Geodesics with Constraints on Heisenberg Manifolds 1 Introduction

Geodesics with constraints on Heisenberg manifolds

Ovidiu Calin and Vittorio Mangione

Abstract We provide a qualitative description for the solutions of Euler- Lagrange equations associated to Lagrangians with linear and quadratic constraints. An important role is played by the natural metric induced by the Heisenberg manifold. In the second part we arrive at a formula which is the analog for the Gauss’ formula for the Heisenberg group.

Key words: Ricci, Riemann, curvature, geodesics, Levi-Civita connection MS Classiﬁcation: (2000) Principal: 37J60; Secondary: 53B21, 70H03

1 Introduction

Riemannian geometry and elliptic operators are closely related. The most simple Riemannian metric is the ﬂat one, and the associated elliptic operator is the usual Laplacian, which is invariant under translations on the additive group (Rn, +). In a similar way, in the sub-Riemannian geometry the associated operator is sub-elliptic (see [BGG1], [BGG2], [BGG3], [BGG4], [Ga1]). Heisenberg group is the prototype model for the sub-Riemannian geometry. It is the simplest noncomutative Lie group. The vector ﬁelds

X1 = ∂x + 2y∂z,X2 = ∂y − 2x∂z (1.1)

are left invariant under the Heisenberg group law on R3 and hence the Heisenberg Laplacian operator 1 ∆ = (X2 + X2), (1.2) H 2 1 2 which is sub-elliptic. For a general survey see [Ho]. A more detailed analysis involving the relationship with equations can be found in [Tr]. The concept of geodesic from Riemannian geometry can be extended in the sub- Riemannian case as the shortest curve between two points subject to some nonholonomic constraints (constraints on the velocity vector). For a study of geodesics and cut locus see [Str]. For a detailed study of sub-Riemannian geodesics on Heisenberg group see [BGG5] and [Ga1]. For a general study of sub-Riemannian manifolds, see [Be]. For variational calculus on Heisenberg manifolds, see [CM]. An important application of the Heisenberg group is the non-commutative geometric analysis approach to clinical MRI (see [Sch]). This is based on the temporal magnetic resonance phenomenon discovered by Bloch (see [Fu]). He detected an induction field of nuclear origin, which indicates a rotation of the total oscillating field around a constant magnetic field. The static magnetic field is described by the missing direction ∂z. This is given by the Lie bracket of the vector fields X1 and X2. The nuclear field acts at resonance over the Larmour orbits. The Larmour equation for the magnetic spin precession describes the geodesics on the Heisenberg group. The vector fields X1,X2 and ∂z span the Lie algebra of the Heisenberg group. The cannonical commutations relations, specific to Quantum Mechanics are

[X1,X2] = −4∂z, [X2, ∂z] = 0. Returning to geometry, we note that the Heisenberg group serves also as a contact manifold. This is the odd dimension analog for a symplectic structure. The 1-form ω = dz − 2ydx + 2xdy, which anihilates X1 and X2 is a contact form, i.e. ω ∧ dω never vanishes. The 2-form Ω = dω = dx ∧ dy is a symplectic form on the distribution generated by X1 and X2. It can be considered also as a magnetic field. Then the Maxwell’s equation can be written as dΩ = 0. More general, a sub-Riemannian geometry can be defined on R3 by the vector fields

X1 = ∂x + A1(y)∂t,X2 = ∂y − A2(x)∂t (1.3) without the condition of having an underlying Lie group structure The vector fields need to satisfy the bracket condition. This is, the vector fields X1 and 3 X2 together with the bracket [X1,X2] span the tangent space of R at each point. This is a natural generalization of the Heisenberg group and is called Heisenberg manifold structure. For calculus on Heisenberg manifolds see [BG]. In the present paper we use the Heisenberg manifold structure to provide a qualitative characterization for geodesics under nonholonomic constraints. For a study of nonholonomic systems, see [BC]. If ϕ : [0, 1] → R3 is the trajectory of a particle of mass m = 1, its energy is given by Z 1 1 |ϕ˙(s)|2 ds, (1.4) 0 2 where ϕ(s) = (x(s), y(s), z(s)) and |ϕ˙(s)|2 = g(ϕ, ˙ ϕ˙), where g is a Rieman- nian metric which will be specified later. In the first part we shall consider the dynamics of a charged particle in a constant magnetic field Ω = dω. In other words, we study the dynamics described by the Lagrangian 1 L = |ϕ˙|2 − λ ω(ϕ ˙)2, (1.5) 2 where λ is a constant, which has the physical significance of charge. This Lagrangian is the difference between a kinetic energy and the square of a magnetic potential. Hence we deal with a variational problem with nonholonomic constraints. In general, the extremizers for such Lagrangians are hard to characterize because of the constraints. The Euler-Lagrange equations are even harder to solve. The idea in this case is to give a qualitative description of the solutions. The goal of this paper is to provide an example of a Riemannian metric g such that the solutions of the Euler-Lagrange equations associated to the Lagrangian (1.5) are geodesics in a certain metric. This metric will be constructed using the structure of the Heisenberg group on R3. In the second part of the paper we consider the more general problem where the Lagrangian has also a nonholonomic linear term 1¯ ¯ L = ¯ϕ˙(s)¯2 − λ ω(ϕ ˙(s))2 + ξ(φ˙)ω(φ˙), 2 h(λ) where ξ is a 1-form which will be defined later. This term is chosen such that the extremizers are geodesics in a certain metric (see Theorem 2.9).

