\begin{document}

\title{On the geometry of non-degenerate surfaces in Lorentzian homogeneous $3$-manifolds}

\footnote{ 
2020 \textit{Mathematics Subject Classification}.
Primary 53C42; Secondary 53A10, 53C20, 53C50.\\
\textit{Key words and phrases}. Non-degenerate surfaces, Homogeneous manifolds, Minimal and maximal surfaces, Extrinsic curvature, Hopf cylinders.\\
$^{\ast}$ Corresponding author.}

\begin{abstract}
In this paper we deal with non-degenerate surfaces $\Sigma^2$ immersed in the $3$-dimensional homogeneous space $\mathbb{L}^{3}(\kappa,\tau)$ endowed with two different metrics, the one induced by the Riemannian metric of $\mathbb{E}^3(\kappa,\tau)$ and the non-degenerate metric inherited by the Lorentzian one of $\mathbb{L}^3(\kappa,\tau)$. Therefore, we have two different geometries on $\Sigma^2$ and we can compare them. In particular, we study the case where the mean curvature functions with respect to both metrics simultaneously vanish, and in this case we show that the surface is ruled. Furthermore, we consider the case where both mean curvature functions coincide but do not necessarily vanish and we also consider the situation where the extrinsic curvatures with respect to both metrics coincide.
\end{abstract}

\section*{\sc Introduction}

In $1983$ Kobayashi~\cite{Kobayashi:83} proved that the only surfaces in the 3-dimensional Lorentz-Minkowski space $\mathbb{L}^3$ which are simultaneously minimal and maximal are open pieces of spacelike planes or of the helicoid, in the region where it is spacelike. On the one hand, it is well known that a minimal surface in $\mathbb{R}^3$ is a surface with zero mean curvature. On the other hand, let us recall that a spacelike surface in the Lorentz-Minkowski space is an immersed surface such that the metric induced from $\mathbb{L}^3$ is a Riemannian one. Furthermore, a maximal surface is a spacelike surface with identically zero mean curvature. The terminology minimal (maximal, respectively) comes from the fact that these surfaces locally minimize (maximize, respectively) area among close enough surfaces. Since any spacelike surface in $\mathbb{L}^3$ can also be endowed with a second Riemannian metric, the one induced from the Euclidean space $\mathbb{R}^3$, the problem studied by Kobayashi makes sense. 

During the last years several authors have generalized the above result both to general dimension and to surfaces in different ambient spaces. As a first extension, in $2009$ Kim, Koh, Shin and Yang~\cite{Kim:09} considered the situation of a simultaneously minimal and maximal surface in a Lorentzian product space of the type $M^2\times\mathbb{R}_1$, where $M^2$ is a Riemannian surface, and they proved that those surfaces are horizontally ruled by geodesics. Besides, they classified those surfaces in $\mathbb{S}^2\times\mathbb{R}_1$ and $\mathbb{H}^2\times\mathbb{R}_1$. More recently, Alarc\'on, Al\'ias and dos Santos~\cite{Eva:19} and Albujer and Caballero~\cite{Alma:preprint} have generalized the problem to general dimension proving independently that the level hypersurfaces of any simultaneously minimal and maximal hypersurface $\Sigma^n$ in a product $M^n\times\mathbb{R}_1$, $M^n$ being a Riemannian manifold, are minimal hypersurfaces of $\Sigma^n$. Furthermore, in~\cite{Eva:19} the authors consider not only spacelike but non-degenerate hypersurfaces. Finally, it is worth mentioning the recent work by de Lima and Roitman~\cite{deLima:21}. Although their work has some overlapping with the results in~\cite{Eva:19} and~\cite{Alma:preprint}, their main goal is different, since they study helicoids and catenoids in a Riemannian product $M^2\times\mathbb{R}$, providing several interesting examples and properties of such surfaces, as well as some characterization results. In particular, they show that vertical helicoids in a Riemannian product $M^2\times\mathbb{R}$, which are spacelike with respect to the metric induced from the Lorentzian product $M^2\times\mathbb{R}_1$ are the only ones which are simultaneously minimal and maximal.

Going a step further, it is possible to consider as ambient space the 3-dimensional Heisenberg space. It is well-known that the 3-dimensional Heisenberg space $\textrm{Nil}^3$ is an homogenous manifold with 4-dimensional isometry group, which is isometric to $\mathbb{R}^3$ endowed with the left-invariant Riemannian metric
\begin{equation}
g_R=dx^2+dy^2+\left(dz+\frac{1}{2}(ydx-xdy)\right)^2.
\end{equation}
When considering instead the left-invariant Lorentzian metric on $\mathbb{R}^3$ given by
\begin{equation}
g_L=dx^2+dy^2-\left(dz+\frac{1}{2}(ydx-xdy)\right)^2,
\end{equation}
we get its Lorentzian counterpart $\mathrm{Nil}^3_1$. Therefore, any spacelike surface, or more generally any non-degenerate surface, in $\mathrm{Nil}^3_1$, can be endowed with two different metrics that we will also denote as usual by $g_R$ and $g_L$. Thus, there are defined two different mean curvature functions in any non-degenerate surface in $\mathrm{Nil}^3_1$: $H_R$ and $H_L$ respectively. In this context, Shin, Kim, Koh, Lee and Yang~\cite{Shin:13} classified non-degenerate surfaces in $\mathrm{Nil}^3_1$ such that $H_R=H_L=0$ as open pieces of a horizontal plane, a vertical plane, a helicoid or a hyperboloid paraboloid.

As a different extension of the problem, Albujer and Caballero have recently studied some geometric properties of spacelike surfaces in $\mathbb{L}^3$,~\cite{Alma:17} and of spacelike hypersurfaces in a product $M^n\times\mathbb{R}_1$,~\cite{Alma:preprint}, such that $H_R=H_L$ non-necessarily zero.

In this manuscript we consider the well-known family of Bianchi-Cartan-Vranceanu spaces $\mathbb{E}^{3}(\kappa,\tau)$. It is well known that $\mathbb{E}^{3}(\kappa,\tau)$, with $\kappa-4\tau^2\neq 0$, model all the simply connected 3-dimensional manifolds with isometry group of dimension 4. Observe that, in particular, $\mathbb{E}^3(\kappa,0)$ model the product spaces $M^2(\kappa)\times\mathbb{R}$, $M^2(\kappa)$ being the 2-dimensional space form with Gaussian curvature $\kappa$, whereas $\mathbb{E}^3(0,\tau)$, $\tau\neq 0$ is isometric to the Heisenberg space. Finally, in the case where $\tau\neq 0$ and $\kappa\neq 0$, $\mathbb{E}^{3}(\kappa,\tau)$ is isometric to a Berger sphere when $\kappa>0$ and to the universal covering of the special linear group if $\kappa<0$. Furthermore, the spaces $\mathbb{E}^{3}(\kappa,\tau)$ have their Lorentzian counterparts $\mathbb{L}^{3}(\kappa,\tau)$ (see for instance~\cite{Lee:11}), being also 3-dimensional Lorentzian manifolds with isometry group of dimension 4 whenever $\kappa+4\tau^2\neq 0$. Therefore, the spaces $\mathbb{E}^{3}(\kappa,\tau)$ and $\mathbb{L}^{3}(\kappa,\tau)$ are a natural generalization of the previously considered ambient spaces. Since both spaces represent the same topological manifold, it makes sense to consider the same topological surface endowed with two different metrics, and to complete the classification of simultaneously minimal and maximal surfaces in those spaces started in~\cite{Kobayashi:83,Kim:09,Shin:13}. Moreover, it is known that $\mathbb{L}^{3}(\kappa,\tau)$ does not admit complete spacelike surfaces if $\kappa+4\tau^2 >0$, see~\cite{Lee:13}. Thus, it seems reasonable to consider not only spacelike, but in general non-degenerate surfaces in $\mathbb{L}^{3}(\kappa,\tau)$ such that both mean curvature functions related to the induced metrics from $\mathbb{E}^{3}(\kappa,\tau)$ and $\mathbb{L}^{3}(\kappa,\tau)$ vanish.

In Section~\ref{sec:ELkt} we present the homogeneous spaces $\mathbb{E}^{3}(\kappa,\tau)$ and $\mathbb{L}^{3}(\kappa,\tau)$. We describe and classify them, and we give some important relations between their intrisic geometries. In particular, in Lemma~\ref{lem:1.1} we relate their first fundamental forms, and in Lemma~\ref{lem:1.2} we relate and compare their Levi-Civita connections. In Section~\ref{Sec:2} we study, from a local point of view, non-degenerate surfaces on $\mathbb{L}^{3}(\kappa,\tau)$. Since such surfaces can be endowed with two different metrics, we have in fact two different surfaces, so we can compare their extrinsic geometry. In particular, when making an appropriate choice of the normal vector fields, we obtain a relation between both normal vector fields, between the related shape operators and between both mean curvature functions. It is worth pointing out that in the case where the surface is timelike we need a further assumption. As an application of the relation between the normal vector fields we prove that a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ has constant angle if and only if has constant angle as a surface in $\mathbb{E}^{3}(\kappa,\tau)$, see Corollary~\ref{cor:constantangle}.

The main results about the characterization of non-degenerate surfaces such that $H_R=H_L=0$ are obtained in Section~\ref{Sec:charact}. In the first main result, Theorem~\ref{main result}, we prove that any non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ with vanishing mean curvature functions with respect to the metrics induced from $\mathbb{E}^{3}(\kappa,\tau)$ and $\mathbb{L}^{3}(\kappa,\tau)$ is (almost everywhere in the timelike case) a ruled surface in both spaces. As a consequence of Theorem~\ref{main result} and of the already known classification of ruled minimal surfaces in a Berger sphere,~\cite{Shin:15}, and in the universal cover of the special linear group,~\cite{Shin:18}, we are able to classify all non-degenerate surfaces in $\mathbb{L}^{3}(\kappa,\tau)$, $\kappa,\tau\neq 0$, such that $H_R=H_L=0$.

In Section~\ref{sec:4} we consider the general situation where a spacelike surface in $\mathbb{L}^{3}(\kappa,\tau)$ has the same mean curvature functions with respect to both metrics, not necessarily zero. After establishing a nice relation between the normal curvatures at any point with respect to both metrics, we prove in Theorem~\ref{teo:4.1} that the shape operator with respect to the metric induced from the one of $\mathbb{E}^{3}(\kappa,\tau)$ is nowhere definite. Furthermore, in the case where the ambient is a Hadamard manifold, Theorem~\ref{teo:4.1} derives a non-convexity result for the surfaces we are considering.

Finally, we close this manuscript by addressing timelike surfaces with the same Gaussian curvatures and/or the same extrinsic curvatures, when the surfaces are considered as immersions into $\mathbb{E}^3(\kappa,\tau)$ and $\mathbb{L}^3(\kappa,\tau)$. Let us recall that  the extrinsic curvature is defined as the product of the principal curvatures, and both in the Euclidean space and in the Lorentz-Minkowski space the Gaussian and the extrinsic curvatures coincide. In this direction, we develop a relation between the extrinsic curvatures of a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ and in $\mathbb{E}^3(\kappa,\tau)$. Using this relation, we 
characterize open pieces of Hopf cylinders as the unique timelike surfaces with the same Gaussian and extrinsic curvature in $\mathbb{L}^{3}(\kappa,\tau)$ for $\kappa+4\tau^{2}>0$ and $\tau\neq0$.

\section{On the geometry of the homogeneous $3$-manifolds}\label{sec:ELkt}

Let $\kappa$ and $\tau$ be real numbers. The region $\mathcal{D}$ of the Euclidean space $\mathbb{R}^{3}$ given by
\begin{equation}\label{eq:1.1}
\mathcal{D}=\left\{
\begin{array}{ccc}
\mathbb{R}^{3}, & \mbox{if} & \kappa\geq0 \\
\mathbb{D}(2/\sqrt{-\kappa})\times\mathbb{R}, & \mbox{if} & \kappa<0
\end{array}
\right. 
\end{equation}
and endowed with the homogeneous Riemannian metric
\begin{equation}\label{eq:1.2}
\langle\,,\rangle_{R}=\lambda^{2}(dx^{2}+dy^{2})+\left(dz+\tau\lambda(ydx-xdy)\right)^{2},\quad\lambda=\dfrac{1}{1+\frac{\kappa}{4}(x^{2}+y^{2})},
\end{equation}
is the so-called {\it Bianchi-Cartan-Vranceanu space} ($BCV$-space) which is usually denoted by $\mathbb{E}^{3}(\kappa,\tau):=\left(\mathcal{D},\langle\,,\rangle_{R}\right)$.

The $BCV$-spaces are the only simply connected $3$-dimensional homogeneous spaces with $4$-dimensional isometry group. They are classified according to the constants $\kappa$ and $\tau$ in the following sense:
\begin{itemize}
	\item when $\tau=0$, $\mathbb{E}^3(\kappa,0)=M^{2}(\kappa)\times\mathbb{R}$ with $M^{2}(\kappa)$ the sphere $\mathbb{S}^{2}(\kappa)$ of curvature $\kappa$, the Euclidean space $\mathbb{R}^2$ or the hyperbolic plane $\mathbb{H}^{2}(\kappa)$ of curvature $\kappa$, when $\kappa>0$, $\kappa=0$ or $\kappa<0$ respectively.
	\item when $\tau\neq0$, $\mathbb{E}^{3}(\kappa,\tau)$ is isometric to the Berger sphere $\mathbb{S}^{3}_{b}(\kappa,\tau)$ if $\kappa>0$, to the Heisenberg group ${\rm Nil}_{3}(\tau)$ if $\kappa=0$ and to the universal cover of $SL(2,\mathbb{R})$ if $\kappa<0$.
\end{itemize}

As it is well known, there exists always a Riemannian submersion $\pi:\mathbb{E}^{3}(\kappa,\tau)\rightarrow M^{2}(\kappa)$, where $M^{2}(\kappa)$ is the $2$-dimensional simply connected space form of constant curvature $\kappa$, such that $\pi$ has bundle curvature $\tau$ and totally geodesic fibers. Furthermore $\xi=\partial_z$ is a unit Killing field on $\mathbb{E}^{3}(\kappa,\tau)$ which is vertical with respect to $\pi$.

On the other hand, if we endow $\mathcal{D}$ with the Lorentzian metric
\begin{equation}\label{eq:1.9}
	\langle\,,\rangle_{L}=\lambda^{2}(dx^{2}+dy^{2})-\left(dz+\tau\lambda(ydx-xdy)\right)^{2},\quad\lambda=\dfrac{1}{1+\frac{\kappa}{4}(x^{2}+y^{2})},
\end{equation}
we obtain the so-called {\it Lorentz-Bianchi-Cartan-Vranceanu spaces} ($LBCV$-spaces), which have been denoted in the literature by $\mathbb{L}^{3}(\kappa,\tau)$ (see for instance~\cite{Lee:11,Lee:13}). In an analogous way to the Riemannian situation, it holds that $\xi=\partial_z$ is a timelike unit Killing vector field in $\mathbb{L}^{3}(\kappa,\tau)$, which is tangent to the fibers of the submersion $\pi$.

