# Double extensions for quadratic Hom-Lie algebras with equivariant twist maps

R. García-Delgado, G. Salgado, O. A. Sánchez-Valenzuela  Unidad Mérida del
CIMAT; Mérida, Yucatán, México<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>Facultad de
Ciencias Universidad Autónoma de San Luis Potosí; San Luis Potosí, México
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Quadratic Hom-Lie algebras with equivariant twist maps are studied. They are
completely characterized in terms of a maximal proper ideal that contains the
kernel of the twist map and a complementary subspace to it that is either
$1$-dimensional, or has the structure of a simple Lie algebra. It is shown how
the analogue of the double extension construction works well for quadratic
Hom-Lie algebras with equivariant twist maps and prove that any indecomposable
and quadratic Hom-Lie algebra with equivariant and nilpotent twist map can be
identified with such a double extension.

###### Key words and phrases:

Quadratic Hom Lie algebras; Double extensions; Simple Lie algebras; Nilpotent
Lie algebras.

###### 1991 Mathematics Subject Classification:

Primary: 17B61, 17B30. Secondary: 17A30, 17B60, 17D30.

## Introduction

Two important tools in the study of Lie algebras $\mathfrak{g}$ equipped with
invariant metrics are the Medina-Revoy theorem on the one hand, and the double
extension procedure on the other [8]. The generalization of these tools to
different algebraic structures defined on $\mathfrak{g}$ —eg, like quadratic
Lie superalgebras as approached in [1] or like quadratic Hom-Lie algebras as
approached in [2]— has some difficulties. It is known, for example, that if
$I$ is a maximal proper ideal of a Lie algebra $\mathfrak{g}$, then the
sequence,

(1)
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{g}/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

splits (see [4], Prop. A.1). It is the splitting of this sequence what makes
it possible to exploit the Medina-Revoy theorem and the double extension
construction. So far, however, it is not known if this sequence splits under
more general hypotheses for $\mathfrak{g}$ (ie, for different algebraic
structures defined on $\mathfrak{g}$). For quadratic Lie superalgebras, we
refer the reader to [1].

The aim of this work is to prove that (1) splits for quadratic Hom-Lie
algebras having an equivariant twist map (see Thm. 2.1). Then we also show how
to generalize the “double extension” procedure in the quadratic Hom-Lie
setting (see Prop. 1.1 and Prop 1.2) and prove that any indecomposable and
quadratic Hom-Lie algebra with equivariant and nilpotent twist map can be
constructed using this double extension mechanism (see Thm. 2.2).

We may refer the reader to [5] as the general background for this work. It is
shown there how to produce a quadratic Hom-Lie algebra with equivariant twist
map on a vector space of the form $\mathfrak{h}\oplus\mathfrak{h}^{\ast}$,
once we are given a Hom-Lie algebra with equivariant twist map defined on
$\mathfrak{h}$. The construction was inspired in the cotangent bundle
associated to a quadratic Lie algebra. In particular, the family of such
examples $\mathfrak{h}\oplus\mathfrak{h}^{\ast}$ exhibited Hom-Lie products
which were not Lie brackets. In this work we start again as in [5] with a
quadratic Hom-Lie algebra $\mathfrak{h}$ whose twist map is equivariant and
produce a quadratic Hom-Lie algebra that can be decomposed in the form
$\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, having an
equivariant twist map, and with $\mathfrak{s}$ being either a one-dimensional
subspace or $\mathfrak{s}$ being the underlying vector space of a quadratic
Lie algebra. In addition, $\mathfrak{s}^{\ast}$ is contained in the kernel of
the corresponding twist map (see Prop. 1.1 and Prop. 1.2). We then show that
any indecomposable quadratic Hom-Lie algebra with equivariant and nilpotent
twist map has this structure, and furthermore, $\mathfrak{s}$ is a simple Lie
algebra (see Thm. 2.2). This goal is achieved by choosing first a maximal
proper ideal $I$ of the Hom-Lie algebra $\mathfrak{g}$ such that $I$ contains
the kernel of the twist map. We then prove that there exists a proper subspace
$\mathfrak{s}\subset\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{s}\oplus
I$. We finally show that $\mathfrak{s}$ is either one-dimensional or that
$\mathfrak{s}$ has a Lie bracket making it into a simple Lie algebra (see Thm
2.1). In particular, we prove the splitting of the sequence (1) under the
given hypotheses for the Hom-Lie algebras addressed in this work.

## Definitions and Notation

The setting of this work is that of [5]. We study _quadratic Hom-Lie algebras_
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ on a finite-dimensional vector
space $\mathfrak{g}$ defined over an algebraically closed field $\mathbb{F}$
of characteristic zero. The Hom-Lie product
$[\,\cdot\,,\,\cdot\,]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$, is a
skew-symmetric bilinear map, whose twist linear map
$T\in\operatorname{End}(\mathfrak{g})$, satisfies,

$[T(x),[y,z]]+[T(y),[z,x]]+[T(z),[x,y]]=0,$

for all $x,y$ and $z$ in $\mathfrak{g}$. If
$T=\operatorname{Id}_{\mathfrak{g}}$, then $[\,\cdot\,,\,\cdot\,]$ is a Lie
algebra bracket on $\mathfrak{g}$ and the triple
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],B)$ is a quadratic Lie algebra. Within
the context of quadratic Hom-Lie algebras
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ addressed in this work, it is
assumed that $B$ is a symmetric non-degenerate bilinear form on $\mathfrak{g}$
which is invariant under $[\,\cdot\,,\,\cdot\,]$; that is,
$B([x,y],z)=B(x,[y,z])$, for all $x,y$ and $z$ in $\mathfrak{g}$. Finally, it
is assumed that the twist map $T$ is $B$-self-adjoint —that is,
$B(T(x),y)=B(x,T(y))$, for all $x,y$ in $\mathfrak{g}$— and that it satisfies
the equivariance property,

$T([x,y])=[T(x),y],\,\text{ for all }x,y\in\mathfrak{g}.$

In this setting, (see [5]), an ideal is a subspace $I$ of $\mathfrak{g}$ such
that $[\mathfrak{g},I]\subset I$ and $T(I)\subset I$. For each $x$ in
$\mathfrak{g}$, the linear map $y\mapsto[x,y]$, shall be denoted by
$\operatorname{ad}(x)$. A linear map $D:\mathfrak{g}\to\mathfrak{g}$,
satisfying, $D([x,y])=[D(x),y]+[x,D(y)]$, for all $x,y$ in $\mathfrak{g}$, is
called a derivation. In general, $\operatorname{ad}(x)$ is not a derivation
because $[\,\cdot\,,\,\cdot\,]$ is not a Lie bracket. We shall denote by
$\operatorname{Der}(\mathfrak{g})$ the vector subspace of
$\operatorname{End}(\mathfrak{g})$ consisting of all derivations. Also denote
by $\mathfrak{o}(B)$ the following subspace of
$\operatorname{End}(\mathfrak{g})$:

$\mathfrak{o}(B)=\\{S\in\operatorname{End}(\mathfrak{g})\,\mid\,B(S(x),y)=-B(x,S(y)),\text{
for all }x,y\in\mathfrak{g}\\}.$

It is trivially seen that if $S_{1}$ and $S_{2}$ belong to $\mathfrak{o}(B)$,
the commutator $S_{1}\circ S_{2}-S_{2}\circ S_{1}$ also belongs to
$\mathfrak{o}(B)$; it is the usual orthogonal Lie algebra defined by $B$.

Convention: Besides Hom-Lie algebras, we shall also be dealing with some Lie
algebras. For a given Lie algebra $\mathfrak{h}$, its Lie bracket will be
denoted by $[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$. Thus, for example, we have
in general that, $[S_{1},S_{2}]_{\mathfrak{gl}(\mathfrak{g})}=S_{1}\circ
S_{2}-S_{2}\circ S_{1}$, for $S_{1}$ and $S_{2}$ in
$\operatorname{End}(\mathfrak{g})$, which restricts precisely to
$[S_{1},S_{2}]_{\mathfrak{o}(B)}$, when $S_{1}$ and $S_{2}$ belong to
$\mathfrak{o}(B)$. In using the notation
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$ we shall in general specify what the
context is. For example, in Prop. 1.1 below,
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$ is not a Lie bracket, but it is the
Hom-Lie algebra product in a quadratic Hom-Lie algebra explicilty specified by
the quadruple
$(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},T,B_{\mathfrak{h}})$.

Let $B^{\flat}:\mathfrak{g}\to\mathfrak{g}^{\ast}$ be the linear map defined
by $B^{\flat}(x)\,(y)=B(x,y)$, for all $x$ and $y$ in $\mathfrak{g}$. In
particular, $S\in\mathfrak{o}(B)$ if and only if, $B^{\flat}\circ
S=-S^{\ast}\circ B^{\flat}$, where $S^{\ast}$ is the dual map
$\mathfrak{g}^{\ast}\to\mathfrak{g}^{\ast}$ given by, $\xi\mapsto
S^{\ast}(\xi)=\xi\circ S$. In particular, if $\mathfrak{s}$ is a Lie algebra
having an invariant metric $B_{\mathfrak{s}}$ for its Lie bracket
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}$, then its coadjoint representation
$\operatorname{ad}_{\mathfrak{s}}^{*}:\mathfrak{s}\to\mathfrak{gl}(\mathfrak{s}^{\ast})$,
defined by
$\operatorname{ad}_{\mathfrak{s}}^{*}(x)(\xi)=-\xi\circ\operatorname{ad}_{\mathfrak{s}}(x)$
for any $x$ in $\mathfrak{s}$ and any $\xi$ in $\mathfrak{s}^{*}$, satisfies,
$B_{\mathfrak{s}}^{\flat}(y)\circ\operatorname{ad}_{\mathfrak{s}}(x)=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(B_{\mathfrak{s}}^{\flat}(y))$.

Let $(\mathfrak{g}_{1},[\,\cdot\,,\,\cdot\,]_{1},T_{1})$ and
$(\mathfrak{g}_{2},[\,\cdot\,,\,\cdot\,]_{2},T_{2})$ be Hom-Lie algebras. A
homomorphism between them, is a linear map
$\psi\\!:\\!\mathfrak{g}_{1}\to\\!\mathfrak{g}_{2}$ such that $\psi\circ
T_{1}=T_{2}\circ\psi$, and,
$\psi(\,[x,y]_{1})\\!\\!=\\!\\![\psi(x),\psi(y)]_{2}$, for all $x$ and $y$ in
$\mathfrak{g}_{1}$.

##  1\. Double extensions of quadratic Hom-Lie algebras with equivariant twist
maps

Let $(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},L,B_{\mathfrak{h}})$
be a quadratic Hom-Lie algebra with equivariant twist map $L$ and let
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}},B_{\mathfrak{s}})$ be a
quadratic Lie algebra. Our first result states what is needed in order to turn
the space
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ into a
quadratic Hom-Lie algebra with equivariant twist map.

######  1.1 Proposition.

Let $(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},L,B_{\mathfrak{h}})$
be a finite-dimensional quadratic Hom-Lie algebra with equivariant $L$ with
respect to its Hom-Lie product $[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$ and let
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}},B_{\mathfrak{s}})$ be an
$r$-dimensional quadratic Lie algebra. Suppose there are linear maps
$f:\mathfrak{s}\to\mathfrak{h}$ and
$\rho:\mathfrak{s}\to\mathfrak{o}(B_{\mathfrak{h}})$ such that:

(i) $f\circ\operatorname{ad}_{\mathfrak{s}}(x)=\rho(x)\circ f$, for all $x$ in
$\mathfrak{s}$.

(ii) $L\circ\rho(x)=\operatorname{ad}_{\mathfrak{h}}(f(x))=\rho(x)\circ
L\in\operatorname{Der}(\mathfrak{h})$, for all $x$ in $\mathfrak{s}$.

(iii) $\rho([x,y]_{\mathfrak{s}})\circ
L=[\rho(x),\rho(y)]_{\mathfrak{gl}(\mathfrak{h})}\circ L$, for all $x$ and $y$
in $\mathfrak{s}$.

Then, the vector space
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ has a
skew-symmetric bilinear Hom-Lie product $[\,\cdot\,,\,\cdot\,]$, an
equivariant twist map $T\in\operatorname{End}(\mathfrak{g})$ and a
nondegenerate symmetric bilinear form $B$ invariant under
$[\,\cdot\,,\,\cdot\,]$, making $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$
into a quadratic Hom-Lie algebra.

