# Analogs of Bol operators on superstrings

Sofiane Bouarroudja, Dimitry Leitesa,b,∗, Irina Shchepochkinac
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>aDivision of Science and Mathematics
New York University Abu Dhabi
Po Box 129188, United Arab Emirates

bDepartment of Mathematics
Stockholm University, Stockholm, Sweden

cIndependent University of Moscow, B. Vlasievsky per., d. 11, RU-119 002
Moscow, Russia

∗ The corresponding author

###### Abstract.

The Bol operators are unary differential operators between spaces of weighted
densities on the 1-dimensional manifold invariant under projective
transformations of the manifold. On the $1|n$-dimensional supermanifolds
(superstrings), we classify analogs of Bol operators invariant under the
simple maximal subalgebra $\mathfrak{h}$ of the same rank as its simple
ambient superalgebra $\mathfrak{g}$ of vector fields and containing all
elements of negative degree of $\mathfrak{g}$ in a $\mathbb{Z}$-grading. We
also consider the Lie superalgebras of vector fields $\mathfrak{g}$ preserving
a contact structure on the superstring. We have discovered many new operators.

###### Key words and phrases:

Lie superalgebra, invariant differential operator, Veblen’s problem,
superstring, Bol operator

###### 2010 Mathematics Subject Classification:

Primary 17B10 Secondary 53B99, 32Wxx

##  1\. Introduction

###  1.1. Veblen’s problem ([GLS, BL, BouL])

On a given $m$-dimensional manifold ${\mathcal{M}}$, let $T_{U}(V)$ or just
$T(V)$ be the space of $V$-valued tensor fields over a domain
$U\subset{\mathcal{M}}$. Let
$\rho:\mathop{\mathrm{{GL}}}\nolimits(m)\longrightarrow\mathop{\mathrm{{GL}}}\nolimits(V)$
be a representation. Let $J_{A}$ be the Jacobi matrix of the diffeomorphism
$A$ computed in coordinates at points $X\in{\mathcal{M}}$ and
$A^{-1}(X)\in{\mathcal{M}}$. The group
$\mathop{\mathrm{Diff}}\nolimits{\mathcal{M}}$ of diffeomorphisms of
${\mathcal{M}}$ acts on $T(V)$ by invertible changes of coordinates as
follows:

$A(t)(X):=\rho(J_{A})(t(A^{-1}(X))\text{~{}~{}for any
$A\in\mathop{\mathrm{Diff}}\nolimits{\mathcal{M}}$, $X\in{\mathcal{M}}$, $t\in
T(V)$}.$

In 1928, at the IMC, O. Veblen formulated a problem (over ${\mathbb{R}}$)
which A. Kirillov later reformulated in more comprehensible terms of modern
differential geometry as follows:

(1) “describe operators $D:T(V_{1})\otimes\cdots\otimes T(V_{k})\longmapsto
T(W)$”,

invariant under all invertible diffeomorphisms of ${\mathcal{M}}$.

For differential operators, J. Bernstein [BL] further reformulated Veblen’s
problem as a purely algebraic problem meaningful over any ground field
${\mathbb{K}}$. Thus interpreted, Rudakov’s solution of Veblen’s problem is
reviewed in [GLS] for several types of unary ($k=1$) operators: invariant
under arbitrary diffeomorphisms, or under diffeomorphisms preserving either a
volume element, or a symplectic form, or a contact structure. Being purely
algebraic and local (hence, ${\mathcal{M}}$ is, actually, any open domain)
Veblen’s problem can be reformulated as follows and its generalization for
supermanifolds and passage to ground fields of characteristic $p>0$ are
immediate. The answers, however, are not that easy to obtain, and there are
many new operators to be discovered: for unary operators in the super case
over ${\mathbb{C}}$, see a review paper [GLS]; for modular cases and
$\dim{\mathcal{M}}=1$, see [BouL]. (In presence of even indeterminates on a
supermanifold, the integral is an example of non-differential invariant
operator from the space of densities with compact support to the space of
functions; for just one more known example of non-differential invariant
operator, first discovered by A.A. Kirillov only by its symbol, see [IM].)

Let the ground field be ${\mathbb{C}}$. Let ${\mathcal{F}}$ be the algebra of
functions (which in our case are polynomials or formal series). Denote the Lie
algebra of vector fields by
${\mathcal{L}}:={\mathfrak{vect}}(m)={\mathfrak{der}}\ {\mathcal{F}}$.

Let ${\mathcal{L}}_{0}$ be the maximal subalgebra of finite codimension (for
${\mathfrak{vect}}(m|0)$ it consists of the vector fields that vanish at the
origin). Define the Weisfeiler filtration

${\mathcal{L}}={\mathcal{L}}_{-d}\supset\dots\mathcal{L}_{-1}\supset{\mathcal{L}}_{0}\supset{\mathcal{L}}_{1}\supset\dots$

by letting ${\mathcal{L}}_{-1}$ be a minimal ${\mathcal{L}}_{0}$-invariant
subspace strictly containing ${\mathcal{L}}_{0}$; the other terms being
defined by the following formula, where $i\geq 1$

(2)
${\mathcal{L}}_{-i-1}=[{\mathcal{L}}_{-1},{\mathcal{L}}_{-i}]+{\mathcal{L}}_{-i}\;\text{
and
}\;{\mathcal{L}}_{i}=\\{D\in{\mathcal{L}}_{i-1}\mid[D,{\mathcal{L}}_{-1}]\subset{\mathcal{L}}_{i-1}\\}.$

Let $V$ be an ${\mathcal{L}}_{0}$-module, such that ${\mathcal{L}}_{1}V=0$,
i.e., $V$ is, actually,
${\mathfrak{gl}}(m)\simeq{\mathcal{L}}_{0}/{\mathcal{L}}_{1}$-module
considered as an ${\mathcal{L}}_{0}$-module. The space of tensor fields of
type $V$ with formal coefficients is defined to be

$T(V):=\text{Hom}_{U({\mathcal{L}}_{0})}(U({\mathcal{L}}),V)\simeq{\mathbb{C}}[[x]]\otimes
V.$

To consider irreducible ${\mathcal{L}}_{0}/{\mathcal{L}}_{1}$-modules $V_{i}$
and $W$, see (1), is a natural first step, especially in the super situation
and over fields of positive characteristics. Obviously, it is natural to
replace ${\mathfrak{g}}={\mathfrak{vect}}(m)$ with any other simple Lie
(super)algebra ${\mathcal{L}}$ of vector fields with polynomial (or formal if
we need closedness with respect to the $x$-adic topology) coefficients; for
the list of known such Lie superalgebras, see [BGLLS]. It is also natural to
consider operators (1) invariant under the action of a maximal subalgebra
${\mathfrak{h}}$ of ${\mathfrak{g}}$.

Here we solve Veblen’s problem for finite-dimensional maximal simple
${\mathbb{Z}}$-graded subalgebras ${\mathfrak{h}}$ of the vectorial Lie
superalgebra ${\mathfrak{g}}={\mathfrak{vect}}(m|n)$ or
${\mathfrak{k}}(2m+1|n)$ or ${\mathfrak{m}}(n)$ in the standard grading of
${\mathfrak{g}}$ and such that
${\mathfrak{h}}_{-}={\mathfrak{g}}_{-}:=\oplus_{i<0}{\mathfrak{g}}_{i}$ and
the maximal tori of ${\mathfrak{h}}$ and ${\mathfrak{g}}$ coincide.

(3) The adjective simple applied to ${\mathfrak{h}}$ and ${\mathfrak{g}}$ in
this problem is not natural; we should (and do) consider various “relatives”
of simple algebras: their central extensions (not only nontrivial) and
algebras of derivations.

###  1.2. Dualization and permutation of arguments

For tensor fields with compact support, we have

(4)
$(T(W))^{*}:=T(W^{*})\otimes_{\mathcal{F}}\mathop{\mathrm{{Vol\/}}}\nolimits.$

Let us define the dual spaces of tensor fields for the fields with polynomial
or formal coefficients via formula (4) as if the fields were with compact
support. For any invariant operator ${D:T(V)\longrightarrow T(W)}$, the dual
operator

$D^{*}:(T(W))^{*}:=T(W^{*})\otimes_{\mathcal{F}}\mathop{\mathrm{{Vol\/}}}\nolimits\longrightarrow(T(V))^{*}:=T(V^{*})\otimes_{\mathcal{F}}\mathop{\mathrm{{Vol\/}}}\nolimits$

is also invariant, and since
$\mathop{\mathrm{{Vol\/}}}\nolimits:=T(\mathop{\mathrm{tr}}\nolimits)$, where
the (super)trace $\mathop{\mathrm{tr}}\nolimits$ denotes both the map
$A\longrightarrow\mathop{\mathrm{tr}}\nolimits(A)$ for any
$A\in{\mathfrak{gl}}(m|n)$ and the ${\mathfrak{gl}}(m|n)$-module it defines,
we can describe $D^{*}$ as

(5) $D^{*}:T(W^{*}\otimes\mathop{\mathrm{tr}}\nolimits)\longrightarrow
T(V^{*}\otimes\mathop{\mathrm{tr}}\nolimits).$

In particular, for the spaces of weighted densities, see Subsection 2.3, let
$D:{\mathcal{F}}_{a}\longrightarrow{\mathcal{F}}_{b}$ be an invariant
operator, and $\mathop{\mathrm{{Vol\/}}}\nolimits={\mathcal{F}}_{v}$, where
$v$ depends on $\mathop{\mathrm{sdim}}\nolimits{\mathcal{M}}=m|n$. Then,
${D^{*}:{\mathcal{F}}_{v-b}\longrightarrow{\mathcal{F}}_{v-a}}$ must be (and
is) invariant. This helps to double-check the answers, see [Gr1, Gr2, BGL,
BouL].

###  1.3. Feasible versions of Veblen’s problem over ${\mathbb{C}}$

Let $k$ be the arity of the operator (1). For $k=1$, Veblen’s problem was
solved for several simple Lie superalgebras, see a review [GLS].

For $k=2$, Grozman solved Veblen’s problem for ${\mathfrak{vect}}(m)$, its
divergence-free subalgebra ${\mathfrak{svect}}(m)$, and partly for the Lie
algebra of Hamiltonian vector fields, see [Gr1, Gr2].

The super versions of Veblen’s problem for $k=2$ are open for any Lie
superalgebra ${\mathcal{L}}$, apart from some occasional results for contact
Lie superalgebras ${\mathfrak{k}}(2m+1|n)$ and ${\mathfrak{m}}(n)$ for small
values of $k$ and $m|n$, see [KLV, Lr].

For $k\geq 3$, Veblen’s problem is out of reach in general, though it is
conceivable to find the operator $D$, see (1), of any prescribed order for any
${\mathcal{L}}$ and any arity $k$ for any given $V_{i}$ and $W$, where “any”
is understood “within reason” depending on computers available.

There are, however, cases where one can get the complete answer even for $k$
“large”, as, e.g., in [Bter, B06, Bc]. These are cases where (A) the spaces
$T(V_{i})$ and $T(W)$ are weighted densities, see §2, i.e., rank 1 modules
over the algebra ${\mathcal{F}}$ of functions, and (B) the operators $D$ are
anti-symmetric, as in [FF].

There is also an exceptional case where the simple Lie superalgebra of vector
fields ${\mathcal{L}}$ is such that ${\mathcal{L}}_{0}/{\mathcal{L}}_{1}$ is
solvable (and hence the irreducible
${\mathcal{L}}_{0}/{\mathcal{L}}_{1}$-modules are finite-dimensional). Over
${\mathbb{C}}$, this only happens if
${\mathcal{L}}\simeq{\mathfrak{vect}}(1|1)$.

###  1.4. The problems we consider over ${\mathbb{C}}$

In various problems of mathematical physics, it is natural to consider a
maximal simple subalgebra ${\mathfrak{s}}$ of a simple vectorial Lie algebra
${\mathfrak{g}}$; for their classification, see [LSh], where “simple” is
understood in the above-mentioned “weak” sense, see (3). Let
$i:{\mathfrak{s}}\longrightarrow{\mathfrak{g}}$ be the corresponding
embedding. Consider examples.

In ${\mathfrak{k}}(1|0)$, there is only one simple maximal subalgebra:
${\mathfrak{sl}}(2)$.

######  Theorem (Bol’s theorem [Bol]).

Let $m|n=1|0$.

_(i)_ For every $k\in{\mathbb{N}}:=\\{1,2,\dots\\}$, there exists a unique
$i(\mathfrak{sl}(2))$-invariant order $k$ linear differential operator
$A^{k}:{\mathcal{F}}_{(1-k)/2}\longrightarrow{\mathcal{F}}_{(1+k)/2}$;
explicitly

$A^{k}(\varphi(dt)^{(1-k)/2})=\varphi^{(k)}(dt)^{(1+k)/2}.$

_(ii)_ If $(\lambda,\mu)\not=\frac{1}{2}(1-k,1+k)$, then there are no nonzero
$i(\mathfrak{sl}(2))$-invariant linear differential operators
${\mathcal{F}}_{\lambda}\longrightarrow{\mathcal{F}}_{\mu}$.

These operators $A^{k}$ are called the Bol operators; they were first
described in [Bol]. For some of their applications, see [BO] and [OT, p.9].
Considering the spaces of polynomial weighted densities on 1-dimensional
manifold as Verma modules with lowest weight vector of weight
$\mu\in{\mathbb{C}}$, the description of Bol operators is a trivial corollary
of the classification of irreducible ${\mathfrak{sl}}(2)$-modules (if
$\mu\in{\mathbb{C}}\setminus{\mathbb{Z}}_{\leq 0}$) and their finite-
dimensional submodules (if $\mu\in{\mathbb{Z}}_{\leq 0}$), see any
classification of irreducible ${\mathfrak{sl}}(2)$-modules with lowest or
highest weight vector in terms of Verma modules, e.g., [Ber, Kir].

