# On the classification of 2-solvable Frobenius Lie algebras

André Diatta(1)111 (1) Aix-Marseille Univ, CNRS, Centrale Marseille, Institut
Fresnel, 13013 Marseille, France.
∗ Corresponding author<EMAIL_ADDRESS><EMAIL_ADDRESS>;
Bakary Manga(2) and Ameth Mbaye(2)222(2) Département de Mathématiques et
Informatique, Université Cheikh Anta Diop de Dakar, BP 5005 Dakar-Fann, Dakar,
Sénégal. Email<EMAIL_ADDRESS><EMAIL_ADDRESS>AD and BM
acknowledge financial support from the Simons Foundation through the NLAGA
project.

###### Abstract

We discuss the classification of $2$-solvable Frobenius Lie algebras. We prove
that every $2$-solvable Frobenius Lie algebra splits as a semidirect sum of an
$n$-dimensional vector space V and an $n$-dimensional maximal Abelian
subalgebra (MASA) of the full space of endomorphisms of $V.$ We supply a
complete classification of $2$-solvable Frobenius Lie algebras corresponding
to nonderogatory endomorphisms, as well as those given by maximal Abelian
nilpotent subalgebras (MANS) of class 2, hence of Kravchuk signature
$(n-1,0,1)$. In low dimensions, we classify all 2-solvable Frobenius Lie
algebras in general up to dimension $8$. We correct and complete the
classification list of MASAs of $\mathfrak{sl}(4,\mathbb{R})$ by Winternitz
and Zassenhaus. As a biproduct, we give a simple proof that every
nonderogatory endormorphism of a real vector space admits a Jordan form and
also provide a new characterization of Cartan subalgebras of
$\mathfrak{sl}(n,\mathbb{R})$.

##  1 Introduction

Throughout this work, $V$ stands for a real vector space of dimension $n$. We
are mainly interested in the classification of 2-solvable Frobenius Lie
algebras. It turns out that such a classification is closely related to that
of $n$-dimensional maximal Abelian subalgebras (MASAs) of the space
$\mathfrak{gl}(V)$ of linear operators (endomorphisms) on $V.$ The sudy of
MASAs traces back to Frobenius and Schur, it has been a vibrant and vivid
subject these last decades, in relation with several subjects in mathematics
and physics such as the classification of Lie algebras and the study of
dynamical systems ([5, 10, 12, 13, 14, 20, 22, 24, 25, 26]). Our interest in
$2$-solvable Frobenius Lie algebras is mainly related to their applications in
symplectic ([3, 7, 9, 15]) and information geometry ([16]), as well as to the
investigation of Lie algebras in general. Frobenius Lie algebras have gained
popularity mainly since the work of Ooms ([18, 19]). Among other results, a
characterization of 2-solvable Frobenius Lie algebras is given in [18]
(Theorem 4.1, Section 4, p. 43). On the other hand, the family of 2-solvable
Frobenius Lie algebras with an Abelian nilradical is studied in [1]. The main
results of the present paper are the following.

$\bullet$ We show that, if $\mathfrak{B}\subset\mathfrak{gl}(V)$ is Abelian
and the semidirect sum $\mathfrak{B}\ltimes V$ is a Frobenius Lie algebra,
then $\mathfrak{B}$ is a MASA of $\mathfrak{gl}(V)$ (Theorem 1). Moreover,
every 2-solvable Frobenius Lie algebra is of that form. Two 2-solvable
Frobenius Lie algebras $\mathfrak{B}_{1}\ltimes V$ and
$\mathfrak{B}_{2}\ltimes V$ are isomorphic if and only if there exists some
$\phi$ in the group GL$(V)$ of invertible linear operators of $V$, such that
$\mathfrak{B}_{2}=\phi\mathfrak{B}_{1}\phi^{-1}$ (Proposition 3.1).

$\bullet$ In Theorem 2, we classify 2-solvable Frobenius Lie algebras
$\mathfrak{B}\ltimes V$, where $\mathfrak{B}\cap\mathfrak{sl}(V)$ is a maximal
Abelian nilpotent subalgebra (MANS) of class 2 and hence of Kravchuk signature
$(n-1,0,1)$. Here $\mathfrak{sl}(V)\subset\mathfrak{gl}(V)$ stands for the
subspace of traceless linear operators.

$\bullet$ We fully classify, in Theorem 4, the family of $2$-solvable
Frobenius Lie algebras $\mathfrak{B}\ltimes V$, where $\mathfrak{B}$ is the
algebra $\mathbb{K}[\phi]$ of the polynomials in some nonderogatory
$\phi\in\mathfrak{gl}(V)$. We will simply write
$\mathcal{G}_{\phi}=\mathbb{K}[\phi]\ltimes V$ instead of $\mathfrak{B}\ltimes
V$.

$\bullet$ We supply a full classification list of all 2-solvable Frobenius Lie
algebras in general, up to and including dimension $8$ (Theorem 6).

$\bullet$ Theorems 3, 5 give a new characterization of Cartan subalgebras of
$\mathfrak{sl}(V).$

$\bullet$ We correct the classification list of MASAs of
$\mathfrak{sl}(4,\mathbb{R})$ of [25] and further complete it with a missing
item.

$\bullet$ We prove that every nonderogatory real matrix admits a Jordan form
(Theorem 7).

The paper is organized as follows. Section 2, is devoted to some preliminaries
and notations. In Section 3, we discuss the classification of $2$-solvable
Frobenius Lie algebras, in general. General examples of MASAs and 2-solvable
Frobenius Lie algebras are discussed in details and a new characterization of
Cartan subalgebras of $\mathfrak{sl}(n,\mathbb{R})$ is given. We supply a full
classification of $2$-solvable Frobenius Lie algebras given by nonderogatory
maps in Section 4. We fully classify all $2$-solvable Frobenius Lie algebras
up to dimension $8$ in Section 5 and furthermore, we correct and complete the
classification list of MASAs of $\mathfrak{sl}(4,\mathbb{R})$ of [25] with a
missing item. The paper ends by a simple direct proof of the Jordanization of
nonderogatory real matrices in Appendix (Section 6).

##  2 Preliminaries

Throughout this work, if $\mathfrak{F}$ is a vector space, $\mathfrak{F}^{*}$
will stand for its (linear) dual, $\mathfrak{gl}(\mathfrak{F})$ the space of
linear maps $\psi:\mathfrak{F}\to\mathfrak{F}$ and GL($\mathfrak{F}$) the
subspace of $\mathfrak{gl}(\mathfrak{F})$ consisting of invertible maps. The
symmetric bilinear form $\langle,\rangle$ stands for the duality pairing
between vectors and linear forms: $\langle u,f\rangle:=f(u),$ for
$u\in\mathfrak{F},$ $f\in\mathfrak{F}^{*}$. If $(e_{j})$ is a basis of
$\mathfrak{F}$, we denote by $(e_{j}^{*})$ its dual basis. Let $\mathfrak{B}$
be a Lie subalgebra of $\mathfrak{gl}(V)$, where $V$ is a vector space. We
consider the natural action $\mathfrak{B}\times V\to V,$
$(a,x)\mapsto\rho(a)x:=ax$. For any $(a,f)\in\mathfrak{B}\times V^{*}$, let
$\rho^{*}(a)f$ be the element of $V^{*}$ defined on elements $x$ of $V$ by
$\langle\rho^{*}(a)f,x\rangle:=-f(ax).$ We recall that $\rho^{*}$ is called
the contragredient representation (or just action) of $\rho$.

######  Definition 2.1.

By abuse of language, we say that the orbit of $\alpha\in V^{*}$, under
$\rho^{*}$, is open, if $V^{*}=\\{\rho^{*}(a)\alpha,\;\;a\in\mathfrak{B}\\}$.
We say that $\alpha$ has a trivial isotropy if the equality
$\rho^{*}(a)\alpha=0$ implies $a=0.$

The vector space $\mathfrak{B}\oplus V$ endowed with the Lie bracket defined,
for every $a,b\in\mathfrak{B}$ and $x,y\in V$, as
$[a,b]=[a,b]_{\mathfrak{B}}\;,\;[x,y]=0,\text{ and }[a,x]=\rho(a)x=ax,$ will
be termed the semidirect sum of $\mathfrak{B}$ and $V$ via $\rho$ and denoted
by $\mathfrak{B}\ltimes_{\rho}V$, or simply $\mathfrak{B}\ltimes V$ if no
confusion is to be made. In the present work, $\mathfrak{B}$ is Abelian,
$[,]_{\mathfrak{B}}=0.$

A Lie algebra $\mathcal{G}$ is called a Frobenius Lie algebra if there exists
$\alpha\in\mathcal{G}^{*}$, called a Frobenius functional, such that
$\partial\alpha$ is nondegenerate, where, for any $u,v\in\mathcal{G},$

$\displaystyle\partial\alpha(u,v):=-\langle\alpha,[u,v]\rangle.$ (1)

Every Frobenius Lie algebra is a codimension 1 subalgebra of some contact Lie
algebra and could be used to construct the latter. The converse is true under
some conditions ([8]). Any Lie group $G$ with Lie algebra $\mathcal{G}$, has a
left invariant symplectic form $\omega^{+}$ with value
$\omega_{\epsilon}^{+}=\partial\alpha$ at the unit (neutral) element
$\epsilon$ of $G.$ Here $\mathcal{G}$ is identified with the tangent space to
$G$ at $\epsilon.$ See for example [7], [9], for more details.

######  Definition 2.2.

A Lie algebra $\mathcal{G}$ is said to be $2$-step solvable (2-solvable, for
short) if its derived ideal $[\mathcal{G},\mathcal{G}]$ is Abelian. A Lie
algebra is said to be indecomposable if it cannot be written as the direct sum
of two of its ideals.

######  Remark 2.1.

Note that two direct sums $\mathcal{G}_{1}\oplus\dots\oplus\mathcal{G}_{k}$
and $\mathcal{G}_{1}^{\prime}\oplus\dots\oplus\mathcal{G}_{k}^{\prime}$ of
$2$-solvable Frobenius Lie algebras, are isomorphic if and only if, up to
rearranging the order of the indices, each $\mathcal{G}_{j}$ is isomorphic to
each $\mathcal{G}_{j}^{\prime}.$ So the classification of $2$-solvable
Frobenius Lie algebras boils down to that of the indecomposable ones.

######  Definition 2.3.

Let $\mathcal{G}$ be a Lie algebra, $\mathcal{A}$ a subalgebra of
$\mathcal{G}.$ The normalizer of $\mathcal{A}$ in $\mathcal{G}$ is
$\mathcal{N}_{\mathcal{G}}(\mathcal{A}):=\\{a\in\mathcal{G},[a,\mathcal{A}]\subset\mathcal{A}\\}.$
The centralizer of $\mathcal{A}$ in $\mathcal{G}$ is
cent$(\mathcal{A},\mathcal{G}):=\\{a\in\mathcal{G},[a,\mathcal{A}]=0\\}.$ We
say that $\mathcal{A}$ is a maximal Abelian subalgebra (MASA) of
$\mathcal{G}$, if $\mathcal{A}=$cent($\mathcal{A},\mathcal{G}).$ Equivalently,
$\mathcal{A}$ is Abelian and is contained in no Abelian subalgebra of higher
dimension of $\mathcal{G}.$ For $\phi\in\mathfrak{gl}(V),$ we use the
convention $\phi^{0}=\mathbb{I}_{V}=$ identity map of $V$ and
$\phi^{p+1}(x)=\phi(\phi^{p}(x)),$ for every $x\in V$ and $p$ an integer. We
let $\mathbb{K}[\phi]$ stand for the space of polynomials in $\phi$ with
coefficients in $\mathbb{K}$ and denote by $cent(\phi,\mathfrak{gl}(V))$ the
set of endomorphisms commuting with $\phi$. Recall that $\phi$ (resp. an
$n\times n$ matrix $M$) is said to be nonderogatory (or cyclic), if its
characteristic polynomial $\chi_{\phi}$ and its minimal polynomial
$\chi_{\min,\phi}$ coincide.

######  Lemma 2.1 (see e.g. [3]).

Let $V$ be a vector space of dimension $n$ over a field $\mathbb{K}$ of
characteristic zero and $\phi\in\mathfrak{gl}(V)$. The following assertions
are equivalent:

there exists $\bar{\alpha}\in V^{*}$ such that
$(\bar{\alpha},\bar{\alpha}\circ\phi,...,\bar{\alpha}\circ\phi^{n-1})$ is a
basis of $V^{*}$,

there exists $\bar{x}\in V$ such that
$(\bar{x},\phi(\bar{x}),...,\phi^{n-1}(\bar{x}))$ is a basis of $V$,

$\dim(cent(\phi,\mathfrak{gl}(V)))=n$ and $cent(\phi,\mathfrak{gl}(V))$ is
commutative,

$cent(\phi,\mathfrak{gl}(V))=\mathbb{K}[\phi]$,

the characteristic and the minimal polynomials of $\phi$ are the same,

In every extension $\bar{\mathbb{K}}$ of $\mathbb{K}$ where $\phi$ admits a
Jordan form, this latter has only one Jordan bloc for each eigenvalue.

In Theorem 7, we go beyond Lemma 2.1 to prove that, for every $n$-dimensional
real vector space $V,$ every nonderogatory $\phi\in\mathfrak{gl}(V)$ admits a
Jordan form. Throughout this work, $E_{i,j}$ stands for the $n\times n$ matrix
with zero in all entries except the $(i,j)$ entry which is equal to $1$.

##  3 On the structure of $2$-solvable Frobenius Lie algebras and
$n$-dimensional MASAs

###  3.1 MASAs of $\mathfrak{gl}(V)$ and maximal Abelian subgroups of $GL(V)$

In this section, we briefly discuss the 1-1 correspondence between MASAs of
$\mathfrak{gl}(V)$ and maximal Abelian subgroups of $GL(V)$. Let
$\mathfrak{B}$ be a MASA of $\mathfrak{gl}(V).$ Then $\mathfrak{B}$ is a
commutative subalgebra of the associative algebra $\mathfrak{gl}(V)$, and it
contains the unit $\mathbb{I}_{V}$. Hence its (open) subset $U(\mathfrak{B})$
consisting of invertible elements is an analytic Lie group and $\dim
U(\mathfrak{B})=\dim\mathfrak{B}$, (see e.g. J. P. Serre [21], p. 103, for the
general case) which, by definition, is also an Abelian Lie subgroup of GL(V).
Furthermore, since $\mathfrak{B}$ is a MASA of $\mathfrak{gl}(V)$, it contains
the exponential of all its elements. Thus $\exp\mathfrak{B}$ coincides with
the connected component of the unit of $U(\mathfrak{B})$. In fact,
$U(\mathfrak{B})$ is a maximal Abelian subgroup of $GL(V)$. Conversely, every
maximal Abelian subgroup of $GL(V)$ is of that form $U(\mathfrak{B})$, for
some MASA $\mathfrak{B}$ of $\mathfrak{gl}(V)$. Two such subgroups are
conjugate in $GL(V)$ if and only if the corresponding MASAs are conjugate in
$\mathfrak{gl}(V)$ (see [22]). So we deduce that a simple way to construct a
2-solvable Lie group with Lie algebra $\mathfrak{B}\ltimes V$ is to simply
take $U(\mathfrak{B})\ltimes V.$

###  3.2 Main result on MASAs and 2-solvable Frobenius Lie algebras

Let $V$ be a vector space of dimension $n$ and $\mathfrak{B}$ an
$n$-dimensional Abelian Lie subalgebra of $\mathfrak{gl}(V).$ Then
$\mathfrak{B}\ltimes V$ is a Frobenius Lie algebra if and only if
$\mathfrak{B}$ has an open orbit in $V^{*}$ and every 2-solvable Frobenius Lie
algebra is of that form ([6]). Equivalently, the Lie group $U(\mathfrak{B})$
of invertible elements of $\mathfrak{B}$, has some open or dense orbit in
$V^{*}$. Here we prove the following.

######  Theorem 1.

Let $V$ be a vector space of dimension $n$ and $\mathfrak{B}$ an
$n$-dimensional Abelian Lie subalgebra of $\mathfrak{gl}(V).$ Suppose
$\mathfrak{B}\ltimes V$ is a Frobenius Lie algebra. Then $\mathfrak{B}$ is a
MASA of $\mathfrak{gl}(V)$. The Lie group $U(\mathfrak{B})$ of invertible
elements of $\mathfrak{B}$, is a maximal Abelian subgroup of $\text{GL}(V).$
Conversely, every 2-solvable Frobenius Lie algebra is isomorphic to one as
above.

###### Proof.

Without loss of generality, set $V=\mathbb{K}^{n}.$ The Lie algebra
$\mathfrak{B}\ltimes\mathbb{K}^{n}$ is Frobenius if and only if the action
$\mathfrak{B}\times(\mathbb{K}^{n})^{*}\to(\mathbb{K}^{n})^{*}$,
$(a,f)\mapsto-f\circ a$, has an open orbit ([6]). Consider some
$\alpha\in(\mathbb{K}^{n})^{*}$ with an open orbit. Thus, any basis
$(a_{1},\dots,a_{n})$ of $\mathfrak{B}$ gives rise to a basis
$(\rho^{*}(a_{1})\alpha,\dots,\rho^{*}(a_{n})\alpha)$ of
$(\mathbb{K}^{n})^{*}$ and the (linear) orbital map
$Q:\mathfrak{B}\to(\mathbb{K}^{n})^{*}$, $Q(a)=\rho^{*}(a)\alpha$ is an
isomorphism between the vector spaces $\mathfrak{B}$ and
$(\mathbb{K}^{n})^{*}$. Just for convenience, we will let
$(\hat{e}_{1},\dots,\hat{e}_{n})$ stand for the basis of $\mathbb{K}^{n}$
whose dual basis is
$(\hat{e}_{1}^{*}:=\rho^{*}(a_{1})\alpha,\dots,\hat{e}_{n}^{*}:=\rho^{*}(a_{n})\alpha).$
Suppose $\tilde{a}\in\mathfrak{gl}(n,\mathbb{K})$ is such that
$[\tilde{a},a]=0$ for any $a\in\mathfrak{B}$ and $\tilde{a}\neq 0$. Assume
$\tilde{a}$ is not an element of $\mathfrak{B}$, then
$\mathfrak{\tilde{B}}:=\mathbb{K}\tilde{a}\oplus\mathfrak{B}$ is an
$(n+1)$-dimensional Abelian subalgebra of $\mathfrak{gl}(n,\mathbb{K})$.
Because $\dim\mathfrak{\tilde{B}}=\dim(\mathbb{K}^{n})^{*}+1$, the orbital map
$\tilde{Q}:\mathfrak{\tilde{B}}\to(\mathbb{K}^{n})^{*}$, also given by
$\tilde{Q}(a)=\tilde{\rho}^{*}(a)\alpha:=-\alpha\circ a$, must have a
$1$-dimensional kernel. So, there exists some $\tilde{b}=k\tilde{a}+a_{0}\neq
0$, with $k\in\mathbb{K}$ and $a_{0}\in\mathfrak{B}$ such that
$\tilde{\rho}^{*}(\tilde{b})\alpha=-\alpha\circ\tilde{b}=0.$ Since
$\tilde{\rho}^{*}(a_{0})\alpha=\rho^{*}(a_{0})\alpha=Q(a_{0})\neq 0$ if
$a_{0}\neq 0$, we must have $k\neq 0$. We then must have $\tilde{b}x=0$, for
any $x\in\mathbb{K}^{n},$ or equivalently $\tilde{b}=0.$ Indeed, for any
$x\in\mathbb{K}^{n}$, the expression of $\hat{b}x$ in the above basis is
$\hat{b}x=\sum\limits_{j=1}^{n}\langle\hat{e}_{j}^{*},\tilde{b}x\rangle\hat{e}_{j}.$
But the components $\langle\hat{e}_{j}^{*},\tilde{b}x\rangle$ are all egal to
zero for any $j=1,\dots,n$, since the following holds true

$\displaystyle\langle\hat{e}_{j}^{*},\tilde{b}x\rangle=\langle\rho^{*}(a_{j})\alpha,\tilde{b}x\rangle=-\langle\alpha,a_{j}\tilde{b}x\rangle=-\langle\alpha,\tilde{b}a_{j}x\rangle=\langle\tilde{\rho}^{*}(\tilde{b})\alpha,a_{j}x\rangle=0.$

Now, the equality $\tilde{b}=k\tilde{a}+a_{0}=0$, contradicts the assumption
that $\tilde{a}$ is not in $\mathfrak{B}$. So $\tilde{a}$ must necessary be in
$\mathfrak{B}$. This proves that $\mathfrak{B}$ is a MASA of
$\mathfrak{gl}(n,\mathbb{K})$. Due to the correspondence between MASAs of
$\mathfrak{gl}(n,\mathbb{K})$ and maximal Abelian subgroups of
GL($n,\mathbb{K}$), see [22] and Section 3.1, $\mathfrak{B}$ is a MASA if and
only if $U(\mathfrak{B})$ is a maximal Abelian subgroup of GL$(n,\mathbb{K}).$
∎

######  Remark 3.1.

From Lemma 2.1, if $\phi\in\mathfrak{gl}(V)$ is nonderogatory, then
$\mathbb{K}[\phi]$ is an $n$-dimensional MASA of $\mathfrak{gl}(V),$ there is
$\bar{\alpha}\in V^{*}$ with an open orbit for the action of
$\mathbb{K}[\phi]$. The 2-solvable Lie algebra
$\mathcal{G}_{\phi}:=\mathbb{K}[\phi]\ltimes V$ is a Frobenius Lie algebra
(Theorem 1). Frobenius Lie algebras of type $\mathcal{G}_{\phi}$ are studied
in details and fully classified in Section 4. Obviously, if
$\phi,\psi\in\mathfrak{gl}(V)$ satisfy $\phi=P\psi P^{-1}$, for some $P\in
GL(V),$ then $\mathbb{K}[\phi]$ and $\mathbb{K}[\psi]$ are conjugate,
$\mathcal{G}_{\phi}$ and $\mathcal{G}_{\psi}$ are isomorphic via the map
$\xi:\mathcal{G}_{\psi}\to\mathcal{G}_{\phi}$, $\xi(a)=PaP^{-1}$, $\xi(x)=Px$,
for any $a\in\mathbb{K}[\psi]$ and $x\in V.$

