Homogeneous Geodesics in Five-Dimensional Generalized Symmetric Spaces

G. Calvaruso and R.A. Marinosci

Dedicated to the Memory of Grigorios TSAGAS (1935-2003), President of Balkan Society of Geometers (1997-2003)

Abstract

O.Kowalski and J.Szenthe [7] proved that every homogeneous Riemannian manifold admits a homogeneous geodesic, that is, a geodesic which is an orbit of a one-parameter group of isometries. Then, several authors investigated the set of all homogeneous geodesics of some homogeneous spaces.

In this paper, we study the set of homogeneous geodesics of five-dimen- sional generalized symmetric spaces and we find several interesting behaviours.

Mathematics Subject Classification: 53C20, 53C22, 53C30. Key words: homogeneous spaces, homogeneous geodesics

1 Introduction

Let (M, g) be a (connected) homogeneous space, that is, a Riemannian manifold admitting a connected group of isometries G, acting transitively and effectively on M. Then, M can be identified with (G/H, g), where H is the isotropy group at a fixed point o of M. The Lie algebra \underline{g} of G has a reductive decomposition \underline{g} = m⊕\underline{h}, where m ⊂ \underline{g} is a subspace of \underline{g} isomorphic to the tangent space T_{o}(M) and \underline{h} is the Lie algebra of H. In general, such decomposition is not unique. A geodesic γ(t) through the origin o of M = G/H is called homogeneous if it is an orbit of a one-parameter subgroup of G, that is γ(t) = exp(tZ)(o), t ∈ IR, (1.1)

where Z is a nonzero vector of \underline{g} .

A homogeneous Riemannian manifold is called a g.o. space if all geodesics are homogeneous with respect to the largest connected group of isometries. All naturally reductive spaces are g.o.spaces, but the converse does not hold. In fact, A. Kaplan [3] proved the existence of g.o. spaces which are in no way naturally reductive; the examples of A.Kaplan are generalized Heisenberg groups with two-dimensional center. In [8], O.Kowalski and L.Vanhecke gave a classification of all g.o.spaces (which are in no way naturally reductive) up to dimension six.

∗ Balkan Journal of Geometry and Its Applications, Vol.8, No.1, 2003, pp. 1-19.

\begin{matrix}c \\ ◯ \end{matrix} Balkan Society of Geometers, Geometry Balkan Press 2003.

About the existence of homogeneous geodesics in a general homogeneous Riemannian manifold, V.V.Kajzer [2] proved that a Lie group endowed with a leftinvariant metric admits at least one homogeneous geodesic. More recently, O.Kowalski and J.Szenthe [7] proved that any homogeneous Riemannian space M = G/H admits at least one homogeneous geodesic through the origin. They also proved that if G is semi-simple, then M admits n = dimM mutually orthogonal homogeneous geodesics through the origin.

A natural problem is then to study the set of all homogeneous geodesics of a homogeneous Riemannian manifold. Several authors investigated the sets of homogeneous geodesics on some types of homogeneous spaces (see [6], [9], [10]).

In this paper, we consider five-dimensional generalized symmetric spaces. The study of homogeneous geodesics in these homogeneous spaces is of particular interest, because their Lie groups in general are not semi-simple. For the ones of order 6 (type 9, see [4]), the Lie group is solvable. In section 2 we shall recall some basic facts about homogeneous geodesics in homogeneous Riemannian manifolds. In sections 3,4,5,6,7 and 8 we study the sets of all homogeneous geodesics of five-dimensional generalized symmetric spaces of type 2, 3, 4, 7, 8 and 9, respectively, which are all the examples of five-dimensional generalized symmetric spaces which are not g.o. spaces. The most interesting results we found concern homogeneous geodesics of fivedimensional generalized symmetric spaces of type 7a and 9. Some other cases, in particular types 3 and 8, are interesting because they present quite complicated sets of geodesic vectors.

2 Preliminaries on homogeneous geodesics and generalized symmetric spaces

Let (M = G/H, g) be a homogeneous Riemannian manifold with a fixed origin o, \underline{g} and \underline{h} the Lie algebras of G and H respectively and

\underline{g} = m ⊕ \underline{h} (2.1)

a reductive decomposition. The canonical projection p : G → G/H induces an isomorphism between the subspace m and the tangent space T_{o}(M). Consequently, the scalar product g_{o} on T_{o}(M) induces a scalar product <, > on m which is Ad(H)- invariant. A non-zero vector Z ∈ \underline{g} is called a geodesic vector if the curve exptX(o) is a geodesic. We recall the following characterization of geodesic vectors: Lemma 2.1 [8] A non-zero vector X ∈ \underline{g} is a geodesic vector if and only if

< [X, Y]_{m}, X_{m} >= 0 (2.2)

for all Y ∈ m (the subscript m denotes the projection into m).

Therefore, looking for all homogeneous geodesics of a homogeneous Riemannian manifold (M = G/H, g), we first calculate the connected component G of the full isometry group I(M), or at least the corresponding Lie algebra \underline{g} . Then, we find a decomposition of the form (2.1) and look for the geodesic vectors in the form

Z = ∑_{i=1}^{r} x_{i}e_{i} + ∑_{j=1}^{s} a_{j}A_{j} (2.3)

where \{e_{i}\}_{i=1,2,...,r} is a convenient basis of m and \{A_{j}\}_{j=1,2,...,s} is a basis of \underline{h}. When we take Y = e_{i} , i = 1, 2, ..., r, the condition (2.2) produces a system of r quadratic equations for the variables x_{i} and a_{j} . Then, we see for which values of x_{1}, x_{2}, ..., x_{r} and a_{1}, a_{2}, ..., a_{s} this system is satisfied. To such solutions, for which x_{1}, x_{2}, ..., x_{r} are not all equal to zero, correspond geodesic vectors (see also [6]).

A finite family {γ_{1}, γ_{2}, ..., γ_{k}} of homogeneous geodesics through o ∈ M is said to be orthogonal (respectively, linearly independent) if the corresponding initial tangent vectors at o are orthogonal (resp., linearly independent). The following result holds: Proposition 2.2 [6] A finite family {γ_{1}, γ_{2}, ..., γ_{k}} of homogeneous geodesics through o ∈ M is orthogonal (respectively, linearly independent) if the m-components of the corresponding geodesic vectors are orthogonal (respectively, linearly independent).

We now recall some basic facts about generalized symmetric spaces. A generalized symmetric space is a connected Riemannian manifold (M, g) admitting a regular sstructure, that is, a family {s_{x} : x ∈ M} of symmetries on M, such that

s_{x} ∘ s_{y} = s_{z} ∘ s_{x}, z = s_{x}(y),

for every points x, y, ∈ M [5]. As it is well-known, every generalized symmetric space is a homogeneous Riemannian space G/H [4]. An s-structure {s_{x} : x ∈ M} is said to be of order k ≥ 2 if (s_{x})^{k} = id for all x ∈ M and (s_{x})^{i} ≠ id for i < k. A Riemannian manifold (M, g) is said to be k-symmetric if it admits a regular s-structure of order k. Each generalized symmetric space is k-symmetric for some k [4]. The order of a generalized symmetric space is the least integer k such that (M, g) is k-symmetric.

Low-dimensional generalized symmetric spaces have been classified [4], [5]. Comparing this classification with the classification of low-dimensional g.o. spaces given in [8], we see that the generalized symmetric spaces which are not g.o. spaces are the ones of type 2, 3, 4, 7, 8 (all of order 4) and 9 (of order 6) in the classification given in [5].