######  Lemma ([LSh]).

The maximal simple subalgebras ${\mathfrak{s}}$ of the simple vectorial Lie
superalgebra ${\mathfrak{g}}$ such that
$\mathop{\mathrm{rk}}\nolimits{\mathfrak{s}}=\mathop{\mathrm{rk}}\nolimits{\mathfrak{g}}$
are as follows

(6) ${\mathfrak{osp}}(n|2m+2)\subset{\mathfrak{k}}(2m+1|n)$
---
${\mathfrak{pgl}}(m|n+2)\simeq{\mathfrak{pgl}}(n+2|m)\subset{\mathfrak{k}}(2n+1|2m)$
${\mathfrak{osp}}_{a}(4|2)\subset{\mathfrak{k}}(1|4)$
${\mathfrak{as}}\subset{\mathfrak{k}}{\mathfrak{as}}\subset{\mathfrak{k}}(1|6)$
${\mathfrak{pgl}}(m+1|n)\simeq{\mathfrak{pgl}}(n|m+1)\subset{\mathfrak{vect}}(m|n)$
for $(m,n)\neq(0,1),(0,2)$
${\mathfrak{pe}}(n+1)\subset{\mathfrak{m}}(n)$
${\mathfrak{pgl}}(m+1|n+1)\simeq{\mathfrak{pgl}}(n+1|m+1)\subset{\mathfrak{m}}(n+m)$
for $m^{2}+n^{2}>0$
${\mathfrak{spe}}_{a,b}(n+1)\subset{\mathfrak{b}}_{a,b}(n)$

Note that in Table (6) we always have projective embeddings (modulo center),
but since ${\mathfrak{pgl}}(a|b)\simeq{\mathfrak{sl}}(a|b)$ if $a\neq b$, we
sometimes write ${\mathfrak{sl}}$ instead of ${\mathfrak{pgl}}$ for clarity of
matrix realization.

We leave the cases where
$\mathop{\mathrm{rk}}\nolimits{\mathfrak{s}}<\mathop{\mathrm{rk}}\nolimits{\mathfrak{g}}$,
also classified in [LSh], for the future.

From Table (6) we see that there are three major types of embeddings:

type ${\mathfrak{k}}$: Embeddings into Lie superalgebras of contact type
${\mathfrak{k}}$;

type ${\mathfrak{m}}$: Embeddings into Lie superalgebras of contact type
${\mathfrak{m}}(n)$ and its subalgebra ${\mathfrak{b}}_{a,b}(n)$; according to
Convention (7), we consider only $n=1$ and do not consider the case of non-
simple ambient algebra ${\mathfrak{b}}_{a,b}(1)$;

type ${\mathfrak{vect}}$: Embeddings into Lie superalgebras of general type
${\mathfrak{vect}}$. They are considered in a separate paper.

(7) In the case of embeddings into ${\mathfrak{k}}(2m+1|n)$, the case of $m=0$
is of particular interest: the supermanifold, on which we consider the
problem, is what physicists call a superstring. Even in this case, we consider
analogs of Bol operators only between spaces of weighted densities: as we will
show elsewhere, the case of operators between spaces of tensor fields on
supermanifolds other than superstrings is out of reach. We leave the case of
the exceptional embedding
${{\mathfrak{as}}\subset{\mathfrak{k}}{\mathfrak{as}}\subset{\mathfrak{k}}(1|6)}$
for the future.

Let us give here an extremely beautiful formula (8) describing invariant
operators in the exceptional case
${\mathfrak{osp}}_{a}(4|2)\subset{\mathfrak{k}}(1|4)$:

######  Theorem (Bol’s analogs of $i({\mathfrak{osp}}_{a}(4|2))$-invariant
operators).

Let $m|n=0|4$.

_(i)_ Let $a\neq-1,\infty$. The only $i({\mathfrak{osp}}_{a}(4|2))$-invariant
differential operators are

$A^{k}:{\mathcal{F}}_{-(k+3)}\longrightarrow{\mathcal{F}}_{k+1}\text{~{}~{}given
by~{}~{}}\varphi\alpha_{1}^{-(k+3)/{2}}\mapsto\bar{A}(\varphi)\alpha_{1}^{(k+1)/{2}}\text{for
any $k\geq 0$},$

where (for the definition of the $D_{f}$ and super anti-commutator, see eqs.
(35) and (34))

(8)
$\bar{A}(\varphi)=[D_{\eta_{1}},D_{\xi_{1}}]_{+}\circ[D_{\eta_{2}},D_{\xi_{2}}]_{+}\circ\partial_{t}^{k}(\varphi)+\mathchoice{\raisebox{0.25pt}{$\dfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{0.25pt}{$\dfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{-0.5pt}{$\tfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{-0.5pt}{$\tfrac{4(1-a)}{(k+2)(1+a)}$}}\partial_{t}^{k+2}(\varphi).$

_(ii)_ The only $i({\mathfrak{osp}}_{-1}(4|2))$-invariant differential
operators are

$A^{k}:{\mathcal{F}}_{\lambda}\longrightarrow{\mathcal{F}}_{\lambda+2k}\text{~{}~{}given
by~{}~{}}\bar{A}:\varphi\alpha_{1}^{\lambda/{2}}\mapsto\partial_{t}^{k}(\varphi)\alpha_{1}^{{\lambda}/{2}+k}\text{for
any $k\geq 0$}.$

_(iii)_ The only $i({\mathfrak{osp}}_{\infty}(4|2))$-invariant differential
operators are

$A^{k}:{\mathcal{F}}_{-(k+3)}\longrightarrow{\mathcal{F}}_{k+1}\text{~{}~{}given
by~{}~{}}\varphi\alpha_{1}^{-(k+3)/{2}}\mapsto\bar{A}(\varphi)\alpha_{1}^{(k+1)/{2}}\text{for
any $k\geq 0$},$

where (the coefficient of $\partial_{t}^{k+2}$ is
$\mathop{\lim}\limits_{a\longrightarrow\infty}\mathchoice{\raisebox{0.25pt}{$\dfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{0.25pt}{$\dfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{-0.5pt}{$\tfrac{4(1-a)}{(k+2)(1+a)}$}}{\raisebox{-0.5pt}{$\tfrac{4(1-a)}{(k+2)(1+a)}$}}$,
see eq. (8))

$\bar{A}(\varphi)=[D_{\eta_{1}},D_{\xi_{1}}]_{+}\circ[D_{\eta_{2}},D_{\xi_{2}}]_{+}\circ\partial_{t}^{k}(\varphi)-\mathchoice{\raisebox{0.25pt}{$\dfrac{4}{(k+2)}$}}{\raisebox{0.25pt}{$\dfrac{4}{(k+2)}$}}{\raisebox{-0.5pt}{$\tfrac{4}{(k+2)}$}}{\raisebox{-0.5pt}{$\tfrac{4}{(k+2)}$}}\partial_{t}^{k+2}(\varphi).$

###  1.5. Comments

In ${\mathfrak{k}}(1|1)$, there is only one simple maximal subalgebra:
${\mathfrak{osp}}(1|2)$, see [G, GT], the result is quoted in Theorem 3.3.
Theorem.

In ${\mathfrak{k}}(1|2)\simeq{\mathfrak{m}}(1)$, there is one simple maximal
subalgebra:
${\mathfrak{osp}}(2|2)\simeq{\mathfrak{sl}}(1|2)\simeq{\mathfrak{sl}}(2|1)$
corresponding to the first 2 lines in Table (6), and one maximal subalgebra
“close to simple”: ${\mathfrak{pe}}(2)\subset{\mathfrak{m}}(1)$ corresponding
to line 6 in Table (6).

In order to explain where this subalgebra ${\mathfrak{pe}}(2)$ comes from,
recall “occasional” isomorphisms, see the bottom line in eq. (9) where the Lie
superalgebras are considered as abstract, whereas as filtered or
${\mathbb{Z}}$-graded they are not isomorphic:

(9)
$\begin{array}[]{l}{\mathfrak{osp}}(2|2)\simeq{\mathfrak{sl}}(1|2)\simeq{\mathfrak{vect}}(0|2),\\\
{\mathfrak{k}}(1|2)\simeq{\mathfrak{vect}}(1|1)\simeq{\mathfrak{m}}(1).\\\
\end{array}$

Considering the Lie superalgebras in the second line of eq. (9) not as
abstract, which is unreasonable in Veblen’s problem, but as filtered, we see
that the $({\mathcal{F}},{\mathfrak{g}})$-bimodules of rank 1, where
${\mathcal{F}}$ is the algebra of functions on ${\mathcal{M}}$ and
${\mathfrak{g}}$ is a Lie superalgebra on ${\mathcal{M}}$, are quite
different: over ${\mathfrak{k}}(1|2)$ and ${\mathfrak{m}}(1)$ we have
2-parameter modules ${\mathcal{F}}_{a,b}$, whereas the modules of weighted
densities over ${\mathfrak{vect}}(1|1)$ can only depend on 1 parameter.

Accordingly, we replace ${\mathfrak{sl}}(2)$ with the Lie superalgebra
${\mathfrak{h}}:={\mathfrak{osp}}(2|2)$, ${\mathfrak{sl}}(1|2)$ and
${\mathfrak{pe}}(2)$, respectively, considered as ${\mathbb{Z}}$-graded, and
embedded in the ${\mathbb{Z}}$-graded ambient algebra ${\mathfrak{g}}$ so that
${\mathbb{Z}}$-grading is preserved; i.e.,
$i({\mathfrak{h}}_{-})={\mathfrak{g}}_{-}$. This forces us to consider
${\mathfrak{pe}}(2)$ as a subalgebra of ${\mathfrak{m}}(1)$.

##  2\. Weighted densities

The ${\mathfrak{vect}}(m|n)$-module $\mathop{\mathrm{{Vol\/}}}\nolimits^{a}$
of weighted densities of weight $a\in{\mathbb{K}}$ is a rank $1$ module over
the algebra of functions ${\mathcal{F}}$ in indeterminates
$x:=(u_{1},\dots,u_{m};\xi_{1},\dots,\xi_{n})$ generated by
$\mathop{\mathrm{{vol\/}}}\nolimits^{a}$, where
$\mathop{\mathrm{{vol\/}}}\nolimits$ is the volume element with constant
coefficients. This $\mathop{\mathrm{{vol\/}}}\nolimits$ is the class of the
element $du_{1}\wedge\dots\wedge
du_{m}\partial_{\xi_{1}}\dots\partial_{\xi_{n}}$ in the quotient of

$E^{m}_{\mathcal{F}}(({\mathfrak{vect}}(m|n))^{*})\otimes_{\mathcal{F}}S^{n}_{\mathcal{F}}({\mathfrak{vect}}(m|n))$

— the tensor product over ${\mathcal{F}}$ of the $m$th exterior power of the
${\mathcal{F}}$-module of covector fields by the $n$th symmetric power of the
${\mathcal{F}}$-module of vector fields — modulo the maximal
${\mathfrak{vect}}(m|n)$-submodule of corank 1. The
${\mathfrak{vect}}(m|n)$-action is given for any $D\in{\mathfrak{vect}}(m|n)$,
$a\in{\mathbb{K}}$, and $f\in{\mathcal{F}}$, by the Lie derivative, where $p$
designates the parity function

(10)
$L_{D}(f\mathop{\mathrm{{vol\/}}}\nolimits^{a})=(D(f)+(-1)^{p(f)p(D)}af\mathop{\text{\rm
div}}\nolimits(D))\mathop{\mathrm{{vol\/}}}\nolimits^{a}.$

###  2.1. Weighted densities on supermanifolds with a contact structure

There are two types of contact structures; in these cases certain modules of
tensor fields $T(V_{i})$ on a supermanifold ${\mathcal{M}}$ form rank 1
modules over the algebra ${\mathcal{F}}$, see [BGLLS]. The corresponding Lie
superalgebras are:

$\bullet$ ${\mathfrak{k}}(2n+1|m)$ preserving the distribution on
${\mathcal{M}}$ given by an odd contact form $\alpha_{1}$ with even “time”;

$\bullet$ ${\mathfrak{m}}(r)$ preserving the distribution on ${\mathcal{M}}$
given by an even contact form $\alpha_{0}$ with odd “time”.

####  2.1.1. Case ${\mathfrak{k}}(2n+1|m)$.

It is more natural to express the ${\mathfrak{k}}(2n+1|m)$-modules of weighted
densities in terms of the contact form $\alpha_{1}$ than in terms of
$\mathop{\mathrm{{vol\/}}}\nolimits$:

(11)
${\mathcal{F}}_{\lambda}:=\begin{cases}{\mathcal{F}}\alpha_{1}^{\lambda/2}&\text{if
$n+m>0$,}\\\ {\mathcal{F}}\alpha_{1}^{\lambda}&\text{if $n=m=0$}.\\\
\end{cases}$

with the ${\mathfrak{k}}(2n+1|m)$-action given by the formula (compare with
eq. (10) and (23))

(12) $L_{K_{f}}(\varphi\alpha_{1}^{\lambda})=(K_{f}(\varphi)+\lambda
K_{1}(f)\varphi)\alpha_{1}^{\lambda}.$

Important: as ${\mathfrak{k}}(2n+1|m)$-modules,
$\mathop{\mathrm{{Vol\/}}}\nolimits^{a}\simeq{\mathcal{F}}_{a(2+2n-m)}$.