######  Proposition 3.1.

Let $V$ be a vector space and $\mathfrak{B}_{1},$ $\mathfrak{B}_{2}$ two MASAs
of $\mathfrak{gl}(V),$ such that $\dim
V=\dim\mathfrak{B}_{1}=\dim\mathfrak{B}_{2}.$ Then $\mathfrak{B}_{1}\ltimes V$
and $\mathfrak{B}_{2}\ltimes V$ are isomorphic if and only if
$\mathfrak{B}_{1}$ and $\mathfrak{B}_{2}$ are conjugate. A linear map
$\psi:\mathfrak{B}_{1}\ltimes V\to\mathfrak{B}_{2}\ltimes V$ is an isomorphism
if and only if there exists $(\phi_{\psi},x_{\psi})\in GL(V)\times V$ such
that $\phi_{\psi}\circ\mathfrak{B}_{1}\circ\phi_{\psi}^{-1}=\mathfrak{B}_{2}$
and for any $(a,x)\in\mathfrak{B}_{1}\ltimes V,$ $\psi(a,x)=(\phi_{\psi}\circ
a\circ\phi_{\psi}^{-1}\;,\;\phi_{\psi}\circ
a\circ\phi_{\psi}^{-1}(x_{\psi})+\phi_{\psi}(x)).$

###### Proof.

Suppose $\mathcal{G}_{1}:=\mathfrak{B}_{1}\ltimes V$ and
$\mathcal{G}_{2}:=\mathfrak{B}_{2}\ltimes V$ are isomorphic and
$\psi:\mathcal{G}_{1}\to\mathcal{G}_{2}$ is an isomorphism. As the derived
ideal $[\mathcal{G}_{1},\mathcal{G}_{1}]=V$ must be mapped to
$[\mathcal{G}_{2},\mathcal{G}_{2}]=V$, there are linear maps
$\psi_{1,1}:\mathfrak{B}_{1}\to\mathfrak{B}_{2}$,
$\psi_{1,2}:\mathfrak{B}_{1}\to V$, $\phi_{\psi}:V\to V$, with $\psi_{1,1}$
and $\phi_{\psi}$ invertible, such that $\psi(a)=\psi_{1,1}(a)+\psi_{1,2}(a)$
and $\psi(x)=\phi_{\psi}(x)$ for any $a\in\mathfrak{B}_{1}$, $x\in V$. We
deduce the equality $\psi_{1,1}(a)=\phi_{\psi}\circ a\circ\phi_{\psi}^{-1}$
from the following

$\displaystyle\phi_{\psi}(ax)=\psi([a,x])=[\psi_{1,1}(a)+\psi_{1,2}(a),\phi_{\psi}(x)]=\Big{(}\psi_{1,1}(a)\circ\phi_{\psi}\Big{)}(x).$

In particular $\phi_{\psi}\circ a\circ\phi_{\psi}^{-1}\in\mathfrak{B}_{2}$,
for any $a\in\mathfrak{B}_{1}$, equivalently
$\phi_{\psi}\circ\mathfrak{B}_{1}\circ\phi_{\psi}^{-1}=\mathfrak{B}_{2}$. Now
set $x_{\psi}:=\psi_{1,2}(b)$, with $b=\mathbb{I}_{V}\in\mathfrak{B}_{1}$. We
have $\psi_{1,1}(\mathbb{I}_{V})=\mathbb{I}_{V}$, since $\psi$ is an
isomorphism. The equalities

$\displaystyle
0=\psi([a,b])=[\psi(a),\psi(b))]=\psi_{1,1}(a)\psi_{1,2}(b)-\psi_{1,1}(b)\psi_{1,2}(a)$

yield $\psi_{1,2}(a)=\phi_{\psi}\circ a\circ\phi_{\psi}^{-1}x_{\psi},$ for any
$a\in\mathfrak{B}_{1}.$ Conversely, suppose there exists some $\phi\in GL(V)$
such that $\mathfrak{B}_{2}=\phi\mathfrak{B}_{1}\phi^{-1}$. Then the map
$\psi:\mathcal{G}_{1}\to\mathcal{G}_{2}$ given for any $a\in\mathfrak{B}_{1}$
and $x\in V$ by $\psi(a+x)=\phi\circ a\circ\phi^{-1}+\phi\circ
a\circ\phi^{-1}(x_{0})+\phi(x)$ is an isomorphism, for any $x_{0}\in V.$ ∎

###  3.3 Different types of MASAs

A MASA of $\mathfrak{gl}(V)$ is said to be decomposable, if $V$ can be written
as the direct sum of subspaces which are all preserved by the MASA. Otherwise,
the MASA is said to be indecomposable. Every MASA of $\mathfrak{gl}(V)$ is a
direct sum of $\mathbb{K}\mathbb{I}_{V}$ and a MASA of $\mathfrak{sl}(V)$.
Without loss of generality, we will sometimes simply take $V=\mathbb{K}^{n}.$
A MASA of $\mathfrak{sl}(n,\mathbb{R})$ is decomposable into a direct sum of
indecomposable MASAs. The latter can either be absolutely indecomposable
(AID), or indecomposable but not absolutely indecomposable (ID $\&$ NAID). The
NAID ones become decomposable after complexification. See e.g. [17]. Let us
recall that a set of matrices is decomposable if and only if it commutes with
some idempotent $Z\neq\mathbb{I}_{\mathbb{K}^{n}}=:\mathbb{I}_{n}$. Further
recall that, an idempotent of a $\mathbb{K}$-algebra, is an element $Z$
satisfying $Z^{2}=Z.$ The NAID MASAs of $\mathfrak{sl}(n,\mathbb{R})$ are
those MASAs all of whose elements commute with $M_{s},$ given in (8). Upon
complexification, $M_{s}$ becomes
$M_{\mathbb{C}}:=$diag($i\mathbb{I}_{\frac{n}{2}}$,
$-i\mathbb{I}_{\frac{n}{2}}$), so that
$Z:=\frac{1}{2}(\mathbb{I}_{n}+iM_{\mathbb{C}})$ and
$Z^{\prime}:=\frac{1}{2}(\mathbb{I}_{n}-iM_{\mathbb{C}})$ are idempotents that
commute with the elements of the NAID MASAs. They exist only for $n$ even and
have the form
$\mathbb{R}M_{s}\oplus$MASA($\mathfrak{sl}(\frac{n}{2},\mathbb{C})$), and in
our case, we only need MASAs of $\mathfrak{sl}(\frac{n}{2},\mathbb{C})$ of
real dimension $n-2.$ An indecomposable MASA of $\mathfrak{sl}(n,\mathbb{C})$
is always a maximal Abelian nilpotent subalgebra (MANS). The AID MASAs are
MANS and they remain indecomposable after field extension. A MANS is
represented by nilpotent matrices in any finite-dimensional representation.
Following [5], a MANS $\mathcal{A}$ of $\mathfrak{sl}(n,\mathbb{K}),$
$\mathbb{K}=\mathbb{R}$ or $\mathbb{C},$ is characterized by a Kravchuk
signature $(\nu,m,\mu),$ where
$1\leq\nu=n-\dim_{\mathbb{K}}\mathcal{A}\mathbb{K}^{n}$,
$1\leq\mu=\dim\ker\mathcal{A}$, where
$\ker\mathcal{A}:=\\{x\in\mathbb{K}^{n},ax=0,\;\forall a\in\mathcal{A}\\}$ and
$m=n-\mu-\nu.$

######  Definition 3.1.

We say that $\mathcal{A}$ is of class $p\geq 2$ if $a_{1}a_{2}\cdots a_{p}=0$
for any p elements $a_{1},\dots,a_{p}\in\mathcal{A}$ and there exists some
$a\in\mathcal{A}$ such that $a^{p-1}\neq 0.$ We simply write
$\mathcal{A}^{p-1}\neq 0$ and $\mathcal{A}^{p}=0.$ Set
Im($\mathcal{A}^{k}$):$=\mathcal{A}^{k}\mathbb{K}^{n},$ $k\geq 1.$

###  3.4 Classification of 2-solvable Frobenius Lie algebras corresponding to
class 2 MANS

In this section, we treat one of the extreme cases, namely the case of MANS of
class 2. The other extreme case, the class $n$, will be treated together with
the more general case of MASAs given by nonderogatory maps, in Section 4.

######  Lemma 3.1.

Let $\mathcal{A}$ be an $(n-1)$-dimensional MASA of
$\mathfrak{sl}(n,\mathbb{R})$ such that $\mathcal{A}^{p+1}=0$ and
$\mathcal{A}^{p}\neq 0,$ for some $p\geq 1.$ If
$\mathcal{G}:=(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is a Frobenius Lie algebra, then $\dim$ Im($\mathcal{A}^{p}$)$=1.$ In
particular if $\mathcal{A}^{2}=0$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is Frobenius, then $\nu=n-1$ and $\dim Im(\mathcal{A})=1$.

###### Proof.

If $\dim$ Im($\mathcal{A}^{p}$)$\geq 2$, then for any
$\alpha\in\mathcal{G}^{*}$, we would have
$Im(\mathcal{A}^{p})\cap\ker\alpha\neq 0.$ Choose $x\in
Im(\mathcal{A}^{p})\cap\ker\alpha$ such that $x\neq 0.$ As any
$v\in\mathcal{G}$ is of the form $v=k\mathbb{I}_{\mathbb{R}^{n}}+a+y,$ where
$k\in\mathbb{R}$, $a\in\mathcal{A}$, $y\in\mathbb{R}^{n}$, we would have
$\partial\alpha(x,v)=\langle\alpha,kx+ax\rangle=0,$ $\forall v\in\mathcal{G}.$
This would mean that $\partial\alpha$ is degenerate, for any
$\alpha\in\mathcal{G}^{*}.$ Thus $\mathcal{G}$ would not be Frobenius. In
particular if $p=1$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is Frobenius, then $\dim Im(\mathcal{A})=1$ and hence $\nu=n-1.$ ∎

Now we are ready to prove the following classification theorem.

######  Theorem 2.

Up to conjugation, $\mathcal{A}:=\mathcal{A}_{n,1}$ as in Example 3.1, is the
unique MANS of class $2$, of $\mathfrak{sl}(n,\mathbb{R})$, such that
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is Frobenius. Equivalently, it is the unique MANS of
$\mathfrak{sl}(n,\mathbb{R})$ with Kravchuk signature $(n-1,0,1)$ such that
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is Frobenius. In other words, $\mathcal{G}_{n,1}$ as in Example 3.1, is the
unique 2-solvable Frobenius Lie algebra given by a MANS of class 2.

###### Proof.

First, note that if $\mathcal{A}\subset\mathfrak{sl}(n,\mathbb{R})$ is an
$(n-1)$-dimensional MASA such that $\mathcal{A}^{2}=0$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is a Frobenius Lie algebra, then from Lemma 3.1, $\dim(Im(\mathcal{A}))=1$.
Hence $\nu=n-1$ and $\mu=1.$ Conversely, if $\nu=n-1$, since $\mathcal{A}$ is
nilpotent, we necessarily have $\mathcal{A}^{2}=0.$ Now, under the assumption
that $\mathcal{A}^{2}=0$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is a Frobenius Lie algebra, let $\hat{e}_{1}\in\mathbb{R}^{n}$ such that
$\mathcal{A}\mathbb{R}^{n}=\mathbb{R}\hat{e}_{1}.$ We have
$\mathcal{A}\hat{e}_{1}=0$ and $\ker\mathcal{A}=\mathbb{R}\hat{e}_{1}.$ Thus,
in any basis of $\mathbb{R}^{n}$ whose first vector is $\hat{e}_{1},$ the
elements of $\mathcal{A}$ are linear combinations of $E_{1,j},$ $j=2,\dots,n$
and so $\mathcal{A}$ assumes the same form as the MASA $\mathcal{A}_{n,1}$ of
$\mathfrak{sl}(n,\mathbb{R})$ of Example 3.1. Consequently, $\mathcal{A}$ is
conjugate to $\mathcal{A}_{n,1}$ in $\mathfrak{gl}(n,\mathbb{R})$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{A})\ltimes\mathbb{R}^{n}$
is isomorphic to the Lie algebra $\mathcal{G}_{n,1}$, see Example 3.1). ∎

###  3.5 Some important examples

We propose several families of pairwise non-isomorphic (resp. non-conjugate)
$2$-solvable Frobenius Lie algebras (resp. MASAs).

######  Example 3.1.

For any integer $n\geq 2$, we construct below a family of ($n-1$) 2-solvable
Frobenius Lie algebras
$\mathcal{G}_{n,p}:=\mathfrak{B}_{n,p}\ltimes\mathbb{R}^{n}$, $p=1,\dots,n-1$.
The $\mathfrak{B}_{n,p}$’s are pairwise non-conjugate commutative algebras of
polynomials in $n-1,n-2,\dots,2,1$ $n\times n$ matrices, respectively. For a
given $p$, with $1\leq p\leq n-1,$ set
$M_{n,p}:=\sum\limits_{l=1}^{p}E_{l,l+1}$, so that
$M_{n,p}^{j-1}=\sum\limits_{i=1}^{p-j+2}E_{i,i+j-1}$, for $2\leq j\leq p+1$,
and $\;M_{n,p}^{p+1}=0$. Let $\mathfrak{B}_{n,p}$ be the Abelian subalgebra of
$\mathfrak{gl}(n,\mathbb{R})$ spanned by $e_{j},$ $j=1,\dots,n$, where
$e_{j}:=M_{n,p}^{j-1}$, $j=1,\dots,p+1$, and for $j=p+2,\dots,n$,
$e_{j}:=E_{1,j}$. In the canonical basis $(\tilde{e}_{1},\dots,\tilde{e}_{n})$
of $\mathbb{R}^{n}$, with dual basis
$(\tilde{e}_{1}^{*},\dots,\tilde{e}_{n}^{*})$, we have

$\displaystyle M_{n,p}\tilde{e}_{1}$ $\displaystyle=$ $\displaystyle
M_{n,p}\tilde{e}_{q}=0,\;\;q\geq p+2,$ $\displaystyle
M_{n,p}^{k}\tilde{e}_{q}$ $\displaystyle=$ $\displaystyle
M_{n,p}^{k-1}\tilde{e}_{q-1}=M_{n,p}^{k-j}\tilde{e}_{q-j},\;\;1\leq
j\leq\text{min}(k,q-1),\forall q,\;2\leq q\leq p+1.$

So if $k\geq q$, then $M_{n,p}^{k}\tilde{e}_{q}=0$, and if $k\leq q-1\leq p$,
then $M_{n,p}^{k}\tilde{e}_{q}=\tilde{e}_{q-k}$. As one can see, we have
$\tilde{e}_{1}^{*}\circ e_{j}=\tilde{e}_{1}^{*}\circ
M_{p}^{j-1}=\tilde{e}_{j}^{*}$, $j=1,\dots,p+1$ and $\tilde{e}_{1}^{*}\circ
e_{j}=\tilde{e}_{1}^{*}\circ E_{1,j}=\tilde{e}_{j}^{*}$, for $j=p+2,\dots,n$.
This shows that $\tilde{e}_{1}^{*}$ has an open orbit for the contragredient
action of $\mathfrak{B}_{n,p}$ on $(\mathbb{R}^{n})^{*}$. For any
$M:=x_{1}e_{1}+\dots+x_{n}e_{n}\in\mathfrak{B}_{n,p},$ we have
$\chi_{M}(\lambda)=(x_{1}-\lambda)^{n}$. But the equality $(x_{1}-M)^{p+1}=0$
implies that $\chi_{\min,M}(\lambda)$ divides $(x_{1}-\lambda)^{p+1}$. So
$\mathfrak{B}_{n,p}$ is not the space of polynomials of a nonderogatory
matrix, unless $p=n-1$. In the basis
$(e_{j},e_{n+j}=\tilde{e}_{j},\;j=1,\dots,n)$ the Lie bracket of the
2-solvable Frobenius Lie algebra
$\mathcal{G}_{n,p}:=\mathfrak{B}_{n,p}\ltimes\mathbb{R}^{n}$, is given by the
following table, $j,k,q\in\\{1,2,\dots,n\\},$

$\displaystyle\;\;\;\;\;\;\;\;[e_{1},e_{n+j}]$ $\displaystyle=$ $\displaystyle
e_{n+j}\;,\;j=1,\dots,n,$ $\displaystyle\;[e_{p},e_{n+q}]$ $\displaystyle=$
$\displaystyle 0,\text{ if }p\geq q+1,$ $\displaystyle\;[e_{k},e_{n+q}]$
$\displaystyle=$ $\displaystyle e_{n+q-k+1}\text{ if }1\leq k\leq q\leq p+1,$
$\displaystyle\;[e_{q},e_{n+q}]$ $\displaystyle=$ $\displaystyle
e_{n+1},\text{ for }q=p+2,\dots,n.$ (2)

The form $e_{n+1}^{*}$ is a Frobenius functional on $\mathcal{G}_{n,p}$ and
$\partial e_{n+1}^{*}=-\sum\limits_{i=1}^{n}e_{j}^{*}\wedge e_{n+j}^{*}$. Note
that $\mathcal{G}_{n,p}$ and $\mathcal{G}_{n,q}$ are isomorphic if and only if
$p=q$. Indeed, although each $\mathcal{G}_{n,p}$ has a codimension $1$
nilradical $\mathcal{N}_{n,p}=\hbox{\rm span}(e_{2},\dots,e_{2n})$, the
derived ideal $[\mathcal{N}_{n,p},\mathcal{N}_{n,p}]=\hbox{\rm
span}(e_{n+1},\dots,e_{n+p})$ is $p$-dimensional and $\mathcal{N}_{n,p}$ is
$(p+1)$-nilpotent. So $\mathcal{N}_{n,p}$ and $\mathcal{N}_{n,q}$ are not
isomorphic whenever $p\neq q$ and hence, neither are the Lie algebras
$\mathcal{G}_{n,p}$ and $\mathcal{G}_{n,q}$. Thus $\mathfrak{B}_{n,p}$ and
$\mathfrak{B}_{n,q}$ are not conjugate. The family
($\mathcal{G}_{n,p}$)1≤p≤n-1, has two special cases. (1) The case
$\mathfrak{B}_{n,n-1}=\mathbb{R}[M_{0}],$ where
$M_{0}=\sum\limits_{j=1}^{n-1}E_{j,j+1}$ is nonderogatory. See Section 4 for a
full classification of 2-solvable Frobenius Lie algebras $\mathbb{R}[M]\ltimes
V$, for nonderogatory $M\in\mathfrak{gl}(V)$. To keep the same notations as in
Section 4, set $\mathfrak{D}_{0}^{n}=\mathcal{G}_{n,n-1}.$ (2) As regards the
case $p=1$, the nilradical $\mathcal{N}_{n,1}$ of $\mathcal{G}_{n,1}$ is the
$(2n-1)$-dimensional Heisenberg Lie algebra
$\mathcal{H}_{2n-1}:=$span($e_{j}:=E_{1,j},j=2,\dots,n,e_{n+k},\;k=1,\dots,n$),
with Lie brackets $[e_{j},e_{n+j}]=e_{n+1}$, $j=2,\dots,n.$ Note also that the
$(n-1)$ spaces $\mathcal{A}_{n,p}=$span($e_{2},e_{3},\dots,e_{n}$) are all
$(n-1)$-dimensional Abelian subalgebras of $\mathfrak{sl}(n,\mathbb{R})$,
which according to Theorem 1 and Proposition 3.1, are pairwise non-conjugate
MASAs of $\mathfrak{sl}(n,\mathbb{R}).$ In particular, $\mathcal{A}_{n,1}$ is
spanned by $E_{1,j},$ $j=2,\dots,n$ and so $\mathcal{A}_{n,1}^{2}=0.$

######  Example 3.2.

For $n\geq 3,$ let $\mathcal{P}_{n,p}$ be the Abelian subalgebra of
$\mathfrak{sl}(n,\mathbb{R})$ defined as
$\mathcal{P}_{n,p}=\\{\sum\limits_{j=2}^{p}m_{j}(E_{1,j}+E_{j,n})+\sum\limits_{j=p+1}^{n}m_{j}E_{1,j},\;m_{2},\dots,m_{n}\in\mathbb{R}\\}$,
$p=2,3,\dots,n-1$. On
$\mathcal{C}_{n,p}:=\mathcal{P}_{n,p}\oplus\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}$,
set $e_{1}:=\mathbb{I}_{\mathbb{R}^{n}}$ and $e_{j}:=E_{1,j}+E_{j,n}$,
$j=2,\dots,p$, $e_{p+i}=E_{1,p+i}$, $i=1,\dots,n-p$. Let $(\tilde{e}_{j})$ be
the canonical basis of $\mathbb{R}^{n}$. For any
$M=m_{1}e_{1}+\dots+m_{n}e_{n}$, we have $\chi_{M}(X)=(m_{1}-X)^{n}$,
$(m_{1}-M)^{2}=(m_{2}^{2}+\dots+m_{p}^{2})E_{1,n}$ and $(m_{1}-M)^{3}=0$. So,
up to a scaling, $\chi_{\min,M}=(m_{1}-X)^{3}$. Thus for any $n\geq 4,$ the
algebra $\mathcal{C}_{n,p}$ contains no nonderogatory matrix. We have
$\tilde{e}_{1}^{*}\circ e_{1}=\tilde{e}_{1}^{*}$, $\tilde{e}_{1}^{*}\circ
e_{j}=\tilde{e}_{1}^{*}\circ(E_{1,j}+E_{j,n})=\tilde{e}_{1}^{*}\circ
E_{1,j}=\tilde{e}_{j}^{*}$, $\tilde{e}_{1}^{*}\circ
e_{p+i}=\tilde{e}_{1}^{*}\circ E_{1,p+i}=\tilde{e}_{p+i}^{*}$,
$i=1,\dots,n-p,$ $j=2,\dots,p.$ So $\tilde{e}_{1}^{*}$ has an open orbit.
Hence $\mathcal{C}_{n,p}$ and $\mathcal{P}_{n,p}$ are MASAs of
$\mathfrak{gl}(n,\mathbb{R})$ and $\mathfrak{sl}(n,\mathbb{R})$, respectively
(Theorem 1). The Lie bracket of
$\mathfrak{h}_{n,p}:=\mathcal{C}_{n,p}\ltimes\mathbb{R}^{n}$, in the basis
$(e_{j},e_{n+j}:=\tilde{e}_{j},j=1,\dots,n),$ is

