# Hom-Lie superalgebras in characteristic 2

Sofiane Bouarroudj  Division of Science and Mathematics, New York University
Abu Dhabi, Po Box 129188, Abu Dhabi, United Arab Emirates.
<EMAIL_ADDRESS>and  Abdenacer Makhlouf  IRIMAS-Mathematics
Department, University of Haute Alsace, Mulhouse, France.
<EMAIL_ADDRESS>
(Date: August 27, 2024, )

###### Abstract.

The main goal of this paper is to develop the structure theory of Hom-Lie
superalgebras in characteristic 2. We discuss their representations,
semidirect product, $\alpha^{k}$-derivations and provide a classification in
low dimension. We introduce another notion of restrictedness on Hom-Lie
algebras in characteristic 2, different from one given by Guan and Chen. This
definition is inspired by the process of queerification of restricted Lie
algebras in characteristic 2. We also show that any restricted Hom-Lie algebra
in characteristic 2 can be queerified to give rise to a Hom-Lie superalgebra.
Moreover, we develop a cohomology theory of Hom-Lie superalgebras in
characteristic 2, which provides a cohomology of ordinary Lie superalgebras.
Furthermore, we establish a deformation theory of Hom-Lie superalgebras in
characteristic 2 based on this cohomology.

Keywords. Hom-Lie superalgebra, modular Lie superalgebra, characteristic 2,
representation, queerification, cohomology, deformation.

###### Key words and phrases:

Hom-Lie superalgebra; Characteristic 2; Deformation

###### 2020 Mathematics Subject Classification:

17B61, 17B05, 17A70

SB was supported by the grant NYUAD-065.

###### Contents

  1. 1 Introduction
    1. 1.1 Lie superalgebras in characteristic 2
    2. 1.2 Hom-Lie superalgebras in characteristic 2
    3. 1.3 The main results
  2. 2 Backgrounds and main definitions
    1. 2.1 Lie superalgebras in characteristic 2.
    2. 2.2 Hom-Lie algebras in characteristic 2
    3. 2.3 Hom-Lie superalgebras in characteristic 2
    4. 2.4 The classification in low dimensions
      1. 2.4.1 The case $\mathrm{sdim}(\mathfrak{g})=1|1$
      2. 2.4.2 The case $\mathrm{sdim}(\mathfrak{g})=1|2$
  3. 3 Representations and semidirect product
  4. 4 $\alpha^{k}$-Derivations
  5. 5 $p$-structures and queerification of Hom-Lie algebras in characteristic 2
  6. 6 Cohomology and Deformations of finite dimensional Hom-Lie superalgebras
    1. 6.1 Cohomology of ordinary Lie superalgebras in characteristic 2
    2. 6.2 Elucidation for $n=2,3$
    3. 6.3 Cohomology of Hom-Lie superalgebras in characteristic 2
    4. 6.4 Deformations of Hom-Lie superalgebras

##  1\. Introduction

Throughout the text, $\mathbb{K}$ stands for an arbitrary field of
characteristic 2. In almost all our constructions, $\mathbb{K}$ is arbitrary.
There are a few instances where $\mathbb{K}$ is required to be infinite. We
will point out at these instances.

###  1.1. Lie superalgebras in characteristic 2

Roughly speaking, a Lie superalgebra in characteristic 2 is a
$\mathbb{Z}/2$-graded vector space that has a Lie algebra structure on the
even part, and endowed with a squaring on the odd part that satisfies a
modified Jacobi identity, see §2.1 for a precise definition. Because we are in
characteristic 2, those Lie superalgebras are sometimes being confused with
${\mathbb{Z}}/2$-graded Lie algebras, though they are totally different
algebras due to the presence of the squaring. They can, however, be considered
as $\mathbb{Z}/2$-graded Lie algebra by forgetting the super structure. The
other way round is not always true in general.

The classification of simple Lie superalgebras in characteristic 2 is still an
open and wide problem. Nevertheless, Lie superalgebras in characteristic 2
admitting a Cartan matrix have been classified in [BGL], with the following
assumption: each Lie superalgebra posses a Dynkin diagram with only one odd
node. The list of non-equivalent Cartan matrices for each Lie superalgebra is
also listed in [BGL]. Moreover, it has recently been showed in [BLLSq] that
each finite-dimensional simple Lie superalgebra in characteristic 2 can be
obtained from a simple finite-dimensional Lie algebra in characteristic 2 by
one of two methods, hence reducing the classification to the classification of
simple Lie algebras which in its own is a very tough problem. As a matter of
fact, there are plenty of (vectorial and non-vectorial) Lie superalgebras in
characteristic 2 that have no analogue in other characteristics, see [BeBou,
BGLLS, BLLSq, BLS, LeD].

It is worth mentioning that the characteristic 2 case is a very tricky case,
due to the presence of the squaring. It does require new ideas and techniques.

###  1.2. Hom-Lie superalgebras in characteristic 2

The first instances of Hom-type algebras appeared in physics literature, see
for example [CKL], where $q$-deformations of some Lie algebras of vector
fields led to a structure in which Jacobi identity is no longer satisfied.
This class of algebras were formalized and studied in [HLS, LS], where they
were called Hom-Lie algebras since the Jacobi identity is twisted by a
homomorphism. The super case were considered in [AM], where Hom-Lie
superalgebras were introduced as a $\mathbb{Z}/2$-graded generalization of the
Hom-Lie algebras. The authors of [AM] characterized Hom-Lie admissible
superalgebras and proved a $\mathbb{Z}/2$-graded version of a Hartwig-Larsson-
Silvestrov Theorem which led to a construction of a $q$-deformed Witt
superalgebra using $\sigma$-derivations. Moreover, they derived a one
parameter family of Hom-Lie superalgebras deforming the orthosymplectic Lie
superalgebra $\mathfrak{osp}(1|2)$. The cohomology of Hom-Lie superalgebras
was defined in [AMS]. Notice that all these studies and results were performed
over a field of characteristic 0.

###  1.3. The main results

The main purpose of this this paper is to tackle the positive characteristic
and provide a study of Hom-Lie superalgebras in characteristic 2. We introduce
the main definitions and some key constructions as well as a cohomology theory
fitting with a deformation theory. In Section 2, we recall some basic
definitions, and introduce Hom-Lie algebras and Hom-Lie superalgebras over
fields of characteristic 2 and some related structures. We show that a Lie
superalgebra in characteristic 2 and an even Lie algebra morphism give rise to
a Hom-Lie superalgebra in characteristic 2. Moreover, we provide a
classification of Hom-Lie superalgebras in characteristic 2 in low dimensions.
In Section 3, we consider representations and semidirect product of Hom-Lie
superalgebras in characteristic 2. The structure map defining a Hom-Lie
superalgebra in characteristic 2 allows a new type of derivations called
$\alpha^{k}$-derivations discussed in Section 4. In Section 5, we introduce
the notion of $p$-structure and discuss queerification of restricted Hom-Lie
algebras in characteristic 2. Section 6 is dedicated to cohomology theory. We
construct a cohomology complex of a Hom-Lie superalgebra $\mathfrak{g}$ in
characteristic 2 with values in a $\mathfrak{g}$-module. This cohomology
complex has no analogue in characteristic $p\not=2$. In the last section, we
provide a deformation theory of Hom-Lie superalgebras in characteristic 2
using the cohomology we constructed.

##  2\. Backgrounds and main definitions

Let $V$ and $W$ be two vector spaces over $\mathbb{K}$. A map $s:V\rightarrow
W$ is called a squaring if

(1) $s(\lambda x)=\lambda^{2}s(x)\quad\text{ for all $\lambda\in\mathbb{K}$
and for all $x\in V$.}$

###  2.1. Lie superalgebras in characteristic 2.

Following [LeD], 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 an ordinary 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 $[\cdot,\cdot]$. For any $x,y\in\mathfrak{g}_{\bar{1}}$, their bracket
is then defined by

$[x,y]:=s(x+y)-s(x)-s(y).$

The bracket is extended to non-homogeneous elements by bilinearity. The Jacobi
identity involving the squaring reads as follows:

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

Such a Lie superalgebra in characteristic 2 will be denoted by
$(\mathfrak{g},[\cdot,\cdot],s)$.

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

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

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

(2) $\displaystyle D([x,y])$ $\displaystyle=$
$\displaystyle[D(x),y]+[x,D(y)]\quad\text{for any $x\in\mathfrak{g}_{\bar{0}}$
and $y\in\mathfrak{g}$ we have}$ (3) $\displaystyle D(s(x))$ $\displaystyle=$
$\displaystyle[D(x),x]\quad\text{for any $x\in\mathfrak{g}_{\bar{1}}$}.$

It is worth noticing that condition (3) implies condition (2) if
$x,y\in\mathfrak{g}_{\bar{1}}$.

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

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ and
$(\mathfrak{h},[\cdot,\cdot]_{\mathfrak{h}},s_{\mathfrak{h}})$ be two Lie
superalgebras in characteristic 2. An even linear map
$\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$ is called a morphism (of Lie
superalgebras) if, in addition to

$\displaystyle\varphi([x,y]_{\mathfrak{g}})$ $\displaystyle=$
$\displaystyle[\varphi(x),\varphi(y)]_{\mathfrak{h}}\quad\text{for any
$x\in\mathfrak{g}_{\bar{0}}$ and $y\in\mathfrak{g}$, we have}$
$\displaystyle\varphi(s_{\mathfrak{g}}(x))$ $\displaystyle=$ $\displaystyle
s_{\mathfrak{h}}(\varphi(x))\quad\text{for any $x\in\mathfrak{g}_{\bar{1}}$}.$

Therefore, morphisms in the category of Lie superalgebras in characteristic 2
preserve not only the bracket but the squaring as well. In particular,
subalgebras and ideals have to be stable under the bracket and the squaring.

An even linear map $\rho:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ is a
representation of the Lie superalgebra $(\mathfrak{g},[\cdot,\cdot],s)$ in the
superspace $V$ called $\mathfrak{g}$-module if

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

######  Remark 2.1.

Associative superalgebras in characteristic 2 leads to Lie superalgebras in
characteristic 2. The bracket is standard and the squaring is defined by
$s(x)=x\cdot x,$ for every odd element $x$.

###  2.2. Hom-Lie algebras in characteristic 2

A Hom-Lie algebra in characteristic 2 is a vector space $\mathfrak{g}$ over
$\mathbb{K}$ and a map $\alpha\in\mathrm{End}(\mathfrak{g})$ together with a
bracket satisfying the following conditions:

$\begin{array}[]{c}[x,x]=0,\;\;\;\alpha[x,y]=[\alpha(x),\alpha(y)]\;\;\text{
and }\;\;[\alpha(x),[y,z]]+\circlearrowleft(x,y,z)=0,\quad\text{for all
$x,y,z\in\mathfrak{g}$.}\end{array}$

Such a Hom-Lie algebra will be denoted by
$(\mathfrak{g},[\cdot,\cdot],\alpha)$.

A representation of a Hom-Lie algebra
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},\alpha)$ is a triplet
$(V,[\cdot,\cdot]_{V},\beta)$ where $V$ is a vector space,
$\beta\in\mathfrak{gl}(V)$, and $[\cdot,\cdot]_{V}$ is the action of
$\mathfrak{g}$ on $V$ such that (for all $x,y\in\mathfrak{g}$ and for all
$v\in V$):

(5)
$\begin{array}[]{l}[\alpha(x),\beta(v)]_{V}=\beta([x,v]_{V}),\quad[[x,y]_{\mathfrak{g}},\beta(v)]_{V}=[\alpha(x),[y,v]_{V}]_{V}+[\alpha(y),[x,v]_{V}]_{V}.\end{array}$

Writing Eq. (5) using the notation of Eq. (4), we put
$\rho_{\beta}:=[\cdot,\cdot]_{V}$ and obtain (for all $x,y\in\mathfrak{g}$):

(6)
$\begin{array}[]{l}\rho_{\beta}(\alpha(x))\circ\beta=\beta\circ\rho(x),\quad\rho([x,y]_{\mathfrak{g}})\circ\beta=\rho(\alpha(x))\rho(y)+\rho(\alpha(y))\rho(x).\end{array}$

###  2.3. Hom-Lie superalgebras in characteristic 2

Our main definition is given below. Due to the presence of the squaring, our
approach to define Hom-Lie superalgebras in characteristic 2 will differ from
that used in characteristics $p\not=2$, see [AM].

######  Definition 2.2.

A Hom-Lie superalgebra in characteristic 2 is a quadruple
$(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ consisting of a $\mathbb{Z}/2$-graded
superspace $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$
over $\mathbb{K}$, a symmetric bracket $[\cdot,\cdot]$, a squaring
$s:\mathfrak{g}_{\bar{1}}\rightarrow\mathfrak{g}_{\bar{0}}$, and an even map
$\alpha\in\mathrm{End}(\mathfrak{g})$ such that

  1. (i)

$(\mathfrak{g}_{\bar{0}},[\cdot,\cdot],\alpha|_{\mathfrak{g}_{\bar{0}}})$ is
an ordinary Hom-Lie algebra,

  2. (ii)

$\mathfrak{g}_{\bar{1}}$ is a $\mathfrak{g}_{\bar{0}}$-module made two-sided
by symmetry, where the action is still denoted by the bracket $[\cdot,\cdot]$.

  3. (iii)

the map

(7)
$\mathfrak{g}_{\bar{1}}\times\mathfrak{g}_{\bar{1}}\rightarrow\mathfrak{g}_{\bar{0}}\quad(x,y)\mapsto
s(x+y)+s(x)+s(y)$

is bilinear, and induces the bracket on odd elements; namely, for any
$x,y\in\mathfrak{g}_{\bar{1}}$:

$[x,y]:=s(x+y)+s(x)+s(y),$

  4. (iv)

the following three conditions hold

(8) $\displaystyle[s(x),\alpha(y)]$ $\displaystyle=$
$\displaystyle[\alpha(x),[x,y]]\;\text{ for any $x\in\mathfrak{g}_{\bar{1}}$
and $y\in\mathfrak{g}$},$ (9) $\displaystyle\alpha([x,y])$ $\displaystyle=$
$\displaystyle[\alpha(x),\alpha(y)]\;\text{ for any }x,y\in\mathfrak{g},$ (10)
$\displaystyle\alpha(s(x))$ $\displaystyle=$ $\displaystyle
s(\alpha(x))\;\text{ for any }x\in\mathfrak{g}_{\bar{1}}.$

######  Remark 2.3.

(i) The Jacobi identity on triples in
$\\{\mathfrak{g}_{0},\mathfrak{g}_{1},\mathfrak{g}_{1}\\}$ and
$\\{\mathfrak{g}_{1},\mathfrak{g}_{1},\mathfrak{g}_{1}\\}$ follow from
condition 8. We, therefore, recover the usual definition of Hom-Lie
superalgebras [AM].

(ii) Since we work over of field of characteristic 2, skewsymmetry and
symmetry coincide since $-1\equiv 1\,(\mod 2)$.

(iii) We may want to consider Hom-Lie superalgebras in characteristic 2
without conditions (9) and (10), which corresponds to the multiplicativity of
the structure map $\alpha$.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ and
$(\mathfrak{g}^{\prime},[\cdot,\cdot]_{\mathfrak{g}^{\prime}},s_{\mathfrak{g}^{\prime}},\alpha^{\prime})$
be two Hom-Lie superalgebras in characteristic 2. A map
$\phi:\mathfrak{g}\rightarrow\mathfrak{g}^{\prime}$ is a morphism of Hom-Lie
superalgebras if the following conditions are satisfied:

(11)
$\phi([\cdot,\cdot]_{\mathfrak{g}})=[\phi(\cdot),\phi(\cdot)]_{\mathfrak{g}^{\prime}},\quad\phi\circ
s_{\mathfrak{g}}=s_{\mathfrak{g}^{\prime}}\circ\phi,\quad\phi\circ\alpha=\alpha^{\prime}\circ\phi.$

Two Hom-Lie superalgebras
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ and
$(\mathfrak{g}^{\prime},[\cdot,\cdot]_{\mathfrak{g}^{\prime}},s_{\mathfrak{g}^{\prime}},\alpha^{\prime})$
are called _isomorphic_ if there exists a homomorphism
$\phi:\mathfrak{g}\rightarrow\mathfrak{g}^{\prime}$ as in (11) that it is
bijective.

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra in
characteristic 2. Let $I$ be a subset of $\mathfrak{g}$. The set $I$ is called
an ideal of $\mathfrak{g}$ if and only if $I$ is closed under addition and
scalar multiplication, together with

$[I,\mathfrak{g}]\subseteq I,\;\alpha(I)\subseteq I\text{ and $s(x)\in I$
whenever $x\in I\cap\mathfrak{g}_{\bar{1}}$.}$

In particular, if the ideal $I$ is homogeneous; namely
$I=I\cap\mathfrak{g}_{\bar{0}}\oplus
I\cap\mathfrak{g}_{\bar{1}}=I_{\bar{0}}\oplus I_{\bar{1}}$ then the condition
involving the squaring reads $s(x)\in I_{\bar{0}}$ for all $x\in I_{\bar{1}}$.
In addition, the superspace $\mathfrak{g}/I$ is also a Hom-Lie superalgebra in
characteristic 2. The bracket and the squaring are defined as follows:

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

while the twist map $\tilde{\alpha}$ on $\mathfrak{g}/I$ is defined by

$\tilde{\alpha}(x+I)=\alpha(x)+I\quad\text{for all $x\in\mathfrak{g}$}.$

We will only show that the squaring is well-defined. Suppose that
$\tilde{x}-x=i\in I_{\bar{1}}$ we have,

$s(\tilde{x})=s(x+i)=s(x)+s(i)+[x,i]=s(x)\mod(I).$

In the following proposition, we will show that an ordinary Lie superalgebra
together with a morphism give rise to a Hom-Lie superalgebra structure on the
underlying vector space.

######  Proposition 2.4.

Let $(\mathfrak{g},[\cdot,\cdot],s)$ be a Lie superalgebra in characteristic
2, and let $\alpha:\mathfrak{g}\rightarrow\mathfrak{g}$ be an even
superalgebra morphism. Then
$(\mathfrak{g},[\cdot,\cdot]_{\alpha},s_{\alpha},\alpha)$, where
$[\cdot,\cdot]_{\alpha}=\alpha\circ[\cdot,\cdot]$ and $s_{\alpha}=\alpha\circ
s$, is a Hom-Lie superalgebra in characteristic 2.

