# On gradings modulo $2$ of simple Lie algebras in characteristic $2$

Andrey Krutova, Alexei Lebedevb aInstitute of Mathematics, Polish Academy of
Sciences, ul. Śniadeckich 8, 00-656 Warszawa, Poland; Independent University
of Moscow, Bolshoi Vlasyevskij Pereulok 11, 119002, Moscow, Russia;
<EMAIL_ADDRESS><EMAIL_ADDRESS>bEqua Simulation AB, Stockholm,
Sweden<EMAIL_ADDRESS>
###### Abstract.

The ground field in the text is of characteristic 2. The classification of
modulo 2 gradings of simple Lie algebras is vital for the classification of
simple finite-dimensional Lie superalgebras: with each grading, a simple Lie
superalgebra is associated, see arXiv:1407.1695. No classification of gradings
was known for any type of simple Lie algebras, bar restricted Zassenhaus
algebras (a.k.a. Witt algebras, i.e., Lie algebras of vector fields with
truncated polynomials as coefficients) on not less than 3 indeterminates.

Here (under a natural assumption about their algebras of derivations) we
completely describe gradings modulo 2 for several series of simple (relatives
of the) Lie algebras: of special linear series (except for $\mathfrak{psl}(4)$
for which a conjecture is given), two inequivalent orthogonal, and
projectivizations of their derived Lie algebras. The classification of
gradings is new but all of the corresponding superizations are known.

For the Zassenhaus algebras on one indeterminate of height $n>1$, there is an
$(n-2)$-parametric family of modulo 2 gradings; all but one of the
corresponding simple Lie superalgebras are new. Our classification yields non-
triviality of a deformation of a simple $(3|2)$-dimensional Lie superalgebra
(new result).

###### Key words and phrases:

modular vectorial Lie algebra, characteristic 2, simple Lie algebra, simple
Lie superalgebra

###### 2010 Mathematics Subject Classification:

Primary 17B50, 17B20, 17B70

##  1\. Introduction

###  1.1. Basic definitions

Hereafter ${\mathbb{K}}$ is an algebraically closed field of characteristic
$p=2$, unless otherwise specified; all algebras are finite-dimensional; for a
review of simple vectorial Lie (super)algebras over ${\mathbb{K}}$ and basic
background, see [BGLLS2].

####  1.1.1. Lie superlagebras in characteristic 2.

A Lie superalgebra is a superspace
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ such
that the even part ${\mathfrak{g}}_{\bar{0}}$ is a Lie algebra, the odd part
${\mathfrak{g}}_{\bar{1}}$ is a ${\mathfrak{g}}_{\bar{0}}$-module, and on
${\mathfrak{g}}_{\bar{1}}$, a squaring (roughly speaking, the halved bracket)
is defined as a map

$\displaystyle x\mapsto x^{2}\quad\text{such that $(ax)^{2}=a^{2}x^{2}$ for
any $x\in{\mathfrak{g}}_{\bar{1}}$ and $a\in{\mathbb{K}}$, and}$
$(x+y)^{2}-x^{2}-y^{2}$ is a bilinear form on ${\mathfrak{g}}_{\bar{1}}$ with
values in ${\mathfrak{g}}_{\bar{0}}$.

Then the bracket of odd elements is defined to be

$[x,y]:=(x+y)^{2}-x^{2}-y^{2}.$

The Jacobi identity involving odd elements takes the following form:

(1) $\begin{array}[]{l}~{}[x^{2},y]=[x,[x,y]]\text{~{}for
any~{}}x\in{\mathfrak{g}}_{\bar{1}},y\in{\mathfrak{g}}_{\bar{0}},\\\
~{}[x^{2},x]=0\;\text{ for any $x\in{\mathfrak{g}}_{\bar{1}}$.}\end{array}$

####  1.1.2. Divided powers.

There are two natural integer bases of the commutative algebra
${\mathbb{C}}[x]$ of polynomials in $m$ indeterminates
$x=(x_{1},\ldots,x_{m})$: the monomial one and the basis of divided powers
constructed as follows. For any multi-index
${\underline{r}}=(r_{1},\ldots,r_{m})$, where $r_{1},\ldots,r_{m}$ are non-
negative integers, we set

$\small u_{i}^{(r_{i})}:=\frac{x_{i}^{r_{i}}}{r_{i}!}\quad\text{and}\quad
u^{({\underline{r}})}:=\prod_{1\leq i\leq m}u_{i}^{(r_{i})}.$

These $u^{({\underline{r}})}$ form an integer basis of ${\mathbb{C}}[x]$.
Their multiplication relations are

$\small u^{({\underline{r}})}\cdot
u^{(\underline{s})}=\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}\Big{)}u^{({\underline{r}}+\underline{s})},\quad\text{where}\quad\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{{\underline{r}}+\underline{s}}{{\underline{r}}}$}}\Big{)}=\prod_{1\leq
i\leq
m}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{r_{i}+s_{i}}{r_{i}}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{r_{i}+s_{i}}{r_{i}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{r_{i}+s_{i}}{r_{i}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{r_{i}+s_{i}}{r_{i}}$}}\Big{)}.$

For an arbitrary field ${\mathbb{K}}$ of characteristic $p>0$, we can consider
the commutative algebra ${\mathbb{K}}[u]$ spanned by the elements
$u^{({\underline{r}})}$ with multiplication relations written above. For any
$m$-tuple $\underline{N}=(N_{1},\ldots,N_{m})$, where $N_{i}$ are either
positive integers or infinity, denote (we set $p^{\infty}:=\infty$)

${\mathcal{O}}(m;\underline{N}):={\mathbb{K}}[u;\underline{N}]=\mathop{\mathrm{{Span}}}\nolimits_{\mathbb{K}}\left(u^{({\underline{r}})}\mid
r_{i}<p^{N_{i}}\right).$

The algebra ${\mathbb{K}}[u]$ and its subalgebras
${\mathbb{K}}[u;\underline{N}]$ are called the algebras of divided powers;
they are analogs of the polynomial algebra. Let $\mathbbmss{1}:=(1,\dots,1)$
denote the shearing vector $\underline{N}$ with the smallest values of heights
$N_{i}$ of the indeterminates.

Since any derivation $D$ of a given algebra is determined by the values of $D$
on the generators, we see that
${\mathfrak{der}}({\mathcal{O}}(m;\underline{N}))$ has more than $m$
functional parameters if $N_{i}\neq 1$ for at least one $i$. Define
distinguished partial derivatives $\partial_{i}:=\partial_{u_{i}}$ by setting

$\partial_{i}(u_{j}^{(k)})=\delta_{ij}u_{j}^{(k-1)}~{}\text{for any $k$}.$

The Lie algebra of all derivations
${\mathfrak{der}}({\mathcal{O}}(m;\underline{N}))$ turns out to be not so
interesting (it is not transitive, and hence can not be simple) as the general
vectorial Lie algebra of distinguished derivations:111For $p>0$, the Lie
algebra ${\mathfrak{vect}}(m;\underline{N})$ is called Zassenhaus algebra if
$\underline{N}\neq\mathbbmss{1}$; it is usually denoted $W(m;\underline{N})$
for any $m$ and $\underline{N}$; if $\underline{N}=\mathbbmss{1}$, it is
called Witt algebra. Zassenhaus algebras are simple if $m>1$; there is no
special name for the simple derived
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, see a discussion in [GZ]. The Lie
algebra of divergence-free, or “special”, vector fields is denoted
${\mathfrak{svect}}(m;\underline{N})$, usually abbreviated to
$S(m;\underline{N})$.

${\mathfrak{vect}}(m;\underline{N})=\mathop{\mathrm{{Span}}}\nolimits_{\mathbb{K}}\left(f\partial_{k}\mid
f\in{\mathcal{O}}(m;\underline{N}),\ k=1,\dots,m\right).$

####  1.1.3. $p$-structure.

For every $x\in{\mathfrak{g}}$, the operator
$(\mathop{\mathrm{ad}}\nolimits_{x})^{p}$ is a derivation of ${\mathfrak{g}}$.
If this derivation is an inner one for every $x\in{\mathfrak{g}}$, then the
Lie algebra ${\mathfrak{g}}$ is said to be restricted or having a
$p$-structure. More specifically, a $p$-structure on ${\mathfrak{g}}$ is a map
$[p]\colon{\mathfrak{g}}\to{\mathfrak{g}}$, $x\mapsto x^{[p]}$ such that

(2)
$\begin{array}[]{l}[x^{[p]},y]=(\mathop{\mathrm{ad}}\nolimits_{x})^{p}(y)\quad\text{for
any $x,y\in{\mathfrak{g}}$,}\\\ (ax)^{[p]}=a^{p}x^{[p]}\quad\text{~{}for
any~{}}a\in{\mathbb{K}},~{}x\in{\mathfrak{g}};\\\
(x+y)^{[p]}=x^{[p]}+y^{[p]}+\mathop{\sum}\limits_{1\leq i\leq
p-1}s_{i}(x,y)\quad\text{~{}for any~{}}x,y\in{\mathfrak{g}},\end{array}$

where $s_{i}(x,y)$ is the coefficient of $\lambda^{i-1}$ in
$(\mathop{\mathrm{ad}}\nolimits_{\lambda x+y})^{p-1}(x)$.

######  Remarks.

1) If the Lie algebra ${\mathfrak{g}}$ is without center, then the last two
conditions of (2) follow from the first one. There might be several
$p$-structures on one Lie algebra; all of them are equal modulo center. Hence,
on any simple Lie algebra, there is at most one $p$-structure.

2) For every $x\in{\mathfrak{g}}$, the operator
$(\mathop{\mathrm{ad}}\nolimits_{x})^{p}$ is a derivation of ${\mathfrak{g}}$.
If this derivation is an inner one for every $x\in{\mathfrak{g}}$, then the
following condition is sufficient for a Lie algebra ${\mathfrak{g}}$ to
possess a $p$-structure: for a basis $\\{g_{i}\\}_{i\in I}$ of
${\mathfrak{g}}$, there exist elements $g_{i}^{[p]}$ such that

$[g_{i}^{[p]},y]=(\mathop{\mathrm{ad}}\nolimits_{g_{i}})^{p}(y)\quad\text{~{}for
any~{}}y\in{\mathfrak{g}}.$

###  1.2. Problem

We consider the following problem, whose solution is well-known for $p=0$; for
a clear exposition of the solution, see the book [H] around p.500:

(3) For any finitely generated commutative group $G$, classify $G$-gradings of
simple finite-dimensional Lie algebras over ${\mathbb{K}}$.

Lie algebras for which a solution of Problem (3) is known. Although we are
interested in a particular case of Problem (3), let us briefly review the
known general results.

For $p\neq 2$, [Ko] is a very clear review of the cases where the solution of
Problem (3) has been found; for further details, see [BK, KPS]; for examples
of applications of certain $G$-gradings, see [Kos].

For $p=2$, the result of [BK] is as follows: all $G$-gradings of the Witt
algebra $W(m;\mathbbmss{1})$ (it is ${\mathfrak{vect}}(m;\mathbbmss{1})$ in
our notation) for $m\geq 3$ are given by $G$-gradings of the corresponding
algebra of divided powers $\mathcal{O}(m;\mathbbmss{1})$ due to an isomorphism
of their automorphism group schemes [Sk01]. The classification of such
gradings is given. Any $G$-grading of ${\mathcal{O}}(m;\mathbbmss{1})$ is
equivalent, up to an algebra automorphism, to one that can be described as
follows. For a given $s$ such that $0\leq s\leq m$, and $a_{1},\ldots,a_{m}\in
G$, set

(4)
${\mathcal{O}}_{g}=\mathop{\mathrm{{Span}}}\nolimits\left\\{(1+x_{1})^{j_{1}}\ldots(1+x_{s})^{j_{s}}x_{s+1}^{j_{s+1}}\ldots
x_{m}^{j_{m}}\mid j_{i}=0\text{~{}~{}or~{}~{}}1,\ \ \sum_{1\leq i\leq
m}j_{i}a_{i}=g\right\\},$

where $a_{1},\ldots,a_{m}\in G$ are the respective degrees of the
indeterminates generating $\mathcal{O}(m;\mathbbmss{1})$, i.e.,

$1+x_{1},\ldots,{1+x_{s}},\ x_{s+1},\ldots,x_{m}.$

For $p>0$ and the restricted Lie algebras considered in [BK] (for $p=2$: Witt
algebras; for $p>2$: Witt algebras, simple relatives of Hamiltonian algebras
and algebras of divergence-free vector fields), the only possible
${\mathbb{Z}}/2$-gradings are generated by those described in eq. (4). For
non-restricted Lie algebras, we know several examples of
${\mathbb{Z}}/2$-gradings of Kaplansky algebras, see [BLLS].

###  1.3. Problem (3) for $G={\mathbb{Z}}/2$ and $p=2$: an application of the
solution

In [BLLS1], there are offered two methods for constructing a simple finite-
dimensional Lie superalgebra from each simple finite-dimensional Lie algebra
over a field of characteristic 2; it is proved that every simple finite-
dimensional Lie superalgebra can be obtained by one of these two methods
(queerification and “method 2”). The “method 2” depends on the
${\mathbb{Z}}/2$-gradings of the simple Lie algebra which is being superized.
Let us recall the method:

Let
${\mathfrak{g}}={\mathfrak{g}}_{{\bar{0}}}\oplus{\mathfrak{g}}_{{\bar{1}}}$ be
a simple Lie algebra with a ${\mathbb{Z}}/2$-grading
$\mathop{\mathrm{gr}}\nolimits$. Let the 1-step restricted closure of
${\mathfrak{g}}$ associated with the grading $\mathop{\mathrm{gr}}\nolimits$
be

(5) $\begin{array}[]{ll}{\mathfrak{g}}^{<1>}:=&\text{the minimal Lie
subalgebra of the restricted closure $\overline{{\mathfrak{g}}}$ containing
${\mathfrak{g}}$}\\\ &\text{ and all the elements $x^{[2]}$, where
$x\in{\mathfrak{g}}_{{\bar{1}}}$.}\end{array}$

Clearly, there is a single way to extend the grading
$\mathop{\mathrm{gr}}\nolimits$ from ${\mathfrak{g}}$ to
${\mathfrak{g}}^{<1>}$; we assume this extension performed. On the space of
${\mathfrak{g}}^{<1>}$, define the structure of a Lie superalgebra denoted (in
what follows, we often omit indicating the grading
$\mathop{\mathrm{gr}}\nolimits$ since it enters the definition of
${\mathfrak{g}}^{<1>}$)

(6) $S({\mathfrak{g}}^{<1>},\mathop{\mathrm{gr}}\nolimits)\text{~{}~{} by
setting $x^{2}:=x^{[2]}$ for any $x\in{\mathfrak{g}}_{{\bar{1}}}$,}$

and retaining the bracket of any even elements with any other element. The Lie
superalgebra $S({\mathfrak{g}}^{<1>})$ is simple, see [BLLS1]. Our strategic
goal is classification of simple Lie (super)algebras, so we formulate our
description of ${\mathbb{Z}}/2$-gradings $\mathop{\mathrm{gr}}\nolimits$ of
${\mathfrak{g}}$ in terms of superizations $S({\mathfrak{g}}^{<1>})$ whenever
we can.

###  1.4. How we seek ${\mathbb{Z}}/2$-gradings

Let ${\mathfrak{g}}$ be a Lie algebra, and
${\mathfrak{g}}={\mathfrak{g}}_{{\bar{0}}}\oplus{\mathfrak{g}}_{{\bar{1}}}$
its ${\mathbb{Z}}/2$-grading. For an arbitrary $x\in{\mathfrak{g}}$, we denote
by $x_{{\bar{0}}}$ its even part and by $x_{{\bar{1}}}$ its odd part. We have

$[x,y]=[x_{{\bar{0}}}+x_{{\bar{1}}},y_{{\bar{0}}}+y_{{\bar{1}}}]=[x_{{\bar{0}}},y_{{\bar{0}}}]+[x_{{\bar{1}}},y_{{\bar{1}}}]+[x_{{\bar{0}}},y_{{\bar{1}}}]+[x_{{\bar{1}}},y_{{\bar{0}}}]\text{~{}~{}for
any $x,y\in{\mathfrak{g}}$}.$

This implies

$([x,y])_{{\bar{0}}}=[x_{{\bar{0}}},y_{{\bar{0}}}]+[x_{{\bar{1}}},y_{{\bar{1}}}]\quad\text{~{}and}\quad([x,y])_{{\bar{1}}}=[x_{{\bar{0}}},y_{{\bar{1}}}]+[x_{{\bar{1}}},y_{{\bar{0}}}].$

Finally, let $U\in\mathop{\mathrm{End}}\nolimits({\mathfrak{g}})$, where
${\mathfrak{g}}$ is considered as a vector space, be the projection on the odd
part, i.e., $Ux=x_{{\bar{1}}}$ and $(I-U)x=x_{{\bar{0}}}$, where $I$ is the
identity operator. A linear operator $U$ is such a projection on the odd part
of ${\mathfrak{g}}$ in some ${\mathbb{Z}}/2$-grading if and only if it
satisfies the following conditions for all $x,y\in{\mathfrak{g}}$:

(7a) $\displaystyle U^{2}={}$ $\displaystyle U,$ (7b) $\displaystyle
U[x,y]={}$ $\displaystyle[Ux,(I-U)y]+[(I-U)x,Uy]=-2[Ux,Uy]+[Ux,y]+[x,Uy].$

Having fixed a basis in ${\mathfrak{g}}$, we express these conditions in terms
of structure constants $c^{ij}_{k}$ as (summation over repeated indices is
assumed)

(8a) $\displaystyle U^{k}_{j}U^{i}_{k}={}$ $\displaystyle U^{i}_{j};$ (8b)
$\displaystyle c^{ij}_{l}U^{l}_{k}={}$ $\displaystyle
c^{il}_{k}U^{j}_{l}+c^{lj}_{k}U^{i}_{l}-2c^{lm}_{k}U^{i}_{l}U^{j}_{m}.$

For $p=2$, eq. (8b) becomes linear and easy (for example, for Mathematica-
based computer package SuperLie, see [Gr]) to solve. There remains, however, a
problem to be solved:

(9a) $\displaystyle\text{condition \eqref{cond2nonlin} is still quadratic};$
(9b) we need equivalence classes of solutions $U$ mod
$\mathop{\mathrm{{Aut\/}}}\nolimits({\mathfrak{g}})$, not individual
operators.

