# Ghost distributions on supersymmetric spaces I: Koszul induced superspaces,
branching, and the full ghost centre

Alexander Sherman

###### Abstract.

Given a Lie superalgebra $\mathfrak{g}$, Gorelik defined the anticentre
$\mathcal{A}$ of its enveloping algebra, which consists of certain elements
that square to the center. We seek to generalize and enrich the anticentre to
the context of supersymmetric pairs $(\mathfrak{g},\mathfrak{k})$, or more
generally supersymmetric spaces $G/K$. We define certain invariant
distributions on $G/K$, which we call ghost distributions, and which in some
sense are induced from invariant distributions on $G_{0}/K_{0}$. Ghost
distributions, and in particular their Harish-Chandra polynomials, give
information about branching from $G$ to a symmetric subgroup $K^{\prime}$
which is related (and sometimes conjugate) to $K$. We discuss the case of
$G\times G/G$ for an arbitrary quasireductive supergroup $G$, where our
results prove the existence of a polynomial which determines projectivity of
irreducible $G$-modules. Finally, a generalization of Gorelik’s ghost centre
is defined called the full ghost centre, $\mathcal{Z}_{full}$. For type I
basic Lie superalgebras $\mathfrak{g}$ we fully describe $\mathcal{Z}_{full}$,
and prove that if $\mathfrak{g}$ contains an internal grading operator,
$\mathcal{Z}_{full}$ consists exactly of those elements in
$\mathcal{U}\mathfrak{g}$ acting by $\mathbb{Z}$-graded constants on every
finite-dimensional irreducible representation.

##  1\. Introduction

Let $\mathfrak{g}$ be a Lie superalgebra over an algebraically closed field
$k$ of characteristic zero. In [Gor00], Gorelik defined a certain natural
’twisted’ adjoint action of a Lie superalgebra $\mathfrak{g}$ on its
enveloping algebra $\mathcal{U}\mathfrak{g}$. The action was originally
considered by [ABF97] for $\mathfrak{o}\mathfrak{s}\mathfrak{p}(1|2n)$, where
a certain element $T$ in the enveloping algebra was constructed, called
Casimir’s ghost, which squared to the center.

The action defined by Gorelik in general is remarkable in that the structure
of $\mathcal{U}\mathfrak{g}$ becomes that of an induced module,
$\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}\mathcal{U}\mathfrak{g}_{\overline{0}}$.
Further the invariants of this action, denoted $\mathcal{A}$ and called the
anticentre in [Gor00], are both a module over the center
$\mathcal{Z}:=\mathcal{Z}(\mathcal{U}\mathfrak{g})$ and multiply into the
center, so that $\tilde{\mathcal{Z}}:=\mathcal{Z}+\mathcal{A}$ is an algebra
which Gorelik called the ghost centre of $\mathfrak{g}$. If $\mathfrak{g}$ is
quasireductive, i.e. $\mathfrak{g}_{\overline{0}}$ is reductive and acts
semisimply on $\mathfrak{g}$, and $\Lambda^{top}\mathfrak{g}_{\overline{1}}$
is a trivial $\mathfrak{g}_{\overline{0}}$-module, Gorelik obtained an
explicit identification as vector spaces of $\mathcal{A}$ with the center of
$\mathcal{U}\mathfrak{g}_{\overline{0}}$. Further, for basic classical Lie
superalgebras, Gorelik gave a complete description of the Harish-Chandra image
of $\mathcal{A}$ and the structure of $\tilde{\mathcal{Z}}$. In particular she
showed that $\tilde{\mathcal{Z}}$ consists of exactly those elements of
$\mathcal{U}\mathfrak{g}$ that act by superconstants on every irreducible
representation. We also note that Gorelik fully computed the $\mathcal{A}$ for
$Q$-type superalgebras.

###  1.1. Generalization to supersymmetric spaces

We seek to understand Gorelik’s original results from the geometric
perspective, and thereby understand how they may be generalized. Consider the
setting of symmetric supervarieties (or supersymmetric spaces, if one prefers
that term) $G/K$ and their corresponding supersymmetric pairs
$(\mathfrak{g},\mathfrak{k})$, corresponding to an involution $\theta$. We
have a decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ into the
$\pm 1$ eigenspaces of $\theta$. We define a new subalgebra

$\mathfrak{k}^{\prime}:=\mathfrak{k}_{\overline{0}}\oplus\mathfrak{p}_{\overline{1}},$

which itself is the fixed points of the involution $\delta\circ\theta$, where
$\delta(x)=(-1)^{\overline{x}}x$ is the grading operator on $\mathfrak{g}$.
Let $K^{\prime}$ be the subgroup of $G$ with $K^{\prime}_{0}=K_{0}$ and
$\operatorname{Lie}(K^{\prime})=\mathfrak{k}^{\prime}$. Then because
$\mathfrak{k}_{\overline{1}}^{\prime}\oplus\mathfrak{k}_{\overline{1}}=\mathfrak{g}_{\overline{1}}$,
we will see that the action of $K^{\prime}$ on $G/K$ enjoys many nice
properties, in particular many of which are not explicitly enjoyed by the
action of $K$ on $G/K$. The first result indicating this is the following.
Recall that for an affine supervariety $X$ with a closed point $x$ and maximal
ideal $\mathfrak{m}_{x}\subseteq k[X]$ in the space of functions, we may
consider the super vector space of distributions

$\operatorname{Dist}(X,x)=\\{\psi:k[X]\to k:\psi(\mathfrak{m}_{x}^{n})=0\text{
for }n\gg 0\\}.$

If we consider the point $eK$ on $G/K$, $K_{0}^{\prime}$ fixes it, and thus
$K^{\prime}$ has a natural action on $\operatorname{Dist}(G/K,eK)$.

######  Theorem 1.1.

We have an isomorphism of $K^{\prime}$-modules

$\operatorname{Dist}(G/K,eK)\cong\operatorname{Ind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}}\operatorname{Dist}(G_{0}/K_{0},eK_{0}).$

In particular, if $K^{\prime}$ is quasireductive and
$\Lambda^{top}\mathfrak{p}_{\overline{1}}$ is a trivial $K_{0}$-module, then
we have an explicit isomorphism of vector spaces

$\operatorname{Dist}(G/K,eK)^{K^{\prime}}\cong\operatorname{Dist}(G_{0}/K_{0},eK_{0})^{K_{0}}.$

For the symmetric supervariety $G\times G/G\cong G$, we have identifications
$G^{\prime}\cong G$ and
$\operatorname{Dist}(G,eG)\cong\mathcal{U}\mathfrak{g}$, and the action we
obtain in this case of $G^{\prime}$ on $\operatorname{Dist}(G,eG)$ is exactly
Gorelik’s twisted adjoint action, thus reproducing her results from this
context.

The proof of 1.1 uses a construction due to Koszul ([Kos82]) which allows one
to take the variety $G_{0}/K_{0}$, which has an action of
$K_{0}^{\prime}=K_{0}$, and induce it to a $K^{\prime}$-supervariety denoted
$(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}$. The latter supervariety has the
special property that the vector fields $\mathfrak{k}_{\overline{1}}$ are
everywhere non-vanishing, and its algebra of functions and spaces of
distributions, respectively, are (co)induced from $G_{0}/K_{0}$. Then by a
general result, $G/K$ and $(G_{0}/K_{0})^{\mathfrak{k}}$ are locally
isomorphic as $K^{\prime}$-supervarieties, implying an isomorphism of their
spaces of distributions as in 1.1.

###  1.2. The Harish-Chandra morphism

Now consider a symmetric supervariety $G/K$ where we assume $G$ is
quasireductive and Cartan-even (i.e. its Lie superalgebra has an even Cartan
subalgebra), $\Lambda^{top}\mathfrak{p}_{\overline{1}}$ is the trivial
$K_{0}$-module, and we have an Iwasawa decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$. We write

$\mathcal{A}_{G/K}:=\operatorname{Dist}(G/K,eK)^{K^{\prime}},$

for the $K^{\prime}$-invariant distributions, and call elements of
$\mathcal{A}_{G/K}$ ghost distributions on $G/K$. One important problem is to
compute the image of $\mathcal{A}_{G/K}$ under the Harish-Chandra homomorphism

$HC:\operatorname{Dist}(G/K,eK)\to S(\mathfrak{a}).$

In this case $G/K$ is also spherical, and admits rational functions
$f_{\lambda}$ indexed by a full rank lattice
$\Lambda\subseteq\mathfrak{a}^{*}$ which are $\mathfrak{n}$-invariant and
eigenfunctions for $\mathfrak{a}$, where we normalize by requiring that
$f_{\lambda}(eK)=1$. If $\lambda$ is a dominant weight, then generically
$f_{\lambda}$ will be a regular function. Then for a distribution $\psi$,

$HC(\psi)(\lambda)=\psi(f_{\lambda}).$

Using this, we have the following representation-theoretic interpretation of
the values of $\mathcal{A}_{G/K}$ on such eigenfunctions.

######  Theorem 1.2.

Let $f_{\lambda}\in k[G/K]$ be a highest weight vector of weight $\lambda$.
Then the $K^{\prime}$-module generated by $f_{\lambda}$ contains a copy of
$I_{K^{\prime}}(k)$ if and only if there exists $\gamma\in\mathcal{A}_{G/K}$
such that $HC(\gamma)(\lambda)\neq 0$.

Here $I_{K^{\prime}}(k)$ denote the injective indecomposable
$K^{\prime}$-module with socle $k$. From this we obtain, as one corollary:

######  Corollary 1.3.

Keep the hypotheses of 1.2, and suppose further that the $G$-module generated
by $f_{\lambda}$, $L$, is irreducible. Then $I_{G}(L)$ is a submodule of
$k[G/K^{\prime}]$.

Thus obtaining $HC(\mathcal{A}_{G/K})$ is of interest, and this will be taken
up in more detail for the case of basic classical Lie superalgebras in a
subsequent article. However even for such Lie superalgebras the answer is not
known in general (outside the case $G\times G/G$, originally done by Gorelik).

###  1.3. The case of $G\times G/G$

For $G\times G/G$, we have the following nice application of 1.3. Note that in
the following we remove the restriction that $G$ be Cartan-even. For this one
needs to generalize the Harish-Chandra morphism to the non-Cartan-even case,
which is done in the appendix.

######  Theorem 1.4.

Let $G$ be a quasireductive supergroup with Cartan subalgebra
$\mathfrak{h}\subseteq\mathfrak{g}$, and choose a Borel subgroup $B$ with Lie
superalgebra $\mathfrak{b}$ containing $\mathfrak{h}$. Then there exists a
polynomial $p_{G,B}\in S(\mathfrak{h})$ of degree less than or equal to
$\operatorname{dim}\mathfrak{b}_{\overline{1}}$, such that for a
$\mathfrak{b}$-dominant weight $\lambda$,

$p_{G,B}(\lambda)\neq 0\ \text{ if and only if }\ L_{B}(\lambda)\text{ is
projective},$

where $L_{B}(\lambda)$ is the irreducible $G$-module of $B$-highest weight
$\lambda$.

In particular the above result implies that if one simple $G$-module is
projective then this is a generic property of simple $G$-modules. However it
is possible that $p_{G,B}$ is the zero polynomial, so that no simple
$G$-modules are projective. Although this was already know for many Lie
superalgebras by direct study, this gives a general explanation for this
phenomenon.

######  Remark 1.5.

In every example the author is aware of, $p_{G,B}$ is a product of linear
polynomials. We do not see an a priori reason for this, and it would be
interesting to find an example where this does not occur (or to prove it
always happens).

Recall that $\mathcal{A}$ denotes the ghost center of
$\mathcal{U}\mathfrak{g}$. Then $\mathcal{A}$ contains an element of least
degree, which we write as $T_{\mathfrak{g}}$. This operator can test whether a
given semisimple $G$-module is projective in the following sense:

######  Proposition 1.6.

Let $L$ be a simple $G$-module. Then $T_{\mathfrak{g}}$ acts by $0$ on $L$ if
and only if $L$ is not projective. If $L$ is projective, then
$T_{\mathfrak{g}}$ acts by one of the following automorphisms:

  * •

if $T_{\mathfrak{g}}$ is even, then up to scalar it acts by $\delta_{L}$, the
parity operator on $L$;

  * •

if $T_{\mathfrak{g}}$ is odd, then it acts by $\delta_{L}\circ\sigma_{L}$,
where $\sigma_{L}:L\to L$ is a $G$-equivariant odd automorphism of $L$.

For a different application of 1.4, we can prove the following general
sufficient criteria for having projective irreducible modules.

######  Theorem 1.7.

Let $G$ be a Cartan-even quasireductive supergroup with a chosen Cartan
subalgebra $\mathfrak{h}$ such that the following conditions hold on its Lie
superalgebra:

  1. (1)

for a root $\alpha\in\mathfrak{h}^{*}$,
$\operatorname{dim}[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]=1$; and

  2. (2)

for a root $\alpha\in\mathfrak{h}^{*}$, the pairing

$[-,-]:\mathfrak{g}_{\alpha}\otimes\mathfrak{g}_{-\alpha}\to[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]$

is nondegenerate.

Choose an arbitrary Borel subgroup $B$ whose Lie superalgebra contains
$\mathfrak{h}$, and let $\alpha_{1},\dots,\alpha_{n}$ denote the odd positive
roots with $r_{i}=\operatorname{dim}\mathfrak{g}_{\alpha_{i}}$. Write
$h_{\alpha_{i}}$ for a nonzero element of
$[\mathfrak{g}_{\alpha_{i}},\mathfrak{g}_{-\alpha_{i}}]$. Then we have (up to
a scalar)

$p_{G,B}=h_{\alpha_{1}}^{r_{1}}\cdots h_{\alpha_{n}}^{r_{n}}+l.o.t.$

In particular $p_{G,B}\neq 0$, so $G$ admits irreducible projective modules.

Note that the above conditions hold in particular for a Cartan-even quadratic
quasireductive supergroup, i.e. a Cartan-even quasireductive supergroup whose
Lie superalgebra admits a nondegenerate, invariant, and even supersymmetric
form. See [Ben00] for a classification of quadratic quasireductive Lie
superalgebras.

###  1.4. The full ghost centre $\mathcal{Z}_{full}$

We may generalize Gorelik’s results in a different way as follows to produce
an interesting subalgebra of $\mathcal{U}\mathfrak{g}$ which contains
Gorelik’s ghost centre. Let
$\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})$ be those
automorphisms of $\mathfrak{g}$ that fix $\mathfrak{g}_{\overline{0}}$
pointwise. For
$\phi\in\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})$, define
the $\phi$-twisted adjoint action $\operatorname{ad}_{\phi}$ of $\mathfrak{g}$
on $\mathcal{U}\mathfrak{g}$ by

$\operatorname{ad}_{\phi}(u)(v)=uv-(-1)^{\overline{u}\overline{v}}v\phi(u).$

When $\phi=\delta$, we obtain the twisted adjoint action studied by Gorelik.
Then using 1.1 we can prove that:

######  Theorem 1.8.

If $\phi(x)=x$ implies that $x\in\mathfrak{g}_{\overline{0}}$, then
$\mathcal{U}\mathfrak{g}\cong\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}\mathcal{U}\mathfrak{g}_{\overline{0}}$
under the $\phi$-twisted adjoint action.

Write $\mathcal{A}_{\phi}\subseteq\mathcal{U}\mathfrak{g}$ for the
$\operatorname{ad}_{\phi}$-invariant elements in $\mathcal{U}\mathfrak{g}$,
for any $\phi\in\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})$.
Then $\mathcal{A}_{\operatorname{id}}=\mathcal{Z}$, and
$\mathcal{A}_{\delta}=\mathcal{A}$. Further, for
$\phi,\psi\in\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})$
multiplication induces a morphism

$\mathcal{A}_{\phi}\otimes\mathcal{A}_{\psi}\to\mathcal{A}_{\psi\phi}.$

Therefore if we set

$\mathcal{Z}_{full}:=\sum\limits_{\phi\in\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})}\mathcal{A}_{\phi},$

we obtain a subalgebra of $\mathcal{U}\mathfrak{g}$, which also contains
$\tilde{\mathcal{Z}}$. For $\mathfrak{g}$ one of the type I basic classical
Lie superalgebras $\mathfrak{g}\mathfrak{l}(m|n)$,
$\mathfrak{s}\mathfrak{l}(m|n)$, $\mathfrak{p}\mathfrak{s}\mathfrak{l}(n|n)$
with $n>2$, or $\mathfrak{o}\mathfrak{s}\mathfrak{p}(2|2n)$) we have an
explicit description of this algebra. Here,
$\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})\cong
k^{\times}$, so we write
$\phi_{c}\in\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})$ for
the automorphism corresponding to $c\in k^{\times}$ and $\mathcal{A}_{c}$ for
the invariants of the $\phi_{c}$-twisted adjoint action. Then $\phi_{c}$
satisfies the conditions of 1.8 exactly if $c\neq 1$.

######  Theorem 1.9.

Let $N=\operatorname{dim}\mathfrak{g}_{\overline{1}}/2$. Then
$HC(\mathcal{A}_{c})=HC(\mathcal{A}_{-1})$ for all $c\neq 1$, and

$\mathcal{Z}_{full}=\bigoplus\limits_{\zeta^{N}=1}\mathcal{A}_{\zeta}.$

Further, for $\mathfrak{g}\mathfrak{l}(m|n)$, $\mathfrak{s}\mathfrak{l}(m|n)$
with $m\neq n$, and $\mathfrak{o}\mathfrak{s}\mathfrak{p}(2|2n)$,
$\mathcal{Z}_{full}$ consists of exactly the set of elements in
$\mathcal{U}\mathfrak{g}$ which act by $\mathbb{Z}$-graded constants on all
finite-dimensional irreducible representations of $\mathfrak{g}$.

We exclude $\mathfrak{s}\mathfrak{l}(n|n)$ and
$\mathfrak{p}\mathfrak{s}\mathfrak{l}(n|n)$ in the last statement due to the
lack of an internal grading operator. Further,
$\mathfrak{p}\mathfrak{s}\mathfrak{l}(2|2)$ is fully excluded because in this
case $\operatorname{Aut}(\mathfrak{g},\mathfrak{g}_{\overline{0}})=SL_{2}$
(see section 5.5 of [Mus12]). For this reason it would be interesting to
compute the full ghost center for this superalgebra.

###  1.5. Future work

This article is the first of two on ghost distributions and their
applications. In the subsequent article we study in more detail two questions
of interest: (1) when is it possible to produce an algebra using ghost
distributions on a general supersymmetric space? In particular can we form an
algebra of differential operators, and if not can we at least produce an
algebra of polynomials using the Harish-Chandra homomorphism? (2) Computing
$HC(\mathcal{A}_{G/K})$ as much as possible in the case when $\mathfrak{g}$ is
an almost simple basic classical Lie superalgebra and the involution under
consideration preserves the form on $\mathfrak{g}$. In this case we give some
general properties of $HC(\mathcal{A}_{G/K})$, and seek to compute it for all
rank one supersymmetric pairs.