$\displaystyle\;[e_{1},e_{n+j}]$ $\displaystyle=$ $\displaystyle
e_{n+j}\;,\;j=1,\dots,n,$ $\displaystyle\;[e_{j},e_{n+j}]$ $\displaystyle=$
$\displaystyle e_{n+1},\;j=2,\dots,n,\;$ $\displaystyle\;[e_{k},e_{2n}]$
$\displaystyle=$ $\displaystyle e_{n+k}\;,\;\;k=2,\dots,p,$

so $\partial e_{n+1}^{*}=-\sum\limits_{j=1}^{n}e_{j}^{*}\wedge e_{n+j}^{*}$ is
nondegenerate, $\mathfrak{h}_{n,p}$ is a Frobenius Lie algebra. The nilradical
$\mathfrak{n}_{n,p}:=$span$(e_{2},\dots,e_{2n})$ of $\mathfrak{h}_{n,p}$ is of
codimension $1.$ It is the semidirect sum
$\mathfrak{n}_{n,p}=\mathbb{R}e_{2n}\ltimes\Big{(}\mathcal{H}_{2n-3}\oplus\mathbb{R}e_{n}\Big{)}$
of the line $\mathbb{R}e_{2n}$ and the so-called Abbena Lie algebra
$\mathcal{H}_{2n-3}\oplus\mathbb{R}e_{n}$ where the former acts on the latter
by nilpotent derivations. Here, $\mathcal{H}_{2n-3}$ is the
($2n-3$)-dimensional Heisenberg Lie algebra
$\mathcal{H}_{2n-3}=\text{span}(e_{2},\dots,e_{n-1},e_{n+1},\dots,e_{2n-1})$.
The derived ideal $[\mathfrak{n}_{n,p},\mathfrak{n}_{n,p}]$ is $p$-dimensional
and spanned by $(e_{n+1},\dots,e_{n+p}).$ So if $p\neq q$, then
$\mathfrak{n}_{n,p}$ and $\mathfrak{n}_{n,q}$ are not isomorphic, hence
$\mathcal{P}_{n,p}$ and $\mathcal{P}_{n,q}$ are not conjugate,
$\mathfrak{h}_{n,p}$ and $\mathfrak{h}_{n,q}$ are not isomorphic. Note that
$\mathfrak{n}_{n,p}$ is $3$-step nilpotent. So, for any $(n,p)$ with $n\geq 4$
and $3\leq p\leq n-1$, $\mathfrak{n}_{n,p}$ is isomorphic to none of the
nilradicals $\mathcal{N}_{n,q}$, $3\leq q\leq n-1$, of Example 3.1, as they
are all $(q+1)$-step nilpotent. When $p=2,$ we note that the linear map
$\psi:\mathcal{G}_{n,2}\to\mathfrak{h}_{n,2}$,

$\displaystyle\psi(e_{m})$ $\displaystyle=$ $\displaystyle
e_{m}^{\prime}\;,\;\;1\leq m\leq 2n\;,\;\;m\notin\\{3,n,n+3,2n\\}\;,$
$\displaystyle\psi(e_{3})$ $\displaystyle=$ $\displaystyle
e_{n}^{\prime}\;,\;\psi(e_{n})=e_{3}^{\prime}\;,\;\psi(e_{n+3})=e_{2n}^{\prime}\;,\;\psi(e_{2n})=e_{n+3}^{\prime}\;,$
(3)

is an isomorphism of Lie algebras, for any $n\geq 3,$ where
$\mathcal{G}_{n,2}$ is as in Example 3.1 and the above basis of
$\mathfrak{h}_{n,2}$ has been renamed
$(e_{1}^{\prime},\dots,e_{2n}^{\prime}).$ Hence, altogether, the Lie algebras
$\mathfrak{h}_{n,p}$ are not isomorphic to any of the Lie algebras
$\mathcal{G}_{n,q}$ of Example 3.1, unless $p=q=2.$ Thus none of the MASAs
$\mathfrak{B}_{n,q}$ of Example 3.1 is conjugate to $\mathcal{C}_{n,p},$
unless $p=q=2.$

######  Example 3.3.

Let $L_{n,n}^{\prime}$ be the following Abelian subspace of
$\mathfrak{sl}(n,\mathbb{R})$,

$\displaystyle
L_{n,n}^{\prime}=\\{M:=\sum\limits_{j=2}^{n}m_{j}(E_{1,j}+E_{n-j+1,n})\;,\;m_{j}\in\mathbb{R},j=2,\dots,n\\}.$
(4)

On $\mathfrak{B}_{n,n}^{\prime}:=\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus
L_{n,n}^{\prime}$, we set $e_{1}:=\mathbb{I}_{\mathbb{R}^{n}},$
$e_{j}:=E_{1,j}+E_{n-j+1,n},$ $j=2,\dots,n$. For any
$M=m_{1}e_{1}+\dots+m_{n}e_{n}\in\mathfrak{B},$ we have
$\chi(X)=(m_{1}-X)^{n}$. We also have $\chi_{\min,M}=(m_{1}-X)^{3}$, since
$(m_{1}-M)^{3}=0$ and $(m_{1}-M)^{2}\neq 0$. Thus, for any $n\geq 4,$
$\mathfrak{B}_{n,n}^{\prime}$ contains no nonderogatory matrix. We have
$\tilde{e}_{1}^{*}\circ e_{j}=\tilde{e}_{j}^{*}$, $j=1,\dots,n-1,$
$\tilde{e}_{1}^{*}\circ e_{n}=2\tilde{e}_{n}^{*}.$ So $\tilde{e}_{1}$ has an
open orbit. According to Theorem 1, $\mathfrak{B}_{n,n}^{\prime}$ and
$L_{n,n}^{\prime}$ are MASAs of $\mathfrak{gl}(n,\mathbb{R})$ and
$\mathfrak{sl}(n,\mathbb{R})$, respectively. In the basis
$(e_{j},e_{n+j}:=\tilde{e}_{j},j=1,\dots,n),$ the Lie bracket of
$\mathcal{G}_{n,n}^{\prime}:=\mathfrak{B}_{n,n}^{\prime}\ltimes\mathbb{R}^{n}$,
is

$\displaystyle\;[e_{1},e_{n+j}]$ $\displaystyle=$ $\displaystyle
e_{n+j}\;,\;j=1,\dots,n,$ $\displaystyle\;[e_{j},e_{n+j}]$ $\displaystyle=$
$\displaystyle
e_{n+1},\;[e_{j},e_{2n}]=e_{2n-j+1},\;j=2,\dots,n-1,\;\;[e_{n},e_{2n}]=2e_{n+1}\;.$

As one sees, $\partial e_{n+1}^{*}=-\sum\limits_{j=1}^{n-1}e_{j}^{*}\wedge
e_{n+j}^{*}-2e_{n}^{*}\wedge e_{2n}^{*}$ is nondegenerate, so
$\mathcal{G}_{n,n}^{\prime}$ is a Frobenius Lie algebra. It also has a
codimension $1$ nilradical
$\mathcal{N}_{n,n}^{\prime}=span(e_{2},\dots,e_{2n})$, which is the semidirect
sum of $\mathbb{R}e_{2n}$ and $\mathcal{H}_{2n-3}\oplus\mathbb{R}e_{n},$ where
the former acts on the latter by nilpotent derivations. However,
$[\mathcal{N}_{n,n}^{\prime},\mathcal{N}_{n,n}^{\prime}]=\hbox{\rm
span}(e_{n+1},\dots,e_{2n-1})$ is $(n-1)$-dimensional. So
$\mathcal{G}_{n,n}^{\prime}$ is not isomorphic to any of the Lie algebras
$\mathcal{G}_{n,p}$ of Example 3.1 and $\mathfrak{h}_{n,p}$ of Example 3.2.
From Theorem 1, none of the MASAs $\mathfrak{B}_{n,p}$ in Example 3.1 or
$\mathcal{C}_{n,p}$ in Example 3.2, is conjugate to
$\mathfrak{B}_{n,n}^{\prime}.$

######  Remark 3.2.

From Theorem 1, the classification of $2$-solvable Frobenius Lie algebras is
equivalent to that of $n$-dimensional MASAs of $\mathfrak{gl}(n,\mathbb{K})$
acting on $(\mathbb{K}^{n})^{*}$ with an open an orbit. However, not all
$n$-dimensional MASAs have open orbit in $(\mathbb{K}^{n})^{*}$. Indeed, for
$n\geq 3$, the algebra
$\mathfrak{B}_{n}:=\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus L_{n}$, is a
MASA of $\mathfrak{gl}(n,\mathbb{R})$, where
$L_{n}:=\Big{\\{}\sum\limits_{i=1}^{n-1}k_{i,n}E_{i,n},\;k_{i,n}\in\mathbb{R},i=1,\dots,n-1\Big{\\}}$
is a MASA of $\mathfrak{sl}(n,\mathbb{R}).$ To see that, consider
$b=\sum\limits_{p,q=1}^{n}t_{p,q}E_{p,q}\in\mathfrak{gl}(n,\mathbb{R})$, with
$t_{p,q}\in\mathbb{R}$, $p,q=1\dots,n.$ We have
$[E_{i,n},b]=\sum\limits_{l=1}^{n-1}t_{n,l}E_{i,l}+(t_{n,n}-t_{i,i})E_{i,n}-\sum\limits_{1\leq
k\leq n,\;k\neq i}t_{k,i}E_{k,n}$, for any $i=1,\dots,n-1.$ So the relation
$[b,a]=0$, for any $a\in\mathfrak{B}_{n}$, is equivalent to the following,
valid for any $i$ with $1\leq i\leq n-1$: $t_{i,i}=t_{n,n}$ and for any $k$
with $1\leq k\leq n$ and $k\neq i$, one has $t_{k,i}=0$. In other words
$b=t_{n,n}\mathbb{I}_{\mathbb{R}^{n}}+t_{1,n}E_{1,n}+\dots+t_{n-1,n}E_{n-1,n}$,
or equivalently, $b$ is an element of $\mathfrak{B}_{n}$. This simply means
that $\mathfrak{B}_{n}$ is a MASA of $\mathfrak{gl}(n,\mathbb{R})$. The orbit
$\\{\alpha\circ a,a\in\mathfrak{B}_{n}\\}$ of any
$\alpha\in(\mathbb{R}^{n})^{*}$ is at most 2-dimensional and spanned by
$\alpha$ and $\tilde{e}_{n}^{*}$. More precisely, let
$(\tilde{e}_{1},\dots,\tilde{e}_{n})$ be the canonical basis of
$\mathbb{R}^{n}$ and let
$\alpha=s_{1}\tilde{e}_{1}^{*}+\dots+s_{n}\tilde{e}_{n}^{*}\in(\mathbb{R}^{n})^{*}$
, where $s_{1},\dots,s_{n}\in\mathbb{R},$ then for any
$k_{1},k_{1,n},\dots,k_{n-1,n}\in\mathbb{R}$ and
$a=k_{1}\mathbb{I}_{\mathbb{R}^{3}}+k_{1,n}E_{1,n}+\dots+k_{n-1,n}E_{n-1,n}\in\mathfrak{B}_{n}$,
one has $\tilde{e}_{i}^{*}\circ
a=k_{1}\tilde{e}_{i}^{*}+k_{i,n}\tilde{e}_{n}^{*}$ , $i=1,\dots,n-1$ and
$\tilde{e}_{n}^{*}\circ a=k_{1}\tilde{e}_{n}^{*}$, so that

$\displaystyle\alpha\circ a$ $\displaystyle=$ $\displaystyle
k_{1}\alpha+(k_{1,n}s_{1}+k_{2,n}s_{2}+\dots+k_{n-1,n}s_{n-1})e_{n}^{*}.$

For $n=3$, $L_{3}$ coincides with $L_{2,4}$ in the list of MASA of
$\mathfrak{sl}(n,\mathbb{R})$ supplied in [26].

###  3.6 Cartan subalgebras of $\mathfrak{sl}(n,\mathbb{R})$

Recall that a Cartan subalgebra $\mathfrak{h}$ of a Lie algebra $\mathcal{G}$
is a nilpotent subalgebra which is equal to its own normalizer
$\mathcal{N}_{\mathcal{G}}(\mathfrak{h})$ in $\mathcal{G}.$ Cartan subalgebras
of semisimple Lie algebras must necessarily be Abelian, more precisely they
are MASAs which contain only semisimple elements. They have been extensively
studied by several authors amongst which E. Cartan, Harish-Chandra ([12]), B.
Kostant ([14]), M. Sugiura ([24]), etc. Further recall that a Cartan
subalgebra $\mathfrak{h}$ of a semisimple Lie algebra $\mathcal{G}$, splits
into a direct sum $\mathfrak{h}=\mathfrak{h^{+}}\oplus\mathfrak{h}^{-}$ of two
subalgebras $\mathfrak{h}^{+}$ and $\mathfrak{h}^{-}$, respectively called its
toroidal and its vector parts, such that $\mathfrak{h}^{+}$ is only made of
elements of $\mathfrak{h}$ whose adjoint operator (as a linear operator of
$\mathcal{G}$) only has purely imaginary eigenvalues and adjoint operators of
elements of $\mathfrak{h}^{-}$ have only real eigenvalues. See e.g. [24]. We
propose the following characterization of Cartan subalgebras of
$\mathfrak{sl}(n,\mathbb{R}).$

######  Theorem 3.

Let $\mathfrak{h}\subset\mathfrak{sl}(n,\mathbb{R})$ be a subalgebra. Set
$\mathfrak{B}_{\mathfrak{h}}:=\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathfrak{h}$.
The following are equivalent. (a) $\mathfrak{h}$ is a Cartan subalgebra of
$\mathfrak{sl}(n,\mathbb{R})$, with toroidal and vector parts of respective
dimensions $p$ and $q.$ (b) $\mathfrak{B}_{\mathfrak{h}}=\mathbb{R}[M]$, for
some nonderogatory $M\in\mathfrak{gl}(n,\mathbb{R})$ with distinct $2p$
complex and q real eigenvalues, where $2p+q=n.$ (c)
$\mathfrak{B}_{\mathfrak{h}}\ltimes\mathbb{R}^{n}$ is the direct sum of $p$
copies of $\mathfrak{aff}(\mathbb{C})$ and $q$ copies of
$\mathfrak{aff}(\mathbb{R})$, where $\mathfrak{aff}(\mathbb{R})$ and
$\mathfrak{aff}(\mathbb{C})$ are as in Examples 4.1.1, 4.1.2.

Theorem 3 is proved in Section 4.6. Theorem 5 gives a complementary
characterization.

##  4 Classification of 2-solvable Frobenius Lie algebras given by
nonderogatory linear maps

###  4.1 Some key examples

####  4.1.1 The Lie algebras $\mathfrak{D}_{0}^{n}$

(1) Consider the simplest nonderogatory linear map
$\psi:=\mathbb{I}_{\mathbb{R}}.$ One gets the Lie algebra
$\mathcal{G}_{\psi}=\mathfrak{aff}(\mathbb{R})$ of the group of affine motions
of $\mathbb{R}.$ In the basis $(e_{1},e_{2})$, with $e_{1}=\psi^{0}$ and
$e_{2}\in\mathbb{R}$, the Lie bracket is $[e_{1},e_{2}]=e_{2}.$ The 2-form
$\partial e_{2}^{*}=-e_{1}^{*}\wedge e_{2}^{*}$ is nondegenerate. (2) In the
canonical basis $(\tilde{e}_{j})$ of $\mathbb{R}^{n}$, let
$M_{0}\in\mathfrak{gl}(n,\mathbb{R})$ be the principal nilpotent matrix

$\displaystyle M_{0}=\sum\limits_{i=1}^{n-1}E_{i,i+1}.$ (5)

It is nonderogatory, as $\chi_{\min,M_{0}}(X)=\chi_{M_{0}}(X)=X^{n}.$ We use
the notation $e_{j}:=M_{0}^{j-1},$ $j=1,\dots,n.$ Set
$e_{n+j}:=M_{0}^{j-1}\bar{x}=\tilde{e}_{n-j+1}$, where
$\bar{x}:=\tilde{e}_{n}$. Then ($e_{n+j}$) is a basis of $\mathbb{R}^{n}$. In
the basis $(e_{1},e_{2},\dots,e_{2n})$, the Lie bracket of
$\mathcal{G}_{M_{0}}=\mathbb{R}[M_{0}]\ltimes\mathbb{R}^{n},$ is

$\displaystyle[e_{i},e_{n+j}]=e_{n+j+i-1},\leavevmode\nobreak\ i,j=1,\dots,n,$
(6)

where we use the convention $e_{2n+k}=0,$ whenever $k\geq 1.$ In particular,
$[e_{1},e_{n+j}]=e_{n+j},$ $j=1,\dots,n,$ and $[e_{j},e_{2n}]=0,$
$j=2,\dots,n.$ We write ${\mathfrak{D}}_{0}^{n}$ instead of
$\mathcal{G}_{M_{0}}.$ The following 2-form on ${\mathfrak{D}}_{0}^{n}$, is
non-degenerate,

$\displaystyle\partial e_{2n}^{*}=-\sum\limits_{j=1}^{n}e_{j}^{*}\wedge
e_{2n-j+1}^{*}=-\sum\limits_{j=1}^{n}e_{n-j+1}^{*}\wedge e_{n+j}^{*}\;.$ (7)

Note that ${\mathfrak{D}}_{0}^{n}$ has an $n$-nilpotent codimension $1$
nilradical $\mathcal{N}=$span$(e_{2},\dots,e_{2n}).$

####  4.1.2 The Lie algebra $\mathfrak{D}_{0,1}^{n}$

(A) Let $\psi\in\mathfrak{gl}(2,\mathbb{R})$ with
$\psi(\tilde{e}_{1})=\tilde{e}_{2}$, $\psi(\tilde{e}_{2})=-\tilde{e}_{1}$. One
has $\chi_{\min,\psi}(X)=\chi_{\psi}(X)=1+X^{2}$. So $\psi$ has the $2$
complex eigenvalues $i$, $-i$. Of course,
$\mathbb{R}[\psi]=\mathbb{R}\mathbb{I}_{\mathbb{R}^{2}}\oplus\mathbb{R}\psi$
is a MASA of $\mathfrak{gl}(2,\mathbb{R})$ and
$\tilde{e}_{1}^{*}\circ\psi^{0}=\tilde{e}_{1}^{*}$,
$\tilde{e}_{1}^{*}\circ\psi=-\tilde{e}_{2}^{*}$. In the basis
$(e_{1},\dots,e_{4})$ of
$\mathcal{G}_{\psi}:=\mathbb{R}[\psi]\ltimes\mathbb{R}^{2}$, with
$e_{1}=\psi^{0}$, $e_{2}=\psi,$ $e_{3}:=\tilde{e}_{1}$,
$e_{4}:=\tilde{e}_{2}$, the Lie bracket reads
$[e_{1},e_{3}]=e_{3},[e_{1},e_{4}]=e_{4},[e_{2},e_{3}]=e_{4},[e_{2},e_{4}]=-e_{3}.$
The 2-form $\partial e_{3}^{*}=-e_{1}^{*}\wedge e_{3}^{*}+e_{2}^{*}\wedge
e_{4}^{*}$ is nondegenerate on $\mathcal{G}_{\psi}.$ Note that
$\mathcal{G}_{\psi}$ is the Lie algebra
$\mathfrak{aff}(\mathbb{C})$=$\Big{\\{}N(z_{1},z_{2}):=\begin{pmatrix}z_{1}&z_{2}\\\
0&0\end{pmatrix},z_{j}\in\mathbb{C},\;j=1,2.\Big{\\}}$ of the group of affine
motions of the complex line $\mathbb{C},$ looked at as a real Lie algebra,
e.g. with the identifications $e_{1}=N(1,0)$, $e_{2}=N(i,0),$ $e_{3}=N(0,1)$,
$e_{4}=N(0,i).$ We set $\mathfrak{D}_{0,1}^{2}:=\mathfrak{aff}(\mathbb{C}).$
(B) Let
$M_{0,1}=E_{2,1}-E_{1,2}+E_{4,3}-E_{3,4}+E_{1,3}+E_{2,4}\in\mathfrak{gl}(4,\mathbb{R})$.
One has $\chi_{\min,M_{0,1}}(X)=\chi_{M_{0,1}}(X)=(X^{2}+1)^{2}$. So $i$ and
$-i$ are the only eigenvalues of $M_{0,1}$ and $\mathbb{R}[M_{0,1}]$ is a MASA
of $\mathfrak{gl}(4,\mathbb{R})$. In the basis $(e_{1},e_{2},\dots,e_{8})$ of
$\mathcal{G}_{M_{0,1}}$, with $e_{1}:=\mathbb{I}_{\mathbb{R}^{4}}$,
$e_{2}:=M_{0,1}$, $e_{3}:=M_{0,1}^{2}$, $e_{4}:=M_{0,1}^{3}$,
$e_{4+j}=\tilde{e}_{j}$ , $j=1,\dots,4$, the Lie bracket reads

$\displaystyle[e_{1},e_{l}]=e_{l}\;,\;\;l=5,6,7,8,$
$\displaystyle[e_{2},e_{5}]=e_{6}\;,\;\;\;\;[e_{2},e_{6}]=-e_{5}\;,\;[e_{2},e_{7}]=e_{5}+e_{8}\;,\;\;\;\;\;[e_{2},e_{8}]=e_{6}-e_{7}\;,$
$\displaystyle[e_{3},e_{5}]=-e_{5}\;,\;[e_{3},e_{6}]=-e_{6}\;,\;[e_{3},e_{7}]=2e_{6}-e_{7}\;,\;\;\;\;[e_{3},e_{8}]=-2e_{5}-e_{8}\;,$
$\displaystyle[e_{4},e_{5}]=-e_{6}\;,\;[e_{4},e_{6}]=e_{5}\;,\;\;\;\;[e_{4},e_{7}]=-3e_{5}-e_{8}\;,\;[e_{4},e_{8}]=-3e_{6}+e_{7}\;.$

Note that both
$(e_{7},M_{0,1}e_{7}=e_{5}+e_{8},M_{0,1}^{2}e_{7}=2e_{6}-e_{7},M_{0,1}^{3}e_{7}=-3e_{5}-e_{8})$
and

