# Manin Triples and non-degenerate anti-symmetric bilinear forms on Lie
Superalgebras in Characteristic $2$

Saïd Benayadi  Laboratoire de Mathématiques IECL UMR CNRS 7502, Université de
Lorraine, 3 rue Augustin Fresnel, BP 45112, F-57073 Metz Cedex 03, FRANCE.
<EMAIL_ADDRESS>and  Sofiane Bouarroudj  Division of Science
and Mathematics, New York University Abu Dhabi, Po Box 129188, Abu Dhabi,
United Arab Emirates<EMAIL_ADDRESS>
###### Abstract.

In this paper, we consider the notions of Manin triples and double extensions
for a Lie superalgebra ${\mathfrak{g}}$ defined over a field of characteristic
$p=2$. We found cohomological necessary conditions for the pair
$({\mathfrak{g}},{\mathfrak{g}}^{*})$ to form a Manin triple. We introduce the
concept of bi-superalgebras for $p=2$ and established a link between Manin
triples and bi-superalgebras. We show that any double extension of any Manin
triple is also a Manin triple.

We introduce the notion of left-symmetric structures – also known as Vinberg
structures – on Lie algebras in characteristic $2$ and show that every Lie
algebra with an invertible derivation gives rise to a left-symmetric structure
on it. We then show that every ${\mathbb{Z}}/2-$graded Lie algebra with an
invertible derivation can be endowed with a Lie superalgebra structure if the
associator of the corresponding left-symmetric structure vanishes. In
particular, if the Lie algebra is 2-step nilpotent, then the associator
vanishes.

Due to the distinguished role of the squaring on the Lie superalgebra in
characteristic $2$, all these notions require special approaches as compared
to their analogs in other characteristics.

Keywords. Manin triple; modular Lie superalgebra; characteristic $2$

The second author was supported by the grant NYUAD-065.

###### Contents

  1. 1 Introduction
    1. 1.1 Manin triples for Lie (super)algebras
    2. 1.2 $r$-Matrices on (super)algebras
    3. 1.3 Double extensions (DEs) of Lie (super)algebras with a non-degenerate invariant symmetric bilinear form (NIS)
    4. 1.4 Summary of our results
  2. 2 Background (see [LSoS, Ch.1])
    1. 2.1 Symmetric and anti-symmetric bilinear forms on superspaces for $p\not=2$
    2. 2.2 Symmetric and anti-symmetric bilinear forms on superspaces for $p=2$
    3. 2.3 Lie superalgebras in characteristic $2$.
    4. 2.4 Non-degenerate anti-symmetric bilinear forms on Lie superalgebras for $p=2$
  3. 3 The Manin triples and matched pairs for Lie superalgebras in characteristic $2$
    1. 3.1 Manin triples
    2. 3.2 Two examples of Manin triples in characteristic $2$
      1. 3.2.1 A Manin triple constructed from the Heisenberg superalgebra
      2. 3.2.2 The Poisson superalgebras [LSh]
    3. 3.3 Review on cohomology of Lie superalgebras in characteristic $2$
    4. 3.4 $r$-matrices and Manin triples
  4. 4 The Matched pairs for Lie superalgebras in characteristic $2$
    1. 4.2 Proof of Theorem 3.1.1. Theorem
  5. 5 Anti-symmetric forms, Manin Triples and double extensions
    1. 5.1 Left-symmetric and left-alternating algebras in characteristic $2$
    2. 5.2 Double extensions (DEs) of even NIS-Lie superalgebras equipped with closed non-degenerate anti-symmetric bilinear forms
    3. 5.3 Double extensions (DEs) of odd NIS-Lie superalgebras with closed non-degenerate anti-symmetric bilinear forms
    4. 5.4 Double extensions of Manin triples in chararcteristic $2$

##  1\. Introduction

Hereafter, $\mathbb{K}$ is an arbitrary field of characteristic $p$. We will
specify when $\mathbb{K}$ has to be algebraically closed. For a list of simple
Lie superalgebras and their “relatives” we use in this paper, see [BGL,
BGLLS].

###  1.1. Manin triples for Lie (super)algebras

Drinfeld [D] proved that a complex Lie algebra ${\mathfrak{g}}$ is also a Lie
bialgebra if and only if it constitutes, together with its dual
${\mathfrak{g}}^{*}$, a Manin triple
$({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{g}}^{*})$, i.e.,
${\mathfrak{h}}={\mathfrak{g}}\oplus{\mathfrak{g}}^{*}$ as vector spaces,
${\mathfrak{h}}$ is a Lie algebra having a non-degenerate invariant symmetric
bilinear form $B$ (NIS for short) such that ${\mathfrak{g}}$ and
${\mathfrak{g}}^{*}$ are isotropic with respect to $B$. A superization of
Drinfeld’s result is due to Olshansky [O]. Further, Leites and Shapovalov
[LSh] listed, à la Olshlansky, important examples of simple (and close to
simple) Lie superalgebras ${\mathfrak{h}}$ finite dimension and of polynomial
growth such that ${\mathfrak{h}}$ splits into Manin-Olshansky triple.

In this paper we construct matched pairs as well as Manin triples for Lie
superalgebras in characteristic $2$, and consider analogs of some of the
examples of [LSh]. Our construction is different from the case where $p\not=2$
because of the squaring, which was equivalent to the bracket for $p\not=2$ but
plays an important role now, and introduce several completely new concepts.

###  1.2. $r$-Matrices on (super)algebras

Belavin and Drinfeld [BD1, BD2] classified $r$-matrices (that describe
quasitriangular Lie bialgebra structures). Their classification uses the
notion of admissible triples that satisfy certain relations. A superization of
[BD1, BD2] is due to Leites and Serganova [LS] with corrections in [Lyb] and
by Karaali [Ka] who used similar constructions using ideas of Belavin and
Drinfeld for Lie superalgebras with a NIS. She then constructed several
examples of $r$-matrices on simple Lie superalgebras; however, the
classification is not reached yet since certain non-degenerate $r$-matrices do
not fit such a description.

For Lie superalgebras over a field of characteristic $2$, we define the notion
of quasi-triangular Lie superalgebra and give a cohomological interpretation.

For Lie algebras, Michaelis [M] found a condition under which the structure of
a triangular, coboundary Lie bialgebra exists on any Lie algebra containing
linearly independent elements $a$ and $b$ satisfying $[a,b]=kb$ for some non-
zero $k\in\mathbb{K}$. This method gives the structure of a Lie bialgebra on
the Virasoro algebra (with or without a central element) and on many other Lie
algebras.

Feldvoss in [F1] showed that for every finite-dimensional Lie algebra over an
algebraically closed field of characteristic $p\not=2$, the following are
equivalent:

(i) the algebra is either abelian or the three dimensional Heisenberg algebra;

(ii) it does not admits any non-trivial triangular Lie bialgebra structure;

(iii) it does not admit any non-trivial quasi-triangular Lie bialgebra
structure.

However, if the characteristic is equal to 2, then any non-abelian Lie algebra
admits a non-trivial quasi-triangular Lie bialgebra structure.

A further generalization to an arbitrary field was carried out by Feldvoss
himself in [F2].

De Smedt ([Sm]) in 1994 (rediscovered in [F2]) showed that every finite-
dimensional non-abelian Lie algebra over the real or complex numbers admits a
non-trivial coboundary Lie bialgebra structure.

###  1.3. Double extensions (DEs) of Lie (super)algebras with a non-degenerate
invariant symmetric bilinear form (NIS)

Medina and Revoy in [MR1] introduced the notion of double extensions for Lie
algebras with NIS in characteristic zero by means of 1-dimensional space on a
simple Lie algebra. They showed that if ${\mathfrak{g}}$ is such a non-simple
Lie algebra, then it can be described inductively in terms of another Lie
algebra ${\mathfrak{a}}$ with NIS; in particular, if the center of
${\mathfrak{g}}$ is not trivial, then ${\mathfrak{g}}$ is a double extension
of a Lie algebra ${\mathfrak{a}}$ with NIS by means of a 1-dimensional space
(see also [FS]). Solvable NIS-Lie algebras ${\mathfrak{g}}$ can be embraced by
this method since the center of ${\mathfrak{g}}$ is non-trivial. Passage to
the ground field of characteristic $p>0$ brings new examples (see [BBH]), but
nothing conceptually new when it comes to the construction of double
extensions, except, perhaps, for inductive descriptions à la Medina and Revoy,
since Lie’s theorem and the Levi decomposition do not hold true anymore if
$p>0$.

The first superization of results by Medina and Revoy was initiated in [BeB]
provided $\mathrm{dim}({\mathfrak{g}}_{\bar{1}})=2$. They showed that every
non-simple NIS-Lie superalgebra can be described by successive double
extensions. The construction have been improved in a series of papers (see
[Bor, B, BBB, ABB]). Superization of the definition of a double extension
introduces completely new possibilities: the bilinear form ${\mathscr{B}}$,
the derivation ${\mathscr{D}}$, the central element $x$ may be odd provided
$p({\mathscr{B}})=p(x)+p(x^{*})$. The first of these possibilities is related
with Manin-Olshansky triples of the type described in [LSh], and loops with
values in ${\mathfrak{p}}{\mathfrak{s}}{\mathfrak{q}}(n)$ and
${\mathfrak{h}}^{(1)}(0|2n+1)$, see [BLS].

The case of characteristic $2$ is completely different and requires new
definitions and methods as shown in [BeBou].

Here we will be studying the notion of double extensions of Lie superalgebras
equipped with a closed ${\bar{1}}$-antisymmetric ortho-orthogonal and
periplectic forms – see Section 2.2 for the terminology – as well as Manin
triples defined over a field of characteristic $p=2$.

###  1.4. Summary of our results

Here is a summary of our results:

(1) We study Manin triples for Lie superalgebras in characteristic $2$ and
provide two examples of such triples in Section 3.2

(2) We generalize the construction to any pair of Lie superalgebras
${\mathfrak{g}}$ and ${\mathfrak{k}}$ satisfying certain conditions. We call a
pair satisfying these conditions a matched pair.

(3) We showed that if there is an $r$-matrix with values in a Lie superalgebra
${\mathfrak{g}}$, then we can naturally define a squaring on
${\mathfrak{g}}^{*}$. We then showed that this squaring satisfies the Jacobi
identity if and only if a certain condition (26) is satisfied.

(4) The existence of an $r$-matrix with values in ${\mathfrak{g}}$ gives rise
to a non-degenerate anti-symmetric form on $F({\mathfrak{g}})$, where $F$ is
the functor forgetting the parity.

(5) The existence of a symplectic form on a Lie algebra induces a left-
symmetric structure on the algebra – also known as Vinberg algebras.

(6) We showed that every Lie algebra with an invertible derivation gives rise
to a left-symmetric structure on it. We then show that every
${\mathbb{Z}}/2$-graded Lie algbera in characteristic $2$ with an invertible
derivation can be turned into a Lie superalgbera provided that associator of
the left-symmetric structure vanishes. Compare with [BLLSq].

(7) We showed that for every Lie superalgebra equipped with either a closed
${\bar{1}}$-antisymmetric ortho-orthogonal or a periplectic form – see Section
2.2 for the terminology – its double extension must also have a closed
${\bar{1}}$-antisymmetric ortho-orthogonal or a periplectic form as well.
Conversely, we showed that the converse is true provided some conditions are
fulfilled. Here there are 4 cases: the NIS is either even or odd and the form
is either ortho-orhtogonal (even) or periplectic (odd).

### Notation

Let ${\mathfrak{g}}$ be a finite-dimensional Lie superalgebra, and let
${\mathfrak{g}}^{*}$ be its dual also endowed with a Lie superalgebra
structure. By identifying ${\mathfrak{g}}\cong{\mathfrak{g}}^{**}$ we view the
map $\mathop{\mathrm{ad}}\nolimits_{f}(-)(x)\in{\mathfrak{g}}^{**}$, for every
$f\in{\mathfrak{g}}^{*}$ and $x\in{\mathfrak{g}}$, as an element of
${\mathfrak{g}}$ and denote it by $x\circ\mathop{\mathrm{ad}}\nolimits_{f}$
for short.

##  2\. Background (see [LSoS, Ch.1])

The general linear Lie superalgebra ${\mathfrak{gl}}(V)$ of all linear
operators in the superspace $V=V_{\bar{0}}\oplus V_{\bar{1}}$ over the ground
field ${\mathbb{K}}$ is denoted by ${\mathfrak{gl}}(\text{Size})$, where
$\text{Size}:=(p_{1},\dots,p_{|\text{Size}|})$ is an ordered collection of
parities of the basis vectors of $V$ for which we take only vectors
homogeneous with respect to parity, and $|\text{Size}|:=\dim V$. The Lie
superalgebra of all supermatrices of size Size is also denoted by
${\mathfrak{gl}}(\text{Size})$. Usually, for the standard (simplest from a
certain point of view) format
$\text{Size}_{st}:=({\bar{0}},\dots,{\bar{0}},{\bar{1}},\dots,{\bar{1}})$, the
notation ${\mathfrak{gl}}(\text{Size}_{st})$ is abbreviated to
${\mathfrak{gl}}(\dim V_{\bar{0}}|\dim V_{\bar{1}})$. Any
$X\in{\mathfrak{gl}}(\text{Size})$ can be uniquely expressed as the sum of its
even and odd parts; in the standard format this is the following block
expression; on non-zero summands the parity is defined:

$X=\text{\small$\begin{pmatrix}A&B\\\
C&D\end{pmatrix}$}=\text{\small$\begin{pmatrix}A&0\\\
0&D\end{pmatrix}$}+\text{\small$\begin{pmatrix}0&B\\\ C&0\end{pmatrix}$},\quad
p\left(\text{\small$\begin{pmatrix}A&0\\\
0&D\end{pmatrix}$}\right)={\bar{0}},\;p\left(\text{\small$\begin{pmatrix}0&B\\\
C&0\end{pmatrix}$}\right)={\bar{1}}.$

###  2.1. Symmetric and anti-symmetric bilinear forms on superspaces for
$p\not=2$

Let $V$ and $W$ be two superspaces defined over an arbtray field
${\mathbb{K}}$ of characteristic $p\not=2$.

Let ${\mathscr{B}}\in\text{Bil}(V,W)$ be a homogenous bilinear form. The Gram
matrix $B=(B_{ij})$ associated to ${\mathscr{B}}$ is given by the formula,

(1)
$B_{ij}=(-1)^{p({\mathscr{B}})p(v_{i})}{\mathscr{B}}(v_{i},v_{j})\text{~{}~{}for
the basis vectors $v_{i}\in V$.}$

This definition can be extended by linearity to non-homegenous forms.
Moreover, it allows us to identify a bilinear form $B(V,W)$ with an element of
$\mathop{\mathrm{Hom}}\nolimits(V,W^{*})$. Consider the upsetting of bilinear
forms
$u\colon\mathop{\mathrm{Bil}}\nolimits(V,W)\longrightarrow\mathop{\mathrm{Bil}}\nolimits(W,V)$
given by the formula

(2) $u({\mathscr{B}})(w,v)=(-1)^{p(v)p(w)}{\mathscr{B}}(v,w)\text{~{}~{}for
any $v\in V$ and $w\in W$.}$

In terms of the Gram matrix $B$ of ${\mathscr{B}}$: the form ${\mathscr{B}}$
is symmetric if and only if

(3) $u(B)=B,\;\text{ where
$u(B)=\text{\small$\begin{pmatrix}R^{t}&(-1)^{p({\mathscr{B}})}T^{t}\\\
(-1)^{p({\mathscr{B}})}S^{t}&-U^{t}\end{pmatrix}$}$ for
$B=\text{\small$\begin{pmatrix}R&S\\\ T&U\end{pmatrix}$}$.}$

Similarly, anti-symmetry of ${\mathscr{B}}$ means that $u(B)=-B$.

Since $p\not=2$, observe that if ${\mathscr{B}}$ is symmetric, then $U$ is
zero-diagonal, and if ${\mathscr{B}}$ is anti-symmetric, then $R$ is zero-
diagonal.

A non-degenerate even (resp. odd) bilinear form is called orthosymplectic
(resp. periplectic).

Most popular normal shapes of the even non-degenerate symmetric forms are the
following matrices in the standard format:

(4)
$\begin{array}[]{l}B^{\prime}_{ev}(m|2n)=\text{\small$\begin{pmatrix}1_{m}&0\\\
0&J_{2n}\end{pmatrix}$},\quad\text{where
$J_{2n}=\text{\small$\begin{pmatrix}0&1_{n}\\\ -1_{n}&0\end{pmatrix}$}$}\\\
B_{ev}(2k|2n)=\text{\small$\begin{pmatrix}\Pi_{2k}&0\\\
0&J_{2n}\end{pmatrix}$},\quad\text{where
$\Pi_{2k}=\text{\small$\begin{pmatrix}0&1_{k}\\\ 1_{k}&0\end{pmatrix}$}$},\\\
B_{ev}(2k+1|2n)=\text{\small$\begin{pmatrix}\Pi_{2k+1}&0\\\
0&J_{2n}\end{pmatrix}$},\quad\text{where
$\Pi_{2k+1}=\text{\footnotesize$\begin{pmatrix}0&0&1_{k}\\\ 0&1&0\\\
1_{k}&0&0\end{pmatrix}$}$}.\end{array}$

The normal shapes of the odd non-degenerate bilinear form are the following
matrices in the standard format (NB: we did not make a mistake below selecting
(anti-)symmetric form)

(5) $B_{odd}(n|n)=\begin{cases}J_{2n}&\text{if the form is symmetric},\\\
\Pi_{2n}&\text{if the form is anti-symmetric}.\\\ \end{cases}$

###  2.2. Symmetric and anti-symmetric bilinear forms on superspaces for $p=2$

Let ${\mathscr{B}}\in\text{Bil}(V,W)$ be a bilinear form. The Gram matrix
$B=(B_{ij})$ associated to ${\mathscr{B}}$ is given by the formula

(6) $B_{ij}={\mathscr{B}}(v_{i},v_{j})\text{~{}~{}for the basis vectors
$v_{i}\in V$.}$

The upsetting of bilinear forms
$u\colon\mathop{\mathrm{Bil}}\nolimits(V,W)\longrightarrow\mathop{\mathrm{Bil}}\nolimits(W,V)$
is given by the formula

(7) $u({\mathscr{B}})(w,v)={\mathscr{B}}(v,w)\text{~{}~{}for any $v\in V$ and
$w\in W$.}$

In terms of the Gram matrix $B$ of ${\mathscr{B}}$, we have

$u(B)=\text{\small$\begin{pmatrix}R^{t}&T^{t}\\\
S^{t}&U^{t}\end{pmatrix}$}\text{ for }B=\text{\small$\begin{pmatrix}R&S\\\
T&U\end{pmatrix}$}.$

We say that the bilinear form ${\mathscr{B}}$ is

  * •

symmetric if $u(B)=B$;

  * •

${\bar{0}}$-antisymmetric (called supersymmetric in [BeBou]) if $u(B)=B$ and
$U$ is zero-diagonal;

  * •

${\bar{1}}$-antisymmetric if $u(B)=B$ and $R$ is zero-diagonal.

In the particular case of a purely even space, we say that $B$ is symmetric if
$u(B)=B$ and antisymmetric if $u(B)=B$ and $R$ is zero-diagonal (recall that
in this case $S=0,T=0,$ and $U=0$.)

A non-degenerate even (resp. odd) bilinear form is called ortho-orthogonal
(resp. periplectic).

Lebedev proved (see [LeD]) that over algebraically closed fields of
characteristic $2$ the equivalence classes of non-degenerate symmetric
bilinear forms ${\mathscr{B}}$ on the $n$-dimensional space $V$ are as
follows:

(i) for $n$ odd, there is just one equivalence class, for the normal shape of
${\mathscr{B}}$ one can take $1_{n}$.

(ii) for $n$ even, there are two equivalence class. For the normal shape of
${\mathscr{B}}$ of one class one can take $1_{n}$, of another class one can
take $\Pi_{n}$.

Following the custom in differential geometry, we will denote a
${\bar{1}}$-antisymmetric bilinear form by $\omega$.

Let us reformulate the condition of ${\bar{1}}$-antisymmetry in terms of the
bilinear form $\omega$. An even bilinear form is ${\bar{1}}$-antisymmetric if
and only if:

(8) $\omega(x,y)=\omega(y,x),\text{ for all $x,y\in V_{\bar{i}}$ where
$\bar{i}={\bar{0}},{\bar{1}}$; and $\omega(x,x)=0$ for all $x\in
V_{\bar{0}}$}.$

An odd bilinear form is ${\bar{1}}$-antisymmetric if and only if:

(9) $\omega(x,y)=\omega(x,y)\text{ for all $x,y\in{\mathfrak{g}}$};$

###  2.3. Lie superalgebras in characteristic $2$.

Following [BGL, LeD, BGLLS, BLLSq], a Lie superalgebra in characteristic $2$
is a superspace
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ over
$\mathbb{K}$ such that ${\mathfrak{g}}_{\bar{0}}$ is a Lie algebra,
${\mathfrak{g}}_{\bar{1}}$ is a ${\mathfrak{g}}_{\bar{0}}$-module made two-
sided by symmetry, and on ${\mathfrak{g}}_{\bar{1}}$ a squaring, denoted by
$s_{\mathfrak{g}}:{\mathfrak{g}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}$,
is given. The bracket on ${\mathfrak{g}}_{\bar{0}}$, as well as the action of
${\mathfrak{g}}_{\bar{0}}$ on ${\mathfrak{g}}_{\bar{1}}$, is denoted by the
same symbol $[\cdot,\cdot]_{\mathfrak{g}}$. For any
$x,y\in{\mathfrak{g}}_{\bar{1}}$, their bracket is then defined by

$[x,y]_{\mathfrak{g}}:=s_{\mathfrak{g}}(x+y)-s_{\mathfrak{g}}(x)-s_{\mathfrak{g}}(y).$

The bracket is extended to non-homogenous elements by bilinearity.