###  1.6. Outline of paper

In section 2 we introduce the basic algebraic supergeometry we need, in
particular the algebra of differential operators and the space of
distributions at a given point. Section 3 recalls basic facts about algebraic
supergroups and their actions, and gives a description of invariant
differential operators on homogeneous spaces. In section 4, we explain the
main technical construction, the induced superspace as defined by Koszul.
Section 5 applies the ideas of section 4 to homogeneous superspaces, and
deduces what we need to generalize the results of [Gor00]. Section 6 studies
the case of $G/G_{0}$, looking in particular at a certain invariant
distribution, $v_{\mathfrak{g}}$, which will play an important role in the
theory of ghost distributions. Section 7 looks at applications to a symmetric
supervariety $G/K$, and section 8 studies more closely the case when an
Iwasawa decomposition is present, giving the definition of the Harish-Chandra
map and its interpretations. In section 9 we take a special look at the case
of $G\times G/G$, where the theory of ghost distributions is most developed,
and prove 1.4. Finally in section 10 we define and study the full ghost centre
$\mathcal{Z}_{full}$, and look especially at the cases of type I algebras. The
appendix proves 1.4 by generalizing the Harish-Chandra homomorphism to the
case when $G$ is not Cartan-even, and studies the image of the Harish-Chandra
homomorphism in this case.

###  1.7. Acknowledgments

The author would like to thank Alexander Alldridge for many enlightening
discussions about this project. Thanks to Vera Serganova for her patience,
tremendous support and guidance throughout my PhD, during which much of this
work was done. Thank you to Maria Gorelik for patiently explaining many
aspects of her relevant papers to me. Finally thank you to Siddartha Sahi,
Johannes Flakes, and Inna Entova-Aizenbud for helpful comments and
suggestions. This research was partially supported by ISF grant 711/18 and
NSF-BSF grant 2019694.

##  2\. Preliminaries from algebraic supergeometry

###  2.1. Linear super algebra notation

We work throughout over an algebraically closed field $k$ of characteristic
zero. For a super vector space $V$ we write $V=V_{\overline{0}}\oplus
V_{\overline{1}}$ for its parity decomposition. Even though we precisely
consider $V_{\overline{0}}$ and $V_{\overline{1}}$ to be even vector spaces,
we will occasionally (and abusively) view them as super vector spaces where
$V_{\overline{0}}$ is purely even and $V_{\overline{1}}$ is purely odd. We
write
$\operatorname{dim}V=\operatorname{dim}V_{\overline{0}}+\operatorname{dim}V_{\overline{1}}$
for the dimension of the underlying vector space of $V$, and the
superdimension of $V$ is given by
$\operatorname{sdim}V=\operatorname{dim}V_{\overline{0}}-\operatorname{dim}V_{\overline{1}}$.

###  2.2. Algebraic supergeometry notation

We will use the symbols $X,Y,\dots$ for supervarieties with even subschemes
$X_{0},Y_{0},\dots.$ We will be considering supervarieties in the sense of
[She19], however all spaces of interest will be smooth and affine. A smooth
affine supervariety is always given by $\Lambda^{\bullet}E$, the exterior
algebra of a vector bundle $E$ on a smooth affine variety $X_{0}$ (see
[VMP90]). Thus one will lose almost nothing if one simply works with
supervarieties of this form in this article. Note that affine supervarieties
and morphisms between them are entirely determined by their spaces of global
functions and maps between them, just as with affine varieties. See [CCF11]
for more on the basics of algebraic supergeometry.

If $X$ is a supervariety, there is a canonical closed embedding
$i_{X}:X_{0}\to X$, and this is a homeomorphism of underlying topological
spaces. The closed points of $X$ are the $k$-points, which we write as $X(k)$,
and they are canonically identified with the closed points of $X_{0}$ via
$i_{X}$. If $x$ is a closed point of $X$ and $\mathcal{F}$ is a sheaf on $X$,
we write $\mathcal{F}_{x}$ for the stalk of $\mathcal{F}$ at $x$. Then
$\mathcal{O}_{X,x}$ is a local superalgebra, and we write $\mathfrak{m}_{x}$
for its corresponding maximal ideal and ${}_{0}\mathfrak{m}_{x}$ for the
maximal ideal in $\mathcal{O}_{X_{0},x}$. For affine supervarieties we will
also write $\mathfrak{m}_{x}$ for the maximal ideal of $k[X]$ corresponding to
$x$, and similarly ${}_{0}\mathfrak{m}_{x}$ for the maximal ideal of
$k[X_{0}]$ corresponding to $x$.

###  2.3. Differential operators and distributions

######  Definition 2.1.

For a supervariety $X$, let $\mathcal{D}_{X}$ denote the sheaf of filtered
algebras which is the subsheaf of $\mathcal{E}nd(\mathcal{O}_{X})$ defined
inductively as follows. We set $\mathcal{D}_{X}^{n}=0$ for $n<0$, and for
$n\geq 0$ and an open subset $U$ of $X$, set

$\Gamma(U,\mathcal{D}_{X}^{n}):=\\{D\in\operatorname{End}(\mathcal{O}_{U})):[D,f]\in\Gamma(U,\mathcal{D}_{X}^{n-1})\text{
for all }f\in\mathcal{O}_{X}(U)\\}$

We call $\mathcal{D}_{X}$ the sheaf of differential operators on $X$, and
refer to its sections as differential operators. If $\mathcal{F}$ is a sheaf
on $X$, we say that it is a left, resp. right $\mathcal{D}_{X}$-module if it
is exactly that.

######  Definition 2.2.

If $x$ is a closed point of $X$, define the super vector space of
distributions at $x$ to be all (not necessarily even) linear maps
$\psi:\mathcal{O}_{X,x}\to k$ such that for some $n\in\mathbb{N}$ we have
$\psi(\mathfrak{m}_{x}^{n})=0$. We denote this super vector space by
$\operatorname{Dist}(X,x)$. Define
$\operatorname{Dist}^{n}(X,x)\subseteq\operatorname{Dist}(X,x)$ to be those
distributions vanishing on $\mathfrak{m}_{x}^{n+1}$ so that
$\operatorname{Dist}(X,x)$ obtains a filtration. Note that
$\operatorname{Dist}^{0}(X,x)$ is one-dimensional and consists of the
distinguished even distribution given by evaluation at $x$, which we denote by
$\operatorname{ev}_{x}$.

We may give $\operatorname{Dist}(X,x)$ the structure of a right
$\mathcal{D}_{X}$-module as follows. We view $\operatorname{Dist}(X,x)$ as a
sheaf on $X$ supported on $x$. Given a differential operator $D$ defined in a
neighborhood of $x$, and a distribution $\psi$, define

$(\psi D)(f):=\psi(Df)$

This action respects the filtration on $\operatorname{Dist}(X,x)$, so it
becomes a filtered right $\mathcal{D}_{X}$-module.

The following lemma is proved in the same way as in the classical setting, so
we omit the proof.

######  Lemma 2.3.

Let $X$ be a supervariety with a closed point $x$.

  1. (1)

Given a map of supervarieties $\phi:X\to Y$, we have a natural map of filtered
super vector spaces
$d\phi_{x}:\operatorname{Dist}(X,x)\to\operatorname{Dist}(Y,\phi(x))$.

  2. (2)

The chain rule holds: if $\phi:X\to Y$ and $\psi:Y\to Z$, then
$d(\psi\circ\phi)=d\psi\circ d\phi$.

  3. (3)

If $X$ is affine, then the natural pairing $\operatorname{Dist}(X,x)\otimes
k[X]\to k$ has the property that if $\psi(f)=0$ for all $f\in k[X]$, then
$\psi=0$.

  4. (4)

There is a natural restriction morphism
$\operatorname{res}_{x}:\Gamma(U,\mathcal{D}_{X})\to\operatorname{Dist}(X,x)$
for any open subscheme $U$ containing $x$, given by
$\operatorname{res}_{x}(D)(f)=D(f)(x)$. This is a morphism of filtered right
$\mathcal{D}_{X}$-modules, where $\mathcal{D}_{X}$ acts on itself by right
multiplication.

######  Remark 2.4.

We have the following identifications:

$\operatorname{Dist}^{n}(X,x)=(\mathcal{O}_{X,x}/\mathfrak{m}_{x}^{n+1})^{*},\
\
\operatorname{Dist}(X,x)=\lim\limits_{\rightarrow}(\mathcal{O}_{X,x}/\mathfrak{m}_{x}^{n+1})^{*}.$

Furthermore when $X$ is affine, we have:

$\operatorname{Dist}^{n}(X,x)=(k[X]/\mathfrak{m}_{x}^{n+1})^{*},\ \
\operatorname{Dist}(X,x)=\lim\limits_{\rightarrow}(k[X]/\mathfrak{m}_{x}^{n+1})^{*}.$

In general we have an isomorphism of $\mathcal{D}_{X}$-modules

$\operatorname{Dist}(X,x)=\Gamma_{\mathfrak{m}_{x}}(\mathcal{O}_{X,x})^{*}$

and for $X$ affine an isomorphism of $\Gamma(X,\mathcal{D}_{X})$-modules

$\operatorname{Dist}(X,x)=\Gamma_{\mathfrak{m}_{x}}k[X]^{*}.$

We recall the definition of the functor $\Gamma_{\mathfrak{m}_{x}}$:

$\Gamma_{\mathfrak{m}_{x}}M=\\{m\in M:\mathfrak{m}_{x}^{n}m=0\text{ for }n\gg
0\\}.$

######  Definition 2.5.

Define the sheaf $\mathcal{T}_{X}$ of vector fields on $X$ by setting
$\Gamma(\text{Spec}A,\mathcal{T}_{X})=\operatorname{Der}(A)$ for an affine
open subscheme $\text{Spec}A$ of $X$. In this way $\mathcal{T}_{X}$ becomes a
subsheaf of $\mathcal{D}_{X}^{1}$, and a sheaf of Lie superalgebras under
supercommutator.

######  Definition 2.6.

For a supervariety $X$ and a closed point $x$ of $X$, we define the tangent
space of $X$ at $x$ to be the super vector space
$T_{x}X:=(\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2})^{*}$. In this way $T_{x}X$ is
exactly the subspace of $\operatorname{Dist}^{1}(X,x)$ given by functionals
$\psi:\mathcal{O}_{X,x}/\mathfrak{m}_{x}^{2}\to k$ such that $\psi(1)=0$.

Observe the restriction morphism
$\operatorname{res}_{x}:\mathcal{D}_{X,x}\to\operatorname{Dist}(X,x)$
restricts to a morphism $\mathcal{T}_{X,x}\to T_{x}X$.

######  Definition 2.7.

A supervariety $X$ is smooth if for all $x\in X(k)$ the morphism
$\operatorname{res}_{x}:\mathcal{T}_{X,x}\to T_{x}X$ is surjective.

See the appendix of [She19] for more equivalent conditions of smoothness.

We have the following standard theorem for $\mathcal{D}_{X}$-modules which we
will use later on. The proof is almost verbatim from the classical case, so we
omit the proof.

######  Proposition 2.8.

Suppose that $X$ is a smooth supervariety. Then if a left (or right)
$\mathcal{D}_{X}$-module $\mathcal{F}$ is coherent over $\mathcal{O}_{X}$,
then it is locally free over $\mathcal{O}_{X}$.

######  Lemma 2.9.

Let $X$ be a supervariety and $x\in X(k)$ a closed point at which $X$ is
smooth. Suppose $V$ is a subspace of vector fields defined in a neighborhood
$U$ of $x$, such that the restriction map $V\to T_{x}X$ is an isomorphism. If
$v_{1},\dots,v_{n}$ is a homogeneous basis of $V$, then the restriction of the
set of all monomials in $v_{1},\dots,v_{n}$, in any order, is a spanning set
of $\operatorname{Dist}(X,x)$

###### Proof.

Choose homogeneous functions $f_{1},\dots,f_{n}$ defined in a neighborhood of
$x$ which vanish at $x$ such that they project to a basis of
$\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}$. Then
$\mathcal{O}_{X,x}/\mathfrak{m}_{x}^{\ell}$ is isomorphic to the set of
polynomials in $f_{1},\dots,f_{n}$ of degree less than or equal to $\ell$. By
applying a linear automorphism, we may assume that
$\operatorname{res}_{x}(v_{i})(f_{j})=\delta_{ij}$. Then we see that

$\operatorname{res}_{x}(v_{1}^{r_{1}}\cdots v_{n}^{r_{n}})(f_{1}^{s_{1}}\cdots
f_{n}^{s_{n}})=\pm s_{1}!\cdots
s_{n}!\delta_{r_{1}s_{1}}\cdots\delta_{r_{n}s_{n}},$

where the sign is determined by the number of parity changes which occurs in
the computation. From this the statement follows. ∎

######  Definition 2.10.

Suppose $\mathcal{F}$ is a quasi-coherent sheaf on $X$. For a closed point $x$
of $X$, define distributions on $\mathcal{F}$ at $x$ to be all linear maps
$\psi:\mathcal{F}_{x}\to k$ such that for some $n\in\mathbb{N}$,
$\psi(\mathfrak{m}_{x}^{n}\mathcal{F}_{x})=0$. We write this as
$\operatorname{Dist}(\mathcal{F},x)$.

######  Remark 2.11.

Notice that if $\mathcal{F}$ has a connection, hence an action by vector
fields, then $\operatorname{Dist}(\mathcal{F},x)$ naturally admits a right
action by vector fields, given by

$(\psi v)(s)=\psi(vs).$

If the connection is flat and $X$ is smooth, then this action extends in the
natural way to a right action by all differential operators.

As before, we have an identification for each $x\in X(k)$:

$\operatorname{Dist}(\mathcal{F},x)\cong\Gamma_{\mathfrak{m}_{x}}(\mathcal{F}_{x})^{*}$

and this identification respects the right action by vector fields when
$\mathcal{F}$ admits a connection.

##  3\. Differential Operators on a $G$-variety

###  3.1. Preliminaries on algebraic supergroups

Recall that an algebraic supergroup is a group object in the category of
algebraic supervarieties. Morphisms of supergroups are those which respect the
multiplication morphisms. We only consider affine algebraic supergroups, or
equivalently those algebraic supergroups that are linear, i.e. have a faithful
finite-dimensional representation. The letters $G,H,\dots$ will denote an
affine algebraic supergroup. To avoid cumbersome language, we will not write
the adjectives affine and algebraic when referring to affine algebraic
supergroups, and instead simply call them supergroups. Similarly, we use the
term subgroup instead of subsupergroup. We refer again to [CCF11] for basics
on algebraic supergroups.

A supergroup has a Lie superalgebra which we will denote with the letters
$\mathfrak{g},\mathfrak{h},\dots$. The Lie superalgebra may be defined as the
super vector space of left-invariant (or right-invariant) vector fields on
$G$. As in the classical case, the Lie superalgebra is canonically identified
with the tangent space of $G$ at the identity, $T_{e}G$. Morphisms of
algebraic supergroups induce morphisms of the corresponding Lie superalgebras.

If $G$ is a supergroup, then $G_{0}$ is an algebraic group, and the morphism
$i_{G}:G_{0}\to G$ is a morphisms of supergroups. Further, $i_{G}$ induces an
isomorphism of Lie algebras
$\operatorname{Lie}(G_{0})\cong\operatorname{Lie}(G)_{\overline{0}}$.

###  3.2. Representations of supergroups

In this article we will be using some basic facts about the representation
theory of quasireductive supergroups. We refer to [Ser11] for further
background.

Recall that for an affine algebraic supergroup $G$, $k[G]$ has the natural
structure of a supercommutative Hopf superalgebra. We define a left $G$-module
to be a left $k[G]$-comodule, and a right $G$-module to be a right
$k[G]$-comodule. It will be necessary for us to consider both left and right
$G$-modules due to the fact that distributions form a right module over the
algebra of differential operators, as we have seen. The category of left
$G$-modules is equivalent to the category of right $G$-modules because $k[G]$
has a Hopf structure. We will sometimes call a left or right $G$-module simply
a representation of $G$, or even a $G$-module, without specifying whether it
has a left or right action. In this case either the type of action is apparent
or is of no importance.

A left (resp. right) $G$-module induces in a natural way a left (resp. right)
representation of the Lie superalgebra. Recall that the category of
representations of $G$ is equivalent to the category of
$(G_{0},\mathfrak{g}=\operatorname{Lie}(G))$-modules such that the action of
$G_{0}$ and $\mathfrak{g}_{\overline{0}}\subseteq\mathfrak{g}$ are compatible
(see [CCF11]).

Given a representation $V$ of a Lie supergroup $G$ such that
$\operatorname{dim}V_{\overline{1}}=n$, we write $\operatorname{Ber}(V)$ for
the Berezinian of $V$, which is the one-dimensional $G$-module with the same
parity as $n$, where the action by $G$ is given by the Berezinian morphism
$G\to GL(V)\xrightarrow{\operatorname{Ber}}\mathbb{G}_{m}$ (see chapter 3 of
[Man13]). If $V$ is purely even (resp. purely odd) then
$\operatorname{Ber}(V)$ coincides with the top exterior power (resp. top
symmetric power) of $V$. We write $\operatorname{ber}_{V}:G\to\mathbb{G}_{m}$
for the character of $G$ determined by $\operatorname{Ber}(V)$, and by abuse
of notation we also write $\operatorname{ber}_{V}$ for the character of
$\mathfrak{g}$ that this determines. Then we have
$\operatorname{ber}_{V}=\det_{V_{\overline{0}}}\cdot\det_{V_{\overline{1}}}^{-1}$
as a character of $G_{0}$, and
$\operatorname{ber}_{V}=\text{tr}_{V_{\overline{0}}}-\text{tr}_{V_{\overline{1}}}$
as a character of $\mathfrak{g}_{\overline{0}}$.

If $\chi:G\to\mathbb{G}_{m}$ is a character of $G$, and $V$ is a
representation of $G$, we will write $V^{\chi}$ for the subspace of $V$ where
$G$ acts by $\chi$. If $\chi$ is the trivial character, we just write
$V^{G}:=V^{\chi}$.

If $V$ is a $G$-module and $v\in V$ is a homogeneous element, the $G$-module
generated by $v$, which we write as $\langle G\cdot v\rangle$, is given by
$\mathcal{U}\mathfrak{g}\cdot\langle G_{0}\cdot v\rangle$. That is, we first
take the $G_{0}$-module generated by $v$, and then take the
$\mathcal{U}\mathfrak{g}$-module which that generates.

Finally, if $V$ is a $G$-module then we will write (when they exist)
$I_{G}(V)$, resp. $P_{G}(V)$ (or $I(V)$, resp. $P(V)$ when the context is
clear) for the injective hull, resp projective cover of $V$.

###  3.3. Actions of supergroups

If $X$ is a supervariety, $G$ a supergroup, and $G$ acts on (the left on) $X$,
then we call $X$ a $G$-supervariety. We will usually reserve the letter
$a_{X}=a$ for the action morphism, i.e. $a:G\times X\to X$. In this case we
will consider $k[X]$ as a right $G$-module via translation. Explicitly, $g\in
G_{0}(k)$ acts by pullback $L_{g}^{*}$ along the left translation morphism
$L_{g}:X\to X$. The Lie superalgebra acts, for $u\in T_{e}G$, by

$u\mapsto(u\otimes 1)\circ a^{*}.$

This induces an map of superalgebras
$\mathcal{U}\mathfrak{g}\to\Gamma(X,\mathcal{D}_{X})^{op}$, and in this way
$\operatorname{Dist}(X,x)$ becomes a left $\mathcal{U}\mathfrak{g}$-module for
any $x\in X(k)$. In general, we will say $\mathfrak{g}$ acts on a supervariety
$X$ if it admits a homomorphism of algebras
$\mathcal{U}\mathfrak{g}\to\Gamma(X,\mathcal{D}_{X})^{op}$ such that
$\mathfrak{g}$ maps into $\Gamma(X,\mathcal{T}_{X})$.

######  Remark 3.1.

Suppose that $G$ acts on $X$, and $x\in X(k)$ is a closed point which is fixed
by $G_{0}$. Then $G_{0}$ acts on $\operatorname{Dist}(X,x)$. However
$\mathfrak{g}$ also acts on $\operatorname{Dist}(X,x)$, and in this way
$\operatorname{Dist}(X,x)$ obtains the structure of a $G$-module. Notice that
this will happen even if $x$ is not stabilized by all of $G$.