####  1.4.1. A comment: gradings and derivations.

For a general discussion of relation between gradings and derivations, and
interesting examples for $p=2$, see [BLLS]. For $p=3$, an example illustrating
the said discussion is the ${\mathbb{Z}}/4$-grading of the Skryabin algebra
${\mathfrak{by}}$, see [GL].

The Lie algebra ${\mathfrak{der}}\,{\mathfrak{vect}}(m;\underline{N})$ of all
derivations of ${\mathfrak{vect}}(m;\underline{N})$ coincides with the
$p$-envelop of ${\mathfrak{vect}}(m;\underline{N})$. Hence, the problem of
classification of equivalence classes of ${\mathbb{Z}}/p$-gradings reduces to
the classification of equivalence classes of toral elements in
${\mathfrak{der}}\,{\mathfrak{vect}}(m;\underline{N})$. The equivalence
classes of maximal tori with respect to the group of automorphisms are
described in [Kuz, T] and the answer depends on $N_{1}-2$ parameters for
$m=1$.

###  1.5. Our results

1) We classify the ${\mathbb{Z}}/2$-gradings of the Lie algebras of series
${\mathfrak{sl}}$, their simple subquotients, and simple derived or
subquotients of both orthogonal series ${\mathfrak{o}}_{I}$ and
${\mathfrak{o}}_{\Pi}$, (under an assumption of the structure of Lie algebras
of their derivations in general case; this assumption is proved for ranks
$\leq 8$ in [BGLL2]) except for ${\mathfrak{psl}}(4)$, see subsec. 1.6.1. In
all cases considered, these gradings yield the known superizations of the
corresponding Lie algebras.

Note that one of the gradings of ${\mathfrak{o}}^{(1)}(3)$ has no analogs
among the ${\mathbb{Z}}/2$-gradings of the simple 3-dimensional Lie algebras
for $p\neq 2$. This unusual grading has analogs among
${\mathbb{Z}}/2$-gradings of the Zassenhaus algebras
${\mathfrak{vect}}^{(1)}(1;\underline{n})$.

2) The classification of ${\mathbb{Z}}/2$-gradings of
${\mathfrak{o}}^{(1)}(3)$ has one more application: it gives the shortest
known proof of the fact that the deform of
${\mathfrak{oo}}_{I\Pi}^{(1)}(1|2)$, one of the two superizations of
${\mathfrak{o}}^{(1)}(3)$, found in [BGL1], is a “true” one, not
“semitrivial”, see Remark 6.3.3. Remarks; this fact is new and unexpected
(although the deform itself, ${\mathfrak{oo}}_{II}^{(1)}(1|2)$, is known).

3) We describe $(n-2)$-parametric collection of ${\mathbb{Z}}/2$-gradings of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$. For $n=2$, these gradings yield
three non-isomorphic Lie superalgebras.

For $n>2$, these ${\mathbb{Z}}/2$-grading yield

a) purely even Lie superalgebra ${\mathfrak{vect}}^{(1)}(1;\underline{n}|0)$;

b) $(n-2)$-parametric family of filtered deforms of
${\mathfrak{k}}(1;\underline{n-1}|1)$;

c) the filtered deform of
${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$.

####  1.5.1. Examples where ${\mathbb{Z}}/2$-gradings of “the same algebra”
for $p\neq 2$ and $p=2$ differ or where previously unknown
${\mathbb{Z}}/2$-gradings have been found.

These are the most interesting of our results.

1) For any $p$, there is only one simple Lie algebra of dimension 3: for
$p\neq 2$, it is ${\mathfrak{o}}(3)\simeq{\mathfrak{sl}}(2)$; for $p=2$, it is
${\mathfrak{o}}^{(1)}(3)$, the derived of ${\mathfrak{o}}(3)$.

For $p\neq 2$, there is only one nontrivial ${\mathbb{Z}}/2$-grading (as
always, up to an automorphism), and hence if the superization procedure had
been defined for $p\neq 2$, the superization of ${\mathfrak{sl}}(2)$ would
have been unique, ${\mathfrak{sl}}(1|1)$.

For $p=2$, there are 2 inequivalent nontrivial ${\mathbb{Z}}/2$-gradings of
${\mathfrak{o}}^{(1)}(3)$, as we will see below. The corresponding non-
isomorphic, see [Leb], Lie superalgebras are
${\mathfrak{oo}}^{(1)}_{I\Pi}(1|2)$ and ${\mathfrak{oo}}^{(1)}_{II}(1|2)$. (It
is interesting to find out if the Lie superalgebra
${\mathfrak{o}}{\mathfrak{o}}_{II}^{(1)}(1|2n)$ is a deform of
${\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2n)$ for $n>1$; this requires to
consider ${\mathbb{Z}}/2$-gradings and deforms of
${\mathfrak{k}}(2n+1;\mathbbmss{1})$ for $n>0$.)

2) For $p\neq 2$, the ${\mathbb{Z}}/2$-gradings of the Zassenhaus algebra
${\mathfrak{vect}}(1;\underline{n})$ were only known for $n=1$, i.e., for the
restricted algebras, see [BK] (and, for $n>1$, obvious
${\mathbb{Z}}/2$-gradings with some of either $x$ or $x+1$ declared odd). We
have found new $(n-2)$-parametric family of ${\mathbb{Z}}/2$-gradings in the
non-restricted case. These gradings yeild superizations which have no known
analogs in the case $p\neq 2$.

###  1.6. Open questions

For $G={\mathbb{Z}}/2$, the gradings (4) with $s=0$ of
${\mathfrak{vect}}(m;\mathbbmss{1})$ for $m>2$ yield
${\mathfrak{vect}}(k;\mathbbmss{1}|m-k)$, where the vectors $\mathbbmss{1}$
have $m$ and $k$ coordinates, respectively, i.e., known superizations. If
$s\neq 0$ in (4) and $a_{i}={\bar{1}}$ for some of the indices $i\leq s$, then
the corresponding indeterminates $1+x_{i}$ are odd; at the moment we are
unable to identify the resulting Lie superalgebra.

The list of simple Lie algebras for which Problem (3) is open:

$\bullet$ For $p>3$, the simple vectorial Lie algebras for the shearing vector
$\underline{N}\neq\mathbbmss{1}$, and the deforms (results of deformations)
thereof (for their classification and description, see [S, Kos, SkH]);

$\bullet$ For $p=3$, the simple vectorial Lie algebras; in particular,
exceptional ones, mainly discovered by Skryabin and lucidly described in [GL],
and their deforms to be described (for a review, see [BGL1]); the deforms of
${\mathfrak{o}}(5)$ and Brown algebras described in [BLW];

$\bullet$ For $p=2$, there are many examples of simple Lie algebras of types
not existing for $p\neq 2$, see [SkT1, Ei, BGLL2, BLLS, GZ, BGLLS1, BGLLS2],
and [BGL, BGL1].

####  1.6.1. Two conjectures (D. Leites).

1) For (the simple derived of) Lie algebras with indecomposable Cartan matrix
not considered in this note, i.e., parametric families of Weisfeiler-Kac
algebras, all their ${\mathbb{Z}}/2$-gradings correspond, conjecturally, to
their known superizations listed in [BGL].

2) Since
${\mathfrak{psl}}(4)\simeq{\mathfrak{h}}_{\Pi}^{(1)}(4;\mathbbmss{1})\simeq{\mathfrak{svect}}^{(1)}(3;\mathbbmss{1})$,
see [ChKu], it is natural to conjecture that all ${\mathbb{Z}}/2$-gradings of
${\mathfrak{psl}}(4)$ correspond to superizations of superdimension $8|6$,
namely

${\mathfrak{svect}}^{(1)}(0|3)\simeq{\mathfrak{psl}}(3|1)$,
${\mathfrak{svect}}^{(1)}(2;\mathbbmss{1}|1)$,
${\mathfrak{h}}_{I\Pi}^{(1)}(1;\mathbbmss{1}|3)$,
${\mathfrak{h}}_{II}^{(1)}(3;\mathbbmss{1}|1)$,

and of superdimension $6|8$, namely

${\mathfrak{h}}_{\Pi}^{(1)}(0|4)\simeq{\mathfrak{psl}}(2|2)$,
${\mathfrak{h}}_{\Pi\Pi}^{(1)}(2;\mathbbmss{1}|2)$,
${\mathfrak{svect}}^{(1)}(1;\mathbbmss{1}|2)$,

and two more, most interesting, superizations of the same type as described,
for any simple vectorial Lie algebra and its standard ${\mathbb{Z}}$-grading,
in §6 of [BGLLS2]; the natural ${\mathbb{Z}}$-gradings of these two
superizations are supported in degrees $-2$ to 1, and their superdimensions
are $10|8$ for ${\mathfrak{h}}_{\Pi}^{(1)}(4;\mathbbmss{1})$ and $11|6$ for
${\mathfrak{svect}}^{(1)}(3;\mathbbmss{1})$.

####  1.6.2. A technical hypothesis.

In §2, §3, §4, and §5, we use the conjectural description of all derivations
of Lie algebras ${\mathfrak{sl}}(n)$, ${\mathfrak{psl}}(2n)$, as well as
${\mathfrak{o}}_{\Pi}(2n)$ and ${\mathfrak{o}}_{I}(n)$ and their simple
relatives, see [BGLL2]. This description is proved if
$\mathop{\mathrm{rk}}\nolimits{\mathfrak{g}}\leq 8$, and sometimes for any
$n$.

##  2\. The ${\mathfrak{sl}}(n)$ and ${\mathfrak{psl}}(2n)$ for $n>2$: same
answer as for $p\neq 2$

Clearly, ${\mathfrak{sl}}(0|n)={\mathfrak{sl}}(n)$ and
${\mathfrak{psl}}(0|2n)={\mathfrak{psl}}(2n)$.

######  Theorem.

All ${\mathbb{Z}}/2$-gradings of the Lie algebras ${\mathfrak{sl}}(n)$ for
$n>2$ and $n\neq 4$ and ${\mathfrak{psl}}(2n)$ for $n>2$ are analogous to
those for $p=0$, i.e., correspond to ${\mathfrak{sl}}(k|n-k)$ for
$k=0,\ldots,\left[\frac{n}{2}\right]$ and ${\mathfrak{psl}}(k|2n-k)$ for
$k=0,\ldots,n$.

####  2.1.1. Proof: the ${\mathfrak{sl}}(2n+1)$ series.

All derivations of ${\mathfrak{sl}}(2n+1)$ are inner ones, see [BGLL2]. Then
the general solution of the linear equation (7b) is
$U=\mathop{\mathrm{ad}}\nolimits_{A}$, where $A\in{\mathfrak{sl}}(2n+1)$. The
nonlinear equation (7a) reduces then to
$\mathop{\mathrm{ad}}\nolimits_{A}=(\mathop{\mathrm{ad}}\nolimits_{A})^{2}$.
Now recall that on ${\mathfrak{sl}}(2n+1)$ there is the 2-structure given by
$A^{[2]}=A^{2}$, see [BGL, BLLS1]. Therefore

$\mathop{\mathrm{ad}}\nolimits_{A}=(\mathop{\mathrm{ad}}\nolimits_{A})^{2}=\mathop{\mathrm{ad}}\nolimits_{A^{[2]}}=\mathop{\mathrm{ad}}\nolimits_{A^{2}};$

and since ${\mathfrak{sl}}(2n+1)$ has no center, we have $A=A^{2}$, i.e., $A$
is a projection. Since its trace is equal to the dimension of the subspace
onto which the projection projects (just look at the normal form of the matrix
of the projection), it follows that

(10) a given projection $A$ belongs to ${\mathfrak{sl}}$ if and only if it is
a projection onto a subspace of even dimension.

If $A\in{\mathfrak{sl}}(2n+1)$ is a projection, then the Lie superalgebra
obtained from it by “method 2” of [BLLS1] is isomorphic to

${\mathfrak{sl}}(\dim\mathop{\mathrm{Ker}}\nolimits A|\dim\mathop{\text{\rm
Im}}\nolimits A)\cong{\mathfrak{sl}}(\dim\mathop{\text{\rm Im}}\nolimits
A|\dim\mathop{\mathrm{Ker}}\nolimits A).$

So we see there are $n+1$ equivalence classes of ${\mathbb{Z}}/2$-gradings of
${\mathfrak{sl}}(2n+1)$ and $n+1$ of its nonisomorphic superizations
(including the trivial purely even one). We can enumerate them either as

${\mathfrak{sl}}(2n+1-2k|2k)\cong{\mathfrak{sl}}(2k|2n+1-2k),\text{~{}~{}where
$k=0,\dots,n$,}$

or, more simply, as ${\mathfrak{sl}}(k|2n+1-k)$, where $k=0,\dots,n$. So the
answer is the same as for $p\neq 2$.

####  2.1.2. Proof: the ${\mathfrak{sl}}(2n)$ series, $n>2$.

The Lie algebra ${\mathfrak{der}}\ {\mathfrak{sl}}(2n)$ can be identified, see
[BGLL2], with ${\mathfrak{pgl}}(2n)$ in the sense that for any
$D\in{\mathfrak{der}}\ {\mathfrak{sl}}(2n)$, there is
${A_{D}\in{\mathfrak{gl}}(2n)}$ such that $D$ coincides with the restriction
of $\mathop{\mathrm{ad}}\nolimits_{A_{D}}$ on ${\mathfrak{sl}}(2n)$. These
elements $A_{D}$ are defined modulo center; in particular, one can take
$A_{D^{2}}$ to be $(A_{D})^{2}$. Therefore $D$ satisfies the condition (7a) if
and only if $(A_{D})^{2}=A_{D}+cI_{2n}$ for some $c\in{\mathbb{K}}$. Let
$d\in{\mathbb{K}}$ be a root of the equation $d^{2}=d+c$ and set
$A^{\prime}_{D}=A_{D}+dI_{2n}$, then $(A^{\prime}_{D})^{2}=A^{\prime}_{D}$. So
we see that an operator $U$ on ${\mathfrak{sl}}(2n)$ satisfies the conditions
(7a) if and only if it can be represented as a restriction of
$\mathop{\mathrm{ad}}\nolimits_{A}$ on ${\mathfrak{sl}}(2n)$ for some
projection $A\in{\mathfrak{gl}}(2n)$, and then the Lie superalgebra obtained
from $U$ by “method 2” is isomorphic to
${\mathfrak{sl}}(\dim\mathop{\mathrm{Ker}}\nolimits A|\dim\mathop{\text{\rm
Im}}\nolimits A)\cong{\mathfrak{sl}}(\dim\mathop{\text{\rm Im}}\nolimits
A|\dim\mathop{\mathrm{Ker}}\nolimits A)$. So there are $n+1$ nonisomorphic
superizations of ${\mathfrak{sl}}(2n)$ (including the trivial purely even
one): ${\mathfrak{sl}}(k|2n-k)$, where $k=0,\dots,n$. Again, the answer is the
same as for $p\neq 2$.