###### Proof.

Let $\\{x_{1},\ldots,x_{r}\\}$ be a basis of $\mathfrak{s}$ and let
$\\{\xi^{1},\ldots,\xi^{r}\\}$ be its corresponding dual basis for
$\mathfrak{s}^{\ast}$, so that, $\xi^{j}(x_{k})=\delta_{jk}$ ($1\leq j,k\leq
r$). Define a bilinear map,
$\Gamma:\mathfrak{h}\times\mathfrak{h}\to\mathfrak{s}^{\ast}$, by means of,

(2)
$\Gamma(u,v)=B_{\mathfrak{h}}(\rho(x_{1})(u),v)\xi^{1}+\ldots+B_{\mathfrak{h}}(\rho(x_{r})(u),v)\xi^{r}.$

Let
$\operatorname{ad}_{\mathfrak{s}}^{*}:\mathfrak{s}\to\mathfrak{gl}(\mathfrak{s}^{*})$
be the coadjoint representation of $\mathfrak{s}$ in $\mathfrak{s}^{*}$.
Define the following skew-symmetric bilinear product $[\,\cdot\,,\,\cdot\,]$
on $\mathfrak{g}$:

If $\displaystyle x,y\in\mathfrak{s},$
$\displaystyle\,[x,y]=[x,y]_{\mathfrak{s}};$ if $\displaystyle
x\in\mathfrak{s},\,v\in\mathfrak{h},$ $\displaystyle\,[x,v]=\rho(x)(v);$ if
$\displaystyle x\in\mathfrak{s},\,\xi\in\mathfrak{s}^{\ast},$
$\displaystyle\,[x,\xi]=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(\xi);$ if
$\displaystyle u,v\in\mathfrak{h},$
$\displaystyle[u,v]\\!=\\![u,v]_{\mathfrak{h}}+\Gamma(u,v),$

Now use the $[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}$-invartiant metric
$B_{\mathfrak{s}}:\mathfrak{s}\times\mathfrak{s}\to\mathbb{F}$, the maps
$f:\mathfrak{s}\to\mathfrak{h}$ and
$\rho:\mathfrak{s}\to\mathfrak{o}(B_{\mathfrak{h}})$ from the hypotheses and a
scalar $\eta\in\mathbb{F}$, to define the linear map
$T:\mathfrak{g}\to\mathfrak{g}$ as follows:

(3) $\begin{array}[]{cccl}\text{If\
}&x\in\mathfrak{s},&&\,T(x)=f(x)+\eta\,B_{\mathfrak{s}}^{\flat}(x);\\\
\text{if\ }&v\in\mathfrak{h},&&\,T(v)=L(v)+R(v);\\\ \text{if\
}&\xi\in\mathfrak{s}^{\ast},&&T(\xi)=0,\end{array}$

where $R(v)=f^{*}\left(B^{\flat}_{\mathfrak{h}}(v)\right)=(f^{*}\circ
B^{\flat}_{\mathfrak{h}})(v)$, for any $v\in\mathfrak{h}$. We shall now show
that $T$ is equivariant for the given skew-symmetric bilinear product on
$\mathfrak{g}$. First, for $x$ and $y$ in $\mathfrak{s}$ we have,

$\displaystyle T([x,y])$ $\displaystyle=f([x,y]_{\mathfrak{s}})+\eta
B_{\mathfrak{s}}^{\flat}([x,y]_{\mathfrak{s}})$
$\displaystyle=f\circ\operatorname{ad}_{s}(x)(y)+\eta
B_{\mathfrak{s}}^{\flat}\circ\operatorname{ad}_{\mathfrak{s}}(x)(y)$
$\displaystyle=\rho(x)(f(y))+\eta\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(B_{\mathfrak{s}}^{\flat}(y))$
$\displaystyle=[x,f(y)]+[x,\eta B_{\mathfrak{s}}^{\flat}(y)]=[x,T(y)].$

Now, for $x$ in $\mathfrak{s}$ and $v$ in $\mathfrak{h}$, we get,

$\displaystyle T([x,v])=T(\rho(x)(v))=L\circ\rho(x)(v)+R\circ\rho(x)(v)$
$\displaystyle=[f(x),v]_{\mathfrak{h}}+\Gamma(f(x),v)=[f(x),v]=[T(x),v],$

where we have used the following fact that follows from (i) in the statement
and the definition (2) of $\Gamma$:

$R\circ\rho(x)=\operatorname{ad}^{\ast}_{\mathfrak{s}}(x)\circ
R=\Gamma(f(x),\,\cdot\,),\quad\text{for all\ }x\text{\ in\ }\mathfrak{s}.$

Also observe that the following is true, which follows again from (2) and the
stated hypotheses:

$R([u,v]_{\mathfrak{h}})=\Gamma(L(u),v)=\Gamma(u,L(v)),\quad\text{for all\
}u,v\text{ in }\mathfrak{h}.$

Now for $u$ and $v$ in $\mathfrak{h}$ we have,

$\displaystyle T([u,v])$
$\displaystyle=L([u,v]_{\mathfrak{h}})+R([u,v]_{\mathfrak{h}})$
$\displaystyle=[L(u),v]_{\mathfrak{h}}+\Gamma(L(u),v)$
$\displaystyle=[L(u),v]=[T(u),v].$

Since $\mathfrak{s}^{\ast}\subset\operatorname{Ker}(T)$,
$\operatorname{Im}(T)\subset\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ and
$[\mathfrak{h},\mathfrak{s}^{\ast}]=\\{0\\}$, we have
$T([\mathfrak{g},\mathfrak{s}^{\ast}])=[T(\mathfrak{g}),\mathfrak{s}^{\ast}]=[\mathfrak{g},T(\mathfrak{s}^{\ast})]=\\{0\\}$.
Therefore, $T$ is indeed equivariant with respect to $[\,\cdot\,,\,\cdot\,]$.

We shall now prove that $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ is a Hom-Lie
algebra. First take $x$, $y$ and $z$ in $\mathfrak{s}$. We shall use the
hypothesis (i), the definition (3) of $T$, and the fact that
$B_{\mathfrak{s}}$ is invariant under $[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}$,
to conclude that,

(4)
$\begin{split}[T(x),[y,z]]&=[T(x),[y,z]_{\mathfrak{s}}]=[f(x),[y,z]_{\mathfrak{s}}]+\eta[B_{\mathfrak{s}}^{\flat}(x),[y,z]_{\mathfrak{s}}]\\\
&=-\rho([y,z]_{\mathfrak{s}})(f(x))-\eta\operatorname{ad}_{\mathfrak{s}}^{\ast}([y,z]_{\mathfrak{s}})(B_{\mathfrak{s}}^{\flat}(x))\\\
&=f([x,[y,z]_{\mathfrak{s}}]_{\mathfrak{s}})+\eta
B_{\mathfrak{s}}^{\flat}(x)\circ\operatorname{ad}_{\mathfrak{s}}([y,z]_{\mathfrak{s}})\\\
&=f([x,[y,z]_{\mathfrak{s}}]_{\mathfrak{s}})+\eta
B_{\mathfrak{s}}^{\flat}([x,[y,z]_{\mathfrak{s}}]_{\mathfrak{s}}).\end{split}$

Using the Jacobi identity for the Lie algebra $\mathfrak{s}$, we obtain,

$[T(x),[y,z]]+[T(y),[z,x]]+[T(z),[x,y]]=0,\quad\text{ for all
}x,y,z\in\mathfrak{s}.$

Next, prove that $[T(x),[y,v]]+[T(y),[v,x]]+[T(v),[x,y]]=0$, for all $x,y$ in
$\mathfrak{s}$ and $v$ in $\mathfrak{h}$. In this case we use the hypothesis
(ii) to get,

(5) $\begin{split}[T(x),[y,v]]&=[T(x),\rho(y)(v)]=[f(x),\rho(y)(v)]\\\
\,&=[f(x),\rho(y)(v)]_{\mathfrak{h}}+\Gamma(f(x),\rho(y)(v))\\\
\,&=\rho(x)\circ\rho(y)(L(v))+\Gamma(f(x),\rho(y)(v)).\end{split}$

Similarly, one has,

(6) $[T(y),[v,x]]=-\rho(y)\circ\rho(x)(L(v))-\Gamma(f(y),\rho(x)(v)).$

Finally, compute $[T(v),[x,y]]$ as follows:

(7)
$\begin{split}[T(v),[x,y]_{\mathfrak{s}}]&=[T(v),[x,y]_{\mathfrak{s}}]=[L(v),[x,y]_{\mathfrak{s}}]+[R(v),[x,y]_{\mathfrak{s}}]\\\
\,&=-\rho([x,y]_{\mathfrak{s}})(L(v))-\operatorname{ad}_{\mathfrak{s}}^{\ast}([x,y]_{\mathfrak{s}})(R(v)).\end{split}$

Adding (5), (6) and (7), and using the hypothesis (iii), we obtain,

$\displaystyle[T(x),[y,v]]+[T(y),[v,x]]+[T(v),[x,y]]$
$\displaystyle=\rho(x)\circ\rho(y)(L(v))-\rho(y)\circ\rho(x)(L(v))-\rho([x,y]_{\mathfrak{s}})(L(v))$
$\displaystyle\quad+\Gamma(f(x),\rho(y)(v))-\Gamma(f(y),\rho(x)(v))-\operatorname{ad}_{\mathfrak{s}}^{\ast}([x,y]_{\mathfrak{s}})(R(v))$
$\displaystyle=\Gamma(f(x),\rho(y)(v))-\Gamma(f(y),\rho(x)(v))-\operatorname{ad}_{\mathfrak{s}}^{\ast}([x,y]_{\mathfrak{s}})(R(v)).$

Also observe that,

(8) $\Gamma(f(x),\rho(y)(v))=\operatorname{ad}^{\ast}_{\mathfrak{s}}(x)\circ
R\circ\rho(y)(v)=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)\circ\operatorname{ad}_{\mathfrak{s}}^{\ast}(y)\circ
R(v).$

And similarly,

(9)
$\Gamma(f(y),\rho(x)(v))=\operatorname{ad}_{\mathfrak{s}}^{\ast}(y)\circ\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)\circ
R(v).$

Then, from (8) and (9), obtain,

$\Gamma(f(x),\rho(y)(v))-\Gamma(f(y),\rho(x)(v))-\operatorname{ad}_{\mathfrak{s}}^{\ast}([x,y]_{\mathfrak{s}})\circ
R(v)=0,$

which implies that $[T(x),[y,v]]+[T(y),[v,x]]+[T(v),[x,y]]=0$.

Now take $u$ and $v$ in $\mathfrak{h}$ and $x$ in $\mathfrak{s}$. We need to
prove that, $[T(x),[u,v]]+[T(u),[v,x]]+[T(v),[u,x]]=0$. Since
$[\mathfrak{h},\mathfrak{s}^{\ast}]=\\{0\\}$, we have,

(10)
$\begin{split}[T(x),[u,v]]&=[T(x),[u,v]_{\mathfrak{h}}]=[f(x),[u,v]_{\mathfrak{h}}]\\\
\,&=[f(x),[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}+\Gamma(f(x),[u,v]_{\mathfrak{h}}).\end{split}$

Since $L$ is equivariant with respect to
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$, we obtain by means of the hypothesis
(ii) that,

(11) $\begin{split}[T(u),[v,x]]&=-[T(u),\rho(x)(v)]=-[L(u),\rho(x)(v)]\\\
\,&=-[L(u),\rho(x)(v)]_{\mathfrak{h}}-\Gamma(L(u),\rho(x)(v))\\\
\,&=-[u,L(\rho(x)(v))]_{\mathfrak{h}}-\Gamma(L(u),\rho(x)(v))\\\
\,&=-[u,[f(x),v]_{\mathfrak{h}}]_{\mathfrak{h}}-\Gamma(L(u),\rho(x)(v)).\end{split}$

Under the same arguments we get,

(12)
$[T(v),[x,u]]=[v,[f(x),u]_{\mathfrak{h}}]_{\mathfrak{h}}+\Gamma(L(v),\rho(x)(u)).$

Using the hypotheses —specially (ii) and (iii)— one proves that for any $x$ in
$\mathfrak{s}$ and any $u$ and $v$ in $\mathfrak{h}$,

$\Gamma\left(f(x),[u,v]_{\mathfrak{h}}\right)=\Gamma\left(L(u),\rho(x)(v)\right)-\Gamma\left(L(v),\rho(x)(u)\right).$

Using this fact, we may now add (10), (11) and (12) to obtain,

$\displaystyle[T(x),[u,v]]+[T(u),[v,x]]+[T(v),[u,x]]$
$\displaystyle=[f(x),[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}-[u,[f(x),v]_{\mathfrak{h}}]_{\mathfrak{h}}+[v,[f(x),u]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle\quad+\Gamma(f(x),[u,v]_{\mathfrak{h}})-\Gamma(L(u),\rho(x)(v))+\Gamma(L(v),\rho(x)(u))=0.$

Finally, let $u,v$ and $w$ be in $\mathfrak{h}$. Using (i) and (iii) this
time, it is a straightforward matter to verify that,

(13)
$\Gamma\left(L(u),[v,w]_{\mathfrak{h}}\right)+\Gamma\left(L(v),[w,u]_{\mathfrak{h}}\right)+\Gamma\left(L(w),[u,v]_{\mathfrak{h}}\right)=0.$

To complete the proof that $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ we need to
check that the cyclic sum of the terms $[T(u),[v,w]]$ vanishes. Since,

$[T(u),[v,w]]=[T(u),[v,w]_{\mathfrak{h}}]=[L(u),[v,w]]_{\mathfrak{h}}+\Gamma(L(u),[v,w]_{\mathfrak{h}}),$

we only need to use the fact that
$(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},L)$ is a Hom-Lie algebra
and (13) to conclude that $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ is indeed a
Hom-Lie algebra.

Finally, let $B:\mathfrak{g}\times\mathfrak{g}\to\mathbb{F}$ be the non-
degenerate bilinear form defined on
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{*}$ by means
of,

(14) $B(x+u+\xi,y+v+\zeta)=\xi(y)+\zeta(x)+B_{\mathfrak{h}}(u,v),$

for any $x$ and $y$ in $\mathfrak{s}$, $\xi$ and $\zeta$ in
$\mathfrak{s}^{*}$, $u$ and $v$ in $\mathfrak{h}$.

We shall first prove that $T$ is $B$-self-adjoint. Take first $x$ and $y$ in
$\mathfrak{s}$, and use the symmetry of the invariant metric
$B_{\mathfrak{s}}$ on $\mathfrak{s}$ to conclude that,

$\displaystyle B(T(x),y)$ $\displaystyle=\eta
B(K^{\flat}_{\mathfrak{s}}(x),y)=\eta B_{\mathfrak{s}}(x,y)=\eta
B_{\mathfrak{s}}^{\flat}(y)(x)$ $\displaystyle=B(\eta
B_{\mathfrak{s}}^{\flat}(y),x)=B(T(y),x)=B(x,T(y)).$

Now take $x$ in $\mathfrak{s}$ and $v$ be in $\mathfrak{h}$. This time use the
definition of $R=f^{*}\circ B^{\flat}_{\mathfrak{h}}$, to obtain,

$\displaystyle
B(T(x),v)=B(f(x),v)=B_{\mathfrak{h}}(v,f(x))=B_{\mathfrak{h}}^{\flat}(v)\left(f(x)\right)$
$\displaystyle=\left(\left(f^{\ast}\circ
B_{\mathfrak{h}}^{\flat}\right)(v)\right)(x)=R(v)(x)=B(x,R(v))=B(x,T(v)).$

Finally, let $u$ and $v$ in $\mathfrak{h}$. Since $L$ is
$B_{\mathfrak{h}}$-self-adjoint,

$B(T(u),v)=B(L(u),v)=B_{\mathfrak{h}}(L(u),v)=B_{\mathfrak{h}}(u,L(v))=B(u,T(v)).$

Whence, $T$ is $B$-self-adjoint.

