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\markboth{Mohammed Guediri}{Left-invariant affine structures on the oscillator group}

\title{Classification of complete left-invariant affine structures on the
oscillator group\thanks{The project was supported by King Saud University, Deanship of Scientific
Research, College of Science Research Center.}}

\author[M.\,Guediri]{Mohammed Guediri\affil{1}\comma\corrauth }

\address{\affilnum{1}\ Department of Mathematics, College of Science, King Saud
University, P.O. Box 2\,455, Riyadh 11\,451, Saudi Arabia}

\email{{\tt mguediri@ksu.edu.sa} (M.\,Guediri)}

\begin{abstract}
The goal of this paper is to provide a method, based on the theory of
extensions of left-symmetric algebras, for classifying left-invariant affine
structures on a given solvable Lie group of low dimension. To illustrate our method
better, we shall apply it to classify all complete
left-invariant affine structures on the oscillator group.
\end{abstract}

\keywords{left-invariant affine structures, left-symmetric algebras,
extensions and cohomologies of Lie algebras and left-symmetric algebras}

\ams{53C50, 53A15}

\maketitle

\section{Introduction}
It is a well known result (see \cite{auslander,milnor}) that a
simply connected Lie group $G$ which admits a complete left-invariant affine
structure, or equivalently $G$ acts simply transitively by affine
transformations on $\mathbb{R}^{n},$ must be solvable. It is also well known
that not every solvable (even nilpotent) Lie group can admit an affine
structure \cite{benoist}.

The goal of this paper is to provide a method for classifying all complete
left-invariant affine structures on a given solvable Lie group of low
dimension. Since the classification has been completely achieved up to
dimension four in the nilpotent case (see \cite{friedgold,kim,kuiper}), we will illustrate our method by applying it to the remarkable
solvable non-nilpotent $4$-dimensional Lie group $O_{4}$ known as the \emph{oscillator group}. Since complete left-invariant affine structures on a Lie
group $G$ are in one-to-one correspondence with complete (in the sense of
\cite{segal}) left-symmetric structures on its Lie algebra $\mathcal{G}$
\cite{kim}, we will carry out the classification in terms of complete
left-symmetric structures on the oscillator algebra $\mathcal{O}_{4}$.

The paper is organized as follows. In Section 2, we will recall the notion
of extensions of Lie algebras and its relationship to the notion of
$\mathcal{G}$-kernels. In Section 3, we will give some necessary definitions
and basic results on left-symmetric algebras and their extensions. In
Section 4, given a complete left-symmetric algebra $A_{4}$ whose associated
Lie algebra is $\mathcal{O}_{4}$, we will use the complexification of $A_{4}$
and some results in \cite{burde} and \cite{kbm} to show first that $A_{4}$
is not simple. Precisely, we will show that $A_{4}$ has a proper two-sided
ideal whose associated Lie algebra is isomorphic to the center $Z\left(
\mathcal{O}_{4}\right) \cong \mathbb{R}$ or the commutator ideal $\left[
\mathcal{O}_{4},\mathcal{O}_{4}\right] \cong \mathcal{H}_{3}$ of $\mathcal{O}
_{4}.$ In the latter case, we will show that the so-called center of $A_{4}$
is nontrivial, and therefore we can get $A_{4}$ as a central extension (in
some sense that we will define later) of a complete $3$-dimensional
left-symmetric algebra $A_{3}$ by the trivial left-symmetric algebra $
\mathbb{R}$ (i.e., the vector space $\mathbb{R}$ with the trivial
left-symmetric product). In Section 5, we will show that in both cases we
have a short exact sequence (which turns out to be central) of
left-symmetric algebras of the form $0\rightarrow \mathbb{R}\overset{i}{
\rightarrow }A_{4}\overset{\pi }{\rightarrow }A_{3}\rightarrow 0,$ where
here $A_{3}$\ is a complete left-symmetric algebra whose Lie algebra is
isomorphic to the Lie algebra$\mathcal{\ E}\left( 2\right) \ $of the group
of Euclidean motions of the plane. We will then show that, up to
left-symmetric isomorphism, there are only two non-isomorphic complete
left-symmetric structures on $\mathcal{E}\left( 2\right) $, and we will use
these to carry out all complete left-symmetric structures on $\mathcal{O}
_{4}.$ We will see that one of these two left-symmetric structures on $
\mathcal{E}\left( 2\right) $ yields exactly one complete left-symmetric
structure on $\mathcal{O}_{4}.$ However, the second one yields a
two-parameter family of complete left-symmetric algebras $A_{4}\left(
s,t\right) $ whose associated Lie algebra is $\mathcal{O}_{4},$ and the
conjugacy class of $A_{4}\left( s,t\right) $ is given as follows: $
A_{4}\left( s^{\prime },t^{\prime }\right) $ is isomorphic to $A_{4}\left(
s,t\right) $ if and only if $\left( s^{\prime },t^{\prime }\right) =\left(
\alpha s,\pm t\right) $ for some $\alpha \in \mathbb{R}^{\ast }.$ By using
the Lie group exponential maps, we will deduce the classification of all
complete left-invariant affine structures on the oscillator group $O_{4}$ in
terms of simply transitive actions of subgroups of the affine group $
Aff\left( \mathbb{R}^{4}\right) =GL\left( \mathbb{R}^{4}\right) \ltimes
\mathbb{R}^{4}$ (see Theorem \ref{thm3}).

Throughout this paper, all vector spaces, Lie algebras, and left-symmetric
algebras are supposed to be over the field $\mathbb{R}$, unless otherwise
specified. We shall also suppose that all Lie groups are connected and
simply connected.

\section{Extensions of Lie algebras}

The notion of extensions of a Lie algebra $\mathcal{G}$ by an abelian Lie
algebra $\mathcal{A}$ is well known (see, for instance, books \cite{de azcarraga} and \cite{jacobson}). In light of \cite{neeb}, we will briefly
describe here the notion of extension $\widetilde{\mathcal{G}}$ of a Lie
algebra $\mathcal{G}$ by a Lie algebra $\mathcal{A}$ which is not
necessarily abelian.

Suppose that a vector space extension $\widetilde{\mathcal{G}}$ of a Lie
algebra $\mathcal{G}$ by another Lie algebra $\mathcal{A}$ is known, and we
want to define a Lie structure on $\widetilde{\mathcal{G}}$ in terms of the
Lie structures of $\mathcal{G}$ and $\mathcal{A}$. Let $\sigma :\mathcal{G}
\rightarrow \widetilde{\mathcal{G}}$ be a section, that is, a linear map
such that $\pi \circ \sigma =id.$ Then the linear map $\Psi :\left(
a,x\right) \mapsto i\left( a\right) +\sigma \left( x\right) $ from $\mathcal{
A\oplus G}$ onto $\widetilde{\mathcal{G}}$ is an isomorphism of vector
spaces. For $\left( a,x\right) $ and $\left( b,y\right) $ in $\mathcal{
A\oplus G}$, a commutator on $\widetilde{\mathcal{G}}$ must satisfy
\begin{eqnarray}
\left[ i\left( a\right) +\sigma \left( x\right) ,i\left( b\right) +\sigma
\left( y\right) \right] &=&i\left( \left[ a,b\right] \right) +\left[ \sigma
\left( x\right) ,i\left( b\right) \right]  \label{eq0} \\
&&+\left[ i\left( a\right) ,\sigma \left( y\right) \right] +\left[ \sigma
\left( x\right) ,\sigma \left( y\right) \right]  \notag
\end{eqnarray}
Now we define a linear map $\phi :\mathcal{G}\rightarrow End\left(\mathcal{A}\right)$
by
\begin{equation}
\phi \left( x\right) a=\left[ \sigma \left( x\right) ,i\left( a\right)
\right]  \label{eq0.5}
\end{equation}
On the other hand, since $\pi \left( \left[ \sigma \left( x\right) ,\sigma
\left( y\right) \right] \right) =\pi \left( \sigma \left( \left[ x,y\right]
\right) \right) ,$ it follows that there exists an alternating bilinear map $
\omega :\mathcal{G\times G\rightarrow A}$ such that $\left[ \sigma \left(
x\right) ,\sigma \left( y\right) \right] =\sigma \left[ x,y\right] +\omega
\left( x,y\right)$.

To sum up, by means of the isomorphism above, $\widetilde{\mathcal{G}}\cong
\mathcal{A\oplus G}$ and its elements may be denoted by $\left( a,x\right) $
with $a\in \mathcal{A}$ and $x$ is simply characterized by its coordinates
in $\mathcal{G}$. The commutator defined by (\ref{eq0}) is now given by
\begin{equation}
\left[ \left( a,x\right) ,\left( b,y\right) \right] =\left( \left[ a,b\right]
+\phi \left( x\right) b-\phi \left( y\right) a+\omega \left( x,y\right),
\left[ x,y\right] \right) ,  \label{eq1}
\end{equation}
for all $\left( a,x\right) \in \widetilde{\mathcal{G}}\cong \mathcal{A\oplus
G}$.

It is easy to see that this is actually a Lie bracket (i.e, it verifies the
Jacobi identity) if and only if the following three conditions are satisfied

\begin{enumerate}
\item $\phi \left( x\right) \left[ b,c\right] =\left[ \phi \left( x\right)
b,c\right] +\left[ b,\phi \left( x\right) c\right] ,$

\item $\left[ \phi \left( x\right) ,\phi \left( y\right) \right] =\phi
\left( \left[ x,y\right] \right) +ad_{\omega \left( x,y\right) },$

\item $\omega \left( \left[x,y\right]\!, z\right) -\omega \left(x,\!\left[y,z
\right] \right) +\omega \left(y,\!\left[ x,z\right] \right)\!=\!\phi \left(
x\right) \omega \left( y,z\right) +\phi \left( y\right) \omega \left(
z,x\right) +\phi \left( z\right) \omega \left( x,y\right)$.
\end{enumerate}

\begin{remark}\label{remark1}
We see that condition (1) above is equivalent to say that
$\phi \left( x\right) $ is a derivation of $\mathcal{A}$. In other words, $
\mathcal{G}$ is actually acting by derivations, that is, $\phi :\mathcal{G}
\rightarrow Der\left( \mathcal{A}\right) .$ Condition (2) indicates
clearly that if $\mathcal{A}$ is supposed to be abelian, then $\mathcal{A}$
becomes a $\mathcal{G}$-module in a natural way, because in this case the
linear map $\phi :\mathcal{G}\rightarrow Der\left( \mathcal{A}\right) $
given by $\phi \left( x\right) a=\left[ \sigma \left( x\right) ,i\left(
a\right) \right] $ is well defined. Condition (3) is equivalent to the
fact that, if $\mathcal{A}$ is abelian, $\omega $ is a $2$-cocycle (i.e., $
\delta _{\phi }\omega =0,$ where $\delta _{\phi }$ refers to the coboundary
operator corresponding to the action $\phi $).
\end{remark}

If now $\sigma ^{\prime }:\mathcal{G}\rightarrow \widetilde{\mathcal{G}}$ is
another section, then $\sigma ^{\prime }-\sigma =\tau $ for some linear map $
\tau :\mathcal{G}\rightarrow \mathcal{A}$, and it follows that the
corresponding morphism and the $2$-cocycle are $\phi ^{\prime
}=\phi +ad\circ \tau $ and $\omega ^{\prime }=\omega +\delta _{\phi }\tau +
\frac{1}{2}\left[ \tau ,\tau \right]$, respectively, where $ad$ stands here and below (if
there is no ambiguity) for the adjoint representation in $\mathcal{A}$, and
where $\left[ \tau ,\tau \right] $ has the following meaning: Given two
linear maps $\alpha ,\beta :\mathcal{G}\rightarrow \mathcal{A}$, we define $
\left[ \alpha ,\beta \right] \left( x,y\right) =\left[ \alpha \left(
x\right) ,\beta \left( y\right) \right] -\left[ \alpha \left( y\right)
,\beta \left( x\right) \right] .$ In particular, we have $\frac{1}{2}\left[
\tau ,\tau \right] \left( x,y\right) =\left[ \tau \left( x\right) ,\tau
\left( y\right) \right] $. Note here that the Lie algebra $\mathcal{A}$ is
not necessarily abelian. Therefore, $\omega ^{\prime }-\omega $ is a $2$
-coboundary if and only if $\left[ \tau \left( x\right) ,\tau \left(
y\right) \right] =0$ for all $x,y\in \mathcal{G}$. Equivalently, $\omega
^{\prime }-\omega $ is a $2$-coboundary if and only if $\tau $ has its range
in the center $Z\left( \mathcal{A}\right) $ of $\mathcal{A}$. In that case,
we get $\omega ^{\prime }-\omega =\delta _{\phi }\tau \in B_{\phi
}^{2}\left( \mathcal{G},Z\left( \mathcal{A}\right) \right) ,$ the group of $
2 $-coboundaries for $\mathcal{G}\ $with values in $Z\left( \mathcal{A}
\right) .$

