Two-step homogeneous geodesics in pseudo-Riemannian manifolds

Abstract

Given a homogeneous pseudo-Riemannian space $(G/H,\langle \ , \ \rangle),$ a geodesic $\gamma :I\rightarrow G/H$ is said to be two-step homogeneous if it admits a parametrization $t=\phi (s)$ (s affine parameter) and vectors X, Y in the Lie algebra ${\mathfrak{g}}$, such that $\gamma (t)=\exp (tX)\exp (tY)\cdot o$, for all $t\in \phi (I)$. As such, two-step homogeneous geodesics are a natural generalization of homogeneous geodesics (i.e., geodesics which are orbits of a one-parameter group of isometries). We obtain characterizations of two-step homogeneous geodesics, both for reductive homogeneous spaces and in the general case, and undertake the study of two-step g.o. spaces, that is, homogeneous pseudo-Riemannian manifolds all of whose geodesics are two-step homogeneous. We also completely determine the left-invariant metrics $\langle \ ,\ \rangle$ on the unimodular Lie group $SL(2,{{\mathbb{R}}})$ such that $\big (SL(2,{{\mathbb{R}}}),\langle \ ,\ \rangle \big )$ is a two-step g.o. space.

1 Introduction

Let (M, g) be a homogeneous pseudo-Riemannian manifold. Denoted by $G \subset I_0 (M,g)$, a connected Lie group of isometries acting transitively on M determines a corresponding realization of the manifold, given by the pseudo-Riemannian homogeneous space (G/H, g). Here, H denotes the isotropy group at a point $o \in M$, chosen the origin.

A geodesic $\gamma :I\rightarrow G/H$ through o is called homogeneous if it is the (reparametrization of an) orbit of a one-parameter subgroup. For homogeneous pseudo-Riemannian manifolds, homogeneous geodesics and related topics have been studied extensively in past years. In particular, a pseudo-Riemannian homogeneous space (G/H, g) is called g.o. space if every geodesic in G/H is homogeneous. The terminology was introduced by Kowalski and Vanhecke in [13] for a Riemannian space. For comprehensive reviews and various results on the subject, we refer to [2, 5,6,7, 9, 15] and references therein.

In the work [3], the first and third author considered a generalization of homogeneous geodesics, namely geodesics of the form

$$\begin{aligned} \gamma (t)=\exp (tX)\exp (tY)\cdot o, \quad X,Y\in {\mathfrak{g}}, \end{aligned}$$

(1.1)

which were called two-step homogeneous geodesics. Geodesics of the form (1.1) had previously appeared in semisimple Lie groups G equipped with a metric induced by a Cartan involution of the Lie algebra ${\mathfrak{g}}$ of G. More specifically, in [19] it was shown that if B is the Killing form and $\theta$ is an involution of ${\mathfrak{g}}$ then the geodesics through $e\in G$, with respect to the metric $\langle X,Y\rangle =-B(X,\theta Y)$, have the form $\gamma (t)=\exp t(-\theta (Z))\exp t(Z+\theta (Z))$, $Z=\dot{\gamma }(0)$. The above result was generalized for Riemannian homogeneous spaces $(G/H,\langle \ ,\ \rangle )$ in [10]; there it was proven that if the tangent space at o decomposes into the orthogonal sum of two spaces ${\mathfrak{m}}_1,{\mathfrak{m}}_2$ such that $\langle [X,Y]_{{\mathfrak{m}}_2},Z\rangle +c\langle X,[Z,Y]\rangle =0$ for $X,Y\in {\mathfrak{m}}_1$, $Z\in {\mathfrak{m}}_2$, and under certain algebraic conditions for ${\mathfrak{m}}_1,{\mathfrak{m}}_2$, then the geodesics have the form $\exp t(X_1+cX_2)\exp t(1-c)X_2\cdot o$. One of those algebraic conditions requires that

$$\begin{aligned}{}[{\mathfrak{m}}_1,{\mathfrak{m}}_2]\subseteq {\mathfrak{m}}_1. \end{aligned}$$

(1.2)

In [3], the first and the third author proved that condition (1.2) is sufficient for a Riemannian homogeneous space to admit two-step homogeneous geodesics. In particular, if $(G/H,\langle \ ,\ \rangle )$ is a compact homogeneous Riemannian space and ${\mathfrak{m}}_1, {\mathfrak{m}}_2\subset T_{o}(G/H)$ are eigenspaces of the metric endomorphism satisfying (1.2), then any geodesic tangent to ${\mathfrak{m}}_1\oplus {\mathfrak{m}}_2$ is two-step homogeneous ([3], Theorem 2.3). A Riemannian homogeneous space G/H such that any geodesic of G/H passing through the origin is two-step homogeneous is called a two-step g.o. space.

As Remark 1.3 shows, form (1.1) is invariant by left translations. The same invariance holds for any curve of the form

$$\begin{aligned} \gamma (t)=\exp (tX_1)\ldots \exp (tX_n)\cdot o,\quad X_1,\ldots ,X_n \in {\mathfrak{g}}. \end{aligned}$$

(1.3)

It is then natural to investigate the cases where geodesics in homogeneous spaces have the general form (1.3).

The aim of the present paper is to initiate a systematic study of two-step homogeneous geodesics and two-step g.o. spaces in the pseudo-Riemannian setting. As it will turn out, the theory is not a direct generalization of the Riemannian case. We start with the following.

Definition 1.1

Let $(G/H,\langle \ ,\ \rangle )$ be a homogeneous pseudo-Riemannian space and consider a point $o\in G/H$. A geodesic $\gamma :I\rightarrow G/H$ through o, with an affine parameter s, is called two-step homogeneous if there exists a parametrization $t=\phi (s)$ of $\gamma$ and vectors X, Y in the Lie algebra ${\mathfrak{g}}$ of G, such that

$$\begin{aligned} \gamma (t)=\exp (tX)\exp (tY)\cdot o \quad {\text{for all}} \;\, t\in \phi (I), \end{aligned}$$

where $\cdot$ denotes the action of G on G/H.

Obviously, setting $X=0$ or $Y=0$, a two-step homogeneous geodesic reduces to a homogeneous geodesic. A g.o. space (“geodesic orbit space”) is a coset representation $(M = G/H,\langle \ ,\ \rangle )$ of a homogeneous pseudo-Riemannian manifold M, so that all geodesics are homogeneous. We extend the concept of g.o. space to the following:

Definition 1.2

A two-step geodesic orbit space (or two-step g.o. space) is a pseudo-Riemannian homogeneous space $(G/H,\langle \ , \ \rangle )$ such that every geodesic through a point $o\in G/H$ is two-step homogeneous.

Remark 1.3

Similarly to the case of g.o. spaces, Definition 1.2 is independent of the choice of the point $o\in G/H$. Indeed, if the curve $\gamma :I\rightarrow G/H$ with $\gamma (t)=\exp (tX)\exp (tY)\cdot o$ is a geodesic through o and $o'=g\cdot o$ is another point in G/H, then $\tau _g\circ \gamma$ is a geodesic through $o'$, where $\tau _g:G/H \rightarrow G/H$ denotes the left translation by g in G/H. Moreover, it satisfies

$$\begin{aligned} (\tau _g\circ \gamma )(t)= & {} g\exp (tX)\exp (tY)\cdot o=\big (g\exp (tX){g}^{-1}\big )\big (g\exp (tY){g}^{-1}\big )(g\cdot o)\\= & {} \exp ({\text{Ad}}(g)tX)\exp ({\text{Ad}}(g)tY)(g\cdot o)=\exp (\widetilde{X})\exp (\widetilde{Y})\cdot o', \end{aligned}$$

where $\widetilde{X}={\text{Ad}}(g)tX$ and $\widetilde{Y}={\text{Ad}}(g)tY$. Therefore, $\tau _g\circ \gamma$ is also a two-step homogeneous geodesic.

It is clear that both the notions of g.o. and two-step g.o. spaces are properties of the specific coset representation of the homogeneous pseudo-Riemannian manifold. For this reason, a pseudo-Riemannian homogeneous manifold (M, g) is said to be a g.o. manifold (respectively, a two-step g.o. manifold) if it admits a coset representation given by a g.o. space (respectively, by a two-step g.o. space). Clearly, not all the representations of a g.o. manifold need to be g.o. spaces, and not all the representations of a two-step g.o. manifold are necessarily two-step g.o..

The paper is organized as follows. In Sect. 2, we provide the appropriate background for homogeneous spaces and reparametrization of geodesics in pseudo-Riemannian homogeneous spaces. The main results of Sect. 3 will provide some criteria to determine whether a geodesic is two-step homogeneous, both for general (not necessarily reductive) homogeneous pseudo-Riemannian spaces, and in the special case of reductive homogeneous pseudo-Riemannian spaces. This leads to an algebraic characterization of two-step homogeneous geodesics, which generalizes the well-known algebraic characterization of homogeneous geodesics for reductive homogeneous spaces (known as “Geodesic Lemma”, cf. [11]). Further characterizations of two-step homogeneous geodesics are given in Sect. 4, with particular regard to the case of left-invariant metrics on Lie groups. In Sect. 5 we turn our attention to two-step g.o. spaces and illustrate some ways to construct such examples. Finally, in Sect. 6 we provide some explicit examples of homogeneous pseudo-Riemannian spaces which are two-step g.o. but not g.o. spaces. In particular, we completely determine the left-invariant (Lorentzian and Riemannian) metrics $\langle$, $\rangle$ on the unimodular Lie group $SL (2,{\mathbb{R}})$ such that $\big (SL(2,{{\mathbb{R}}}),\langle \ ,\ \rangle \big )$ is a two-step g.o. space.

2 Preliminaries

2.1 Invariant metrics and killing vector fields in homogeneous spaces

Consider a homogeneous pseudo-Riemannian manifold $(M=G/H,\langle \ , \ \rangle )$. Let $\pi :G\rightarrow G/H$ denote the projection and $o=\pi (e)$ be the origin of G/H. For $g\in G$, let $\tau _g:G/H\rightarrow G/H$ be the left translation by g, i.e., $\tau _g(g^{\prime }H)=(gg^{\prime })H$. For $g\in G$, denote by $L_g,R_g:G\rightarrow G$ the left and the right translations by g and let ${\text{Ad}}:G\rightarrow {\text{Aut}}({\mathfrak{g}})$ denote the adjoint representation of G. Recall also the relation $\pi \circ L_g=\tau _g\circ \pi$. For $X\in {\mathfrak{g}}$ let $X^L$ (resp. $X^R$) be the left-invariant (resp. right-invariant) vector field in G induced by X. In other words, $X^L_{g}:=(L_g)_*(X)$ and $X^R_{g}:=(R_g)_*(X)$.

