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Volume 16, Number 2, 2006

Geometric Analysis on Quaternion H -Type

Groups

ByDer-Chen Chang and Irina Markina

ABSTRACT. We construct some examples ofH-types Carnot groups related to quaternion numbers and study their geometric properties. We involve the Hamiltonian formalism to obtain the equations of geodesics and calculate the cardinality of geodesics joining two different points on these groups. We prove Kepler’s law and give a nice geometric interpretation of the length of geodesics.

1. Introduction

Since their introduction by Kaplan [11] in 1980, generalized Heisenberg groups, also known as groups of Heisenberg type orH-type groups, have provided a framework in which to construct interesting examples in geometry and analysis [8, 12, 13, 17]. Analysis on homogeneous groups, and in particular, on theH-type groups, is a good test ground for the study of general sub-elliptic problems arising from vector fieldsX1, . . . , Xk, satisfying the Chow’s bracket generating condi- tion [7]. More precisely, vector fieldsX1, . . . , Xk together with a finite number of Lie brackets of X1, . . . , Xk generate the tangent bundle of the group. The information ofH-type groups can be also applied to the study of pseudoconvex domains in complex analysis, subRiemannian geometry, control theory, and semiclassical analysis of quantum mechanics.

Recently, the real and complex Hamiltonian mechanics has been involved to study sub- Riemannian manifolds [1, 2, 4, 5]. It is well known that the elliptic Laplace-Beltrami operator induces the Riemannian geometry. In the same way the sub-elliptic Laplacian, the sum of squares of vector fields, which jointly with their commutators generate the tangent bundle on the groups, responds the geometry on subRiemannian manifolds. Thus, the study of the Hamiltonian function associated with the sub-Laplacian plays a crucial role.

In this article we consider seven examples ofH-type groups, which are related to the quater- nion numbers. Three of them have a one-dimensional center, and actually, are isomorphic to the H2Heisenberg group. Another three have a two-dimensional center and the last one has a three- dimensional center which is isomorphic to imaginary quaternions. In each case we construct the

Math Subject Classifications.53C17, 53C22, 35H20.

Key Words and Phrases.Quaternion numbers, Carnot-Carathéodory metrics, nilpotent Lie groups, Hamiltonian formalism.

Acknowledgements and Notes.This work was supported by Projects FONDECYT (Chile) # 7040027, #1040333, and #1030373. Part of this article was finished while the authors visited Universidad Técnica Federico Santa María, Valparaíso, Chile, in March 2005, under the grant Project FONDECYT (Chile) #7040027.

©2006 The Journal of Geometric Analysis ISSN 1050-6926

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Hamiltonian function associated with the sub-Laplacian. We solve the Hamiltonian system of differential equations and give exact solutions that describe the geodesics on the groups. We con- sider different positions of points and study a number of geodesics connecting these points. We prove Kepler’s law, that gives a nice geometric interpretation of the length of a part of a horizontal curve belonging to a center of the group and the area of certain surface. We also consider complex geodesics and find a relation between the complex action function and the Carnot-Carathéodory length of geodesics.

2. Definitions

A quaternion is a mathematical concept introduced by William Rowan Hamilton from Ireland in 1843. The idea captured the popular imagination for a time because it involves relatively simple calculations that abandon the commutative law, one of the basic rules of arithmetic. Specifically, a quaternion is a noncommutative extension of the complex numbers. As a vector space over the real numbers, the quaternions have dimension 4, whereas the complex numbers have dimension 2. While the complex numbers which satisfiesi2= −1, the quaternions are obtained by adding the elementsi,j, andkto the real numbers which satisfy the following relations

i2=j2=k2=ijk= −1.

Unlike real or complex numbers, multiplication of quaternions is not commutative, e.g., ij=k, ji= −k, jk=i, kj= −i, ki=j, ik= −j. (2.1) The quaternions are an example of a division ring, an algebraic structure similar to a field except for commutativity of multiplication. In particular, multiplication is still associative and every nonzero element has a unique inverse.

