq
n}\operatorname{Hom}_{\mathbb{C}}(\mathbb{C}e_{i}\oplus\mathbb{C}f_{i},\mathbb{C}e_{j}\oplus\mathbb{C}f_{j}).$

Define a basis $u_{ij}^{\pm},v_{ij}^{\pm}$ in $T_{p}\mathcal{M}$ by

$u_{ij}^{\pm}=e_{i}\otimes e_{j}^{*}\pm f_{i}\otimes f_{j}^{*},\quad
v_{ij}^{\pm}=f_{i}\otimes e_{j}^{*}\pm e_{i}\otimes f_{j}^{*}.$

Then we have

$Qu_{ij}^{\pm}=(1+i)(a_{i}\mp a_{j})v_{ji}^{\pm},\quad
Qv_{ij}^{\pm}=(1+i)(a_{i}\pm a_{j})u_{ji}^{\pm}.$

Further $\theta u_{ij}^{\pm}=-u_{ji}^{\pm}$, $\theta v_{ij}^{\pm}=\pm
iv_{ji}^{\pm}$. Thus we have

$\langle u_{ij}^{\pm},u_{ij}^{\mp}\rangle=-2,\ \ \ \langle
iu_{ij}^{\pm},iu_{ij}^{\mp}\rangle=2.$ $\langle
iv_{ij}^{\mp},v_{ij}^{\pm}\rangle=2.$

Let $z_{ij}^{\pm}=x_{ij}^{\pm}\pm iy_{ij}^{\pm}$ be the complex coordinate
dual to $u_{ij}^{\pm}$, and let $\zeta_{ij}=\xi_{ij}^{\pm}+i\eta_{ij}^{\pm}$
be the complex coordinate dual to $v_{ij}^{\pm}$. Then the bilinear form (as
an element of $S^{2}(\mathcal{T}^{*}\mathcal{M}_{p})$) at this points looks as
follows:

$2\sum\limits_{i,j}\left(-dx_{ij}^{+}dx_{ij}^{-}+dy_{ij}^{+}dy_{ij}^{-}\right)-2\sum\limits_{i,j}\left(d\xi_{ij}^{+}d\eta_{ij}^{-}+d\xi_{ij}^{-}d\eta_{ij}^{+}\right).$

Thus the volume form at this point is given by:

$\frac{\prod\limits_{1\leq i\leq r<j\leq
n}dx_{ij}^{+}dy_{ij}^{+}dx_{ij}^{-}dy_{ij}^{-}}{\prod\limits_{1\leq i\leq
r<j\leq
n}d\xi_{ij}^{+}d\eta_{ij}^{-}d\xi_{ij}^{+}d\eta_{ij}^{-}}=\frac{\prod\limits_{1\leq
i\leq r<j\leq
n}dz_{ij}^{+}d\overline{z_{ij}^{+}}dz_{ij}^{-}d\overline{z_{ij}^{-}}}{\prod\limits_{1\leq
i\leq r<j\leq
n}d\zeta_{ij}^{+}d\overline{\zeta_{ij}^{+}}d\zeta_{ij}^{-}d\overline{\zeta_{ij}^{-}}}$

Therefore by Lemma 5.4 we find that:

$\alpha(T_{p}\mathcal{M},\omega_{p})=\prod_{i\leq
r,j>r}\frac{a_{i}+a_{j}}{a_{i}-a_{j}}\frac{a_{i}-a_{j}}{a_{i}+a_{j}}=1.$

In particular we see that $\alpha(T_{p}\mathcal{M},\omega_{p})$ does not
depend on a choice of $a_{1},\dots,a_{n}$.