###  3.4. Differential operators on a $G$-supervariety

Let $X$ be an affine $G$-supervariety, with action morphism $a:G\times X\to
X$, and consider $D_{X}=\Gamma(X,\mathcal{D}_{X})$.

######  Lemma 3.2.

For a differential operator $D\in D_{X}$, the following are equivalent:

  1. (1)

The map $D:k[X]\to k[X]$ is $G$-equivariant;

  2. (2)

We have $a^{*}\circ D=\operatorname{id}\otimes D\circ a^{*}$;

and in the case when $G$ is connected, we have the third equivalent condition:

  * (3)

For all $u\in\mathfrak{g}$, $[u,D]=0$.

In this case we say that $D$ is $G$-invariant.

###### Proof.

(1)$\iff$(3) is obvious when $G$ is connected. And (2) says that $D$ is a
$k[G]$-comodule homomorphism, equivalently a $G$-module homomorphism giving
(1)$\iff$(2). ∎

######  Definition 3.3.

Write $D_{X}^{G}$ for the superalgebra of $G$-invariant differential operators
on $X$.

For the meaning of an open orbit of an algebraic supergroup, see [She19].

######  Proposition 3.4.

Let $X$ be a $G$-supervariety, and $x$ a point of $X$ with stabilizer subgroup
$K\subseteq G$. If $G$ has an open orbit at $x$, then the morphism
$\operatorname{res}_{x}:D_{X}\to\operatorname{Dist}(X,x)$ restricts to an
injection $D_{X}^{G}\to\operatorname{Dist}(X,x)^{K}$.

###### Proof.

The map $\operatorname{res}_{x}$ is $K$-equivariant, so certain
$\operatorname{res}_{x}(D_{X}^{G})\subseteq\operatorname{Dist}(X,x)^{K}$. To
see that it is injective, let $a_{x}:G\to X$ be the orbit map at $x$, and
$D\in D_{X}^{G}$. Then by
$a_{x}^{*}:\mathcal{O}_{X}\to(a_{x})_{*}\mathcal{O}_{G}$ is an injective
morphism of sheaves by Prop. 3.11 of [She19]. Therefore, if
$\operatorname{res}_{x}(D)=0$, we have $D(f)(x)=0$ for all $f$ defined in an
open neighborhood of $x$. Equivalently, $a_{x}^{*}(D(f))(e)=0$ for all such
$f$. But we have the factorization $a_{x}=a\circ(\operatorname{id}_{G}\times
i_{x})$, so this says that

$\displaystyle(\operatorname{id}_{G}\times i_{x})^{*}\circ a^{*}(D(f))$
$\displaystyle=$ $\displaystyle(\operatorname{id}_{G}\times
i_{x})^{*}(\operatorname{id}\otimes D)(a^{*}(f))$ $\displaystyle=$
$\displaystyle(\operatorname{id}\otimes\operatorname{res}_{x}(D))(a^{*}(f))=0.$

This implies $D(f)=0$ for all $f$ defined in an open neighborhood of $X$.
Since by definition the restriction morphism on functions is injective for a
supervariety, this implies that $D=0$. ∎

######  Proposition 3.5.

In the context of the previous proposition, if $X\cong G/K$ via the orbit map
at $x$ then the map $D_{X}^{G}\to\operatorname{Dist}(X,x)^{K}$ is an
isomorphism.

###### Proof.

It remains to show it is surjective. For this, if
$\psi\in\operatorname{Dist}(X,x)^{K}$, define $D_{\psi}$ by
$f\mapsto(\operatorname{id}\otimes\psi)\circ a^{*}(f)$. Indeed,

$\displaystyle a^{*}\circ D_{\psi}$ $\displaystyle=$ $\displaystyle
a^{*}\circ(\operatorname{id}\otimes\psi)\circ a^{*}$ $\displaystyle=$
$\displaystyle(\operatorname{id}\otimes\operatorname{id}\otimes\psi)\circ(a^{*}\otimes\operatorname{id})\circ
a^{*}$ $\displaystyle=$
$\displaystyle(\operatorname{id}\otimes\operatorname{id}\otimes\psi)\circ(\operatorname{id}\otimes
a^{*})\circ a^{*}$ $\displaystyle=$
$\displaystyle[\operatorname{id}\otimes((\operatorname{id}\otimes\psi)\circ
a^{*})]\circ a^{*}$ $\displaystyle=$ $\displaystyle(\operatorname{id}\otimes
D_{\psi})\circ a^{*}.$

∎

##  4\. Induced supervarieties in the sense of Koszul

We present a construction that is originally due to Koszul in [Kos82].
Although it may be defined for non-affine supervarieties, for simplicity we
stick to the affine case since that is all we need. Throughout, we let $H$ be
a supergroup and write $\mathfrak{h}=\operatorname{Lie}(H)$.

###  4.1. Induced and coinduced modules

######  Definition 4.1.

Let $V_{0}$ be an $\mathfrak{h}_{\overline{0}}$-module, and define the
$\mathfrak{h}$-module
$\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}$ to be

$\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}:=\mathcal{U}\mathfrak{h}\otimes_{\mathcal{U}\mathfrak{h}_{\overline{0}}}V_{0}.$

The action by $\mathfrak{h}$ is left multiplication. If the action of
$\mathfrak{h}_{\overline{0}}$ on $V_{0}$ integrates to an action of $H_{0}$,
then the $\mathfrak{h}$ action on
$\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}$
integrates to an action of $H$, where $H_{0}$ acts by

$h\cdot(u\otimes v)=\operatorname{Ad}(h)(u)\otimes h\cdot v.$

Similarly we define the $\mathfrak{h}$-module
$\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}$ by

$\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}:=\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},V_{0}).$

Here we consider $\mathcal{U}\mathfrak{h}$ as a left
$\mathcal{U}\mathfrak{h}_{\overline{0}}$-module. The action by $\mathfrak{h}$
on this module is

$(u\eta)(v)=(-1)^{\overline{u}(\overline{\eta}+\overline{v})}\eta(vu).$

Once again, if $V_{0}$ is actually an $H_{0}$-module then
$\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0}$ will
be an $H$-module, where the action by $H_{0}$ is given by

$(h\cdot\eta)(v)=h\cdot\eta(\operatorname{Ad}(h^{-1})(v)).$

######  Remark 4.2.

If $A_{0}$ is an algebra on which $\mathfrak{h}_{\overline{0}}$ acts by
derivations, then
$\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}A_{0}$ is
naturally a superalgebra with multiplication given by

$(\eta\xi)(u)=m_{A_{0}}\circ(\eta\otimes\xi)(\Delta(u)).$

In this case, the action of $\mathfrak{h}$ on
$\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}A_{0}$ is by
super derivations. A similar statement holds if $A_{0}$ is a Hopf algebra
which $\mathfrak{h}_{\overline{0}}$ acts on by derivations preserving the Hopf
algebra structure.

######  Lemma 4.3.

For an $\mathfrak{h}_{\overline{0}}$-module $V_{0}$, we have a canonical
isomorphism of $\mathfrak{h}$-modules

$(\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0})^{*}\cong\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}(V_{0})^{*}.$

This extends to an isomorphism of $H$-modules when $V_{0}$ is an
$H_{0}$-module.

###### Proof.

We always have a canonical map of $\mathfrak{h}$-modules (or $H$-modules when
applicable)

$\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}(V_{0})^{*}\to(\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}V_{0})^{*}$

given by

$(u\otimes\varphi)(\eta)=(-1)^{\overline{u}(\overline{\varphi}+\overline{\eta})}\varphi(\eta(s(u))),$

where $s:\mathcal{U}\mathfrak{h}\to\mathcal{U}\mathfrak{h}$ is the antipode.
Here it is an isomorphism by the PBW theorem, since $\mathcal{U}\mathfrak{h}$
is a finitely generated free $\mathcal{U}\mathfrak{h}_{\overline{0}}$-module.

∎

###  4.2. Koszul’s induced superspace

######  Definition 4.4.

For an affine variety $X_{0}$ with an action of $\mathfrak{h}_{\overline{0}}$,
define $(X_{0})^{\mathfrak{h}}$ to be the affine variety with coordinate ring

$k[(X_{0})^{\mathfrak{h}}]:=\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},k[X_{0}])=\operatorname{Coind}_{\mathfrak{h}_{0}}^{\mathfrak{h}}k[X_{0}].$

By remark 4.2, $(X_{0})^{\mathfrak{h}}$ is an affine supervariety with an
action by $\mathfrak{h}$. If the action of $\mathfrak{h}_{\overline{0}}$ on
$X_{0}$ comes from an action of $H_{0}$ on $X_{0}$, then
$(X_{0})^{\mathfrak{h}}$ will be an $H$-supervariety. We have that
$((X_{0})^{\mathfrak{h}})_{0}=X_{0}$, and the natural projection
$k[(X_{0})^{\mathfrak{h}}]\to k[X_{0}]$ is given by $\eta\mapsto\eta(1)$.

######  Remark 4.5.

The construction of the induced superspace given above can be done for any
supervariety $X_{0}$ with an $\mathfrak{h}_{\overline{0}}$-action as follows.
We define $(X_{0})^{\mathfrak{h}}$ to have underlying space $X_{0}$ and sheaf
of functions given by, for an open subset $U_{0}\subseteq X_{0}$,

$\Gamma(U_{0},\mathcal{O}_{(X_{0})^{\mathfrak{h}}})=\operatorname{Coind}_{\mathfrak{h}_{0}}^{\mathfrak{h}}k[U_{0}]=k[(U_{0})^{\mathfrak{h}}].$

One can check this construction respects localization and thus gives a well-
defined supervariety.

######  Remark 4.6.

If $X_{0}$ is smooth, then $(X_{0})^{\mathfrak{h}}$ is also smooth. In any
case, $\mathfrak{h}_{\overline{1}}$ defines everywhere non-vanishing vector
fields on $(X_{0})^{\mathfrak{h}}$ so that $\mathfrak{h}_{\overline{1}}\to
T_{x}(X_{0})^{\mathfrak{h}}$ is an isomorphism for all $x\in X_{0}(k)$.

######  Remark 4.7.

In [Kos82], it was shown that $G=(G_{0})^{\mathfrak{g}}$. Thus we have an
explicit description of the algebra of functions on $k[G]$ given by

$k[G]=\operatorname{Hom}_{\mathfrak{g}_{\overline{0}}}(\mathcal{U}\mathfrak{g},k[G_{0}]).$

From this one can also determine the Hopf algebra structure on $k[G]$ from the
Hopf algebra structures on $\mathcal{U}\mathfrak{g}$ and $k[G_{0}]$.

######  Proposition 4.8.

Let $X_{0}$ be an affine variety with an action of
$\mathfrak{h}_{\overline{0}}$. Then $(X_{0})^{\mathfrak{h}}$ has the following
universal property: given an affine supervariety $Y$ with an action of
$\mathfrak{h}$, and an $\mathfrak{h}_{\overline{0}}$-equivariant map
$\overline{\phi}:X_{0}\to Y_{0}$, there exists a unique
$\mathfrak{h}$-equivariant map $\phi:(X_{0})^{\mathfrak{h}}\to Y$ such that
$\phi_{0}=\overline{\phi}$ and the following diagram commutes:

$\textstyle{(X_{0})^{\mathfrak{h}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{Y}$$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{X_{0}}}$$\scriptstyle{\overline{\phi}}$$\textstyle{Y_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces.}$$\scriptstyle{i_{Y_{0}}}$

If the actions of $\mathfrak{h}_{\overline{0}}$ and $\mathfrak{h}$ integrate
to actions of $H_{0}$ and $H$, respectively, then $\phi$ is a morphism of
$H$-supervarieties.

###### Proof.

Since everything is affine, we may work instead with the algebras of
functions. Then the commutativity of the square says that for such a map
$\phi$, we must have that for $f\in k[Y]$,

$\phi^{*}(f)(1)=(\overline{\phi})^{*}\circ(i_{Y_{0}})^{*}(f)$

The property of begin an $\mathfrak{h}$-equivariant map then forces, for
$u\in\mathcal{U}\mathfrak{h}$,

$\phi^{*}(f)(u)=(-1)^{\overline{u}\overline{f}}(u\phi^{*}(f))(1)=(-1)^{\overline{u}\overline{f}}\phi^{*}(uf)(1)=(-1)^{\overline{u}\overline{f}}(\overline{\phi})^{*}((i_{Y_{0}})^{*}(uf))$

so the definition of $\phi^{*}$ is forced on us. One can check that the above
definition defines an algebra homomorphism, and so we are done. ∎

For the following, observe that if $X_{0}$ is an $H_{0}$-variety and $x$ is a
closed point of $X_{0}$ which is fixed by $H_{0}$, then $H_{0}$ preserves
${}_{0}\mathfrak{m}_{x}$ and thus acts on $\operatorname{Dist}(X_{0},x)$.

######  Proposition 4.9.

Let $X_{0}$ be affine $H_{0}$-variety, and $x\in X_{0}(k)$ a closed point
which is fixed by $H_{0}$. Then we have an isomorphism of $H$-modules

$\operatorname{Dist}((X_{0})^{\mathfrak{h}},x)\cong\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\operatorname{Dist}(X_{0},x).$

(See remark 3.1 for the $H$-module structure on
$\operatorname{Dist}((X_{0})^{\mathfrak{h}},x)$.)

###### Proof.

We have the following string of isomorphisms of $H$-modules:

$\displaystyle\operatorname{Dist}((X_{0})^{\mathfrak{h}},x)$
$\displaystyle\cong$
$\displaystyle\Gamma_{\mathfrak{m}_{x}}(k[(X_{0})^{\mathfrak{h}}])^{*}$
$\displaystyle=$
$\displaystyle\Gamma_{\mathfrak{m}_{x}}(\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}k[X_{0}])^{*}$
$\displaystyle\cong$
$\displaystyle\Gamma_{\mathfrak{m}_{x}}\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}(k[X_{0}])^{*}$
$\displaystyle=$
$\displaystyle\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\Gamma_{{}_{0}\mathfrak{m}_{x}}(k[X_{0}])^{*}$
$\displaystyle=$
$\displaystyle\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\operatorname{Dist}(X_{0},x)$

The non-trivial step here is that we can move $\Gamma$ past
$\operatorname{Ind}$. To see this, observe that

$\mathfrak{m}_{x}=\\{\eta\in\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},k[X_{0}]):\eta(1)\in{{}_{0}\mathfrak{m}_{x}}\\}.$

It follows that for any $d\in\mathbb{N}$, for $n\gg d$, if
$\eta\in\mathfrak{m}_{x}^{n}$, then $\eta(u)\in$ ${}_{0}\mathfrak{m}_{x}^{d}$
for all $u\in\mathcal{U}\mathfrak{h}$, because $\mathfrak{h}_{\overline{0}}$
preserves the ideal ${}_{0}\mathfrak{m}_{x}$. The claim now follows. ∎

###  4.3. Induced vector bundles

Let us extend the above framework to more general vector bundles. Note that
this subsection can largely be skipped, as it is only used to prove the
$\operatorname{Ind}-\operatorname{Coind}$ isomorphism in section 6.1, which is
already known via other methods. See section 4.6 of [She20a] for the
definition of $\mathfrak{h}$-equivariant vector bundle.

Again assume that $X_{0}$ is affine, and suppose that $F_{0}$ is an
$\mathfrak{h}_{\overline{0}}$-equivariant vector bundle over $X_{0}$. Then
$\mathfrak{h}_{\overline{0}}$ acts on $\Gamma(X_{0},F_{0})$ and satisfies, for
$f_{0}\in k[X_{0}]$, $s_{0}\in\Gamma(X_{0},F_{0})$, and
$u_{0}\in\mathfrak{h}_{\overline{0}}$,

$u_{0}\cdot(f_{0}s_{0})=u_{0}(f_{0})s_{0}+f_{0}u_{0}(s_{0}),$

where the action of $u_{0}$ on $f_{0}$ is coming from the action on
$k[X_{0}]$. If $F_{0}$ is actually an $H_{0}$-equivariant vector bundle, then
for $h\in H_{0}$ we have

$h\cdot(f_{0}s_{0})=L_{h}^{*}(f_{0})h\cdot(s_{0}).$

Then we define an $\mathfrak{h}$-equivariant vector bundle
$(F_{0})^{\mathfrak{h}}$ over $(X_{0})^{\mathfrak{h}}$ to have the space of
sections given by

$\Gamma((X_{0})^{\mathfrak{h}},(F_{0})^{\mathfrak{h}}):=\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\Gamma(X_{0},F_{0})=\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},\Gamma(X_{0},F_{0})).$

Then this admits a natural $\mathfrak{h}$-action by virtue of being a
coinduced module, and if $F_{0}$ is $H_{0}$-equivariant,
$(F_{0})^{\mathfrak{h}}$ will be $H$-equivariant. The
$k[(X_{0})^{\mathfrak{h}}]$-module structure is defined by

$(\varphi s)(u)=a_{0,F_{0}}^{*}\circ(\varphi\otimes s)(\Delta(u))$

where
$a_{0,F_{0}}^{*}:k[X_{0}]\otimes\Gamma(X_{0},F_{0})\to\Gamma(X_{0},F_{0})$ is
the action map. We check that this makes sense: for
$v_{0}\in\mathfrak{h}_{\overline{0}}$, writing $\Delta(u)=\sum u_{i}\otimes
u^{i}$,

$\displaystyle(\varphi s)(v_{0}u)$ $\displaystyle=$ $\displaystyle
a_{0}\circ(\varphi\otimes s)((v_{0}\otimes 1+1\otimes v_{0})\Delta(u))$
$\displaystyle=$
$\displaystyle(-1)^{\overline{s}\overline{u_{i}}}(\varphi(v_{0}u_{i})s(u^{i})+\varphi(u_{i})s(v_{0}u^{i}))$
$\displaystyle=$
$\displaystyle(-1)^{\overline{s}\overline{u_{i}}}((v_{0}\varphi)(u_{i})s(u^{i})+\varphi(u_{i})(v_{0}s)(u^{i}))$
$\displaystyle=$ $\displaystyle v_{0}((\varphi s)(u)).$

It’s also straightforward to show that for $v\in\mathfrak{h}$ we have

$v(\varphi s)=v(\varphi)s+(-1)^{\overline{v}\overline{\varphi}}\varphi v(s),$

and (when applicable) for $h\in H_{0}$ we have

$h\cdot(\varphi s)=L_{h}^{*}(\varphi)h\cdot s.$

This construction is local in the sense that we have, for an open subvariety
$U_{0}\subseteq X_{0}$,

$\Gamma(U_{0},(F_{0})^{\mathfrak{h}})=\operatorname{Coind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\Gamma(U_{0},F_{0})=\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},\Gamma(U_{0},F_{0})).$

From this it is not hard to show that $(F_{0})^{\mathfrak{h}}$ is an even rank
vector bundle on $(X_{0})^{\mathfrak{h}}$ of the same rank as $F_{0}$.

######  Proposition 4.10.

There is a natural bijection between even rank $\mathfrak{h}$-equivariant
vector bundles on $(X_{0})^{\mathfrak{h}}$ and
$\mathfrak{h}_{\overline{0}}$-equivariant vector bundles on $X_{0}$, given
explicitly as follows:

  * •

given an $\mathfrak{h}_{\overline{0}}$-equivariant vector bundle $F_{0}$ on
$X_{0}$, we may produce the $\mathfrak{h}$-equivariant vector bundle
$(F_{0})^{\mathfrak{h}}$ on $X$; and

  * •

given an $\mathfrak{h}$-equivariant vector bundle $F$ on
$(X_{0})^{\mathfrak{h}}$ we may take the
$\mathfrak{h}_{\overline{0}}$-equivariant vector bundle on $X_{0}$ gotten by
pulling back along embedding $i:X_{0}\hookrightarrow(X_{0})^{\mathfrak{h}}$.