For all these spaces, in [4] it is given a basis of \underline{g} , containing a basis \{X_{1}, Y_{1}, X_{2}, Y_{2}, W \} of m , with respect to which the Lie bracket [,] of \underline{g} and the scalar product <, > of m are explicitly described.

According to Lemma 2.1, a non-zero vector X ∈ \underline{g} is geodesic if and only if

< [X, Y]_{m}, X_{m} >= 0

for all Y ∈ m or, equivalently, if and only if

< [X, X_{1}]_{m}, X_{m} >= 0,
< [X, X_{2}]_{m}, X_{m} >= 0,
< [X, Y_{1}]_{m}, X_{m} >= 0,
< [X, Y_{2}]_{m}, X_{m} >= 0,
< [X, W]_{m}, X_{m} >= 0.

(2.4)

3 Homogeneous geodesics of generalized symmetric spaces of type 2

A five-dimensional generalized symmetric space M of type 2 is IR^{5}(x, y, z, w, t), equipped with the Riemannian metric

g = e^{−2λ_{1}t}dx^{2} + e^{2λ_{1}t}dy^{2} + e^{−2λ_{2}t}dz^{2} + e^{2λ_{2}t}dw^{2} + dt^{2} +
2α[e^{−(λ_{1}+λ_{2})t}dxdz + e^{(λ_{1}+λ_{2})t}dydw] +
+2β[e^{(λ_{1}−λ_{2})t}dydz − e^{(λ_{2}−λ_{1})t}dxdw],

where either λ_{1} > λ_{2} > 0, α^{2} + β^{2} < 1, or λ_{1} = λ_{2} > 0, α = 0 and 0 ≤ β < 1, or λ_{1} < 0, λ_{2} = 0, α = 0 and 0 < β < 1. As homogeneous space, M = G/H, where G is the group of all matrices of the form

( \begin{matrix}e^{λ_{1}t} & 0 & 0 & 0 & x \\ 0 & e^{−λ_{1}t} & 0 & 0 & y \\ 0 & 0 & e^{λ_{2}t} & 0 & z \\ 0 & 0 & 0 & e^{−λ_{2}t} & w \\ 0 & 0 & 0 & 0 & 1 \end{matrix} )

The linear subspace m of \underline{g} admits a basis \{X_{1}, Y_{1}, X_{2}, Y_{2}, W \} such that

[X_{j}, W] = −λ_{j}X_{j}, [Y_{j}, W] = λ_{j}Y_{j}, [·, ·] = 0 otherwise.

2a): λ_{1} > λ_{2} > 0 and α^{2} + β^{2} < 1.

In this case, \underline{h} = 0 [4]. The Lie bracket [,] and the Riemannian metric <, > are respectively determined by

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 0 & 0 & 0 & 0 & −λ_{1}X_{1} \\ X_{2} & 0 & 0 & 0 & 0 & −λ_{2}X_{2} \\ Y_{1} & 0 & 0 & 0 & 0 & λ_{1}Y_{1} \\ Y_{2} & 0 & 0 & 0 & 0 & λ_{2}Y_{2} \\ W & λ_{1}X_{1} & λ_{2}X_{2} & −λ_{1}Y_{1} & −λ_{2}Y_{2} & 0 \end{matrix} (3.1)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 1 & α & 0 & −β & 0 \\ X_{2} & α & 1 & β & 0 & 0 \\ Y_{1} & 0 & β & 1 & α & 0 \\ Y_{2} & −β & 0 & α & 1 & 0 \\ W & 0 & 0 & 0 & 0 & 1 \end{matrix} (3.2)

(see [4]). Using (3.1) and (3.2) to compute (2.4), we obtain that X ∈ \underline{g} = m is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w) with respect to {X_{1}, X_{2}, Y_{1}, Y_{2}, W } satisfy

\{ \begin{matrix}w(x_{1} + αx_{2} − βy_{2}) = 0, \\ w(αx_{1} + x_{2} + βy_{1}) = 0, \\ w(βx_{2} + y_{1} + αy_{2}) = 0, \\ w(−βx_{1} + αy_{1} + y_{2}) = 0, \\ −λ_{1}x_{1}(x_{1} + αx_{2} − βy_{2}) − λ_{2}x_{2}(αx_{1} + x_{2} + βy_{1})+ \\ +λ_{1}y_{1}(βx_{2} + y_{1} + αy_{2}) + λ_{2}y_{2}(−βx_{1} + αy_{1} + y_{2}) = 0, \end{matrix} (3.3)

where we took into account that λ_{1} > λ_{2} > 0.

If w ≠ 0, (3.4) gives easily x_{1} = x_{2} = y_{1} = y_{2} = 0. If w = 0, the solutions of (3.4) are given by (x_{1}, x_{2}, y_{1}, y_{2}, 0), satisfying

−λ_{1}x_{1}(x_{1} + αx_{2} − βy_{2}) − λ_{2}x_{2}(αx_{1} + x_{2} + βy_{1}) + (3.4)

+λ_{1}y_{1}(βx_{2} + y_{1} + αy_{2}) + λ_{2}y_{2}(−βx_{1} + αy_{1} + y_{2}) = 0,

Hence, we proved that X is a geodesic vector of a generalized symmetric space of type 2a if and only if

(i) X = wW, or
(ii) X = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and (3.4) holds.

Geometrically speaking, geodesic vectors of a generalized symmetric space of type 2a form a straight line (i) and a hypercone (of equation (3.4)) in the orthogonal complement of such line (ii).

Note that W is a geodesic vector of type (i), while X_{1} +Y_{1}, X_{1} −Y_{1}, −(α−β)X_{1} + X_{2} − (α + β)Y_{1} + Y_{2} and −(α + β)X_{1} + X_{2} + (α − β)Y_{1} − Y_{2} are geodesic vectors of type (ii) and all together they form an orthogonal basis of geodesic vectors. 2b): λ_{1} = λ_{2} > 0, α = 0 and β > 0.

In this case, \underline{h} = so(2) = span(A), where A is determined by AX_{1} = X_{2}, AX_{2} = −X_{1}, AY_{1} = Y_{2}, AY_{2} = −Y_{1}, AW = 0 (see [4]). Since [A, X] = AX for all X ∈ \underline{g} , we get

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A \\ X_{1} & 0 & 0 & 0 & 0 & −λ_{1}X_{1} & −X_{2} \\ X_{2} & 0 & 0 & 0 & 0 & −λ_{2}X_{2} & X_{1} \\ Y_{1} & 0 & 0 & 0 & 0 & λ_{1}Y_{1} & −Y_{2} \\ Y_{2} & 0 & 0 & 0 & 0 & λ_{2}Y_{2} & Y_{1} \\ W & λ_{1}X_{1} & λ_{2}X_{2} & −λ_{1}Y_{1} & −λ_{2}Y_{2} & 0 & 0 \\ A & X_{2} & −X_{1} & Y_{2} & −Y_{1} & 0 & 0 \end{matrix} (3.5)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 1 & 0 & 0 & −β & 0 \\ X_{2} & 0 & 1 & β & 0 & 0 \\ Y_{1} & 0 & β & 1 & 0 & 0 \\ Y_{2} & −β & 0 & 0 & 1 & 0 \\ W & 0 & 0 & 0 & 0 & 1 \end{matrix} (3.6)