$\bullet$ Let $u=(t,p_{1},\dots,p_{n},q_{1},\dots,q_{n})$ be even
indeterminates, $\Theta=\begin{cases}(\xi,\eta)&\text{ if }\ m=2k\\\
(\xi,\eta,\theta)&\text{ if }\ m=2k+1,\end{cases}$ be odd indeterminates and
let

(13) $\alpha_{1}=dt+\mathop{\sum}\limits_{1\leq i\leq
n}(p_{i}dq_{i}-q_{i}dp_{i})+\mathop{\sum}\limits_{1\leq j\leq
k}(\xi_{j}d\eta_{j}+\eta_{j}d\xi_{j})+\begin{cases}0&\text{ if }\ m=2k\\\
\theta d\theta&\text{ if }\ m=2k+1.\end{cases}$

Consider the distribution singled out by the Pfaff equation

(14) $\alpha_{1}(X)=0$ for $X\in{\mathfrak{vect}}(2n+1|m)$.

The Lie superalgebra, called contact Lie superalgebra, that preserves this
distribution is

(15) ${\mathfrak{k}}(2n+1|m)=\\{D\in{\mathfrak{vect}}(2n+1|m)\mid
L_{D}\alpha_{1}=f_{D}\alpha_{1}\text{ for some
}f_{D}\in{\mathbb{C}}[t,p,q,\theta]\\}.$

####  2.1.2. Case ${\mathfrak{m}}(n)$.

Similarly, set $u=q=(q_{1},\dots,q_{n})$, let
$\theta=(\xi_{1},\dots,\xi_{n};\tau)$ be odd. Set

(16)
$\begin{array}[]{c}\alpha_{0}=d\tau+\mathop{\sum}\limits_{i}(\xi_{i}dq_{i}+q_{i}d\xi_{i}),\end{array}$

and call this forms, the pericontact.

Consider the distribution singled out by the Pfaff equation

(17) $\alpha_{0}(X)=0$ for $X\in{\mathfrak{vect}}(n|n+1)$.

The Lie superalgebra, called pericontact Lie superalgebra, preserves this
distribution is

(18) ${\mathfrak{m}}(n)=\\{D\in{\mathfrak{vect}}(n|n+1)\mid
L_{D}\alpha_{0}=f_{D}\cdot\alpha_{0}\text{ for some
}\;f_{D}\in{\mathbb{C}}[q,\xi,\tau]\\}.$

To call ${\mathfrak{m}}$ “odd contact superalgebra” is misleading since the
even part of ${\mathfrak{m}}$ is nonzero. To say “the “odd” contact
superalgebra” is OK, though a bit long.

####  2.1.3. Generating functions: contact series and their subalgebras

A laconic way to describe ${\mathfrak{k}}$, ${\mathfrak{m}}$ and their
subalgebras is via generating functions. We express the contact fields of
series ${\mathfrak{k}}$ (resp. ${\mathfrak{m}}$) in natural bases $K_{f}$
(resp. $M_{f}$).

$\bullet$ Odd form $\alpha_{1}$. For any $f\in{\mathbb{C}}[t,p,q,\theta]$,
set:

(19)
$K_{f}=(2-E)(f){\frac{\partial}{\partial{t}}}-H_{f}+{\frac{\partial{f}}{\partial{t}}}E,$

where $E=\sum_{i}y_{i}{\frac{\partial}{\partial{y_{i}}}}$ (here the $y_{i}$
are all the coordinates except $t$) is the Euler operator, and $H_{f}$ is the
hamiltonian field with Hamiltonian $f$ that preserves $d\alpha_{1}$:

(20) $H_{f}=\mathop{\sum}\limits_{i\leq
n}\left({\frac{\partial{f}}{\partial{p_{i}}}}{\frac{\partial}{\partial{q_{i}}}}-{\frac{\partial{f}}{\partial{q_{i}}}}{\frac{\partial}{\partial{p_{i}}}}\right)-(-1)^{p(f)}\left({\frac{\partial{f}}{\partial{\theta}}}{\frac{\partial}{\partial{\theta}}}+\mathop{\sum}\limits_{j\leq
k}\left({\frac{\partial{f}}{\partial{\xi_{j}}}}{\frac{\partial}{\partial{\eta_{j}}}}+{\frac{\partial{f}}{\partial{\eta_{j}}}}{\frac{\partial}{\partial{\xi_{j}}}}\right)\right).$

$\bullet$ Even form $\alpha_{0}$. For any $f\in{\mathbb{C}}[q,\xi,\tau]$, set:

(21)
$M_{f}=(2-E)(f){\frac{\partial}{\partial{\tau}}}-Le_{f}-(-1)^{p(f)}{\frac{\partial{f}}{\partial{\tau}}}E,$

where $E=\sum_{i}y_{i}{\frac{\partial}{\partial{y_{i}}}}$ (here the $y_{i}$
are all the coordinates except $\tau$), and

(22) $Le_{f}=\mathop{\sum}\limits_{i\leq
n}\left({\frac{\partial{f}}{\partial{q_{i}}}}\
{\frac{\partial}{\partial{\xi_{i}}}}+(-1)^{p(f)}{\frac{\partial{f}}{\partial{\xi_{i}}}}\
{\frac{\partial}{\partial{q_{i}}}}\right).$

Since

(23)
$\begin{array}[]{l}L_{K_{f}}(\alpha_{1})=2{\frac{\partial{f}}{\partial{t}}}\alpha_{1}=K_{1}(f)\alpha_{1},\\\
L_{M_{f}}(\alpha_{0})=-(-1)^{p(f)}2{\frac{\partial{f}}{\partial{\tau}}}\alpha_{0}=-(-1)^{p(f)}M_{1}(f)\alpha_{0},\end{array}$

it follows that $K_{f}\in{\mathfrak{k}}(2n+1|m)$ and
$M_{f}\in{\mathfrak{m}}(n)$. Observe that

$p(Le_{f})=p(M_{f})=p(f)+{\bar{1}}.$

$\bullet$ To the (super)commutators $[K_{f},K_{g}]$ or $[M_{f},M_{g}]$ there
correspond contact brackets of the generating functions:

$\begin{array}[]{l}{}[K_{f},K_{g}]=K_{\\{f,\;g\\}_{k.b.}};\qquad{}[M_{f},M_{g}]=M_{\\{f,\;g\\}_{m.b.}}.\end{array}$

The explicit formulas for the contact brackets are as follows. Let us first
define the brackets on functions that do not depend on $t$ (resp. $\tau$).

The Poisson bracket $\\{-,-\\}_{P.b.}$ in the realization with the form
$\omega_{0}$ is given by the formula

(24) $\begin{array}[]{ll}\\{f,g\\}_{P.b.}&=\mathop{\sum}\limits_{i\leq n}\
\bigg{(}\displaystyle{\frac{\partial{f}}{\partial{p_{i}}}}\
{\frac{\partial{g}}{\partial{q_{i}}}}-\ {\frac{\partial{f}}{\partial{q_{i}}}}\
{\frac{\partial{g}}{\partial{p_{i}}}}\bigg{)}\\\
&-(-1)^{p(f)}\bigg{(}\displaystyle{\frac{\partial{f}}{\partial{\theta}}}\
{\frac{\partial{g}}{\partial{\theta}}}+\mathop{\sum}\limits_{j\leq
m}\left(\displaystyle{\frac{\partial{f}}{\partial{\xi_{j}}}}\
{\frac{\partial{g}}{\partial{\eta_{j}}}}+{\frac{\partial{f}}{\partial{\eta_{j}}}}\
{\frac{\partial{g}}{\partial{\xi_{j}}}}\right)\bigg{)}\\\ &\text{ for any
}f,g\in{\mathbb{C}}[p,q,\xi,\eta,\theta].\end{array}$

The Buttin bracket $\\{-,-\\}_{B.b.}$ is given by the formula

(25) $\\{f,g\\}_{B.b.}=\mathop{\sum}\limits_{i\leq n}\
\bigg{(}{\frac{\partial{f}}{\partial{q_{i}}}}\
{\frac{\partial{g}}{\partial{\xi_{i}}}}+(-1)^{p(f)}\
{\frac{\partial{f}}{\partial{\xi_{i}}}}\
{\frac{\partial{g}}{\partial{q_{i}}}}\bigg{)}\text{ for any
}f,g\in{\mathbb{C}}[q,\xi].$

In terms of the Poisson and Buttin brackets, respectively, the contact
brackets are

(26)
$\\{f,g\\}_{k.b.}=(2-E)(f){\frac{\partial{g}}{\partial{t}}}-{\frac{\partial{f}}{\partial{t}}}(2-E)(g)-\\{f,g\\}_{P.b.}$

and

(27)
$\\{f,g\\}_{m.b.}=(2-E)(f){\frac{\partial{g}}{\partial{\tau}}}+(-1)^{p(f)}{\frac{\partial{f}}{\partial{\tau}}}(2-E)(g)-\\{f,g\\}_{B.b.}.$

###  2.2. Divergences

####  2.2.1. Divergences on ${\mathfrak{k}}(2n+1|m)$.

For $m\neq 2n+2$, the restriction of the divergence from
${\mathfrak{vect}}(2n+1|n)$ to ${\mathfrak{k}}(2n+1|m)$ is equal to

$\mathop{\text{\rm div}}\nolimits(K_{f})=(2n+2-m)\partial_{t}(f).$

Therefore, ${\mathfrak{k}}(2n+1|2n+2)\subset{\mathfrak{svect}}(2n+1|2n+2)$. If
$m\neq 2n+2$, then the divergence-free subalgebra of ${\mathfrak{k}}(2n+1|m)$
is the Poisson Lie superalgebra of functions independent of $t$, spanned by
$K_{f}$ such that $\partial_{t}(f)=0$.

The standard grading in ${\mathcal{F}}$, where $\deg t=2$ and the degree of
any other indeterminate being equal to 1, induces the standard grading in
${\mathfrak{k}}(2n+1|m)$: we set $\deg(K_{f})=\deg(f)-2$.

Since codimension of $({\mathfrak{k}}(2n+1|m)_{0})^{\prime}$ in
${\mathfrak{k}}(2n+1|m)_{0}$ is equal to 1, there is one (up to a non-zero
factor) trace on ${\mathfrak{k}}(2n+1|m)_{0}$. Hence, there should be one (up
to a non-zero factor) divergence on ${\mathfrak{k}}(2n+1|m)$, see [BGLLS]. Set
$\mathop{\mathrm{tr}}\nolimits(K_{t}|_{{\mathfrak{g}}_{-1}})=1$, then the
divergence is given by $\partial_{t}$ for any $n$ and $m$.

##### The case ${\mathfrak{k}}(1|2)$

This case is exceptional. The algebra ${\mathfrak{k}}(1|2)_{0}$ of degree 0
elements is commutative and $\dim{\mathfrak{k}}(1|2)_{0}=2$, so there are 2
traces on ${\mathfrak{k}}(1|2)_{0}$, and hence 2 divergences on
${\mathfrak{k}}(1|2)$. Let $\alpha_{1}:=dt+\xi d\eta+\xi d\eta$. To the
operators

$\tilde{K}_{\xi}(f)=(-1)^{p(f)}(\partial_{\eta}-\xi\partial_{t})(f),\
\tilde{K}_{\eta}(f)=(-1)^{p(f)}(\partial_{\xi}-\eta\partial_{t})(f)\text{~{}~{}and~{}~{}}\tilde{K}_{1}=\partial_{t}$

there corresponds the second divergence

(28) $\mathop{\text{\rm
div}}\nolimits_{2}:=\tilde{K}_{\eta}\tilde{K}_{\xi}-\tilde{K}_{1}.$

Let $\beta$ be the symbol of the class of a pseudo-differential form that can
be “visualized” as $\beta:=\sqrt{[d\xi(d\eta)^{-1}]}$. Its actual expression
is irrelevant, we need to know only how the Lie derivative acts on it. Let wht
be the weight with respect to $K_{t}$ and $K_{\xi\eta}$; then

(29) ${\text{wht}}(\xi)=(1,1)$ and ${\text{wht}}(\eta)=(1,-1)$, so
${\text{wht}}([d\xi(d\eta)^{-1}])=(0,2)$.

Since it is natural to have a unit of weight be equal to 1, not 2, we take the
square root.

The Lie derivative acts as follows

(30)
$\begin{array}[]{l}L_{K_{f}}(\alpha_{1}^{a}\beta^{b})=(2a\partial_{t}(f)+(-1)^{p(f)}b\mathop{\text{\rm
div}}\nolimits_{2}(K_{f}))(\alpha_{1}^{a}\beta^{b})\\\
=((2a-(-1)^{p(f)}b)\partial_{t}(f)+(-1)^{p(f)}b\tilde{K}_{\eta}\tilde{K}_{\xi}(f))(\alpha_{1}^{a}\beta^{b}).\end{array}$

The space ${\mathcal{F}}_{a,b}$ of weighted densities over
${\mathfrak{k}}(1|2)$ is a rank 1 module over the algebra of functions
generated by $\alpha^{a/2}_{1}\beta^{b}$. (The above description is OK unless
the characteristic of the ground field is equal to $2$, cf. [BGLLS].)