A metric $\langle \ ,\rangle$ on G/H is called G-invariant if the left translations are isometries of $(G/H,\langle \ ,\rangle )$. The G-invariant metrics on G/H are in one-to-one correspondence with ${\text{Ad}}(H)$-invariant scalar products in $T_o(G/H)$. Let ${\mathfrak{g}},{\mathfrak{h}}$ be the Lie algebras of G, H, respectively. The space G/H is called reductive if there exists a decomposition

$$\begin{aligned} {\mathfrak{g}}={\mathfrak{h}}\oplus {{\mathfrak{m}}}, \end{aligned}$$

(2.1)

such that ${\text{Ad}}(H){\mathfrak{m}}\subseteq {\mathfrak{m}}$. The decomposition (2.1) is also called reductive. When G/H is reductive, we naturally identify ${\mathfrak{m}}$ with the tangent space $T_o(G/H)=\pi _*({\mathfrak{g}})$, where $\pi _*:{\mathfrak{g}}\rightarrow T_o(G/H)$ is the differential of the projection at e.

For any $W\in {\mathfrak{g}}$, the correspondence $W^*:G/H\rightarrow T(G/H)$, with

$$\begin{aligned} W^*_{aH}=\left. \frac{\mathrm{d}}{\mathrm{d}t}\right| _{t=0}\exp (tW)aH=(\pi \circ R_a)_*W,\quad aH\in G/H, \end{aligned}$$

(2.2)

is a well-defined vector field in G/H which is a Killing vector field for all G-invariant metrics in G/H. Moreover, since $\pi$ is a submersion, the tangent space of G/H at each point aH is spanned by the vectors $W^*_{aH}$, $W\in {\mathfrak{g}}$.

2.2 Reparametrizations of geodesics in homogeneous spaces

Let $\big (M=G/H,\langle \ ,\rangle \big )$ be a homogeneous pseudo-Riemannian manifold with the Levi-Civita connection $\nabla$. A curve $\gamma : J \rightarrow M$ is called a geodesic up to a reparametrization if its tangent vector field $\dot{\gamma }$ is parallel along $\gamma$, that is

$$\begin{aligned} \nabla _{\dot{\gamma }}\dot{\gamma }=k\dot{\gamma }, \end{aligned}$$

(2.3)

where k is a real function of the affine parameter t of $\gamma$ (see, for example, [8, p. 14]). It is always possible to find a new parameter s for which $k=0$ along $\gamma$, so that the geodesic equation reduces to $\nabla _{\dot{\gamma }}\dot{\gamma }= 0$.

Given a curve $\gamma :J\rightarrow G/H$, a vector $W\in {\mathfrak{g}}$ and a real function k, we introduce the function ${\mathcal{G}}_{k}^W:J\rightarrow {{\mathbb{R}}}$ defined by

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)=\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)}, \end{aligned}$$

where $\dot{\gamma }$ denotes a local extension of the vector field $\dot{\gamma }(t)$ along $\gamma$ and $W^*$ is the vector field defined by Eq. (2.2). We have the following.

Proposition 2.1

Let$(G/H,\langle \ ,\rangle )$ be a homogeneous pseudo-Riemannian space and let $\gamma :J\rightarrow G/H$ be a curve. Then, $\gamma$ is a geodesic up to reparametrization if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that ${\mathcal{G}}_{k}^W(t)=0$ for any $W\in {\mathfrak{g}}$ and for any$t\in J$.

Proof

Using the nondegeneracy of $\langle \ , \ \rangle$, we have that Eq. (2.3) holds, if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

$$\begin{aligned} \langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },V\rangle _{\gamma (t)}=0, \end{aligned}$$

(2.4)

for any vector field V in G/H and for any $t\in J$. Since the tangent space of G/H at each point aH is spanned by the vectors $W^*_{aH}$, $W\in {\mathfrak{g}}$, we have $T_{\gamma (t)}(G/H)={\text{span}}\{(W_1^*)_{\gamma (t)},\dots ,(W_n^*)_{\gamma (t)}\}$ for some $W_i\in {\mathfrak{g}}$. Write $V_{\gamma (t)}=\sum _{i=1}^{n}c_i(W_i)^*_{\gamma (t)}$, $c_i\in {{\mathbb{R}}}$. Then, $\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },V\rangle _{\gamma (t)}=\sum _{i=1}^{n}c_i\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },(W_i)^*\rangle _{\gamma (t)}$. Hence, if $\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)}=0$ for any $W\in {\mathfrak{g}}$ then $\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },V\rangle _{\gamma (t)}=0$ for any vector field V in G/H. Conversely, if $\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },V\rangle _{\gamma (t)}=0$ for any vector field V in G/H, then $\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)}=0$ for any $W\in {\mathfrak{g}}$. Therefore, it suffices to replace V in (2.4) with any vector field $W^*$, $W\in {\mathfrak{g}}$.

We conclude that $\gamma$ is a geodesic in G/H if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)=\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)}=0, \end{aligned}$$

for any $W\in {\mathfrak{g}}$ and for any $t\in J$. $\square$

3 The generalized geodesic lemma

We start with the following general characterization of two-step homogeneous geodesics.

Theorem 3.1

Let $(G/H,\langle \ ,\rangle )$ be a homogeneous pseudo-Riemannian space with the natural projection $\pi :G\rightarrow G/H$, and let $o=\pi (e)$ be the origin in G/H. Let $\gamma :J\rightarrow G/H$ be the curve

$$\begin{aligned} \gamma (t)=\pi \big (\exp (tX)\exp (tY)\big ), \quad X,Y \in {\mathfrak{g}}. \end{aligned}$$

(3.1)

Moreover, let $T: J\rightarrow {\text{Aut}}({\mathfrak{g}})$ be the map

$$\begin{aligned} T(t)={\text{Ad}}(\exp (-tY))=\sum _{n=0}^{\infty }{\frac{t^n}{n!}{\text{ad}}^n(-Y)}. \end{aligned}$$

(3.2)

Then, $\gamma$ is a geodesic up to reparametrization (i.e., a two-step homogeneous geodesic) if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$, such that the function ${\mathcal{G}}_{k}^W(t)$ defined by the formula

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle \pi _*(T(t)X+Y),\pi _*([W,T(t)X+Y])\rangle _o+\langle \pi _*(W),\pi _*([T(t)X,Y])\rangle _o\nonumber \\&\ \ -k(t)\langle \pi _*(W) , \pi _*(T(t)X+Y)\rangle _o=0, \end{aligned}$$

(3.3)

for any $W\in {\mathfrak{g}}$ and for any $t\in J$.

To prove Theorem 3.1 we need the following lemma.

Lemma 3.2

Let $X,Y\in {\mathfrak{g}}$, let $\gamma$ be the curve described by (3.1) and let $\alpha :J\rightarrow G$ be the curve defined by

$$\begin{aligned} \alpha (t)=\exp (tX)\exp (tY). \end{aligned}$$

Then, the velocity of $\gamma$ is given by

$$\begin{aligned} \dot{\gamma }(t)=(\pi _*)_{\alpha (t)}\big (\big (X^R+Y^L\big )_{\alpha (t)}\big )=((\tau _{\alpha (t)}\circ \pi )_*)_e(T(t)X+Y). \end{aligned}$$

(3.4)

Proof

We have that $\gamma =\pi \circ \alpha .$ Therefore,

$$\begin{aligned} \dot{\gamma }(t)= & {} \pi _*(\dot{\alpha }(t))=\pi _*\left( \displaystyle {\left. \frac{\mathrm{d}}{\mathrm{d}s}\right| _{s=0}\alpha (t+s)}\right) =\pi _*\left( \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp (t+s)X \exp (t+s)Y}\right) \\= & {} \pi _*\left( \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp (t+s)X \exp tY} +\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp tX \exp (t+s)Y}\right) \\= & {} \pi _*\left( \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp sX \exp tX \exp tY} + \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp tX \exp tY \exp sY}\right) \\= & {} \pi _*\left( \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\exp sX \alpha (t)}+ \displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\alpha (t) \exp sY}\right) =\pi _*\left( \displaystyle {(R_{\alpha (t)})_*(X)+(L_{\alpha (t)})_*Y}\right) \\= & {} (\pi _*)_{\alpha (t)}\big (\big (X^R+Y^L\big )_{\alpha (t)}\big ), \end{aligned}$$

which proves the first equality of (3.4). Moreover, it equals

$$\begin{aligned}(\pi _*)_{\alpha (t)}\big (\big (X^R+Y^L\big )_{\alpha (t)}\big )= & {} \displaystyle {((\pi \circ L_{\alpha (t)}\circ L_{\alpha (t)^{-1}})_*)_{\alpha (t)}(\displaystyle {(R_{\alpha (t)})_*(X)+(L_{\alpha (t)})_*Y})}\\= & {} \displaystyle {((\pi \circ L_{\alpha (t)})_*)_{\alpha (t)}({\text{Ad}}(\alpha (t)^{-1})X+Y)}\\= & {} \displaystyle {((\tau _{\alpha (t)}\circ \pi )_*)_e({\text{Ad}}(\alpha (t)^{-1})X+Y)}\\= & {} \displaystyle {((\tau _{\alpha (t)}\circ \pi )_*)_e({\text{Ad}}(\exp (-tY)\exp (-tX))X+Y)}\\= & {} \displaystyle {((\tau _{\alpha (t)}\circ \pi )_*)_e({\text{Ad}}(\exp (-tY)){\text{Ad}}(\exp (-tX))X+Y)}\\= & {} \displaystyle {((\tau _{\alpha (t)}\circ \pi )_*)_e({\text{Ad}}(\exp (-tY))X+Y)}\\= & {} \displaystyle {((\tau _{\alpha (t)}\circ \pi )_*)_e(T(t)X+Y)}, \end{aligned}$$

which proves the second equality of (3.4). $\square$

We now proceed to the proof of Theorem 3.1.