We computed <, > as in case 2a), taking into account α = 0. X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a) with respect to the basis \{X_{1}, Y_{1}, X_{2}, Y_{2}, W, A\} of \underline{g} satisfy

\{ \begin{matrix}λ_{1}w(x_{1} − βy_{2}) + a(x_{2} + βy_{1}) = 0, \\ λ_{2}w(x_{2} + βy_{1}) − a(x_{1} − βy_{2}) = 0, \\ −λ_{1}w(βx_{2} + y_{1}) + a(−βx_{1} + y_{2}) = 0, \\ −λ_{2}w(−βx_{1} + y_{2}) − a(βx_{2} + y_{1}) = 0, \\ −λ_{1}x_{1}(x_{1} − βy_{2}) − λ_{2}x_{2}(x_{2} + βy_{1})+ \\ +λ_{1}y_{1}(βx_{2} + y_{1}) + λ_{2}y_{2}(−βx_{1} + y_{2}) = 0. \end{matrix} (3.7)

If a ≠ 0, then the solutions of (3.7) are given by x_{1} = x_{2} = y_{1} = y_{2} = 0. For a = 0, we put a_{1} = x_{1} − βy_{2}, a_{2} = x_{2} + βy_{1}, a_{3} = βx_{2} + y_{1} and a_{4} = −βx_{1} + y_{2}. The system (3.7) reduces to

\{ \begin{matrix}wa_{i} = 0, i = 1, 2, 3, 4, \\ −λ_{1}x_{1}a_{1} − λ_{2}a_{2}λ_{1}y_{1}a_{3} + λ_{2}y_{2}a_{4} = 0. \end{matrix} (3.8)

If w ≠ 0, (3.8) gives a_{i} = 0, i = 1, .., 4, from which it follows easily x_{1} = x_{2} = y_{1} = y_{2} = 0. If w = 0, then (3.8) reduces to the last equation, which gives x_{1}^{2} +x_{2}^{2} = y_{1}^{2} +y_{2}^{2}. Hence, X is a geodesic vector of a generalized symmetric space of type 2b if and only if its m-component is

\begin{matrix}(i) & X_{m} = wW, or \\ (ii) & X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2}. \end{matrix}

We can check easily that {W, X_{1} + Y_{2}, X_{1} − Y_{2}, X_{2} + Y_{1}, X_{2} − Y_{1}} is an orthogonal basis of geodesic vectors of m. 2c): λ_{1} > 0, λ_{2} = 0, α = 0 and 0 < β < 1.

In this case, \underline{h} = so(2) ⨁ so(2) = span(A_{1}, A_{2}), where A_{1} = A of case 2b), while A_{2} is determined by A_{2}X_{1} = X_{2}, A_{2}X_{2} = −X_{1}, A_{2}Y_{1} = −Y_{2}, A_{2}Y_{2} = Y_{1}, A_{2}W = 0 (see [4]). Similarly to case 2b), we get

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A_{1} & A_{2} \\ X_{1} & 0 & 0 & 0 & 0 & −λ_{1}X_{1} & −X_{2} & −X_{2} \\ X_{2} & 0 & 0 & 0 & 0 & 0 & X_{1} & X_{1} \\ Y_{1} & 0 & 0 & 0 & 0 & λ_{1}Y_{1} & −Y_{2} & Y_{2} \\ Y_{2} & 0 & 0 & 0 & 0 & 0 & Y_{1} & −Y_{1} \\ W & λ_{1}X_{1} & 0 & −λ_{1}Y_{1} & 0 & 0 & 0 & 0 \\ A_{1} & X_{2} & −X_{1} & Y_{2} & −Y_{1} & 0 & − & − \\ A_{2} & X_{2} & −X_{1} & −Y_{2} & Y_{1} & 0 & − & − \end{matrix} (3.9)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 1 & 0 & 0 & −β & 0 \\ X_{2} & 0 & 1 & β & 0 & 0 \\ Y_{1} & 0 & β & 1 & 0 & 0 \\ Y_{2} & −β & 0 & 0 & 1 & 0 \\ W & 0 & 0 & 0 & 0 & 1 \end{matrix} (3.10)

X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a, b) with respect to {X_{1}, Y_{1}, X_{2}, Y_{2}, W, A_{1}, A_{2}} satisfy

\{ \begin{matrix}λ_{1}w(x_{1} − βy_{2}) + (a + b)(x_{2} + βy_{1}) = 0, \\ −(a + b)(x_{1} − βy_{2}) = 0, \\ −λ_{1}w(βx_{2} + y_{1}) + (a − b)(−βx_{1} + y_{2}) = 0, \\ −(a − b)(βx_{2} + y_{1}) = 0, \\ −λ_{1}x_{1}(x_{1} − βy_{2}) + λ_{1}y_{1}(βx_{2} + y_{1}) = 0. \end{matrix} (3.11)

If a ≠ ±b, then x_{1} = x_{2} = y_{1} = y_{2} = 0. If a = b = 0, then either w = 0 and x_{1}(x_{1} − βy_{2}) + y_{1}(βx_{2} + y_{1}) = 0, or w ≠ 0 and x_{1} = βy_{2}, y_{1} = −βx_{2}. If a = b ≠ −b, then x_{1} = βy_{2} and x_{2} = y_{1} = 0, while for a = −b ≠ b, we obtain x_{1} = y_{2} = 0 and y_{1} = −βx_{2}. Hence, if X is a geodesic vector of a generalized symmetric space of type 2c, then its m-component is

(i) X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} with x_{1}(x_{1} − βy_{2}) + y_{1}(βx_{2} + y_{1}) = 0,

or

(ii) X_{m} = βy_{2}X_{1} + x_{2}X_{2} − βx_{2}Y_{1} + y_{2}Y_{2} + wW.

We can check easily that X_{1} + Y_{1}, X_{1} − Y_{1} (type (i)), together with X_{2} − βY_{1} + W, X_{2} − βY_{1} − (1 − β^{2})W and βX_{1} + Y_{2} (type (ii)), form an orthogonal basis of m.

From the study of cases 2a), 2b) and 2c) and taking into account Proposition 2.2, we can conclude that all five-dimensional generalized symmetric spaces of type 2 admit five mutually orthogonal homogeneous geodesics through the origin.

4 Homogeneous geodesics of generalized symmetric spaces of type 3

A five-dimensional generalized symmetric space M of type 3 is the homogeneous space M = SO(3, C)/SO(2), where SO(3, C) is the special complex orthogonal group and the Riemannian metric of M is induced by a real invariant positive semidefinite form of GL(3, C) (see [4]). The subalgebra is \underline{h} = so(2) = span(A), where AX_{1} = X_{2}, AX_{2} = −X_{1}, AY_{1} = Y_{2}, AY_{2} = −Y_{1}, AW = 0, with respect to a basis {X_{1}, X_{2}, Y_{1}, Y_{2}, W } of m . The Lie bracket [,] and the Riemannian metric <, > are respectively given by

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A \\ X_{1} & 0 & 0 & 0 & −W & −X_{1} & −X_{2} \\ X_{2} & 0 & 0 & W & 0 & −X_{2} & X_{1} \\ Y_{1} & 0 & −W & 0 & 0 & Y_{1} & −Y_{2} \\ Y_{2} & W & 0 & 0 & 0 & Y_{2} & Y_{1} \\ W & X_{1} & X_{2} & −Y_{1} & −Y_{2} & 0 & 0 \\ A & X_{2} & −X_{1} & Y_{2} & −Y_{1} & 0 & 0 \end{matrix} (4.1)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & a^{2} & 0 & 0 & −γ & 0 \\ X_{2} & 0 & a^{2} & γ & 0 & 0 \\ Y_{1} & 0 & γ & a^{2} & 0 & 0 \\ Y_{2} & −γ & 0 & 0 & a^{2} & 0 \\ W & 0 & 0 & 0 & 0 & b^{2} \end{matrix} (4.2)

where a, b > 0, γ are real numbers, a^{2} > |γ| (see [4]).