The Jacobi identity involving the squaring reads as follows:

$[s_{\mathfrak{g}}(x),y]_{\mathfrak{g}}=[x,[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}\;\text{
for any $x\in{\mathfrak{g}}_{\bar{1}}$ and $y\in{\mathfrak{g}}$}.$

Let ${\mathscr{B}}_{\mathfrak{g}}$ be a bilinear form on
${\mathfrak{g}}:=({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$.
Following [BeBou], we say that ${\mathscr{B}}_{\mathfrak{g}}$ is invariant if
${\mathscr{B}}_{\mathfrak{g}}([x,y]_{\mathfrak{g}},z)={\mathscr{B}}_{\mathfrak{g}}(x,[y,z]_{\mathfrak{g}})$
for any $x,y,z\in{\mathfrak{g}}$. Observe that the squaring is not involved
here.

We call the Lie superalgebra
${\mathfrak{g}}:=({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$
a NIS-superalgebra (sometimes called quadratics, see [BeB]) if it has a non-
degenerate invariant ${\bar{0}}$-antisymmetric bilinear form
${\mathscr{B}}_{\mathfrak{g}}$. A NIS-Lie superalgebra is called irreducible
if it cannot be decomposed into direct sum of mutually orthogonal ideals.

For any Lie superalgebra ${\mathfrak{g}}$ in characteristic $2$, its derived
algebras are defined to be (for $i\geq 0$)

${\mathfrak{g}}^{(0)}:={\mathfrak{g}},\quad{\mathfrak{g}}^{(i+1)}=[{\mathfrak{g}}^{(i)},{\mathfrak{g}}^{(i)}]_{\mathfrak{g}}+\mathop{\mathrm{{Span}}}\nolimits\\{s_{\mathfrak{g}}(x)\mid
x\in({\mathfrak{g}}^{(i)})_{\bar{1}}\\}.$

A linear map ${\mathscr{D}}:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ is called
a derivation of the Lie superalgebra ${\mathfrak{g}}$ if, in addition to

(10)
${\mathscr{D}}([x,y]_{\mathfrak{g}})=[{\mathscr{D}}(x),y]_{\mathfrak{g}}+[x,{\mathscr{D}}(y)]_{\mathfrak{g}}\quad\text{for
any $x\in{\mathfrak{g}}_{\bar{0}}$ and $y\in{\mathfrak{g}}$},$

we have

(11)
${\mathscr{D}}(s_{\mathfrak{g}}(x))=[{\mathscr{D}}(x),x]_{\mathfrak{g}}\quad\text{for
any $x\in{\mathfrak{g}}_{\bar{1}}$}.$

It is worth noticing that condition (11) implies condition (10) if
$x,y\in{\mathfrak{g}}_{\bar{1}}$.

We denote the space of all derivations of ${\mathfrak{g}}$ by
${\mathfrak{der}}({\mathfrak{g}})$.

###  2.4. Non-degenerate anti-symmetric bilinear forms on Lie superalgebras
for $p=2$

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a Lie superalgebra in characteristic $2$.

Let $\omega:{\mathfrak{g}}\times{\mathfrak{g}}\rightarrow{\mathbb{K}}$ be a
bilinear form. We say that $\omega$ is closed if the following cocycle
conditions are satisfied:

(12) $\displaystyle\omega([x,y],z)+\omega([z,x],y)+\omega([y,z],x)$
$\displaystyle=$ $\displaystyle 0\quad\text{ for all
$x,y,z\in{\mathfrak{g}}$};$ (13) $\displaystyle\omega(s_{\mathfrak{g}}(x),z)$
$\displaystyle=$ $\displaystyle\omega(x,[x,z])\quad\text{ for all
$x\in{\mathfrak{g}}_{\bar{1}}$ and $z\in{\mathfrak{g}}$}.$

Observe that the squaring has to be taken into account in defining the cocycle
condition. For more details, see [BGL2].

######  Proposition (Link NIS’s with anti-symmetric forms).

Let $({\mathfrak{g}},\mathscr{B})$ be a NIS Lie superalgebra with
$p({\mathscr{B}})=\bar{0}$ (resp. $p({\mathscr{B}})=\bar{1}$). Then
${\mathfrak{g}}$ admits an ${\bar{1}}$-antisymmetric ortho-orthogonal (resp.
periplectic) form $\omega$ if and only if there exists an invertible
$\Delta\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ such that
${\mathscr{B}}$ is $\Delta$-invariant and

$\omega(x,y)=\mathscr{B}(\Delta(x),y)\quad\text{for all
$x,y\in{\mathfrak{g}}$}.$

###### Proof.

Let us prove the “only if” part. Suppose that ${\mathfrak{g}}$ admits a
${\bar{1}}$-antisymmetric ortho-orthogonal (resp. periplectic) form $\omega$.
Since $\omega$ and ${\mathscr{B}}$ are non-degenerate, there exists
$\Delta\in\mathop{\mathrm{End}}\nolimits({\mathfrak{g}})_{\bar{0}}$ such that

$\omega(x,y)={\mathscr{B}}(\Delta(x),y)\quad\text{for all
$x,y\in{\mathfrak{g}}$}.$

Let us prove that $\Delta$ is an invertible derivation. Indeed, the fact that
$\Delta$ is invertible follows from the fact that $\omega$ is non-degenerate.
Moreover, the fact that ${\mathscr{B}}$ is $\Delta$-invariant follows from
condition 8 or 9 above. For all $x,y,z\in{\mathfrak{g}}$, we have

$\displaystyle{\mathscr{B}}(\Delta([x,y]),z)=\omega([x,y],z)=\omega([z,x],y)+\omega([y,z],x)={\mathscr{B}}(\Delta([z,x]),y)+{\mathscr{B}}(\Delta([y,z]),x)$
$\displaystyle={\mathscr{B}}([z,x],\Delta(y))+{\mathscr{B}}([y,z],\Delta(x))={\mathscr{B}}(z,[x,\Delta(y)])+{\mathscr{B}}(z,[y,\Delta(x)])$
$\displaystyle={\mathscr{B}}(z,[x,\Delta(y)]+[y,\Delta(x)]).$

Therefore, $\Delta([x,y])=[\Delta(x),y]+[x,\Delta(y)]$ since ${\mathscr{B}}$
is non-degenerate and ${\bar{0}}$-antisymmetric. On the other hand, (for all
$x\in{\mathfrak{g}}_{\bar{1}}$ and $z\in{\mathfrak{g}}$):

$\begin{array}[]{l}{\mathscr{B}}(\Delta(s(x)),z)=\omega(s(x),z)=\omega(x,[x,z])={\mathscr{B}}(\Delta(x),[x,z])={\mathscr{B}}([\Delta(x),x],z).\end{array}$

It follows that $\Delta(s(x))=[x,\Delta(x)]$ since ${\mathscr{B}}$ is non-
degenerate. Therefore, $\Delta$ is indeed a derivation.

For the “if” part, the proof results from the fact that $\Delta$ is a
derivation and the bilinear form ${\mathscr{B}}$ is $\Delta$-invariant. ∎

##  3\. The Manin triples and matched pairs for Lie superalgebras in
characteristic $2$

Throughout this section $\mathbb{K}$ is an arbitray field of characteristic
$2$.

###  3.1. Manin triples

Let ${\mathfrak{h}}$ be a Lie superalgebra and ${\mathfrak{g}},{\mathfrak{k}}$
be two Lie sub-superalgebras of ${\mathfrak{h}}$. The triple
$({\mathfrak{h}},{\mathfrak{g}},{\mathfrak{k}})$ is called a Manin triple if
it satisfies the following two conditions:

  1. (1)

${\mathfrak{h}}={\mathfrak{g}}\oplus{\mathfrak{k}}$, the direct sum of vector
subspaces ${\mathfrak{g}}$ and ${\mathfrak{k}}$;

  2. (2)

there exists a NIS $\mathscr{B}$ on ${\mathfrak{h}}$ such that
${\mathfrak{g}}$ and ${\mathfrak{k}}$ are isotropic with respect to
$\mathscr{B}$

For a comprehensive construction of Manin triples for complex Lie algebras,
see [CP]. For Lie superalgebras in chararcteristic 2, the first example of a
Manin triple was given in [BeBou]. Here we generalize the construction
developed in [BeBou].

######  Theorem (On Manin Triples).

Let $({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ be a Lie
superalgebra in characteristic $2$, and suppose that on the dual space
${\mathfrak{g}}^{*}$ a Lie superalgebra structure
$({\mathfrak{g}}^{*},[\cdot,\cdot]_{{\mathfrak{g}}^{*}},s_{{\mathfrak{g}}^{*}})$
is also given. On the space
${\mathfrak{h}}:={\mathfrak{g}}\oplus{\mathfrak{g}}^{*}$, we consider a
natural bilinear form defined by

${\mathscr{B}}_{\mathfrak{h}}(x+f,y+g)=g(x)+f(y)\quad\text{ for every
$x,y\in{\mathfrak{g}}$ and $f,g\in{\mathfrak{g}}^{*}$.}$

There exists a unique NIS-superalgebra structure on
$({\mathfrak{h}},{\mathscr{B}}_{\mathfrak{h}})$ provided the conditions (14),
(15), and (16) are fulfilled:

(14) $\displaystyle s_{\mathfrak{g}}(x)\circ\mathop{\mathrm{ad}}\nolimits_{h}$
$\displaystyle=$
$\displaystyle[x,x\circ\mathop{\mathrm{ad}}\nolimits_{h}]_{\mathfrak{g}}+x\circ\mathop{\mathrm{ad}}\nolimits_{h\circ\mathop{\mathrm{ad}}\nolimits_{x}}\;\text{
for all $x\in{\mathfrak{g}}_{\bar{1}}$ and $h\in{\mathfrak{g}}^{*}$};$ (15)
$\displaystyle
s_{{\mathfrak{g}}^{*}}(f)\circ\mathop{\mathrm{ad}}\nolimits_{y}$
$\displaystyle=$
$\displaystyle[f,f\circ\mathop{\mathrm{ad}}\nolimits_{y}]_{{\mathfrak{g}}^{*}}+f\circ\mathop{\mathrm{ad}}\nolimits_{y\circ\mathop{\mathrm{ad}}\nolimits_{f}}\;\text{
for all $y\in{\mathfrak{g}}$ and $f\in{\mathfrak{g}}^{*}_{\bar{1}}$};$ (16)
$\displaystyle[x\circ\mathop{\mathrm{ad}}\nolimits_{f},y]_{\mathfrak{g}}$
$\displaystyle=$ $\displaystyle
y\circ\mathop{\mathrm{ad}}\nolimits_{f\circ\mathop{\mathrm{ad}}\nolimits_{x}}+[x,y\circ\mathop{\mathrm{ad}}\nolimits_{f}]_{\mathfrak{g}}+x\circ\mathop{\mathrm{ad}}\nolimits_{f\circ\mathop{\mathrm{ad}}\nolimits_{y}}+[x,y]_{\mathfrak{g}}\circ\mathop{\mathrm{ad}}\nolimits_{f},$
$\begin{array}[]{l}\text{for all $x,y\in{\mathfrak{g}}_{\bar{0}}$, and
$f\in{\mathfrak{g}}^{*}$};\\\ \text{or for all $x\in{\mathfrak{g}}_{\bar{0}}$,
$y\in{\mathfrak{g}}$ and $f\in{\mathfrak{g}}^{*}_{\bar{0}}$};\\\ \text{or for
all $x\in{\mathfrak{g}}_{\bar{1}}$, $y\in{\mathfrak{g}}$ and
$f\in{\mathfrak{g}}^{*}_{\bar{1}}$}.\end{array}$

Explicitly,

  * •

The bracket is given by (for all $x,y\in{\mathfrak{g}}$ and for all
$f,h\in{\mathfrak{g}}^{*}$)

$[x+f,y+h]_{\mathfrak{h}}:=[x,y]_{\mathfrak{g}}+y\circ\mathop{\mathrm{ad}}\nolimits_{f}+x\circ\mathop{\mathrm{ad}}\nolimits_{h}+[f,h]_{{\mathfrak{g}}^{*}}+f\circ\mathop{\mathrm{ad}}\nolimits_{y}+h\circ\mathop{\mathrm{ad}}\nolimits_{x}.$

  * •

The squaring is given by

$s_{\mathfrak{h}}(x+f)=s_{\mathfrak{g}}(x)+s_{{\mathfrak{g}}^{*}}(f)+f\circ\mathop{\mathrm{ad}}\nolimits_{x}+x\circ\mathop{\mathrm{ad}}\nolimits_{f}\;\text{
for all $x\in{\mathfrak{g}}_{\bar{1}}$ and
$f\in{\mathfrak{g}}^{*}_{\bar{1}}$.}$

The bilinear form ${\mathscr{B}}_{\mathfrak{h}}$ is
$\mathop{\mathrm{ad}}\nolimits_{\mathfrak{h}}$-invariant.

Clearly,

$[\cdot,\cdot]_{\mathfrak{h}}|_{\mathfrak{g}}=[\cdot,\cdot]_{\mathfrak{g}},\quad[\cdot,\cdot]_{\mathfrak{h}}|_{{\mathfrak{g}}^{*}}=[\cdot,\cdot]_{{\mathfrak{g}}^{*}},\quad
s_{\mathfrak{h}}|_{\mathfrak{g}}=s_{\mathfrak{g}},\quad
s_{\mathfrak{h}}|_{{\mathfrak{g}}^{*}}=s_{{\mathfrak{g}}^{*}},$

and therefore
$({\mathfrak{h}}:={\mathfrak{g}}\oplus{\mathfrak{g}}^{*},{\mathfrak{g}},{\mathfrak{g}}^{*})$
is a Manin triple.

For the proof, see Section 4.

The reader may wonder why there is no analogue of Eq. (16) on
${\mathfrak{g}}^{*}$. The answer is yes indeed it should be, but Lemma 3.1.2.
Lemma shows that it is equivalent to Eq. (16).

######  Lemma (Technical).

(i) For all $x,y\in{\mathfrak{g}}$ and $f,h\in{\mathfrak{g}}^{*}$, we have

$h([x,y\circ\mathop{\mathrm{ad}}\nolimits_{f}]_{\mathfrak{g}})=[f,h\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}}(y).$

(ii) For all $x,y\in{\mathfrak{g}}$ and $f,h\in{\mathfrak{g}}^{*}$, the
following two conditions are equivalent

$\begin{array}[]{lcl}[x\circ\mathop{\mathrm{ad}}\nolimits_{f},y]_{\mathfrak{g}}+y\circ\mathop{\mathrm{ad}}\nolimits_{f\circ\mathop{\mathrm{ad}}\nolimits_{x}}&=&[x,y\circ\mathop{\mathrm{ad}}\nolimits_{f}]_{\mathfrak{g}}+x\circ\mathop{\mathrm{ad}}\nolimits_{f\circ\mathop{\mathrm{ad}}\nolimits_{y}}+[x,y]_{\mathfrak{g}}\circ\mathop{\mathrm{ad}}\nolimits_{f},\\\\[5.69054pt]
[f\circ\mathop{\mathrm{ad}}\nolimits_{x},h]_{{\mathfrak{g}}^{*}}+h\circ\mathop{\mathrm{ad}}\nolimits_{x\circ\mathop{\mathrm{ad}}\nolimits_{f}}&=&[f,h\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}}+f\circ\mathop{\mathrm{ad}}\nolimits_{x\circ\mathop{\mathrm{ad}}\nolimits_{h}}+[f,h]_{{\mathfrak{g}}^{*}}\circ\mathop{\mathrm{ad}}\nolimits_{x}.\end{array}$

###### Proof.

For Part (i), we have

$\begin{array}[]{rcl}h([x,y\circ\mathop{\mathrm{ad}}\nolimits_{f}]_{\mathfrak{g}})&=&h\circ\mathop{\mathrm{ad}}\nolimits_{x}(y\circ\mathop{\mathrm{ad}}\nolimits_{f})=y\circ\mathop{\mathrm{ad}}\nolimits_{f}(h\circ\mathop{\mathrm{ad}}\nolimits_{x})\quad\text{(we
identify ${\mathfrak{g}}$ with ${\mathfrak{g}}^{**}$)}\\\\[5.69054pt]
&=&y([f,h\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}})=[f,h\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}}(y).\end{array}$

Part (ii) follows by evaluating the first equation at an arbitrary
$h\in{\mathfrak{g}}^{*}$, evaluating the second equation at an arbitrary
$y\in{\mathfrak{g}}$, and then using Part (i).∎

###  3.2. Two examples of Manin triples in characteristic $2$

Here we provide with two examples of Manin triple of Lie superalgebras in
characteristic $2$.

####  3.2.1. A Manin triple constructed from the Heisenberg superalgebra

Consider the Heisenberg superalgebra ${\mathfrak{g}}:={\mathfrak{hei}}(0|2)$
spanned by $p,q$ (odd) and $z$ (even), with
$s_{\mathfrak{g}}(p)=s_{\mathfrak{g}}(q)=0$ and the only nonzero bracket:
$[p,q]=z$. We endow ${\mathfrak{g}}^{*}$ with a Lie superalgebra structure
given by:

$[p^{*},z^{*}]=sq^{*}+tp^{*},\quad[q^{*},z^{*}]=uq^{*}+vp^{*}\text{ for any
fixed $s,t,u,v\in\mathbb{K}$}.$

A NIS-Lie superalgebra
${\mathfrak{h}}:={\mathfrak{hei}}(0|2)\oplus{\mathfrak{hei}}(0|2)^{*}$ exists.
A direct computation shows the following:

  * •

The bracket on ${\mathfrak{h}}$ is given by

$\begin{array}[]{l}[p,p^{*}]=tz,\quad[p,q^{*}]=vz,\quad[p,z^{*}]=tp+vq+q^{*},\\\\[5.69054pt]
[q,p^{*}]=sz,\quad[q,q^{*}]=uz,\quad[q,z^{*}]=sp+uq+p^{*}.\end{array}$

  * •

The squaring on ${\mathfrak{h}}$ is given by (where
$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{K}$):

$\begin{array}[]{lcl}s_{\mathfrak{h}}(\alpha_{1}p+\alpha_{2}q+\alpha_{3}p^{*}+\alpha_{4}q^{*})&=&(\alpha_{1}\alpha_{2}+\alpha_{1}\alpha_{3}t+\alpha_{1}\alpha_{4}v+\alpha_{2}\alpha_{3}s+\alpha_{2}\alpha_{4}u)z.\\\\[5.69054pt]
\end{array}$

We give an example of an outer derivation ${\mathscr{D}}$ on ${\mathfrak{h}}$
that is invertible. This example will justify the studies in Section 5. The
example we provide here was borrowed from [BeBou] in the case where
${\mathfrak{hei}}(0|2)^{*}$ has an abelian structure; namely, $s=t=u=v=0$. The
ground field ${\mathbb{K}}$ is assumed to be algebraically closed, hence
infinite. The invertible derivation is as follows (for
$\alpha_{9},\alpha_{10}\not=0,1$):

${\mathscr{D}}=\alpha_{2}D_{2}+\alpha_{4}D_{4}+\alpha_{9}D_{9}+\alpha_{10}D_{10}+(\alpha_{9}+\alpha_{10})D_{11},$

where

$\begin{array}[]{lcllcllcllcl}D_{2}&=&q^{*}\otimes\widehat{q},&D_{4}&=&p^{*}\otimes\widehat{p},\\\\[5.69054pt]
D_{9}&=&p\otimes\widehat{p}+q^{*}\otimes\widehat{q^{*}}+z\otimes\widehat{z},&D_{10}&=&q\otimes\widehat{q}+p^{*}\otimes\widehat{p^{*}}+z\otimes\widehat{z},\\\\[5.69054pt]
D_{11}&=&q^{*}\otimes\widehat{q^{*}}+p^{*}\otimes\widehat{p^{*}}+z^{*}\otimes\widehat{z^{*}}.\end{array}$

####  3.2.2. The Poisson superalgebras [LSh]

Consider ${\mathfrak{g}}={\mathfrak{po}}(0|2n-1)$. Let
${\mathfrak{h}}={\mathfrak{po}}(0|2n)$ be generated by
$\mathbb{K}[\theta_{1},...,\theta_{2n}]$ and ${\mathfrak{g}}$ by
$\mathbb{K}[\theta_{1},...,\theta_{2n-1}]$. The form
${\mathscr{B}}_{\mathfrak{h}}$ on ${\mathfrak{h}}$ is given by the Berezin
integral ${\mathscr{B}}_{\mathfrak{h}}(x,y)=\int xy\;\mathrm{vol}(\theta)$.
Let ${\mathfrak{g}}^{*}={\mathfrak{g}}\cdot\zeta$, where
$\zeta=\theta_{2n}+\sum_{i=1}^{2n-1}k_{i}\theta_{i}$ and
$\sum_{i=1}^{2n-1}{k_{i}}^{2}=1.$

###  3.3. Review on cohomology of Lie superalgebras in characteristic $2$

Recall basic facts concerning the cohomology of Lie superalgebras in
characteristic $2$; for more details, see [BGL2]. 1-cochains are just linear
functions on ${\mathfrak{g}}$ with values in a ${\mathfrak{g}}$-module $M$. In
our context, the differential of a 1-cochain $c$ is defined as follows:

(17) $\begin{array}[]{lcl}{\mathfrak{d}}c(x,y)&=&c([x,y])+x\cdot c(y)+y\cdot
c(x)\quad\text{for all $x,y\in{\mathfrak{g}}$};\\\\[2.84526pt]
{\mathfrak{d}}c(x,x)&=&c(s_{\mathfrak{g}}(x))+x\cdot c(x)\quad\text{for all
$x\in{\mathfrak{g}}_{\bar{1}}$}.\end{array}$

A 1-cocycle $c$ on ${\mathfrak{g}}$ with values in a ${\mathfrak{g}}$-module
$M$ must satisfy the following conditions:

(18) $\displaystyle x\cdot c(y)+y\cdot c(x)+c([x,y]_{\mathfrak{g}})$
$\displaystyle=$ $\displaystyle 0\quad\text{for all $x,y\in{\mathfrak{g}}$},$
(19) $\displaystyle x\cdot c(x)+c(s_{\mathfrak{g}}(x))$ $\displaystyle=$
$\displaystyle 0\quad\text{for all $x\in{\mathfrak{g}}_{\bar{1}}$}.$

The space of all 1-cocycles is denoted by $Z^{1}({\mathfrak{g}};M)$.

