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DER-CHEN CHANG, IRINA MARKINA, AND ALEXANDER VASIL’EV

Abstract. Sub-Riemannian Geometry is proved to play an important role in many applications, e.g., Mathematical Physics and Control Theory. The simplest example of sub-Riemannian structure is provided by the 3-D Heisenberg group. Sub-Riemannian Geometry enjoys major differences from the Riemannian being a generalisation of the latter at the same time, e.g., geodesics are not unique and may be singular, the Haus-dorff dimension is larger than the manifold topological dimension. There exists a large amount of literature developing sub-Riemannian Geometry. However, very few is known about its natural extension to pseudo-Riemannian analogues. It is natural to begin such a study with some low-dimensional manifolds. Based on ideas from sub-Riemannian geometry we develop sub-Lorentzian geometry over the classical 3-D anti-de Sitter space. Two different distributions of the tangent bundle of anti-de Sitter space yield two different geometries: sub-Lorentzian and sub-Riemannian. It is shown that the set of timelike and spacelike ‘horizontal’ curves is non-empty and we study the problem of horizontal connectivity in anti-de Sitter space. We also use Lagrangian and Hamiltonian formalisms for both sub-Lorentzian and sub-Riemannian geometries to find geodesics.

1. Introduction

Many interesting studies of anticommutative algebras and sub-Riemannian structures may be seen in a general setup of Clifford algebras and spin groups. Among others we dis-tinguish the following example. The unit 3-dimensional sphere S3 being embedded into

the Euclidean space R4 possesses a clear manifold structure with the Riemannian

met-ric. It is interesting to consider the sphere S3 as an algebraic object S3 = SO(4)/ SO(3)

where the group SO(4) preserves the global Euclidean metric of the ambient space R4

and SO(3) preserves the Riemannian metric on S3. The quotient SO(4)/ SO(3) can

be realised as the group SU(2) acting on S3 as on the space of complex vectors z 1, z2

of unit norm |z1|2 + |z2|2 = 1. It is isomorphic to the group of unit quaternions with

the group operation given by the quaternion multiplication. It is natural to make the correspondence between S3 as a smooth manifold and S3 as a Lie group acting on this

manifold. The corresponding Lie algebra is given by left-invariant vector fields with non-vanishing commutators. This leads to construction of a sub-Riemannian structure

2000 Mathematics Subject Classification. Primary: 53C50, 53C27; Secondary: 83C65.

Key words and phrases. Sub-Riemannian and sub-Lorentzian geometries, geodesic, anti-de Sitter space, Hamiltonian system, Lagrangian, spin group, spinors.

The first author has been supported by a research grant from the United States Army Research Office and by a competitive research grant of the Georgetown University.

The second and the third authors have been supported by the grant of the Norwegian Research Council # 177355/V30 and by the European Science Foundation Networking Programme HCAA.

All authors were partially supported by the grant of the Norwegian Research Council #180275/D15. 1

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on S3, see [4] (more about sub-Riemannian geometry see, for instance, [11, 19, 20, 21]).

The commutation relations for vector fields on the tangent bundle of S3 come from the

non-commutative multiplication for quaternions. Unit quaternions, acting by conjuga-tion on vectors from R3 (and R4), define rotation in R3 (and R4), thus preserving the

positive-definite metric in R4. At the same time, the Clifford algebra over the vector

space R3with the standard Euclidean metric gives rise to the spin group Spin(3) = SU(2)

that acts on the group of unit spinors in the same fashion leaving some positive-definite quadratic form invariant. Two models are equivalent but the latter admits various gen-eralisations. We are primary aimed at switching the Euclidean world to the Lorentzian one and sub-Riemannian geometry to sub-Lorentzian following a simple example similar to the above of a low-dimensional space that leads us to sub-Lorentzian geometry over the pseudohyperbolic space H1,2 in R2,2. In General Relativity the simply connected

covering manifold of H1,2 is called the universal anti-de Sitter (AdS) space [15, 16, 22].

We start with some more rigorous explanations. A real Clifford algebra is associated with a vector space V equipped with a quadratic form Q(·, ·). The multiplication (let us denote it by ⊗) in the Clifford algebra satisfies the relation

v ⊗ v = −Q(v, v)1,

for v ∈ V , where 1 is the unit element of the algebra. We restrict ourselves to V = R3

with two different quadratic forms:

QE(v, v) = Ev · v, E =   1 0 00 1 0 0 0 1   , and Q(v, v) = Iv · v, I =   −1 0 00 1 0 0 0 1   .

The first case represents the standart inner product in the Euclidean space R3. The

second case corresponds to the Lorentzian metric in R3 given by the diagonal metric

tensor with the signature (−, +, +). The corresponding Clifford algebras we denote by Cl(0, 3) = Cl(3) and Cl(1, 2). The basis of the Clifford algebra Cl(3) consists of the elements

{1, i1, i2, i3, i1 ⊗ i2, i1⊗ i3, i2⊗ i3, i1⊗ i2⊗ i3}, with i1⊗ i1 = i2 ⊗ i2 = i3⊗ i3 = −1.

The algebra Cl(3) can be associated with the space H × H, where H is the quaternion algebra. The basis of the Clifford algebra Cl(1, 2) is formed by

{1, e, i1, i2, e ⊗ i1, e ⊗ i2, i1⊗ i2, e ⊗ i1⊗ i2}, with e ⊗ e = 1, i1⊗ i1 = i2 ⊗ i2 = −1.

In this case the algebra is represented by 2 × 2 complex matrices.

Spin groups are generated by quadratic elements of Clifford algebras. We obtain the spin group Spin(3) in the case of the Clifford algebra Cl(3), and the group Spin(1, 2) in the case of the Clifford algebra Cl(1, 2). The group Spin(3) is represented by the group SU(2) of unitary 2 × 2 complex matrices with determinant 1. The elements of SU(2)

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can be written as 

a b −¯b ¯a



, a, b ∈ C, |a|2+ |b|2 = 1.

The group Spin(3) = SU(2) forms a double cover of the group of rotations SO(3). In this case the Euclidean metric in R3 is preserved under the actions of the group

SO(3). The group Spin(3) = SU(2) acts on spinors similarly to how SO(3) acts on vectors from R3. Indeed, given an element R ∈ SO(3) the rotation is performed by

the matrix multiplication RvR−1, where v ∈ R3. An element U ∈ SU(2) acts over

spinors regarded as 2 component vectors z = (z1, z2) with complex entries in the same

way UzU−1. This operation defines a ‘half-rotation’ and preserves the positive-definite

metric for spinors. Restricting ourselves to spinors of length 1, we get the manifold {(z1, z2) ∈ C2 : |z1|2+ |z2|2 = 1} which is the unit sphere S3.

Now we turn to the Lorentzian metric and to the Clifford algebra Cl(1, 2). The spin group Spin+(1, 2) is represented by the group SU+(1, 1) whose elements are

 a b ¯b ¯a



, a, b ∈ C, |a|2− |b|2 = 1.

The group Spin+(1, 2) = SU+(1, 1) forms a double cover of the group of Lorentzian rotations SO(1, 2) preserving the Lorentzian metric Q(v, v). Acting on spinors, the group Spin+(1, 2) = SU+(1, 1) preserves the pseudo-Riemannian metric for spinors. Unit spinors (z1, z2), |z1|2 − |z2|2 = 1, are invariant under the actions of the corresponding

group Spin+(1, 2) = SU+(1, 1). The manifold H1,2 = {(z

1, z2) ∈ C2 : |z1|2 − |z2|2 = 1}

is a 3-dimensional Lorentzian manifold known as a pseudohyperbolic space in Geometry and as the anti-de Sitter space AdS3in General Relativity. In fact, AdSnis the maximally

symmetric, simply connected, Lorentzian manifold of constant negative curvature. It is one of three maximally symmetric cosmological constant solutions to Einstein’s field equation: de Sitter space with a positive cosmological constant Λ, anti-de Sitter space with a negative cosmological constant −Λ, and the flat space. Both de Sitter dS3

and anti-de Sitter AdS3 spaces may be treated as non-compact hypersurfaces in the

corresponding pseudo-Euclidean spaces R1,3 and R2,2. Sometimes de Sitter space dS 3 or

the hypersphere is used as a direct analogue to the sphere S3 given its positive curvature.

However, AdS3 geometrically is a natural object for us to work with. We reveal the

analogy between S3 and AdS

3 as follows. The group of rotations SO(4) in the usual

Euclidean 4-dimensional space acts as translations on the Euclidean sphere S3 leaving it

invariant. As it has been mentioned at the beginning, the sphere S3 can be thought of as

the Lie group S3 = SO(4)/ SO(3) endowed with the group law given by the multiplication

of matrices from SU(2) which is the multiplication law for unit quaternions. The Lie algebra is identified with the left-invariant vector fields from the tangent space at the unity. The tangent bundle admits the natural sub-Riemannian structure and S3 can be

considered as a sub-Riemannian manifold. This geometric object was studied in details in [4]. It appears throughout celestial mechanics in works of Feynman and Vernon who described it in the language of two-level systems, in Berry’s phase in quantum mechanics or in the Kustaaheimo-Stifel transformation for regularizing binary collision.

Instead of R4, we consider now the space

R2,2 = { (x1, x2, x3, x4) ∈ R4 with a pseudo-metric dx2 = −dx2

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The group SO(2, 2) acting on R2,2 is a direct analog of the rotation group SO(4) acting

on R4. We consider AdS

3 as a manifold H1,2 = SO(2, 2)/ SO(1, 2) with the Lorentzian

metric induced from R2,2. The group SO(2, 2) acts as translations on H1,2. We define

the group law on H1,2 by the multiplication of elements from SU+(1, 1). Under this

rule the manifold H1,2 can be considered as a Lie group. The reader can find more

information about the group actions and relation to General Relativity, e. g. [12, 17]. Left-invariant vector fields on the tangent bundle are not commutative and this gives us an opportunity to consider an analogue of sub-Riemannian geometry, that is called sub-Lorentzian geometry on H1,2. The geometry of anti-de Sitter space was studied in

numerous works, see, for example, [1, 5, 10, 13, 18].