###### Proof.

The first part of the proof is given in [AM]. We have to check Eqns. (1) and
(8). Indeed, let $\lambda\in\mathbb{K}$ and let $x\in\mathfrak{g}_{\bar{1}}$.
We have

$\displaystyle s_{\alpha}(\lambda x)=\alpha(s(\lambda
x))=\alpha(\lambda^{2}s(x))=\lambda^{2}\alpha(s(x))=\lambda^{2}s_{\alpha}(x).$

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

$\displaystyle[s_{\alpha}(x),\alpha(y)]_{\alpha}=\alpha([\alpha(s(x)),\alpha(y)])=\alpha^{2}([s(x),y])=\alpha^{2}([x,[x,y]])$
$\displaystyle=\alpha([\alpha(x),\alpha([x,y])])=[\alpha(x),[x,y]_{\alpha}]_{\alpha}.\qed$

More generally, let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie
superalgebra in characteristic 2 and let
$\beta:\mathfrak{g}\rightarrow\mathfrak{g}$ be an even weak superalgebra
morphism (the third condition of (11) is not necessary satisfied). Then
$(\mathfrak{g},[\cdot,\cdot]_{\beta}:=\beta\circ[\cdot,\cdot],s_{\beta}:=\beta\circ
s,\alpha\circ\beta)$ is a Hom-Lie superalgebra in characteristic 2. The proof
is similar to that of Proposition 2.4.

######  Example 2.5.

Consider the ortho-orthogonal Lie superalgebra
$\mathfrak{g}:={\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2)$ (see [BGL,
LeD]) spanned by the even vectors $h,x_{2},y_{2}$ and the odd vectors
$x_{1},y_{1}$ with the non-zero brackets:

$[x_{1},y_{1}]=[x_{2},y_{2}]=h,\quad[h,x_{1}]=x_{1},\quad[h,y_{1}]=y_{1},\quad[x_{2},y_{1}]=x_{1},\quad[y_{2},x_{1}]=y_{1},$

and the squaring

$s(x_{1})=x_{2},\quad s(y_{1})=y_{2}.$

Let us define the map $\alpha$ on the vector space underlying
${\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2)$:

$\begin{array}[]{lll}\alpha(x_{1})=\delta_{1}x_{1}+\delta_{2}y_{1},&\alpha(y_{1})=\varepsilon_{1}x_{1}+\varepsilon_{2}y_{1},&\alpha(x_{2})=\lambda_{1}h+\lambda_{2}x_{2}+\lambda_{3}y_{2},\\\\[5.69054pt]
\alpha(y_{2})=\beta_{1}h+\beta_{2}x_{2}+\beta_{3}y_{2},&\alpha(h)=\gamma_{1}h.&\end{array}$

A direct computation shows that the map $\alpha$ is a morphism of Lie
superalgebras if and only if (where we have put for simplicity
$T:=1+\delta_{2}\varepsilon_{1}+\delta_{1}\varepsilon_{2}$):

$\gamma_{1}=(1+T)^{2},\quad\beta_{1}=\varepsilon_{1}\varepsilon_{2},\quad\beta_{2}=\varepsilon_{1}^{2},\quad\beta_{3}=\varepsilon_{2}^{2},\quad\lambda_{1}=\delta_{1}\delta_{2},\quad\lambda_{2}=\delta_{1}^{2},\quad\lambda_{3}=\delta_{2}^{2};$

together with

(12)
$\varepsilon_{1}\,T=\varepsilon_{1}\,T^{2}=\varepsilon_{2}\,T=\varepsilon_{2}\,T^{2}=\delta_{1}\,T=\delta_{1}\,T^{2}=\delta_{2}\,T=\delta_{2}\,T^{2}=T(1+T)=0.$

The only solutions to Eqns. (12) that do not produce the zero map are given by
$T=0$.

We can, therefore, construct a Hom-Lie superalgebra by means of the map
$\alpha$, depending on three parameters, as in Proposition 2.4. So, we have

$\begin{array}[]{lll}\alpha(x_{1})=\delta_{1}x_{1}+\delta_{2}y_{1},&\alpha(y_{1})=\varepsilon_{1}x_{1}+\varepsilon_{2}y_{1},&\alpha(x_{2})=\delta_{1}\delta_{2}h+\delta_{1}^{2}x_{2}+\delta_{2}^{2}y_{2},\\\\[5.69054pt]
\alpha(y_{2})=\varepsilon_{1}\varepsilon_{2}h+\varepsilon_{1}^{2}x_{2}+\varepsilon_{2}^{2}y_{2},&\alpha(h)=h.&\end{array}$

such that $\left(\begin{array}[]{cc}\varepsilon_{2}&\varepsilon_{1}\\\
\delta_{2}&\delta_{1}\end{array}\right)\in SL_{2}(\mathbb{K})$.

In particular, we have the following Hom-Lie superalgebra in characteristic 2,
which we denote by ${\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2)_{\alpha}$,
defined by the brackets:

$[x_{1},y_{1}]_{\alpha}=[x_{2},y_{2}]_{\alpha}=h,\;[h,x_{1}]_{\alpha}=x_{1},\;[h,y_{1}]_{\alpha}=\varepsilon
x_{1}+y_{1},\;[x_{2},y_{1}]_{\alpha}=x_{1},\;[y_{2},x_{1}]_{\alpha}=\varepsilon
x_{1}+y_{1},$

with the corresponding squaring:

$s(x_{1})=x_{2},\quad s(y_{1})=\varepsilon h+\varepsilon^{2}x_{2}+y_{2},$

and the twist map

$\displaystyle\alpha(x_{1})=x_{1},\quad\alpha(y_{1})=\varepsilon
x_{1}+y_{1},\quad\alpha(x_{2})=x_{2},\quad\alpha(y_{2})=\varepsilon
h+\varepsilon^{2}x_{2}+y_{2},\quad\alpha(h)=h,$

where $\varepsilon$ is a parameter in $\mathbb{K}$. We recover the Lie
superalgebra ${\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2)$ for
$\varepsilon=0$.

###  2.4. The classification in low dimensions

Let us assume here that the field $\mathbb{K}$ is infinite (for instance,
algebraically closed). For the classification of Hom-Lie algebras and
superalgebras in low dimensions, see [MS, GSS, GSSc, LL, ORS1, ORS2, R, WZW].

####  2.4.1. The case $\mathrm{sdim}(\mathfrak{g})=1|1$

Assume that $\mathfrak{g}_{\bar{0}}=\mathrm{Span}\\{e\\}$ and
$\mathfrak{g}_{\bar{1}}=\mathrm{Span}\\{f\\}$. We set

$\displaystyle\alpha(e)=\lambda_{1}e,\quad\alpha(f)=\lambda_{2}f,\quad
s_{\mathfrak{g}}(f)=\rho\,e,\quad[e,e]=0,\quad[e,f]=\gamma f.$

It follows that $[f,f]=s(2f)-2s(f)=2s(f)=2\rho e=0.$ By straightforward
computations on the conditions, one gets that the only non-trivial case is
given by $\lambda_{1}=1$ and $\gamma=\rho\lambda_{2}$. Therefore, any
$(1|1)$-dimensional Hom-Lie superalgebra in characteristic 2 is isomorphic to
the Hom-Lie superalgebra given, with respect to basis $\\{e,f\\}$, by

$\displaystyle\alpha(e)=e,\quad\alpha(f)=\lambda f,\quad
s_{\mathfrak{g}}(f)=\rho\,e,$ $\displaystyle[e,e]=0,\quad[e,f]=\rho\lambda
f,\quad[f,f]=0,$

where $\lambda$ and $\rho$ are non-zero parameters. As the field $\mathbb{K}$
is infinite, we have a family of Hom-Lie superalgebras that depends on
parameters $\lambda$ and $\rho$.

####  2.4.2. The case $\mathrm{sdim}(\mathfrak{g})=1|2$

Assume that $\mathfrak{g}_{\bar{0}}=\mathrm{Span}\\{e\\}$ and
$\mathfrak{g}_{\bar{1}}=\mathrm{Span}\\{f_{1},f_{2}\\}$. We define the
brackets as (where $a_{i},b_{i}\in\mathbb{K}$ for $i,j=1,2$):

$\displaystyle[e,f_{1}]=a_{1}f_{1}+a_{2}f_{2},\quad[e,f_{2}]=b_{1}f_{1}+b_{2}f_{2},$

and finally the squaring as (where $\rho_{i}\in\mathbb{K}$ for $i=1,2,3$):

$s(f_{1})=\rho_{1}\,e,\quad s(f_{2})=\rho_{2}\,e,\quad
s(f_{1}+f_{2})=\rho_{3}\,e.$

Let us consider a linear map $\alpha$ by which we will construct the Hom-
structure. As $\alpha$ preserves the $\mathbb{Z}/2\mathbb{Z}$-grading, and by
using the Jordan decomposition we distinguish two cases:

Case 1: Suppose that $\alpha$ is given by (where
$s,t_{1},r_{2}\in\mathbb{K}$):

$\displaystyle\alpha(e)=se,\quad\alpha(f_{1})=t_{1}f_{1},\quad\quad\alpha(f_{2})=r_{2}f_{2}.$

A direct computation shows that there are only the following sub-cases:

Sub-case 1a): We have
$\rho_{3}=\rho_{1}+\rho_{2},\;\rho_{1}\not=0,\;s=t_{1}^{2},\;\rho_{2}\,(s+r_{2}^{2})=0$
and $a_{i}=b_{i}=0$ for $i=1,2$. Here are the two possible cases:

$\begin{array}[]{l}\rho_{3}=\rho_{1}+\rho_{2},\;\rho_{1}\not=0,\;\rho_{2}=0,\;s=t_{1}^{2},\;a_{1}=b_{1}=a_{2}=b_{2}=0,\;r_{2}\text{
arbitrary};\text{ or }\\\\[5.69054pt]
\rho_{3}=\rho_{1}+\rho_{2},\;\rho_{1},\rho_{2}\not=0,\;s=t_{1}^{2},\;t_{1}=r_{2},\;a_{1}=b_{1}=a_{2}=b_{2}=0.\end{array}$

Sub-case 1b): We have $\rho_{1}=\rho_{2}=\rho_{3}=0$ together with

$b_{1}(t_{1}+sr_{2})=0,\;b_{2}r_{2}(1+s)=0,\;a_{1}t_{1}(1+s)=0,\;a_{2}(r_{2}+st_{1})=0.$

We can disregard this case, because it produces a Lie algebra instead of a Lie
superalgebra.

Sub-case 1c): We have $\rho_{1}+\rho_{2}+\rho_{3}\not=0,\;\rho_{1}\not=0,$
together with

$s=t_{1}^{2}=t_{1}r_{2},\;\rho_{2}t_{1}^{2}=r_{2}^{2}\rho_{2},\;a_{1}=b_{1}=a_{2}=b_{2}=0.$

Here are the two possible cases:

$\begin{array}[]{l}\rho_{1}+\rho_{2}+\rho_{3}\not=0,\;\rho_{1}\not=0,\;\rho_{2}=0,\;s=t_{1}^{2}=t_{1}r_{2},\;r_{2}\not=0,\;a_{1}=b_{1}=a_{2}=b_{2}=0;\text{
or }\\\\[5.69054pt]
\rho_{1}+\rho_{2}+\rho_{3}\not=0,\;\rho_{1},\rho_{2}\not=0,\;s=t_{1}^{2},\;t_{1}=r_{2}\not=0,\;a_{1}=b_{1}=a_{2}=b_{2}=0.\end{array}$

Case 2: Suppose that $\alpha$ is given by (where $s,t_{1}\in\mathbb{K}$):

$\displaystyle\alpha(e)=se,\quad\alpha(f_{1})=t_{1}f_{1},\quad\quad\alpha(f_{2})=f_{1}+t_{1}f_{2}.$

A direct computation shows that there are only the following sub-cases:

Subcase 2a): We have $\rho_{1},\rho_{2}\not=0$ but $\rho_{3}$ arbitrary,
together with

$a_{1}=a_{2}=b_{1}=b_{2}=0,\;s=t_{1}^{2},\;\rho_{1}(1+t_{1})=t_{1}(\rho_{2}+\rho_{3}).$

Subcase 2b): We have $\rho_{1},\rho_{3}\not=0$ but $\rho_{2}=0$ arbitrary,
together with

$a_{1}=a_{2}=b_{1}=b_{2}=0,\;s=t_{1}^{2},\;\rho_{1}(1+t_{1})=t_{1}\rho_{3}.$

Subcase 2c): We have $\rho_{1}\not=0$ but $\rho_{2}=\rho_{3}=0$, together with

$a_{1}=a_{2}=b_{1}=b_{2}=0,\;s=1,\;t_{1}=1.$

Subcase 2d): We have $\rho_{1}=0,\;\rho_{2}\not=0$ but $\rho_{3}$ arbitrary,
together with

$a_{1}=a_{2}=b_{1}=b_{2}=0,\;s=t_{1}^{2},\;t_{1}(\rho_{2}+\rho_{3})=0.$

Subcase 2e): We have $\rho_{1}=\rho_{2}=0,$ but $\rho_{3}\not=0$, together
with

$a_{1}=a_{2}=b_{1}=b_{2}=0,\;s=0,\;t_{1}=0.$

The tables below summarize our finding. We find it convenient to order the
Hom-Lie superalgebras into two groups: (i) of type I are those for which the
$\mathfrak{g}_{0}$-module structure on $\mathfrak{g}_{\bar{1}}$ is trivial;
(ii) of type II are those for which the $\mathfrak{g}_{0}$-module structure on
$\mathfrak{g}_{\bar{1}}$ is not trivial.

The HLSA | The squaring $s$ | The conditions
---|---|---
$A_{1}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&0,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\rho_{1}e\end{array}$ | $\begin{array}[]{l}\rho_{1}\not=0,\;s=t_{1}^{2},\\\\[2.84526pt] r_{2}\text{ arbitrary}\end{array}$
$A_{2}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&\rho_{2}e,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&(\rho_{1}+\lambda^{2}\rho_{2})e\end{array}$ | $\begin{array}[]{l}\rho_{1},\rho_{2}\not=0,\\\\[2.84526pt] s=t_{1}^{2},r_{2}=t_{1}\end{array}$
$A_{3}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&0,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&((1+\lambda)\rho_{1}+\lambda\rho_{3})e\end{array}$ | $\begin{array}[]{l}\rho_{1}\not=0,\;\rho_{1}+\rho_{3}\not=0,\\\\[2.84526pt] s=t_{1}^{2}=t_{1}r_{2},\;r_{2}\not=0\end{array}$
$A_{4}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&\rho_{2}e,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\lambda(\rho_{1}+(1+\lambda)\rho_{2}+\rho_{3})e\\\\[2.84526pt] &&+\rho_{1}e\end{array}$ | $\begin{array}[]{l}\rho_{1},\rho_{2}\not=0,\;\rho_{1}+\rho_{2}+\rho_{3}\not=0,\\\\[2.84526pt] s=t_{1}^{2},\;t_{1}=r_{2},\;r_{2}\not=0\end{array}$
Table 1. Type I (i.e. $[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]=\\{0\\}$) with $\alpha(e)=se,\;\alpha(f_{1})=t_{1}f_{1},\;\alpha(f_{2})=r_{2}f_{2}$ The HLSA | The squaring $s$ | The conditions
---|---|---
$A_{5}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&\rho_{2}f_{2},\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\lambda(\rho_{1}+(1+\lambda)\rho_{2})e\\\\[2.84526pt] &&+(\lambda\rho_{3}+\rho_{1})e\end{array}$ | $\begin{array}[]{l}\rho_{1},\rho_{2}\not=0,\;\displaystyle\rho_{3}=\frac{1+t_{1}}{t_{1}}\rho_{1}+\rho_{2},\\\\[2.84526pt] s=t_{1}^{2}\end{array}$
$A_{6}$ | $\begin{array}[]{lcllcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&0,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&(\lambda(\rho_{1}+\rho_{3})+\rho_{1})e&&&\end{array}$ | $\begin{array}[]{l}\rho_{1},\rho_{3}\not=0,\;\displaystyle\rho_{3}=\frac{1+t_{1}}{t_{1}}\rho_{1},\\\\[2.84526pt] s=t_{1}^{2}\end{array}$
$A_{7}$ | $\begin{array}[]{lcl}s(f_{1})&=&\rho_{1}e,\\\\[2.84526pt] s(f_{2})&=&0,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\rho_{1}(1+\lambda)e\end{array}$ | $\begin{array}[]{l}\rho_{1}\not=0,\\\\[2.84526pt] s=t_{1}=1\end{array}$
$A_{8}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\\[2.84526pt] s(f_{2})&=&\rho_{2}e,\\\ s(f_{1}+\lambda f_{2})&=&\lambda(\rho_{2}+\rho_{3}+\lambda\rho_{2})e\end{array}$ | $\begin{array}[]{l}\rho_{2}\not=0,\;\rho_{3}\text{ arbitrary}\\\\[2.84526pt] s=t_{1}=0,\end{array}$
$A_{9}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\\[2.84526pt] s(f_{2})&=&\rho_{2}e,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\lambda^{2}\rho_{2}e\end{array}$ | $\begin{array}[]{l}\rho_{2}\not=0,\\\\[2.84526pt] s=t_{1}^{2},\;t_{1}\not=0\end{array}$
$A_{10}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\\[2.84526pt] s(f_{2})&=&0,\\\\[2.84526pt] s(f_{1}+\lambda f_{2})&=&\lambda\rho_{3}e\end{array}$ | $\begin{array}[]{l}\rho_{3}\not=0,\\\\[2.84526pt] s=t_{1}=0\end{array}$
Table 2. Type I (i.e. $[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]=\\{0\\}$) with $\alpha(e)=se,\;\alpha(f_{1})=t_{1}f_{1},\;\alpha(f_{2})=f_{1}+t_{1}f_{2}$ The HLSA | The squaring $s$ | $[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]$ | The conditions
---|---|---|---
$B_{1}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{1}f_{1}+a_{2}f_{2},\\\\[2.84526pt] [e,f_{2}]&=&b_{1}f_{1}+b_{2}f_{2}\end{array}$ | $\begin{array}[]{l}s=1,\\\\[2.84526pt] t_{1}=r_{2}\\\\[2.84526pt] a_{1},a_{2},b_{1},b_{2}\text{ arbitrary}\end{array}$
$B_{2}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{1}f_{1},\\\\[2.84526pt] [e,f_{2}]&=&b_{2}f_{2}\end{array}$ | $\begin{array}[]{l}s=1,\\\\[2.84526pt] t_{1}\not=r_{2}\\\\[2.84526pt] a_{1},b_{2}\text{ arbitrary}\end{array}$
$B_{3}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{1}f_{1}+a_{2}f_{2},\\\\[2.84526pt] [e,f_{2}]&=&b_{1}f_{1}+b_{2}f_{2}\end{array}$ | $\begin{array}[]{l}s\not=0,1,\\\\[2.84526pt] t_{1}=r_{2}=0\\\\[2.84526pt] a_{1},a_{2},b_{1},b_{2}\text{ arbitrary}\end{array}$
$B_{4}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&0,\\\\[2.84526pt] [e,f_{2}]&=&b_{2}f_{2}\end{array}$ | $\begin{array}[]{l}s\not=0,1,\\\\[2.84526pt] t_{1}\not=0,r_{2}=0\\\\[2.84526pt] b_{2}\text{ arbitrary}\end{array}$
$B_{5}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&0,\\\\[2.84526pt] [e,f_{2}]&=&b_{1}f_{1}\end{array}$ | $\begin{array}[]{l}s\not=0,1,\\\\[2.84526pt] t_{1}=rs_{2},\\\\[2.84526pt] b_{1}\not=0\text{ and arbitrary}\end{array}$
$B_{6}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{2}f_{2},\\\\[2.84526pt] [e,f_{2}]&=&0\end{array}$ | $\begin{array}[]{l}s\not=0,1,\\\\[2.84526pt] t_{1}\not=rs_{2},\\\\[2.84526pt] a_{2}\not=0\text{ and arbitrary}\end{array}$
$B_{7}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{2}f_{2},\\\\[2.84526pt] [e,f_{2}]&=&b_{2}f_{2}\end{array}$ | $\begin{array}[]{l}s\not=0,1,\\\\[2.84526pt] r_{2}=0,\\\\[2.84526pt] a_{2},b_{2}\text{ arbitrary}\end{array}$
Table 3. Type II (i.e. $[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]\not=\\{0\\}$) with $\alpha(e)=se,\;\alpha(f_{1})=t_{1}f_{1},\;\alpha(f_{2})=r_{2}f_{2}$ The HLSA | The squaring $s$ | $[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]$ | The conditions
---|---|---|---
$B_{8}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{1}f_{1},\\\\[2.84526pt] [e,f_{2}]&=&b_{1}f_{1}+a_{1}f_{2}\end{array}$ | $\begin{array}[]{l}s=1,\\\\[2.84526pt] t_{1}=0\\\\[2.84526pt] a_{1},b_{1}\text{ arbitrary}\end{array}$
$B_{9}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&0,\\\\[2.84526pt] [e,f_{2}]&=&b_{1}f_{1}\end{array}$ | $\begin{array}[]{l}s\not=1,\\\\[2.84526pt] t_{1}=0\\\\[2.84526pt] b_{1}\not=0\text{ and arbitrary}\end{array}$
$B_{10}$ | $\begin{array}[]{lcl}s(f_{1})&=&0,\\\ s(f_{2})&=&0,\\\ s(f_{1}+\lambda f_{2})&=&0\end{array}$ | $\begin{array}[]{lcl}[e,f_{1}]&=&a_{1}f_{1},\\\\[2.84526pt] [e,f_{2}]&=&a_{1}f_{2}\end{array}$ | $\begin{array}[]{l}s=1,\\\\[2.84526pt] t_{1}\not=0\\\\[2.84526pt] a_{1}\not=0\text{ and arbitrary}\end{array}$
Table 4. Type II (i.e.
$[\mathfrak{g}_{\bar{0}},\mathfrak{g}_{\bar{1}}]\not=\\{0\\}$) with
$\alpha(e)=se,\;\alpha(f_{1})=t_{1}f_{1},\;\alpha(f_{2})=f_{1}+t_{1}f_{2}$