Our discussion will involve a description of the quaternions in different forms. One of them is a combination of a scalar and a vector in analogy with the complex numbers being representable as a sum of real and imaginary parts,a·1+b·i. For a quaternionH=a+bi+cj+dkwe call a scalarathereal partand the three-dimensional vectoru=bi+cj+dkis called theimaginary part.The quaternions can be represented using complex 2×2 matrices

H = a+ib c+id

c+id aib

=aU+bI+cJ+dK ,

where

U = 1 0 0 1

, I = i 0 0 −i

, J = 0 1

−1 0

, K = 0 i i 0

. InR4, the basis of quaternion numbers can be given by real matrices

U =



1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1



, I=



0 1 0 0

−1 0 0 0

0 0 0 1

0 0 −1 0



,

J =



0 0 0 −1

0 0 −1 0

0 1 0 0

1 0 0 0



, K=



0 0 −1 0

0 0 0 1

1 0 0 0

0 −1 0 0



.

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We have

H =



a bdc

b ac d

d c a b

cdb a



=aU+bI+cJ +dK.

Similarly to complex numbers, vectors, and matrices, the addition of two quaternions is equivalent to summing up the elements. SetH =a+uandQ=t+xi+yj+zk=t+v. Then

H+Q=(a+t )+(u+v)=(a+t )+(b+x)i+(c+y)j+(d+z)k.

Addition satisfies all the commutation and association rules of real and complex numbers. The quaternion multiplication (the Grassmanian product) is defined by

H Q=(atu·v)+(av+tu+u×v) ,

whereu·vis the scalar product andu×vis the vector product ofuandv. The multiplication is not commutative because of the noncommutative vector product. The noncommutativity of mul- tiplication has some unexpected consequences, e.g., polynomial equations over the quaternions may have more distinct solutions then the degree of a polynomial. The equationH2+1=0, for instance, has infinitely many quaternion solutionsH =a+bi+cj+dkwithb2+c2+d2=1.

The conjugateof a quaternionH=a+bi+cj+dk, is defined asH=abicjdkand theabsolute valueofHis defined as|H| =√

H H=√

a2+b2+c2+d2.

In this article we will construct 2-step homogeneous groups ofH-type related to quaternion numbers. We will call these groupsquaternionH-type groups, or, simply,quaternion groups.

Let us start with some definitions.

H-type homogeneous groups are simply connected 2-step Lie groupsGwhose algebrasG are graded and carry an inner product such that

(i) Gis the orthogonal direct sum of the generating subspaceV1and the centerV2: G = V1V2, V2= [V1, V1], [V1, V2] =0,

(ii) the homomorphismsJZ:V1V1,ZV2, defined by JZX, X

= Z, X, X

, X, XV1, satisfy the equation

JZ2= −|Z|2I, ZV2.

Here·,·is a positively definite nondegenerating quadratic form onG,,·]is a commutator and I is the identity. These groups are generated by their algebras by exponentiation.

To construct the quaternionH-type groups we take the space of quaternions orR4 asV1 with the corresponding structures. The matricesI,J, andKdefine the homomorphismsJZ. We construct the quaternionH-type groups with centersV2of different dimensions. The following notations

HI,0,0=HI, H0,J,0=HJ, H0,0,K=HK

will stand for the groups with a one-dimensional centerV2. In this case, only one of the matrices I,J, orK, is involved. Essentially, these groups are isomorphic toH2Heisenberg groups which were studied intensively (see, for instance, [2, 3, 12, 13, 17]). The groups

HI,J,0=HIJ, HI,0,K=HIK, H0,J,K=HJ K

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with a two-dimensional centerV2are obtained by making use of two of the matricesI,J, or K. The groupHIJ K (we will denote it simply by H) has a three-dimensional center. The corresponding algebras

HI, HJ, HK, HIJ, HIK, HJ K, H are the two-step algebrasV1V2. The topological dimensions of the groups are

n(HI)=n(HJ)=n(HK)=5, n(HIJ)=n(HJ K)=n(HIK)=6, n(H)=7. The homogeneous dimension defined by the formulaν=dimV1+2 dimV2plays an important role in analysis on homogeneous groups. We see that the homogeneous dimensions are greater than the topological dimensions and equal to