####  2.2.2. Case ${\mathfrak{m}}(n)$.

On ${\mathfrak{m}}(n)$, there are two analogs of the divergence on
${\mathfrak{vect}}$. One is the operator $\mathop{\text{\rm div}}\nolimits$ —
the restriction of the divergence from ${\mathfrak{vect}}(n|n+1)$ to
${\mathfrak{m}}(n)$:

$L_{M_{f}}\mathop{\mathrm{{vol\/}}}\nolimits=\mathop{\text{\rm
div}}\nolimits(M_{f})\mathop{\mathrm{{vol\/}}}\nolimits;$

the other one, $\mathop{\text{\rm
div}}\nolimits_{\tau}:=2\partial_{\tau}\circ\text{sign}$, arises from the Lie
derivative acting on $\alpha_{0}$:

(31) $L_{M_{f}}\alpha_{0}=2(-1)^{p(f)}\partial_{\tau}(f)\alpha_{0}.$

Explicitly, (here $E=\sum y_{i}\partial_{y_{i}}$ the sum of all the
coordinates except $\tau$)

(32) $\mathop{\text{\rm
div}}\nolimits(M_{f})=2(-1)^{p(f)}\Delta^{\mathfrak{m}}(f),\text{where
$\Delta^{\mathfrak{m}}:=(1-E)\partial_{\tau}(f)-\sum_{i}\frac{\partial^{2}}{\partial_{\xi_{i}}\partial_{q_{i}}}$}.$

Rank 1 modules ${\mathcal{F}}_{a,b}$, where $a,b\in{\mathbb{C}}$, over the
algebra ${\mathcal{F}}$ are generated by
$\alpha_{0}^{a}\mathop{\mathrm{{vol\/}}}\nolimits^{b}$, on which
${\mathfrak{m}}(n)$ acts as follows:

(33)
$L_{M_{f}}(\alpha_{0}^{a}\mathop{\mathrm{{vol\/}}}\nolimits^{b})=2(-1)^{p(f)}(a\partial_{\tau}(f)+b~{}\Delta^{\mathfrak{m}}(f))(\alpha_{0}^{a}\mathop{\mathrm{{vol\/}}}\nolimits^{b}).$

###  2.3. The volume elements and weighted densities

In this subsection all isomorphisms are up to the change of parity functor
$\Pi$ because (a) there are two natural choices of parity on the space
$\mathop{\mathrm{{Vol\/}}}\nolimits$, see [MaG], and (b) neither choice
affects the invariance.

Over ${\mathfrak{vect}}(m|n)$, we have
$\mathop{\mathrm{{Vol\/}}}\nolimits:={\mathcal{F}}\mathop{\mathrm{{vol\/}}}\nolimits$.
Since $\mathop{\mathrm{{vol\/}}}\nolimits=[dx_{1}\dots
dx_{m}\partial_{\xi_{1}}\dots\partial_{\xi_{n}}]$, then
$\mathop{\mathrm{{Vol\/}}}\nolimits\simeq{\mathcal{F}}_{m-n}$, where the
weight is taken relative the Euler operator $E=\sum
x_{i}\partial_{x_{i}}+\sum\xi_{j}\partial_{\xi_{j}}$.

Over ${\mathfrak{k}}(2m+1|n)$, we have
$\mathop{\mathrm{{Vol\/}}}\nolimits\simeq{\mathcal{F}}_{2m+2-n}$, where the
weight is taken relative $K_{t}$.

Over ${\mathfrak{m}}(n)$, we have
$\mathop{\mathrm{{Vol\/}}}\nolimits\simeq{\mathcal{F}}_{-2}$, where the weight
is taken relative $M_{\tau}$.

##  3\. The ${\mathfrak{osp}}(n|2m+2)\subset{\mathfrak{k}}(2m+1|n)$ cases

According to (7), in this paper we consider only the case $m=0$.

###  3.1. Preliminaries

The super commutator and the super anti-commutator of $D_{\eta}$ and $D_{\xi}$
are, respectively,

(34) $[D_{\eta},D_{\xi}]:=D_{\eta}D_{\xi}+D_{\xi}D_{\eta}$ and
$[D_{\eta},D_{\xi}]_{+}:=D_{\eta}D_{\xi}-D_{\xi}D_{\eta}$.

Recall the definition of $K_{f}$ and introduce the operators $D_{f}$ for
$f=1,p,q$ (even) and $\xi,\eta,\theta$ (odd):

(35) $\begin{array}[]{l}\text{$D_{1}=\partial_{t}$,\ \
$K_{1}=2\partial_{t}$};\\\
\text{$D_{\eta_{i}}:=\xi_{i}\partial_{t}+\partial_{\eta_{i}}$,\
$D_{\xi_{i}}:=\eta_{i}\partial_{t}+\partial_{\xi_{i}}$, \
$D_{\theta}:=\theta\partial_{t}+\partial_{\theta}$; \
$D_{p_{j}}=p_{j}\partial_{t}+\partial_{q_{j}}$, \
$D_{q_{j}}=q_{j}\partial_{t}-\partial_{p_{j}}$;}\\\\[5.69054pt]
\text{$K_{\xi_{i}}=\xi_{i}\partial_{t}-\partial_{\eta_{i}}$,\
$K_{\eta_{i}}=\eta_{i}\partial_{t}-\partial_{\xi_{i}}$, \
$K_{\theta}=\theta\partial_{t}-\partial_{\theta}$; \
$K_{p_{j}}=p_{j}\partial_{t}-\partial_{q_{j}}$, \
$K_{q_{j}}=q_{j}\partial_{t}+\partial_{p_{j}}$}.\end{array}$

The cases ${\mathfrak{k}}(1|n)$ for $n=2$ and $n=4$ are exceptional: if $n=2$,
the weight of densities acquires an extra parameter and new Bol-type operators
associated with this parameter; if $n=4$, the embedded algebra can be deformed
and there are new Bol-type operators depending on the parameter of
deformation.

The invariance of the operator
$A:{\mathcal{F}}_{a}\longrightarrow{\mathcal{F}}_{b}$ or, equivalently, of the
map $A(f\alpha^{a})=\bar{A}(f)\alpha^{b}$, where $\alpha$ denotes the
generator of the ${\mathcal{F}}$-module of weighted densities (i.e.,
$\alpha_{1}$, or $\alpha_{0}\mathop{\mathrm{{vol\/}}}\nolimits$, or
$\mathop{\mathrm{{vol\/}}}\nolimits$) over ${\mathfrak{vect}}({\mathcal{M}})$
or any its subalgebra ${\mathfrak{g}}$ means that

(36)
$L_{D}(A(f\alpha^{a}))=(-1)^{p(A)p(D)}A(L_{D}(f\alpha^{a}))\text{~{}~{}for any
$D\in{\mathfrak{g}}$ and $f\in{\mathcal{F}}$}.$

Now, let ${\mathfrak{h}}$ be a subalgebra of
${\mathfrak{g}}:={\mathfrak{vect}}({\mathcal{M}})$, or
${\mathfrak{k}}({\mathcal{M}})$ or ${\mathfrak{m}}({\mathcal{M}})$ on the
supermanifold ${\mathcal{M}}$ such that
${\mathfrak{h}}_{-}={\mathfrak{g}}_{-}$. Then,

$L_{D}(\alpha)=0\text{~{}~{}for any $D\in{\mathfrak{g}}_{-}$},$

and therefore if condition (36) is satisfied, then the action of the operators
$\bar{A}$ and $L_{D}$ on functions supercommute.

For the elements of ${\mathfrak{m}}(n)$, recall the definition of $M_{f}$ and
introduce the operators $D_{f}$ for $f=1,q$ (even) and $\xi$ (odd):

(37) $\begin{array}[]{l}\text{$D_{1}=\partial_{\tau}$,\ \
$M_{1}=2\partial_{\tau}$};\\\
\text{$M_{\xi_{i}}:=\xi_{i}\partial_{\tau}+\partial_{q_{i}}$,\
$M_{q_{i}}:=q_{i}\partial_{\tau}-\partial_{\xi_{i}}$;}\\\\[5.69054pt]
\text{$D_{\xi_{i}}:=\xi_{i}\partial_{\tau}-\partial_{q_{i}}$,\
$D_{q_{i}}:=q_{i}\partial_{\tau}+\partial_{\xi_{i}}$}.\end{array}$

######  Theorem (An analog of J. Peetre’s theorem [Pe]).

Let ${\mathfrak{h}}$ be a subalgebra in a vectorial Lie superalgebra
${\mathfrak{g}}$, and ${\mathfrak{h}}_{-}={\mathfrak{g}}_{-}$.

Let ${\mathfrak{g}}:={\mathfrak{k}}(2m+1|n)$. Any
${\mathfrak{h}}_{-}$-invariant differential operator $\bar{A}$ is a polynomial
with constant coefficients in the operators $D_{f}$, see eq. (35).

If ${\mathfrak{g}}={\mathfrak{m}}(n)$, then any ${\mathfrak{h}}_{-}$-invariant
differential operator $\bar{A}$ is a polynomial with constant coefficients in
the operators $D_{f}$, see eq. (37).

If ${\mathfrak{g}}={\mathfrak{vect}}(m|n)$, then any
${\mathfrak{h}}_{-}$-invariant differential operator $\bar{A}$ is a polynomial
with constant coefficients in partial derivatives.

###### Proof.

Let us arrange monomials in $D_{\xi_{i}}$, $D_{\eta_{i}}$, and
$D_{1}=\partial_{t}$ in this order to form a basis over ${\mathcal{F}}$ in the
space of all differential operators. (Since $[D_{\xi_{i}},D_{\eta_{i}}]=\pm
D_{1}$, any basis contains $D_{1}$ and one of the two: either
$D_{\xi_{i}}D_{\eta_{i}}$ or $D_{\eta_{i}}D_{\xi_{i}}$.) All operators
$D_{f}$, where $\deg f<2$, commute with ${\mathfrak{g}}_{-}$. Then for any
monomial $F$ in $D_{f}$, any $\varphi\in{\mathcal{F}}$ and
$K_{g}\in{\mathfrak{g}}_{-}$, we have

$[K_{g},\varphi F]=K_{g}(\varphi)F.$

Hence, any ${\mathfrak{g}}_{-}$-invariant differential operator must have
constant coefficients ($K_{g}(\varphi)=0$). ∎

(38) Since the condition (36) should hold for any function
$f\in{\mathcal{F}}$, then having selected a basis of ${\mathcal{F}}$
homogeneous with respect to a grading, we see that all components of $\bar{A}$
homogeneous with respect to the same grading should be invariant.

Thanks to (38), we can assume that $\bar{A}$ is homogeneous of degree $s$.
Since

$K_{t}(f)=(\deg f)f,\text{~{}and~{}}K_{t}(\alpha_{1}^{a})=2a\alpha_{1}^{a},$

then (36) for $D=K_{t}$ turns into

$\begin{array}[]{ll}&AK_{t}(f\alpha_{1}^{a})=(\deg
f+2a)\bar{A}(f)\alpha_{1}^{b}\\\
=&K_{t}A(f\alpha_{1}^{a})=(\deg\bar{A}(f)+2b)\bar{A}(f)\alpha_{1}^{b}=(s+\deg
f+2b)\bar{A}(f)\alpha_{1}^{b},\end{array}$

i.e.,

$2a=s+2b.$

Analogously,

(39) $L_{D}(\mathop{\mathrm{{vol\/}}}\nolimits)=(\mathop{\text{\rm
div}}\nolimits D)\mathop{\mathrm{{vol\/}}}\nolimits.$

Over ${\mathfrak{k}}(2m+1|n)$, we define the spaces of weighted densities
generated, as modules over the algebra of functions ${\mathcal{F}}$, by
$\alpha_{1}^{a}$ (or $\alpha_{1}^{a}\beta^{b}$ in dimension $1|2$), see (12),
NOT by $\mathop{\mathrm{{vol\/}}}\nolimits^{a}$, as in [BaL].

Observe also that $L_{K_{\xi_{i}\eta_{i}}}(\alpha_{1})=0$ for all $i$, and
hence the actions of the operators $\bar{A}$ and $K_{\xi_{i}\eta_{i}}$ on the
space of functions must commute. From commutation relations

$\begin{array}[]{l}{}[K_{\xi_{i}\eta_{i}},D_{\xi_{i}}]=D_{\xi_{i}},\ \
[K_{\xi_{i}\eta_{i}},D_{\eta_{i}}]=-D_{\eta_{i}},\\\
{}[K_{\xi_{i}\eta_{i}},D_{\xi_{j}}]=[K_{\xi_{i}\eta_{i}},D_{\eta_{j}}]=0\text{~{}~{}for
$i\neq j$}\end{array}$

it follows that $\bar{A}$ should contain products $D_{\xi_{i}}D_{\eta_{i}}$
and $D_{\eta_{i}}D_{\xi_{i}}$.

Thus, $\bar{A}$ should be a sum of monomials in
$[D_{\xi_{i}},D_{\eta_{i}}]_{+}$ and $\partial_{t}$ with constant
coefficients. This argument is not applicable to the
${\mathfrak{k}}(1|2)$-invariant operators
${\mathcal{F}}_{a,b}\longrightarrow{\mathcal{F}}_{c,d}$ with $b,d\neq 0$. Odd
Bols can and do appear. Actually, there are odd operators
${\mathcal{F}}_{a,b}\longrightarrow{\mathcal{F}}_{c,d}$ invariant with respect
to the whole ${\mathfrak{k}}(1|2)$, not only Bols, see [Lr].

######  Theorem (Bol’s analogs for $i({\mathfrak{osp}}(1|2))$-invariant
operators, [G, GT]).

Let $m|n=0|1$.

_(i)_ For every $k\in{\mathbb{Z}}_{+}:={\mathbb{N}}\cup\\{0\\}$, there is an
$i({\mathfrak{osp}}(1|2))$-invariant differential operator

$A^{k}:{\mathcal{F}}_{-k}\longrightarrow{\mathcal{F}}_{k+1}\text{~{}~{}given
by~{}~{}}\varphi\alpha_{1}^{-\frac{k}{2}}\mapsto
D_{\theta}\circ\partial_{t}^{k}(\varphi)\alpha_{1}^{\frac{k+1}{2}}.$

_(ii)_ There are no other nonzero $i({\mathfrak{osp}}(1|2))$-invariant
differential operators ${\mathcal{F}}_{a}\longrightarrow{\mathcal{F}}_{b}$.