This restricts to a bijection of even rank $H$-equivariant vector bundles on
$(X_{0})^{\mathfrak{h}}$ and $H_{0}$-equivariant vector bundles on $X_{0}$,
when applicable.

###### Proof.

If we start with an $\mathfrak{h}_{\overline{0}}$-equivariant vector bundle
$F_{0}$, then we have a natural surjection

$\Gamma(X_{0},i^{*}(F_{0})^{\mathfrak{h}})\to\Gamma(X_{0},F_{0})$

given by evaluating an element of
$\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},\Gamma(X_{0},F_{0}))$
at $1\in\mathcal{U}\mathfrak{h}$. It is an
$\mathfrak{h}_{\overline{0}}$-equivariant map, and since the vector bundles
have the same rank, it must be an isomorphism.

In the other direction, suppose we have an $\mathfrak{h}$-equivariant bundle
$F$. Then we have a natural map of $\mathfrak{h}$-equivariant vector bundles
$F\to(i^{*}F)^{\mathfrak{h}}$, given on sections

$\Gamma(X,F)\to\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},\Gamma(X,i^{*}F))$

by

$s\mapsto(u\mapsto i^{*}(us))$

To see this is an isomorphism, it suffices to look at the map on fibers, which
are isomorphisms. ∎

######  Proposition 4.11.

Under the same hypothesis of 4.9, let $F_{0}$ be an $H_{0}$-equivariant vector
bundle on $X_{0}$. Then we have a natural isomorphism of $H$-modules

$\operatorname{Dist}((F_{0})^{\mathfrak{h}},x)\cong\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\operatorname{Dist}(F_{0},x)$

###### Proof.

The proof extends almost verbatim from 4.9. ∎

##  5\. Induced spaces and homogeneous spaces

###  5.1. A local isomorphism for certain homogeneous supervarieties

Given a supergroup $G$ and a closed algebraic subgroup $K$, we may form the
homogeneous supervariety $G/K$. For technical aspects of such spaces see
[MT18] and [MZ10]. In particular $G/K$ is always smooth and
$(G/K)_{0}=G_{0}/K_{0}$, so that $G/K$ is affine if and only if $G_{0}/K_{0}$
is.

Let $X=G/K$ be a homogeneous affine supervariety, and suppose that there
exists a subgroup $H$ of $G$ such that
$\mathfrak{h}_{\overline{1}}\oplus\mathfrak{k}_{\overline{1}}=\mathfrak{g}_{\overline{1}}$.
Then consider the $H$-supervariety $(G_{0}/K_{0})^{\mathfrak{h}}$. By its
universal property it admits a canonical $H$-equivariant morphism to $G/K$.

######  Proposition 5.1.

The canonical $H$-equivariant map $\phi:(G_{0}/K_{0})^{\mathfrak{h}}\to G/K$
induces an isomorphism of supervarieties in a Zariski open neighborhood of
$eK_{0}$. In particular, the map on functions

$\phi^{*}:k[G/K]\to
k[(G_{0}/K_{0})^{\mathfrak{h}}]=\operatorname{Hom}_{\mathfrak{h}_{\overline{0}}}(\mathcal{U}\mathfrak{h},k[G_{0}/K_{0}])$

is an injective $H$-module homomorphism

First we need a lemma.

######  Lemma 5.2.

Suppose that $f:X\to Y$ is a morphism of smooth affine supervarieties such
that $f_{0}:X_{0}\to Y_{0}$ is an isomorphism and $df_{x}$ is an isomorphism
for all closed points $x\in X(k)$. Then $f$ is an isomorphism.

###### Proof.

Because $X$ and $Y$ are smooth and affine, we may present them as exterior
algebras of vector bundles $E_{X_{0}}$, $E_{Y_{0}}$ on $X_{0}$, $Y_{0}$.
Working on a cover, we may assume these vector bundles are trivial.

Let $\xi_{1},\dots,\xi_{n}\in\Gamma(Y_{0},E_{Y_{0}})\subseteq
k[Y]_{\overline{1}}$ be a $k[Y_{0}]$-basis for $\Gamma(Y_{0},E_{Y_{0}})$ so
these elements project to a basis of $(T_{y}Y)_{\overline{1}}$ for each $y\in
Y(k)$. Then $f^{*}(\xi_{1}),\dots,f^{*}(\xi_{n})\in k[X]_{\overline{1}}$ must
project to a basis of $T_{x}X$ for all $x\in X(k)$. It follows that
$f^{*}(\xi_{1}),\dots,f^{*}(\xi_{n})$ project to a $k[X_{0}]$-basis of
$\Gamma(X_{0},E_{X_{0}})$ in the associated graded of $k[X]$. Hence the
associated graded morphism of $f^{*}$ is an isomorphism, which implies that
$f^{*}$ is an isomorphism, and we are done. ∎

###### Proof of 5.1.

Each space admits an action by $\mathfrak{h}$ as vector fields acting on
functions, and since $\phi^{*}$ is a $\mathfrak{h}$-homomorphism we have

$\phi^{*}(uf)=u\phi^{*}(f)$

for $u\in\mathfrak{h},f\in k[G/K]$. Hence for any closed point $x$ of $G/K$
the following diagram is commutative:

$\textstyle{\mathfrak{h}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{x}(G_{0}/K_{0})^{\mathfrak{h}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{df_{x}}$$\textstyle{T_{x}(G/K)}$

In particular, wherever $\mathfrak{h}_{\overline{1}}$ spans the odd part of
the tangent space of $G/K$ at $x$, $df_{x}$ will be an isomorphism of vector
spaces when restricted to the odd part. However, we see that $f_{0}^{*}$ is
the identity map by construction, hence we get a commutative diagram:

$\textstyle{(G_{0}/K_{0})^{\mathfrak{h}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{G/K}$$\textstyle{G_{0}/K_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{0}=\operatorname{id}}$$\textstyle{G_{0}/K_{0}.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

If we restrict to the open neighborhood of $eK_{0}$ upon which
$\mathfrak{h}_{\overline{1}}$ spans the odd tangent space of $G/K$, $f$ will
be an isomorphism by lemma 5.2. ∎

######  Corollary 5.3.

Maintain the assumptions of 5.1. and suppose further that $H_{0}\subseteq
K_{0}$. Then we have a canonical isomorphism of $H$-modules

$\operatorname{Dist}(G/K,eK)\cong\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\operatorname{Dist}(G_{0}/K_{0},eK_{0}).$

Or in terms of enveloping algebras,

$\mathcal{U}\mathfrak{g}/(\mathcal{U}\mathfrak{g}\mathfrak{k})\cong\operatorname{Ind}_{\mathfrak{h}_{\overline{0}}}^{\mathfrak{h}}\mathcal{U}\mathfrak{g}_{\overline{0}}/(\mathcal{U}\mathfrak{g}_{\overline{0}}\mathfrak{k}_{\overline{0}})$

###### Proof.

The first statement follows from 4.9. The second statement follows from the
usual identification of $\mathcal{U}\mathfrak{g}$-modules

$\operatorname{Dist}(G/K,eK)\cong\mathcal{U}\mathfrak{g}/(\mathcal{U}\mathfrak{g}\mathfrak{k}).$

which comes from the composition

$\mathcal{U}\mathfrak{g}\to\Gamma(G/K,\mathcal{D}_{G/K})\xrightarrow{\operatorname{res}_{eK}}\operatorname{Dist}(G/K,eK).$

∎

###  5.2. Supersymmetric spaces

From now on, we assume that $G$ is a connected supergroup, i.e. an affine
algebraic supergroup such that $G_{0}$ is connected. Let $\theta$ be an
involution of $G$, and let $K$ be a closed subgroup of $G$ such that
$(G^{\theta})^{0}\subseteq K\subseteq G^{\theta}$. In particular $K$ need not
be connected. From this we may consider the homogeneous supervariety $G/K$,
and we call $G$-supervarieties of this form symmetric supervarieties (or
supersymmetric spaces).

On the level of Lie superalgebras, $\theta$ induces an involution on
$\mathfrak{g}$, which by abuse of notation we again write as $\theta$, giving
rise to the decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$, where
$\mathfrak{k}=\operatorname{Lie}K$ is the fixed subspace of $\theta$ and
$\mathfrak{p}$ the $(-1)$-eigenspace of $\theta$.

######  Definition 5.4.

Define
$\mathfrak{k}^{\prime}:=\mathfrak{k}_{\overline{0}}\oplus\mathfrak{p}_{\overline{1}}$,
which is the fixed points of the involution $\delta\circ\theta$, where
$\delta(x)=(-1)^{\overline{x}}x$ is the grading operator on $\mathfrak{g}$.
Let $K^{\prime}$ denote the closed algebraic subgroup of $G$ such that
$K^{\prime}_{0}=K_{0}$ and
$\operatorname{Lie}K^{\prime}=\mathfrak{k}^{\prime}$.

Notice that $\delta\circ\theta$ is an involution on $G$ such that
$(G^{\delta\circ\theta})^{0}\subseteq K^{\prime}\subseteq
G^{\delta\circ\theta}$, hence $G/K^{\prime}$ is another symmetric
supervariety.

######  Proposition 5.5.

We have a $K^{\prime}$-equivariant morphism

$(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}\to G/K$

which is an isomorphism in a neighborhood of $eK_{0}$. In particular, the
pullback morphism of functions

$k[G/K]\to
k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]=\operatorname{Coind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}^{\prime}}k[G_{0}/K_{0}]$

is injective.

###### Proof.

This follows immediately from 5.1. ∎

######  Corollary 5.6.

We have a natural injective morphism of algebras

$k[K^{\prime}\backslash G/K]=k[G/K]^{K^{\prime}}\to k[K_{0}\backslash
G_{0}/K_{0}]=k[G_{0}/K_{0}]^{K_{0}}.$

In particular, $k[K^{\prime}\backslash G/K]$ is an integral domain.

###### Proof.

Taking $K^{\prime}$ invariants of the pullback morphism we obtain an injection

$k[G/K]^{K^{\prime}}\to\left(\operatorname{Coind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}^{\prime}}k[G_{0}/K_{0}]\right)^{K^{\prime}}.$

Now one may use Frobenius reciprocity to identify
$\left(\operatorname{Coind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}^{\prime}}k[G_{0}/K_{0}]\right)^{K^{\prime}}$
with $k[G_{0}/K_{0}]^{K_{0}}$ as an algebra. ∎

The following result, which now is proven easily from 5.3, is an appropriate
generalization of the fundamental observation made in [Gor00].

######  Proposition 5.7.

We have a canonical isomorphism of $K^{\prime}$-modules

$\operatorname{Dist}(G/K,eK)\cong\operatorname{Ind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}^{\prime}}\operatorname{Dist}(G_{0}/K_{0},eK_{0}).$

##  6\. The symmetric space $G/G_{0}$

Consider the involution $\theta=\delta$, the canonical grading operator on
$\mathfrak{g}$ which is defined by $\delta(x)=-x$. In this case
$G^{\delta}=G_{0}$, and the local isomorphism of 5.5 becomes a global
isomorphism of $G$-supervarieties (both consisting of just one point):

$G/G_{0}\cong(G_{0}/G_{0})^{\mathfrak{g}}.$

###  6.1. $\operatorname{Ind}$-$\operatorname{Coind}$ isomorphism

The homogeneous space $G/G_{0}$ has one closed point, which we will call $e$.
Let $V$ be a $G$-equivariant vector bundle on $G/G_{0}$, and write $V$ again
for its space of sections. Since $\mathfrak{m}_{e}$ is nilpotent, we have the
identification $\operatorname{Dist}(V,e)=V^{*}$.

A $G_{0}$-equivariant vector bundle on $G_{0}/G_{0}$ is the same data as a
finite-dimensional $G_{0}$-representation $V_{0}$, and its sections are again
$V_{0}$. Then 4.11 tells us that

$\operatorname{Dist}((V_{0})^{\mathfrak{g}},e)=(\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}V_{0})^{*}\cong\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}(V_{0})^{*},$

which we already showed in lemma 4.3.

We will write $(V_{0})^{\mathfrak{g}}$ once again for the sections of
$(V_{0})^{\mathfrak{g}}$ on $G/G_{0}$ since this space has only one point.
Since $D_{G/G_{0}}$ is generated by $\mathcal{U}\mathfrak{g}$ and $k[G/G_{0}]$
as an algebra, $(V_{0})^{\mathfrak{g}}$ will be a $D_{G/G_{0}}$-module on
$G/G_{0}$.

Thus $\operatorname{Dist}((V_{0})^{\mathfrak{g}},e)$ is a
$D_{G/G_{0}}$-module, and, being finite-dimensional, must also be coherent. By
2.8, $\operatorname{Dist}((V_{0})^{\mathfrak{g}},e)$ must be a vector bundle
on $G/G_{0}$, and is $G$-equivariant via its $\mathcal{D}_{G/G_{0}}$-module
structure. Further it is of even rank since since $(V_{0})^{\mathfrak{g}}$ is
of even rank, and therefore by 4.10 there exists a $G_{0}$-equivariant vector
bundle on $G_{0}/G_{0}$, that is a $G_{0}$-representation $W_{0}$, such that

$\operatorname{Dist}((V_{0})^{\mathfrak{g}},e)\cong(W_{0})^{\mathfrak{g}}.$

By 4.10, $W_{0}$ is obtained by taking the sections of the restriction of
$(V_{0})^{\mathfrak{g}}$ to $G_{0}/G_{0}$, i.e.

$W_{0}\cong\operatorname{Dist}((V_{0})^{\mathfrak{g}},e)/\mathfrak{m}_{e}\operatorname{Dist}((V_{0})^{\mathfrak{g}},e).$

Now let $n=\operatorname{dim}\mathfrak{g}_{\overline{1}}$ so that
$\mathfrak{m}_{e}^{n+1}=0$ but $\mathfrak{m}_{e}^{n}\neq 0$. Then
$\operatorname{Dist}(V,e)\cong((V_{0})^{\mathfrak{g}}/\mathfrak{m}_{e}^{n+1}(V_{0})^{\mathfrak{g}})^{*}$,
and therefore

$((V_{0})^{\mathfrak{g}}/\mathfrak{m}_{e}^{n+1}(V_{0})^{\mathfrak{g}})^{*}/\mathfrak{m}_{e}((V_{0})^{\mathfrak{g}}/\mathfrak{m}_{e}^{n+1}(V_{0})^{\mathfrak{g}})^{*}\cong(\mathfrak{m}_{e}^{n}(V_{0})^{\mathfrak{g}})^{*}.$

Now as a $G_{0}$-module,

$(V_{0})^{\mathfrak{g}}=\operatorname{Hom}_{\mathfrak{g}_{\overline{0}}}(\mathcal{U}\mathfrak{g},V_{0})\cong\operatorname{Hom}(\Lambda\mathfrak{g}_{\overline{1}},V_{0})\cong\Lambda\mathfrak{g}_{\overline{1}}^{*}\otimes
V_{0}$

and $\mathfrak{m}_{e}^{n}V$ sits inside as the
$\mathfrak{g}_{\overline{0}}$-submodule
$\Pi^{n}\Lambda^{n}\mathfrak{g}_{\overline{1}}^{*}\otimes V_{0}$. Therefore,
$W_{0}\cong\Pi^{n}\Lambda^{n}\mathfrak{g}_{\overline{1}}\otimes
V_{0}^{*}=\operatorname{Ber}(\mathfrak{g}_{\overline{1}})\otimes V_{0}^{*}$.

Putting everything together and replacing $V_{0}$ by $V_{0}^{*}$, we have
reproduced the following well-known result (see for instance section 9 of
[Ser11] for an algebraic proof):

######  Proposition 6.1.

For a $G_{0}$-module $V_{0}$, we have a canonical isomorphism of $G$-modules

$\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}V_{0}\cong\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}(\operatorname{Ber}(\mathfrak{g}_{\overline{1}})\otimes
V_{0}).$

###  6.2. The invariant differential ghost operator

Now we study differential operators on $G/G_{0}$. By 3.5, the invariant
differential operators are given by $G_{0}$-invariant distributions, i.e.
$(\mathcal{U}\mathfrak{g}/(\mathcal{U}\mathfrak{g})\mathfrak{g}_{\overline{0}})^{G_{0}}$
which can be identified, via symmetrization, with
$S(\mathfrak{g}_{\overline{1}})^{G_{0}}$. However, we also have
$\operatorname{Dist}(G/G_{0},e)\cong\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k$
as a $\mathfrak{g}$-module. Therefore by 6.1 we have

$\operatorname{Dist}(G/G_{0},e)\cong\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}(\operatorname{Ber}(\mathfrak{g}_{\overline{1}}))$

Recall that $\operatorname{ber}_{\mathfrak{g}}$ is the character of $G$
determined determined by $\operatorname{Ber}(\mathfrak{g})$. If $V$ is a
$G$-module, we write $V^{\operatorname{ber}_{\mathfrak{g}}}$ for the subspace
of eigenvectors for $G$ of weight $\operatorname{ber}_{\mathfrak{g}}$.

######  Corollary 6.2.

Suppose that $\Lambda^{top}\mathfrak{g}_{\overline{0}}$ is a trivial
$G_{0}$-module. Then the subspace
$\operatorname{Dist}(G/G_{0},e)^{\operatorname{ber}_{\mathfrak{g}}}$ is one-
dimensional.

###### Proof.

Under this assumption,
$\operatorname{Ber}(\mathfrak{g})=\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$,
and so the result follows by Frobenius reciprocity. ∎

Observe that the conditions of 6.2 hold if $G_{0}$ is reductive. When it
exists, we write
$v_{\mathfrak{g}}\in\operatorname{Dist}(G/G_{0},e)^{\operatorname{ber}_{\mathfrak{g}}}$
for a chosen non-zero element. If $\beta=0$, i.e.
$\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is a trivial $G_{0}$-module,
then
$v_{\mathfrak{g}}\in\operatorname{Dist}(G/G_{0},e)^{G}\subseteq\operatorname{Dist}(G/G_{0},e)^{G_{0}}\cong
D_{G/G_{0}}^{G}$. Therefore in this case $v_{\mathfrak{g}}$ corresponds to a
$G$-invariant differential operator on $G/G_{0}$, which we write as
$D_{\mathfrak{g}}$.

######  Definition 6.3.

We say a supergroup $G$ is quasireductive if $G_{0}$ is reductive. We say a
Lie superalgebra $\mathfrak{g}$ is quasireductive if it is the Lie
superalgebra of a quasireductive supergroup.

Assumption: We assume for the rest of the section that $G$ is quasireductive
and
$\operatorname{Ber}(\mathfrak{g})=\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$
is the trivial $G_{0}$-module.

Under this assumption, $v_{\mathfrak{g}}$ corresponds, as we said, to a
certain $G$-invariant differential operator. We now determine what it is.

######  Lemma 6.4.

$D_{G/G_{0}}=\operatorname{End}_{k}(k[G/G_{0}])$.