Now consider $Q$ as a vector field on $\mathcal{M}$ and let us compute $Z(Q)$.
As in the proof of Proposition 5.13 we should describe

$S:=\\{g\in\mathcal{U}_{0}\mid\operatorname{Ad}_{g}^{-1}(Q)\in\operatorname{Lie}\mathcal{K}\\}.$

Then $Z(Q)$ is in bijection $S/\mathcal{K}$. Note that the projection of
$Q^{2}$ to either factor of $GL(n)\times GL(n)=GL(n|n)_{0}$ is a generic
element of the Cartan subalgebra, so we can reduce the situation to
computation elements $w\in W/W_{K}$ such that

$w(Q)\in\operatorname{Lie}\mathcal{K},$ (5.3)

where $W$ is the Weyl group of $\mathcal{U}$ and $W_{K}$ is the Weyl group of
$\mathcal{K}$. Note that $W\simeq S_{n}\times S_{n}$. Let $w=(\sigma,\tau)$
for some $\sigma,\tau\in S_{n}$. Then $w$ satisfies (5.3) iff for each $i$
either $\sigma(i),\tau(i)\in\\{1,\dots,r\\}$ or
$\sigma(i),\tau(i)\in\\{r+1,\dots,n\\}$. Therefore
$w=(\sigma,\sigma)w^{\prime}$ for some $w^{\prime}\in W_{K}$. Thus,
$|Z(Q)|=\binom{n}{r}$. Moreover, $Z(Q)$ is on the orbit of the diagonal
$S_{n}$ in $W$. Furthermore,

$\alpha(T_{wp}\mathcal{M},\omega_{wp})=\alpha(T_{p}\mathcal{M},\omega_{p})=1,$

since $w^{-1}Q$ has the same form as $Q$ with permuted $a_{1},\dots,a_{n}$.
Thus, Theorem 5.5 gives $\int_{\mathcal{M}}\omega=\pi^{r(n-r)}\binom{n}{r}$. ∎

######  Proposition 5.15.

Let $m>n$. Then we have

$V(m-n|0,m|n))=(-1)^{n(m-n)}V(n|n,m|n)=(2\pi)^{n(m-n)}.$

###### Proof.

We show that $V(m-n|0,m|n)=(2\pi)^{n(m-n)}$, with the second equality
following from Lemma 5.10.

Choose $Q\in\operatorname{Lie}U(n|n)\subset\operatorname{Lie}{\mathcal{K}}$ in
the same way as in the proof of Lemma 5.14. One can easily see $Q$ has exactly
one zero point $p$ with stabilizer $\mathcal{K}$; it remains to compute
$\alpha(T_{p}Gr(m-n|0,m|n),\omega_{p})$. In the first case,
$T_{p}Gr(m-n|0,m|n)=\operatorname{Hom}(\mathbb{C}^{m-n|0},\mathbb{C}^{n|n})$,
and we may take as basis

$u_{ij}=e_{i}\otimes e_{j}^{*},\ \ \ \ v_{ij}=f_{i}\otimes e_{j}^{*},\ \ \
1\leq i\leq n,\ n<j\leq m.$

We see that

$Qu_{ij}=(1+i)a_{i}v_{ij},\ \ \ \ Qv_{ij}=(1+i)a_{i}u_{ij}.$

Thus following the same reasoning as in Lemma 5.14, we find that

$\alpha(T_{p}\mathcal{M},\omega_{p})=\prod\limits_{i,j}\frac{a_{ij}}{a_{ij}}=1.$

∎

In what follows, we set $V(a,b):=V(a|0,b|0)=V(0|a,0|b)$ for $0\leq a\leq b$,
and use the notation $I(\mathcal{M},\omega):=\int_{\mathcal{M}}\omega$.

######  Corollary 5.16.

Let $a\leq b\leq c$. Then

$V(b|a,c|a)=\frac{V(b-a,c-a)V(a|a,c|a)}{V(a|a,b|a)}.$

###### Proof.

Consider the partial flag variety $Fl(a|a,b|a,c|a)$ with invariant form
$\tilde{\omega}$. We use the double bundle

$Gr(b|a,c|a)\xleftarrow{Gr(a|a,b|a)}Fl(a|a,b|a,c|a)\xrightarrow{Gr(b-a,c-a)}Gr(a|a,c|a).$

The fibers are marked on the arrows. By $\mathcal{U}$-equivariance we can
choose an invariant volume form $\tilde{\omega}$ on $Fl(a|a,b|a,c|a)$ such
that

$\tilde{\omega}=\omega_{f}\omega_{b}=\omega^{\prime}_{f}\omega^{\prime}_{b},$

where $\omega_{f}$ ( respectively, $\omega^{\prime}_{f}$) are is a volume form
on the fiber of the left (respectively, right) bundle and $\omega_{b}$
(respectively, $\omega^{\prime}_{b}$) is a volume form on the base of the left
(respectively, right) bundle.