##  3\. Representations and semidirect product

######  Definition 3.1.

A representation of a Hom-Lie superalgebra
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ is a
triple $(V,[\cdot,\cdot]_{V},\beta)$, where $V$ is a superspace, $\beta$ an
even map in $\mathfrak{gl}(V)$, and $[\cdot,\cdot]_{V}$ is the action of
$\mathfrak{g}$ on $V$ such that

(13) $\begin{array}[]{lcl}[\alpha(x),\beta(v)]_{V}&=&\beta([x,v]_{V})\;\text{
for any $x\in\mathfrak{g}$ and $v\in V$},\\\\[5.69054pt]
[[x,y]_{\mathfrak{g}},\beta(v)]_{V}&=&[\alpha(x),[y,v]_{V}]_{V}+[\alpha(y),[x,v]_{V}]_{V}\;\text{
for any $x,y\in\mathfrak{g}$ and $v\in V$,}\\\\[5.69054pt]
[s_{\mathfrak{g}}(x),\beta(v)]_{V}&=&[\alpha(x),[x,v]_{V}]_{V}\;\text{ for any
$x\in\mathfrak{g}_{\bar{1}}$ and $v\in V$}.\end{array}$

We say that $V$ is a $\mathfrak{g}$-module.

Sometimes it is more convenient to use the notation
$\rho_{\beta}=[\cdot,\cdot]_{V}$ and write:

(14)
$\begin{array}[]{lcl}\rho_{\beta}\circ\alpha(x)&=&\beta\circ\rho_{\beta}(x)\;\text{
for any $x\in\mathfrak{g}$,}\\\\[5.69054pt]
\rho_{\beta}([x,y]_{\mathfrak{g}})\circ\beta&=&\rho_{\beta}(\alpha(x))\rho(y)+\rho_{\beta}(\alpha(y))\rho(x)\;\text{
for any $x,y\in\mathfrak{g}$,}\\\\[5.69054pt] \rho_{\beta}\circ
s_{\mathfrak{g}}(x)\circ\beta&=&\rho_{\beta}(\alpha(x))\circ\rho_{\beta}(x)\;\text{
for any $x\in\mathfrak{g}_{\bar{1}}$}.\end{array}$

######  Theorem 3.2.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ be a
Hom-Lie superalgebra and $(V,[\cdot,\cdot]_{V},\beta)$ be a representation.
With the above notation, we define a Hom-Lie superalgebra structure on the
superspace $\mathfrak{g}\oplus
V=(\mathfrak{g}_{\bar{0}}+V_{\bar{0}})\oplus(\mathfrak{g}_{\bar{1}}+V_{\bar{1}})$,
where the bracket is defined by

$[x+v,y+w]_{\mathfrak{g}\oplus
V}=[x,y]_{\mathfrak{g}}+[x,w]_{V}+[y,v]_{V}\text{ for any
}x,y\in\mathfrak{g}\text{ and }v,w\in V,$

the squaring
$s_{\mathfrak{g}+V}:\mathfrak{g}_{\bar{1}}+V_{\bar{1}}\rightarrow\mathfrak{g}_{\bar{0}}+V_{\bar{0}}$
is defined by

$s_{\mathfrak{g}+V}(x+v)=s_{\mathfrak{g}}(x)+[x,v]_{V}\text{ for any
}x\in\mathfrak{g}_{\bar{1}}\text{ and }v\in V_{\bar{1}},$

and the structure map $\alpha_{\mathfrak{g}\oplus V}:\mathfrak{g}\oplus
V\rightarrow\mathfrak{g}\oplus V$ defined by

$\alpha_{\mathfrak{g}\oplus V}(x+v)=\alpha(x)+\beta(v)\text{ for any
}x\in\mathfrak{g}\text{ and }v\in V.$

The Hom-Lie superalgebra $(\mathfrak{g}\oplus
V,[\cdot,\cdot]_{\mathfrak{g}\oplus
V},s_{\mathfrak{g}+V},\alpha_{\mathfrak{g}\oplus V})$ is called the semidirect
product of
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ by the
representation $(V,[\cdot,\cdot]_{V},\beta)$.

###### Proof.

Checking Axioms (i) and (ii) of Definition 2.2 is a routine; we can refer to
[AM]. We should check the conditions relative to the squaring. Let us first
check that the map $s_{\mathfrak{g}}\oplus V$ is indeed a squaring. For all
$x+v\in\mathfrak{g}_{\bar{1}}\oplus V_{\bar{1}}$ and for all
$\lambda\in\mathbb{K}$, we have

$s_{\mathfrak{g}\oplus V}(\lambda(x+v))=s_{\mathfrak{g}}(\lambda x)+[\lambda
x,\lambda
v]_{V}=\lambda^{2}s_{\mathfrak{g}}(x)+\lambda^{2}[x,v]_{V}=\lambda^{2}s_{\mathfrak{g}\oplus
V}(x+v).$

Now, for all $x+v\in\mathfrak{g}_{\bar{1}}\oplus V_{\bar{1}}$ and for all
$y+w\in\mathfrak{g}\oplus V$, we have

$\displaystyle[s_{\mathfrak{g}\oplus V}(x+v),\alpha_{\mathfrak{g}\oplus
V}(y+w)]_{\mathfrak{g}\oplus
V}=[s_{\mathfrak{g}}(x)+[x,v]_{V},\alpha(y)+\beta(w)]_{\mathfrak{g}\oplus V}$
$\displaystyle=[s_{\mathfrak{g}}(x),\alpha(y)]_{\mathfrak{g}}+[s_{\mathfrak{g}}(x),\beta(w)]_{V}+[\alpha(y),[x,v]_{V}]_{V}$
$\displaystyle=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}+[\alpha(x),[x,w]_{V}]_{V}+[\alpha(y),[x,v]_{V}]_{V}.$

On the other hand,

$\displaystyle[\alpha_{\mathfrak{g}\oplus
V}(x+v),[x+v,y+w]_{\mathfrak{g}\oplus V}]_{\mathfrak{g}\oplus
V}=[\alpha(x)+\beta(v),[x,y]_{\mathfrak{g}}+[x,w]_{V}+[y,v]_{V}]_{\mathfrak{g}\oplus
V}$
$\displaystyle=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}+[\alpha(x),[x,w]_{V}+[y,v]_{V}]_{V}+[[x,y]_{\mathfrak{g}},\beta(v)]_{V}$
$\displaystyle=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}+[\alpha(x),[x,w]_{V}+[y,v]_{V}]_{V}+[\alpha(x),[y,v]_{V}]_{V}+[\alpha(y),[x,v]_{V}]_{V}$
$\displaystyle=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}+[\alpha(x),[x,w]_{V}]_{V}+[\alpha(y),[x,v]_{V}]_{V}.$

Therefore, Eq. (8) is satisfied. Now,

$\displaystyle\alpha_{\mathfrak{g}\oplus V}(s_{\mathfrak{g}\oplus
V}(x+v))=\alpha_{\mathfrak{g}\oplus
V}(s_{\mathfrak{g}}(x)+[x,v]_{V})=\alpha(s_{\mathfrak{g}}(x))+\beta([x,v]_{V})$
$\displaystyle=\alpha(s_{\mathfrak{g}}(x))+[\alpha(x),\beta(v)]_{V}=s_{\mathfrak{g}}(\alpha(x))+[\alpha(x),\beta(v)]_{V}=s_{\mathfrak{g}\oplus
V}(\alpha(x)+\beta(v))$ $\displaystyle=s_{\mathfrak{g}\oplus
V}(\alpha_{\mathfrak{g}\oplus V}(x+v)).$

Therefore, Eq. (10) is satisfied. ∎

In the following proposition, we show how to twist a Lie superalgebra and its
representation into a Hom-Lie superalgebra together with a representation in
characteristic 2.

######  Proposition 3.3.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}})$ be a Lie
superalgebra and $(V,\rho)$ a representation. Let
$\alpha:\mathfrak{g}\rightarrow\mathfrak{g}$ be an even superalgebra morphism
and $\beta\in\mathfrak{gl}(V)$ be a linear map such that
$\rho(\alpha(x))\circ\beta=\beta\circ\rho(x)$. Then
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g},\alpha},s_{\mathfrak{g},\alpha},\alpha)$,
where
$[\cdot,\cdot]_{\mathfrak{g},\alpha}=\alpha\circ[\cdot,\cdot]_{\mathfrak{g}}$
and $s_{\mathfrak{g},\alpha}=\alpha\circ s_{\mathfrak{g}}$, is a Hom-Lie
superalgebra and $(V,\rho_{\beta},\beta)$, where
$\rho_{\beta}=\beta\circ\rho$, is a representation.

###### Proof.

We have already proved in Proposition 2.4 that
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g},\alpha},s_{\mathfrak{g},\alpha},\alpha)$
is a Hom-Lie superalgebra. Let us check that $(V,\rho_{\beta},\beta)$ is a
representation with respect to
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g},\alpha},s_{\mathfrak{g},\alpha},\alpha)$.
Indeed, the first condition is provided by the hypothesis while the second and
the third ones are straightforward. Let us check the last one. For any
$x\in\mathfrak{g}_{\bar{1}}$ and $v\in V$, we have

$\displaystyle[s_{\mathfrak{g},\alpha}(x),\beta(v)]_{V,\beta}=\beta([\alpha(s_{\mathfrak{g}}(x)),\beta(v)]_{V})=\beta^{2}([s_{\mathfrak{g}}(x),v]_{V}),$

and

$\displaystyle[\alpha(x),[x,v]_{V,\beta}]_{V,\beta}=\beta[\alpha(x),\beta([x,v]_{V}]_{V})=\beta^{2}([x,[x,v]_{V}]_{V}).$

The equality follows from the fact that
$[s_{\mathfrak{g}}(x),v]_{V}=[x,[x,v]_{V}]_{V}$. ∎

######  Example 3.4.

The classification of irreducible modules over
$\mathfrak{oo}_{I\Pi}^{(1)}(1|2)$ having highest weight vectors has been
carried out in [BGKL]. We will borrow here the simplest example. Consider the
Hom-Lie superalgebra $\mathfrak{oo}_{I\Pi}^{(1)}(1|2)$ with the twist $\alpha$
given as in Example (2.5). We consider the
$\mathfrak{oo}_{I\Pi}^{(1)}(1|2)$-module $M$ with basis: (even $|$ odd)

$m_{1},m_{3}\quad|\quad m_{2}.$

The vector $m_{1}$ is a highest weight vector with $\text{weight
}(m_{1})=(1)$. The map $\beta$ is given as follows:

$\beta(m_{1})=\delta_{1}m_{1}+\delta_{2}m_{3},\quad\beta(m_{3})=\varepsilon_{1}m_{1}+\varepsilon_{2}m_{3},\quad\beta(m_{2})=m_{2},$

where the coefficients $\delta_{1},\delta_{2},\varepsilon_{1},\varepsilon_{2}$
are given as in Example 2.5.

Here we will introduce another point of view concerning the representations of
Hom-Lie superalgebras in characteristic 2, inspired by [Sh].

Let $V=V_{\bar{0}}\oplus V_{\bar{1}}$ be a vector superspace, and let
$\beta\in GL(V)$ be even map. We will define a bracket on $\mathfrak{gl}(V)$
as well as a product as follows: (where $\beta^{-1}$ is the inverse of
$\beta$):

(15) $\displaystyle[f,g]_{\mathfrak{gl}(V)}$ $\displaystyle:=$
$\displaystyle\beta\circ f\circ\beta^{-1}g\circ\beta^{-1}+\beta\circ
g\circ\beta^{-1}f\circ\beta^{-1}\quad\text{ for all
$f,g\in\mathfrak{gl}(V)$},$ (16) $\displaystyle s_{\mathfrak{gl}(V)}(f)$
$\displaystyle:=$ $\displaystyle\beta\circ
f\circ\beta^{-1}f\circ\beta^{-1}\quad\text{ for all
$f\in\mathfrak{gl}(V)_{\bar{1}}$}.$

Obviously, $s_{\mathfrak{gl}(V)}(\lambda
f)=\lambda^{2}s_{\mathfrak{gl}(V)}(f)$ for all $\lambda\in{\mathbb{K}}$ and
for all $f\in\mathfrak{gl}(V)_{\bar{1}}$. Now, the map

$(f,g)\mapsto
s_{\mathfrak{gl}(V)}(f+g)+s_{\mathfrak{gl}(V)}(f)+s_{\mathfrak{gl}(V)}(g)=\beta\circ
f\circ\beta^{-1}g\circ\beta^{-1}+\beta\circ g\circ\beta^{-1}f\circ\beta^{-1}$

is obviously bilinear on $\mathfrak{gl}(V)_{\bar{1}}$ as well.

Denote by $\mathrm{Ad}_{\beta}:\mathfrak{gl}(V)\rightarrow\mathfrak{gl}(V)$
the adjoint action on $\mathfrak{gl}(V)$, i.e.
$\mathrm{Ad}_{\beta}(f)=\beta\circ f\circ\beta^{-1}$.

######  Proposition 3.5.

The brackets and the squaring defined in Eqns. (15) and (16) make
$(\mathfrak{gl}(V),[\cdot,\cdot]_{\mathfrak{gl}(V)},s_{\mathfrak{gl}(V)},\mathrm{Ad}_{\beta})$
a Hom-Lie superalgebra in characteristic 2.