Let us assume along the rest of the section that $\tau\neq 0$ unless otherwise stated. In this case, it is a standard fact that the following canonical frame $\{E_{1},E_{2},E_{3}\}$ on $\mathbb{E}^{3}(\kappa,\tau)$, defined as
\begin{equation}\label{eq:1.3}
\left.
\begin{array}{ccl}
E_{1}\!\!\!\!&=&\!\!\!\!\lambda^{-1}\left(\cos(\sigma z)\partial_{x}+\sin(\sigma z)\partial_{y}\right)+\tau\left(x\sin(\sigma z)-y\cos(\sigma z)\right)\partial_{z},\\
E_{2}\!\!\!\!&=&\!\!\!\!\lambda^{-1}\left(-\sin(\sigma z)\partial_{x}+\cos(\sigma z)\partial_{y}\right)+\tau\left(x\cos(\sigma z)+y\sin(\sigma z)\right)\partial_{z},\\
E_{3}\!\!\!\!&=&\!\!\!\!\partial_{z},
\end{array}
\right.
\end{equation}
where $\sigma=\kappa/2\tau$, is an orthonormal frame with respect to $\langle\,,\rangle_R$, that is,
\begin{equation}\label{eq:1.4}
\langle E_{i},E_{i}\rangle_{R}=\langle E_{2},E_{2}\rangle_{R}=\langle E_{3},E_{3}\rangle_{R}=1\quad\mbox{and}\quad\langle E_{i},E_{j}\rangle_{R}=0\quad\mbox{for}\quad i\neq j.
\end{equation}
Moreover, it also satisfies
\begin{equation}\label{eq:1.5}
[E_{1},E_{2}]=2\tau E_{3},\quad[E_{2},E_{3}]=\sigma E_{1}\quad\mbox{and}\quad[E_{3},E_{1}]=\sigma E_{2}.
\end{equation}

Observe that since $E_3=\xi$ is vertical to the fibration, $E_1$ and $E_2$ are horizontal vector fields on $\mathbb{E}^{3}(\kappa,\tau)$. Moreover, the $BCV$-spaces are oriented, and then we can define a vectorial product $\wedge_{R}$ on $\mathfrak{X}(\mathcal{D})$ given by the formula $\langle U \wedge_R V,W\rangle_R=\textrm{det}(U,V,W)$ for any $U,V,W\in\mathfrak{X}(\mathcal{D})$. In other words,
\begin{equation}\label{eq:1.6}
U\wedge_{R}V=(u_2v_3-u_3v_2)E_1-(u_1v_3-u_3v_1)E_2-(u_2v_1-u_1v_2)E_3,
\end{equation}
for $U=(u_,u_2,u_3)=u_1E_1+u_2E_2+u_3E_3$ and $V=(v_1,v_2,v_3)$ tangent vector fields in $\mathbb{E}^{3}(\kappa,\tau)$.

Since $\xi$ is Killing, it is not difficult to check (see~\cite{Daniel:07}) that for any vector field $X$ on $\mathbb{E}^{3}(\kappa,\tau)$ the following identity holds
\begin{equation}\label{eq:1.7}
	\overline{\nabla}^{R}_{X}\xi=\tau(X\wedge_{R}\xi),
\end{equation}
where $\overline{\nabla}^R$ stands for the Levi-Civita connection in $\mathbb{E}^{3}(\kappa,\tau)$. From the above equation and Koszul's formula, we easily get the following expressions
\begin{equation}\label{eq:1.8}
\left.
\begin{array}{ccccccccccccc}
\overline{\nabla}^{R}_{E_{1}}E_{1}\!\!\!\!&=&\hspace{-1,5cm}0,  &\quad & \overline{\nabla}^{R}_{E_{1}}E_{2}\!\!\!\!&=&\hspace{-1,5cm}\tau E_{3}, &\quad & \overline{\nabla}^{R}_{E_{1}}E_{3}\!\!\!\!&=&\!\!\!\!-\tau E_{2},\\
\overline{\nabla}^{R}_{E_{2}}E_{1}\!\!\!\!&=&\hspace{-0,8cm}-\tau E_{3},  &\quad & \overline{\nabla}^{R}_{E_{2}}E_{2}\!\!\!\!&=&\hspace{-1,8cm}0, &\quad & \overline{\nabla}^{R}_{E_{2}}E_{3}\!\!\!\!&=&\hspace{-0,5cm}\tau E_{1},\\
\overline{\nabla}^{R}_{E_{3}}E_{1}\!\!\!\!&=&\!\!\!\!\dfrac{\kappa-2\tau^{2}}{2\tau}E_{2},  &\quad & \overline{\nabla}^{R}_{E_{3}}E_{2}\!\!\!\!&=&\!\!\!\!-\dfrac{\kappa-2\tau^{2}}{2\tau}E_{1}, &\quad & \overline{\nabla}^{R}_{E_{3}}E_{3}\!\!\!\!&=&\hspace{-0,8cm}0.
\end{array}
\right.
\end{equation}

Let us refer the reader to~\cite{Daniel:07} and~\cite{Espinar:11} for a deeper study of the geometry of the $BCV$-spaces.

With respect to the $LBCV$-spaces, it is easy to show that the frame $\{E_1,E_2,E_3\}$ defined as in~\eqref{eq:1.3} is orthonormal with respect to the Lorentzian metric~\eqref{eq:1.9} as well, that is,
\begin{equation}\label{eq:1.10}
\langle E_{1},E_{1}\rangle_{L}=\langle E_{2},E_{2}\rangle_{L}=1,\quad\langle E_{3},E_{3}\rangle_{L}=-1\quad\mbox{and}\quad\langle E_{i},E_{j}\rangle_{L}=0\quad\mbox{for}\quad i\neq j.
\end{equation}
In an analogous way to the Riemannian situation, it holds that the timelike unit Killing vector field $\xi$ satisfies the following identity,
\begin{equation}\label{eq:1.11}
\overline{\nabla}^{L}_{X}\xi=-\tau(X\wedge_{L}\xi),
\end{equation}
where $\overline{\nabla}^{L}$ denotes the Levi-Civita connection in $\mathbb{L}^{3}(\kappa,\tau)$ and $\wedge_{L}$ stands for the vectorial product in $\mathfrak{X}(\mathcal{D})$ defined by the formula  $\langle U \wedge_L V,W\rangle_L=\textrm{det}(U,V,W)$ for any $U,V,W\in\mathfrak{X}(\mathcal{D})$. Therefore,
\begin{equation}\label{eq:1.12}
U\wedge_{L}V=(u_2v_3-u_3v_2)E_1-(u_1v_3-u_3v_1)E_2-(u_1v_2-u_2v_1)E_3,
\end{equation}
for $U=(u_1,u_2,u_3)$ and $V=(v_1,v_2,v_3)$ tangent vector fields in $\mathbb{L}^{3}(\kappa,\tau)$. Furthermore, from~\eqref{eq:1.5} and~\eqref{eq:1.11}, we also get
\begin{equation}\label{eq:1.13}
\left.
\begin{array}{cccccccccccc}
\overline{\nabla}^{L}_{E_{1}}E_{1}\!\!\!\!&=&\hspace{-1.5cm}0,  &\quad & \overline{\nabla}^{L}_{E_{1}}E_{2}\!\!\!\!&=&\hspace{-1.5cm}\tau E_{3}, &\quad & \overline{\nabla}^{L}_{E_{1}}E_{3}\!\!\!\!&=&\!\!\!\!\tau E_{2},\\
\overline{\nabla}^{L}_{E_{2}}E_{1}\!\!\!\!&=&\hspace{-1cm}-\tau E_{3},  &\quad & \overline{\nabla}^{L}_{E_{2}}E_{2}\!\!\!\!&=&\hspace{-2cm}0, &\quad & \overline{\nabla}^{L}_{E_{2}}E_{3}\!\!\!\!&=&\!\!\!\!-\tau E_{1},\\
\overline{\nabla}^{L}_{E_{3}}E_{1}\!\!\!\!&=&\!\!\!\!\dfrac{\kappa+2\tau^{2}}{2\tau}E_{2},  &\quad & \overline{\nabla}^{L}_{E_{3}}E_{2}\!\!\!\!&=&\!\!\!\!-\dfrac{\kappa+2\tau^{2}}{2\tau}E_{1}, &\quad & \overline{\nabla}^{L}_{E_{3}}E_{3}\!\!\!\!&=&\!\!\!\!0.
\end{array}
\right.
\end{equation}
Given $X\in\mathfrak{X}(\mathcal{D})$, we can write $X=X^{h}+X^{v}$ where $X^{h}\,\in\,\textrm{span}\{E_1,E_2\}$ and $X^{v}=\alpha E_3$ denote, respectively, the horizontal and vertical components of a vector field $X$ in $\mathcal{D}$. It follows immediately that, for all $X,Y\in\mathfrak{X}(\mathcal{D})$,
\begin{equation}\label{eq:1.7.bis}
\alpha =\langle X,E_{3}\rangle_{R}=-\langle X,E_{3}\rangle_{L}\quad\mbox{and}\quad\langle X^{h},Y^{h}\rangle_{R}=\langle X^{h},Y^{h}\rangle_{L}
\end{equation}

From this decomposition, the two following results provide interesting relations between the Riemannian and Lorentzian metrics and the Levi-Civita connections $\overline\nabla^R$ and $\overline\nabla^L$.
\begin{lemma}\label{lem:1.1}
Given $X,Y$ tangent vector fields to $\mathcal{D}$, we have
\begin{equation}\label{eq:1.18}
\langle X,Y\rangle_{R}+\langle X,Y\rangle_{L}=2\langle X^{h},Y^{h}\rangle_{R}=2\langle X^{h},Y^{h}\rangle_{L}
\end{equation}
and
\begin{equation}\label{eq:1.19}
\langle X,Y\rangle_{R}-\langle X,Y\rangle_{L}=2\langle X, E_{3}\rangle_{R}\langle Y, E_{3}\rangle_{R}=2\langle X, E_{3}\rangle_{L}\langle Y, E_{3}\rangle_{L}.
\end{equation}
\end{lemma}

\begin{proof}
The proof follows immediately from~\eqref{eq:1.7.bis} and from the already mentioned fact that the canonical frame on $\mathfrak{X}(\mathcal{D})$, $\{E_1,E_2,E_3\}$, is orthonormal with respect to both metrics (see equations~\eqref{eq:1.4} and~\eqref{eq:1.10}). 
\end{proof}

\begin{lemma}\label{lem:1.2}
The Levi-Civita connections of the homogeneous spaces $\mathbb{E}^{3}(\kappa,\tau)$ and $\mathbb{L}^{3}(\kappa,\tau)$ are related by
\begin{equation*}
\overline{\nabla}^{R}_{Y}X-\overline{\nabla}^{L}_{Y}X=W(X,Y),
\end{equation*}
where
\begin{equation}\label{eq:1.22}
W(X,Y)=2\tau\left(X^{h}\wedge_{R}Y^{v}-X^{v}\wedge_{R}Y^{h}\right)=2\tau\left(X^{h}\wedge_{L}Y^{v}-X^{v}\wedge_{L}Y^{h}\right)
\end{equation}
for all $X,Y\in\mathfrak{X}(\mathcal{D})$.
\end{lemma}

\begin{proof}
As above let us write $X=X^{h}+X^{v}$ and $Y=Y^{h}+Y^{v}$, where $X^{h}=x_{1}E_{1}+x_{2}E_{2}$, $Y^{h}=y_{1}E_{1}+y_{2}E_{2}$, $X^{v}=x_{3}E_{3}$ and $Y^{v}=y_{3}E_{3}$, with $x_{i}=\langle X,E_{i}\rangle_{R}$ and $y_{i}=\langle Y,E_{i}\rangle_{R}$ for $i=1,2,3$. From~\eqref{eq:1.8} and~\eqref{eq:1.13} it holds 
\begin{equation*}
\overline{\nabla}^{R}_{E_{i}}E_{j}=\overline{\nabla}^{L}_{E_{i}}E_{j},\,\, i,j=1,2 \quad \textrm{and} \quad \overline{\nabla}^{R}_{E_{3}}E_{3}=\overline{\nabla}^{L}_{E_{3}}E_{3}.
\end{equation*}
Thus, we immediately get
\begin{equation*}
\overline{\nabla}^{R}_{Y^h}X^h-\overline{\nabla}^{L}_{Y^h}X^h=0 \quad \textrm{and} \quad \overline{\nabla}^{R}_{Y^v}X^v-\overline{\nabla}^{L}_{Y^v}X^v=0,
\end{equation*}
and consequently
\begin{equation}\label{eq:1.23}
\overline{\nabla}^{R}_{Y}X-\overline{\nabla}^{L}_{Y}X= \left(\overline{\nabla}^{R}_{Y^h}X^v-\overline{\nabla}^{L}_{Y^h}X^v\right)+\left(\overline{\nabla}^{R}_{Y^v}X^h-\overline{\nabla}^{L}_{Y^v}X^h\right).
\end{equation}

Let us now consider the first term in the right hand side of~\eqref{eq:1.23}. Using again~\eqref{eq:1.8} and~\eqref{eq:1.13} we get
\begin{equation}\label{eq:1.24}
\overline{\nabla}^{R}_{Y^h}X^v-\overline{\nabla}^{L}_{Y^h}X^v=2\tau\langle X^v,E_3\rangle_R\left(\langle Y^h,E_2\rangle_R E_1-\langle Y^h,E_1\rangle_R E_2\right).
\end{equation}
Furthermore, from~\eqref{eq:1.6} it yields that  $E_3\wedge_R Y^h=\langle Y^h,E_1\rangle_R E_2-\langle Y^h,E_2\rangle_R E_1$. Therefore,
\begin{equation}\label{eq:1.25}
\overline{\nabla}^{R}_{Y^h}X^v-\overline{\nabla}^{L}_{Y^h}X^v=-2\tau(X^v\wedge_RY^h).
\end{equation}
Analogously, considering the last term in~\eqref{eq:1.23} we obtain
\begin{equation}\label{eq:1.26}
\overline{\nabla}^{R}_{Y^v}X^h-\overline{\nabla}^{L}_{Y^v}X^h=2\tau(X^h\wedge_R Y^v).
\end{equation}

Finally, taking into account~\eqref{eq:1.25} and~\eqref{eq:1.26}, equation~\eqref{eq:1.23} reads
\begin{equation}\label{eq:1.27}
\overline{\nabla}^{R}_{Y}X-\overline{\nabla}^{L}_{Y}X=2\tau\left(X^h\wedge_R Y^v-X^v\wedge_R Y^h\right),
\end{equation}
and the proof follows by observing that $X^h\wedge_RY^v=X^h\wedge_LY^v$ for any $X,Y\in\mathfrak{X}(\mathcal{D})$.
\end{proof}

\begin{remark}
Observe that in the case $\tau=0$, $\mathbb{E}^3(\kappa,0)$ and $\mathbb{L}(\kappa,0)$ are product spaces and it is possible to reproduce the above study by considering the orthonormal frame $\{\partial_x,\partial_y,\partial_z\}$ on $\mathfrak{X}(\mathcal{D})$ instead of $\{E_1,E_2,E_3\}$. In particular, $\partial_x$ and $\partial_y$ are horizontal vectors and it is possible also to consider the vectorial products $\wedge_R$ and $\wedge_L$, giving their expressions in terms of this frame. This case has been previously study in~\cite{Eva:19}. Specifically in~\cite{Eva:19} Lemma~\ref{lem:1.1} has been obtained for $\tau=0$, and it has been proved that in this situation the Riemannian and the Lorentzian Levi-Civita connection coincide, so Lemma~\ref{lem:1.2} also holds.
\end{remark}

\section{Surfaces in homogeneous spaces}\label{Sec:2}

A $2$-dimensional connected surface $\Sigma^{2}$ is said to be a {\em non-degenerate} surface in $\mathbb{L}^{3}(\kappa,\tau)$ if there exists a smooth immersion $\psi:\Sigma^{2}\longrightarrow\mathbb{L}^{3}(\kappa,\tau)$ such that $\psi$ induces, from the Lorentzian metric of $\mathbb{L}^{3}(\kappa,\tau)$, a non-degenerate metric on $\Sigma^{2}$ which, as usual, is also denoted by $\langle\,,\rangle_{L}$. In this case, there are two possibilities:
\begin{itemize}
\item[(a)] either the induced metric on $\Sigma^{2}$ is Riemannian, and we say that $\Sigma^{2}$ is a {\em spacelike} surface of $\mathbb{L}^{3}(\kappa,\tau)$, or
\item[(b)] the induced metric on $\Sigma^{2}$ is Lorentzian, and $\Sigma^{2}$ is said to be a {\em timelike} surface.
\end{itemize}
Equivalently, in case (a) the tangent plane $\psi_{\ast}(T_{p}\Sigma)$ is a spacelike plane for every $p\in\Sigma$ while in case (b) it is timelike.