###### Proof.

This easily follows from the definition of differential operators. ∎

Recall that since $G_{0}$ is reductive,
$k[G/G_{0}]=\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k$
is projective, and hence is a sum of injective indecomposable modules $I(L)$
for $L$ a simple $G$-module in the socle of
$\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k$. We have
that the trivial module, $k$, shows up exactly once in
$\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k$, hence we
can write:

$k[G/G_{0}]=\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k=I(k)\oplus
V.$

Since $\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is trivial, we have
that $I(k)=P(k)$ (see [Ser11]), and thus the head and tail of $I(k)$ are both
the trivial module. It follows that $\operatorname{End}(I(k))$ contains a
unique up to scalar endomorphism $\phi$ taking the head to the tail, which is
nilpotent exactly if $G$ does not have semisimple representations (i.e. $k$ is
not projective).

######  Proposition 6.5.

Up to a non-zero scalar, $D_{\mathfrak{g}}$ is the endomorphism of
$k[G/G_{0}]=I(k)\oplus V$ given by $\phi\oplus 0_{V}$.

###### Proof.

Since $D_{\mathfrak{g}}$ is $G$-invariant by construction, it suffices to show
that $\operatorname{res}_{e}(D_{\mathfrak{g}})$ is $\mathfrak{g}$-invariant.
However, we observe that $uD_{\mathfrak{g}}=D_{\mathfrak{g}}u=0$ for all
$u\in\mathfrak{g}$, so since $D_{G/G_{0}}\to\operatorname{Dist}(G/G_{0},e)$ is
right $D_{G/G_{0}}$-equivariant, we are done. ∎

The following is now easy to show:

######  Corollary 6.6.

For a $G_{0}$-module $V_{0}$, $D_{\mathfrak{g}}$ acts on
$\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}V_{0}$ by
$\phi$ on each summand isomorphic to $I(k)$, and zero otherwise.

We have the following characterization of $v_{\mathfrak{g}}$:

######  Corollary 6.7.

Let $\mathfrak{g}$ be a quasireductive Lie superalgebra such that
$\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is the trivial
$G_{0}$-module, and write $v_{\mathfrak{g}}$ for a non-zero element of
$(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}})^{G}$.
Then $v_{\mathfrak{g}}$ is the unique non-zero element up to scalar with the
property that
$uv_{\mathfrak{g}}\in\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}$ for
all $u\in\mathfrak{g}$.

###  6.3. Relation to Gorelik’s element $v_{\emptyset}$

######  Proposition 6.8.

Let $\mathfrak{g}$ be a Lie superalgebra such that
$\Lambda^{top}\mathfrak{g}_{\overline{1}}$ is the trivial $G_{0}$-module. Then
for a $G_{0}$-module $V_{0}$ we have a natural isomorphism

$(V_{0})^{G}\to(\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}V_{0})^{\operatorname{ber}_{\mathfrak{g}}}$

given by

$z\mapsto v_{\mathfrak{g}}z.$

###### Proof.

This easily follows from the work in already done in this section. ∎

In [Gor00], it was proven that if
$\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is trivial then there exists
an element $v_{\emptyset}\in\mathcal{U}\mathfrak{g}$ with the property that
for a $\mathfrak{g}_{\overline{0}}$-module $V$, the map

$z\mapsto v_{\emptyset}z$

defines an isomorphism

$V^{\mathfrak{g}_{\overline{0}}}\to(\operatorname{Ind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}V)^{\mathfrak{g}}.$

######  Corollary 6.9.

The element $v_{\mathfrak{g}}$ agrees with the construction of such an element
given by Gorelik in [Gor00].

###### Proof.

Gorelik’s element has the property that it defined a nonzero element of
$(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}})^{\mathfrak{g}}$,
so we are done. ∎

###  6.4. Computations of $v_{\mathfrak{g}}$ for some Lie superalgebras

We compute explicitly the element
$v_{\mathfrak{g}}\in\left(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}\right)^{\operatorname{ber}_{\mathfrak{g}}}$
for certain quasireductive Lie superalgebras. First, an easy example:

######  Lemma 6.10.

If $[\mathfrak{g}_{\overline{1}},\mathfrak{g}_{\overline{1}}]$ is central in
$\mathfrak{g}$, then $v_{\mathfrak{g}}=v_{1}\cdots v_{n}$ for any choice of
basis $v_{1},\dots,v_{n}$ of $\mathfrak{g}_{\overline{1}}$.

###### Proof.

This element is acted on by $\mathfrak{g}_{\overline{0}}$ according to its
action on $\operatorname{Ber}(\mathfrak{g})$. Observe that

$v_{i}v_{\mathfrak{g}}=(-1)^{i-1}v_{1}\cdots v_{i}^{2}\cdots
v_{n}+\sum\limits_{1\leq j<i}(-1)^{j-1}v_{1}\dots
v_{j-1}[v_{i},v_{j}]v_{j+1}\cdots v_{n}.$

Since both $[v_{j},v_{i}]$ and $v_{i}^{2}$ are central, we may rewrite this
sum as

$(-1)^{i-1}v_{1}\cdots\widehat{v_{i}}\cdots
v_{n}v_{i}^{2}+(-1)^{j-1}v_{1}\dots v_{j-1}\widehat{v_{j}}v_{j+1}\cdots
v_{n}[v_{j},v_{i}]\in\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}.$

∎

For type I algebras, we have the following:

######  Proposition 6.11.

Suppose that $\mathfrak{g}$ is a quasireductive Lie superalgebra with a
$\mathbb{Z}$-grading
$\mathfrak{g}=\mathfrak{g}_{-1}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{1}$
such that $[\mathfrak{g}_{i},\mathfrak{g}_{j}]\subseteq\mathfrak{g}_{i+j}$,
$\mathfrak{g}_{\overline{0}}=\mathfrak{g}_{0}$ and
$\mathfrak{g}_{\overline{1}}=\mathfrak{g}_{-1}\oplus\mathfrak{g}_{1}$. Suppose
further that for an odd weight $\alpha$, of a Cartan subalgebra of
$\mathfrak{g}_{\overline{0}}$,
$[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]$ acts trivially on
$\Lambda^{top}\mathfrak{g}_{1}$. Let $v_{1},\dots v_{n}$ be any basis of
$\mathfrak{g}_{-1}$ and $w_{1},\dots,w_{m}$ any basis of $\mathfrak{g}_{1}$.
Then

$v_{\mathfrak{g}}=(v_{1}\cdots v_{n})\cdot(w_{1}\cdots w_{m}).$

###### Proof.

Clearly $\mathfrak{g}_{\overline{0}}$ acts on it as the top exterior power of
$\mathfrak{g}_{\overline{1}}$, and $\mathfrak{g}_{-1}$ annihilates it.
Therefore it remains to show that $\mathfrak{g}_{1}$ also annihilates it. Let
$w\in\mathfrak{g}_{1}$ be a weight vector, of weight $\alpha$ (note that
$\alpha$ could be 0). The above expression is seen to be independent of the
choice of bases up to a nonzero scalar, so let us assume that
$v_{1},\dots,v_{n}$ are weight vectors and that $v_{1},\dots,v_{i}$ are a
basis of $\mathfrak{g}_{-\alpha}$ (and if $\mathfrak{g}_{-\alpha}=0$ then this
condition is vacuous). We see that (working up to
$\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}$)

$\displaystyle w(v_{1}\cdots v_{n})\cdot(w_{1}\cdots w_{m})$ $\displaystyle=$
$\displaystyle\sum\limits_{j}(-1)^{j-1}v_{1}\cdots
v_{j-1}[w,v_{j}]v_{j+1}\cdots v_{n}w_{1}\cdots w_{m}$ $\displaystyle=$
$\displaystyle\sum\limits_{j\leq i}(-1)^{j-1}v_{1}\cdots
v_{j-1}[[w,v_{j}],v_{j+1}\cdots v_{n}w_{1}\cdots w_{m}]$ $\displaystyle+$
$\displaystyle\sum\limits_{j>i,k>j}(-1)^{j-1}v_{1}\cdots v_{j-1}v_{j+1}\cdots
v_{k-1}[[w,v_{j}],v_{k}]v_{k+1}\cdots v_{n}w_{1}\cdots w_{m}$ $\displaystyle+$
$\displaystyle\sum\limits_{j>i}(-1)^{j-1}v_{1}\cdots v_{j-1}v_{j+1}\cdots
v_{n}[[w,v_{j}],w_{1}\cdots w_{m}].$

In the first sum, we have that action of
$[w,v_{j}]\in[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]$ on the vector
$v_{j+1}\cdots v_{n}w_{1}\cdots w_{m}$ which has weight

$j\alpha+\sum\limits_{\beta\in\Delta}\beta.$

By assumption,
$\sum\limits_{\beta\in\Delta}\beta([\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}])=0$,
and by the Jacobi identity we have
$\alpha([\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}])=0$ as well. Therefore
this first term is zero.

For the second sum, the weight of $[[w,v_{j}],v_{k}]$ is
$\alpha+\alpha_{j}+\alpha_{k}$, where $\alpha_{j}$, resp. $\alpha_{k}$ is the
weight of $v_{j}$, resp. $v_{k}$. Since $\alpha\neq-\alpha_{j},-\alpha_{k}$,
we have that $\alpha+\alpha_{j}+\alpha_{k}\neq\alpha_{j},\alpha_{k}$, and
therefore $[[w,v_{j}],v_{k}]$ is either zero or a root vector in
$\mathfrak{g}_{-1}$ which we may assume already appears in the product
$v_{1}\cdots v_{j-1}v_{j+1}\cdots v_{k-1}v_{k+1}\cdots v_{n}$, giving zero.

For the final sum, we know that $[w,v_{j}]$ is a nilpotent element of
$\mathfrak{g}_{\overline{0}}$ and thus acts trivially on
$\Lambda^{top}\mathfrak{g}_{1}$, so we once again get zero. ∎

######  Corollary 6.12.

For $\mathfrak{g}\mathfrak{l}(m|n)$,
$(\mathfrak{p})\mathfrak{s}\mathfrak{l}(m|n)$,
$\mathfrak{o}\mathfrak{s}\mathfrak{p}(2|2n)$, and
$(\mathfrak{s})\mathfrak{p}(n)$ the formula for $v_{\mathfrak{g}}$ is given by
6.11.

###### Proof.

It is straightforward to check the conditions of 6.11 for these superalgebras.
∎

######  Proposition 6.13.

Let $\pm\delta_{1},\dots,\pm\delta_{n}$ denote the odd roots of
$\mathfrak{g}=\mathfrak{o}\mathfrak{s}\mathfrak{p}(1|2n)$ for the standard
presentation of $\mathfrak{o}\mathfrak{s}\mathfrak{p}(1|2n)$ as given for
example in [Mus12]. Choose elements $u_{1},\dots,u_{n}$ of weight
$\delta_{1},\dots,\delta_{n}$ and $v_{1},\dots,v_{n}$ of weight
$-\delta_{1},\dots,-\delta_{n}$ such that if $h_{i}=[u_{i},v_{i}]$, then
$\delta_{i}(h_{i})=1$. Write $t_{i}=u_{i}v_{i}$. Then we have

$v_{\mathfrak{g}}=(1+t_{1})(3+t_{2})\cdots((2n-1)+t_{n}).$

###### Proof.

This is in fact proven in section 4 of [DH+76]. There they prove that the
above element is equal, mod
$\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}$, to

$(1+t_{\sigma(1)})(3+t_{\sigma(2)})\cdots(2n-1+t_{\sigma(n)}).$

for any permutation $\sigma$. Now

$u_{i}(1+t_{i})=-v_{i}u_{i}^{2},\ \ \ \ \
v_{i}(1+t_{i})=-u_{i}v_{i}^{2}+v_{i}h_{i}.$

Since $u_{i}^{2},v_{i}^{2},$ and $h_{i}$ commute with $t_{j}$ for $j\neq i$,
we may move them all the way to the right and obtain elements of
$\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}$. ∎

It would be interesting to obtain formulas for $v_{\mathfrak{g}}$ when
$\mathfrak{g}=\mathfrak{o}\mathfrak{s}\mathfrak{p}(m|2n)$ with $m>2$,
$\mathfrak{g}=\mathfrak{q}(n)$, or when $\mathfrak{g}$ is exceptional simple.
Explicit formulas are important in computing the image of ghost distributions
under the Harish-Chandra homomorphism, which we discuss later.

###  6.5. Semisimplicity criteria

We give a brief application of the ideas above.

######  Theorem 6.14.

Let $G$ be a quasireductive supergroup. Then the following are equivalent:

  1. (1)

The category $\operatorname{Rep}(G)$ of representations of $G$ is semisimple;

  2. (2)

$\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is trivial and
$D_{\mathfrak{g}}$ is not nilpotent.

  3. (3)

$\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is trivial and
$\varepsilon(v_{\mathfrak{g}})\neq 0$, where $\varepsilon$ is the counit on
$\mathcal{U}\mathfrak{g}$ (and is well-defined on
$\mathcal{U}\mathfrak{g}/\mathfrak{g}_{\overline{0}}\mathcal{U}\mathfrak{g}$).

Note that the condition that $G$ be quasireductive is necessary in order for
$\operatorname{Rep}(G)$ to be semisimple.

###### Proof.

The equivalence $(3)\iff(2)$ is clear, so we show $(2)\iff(1)$. Since $G$ is
quasireductive,
$k[G/G_{0}]=\operatorname{Coind}_{\mathfrak{g}_{\overline{0}}}^{\mathfrak{g}}k$
is projective. Since $\operatorname{Rep}(G)$ is semisimple if and only if the
trivial module $k$ is projective, it is equivalent to show that $k$ splits off
of $k[G/G_{0}]$. For this it is equivalent that $D_{\mathfrak{g}}(1)\neq 0$,
which gives $(2)\iff(1)$. ∎

It is well-known, going back to a result of Djokovic and Hochschild in
[DH+76], that if $G$ is a connected algebraic supergroup such that
$\operatorname{Rep}(G)$ is semisimple, then $G\cong K\times
SOSp(1|2n_{1})\times\dots\times SOSp(1|2n_{k})$, where $K$ is a reductive Lie
group. Using 6.14 and 6.13 one can obtain a simple proof of this statement,
and this has been carried out in [She20b].

##  7\. General symmetric space $G/K$

We come back to the general case of symmetric supervarieties $G/K$. For the
rest of the article we assume that $G$ is quasireductive, so that the
connected component of the identity of $K$ is quasireductive, and in
particular $K_{0}$ has semisimple representation theory.

###  7.1. Ghost distributions $\mathcal{A}_{G/K}$

We know by 5.7 that

$\operatorname{Dist}(G/K,eK)\cong\operatorname{Ind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}\operatorname{Dist}(G_{0}/K_{0},eK_{0})$

as $K^{\prime}$-modules. Since $K_{0}$ has semisimple representation theory,
$\operatorname{Dist}(G_{0}/K_{0},eK_{0})$ is a sum of finite-dimensional
$K_{0}$-modules. Hence by 6.1 we have that

$\operatorname{Ind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}\operatorname{Dist}(G_{0}/K_{0},eK_{0})\cong\operatorname{Coind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}(\operatorname{Dist}(G_{0}/K_{0},eK_{0})\otimes\operatorname{Ber}(\mathfrak{k}^{\prime}))$

where we have used that $\mathcal{U}\mathfrak{k}^{\prime}$ is a finite rank,
free left $\mathcal{U}\mathfrak{k}_{0}$-module. Now we have

$\displaystyle\operatorname{Dist}(G/K,ek)^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$
$\displaystyle\cong$
$\displaystyle(\operatorname{Coind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}(\operatorname{Dist}(G_{0}/K_{0},eK_{0})\otimes\operatorname{Ber}(\mathfrak{k}^{\prime}))^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$
$\displaystyle\cong$
$\displaystyle(\operatorname{Dist}(G_{0}/K_{0},eK_{0})\otimes\operatorname{Ber}(\mathfrak{k}^{\prime}))^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$
$\displaystyle\cong$
$\displaystyle\operatorname{Dist}(G_{0}/K_{0},eK_{0})^{K_{0}}$

Write $v_{\mathfrak{k}^{\prime}}$ for an element of
$\mathcal{U}\mathfrak{k}^{\prime}$ which projects to a nonzero element of
$(\mathcal{U}\mathfrak{k}^{\prime}/\mathcal{U}\mathfrak{k}^{\prime}\mathfrak{k}_{\overline{0}})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$.
Then by 6.8 we have that

######  Proposition 7.1.

The isomorphism

$\eta:\operatorname{Dist}(G_{0}/K_{0},eK_{0})^{K_{0}}\to\operatorname{Dist}(G/K,eK)^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$

is given by

$z\mapsto v_{\mathfrak{k}^{\prime}}\cdot z.$

######  Remark 7.2.

Recall we have an identification
$\operatorname{Dist}(G_{0}/K_{0},eK_{0})^{K_{0}}\cong D^{G_{0}}(G_{0}/K_{0})$,
and this algebra is identified with $S(\mathfrak{a})^{W_{\mathfrak{a}}}$,
where $\mathfrak{a}\subseteq\mathfrak{p}_{\overline{0}}$ is a Cartan subspace
and $W_{\mathfrak{a}}$ is the little Weyl group associated to the symmetric
space.

Stated in terms of enveloping algebras, we have shown:

######  Corollary 7.3.

We have an isomorphism

$(\mathcal{U}\mathfrak{g}_{\overline{0}}/\mathcal{U}\mathfrak{g}_{\overline{0}}\mathfrak{k}_{\overline{0}})^{K_{0}}\to(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$

given by $\phi(z)=v_{\mathfrak{k}^{\prime}}z$.

######  Definition 7.4.

We define $\mathcal{A}_{G/K}$ to be
$\operatorname{Dist}(G/K,eK)^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$,
and refer to elements of $\mathcal{A}_{G/K}$ as ghost distributions on $G/K$.
We will often use the letter $\gamma$ to denote such a distribution.

######  Remark 7.5 (Caution).

In [Gor00], $\mathcal{A}$ is used to denote the $G^{\prime}$ invariants in
$\operatorname{Dist}(G\times G/G,eG)$ as we will see later on. However in our
notation, $\mathcal{A}_{G\times G/G}$ denotes the
$\operatorname{ber}_{\mathfrak{g}^{\prime}}$ semi-invariants of $G^{\prime}$
acting on $\operatorname{Dist}(G\times G/G,eG)$. Thus these will agree only
when $\operatorname{Ber}(\mathfrak{g})$ is the trivial module.

In section 10, we will introduce another object, $\mathcal{A}_{\phi}$, which
is a subspace of $\mathcal{U}\mathfrak{g}$ that is invariant under a certain
twisted adjoint action depending on an automorphism $\phi$. For this notation,
we have that $\mathcal{A}=\mathcal{A}_{\delta}$.

###  7.2. Module structure of $\mathcal{A}_{G/K}$

Write $\mathcal{Z}_{G/K}$ for
$\operatorname{Dist}(G/K,eK)^{K}=(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{K}$.
This is identified with the algebra of invariant differential operators on
$G/K$, as explained in 3.5.

######  Proposition 7.6.

We have a natural map

$\mathcal{A}_{G/K}\otimes\mathcal{Z}_{G/K}=(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}\otimes(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{K}\to(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}=\mathcal{A}_{G/K}$

making $\mathcal{A}_{G/K}$ into a right module over $\mathcal{Z}_{G/K}$.