$(e_{8},M_{0,1}e_{8}=e_{6}-e_{7},\;M_{0,1}^{2}e_{8}=-2e_{5}-e_{8},M_{0,1}^{3}e_{8}=-3e_{5}+e_{7})$
are bases of $\mathbb{R}^{4}.$ The 2-form $\partial e_{5}^{*}=-e_{1}^{*}\wedge
e_{5}^{*}+e_{2}^{*}\wedge(e_{6}^{*}-e_{7}^{*})+e_{3}^{*}\wedge(e_{5}^{*}+2e_{8}^{*})-e_{4}^{*}\wedge(e_{6}^{*}-3e_{7}^{*})$
is nondegenerate. (C) This generalizes (see Section 4.4.3) to a
$2n$-dimensional 2-solvable Frobenius Lie algebra
$\mathfrak{D}_{0,1}^{n}:=\mathcal{G}_{M_{0,1}}=\mathbb{R}[M_{0,1}]\ltimes\mathbb{R}^{n},$
for any even $n\geq 4$, where the nonderogatory
$M_{0,1}\in\mathfrak{gl}(n,\mathbb{R})$ is defined, in the canonical basis
$(\tilde{e}_{1},\dots,\tilde{e}_{n})$ of $\mathbb{R}^{n}$, as

$\displaystyle M_{0,1}=M_{s}+M_{n},\text{ where
}M_{s}=-\sum\limits_{j=0}^{\frac{n}{2}-1}(E_{2j+1,2j+2}-E_{2j+2,2j+1}),\;M_{n}=\sum\limits_{j=1}^{n-2}E_{j,j+2}.$
(8)

Hence, $\chi_{\min,M_{0,1}}(X)=\chi_{M_{0,1}}(X)=(X^{2}+1)^{\frac{n}{2}}$, so
$i$ and $-i$ are the only (${\frac{n}{2}}$ times repeated complex conjugate)
eigenvalues of $M_{0,1}$ and $\mathbb{R}[M_{0,1}]$ is a MASA of
$\mathfrak{gl}(n,\mathbb{R})$. In the basis $(e_{1},e_{2},\dots,e_{2n})$ of
$\mathfrak{D}_{0,1}^{n}$, with $e_{j}:=(M_{0,1})^{j-1}$,
$e_{n+j}=\tilde{e}_{j}$, $j=1,\dots,n$, the Lie bracket reads, for any
$j,l=1,\dots,n,$

$\displaystyle[e_{j},e_{n+l}]=(M_{0,1})^{j-1}\tilde{e}_{l}\;.$ (9)

###  4.2 The classification Theorem

In Theorem 4, we completely classify all $2$-solvable Frobenius Lie algebras
of the form $\mathcal{G}_{\phi}:=\mathbb{R}[\phi]\ltimes V$, where
$\phi\in\mathfrak{gl}(V)$ is nonderogatory and $V$ a real vector space of
dimension $n$. In particular, we show that the Lie algebras
$\mathfrak{D}_{0}^{p}$, $\mathfrak{D}_{0,1}^{2p}$, where $p\geq 1$ is an
integer, discussed in Section 4.1, are the building blocks that make up, in a
trivial way (direct sums), the Lie algebras $\mathcal{G}_{\phi}$. As in
Examples 4.1.1, 4.1.2, the notations
$\mathfrak{D}_{0}^{1}:=\mathfrak{aff}(\mathbb{R})$ and
$\mathfrak{D}_{0,1}^{2}:=\mathfrak{aff}(\mathbb{C})$, are implicitly adopted.
It is obvious that, if $z,\bar{z}$ are two complex conjugate eigenvalues of
$\phi$, then $(X-Re(z))^{2}+Im(z)^{2}$ divides $\chi_{\phi}(X)$ (Lemma 4.1).

######  Definition 4.1.

We say that $2$ complex conjugate eigenvalues $z,\bar{z}$ of
$\phi\in\mathfrak{gl}(V)$ are of multiplicity $m$, if $m$ is the largest
integer such that $\Big{(}(X-Re(z))^{2}+Im(z)^{2}\Big{)}^{m}$ is a factor of
$\chi_{\phi}(X)$.

We call the following, the Factorization Lemma.

######  Lemma 4.1 (Factorization Lemma).

Let $V$ be a real vector space, $\dim V=n$. Suppose a nonderogatory
$\phi\in\mathfrak{gl}(V)$ has $p$ distinct real eigenvalues $\lambda_{j},$
with multiplicity $k_{j}$, $j=1,\dots,p$ and $2q$ complex eigenvalues
$\lambda_{p+l},\bar{\lambda}_{p+l}$, with multiplicity $k_{p+{l}}$,
$l=1,\dots,q,$ where the equality
$n=k_{1}+\dots+k_{p}+2(k_{p+1}+\dots+k_{p+q})$ holds. Then the characteristic
polynomial of $\phi$ factorizes as

$\displaystyle\chi_{\phi}(X)=\displaystyle\prod_{j=1}^{p}(X-\lambda_{j})^{k_{j}}\prod_{l=1}^{q}\Big{(}(X-Re(\lambda_{p+l}))^{2}+Im(\lambda_{p+l})^{2}\Big{)}^{k_{p+l}}.$
(10)

In particular, if $\phi$ has only $2$ eigenvalues which are both complex
$\lambda,$ $\bar{\lambda}$, then

$\displaystyle\chi_{\phi}(X)=\Big{(}(Re(\lambda)-X)^{2}+Im(\lambda)^{2}\Big{)}^{\frac{n}{2}}.$
(11)

###### Proof.

Suppose a nonderogatory $\phi\in\mathfrak{gl}(V)$ has only $2$ eigenvalues
which are both complex $\lambda,$ $\bar{\lambda}$. Since
$(X-\lambda)(X-\bar{\lambda})=(X-Re(\lambda))^{2}+Im(\lambda)^{2}$ divides
$\chi_{\phi}(X),$ we therefore write
$\chi_{\phi}(X)=\Big{(}(X-Re(\lambda))^{2}+Im(\lambda)^{2}\Big{)}P_{1}(X)$
where $P_{1}(X)$ is a polynomial of degree $n-2,$ with real coefficients. As
$\mathbb{C}$ is a closed field, $P_{1}(X)$ admits some complex zeros, bound to
be $\lambda,$ $\bar{\lambda}$ since they are the only zeros of
$\chi_{\phi}(X)$, by hypothesis. We now re-write $\chi_{\phi}(X)$ as
$\chi_{\phi}(X)=\Big{(}(X-Re(\lambda))^{2}+Im(\lambda)^{2}\Big{)}^{2}P_{2}(X)$
where $P_{2}(X)$ is a polynomial of degree $n-4$, with real coefficients. The
result follows by inductively reapplying the same process to $P_{2}$. The
proof of the general case immediately follows by applying the Primary
Decomposition Theorem to $\chi_{\phi}(X)$ to reduce the problem to the cases
where $\phi$ admits a unique eigenvalue or only two eigenvalues which are both
complex and conjugate, as above. ∎

Note that Lemma 4.1 is still valid even if $\phi$ is not nonderogatory. In
Theorem 4 below, $V$ is a real vector space of dimension $n$.

######  Theorem 4.

Let $\phi\in\mathfrak{gl}(V)$ be nonderogatory with $p$ real distinct
eigenvalues $\lambda_{j}$ with multiplicities $k_{j},$ $j=1,\dots,p$ and $2q$
distinct complex eigenvalues $z_{l},\bar{z}_{l},$ with multiplicities $m_{l},$
$l=1,\dots,q$ where $n=k_{1}+\dots+k_{p}+2(m_{1}+\dots+m_{q})$.

Then the Lie algebra $\mathcal{G}_{\phi}:=\mathbb{R}[\phi]\ltimes V$ is
isomorphic to the direct sum of the Lie algebras ${\mathfrak{D}}_{0}^{k_{1}},$
$\dots,$ ${\mathfrak{D}}_{0}^{k_{p}}$, ${\mathfrak{D}}_{0,1}^{2m_{1}},$
$\dots,$ ${\mathfrak{D}}_{0,1}^{2m_{q}}.$ In particular,

  * •

(a) if $p=n,$ then $\mathcal{G}_{\phi}$ is isomorphic to the direct sum of $n$
copies of $\mathfrak{aff}(\mathbb{R})$,

  * •

(b) if $p=1$ and $q=0,$ then $\mathcal{G}_{\phi}$ is isomorphic to
${\mathfrak{D}}_{0}^{n}$,

  * •

(c) if $2q=n,$ then $\mathcal{G}_{\phi}$ is isomorphic to the direct sum of
$\frac{n}{2}$ copies of $\mathfrak{aff}(\mathbb{C})$,

  * •

(d) if $p=0$ and $q=1,$ then $\mathcal{G}_{\phi}$ is isomorphic to
${\mathfrak{D}}_{0,1}^{n}$.

The following is a direct corollary of Theorem 4.

######  Corollary 4.1.

Let $\phi\in\mathfrak{gl}(V)$ be nonderogatory. The Lie algebra
$\mathcal{G}_{\phi}$ is indecomposable if and only if one of the following
holds true:

(a) $\phi$ has a unique eigenvalue, in which case $\mathcal{G}_{\phi}$ is
completely solvable, or

(b) $\phi$ has only 2 eigenvalues which are both complex.

The rest of this section and Sections 4.3, 4.4 are mainly concerned with
discussions and the proof of Theorem 4. Lemma 4.2 allows us to split the proof
into two main cases discussed in Propositions 4.1, 4.3, 4.4, 4.5. Part (a) of
Theorem 4 is obtained by taking, in the general case, $p=n$, $q=0$, with the
identification $\mathfrak{D}_{0}^{1}=\mathfrak{aff}(\mathbb{R})$. A direct
proof (for nonzero eigenvalues) is also presented in Proposition 4.2. Part (b)
can be directly found in Lemma 4.4, whereas parts (c) and (d) are directly
proved in Propositions 4.3 and 4.4, respectively.

######  Lemma 4.2.

Suppose $V=\mathcal{E}_{1}\oplus\mathcal{E}_{2}$ and
$\phi\mathcal{E}_{j}\subset\mathcal{E}_{j},$ $j=1,2,$ where
$\phi\in\mathfrak{gl}(V)$ is nonderogatory. Let
$\mathcal{G}_{\phi_{j}}:=\mathbb{R}[\phi_{j}]\ltimes\mathcal{E}_{j}$, where
$\phi_{j}$ is the restriction of $\phi$ to $\mathcal{E}_{j}$. Then both
$\mathcal{G}_{\phi_{1}}$, $\mathcal{G}_{\phi_{2}}$ are ideals of
$\mathcal{G}_{\phi}$ and, in fact,
$\mathcal{G}_{\phi}=\mathcal{G}_{\phi_{1}}\oplus\mathcal{G}_{\phi_{2}}$.

###### Proof.

Let us extend $\phi_{j}$ to the linear map $\tilde{\phi}_{j}$ on $V$ such that
$\tilde{\phi}_{j}(\mathcal{E}_{p})=0$ if $p\neq j.$ So
$[\tilde{\phi}_{1},\phi]=[\tilde{\phi}_{2},\phi]=0$ and hence
$\tilde{\phi}_{1},$ $\tilde{\phi}_{2}\in\mathbb{R}[\phi]$. More precisely
$\mathbb{R}[\tilde{\phi}_{j}]$ is a subalgebra of $\mathbb{R}[\phi]$. We thus
see that $\mathcal{G}_{\phi_{j}}$, identified with
$\mathcal{G}_{{\tilde{\phi}}_{j}}=\mathbb{R}[\tilde{\phi}_{j}]\ltimes\mathcal{E}_{j}$,
is a Lie subalgebra of $\mathcal{G}_{\phi}:=\mathbb{R}[\phi]\ltimes V.$ As a
matter of fact, each $\mathcal{G}_{\phi_{j}}$ is an ideal of
$\mathcal{G}_{\phi}$. This directly follows from the combination of the
following properties $\phi=\tilde{\phi}_{1}+\tilde{\phi}_{2},$
$\mathbb{R}^{n}=\mathcal{E}_{1}\oplus\mathcal{E}_{2},$
$\phi\mathcal{E}_{j}\subset\mathcal{E}_{j},$ $j=1,2$ and
$\tilde{\phi}_{j}(\mathcal{E}_{p})=0$ if $p\neq j.$ Thus, as the ideals
$\mathcal{G}_{\phi_{j}}$ form a direct sum (they only meet at $\\{0\\}$,
unless they are identical), we get
$\mathcal{G}:=\mathcal{G}_{{\tilde{\phi}}_{1}}\oplus\mathcal{G}_{{\tilde{\phi}}_{2}}$.
∎

Consider the general case where $\phi\in\mathfrak{gl}(V)$ admits $p$ real and
$2q$ complex eigenvalues. We write $\chi_{\phi}(X)$ as the product
$\chi_{\phi}(X)=Q_{1}(X)Q_{2}(X)$ where $Q_{1}(X)$ has only complex (nonreal)
zeros, whereas the zeros of $Q_{2}(X)$ are all real. Of course as $Q_{1}(X)$
and $Q_{2}(X)$ must be relatively prime, the Primary Decomposition Theorem
combined with Cayley-Hamilton theorem imply the following

$\displaystyle
V=\ker(\chi_{\phi}(\phi))=\ker(Q_{1}(\phi))\oplus\ker(Q_{2}(\phi)).$ (12)

As $\ker Q_{1}(\phi)$ and $\ker Q_{2}(\phi)$ are both stable by $\phi$, Lemma
4.2 reduces the proof of Theorem 4 to two cases: the case where all the
eigenvalues of $\phi$ are real and the case where all the eigenvalues of
$\phi$ are complex.

###  4.3 Nonderogatory $\phi\in\mathfrak{gl}(V)$ with only real eigenvalues

If a nonderogatory $\phi\in\mathfrak{gl}(V)$ has only real eigenvalues
$\lambda_{j}$ of respective multiplicity $k_{j}$, $j=1,\dots,p$, then its
restriction $\phi_{j}$ to each subspace
$\mathcal{E}_{j}:=\ker(\phi-\lambda_{j})^{k_{j}}$ is again nonderogatory with
a unique eigenvalue $\lambda_{j}$ of multiplicity $k_{j}$ and $V$ splits as
$V=\mathcal{E}_{1}\oplus\cdots\oplus\mathcal{E}_{p}.$

######  Definition 4.2 (Notation).

When a nonderogatory $\phi\in\mathfrak{gl}(V)$ has a unique real eigenvalue
$\lambda$ of multiplicity $n$, we set
$\mathfrak{D}_{\lambda}^{n}:=\mathcal{G}_{\phi}=\mathbb{R}[\phi]\ltimes V.$

######  Lemma 4.3.

Suppose a nonderogatory $\phi\in\mathfrak{gl}(V)$ has $p$ distinct eigenvalues
$\lambda_{1},\dots,\lambda_{p},$ all of which are real and of respective
multiplicity $k_{1},\dots,k_{p}$, where $k_{1}+\cdots+k_{p}=n.$ Then
$\mathcal{G}_{\phi}$ is isomorphic to the direct sum
$\mathfrak{D}_{\lambda_{1}}^{k_{1}}\oplus\cdots\oplus\mathfrak{D}_{\lambda_{p}}^{k_{p}}$.

###### Proof.

Applying Lemma 4.2 to Equality (19), where
$\ker\chi_{j}(\phi)=\ker(\phi-\lambda_{j}\mathbb{I}_{V})^{k_{j}}$
$=:\mathcal{E}_{j},$ $j=1,\dots,p,$ leads to
$\mathcal{G}_{\phi}=\mathcal{G}_{\phi_{1}}\oplus\dots\oplus\mathcal{G}_{\phi_{p}}$,
where $\phi_{j}$ is the restriction of $\phi$ to $\mathcal{E}_{j}.$ Each
$\phi_{j}$ is a nonderogatory element of $\mathfrak{gl}(\mathcal{E}_{j})$,
with a unique eigenvalue $\lambda_{j}$ of multiplicity $k_{j}$, so
$\mathcal{G}_{\phi_{j}}=\mathfrak{D}_{\lambda_{j}}^{k_{j}},$ by Definition
4.2. Thus
$\mathcal{G}_{\phi}=\mathfrak{D}_{\lambda_{1}}^{k_{1}}\oplus\dots\oplus\mathfrak{D}_{\lambda_{p}}^{k_{p}}$.
∎

######  Lemma 4.4.

If a nonderogatory $\phi\in\mathfrak{gl}(V)$ has a unique real eigenvalue
$\lambda$, then the Lie algebra
$\mathcal{G}_{\phi}=:\mathfrak{D}_{\lambda}^{n}$ is isomorphic to
${\mathfrak{D}}_{0}^{n}$ as in Example 4.1.1.

###### Proof.

Following Theorem 7, consider a basis in which the matrix of $\phi$ has the
form $M_{\lambda}=\lambda\mathbb{I}_{V}+M_{0},$ with $M_{0}$ as in (5). Both
$M_{0}$ and $M_{\lambda}$ being nonderogatory and $M_{\lambda}$ being a
polynomial in $M_{0},$ imply the following: cent$(M_{0},\mathfrak{gl}(V))=$
cent$(M_{\lambda},\mathfrak{gl}(V))=$ $\mathbb{K}[M_{0}]$. Thus we have
${\mathfrak{D}}_{0}^{n}=\mathcal{G}_{M_{0}}=\mathcal{G}_{M_{\lambda}}=\mathfrak{D}_{\lambda}^{n}.$
∎

Lemmas 4.3 and 4.4 prove the following.

######  Proposition 4.1.

Suppose a nonderogatory $\phi\in\mathfrak{gl}(V)$ has $p$ distinct eigenvalues
$\lambda_{1},\dots,\lambda_{p},$ all of which are real and of respective
multiplicity $k_{1},\dots,k_{p}$, where $k_{1}+\cdots+k_{p}=n.$ Then
$\mathcal{G}_{\phi}$ is isomorphic to the direct sum
$\mathfrak{D}_{0}^{k_{1}}\oplus\cdots\oplus\mathfrak{D}_{0}^{k_{p}}$.

###### Proof.

Indeed, if a nonderogatory $\phi\in\mathfrak{gl}(V)$ has $p$ real eigenvalues
$\lambda_{1},\dots,\lambda_{p},$ of respective multiplicity
$k_{1},\dots,k_{p}$, where $k_{1}+\cdots+k_{p}=n,$ then Lemma 4.3 ensures that
$\mathcal{G}_{\phi}$ is isomorphic to the direct sum
$\mathfrak{D}_{\lambda_{1}}^{k_{1}}\oplus\cdots\oplus\mathfrak{D}_{\lambda_{p}}^{k_{p}}$
and Lemma 4.4 further proves that each $\mathfrak{D}_{\lambda_{j}}^{k_{j}}$ is
isomorphic to $\mathfrak{D}_{0}^{k_{j}}$, for $j=1,\dots,p.$ ∎

This concludes the proof of Theorem 4 in the case of nonderogatory
$\phi\in\mathfrak{gl}(V)$ with only real eigenvalues. Proposition 4.2 supplies
another rather direct proof (valid only when all the eigenvalues are nonzero)
of Theorem 4 (a).

######  Proposition 4.2.

Let $\mathcal{G}$ be the direct sum of $n$ copies of the Lie algebra
$\mathfrak{aff}(\mathbb{R})$ and $\lambda_{1},\dots,\lambda_{n},$ $n$ distinct
nonzero real numbers. Suppose $\phi\in\mathfrak{gl}(\mathbb{R}^{n})$ is
nonderogatory with eigenvalues $\lambda_{1},\dots,\lambda_{n}.$ Then
$\mathcal{G}$ and $\mathcal{G}_{\phi}$ are isomorphic.