To every Lie superalgebra ${\mathfrak{g}}$ in characteristic $2$, we can
assign its desuperization denoted by $F({\mathfrak{g}})$, where $F$ is the
functor of forgetting the super structure. The space $F({\mathfrak{g}})$ turns
into an ordinary $\mathbb{Z}/2$-graded Lie algebra; for more details, see
[BLLSq]. Now, any 1-cocycle $c$ on the Lie algebra $F({\mathfrak{g}})$ is
captured only by the condition (18), and therefore the cohomology of
$F({\mathfrak{g}})$ viewed as a Lie algebra and that of ${\mathfrak{g}}$
viewed as a Lie superalgebra are not necessarily isomorphic. We refer to
[BGL2] for illustrative examples.

Let us define a linear from
$\langle-,-\rangle:{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}\rightarrow\mathbb{K}$
by $\langle f,x\rangle=f(x)$ for all $x\in{\mathfrak{g}}$ and for all
$f\in{\mathfrak{g}}^{*}$. Since ${\mathfrak{g}}$ is finite-dimensional, we
have
$({\mathfrak{g}}\otimes{\mathfrak{g}})^{*}={\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*}$
so we also denote by $\langle-,-\rangle$ the linear form on
$({\mathfrak{g}}\otimes{\mathfrak{g}})^{*}\otimes({\mathfrak{g}}\otimes{\mathfrak{g}})$
defined by

$\langle f\otimes h,x\otimes y\rangle=f(x)h(y)\text{ for all
$x,y\in{\mathfrak{g}}$ and $f,h\in{\mathfrak{g}}$},$

then extend the definition by linearity. Given the bracket
$[\cdot,\cdot]_{{\mathfrak{g}}^{*}}:{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*}\rightarrow{\mathfrak{g}}^{*}$,
its dual is the map
$c_{\mathfrak{g}}:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$.
Similarly, given the bracket
$[\cdot,\cdot]_{\mathfrak{g}}:{\mathfrak{g}}\otimes{\mathfrak{g}}\rightarrow{\mathfrak{g}}$,
its dual is the map
$c_{{\mathfrak{g}}^{*}}:{\mathfrak{g}}^{*}\rightarrow{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*}$.
More precisely,

(20) $\displaystyle\langle[f,h]_{{\mathfrak{g}}^{*}},x\rangle$
$\displaystyle=$ $\displaystyle\langle f\otimes
h,c_{\mathfrak{g}}(x)\rangle\quad\text{ for all $x\in{\mathfrak{g}}$ and
$f,h\in{\mathfrak{g}}^{*}$},$ (21) $\displaystyle\langle
c_{{\mathfrak{g}}^{*}}(f),x\otimes y\rangle$ $\displaystyle=$
$\displaystyle\langle f,[x,y]_{\mathfrak{g}}\rangle\quad\text{ for all
$x,y\in{\mathfrak{g}}$ and $f\in{\mathfrak{g}}^{*}$}.$

We observe here that both $c_{\mathfrak{g}}$ and $c_{{\mathfrak{g}}^{*}}$ have
images generated by symmetric 2-tensors.

######  Proposition (A cohomological interpretation).

The conditions (14), (15) and (16) in Theorem 3.1.1. Theorem are
cohomological. Namely, they are equivalent to the fact that
$c_{\mathfrak{g}}\in
Z^{1}({\mathfrak{g}};{\mathfrak{g}}\otimes{\mathfrak{g}})$ and
$c_{{\mathfrak{g}}^{*}}\in
Z^{1}({\mathfrak{g}}^{*};{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*})$.

###### Proof.

We will only be dealing with the map $c_{\mathfrak{g}}$. Studying the
properties of the map $c_{{\mathfrak{g}}^{*}}$ is absolutely the same.

Now, checking that the cocycle condition (18) is equivalent to (16) is a
routine, so we omit this part. Let us check the conditions involving the
squaring. We have

$\displaystyle\langle f\otimes h,x\cdot
c_{\mathfrak{g}}(x)+c_{\mathfrak{g}}(s_{\mathfrak{g}}(x))\rangle=\langle
f\otimes h,x\cdot c_{\mathfrak{g}}(x)\rangle+\langle f\otimes
h,c_{\mathfrak{g}}(s_{\mathfrak{g}}(x))\rangle$ $\displaystyle=x\cdot(f\otimes
h)(c_{\mathfrak{g}}(x))+(f\otimes
h)(c_{\mathfrak{g}}(s_{\mathfrak{g}}(x)))=((x\cdot f)\otimes h+f\otimes(x\cdot
h))(c_{\mathfrak{g}}(x))+[f,h]_{{\mathfrak{g}}^{*}}(s_{\mathfrak{g}}(x))$
$\displaystyle=\langle[x\cdot
f,h]_{{\mathfrak{g}}^{*}},x\rangle+\langle[f,x\cdot
h]_{{\mathfrak{g}}^{*}},x\rangle+[f,h]_{{\mathfrak{g}}^{*}}(s_{\mathfrak{g}}(x))$
$\displaystyle=\langle[f\circ\mathop{\mathrm{ad}}\nolimits_{x},h]_{{\mathfrak{g}}^{*}},x\rangle+\langle
f,h\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}},x\rangle+\langle[f,h]_{{\mathfrak{g}}^{*}},s_{\mathfrak{g}}(x)\rangle.$

On the other hand, let us consider Eq. (14) applied to $f$. We get:

$\displaystyle\left(s_{\mathfrak{g}}(x)\circ\mathop{\mathrm{ad}}\nolimits_{h}+[x,x\circ\mathop{\mathrm{ad}}\nolimits_{h}]_{\mathfrak{g}}+x\circ\mathop{\mathrm{ad}}\nolimits_{h\circ\mathop{\mathrm{ad}}\nolimits_{x}}\right)(f)$
$\displaystyle=[h,f]_{{\mathfrak{g}}^{*}}(s_{\mathfrak{g}}(x))+f([x,x\circ\mathop{\mathrm{ad}}\nolimits_{h}]_{\mathfrak{g}})+\mathop{\mathrm{ad}}\nolimits_{h\circ\mathop{\mathrm{ad}}\nolimits_{x}}(f)(x)\quad\text{
using Lemma (\ref{EqBra0})}$
$\displaystyle=\langle[h,f]_{{\mathfrak{g}}^{*}},s_{\mathfrak{g}}(x)\rangle+\langle[h,f\circ\mathop{\mathrm{ad}}\nolimits_{x}]_{{\mathfrak{g}}^{*}},x\rangle+\langle[h\circ\mathop{\mathrm{ad}}\nolimits_{x},f]_{{\mathfrak{g}}^{*}},x\rangle.$

It follows that Eq. (14) is equivalent to Eq. (19).∎

######  Proposition (Relation between $c_{\mathfrak{g}}$ and
$c_{{\mathfrak{g}}^{*}}$).

For all $f,g\in{\mathfrak{g}}^{*}$ and $x,y\in{\mathfrak{g}}$, we have

$\langle f\cdot c_{{\mathfrak{g}}^{*}}(h)+h\cdot
c_{{\mathfrak{g}}^{*}}(f)+c_{{\mathfrak{g}}^{*}}([f,h]),x\otimes
y\rangle=\langle f\otimes h,x\cdot c_{\mathfrak{g}}(y)+y\cdot
c_{\mathfrak{g}}(x)+c_{\mathfrak{g}}([x,y])\rangle.$

Thus, the map $c_{\mathfrak{g}}$ satisfies condition (18) if and only if the
map $c_{{\mathfrak{g}}^{*}}$ satisfies condition (18).

######  Remark.

For $p\not=2$, the 1-cocycle conditions of the maps $c_{\mathfrak{g}}$ and
$c_{{\mathfrak{g}}^{*}}$ are actually equivalent thanks to an analogue of
Proposition 3.3.2. Proposition. For $p=2$, they are not necessarily equivalent
because of the different role of the squaring.

In order to prepare for the proof of Proposition 3.3.2. Proposition, we will
be using Sweedler’s notations. Let us write

$c_{{\mathfrak{g}}^{*}}(f):=\sum_{(f)}f_{(1)}\otimes f_{(2)}\text{ for all
$f\in{\mathfrak{g}}^{*}$ and $c_{{\mathfrak{g}}}(x):=\sum_{(x)}x_{(1)}\otimes
x_{(2)}$ for all $x\in{\mathfrak{g}}$}.$

######  Lemma (Technical).

For all $x,y\in{\mathfrak{g}}$ and $f,h\in{\mathfrak{g}}^{*}$, we have

$\langle f\cdot c_{{\mathfrak{g}}^{*}}(h),x\otimes y\rangle=\langle f\otimes
h,\sum_{(x)}x_{(1)}\otimes[x_{(2)},y]+\sum_{(y)}y_{(1)}\otimes[x,y_{(2)}]\rangle.$

###### Proof.

For all $f,h\in{\mathfrak{g}}^{*}$ and $x,y\in{\mathfrak{g}}$, we have

$\displaystyle\langle f\cdot c_{{\mathfrak{g}}^{*}}(h),x\otimes
y\rangle=\langle\sum_{(h)}([f,h_{(1)}]_{{\mathfrak{g}}^{*}}\otimes h_{(2)}+\
h_{(1)}\otimes[f,h_{(2)}]_{{\mathfrak{g}}^{*}},x\otimes y\rangle$
$\displaystyle=\sum_{(h)}(f\otimes h_{(1)})(c_{\mathfrak{g}}(x))\,h_{(2)}(y)+\
h_{(1)}(x)\,(f\otimes h_{(2)})(c_{\mathfrak{g}}(y))$
$\displaystyle=\sum_{(h)}\sum_{(x)}f(x_{(1)})h_{(1)}(x_{(2)})\,h_{(2)}(y)+\sum_{(h)}\sum_{(y)}h_{(1)}(x)\,f(y_{(1)})h_{(2)}(y_{(2)})$
$\displaystyle=f(\sum_{(x)}c_{{\mathfrak{g}}^{*}}(h)(x_{(2)}\otimes
y)\,x_{(1)})+f(\sum_{(y)}c_{{\mathfrak{g}}^{*}}(h)(x\otimes
y_{(2)})\,y_{(1)})$
$\displaystyle=f(\sum_{(x)}h([x_{(2)},y])\,x_{(1)})+\sum_{(y)}h([x,y_{(2)}])\,y_{(1)}])$
$\displaystyle=\langle f\otimes
h,\sum_{(x)}x_{(1)}\otimes[x_{(2)},y]+\sum_{(y)}y_{(1)}\otimes[x,y_{(2)}]\rangle.\qed$

###### (Proof of Proposition 3.3.2. Proposition).

Using Lemma 3.3.3. Lemma and the fact that $c_{\mathfrak{g}}$ takes values in
the space of symmetric 2-tensors, we have

$\displaystyle\langle f\cdot c_{{\mathfrak{g}}^{*}}(h)+h\cdot
c_{{\mathfrak{g}}^{*}}(f)+c_{{\mathfrak{g}}^{*}}([f,h]),x\otimes y\rangle$
$\displaystyle=(f\otimes
h)(\sum_{(x)}x_{(1)}\otimes[x_{(2)},y]+\sum_{(y)}y_{(1)}\otimes[x,y_{(2)}])$
$\displaystyle+(f\otimes h)(\sum_{(x)}[x_{(2)},y]\otimes
x_{(1)}+\sum_{(y)}[x,y_{(2)}]\otimes y_{(1)})+(f\otimes
h)(c_{\mathfrak{g}}([x,y]))$ $\displaystyle=(f\otimes
h)(\sum_{(x)}x_{(1)}\otimes[x_{(2)},y]+\sum_{(x)}[x_{(2)},y]\otimes x_{(1)}$
$\displaystyle+(f\otimes
h)(\sum_{(y)}y_{(1)}\otimes[x,y_{(2)}]+\sum_{(y)}[x,y_{(2)}]\otimes
y_{(1)})+(f\otimes h)(c_{\mathfrak{g}}([x,y]))$ $\displaystyle=(f\otimes
h)(\sum_{(x)}\otimes[y,x_{(1)}]\otimes
x_{(2)}+\sum_{(x)}x_{(1)}\otimes[x_{(2)},y]$ $\displaystyle+(f\otimes
h)(\sum_{(y)}[x,y_{(1)}]\otimes
y_{(2)}+\sum_{(y)}y_{(1)}\otimes[x,y_{(2)}])+(f\otimes
h)(c_{\mathfrak{g}}([x,y]))$ $\displaystyle=(f\otimes h)(x\cdot
c_{\mathfrak{g}}(y)+y\cdot c_{\mathfrak{g}}(x)+c_{\mathfrak{g}}([x,y])).\qed$

###  3.4. $r$-matrices and Manin triples

This section is devoted to the study of triangular Lie bi-superalgebras in
characteristic $2$. We refer to [CP, D, GZB] in the case of Lie
(super)algebras defined over a field of characteristic zero, and to [F1, F2]
in the case of modular Lie algebras.

A Lie coalgebra over a field $\mathbb{K}$ of characteristic $p\not=2$ is a
vector space ${\mathfrak{g}}$ over $\mathbb{K}$ together with a linear mapping
$c:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$, such that

  * (i)

$\mathop{\text{\rm Im}}\nolimits(c)\subseteq\mathop{\text{\rm
Im}}\nolimits(1-\tau)$, and

  * (ii)

$(1+\xi+\xi^{2})\circ(1\otimes c)\otimes c=0,$ where
$\tau:{\mathfrak{g}}\otimes{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$
denotes the switch mapping sending $x\otimes y$ to $y\otimes x$ for all
$x,y\in{\mathfrak{g}}$ while
$\xi:{\mathfrak{g}}\otimes{\mathfrak{g}}\otimes{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}\otimes{\mathfrak{g}}$
denotes the cycle mapping sending $x\otimes y\otimes z$ to $y\otimes z\otimes
x$ for all $x,y,z\in{\mathfrak{g}}$.

The mapping $c$ is called the cobracket of ${\mathfrak{g}}$, (i) is called co-
anticommutativity, and (ii) is called the co-Jacobi identity.

For $p=2$, the co-anticommutativity condition (i) has to be changed with the
following condition:

$c^{t}(f\otimes f)=0\text{ for all $f\in{\mathfrak{g}}^{*}$},$

where
$c^{t}:{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*}\rightarrow{\mathfrak{g}}^{*}$
is the dual of $c$. Our choice is based on the fact that a coalgebra structure
on ${\mathfrak{g}}$ will define a Lie algebra structure on the dual space
${\mathfrak{g}}^{*}$. Recall that for $p=2$, the anti-commutativity of the
bracket is expressed as $[a,a]=0.$

A $\mathbb{Z}/2$-graded Lie coalgebra is a $\mathbb{Z}/2$-graded vector space
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ over
$\mathbb{K}$ endowed with a cobracket $c$ such that
$c({\mathfrak{g}}_{\bar{i}})\subset({\mathfrak{g}}\otimes{\mathfrak{g}})_{\bar{i}}$,
where $\bar{i}={\bar{0}},{\bar{1}}$.

A Lie bialgebra over $\mathbb{K}$ is a vector space ${\mathfrak{g}}$ over
$\mathbb{K}$ together with $\mathbb{K}$-linear mappings
$[\cdot,\cdot]:{\mathfrak{g}}\otimes{\mathfrak{g}}\rightarrow{\mathfrak{g}}$
and $c:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$ such that
$({\mathfrak{g}},[\cdot,\cdot])$ is a Lie algebra, $({\mathfrak{g}},c)$ is a
Lie coalgebra and $c$ is a derivation of the Lie algebra ${\mathfrak{g}}$ to
the ${\mathfrak{g}}$-module ${\mathfrak{g}}\otimes{\mathfrak{g}}$. Namely,

$c([x,y))=x\cdot c(y)-y\cdot(x)\text{ for all $x,y\in{\mathfrak{g}}$}.$

A coboundary Lie bialgebra is a Lie bialgebra such that the cobracket $c$ is
an inner derivation, i.e., there exists an element
$r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$ such that $\delta(x)=x\cdot r$ for
all $x\in{\mathfrak{g}}$.

Let ${\mathfrak{g}}$ be a Lie algebra and
$r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$ be defined by
$r=\sum_{i=1}^{n}a_{i}\otimes b_{i}$, with $a_{i}\in{\mathfrak{g}},$ and
$b_{i}\in{\mathfrak{g}}$ for all $i\in\\{1,\ldots,n\\}$. Consider
${\mathfrak{g}}$ embedded in its universal enveloping superalgebra
$U({\mathfrak{g}})$. We put

$r^{12}=\sum_{1\leq i\leq n}a_{i}\otimes b_{i}\otimes 1,\quad
r^{13}=\sum_{1\leq i\leq n}a_{i}\otimes 1\otimes b_{i},\quad\text{ and }\quad
r^{23}=\sum_{1\leq i\leq n}\otimes a_{i}\otimes b_{i}.$

The mapping
$\mathrm{CBY}:{\mathfrak{g}}\otimes{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}\otimes{\mathfrak{g}}$
given by

$r\rightarrow[r^{12},r^{13}]+[r^{12},r^{13}]+[r^{13},r^{23}]$

is called the classical Yang-Baxter operator (CYBO), the equation
$\mathrm{CYB}(r)=0$ is called the classical Yang-Baxter equation (CYBE), and a
solution of the CYBE is called a classical $r$-matrix.

If ${\mathfrak{g}}$ is a Lie algebra and $V$ is a ${\mathfrak{g}}$-module,
then the set of ${\mathfrak{g}}$-invariant elements of $V$ is defined as

$V^{\mathfrak{g}}:=\\{v\in V\,|\,x\cdot v=0\text{ for all
}x\in{\mathfrak{g}}\\}.$

In the case of Lie algebras defined over a field of characteristic zero,
Drinfeld proved the following theorem.

######  Theorem (Lie bialgebra structure, see[D]).

Let $r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$ and define
$\delta_{r}(x):=x\cdot r$ for every $x\in{\mathfrak{g}}$. Then, $r$ defines a
Lie bialgebra structure on ${\mathfrak{g}}$ if and only if

(i) $r+\tau(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$, and

(ii)
$\mathrm{CYB}(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$.

Therefore, every solution $r$ of the CYBE satisfying
$r+\tau(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$ gives rise
to a coboundary Lie bialgebra structure on ${\mathfrak{g}}$ by means of
$\delta_{r}$. Following [D], such a Lie bialgebra structure is called quasi-
triangular. The quasi-triangular Lie bialgebra structures arising from anti-
symmetric classical $r$-matrices are called triangular.

Following [F1, F2], coboundary Lie bialgebra structures defined over a field
of characteristic $2$ are called quasi-triangular if
$\mathrm{Im}(\delta_{r})\subseteq\mathrm{Im}(1+\tau)$, and called triangular
if $r\in\mathrm{Im}(1-\tau)$.

For $p=2$, the condition (i) in Theorem 3.4.1. Theorem is too weak. Feldvoss
in [F1] has replaced it with the condition
$\mathrm{Im}(\delta_{r})\subseteq\mathrm{Im}(1+\tau)$. We realized that the
condition

(22) $(\delta_{r})^{t}(f\otimes f)=0\text{ for all $f\in{\mathfrak{g}}^{*}$},$

which is weaker, also works.

######  Proposition (Eq. (22) $\Rightarrow
r+\tau(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$).

(i) We have: Eq. (22) holds if and only if

$\sum_{1\leq i\leq
n}\left(f(b_{i})f\circ\mathop{\mathrm{ad}}\nolimits_{a_{i}}+f(a_{i})f\circ\mathop{\mathrm{ad}}\nolimits_{b_{i}}\right)=0\text{
for all $f\in{\mathfrak{g}}^{*}$.}$

(ii) If Eq. (22) holds, then
$r+\tau(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$.

###### Proof.

Let us just prove part (ii). Observe that if Eq. (22) holds, then

$(\delta_{r})^{t}(f\otimes g+g\otimes f)=0\text{ for all
$f,g\in{\mathfrak{g}}.$}$

Now, for all $x\in{\mathfrak{g}}$, we have

$\displaystyle 0=$ $\displaystyle\langle f\otimes g+g\otimes
f,\delta_{r}(x)\rangle=\langle f\otimes
g,\delta_{r}(x)+\tau\circ\delta_{r}(x)\rangle.$

It remains to show that the condition

$\delta_{r}(x)+\tau\circ\delta_{r}(x)=0\text{ for all $x\in{\mathfrak{g}}$}$

is equivalent to
$r+\tau(r)\in({\mathfrak{g}}\otimes{\mathfrak{g}})^{\mathfrak{g}}$. Indeed,

$\delta_{r}(x)+\tau\circ\delta_{r}(x)=x\cdot r+\tau\circ x\cdot
r=x\cdot(r+\tau(r)).\qed$

######  Theorem (Quasi-triangular Lie bialgebra structure, see[F1]).

If ${\mathfrak{g}}$ is a finite-dimensional non-abelian Lie algebra over an
algebraically closed field of characteristic $2$, then ${\mathfrak{g}}$ admits
a non-trivial quasi-triangular Lie bialgebra structure.

######  Remark.

(i) If the Lie algebra is $\mathbb{Z}/2$-graded and the quasi-triangular
structure given by $r$ satisfies
$r\in({\mathfrak{g}}\otimes{\mathfrak{g}})_{\bar{0}}$ then we say that
${\mathfrak{g}}$ admits a non-trivial quasi-triangular $\mathbb{Z}/2$-graded
Lie bialgebra structure.

(ii) If ${\mathfrak{g}}$ is a Lie superalgebra in charcteristic 2 where
${\mathfrak{g}}_{\bar{0}}$ is a non-abelian Lie algebra, then it is always
possible to find a non-trivial quasi-triangular $r$-matrix on
$F({\mathfrak{g}})$. Indeed, just apply Theorem 3.5.1. Theorem to
${\mathfrak{g}}_{\bar{0}}$.

(iii) In the case of Lie superalgebras over a field of characteristic zero,
and under certain conditions, several $r$-matrices have been investigated in
[Ka].

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a $\mathbb{Z}/2$-graded vector space over $\mathbb{K}$ and
$c:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$ be an even
linear map. We say that $({\mathfrak{g}},c)$ is a Lie supercoalgebra if:

(i) $({\mathfrak{g}},c)$ is a $\mathbb{Z}/2$-graded coalgebra;

(ii) There exsits a squaring $s_{{\mathfrak{g}}^{*}}$ on ${\mathfrak{g}}^{*}$
such that
$({\mathfrak{g}}^{*},[\cdot,\cdot]_{{\mathfrak{g}}^{*}},s_{{\mathfrak{g}}^{*}})$,
where $[\cdot,\cdot]_{{\mathfrak{g}}^{*}}$ is given as in (20), is a Lie
superalgebra.