ν(HI)=ν(HJ)=ν(HK)=6, ν(HIJ)=ν(HJ K)=ν(HIK)=8, ν(H)=10. We will give calculations for the groupHIJ K=H. The results for other groups are obtained by vanishing the corresponding index. We set the standard orthonormal systems{X1, X2, X3, X4} ∈V1,{ZI, ZJ, ZK} ∈V2. Then the matricesI,J,Ktransform the basis vectors in the following form

IX1 = −X2, IX2=X1, IX3 = −X4, IX4 =X3, JX1=X4, JX2=X3, JX3= −X2, JX4= −X1

KX1=X3, KX2= −X4, KX3= −X1, KX4=X2.

(2.2)

We use the normal coordinates(x, z)=(x1, . . . , x4, zI, zJ, zK)for the elements exp

4

α=1

xαXα+zIZI+zJZJ +zKZK

HIJ K =H.

If one or two indices of the corresponding algebraHIJ Kis equal to zero, then the corresponding coordinates vanish. For example, we write q = (x, zI, zK)for coordinatesqHIK. The Baker-Campbell-Hausdorff formula

exp(X+Z)exp

X+Z

=exp

X+X, Z+Z+1 2

X, X , forX, XV1,Z, ZV2defines the multiplication law onH. Precisely, we have

Lq

q

=L(x,z)

x, z

=(x, zI, zJ, zK)

x, zI, zJ, zK

=

x+x, zI+zI+1 2

Ix, x

, zJ +zJ +1 2

Jx, x

, zK+zK+1 2

Kx, x , forq =(x, z)andq=(x, z), wherex, x ∈R4and(Ix, x),(Jx, x),(Kx, x)are the usual scalar product of the vectorsIx,Jx,Kx belonging toR4byx ∈R4. This scalar product(·,·) defined onV1G, can be continued up to the quadratic form·,·onG. The multiplication “◦” defines the left translationLqofqby the elementq=(x, z)Hon the groupH.

We associate the Lie algebraHof the groupHwith the set of all left invariant vector fields of the tangent bundleTH. The tangent bundle contains a natural subbundleThHconsisting of

“horizontal” vectors. We callThHthe horizontal bundle. The horizontal bundle is spanned by the left-invariant vector fieldsX1(x, z), . . . ,X4(x, z)withXα(0,0)=XαV1,α=1, . . . ,4

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(see, for example, [16]). In coordinates of the standard Euclidean basis∂x

α,α=1, . . . ,4,∂z

I,

∂zJ,∂z

K, these vector fields are expressed as Xα(x, z)=

∂xα −1 2

(IXα, X)

∂zI +(JXα, X)

∂zJ +(KXα, X)

∂zK

where we setX=4

β=1xβXβ. Explicitly, X1(x, z)=

∂x1+1 2

+x2

∂zIx4

∂zJx3

∂zK

, X2(x, z)=

∂x2+1 2

x1

∂zIx3

∂zJ +x4

∂zK

, X3(x, z)=

∂x3+1 2

+x4

∂zI +x2

∂zJ +x1

∂zK

, X4(x, z)=

∂x4+1 2

x3

∂zI +x1

∂zJx2

∂zK

.

(2.3)

The left invariant vector fieldsZβ(x, z)withZβ(0,0)=ZβV2,β =I,J,K, are simply the vector fields

Zβ(x, z)=

∂zβ .

We write simplyXα andZβ instead ofXα(x, z)andZβ(x, z), if no confusion may arise. Note that in the case of the groupsHI,HJ,andHK, the vector fields (2.3) are reduced to the standard vector fields onH2Heisenberg group. We also use the notationX=(X1, . . . , X4), and call it thehorizontal gradient. The horizontal gradient can be written in the form

X= ∇x+1 2

Ix

∂zI +Jx

∂zJ +Kx

∂zK

, withx =(x1, . . . , x4)and∇x=

∂x1, . . . ,∂x

4

. Any vector fieldY belonging toThHis called thehorizontal vector field.In particular, the horizontal gradientXis a horizontal vector field.