Very few is known about extension of sub-Riemannian geometry to its pseudo-Rieman-nian analogues. The simplest example of a sub-Riemanpseudo-Rieman-nian structure is provided by the 3-D Heisenberg group. Let us mention that recently Grochowski studied its sub-Lorenzian analogue [7, 8]. Our approach deals with non-nilpotent groups over S3 and

AdS3.

The paper is organised in the following way. In Section 2 we give the precise form of left-invariant vector fields defining sub-Lorentzian and sub-Riemannian structures on anti-de Sitter space. In Sections 3 and 4 the question of existence of smooth horizontal curves in the sub-Lorentzian manifold is studied. The Lagrangian and Hamiltonian formalisms are applied to find sub-Lorentzian geodesics in Sections 5 and 6. Section 7 is devoted to the study of a sub-Riemannian geometry defined by the distribution generated by spacelike vector fields of anti-de Sitter space. In both sub-Lorentzian and sub-Riemannian cases we find geodesics explicitly.

2. Left-invariant vector fields

We consider the space AdS3 as a 3-dimensional manifold H1,2 in R2,2

H1,2= {(x1, x2, x3, x4) ∈ R2,2 : −x21− x22+ x23+ x24 = −1},

and the group law is given by the multiplication of the matrices from SU+(1, 1). We write a = x1 + ix2, b = x3 + ix4, where i is the complex unity. For each matrix

 a b ¯b ¯a



∈ SU+(1, 1) we associate its coordinates to the complex vector p = (a, b). Then the multiplication law between p = (a, b) and q = (c, d) written in coordinates is (2.1) pq = (a, b)(c, d) = (ac + b ¯d, ad + b¯c).

The manifold H1,2 with the multiplication law (2.1) is the Lie group with the unity

(1, 0), with the inverse to p = (a, b) element p−1

= (¯a, −b), and with the left translation Lp(q) = pq. The Lie algebra is associated with the left-invariant vector fields at the

identity of the group. To calculate the real left-invariant vector fields, we write the multiplication law (2.1) in real coordinates, setting c = y1+ iy2, d = y3+ iy4. Then

pq = (x1, x2, x3, x4)(y1, y2, y3, y4)

= (x1y1 − x2y2+ x3y3+ x4y4, x2y1+ x1y2+ x4y3− x3y4,

x3y1+ x4y2+ x1y3− x2y4, x4y1− x3y2+ x2y3+ x1y4).

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The tangent map (Lp)∗ corresponding to the left translation Lp(q) is (Lp)∗ =     x1 −x2 x3 x4 x2 x1 x4 −x3 x3 x4 x1 −x2 x4 −x3 x2 x1     .

The left-invariant vector fields are the left translations of vectors at the unity by the tangent map (Lp)∗: eX = (Lp)∗X(0). Letting X(0) be the vectors of the standard basis

in R2,2 (that coincides with the Euclidean basis in R4), we get the left-invariant vector

fields e X1 = x1∂x1 + x2∂x2 + x3∂x3 + x4∂x4, e X2 = −x2∂x1 + x1∂x2 + x4∂x3 − x3∂x4, e X3 = x3∂x1 + x4∂x2 + x1∂x3 + x2∂x4, e X4 = x4∂x1 − x3∂x2 − x2∂x3 + x1∂x4

in the basis ∂x1, ∂x2, ∂x3, ∂x4. Let us introduce the matrices

U =     1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1     , J =     0 1 0 0 −1 0 0 0 0 0 0 −1 0 0 1 0     , E1 =     0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0     , E2 =     0 0 0 1 0 0 −1 0 0 −1 0 0 1 0 0 0     .

Then the left-invariant vector fields can be written in the form e

X1 = xU · ∇x, Xe2 = xJ · ∇x, Xe3 = xE1· ∇x, Xe4 = xE2· ∇x,

where x = (x1, x2, x3, x4), ∇x = (∂x1, ∂x2, ∂x3, ∂x4) and ”·” is the dot-product in R

4. The

matrices possess the following properties:

• Anti-commutative rule or the Clifford algebra condition:

(2.3) JE1 + E1J = 0, E2E1+ E1E2 = 0, JE2+ E2J = 0. • Non-commutative rule: (2.4) [1 2J, 1 2E1] = 1 4(JE1− E1J) = 1 2E2, [ 1 2E2, 1 2E1] = 1 2J, [ 1 2J, 1 2E2] = − 1 2E1. • Transpose matrices: (2.5) JT = −J, E2T = E2, E1T = E1. • Square of matrices: (2.6) J2 = −U, E22 = U, E12 = U. As a consequence we obtain

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• Product of matrices:

(2.7) JE1 = E2, E2E1 = J, JE2 = −E1.

The inner h·, ·i product in R2,2 is given by

(2.8) hx, yi = Ix · y, with I =     −1 0 0 0 0 −1 0 0 0 0 1 0 0 0 0 1     .

Given the inner product (2.8) we have

(2.9) hx, xE1i = hx, xJi = hx, xE2i = 0,

(2.10) hxJ, xE1i = hxE2, xE1i = hxJ, xE2i = 0,

(2.11) hxJ, xJi = −1, hxE2, xE2i = hxE1, xE1i = 1.

The vector field eX1 is orthogonal to H1,2. Indeed, if we write H1,2 as a hypersurface

F (x1, x2, x3, x4) = −x12− x22+ x23+ x24+ 1 = 0, then dF (c(s)) ds = 2  − x1 dx1 ds − x2 dx2 ds + x3 dx3 ds + x4 dx4 ds  = h eX1, dc(s) ds i = 0

for any smooth curve c(s) = (x1(s), x2(s), x3(s), x4(s)) on H1,2. From now on we denote

the vector field eX1 by N. Observe, that |N|2 = hN, Ni = −1. Up to certain ambiguity

we use the same notation | · | as the norm (whose square is not necessary positive) of a vector and as the absolute value (non-negative) of a real/complex number. Other vector fields are orthogonal to N with respect to the inner product h·, ·i in R2,2:

hN, eX2i = hN, eX3i = hN, eX4i = 0.

We conclude that the vector fields eX2, eX3, eX4 are tangent to H1,2. Moreover, they are

mutually orthogonal with

| eX2|2 = h eX2, eX2i = −1, | eX3|2 = | eX4|2 = 1.

We denote the vector field eX2 by T providing time orientation (for the terminology see

the end of the present section). The spacelike vector fields eX3 and eX4 will be denoted

by X and Y respectively. We conclude that T, X, Y is the basis of the tangent bundle of H1,2. In Table 1 the commutative relations between T, X, and Y are presented.

We see that if we fix two of the vector fields, then they generate, together with their Table 1. Commutators of left-invariant vector fields

T X Y

T 0 2Y −2X

X −2Y 0 −2T

Y 2X 2T 0

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Definition 1. Let M be a smooth n-dimensional manifold, D be a smooth k-dimensional, k < n, bracket generating distribution on T M, and h·, ·iDbe a smooth Lorentzian metric

on D. Then the triple (M, D, h·, ·iD) is called the sub-Lorentzian manifold.

We deal with two following cases in Sections 3–6 and Section 7 respectively:

1. The horizontal distribution D is generated by the vector fields T and X: D = span{T, X}. In this case T provides the time orientation and X gives the spatial direction. The direction Y = 1

2[T, X], orthogonal to the distribution D, is the

second spatial direction. The metric h·, ·iDis given by the restriction of h·, ·i from

R2,2. This case corresponds to the sub-Lorentzian manifold (H1,2, D, h·, ·iD). 2. The horizontal distribution D is generated by the vector fields X and Y : D =

span{X, Y }. In this case both of the directions are spatial. The direction T = 1

2[Y, X], orthogonal to the distribution D, is time. In this case, the triple

(H1,2, D, h·, ·i

D) is a sub-Riemannian manifold.

The ambient metric with the signature (−, −, +, +) of R2,2 restricted to the tangent

bundle T H1,2of H1,2is the Lorentzian metric with the signature (−, +, +), and therefore,

H1,2 is a Lorentzian manifold. The vector fields T, X, Y form an orthonormal basis of

each tangent space TpH1,2 at p ∈ H1,2. We introduce a time orientation on H1,2. A

vector v ∈ TpH1,2 is said to be timelike if hv, vi < 0, spacelike if hv, vi > 0 or v = 0, and

lightlike if hv, vi = 0 and v 6= 0. By previous consideration we have T as a timelike vector field and X, Y as spacelike vector fields at each p ∈ H1,2. A timelike vector v ∈ T

pH1,2

is said to be future-directed if hv, T i < 0 or past-directed if hv, T i > 0. A smooth curve γ : [0, 1] → H1,2 with γ(0) = p and γ(1) = q is called timelike (spacelike, lightlike) if the tangent vector ˙γ(t) is timelike (spacelike, lightlike) for any t ∈ [0, 1]. If Ωp,q is the

non-empty set of all timelike, future-directed smooth curves γ(t) connecting the points p and q on H1,2, then the distance between p and q is defined as

dist := sup γ∈Ωp,q 1 Z 0 p −h ˙γ(t), ˙γ(t)idt.

A geodesic in any manifold M is a curve γ : [0, 1] → M whose vector field is parallel, or equivalently, geodesics are the curves of acceleration zero. A manifold M is called geodesically connected if, given two points p, q ∈ M, there is a geodesic curve γ(t) connecting them. Anti-de Sitter space H1,2 is not geodesically connected, see [9, 14].

The concept of causality is important in the study of Lorentz manifolds. We say that p ∈ M chronologically (causally) precedes q ∈ M if there is a timelike (non-spacelike) future-directed (if non-zero) curve starting at p and ending at q. For each p ∈ M we define the chronological future of p as

I+(p) = {q ∈ M : p chronologically precedes q}, and the causal future of p as

J+(p) = {q ∈ M : p causally precedes q}.

The conformal infinity due to Penrose is timelike. One can make analogous definitions replacing ‘future’ by ‘past’.

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From the mathematical point of view the spacelike curves have the same right to be studied as timelike or lightlike curves. Nevertheless, the timelike curves and lightlike curves possess an additional physical meaning as the following example shows.