###### Proof.

Choose a basis $(a_{i},b_{i})$ of the ith copy of
$\mathfrak{aff}(\mathbb{R})$, so that $[a_{i},b_{i}]=b_{i}$ and set
$\mathfrak{B}=\hbox{\rm span}(a_{1},\dots,a_{n})$. Let
$V:=[\mathcal{G},\mathcal{G}]=\hbox{\rm span}(b_{1},\dots,b_{n})$. For
$a\in\mathfrak{B}$, denote by $\rho(a)$ the restriction to $V$ of the adjoint
operator $[a,\cdot]$ of $a.$ So one has
$\mathcal{G}=\mathfrak{B}\ltimes_{\rho}V.$ Further choose
$a_{0}=\sum\limits_{i=1}^{n}\lambda_{i}a_{i}$, with $\lambda_{i}$ as in the
hypotheses, so that the $\lambda_{i}$’s are the eigenvalues of $\rho(a_{0}).$
By hypothesis, the characteristic and minimal polynomials of $\rho(a_{0})$
coincide. Hence $\mathbb{R}[\rho(a_{0})]$ and $\mathbb{R}[\phi]$ are conjugate
(with the identification $V=\mathbb{R}^{n}$). Consider the linear map
$\psi:\mathcal{G}\to\mathbb{R}[\rho(a_{0})]\ltimes V=\mathcal{G}_{\phi}$
defined by $\psi(b)=b$, for every $b\in V$ and
$\psi(a)=\sum\limits_{s=1}^{n}v_{s}(\rho(a_{0}))^{s}$ for any
$a=\sum\limits_{i=1}^{n}k_{i}a_{i}$, where $v=(v_{1},\dots,v_{n})$ is the
solution of the equation $Nv^{T}=K^{T}$, with $K=(k_{1},\dots,k_{n})$ and $N$
is the $n\times n$ matrix with coefficients $N_{ij}=\lambda_{i}^{j}$. One sees
that, for any $b=\sum\limits_{j}^{n}t_{j}b_{j},$ $t_{j}\in\mathbb{R},$
$\psi[a,b]=[a,b]=\sum\limits_{j=1}^{n}t_{j}k_{j}b_{j}=\sum\limits_{j=1}^{n}t_{j}\sum\limits_{s=1}^{n}N_{js}v_{s}b_{j}=\sum\limits_{j=1}^{n}t_{j}\sum\limits_{s=1}^{n}\lambda_{j}^{s}v_{s}b_{j}$
and
$[\psi(a),\psi(b)]=\sum\limits_{j=1}^{n}t_{j}\sum\limits_{s=1}^{n}v_{s}(\rho(a_{0}))^{s}b_{j}=\sum\limits_{j=1}^{n}t_{j}\sum\limits_{s=1}^{n}v_{s}\lambda_{j}^{s}b_{j}=\psi[a,b]$
and
$[\psi(a),\psi(a^{\prime})]=\psi[a,a^{\prime}]=[\psi(b),\psi(b^{\prime})]=\psi[b,b^{\prime}]=0,$
for any $a,a^{\prime}\in\mathfrak{B}$, $b,b^{\prime}\in V$. We further have

$\det(N)=\left(\prod\limits_{i=1}^{n}\lambda_{i}\right)\left(\prod\limits_{1\leq
i<j\leq n}(\lambda_{j}-\lambda_{i})\right)\neq 0.$

Thus $\psi$ is a Lie algebra isomorphism. ∎

######  Example 4.1.

Let $\mathcal{G}$ be the $6$-dimensional rank-three Kähler-Einstein solvable
Lie algebra in [11], with Lie bracket $[e_{4},e_{1}]=se_{1}$,
$[e_{4},e_{2}]=se_{2}$, $[e_{4},e_{3}]=se_{3}$,
$[e_{5},e_{1}]=-s\frac{\sqrt{6}}{2}e_{1}$,
$[e_{5},e_{2}]=s\frac{\sqrt{6}}{2}e_{2}$,
$[e_{6},e_{1}]=\frac{\sqrt{2}}{2}e_{1}$,
$[e_{6},e_{2}]=s\frac{\sqrt{2}}{2}e_{2}$, $[e_{6},e_{3}]=-s\sqrt{2}e_{3},$ in
a basis $(e_{1},\dots,e_{6}),$ with $s\in\mathbb{R}.$ Consider
$\mathcal{B}=\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})$
and a basis $(a_{1},b_{1},a_{2},b_{2},a_{3},b_{3})$ in which its Lie bracket
reads $[a_{j},b_{j}]=b_{j},$ $j=1,2,3$. The linear map
$\psi:\mathcal{G}\to\mathcal{B}$, $\psi(e_{1})=b_{1}$, $\psi(e_{2})=b_{2}$,
$\psi(e_{3})=b_{3}$, $\psi(e_{4})=s(a_{1}+a_{2}+a_{3})$,
$\psi(e_{5})=s\frac{\sqrt{6}}{2}(a_{2}-a_{1})$,
$\psi(e_{6})=\frac{\sqrt{2}}{2}(a_{1}+sa_{2}-2sa_{3})$, is a homomorphism and
$\det(\psi)=-\frac{\sqrt{3}}{2}s^{2}(5s+1)$. So, $\psi$ is an isomorphism
between the Lie algebras $\mathcal{G}$ and $\mathcal{B}$, except when
$s\in\\{0,-\frac{1}{5}\\}$. Thus, any $\phi\in\mathfrak{gl}(3,\mathbb{R})$
with $3$ distinct real eigenvalues gives rise to
$\mathcal{G}_{\phi}=\mathcal{G},$ when $s\notin\\{0,-\frac{1}{5}\\}$. Note
that $\mathcal{G}$ is Frobenius if and only if $s\notin\\{0,-\frac{1}{5}\\}$.

###  4.4 Nonderogatory $\phi\in\mathfrak{gl}(V)$ all of whose eigenvalues are
complex

####  4.4.1 Nonderogatory $\phi\in\mathfrak{gl}(V)$ diagonalizable in
$\mathbb{C}$

Let $\phi\in\mathfrak{gl}(V)$ be nonderogatory. Suppose all the eigenvalues
$\lambda_{1},\bar{\lambda}_{1},\dots,\lambda_{\frac{n}{2}},\bar{\lambda}_{\frac{n}{2}}$
of $\phi$ are complex (nonreal) and distinct, in which case $n$ is even and
$\phi$ is diagonalizable in $\mathbb{C}$. From Theorem 7 (B), there is a basis
of $V$ in which the matrix $\phi$ is of the form
diag($J_{1},\dots,J_{\frac{n}{2}}$), where each block $J_{j}$ is of the form
$\begin{pmatrix}Re(\lambda_{j})&-Im(\lambda_{j})\\\
Im(\lambda_{j})&Re(\lambda_{j})\end{pmatrix},$ $Im(\lambda_{j})\neq 0.$ Note
that each $J_{j}$ is nonderogatory with characteristic polynomial
$\chi_{J_{j}}(X)=(X-Re(\lambda_{j}))^{2}+Im(\lambda_{j})^{2}.$ Let $v_{j}$ be
an eigenvector of $\phi$ with corresponding eigenvalue $\lambda_{j}$. Both
$Re(v_{j})$ and $Im(v_{j})$ are in $\ker\chi_{J_{j}}(\phi)=:\mathcal{E}_{j}.$
Obviously, one has $\phi\mathcal{E}_{j}\subset\mathcal{E}_{j},$
$j=1,\dots,\frac{n}{2}.$ Set $\phi_{j}:=\phi_{|_{\mathcal{E}_{j}}}.$ Of
course,
$\mathbb{R}[\phi_{j}]=\mathbb{R}\mathbb{I}_{\mathcal{E}_{j}}\oplus\mathbb{R}\phi_{j}$.
In the basis $e_{1}=\mathbb{I}_{\mathcal{E}_{j}},$ $e_{2}=\phi_{j},$
$e_{3}=Re(v_{j}),$ $e_{4}=Im(v_{j})$, the Lie bracket of
$\mathcal{G}_{{\phi}_{j}}:=\mathbb{R}[\phi_{j}]\ltimes\mathcal{E}_{j}$, is
$[e_{1},e_{3}]=e_{3},$ $[e_{1},e_{4}]=e_{4},$
$[e_{2},e_{3}]=Re(\lambda_{j})e_{3}+Im(\lambda_{j})e_{4}$,
$[e_{2},e_{4}]=-Im(\lambda_{j})e_{3}+Re(\lambda_{j})e_{4}.$ Lemma 4.2 applied
to the direct sum
$V=\mathcal{E}_{1}\oplus\dots\oplus\mathcal{E}_{{\frac{n}{2}}}$, leads to
$\mathcal{G}_{\phi}=\mathcal{G}_{{\phi}_{1}}\oplus\cdots\oplus\mathcal{G}_{{\phi}_{\frac{n}{2}}}.$
We identify $\mathcal{E}_{j}$ with $\mathbb{R}^{2}$ and show below that each
Lie ideal $\mathcal{G}_{{\phi}_{j}}$ is isomorphic to
$\mathfrak{aff}(\mathbb{C}).$

######  Lemma 4.5.

Let $\phi\in\mathfrak{gl}(2,\mathbb{R})$ with complex eigenvalues
$\lambda,\;\bar{\lambda},$ where $\lambda=r-is$ and $r,\;s\in\mathbb{R},$ with
$s\neq 0$. Then $\mathcal{G}_{\phi}$ is isomorphic to
$\mathfrak{aff}(\mathbb{C}),$ as in Example 4.1.2.

###### Proof.

Without loss of generality, set
$\phi(\tilde{e}_{1})=r\tilde{e}_{1}+s\tilde{e}_{2}$,
$\phi(\tilde{e}_{2})=-s\tilde{e}_{1}+r\tilde{e}_{2}$, in the canonical basis
$(\tilde{e}_{1},\tilde{e}_{2})$ of $\mathbb{R}^{2}.$ In the basis
$e_{1}^{\prime}=\mathbb{I}_{\mathbb{R}^{2}},\;e_{2}^{\prime}=\phi$,
$e_{3}^{\prime}=\tilde{e}_{1},\;e_{4}^{\prime}=\tilde{e}_{2}$, the Lie bracket
of $\mathcal{G}_{\phi}$ reads
$[e_{1}^{\prime},e_{3}^{\prime}]=e_{3}^{\prime}$,
$[e_{1}^{\prime},e_{4}^{\prime}]=e_{4}^{\prime}$,
$[e_{2}^{\prime},e_{3}^{\prime}]=re_{3}^{\prime}+se_{4}^{\prime}$,
$[e_{2}^{\prime},e_{4}^{\prime}]=-se_{3}^{\prime}+re_{4}^{\prime}$. In the new
basis $X_{1}:=e_{1}^{\prime}$,
$X_{2}:=-\frac{r}{s}e_{1}^{\prime}+\frac{1}{s}e_{2}^{\prime}$,
$X_{3}:=pe_{3}^{\prime}-qe_{4}^{\prime}$,
$X_{4}:=qe_{3}^{\prime}+pe_{4}^{\prime}$, with $p^{2}+q^{2}\neq 0$, we now
have $[X_{1},X_{3}]=X_{3}$, $[X_{1},X_{4}]=X_{4}$, $[X_{2},X_{3}]=X_{4}$,
$[X_{2},X_{4}]=-X_{3}$, which is the Lie bracket of
$\mathfrak{aff}(\mathbb{C}).$ In other words, the invertible linear map
$\psi:\mathfrak{aff}(\mathbb{C})\to\mathcal{G}_{\phi}$, $\psi(e_{j})=X_{j}$,
$j=1,2,3,4,$ is an isomorphism between the Lie algebras
$\mathfrak{aff}(\mathbb{C})$ and $\mathcal{G}_{\phi}$. ∎

######  Proposition 4.3.

Suppose a nonderogatory $\phi\in\mathfrak{gl}(V)$ has $n$ distinct complex
eigenvalues $\lambda_{j},\bar{\lambda}_{j},$ $j=1,\dots,\frac{n}{2}.$ Then the
Lie algebra $\mathcal{G}_{\phi}:=\mathbb{R}[\phi]\ltimes V$ is isomorphic to
the direct sum
$\mathfrak{aff}(\mathbb{C})\oplus\cdots\oplus\mathfrak{aff}(\mathbb{C})$ of
$\frac{n}{2}$ copies of the Lie algebra $\mathfrak{aff}(\mathbb{C})$.

###### Proof.

Suppose a nonderogatory $\phi\in\mathfrak{gl}(V)$ has $n$ distinct complex
(nonreal) eigenvalues $\lambda_{j},\bar{\lambda}_{j},$
$j=1,\dots,\frac{n}{2}.$ Then $\mathcal{G}_{\phi}$ is isomorphic to the direct
sum of the ideals
$\mathcal{G}_{{\phi}_{j}}:=\mathbb{R}[\phi_{j}]\ltimes\mathcal{E}_{j}$,
$j=1,\dots,\frac{n}{2}$, where
$\ker\chi_{\phi_{j}}(\phi_{j})=:\mathcal{E}_{j},$ as above. From Lemma 4.5,
each $\mathcal{G}_{{\phi}_{j}}$ is isomorphic to $\mathfrak{aff}(\mathbb{C})$.
∎

####  4.4.2 Example: the circular permutation of the vectors of a basis

Here is a typical example of a nonderogatory
$\phi\in\mathfrak{gl}(n,\mathbb{R})$ with $n$ real and complex eigenvalues,
hence diagonalizable in $\mathbb{C}$. It is given by the circular permutation
of the canonical basis $\phi(\tilde{e}_{i})=\tilde{e}_{i+1}$, for
$i=1,\dots,n-1$ and $\phi(\tilde{e}_{n})=\tilde{e}_{1}$. Any vector
$\tilde{e}_{i}$ of this basis is such that
$(\tilde{e}_{i},\phi(\tilde{e}_{i}),\dots,\phi^{n-1}(\tilde{e}_{i}))$ is again
a basis of $\mathbb{R}^{n}$. The map $\phi$ is nonderogatory and its matrix in
the above basis reads $[\phi]=E_{1,n}+\sum\limits_{i=1}^{n-1}E_{i+1,i}.$ Its
characteristic polynomial is $\chi_{\phi}(X)=X^{n}-1$, up to a sign. So the
eigenvalues are the complex nth roots of $1.$ They are
$z_{k}=e^{ik\frac{2\pi}{n}}$, where $k=1,2,\dots,n.$ When $n=2,$ then it reads
$[\phi]=E_{1,2}+E_{2,1}$ and has the two distinct real eigenvalues $z_{1}=-1$
and $z_{2}=1.$ So $\mathcal{G}_{\phi}$ is isomorphic to
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R}).$ For $n=3,$ the
eigenvalues are $z_{1}=1,$ $z_{2}=-e^{i\frac{\pi}{3}}$,
$z_{3}=-e^{-i\frac{\pi}{3}}.$ So $\mathcal{G}_{\phi}$ is isomorphic to
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{C})$. For $n=4,$ the
eigenvalues are $z_{1}=-1,$ $z_{2}=1,$ $z_{3}=i,$ $z_{4}=-i$, thus
$\mathcal{G}_{\phi}$ is isomorphic to
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{C}).$
When $n=5$, there are one real and $4$ complex eigenvalues and hence we get
$\mathcal{G}_{\phi}=\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{C}).$
For $n=7$, we have one real and six complex eigenvalues, namely
$1,-e^{i\frac{\pi}{7}},-e^{-i\frac{\pi}{7}},e^{i\frac{2\pi}{7}},e^{-i\frac{2\pi}{7}},-e^{i\frac{3\pi}{7}},-e^{-i\frac{3\pi}{7}}.$
This gives rise to
$\mathcal{G}_{\phi}=\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{C}).$

####  4.4.3 Nonderogatory $\phi\in\mathfrak{gl}(V)$ non-diagonalizable in
$\mathbb{C}$

As above, $V$ is a real vector space with $\dim V=n.$ In this section, we
discuss the case of nonderogatory $\phi\in\mathfrak{gl}(V)$ all of whose
eigenvalues are complex (nonreal, in which case, $n$ is even), but which are
not diagonalizable in $\mathbb{C}.$ First, consider the case where $\phi$ has
only $2$ eigenvalues, say $z=r+is$ and $r-is$, with $s\neq 0$. In the same
basis in which the matrix of $\phi$ is in the form $M_{z}$ as in (18),
consider the nonderogatory $M_{0,1}=M_{s}+M_{n}$, where $M_{s}$, $M_{n}$ are
as in (8). We have

$\displaystyle M_{s}M_{n}$ $\displaystyle=$
$\displaystyle-\sum\limits_{j=0}^{\frac{n}{2}-2}\Big{(}E_{2j+1,2j+4}-E_{2j+2,2j+3}\Big{)}=M_{n}M_{s}\;,$
(13)

and, of course $[M_{0,1},M_{s}]=[M_{n},M_{s}]=[M_{n},M_{0,1}]=0$. In
particular, $M_{s}$ and $M_{n}$ are respectively the semisimple and the
nilpotent parts of $M_{0,1}.$ So, $M_{s}$ and $M_{n}$ are both polynomials in
$M_{0,1}$. Thus, $M_{z}=r\mathbb{I}_{\mathbb{R}^{n}}+sM_{s}+M_{n}$ is also a
polynomial in $M_{0,1}$. This induces the following equalities

$\displaystyle\mathbb{R}[M_{0,1}]=\mathbb{R}[M_{z}]\text{ and
}\mathcal{G}_{M_{z}}=\mathcal{G}_{M_{0,1}}=\mathfrak{D}_{0,1}^{n}.$ (14)

We have thus proved the

######  Proposition 4.4.

If a nonderogatory $\phi\in\mathfrak{gl}(V)$ has only $2$ eigenvalues which
are complex (thus conjugate), and if $\phi$ is non-diagonalizable in
$\mathbb{C}$, then $\mathcal{G}_{\phi}=\mathfrak{D}_{0,1}^{n}.$

Finally, we deduce the following.

######  Proposition 4.5.

If all the eigenvalues
$\lambda_{1},\bar{\lambda}_{1},\dots,\lambda_{p},\bar{\lambda}_{p}$ of a
nonderogatory $\phi\in\mathfrak{gl}(V),$ are complex, then
$\mathcal{G}_{\phi}$ is isomorphic to the direct sum
$\mathfrak{D}_{0,1}^{2k_{1}}\oplus\dots\oplus\mathfrak{D}_{0,1}^{2k_{p}}$ of
copies of $\mathfrak{D}_{0,1}^{2k_{j}}$, where $k_{j}$ is the multiplicity of
the eigenvalues $\lambda_{j}$, $\bar{\lambda}_{j}$.

###### Proof.

Factorizing the characteristic polynomial of $\phi$, via Lemma 4.1, then
successively applying the Primary Decomposition Theorem, Lemma 4.2, together
with Proposition 4.4, yield the result. ∎

####  4.4.4 More on the Lie algebra $\mathfrak{D}_{0,1}^{n}$

Without loss of generality, we suppose that the form (18) is achieved in the
canonical basis $(\tilde{e}_{1},\dots,\tilde{e}_{n})$ of $\mathbb{R}^{n}.$
Note that $M_{s}$ and $M_{n}$ in (8) satisfy
$M_{s}^{2}=-\mathbb{I}_{\mathbb{R}^{n}},$ $M_{n}^{\frac{n}{2}}=0$,
$(M_{s}M_{n})^{\frac{n}{2}}=0$ and for any $j=1,\dots,\frac{n}{2}-1,$

$\displaystyle\Big{(}M_{n}\Big{)}^{j}=\sum\limits_{p=1}^{n-2j}E_{p,p+2j}\text{
and
}M_{s}\Big{(}M_{n}\Big{)}^{j}=-\sum\limits_{p=0}^{\frac{n}{2}-j-1}\Big{(}E_{2p+1,2p+2j+2}-E_{2p+2,2p+2j+1}\Big{)}.$

Each
$M_{0,1}^{p}=(M_{s}+M_{n})^{p}=\sum\limits_{j=0}^{p}\complement_{p}^{j}\;(M_{s})^{j}\;(M_{n})^{p-j},$
$p=0,1,2,\dots,n-1$, is a linear combination of $\mathbb{I}_{\mathbb{R}^{n}}$,
$M_{s}$, $(M_{n})^{j},$ $M_{s}(M_{n})^{j}$, $j=1,\dots,\frac{n}{2}-1,$ where
as above, $\complement_{p}^{j}=\frac{p!}{j!(p-j)!}$, $p\geq j$. Thus
$e_{1}:=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}:=M_{s},$ $e_{2j+1}:=(M_{n})^{j},$
$e_{2j+2}:=-M_{s}(M_{n})^{j},$ $j=1,\dots,\frac{n}{2}-1,$ is another basis of
$\mathbb{R}[M_{0,1}]$. In the basis ($e_{s},\tilde{e}_{s}$, $s=1,\dots,n$),
the Lie bracket of $\mathfrak{D}_{0,1}^{n}$ reads, for any
$q=0,1,\dots,\frac{n}{2}-1$ and $i=1,\dots,n,$

$[e_{2q+1},\tilde{e}_{i}]=\tilde{e}_{i-2q},\;\;\text{ if }2q+1\leq i\leq
n,\text{ and }[e_{2q+1},\tilde{e}_{i}]=0\text{ otherwise, }$

$[e_{2q+2},\tilde{e}_{i}]=\left\\{\begin{array}[]{cll}-\tilde{e}_{i-2q+1},&\text{
if $i$ odd and }2q+1\leq i\leq n-1,\\\ \tilde{e}_{i-2q-1},&\;\text{ if $i$
even and }2q+2\leq i\leq n,\\\ 0,&\;\text{ otherwise.}\end{array}\right.$

So the 2-form
$\partial\tilde{e}_{1}^{*}=-\sum\limits_{k=1}^{n}e_{k}^{*}\wedge\tilde{e}_{k}^{*}$
is non-degenerate on $\mathfrak{D}_{0,1}^{n}$. The codimension $2$ subspace
$\mathcal{N}:=span\Big{(}(M_{n})^{j},M_{s}(M_{n})^{j},\;j=1,\dots,\frac{n}{2}-1\Big{)}\ltimes\mathbb{R}^{n}$
is an ideal of $\mathfrak{D}_{0,1}^{n}$, as it contains the derived ideal
$\mathbb{R}^{n}=[\mathfrak{D}_{0,1}^{n},\mathfrak{D}_{0,1}^{n}]$. The
equalities $(M_{n})^{\frac{n}{2}}=(M_{s}M_{n})^{\frac{n}{2}}=0$, show that
$\mathcal{N}$ is $\frac{n}{2}$-step nilpotent, and is in fact the nilradical
of $\mathfrak{D}_{0,1}^{n}$. Indeed, if we set
$\mathfrak{F}:=\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathbb{R}M_{0,1}$,
then the vector space underlying $\mathfrak{D}_{0,1}^{n}$ splits as the direct
sum $\mathfrak{D}_{0,1}^{n}=\mathfrak{F}\oplus\mathcal{N}$, so that, any
subspace of dimension higher than $n-2$, must meet $\mathfrak{F}$ non-
trivially and hence cannot be a nilpotent subalgebra of
$\mathfrak{D}_{0,1}^{n}$. Thus $\mathcal{N}$ is the biggest nilpotent ideal of
$\mathfrak{D}_{0,1}^{n}$. Altogether, $\mathfrak{D}_{0,1}^{n}$ is an
indecomposable, non-completely solvable (the adjoint of $M_{s}$ has complex
eigenvalues) $2$-solvable Frobenius Lie algebra with a codimension 2 non-
Abelian nilradical $\mathcal{N}$. One sees that $\mathfrak{D}_{0,1}^{n}$ is
not isomorphic to $\mathfrak{D}_{0}^{n}$ which is completely solvable and has
a codimension $1$ non-Abelian nilradical (except for
$\mathfrak{D}_{0}^{1}=\mathfrak{aff}(\mathbb{R})$).