Let $({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ be a Lie
superalgebra in characteristic $2$. Let
$c_{\mathfrak{g}}:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$
be an even linear map such that
$({\mathfrak{g}},c_{\mathfrak{g}},s_{{\mathfrak{g}}^{*}})$ is a supercoalgebra
in characteristic $2$. We say that
$({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},c_{\mathfrak{g}},s_{\mathfrak{g}},s_{{\mathfrak{g}}^{*}})$
is a superbialgebra if:

(i) $c_{\mathfrak{g}}$ is a 1-cocyle on ${\mathfrak{g}}$ with values in
${\mathfrak{g}}\otimes{\mathfrak{g}}$; namely, it satisfies Eqs. (18) and
(19).

(ii) $s_{{\mathfrak{g}}^{*}}$ satisfies the condition

(23) $s_{{\mathfrak{g}}^{*}}(f)\circ[\cdot,\cdot]_{\mathfrak{g}}=(f\otimes
f)\circ(\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}\otimes[\cdot,\cdot]_{\mathfrak{g}})\circ(c_{{\mathfrak{g}}}\otimes\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}+\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}\otimes
c_{{\mathfrak{g}}})\text{ for all $f\in{\mathfrak{g}}_{1}^{*}$}.$

As in the classical case, we will show that with our definitions the notion of
Manin triples is equivalent to the notion of superbialgebra. We have already
shown in Proposition 3.3.2. Proposition that if one of the maps
$c_{\mathfrak{g}}$ and $c_{{\mathfrak{g}}^{*}}$ satisfies the 1-cocycle
condition (18), then the other one must automatically satisfy this condition.
It remains to show that the condition (23) is equivalent to the 1-cocycle
condition (19) for the map $c_{{\mathfrak{g}}^{*}}$. We have

######  Proposition (Eq. (23) is cohomological).

The squaring $s_{{\mathfrak{g}}^{*}}$ satisfies Eq. (23) if and and only if
the map $c_{{\mathfrak{g}}^{*}}$ satisfies Eq. (19) of the $1$-cocycle
condition.

###### Proof.

For all $f\in{\mathfrak{g}}_{\bar{1}}^{*}$, let us write
$c_{{\mathfrak{g}}^{*}}(f)=\sum_{(f)}f_{(1)}\otimes f_{(2)}$. Now, for all
$x,y\in{\mathfrak{g}}$ we have:

$\displaystyle\langle f\cdot c_{{\mathfrak{g}}^{*}}(f),x\otimes
y\rangle=\langle\sum_{(f)}f\cdot f_{(1)}\otimes f_{(2)}+f_{(1)}\otimes f\cdot
f_{(2)},x\otimes y\rangle$ $\displaystyle=\sum_{(f)}f\cdot
f_{(1)}(x)f_{(2)}(y)+f_{(1)}(x)(f\cdot f_{(2)})(y)$
$\displaystyle=\sum_{(f)}(f\otimes
f_{(1)})(c_{\mathfrak{g}}(x))f_{(2)}(y)+f_{(1)}(x)(f\otimes
f_{(2)})(c_{\mathfrak{g}}(y))$ $\displaystyle=\sum_{(f)}\langle f\otimes
f_{(1)}\otimes f_{(2)},c_{\mathfrak{g}}(x)\otimes y+x\otimes
c_{\mathfrak{g}}(y)\rangle$ $\displaystyle=\langle(f\otimes
c_{{\mathfrak{g}}^{*}}(f))\circ(c_{\mathfrak{g}}\otimes\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}+\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}\otimes
c_{\mathfrak{g}}),x\otimes y\rangle$ $\displaystyle=\langle(f\otimes
f)\circ(\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}\otimes[\cdot,\cdot]_{\mathfrak{g}})\circ(c_{\mathfrak{g}}\otimes\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}+\mathop{\mathrm{id}}\nolimits_{\mathfrak{g}}\otimes
c_{\mathfrak{g}}),x\otimes y\rangle.$

On the other hand

$\langle c_{{\mathfrak{g}}^{*}}(s_{{\mathfrak{g}}^{*}}(f)),x\otimes
y\rangle=s_{{\mathfrak{g}}^{*}}(f)([x,y]_{\mathfrak{g}})=\langle
s_{{\mathfrak{g}}^{*}}(f)\otimes[\cdot,\cdot]_{\mathfrak{g}},x\otimes
y\rangle.\qed$

Let ${\mathfrak{g}}$ be a non-abelian Lie superalgebra in characteristic $2$.
Denote by $F({\mathfrak{g}})$ the Lie algebra structure on the underlying
vector space obtained by forgetting the super structure. The Lie algebra
$F({\mathfrak{g}})$ is $\mathbb{Z}/2$-graded, the grading being inherited from
the super structure. Theorem 3.5.1. Theorem guaranteed the existence of a non-
trivial quasi-triangle Lie bialgebra structure on $F({\mathfrak{g}})$. Let us
suppose further that this structure is $\mathbb{Z}/2\mathbb{Z}$-graded. Denote
the corresponding $r-$matrix by $r$ and let us write

$r=\sum_{1\leq i\leq n}a_{i}\otimes b_{i},\quad\text{where
\quad$\mathrm{gr}(a_{i})=\mathrm{gr}(b_{i})$.}$

Let us define the map
$c:{\mathfrak{g}}\rightarrow{\mathfrak{g}}\otimes{\mathfrak{g}}$ by setting

(24) $c_{\mathfrak{g}}(x):=\sum_{1\leq i\leq
n}\left([x,a_{i}]_{\mathfrak{g}}\otimes
b_{i}+a_{i}\otimes[x,b_{i}]_{\mathfrak{g}}\right)\quad\text{for all
$x\in{\mathfrak{g}}$}.$

######  Lemma ($c_{\mathfrak{g}}$ is a 1-cocycle).

For all $x\in{\mathfrak{g}}_{\bar{1}}$, we have

$x\cdot c_{\mathfrak{g}}(x)=c_{\mathfrak{g}}(s_{\mathfrak{g}}(x)).$

###### Proof.

For all $x\in{\mathfrak{g}}_{\bar{1}}$, we have

$\displaystyle x\cdot c_{\mathfrak{g}}(x)=x\cdot\left(\sum_{1\leq i\leq
n}\left([x,a_{i}]\otimes b_{i}+a_{i}\otimes[x,b_{i}]\right)\right)$
$\displaystyle=\sum_{1\leq i\leq n}\left([x,[x,a_{i}]]\otimes
b_{i}+[x,a_{i}]\otimes[x,b_{i}]\right)+\sum_{1\leq i\leq
n}\left([x,a_{i}]\otimes[x,b_{i}]+a_{i}\otimes[x,[x,b_{i}]]\right)$
$\displaystyle=\sum_{1\leq i\leq n}\left([s_{\mathfrak{g}}(x),a_{i}]\otimes
b_{i}+a_{i}\otimes[s_{\mathfrak{g}}(x),b_{i}]\right)=c_{\mathfrak{g}}(s_{\mathfrak{g}}(x)).\qed$

This Lemma shows that the map $c_{\mathfrak{g}}$ defined by Eq. (24) is not
only a 1-cocycle on the Lie algebra $F({\mathfrak{g}})$ but also a 1-cocyle on
the Lie superalgebra ${\mathfrak{g}}$ as well. Therefore, we can write
$c_{\mathfrak{g}}={\mathfrak{d}}r$, where $\mathfrak{d}$ is the coboundary
operator, see Eq. (17).

######  Proposition (Existence of a squaring on ${\mathfrak{g}}^{*}$).

Let ${\mathfrak{g}}$ be a Lie superalgebra with an $r$-matrix. There exists a
squaring on ${\mathfrak{g}}^{*}$ given as follows:

(25)
$s_{{\mathfrak{g}}^{*}}:{\mathfrak{g}}^{*}_{\bar{1}}\rightarrow{\mathfrak{g}}^{*}_{\bar{0}}\quad\quad
f\mapsto s_{{\mathfrak{g}}^{*}}(f):=\sum_{1\leq i\leq
n}f([-,a_{i}]_{\mathfrak{g}})f(b_{i}).$

###### Proof.

First, for all $\lambda\in{\mathbb{K}}$ and $f\in{\mathfrak{g}}^{*}_{\bar{1}}$
we have

$s_{{\mathfrak{g}}^{*}}(\lambda f)=\sum_{1\leq i\leq n}\lambda
f([-,a_{i}]_{\mathfrak{g}})\lambda f(b_{i})=\lambda^{2}\sum_{1\leq i\leq
n}f([-,a_{i}]_{\mathfrak{g}})f(b_{i})=\lambda^{2}s_{{\mathfrak{g}}^{*}}(f).$

Second, the map (for all $f,g\in{\mathfrak{g}}^{*}_{\bar{1}}$)

$\begin{array}[]{lcl}(f,g)&\mapsto&s_{{\mathfrak{g}}^{*}}(f+g)+s_{{\mathfrak{g}}^{*}}(f)+s_{{\mathfrak{g}}^{*}}(g)\\\\[2.84526pt]
&&\displaystyle=\sum_{1\leq i\leq
n}\left(f([-,a_{i}]_{\mathfrak{g}})g(b_{i})+g([-,a_{i}]_{\mathfrak{g}})f(b_{i})\right).\end{array}$

is obviously bilinear. ∎

The existence of the squaring on ${\mathfrak{g}}^{*}$ does not endow
${\mathfrak{g}}^{*}$ with a Lie superalgebra structure in characteristic $2$.
To do that we will have to show that the Jacobi identity involving the
squaring is satisfied.

######  Theorem (Condition on the squaring to satisfy the Jacobi identity).

Let ${\mathfrak{g}}$ be a non-trivial Lie superalgebra such that
$F({\mathfrak{g}})$ admits a quasi-triangular Lie bialgebra structure defined
by a $\mathbb{Z}/2$-graded tensor $r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$.
There exists a Lie superalgebra structure on the dual space
${\mathfrak{g}}^{*}$, where the squaring is given by Eq. (25), if and only if
the following condition is satisfied (for all $x\in{\mathfrak{g}}$):

$\displaystyle\sum_{i,j}f(b_{j})\left(f([[x,a_{i}]_{\mathfrak{g}},a_{j}]_{\mathfrak{g}})h(b_{i})+f([a_{i},a_{j}]_{\mathfrak{g}})h([x,b_{i}]_{\mathfrak{g}})\right)$
(26)
$\displaystyle=\sum_{i,j}f([x,a_{i}]_{\mathfrak{g}})(f([b_{i},a_{j}]_{\mathfrak{g}}h(b_{j})+f(a_{j})h([b_{i},b_{j}]_{\mathfrak{g}})+f(a_{i})f([[x,b_{i}]_{\mathfrak{g}},a_{j}]_{\mathfrak{g}})h(b_{j})$
$\displaystyle+\sum_{i,j}f(a_{i})f(a_{j})h([[x,b_{i}]_{\mathfrak{g}},b_{j}]_{\mathfrak{g}}).$

###### Proof.

We have already a Lie algebra structure on $F({\mathfrak{g}}^{*})$ with the
bracket given by

$\begin{array}[]{lcl}\langle[f,h]_{{\mathfrak{g}}^{*}},x\rangle&=&\langle
f\otimes h,c_{\mathfrak{g}}(x)\rangle\quad\text{ for all $x\in{\mathfrak{g}}$
and $f,h\in{\mathfrak{g}}^{*}$}.\end{array}$

The Jacobi identity involving the squaring reads

$\displaystyle\langle[s_{{\mathfrak{g}}^{*}}(f),h]_{{\mathfrak{g}}^{*}},x\rangle=\langle
s_{{\mathfrak{g}}^{*}}(f)\otimes h,c_{\mathfrak{g}}(x)\rangle=\langle
s_{{\mathfrak{g}}^{*}}(f)\otimes h,\sum_{1\leq i\leq
n}\left([x,a_{i}]_{\mathfrak{g}}\otimes
b_{i}+a_{i}\otimes[x,b_{i}]_{\mathfrak{g}}\right)\rangle$
$\displaystyle=\sum_{1\leq i\leq
n}\left(s_{{\mathfrak{g}}^{*}}(f)([x,a_{i}]_{\mathfrak{g}})h(b_{i})+s_{{\mathfrak{g}}^{*}}(f)(a_{i})h([x,b_{i}]_{\mathfrak{g}})\right)$
$\displaystyle=\sum_{1\leq i,j\leq
n}\left(f([[x,a_{i}]_{\mathfrak{g}},a_{j}]_{\mathfrak{g}})f(b_{j})h(b_{i})+f([a_{i},a_{j}]_{\mathfrak{g}})h([x,b_{i}]_{\mathfrak{g}})\right).$

On the other hand,

$\displaystyle\langle[f,[f,h]_{{\mathfrak{g}}^{*}}]_{{\mathfrak{g}}^{*}},x\rangle=\langle
f\otimes[f,h]_{{\mathfrak{g}}^{*}},c_{\mathfrak{g}}(x)\rangle=\langle
f\otimes[f,h]_{{\mathfrak{g}}^{*}},\sum_{1\leq i\leq
n}\left([x,a_{i}]_{\mathfrak{g}}\otimes
b_{i}+a_{i}\otimes[x,b_{i}]_{\mathfrak{g}}\right)\rangle$
$\displaystyle=\sum_{1\leq i\leq
n}\left(f([x,a_{i}]_{\mathfrak{g}})[f,h]_{{\mathfrak{g}}^{*}}(b_{i})+f(a_{i})[f,h]_{{\mathfrak{g}}^{*}}([x,b_{i}]_{\mathfrak{g}})\right)$
$\displaystyle=\sum_{1\leq i\leq n}\left(f([x,a_{i}]_{\mathfrak{g}})\langle
f\otimes h,c_{\mathfrak{g}}(b_{i})\rangle+f(a_{i})\langle f\otimes
h,c_{\mathfrak{g}}([x,b_{i}]_{\mathfrak{g}})\rangle\right)$
$\displaystyle=\sum_{1\leq i,j\leq
n}f([x,a_{i}]_{\mathfrak{g}})(f([b_{i},a_{j}]_{\mathfrak{g}})h(b_{j})+f(a_{j})h([b_{i},b_{j}]_{\mathfrak{g}}))$
$\displaystyle+\sum_{1\leq i,j\leq
n}f(a_{i})\left(f([[x,b_{i}]_{\mathfrak{g}},a_{j}]_{\mathfrak{g}})h(b_{j})+f(a_{j})h([[x,b_{i}]_{\mathfrak{g}},b_{j}]_{\mathfrak{g}})\right).\qed$

Let us now study the property of the map $c_{{\mathfrak{g}}^{*}}$ as given in
Eq. (21). We will show that it satsifies the 1-cocycle condition involving the
squaring. First, we need the following lemma.

######  Lemma (Technical).

For all $x,y\in{\mathfrak{g}}$ and for all $f\in{\mathfrak{g}}$, we have

$f\cdot c_{{\mathfrak{g}}^{*}}(f)(x\otimes y)=(f\otimes y\cdot
f)(c_{\mathfrak{g}}(x))+(f\otimes x\cdot f)(c_{\mathfrak{g}}(y)).$

###### Proof.

Indeed,

$\displaystyle f\cdot c_{{\mathfrak{g}}^{*}}(f)(x\otimes
y)=c_{{\mathfrak{g}}^{*}}(f)(f\cdot x\otimes
y)+c_{{\mathfrak{g}}^{*}}(f)(x\otimes f\cdot y)=f([f\cdot
x,y]_{\mathfrak{g}})+f([x,f\cdot y]_{\mathfrak{g}})$
$\displaystyle=f([x\circ\mathop{\mathrm{ad}}\nolimits_{f},y]_{\mathfrak{g}})+f([x,y\circ\mathop{\mathrm{ad}}\nolimits_{f}]_{\mathfrak{g}})=(y\cdot
f)(x\circ\mathop{\mathrm{ad}}\nolimits_{f})+(x\cdot
f)(y\circ\mathop{\mathrm{ad}}\nolimits_{f})$ $\displaystyle=[f,y\cdot
f]_{{\mathfrak{g}}^{*}}(x)+[f,x\cdot f]_{{\mathfrak{g}}^{*}}(y)=(f\otimes
y\cdot f)(c_{\mathfrak{g}}(x))+(f\otimes x\cdot f)(c_{\mathfrak{g}}(y)).\qed$

######  Proposition ($c_{{\mathfrak{g}}^{*}}$ is a 1-cocycle).

We have $c_{{\mathfrak{g}}^{*}}\in
Z^{1}({\mathfrak{g}}^{*},{\mathfrak{g}}^{*}\otimes{\mathfrak{g}}^{*})$.

###### Proof.

Let us check that condition Eq. (19) is satisfied. Using Lemma 3.5.7. Lemma we
get (for all $x,y\in{\mathfrak{g}}$ and for all
$f\in{\mathfrak{g}}^{*}_{\bar{1}}$):

$\begin{array}[]{lcl}f\cdot c_{{\mathfrak{g}}^{*}}(f)(x\otimes y)&=&(f\otimes
y\cdot f)(\sum_{1\leq i\leq n}[x,a_{i}]_{\mathfrak{g}}\otimes
b_{i}+a_{i}\otimes[x,b_{i}]_{\mathfrak{g}})+(x\leftrightarrow
y)\\\\[5.69054pt] &=&\sum_{1\leq i\leq
n}f([x,a_{i}]_{\mathfrak{g}})f([y,b_{i}])+f(a_{i})f([y,[x,b_{i}]_{\mathfrak{g}}]_{\mathfrak{g}})+(x\leftrightarrow
y)\\\\[5.69054pt] &=&\sum_{1\leq i\leq
n}f(a_{i})f([[x,y]_{\mathfrak{g}},b_{i}]_{\mathfrak{g}})=c_{{\mathfrak{g}}^{*}}(s_{{\mathfrak{g}}^{*}}(f))(x\otimes
y).\qed\end{array}$

Consider $r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$ given by $r=\sum_{1\leq
i\leq n}a_{i}\otimes b_{i}$, and define a linear map
$R:{\mathfrak{g}}^{*}\rightarrow{\mathfrak{g}}$ by setting

(27) $R(f):=(f\otimes 1)(r)=\sum_{1\leq i\leq n}f(a_{i})b_{i}\quad\text{ for
all $f\in{\mathfrak{g}}^{*}$.}$

######  Proposition (Technical).

Let ${\mathfrak{g}}$ be a Lie superalgebra in characteristic $2$, and let $R$
be the map (27) defined by means of $r\in{\mathfrak{g}}\otimes{\mathfrak{g}}$.
Then, $r$ is an even and symmetric classical $r$-matrix if and only if $R$ is
even and satisfies the following two conditions:

(i) $\langle f,R(h)\rangle=\langle h,R(f)\rangle\quad\text{for all
$f,h\in{\mathfrak{g}}^{*}$};$

(ii) $\langle f,[R(h),R(l)]_{\mathfrak{g}}\rangle+\langle
l,[R(f),R(h)]_{\mathfrak{g}}\rangle+\langle
h,[R(l),R(f)]_{\mathfrak{g}}\rangle=0\quad\text{for all
$f,h,l\in{\mathfrak{g}}^{*}$.}$

###### Proof.

Obviously, $r$ is even if and only if $R$ is even. Also $r$ is symmetric if
and only if Part (i) is satisfied. Now, a direct computation shows that

$\begin{array}[]{l}\langle f,[R(h),R(l)]_{\mathfrak{g}}\rangle+\langle
l,[R(f),R(h)]_{\mathfrak{g}}\rangle+\langle
h,[R(l),R(f)]_{\mathfrak{g}}\rangle\\\\[2.84526pt] =\langle f\otimes h\otimes
l,[r_{12},r_{13}]+[r_{12},r_{23}]+[r_{13},r_{23}]\rangle.\qed\end{array}$

######  Theorem (A bracket and a squaring on ${\mathfrak{g}}^{*}$).

Let ${\mathfrak{g}}$ be a Lie superalgeba in characteristic $2$. Let
$F({\mathfrak{g}})$ admit a triangular Lie bi-algebra structure such that the
co-multiplication on $F({\mathfrak{g}})$ is the $1$-coboundary of $r$ (where
$r$ is an even symmetric $r$-matrix). Then, the multiplication on
${\mathfrak{g}}^{*}$ defined by Eq. (21) is explicitly determined by the
formula

$[f,h]_{{\mathfrak{g}}^{*}}=h\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}+f\circ\mathop{\mathrm{ad}}\nolimits_{R(h)}\quad\text{for
all $f,h\in{\mathfrak{g}}^{*}$.}$

and the squaring given by Eq. (25) is explicitly given by the formula

(28) $s_{{\mathfrak{g}}^{*}}(f)=f\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}.$

###### Proof.

Let us first deal with the bracket. The LHS reads

$\displaystyle\langle[f,h]_{{\mathfrak{g}}^{*}},x\rangle=\langle f\otimes
h,c_{\mathfrak{g}}(x)\rangle=\langle f\otimes h,\sum_{1\leq i\leq
n}([x,a_{i}]\otimes b_{i}+a_{i}\otimes[x,b_{i}])\rangle$
$\displaystyle=\sum_{1\leq i\leq n}f([x,a_{i}])h(b_{i})+f(a_{i})h([x,b_{i}]).$

The RHS reads

$\displaystyle\langle
h\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}+f\circ\mathop{\mathrm{ad}}\nolimits_{R(h)}\,,x\rangle=h\circ[R(f),x]_{\mathfrak{g}}+f\circ[R(h),x]_{\mathfrak{g}}=\sum_{1\leq
i\leq n}f([x,a_{i}])h(b_{i})+f(a_{i})h([x,b_{i}]).$

When it comes to the squaring, and using the fact that $r$ is symmetric, we
get (for all $x\in{\mathfrak{g}}$):

$\displaystyle\langle
f\circ\mathop{\mathrm{ad}}\nolimits_{R(f)},x\rangle=f([R(f),x]_{\mathfrak{g}})=f([\sum_{1\leq
i\leq n}f(a_{i})b_{i},x]_{\mathfrak{g}})$ $\displaystyle=\sum_{1\leq i\leq
n}f(a_{i})f([b_{i},x]_{\mathfrak{g}})=\sum_{1\leq i\leq
n}f(b_{i})f([a_{i},x]_{\mathfrak{g}})=\langle
s_{{\mathfrak{g}}^{*}}(f),x\rangle.\qed$