Example 1. Interpreting the x1-coordinate of H1,2 as time measured in some inertial

frame (x1 = t), the timelike curves represent motions of particles such that

dx2 dt 2 +dx3 dt 2 < 1.

It is assumed that units have been chosen so that 1 is the maximal allowed velocity for a matter particle (the speed of light). Therefore, timelike curves represents motions of matter particles. Timelike geodesics represent motions with constant speed. In addition, the length τ (γ) = 1 Z 0 p −h ˙γ(t), ˙γ(t)i dt,

of a timelike curve γ : [0, 1] → H1,2 is interpreted as the proper time measured by a

particle between events γ(0) and γ(1).

Lightlike curves represent motions at the speed of light and the lightlike geodesics represent motions along the light rays.

3. Horizontal curves with respect to the distribution D = span{T, X} Up to Section 7 we shall work with the horizontal distribution D = span{T, X} and the Lorentzian metric on D, which is the restriction of the metric h·, ·i from R2,2. We

say that an absolutely continuous curve c(s) : [0, 1] → H1,2 is horizontal if the tangent

vector ˙c(s) satisfies the relation ˙c(s) = α(s)T (c(s)) + β(s)X(c(s)).

Lemma 1. A curve c(s) = (x1(s), x2(s), x3(s), x4(s)) is horizontal with respect to the

distribution D = span{T, X}, if and only if,

(3.1) −x4˙x1+ x3˙x2− x2˙x3 + x1˙x4 = 0 or hxE2, ˙ci = 0.

Proof. The tangent vector to the curve c(s) = (x1(s), x2(s), x3(s), x4(s)) written in the

left-invariant basis (T, X, Y ) admits the form

˙c(s) = αT + βX + γY. Then

γ = h ˙c, Y i = I ˙c · Y = −x4˙x1+ x3˙x2− x2˙x3+ x1˙x4 = hxE2, ˙ci.

We conclude that

γ = 0,

if and only if, the condition (3.1) holds. 

In other words, a curve c(s) is horizontal, if and only if, its velocity vector ˙c(s) is orthogonal to the missing direction Y . The left-invariant coordinates α(s) and β(s) of a horizontal curve c(s) = (x1(s), x2(s), x3(s), x4(s)) are

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(3.3) β = h ˙c, Xi = −x3˙x1− x4˙x2+ x1˙x3+ x2˙x4 = hxE1, ˙ci.

Let us write the definition of the horizontal distribution D = span{T, X} using the contact form. We define the form ω = −x4dx1+ x3dx2 − x2dx3 + x1dx4 = hxE2, dxi.

Then,

ω(N) = 0, ω(T ) = 0, ω(X) = 0, ω(Y ) = 1,

and ker ω = span{N, T, Y }, The horizontal distribution can be defined as follows D = {V ∈ T H1,2: ω(V ) = 0}, or D = ker ω ∩ T H1,2,

where T H1,2 is the tangent bundle of H1,2.

The length l(c) of a horizontal curve c(s) : [0, 1] → H1,2 is defined by the following

formula

l(c) = Z 1

0 |h ˙c(s), ˙c(s)i| 1/2ds.

Using the orthonormality of the vector fields T and X, we deduce that l(c) =

Z 1 0

− α2(s) + β2(s) 1/2ds.

We see that the restriction onto the horizontal distribution D ⊂ T H1,2 of the

non-degenerate metric h·, ·i defined on T H1,2 gives the Lorentzian metric which is

non-degenerate. The definitions of timelike (spacelike, lightlike) horizontal vectors v ∈ Dp

are the same as for the vectors v ∈ TpH1,2. A horizontal curve c(s) is timelike (spacelike,

lightlike) if its velocity vector ˙c(s) is horizontal timelike (spacelike, lightlike) vector at each point of this curve.

Lemma 2. Let γ(s) = (y1(s), y2(s), y3(s), y4(s)) be a horizontal timelike future-directed

(or past-directed) curve and c(s) = Lp(γ(s)) be its left translation by p = (p1, p2, p3, p4),

p ∈ H1,2. Then the curve c(s) is horizontal timelike and future-directed (or past-directed).

Proof. Let us denote by (c1(s), c2(s), c3(s), c4(s)) the coordinates of the curve c(s). Then,

by (2.2) we have c1(s) = p1y1(s) − p2y2(s) + p3y3(s) + p4y4(s), c2(s) = p2y1(s) + p1y2(s) + p4y3(s) − p3y4(s), (3.4) c3(s) = p3y1(s) + p4y2(s) + p1y3(s) − p2y4(s), c4(s) = p4y1(s) − p3y2(s) + p2y3(s) + p1y4(s).

Differentiating with respect to s, we calculate the horizontality condition (3.1) for the curve c(s). Since −p2

1− p22+ p23 + p24 = −1, straightforward simplifications lead to the

relation

h ˙c, Y i = −c4˙c1+ c3˙c2− c2˙c3+ c1˙c4 = (−p21− p22+ p23+ p24)(−y4˙y1+ y3˙y2− y2˙y3+ y1˙y4) = 0,

and the curve γ is horizontal.

Let us show that the curve c(s) is timelike and future-directed provided γ(s) is such. We calculate

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and

h ˙c, Xi = −c3˙c1−c4˙c2+c1˙c3+c2˙c4 = (p21+p22−p23−p24)(−y3˙y1−y4˙y2+y1˙y3+y2˙y4) = h ˙γ, Xi

from (3.2), (3.3), and (3.4). Since the horizontal coordinates are not changed, we con-clude that the property timelikeness and future-directness is preserved under the left

translations. 

In view that the left-invariant coordinates of the velocity vector to a horizontal curve do not change under left translations, we conclude the following analogue of the preceding lemma.

Lemma 3. Let γ(s) = (y1(s), y2(s), y3(s), y4(s)) be a horizontal spacelike (or lightlike)

curve and c(s) = Lp(γ(s)) be its left translation by p = (p1, p2, p3, p4), p ∈ H1,2. Then

the curve c(s) is horizontal spacelike (or lightlike).

4. Existence of smooth horizontal curves on H1,2

The question of the connectivity by geodesics of two arbitrary points on a Lorentzian manifold is not trivial, because we have to distinguish timelike and spacelike curves. The problem becomes more difficult if we study connectivity for sub-Lorentzian geom-etry. In the classical Riemannian geometry all geodesics can be found as solutions to the Euler-Lagrange equations and they coincide with the solutions to the correspond-ing Hamiltonian system obtained by the Legandre transform. In the sub-Riemannian geometry, any solution to the Hamilton system is a horizontal curve and satisfies the Euler-Lagrange equations. However, a solution to the Euler-Lagrange equations is a solution to the Hamiltonian system only if it is horizontal.

In the case of sub-Lorentzian geometry we have no information about such a correspon-dence. As it will be shown in Sections 6 and 7 the solutions to the Hamiltonian system are horizontal. It is a rather expectable fact given the corresponding analysis of sub-Riemannian structures, e. g., on nilpotent groups, see [2, 3]. Since {T, X, Y = 1/2[T, X]} span the tangent space at each point of H1,2 the existence of horizontal curves is guar-anteed by Chow’s theorem [6]. So as the first step, in this section we study connectivity by smooth horizontal curves. The main results states that any two points can be con-nected by a smooth horizontal curve. A naturally arisen question is whether the found horizontal curve is timelike (spacelike, lightlike)?

First, we introduce a parametrisation of H1,2 and present the horizontality condition

and the horizontal coordinates in terms of this parametrisation. The manifold H1,2 can be parametrised by

x1 = cos a cosh θ,

x2 = sin a cosh θ,

x3 = cos b sinh θ,

(4.1)

x4 = sin b sinh θ,

with a, b ∈ (−π, +π], θ ∈ (−∞, ∞). Setting ψ = a − b, ϕ = a + b, we formulate the following lemma.

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Lemma 4. Let c(s) = (ϕ(s), ψ(s), θ(s)) be a curve on H1,2. The curve is horizontal, if

and only if,

(4.2) ϕ cos ψ sinh 2θ − 2 ˙θ sin ψ = 0.˙ The horizontal coordinates α and β of the velocity vector are

(4.3) α = −1

2( ˙ϕ cosh 2θ + ˙ψ) = − ˙a cosh

2

θ − ˙b sinh2θ,

(4.4) β = 1

2( ˙ϕ sin ψ sinh 2θ + 2 ˙θ cos ψ). Proof. Using the parametrisation (4.1) of H1,2, we calculate

˙x1 = − ˙a sin a cosh θ + ˙θ cos a sinh θ,

˙x2 = ˙a cos a cosh θ + ˙θ sin a sinh θ,

˙x3 = −˙b sin b sinh θ + ˙θ cos b cosh θ,

(4.5)

˙x4 = ˙b cos b sinh θ + ˙θ sin b cosh θ.

Substituting the expressions for xk and ˙xk, k = 1, 2, 3, 4, in (3.1), (3.2), and (3.3), in

terms of ϕ, ψ and θ, we get the necessary result. 

We also need the following obvious technical lemma formulated without proof. Lemma 5. Given q0, q1, I ∈ R, there is a smooth function q : [0, 1] → R, such that

q(0) = q0, q(1) = q1,

Z 1 0

q(u) du = I.

Theorem 1. Let P and Q be two arbitrary points in H1,2. Then there is a smooth

horizontal curve joining P and Q.

Proof. Let P = P (ϕ0, ψ0, θ0) and Q = Q(ϕ1, ψ1, θ1) be coordinates of the points P

and Q. In order to find a horizontal curve c(s) we must solve equation (4.2) with the boundary conditions

c(0) = P, or ϕ(0) = ϕ0, ψ(0) = ψ0, θ(0) = θ0,

c(1) = Q, or ϕ(1) = ϕ1, ψ(1) = ψ1, θ(1) = θ1.

Assume that sin ψ 6= 0 we rewrite the equation (4.2) as

(4.6) 2 ˙θ = ˙ϕ cot ψ sinh 2θ.