To overcome all these difficulties, we proceed as follows. Let $C^{2}\left(
\mathcal{G},\mathcal{A}\right) $ be the abelian group of all $2$-cochains,
i.e., alternating bilinear mappings $\mathcal{G}\times \mathcal{G}\rightarrow
\mathcal{A}$. For a given $\phi :\mathcal{G}\rightarrow Der\left( \mathcal{A}
\right) ,$ let $T_{\phi }\in C^{2}\left( \mathcal{G},\mathcal{A}\right) $ be
defined by
\[
T_{\phi }\left( x,y\right) =\left[ \phi \left( x\right) ,\phi \left(
y\right) \right] -\phi \left( \left[ x,y\right] \right) ,\ \ \ \text{for all
}x,y\in \mathcal{G}.
\]
If there exists some $\omega \in C^{2}\left( \mathcal{G},\mathcal{A}\right) $
such that $T_{\phi }=ad\circ \omega ~$and $\delta _{\phi }\omega =0,$ then
the pair $\left( \phi ,\omega \right) $ is called a \textsl{factor system}
for\textsl{\ }$\left( \mathcal{G},\mathcal{A}\right) .$ \textsl{\ }Let $
Z^{2}\left( \mathcal{G},\mathcal{A}\right) $ be the set of all factor
systems for $\left( \mathcal{G},\mathcal{A}\right)$. It is well known that
the equivalence classes of extensions of a Lie algebra $\mathcal{G}$ by a
Lie algebra $\mathcal{A}$ are in one-to-one correspondence with the elements
of the quotient space $Z^{2}\left( \mathcal{G},\mathcal{A}\right)
/C^{1}\left( \mathcal{G},\mathcal{A}\right) ,$ where $C^{1}\left( \mathcal{G}
,\mathcal{A}\right) $ is the space of linear maps from $\mathcal{G}$ into $
\mathcal{A}$ (see, for instance, \cite{neeb}, Theorem II.7). Note that if we
assume that $\mathcal{A}$ is abelian, then we meet the well known result
(see, for instance, \cite{chevalley}) stating that for a given action$\ \ \phi
:\mathcal{G}\rightarrow End\left( \mathcal{A}\right) ,$ the equivalence
classes of extensions of $\mathcal{G}$ by $\mathcal{A}$ are in one-to-one
correspondence with the elements of the second cohomology group
\[
H_{\phi }^{2}\left( \mathcal{G},\mathcal{A}\right) =Z_{\phi }^{2}\left(
\mathcal{G},\mathcal{A}\right) /B_{\phi }^{2}\left( \mathcal{G},\mathcal{A}
\right) .
\]
In the present paper, we will be concerned with the special case where $
\mathcal{A}$ is non-abelian and $\mathcal{G}$ is $\mathbb{R}$, and
henceforth the cocycle $\omega $ is identically zero.

\begin{remark}
It is worth noticing that the construction above is closely related to the
notion of $\mathcal{G}$-kernels considered for Lie algebras firstly in \cite{mori}.
\end{remark}

\section{Left-symmetric algebras}
The notion of a \emph{left-symmetric algebra} arises naturally in various
areas of mathematics and physics. It originally appeared in the works of
Vinberg \cite{vinberg} and Koszul \cite{koszul} concerning convex
homogeneous cones and bounded homogeneous domains, respectively. It also
appears, for instance, in connection with Yang-Baxter equation and
integrable hydrodynamic systems (cf. \cite{bordemann,gol-sokolov,kupershmidt}). A left-symmetric algebra $\left( A,.\right) $ is a
finite-dimensional algebra $A$ in which the products, for all $x,y,z\in A,$
satisfy the identity
\begin{equation}
\left( x\cdot y\right) \cdot z-x\cdot \left( y\cdot z\right) =\left( y\cdot
x\right) \cdot z-y\cdot \left( x\cdot z\right)  \label{eq2}
\end{equation}
It is clear that an associative algebra is a left-symmetric algebra.
Actually, if $A$ is a left-symmetric algebra and $\left( x,y,z\right)
=\left( x\cdot y\right) \cdot z-x\cdot \left( y\cdot z\right) $ is the
associator of $x,y,z$, then we can see that (\ref{eq2}) is equivalent to $
\left( x,y,z\right) =\left( y,x,z\right) .$ This means that the notion of a
left-symmetric algebra is a natural generalization of the notion of an
associative algebra. If $A$ is a left-symmetric algebra, then the commutator
\begin{equation}
\left[ x,y\right] =x\cdot y-y\cdot x  \label{eq3}
\end{equation}
defines the structure of a Lie algebra on $A,$ called the \emph{associated Lie
algebra}. Conversely, if $\mathcal{G}$ is a Lie algebra with a
left-symmetric product $\cdot $ satisfying (\ref{eq3}), then we say that the
left-symmetric structure is \emph{compatible }with the Lie structure on $
\mathcal{G}$.

On the other hand, let $G$ be a Lie group with a left-invariant flat affine
connection $\nabla ,$ and define a product $\cdot $ on the Lie algebra $
\mathcal{G}$ of $G$ by
\begin{equation}
x\cdot y=\nabla _{x}y, \text{ for all } x,y\in \mathcal{G}.  \label{eq4}
\end{equation}
Then, conditions on the connection $\nabla $ for being flat and
torsion-free are now equivalent to conditions (\ref{eq2}) and (\ref{eq3}), respectively. Conversely, suppose that $\mathcal{G}$ is endowed with a
left-symmetric product $\cdot $ which is compatible with the Lie bracket of $
\mathcal{G}$. In this case, in order to obtain a left-invariant flat affine
structure on $G,$ we can define an operator $\nabla $ on $\mathcal{G}$
according to identity (\ref{eq4}) and then extend it by left-translations to
the whole Lie group $G.$ To sum up, the left-invariant flat affine
structures on $G$ are in one-to-one correspondence with the left-symmetric
structures on $\mathcal{G}$ compatible with the Lie structure.

Let now $A$ be a left-symmetric algebra, and let $L_{x}$ and $R_{x}$ be the
left and right multiplications by the element $x$, that is, $L_{x}y=x\cdot y$
and $R_{x}y=y\cdot x.$ We say that $A$ is \emph{complete} if $R_{x}$ is a
nilpotent operator, for all $x\in A.$ It turns out that, for a given simply
connected Lie group $G$ with Lie algebra $\mathcal{G}$, the complete
left-invariant flat affine structures on $G$ are in one-to-one
correspondence with the complete left-symmetric structures on $\mathcal{G}$
compatible with the Lie structure (see, for example, \cite{kim}). It is also
known that an $n$-dimensional simply connected Lie group admits a complete
left-invariant flat affine structure if and only if it acts simply
transitively on $\mathbb{R}^{n}$ by affine transformations (see \cite{kim}).
A simply connected Lie group acting simply transitively on $\mathbb{
R}^{n}$ by affine transformations must be solvable according to \cite{auslander}, but it is worth noticing that there exist solvable (even
nilpotent) Lie groups which do not admit affine structures (see \cite{benoist}).

We close this section by fixing some notations which we will use
in what follows. For a left-symmetric algebra $A$, we can easily check that
the subset
\begin{equation}
T\left( A\right) =\left\{ x\in A:L_{x}=0\right\}  \label{T(A)}
\end{equation}
is a two-sided ideal in $A$. Geometrically, if $G$ is a Lie group which acts
simply transitively on $\mathbb{R}^{n}$ by affine transformations, then $ T\left( \mathcal{G}\right) $ corresponds to the set of translational
elements in $G,$ where $\mathcal{G}$ is endowed with the complete
left-symmetric product corresponding to the action of $G$ on $\mathbb{R} ^{n}$.
It has been conjectured in \cite{auslander} that every nilpotent Lie
group $G$ which acts simply transitively on $\mathbb{R}^{n}$ by affine
transformations contains a translation which lies in the center of $G$, but
this conjecture turned out to be false (see \cite{fried}).

\subsection{Extensions of left-symmetric algebras}
In this section, we will briefly discuss the problem of an extension of a
left-symmetric algebras. To our knowledge, this notion has been considered
for the first time in \cite{kim}. Suppose we are given a vector space $A$ as
an extension of a left-symmetric algebra $K$ by another left-symmetric
algebra $E$. We want to define a left-symmetric structure on $A$ in terms
of the left-symmetric structures given on $K$ and $E.$ In other words, we
want to define a left-symmetric product on $A$ for which $E$ becomes a
two-sided ideal in $A$ such that $A/E\cong K;$ or equivalently,
$0\rightarrow E\rightarrow A\rightarrow K\rightarrow 0$ becomes a short exact
sequence of left-symmetric algebras.

\begin{theorem}[See \protect\cite{kim}]\label{thm1}
There exists a left-symmetric structure on $A$ extending a
left-symmetric algebra $K$ by a left-symmetric algebra $E$ if and only if
there exist two linear maps $\lambda ,$ $\rho :K\rightarrow End\left(
E\right) $ and a bilinear map $g:K\times K\rightarrow E$ such that, for all
$x,y,z\in K$ and $a,b\in E,$ the following conditions are satisfied.

\begin{itemize}
\item[$($i$)$] $\lambda _{x}\left( a\cdot b\right) =\lambda _{x}\left( a\right)
\cdot b+a\cdot \lambda _{x}\left( b\right) -\rho _{x}\left( a\right) \cdot
b, $

\item[$($ii$)$] $\rho _{x}\left( \left[ a,b\right] \right) =a\cdot \rho
_{x}\left( b\right) -b\cdot \rho _{x}\left( a\right) ,$

\item[$($iii$)$] $\left[ \lambda _{x},\lambda _{y}\right] =\lambda _{\left[ x,y
\right] }+L_{g\left( x,y\right) -g\left( y,x\right) },$ where $L_{g\left(
x,y\right) -g\left( y,x\right) }$ denotes the left multiplication in $E$ by $
g\left( x,y\right) -g\left( y,x\right)$,

\item[$($iv$)$] $\left[ \lambda _{x},\rho _{y}\right] =\rho _{x\cdot y}-\rho
_{y}\circ \rho _{x}+R_{g\left( x,y\right) },$ where $R_{g\left( x,y\right) }$
denotes the right multiplication in $E$ by $g\left( x,y\right)$,

\item[$($v$)$]
$g( x,y\cdot z) -g( y,x\cdot z) +\lambda_{x}( g( y,z) ) -\lambda _{y}( g( x,z)) -g([ x,y] ,z) -\rho _{z}( g(x,y)-g( y,x)) =0$.
\end{itemize}
\end{theorem}

If the conditions of Theorem \ref{thm1} are fulfilled, then the extended
left-symmetric product on $A\cong K\times E$ is given by
\begin{equation}
\left( x,a\right) \cdot \left( y,b\right) =\left( x\cdot y,a\cdot b+\lambda
_{x}\left( b\right) +\rho _{y}\left( a\right) +g\left( x,y\right) \right) .
\label{eq5}
\end{equation}
It is remarkable that if the left-symmetric product of $E$ is trivial, then
the conditions of Theorem \ref{thm1} simplify to the following three
conditions:

\begin{itemize}
\item[$($i$)$] $\left[ \lambda _{x},\lambda _{y}\right] =\lambda _{\left[ x,y
\right] },$ i.e., $\lambda $ is a representation of Lie algebras,

\item[$($ii$)$] $\left[ \lambda _{x},\rho _{y}\right] =\rho _{x\cdot y}-\rho
_{y}\circ \rho _{x}$,

\item[$($iii$)$] $g( x,y\cdot z) -g( y,x\cdot z) +\lambda_{x}( g( y,z)) -\lambda _{y}( g( x,z))
-g([ x,y] ,z) -\rho _{z}( g( x,y) -g( y,x) ) =0.$
\end{itemize}

In this case, $E$ becomes a $K$-bimodule and the extended product
given in (\ref{eq5}) simplifies, too. Recall that if $K$ is a left-symmetric
algebra and $V$ is a vector space, then we say that $V$ is a $K$-bimodule if
there exist two linear maps $\lambda ,$ $\rho :K\rightarrow End\left(
V\right) $ which satisfy conditions (i) and (ii) stated above.