###  4.5 Derivations and automorphisms of $\mathcal{G}_{\phi}$

######  Proposition 4.6.

Let $\phi\in\mathfrak{gl}(V)$ be nonderogatory. Set

$\displaystyle\mathfrak{N}:=\\{N\in\mathfrak{gl}(V),\text{ such that
}[N,\mathbb{K}[\phi]]\subset\mathbb{K}[\phi]\\}=\mathcal{N}_{\mathfrak{gl}(V)}(\mathbb{K}[\phi]).$
(15)

Up to isomorphism, the Lie algebra $Der(\mathcal{G}_{\phi})$ of derivations of
$\mathcal{G}_{\phi}$ is given by

$\displaystyle Der(\mathcal{G}_{\phi})=\mathfrak{N}\ltimes V.$ (16)

More precisely, $D\in Der(\mathcal{G}_{\phi})$ if and only if there are
$x_{D}\in V$ and $h\in\mathfrak{N},$ such that $D$ is of the form
$D(a+x)=[h,a]+ax_{D}+h(x),$ for every $a\in\mathbb{K}[\phi]$ and $x\in V.$

###### Proof.

Let $D\in Der(\mathcal{G}_{\phi})$. As the derived ideal
$V=[\mathcal{G}_{\phi},\mathcal{G}_{\phi}]$ is preserved by $D$, we write
$D(x)=h(x)$ and $D(a)=D_{1,1}(a)+D_{1,2}(a),$ for $a\in\mathbb{K}[\phi]$,
$x\in V,$ where $D_{1,1}\in\mathfrak{gl}(\mathbb{K}[\phi])$,
$h\in\mathfrak{gl}(V)$ and $D_{1,2}:\mathbb{K}[\phi]\to V$ is linear. Now,
setting $e_{1}:=\mathbb{I}_{V}$ and $x_{D}:=D_{12}(e_{1})$, the equality
$D_{12}(a)=ax_{D},$ for any $a\in\mathbb{K}[\phi]$, follows:
$0=D([e_{1},a])$$=[D(e_{1}),a]+[e_{1},D(a)]$$=[x_{D},a]+[e_{1},D_{12}(a)]$$=D_{12}(a)-ax_{D}.$
We also have $h(ax)=D[a,x]=[D_{11}(a),x]+[a,h(x)]=D_{11}(a)x+ah(x)$, thus
entailing $[h,a]=D_{11}(a)\in\mathbb{K}[\phi],$ for any
$a\in\mathbb{K}[\phi].$ So the equality $D_{11}(a):=[h,a]$ stands as the
definition of $D_{11}(a)$ and implies $h\in\mathfrak{N}$. Thus for any $D$ in
$Der(\mathcal{G}_{\phi})$, there are $x_{D}\in V$, $h\in\mathfrak{N}$ such
that $D(a+x)=[h,a]+h(x)+ax_{D}.$ Conversely, any
$(h,\hat{x})\in\mathfrak{N}\times V$ defines a unique $D\in
Der(\mathcal{G}_{\phi})$ by the formula $D(a+x)=[h,a]+h(x)+a\hat{x}.$ We get
the invertible linear $\psi:Der(\mathcal{G}_{\phi})\to\mathfrak{N}\ltimes V$,
$D\mapsto(h,x_{D})$. As easily seen, for any $D_{j}\in
Der(\mathcal{G}_{\phi})$ given by $(h_{j},x_{D_{j}})$, $j=1,2,$ we have
$\psi([D_{1},D_{2}])=([h_{1},h_{2}],h_{1}(x_{D_{2}})-h_{2}(x_{D_{1}}))=[\psi(D_{1}),\psi(D_{2})]$.
Hence $\psi$ is an isomorphism between the Lie algebras
$Der(\mathcal{G}_{\phi})$ and $\mathfrak{N}\ltimes V.$ ∎

Note that in Proposition 4.6, the inner derivations of $\mathcal{G}_{\phi}$
are those for which the component $h$ belongs to $\mathbb{K}[\phi]$.

######  Example 4.2.

The space $Der(\mathfrak{D}_{0}^{n})$ of derivations of
$\mathfrak{D}_{0}^{n}$.

For $n=2,$ the normalizer of $\mathbb{R}[E_{1,2}]$ in
$\mathfrak{gl}(2,\mathbb{R}),$ is the 3-dimensional real algebra spanned by
$E_{1,1},E_{1,2},E_{22}$, which is $1$ dimension higher than
$\mathbb{R}[E_{1,2}]$. For example, $E_{2,2}$ is not in $\mathbb{R}[E_{1,2}]$
and will thus act as an outer derivation on $\mathfrak{D}_{0}^{2}.$
Considering elements of $\mathbb{R}^{2}$ as inner derivations, we see that
$Der(\mathfrak{D}_{0}^{2})$ is spanned by
$E_{1,1},E_{1,2},E_{22},\tilde{e}_{1},\tilde{e}_{2}.$ For $n=3,$ the
normalizer of $\mathbb{R}[E_{1,2}+E_{2,3}]$ in $\mathfrak{gl}(3,\mathbb{R}),$
is the $5$-dimensional algebra of matrices spanned by $E_{1,1}-E_{3,3},$
$E_{22}+2E_{33}$, $E_{1,2}$, $E_{2,3}$, $E_{1,3}$. For example,
$E_{1,1}-E_{3,3},$ $E_{22}+2E_{33}$, are not in $\mathbb{R}[E_{1,2}+E_{2,3}]$,
so they represent outer derivations of $\mathfrak{D}_{0}^{3}.$ Counting
elements of $\mathbb{R}^{3}$ in, as inner derivations, we get $\dim
Der(\mathfrak{D}_{0}^{3})=8.$ More generally, for a given $n\geq 4$, if we set
$M_{0}:=E_{1,2}+E_{2,3}+\dots+E_{n-1,n}$ as in (5), then the normalizer of
$\mathbb{R}[M_{0}]$ in $\mathfrak{gl}(n,\mathbb{R}),$ is of dimension $2n-1$
and is spanned by $D_{k}:=E_{1,k}-\sum\limits_{j=3}^{n-k+1}(j-2)E_{j,j+k-1}$,
$D_{k}^{\prime}:=E_{2,k+1}+\sum\limits_{j=3}^{n-k+1}(j-1)E_{j,j+k-1}$,
$k=1,\dots,n-2$ and $D_{n-1}:=E_{1,n-1}$, $D_{n-1}^{\prime}:=E_{2,n}$,
$D_{n}:=E_{1,n}.$ For example, $D_{1}$ and $D_{1}^{\prime}$ are not in
$\mathbb{R}[M_{0}]$ and will act as non-trivial outer derivations of
$\mathfrak{D}_{0}^{n}$. In particular $\mathbb{R}[M_{0}]$ is not a Cartan
subalgebra of $\mathfrak{gl}(n,\mathbb{R})$ and $\dim
Der(\mathfrak{D}_{0}^{n})=3n-1.$

######  Example 4.3.

On the space $Der(\mathfrak{D}_{0,1}^{n})$ of derivations of
$\mathfrak{D}_{0,1}^{n}$.

For $n=4$, consider the $n\times n$ matrix
$M_{0,1}=E_{2,1}-E_{1,2}+E_{4,3}-E_{3,4}+E_{1,3}+E_{2,4}$ as in Example 4.1.2.
The normalizer of $\mathbb{R}[M_{0,1}]$ in $\mathfrak{gl}(4,\mathbb{R}),$ is
of dimension $6$ and is spanned by the matrices $E_{1,1}+E_{2,2},$
$E_{3,3}+E_{4,4},$ $E_{1,2}-E_{2,1},$ $E_{3,4}-E_{4,3},$ $E_{1,3}+E_{2,4},$
$E_{2,3}-E_{1,4}$. In particular the two matrices $E_{1,1}+E_{2,2},$
$E_{3,3}+E_{4,4},$ are not elements of $\mathbb{R}[M_{0,1}]$. So they both
represent non-trivial outer derivations of $\mathfrak{D}_{0,1}^{4}$. More
generally, for any $n\geq 6$, we let again $M_{0,1}$ stand for the $n\times n$
matrix $M_{0,1}=M_{s}+M_{n}$ as in (8). The $n\times n$ matrix
$Z_{1}:=E_{1,1}+E_{2,2}-\sum\limits_{j=2}^{\frac{n}{2}-1}(j-1)(E_{2j+1,2j+1}+E_{2j+2,2j+2})$
is in the normalizer of $\mathbb{R}[M_{0,1}]$ in
$\mathfrak{gl}(n,\mathbb{R})$, but not in $\mathbb{R}[M_{0,1}]$. Indeed, for
any $s\geq 1,$ we have
$[Z_{1},\Big{(}M_{n}\Big{)}^{s}]=s\sum\limits_{j=1}^{n-2s}E_{j,j+2s}=s\Big{(}M_{n}\Big{)}^{s}$,
$[Z_{1},M_{s}]=0$ and
$[Z_{1},M_{s}\Big{(}M_{n}\Big{)}^{s}]=M_{s}[Z_{1},\Big{(}M_{n}\Big{)}^{s}]+[Z_{1},M_{s}]\Big{(}M_{n}\Big{)}^{s}=sM_{s}\Big{(}M_{n}\Big{)}^{s}.$
So the linear map $M\mapsto[Z_{1},M]$ preserves $\mathbb{R}[M_{0,1}]$ and has
a diagonal matrix diag($0,0,1,2,\dots,\frac{n}{2}-1,1,2,\dots,\frac{n}{2}-1$)
in the basis
$(\mathbb{I}_{\mathbb{R}^{n}},M_{s},M_{n},(M_{n})^{2},\dots,(M_{n})^{\frac{n}{2}-1},M_{s}M_{n},M_{s}(M_{n})^{2},\dots,M_{s}(M_{n})^{\frac{n}{2}-1})$
and $Z_{1}\notin\mathbb{R}[M_{0,1}]$, given that $\mathbb{R}[M_{0,1}]$ is
Abelian. In particular $\mathbb{R}[M_{0,1}]$ is not a Cartan subalgebra of
$\mathfrak{gl}(n,\mathbb{R})$. Also, $Z_{1}$ will act as an outer derivation
of $\mathfrak{D}_{0,1}^{n}$.

We have the following.

######  Theorem 5.

Let $\phi\in\mathfrak{gl}(n,\mathbb{R})$ be nonderogatory. The following are
equivalent.

(1) Every derivation of $\mathcal{G}_{\phi}$ is an inner derivation.

(2) $\mathbb{R}[\phi]$ is a Cartan subalgebra of
$\mathfrak{gl}(n,\mathbb{R})$.

(3) $\phi$ has $n$ distinct (real or complex) eigenvalues.

(4) The nilradical of $\mathcal{G}_{\phi}$ is Abelian.

(5) $\mathcal{G}_{\phi}$ is the direct sum of only copies of
$\mathfrak{aff}(\mathbb{R})$ and $\mathfrak{aff}(\mathbb{C})$.

###### Proof.

The equivalence between (1) and (2) directly follows from Proposition 4.6.
Indeed, every derivation of $\mathcal{G}_{\phi}$ is inner, if and only if the
normalizer of $\mathbb{R}[\phi]$ in $\mathfrak{gl}(n,\mathbb{R}),$ coincides
with $\mathbb{R}[\phi]$. The equivalence between (2), (3) and (5) has been
shown in Theorem 3. The proof that (5) implies (4) directly follows from the
fact that each copy, of either $\mathfrak{aff}(\mathbb{R})$ or
$\mathfrak{aff}(\mathbb{C})$, has an Abelian nilradical. We now prove that (4)
implies (5). In the decomposition of $\mathcal{G}_{\phi}$ (Theorem 4), the
nilradical of $\mathcal{G}_{\phi}$ is the sum of the nilradicals of its
ideals. From Examples 4.2, 4.3, the nilradicals of $\mathcal{D}_{0}^{k}$ and
$\mathcal{D}_{0,1}^{k}$ are not Abelian, for any $k\geq 2$. So in order for
$\mathcal{G}_{\phi}$ to have an Abelian nilradical, it must not contain an
ideal isomorphic $\mathcal{D}_{0}^{k}$ or $\mathcal{D}_{0,1}^{k}$, $k\geq 2$.
∎

Note that Theorem 5 is in agreement with the classification of Cartan
subalgebras of $\mathfrak{gl}(n,\mathbb{R})$ supplied by Kostant ([14]) and
Sugiura ([24]). Theorem 5 also implies that when $\mathbb{R}[\phi]$ is a
Cartan subalgebra of $\mathfrak{gl}(n,\mathbb{R})$, then $\mathcal{G}_{\phi}$
is a part of the 2-solvable Lie algebras studied in [1], [17].

###  4.6 Proof of Theorem 3

From Proposition 5, for a given $n$, the number of isomorphism classes of
$2$-solvable Frobenius Lie algebras of dimension $2n$ of the form
$\mathcal{G}_{\phi}:=\mathbb{R}[\phi]\ltimes\mathbb{R}^{n}$, where
$\mathbb{R}[\phi]$ is a Cartan subalgebra of $\mathfrak{gl}(n,\mathbb{R}),$ is
exactly $[\frac{n}{2}]+1$. Indeed, one can look at $[\frac{n}{2}]+1$ as the
number (counting from zero) of possible copies of $\mathfrak{aff}(\mathbb{C})$
that one can count in a decomposable Lie algebra containing only copies of
either $\mathfrak{aff}(\mathbb{C})$ or $\mathfrak{aff}(\mathbb{R})$. On the
other hand, from e.g. [24], there are exactly $[\frac{n}{2}]+1$ non-conjugate
Cartan subalgebras of $\mathfrak{gl}(n,\mathbb{R}).$ So we have derived
Theorem 3, in a simple and direct way. More precisely we prove it as follows.

###### Proof.

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{sl}(n,\mathbb{R})$
with a $k$-dimensional toroidal part. From [24], up to conjugacy under an
element of the Weyl group, there is a basis
($\tilde{e}_{1},\dots,\tilde{e}_{n}$), considered here as the canonical basis
of $\mathbb{R}^{n}$, in which $\mathfrak{h}$ is of the form
$\mathfrak{h}=\Big{\\{}\begin{pmatrix}D_{1}&-D_{2}&{\mathbf{0}}\\\
D_{2}&D_{1}&{\mathbf{0}}\\\
{\mathbf{0}}&{\mathbf{0}}&D_{3}\end{pmatrix}\Big{\\}}$, where
$D_{1}=\text{diag}(h_{1},\dots,h_{k})$,
$D_{2}=\text{diag}(h_{k+1},\dots,h_{2k}),$
$D_{3}=\text{diag}(h_{2k+1},\dots,h_{n})$ with
$h_{j}\in\mathbb{R},j=1,\dots,n$. So that, $\mathfrak{h}$ is conjugate to
$\mathfrak{h}^{\prime}=\Big{\\{}M:=\text{diag}(D_{1}^{\prime},\dots,D_{k}^{\prime},D_{3}),$
with $D_{j}^{\prime}=\begin{pmatrix}{h_{j}}&-h_{k+j}\\\
h_{k+j}&{h_{j}}\end{pmatrix}$ and
$D_{3}=\text{diag}(h_{2k+1},\dots,h_{n}),h_{j}\in\mathbb{R},j=1,\dots,n\;\Big{\\}}$,
obtained by reordering the canonical basis of $\mathbb{R}^{n}$ into
$(\tilde{e}_{1},\tilde{e}_{k+1},\tilde{e}_{2},\tilde{e}_{k+2},\dots,\tilde{e}_{k},\tilde{e}_{2k},\tilde{e}_{2k+1},\tilde{e}_{2k+2},\dots,\tilde{e}_{n}).$
An $M$ in $\mathfrak{h}^{\prime}$ is nonderogatory if and only if
$(h_{i},h_{k+i})\neq(h_{j},h_{k+j})$, whenever $i\neq j$ and $h_{2k+s}\neq
h_{2k+l}$ whenever $s\neq l.$ But the existence of an (regular) element
satisfying these conditions is guaranteed by the fact that $\mathfrak{h}$ is a
Cartan subalgebra. Hence, there exists a nonderogatory $M$ with the $n$
distinct eigenvalues
$h_{1}+ih_{k+1},h_{1}-ih_{k+1},\dots,h_{k}+ih_{2k},h_{k}-ih_{2k},$
$h_{2k+1},\dots,h_{n}$, such that, up to a conjugation,
$\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathfrak{h}=\mathbb{R}[M]=\mathfrak{B}_{\mathfrak{h}}$.
So (a) implies (b). Furthermore, Theorem 4 ensures that the Lie algebra
$\mathfrak{B}_{\mathfrak{h}}\ltimes\mathbb{R}^{n}$ is isomorphic to the direct
sum
$\mathfrak{aff}(\mathbb{C})\oplus\dots\oplus\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{R})\oplus\dots\oplus\mathfrak{aff}(\mathbb{R})$
of $k$ copies of $\mathfrak{aff}(\mathbb{C})$ and $(n-k)$ of
$\mathfrak{aff}(\mathbb{R})$. In fact, the equivalence between (b) and (c) has
already been proved by Theorem 4. Conversely, suppose
$M\in\mathfrak{gl}(n,\mathbb{R})$ is nonderogatory and has $n$ distinct
eigenvalues, 2k of which are complex. From Proposition 5, $\mathbb{R}[M]$ is a
Cartan subalgebra of $\mathfrak{gl}(n,\mathbb{R})$ and so (b) implies (a).
Furthermore, using the Primary Decomposition Theorem and results from Section
4.4, we put $\mathbb{R}[M]$ in the form $\mathfrak{h}^{\prime}$. ∎

##  5 Classification of low dimensional 2-solvable Frobenius Lie algebras

Applying Theorem 4 and the analysis done above, we get Theorem 6 below which
provides, up to isomorphism, a complete list of all 2-solvable Frobenius Lie
algebras of dimension $2$, $4,$ $6$ or $8.$ We note that
$\mathfrak{D}_{0}^{2}$, $\mathfrak{aff}({\mathbb{C}})$ and
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$ correspond to
the family PHC7 of [2] and to S11, S8 and S10 respectively in [23]. We also
note that $\mathfrak{D}_{0}^{2}\oplus\mathfrak{D}_{0}^{2}$ is missing from the
list in [25], see Section 5.2. It is also worth recalling here, that the Lie
algebras $\mathcal{G}_{4,2}$ of Example 3.1 and $\mathfrak{h}_{4,2}$ of
Example 3.2 are isomorphic via the isomorphism (3.2).

######  Theorem 6.

A 2-solvable Frobenius Lie algebra of dimension $\leq 6$ is either isomorphic
to $\mathfrak{aff}({\mathbb{R}})$, $\mathfrak{D}_{0}^{2}$,
$\mathfrak{aff}({\mathbb{C}})$, $\mathfrak{D}_{0}^{3}$, $\mathcal{G}_{3,1}$ as
in Example 3.1, or to one of the direct sums
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$
of their copies. In dimension $8$, there are $14$ non-isomorphic $2$-solvable
Frobenius Lie algebras, $9$ of which are of the form $\mathcal{G}_{M}$, for
some nonderogatory $M\in\mathfrak{gl}(4,\mathbb{R})$, namely
${\mathfrak{D}}_{0,1}^{4}$, $\mathfrak{D}_{0}^{4}$, the direct sums
$\mathfrak{D}_{0}^{3}\oplus\mathfrak{aff}({\mathbb{R}}),$
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{D}_{0}^{2}$,
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{C}})$,
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{C}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
and $5$ of which are not given by a nonderogatory
$M\in\mathfrak{gl}(4,\mathbb{R})$, namely $\mathcal{G}_{4,1}$ of Example 3.1,
$\mathfrak{h}_{4,2}$ and $\mathfrak{h}_{4,3}$ of Example 3.2,
$\mathcal{G}_{4,4}^{\prime}$ of Example 3.3,
$\mathcal{G}_{3,1}\oplus\mathfrak{aff}({\mathbb{R}})$.

###  5.1 Proof of Theorem 6

First, let us recall the classification list of nilpotent 3-dimensional
associative real algebras $A_{4,j},$ $j=1,2,\dots,6,$ given in [4]. We will
need the commutative ones for the proof of Theorem 6. They have a basis
$(a,b,c)$ in which the (non-zero) products are as follows.

$\displaystyle A_{3,1}:$ $\displaystyle(A_{3,1})^{2}=0,$ $\displaystyle
A_{3,2}:$ $\displaystyle a^{2}=c,$ $\displaystyle A_{3,3}^{s}:$ $\displaystyle
a^{2}=sc,\;\;b^{2}=c,\;\;s\in\mathbb{R},\;\;s\neq 0,$ $\displaystyle
A_{3,4}^{s}:$ $\displaystyle
a^{2}=sc,\;\;b^{2}=c,\;\;ab=c,\;\;s\in\mathbb{R},$ $\displaystyle A_{3,5}:$
$\displaystyle ab=c,\;\;ba=-c,$ $\displaystyle A_{3,6}:$ $\displaystyle
a^{2}=b,\;\;ab=ba=c.$ (17)

Note that $A_{3,4}^{s}$ and $A_{3,5}$ are not commutative, since $ab\neq ba.$
As noted in [4], $A_{3,3}^{s}$ and $A_{3,3}^{t}$ are isomorphic if and only if
$t=\epsilon^{2}s$ for some $\epsilon\neq 0.$ In our case where the ground
field is $\mathbb{R},$ there are only two isomorphism classes corresponding to
$s>0$, say, $A_{3,3}^{1}$ and $s<0$, say $A_{3,3}^{-1}$.

The above being said, let us now dive into the proof of Theorem 3.