To simplify matters, let us introduce two new smooth functions p(s) and q(s) by 2θ(s) = arcsinh p(s), ψ(s) = arccot q(s),

and let the function ϕ(s) is set as ϕ(s) = ϕ0 + s(ϕ1 − ϕ0). Then we will define the

smooth functions p(s) and q(s) satisfying the horizontality condition (4.6) for c = c(s). Let k = ϕ1− ϕ0. Then equation (4.6) admits the form

˙p(s) p

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Separation of variables leads to the equation dp

pp1 + p2 = kq(s) ds,

that after integrating gives

− arctanhp 1

1 + p2(s) = k

 Z s 0

q(τ ) dτ + C

To define the constant C, we use the boundary conditions at s = 0. Observe that 1 p 1 + p2(0) = 1 cosh 2θ0 and p 1 1 + p2(1) = 1 cosh 2θ1 . Then C = −k1arctanh 1 cosh 2θ0 .

Applying the boundary condition at s = 1 we find the value of R01q(τ ) dτ as Z 1 0 q(τ ) dτ = − 1 k  arctanh 1 cosh 2θ1 + arctanh 1 cosh 2θ0  .

Since, moreover, q(0) = cot ψ0, q(1) = cot ψ1, Lemma 5 implies the existence of a smooth

function q(s) satisfying the above relation. The function p(s) can be defined by

1 p 1 + p2(s) = − tanh h k Z s 0 q(τ ) dτ − arctanh 1 cosh 2θ0 i .

The curve c(s) = ϕ(s), ψ(s), θ(s)) = (ϕ0+ s(ϕ1− ϕ0), arccot q(s)),12arcsinh p(s)

 is the desired horizontal curve.

 Remark 1. Of course, the proof is given for a particular parametrisation by a linear func-tion ϕ. One may easily modify this proof for an arbitrary smooth funcfunc-tion ϕ obtaining a wider class of smooth horizontal curves.

Some of the points on H1,2 can be connected by a curve that maintain one of the

coordinate constant.

Theorem 2. If P = P (ϕ0, ψ, θ0) and Q = Q(ϕ1, ψ, θ1) with

(4.7) ψ = arccotlntanh θ1

tanh θ0

/ ϕ0− ϕ1



are two points that can be connected, then there is a smooth horizontal curve joining P and Q with the constant ψ-coordinate given by (4.7).

Proof. Let c = c(ϕ, ψ, θ) be a horizontal curve with the constant ψ-coordinate. Then it satisfies the equation (4.2) that in this case we write as

cot ψ dϕ = d(2θ) sinh 2θ.

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Integrating yields cot ψ Z θ θ0 dϕ = Z θ θ0 d(2θ) sinh 2θ ⇒ (4.8) cot ψ ϕ(θ) − ϕ(θ0)  = ln tanh θ − ln tanh θ0.

For θ = θ1 we get formula (4.7) for the value of ψ. Solving (4.8) with respect to ϕ(θ) we

get

ϕ(θ) = ϕ0+

ln tanh θ/ tanh θ0

 cot ψ

with ψ given by (4.7). Finally, the horizontal curve joining the points P and Q satisfies the equation (ϕ, ψ, θ) =ϕ0+ ln tanh θ/ tanh θ0  cot ψ , ψ, θ  .  Upon solving the problem of the connectivity of two arbitrary points by a horizontal curve we are interested in determining its character: timelikeness (spacelikeness or light-likeness). It is not an easy problem. We are able to present some particular examples showing its complexity. Let us start with the following remark.

Remark 2. If P, Q ∈ H1,2 are two points connectable only by a family of smooth timelike

(spacelike, lightlike) curves, then smooth horizontal curves (its existence is known by the preceding theorem) joining P and Q are timelike (spacelike, lightlike).

Indeed, let ΩP,Qbe a family of smooth timelike (lightlike) curves connecting P and Q.

If δ(s) ∈ ΩP,Q, then its velocity vector ˙δ(s) can be written in the left-invariant basis

T, X, Y as

˙δ(s) = α(s)T (δ(s)) + β(s)X(δ(s)) + γ(s)Y (δ(s))

with h ˙δ(s), ˙δ(s)i = −α2 + β2+ γ2 < 0(= 0). If moreover, it is horizontal, then γ = 0.

Therefore, −α2+ β2 < 0(= 0), and the horizontal curve connecting P and Q is timelike

(lightlike).

If the points P and Q are connectable only by a family of spacelike curves, then the inequality −α2+ β2 > γ2 holds for them. It implies −α2+ β2 > 0 for a horizontal curve.

We conclude that in this case the horizontal curve is still spacelike.

Making use of (4.3) and (4.4) as well as parametrisation (4.1) we calculate the square of the velocity vector for a horizontal curve in terms of the variables ϕ, ψ, θ as

(4.9) −α2+ β2= − ˙ϕ2− ˙ψ2+ 4 ˙θ2− 2 ˙ϕ ˙ψ cosh 2θ.

We present some particular timelike, spacelike, and lightlike solutions of (4.2).

Example 2. Let ˙ϕ = 0. Then, ϕ ≡ ϕ0 is constant. In order to satisfy (4.2) we have

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2.1 ˙θ = 0 =⇒ θ ≡ θ0 is constant. Then | ˙c|2 = − ˙ψ2 ≤ 0. We conclude

that all non-constant horizontal curves c(s) = (ϕ0, ψ(s), θ0) are timelike. The

projections of these curves onto the (x1, x2)- and (x3, x4)-planes are circles. All

lightlike horizontal curves are only constant ones.

2.2 ψ = πn, n ∈ Z. Then | ˙c|2 = 4 ˙θ2 ≥ 0. We conclude that all non-constant

horizontal curves c(s) = (ϕ0, πn, θ(s)), n ∈ Z are spacelike. The projections of

these curves onto the (x1, x3)- and (x2, x4)-planes are hyperbolas. All lightlike

horizontal curves are only constant ones.

Example 3. Let ˙ϕ 6= 0. We choose ϕ as a parameter. Then the square of the norm of the velocity vector is

(4.10) −α2 + β2 = −1 − ˙ψ2+ 4 ˙θ2− 2 ˙ψ cosh 2θ,

where the derivatives are taken with respect to the parameter ϕ. The horizontality condition becomes

(4.11) 2 ˙θ sin ψ = cos ψ sinh 2θ.

As in the previous example we consider different cases.

3.1 Suppose ˙θ = 0 and assume that θ = θ0 6= 0. Then the horizontal curves are

parametrised by c(s) = (ϕ,π2 + πn, θ0), n ∈ Z. All these curves are timelike,

since | ˙c|2 = −1. There are no lightlike or spacelike horizontal curves.

3.2 If θ0 = 0, then any curve in the (ϕ, ψ)-plane is horizontal and timelike since

| ˙c|2 = −(1 + ˙ψ)2.

3.3 Suppose that ˙ψ = 0 and ψ ≡ ψ0 6= πk2 , k ∈ Z. Then (4.10) and (4.11) are

simplified to

(4.12) −α2+ β2 = −1 + 4 ˙θ2,

(4.13) ˙θ = K sinh 2θ with K = cot ψ0 2 .

Let θ = θ(ϕ) solves equation (4.13). Then the horizontal curve

(4.14) c(s) = (ϕ, ψ0, θ(ϕ))

is timelike when |θ| < 12arcsinh 1

2K. If |θ| > (=) 1

2arcsinh 1

2K, then the horizontal

curve (4.14) is spacelike (lightlike).

Thus any two points P (ϕ0, ψ0, θ0), Q(ϕ1, ψ1, θ0), can be connected by a piecewise smooth

timelike horizontal curve. This curve consists of straight segments with constant ϕ-coordinates or with coordinate ψ = π2 + πn, n ∈ Z. In the case θ0 = 0, this horizontal

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5. Sub-Lorentzian geodesics

In Lorentzian geometry there are no curves of minimal length because two arbitrary points can be connected by a piecewise lightlike curve. However, there do exist timelike curves with maximal length which are timelike geodesics [14]. By this reason, we are looking for the longest curve among all horizontal timelike ones. It will be given by extremizing the action integral S = 12R01 − α2(s) + β2(s)ds under the non-holonomic

constrain hxE2, ˙ci = 0. The extremal curve will satisfy the Euler-Lagrange system

(5.1) d ds ∂L ∂ ˙c = ∂L ∂c with the Lagrangian

L(c, ˙c) = 1 2(−α

2+ β2

) + λ(s)hxE2, ˙ci.

The function λ(s) is the Lagrange multiplier function and the values of α and β are given by (3.2) and (3.3). The Euler-Lagrange system (5.1) can be written in the form

− ˙αx2− ˙βx3 = 2(α ˙x2+ β ˙x3− λ ˙x4) − ˙λx4,

˙αx1− ˙βx4 = 2(−α ˙x1+ β ˙x4+ λ ˙x3) + ˙λx3,

− ˙αx4+ ˙βx1 = 2(α ˙x4− β ˙x1− λ ˙x2) − ˙λx2,

˙αx3+ ˙βx2 = 2(−α ˙x3− β ˙x2+ λ ˙x1) + ˙λx4.

for the extremal curve c(s) = (x1(s), x2(s), x3(s), x4(s)). Multiplying these equations by

x2, −x1, −x4, x3, respectively and then, summing them up we obtain

− ˙α = 2(−αh ˙c, Ni − βh ˙c, Y i − λβ) = −2λβ

because h ˙c, Y i = h ˙c, Ni = 0. Now, multiplying the equations by x3, x4, x1, x2,

respec-tively and then, summing them up we get

− ˙β = 2(αh ˙c, Y i + βh ˙c, Ni + λα) = 2λα

in a similar way. The values of α and β are concluded to satisfy the system ˙α(s) = 2λβ(s),

(5.2)

˙

β(s) = 2λα(s).

Case λ(s) = 0. In the Riemannian geometry the Schwartz inequality allows us to define the angle ϑ between two vectors v and w as a unique number 0 ≤ ϑ ≤ π, such that

cos ϑ = v · w |v||w|.