Using (4.1) and (4.2) to compute (2.4), we obtain that X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, r) with respect to {X_{1}, X_{2}, Y_{1}, Y_{2}, W, A} satisfy

\{ \begin{matrix}w(a^{2}x_{1} + (b^{2} − γ)y_{2}) + r(a^{2}x_{2} + γy_{1}) = 0, \\ w(a^{2}x_{2} − (b^{2} − γ)y_{1}) − r(a^{2}x_{1} − γy_{2}) = 0, \\ w(−a^{2}y_{1} + (b^{2} − γ)x_{2}) + r(a^{2}y_{2} − γx_{1}) = 0, \\ w(−a^{2}y_{2} − (b^{2} − γ)x_{1}) − r(a^{2}y_{1} + γx_{2}) = 0, \\ x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2}, \end{matrix} (4.3)

taking into account that a^{2} ≠ 0.

We can now find all the solutions of (4.3). If r = w = 0, then (4.3) reduces to x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2}. If r = 0 ≠ w, the solutions of (4.3) are x_{1} = x_{2} = y_{1} = y_{2} = 0 if a^{2} − b^{2} + γ ≠ 0, and x_{1} + y_{2} = x_{2} − y_{1} = 0 if a^{2} − b^{2} + γ = 0.

The case r ≠ 0 is much more complicated. After some standard but quite long calculations, we eventually find that, when r ≠ 0, the solutions of (4.3) are given by

\{ \begin{matrix}x_{1} = −\frac{b^{2}w(ry_{1} + wy_{2})}{a^{2}(r^{2} + w^{2})} + \frac{γ}{a^{2}} y_{2}, \\ x_{2} = \frac{b^{2}w(wy_{1} − ry_{2})}{a^{2}(r^{2} + w^{2})} − \frac{γ}{a^{2}} y_{1}, with r^{2} = \frac{b^{2}w^{2}(b^{2} − 2γ)}{a^{4} − γ^{2}} − w^{2}. \end{matrix}

Thus, if X is a geodesic vector of a generalized symmetric space of type 3, then its m-component is

(i) X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2}, or
(ii) X_{m} = wW if a^{2} − b^{2} + γ ≠ 0, while
X_{m} = x_{2}X_{2} + x_{2}Y_{1} + wW if a^{2} − b^{2} + γ = 0, or
(iii) X_{m} = (−\frac{b^{2}w(ry_{1} + wy_{2})}{a^{2}(r^{2} + 1)} + \frac{γ}{a^{2}} y_{2})X_{1} + (\frac{b^{2}w(wy_{1} − ry_{2})}{a^{2}(r^{2} + 1)} − \frac{γ}{a^{2}} y_{1})X_{2} + y_{1}Y_{1}+
+y_{2}Y_{2} + wW, where r^{2} = \frac{b^{2}w^{2}(b^{2}−2γ)}{a^{4}−γ^{2}} − w^{2} ≠ 0.

One can check that W is a vector of type (ii), while V_{1} = X_{1} + Y_{1}, V_{2} = X_{1} − Y_{1}, V_{3} = γX_{1} +a^{2}X_{2} −γY_{1} +a^{2}Y_{2} and V_{4} = γX_{1} −a^{2}X_{2} +γY_{1} +a^{2}Y_{2} are of type (i), and {V_{1}, V_{2}, V_{3}, V_{4}, W } is an orthogonal basis of m. Therefore, Proposition 2.2 implies that five-dimensional generalized symmetric spaces of type 3 admit five mutually orthogonal homogeneous geodesics through the origin o.

5 Homogeneous geodesics of generalized symmetric spaces of type 4

Five-dimensional generalized symmetric spaces M of type 4 are complex matrix groups

( \begin{matrix}e^{λt} & 0 & z \\ 0 & e^{−λt} & w \\ 0 & 0 & 1 \end{matrix} )

where z, w ∈ C and t ∈ IR. M is also the space C^{2}(z, w) × IR(t), equipped with a Riemannian metric

g = e^{−(λ+λ̄)t}dzdz̄ + e^{(λ+λ̄)t}dwdw̄ + dt^{2} + 2c[e^{(λ̄−λ)t}dzdw̄ + e^{(λ−λ̄)t}dz̄dw] +
+γe^{−2λt}dz^{2} + γ̄e^{−2λ̄t}dz̄^{2} − γe^{2λt}dw^{2} − γ̄e^{2λ̄t}dw̄^{2},

with λ, γ ∈ C, c ∈ IR, γγ̄ + c^{2} < 1/4 [4]. Put ν = (1 + b^{2})γ, where c = \frac{1 − b^{2}}{2(1 + b^{2})}.

Then, γγ̄ + c^{2} < 1/4 is equivalent to νν̄ < b^{2}. 2a): λ + λ̄ ≠ 0 and ν ≠ 0.

In this case, \underline{h} = 0 [4]. With respect to a suitable basis \{X_{1}, X_{2}, Y_{1}, Y_{2}, W \} of \underline{g}, the Lie bracket [,] and the Riemannian metric <, > are given by (see [4])

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 0 & 0 & 0 & 0 & −ηX_{2} − μY_{2} \\ X_{2} & 0 & 0 & 0 & 0 & ηX_{1} − μY_{1} \\ Y_{1} & 0 & 0 & 0 & 0 & −μX_{2} + ηY_{2} \\ Y_{2} & 0 & 0 & 0 & 0 & −μX_{1} − ηY_{1} \\ W & ηX_{2} + μY_{2} & −ηX_{1} + μY_{1} & μX_{2} − ηY_{2} & μX_{1} + ηY_{1} & 0 \end{matrix} (5.1)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 1 & α & 0 & −β & 0 \\ X_{2} & α & b^{2} & β & 0 & 0 \\ Y_{1} & 0 & β & 1 & α & 0 \\ Y_{2} & −β & 0 & α & b^{2} & 0 \\ W & 0 & 0 & 0 & 0 & 1 \end{matrix} (5.2)

where λ = η +iμ and ν = α+iβ. Using (5.1) and (5.2) in (2.4), we obtain that X ∈ \underline{g} is geodesic if and only if its components satisfy

\{ \begin{matrix}wη(αx_{1} + b^{2}x_{2} + βy_{1}) + wμ(−βx_{1} + αy_{1} + b^{2}y_{2}) = 0, \\ −wη(x_{1} + αx_{2} − βy_{2}) + wμ(βx_{2} + y_{1} + αy_{2}) = 0, \\ wμ(αx_{1} + b^{2}x_{2} + βy_{1}) − wη(−βx_{1} + αy_{1} + b^{2}y_{2}) = 0, \\ wμ(x_{1} + αx_{2} − βy_{2}) + wη(βx_{2} + y_{1} + αy_{2}) = 0, \\ (ηx_{1} + μy_{1})(αx_{1} + b^{2}x_{2} + βy_{1}) + (μx_{1} − ηy_{1})(−βx_{1} + αy_{1} + b^{2}y_{2})+ \\ +(−ηx_{2} + μy_{2})(x_{1} + αx_{2} − βy_{2}) + (μx_{2} + ηy_{2})(βx_{2} + y_{1} + αy_{2}) = 0 \end{matrix} (5.3)