Given a non-degenerate surface $\Sigma^2$ in $\mathbb{L}^{3}(\kappa,\tau)$, we can define the open region given by
\begin{equation}\label{eq:IWOK}
\widehat{\Sigma}^2=\{p\in\Sigma^2\,;\,\exists\,\, \mathcal{U}_p, N_L \textrm{ s.t. } \langle N_L,\xi\rangle_L \leq 0 \textrm{ on } \mathcal{U}_p\},
\end{equation}
where $\mathcal{U}_p\subseteq \Sigma^2$ is a neighbourhood of $p$ and $N_L$ a unit normal vector field (locally) defined on $\widehat{\Sigma}^2$. On the one hand, observe that when $\Sigma^{2}$ is spacelike, since $\xi$ is a unit timelike vector field globally defined on $\mathbb{L}^{3}(\kappa,\tau)$, there exists a unique global timelike unit normal vector field on $\Sigma^{2}$, $N_{L}$, which is in the same time-orientation as $\xi$, and hence we may assume that $\Sigma^{2}$ is oriented by $N_{L}$. As a consequence of the backwards Schwarz inequality in the Lorentzian context, it follows that
\begin{equation}\label{eq:2.1}
\langle N_{L},\xi\rangle_{L}\leq-1<0,
\end{equation}
with equality if and only if $N_{L}=\xi$, so $\widehat{\Sigma}^2=\Sigma^2$. In particular, there exists a unique number $\varphi\geq0$, called the {\em hyperbolic angle} between $N_{L}$ and $\xi$, such that
\begin{equation}\label{eq:2.2}
\langle N_{L},\xi\rangle_{L}=-\cosh\varphi.
\end{equation}
On the other hand, in the case where $\Sigma^{2}$ is timelike we can choose in any connected component of $\widehat{\Sigma}^2$ a globally defined unique spacelike unit normal vector field $N_L$ satisfying the inequality in~\eqref{eq:IWOK}. Let us observe in that case that $\widehat{\Sigma}^2$ is an open and dense subset of $\Sigma^2$. In order to unify our notation, we will denote by $\varepsilon$ the sign of $\langle N_{L},N_{L}\rangle_{L}$, that is,
\begin{equation}\label{eq:2.4}
\varepsilon=\left\{
\begin{array}{rl}
 -1, & \mbox{if}\,\, \Sigma^{2}\,\, \mbox{is spacelike.}  \\
 1, & \mbox{if}\,\, \Sigma^{2}\,\, \mbox{is timelike.}
\end{array}
\right.
\end{equation}

Given $\psi:\Sigma^{2}\longrightarrow \mathbb{L}^{3}(\kappa,\tau)$ a non-degenerate surface, $\psi:\Sigma^{2}\longrightarrow\mathbb{E}^{3}(\kappa,\tau)$ also defines an immersed surface into the Riemannian homogeneous space $\mathbb{E}^{3}(\kappa,\tau)$. Our first objective here is to relate the extrinsic geometry of both surfaces. We will do it by using the following essential proposition.

\begin{proposition}\label{prop:2.1}
Let $\psi:\Sigma^{2}\longrightarrow \mathbb{L}^{3}(\kappa,\tau)$ be a non-degenerate surface immersed into the Lorentzian homogeneous space $\mathbb{L}^{3}(\kappa,\tau)$, and consider its restriction to any connected component of $\widehat{\Sigma}^2$, $\widehat{\Sigma}^2_0$. Let $N_L$ be the globally defined unit normal vector field on $\widehat{\Sigma}^2_0$ such that $\langle N_{L},\xi\rangle_{L}\leq0$. Then
\begin{equation}\label{normal R}
N_{R}=\dfrac{1}{\omega_{L}}\left(\sqrt{2(\omega_{L}^{2}-\varepsilon)}\xi-N_{L}\right)
\end{equation}
globally defines the unique unit normal vector field on  $\widehat{\Sigma}^2_0$ with respect to the Riemannian metric $\langle\,,\rangle_{R}$ induced from $\mathbb{E}^{3}(\kappa,\tau)$ such that $\langle N_{R},\xi\rangle_{R}\geq0$, where
\begin{equation}\label{eq:2.5}
\omega_{L}=\sqrt{\varepsilon+2\langle N_{L},\xi\rangle_{L}^{2}}\geq1>0.
\end{equation}
\end{proposition}

\begin{proof}
First of all, let us compute
\begin{equation}\label{eq:2.6}
\langle N_{R},N_{R}\rangle_{R}=\dfrac{1}{\omega_{L}^{2}}\left(2(\omega_{L}^{2}-\varepsilon)-2\sqrt{2(\omega_{L}^{2}-\varepsilon)}\langle N_{L},\xi\rangle_{R}+\langle N_{L},N_{L}\rangle_{R}\right).
\end{equation}
Since $\langle N_{L},\xi\rangle_{L}\leq0$, it follows from~\eqref{eq:2.5} that
\begin{equation}\label{eq:2.7}
\langle N_{L},\xi\rangle_{L}=-\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2}
\end{equation}
and using~\eqref{eq:1.18} we get
\begin{equation}\label{eq:2.8}
\langle N_{L},\xi\rangle_{R}=-\langle N_{L},\xi\rangle_{L}=\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2}.
\end{equation}
Thus
\begin{equation}\label{eq:2.9}
-2\sqrt{2(\omega_{L}^{2}-\varepsilon)}\langle N_{L},\xi\rangle_{R}=-2(\omega_{L}^{2}-\varepsilon).
\end{equation}
On the other hand,~\eqref{eq:1.19} yields
\begin{equation}\label{eq:2.13}
\langle N_{L},N_{L}\rangle_{R}=\varepsilon+2\langle N_{L},\xi\rangle_{L}^{2}=\omega_{L}^{2}.
\end{equation}
Taking into account~\eqref{eq:2.9} and~\eqref{eq:2.13}, we can immediately check from~\eqref{eq:2.6} that $\langle N_{R},N_{R}\rangle_{R}=1$.

Furthermore, from~\eqref{normal R} we also obtain
\begin{equation}\label{eq:2.15.bis}
\langle N_{R},\xi\rangle_{R}=\dfrac{1}{\omega_{L}}\left(\sqrt{2(\omega_{L}^{2}-\varepsilon)}-\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2}\right)=\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2\omega_{L}}\geq0.
\end{equation}
Finally, from~\eqref{normal R},~\eqref{eq:2.7} and Lemma~\ref{lem:1.1}, for every $X\in\mathfrak{X}(\Sigma)$ we have
\begin{equation}\label{eq:2.16}\begin{split}
\langle N_{R},X\rangle_{R}=&\dfrac{1}{\omega_{L}}\left(\sqrt{2(\omega^{2}_{L}-\varepsilon)}\langle \xi,X\rangle_{R}-\langle N_{L},X\rangle_{R}\right)\\
=&\dfrac{1}{\omega_{L}}\left(-\sqrt{2(\omega^{2}_{L}-\varepsilon)}-2\langle N_{L},\xi\rangle_{L}\right)\langle\xi, X\rangle_L=0\nonumber
\end{split}
\end{equation}
which finishes the proof.
\end{proof}

The last result expresses how to write the vector field $N_{R}$ in terms of $N_{L}$, but the opposite is also possible. Indeed, from~\eqref{normal R} we write
\begin{equation}\label{eq:2.17}
N_{L}=\sqrt{2(\omega_{L}^{2}-\varepsilon)} \xi-\omega_{L}N_{R}=\dfrac{1}{\omega_{R}}\left(\sqrt{2(1-\varepsilon\omega_{R}^{2})} \xi-N_{R}\right),
\end{equation}
where, by definition $\omega_{R}:=\dfrac{1}{\omega_{L}}$. Observe also that by~\eqref{eq:2.15.bis} we have $2\langle N_{R},\xi\rangle_{R}^{2}=1-\varepsilon\omega_{R}^{2}$ and hence
\begin{equation}\label{eq:2.19}
\omega_{R}=\sqrt{\varepsilon\left(1-2\langle N_{R},\xi\rangle_{R}^{2}\right)}.
\end{equation}

In what follows, we will always assume that given a non-degenerate surface $\Sigma^{2}$ in $\mathbb{L}^{3}(\kappa,\tau)$, any connected component $\widehat{\Sigma}^2_0$ of $\widehat{\Sigma}$ is oriented by the unique unit normal field $N_{L}$ such that $\langle N_{L},\xi\rangle_{L}\leq0$. Doing so, as a surface of the space $\mathbb{E}^{3}(\kappa,\tau)$, $\widehat\Sigma^{2}_0$ will be always oriented by the unit normal vector field $N_{R}$ given by~\eqref{normal R}, with $\langle N_{R},\xi\rangle_{R}\geq0$. We observe that, given any $V\in\mathfrak{X}(D)$,~\eqref{normal R} yields
\begin{equation}\label{eq:2.24_aux}
\begin{split}
\langle N_{R},V\rangle_{R}=&\dfrac{1}{\omega_{L}}\sqrt{2(\omega_{L}^{2}-\varepsilon)}\langle \xi,V\rangle_{R}-\dfrac{1}{\omega_{L}}\langle N_{L},V\rangle_{R}\\
=&-\dfrac{1}{\omega_{L}}\sqrt{2(\omega_{L}^{2}-\varepsilon)}\langle  \xi,V\rangle_{L}-\dfrac{1}{\omega_{L}}\langle N_{L},V\rangle_{R}.
\end{split}
\end{equation}
Now, from~\eqref{eq:1.19} and~\eqref{eq:2.7} it holds
\begin{equation}\label{eq:2.24_aux2}
\langle N_{L},V\rangle_{R}=\langle N_{L},V\rangle_{L}+2\langle N_{L},\xi\rangle_{L}\langle \xi,V\rangle_{L}=\langle N_{L},V\rangle_{L}-\sqrt{2(\omega_{L}^{2}-\varepsilon)}\langle \xi,V\rangle_{L}.
\end{equation}
Hence,
\begin{equation}\label{eq:2.24}
\langle N_{R},V\rangle_{R}=-\dfrac{1}{\omega_{L}}\langle N_{L},V\rangle_{L} \quad \textrm{on} \quad \widehat{\Sigma}^2_0.
\end{equation}

At this moment, we recall that an orientable surface $\Sigma^{2}$ in $\mathbb{E}^{3}(\kappa,\tau)$ and/or $\mathbb{L}^{3}(\kappa,\tau)$ is said to have {\em constant angle} if its normal vector field makes a constant angle with respect to $\xi$. Thus, taking $V=\xi$ in~\eqref{eq:2.24}, we get

\begin{corollary}\label{cor:constantangle}
A non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ has constant angle if and only if has constant angle as a surface in $\mathbb{E}^{3}(\kappa,\tau)$.
\end{corollary}
\begin{proof}
The corollary is immediate from~\eqref{eq:2.24} and~\eqref{eq:2.5} in any connected component of $\widehat{\Sigma}^2$. Since $\widehat{\Sigma}^2$ is dense on the (connected) surface $\Sigma^2$, the result follows by a continuity argument.
\end{proof}

From now on, and unless the opposite is explicitely mentioned, we will assume we are working on a connected component of $\widehat{\Sigma}^2$, $\widehat{\Sigma}^2_0$. Let us denote by $T_{R}$ and $T_{L}$ the tangential components of $\xi$ along $\widehat\Sigma^{2}_0$ with respect to the metrics $\langle\,,\rangle_{R}$ and $\langle\,,\rangle_{L}$, respectively. In this setting, we have the following splittings
\begin{equation}\label{eq:2.25}
\xi=T_{R}+\langle N_{R},\xi\rangle_{R}N_{R}\quad\mbox{and}\quad\xi=T_{L}+\varepsilon\langle N_{L},\xi\rangle_{L}N_{L}.
\end{equation}
Therefore,
\begin{equation}\label{eq:2.26}
1=\langle\xi,\xi\rangle_{R}=|T_{R}|_{R}^{2}+\langle N_{R},\xi\rangle_{R}^{2}\quad\mbox{and}\quad-1=\langle\xi,\xi\rangle_{L}=|T_{L}|^2_{L}+\varepsilon\langle N_{L},\xi\rangle_{L}^{2},
\end{equation}
where $|\cdot|_R=\sqrt{\langle\cdot,\cdot\rangle_R}$ and $|\cdot|_L=\sqrt{\varepsilon\langle\cdot,\cdot\rangle_{L}}$ denote the norm on $(\Sigma^2,\langle,\rangle_R)$ and $(\Sigma^2,\langle,\rangle_L)$, respectively. 