There is an analogous result in Lorentzian geometry which is formulated as follows. Proposition 1. [14] Let v and w be timelike vectors. Then,

1. |hv, wi| ≥ |v||w| where the equality is attained if and only if v and w are collinear. 2. If hv, wi < 0, there is a unique number ϑ ≥ 0, called the hyperbolic angle between

v and w, such that

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Theorem 3. The family of timelike future-directed horizontal curves contains horizontal timelike future-directed geodesics c(s) with the following properties

1. The length | ˙c| is constant along the geodesic.

2. The inner products hT, ˙ci = α, hX, ˙ci = β, hY, ˙ci = 0 are constant along the geodesic.

3. The hyperbolic angle between the horizontal time vector field T and the velocity vector ˙c is constant.

Proof. The system (5.2) implies

˙α(s) = 0 β(s) = 0.˙

The existence of a geodesic follows from the general theory of ordinary differential equa-tions, employing, for example, the parametrisation given for α, β, γ in the preceding section. Since the horizontal coordinates α(s) and β(s) are constant along the curve c we conclude that c is geodesic. We denote by α and β its respective horizontal coordinates. The length of the velocity vector ˙c is | − α2 + β2|1/2 and it is constant along the

geodesic.

The second statement is obvious. Since c(s) is a future-directed geodesic, we have hT, ˙ci < 0, and cosh(∠T, ˙c) = −hT, ˙ci |T || ˙c| = −α p | − α2+ β2| is constant.  Case λ(s) 6= 0. We continue to study the extremals given by the solutions of the Euler-Lagrange equation (5.1).

Lemma 6. Let c(s) be a timelike future-directed solution of the Euler-Lagrange sys-tem (5.1) with λ(s) 6= 0. Then,

1. The length | − α2(s) + β2(s)|1/2 of the velocity vector ˙c(s) is constant along the

solution.

2. The hyperbolic angle between the curve c(s) and the integral curve of the time vector field T is given by

ϑ = ∠( ˙c, T ) = −2Λ(s) + θ0,

where Λ is the primitive of λ.

Proof. Multiplying the first equation of (5.2) by α, the second one by β and subtracting, we deduce that α ˙α − β ˙β = 0. This implies that −α2 + β2 = h ˙c, ˙ci is constant. The

horizontal solution is timelike if the initial velocity vector is timelike. The first assertion is proved.

Set r = p| − α2+ β2|. Using the hyperbolic functions we write

α(s) = −r cosh θ(s), β(s) = r sinh θ(s). Substituting α and β in (5.2), we have

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Denote Λ(s) = R0sλ(s) ds and write the solution of the latter equation as θ = −2Λ(s)+θ0.

Thus,

α(s) = −r cosh(−2Λ(s) + θ0), β(s) = r sinh(−2Λ(s) + θ0).

(5.3)

In order to find the value of the constant θ0 we put s = 0 and get θ0 = arctanhβ(0)α(0).

Let c(s) be a horizontal timelike future-directed solution of (5.1). Then h ˙c, T i < 0 and

α = h ˙c, T i = −| ˙c||T | cosh ϑ = −r cosh(∠( ˙c, T )).

Comparing with (5.3) finishes the proof of the theorem.  There is no counterpart of Proposition 1 for spacelike vectors. Nevertheless, we obtain the following analogue of Lemma 6 .

Lemma 7. Let c(s) be a spacelike solution of the Euler-Lagrange system (5.1) with λ(s) 6= 0. Then,

1. The length of the velocity vector ˙c(s) is constant along the solution; 2. The horizontal coordinates are expressed by (5.3).

As the next step, we shall study the function Λ(s). First, let us prove some useful facts.

Proposition 2. Let c(s) = (x1(s), x2(s), x3(s), x4(s)) be a horizontal timelike (spacelike)

curve. Then, 1. − ˙x2

1(s) − ˙x22(s) + ˙x23(s) + ˙x24(s) = −α2(s) + β2(s);

2. ¨c = a(s)T + b(s)X + ω(s)Y + w(s)N, with a = ˙α, b = ˙β, ω = 0, w = α2− β2.

Proof. Let us write the coordinates of ˙c(s) in the basis T, X, Y, N as ˙c(s) = α(s)T + β(s)X + γ(s)Y + δ(s)N, where α = h ˙c, T i = x2˙x1− x1˙x2+ x4˙x3 − x3˙x4, β = h ˙c, Xi = −x3˙x1 − x4˙x2 + x1˙x3+ x2˙x4, 0 = γ = h ˙c, Y i = x4˙x1 − x3˙x2+ x2˙x3− x1˙x4, 0 = δ = h ˙c, Ni = −x1˙x1− x2˙x2+ x3˙x3+ x4˙x4.

By the direct calculation we get

−α2+ β2 = −α2− δ2+ β2+ γ2 = − ˙x21− ˙x22+ ˙x23+ ˙x24. In order to prove the second statement of the proposition we calculate

˙α = x2x¨1− x1x¨2+ x4x¨3− x3x¨4 = h¨c, T i = a,

˙

β = −x3x¨1− x4x¨2+ x1x¨3+ x2x¨4 = h¨c, Xi = b.

Differentiating the horizontality condition (3.1), we find 0 = d dsh ˙c, Y i = d ds x4˙x1−x3˙x2+ x2˙x3−x1˙x4  = x4x¨1−x3x¨2+ x2x¨3−x1x¨4 = h¨c, Y i = ω.

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Then, 0 = dsdh ˙c, Ni = dsd − x1˙x1− x2˙x2+ x3˙x3+ x4˙x4  = −x1x¨1− x2x¨2 + x3x¨3+ x4x¨4 +(− ˙x2 1− ˙x22+ ˙x23 + ˙x42) = h¨c, Ni + (−α2+ β2) = w − α2+ β2,

by the first statement. The proof is finished. 

Theorem 4. The Lagrange multiplier λ(s) is constant along the horizontal timelike (spacelike, lightlike) solution of the Euler-Lagrange system (5.1).

Proof. We consider the equivalent Lagrangian function bL(x, ˙x), changing the length function −α2+ β2 to − ˙x2

1− ˙x22 + ˙x23+ ˙x24. The solutions of the Euler-Lagrange system

for both Lagrangians give the same curve. Thus, the new Lagrangian is b L(x, ˙x) = 1 2 − ˙x 2 1− ˙x22 + ˙x23+ ˙x24  + λ(s) ˙x1x4− ˙x4x1 − ˙x2x3 + ˙x3x2  . The corresponding Euler-Lagrange system is

−¨x1 = − ˙λx4− 2λ ˙x4, −¨x2 = ˙λx3+ 2λ ˙x3, ¨ x3 = − ˙λx2− 2λ ˙x2, ¨ x4 = − ˙λx1+ 2λ ˙x1.

We multiply the first equation by −x4, the second equation by x3, the third one by x2,

and the last one by −x1, finally, sum them up. This yields

¨

x1x4− ¨x2x3+ ¨x3x3− ¨x4x1 = ˙λ(x24+ x32− x22− x21) + 2λ ˙x4x4+ ˙x3x3− ˙x2x2− ˙x1x1

 ⇒ h¨c, Y i = − ˙λ + 2λh ˙c, Ni ⇒ ˙λ = 0.

We conclude that λ is constant along the solution. 

We see from the proof of Lemma 6 that the function Λ(s) is just a linear function. This leads to the following property of horizontal timelike future-directed solutions of the Euler-Lagrange system (5.1).

Corollary 1. If c(s) is a horizontal timelike future-directed solution of (5.1), then the hyperbolic angle between its velocity and the time vector field T increases linearly in s.

6. Hamiltonian formalism

The sub-Laplacian, which is the sum of the squares of the horizontal vector fields plays the fundamental role in Riemannian geometry. The counterpart of the sub-Laplacian in the Lorentz setting is the operator

L = 12(−T2+ X2) = 1 2  − − x2∂x1 + x1∂x2 + x4∂x3 − x3∂x4 2 + x3∂x1 + x4∂x2 + x1∂x3 + x2∂x4 2 . (6.1)

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Then the Hamiltonian function corresponding to the operator (6.1) is H(x, ξ) = 1 2  − − x2ξ1+ x1ξ2+ x4ξ3− x3ξ4 2 + x3ξ1+ x4ξ2+ x1ξ3+ x2ξ4 2 = 1 2 − τ 2 + ς2, (6.2)

where we use the notations ξk = ∂xk, τ = −x2ξ1 + x1ξ2+ x4ξ3 − x3ξ4, and ς = x3ξ1+

x4ξ2+ x1ξ3+ x2ξ4. There are close relations between the solutions of the Euler-Lagrange

equation and the solutions of the Hamiltonian system ˙x = ∂H

∂ξ , ˙ξ = − ∂H

∂x.

The solutions of the Euler-Lagrange system (5.1) coincide with the projection of the solu-tions of the Hamiltonian system onto the Riemannian manifold. In the sub-Riemannian case the solutions coincide, if and only if, the solution of the Euler-Lagrange system is a horizontal curve. We are interested in relations of the solutions of these two systems in our situation. The Hamilton system admits the form



˙x = ∂H∂ξ = −τxJ + ςxE1,

˙ξ = −∂H

∂x = −τξJ − ςξE1.

(6.3)

Lemma 8. The solution of the Hamiltonian system (6.3) is a horizontal curve and

(6.4) τ = α, ς = β,

where α and β are given by (3.2) and (3.3) respectively. Proof. Let c(s) = x1(s), x2(s), x3(s), x4(s)



be a solution of (6.3). In order to prove its horizontality we need to show that the inner product h ˙x, xE2i vanishes. We substitute

˙x from (6.3) and get

h ˙x, xE2i = −τhxJ, xE2i + ςhxE1, xE2i = 0

by (2.10).

Using the first line in the Hamiltonian system and the definitions of horizontal coor-dinates (3.2) and (3.3), we get

α = h ˙x, xJi = −τhxJ, xJi + ςhxE1, xJi = τ,

β = h ˙x, xE1i = −τhxJ, xE1i + ςhxE1, xE1i = ς

from (2.10) and (2.11). 

Lemma 8 implies the following form of the Hamiltonian system (6.3) ˙x1 = −α(−x2) + βx3,

˙x2 = −αx1+ βx4,

(6.5)

˙x3 = −αx4+ βx1,

˙x4 = −α(−x3) + βx2.