The commutation relations are as follows:

[X1, X2] = −ZI, [X1, X3] =ZK, [X1, X4] =ZJ , [X2, X3] =ZJ, [X2, X4] = −ZK, [X3, X4] = −ZI.

A basis of one-forms dual toX1, . . . , X4, ZI, ZJ, ZKis given bydx1, . . . , dx4,ϑI,ϑJ, ϑKwith

ϑI =dzI−1

2(+x2dx1x1dx2+x4dx3x3dx4)=dzI−1 2

Ix, dx , ϑJ =dzJ −1

2(x4dx1x3dx2+x2dx3+x1dx4)=dzJ −1 2

Jx, dx , ϑK=dzK−1

2(x3dx1+x4dx2+x1dx3x2dx4)=dzK−1 2

Kx, dx .

(2.4)

Since the exterior productϑβ(Xα)vanishes for allβ = I,J,K,α = 1, . . . ,4, we have the productϑβ(Y )vanishing on all horizontal vector fieldsY.

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3. Horizontal curves and their geometric characteristics

The geometry of subRiemannian manifolds, examples of which are H-type quaternion groups, is quite different from Riemannian manifolds. The definitions and basic notations of subRiemanian geometry can be found in, e.g., [18]. The velocity and the distance should respect the horizontal bundleThH. Since [Xαi, Xαj] ∈/ ThHfor i, j = 1, . . . ,4, hence the horizon- tal bundle is not integrable, i.e., there is no surface locally tangent to it. A continuous map c(s): [0,1] →His called a curve. We say that a curvec(s)ishorizontalif its tangent vector

˙

c(s)belongs toThHat each pointc(s). In other words, there are (measurable) functionsaα(s) such thatc(s)˙ =4

α=1aα(s)Xα(c(s)). We present some simple propositions that describe the geometry ofH-type quaternion groups.

Proposition 3.1.

A curvec(s)=(x1(s), . . . , x4(s), zI(s), zJ(s), zK(s))is horizontal if and only if

˙ zI =1

2(+x2x˙1x1x˙2+x4x˙3x3x˙4)= 1

2(Ix,x) ,˙

˙ zJ =1

2(x4x˙1x3x˙2+x2x˙3+x1x˙4)= 1

2(Jx,x) ,˙

˙ zK=1

2(x3x˙1+x4x˙2+x1x˙3x2x˙4)= 1

2(Kx,x) ,˙

(3.1)

wherex˙ =(x˙1, . . . ,x˙4).

Proof. We can write the tangent vectorc(s)˙ in the form

˙ c(s)=

˙

x1(s), . . . ,x˙4(s),z˙I(s),z˙J(s),z˙K(s)

= 4 α=1

˙ xα(s)

∂xα + ˙zI(s)

∂zI + ˙zJ(s)

∂zJ + ˙zK(s)

∂zK

=

˙

x(s),x+1 2

Ix

∂zI +Jx

∂zJ +Kx

∂zK

−1 2

x(s),˙ Ix

∂zI +Jx

∂zJ +Kx

∂zK + ˙zI(s)

∂zI + ˙zJ(s)

∂zJ + ˙zK(s)

∂zK

=

˙ x(s), X

+

˙

zI(s)−1 2(Ix,x)˙

∂zI +

˙

zJ(s)−1 2(Jx,x)˙

∂zJ +

˙

zK(s)−1

2(Kx,x)˙

∂zK . It is clear thatc(s)˙ is horizontal if and only if the coefficients in front of∂z

I,∂z

J, and∂z

Kvanish.

This proves Proposition 3.1.

Corollary 3.2.

If a curvec(s)=(x1(s), . . . , x4(s), zI(s), zJ(s), zK(s))is horizontal, then

˙ c(s)=

˙ x(s), X

= 4 α=1

˙

xα(s)Xα .

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Proposition 3.3.