To state our next result, we recall some classical surfaces in $\mathbb{E}^{3}(\kappa,\tau)$ which can be constructed in the following way. On the one hand, given any regular curve $\alpha$ in $\mathbb{M}^{2}(\kappa)$, $\pi^{-1}(\alpha)$ is a surface in $\mathbb{E}^{3}(\kappa,\tau)$ which has $\xi$ as a tangent vector field. $\pi^{-1}(\alpha)$ is usually called a {\em Hopf surface} of $\mathbb{E}^{3}(\kappa,\tau)$. Hopf surfaces are timelike surfaces in $\mathbb{L}^3(\kappa,\tau)$ characterized by $\langle N_R,\xi\rangle_R=0$. On the other hand, $\Sigma^2$ is said to be a {\em horizontal surface} in $\mathbb{E}^{3}(\kappa,\tau)$ if for every $p\in\Sigma^{2}$, the tangent plane $T_p\Sigma$ is horizontal. Equivalently, if $N_R=N_L=\xi$. Let us observe that a horizontal surface is always spacelike. The last result of this section gives us a characterization of those surfaces in terms of $\omega_L$ (or $\omega_R$).

\begin{proposition}\label{prop:1.1}
Let $\Sigma^2$ be a non-degenerate surface immersed in $\mathbb{L}^{3}(\kappa,\tau)$ with unit normal vector field $N_L$ on any connected component $\widehat{\Sigma}^2_0$ of $\widehat{\Sigma}^2$ such that $\langle N_L,\xi\rangle_L\leq 0$. Then $\omega_{L}=1$ in $\Sigma^2$ if and only if the surface is locally isometric to a Hopf surface $\pi^{-1}(\alpha)$ for some $\alpha:I\rightarrow M^2(\kappa)$ or to a horizontal surface.
\end{proposition}

\begin{proof}
Let us assume first that $\omega_{L}=1$. From~\eqref{eq:2.5} we have in any $\widehat{\Sigma}^2_0$ that
\begin{equation}\label{eq:2.27}
\langle N_{L},\xi\rangle_{L}^{2}=\dfrac{1-\varepsilon}{2}.
\end{equation}
Then, if $\varepsilon=-1$ we have $\langle N_L,\xi\rangle_L^2=1$ and consequently $N_L=N_R=\xi$, so $\widehat{\Sigma}^2_0$ is isometric to an open region of a horizontal surface. In the case $\varepsilon=1$,~\eqref{eq:2.27} yields $\langle N_L,\xi\rangle_L=0$. Thus, from~\eqref{eq:2.24} we get $\langle N_R,\xi\rangle_R=0$, and $\widehat{\Sigma}^2_0$ is isometric to an open region of a Hopf surface. The result follows again from a continuity argument. The converse follows immediately from~\eqref{eq:2.5} and~\eqref{eq:2.25}.
\end{proof}

\subsection{The integrability equations}\label{Sec:3}

Let $\psi:\Sigma^{2}\longrightarrow\mathbb{L}^{3}(\kappa,\tau)$ be a non-degenerate surface with unit normal field $N_{L}$, defined on any connected component of $\widehat{\Sigma}^2$, such that $\langle N_{L},\xi\rangle_{L}\leq0$, and let $N_{R}$ be given by~\eqref{normal R}. Recall that the Levi-Civita connections of $\mathbb{L}^{3}(\kappa,\tau)$ and $\mathbb{E}^{3}(\kappa,\tau)$ are related by Lemma~\ref{lem:1.2}. Throughout this paper we will denote by $\nabla^{L}$ the Levi-Civita connection related to $(\Sigma^2,\langle,\rangle_L)$ and by $\nabla^{R}$ the one related to $(\Sigma^{2},\langle,\rangle_R)$. With this notation, the Gauss and Weingarten formulas for $\psi:\Sigma^{2}\longrightarrow\mathbb{L}^{3}(\kappa,\tau)$ are given, respectively, by
\begin{equation}\label{eq:3.1}
\overline{\nabla}^{L}_{X}Y=\nabla_{X}^{L}Y+\varepsilon\langle A_{L}(X),Y\rangle_{L}N_{L}
\end{equation}
and
\begin{equation}\label{eq:3.2}
A_{L}(X)=-\overline{\nabla}^{L}_{X}N_{L},
\end{equation}
for every tangent vector fields $X,Y\in\mathfrak{X}(\Sigma)$, where $A_{L}:\mathfrak{X}(\Sigma)\rightarrow\mathfrak{X}(\Sigma)$ stands for the shape operator (or second fundamental form) of $(\Sigma^{2},\langle,\rangle_L)$ with respect to $N_{L}$. From the above Gauss and Weingarten formulas, jointly with~\eqref{eq:1.11} and~\eqref{eq:2.25}, we obtain
\begin{equation}\label{eq:3.8}
\begin{split}
-\tau(X\wedge_{L}\xi)=&\overline{\nabla}^{L}_{X}\xi=\overline{\nabla}^{L}_{X}\left(T_{L}+\varepsilon\langle N_{L},\xi\rangle_{L}N_{L}\right)\\
=&\overline{\nabla}^{L}_{X}T_{L}+\varepsilon X\left(\langle N_{L},\xi\rangle_{L}\right)N_{L}+\varepsilon\langle N_{L},\xi\rangle_{L}\overline{\nabla}^{L}_{X}N_{L}\\
=&\nabla^{L}_{X} T_{L}+\varepsilon\langle A_{L}(X), T_{L}\rangle_{L}N_{L}+\varepsilon X\left(\langle N_{L},\xi\rangle_{L}\right)N_{L}-\varepsilon\langle N_{L},\xi\rangle_{L}A_{L}(X)
\end{split}
\end{equation}
On the other hand,
\begin{equation}\label{eq:3.9}
	\begin{split}
\tau(X\wedge_{L}\xi)=&\tau\left(X\wedge_{L}\left(T_{L}+\varepsilon\langle N_{L},\xi\rangle_{L}N_{L}\right)\right)\\
=&\tau(X\wedge_{L} T_{L})+\varepsilon\tau\langle N_{L},\xi\rangle_{L}(X\wedge_{L}N_{L})
\\
=&-\varepsilon\tau\langle N_{L},\xi\rangle_{L}J_{L}(X)-\varepsilon\tau\langle J_{L}(T_{L}),X\rangle_{L}N_{L},
\end{split}
\end{equation}
where $J_{L}(X)=N_{L}\wedge_{L}X$ for all $X\in\mathfrak{X}(\Sigma)$, so that $X\wedge_{L} T_{L}=-\varepsilon\langle J_{L}(T_{L}),X\rangle_{L}N_{L}$. Therefore, comparing tangent and normal components, we must have
\begin{equation}\label{eq:int1_L}
\nabla^{L}_{X} T_{L}=\varepsilon\langle N_{L},\xi\rangle_{L}\left(A_{L}(X)+\tau J_{L}(X)\right)
\end{equation}
and
\begin{equation}\label{eq:int2_L}
X\left(\langle N_{L},\xi\rangle_{L}\right)=-\langle(A_{L}-\tau J_{L})(T_{L}),X\rangle_L,
\end{equation}
for all $X\in\mathfrak{X}(\Sigma)$. Finally, the mean curvature function of $\psi$ is defined by $2H_{L}=\varepsilon{\rm tr}_{L}(A_{L})$, where ${\rm tr}_{L}$ denotes the trace in $(\Sigma^2,\langle\,,\rangle_{L})$.

Similarly, the Gauss and Weingarten formulas of the surface $\psi:\Sigma^{2}\longrightarrow\mathbb{E}^{3}(\kappa,\tau)$ are given, respectively, by
\begin{equation}\label{eq:3.3}
\overline{\nabla}^{R}_{X}Y=\nabla_{X}^{R}Y+\langle A_{R}(X),Y\rangle_{R}N_{R}
\end{equation}
and
\begin{equation}\label{eq:WeingR}
A_{R}(X)=-\overline{\nabla}^{R}_{X}N_{R},
\end{equation}
for every tangent vector fields $X,Y\in\mathfrak{X}(\Sigma)$, where $A_{R}:\mathfrak{X}(\Sigma)\rightarrow\mathfrak{X}(\Sigma)$ stands for the shape operator (or second fundamental form) of $(\Sigma^{2},\langle,\rangle_R)$ with respect to $N_{R}$.

Therefore,
\begin{equation}\label{eq:3.10}
\nabla^{R}_{X}T_{R}=\langle N_{R},\xi\rangle_{R}\left(A_{R}(X)-\tau J_{R}(X)\right)
\end{equation}
and
\begin{equation}\label{eq:3.11}
X(\langle N_{R},\xi\rangle_{R})=-\langle(A_{R}+\tau J_{R})(T_{R}),X\rangle_{R},
\end{equation}
for all $X\in\mathfrak{X}(\Sigma)$, where $J_R(X)=N_R\wedge_R X$. We also define the mean curvature of $\Sigma^{2}$ by $2H_{R}={\rm tr}_{R}(A_{R}),$ where ${\rm tr}_{R}$ denotes the trace with respect to the Riemannian metric $\langle\,,\rangle_{R}$.

In our next result we establish the relation between both shape operators.
\begin{proposition}\label{prop:3.1}
Let $\Sigma^{2}$ be a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ and let $\widehat{\Sigma}^2_0$ be a connected component of $\widehat{\Sigma}^2$. Then, the corresponding shape operators of $(\Sigma^2,\langle,\rangle_R)$ and $(\Sigma^2,\langle,\rangle_L)$ with respect to $N_R$ and $N_L$ in $\widehat{\Sigma}^2_0$ are related by
\begin{equation}\label{shape R}
A_{R}(X)=-\dfrac{1}{\omega_{L}}A_{L}(X)-\dfrac{2\varepsilon}{\omega_{L}^{3}}\langle(A_{L}-\tau J_{L})(T_{L}),X\rangle_{L}T_{L}-\dfrac{2\tau}{\omega_{L}}\langle T_{L},X\rangle_{L}J_{L}(T_{L}).
\end{equation}
Equivalently,
\begin{equation}\label{shape L}
A_{L}(X)=-\dfrac{1}{\omega_{R}}A_{R}(X)+\dfrac{2\varepsilon}{\omega_{R}^{3}}\langle (A_{R}+\tau J_{R})(T_{R}),X\rangle_{R}T_{R}-\dfrac{2\tau}{\omega_{R}}\langle T_{R},X\rangle_{R}\,J_{R}(T_{R}),
\end{equation}
for all $X\in\mathfrak{X}(\widehat{\Sigma}_0)$.
\end{proposition}

\begin{proof}
Let us prove~\eqref{shape R} since the second relation is analogous. From Lemma~\ref{lem:1.2} and using~\eqref{normal R}, we have
\begin{equation}\label{eq:3.12}\begin{split}
A_{R}(X)=&-\overline{\nabla}^{R}_{X}N_{R}=-\overline{\nabla}^{L}_{X}N_{R}-W(N_{R},X)\\
=&-\overline{\nabla}^{L}_{X}\left(\dfrac{1}{\omega_{L}}\left(\sqrt{2(\omega^{2}_{L}-\varepsilon)} \xi-N_{L}\right)\right)-W(N_{R},X)\\
=&\dfrac{X(\omega_{L})}{\omega^{2}_{L}}\left(\sqrt{2(\omega^{2}_{L}-\varepsilon)} \xi-N_{L}\right)+\dfrac{1}{\omega_{L}}\overline{\nabla}^{L}_{X}N_{L}-\dfrac{1}{\omega_{L}}X\left(\sqrt{2(\omega^{2}_{L}-\varepsilon)}\right)\xi\\
&-\dfrac{1}{\omega_{L}}\sqrt{2(\omega^{2}_{L}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi-W(N_{R},X),
\end{split}
\end{equation}
for every $X\in\mathfrak{X}(\widehat{\Sigma}_0)$. On the one hand, from~\eqref{eq:2.7}, we have
\begin{equation}\label{eq:3.13}
X\left(\langle N_{L},\xi\rangle_{L}^{2}\right)=2\langle N_{L},\xi\rangle_{L}X\left(\langle N_{L},\xi\rangle_{L}\right)=-X\left(\langle N_{L},\xi\rangle_{L}\right)\sqrt{2(\omega^{2}_{L}-\varepsilon)},
\end{equation}
so that
\begin{equation}\label{eq:3.14}
X(\omega_{L})=X\left(\sqrt{\varepsilon+2\langle N_{L},\xi\rangle_{L}^{2}}\right)=-\dfrac{\sqrt{2(\omega^{2}_{L}-\varepsilon)}}{\omega_{L}}X\left(\langle N_{L},\xi\rangle_{L}\right).
\end{equation}
Consequently,
\begin{equation}\label{eq:3.15}
\dfrac{X(\omega_{L})}{\omega^{2}_{L}}=-\dfrac{\sqrt{2(\omega^{2}_{L}-\varepsilon)}}{\omega_{L}^{3}}X\left(\langle N_{L},\xi\rangle_{L}\right)\quad\mbox{and}\quad X\left(\sqrt{2(\omega^{2}_{L}-\varepsilon)}\right)=-2X\left(\langle N_{L},\xi\rangle_{L}\right).
\end{equation}
Hence, inserting~\eqref{eq:3.15} in~\eqref{eq:3.12} we obtain
\begin{equation}\label{eq:3.17}\begin{split}
A_{R}(X)=&-\dfrac{1}{\omega_{L}}A_{L}(X)+\dfrac{1}{\omega_{L}^{3}}X\left(\langle N_{L},\xi\rangle_{L}\right)\left(-2(\omega^{2}_{L}-\varepsilon)\xi+\sqrt{2(\omega_{L}^{2}-\varepsilon)}N_{L}+2\omega^{2}_{L}\xi\right)\\
&-\,\dfrac{1}{\omega_{L}}\sqrt{2(\omega^{2}_{L}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi-W(N_{R},X).
\end{split}
\end{equation}
On the other hand, using one more time~\eqref{eq:2.7} and~\eqref{eq:2.25},
\begin{equation}\label{eq:3.18}
 \xi=T_{L}-\dfrac{\varepsilon}{2}\sqrt{2(\omega_{L}^{2}-\varepsilon)}N_{L}.
\end{equation}
Therefore, inserting~\eqref{eq:3.18} in~\eqref{eq:3.17} we obtain
\begin{equation}\label{eq:3.19}
A_{R}(X)=-\dfrac{1}{\omega_{L}}A_{L}(X)+\dfrac{2\varepsilon}{\omega_{L}^{3}}X\left(\langle N_{L},\xi\rangle_{L}\right)T_{L}-\dfrac{1}{\omega_{L}}\sqrt{2(\omega^{2}_{L}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi-W(N_{R},X).
\end{equation}
From~\eqref{eq:1.22},~\eqref{normal R} and~\eqref{eq:2.7}, a direct computation yields
\begin{equation}\label{eq:3.20}
\begin{split}
W(N_{R},X)&=-\dfrac{2}{\omega_{L}}\left(\langle X, \xi\rangle_{L}\overline{\nabla}^{L}_{N_{L}} \xi+\sqrt{2(\omega_{L}^{2}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi+\langle N_{L},\xi\rangle_{L}\overline{\nabla}^{L}_{X}\xi\right)\\
&=-\dfrac{2}{\omega_{L}}\left(\langle X, \xi\rangle_{L}\overline{\nabla}^{L}_{N_{L}} \xi+\sqrt{2(\omega_{L}^{2}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi-\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2}\overline{\nabla}^{L}_{X}\xi\right)\\
&=-\dfrac{2}{\omega_{L}}\left(\langle X, \xi\rangle_{L}\overline{\nabla}^{L}_{N_{L}} \xi+\dfrac{\sqrt{2(\omega_{L}^{2}-\varepsilon)}}{2}\overline{\nabla}^{L}_{X}\xi\right),
\end{split}
\end{equation}
and then
\begin{equation}\label{eq:3.21}
-\dfrac{1}{\omega_{L}}\sqrt{2(\omega^{2}_{L}-\varepsilon)}\overline{\nabla}^{L}_{X}\xi-W(N_{R},X)=\dfrac{2}{\omega_{L}}\langle X, \xi\rangle_{L}\overline{\nabla}^{L}_{N_{L}}\xi,
\end{equation}
for all $X\in\mathfrak{X}(\Sigma)$. Furthermore, from~\eqref{eq:1.11} and~\eqref{eq:2.25} we have
\begin{equation}\label{eq:3.22}
\overline{\nabla}^{L}_{N_{L}} \xi=-\tau(N_{L}\wedge_{L} \xi)=-\tau(N_{L}\wedge_{L}(T_{L}+\varepsilon \langle N_{L},\xi\rangle_{L}N_{L}))=-\tau J_{L}(T_{L}).
\end{equation}
The results follows now easily from~\eqref{eq:int2_L},~\eqref{eq:3.19},~\eqref{eq:3.21} and~\eqref{eq:3.22}.
\end{proof}