Proof of Theorem 3.1

Using Koszul formula, we have that

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)} \nonumber \\= & {} \bigg (\dot{\gamma }\langle W^*,\dot{\gamma }\rangle +\langle \dot{\gamma },[W^*,\dot{\gamma }]\rangle -\frac{1}{2}W^*\langle \dot{\gamma },\dot{\gamma }\rangle -k\langle W^*,\dot{\gamma }\rangle \bigg )_{\gamma (t)}. \end{aligned}$$

(3.5)

Moreover, using the compatibility of the Levi-Civita connection $\nabla$ with the metric along with its torsion-free property (see [16]), we have the following:

$$\begin{aligned} W^*\langle \dot{\gamma },\dot{\gamma }\rangle =2\langle \nabla _{W^*}{\dot{\gamma }},\dot{\gamma }\rangle \end{aligned}$$

(3.6)

$$\begin{aligned} \nabla _{W^*}{\dot{\gamma }}-\nabla _{\dot{\gamma }}{W^*}=[W^*,\dot{\gamma }]. \end{aligned}$$

(3.7)

Furthermore, since W is a Killing vector field we have that

$$\begin{aligned} \langle \nabla _{\dot{\gamma }}{W^*},\dot{\gamma })=0 \end{aligned}$$

(3.8)

(see [16]). By taking into account Eqs. (3.6)–(3.8), we see that $\langle \dot{\gamma },[W^*,\dot{\gamma }]\rangle -\frac{1}{2}W^*\langle \dot{\gamma },\dot{\gamma }\rangle =0$ and thus Eq. (3.5) is equivalent to

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)=(\dot{\gamma }\langle W^*, \dot{\gamma }\rangle )_{\gamma (t)}-k(t)\langle W^*,\dot{\gamma }\rangle _{\gamma (t)}. \end{aligned}$$

(3.9)

We will describe explicitly each term of the right hand side of Eq. (3.9). Using Eq. (2.2), the G-invariance of the metric as well as Lemma 3.2, the first term of the right-hand side of Eq. (3.9) becomes

$$\begin{aligned}&(\dot{\gamma }\langle W^*,\dot{\gamma }\rangle )_{\gamma (t)}=\dot{\gamma }_{\gamma (t)}\langle W^*,\dot{\gamma }\rangle _{\gamma (t)} =\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\langle W^*_{\gamma (t+s)},\dot{\gamma }_{\gamma (t+s)}\rangle _{\gamma (t+s)}} \nonumber \\&\quad =\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\big \langle (\pi \circ R_{\alpha (t+s)})_*W,(\tau _{\alpha (t+s)}\circ \pi )_*(T(t+s)X+Y)\big \rangle _{\gamma _{(t+s)}}}\nonumber \\&\quad =\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\big \langle (\pi \circ L_{\alpha (t+s)}\circ L_{\alpha ^{-1}(t+s)}\circ R_{\alpha (t+s)})_*W,(\tau _{\alpha (t+s)}\circ \pi )_*(T(t+s)X+Y)\big \rangle _{\gamma _{(t+s)}}}\nonumber \\&\quad =\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\big \langle (\tau _{\alpha (t+s)}\circ \pi )_*\big ({\text{Ad}}(\alpha ^{-1}(t+s))W\big ),(\tau _{\alpha (t+s)}\circ \pi )_*(T(t+s)X+Y)\big \rangle _{\gamma _{(t+s)}}}\nonumber \\&\quad =\displaystyle {\left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\big \langle \pi _*({\text{Ad}}(\alpha ^{-1}(t+s))W),\pi _*(T(s)T(t)X+Y)\big \rangle _o}, \end{aligned}$$

(3.10)

By Lemma 3.2, we obtain $\dot{\alpha }(t)=(L_{\alpha (t)})_*(T(t)X+Y)$ [one can see this for example by assuming that $\gamma$ is a curve in G by setting $\pi _*:={\text{id}}$ in Eq. (3.4)]. So by differentiating the relation $\alpha ^{-1}(t)\alpha (t)=e$ we have $(R_{\alpha (t)})_*\dot{\alpha }^{-1}(t)=-(L_{\alpha ^{-1}(t)})_*\dot{\alpha }(t)=-(T(t)X+Y)$ (here by $\dot{\alpha }^{-1}(t)$ we denote the quantity $\frac{{\mathrm{d}}}{{\mathrm{d}}t}\alpha ^{-1}(t)$). Along with the fact that ${\text{Ad}}:G\rightarrow {\text{Aut}}({\mathfrak{g}})$ is a homomorphism and by setting

$$\begin{aligned} \widetilde{W}:={\text{Ad}}(\alpha ^{-1}(t))W, \end{aligned}$$

we obtain

$$\begin{aligned} \left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}{\text{Ad}}(\alpha ^{-1}(t+s))W= & {} \left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}{\text{Ad}}(\alpha ^{-1}(t+s)\alpha (t)){\text{Ad}}(\alpha ^{-1}(t))W\nonumber \\= & {} ({\text{Ad}}_*)_e\left( \left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\alpha ^{-1}(t+s)\alpha (t)\right) {\text{Ad}}(\alpha ^{-1}(t))W\nonumber \\= & {} ({\text{Ad}}_*)_e\big ((R_{\alpha (t)})_*\dot{\alpha }^{-1}(t)\big )\widetilde{W}\nonumber \\= & {} ({\text{Ad}}_*)_e\big (-(L_{\alpha ^{-1}(t)})_*\dot{\alpha }(t)\big )\widetilde{W}\nonumber \\= & {} ({\text{Ad}}_*)_e(-(T(t)X+Y))\widetilde{W}=[\widetilde{W},T(t)X+Y]. \end{aligned}$$

(3.11)

Moreover, by taking into account Eq. (3.2) we have

$$\begin{aligned} \left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}\pi _*(T(s)T(t)X+Y)=\pi _*\left( \left. \frac{{\mathrm{d}}}{{\mathrm{d}}s}\right| _{s=0}{\text{Ad}}(\exp (-sY))T(t)X\right) =[T(t)X,Y]. \end{aligned}$$

(3.12)

Using Eqs. (3.11) and (3.12), Eq. (3.10) implies that the first term of the right-hand side Eq. (3.9) becomes

$$\begin{aligned} (\dot{\gamma }\langle W^*,\dot{\gamma }\rangle )_{\gamma (t)}=\langle \pi _*([\widetilde{W},T(t)X+Y]),\pi _*(T(t)X+Y)\rangle _o+\langle \pi _*(\widetilde{W}),\pi _*([T(t)X,Y])\rangle _o. \end{aligned}$$

(3.13)

Finally, the second term at the right-hand side of Eq. (3.9) becomes

$$\begin{aligned} -k(t)\langle W^*_{\gamma (t)},\dot{\gamma }(t)\rangle _{\gamma (t)}= & {} \displaystyle {-\,k(t)\langle (\tau _{\alpha (t)}\circ \pi )_*({\text{Ad}}(\alpha ^{-1}(t))W),(\tau _{\alpha (t)}\circ \pi )_*(T(t)X+Y)\rangle _{\gamma (t)}}\nonumber \\= & {} \displaystyle {-\,k(t)\langle \pi _*(\widetilde{W}), \pi _*(T(t)X+Y)\rangle _o}. \end{aligned}$$

(3.14)

We substitute (3.13) and (3.14) into (3.9) to obtain that $\gamma$ is a geodesic if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

$$\begin{aligned} {\mathcal{G}}_{k}^{\widetilde{W}}(t)= & {} \langle \pi _*(T(t)X+Y),\pi _*([\widetilde{W},T(t)X+Y])\rangle _o+\langle \pi _*(\widetilde{W}),\pi _*([T(t)X,Y])\rangle _o\nonumber \\&\ \ -k(t)\langle \pi _*(\widetilde{W}) , \pi _*(T(t)X+Y)\rangle _o=0, \end{aligned}$$

for any $t\in J$ and for any $\widetilde{W}={\text{Ad}}(\alpha ^{-1}(t))W$, $W \in {\mathfrak{g}}$. But ${\text{Ad}}(\alpha ^{-1}(t))$ is an automorphism of ${\mathfrak{g}}$ and thus we may substitute “for any $\widetilde{W} \in {\mathfrak{g}}$” with “for any $W\in {\mathfrak{g}}$”, which concludes the proof of the theorem. $\square$

For the rest of this paper, we will use the notation $\langle \ ,\ \rangle$ to denote both the metric on G/H and the corresponding inner product on $T_o(G/H)$. For the reductive case, Theorem 3.1 is simplified in the following way.

Corollary 3.3

(Generalized Geodesic Lemma) Let $(G/H,\langle \ ,\rangle )$ be a pseudo-Riemannian reductive homogeneous space with reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$. Then, the curve $\gamma$ in Theorem 3.1 is a geodesic up to reparametrization if and only if there exists a function $k:J\rightarrow {{\mathbb{R}}}$, such that

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle (T(t)X+Y)_{{\mathfrak{m}}},[W,T(t)X+Y]_{{\mathfrak{m}}}\rangle +\langle W,[T(t)X,Y]_{{\mathfrak{m}}}\rangle \nonumber \\&-\,k(t)\langle W , (T(t)X+Y)_{{\mathfrak{m}}}\rangle =0, \end{aligned}$$

(3.15)

for any $W\in {\mathfrak{m}}$ and for any $t\in J$.