###### Proof.

The map $\mathrm{Ad}_{\beta}$ is invertible with inverse
$\mathrm{Ad}_{\beta^{-1}}$. Let us check the multiplicativity conditions:

$\displaystyle[\mathrm{Ad}_{\beta}(f),\mathrm{Ad}_{\beta}(g)]_{\mathfrak{gl}(V)}=[\beta\circ
f\circ\beta^{-1},\beta\circ g\circ\beta^{-1}]_{\mathfrak{gl}(V)}$
$\displaystyle=\beta\circ(\beta\circ
f\circ\beta^{-1})\circ\beta^{-1}(\beta\circ
g\circ\beta^{-1})\circ\beta^{-1}+\beta\circ(\beta\circ
g\circ\beta^{-1})\circ\beta^{-1}(\beta\circ f\circ\beta^{-1})\circ\beta^{-1}$
$\displaystyle=\beta\circ(\beta\circ f\circ\beta^{-1}\circ
g\circ\beta^{-1})\circ\beta^{-1}+\beta\circ(\beta\circ g\circ\beta^{-1}\circ
f\circ\beta^{-1})\circ\beta^{-1}$
$\displaystyle=\mathrm{Ad}_{\beta}([f,g]_{\mathfrak{gl}(V)}).$

Similarly,

$\displaystyle
s_{\mathfrak{gl}(V)}(\mathrm{Ad}_{\beta}(f))=s_{\mathfrak{gl}(V)}(\beta\circ
f\circ\beta^{-1})=\beta\circ(\beta\circ
f\circ\beta^{-1})\circ\beta^{-1}\circ(\beta\circ
f\circ\beta^{-1})\circ\beta^{-1}$ $\displaystyle=\beta\circ(\beta\circ
f\circ\beta^{-1}\circ
f\circ\beta^{-1})\circ\beta^{-1}=\mathrm{Ad}_{\beta}(s_{\mathfrak{gl}(V)}).$

For the Jacobi identity, let us just deal with the squaring. The LHS of the JI
reads (for all $f\in\mathfrak{gl}(V)_{\bar{1}}$ and for all
$g\in\mathfrak{gl}(V)$)

$\begin{array}[]{lcl}[s_{\mathfrak{gl}(V)}(f),\mathrm{Ad}_{\beta}(g)]_{\mathfrak{gl}(V)}&=&\beta\circ
s_{\mathfrak{gl}(V)}(f)\circ\beta^{-1}\circ\beta\circ
g\circ\beta^{-1}\circ\beta^{-1}+\\\\[5.69054pt] &&+\beta\circ\beta\circ
g\circ\beta^{-1}\circ\beta^{-1}\circ
s_{\mathfrak{gl}(V)}(f)\circ\beta^{-1}\\\\[5.69054pt]
&=&\beta^{2}\circ(f\circ\beta^{-1}\circ f\circ\beta^{-1}\circ
g+g\circ\beta^{-1}\circ f\circ\beta^{-1}\circ f)\circ\beta^{-2}.\end{array}$

While the RHS reads

$\begin{array}[]{lcl}[\mathrm{Ad}_{\beta}(f),[f,g]_{\mathfrak{gl}(V)}]_{\mathfrak{gl}(V)}&=&\beta^{2}\circ
f\circ\beta^{-2}\circ[f,g]_{\mathfrak{gl}(V)}\circ\beta^{-1}+\beta\circ[f,g]_{\mathfrak{gl}(V)}\circ
f\circ\beta^{-1}\circ\beta^{-1}\\\\[5.69054pt] &=&\beta^{2}\circ
f\circ\beta^{-2}\circ(\beta\circ f\circ\beta^{-1}g\circ\beta^{-1}+\beta\circ
g\circ\beta^{-1}f\circ\beta^{-1})\circ\beta^{-1}\\\\[5.69054pt]
&&+\beta\circ(\beta\circ f\circ\beta^{-1}g\circ\beta^{-1}+\beta\circ
g\circ\beta^{-1}f\circ\beta^{-1})\circ f\circ\beta^{-2}\\\\[5.69054pt]
&=&\beta^{2}\circ(f\circ\beta^{-1}\circ f\circ\beta^{-1}\circ
g+g\circ\beta^{-1}\circ f\circ\beta^{-1}\circ
f)\circ\beta^{-2}.\qed\end{array}$

######  Theorem 3.6.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ be a
Hom-Lie superalgebra in characteristic 2. Let $V$ be a vector superspace, and
let $\beta\in GL(V)$ be even. Then, the map
$\rho_{\beta}:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ is a representation of
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ on $V$
with respect to $\beta$ if and only if the map
$\rho_{\beta}:(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)\rightarrow(\mathfrak{gl}(V),[\cdot,\cdot]_{\mathfrak{gl}(V)},s_{\mathfrak{gl}(V)},\mathrm{Ad}_{\beta})$
is a morphism of Hom-Lie superalgebras.

###### Proof.

Let us only proof one direction. Suppose that
$\rho_{\beta}:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ is a representation of
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ on $V$
with respect to $\beta$. Since
$\rho_{\beta}(\alpha(x))\circ\beta=\beta\circ\rho(x)$, for all
$f\in\mathfrak{g}$, it follows that

$\rho_{\beta}(x)\circ\alpha=\beta\circ\rho(x)\circ\beta^{-1}=\mathrm{Ad}_{\beta}\circ\rho_{\beta}(x).$

Now,

$\begin{array}[]{lcl}\rho_{\beta}([x,y]_{\mathfrak{g}})&=&\rho_{\beta}(\alpha(x))\circ\rho(y)\circ\beta^{-1}+\rho_{\beta}(\alpha(y))\circ\rho(x)\circ\beta^{-1}\\\\[5.69054pt]
&=&\rho_{\beta}(\alpha(x))\circ\beta\circ\beta^{-1}\circ\rho_{\beta}(y)\circ\beta^{-1}+\rho_{\beta}(\alpha(y))\circ\beta\circ\beta^{-1}\circ\rho_{\beta}(x)\circ\beta^{-1}\\\\[5.69054pt]
&=&\beta\circ\rho_{\beta}(x)\circ\beta^{-1}\circ\rho_{\beta}(y)\circ\beta^{-1}+\beta\circ\rho_{\beta}(y)\circ\beta^{-1}\circ\rho_{\beta}(x)\circ\beta^{-1}\\\\[5.69054pt]
&=&[\rho_{\beta}(x),\rho_{\beta}(y)]_{\mathfrak{gl}(V)}\end{array}$

For the squaring, we have

$\begin{array}[]{lcl}\rho_{\beta}(s_{\mathfrak{g}}(x))&=&\rho_{\beta}(\alpha(x))\circ\rho_{\beta}(x)\circ\beta^{-1}\\\\[5.69054pt]
&=&\beta\circ\rho_{\beta}(x)\circ\beta^{-1}\circ\rho_{\beta}(x)\circ\beta^{-1}\\\\[5.69054pt]
&=&s_{\mathfrak{gl}(V)}(\rho_{\beta}(x)).\end{array}$

It follows that $\rho_{\beta}$ is a homomorphism of Hom-Lie superalgebras in
characteristic 2. ∎

######  Corollary 3.7.

Let $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},s_{\mathfrak{g}},\alpha)$ be a
Hom-Lie superalgebra in characteristic 2. Then, the adjoint representation
$\mathrm{ad}:\mathfrak{g}\rightarrow\mathfrak{gl}(\mathfrak{g})$, which is
defined by $\mathrm{ad}_{x}(y)=[x,y]_{\mathfrak{g}}$, is a morphism from
$(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}},\alpha)$ to
$(\mathfrak{gl}(\mathfrak{g}),[\cdot,\cdot]_{\mathfrak{gl}(\mathfrak{g})},s_{\mathfrak{gl}(\mathfrak{g})},\mathrm{Ad}_{\alpha})$.

##  4\. $\alpha^{k}$-Derivations

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra in
characteristic 2. We denote by $\alpha^{k}$ the $k$-times composition of
$\alpha$, where $\alpha^{0}$ is the identity map. We will be needing the
following linear map

(17) $\mathrm{ad}_{\alpha^{s,k}}(x):y\mapsto[\alpha^{s}(x),\alpha^{k}(y)].$

######  Definition 4.1.

A linear map $D:\mathfrak{g}\rightarrow\mathfrak{g}$ is called an
$\alpha^{k}$-derivation of the Hom-Lie superalgebra $\mathfrak{g}$ if

(18) $\displaystyle D\circ\alpha$ $\displaystyle=$ $\displaystyle\alpha\circ
D,\text{ namely $D$ and $\alpha$ commutes.}$ (19) $\displaystyle D([x,y])$
$\displaystyle=$
$\displaystyle[D(x),\alpha^{k}(y)]+[\alpha^{k}(x),D(y)]\quad\text{for any
$x\in\mathfrak{g}_{\bar{0}}$ and $y\in\mathfrak{g}$ }.$ (20) $\displaystyle
D(s(x))$ $\displaystyle=$ $\displaystyle[D(x),\alpha^{k}(x)]\quad\text{for any
$x\in\mathfrak{g}_{\bar{1}}$}.$

######  Remark 4.2.

Notice that condition (20) implies condition (19) if
$x,y\in\mathfrak{g}_{\bar{1}}$.

Let us give an example. Let $x\in\mathfrak{g}$ such that $\alpha(x)=x$. The
linear map $\mathrm{ad}_{\alpha^{0,k}}(x):y\mapsto[x,\alpha^{k}(y)]$ (see Eq.
(17)) is an $\alpha^{k}$-derivation. Let us just check the condition related
to the squaring. Indeed,

$\mathrm{ad}_{\alpha^{0,k}}(x)(s(y))=[x,\alpha^{k}(s(y))]=[x,s(\alpha^{k}(y))]=[[x,\alpha^{k}(y)],\alpha^{k}(y)]=[\mathrm{ad}_{\alpha^{0,k}}(x)(y),\alpha^{k}(y)]$

Let us denote the space of $\alpha^{k}$-derivations by
$\mathfrak{der}^{\alpha}(\mathfrak{g})$. We have the following proposition.

######  Proposition 4.3.

The space $\mathfrak{der}^{\alpha}(\mathfrak{g})$ can be endowed with a Lie
superalgebra structure in characteristic 2. The bracket is the usual
commutator, and the squaring is given by

$s_{\mathfrak{der}^{\alpha}(\mathfrak{g})}(D):=D^{2}\quad\text{ for all
$D\in\mathfrak{der}^{\alpha}_{\bar{1}}(\mathfrak{g})$.}$

###### Proof.

As we did before, we only prove the requirements when the squaring is
involved. Let us first show that $D^{2}$ is an $\alpha^{2k}$-derivation.
Checking the bracket is a routine. For the squaring, we have (for all
$x\in\mathfrak{g}_{\bar{1}}$):

$\displaystyle
D^{2}(s_{\mathfrak{g}}(x))=D([D(x),\alpha^{k}(x)]_{\mathfrak{g}})=[D^{2}(x),\alpha^{k}(\alpha^{k}(x))]_{\mathfrak{g}}+[\alpha^{k}(D(x)),D(\alpha^{k}(x))]_{\mathfrak{g}}$
$\displaystyle=[D^{2}(x),\alpha^{2k}(x)]_{\mathfrak{g}}+[\alpha^{k}(D(x)),\alpha^{k}(D(x))]_{\mathfrak{g}}=[D^{2}(x),\alpha^{2k}(x)]_{\mathfrak{g}}.$

Before we proceed with the proof, let us re-denote the space
$\mathfrak{der}^{\alpha}(\mathfrak{g})$ by $\mathfrak{h}$ for simplicity.

Now, for all $D\in{\mathfrak{h}}_{\bar{1}}$ and for all
$E\in{\mathfrak{h}}_{\bar{1}}$, we have (for all $x\in\mathfrak{g}$):

$[s_{\mathfrak{h}}(D),E]_{\mathfrak{h}}(x)=[D^{2},E]_{\mathfrak{h}}(x)=D^{2}\circ
E(x)+E\circ D^{2}(x).$

On the other hand,

$\displaystyle[D,[D,E]_{\mathfrak{h}}]_{\mathfrak{h}}(x)=[D,D\circ E+E\circ
D]_{\mathfrak{h}}(x)$ $\displaystyle=D\circ(D\circ E+E\circ D)(x)+(D\circ
E+E\circ D)\circ D(x)$ $\displaystyle=D^{2}\circ E(x)+E\circ D^{2}(x).$

Therefore,
$[s_{\mathfrak{h}}(D),E]_{\mathfrak{h}}=[D,[D,E]_{\mathfrak{h}}]_{\mathfrak{h}}$.∎

The space $\mathfrak{der}^{\alpha}(\mathfrak{g})$ is actually graded as
$\mathfrak{der}^{\alpha}(\mathfrak{g})=\oplus\mathfrak{der}^{\alpha}_{k}(\mathfrak{g})$
where $\mathfrak{der}^{\alpha}_{k}(\mathfrak{g})$ is the space of
$\alpha^{k}$-derivations where $k$ is fixed. Indeed, we have

$[\mathfrak{der}^{\alpha}_{k}(\mathfrak{g}),\mathfrak{der}^{\alpha}_{l}(\mathfrak{g})]\subseteq\mathfrak{der}^{\alpha}_{k+l}(\mathfrak{g})\quad\text{
and }\quad
s(\mathfrak{der}^{\alpha}_{k}(\mathfrak{g})_{\bar{1}})\subseteq\mathfrak{der}^{\alpha}_{2k}(\mathfrak{g}).$

######  Example 4.4.

We will describe all $\alpha^{k}$-derivations of the Hom-Lie superalgebra
${\mathfrak{o}}{\mathfrak{o}}_{I\Pi}^{(1)}(1|2)_{\alpha}$ introduced in
Example 2.5. First, observe that

$\alpha^{2k}=\alpha^{0}=\mathrm{Id},\quad\alpha^{2k+1}=\alpha,\quad\text{for
all $k\geq 0$}.$

The case of $\alpha^{0}$-derivations:

$\begin{array}[]{llcl}\text{(Even)}&D_{1}^{0}&=&h_{1}\otimes
y_{2}^{*}+x_{1}\otimes y_{1}^{*},\\\\[5.69054pt]
\text{(Even)}&D_{2}^{0}&=&x_{1}\otimes x_{1}^{*}+y_{1}\otimes
y_{1}^{*},\\\\[5.69054pt] \text{(Odd)}&D_{3}^{0}&=&x_{1}\otimes
h^{*}_{1}+h_{1}\otimes y_{1}^{*}+y_{1}\otimes y_{2}^{*}.\end{array}$

The case of $\alpha$-derivations:

$\begin{array}[]{llcl}\text{(Even)}&D_{1}^{1}&=&h_{1}\otimes
y_{2}^{*}+x_{1}\otimes y_{1}^{*},\\\\[5.69054pt]
\text{(Even)}&D_{2}^{1}&=&\epsilon\,x_{1}\otimes y_{1}^{*}+x_{1}\otimes
x_{1}^{*}+y_{1}\otimes y_{1}^{*},\\\\[5.69054pt]
\text{(Odd)}&D_{3}^{1}&=&\epsilon\,x_{1}\otimes y_{2}^{*}+x_{1}\otimes
h_{1}^{*}+h_{1}\otimes y_{1}^{*}+y_{1}\otimes y_{2}^{*}.\end{array}$

##  5\. $p$-structures and queerification of Hom-Lie algebras in
characteristic 2

We will first introduce the concept of $p$-structures on Hom-Lie algebras. In
the case of Lie algebras, the definition is due to Jacobson [J]. Roughly
speaking, one requires the existence of an endomorphism on the modular Lie
algebra that resembles the pth power mapping $x\mapsto x^{p}$ in associative
algebras. In the case of Hom-Lie algebra, there is a definition proposed in
[GC] but it turns out that this definition is not appropriate to queerify a
restricted Hom-Lie algebras in characteristic two, as done in [BLLSq] in the
case of ordinary restricted Lie algebras. Here, we will give an alternative
definition and justify the construction.

######  Definition 5.1.

Let $\mathfrak{g}$ be a Hom-Lie algebra in characteristic $p$ with a twist
$\alpha$. A mapping
$[p]_{\alpha}:\mathfrak{g}\rightarrow\mathfrak{g},\;a\mapsto a^{[p]_{\alpha}}$
is called a $p$-structure of $\mathfrak{g}$ and $\mathfrak{g}$ is said to be
restricted if

  1. (R1)

$\mathrm{ad}(x^{[p]_{\alpha}})\circ\alpha^{p-1}=\mathrm{ad}(\alpha^{p-1}(x))\circ\mathrm{ad}(\alpha^{p-2}(x))\circ\cdots\circ\mathrm{ad}(x)$
for all $x\in\mathfrak{g}$;

  2. (R2)

$(\lambda x)^{[p]_{\alpha}}=\lambda^{p}x^{[p]_{\alpha}}$ for all
$x\in\mathfrak{g}$ and for all $\lambda\in\mathbb{K}$;

  3. (R3)

$(x+y)^{[p]_{\alpha}}=x^{[p]_{\alpha}}+y^{[p]_{\alpha}}+\displaystyle\sum_{1\leq
i\leq p-1}s_{i}(x,y)$, where the $s_{i}(x,y)$ can be obtained from

$\mathrm{ad}({\alpha^{p-2}}(\lambda
x+y))\circ\mathrm{ad}({\alpha^{p-3}}(\lambda
x+y))\circ\cdots\circ\mathrm{ad}(\lambda x+y)(x)=\sum_{1\leq i\leq
p-1}is_{i}(x,y)\lambda^{i-1}.$

Let us exhibit this $p$-structure in the case where $p=2$. The conditions (R2)
and (R3) read, respectively, as

$[x^{[2]_{\alpha}},\alpha(y)]=[\alpha(x),[x,y]]\;\text{ and
}\;(x+y)^{[2]_{\alpha}}=x^{[2]_{\alpha}}+y^{[2]_{\alpha}}+[x,y].$

######  Proposition 5.2.

Twisting with a morphism $\alpha$ an ordinary Lie algebra with a $p$-structure
gives rise to a Hom-Lie algebra with a $p$-structure. More precisely, given an
ordinary Lie algebra $(\mathfrak{g},[\cdot,\cdot])$ and a Lie algebra morphism
$\alpha$. Then $(\mathfrak{g},[\cdot,\cdot]_{\alpha},\alpha)$, where
$[\cdot,\cdot]_{\alpha}:=\alpha\circ[\cdot,\cdot]$, is a Hom-Lie algebra with
a $p$-structure given by

$x^{[p]_{\alpha}}:=\alpha^{p-1}(x^{[p]}).$

###### Proof.

It has been shown in [Y] that if $(\mathfrak{g},[\cdot,\cdot])$ is an ordinary
Lie algebra, then $(\mathfrak{g},[\cdot,\cdot]_{\alpha})$ where
$[\cdot,\cdot]_{\alpha}:=\alpha\circ[\cdot,\cdot]$, is a Hom-Lie algebra. Now,
let us show that the map $[p]_{\alpha}$ defines a $p$-structure on the Hom-Lie
algebra $(\mathfrak{g},[\cdot,\cdot]_{\mathfrak{g}})$. Indeed, let us check
Axiom (R1). The LHS reads

$\mathrm{ad}(x^{[p]_{\alpha}})\circ\alpha^{p-1}(y)=[x^{[p]_{\alpha}},\alpha^{p-1}(y)]_{\alpha}=\alpha([\alpha^{p-1}(x^{[p]}),\alpha^{p-1}(y)])=\alpha^{p}([x^{[p]},y]).$

The RHS reads

$\displaystyle\mathrm{ad}(\alpha^{p-1}(x))\circ\mathrm{ad}(\alpha^{p-2}(x))\circ\cdots\circ\mathrm{ad}(x)(y)=[\alpha^{p-1}(x),[\alpha^{p-2}(x),\ldots,[x,y]_{\alpha}]_{\alpha}$
$\displaystyle=\alpha([\alpha^{p-1}(x),\alpha([\alpha^{p-2}(x),[\ldots,\alpha([x,y]))])=\alpha^{p}([x,[x,\ldots,[x,y]])=\alpha^{p}([x^{[p]},y]).$

Axiom (R2) is obviously satisfied. Let us check Axiom (R3). Indeed,

$\displaystyle(x+y)^{[p]_{\alpha}}=\alpha^{p-1}((x+y)^{[p]})=\alpha^{p-1}\left(x^{[p]}+y^{[p]}+\sum_{1\leq
i\leq p-2}s_{i}(x,y)\right)$
$\displaystyle=x^{[p]_{\alpha}}+y^{[p]_{\alpha}}+\left(\sum_{1\leq i\leq
p-2}\alpha^{p-1}(s_{i}(x,y))\right).$