It is not difficult to show that if w ≠ 0, the solutions of (5.3) are given by (0, 0, 0, 0, w), while if w = 0, (5.3) reduces to the fifth equation. So, in this case the solutions are (x_{1}, x_{2}, y_{1}, y_{2}, 0) such that

(ηx_{1} + μy_{1})(αx_{1} + b^{2}x_{2} + βy_{1}) + (μx_{1} − ηy_{1})(−βx_{1} + αy_{1} + b^{2}y_{2}) + (5.4)

+(−ηx_{2} + μy_{2})(x_{1} + αx_{2} − βy_{2}) + (μx_{2} + ηy_{2})(βx_{2} + y_{1} + αy_{2}) = 0

We then proved that X is a geodesic vector of a generalized symmetric space of type 4a if and only if

(i) X = wW, or
(ii) X = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and (5.4) holds.

Then, for generalized symmetric spaces of type 4a, geodesic vectors form a straight line (i) and a hypercone (of equation (5.4)) in the orthogonal complement of such line (ii).

About the existence of an orthogonal basis of geodesic vectors, we found that such basis always exists, but it strongly varies with the different values of α, β, η and μ. Taking into account that η ≠ 0 and α+iβ ≠ 0, an orthogonal basis of geodesic vectors is given by:

a) {X_{1}, Y_{1}, X_{2} − βY_{1}, βX_{1} + Y_{2}, W } when α = μ = 0;

b) \{X_{1} + \frac{αμ + βη + \sqrt{Δ}}{αη − βμ}Y_{1}, X_{1} + \frac{αμ + βη − \sqrt{Δ}}{αη − βμ}Y_{1}, (βk − α)X_{1} + X_{2} − (αk +

β)Y_{1} + kY_{2}, (β + αk)X_{1} − kX_{2} + (βk − α)Y_{1} + Y_{2}, W }, with Δ = (α^{2} + β^{2})(η^{2} + μ^{2})

and k = \frac{αμ − βη + \sqrt{Δ}}{αη + βμ}, when α ≠ ±\frac{μ}{η}β;

c) \{X_{1}, Y_{1}, X_{2}−(1+\frac{μ^{2}}{η^{2}} )βY_{1}+\frac{μ}{η}Y_{2}, −\frac{η^{2} + μ^{2}}{ημ} X_{1}+X_{2}−\frac{η}{μ}Y_{2}, W \} when α = \frac{μ}{η}β ≠ 0;

d) \{X_{1}+ \frac{μ}{η}Y_{1}, X_{1}− \frac{η}{μ}Y_{1}, \frac{μ}{η}βX_{1}+X_{2}−βY_{1}, βX_{1}+ \frac{μ}{η}βY_{1}+Y_{2}, W \} if α = −\frac{μ}{η}β ≠ 0.

Therefore, five-dimensional generalized symmetric spaces of type 4a admit five mutually orthogonal homogeneous geodesics through the origin. 2b): λ + λ̄ = 0, ν = 0 and b^{2} ≠ 1.

Since λ + λ̄ = 0, we have λ = iμ, μ ∈ IR. In this case, \underline{h} = so(2) = span(A), where A is determined by AX_{1} = −Y_{1}, AX_{2} = Y_{2}, AY_{1} = X_{1}, AY_{2} = −X_{2}, AW = 0 (see [4]). Computing [,] and <, >, taking into account that now ν = 0 and λ = iμ, we get

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A \\ X_{1} & 0 & 0 & 0 & 0 & −μY_{2} & Y_{1} \\ X_{2} & 0 & 0 & 0 & 0 & −μY_{1} & −Y_{2} \\ Y_{1} & 0 & 0 & 0 & 0 & −μX_{2} & −X_{1} \\ Y_{2} & 0 & 0 & 0 & 0 & −μX_{1} & X_{2} \\ W & μY_{2} & μY_{1} & μX_{2} & μX_{1} & 0 & 0 \\ A & −Y_{1} & Y_{2} & X_{1} & −X_{2} & 0 & 0 \end{matrix} (5.5)

and {X_{1}, X_{2}, Y_{1}, Y_{2}, W } is an orthogonal basis, with < X_{1}, X_{1} >=< Y_{1}, Y_{1} >= < W, W >= 1 and < X_{2}, X_{2} >=< Y_{2}, Y_{2} >= b^{2} ≠ 1.

Therefore, X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a) with respect to the basis {X_{1}, Y_{1}, X_{2}, Y_{2}, W, A} satisfy

\{ \begin{matrix}μwb^{2}y_{2} − ay_{1} = 0, \\ μwy_{1} + ab^{2}y_{2} = 0, \\ μwb^{2}x_{2} + ax_{1} = 0, \\ μwx_{1} − ab^{2}x_{2} = 0, \\ 2μ(x_{1}y_{2} + x_{2}y_{1}) = 0. \end{matrix} (5.6)

It is easy to show that if μ = 0, then any vector of m is the component of a geodesic vector. Hence, we now focus on the case μ ≠ 0. When w ≠ 0, from (5.6) we get x_{1} = x_{2} = y_{1} = y_{2} = 0, while if w = 0, then (5.6) reduces to x_{1}y_{2} + x_{2}y_{1} = 0. Therefore, if μ ≠ 0, X is a geodesic vector of a generalized symmetric space of type 4b if and only if

\begin{matrix}(i) & X_{m} = wW, or \\ (ii) & X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}y_{2} + x_{2}y_{1} = 0. \end{matrix}

We can check easily that X_{1}, X_{2}, Y_{1}, Y_{2} and W are mutually orthogonal geodesic vectors. Thus, five-dimensional generalized symmetric spaces of type 4b admit five linearly independent homogeneous geodesics through the origin o. 2c): λ + λ̄ ≠ 0, ν = 0 and b^{2} = 1.

In this case, \underline{h} = so(2) ⨁ so(2) = span(A_{1}, A_{2}), where A_{1} = A of type 4b, while A_{2} is determined by A_{2}X_{1} = X_{2}, A_{2}X_{2} = −X_{1}, A_{2}Y_{1} = −Y_{2}, AY_{2} = Y_{1}, AW = 0 (see [4]). Hence, computing the Lie bracket, we get

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A_{1} & A_{2} \\ X_{1} & 0 & 0 & 0 & 0 & −ηX_{2} − μY_{2} & Y_{1} & −X_{2} \\ X_{2} & 0 & 0 & 0 & 0 & ηX_{1} − μY_{1} & −Y_{2} & X_{1} \\ Y_{1} & 0 & 0 & 0 & 0 & −μX_{2} + ηY_{2} & −X_{1} & Y_{2} \\ Y_{2} & 0 & 0 & 0 & 0 & −μX_{1} − ηY_{1} & X_{2} & −Y_{1} \\ W & ηX_{2} + μY_{2} & −ηX_{1} + μY_{1} & μX_{2} − ηY_{2} & μX_{1} + ηY_{1} & 0 & 0 & 0 \\ A_{1} & −Y_{1} & Y_{2} & X_{1} & −X_{2} & 0 & − & − \\ A_{2} & X_{2} & −X_{1} & −Y_{2} & Y_{1} & 0 & − & − \end{matrix}

Note that in this case {X_{1}, X_{2}, Y_{1}, Y_{2}, W } is an orthonormal basis of m .