Left translationLqof a horizontal curvec(s)is a horizontal curvec(s)˜ = Lq(c(s))with the velocity

˙˜

c(s)=(Lq)c(s)˙ = 4 α=1

˙ cα(s)Xα

˜ c(s)

=

˙

cα(s), X

˜ c(s)

. (3.2)

Proof. Since the left translationLq(p) by an elementqHis a conformal mapping, it preserves the horizontality [14, 15]. Let us show this by the direct calculation. Let c(s) = (x1(s), . . . , x4(s), zI(s), zJ(s), zK(s))be a horizontal curve and

˜ c(s)=

˜

x1(s), . . . ,x˜4(s),z˜I(s),z˜J(s),z˜K(s)

=Lq(c(s)) be its left translation by an elementqH. Let

q=(p1, . . . , p4, wI, wJ, wK)=(p, wI, wJ, wK) . Sincex˜α(s)=pα+xα(s), we have

˙˜

xα(s)= ˙xα(s) for α=1,2,3,4. (3.3) Differentiatingz˜I(s)=wI+zI(s)+21(Ip, x(s)), making use of the horizontality condition (3.1) forc(s), and (3.3), we deduce

˙˜

zI(s)= ˙zI(s)+1

2(Ip,x(s))˙ =1

2(Ix(s),x(s))˙ +1

2(Ip,x(s))˙

= 1

2(I(p+x(s)),x(s))˙ =1 2

Ix(s),˜ x(s)˙˜

.

(3.4)

Similarly, we obtain

˙˜

zJ(s)= 1 2

Jx(s),˜ x(s)˙˜

, z˙˜K(s)= 1 2

Kx(s),˜ x(s)˙˜

, (3.5)

from what it follows that the curvec(s)˜ is horizontal.

Let us show (3.2). Sincec(s)˜ is horizontal, we get

˙˜

c(s)= 4 α=1

˙˜

cα(s)Xα

˜ c(s)

= 4 α=1

˙ cα(s)Xα

˜ c(s)

= 4 α=1

˙

cα(s)(Lq)Xα(c(s))

=(Lq) 4

α=1

˙

cα(s)Xα(c(s))

=(Lq)c(s) .˙

This completes the proof of the proposition.

The next properties of the matricesI,J,Kare obvious:

(i) I2=J2=K2= −U2,

(ii) IJ = −J I=K, J K= −KJ =I, KI= −IK=J,

(iii) I1= −I,J1= −J,K1 = −K, whereI1,J1,K1are the inverse matrices ofI,J,K, respectively.

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(iv) IT = −I,JT = −J,KT = −K, whereIT,JT,KT are the transpose matrices of I,J,K, respectively.

(v) (Ix, x)=0,(Jx, x)=0,(Kx, x)=0 for anyx ∈R4.

Let us study the osculator plane. Letc(s)be a curve. Theosculatorplane atc(s)is defined asT =span{˙c(s),c(s)¨ }.

Proposition 3.4.

A curvec(s)is horizontal if and only if the osculator planeT atc(s)belongs to the horizontal spaceThHc(s)at the pointc(s).

Proof. Obviously, ifT =span{˙c(s),c(s)¨ } ∈ThHc(s), thenc(s)˙ ∈ThHc(s), and the curvec(s) is horizontal.

Letc(s)be a horizontal curve. Thenc(s)˙ ∈ ThHc(s). Let us show thatc(s)¨ ∈ ThHc(s). Differentiating equalities (3.1) of horizontality condition and making use of the property (v), we deduce that

¨

zI(s)=1 2

Ix(s),˙ x(s)˙ +1

2

Ix(s),x(s)¨

= 1 2

Ix(s),x(s)¨ ,

¨

zJ(s)=1 2

Jx(s),˙ x(s)˙ +1

2

Jx(s),x(s)¨

=1 2

Jx(s),x(s)¨ ,

¨

zK(s)=1 2

Kx(s),˙ x(s)˙ +1

2

Kx(s),x(s)¨

= 1 2

Kx(s),x(s)¨ . The acceleration vector alongc(s)is

¨ c(s)=

4 α=1

¨ xα(s)

∂xα + ¨zI(s)

∂zI + ¨zJ(s)

∂zJ + ¨zK(s)

∂zK

=

¨ x(s),x

+

¨ z(s),z

=

¨

x(s),x+1 2

Ix(s)

∂zI +Jx(s)

∂zJ +Kx(s)

∂zK +

¨

zI(s)−1 2

x(s),¨ Ix(s)

∂zI +

¨

zJ(s)−1 2

x(s),¨ Jx(s)

∂zJ +

¨

zK(s)−1 2

x(s),¨ Kx(s)

∂zK

=

¨

x(s), X(c(s))

= 4 α=1

¨

xα(s)Xα.