###### Proof.

We define

$(\gamma+\mathcal{U}\mathfrak{g}\mathfrak{k})(z+\mathcal{U}\mathfrak{g}\mathfrak{k}):=\gamma
z+\mathcal{U}\mathfrak{g}\mathfrak{k}$

Since $z$ is $K$-invariant, it is easy to check this is well-defined. ∎

######  Lemma 7.7.

Suppose that $\operatorname{Ber}(\mathfrak{k}_{\overline{1}})$ and
$\operatorname{Ber}(\mathfrak{p}_{\overline{1}})$ are trivial $K_{0}$-modules.
Then we have a natural map
$\mathcal{A}_{G/K}\otimes\mathcal{A}_{G/K^{\prime}}\to\mathcal{Z}_{G/K}$, or

$(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k})^{K^{\prime}}\otimes(\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k}^{\prime})^{K}\to(\mathcal{U}\mathfrak{g}/\mathfrak{k}\mathcal{U}\mathfrak{g})^{K},$

given by

$(\gamma+\mathcal{U}\mathfrak{g}\mathfrak{k})\otimes(\gamma^{\prime}+\mathcal{U}\mathfrak{g}\mathfrak{k}^{\prime})\mapsto\gamma\gamma^{\prime}+\mathfrak{k}\mathcal{U}\mathfrak{g}.$

###### Proof.

The proof is straightforward. ∎

###  7.3. $k[G/K]$ as a $K^{\prime}$-module

Observe that via pullback and the isomorphism of distributions we have the
commutative diagram

$\textstyle{\operatorname{Dist}(G/K,eK)^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}\otimes
k[G/K]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Dist}((G_{0}/K_{0})^{\mathfrak{k}^{\prime}},eK_{0})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}\otimes
k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k.}$

Now let $\gamma$ be an element of
$\operatorname{Dist}((G_{0}/K_{0})^{\mathfrak{k}^{\prime}},eK_{0})^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$,
and suppose that $\gamma(f)\neq 0$ for $f\in
k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$. Then necessarily the
$K^{\prime}$-module generated by $f$ generates a copy of $I_{K^{\prime}}(k)$,
the injective hull of $k$ for $K^{\prime}$. It follows that the same must be
true for the space $G/K$, and so we have:

######  Proposition 7.8.

Let $\gamma\in\mathcal{A}_{G/K}$, and suppose that $f\in k[G/K]$ is such that
$\gamma(f)\neq 0$. Then the $K^{\prime}$-module generated by $f$ contains a
copy of $I_{K^{\prime}}(k)$. In particular, if
$\mathfrak{k}_{\overline{0}}\neq\mathfrak{g}_{\overline{0}}$ then $k[G/K]$
contains $I_{K^{\prime}}(k)$ with infinite multiplicity.

###### Proof.

For the last statement, we observe that since $G/K$ is affine the pairing of
distributions with functions is nondegenerate. Using the fact that the
$K^{\prime}$-semi-invariant distributions form an infinite-dimensional vector
space (being isomorphic, as a vector space, to
$S(\mathfrak{a})^{W_{\mathfrak{a}}}$), it is not hard to prove that the
multiplicity must be infinite. ∎

######  Remark 7.9.

We observe that since $k[G_{0}/K_{0}]^{K_{0}}$ is a subalgebra of
$k[G_{0}/K_{0}]$,

$A:=\operatorname{Coind}_{\mathfrak{k}_{\overline{0}}}^{\mathfrak{k}^{\prime}}k[G_{0}/K_{0}]^{K_{0}}$

is a subalgebra of $k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$. In particular
$A$ is the sum of all copies of $I_{K^{\prime}}(k)$ appearing in
$k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$. It follows that $k[G/K]\cap A$ is
a subalgebra of $k[G/K]$ which contains all copies of $I_{K^{\prime}}(k)$ in
$k[G/K]$, as well as all of $k[G/K]^{K^{\prime}}$. The author has a rather
limited understanding of $A\cap k[G/K]$.

Further we do not have a good answer, or even good formulation of the question
of ‘how many’ copies of $I_{K^{\prime}}(k)$ are within $k[G/K]$. The copies of
$I_{K^{\prime}}(k)$ in $k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$ may be
indexed by a basis of $k[G_{0}/K_{0}]^{K_{0}}$, which itself is indexed by the
irreducible summands of $k[G_{0}/K_{0}]$, i.e. by certain dominant weights in
$\mathfrak{a}$. For each copy of $I_{K^{\prime}}(k)$ in $k[G/K]$, one could
record which dominant weights of $\mathfrak{a}$ it is supported on in
$k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$. One could then look at all weights
that appear in the supports of such copies of $I_{K^{\prime}}(k)$ in $k[G/K]$.
This collection of weights would be infinite and could not, for example, lie
within any hyperplane.

We also observe that since each copy of $I_{K^{\prime}}(k)$ contains a
$K^{\prime}$-invariant function, we can deduce there are many
$K^{\prime}$-invariant functions on $k[G/K]$ (recall we already know from 5.6
that $k[G/K]^{K^{\prime}}$ is an integral domain). Again the structure of
$K[G/K]^{K^{\prime}}$ is not generally understood, although it has been
partially computed in certain examples, i.e. $G\times G/G$ and the
superspheres $OSP(m|2n)/OSP(m-1|2n)$.

######  Remark 7.10.

Suppose again the
$\mathfrak{k}_{\overline{0}}\neq\mathfrak{g}_{\overline{0}}$. Following a
similar argument to the one above, one may deduce the existence of many
projective $K^{\prime}$-submodules of $k[G/K]$ as follows: given an
irreducible $K^{\prime}$-submodule $L$ of $\operatorname{Dist}(G/K,eK)$, it
defines an irreducible submodule of
$\operatorname{Dist}((G_{0}/K_{0})^{\mathfrak{k}^{\prime}},eK_{0})$ via our
isomorphism. Thus $L$ must pair nontrivially with some projective
indecomposable summand $P=P_{K^{\prime}}(V)$ of
$k[(G_{0}/K_{0})^{\mathfrak{k}^{\prime}}]$. However this is only possible if
$V\cong L^{*}$, and $L$ pairs with $P$ via $L\otimes P\to L\otimes L^{*}\to
k$. It follows that $k[G/K]$ must contain a copy of $P_{K^{\prime}}(L^{*})$ as
well.

##  8\. Pairs that have an Iwasawa decomposition

###  8.1. The Iwasawa decomposition

We continue to assume $G$ is quasireductive. Due to technical difficulties
with Lie superalgebras like $\mathfrak{q}(n)$, which have Cartan subalgebras
that are not purely even, we will from now on also assume $G$ is Cartan-even.

######  Definition 8.1.

A quasireductive supergroup $G$ is Cartan-even if for a Cartan subalgebra
$\mathfrak{h}\subseteq\mathfrak{g}$ we have
$\mathfrak{h}=\mathfrak{h}_{\overline{0}}$ (i.e. $\mathfrak{h}_{\overline{0}}$
is self-centralizing in $\mathfrak{g}$).

Let $(\mathfrak{g},\mathfrak{k})$ be a supersymmetric pair, and let
$\mathfrak{a}\subseteq\mathfrak{p}_{\overline{0}}$ be a Cartan subspace, i.e.
a maximal subalgebra of $\mathfrak{p}_{\overline{0}}$ consisting only of
semisimple elements. Then we may decompose $\mathfrak{g}$ into weight spaces
according to the action of $\mathfrak{a}$ as

$\mathfrak{g}=\mathfrak{m}\oplus\bigoplus\limits_{\overline{\alpha}\in\mathfrak{a}^{*}}\mathfrak{g}_{\overline{\alpha}}$

where $\overline{\alpha}\neq 0$ and $\mathfrak{m}$ is the centralizer of
$\mathfrak{a}$ in $\mathfrak{g}$. We write $\overline{\Delta}$ for the set of
nonzero weights of this action, and call them restricted roots. Note that the
weights of the action are exactly the restriction to $\mathfrak{a}$ of the
roots under the action of a maximal torus in $\mathfrak{g}_{\overline{0}}$
which contains $\mathfrak{a}$. We say that the pair
$(\mathfrak{g},\mathfrak{k})$ admits an Iwasawa decomposition if there is some
choice of positive roots in $\overline{\Delta}$,
$\overline{\Delta}=\overline{\Delta}^{+}\sqcup\overline{\Delta^{-}}$, such
that

$\text{if }\
\mathfrak{n}=\bigoplus\limits_{\alpha\in\overline{\Delta}^{+}}\mathfrak{g}_{\alpha},\
\text{ then }\ \mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}.$

In this case, we may extend the choice of positive restricted roots to a
choice of positive roots $\Delta=\Delta^{+}\sqcup\Delta^{-}$ for
$\mathfrak{g}$. In that case, if $\mathfrak{b}$ is the Borel subalgebra
determined by this choice of positive roots then we have
$\mathfrak{a}\oplus\mathfrak{n}\subseteq\mathfrak{b}$. In particular,
$\mathfrak{b}+\mathfrak{k}=\mathfrak{g}$, i.e. $\mathfrak{k}$ has a
complimentary Borel subalgebra.

######  Definition 8.2.

We call a Borel subalgebra arising in this way an Iwasawa Borel subalgebra for
$\mathfrak{k}$ with respect to $\mathfrak{a}$. If $B\subseteq G$ integrates
$\mathfrak{b}$, then we call $B$ an Iwasawa Borel subgroup of $G$.

###  8.2. $G/K$ is spherical; the rational functions $f_{\lambda}$

Recall that a $G$-supervariety $X$ is spherical if $X$ admits an open
$B$-orbit for some Borel $B$. For $G/K$, this is equivalent to $\mathfrak{k}$
admitting a complementary Borel subalgebra. Hence if $\mathfrak{g}$ admits an
Iwasawa decomposition, $G/K$ is spherical. Choose a Cartan subspace
$\mathfrak{a}$ and let $B$ be an Iwasawa Borel subgroup with respect to
$\mathfrak{a}$.

In this case, $G_{0}/K_{0}$ is spherical with respect to $B_{0}$. Write
$\Lambda_{0}^{+}\subseteq\mathfrak{a}^{*}$ for the dominant weights that
appear as $B_{0}$-highest weights in $k[G_{0}/K_{0}]$. Then the lattice
generated by $\Lambda_{0}^{+}$ is a full rank lattice in $\mathfrak{a}^{*}$
(see, for instance, [Tim11]).

By the general theory of spherical supervarieties (see [She19]), the
$B$-highest weights in $k[G/K]$, which we will call $\Lambda^{+}$, are a
subset of $\Lambda_{0}^{+}$, and there is at most one function $f_{\lambda}\in
k[G/K]$ of highest weight $\lambda$ for any $\lambda\in\Lambda_{0}^{+}$.
Further, $f_{\lambda}$ is even and non-nilpotent, hence it maps down to a non-
zero function of highest weight $\lambda$ on $G_{0}/K_{0}$. Further,
$\Lambda^{+}$ generates the same lattice as $\Lambda_{0}^{+}$ in
$\mathfrak{a}^{*}$, which we call $\Lambda$, and in particular is Zariski
dense. It follows that by inverting the functions $f_{\lambda}$ for
$\lambda\in\Lambda^{+}$ and taking arbitrary products, we obtain rational
functions $f_{\lambda}$ for all $\lambda\in\Lambda$ which are
$\mathfrak{a}\oplus\mathfrak{n}$-eigenfunctions and are regular in a
neighborhood of $eK$.

###  8.3. A different perspective via induced spaces

We may gain another perspective on the sphericity of $G/K$ using the notion of
induced spaces. Write $A$ and $N$ for the subgroups of $G$ integrating
$\mathfrak{a}$ and $\mathfrak{n}$, respectively. Then $AN$ is a subgroup of
$G$ which integrates $\mathfrak{a}\oplus\mathfrak{n}$. Since
$(\mathfrak{a}\oplus\mathfrak{n})_{\overline{1}}$ is complimentary to
$\mathfrak{k}_{\overline{1}}$ by the Iwasawa decomposition, we may apply 5.1
to obtain the existence of a canonical map of $AN$-supervarieties

$(G_{0}/K_{0})^{\mathfrak{a}\oplus\mathfrak{n}}\to G/K$

which is an isomorphism in a neighborhood of $eK_{0}$. In particular the map
on functions

$k[G/K]\to k[(G_{0}/K_{0})^{\mathfrak{a}\oplus\mathfrak{n}}]$

is injective map of $AN$-modules. It follows that we have an injective
morphism

$k[G/K]^{N}\to k[(G_{0}/K_{0})^{\mathfrak{a}\oplus\mathfrak{n}}]^{N},$

and the right hand side is isomorphic to the subalgebra of highest weight
vectors in $k[G_{0}/K_{0}]$. Further, in a neighborhood of the $eK_{0}$ in
$(G_{0}/K_{0})^{\mathfrak{a}\oplus\mathfrak{n}}$, we have functions
$f_{\lambda}$ annihilated by $\mathfrak{n}$ and of weight
$\lambda\in\mathfrak{a}^{*}$ for every $\lambda\in\Lambda$. Using the local
isomorphism, we find that such functions also exist in a neighborhood of $eK$
in $G/K$, and in particular they are the highest weight functions of weight
$\lambda\in\Lambda$. .

######  Definition 8.3.

In the setup above, for $\lambda\in\Lambda$ we let $f_{\lambda}$ denote the
rational $\mathfrak{a}\oplus\mathfrak{n}$-eigenfunction on $G/K$ such that
$f_{\lambda}(eK)=1$.

Note we can always normalize as above, because the
$\mathfrak{a}\oplus\mathfrak{n}$ eigenvectors are all non-nilpotent and if
they vanished at $eK$ then necessarily they would vanish everywhere.

###  8.4. Highest weight submodules of $k[G/K]$

Recall that for a homogeneous vector $v$ we write $\langle K^{\prime}\cdot
v\rangle$ for the $K^{\prime}$-module generated by $v$.

######  Lemma 8.4.

If $\lambda\in\Lambda^{+}$, then $\mathcal{U}\mathfrak{k}^{\prime}f_{\lambda}$
is stable under $K^{\prime}$, and thus

$\langle K^{\prime}\cdot
f_{\lambda}\rangle=\mathcal{U}\mathfrak{k}^{\prime}f_{\lambda}.$

###### Proof.

It suffices to prove that $\mathcal{U}\mathfrak{k}^{\prime}f_{\lambda}$ is
stable under $K_{0}$. For this we notice that by the Iwasawa decomposition for
$\mathfrak{g}_{\overline{0}}$ and the fact that $G_{0}$ is connected,

$\langle G_{0}\cdot
f_{\lambda}\rangle=\mathcal{U}\mathfrak{g}_{\overline{0}}f_{\lambda}=\mathcal{U}\mathfrak{k}_{\overline{0}}f_{\lambda}\subseteq\langle
K_{0}\cdot f_{\lambda}\rangle.$

Since $\langle K_{0}\cdot f_{\lambda}\rangle\subseteq\langle G_{0}\cdot
f_{\lambda}\rangle$ it follows that $\langle K_{0}\cdot
f_{\lambda}\rangle=\mathcal{U}\mathfrak{k}_{\overline{0}}f_{\lambda}$. From
here it is easy to check that $\mathcal{U}\mathfrak{k}^{\prime}f_{\lambda}$ is
$K_{0}$-stable. ∎

######  Lemma 8.5.

For $\lambda\in\Lambda^{+}$, $\langle K^{\prime}\cdot f_{\lambda}\rangle$
contains at most one summand isomorphic to $I_{K^{\prime}}(k)$.

###### Proof.

By the classical Iwasawa decomposition, we have $\langle K_{0}\cdot
f_{\lambda}\rangle=\mathcal{U}\mathfrak{g}_{\overline{0}}f_{\lambda}\cong
L_{0}(\lambda)$, since $f_{\lambda}$ is a $B_{0}$-highest weight vector. Hence
we have a surjection of $K^{\prime}$-modules

$\operatorname{Ind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}L_{0}(\lambda)\to\langle
K^{\prime}\cdot f_{\lambda}\rangle.$

Each copy of $I_{K^{\prime}}(k)$ showing up in $\langle K^{\prime}\cdot
f_{\lambda}\rangle$ would have to split disjointly back into
$\operatorname{Ind}_{\mathfrak{k}_{0}}^{\mathfrak{k}^{\prime}}L_{0}(\lambda)$.
However, since $K_{0}$ is a spherical subgroup of $G_{0}$,
$L_{0}(\lambda)^{K_{0}}$ is one-dimensional, so the induced module has only
one copy of $I_{K^{\prime}}(k)$. ∎

######  Remark 8.6.

If $(\mathfrak{g},\mathfrak{k}^{\prime})$ also satisfies the Iwasawa
decomposition then we have $\langle K^{\prime}\cdot f_{\lambda}\rangle=\langle
G\cdot f_{\lambda}\rangle$. Also, it will be shown in the subsequent article
that if $\mathfrak{g}$ is basic classical and the involution $\theta$
preserves the nondegenerate form on $\mathfrak{g}$, then we always have
$\langle K^{\prime}\cdot f_{\lambda}\rangle=\langle G\cdot
f_{\lambda}\rangle$.

###  8.5. Harish-Chandra Homomorphism

The Iwasawa decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$ implies that
$\mathcal{U}\mathfrak{g}=S(\mathfrak{a})\oplus(\mathfrak{n}\mathcal{U}\mathfrak{g}+\mathcal{U}\mathfrak{g}\mathfrak{k})$.
Thus we have a vector space isomorphism

$\operatorname{Dist}(G/K,eK)\cong\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{k}\cong
S(\mathfrak{a})\oplus\mathfrak{n}\mathcal{U}\mathfrak{g}/(\mathfrak{n}\mathcal{U}\mathfrak{g}\cap\mathcal{U}\mathfrak{g}\mathfrak{k}).$

######  Definition 8.7.

The Harish-Chandra morphism $HC:\operatorname{Dist}(G/K,eK)\to
S(\mathfrak{a})$ is given by the projection onto $S(\mathfrak{a})$ along
$\mathfrak{n}\mathcal{U}\mathfrak{g}/(\mathfrak{n}\mathcal{U}\mathfrak{g}\cap\mathcal{U}\mathfrak{g}\mathfrak{k})$.

Choose a homogeneous basis $p_{1},\dots,p_{r}$ of $\mathfrak{p}$. Recall from
lemma 2.9 that $\operatorname{Dist}(G/K,eK)$ is spanned by monomials

$p_{1}^{k_{1}}\cdots p_{r}^{k_{r}},$

where we abusively identify them with their restrictions to
$\operatorname{Dist}(G/K,eK)$. Define a filtration $F^{\bullet}$ on
$\operatorname{Dist}(G/K,eK)$ by setting
$F^{0}=\langle\operatorname{ev}_{eK}\rangle$, and set $F^{i}$ to be the span
of all monomials as above where the degree of $p_{i}$ is 2 if $p_{i}$ is even
and 1 if $p_{i}$ is odd. The following is a generalization of 4.2.2 in
[Gor00].

######  Lemma 8.8.

$HC(F^{r})\subseteq\sum\limits_{j\leq r/2}S^{j}\mathfrak{a}$.