X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a, b) with respect to {X_{1}, Y_{1}, X_{2}, Y_{2}, W, A_{1}, A_{2}} satisfy

\{ \begin{matrix}w(ηx_{2} + μy_{2}) − ay_{1} + bx_{2} = 0, \\ w(ηx_{1} − μy_{1}) − ay_{2} + bx_{1} = 0, \\ w(μx_{2} − ηy_{2}) + ax_{1} − by_{2} = 0, \\ w(μx_{1} + ηy_{1}) − ax_{2} + by_{1} = 0, \\ 2μ(x_{1}y_{2} + y_{1}x_{2}) = 0. \end{matrix} (5.7)

When w ≠ 0, (5.7) gives x_{1} = x_{2} = y_{1} = y_{2} = 0. For w = 0, (5.7) reduces to its last equation x_{1}y_{2} + y_{1}x_{2} = 0. Thus, if X is a geodesic vector of a generalized symmetric space of type 4c, then its m-component is

\begin{matrix}(i) & X_{m} = wW, or \\ (ii) & X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}y_{2} + x_{2}y_{1} = 0. \end{matrix}

Note that {X_{1}, X_{2}, Y_{1}, Y_{2}, W } is an orthonormal basis of m, where X_{1}, X_{2}, Y_{1}, Y_{2} are of type (i) while W is of type (ii). Therefore, from Proposition 2.2 it follows that five-dimensional generalized symmetric spaces of type 4c admit five mutually orthogonal homogeneous geodesics through the origin.

6 Homogeneous geodesics of generalized symmetric spaces of type 7

As homogeneous spaces, five-dimensional generalized symmetric spaces M of type 7 are real matrix groups

( \begin{matrix}e^{λt} & 0 & 0 & 0 & x \\ 0 & e^{−λt} & 0 & 0 & y \\ te^{λt} & 0 & e^{λt} & 0 & u \\ 0 & −te^{−λt} & 0 & e^{−λt} & v \\ 0 & 0 & 0 & 0 & 1 \end{matrix} )

M is also IR^{5}(x, y, u, v, t), equipped with a Riemannian metric

g = dt^{2} + e^{−2λt}(tdx − du)^{2} + e^{2λt}(tdy + dv)^{2} + a^{2}(e^{−2λt}dx^{2} + e^{2λt}dy^{2}) +
+2γ(dydu − dxdv),

where λ, a, γ ∈ IR, λ ≥ 0, a > 0 and γ^{2} < a^{2}. 2a): λ ≠ 0.

In this case, \underline{h} = 0 [4]. Following [4], there exists a basis \{X_{1}, X_{2}, Y_{1}, Y_{2}, W \} of \underline{g} such that

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 0 & 0 & 0 & 0 & −λX_{1} − X_{2} \\ X_{2} & 0 & 0 & 0 & 0 & −λX_{2} \\ Y_{1} & 0 & 0 & 0 & 0 & λY_{1} + Y_{2} \\ Y_{2} & 0 & 0 & 0 & 0 & λY_{2} \\ W & λX_{1} + X_{2} & λX_{2} & −λY_{1} − Y_{2} & −λY_{2} & 0 \end{matrix} (6.1)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & a^{2} & 0 & 0 & −γ & 0 \\ X_{2} & 0 & 1 & γ & 0 & 0 \\ Y_{1} & 0 & γ & a^{2} & 0 & 0 \\ Y_{2} & −γ & 0 & 0 & 1 & 0 \\ W & 0 & 0 & 0 & 0 & 1 \end{matrix} (6.2)

We now use (6.1) and (6.2) to compute (2.4). Taking into account that λ ≠ 0, we get

\{ \begin{matrix}a^{2}x_{1} − γy_{2} = 0, \\ x_{2} + γy_{1} = 0, \\ γx_{2} + a^{2}y_{1} = 0, \\ y_{2} − γx_{1} = 0, \end{matrix} (6.3)

which only admits the solutions (0, 0, 0, 0, w). So, X is a geodesic vector of a generalized symmetric space of type 7a if and only if X is parallel to W. In other words, geodesic vectors of a five-dimensional generalized symmetric space of type 7a form a straight line. As a consequence, we clearly have the following Theorem 6.1 Five-dimensional generalized symmetric spaces of type 7a only admit one homogeneous geodesic through the origin. 2b): λ = 0.

In this case, \underline{h} = so(2) = span(A), where A is determined by AX_{1} = −Y_{1}, AX_{2} = Y_{2}, AY_{1} = X_{1}, AY_{2} = −X_{2}, AW = 0 (see [4]). Computing [,] and <, >, taking into account that now λ = 0 and γ = 0 [4], we get

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A \\ X_{1} & 0 & 0 & 0 & 0 & −X_{2} & Y_{1} \\ X_{2} & 0 & 0 & 0 & 0 & 0 & −Y_{2} \\ Y_{1} & 0 & 0 & 0 & 0 & Y_{2} & −X_{1} \\ Y_{2} & 0 & 0 & 0 & 0 & 0 & X_{2} \\ W & X_{2} & 0 & −Y_{2} & 0 & 0 & 0 \\ A & −Y_{1} & Y_{2} & X_{1} & −X_{2} & 0 & 0 \end{matrix} (6.4)

and \{X_{1}, X_{2}, Y_{1}, Y_{2}, W \} is an orthogonal basis of \underline{g} , with < X_{1}, X_{1} >=< Y_{1}, Y_{1} >= a^{2} and < X_{2}, X_{2} >=< Y_{2}, Y_{2} >=< W, W >= 1.

Using (6.4) to compute (2.4), it is easy to show that X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a) with respect to the basis {X_{1}, Y_{1}, X_{2}, Y_{2}, W, A} satisfy

\{ \begin{matrix}ay_{1} − wx_{2} = 0, \\ ay_{2} = 0, \\ ax_{1} − wy_{2} = 0, \\ ax_{2} = 0, \\ x_{1}x_{2} − y_{1}y_{2} = 0. \end{matrix} (6.5)

For a ≠ 0, (6.5) gives x_{1} = x_{2} = y_{1} = y_{2} = 0. For a = 0, we have either w = 0 and x_{1}x_{2} − y_{1}y_{2} = 0, or w ≠ 0 and x_{2} = y_{2} = 0. Therefore, if X is a geodesic vector of a generalized symmetric space of type 7b, then its m-component is given by

\begin{matrix}(i) & X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}x_{2} − y_{1}y_{2} = 0, or \\ (ii) & X_{m} = x_{1}X_{1} + y_{1}Y_{1} + wW. \end{matrix}

Finally, since all the vectors of the orthogonal basis {X_{1}, X_{2}, Y_{1}, Y_{2}, W } are of type (i) or (ii), we can conclude that five-dimensional generalized symmetric spaces of type 7b admit five mutually orthogonal homogeneous geodesics through the origin.