Using the right bundle we get

$I({Fl(a|a,b|a,c|a)},\tilde{\omega})=I(Gr(b-a,c-a),\omega^{\prime}_{f})I(Gr(a|a,c|a),\omega^{\prime}_{b}).$

Applying now the left bundle we obtain

$I({Fl(a|a,b|a,c|a)},\tilde{\omega})=I({Gr(a|a,b|a)},\omega_{f})I({Gr(b|a,c|a)},\omega_{b}).$

That gives the desired formula. ∎

######  Proposition 5.17.

Assume that $\operatorname{sdim}Gr(r|s,m|n)\geq 0$ and $r\geq s$ then

$V(r|s,m|n)=(-1)^{(r-s)(n-s)}\frac{V(s|s,n|n)V(n|n,m|n)V(r-s,m-n)}{V(s|s,r|s)V(n-s|n-s,m-r|n-s)}.$

###### Proof.

We start with the case $r=s$ and use again the same method as in the proof of
Corollary 5.16 and we use the same notations. Note that $m>n$. Consider the
partial flag variety $Fl(s|s,n|n,m|n)$ and the double bundle

$Gr(s|s,m|n)\xleftarrow{Gr(n-s|n-s,m-s|n-s)}Fl(s|s,n|n,m|n)\xrightarrow{Gr(s|s,n|n)}Gr(n|n,m|n).$

Then by Proposition 5.14 and Proposition 5.15 we obtain

$I({Fl(s|s,n|n,m|n)},\tilde{\omega})=V(s|s,n|n)V(n|n,m|n)\neq 0.$

From the left bundle we get

$I({Fl(s|s,n|n,m|n)},\tilde{\omega})=V(n-s|n-s,m-s|n-s)V(s|s,m|n).$

That gives

$V(s|s,m|n)=\frac{V(s|s,n|n)V(n|n,m|n)}{V(n-s|n-s,m-s|n-s)}.$

For the general statement consider the double bundle

$Gr(r|s,m|n)\xleftarrow{Gr(s|s,r|s)}Fl(s|s,r|s,m|n)\xrightarrow{Gr(r-s|0,m-s|n-s)}Gr(s|s,m|n).$

Then

$I({Fl(s|s,r|s,m|n)},\tilde{\omega})=V(r-s|0,m-s|n-s)V(s|s,m|n).$

On the other hand, from the left bundle

$I({Fl(s|s,r|s,m|n)},\tilde{\omega})=V(s|s,r|s)V(r|s,m|n),$

and using Lemma 5.10 we get

$V(r|s,m|n)=(-1)^{(r-s)(n-s)}\frac{V(s|s,n|n)V(n|n,m|n)V(m-r|n-s,m-s|n-s)}{V(s|s,r|s)V(n-s|n-s,m-s|n-s)}.$

Using Corollary 5.16 for $a=n-s,b=m-r,c=m-s$ we have

$V(m-r|n-s,m-s|n-s)=\frac{V(r-s,m-n)V(n-s|n-s,m-s|n-s)}{V(n-s|n-s,m-r|n-s)},$

and substitution gives the desired formula. ∎

######  Corollary 5.18.

Assume that $r\geq s$ and $\operatorname{sdim}Gr(r|s,m|n)\geq 0$. Then

$\int_{Gr(r|s,m|n)}\omega=(-1)^{s(m+n+r+s)}\binom{n}{s}(2\pi)^{A}\int_{Gr(r-s,m-n)}\omega,$

where $A=(m-r)s+(n-s)r=\operatorname{dim}Gr(r|s,m|n)_{1}$.