Proof

From the reductive decomposition, the tangent space $T_o(G/H)$ is naturally identified with ${\mathfrak{m}}$ via the differential $\pi _*$. In particular, for $W\in {\mathfrak{g}}$ the vector $\pi _*(W)$ is identified with $W_{{\mathfrak{m}}}\in {\mathfrak{m}}$. Under the above identification, Eq. (3.3) is equivalent to

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle (T(t)X+Y)_{{\mathfrak{m}}},[W,T(t)X+Y]_{{\mathfrak{m}}}\rangle +\langle W_{{\mathfrak{m}}},[T(t)X,Y]_{\mathfrak{m}}\rangle \nonumber \\&-k(t)\langle W_{\mathfrak{m}} , (T(t)X+Y)_{\mathfrak{m}}\rangle =0, \end{aligned}$$

(3.16)

for any $W\in {\mathfrak{g}}$ and for any $t\in J$. Moreover, using the ${\text{ad}}({\mathfrak{h}})$-invariance of $\langle \ ,\rangle$, for any $Z\in {\mathfrak{g}}$ and $a\in {\mathfrak{h}}$ we obtain that

$$\begin{aligned} \langle Z_{\mathfrak{m}}, [a,Z]_{\mathfrak{m}}\rangle= & {} \langle Z_{\mathfrak{m}}, [a,Z_{\mathfrak{h}}+Z_{\mathfrak{m}}]_{\mathfrak{m}}\rangle =\langle Z_{\mathfrak{m}}, [a, Z_{\mathfrak{m}}]_{\mathfrak{m}}\rangle =\langle Z_{\mathfrak{m}}, [a, Z_{\mathfrak{m}}]\rangle \\= & {} -\langle [a,Z_{\mathfrak{m}}], Z_{\mathfrak{m}}\rangle , \end{aligned}$$

and thus $\langle Z_{\mathfrak{m}}, [a,Z]_{\mathfrak{m}}\rangle =0$. Using the above relation, it follows that

$$\begin{aligned} \langle Z_{\mathfrak{m}},[W,Z]_{\mathfrak{m}}\rangle= & {} \langle Z_{\mathfrak{m}}, [W_{\mathfrak{h}}+W_{\mathfrak{m}},Z]_{\mathfrak{m}}\rangle =\langle Z_{\mathfrak{m}},[W_{\mathfrak{h}},Z]_{\mathfrak{m}}\rangle +\langle Z_{\mathfrak{m}},[W_{\mathfrak{m}},Z]_{\mathfrak{m}}\rangle \\= & {} \langle Z_{\mathfrak{m}},[W_{\mathfrak{m}},Z]_{\mathfrak{m}}\rangle , \quad \text{for all} \;\, W\in {\mathfrak{g}}. \end{aligned}$$

For $Z=T(t)X+Y$, the above equation yields

$$\begin{aligned} \langle (T(t)X+Y)_{\mathfrak{m}},[W,T(t)X+Y]_{\mathfrak{m}}\rangle =\langle (T(t)X+Y)_{\mathfrak{m}},[W_{\mathfrak{m}},T(t)X+Y]_{\mathfrak{m}}\rangle . \end{aligned}$$

Substituting the above into the first term of Eq. (3.16), we obtain

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle (T(t)X+Y)_{\mathfrak{m}},[W_{\mathfrak{m}},T(t)X+Y]_{\mathfrak{m}}\rangle +\langle W_{\mathfrak{m}},[T(t)X,Y]_{\mathfrak{m}}\rangle \\&-k(t)\langle W_{\mathfrak{m}} , (T(t)X+Y)_{\mathfrak{m}}\rangle =0, \end{aligned}$$

for any $t\in J$ and $W_{\mathfrak{m}}\in {\mathfrak{m}}$, $W\in {\mathfrak{g}}$. Hence, we may assume without any loss of generality that $W\in {\mathfrak{m}}$, and thus the above equation is equivalent to

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)= & {} \langle (T(t)X+Y)_{\mathfrak{m}},[W,T(t)X+Y]_{\mathfrak{m}}\rangle +\langle W,[T(t)X,Y]_{\mathfrak{m}}\rangle \\&-k(t)\langle W , (T(t)X+Y)_{\mathfrak{m}}\rangle =0, \end{aligned}$$

for any $W\in {\mathfrak{m}}$ and $t\in J$. $\square$

Remark 3.4

By setting $X=0,$ Eq. (3.15) reduces to

$$\begin{aligned} \langle Y_{\mathfrak{m}},[W,Y]_{\mathfrak{m}}\rangle =k(t)\langle W,Y_{\mathfrak{m}}\rangle , \quad \text{for all} \;\, W\in {\mathfrak{m}}, \ t\in J. \end{aligned}$$

The above equation implies that k(t) is independent of t and so, $k(t)=k$ is a constant. Hence, for $X=0$, Corollary 3.3 implies that the curve $\gamma$ with $\gamma (t)=\exp (tY)\cdot o$ is a geodesic up to some parameter if and only if there exists a constant k such that

$$\begin{aligned} \langle Y_{\mathfrak{m}},[W,Y]_{\mathfrak{m}}\rangle =k\langle W,Y_{\mathfrak{m}}\rangle \quad \text{for all} \;\, W\in {\mathfrak{m}}. \end{aligned}$$

This is exactly the Geodesic Lemma proved in [11]. For this reason, we called Lemma 3.3 “Generalized Geodesic Lemma”.

4 Two-step homogeneous geodesics in pseudo-Riemannian spaces

We shall now obtain various characterizations of two-step homogeneous geodesics in pseudo-Riemannian homogeneous spaces. In particular, we describe such geodesics in pseudo-Riemannian Lie groups. We start with the following.

Proposition 4.1

Let $(G/H,\langle \ , \ \rangle )$ be a pseudo-Riemannian space with the reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$, and let $X,Y\in {\mathfrak{g}}$ with $Y=a+\widetilde{Y}$, $a\in {\mathfrak{h}},\widetilde{Y}\in {\mathfrak{m}}$. The following properties are equivalent.

• (1) The curve $\gamma :J\rightarrow G/H$ with $\gamma (t)=\exp (tX)\exp (tY)\cdot o$ is a geodesic up to reparametrization.

• (2) There exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

  $$\begin{aligned} \langle (T(t)X+Y)_{\mathfrak{m}},[W,T(t)X+Y]_{\mathfrak{m}}\rangle +\langle W,[T(t)X,Y]_{\mathfrak{m}}\rangle =k(t)\langle W , (T(t)X+Y)_{\mathfrak{m}}\rangle , \end{aligned}$$

  for any $W\in {\mathfrak{m}}$, $t\in J$.

• (3) There exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

  $$\begin{aligned} \langle (T(t)X)_{\mathfrak{m}}+\widetilde{Y},[W,T(t)X+\widetilde{Y}]_{\mathfrak{m}}\rangle +\langle W,[T(t)X,\widetilde{Y}]_{\mathfrak{m}}\rangle +\langle W,[a,\widetilde{Y}]\rangle =k(t)\langle W , (T(t)X)_{\mathfrak{m}}+\widetilde{Y}\rangle , \end{aligned}$$

  for any $W\in {\mathfrak{m}}$, $t\in J$. Moreover, assume that the following property is satisfied: (P) There exists an ${\text{Ad}}$-invariant inner product B in ${\mathfrak{g}}$ such that the reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$ is B-orthogonal and let $\Lambda :{\mathfrak{m}}\rightarrow {\mathfrak{m}}$ be the symmetric, nondegenerate, ${\text{Ad}}(H)$-equivariant operator determined by the metric $\langle \ ,\ \rangle$, i.e.,

  $$\begin{aligned} \langle Z,W\rangle =B(\Lambda (Z),W), \quad Z,W\in {\mathfrak{m}}. \end{aligned}$$

  Then, any of the above properties (1)-(3) are equivalent to the following.

• (4) There exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that

  $$\begin{aligned}{}[T(t)X+Y,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )]+\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )=k(t)\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big ), \end{aligned}$$

  for any $t\in J$.

Proof

The equivalence of (1) and (2) follows from Theorem 3.1. Setting $Y=a+\widetilde{Y}$, Eq. (3.15) becomes

$$\begin{aligned} \langle (T(t)X)_{\mathfrak{m}}+ & {} \widetilde{Y},[W,T(t)X+\widetilde{Y}]_{\mathfrak{m}}\rangle +\langle (T(t)X)_{\mathfrak{m}}+\widetilde{Y},[W,a]_{\mathfrak{m}}\rangle +\langle W,[T(t)X,a]_{\mathfrak{m}}\rangle \nonumber \\+ & {} \langle W,[T(t)X,\widetilde{Y}]_{\mathfrak{m}}\rangle =k(t)\langle W , (T(t)X)_{\mathfrak{m}}+\widetilde{Y}\rangle , \end{aligned}$$

where $W\in {\mathfrak{m}}$, $t\in J$. Moreover, using the ${\text{Ad}}(H)$-invariance of $\langle \ ,\rangle$, the second and third terms of the left-hand side of the above equation add to

$$\begin{aligned}&\langle (T(t)X)_{\mathfrak{m}}+\widetilde{Y},[W,a]_{\mathfrak{m}}\rangle +\langle W,[T(t)X,a]_{\mathfrak{m}}\rangle =\langle (T(t)X)_{\mathfrak{m}}+\widetilde{Y},[W,a]\rangle +\langle W,[T(t)X,a]\rangle \nonumber \\&\quad =\langle (T(t)X)_{\mathfrak{m}}+\widetilde{Y},[W,a]\rangle -\langle (T(t)X)_{\mathfrak{m}},[W,a]\rangle \nonumber \\&\quad =\langle \widetilde{Y},[W,a]\rangle =\langle W, [a,\widetilde{Y}]\rangle , \end{aligned}$$

which implies the equivalence of (2) and (3). Finally, we will prove the equivalence of (2) and (4) under the additional assumption (P). Eq. (3.15) is equivalent to

$$\begin{aligned}&B\big (\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big ),[W,T(t)X+Y]_{\mathfrak{m}}\big )+B\big (W,\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )\big )\\&\quad =k(t)B\big (W,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )\big ),\\&\quad \text{ or } \ B\big (\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big ),[W,T(t)X+Y]\big )+B\big (W,\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )\big )\\&\quad =k(t)B\big (W,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )\big ), \end{aligned}$$

for any $W\in {\mathfrak{m}}$, $t\in J$. By the ${\text{Ad}}$-invariance and the bilinearity of B, the above is in turn equivalent to

$$\begin{aligned} B\Big ([T(t)X+Y,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )])+\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )-k(t)\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big ),W\Big )=0, \end{aligned}$$

for any $W\in {\mathfrak{m}}$, $t\in J$. Taking into account the B-orthogonality of the reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$, the above equation is equivalent to

$$\begin{aligned}{}[T(t)X+Y,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )]+\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )-k(t)\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )\in {\mathfrak{h}}. \end{aligned}$$

(4.1)