###  3.4. Super Bol operators
${\mathcal{F}}_{\lambda}\longrightarrow{\mathcal{F}}_{\mu}$ in terms of
$\theta$’s

In [BaL], the authors considered the realization of ${\mathfrak{k}}(1|n)$
preserving the distribution singled out by the contact form

(40) $dt+\sum\theta_{i}d\theta_{i}.$

Then,

$K_{\theta_{i}}=\theta_{i}\partial_{t}-\partial_{\theta_{i}},\ \
D_{\theta_{i}}=\partial_{\theta_{i}}+\theta_{i}\partial_{t}.$

######  Theorem (Bol’s analogs for $i({\mathfrak{osp}}(1|2n))$-invariant
operators, [BaL]).

The only $i({\mathfrak{osp}}(1|2n))$-invariant differential operators
${\mathcal{F}}_{\lambda}\longrightarrow{\mathcal{F}}_{\mu}$ are the above-
described for $n=0$ and $1$, and

$A^{k}:\varphi\alpha_{1}^{\lambda/2}\mapsto D_{\theta_{1}}\circ\dots\circ
D_{\theta_{n}}\partial_{t}^{k}(\varphi)\alpha_{1}^{\mu/2}\text{~{}~{}for any
$k\in{\mathbb{Z}}_{+}$},$

where

(41) $\lambda=1-k-n\text{ and }\mu=k+1.$

Observe that in [BaL], the authors define ${\mathcal{F}}_{a}$ not as we do
(11), but as ${\mathcal{F}}\mathop{\mathrm{{vol\/}}}\nolimits^{a}$, but denote
$\mathop{\mathrm{{vol\/}}}\nolimits$ by $\alpha$, hence the values of
$\lambda$ and $\mu$ in [BaL] differ from the above.

###  3.5. The realization of ${\mathfrak{k}}(1|n)$ preserving the distribution
given by the contact form (13)

######  Theorem (A $\xi\eta$-version of [BaL], Eq. (2.1), see Th. 3.4.1.
Theorem).

For $n>2$, the only $i({\mathfrak{osp}}(2|2n))$-invariant operators
$A^{k}:{\mathcal{F}}_{\lambda}\longrightarrow{\mathcal{F}}_{\mu}$ of parity
$n\pmod{2}$ are as follows for any $k\in{\mathbb{Z}}_{+}$

(42) $\begin{array}[]{ll}\varphi\mapsto
D_{\theta}\circ[D_{\eta_{1}},D_{\xi_{1}}]_{+}\circ\cdots\circ[D_{\eta_{\tilde{n}}},D_{\xi_{\tilde{n}}}]_{+}\circ\partial_{t}^{k}(\varphi)&\text{for
$n=2\tilde{n}+1$ odd},\\\
\varphi\mapsto[D_{\eta_{1}},D_{\xi_{1}}]_{+}\circ\cdots\circ[D_{\eta_{\tilde{n}}},D_{\xi_{\tilde{n}}}]_{+}\circ\partial_{t}^{k}(\varphi)&\text{for
$n=2\tilde{n}$ even},\end{array}$

where

(43) $\lambda=1-k-n\text{ and }\mu=k+1.$

######  Theorem (Bol’s analogs for $i({\mathfrak{osp}}(2|2))$-invariant
operators).

Let $m|n=0|2$.

_(i)_ For every $k\in{\mathbb{Z}}_{+}$, the only even
$i({\mathfrak{osp}}(2|2))$-invariant differential operators

$A:\varphi\alpha^{\lambda}\beta^{\mu}\mapsto\bar{A}(\varphi)\alpha^{\nu}\beta^{\omega}$

where

(44)
$\lambda=-\nu=\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}},\
\ \mu=\omega=-c(k+1),$

are of the form

$\bar{A}=[D_{\xi},D_{\eta}]_{+}D_{1}^{k}+cD_{1}^{k+1},\text{~{}~{}where
$c\in{\mathbb{C}}$}.$

(For $\mu=\omega=c=0$, we recover the result of [BaL].)

_(ii)_ There are only two $1$-parameter families of odd
$i({\mathfrak{osp}}(2|2))$-invariant operators:

$\begin{array}[]{ll}A_{\xi,\lambda}:{\mathcal{F}}_{\lambda,-\lambda}\longrightarrow{\mathcal{F}}_{1+\lambda,-1-\lambda},&\bar{A}_{\xi,\lambda}(\varphi)=D_{\xi}(\varphi);\\\\[5.69054pt]
A_{\eta,\lambda}:{\mathcal{F}}_{\lambda,\lambda}\longrightarrow{\mathcal{F}}_{1+\lambda,1+\lambda},&\bar{A}_{\eta,\lambda}(\varphi)=D_{\eta}(\varphi).\end{array}$

####  3.5.3. Recapitulation from [Lr]: elucidation of Theorem 3.5.2. Theorem.

For the a description of ${\mathfrak{g}}={\mathfrak{k}}(1|2)$-invariant
differential operators $D:T(V)\longrightarrow T(W)$ with irreducible
${\mathfrak{g}}_{0}$-modules $V$ and $W$, see [Lr]. In this case, all modules
$T(V)$ are of the form ${\mathcal{F}}_{a,b}$, so the answers in [Lr] should be
among our Bols. Let us reproduce the results from [Lr] and amend them to get
Bols.

On the $n$-dimensional manifold $M$, we have $(\Omega^{i})^{*}=\Omega^{n-i}$.
Recall that on the supermanifold ${\mathcal{M}}$, the dual to the space of
differential forms $\Omega^{i}:=T(E^{i}(V))$, where $E^{i}$ is the operator of
raising to the $i$th exterior power and $V$ is the tautological
${\mathfrak{gl}}(V)={\mathfrak{gl}}(\dim({\mathcal{M}}))$-module, is the space
of integrable forms

$\Sigma_{-i}({\mathcal{M}}):=(\Omega^{i}({\mathcal{M}}))^{*}\otimes_{\mathcal{F}}\mathop{\mathrm{{Vol\/}}}\nolimits({\mathcal{M}})=T(E^{i}(V^{*})\otimes\mathop{\mathrm{tr}}\nolimits),$

where $\mathop{\mathrm{{Vol\/}}}\nolimits({\mathcal{M}})$ is the
${\mathcal{F}}$-module of volume forms (aka Berezinian). There are two
complexes (series of mappings such that the product of two consecutive
mappings is equal to 0)

(45) $\begin{array}[]{l}\Omega^{0}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\Omega^{1}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\dots\Omega^{i}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\Omega^{i+1}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}\dots;\\\ \dots\stackrel{{\scriptstyle
d^{*}}}{{\longrightarrow}}\Sigma_{j}\stackrel{{\scriptstyle
d^{*}}}{{\longrightarrow}}\Sigma_{j+1}\stackrel{{\scriptstyle
d^{*}}}{{\longrightarrow}}\dots\Sigma_{-1}\stackrel{{\scriptstyle
d^{*}}}{{\longrightarrow}}\Sigma_{0}.\end{array}$

In [Lr], following Rudakov, instead of modules
$T(V):=\mathop{\mathrm{Hom}}\nolimits_{U({\mathfrak{g}}_{0}\oplus{\mathfrak{g}}_{+})}(U({\mathfrak{g}}),V)$,
Leites considered the induced modules

$I(V):=U({\mathfrak{g}})\otimes_{U({\mathfrak{g}}_{0}\oplus{\mathfrak{g}}_{+})}V\simeq{\mathbb{C}}[{\mathfrak{g}}_{-}]\otimes
V=(T(V^{*}))^{*},$

where ${\mathfrak{g}}_{-}:=\oplus_{i<0}{\mathfrak{g}}_{i}$ and
${\mathfrak{g}}_{+}:=\oplus_{i>0}{\mathfrak{g}}_{i}$, and $V$ is considered as
a ${\mathfrak{g}}_{0}\oplus{\mathfrak{g}}_{+}$-module such that
${\mathfrak{g}}_{+}V=0$. Since ${\mathfrak{g}}_{0}$ is commutative, then any
irreducible ${\mathfrak{g}}_{0}$-module is 1-dimensional. Observe that

(46)
$(K_{\xi})^{2}=\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}[K_{\xi},K_{\xi}]=0,\
\
(K_{\eta})^{2}=\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}[K_{\eta},K_{\eta}]=0,\
\ [K_{\xi},K_{\eta}]=K_{1}.$

The invariant operators $D:T(V)\longrightarrow T(W)$ are in one-to-one
correspondence with invariant operators

$D^{*}:T(W)^{*}=I(W^{*})\longrightarrow T(V)^{*}=I(V^{*}),$

and hence singular vectors $f\in I(V^{*})$, i..e., such that $\deg f<0$ and
${\mathfrak{g}}_{+}f=0$. Since ${\mathfrak{g}}_{+}$ is generated by 3
operators $K_{t\xi}$, $K_{t\eta}$, and $K_{t\xi\eta}$, it suffices to solve
systems of 3 equations for every $\deg f\leq-2$ and only 2 equations if $\deg
f=-1$ since in this case $K_{t\xi\eta}f=0$ holds automatically.

In what follows, let $K_{t}v=Ev$ and $K_{\xi\eta}v=Hv$ for some
$E,H\in{\mathbb{C}}$ and $v\in V$.

$\bullet$ $\deg f=-1$. Then $f=(aK_{\xi}+bK_{\eta})v$, where
$a,b\in{\mathbb{C}}$, and $ab=0$ because $f$ is supposed to be a weight vector
with respect to $K_{t}$ and $K_{\xi\eta}$. Recall that
${\mathfrak{g}}_{+}v=0$. Let us solve the system

(47) $\begin{array}[]{ll}(A)&K_{t\xi}f=0;\\\ (B)&K_{t\eta}f=0.\end{array}$

The system (47) is equivalent to

(48) $\begin{array}[]{ll}(A)&b(K_{\xi\eta}-K_{t})v=0,\\\
(B)&a(K_{\xi\eta}+K_{t})v=0.\end{array}$

Therefore, recall (29),

(49) $\begin{array}[]{ll}(A)&\text{$H=E=\lambda$ and
${\text{wht}}(f)=(\lambda-1,\lambda-1)$},\\\ (B)&\text{$H=-E=\lambda$ and
${\text{wht}}(f)=(\lambda-1,-\lambda+1)$}.\end{array}$

Let us try to understand the nature of these operators. We know that the
exterior differential $d:{\mathcal{F}}_{0,0}\longrightarrow\Omega^{1}$ is
invariant with respect to any changes of variables, not only
${\mathfrak{g}}$-invariant. Since ${\mathcal{F}}\alpha_{1}$ is a
${\mathfrak{g}}$-invariant submodule in $\Omega^{1}$, we have projections

$\begin{array}[]{l}A_{\xi,0}:=\mathop{\mathrm{pr}}\nolimits_{\xi}\circ
d:{\mathcal{F}}_{0,0}\longrightarrow{\mathcal{F}}[d\xi]={\mathcal{F}}_{1,1};\\\
A_{\eta,0}:=\mathop{\mathrm{pr}}\nolimits_{\eta}\circ
d:{\mathcal{F}}_{0,0}\longrightarrow{\mathcal{F}}[d\eta]={\mathcal{F}}_{1,-1},\end{array}$

where $[d\xi]$ and $[d\eta]$ are classes of $d\xi$ and $d\eta$, respectively,
modulo ${\mathcal{F}}\alpha_{1}$. Calculations (47) and (49) show that there
are ${\mathfrak{k}}(1|2)$-invariant analogs of these operators

$A_{\xi,a}:={\mathcal{F}}_{a,a}\longrightarrow{\mathcal{F}}_{a+1,a+1}$ and
$A_{\eta,a}:=\mathop{\mathrm{pr}}\nolimits_{\eta}\circ
d:{\mathcal{F}}_{a,a}\longrightarrow{\mathcal{F}}_{a+1,a-1}$.

Since $i({\mathfrak{osp}}(2|2))$ contains $K_{t\xi}$ and $K_{t\eta}$, the
operators $A_{\xi,a}$ and $A_{\eta,a}$ are Bols.

Several ${\mathfrak{k}}(1|2)$-invariant unary differential operators of the
degree 2 found in [Lr] are obvious: consider the restrictions of the
composition

(50) $\Sigma_{-1}\stackrel{{\scriptstyle
d}}{{\to}}\Sigma_{0}\stackrel{{\scriptstyle\phantom{aa}\mathop{\mathrm{{vol\/}}}\nolimits^{-1}}}{{\simeq}}\Omega_{0}\stackrel{{\scriptstyle
d}}{{\to}}\Omega_{1}$

acting from submodules of the form ${\mathcal{F}}_{a,b}$ to the quotients of
the same form.

There are many more Bol operators than restrictions of operators (50) to rank
1 submodules of weighted densities, see (44). Unlike operators (50), we can
not interpret them.

$\bullet$ $\deg f=-3$. Then $f=(aK_{\xi}K_{1}+bK_{\eta}K_{1})v$, where
$a,b\in{\mathbb{C}}$, and $ab=0$. It is easy to check that $f$ can not be
annihilated by both $K_{t\xi}$ and $K_{t\eta}$. The same is true (by
induction) for any vector of degree $-(2n+1)<-1$.