####  2.1.3. Proof: the ${\mathfrak{psl}}(2n)$ series, $n>2$.

The algebra ${\mathfrak{der}}\ {\mathfrak{psl}}(2n)$ is isomorphic to
${\mathfrak{der}}\ {\mathfrak{sl}}(2n)$, see [BGLL2]. So the arguments from
the previous subsection apply. There are $n+1$ non-isomorphic superizations of
${\mathfrak{psl}}(2n)$ (including the purely even one), which can be
enumerated as ${\mathfrak{psl}}(k|2n-k)$, where $k=0,\dots,n$. ∎

####  2.1.4. Remark.

A ${\mathbb{Z}}/2$-grading of ${\mathfrak{gl}}(2n+1)$ for $n>2$ can produce a
Lie superalgebra which is not isomorphic to any superalgebra of the form
${\mathfrak{gl}}(k|2n+1-k)$. More specifically, such a superalgebra would be
isomorphic to the direct sum of a Lie superalgebra of the form
${{\mathfrak{sl}}(k|2n+1-k)}$, where $0\leq k\leq n$, and $0|1$-dimensional
center (i.e., 1-dimensional odd center). But since it does not result in any
new simple Lie superalgebra, we do not consider gradings of ${\mathfrak{gl}}$
here.

##  3\. The ${\mathfrak{o}}^{(1)}(2n+1)$ series

Clearly,
${\mathfrak{o}}{\mathfrak{o}}_{B_{1}B_{2}}(0|n)={\mathfrak{o}}_{B_{2}}(n)$ and
${\mathfrak{o}}{\mathfrak{o}}_{B_{1}B_{2}}(n|0)={\mathfrak{o}}_{B_{1}}(n)$.

###  3.1. Example: ${\mathfrak{o}}^{(1)}(3)$

Observe the occasional isomorphism
${\mathfrak{o}}^{(1)}(3)\simeq{\mathfrak{vect}}^{(1)}(1;\underline{2})$, so
this case is considered in Subsec. 6.3. In [BGL], the deforms of the
superization of ${\mathfrak{o}}^{(1)}(3)$ associated with one of the three
${\mathbb{Z}}/2$-gradings of ${\mathfrak{o}}^{(1)}(3)$ are described; but
until now it was unclear if these deforms are true ones, i.e., the deformed
algebra is not isomorphic to the initial algebra, as is the case for
semitrivial deformations corresponding to certain integrable cocycles from
nontrivial cohomology class, for examples, see [BLLS, Ri]. The results of
Subsec. 6.3 prove that one of the deforms found in [BGL] is a true one.

######  Theorem.

For any $n\geq 1$, all ${\mathbb{Z}}/2$-gradings of the Lie algebra
${\mathfrak{o}}^{(1)}(2n+1)$ correspond to Lie superalgebras
${\mathfrak{oo}}^{(1)}_{II}(2n+1-2k|2k)$, where $0\leq k\leq n$ (or, which is
the same but looks simpler, ${\mathfrak{oo}}^{(1)}_{II}(k|2n+1-k)$, where
$0\leq k\leq n$), and ${\mathfrak{oo}}^{(1)}_{I\Pi}(2n+1-2k|2k)$, where $1\leq
k\leq n$.

###### Proof.

The algebra of derivations of ${\mathfrak{o}}^{(1)}(2n+1)$ is
${\mathfrak{o}}(2n+1)/{\mathfrak{c}}$, the quotient of ${\mathfrak{o}}(2n+1)$
modulo center, or the subalgebra of traceless elements of
${\mathfrak{o}}(2n+1)$, see [BGLL2].

Again, an element $A\in{\mathfrak{der}}\ ~{}{\mathfrak{o}}^{(1)}(2n+1)$
describes a ${\mathbb{Z}}/2$-grading of ${\mathfrak{o}}^{(1)}(2n+1)$ if and
only if $A^{2}=A$, i.e., $A$ is a projection.

By definition, an operator $A\in{\mathfrak{gl}}(V)$ belongs to
${\mathfrak{o}}_{B}(V)$, where $B$ is a nondegenerate symmetric bilinear form,
if and only if

(11) $B(Ax,y)+B(x,Ay)=0\text{~{}~{}for all $x,y\in V$.}$

In what follows, we assume that $A$ is a projection. It is easy to check that
(11) is automatically satisfied if $x,y\in\mathop{\text{\rm Im}}\nolimits A$
or if $x,y\in\mathop{\mathrm{Ker}}\nolimits A$. If $x\in\mathop{\text{\rm
Im}}\nolimits A$ and $y\in\mathop{\mathrm{Ker}}\nolimits A$, then (11) is
equivalent to $B(x,y)=0$. So $A\in{\mathfrak{o}}_{B}(V)$ if and only if
$\mathop{\text{\rm Im}}\nolimits A$ and $\mathop{\mathrm{Ker}}\nolimits A$ are
orthogonal with respect to $B$. For $A$ to be traceless,
$\dim\mathop{\text{\rm Im}}\nolimits A$ must be even, see condition (10).

Let $A$ be a projection belonging to ${\mathfrak{der}}\
~{}{\mathfrak{o}}^{(1)}_{B}(2n+1)$ with $\dim\mathop{\text{\rm Im}}\nolimits
A=2k>0$. Denote restrictions of $B$ to $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ by $B_{\mathop{\text{\rm Im}}\nolimits A}$
and $B_{\mathop{\mathrm{Ker}}\nolimits A}$, respectively.

Since $B$ is nondegenerate, $\mathop{\text{\rm Im}}\nolimits
A\oplus\mathop{\mathrm{Ker}}\nolimits A=V$ and $\mathop{\text{\rm
Im}}\nolimits A\perp\mathop{\mathrm{Ker}}\nolimits A$ with respect to $B$, it
follows that these restrictions are nondegenerate.

Since $\dim\mathop{\mathrm{Ker}}\nolimits A=2n+1-2k$ is odd,
$B_{\mathop{\mathrm{Ker}}\nolimits A}$ is equivalent to $I_{2n+1-2k}$. As
$\dim\mathop{\text{\rm Im}}\nolimits A=2k$ is even, $B_{\mathop{\text{\rm
Im}}\nolimits A}$ may be equivalent to either $I_{2k}$ or $\Pi_{2k}$. In the
latter case, the resulting Lie superalgebra is isomorphic to
${\mathfrak{oo}}^{(1)}_{I\Pi}(2n+1-2k|2k)$; in the former case, it is
isomorphic to ${\mathfrak{oo}}^{(1)}_{II}(2k|2n+1-2k)$. Note that the
collection “${\mathfrak{oo}}^{(1)}_{II}(2k|2n+1-2k)$, where $1\leq k\leq n$”
can be described more simply as “${\mathfrak{oo}}^{(1)}_{II}(k|2n+1-k)$, where
$1\leq k\leq n$”.

So we get $2n+1$ nonisomorphic Lie superalgebras from
${\mathfrak{o}}^{(1)}(2n+1)$, including the purely even case. To see that all
those cases are really realizable with some projection $A$, the following
approach can be used. Recall that in an odd-dimensional space, all
nondegenerate symmetric forms are equivalent, so for any such form $B$ and any
symmetric invertible matrix, there is a basis in which the matrix of $B$ is
equal to the given one. In particular, for a given $k$, there are bases in
which the Gram matrix of $B$ is either $I_{2n+1}$ (call it the first basis) or
$\mathop{\mathrm{diag}}\nolimits(I_{2n+1-2k},\Pi_{2k})$ (call it the second
basis for this value of $k$). Take the operator $A$ whose matrix in the first
basis is $\mathop{\mathrm{diag}}\nolimits(0_{2n+1-2k},I_{2k})$; clearly,
$A^{2}=A$ and $\dim\mathop{\text{\rm Im}}\nolimits A=2k$. Relative the form
$B$, the space $\mathop{\text{\rm Im}}\nolimits A$, which is spanned by the
last $2k$ vectors of the basis, is, clearly, orthogonal to
$\mathop{\mathrm{Ker}}\nolimits A$, which is spanned by the first $2n+1-2k$
vectors of the basis. So $A\in{\mathfrak{der}}\ {\mathfrak{o}}_{B}$, and hence
determines a grading of ${\mathfrak{o}}^{(1)}_{B}$. It is easy to see that the
Lie superalgebra obtained by “method 2” of [BLLS1] is
${\mathfrak{oo}}^{(1)}_{II}(2n+1-2k|2k)$.

Analogously, an operator $A$ with the same matrix
$\mathop{\mathrm{diag}}\nolimits(0_{2n+1-2k},I_{2k})$ in the second basis (it
does not matter what is the matrix of $A$ in the first basis) determines a
grading that corresponds to the Lie superalgebra
${\mathfrak{oo}}^{(1)}_{I\Pi}(2n+1-2k|2k)$.

It is possible to show that there are $2n+1$ equivalence classes of
${\mathbb{Z}}/2$-gradings. It could have happened that there were more
gradings than nonisomorphic superizations if two inequivalent gradings yielded
isomorphic Lie superalgebras. This, however, does not happen; we skip the
details.∎

####  3.2.1. Remark.

Similarly to ${\mathfrak{gl}}(2n+1)$, it is clear that
${\mathbb{Z}}/2$-gradings of ${\mathfrak{o}}(2n+1)$ can produce Lie
superalgebras not isomorphic to any superalgebras of the form
${\mathfrak{oo}}_{II}(2n+1-2k|2k)$ or ${\mathfrak{oo}}_{I\Pi}(2n+1-2k|2k)$:
for example, a direct sum of ${\mathfrak{oo}}^{(1)}_{II}(2n+1-2k|2k)$ and
$0|1$-dimensional center. But again, these gradings do not produce any new
simple Lie superalgebras, so we do not consider such gradings of
${\mathfrak{o}}(2n+1)$ here.

##  4\. The ${\mathfrak{o}}^{(1)}_{I}(2n)$ series

######  Theorem.

For $n>2$, all ${\mathbb{Z}}/2$-gradings of the Lie algebra
${\mathfrak{o}}^{(1)}_{I}(2n)$ correspond to Lie superalgebras
${\mathfrak{oo}}^{(1)}_{II}(m|2n-m)$, where $1\leq m\leq n$, and
${\mathfrak{oo}}^{(1)}_{I\Pi}(2n-2k|2k)$, where $0\leq k\leq n$.

###### Proof.

If a bilinear form $B$ on vector space $V$ of dimension $2n$ is equivalent to
$I_{2n}$, then the algebra ${\mathfrak{der}}\ {\mathfrak{o}}^{(1)}_{B}(2n)$
can be identified, see [BGLL2], with ${\mathfrak{o}}_{B}(2n)/{\mathfrak{c}}$
in the sense that for any $D\in{\mathfrak{der}}\
{\mathfrak{o}}^{(1)}_{B}(2n)$, there is $A_{D}\in{\mathfrak{o}}_{B}(2n)$ such
that $D$ coincides with the restriction of
$\mathop{\mathrm{ad}}\nolimits_{A_{D}}$ on ${\mathfrak{o}}^{(1)}_{B}(2n)$.
These $A_{D}$ are defined up to a central element; in particular, one can take
$A_{D^{2}}$ to be $(A_{D})^{2}$. By arguments similar to the ones we used in
the case of ${\mathfrak{sl}}(2n)$, one can show that an operator $U$ on
${\mathfrak{o}}^{(1)}_{B}(2n)$ satisfies the conditions (7a) if and only if it
can be represented as a restriction of $\mathop{\mathrm{ad}}\nolimits_{A}$ on
${\mathfrak{o}}^{(1)}_{B}(2n)$ for some projection
$A\in{\mathfrak{o}}_{B}(2n)$.

By arguments we used in the previous section, a projection $A$ belongs to
${\mathfrak{o}}_{B}(2n)$ if and only if $\mathop{\text{\rm Im}}\nolimits A$
and $\mathop{\mathrm{Ker}}\nolimits A$ are orthogonal with respect to $B$. We
denote restrictions of $B$ to $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ as $B_{\mathop{\text{\rm Im}}\nolimits A}$
and $B_{\mathop{\mathrm{Ker}}\nolimits A}$, respectively. These restrictions
have to be nondegenerate.

If $\dim\mathop{\text{\rm Im}}\nolimits A=2k+1$, where $0\leq k\leq n-1$, then
these restrictions are equivalent to $I_{2k+1}$ and $I_{2n-2k-1}$,
respectively, and the resulting Lie superalgebra is isomorphic to

${\mathfrak{o}}^{(1)}_{II}(2k+1|2n-2k-1)\simeq{\mathfrak{o}}^{(1)}_{II}(2n-2k-1|2k+1).$

If $\dim\mathop{\text{\rm Im}}\nolimits A=2k$, where $1\leq k\leq n-1$, then
$B_{\mathop{\text{\rm Im}}\nolimits A}$ can be equivalent to either $I_{2k}$
or $\Pi_{2k}$, and $B_{\mathop{\mathrm{Ker}}\nolimits A}$ can be equivalent to
either $I_{2n-2k}$ or $\Pi_{2n-2k}$. However, it is impossible for
$B_{\mathop{\text{\rm Im}}\nolimits A}$ to be equivalent to $\Pi_{2k}$ while
$B_{\mathop{\mathrm{Ker}}\nolimits A}$ is equivalent to $\Pi_{2n-2k}$ at the
same time, because the direct sum of these two forms is equivalent to
$\Pi_{2n}$. All the other combinations are possible, and, depending on them,
the resulting Lie superalgebra can be isomorphic to either
${\mathfrak{o}}^{(1)}_{II}(2k|2n-2k)$, or
${\mathfrak{o}}^{(1)}_{I\Pi}(2k|2n-2k)$, or ${\mathfrak{o}}^{(1)}_{\Pi
I}(2k|2n-2k)\simeq{\mathfrak{o}}^{(1)}_{I\Pi}(2n-2k|2k)$.

If $\dim\mathop{\text{\rm Im}}\nolimits A=0$ or $2n$, then the resulting Lie
superalgebra is purely even. So, as described above,
${\mathfrak{o}}^{(1)}_{B}(2n)$ has $2n+1$ nonisomorphic superizations.∎

####  4.1.1. Remark.

The algebras ${\mathfrak{o}}^{(1)}_{I}(2)$ and ${\mathfrak{o}}^{(1)}_{I}(4)$
are not simple, so we do not consider them here.

##  5\. The simple relatives of ${\mathfrak{o}}_{\Pi}(2n)$ series

The algebras ${\mathfrak{o}}_{\Pi}(2)$ and ${\mathfrak{o}}_{\Pi}(4)$ do not
have simple relatives, so we do not consider them.

######  Theorem.

For $n>2$, all ${\mathbb{Z}}/2$-gradings of the simple relative of
${\mathfrak{o}}_{\Pi}(2n)$ (that is, of ${\mathfrak{o}}^{(2)}_{\Pi}(2n)$, if
$n$ is odd, or of ${\mathfrak{o}}^{(2)}_{\Pi}(2n)/{\mathfrak{c}}$, if $n$ is
even) correspond to the simple relatives of the corresponding superizations,
i.e., ${\mathfrak{oo}}^{(2)}_{\Pi\Pi}(2k|2n-2k)$, where $0\leq
k\leq\left[\frac{n}{2}\right]$, and ${\mathfrak{pe}}^{(2)}(n)$, if $n$ is odd,
or ${\mathfrak{oo}}^{(2)}_{\Pi\Pi}(2k|2n-2k)/{\mathfrak{c}}$, where $0\leq
k\leq\left[\frac{n}{2}\right]$, and ${\mathfrak{pe}}^{(2)}(n)/{\mathfrak{c}}$,
if $n$ is even.

###### Proof.

If $n$ is odd, then the center of ${\mathfrak{o}}^{(2)}_{\Pi}(2n)$ is trivial,
so in this case
${\mathfrak{o}}^{(2)}_{\Pi}(2n)/{\mathfrak{c}}\equiv{\mathfrak{o}}^{(2)}_{\Pi}(2n)$.
Keeping this in mind, we will denote the simple relative of
${\mathfrak{o}}_{\Pi}(2n)$ as ${\mathfrak{o}}^{(2)}_{\Pi}(2n)/{\mathfrak{c}}$
for both even and odd values of $n$. Similarly, we will denote the simple
relatives of ${\mathfrak{oo}}_{\Pi\Pi}(2k|2n-2k)$ and
${\mathfrak{pe}}^{(2)}(n)$ as
${\mathfrak{oo}}^{(2)}_{\Pi\Pi}(2k|2n-2k)/{\mathfrak{c}}$ and
${\mathfrak{pe}}^{(2)}(n)/{\mathfrak{c}}$ for both even and odd values of $n$.