It suffices to show that the left-hand side of Eq. (4.1) is also an element of ${\mathfrak{m}}$. Indeed, this will imply that the right-hand side of Eq. (4.1) is zero, which will yield the equivalence of (2) and (4). Since $\Lambda ({\mathfrak{m}})\subset {\mathfrak{m}}$, it suffices to show that

$$\begin{aligned}{}[T(t)X+Y,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )]\in {\mathfrak{m}}. \end{aligned}$$

(4.2)

Indeed, using the symmetry of $\Lambda$, the B-orthogonality of ${\mathfrak{h}}$ and ${\mathfrak{m}}$, the ${\text{Ad}}$-invariance of B, the fact that $[{\mathfrak{h}},{\mathfrak{m}}]\subset {\mathfrak{m}}$ as well as the ${\text{ad}}({\mathfrak{h}})$-equivariance of $\Lambda$, for any $Z\in {\mathfrak{g}}$ and $b\in {\mathfrak{h}}$ we have

$$\begin{aligned} B([Z,\Lambda (Z_{\mathfrak{m}})],b)= & {} B(\Lambda (Z_{\mathfrak{m}}),[b,Z])=B(\Lambda (Z_{\mathfrak{m}}),[b,Z]_{\mathfrak{m}})\\= & {} B(\Lambda (Z_{\mathfrak{m}}),[b,Z_{\mathfrak{m}}])=B(Z_{\mathfrak{m}},\Lambda ([b,Z_{\mathfrak{m}}]))\\= & {} B(Z_{\mathfrak{m}},[b,\Lambda (Z_{\mathfrak{m}})])=B(Z,[b,\Lambda (Z_{\mathfrak{m}})])\\= & {} -B([Z,\Lambda (Z_{\mathfrak{m}})],b), \end{aligned}$$

which implies that $B([Z,\Lambda (Z_{\mathfrak{m}})],b)=0$. Therefore,

$$\begin{aligned}{}[Z,\Lambda (Z_{\mathfrak{m}})]\in {\mathfrak{m}},\quad \text{ for } \text{ any } \quad Z\in {\mathfrak{g}}, \end{aligned}$$

which verifies Eq. (4.2) and this concludes the proof of the Proposition. $\square$

By setting $X=0$ in Proposition 4.1 we obtain the following conditions for homogeneous geodesics (see also [1]).

Corollary 4.2

Let $(G/H,\langle \ , \ \rangle )$ be a pseudo-Riemannian space with reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$, and let $X,Y\in {\mathfrak{g}}$ with $Y=a+\widetilde{Y}$, $a\in {\mathfrak{h}},\ \widetilde{Y}\in {\mathfrak{m}}$. The following properties are equivalent.

• (1) The curve $\gamma :J\rightarrow G/H$ with $\gamma (t)=\exp t(a+\widetilde{Y})\cdot o$ is a geodesic up to reparametrization.

• (2) There exists a constant k, such that $\langle \widetilde{Y},[W,Y]_{\mathfrak{m}}\rangle =k\langle W , \widetilde{Y}\rangle ,$ for any $W\in {\mathfrak{m}}$.

• (3) There exists a constant k, such that $\langle \widetilde{Y},[W,\widetilde{Y}]_{\mathfrak{m}}\rangle +\langle W,[a,\widetilde{Y}]\rangle =k\langle W ,\widetilde{Y}\rangle ,$ for any $W\in {\mathfrak{m}}$. Moreover, if Property (P) is satisfied, then any of the above properties (1)-(3) are equivalent to the following.

• (4) There exists a constant k, such that $[a+\widetilde{Y},\Lambda (\widetilde{Y})]=k\Lambda (\widetilde{Y}).$

The function $T(t)=\sum _{n=0}^{\infty }{\frac{t^n}{n!}{\text{ad}}^n(-Y)}$ is in general hard to compute. The following proposition simplifies the conditions (2) and (4) of Proposition 4.1 when $H=\{e\}$ and ${\text{ad}}(Y)$ is skew-symmetric.

Proposition 4.3

Let $(G,\langle \ , \ \rangle )$ be a pseudo-Riemannian Lie group, where $\langle \ , \ \rangle$ is a left-invariant metric. Assume that for some $Y\in {\mathfrak{g}}$ the endomorphism ${\text{ad}}(Y)$ is skew-symmetric with respect to $\langle \ , \ \rangle$. Let k be a real function and $\gamma (t)=\exp (tX)\exp (tY)$. Then, ${\mathcal{G}}_{k}^W(t)=0$ for all t if and only if ${\mathcal{G}}_{k}^W(0)=0$. In particular, the following properties are equivalent.

• (1) The curve $\gamma :J\rightarrow G$ with $\gamma (t)= \exp (tX)\exp (tY)$, $X,Y\in {\mathfrak{g}}$, is a geodesic up to reparametrization.

• (2) There exists a constant function $k:J \rightarrow {{\mathbb{R}}}$ such that

  $$\begin{aligned} {\mathcal{G}}_{k}^W(0)=\langle X+Y,[W,X]-kW \rangle =0, \end{aligned}$$

for any $W\in {\mathfrak{g}}$. Moreover, assume that there exists an ${\text{Ad}}$ -invariant inner product on ${\mathfrak{g}}$. Then, properties (1) and (2) are equivalent to the following. (3) There exists a constant k such that

$$\begin{aligned}{}[X+Y,\Lambda (X+Y))]+\Lambda ([X,Y])=k\Lambda (X+Y). \end{aligned}$$

(4.3)

Proof

We have that ${\mathfrak{m}}={\mathfrak{g}}$. Moreover, since ${\text{Ad}}(g)\in {\text{Aut}}({\mathfrak{g}})$, $g\in G$, we may replace W by T(t)W in Eq. (3.15). Finally, by the definition of T(t), we have that $T(t)Y=Y$. Taking into account Eqs. (4.3), (3.15) can be rewritten as

$$\begin{aligned} {\mathcal{G}}_{k}^W(t)&=\langle T(t)(X+Y),T(t)[W,X+Y]\rangle \nonumber \\&\quad +\langle T(t)W,T(t)[X,Y]\rangle -k(t)\langle T(t)W,T(t)(X+Y)\rangle =0, \end{aligned}$$

(4.4)

for $W\in {\mathfrak{g}}$ and $t\in J$.

Using the skew-symmetry of ${\text{ad}}(Y)$ with respect to $\langle \ , \ \rangle$, for any $Z,Z^{\prime }\in {\mathfrak{g}}$ we obtain

$$\begin{aligned} \langle T(t)Z,T(t)Z^{\prime }\rangle= & {} \sum _{n=0}^{\infty }{\frac{t^n}{n!}\langle {\text{ad}}^n(-Y)Z,T(t)Z^{\prime }\rangle }=\sum _{n=0}^{\infty }{\frac{t^n}{n!}\langle Z, {\text{ad}}^n(Y)T(t)Z^{\prime }\rangle }\\= & {} \left\langle Z,\sum _{n=0}^{\infty }{\frac{t^n}{n!}{\text{ad}}^n(Y)T(t)Z^{\prime }}\right\rangle =\langle Z,T(-t)T(t)Z^{\prime }\rangle \\= & {} \langle Z,T(t)^{-1}T(t)Z^{\prime }\rangle =\langle Z,Z^{\prime }\rangle . \end{aligned}$$

Using the above equality, Eq. (4.4) is equivalent to

$$\begin{aligned} {\mathcal{G}}_{k}^W(0)=\langle X+Y,[W,X+Y]\rangle +\langle W,[X,Y]\rangle -k(t)\langle W,X+Y\rangle =0, \end{aligned}$$

(4.5)

for any $W\in {\mathfrak{g}}$ and $t\in J$. We deduce that $k(t)=k(0)$ is a constant, and the equivalence between equations ${\mathcal{G}}_{k}^W(t)=0$ for all t and ${\mathcal{G}}_{k}^W(0)=0$ follows. Moreover, writing $[X,Y]=[X+Y,Y]$ in the second term of the right-hand side of Eq. (4.5) and then using the skew-symmetry of ${\text{ad}}(Y)$ with respect to $\langle \ ,\ \rangle$, Eq. (4.5) becomes

$$\begin{aligned} {\mathcal{G}}_{k}^W(0)=\langle X+Y,[W,X+Y]\rangle -\langle [W,Y],X+Y\rangle -k\langle W,X+Y\rangle =\langle X+Y,[W,X]-kW \rangle =0. \end{aligned}$$

Therefore, (1) and (2) are equivalent. The equivalence of (2) and (3) is obtained by the equivalence of (2) and (4) in Proposition 4.1, setting $t=0$. $\square$

5 Two-step g.o. spaces

In the present section, we characterize two-step g.o. spaces and obtain large classes of such spaces. We start with the following characterization.

Proposition 5.1

A pseudo-Riemannian space $(G/H,\langle \ , \ \rangle )$ with origin $o=\pi (e)\in G/H$ is a two-step g.o. space if and only if for any $V\in T_o(G/H)$ there exist $X,Y\in {\mathfrak{g}}$ such that

• (1) $\pi _*(X+Y)=V$ and

• (2) there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that the curve $\gamma :J\rightarrow G/H$, defined by $\gamma (t)=\exp (tX)\exp (tY)\cdot o$, satisfies the equation

  $$\begin{aligned} {\mathcal{G}}_{k}^W(t)=\langle \nabla _{\dot{\gamma }}\dot{\gamma }-k\dot{\gamma },W^*\rangle _{\gamma (t)}=0, \end{aligned}$$

  for all $W\in {\mathfrak{g}}$ and $t\in J$. In particular, if $(G/H,\langle \ , \ \rangle )$ is reductive and ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}$ is a reductive decomposition of its Lie algebra, then the space is two-step g.o. if and only if for any $V\in {\mathfrak{m}}$ there exist $X,Y\in {\mathfrak{g}}$ such that

$(1)^\prime$ :

$(X+Y)_{\mathfrak{m}}=V$ and

$(2)^\prime$ :

one of the equivalent conditions (1)-(4) of Proposition 4.1 is satisfied, where Condition (4) is satisfied if Property (P) holds.