\begin{corollary}\label{cor:3.1}
Following the notation of Proposition~\ref{prop:3.1}, the following relation holds
\begin{equation}\label{eq:3.23}
\langle A_{R}(X),Y\rangle_{R}=-\dfrac{1}{\omega_{L}}\left(\langle A_{L}(X),Y\rangle_{L}-\tau\left(\langle J_{L}(X),Y\rangle_{R}+\langle J_{L}(Y),X\rangle_{R}\right)\right).
\end{equation}
Equivalently,
\begin{equation}\label{eq:3.24}
\langle A_{L}(X),Y\rangle_{L}=-\dfrac{1}{\omega_{R}}\left(\langle A_{R}(X),Y\rangle_{R}+\tau\left(\langle J_{R}(X),Y\rangle_{L}+\langle J_{R}(Y),X\rangle_{L}\right)\right),
\end{equation}
for every $X,Y\in\mathfrak{X}(\widehat\Sigma_0)$, on any connected component $\widehat{\Sigma}^2_0$ of $\widehat{\Sigma}^2$.
\end{corollary}

\begin{proof}
Given $X,Y\in\mathfrak{X}(\widehat{\Sigma}_0),$ from~\eqref{shape R} we have
\begin{equation}\label{eq:3.25}
	\begin{split}
\langle A_{R}(X),Y\rangle_{R}=&-\dfrac{1}{\omega_{L}}\langle A_{L}(X),Y\rangle_{R}-\dfrac{2\varepsilon}{\omega_{L}^{3}}\langle(A_{L}-\tau J_{L})(T_{L}),X\rangle_{L}\langle T_{L},Y\rangle_{R}\\
&-\dfrac{2\tau}{\omega_{L}}\langle T_{L},X\rangle_{L}\langle J_{L}(T_{L}),Y\rangle_{R}.
\end{split}
\end{equation}
On the other hand, from Lemma~\ref{lem:1.1},~\eqref{eq:2.7} and~\eqref{eq:2.25}, we collect 
\begin{equation}\label{eq:3.26}
\left\{
\begin{array}{ccc}
\langle A_{L}(X),Y\rangle_{R}\!\!\!\!&=&\!\!\!\!\langle A_{L}(X),Y\rangle_{L}+2\langle A_{L}(X),T_{L}\rangle_{L}\langle T_{L},Y\rangle_{L} \\
\langle J_{L}(T_L),X\rangle_{R}\!\!\!\!&=&\hspace{-4.7cm}\langle J_{L}(T_L),X\rangle_{L}\\
\langle T_{L},X\rangle_{R}\!\!\!\!\!\!\!\!\!\!\!\!&=&\hspace{-4.4cm}-\varepsilon\omega_{L}^{2}\langle T_{L},X\rangle_{L}
\end{array}
\right. .
\end{equation}
Hence, inserting these three last expressions in~\eqref{eq:3.25},
\begin{equation}\label{eq:3.27}
	\begin{split}
\langle A_{R}(X),Y\rangle_{R}=&-\dfrac{2\tau}{\omega_{L}}\left(\langle J_{L}(T_{L}),X\rangle_{L}\langle T_{L},Y\rangle_{L}+\langle J_{L}(T_{L}),Y\rangle_{L}\langle T_{L},X\rangle_{L}\right)\\
&-\dfrac{1}{\omega_{L}}\langle A_{L}(X),Y\rangle_{L}.
\end{split}
\end{equation}
By Lemma~\ref{lem:1.1}, the first term in the right-hand side of~\eqref{eq:3.27} can be computed as follows,
\begin{equation}\label{eq:3.28}
\langle J_{L}(T_{L}),X\rangle_{L}\langle T_{L},Y\rangle_{L}+\langle J_{L}(T_{L}),Y\rangle_{L}\langle T_{L},X\rangle_{L}=-\dfrac{1}{2}\left(\langle J_{L}(X),Y\rangle_{R}+\langle J_{L}(Y),X\rangle_{R}\right).
\end{equation}
Therefore, from~\eqref{eq:3.27} and~\eqref{eq:3.28} we obtain~\eqref{eq:3.23}.

With an analogous computation, we obtain~\eqref{eq:3.24}.
\end{proof}

Considering now the traces of the shape operators, we obtain the relation between the mean curvature functions $H_L$ and $H_R$.
\begin{corollary}\label{cor:3.2}
Let $\Sigma^{2}$ be a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$. Then, the mean curvatures functions $H_L$ and $H_R$ are related in any connected component $\widehat{\Sigma}^2_0$ of $\widehat{\Sigma}^2$ by
\begin{equation}\label{R-M.C}
H_{L}=-\dfrac{\varepsilon}{\omega_{R}}H_{R}+\dfrac{1}{\omega^{3}_{R}}\langle A_{R}(T_{R}),T_{R}\rangle_{R}.
\end{equation}
Equivalently,
\begin{equation}\label{L-M.C}
H_{R}=-\dfrac{\varepsilon}{\omega_{L}}H_{L}-\dfrac{\varepsilon}{\omega^{3}_{L}}\langle A_{L}(T_{L}),T_{L}\rangle_{L}.
\end{equation}
\end{corollary}

\begin{proof}
Fix $p\in \widehat{\Sigma}^{2}_0$ and let $\{e_{1},e_{2}\}$ be an orthonormal frame on a neighborhood of $p$ with respect to the metric $\langle\,,\rangle_{R}$ and such that $A_{R}(e_{i})=\lambda^{R}_{i}e_{i}$ for $i=1,2$. Thus, from Proposition~\ref{prop:3.1} we write
\begin{equation}\label{eq:3.29}
A_{L}(e_{i})=-\dfrac{\lambda^{R}_{i}}{\omega_{R}}e_{i}+\dfrac{2\varepsilon\lambda_{i}^{R}}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}T_{R}+\dfrac{2\varepsilon\tau}{\omega_{R}^{3}}\langle J_{R}(T_{R}),e_{i}\rangle_{R}T_{R}-\dfrac{2\tau}{\omega_{R}}\langle T_{R},e_{i}\rangle_{R}\,J_{R}(T_{R})
\end{equation}
Since $A_{L}(e_{i})=\sum_{j=1}^{2}a_{ij}^{L}e_{j}$ we have
\begin{equation}\label{eq:3.30}
a_{ii}^{L}=-\dfrac{\lambda^{R}_{i}}{\omega_{R}}+\dfrac{2\varepsilon\lambda^{R}_{i}}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}^{2}+\dfrac{2\tau(\varepsilon-\omega_{R}^{2})}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}\langle J_{R}(T_{R}),e_{i}\rangle_{R}.
\end{equation}\label{eq:3.31}
So, taking into account that $2\varepsilon H_{L}=\sum_{i=1}^{2}a_{ii}^{L}$ and that
\begin{equation}\label{eq:3.32}
\sum_{i=1}^{2}\langle J_{R}(T_{R}),\langle T_{R},e_{i}\rangle_{R}e_{i}\rangle_{R}=\langle J_{R}(T_{R}),T_{R}\rangle_{R}=0,
\end{equation}
we get
\begin{equation}\label{eq:3.33}
\varepsilon H_{L}=-\dfrac{1}{\omega_{R}}H_{R}+\dfrac{\varepsilon}{\omega^{3}_{R}}\langle A_{R}(T_{R}),T_{R}\rangle_{R}.
\end{equation}

For the proof of~\eqref{L-M.C}, we observe that~\eqref{R-M.C} implies
\begin{equation}\label{eq:3.34}
H_{R}=-\varepsilon\omega_{R}H_{L}+\dfrac{\varepsilon}{\omega^{2}_{R}}\langle A_{R}(T_{R}),T_{R}\rangle_{R}.
\end{equation}
Now, we claim that 
\begin{equation}\label{eq:3.35}
T_{R}=\dfrac{\varepsilon}{\omega_{L}^{2}}T_{L}.
\end{equation}
Indeed, from~\eqref{normal R},~\eqref{eq:2.8} and~\eqref{eq:2.24} (making $V=\xi$) we get
\begin{equation}\label{eq:3.35.1}
\begin{split}
\langle N_{R},\xi\rangle_{R}N_{R}&=\dfrac{1}{\omega_{L}}\langle N_{R},\xi\rangle_{R}\left(\sqrt{2(\omega_{L}^{2}-\varepsilon)}\xi-N_{L}\right)\\
&=\dfrac{1}{\omega_{L}^{2}}\left((\omega_{L}^{2}-\varepsilon)\xi+\langle N_{L},\xi\rangle_{L}N_{L}\right)\\
&=\xi-\dfrac{\varepsilon}{\omega_{L}^{2}}\left(\xi-\varepsilon\langle N_{L},\xi\rangle_{L}N_{L}\right)\\
&=\xi-\dfrac{\varepsilon}{\omega_{L}^{2}}T_{L},
\end{split}
\end{equation}
where in the last equality the right side of~\eqref{eq:2.25} has been used. Hence the claim follows from~\eqref{eq:2.25}.
So, inserting~\eqref{eq:3.35} in~\eqref{eq:3.23},
\begin{equation}\label{eq:3.36}
\langle A_{R}(T_{R}),T_{R}\rangle_{R}=-\dfrac{1}{\omega_{L}^{5}}\langle A_{L}(T_{L}),T_{L}\rangle_{L}.
\end{equation}
Therefore, using that $\omega_{L}\omega_{R}=1$, from~\eqref{eq:3.34} and~\eqref{eq:3.36} we conclude that
\begin{equation}
H_{R}=-\varepsilon\omega_{R}H_{L}-\dfrac{\varepsilon}{\omega^{3}_{L}}\langle A_{L}(T_{L}),T_{L}\rangle_{L}.
\end{equation}
\end{proof}

\section{Characterization of surfaces with $H_{R}=H_{L}=0$}\label{Sec:charact}

In this section we consider non-degenerate surfaces in $\mathbb{L}^{3}(\kappa,\tau)$ whose mean curvature functions with respect to both metrics $\langle\,,\rangle_{R}$ and $\langle\,,\rangle_{L}$ are identically zero.

\begin{theorem}\label{main result}
Let $\Sigma^{2}$ be an immersed surface in the Riemannian homogeneous space $\mathbb{E}^{3}(\kappa,\tau)$ which is non-degenerate with respect to the metric induced from $\mathbb{L}^3(\kappa,\tau)$. If\, $\Sigma^{2}$ has zero mean curvature with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$ then it is ruled both in $\mathbb{E}^{3}(\kappa,\tau)$ and in $\mathbb{L}^{3}(\kappa,\tau)$.
\end{theorem}

\begin{proof}
Let $\widehat{\Sigma}^2_0$ be any connected component of $\widehat{\Sigma}^2$ as defined in~\eqref{eq:IWOK}, and let us consider a point $p\in\widehat{\Sigma}^2_{0,\ast}$, where
$$\widehat{\Sigma}^2_{0,\ast}=\{p\in\widehat{\Sigma}^2_0\,;\,T_R(p)\neq 0\}.$$ Then, we can consider the orthonormal basis in $(\widehat{\Sigma}^2_{0,\ast},\langle,\rangle_R)$, $\{e_1,e_2\}\subseteq T_p\widehat{\Sigma}_{0,\ast}$, given by $e_1=T_R(p)/|T_R(p)|_R$ and $e_2=J_R(e_1)$. Let us see that the integral curve through $e_2$ at $p$ is a geodesic in $\mathbb{E}^{3}(\kappa,\tau)$. In the following we will omit the point $p$ for the sake of simplicity. 

Since $H_{R}=H_{L}=0$, from Corollary~\ref{cor:3.2} we get $\langle A_{R}(T_{R}),T_{R}\rangle_{R}=0$ and so $\langle A_{R}(e_{1}),e_{1}\rangle_{R}=0$. Thus, since $\mathrm{tr}_R(A_R)=0$ it also holds $\langle A_R(e_2),e_2\rangle_R=0$ and
\begin{equation}\label{eq:4.1}
A_{R}(e_{1})=\lambda e_{2}\qquad\mbox{and}\qquad A_{R}(e_{2})=\lambda e_{1},
\end{equation}
where $\lambda=\langle A_R(e_1),e_2\rangle_R=\langle A_R(e_2),e_1 \rangle_R$. Besides, the following identity holds,
\begin{equation}\label{eq:4.2}
\overline{\nabla}^{R}_{J_R(T_R)}J_R(T_R)=\overline{\nabla}^{R}_{J_R(T_R)}(N_{R}\wedge_{R}T_R)=\overline{\nabla}^{R}_{J_R(T_R)}N_{R}\wedge_{R}T_R+N_{R}\wedge_{R}\overline{\nabla}^{R}_{J_R(T_R)}T_R.
\end{equation}
On the one hand, since $\overline{\nabla}^{R}_{J_R(T_R)}N_{R}=-A_{R}(J_R(T_R))=-\lambda T_R$, we have
\begin{equation}\label{eq:4.3}
\overline{\nabla}^{R}_{J_R(T_R)}N_{R}\wedge_R T_R=-\lambda T_R\wedge_{R}T_R=0,
\end{equation}
and hence
\begin{equation}\label{eq:4.4}
\overline{\nabla}^{R}_{J_R(T_R)}J_R(T_R)=-\overline{\nabla}^{R}_{J_R(T_R)}T_R\wedge_{R}N_{R}.
\end{equation}
On the other hand, from~\eqref{eq:3.10} and~\eqref{eq:4.1}, we get
\begin{equation}\label{eq:4.5}
\nabla^{R}_{J_R(T_R)}T_R=\langle N_{R},\xi\rangle_{R}\left(A_{R}(J_R(T_R))+\tau T_R\right)=\langle N_{R},\xi\rangle_{R}(\lambda+\tau)T_R,
\end{equation}
so from the Gauss formula~\eqref{eq:3.3},
\begin{equation}\label{eq:4.9}
\overline{\nabla}^{R}_{J_R(T_R)}T_R=\nabla^{R}_{J_R(T_R)}T_R+\langle A_{R}(T_R),J_R(T_R)\rangle_{R}N_{R}=\langle N_{R},\xi\rangle_{R}(\lambda+\tau)T_R+\lambda |T_R|_R^2 N_{R}.
\end{equation}
Hence,
\begin{equation}\label{eq:4.10}
\overline{\nabla}^{R}_{J_R(T_R)}J_R(T_R)=-(\langle N_{R},\xi\rangle_{R}\left(\lambda+\tau)T_R+\lambda |T_R|_ R^2N_{R}\right)\wedge_{R}N_{R}=\langle N_{R},\xi\rangle_{R}(\lambda+\tau)J_R(T_R),
\end{equation}
which means that the integral curve of $J_R(T_R)$ passing through a point in $\Sigma^\ast$ is a pre-geodesic in $\mathbb{E}^{3}(\kappa,\tau)$. Observe that, since $J_R(T_R)$ is horizontal, $$\overline{\nabla}^{R}_{J_R(T_R)}J_R(T_R)=\overline{\nabla}^{L}_{J_R(T_R)}J_R(T_R),$$ so the same holds in $\mathbb{L}^{3}(\kappa,\tau)$. 