Now,

$\displaystyle\mathrm{ad}({\alpha^{p-2}}(\lambda
x+y))\circ\mathrm{ad}({\alpha^{p-3}}(\lambda
x+y))\circ\cdots\circ\mathrm{ad}(\lambda x+y)(x)$
$\displaystyle=[\alpha^{p-2}(\lambda x+y),[\alpha^{p-3}(\lambda
x+y),\ldots,[\lambda x+y,x]_{\alpha}]_{\alpha}$
$\displaystyle=\alpha([\alpha^{p-2}(\lambda x+y),\alpha([\alpha^{p-3}(\lambda
x+y),[\ldots,\alpha([\lambda x+y,x]))])$ $\displaystyle=\alpha^{p-1}([\lambda
x+y,[\lambda x+y,[\ldots,[\lambda x+y,x]])$
$\displaystyle=\alpha^{p-1}\left(\sum_{1\leq i\leq
p-1}is_{i}(x,y)\lambda^{i-1}\right)=\sum_{1\leq i\leq
p-1}i\alpha^{p-1}(s_{i}(x,y))\lambda^{i-1}.$

The proof is now complete. ∎

######  Proposition 5.3.

Let $\mathfrak{g}$ be a restricted Hom-Lie algebra in characteristic 2 with a
twist map $\alpha$. On the superspace
$\mathfrak{h}:=\mathfrak{g}\oplus\Pi(\mathfrak{g})$ there exists a Hom-Lie
superalgebra structure defined as follows (for all $x,y\in\mathfrak{g}$):

$[x,y]_{\mathfrak{h}}:=[x,y]_{\mathfrak{g}},\quad[\Pi(x),y]_{\mathfrak{h}}:=\Pi([x,y]_{\mathfrak{g}}),\quad
s_{\mathfrak{h}}(\Pi(x))=x^{[2]_{\alpha}}.$

###### Proof.

Let us check that the map $s_{\mathfrak{h}}$ is indeed a squaring on
$\mathfrak{h}$. The condition
$s_{\mathfrak{h}}(\lambda\Pi(x))=\lambda^{2}s_{\mathfrak{h}}(\Pi(x))$, for all
$\lambda\in\mathbb{K}$ and for all $x\in\mathfrak{g}$, is an immediate
consequence of condition (R2). Moreover, the map

$(\Pi(x),\Pi(y)\mapsto
s_{\mathfrak{h}}(\Pi(x)+\Pi(y))+s_{\mathfrak{h}}(\Pi(x))+s_{\mathfrak{h}}(\Pi(y))=(x+y)^{[2]_{\alpha}}+x^{[2]_{\alpha}}+y^{[2]_{\alpha}}=[x,y]_{\mathfrak{g}}$

is obviously bilinear because it coincides with the Lie bracket on
$\mathfrak{g}$.

Let us check the Jacobi identity involving the squaring. Indeed, for all
$y\in\mathfrak{h}_{\bar{0}}$ and for all $\Pi(x)\in\mathfrak{h}_{\bar{1}}$, we
have

$\displaystyle[s_{\mathfrak{h}}(\Pi(x)),\alpha(y)]_{\mathfrak{h}}=[x^{[2]_{\alpha}},\alpha(y)]_{\mathfrak{h}}=[x^{[2]_{\alpha}},\alpha(y)]_{\mathfrak{g}}=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}.$

On the other hand

$\displaystyle[\alpha(\Pi(x)),[\Pi(x),y]_{\mathfrak{h}}]_{\mathfrak{h}}=[\Pi(\alpha(x)),\Pi([x,y]_{\mathfrak{g}})]_{\mathfrak{h}}=\Pi([\Pi(\alpha(x)),[x,y]_{\mathfrak{g}}]_{\mathfrak{h}})=\Pi^{2}([\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{h}})$
$\displaystyle=[\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}.$

For all $\Pi(y)\in\mathfrak{h}_{\bar{1}}$ and for all
$\Pi(x)\in\mathfrak{h}_{\bar{1}}$, we have

$\displaystyle[s_{\mathfrak{h}}(\Pi(x)),\alpha(\Pi(y))]_{\mathfrak{h}}=[x^{[2]_{\alpha}},\alpha(\Pi(y))]_{\mathfrak{h}}=\Pi([x^{[2]_{\alpha}},\alpha(y)]_{\mathfrak{g}})=\Pi([\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}).$

On the other hand

$\displaystyle[\alpha(\Pi(x)),[\Pi(x),\Pi(y)]_{\mathfrak{h}}]_{\mathfrak{h}}=[\Pi(\alpha(x)),s_{\mathfrak{h}}(\Pi(x)+\Pi(y))+s_{\mathfrak{h}}(\Pi(x))+s_{\mathfrak{h}}(\Pi(x)]_{\mathfrak{h}}$
$\displaystyle=[\Pi(\alpha(x)),(x+y)^{[2]_{\alpha}}+x^{[2]_{\alpha}}+y^{[2]_{\alpha}}]_{\mathfrak{h}}=[\Pi(\alpha(x)),[x,y]_{\mathfrak{g}}]_{\mathfrak{h}}=\Pi([\alpha(x),[x,y]_{\mathfrak{g}}]_{\mathfrak{g}}).\qed$

######  Proposition 5.4.

Let $\mathfrak{g}$ be a restricted Lie algebra in characteristic 2 and
$\mathfrak{h}:=\mathfrak{g}\oplus\Pi(\mathfrak{g})$ be its queerification, see
[BLLSq], defined as follows (for all $x,y\in\mathfrak{g}$):

$[x,y]_{\mathfrak{h}}:=[x,y]_{\mathfrak{g}},\quad[\Pi(x),y]_{\mathfrak{h}}:=\Pi([x,y]_{\mathfrak{g}}),\quad
s_{\mathfrak{h}}(\Pi(x))=x^{[2]}.$

Let $\alpha:\mathfrak{g}\rightarrow\mathfrak{g}$ be a Lie algebra morphism.
Let us extend it to $\tilde{\alpha}$ on $\mathfrak{h}$ by declaring
$\alpha(\Pi(x)):=\Pi(\alpha(x))$ for all $x\in\mathfrak{g}$. Then twisting the
Lie superalgebra $\mathfrak{h}$ along $\tilde{\alpha}$ is exactly the
queerification of the Hom-Lie algebra $\mathfrak{g}_{\alpha}$ obtained by
twisting $\mathfrak{g}$ along $\alpha$. Namely,

$\mathfrak{h}_{\tilde{\alpha}}=(\mathfrak{g}\oplus\Pi(\mathfrak{g}))_{\tilde{\alpha}}=\mathfrak{g}_{\alpha}\oplus\Pi(\mathfrak{g}_{\alpha}).$

###### Proof.

Let $x,y\in\mathfrak{g}$. We have

$[x,y]_{\mathfrak{h}_{\tilde{\alpha}}}=\tilde{\alpha}([x,y]_{\mathfrak{h}})=\alpha([x,y]_{\mathfrak{g}}).$

On the other hand,

$[x,y]_{\mathfrak{g}_{\alpha}\oplus\Pi(\mathfrak{g}_{\alpha})}=[x,y]_{\mathfrak{g}_{\alpha}}=\alpha([x,y]_{\mathfrak{g}}).$

Similarly, one can easily prove that

$[\Pi(x),y]_{\mathfrak{h}_{\tilde{\alpha}}}=[\Pi(x),y]_{\mathfrak{g}_{\alpha}\oplus\Pi(\mathfrak{g}_{\alpha})}.$

Let us only prove that their squarings coincide. Indeed, for all
$x\in\mathfrak{g}$ we have

$s_{\mathfrak{h}_{\tilde{\alpha}}}(\Pi(x))=\alpha\circ
s_{\mathfrak{h}}(\Pi(x))=\alpha(x^{[2]}).$

On the other hand,

$s_{\mathfrak{g}_{\alpha}\oplus\Pi(\mathfrak{g}_{\alpha})}(\Pi(x))=x^{[2]_{\alpha}}=\alpha(x^{[2]}).\qed$

##  6\. Cohomology and Deformations of finite dimensional Hom-Lie
superalgebras

###  6.1. Cohomology of ordinary Lie superalgebras in characteristic 2

In this section we define a cohomology theory of Lie superalgebras in
characteristic 2. The first instances can be found in [BGLL1]. Let
$\mathfrak{g}$ be a Lie superalgebra in characteristic 2 and $M$ be a
$\mathfrak{g}$-module. Let us introduce a map

(21)
$\mathfrak{p}:\mathfrak{g}_{\bar{1}}\times\wedge^{n}\mathfrak{g}\rightarrow
M,$

with the following properties:

(i) $\mathfrak{p}(\lambda x,z)=\lambda^{2}\mathfrak{p}(x,z)$ for all
$x\in\mathfrak{g}_{\bar{1}}$, for all $z\in\wedge^{n}\mathfrak{g}$ and for all
$\lambda\in\mathbb{K}$;

(ii) For all $x\in\mathfrak{g}_{\bar{1}}$, the map
$z\mapsto{\mathfrak{p}}(x,z)$ is multi-linear.

For $n=0$, the map ${\mathfrak{p}}$ should be understood as a quadratic form
on $\mathfrak{g}_{\bar{1}}$ with values in $M$.

We are now ready to define the space of cochains on $\mathfrak{g}$ with values
in $M$. We set ($n>1$)

(22) $\begin{array}[]{lcl}XC^{-1}(\mathfrak{g};M)&:=&\\{0\\},\\\\[5.69054pt]
XC^{0}(\mathfrak{g};M)&:=&M,\\\\[5.69054pt]
XC^{1}(\mathfrak{g};M)&:=&\\{c\;|\text{ where $c:\mathfrak{g}\rightarrow M$ is
linear}\\},\\\\[5.69054pt]
XC^{n}(\mathfrak{g};M)&:=&\\{(c,\mathfrak{p})\;|\text{ where
}c:\wedge^{n}\mathfrak{g}\rightarrow M\text{ is a multi-linear map and
}\mathfrak{p}:\mathfrak{g}_{\bar{1}}\times\wedge^{n-2}\mathfrak{g}\rightarrow
M\\\\[2.84526pt] &&\text{ is a map as in \eqref{mapp} }\text{ such that
}\mathfrak{p}(x+y,z)+\mathfrak{p}(x,z)+\mathfrak{p}(y,z)=c(x,y,z)\\\\[2.84526pt]
&&\text{ for all }x,y\in\mathfrak{g}_{\bar{1}}\text{ and
}z\in\wedge^{n-2}\mathfrak{g}\\}.\end{array}$

We define the differential
${\mathfrak{d}}^{-1}:XC^{-1}(\mathfrak{g},M)\rightarrow
XC^{0}(\mathfrak{g},M)$ to be the trivial map. The differential
${\mathfrak{d}}^{0}$ is given by

${\mathfrak{d}}^{0}:XC^{0}(\mathfrak{g},M)\rightarrow
XC^{1}(\mathfrak{g},M)\quad m\mapsto{\mathfrak{d}}^{0}(m),$

where ${\mathfrak{d}}^{0}(m)(x)=x\cdot m.$ The differential
${\mathfrak{d}}^{1}$ is given by

${\mathfrak{d}}^{1}:XC^{1}(\mathfrak{g},M)\rightarrow
XC^{2}(\mathfrak{g},M)\quad c\mapsto(dc,\mathfrak{q}),$

where

(23) $\begin{array}[]{lcl}dc(x,z)&=&c([x,z])+x\cdot c(z)+z\cdot
c(x)\quad\text{for all $x,z\in\mathfrak{g}$};\\\\[2.84526pt]
\mathfrak{q}(x)&=&c(s(x))+x\cdot c(x)\quad\text{for all
$x\in\mathfrak{g}_{\bar{1}}$}.\end{array}$

Now, for $n\geq 2$ the differential ${\mathfrak{d}}^{n}$ is given by

${\mathfrak{d}}^{n}:XC^{n}(\mathfrak{g},M)\rightarrow
XC^{n+1}(\mathfrak{g},M)\quad(c,\mathfrak{p})\mapsto(d^{n}c,d^{n}\mathfrak{p}),$

where

(24)
$\begin{array}[]{lcl}d^{n}c(z_{1},\ldots,z_{n+1})&=&\displaystyle\sum_{1\leq
i\leq n+1}z_{i}\cdot
c(z_{1},\ldots,\hat{z_{i}},\ldots,z_{n+1})\\\\[11.38109pt]
&&\displaystyle+\sum_{1\leq i<j\leq
n+1}c([z_{i},z_{j}],z_{1},\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots
z_{n+1}),\\\\[5.69054pt]
d^{n}{\mathfrak{p}}(x,z_{1},\ldots,z_{n-1})&=&\displaystyle x\cdot
c(x,z_{1},\ldots,z_{n-1})+\sum_{1\leq i\leq
n-1}z_{i}\cdot{\mathfrak{p}}(x,z_{1},\ldots,\hat{z_{i}},\ldots,z_{n-1})\\\\[8.53581pt]
&&+c(s(x),z_{1},\ldots,z_{n-1})+\displaystyle\sum_{1\leq i\leq
n-1}c([x,z_{i}],x,z_{1},\ldots,\hat{z_{i}},\ldots,z_{n-1})\\\\[5.69054pt]
&&\displaystyle+\sum_{1\leq i<j\leq
n-1}{\mathfrak{p}}(x,[z_{i},z_{j}],z_{1},\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots,z_{n-1}).\end{array}$

######  Theorem 6.1.

The maps ${\mathfrak{d}}^{n}$ is well defined. Moreover, for all integers $n$

${\mathfrak{d}}^{n+1}\circ{\mathfrak{d}}^{n}=0.$

Hence, the pair $(XC^{*}(\mathfrak{g},M),{\mathfrak{d}}^{*})$ defines a
cohomology complex for Lie superalgebras in characteristic 2.

The proof of the theorem will be given next when considering the cohomology of
Hom-Lie superalgebras that reduce to ordinary Lie superalgebras when the
structure map is the identity.

###  6.2. Elucidation for $n=2,3$

Let us first exhibit the sets of cochains in the case where $n=2,3$.

If $v\in M$ and $a,b\in\mathfrak{g}^{*}_{\bar{1}}$, we can define the cochain
$(v\otimes a\wedge b,\mathfrak{q})\in XC^{2}(\mathfrak{g},M)$ such that the
quadratic form is ${\mathfrak{q}}(x)=a(x)b(x)\,v$ for all
$x\in\mathfrak{g}_{\bar{1}}$. The polar form111Recall that to each quadratic
from $\mathfrak{q}$ with values in a space $M$, its polar form is the bilinear
form with values in $M$ given by
$B_{\mathfrak{q}}(x,y):=\mathfrak{q}(x+y)+\mathfrak{q}(x)+\mathfrak{q}(y).$
associated to $\mathfrak{q}$ is

$B_{\mathfrak{q}}(x,y)=(a(x)b(y)+a(y)b(x))\,v\text{ for all
$x,y\in\mathfrak{g}_{\bar{1}}$}.$

In particular, we can define the cochain $c=v\otimes a\wedge a$, where
$q(x)=v(a(x))^{2}$ for all $x\in\mathfrak{g}_{\bar{1}}$ and $c(x,y)=0$ for all
$x,y\in\mathfrak{g}$.

Similarly, if $v\in M$, and $a,b\in\mathfrak{g}^{*}_{\bar{1}}$ but
$c\in\mathfrak{g}$, we can define the cochain $(v\otimes a\wedge b\wedge
c,{\mathfrak{p}})\in XC^{2}(\mathfrak{g},M)$ such that the map $\mathfrak{p}$
is

${\mathfrak{p}}(x,z)=(a(x)b(x)c(z)+a(z)b(x)c(x)+a(x)b(z)c(x))\,v\quad\text{for
all $x\in\mathfrak{g}_{\bar{1}}$ and $z\in\mathfrak{g}$}$

Now, a direct computation shows that

$\begin{array}[]{lcl}\mathfrak{p}(x+y,z)+\mathfrak{p}(x,z)+\mathfrak{p}(y,z)&=&(a(x)b(y)c(z)+a(y)b(x)c(z)+a(z)b(x)c(y))\,v\\\\[5.69054pt]
&&+(a(z)b(y)c(x)+b(z)a(x)c(y)+b(z)a(y)c(x))v\\\\[5.69054pt]
&=&v\otimes(a\wedge b\wedge c)(x,y,z).\end{array}$

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

(25) $\displaystyle x\cdot c(z)+z\cdot c(x)+c([x,z])$ $\displaystyle=$
$\displaystyle 0\quad\text{for all $x,z\in\mathfrak{g}$},$ (26) $\displaystyle
x\cdot c(x)+c(s(x))$ $\displaystyle=$ $\displaystyle 0\quad\text{for all
$x\in\mathfrak{g}_{\bar{1}}$}.$

A 2-cocycle $(c,\mathfrak{q})$ on $\mathfrak{g}$ with values in $M$ must
satisfy the following conditions:

(27) $\displaystyle 0$ $\displaystyle=$ $\displaystyle x\cdot
c(y,z)+c([x,y],z)+\circlearrowleft(x,y,z)\quad\text{for all
$x,y,z\in\mathfrak{g}$},$ $\displaystyle 0$ $\displaystyle=$ $\displaystyle
x\cdot c(x,z)+z\cdot\mathfrak{q}(x)+c(s(x),z)+c([x,z],x]$
$\displaystyle\text{for all $x\in\mathfrak{g}_{\bar{1}}$ and for all
$z\in\mathfrak{g}$},$

###  6.3. Cohomology of Hom-Lie superalgebras in characteristic 2

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra in
characteristic 2 and $(M,\beta)$ be a $\mathfrak{g}$-module, see Definition
3.1. The space of $n$-cochains are defined similarly to (22) with a slight
difference with respect to degree 0 space and an extra condition that is

(29) $\beta\circ c=c\circ(\alpha\wedge\cdots\wedge\alpha),\quad\text{ and
}\quad\beta\circ{\mathfrak{p}}={\mathfrak{p}}\circ(\alpha\wedge\cdots\wedge\alpha).$
(30)
$\begin{array}[]{lcl}XC_{\alpha}^{-1}(\mathfrak{g};M)&:=&\\{0\\},\\\\[5.69054pt]
XC_{\alpha}^{0}(\mathfrak{g};M)&:=&\\{m\in M\;|\;\beta(m)=m\text{ and
}\alpha(x)\cdot(y\cdot m)=x\cdot(y\cdot m)\text{ for all
$x,y\in\mathfrak{g}$}\\},\\\\[5.69054pt]
XC_{\alpha}^{1}(\mathfrak{g};M)&:=&\\{c\;|\text{ where
$c:\mathfrak{g}\rightarrow M$ is linear and satisfies Eq.
\ref{cohalphabeta}}\\},\\\\[5.69054pt]
XC_{\alpha}^{n}(\mathfrak{g};M)&:=&\\{(c,\mathfrak{p})\;|\text{ where
}c:\wedge^{n}\mathfrak{g}\rightarrow M\text{ is a multi-linear map satisfying
Eq.\ref{cohalphabeta} and }\\\\[2.84526pt]
&&\mathfrak{p}:\mathfrak{g}_{\bar{1}}\times\wedge^{n-2}\mathfrak{g}\rightarrow
M\text{ is a map as in \eqref{mapp} satisfying Eq. \ref{cohalphabeta} }\text{
such that }\\\\[2.84526pt]
&&\mathfrak{p}(x+y,z)+\mathfrak{p}(x,z)+\mathfrak{p}(y,z)=c(x,y,z)\text{ for
all }x,y\in\mathfrak{g}_{\bar{1}}\text{ and
}z\in\wedge^{n-2}\mathfrak{g}\\}.\end{array}$

1-cochains are just linear functions $c$ on $\mathfrak{g}$ with values in a
$\mathfrak{g}$-module $M$ such that $\beta\circ c=c\circ\alpha$. Let us define
the differentials in our context. First, let us define define
$\mathfrak{d}_{\alpha}^{0}$ and $\mathfrak{d}_{\alpha}^{1}$.