####  3.5.4. Proof of Theorem 3.5.2. Theorem.

(i) Any even operator commuting with ${\mathfrak{g}}_{-}$ is of the form

(51)
$A:\varphi\alpha^{\lambda}\beta^{\mu}\longrightarrow\bar{A}(\varphi)\alpha^{\nu}\beta^{\omega},\text{
where $\bar{A}=aD_{\xi}D_{\eta}\partial_{t}^{k}+b\partial_{t}^{k+1}$ and
$a,b\in{\mathbb{C}}$}.$

The operator $K_{f}$ commutes with the operator (51) if and only if for any
$\varphi$ we have

(52) $K_{f}\bar{A}(\varphi)+(2\partial_{t}(f)\nu+\omega\mathop{\text{\rm
div}}\nolimits_{2}(f))\bar{A}(\varphi)=\bar{A}K_{f}(\varphi)+2\lambda\bar{A}(\partial_{t}(f)\varphi)+\mu\bar{A}(\mathop{\text{\rm
div}}\nolimits_{2}(f)\varphi).$

To find out when the operator (51) commutes with the subalgebra
${\mathfrak{osp}}(2|2)\subset{\mathfrak{k}}(1|2)$, it suffices to find when
(51) commutes with $K_{t^{2}}$ and
${\mathfrak{osp}}(2|2)_{-}={\mathfrak{k}}(1|2)_{-}$ because they generate
${\mathfrak{osp}}(2|2)$. The corresponding calculations are not that easy
(especially by bare hands), so let us first consider commutations with $K_{t}$
and $K_{\xi\eta}$. This will help in further calculations.

Set

$\mathop{\mathrm{gr}}\nolimits(t)=2,\quad\mathop{\mathrm{gr}}\nolimits(\xi)=\mathop{\mathrm{gr}}\nolimits(\eta)=1,$

let $\deg(x)=1$ for any $x=t,\xi$ or $\eta$.

The action of $K_{t}$:

$\begin{array}[]{l}K_{t}=2t\partial_{t}+\xi\partial_{\xi}+\eta\partial_{\eta},\;K_{t}\alpha=2\alpha,\;K_{t}\beta=0,\;K_{t}(\varphi)=\mathop{\mathrm{gr}}\nolimits(\varphi)\varphi,\\\
K_{t}(\bar{A}\varphi)=(\mathop{\mathrm{gr}}\nolimits(\bar{A})+\mathop{\mathrm{gr}}\nolimits(\varphi))\bar{A}\varphi=(-2k-2+\mathop{\mathrm{gr}}\nolimits(\varphi))\bar{A}\varphi.\end{array}$

Thus, the equation (52) for $f=t$ is equivalent to the equation

(53)
$\mathop{\mathrm{gr}}\nolimits(\bar{A})+\mathop{\mathrm{gr}}\nolimits(\varphi)+2\nu=\mathop{\mathrm{gr}}\nolimits(\varphi)+2\lambda\
{\Longleftrightarrow}\ 2\nu+2k+2=2\lambda\ {\Longleftrightarrow}\
\nu=\lambda+k+1.$

Now, consider the action of
$K_{\xi\eta}=-\xi\partial_{\xi}+\eta\partial_{\eta}$:

$\begin{array}[]{l}K_{\xi\eta}\alpha=0,\;K_{\xi\eta}\beta=\beta,\\\
{}[K_{\xi\eta},D_{\xi}]=D_{\xi},\;[K_{\xi\eta},D_{\eta}]=-D_{\eta},\;[K_{\xi\eta},D_{1}]=0{\Longrightarrow}[K_{\xi\eta},\bar{A}]=0.\end{array}$

Hence for $f=\xi\eta$ the equation (52) is of the form

(54) $\omega\bar{A}(\varphi)=\mu\bar{A}(\varphi)\ {\Longleftrightarrow}\
\omega=\mu.$

Now, pass to $K_{t^{2}}$ and introduce $T$:

(55)
$2T:=K_{t^{2}}=2t^{2}\partial_{t}+2t(\xi\partial_{\xi}+\eta\partial_{\eta}),\;K_{t^{2}}\alpha=4t\alpha,\;K_{t^{2}}\beta=2\xi\eta\beta,\;T(\varphi)=(\deg\varphi)t\varphi.$

Accordingly, the equations (52) for $f=t^{2}$ with equality (54) taken into
account take the form

$T\bar{A}(\varphi)+(2t\nu+\omega\xi\eta)\bar{A}(\varphi)=\bar{A}T(\varphi)+2\lambda\bar{A}(t\varphi)+\omega\bar{A}(\xi\eta\varphi),$

which taking (55) into account, is equivalent to

(56)
$T\bar{A}(\varphi)+(2t\nu+\omega\xi\eta)\bar{A}(\varphi)=(\deg\varphi+2\lambda)\bar{A}(t\varphi)+\omega\bar{A}(\xi\eta\varphi).$

Consider the action of $\bar{A}$ on monomials:

(57) $\bar{A}(t^{s})=\begin{cases}0&\text{for $s\leq k$}\\\
(a+b)(k+1)!&\text{for $s=k+1$}\\\
\frac{s!}{(s-k-1)!}\left((a+b)t^{s-k-1}-a(s-k-1)\xi\eta
t^{s-k-2}\right)&\text{for $s>k+1$}\\\ \end{cases}$

(58) $\begin{array}[]{l}\bar{A}(\xi t^{s})=\begin{cases}0&\text{for $s\leq
k$}\\\ b\frac{s!}{(s-k-1)!}\xi t^{s-k-1}&\text{for $s>k$}\\\ \end{cases}\\\
\bar{A}(\eta t^{s})=\begin{cases}0&\text{for $s\leq k$}\\\
(2a+b)\frac{s!}{(s-k-1)!}\eta t^{s-k-1}&\text{for $s>k$}\\\
\end{cases}\end{array}$

(59) $\bar{A}(\xi\eta t^{s})=\begin{cases}0&\text{for $s<k$}\\\ -ak!&\text{for
$s=k$}\\\ \frac{s!}{(s-k)!}\left(-at^{s-k}+(a+b)(s-k)\xi\eta
t^{s-k-1}\right)&\text{for $s>k$}\\\ \end{cases}$

Let us write the condition (56) for $\varphi=t^{k}$ taking (57) into account.
We get

$(k+2\lambda)(a+b)(k+1)!+\omega(-ak!)=0,$

implying the condition for $\omega$:

(60) $a\omega=(k+1)(a+b)(k+2\lambda)$

Now, let us write the condition (56) for $\varphi=t^{k+1}$ taking (57) into
account:

$\begin{array}[]{l}(2t\nu+\omega\xi\eta)(a+b)(k+1)!\\\
=(k+1+2\lambda)(k+2)!\left((a+b)t-a\xi\eta\right)+\omega(k+1)!(-at+(a+b)\xi\eta)\end{array}$

After simplification we get

$t\left((a+b)(2\nu-(k+2)(k+1+2\lambda))+\omega
a\right)+a(k+1+2\lambda)(k+2)\xi\eta=0,$

which is equivalent to the system of two conditions

$a\omega=(a+b)((k+2)(k+1+2\lambda)-2\nu)\text{ and }a(k+1+2\lambda)=0.$

Substituting $\nu=\lambda+k+1$ in these conditions and taking (53) into
account we see that the first condition is equivalent to (60), while the
second one is equivalent to the following condition

$a=0$ or $\lambda=\nu$.

If $a\neq 0$, we can set $a=1$. Then the condition $\lambda=\nu$ with (53),
(54) and (60) taken into account yields the following values of parameters:

(61)
$\lambda=-\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}},\quad\nu=\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}},\quad\mu=\omega=-(k+1)(1+b).$

Actually, this is the answer desired. To be absolutely sure, we have to

1) check that for $a=0$ the operator $A$ is not invariant;

2) check that for $a=1$ if conditions (61) are satisfied, then $A$ is indeed
invariant, i.e., conditions (56) hold for any function $\varphi$.

Let $a=0$. Then, we can assume that $\bar{A}=\partial_{t}^{k}$. In this case,
equalities $\xi\eta\bar{A}(\varphi)=\bar{A}(\xi\eta\varphi)$ and (56) turn
into

(62) $T\bar{A}(\varphi)+2\nu
t\bar{A}(\varphi)=(\deg\varphi+2\lambda)\bar{A}(t\varphi).$

Let $\varphi=\varphi_{0}t^{k}$, where the function $\varphi_{0}$ depends only
on $\xi,\eta$ and $\deg\varphi_{0}=d$. Then, the equation (62) takes the form

$\begin{array}[]{l}k!(d+2\nu)\varphi_{0}t=(d+k+2\lambda)(k+1)!\varphi_{0}t\\\
\ {\Longleftrightarrow}\ d+2\nu=(d+k+2\lambda)(k+1)\ {\Longleftrightarrow}\
dk=2\nu-(k+2\lambda)(k+1).\end{array}$

This equality should hold for any $d=0,1,2$, which is impossible since RHS is
a constant. Thus, the case $a=0$ is ruled out.

It only remains now to verify the invariance of $A$ for parameters given by
relations (61). Just plug (61) in (56) and verify for all monomials $\varphi$
using formulas (57)–(59).

Thus for $\varphi=t^{s}$ and $s>k+1$ we see that the LHS of (56) is of the
form

$\begin{array}[]{l}\frac{s!}{(s-k-1)!}((1+b)(s-k-1)t^{s-k}-(s-k-1)(s-k)\xi\eta
t^{s-k-1})\\\ +t(k+1)\frac{s!}{(s-k-1)!}((1+b)t^{s-k-1}-(s-k-1)\xi\eta
t^{s-k-2})-\\\ -(k+1)(1+b)\xi\eta\frac{s!}{(s-k-1)!}(1+b)t^{s-k-1}\\\
=\frac{s\cdot
s!}{(s-k-1)!}(1+b)t^{s-k}-\left(\frac{(s+1)!}{(s-k-2)!}+\frac{s!(k+1)(1+b)^{2}}{(s-k-1)!}\right)\xi\eta
t^{s-k-1}\end{array}$

whereas the RHS of (56) is of the form

$\begin{array}[]{l}(s-k-1)\frac{(s+1)!}{(s-k)!}((1+b)t^{s-k}-(s-k)\xi\eta
t^{s-k-1})-\\\ -(k+1)(1+b)\frac{s!}{(s-k)!}(-t^{s-k}+(1+b)(s-k)\xi\eta
t^{s-k-1})=\\\
=(1+b)\frac{s!}{(s-k)!}((s-k-1)(s+1)+k+1)t^{s-k}-\left(\frac{(s+1)!}{(s-k-2)!}+\frac{s!(k+1)(1+b)^{2}}{(s-k-1)!}\right)\xi\eta
t^{s-k-1}.\end{array}$

Since $(s-k-1)(s+1)+k+1=s(s-k)$, then LHS=RHS. Thus, on monomials
$\varphi=t^{s}$ the equality (56) holds.

Now, check the equality (56) for $\varphi=\xi t^{s}$. In this case,
$\xi\eta\bar{A}(\varphi)=\bar{A}(\xi\eta\varphi)=0$. The LHS of (56) is of the
form

$(s-k+k+1)b\mathchoice{\raisebox{0.25pt}{$\dfrac{s!}{(s-k-1)!}$}}{\raisebox{0.25pt}{$\dfrac{s!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{s!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{s!}{(s-k-1)!}$}}\xi
t^{s-k}=\mathchoice{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k-1)!}$}}b\xi
t^{s-k},$

whereas the RHS is of the form

$(s+1-k-1)b\mathchoice{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k)!}$}}{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k)!}$}}\xi
t^{s-k}=\mathchoice{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{0.25pt}{$\dfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k-1)!}$}}{\raisebox{-0.5pt}{$\tfrac{(s+1)!}{(s-k-1)!}$}}b\xi
t^{s-k}.$

Thus, the equality (56) holds for such monomials as well.

Calculations for $\varphi=\eta t^{s}$ are analogous.

It remains to verify equality (56) for $\varphi=\xi\eta t^{s}$. In this case,
$\bar{A}(\xi\eta\varphi)=0$, but $\xi\eta\bar{A}(\varphi)\neq 0$. The RHS is
equal to

$\begin{array}[]{l}(s+2-k-1)\frac{(s+1)!}{(s-k+1)!}(-t^{s-k+1}+(s-k+1)(1+b)\xi\eta
t^{s-k})\\\ =\frac{(s+1)!}{(s-k)!}(-t^{s-k+1}+(s-k+1)(1+b)\xi\eta
t^{s-k}).\end{array}$

Let us calculate the LHS:

$\begin{array}[]{l}\frac{s!}{(s-k)!}(-(s-k)t^{s-k+1}+(s-k+1)(s-k)(1+b)\xi\eta
t^{s-k})\\\ +(k+1)\frac{s!}{(s-k)!}(-t^{s-k+1}+(s-k)(1+b)\xi\eta
t^{s-k})-\omega\frac{s!}{(s-k)!}\xi\eta t^{s-k}\\\
=-(s+1)\frac{s!}{(s-k)!}t^{s-k+1}+\left(\frac{s!}{(s-k-1)!}(1+b)(s+2)+\frac{s!}{(s-k)!}(1+b)(k+1)\right)\xi\eta
t^{s-k}\\\
=-\frac{(s+1)!}{(s-k)!}t^{s-k+1}+\frac{s!}{(s-k)!}(1+b)((s+2)(s-k)+k+1)\xi\eta
t^{s-k}.\end{array}$

LHS=RHS since

$(s+2)(s-k)+k+1=(s+1)(s-k+1).$

Thus, the equality (56) is verified for all monomials $\varphi$, and hence the
operator

$A:\varphi\alpha^{\lambda}\beta^{\mu}\longrightarrow\bar{A}(\varphi)\alpha^{\nu}\beta^{\omega},$

where

$\bar{A}=D_{\xi}D_{\eta}\partial_{t}^{k}+b\partial_{t}^{k+1},\quad\nu=-\lambda=\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}},\quad\mu=\omega=-(k+1)(1+b)$

is invariant.