Let $K$ be a left-symmetric algebra, and let $V$ be a $K$-bimodule. Let $
L^{p}\left( K,V\right) $ be the space of all $p$-linear maps from $K$ to $V,$
and define two coboundary operators $\delta _{1}:L^{1}\left( K,V\right)
\rightarrow L^{2}\left( K,V\right) $ and $\delta _{2}:L^{2}\left( K,V\right)
\rightarrow L^{3}\left( K,V\right) $ as follows: For a linear map $h\in
L^{1}\left( K,V\right) $ we set
\begin{equation}
\delta _{1}h\left( x,y\right) =\rho _{y}\left( h\left( x\right) \right)
+\lambda _{x}\left( h\left( y\right) \right) -h\left( x\cdot y\right) ,
\label{delta1}
\end{equation}
and for a bilinear map $g\in L^{2}\left( K,V\right) $ we set
\begin{eqnarray}
\delta _{2}g\left( x,y,z\right) &=&g\left( x,y\cdot z\right) -g\left(
y,x\cdot z\right) +\lambda _{x}\left( g\left( y,z\right) \right) -\lambda
_{y}\left( g\left( x,z\right) \right)  \label{delta2} \\
&&-g\left( \left[ x,y\right] ,z\right) -\rho _{z}\left( g\left( x,y\right)
-g\left( y,x\right) \right) .  \notag
\end{eqnarray}

It is straightforward to check that $\delta _{2}\circ \delta _{1}=0.$
Therefore, if we set $Z_{\lambda ,\rho }^{2}\left( K,V\right) =\ker \delta
_{2}$ and $B_{\lambda ,\rho }^{2}\left( K,V\right) =\mathop{\rm Im}\delta
_{1}$, we can define a notion of second cohomology for the actions $\lambda $
and $\rho $ by simply setting $H_{\lambda ,\rho }^{2}\left( K,V\right)
=Z_{\lambda ,\rho }^{2}\left( K,V\right) /B_{\lambda ,\rho }^{2}\left(
K,V\right) .$ As in the case of extensions of Lie algebras, we can prove
that for given linear maps $\lambda ,$ $\rho :K\rightarrow End\left(
V\right) $, the equivalence classes of extensions $0\rightarrow V\rightarrow
A\rightarrow K\rightarrow 0$ of $K$ by $V$ are in one-to-one correspondence
with the elements of the second cohomology group $H_{\lambda ,\rho
}^{2}\left( K,V\right) .$ We close this subsection with the following lemma
on completeness of left-symmetric algebras (see \cite[Proposition 3.4]{changkimlee}).

\begin{lemma}\label{completelemma}
Let $0\rightarrow E\rightarrow A\rightarrow
K\rightarrow 0$ be a short exact sequence of left-symmetric algebras. Then, $A$ is complete if and only if $E$ and $K$ are so.
\end{lemma}

\subsection{Central extensions of left-symmetric algebras}
The notion of central extensions known for Lie algebras may analogously be
defined for left-symmetric algebras. Let $A$ be a left-symmetric extension
of a left-symmetric algebra $K$ by another left-symmetric algebra $E$, and
let $\mathcal{G}$ be the Lie algebra associated to $A.$ Define \textsl{the
center} of $A$ to be $C\left( A\right) =T\left( A\right) \cap Z\left(
\mathcal{G}\right)$, that is,
\begin{equation}
C\left( A\right) =\left\{ x\in A:x\cdot y=y\cdot x=0, \text{ for all }
y\in A\right\} ,  \label{l.s. center}
\end{equation}
where $Z\left( \mathcal{G}\right) $ is the center of the Lie algebra $
\mathcal{G}$ and $T\left( A\right) $ is the two-sided ideal of $A$ defined
by (\ref{T(A)}).

\begin{definition}
The extension $0\rightarrow E\overset{i}{\rightarrow }A\overset{\pi }{\rightarrow }K\rightarrow 0$ of left-symmetric algebras is said to be
\textsl{central} (resp. \textsl{exact}) if $i\left( E\right) \subseteq
C\left( A\right) $ (resp. $i\left( E\right) =C\left( A\right) $).\textsl{\ }
\end{definition}

\begin{remark}
It is not difficult to show that if the extension $0\rightarrow E\overset{i}{
\rightarrow }A\overset{\pi }{\rightarrow }K\rightarrow 0$ is central, then
both the left-symmetric product and the $K$-bimodule on $E$ are trivial
(i.e., $a\cdot b=0$ for all $a,b\in E,$ and $\lambda =\rho =0$). It is also
easy to show that if $\left[ g\right] $ is the cohomology class associated
to this extension, and if
\[
I_{\left[ g\right] }=\left\{ x\in K:x\cdot y=y\cdot x=0 \text{ and }g\left(
x,y\right) =g\left( y,x\right) =0, \text{ for all }y\in K\right\} ,
\]
then the extension is exact if and only if $I_{\left[ g\right] }=0$ (see
\cite{kim}). We note here that $I_{\left[ g\right] }$ is well defined
because any other element in $\left[ g\right] $ takes the form $g+\delta
_{1}h,$ with $\delta _{1}h\left( x,y\right) =-h\left( x\cdot y\right) .$
\end{remark}

Let now $K$ be a left-symmetric algebra, and $E$ a trivial $K$-bimodule.
Denote by $\left( A,\left[ g\right] \right) $ the central extension $0\rightarrow E\rightarrow A\rightarrow K\rightarrow 0$ corresponding to the
cohomology class $\left[ g\right] \in H^{2}\left( K,E\right) .$ Let $\left(
A,\left[ g\right] \right) $ and $\left( A^{\prime },\left[ g^{\prime }\right]
\right) $ be two central extensions of $K$ by $E,$ and let $\mu \in
Aut\left( E\right) =GL\left( E\right) $ and $\eta \in Aut\left( K\right),$
where $Aut\left( E\right) $ and $Aut\left( K\right) $ are the groups of
left-symmetric automorphisms of $E$ and $K,$ respectively. It is clear that
if $h\in L^{1}\left( K,E\right) ,$ then the linear mapping $\psi
:A\rightarrow A^{\prime }$ defined by $\psi \left( x,a\right) =\left( \eta
\left( x\right) ,\mu \left( a\right) +h\left( x\right) \right) $ is an
isomorphism provided $g^{\prime }\left( \eta \left( x\right) ,\eta \left(
y\right) \right) =\mu \left( g\left( x,y\right) \right) -\delta _{1}h\left(
x,y\right) $ for all $\left( x,y\right) \in K\times K,$ i.e. $\eta ^{\ast }
\left[ g^{\prime }\right] =\mu _{\ast }\left[ g\right] .$ This allows us to
define an action of the group $G=Aut\left( E\right) \times Aut\left(
K\right) $ on $H^{2}\left( K,E\right) $ by setting

\begin{equation}
\left( \mu ,\eta \right) .\left[ g\right] =\mu _{\ast }\eta ^{\ast }\left[ g
\right] ,  \label{action1}
\end{equation}
or equivalently, $\left( \mu ,\eta \right) .g\left( x,y\right) =\mu \left(
g\left( \eta \left( x\right) ,\eta \left( y\right) \right) \right) $ for all
$x,y\in K.$

Denoting the set of all exact central extensions of $K$ by $E$ by
\[
H_{ex}^{2}\left( K,E\right) =\left\{ \left[ g\right] \in H^{2}\left(
K,E\right) :I_{\left[ g\right] }=0\right\} ,
\]
and the orbit of $\left[ g\right] $ by $G_{\left[ g\right] },$ it turns out
that the following result is valid (see \cite{kim}).

\begin{proposition}
\label{prop3} Let $\left[ g\right] $ and $\left[ g^{\prime }\right] $ be two
classes in $H_{ex}^{2}\left( K,E\right) .$ Then, the central extensions $
\left( A,\left[ g\right] \right) $ and $\left( A^{\prime },\left[ g^{\prime}
\right] \right) $ are isomorphic if and only if $G_{\left[ g\right] }=G_{\left[ g^{\prime }\right] }.$ In other words, the classification of the
exact central extensions of $K$ by $E$ is, up to left-symmetric isomorphism,
the orbit space of $H_{ex}^{2}\left( K,E\right) $ under the natural action
of $G=Aut\left( E\right) \times Aut\left( K\right) .$
\end{proposition}

\subsection{Complexification of a real left-symmetric algebra}

Let $A$ be a real left-symmetric algebra of dimension $n,$ and let $A^{\mathbb{C}}$ denote the real vector space $A\oplus A.$ Let $J:A\oplus
A\rightarrow A\oplus A$ be the linear map on $A\oplus A$ defined by $J\left(
x,y\right) =\left( -y,x\right) .$ For $\alpha +i\beta \in \mathbb{C}$ and $x,x^{\prime },y,y^{\prime }\in A,$ we define
\begin{eqnarray}
\left( \alpha +i\beta \right) \left( x,y\right) &=&\left( \alpha x-\beta
y,\alpha y+\beta x\right), \label{cmplx1}
\\
\left( x,y\right) \cdot \left( x^{\prime },y^{\prime }\right) &=&\left(
xx^{\prime }-yy^{\prime },xy^{\prime }+yx^{\prime }\right).  \label{cmplx2}
\end{eqnarray}
We endow the set $A^{\mathbb{C}}$ with the componentwise addition,
multiplication by complex numbers defined by (\ref{cmplx1}), and the product
defined by (\ref{cmplx2}). It is then straightforward to verify that $A^{\mathbb{C}},$ when endowed with the product defined by (\ref{cmplx2}),
becomes a complex left-symmetric algebra called \textsl{the complexification
}of $A.$ The left-symmetric algebra $A$ can be identified with the set of
elements in $A^{\mathbb{C}}$ of the form $\left( x,0\right) ,$ where $x\in
A. $ If $e_{1},\ldots ,e_{n}$ is a basis of $A,$ then the elements $\left(
e_{1},0\right) ,\ldots ,\left( e_{n},0\right) $ form a basis of the complex
vector space $A^{\mathbb{C}}.$ It follows that $\dim _{\mathbb{C}}\left( A^{\mathbb{C}}\right) =\dim _{\mathbb{R}}\left( A\right) .$

Since $A^{\mathbb{C}}$ is a left-symmetric algebra, we know that the
commutator $\left[\left(x,y\right), \!\left( x^{\prime}\!,y^{\prime }\right)
\right]$ $=\left( x,y\right) \cdot \left( x^{\prime },y^{\prime }\right)
-\left( x^{\prime },y^{\prime }\right) \cdot \left( x,y\right) $ defines a
Lie algebra $\mathcal{G}^{\mathbb{C}}$ on $A^{\mathbb{C}}.$ Computing this
commutator, we get the following lemma.

\begin{lemma}
\label{cmplxlemma}The complex Lie algebra $\mathcal{G}^{\mathbb{C}}$
associated to the complex left-symmetric algebra $A^{\mathbb{C}}$ is
isomorphic to the complexification of the Lie algebra $\mathcal{G}$
associated to the left-symmetric algebra $A.$
\end{lemma}

Therefore, if $e_{1},\ldots ,e_{n}$ is a basis of $A,$ then the elements $\left( e_{1},0\right) ,\ldots ,\left( e_{n},0\right) $ form a basis of $
\mathcal{G}^{\mathbb{C}},$ and the structural constants of $\mathcal{G}^{
\mathbb{C}}$ are real since they coincide with the structural constants of $
\mathcal{G}$ in the basis $e_{1},\ldots ,e_{n}.$