Proof

Assume that $(G/H,\langle \ , \ \rangle )$ is a two-step g.o. space and consider $V\in T_o(G/H)$. Then, there exists a unique geodesic $\gamma :I\rightarrow G/H$ with $\gamma (0)=o$ and $\dot{\gamma }(0)=V$. Let s be the (affine) parameter of the geodesic $\gamma$. By assumption $\gamma$ is two-step homogeneous. Therefore, there exists a parameter $t=\phi (s)$, $\phi :I\rightarrow J$, as well as vectors $X,Y\in {\mathfrak{g}}$, such that

$$\begin{aligned} \gamma (t)=\gamma (\phi (s))=\pi \big (\exp (tX)\exp (tY)\big ), \quad t\in J. \end{aligned}$$

(5.1)

Without loss of generality, we may assume that $\phi (0)=0$. By virtue of Theorem 3.1, there exists a function $k:J\rightarrow {{\mathbb{R}}}$ such that the curve $\gamma$ satisfies the condition ${\mathcal{G}}_{k}^W(t)=0$ for any $W\in {\mathfrak{g}}$, $t\in J$. On the other hand, by differentiating Eq. (5.1) at $s=0,$ we obtain

$$\begin{aligned} \dot{\gamma }(\phi (0))\phi ^{\prime }(0)=\phi ^{\prime }(0)\pi _*(X+Y), \end{aligned}$$

which is equivalent to $\pi _*(X+Y)=V$.

Conversely, assume that for any $V\in T_o(G/H)$, there exist $X,Y\in {\mathfrak{g}}$ such that conditions (1) and (2) are satisfied. We will show that the unique geodesic through the arbitrary V is two-step homogeneous. By Theorem 3.1 the curve $\gamma :J\rightarrow {{\mathbb{R}}}$ with $\gamma (t)=\pi \big (\exp (tX)\exp (tY)\big )$ is a geodesic with respect to some parameter $s=\phi (t)$, $\phi :J\rightarrow I$.

Again, we can use an affine transformation so that $\phi (0)=0$ and $\phi ^{\prime }(0)=1$. Then, the geodesic $\widetilde{\gamma }=\gamma \circ \phi$ passes through o, it is two-step homogeneous and $\left. \frac{{\mathrm{d}}}{{\mathrm{d}}t}\right| _{t=0}\widetilde{\gamma }(t)=\dot{\gamma }(\phi (0))\phi ^{\prime }(0)=V$, which concludes the proof. $\square$

Large classes of pseudo-Riemannian two-step g.o. spaces can be obtained by applying the following proposition.

Proposition 5.2

Let G be a compact Lie group and let G/H be a homogeneous space with origin o. Consider the homogeneous fibration $K/H\rightarrow G/H \rightarrow G/K$, where K is a closed subgroup of G such that $H\subset K\subset G$. We assume that B is an ${\text{Ad}}$ -invariant inner product in ${\mathfrak{g}}$, and we endow G/H with a 1-parameter family of pseudo-Riemannian metrics $\langle \ , \ \rangle _{\lambda }$, $\lambda \in {{\mathbb{R}}}^*$, constructed as deformations

$$\begin{aligned} \langle \ , \ \rangle _\lambda =B|_{{ T}(G/K) \times {T}(G/K)}+\lambda B|_{{ T}(K/H) \times T (K/H)} \end{aligned}$$

of B along the fiber K/H. Then, $(G/H,\langle \ ,\ \rangle _{\lambda })$ is a two-step g.o. space.

Proof

Let ${\mathfrak{k}}$ and ${\mathfrak{h}}$ be the Lie algebras of K and H respectively. We consider the B-orthogonal reductive decompositions ${\mathfrak{g}}=\mathfrak{k}\oplus {\mathfrak{m}}_1$ and ${\mathfrak{k}}={\mathfrak{h}}\oplus {\mathfrak{m}}_2$. Then, we have the reductive decomposition ${\mathfrak{g}}={\mathfrak{h}}\oplus {\mathfrak{m}}_1\oplus {\mathfrak{m}}_2$ with ${\mathfrak{m}}_1=T_{eK}(G/K)$, ${\mathfrak{m}}_2=T_o(K/H)$, and ${\mathfrak{m}}_1\oplus {\mathfrak{m}}_2={\mathfrak{m}}=T_o(G/H)$. Moreover, we have

$$\begin{aligned}{}[{\mathfrak{m}}_1,{\mathfrak{m}}_2]\subseteq {\mathfrak{m}}_1. \end{aligned}$$

(5.2)

The deformation metric is induced by the scalar product $\langle \ , \ \rangle =\left. B\right| _{{\mathfrak{m}}_1\times {\mathfrak{m}}_1}+\left. \lambda B\right| _{\mathfrak{m}_2\times {\mathfrak{m}}_2}$, $\lambda \in {{\mathbb{R}}}^*$.

Let $V=X_1+X_2\in {\mathfrak{m}}$, $X_i\in {\mathfrak{m}}_i$. Then, the metric endomorphism $\Lambda :{\mathfrak{m}}\rightarrow {\mathfrak{m}}$ corresponding to the deformation metric has the form $\Lambda (V)=X_1+\lambda X_2$. Consider the vectors $X=X_1+\lambda X_2$ and $Y=(1-\lambda )X_2$. Then, $X+Y=X_1+X_2=V$. Moreover, by virtue of relation (5.2), we have

$$\begin{aligned} T(t)X_1=\sum _{n=0}^{\infty }{\frac{t^n}{n!}{\text{ad}}^n((\lambda -1)X_2)X_1}\in \sum _{n=0}^{\infty }{{\text{ad}}^n({\mathfrak{m}}_2)\mathfrak{m}_1}\subseteq {\mathfrak{m}}_1. \end{aligned}$$

Also, we have $T(t)X_2=\sum _{n=0}^{\infty }{\frac{t^n}{n!}{\text{ad}}^n((\lambda -1)X_2)X_2}=X_2$. Therefore, $T(t)X+Y=T(t)(X+Y)$. Taking into account the above facts and setting $k(t)=0$, the left-hand side of the equation in part (4) of Proposition 4.1 becomes

$$\begin{aligned}&[T(t)X+Y,\Lambda \big ((T(t)X+Y)_{\mathfrak{m}}\big )]+\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )\\&\quad =[T(t)(X+Y),\Lambda \big ((T(t)(X+Y))_{\mathfrak{m}}\big )]+\Lambda \big ([T(t)X,Y]_{\mathfrak{m}}\big )\\&\quad =[T(t)(X_1+X_2),\Lambda \big ((T(t)(X_1+X_2))_{\mathfrak{m}}\big )]+ (1-\lambda )\Lambda \big ([T(t)(X_1+\lambda X_2),X_2]_{\mathfrak{m}}\big )\\&\quad =[T(t)X_1+X_2,\Lambda \big (T(t)X_1+X_2\big )]+(1-\lambda )\Lambda \big ([T(t)X_1+\lambda X_2),X_2]_{\mathfrak{m}}\big )\\&\quad =[T(t)X_1,\Lambda (T(t)X_1)]+[T(t)X_1,\Lambda (X_2)]+[X_2,\Lambda (T(t)X_1)]\\&\qquad +[X_2,\Lambda (X_2)] +(1-\lambda )\Lambda \big ([T(t)X_1,X_2]_{\mathfrak{m}}\big )\\&\quad =[T(t)X_1,T(t)X_1]+[T(t)X_1,\lambda X_2]+[X_2,T(t)X_1]+[X_2,\lambda X_2]\\&\qquad + (1-\lambda )\Lambda \big ([T(t)X_1,X_2]_{\mathfrak{m}_1}\big )\\&\quad =(\lambda -1)[T(t)X_1,X_2]+(1-\lambda )[T(t)X_1,X_2]_{\mathfrak{m}_1}\\&\quad =(\lambda -1)[T(t)X_1,X_2]+(1-\lambda )[T(t)X_1,X_2]=0. \end{aligned}$$

The result then follows from Propositions 5.1 and 4.1. $\square$

6 Lorentzian two-step g.o. Lie groups

In the present section, we discuss a concrete example of a two-step g.o. space, namely a three-dimensional unimodular Lorentzian Lie group. We recall that a three-dimensional Lorentzian Lie group is a three-dimensional Lie group G endowed with a Lorentzian left-invariant metric $\langle \ ,\ \rangle$. We also recall the following classification result.

Theorem 6.1

[17] A three-dimensional simply connected unimodular Lorentzian Lie group G admits a pseudo-orthonormal frame field $\{ e_1,e_2,e_3\}$, with $e_3$ time-like, such that the Lie algebra ${\mathfrak{g}}$ of G is one of the following:

$$\begin{aligned}&\left[ e_1,e_2 \right] =\alpha e_1-\beta e_3, \\ {\mathfrak{g}} _1 :&\left[ e_1,e_3\right] =-\alpha e_1-\beta e_2, \\&\left[ e_2,e_3\right] =\beta e_1 +\alpha e_2 +\alpha e_3 \qquad \alpha \ne 0. \end{aligned}$$

If $\beta \ne 0$, then $G=\widetilde{SL (2,{{\mathbb{R}}})}$, while for $\beta =0$ it is $G=E(1,1)$, the group of rigid motions of the Minkowski two-space.

$$\begin{aligned}&\left[ e_1,e_2 \right] =-\gamma e_2-\beta e_3, \\ {\mathfrak{g}} _2:&\left[ e_1,e_3\right] =-\beta e_2+\gamma e_3, \qquad \gamma \ne 0, \\&\left[ e_2,e_3\right] =\alpha e_1 . \end{aligned}$$

In this case, $G=\widetilde{SL (2,{{\mathbb{R}}})}$ if $\alpha \ne 0$, while $G=E(1,1)$ if $\alpha =0$.

$$\begin{aligned} {\mathfrak{g}} _3: \quad \left[ e_1,e_2 \right] =-\gamma e_3, \quad \left[ e_1,e_3\right] =-\beta e_2, \quad \left[ e_2,e_3\right] =\alpha e_1 . \end{aligned}$$

Table 1 (where $\widetilde{E}(2)$ and $H_3,$ respectively, denote the universal covering of the group of rigid motions in the Euclidean two-space and the Heisenberg group) lists all the Lie groups G which admit a Lie algebra ${\mathfrak{g}}_3$, according to the different possibilities for the signs of $\alpha$, $\beta$ and $\gamma$:

Table 1 3D Lorentzian Lie groups with Lie algebra ${\mathfrak{g}}_3$

$$\begin{aligned}&\left[ e_1,e_2 \right] =- e_2 + (2 \varepsilon - \beta ) e_3, \qquad \varepsilon = \pm 1, \\ {\mathfrak{g}} _4:&\left[ e_1,e_3\right] =-\beta e_2 + e_3, \\&\left[ e_2,e_3\right] =\alpha e_1 . \end{aligned}$$

Table 2describes all Lie groups G admitting a Lie algebra ${\mathfrak{g}}_4$:

Table 2 3D Lorentzian Lie groups with Lie algebra ${\mathfrak{g}}_4$

We observe that it would be possible to unify the two cases described in Table 2. However, in order to unify them, the second column should list conditions for $\varepsilon a$ instead of a, which would make the whole table less readable.