Let us assume now that $p\in\widehat{\Sigma}^2_0\setminus\widehat{\Sigma}^2_{0,\ast}$. The case $p\in\mathrm{int}(\widehat{\Sigma}_0\setminus\widehat{\Sigma}_{0,\ast})$ can only hold in the case where $\tau=0$, since from~\eqref{eq:1.5} in the case $\tau\neq 0$ the orthogonal distribution to $\xi$ is not (even locally) integrable in $\mathbb{E}^{3}(\kappa,\tau)$, so $\mathrm{int}(\widehat{\Sigma}_0\setminus\widehat{\Sigma}_{0,\ast})=\emptyset$. Therefore, in this case $\mathbb{E}^3(\kappa,0)=M^2(\kappa)\times\mathbb{R}$ and any connected component of $\mathrm{int}(\widehat{\Sigma}_0\setminus\widehat{\Sigma}_{0,\ast})$ is an open subset of a slice $M^2(\kappa)\times\{t_0\}$, for a certain $t_0\in\mathbb{R}$, which is ruled. Otherwise $p\in\partial\widehat{\Sigma}_{0,\ast}$ and a continuity argument shows that $\widehat{\Sigma}^2_0$ is ruled. Since any connected component of $\widehat{\Sigma}^2$ is ruled, $\widehat{\Sigma}^2$ is itself ruled.

The result follows again from a continuity argument. Given any $p\in\Sigma^2\setminus\widehat{\Sigma}^2$, either it is an endpoint of a horizontal ruling, or $\Sigma^2\setminus\widehat{\Sigma}^2$ is locally the limit of two families of horizontal rulings, and so it is a horizontal ruling itself. 
\end{proof}

As a consequence of Theorem~\ref{main result}, we can complete the classification of non-degenerate simultaneously minimal and maximal surface in $\mathbb{L}^{3}(\kappa,\tau)$ for any real values $\kappa$ and $\tau$. As stated in the introduction, such classification is already known for the case where $\tau=0$, see~\cite{Eva:19}, and for the case $\kappa=0$ and $\tau\neq 0$, so the ambient is the Heisenberg space, see~\cite{Shin:13}. Let us complete it by considering the case where $\mathbb{E}^{3}(\kappa,\tau)$ is either a Berger sphere or the universal cover of $SL_2(\mathbb{R})$.

\subsection{The Berger spheres $\mathbb{S}^{3}_{b}(\kappa,\tau)$}\label{Sec: Berger}

In the case when $\tau\neq0$ and $\kappa>0$, the homogeneous Riemannian space $\mathbb{E}^{3}(\kappa,\tau)$ is known as a Berger sphere and it is usually denoted by $\mathbb{S}^3_b(\kappa,\tau)$. It is possible to define the Berger spheres in an alternative way as follows. Taking complex coordinates $z=x+iy$ and $w=u+iv$ in $\mathbb{C}^{2}=\mathbb{R}^{4}$, we define the unit sphere $\mathbb{S}^{3}\subset\mathbb{R}^{4}$ as
\begin{equation}
\mathbb{S}^{3}=\{(z,w)\in\mathbb{C}^{2}\,;\,|z|^{2}+|w|^{2}=1\},
\end{equation}
which can be identified in a natural way with the special unitary group
\begin{equation}
SU_{2}(\mathbb{C})=\left\{B=
\left(
\begin{array}{cc}
 z & w   \\
 -\overline{w} & \overline{z} 
\end{array}
\right)
\in GL_2(\mathbb{C})\,;\,z\overline{z}+w\overline{w}=1\right\}.
\end{equation}
The Lie algebra $\mathfrak{su}_{2}$ is spanned by
\begin{equation}
X_{1}=
\left(
\begin{array}{ccc}
0 & 1\\
-1& 0
\end{array}
\right),\quad X_{2}=
\left(
\begin{array}{ccc}
0 & i\\
i & 0
\end{array}
\right)\quad\mbox{and}\quad X_{3}=
\left(
\begin{array}{ccc}
i & 0\\
0 & -i
\end{array}
\right),
\end{equation}
with the corresponding left invariant vector fields given by
\begin{equation*}
X_{1}(z,w)=
\left(
\begin{array}{ccc}
-w & z\\
-\overline{z} & -\overline{w}
\end{array}
\right),\quad X_{2}(z,w)=
\left(
\begin{array}{ccc}
iw & iz\\
i\overline{z} & -i\overline{w}
\end{array}
\right)\quad\mbox{and}\quad X_{3}(z,w)=
\left(
\begin{array}{ccc}
iz & -iw\\
-i\overline{w} & -i\overline{z}
\end{array}
\right),
\end{equation*}
which seen as tangent vector fields on $\mathbb{S}^{3}$ become
\begin{equation}\label{eq:5.18}
X_{1}(z,w)=(-w,z),\quad X_{2}(z,w)=(iw,iz)\quad\mbox{and}\quad X_{3}(z,w)=(iz,-iw).
\end{equation}

Associated to the left invariant vector fields~\eqref{eq:5.18}, we can consider the left invariant metric in $\mathbb{S}^{3}$ given by (see~\cite{Torralbo:10})
\begin{equation}\label{eq:metricBerger_R}
\langle X,Y\rangle_{R}=\dfrac{4}{\kappa}\left(\langle X,Y\rangle_{\mathbb{S}^{3}}+\left(\dfrac{4\tau^{2}}{\kappa}-1\right)\langle X,X_3\rangle_{\mathbb{S}^{3}}\langle Y,X_3\rangle_{\mathbb{S}^{3}}\right),
\end{equation}
where $\langle,\rangle_{\mathbb{S}^{3}}$ stands for the usual metric on the sphere. Then, the Berger sphere $\mathbb{S}^3_b(\kappa,\tau)$ is obtained as the usual 3-sphere $\mathbb{S}^3$ endowed with the metric~\eqref{eq:metricBerger_R}. We note that if $\kappa=4\tau^{2}$ then $\mathbb{S}_{b}^{3}(\kappa,\tau)$ is, up to homotheties, the round sphere. 

Analogously, the Lorentzian Berger sphere $\mathbb{S}^3_{b,1}(\kappa,\tau)$ can be constructed as $(\mathbb{S}^3,\langle ,\rangle_L)$ where
\begin{equation}\label{eq:metricBerger_L}
\langle X,Y\rangle_{L}=\dfrac{4}{\kappa}\left(\langle X,Y\rangle_{\mathbb{S}^{3}}-\left(\dfrac{4\tau^{2}}{\kappa}+1\right)\langle X,X_3\rangle_{\mathbb{S}^{3}}\langle Y,X_3\rangle_{\mathbb{S}^{3}}\right).
\end{equation}

It is well known that the Hopf fibration $\pi:\mathbb{S}^{3}_b(\kappa,\tau)\rightarrow\mathbb{S}^{2}(\kappa)$, where $\mathbb{S}^{2}(\kappa)$ stands for the $2$-sphere of radius $1/\sqrt{\kappa}$, defined by
\begin{equation}
\pi(z,w)=\dfrac{1}{\sqrt{\kappa}}\left(zw,\dfrac{1}{2}\left(|z|^{2}-|w|^{2}\right)\right),
\end{equation}
is a Riemannian submersion whose fibers are geodesics. Furthermore, the vertical unitary Killing vector field is given by $\xi=\dfrac{\kappa}{4\tau}X_{3}$. Let us observe that the fiber over $(z,w)$ is $\{(e^{i\theta}z,e^{-i\theta}w)\}$, to which the vector field $\xi$ is tangent.

\begin{proposition}
\label{ex:berger_alma}
The only non-degenerate surfaces in $\mathbb{S}^3_{b,1}(\kappa,\tau)$ having zero mean curvature with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$ are locally isometric to an open subset of a helicoid
\begin{equation}
\label{eq:par_berger}
X(s,t)=(e^{i\alpha s}\cos t,e^{i s}\sin t), \quad \alpha\in\mathbb{R}.
\end{equation}
\end{proposition}
\begin{proof}
Let $\Sigma^2$ be a non-degenerate surface immersed into $\mathbb{S}^3_{b,1}(\kappa,\tau)$ having zero mean curvature with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$. By Theorem~\ref{main result} $\Sigma^2$ is a ruled surface in the Berger sphere $\mathbb{S}^3_b(\kappa,\tau)$. Then, taking into account~\cite{Shin:15} $\Sigma^2$ is horizontally ruled by geodesics, it is congruent to an open subset of a helicoid parametrized by~\eqref{eq:par_berger} and it satisfies
\begin{equation}\label{eq:ruled_minimal}
\langle \overline\nabla^R_{X_s} X_s, X_s\times X_t \rangle_R=0.
\end{equation}
Furthermore, $X_t$ is a horizontal field, since it is tangent to the horizontal geodesics, and $X_s\perp X_t$. Therefore, $T_R$ is proportional to $X_s$. It remains to show that~\eqref{eq:par_berger} determines a surface with $H_L=0$ in $\mathbb{S}^3_{b,1}(\kappa,\tau)$.
	
From~\eqref{eq:ruled_minimal} we get
\begin{equation}\label{eq:aux_ex_berger} 
\langle A_R(X_s),X_s\rangle_R=-\langle \overline\nabla^R_{X_s} N_R,X_s\rangle_R=\langle\overline\nabla^R_{X_s}X_s,N_R\rangle_R=0,
\end{equation}
so it follows that $\langle A_R (T_R), T_R \rangle_R=0$. Thus from Corollary~\ref{cor:3.2} $H_L=0$ and the result follows.
\end{proof}

\subsection{The linear special group $SL_2(\mathbb{R})$}

When $\tau\neq 0$ and $\kappa <0$, $\mathbb{E}^{3}(\kappa,\tau)$ has the isometry group of the universal cover of the special linear group 
\begin{equation}\label{eq:SL2R}
SL_2(\mathbb{R})=\left\{\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\in GL_2(\mathbb{R}); ad-bc=1\right\}
\end{equation}
and of the special unitary group 
\begin{equation}\label{eq:SU11}
SU_{(1,1)}(\mathbb{C})=\left\{
\left(
\begin{array}{ccc}
z & w   \\
\overline{w} & \overline{z} 
\end{array}
\right)
\in GL_2(\mathbb{C})\,;\,z\overline{z}-w\overline{w}=1\right\}.
\end{equation}
Let us observe that the Lie algebra $\mathfrak{su}(1,1)$ is spanned by
\begin{equation}
X_{1}=
\left(
\begin{array}{ccc}
0 & 1\\
1& 0
\end{array}
\right),\quad X_{2}=
\left(
\begin{array}{ccc}
0 & i\\
-i & 0
\end{array}
\right)\quad\mbox{and}\quad X_{3}=
\left(
\begin{array}{ccc}
i & 0\\
0 & -i
\end{array}
\right),
\end{equation}
with the corresponding left invariant vector fields given by
\begin{equation}
X_{1}(z,w)=
\left(
\begin{array}{ccc}
\overline{w} & \overline{z}\\
z & w
\end{array}
\right),\quad X_{2}(z,w)=
\left(
\begin{array}{ccc}
i\overline{w} & i\overline{z}\\
-iz & -iw
\end{array}
\right)\quad\mbox{and}\quad X_{3}(z,w)=
\left(
\begin{array}{ccc}
iz & iw\\
-i\overline{w} & -i\overline{z}
\end{array}
\right).
\end{equation}

Then, the Riemannian universal cover of $SL_2(\mathbb{R})$, $\widetilde{SL}_2(\mathbb{R})$, can be modelled as $(SU_{1,1}(\mathbb{C}),\langle,\rangle_R)$, where $\langle,\rangle_R$ is the unique Riemannian metric on $SU_{1,1}(\mathbb{C})$ such that $\{X_1,X_2,X_3\}$ is an orthogonal frame with $\langle X_1, X_1\rangle_R=\langle X_2,X_2\rangle_R=4/\kappa$ and $\langle X_3,X_3\rangle_R=16\tau^2/\kappa^2$. Analogously, the Lorentzian universal cover of $SL_2(\mathbb{R})$, $\widetilde{SL}_{2,1}(\mathbb{R})$, can be identifed with $(SU_{1,1}(\mathbb{C}),\langle,\rangle_L)$, $\langle,\rangle_L$ being the unique Lorentzian metric on $SU_{1,1}(\mathbb{C})$ for which $\{X_1,X_2,X_3\}$ is an orthogonal frame with $\langle X_1, X_1\rangle_L=\langle X_2,X_2\rangle_L=4/\kappa$ and $\langle X_3,X_3\rangle_L=-16\tau^2/\kappa^2$.