This means that the vectorc(s)¨ is horizontal. The proposition is proved.

Let us use the hyperspherical coordinates(r, θ, ξ1, ξ2)to obtain a geometric interpretation ofz-coordinates. To do this we introduce the complex coordinates(ω1, ω2)∈C2=R4by

ω1=x1+ix3=re1cosθ

2 and ω2=x2+ix4=re2sinθ 2 .

Herer ∈ [0,∞),θ ∈ [0, π],ξ1, ξ2∈ [0,2π]. Note that for any fixed valuesrandθ, the pair 1, ξ2)parameterizes a two-dimensional torus, except for the degenerate casesθ=0 orθ=π, that describe a circle. The round metric on the three-dimensional unite sphere in these coordinates is given by

dlx= 1

42+cos2 θ

2

12+sin2 θ

2

22ds .

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The volume form on a 3-sphere is dvx= r4

16sinθ dθ12. The horizontality condition (3.1) has the form

˙ zI= 1

2

Reω2ω˙1−Reω1ω˙2 ,

˙ zJ = 1

2

−Imω2ω˙1−Imω1ω˙2

,

˙ zK= 1

2

−Imω1ω˙1+Imω2ω˙2

,

(3.6)

in coordinates1, ω2). We calculate ω2ω˙1= rr˙

2ei(ξ2ξ1)sinθir2ξ˙1

2 ei(ξ2ξ1)sinθr2θ˙

2 ei(ξ2ξ1)sin2θ 2 , ω1ω˙2= rr˙

2ei(ξ2+ξ1)sinθir2ξ˙2

2 ei(ξ2+ξ1)sinθ+r2θ˙

2 ei(ξ2+ξ1)cos2θ 2 , ω1ω˙1=rr˙cos2θ

2 −ir2ξ˙1cos2θ 2 +r2θ˙

4 sinθ , ω2ω˙2=rr˙sin2θ

2 −ir2ξ˙2sin2θ 2 +r2θ˙

4 sinθ . The conditions (3.6) become

˙ zI = r2

4

ξ˙1+ ˙ξ2

sin(ξ2ξ1)sinθ− ˙θcos(ξ2ξ1)

,

˙ zJ = r2

4

ξ˙1+ ˙ξ2

cos(ξ2ξ1)sinθ+ ˙θsin(ξ2ξ1)

,

˙ zK= r2

4

2ξ˙1cos2θ

2 −2ξ˙2sin2θ 2

.

(3.7)

Then,

˙

z2I+ ˙zJ2 + ˙z2K=r4 4

θ˙2 4 +cos2

θ 2

ξ˙12+sin2 θ

2 ξ˙22

. This means that the element of lengthdz=

˙

z2I+ ˙zJ2 + ˙z2Kdsofz-components is equal to the round metric on the three-dimensional sphere multiplied byr22.

We rewrite conditions (3.7) in the differential form as dzI= r2

4

1+2

sin(ξ2ξ1)sinθcos(ξ2ξ1)

, dzJ = r2

4

1+2

cos(ξ2ξ1)sinθ+sin(ξ2ξ1)

, dzK= r2

4

12+(dξ1+2)cosθ ,

(3.8)

and calculate the element of volume inz-components asdvz=dzIdzJdzK, that yields dvz=dzIdzJdzK=r6

43(2 sinθ ) dθ12=r2 2 dvx.