We see from the above classification that the Lie algebra ${\mathfrak{sl}}_2$ of $SL (2,{{\mathbb{R}}})$ is the one occurring most frequently in the classification of three-dimensional unimodular Lorentzian Lie algebras. Observe that $SL (2,{{\mathbb{R}}})$ has been an important source of examples for very different topics (see, for example, [4, 18]).

The main result of this section is the following.

Theorem 6.2

The Lorentzian Lie group $(SL (2,{{\mathbb{R}}}),\langle \ , \ \rangle )$ is a two-step g.o. space if and only if one of the following properties holds for its Lie algebra ${\mathfrak{sl}}_2$:

• (a) ${\mathfrak{sl}}_2={\mathfrak{g}}_3$ with either $\alpha =\beta$ or $\alpha =\gamma$ or $\beta =\gamma$.

• (b) ${\mathfrak{sl}}_2={\mathfrak{g}}_4$ with $\alpha =\beta -\varepsilon$.

To prove our main result, we need the following lemma.

Lemma 6.3

Let $(G,\langle \ ,\ \rangle )$ be a Lie group such that ${\mathfrak{g}}={\text{span}}\{e_1,\dots ,e_n\}$. If $(G,\langle \ ,\ \rangle )$ is a two-step g.o. space, then for any $V\in {\mathfrak{g}}$ there exist $X,Y\in {\mathfrak{g}}$, such that

• (i) $X+Y=V$ and

• (ii) there exists a function k such that ${\mathcal{G}}_{k}^{e_i}(0)=0$ for all indices $i=1,\dots , n$, where ${\mathcal{G}}_{k}^W(t)$ is the function

  $$\begin{aligned} {\mathcal{G}}_{k}^W(t)=\langle T(t)X+Y,[W,T(t)X+Y]\rangle +\langle W,[T(t)X,Y]\rangle -k(t)\langle W,T(t)X+Y\rangle . \end{aligned}$$

  Conversely, if for any $V\in {\mathfrak{g}}$ there exist $X,Y\in {\mathfrak{g}}$ satisfying conditions (i), (ii) and

• (iii) ${\text{ad}}(Y)$ is skew-symmetric with respect to $\langle \ , \ \rangle$, then $(G,\langle \ , \ \rangle )$ is a two-step g.o. space.

Proof

If $(G,\langle \ , \ \rangle )$ is a two-step g.o. space, then conditions (i) and (ii) follow immediately by Proposition 5.1. To prove the converse, we observe that the quantity ${\mathcal{G}}_{k}^W(t)$ is linear with respect to W. Since ${\mathfrak{g}}={\text{span}}\{e_1,\dots e_n\}$, condition (ii) implies that ${\mathcal{G}}_{k}^W(0)=0$ for any $W\in {\mathfrak{g}}$. By taking into account condition (iii) and Proposition 4.3, we obtain that ${\mathcal{G}}_{k}^W(t)=0$, for all $W\in {\mathfrak{g}}$ and for all $t\in J$. Also, by taking into account condition (i) and Proposition 5.1 we conclude that $(G,\langle \ , \ \rangle )$ is a two-step g.o. space.

Proof of Theorem 6.2

We assume that there exists a left-invariant Lorentzian two-step g.o. metric $\langle \ , \ \rangle$ in $SL (2,{{\mathbb{R}}})$. We will use the classification of three-dimensional Lorentzian Lie algebras mentioned in Theorem 6.1, and we will examine each of the four cases in which the Lie algebra ${\mathfrak{sl}}_2$ occurs. We recall that for $X,Y\in {\mathfrak{g}}$ we have

$$\begin{aligned} {\mathcal{G}}_{k}^W(0)=\langle X+Y,[W,X+Y]\rangle +\langle W,[X,Y]\rangle -k\langle W,X+Y\rangle . \end{aligned}$$

Case 1 ${\mathfrak{sl}}_2={\mathfrak{g}}_1$ with $\beta \ne 0$. Let $V=2\alpha e_1+\beta e_2-\beta e_3\in {\mathfrak{g}}_1.$

By Lemma 6.3, there exist $X,Y\in {\mathfrak{g}}$ such that $X+Y=V$ and ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$. We set

$$\begin{aligned} X=x_1e_1+x_2e_2+x_3e_3, \end{aligned}$$

so that

$$\begin{aligned} Y=(2\alpha -x_1)e_1+(\beta -x_2)e_2+(-\beta -x_3)e_3. \end{aligned}$$

Taking into account the Lie bracket relations for the algebra ${\mathfrak{g}}_1$ as well as the expressions of X, Y, then system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$, is equivalent to

$$\begin{aligned}&2\alpha \beta x_1-\big (2\alpha ^2+\beta ^2\big )x_2+\big (2\alpha ^2-\beta ^2\big )x_3=2k\alpha \\&\beta ^2x_1-\alpha \beta x_2+\alpha \beta x_3=k\beta +2\alpha \big (2\alpha ^2+\beta ^2\big )\\&\beta ^2x_1-\alpha \beta x_2+\alpha \beta x_3=k\beta -2\alpha \big (2\alpha ^2+\beta ^2\big ). \end{aligned}$$

The last two equations imply that $\alpha =0$, which is a contradiction. Therefore, ${\mathfrak{sl}}_2\ne {\mathfrak{g}}_1$.

Case 2 ${\mathfrak{sl}}_2={\mathfrak{g}}_2$ with $\alpha \ne 0$. Let $V=v_1 e_1+v_2 e_2+v_3 e_3\in {\mathfrak{g}}_2$.

By Lemma 6.3, there exist $X,Y\in {\mathfrak{g}}$, such that $X+Y=V$ and ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$. We consider again $X=x_1e_1+x_2e_2+x_3e_3$, so that

$$\begin{aligned} Y=(v_1-x_1)e_1+(v_2-x_2)e_2+(v_3-x_3)e_3. \end{aligned}$$

Taking into account the Lie bracket relations for the Lie algebra ${\mathfrak{g}}_2$ as well as the expressions of X, Y we deduce that the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$, is equivalent to

$$\begin{aligned}&\alpha v_3x_2-\alpha v_2x_3=kv_1+\gamma (v_2^2+v_3^2)\\&-(\gamma v_2+\beta v_3)x_1+\gamma v_1 x_2+\beta v_1 x_3=kv_2+(\beta -\alpha )v_1v_3-\gamma v_1v_2\\&(\beta v_2-\gamma v_3)x_1-\beta v_1 x_2+\gamma v_1 x_3=-kv_3+(\alpha -\beta )v_1v_2-\gamma v_1v_3 \end{aligned}$$

with unknowns $x_1,x_2,x_3$ and parameters $v_1,v_2,v_3,k$. The determinant D of the above system is zero. The determinant $D_{x_1}$ is given by $D_{x_1}=Ak+B$, where $A=v_1\big ((\beta ^2+\gamma ^2)v_1^2-2\alpha \gamma v_2v_3-\alpha \beta v_3^2+\alpha \beta v_2^2\big )$ and $B=v_1^2(v_2^2+v_3^2)\gamma ((\alpha -\beta )^2+\gamma ^2)$. Here, $D_{x_1}$ is the determinant obtained by replacing first column of the above system, by the column of constant terms.

If $v_1,v_2,v_3\ne 0$ then $B\ne 0$, because $\gamma \ne 0$. Moreover, as $\alpha \gamma \ne 0$, there exist non zero real numbers $v_1,v_2,v_3$, such that $A=0$. Indeed, it suffices to choose

$$\begin{aligned} v_1=\sqrt{\frac{2\varepsilon \alpha \gamma }{\beta ^2+\gamma ^2}},\quad v_2=\varepsilon , \quad v_3=1. \end{aligned}$$

Here, $\varepsilon =1$ if $\alpha \gamma >0$ and $\varepsilon =-1$ if $\alpha \gamma <0$. With the above choices for $v_1,v_2,v_3,$ we obtain that $D_{x_1}\ne 0$. Since $D=0$, this implies that there exist no solutions $x_i$ for the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$, which is a contradiction. Therefore, ${\mathfrak{sl}}_2\ne {\mathfrak{g}}_2$

Case 3 ${\mathfrak{sl}}_2={\mathfrak{g}}_3$ with $\alpha ,\beta ,\gamma >0$ or $\alpha >0$ and $\beta ,\gamma <0$. We set $V=v_1 e_1+v_2 e_2+v_3 e_3\in {\mathfrak{g}}_3$.

Assume that $(SL (2,{{\mathbb{R}}}),\langle \ , \ \rangle )$ is two-step g.o. and $\alpha ,\beta ,\gamma$ are all distinct. By Lemma 6.3, there exist $X,Y\in {\mathfrak{g}}_3$ such that $X+Y=V$ and ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$. We set $X=x_1e_1+x_2e_2+x_3e_3$, so that

$$\begin{aligned} Y=(v_1-x_1)e_1+(v_2-x_2)e_2+(v_3-x_3)e_3. \end{aligned}$$

Taking into account the Lie bracket relations for the algebra ${\mathfrak{g}}_3$ as well as the expressions of X, Y, the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$ is equivalent to

$$\begin{aligned}&\alpha v_3x_2-\alpha v_2 x_3=(\beta -\gamma )v_2v_3+kv_1 \end{aligned}$$

(6.1)

$$\begin{aligned}&-\beta v_3x_1+\beta v_1x_3=(\gamma -\alpha )v_1v_3-kv_2 \end{aligned}$$

(6.2)

$$\begin{aligned}&\gamma v_2x_1-\gamma v_1x_2=(\alpha -\beta )v_1v_2-kv_3, \end{aligned}$$

(6.3)

with unknowns $x_1,x_2,x_3$ and parameters $v_1,v_2,v_3,k$.