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6.1.1. Timelike case. In this section we are aimed at finding geodesics corresponding to the extremals (Section 5) with constant horizontal coordinates α and β giving the vanishing value to the Lagrangian multiplier λ. We give an explicit picture for the base point (1, 0, 0, 0). Left shifts transport it to any other point of H1,2. Without lost of

generality, let us assume that −α2+ β2 = −1, α = cosh ψ, β = sinh ψ.

The Hamiltonian system (6.5) written for constant α and β is reduced to a second-order differential equation

(6.6) x¨k = −xk, k = 1, . . . 4.

The general solution is given in the trigonometric basis as xk= Akcos s + Bksin s. The

initial condition x(0) = (1, 0, 0, 0) defines the coefficients Ak by A1 = 1, A2 = A3 =

Ak = 0. Returning back to the first-order system (6.5) we calculate the coefficients Bk

as B1 = 0, B2 = −α, B3 = β, B4 = 0. Finally, the solution is

(6.7) x1 = cos s, x2 = − cosh ψ sin s, x3 = sinh ψ sin s, x4 ≡ 0.

These timelike geodesics are closed. Varying ψ they sweep out the one-sheet hyperboloid x2

1+ x22− x23 = 1 in R3.

Let us calculate the vertical line Γ, the line corresponding to the vanishing horizontal velocity (α, β) and with the constant value γ = 1 6= 0, passing the base point (1, 0, 0, 0). Its parametric representation Γ = Γ(s) satisfies the system

α = x2˙x1− x1˙x2+ x4˙x3− x3˙x4 = 0,

β = −x3˙x1− x4˙x2+ x1˙x3+ x2˙x4 = 0,

γ = x4˙x1− x3˙x2+ x2˙x3− x1˙x4 = 1,

δ = x1˙x1+ x2˙x2 − x3˙x3− x4˙x4 = 0.

The discriminant of this system calculated with respect the derivatives as variables is (-1), and we reduce the system to a simple one

˙x1 = −x4, ˙x2 = x3, ˙x3 = x2, ˙x4 = −x1,

with the initial condition Γ(0) = x(0) = (1, 0, 0, 0). The solution is Γ(s) = (cosh s, 0, 0, − sinh s).

The vertical line (hyperbola) Γ meets the surface (6.7) at the point (1,0,0,0) orthogonally with respect to the scalar product in R2,2. Comparing this picture with the classical

sub-Riemannian case of the Heisenberg group, we observe that in the Heisenberg case all sraight line geodesics lie on the horizontal plane R2 and the center is the third vertical

axis. In our case the surface (6.7) corresponds to the horizontal plane, timelike geodesics correspond to the sraight line Heisenberg geodesics, and Γ corresponds to the vertical center.

6.1.2. Spacelike/lightlike case. Again we consider constant horizontal coordinates α and β, and let us assume that −α2+ β2 = 1, α = sinh ψ, β = cosh ψ.

The Hamiltonian system (6.5) is reduced to the second-order differential equation (6.8) x¨k = xk, k = 1, . . . 4.

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Arguing as in the previous case we deduce the solution passing the point (1,0,0,0) as x1 = cosh s, x2 = − sinh ψ sinh s, x3 = cosh ψ sinh s, x4 ≡ 0.

These non-closed spacelike geodesics sweep the same hyperboloid of one sheet in R3.

The vertical line Γ meets orthogonally each spacelike geodesic on this hyperboloid at the point (1,0,0,0).

In the lightlike case α2 = β2 = 1 the Hamiltonian system (6.5) has a linear solution

given by

x1 ≡ 1, x2 = −αs, x3 = βs, x4 ≡ 0,

which are two straight lines on the hyperboloid, and again Γ meets them orthogonally at the unique point (1,0,0,0).

6.2. Geodesics with non-constant horizontal coordinates. If the horizontal coor-dinates α(s) and β(s) are constant, then (6.5) is a linear system of ordinary differential equations with constant coefficients. If α(s) and β(s) are not constant, then using the expression (3.2) and (3.3) for α(s) and β(s), we get the homogeneous system of ordinary differential equations which is linear with respect to derivatives

˙x1 1 − x22+ x23  + ˙x2 x1x2+ x3x4  − ˙x3 x2x4+ x1x3  = 0 ˙x1 x1x2+ x3x4  + ˙x2 1 − x21+ x24  − ˙x4 x2x4+ x1x3  = 0 ˙x1 x2x4+ x1x3  + ˙x3 1 − x21+ x24  − ˙x4 x1x2+ x3x4  = 0 (6.9) ˙x2 x2x4+ x1x3  − ˙x3 x1x2+ x3x4  + ˙x4 1 − x22+ x23  = 0.

The determinant of the system vanishes. The direct calculations show that the rank of the system is equal to 2.

Fix an initial point x(0). We shall give two approaches to solve this Hamiltonian

system based on a direct solution and on a parametrisation of H1,2.

Direct solution. Subtracting and summing the first and the last equations from (6.9) are equivalent to subtracting and summing the second and the third equations from (6.9). These simple linear combinations yield the following system

(x1+ x4)(( ˙x1+ ˙x4)(x1− x4) + ( ˙x2− ˙x3)(x2+ x3)) = 0,

(x1− x4)(( ˙x1− ˙x4)(x1+ x4) + ( ˙x2+ ˙x3)(x2− x3)) = 0.

The degenerating case x1± x4 = 0 implies a straight line solution x2 = (1 − C2)−1/2,

x3 = C(1 − C2)−1/2, where C is a constant, −1 < C < 1. The square of the velocity

vector is | ˙c|2 = − ˙x2

1 − ˙x22 + ˙x23 + ˙x24, and it vanishes on this solution. Therefore, the

degenerating case gives lightlike geodesics.

The non-degenerating case x1 ± x4 6= 0 gives the following solution

x1+ x4 = (x(0)1 + x (0) 4 ) exp t Z 0 ( ˙x3− ˙x2)(x3+ x2) 1 + x2 3− x22 dt, x1− x4 = (x(0)1 − x (0) 4 ) exp t Z 0 ( ˙x3+ ˙x2)(x3− x2) 1 + x2 3− x22 dt.

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One verifies that given an initial point (x(0)1 , x(0)2 , x(0)3 , x(0)4 ) at H1,2, the whole solution

trajectory (x1(s), x2(s), x3(s), x4(s)) lies on H1,2.

Let us check the direction properties of the non-degenerating solution. In this case the square of the velocity vector is given by | ˙c|2 = ( ˙x2

3 − ˙x22)(x21 − x24) ≡ (( ˙x (0) 3 )2 − ( ˙x(0)2 )2)((x(0) 1 )2 − (x (0)

4 )2). The time-(space-, light-)ness of the solution is completely

defined by the choice of ˙x3, ˙x2 at the initial point. The second factor is non-zero,

therefore, lightlikeness is provided by vanishing ( ˙x2

3− ˙x22) that gives also a straight line

solution as in the degenerating case. Thus, combining these two cases we conclude that lightlike solutions of the Hamiltonian system (6.9) are only straight lines.

Parametric solution. Here it is more convenient to use the following parametrisation different from (4.1) x1 = cos φ cosh ξ, x2 = sin φ cosh χ, x3 = sin φ sinh χ, (6.10) x4 = cos φ sinh ξ.

The coefficients in system (6.9) become

a = 1 + x23− x22 = cos2φ, b = 1 − x21 + x24 = sin2φ,

c = x2x4+ x1x3 = cos φ sin φ sinh(ξ + χ),

(6.11)

d = x1x2+ x3x4 = cos φ sin φ cosh(ξ + χ).

Let us use the parametrisation (6.10) to solve the system (6.9). Since the system (6.9) is of rank 2, we can assume that ˙x2 and ˙x3 are arbitrary smooth functions. We are

looking for a horizontal curve c(s) = φ(s), χ(s), ξ(s), s ∈ [0, 1], starting from a point P = (φ0, χ0, ξ0) ∈ H1,2. Let φ(s) and χ(s) be smooth functions such that

φ(0) = φ0, χ(0) = χ0,

Let us set

x2 = K(s) = sin φ(s) cosh χ(s), x3 = L(s) = sin φ(s) sinh χ(s),

and let us denote ˙x2(s) = k(s), ˙x3(s) = l(s). Then ,

(6.12) x2 = sin φ cosh χ ⇒ sin φ =

K cosh χ = K |K| √ K2− L2.

We need only to determine the function ξ. The system (6.9) yields ˙x1(s) = l(s)c(s) − k(s)d(s)

a(s) , ˙x4(s) =

k(s)c(s) + l(s)d(s)

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Making use of the parametrisation (6.10), we write the latter equations in the form (6.13)            d cos φ(s)

ds cosh ξ(s) + ˙ξ(s) cos φ(s) sinh ξ(s)

= tan φ(s)l sinh ξ(s) + χ(s)− k cosh ξ(s) + χ(s),

d cos φ(s)

ds sinh ξ(s) + ˙ξ(s) cos φ(s) cosh ξ(s)

= tan φ(s)k sinh ξ(s) + χ(s)+ l cosh ξ(s) + χ(s). Multiplying both equations by cos φ cosh ξ1 we express tanh ξ from the second equation as

tanh ξ = − ˙ξ +

tan φ

cos φ k sinh χ + l cosh χ

 1 cos φ d cos φ ds − tan φ

cos φ k cosh χ + l sinh χ

.

Substituting tanh ξ in the first equation of (6.13), we get the equation for ˙ξ ˙ξ2

− 2 ˙ξtan φcos φl cosh χ − 1cos φd cos φds 2+ 2d cos φ ds tan φ cos2φl sinh χ + tan2φ cos2φ(k 2+ l2) = 0.