Consider the Lie algebra

(12)
$\begin{array}[]{ll}\tilde{\mathfrak{o}}_{B}(V):=&\\{M\in{\mathfrak{gl}}(V)\mid\text{there
is $c_{M}\in{\mathbb{K}}$ such that~{}~{}}\\\ &\hskip
65.44142ptB(Mx,y)+B(x,My)=c_{M}B(x,y)\text{~{}~{}for all $x,y\in V$}\\}.\\\
\end{array}$

As is not difficult to see, for $B\sim\Pi_{2n}$ we have

$\tilde{\mathfrak{o}}_{\Pi_{2n}}(V)={\mathfrak{o}}_{\Pi_{2n}}(V)\ltimes{\mathbb{K}}d_{n},\text{~{}~{}where
$d_{n}=\mathop{\mathrm{diag}}\nolimits(0_{n},1_{n})$ or
$\mathop{\mathrm{diag}}\nolimits(1_{n},0_{n})$}.$

Then, according to [BGLL2]

(13) ${\mathfrak{der}}\
({\mathfrak{o}}^{(2)}_{\Pi_{2n}}(V)/{\mathfrak{c}})\simeq\tilde{\mathfrak{o}}_{\Pi_{2n}}(V)/{\mathfrak{c}}.$

(To compare with our previous results, observe that if $B\sim I_{2n}$, then
$\tilde{\mathfrak{o}}_{I_{2n}}(V)={\mathfrak{o}}_{I_{2n}}(V)$.)

As before, a given operator $U\in{\mathfrak{der}}\
({\mathfrak{o}}^{(2)}_{B}(V)/{\mathfrak{c}})\simeq\tilde{\mathfrak{o}}_{B}(V)/{\mathfrak{c}}$
satisfies the condition $U^{2}=U$ if and only if the corresponding equivalence
class of $\tilde{\mathfrak{o}}_{B}(V)$ contains a projection $A$. Let us
consider the values $c_{A}$ can take.

$c_{A}={\bar{0}}$: In this case, the definition (12) is equivalent to the
statement that $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ are orthogonal with respect to $B$. It
means that the restrictions of $B$ on $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ have to be nondegenerate. Since $B$ is
anti-symmetric (i.e., $B(x,x)=0$ for any $x\in V$), these restrictions are
anti-symmetric as well, which means that dimensions of $\mathop{\text{\rm
Im}}\nolimits A$ and $\mathop{\mathrm{Ker}}\nolimits A$ are even. The
resulting Lie superalgebra in this case is isomorphic to

${\mathfrak{oo}}^{(2)}_{\Pi\Pi}(\dim\mathop{\text{\rm Im}}\nolimits
A|\dim\mathop{\mathrm{Ker}}\nolimits
A)/{\mathfrak{c}}\simeq{\mathfrak{oo}}^{(2)}_{\Pi\Pi}(\dim\mathop{\mathrm{Ker}}\nolimits
A|\dim\mathop{\text{\rm Im}}\nolimits A)/{\mathfrak{c}}.$

$c_{A}={\bar{1}}$: In this case, the condition (12) is equivalent to the
statement that both $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ are isotropic with respect to $B$. Since
$\mathop{\text{\rm Im}}\nolimits A\oplus\mathop{\mathrm{Ker}}\nolimits A=V$,
this means that

$\dim\mathop{\text{\rm Im}}\nolimits A=\dim\mathop{\mathrm{Ker}}\nolimits A=n$

and there is an invertible linear map $f:\mathop{\mathrm{Ker}}\nolimits
A\to(\mathop{\text{\rm Im}}\nolimits A)^{*}$ such that

$B(x,y)=(f(y))(x)\text{~{}~{} for all $x\in\mathop{\text{\rm Im}}\nolimits
A,y\in\mathop{\mathrm{Ker}}\nolimits A$.}$

The resulting Lie superalgebra in this case is isomorphic to
${\mathfrak{pe}}^{(2)}(n)/{\mathfrak{c}}$.

$c_{A}\neq{\bar{0}},{\bar{1}}$: This is impossible, because in this case, (12)
would mean that $\mathop{\text{\rm Im}}\nolimits A$ and
$\mathop{\mathrm{Ker}}\nolimits A$ are both isotropic and orthogonal to each
other with respect to $B$, i.e., $B(x,y)=0$ for all $x,y\in V$.

So we get $\left[\frac{n}{2}\right]+2$ nonisomorphic superizations of
${\mathfrak{o}}^{(2)}_{\Pi}(2n)/{\mathfrak{c}}$ (including the purely even
one), which are ${\mathfrak{oo}}^{(2)}_{\Pi\Pi}(2k|2n-2k)/{\mathfrak{c}}$,
where $0\leq k\leq\left[\frac{n}{2}\right]$, and
${\mathfrak{pe}}^{(2)}(n)/{\mathfrak{c}}$.∎

##  6\. The ${\mathfrak{vect}}^{(1)}(1;\underline{n})$ series

We describe the ${\mathbb{Z}}/2$-gradings of
${\mathfrak{vect}}^{(1)}(1;\underline{n-1})$ in Theorem 6.2. Theorem. First,
we need some definitions.

Let ${\mathfrak{v}}_{n}$ be the minimal Lie subalgebra of the restricted
closure of ${\mathfrak{vect}}^{(1)}(1;\underline{n-1})$ that contains
${\mathfrak{vect}}^{(1)}(1;\underline{n-1})$ and all the elements
$\mathcal{X}^{[2]}$, where
$\mathcal{X}\in{\mathfrak{vect}}^{(1)}(1;\underline{n-1})$. As a vector space,
${\mathfrak{v}}_{n}$ can be written as a direct sum

(14)
${\mathfrak{v}}_{n}={\mathfrak{vect}}(1;\underline{n-1})\oplus{\mathbb{K}}\partial^{2}.$

Thus, $\dim{\mathfrak{v}}_{n}=2^{n-1}+1$. Observe that
${\mathfrak{v}}_{n}=({\mathfrak{vect}}^{(1)}(1;\underline{n-1}))^{<1>}$, see
eq. (5), associated with the ${\mathbb{Z}}/2$-grading induced by the standard
${\mathbb{Z}}$-grading of ${\mathfrak{vect}}^{(1)}(1;\underline{n-1})$.

For $n>2$, consider the Lie superalgebra
${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$. As a vector space,
this Lie superalgebra is isomorphic to
${\mathfrak{v}}_{n}\oplus\Pi({\mathfrak{vect}}(1;\underline{n-1}))$, where
$\Pi$ is the change of parity functor. Consider the Lie algebra
${\mathfrak{vect}}(1;\underline{n})$ in indeterminate $z$ in order to
distinguish from indeterminate $x$ in eq. (16). Let

$X_{-2}=\partial^{2},\ X_{-1}=\partial,\ X_{0}=z\partial,\ \ldots,\
X_{2^{n-1}-1}=z^{(2^{n-1})}\partial,$

be the basis in the even part ${\mathfrak{v}}_{n}$ of
${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$ and

$Y_{-1}=\Pi\partial,\ Y_{0}=\Pi(z\partial),\ \ldots,\
Y_{2^{n-1}-2}=\Pi(z^{(2^{n-1}-1)}\partial),$

be the basis in the odd part of
${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$. The Lie superalgebra
structure in ${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$ is given
by following formulas

$\displaystyle[X_{-2},X_{k}]={}$
$\displaystyle[\partial^{2},z^{(k-1)}\partial]=X_{k-2}\quad\text{for
$k=1,\ldots,2^{n-1}-1$},$ $\displaystyle[X_{-2},Y_{k}]={}$
$\displaystyle\Pi[\partial^{2},z^{(k-1)}\partial]=Y_{k-2}\quad\text{for
$k=1,\ldots,2^{n-1}-1$},$ $\displaystyle[X_{-2},X_{-1}]={}$
$\displaystyle[X_{-2},X_{0}]=[X_{-2},Y_{-1}]=[X_{-2},Y_{0}]=[X_{-1},Y_{-1}]=0,$
$\displaystyle[X_{-1},X_{m}]={}$ $\displaystyle
X_{m-1}\quad\text{and}\quad[X_{-1},Y_{m}]=Y_{m-1}\quad\text{for
$m=0,\ldots,2^{n-1}-1$},$ $\displaystyle[X_{k},X_{m}]={}$
$\displaystyle[z^{(k-1)}\partial,z^{(m-1)}\partial]=\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}\Big{)}X_{k+m}\quad\text{for
$k,m=0,\ldots,2^{n-1}-1$},$ $\displaystyle[X_{k},Y_{s}]={}$
$\displaystyle[z^{(k-1)}\partial,z^{(s-1)}\partial]=\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+s+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+s+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+s+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+s+2}{k+1}$}}\Big{)}Y_{k+s}\quad\text{for
$k=0,\ldots,2^{n-1}-1$, $s=0,\ldots,2^{n-1}-2$}$
$\displaystyle[Y_{t},Y_{s}]={}$
$\displaystyle[z^{(t-1)}\partial,z^{(s-1)}\partial]=\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{t+s+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{t+s+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{t+s+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{t+s+2}{k+1}$}}\Big{)}X_{t+s}\quad\text{for
$t,s=0,\ldots,2^{n-1}-2$},$ $\displaystyle(Y_{-1})^{2}={}$ $\displaystyle
X_{-2},\quad(Y_{k})^{2}=(z^{(k+1)}\partial)^{[2]}=\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2k+1}{k}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2k+1}{k}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2k+1}{k}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2k+1}{k}$}}\Big{)}X_{2k}\quad\text{for
$k=0,\ldots,2^{n-1}-2$}.$

On the superspace of ${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$,
define another Lie superalgebra structure, call it
${\widetilde{\mathfrak{qv}}}_{n-1}$, by the following formulas:

$\displaystyle[X_{-2},X_{-1}]:={}$ $\displaystyle 0,\quad[X_{-2},X_{0}]:=0,$
$\displaystyle[X_{-2},X_{m}]:={}$ $\displaystyle
X_{m-2}\quad\text{for~{}$m=0,\ldots,2^{n-1}-1$},$
$\displaystyle[X_{k},X_{m}]:={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}\Big{)}X_{k+m}\quad\text{for~{}$m,k=-2,\ldots,2^{n-1}-1$,
except for $m=k=-1$,}$ $\displaystyle[Y_{k},Y_{m}]:={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}\Big{)}X_{k+m}\quad\text{for~{}$m,k=-2,\ldots,2^{n-1}-2$,
except for $m=k=-1$,}$ $\displaystyle[X_{-1},Y_{-1}]:={}$ $\displaystyle
Y_{-1},$ $\displaystyle[X_{k},Y_{m}]:={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}\Big{)}Y_{k+m}+\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{k+m+2}{k+1}$}}\Big{)}Y_{k+m+1}\quad\text{for
$m=-2,\ldots,2^{n-1}-1$}$ and $k=-2,\ldots,2^{n-1}-2$, except for $m=k=-1$,
$\displaystyle(Y_{-1})^{2}:={}$ $\displaystyle
X_{-2}+X_{0},\quad(Y_{k})^{2}:={}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2k+1}{k}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2k+1}{k}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2k+1}{k}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2k+1}{k}$}}\Big{)}X_{2k}\quad\text{for
$k=0,\ldots,2^{n-1}-2$}.$

I. Shchepochkina observed that the Lie superalgebra
${\widetilde{\mathfrak{qv}}}_{n-1}$ is a filtered deform of the Lie
superalgebra ${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$.

######  Lemma.

Lie superalgebras ${\widetilde{\mathfrak{qv}}}_{n-1}$ and
${\mathfrak{q}}({\mathfrak{vect}}(1;\underline{n-1}))$ are not isomorphic.

###### Proof.

Let us prove that ${\widetilde{\mathfrak{qv}}}_{n-1}$ is not isomorphic to any
Lie superalgebra of the form ${\mathfrak{q}}({\mathfrak{g}})$, where
${\mathfrak{g}}$ is a Lie algebra. A superalgebra of the form
${\mathfrak{q}}({\mathfrak{g}})$ possesses the following property: if
$x\in{\mathfrak{q}}({\mathfrak{g}})_{\bar{0}}$ acts nilpotently on
${\mathfrak{q}}({\mathfrak{g}})_{\bar{0}}$, then it acts nilpotently on
${\mathfrak{q}}({\mathfrak{g}})_{\bar{1}}$ as well. Indeed, if
$(\mathop{\mathrm{ad}}\nolimits_{x})^{k}|_{{\mathfrak{q}}({\mathfrak{g}})_{\bar{0}}}=0$
for some positive integer $k$, then
$(\mathop{\mathrm{ad}}\nolimits_{x})^{k}\Pi
y=\Pi\left((\mathop{\mathrm{ad}}\nolimits_{x})^{k}y\right)=0$ for any
$y\in{\mathfrak{g}}$.

On the other hand, one can see from the relations above that $X_{-1}$ acts
nilpotently on $({\widetilde{\mathfrak{qv}}}_{n-1})_{\bar{0}}$ (more
specifically,
$(\mathop{\mathrm{ad}}\nolimits_{X_{-1}})^{2^{n-1}+1}|_{({\widetilde{\mathfrak{qv}}}_{n-1})_{\bar{0}}}=0$),
but $[X_{-1},Y_{-1}]=Y_{-1}$, so the action of $X_{-1}$ on
$({\widetilde{\mathfrak{qv}}}_{n-1})_{\bar{1}}$ is not nilpotent. ∎

Consider the Lie algebra ${\mathfrak{vect}}(1;\underline{n})$ in indeterminate
$x$. For a basis of ${\mathfrak{vect}}(1;\underline{n})$ we take
$\boldsymbol{e}_{i}:=x^{(i+1)}\partial$ for $i=-1,0,\ldots,2^{n}-2$.

######  Theorem.

All ${\mathbb{Z}}/2$-gradings of the Lie algebra
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ correspond to the one of the
following Lie superalgebras:

_1)_ purely even ${\mathfrak{vect}}^{(1)}(1;\underline{n}|0)$;

_2)_ ${\mathfrak{k}}(1;\underline{n-1}|1)$, described in [BGLLS2], and its
$(n-2)$-parametric family of filtered deforms described below;

_3)_ for $n>2$, the Lie superalgebra ${\widetilde{\mathfrak{qv}}}_{n-1}$.

###### Proof.

Let $u\in{\mathcal{O}}(1;\underline{n})$ be a linear combination of only even
powers of the indeterminate $x$, and let $a$ be its constant term222The
notation $u(0)$ we use for brevity in what follows is meaningless, strictly
speaking, because the divided power polynomials can not be evaluated at any
$x\in{\mathbb{K}}$ for $p>0$; by writing like this we mean
$a\in{\mathbb{K}}\cdot 1$ which is the value of $u$ modulo the maximal ideal
of ${\mathcal{O}}(1;\underline{n})$; this is a standard abuse of notation.. In
other words, let $u,a\in{\mathcal{O}}(1;\underline{n})$ be such that

(15) $\displaystyle x\cdot(\partial u)={}$ $\displaystyle 0;\ \ \ \partial
a={}0;\ \ \ u^{2}={}a^{2}.$

Consider derivation $D_{u}\in{\mathfrak{der}}\ {\mathcal{O}}(1;\underline{n})$
of the form

$D_{u}=\left(u+x+xu\sum\limits_{1\leq i\leq
n-1}a^{2^{i}-2}(\partial^{2^{i}}u)\right)\partial+\sum\limits_{1\leq i\leq
n-1}a^{2^{i}}\partial^{2^{i}}.$

Properties (15) imply (after rather lengthy calculations we omit) that
$(D_{u})^{2}=D_{u}$, so $D_{u}$ describes a ${\mathbb{Z}}/2$-grading of
${\mathcal{O}}(1;\underline{n})$.

Consider the linear map $\mathop{\mathrm{ad}}\nolimits_{D_{u}}$ on the Lie
algebra ${\mathfrak{der}}\ {\mathcal{O}}(1;\underline{n})$ given by
$D\mapsto[D_{u},D]$. Due to the Jacobi identity, we see that
$\mathop{\mathrm{ad}}\nolimits_{D_{u}}\in{\mathfrak{der}}({\mathfrak{der}}\
{\mathcal{O}}(1;\underline{n}))$, where the outer ${\mathfrak{der}}$ is for
derivations of the Lie algebra ${\mathfrak{der}}\
{\mathcal{O}}(1;\underline{n})$. Since the Lie algebra ${\mathfrak{der}}\
{\mathcal{O}}(1;\underline{n})$ has a $2$-structure given by $D^{[2]}=D^{2}$,
we have
$(\mathop{\mathrm{ad}}\nolimits_{D_{u}})^{2}=\mathop{\mathrm{ad}}\nolimits_{D_{u}}$.
Thus, $\mathop{\mathrm{ad}}\nolimits_{D_{u}}$ describes a
${\mathbb{Z}}/2$-grading of ${\mathfrak{der}}\
{\mathcal{O}}(1;\underline{n})$.