${\mathfrak{d}}^{0}_{\alpha}:XC^{0}_{\alpha}(\mathfrak{g},M)\rightarrow
XC^{1}_{\alpha}(\mathfrak{g},M)\quad m\mapsto d^{0}_{\alpha}m,$

where $d^{0}_{\alpha}m(x)=x\cdot m$ for all $x\in\mathfrak{g}$. Additionally,

${\mathfrak{d}}^{1}_{\alpha}:XC^{1}_{\alpha}(\mathfrak{g},M)\rightarrow
XC^{2}_{\alpha}(\mathfrak{g},M)\quad c\mapsto(d^{1}_{\alpha}c,\mathfrak{q}),$

where

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

Note that these definitions are consistent as showed by the following Lemma.

######  Proposition 6.2.

The differentials ${\mathfrak{d}}^{0}_{\alpha}$ and
${\mathfrak{d}}^{1}_{\alpha}$ are indeed well-defined; namely,
$\text{\rm{Im}}\,({\mathfrak{d}}^{0}_{\alpha})\subseteq
XC^{1}_{\alpha}(\mathfrak{g},M)$ and
$\text{\rm{Im}}\,({\mathfrak{d}}^{1}_{\alpha})\subseteq
XC^{2}_{\alpha}(\mathfrak{g},M)$.

###### Proof.

Let us first deal with ${\mathfrak{d}}^{0}_{\alpha}$. We have

$d^{0}_{\alpha}m(\alpha(x))=\alpha(x)\cdot
m=\alpha(x)\cdot\beta(m)=\beta(x\cdot m)=\beta(d^{0}_{\alpha}m(x)).$

Therefore, Eq. 29 is satisfied. Let us now deal deal with
${\mathfrak{d}}^{1}_{\alpha}$. We will only prove that $\mathfrak{q}$
satisfies Eq. 29. Indeed,

$\begin{array}[]{lcl}\mathfrak{q}(\alpha(x))&=&c(s(\alpha(x)))+\alpha(x)\cdot
c(\alpha(x))=c(\alpha(s(x)))+\alpha(x)\cdot\beta(c(x))\\\\[5.69054pt]
&=&\beta(c(s(x)))+\beta(x\cdot c(x))=\beta({\mathfrak{q}}(x)).\end{array}$

On the other hand, we have

$\begin{array}[]{lcl}\mathfrak{q}(x+y)+\mathfrak{q}(x)+\mathfrak{q}(y)&=&c(s(x+y))+(x+y)\cdot
c(x+y)+c(s(x))\\\\[5.69054pt] &&+x\cdot c(x)+c(s(y))+y\cdot
c(y)\\\\[5.69054pt] &=&c([x,y])+y\cdot c(x)+x\cdot
c(y)=d^{1}_{\alpha}c(x,y).\qed\end{array}$

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

(32) $\displaystyle x\cdot c(y)+y\cdot c(x)+c([x,y])$ $\displaystyle=$
$\displaystyle 0\quad\text{for all $x,y\in\mathfrak{g}$},$ (33) $\displaystyle
x\cdot c(x)+c(s(x))$ $\displaystyle=$ $\displaystyle 0\quad\text{for all
$x\in\mathfrak{g}_{\bar{1}}$}.$

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

Now, for $n\geq 2$ the differential ${\mathfrak{d}}^{n}_{\alpha}$ is given by

${\mathfrak{d}}^{n}:XC^{n}_{\alpha}(\mathfrak{g},M)\rightarrow
XC^{n+1}_{\alpha}(\mathfrak{g},M)\quad(c,\mathfrak{p})\mapsto(d^{n}_{\alpha}c,d^{n}_{\alpha}\mathfrak{p}),$

where

$\displaystyle d^{n}_{\alpha}c(z_{1},\ldots,z_{n+1})=\displaystyle\sum_{1\leq
i\leq n+1}\alpha^{n-1}(z_{i})\cdot c(z_{1},\ldots,\hat{z_{i}},\ldots,z_{n+1})$
$\displaystyle\displaystyle+\sum_{1\leq i<j\leq
n+1}c([z_{i},z_{j}],\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots\alpha(z_{n+1})),$
$\displaystyle
d^{n}_{\alpha}{\mathfrak{p}}(x,z_{1},\ldots,z_{n-1})=\displaystyle\alpha^{n-1}(x)\cdot
c(x,z_{1},\ldots,z_{n-1})+c(s(x),\alpha(z_{1}),\ldots,\alpha(z_{n-1}))$
$\displaystyle\displaystyle+\sum_{1\leq i\leq
n-1}\alpha^{n-1}(z_{i})\cdot{\mathfrak{p}}(x,z_{1},\ldots,\hat{z_{i}},\ldots,z_{n-1})$
$\displaystyle+\displaystyle\sum_{1\leq i\leq
n-1}c([x,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\alpha(z_{n-1}))$
$\displaystyle\displaystyle+\sum_{1\leq i<j\leq
n-1}{\mathfrak{p}}(\alpha(x),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots,\alpha(z_{n-1})).$

In particular for $n=2$, the differential is given by

${\mathfrak{d}}^{2}_{\alpha}:XC^{2}_{\alpha}(\mathfrak{g},M)\rightarrow
XC^{3}_{\alpha}(\mathfrak{g},M)\quad(c,{\mathfrak{p}})\mapsto(d^{2}_{\alpha}c,d^{2}_{\alpha}\mathfrak{p}),$

where

$\begin{array}[]{lcl}d^{2}_{\alpha}c(z_{1},z_{2},z_{3})&=&\alpha(z_{1})\cdot
c(z_{2},z_{3})+c([z_{1},z_{2}],\alpha(z_{3}))+\circlearrowleft(z_{1},z_{2},z_{3})\quad\text{for
all $z_{1},z_{2},z_{3}\in\mathfrak{g}$};\\\\[2.84526pt]
d^{2}_{\alpha}{\mathfrak{p}}(x,z_{1})&=&\alpha(x)\cdot
c(x,z_{1})+\alpha(z_{1})\cdot{\mathfrak{p}}(x)+c(s(x),\alpha(z_{1}))+c([x,z_{1}],\alpha(x))\\\\[5.69054pt]
&&\text{for all $x\in\mathfrak{g}_{\bar{1}},$ and for all
$z_{1}\in\mathfrak{g}$}.\end{array}$

A 2-cocycle is 2-tuple $(c,\mathfrak{p})$ satisfying the following conditions:

(34) $\displaystyle 0$ $\displaystyle=$ $\displaystyle\alpha(z_{3})\cdot
c(z_{1},z_{2})+c([z_{1},z_{2}],\alpha(z_{3}))+\circlearrowleft(z_{1},z_{2},z_{3})\quad\text{for
all $z_{1},z_{2},z_{3}\in\mathfrak{g}$},$ $\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha(x)\cdot
c(x,z_{1})+\alpha(z_{1})\cdot{\mathfrak{p}}(x)+c(s(x),\alpha(z_{1}))+c([x,z_{1}],\alpha(x)]$
$\displaystyle\text{for all $x\in\mathfrak{g}_{\bar{1}}$ and for all
$z_{1}\in\mathfrak{g}$},$

The first step here is to show that the map ${\mathfrak{d}}_{\alpha}^{n}$ is
well defined, for every twist $\alpha$. By doing so, we give a proof to the
first part of Theorem 6.1 in the case where $\alpha=\mathrm{id}$.

######  Proposition 6.3.

The map $\mathfrak{d}^{n}_{\alpha}$ is well-defined; namely,
$\text{\rm{Im}}\,({\mathfrak{d}}^{n}_{\alpha})\subseteq
XC^{n+1}_{\alpha}(\mathfrak{g},M)$.

###### Proof.

For all $x,y\in\mathfrak{g}_{\bar{1}}$ and for all
$z_{1},\ldots,z_{n}\in\mathfrak{g}$, we have

$\displaystyle d^{n}_{\alpha}{\mathfrak{p}}(x+y,z_{1},\ldots,z_{n})$
$\displaystyle=\displaystyle\alpha^{n-1}(x+y)\cdot
c(x+y,z_{1},\ldots,z_{n-1})+\sum_{1\leq i\leq
n-1}\alpha^{n-1}(z_{i})\cdot{\mathfrak{p}}(x+y,z_{1},\ldots,\hat{z_{i}},\ldots,z_{n-1})$
$\displaystyle+c(s(x+y),\alpha(z_{1}),\ldots,\alpha(z_{n-1}))+\displaystyle\sum_{1\leq
i\leq
n-1}c([x+y,z_{i}],\alpha(x+y),\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\alpha(z_{n-1}))$
$\displaystyle\displaystyle+\sum_{1\leq i<j\leq
n-1}{\mathfrak{p}}(\alpha(x+y),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots,\alpha(z_{n-1})).$
$\displaystyle=d^{n}_{\alpha}{\mathfrak{p}}(x,z_{1},\ldots,z_{n})+d^{n}_{\alpha}{\mathfrak{p}}(y,z_{1},\ldots,z_{n})+\alpha^{n-1}(x)\cdot
c(y,z_{1},\ldots,z_{n-1})$ $\displaystyle+\alpha^{n-1}(y)\cdot
c(x,z_{1},\ldots,z_{n-1})+\sum_{1\leq i\leq n-1}\alpha^{n-1}(z_{i})\cdot
c(x,y,z_{1},\ldots,\hat{z_{i}},\ldots,z_{n-1})$
$\displaystyle+c([x,y],\alpha(z_{1}),\ldots,\alpha(z_{n-1}))+\displaystyle\sum_{1\leq
i\leq
n-1}c([x,z_{i}],\alpha(y),\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\displaystyle\sum_{1\leq i\leq
n-1}c([y,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\alpha(z_{n-1}))$
$\displaystyle\displaystyle+\sum_{1\leq i<j\leq
n-1}c(\alpha(x),\alpha(y),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\hat{z_{i}},\ldots,\hat{z_{j}},\ldots,\alpha(z_{n-1}))$
$\displaystyle=d^{n}_{\alpha}{\mathfrak{p}}(x,z_{1},\ldots,z_{n})+d^{n}_{\alpha}{\mathfrak{p}}(y,z_{1},\ldots,z_{n})+d^{n}_{\alpha}c(x,y,z_{1},\ldots,z_{n}),$

where we have used the fact that $s(x+y)=s(x)+s(y)+[x,y]$ and

$\begin{array}[]{lcl}{\mathfrak{p}}(x+y,z_{1},\ldots,z_{n-1})+{\mathfrak{p}}(x,z_{1},\ldots,z_{n-1})+{\mathfrak{p}}(y,z_{1},\ldots,z_{n-1})&=&c(x,y,z_{1},\ldots,z_{n-1}).\qed\end{array}$

######  Theorem 6.4.

For all $n\geq 1$, we have
$\mathfrak{d}_{\alpha}^{n}\circ\mathfrak{d}_{\alpha}^{n-1}=0$. Hence, the pair
$(XC^{*}_{\alpha}(\mathfrak{g},M),{\mathfrak{d}}_{\alpha}^{*})$ defines a
cohomology complex for Hom-Lie superalgebras in characteristic 2.

In order to prove this theorem, we will need the following Lemma.

######  Lemma 6.5.

If $(c,{\mathfrak{p}})\in XC^{n}_{\alpha}(\mathfrak{g},M)$, then

(i) $\alpha^{n-1}(x)\cdot(\alpha^{n-2}(x)\cdot
c(z_{1},\ldots,z_{n}))=\alpha^{n-2}(s(x))\cdot
c(\alpha(z_{1}),\ldots,\alpha(z_{n}))$ for all $x\in\mathfrak{g}_{\bar{1}}$
and for all $z_{1},\ldots,z_{n}\in\mathfrak{g}$.

(ii) $\alpha(z_{i})\cdot(z_{j}\cdot
c(z_{1},\ldots,z_{n}))+\alpha(z_{j})\cdot(z_{i}\cdot
c(z_{1},\ldots,z_{n}))=[z_{i},z_{j}]\cdot
c(\alpha(z_{1}),\ldots\alpha(z_{n}))$ for all $z_{1},\ldots
z_{n}\in\mathfrak{g}$.

###### Proof.

Let us only prove Part (i). Using the fact that $\beta\circ
c=c\circ(\alpha\wedge\cdots\wedge\alpha)$ we get

$\displaystyle\alpha^{n-1}(x)\cdot(\alpha^{n-2}(x)\cdot
c(z_{1},\ldots,z_{n}))=s(\alpha^{n-2}(x))\cdot\beta(c(z_{1},\ldots,z_{n}))$
$\displaystyle=\alpha^{n-2}(s(x))\cdot(c(\alpha(z_{1}),\ldots,\alpha(z_{n}))).\qed$

###### Proof of Theorem 6.4.

Lert us first show that
$\mathfrak{d}^{1}_{\alpha}\circ\mathfrak{d}^{0}_{\alpha}=0$. Indeed, for all
$x,y\in\mathfrak{g}$ and $m\in XC_{\alpha}^{0}(\mathfrak{g};M)$ we have

$\begin{array}[]{lcl}d^{1}_{\alpha}\circ d^{0}_{\alpha}m(x,y)&=&x\cdot
d^{0}_{\alpha}m(y)+y\cdot
d^{0}_{\alpha}m(x)+d^{0}_{\alpha}m([x,y])=x\cdot(y\cdot m)+y\cdot(x\cdot
m)+[x,y]\cdot m\\\\[2.84526pt] &=&\alpha(x)\cdot(y\cdot
m)+\alpha(y)\cdot(x\cdot m)+[x,y]\cdot\beta(m)=0.\end{array}$

On the other hand, for all $x\in\mathfrak{g}_{\bar{1}}$ and $m\in
XC_{\alpha}^{0}(\mathfrak{g};M)$, we have

${\mathfrak{q}}(x)=d^{0}_{\alpha}m(s(x))+x\cdot d^{0}_{\alpha}m(x)=s(x)\cdot
m+x\cdot(x\cdot m)=s(x)\cdot\beta(m)+\alpha(x)\cdot(x\cdot m)=0.$

Let us now show that
$\mathfrak{d}^{n}_{\alpha}\circ\mathfrak{d}^{n-1}_{\alpha}=0$ for all $n>1$.
To show that $d^{n}_{\alpha}\circ d^{n-1}_{\alpha}(c)=0$ is a routine, see for
instance [AMS]. Let us show that $d^{n}_{\alpha}\circ
d^{n-1}_{\alpha}({\mathfrak{p}})=0$. This would imply that
${\mathfrak{d}}^{n}_{\alpha}\circ{\mathfrak{d}}^{n-1}_{\alpha}(c,{\mathfrak{p}})=0$.
Actually, the computation is very cumbersome so we will break it into small
pieces. First, we compute:

$\displaystyle d^{n}_{\alpha}\circ
d^{n-1}_{\alpha}({\mathfrak{p}})(x,z_{1},\ldots,z_{n-1})=\alpha^{n-1}(x)\cdot
d^{n-1}_{\alpha}c(x,z_{1},\ldots,z_{n-1})$
$\displaystyle+\sum_{i=1}^{n-1}\alpha^{n-1}(z_{i})d^{n-1}_{\alpha}{\mathfrak{p}}(x,z_{1},\ldots,\widehat{z_{i}},\ldots,z_{n-1})+d^{n-1}_{\alpha}c(s(x),\alpha(z_{1}),\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{i=1}^{n-1}d^{n-1}_{\alpha}c([x,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq i<j\leq
n-1}d^{n-1}_{\alpha}{\mathfrak{p}}(\alpha(x),[z_{i},z_{j}],\alpha(z_{1}),\ldots\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1})).$

There are five terms in the expression above. We will compute each term
separately.