######  Proposition (A Lie superalgebra structure on ${\mathfrak{g}}^{*}$).

Let ${\mathfrak{g}}$ be a Lie superalgeba in characteristic $2$. Let
$F({\mathfrak{g}})$ admit a triangular Lie bi-algebra structure such that the
co-multiplication on $F({\mathfrak{g}})$ is the $1$-coboundary of $r$ (where
$r$ is an even symmetric $r$-matrix). Then, there exists a Lie superalgebra
structure on the dual space ${\mathfrak{g}}^{*}$, where the squaring is given
by (28), if and only if the following condition is satisfied (for all
$x\in{\mathfrak{g}}$, for all $h\in{\mathfrak{g}}^{*}$ and for all
$f\in{\mathfrak{g}}^{*}_{\bar{1}}$):

$\displaystyle
h([R(f\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}),x]_{{\mathfrak{g}}})+f([[R(h),R(f)]_{\mathfrak{g}},x]_{{\mathfrak{g}}})$
$\displaystyle=h([R(f),[R(f),x]_{\mathfrak{g}}]_{\mathfrak{g}})+f([R(h),[R(f),x]_{\mathfrak{g}}]_{\mathfrak{g}})+f\circ
h([R(f),x]_{\mathfrak{g}})+f\circ f([R(h),x]_{\mathfrak{g}}).$

###### Proof.

For the proof of the Jacobi identity on ${\mathfrak{g}}^{*}_{\bar{0}}$ and the
fact that ${\mathfrak{g}}_{\bar{1}}$ is a ${\mathfrak{g}}_{\bar{0}}$-module,
we refer to [BaB, Prop. 3.1]. Let us deal with the squaring

$\displaystyle\langle[s_{{\mathfrak{g}}^{*}}(f),h],x\rangle=\langle
h\circ\mathop{\mathrm{ad}}\nolimits_{R(s_{{\mathfrak{g}}^{*}}(f))}+s_{{\mathfrak{g}}^{*}}(f)\circ\mathop{\mathrm{ad}}\nolimits_{R(h)},x\rangle$
$\displaystyle=h([R(f\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}),x]_{\mathfrak{g}})+f([R(f),[R(h),x]_{\mathfrak{g}}]_{\mathfrak{g}}).$

On the other hand,

$\displaystyle\langle[f,[f,h]_{{\mathfrak{g}}*}]_{{\mathfrak{g}}*},x\rangle=\langle[f,h\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}+f\circ\mathop{\mathrm{ad}}\nolimits_{R(h)}]_{{\mathfrak{g}}*},x\rangle$
$\displaystyle=\langle
h\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}+f\circ\mathop{\mathrm{ad}}\nolimits_{R(h)})\circ\mathop{\mathrm{ad}}\nolimits_{R(f)},x\rangle+\langle
f\circ\mathop{\mathrm{ad}}\nolimits_{h\circ\mathop{\mathrm{ad}}\nolimits_{R(f)}+f\circ\mathop{\mathrm{ad}}\nolimits_{R(h)}},x\rangle$
$\displaystyle=h([R(f),[R(f),x]_{\mathfrak{g}}]_{\mathfrak{g}})+f([R(h),[R(f),x]_{\mathfrak{g}}]_{\mathfrak{g}})+f\circ
h([R(f),x]_{\mathfrak{g}})+f\circ f([R(h),x]_{\mathfrak{g}})\qed$

Consider the space

$\mathrm{Im}(R):=\\{R(f)\;|\;f\in{\mathfrak{g}}^{*}\\},$

and consider the following bilinear form

(29)
$\omega:\mathrm{Im}(R)\times\mathrm{Im}(R)\rightarrow\mathbb{K}\quad(R(f),R(g))\mapsto\langle
g,R(f)\rangle.$

The form $\omega$ is well-defined. Indeed, if $R(f)=R(f^{\prime})$ and
$R(h)=R(h^{\prime})$, then

$\langle f,R(h)\rangle=\langle R(f),h\rangle=\langle
R(f^{\prime}),h\rangle=\langle f^{\prime},R(h)\rangle=\langle
f^{\prime},R(h^{\prime})\rangle.$

######  Proposition (A closed non-degenerate antisymmetric form on
$\mathrm{Im}(R)$).

The space $\mathrm{Im}(R)$ is a $\mathbb{Z}/2$-graded Lie algebra (not super!)
in characteristic $2$ equipped with the form (29) that is closed, non-
degenerate and antisymmetric.

###### Proof.

We should prove first that $\mathrm{Im}(R)$ is closed under the bracket, and
hence $\mathrm{Im}(R)$ is a Lie sub superalgebra of $F({\mathfrak{g}})$.
Indeed, let $R(f),R(h)\in\mathrm{Ker}(R)$. For all $l\in\mathrm{Ker}(R)$,
using Part (ii) of Prop. (3.5.9. Proposition), we see that

$\langle l,[R(f),R(h)]_{\mathfrak{g}}\rangle=\langle
h,[R(l),R(f)]_{\mathfrak{g}}\rangle+\langle
f,[R(h),R(l)]_{\mathfrak{g}}\rangle=0.$

It follows that $[R(f),R(h)]_{\mathfrak{g}}\in(\mathrm{Ker}(R))^{\perp}$,
where here the orthogonality should be understood with respect to the bilinear
form $\langle-,-\rangle$. It follows that
$[R(f),R(h)]_{\mathfrak{g}}\in\mathrm{Im}(R)$.

It is easy to see that the map $\omega$ is bilinear. Let us check the
2-cocycle condition. Indeed, using Part (ii) of Proposition 3.5.9.
Proposition, we get

$\displaystyle\omega(R(l),[R(f),R(h)]_{\mathfrak{g}})+\omega(R(h),[R(l),R(f)]_{\mathfrak{g}})+\omega(R(f),[R(h),R(l)]_{\mathfrak{g}})$
$\displaystyle=\langle l,[R(f),R(h)]_{\mathfrak{g}})\rangle+\langle
h,[R(l),R(f)]_{\mathfrak{g}})\rangle+\langle
f,[R(h),R(l)]_{\mathfrak{g}})\rangle=0.\qed$

Suppose now that ${\mathfrak{g}}$ has a NIS ${\mathscr{B}}$. Consider the
natural isomorphism

$\Phi:{\mathfrak{g}}\rightarrow{\mathfrak{g}}^{*}\quad
x\mapsto\Phi(x):={\mathscr{B}}(x,\cdot).$

We define a new bracket and a new squaring on ${\mathfrak{g}}$ as follows:

$\begin{array}[]{lcll}\widetilde{[\cdot,\cdot]}&:&{\mathfrak{g}}\times{\mathfrak{g}}\rightarrow{\mathfrak{g}}&\widetilde{[x,y]}=\Phi^{-1}([\Phi(x),\Phi(y)]_{{\mathfrak{g}}^{*}})\text{
for all $x,y\in{\mathfrak{g}}$}\\\\[2.84526pt]
\widetilde{s_{\mathfrak{g}}(\cdot)}&:&{\mathfrak{g}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}&\widetilde{s_{\mathfrak{g}}(x)}=\Phi^{-1}(s_{{\mathfrak{g}}^{*}}(\Phi(x)))\text{
for all $x\in{\mathfrak{g}}_{\bar{1}}$}.\end{array}$

######  Theorem (Deforming the Lie structure).

Let $({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}}(\cdot))$ be
a Lie superalgebra in characteristic $2$ with a NIS. Suppose that
${\mathfrak{g}}$ has an invertible and symmetric $r$-matrix. If we denote
$U:=R\circ\Phi:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$, then

$\begin{array}[]{lcl}\widetilde{[x,y]}&=&[U(x),y]_{\mathfrak{g}}+[x,U(y)]_{\mathfrak{g}}\quad\text{for
all $x,y\in{\mathfrak{g}}$,}\\\\[5.69054pt]
\widetilde{s}_{\mathfrak{g}}(x)&=&[U(x),x]_{\mathfrak{g}}\quad\text{for all
$x\in{\mathfrak{g}}_{\bar{1}}$}.\end{array}$

Moreover,
$({\mathfrak{g}},\widetilde{[\cdot,\cdot]},\widetilde{s}_{\mathfrak{g}}(\cdot))$
is a Lie superalgebra.

###### Proof.

We will only be dealing with the squaring. Indeed, for all
$x\in{\mathfrak{g}}_{\bar{1}}$.

$\displaystyle\widetilde{s_{\mathfrak{g}}(x)}=\Phi^{-1}(s_{{\mathfrak{g}}^{*}}(\Phi(x)))=\Phi^{-1}(\Phi(x)\circ\mathop{\mathrm{ad}}\nolimits_{R(\Phi(x))})=\Phi^{-1}({\mathscr{B}}(x,[R(\Phi(x))]_{\mathfrak{g}},\cdot])$
$\displaystyle=\Phi^{-1}({\mathscr{B}}([x,R(\Phi(x))]_{\mathfrak{g}},\cdot))=[x,R(\Phi(x))]_{\mathfrak{g}}=[x,U(x)]_{\mathfrak{g}}.$

The JI pertaining to the squaring reads: (for all
$x\in{\mathfrak{g}}_{\bar{1}}$ and for all $y\in{\mathfrak{g}}$)

$\displaystyle\widetilde{[\widetilde{s}_{\mathfrak{g}}(x),y]}=\Phi^{-1}([s_{\mathfrak{g}}(\Phi(x)),\Phi(y)]_{\mathfrak{g}})=\Phi^{-1}([\Phi(x),[\Phi(x),\Phi(y)]_{\mathfrak{g}}]_{\mathfrak{g}})=\Phi^{-1}([\Phi(x),\Phi(\widetilde{[x,y]})]_{\mathfrak{g}})$
$\displaystyle=\widetilde{[x,\widetilde{[x,y]}]}.\qed$

######  Proposition ($U$-invariance).

The map
$U:({\mathfrak{g}},\widetilde{[\cdot,\cdot]})\rightarrow({\mathfrak{g}},[\cdot,\cdot])$
is a morphism of Lie superalgebras; moreover,

${\mathscr{B}}(U(x),y)={\mathscr{B}}(x,U(y)).$

Conversely, if $U\in\mathop{\mathrm{End}}\nolimits({\mathfrak{g}})$ is even
and ${\mathscr{B}}(U(x),y)={\mathscr{B}}(x,U(y))$, then

$\begin{array}[]{lcl}\widetilde{[x,y]}_{\mathfrak{g}}&:=&[U(x),y]_{\mathfrak{g}}+[x,U(y)]_{\mathfrak{g}}\quad\text{for
all $x,y\in{\mathfrak{g}}$}\\\\[5.69054pt]
\widetilde{s}_{\mathfrak{g}}(x)&:=&[x,U(x)]_{\mathfrak{g}}\quad\text{for all
$x\in{\mathfrak{g}}_{\bar{1}}$}\end{array}$

defines a Lie superalgebra structure in characteristic $2$; moreover,

$R:=U\circ\Phi^{-1}:{\mathfrak{g}}^{*}\rightarrow{\mathfrak{g}}\text{ is
a~{}associated with an $r$-matrix.}$

###### Proof.

Similar to that of [BaB, Prop 3.9].∎

Recall that in characteristic $2$ the morphism $R$ is a morphism of Lie
superalgebras if in addition to the bracket we have

$R(s_{{\mathfrak{g}}^{*}}(f))=s_{\mathfrak{g}}(R(f))\quad\text{for all
$f\in{\mathfrak{g}}^{*}_{\bar{1}}$.}$

Therefore,

$R(s_{{\mathfrak{g}}^{*}}(f))=\sum_{i=1}^{n}s_{{\mathfrak{g}}^{*}}(f)(a_{i})b_{i}=s_{\mathfrak{g}}\left(\sum_{i=1}^{n}f(a_{i})b_{i}\right).$

If $R$ is invertible, then
$s_{{\mathfrak{g}}^{*}}(f)=R^{-1}(s_{\mathfrak{g}}(R(f)))$ defines a squaring
on ${\mathfrak{g}}^{*}$.

##  4\. The Matched pairs for Lie superalgebras in characteristic $2$

The goal of this section is to look at Manin triples in a more general setting
by introducing the concept of matched pairs. As far as we know, this concept
was introduced by Majid in [Maj].

Let $({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ and
$({\mathfrak{k}},[\cdot,\cdot]_{\mathfrak{k}},s_{\mathfrak{k}})$ be two Lie
superalgebras in characteristic $2$ – in the case where
${\mathfrak{k}}={\mathfrak{g}}^{*}$, the triple
$({\mathfrak{h}}:={\mathfrak{g}}\oplus{\mathfrak{g}}^{*},{\mathfrak{g}},{\mathfrak{g}}^{*})$
is a Manin triple. Let us suppose they are both endowed with the
representations

$\pi:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{k}}_{\bar{0}}),\quad\lambda:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{k}}_{\bar{1}}),\quad\rho:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{g}}_{\bar{0}}),\quad\mu:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{g}}_{\bar{1}}),$

together with the following maps

$\tilde{\lambda}:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{Hom}({\mathfrak{k}}_{\bar{1}},{\mathfrak{g}}_{\bar{1}}),\quad\tilde{\mu}:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{Hom}({\mathfrak{g}}_{\bar{1}},{\mathfrak{k}}_{\bar{1}}).$

Now suppose that these maps satisfy the following conditions: (for all
$x_{0},y_{0}\in{\mathfrak{g}}_{\bar{0}}$, $x_{1}\in{\mathfrak{g}}_{\bar{1}}$,
$a_{0},b_{0}\in{\mathfrak{k}}_{\bar{0}}$, $a_{1}\in{\mathfrak{k}}_{\bar{1}}$):

(30)
$\begin{array}[]{l}\rho(a_{0})([x_{0},y_{0}]_{\mathfrak{g}})+[\rho(a_{0})(x_{0}),y_{0}]_{\mathfrak{g}}+[x_{0},\rho(a_{0})(y_{0})]_{\mathfrak{g}}\\\\[2.84526pt]
=\rho(\pi(y_{0})(a_{0}))(x_{0})+\rho(\pi(x_{0})(a_{0}))(y_{0}),\\\\[5.69054pt]
\pi(x_{0})([a_{0},b_{0}]_{\mathfrak{k}})+[\pi(x_{0})(a_{0}),b_{0}]_{\mathfrak{k}}+[a_{0},\pi(x_{0})(b_{0})]_{\mathfrak{k}}\\\\[2.84526pt]
=\pi(\rho(b_{0})(x_{0}))(a_{0})+\pi(\rho(a_{0})(x_{0}))(b_{0}),\end{array}$

and

(31)
$\begin{array}[]{lcl}\mu(b_{0})([x_{0},x_{1}]_{\mathfrak{g}})+[x_{0},\mu(b_{0})(x_{1})]_{\mathfrak{g}}&=&[\rho(b_{0})(x_{0}),x_{1}]_{\mathfrak{g}}+\mu(\pi(x_{0})(b_{0}))(x_{1})+\tilde{\lambda}(x_{0})(\tilde{\mu}(b_{0})(x_{1})),\\\\[8.53581pt]
\lambda(x_{0})([b_{0},a_{1}]_{\mathfrak{k}})+[b_{0},\lambda(x_{0})(a_{1})]_{\mathfrak{k}}&=&[\pi(x_{0})(b_{0}),a_{1}]_{\mathfrak{k}}+\lambda(\rho(b_{0})(x_{0}))(a_{1})+\tilde{\mu}(b_{0})(\tilde{\lambda}(x_{0})(a_{1})),\\\\[8.53581pt]
\tilde{\lambda}(\rho(b_{0})(x_{0}))(a_{1})+\tilde{\lambda}(x_{0})([b_{0},a_{1}]_{\mathfrak{k}})&=&\mu(b_{0})(\tilde{\lambda}(x_{0})(a_{1})),\\\\[8.53581pt]
\tilde{\mu}(\pi(x_{0})(b_{0}))(x_{1})+\tilde{\mu}(b_{0})([x_{0},x_{1}]_{\mathfrak{g}})&=&\lambda(x_{0})(\tilde{\mu}(b_{0})(x_{1})),\\\\[8.53581pt]
\tilde{\lambda}([x_{0},y_{0}]_{\mathfrak{g}})(a_{1})+[x_{0},\tilde{\lambda}(y_{0})(a_{1})]_{\mathfrak{g}}&=&[y_{0},\tilde{\lambda}(x_{0})(a_{1})]_{\mathfrak{g}}+\tilde{\lambda}(x_{0})\lambda(y_{0})(a_{1})+\tilde{\lambda}(y_{0})\lambda(x_{0})(a_{1}),\\\\[8.53581pt]
\tilde{\mu}([a_{0},b_{0}]_{\mathfrak{k}})(x_{1})+[a_{0},\tilde{\mu}(b_{0})(x_{1})]_{\mathfrak{k}}&=&[b_{0},\tilde{\mu}(a_{0})(x_{1})]_{\mathfrak{k}}+\tilde{\mu}(a_{0})\mu(b_{0})(x_{1})+\tilde{\mu}(b_{0})\mu(a_{0})(x_{1}).\end{array}$

Suppose further that there are two bilinear maps
$r_{\mathfrak{g}}:{\mathfrak{g}}_{\bar{1}}\odot{\mathfrak{k}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}$
and
$r_{\mathfrak{k}}:{\mathfrak{g}}_{\bar{1}}\odot{\mathfrak{k}}_{\bar{1}}\rightarrow{\mathfrak{k}}_{\bar{0}}$,
satifying the following conditions (for all $b\in{\mathfrak{k}}_{\bar{0}}$ and
for all $y\in{\mathfrak{g}}_{\bar{0}}$):

(32)
$\begin{array}[]{lcl}[r_{\mathfrak{g}}(x_{1},a_{1}),y]_{\mathfrak{g}}+\rho(r_{\mathfrak{k}}(x_{1},a_{1}))(y)&=&r_{\mathfrak{g}}(x_{1},\lambda(y)(a_{1}))+[x_{1},\tilde{\lambda}(y)(a_{1})]_{\mathfrak{g}}+r_{\mathfrak{g}}(a_{1},[x_{1},y]_{\mathfrak{g}}),\\\\[8.53581pt]
\rho(b)(r_{\mathfrak{g}}(x_{1},a_{1}))+r_{\mathfrak{g}}(a_{1},\mu(b)(x_{1}))&=&r_{\mathfrak{g}}(x_{1},[a_{1},b]_{\mathfrak{k}}),\\\\[8.53581pt]
\rho(b)(s_{\mathfrak{g}}(x_{1}))+[x_{1},\mu(b)(x_{1})]_{\mathfrak{g}}&=&r_{\mathfrak{g}}(x_{1},\tilde{\mu}(b)(x_{1})),\\\\[8.53581pt]
\rho(s_{\mathfrak{k}}(a_{1}))(y)&=&r_{\mathfrak{g}}(a_{1},\tilde{\lambda}(y)(a_{1})),\end{array}$

and also

(33)
$\begin{array}[]{lcl}[r_{\mathfrak{k}}(x_{1},a_{1}),b]_{\mathfrak{k}}+\pi(r_{\mathfrak{g}}(x_{1},a_{1}))(b)&=&r_{\mathfrak{k}}(a_{1},\mu(b)(x_{1}))+[a_{1},\tilde{\mu}(b)(x_{1})]_{\mathfrak{k}}+r_{\mathfrak{k}}(x_{1},[a_{1},b]_{\mathfrak{k}}),\\\\[8.53581pt]
\pi(y)(r_{\mathfrak{k}}(x_{1},a_{1}))+r_{\mathfrak{k}}(x_{1},\lambda(y)(a_{1}))&=&r_{\mathfrak{k}}(a_{1},[x_{1},y]_{\mathfrak{g}}),\\\\[8.53581pt]
\pi(y)(s_{\mathfrak{k}}(a_{1}))+[a_{1},\lambda(y)(a_{1})]_{\mathfrak{k}}&=&r_{\mathfrak{k}}(a_{1},\tilde{\lambda}(y)(a_{1})),\\\\[8.53581pt]
\pi(s_{\mathfrak{g}}(x_{1}))(b)&=&r_{\mathfrak{k}}(x_{1},\tilde{\mu}(b)(x_{1})).\end{array}$

Moreover, (for all $b\in{\mathfrak{k}}_{\bar{1}}$ and for all
$y\in{\mathfrak{g}}_{\bar{1}}$)

(34)
$\begin{array}[]{lcl}\tilde{\mu}(r_{\mathfrak{k}}(x_{1},a_{1}))(y)&=&\lambda([x_{1},y]_{\mathfrak{g}})(a_{1})+\tilde{\mu}(r_{\mathfrak{k}}(a_{1},y))(x_{1}),\\\\[8.53581pt]
\tilde{\mu}(s_{\mathfrak{k}}(a_{1}))(y)&=&[r_{\mathfrak{k}}(a_{1},y),a_{1}]_{\mathfrak{k}}+\lambda(r_{\mathfrak{g}}(a_{1},y))(a_{1}),\\\\[8.53581pt]
\lambda(s_{\mathfrak{g}}(x_{1}))(b)&=&\tilde{\mu}(r_{\mathfrak{k}}(x_{1},b))(x_{1}),\\\\[8.53581pt]
\lambda(r_{\mathfrak{g}}(x_{1},a_{1}))(b)&=&[r_{\mathfrak{k}}(x_{1},a_{1}),b]_{\mathfrak{k}}+\lambda(r_{\mathfrak{g}}(x_{1},b))(a_{1})+\tilde{\mu}([a,b]_{\mathfrak{k}})(x_{1})+[r_{\mathfrak{k}}(x_{1},b),a_{1}]_{\mathfrak{k}},\end{array}$

also

(35)
$\begin{array}[]{lcl}\tilde{\lambda}(r_{\mathfrak{g}}(x_{1},a_{1}))(b)&=&\mu([a,b]_{\mathfrak{k}})(x_{1})+\tilde{\lambda}(r_{\mathfrak{g}}(x_{1},b))(a),\\\\[8.53581pt]
\tilde{\lambda}(s_{\mathfrak{g}}(x_{1}))(b)&=&[r_{\mathfrak{g}}(x_{1},b),x_{1}]_{\mathfrak{g}}+\mu(r_{\mathfrak{k}}(x_{1},b))(x),\\\\[8.53581pt]
\mu(s_{\mathfrak{k}}(a_{1}))(y)&=&\tilde{\lambda}(r_{\mathfrak{g}}(a_{1},y))(a_{1}),\\\\[8.53581pt]
\mu(r_{\mathfrak{k}}(x_{1},a_{1}))(y)&=&[r_{\mathfrak{g}}(x_{1},a_{1}),y]_{\mathfrak{g}}+\mu(r_{\mathfrak{k}}(a_{1},y))(x_{1})+\tilde{\lambda}(a_{1})([x_{1},y]_{\mathfrak{g}})+[r_{\mathfrak{g}}(a_{1},y),x_{1}]_{\mathfrak{g}}.\end{array}$

######  Theorem (Matched pairs).

Let $({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ and
$({\mathfrak{k}},[\cdot,\cdot]_{\mathfrak{k}},s_{\mathfrak{k}})$ be two Lie
superalgebras in characteristic $2$, endowed with

  * •

Representation maps

$\pi:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{k}}_{\bar{0}}),\quad\lambda:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{k}}_{\bar{1}}),\quad\rho:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{g}}_{\bar{0}}),\quad\mu:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{End}({\mathfrak{g}}_{\bar{1}});$

  * •

Linear maps
$\tilde{\lambda}:{\mathfrak{g}}_{\bar{0}}\rightarrow\mathrm{Hom}({\mathfrak{k}}_{\bar{1}},{\mathfrak{g}}_{\bar{1}})$
and
$\tilde{\mu}:{\mathfrak{k}}_{\bar{0}}\rightarrow\mathrm{Hom}({\mathfrak{g}}_{\bar{1}},{\mathfrak{k}}_{\bar{1}})$
satisfying Eqns (30), (31);

  * •

Bilinear maps
$r_{\mathfrak{g}}:{\mathfrak{g}}_{\bar{1}}\odot{\mathfrak{k}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}$
and
$r_{\mathfrak{k}}:{\mathfrak{g}}_{\bar{1}}\odot{\mathfrak{k}}_{\bar{1}}\rightarrow{\mathfrak{k}}_{\bar{0}}$,
satifying Eqns (32), (33), (34), (35).