The linear map $\mathop{\mathrm{ad}}\nolimits_{D_{u}}$ sends elements of
${\mathfrak{vect}}(1;\underline{n})$ to
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ because $D_{u}$ is a linear
combination of a vector field and elements of the form $\partial^{2^{i}}$, and

$[\partial^{2^{i}},f\partial]=(\partial^{2^{i}}f)\partial=[\partial,(\partial^{2^{i}-1}f)\partial]\in{\mathfrak{vect}}^{(1)}(1;\underline{n})\text{~{}~{}for
any $f\in{\mathcal{O}}(1;\underline{n})$.}$

This means that the restriction
$(\mathop{\mathrm{ad}}\nolimits_{D_{u}})|_{{\mathfrak{vect}}(1;\underline{n})}$
describes a ${\mathbb{Z}}/2$-grading of ${\mathfrak{vect}}(1;\underline{n})$
and the restriction
$(\mathop{\mathrm{ad}}\nolimits_{D_{u}})|_{{\mathfrak{vect}}^{(1)}(1;\underline{n})}$
describes a ${\mathbb{Z}}/2$-grading of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$. In what follows, we will denote
the latter restriction by $D_{u}$ as well. Different polynomials $u$
correspond to different gradings, as $[D_{u},x\partial]=u\partial$. We need,
however, not individual gradings, but their equivalence classes. We were
unable to solve this problem completely so far.

In notation of [T], any torus of the restricted closure of
${\mathfrak{vect}}(1;\underline{n})$ lying in the maximal subalgebra of
elements of non-negative degree (assuming $\deg x=1$) is called an inner one,
the other tori are called outer ones.

If $u(0)=a\neq 0$, the derivation $D_{u}$ is an outer toroidal derivation of
${\mathcal{O}}(1;\underline{n})$ (i.e., it spans an outer torus of $2$-closure
of ${\mathfrak{vect}}(1;\underline{n})$). The automorphism group of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ acts transitively on the set of
outer toroidal derivations, see [T].333In [T], Tyurin proved that, for any two
outer toroidal derivations $O_{1}$ and $O_{2}$ of
${\mathfrak{vect}}(1;\underline{n})$, there exists an $A\in{\rm
Aut}({\mathfrak{vect}}(1;\underline{n}))$ such that $O_{2}=A^{-1}O_{1}A$.
Instead we need “for any two outer toroidal derivations $O_{1}$ and $O_{2}$ of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, there exists an $A\in{\rm
Aut}({\mathfrak{vect}}^{(1)}(1;\underline{n}))$ such that
$O_{2}=A^{-1}O_{1}A$”. This is so because “our”
$D_{u}=(D^{T}_{u})|_{{\mathfrak{vect}}^{(1)}}\in{\mathfrak{der}}~{}{\mathfrak{vect}}^{(1)}(1;\underline{n})$
is the restriction of the corresponding “Tyurin’s”
$D^{T}_{u}\in{\mathfrak{der}}~{}{\mathfrak{vect}}(1;\underline{n})$ to
${\mathfrak{der}}~{}{\mathfrak{vect}}^{(1)}(1;\underline{n})$. If $D_{u}$ and
$D_{v}$ are outer toroidal derivations of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, then $D^{T}_{u}$ and $D^{T}_{v}$
are outer toroidal derivations of ${\mathfrak{vect}}(1;\underline{n})$, and
therefore there exists an $A\in{\rm Aut}({\mathfrak{vect}}(1;\underline{n}))$
such that $D^{T}_{v}=A^{-1}D^{T}_{u}A$. Any automorphism of
${\mathfrak{vect}}(1;\underline{n})$ preserves
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, and hence
${D_{v}=(\bar{A})^{-1}D_{u}\bar{A}}$, where
$\bar{A}:=A|_{{\mathfrak{vect}}^{(1)}}(1;\underline{n})$ is an automorphism of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$. Hence, any derivation $D_{u}$ with
$a\neq 0$ is conjugate by an automorphism of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ to the derivation

${D_{1}=(1+x)\partial+\mathop{\sum}\limits_{1\leq i\leq
n-1}\partial^{2^{i}}}.$

Recall that

$(u+v)^{(k)}=\mathop{\sum}\limits_{0\leq i\leq k}u^{(k-i)}v^{(i)}$

and for $k=-2,\ldots,2^{n-1}-1$, define
$e_{k}\in({\mathfrak{vect}}^{(1)}(1;\underline{n});D_{1})^{<1>}$ by setting

(16) $\begin{array}[]{l}e_{-2}=\partial^{2}+(1+x)\partial,\\\
e_{k}=\partial\left((x+x^{(2)})^{(k+2)}\right)\partial=(1+x)(x+x^{(2)})^{(k+1)}\partial\text{~{}~{}for
$k>-2$},\end{array}$

and for $k=-1,\ldots,2^{n-1}-2$, define $o_{k}$ by formulas

(17)
$o_{k}=(x+x^{(2)})^{(k+1)}\partial,\text{~{}~{}where~{}~{}}k=-1,\dots,2^{n-1}-2.$

Direct computations show that $D_{1}(e_{k})=0$ and $D_{1}(o_{m})=o_{m}$. Let
$e_{-2},\ldots,e_{2^{n-1}-2}$ and $\Pi o_{-1},\ldots,\Pi o_{2^{n}-2}$ form a
basis in the Lie superalgebra corresponding to the ${\mathbb{Z}}/2$-grading
defined by $D_{1}$. The formulas

$\begin{array}[]{l}e_{k}\mapsto
X_{k},\text{~{}~{}where~{}~{}}k=-2,\ldots,2^{n-1}-2,\\\ o_{m}\mapsto
Y_{m},\text{~{}~{}where~{}~{}}m=-1,\ldots,2^{n-1}-2,\end{array}$

establish an isomorphism between this Lie superalgebra, spanned by the $e_{i}$
and $o_{j}$, and ${\widetilde{\mathfrak{qv}}}_{n-1}$.

If $u(0)=a=0$, the derivation $D_{u}$ spans an inner torus. Hence, see [T],
the derivation $D_{u}$ is conjugate to the derivation $D_{f}$, where
$f=\mathop{\sum}\limits_{1\leq i\leq n-1}c_{i}x^{(2^{i})}$ and the summands
$c_{i}x^{(2^{i})}$ are the corresponding terms of $u$. This gives us an
$(n-1)$-parametric family of ${\mathbb{Z}}/2$-gradings. Let the automorphism
$\sigma_{\varepsilon}$ of ${\mathcal{O}}(1;\underline{n})$ be given on its
generators by the formulas

$\sigma_{\varepsilon}(x)=\varepsilon
x,\quad\sigma_{\varepsilon}(x^{(2)})=\varepsilon^{2}x^{(2)},\quad\ldots,\quad\sigma_{\varepsilon}(x^{(2^{n-1})})=\varepsilon^{2^{n-1}}x^{(2^{n-1})},$

where $\varepsilon\in{\mathbb{K}}$ and $\varepsilon\neq 0$. Then

$\sigma_{\varepsilon}\left(\left(\mathop{\sum}\limits_{1\leq i\leq
n-1}c_{i}x^{(2^{i})}\right)\partial\right)=\left(\mathop{\sum}\limits_{1\leq
i\leq n-1}c_{i}\varepsilon^{2^{i}-1}x^{(2^{i})}\right)\partial.$

Hence, the conjugacy class of the derivation $D_{u}$ is defined by a tuple of
parameters $(c_{1},c_{2},\ldots,c_{n-1})$ up to equivalence

(18) $(c_{1},c_{2},\ldots,c_{n-1})\sim(\varepsilon
c_{1},\varepsilon^{3}c_{2},\ldots,\varepsilon^{2^{n-1}-1}c_{n-1}).$

Therefore, the number of parameters of ${\mathbb{Z}}/2$-gradings reduces to
$n-2$. And, indeed, for $n=2$, we do not have parametric families of
${\mathbb{Z}}/2$-gradings, see Subsection 6.3.

Observe that $x\cdot\partial u=0$, so

$0=\partial(x\partial u)=\partial u+x\partial^{2}u,\text{~{}which
implies~{}}x\partial^{2}u=\partial u.$

Then

$D_{u}=(u+x+xu\partial^{2}u)\partial=(u+x+u\partial u)\partial.$

Desuperization of any superization of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ can be considered as a Lie
subalgebra in
${\mathfrak{v}}_{n+1}:=({\mathfrak{vect}}^{(1)}(1;\underline{n}))^{<1>}$, see
(14). The operator $\mathop{\mathrm{ad}}\nolimits_{D_{u}}$ preserves
${\mathfrak{v}}_{n+1}$, so the grading it defines on
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ can be extended to
${\mathfrak{v}}_{n+1}$ with the same definition: $D\mapsto[D_{u},D]$. This
extended operator describes a ${\mathbb{Z}}/2$-grading of
${\mathfrak{v}}_{n+1}$. The elements of ${\mathfrak{v}}_{n+1}$ are said to be
$u$-even or $u$-odd if they are, respectively, even or odd in this grading.

In particular, for the grading on ${\mathfrak{v}}_{n+1}$ given by the function
$u=0$, we have $D_{0}=x\partial$, and the resulting superization is isomorphic
to ${\mathfrak{k}}(1;\underline{n-1}|1)$, see [BLLS1, §7].

Consider the linear maps $T_{u},\
A_{u}:{\mathfrak{v}}_{n+1}\longrightarrow{\mathfrak{v}}_{n+1}$ defined as
follows (their action on other elements being extended by linearity):

(19) $\begin{array}[]{l}T_{u}(f\partial)=(f+u(1+\partial u)\partial
f)(1+\partial u+u\partial^{2}u)\partial\text{~{}for
any~{}}f\in{\mathcal{O}}(1;\underline{n});\\\
T_{u}\partial^{2}=(T_{u}\partial)^{2}=\partial^{2}+(\partial^{2}u+u\partial^{3}u+u(\partial^{2}u)^{2}+u\partial
u(\partial^{2}u)^{2})\partial;\\\ A_{u}(f\partial)=(f+u\partial u\partial
f)\partial\text{~{}for any~{}}f\in{\mathcal{O}}(1;\underline{n});\\\
A_{u}\partial^{2}=\partial^{2}.\end{array}$

For any $X\in{\mathfrak{v}}_{n+1}$, one can check that $T_{u}X$ is $u$-even
(resp. $u$-odd) if and only if $X$ is $0$-even (resp. $0$-odd). We omit the
calculations that show it, but the idea is as follows: $\partial$ is $0$-odd,
while $(1+\partial u+u\partial^{2}u)\partial$ is $u$-odd. Besides, if we
extend the concept of $u$-evenness/$u$-oddness to
${\mathcal{O}}(1;\underline{n})$, then $f+u(1+\partial u)\partial f$ is
$u$-even (resp. $u$-odd) for any $f\in{\mathcal{O}}(1;\underline{n})$ if and
only if $f$ is $0$-even (resp. $0$-odd).

Also, the following is true for any
$X,Y\in{\mathfrak{vect}}(1;\underline{n})$:

$\begin{array}[]{l}{}[T_{u}X,T_{u}Y]=T_{u}A_{u}[X,Y];\\\
(T_{u}X)^{2}=T_{u}A_{u}X^{2}\text{~{}for any $0$-odd~{}}X;\\\
{}[T_{u}\partial^{2},T_{u}X]=T_{u}((1+u\partial^{2}u)\cdot[\partial^{2},X]).\end{array}$

Note also that the operator $T_{u}$ is invertible, so its image is the whole
${\mathfrak{v}}_{n+1}$. The invertibility follows from the fact that the
matrix of $T_{u}$ is upper triangular with ${\bar{1}}$’s on the main diagonal
in the basis $\partial^{2},\partial,x\partial,\dots,x^{(2^{n}-1)}\partial$.

This means that the superization given by any function $u$ such that $u(0)=0$
can be considered as a deform of the superization given by $u=0$ with the
deformation parametrized by the polynomial $u$ or, which is the same, the
coefficients of $u$ as follows:

(20)
$\begin{array}[]{l}{}[X,Y]_{u}=\begin{cases}A_{u}[X,Y]&\text{if~{}~{}}X,Y\in{\mathfrak{vect}}(1;\underline{n});\\\
(1+u\partial^{2}u)\cdot[X,Y]&\text{if~{}~{}}X=\partial^{2}\text{~{}~{}and~{}~{}}Y\in{\mathfrak{vect}}(1;\underline{n});\\\
\text{defined by linearity and anti-symmetry}&\text{in other
cases};\end{cases}\\\ (X^{2})_{u}=A_{u}X^{2}\text{~{}~{}for 0-odd
$X$}.\end{array}$

This deformation is filtered relative the decreasing filtration in which
${\mathcal{L}}_{-2}=S({\mathfrak{v}}_{n+1})$, see (6), while
${\mathcal{L}}_{k-1}$ for $k\geq-1$ consists of vector fields of the form
$f\partial$, where $f\in{\mathcal{O}}(1;\underline{n})$ does not contain term
of degree $<k$. Such superizations are listed in heading 2) of Theorem.

Finally, observe that these formulas do not capture the trivial
${\mathbb{Z}}/2$-grading given by the derivation $U=0$; the even part
${\mathfrak{g}}_{{\bar{0}}}$ of this grading is the whole Lie algebra
${\mathfrak{vect}}^{(1)}(1;\underline{n})$. Since any torus in
${\mathfrak{der}}({\mathfrak{vect}}^{(1)}(1;\underline{n}))$ has a form
$D_{u}$, see [Kuz, T] and footnote 3 on p. 3, we completely described all
${\mathbb{Z}}/2$-gradings of ${\mathfrak{vect}}^{(1)}(1;\underline{n})$ and
the corresponding superizations. ∎

####  6.2.1. Remark.

Observe an unpredicted fact: according to (20), if
$X,Y\in{\mathfrak{vect}}(1;\underline{n})$, then $[X,Y]_{u}$ can be expressed
in terms of $[X,Y]$ and $u$. In particular, concerning the $0$-even part of
${\mathfrak{vect}}(1;\underline{n})$, this implies the following fact:

######  Corollary.

If $u(0)=0$, then the even part of the superization corresponding to $D_{u}$
is a solvable Lie algebra.

For $n=2$, any outer torus is conjugate to an inner one, see (22). This is not
so for $n>2$.

######  Lemma.

Let $n>2$, consider the superization
${\widetilde{\mathfrak{qv}}}_{n-1}={\mathfrak{g}}_{{\bar{0}}}\oplus{\mathfrak{g}}_{{\bar{1}}}$
corresponding to the outer torus $D_{1}$. Its even part
${\mathfrak{g}}_{{\bar{0}}}\simeq{\mathfrak{v}}_{n}$ is spanned by the
elements (16), and ${\mathfrak{g}}_{{\bar{1}}}$ is spanned by the elements
(17). The odd part ${\mathfrak{g}}_{{\bar{1}}}$ is a reducible
${\mathfrak{g}}_{{\bar{0}}}$-module with no lowest weight vector and with the
two highest weight vectors $\Pi o_{2^{n-1}-3}$ and $\Pi o_{2^{n-1}-2}$ with
respect to the standard ${\mathbb{Z}}$-grading of
${\mathfrak{g}}_{{\bar{0}}}$, namely
$({\mathfrak{g}}_{{\bar{0}}})_{k}={\mathbb{K}}e_{k}$ for
$k=-2,\ldots,2^{n}-1$.

###### Proof.

Using (16) and (17), for positive generators $e_{2^{k}-1}$, where
$k=1,\ldots,n-2$, of ${\mathfrak{g}}_{{\bar{0}}}$ we have

$\displaystyle[e_{2^{k}-1},\Pi o_{2^{n-1}-2}]={}$ $\displaystyle\Pi[(\partial
w)w^{(2^{k})}\partial,w^{(2^{n-1}-1)}\partial]$ $\displaystyle{}={}$
$\displaystyle\Pi\left(w^{(2^{k})}w^{(2^{n-1}-2)}+w^{(2^{k})}w^{(2^{n-1}-1)}+w^{(2^{k}-1)}w^{(2^{n-1}-1)}\right)\partial,$

where $w=(x+x^{(2)})$. Note that

$\displaystyle 2^{n-1}-1={}$ $\displaystyle 1\cdot 2^{n-2}+1\cdot
2^{n-3}+\ldots+1\cdot 2+1,$ $\displaystyle 2^{n-1}-2={}$ $\displaystyle 1\cdot
2^{n-2}+1\cdot 2^{n-3}+\ldots+1\cdot 2+0.$

Then by Lucas’s Theorem for $k<n-1$, we have

$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-2+2^{k}}{2^{k}}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-2+2^{k}}{2^{k}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-2+2^{k}}{2^{k}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-2+2^{k}}{2^{k}}$}}\Big{)}\mod
2={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\underbrace{\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}\Big{)}}_{\text{at
$k$th
place}}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}=0,$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-1+2^{k}}{2^{k}}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-1+2^{k}}{2^{k}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-1+2^{k}}{2^{k}}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-1+2^{k}}{2^{k}}$}}\Big{)}\mod
2={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\underbrace{\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}\Big{)}}_{\text{at
$k$th
place}}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}=0,$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-1+2^{k}-1}{2^{k}-1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{2^{n-1}-1+2^{k}-1}{2^{k}-1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-1+2^{k}-1}{2^{k}-1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{2^{n-1}-1+2^{k}-1}{2^{k}-1}$}}\Big{)}\mod
2={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{0}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}\underbrace{\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{0}$}}\Big{)}}_{\text{at
$k$th
place}}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{1}$}}\Big{)}\cdots\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{1}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{1}{1}$}}\Big{)}\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{0}{1}$}}\Big{)}=0.$

Thus, we have $w^{(2^{k})}w^{(2^{n-1}-2)}=0$, $w^{(2^{k})}w^{(2^{n-1}-1)}=0$,
and $w^{(2^{k}-1)}w^{(2^{n-1}-1)}=0$. Finally, we see that $[e_{2^{k}-1},\Pi
o_{2^{n-1}-2}]=0$ for $k=1,\ldots,n-2$. Computations of the same kind show
that $[e_{2^{k}-1},\Pi o_{2^{n-1}-3}]=0$ for $k=1,\ldots,n-2$. Therefore,
$o_{2^{n-1}-2}$ and $o_{2^{n-1}-3}$ are highest weight vectors in
${\mathfrak{g}}_{{\bar{1}}}$.