\section{Left-symmetric structures on the oscillator algebra\label{oscillator section}}
Recall that the Heisenberg group $H_{3}$ is the 3-dimensional Lie
group diffeomorphic to $\mathbb{R\times C}$ with the group law
\[
(v_{1},z_{1})\cdot (v_{2},z_{2})=(v_{1}+v_{2}+\frac{1}{2}\mathop{\rm Im}(\overline{z_{1}}z_{2}),z_{1}+z_{2}),
\]
for all $v_{1},v_{2}\in \mathbb{R}$ and $z_{1},z_{2}\in \mathbb{C}$. Let $\lambda >0,$ and let $G=\mathbb{Rn}H_{3}$ be equipped with the group law
\[
(t_{1},v_{1},z_{1})\cdot (t_{2},v_{2},z_{2})=(t_{1}+t_{2},v_{1}+v_{2}+\frac{1}{2}\mathop{\rm Im}(\overline{z_{1}}z_{2}e^{i\lambda
t_{1}}),z_{1}+z_{2}e^{i\lambda t_{1}}),
\]
for all $t_{1},t_{2}\in \mathbb{R}$ and $(v_{1},z_{1}),(v_{2},z_{2})\in
H_{3}.$ This is a 4-dimensional Lie group with Lie algebra $\mathcal{G}$
having a basis $\{e_{1},e_{2},e_{3},e_{4}\}$ such that
\[
\lbrack e_{1},e_{2}]=e_{3},\ [e_{4},e_{1}]=\lambda e_{2},\ \left[ e_{4},e_{2}\right] =-\lambda e_{1},
\]
and all the other brackets are zero. It follows that the derived series is
given by
\[
\mathcal{D}^{1}\mathcal{G}=[\mathcal{G},\mathcal{G}]=span\{e_{1},e_{2},e_{3}\},\ \mathcal{D}^{2}\mathcal{G}=span\{e_{3}\},\ \mathcal{D}^{3}\mathcal{G}=\{0\},
\]
and therefore $\mathcal{G}$ is a (non-nilpotent) 3-step solvable Lie
algebra. When $\lambda =1,$ $G$ is known as the \emph{oscillator group. }We
will denote it by $O_{4}$, and we shall denote its Lie algebra by $\mathcal{O}_{4}$ and call it the \emph{oscillator algebra.}

From now on, $A_{4}$ will be a complete real left-symmetric algebra whose
associated Lie algebra is $\mathcal{O}_{4}.$ We begin by proving the
following proposition which will be crucial to the classification of
complete left-symmetric structures on $\mathcal{O}_{4}.$

\begin{proposition}
$A_{4}$ is not simple (i.e., $A_{4}$ contains a proper two-sided ideal).
\end{proposition}

\begin{proof}
Assume to the contrary that $A_{4}$ is simple, and let $A_{4}^{\mathbb{C}}$
be its complexification. By \cite{kbm}, Lemma 2.10, it follows that $A_{4}^{\mathbb{C}}$ is either simple or a direct sum of two simple ideals having
the same dimension. If $A_{4}^{\mathbb{C}}$ is simple, then we can apply
Proposition 5.1 in \cite{burde} to deduce that, being simple and complete, $A_{4}^{\mathbb{C}}$ is necessarily isomorphic to the complex left-symmetric
algebra $B_{4}$ having a basis $\left\{ e_{1},e_{2},e_{3},e_{4}\right\} $
such that
\begin{eqnarray*}
e_{1}\cdot e_{2} &=&e_{2}\cdot e_{1}=e_{4},~e_{2}\cdot e_{3}=2e_{1},~ \\
e_{3}\cdot e_{2} &=&e_{4}\cdot e_{1}=e_{1},~e_{4}\cdot
e_{2}=-e_{2},~e_{4}\cdot e_{3}=2e_{3},
\end{eqnarray*}
and all other products are zero. It follows that the Lie algebra $\mathcal{G}_{4}$ associated to $B_{4}$ admits a basis $\left\{
e_{1},e_{2},e_{3},e_{4}\right\} $ such that
\[
\left[ e_{2},e_{3}\right] =\left[ e_{4},e_{1}\right] =e_{1},~\left[
e_{2},e_{4}\right] =e_{2},~\left[ e_{3},e_{4}\right] =-2e_{3}.
\]
This leads to a contradiction since, according to Lemma \ref{cmplxlemma}, $\mathcal{G}_{4}$ should be isomorphic to the complexification of the Lie
algebra $\mathcal{O}_{4},$ but this is obviously not the case. This
contradiction shows that $A_{4}^{\mathbb{C}}$ cannot be simple.

If $A_{4}^{\mathbb{C}}$ is a direct sum of two simple ideals having the same
dimension, say $A_{4}^{\mathbb{C}}=A_{1}\oplus A_{2},$ it follows that $\dim
A_{1}=\dim A_{2}=\frac{1}{2}\dim A_{4}^{\mathbb{C}}=2.$ In this case, by
Corollary 4.1 in \cite{burde}, $A_{1}$ and $A_{2}$ are both isomorphic to
the unique two-dimensional complex simple left-symmetric algebra having a
basis
\[
B_{2}=\left\langle e_{1},e_{2}:e_{1}\cdot e_{1}=2e_{1},~e_{1}\cdot
e_{2}=e_{2},~e_{2}\cdot e_{2}=e_{1}\right\rangle .
\]
This is a contradiction, since $A_{1}$ and $A_{2}$ are complete but $B_{2}$ is
not. This contradiction shows that $A_{4}^{\mathbb{C}}$ cannot be
direct sum of two simple ideals. We deduce that $A_{4}$ is not simple, and
this completes the proof of the proposition.
\end{proof}

Before we return to the algebra $A_{4},$ we need to give the following
lemmas.

\begin{lemma}
\label{lemma0} Let $A$ be a left-symmetric algebra with Lie algebra $
\mathcal{G}$, and $R$ a two-sided ideal in $A.$ Then, the Lie algebra $
\mathcal{R}$ associated to $R$ is an ideal in $\mathcal{G}$.
\end{lemma}

\begin{proof}
Let $x\in \mathcal{R}$ and $y\in \mathcal{G}.$ Since $R$ is a two-sided ideal, then $x\cdot y$
and $y\cdot x$ belong to $R.$ It follows that $\left[ x,y\right] =x\cdot
y-y\cdot x\in R,$ and therefore $\mathcal{R}$ is an ideal in $\mathcal{G}.$
\end{proof}

\begin{lemma}
\label{lemma1} The oscillator algebra $\mathcal{O}_{4}$ contains only two
proper ideals which are $Z\left( \mathcal{O}_{4}\right) \cong \mathbb{R}$
and $\left[ \mathcal{O}_{4},\mathcal{O}_{4}\right] \cong \mathcal{H}_{3}.$
\end{lemma}

\begin{proof}
It is clear that $\mathcal{Z}\left( \mathcal{O}_{4}\right) \cong \mathbb{R}$
and $\left[ \mathcal{O}_{4},\mathcal{O}_{4}\right] \cong \mathcal{H}_{3}$
are proper ideals in $\mathcal{O}_{4}.$ If $\mathcal{I}$ is a proper ideal
in $\mathcal{O}_{4},$ then $\mathcal{I}$ should be unimodular. If $\dim
\left( \mathcal{I}\right) =1,$ then $\mathcal{I}$ is isomorphic to $\mathcal{
Z}\left( \mathcal{O}_{4}\right) \cong \mathbb{R}.$ If $\dim \left( \mathcal{I
}\right) =2$, then being unimodular, $\mathcal{I}$ is isomorphic to $\mathbb{
R}^{2}.$ In particular, $\mathcal{I}$ contains $\mathcal{Z}\left( \mathcal{O}
_{4}\right)$ and thus $\mathcal{O}_{4}/\mathcal{I}$ is abelian, a
contradiction since $\mathcal{O}_{4}$ is not nilpotent. Hence, $\mathcal{O}
_{4}$ contains no two-dimensional ideals. If $\dim \left( \mathcal{I}\right)
=3$, then being unimodular and solvable, $\mathcal{I}$ is isomorphic to
either $\mathcal{H}_{3},$ the Lie algebra $\mathcal{E}\left( 2\right) $ of
the group of the rigid motions of the plane, or the Lie algebra $\mathcal{E}
\left( 1,1\right) $ of the group of the rigid motions of the Minkowski
plane. However, it is straightforward to show that $\mathcal{O}_{4}$ cannot
be obtained as an extension of $\mathcal{E}\left( 2\right) $ or $\mathcal{E}
\left( 1,1\right) .$ We have therefore proved the lemma.
\end{proof}

By the above proposition, $A_{4}$ is not simple and hence it has a proper
two-sided ideal $I,$ so we get a short exact sequence of complete
left-symmetric algebras
\begin{equation}
0\rightarrow I\overset{i}{\rightarrow }A_{4}\overset{\pi }{\rightarrow }
J\rightarrow 0.  \label{seq1.5}
\end{equation}
In fact, according to Lemma \ref{completelemma}, the completeness of $I$ and $J$ comes
from that of $A_{4}$. If $\mathcal{I}$ is the Lie
subalgebra associated to $I$ then, by Lemma \ref{lemma0}, $\mathcal{I}$ is
an ideal in $\mathcal{O}_{4}$. From Lemma \ref{lemma1}, it follows that
there are two cases to consider according to whether $\mathcal{I}$ is
isomorphic to $\mathcal{H}_{3}$ or $\mathbb{R}$. Next, we will focus on the
case where\textbf{\ }$\mathcal{I}$\textbf{\ }is isomorphic to $\mathcal{H}
_{3}\cong \left[ \mathcal{O}_{4},\mathcal{O}_{4}\right] .$ In this case,$\ $
the short exact sequence (\ref{seq1.5}) becomes
\begin{equation}
0\rightarrow I_{3}\overset{i}{\rightarrow }A_{4}\overset{\pi }{\rightarrow }
I_{0}\rightarrow 0,  \label{seq2}
\end{equation}
where $I_{3}$ is a complete $3$-dimensional left-symmetric algebra whose Lie
algebra is $\mathcal{H}_{3},$ and $I_{0}=\left\{ e_{0}:e_{0}\cdot
e_{0}=0\right\} $ the trivial one-dimensional real left-symmetric algebra.
It is easy to prove the following proposition (cf. \cite[Theorem 3.5]{friedgold}).

\begin{proposition}
\label{prop1}Up to left-symmetric isomorphism, the complete left-symmetric
structures on the Heisenberg algebra $\mathcal{H}_{3}$ are classified as
follows: There is a basis $\left\{ e_{1},e_{2},e_{3}\right\} $ of $\mathcal{H
}_{3}$ relative to which the left-symmetric product is given by one of the
following classes:

\begin{itemize}
\item[$($i$)$] $e_{1}\cdot e_{1}=pe_{3},$ $e_{2}\cdot e_{2}=qe_{3},$ $e_{1}\cdot
e_{2}=\frac{1}{2}e_{3},$ $e_{2}\cdot e_{1}=-\frac{1}{2}e_{3},$ where $p,q\in
\mathbb{R}.$

\item[$($ii$)$] $e_{1}\cdot e_{2}=me_{3},$ $e_{2}\cdot e_{1}=\left( m-1\right)
e_{3},$ $e_{2}\cdot e_{2}=e_{1},$ where $m\in \mathbb{R}.$
\end{itemize}
\end{proposition}

\begin{remark}
It is noticeable that the left-symmetric products on $\mathcal{H}_{3}$
belonging to class (i) in Proposition \ref{prop1} are obtained by central
extensions (in the sense of fixed in Subsection 3.1) of $\mathbb{R}^{2}$
endowed with some complete left-symmetric structure by $I_{0}.$ However, the
left-symmetric products on $A_{3}$ belonging to class (ii) are obtained by
central extensions of the non-abelian two-dimensional Lie algebra $\mathcal{G
}_{2}$ endowed with its unique complete left-symmetric structure by $I_{0}.$
\end{remark}