Corollary 5.18 and Proposition 5.13 imply Theorem 5.11.

###  5.6. The volume of $Q$-Grassmannian

In this subsection we assume that $G=Q(n)$ and $P$ is the parabolic subgroup
with the Levi subgroup $K=Q(r)\times Q(n-r)$. The homogeneous space
$QGr(r,n)=G/P$ is called a $Q$-Grassmannian. It represents the functor of all
$r|r$-dimensional subspaces in $\mathbb{C}^{n|n}$ invariant under the fixed
odd linear operator $F:\mathbb{C}^{n|n}\to\mathbb{C}^{n|n}$. As in the case of
$GL$ the complex quasireductive group $Q(n)$ has a compact form
$\mathcal{U}=UQ(n)$ and the Lie algebra of $\mathcal{U}$ is the set of fixed
of the involution $\theta$ restricted to $\mathfrak{q}(n)$ defined in the
previous section. It is well defined as
$\mathfrak{q}(n)\subset\mathfrak{gl}(n|n)$ is $\theta$-stable. We set
$\mathcal{K}=\mathcal{U}\cap K\simeq UQ(r)\times UQ(n-r)$. Here is an analogue
of Proposition 5.8.

######  Proposition 5.19.

There is an isomorphism $QGr(r,n)\simeq G/P\simeq\mathcal{U}/\mathcal{K}$.
Furthermore, there exists an unique up to normalization
$\mathcal{U}$-invariant volume form $\omega$ on $\mathcal{M}:=QGr(r,n)$.

###### Proof.

The first assertion can be proven exactly as in Proposition 5.8. To prove the
existence of invariant form let us consider a point $p\in\mathcal{M}$ with
stabilizer $\mathcal{K}$. There is an obvious identification of the tangent
space $T_{p}\mathcal{M}$ with
$\operatorname{Hom}_{\mathbb{C}[F]}(\mathbb{C}^{r|r},\mathbb{C}^{n-r|n-r})$.
Note that in contrast with the case of undecorated supergrassmannians the
$\mathcal{K}$-module $T_{p}\mathcal{M}$ is not self-dual. However, since
$\mathcal{K}$ admits no nontrivial characters, we have that
$\mathcal{K}\subset SL(T_{p}\mathcal{M})$. It will be useful to choose a
precise coordinate system in $T_{p}\mathcal{M}$. If we represent a tangent
vector by a matrix of the form $\left(\begin{matrix}Z&W\\\
W&Z\end{matrix}\right)$ where $Z,W$ are $r\times(n-r)$-matrices, then

$\omega_{p}=\prod_{i\leq
r,j>r}\frac{dz_{ij}d\bar{z}_{ij}}{dw_{ij}d\bar{w}_{ij}}.$

∎

######  Proposition 5.20.

Let $\mathcal{M}=QGr(n,r)$. Then
$\int_{\mathcal{M}}\omega=(2\pi)^{2r(n-r)}C(r,n)$, where

$C(r,n)=\begin{cases}\binom{m}{l}\ \text{if}\ n=2m,r=2l,n=2m+1,r=2l+1,2l\\\ 0\
\ \ \ \text{if}\ n=2m,r=2l+1\end{cases}.$

In particular, $\int_{\mathcal{M}}\omega\neq 0$ if and only if $r(n-r)$ is
even.

###### Proof.

The idea of the proof is to use again the Schwarz–Zaboronsky formula for a
suitably chosen odd vector field. We observe that $Q$ constructed in the proof
of proposition 5.14 is an element of $\operatorname{Lie}UQ(n)$ via the
embedding $UQ(n)\subset U(n|n)$. It is straightforward to check that $Z(Q)$
has $\binom{n}{r}$ points and all these points are obtained from the point $p$
by the action of the Weyl group $S_{n}$. Every such point represents the
coordinate subspace of $\mathbb{C}^{n|n}$ with basis
$u_{i_{1}},\dots,u_{i_{r}},v_{i_{1}},\dots,v_{i_{r}}$ if
$u_{1},\dots,u_{n},v_{1},\dots,v_{n}$ is the standard basis. Thus, we
enumerate the points of $Z(Q)$ by $r$ element subsets $S$ of $\\{1,\dots,n\\}$
and use the notation $p_{S}$. Choosing exactly half of the basis
$u_{ij}^{+},v_{ij}^{+}$ defined in the proof Proposition 5.14 we obtain