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Let r = r(θ (s), ξ1(s), ξ2(s)) be the equation in the hyperspherical coordinates of the projection of a horizontal curve on the x-space. Let us consider the surface swept by the radius-vector if a point runs along the projection r. The area dA of an infinitesimal trian- gle lying on this surface with the vertices at the origin, at

r(s0), θ (s0), ξ1(s0), ξ2(s0) , and at r(s0+ds), θ (s0+ds), ξ1(s0+ds), ξ2(s0+ds)

is approximately equal to r22dlx. Integrat- ing, we obtain the area of a conic surface swept by the vectorial radius between the initial point r(s0), θ (s0), ξ1(s0), ξ2(s0)

and a point

r(s), θ (s), ξ1(s), ξ2(s) : A= 1

2 s

s0

r2dlx.

We see thatdlz= r22 dlx =dA. Integrating this equality we can say that the part of the length of a horizontal curvelzinz-space is equal to the area of a conic surface swept by the radius-vector of the projection of this curve onto thex-space.

The rate of the area changeA˙is defined asA˙= dAds. We proved the following theorem which can be considered as a generalization of Kepler’s law.

Theorem 3.5.

If a curvecis horizontal, then the rate of change of itsz-components is equal to the absolute value of the rate of the area change| ˙A|, i.e.,z| = | ˙A|.

Theorem 3.6.

A smooth curvec(s)is horizontal with constantz-coordinates if and only if c(s)=(a1s, . . . , a4s, z1, z2, z3)witha1, . . . , a4∈Randa12+. . .+a42=0.

Proof. Letc(s)be a horizontal curve with constantz-componentszI =z1,zJ =z2,zK=z3. Then the equation|˙z|2=0= r44θ˙2

4 +cos2θ

2

ξ˙12+sin2θ

2

ξ˙22 implies θ˙=0, cos

θ 2

ξ˙1=0, and sin θ

2

ξ˙2=0.

From the first equation we conclude thatθ =θ0 is constant. In the caseθ0 = π+2π k, and θ0=2π n,k, n∈Z, we see thatξ1 =ξ10, andξ2=ξ20. The parameterization of the curvec(s) has the form

c(s)=

scosξ10cos θ0

2

, scosξ20sin θ0

2

, ssinξ10cos θ0

2

, ssinξ20sin θ0

2 b

, z1, z2, z3

. Ifθ0=π+2π k,k∈Z, then

c(s)=

0,±scosξ20,0,±ssinξ20, z1, z2, z3

by the formulas of hyperspherical coordinates. Ifθ0=2π n,n∈Z, then

c(s)=

±scosξ10,0,±ssinξ10,0, z1, z2, z3 .

Now, let us assume thatc(s)=(a1s, . . . , a4s, z1, z2, z3)with constantz-components. Set as = (a1s, . . . , a4s). Observe that (Ia, a) = (Ja, a) = (Ka, a) = 0 for any vectora = (a1, . . . , a4). Then,

˙

zI =0=1 2

I(as),(as)˙

= s

2(Ia, a), z˙J =0=1 2

J(as),(as)˙

= s

2(Ja, a) ,

˙

zK=0=1 2

K(as),(as)˙

= s

2(Ka, a) .

The horizontal condition (3.1) holds for all threez-components.

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4. Hamiltonian formalism

In this section we study the geometry of quaternionH-type groups making use of the Hamil- tonian formalism. The geometry of the group is induced by the sub-Laplacianh =4

α=1Xα2. Since the vector fieldsX1, . . . , X4satisfying the Chow’s condition, by a theorem of Hörman- der [10], the operatorhis hypoelliptic. Explicitly,

4 α=1

Xα2= 2

∂x12 + 2

∂x22 + 2

∂x32+ 2

∂x24 +1

4

x12+x22+x32+x42 2

∂z2I + 2

∂zJ2 + 2

∂z2K +

x2

∂x1x1

∂x2 +x4

∂x3x3

∂x4

∂zI +

x4

∂x1x3

∂x2+x2

∂x3 +x1

∂x4

∂zJ +

x3

∂x1+x4

∂x2+x1

∂x3x2

∂x4

∂zK

=x+1

4|x|2z+ Ix,x

∂zI + Jx,x

∂zJ + Kx,x

∂zK where∇x=(∂x

1, . . . ,∂x

4),x=4

α=1

2

∂x2α,z= ∂z22 I +∂z22

J +∂z22 K. The associated Hamiltonian functionH (ξ, θ, x, z)is of the form

H (ξ, θ, x, z)= 4 α=1

ξα−1

2

θII+θJJ +θKK Xα, x

2

= |ξ|2+1

4|x|2|θ|2+

θII+θJJ +θKK x, ξ

,

(4.1)