The square of the discriminant D of this quadratic equation is D2 =tan φ

cos φ 2

l sinh χ + ˙φ cos φ2− k22. Since tanh χ(s) = K(s)L(s), we get

(6.14) l = k tanh χ + K cosh2χ ˙χ. From (6.12) (6.15) φ cos φ =˙ s ds sin φ = k cosh χ − K sinh χ cosh2χ ˙χ. Substituting (6.14) and (6.15) in the expression of D2 we deduce that

D2 =tan φ cos φ 2 k cosh χ(sinh 2 χ + 1)2− k2=tan φ cos φ 2 k2sinh2χ. The solution of the quadratic equation gives

˙ξ(s) = tan φ

cos φ l cosh χ ± k sinh χ). Differentiating the functions K(s) and L(s) leads to

(6.16) ˙ξ−(s) = tan2φ(s) ˙χ(s)

for the sign minus in the solution of the quadratic equation. For the sign plus we get (6.17) ˙ξ+(s) = ˙φ(s) tan φ(s) sinh 2χ(s) + ˙χ(s) tan2φ(s) cosh 2χ(s).

Equation (6.16) is just a horizontality condition written in the parametrization (6.10) and we come to the conclusion of Lemma 8. Since we have to satisfy both of the equations (6.16) and (6.17) we get ξ− = ξ+. This leads to the equation

˙ φ

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that is equivalent to the condition

D = 0 =⇒ k = 0 =⇒ sin φ cosh χ = C0 is constant.

The value of C0 is defined from the initial data, so C0 = sin φ0cosh χ0.

Take the function χ as a parameter. Then φ can be found from the equation

(6.18) sin φ = C0

cosh χ. From the equation (6.18) we find that tan2φ = C02

cosh2

χ−C2

0. We substitute tan

2φ into

equation (6.16) and get ξ − ξ0 = C02 Z dχ cosh2χ − C2 0 = C02 Z d2χ cosh(2χ) + 1 − 2C2 0 .

The value of the last integral depends on C0 and leads to the following three cases:

1. if C2 0 − 1 > 0, then (6.19) ξ(s) = ξ0+ C2 0 2|C0| p C2 0 − 1 ln 4χ + 1 − 2C 2 0 − 2|C0| p C2 0 − 1 4χ + 1 − 2C2 0 + 2|C0| p C2 0 − 1 χ χ0 , 2. if C2 0 − 1 < 0, then (6.20) ξ(s) = ξ0+ C2 0 |C0| p 1 − C2 0 arctan 4χ + 1 − 2C 2 0 2|C0| p 1 − C2 0 χ χ0 , 3. if C2 0 = 1, then (6.21) ξ(s) = ξ0+ 4 4χ − 1 χ χ0 .

Since the velocity vector is constant along the geodesic, the timelike, spacelike or lightlike properties are defined by the nature of the initial velocity vector. We proved the following theorem.

Theorem 5. Given a point P (φ0, χ0, ξ0), there is a timelike (spacelike, lightlike) geodesic

c(s) = (φ(χ), χ, ξ(χ)) with φ(χ) satisfying to (6.18) and ξ(χ) satisfying to (6.19), (6.20) or (6.21) according to the position of the point P . The property to be timelike, spacelike or lightlike depends on the choice of the function χ.

7. Geodesics with respect to the distribution D = span{X, Y }

This case reveals the sub-Riemannian nature of such a distribution. In principle, one can easily modify the classical results from sub-Riemannian geometry (Chow-Rashevskii theorem, in particular). However we prefer to modify our own results proved in previous sections to show some particular features and to compare with the sub-Lorentzian case defined by the distribution D = span{T, X}.

Lemma 9. A curve c(s) = (x1(s), x2(s), x3(s), x4(s)) is horizontal with respect to the

distribution D = span{X, Y }, if and only if,

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Proof. The tangent vector to a curve c(s) = (x1(s), x2(s), x3(s), x4(s)) written in the

left-invariant basis is of the form

˙c(s) = αT + βX + γY. Then

α = h ˙c, T i = I ˙c · T = ˙x1x2− ˙x2x1+ ˙x3x4− ˙x4x3.

We conclude that α = 0, if and only if, (7.1) holds. 

In this case a curve is horizontal, if and only if, its velocity vector is orthogonal to the vector field T . The left-invariant coordinates β(s) and γ(s) of a horizontal curve c(s) = (x1(s), x2(s), x3(s), x4(s)) are

(7.2) β = h ˙c, Xi = −x3˙x1− x4˙x2+ x1˙x3+ x2˙x4 = hxE1, ˙ci.

(7.3) γ = h ˙c, Y i = −x4˙x1+ x3˙x2− x2˙x3+ x1˙x4 = hxE2, ˙ci.

The form w = −x2dx1+ x1dx2− x4dx3+ x3dx4 = −hxJ, dxi is a contact form for the

horizontal distribution D = span{X, Y }. Indeed

w(N) = 0, w(T ) = 1, w(X) = 0, w(Y ) = 0.

Thus, ker w = span{N, X, Y }, The horizontal distribution can be defined as follows D = {V ∈ T H1,2 : w(V ) = 0}, or D = ker w ∩ T H1,2.

The length l(c) of a horizontal curve c(s) : [0, 1] → H1,2 is given by

l(c) = Z 1 0 h ˙c(s), ˙c(s)i 1/2ds = Z 1 0 β2(s) + γ2(s)1/2ds.

The restriction of the non-degenerate metric h·, ·i onto the horizontal distribution D ⊂ T H1,2 gives a positive-definite metric that we still denote by h·, ·i

D. Thus from now

on, we shall work only with one type of the curves (that we shall call simply horizontal curves), since the horizontality condition requires the vanishing coordinate function of the vector field T .

7.1. Existence of horizontal curves. The following theorem is an analogue to The-orem 1 proved for the distribution D = span{T, X} in Section 4.

Theorem 6. Let P , Q ∈ H1,2 be arbitrary given points. Then there is a smooth hori-zontal curve connecting P with Q.

Proof. We use parametrisation (4.1), in which the horizontality condition for a curve c(s) is expressed by (4.3) as

˙

ψ + ˙ϕ cosh 2θ = 0. This equation is to be sold for the initial conditions

c(0) = P, or ϕ(0) = ϕ0, ψ(0) = ψ0, θ(0) = θ0,

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Let ψ = ψ(s) be a smooth arbitrary function with ˙ψ(0) = lim s→0+ ˙ ψ(s) and ˙ψ(1) = lim s→1− ˙

ψ(s). Set 2θ(s) = arccosh p(s). Then the equation (4.3) admits the form

˙ ϕ = − ψ˙ cosh 2θ = − ˙ ψ p(s) ⇒ ϕ(s) = − Z s 0 ˙ ψ(s) ds p(s) + ϕ(0).

Denote q(s) = ψ(s)p(s)˙ . Since q(0) = cosh 2θψ(0)˙ 0, q(1) = cosh 2θψ(1)˙ 1, and R01q(s) ds = ϕ0 − ϕ1

applying Lemma 5 we conclude that there exists such a smooth function q(s). The function p(s) is found as p(s) = ψ(s)q(s)˙ . We get a curve c(s) = (ϕ(s), ψ(s), θ(s)) with

ψ = ψ(s), (7.4) ϕ(s) = − Z s 0 ˙ ψ(s) ds p(s) + ϕ(0), (7.5) θ(s) = 1 2arccosh p(s). (7.6)  Remark 3. Observe that in the general Chow-Rashevskii theorem smoothness was not concluded.

Theorem 7. Given two arbitrary points P = P (ϕ0, ψ0, θ0) and Q = Q(ϕ1, ψ1, θ0) with

2θ0 = arccoshψ1 −ψ0

ϕ0−ϕ1, there is a horizontal curve with the constant θ-coordinate connecting

P with Q.

Proof. If the θ-coordinate is constant, then the governing equation is ˙

ψ = − ˙ϕ cosh 2θ0 ⇒ ψ(s) = −ϕ(s) cosh 2θ0+ C.

Applying the initial conditions

c(0) = ϕ0, ψ0, θ0  , and c(1) = ϕ1, ψ1, θ0  , we find 2θ0 = arccosh ψ1− ψ0 ϕ0− ϕ1  , C = ψ0+ ϕ0 ψ1− ψ0 ϕ0− ϕ1 . Therefore, for any parameter ϕ, the horizontal curve

c(s) =ϕ, ψ0+ ϕ(0) − ϕ ψ1− ψ0 ϕ0− ϕ1 , θ0  , 2θ0 = arccosh ψ1− ψ0 ϕ0− ϕ1 ,

joins the points P = P (ϕ0, ψ0, θ0) and Q = Q(ϕ1, ψ1, θ0). 

7.2. Lagrangian formalism. Dealing with D = span{X, Y } and a positive-definite metric h·, ·iDon it, one might compare with the geometry generated by the sub-Riemannian

distribution on sphere S3in [4]. The minimising length curve can be found by minimising

the action integral

S = 1 2

Z 1 0

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under the non-holonomic constrain α = h ˙c, xJi = 0. The corresponding Lagrangian is

(7.7) L(c, ˙c) = 1

2 β

2(s) + γ2(s)+ λ(s)α(s).

The extremal curve is given by the solution of the Euler-Lagrange system (5.1) with the Lagrangian (7.7).

Let us make some preparatory calculations. Write the system (5.1) for the La-grangian (7.7) as the follows

2β ˙x3+ 2γ ˙x4 − 2λ ˙x2 + ˙βx3+ ˙γx4− ˙λx2 = 0,

2β ˙x4− 2γ ˙x3+ 2λ ˙x1 + ˙βx4− ˙γx3+ ˙λx1 = 0,

−2β ˙x1+ 2γ ˙x2− 2λ ˙x4− ˙βx1+ ˙γx2− ˙λx4 = 0,

−2β ˙x2− 2γ ˙x1 + 2λ ˙x3− ˙βx2− ˙γx1+ ˙λx3 = 0.

Multiply the equations by x3, x4, x1, and x2, respectively and sum them up. We get

2βh ˙c, Ni − 2γh ˙c, T i − 2λh ˙c, Y i − ˙β + 0 ˙γ + 0 ˙λ = 0 ⇒ β = 2λγ,˙ (7.8)

2βh ˙c, T i − 2γh ˙c, Ni + 2λh ˙c, Xi + 0 ˙β − ˙γ + 0 ˙λ = 0 ⇒ ˙γ = 2λβ. (7.9)

Let us consider two cases.