Now we return to the short exact sequence (\ref{seq2}). First, let $\sigma
:I_{0}\rightarrow A_{4}$ be a section, and set $\sigma \left( e_{0}\right)
=x_{0}\in A_{4}.$ Define two linear maps $\lambda ,$ $\rho \in End\left(
I_{3}\right) $ by putting $\lambda \left( y\right) =x_{0}\cdot y$ and $\rho
\left( y\right) =y\cdot x_{0},$ and put $e=x_{0}\cdot x_{0}$ (clearly $e\in
I_{3}$). Let $g:I_{0}\times I_{0}\rightarrow I_{3}$ be the bilinear map
defined by $g\left( e_{0},e_{0}\right) =e.$ It is obvious, using the
notation of Subsection 3.1, to verify that $\delta _{2}g=0,$ i.e. $g\in
Z_{\lambda ,\rho }^{2}\left( I_{0},I_{3}\right) $. The extended
left-symmetric product on $I_{3}\oplus I_{0}$ given by (\ref{eq5}) turns out
to take the simplified form $\left( x,ae_{0}\right) \cdot \left(
y,be_{0}\right) =\left( x\cdot y+a\lambda \left( y\right) +b\rho \left(
x\right) +abe,0\right) ,$ for all $x,y\in I_{3}$ and $a,b\in \mathbb{R}.$
The conditions in Theorem \ref{thm1} can be simplified to the following
conditions:
\begin{eqnarray}
\lambda \left( x\cdot y\right) &=&\lambda \left( x\right) \cdot y+x\cdot
\lambda \left( y\right) -\rho \left( x\right) \cdot y  \label{eq6.1} \\
\rho \left( \left[ x,y\right] \right) &=&x\cdot \rho \left( y\right) -y\cdot
\rho \left( x\right)  \label{eq6.2} \\
\left[ \lambda ,\rho \right] +\rho ^{2} &=&R_{e}  \label{eq6.3}
\end{eqnarray}

Let $\phi :\mathbb{R}\rightarrow End\left( \mathcal{H}_{3}\right) $ be the
linear map defined by formula (\ref{eq0.5}). As we mentioned in Remark \ref{remark1}, $\mathbb{R}$ acts on $\mathcal{H}_{3}$ by derivations, that is, $
\phi :\mathbb{R}\rightarrow Der\left( \mathcal{H}_{3}\right) .$ In
particular, we deduce in view of (\ref{eq1}) that $\lambda =D+\rho $ for
some derivation $D$ of $\mathcal{H}_{3}.$ The derivations of $\mathcal{H}
_{3} $ are given by the following lemma, whose proof is straightforward and
is therefore omitted.

\begin{lemma}
\label{lemma2}In a basis $\left\{ e_{1},e_{2},e_{3}\right\} $ of $\mathcal{H}
_{3}$ satisfying $\left[ e_{1},e_{2}\right] =e_{3}$, a derivation $D$ of $
\mathcal{H}_{3}$ takes the form
\[
D=\left(
\begin{array}{ccc}
a_{1} & b_{1} & 0 \\
a_{2} & b_{2} & 0 \\
a_{3} & b_{3} & a_{1}+b_{2}
\end{array}
\right).
\]
\end{lemma}

On the other hand, observe that $\left( x,ae_{0}\right) \in T\left(
A_{4}\right) $ if and only if $\left( x,ae_{0}\right) \cdot \left(
y,be_{0}\right) =\left( 0,0\right) $ for all $\left( y,be_{0}\right) \in
I_{3}\oplus I_{0}$, or equivalently, $x\cdot y+a\lambda \left( y\right)
+b\rho \left( x\right) +abe=0$ for all $\left( y,be_{0}\right) \in
I_{3}\oplus I_{0}.$ Since $y\ $and $b$ are arbitrary, we conclude that this
is also equivalent to say that $\left( L_{x}\right) _{\mid
_{A_{3}}}=-a\lambda $ and $\rho \left( x\right) =-ae.$ In particular, an
element $x\in I_{3}$ belongs to $T\left( A_{4}\right) $ if and only if $
\left( L_{x}\right) _{\mid _{I_{3}}}=0$ and $\rho \left( x\right) =0,$ or
equivalently,
\begin{equation}
I_{3}\cap T\left( A_{4}\right) =T\left( I_{3}\right) \cap \ker \rho .
\label{eq7}
\end{equation}

The following lemma will be crucial for the next section.

\begin{lemma}
\label{lemma3}The center $C\left( A_{4}\right) =T\left( A_{4}\right) \cap
Z\left( \mathcal{O}_{4}\right) $ is non-trivial.
\end{lemma}

\begin{proof}
In view of Proposition \ref{prop1}, we have to consider two cases.

\textbf{Case 1.} Assume that there is a basis $\left\{
e_{1},e_{2},e_{3}\right\} $ of $\mathcal{H}_{3}$ relative to which the
left-symmetric product of $I_{3}$ is given by : $e_{1}\cdot e_{1}=pe_{3},$ $
e_{2}\cdot e_{2}=qe_{3},$ $e_{1}\cdot e_{2}=\frac{1}{2}e_{3},$ $e_{2}\cdot
e_{1}=-\frac{1}{2}e_{3},$ where $p,q\in \mathbb{R}.$
Substituting $x=e_{1}$ and $y=e_{2}$ into (\ref{eq6.2}), we find that the
operator $\rho $ takes the form
\[
\rho =\left(
\begin{array}{ccc}
\alpha _{1} & \beta _{1} & 0 \\
\alpha _{2} & \beta _{2} & 0 \\
\alpha _{3} & \beta _{3} & \gamma _{3}
\end{array}
\right) ,
\]
with $\gamma _{3}=p\beta _{1}-q\alpha _{2}+\frac{1}{2}\left( \alpha
_{1}+\beta _{2}\right) .$ Since $\lambda =D+\rho $ for some $D\in \mathcal{H}
_{3}$, we use Lemma~\ref{lemma2} to deduce that
\[
\lambda =\left(
\begin{array}{ccc}
\alpha _{1}+a_{1} & \beta _{1}+b_{1} & 0 \\
\alpha _{2}+a_{2} & \beta _{2}+b_{2} & 0 \\
\alpha _{3}+a_{3} & \beta _{3}+b_{3} & \gamma _{3}+a_{1}+b_{2}
\end{array}
\right) .
\]
Since $\left( L_{e_{3}}\right) _{\mid _{I_{3}}}=0$ and $e\in I_{3}$, then (\ref{eq6.3}), when applied to $e_{3}$, gives
\[
\gamma _{3}^{2}e_{3}=e_{3}\cdot e=0,
\]
from which we get $\gamma _{3}=0,$ i.e., $\rho \left( e_{3}\right) =0.$ It
follows from (\ref{eq7}) that $e_{3}\in T\left( A_{4}\right) .$ Since $
Z\left( \mathcal{O}_{4}\right) =\mathbb{R}e_{3},$ we deduce that $C\left(
A_{4}\right) =T\left( A_{4}\right) \cap Z\left( \mathcal{O}_{4}\right) \neq
0,$ as required.

\textbf{Case 2.} Assume now that there is a basis $\left\{
e_{1},e_{2},e_{3}\right\} $ of $\mathcal{H}_{3}$ relative to which the
left-symmetric product of $I_{3}$ is given by : $e_{1}\cdot e_{2}=me_{3},$ $
e_{2}\cdot e_{1}=\left( m-1\right) e_{3},$ $e_{2}\cdot e_{2}=e_{1},$ where $
m $ is a real number.

Substituting successively $x=e_{1},~y=e_{2}$ and $x=e_{2},~y=e_{3}$ into
equation (\ref{eq6.2}), we find that the operator $\rho $ takes the form
\begin{equation}
\rho =\left(
\begin{array}{ccc}
\alpha _{1} & \beta _{1} & -\alpha _{2} \\
\alpha _{2} & \beta _{2} & 0 \\
\alpha _{3} & \beta _{3} & m\beta _{2}-\left( m-1\right) \alpha _{1}
\end{array}
\right) ,  \label{eq8}
\end{equation}
with~$\left( m-1\right) \alpha _{2}=0.$

We claim that\textbf{\ }$\alpha _{2}=0.$ To prove this, let us assume to the
contrary that $\alpha _{2}\neq 0.$ It follows that $m=1,$ and therefore
\begin{eqnarray*}
\rho \left( e_{3}\right) &=&-\alpha _{2}e_{1}+\beta _{2}e_{3} \\
\rho ^{2}\left( e_{3}\right) &=&-\alpha _{2}\left( \alpha _{1}+\beta
_{2}\right) e_{1}-\alpha _{2}^{2}e_{2}+\left( \beta _{2}^{2}-\alpha
_{2}\alpha _{3}\right) e_{3}
\end{eqnarray*}

Since $\alpha _{2}\neq 0,$ we deduce that $e_{3},$ $\rho \left( e_{3}\right)
,$ $\rho ^{2}\left( e_{3}\right) $ form a basis of $I_{3}.$ Since $\rho $ is
nilpotent (by completeness of the left-symmetric structure), it follows that
$\rho ^{3}\left( e_{3}\right) =0.$ In other words, $\rho $ has the form
\[
\rho =\left(
\begin{array}{rcc}
0 & 0 & 1 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right) ,
\]
with respect to the basis $e_{1}^{\prime }=-\rho \left( e_{3}\right) ,$ $
e_{2}^{\prime }=\rho ^{2}\left( e_{3}\right) ,$ $e_{3}^{\prime }=-e_{3}.$

Using the fact that $\alpha _{1}+2\beta _{2}=0$ which follows from the
identity $\rho ^{3}\left( e_{3}\right) =0,$ we see that $e_{1}^{\prime
}\cdot e_{2}^{\prime }=\alpha _{2}^{3}e_{3}^{\prime },$ $e_{2}^{\prime
}\cdot e_{2}^{\prime }=\alpha _{2}^{3}e_{1}^{\prime },$ and all other
products are zero.

For simplicity, assume without loss of generality that $\alpha _{2}=1.$
Since $\lambda =D+\rho $ for some $D\in \mathcal{H}_{3}$, Lemma \ref{lemma2}
tells us that, with respect to the basis $e_{1}^{\prime },e_{2}^{\prime },$ $
e_{3}^{\prime }$, the operator $\lambda $ takes the form
\[
\lambda =\left(
\begin{array}{ccc}
a_{1} & b_{1} & 1 \\
a_{2}-1 & b_{2} & 0 \\
a_{3} & b_{3} & a_{1}+b_{2}
\end{array}
\right) .
\]
Applying formula (\ref{eq6.3}) to $e_{3}^{\prime }$ and recalling that $
e_{3}^{\prime }\cdot e=0$ since $\mathbf{e}\in I_{3},$ we deduce that $
a_{2}=1$ and $b_{2}=a_{3}=0.$ Then, substituting $x=y=e_{2}^{\prime }$ into
equation (\ref{eq6.1}), we get $a_{1}=b_{1}=0.$ Thus, the form of $\lambda $
reduces to
\[
\lambda =\left(
\begin{array}{ccc}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & b_{3} & 0
\end{array}
\right) .
\]
Now, by setting $\mathbf{e}=ae_{1}+be_{2}+ce_{3}$ and applying (\ref{eq6.3})
to $e_{1},$ we get that $b_{3}=-b.$ By using (\ref{eq5}), we deduce that
the nonzero left-symmetric products are
\begin{eqnarray*}
e_{1}^{\prime }\cdot e_{2}^{\prime } &=&e_{3}^{\prime },\ \ e_{2}^{\prime
}\cdot e_{2}^{\prime }=e_{1}^{\prime }, \\
e_{1}^{\prime }\cdot e_{4}^{\prime } &=&-e_{2}^{\prime },\ \ e_{4}^{\prime
}\cdot e_{2}^{\prime }=-be_{3}^{\prime } \\
e_{3}^{\prime }\cdot e_{4}^{\prime } &=&e_{4}^{\prime }\cdot e_{3}^{\prime
}=e_{1}^{\prime },\ \ e_{4}^{\prime }\cdot e_{4}^{\prime }=\mathbf{e}.
\end{eqnarray*}

This implies, in particular, that $\dim \left[ \mathcal{O}_{4},\mathcal{O}
_{4}\right] =\dim \left[ A_{4},A_{4}\right] =2,$ a contradiction. It follows
that $\alpha _{2}=0,$ as desired.