Finally, we need to prove that $B$ is invariant under $[\,\cdot\,,\,\cdot\,]$.
Take first $x$ in $\mathfrak{s}$ and $u$ and $v$ in $\mathfrak{h}$. Since
$\rho(x)$ belongs to $\mathfrak{o}(B_{\mathfrak{h}})$, we have,

$\displaystyle B([x,u],v)$
$\displaystyle=B(\rho(x)(u),v)=B_{\mathfrak{h}}(\rho(x)(u),v)=-B_{\mathfrak{h}}(u,\rho(x)(v))$
$\displaystyle=B(u,\rho(x)(v))=-B(u,[x,v]).$

Now, for any element $x_{j}$ in the given basis of $\mathfrak{s}$ ($1\leq
j\leq r$), we have,

$\displaystyle B(x_{j},[u,v])$
$\displaystyle=B(x_{j},\Gamma(u,v))=\sum_{k=1}^{r}B_{\mathfrak{h}}(\rho(x_{k})(u),v)\xi^{k}(x_{j})$
$\displaystyle=\sum_{k=1}^{r}B_{\mathfrak{h}}(\rho(x_{k})(u),v)\delta_{jk}=B_{\mathfrak{h}}(\rho(x_{j})(u),v)$
$\displaystyle=B(\rho(x_{j})(u),v)=B([x_{j},u],v).$

Therefore, $B(x,[u,v])=B([x,u],v)=-B(u,[x,v])$. Since $B_{\mathfrak{h}}$ is
already invariant under $[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$ it follows that
$B(u,[v,w])=B([u,v],w)$, for all $u$, $v$ and $w$ in $\mathfrak{h}$. And
finally, for $x$ and $y$ in $\mathfrak{s}$ and $\xi$ in $\mathfrak{s}^{\ast}$
we use the coadjoint representation to conclude that,

$\displaystyle B([x,y],\xi)$
$\displaystyle=\xi([x,y]_{\mathfrak{s}})=-\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(\xi)(y)$
$\displaystyle=-B(y,\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(\xi))=-B(y,[x,\xi]).$

∎

Given a quadratic Hom-Lie algebra $(\mathfrak{h},[\,\cdot\,,\,\cdot\,],B,L)$
with equivariant $L$, the next result states what is needed to produce a
quadratic Hom-Lie algebra with equivariant twist map in a vector space of the
form $\mathbb{F}d\oplus\mathfrak{h}\oplus\mathbb{F}c$, where $\mathbb{F}c$ is
a one-dimensional ideal and
$(\mathbb{F}c)^{\perp}=\mathfrak{h}\oplus\mathbb{F}c$.

######  1.2 Proposition.

Let $(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},B,L)$ be a quadratic
Hom-Lie algebra as in Prop. 1.1 and let $D\in\mathfrak{o}(B_{\mathfrak{h}})$
and $v^{\prime}\in\mathfrak{h}$ be such that:

(i) $L\circ D=D\circ L=\operatorname{ad}_{\mathfrak{h}}(v^{\prime})$.

(ii) $\operatorname{ad}_{\mathfrak{h}}(v^{\prime})$ is a derivation of
$\mathfrak{h}$.

(iii) $v^{\prime}$ belongs to $\operatorname{Ker}(D)$.

Then, the vector space
$\mathfrak{g}=\mathbb{F}d\oplus\mathfrak{h}\oplus\mathbb{F}c$, has a Hom-Lie
product $[\,\cdot\,,\,\cdot\,]$, an equivariant twist map
$T\in\operatorname{End}(\mathfrak{g})$ and a non-degenerate symmetric bilinear
form $B$ invariant under $[\,\cdot\,,\,\cdot\,]$, making
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ into a quadratic Hom-Lie algebra,
with a one-dimensional ideal $\mathbb{F}c$ satisfying,
$(\mathbb{F}c)^{\perp}=\mathfrak{h}\oplus\mathbb{F}c$, and
$\mathbb{F}d\simeq(\mathbb{F}c)^{*}$.

###### Proof.

Define the skew-symmetric bilinear map $[\,\cdot\,,\,\cdot\,]$ on
$\mathfrak{g}$ by means of:

If $\displaystyle u,v\in\mathfrak{h},$
$\displaystyle[u,v]=[u,v]_{\mathfrak{h}}+B_{\mathfrak{h}}(D(u),v)c;$ if
$\displaystyle u\in\mathfrak{h},$ $\displaystyle[d,u]=D(u),\,\text{ and
}\,[c,\mathfrak{g}]=0.$

Now define the linear map $T:\mathfrak{g}\to\mathfrak{g}$ by,

If $\displaystyle T(d)=v^{\prime}+\lambda^{\prime}c;$ if $\displaystyle
T(u)=L(u)+B_{\mathfrak{h}}(v^{\prime},u)c,\,\text{ for all }u\in\mathfrak{h},$
and $\displaystyle T(c)=0.$

First we shall prove that $T$ is equivariant. Let $u$ be in $\mathfrak{h}$;
from (i)-(iii), we have:

$\displaystyle T([d,u])$
$\displaystyle=T(D(u))=L(D(u))+B_{\mathfrak{h}}(v^{\prime},D(u))c=L(D(u)),$
$\displaystyle\,[T(d),u]$
$\displaystyle=[v^{\prime},u]=[v^{\prime},u]_{\mathfrak{h}}+B_{\mathfrak{h}}(D(v^{\prime}),u)c=\operatorname{ad}_{\mathfrak{h}}(v^{\prime})(u),$
$\displaystyle\,[d,T(u)]$ $\displaystyle=[d,L(u)]=D(L(u)).$

From (i), $T([d,u])=[T(d),u]=[d,T(u)]$. Now let $u$ and $v$ be in
$\mathfrak{h}$. Since $T(c)=0$ and $[c,\mathfrak{g}]=\\{0\\}$, we get,

$\displaystyle T([u,v])$
$\displaystyle=T([u,v]_{\mathfrak{h}})=L([u,v]_{\mathfrak{h}})+B_{\mathfrak{h}}(v^{\prime},[u,v]_{\mathfrak{h}})c,$
$\displaystyle\,[T(u),v]$
$\displaystyle=[L(u),v]=[L(u),v]_{\mathfrak{h}}+B_{\mathfrak{h}}(D(L(u)),v)c.$

Now use the fact that $L$ is equivariant with respect to
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$, together with the fact that
$B_{\mathfrak{h}}$ is invariant under $[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$,
in conjunction with the hypothesis (i) in the statement to conclude that
$T([u,v])=[T(u),v]$. Thus, $T$ is equivariant with respect to
$[\,\cdot\,,\,\cdot\,]$.

We shall prove next that $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ is a Hom-Lie
algebra. Let $u,v$ and $w$ be in $\mathfrak{h}$. Since,

$[T(u),[v,w]]=[L(u),[v,w]_{\mathfrak{h}}]=[L(u),[v,w]_{\mathfrak{h}}]_{\mathfrak{h}}+B_{\mathfrak{h}}(v^{\prime},[u,[v,w]_{\mathfrak{h}}]_{\mathfrak{h}})\,c,$

using (i) and the $[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}$-invariance of
$B_{\mathfrak{h}}$, one concludes that
$[T(u),[v,w]]+[T(v),[w,u]]+[T(w),[u,v]]=0$. Finally, observe that,

$\displaystyle[T(d),[u,v]]+[T(u),[v,d]]+[T(v),[d,u]]$
$\displaystyle=[v^{\prime},[u,v]_{\mathfrak{h}}]-[L(u),D(v)]+[L(v),D(u)]$
$\displaystyle=[v^{\prime},[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}-[L(u),D(v)]_{\mathfrak{h}}-B_{\mathfrak{h}}(D(L(u)),D(v))c$
$\displaystyle\quad+[L(v),D(u)]_{\mathfrak{h}}+B_{\mathfrak{h}}(D(L(v)),D(u))c$
$\displaystyle=[v^{\prime},[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}-[u,[v^{\prime},v]_{\mathfrak{h}}]_{\mathfrak{h}}+[v,[v^{\prime},u]_{\mathfrak{h}}]_{\mathfrak{h}}=0,$

where we have used the fact that $v^{\prime}$ lies in $\operatorname{Ker}(D)$
and the hypotheses (i) and (ii).

To finish the proof, define the non-degenerate symmetric bilinear form
$B:\mathfrak{g}\times\mathfrak{g}\to\mathbb{F}$, by means of,

$B(\lambda\,d+u+\zeta\,c,\lambda^{\prime}\,d+v+\zeta^{\prime}\,c)=\lambda\,\zeta^{\prime}+\lambda^{\prime}\,\zeta+B_{\mathfrak{h}}(u,v),$

for any $\lambda$, $\lambda^{\prime}$, $\zeta$ and $\zeta^{\prime}$ in
$\mathbb{F}$ and any $u$ and $v$ in $\mathfrak{h}$. To see that $B$ is
invariant under $[\,\cdot\,,\,\cdot\,]$ one only needs the fact that
$D\in\mathfrak{o}(B_{\mathfrak{h}})$ and observe that
$B|_{\mathfrak{h}\times\mathfrak{h}}=B_{\mathfrak{h}}$. The proof that $T$ is
$B$-self-adjoint is similar to its counterpart in Prop. 1.1 but easier, and we
may safely omit it. ∎

##  2\. Indecomposable quadratic Hom-Lie algebras with equivariant twist map

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ be a Hom-Lie algebra with
equivariant twist map under $[\,\cdot\,,\,\cdot\,]$. By Fitting’s lemma, there
exists a maximal positive integer $\ell$ such that
$\mathfrak{g}=\operatorname{Im}\left(T^{\ell}\right)\oplus\operatorname{Ker}\left(T^{\ell}\right)$,
where $\operatorname{Im}(T^{\ell})=\operatorname{Im}(T^{\ell+1})$ and
$\operatorname{Ker}(T^{\ell})=\operatorname{Ker}(T^{\ell+1})$. The subspaces
$\operatorname{Im}(T^{\ell})$ and $\operatorname{Ker}(T^{\ell})$ are easily
seen to be Hom-Lie ideals of $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$.
Moreover, since $T|_{\operatorname{Im}\left(T^{\ell}\right)}$ is invertible,
the Hom-Lie algebra structure inherited by $\operatorname{Im}(T^{\ell})$ under
the $[\,\cdot\,,\,\cdot\,]$-equivariant hypotheses of $T$, is that of a Lie
algebra. On the other hand, for $\ell\geq 1$,
$T|_{\operatorname{Ker}\left(T^{\ell}\right)}$ is nilpotent. Thus, Hom-Lie
algebras with equivariant and non-invertible $T$ may be reduced, under
restriction to $\operatorname{Ker}(T^{\ell})$, to the case when $T$ is
nilpotent. Having these facts in mind, we may now state and prove the
following result:

######  2.1 Theorem.

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ be a Hom-Lie algebra with
equivariant and nilpotent twist map $T$. If $I$ is a maximal proper ideal of
$\mathfrak{g}$, then there exists a subspace $\mathfrak{s}$ of $\mathfrak{g}$,
such that, $\mathfrak{g}=\mathfrak{s}\oplus I$,
$[\mathfrak{s},\mathfrak{s}]\subset\mathfrak{s}$ and $\mathfrak{s}$ is either
one dimensional or
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]\mid_{\mathfrak{s}\times\mathfrak{s}})$ is
a simple Lie algebra.

###### Proof.

Consider the Hom-Lie algebra structure in the quotient $\mathfrak{g}/I$
induced in the obvious way and let $\pi_{I}:\mathfrak{g}\to\mathfrak{g}/I$ be
its corresponding canonical projection. Let $\tilde{J}$ be an ideal of
$\mathfrak{g}/I$ and let $J=\pi_{I}^{-1}(\tilde{J})$. Then, $\tilde{J}=J/I$
and $J$ is an ideal of $\mathfrak{g}$ containing $I$. Since $I$ is maximal,
either $J=I$ or $J=\mathfrak{g}$. Let
$\widetilde{T}:\mathfrak{g}/I\to\mathfrak{g}/I$ be the twist map of
$\mathfrak{g}/I$ induced by $T$, so that $\widetilde{T}(x+I)=T(x)+I$, for all
$x$ in $\mathfrak{g}$. Since $\widetilde{T}$ is equivariant, both
$\operatorname{Ker}(\widetilde{T})$ and $\operatorname{Im}(\widetilde{T})$ are
ideals of $\mathfrak{g}/I$. Now, $\widetilde{T}$ nilpotent and
$I\subset\mathfrak{g}$ maximal imply that
$\operatorname{Im}(\widetilde{T})=\\{0\\}$, which in turn means that
$\operatorname{Im}(T)\subset I$. And since $T$ is nilpotent,
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)\neq\\{0\\}$.

Now consider the Hom-Lie algebra structure inherited by
$\mathfrak{g}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$ in the obvious
way. Observe that $I/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$ is a
maximal proper ideal of
$\mathfrak{g}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. Proceeding by
induction on $\operatorname{dim}\mathfrak{g}$, we shall deduce that,

(15)
$\mathfrak{g}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)=\widetilde{\mathfrak{h}}\oplus\left(I/\operatorname{Ker}(T)\cap\operatorname{Im}(T)\right),$

where $\widetilde{\mathfrak{h}}$ is either one-dimensional or, if $\mu$
denotes the Hom-Lie bracket on
$\mathfrak{g}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$, then
$\mu(\widetilde{\mathfrak{h}},\widetilde{\mathfrak{h}})\subset\widetilde{\mathfrak{h}}$.

Let
$\pi:\mathfrak{g}\to\mathfrak{g}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$
be the canonical projection onto the quotient and let
$\mathfrak{h}=\pi^{-1}(\widetilde{\mathfrak{h}})$, so that
$\widetilde{\mathfrak{h}}=\mathfrak{h}/\operatorname{Ker}(T)\cap\operatorname{Im}(T)$.
It then follows from (15) that $\mathfrak{h}\cap
I=\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. The induction hypothesis
states that
$\mu(\widetilde{\mathfrak{h}},\widetilde{\mathfrak{h}})\subset\widetilde{\mathfrak{h}}$.
Therefore, $[\mathfrak{h},\mathfrak{h}]\subset\mathfrak{h}$. In addition, it
follows from (15) that $\mathfrak{g}=\mathfrak{h}+I$.

Now, the equivariance of $T$ implies that
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)$ is an Abelian ideal of
$\mathfrak{g}$. Indeed, let $u$ and $v$ be in
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. Obviously, there exists an
element $u^{\prime}$ in $\mathfrak{g}$ such that $u=T(u^{\prime})$. Therefore,
$[u,v]=[T(u^{\prime}),v]=[u^{\prime},T(v)]=0$.