###### Proof.

Since the Harish-Chandra projection is linear, it suffices to prove this on
monomials. We induct on the length of the monomial. If the monomial is length
zero, the result is clear. Suppose we have a monomial $b_{1}\cdots b_{t}\in
F^{r}$. Using the Iwasawa decomposition, we may write $b_{1}=k+a+n$ where
$k\in\mathfrak{k}$, $a\in\mathfrak{a}$, and $n\in\mathfrak{n}$. Since
$nb_{2}\cdots b_{t}\in\mathfrak{n}\mathcal{U}\mathfrak{g}$ it vanishes under
the Harish-Chandra projection, so we have

$\displaystyle HC(b_{1}\dots b_{t})$ $\displaystyle=$ $\displaystyle
HC(ab_{2}\cdots b_{t})+HC(kb_{2}\cdots b_{t})$ $\displaystyle=$ $\displaystyle
HC(ab_{2}\cdots b_{t})+\sum\limits_{i}HC(b_{2}\cdots[k,b_{i}]\cdots
b_{t})+HC(b_{2}\cdots b_{t}k).$

Since $b_{2}\cdots b_{t}k\in\mathcal{U}\mathfrak{g}\mathfrak{k}$, the last
term vanishes. Now notes that $a\neq 0$ only if $b_{t}$ is even, and in that
case $b_{2}\cdots b_{t}\in F^{r-2}$, and $HC(ab_{2}\cdots
b_{t})=HC(a)HC(b_{2}\cdots b_{t})$. Thus by induction,

$\deg HC(ab_{2}\cdots b_{t})\leq\deg HC(b_{2}\cdots
b_{t})+1\leq\frac{r-2}{2}+1=\frac{r}{2}.$

As for $HC(b_{2}\cdots[k,b_{i}]\cdots b_{t})$, if $b_{1}$ is even then the
parity of $[k,b_{i}]$ is the same as the parity of $b_{i}$, and thus
$b_{2}\cdots[k,b_{i}]\cdots b_{t}\in F^{r-2}$. Therefore

$\deg HC(b_{2}\cdots[k,b_{i}]\cdots b_{t})\leq(r-2)/2\leq r/2.$

If $b_{1}$ is odd, then the parity of $[k,b_{i}]$ is opposite the parity of
$b_{i}$, so $b_{2}\cdots[k,b_{i}]\cdots b_{t}\in F^{r}$, and by induction we
have

$HC(b_{2}\cdots[k,b_{i}]\cdots b_{t})\leq r/2.$

∎

######  Corollary 8.9.

Let $z\in\operatorname{Dist}^{r}(G_{0}/K_{0},eK_{0})^{K_{0}}$ lie in the $r$th
part of the standard filtration on $\operatorname{Dist}(G_{0}/K_{0},eK_{0})$
defined in definition 2.2. Then $v_{\mathfrak{k}^{\prime}}\cdot
z\in\mathcal{A}_{G/K}$ has

$HC(v_{\mathfrak{k}^{\prime}}\cdot z)\leq
r+\operatorname{dim}\mathfrak{p}_{\overline{1}}/2.$

###### Proof.

We have that $z\in F^{2r}$ and $v_{\mathfrak{k}^{\prime}}\in
F^{\operatorname{dim}\mathfrak{p}_{\overline{1}}}$, so
$v_{\mathfrak{k}^{\prime}}\cdot z=zv_{\mathfrak{k}^{\prime}}\in
F^{2r+\operatorname{dim}\mathfrak{p}_{\overline{1}}}$. ∎

######  Lemma 8.10.

$HC(\mathcal{A}_{G/K})$ is naturally a module over $HC(\mathcal{Z}_{G/K})$
such that $HC:\mathcal{A}_{G/K}\to S(\mathfrak{a})$ induces a morphism of
$\mathcal{Z}_{G/K}$-modules.

###### Proof.

Let $\gamma\in\mathcal{A}_{G/K}$ and $z\in\mathcal{Z}_{G/K}$. Write
$z=n+HC(z)+\mathcal{U}\mathfrak{g}\mathfrak{k}$ and
$\gamma=n^{\prime}+HC(\gamma)+\mathcal{U}\mathfrak{g}\mathfrak{k}$. Then we
see that

$\gamma
z=(n^{\prime}+HC(\gamma)+\mathcal{U}\mathfrak{g}\mathfrak{k})(n+HC(z)+\mathcal{U}\mathfrak{g}\mathfrak{k})=n^{\prime}(n+HC(z))+HC(\gamma)n+HC(z)HC(\gamma)+\mathcal{U}\mathfrak{g}\mathfrak{k}.$

Clearly $n^{\prime}(n+HC(z))\in\mathfrak{n}\mathcal{U}\mathfrak{g}$. And since
$HC(\gamma)\in S(\mathfrak{a})$, it preserves
$\mathfrak{n}\mathcal{U}\mathfrak{g}$ under commutator and thus
$HC(\gamma)n\in\mathfrak{n}\mathcal{U}\mathfrak{g}$ as well. Hence we find
that

$HC(\gamma z)=HC(\gamma)HC(z),$

as desired. ∎

######  Lemma 8.11.

For $\gamma\in\operatorname{Dist}(G/K,eK)$ and $\lambda\in\Lambda$ we have
that $HC(\gamma)(\lambda)=\gamma(f_{\lambda})$.

###### Proof.

Since $\mathcal{U}\mathfrak{g}\mathfrak{n}$ annihilates $f_{\lambda}$, the
result follows from our normalization of $f_{\lambda}$. ∎

The following illustrates the importance of understanding
$HC(\mathcal{A}_{G/K})$, especially in light of remark 8.6.

######  Corollary 8.12.

Let $\lambda\in\Lambda^{+}$. Then $\langle K^{\prime}\cdot f_{\lambda}\rangle$
contains a copy of $I_{K^{\prime}}(k)$ if and only if there exists
$\gamma\in\mathcal{A}_{G/K}$ such that
$HC(\gamma)(\lambda)=\gamma(f_{\lambda})\neq 0$.

###### Proof.

These results follow from the work in section 7.3. ∎

For the next lemma, let $\mathfrak{h}\subseteq\mathfrak{g}$ be a
$\theta$-stable Cartan subalgebra which contains $\mathfrak{a}$, so that
$\mathfrak{h}=\mathfrak{t}\oplus\mathfrak{a}$, where $\mathfrak{t}$ are the
fixed points of $\theta$ in $\mathfrak{h}$. In particular,
$\mathfrak{t}\subseteq\mathfrak{k}_{\overline{0}}$.

######  Proposition 8.13.

$HC(\mathcal{A}_{G/K})=0$ if one of the following two conditions holds.

  1. (1)

$\operatorname{dim}\mathfrak{p}_{\overline{1}}$ is odd;

  2. (2)

The restriction of $\operatorname{ber}_{\mathfrak{k}^{\prime}}$ to
$\mathfrak{t}$ is non-zero. Note this holds if
$\operatorname{ber}_{\mathfrak{k}^{\prime}}\neq 0$ and the map
$\mathfrak{t}\to\mathfrak{k}^{\prime}/[\mathfrak{k}^{\prime},\mathfrak{k}^{\prime}]$
is surjective.

###### Proof.

If $\mathfrak{p}_{\overline{1}}$ is odd-dimensional then
$v_{\mathfrak{k}^{\prime}}$ is odd, and thus all elements of
$\mathcal{A}_{G/K}$ are odd. Since $S(\mathfrak{a})$ is purely even and $HC$
is an even map, we must have $HC(\mathcal{A}_{G/K})=0$.

Now suppose condition (2) holds. Let
$\gamma\in\mathcal{A}_{G/K}=\operatorname{Dist}(G/K,eK)^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$,
and let $\lambda\in\Lambda^{+}$. Then since $\lambda\in\mathfrak{a}^{*}$,
$\lambda(\mathfrak{t})=0$. If $t\in\mathfrak{t}$ we see that

$\operatorname{ber}_{\mathfrak{k}^{\prime}}(t)\gamma(f_{\lambda})=(t\cdot\gamma)(f_{\lambda})=\gamma
tf_{\lambda}=\lambda(t)\gamma(f_{\lambda})=0.$

However by assumption there exists $t\in\mathfrak{t}$ such that
$\operatorname{ber}_{\mathfrak{k}^{\prime}}(t)\neq 0$, so if we use this $t$
we find that $\gamma(f_{\lambda})=0$ for all $\lambda$, so that
$HC(\gamma)=0$. ∎

###  8.6. The polynomial $p_{G/K,B}$

We now interpret the polynomial
$p_{G/K,B}:=HC(\operatorname{ev}_{eK}v_{\mathfrak{k}^{\prime}})$, noting that
it depends on our Iwasawa Borel subgroup $B$. Recall that
$\operatorname{ev}_{eK}v_{\mathfrak{k}^{\prime}}=v_{\mathfrak{k}^{\prime}}\cdot\operatorname{ev}_{eK}$
is a ghost distribution, and is the one of minimal degree. Observe that we
have a $K^{\prime}$-equivariant embedding

$i:K^{\prime}/K_{0}\to G/K.$

This induces a $K^{\prime}$-equivariant surjective morphism on functions

$i^{*}:k[G/K]\to k[K^{\prime}/K_{0}].$

Write $\operatorname{ev}_{eK}$ for the evaluation distribution at $eK$ on
$G/K$. Then on $K^{\prime}$ distributions we obtain a map which sends

$v_{\mathfrak{k}^{\prime}}\mapsto
v_{\mathfrak{k}^{\prime}}\cdot\operatorname{ev}_{eK}=\operatorname{ev}_{eK}v_{\mathfrak{k}^{\prime}}.$

Thus for $\lambda\in\Lambda^{+}$ we have

$p_{G/K,B}(\lambda)=(\operatorname{ev}_{eK}v_{\mathfrak{k}^{\prime}})f_{\lambda}=v_{\mathfrak{k}^{\prime}}i^{*}(f_{\lambda})=\operatorname{ev}_{eK_{0}}\circ
D_{\mathfrak{k}^{\prime}}\circ i^{*}(f_{\lambda}).$

It follows that:

######  Lemma 8.14.

For $\lambda\in\Lambda^{+}$, $p_{G/K,B}(\lambda)\neq 0$ if and only if
$\langle K^{\prime}\cdot i^{*}f_{\lambda}\rangle$ contains
$I_{K^{\prime}}(k)$.

######  Corollary 8.15.

If $\lambda\in\Lambda^{+}$ and $\langle K^{\prime}\cdot f_{\lambda}\rangle$
contains a copy of $I_{K^{\prime}}(k)$, then $p_{G/K,B}(\lambda)\neq 0$ if and
only if the $K^{\prime}$-invariant in the copy of $I_{K^{\prime}}(k)$ is non-
zero at $eK$.

We have now shown that:

######  Corollary 8.16.

We have $p_{G/K,B}(\lambda)\neq 0$ if and only if $\langle K^{\prime}\cdot
f_{\lambda}\rangle$ contains a copy of $I_{K^{\prime}}(k)$ and the
$K^{\prime}$-invariant of $I_{K^{\prime}}(k)$ does not vanish at $eK$.

The following gives one reason for the importance of the polynomial
$p_{G/K,B}$.

######  Proposition 8.17.

Suppose that whenever $\langle K^{\prime}\cdot f_{\lambda}\rangle$ contains a
copy of $I_{K^{\prime}}(k)$ for $\lambda\in\Lambda^{+}$, the
$K^{\prime}$-invariant in it does not vanish at $eK$. Then,

  1. (1)

For $\lambda\in\Lambda^{+}$, $p_{G/K,B}(\lambda)\neq 0$ if and only if
$\langle K^{\prime}\cdot f_{\lambda}\rangle$ contains a copy of
$I_{K^{\prime}}(k)$.

  2. (2)

For any $\lambda\in\Lambda^{+}$, if $p_{G/K,B}(\lambda)=0$, then
$HC(\gamma)(\lambda)=0$ for all $\gamma\in\mathcal{A}_{G/K}$.

###### Proof.

(1) follows from 8.16.

(2) follows from 8.12. ∎

###  8.7. Relationship between $G/K$ and $G/K^{\prime}$

We continue to suppose that $(\mathfrak{g},\mathfrak{k})$ satisfies the
Iwasawa decomposition.

######  Proposition 8.18.

For $\lambda\in\Lambda^{+}$, suppose that

  1. (1)

$\langle K^{\prime}\cdot f_{\lambda}\rangle$ contains a copy of
$I_{K^{\prime}}(k)$ (equivalently $HC(\gamma)(\lambda)\neq 0$ for some
$\gamma\in\mathcal{A}_{G/K}$); and

  2. (2)

$\langle G\cdot f_{\lambda}\rangle\cong L(\lambda)$ is an irreducible
$G$-module.

Then $I_{G}(L(\lambda))$ is a submodule of
$k[G]^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$.

Note that $k[G]^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}$ is equal to the
induced bundle
$\operatorname{Ind}_{K^{\prime}}^{G}\operatorname{Ber}(\mathfrak{k}^{\prime})$,
and if $\operatorname{Ber}(\mathfrak{k}^{\prime})$ is trivial then
$k[G]^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}=k[G/K^{\prime}]$.

###### Proof.

Since $I_{K^{\prime}}(k)\cong
P_{K^{\prime}}(\operatorname{Ber}(\mathfrak{k}^{\prime}))$,
$I_{K^{\prime}}(k)$ has a morphism
$\varphi:I_{K^{\prime}}(k)\to\operatorname{Ber}(\mathfrak{k}^{\prime})$
determined by the projection onto its head. Since $I_{K^{\prime}}(k)$ splits
off $L(\lambda)$ as a $K^{\prime}$-module, it also must split off
$I_{G}(L(\lambda))$ as a $K^{\prime}$-module. Therefore we may extend
$\varphi$ to a $K^{\prime}$-equivariant morphism
$\phi:I_{G}(L(\lambda))\to\operatorname{Ber}(\mathfrak{k}^{\prime})$. By
construction $\phi$ is non-zero on $L(\lambda)$, the socle of
$I_{G}(L(\lambda))$. Thus by Frobenius reciprocity, $\phi$ defines an
injective morphism of $G$-modules

$\Phi:I_{G}(L(\lambda))\to k[G]^{\operatorname{ber}_{\mathfrak{k}^{\prime}}}.$

∎

######  Remark 8.19.

The above theorem is especially useful when $G/K\cong G/K^{\prime}$, as it
helps to determine the structure of $k[G/K]$ as a $G$-module. We will see one
nice application of this in the next section.

##  9\. The Group Case $G\times G/G$

We now consider the case of the supersymmetric space $G\times G/G$, where $G$
is embedded diagonally. This space is isomorphic to $G$ with the $G\times G$
action given by left and right translation. Since $G$ is assumed to be
quasireductive, it must be connected in particular, so we can work with the
Lie superalgebra without losing much. The involution we take in this case,
$\theta$, is given on the Lie superalgebra $\mathfrak{g}\times\mathfrak{g}$ by
$\theta(x,y)=(y,x)$.

######  Lemma 9.1.

We have a $\mathfrak{g}\times\mathfrak{g}$-module isomorphism

$\mathcal{U}\mathfrak{g}\cong\operatorname{Dist}(G\times
G/G,eG)=\mathcal{U}(\mathfrak{g}\times\mathfrak{g})/\mathcal{U}(\mathfrak{g}\times\mathfrak{g})\mathfrak{g}$

given by

$u\mapsto u\otimes 1.$

###### Proof.

Let $(v_{1},v_{2})\in\mathfrak{g}\times\mathfrak{g}$ and
$u\in\mathcal{U}\mathfrak{g}$. We see that $(v_{1},v_{2})\cdot
u=v_{1}u-(-1)^{\overline{u}\overline{v_{2}}}uv_{2}$, and this will map to
(before modding out by
$\mathcal{U}(\mathfrak{g}\times\mathfrak{g})\mathfrak{g}$)

$\displaystyle v_{1}u\otimes
1-(-1)^{\overline{u}\overline{v_{2}}}uv_{2}\otimes 1$ $\displaystyle=$
$\displaystyle v_{1}u\otimes 1+(-1)^{\overline{v_{2}}\overline{u}}u\otimes
v_{2}-(-1)^{\overline{v_{2}}\overline{u}}uv_{2}\otimes
1-(-1)^{\overline{v_{2}}\overline{u}}u\otimes v_{2}$ $\displaystyle=$
$\displaystyle(v_{1}\otimes 1+1\otimes v_{2})(u\otimes
1)-(-1)^{\overline{v_{2}}\overline{u}}(u\otimes 1)(v_{2}\otimes 1+1\otimes
v_{2})$

which shows the $\mathfrak{g}\times\mathfrak{g}$-equivariance of the map. It
is an isomorphism by the PBW theorem. ∎

From now on we work with $\mathcal{U}\mathfrak{g}$ as our space of
distributions. Observe that in this case,

$\mathfrak{g}^{\prime}=\\{(u,\delta(u)):u\in\mathfrak{g}\\}\cong\mathfrak{g},$

and so we can identify $\mathfrak{g}^{\prime}$ with $\mathfrak{g}$ as a Lie
superalgebra. With this setup, the action of $\mathfrak{g}^{\prime}$ on
$\operatorname{Dist}(G\times G/G,eG)$ is given by, for $u\in\mathfrak{g}$ and
$v\in\mathcal{U}\mathfrak{g}$,

$u\cdot v=uv-(-1)^{\overline{u}\overline{v}+\overline{u}}vu.$

This is exactly Gorelik’s twisted adjoint action defined in [Gor00]. There,
she proved algebraically that $\mathcal{U}\mathfrak{g}$ is an induced module
from $\mathfrak{g}_{\overline{0}}$ under this action. However this follows
from our geometric perspective via 5.7. Gorelik defined
$\mathcal{A}\subseteq\mathcal{U}\mathfrak{g}$, the anticentre of
$\mathcal{U}\mathfrak{g}$, to be the $\mathfrak{g}^{\prime}$-invariant
distributions on $G$. However please note remark 7.5 regarding notation.

###  9.1. Structure of $k[G]$ as a $G\times G$-module

We assume that $G$ is Cartan-even. The following is proven in [She19].

######  Proposition 9.2.

  1. (1)

As a $G\times G$-module, we may write

$k[G]=\bigoplus\limits_{\mathcal{B}}M_{\mathcal{B}},$

where each $M_{\mathcal{B}}$ is an indecomposable $G\times G$-module, and the
sum runs over all blocks of $\operatorname{Rep}(G)$ up to parity.

  2. (2)

The socle of $k[G]$ is the submodule generated by all highest weight vectors
with respect to an Iwasawa Borel, and is equal to
$\bigoplus\limits_{L}L^{*}\boxtimes L$, where $L$ runs over all simple modules
of $G$ up to parity.

  3. (3)

The socle of $M_{\mathcal{B}}$ is $\bigoplus\limits_{L}L^{*}\boxtimes L$,
where the sum runs over all simple $G$-modules $L$ in $\mathcal{B}$, up to
parity. In particular $M_{\mathcal{B}}$ is simple if and only if its socle is
simple.

  4. (4)

For a simple $G$-module $L$, the submodule $L^{*}\boxtimes L$ is given by the
image of the $\operatorname{End}(L)^{*}$ in $k[G]$ under the pullback morphism

$k[\operatorname{End}(L)]\to k[G]$

coming from the representation $G\to\operatorname{End}(L)$.

######  Proposition 9.3.

Let $L$ be a simple $G$-module. Then $\operatorname{End}(L)$, as a $G\times
G$-module, admits a unique, up to scalar, $G^{\prime}$-invariant given by
$\delta_{L}$, the parity involution. Consequently,
$\text{tr}_{L}\in\operatorname{End}(L)^{*}$ defines a nonzero
$G^{\prime}$-invariant function on $G$.

###### Proof.

Clearly $\delta_{L}$ is $\mathfrak{g}_{\overline{0}}$-invariant. For
$u\in\mathfrak{g}_{1}$, we see that

$((u,-u)\delta)(v)=u\delta(v)+\delta(uv)=(-1)^{\overline{v}}uv+(-1)^{\overline{u}+\overline{v}}uv=0$

so $\gamma$ is $\mathfrak{g}^{\prime}$-invariant. Now as a tensor,
$\delta_{L}=\sum\limits_{i}(-1)^{\overline{e_{i}}}e_{i}\otimes\varphi_{i}$,
where $\varphi_{i}(e_{i})=1$, so after applying the braiding to switch the
order of tensors, we obtain the trace in $\operatorname{End}(L)^{*}$. ∎

######  Corollary 9.4.

If $L$ is a simple $G$-module, $L^{*}\boxtimes L\subseteq k[G]$ has a unique
$\mathfrak{g}^{\prime}$-invariant, given by $\text{tr}_{L}$. In particular,
$\text{tr}_{L}(eG)=\operatorname{dim}L\neq 0$, so that the hypotheses of 8.17
apply.