Case λ(s) = 0. In this case equation (7.8) admits the form

(7.10) β = 0,˙ ˙γ = 0,

and we deduce the following theorem.

Theorem 8. There are horizontal geodesics with the following properties: 1. The coordinates α = h ˙c, T i = 0, β = h ˙c, Xi, and γ = h ˙c, Y i are constant; 2. The length | ˙c| along the geodesics;

3. The angles between the velocity vector and horizontal frame is constant along along the geodesic.

Proof. Taking into account the solution of (7.10), we denote β(s) = β and γ(s) = γ. Then the length of the velocity vector | ˙c| =pβ2+ γ2 is constant.

Since h ˙c, Xi = h ˙c, XiD = | ˙c|D|X|Dcos(∠ ˙c, X), h ˙c, Y i = h ˙c, Y iD = | ˙c|D|Y |Dcos(∠ ˙c, Y ),

we have cos(∠ ˙c, X) = p β β2+ γ2, cos(∠ ˙c, Y ) = γ p β2+ γ2,

that proves the third assertion. 

Case λ(s) 6= 0.

Theorem 9. There are horizontal geodesics with the following properties: 1. The velocity vector | ˙c| of a geodesic is constant along the geodesic;

2. The angles between the velocity vector and the horizontal frame are given by ∠ ˙c, X = cs + θ0, ∠ ˙c, Y =

π

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Proof. Since

(7.11) β = 2λγ,˙ ˙γ = 2λβ

implies dsd β2+ γ2 = 0, we conclude, that the length of the velocity vector | ˙c| is

con-stant. Taking into account positivity of β2+ γ2 let us denote it by r2. Set β = r cos θ(s)

and γ = r sin θ(s). Substituting them in (7.11), we get

˙θ(s) = 2λ(s) ⇒ θ(s) = 2Z λ(s) ds + θ0.

Let us find the function λ(s). Observe that

β2+ γ2 = − ˙x21− ˙x22+ ˙x23+ ˙x24.

It can be shown similarly to the proof of Proposition 2, having α = δ = 0. By the direct calculation (see also Proposition 2) we show that

h¨c, T i = dsd h ˙c, T i = 0.

Now, we consider an equivalent to (7.7) extremal problem with the Lagrangian (7.12) L(c, ˙c) =b 1 2 − ˙x 2 1− ˙x22+ ˙x23+ ˙x24  + λ(s)h ˙c, T i. The Euler-Lagrange system admits the form

−¨x1 = −2λ ˙x2− ˙λx2, −¨x2 = 2λ ˙x1+ ˙λx1, ¨ x3 = −2λ ˙x4− ˙λx4, ¨ x4 = 2λ ˙x3+ ˙λx3.

Multiplying these equations by x2, −x1, −x4, x3 respectively and then, summing them

up, we obtain

−h¨c, T i = 2λh ˙c, Ni − ˙λ.

This allows us to conclude, that the function λ(s) is constant along the solution of the Euler-Lagrange equation that yields the second assertion of the theorem.  7.3. Hamiltonian formalism. The sub-Laplacian is L = X2+ Y2 and the

correspond-ing Hamiltonian function is H(x, ξ) = 1 2  x3ξ1+ x4ξ2+ x1ξ3+ x2ξ4 2 + x4ξ1− x3ξ2− x2ξ3+ x1ξ4 2 = 1 2(ς 2+ κ2).

The Hamilton system is written as ˙x = ∂H

∂ξ = ςxE1+ κxE2, ˙ξ = −∂H

∂x = −ςξE1− κξE2, (7.13)

As in the previous section we are able to prove the following proposition. Proposition 3. The solution of the Hamilton system is a horizontal curve and

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Corollary 2. The Hamilton function is the energy H(x, ξ) = 1 2(β

2+ γ2).

Making use of Proposition 3 we write the first line of the Hamiltonian system (7.13) in the form. ˙x1 = βx3+ γx4, ˙x2 = βx4− γx3, ˙x3 = βx1− γx2, (7.14) ˙x4 = βx2+ γx1,

7.4. Geodesics with constant horizontal coordinates. In this section we consider constant horizontal coordinates β and γ. We give an explicit picture for the base point (1, 0, 0, 0). Without lost of generality, let us assume that β2 + γ2 = 1, β = cos ψ,

γ = sin ψ.

The Hamiltonian system (7.14) written for constant β and γ is reduced to a second-order differential equation

(7.15) x¨k = xk, k = 1, . . . 4.

The general solution is given in the trigonometric basis as xk = Akcosh s + Bksinh s.

The initial condition x(0) = (1, 0, 0, 0) defines the coefficients Ak by A1 = 1, A2 = A3 =

Ak= 0. Returning back to the first-order system (7.14) we calculate the coefficients Bk

as B1 = 0, B2 = 0, B3 = β, B4 = γ. Finally, the solution is

(7.16) x1 = cosh s, x2 ≡ 0, x3 = cos ψ sinh s, x4 = sin ψ sinh s.

Varying ψ they sweep out the two-sheet hyperboloid x2

1−x23−x24 = 1 in R3. We use only

one sheet containing the point (1, 0, 0, 0). Geodesics are hyperbolas passing this point. The vertical line corresponds to the vanishing horizontal velocity (β, γ) and with the constant value α = 1 6= 0, passing the base point (1, 0, 0, 0). The solution is

Γ(s) = (cos s, sin s, 0, 0).

The vertical line (circle) Γ meets the surface (7.16) at the point (1,0,0,0) orthogonally with respect to the scalar product in R2,2.

7.5. Geodesics with non-constant horizontal coordinates. Using expressions (7.2) and (7.3) for β(s) and γ(s) we get a homogeneous system of ordinary differential equa-tions which is linear with respect to derivatives

˙x1 1 + x23+ x24  − ˙x3 x1x3− x2x4  − ˙x4 x1x4+ x2x3  = 0 ˙x2 1 + x23+ x24  − ˙x3 x1x4+ x2x3  + ˙x4 x1x3− x2x4  = 0 ˙x1 x1x3− x2x4  + ˙x2 x1x4+ x2x3  + ˙x3 1 − x21− x22  = 0 (7.17) ˙x1 x1x4 + x2x3  − ˙x2 x1x3− x2x4  + ˙x4 1 − x21− x22  = 0.

The determinant of the system vanishes. The direct calculations show that the rank of the system is equal to 2.

Direct solution. As one can observe, the first and the second lines in the system (7.17) are the real and the imaginary parts of the equation

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The third and the last lines yield the same equation because of the rank of the system. The solution of the system must lie on H1,2, hence the functions x

1 and x2 never vanish

simultaneously. So this equation is reduced to

( ˙x1+ i ˙x2)(x1− ix2) = ( ˙x3− i ˙x4)(x3+ ix4).

Dividing both parts of this equation by x2

1 + x22 and by 1 + (x23 + x24) respectively, and

integrating we get the solution in the form x1 + ix2 = (x(0)1 + ix (0) 2 ) exp t Z 0 ( ˙x3− i ˙x4)(x3+ ix4) 1 + (x2 3 + x24) dt.

One verifies that given an initial point (x(0)1 , x(0)2 , x(0)3 , x(0)4 ) at H1,2, the whole solution

trajectory (x1(s), x2(s), x3(s), x4(s)) lies on H1,2. Let us remark that we work only with

spacelike curves and there is no degeneration as in the previous sub-Lorentzian case, where it corresponds to lightlike solutions. Thus, we got a family of solutions starting from the point (x(0)1 , x

(0) 2 , x

(0) 3 , x

(0)

4 ) ∈ H1,2, parametrised by two functions x3, x4.

8. Acknowledgement

The paper was initiated when the authors visited the National Center for Theoretical Sciences and National Tsing Hua University during January 2007. They would like to express their profound gratitude to Professors Jing Yu and Shu-Cheng Chang for their invitation and for the warm hospitality extended to them during their stay in Taiwan.

References

[1] I. Bengtsson, P. Sandin, Anti-de Sitter space, squashed and stretched. Classical Quantum Gravity 23(2006), no. 3, 971–986.

[2] R. Beals, B. Gaveau, P. C. Greiner, Complex Hamiltonian mechanics and parametrices for subel-liptic Laplacians, I, II, III. Bull. Sci. Math., 21 (1997), no. 1–3, 1–36, 97–149, 195–259.

[3] O. Calin, D. C. Chang, P. C. Greiner, Geometric Analysis on the Heisenberg Group and Its Gen-eralizations, AMS/IP Ser. Adv. Math., 40, International Press, Cambridge, MA, 2007.

[4] O. Calin, D.-Ch. Chang, I. Markina, Sub-Riemannian geometry of the sphere S3

, submitted. [5] S. Carlip, Conformal field theory, (2 + 1)-dimensional gravity and the BTZ black hole. Classical

Quantum Gravity 22 (2005), no. 12, R85–R123.

[6] W. L. Chow. Uber Systeme von linearen partiellen Differentialgleichungen erster Ordnung, Math. Ann., 117 (1939), 98-105.

[7] M. Grochowski, Geodesics in the sub-Lorentzian geometry. Bull. Polish Acad. Sci. Math. 50 (2002), no. 2, 161–178.

[8] M. Grochowski, On the Heisenberg sub-Lorentzian metric on R3

. Geometric Singularity Theory, Banach Center Publ., Polish Acad. Sci. 65 (2004), 57–65.

[9] S. W. Hawking, G. F. Ellis, The large scale structure of space time, Cambridge Univ. Press, 1973. [10] Q. K. Lu, Heisenberg group and energy-momentum conservative law in de-Sitter spaces. Commun.

Theor. Phys. (Beijing) 44 (2005), no. 3, 389–392.

[11] R. Montgomery, A tour of subriemannian geometries, their geodesics and applications. Mathe-matical Surveys and Monographs, 91. American MatheMathe-matical Society, Providence, RI, 2002. 259 pp.

[12] G. L. Naber, The geometry of Minkowski spacetime. An introduction to the mathematics of the special theory of relativity. Applied Mathematical Sciences, 92. Springer-Verlag, New York, 1992. 257 pp.

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