If $\operatorname{dim}\widetilde{\mathfrak{h}}=1$, we may write
$\mathfrak{h}=\mathbb{F}d\oplus(\operatorname{Ker}(T)\cap\operatorname{Im}(T))$;
whence $\mathfrak{g}=\mathfrak{h}+I=\mathbb{F}d\oplus I$, with $\mathbb{F}d$
being a subspace of $\mathfrak{g}$ complementary to $I$, such that
$[\mathbb{F}d,\mathbb{F}d]=\\{0\\}\subset\mathbb{F}d$.

Now suppose $\operatorname{dim}\widetilde{\mathfrak{h}}>1$. By the induction
hypothesis, the Hom-Lie bracket $\mu$ restricted to $\widetilde{\mathfrak{h}}$
yields a simple Lie algebra. Let $\mathfrak{s}$ be a subspace of
$\mathfrak{h}$ such that
$\mathfrak{h}=\mathfrak{s}\oplus(\operatorname{Ker}(T)\cap\operatorname{Im}(T))$.
For each pair of elements $x$ and $y$ in $\mathfrak{s}$, write,

(16) $[x,y]=[x,y]_{\mathfrak{s}}+\Lambda(x,y),$

where $[x,y]_{\mathfrak{s}}$ and $\Lambda(x,y)$ are the components of $[x,y]$
along $\mathfrak{s}$ and $\operatorname{Ker}(T)\cap\operatorname{Im}(T)$,
respectively, thus obtaining the following skew-symmetric and bilinear maps:

$\displaystyle[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}:\mathfrak{s}\times\mathfrak{s}\to\mathfrak{s},\quad$
$\displaystyle(x,y)\mapsto[x,y]_{\mathfrak{s}}$
$\displaystyle\Lambda:\mathfrak{s}\times\mathfrak{s}\to\operatorname{Ker}(T)\cap\operatorname{Im}(T),\quad$
$\displaystyle(x,y)\mapsto\Lambda(x,y).$

Let $p:\mathfrak{h}\to\mathfrak{s}$ be the projection onto $\mathfrak{s}$;
that is, $p(x+v)=x$, whenever $x$ belongs to $\mathfrak{s}$ and $v$ lies in
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. Observe that,

$p([x+u,y+v])=[p(x+u),p(y+v)]_{\mathfrak{s}},\,\text{ and
}\operatorname{Ker}(p)=\operatorname{Ker}(T)\cap\operatorname{Im}(T).$

This shows that $p$ induces an algebra isomorphism between
$\widetilde{\mathfrak{h}}$ and
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}})$. Then
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}})$ is a simple Lie algebra.

Let $\\{v_{1},\ldots,v_{r}\\}$ be a basis of
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)$ and let $K_{\mathfrak{s}}$ be
the Cartan-Killing form of $\mathfrak{s}$. Since $K_{\mathfrak{s}}$ is non-
degenerate and $\Lambda(x,y)$ is a linear combination of $v_{1},\ldots,v_{r}$,
$\Lambda(x,y)$ may be written as follows:

(17)
$\Lambda(x,y)=K_{\mathfrak{s}}(D_{1}(x),y)v_{1}+\ldots+K_{\mathfrak{s}}(D_{r}(x),y)v_{r},$

where $D_{i}:\mathfrak{s}\to\mathfrak{s}$ is a linear map. Since $\Lambda$ is
skew-symmetric, $D_{i}$ actually belongs to $\mathfrak{o}(K_{\mathfrak{s}})$,
for all, $1\leq i\leq r$. We may use the fact that $\mathfrak{s}$ is a simple
Lie algebra if and only if
$\operatorname{ad}(\mathfrak{s})=\mathfrak{o}(\mathfrak{s})$ (see Thm. A in
the Appendix) to further conclude that for each $1\leq i\leq r$, there exists
some $x_{i}$ in $\mathfrak{s}$ such that
$D_{i}=\operatorname{ad}_{\mathfrak{s}}(x_{i})\in\operatorname{Der}(\mathfrak{s})$.
Therefore, (17) can be written in the form,

(18)
$\Lambda(x,y)\\!=\\!K_{\mathfrak{s}}(x_{1},[x,y]_{\mathfrak{s}})v_{1}\\!+\\!\ldots\\!+\\!K_{\mathfrak{s}}(x_{r},[x,y]_{\mathfrak{s}})v_{r},$

with $x_{i}$ in $\mathfrak{s}$ ($1\leq i\leq r$), for any given pair of
elements $x$ and $y$ in $\mathfrak{s}$. Now define a skew-symmetric product,
$[\,\cdot\,,\,\cdot\,]^{\prime}$, in the subspace
$\mathfrak{s}\oplus\left(\operatorname{Ker}(T)\cap\operatorname{Im}(T)\right)$,
by means of,

$[x+u,y+v]^{\prime}=[x,y]_{\mathfrak{s}}+[x,v]-[y,u],$

for all $x$ and $y$ in $\mathfrak{s}$, and all $u$ and $v$ in
$\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. Observe that
$[\mathfrak{s},\mathfrak{s}]^{\prime}\subset\mathfrak{s}$. Let
$\Phi:\mathfrak{h}\to\mathfrak{s}\oplus(\operatorname{Ker}(T)\cap\operatorname{Im}(T))$
be the linear map defined by:

$\displaystyle\Phi(x)$ $\displaystyle=x-K_{\mathfrak{s}}(x_{1},x)v_{1}-\ldots-
K_{\mathfrak{s}}(x_{r},x)v_{r},\,\text{ for all }x\in\mathfrak{s},$
$\displaystyle\Phi(v)$ $\displaystyle=v,\,\,\text{ for all
}v\in\operatorname{Ker}(T)\cap\operatorname{Im}(T).$

From (16) and (18), $\Phi$ is an isomorphism between
$(\mathfrak{h},[\,\cdot\,,\,\cdot\,]\mid_{\mathfrak{h}\times\mathfrak{h}})$
and
$(\mathfrak{s}\oplus(\operatorname{Ker}(T)\cap\operatorname{Im}(T)),[\,\cdot\,,\,\cdot\,]^{\prime})$.
Since $[\mathfrak{h},\mathfrak{h}]\subset\mathfrak{h}$, we might as well
assume that, $[\mathfrak{s},\mathfrak{s}]\subset\mathfrak{s}$. We now claim
that $\mathfrak{g}=\mathfrak{s}\oplus I$. Indeed, let $x$ be in
$\mathfrak{s}\cap I$, then $x$ lies in $\mathfrak{h}\cap
I=\operatorname{Ker}(T)\cap\operatorname{Im}(T)$. Thus, $x$ belongs to
$\mathfrak{s}\cap(\operatorname{Ker}(T)\cap\operatorname{Im}(T))=\\{0\\}$.
From $\mathfrak{g}=\mathfrak{h}+I$, we get
$\mathfrak{g}=\mathfrak{s}+I=\mathfrak{s}\oplus I$.

Finally, observe that the inclusion map $\iota:\mathfrak{s}\to\mathfrak{g}$
satisfies $\iota([x,y]_{\mathfrak{s}})=[\iota(x),\iota(y)]$, for all $x,y$ in
$\mathfrak{s}$, and $q\circ\iota=\operatorname{Id}_{\mathfrak{s}}$, where
$q:\mathfrak{g}\to\mathfrak{s}$ is the projection onto $\mathfrak{s}$ with
kernel $I$. Since $\operatorname{Im}(T)\subset I$, one may regard the zero map
as a twist map for $\mathfrak{s}$. In this way, $\iota$ becomes a morphism of
Hom-Lie algebras under which the following short exact sequence of Hom-Lie
algebras actually splits:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathfrak{s}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

∎

We are now prepared to state and prove one of the main results of this work;
namely, that any indecomposable quadratic Hom-Lie algebra with equivariant and
nilpotent twist map, can be identified with one of the constructions given in
either Prop. 1.1 or Prop. 1.2.

######  2.2 Theorem.

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ be an indecomposable quadratic
Hom-Lie algebra with equivariant and nilpotent twist map $T$. Then, either
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, where
$\mathfrak{s}$ is a simple Lie algebra and its structure is described by Prop.
1.1, or $\mathfrak{g}=\mathbb{F}d\oplus\mathfrak{h}\oplus\mathbb{F}c$, and its
structure is described by Prop. 1.2 .

###### Proof.

Let $\mathcal{F}$ be the family of ideals of $\mathfrak{g}$ defined by:

$\mathcal{F}=\\{I\subset\mathfrak{g}\mid I\text{ is a proper ideal of
}\mathfrak{g}\text{ such that }I\supset\operatorname{Ker}(T)\\}.$

Let $I$ be a maximal element of $\mathcal{F}$. Arguing as in Thm. 2.1, we have
$\operatorname{Im}(T)\subset I$. Thus,
$\operatorname{Ker}(T)+\operatorname{Im}(T)\subset I$ and
$I^{\perp}\subset\operatorname{Ker}(T)\cap\operatorname{Im}(T)\subset I$.
Using the Witt decomposition of $\mathfrak{g}$ associated to $I^{\perp}$, we
conclude that $\mathfrak{g}$ has a vector space decomposition of the form,
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus I^{\perp}$, where
$I=\mathfrak{h}\oplus I^{\perp}$, with $\mathfrak{h}$ being a non-degenerate
subspace of $\mathfrak{g}$, $\mathfrak{h}^{\perp}=\mathfrak{s}\oplus
I^{\perp}$ and $B(\mathfrak{s},\mathfrak{s})=\\{0\\}$.

We shall only prove the case for which $\operatorname{dim}\mathfrak{s}>1$,
condition which will be assumed throughout this proof. The proof of the case
$\operatorname{dim}\mathfrak{s}=1$ is completely analogous, but much simpler
and may be safely omitted. From Thm. 2.1, we may further assume that
$[\mathfrak{s},\mathfrak{s}]\subset\mathfrak{s}$ and that
$(\mathfrak{s},[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}})$ is a simple Lie algebra,
with
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}=[\,\cdot\,,\,\cdot\,]\mid_{\mathfrak{s}\times\mathfrak{s}}$.
Now observe that $[I,I^{\perp}]=\\{0\\}$ and since $B$ is invariant under
$[\,\cdot\,,\,\cdot\,]$, it follows that
$[\mathfrak{s},\mathfrak{h}]\subset\mathfrak{h}$. Therefore, the Hom-Lie
bracket $[\,\cdot\,,\,\cdot\,]$ of $\mathfrak{g}$ has the following
decomposition:

If $\displaystyle x,y\in\mathfrak{s},$
$\displaystyle\,[x,y]=[x,y]_{\mathfrak{s}},$ if $\displaystyle
x\in\mathfrak{s},v\in\mathfrak{h},$ $\displaystyle\,[x,v]=\rho(x)(v),$ if
$\displaystyle x\in\mathfrak{s},\,\alpha\in I^{\perp},$
$\displaystyle\,[x,\alpha]=\sigma(x)(\alpha),$ if $\displaystyle
u,v\in\mathfrak{h},$ $\displaystyle\,[u,v]=[u,v]_{\mathfrak{h}}+\Gamma(u,v),$

where $\rho:\mathfrak{s}\to\mathfrak{gl}_{\mathbb{F}}(\mathfrak{h})$, and
$\sigma:\mathfrak{s}\to\mathfrak{gl}_{\mathbb{F}}(I^{\perp})$ are linear maps.
In addition,
$[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}}:\mathfrak{h}\times\mathfrak{h}\to\mathfrak{h}$,
and $\Gamma:\mathfrak{h}\times\mathfrak{h}\to I^{\perp}$, are skew-symmetric
bilinear maps.

Since $\operatorname{Im}(T)\subset I=\mathfrak{h}\oplus I^{\perp}$, the twist
map $T$ has the following decomposition:

(19) $\begin{array}[]{rccl}\text{For,\ \
}&x\in\mathfrak{s},&&T(x)=f(x)+g(x);\\\ \text{and for,\ \
}&v\in\mathfrak{h},&&T(v)=L(v)+R(v),\end{array}$

where,

$f:\mathfrak{s}\to\mathfrak{h},\quad g:\mathfrak{s}\to I^{\perp},\quad
L:\mathfrak{h}\to\mathfrak{h},\quad\text{ and }\quad R:\mathfrak{h}\to
I^{\perp},$

are all linear. Now, take $x$ and $y$ in $\mathfrak{s}$. On the one hand,

$T([x,y])=T([x,y]_{\mathfrak{s}})=f([x,y]_{\mathfrak{s}})+g([x,y]_{\mathfrak{s}}),$

whereas,

$[x,T(y)]=[x,f(y)]+[x,g(y)]=\rho(x)(f(y))+\sigma(x)(g(y)).$

Then, from $T([x,y])=[T(x),y]$, conclude that,

(20) $\displaystyle f\circ\operatorname{ad}_{\mathfrak{s}}(x)$
$\displaystyle=\rho(x)\circ f,\,\,\text{ and,}$ (21) $\displaystyle
g\circ\operatorname{ad}_{\mathfrak{s}}(x)$ $\displaystyle=\sigma(x)\circ
g,\,\text{ for all }x\in\mathfrak{s}.$

In particular, (20) states that (i) in Prop. 1.1 is satisfied. Taking $x$ in
$\mathfrak{s}$ and $v$ in $\mathfrak{h}$, the equivariance condition
$T([x,v])=[T(x),v]=[x,T(v)]$, yields,

(22) $L\circ\rho(x)=\operatorname{ad}_{\mathfrak{h}}(f(x))=\rho(x)\circ
L,\quad\text{for all }x\in\mathfrak{s}.$

This time (22) states that condition (ii) of Prop. 1.1 is satisfied. For $u$
and $v$ in $\mathfrak{h}$, the equivariance condition $T([u,v])=[T(u),v]$
leads to,

(23) $L([u,v]_{\mathfrak{h}})=[L(u),v]_{\mathfrak{h}},\quad\text{for all
}u,v\in\mathfrak{h}.$

Let $x,y$ be in $\mathfrak{s}$ and $v$ be in $\mathfrak{h}$. From the Hom-Lie
Jacobi identity on $[T(x),[y,v]]+[T(y),[v,x]]+[T(v),[x,y]]=0$, we obtain:

(24)
$\rho([x,y]_{\mathfrak{s}})(L(v))=[\rho(x),\rho(y)]_{\mathfrak{gl}(\mathfrak{h})}\circ
L(v).$

Now let $x$ be in $\mathfrak{s}$ and let $u$ and $v$ be in $\mathfrak{h}$. The
Hom-Lie Jacobi identity $[T(x),[u,v]]+[T(u),[v,x]]+[T(v),[x,u]]=0$, leads to,

(25)
$[f(x),[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}=[[f(x),u]_{\mathfrak{h}},v]_{\mathfrak{h}}+[u,[f(x),v]_{\mathfrak{h}}]_{\mathfrak{h}}.$

That is, $\operatorname{ad}_{\mathfrak{h}}(f(x))$ is a derivation of
$(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}})$, for all $x$ in
$\mathfrak{s}$. Then (24) and (25) state that condition (iii) of Prop. 1.1 is
satisfied.