Then, there exists a Lie superalgebra structure on
${\mathfrak{h}}:={\mathfrak{g}}\oplus{\mathfrak{k}}$ given as follows.

  * •

The Lie algebra structure on ${\mathfrak{h}}_{\bar{0}}$: The Lie bracket is
given by:

$[x_{0}+a_{0},y_{0}+b_{0}]_{\mathfrak{h}}:=[x_{0},y_{0}]_{\mathfrak{g}}+\rho(b_{0})(x_{0})+\rho(a_{0})(y_{0})+[a_{0},b_{0}]_{\mathfrak{k}}+\pi(x_{0})(b_{0})+\pi(y_{0})(a_{0}),$

for all $x_{0},y_{0}\in{\mathfrak{g}}_{\bar{0}}$ and
$a_{0},b_{0}\in{\mathfrak{k}}_{\bar{0}}$.

  * •

The ${\mathfrak{h}}_{\bar{0}}$-module structure on ${\mathfrak{h}}_{\bar{1}}$:
The action is given by:

$[x_{0}+a_{0},x_{1}+a_{1}]:=[x_{0},x_{1}]_{\mathfrak{g}}+\mu(a_{0})(x_{1})+\tilde{\lambda}(x_{0})(a_{1})+[a_{0},a_{1}]_{\mathfrak{k}}+\lambda(x_{0})(a_{1})+\tilde{\mu}(a_{0})(x_{1}),$

for all $x_{0}\in{\mathfrak{g}}_{\bar{0}},x_{1}\in{\mathfrak{g}}_{\bar{1}}$
and $a_{0}\in{\mathfrak{k}}_{\bar{0}},b_{1}\in{\mathfrak{k}}_{\bar{1}}$.

  * •

The squaring on ${\mathfrak{h}}_{\bar{1}}$ (where we have put
$r:=r_{\mathfrak{g}}+r_{\mathfrak{k}}$):

$S_{\mathfrak{h}}(x_{1}+a_{1}):=s_{\mathfrak{g}}(x_{1})+s_{\mathfrak{k}}(a_{1})+r(x_{1},a_{1}),$

for all $x_{1}\in{\mathfrak{g}}_{\bar{1}}$ and
$a_{1}\in{\mathfrak{k}}_{\bar{1}}$.

###### Proof.

The Jacobi identity on ${\mathfrak{h}}_{\bar{0}}$ is equivalent to the fact
that $\pi$ and $\rho$ are representations satisfying conditions (30). We omit
the details since the proof is similar to the case of a Lie algebra, see [D].

The ${\mathfrak{h}}_{\bar{0}}$-module structure ${\mathfrak{h}}_{\bar{1}}$ is
equivalent to the fact that $\lambda$ and $\mu$ are representations satisfying
conditions (31). Here also we omit the details of the computations.

Let us study the squaring on ${\mathfrak{h}}_{\bar{1}}$. We have (for all
$\lambda\in\mathbb{K},$ for all $x_{1}\in{\mathfrak{g}}_{\bar{1}}$ and
$a_{1}\in{\mathfrak{k}}_{\bar{1}}$):

$\begin{array}[]{lcl}S_{\mathfrak{h}}(\lambda(x_{1}+a_{1}))&=&r(\lambda
x_{1},\lambda a_{1})+s_{\mathfrak{g}}(\lambda x_{1})+s_{\mathfrak{k}}(\lambda
a_{1})\\\\[8.53581pt]
&=&\lambda^{2}r(x_{1},a_{1})+\lambda^{2}s_{\mathfrak{g}}(x_{1})+\lambda^{2}s_{\mathfrak{k}}(a_{1})=\lambda^{2}S_{\mathfrak{h}}(x_{1}+a_{1}),\end{array}$

Besides, the map

$(x_{1}+a_{1},y_{1}+b_{1})\mapsto
B(x_{1}+a_{1},y_{1}+b_{1}):=S_{\mathfrak{h}}(x_{1}+y_{1}+a_{1}+b_{1})-S_{\mathfrak{h}}(x_{1}+a_{1})-S_{\mathfrak{h}}(y_{1}+b_{1})$

is a bilinear map on ${\mathfrak{h}}_{\bar{1}}$ with values in
${\mathfrak{h}}_{\bar{0}}$. Indeed,

$\begin{array}[]{lcl}B(x_{1}+a_{1},y_{1}+b_{1})&=&r(x_{1}+y_{1},a_{1}+b_{1})+s_{\mathfrak{g}}(x_{1}+y_{1})+s_{\mathfrak{k}}(a_{1}+b_{1})+r(x_{1},a_{1})+s_{\mathfrak{g}}(x_{1})\\\\[8.53581pt]
&&+s_{\mathfrak{k}}(a_{1})+r(y_{1},b_{1})+s_{\mathfrak{g}}(y_{1})-s_{\mathfrak{k}}(b_{1})\\\\[8.53581pt]
&=&r(x_{1},b_{1})+r(y_{1},a_{1})+[x_{1},y_{1}]_{\mathfrak{g}}+[a_{1},b_{1}]_{\mathfrak{k}}.\end{array}$

Let us study the Jacobi identity involving to the squaring; for all
$h=x_{1}+a_{1}\in{\mathfrak{h}}_{\bar{1}}$ and $k\in{\mathfrak{h}}$ we have

  * •

The case where $k=y_{0}+b_{0}\in{\mathfrak{h}}_{\bar{0}}$. The LHS reads

$\begin{array}[]{lcl}[S_{\mathfrak{h}}(h),k]_{\mathfrak{h}}&=&[r(x_{1},a_{1})+s_{\mathfrak{g}}(x_{1})+s_{\mathfrak{k}}(a_{1}),y_{0}+b_{0}]_{\mathfrak{h}}+[r_{\mathfrak{g}}(x_{1},a_{1}),y_{0}]_{\mathfrak{g}}\\\\[8.53581pt]
&&+\rho(r_{\mathfrak{k}}(x_{1},a_{1}))(y_{0})+\pi(r_{\mathfrak{g}}(x_{1},a_{1}))(b_{0})+\pi(y)(r_{\mathfrak{k}}(x_{1},a_{1}))+\rho(b)(r_{\mathfrak{g}}(x_{1},a_{1}))\\\\[5.69054pt]
&&+[r_{k}(x_{1},a_{1}),b_{0}]_{\mathfrak{k}}+[x_{1},[x_{1},y_{0}]_{\mathfrak{g}}]_{\mathfrak{g}}+\pi(s_{\mathfrak{g}}(x_{1}))(b_{0})+\rho(b)(s_{\mathfrak{g}}(x_{1}))+\rho(s_{\mathfrak{k}}(a_{1}))(y_{0})\\\\[8.53581pt]
&&+\pi(y)(s_{\mathfrak{k}}(a_{1}))+[a_{1},[a_{1},b_{0}]_{\mathfrak{k}}]_{\mathfrak{k}}.\end{array}$

The RHS reads

$\begin{array}[]{lcl}[h,[h,k]_{\mathfrak{h}}]_{\mathfrak{h}}&=&[x_{1}+a_{1},[x_{1},y_{0}]_{\mathfrak{g}}+\mu(b_{0})(x_{1})+\tilde{\mu}(b_{0})(x_{1})+\lambda(y_{0})(a_{1})+\tilde{\lambda}(y_{0})(a_{1})+[a_{1},b_{0}]_{\mathfrak{k}}]_{\mathfrak{h}}\\\\[8.53581pt]
&=&[x_{1},[x_{1},y_{0}]_{\mathfrak{g}}]_{\mathfrak{g}}+[x_{1},\mu(b_{0})(x_{1})]_{\mathfrak{g}}+r(x_{1},\tilde{\mu}(b)(x_{1}))+r(x_{1},\lambda(y)(a_{1}))\\\\[8.53581pt]
&&+[x_{1},\tilde{\lambda}(y)(a_{1})]_{\mathfrak{g}}+r(a_{1},[x_{1},y]_{\mathfrak{g}})+r(a_{1},\mu(b)(x_{1}))+[a_{1},\tilde{\mu}(b)(x_{1})]_{\mathfrak{k}}\\\\[8.53581pt]
&&+[a_{1},\lambda(y)(a_{1})]_{\mathfrak{k}}+r(a_{1},\tilde{\lambda}(y)(a_{1}))+r(x_{1},[a_{1},b]_{\mathfrak{k}})+[a_{1},[a_{1},b_{0}]_{\mathfrak{k}}]_{\mathfrak{k}}.\end{array}$

Therfore, LHS=RHS is equivalent to conditions (32) and (33).

  * •

The case where $k=y_{1}+b_{1}\in{\mathfrak{h}}_{\bar{1}}$. The LHS reads

$\begin{array}[]{lcl}[S_{\mathfrak{h}}(h),k]_{\mathfrak{h}}&=&[r(x_{1},a_{1})+s_{\mathfrak{g}}(x_{1})+s_{\mathfrak{k}}(a_{1}),y_{1}+b_{1}]_{\mathfrak{h}}\\\\[8.53581pt]
&=&[r_{\mathfrak{g}}(x_{1},a_{1}),y_{1}]_{\mathfrak{g}}++\lambda(r_{\mathfrak{g}}(x_{1},a_{1}),b)+\tilde{\lambda}(r_{\mathfrak{g}}(x_{1},a_{1}),b)+\mu(r_{\mathfrak{k}}(x_{1},a_{1}))(y_{1})\\\\[8.53581pt]
&&+\tilde{\mu}(r_{\mathfrak{k}}(x_{1},a_{1}))(y_{1})+[s_{\mathfrak{g}}(x_{1}),y_{1}]_{\mathfrak{g}}+\mu(s_{\mathfrak{k}}(a_{1}))(y_{1})+\tilde{\mu}(s_{\mathfrak{k}}(a_{1}))(y_{1})\\\\[8.53581pt]
&&+[r_{\mathfrak{k}}(x_{1},a_{1}),b_{1}]_{\mathfrak{k}}+\lambda(s_{\mathfrak{g}}(x_{1}))(b_{1})+\tilde{\lambda}(s_{\mathfrak{g}}(x_{1}))(b_{1})+[s_{\mathfrak{k}}(a_{1}),b_{1}]_{\mathfrak{k}}.\end{array}$

The RHS reads

$\begin{array}[]{lcl}[h,[h,k]_{\mathfrak{h}}]_{\mathfrak{h}}&=&[x_{1}+a_{1},[x_{1},y_{1}]_{\mathfrak{g}}+r(y_{1},a_{1})+r(x_{1},b_{1})+[a_{1},b_{1}]_{\mathfrak{k}}]_{\mathfrak{h}}\\\\[8.53581pt]
&=&[x_{1},[x_{1},y_{1}]_{\mathfrak{g}}]_{\mathfrak{g}}+\lambda([x_{1},y_{1}]_{\mathfrak{g}})(a_{1})+\tilde{\lambda}([x_{1},y_{1}]_{\mathfrak{g}})(a_{1})+[a_{1},[a_{1},b]_{\mathfrak{k}}]_{\mathfrak{k}}\\\\[8.53581pt]
&&+\mu([a,b]_{\mathfrak{k}})(x_{1})+\tilde{\mu}([a,b]_{\mathfrak{k}})(x_{1})+\mu(r_{\mathfrak{k}}(x_{1},a_{1}))(x_{1})+\tilde{\mu}(r_{\mathfrak{k}}(x_{1},a_{1}))(x_{1})\\\\[8.53581pt]
&&+[r_{\mathfrak{k}}(x_{1},a_{1}),a_{1}]_{\mathfrak{k}}+[r_{\mathfrak{g}}(x_{1},b),x_{1}]_{\mathfrak{g}}+\lambda(r_{\mathfrak{g}}(x_{1},b))(a_{1})+\tilde{\lambda}(r_{\mathfrak{g}}(x_{1},b))(a_{1})\\\\[8.53581pt]
&&+\mu(r_{\mathfrak{k}}(a_{1},y))(x_{1})+\tilde{\mu}(r_{\mathfrak{k}}(a_{1},y))(x_{1})+[r_{\mathfrak{k}}(a_{1},y),a_{1}]_{\mathfrak{k}}+[x_{1},r_{\mathfrak{g}}(a_{1},y)]_{\mathfrak{g}}\\\\[8.53581pt]
&&+\lambda(r_{\mathfrak{g}}(a_{1},y))(a_{1})+\tilde{\lambda}(r_{\mathfrak{g}}(a_{1},y))(a_{1}).\end{array}$

Therfore, LHS=RHS is equivalent to conditions (34) and (35).∎

###  4.2. Proof of Theorem 3.1.1. Theorem

###### Proof.

The proof will be a consequence of Theorem 4.1. Theorem. Let us explicitly
describe the maps $\rho,\pi,\lambda,\tilde{\lambda},\mu,$ and $\tilde{\mu}$:

$\begin{array}[]{lcllcl}\pi(x)(f)&=&f\circ\mathop{\mathrm{ad}}\nolimits_{x}\;\text{($x\in{\mathfrak{g}}_{\bar{0}}$
and
$f\in{\mathfrak{g}}_{\bar{0}}^{*}$)},&\lambda(x)(f)&=&f\circ\mathop{\mathrm{ad}}\nolimits_{x}\;\text{($x\in{\mathfrak{g}}_{\bar{0}}$
and $f\in{\mathfrak{g}}_{\bar{1}}^{*}$),}\\\\[5.69054pt]
\rho(f)(x)&=&x\circ\mathop{\mathrm{ad}}\nolimits_{f},\;\text{($x\in{\mathfrak{g}}_{\bar{0}}$
and
$f\in{\mathfrak{g}}_{\bar{0}}^{*}$)},&\mu(f)(x)&=&x\circ\mathop{\mathrm{ad}}\nolimits_{f},\;\text{($x\in{\mathfrak{g}}_{\bar{1}}$
and $f\in{\mathfrak{g}}_{\bar{0}}^{*}$),}\\\\[5.69054pt]
\tilde{\lambda}(x)(f)&=&x\circ\mathop{\mathrm{ad}}\nolimits_{f},\text{($x\in{\mathfrak{g}}_{\bar{0}}$
and
$f\in{\mathfrak{g}}_{\bar{1}}^{*}$)},&\tilde{\mu}(f)(x)&=&f\circ\mathop{\mathrm{ad}}\nolimits_{x},\;\text{($x\in{\mathfrak{g}}_{\bar{0}}$
and $f\in{\mathfrak{g}}_{\bar{1}}$).}\end{array}$

On the other hand, the maps $r_{\mathfrak{g}}$ and $r_{{\mathfrak{g}}^{*}}$
are given by (where $f\in{\mathfrak{g}}_{\bar{1}}^{*}$ and
$x\in{\mathfrak{g}}_{\bar{1}}$)

$r_{\mathfrak{k}}(x,f)=f\circ\mathop{\mathrm{ad}}\nolimits_{x},\quad
r_{\mathfrak{g}}(x,f)=x\circ\mathop{\mathrm{ad}}\nolimits_{f}.$

It follows that the squaring on ${\mathfrak{h}}$ is given by

$s_{\mathfrak{h}}(x+f)=s_{\mathfrak{g}}(x)+s_{{\mathfrak{g}}^{*}}(f)+f\circ\mathop{\mathrm{ad}}\nolimits_{x}+x\circ\mathop{\mathrm{ad}}\nolimits_{f},\;\text{for
all $x\in{\mathfrak{g}}_{\bar{1}}$ and $f\in{\mathfrak{g}}_{\bar{1}}^{*}$}.$

A direct computation shows that the conditions (14), (15) and (16) are
equivalent to (30), (31), (32), (33), (34) and (35). The proof of Theorem
3.1.1. Theorem is now complete. ∎

##  5\. Anti-symmetric forms, Manin Triples and double extensions

Let ${\mathfrak{g}}$ be a Lie superalgebra in characteristic $2$ with a closed
${\bar{1}}$-antisymmetric ortho-orthogonal form $\omega$.

###  5.1. Left-symmetric and left-alternating algebras in characteristic $2$

The existence of a closed ${\bar{1}}$-antisymmetric ortho-orthogonal form (see
Section 2.2) on a Lie superalgebra ${\mathfrak{g}}$ will help us to naturally
define the notion of left-symmetric (resp. left-alternating) structure on the
Lie algebra $F({\mathfrak{g}})$, where $F$ is the functor of forgetting the
super structure. These structures are also known as Vinberg algebras as they
were used in [V]. For a thourough study of these structures, see [N].

Let $({\mathfrak{g}},\mathscr{B})$ be a NIS Lie superalgebra with a close
${\bar{1}}$-antisymmetric ortho-orthogonal form $\omega$.

For all $x,y\in{\mathfrak{g}}$, the map $z\mapsto\omega(y,[x,z])$ is a linear
form on ${\mathfrak{g}}$. Moreover, the map $z\mapsto\omega(z,\cdot)$
identifies ${\mathfrak{g}}$ with ${\mathfrak{g}}^{*}$. It follows that there
exists $x\star y\in{\mathfrak{g}}$ such that

$\omega(x\star y,z)=\omega(y,[x,z]).$

For all $x,y,z\in{\mathfrak{g}}$, we have

$0=\omega(x,[y,z])+\omega(y,[z,x])+\omega(z,[x,y])=\omega(y\star
x,z)+\omega(x\star y,z)+\omega(z,[x,y]).$

Therefore,

(36) $[x,y]=x\star y+y\star x\;\text{ for all $x,y\in{\mathfrak{g}}$}.$

On the other hand, for all $x\in{\mathfrak{g}}_{\bar{1}}$ and
$y\in{\mathfrak{g}}$, we have

$0=\omega(s(x),y)+\omega(x,[x,y])=\omega(x\star x,y).$

Therefore,

(37) $s(x)=x\star x\;\text{ for all $x\in{\mathfrak{g}}_{\bar{1}}$}.$

Now, (for all $x,y,z,t\in{\mathfrak{g}}$)

$\begin{array}[]{lcl}0&=&\omega(t,[x,[y,z]]+[y,[z,x]]+[z,[x,y]])\\\\[5.69054pt]
&=&\omega(t,[x,[y,z]])+\omega(t,[y,[z,x]])+\omega(t,[z,[x,y]])\\\\[5.69054pt]
&=&\omega(y\star(x\star t),z)+\omega(x\star(y\star t)t,z)+\omega([x,y]\star
t,z).\end{array}$

Now Eq. (36) and the fact that $\omega$ is non-degenerate imply that

$\mathrm{Asso}(x,y,t)=\mathrm{Asso}(y,x,t)\;\text{ for all
$x,y,t\in{\mathfrak{g}}$},$

where

$\begin{array}[]{lcl}\mathrm{Asso}(x,y,t)&:=&(x\star y)\star t+x\star(y\star
t)\;\text{ for all $x,y,t\in{\mathfrak{g}}$}\\\\[5.69054pt] \end{array}$

On the other hand,

$\begin{array}[]{lcl}0&=&\omega([s(x),y]+[x,[x,y]],t)\\\\[5.69054pt]
&=&\omega(t,[s(x),y])+\omega(t,[x,[x,y]])\\\\[5.69054pt] &=&\omega(s(x)\star
t,y)+\omega(x\star(x\star t),y).\end{array}$

Now Eq. (37) and the fact that $\omega$ is non-degenerate imply that

$\mathrm{Asso}(x,x,t)=0\;\text{ for all $x\in{\mathfrak{g}}_{\bar{1}}$ and
$t\in{\mathfrak{g}}$}.$

Let us describe $\star$ in terms of the derivation $\Delta$. For all
$x,y,z\in{\mathfrak{g}}$, and using Proposition 2.4.1. Proposition we have

$0=\omega(x\star y,z)-\omega(y,[x,z])={\mathscr{B}}(\Delta(x\star
y),z)-{\mathscr{B}}(\Delta(y),[x,z])={\mathscr{B}}(\Delta(x\star
y)-[\Delta(y),x],z).$

Since ${\mathscr{B}}$ is non-degenerate and $\Delta$ is invertible, it follows
that

$x\star y=\Delta^{-1}([x,\Delta(y)])\;\text{ for all $x,y\in{\mathfrak{g}}$}.$

In particular, $s(x)=x\star x=\Delta^{-1}([x,\Delta(x)])$. We then arrive at
the following definition.