Now, let us prove that there are no lowest weight vectors in
${\mathfrak{g}}_{{\bar{1}}}$. Consider the action of $e_{-1}$ on elements of
${\mathfrak{g}}_{{\bar{1}}}$. Using (16) and (17), we obtain

(21) $[e_{-1},\Pi o_{-1}]=o_{1},\quad\text{~{}and~{}}\quad[e_{-1},\Pi
o_{k}]=\Pi o_{k}+\Pi o_{k-1}\text{~{}for $k=0,\ldots,2^{n-1}-2$}.$

Therefore, the equation

${}[e_{-1},\mathop{\sum}\limits_{-1\leq k\leq 2^{n-1}-2}q_{k}\Pi o_{k}]=0$

for lowest weight vectors reduces to the following system of linear equations

$\displaystyle q_{-1}+q_{0}={}$ $\displaystyle 0,$ $\displaystyle
q_{0}+q_{1}={}$ $\displaystyle 0,$ $\displaystyle\ldots,$ $\displaystyle
q_{2^{n-1}-3}+q_{2^{n-1}-2}={}$ $\displaystyle 0,$ $\displaystyle
q_{2^{n-1}-2}=0.$

Clearly, this system has only one solution: 0. Hence, there are no lowest
weight vectors in ${\mathfrak{g}}_{{\bar{1}}}$.

Observe that the highest weight vector $\Pi o_{2^{n-1}}$ generates the whole
${\mathfrak{g}}_{{\bar{0}}}$-module ${\mathfrak{g}}_{{\bar{1}}}$ and the
highest weight vector $\Pi o_{2^{n-1}-1}$ generates a
${\mathfrak{g}}_{{\bar{0}}}$-submodule of codimesion 1 in
${\mathfrak{g}}_{{\bar{1}}}$, as follows from (21). ∎

######  Remark.

Put $v=\alpha\Pi o_{2^{n-1}-3}+\beta\Pi o_{2^{n-1}-2}$, where
$\alpha,\beta\in{\mathbb{K}}$. Construct a new basis in
${\mathfrak{g}}_{{\bar{1}}}$ by the rule

$o^{\prime}_{1}=v,\quad o_{k}^{\prime}=[e_{-1},o_{k-1}^{\prime}].$

For $n=4$, the coordinates of $o_{k}^{\prime}$ in the old basis $o_{k}$ are
defined by the following matrix

${\tiny\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&\alpha&\alpha+\beta\\\
0&0&0&0&0&\alpha&\beta&\alpha+\beta\\\
0&0&0&0&\alpha&\alpha+\beta&\alpha&\alpha+\beta\\\
0&0&0&\alpha&\beta&\beta&\beta&\alpha+\beta\\\
0&0&\alpha&\alpha+\beta&0&0&\alpha&\alpha+\beta\\\
0&\alpha&\beta&\alpha+\beta&0&\alpha&\beta&\alpha+\beta\\\
\alpha&\alpha+\beta&\alpha&\alpha+\beta&\alpha&\alpha+\beta&\alpha&\alpha+\beta\\\
\beta&\beta&\beta&\beta&\beta&\beta&\beta&\beta\\\ \end{array}\right).}$

Notice that $[e_{-1},o_{8}^{\prime}]=o_{1}^{\prime}$. For $\beta=0$, we have
$o_{8}^{\prime}=o_{1}^{\prime}+o_{2}^{\prime}+\ldots+o_{7}^{\prime}$. For
$\alpha=0$, the ${\mathfrak{g}}_{{\bar{0}}}$-action on
${\mathfrak{g}}_{{\bar{1}}}$ in the new basis is defined by the following
table

${\tiny\begin{array}[]{c|cccccccccccccccc}{}\hfil&e_{-1}&e_{0}&e_{1}&e_{2}&e_{3}&e_{4}&e_{5}&e_{6}&e_{7}&e_{8}&e_{9}&e_{10}&e_{11}&e_{12}&e_{13}&e_{14}\\\
\hline\cr o^{\prime}_{1}&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{2}&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{3}&o^{\prime}_{4}&0&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{4}&o^{\prime}_{5}&o^{\prime}_{4}&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{5}&o^{\prime}_{6}&0&0&0&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{6}&o^{\prime}_{7}&o^{\prime}_{6}&0&0&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{7}&o^{\prime}_{8}&0&o^{\prime}_{6}&0&o^{\prime}_{4}&0&o^{\prime}_{2}&0&0&0&0&0&0&0&0&0\\\
o^{\prime}_{8}&o^{\prime}_{9}&o^{\prime}_{8}&o^{\prime}_{7}&o^{\prime}_{6}&o^{\prime}_{5}&o^{\prime}_{4}&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0&0&0&0&0\\\
o^{\prime}_{9}&o^{\prime}_{10}&0&0&0&0&0&0&0&o^{\prime}_{2}&0&0&0&0&0&0&0\\\
o^{\prime}_{10}&o^{\prime}_{11}&o^{\prime}_{10}&0&0&0&0&0&0&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0&0&0\\\
o^{\prime}_{11}&o^{\prime}_{12}&0&o^{\prime}_{10}&0&0&0&0&0&o^{\prime}_{4}&0&o^{\prime}_{2}&0&0&0&0&0\\\
o^{\prime}_{12}&o^{\prime}_{13}&o^{\prime}_{12}&o^{\prime}_{11}&o^{\prime}_{10}&0&0&0&0&o^{\prime}_{5}&o^{\prime}_{4}&o^{\prime}_{3}&o^{\prime}_{2}&0&0&0&0\\\
o^{\prime}_{13}&o^{\prime}_{14}&0&0&0&o^{\prime}_{10}&0&0&0&o^{\prime}_{6}&0&0&0&o^{\prime}_{2}&0&0&0\\\
o^{\prime}_{14}&o^{\prime}_{15}&o^{\prime}_{14}&0&0&o^{\prime}_{11}&o^{\prime}_{10}&0&0&o^{\prime}_{7}&o^{\prime}_{6}&0&0&o^{\prime}_{3}&o^{\prime}_{2}&0&0\\\
o^{\prime}_{15}&o^{\prime}_{16}&0&o^{\prime}_{14}&0&o^{\prime}_{12}&0&o^{\prime}_{10}&0&o^{\prime}_{8}&0&o^{\prime}_{6}&0&o^{\prime}_{4}&0&o^{\prime}_{2}&0\\\
o^{\prime}_{16}&o^{\prime}_{1}&o^{\prime}_{16}&o^{\prime}_{15}&o^{\prime}_{14}&o^{\prime}_{13}&o^{\prime}_{12}&o^{\prime}_{11}&o^{\prime}_{10}&o^{\prime}_{9}&o^{\prime}_{8}&o^{\prime}_{7}&o^{\prime}_{6}&o^{\prime}_{5}&o^{\prime}_{4}&o^{\prime}_{3}&o^{\prime}_{2}\\\
\end{array}}$

In what follows we consider only generating functions of the form

$u(c)=c_{0}+\mathop{\sum}\limits_{1\leq k\leq
n-1}c_{k}x^{(2^{k})},\text{~{}~{}where $c=(c_{0},c_{1},\ldots,c_{n-1})$ is a
set of free parameters.}$

Let us consider the classification of ${\mathbb{Z}}/2$-gradings of
${\mathfrak{vect}}^{(1)}(1;\underline{2})$ in more detail.

###  6.3. ${\mathfrak{g}}={\mathfrak{vect}}^{(1)}(1;\underline{2})$

The general solution $U_{\text{lin}}$ of linear equations (8b), where
$c_{1,1}$, $c_{1,2}$, $c_{1,3}$, $c_{2,1}$, $c_{3,1}$ are free parameters, and
the general solution $U$ of eq. (8), where $c_{0}$, $c_{1}$ are free
parameters, and the schematic form of $U$ are as follows:

$U_{\text{lin}}=\begin{pmatrix}c_{1,1}&c_{1,2}&c_{1,3}\\\ c_{2,1}&0&c_{1,2}\\\
c_{3,1}&c_{2,1}&c_{1,1}\\\ \end{pmatrix},\qquad
U=\left(\begin{array}[]{ccc}c_{0}c_{1}+1&c_{0}&c_{0}^{2}\\\ c_{1}&0&c_{0}\\\
c_{1}^{2}&c_{1}&c_{0}c_{1}+1\\\
\end{array}\right),\qquad\begin{pmatrix}*&*&*\\\ *&&*\\\ *&*&*\end{pmatrix}.$

This solution corresponds to the derivation $D_{u}$ with
$u(c)=c_{0}+c_{1}x^{(2)}$.

####  6.3.1. Parameters and $\mathop{\mathrm{{Aut\/}}}\nolimits{\mathfrak{g}}$

We denote by ${\mathfrak{g}}_{{\bar{0}}}(c)$ the kernel of the projection
$D_{u(c)}$, the even part of the corresponding ${\mathbb{Z}}/2$-grading, e.g.,
${\mathfrak{g}}_{{\bar{0}}}(c_{0},c_{1})$, where $u=c_{0}+c_{1}x^{(2)}$.

Certain automorphisms and their actions:

${\tiny\begin{pmatrix}\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{c_{0}}$}}{\raisebox{0.25pt}{$\dfrac{1}{c_{0}}$}}{\raisebox{-0.5pt}{$\tfrac{1}{c_{0}}$}}{\raisebox{-0.5pt}{$\tfrac{1}{c_{0}}$}}&0&0\\\
0&1&0\\\
\mathchoice{\raisebox{0.25pt}{$\dfrac{c_{1}+c_{0}c_{1}}{c_{0}}$}}{\raisebox{0.25pt}{$\dfrac{c_{1}+c_{0}c_{1}}{c_{0}}$}}{\raisebox{-0.5pt}{$\tfrac{c_{1}+c_{0}c_{1}}{c_{0}}$}}{\raisebox{-0.5pt}{$\tfrac{c_{1}+c_{0}c_{1}}{c_{0}}$}}&0&c_{0}\end{pmatrix}}\text{~{}~{}
maps ${\mathfrak{g}}_{{\bar{0}}}(c_{0},c_{1})$ to
${\mathfrak{g}}_{{\bar{0}}}(1,c_{1})$ for all $c_{0}\neq 0$.}$
${\tiny\begin{pmatrix}1+c_{1}&0&1\\\ 0&1&0\\\ 1&0&0\end{pmatrix}}\text{~{}~{}
maps ${\mathfrak{g}}_{{\bar{0}}}(1,c_{1})$ to
${\mathfrak{g}}_{{\bar{0}}}(1,1)$ for all $c_{1}\in{\mathbb{K}}$.}$
${\tiny\begin{pmatrix}c_{1}&0&0\\\ 0&1&0\\\
0&0&\mathchoice{\raisebox{0.25pt}{$\dfrac{1}{c_{1}}$}}{\raisebox{0.25pt}{$\dfrac{1}{c_{1}}$}}{\raisebox{-0.5pt}{$\tfrac{1}{c_{1}}$}}{\raisebox{-0.5pt}{$\tfrac{1}{c_{1}}$}}\end{pmatrix}}\text{~{}~{}
maps ${\mathfrak{g}}_{{\bar{0}}}(0,c_{1})$ to
${\mathfrak{g}}_{{\bar{0}}}(0,1)$ for all $c_{1}\neq 0$.}$ (22)
${\tiny\begin{pmatrix}1&0&1\\\ 0&1&0\\\
0&0&1\end{pmatrix}}\text{~{}~{}establishes equivalence of
${\mathfrak{g}}_{{\bar{0}}}(1,1)$ and~{}${\mathfrak{g}}_{{\bar{0}}}(0,1)$}.$

Observe that the automorphism (22) sends the outer derivation $D_{1+x^{(2)}}$
to the inner derivation $D_{x^{(2)}}$.

The final answer: there are three inequivalent ${\mathbb{Z}}/2$-gradings;
their ${\mathfrak{g}}_{{\bar{0}}}$’s are:

(23)
$\begin{array}[]{l}{\mathfrak{g}}_{{\bar{0}}}(0,0)={\mathbb{K}}\boldsymbol{e}_{0}\text{~{}~{}and~{}~{}}{\mathfrak{g}}_{{\bar{0}}}(1,1)={\mathbb{K}}(\boldsymbol{e}_{-1}+\boldsymbol{e}_{0}+\boldsymbol{e}_{1}),\\\
\text{and the trivial ${\mathbb{Z}}/2$\/-\/grading with
}{\mathfrak{g}}_{{\bar{0}}}={\mathfrak{vect}}^{(1)}(1;\underline{2}),\end{array}$

not two as is the case for ${\mathbb{Z}}/2$-gradings of
${\mathfrak{o}}^{(1)}(3)={\mathfrak{o}}(3)$ for any $p\neq 2$.

####  6.3.2. Lie superalgebras corresponding to the gradings (23)

##### Grading $(0,0)$

The minimal Lie subsuperalgebra

(24)
${\mathfrak{g}}^{<1>}:=({\mathfrak{g}}^{<1>})_{{\bar{0}}}\oplus({\mathfrak{g}}^{<1>})_{{\bar{1}}}$

of the restricted closure $\bar{\mathfrak{g}}$ containing all elements
$v^{[2]}$, where
$v\in{\mathfrak{g}}_{{\bar{1}}}=({\mathfrak{g}}^{<1>})_{{\bar{1}}}$, see eq.
(5), is as follows:

$({\mathfrak{g}}^{<1>})_{{\bar{0}}}=\mathop{\mathrm{{Span}}}\nolimits(e_{-2}=\partial^{2},\
e_{0}=x\partial,\
e_{2}=x^{(3)}\partial),\qquad({\mathfrak{g}}^{<1>})_{{\bar{1}}}=\mathop{\mathrm{{Span}}}\nolimits(o_{-1}=\partial,\
o_{1}=x^{(2)}\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$o_{-1}^{2}=(\partial)^{2}=e_{-2}=\partial^{2},\qquad
o_{1}^{2}=(x^{(2)}\partial)^{2}=e_{2}=x^{(3)}\partial.$

The commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are defined
by the following formulas:

$[e_{0},e_{-2}]=0,\quad[e_{0},e_{2}]=0,\quad[e_{-2},e_{2}]=e_{0}.$

Observe that $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ is the Heisenberg algebra.
The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{-2},o_{-1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{0},o_{-1}]={}$ $\displaystyle o_{-1},$
$\displaystyle[e_{2},o_{-1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{-2},o_{1}]={}$ $\displaystyle o_{-1},$
$\displaystyle[e_{0},o_{1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle 0.$

It is easy to see that
$S({\mathfrak{g}}^{<1>})\simeq{\mathfrak{oo}}^{(1)}_{I\Pi}(1|2)\simeq{\mathfrak{k}}(1;\underline{1}|1)$;
for the (non-obvious) definition of the latter, see [BGLLS2]. The Lie
superalgebra ${\mathfrak{oo}}^{(1)}_{I\Pi}(1|2)$ is given by supermatrices of
the form

${\tiny\begin{pmatrix}0&b_{2}&b_{1}\\\ b_{1}&a_{1}&a_{2}\\\
b_{2}&a_{3}&a_{1}\end{pmatrix},}\text{~{}where~{}}a_{i},b_{i}\in{\mathbb{K}}.$

The correspondence between the abstract and matrix representations of the
elements of ${\mathfrak{oo}}^{(1)}_{I\Pi}(1|2)$ is as follows

$\begin{array}[]{l}e_{-2}=e_{2,3},\quad e_{0}=e_{2,2}+e_{3,3},\quad
e_{2}=e_{3,2},\\\ o_{-1}=e_{1,3}+e_{2,1},\qquad
o_{1}=e_{1,2}+e_{3,1}.\end{array}$

##### Grading $(1,1)\simeq(1,0)$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{1}=(1+x+x^{(2)})\partial,\
e_{2}=\partial^{2}+(1+x)\partial,\ e_{3}=\partial^{2}+(x+x^{(3)})\partial),$
$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{1}=(1+x)\partial,\
o_{2}=(1+x^{(2)})\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$o_{1}^{2}=e_{2},\qquad o_{2}^{2}=e_{3}.$

The commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are defined
by the following formulas:

(25) $[e_{1},e_{2}]=e_{1},\quad[e_{1},e_{3}]=0,\quad[e_{2},e_{3}]=e_{1}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{1},o_{1}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{1}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{1},o_{2}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{2},o_{2}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{3},o_{2}]={}$ $\displaystyle 0.$

Let us show that
$S({\mathfrak{g}}^{<1>})\simeq{\mathfrak{oo}}^{(1)}_{II}(1|2)$. The Lie
superalgebra ${\mathfrak{oo}}^{(1)}_{II}(1|2)$ consists of symmetric
supermatrices with supertrace 0. For a basis of its even part we take

$E_{1}=e_{1,1}+e_{2,2},\ \ E_{2}=e_{2,2}+e_{3,3},\ \ E_{3}=e_{2,3}+e_{3,2}.$

For a basis of its odd part we take $O_{1}=e_{1,2}+e_{2,1}$,
$O_{2}=e_{1,3}+e_{3,1}$. The commutation relations are given by the following
formulas:

$[E_{1},E_{2}]=0,\qquad[E_{1},E_{3}]=E_{3},\quad[E_{2},E_{3}]=0.$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$O_{1}^{2}=E_{1},\qquad O_{2}^{2}=E_{1}+E_{2}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure is as follows:

$[E_{2},O_{1}]=O_{1},\qquad[E_{2},O_{2}]=O_{2}.$

The multiplication by elements $E_{1}$ and $E_{3}$ is given by the following
formulas:

$[E_{1},O_{1}]=0,\quad[E_{1},O_{2}]=O_{2},\qquad[E_{3},O_{1}]=O_{2},\quad[E_{3},O_{2}]=O_{1}.$

The isomorphism between ${\mathfrak{oo}}^{(1)}_{II}(1|2)$ and
${\mathfrak{g}}^{<1>}$ is given by the following formulas:

$e_{1}=E_{3},\quad e_{2}=E_{1},\quad e_{3}=E_{2}+E_{3},\quad
o_{1}=O_{1},\qquad o_{2}=O_{1}+O_{2}.$

######  Remarks.

1) Observe that the even parts of both superizations are solvable Lie
algebras, but the corresponding Lie superalgebras are simple. For more details
and examples of this phenomenon indigenous to $p=2$, see Shchepochkina’s
comment (the last section) in [BGL].