Set $c:=1+b$. Then

$\bar{A}=(D_{\xi}D_{\eta}-\partial_{t})\partial_{t}^{k}+c\partial_{t}^{k+1}.$

Since $[D_{\xi},D_{\eta}]=2\partial_{t}$, we have

$-D_{\eta}D_{\xi}=D_{\xi}D_{\eta}-2\partial_{t}~{}{\Longrightarrow}~{}[D_{\xi},D_{\eta}]_{+}=2(D_{\xi}D_{\eta}-\partial_{t}).$

Since $A$ is defined up to a non-zero scalar factor, having denoted $2A$ by
$A$ and $2c$ by $c$, we get the final form of $A$:

$A:\varphi\alpha^{-\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}}\beta^{-c(k+1)}\longrightarrow([D_{\xi},D_{\eta}]_{+}\partial_{t}^{k}+c\partial_{t}^{k+1})(\varphi)\alpha^{\mathchoice{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{0.25pt}{$\dfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{k+1}{2}$}}}\beta^{-c(k+1)}.$

(ii) Odd operators. For the classification of ${\mathfrak{k}}(1|2)$-invariant
differential operators of degree 1, see [Lr] and the proof reproduced in
Subsection 3.5.3. There are no ${\mathfrak{k}}(1|2)$-invariant differential
operators of order $>2$, see [Lr]. There are no Bols either, because already
invariance with respect to $K_{t\xi}$ and $K_{t\eta}$ is violated. ∎

####  3.5.5. $i({\mathfrak{osp}}_{a}(4|2))\subset{\mathfrak{k}}(1|4)$.

In Section 3.5.5 we explain the difference between the parametric family
${\mathfrak{osp}}(4|2;\alpha)$ considered in [BGL] and the family of embedded
algebras ${\mathfrak{osp}}_{a}(4|2)$ considered here.

The following generating functions form a basis of
$i({\mathfrak{osp}}_{a}(4|2))\subset{\mathfrak{k}}(1|4)$, see [LSh]:

(63) $\begin{array}[]{ll}{\mathfrak{osp}}_{a}(4|2)_{\leq
0}&={\mathfrak{k}}(1|4)_{\leq 0};\\\
{\mathfrak{osp}}_{a}(4|2)_{1}:&(1+a)t\xi_{i}+(a-1)\partial_{\eta_{i}}\Xi\text{
and }(1+a)t\eta_{i}+(a-1)\partial_{\xi_{i}}\Xi,\text{ where }i=1,2;\\\
{\mathfrak{osp}}_{a}(4|2)_{2}:&\frac{1}{2}(1+a)t^{2}-(a-1)\Xi,\text{ where
}\Xi=\xi_{1}\xi_{2}\eta_{1}\eta_{2}.\end{array}$

Observe that
$i({\mathfrak{osp}}_{1}(4|2))\simeq{\mathfrak{osp}}(4|2;1)\simeq{\mathfrak{osp}}(4|2)$,
so the list of Bol operators in this case should be as in Theorem 3.5.1.
Theorem. It is, but
${\mathfrak{osp}}(4|2;1)\simeq{\mathfrak{osp}}(4|2;-2)\simeq{\mathfrak{osp}}(4|2;-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}})$,
see [BGL], whereas the answers for $a=-2$ and
$-\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{0.25pt}{$\dfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}{\raisebox{-0.5pt}{$\tfrac{1}{2}$}}$
are different, see Theorem 1.4.3. Theorem!

This is, however, correct: We will see that the corresponding spaces of
parameters $\alpha$ and $a$ have non-isomorphic symmetry groups.

Observe also that ${\mathfrak{osp}}(4|2;\alpha)$ is not simple for
$\alpha=0,-1$, and the list of Bols might differ in these cases. But we are
working with ${\mathfrak{osp}}_{a}(4|2)$, not with $\alpha$, and for it the
Bols are described by the general formula (8) evaluated at $a=0$ and as
$a\to\infty$ as well, which we gave separately for clarity. The algebra
${\mathfrak{osp}}_{-1}(4|2)$ is not simple and the Bols are predictably
different.

##### An explanation of the results of Theorem 3.5.2. Theorem

Recall that a parametric family Kaplansky discovered and denoted
$\Gamma(\sigma_{1},\sigma_{2},\sigma_{3})$, see [Kapp], contains deformations
of ${\mathfrak{osp}}(4|2)$, denoted here by ${\mathfrak{osp}}_{a}(4|2)$ or by
${\mathfrak{osp}}(4|2;\alpha)$, where the parameter $a$ was introduced in
formulas (63) whereas $\alpha$ is described in what follows.

For ${\mathfrak{g}}={\mathfrak{osp}}(4|2;\alpha)$, the bracket
$S^{2}({\mathfrak{g}}_{\bar{1}})\longrightarrow{\mathfrak{g}}_{\bar{0}}$,
where
${\mathfrak{g}}_{\bar{0}}={\mathfrak{sl}}_{1}(2)\oplus{\mathfrak{sl}}_{2}(2)\oplus{\mathfrak{sl}}_{3}(2)$
and
${\mathfrak{g}}_{\bar{1}}:=\mathop{\mathrm{id}}\nolimits_{1}\otimes\mathop{\mathrm{id}}\nolimits_{2}\otimes\mathop{\mathrm{id}}\nolimits_{3}$,
where the isomorphic copies of ${\mathfrak{sl}}(2)$ and their respective
tautological modules are numbered for convenience. The Lie superalgebra
structure on ${\mathfrak{g}}$ is completely defined by the bracket of any two
odd elements, i.e., by the mapping
$S^{2}({\mathfrak{g}}_{\bar{1}})\longrightarrow{\mathfrak{g}}_{\bar{0}}$
defined up to a permutation of the isomorphic summands ${\mathfrak{sl}}(2)$ of
${\mathfrak{g}}_{\bar{0}}$ or, which is the same, on non-vanishing
simultaneously coefficients $\sigma_{i}$, where $i=1,2,3$, with which the
projections
$S^{2}({\mathfrak{g}}_{\bar{1}})\longrightarrow{\mathfrak{sl}}_{i}(2)$ enter
the bracket. The Jacobi identity is satisfied if and only if
$\sigma_{1}+\sigma_{2}+\sigma_{3}=0$, see [Kapp]. Since the $\sigma_{i}$ are
defined up to a common non-zero multiple and at least two of them must be non-
zero, let $\sigma_{1}=1$ and
$\alpha:=\mathchoice{\raisebox{0.25pt}{$\dfrac{\sigma_{i}}{\sigma_{j}}$}}{\raisebox{0.25pt}{$\dfrac{\sigma_{i}}{\sigma_{j}}$}}{\raisebox{-0.5pt}{$\tfrac{\sigma_{i}}{\sigma_{j}}$}}{\raisebox{-0.5pt}{$\tfrac{\sigma_{i}}{\sigma_{j}}$}}$,
where $\sigma_{j}\neq 0$ for $\\{i,j\\}=\\{2,3\\}$. Thus, $S_{3}$ acts on the
space of parameter $\alpha$. For explicit conditions when
${\mathfrak{osp}}(4|2;\alpha_{1})\simeq{\mathfrak{osp}}(4|2;\alpha_{2})$, see
[BGL]. This family is often called $D(2,1;\alpha)$. Unlike
${\mathfrak{osp}}(4|2;\alpha)$ or ${\mathfrak{osp}}_{\alpha}(4|2)$, the
notation $D(2,1;\alpha)$ is ill-chosen as explained, e.g., in [CCLL].

We know, see [BGL], that $S_{3}$ acts on the parameter $\alpha$ of
${\mathfrak{osp}}(4|2;\alpha)$ with 6 elements in the orbit of general
position and 3 elements in the exceptional cases. The reader might wonder if
we lost some of the operators corresponding to the other values of parameter
of the isomorphic algebras. Let us show that we did not lose anything. For
brevity, we say “$f$ acts” meaning “$K_{f}$ acts”, etc.

The case of $a=-1$ is exceptional. Indeed, the space of
$i({\mathfrak{osp}}_{-1}(4|2))$ is isomorphic to
$\Lambda(\xi_{1},\xi_{2},\eta_{1},\eta_{2})\oplus{\mathbb{C}}t$.

In ${\mathfrak{g}}$, the subalgebra generated by
$\\{f\in\Lambda(\xi,\eta)\mid\int
f\mathop{\mathrm{{vol\/}}}\nolimits(\xi,\eta)=0\\}$ is an ideal isomorphic to
${\mathfrak{po}}^{\prime}(0|4)$; the elements $t$ and $\Xi$, see eq. (63), act
on the ideal by outer derivations. The Lie superalgebra
${\mathfrak{po}}^{\prime}(0|4)$ has an ideal generated by ${\mathbb{C}}\cdot
1$, the quotient is ${\mathfrak{h}}^{\prime}(0|4)\simeq{\mathfrak{psl}}(2|2)$.
The “additional” to ${\mathfrak{sl}}(2|2)$ elements $1,t,\Xi$ form a
subalgebra NOT isomorphic to ${\mathfrak{sl}}(2)$, because $\\{1,\Xi\\}=0$.

If $a\neq-1$, set

$b:=\mathchoice{\raisebox{0.25pt}{$\dfrac{a-1}{a+1}$}}{\raisebox{0.25pt}{$\dfrac{a-1}{a+1}$}}{\raisebox{-0.5pt}{$\tfrac{a-1}{a+1}$}}{\raisebox{-0.5pt}{$\tfrac{a-1}{a+1}$}},\quad
A_{i}:=t\xi_{i}+b\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial\Xi}{\partial\eta_{i}}$}}{\raisebox{0.25pt}{$\dfrac{\partial\Xi}{\partial\eta_{i}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial\Xi}{\partial\eta_{i}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial\Xi}{\partial\eta_{i}}$}},\;B_{i}:=t\eta_{i}+b\mathchoice{\raisebox{0.25pt}{$\dfrac{\partial\Xi}{\partial\xi_{i}}$}}{\raisebox{0.25pt}{$\dfrac{\partial\Xi}{\partial\xi_{i}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial\Xi}{\partial\xi_{i}}$}}{\raisebox{-0.5pt}{$\tfrac{\partial\Xi}{\partial\xi_{i}}$}}\;(i=1,2),\quad
F:=-t^{2}+2b\Xi.$

Then, the non-zero brackets of $[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{-1}]$ are
only $\\{\xi_{i},\eta_{i}\\}=1$, and the non-zero brackets of
$[{\mathfrak{g}}_{1},{\mathfrak{g}}_{1}]$ are only $\\{A_{i},B_{i}\\}=F$.
Besides,

$\\{1,A_{i}\\}=2\xi_{i},\;\\{1,B_{i}\\}=2\eta_{i}\quad\text{ and
}\\{F,\xi_{i}\\}=2A_{i},\;\\{F,\eta_{i}\\}=2B_{i}.$

Let us now look how
${\mathfrak{g}}_{0}={\mathfrak{o}}(4)\oplus{\mathbb{C}}t\simeq{\mathfrak{sl}}(2)_{1}\oplus{\mathfrak{sl}}(2)_{2}\oplus{\mathbb{C}}t$
acts. The element $t$ is a grading operator
$\mathop{\mathrm{ad}}\nolimits_{t}|_{{\mathfrak{g}}_{k}}=k\cdot\mathop{\mathrm{id}}\nolimits$.

The elements

$x=u(\xi_{1}\eta_{1}-\xi_{2}\eta_{2})+v\xi_{1}\eta_{2}+w\xi_{2}\eta_{1}$,
where $u,v,w\in{\mathbb{C}}$

span an ideal in ${\mathfrak{g}}_{0}$ isomorphic to ${\mathfrak{sl}}(2)$. We
denote it ${\mathfrak{sl}}(2)_{1}$ for definiteness. The matrix of the
restriction of $\mathop{\mathrm{ad}}\nolimits_{x}$ to ${\mathfrak{g}}_{-1}$ in
the basis $\xi_{1},\xi_{2},\eta_{1},\eta_{2}$, and to ${\mathfrak{g}}_{1}$ in
the basis $A_{1},A_{2},B_{1},B_{2}$ is

$\begin{pmatrix}-U^{t}&0\\\ 0&U\\\ \end{pmatrix},\text{
where~{}~{}}U=\begin{pmatrix}u&w\\\ v&-u\\\ \end{pmatrix}.$

The elements

$y=\lambda(\xi_{1}\eta_{1}+\xi_{2}\eta_{2})+\mu\xi_{1}\xi_{2}+\nu\eta_{1}\eta_{2}$,
where $\lambda,\mu,\nu\in{\mathbb{C}}$

also span an ideal in ${\mathfrak{g}}_{0}$ isomorphic to ${\mathfrak{sl}}(2)$.
We denote it ${\mathfrak{sl}}(2)_{2}$.