Finally, take $u$, $v$ and $w$ in $\mathfrak{h}$. From the Hom-Lie Jacobi
identity and the decomposition of $T$ given in (19), conclude that,

$[L(u),[v,w]_{\mathfrak{h}}]_{\mathfrak{h}}+[L(v),[w,u]_{\mathfrak{h}}]_{\mathfrak{h}}+[L(w),[u,v]_{\mathfrak{h}}]_{\mathfrak{h}}=0.$

It follows that $(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},L)$ is a
Hom-Lie algebra and (23) says that $L$ is equivariant.

Let $B_{\mathfrak{h}}=B|_{\mathfrak{h}\times\mathfrak{h}}$. Since
$\mathfrak{h}\subset I$, it follows that
$B_{\mathfrak{h}}([u,v]_{\mathfrak{h}},w)=B_{\mathfrak{h}}(u,[v,w]_{\mathfrak{h}})$.
On the other hand, from the fact that $T$ is $B$-self-adjoint, and the
decomposition given in (19), we obtain,
$B_{\mathfrak{h}}(L(u),v)=B_{\mathfrak{h}}(u,L(v))$, for all $u$, $v$ and $w$
in $\mathfrak{h}$. Therefore,
$(\mathfrak{h},[\,\cdot\,,\,\cdot\,]_{\mathfrak{h}},L,B_{\mathfrak{h}})$ is a
quadratic Hom-Lie algebra with equivariant and nilpotent twist map $L$.

Now use the fact that $B$ is non-degenerate and invariant to conclude that
there is an isomorphism $\psi:I^{\perp}\to\mathfrak{s}^{\ast}$, given by
$\psi(\alpha)(x)=B(\alpha,x)$, for all $\alpha$ in $I^{\perp}$ and any $x$ in
$\mathfrak{s}$, such that,

(26)
$\psi\circ\sigma(x)=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)\circ\psi,\,\text{
for all }x\in\mathfrak{s}.$

This implies that $\sigma:\mathfrak{s}\to\mathfrak{gl}(I^{\perp})$ is a
representation of the Lie algebra $\mathfrak{s}$ in $I^{\perp}$ and that
$\psi$ is an isomorphism of $\mathfrak{s}$-modules. Now, using again the
decomposition of $T$ given in (19) together with the fact that it is $B$-self-
adjoint, one conlcudes that $\psi(g(x))(y)=\psi(g(y))(x)$. Let $g_{\psi}$ be
the bilinear form on $\mathfrak{s}$ defined by $g_{\psi}(x,y)=\psi(g(x))(y)$,
for all $x,y$ in $\mathfrak{s}$. It is clear that $g_{\psi}$ is symmetric.
Using (21) and (26), we see that,

$\displaystyle g_{\psi}([x,y]_{\mathfrak{s}},z)$
$\displaystyle=\psi(g([x,y]_{\mathfrak{s}}))(z)=\psi(\sigma(x)(g(y)))(z)$
$\displaystyle=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)\circ\psi(g(y))(z)=-\psi(g(y))([x,z]_{\mathfrak{s}})$
$\displaystyle=g_{\psi}(y,[x,z]_{\mathfrak{s}}),\,\text{ for all
}x,y,z\in\mathfrak{s}.$

That is, $g_{\psi}$ is invariant under $[\,\cdot\,,\,\cdot\,]_{\mathfrak{s}}$.
Since $\mathfrak{s}$ is a simple Lie algebra, there exists a non-zero scalar
$\eta$ such that $g_{\psi}=\eta K_{\mathfrak{s}}$, where $K_{\mathfrak{s}}$ is
the Cartan-Killing form of $\mathfrak{s}$. Finally, for $x$ in $\mathfrak{s}$
and $u$ in $\mathfrak{h}$, the condition $B(T(x),u)=B(x,T(u))$, yields,

$\displaystyle B_{\mathfrak{h}}(f(x),v)$
$\displaystyle=B(x,R(v))=\psi(R(v))(x)$
$\displaystyle=B_{\mathfrak{h}}(v)^{\flat}(f(x))=\psi(R(v))(x)$
$\displaystyle=f^{\ast}(B_{\mathfrak{h}}^{\flat}(v))(x)=\psi(R(v))(x).$

Then, $f^{\ast}\circ B_{\mathfrak{h}}^{\flat}(v)=\psi(R(v))$; that is
$f^{\ast}\circ B_{\mathfrak{h}}^{\flat}=\psi\circ R$. Let
$R_{\psi}:\mathfrak{h}\to\mathfrak{s}^{\ast}$ be the map defined by
$R_{\psi}=\psi\circ R$. Then, $R_{\psi}=f^{\ast}\circ
B_{\mathfrak{h}}^{\flat}$ (compare with the definition of the map $R$ given in
(3)).

Now, introduce the skew-symmetric bilinear map $[\,\cdot\,,\,\cdot\,]_{\psi}$
defined in the vector space
$\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, by,

if $\displaystyle x,y\in\mathfrak{s},$
$\displaystyle\,[x,y]_{\psi}=[x,y]_{\mathfrak{s}},$ if $\displaystyle
x\in\mathfrak{s},v\in\mathfrak{h},$ $\displaystyle\,[x,v]_{\psi}=\rho(x)(v),$
if $\displaystyle x\in\mathfrak{s},\,\,\xi\in\mathfrak{s}^{\ast},$
$\displaystyle\,[x,\alpha]_{\psi}=\operatorname{ad}_{\mathfrak{s}}^{\ast}(x)(\xi),$
and if $\displaystyle u,v\in\mathfrak{h},$
$\displaystyle\,[u,v]_{\psi}=[u,v]_{\mathfrak{h}}+\psi(\Gamma(u,v)).$

Similarly, consider the linear map
$T_{\psi}:\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}\to\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$
as follows:

for $\displaystyle x\in\mathfrak{s},\,$ $\displaystyle T(x)=f(x)+\eta
K_{\mathfrak{s}}(x),$ and for $\displaystyle v\in\mathfrak{h},\,$
$\displaystyle T(v)=L(v)+R_{\psi}(v).$

Finally, let $B_{\psi}$ on
$\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ be the bilinear form
defined by,

$B(x+u+\xi,y+v+\zeta)=\xi(y)+\zeta(x)+B_{\mathfrak{h}}(u,v),$

for all $x,y$ in $\mathfrak{s}$, $u,v$ in $\mathfrak{h}$ and $\xi,\zeta$ in
$\mathfrak{s}^{\ast}$. From (20), (21), (22), (24) and Prop. 1.1,
$\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ has a quadratic Hom-
Lie algebra structure, and the linear map
$\Psi:\mathfrak{g}\to\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$
defined by $\Psi(x+u+\alpha)=x+u+\psi(\alpha)$, is an isometric isomorphism. ∎

##  3\. Quadratic Lie algebras over quadratic Hom-Lie algebras with
equivariant twist maps

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ be a Hom-Lie algebra with
equivariant twist map $T$. It has been shown in [5] that
$[\,\cdot\,,\,\cdot\,]_{\text{Lie}}=T\circ[\,\cdot\,,\,\cdot\,]$, is a Lie
bracket in $\mathfrak{g}$, satisfying
$[\operatorname{Ker}(T),\mathfrak{g}]_{\text{Lie}}=\\{0\\}$. This fact is used
in the proof of our next result:

######  3.1 Proposition.

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ be a nilpotent Hom-Lie algebra
with equivariant twist map $T$. Define the Lie bracket
$[\,\cdot\,,\,\cdot\,]_{\text{Lie}}=T\circ[\,\cdot\,,\,\cdot\,]$ in
$\mathfrak{g}$ as in [5]. Then,
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$ is a nilpotent Lie
algebra.

###### Proof.

Define recursively, $\mathfrak{g}^{n+1}:=[\mathfrak{g},\mathfrak{g}^{n}]$,
with $\mathfrak{g}^{2}:=[\mathfrak{g},\mathfrak{g}]$. The Hom-Lie algebra
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ is nilpotent if there is some
$m\in\mathbb{N}$ such that, $\mathfrak{g}^{m}=0$. The equivariance property of
$T$ implies that for any pair of subspaces $S$ and $S^{\prime}$ of
$\mathfrak{g}$,

$\left[S,S^{\prime}\right]_{\text{Lie}}=[T(S),S^{\prime}]=[S,T(S^{\prime})]=T([S,S^{\prime}]).$

Let $\mathfrak{g}^{1}_{\text{Lie}}=[\mathfrak{g},\mathfrak{g}]_{\text{Lie}}$,
and for $\ell>1$, inductively set
$\mathfrak{g}^{\ell}_{\text{Lie}}=[\mathfrak{g},\mathfrak{g}^{\ell-1}_{\text{Lie}}]_{\text{Lie}}$.
Then,
$\mathfrak{g}_{\text{Lie}}^{1}=T\left([\mathfrak{g},\mathfrak{g}]\right)=[\mathfrak{g},T(\mathfrak{g})]\subset[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}^{1}$.
Inductively assume that
$\mathfrak{g}_{\text{Lie}}^{\ell}\subset\mathfrak{g}^{\ell}$. Then, using the
equivariance property of $T$, it follows that,

$\mathfrak{g}_{\text{Lie}}^{\ell+1}=[\mathfrak{g},\mathfrak{g}^{\ell}]_{\text{Lie}}\subset[\mathfrak{g},\mathfrak{g}^{\ell}]_{\text{Lie}}=[\mathfrak{g},T(\mathfrak{g}^{\ell})]=[T(\mathfrak{g}),\mathfrak{g}^{\ell}]\subset[\mathfrak{g},\mathfrak{g}^{\ell}]=\mathfrak{g}^{\ell+1}.$

This proves that
$\mathfrak{g}_{\text{Lie}}^{\ell+1}\subset\mathfrak{g}^{\ell+1}$, for any
$\ell\in\mathbb{N}$. Under the hypothesis that
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ is nilpotent, it follows that
$\mathfrak{g}_{\text{Lie}}^{m}=[\mathfrak{g},\mathfrak{g}^{m-1}]_{\text{Lie}}=\\{0\\}$.
Therefore, $(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$ is a nilpotent
Lie algebra. ∎

It has been proved in [5] that there is a cocycle with values in
$V\simeq\operatorname{Ker}(T)$ (vector space isomorphism only) associated to
the Lie algebra $(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$, such
that, if it is not a coboundary, there exists a non-associative simple algebra
from which the Hom-Lie algebra structure of
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ can be recovered (see Thm. 2.2 in
[5]). Moreover, from the results in [5], we know how to associate a quadratic
Lie algebra to a quadratic Hom-Lie algebra with equivariant twist map. Since
we also know how to produce a central extension of it in terms of an
explicitly produced 2-cocycle, one can say exactly when this Hom-Lie algebra
is of the form
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, with
$\mathfrak{s}$ being a simple Lie algebra as described in Thm. 2.2 above;
namely, when the cocycle is not a coboundary. In addition, the following
result shows that the Lie algebra
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$ can be nilpotent, even if
the Hom-Lie algebra $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ satisfies
$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$.

######  3.2 Proposition.

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ be an indecomposable quadratic
Hom-Lie algebra with equivariant and nilpotent twist map $T$. For any $x$ and
$y$ in $\mathfrak{g}$, let $[x,y]_{\text{Lie}}=T([x,y])$. Then,

(i) $(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}},B)$ is a nilpotent
quadratic Lie algebra.

(ii) Let $V$ be a trivial module of
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$, with
$\operatorname{dim}V=\operatorname{dim}\operatorname{Ker}(T)$ $=d$. Let
$\mathfrak{a}$ be a subspace of $\mathfrak{g}$ such that
$\mathfrak{g}=\mathfrak{a}\oplus\operatorname{Im}(T)$. Let
$\\{a_{1},\ldots,a_{d}\\}$ be a basis of $\mathfrak{a}$ and let
$\\{v_{1},\ldots,v_{d}\\}$ be a basis of $V$. Let
$\theta:\mathfrak{g}\times\mathfrak{g}\to V$ be the skew-symmetric bilinear
map defined by,

(27) $\theta(x,y)=B([a_{1},x],y)v_{1}+\ldots+B([a_{d},x],y)v_{d},\,\text{ for
all }x,y\in\mathfrak{g}.$

Then $\theta$ is a 2-cocycle of the Lie algebra
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$, with values in $V$.

(iii) If $\theta$ is not a coboundary,
$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ and the underlying space of
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ is of the form
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$ as
given in Thm. 2.2, whose structure is described in Prop. 1.1.

###### Proof.

From Prop. 1.1 of [5] we know that
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}},B)$ is a quadratic Lie
algebra. Moreover, the definition of the Lie bracket
$[\,\cdot\,,\,\cdot\,]_{\text{Lie}}$, makes it clear that
$\mathfrak{g}^{1}_{\text{Lie}}=[\mathfrak{g},\mathfrak{g}]_{\text{Lie}}\subset\operatorname{Im}(T)$.
Since $T$ is equivariant with respect to $[\,\cdot\,,\,\cdot\,]$, we conclude
that,

$\mathfrak{g}^{2}_{\text{Lie}}=[\mathfrak{g},\mathfrak{g}^{1}_{\text{Lie}}]_{\text{Lie}}\subset[\mathfrak{g},\operatorname{Im}(T)]_{\text{Lie}}\subset\operatorname{Im}(T^{2}).$

By induction, we may further conclude that
$\mathfrak{g}^{\ell}_{\text{Lie}}\subset\operatorname{Im}(T^{\ell})$, for all
$\ell$ in $\mathbb{N}$. Since $T$ is nilpotent,
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})$ is a nilpotent Lie
algebra.