$\alpha(S):=\alpha(T_{p_{S}}\mathcal{M})=\prod_{i\in S,j\notin
S}\frac{a_{i}+a_{j}}{a_{i}-a_{j}}.$

Thus, we have reduced the proof of the theorem to computation of the
expression

$C(r,n):=\sum_{|S|=r}\alpha(S).$

First, note that $C(r,n)$ is a symmetric rational function in
$a_{1},\dots,a_{n}$ of homogeneous degree $0$ with denominator given by the
Vandermonde determinant. That actually means that it is constant which does
not depend on $a_{1},\dots,a_{n}$ as it should be by Theorem 5.5. Next, the
substitution $a_{n}=0$ gives the recursions:

$C(r,n)=C(r,n-1)+(-1)^{n-r}C(r-1,n-1),\quad C(r,n)=(-1)^{r(n-r)}C(n-r,n),\quad
C(1,2)=0.$

From this one may show that $C(r,n)$ has the desired values. ∎

##  6\. Splitting subgroups for $GL(m|n)$ and $Q(n)$

######  Theorem 6.1.

(a) Let $G=GL(m|n)$, $K=GL(r|s)\times GL(m-r|n-s)$. Then $K$ is a splitting
subgroup of $G$ if and only if $\operatorname{sdim}Gr(r|s,m|n)\geq 0$.

(b) Let $G=Q(n)$, $K=Q(r)\times Q(n-r)$. Then $K$ is a splitting subgroup of
$G$ if and only if $r(n-r)$ is even.

###### Proof.

If $\operatorname{sdim}Gr(r|s,m|n)<0$ or $r(n-r)$ is odd, the necessary
condition for $K$ does not hold. Therefore it remains to show that if
$\operatorname{sdim}Gr(r|s,m|n)\geq 0$ or $r(n-r)$ is even the trivial
$G$-submodule splits in $\mathcal{O}(G/K)$. We use unitary trick. Consider the
embedding $\mathcal{M}=\mathcal{U}/\mathcal{K}\to G/K$ as the closed
$\mathcal{U}$-orbit of $p\in G/K$ with stabilizer $K$. Define the map
$I:\mathcal{O}(G/K)\to\mathbb{C}$ by

$I(f):=\int_{\mathcal{M}}f\omega.$

Note that $I$ is $\mathcal{U}$-invariant and hence $G$-invariant. Theorem 5.11
and Theorem 5.20 imply $I(1)\neq 0$. Therefore $\mathbb{C}$ splits. ∎

######  Corollary 6.2.

(a) Let $G=GL(m|n)$ with defect $d=\min(m,n)$, then the defect subgroup
$D\simeq SL(1|1)^{d}$ is splitting in $G$ and hence any subgroup of $G$
containing $D$ is splitting.

(b) Let $G=Q(2d)$ (respectively, $Q(2d+1)$) then the subgroup $Q(2)^{d}$
(respectively, $Q(2)^{d}\times Q(1)$) is splitting.

###### Proof.

For (b) we just use Theorem 6.1 and transitivity Corollary 2.7(2). For (a)
assume that $d=n$, then by the same argument $GL(1|1)^{d}\times GL(m-n)$ is
splitting. Furthermore, $H:=GL(1|1)^{d}$ is splitting in $GL(1|1)^{d}\times
GL(m-n)$. Therefore $H$ is splitting in $G$. Finally $D$ is splitting in $H$
by Lemma 2.11. ∎

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Vera Serganova, Dept. of Mathematics, University of California at Berkeley,
Berkeley, CA 94720

Email address<EMAIL_ADDRESS>
Alexander Sherman, Dept. of Mathematics, Ben Gurion University, Beer-Sheva,
Israel

Email address<EMAIL_ADDRESS>