\begin{example}\label{E.H. PSL}
Let us define the following curves on $SU_{1,1}(\mathbb{C})$:	
\begin{equation}\label{eq:lambda_esp1}
\Lambda_{e}(s)=
\left(
\begin{array}{ccc}
e^{is}  & 0 \\
0  & e^{-is}
\end{array}
\right),\,\Lambda_{p}(s)=
\left(
\begin{array}{ccc}
1+is  & -is \\
is & 1-is
\end{array}
\right),\,\Lambda_{h}(s)=
\left(
\begin{array}{ccc}
\cosh(s)  & \sinh(s) \\
\sinh(s) & \cosh(s)
\end{array}
\right),
\end{equation}
\begin{equation}\label{eq:lambda_esp2}
\Lambda_{p1}(s)=
\left(
\begin{array}{ccc}
1+is  & is \\
-is & 1-is
\end{array}
\right)\quad\mbox{and}\quad\Lambda_{h1}(s)=
\left(
\begin{array}{ccc}
\cosh(s) & i\sinh(s) \\
-i\sinh(s) & \cosh(s)
\end{array}
\right),
\end{equation}
and let $\Lambda(s)$ denote any of 	
\begin{equation}\label{eq:lambda}
\Lambda_{e}(-2\alpha s),\quad\Lambda_{h1}(2\alpha s),\quad\Lambda_{p1}(bs)\quad{and}\quad\Lambda_{p}(-as),\quad\alpha,a,b\in\mathbb{R}.
\end{equation}	
Then, for any arbitrary $A\in SU_{1,1}(\mathbb{C})$ the following map can be defined
\begin{equation}\label{H. PSL}
\left.
\begin{array}{cccccc}
X:\!\!\!\!\!\!&\mathcal{U}&\longrightarrow &SU_{1,1}(\mathbb{C})&\\
  &(s,t)&\longmapsto&X(s,t)&\!\!\!\!\!=&\!\!\!\!A\,\Lambda(s)\Lambda_{h}(\kappa^2t/2)\Lambda_{e}(s/2)
\end{array}
\right. , 
\end{equation}
where
\begin{equation}
\mathcal{U}=
\left\{
\begin{array}{ccccccccc}
\mathbb{R}\times\mathbb{R}^{+} & or & \mathbb{R}\times\mathbb{R}^{-},&\mbox{if}\,\,\,\Lambda(s)&\!\!\!\!\!\!=\Lambda_{e}(-2\alpha s)\,\,\mbox{and}\,\,\alpha=1/4\\
\!\!\!\mathbb{R}\times\mathbb{R}, & & &  \mbox{otherwise}&
\end{array}
\right. .
\end{equation}

In~\cite{Shin:18}, Y. W. Kim et al., proved that~\eqref{H. PSL} determines a ruled minimal surface in $SU_{1,1}(\mathbb{C})$. Furthermore, they showed that every ruled minimal surface in $SU_{1,1}(\mathbb{C})$ can be parametrized as in~\eqref{H. PSL} and it is a helicoid in the sense of~\cite[Definition 3.4]{Shin:18}.
\end{example}

\begin{proposition}\label{prop_SL2R}
The only non-degenerate surfaces in the universal cover of the Lorentzian special linear group $\widetilde{SL}_{2,1}(\mathbb{R})$ having zero mean curvature with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$ are locally isometric to an open subset of a helicoid as described in Example~\ref{E.H. PSL}.
\end{proposition}
\begin{proof}
The proof is analogous to the one of Proposition~\ref{ex:berger_alma}. In fact, let $\Sigma^2$ be a non-degenerate surface immersed into $SU_{1,1}(\mathbb{C})$ having zero mean curvature with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$. By Theorem~\ref{main result} $\Sigma^2$ is a ruled surface. Then, taking into account~\cite{Shin:18} $\Sigma^2$ is horizontally ruled by geodesics, it is congruent to an open subset of~\eqref{H. PSL} and it satisfies
\begin{equation}\label{eq:ruled_minimal_PSL}
\langle \overline\nabla^R_{X_s} X_s, X_s\times X_t \rangle_R=0.
\end{equation}
Then, the proof concludes as in Proposition~\ref{ex:berger_alma} since again in~\cite{Shin:18} it is shown that $X_t$ is a horizontal field and $X_s\perp X_t$.
\end{proof}

\section{Characterization of spacelike surfaces with $H_{R}=H_{L}$}\label{sec:4}

In this section we will consider the general situation where $\Sigma^2$ is a spacelike surface in $\mathbb{L}^{3}(\kappa,\tau)$ such that $H_R=H_L$ not necessarily zero. Before proving the main result of this secion, we previously need some general computations related to the normal curvatures of $\Sigma^2$. Those first computations are true even in the case of a non-degenerate surface.

Given $\Sigma^{2}$ a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$, $p\in\Sigma^2$ and $v\in T_p\Sigma$, let $\lambda_v^R(p)$ and $\lambda_v^L(p)$ denote the normal curvatures at $p$ in the direction of $v$ with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$, respectively. From here on we will work at $p$, but for simplicity we will omit it. By definition, given $\alpha$ a smooth curve on $\Sigma^2$ such that $\alpha(0)=p$ and $\alpha'(0)=v$ we have
\begin{equation}\label{eq:4.1_normcurv}
\lambda_v^R=\langle A_R(t_R),t_R\rangle_R
\quad \textrm{and} \quad
\lambda_v^L=\varepsilon_v\langle A_L(t_L),t_L\rangle_L,
\end{equation}
where $t_L=\frac{\alpha'}{|\alpha'|_L}$, $t_R=\frac{\alpha'}{|\alpha'|_R}$, $\varepsilon_v=1$ if $v$ is spacelike with respect to $\langle,\rangle_L$ and $\varepsilon_v=-1$ if $v$ is timelike.

Considering $X=Y=t_L$ in Corollary~\ref{cor:3.1}, we have
\begin{equation}\label{eq:4.11}
\langle A_{L}(t_{L}),t_{L}\rangle_{L}=-\dfrac{1}{\omega_{R}}\langle A_{R}(t_{L}),t_{L}\rangle_{R}-\dfrac{2\tau}{\omega_{R}}\langle J_{R}(t_{L}),t_{L}\rangle_{L}.
\end{equation}
Furthermore,
\begin{equation}\label{eq:4.12}
\langle A_{R}(t_{L}),t_{L}\rangle_{R}=\dfrac{|\alpha'|_{R}^{2}}{|\alpha'|_{L}^{2}}\langle A_{R}(t_{R}),t_{R}\rangle_{R}=\dfrac{|\alpha'|_{R}^{2}}{|\alpha'|_{L}^{2}}\lambda_{v}^{R}
\end{equation}
and
\begin{equation}\label{eq:4.13}
\langle J_{R}(t_{L}),t_{L}\rangle_{L}=\dfrac{|\alpha'|_{R}^{2}}{|\alpha'|_{L}^{2}}\langle J_{R}(t_{R}),t_{R}\rangle_{L}.
\end{equation}
Hence, from~\eqref{eq:4.1_normcurv} it follows
\begin{equation}\label{eq:relacion_lambda}
\varepsilon_v\lambda_v^L=-\frac{|\alpha'|^2_R}{\omega_R|\alpha'|^2_L}\,\left(\lambda_v^R+2\tau\left\langle\frac{\alpha'}{|\alpha'|_R},J_R\left(\frac{\alpha'}{|\alpha'|_R}\right)\right\rangle_L\right).
\end{equation}

It is worth pointing out that~\eqref{eq:relacion_lambda} holds for any non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$, without any assumption on its mean curvature functions.

We are now in condition to state and prove our last main result for spacelike surfaces satisying $H_R=H_L$.

\begin{theorem}\label{teo:4.1}
Let $\Sigma^2$ be a non-degenerate surface in $\mathbb{L}^{3}(\kappa,\tau)$ such that $H_R=H_L$, then the shape operator $A_R$ is nowhere definite. That is, not all the principal curvatures of $(\Sigma^{2},\langle\,,\rangle_R)$ have the same sign.
\end{theorem}

\begin{proof}
Given $p\in\Sigma^2$, let $\{v_1,w_1\}$ and $\{v_2,w_2\}$ be two orthogonal basis of $T_p\Sigma$ with respect to $\langle,\rangle_R$ and $\langle,\rangle_L$, respectively, then
\begin{equation}\label{eq:4.6}
H_R=\frac{1}{2}\left(\lambda_{v_1}^R+\lambda_{w_1}^R\right) \qquad \textrm{and} \qquad H_L=-\frac{1}{2}\left(\lambda_{v_2}^L+\lambda_{w_2}^L\right).
\end{equation}
The idea of the proof is to consider a basis of $T_p\Sigma$  which is simultaneously orthogonal with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$. To do that, let us first consider a point $p$ in the open subset of $\Sigma^2$ given by $\Sigma^2_\ast=\{p\in\Sigma^2 \,;\, T_R(p)\neq 0\}$. Then, if $v=T_R/|T_R|_R$, $\{v,J_R (v)\}$ is an orthogonal basis with respect to both metrics $\langle,\rangle_R$ and $\langle,\rangle_L$ satisfying $$\langle J_R(v),J_R(v)\rangle_R=\langle J_R(v),J_R(v)\rangle_L=1,\quad \langle v,v\rangle_R=1$$ and, taking into account~\eqref{eq:1.18} and~\eqref{eq:2.15.bis},
\begin{equation}\label{eq:4.7}
\langle v,v \rangle_L=2\langle v^h,v^h\rangle_R-1=2\langle N_R,\xi\rangle_R^2-1=\frac{1}{\omega^2_L},
\end{equation}
where we are omitting the point $p$ for the sake of simplicity.
Therefore, from~\eqref{eq:relacion_lambda} we get the following relations,
\begin{equation}\label{eq:4.8}
\lambda_v^L=-\omega_L^3\lambda_v^R \qquad \textrm{and} \qquad \lambda_{J_R(v)}^L=- \omega_L\lambda_{J_R(v)}^R.
\end{equation}
Since $H_R=H_L$ by assumption, from~\eqref{eq:4.6} we get
\begin{equation}\label{eq:4.9bis}
\lambda_v^R+\lambda_{J_R(v)}^R=-\lambda_v^L-\lambda_{J_R(v)}^L,
\end{equation}
so from a straightforward computation using~\eqref{eq:4.8} we finally obtain
\begin{equation}\label{eq:4.10bis}
\lambda_{v}^R=-\left(\frac{1-\omega_L}{1-\omega_L^3}\right)\lambda_{J_R(v)}^R,
\end{equation}
so $\lambda_v^R\lambda_{J_R(v)}^R\leq 0$ since from~\eqref{eq:2.5} it holds $\omega_L\geq 1$. Consequently the same inequality holds for the principal curvatures on $\Sigma^2_\ast$.
	
Consider now $p\in\mathrm{int}(\Sigma^2\setminus\Sigma^2_\ast)$, then $T_R$ vanishes on a neighbourhood at $p$, so $N_R=\xi$ and $\Sigma^2$ is locally a horizontal surface. We can easily observe from~\eqref{eq:1.8} that in this case $\langle A_R(E_1),E_1\rangle_R=\langle A_R(E_2),E_2\rangle_R=0$ and $\langle A_L(E_1),E_1\rangle_L=\langle A_L(E_2),E_2\rangle_L=0$, so $H_R=H_L=0$. Otherwise, $p\in\partial\Sigma_\ast$ and the result follows by a continuity argument.	
\end{proof}

\begin{remark}\label{rem:4.1}
In the case where the ambient space is a Hadamard manifold, it is possible to deduce from Theorem~\ref{teo:4.1} a non-convexity result for the surface $\Sigma^{2}$. This is the case, for instance, when $\kappa=\tau=0$, so $\mathbb{E}^{3}(0,0)=\mathbb{R}^3$ and $\mathbb{L}^{3}(0,0)=\mathbb{L}^3$, which was previously studied in~\cite{Alma:17}, and when $\kappa<0$ and $\tau=0$. This is the case where $\mathbb{E}^3(\kappa,0)=\mathbb{H}^2(\kappa)\times\mathbb{R}$ and $\mathbb{L}^3(\kappa,0)=\mathbb{H}^2(\kappa)\times\mathbb{R}_1$ and has been studied in~\cite{Alma:preprint}.
\end{remark}

\section{Characterization of timelike surfaces with the same Gaussian and extrinsic curvature}\label{sec:5}

In this last section we will consider non-degenerate surfaces in the homogeneous space $\mathbb{L}^{3}(\kappa,\tau)$ such that $\kappa\neq 4\tau^2$ and $\kappa \neq -4\tau^2$, with the same extrinsic and Gaussian curvature when considered as immersions into $\mathbb{L}^3(\kappa,\tau)$ and into $\mathbb{E}^3(\kappa,\tau)$, and as a consequence we will characterize timelike surfaces satisfying this property. 

Given $\Sigma^2$ a non-degenerate surface in $\mathbb{L}^3(\kappa,\tau)$ with $\kappa\neq 4\tau^ 2$ and $\kappa \neq -4\tau^2$, $\overline{K}_{R}$ and $\overline{K}_{L}$ will denote the sectional curvatures in $\mathbb{L}^{3}(\kappa,\tau)$ and $\mathbb{E}^3(\kappa,\tau)$, respectively, of the non-degenerate tangent plane to $\Sigma^{2}$. Following~\cite{Daniel:07}, the curvature tensor\footnote{We adopt for the $(1,3)$-curvature tensor of the spacetime the following definition (\cite[Chapter 3]{O'Neill:83}): $\overline{R}(X,Y)Z=\overline{\nabla}_{[X,Y]}Z-[\overline\nabla_X,\overline\nabla_Y]Z$.} of $\mathbb{E}^3(\kappa,\tau)$ is given by
\begin{equation}\label{eq:5.2}
	\begin{split}
\overline{R}_{R}(X,Y)Z=&(\kappa-3\tau^{2})(\langle X,Z\rangle_{R}Y-\langle Y,Z\rangle_{R}X)\\
&+(\kappa-4\tau^{2})\langle Z,\xi\rangle_{R}(\langle Y,\xi\rangle_{R}X-\langle X,\xi\rangle_{R}Y)\\
&+(\kappa-4\tau^{2})(\langle Y,Z\rangle_{R}\langle X,\xi\rangle_{R}-\langle X,Z\rangle_{R}\langle Y,\xi\rangle_{R})\xi,
\end{split}
\end{equation}
for any $X,Y,Z\in\mathfrak{X}(\mathcal{D})$. Analogously, it is easy to check that the curvature tensor of $\mathbb{L}^{3}(\kappa,\tau)$ is given by
\begin{equation}\label{eq:5.2.1}
	\begin{split}
\overline{R}_{L}(X,Y)Z=&(\kappa+3\tau^{2})(\langle X,Z\rangle_{L}Y-\langle Y,Z\rangle_{L}X)\\
&-(\kappa+4\tau^{2})\langle Z,\xi\rangle_{L}(\langle Y,\xi\rangle_{L}X-\langle X,\xi\rangle_{L}Y)\\
&-(\kappa+4\tau^{2})(\langle Y,Z\rangle_{L}\langle X,\xi\rangle_{L}-\langle X,Z\rangle_{L}\langle Y,\xi\rangle_{L})\xi,
\end{split}
\end{equation}
where $X,Y,Z\in\mathfrak{X}(\mathcal{D})$.