whereξα= ∂xα,α=1, . . . ,4,θ=I, θJ, θK)=

∂zI,∂z

J,∂z

K

, and

θII+θJJ +θKK=



0 θIθKθJ

θI 0 −θJ θK

θK θJ 0 θI

θJθKθI 0



.

For simplicity, we introduce the notationM=θII+θJJ +θKK. The corresponding Hamil- tonian system is







































˙

x =∂H

∂ξ =2ξ+Mx

˙

zI = ∂H

∂θI = 1

2|x|2θI+ Ix, ξ

˙

zJ = ∂H

∂θJ = 1

2|x|2θJ + Jx, ξ

˙

zK = ∂H

∂θK = 1

2|x|2θK+ Kx, ξ ξ˙ = −∂H

∂x = −1

2|θ|2x+ θ˙ = −∂H

∂z =0.

(4.2)

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The solutionsγ (s)=(x(s), z(s), ξ(s), θ (s))of the system (4.2) are calledbicharacteristics.

Definition 4.1. LetP (x0, z0),Q(x1, z1)H. A geodesic betweenP andQis the projection of a bicharacteristicγ (s),s∈ [0, τ], onto the(x, z)-space, that satisfies the boundary conditions

x(0), z(0)

=(x0, z0),

x(τ ), z(τ )

= xτ, zτ

.

Lemma 4.2.

Any geodesic is a horizontal curve.

Proof. Letc(s)=(x1(s), . . . , x4(s), zI(s), zJ(s), zK(s))be a geodesic. The second equa- tion of the system (4.2) implies

˙ zI =θI

2 |x|2+1 2

Ix,

= θI

2 |x|2+1 2

Ix,x˙ +1

2

Ix,2ξ− ˙x

. (4.3)

Substituting the first equation of (4.2) in the last term of (4.3) we obtain 1

2

Ix,2ξ − ˙x

= −θI

2 (Ix,Ix)θJ

2 (Ix,Jx)θK

2 (Ix,Kx)

= −θI

2 |x|2+θJ

2 (J Ix, x)+θK

2 (KIx, x)= −θI 2 |x|2.

(4.4)

Here we use the properties (iv) and (v) of matricesI,J,K. Combining (4.3) and (4.4) we deduce

˙ zI= θI

2 |x|2+1 2

Ix,

=1 2

Ix,x˙ . Similar calculations show that

˙ zJ = θJ

2 |x|2+1 2

Jx,

= 1 2

Jx,x˙

, z˙K=θK

2 |x|2+1 2

Kx,

= 1 2

Kx,x˙ . Therefore,c(s)is a horizontal curve by Proposition 3.1.

Properties (i) and (ii) of the matricesI,J,Kgive us the following identities for the matrix M

M2= −|θ|2, M3= −|θ|2M, M4= |θ|4, M5= |θ|4M, . . . . (4.5)

Lemma 4.3.

The exponentexp

2sM

is a rotation by the angle4s|θ|about the unit vector (0,θ|θI|,θ|θJ|,θ|θK|)in thex-space.

Proof. We observe that exp

2sM)= n=0

(2s)n

n! Mn=U k=0

(2s|θ|)4k (4k)! +M

|θ| k=0

(2s|θ|)4k+1 (4k+1)!

U

k=0

(2s|θ|)4k+2 (4k+2)! −M

|θ| k=0

(2s|θ|)4k+3 (4k+3)! by (4.5). We conclude that the matricesMand exp

2sM

commute. Note that

k=0

(2s|θ|)4k (4k)! −

k=0

(2s|θ|)4k+2

(4k+2)! =cos(2s|θ|)

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