The case $n=1$ is trivial, as every 2-dimensional non-Abelian Lie algebra is
isomorphic to $\mathfrak{aff}(\mathbb{R})=\mathcal{G}_{\psi}$, where
$\psi=\mathbb{I}_{\mathbb{R}}$, as in Example 4.1.1. Let $\mathcal{G}$ be a
2-solvable Frobenius Lie algebra of dimension $2n$, with $2\leq n\leq 3.$
Write $\mathcal{G}=\mathfrak{B}\ltimes\mathbb{R}^{n}$ where $\mathfrak{B}$ is
an $n$-dimensional MASA of $\mathfrak{gl}(n,\mathbb{R})$. Note that, when
$n\leq 3,$ then $[\frac{n^{2}}{4}]+1=n$, so that from Jacobson’s theorem
([13]), every $n$-dimensional Abelian subalgebra of
$\mathfrak{gl}(n,\mathbb{R})$ is a MASA. Further set
$\mathfrak{B}=\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus L$, where $L$ is a
MASA of $\mathfrak{sl}(n,\mathbb{R})$. When $n=2,$ then $L=\mathbb{R}M$, for
some nonzero $M\in\mathfrak{sl}(2,\mathbb{R})$. But every nonzero
$M\in\mathfrak{sl}(2,\mathbb{R})$ is nonderogatory. Thus,
$\mathfrak{B}=\mathbb{R}[M]$ and
$\mathfrak{B}\ltimes\mathbb{R}^{2}=\mathcal{G}_{M}$. From Theorem 4,
$\mathcal{G}_{M}$ is isomorphic to $\mathfrak{D}_{0}^{2}$,
$\mathfrak{aff}(\mathbb{C})$, or
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})$. For $n=3,$ using
the Lie algebras of lower dimensions listed above, the list of decomposable
2-solvable Frobenius Lie algebras of dimension $6$ simply reads
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$.
As regards the non-decomposable ones, they are all of the form
$\mathfrak{g}=(\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus\mathcal{A})\ltimes\mathbb{R}^{3}$
with $\mathcal{A}$ a $2$-dimensional AID MASA of
$\mathfrak{sl}(3,\mathbb{R}),$ since $n$ is odd. We have $2$ possibilities.
The first is $\mathcal{A}^{2}\neq 0$, in which case
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus\mathcal{A}=\mathbb{R}[M],$
yielding the only solution $M=M_{0}$ and $\mathfrak{g}=\mathfrak{D}_{0}^{3}.$
The second possibility is $\mathcal{A}^{2}=0.$ From Theorem 2, $\mathcal{A}$
is unique and the corresponding Lie algebra is
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus\mathcal{A}=\mathfrak{g}_{3,1},$
as in Example 3.1. For the case $n=4,$ again using the Lie algebras of lower
dimensions, we get the following list of decomposable 2-solvable Frobenius Lie
algebras of dimension $8$:
$\mathfrak{D}_{0}^{3}\oplus\mathfrak{aff}({\mathbb{R}}),$
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{D}_{0}^{2}$,
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{C}})$,
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{C}})$,
$\mathfrak{aff}({\mathbb{C}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$,
$\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})\oplus\mathfrak{aff}({\mathbb{R}})$
and $\mathcal{G}_{3,1}\oplus\mathfrak{aff}({\mathbb{R}}).$ We put the non-
decomposable ones in the form
$\mathfrak{g}=(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A})\ltimes\mathbb{R}^{4}$
where $\mathcal{A}$ is either an AID or NAID $3$-dimensional MASA of
$\mathfrak{sl}(4,\mathbb{R}).$ First, in the case where $\mathcal{A}$ is a
NAID MASA of $\mathfrak{sl}(4,\mathbb{R}),$ we write
$\mathcal{A}=\mathbb{R}M_{s}\oplus\mathcal{A}^{\prime}$, where
$\mathcal{A}^{\prime}$ a $2$-dimensional MANS of
$\mathfrak{sl}(2,\mathbb{C}).$ Using a direct approach and writing elements of
$\mathcal{A}^{\prime}$ as upper triangular $4\times 4$ matrices commuting with
$M_{s}$, one easily sees that
$\mathcal{A}^{\prime}=\\{m_{1}(E_{1,3}+E_{2,4})+m_{2}(E_{2,3}-E_{1,4}),m_{1},m_{2}\in\mathbb{R}\\}$.
This leads to
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A}=\mathbb{R}[M_{0,1}]$
and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A})\ltimes\mathbb{R}^{4}=\mathfrak{D}_{0,1}^{4},$
as in Example 4.1.2. Note that using lower triangular matrices leads to
$\mathbb{R}[(M_{0,1})^{T}]$ which is conjugate to $\mathbb{R}[M_{0,1}]$. Now
in the case where $\mathcal{A}$ is an AID, there are $3$ possibilities. (A)
$\mathcal{A}^{3}\neq 0$ and $\mathcal{A}^{4}=0$, there exists
$a\in\mathcal{A}$ such that $a^{3}\neq 0$ and $a^{4}=0,$ so that
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A}=\mathbb{R}[a].$ From
Theorem 4, we have
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A}=\mathbb{R}[M_{0}]$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A})\ltimes\mathbb{R}^{4}=\mathfrak{D}_{0}^{4}$.
This corresponds to the case $A_{3,6}$ in [4], as
$A_{3,6}=span(a,a^{2}=b,a^{3}=c).$ (B) The second possibility is
$\mathcal{A}^{3}=0$ and $\mathcal{A}^{2}\neq 0.$ From Lemma 3.1, we have $\dim
Im(\mathcal{A}^{2})=1.$ This splits into 3 cases (5.1): (a) $A_{3,2}$: with
$(A_{3,2})^{2}=\mathbb{R}c$. A representative of this class is
$\mathcal{P}_{4,2}$ as in Example 3.2 with $a=E_{1,2}+E_{2,4}$, $b=E_{1,3}$,
$c=E_{1,4}$, so the corresponding Lie algebra is
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus
A_{3,2})\ltimes\mathbb{R}^{4}=\mathfrak{h}_{4,2}$. (b) $A_{3,3}^{1}$, with
$(A_{3,3}^{1})^{2}=\mathbb{R}c.$ A representative of this class is
$\mathcal{P}_{4,3}$ as in Example 3.2 with $a=E_{1,2}+E_{2,4}$,
$b=E_{1,3}+E_{3,4}$, $c=E_{1,4}$, and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus
A_{3,3}^{1})\ltimes\mathbb{R}^{4}=\mathfrak{h}_{4,3}$. (c) $A_{3,3}^{-1}$ : A
representative of this class is $L_{4,4}^{\prime}$ as in Example 3.3 with
$a=E_{1,2}+E_{3,4}-E_{1,3}-E_{2,4},$ $b=E_{1,2}+E_{3,4}+E_{1,3}+E_{2,4},$
$c=2E_{1,4}$, so that $(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus
A_{3,3}^{-1})\ltimes\mathbb{R}^{4}=\mathcal{G}_{4,4}^{\prime}$. (C) The last
case is given by the condition $\mathcal{A}^{2}=0$, which, from Theorem 2,
provides a unique $\mathcal{A}$ and
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{A})\ltimes\mathbb{R}^{4}=\mathfrak{g}_{4,1},$
as in Example 3.1. Note that this corresponds to $\mathcal{A}=A_{3,1}$ in [4].
∎

###  5.2 A discussion on the lists by Winternitz and Zassenhaus

We revisit the classification lists of MASAs of $\mathfrak{sl}(n,\mathbb{R})$,
$n=3,4$ provided in [26] and [25]. A systematic comparison shows a match with
our classification, except for one missing item which we complete and correct
some misprint on another item.

For $n=3,$ according to [26], there are six classes of non-mutually conjugate
MASAs of $\mathfrak{sl}(3,\mathbb{R})$. With the same notation as in [26], we
denote them by $L_{2,i}$, $i=1,\dots,6.$

(1) The first is
$L_{2,1}:=\\{diag(k_{1}+k_{2},-k_{1}+k_{2},-2k_{2}),\;k_{1},k_{2}\in\mathbb{R}\\}.$

We note that
$L_{2,1}:=\\{(k_{2}-k_{1})S_{2,1}^{0}+\frac{1}{2}(k_{1}+3k_{2})S_{2,1}+\frac{3}{2}(k_{1}-k_{2})S_{2,1}^{2},\;k_{1},k_{2}\in\mathbb{R}\\}$$\subset\mathbb{R}[S_{2,1}],$
where $S_{2,1}=diag(1,0,-1),$ and $\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus
L_{2,1}=\mathbb{R}[S_{2,1}].$ So $(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus
L_{2,1})\ltimes\mathbb{R}^{3}$ is isomorphic to
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})$,
as $S_{2,1}$ is nonderogatory with $3$ eigenvalues.

(2) In
$L_{2,2}:=\Big{\\{}k_{1}(E_{1,1}+E_{2,2}-2E_{3,3})+k_{2}(E_{1,2}-E_{2,1}),k_{1},k_{2}\in\mathbb{R}\Big{\\}},$
every element is a nonderogatory matrix with one real eigenvalue $-2k_{1}$ and
two complex conjugate eigenvalues $k_{1}+ik_{2}$ and $k_{1}-ik_{2},$ except
when $k_{2}=0.$ For instance, $S_{2,2}:=E_{1,2}-E_{2,1}\in L_{2,2}$ is
nonderogatory, so $\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus
L_{2,2}=\mathbb{R}[S_{2,2}].$ From Theorem 4,
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus L_{2,2})\ltimes\mathbb{R}^{3}$
is isomorphic to $\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{R}).$

(3) In
$L_{2,3}:=\Big{\\{}k_{1}(E_{1,1}+E_{2,2}-2E_{3,3})+k_{2}E_{1,2},k_{1},k_{2}\in\mathbb{R}\Big{\\}}$,
each element is of the form
$k_{1}S_{2,3}^{0}+k_{2}S_{2,3}-(3k_{1}+k_{2})S_{2,3}^{2}$ where
$S_{2,3}:=E_{1,2}+E_{3,3}$ is nonderogatory with a double eigenvalue $0$ and a
simple eigenvalue 1. Hence, $\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus
L_{2,3}=\mathbb{R}[S_{2,3}]$ and $(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus
L_{2,3})\ltimes\mathbb{R}^{3}=\mathcal{G}_{S_{2,3}}$ is isomorphic to
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{D}_{0}^{2}$ (Theorem 4).

(4) As regards
$L_{2,5}:=\Big{\\{}k_{1}E_{1,2}+k_{2}E_{1,3},k_{1},k_{2}\in\mathbb{R}\Big{\\}}$,
the algebra $\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus L_{2,5}$ is the same
as $\mathfrak{B}_{3,1}$ in Example 3.1 and
$\mathfrak{B}_{3,1}\ltimes\mathbb{R}^{3}=\mathcal{G}_{3,1}$ is indecomposable.

(5)
$L_{2,6}:=\Big{\\{}k_{1}(E_{1,2}+E_{2,3})+k_{2}E_{1,3},k_{1},k_{2}\in\mathbb{R}\Big{\\}}$
is such that $\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus
L_{2,6}=\mathbb{R}[M_{0}]$, where the nonderogatory matrix
$M_{0}=E_{1,2}+E_{2,3}$ has $0$ as a unique eigenvalue of multiplicity 3.
Thus, $(\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus
L_{2,6})\ltimes\mathbb{R}^{3}=\mathfrak{D}_{0}^{3},$ (Theorem 4).

(6) For the remaining algebra
$L_{2,4}:=\Big{\\{}k_{1,3}E_{1,3}+k_{2,3}E_{2,3},k_{1,3},k_{2,3}\in\mathbb{R}\Big{\\}}$
in the list in [26], it is easy to see that
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{3}}\oplus L_{2,4})\ltimes\mathbb{R}^{3}$
is not a Frobenius Lie algebra, as every linear form $\alpha$ satisfies
$(\partial\alpha)^{3}=0.$ See also Remark 3.2.

For $n=4$, we treat the pair-wise non-conjugate $16$ MASAs of
$\mathfrak{sl}(4,\mathbb{R})$ in the order in which they appear in the
classification list in [25] and name them $\mathcal{Y}_{j}$, $j=1,\dots,16.$
We correct the algebra $\mathcal{Y}_{8}$ from its original expression in [25]
which was not commutative and complete the list of MASAs from which the MASA
$\mathcal{Y}_{17}$ giving rise to
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{D}_{0}^{2}$ is missing.

  1. 1.

$\mathcal{Y}_{1}:=\\{k_{1,3}E_{1,3}+k_{1,4}E_{1,4}+k_{2,3}E_{2,3}+k_{2,4}E_{2,4}$,
$k_{1,3},k_{1,4},k_{2,3},k_{2,4}\in\mathbb{R}\\}$, is $4$-dimensional, thus it
is not relevant to our study.

  2. 2.

$\mathcal{Y}_{2}:=\\{k_{1,2}E_{1,2}+k_{1,3}E_{1,3}+k_{1,4}E_{1,4}$,
$k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$ is the algebra $L_{4,1}$ in Example
3.1, so the $2$-solvable Frobenius Lie algebra
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{2})\ltimes\mathbb{R}^{4}$ is
isomorphic to $\mathcal{G}_{4,1}.$

  3. 3.

$\mathcal{Y}_{3}:=\\{k_{1,4}E_{1,4}+k_{2,4}E_{2,4}+k_{3,4}E_{3,4}$,
$k_{1,4},k_{2,4},k_{3,4}\in\mathbb{R}\\}$ is the algebra $L_{n}$ in Remark
3.2, when $n=4$. So
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{3})\ltimes\mathbb{R}^{4}$
is not a Frobenius Lie algebra.

  4. 4.

$\mathcal{Y}_{4}:=\\{k_{1,3}(E_{1,3}+E_{3,4})+k_{1,4}E_{1,4}+k_{2,4}E_{2,4}$,
$k_{1,3},k_{1,4},k_{2,4}\in\mathbb{R}\\}$ yields a 2-solvable Lie algebra
which is not a Frobenius Lie algebra. Indeed, if we set
$e_{1}:=\mathbb{I}_{\mathbb{R}^{4}},$ $e_{2}:=E_{1,3}+E_{3,4},$
$e_{3}:=E_{1,4},$ $e_{4}=E_{2,4},$ the Lie bracket of
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{4})\ltimes\mathbb{R}^{4}$ is
$[e_{1},\tilde{e}_{j}]=\tilde{e}_{j},$ $j=1,\dots,4,$
$[e_{2},\tilde{e}_{3}]=\tilde{e}_{1},$ $[e_{2},\tilde{e}_{4}]=\tilde{e}_{3},$
$[e_{3},\tilde{e}_{4}]=\tilde{e}_{1},$ $[e_{4},\tilde{e}_{4}]=\tilde{e}_{2}.$

Any linear form
$\alpha=k_{1}e_{1}^{*}+k_{2}e_{2}^{*}+k_{3}e_{3}^{*}+k_{4}e_{4}^{*}+s_{1}\tilde{e}_{1}^{*}+s_{2}\tilde{e}_{2}^{*}+s_{3}\tilde{e}_{3}^{*}+s_{4}\tilde{e}_{4}^{*}$
satisfies $(\partial\alpha)^{4}=0$, as we have
$\partial\alpha=-e_{1}^{*}\wedge\eta_{1}-e_{2}^{*}\wedge\eta_{2}-\eta_{3}\wedge\tilde{e}_{4}^{*}$,
where
$\eta_{1}^{*}:=s_{1}\tilde{e}_{1}^{*}+s_{2}\tilde{e}_{2}^{*}+s_{3}\tilde{e}_{3}^{*}+s_{4}\tilde{e}_{4}^{*},$
$\eta_{2}:=s_{1}\tilde{e}_{3}^{*}+s_{3}\tilde{e}_{4}^{*},$
$\eta_{3}:=s_{1}e_{3}^{*}+s_{2}e_{4}^{*}.$

  5. 5.

$\mathcal{Y}_{5}:=\\{k_{1,2}(E_{1,2}+E_{2,4})+k_{1,3}E_{1,3}+k_{1,4}E_{1,4},k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$
is the MASA $\mathcal{P}_{4,2}$ of $\mathfrak{sl}(4,\mathbb{R})$ given in
Example 3.2. So
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{5})\ltimes\mathbb{R}^{4}$ is
the $2$-solvable Frobenius Lie algebra $\mathfrak{h}_{4,2}$ in Example 3.2.

  6. 6.

$\mathcal{Y}_{6,\varepsilon}:=\\{k_{1,2}(E_{1,2}+E_{2,4})+k_{1,3}(E_{1,3}+\varepsilon
E_{3,4})+k_{1,4}E_{1,4},\;k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$,
$\varepsilon=\pm 1$, yields a 2-solvable Frobenius Lie algebra
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{6,\varepsilon})\ltimes\mathbb{R}^{4}$
with the following Lie brackets, in the basis
$e_{1}:=\mathbb{I}_{\mathbb{R}^{4}},$ $e_{2}:=E_{1,2}+E_{2,4},$
$e_{3}:=E_{1,3}+\varepsilon E_{3,4},$ $e_{4}=E_{1,4},$ $\tilde{e}_{j},$ $1\leq
j\leq 4:$ $[e_{1},\tilde{e}_{j}]=\tilde{e}_{j},$ $j=1,\dots,4,$
$[e_{2},\tilde{e}_{2}]=\tilde{e}_{1},$ $[e_{2},\tilde{e}_{4}]=\tilde{e}_{2},$
$[e_{3},\tilde{e}_{3}]=\tilde{e}_{1},$
$[e_{3},\tilde{e}_{4}]=\varepsilon\tilde{e}_{3},$
$[e_{4},\tilde{e}_{4}]=\tilde{e}_{1}.$ Furthermore
$\partial\tilde{e}_{1}^{*}=-e_{1}^{*}\wedge\tilde{e}_{1}^{*}-e_{2}^{*}\wedge\tilde{e}_{2}^{*}-e_{3}^{*}\wedge\tilde{e}_{3}^{*}-e_{4}^{*}\wedge\tilde{e}_{4}^{*}$
is nondegenerate. Note that $\mathcal{Y}_{6,1}$ coincides with
$\mathcal{P}_{4,3}$ in Example 3.2. So
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{6,1})\ltimes\mathbb{R}^{4}$ is
the same as $\mathfrak{h}_{4,3}$ in Example 3.2. When $\varepsilon=-1$, the
linear map defined by
$\psi:(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{6,-1})\ltimes\mathbb{R}^{4}\to\mathcal{G}_{4,4}^{\prime},$
$\psi(e_{1})=e_{1}^{\prime},$ $\psi(e_{2})=e_{2}^{\prime}-e_{3}^{\prime},$
$\psi(e_{3})=-e_{2}^{\prime}-e_{3}^{\prime},$ $\psi(e_{4})=-2e_{4}^{\prime},$
$\psi(\tilde{e}_{2})=-\tilde{e}_{2}^{\prime}+\tilde{e}_{3}^{\prime},$
$\psi(\tilde{e}_{3})=\tilde{e}_{2}^{\prime}+\tilde{e}_{3}^{\prime},$
$\psi(\tilde{e}_{1})=-4\tilde{e}_{1}^{\prime},$ is a Lie algebra isomorphism,
where $\mathcal{G}_{4,4}^{\prime}$ is as in Example 3.3. So
$\mathcal{Y}_{6,-1}$ is conjugate to $L_{4,4}^{\prime}.$

  7. 7.

$\mathcal{Y}_{7}:=\\{k_{1,2}(E_{1,2}+E_{2,3}+E_{3,4})+k_{1,3}(E_{1,3}+E_{2,4})+k_{1,4}E_{1,4},\;k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$
satisfies
$\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{7}=\mathbb{R}[M_{0}],$ with
$M_{0}=E_{1,2}+E_{2,3}+E_{3,4}.$ So
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{7})\ltimes\mathbb{R}^{4}=\mathfrak{D}_{0}^{4}$.

  8. 8.

$\mathcal{Y}_{8}:=\\{k_{1,2}(E_{1,2}-E_{2,1}+E_{3,4}-E_{4,3})+k_{1,3}(E_{1,3}+E_{2,4})+k_{1,4}(E_{1,4}-E_{2,3}),$
$k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$ is such that
$\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{8}=\mathbb{R}[M_{0,1}]$, where
$M_{0,1}=E_{1,2}-E_{2,1}+E_{3,4}-E_{4,3}+E_{1,3}+E_{2,4}$. Thus
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{8})\ltimes\mathbb{R}^{4}=\mathfrak{D}_{0,1}^{4}$.
Note that, in the original list in [25], this item was listed with some
mistakes as
$\\{k_{1,2}(E_{1,2}-E_{2,1}+E_{2,4}-E_{4,3})+k_{1,3}(E_{1,3}+E_{2,4})+k_{1,4}E_{1,4},$
$k_{1,2},k_{1,3},k_{1,4}\in\mathbb{R}\\}$ which is not commutative.

  9. 9.

$\mathcal{Y}_{9}:=\\{k_{1,1}(E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4})+k_{1,2}E_{1,2}+k_{1,3}E_{1,3},$
$k_{1,1},k_{1,2},k_{1,3}\in\mathbb{R}\\}$, consider the basis
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4},$
$e_{3}=E_{1,2},$ $e_{4}:=E_{1,3}$ of
$\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{9}$ and change it into
$e_{1}^{\prime}:=\frac{1}{4}(3e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=e_{4},$ $e_{4}^{\prime}:=\frac{1}{4}(e_{1}-e_{2}).$ The Lie
bracket of
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{9})\ltimes\mathbb{R}^{4}$
reads $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{1}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3},\tilde{e}_{3}]=\tilde{e}_{1},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4}.$ That is clearly the Lie
bracket of $\mathcal{G}_{3,1}\oplus\mathfrak{aff}(\mathbb{R})$, where
$\mathcal{G}_{3,1}=span(e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime},\tilde{e}_{1},\tilde{e}_{2},\tilde{e}_{3})$
corresponds to $n=3,p=1$ in Example 3.1.

  10. 10.

$\mathcal{Y}_{10}:=\\{k_{1,1}(E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4})+k_{1,3}E_{1,3}+k_{2,3}E_{2,3},\;k_{1,1},k_{1,3},k_{2,3}\in\mathbb{R}\\}$,
a basis of $\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{10}$ is
$e_{1}=\mathbb{I}_{\mathbb{R}^{4}},$ $e_{2}=E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4},$
$e_{3}=E_{1,3},$ $e_{4}:=E_{2,3}$ which we change into
$e_{1}^{\prime}:=\frac{1}{4}(3e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=e_{4},$ $e_{4}^{\prime}:=\frac{1}{4}(e_{1}-e_{2}).$ The Lie
bracket of
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{10})\ltimes\mathbb{R}^{4}$
reads $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{1}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{2}^{\prime},\tilde{e}_{3}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{2},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4},$ which is the same as that of
$(\mathfrak{B}_{3}\ltimes\mathbb{R}^{3})\oplus\mathfrak{aff}(\mathbb{R})$,
where $\mathfrak{B}_{3}=$span($e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime})$
is as in Remark 3.2, when $n=3$ and
$\mathbb{R}^{3}=$span($\tilde{e}_{1},\tilde{e}_{2},\tilde{e}_{3})$. As
$\mathfrak{B}_{3}\ltimes\mathbb{R}^{3}$ is not a Frobenius Lie algebra (Remark
3.2), neither is
$(\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{10})\ltimes\mathbb{R}^{4}$.

  11. 11.

$\mathcal{Y}_{11}:=\\{k_{1,1}(E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4})+k_{1,2}(E_{1,2}+E_{2,3})+k_{1,3}E_{1,3},$
$k_{1,1},$ $k_{1,2},k_{1,3}\in\mathbb{R}\\}$, a basis of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{Y}_{11}$ is
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4},$
$e_{3}=E_{1,2}+E_{2,3},$ $e_{4}:=E_{1,3}=e_{2}^{2}$, which we change into
$e_{1}^{\prime}:=\frac{1}{4}(3e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=e_{4},$ $e_{4}^{\prime}:=\frac{1}{4}(e_{1}-e_{2}),$ so that
the Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{Y}_{11})\ltimes\mathbb{R}^{4}$
reads $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{1}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{2}^{\prime},\tilde{e}_{3}]=\tilde{e}_{2},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{1},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4}.$ This is the Lie bracket of
$\mathfrak{D}_{0}^{3}\oplus\mathfrak{aff}(\mathbb{R})$, where
$\mathfrak{D}_{0}^{3}=span(e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime},\tilde{e}_{1},\tilde{e}_{2},\tilde{e}_{3})=\mathbb{R}[e_{2}^{\prime}]\ltimes\mathbb{R}^{3}$.