An algebra ${\mathfrak{g}}$ in characteristic $2$ is called left-symmetric
(resp. left-alternating) if it is endowed with a bilinear operation $\star$
satisfying (for all $x,y,z\in{\mathfrak{g}}$):

$\mathrm{Asso}(x,y,z)=\mathrm{Asso}(y,x,z)\quad\text{(resp.
$\mathrm{Asso}(x,x,z)=0$)},$

where

$\begin{array}[]{lcl}\mathrm{Asso}(x,y,z)&:=&(x\star y)\star z+x\star(y\star
z)\text{ for all $x,y,z\in{\mathfrak{g}}$},\end{array}$

is called the associator.

######  Proposition (left-alternating implies left-symmetric).

In characteristic $2$, every left-alternating algebra is left-symmetric.

###### Proof.

Indeed, for all $x,y,z\in{\mathfrak{g}}$ we have

$\begin{array}[]{l}\mathrm{Asso}(x+y,x+y,z)=\mathrm{Asso}(x,x,z)+\mathrm{Asso}(x,y,z)+\mathrm{Asso}(y,x,z)+\mathrm{Asso}(y,y,z).\qed\end{array}$

######  Theorem (Constructing Lie superalgebras from Lie algebras).

Let $p=2$. (i) Every Lie algebra ${\mathfrak{g}}$ with an invertible
derivation $\Delta\in{\mathfrak{der}}({\mathfrak{g}})$ admits a left-symmetric
structure. Explicitly,

(38) $x\star y:=\Delta^{-1}([x,\Delta(y)])\;\text{ for all
$x,y\in{\mathfrak{g}}$}.$

(ii) Every ${\mathbb{Z}}/2$-graded Lie algebra
${\mathfrak{g}}:={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
an invertible derivation $\Delta$ preserving the ${\mathbb{Z}}/2$-grading can
be turned into a Lie superalgebra provided that

$\mathrm{Ass}(x,x,y)=0\;\text{ for all $x\in{\mathfrak{g}}_{\bar{1}}$ and
$y\in{\mathfrak{g}}$},$

where the assciator $\mathrm{Asso}$ is associated with the operation (38).
Explicitly, the map

$s:{\mathfrak{g}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}\quad x\mapsto
x\star x$

is a squaring on ${\mathfrak{g}}$.

###### Proof.

(i) Let us first compute the associator. For all $x,y,z\in{\mathfrak{g}}$, we
have

$\displaystyle\mathrm{Asso}(x,y,z)=(x\star y)\star z+x\star(y\star
z)=\Delta^{-1}\left([x\star y,\Delta(z)]+[x,\Delta(y\star z)]\right)$
$\displaystyle=\Delta^{-1}\left([\Delta^{-1}[x,\Delta(y)],\Delta(z)]+[x,[y,\Delta(z)]\right)$
$\displaystyle=\Delta^{-1}\left([[x,y],\Delta(z)]+[\Delta^{-1}[\Delta(x),y],\Delta(z)]+[x,[y,\Delta(z)]]\right).$

Now, using the Jacobi identity and the fact that $\Delta$ is a derivation we
see that

$\begin{array}[]{lcl}\mathrm{Asso}(x,y,z)+\mathrm{Asso}(y,x,z)&=&\Delta^{-1}\left([\Delta^{-1}[\Delta(x),y],\Delta(z)]+[x,[y,\Delta(z)]]\right)\\\\[5.69054pt]
&&+\Delta^{-1}\left([\Delta^{-1}[\Delta(y),x],\Delta(z)]+[y,[x,\Delta(z)]]\right)\\\\[5.69054pt]
&=&\Delta^{-1}\left([[x,y],\Delta(z)]+[\Delta(z),[x,y]]\right)=0.\end{array}$

Therefore, the operation (38) defines a left-symmetric structure on
${\mathfrak{g}}$.

(ii) Suppose now that
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ is
${\mathbb{Z}}/2$-graded and that $\Delta$ preserves the parity. Let us check
that the map

$s:{\mathfrak{g}}_{\bar{1}}\rightarrow{\mathfrak{g}}_{\bar{0}}\quad x\mapsto
x\star x$

is indeed a squaring. For all $\lambda\in{\mathbb{K}}$ and
$x\in{\mathfrak{g}}_{\bar{1}}$, we have

$s(\lambda x)=(\lambda x)\star(\lambda x)=\lambda^{2}(x\star
x)=\lambda^{2}s(x).$

On the other hand, for all $x,y\in{\mathfrak{g}}_{\bar{1}}$:

$\displaystyle s(x+y)-s(x)-s(y)=(x+y)\star(x+y)-x\star x-y\star y=x\star
y+y\star x$
$\displaystyle=\Delta^{-1}[x,\Delta(y)]+\Delta^{-1}[y,\Delta(x)]=\Delta^{-1}(\Delta([x,y]))=[x,y].$

Moreover (for all $x\in{\mathfrak{g}}_{\bar{1}}$ and $y\in{\mathfrak{g}}$):

$\begin{array}[]{lcl}[s(x),y]+[x,[x,y]]&=&s(x)\star y+y\star
s(x)+x\star[x,y]+[x,y]\star x\\\\[5.69054pt] &=&(x\star x)\star
y+y\star(x\star x)+x\star(x\star y+y\star x)+(x\star y+y\star x)\star
x\\\\[5.69054pt] &=&(x\star x)\star y+x\star(x\star y)+y\star(x\star
x)+(y\star x)\star x+x\star(y\star x)\\\\[5.69054pt] &&+(x\star y)\star
x\\\\[5.69054pt]
&=&\mathrm{Asso}(x,x,y)+\mathrm{Asso}(y,x,x)+\mathrm{Asso}(x,y,x)=\mathrm{Asso}(x,x,y).\end{array}$

It follows that the Jacobi identity involving the squaring holds if and only
if

$\mathrm{Asso}(x,x,y)=0\text{ for all $x\in{\mathfrak{g}}_{\bar{1}}$ and
$y\in{\mathfrak{g}}$}.\qed$

######  Corollary (The case of $2$-step nilpotent algebras).

The space of every ${\mathbb{Z}}/2$-graded $2$-step nilpotent Lie algebra in
characateristic $2$ with an invertible derivation can naturally be endowed
into a Lie superalgebra.

###### Proof.

Let us first observe that since $\Delta$ is invertible, then
$\Delta([{\mathfrak{g}},{\mathfrak{g}}])=[{\mathfrak{g}},{\mathfrak{g}}]$.
Now, for all $x\in{\mathfrak{g}}_{\bar{1}}$ and $y\in{\mathfrak{g}}$:

$\displaystyle\mathrm{Asso}(x,x,y)=x\star(x\star y)+(x\star x)\star
y=x\star\Delta^{-1}[x,\Delta(y)]+\Delta^{-1}[x,\Delta(x)]\star y$
$\displaystyle=\Delta^{-1}\left([x,[x,\Delta(y)]]+[\Delta^{-1}[x,\Delta(x)],\Delta(y)]\right)=\Delta^{-1}(0)=0.\qed$

###  5.2. Double extensions (DEs) of even NIS-Lie superalgebras equipped with
closed non-degenerate anti-symmetric bilinear forms

The notion of double extensions of Lie superalgebras in characteristic $2$ has
been introduced and studied in [BeBou]. For interesting examples of DEs of
simple Lie superalgebras classified up to isomorphisms, see [BeBou, BLS]. Here
we extend this notion to NIS-Lie supearlgebras equipped with a
${\bar{1}}$-antisymmetric ortho-orthogonal or periplectic forms as well as to
Manin triples.

The first two theorems below have been proved in [BeBou].

######  Theorem (${\mathscr{D}}_{\bar{0}}$-extension – NIS even).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ be a NIS-Lie superalgebra
in characteristic $2$ such that ${\mathscr{B}}_{\mathfrak{a}}$ is even. Let
${\mathscr{D}}\in{\mathfrak{der}}_{\bar{0}}({\mathfrak{a}})$ be a derivation
satisfying the following conditions:

(39)
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b))=0\text{
for any
}a,b\in{\mathfrak{a}};\;{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),a)=0\text{
for any }a~{}\in{\mathfrak{a}}_{\bar{0}}.$

Let $\alpha:{\mathfrak{a}}_{\bar{1}}\rightarrow\mathbb{K}$ be a quadratic form
and ${\mathscr{B}}_{\alpha}$ its polar form111In characteristic $2$, the polar
form ${\mathscr{B}}_{\alpha}$ associated with a quadratique form $\alpha$ is
defined by
$(a,b)\mapsto{\mathscr{B}}_{\alpha}(a,b):=\alpha(a+b)+\alpha(a)+\alpha(b).$
which satisfies

(40) $\displaystyle{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b))$
$\displaystyle=$ $\displaystyle{\mathscr{B}}_{\alpha}(a,b)\text{ for any
$a,b\in{\mathfrak{a}}_{\bar{1}}$}.$

Then, there exists a NIS-Lie superalgebra structure on
${\mathfrak{g}}:=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{K}^{*}$, where
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ for $x$ even, defined
as follows. The squaring is given by

$s_{\mathfrak{g}}(a):=s_{\mathfrak{a}}(a)+\alpha(a)x\qquad\text{ for any
}a\in{\mathfrak{g}}_{\bar{1}}\;(={\mathfrak{a}}_{\bar{1}}).$

The bracket on ${\mathfrak{g}}$ is defined as follows:

(41)
$[\mathscr{K},{\mathfrak{g}}]_{\mathfrak{g}}=0,\qquad[a,b]_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)x+[a,b]_{\mathfrak{a}},\qquad[x^{*},a]_{\mathfrak{g}}:={\mathscr{D}}(a)\text{
for any }a,b\in{\mathfrak{a}}.$

The non-degenerate ${\bar{0}}$-antisymmetric bilinear form
${\mathscr{B}}_{\mathfrak{g}}$ on ${\mathfrak{g}}$ is defined as follows:

$\displaystyle{{\mathscr{B}}_{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}:={\mathscr{B}}_{\mathfrak{a}},\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}):=0,\quad{\mathscr{B}}_{\mathfrak{g}}(x,x^{*}):=1,\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}^{*}):=0,$
$\displaystyle{\mathscr{B}}_{\mathfrak{g}}(x,x):=0,\
{\mathscr{B}}_{\mathfrak{g}}(x^{*},x^{*})\;\text{ arbitrary}.$

Moreover, the form ${\mathscr{B}}_{\mathfrak{g}}$, obviously even, is
invariant on ${\mathfrak{g}}$.

Therefore, $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a NIS-Lie
superalgebra.

######  Theorem (${\mathscr{D}}_{\bar{1}}$-extension – NIS even).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ be a NIS-Lie superalgebra
in characteristic $2$ such that ${\mathscr{B}}_{\mathfrak{a}}$ is even. Let
${\mathscr{D}}\in{\mathfrak{der}}_{\bar{1}}({\mathfrak{a}})$ and
$a_{0}\in{\mathfrak{a}}_{\bar{0}}$ satisfy the following conditions:

(42) $\displaystyle{\mathscr{B}}_{\mathfrak{a}}(\mathscr{D}(a),b)$
$\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}(a,\mathscr{D}(b))\text{ for any
}a,b\in{\mathfrak{a}};$ (43) $\displaystyle{\mathscr{D}}^{2}$ $\displaystyle=$
$\displaystyle\mathop{\mathrm{ad}}\nolimits_{a_{0}};$ (44)
$\displaystyle{\mathscr{D}}(a_{0})$ $\displaystyle=$ $\displaystyle 0.$

Then there exists a NIS-Lie superalgebra structure on
${\mathfrak{g}}:=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{K}^{*}$, where
$\mathscr{K}:=~{}\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ and $x$ odd,
defined as follows. The squaring is given by

$s_{\mathfrak{g}}(rx+a+tx^{*}):=s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a)\qquad\text{
for any }\;rx+a+tx^{*}\in{\mathfrak{g}}_{\bar{1}}.$

The bracket is given by:

$\begin{array}[]{l}[x,{\mathfrak{g}}]_{\mathfrak{g}}:=0;\qquad[a,b]_{\mathfrak{g}}:=[a,b]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}(\mathscr{D}(a),b)x\quad\text{~{}~{}for
any $a,b\in{\mathfrak{a}}$};\\\\[5.69054pt]
[x^{*},a]_{\mathfrak{g}}:={\mathscr{D}}(a)\quad\text{ for any
}a~{}\in{\mathfrak{a}}.\end{array}$

The bilinear form ${\mathscr{B}}_{\mathfrak{g}}$ on ${\mathfrak{g}}$ defined
by:

$\displaystyle{{\mathscr{B}}_{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}:={\mathscr{B}}_{\mathfrak{a}},\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}):=0,\quad{\mathscr{B}}_{\mathfrak{g}}({\mathfrak{a}},\mathscr{K}^{*}):=0,$
$\displaystyle{\mathscr{B}}_{\mathfrak{g}}(x,x^{*}):=1,\quad{\mathscr{B}}_{\mathfrak{g}}(x,x):={\mathscr{B}}_{\mathfrak{g}}(x^{*},x^{*}):=0.$

is even, non-degenerate, ${\bar{0}}$-antisymmetric and invariant on
$({\mathfrak{g}},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$.

Therefore $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a NIS-Lie
superalgebra.

######  Theorem (Ortho-orthogonal forms for
${\mathscr{D}}_{\bar{0}}$-extensions).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ be a
NIS Lie superalgebra equipped with a closed ${\bar{1}}$-antisymmetric ortho-
orthogonal form $\omega_{\mathfrak{a}}$, and let
$\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{0}}$ be the unique
and invertible derivation such that
$\omega_{\mathfrak{a}}(\cdot,\cdot)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot)$.
Let $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ be the double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ by means of a derivation
${\mathscr{D}}\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ as in Theorem
5.2.1. Theorem such that ${\mathscr{B}}_{\mathfrak{g}}(x^{*},x^{*})=0$.
Suppose there exist $\lambda\in{\mathbb{K}}\backslash\\{0\\}$ and
$a_{0}\in{\mathfrak{a}}_{\bar{0}}$ such that

(45) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{a_{0}},$
(46)
$\displaystyle\lambda\alpha(a)+{\mathscr{B}}_{\alpha}({\mathscr{D}}\circ\tilde{\Delta}(a),a)$
$\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a_{0})\text{
for all $a\in{\mathfrak{a}}_{\bar{1}}$}.$

The linear map $\Delta:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ defined by

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(x^{*})&=&\lambda
x^{*}+a_{0},\\\
\Delta(a)&=&\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{0})x.\end{array}$

is an invertible derivation on ${\mathfrak{g}}$ for which
${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Consequently,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is a
NIS-Lie superalgebra with a closed ${\bar{1}}$-antisymmetric ortho-orthogonal
form
$\omega_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.

###### Proof.

First, let us observe that $\Delta$ is obviously invertible since
$\tilde{\Delta}$ is invertible. Let us prove that $\Delta$ is a derivation. We
will check the condition on the squaring. Indeed, (where
$a\in{\mathfrak{a}}_{\bar{1}}$)

$\Delta(s_{\mathfrak{g}}(a))=\Delta(s_{\mathfrak{a}}(a)+\alpha(a)x)=\Delta(s_{\mathfrak{a}}(a))+\lambda\alpha(a)x=\tilde{\Delta}(s_{\mathfrak{a}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a_{0})x+\lambda\alpha(a)x.$

On the other hand, we have (where $a\in{\mathfrak{a}}_{\bar{1}}$)

$[\Delta(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{g}}=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+{\mathscr{B}}_{\alpha}({\mathscr{D}}\circ\tilde{\Delta}(a),a)x.$

The fact that $\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{0}}$
and Eq. (46) implies that
$\Delta(s_{\mathfrak{g}}(a))=[\Delta(a),a]_{\mathfrak{g}}$.

Now let us check the condition

$\Delta([f,g]_{\mathfrak{g}})=[\Delta(f),g]_{\mathfrak{g}}+[f,\Delta(g)]_{\mathfrak{g}}\text{
for all $f,g\in{\mathfrak{g}}$}.$

We will only check the case where $(f,g)=(a,x^{*})$; checking the other cases
is a routine. Indeed,

$\begin{array}[]{lcl}\Delta([a,x^{*}]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),a_{0})x.\end{array}$

On the other hand,

$\begin{array}[]{lcl}[\Delta(a),x^{*}]_{\mathfrak{g}}+[a,\Delta(x^{*})]_{\mathfrak{g}}&=&[\tilde{\Delta}(a),x^{*}]_{\mathfrak{g}}+[a,\lambda
x^{*}+a_{0}]_{\mathfrak{g}}\\\\[5.69054pt]
&=&{\mathscr{D}}(\tilde{\Delta}(a))+\lambda{\mathscr{D}}(a)+[a,a_{0}]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),a_{0})x.\end{array}$

Therefore,
$\Delta([a,x^{*}]_{\mathfrak{g}})=[\Delta(a),x^{*}]_{\mathfrak{g}}+[a,\Delta(x^{*})]_{\mathfrak{g}}$
is equivalent to Eq. (45).

Let us show that ${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Indeed,
for all $a\in{\mathfrak{a}}_{\bar{0}}$ (recall that $x$ is orthogonal to
${\mathfrak{a}}$):

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),a)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{0})x,a)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),a)=\omega_{\mathfrak{a}}(a,a)=0.$

Similarly,

${\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),x^{*})={\mathscr{B}}_{\mathfrak{g}}(\lambda
x^{*}+a_{0},x^{*})=0\text{ and
}{\mathscr{B}}_{\mathfrak{g}}(\Delta(x),x)={\mathscr{B}}_{\mathfrak{g}}(\lambda
x,x)=0.$

For all $a,b\in{\mathfrak{a}}$, we have

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),b)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{0})x,b)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)={\mathscr{B}}_{\mathfrak{a}}(a,\tilde{\Delta}(b))={\mathscr{B}}_{\mathfrak{g}}(a,\Delta(b)).$

Moreover,

$\begin{array}[]{lcl}{\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),a)+{\mathscr{B}}_{\mathfrak{g}}(x^{*},\Delta(a))&=&{\mathscr{B}}_{\mathfrak{g}}(\lambda
x^{*}+a_{0},a)+{\mathscr{B}}_{\mathfrak{g}}(x^{*},\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{0})x.)\\\\[5.69054pt]
&=&{\mathscr{B}}_{\mathfrak{a}}(a_{0},a)+{\mathscr{B}}_{\mathfrak{a}}(a,a_{0})=0.\end{array}$

The $\Delta$-invariant of ${\mathscr{B}}_{\mathfrak{g}}$ can easily be checked
for the remaining elements of the basis of ${\mathfrak{g}}$.∎

In proving the converse of Theorem 5.2.1. Theorem in [BeBou], the following
object, called special center, has been introduced

${\mathfrak{z}}_{s}({\mathfrak{g}}):={\mathfrak{z}}({\mathfrak{g}})\cap
s_{\mathfrak{g}}({\mathfrak{g}}_{\bar{1}})^{\perp}.$

######  Lemma (Description of the special center).

If the Lie superalgebra $({\mathfrak{g}},{\mathscr{B}})$ admits an invertible
$\Delta$ derivation such that ${\mathscr{B}}$ is $\Delta$-invariant, then
${\mathfrak{z}}_{s}({\mathfrak{g}})={\mathfrak{z}}({\mathfrak{g}})$.

###### Proof.

Let $x\in{\mathfrak{z}}({\mathfrak{g}})$. There exists
$\tilde{x}\in{\mathfrak{z}}({\mathfrak{g}})$ such that $x=\Delta(\tilde{x})$.
We have

${\mathscr{B}}_{\mathfrak{g}}(x,s_{\mathfrak{g}}(y))={\mathscr{B}}_{\mathfrak{g}}(\Delta(\tilde{x}),s_{\mathfrak{g}}(y))={\mathscr{B}}_{\mathfrak{g}}(\tilde{x},\Delta(s_{\mathfrak{g}}(y)))={\mathscr{B}}_{\mathfrak{g}}(\tilde{x},[y,\Delta(y)]_{\mathfrak{g}})={\mathscr{B}}_{\mathfrak{g}}([\tilde{x},y]_{\mathfrak{g}},\Delta(y))=0.$

∎

######  Theorem (Converse of Theorem 5.2.3. Theorem).

Let $({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ be an
irreducible NIS-Lie superalgebra equipped with a closed
${\bar{1}}$-antisymmetric ortho-orthogonal form $\omega_{\mathfrak{g}}$.
Suppose that ${\mathfrak{z}}_{s}({\mathfrak{g}})_{\bar{0}}\not=\\{0\\}.$ Then,
$({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is
obtained as a $\mathscr{D}$-extension of a NIS-Lie superalgebra
$({\mathfrak{a}},\mathscr{B}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ with a
closed ${\bar{1}}$-antisymmetric ortho-orthogonal form
$\omega_{\mathfrak{a}}$.

###### Proof.

Let $\Delta\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ be the unique
invertible derivation such that
$\omega_{\mathfrak{g}}(\cdot,\cdot)=\mathscr{B}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.
It follows that
$\Delta({\mathfrak{z}}_{s}({\mathfrak{g}})_{\bar{0}})={\mathfrak{z}}_{s}({\mathfrak{g}})_{\bar{0}}$.
Since ${\mathbb{K}}$ is algebraically closed, let us choose
$x\in{\mathfrak{z}}_{s}({\mathfrak{g}})_{\bar{0}}$ such that
$\Delta(x)=\lambda x$ for some $\lambda\in{\mathbb{K}}\backslash\\{0\\}$. It
has been shown in [BeBou, Prop. 3.1.3], that the subspaces
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ and
$\mathscr{K}^{\perp}$ are ideals in
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$. Besides,
$\Delta(\mathscr{K}^{\perp})\subset\mathscr{K}^{\perp}$. Indeed, (for all
$f\in\mathscr{K}^{\perp}$) we have

${\mathscr{B}}_{\mathfrak{g}}(\Delta(f),x)={\mathscr{B}}_{\mathfrak{g}}(f,\Delta(x))={\mathscr{B}}_{\mathfrak{g}}(f,\lambda
x)=0.$

Now, since $\mathscr{K}$ is 1-dimensional, then either
$\mathscr{K}\cap\mathscr{K}^{\perp}=\\{0\\}$ or
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$. The first case is to be
disregarded because otherwise
${\mathfrak{g}}=\mathscr{K}\oplus\mathscr{K}^{\perp}$ and the Lie superalgebra
${\mathfrak{g}}$ is not irreducible. Hence,
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$. It follows that
$\mathscr{K}\subset\mathscr{K}^{\perp}$ and
$\dim(\mathscr{K}^{\perp})=\dim({\mathfrak{g}})-1$. Therefore, there exists a
non-zero $x^{*}\in{\mathfrak{g}}_{\bar{0}}$ such that

${\mathfrak{g}}=\mathscr{K}^{\perp}\oplus\mathscr{K}^{*},\quad\text{ where
$\mathscr{K}^{*}:=\mathop{\mathrm{{Span}}}\nolimits\\{x^{*}\\}$}.$

This $x^{*}$ can be normalized such that
${\mathscr{B}}_{\mathfrak{g}}(x,x^{*})=1$. Besides,
${\mathscr{B}}_{\mathfrak{g}}(x,x)=0$ since
$\mathscr{K}\cap\mathscr{K}^{\perp}=\mathscr{K}$.