The determinant D of the above system is zero. The determinant $D_{x_1}$ is equal to

$$\begin{aligned} D_{x_1}=Ak+B, \end{aligned}$$

where $A=v_1(\beta \gamma v_1^2-\alpha \beta v_3^2+\alpha \gamma v_2^2)$ and $B=v_1^2v_2v_3(\alpha -\gamma )(\alpha -\beta )(\beta -\gamma )$. If $\alpha ,\beta ,\gamma >0,$ we set

$$\begin{aligned} v_1=v_2=1,\quad v_3=\sqrt{\frac{\gamma (\alpha +\beta )}{\alpha \beta }}, \end{aligned},$$

whereas if $\alpha >0$ and $\beta ,\gamma <0$ we set

$$\begin{aligned} v_1=v_3=1,\quad v_2=\sqrt{\frac{\beta (\gamma -\alpha )}{-\alpha \gamma }}. \end{aligned}$$

With the above choices for $v_1,v_2,v_3$ and since $\alpha ,\beta ,\gamma$ are all distinct, we obtain that $A=0$ and $B\ne 0$. Therefore, $D_{x_1}\ne 0$. Since $D=0$, this implies that there exist no solutions $x_i$ for the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$, which is a contradiction. Hence, at least two of the structure constants $\alpha ,\beta ,\gamma$ are equal.

Conversely, assume that at least two of the structure constants $\alpha ,\beta ,\gamma$ coincide, so that we have one of the cases below:

If $\alpha =\beta,$ we set $x_1=v_1$, $x_2=v_2$, $x_3=\frac{\gamma }{\alpha }v_3$, so that

$$\begin{aligned} X=v_1e_1+v_2e_2+\frac{\gamma }{\alpha }v_3e_3\quad \text{ and } \quad Y=\left( 1-\frac{\gamma }{\alpha }\right) v_3e_3. \end{aligned}$$

If $\alpha =\gamma,$ we set $x_1=v_1$, $x_3=v_3$, $x_2=\frac{\beta }{\alpha }v_2$, so that

$$\begin{aligned} X=v_1e_1+\frac{\beta }{\alpha }v_2e_2+v_3e_3\quad \text{ and } \quad Y=\left( 1-\frac{\beta }{\alpha }\right) v_2e_2. \end{aligned}$$

Finally, if $\beta =\gamma,$ we set $x_2=v_2$, $x_3=v_3$, $x_1=\frac{\alpha }{\beta }v_1$, so that

$$\begin{aligned} X=\frac{\alpha }{\beta }v_1e_1+v_2e_2+v_3e_3\quad \text{ and } \quad Y=\left( 1-\frac{\alpha }{\beta }\right) v_1e_1. \end{aligned}$$

In any of the above cases, the following conditions are satisfied:

• (i) $X+Y=V$.

• (ii) Equations (6.1)–(6.3) are satisfied for $k=0$; therefore, ${\mathcal{G}}_{0}^{e_i}(0)=0$, $i=1,2,3$.

• (iii) The endomorphism ${\text{ad}}(Y)$ is skew symmetric with respect to $\langle$, $\rangle$.

By virtue of Lemma 6.3, we then conclude that $(SL (2,{\mathbb{R}}),\langle \ , \ \rangle )$ is a two-step g.o. space. $\square$

Case 4 ${\mathfrak{sl}}_2 ={\mathfrak{g}}_4$ with $\alpha \ne 0$ and $\beta \ne \varepsilon$.

Assume that $(SL (2, {{\mathbb{R}}}),\langle \ , \ \rangle )$ is two-step g.o. and $\alpha \ne \beta -\varepsilon$. By virtue of Lemma 6.3, there exist $X,Y\in {\mathfrak{g}}_4$ and a function k, such that $X+Y=V$ and ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$. We set $X=x_1e_1+x_2e_2+x_3e_3$, so that

$$\begin{aligned} Y=(v_1-x_1)e_1+(v_2-x_2)e_2+(v_3-x_3)e_3. \end{aligned}$$

Taking into account the Lie bracket relations for the algebra ${\mathfrak{g}}_4$ as well as the expressions of X, Y, the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$ is equivalent to

$$\begin{aligned}&\alpha v_3x_2-\alpha v_2 x_3=kv_1+(v_2+\varepsilon v_3^2) \end{aligned}$$

(6.4)

$$\begin{aligned}&(-v_2-\beta v_3)x_1+v_1x_2+\beta v_1x_3=kv_2-v_1v_2-v_1v_2(\alpha +2\varepsilon - \beta ) \end{aligned}$$

(6.5)

$$\begin{aligned}&((\beta -2\varepsilon )v_2-v_3)x_1-(\beta -2\varepsilon )v_1x_2+v_1x_3=-kv_3+(\alpha -\beta )v_1v_2-v_1v_3, \end{aligned}$$

(6.6)

with unknowns $x_1,x_2,x_3$ and parameters $v_1,v_2,v_3,k$. The determinant D of the above system is zero. The determinant $D_{x_1}$ is given by

$$\begin{aligned} D_{x_1}=Ak+B, \end{aligned}$$

where $A=kv_1\big (v_1^2(\beta -\varepsilon )^2-2\alpha v_2v_3+\alpha \beta (v_2^2-v_3^2)+2\varepsilon v_2\big )$ and $B=v_1^2(v_2+\varepsilon v_2)^2(\alpha -\beta +\varepsilon )^2$.

If $\alpha <0$ and $\beta \le,0$ we set

$$\begin{aligned} v_2=-\varepsilon ,\quad v_3=2\varepsilon , \quad v_1=\sqrt{\frac{-4\alpha +3\alpha \beta +2}{(\beta -\varepsilon )^2}}. \end{aligned}$$

If $\alpha <0$ and $\beta >0,$ we set

$$\begin{aligned} v_2=-2\varepsilon ,\quad v_3=\varepsilon , \quad v_1=\sqrt{\frac{-4\alpha -3\alpha \beta +4}{(\beta -\varepsilon )^2}}. \end{aligned}$$

If $\alpha >0$ and $\beta \le 0,$ we set

$$\begin{aligned} v_2=-2\varepsilon ,\quad v_3=-\varepsilon , \quad v_1=\sqrt{\frac{4\alpha -3\alpha \beta +4}{(\beta -\varepsilon )^2}}. \end{aligned}$$

If $\alpha >0$ and $\beta >0,$ we set

$$\begin{aligned} v_2=-\varepsilon ,\quad v_3=-2\varepsilon , \quad v_1=\sqrt{\frac{4\alpha +3\alpha \beta +2}{(\beta -\varepsilon )^2}}. \end{aligned}$$

With the above choices for $v_1,v_2,v_3,$ we obtain that $A=0$ and $B\ne 0$. Therefore, $D_{x_1}\ne 0$. Since $D=0$, this implies that there exist no solutions $x_i$ for the system ${\mathcal{G}}_{k}^{e_i}(0)=0$, $i=1,2,3$, which is a contradiction. Hence, $\alpha =\beta -\varepsilon$. Conversely, assume that $\alpha =\beta -\varepsilon$. We set

$$\begin{aligned} x_1=v_1,\quad x_2=\frac{\beta v_2+v_3}{\beta -\varepsilon },\quad x_3=\frac{-v_2+(\beta -\varepsilon )v_3}{\beta -\varepsilon }, \end{aligned}$$

so that

$$\begin{aligned} X=v_1e_1+\frac{\beta v_2+v_3}{\beta -\varepsilon }e_2-\frac{-v_2+(\beta -\varepsilon )v_3}{\beta -\varepsilon }e_3,\quad Y=-\frac{\varepsilon v_2+v_3}{\beta -\varepsilon }e_2+\frac{v_2+\varepsilon v_3}{\beta -\varepsilon }e_3. \end{aligned}$$

Then, the following conditions are satisfied:

• (i) $X+Y=V$.

• (ii) Equations (6.4)–(6.6) are satisfied for $k=0$; therefore ${\mathcal{G}}_0^{e_i}(0)=0$, $i=1,2,3$.

• (iii) The endomorphism ${\text{ad}}(Y)$ is skew symmetric with respect to $\langle \ , \ \rangle$. By virtue of Lemma 6.3, we obtain that $(SL (2,{{\mathbb{R}}}),\langle \ , \ \rangle )$ is a two-step g.o. space, which concludes Case 4. $\square$

Remark 6.4

Observe that for the trivial coset realization $(SL (2,{{\mathbb{R}}}), \langle , \ \rangle )$ not every geodesic is homogeneous ([5]). Hence, the geodesics obtained in Theorem 6.2 are examples of proper two-step g.o. spaces. On the other hand, with respect to the full isometry group, those examples are g.o. (in fact, they are naturally reductive, cf. [5]).

For the Riemannian case, a similar argument proves the following.

Theorem 6.5

The Riemannian Lie group $(SL (2,{{\mathbb{R}}}),\langle \ , \ \rangle )$ is a two-step g.o. space if and only its Lie algebra ${\mathfrak{sl}}_2$ admits an orthonormal basis $\{e_1,e_2,e_3\}$, such that

$$\begin{aligned}{}[e_{1},e_{2}]=\gamma e_{3},\qquad [e_{2},e_{3}]=\alpha e_{1} ,\qquad [e_{3},e_{1}]=\beta e_{2}, \end{aligned}$$

with either $\alpha =\beta$ or $\alpha =\gamma$ or $\beta =\gamma$.

Remark 6.6

By exactly the same argument used in the proof of the “only if ”part of Theorem 6.2, the result remains true if we start from any three-dimensional unimodular Lorentzian Lie algebra (as described in Theorem 6.1). Hence, we have the following.

Theorem 6.7

A unimodular Lorentzian Lie group $(G,\langle \ , \ \rangle )$ is a two-step g.o. space, if one of the following properties holds for its Lie algebra ${\mathfrak{g}}$:

• (a) either ${\mathfrak{g}}=\mathfrak{g}_3$ and at least two among $\alpha$, $\beta$ and $\gamma$ coincide, or

• (b) ${\mathfrak{g}}={\mathfrak{g}}_4$ with $\alpha =\beta -\varepsilon$.

In particular, $G=Sl(2,{{\mathbb{R}}})$.