2) Observe that due to Theorem 6.2. Theorem the Lie superalgebra
${\mathfrak{oo}}_{II}^{(1)}(1|2)$ is a filtered deform of the Lie superalgebra
${\mathfrak{oo}}^{(1)}_{I\Pi}(1|2)$.

###  6.4. Superizations corresponding to inner tori

Consider a superization of ${\mathfrak{vect}}^{(1)}(1;\underline{n})$ given by
a generating function $u$ such that $u(0)=0$. For such a function $u$, the
derivation $D_{u}$ spans an inner torus. According to [T], the derivation
$D_{u}$ is conjugate to the derivation $D_{f}$ in
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, where
$f=\mathop{\sum}\limits_{1\leq i\leq n-1}c_{i}x^{(2^{i})}$ and the
$c_{i}x^{(2^{i})}$ are the corresponding terms of $u$, so the superizations
corresponding to $D_{u}$ and $D_{f}$ are isomorphic. For this reason, in the
rest of this Subsection we consider only generating functions of the form
$\mathop{\sum}\limits_{1\leq i\leq n-1}c_{i}x^{(2^{i})}$. This gives us an
$(n-1)$-parametric family of ${\mathbb{Z}}/2$-gradings.

######  Conjecture.

Let $u_{1}=\mathop{\sum}\limits_{1\leq k\leq n-1}c_{k}x^{(2^{k})}$ and
$u_{2}=\mathop{\sum}\limits_{1\leq k\leq n-1}b_{k}x^{(2^{k})}$ be generating
functions, then the superizations corresponding to $D_{u_{1}}$ and $D_{u_{2}}$
are isomorphic if and only if there exists
$\varepsilon\in{\mathbb{K}}^{\times}$ such that
$c_{k}=\varepsilon^{2^{k}-1}b_{k}$ for all $k=1,\dots,n-1$.

The following is a sketch of a proof of this conjecture, including a complete
computer-aided proof for $2\leq n\leq 6$.

First of all, if there is an $\varepsilon\neq 0$ such that
$c_{k}=\varepsilon^{2^{k}-1}b_{k}$ for all $k=1,\dots,n-1$, then $D_{u_{1}}$
and $D_{u_{2}}$ are conjugate by the automorphism of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ given by
$x^{(m)}\partial\mapsto\varepsilon^{m-1}x^{(m)}\partial$ (this automorphism is
generated by the automorphism of ${\mathcal{O}}(1;\underline{n})$ given by
$x^{(m)}\mapsto\varepsilon^{m}x^{(m)}$).

By Theorem 6.2. Theorem, the superization given by $u_{1}$ is isomorphic to a
deform of ${\mathfrak{k}}(1;\underline{n-1}|1)$. Consider the even part of of
such a deform. According to (20), it contains a commutative subalgebra of
codimension $1$, which is the $0$-even part of
${\mathfrak{vect}}(1;\underline{n})$. The element $\partial^{2}$ acts on this
commutative subalgebra, and its action is given by

(26) $v\partial\mapsto(1+u\partial^{2}u)\partial^{2}v\partial\text{~{}~{}for
any $0$-odd $v\in{\mathcal{O}}(1;\underline{n})$}.$

######  Conjecture (Proved by computer for $n=2,3,4,5,6$).

The characteristic polynomial of the linear operator (26) on the $0$-even part
of ${\mathfrak{vect}}(1;\underline{n})$ for $n\geq 2$ is defined by the
formula

$\lambda^{2^{n-1}}+\mathop{\sum}\limits_{0\leq k\leq
n-2}c_{n-1-k}^{2^{k+1}}\lambda^{2^{k}}.$

For two deforms given by $u_{1}$ and $u_{2}$ to be isomorphic to each other,
their even parts have to be isomorphic, which means that the two actions of
$\partial^{2}$ on the $0$-even part of ${\mathfrak{vect}}(1;\underline{n})$
have to be conjugate (similar) up to a non-zero scalar factor, i.e., if $A$
and $A^{\prime}$ are such operators, then $A^{\prime}=\alpha MAM^{-1}$ for
some non-zero $\alpha\in{\mathbb{K}}$ and an invertible linear map $M$. If two
operators are conjugate up to a scalar multiple $\alpha$, the roots of their
characteristic polynomials differ by the same multiple, so if the dimension of
the space they act on is $d$ (in our case, $d=2^{n-1}$) and one polynomial is
equal to $\mathop{\sum}\limits_{0\leq i\leq d}z_{i}\lambda^{i}$, where
$z_{i}\in{\mathbb{K}}$ and $z_{d}=1$, then the other one would have the form
$\mathop{\sum}\limits_{0\leq i\leq k}z_{i}\alpha^{k-i}\lambda^{i}$. So, if
Conjecture 6.4.2. Conjecture is correct, then for the superizations given by
$u_{1}$ and $u_{2}$ to be isomorphic, there must exist non-zero
$\alpha\in{\mathbb{K}}$ such that

$\lambda^{2^{n-1}}+\mathop{\sum}\limits_{0\leq k\leq
n-2}c_{n-1-k}^{2^{k+1}}\lambda^{2^{k}}=\lambda^{2^{n-1}}+\mathop{\sum}\limits_{0\leq
k\leq n-2}\alpha^{2^{n-1}-2^{k}}b_{n-1-k}^{2^{k+1}}\lambda^{2^{k}},$

or, equivalently, $c_{k}=\varepsilon^{2^{k}-1}b_{k}$, where
$\varepsilon=\sqrt{\alpha}$, for all $k=1,\dots,n-1$.

So the set of equivalence classes of superizations of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$ corresponding to generating
functions $u$ such that $u(0)=0$ consist of the following two types:

A) the superization corresponding to $u=0$, which is isomorphic to
${\mathfrak{k}}(1;\underline{n-1}|1)$,

B) an $(n-2)$-parametric family of its pairwise non-isomorphic deforms. Note,
though, that it is not a result of $(n-2)$-parametric deformation of
${\mathfrak{k}}(1;\underline{n-1}|1)$; to obtain all these deforms, an
$(n-1)$-parametric deformation is needed, but the deforms obtained from some
sets of parameters are isomorphic: parameters $(c_{1},\dots,c_{n-1})$ and
$(b_{1},\dots,b_{n-1})$ produce isomorphic deformations if and only if there
exists $\varepsilon\in{\mathbb{K}}^{\times}$ such that
$c_{k}=\varepsilon^{2^{k}-1}b_{k}$ for all $k\in\overline{1,n-1}$.

###  6.5. Superizations of ${\mathfrak{vect}}^{(1)}(1;\underline{3})$

Let ${\mathfrak{g}}={\mathfrak{vect}}^{(1)}(1;\underline{3})$.

####  6.5.1. Grading $(0,0,0)$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{-2}=\partial^{2},\
e_{0}=x\partial,\ e_{2}=x^{(3)}\partial,\ e_{4}=x^{(5)}\partial,\
e_{6}=x^{(7)}\partial),$ $\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{-1}=\partial,\
o_{1}=x^{(2)}\partial,\ o_{3}=x^{(4)}\partial,\ o_{5}=x^{(6)}\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$\displaystyle o_{-1}^{2}={}$ $\displaystyle e_{-2},$ $\displaystyle
o_{1}^{2}={}$ $\displaystyle e_{2},$ $\displaystyle o_{3}^{2}={}$
$\displaystyle e_{6},$ $\displaystyle o_{5}^{2}={}$ $\displaystyle 0.$

The nonzero commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are
given by the following formulas:

$\displaystyle[e_{2},e_{-2}]={}$ $\displaystyle e_{0},$
$\displaystyle[e_{4},e_{-2}]={}$ $\displaystyle e_{2},$
$\displaystyle[e_{-2},e_{6}]={}$ $\displaystyle e_{4}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{-2},o_{-1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{0},o_{-1}]={}$ $\displaystyle o_{-1},$
$\displaystyle[e_{2},o_{-1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{4},o_{-1}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{6},o_{-1}]={}$ $\displaystyle o_{5},$
$\displaystyle[e_{-2},o_{1}]={}$ $\displaystyle o_{-1},$
$\displaystyle[e_{0},o_{1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{1}]={}$ $\displaystyle o_{5},$
$\displaystyle[e_{6},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{-2},o_{3}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{0},o_{3}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{2},o_{3}]={}$ $\displaystyle o_{5},$
$\displaystyle[e_{4},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{6},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{-2},o_{5}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{0},o_{5}]={}$ $\displaystyle o_{5},$
$\displaystyle[e_{2},o_{5}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{5}]={}$ $\displaystyle 0,$
$\displaystyle[e_{6},o_{5}]={}$ $\displaystyle 0.$

Let $L_{k}$ and $L^{(k)}$ be given by the formulas

(27) $L_{0}=L^{0}=({\mathfrak{g}}^{<1>})_{{\bar{0}}};\quad\quad
L_{k}=[L_{0},L_{k-1}]\text{~{}~{}and~{}~{}}L^{(k)}=[L^{(k-1)},L^{(k-1)}]\text{~{}~{}for
$k>0$}.$

We have

$L_{1}=\mathop{\mathrm{{Span}}}\nolimits(e_{0},\ e_{2},\ e_{4}),\ \
L_{2}=\mathop{\mathrm{{Span}}}\nolimits(e_{0},\ e_{2}),\ \
L_{3}={\mathbb{K}}e_{0},\ \ L_{4}=0.$

####  6.5.2. Grading $(0,1,0)$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{1}=(x+x^{(2)})\partial,\
e_{2}=x^{(3)}\partial,\ e_{3}=(x^{(5)}+x^{(6)})\partial,\
e_{4}=\partial^{2}+(1+x)\partial,\ e_{5}=x^{(7)}\partial),$
$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{1}=(1+x)\partial,\
o_{2}=x^{(2)}\partial,\ o_{3}=(x^{(4)}+x^{(5)})\partial,\
o_{4}=x^{(6)}\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$\displaystyle o_{1}^{2}={}$ $\displaystyle e_{4},$ $\displaystyle
o_{2}^{2}={}$ $\displaystyle e_{2},$ $\displaystyle o_{3}^{2}={}$
$\displaystyle e_{5},$ $\displaystyle o_{4}^{2}={}$ $\displaystyle 0.$

The commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are defined
by the following formulas:

$\displaystyle[e_{1},e_{2}]=0,\ [e_{1},e_{3}]=0,\ [e_{1},e_{4}]=e_{1},\
[e_{1},e_{5}]=0,\ [e_{2},e_{3}]=0,\ [e_{2},e_{4}]=e_{1},$
$\displaystyle[e_{2},e_{5}]=0,\ [e_{3},e_{4}]=e_{2}+e_{3},\ [e_{3},e_{5}]=0,\
[e_{4},e_{5}]=e_{3}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{1},o_{1}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{3},o_{1}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{4},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{5},o_{1}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{1},o_{2}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{2},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{2}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{4},o_{2}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{5},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{3}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{2},o_{3}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{3},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{3}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{5},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{4}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{2},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{4}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{5},o_{4}]={}$ $\displaystyle 0.$

####  6.5.3. Grading $(0,1,\beta)$, where $\beta\neq 0$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{1}=(x+x^{(2)}+\beta
x^{(4)})\partial,\ e_{2}=(x^{(3)}+\beta x^{(5)})\partial,\
e_{3}=(x^{(5)}+x^{(6)})\partial,$ $\displaystyle e_{4}=\partial^{2}+(1+x+\beta
x^{(2)})\partial,\ e_{5}=(x^{(3)}+\beta x^{(5)}+\beta^{2}x^{(7)})\partial),$
$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{1}=(1+x+\beta
x^{(3)})\partial,\ o_{2}=(x^{(2)}+\beta x^{(4)})\partial,\
o_{3}=(x^{(4)}+x^{(5)})\partial,\ o_{4}=x^{(6)}\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$\displaystyle o_{1}^{2}={}$ $\displaystyle e_{4},$ $\displaystyle
o_{2}^{2}={}$ $\displaystyle e_{5},$ $\displaystyle o_{3}^{2}={}$
$\displaystyle\tfrac{1}{\beta^{2}}e_{2}+\tfrac{1}{\beta^{2}}e_{5},$
$\displaystyle o_{4}^{2}={}$ $\displaystyle 0.$

The commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are defined
by the following formulas:

$\displaystyle[e_{1},e_{2}]=0,\quad[e_{1},e_{3}]=0,\quad[e_{1},e_{4}]=e_{1}+\beta
e_{2},\quad[e_{1},e_{5}]=0,$
$\displaystyle[e_{2},e_{3}]=0,\quad[e_{2},e_{4}]=e_{1}+\beta
e_{2}+\beta^{2}e_{3},\quad[e_{2},e_{5}]=0,$
$\displaystyle[e_{3},e_{4}]=e_{2}+(1+\beta)e_{3},\quad[e_{3},e_{5}]=0,\quad[e_{4},e_{5}]=e_{1}+\beta
e_{2}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{1},o_{1}]={}$ $\displaystyle o_{1}+o_{2}+\beta^{2}o_{4},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{3},o_{1}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{4},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{5},o_{1}]={}$ $\displaystyle o_{2}+\beta^{2}o_{4},$
$\displaystyle[e_{1},o_{2}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{2},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{2}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{4},o_{2}]={}$ $\displaystyle
o_{1}+(1+\beta)o_{2}+\beta^{2}o_{3},$ $\displaystyle[e_{5},o_{2}]={}$
$\displaystyle 0,$ $\displaystyle[e_{1},o_{3}]={}$ $\displaystyle
o_{3}+o_{4},$ $\displaystyle[e_{2},o_{3}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{3},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{3}]={}$ $\displaystyle o_{2}+\beta(o_{3}+o_{4}),$
$\displaystyle[e_{5},o_{3}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{1},o_{4}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{2},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{4}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{5},o_{4}]={}$ $\displaystyle 0.$

####  6.5.4. Grading $(0,0,1)$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{1}=(x+x^{(4)})\partial,\
e_{2}=x^{(5)}\partial,\ e_{3}=(x^{(3)}+x^{(6)})\partial,\
e_{4}=\partial^{2}+x^{(2)}\partial,\ e_{5}=x^{(7)}\partial),$
$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{1}=(1+x^{(3)})\partial,\
o_{2}=x^{(4)}\partial,\ o_{3}=(x^{(2)}+x^{(5)})\partial,\
o_{4}=x^{(6)}\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$\displaystyle o_{1}^{2}={}$ $\displaystyle e_{4},$ $\displaystyle
o_{2}^{2}={}$ $\displaystyle e_{5},$ $\displaystyle o_{3}^{2}={}$
$\displaystyle e_{3},$ $\displaystyle o_{4}^{2}={}$ $\displaystyle 0.$