Let us define ${\mathfrak{a}}:=(\mathscr{K}+\mathscr{K}^{*})^{\perp}$. We then
have a decomposition
${\mathfrak{g}}=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{K}^{*}$.

It has been shown in [BeBou, Prop. 3.1.3] that
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$, where
${\mathscr{B}}_{\mathfrak{a}}={\mathscr{B}}_{{\mathfrak{g}}_{|_{{\mathfrak{a}}\times{\mathfrak{a}}}}}$,
by means of a derivation $\mathscr{D}$ and a quadratic form $\alpha$.

Let us write $\Delta$ as (where
$a_{0}\in{\mathfrak{a}}_{\bar{0}},\beta(a),\gamma(a),\mu,\nu\in{\mathbb{K}}$):

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(x^{*})&=&\nu
x^{*}+a_{0}+\mu x,\\\
\Delta(a)&=&\tilde{\Delta}(a)+\beta(a)x+\gamma(a)x^{*}.\end{array}$

The condition
${\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),x)={\mathscr{B}}_{\mathfrak{g}}(x^{*},\Delta(x))$
implies that $\nu=\lambda$. Similarly, the condition
${\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),x^{*})=0$ implies that $\mu=0$ and
the condition
${\mathscr{B}}_{\mathfrak{a}}(\Delta(a),x)={\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))$
implies that $\gamma(a)=0$ for every $a\in{\mathfrak{a}}$.

In addition, the conditions

$\begin{array}[]{lcl}{\mathscr{B}}_{\mathfrak{g}}(\Delta(a),x^{*})&=&{\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+\beta(a)x,x^{*})=\beta(a),\\\\[2.84526pt]
{\mathscr{B}}_{\mathfrak{g}}(a,\Delta(x^{*}))&=&{\mathscr{B}}_{\mathfrak{g}}(a,\lambda
x^{*}+a_{0}+\mu x)={\mathscr{B}}_{\mathfrak{a}}(a,a_{0}),\end{array}$

imply that

(47) $\beta(a)={\mathscr{B}}_{\mathfrak{a}}(a,a_{0})\text{ for all
$a\in{\mathfrak{a}}$}.$

Now, we have

$\begin{array}[]{lcl}\Delta([x^{*},a]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+\beta({\mathscr{D}}(a))x\;\text{
for all $a\in{\mathfrak{a}}$}.\end{array}$

On the other hand, we have

$[\Delta(x^{*}),a]_{\mathfrak{g}}+[x^{*},\Delta(a)]_{\mathfrak{g}}=\lambda{\mathscr{D}}(a)+[a_{0},a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a_{0}),a)x+{\mathscr{D}}(\tilde{\Delta}(a))\text{
for all $a\in{\mathfrak{a}}$}.$

Since $\Delta$ is a derivation on ${\mathfrak{g}}$, it follows that

(48)
$[\tilde{\Delta},{\mathscr{D}}]=\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{a_{0}}.$

Following the same computation as in [BaB], the condition

$\Delta([a,b]_{\mathfrak{g}})=[\Delta(a),b]_{\mathfrak{g}}+[a,\Delta(b)]_{\mathfrak{g}}\;\text{
for all $a,b\in{\mathfrak{a}}$,}$

implies that

(49) $\displaystyle\tilde{\Delta}([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle[\tilde{\Delta}(a),b]_{\mathfrak{a}}+[a,\tilde{\Delta}(b)]_{\mathfrak{a}}$
(50) $\displaystyle\beta([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(\tilde{\Delta}(a)),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),\tilde{\Delta}(b)).$

Eq. (50) is certainly satisfied, thanks to Eqs. (47), (48) and the fact that
${\mathscr{B}}_{\mathfrak{a}}$ is $\tilde{\Delta}$-invariant.

For all $a\in{\mathfrak{a}}$, the condition
$\Delta(s_{\mathfrak{g}}(a))=[a,\Delta(a)]_{\mathfrak{g}}$ implies that

(51) $\displaystyle\tilde{\Delta}(s_{\mathfrak{a}}(a))$ $\displaystyle=$
$\displaystyle[a,\tilde{\Delta}(a)]_{\mathfrak{a}},$ (52)
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a),a_{0})$
$\displaystyle=$
$\displaystyle\lambda\alpha(a)+{\mathscr{B}}_{\alpha}(a,\tilde{\Delta}(a)).$

Eqs. (49), (51) imply that $\tilde{\Delta}$ is a derivation on
${\mathfrak{a}}$. Since $\Delta$ is invertible, then $\tilde{\Delta}$ is
invertible as well. Define

$\omega_{\mathfrak{a}}(a,b):={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)\;\text{
for all $a,b\in{\mathfrak{a}}$}.$

The form $\omega_{\mathfrak{a}}$ is a closed, ${\bar{1}}$-antisymmetric and
ortho-orthogonal over ${\mathfrak{a}}$.∎

######  Theorem (Ortho-orthogonal forms for
${\mathscr{D}}_{\bar{1}}$-extensions).

Let $({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}},\omega_{\mathfrak{a}})$ be a
NIS Lie superalgebra equipped with a closed ${\bar{1}}$-antisymmetric ortho-
orthogonal form $\omega_{\mathfrak{a}}$, and let
$\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ be the unique
and invertible derivation such that
$\omega_{\mathfrak{a}}(\cdot,\cdot)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(\cdot),\cdot)$.
Let $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ be the double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$ by means of a derivation
${\mathscr{D}}\in{\mathfrak{der}}_{\bar{1}}({\mathfrak{a}})$ satisfying (42),
(43), (44). Suppose there exist $\lambda\in{\mathbb{K}}\backslash\\{0\\}$ and
$b_{0}\in{\mathfrak{a}}_{\bar{1}}$ such that

(53) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{b_{0}},$
(54) $\displaystyle\tilde{\Delta}(a_{0})$ $\displaystyle=$
$\displaystyle{\mathscr{D}}(b_{0}).$

The linear map $\Delta:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ defined by
(where $\mu\in{\mathbb{K}}$ is arbitrary)

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(x^{*})&=&\lambda
x^{*}+b_{0}+\mu x,\\\
\Delta(a)&=&\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x.\end{array}$

is an invertible derivation on ${\mathfrak{g}}$ such that
${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Consequently,
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is a
NIS-Lie superalgebra with a closed ${\bar{1}}$-antisymmetric ortho-orthogonal
form
$\omega_{\mathfrak{g}}:={\mathscr{B}}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.

###### Proof.

Obviously, the map $\Delta$ is invertible since $\tilde{\Delta}$ is
invertible. Let us check that $\Delta$ is a derivation ${\mathfrak{g}}$. We
start be checking the condition on the squaring. Indeed, (where
$a\in{\mathfrak{a}}_{\bar{1}}$ and $s,t\in{\mathbb{K}}$)

$\begin{array}[]{l}\Delta(s_{\mathfrak{g}}(rx+a+tx^{*}))\\\\[5.69054pt]
=\Delta(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a))\\\\[5.69054pt]
=\tilde{\Delta}(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}(s_{\mathfrak{a}}(a)+t^{2}a_{0}+t{\mathscr{D}}(a),b_{0})x\\\\[5.69054pt]
=\tilde{\Delta}(s_{\mathfrak{a}}(a))+t^{2}\tilde{\Delta}(a_{0})+t\tilde{\Delta}({\mathscr{D}}(a)).\end{array}$

On the other hand,

$\begin{array}[]{l}[\Delta(rx+a+tx^{*}),rx+a+tx^{*}]_{\mathfrak{g}}\\\\[5.69054pt]
=[r\lambda
x+\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x+t(\lambda
x^{*}+b_{0}+\mu x),rx+a+tx^{*}]_{\mathfrak{g}}\\\\[5.69054pt]
=[\tilde{\Delta}(a)+t(\lambda
x^{*}+b_{0}),a+tx^{*}]_{\mathfrak{g}}\\\\[5.69054pt]
=[\tilde{\Delta}(a),a]_{\mathfrak{a}}+t{\mathscr{D}}(\tilde{\Delta}(a))+t\lambda{\mathscr{D}}(a)+t[b_{0},a]_{\mathfrak{a}}+t^{2}{\mathscr{D}}(b_{0}).\end{array}$

The fact that $\tilde{\Delta}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{1}}$
and Eqs. (53), (54) imply that
$\Delta(s_{\mathfrak{g}}(f))=[\Delta(f),f]_{\mathfrak{g}}$ for all
$f\in{\mathfrak{g}}_{\bar{1}}$.

Now let us check the condition

$\Delta([f,g]_{\mathfrak{g}})=[\Delta(f),g]_{\mathfrak{g}}+[f,\Delta(g)]_{\mathfrak{g}}\text{
for all $f,g\in{\mathfrak{g}}$}.$

We will only check the case where $(f,g)=(a,x^{*})$; checking the other cases
is a routine. Indeed, using Eq. (53) we have

$\begin{array}[]{lcl}\Delta([a,x^{*}]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x.\end{array}$

On the other hand,

$\begin{array}[]{lcl}[\Delta(a),x^{*}]_{\mathfrak{g}}+[a,\Delta(x^{*})]_{\mathfrak{g}}&=&[\tilde{\Delta}(a)+{\mathscr{B}}(a,b_{0})x,x^{*}]_{\mathfrak{g}}+[a,\lambda
x^{*}+b_{0}+\mu x]_{\mathfrak{g}}\\\\[5.69054pt]
&=&{\mathscr{D}}(\tilde{\Delta}(a))+\lambda{\mathscr{D}}(a)+[a,b_{0}]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x\\\\[5.69054pt]
\end{array}$

Therefore, the condition

$\Delta([a,x^{*}]_{\mathfrak{g}})=[\Delta(a),x^{*}]_{\mathfrak{g}}+[a,\Delta(x^{*})]_{\mathfrak{g}}$

is equivalent to Eq. (53).

Let us show that ${\mathscr{B}}_{\mathfrak{g}}$ is $\Delta$-invariant. Indeed,
for all $a\in{\mathfrak{a}}_{\bar{0}}$ (recall that $x$ is orthogonal to
${\mathfrak{a}}$):

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),a)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x,a)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),a)=0.$

For all $a,b\in{\mathfrak{a}}$, we have

${\mathscr{B}}_{\mathfrak{g}}(\Delta(a),b)={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x,b)={\mathscr{B}}_{\mathfrak{a}}(\tilde{\Delta}(a),b)={\mathscr{B}}_{\mathfrak{a}}(a,\tilde{\Delta}(b))={\mathscr{B}}_{\mathfrak{g}}(a,\Delta(b)).$

Similarly,

$\begin{array}[]{lcl}{\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),a)+{\mathscr{B}}_{\mathfrak{g}}(x^{*},\Delta(a))&=&{\mathscr{B}}_{\mathfrak{g}}(\lambda
x^{*}+b_{0}+\mu
x,a)+{\mathscr{B}}_{\mathfrak{g}}(x^{*},\tilde{\Delta}(a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})x)\\\\[5.69054pt]
&=&{\mathscr{B}}_{\mathfrak{a}}(b_{0},a)+{\mathscr{B}}_{\mathfrak{a}}(a,b_{0})=0.\end{array}$

The $\Delta$-invariance of ${\mathscr{B}}_{\mathfrak{g}}$ can easily be
checked for the remaining elements of the basis of ${\mathfrak{g}}$. We define
an orthosymplectic form on ${\mathfrak{g}}$ as follows

$\omega(\cdot,\cdot):={\mathscr{B}}_{\mathfrak{g}}(\Delta(\cdot),\cdot).\qed$

In proving the converse of Theorem 5.2.2. Theorem in [BeBou], we had to
introduce the cone

$\mathscr{C}({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}}):=\\{x\in{\mathfrak{g}}_{\bar{1}}\;|\;{\mathscr{B}}_{\mathfrak{g}}(s_{\mathfrak{g}}(x),s_{\mathfrak{g}}(t))=0\quad\text{
for any }t\in{\mathfrak{g}}_{\bar{1}}\\},$

Clearly, if $x\in\mathscr{C}({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$,
then $\lambda x\in\mathscr{C}({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$
for any $\lambda\in{\mathbb{K}}$, hence the terminology.

######  Lemma (Description of the cone).

If the Lie superalgebra $({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ admits
an invertible derivation $\Delta$ such that ${\mathscr{B}}_{\mathfrak{g}}$ is
$\Delta$-invariant, then

${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\cap~{}\mathscr{C}({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})={\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}.$

###### Proof.

Let $x\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$. For every
$t\in{\mathfrak{g}}_{\bar{1}}$, there exists a
$\tilde{t}\in{\mathfrak{g}}_{\bar{0}}$ such that
$\Delta(\tilde{t})=s_{\mathfrak{g}}(t)$. We have

${\mathscr{B}}_{\mathfrak{g}}(s_{\mathfrak{g}}(x),s_{\mathfrak{g}}(t))={\mathscr{B}}_{\mathfrak{g}}(s_{\mathfrak{g}}(x),\Delta(\tilde{t}))={\mathscr{B}}_{\mathfrak{g}}(\Delta(s_{\mathfrak{g}}(x)),\tilde{t})={\mathscr{B}}_{\mathfrak{g}}([x,\Delta(x)]_{\mathfrak{g}},\tilde{t})=0.$

Therefore, $x\in\mathscr{C}({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ and
we are done. ∎

######  Theorem (Converse of Theorem 5.2.6. Theorem).

Let $({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ be an
irreducible NIS-Lie superalgebra with
$p(\mathscr{B}_{\mathfrak{g}})={\bar{1}}$ and equipped with a closed
${\bar{1}}$-antisymmetric ortho-orthogonal form $\omega_{\mathfrak{g}}$.
Suppose that ${\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\not=\\{0\\}$. Then
$({\mathfrak{g}},\mathscr{B}_{\mathfrak{g}},\omega_{\mathfrak{g}})$ is
obtained as an $\mathscr{D}$-extension from a NIS-Lie superalgebra
$({\mathfrak{a}},\mathscr{B}_{\mathfrak{a}},\omega_{\mathfrak{a}})$, where
$\omega_{\mathfrak{a}}$ is closed, ${\bar{1}}$-antisymmetric and ortho-
orthogonal form on ${\mathfrak{a}}$.

###### Proof.

Let $\Delta\in{\mathfrak{der}}({\mathfrak{g}})_{\bar{0}}$ be the unique
invertible derivation such that
$\omega_{\mathfrak{g}}(\cdot,\cdot)=\mathscr{B}_{\mathfrak{g}}(\Delta(\cdot),\cdot)$.
It follows that
$\Delta({\mathfrak{z}}({\mathfrak{g}})_{\bar{1}})={\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}$.
Since ${\mathbb{K}}$ is algebraically closed, let us choose
$x\in{\mathfrak{z}}({\mathfrak{g}})_{\bar{1}}\cap~{}\mathscr{C}({\mathfrak{g}},B)$
such that $\Delta(x)=\lambda x$ for some
$\lambda\in{\mathbb{K}}\backslash\\{0\\}$. We assume that
$s_{\mathfrak{g}}(x)=0$; otherwise, $s_{\mathfrak{g}}(x)$ will be a non-zero
element of ${\mathfrak{z}}_{\bar{0}}({\mathfrak{g}})$ and hence Theorem 5.2.5.
Theorem can be recalled. Besides,
$\Delta(\mathscr{K}^{\perp})\subset\mathscr{K}^{\perp}$. Indeed, (for all
$f\in\mathscr{K}^{\perp}$)

${\mathscr{B}}_{\mathfrak{g}}(\Delta(f),x)={\mathscr{B}}_{\mathfrak{g}}(f,\Delta(x))={\mathscr{B}}_{\mathfrak{g}}(f,\lambda
x)=0.$

It has been shown in [BeBou, Prop. 3.2.2] that the subspace
$\mathscr{K}:=\mathop{\mathrm{{Span}}}\nolimits\\{x\\}$ is an ideal in
${\mathfrak{g}}$; moreover, the subspace $\mathscr{K}^{\perp}$ is also an
ideal containing $\mathscr{K}$. Same arguments as in Theorem 5.2.5. Theorem
can be used to construct the ideal
${\mathfrak{a}}=(\mathscr{K}\oplus\mathscr{K}^{*})^{\perp},$ and a
decomposition
${\mathfrak{g}}=\mathscr{K}\oplus{\mathfrak{a}}\oplus\mathscr{K}^{*}$, where
the generator $x^{*}$ of $\mathscr{K}^{*}$ can be normalized so that
${\mathscr{B}}_{\mathfrak{g}}(x,x^{*})=1$.

It has been shown in [BeBou, Prop. 3.2.2] that
$({\mathfrak{g}},{\mathscr{B}}_{\mathfrak{g}})$ is a double extension of
$({\mathfrak{a}},{\mathscr{B}}_{\mathfrak{a}})$, where
${\mathscr{B}}_{\mathfrak{a}}={\mathscr{B}}_{{\mathfrak{g}}}|_{{\mathfrak{a}}\times{\mathfrak{a}}}$,
by means of a derivation
$\mathscr{D}\in{\mathfrak{der}}({\mathfrak{a}})_{\bar{1}}$,
$a_{0}\in{\mathfrak{a}}_{\bar{0}}$ and satisfying Eqs. (42), (43), (44).

Let us write $\Delta$ as (where $b_{0}\in{\mathfrak{a}}_{\bar{1}}$, and
$\beta(a),\mu,\nu,\lambda\in{\mathbb{K}}$):

$\begin{array}[]{lcl}\Delta(x)&=&\lambda x,\\\ \Delta(x^{*})&=&\nu
x^{*}+b_{0}+\mu x,\\\
\Delta(a)&=&\tilde{\Delta}(a)+\beta(a)x+\gamma(a)x^{*}.\end{array}$

Since
${\mathscr{B}}_{\mathfrak{g}}(\Delta(x^{*}),x)={\mathscr{B}}_{\mathfrak{g}}(x^{*},\Delta(x))$,
then $\nu=\lambda$, and since
${\mathscr{B}}_{\mathfrak{a}}(\Delta(a),x)={\mathscr{B}}_{\mathfrak{a}}(a,\Delta(x))$,
then $\gamma(a)=0$ for all $a\in{\mathfrak{a}}$.

In addition, the conditions

$\begin{array}[]{l}{\mathscr{B}}_{\mathfrak{g}}(\Delta(a),x^{*})={\mathscr{B}}_{\mathfrak{g}}(\tilde{\Delta}(a)+\beta(a)x,x^{*})=\beta(a)\;\text{
for all $a\in{\mathfrak{a}}$},\\\\[5.69054pt]
{\mathscr{B}}_{\mathfrak{g}}(a,\Delta(x^{*}))={\mathscr{B}}_{\mathfrak{g}}(a,\lambda
x^{*}+b_{0}+\mu x)={\mathscr{B}}_{\mathfrak{a}}(a,b_{0})\;\text{ for all
$a$}\end{array}$

imply that

(55) $\beta(a)={\mathscr{B}}_{\mathfrak{a}}(a,b_{0})\text{ for all
$a\in{\mathfrak{a}}$}.$

Now, we have

$\begin{array}[]{lcl}\Delta([x^{*},a]_{\mathfrak{g}})&=&\Delta({\mathscr{D}}(a))=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b_{0})x=\tilde{\Delta}({\mathscr{D}}(a))+{\mathscr{B}}_{\mathfrak{a}}(a,{\mathscr{D}}(b_{0}))x.\end{array}$

On the other hand, we have

$\begin{array}[]{lcl}[\Delta(x^{*}),a]_{\mathfrak{g}}+[x^{*},\Delta(a)]_{\mathfrak{g}}&=&\lambda{\mathscr{D}}(a)+[b_{0},a]_{\mathfrak{a}}+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(b_{0}),a)x+{\mathscr{D}}(\tilde{\Delta}(a)).\end{array}$

Since $\Delta$ is a derivation, it follows that

(56) $\displaystyle[\tilde{\Delta},{\mathscr{D}}]$ $\displaystyle=$
$\displaystyle\lambda{\mathscr{D}}+\mathop{\mathrm{ad}}\nolimits_{a_{0}}.$

Following the same computation of [BaB], the condition

$\Delta([a,b]_{\mathfrak{g}})=[\Delta(a),b]_{\mathfrak{g}}+[a,\Delta(b)]_{\mathfrak{g}}$

implies that

(57) $\displaystyle\tilde{\Delta}([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle[\tilde{\Delta}(a),b]_{\mathfrak{a}}+[a,\tilde{\Delta}(b)]_{\mathfrak{a}},$
(58) $\displaystyle\beta([a,b]_{\mathfrak{a}})$ $\displaystyle=$
$\displaystyle{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(\tilde{\Delta}(a)),b)+{\mathscr{B}}_{\mathfrak{a}}({\mathscr{D}}(a),\tilde{\Delta}(b)).$

Eq. (58) is certainly satisfied, thanks to Eqs. (55), (56) and the fact that
${\mathscr{B}}_{\mathfrak{a}}$ is $\tilde{\Delta}$-invariant.

Let us study the condition

$\Delta(s_{\mathfrak{g}}(f))=[f,\Delta(f)]_{\mathfrak{g}}\text{ for all
$f\in{\mathfrak{g}}_{\bar{1}}$}.$

The LHS (for $f=rx+a+tx^{*}$):