We now return to (\ref{eq8}). Since $\alpha _{2}=0,$ we have
\[
\rho =\left(
\begin{array}{ccc}
\alpha _{1} & \beta _{1} & 0 \\
0 & \beta _{2} & 0 \\
\alpha _{3} & \beta _{3} & m\beta _{2}-\left( m-1\right) \alpha _{1}
\end{array}
\right) ,
\]
and since $\lambda =D+\rho $ for some $D\in \mathcal{H}_{3}$ then, in view
of Lemma \ref{lemma2}, the operator $\lambda $ takes the form
\[
\lambda =\left(
\begin{array}{ccc}
\alpha _{1}+a_{1} & \beta _{1}+b_{1} & 0 \\
a_{2} & \beta _{2}+b_{2} & 0 \\
\alpha _{3}+a_{3} & \beta _{3}+b_{3} & a_{1}+b_{2}+m\beta _{2}-\left(
m-1\right) \alpha _{1}
\end{array}
\right) .
\]
Once again, by applying (\ref{eq6.3}) to $e_{3}$ and recalling that $
e_{3}\cdot e=0$ since $\mathbf{e}\in I_{3},$ we deduce that $\left( m\beta
_{2}-\left( m-1\right) \alpha _{1}\right) ^{2}=0,$ thereby showing that $
\rho \left( e_{3}\right) =0.$ Now, in view of (\ref{eq7}) we get $e_{3}\in
T\left( A_{4}\right) ,$ and since $Z\left( \mathcal{O}_{4}\right) =\mathbb{R}
e_{3}$ we deduce that $C\left( A_{4}\right) =T\left( A_{4}\right) \cap
Z\left( \mathcal{O}_{4}\right) \neq 0,$ as desired. This completes the proof
of the lemma.
\end{proof}

\section{Classification}

We know from Section 4 that $A_{4}$\ has a proper two-sided ideal $I$\ which
is isomorphic to either the trivial one-dimensional real left-symmetric
algebra $I_{0}=\left\{ e_{0}:e_{0}\cdot e_{0}=0\right\} $ or a $3$
-dimensional left-symmetric algebra $I_{3}$\ (as described in Proposition
\ref{prop1}) whose associated Lie algebra is the Heisenberg algebra $
\mathcal{H}_{3}$. In the case where $I\cong I_{3},$ we know by Lemma \ref{lemma3}
that $C\left( A_{4}\right) \neq \left\{ 0\right\} .$\ Since in our
situation $\dim Z\left( \mathcal{O}_{4}\right) =1,$\ it follows that $
C\left( A_{4}\right) \cong I_{0},$ so that we have a central short exact
sequence of left-symmetric algebras of the form
\begin{equation}
0\rightarrow I_{0}\rightarrow A_{4}\rightarrow I_{3}\rightarrow 0.
\label{seq3}
\end{equation}
In general, one has that the center of a left-symmetric algebra is a part of
the center of the associated Lie algebra, and therefore the following lemma
is proved.

\begin{lemma}
\label{lemma5}The Lie algebra associated to $I_{3}$ is isomorphic to the Lie
algebra $\mathcal{E}\left( 2\right) $ of the group of Euclidean motions of
the plane.
\end{lemma}

Recall that $\mathcal{E}\left( 2\right) $ is solvable non-nilpotent and has
a basis$\ \left\{ e_{1},e_{2},e_{3}\right\} $ which satisfies $\left[
e_{1},e_{2}\right] =e_{3}$ and $\left[ e_{1},e_{3}\right] =-e_{2}$.

In the case where $I\cong I_{0}$, we know by Lemma \ref{lemma0} that the
associated Lie algebra is $\mathcal{I}\cong \mathbb{R}.$ Since, by Lemma \ref{lemma1}, $\mathcal{O}_{4}$ has only two proper ideals which are $Z\left(
\mathcal{O}_{4}\right) \cong \mathbb{R}$ and $\left[ \mathcal{O}_{4},
\mathcal{O}_{4}\right] \cong \mathcal{H}_{3},$ it follows that $\mathcal{I}
\cong \mathbb{R}$ coincides with the center $Z\left( \mathcal{O}_{4}\right)
. $ We deduce from this that, if $\mathcal{J}$ denotes the Lie algebra of
the left-symmetric algebra $J$ in the short exact sequence (\ref{seq1.5}),
then $\mathcal{J}$ is isomorphic to $\mathcal{E}\left( 2\right) .$
Therefore, we have a short sequence of left-symmetric algebras which looks
like (\ref{seq3}), except that it would not necessarily be central. But, as
we will see a little later, this is necessarily a central extension (i.e., $
I\cong C\left( A_{4}\right) \cong I_{0}$).

To summarize, each complete left-symmetric structure on $\mathcal{O}_{4}$\
may be obtained by an extension of a complete $3$-dimensional left-symmetric
algebra $A_{3}$ whose associated Lie algebra is $\mathcal{E}\left( 2\right) $
by $I_{0}.$ Next, we shall determine all the complete left-symmetric
structures on $\mathcal{E}\left( 2\right) .$ These are described by the
following lemma that we state without proof (see \cite{friedgold}, Theorem
4.1).

\begin{lemma}
\label{lemma6}Up to left-symmetric isomorphism, any complete left-symmetric
structure on $\mathcal{E}\left( 2\right) $ is isomorphic to the following
one which is given in a basis $\left\{ e_{1},e_{2},e_{3}\right\} $ of $
\mathcal{E}\left( 2\right) $ by the relations
$ e_{1}\cdot e_{2}=e_{3},~e_{1}\cdot e_{3}=-e_{2},~e_{2}\cdot e_{2}=e_{3}\cdot
e_{3}=\varepsilon e_{1}.$

There are exactly two non-isomorphic conjugacy classes according to whether $
\varepsilon =0$ or $\varepsilon \neq 0$.
\end{lemma}

From now on, $A_{3}$ will denote the vector space $\mathcal{E}\left(
2\right) $ endowed with one of the complete left-symmetric structures
described in Lemma \ref{lemma6}. The extended Lie bracket on $\mathcal{E}
\left( 2\right) \oplus \mathbb{R}$ is given by
\begin{equation}
\left[ \left( x,a\right) ,\left( y,b\right) \right] =\left( \left[ x,y\right]
,\omega \left( x,y\right) \right) ,  \label{eq10}
\end{equation}
with $\omega \in Z^{2}\left( \mathcal{E}\left( 2\right) ,\mathbb{R}\right) .$
The extended left-symmetric product on $A_{3}\oplus I_{0}$ is given by
\begin{equation}
\left( x,ae_{0}\right) \cdot \left( y,be_{0}\right) =\left( x\cdot
y,b\lambda _{x}\left( e_{0}\right) +a\rho _{y}\left( e_{0}\right) +g\left(
x,y\right) \right) ,  \label{eq11}
\end{equation}
with $\lambda ,$ $\rho :A_{3}\rightarrow End\left( I_{0}\right) $ and $g\in
Z_{\lambda ,\rho }^{2}\left( A_{3},I_{0}\right)$.

As we have noticed in Section 3, $I_{0}$ is an $A_{3}$-bimodule, or
equivalently, the conditions in Theorem \ref{thm1} simplify to the following
conditions:
\begin{itemize}
\item[$($i$)$] $\lambda _{\left[ x,y\right] }=0,$

\item[$($ii$)$] $\rho _{x\cdot y}=\rho _{y}\circ \rho _{x},$

\item[$($iii$)$] $g( x,y\cdot z) -g( y,x\cdot z) +\lambda_{x}( g( y,z) ) -\lambda _{y}( g( x,z)) -g( [ x,y] ,z) -\rho _{z}( g( x,y) -g( y,x) ) =0.$
\end{itemize}

By using (\ref{eq10}) and (\ref{eq11}), we deduce from $\left[ \left(
x,a\right) ,\left( y,b\right) \right] =\left( x,ae_{0}\right) \cdot \left(
y,be_{0}\right) -\left( y,be_{0}\right) \cdot \left( x,ae_{0}\right) $ that
\begin{equation}
\omega \left( x,y\right) =g\left( x,y\right) -g\left( y,x\right) \text{ and } \lambda =\rho .  \label{eq12}
\end{equation}
By applying identity (ii) above to $e_{i}\cdot e_{i},$ $1\leq i\leq 3,$ we
deduce that $\rho =0,$ and a fortiori $\lambda =0.$ In other words, the
extension $A_{4}$ is always central (i.e., $I\cong C\left( A_{4}\right) $
even in the case where $\mathcal{I}\cong \mathbb{R}$). In fact, we have

\begin{claim}
\label{claim2}The extension $0\rightarrow I_{0}\rightarrow A_{4}\rightarrow
A_{3}\rightarrow 0$ is exact.
\end{claim}

\begin{proof}
We recall from Subsection 3.1 that the extension given by the short sequence
(\ref{seq3}) is exact, i.e., $i\left( I_{0}\right) =C\left( A_{4}\right) $,
if and only if $I_{\left[ g\right] }=0,$ where
\[
I_{\left[ g\right] }=\left\{ x\in A_{3}:x\cdot y=y\cdot x=0 \text{ and }
g\left( x,y\right) =g\left( y,x\right) =0, \text{ for all }y\in
A_{3}\right\} .
\]
To show that $I_{\left[ g\right] }=0$, let $x$ be an arbitrary element in $
I_{\left[ g\right] }$, and put $x=ae_{1}+be_{2}+ce_{3}\in I_{\left[ g\right]
}.$ Now, by computing all the products $x\cdot e_{i}=e_{i}\cdot x=0,$ $1\leq
i\leq 3,$ we easily deduce that $x=0.$
\end{proof}

Our aim is to classify complete left-symmetric structures on $\mathcal{O}
_{4},$ up to left-symmetric isomorphisms. By Proposition \ref{prop3}, the
classification of exact central extensions of $A_{3}$ by $I_{0}$ is nothing
but the orbit space of $H_{ex}^{2}\left( A_{3},I_{0}\right) $ under the
natural action of $G=Aut\left( I_{0}\right) \times Aut\left( A_{3}\right) .$
Accordingly, we must compute $H_{ex}^{2}\left( A_{3},I_{0}\right) .$ Since $
I_{0}$ is a trivial $A_{3}$-bimodule, we see first from (\ref{delta1}) and (\ref{delta2}) that the coboundary operator $\delta $ simplifies as follows:
\[
\delta _{1}h\left( x,y\right) =-h\left( x\cdot y\right), \quad \delta
_{2}g\left( x,y,z\right) =g\left( x,y\cdot z\right) -g\left( y,x\cdot
z\right) -g\left( \left[ x,y\right] ,z\right) ,
\]
where $h\in L^{1}\left( A_{3},I_{0}\right) $ and $g\in L^{2}\left(
A_{3},I_{0}\right) .$

In view of Lemma \ref{lemma6}, there are two cases to be
considered.