7 Homogeneous geodesics of generalized symmetric spaces of type 8

As homogeneous spaces, five-dimensional generalized symmetric spaces M of type 8 are I^{e}(IR^{3})/SO(2) or I^{h}(IR^{3})/SO(2), where I^{e} (respectively, I^{h}) denotes the group of all positive affine transformations of IR^{3} that preserve dx^{2} + dy^{2} + dz^{2} (respectively, dx^{2} + dy^{2} − dz^{2}). M is also described as submanifold of IR^{6}(x, y, z, α, β, γ), such that α^{2} + β^{2} ± γ^{2} = ±1. The Riemannian metric of M is induced by the regular invariant quadratic form

ḡ = dx^{2} + dy^{2} ± dz^{2} + λ^{2}(dα^{2} + dβ^{2} ± dγ^{2}) + [μ ± (−1)](αdx + βdy ± γdz)^{2},

where λ, μ > 0 [4]. The five-dimensional generalized symmetric spaces of type 8a (respectively, 8b) are obtained when we have the sign ” + ” (respectively, ” − ”) in the previous formulas. Here we analyze the case 8a, the case 8b can be treated similarly and it leads to the same conclusions. In both cases, \underline{h} = so(2) = span(A), where A is determined by AX_{1} = −Y_{1}, AX_{2} = Y_{2}, AY_{1} = X_{1}, AY_{2} = −X_{2}, AW = 0, with {X_{1}, X_{2}, Y_{1}, Y_{2}, W } a basis of m . The Lie bracket on M is determined by

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W & A \\ X_{1} & 0 & W & 0 & 0 & −X_{2} & Y_{1} \\ X_{2} & −W & 0 & 0 & 0 & 0 & −Y_{2} \\ Y_{1} & 0 & 0 & 0 & −W & Y_{2} & −X_{1} \\ Y_{2} & 0 & 0 & W & 0 & 0 & X_{2} \\ W & X_{2} & 0 & −Y_{2} & 0 & 0 & 0 \\ A & −Y_{1} & Y_{2} & X_{1} & −X_{2} & 0 & 0 \end{matrix} (7.1)

Moreover, {X_{1}, X_{2}, Y_{1}, Y_{2}, W } is an orthogonal basis of m , with < X_{1}, X_{1} >= < Y_{1}, Y_{1} >= b^{2}, < X_{2}, X_{2} >=< Y_{2}, Y_{2} >= 1 and < W, W >= c^{2}, where b, c > 0 [4]. Using (7.1) to compute (2.4), we obtain that X ∈ \underline{g} is geodesic if and only if its components (x_{1}, x_{2}, y_{1}, y_{2}, w, a) with respect to the basis {X_{1}, Y_{1}, X_{2}, Y_{2}, W, A} satisfy

\{ \begin{matrix}x_{2}w(1 − c^{2}) − y_{1}ab^{2} = 0, \\ x_{1}wc^{2} + ay_{2} = 0, \\ −y_{2}w(1 − c^{2}) + x_{1}ab^{2} = 0, \\ y_{1}wc^{2} + ax_{2} = 0, \\ x_{1}x_{2} − b^{2}y_{1}y_{2} = 0. \end{matrix} (7.2)

For a = w = 0, (7.2) reduces to x_{1}x_{2} − b^{2}y_{1}y_{2} = 0. If a = 0 ≠ w, we get x_{1} = x_{2} = y_{1} = y_{2} = 0 when c^{2} ≠ 1, while if c^{2} = 1 we only have x_{1} = y_{1} = 0. If a ≠ 0, we must distinguish different cases. We eventually obtain:

1) If c^{2} − 1 < 0, then x_{1} = x_{2} = y_{1} = y_{2} = 0.

2) If c^{2} − 1 > 0 and b^{2} = 1, then, in addition to x_{1} = x_{2} = y_{1} = y_{2} = 0, we also have

the solutions x_{2} = −\frac{wc^{2}}{a} y_{1}, y_{2} = −\frac{wc^{2}}{a} x_{1} and w = ±\frac{ab}{c\sqrt{c^{2} − 1}}.

3) If c^{2} − 1 > 0 and b^{2} ≠ 1, the solutions are either x_{1} = x_{2} = y_{1} = y_{2} = 0, or

x_{2} = y_{1} = 0, y_{2} = −\frac{wc^{2}}{a} x_{1} and w = ±\frac{ab}{c\sqrt{c^{2} − 1}}, or x_{1} = y_{2} = 0, x_{2} = −\frac{wc^{2}}{a} y_{1} and

w = ±\frac{ab}{c\sqrt{c^{2} − 1}}.

In this way, we proved that the if X is a geodesic vectors of a five-dimensional generalized symmetric space of type 8a, then its m-component is:

(i) X_{m} = wW, or
(ii) X_{m} = x_{1}X_{1} + x_{2}X_{2} + y_{1}Y_{1} + y_{2}Y_{2} and x_{1}x_{2} − b^{2}y_{1}y_{2} = 0, or
(iii) X_{m} = x_{2}X_{2} + y_{2}Y_{2} + wW (only when c^{2} = 1), or
(iv) X_{m} = x_{1}X_{1} − \frac{wc^{2}}{a} y_{1}X_{2} + y_{1}Y_{1} − \frac{wc^{2}}{a} x_{1}Y_{2} ± \frac{ab}{c\sqrt{c^{2} − 1}}W (only when c^{2} > 1

and b^{2} = 1), or

(v) X_{m} = x_{1}X_{1} − \frac{wc^{2}}{a} x_{1}Y_{2} ± \frac{ab}{c\sqrt{c^{2} − 1}}W and

X_{m} = −\frac{wc^{2}}{a} y_{1}X_{2} + y_{1}Y_{1} ± \frac{ab}{c\sqrt{c^{2} − 1}}W (only when c^{2} > 1 and b^{2} ≠ 1).

The calculations for spaces of type 8b are similar. Since all the vectors of the orthogonal basis {X_{1}, X_{2}, Y_{1}, Y_{2}, W } are geodesic vectors of type (i) or (ii), which exist for all values of b and c, we can conclude that five-dimensional generalized symmetric spaces of type 8 admit five mutually orthogonal homogeneous geodesics through the origin.

8 Homogeneous geodesics of generalized symmetric spaces of order 6 (type 9)

Five-dimensional generalized symmetric spaces of order 6 can be described in the following way. The underlying manifold is IR^{5}(x, y, z, u, v), equipped with the Riemannian metric

(8.1) g = \frac{2}{3}a^{2}(du^{2} + dudv + dv^{2}) + (2b^{2} + 1)(e^{2(u+v)}dx^{2} + e^{−2u}dy^{2} +
+e^{−2v}dz^{2}) + 2(b^{2} − 1)(e^{v}dxdy + e^{u}dxdz − e^{−(u+v)}dydz),

where a > 0 and b > 0 are real numbers. The space (IR^{5}, g) can be identified with the homogeneous space G/H, where G is the group of all matrices of the form

( \begin{matrix}e^{−(u+v)} & 0 & 0 & x \\ 0 & e^{u} & 0 & y \\ 0 & 0 & e^{v} & y \\ 0 & 0 & 0 & 1 \end{matrix} )

Generalized symmetric spaces of type 9 have a special interest because they are of solvable type, that is, G is a solvable Lie group. In a forthcoming paper [1], the authors and O. Kowalski will study homogeneous geodesics in some examples of generalized symmetric spaces of solvable type of arbitrary odd dimension.