It was also proved in Cor. 1.6 of [5] that there exists a surjective linear
map $h:\mathfrak{g}\oplus V\to\mathfrak{g}$, such that,

(28) $\displaystyle[x,y]$
$\displaystyle=h\left([x,y]_{\text{Lie}}+\theta(x,y)\right),\quad\text{ for
all }x,y\in\mathfrak{g},$ $\displaystyle h(V)$
$\displaystyle=\operatorname{Ker}(T),\quad
h(\mathfrak{g})=\mathfrak{a}^{\perp},\quad\text{and,}\quad\operatorname{Ker}(h)=\mathfrak{a}.$

Let $x$ be in $\mathfrak{g}$ be such that $[x,\mathfrak{g}]=\\{0\\}$. We shall
prove that $x=0$. Indeed, by (28), we have:

$h([x,\mathfrak{g}]_{\text{Lie}})=-h(\theta(x,\mathfrak{g}))\in
h(\mathfrak{g})\cap
h(V)=\mathfrak{a}^{\perp}\cap\operatorname{Ker}(T)=\\{0\\},$

as
$\mathfrak{g}=\mathfrak{a}^{\perp}\oplus\operatorname{Ker}(T)=\mathfrak{a}\oplus\operatorname{Im}(T)$.
Then, $\theta(x,\mathfrak{g})$ lies in $V\cap\mathfrak{a}=\\{0\\}$. From (27),
this says that $[a_{j},x]=0$ for all $1\leq j\leq d$; that is, $x$ belongs to
$\operatorname{Ker}([a_{j},\,\cdot\,])$, for all $1\leq j\leq d$.

On the other hand, it was proved in [3] (see Prop. 2.5 (ii) there) that if
$\theta$ is not a coboundary,

$\bigcap_{j=1}^{d}\operatorname{Ker}([a_{j},\,\cdot\,])=\\{0\\}.$

Thus, $x=0$. Altogether, this shows that
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T)$ has a trivial center. Since $B$ is
invariant with respect to $[\,\cdot\,,\,\cdot\,]$, one concludes that
$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ and that $\mathfrak{g}$ cannot have
one-dimensional ideals. Therefore, $\mathfrak{g}$ is of the form
$\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, with $\mathfrak{s}$
being a simple Lie algebra as in Thm. 2.2. ∎

##  4\. Example

Let
$\mathfrak{s}=\mathfrak{sl}_{2}(\mathbb{F})=\operatorname{Span}_{\mathbb{F}}\\{x_{1},x_{2},x_{3}\\}$,
where the basis is chosen in such a way that
$[x_{1},x_{2}]_{\mathfrak{s}}=x_{3}$, $[x_{2},x_{3}]_{\mathfrak{s}}=x_{1}$,
$[x_{3},x_{1}]_{\mathfrak{s}}=x_{2}$ and
$K_{\mathfrak{s}}(x_{j},x_{k})=\delta_{jk}$, for all $1\leq j,k\leq 3$. Put,
$\mathfrak{h}=\operatorname{Span}_{\mathbb{F}}\\{v_{1},v_{2},v_{3}\\}$, and
consider the non-degenerate symmetric bilinear form $B_{\mathfrak{h}}$ defined
on the basis elements of $\mathfrak{h}$ by,
$B_{\mathfrak{h}}(v_{j},v_{k})=\delta_{jk}$ ($1\leq j,k\leq 3$). We shall
consider $\mathfrak{h}$ as an Abelian quadratic Hom-Lie algebra with
equivariant twist map $L=0$. Let $f:\mathfrak{s}\to\mathfrak{h}$ be the zero
map and let $\rho:\mathfrak{s}\to\mathfrak{o}(B_{\mathfrak{h}})$ be the linear
map defined by:

$\displaystyle\rho(x_{1})(v_{1})$ $\displaystyle=-v_{2},\quad$
$\displaystyle\,\rho(x_{2})(v_{1})$ $\displaystyle=v_{2},\quad$
$\displaystyle\,\rho(x_{3})(v_{1})$ $\displaystyle=-v_{3},$
$\displaystyle\rho(x_{1})(v_{2})$ $\displaystyle=v_{1},\quad$
$\displaystyle\,\rho(x_{2})(v_{2})$ $\displaystyle=-v_{1},\quad$
$\displaystyle\,\rho(x_{3})(v_{2})$ $\displaystyle=0,$
$\displaystyle\rho(x_{1})(v_{3})$ $\displaystyle=0\quad$
$\displaystyle\,\rho(x_{2})(v_{3})$ $\displaystyle=0,\,\quad$
$\displaystyle\rho(x_{3})(v_{3})$ $\displaystyle=v_{1}.$

Since $f=0$, $L=0$ and $\mathfrak{h}$ is Abelian, the conditions given in (i),
(ii) and (iii), of Prop. 1.1, are satisfied. Therefore, the vector space
$\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{h}\oplus\mathfrak{s}^{\ast}$, has a
skew-symmetric bilinear map $[\,\cdot\,,\,\cdot\,]$, a linear endomorphism $T$
and a symmetric bilinear form $B$ defined on it, making
$(\mathfrak{g},[\,\cdot\,,\,\cdot\,],T,B)$ into a quadratic Hom-Lie algebra
with equivariant twist map $T$. Let $\xi^{j}=K_{\mathfrak{s}}^{\flat}(x_{j})$
($1\leq j\leq 3$), be such that $\xi^{j}(x_{k})=\delta_{jk}$. According to
Prop. 1.1, the Hom-Lie bracket $[\,\cdot\,,\,\cdot\,]$ in $\mathfrak{g}$ is
given by,

$\displaystyle=x_{3},\quad$ $\displaystyle\ \ [x_{2},x_{3}]$
$\displaystyle=x_{1},\quad$ $\displaystyle[x_{3},x_{1}]$
$\displaystyle=x_{2},$ $\displaystyle[x_{1},v_{1}]$
$\displaystyle=-v_{2},\quad$ $\displaystyle\ \ [x_{2},v_{1}]$
$\displaystyle=v_{2},\quad$ $\displaystyle[x_{3},v_{1}]$
$\displaystyle=-v_{3},$ $\displaystyle[x_{1},v_{2}]$
$\displaystyle=v_{1},\quad$ $\displaystyle\ \ [x_{2},v_{2}]$
$\displaystyle=-v_{1},\quad$ $\displaystyle[x_{3},v_{2}]$ $\displaystyle=0,$
$\displaystyle[x_{1},v_{3}]$ $\displaystyle=0,\quad$ $\displaystyle\ \
[x_{2},v_{3}]$ $\displaystyle=0,\,\quad$ $\displaystyle[x_{3},v_{3}]$
$\displaystyle=v_{1}.$ $\displaystyle[v_{1},v_{2}]$
$\displaystyle=-\xi^{1}+\xi^{2}-\xi^{3},\quad$ $\displaystyle\ \
[v_{2},v_{3}]$ $\displaystyle=0,\quad\quad\quad$ $\displaystyle[v_{3},v_{1}]=$
$\displaystyle\xi^{3},$

and

$\displaystyle\,[x_{1},\xi^{2}]$ $\displaystyle=-[x_{2},\xi^{1}]=\xi^{3},$
$\displaystyle\,[x_{2},\xi^{3}]$ $\displaystyle=-[x_{3},\xi^{2}]=\xi^{1},$
$\displaystyle\,[x_{3},\xi^{1}]$ $\displaystyle=-[x_{1},\xi^{3}]=\xi^{2}.$

Observe that $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ and
$[\mathfrak{h},\mathfrak{h}]\subset\mathfrak{s}^{\ast}$. In addition,
$[\,\cdot\,,\,\cdot\,]$ is not a Lie bracket, because,

$\displaystyle\,[x_{1},[v_{1},v_{2}]]+[v_{1},[v_{2},x_{1}]]$
$\displaystyle+[v_{2},[x_{1},v_{1}]]=$
$\displaystyle=[x_{1},-\xi^{1}+\xi^{2}-\xi^{3}]-[v_{1},v_{1}]-[v_{2},v_{2}]$
$\displaystyle=\xi^{3}+\xi^{2}\neq 0.$

Let $\eta$ be an scalar in $\mathbb{F}$. The linear map
$T:\mathfrak{g}\to\mathfrak{g}$ that works as an equivariant and twist map for
$[\,\cdot\,,\,\cdot\,]$, is given by:

$T(x_{1})=\eta\,\xi^{1},\quad T(x_{2})=\eta\,\xi^{2},\quad
T(x_{3})=\eta\,\xi^{3},\quad
T(\mathfrak{h}\oplus\mathfrak{s}^{\ast})=\\{0\\}.$

Then, $T^{2}=0$. The Lie algebra bracket $[\,\cdot\,,\,\cdot\,]_{\text{Lie}}$,
described in Thm. 3.2, is given by:

$\displaystyle[x_{1},x_{2}]_{\text{Lie}}=\eta\,\xi^{3},\quad[x_{2},x_{3}]_{\text{Lie}}=\eta\,\xi^{1},\quad[x_{3},x_{1}]_{\text{Lie}}=\eta\,\xi^{1},$
$\displaystyle[\mathfrak{g},\mathfrak{h}\oplus\mathfrak{s}^{\ast}]_{\text{Lie}}=\\{0\\},\quad\mathfrak{h}^{\perp}=\mathfrak{s}\oplus\mathfrak{s}^{\ast}.\quad\mathfrak{h}\text{
is an ideal of }(\mathfrak{g},[\,\cdot\,,\cdot\,]_{\text{Lie}}),$
$\displaystyle(\mathfrak{g},[\,\cdot\,,\,\cdot\,]_{\text{Lie}})\text{ is
2-step nilpotent. }$

## Appendix

######  Theorem A.

Let $(\mathfrak{g},[\,\cdot\,,\,\cdot\,],B)$ be a quadratic Lie algebra. The
following statements are equivalent:

  1. (1)

$\mathfrak{g}$ is a simple Lie algebra.

  2. (2)

$\operatorname{ad}(\mathfrak{g})=\mathfrak{o}(B)$.

###### Proof.

$(1)\Rightarrow(2)$ With no loss of generality, we may assume that $B$ is the
Cartan-Killing form $K$ of $\mathfrak{g}$. Since $K$ is invariant, it follows
that $\operatorname{ad}(\mathfrak{g})$ is contained in $\mathfrak{o}(K)$. Let
$f$ be in $\mathfrak{o}(K)$ and consider the skew-symmetric bilinear map
$\Delta_{f}:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$, defined by,

$\Delta_{f}(x,y)=f([x,y])-[f(x),y]-[x,f(y)],\,\text{ for all
}x,y\in\mathfrak{g}.$

Let
$d:\mathfrak{g}\otimes\wedge^{2}\mathfrak{g}^{*}\to\mathfrak{g}\otimes\wedge^{3}\mathfrak{g}^{*}$
be the differential map. Then,

$\displaystyle d\Delta_{f}(x,y,z)$
$\displaystyle=\operatorname{ad}(x)(\Delta_{f}(y,z))+\operatorname{ad}(y)(\Delta_{f}(z,x))+\operatorname{ad}(z)(\Delta(x,y))$
$\displaystyle-\Delta_{f}([x,y],z)-\Delta_{f}([y,z],x)-\Delta_{f}([z,x],y).$

It is a straightforward matter to see that $d\Delta_{f}(x,y,z)=0$, for all
$x,y,z$ in $\mathfrak{g}$. Thus, there exists a linear map
$g:\mathfrak{g}\to\mathfrak{g}$ such that $\Delta_{f}(x,y)=-g([x,y])$, for all
$x,y$ in $\mathfrak{g}$. Then, $(f+g)([x,y])=[f(x),y]+[x,f(y)]$. This says
that $f$ fits into the concept of quasiderivation (see [7]); namely, a
quasiderivation $f:\mathfrak{g}\to\mathfrak{g}$ is a linear map for which
there exists a linear map $f^{\prime}:\mathfrak{g}\to\mathfrak{g}$,
satisfying,

$f^{\prime}([x,y])=[f(x),y]+[x,f(y)],\,\text{ for all }x,y\in\mathfrak{g}.$

Using Thm. 4.12 of [7], one may find an element $x^{\prime}$ in $\mathfrak{g}$
and a scalar $\eta^{\prime}$ in $\mathbb{F}$ such that,
$f=\operatorname{ad}_{\mathfrak{g}}(x^{\prime})+\eta^{\prime}\operatorname{Id}_{\mathfrak{g}}$.
Since both, $f$ and $\operatorname{ad}_{\mathfrak{g}}(x^{\prime})$ belong to
$\mathfrak{o}(K)$, it follows that $2\eta K=0$; whence, $\eta=0$ and
$f=\operatorname{ad}(x^{\prime})$. Therefore,
$\operatorname{ad}(\mathfrak{g})=\mathfrak{o}(K)$.

$(2)\Rightarrow(1)$ Since $B$ is non-degenerate, $\mathfrak{o}(B)$ is
semisimple (see [6]). On the other hand, since
$\operatorname{ad}:\mathfrak{g}\to\mathfrak{o}(B)$ is a faithful
representation, it is an isomorphism. Therefore, $\mathfrak{g}$ is a
semisimple Lie algebra. We claim that if $\mathfrak{g}$ is not simple, then
$\operatorname{dim}\mathfrak{g}<\operatorname{dim}\mathfrak{o}(B)$. Indeed,
let $\mathfrak{g}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}$ be a semisimple Lie
algebra, where both, $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ are semisimple.
Let $K=K_{1}\oplus K_{2}$ be the symmmetric bilinear form on $\mathfrak{g}$
defined by the Cartan-Killing forms $K_{1}$ and $K_{2}$ of $\mathfrak{g}_{1}$
and $\mathfrak{g}_{2}$, respectively. Since each $\mathfrak{g}_{i}$ is an
ideal of $\mathfrak{g}$, we also have,

$\operatorname{ad}(x)=\begin{pmatrix}\operatorname{ad}_{1}(x)&0\\\
0&0\end{pmatrix},\quad\text{if}\quad x\in\mathfrak{g}_{1},$

where $\operatorname{ad}_{1}(x)$ is the adjoint representation of
$\mathfrak{g}_{1}$. Similarly,

$\operatorname{ad}(x)=\begin{pmatrix}0&0\\\
0&\operatorname{ad}_{2}(x)\end{pmatrix},\quad\text{if}\quad
x\in\mathfrak{g}_{2}.$

This implies that $\operatorname{dim}\mathfrak{g}$ is lesser or equal than the
dimension of the space where the bilinear form $K$ is defined. Now, let
$L:\mathfrak{g}\to\mathfrak{g}$ be a linear map such that
$K(L(x),y)+K(x,L(y))=0$ for all $x$ and $y$ in $\mathfrak{g}$. This equation
is equivalent to,

$\begin{pmatrix}L_{11}&L_{12}\\\
L_{21}&L_{22}\end{pmatrix}^{T}\begin{pmatrix}K_{1}&0\\\
0&K_{2}\end{pmatrix}+\begin{pmatrix}K_{1}&0\\\
0&K_{2}\end{pmatrix}\begin{pmatrix}L_{11}&L_{12}\\\
L_{21}&L_{22}\end{pmatrix}=0.$

In particular, $L$ belongs to $\mathfrak{o}(K)$ if and only if,

$L_{11}^{T}K_{1}+K_{1}L_{11}=0,\quad L_{21}^{T}K_{2}+K_{1}L_{12}=0,\quad
L_{22}^{T}K_{2}+K_{2}L_{22}=0.$

The first and third equations state $L_{11}\in\mathfrak{o}(K_{1})$ and
$L_{22}\in\mathfrak{o}(K_{2})$, respectively. Since $K_{1}$ and $K_{2}$ are
non-degenerate, their corresponding matrices are invertible, so we may choose,
$L_{12}=(K_{1})^{-1}L_{21}^{T}K_{2}$, where $L_{12}\neq 0$. Therefore,
$\operatorname{dim}\mathfrak{g}<\operatorname{dim}\mathfrak{o}(K)$, and
$\mathfrak{g}$ is a simple Lie algebra. ∎

## Acknowledgements

The authors acknowledge the support received through CONACYT Grant $\\#$
A1-S-45886. The author RGD would also like to thank the support provided by
CONACYT post-doctoral fellowship 000153. Finally, GS acknowledges the support
provided by PROMEP grant UASLP-CA-228 and ASV acknowledges the support given
by MB1411.