\textbf{Case 1.} $A_{3}=\left\langle e_{1},e_{2},e_{3}:e_{1}\cdot
e_{2}=e_{3},~e_{1}\cdot e_{3}=-e_{2}\right\rangle .$

In this case, using the first formula above for $\delta _{1}$, we get
\[
\delta _{1}h=\left(
\begin{array}{ccc}
0 & h_{12} & h_{13} \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right),
\]
where $h_{12}=-h\left( e_{3}\right) $ and $h_{13}=h\left( e_{2}\right) .$
Similarly, using the second formula above for $\delta _{2}$, we verify
easily that if $g$ is a cocycle (i.e. $\delta _{2}g=0$) and $g_{ij}=g\left(
e_{i},e_{j}\right) $, then
\[
g=\left(
\begin{array}{ccc}
g_{11} & g_{12} & g_{13} \\
0 & g_{22} & g_{23} \\
0 & -g_{23} & g_{22}
\end{array}
\right),
\]
that is, $g_{21}=g_{31}=0,$ $g_{32}=-g_{23},$ and $g_{33}=g_{22}.$ We deduce
that, in the basis above, the class $\left[ g\right] \in H^{2}\left( A_{3},
\mathbb{R}\right) $ of a cocycle $g$ takes the simplified form
\[
g=\left(
\begin{array}{ccc}
\alpha & 0 & 0 \\
0 & \beta & \gamma \\
0 & -\gamma & \beta
\end{array}
\right) .
\]
We can now determine the extended left-symmetric structure on $A_{4}.$ By
setting $\widetilde{e}_{i}=\left( e_{i},0\right) $, $1\leq i\leq 3,$ and $
\widetilde{e}_{4}=\left( 0,1\right) ,$ and using formula (\ref{eq11}) which
(since $\lambda =\rho =0$) reduces to
\begin{equation}
\left( x,ae_{0}\right) \cdot \left( y,be_{0}\right) =\left( x\cdot y,g\left(
x,y\right) \right) ,  \label{eq11bis}
\end{equation}
we obtain
\begin{eqnarray}
\widetilde{e}_{1}\cdot \widetilde{e}_{1} &=&\alpha \widetilde{e}_{4},\
\widetilde{e}_{2}\cdot \widetilde{e}_{2}=\widetilde{e}_{3}\cdot \widetilde{e}
_{3}=\beta \widetilde{e}_{4}\   \notag \\
\widetilde{e}_{1}\cdot \widetilde{e}_{2} &=&\widetilde{e}_{3},\ \ \widetilde{
e}_{1}\cdot \widetilde{e}_{3}=-\widetilde{e}_{2},  \label{eq16} \\
\widetilde{e}_{2}\cdot \widetilde{e}_{3} &=&\gamma \widetilde{e}_{4},\ \
\widetilde{e}_{3}\cdot \widetilde{e}_{2}=-\gamma \widetilde{e}_{4},  \notag
\end{eqnarray}
and all the other products are zero. We observe here that we should have $
\gamma \neq 0,$ given that the underlying Lie algebra is $\mathcal{O}_{4}.$
We denote by $A_{4}\left( \alpha ,\beta ,\gamma \right) $ the Lie algebra $
\mathcal{O}_{4}$ endowed with the above complete left-symmetric product.

Let now $A_{4}\left( \alpha ,\beta ,\gamma \right) $ and $A_{4}\left( \alpha
^{\prime },\beta ^{\prime },\gamma ^{\prime }\right) $ be two arbitrary
left-symmetric structures on $\mathcal{O}_{4}$ given as above, and let $
\left[ g\right] $ and $\left[ g^{\prime }\right] $ be the corresponding
classes in $H_{ex}^{2}\left( A_{3},I_{0}\right) $. By Proposition \ref{prop3},
we know that $A_{4}\left( \alpha ,\beta ,\gamma \right) $ is isomorphic to
$A_{4}\left( \alpha ^{\prime },\beta ^{\prime },\gamma ^{\prime }\right) $
if and only if the exists $\left( \mu ,\eta \right) \in Aut\left(
I_{0}\right) \times Aut\left( A_{3}\right) $ such that for all $x,y\in
A_{3}, $ we have
\begin{equation}
g^{\prime }\left( x,y\right) =\mu \left( g\left( \eta \left( x\right) ,\eta
\left( y\right) \right) \right) .  \label{action2}
\end{equation}

We shall first determine $Aut\left( I_{0}\right) \times Aut\left(
A_{3}\right) .$ We have $Aut\left( I_{0}\right) \cong \mathbb{R}^{\ast },$
and it is easy too to determine $Aut\left( A_{3}\right) .$ Indeed, recall
that the unique left-symmetric structure of $A_{3}$ is given by $e_{1}\cdot
e_{2}=e_{3},\ \ e_{1}\cdot e_{3}=-e_{2},$ and let $\eta \in Aut\left(
A_{3}\right) $ be given in the basis $\left\{ e_{1},e_{2},e_{3}\right\}$
by
\[
\eta =\left(
\begin{array}{ccc}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}
\right) .
\]
From the identity $\eta \left( e_{3}\right) =\eta \left( e_{1}\cdot
e_{2}\right) =\eta \left( e_{1}\right) \cdot \eta \left( e_{2}\right) ,$ we
get $c_{1}=0,$ $c_{2}=-a_{1}b_{3},$ and $c_{3}=a_{1}b_{2}.$ From the
identity $-\eta \left( e_{2}\right) =\eta \left( e_{1}\cdot e_{3}\right)
=\eta \left( e_{1}\right) \cdot \eta \left( e_{3}\right) $ we get $b_{1}=0,$
$b_{2}=a_{1}c_{3},$ and $b_{3}=-a_{1}c_{2}.$ Since $\det \eta \neq 0,$ we
deduce that $a_{1}=\pm 1.$ It follows, by setting $\varepsilon =\pm 1,$ that
$b_{3}=-\varepsilon c_{2}$ and $c_{3}=\varepsilon b_{2}.$ From the identity $
\eta \left( e_{1}\right) \cdot \eta \left( e_{1}\right) =\eta \left(
e_{1}\cdot e_{1}\right) =0,$ we obtain $a_{2}=a_{3}=0.$ Therefore, $\eta $
takes the form
\[
\eta =\left(
\begin{array}{ccc}
\varepsilon & 0 & 0 \\
0 & b_{2} & c_{2} \\
0 & -\varepsilon c_{2} & \varepsilon b_{2}
\end{array}
\right), \quad b_{2}^{2}+c_{2}^{2}\neq 0.
\]
We now apply formula (\ref{action2}). For this we recall first that in the
basis above the classes $\left[ g\right] $ and $\left[ g^{\prime }\right] $
corresponding to $A_{4}\left( \alpha ,\beta ,\gamma \right) $
and $A_{4}\left( \alpha ^{\prime },\beta ^{\prime },\gamma ^{\prime }\right)$,
respectively, have the forms
\[
g=\left(
\begin{array}{ccc}
\alpha & 0 & 0 \\
0 & \beta & \gamma \\
0 & -\gamma & \beta
\end{array}
\right) \quad \text{and} \quad g^{\prime }=\left(
\begin{array}{ccc}
\alpha ^{\prime } & 0 & 0 \\
0 & \beta ^{\prime } & \gamma ^{\prime } \\
0 & -\gamma ^{\prime } & \beta ^{\prime }
\end{array}
\right),
\]
respectively. From $g^{\prime }\left( e_{1},e_{1}\right) =\mu g\left( \eta \left(
e_{1}\right) ,\eta \left( e_{1}\right) \right) ,$ we get
\begin{equation}
\alpha ^{\prime }=\mu \alpha ,  \label{eq13}
\end{equation}
and from $g^{\prime }\left( e_{2},e_{2}\right) =\mu g\left( \eta \left(
e_{2}\right) ,\eta \left( e_{2}\right) \right) $, we get
\begin{equation}
\beta ^{\prime }=\mu \left( b_{2}^{2}+c_{2}^{2}\right) \beta .  \label{eq14}
\end{equation}

Similarly, from $g^{\prime }\left( e_{2},e_{3}\right) =\mu g\left( \eta
\left( e_{2}\right) ,\eta \left( e_{3}\right) \right) $ we get
\begin{equation}
\gamma ^{\prime }=\mu \varepsilon \left( b_{2}^{2}+c_{2}^{2}\right) \gamma .
\label{eq15}
\end{equation}

Recall here that $\mu \neq 0$, $\gamma \neq 0,$ and $b_{2}^{2}+c_{2}^{2}\neq
0.$

\begin{claim}
\label{claim}
Each $A_{4}\left( \alpha ,\beta ,\gamma \right) $ is
isomorphic to some $A_{4}\left(\alpha ^{\prime },\beta ^{\prime },1\right)$. Precisely, $A_{4}\!\left(\alpha ,\beta ,\gamma\!\right) $ is isomorphic to
$A_{4}\left( \varepsilon \frac{\alpha }{\gamma },\varepsilon \frac{\beta }{
\gamma },1\right) .$
\end{claim}

\begin{proof}
By (\ref{eq13}), (\ref{eq14}), and (\ref{eq15}), $A_{4}\left( \alpha ,\beta
,\gamma \right) $ is isomorphic to $A_{4}\left( \alpha ^{\prime },\beta
^{\prime },1\right) $ if and only if there exists $\mu \in \mathbb{R}^{\ast
} $ and $b,c\in \mathbb{R}$, with $b^{2}+c^{2}\neq 0,$ such that
\begin{eqnarray*}
\alpha ^{\prime } &=&\mu \alpha , \beta ^{\prime }\\ &=&\mu \left( b^{2}+c^{2}\right) \beta ,
1\\
&=&\mu \varepsilon \left( b^{2}+c^{2}\right) \gamma .
\end{eqnarray*}

Now, by taking $b^{2}+c^{2}=1$ (for instance, $b=\cos \theta _{0}$ and $
c=\sin \theta _{0}$ for some $\theta _{0}$), the third equation yields $\mu =
\frac{\varepsilon }{\gamma }.$ Substituting the value of $\mu $ in the two
first equations, we deduce that $\alpha ^{\prime }=\varepsilon \frac{\alpha
}{\gamma }$ and $\beta ^{\prime }=\varepsilon \frac{\beta }{\gamma }.$
Consequently, each $A_{4}\left( \alpha ,\beta ,\gamma \right) $ is
isomorphic to $A_{4}\left( \varepsilon \frac{\alpha }{\gamma },\varepsilon
\frac{\beta }{\gamma },1\right) .$
\end{proof}


\textbf{Case 2.} $A_{3}=\left\langle e_{1},e_{2},e_{3}:e_{1}\cdot
e_{2}=e_{3},~e_{1}\cdot e_{3}=-e_{2},~e_{2}\cdot e_{2}=e_{3}\cdot
e_{3}=e_{1}\right\rangle .$

Similarly to the first case, we get
\[
\delta _{1}h=\left(
\begin{array}{ccc}
0 & h_{12} & h_{13} \\
0 & h_{22} & 0 \\
0 & 0 & h_{22}
\end{array}
\right) \quad \text{and}\quad g=\left(
\begin{array}{ccc}
0 & g_{12} & g_{13} \\
0 & g_{22} & g_{23} \\
0 & -g_{23} & g_{22}
\end{array}
\right) ,
\]
where $h_{12}=-h\left( e_{3}\right) ,~h_{13}=h\left( e_{2}\right)
,~h_{22}=-h\left( e_{1}\right) ,$ and $g_{ij}=g\left( e_{i},e_{j}\right) .$
It follows that in this case the class $\left[ g\right] \in H^{2}\left(
A_{3},\mathbb{R}\right) $ of a cocycle $g$ takes the reduced form
\[
g=\left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & \gamma \\
0 & -\gamma & 0
\end{array}
\right), \quad \gamma \neq 0.
\]
By setting $\widetilde{e}_{i}=\left( e_{i},0\right) $, $1\leq i\leq
3,$ and $\widetilde{e}_{4}=\left( 0,1\right) ,$ and using formula (\ref
{eq11bis}) we find that the nonzero relations are
\begin{eqnarray}
\widetilde{e}_{1}\cdot \widetilde{e}_{2} &=&\widetilde{e}_{3},~\widetilde{e}
_{1}\cdot \widetilde{e}_{3}=-\widetilde{e}_{2},~\widetilde{e}_{2}\cdot
\widetilde{e}_{2}=\widetilde{e}_{3}\cdot \widetilde{e}_{3}=\widetilde{e}
_{1}\   \label{eq16bis} \\
\widetilde{e}_{2}\cdot \widetilde{e}_{3} &=&\gamma \widetilde{e}_{4},\ \
\widetilde{e}_{3}\cdot \widetilde{e}_{2}=-\gamma \widetilde{e}_{4},~~\gamma
\neq 0.  \notag
\end{eqnarray}

We can now state the main result of this paper.

\begin{theorem}
\label{thm2}Let $A_{4}$ be a complete non-simple real left-symmetric algebra
whose associated Lie algebra is $\mathcal{O}\left( 4\right) .$ Then $A_{4}$
is isomorphic to one of the following left-symmetric algebras:

\begin{itemize}
\item[$($i$)$] $A_{4}\left( s,t\right)$: There exist real numbers $s,t$, and a
basis $\left\{ e_{1},e_{2},e_{3},e_{4}\right\} $ of $\mathcal{O}\left(
4\right) $ relative to which the nonzero left-symmetric relations are
\begin{eqnarray*}
e_{1}\cdot e_{1} &=&se_{4},\ \ e_{2}\cdot e_{2}=e_{3}\cdot e_{3}=te_{4} \\
e_{1}\cdot e_{2} &=&e_{3},\ \ e_{1}\cdot e_{3}=-e_{2}, \\
e_{2}\cdot e_{3} &=&\frac{1}{2}e_{4},\ \ e_{3}\cdot e_{2}=-\frac{1}{2}e_{4}.
\end{eqnarray*}

The conjugacy class of $A_{4}\left( s,t\right) $ is given as follows: $
A_{4}\left( s^{\prime },t^{\prime }\right) $ is isomorphic to $A_{4}\left(
s,t\right) $ if and only if $\left( s^{\prime },t^{\prime }\right) =\left(
\alpha s,\pm t\right) $ for some $\alpha \in \mathbb{R}^{\ast }.$

\item[$($ii$)$] $B_{4}$: There is a basis $\left\{
e_{1},e_{2},e_{3},e_{4}\right\} $ of $\mathcal{O}\left( 4\right) $ relative
to which the nonzero left-symmetric relations are
\begin{eqnarray*}
e_{1}\cdot e_{2} &=&e_{3},\ \ e_{1}\cdot e_{3}=-e_{2},\ e_{2}\cdot
e_{2}=e_{3}\cdot e_{3}=e_{1} \\
e_{2}\cdot e_{3} &=&\frac{1}{2}e_{4},\ \ e_{3}\cdot e_{2}=-\frac{1}{2}e_{4}.
\end{eqnarray*}
\end{itemize}
\end{theorem}

\begin{proof}
According to the discussion above, there are two cases to be considered.