Let $\{e_1,e_2\}$ be a local orthonormal frame on $(\Sigma^{2},\langle,\rangle_L)$ such that $\langle e_{1},e_{1}\rangle_{L}=1$ and $\langle e_{2},e_{2}\rangle_{L}=-\varepsilon$, where we recall that $\varepsilon=\langle N_L,N_L\rangle_L$. Thus, from~\eqref{eq:2.26} and~\eqref{eq:5.2},
\begin{equation}\label{eq:5.3}
\overline{K}_{L}=\langle\overline{R}_{L}(e_{1},e_{2})e_{1},e_{2}\rangle_{L}=\varepsilon\tau^{2}+(\kappa+4\tau^{2})\langle N_{L},\xi\rangle_{L}^{2}.
\end{equation}
Similarly, if $\{f_1,f_2\}$ is a local orthonormal frame on $(\Sigma^{2},\langle,\rangle_R)$, we can also show that
\begin{equation}\label{eq:5.4}
\overline{K}_{R}=\langle\overline{R}_{R}(f_{1},f_{2})f_{1},f_{2}\rangle_{R}=\tau^{2}+(\kappa-4\tau^{2})\langle N_{R},\xi\rangle_{R}^{2}.
\end{equation}
Hence from~\eqref{eq:2.24}, \eqref{eq:5.3} and~\eqref{eq:5.4}, we obtain the following relation between both sectional curvatures,
\begin{equation}\label{eq:5.5}
\overline{K}_{R}=\dfrac{1}{\omega_{L}^{2}}\left(\tau^{2}\left(\omega_{L}^{2}-\varepsilon A\right)+A\overline{K}_{L}\right),
\end{equation}
with
\begin{equation}\label{eq:5.5.1}
A=\dfrac{\kappa-4\tau^{2}}{\kappa+4\tau^{2}}\neq0.
\end{equation}

From now on, we will denote by $K_{e}^{R}:=\textrm{det}(A_R)$ the extrinsic curvature of $\psi:\Sigma^2\rightarrow \mathbb{E}^3(\kappa,\tau)$ and by $K_{e}^{L}:=\textrm{det}(A_L)$ the extrinsic curvature of $\psi:\Sigma^2\rightarrow\mathbb{L}^3(\kappa,\tau)$. The next result guarantees a relationship between them.

\begin{proposition}\label{prop:5.1}
Let $\Sigma^{2}$ be a non-degenerate surface immersed into the Lorentzian homogeneous space $\mathbb{L}^{3}(\kappa,\tau)$. Then, the extrinsic curvatures are related as follows
\begin{equation}\label{eq:5.6}
K_{e}^{L}=-\dfrac{\varepsilon}{\omega_{R}^{4}}K_{e}^{R}+\dfrac{4\tau\varepsilon}{\omega_{R}^{4}}\left(\langle A_R(T_R),J_R(T_R)\rangle_R+\tau|T_{R}|_{R}^{2}\right).
\end{equation}
\end{proposition}

\begin{proof}
Firstly, let us recall that
\begin{equation}\label{eq:5.7}
a_{ii}^{L}=-\dfrac{\lambda^{R}_{i}}{\omega_{R}}+\dfrac{2\varepsilon\lambda^{R}_{i}}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}^{2}+\dfrac{2\tau(\varepsilon-\omega_{R}^{2})}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}\langle J_{R}(T_{R}),e_{i}\rangle_{R}
\end{equation}
and
\begin{equation}\label{eq:5.8}
a_{ij}^{L}=\dfrac{2\varepsilon\lambda^{R}_{i}}{\omega_{R}^{3}}\langle T_{R},e_{i}\rangle_{R}\langle T_{R},e_{j}\rangle_{R}+\dfrac{2\tau\varepsilon}{\omega_{R}^{3}}\langle J_{R}(T_{R}),e_{i}\rangle_{R}\langle T_{R},e_{j}\rangle_{R}-\dfrac{2\tau}{\omega_{R}}\langle T_{R},e_{i}\rangle_{R}\langle J_{R}(T_{R}),e_{j}\rangle_{R},
\end{equation}
for $i\neq j$, where $\{e_{1},e_{2}\}$ is an orthonormal frame on $\Sigma^{2}$ adapted to the Riemannian metric $\langle\,,\rangle_{R}$ such that $J_R(e_1)=e_2$,  $A_{R}(e_{1})=\lambda_{1}^{R}e_{1}$ and $A_{R}(e_{2})=\lambda_{2}^{R}e_{2}$. Therefore, after a straightforward computation we get
\begin{equation}\label{eq:5.9}
\begin{split}
{\rm det}(A_{L})=&\dfrac{\lambda^{R}_{1}\lambda^{R}_{2}}{\omega_{R}^{2}}-\dfrac{2\varepsilon\lambda_{1}^{R}\lambda_{2}^{R}}{\omega_{R}^{4}}|T_{R}|_{R}^{2}-\dfrac{2\tau(\varepsilon-\omega_{R}^{2})}{\omega_{R}^{4}}(\lambda_{1}^{R}-\lambda_{2}^{R})\langle T_{R},e_{1}\rangle_{R}\langle T_{R},e_{2}\rangle_{R}\\
&-\dfrac{4\tau\varepsilon}{\omega_{R}^{4}}(\lambda_{1}^{R}-\lambda_{2}^{R})\langle T_{R},e_{1}\rangle_{R}\langle T_{R},e_{2}\rangle_{R}|T_{R}|^{2}_{R}
+\dfrac{4\tau^{2}\varepsilon}{\omega_{R}^{4}}|T_{R}|_{R}^{4}.
\end{split}
\end{equation}

On the one hand, since $T_{R}=\langle T_{R},e_{1}\rangle_{R}e_{1}+\langle T_{R},e_{2}\rangle_{R}e_{2}$, we have
\begin{equation}\label{eq:5.10}
\langle A_{R}(T_{R}),J_{R}(T_{R})\rangle_{R}=-(\lambda_{1}^{R}-\lambda_{2}^{R})\langle T_{R},e_{1}\rangle_{R}\langle T_{R},e_{2}\rangle_{R}.
\end{equation}
Thus,
\begin{equation}\label{eq:5.11}
\begin{split}
{\rm det}(A_{L})=&\dfrac{\varepsilon}{\omega_{R}^{4}}\left(\varepsilon\omega_R^{2}-2|T_{R}|_{R}^{2}\right){\rm det}(A_{R})+\dfrac{4\tau^{2}\varepsilon}{\omega_{R}^{4}}|T_{R}|_{R}^{4}\\
&+\dfrac{2\tau}{\omega_{R}^{4}}\langle A_{R}(T_{R}),J_{R}(T_{R})\rangle_{R}\left(\varepsilon-\omega_{R}^{2}+2\varepsilon|T_{R}|^{2}_{R}\right).
\end{split}
\end{equation}
On the other hand, equations~\eqref{eq:2.19} and~\eqref{eq:2.26} give
\begin{equation}\label{eq:5.12}
\varepsilon\omega_{R}^{2}-2|T_{R}|_{R}^{2}=-1 \quad\mbox{and}\quad\varepsilon-\omega_{R}^{2}+2\varepsilon|T_{R}|_{R}^{2}=2\varepsilon,
\end{equation}
so that
\begin{equation}\label{eq:5.13}
{\rm det}(A_{L})=-\dfrac{\varepsilon}{\omega_{R}^{4}}{\rm det}(A_{R})+\dfrac{4\tau\varepsilon}{\omega_{R}^{4}}\langle A_{R}(T_{R}),J_{R}(T_{R})\rangle_{R}+\dfrac{4\tau^{2}\varepsilon}{\omega_{R}^{4}}|T_{R}|_{R}^{4}.
\end{equation}
Hence, we get the desired result.
\end{proof}

For the next results, we recall that the Gauss equation of $(\Sigma^{2},\langle\,,\rangle_R)$ and $(\Sigma^2,\langle\,,\rangle_L)$ are given, respectively, by
\begin{equation}\label{eq:5.14}
K_{R}=\overline{K}_{R}+K_{e}^{R}\quad\mbox{and}\quad K_{L}=\overline{K}_{L}+\varepsilon K_{e}^{L},
\end{equation}
where $K_R$ and $K_L$ stand for the Gaussian curvature of $(\Sigma^{2},\langle,\rangle_R)$ and $(\Sigma^2,\langle,\rangle_L)$, respectively. So, from~\eqref{eq:5.3} and~\eqref{eq:5.4} we obtain
\begin{equation}\label{eq:5.18.1}
K_{R}=\tau^{2}+(\kappa-4\tau^{2})\langle N_{R},\xi\rangle_{R}^{2}+K_{e}^{R}
\end{equation}
and
\begin{equation}\label{eq:5.18.2}
K_{L}=\varepsilon\tau^{2}+(\kappa+4\tau^{2})\langle N_{L},\xi\rangle_{L}^{2}+\varepsilon K_{e}^{L}.
\end{equation}
Furthermore, with a straightforward computation, by~\eqref{eq:5.5} and~\eqref{eq:5.14}, we have
\begin{equation}\label{eq:5.15}
\begin{split}
K_{R}=&\dfrac{1}{\omega_{L}^{2}}\left(\tau^{2}(\omega_{L}^{2}-\varepsilon A)+A\overline{K}_{L}\right)+K_{e}^{R}\\
=&\dfrac{A}{\omega_{L}^{2}}\overline{K}_{L}+\dfrac{(\omega_{L}^{2}-\varepsilon A)\tau^{2}}{\omega_{L}^{2}}+K_{e}^{R}\\
=&\dfrac{A}{\omega_{L}^{2}}\left(K_{L}-\varepsilon K_{e}^{L}\right)+\dfrac{(1+\omega_{L}^{2})\tau^{2}}{\omega_{L}^{2}}+K_{e}^{R}\\
=&\dfrac{A}{\omega_{L}^{2}}K_{L}+\dfrac{(\omega_{L}^{2}-\varepsilon A)\tau^{2}}{\omega_{L}^{2}}+\dfrac{1}{\omega_{L}^{2}}\left(-\varepsilon A K_{e}^{L}+\omega_{L}^{2}K_{e}^{R}\right).
\end{split}
\end{equation}
Equivalently,
\begin{equation}\label{eq:5.16}
\omega_{L}^{2}K_{R}-A K_{L}=(\omega_{L}^{2}-\varepsilon A)\tau^{2}-\varepsilon A K_{e}^{L}+\omega_{L}^{2}K_{e}^{R}.
\end{equation}

Taking into account the previous digression, we can provide a first result about the geometry of timelike surfaces such that $K_{R}=K_{L}=K$ and $K^{R}_{e}=K^{L}_{e}=K_{e}$.

\begin{theorem}\label{teo:5.1}
Any timelike surface immersed into $\mathbb{L}^{3}(\kappa,\tau)$ with $\kappa+4\tau^{2}>0$ and $\tau\neq0$ satisfying $K_{R}=K_{L}$ and $K^{R}_{e}=K^{L}_{e}$ are open pieces of Hopf surfaces.
\end{theorem}

\begin{proof}
Let us consider $\Sigma^{2}$ a timelike surface in $\mathbb{L}^{3}(\kappa,\tau)$ such that $\kappa+4\tau^2>0$. Then $\varepsilon=1$ and since $K_{R}=K_{L}=K$ and $K^{L}_{e}=K^{L}_{e}=K_{e}$ we get
\begin{equation}\label{eq:5.16_b}
(\omega_{L}^{2}-A)K=(\omega_{L}^{2}-A)\tau^{2}+(\omega_{L}^{2}-A)K_{e}.
\end{equation}
Therefore, either $\omega_{L}^{2}=A$ or $K=\tau^{2}+K_{e}$. 

Observe that, if $\omega_{L}^{2}=A$, from~\eqref{eq:5.5.1} and~\eqref{eq:2.5}, we have
\begin{equation}
\langle N_{L},\xi\rangle_{L}^{2}=-\dfrac{4\tau^{2}}{\kappa+4\tau^{2}}\leq0.
\end{equation}
Consequently $\tau=0$, which is a contradiction. Thus $\omega_{L}^{2}\neq A$, and therefore $K=\tau^{2}+K_{e}$. Then~\eqref{eq:5.18.2} yields $\langle N_{L},\xi\rangle_{L}=0$. Hence, from~\eqref{eq:2.5} it holds $\omega_{L}=1$ and Proposition~\ref{prop:1.1} implies that $\Sigma^{2}$ is locally isometric to a Hopf surface.
\end{proof}

In the case where $\mathbb{L}^3(\kappa,\tau)$ is a Lorentzian Berger sphere, the condition $\kappa+4\tau^{2}>0$ and $\tau\neq0$ is trivially satisfied. Therefore,
\begin{corollary}\label{cor:5.1}
The only timelike surfaces immersed in the Lorentz Berger sphere $\mathbb{S}^{3}_{b,1}(\kappa,\tau)$ satisfying $K_{R}=K_{L}$ and $K^{R}_{e}=K^{L}_{e}$ are open pieces of Hopf surfaces.
\end{corollary}

We observe that in the Heisenberg space $A=-1$, hence, from~\eqref{eq:5.16_b} and following the same steps as in the proof of Theorem~\ref{teo:5.1} we obtain the following consequence,

\begin{corollary}\label{cor:5.2}
The only timelike surfaces immersed in the Lorentz Heisenberg space $Nil^{3}_{1}(\tau)$ satisfying $K_{R}=K_{L}$ and $K^{R}_{e}=K^{L}_{e}$ are open pieces of Hopf surfaces.
\end{corollary}

We close this paper with the case $\tau=0$.
\begin{proposition}\label{prop:5.2}
The only timelike surfaces immersed in the Lorentz product space $M^{3}(\kappa)\times\mathbb{R}_{1}$ satisfying $K_{R}=K_{L}$ and $K^{R}_{e}=K^{L}_{e}$ are open pieces of cylinders over a regular curve of $M^{2}(\kappa)$.
\end{proposition}

\begin{proof}
Making $\tau=0$ in Proposition~\ref{prop:5.1}, we get
\begin{equation}
(\omega_{R}^{4}+1)K_{e}=0,
\end{equation}
which implies $K_{e}=0$. On the other hand, from~\eqref{eq:5.5.1} we have $A=1$. So, from~\eqref{eq:2.5}, \eqref{eq:5.18.1} (or~\eqref{eq:5.18.2}) and~\eqref{eq:5.16},
\begin{equation}
\omega_{L}^{2}=1\quad\mbox{and}\quad K=0.
\end{equation}
Hence, $\Sigma^{2}$ is locally isometric to a cylinder $\gamma\times\mathbb{R}$ where $\gamma:I\subset\mathbb{R}\rightarrow M^{2}(\kappa)$ is a regular curve.
\end{proof}

\begin{remark}
In the case where $\kappa=\tau=0$, $\mathbb{E}^{3}(0,0)\equiv\mathbb{R}^{3}$ and $\mathbb{L}^{3}(0,0)\equiv\mathbb{L}^{3}$, so for any non-degenerate surface $\Sigma^2$ it holds $K_{R}=K_{e}^{R}$ and $K_{L}=K_{e}^{L}$. Consequently, if $K_{R}=K_{L}$ it follows from Proposition~\ref{prop:5.1} that $K_{R}=K_{L}=0$. Therefore, from Theorem~$4.11$ of~\cite{Eva:19}, $\Sigma^{2}$ is isometric to an open piece of one of the following surfaces: planes, cones, cylinders and tangent developable surfaces.
\end{remark}

\address{
Departamento de Matem\'aticas \\
Universidad de C\'ordoba \\
14071 Campus de Rabanales, C\'ordoba\\
Spain
}
{alma.albujer@uco.es}

\address{
Departamento de Matem\'atica \\
Universidade Federal de Pernambuco\\
50.740-540 Recife, Pernambuco \\
Brazil
}
{fabio.reis@ufpe.br}

\end{document}