The commutation relations in $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$ are defined
by the following formulas:

$\displaystyle[e_{1},e_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},e_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},e_{4}]={}$ $\displaystyle e_{2},$
$\displaystyle[e_{1},e_{5}]={}$ $\displaystyle 0,$
$\displaystyle[e_{2},e_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{2},e_{4}]={}$ $\displaystyle e_{3},$
$\displaystyle[e_{2},e_{5}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},e_{4}]={}$ $\displaystyle e_{1},$
$\displaystyle[e_{3},e_{5}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},e_{5}]={}$ $\displaystyle e_{2}.$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{1},o_{1}]={}$ $\displaystyle o_{1}+o_{4},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{3},o_{1}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{4},o_{1}]={}$ $\displaystyle 0,$
$\displaystyle[e_{5},o_{1}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{1},o_{2}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{2},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{2}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{4},o_{2}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{5},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{3}]={}$ $\displaystyle o_{3},$
$\displaystyle[e_{2},o_{3}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{3},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{3}]={}$ $\displaystyle o_{1}+o_{4},$
$\displaystyle[e_{5},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{4}]={}$ $\displaystyle o_{4},$
$\displaystyle[e_{2},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{3},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{4},o_{4}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{5},o_{4}]={}$ $\displaystyle 0,$

####  6.5.5. Grading $(1,0,0)$

The Lie subsuperalgebra ${\mathfrak{g}}^{<1>}$, see eq. (24), is as follows:

$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{0}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(e_{-2}=\partial^{2}+(1+x)\partial,\
e_{-1}=(1+x)\partial,\ e_{0}=(x+x^{(2)}+x^{(3)})\partial,$ $\displaystyle
e_{1}=(x^{(2)}+x^{(4)}+x^{(5)})\partial,\
e_{2}=(x^{(3)}+x^{(5)}+x^{(6)}+x^{(7)})\partial),$
$\displaystyle({\mathfrak{g}}^{<1>})_{{\bar{1}}}={}$
$\displaystyle\mathop{\mathrm{{Span}}}\nolimits(o_{1}=\partial,\
o_{2}=(x+x^{(2)})\partial,\ o_{3}=(x^{(2)}+x^{(3)}+x^{(4)})\partial,\
o_{4}=(x^{(3)}+x^{(5)}+x^{(6)})\partial).$

The squaring in $S({\mathfrak{g}}^{<1>})$ is given by the following formulas:

$\displaystyle o_{1}^{2}={}$ $\displaystyle e_{-2},$ $\displaystyle
o_{2}^{2}={}$ $\displaystyle e_{0},$ $\displaystyle o_{3}^{2}={}$
$\displaystyle e_{2},$ $\displaystyle o_{4}^{2}={}$ $\displaystyle 0.$

The commutation relations in
$({\mathfrak{g}}^{<1>})_{{\bar{0}}}\simeq({\mathfrak{vect}}(1;\underline{3}))^{<1>}$
are defined by the following formulas:

$\displaystyle[e_{-1},e_{0}]={}$ $\displaystyle e_{-1},$
$\displaystyle[e_{-1},e_{1}]={}$ $\displaystyle e_{0},$
$\displaystyle[e_{-1},e_{2}]={}$ $\displaystyle e_{1},$
$\displaystyle[e_{-1},e_{-2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{0},e_{1}]={}$ $\displaystyle e_{1},$
$\displaystyle[e_{0},e_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{0},e_{-2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},e_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},e_{-2}]={}$ $\displaystyle e_{-1},$
$\displaystyle[e_{2},e_{-2}]={}$ $\displaystyle e_{0}$

The $({\mathfrak{g}}^{<1>})_{{\bar{0}}}$-module structure in
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is given by the following formulas:

$\displaystyle[e_{-1},o_{1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{0},o_{1}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{1},o_{1}]={}$ $\displaystyle o_{2}+o_{3},$
$\displaystyle[e_{2},o_{1}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{-2},o_{1}]={}$ $\displaystyle o_{1},$
$\displaystyle[e_{-1},o_{2}]={}$ $\displaystyle o_{1}+o_{2},$
$\displaystyle[e_{0},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{2}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{2},o_{2}]={}$ $\displaystyle 0,$
$\displaystyle[e_{-2},o_{2}]={}$ $\displaystyle o_{2},$
$\displaystyle[e_{-1},o_{3}]={}$ $\displaystyle o_{2}+o_{3},$
$\displaystyle[e_{0},o_{3}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{1},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{2},o_{3}]={}$ $\displaystyle 0,$
$\displaystyle[e_{-2},o_{3}]={}$ $\displaystyle o_{1}+o_{3},$
$\displaystyle[e_{-1},o_{4}]={}$ $\displaystyle o_{3}+o_{4},$
$\displaystyle[e_{0},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{1},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{2},o_{4}]={}$ $\displaystyle 0,$
$\displaystyle[e_{-2},o_{4}]={}$ $\displaystyle o_{2}+o_{4}.$

We have

$[({\mathfrak{g}}^{<1>})_{{\bar{0}}},({\mathfrak{g}}^{<1>})_{{\bar{0}}}]=\mathop{\mathrm{{Span}}}\nolimits(e_{-1},\
e_{0},\ e_{1}),\ \
[({\mathfrak{g}}^{<1>})_{{\bar{0}}},[({\mathfrak{g}}^{<1>})_{{\bar{0}}},({\mathfrak{g}}^{<1>})_{{\bar{0}}}]]=[({\mathfrak{g}}^{<1>})_{{\bar{0}}},({\mathfrak{g}}^{<1>})_{{\bar{0}}}].$

The ideal
$[({\mathfrak{g}}^{<1>})_{{\bar{0}}},({\mathfrak{g}}^{<1>})_{{\bar{0}}}]$ is
isomorphic to ${\mathfrak{o}}^{(1)}(3)$. The ${\mathfrak{o}}^{(1)}(3)$-module
$({\mathfrak{g}}^{<1>})_{{\bar{1}}}$ is irreducible and has no lowest weight
vectors and with highest vectors $o_{3}$ and $o_{4}$, cf. [Do],

###  6.6. Summary of computer-aided experiments

Let $L_{k}$ and $L^{(k)}$ be given by eq. (27); in tables below, $k=1,2,\dots$
up to first stable term. Theorem 6.2. Theorem implies the following fact

$\mathop{\mathrm{sdim}}\nolimits\
S({\mathfrak{g}},D_{u(c)})=(2^{n-1}+1|2^{n-1})\text{~{}~{} and
$\dim[({\mathfrak{g}}^{<1>})_{{\bar{0}}},({\mathfrak{g}}^{<1>})_{{\bar{0}}}]=2^{n-1}-1$.}$

Computer-aided experiments show that (recall that ${\mathfrak{v}}_{n}$ is
defined by (14); the symbol“solv” means that the corresponding Lie algebra is
solvable, see Corollary 6.2.2. Corollary)

${\mathfrak{vect}}^{(1)}(1;\underline{2})\quad\footnotesize\begin{tabular}[]{|l|l|l|l|}\hline\cr$({\mathfrak{g}}^{<1>})_{{\bar{0}}}$&$\dim
L_{k}$&$\dim L^{(k)}$&Parameters\\\\[2.84526pt] \hline\cr
Heisenberg&$1,0$&$1,0$&$(00)$\\\ \hline\cr{solv}, see
\eqref{neHei}&$1$&$1,0$&$(01)$, $(10)$, $(11)$\\\ \hline\cr\end{tabular}$
${\mathfrak{vect}}^{(1)}(1;\underline{3})\quad\footnotesize\begin{tabular}[]{|l|l|l|l|}\hline\cr$({\mathfrak{g}}^{<1>})_{{\bar{0}}}$&$\dim
L_{k}$&$\dim L^{(k)}$&Parameters\\\\[2.84526pt]
\hline\cr${\mathfrak{o}}(3)/{\mathfrak{c}}\simeq{\mathfrak{v}}_{3}$&$3$&$3$&$(1ab)$\\\\[2.84526pt]
\hline\cr{solv}&$3,2,1,0$&$3,0$&$(000)$\\\\[2.84526pt]
\hline\cr{solv}&$3,2$&$3,0$&$(010)$, $(010)$, $(0\alpha 0)$,\text{~{}where
$\alpha\neq 0$}\\\\[2.84526pt] \hline\cr{solv}&$3$&$3,0$&$(001)$, $(011)$,
$(0\alpha\beta)$,\text{where~{}$\beta\neq 0$}\\\\[2.84526pt]
\hline\cr\end{tabular}$
${\mathfrak{vect}}^{(1)}(1;\underline{4})\quad{\footnotesize\begin{tabular}[]{|l|l|l|l|}\hline\cr${\mathfrak{g}}_{{\bar{0}}}^{<1>}$&$\dim
L_{k}$&$\dim L^{(k)}$&Parameters\\\\[2.84526pt]
\hline\cr${\mathfrak{v}}_{4}$&$7$&$7$&$(1000)$ and $(1abc)$\\\
\hline\cr{solv}&$7,6,5,4,3,2,1,0$&$7,0$&$(0000)$\\\\[2.84526pt]
\hline\cr{solv}&$7,6,4$&$7,0$&$(0100)$\\\\[2.84526pt]
\hline\cr{solv}&$7,6$&$7,0$&$(0010)$, $(0110)$\\\\[2.84526pt]
\hline\cr{solv}&$7$&$7,0$&$(0001)$, $(0011)$, $(0101)$,
$(0111)$\\\\[2.84526pt] \hline\cr\end{tabular}}$
${\mathfrak{vect}}^{(1)}(1;\underline{5})\quad{\footnotesize\begin{tabular}[]{|l|l|l|l|}\hline\cr${\mathfrak{g}}_{{\bar{0}}}^{<1>}$&$\dim
L_{k}$&$\dim L^{(k)}$&Parameters\\\\[2.84526pt]
\hline\cr${\mathfrak{v}}_{5}$&$15$&$15$&$(10000)$ and $(1abcd)$\\\
\hline\cr{solv}&$15,14,13,12,11,10,9,8,$&$15,0$&$(00000)$\\\
{}&$7,6,5,4,3,2,1,0$&{}&{}\\\\[2.84526pt]
\hline\cr{solv}&$15,14,13,12,11,10,9,8$&$15,0$&$(01000)$\\\\[2.84526pt]
\hline\cr{solv}&$15,14,13,12$&$15,0$&$(00100)$, $(01100)$\\\\[2.84526pt]
\hline\cr{solv}&$15,14$&$15,0$&$(00010)$, $(00110)$, $(01010)$,
$(01110)$\\\\[2.84526pt] \hline\cr{solv}&$15$&$15,0$&$(00001)$, $(00011)$,
$(00101)$, $(00111)$, $(01001)$,\\\ {}&{}&{}&$(01011)$, $(01101)$,
$(01111)$\\\\[2.84526pt] \hline\cr\end{tabular}}$

Let ${\mathfrak{i}}$ be the last term of the sequence $L_{k}$, see (27), for
the even part ${\mathfrak{g}}_{{\bar{0}}}$ of the corresponding Lie
superalgebra ${\mathfrak{g}}$. Set

$c(i):=(0\ldots 010\ldots 0)\text{~{}~{}with a 1 in the $(i+1)$st slot}.$

######  Conjecture.

Grading $c(k)$, where $k=1,\ldots,n-1$, yields the Lie superalgebras such that
$\dim{\mathfrak{i}}=\mathop{\sum}\limits_{1\leq i\leq k}2^{n-1-i}$. The
corresponding generating function $u$ is $x^{(2^{k})}$.

######  Conjecture.

Let $u=\mathop{\sum}\limits_{1\leq i\leq n-1}c_{i}x^{(2^{i})}$, i.e.,
$c_{0}=0$. Let $k$ be the maximal number such that $c_{k}\neq 0$ and
$c_{k+1}=\ldots=c_{n-1}=0$. Then the grading given by $D_{u}$ yields the Lie
superalgebras such that $\dim{\mathfrak{i}}=\mathop{\sum}\limits_{1\leq i\leq
k}2^{n-1-i}$.

###  6.7. The derivations of ${\mathfrak{vect}}^{(1)}(1;\underline{n})$

The result of this subsection is probably known, but we’d like to draw
attention to Sierpiński sieves here. Observe that solutions of the linear
equations (8b), i.e., derivations of
${\mathfrak{vect}}^{(1)}(1;\underline{n})$, form the Sierpiński sieve of order
$\underline{n}$ under the main diagonal, i.e., for $i\geq j$, we have

$\displaystyle c_{i,j}={}$
$\displaystyle\Big{(}\mathchoice{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{i}{j-1}$}}{\raisebox{0.25pt}{$\genfrac{}{}{0.0pt}{0}{i}{j-1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{i}{j-1}$}}{\raisebox{-0.5pt}{$\genfrac{}{}{0.0pt}{1}{i}{j-1}$}}\Big{)}c_{i-j+1,1},\text{~{}which
corresponds to the
derivation~{}$\mathop{\mathrm{ad}}\nolimits_{x^{(i-j+1)}\partial}$},$ and
nonzero diagonals with parameters $c_{1,2}$, $c_{1,3}$, $c_{1,5}$, …,
$c_{1,2^{n}+1}$, e.g., for $i>j$, we have $\displaystyle c_{1,j}={}$
$\displaystyle 0\quad\text{if $j\neq 2^{k}+1$, where $k$ is a non-negative
integer,}$ $\displaystyle c_{i,j}={}$ $\displaystyle
c_{i-1,j-1}\text{~{}~{}for $i>1$, which corresponds to the
derivation~{}$\mathop{\mathrm{ad}}\nolimits_{\partial^{k}}$}.$

Namely, for $\underline{n}=3$, we have the following solution of linear
equations (8b), and its schematic form:

$U_{\text{lin}}=\begin{pmatrix}c_{1,1}&c_{1,2}&c_{1,3}&0&c_{1,5}&0&0\\\
c_{2,1}&0&c_{1,2}&c_{1,3}&0&c_{1,5}&0\\\
c_{3,1}&c_{2,1}&c_{1,1}&c_{1,2}&c_{1,3}&0&c_{1,5}\\\
c_{4,1}&0&0&0&c_{1,2}&c_{1,3}&0\\\
c_{5,1}&c_{4,1}&0&0&c_{1,1}&c_{1,2}&c_{1,3}\\\
c_{6,1}&0&c_{4,1}&0&c_{2,1}&0&c_{1,2}\\\
c_{7,1}&c_{6,1}&c_{5,1}&c_{4,1}&c_{3,1}&c_{2,1}&c_{1,1}\\\
\end{pmatrix},\qquad\begin{pmatrix}*&*&*&&*&&\\\ *&&*&*&&*&\\\ *&*&*&*&*&&*\\\
*&&&&*&*&\\\ *&*&&&*&*&*\\\ *&&*&&*&&*\\\ *&*&*&*&*&*&*\\\ \end{pmatrix}.$

For $\underline{n}=4$, we have the following schematic form of the solution of
linear equations (8b):

$\left(\begin{array}[]{ccccccccccccccc}*&*&*&&*&&&&*&&&&&&\\\
&&*&*&&*&&&&*&&&&&\\\ &*&*&*&*&&*&&&&*&&&&\\\ &&&&*&*&&*&&&&*&&&\\\
&*&&&*&*&*&&*&&&&*&&\\\ &&*&&*&&*&*&&*&&&&*&\\\ &*&*&*&*&*&*&*&*&&*&&&&*\\\
&&&&&&&&*&*&&*&&&\\\ &*&&&&&&&*&*&*&&*&&\\\ &&*&&&&&&*&&*&*&&*&\\\
&*&*&*&&&&&*&*&*&*&*&&*\\\ &&&&*&&&&*&&&&*&*&\\\ &*&&&*&*&&&*&*&&&*&*&*\\\
&&*&&*&&*&&*&&*&&*&&*\\\ &*&*&*&*&*&*&*&*&*&*&*&*&*&*\\\ \end{array}\right)$

### Acknowledgements

We are thankful to D. Leites, who raised the problem, I. Shchepochkina, and S.
Skryabin for help, to P. Grozman, whose code SuperLie, see [Gr], we used in
our computer experiments. Thanks are due to S. Bouarroudj, and V. Grandjean
for discussion and useful comments. The first author thanks the Organising
committee of the symposium “Groningen Deformation Day” (Oct.7, 2016,
Groningen, The Netherlands), where the results of this note were delivered,
for hospitality and financial support; his research was partly supported by
WCMCS post-doctoral fellowship and the grant AD 065 NYUAD during his visits of
NYUAD. For the possibility to perform the difficult computations of this
research we are grateful to M. Al Barwani, Director of the High Performance
Computing resources at New York University Abu Dhabi.

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