Following [4], the Lie algebra \underline{g} of G admits a basis \{X_{1}, X_{2}, Y_{1}, Y_{2}, W \} such that

\begin{matrix}[ , ] & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & 0 & −X_{2} & 0 & 0 & W \\ X_{2} & X_{2} & 0 & X_{2} & 0 & 0 \\ Y_{1} & 0 & −X_{2} & 0 & Y_{2} & 0 \\ Y_{2} & 0 & 0 & −Y_{2} & 0 & 0 \\ W & −W & 0 & 0 & 0 & 0 \end{matrix} (8.2)

and

\begin{matrix}< , > & X_{1} & X_{2} & Y_{1} & Y_{2} & W \\ X_{1} & \frac{2}{3}a^{2} & 0 & \frac{1}{3}a^{2} & 0 & 0 \\ X_{2} & 0 & 2b^{2} + 1 & 0 & b^{2} − 1 & b^{2} − 1 \\ Y_{1} & \frac{1}{3}a^{2} & 0 & \frac{2}{3}a^{2} & 0 & 0 \\ Y_{2} & 0 & b^{2} − 1 & 0 & 2b^{2} + 1 & −(b^{2} − 1) \\ W & 0 & b^{2} − 1 & 0 & −(b^{2} − 1) & 2b^{2} + 1 \end{matrix} (8.3)

Since \underline{h} = 0 [4], each geodesic vector must be an element of \underline{g}. We use (8.2) and (8.3) to compute (2.4). Putting

h = \frac{b^{2} − 1}{2b^{2} + 1},

we get easily that X ∈ \underline{g} is geodesic if and only if its components satisfy

\{ \begin{matrix}(x_{2} + w)(x_{2} + hy_{2} − w) = 0, \\ (x_{1} + y_{1})(x_{2} + hy_{2} + hw) = 0, \\ (x_{2} + y_{2})(x_{2} − y_{2} + hw) = 0, \\ y_{1}(hx_{2} + y_{2} − hw) = 0, \\ x_{1}(hx_{2} − hy_{2} + w) = 0. \end{matrix} (8.4)

By the definition of h, it follows easily that h − 1 ≠ 0, h + 1 ≠ 0 and 2h − 1 ≠ 0, for all b > 0. We now find all the solutions of (8.4).

a) If x_{1} = 0, (8.4) reduces to

\{ \begin{matrix}(x_{2} + w)(x_{2} + hy_{2} − w) = 0, \\ y_{1}(x_{2} + hy_{2} + hw) = 0, \\ (x_{2} + y_{2})(x_{2} − y_{2} + hw) = 0, \\ y_{1}(hx_{2} + y_{2} − hw) = 0. \end{matrix} (8.5)

Adding the second and the fourth equation of (8.5) and taking into account that h + 1 ≠ 0, we get that either y_{1} = 0 or x_{2} + y_{2} = 0.

If also y_{1} = 0, then (8.5) reduces to

\{ \begin{matrix}(x_{2} + w)(x_{2} + hy_{2} − w) = 0, \\ (x_{2} + y_{2})(x_{2} − y_{2} + hw) = 0, \end{matrix}

whose solutions are (0, −w, 0, w, w), (0, −w, 0, (h − 1)w, w), (0, −y_{2}, 0, y_{2}, (h − 1)y_{2}), (0, (1 − h)w, 0, w, w).

When y_{1} ≠ 0, the only solutions are (0, 0, y_{1}, 0, 0).

b) If x_{1} ≠ 0, from the last equation of (8.4) we get hx_{2} − hy_{2} + w = 0.

If y_{1} = 0, the case is similar to the case x_{1} = 0, y_{1} ≠ 0. Proceeding in the same way, we get the solutions (x_{1}, 0, 0, 0, 0).

If y_{1} ≠ 0, (8.4) gives

\{ \begin{matrix}hx_{2} − hy_{2} + w = 0, \\ hx_{2} + y_{2} − hw) = 0, \\ (x_{2} + w)(x_{2} + hy_{2} − w) = 0, \\ (x_{1} + y_{1})(x_{2} + hy_{2} + hw) = 0, \\ (x_{2} + y_{2})(x_{2} − y_{2} + hw) = 0. \end{matrix} (8.6)

whose solutions are (x_{1}, 0, y_{1}, 0, 0).

So, we proved the following Proposition 8.1 X is a geodesic vector of a five-dimensional generalized symmetric space of type 9 if and only if

(i) X = −wX_{2} + wY_{2} + wW, or
(ii) X = −wX_{2} + (h − 1)wY_{2} + wW, or
(iii) X = −y_{2}X_{2} + y_{2}Y_{2} + (h − 1)y_{2}W, or
(iv) X = (1 − h)wX_{2} + wY_{2} + wW, or
(v) X = x_{1}X_{1} + y_{1}Y_{1}.

Note that two geodesic vectors of two distinct types among (i), (ii), (iii) and (iv) are always distinct, since h − 1 ≠ 1.

If we add to {X_{1}, Y_{1}} a triplet of vectors chosen in {V_{1} = −X_{2} + Y_{2} + W, V_{2} = −X_{2} + (h − 1)Y_{2} + W, V_{3} = −X_{2} + Y_{2} + (h − 1)W, V_{4} = −(h − 1)X_{2} + Y_{2} + W }, we then always get five linearly independent geodesic vectors, taking into account that h − 2 ≠ 0. Hence, there exist five linearly independent geodesic vectors in M.

About the orthogonality, it is easy to prove that a geodesic vector of type (v) is orthogonal to all geodesic vectors of type (i), (ii), (iii) or (iv). Moreover, for example X_{1} and V = X_{1} − 2Y_{1} are two orthogonal geodesic vectors of type (v). Finally, two geodesic vectors chosen in two different types among (i), (ii), (iii) and (iv) are never mutually orthogonal. In fact, < V_{1}, V_{2} >=< V_{1}, V_{3} >= < V_{1}, V_{4} >= 3(h + 1) ≠ 0, while < V_{2}, V_{3} >=< V_{2}, V_{4} >=< V_{3}, V_{4} >= (2b^{2} +1)(2h−1)−(b^{2} −1)(h^{2} +2) ≠ 0. So, we can conclude that there are at most three geodesic vectors mutually orthogonal, for example taking X_{1}, V = X_{1} − 2Y_{1} and V_{1}. So, from Proposition 8.1 it follows Corollary 8.2 Geodesic vectors of five-dimensional generalized symmetric spaces of type 9 form

(a) a plane (2-dimensional vector subspace) P of \underline{g} (type (v)), and

(b) four straight lines, respectively generated by the geodesic vectors −X_{2}+Y_{2}+W, −X_{2} + (h − 1)Y_{2} + W, −X_{2} + Y_{2} + (h − 1)W and (1 − h)X_{2} + Y_{2} + W (types (i), (ii), (iii) and (iv), respectively), in the orthogonal complement of P.

Finally, we can conclude with the following result. Theorem 8.3 Five-dimensional generalized symmetric space of type 9 admit five linearly independent homogeneous geodesics through the origin o, but never five orthogonal ones. There are at most three mutually orthogonal homogeneous geodesics through o. Acknowledgments. The authors wish to express their gratitude towards Dr. E. Boeckx for his helpful hints and comments during the preparation of this paper.

Supported by funds of the University of Lecce and the M.U.R.S.T.

References

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Universitá degli Studi di Lecce, Dipartimento di Matematica ”E. de Giorgi”, Via Provinciale Lecce-Arnesano, 73100 Lecce, Italy. e-mail:giovanni.calvaruso@unile.it; rosanna@ilenic.unile.it