## References

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# Prediction of neutron electromagnetic form factors behaviors just by the
proton electromagnetic form factors data

Anna Z. Dubničková  Department of Theoretical Physics, Comenius University,
Bratislava, Slovak Republic  Stanislav Dubnička  Institute of Physics, Slovak
Academy of Sciences, Bratislava, Slovak Republic

###### Abstract

It is demonstrated how the neutron electromagnetic form factors behaviors in
the whole region of their definition can be predicted theoretically just by
using the proton form factors data to be described by the advanced proton
electromagnetic structure Unitary and Analytic model.

neutrons, protons, vector mesons, electromagnetic form factors, analyticity

##  I Introduction

Dominant building particles of the whole universe are protons $p$ and
electrons $e$. While the electron is always stable and a point-like particle
up to now, the proton, though also to be stable, is compound of three quarks
and as a result in interaction with electrons it appears as an extended object
in the space. The latter property of the proton is called the proton
electromagnetic (EM) structure, for the first time revealed in the middle of
the last century, before the quark model of strongly interacting particles,
hadrons, has been established. Later on this non-point-like nature of the
proton has been generalized also to all other hadrons.

The EM structure of any hadron is completely described by scalar functions to
be called EM form factors (FFs) of one variable, in the spacelike region it is
the momentum transfer squared $t=-Q^{2}$ and in timelike region it is the
total c.m. energy squared $s=w^{2}$.

The number of EM FFs depends on the spin of the hadron under consideration.
The EM structure of a hadron with spin 0 is completely described by one EM FF,
the EM structure of a hadron with spin 1/2 is completely described by two EM
FFs, the EM structure of a hadron with spin 1 is completely described by three
EM FFs. In the case of the EM structure of hadrons with spin higher than one a
situation is even more complicated and generally it is not solved
satisfactorily up to now.

Further our interest will be concentrated to the $1/2^{+}$ octet baryons
consisting of the following particles:
$p,n,\Lambda^{0},\Sigma^{+},\Sigma^{0},\Sigma^{-},\Xi^{0}$ and $\Xi^{-}$.

To know their EM structure means to have adequate multitude of as precise as
possible experimental information on their EM FFs in the whole region of their
definition and to dispose with reliable theoretical models to be able predict
behaviors of these EM FFs consistently with data.

However, from all members of the $1/2^{+}$ octet baryons the most reliable and
dense experimental data exist on the proton EM structure just due to its
stability. They have been obtained by the processes like $e^{-}p\to e^{-}p$,
$e^{+}p\to e^{+}p$, $e^{+}e^{-}\to p\bar{p}$, $p\bar{p}\to e^{+}e^{-}$ and
polarizations there. And thanks to a progress with more accomplished
technologies one can expect even improved situation with these data in near
future. Moreover the advanced proton EM structure $Unitary\&Analytic$ model
ABDD  has been constructed recently, which reproduces all existing data quite
well.

The next member of the $1/2^{+}$ octet baryons is the neutron $n$, for which
there are no very precise and reliable experimental data just due to its
instability. One is unable to prepare neither neutron targets nor antineutron
beams for realization of similar experiments as they have been carried out
with protons. In obtaining of the neutron EM FFs data a lot of model
assumptions have been utilized and therefore finally the neutron data are not
very reliable.

Heaving in mind the latter, in this paper we would like to demonstrate how the
neutron EM FFs behaviors are predicted theoretically in the whole region of
their definition by using the reliable proton EM FFs data, for a description
of which the advanced proton EM structure $Unitary\&Analytic$ model ABDD  is
applied.

In the next papers a similar approach is planned to be realized also for other
members of the $1/2^{+}$ octet baryons, so-called hyperons, starting with
$\Lambda^{0}$ and ending with $\Xi$ hyperons, however for numerical estimation
of free physically interpretable parameters of their $Unitary\&Analytic$
models a completely different and more laborious approach is developed.

##  II Proton electromagnetic structure model

The EM structure of the proton is completely described theoretically by two
independent functions, the proton Dirac $F^{p}_{1}(t)$ and Pauli
$F^{p}_{2}(t)$ FFs, which naturally appear in a decomposition of the proton
matrix element of the EM current $J^{EM}_{\mu}(0)$ in the form of coefficients
of two linearly independent covariants constructed from the four momenta
$p,p^{\prime}$, $\gamma$-matrices and Dirac bi-spinors

$\displaystyle<p|J^{EM}_{\mu}(0)|p>=e\bar{u}(p^{\prime})\\{\gamma_{\mu}F^{p}_{1}(t)+\frac{i}{2m_{p}}\sigma_{\mu\nu}(p^{\prime}-p)_{\mu}F^{p}_{2}(t)\\}u(p),$
(1)

with $m_{p}$ to be the proton mass.

Description of the proton is even improved if the following mixed
transformation properties of the EM current $J^{EM}_{\mu}(0)$ under the
rotation in the isospin space are utilized. One its part transforms like
isoscalar and another like the third component of isovector. These
transformation properties lead to a splitting of the proton Dirac and Pauli EM
FFs to flavour-independent isoscalar $F^{N}_{1s}(t),F^{N}_{2s}(t)$ and
isovector $F^{N}_{1v}(t),F^{N}_{2v}(t)$ parts as follows

$\displaystyle F^{p}_{1}(t)=[F^{N}_{1s}(t)+F^{N}_{1v}(t)]$ $\displaystyle
F^{p}_{2}(t)=[F^{N}_{2s}(t)+F^{N}_{2v}(t)]$ (2)

whereby the sign between them is specified by the sign of the third component
of the proton isospin to be $+1/2$.

The FFs $F^{N}_{1s}(t),F^{N}_{1v}(t),F^{N}_{2s}(t),F^{N}_{2v}(t)$ are analytic
in the whole complex t-plane besides the cuts on the positive real axis
starting for isovector FFs at the two-pion threshold and for isoscalar FFs at
three-pion threshold.

In the paper ABDD  the advanced 9 vector-meson resonance
$\rho(770),\omega(782),\phi(1020);\rho^{\prime}(1450),\omega^{\prime}(1420),\phi^{\prime}(1680);\rho^{\prime\prime}(1700),\omega^{\prime\prime}(1650),\phi^{\prime\prime}(2170);$
PDG
Unitary and Analytic $(U\&A)$ model for proton isoscalar and isovector Dirac
and Pauli FFs has been constructed

$\displaystyle
F^{N}_{1s}[V(t)]=\Bigg{(}\frac{1-V^{2}}{1-V^{2}_{N}}\Bigg{)}^{4}\Bigg{\\{}\frac{1}{2}H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime\prime}}(V)$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(V)H_{\omega^{\prime}}(V)\frac{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega^{\prime}})}{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega^{\prime\prime}})}+H_{\omega^{\prime\prime}}(V)H_{\omega^{\prime}}(V)\frac{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\omega^{\prime}})}{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi^{\prime\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime\prime}}(V)\Bigg{]}(f^{(1)}_{\omega^{\prime}NN}/f_{\omega^{\prime}})$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(V)H_{\phi^{\prime}}(V)\frac{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\phi^{\prime}})}{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega^{\prime\prime}})}+H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime}}(V)\frac{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi^{\prime}})}{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi^{\prime\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime\prime}}(V)\Bigg{]}(f^{(1)}_{\phi^{\prime}NN}/f_{\phi^{\prime}})$
(3)
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(V)L_{\omega}(V)\frac{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega})}{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega^{\prime\prime}})}+H_{\omega^{\prime\prime}}(V)L_{\omega}(V)\frac{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\omega})}{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi^{\prime\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime\prime}}(V)\Bigg{]}(f^{(1)}_{\omega
NN}/f_{\omega})$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(V)L_{\phi}(V)\frac{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\phi})}{(C^{1s}_{\phi^{\prime\prime}}-C^{1s}_{\omega^{\prime\prime}})}+H_{\omega^{\prime\prime}}(V)L_{\phi}(V)\frac{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi})}{(C^{1s}_{\omega^{\prime\prime}}-C^{1s}_{\phi^{\prime\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(V)H_{\phi^{\prime\prime}}(V)\Bigg{]}(f^{(1)}_{\phi
NN}/f_{\phi})\Bigg{\\}}$

with 5 free parameters
$(f^{(1)}_{\omega^{\prime}NN}/f_{\omega^{\prime}}),(f^{(1)}_{\phi^{\prime}NN}/f_{\phi^{\prime}}),(f^{(1)}_{\omega
NN}/f_{\omega}),(f^{(1)}_{\phi NN}/f_{\phi}),t^{1s}_{in}$

$\displaystyle
F^{N}_{1v}[W(t)]=\Bigg{(}\frac{1-W^{2}}{1-W^{2}_{N}}\Bigg{)}^{4}\Bigg{\\{}\frac{1}{2}L_{\rho}(W)L_{\rho^{\prime}}(W)$
$\displaystyle+\Bigg{[}L_{\rho^{\prime}}(W)L_{\rho^{\prime\prime}}(W)\frac{(C^{1v}_{\rho^{\prime}}-C^{1v}_{\rho^{\prime\prime}})}{(C^{1v}_{\rho^{\prime}}-C^{1v}_{\rho})}+L_{\rho}(W)L_{\rho^{\prime\prime}}(W)\frac{(C^{1v}_{\rho}-C^{1v}_{\rho^{\prime\prime}})}{(C^{1v}_{\rho}-C^{1v}_{\rho^{\prime}})}$
(4) $\displaystyle-L_{\rho}(W)L_{\rho^{\prime}}(W)\Bigg{]}(f^{(1)}_{\rho
NN}/f_{\rho})\Bigg{\\}}$

with 2 free parameters $(f^{(1)}_{\rho NN}/f_{\rho})$ and $t^{1v}_{in}$,

$\displaystyle
F^{N}_{2s}[U(t)]=\Bigg{(}\frac{1-U^{2}}{1-U^{2}_{N}}\Bigg{)}^{6}\Bigg{\\{}\frac{1}{2}(\mu_{p}+\mu_{n}-1)H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)H_{\phi^{\prime}}(U)\frac{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\phi^{\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi^{\prime}})}{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\omega^{\prime\prime}})}$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\omega^{\prime}}(U)H_{\phi^{\prime}}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi^{\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi^{\prime}})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi^{\prime\prime}})}$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)H_{\phi^{\prime}}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi^{\prime}})(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\phi^{\prime}})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\omega^{\prime}})(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)\Bigg{]}(f^{(2)}_{\phi^{\prime}NN}/f_{\phi^{\prime}})$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)L_{\omega}(U)\frac{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\omega})}{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\omega^{\prime\prime}})}$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\omega^{\prime}}(U)L_{\omega}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\omega})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\omega})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi^{\prime\prime}})}+$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)L_{\omega}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\omega})(C^{2s}_{\phi^{\prime}}-C^{2s}_{\omega})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\omega^{\prime}})(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime}})}$
(5) $\displaystyle-
H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)\Bigg{]}(f^{(2)}_{\omega
NN}/f_{\omega})$
$\displaystyle+\Bigg{[}H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)L_{\phi}(U)\frac{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\phi})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi})}{(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\omega^{\prime\prime}})}$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\omega^{\prime}}(U)L_{\phi}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi^{\prime\prime}})(C^{2s}_{\omega^{\prime}}-C^{2s}_{\phi^{\prime\prime}})}$
$\displaystyle+H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)L_{\phi}(U)\frac{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\phi})(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\phi})}{(C^{2s}_{\omega^{\prime\prime}}-C^{2s}_{\omega^{\prime}})(C^{2s}_{\phi^{\prime\prime}}-C^{2s}_{\omega^{\prime}})}$
$\displaystyle-
H_{\omega^{\prime\prime}}(U)H_{\phi^{\prime\prime}}(U)H_{\omega^{\prime}}(U)\Bigg{]}(f^{(2)}_{\phi
NN}/f_{\phi})\Bigg{\\}}$

with 4 free parameters $(f^{(2)}_{\phi^{\prime}NN}/f_{\phi^{\prime}})$,
$(f^{(2)}_{\omega NN}/f_{\omega})$, $(f^{(2)}_{\phi
NN}/f_{\phi}),t^{2s}_{in}$, and