\textbf{Case 1.} $A_{3}=\left\langle e_{1},e_{2},e_{3}:e_{1}\cdot
e_{2}=e_{3},~e_{1}\cdot e_{3}=-e_{2}\right\rangle .$

In this case, Claim \ref{claim} asserts that $A_{4}$ is isomorphic to some $
A_{4}\left( \alpha ,\beta ,1\right) $; and according to equations (\ref{eq16}),
we know that there is a basis $\left\{ \widetilde{e}_{1},\widetilde{e}
_{2},\widetilde{e}_{3},\widetilde{e}_{4}\right\} $ of $\mathcal{O}_{4}$
relative to which the nonzero relations for $A_{4}\left( \alpha ,\beta
,1\right) $ are:
\begin{eqnarray*}
\widetilde{e}_{1}\cdot \widetilde{e}_{1} &=&\alpha \widetilde{e}_{4},\
\widetilde{e}_{2}\cdot \widetilde{e}_{2}=\widetilde{e}_{3}\cdot \widetilde{e}
_{3}=\beta \widetilde{e}_{4}\  \\
\widetilde{e}_{1}\cdot \widetilde{e}_{2} &=&\widetilde{e}_{3},\ \ \widetilde{
e}_{1}\cdot \widetilde{e}_{3}=-\widetilde{e}_{2}, \\
\widetilde{e}_{2}\cdot \widetilde{e}_{3} &=&\widetilde{e}_{4},\ \ \widetilde{
e}_{3}\cdot \widetilde{e}_{2}=-\widetilde{e}_{4}.
\end{eqnarray*}

Now, it is clear that by setting $s=\frac{\alpha }{2}$, $t=\frac{\beta }{2},$
$e_{i}=\widetilde{e}_{i}$ for $1\leq i\leq 3,$ and $e_{4}=2\widetilde{e}
_{4}, $ we get the desired two-parameter family $A_{4}\left( s,t\right) .$
On the other hand, we see from (\ref{eq13}), (\ref{eq14}), and (\ref{eq15})
that $A_{4}\left( s^{\prime },t^{\prime }\right) $ is isomorphic
to $A_{4}\left( s,t\right) $ if and only if there exists $\alpha \in \mathbb{R}
^{\ast }$ and $b,c\in \mathbb{R}$, with $b^{2}+c^{2}\neq 0,$ such that
\begin{eqnarray*}
s^{\prime } &=&\alpha s, \\
t^{\prime } &=&\alpha \left( b^{2}+c^{2}\right) t, \\
1 &=&\alpha \varepsilon \left( b^{2}+c^{2}\right) .
\end{eqnarray*}

From the third equation, we get $b^{2}+c^{2}=\frac{\varepsilon }{\alpha };$
and by substituting the latter into the second equation, we
get $t^{\prime }=\varepsilon t.$ In other words, we have shown that $
A_{4}\left( s^{\prime },t^{\prime }\right) $ and $A_{4}\left(
s,t\right) $ are isomorphic if and only if there exists $\alpha \in \mathbb{R}^{\ast }$ such that
$s^{\prime }=\alpha s$ and $t^{\prime }=\pm t$.

\textbf{Case 2.} $A_{3}=\left\langle e_{1},e_{2},e_{3}:e_{1}\cdot
e_{2}=e_{3},~e_{1}\cdot e_{3}=-e_{2},~e_{2}\cdot e_{2}=e_{3}\cdot
e_{3}=e_{1}\right\rangle .$

In this case, by (\ref{eq16bis}), there is a basis $\left\{ \widetilde{e}
_{1},\widetilde{e}_{2},\widetilde{e}_{3},\widetilde{e}_{4}\right\} $ of $
\mathcal{O}_{4}$ relative to which the nonzero relations in $A_{4}$ are:
\begin{eqnarray*}
\widetilde{e}_{1}\cdot \widetilde{e}_{2} &=&\widetilde{e}_{3},~\widetilde{e}
_{1}\cdot \widetilde{e}_{3}=-\widetilde{e}_{2},~\widetilde{e}_{2}\cdot
\widetilde{e}_{2}=\widetilde{e}_{3}\cdot \widetilde{e}_{3}=\widetilde{e}
_{1}\  \\
\widetilde{e}_{2}\cdot \widetilde{e}_{3} &=&\gamma \widetilde{e}_{4},\ \
\widetilde{e}_{3}\cdot \widetilde{e}_{2}=-\gamma \widetilde{e}_{4}, \gamma \neq 0.
\end{eqnarray*}

By setting $e_{i}=\widetilde{e}_{i}$ for $1\leq i\leq 3,$ and $e_{4}=2\gamma
\widetilde{e}_{4},$ we see that $A_{4}$ is isomorphic to $B_{4}.$ This
finishes the proof of the main theorem.
\end{proof}

\begin{remark}
Recall that a left-symmetric algebra $A$ is called \emph{Novikov} if it
satisfies the condition $\left( x\cdot y\right) \cdot z=\left( x\cdot z\right) \cdot y $,
for all $x,y,z\in A.$

Novikov left-symmetric algebras were introduced in
\cite{novikov} (see also \cite{zelmanov} for some important results
concerning this). We note here that $A_{4}\left( s,0\right) $ is Novikov and
that $B_{4}$ is not.
\end{remark}

We can explicitly compute the exponential map $\exp :\mathcal{O}
_{4}\rightarrow O_{4}$ of the oscillator group in the parametrization given
in Section \ref{oscillator section}. Details of the argument are left to the
reader (see \cite{dorr}). It is given by
\[
\exp \left(v,z,t\right) =\left\{\begin{array}{ll}
\left(v+\frac{\left\vert z\right\vert ^{2}}{4t}\left( 1-\frac{\sin 2t}{2t}
\right) ,z\frac{\sin t}{t},t\right),& t\neq 0
\\
\left(v,z,0\right),& t=0
\end{array}
\right.
\]
On the other hand, we note that the mapping $X\mapsto \left( L_{X},X\right) $
is a Lie algebra representation of $\mathcal{O}_{4}$ in $\mathfrak{aff}
\left( \mathbb{R}^{4}\right) =End\left( \mathbb{R}^{4}\right) \oplus \mathbb{
R}^{4}.$ By using the exponential map of the affine group $Aff\left( \mathbb{
R}^{4}\right) =GL\left( \mathbb{R}^{4}\right) \ltimes \mathbb{R}^{4},$
Theorem \ref{thm2} can now be stated, in terms of simply transitive actions
of subgroups of $Aff\left( \mathbb{R}^{4}\right) ,$ as follows. To state it,
define the continuous functions
\begin{eqnarray*}
f\left( x\right)  &=&\left\{
\begin{array}{ll}
\frac{\sin x}{x},& x\neq 0 \\
1,& x=0
\end{array}
\right., \quad g\left( x\right) =\left\{
\begin{array}{ll}
\frac{1-\cos x}{x},& x\neq 0 \\
0,& x=0
\end{array}
\right. , \\
h\left( x\right)  &=&\left\{
\begin{array}{ll}
\frac{x-\sin x}{x^{2}},& x\neq 0 \\
0,& x=0
\end{array}
\right.,\quad k\left( x\right) =\left\{
\begin{array}{ll}
\frac{1-\cos x}{x^{2}},& x\neq 0 \\
0,& x=0
\end{array}
\right. ,
\end{eqnarray*}
and set
\begin{eqnarray*}
\Phi _{t}\left( x\right)  &=&\left( \frac{y}{2}+tz\right) g\left( x\right)
-\left( \frac{z}{2}-ty\right) f\left( x\right) , \\
\Psi _{t}\left( x\right)  &=&\left( \frac{y}{2}+tz\right) f\left( x\right)
+\left( \frac{z}{2}-ty\right) g\left( x\right) .
\end{eqnarray*}
\begin{theorem}
\label{thm3}Suppose that the oscillator group $O_{4}$ acts simply
transitively by affine transformations on $\mathbb{R}^{4}.$ Then, as a
subgroup of $Aff\left( \mathbb{R}^{4}\right) =GL\left( \mathbb{R}^{4}\right)
\ltimes \mathbb{R}^{4},$ $O_{4}$ is conjugate to one of the following
subgroups:
\begin{itemize}
\item[$($i$)$]
\[\hspace*{-16mm}
G_{4}=\left\{
\begin{array}{l}
\left[
\begin{array}{cccc}
1 & yf\left( x\right) +zg\left( x\right) & zf\left( x\right) -yg\left(
x\right) & 0 \\[1ex]
0 & \cos x & -\sin x & 0 \\
0 & \sin x & \cos x & 0 \\
0 & \Phi _{0}\left( x\right) & \Psi _{0}\left( x\right) & 1
\end{array}
\right] \times \left[
\begin{array}{c}
x+\left( y^{2}+z^{2}\right) k\left( x\right) \\
yf\left( x\right) -zg\left( x\right) \\
zf\left( x\right) +yg\left( x\right) \\
w+\frac{\left( y^{2}+z^{2}\right) }{2}h\left( x\right)
\end{array}
\right] \\[2ex]
:x,y,z,w\in \mathbb{R}
\end{array}
\right\} ,
\]
\item[$($ii$)$]
\[
\hspace*{-6mm}
G_{4}\left( s,t\right) =\left\{
\begin{array}{l}
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \cos x & -\sin x & 0 \\
0 & \sin x & \cos x & 0 \\
sx & \Phi _{t}\left( x\right) & \Psi _{t}\left( x\right) & 1
\end{array}
\right]
\times \left[
\begin{array}{c}
x \\
yf\left( x\right) -zg\left( x\right) \\
zf\left( x\right) +yg\left( x\right) \\
w+\frac{s}{2}x^{2}+\left( y^{2}+z^{2}\right) \left( \frac{h\left( x\right) }{
2}+tk\left( x\right) \right)
\end{array}
\right] \\[2ex]
:x,y,z,w\in \mathbb{R}
\end{array}
\right\} ,
\]
where $s,t\in \mathbb{R}.$ The only pairs of conjugate subgroups in $
Aff\left( \mathbb{R}^{4}\right) $ are $G_{4}\left( s,t\right) $ and $
G_{4}\left( \alpha s,\pm t\right) $ where $\alpha \in \mathbb{R}^{\ast }$.
\end{itemize}
\end{theorem}

%%%% Acknowledgment %%%%%%%%
%\section*{Acknowledgement}

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\end{document}