The matrix of restriction of the operator $\mathop{\mathrm{ad}}\nolimits_{y}$
to ${\mathfrak{g}}_{-1}$ in the basis $\xi_{1},\xi_{2},\eta_{1},\eta_{2}$, and
to ${\mathfrak{g}}_{1}$ in the basis $A_{1},A_{2},B_{1},B_{2}$ is

$\begin{pmatrix}-\lambda&0&0&-\mu\\\ 0&-\lambda&\mu&0\\\ 0&-\nu&\lambda&0\\\
\nu&0&0&\lambda\\\ \end{pmatrix}.$

We see that the parameter $b$ arises only in commutators in
$[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{1}]$. Explicitly, they are (for
$i,j=1,2$):

$\begin{array}[]{l}\\{\xi_{i},A_{i}\\}=\\{\eta_{i},B_{i}\\}=0,\\\
\\{\xi_{i},A_{j}\\}=(1+b)\xi_{i}\xi_{j},\ \
\\{\xi_{i},B_{j}\\}=(1-b)\xi_{i}\eta_{j},\\\
\\{\eta_{i},B_{j}\\}=(1+b)\eta_{i}\eta_{j},\ \
\\{\eta_{i},A_{j}\\}=(1-b)\eta_{i}\xi_{j}\text{~{}~{}for $i\neq
j$}.\end{array}$

Finally,

$\begin{array}[]{ll}\\{\xi_{1},B_{1}\\}=-t+\xi_{1}\eta_{1}+b\xi_{2}\eta_{2},&\\{\xi_{2},B_{2}\\}=-t+b\xi_{1}\eta_{1}+\xi_{2}\eta_{2},\\\
\\{\eta_{1},A_{1}\\}=-t-\xi_{1}\eta_{1}-b\xi_{2}\eta_{2},&\\{\eta_{2},A_{2}\\}=-t-b\xi_{1}\eta_{1}-\xi_{2}\eta_{2}.\end{array}$

We see that for $b=-1$, which is equivalent to $a=0$, we have
$[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{1}]\simeq{\mathfrak{sl}}(2)_{1}\oplus{\mathbb{C}}t$.
So, ${\mathfrak{osp}}_{0}(4|2)$ contains an ideal
${\mathfrak{g}}_{-2}\oplus{\mathfrak{g}}_{-1}\oplus[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{1}]\oplus{\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{2}$,
isomorphic to ${\mathfrak{psl}}(2|2)$, on which ${\mathfrak{sl}}(2)_{2}$ acts
by outer derivations.

For $b=1$, which is equivalent to $a=\infty$, we have
$[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{1}]={\mathfrak{sl}}(2)_{2}\oplus{\mathbb{C}}t$.
So, ${\mathfrak{osp}}_{\infty}(4|2)$ contains an ideal
${\mathfrak{g}}_{-2}\oplus{\mathfrak{g}}_{-1}\oplus[{\mathfrak{g}}_{-1},{\mathfrak{g}}_{1}]\oplus{\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{2}$,
isomorphic to a copy of ${\mathfrak{psl}}(2|2)$ on which
${\mathfrak{sl}}(2)_{1}$ acts by outer derivations.

Observe immediately that possibility $a=\infty$ shows that the values of
parameter $a$ do not coincide with the same values of $\alpha$ from [BGL],
except occasionally, e.g. for $a=\alpha=1$.

Furthermore, the group acting on the space of parameter $a$ and establishing
an isomorphism
${\mathfrak{osp}}_{a_{1}}(4|2)\simeq{\mathfrak{osp}}_{a_{2}}(4|2)$ is not
isomorphic to the symmetry group of the parameter $\alpha$. Indeed, having
embedded ${\mathfrak{osp}}_{a}(4|2)$ in ${\mathfrak{k}}(1|4)$ in the case
where $a\neq-1$, we distinguish one ${\mathfrak{sl}}(2)$, spanned by $1,t$ and
$\frac{1}{2}(1+a)t^{2}-(a-1)\Xi$, see eq. (63). This embedding endows
${\mathfrak{g}}_{\bar{1}}$ with a ${\mathbb{Z}}$-grading which shows how the
brackets of two odd elements with values in this distinguished
${\mathfrak{sl}}(2)$ should look.

The parameter $a$ arises only in commuting ${\mathfrak{g}}_{-1}$ with
${\mathfrak{g}}_{1}$ with values in ${\mathfrak{g}}_{0}$ which is now
isomorphic to ${\mathfrak{sl}}(2)\oplus{\mathfrak{sl}}(2)\oplus{\mathbb{C}}$.
So now, we can only permute two copies of ${\mathfrak{sl}}(2)$, and hence the
group establishing isomorphisms is now $S_{2}$, not $S_{3}$.

The formulas for commutators show that $S_{2}$ is generated by the mapping
$b\mapsto-b$, or equivalently, $a\mapsto\frac{1}{a}$. The values $a=\pm 1$ are
exceptional points of this action: their orbits consist of one point, not of
two.

Explicitly, the isomorphism is given by the map

$f:\varphi(\xi_{1},\xi_{2},\eta_{1},\eta_{2})\mapsto\varphi(\xi_{1},\eta_{2},\eta_{1},\xi_{2}).$

######  Remarks.

1) In [BaL], Basdouri and Laraiedh described Bol’s operator in the case of
${\mathfrak{k}}(1|n)$ in realization preserving the distribution singled out
by the form $dt+\sum\theta_{i}d\theta_{i}$. This realization has advantage as
compared with our realization in terms of $\xi$ and $\eta$: the formulas in
[BaL] is uniform for all $n$. Basdouri and Laraiedh used this uniformity to
give an analytic proof of their result, which is also an advantage as compared
with computer-assisted proof which is justly considered less solid than
theoretical.

On the other hand, the $\xi\eta$ realization of ${\mathfrak{k}}(1|n)$ we
considered makes it possible to use weight decomposition and facilitates
computations (either by pencil or by computer).

2) Basdouri and Laraiedh did not consider the second parameter of the weight
densities when $n=2$, and therefore missed two odd Bol operators because the
main object of interest in [BaL] was
$H^{1}({\mathfrak{k}}(1|n),{\mathfrak{osp}}(n|2);\mathcal{F}_{\lambda})$, not
analogs of Bol operators.

3) The ground field in [BaL] is ${\mathbb{R}}$, but the answer is true for any
field of characteristic zero.

##  4\. The
${\mathfrak{pgl}}(m|n+2)\simeq{\mathfrak{pgl}}(n+2|m)\subset{\mathfrak{k}}(2n+1|2m)$
cases

Let $x$ be indeterminates $m$ of which are even and $n$ are odd, and two more
indeterminates $\xi$ and $\eta$ of the same parity. On
${\mathfrak{pgl}}(m|n+2)$ realized by linear differential operators, consider
the grading given by

$\deg x_{i}=1\text{~{}~{}for all~{}~{}}i;\ \ \deg\xi=0;\ \ \deg\eta=2.$

By convention (7), we do not consider the case where $\xi$ and $\eta$ are both
even. Then, the elements of ${\mathfrak{g}}={\mathfrak{pgl}}(m|n+2)$, where
$\xi$ and $\eta$ are both odd, are represented by the vector fields

(64)
$\begin{array}[]{l}{\mathfrak{g}}_{-2}:={\mathbb{C}}\xi\partial_{\eta};\\\
{\mathfrak{g}}_{-1}:=\mathop{\mathrm{{Span}}}\nolimits\\{\xi\partial_{x_{i}}\text{~{}~{}and~{}~{}}x_{i}\partial_{\eta}\\};\\\
{\mathfrak{g}}_{0}:=\mathop{\mathrm{{Span}}}\nolimits\\{x_{i}\partial_{x_{j}},\xi\partial_{\xi};\eta\partial_{\eta}\\}\text{~{}~{}modulo
center}.\\\ \end{array}$

By convention (7), we also have $n=0$, so in terms of supermatrices, we
consider the following ${\mathbb{Z}}$-grading of
${\mathfrak{h}}:={\mathfrak{gl}}(m|2)$: the main block-diagonal elements are
of degree 0, the block-diagonal elements just above (resp. below) the main
diagonal are of degree 1 (resp. $-1$), etc.

(65) $\begin{pmatrix}a&r&b\\\ s&A&z\\\ c&u&d\end{pmatrix},\text{where
$A\in{\mathfrak{gl}}(m)$, $a,b,c,d\in{\mathbb{C}}$}.$

Since ${\mathfrak{sl}}(1|2)\simeq{\mathfrak{osp}}(2|2)$, the case where $n=0$,
$m=1$ is already considered in Theorem 3.5.2. Theorem.

In terms of $K_{f}\in{\mathfrak{k}}(1|2m)$, where
$f\in{\mathbb{C}}[t;\xi_{j},\eta_{j}]_{j=1,2,\dots,m}$, see eq. (13), we
should consider the following cases.

We realize the elements (65) of ${\mathfrak{h}}_{\geq 0}$ by means of contact
vector fields where $F:=\sum\xi_{i}\eta_{i}$

$\begin{array}[]{l}{\mathfrak{h}}_{-2}=\text{$c\mapsto cK_{1}$};\\\
{\mathfrak{h}}_{-1}=\text{$s_{i}\mapsto s_{i}K_{\xi_{i}}$, $u_{j}\mapsto
u_{j}K_{\eta_{j}}$};\\\
{\mathfrak{h}}_{0}=\mathop{\mathrm{{Span}}}\nolimits(K_{t},K_{\xi_{i}\eta_{j}})_{i,j=1,2,\dots,m};\\\
{\mathfrak{h}}_{1}=\mathop{\mathrm{{Span}}}\nolimits\\{\eta_{i}(t+F),\
\xi_{i}(t-F)\\}_{i=1,2,\dots,m};\\\
{\mathfrak{h}}_{2}=\mathop{\mathrm{{Span}}}\nolimits\\{(t+F)(t-F)\\}\end{array}$

######  Theorem (Bol’s analogs of $i({\mathfrak{pgl}}(2|2))$-invariant
operators).

The only $i({\mathfrak{pgl}}(2|2))$-invariant differential operators are

$A^{k}:{\mathcal{F}}_{-(k+3)}\longrightarrow{\mathcal{F}}_{k+1}\text{~{}~{}given
by~{}~{}}\varphi\alpha_{1}^{-(k+3)/{2}}\mapsto\bar{A}(\varphi)\alpha_{1}^{(k+1)/{2}}\text{for
any $k\geq 0$},$

where (for the definition of the super anti-commutator, see (34))

(66)
$\bar{A}(\varphi)=[D_{\eta_{1}},D_{\xi_{1}}]_{+}\circ[D_{\eta_{2}},D_{\xi_{2}}]_{+}\circ\partial_{t}^{k}(\varphi)+\mathchoice{\raisebox{0.25pt}{$\dfrac{4}{(k+2)}$}}{\raisebox{0.25pt}{$\dfrac{4}{(k+2)}$}}{\raisebox{-0.5pt}{$\tfrac{4}{(k+2)}$}}{\raisebox{-0.5pt}{$\tfrac{4}{(k+2)}$}}\partial_{t}^{k+2}(\varphi),$

and one more operator:

$B:{\mathcal{F}}_{-2}\longrightarrow{\mathcal{F}}_{0}:\ \
\varphi\alpha_{1}^{-1}\mapsto\bar{B}(\varphi),$

where

(67)
$\bar{B}(\varphi)=[D_{\eta_{1}},D_{\xi_{1}}]_{+}(\varphi)+[D_{\eta_{2}},D_{\xi_{2}}]_{+}(\varphi).$

###### Proof.

The proof follows the same lines as in the proof of Theorem 3.5.2. Theorem.
First, observe that since $i({\mathfrak{pgl}}(2|2))$ is a Lie subsuperalgebra
of $i({\mathfrak{osp}}(4|2;0))$, then the operator $\bar{A}$ must be
invariant. Proving that the operator $B$ is invariant is a matter of a direct
computation. We omit the details. ∎

######  Theorem (Bol’s analogs of $i({\mathfrak{sl}}(m|2))$-invariant
operators for $m\geq 3$).

There is only one operator:

$B:{\mathcal{F}}_{-m}\longrightarrow{\mathcal{F}}_{2-m}:\ \
\varphi\alpha_{1}^{-m/2}\mapsto\bar{B}(\varphi)\alpha_{1}^{(2-m)/2}$

where

$\bar{B}(\varphi)=\sum_{1\leq i\leq
m}[D_{\eta_{i}},D_{\xi_{i}}]_{+}(\varphi).$

###### Proof.

Similar to the one given in Theorem 3.5.2. Theorem. ∎

##  5\. The ${\mathfrak{pe}}(n+1)\subset{\mathfrak{m}}(n)$ cases

Since the Lie superalgebras ${\mathfrak{spe}}(n)$ are simple only for $n\geq
3$, we consider the case of subalgebra ${\mathfrak{pe}}(2)$ separately.
Besides, the case of ${\mathfrak{m}}(1)$ is exceptional, see (9). Moreover, by
convention (7), we consider only he case $n=1$. In the next Theorem, do not
confuse the order of the operator $\partial_{\tau}$ which is equal to 1 with
its degree which is equal to $-2$. Recall expressions (37).

######  Theorem (Bol’s analogs for $i({\mathfrak{pe}}(2))$-invariant
operators).

The only $i(\mathfrak{pe}(2))$-invariant linear differential operators between
spaces of weighted densities are as follows. For every $k\in{\mathbb{N}}$,
there exists an order $k$ odd $i(\mathfrak{pe}(2))$-invariant linear
differential operator

$\begin{array}[]{l}A^{k}:{\mathcal{F}}_{\frac{k-1}{2},\
\nu}\longrightarrow{\mathcal{F}}_{\frac{1-k}{2},\ \nu-1+\frac{k}{2}}\text{,
where}\\\
\bar{A}^{k}=D_{q}(-D_{\xi})^{k-1}+(k-1)(-D_{\xi})^{k-2}\partial_{\tau}.\end{array}$

###### Proof.

Same routine as above. ∎

Acknowledgments. S.B. and D.L. were supported by the grant NYUAD 065.

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