$\displaystyle
F^{N}_{2v}[X(t)]=\Bigg{(}\frac{1-X^{2}}{1-X^{2}_{N}}\Bigg{)}^{6}\Bigg{\\{}\frac{1}{2}(\mu_{p}-\mu_{n}-1)L_{\rho}(X)L_{\rho^{\prime}}(X)H_{\rho^{\prime\prime}}(X)\Bigg{\\}}$
(6)

dependent on only 1 free parameter $t^{2v}_{in}$ with

$\displaystyle
L_{r}(V)=\frac{(V_{N}-V_{r})(V_{N}-V^{*}_{r})(V_{N}-1/V_{r})(V_{N}-1/V^{*}_{r})}{(V-V_{r})(V-V^{*}_{r})(V-1/V_{r})(V-1/V^{*}_{r})},$
(7) $\displaystyle
C^{1s}_{r}=\frac{(V_{N}-V_{r})(V_{N}-V^{*}_{r})(V_{N}-1/V_{r})(V_{N}-1/V^{*}_{r})}{-(V_{r}-1/V_{r})(V_{r}-1/V^{*}_{r})},r=\omega,\phi$
$\displaystyle
H_{l}(V)=\frac{(V_{N}-V_{l})(V_{N}-V^{*}_{l})(V_{N}+V_{l})(V_{N}+V^{*}_{l})}{(V-V_{l})(V-V^{*}_{l})(V+V_{l})(V+V^{*}_{l})},$
(8) $\displaystyle
C^{1s}_{l}=\frac{(V_{N}-V_{l})(V_{N}-V^{*}_{l})(V_{N}+V_{l})(V_{N}+V^{*}_{l})}{-(V_{l}-1/V_{l})(V_{l}-1/V^{*}_{l})},l=\omega^{\prime\prime},\phi^{\prime\prime},\omega^{\prime},\phi^{\prime}$
$\displaystyle
L_{k}(W)=\frac{(W_{N}-W_{k})(W_{N}-W^{*}_{k})(W_{N}-1/W_{k})(W_{N}-1/W^{*}_{k})}{(W-W_{k})(W-W^{*}_{k})(W-1/W_{k})(W-1/W^{*}_{k})},$
(9) $\displaystyle
C^{1v}_{k}=\frac{(W_{N}-W_{k})(W_{N}-W^{*}_{k})(W_{N}-1/W_{k})(W_{N}-1/W^{*}_{k})}{-(W_{k}-1/W_{k})(W_{k}-1/W^{*}_{k})},k=\rho^{\prime\prime},\rho^{\prime},\rho$
$\displaystyle
L_{r}(U)=\frac{(U_{N}-U_{r})(U_{N}-U^{*}_{r})(U_{N}-1/U_{r})(U_{N}-1/U^{*}_{r})}{(U-U_{r})(U-U^{*}_{r})(U-1/U_{r})(U-1/U^{*}_{r})},$
(10) $\displaystyle
C^{2s}_{r}=\frac{(U_{N}-U_{r})(U_{N}-U^{*}_{r})(U_{N}-1/U_{r})(U_{N}-1/U^{*}_{r})}{-(U_{r}-1/U_{r})(U_{r}-1/U^{*}_{r})},r=\omega,\phi$
$\displaystyle
H_{l}(U)=\frac{(U_{N}-U_{l})(U_{N}-U^{*}_{l})(U_{N}+U_{l})(U_{N}+U^{*}_{l})}{(U-U_{l})(U-U^{*}_{l})(U+U_{l})(U+U^{*}_{l})},$
(11) $\displaystyle
C^{2s}_{l}=\frac{(U_{N}-U_{l})(U_{N}-U^{*}_{l})(U_{N}+U_{l})(U_{N}+U^{*}_{l})}{-(U_{l}-1/U_{l})(U_{l}-1/U^{*}_{l})},l=\omega^{\prime\prime},\phi^{\prime\prime},\omega^{\prime},\phi^{\prime}$
$\displaystyle
L_{k}(X)=\frac{(X_{N}-X_{k})(X_{N}-X^{*}_{k})(X_{N}-1/X_{k})(X_{N}-1/X^{*}_{k})}{(X-X_{k})(X-X^{*}_{k})(X-1/X_{k})(X-1/X^{*}_{k})},$
(12) $\displaystyle
C^{2v}_{k}=\frac{(X_{N}-X_{k})(X_{N}-X^{*}_{k})(X_{N}-1/X_{k})(X_{N}-1/X^{*}_{k})}{-(X_{k}-1/X_{k})(X_{k}-1/X^{*}_{k})},k=\rho^{\prime},\rho$
$\displaystyle
H_{\rho^{\prime\prime}}(X)=\frac{(X_{N}-X_{\rho^{\prime\prime}})(X_{N}-X^{*}_{\rho^{\prime\prime}})(X_{N}+X_{\rho^{\prime\prime}})(X_{N}+X^{*}_{\rho^{\prime\prime}})}{(X-X_{\rho^{\prime\prime}})(X-X^{*}_{\rho^{\prime\prime}})(X+X_{\rho^{\prime\prime}})(X+X^{*}_{\rho^{\prime\prime}})},$
(13) $\displaystyle
C^{2v}_{\rho^{\prime\prime}}=\frac{(X_{N}-X_{\rho^{\prime\prime}})(X_{N}-X^{*}_{\rho^{\prime\prime}})(X_{N}+X_{\rho^{\prime\prime}})(X_{N}+X^{*}_{\rho^{\prime\prime}})}{-(X_{\rho^{\prime\prime}}-1/X_{\rho^{\prime\prime}})(X_{\rho^{\prime\prime}}-1/X^{*}_{\rho^{\prime\prime}})}.$

Whereas the Dirac $F^{p}_{1}(t)$ and Pauli $F^{p}_{2}(t)$ proton FFs, as it is
seen from the previous, are very effective for theoretical considerations of
the proton EM structure, an extraction of experimental information on it from
measured cross sections and polarizations, by the proton electric
$G^{p}_{E}(t)$ and proton magnetic $G^{p}_{M}(t)$ FFs is more comfortable.
They appear e.g. in the total cross section of $e^{+}e^{-}\to p\bar{p}$
process BPZZ

$\displaystyle\sigma_{tot}(e^{+}e^{-}\to
p\bar{p})=\frac{4\pi\alpha^{2}C\beta_{p}(t)}{3t}[|G^{p}_{M}(t)|^{2}+\frac{2m_{p}}{t}|G^{p}_{E}(s)|^{2}]$
(14)

with $\beta_{p}(t)=\sqrt{1-\frac{4m^{2}_{p}}{t}}$ and $C$ to be the Coulomb
enhancement factor, without interference term, unlike the case if Dirac and
Pauli FFs appear in the same total cross section.

The relations between both sets of FFs are

$\displaystyle
G^{p}_{E}(t)=[F^{N}_{1s}(t)+F^{N}_{1v}(t)]+\frac{t}{4m^{2}_{p}}[F^{N}_{2s}(t)+F^{N}_{2v}(t)]$
(15) $\displaystyle
G^{p}_{M}(t)=[F^{N}_{1s}(t)+F^{N}_{1v}(t)]+[F^{N}_{2s}(t)+F^{N}_{2v}(t)]$

with normalizations

$\displaystyle G^{p}_{E}(0)=1;\quad G^{p}_{M}(0)=\mu_{p};$ (16)

and

$\displaystyle F^{N}_{1s}(0)=F^{N}_{1v}(0)=\frac{1}{2};\quad
F^{N}_{2s}(0)=\frac{1}{2}(\mu_{p}+\mu_{n}-1);\quad
F^{N}_{2v}(0)=\frac{1}{2}(\mu_{p}-\mu_{n}-1),$ (17)

where $\mu_{p}$ and $\mu_{n}$ are the magnetic moments of the proton and
neutron, respectively.

The model is in fact a well-matched unification of pole contributions of
unstable vector mesons with cut structure in the complex plane $t$, whereby
these cuts represent so-called continua contributions generated by exchange of
more than one particle in the corresponding Feynman diagrams and they secure
FFs imaginary parts to be different from zero just beyond the lowest possible
thresholds on the positive real axis, as it is required by the FFs unitarity
conditions.

There are approximately 460 more or less reliable experimental points on the
proton EM FFs $G^{p}_{E}(t),G^{p}_{M}(t)$ jones -Ab1  to be graphically
presented in Figs. 1-2.

Figure 1: Experimental data on the ratios of the proton electric to magnetic
FFs in space-like and time-like regions.

The results of their simultaneous analysis by the proton EM structure $U\&A$
model given by (II)-(6) and (15) with 12 free parameters with a clear physical
meaning are given in TABLE.

Figure 2: Experimental data on the proton electric and magnetic FFs in space-
like and time-like regions.

TABLE: Values of free parameters of the proton EM structure $U\&A$ model
evaluated in the analysis of the proton EM FFs data.
$\chi^{2}/ndf=1.74;$ $s^{1s}_{in}=(1.6750\pm
0.0363)GeV^{2};s^{1v}_{in}=(2.9683\pm 0.0091)GeV^{2};$
$s^{2s}_{in}=(1.8590\pm 0.0023)GeV^{2};s^{2v}_{in}=(2.4425\pm 0.0208)GeV^{2};$
$(f^{(1)}_{\omega^{\prime}NN}/f_{\omega^{\prime}})=-0.2937\pm
0.0015;(f^{(1)}_{\phi^{\prime}NN}/f_{\phi^{\prime}})=-0.5298\pm 0.0027;\\\
(f^{(1)}_{\omega NN}/f_{\omega})=0.6384\pm 0.0025;(f^{(1)}_{\phi
NN}/f_{\phi})=-0.0271\pm 0.0005;\\\
(f^{(2)}_{\phi^{\prime}NN}/f_{\phi^{\prime}})=0.3075\pm
0.0156;(f^{(2)}_{\omega NN}/f_{\omega})=0.1676\pm 0.0377;\\\ (f^{(2)}_{\phi
NN}/f_{\phi})=0.1226\pm 0.0035;(f^{(1)}_{\rho NN}/f_{\rho})=-0.0802\pm
0.0014;$

The corresponding behaviors of absolute values of the proton electric and
magnetic FFs on the base of these results are given by full lines in Fig. 3,
where also their comparison with exiting data is presented.

Figure 3: The proton EM structure $U\&A$ model given by (II)-(6) and (15) with
parameters of TABLE reproduce existing data quite well.

##  III Theoretical predictions of neutron electromagnetic form factors
behaviors

Similarly to the proton one can define the neutron Dirac $F^{n}_{1}(t)$ and
Pauli $F^{n}_{2}(t)$ FFs by means of the neutron matrix element of the EM
current $J^{EM}_{\mu}(0)$

$\displaystyle<n|J^{EM}_{\mu}(0)|n>=e\bar{u}(p^{\prime})\\{\gamma_{\mu}F^{n}_{1}(t)+\frac{i}{2m_{n}}\sigma_{\mu\nu}(p^{\prime}-p)_{\mu}F^{n}_{2}(t)\\}u(p),$
(18)

with $m_{n}$ to be the neutron mass. Again mixed transformation properties of
the EM current $J^{EM}_{\mu}(0)$ under the rotation in the isospin space is
utilized, which leads to a splitting of the neutron Dirac $F^{n}_{1}(t)$ and
Pauli $F^{n}_{2}(t)$ FFs to the same flavour-independent isoscalar
$F^{N}_{1s}(t),F^{N}_{2s}(t)$ and isovector $F^{N}_{1v}(t),F^{N}_{2v}(t)$
parts like in the case of the proton as follows

$\displaystyle F^{n}_{1}(t)=[F^{N}_{1s}(t)-F^{N}_{1v}(t)]$ $\displaystyle
F^{n}_{2}(t)=[F^{N}_{2s}(t)-F^{N}_{2v}(t)]$ (19)

however, now with the minus sign between them as the sign of the third
component of the neutron isospin is -1/2.

Figure 4: Predicted absolute value of the neutron electric and magnetic FFs
behaviors without the use of any experimental point hanson -riordan  seen in
this figures.

Then the relations between neutron EM FFs and the flavour-independent neutron
Dirac and Pauli FFs are

$\displaystyle
G^{n}_{E}(t)=[F^{N}_{1s}(t)-F^{N}_{1v}(t)]+\frac{t}{4m^{2}_{n}}[F^{N}_{2s}(t)-F^{N}_{2v}(t)]$
(20) $\displaystyle
G^{n}_{M}(t)=[F^{N}_{1s}(t)-F^{N}_{1v}(t)]+[F^{N}_{2s}(t)-F^{N}_{2v}(t)]$

with normalizations

$\displaystyle G^{n}_{E}(0)=0;\quad G^{n}_{M}(0)=\mu_{n}.$ (21)

Since the FFs $F^{N}_{1s}(t),F^{N}_{2s}(t)$ and $F^{N}_{1v}(t),F^{N}_{2v}(t)$
are completely known by the expressions (II)-(6) with free parameters from
TABLE to be evaluated in the analysis of only the proton EM FFs data, in
combination with relations (20) they represent the advanced neutron
electromagnetic structure $U\&A$ model, which enables to predict behaviors of
the absolute values of the neutron electric $G^{n}_{E}(t)$ and magnetic
$G^{n}_{M}(t)$ FFs theoretically, as they are presented by the full lines in
Fig. 4, without the use of any experimental point hanson -riordan  on the
neutron EM FFs data also shown there for a comparison. The results are
encouraging.

##  IV Conclusions

The absolute values of the neutron EM FFs, $G^{n}_{E}(t),G^{n}_{M}(t)$, are
predicted theoretically in the whole region of their definition without
exploiting of any experimental point on them. This could be carried out by the
fact that the proton EM structure $U\&A$ model given by (II)-(6) and (15) and
the neutron EM structure $U\&A$ model given by (II)-(6) and (20) are compound
of the same flavour-independent isoscalar $F^{N}_{1s}(t),F^{N}_{2s}(t)$ and
isovector $F^{N}_{1v}(t),F^{N}_{2v}(t)$ parts of the corresponding Dirac and
Pauli FFs, and it does not matter that in the case of the proton with "+" sign
and in the case of the neutron with "-" sign, however, they depend on the same
free parameters.

So, these parameters can be numerically evaluated either in the analysis of
the neutron EM FFs data by means of the neutron EM structure $U\&A$ model and
then one predicts absolute values of the proton EM FFs behaviors, or in the
analysis of the proton EM FFs data by means of the proton EM structure $U\&A$
model and then one predicts absolute values of the neutron EM FFs behaviors.
Naturally, as at present the proton EM FFs are more reliable and more precise,
we have preferred the second case.

On the conclusion we would like to emphasize that if more precise and more
reliable proton EM FFs data will appear in as wide region of their definition
as possible, the more reliable predictions for the absolute values of the
neutron EM FFs will be achieved by the approach discussed in this paper.

The authors would like to thank Eric Bartos for valuable discussions. The
support of the Slovak Grant Agency for Sciences VEGA, grant No.2/0153/17, is
acknowledged.

## References

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