$\displaystyle\text{\text@underline{Part 1: }}\sum_{1\leq i\leq
n-1}\alpha^{n-1}_{\alpha}(z_{i})d^{n-1}{\mathfrak{p}}(x,z_{1},\ldots,\widehat{z_{i}},\ldots,z_{n-1})=$
$\displaystyle+\sum_{1\leq i\leq
n-1}\alpha^{n-1}(z_{i})\cdot[\alpha^{n-2}(x)\cdot
c(x,z_{1},\ldots,\widehat{z_{i}},\ldots,z_{n-1})$ $\displaystyle+\sum_{1\leq
j\leq
n-2}\alpha^{n-2}(z_{j}){\mathfrak{p}}(x,z_{1},\ldots,\widehat{z_{i}},\ldots,\widehat{z_{j}},\ldots,z_{n-1})+c(s(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq j\leq
n-2}c([x,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq l<j\leq
n-2}{\mathfrak{p}}(\alpha(x),[z_{l},z_{j}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{l})},\ldots,\widehat{\alpha(z_{j})},\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))].$
$\displaystyle\text{\text@underline{Part 2:
}}\alpha^{n-1}(x)d^{n-1}_{\alpha}c(x,z_{1},\ldots,z_{n-1})=\alpha^{n-1}(x)\cdot[\alpha^{n-2}(x)c(z_{1},\ldots,z_{n-1})$
$\displaystyle+\sum_{1\leq i\leq
n-1}\alpha^{n-2}(z_{i})c(x,z_{1},\ldots,\widehat{z_{i}},\ldots,z_{n-1})+\sum_{1\leq
i\leq
n-1}c([x,z_{i}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq i<j\leq
n-1}c([z_{i},z_{j}],\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))].$
$\displaystyle\text{\text@underline{Part 3:
}}d^{n-1}_{\alpha}c(s(x),\alpha(z_{1}),\ldots,\alpha(z_{n-1}))=\alpha^{n-2}(s(x))\cdot
c(\alpha(z_{1}),\ldots,\alpha(z_{n-1}))$ $\displaystyle+\sum_{1\leq i\leq
n-1}\alpha^{n-1}(z_{i})\cdot
c(s(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle\sum_{1\leq i<j\leq
n-1}c([\alpha(z_{i}),\alpha(z_{j})],\alpha(s(x)),\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots,\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq i\leq
n-1}c([s(x),\alpha(z_{i})],\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\alpha^{2}(z_{n-1})).$
$\displaystyle\text{\text@underline{Part 4: }}\sum_{1\leq i\leq
n-1}d^{n-1}_{\alpha}c([x,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle=\sum_{1\leq i\leq n-1}[\alpha^{n-2}([x,z_{i}])\cdot
c(\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\alpha^{n-1}(x)\cdot
c([x,z_{i}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq l\leq n-2}\alpha^{n-1}(z_{l})\cdot
c([x,z_{i}],\alpha(x),\alpha(z_{1}),\ldots,\widehat{\alpha(z_{l})},\ldots,\widehat{\alpha(z_{i})},\dots,\alpha(z_{n-1}))$
$\displaystyle+c([[x,z_{i}],\alpha(x)],\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq j\leq
n-1}c([[x,z_{i}],\alpha(z_{j})],\alpha^{2}(x),\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq l\leq
n-1}c([\alpha(x),\alpha(z_{l})],\alpha([x,z_{i}]),\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{l})},\ldots\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq u<v\leq
n-1}c(\alpha([x,z_{i}]),\alpha^{2}(x),[\alpha(z_{u}),\alpha(z_{v})],\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{u})},\ldots\widehat{\alpha^{2}(z_{v})},\ldots\alpha^{2}(z_{n-1}))].$
$\displaystyle\text{\text@underline{Part 5 :} }\sum_{1\leq i<j\leq
n-1}d^{n-1}_{\alpha}{\mathfrak{p}}(\alpha(x),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))$
$\displaystyle=\sum_{1\leq i<j\leq n-1}[\alpha^{n-1}(x)\cdot
c(\alpha(x),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\sum_{1\leq l\leq
n-1}\alpha^{n-1}(z_{l}){\mathfrak{p}}(\alpha(x),[z_{i},z_{j}],\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\widehat{\alpha(z_{l})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+\alpha^{n-2}([z_{i},z_{j}])\cdot{\mathfrak{p}}(x,\alpha(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))$
$\displaystyle+c(s(\alpha(x)),\alpha([z_{i},z_{j}]),\alpha^{2}(z_{1}),\ldots,\widehat{\alpha(z_{i})},\ldots,\widehat{\alpha(z_{j})},\ldots,\alpha(z_{n-1}))$
$\displaystyle\sum_{1\leq l\leq
n-1}c([\alpha(x),\alpha(z_{l})],\alpha([z_{i},z_{j}]),\alpha^{2}(x),\alpha^{2}(z_{1}),\ldots,\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots,\alpha^{2}(z_{n-1}))$
$\displaystyle+c([\alpha(x),[z_{i},z_{j}]],\alpha^{2}(z_{1}),\ldots\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots,\widehat{\alpha^{2}(z_{l})},\ldots,\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq l\leq
n-1}{\mathfrak{p}}(\alpha^{2}(x),[\alpha(z_{l}),[z_{i},z_{j}]],\alpha^{2}(z_{1}),\ldots\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots,\widehat{\alpha^{2}(z_{l})},\ldots\alpha^{2}(z_{n-1}))$
$\displaystyle+\sum_{1\leq u<v\leq n-1}T_{ijuv}],$

where

$T_{ijuv}={\mathfrak{p}}(\alpha^{2}(x),\alpha([z_{i},z_{j}]),[\alpha(z_{u}),\alpha(z_{v})],\alpha^{2}(z_{1}),\ldots\widehat{\alpha^{2}(z_{i})},\ldots,\widehat{\alpha^{2}(z_{j})},\ldots,\widehat{\alpha^{2}(z_{u})},\ldots,\widehat{\alpha^{2}(z_{v})},\ldots).$

Now, using Lemma 6.5 a direct computation shows that

$\text{Part 1+Part 2+Part 3+Part 4+Part 5}=0.\qed$

Now, we are ready to define a cohomology of Hom-Lie superalgebras in
characteristic 2. The kernel of the map ${\mathfrak{d}}^{n}_{\alpha}$, denoted
by $Z^{n}_{\alpha}(\mathfrak{g};M)$, is the space of $n$-cocycles. The range
of the map ${\mathfrak{d}}^{n-1}_{\alpha}$, denoted by
$B^{n}_{\alpha}(\mathfrak{g};M)$, is the space of coboundaries.

We define the $n^{th}$ cohomology space as

${\mathrm{H}}^{n}_{\alpha}(\mathfrak{g};M):=Z^{n}_{\alpha}(\mathfrak{g};M)/B^{n}_{\alpha}(\mathfrak{g};M).$

######  Remark 6.6.

The cohomology defined above coincides when
$\alpha=\mathrm{id}_{\mathfrak{g}}$ and $\beta=\mathrm{id}_{M}$, with the
cohomology of Lie superalgebras in characteristic 2 defined in the previous
section.

######  Example 6.7.

We compute the second cohomology of the Hom-Lie superalgebra
$\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha}$ defined in Example 2.5. We will
assume here that the field $\mathbb{K}$ is infinite.

(i) The cohomology space
$\mathrm{H}^{2}_{\alpha}(\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha};\mathbb{K})$
is trivial. Recall that in this case the map $\beta=\mathrm{Id}$.

Let us first show that cocycles of the form $(0,\mathfrak{p})$ are necessarily
trivial. In fact, the condition $\mathfrak{p}=\mathfrak{p}\circ\alpha$ and
$\varepsilon\neq 0$ implie that

${\mathfrak{p}}(x_{1})=0\quad\text{and}\quad{\mathfrak{p}}(y_{1})=m\;\text{
(arbitrary)}.$

Choose $B_{m}=m\,y_{2}^{*}$, where $m\in\mathbb{K}$. A direct computation
shows that $d^{2}_{\alpha}B_{m}=0$. Let us compute the corresponding
${\mathfrak{q}}_{m}$. Indeed,

${\mathfrak{q}}_{m}(x_{1})=B_{m}(s(x_{1}))=B_{m}(x_{2})=0,$

and

${\mathfrak{q}}_{m}(y_{1})=B_{m}(s(y_{1}))=B_{m}(\varepsilon
h_{1}+\varepsilon^{2}x_{2}+y_{2})=m.$

It follows that $(0,\mathfrak{p})=(d^{2}_{\alpha}B_{m},\mathfrak{q}_{m})$ and
hence its cohomlogy class is trivial.

Let us now describe 2-cocycles of the form $(c,\mathfrak{p})$. A direct
computation shows that

$c_{1}=x_{1}^{*}\wedge y_{1}^{*}+x_{2}^{*}\wedge y_{2}^{*},\quad
c_{2}=h_{1}^{*}\wedge y_{1}^{*}+x_{1}^{*}\wedge y_{2}^{*},$

are the only cochains verifying both conditions
$c_{i}\circ(\alpha\wedge\alpha)=c_{i}$ and $d_{\alpha}^{2}c_{i}=0$ for
$i=1,2$. Let us describe the corresponding $\mathfrak{p}$’s. We have

$\begin{array}[]{lcllcl}\mathfrak{p}_{1}(x_{1})&=&\varepsilon^{-1},&\mathfrak{p}_{1}(y_{1})&=&m_{1}\quad\text{(arbitrary)}\\\\[5.69054pt]
\mathfrak{p}_{2}(x_{1})&=&0&\mathfrak{p}_{2}(y_{1})&=&m_{2}\quad\text{(arbitrary)}\end{array}$

We then get that
$\mathfrak{d}^{2}_{\alpha}(c_{1},\mathfrak{p}_{1})=\mathfrak{d}^{2}_{\alpha}(c_{2},\mathfrak{p}_{2})=0$.

Let us now describe the coboundaries. Choose
$b_{1}=h_{1}^{*}+\varepsilon^{-1}x_{2}^{*}$. It follows that

$d^{1}_{\alpha}b_{1}=x_{1}^{*}\wedge y_{1}^{*}+x_{2}^{*}\wedge y_{2}^{*}.$

Now,

$\mathfrak{q}_{1}(x_{1})=b_{1}(s(x_{1}))=b_{1}(x_{2})=\varepsilon^{-1},$

and

$\mathfrak{q}_{1}(y_{1})=b_{1}(s(y_{1}))=b_{1}(\varepsilon
h+\varepsilon^{2}x_{2}+y_{2})=0.$

Choose $b_{2}=y_{1}^{*}$. A direct computation shows that

$d^{1}_{\alpha}b_{2}=h_{1}^{*}\wedge y_{1}^{*}+x_{1}^{*}\wedge
y_{2}^{*},\quad\text{and $\mathfrak{q}_{2}\equiv 0$. }$

It follows that

$(c_{1},\mathfrak{p}_{1})=(d^{2}_{\alpha}b_{1},\mathfrak{q}_{1})+(d^{2}_{\alpha}B_{m_{1}}=0,\mathfrak{q}_{m_{1}}),\quad\text{
and
}\quad(c_{2},\mathfrak{p}_{2})=(d^{2}_{\alpha}b_{2},\mathfrak{q}_{2})+(d^{2}_{\alpha}B_{m_{2}}=0,\mathfrak{q}_{m_{2}}).$

Therefore, the cohomology space
$\mathrm{H}^{2}_{\alpha}(\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha};\mathbb{K})$
is trivial.

(ii) Let us now compute the cohomology space:
$\mathrm{H}^{2}_{\alpha}(\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha};\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha})$.
Recall that in the case where $\alpha=\mathrm{Id}$, this cohomology space has
only two non-trivial 2-cocycles.

The case where $\varepsilon\not=1$: the space
$\mathrm{H}^{2}_{\alpha}(\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha};\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha}$
is generated by the non-trivial 2-cocycles:

$(c_{4},\mathfrak{p}_{4}),\quad(c_{5},\mathfrak{p}_{5}),\quad(c_{9},\mathfrak{p}_{9}),\quad(c_{10},\mathfrak{p}_{10}),\quad(c_{11},\mathfrak{p}_{11}),\quad(c_{12},\mathfrak{p}_{12}),$

where

$\begin{array}[]{lcl}c_{4}&=&\varepsilon x_{1}\otimes h_{1}^{*}\wedge
x_{1}^{*}+\varepsilon^{2}x_{1}\otimes h_{1}^{*}\wedge y_{1}^{*}+x_{1}\otimes
x_{1}^{*}\wedge x_{2}^{*}+\varepsilon^{2}x_{1}\otimes x_{1}^{*}\wedge
y_{2}^{*}+\varepsilon^{2}x_{2}\otimes x_{1}^{*}\wedge y_{1}^{*}\\\\[5.69054pt]
&&\varepsilon^{2}x_{2}\otimes x_{2}^{*}\wedge y_{2}^{*}+\varepsilon
y_{1}\otimes h_{1}^{*}\wedge y_{1}^{*}+y_{1}\otimes x_{2}^{*}\wedge
y_{1}^{*},\\\\[5.69054pt] c_{5}&=&h_{1}\otimes h_{1}^{*}\wedge
y_{2}^{*}+x_{1}\otimes h_{1}^{*}\wedge y_{1}^{*}+x_{1}\otimes x_{1}^{*}\wedge
y_{2}^{*},\\\\[5.69054pt] c_{9}&=&h_{1}\otimes h_{1}^{*}\wedge
x_{1}^{*}+\varepsilon h_{1}\otimes h_{1}^{*}\wedge y_{1}^{*}+\varepsilon
x_{1}\otimes x_{1}^{*}\wedge y_{1}^{*}+\varepsilon x_{2}\otimes
h_{1}^{*}\wedge x_{1}^{*}+\varepsilon^{2}x_{2}\otimes h_{1}^{*}\wedge
y_{1}^{*}\\\\[5.69054pt] &&+\varepsilon^{2}x_{2}\otimes x_{1}^{*}\wedge
y_{2}^{*}+\varepsilon x_{2}\otimes x_{2}^{*}\wedge y_{1}^{*}+y_{2}\otimes
h_{1}^{*}\wedge y_{1}^{*}+y_{2}\otimes x_{1}^{*}\wedge
y_{2}^{*},\\\\[5.69054pt] c_{10}&=&h_{1}\otimes x_{1}^{*}\wedge
y_{1}^{*},\\\\[5.69054pt] c_{11}&=&h_{1}\otimes x_{2}^{*}\wedge
y_{2}^{*}+x_{1}\otimes x_{1}^{*}\wedge y_{2}^{*}+x_{1}\otimes x_{2}^{*}\wedge
y_{1}^{*}+\varepsilon x_{2}\otimes x_{1}^{*}\wedge y_{1}^{*}+\varepsilon
x_{2}\otimes x_{2}^{*}\wedge y_{2}^{*},\\\\[5.69054pt] c_{12}&=&x_{1}\otimes
y_{1}^{*}\wedge x_{1}^{*}+\varepsilon x_{1}\otimes h_{1}^{*}\wedge
y_{1}^{*}+\varepsilon x_{2}\otimes x_{1}^{*}\wedge y_{1}^{*}+\varepsilon
x_{2}\otimes x_{2}^{*}\wedge y_{2}^{*}\\\\[5.69054pt] &&+y_{1}\otimes
h_{1}^{*}\wedge y_{1}^{*}+y_{1}\otimes x_{1}^{*}\wedge y_{2}^{*},\end{array}$

and

$\begin{array}[]{lcllcl}\mathfrak{p}_{4}(x_{1})&=&h_{1},&\mathfrak{p}_{4}(y_{1})&=&\varepsilon^{2}h_{1}+\varepsilon^{3}x_{2}+\varepsilon
y_{2},\\\\[5.69054pt]
\mathfrak{p}_{5}(x_{1})&=&0,&\mathfrak{p}_{5}(y_{1})&=&0,\\\\[5.69054pt]
\mathfrak{p}_{9}(x_{1})&=&x_{1},&\mathfrak{p}_{9}(y_{1})&=&0,\\\\[5.69054pt]
\mathfrak{p}_{10}(x_{1})&=&x_{2},&\mathfrak{p}_{10}(y_{1})&=&\varepsilon
h_{1}+\varepsilon^{2}x_{2}+y_{2},\\\\[5.69054pt]
\mathfrak{p}_{11}(x_{1})&=&x_{2},&\mathfrak{p}_{11}(y_{1})&=&0,\\\\[5.69054pt]
\mathfrak{p}_{12}(x_{1})&=&x_{2},&\mathfrak{p}_{12}(y_{1})&=&0.\end{array}$

The case where $\varepsilon=1$: the space
$\mathrm{H}^{2}_{\alpha}(\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha};\mathfrak{oo}_{I\Pi}^{(1)}(1|2)_{\alpha})$
is generated by the non-trivial 2-cocycles:

$(c_{4},\mathfrak{p}_{4}),\quad(c_{5},\mathfrak{p}_{5}),\quad(c_{10},\mathfrak{p}_{10}),\quad(c_{11},\mathfrak{p}_{11}).$

###  6.4. Deformations of Hom-Lie superalgebras

The deformation theory of Hom-Lie superalgebras in characteristic 2 will be
discussed here. As a result, we also cover the Lie case, namely
$\alpha=\mathrm{Id}_{\mathfrak{g}}$. Over a field of characteristic zero, the
study has been carried out in [AMS, TR].

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra over a
field $\mathbb{K}$ of characteristic 2. A deformation of $\mathfrak{g}$ is a
family of Hom-Lie superalgebras $\mathfrak{g}_{t}$ specializing in
$\mathfrak{g}$ when the parameter $t$ equals $0$ and where the Hom-Lie
superalgebra structure is defined on the tensor product
$\mathfrak{g}\otimes{\mathbb{K}}[[t]]$ when $\mathfrak{g}$ is finite
dimensional. The bracket in the deformed Hom-Lie superalgebra
$\mathfrak{g}_{t}$ is a $\mathbb{K}[[t]]$-bilinear map of the form (for all
$x,y\in\mathfrak{g}$):

$\begin{array}[]{lcl}\displaystyle[x,y]_{t}&=&\displaystyle[x,y]+\sum_{i\geq
1}c_{i}(x,y)t^{i},\end{array}$

while the squaring $s_{t}$, with respect to $\mathbb{K}[[t]]$, on the Hom-Lie
superalgebra $\mathfrak{g}_{t}$ is given by (for all
$x\in\mathfrak{g}_{\bar{1}}$):

$\begin{array}[]{lcl}\displaystyle s_{t}(x)&=&\displaystyle s(x)+\sum_{i\geq
1}{\mathfrak{p}}_{i}(x)t^{i},\end{array}$

where $(c_{i},\mathfrak{p}_{i})\in
XC^{2}_{\alpha,{\bar{0}}}(\mathfrak{g};\mathfrak{g})$ for all $i\geq 1$. We
will assume that $c_{0}(\cdot,\cdot)=[\cdot,\cdot]$ and
${\mathfrak{p}}_{0}(\cdot)=s(\cdot)$.

According to deformation theory, we call a deformation infinitesimal if the
bracket $[\cdot,\cdot]_{t}$ and the squaring $s_{t}(\cdot)$ define a Hom-Lie
superalgebra structure$\mod(t^{2})$ (degree 1), that is
$[\cdot,\cdot]_{t}=[\cdot,\cdot]+c_{1}(\cdot,\cdot)t$ and
$s_{t}(\cdot)=s(\cdot)+{\mathfrak{p}}_{1}(\cdot)t$. A deformation is said to
be of order $n$ if the bracket $[\cdot,\cdot]_{t}$ and the squaring
$s_{t}(\cdot)$ define a Hom-Lie superalgebra structure$\mod(t^{n+1})$, that is

$[\cdot,\cdot]_{t}=[\cdot,\cdot]+\sum_{1\leq i\leq
n}c_{i}(\cdot,\cdot)t^{i}\text{ and }s_{t}(\cdot)=s(\cdot)+\sum_{1\leq i\leq
n}{\mathfrak{p}}_{i}(\cdot)t^{i}.$

Afterwards, one tries to extend a deformation of order $n$ to a deformation of
order $n+1$. All obstructions are cohomological, as we will see.