The following is well-known:

######  Lemma 9.5.

Write $\mathfrak{a}=\\{(h,-h):h\in\mathfrak{h}\\}$. Then $\mathfrak{a}$ is a
Cartan subspace for $(\mathfrak{g}\times\mathfrak{g},\mathfrak{g})$.

Now choose a positive system for $\mathfrak{g}$, say $\Delta^{+}$, with Borel
$\mathfrak{b}^{+}$ and opposite Borel $\mathfrak{b}^{-}$. Write $\rho$ for the
Weyl vector. Then we obtain a positive system for
$\mathfrak{g}\times\mathfrak{g}$ with Borel subalgebra
$\mathfrak{b}^{-}\times\mathfrak{b}^{+}$.

######  Corollary 9.6.

The above choice of Borel subalgebra is an Iwasawa Borel subalgebra. Further,
we have that $\mathfrak{n}=\mathfrak{u}^{-}\times\mathfrak{u}^{+}$, where
$\mathfrak{u}^{\pm}\subseteq\mathfrak{b}^{\pm}$ are the nilpotent radicals of
$\mathfrak{b}^{\pm}$, and the Iwasawa decompositions:

$\mathfrak{g}\times\mathfrak{g}=\mathfrak{g}\oplus\mathfrak{a}\oplus\mathfrak{n}=\mathfrak{g}^{\prime}\oplus\mathfrak{a}\oplus\mathfrak{n}$

######  Lemma 9.7.

For the above choice of $\mathfrak{a}$ and positive system,

$\Lambda^{+}=\\{(-\lambda,\lambda):\lambda\text{ is a
}\mathfrak{b}\text{-dominant weight}\\}.$

Given this lemma, we identify $\Lambda^{+}$ with the set of
$\mathfrak{b}$-dominant weights, and write $f_{\lambda}\in k[G]$ for the
function of highest weight $(-\lambda,\lambda)$ such that $f_{\lambda}(eG)=1$.

######  Proposition 9.8.

We have
$\mathcal{U}(\mathfrak{g}\times\mathfrak{g})f_{\lambda}=\mathcal{U}\mathfrak{g}f_{\lambda}=\mathcal{U}\mathfrak{g}^{\prime}f_{\lambda}=L_{\lambda}^{*}\boxtimes
L_{\lambda}$.

###### Proof.

This follows from the Iwasawa decompositions. ∎

Recall that for this supersymmetric pair, the Harish-Chandra morphism on
distributions corresponds to the usual Harish-Chandra morphism
$HC:\mathcal{U}\mathfrak{g}\to S(\mathfrak{h})$ on the enveloping algebra.

######  Corollary 9.9.

$p_{G\times G/G,B}(\lambda)\neq 0$ if and only if $L_{\lambda}^{*}\boxtimes
L_{\lambda}$ has a copy of $I_{G^{\prime}}(k)$ as a $G^{\prime}$-module.

###### Proof.

This follows from 9.8, 9.4, and 8.17. ∎

###  9.2. Projectivity criteria for irreducible modules

For this section we work with a fixed Borel subalgebra $\mathfrak{b}$ of
$\mathfrak{g}$, which as stated above determines a fixed Iwasawa Borel
subalgebra of $\mathfrak{g}\times\mathfrak{g}$.

We observe that $\operatorname{id}\times\delta$ defines an automorphism of
$G\times G$ which takes $G$ to $G^{\prime}$ and vice-versa. In particular it
defines an isomorphism

$G\times G/G^{\prime}\to G\times G/G$

which is $\operatorname{id}\times\delta$-equivariant, meaning that the
pullback morphism defines a $G\times G$-equivariant isomorphism

$k[G]=k[G\times G/G]\to k[G\times
G/G^{\prime}]^{\operatorname{id}\times\delta}.$

Here we use the notation $V^{\phi}$ for the $\phi$ twist of a $G$-module $V$,
where $\phi$ is an automorphism of $G$. Notice that since
$\operatorname{id}\times\delta$ is the identity on $G_{0}\times G_{0}$, we
have that $(L^{*}\boxtimes L)^{\operatorname{id}\times\delta}\cong
L^{*}\boxtimes L$ for a simple $G$-module $L$, by highest weight theory.

######  Proposition 9.10.

Let $(-\lambda,\lambda)\in\Lambda^{+}$. Then $L(\lambda)^{*}\boxtimes
L(\lambda)$ contains a copy of $I_{G^{\prime}}(k)$ if and only if $L(\lambda)$
is a projective $G$-module.

###### Proof.

For the backward direction, if $L(\lambda)$ is projective then
$L(\lambda)^{*}\boxtimes L(\lambda)$ is too, thus its restriction to
$G^{\prime}$ must be projective since $G$ is quasireductive. By 9.4,
$L(\lambda)^{*}\boxtimes L(\lambda)$ contains a $G^{\prime}$-invariant, and
thus it must contain a copy of $I_{G^{\prime}}(k)$.

Conversely, suppose that $L(\lambda)^{*}\boxtimes L(\lambda)$ contains a copy
of $I_{G^{\prime}}(k)$. Since $L(\lambda)^{*}\boxtimes L(\lambda)$ is
irreducible, we may apply 8.18 to obtain that $I_{G\times
G}(L(\lambda)^{*}\boxtimes L(\lambda))$ is a $G\times G$-submodule of
$k[G\times G/G^{\prime}]$. Using the isomorphism $k[G\times
G/G^{\prime}]^{\operatorname{id}\times\delta}\to k[G\times G/G]$, we obtain
that $I_{G\times G}(L(\lambda)^{*}\boxtimes L(\lambda))$ is a submodule
$k[G]$.

Since $I_{G\times G}(L(\lambda)^{*}\boxtimes L(\lambda))$ is projective, it
splits off $k[G]$ as a $G\times G$-module. By 9.2, we find that $L(\lambda)$
must be in its own block, and is therefore projective. ∎

From this, we obtain the following remarkable result. We write

$p_{G,B}:=p_{G\times
G/G,B}=HC(\operatorname{ev}_{eG}v_{\mathfrak{g}^{\prime}}).$

######  Theorem 9.11.

For a $\mathfrak{b}$-dominant weight $\lambda$, the simple $G$-module
$L(\lambda)$ is projective if and only if $p_{G,B}(\lambda)\neq 0$. In
particular, the dominant weights $\lambda$ for which $L(\lambda)$ is
projective is given by the intersection of $\Lambda^{+}$ with a Zariski open
subset of $\mathfrak{h}^{*}$ given by the non-vanishing set of the polynomial
$p_{G,B}$. In particular if $\operatorname{dim}\mathfrak{g}_{\overline{1}}$ is
odd or $\operatorname{Ber}(\mathfrak{g}_{\overline{1}})$ is not the trivial
$\mathfrak{g}$-module then no irreducible $G$-modules are projective.

###### Proof.

The first statement follows from 9.10 and 9.9. The last statement follows from
8.13. Note that the last statement could be proven more easily using the
result in [Ser11] that $I_{G}(L)\cong
P_{G}(L\otimes\operatorname{Ber}(\mathfrak{g}_{\overline{1}}))$. ∎

In the appendix, we will prove 9.11 in full generality, i.e. without the
assumption the $G$ is Cartan-even.

######  Corollary 9.12.

If $HC(\mathcal{Z})=k$ (e.g. if $\mathcal{Z}=k$) and $\mathfrak{g}\neq 0$,
then no finite-dimensional irreducible $G$-modules are projective.

###### Proof.

By 9.11, if there exists a finite-dimensional irreducible projective
$G$-module, then $p_{G,B}\in S(\mathfrak{h})$ is a nonzero polynomial. By
9.11, $\operatorname{Ber}(\mathfrak{g})$ must be trivial and thus
$\mathcal{A}_{G\times G/G}=\mathcal{A}$. Thus we know from [Gor00] that
$p_{G,B}^{2}\in HC(\mathcal{Z})=k$, and is nonzero, so $p_{G,B}$ must be a
constant polynomial. Therefore $\operatorname{Rep}(G)$ is semisimple, and thus
$HC(\mathcal{Z})$ must be larger than constants by the classification of
algebraic supergroups with semisimple representation theory (see [She20b]). ∎

######  Corollary 9.13.

If there are no finite-dimensional irreducible projective $G$-modules, then
$HC(\mathcal{A}_{G/K})=0$.

###### Proof.

If for $\lambda\in\Lambda^{+}$ we have $HC(\gamma)(\lambda)\neq 0$, then
$L(\lambda)^{*}\boxtimes L(\lambda)$ contains a copy of $I_{G^{\prime}}(k)$,
so that $L(\lambda)$ is projective, a contradiction. ∎

We now can prove a nice sufficient criteria for when $G$ admits projective
irreducible modules.

######  Theorem 9.14.

Suppose that $G$ is a Cartan-even quasireductive supergroup such that the
following conditions on its Lie superalgebra, $\mathfrak{g}$, hold:

  1. (1)

$\operatorname{dim}[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]=1$ for all
$\alpha\in\Delta$; and

  2. (2)

for all $\alpha\in\Delta$, the pairing

$[-,-]:\mathfrak{g}_{\alpha}\otimes\mathfrak{g}_{-\alpha}\to[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]$

is nondegenerate.

Make a choice of positive roots and thus a Borel subalgebra $\mathfrak{b}$
containing $\mathfrak{h}$, and let
$\Delta_{\overline{1}}^{+}=\\{\alpha_{1},\dots,\alpha_{n}\\}$. Let
$h_{\alpha_{i}}\in[\mathfrak{g}_{\alpha_{i}},\mathfrak{g}_{-\alpha_{i}}]$ be a
chosen nonzero element, and write
$r_{i}=\operatorname{dim}\mathfrak{g}_{\alpha_{i}}$. Then up to a scalar we
have:

$p_{G,B}=h_{\alpha_{1}}^{r_{1}}\cdots h_{\alpha_{n}}^{r_{n}}+l.o.t.$

In particular $p_{G,B}\neq 0$, so that $G$ admits projective irreducible
modules.

###### Proof.

We may choose a weight basis of $\mathfrak{g}_{\overline{1}}$ as follows. Let
$u_{0},\dots,u_{m}$ be a root vector basis of $\mathfrak{n}_{\overline{1}}$,
where $u_{i}$ has weight $\beta_{i}$, and the $\beta_{i}$’s need not be
distinct. Then there exists a root vector basis $v_{0},\dots,v_{m}$ of
$\mathfrak{n}^{-}_{\overline{1}}$ such that $v_{i}$ has weight $-\beta_{i}$
and $[u_{i},v_{j}]=\delta_{ij}h_{\beta_{i}}$ whenever $\beta_{i}=\beta_{j}$.
Write $u_{I}$ and $v_{I}$ for the corresponding ordered products when
$I\subseteq\\{0,\dots,m\\}$, and set $u_{\emptyset}=v_{\emptyset}=1$. For a
subset $J\subseteq\\{0,\dots,m\\}$ with $J=\\{j_{1}<\dots<j_{l}\\}$ we set

$\widetilde{v_{J}}=v_{j_{l}}\cdots v_{j_{1}}.$

Then we may take as a basis for
$\mathcal{U}\mathfrak{g}/\mathcal{U}\mathfrak{g}\mathfrak{g}_{\overline{0}}$
the elements $u_{I}\widetilde{v_{J}}$ for $I,J\subseteq\\{0,\dots,m\\}$.
Therefore we may write

$v_{\mathfrak{g}}=u_{0}\cdots u_{m}v_{m}\cdots v_{0}+v^{\prime},$

where
$v^{\prime}\in\mathcal{F}^{\operatorname{dim}\mathfrak{g}_{\overline{1}}-2}$.
Thus

$\operatorname{ad}^{\prime}(v_{\mathfrak{g}})(1)=\operatorname{ad}^{\prime}(u_{0}\cdots
u_{m}v_{m}\cdots v_{0})(1)+\operatorname{ad}^{\prime}(v^{\prime})(1).$

By lemma 8.8 $HC(\operatorname{ad}^{\prime}(v^{\prime})(1))\leq m$, so it
suffices to show that (up to a scalar)

$\operatorname{ad}^{\prime}(u_{0}\cdots u_{m}v_{m}\cdots
v_{0})(1)=h_{\beta_{0}}\cdots h_{\beta_{m}}+l.o.t.$

Observe that

$\operatorname{ad}^{\prime}(v_{n}\cdots
v_{0})(1)=\sum\limits_{I\subseteq\\{1,\dots,m\\}}(-1)^{i_{1}+\dots+i_{j}}\widetilde{v_{I^{c}}}v_{I},$

(note the $\\{1,\dots,m\\}$ in the sum is not a typo) where
$I=\\{i_{1},\dots,i_{j}\\}$ and $I^{c}$ denotes the complement of $I$. Now if
we apply $\operatorname{ad}^{\prime}(u_{0}\cdots u_{m})$ to
$\operatorname{ad}^{\prime}(v_{m}\cdots v_{0})(1)$, the only term that does
not vanish under the Harish-Chandra morphism is

$\sum\limits_{I\subseteq\\{1,\dots,m\\}}(-1)^{i_{1}+\dots+i_{j}}u_{0}\cdots
u_{m}\widetilde{v_{I^{c}}}v_{I}.$

Now it suffices to show that

$HC((-1)^{i_{1}+\dots+i_{j}}u_{0}\cdots
u_{m}\widetilde{v_{I^{c}}}v_{I})=h_{\beta_{0}}\cdots h_{\beta_{m}}+l.o.t.$

The proof of this is inductive. If $m\notin I$, then we may write

$u_{0}\cdots u_{m}\widetilde{v_{I^{c}}}v_{I}=u_{0}\cdots
u_{m}v_{m}\widetilde{v_{I^{c}\setminus\\{m\\}}}v_{I}.$

This is equal to (after removing the term in $v_{m}\mathcal{U}\mathfrak{g}$)

$\displaystyle\sum\limits_{j}(-1)^{m-j}u_{0}\cdots[u_{j},v_{m}]\cdots
u_{m}\widetilde{v_{I^{c}\setminus\\{n\\}}}v_{I\setminus\\{m\\}}$
$\displaystyle+$ $\displaystyle u_{0}\cdots
u_{m-1}\widetilde{v_{I^{c}\setminus\\{m\\}}}v_{I}h_{\beta_{m}}$
$\displaystyle+$ $\displaystyle cu_{0}\cdots
u_{m-1}\widetilde{v_{I^{c}\setminus\\{m\\}}}v_{I}$

In the last summand $c$ is the constant given by
$(-\beta_{0}-\dots-\beta_{m-1})(h_{\beta_{m}})$, but its precise value does
not matter as this term will have Harish-Chandra projection of degree $m$ or
less. For the first summand, let
$e_{j,m}=[u_{j},v_{m}]\in\mathfrak{g}_{\overline{0}}$, and suppose that
$e_{j,m}$ lies in $\mathfrak{n}^{-}$ (note that it cannot lie in
$\mathfrak{h}$ by our choice of basis). The argument for when it lies in
$\mathfrak{n}^{+}$ is entirely analogous. Then we have (after removing the
term in $e_{j,m}\mathcal{U}\mathfrak{g}$.)

$u_{0}\cdots e_{j,m}\cdots
u_{m}\widetilde{v_{I^{c}\setminus\\{m\\}}}v_{I}=\sum\limits_{l}u_{0}\cdots[u_{l},e_{j,m}]\cdots\hat{u_{j}}\cdots
u_{m}\widetilde{v_{I^{c}\setminus\\{m\\}}}v_{I}$

Thus we end up with an element in $\mathcal{F}^{2m}$ so that its Harish-
Chandra projection is of degree at most $m$, and so does not contribute to the
top degree.

Now suppose instead $m\in I$, so that we can write

$u_{0}\cdots u_{m}\widetilde{v_{I^{c}}}v_{I}=u_{0}\cdots
u_{m}\widetilde{v_{I^{c}}}v_{I\setminus\\{m\\}}v_{m}.$

Write $I=\\{i_{1},\dots,i_{s},m\\}$ and $I^{c}=\\{j_{1},\dots,j_{t}\\}$. Then
the above expression is equal to

$\displaystyle(-1)^{n}u_{0}\cdots
u_{m-1}\widetilde{v_{I^{c}}}v_{I\setminus\\{m\\}}h_{\beta_{m}}$
$\displaystyle+$ $\displaystyle\sum\limits_{p}u_{0}\cdots
u_{m-1}v_{j_{t}}\cdots[u_{m},v_{j_{p}}]\cdots v_{j_{t}}v_{I}$ $\displaystyle+$
$\displaystyle\sum\limits_{q}u_{0}\cdots
u_{m-1}\widetilde{v_{I^{c}}}v_{i_{1}}\cdots[u_{m},v_{i_{q}}]\cdots v_{n}.$

Now in the last two summands we may apply a similar argument as above to prove
that their Harish-Chandra projections are of degree less than or equal to $m$.
Therefore, we have shown that the following two elements have the same top
degree term:

$HC((-1)^{i_{1}+\dots+i_{j}}u_{0}\cdots u_{m}\widetilde{v_{I^{c}}}v_{I}),\ \ \
\ HC((-1)^{j_{1}+\cdot+j_{l}}u_{0}\cdots
u_{m-1}\widetilde{v_{J^{c}}}v_{J})h_{\alpha_{m}},$

where $\\{j_{1},\dots,j_{l}\\}=J=I\cap\\{0,\dots,m-1\\}$ and the complement of
$J$ is taken in $\\{0,\dots,m-1\\}$. Now we may apply an induction argument to
show that

$HC((-1)^{i_{1}+\dots+i_{j}}u_{0}\cdots
u_{m}\widetilde{v_{I^{c}}}v_{I})=h_{\beta_{0}}\cdots h_{\beta_{m}}+l.o.t.$

which completes the proof. ∎

We say a Lie superalgebra is quadratic if it admits a nondegenerate,
invariant, and even supersymmetric form $(-,-)$. The following is
straightforward to show from the properties of being quadratic.

######  Corollary 9.15.

Let $G$ be a Cartan-even quasireductive supergroup whose Lie superalgebra is
quadratic. Then the conditions of 9.14 hold. In particular $G$ admits
projective irreducible modules.

###  9.3. $\mathfrak{g}$ basic classical

We now additionally assume that $\mathfrak{g}$ is almost simple and basic
classical, by which we mean it is either basic simple,
$\mathfrak{s}\mathfrak{l}(n|n)$, or $\mathfrak{g}\mathfrak{l}(m|n)$. The
following two results could be proven more easily using 9.11 and facts about
the representation theory of basic Lie superalgebras, but we present the
proofs below as they demonstrate a different approach which generalizes to
other supersymmetric pairs.

######  Lemma 9.16.

Suppose that $\alpha\in\Delta^{+}$ is a simple isotropic root, and that we
have $(\lambda,\alpha)=0$. Then $p_{G,B}(\lambda)=0$.

###### Proof.

Let $u_{\alpha}=(e_{\alpha},0)-(0,e_{\alpha})\in\mathfrak{g}^{\prime}$. Then
$u_{\alpha}f_{\lambda}=0$ because $\alpha$ is simple isotropic and
$(\lambda,\alpha)=0$.

Now suppose that $\mathcal{U}\mathfrak{g}^{\prime}f_{\lambda}$ contained a
copy of $I_{G^{\prime}}(k)$, so that we may write
$\mathcal{U}\mathfrak{g}^{\prime}f_{\lambda}=I(k)\oplus M$, for some
complimentary module $M$. Then with respect the this direct sum decomposition