  12. 12.

$\mathcal{Y}_{12}:=\\{k_{1,1}(E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4})+k_{1,2}(E_{1,2}-E_{2,1})+k_{3,4}(E_{3,4}-E_{4,3}),$
$k_{1,1},k_{1,2},$ $k_{3,4}\in\mathbb{R}\\}$, a basis of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{12}$ is
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4},$
$e_{3}=E_{1,2}-E_{2,1},$ $e_{4}:=E_{3,4}-E_{4,3},$ we change it into
$e_{1}^{\prime}=\frac{1}{2}(e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=\frac{1}{2}(e_{1}-e_{2}),$ $e_{4}^{\prime}=e_{4}.$ The Lie
bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{n}}\oplus\mathcal{Y}_{12})\ltimes\mathbb{R}^{4}$
is given by $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{1}]=-\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{3}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4},$
$[e_{4}^{\prime},\tilde{e}_{3}]=-\tilde{e}_{4},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{3}.$ This is the direct sum
$\mathcal{I}_{1}\oplus\mathcal{I}_{2}$ of two ideals
$\mathcal{I}_{1}=span(e_{1}^{\prime},e_{2}^{\prime},\tilde{e}_{1},\tilde{e}_{2})$
and
$\mathcal{I}_{2}=span(e_{3}^{\prime},e_{4}^{\prime},\tilde{e}_{3},\tilde{e}_{4}).$
Both $\mathcal{I}_{1}$ and $\mathcal{I}_{2}$ are isomorphic to
$\mathfrak{aff}(\mathbb{C}).$ So
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{12})\ltimes\mathbb{R}^{4}$
is isomorphic to $\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{C}).$

  13. 13.

$\mathcal{Y}_{13}:=\\{k_{1,1}(E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4})+k_{1,2}(E_{1,2}-E_{2,1})+k_{3,4}E_{3,4},$
$k_{1,1},k_{1,2},$ $k_{3,4}\in\mathbb{R}\\}$, consider the following basis of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{13}$:
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4},$
$e_{3}=E_{1,2}-E_{2,1},$ $e_{4}:=E_{3,4}$. In the new basis
$e_{1}^{\prime}=\frac{1}{2}(e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=\frac{1}{2}(e_{1}-e_{2}),$ $e_{4}^{\prime}=e_{4},$ we get the
following Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{13})\ltimes\mathbb{R}^{4}$:
$[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{1}]=-\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{3}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{3}.$ This is the Lie bracket of
$\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{D}_{0}^{2},$ with
$\mathfrak{aff}(\mathbb{C})=span(e_{1}^{\prime},e_{2}^{\prime},\tilde{e}_{1},\tilde{e}_{2})$
and
$\mathfrak{D}_{0}^{2}=span(e_{3}^{\prime},e_{4}^{\prime},\tilde{e}_{3},\tilde{e}_{4}).$

  14. 14.

$\mathcal{Y}_{14}:=\\{k_{1,1}(E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4})+k_{1,2}(E_{1,2}-E_{2,1})+k_{3,3}(E_{3,3}-E_{3,4}),$
$k_{1,1},$ $k_{1,2},$ $k_{3,3}\in\mathbb{R}\\}$, consider the following basis
of $\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{14}$:
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4},$
$e_{3}=E_{1,2}-E_{2,1},$ $e_{4}:=E_{3,3}-E_{4,4}$. In the basis
$e_{1}^{\prime}=\frac{1}{2}(e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=\frac{1}{4}(e_{1}-e_{2}+2e_{4}),$
$e_{4}^{\prime}=\frac{1}{4}(e_{1}-e_{2}-2e_{4}),$ $\tilde{e}_{j}$,
$j=1,2,3,4,$ the Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{14})\ltimes\mathbb{R}^{4}$
is: $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{1}]=-\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4}.$ This is the Lie bracket of
$\mathfrak{aff}(\mathbb{C})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R}),$
where
$\mathfrak{aff}(\mathbb{C})=$span($e_{1}^{\prime},e_{2}^{\prime},\tilde{e}_{1},\tilde{e}_{2}$)
and the two copies of $\mathfrak{aff}(\mathbb{R})$ are
span($e_{3}^{\prime},\tilde{e}_{3}$), span($e_{4}^{\prime},\tilde{e}_{4}$).

  15. 15.

$\mathcal{Y}_{15}:=\\{k_{1,1}(E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4})+k_{1,2}E_{1,2}+k_{3,3}(E_{3,3}-E_{4,4}),\;k_{1,1},k_{1,2},k_{3,3}\in\mathbb{R}\\}$,
change the basis $e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$
$e_{2}=E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4},$ $e_{3}=E_{1,2},$
$e_{4}=E_{3,3}-E_{4,4}$ of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{15}$, into
$e_{1}^{\prime}=\frac{1}{2}(e_{1}+e_{2}),$ $e_{2}^{\prime}=e_{3},$
$e_{3}^{\prime}=\frac{1}{4}(e_{1}-e_{2}+2e_{4}),$
$e_{4}^{\prime}=\frac{1}{4}(e_{1}-e_{2}-2e_{4}).$ The Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{15})\ltimes\mathbb{R}^{4}$
is $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4}.$ This is the Lie bracket of
$\mathfrak{D}_{0}^{2}\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R}),$
where
$\mathfrak{D}_{0}^{2}=$span($e_{1}^{\prime},e_{2},\tilde{e}_{1},\tilde{e}_{2}$).

  16. 16.

$\mathcal{Y}_{16}:=\\{k_{1}(E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4})+k_{2}(E_{1,1}+E_{2,2}-2E_{3,3})+k_{3}(E_{1,1}$$-E_{2,2})$,
$k_{1},$$k_{2},$ $k_{3}\in\mathbb{R}\\}$, we consider the basis
$e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$ $e_{2}=E_{1,1}+E_{2,2}+E_{3,3}-3E_{4,4},$
$e_{3}=E_{1,1}+E_{2,2}-2E_{3,3},$ $e_{4}:=E_{1,1}-E_{2,2}$ of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{16}$, then change it
to $e_{1}^{\prime}=\frac{1}{12}(3e_{1}+e_{2}+2e_{3}+6e_{4}),$
$e_{2}^{\prime}=\frac{1}{12}(3e_{1}+e_{2}+2e_{3}-6e_{4}),$
$e_{3}^{\prime}=\frac{1}{4}(3e_{1}+e_{2}-4e_{3}),$
$e_{4}^{\prime}=\frac{1}{4}(e_{1}-e_{2}).$ The Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{16})\ltimes\mathbb{R}^{4}$
is: $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4}.$ This is the Lie bracket of
$\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})\oplus\mathfrak{aff}(\mathbb{R})$,
where the copies of $\mathfrak{aff}(\mathbb{R})$ are
$span(e_{j}^{\prime},\tilde{e}_{j})$, $j=1,2,3,4,$ respectively.

  17. 17.

$\mathcal{Y}_{17}:=\\{k_{1,1}(E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4})+k_{1,2}E_{1,2}+k_{3,4}E_{3,4},\;k_{1,1},k_{1,2},k_{3,4}\in\mathbb{R}\\}$.
Consider the basis $e_{1}=\mathbb{I}_{\mathbb{R}^{n}},$
$e_{2}=E_{1,1}+E_{2,2}-E_{3,3}-E_{4,4},$ $e_{3}=E_{1,2},$ $e_{4}:=E_{3,4}$ of
$\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{17}$ and further
change it into $e_{1}^{\prime}=\frac{1}{2}(e_{1}+e_{2}),$
$e_{2}^{\prime}=e_{3},$ $e_{3}^{\prime}=\frac{1}{2}(e_{1}-e_{2}),$
$e_{4}^{\prime}=e_{4}.$ The Lie bracket of
$(\mathbb{R}\mathbb{I}_{\mathbb{R}^{4}}\oplus\mathcal{Y}_{17})\ltimes\mathbb{R}^{4}$
reads: $[e_{1}^{\prime},\tilde{e}_{1}]=\tilde{e}_{1},$
$[e_{1}^{\prime},\tilde{e}_{2}]=\tilde{e}_{2},$
$[e_{2}^{\prime},\tilde{e}_{2}]=\tilde{e}_{1},$
$[e_{3}^{\prime},\tilde{e}_{3}]=\tilde{e}_{3},$
$[e_{3}^{\prime},\tilde{e}_{4}]=\tilde{e}_{4},$
$[e_{4}^{\prime},\tilde{e}_{4}]=\tilde{e}_{3}.$ This is the Lie bracket of the
direct sum $\mathfrak{D}_{0}^{2}\oplus\mathfrak{D}_{0}^{2}$, where the copies
of $\mathfrak{D}_{0}^{2}$ are
span$(e_{1}^{\prime},e_{2}^{\prime},\tilde{e}_{1},\tilde{e}_{2})$ and
$span(e_{3}^{\prime},e_{4}^{\prime},\tilde{e}_{3},\tilde{e}_{4})$. The present
item is missing in the list in [25].

## References

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##  6 Appendix: Jordan form of nonderogatory real Matrices

######  Theorem 7 (Jordan form of real endomorphisms).

Let $V$ be a real vector space, $\dim V=n.$ Let $\phi\in\mathfrak{gl}(V)$ be
nonderogatory. There exists a basis of $V$ in which the matrix $[\phi]$ of
$\phi$ has a Jordan form, more precisely, the following hold.

(A) If $\phi$ has a unique eigenvalue $\lambda$ with multiplicity $n$, then
$[\phi]=\lambda\mathbb{I}_{V}+\sum\limits_{i=1}^{n-1}E_{i,i+1}.$

(B) If $\phi$ has $n$ distinct complex eigenvalues $\lambda_{j}$,
$\bar{\lambda}_{j}$, $j=1,\dots,\frac{n}{2}$ then $[\phi]$$=$
diag($J_{1},\dots,J_{\frac{n}{2}}$), where
$J_{j}=\begin{pmatrix}Re(\lambda_{j})&-Im(\lambda_{j})\\\
Im(\lambda_{j})&Re(\lambda_{j})\end{pmatrix},$ $Im(\lambda_{j})\neq 0.$

(C) If $\phi$ has only two eigenvalues which are both complex $z=r+is$ and
$\bar{z},$ where $r,s\in\mathbb{R},$ $s\neq 0$, and if $n\geq 4$ (in which
case $\phi$ is not diagonalizable in $\mathbb{C}$), then $[\phi]$ has the
form, with $M_{s}$, $M_{n}$ as in (8):

$\displaystyle[\phi]=r\mathbb{I}_{V}+sM_{s}+M_{n}=:M_{z}\;.$ (18)

(D) In the general case, $[\phi]$ is a block diagonal matrix with blocks in
the form (A), (B), (C) or diagonal with distinct diagonal real entries.

###  6.1 Proof of Theorem 7

Let $\phi\in\mathfrak{gl}(V)$ be nonderogatory, where $V$ is a real
$n$-dimensional vector space. We write $\chi_{\phi}(X)=Q_{1}(X)Q_{2}(X)$ where
$Q_{1}(X)$ has only complex (nonreal) zeros and the zeros of $Q_{2}(X)$ are
all real, so that
$V=\ker(\chi_{\phi}(\phi))=\ker(Q_{1}(\phi))\oplus\ker(Q_{2}(\phi)).$ The
result follows by noting again that $\phi$ preserves both $\ker(Q_{1}(\phi))$
and $\ker(Q_{2}(\phi))$ and its restriction to each of them has a Jordan form,
as discussed below. In fact, $Q_{1}(X)$ factorizes as
$Q_{1}(X)=Q_{1a}(X)Q_{1b}(X)$ where all the zeros of $Q_{1a}(X)$ have
multiplicity $1$ and we can further factorize $Q_{1b}(X)$ into factors, each
having only $2$ (complex conjugate) zeros with multiplicity greater than $1.$
So the latter boils down to the case where $\phi$ has only two complex
conjugate eigenvalues, as discussed in (C) below.

(A) Suppose $\phi$ has p distinct eigenvalues $\lambda_{1},\dots,\lambda_{p},$
all of which are real and of respective multiplicity $k_{1},\dots,k_{p}$, with
$k_{1}+\dots+k_{p}=n$, so that its characteristic polynomial factorizes as
$\chi_{\phi}(X)=\chi_{1}(X)\chi_{2}(X)\cdots\chi_{p}(X)$, where
$\chi_{j}(X)=(X-\lambda_{j})^{k_{j}}$, for $j=1,\dots,p$. The polynomials
$\chi_{j}(X)$ are pairwise relatively prime. By the Primary Decomposition
Theorem and Cayley-Hamilton theorem, we have

$\displaystyle
V=\ker\chi_{\phi}(\phi)=\ker\chi_{1}(\phi)\oplus\ker\chi_{2}(\phi)\oplus\cdots\oplus\ker\chi_{p}(\phi).$
(19)

Of course the subspaces $\mathcal{E}_{j}:=\ker(\phi-\lambda_{j})^{k_{j}},$
$j=1,\dots,p,$ are all stable by $\phi$ and the restriction of $\phi$ to
$\mathcal{E}_{j}$ is again a nonderogatory endomorphism of $\mathcal{E}_{j}$
with a unique eigenvalue, for every $j=1,\dots,p.$ So this reduces to the case
where $\phi$ has $n$ distinct real eigenvalues and is hence diagonalizable, or
$\phi$ has a unique real eigenvalue $\lambda$ with multiplicity $n.$ We now
assume the latter case. Up to a sign,
$\chi_{\phi}(X)=(X-\lambda)^{n}=\sum\limits_{j=0}^{n}\complement_{n}^{j}(-1)^{n-j}\lambda^{n-j}X^{j},$
with $\complement_{p}^{q}=\frac{p!}{q!(p-q)!}$, for $p\geq q$. By Cayley-
Hamilton’s Theorem
$\phi^{n}=\sum\limits_{j=0}^{n-1}\complement_{n}^{j}(-1)^{n-j+1}\lambda^{n-j}\phi^{j}$.
Choose $\bar{x}\in V$ such that
$(\bar{e}_{1},\dots,\bar{e}_{n})=$$(\phi^{n-1}\bar{x},\;\phi^{n-2}\bar{x},\;\dots,\;\phi^{n-j}\bar{x},\dots,\;\bar{x})$
is a basis of $V.$ We thus have
$\phi\bar{e}_{1}=\phi^{n}\bar{x}=\sum\limits_{j=0}^{n-1}\complement_{n}^{j}(-1)^{n-j+1}\lambda^{n-j}\phi^{j}\bar{x}=\sum\limits_{j=1}^{n}\complement_{n}^{j}(-1)^{j+1}\lambda^{j}\bar{e}_{j}$
and $\phi\bar{e}_{j}=\bar{e}_{j-1}$, for $j=2,\dots,n$. So, the matrix of
$\phi$ has the form
$\tilde{M}_{\lambda}=\sum\limits_{j=1}^{n}\tilde{k}_{j}E_{j,1}+\sum\limits_{j=1}^{n-1}E_{j,j+1},$
with $\tilde{k}_{j}=\complement_{n}^{j}(-1)^{j+1}\lambda^{j}.$ If $\lambda=0$,
then we are done. If $\lambda\neq 0,$ we use a direct approach by looking for
the coefficients $p_{l,j}$ of a matrix $P$ such that
$\tilde{M}_{\lambda}P=PM_{\lambda},$ where
$M_{\lambda}=\lambda\mathbb{I}_{V}+\sum\limits_{l=1}^{n-1}E_{l,l+1}.$ We get
$p_{k,l}=\sum\limits_{j=0}^{l-1}(-1)^{k-l+j}\lambda^{k-l+j}\complement_{n-l+j}^{k-l+j}p_{1,j+1}$,
where the numbers $p_{1,j+1}$, $j=0,\dots,n-1,$ are seen as parameters. In
particular, the matrix $P$ whose coefficients in the above basis
($\bar{e}_{s}$) are
$p_{kl}=\left\\{\begin{array}[]{cll}(-1)^{k-l}\lambda^{k-l}\complement_{n-l}^{k-l}&\text{
if }&k\geq l,\\\ 0&\text{ if }&k<l\end{array}\right.$ is a solution.

(B) Suppose all the eigenvalues of $\phi$ are complex (and nonreal) and $\phi$
is diagonalizable in $\mathbb{C}$. Equivalently, $\phi$ has $n$ distinct
complex eigenvalues, say, $\lambda_{j},$ $\bar{\lambda}_{j},$
$j=1,\dots,\frac{n}{2},$ where of course, here, $\bar{\lambda}_{j}$ is the
complex conjugate of $\lambda_{j}.$ Let $\lambda=\lambda_{R}-i\lambda_{I}$ be
an eigenvalue of $\phi$, where $\lambda_{R},\lambda_{I}$ are real numbers and
$\lambda_{I}\neq 0$. Consider an eigenvector $v$ (with complex components) of
$\phi$ with corresponding eigenvalue $\lambda$. Further set
$\mathcal{E}_{\lambda}:=\\{zv+\overline{zv},z\in\mathbb{C}\\}.$ Note that, as
$\bar{v}$ is also an eigenvector of $\phi$ with eigenvalue $\bar{\lambda}$, we
also have $\mathcal{E}_{\lambda}=\mathcal{E}_{\bar{\lambda}}.$ Furthermore,
$\mathcal{E}_{\lambda}$ is a real 2-dimensional vector subspace of $V$ which
is stable by $\phi$ and the vectors $Re(v)$ and $Im(v)$ form a basis of
$\mathcal{E}_{\lambda}$ in which the matrix of the restriction of $\phi$ is
$M_{\lambda}:=\begin{pmatrix}\lambda_{R}&-\lambda_{I}\\\
\lambda_{I}&\lambda_{R}\end{pmatrix}.$ Indeed, any element of
$\mathcal{E}_{\lambda}$ is of the form
$zv+\overline{zv}=2Re(z)\;Re(v)-2Im(z)\;Im(v)$ and taking the real and
imaginary parts of both sides of the equation

$\displaystyle\phi v$ $\displaystyle=$ $\displaystyle\lambda
v=\Big{(}\lambda_{R}-i\;\lambda_{I}\Big{)}\;\Big{(}Re(v)+i\;Im(v)\Big{)}$ (20)
$\displaystyle=$
$\displaystyle\Big{(}\lambda_{R}\;Re(v)+\lambda_{I}\;Im(v)\Big{)}+i\;\Big{(}\lambda_{R}\;Im(v)-\lambda_{I}\;Re(v)\Big{)},$

one gets

$\displaystyle\phi Re(v)=\lambda_{R}\;Re(v)+\lambda_{I}\;Im(v)\;,\;\;\phi
Im(v)=-\lambda_{I}\;Re(v)+\lambda_{R}\;Im(v).$ (21)

Note that $M_{\lambda}$ is nonderogatory with
$\chi_{M_{\lambda}}(X)=(X-\lambda_{R})^{2}+\lambda_{I}^{2}$ and both $Re(v)$
and $Im(v)$ are in $\ker\chi_{M_{\lambda}}(\phi),$ more precisely,
$\ker\chi_{M_{\lambda}}(\phi)=\mathcal{E}_{\lambda}.$

In the basis $(Re(v_{j}),Im(v_{j})),$ $j=1,\dots,\frac{n}{2},$ of $V,$ the map
$\phi$ assumes the following Jordan form
diag($M_{\lambda_{1}},\dots,M_{\lambda_{\frac{n}{2}}}$), where $v_{j}$ is an
eigenvector of $\phi$, as above, with eigenvalue $\lambda_{j}$.

(C) Suppose $\phi$ has only 2 eigenvalues which are complex $z=r+is$ and
$\bar{z},$ and $n\geq 4$. So $\phi$ is not diagonalizable and
$\chi_{\phi}(X)=\Big{(}(X-r)^{2}+s^{2}\Big{)}^{\frac{n}{2}}=\sum\limits_{j=0}^{\frac{n}{2}}\complement_{\frac{n}{2}}^{j}s^{n-2j}(X-r)^{2j}.$
We write the latter as
$\chi_{\phi}(X)=\sum\limits_{k=0}^{n}D_{n,k}(z)X^{k}\;,$ where, by a direct
calculation, we obtain the coefficients $D_{n,j}:\mathbb{C}\to\mathbb{R}$
which are given by

$\displaystyle
D_{n,j}(z)=(-1)^{j}\sum\limits_{p=\varepsilon_{j}}^{\frac{n}{2}}\complement_{\frac{n}{2}}^{p}\complement_{2p}^{j}\Big{(}\frac{z-\bar{z}}{2i}\Big{)}^{n-2p}\Big{(}\frac{z+\bar{z}}{2}\Big{)}^{2p-j}\;,$
(22)

where $2\varepsilon_{j}=j+\frac{1}{2}(1-(-1)^{j}).$ By Cayley-Hamilton’s
Theorem, $\phi^{n}=-\sum\limits_{j=0}^{n-1}D_{n,j}(z)\phi^{j}.$ Following
Lemma 2.1, let $\bar{x}\in V$ such that
$(\bar{e}_{j}:=\phi^{n-j}\bar{x}\;,\;j=1,\dots,n)$ is a basis of $V$. We have

$\displaystyle\phi\bar{e}_{j}=\bar{e}_{j-1},\;2\leq j\leq
n,\;\;\phi\bar{e}_{1}=\phi^{n}\bar{x}=-\sum\limits_{j=0}^{n-1}D_{n,j}(z)\phi^{j}\bar{x}=-\sum\limits_{j=0}^{n-1}D_{n,j}(z)\bar{e}_{n-j},$

which in matrix form, say $\tilde{M}_{z}$, reads
$\tilde{M}_{z}=\sum\limits_{j=1}^{n-1}E_{j,j+1}-\sum\limits_{j=1}^{n}D_{n,n-j}(z)E_{j,1}.$
Now one may use a direct approach by looking for the explicit expressions of