######  Theorem 6.8.

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra in
characteristic 2 and $(\mathfrak{g}_{t},[\cdot,\cdot]_{t},s_{t},\alpha)$ be a
deformation. Assume that $(c_{1},\mathfrak{p}_{1})\neq(0,0)$. Then

(i) $(c_{1},\mathfrak{p}_{1})$ is a 2-cocycle, i.e.
$(c_{1},\mathfrak{p}_{1})\in Z_{\alpha}^{2}(\mathfrak{g};\mathfrak{g})$.

(ii)  For $n>1$, consider the following maps:

$\begin{array}[]{lcl}C_{n}(x,y,z)&:=&\displaystyle\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{i}(c_{j}(x,y),\alpha(z))+\circlearrowleft(x,y,z),\quad\text{for
all $x,y,z\in\mathfrak{g}$,}\\\\[14.22636pt]
Q_{n}(x,y)&:=&\displaystyle\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{i}(\mathfrak{p}_{j}(x),\alpha(y))+\displaystyle\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{i}(c_{j}(x,y),\alpha(x)),\;\;\text{for all
$x\in\mathfrak{g}_{\bar{1}}$ and $y\in\mathfrak{g}$.}\end{array}$

A deformation of order $n-1$ can be extended to a deformation of order $n$ if
and only there exists $(c_{n},\mathfrak{p}_{n})$

$(C_{n},Q_{n})=\mathfrak{d}_{\alpha}^{2}(c_{n},\mathfrak{p}_{n}).$

###### Proof.

(i) Checking that $c_{1}$ satisfies the condition (34) is a routine, see
[AMS]. Let us deal with the squaring $s_{t}$. We have

(36)
$\begin{array}[]{l}[s_{t}(x),\alpha(y)]_{t}=\displaystyle[s(x)+\sum_{i\geq
1}{\mathfrak{p}}_{i}(x)t^{i},\alpha(y)]_{t}\\\\[2.84526pt]
=\displaystyle[s(x),\alpha(y)]+\sum_{i\geq
1}c_{i}(s(x),\alpha(y))t^{i}+\sum_{i\geq
1}[{\mathfrak{p}}_{i}(x),\alpha(y)]t^{i}+\sum_{i,j\geq
1}c_{j}({\mathfrak{p}}_{i}(x),\alpha(y))t^{i+j}.\end{array}$

On the other hand,

(37)
$\begin{array}[]{lcl}[\alpha(x),[x,y]_{t}]_{t}&=&\displaystyle[\alpha(x),[x,y]+\sum_{i\geq
1}c_{i}(x,y)t^{i}]_{t}\\\\[5.69054pt]
&=&\displaystyle[\alpha(x),[x,y]]+\sum_{i\geq
1}c_{i}(\alpha(x),[x,y])t^{i}+\sum_{i\geq
1}[\alpha(x),c_{i}(x,y)]t^{i}+\displaystyle\\\\[5.69054pt] &&+\sum_{i,j\geq
1}c_{i}(\alpha(x),c_{j}(x,y))t^{i+j}.\end{array}$

Collecting the coefficient of $t$ in the condition
$[s_{t}(x),\alpha(y)]_{t}=[\alpha(x),[x,y]_{t}]_{t},$ we get

$c_{1}(s(x),\alpha(y))+[\alpha(y),{\mathfrak{p}}_{i}(x)]+c_{1}(\alpha(x),[x,y])+[\alpha(x),c_{1}(x,y)]=0,$

which corresponds to Condition (6.3). Therefore, $(c_{1},{\mathfrak{p}_{1}})$
is a 2-cocycle on $\mathfrak{g}$ with values in the adjoint representation.

(ii) Let us first show that the pair $(C_{n},Q_{n})$ is a cochain in
$XC^{3}(\mathfrak{g},\mathfrak{g})$. Indeed,

$\displaystyle
Q_{n}(x_{1}+x_{2},y)=\sum_{i,j}c_{i}(\mathfrak{p}_{j}(x_{1}+x_{2}),\alpha(y))+c_{i}(c_{j}(x_{1}+x_{2},y),\alpha(x_{1}+x_{2}))$
$\displaystyle=\sum_{i,j}c_{i}(\mathfrak{p}_{j}(x_{1})+\mathfrak{p}_{j}(x_{2})+c_{j}(x_{1},x_{2}),\alpha(y))+c_{i}(c_{j}(x_{1},y)+c_{j}(x_{2},y),\alpha(x_{1})+\alpha(x_{2}))$
$\displaystyle=Q_{n}(x_{1},y)+Q_{n}(x_{2},y)+\sum_{i,j}(c_{i}(c_{j}(x_{1},x_{2}),\alpha(y))+c_{i}(c_{j}(x_{1},y),\alpha(x_{2}))+c_{i}(c_{j}(x_{2},y),\alpha(x_{1})))$
$\displaystyle=Q_{n}(x_{1},y)+Q_{n}(x_{2},y)+C_{n}(x_{1},x_{2},y).$

Collecting coefficients of $t^{n}$ in (34) leads to
$C_{n}=d^{2}_{\alpha}c_{n}$, see [AMS]. Let us deal with the squaring.
Consider the coefficient of $t^{n}$ in the condition
$[s_{t}(x),\alpha(y)]_{t}=[\alpha(x),[x,y]_{t}]_{t},$ and using Eqns. (36,
37), we get (for all $x\in\mathfrak{g}_{\bar{1}}$ and $y\in\mathfrak{g}$):

$\begin{array}[]{lcl}c_{n}(s(x),\alpha(y))+[{\mathfrak{p}}_{n}(x),\alpha(y)]+\displaystyle\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{j}({\mathfrak{p}}_{i}(x),\alpha(y))&=&c_{n}(\alpha(x),[x,y])+[\alpha(x),c_{n}(x,y)]\\\\[5.69054pt]
&&\displaystyle+\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{i}(\alpha(x),c_{j}(x,y)).\end{array}$

Let us rewrite this expression. We get

$d^{2}_{\alpha}{\mathfrak{p}}_{n}(x,y)=\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{i}(\alpha(x),c_{j}(x,y))+\displaystyle\sum_{\begin{subarray}{c}i+j=n\\\
i,j<n\end{subarray}}c_{j}({\mathfrak{p}}_{i}(x),\alpha(y)).$

Therefore,
$(C_{n},Q_{n})=(d^{2}_{\alpha}c_{n},d^{2}_{\alpha}\mathfrak{p}_{n})={\mathfrak{d}}^{2}_{\alpha}(c_{n},\mathfrak{p}_{n}).$
∎

Now, we discuss equivalent deformations.

######  Definition 6.9.

Let $(\mathfrak{g},[\cdot,\cdot],s,\alpha)$ be a Hom-Lie superalgebra in
characteristic 2. Let $(\mathfrak{g}_{t},[\cdot,\cdot]_{t},s_{t},\alpha)$ and
$(\tilde{\mathfrak{g}}_{t},\widetilde{[\cdot,\cdot]}_{t},\tilde{s}_{t},\alpha)$
be two deformations of $\mathfrak{g}$, such that
$\tilde{[\cdot,\cdot]}_{0}=[\cdot,\cdot]_{0}=[\cdot,\cdot]$ and
$\tilde{s}_{0}=s_{0}=s$. We say that the two deformations $\mathfrak{g}_{t}$
and $\tilde{\mathfrak{g}}_{t}$ are equivalent if there exists a
$\mathbb{K}[[t]]$-linear map
$\tau:\mathfrak{g}_{t}\rightarrow\tilde{\mathfrak{g}}_{t}$ of the form
$\tau(a)=\mathrm{id}_{\mathfrak{g}}(a)+\sum_{i\geq 1}\tau_{i}(a)t^{i}$ for all
$a\in\mathfrak{g}$, that is an isomorphism of Hom-Lie superalgebras.

######  Theorem 6.10.

Two 1-parameter formal deformations $\mathfrak{g}_{t}$ and
$\tilde{\mathfrak{g}}_{t}$ of $\mathfrak{g}$ given by the collections
$(c,{\mathfrak{p}})$ and $(\tilde{c},\tilde{\mathfrak{p}})$ are equivalent
through an isomorphism of the form
$\tau=\mathrm{id}_{\mathfrak{g}}+\sum_{i\geq 1}t^{i}\tau_{i}$ if and only if
$\tau$ links $(c,{\mathfrak{p}})$ and $(\tilde{c},\tilde{\mathfrak{p}})$ by
the following formulae (for all $n>0$):

(38)
$\sum_{i+j=n}\tau_{i}(\tilde{c}_{j}(x,y))=\sum_{i+j+k=n}c_{i}(\tau_{j}(x),\tau_{k}(y)),\quad\text{for
all $x,y\in\mathfrak{g}$},$

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

(39)
$\displaystyle\sum_{i+j=n}\tau_{i}(\tilde{\mathfrak{p}}_{j}(x))=\displaystyle\sum_{i+j=n}c_{i}(x,\tau_{j}(x))+\sum_{2i+j=n}{\mathfrak{p}}_{j}(\tau_{i}(x))+\displaystyle\sum_{\begin{subarray}{c}u+v+j=n\\\
1\leq u<v\\\ 1\leq j\end{subarray}}c_{j}([\tau_{u}(x),\tau_{v}(x)]).$

In particular, if $n=1$ we get

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

Hence, $({\tilde{c}}_{1},\tilde{\mathfrak{p}}_{1})$ and
$(c_{1},\mathfrak{p}_{1})$ are in the same cohomology class.

######  Corollary 6.11.

Infinitesimal deformations are classified by the cohomology group
$\mathrm{H}^{2}_{\alpha}(\mathfrak{g};\mathfrak{g})$.

###### Proof of Theorem 6.10.

Checking Eq. (38) is a routine, see [AMS]. Let us check Eq. (39).

We have

$\displaystyle\tau(\tilde{s}_{t}(x))=\tau\left(s(x)+\sum_{i\geq
1}\tilde{\mathfrak{p}}_{i}(x)t^{i}\right)=s(x)+\sum_{i\geq
1}\tilde{\mathfrak{p}}_{i}(x)t^{i}+\sum_{i,j\geq
1}\tau_{j}(\tilde{\mathfrak{p}}_{i}(x))t^{i+j}+\sum_{j\geq
1}\tau_{j}(s(x))t^{j}$ $\displaystyle=\sum_{i,j\geq
0}\tau_{j}(\tilde{\mathfrak{p}}_{i}(x))t^{i+j}$

On the other hand, we get

$\begin{array}[]{lcl}s_{t}(\tau(x))&=&\displaystyle s\left(x+\sum_{i\geq
1}\tau_{i}(x)t^{i}\right)+\sum_{i\geq 1}{\mathfrak{p}}_{i}\left(x+\sum_{i\geq
1}\tau_{i}(x)t^{i}\right)t^{j}\\\\[8.53581pt] &=&\displaystyle
s(x)+\sum_{i\geq 1}s(\tau_{i}(x))t^{2i}+\sum_{i\geq
1}[x,\tau_{i}(x)]t^{i}+\sum_{i\geq 1}{\mathfrak{p}}_{j}\left(x+\sum_{i\geq
1}\tau_{i}(x)t^{i}\right)t^{j}\\\\[5.69054pt] &=&\displaystyle
s(x)+\sum_{i\geq 1}s(\tau_{i}(x))t^{2i}+\sum_{i\geq
1}[x,\tau_{i}(x)]t^{i}+\sum_{i\geq 1}{\mathfrak{p}}_{j}(x)t^{j}+\sum_{i\geq
1}{\mathfrak{p}}_{j}(\tau_{i}(x))t^{2i+j}\\\\[5.69054pt]
&&\displaystyle+\sum_{i,j\geq
1}c_{j}(x,\tau_{i}(x))t^{i+j}+\displaystyle\sum_{1\leq
j}\sum_{\begin{subarray}{c}1\leq
u<v\end{subarray}}c_{j}(\tau_{u}(x),\tau_{v}(x))t^{j+u+v}\\\\[8.53581pt]
&=&\displaystyle\sum_{i,j\geq
0}{\mathfrak{p}}_{j}(\tau_{i}(x))t^{2i+j}+\displaystyle+\sum_{i,j\geq
0}c_{j}(x,\tau_{i}(x))t^{i+j}+\displaystyle\sum_{1\leq
j}\sum_{\begin{subarray}{c}1\leq
u<v\end{subarray}}c_{j}(\tau_{u}(x),\tau_{v}(x))t^{j+u+v}\end{array}$

The result follows by identification. ∎

## References

  * [AM] Ammar F. and Makhlouf A., Hom-Lie superalgebras and Hom-Lie admissible superalgebras. J. Algebra 324 (2010), no. 7, 1513–1528.
  * [AMS] Ammar F., Makhlouf A. and Saadaoui N., Cohomology of Hom-Lie superalgebras and $q$-deformed Witt superalgebra. Czechoslovak Math. J. 63(138) (2013), no. 3, 721–761.
  * [BeBou] Benayadi S. and Bouarroudj S., Double extensions of Lie superalgebras in characteristic 2 with non-degenerate invariant supersymmetric bilinear forms. J. of Algebra, 510 (2018), 141–179.
  * [BGKL] Bouarroudj S., Grozman P., Krutov A. and Leites D., Irreducible modules over exceptional modualr Lie superalgebras with indecomposable Cartan matrix. In preparation.
  * [BGLL1] Bouarroudj S., Grozman P., Lebedev A. and Leites D., Derivations and central extensions of simple modular Lie algebras and superalgebras; arXiv:1307.1858
  * [BGLLS] Bouarroudj S., Grozman P., Lebedev A., Leites D. and Shchepochkina I., Simple vectorial Lie algebras in characteristic $2$ and their superizations. Symmetry, Integrability and Geometry: Methods and Applications (SIGMA) 16 (2020), 089, 101 pages.
  * [BGL] Bouarroudj S., Grozman P., Leites D., Classification of finite-dimensional modular Lie superalgebras with indecomposable Cartan matrix. Symmetry, Integrability and Geometry: Methods and Applications (SIGMA), 5 (2009), 060, 63 pages.
  * [BLLSq] Bouarroudj S., Lebedev A., Leites D. and Shchepochkina I., Classifications of simple Lie superalgebras in characteristic 2, https://doi.org/10.1093/imrn/rnab265; arXiv:1407.1695
  * [BLS] Bouarroudj S., Leites D. and Shang J., Computer-aided study of double extensions of restricted Lie superalgebras preserving the non-degenerate closed 2-forms in characteristic 2. Experimental Mathematics, 31, No. 2, (2022), 676-688; https://doi.org/10.1080/10586458.2019.1683102.
  * [CKL] Chaichian M., Kulish K. and Lukierski J., $q$-Deformed Jacobi identity, $q$-oscillators and $q$-deformed infinite-dimensional algebras, Phys. Lett. B 237, no. 3-4 (1990), 401–406.
  * [Fu] Fuks (Fuchs) D., Cohomology of infinite dimensional Lie algebras, Consultants Bureau, NY, 1986.
  * [GSS] Garcia-Delgado R., Salgado G. and Sanchez-Valenzuela O.A., On 3-dimensional complex Hom-Lie algebras, J. Algebra 555 (2020), 361–385.
  * [GSSc] Garcia-Delgado R., Salgado G. and Sanchez-Valenzuela O.A., Corrigendum to “On 3-dimensional complex Hom-Lie algebras” [J. Algebra 555 (2020) 361–385], J. Algebra 562 (2020), 286–289.
  * [GC] Guan B. and Chen L., Restricted hom-Lie algebras, Hacettepe Journal of Mathematics and Statistics, 2015, 44(4), pp. 823–837.
  * [HLS] Hartwig J.T., Larsson D. and Silvestrov S., Deformations of Lie algebras using $\sigma$-derivations, J. Algebra 295 (2006), no. 2, 314–361.
  * [J] Jacobson, N., Lie algebras. Interscience, New York (1962).
  * [LS] Larsson D. and Silvestrov S.D., Quasi-hom-Lie algebras, central extensions and $2$-cocycle-like identities, J. Algebra 288 (2005), no. 2, 321–344.
  * [LS2] Larsson D. and Silvestrov S., Graded quasi-Lie algebras, Czech. J. Phys. 55 (2005), 1473–1478.
  * [LeD] Lebedev A., Analogs of the orthogonal, Hamiltonian, Poisson, and contact Lie superalgebras in characteristic 2, J. Nonlinear Math. Phys. 17 (2010), suppl. 1, 217–251.
  * [LL] Li X. and Li Y., Classification of 3-dimensional multiplicative Hom-Lie algebras, J. Xinyang Norm. Univ. Nat. Sci. Ed. 25, 455–475 (2012).
  * [MS] Makhlouf A. and Silvestrov S. D., Hom-algebra structures, J. Gen. Lie Theory Appl. 2(2), 51–64 (2008).
  * [ORS1] Ongong’a E., Richter J. Silvestrov S., Classification of 3-dimensional Hom-Lie algebras, IOP Conf. Series: Journal of Physics: Conf. Series 1194 (2019) 012084 IOP Publishing doi:10.1088/1742-6596/1194/1/012084
  * [ORS2] Ongong’a E., Richter J. Silvestrov S., Classification of Low-Dimensional Hom-Lie Algebras, Classification of Low-Dimensional Hom-Lie Algebras. In: Silvestrov, S., Malyarenko, A., Rancic, M. (eds) Algebraic Structures and Applications. SPAS 2017. Springer Proceedings in Mathematics $\&$ Statistics, vol 317. Springer, Cham. https://doi.org/10.1007/978-3-030-41850-2_9 (2020).
  * [R] Remm E., 3-Dimensional Skew-symmetric Algebras and the Variety of Hom-Lie Algebras, Algebra Colloquium Vol. 25, No. 04, pp. 547–566 (2018).
  * [Sh] Sheng Y. and Xiong Z., On Hom-Lie algebras, Linear and Multilinear Algebra, 63, (2015), no. 12, 2379–2395.
  * [TR] Tripathy K. C. and Patra M. K., Cohomology theory and deformations of $\mathbb{Z}_{2}$-graded Lie algebras, J. Math. Phys. 31, 2822 (1990).
  * [WZW] Wang C., Zhang Q. and Wei Z, A classification of low dimensional multiplicative Hom-Lie superalgebras, Open Mathematics 14(1), 613–628 (2016)
  * [XY] Xiao-chao L. and Yong-feng L., Classification of 3-dimensional multiplicative Hom-Lie algebras, J. Xinyang Norm. Univ., 25 (2012), 427–455.
  * [Y] Yau D., Hom-algebras and homology, Journal of Lie Theory 19 (2009) 409–421.