The Heisenberg group Consider the following group law on R3 (x, y, z) ◦ (x0, y0, z0) = (x + x0, y + y0, z + z0 + 2yx0 − 2xy0). (1.6) The pair (R3, ◦) is a Lie group denoted by H1. The vector ﬁelds

X1 = ∂x + 2y∂z,X2 = ∂y − 2x∂z,T = ∂z (1.7) span the Lie algebra of the Heisenberg group H1. We can check easily that

ω(X1) = ω(X2) = 0, ω(T ) = 1 and [X1,X2] = −4T, (1.8) so that the vector fields X1 and X2 span a 2-dimensional nonintegrable distribution H = ker ω called the horizontal distribution. A curve ϕ = (x, y, z) is called horizontal if ω(ϕ ˙) = 0, or z˙ − 2yx˙ + 2xy˙ = 0. (1.9) The vector fields (1.17) define a unique Riemannian metric h such that h(Xi,Xj) = δij, h(Xi,T ) = 0, h(T,T ) = λ. The coefficients are given by   1 + 4λy2 −4λxy −2λy (λ)  2  hij = −4λxy 1 + 4λx 2λx . (1.10) −2λy 2λx λ To avoid confusion we shall denote the above metric by h(λ). This is the metric in which the energy (1.4) is computed.

2 Main results

2.1 Heisenberg group case We shall construct the Euler-Lagrange equation for the Lagrangian (1.5) in the Levi-Civita connection form.

(λ) Lemma 2.1 If Rij are the components of the Ricci tensor with respect to (λ) 1 the metric hij on H , then

(λ) (−λ) Rij = 8λhij . (2.11) Proof: A computation shows that the non-zero components of the 4-covariant Riemann tensor of curvature are 2 2 R2323 = 4λ ,R1213 = 8λ x, 2 2 R1223 = 8λ y, R1313 = 4λ ,

2 2 2 R1212 = −12λ + 16λ (x + y ). kl Using the contraction formula Rij = Rijklh , the non-zero components of the Ricci tensor are 2 2 2 2 R11 = 8λ − 32λ y ,R12 = 32λ xy, R13 = 16λ y, 2 2 2 2 R22 = 8λ − 32λ x ,R23 = −16λ x, R33 = −8λ . Using (1.10) where we swap λ with −λ, we get the equation (2.11) on components. ¤

Corollary 2.2 If ϕ is a horizontal curve, then

R(λ)(ϕ, ˙ ϕ˙) = 8λ|ϕ˙|2, where |ϕ˙|2 = hλ(ϕ, ˙ ϕ˙) does not depend on λ. Proof: Ifϕ ˙ is horizontal,ϕ ˙ =xX ˙ 1 +yX ˙ 2. In this case h(−λ)(ϕ, ˙ ϕ˙) = h(λ)(ϕ, ˙ ϕ˙) =x ˙ 2 +y ˙2, and apply Lemma 2.1. ¤ The next proposition is a generalization of the previous Corollary to any vector ﬁeld.

Proposition 2.3 For any vector ﬁeld V we have ¡ ¢ R(λ)(V,V ) = 8λ h(λ)(V,V ) − 2λω(V )2 . (2.12)

Proof: Using (1.10) we can write

(−λ) (λ) hij = hij + 2λKij, (2.13) where   −4y2 4xy 2y 2 Kij =  4xy −4x −2x . (2.14) 2y −2x −1 A straightforward computation shows that

K(V,V ) = −ω(V )2, (2.15) and (2.13) becomes

h(−λ)(V,V ) = h(λ)(V,V ) − 2λω(V )2. (2.16)

Using Lemma 2.1 we complete the proof. ¤

Corollary 2.4 The actions Z Z 1 R(λ)(ϕ ˙(s), ϕ˙(s)) ds and |ϕ˙(s)|2 − λ ω(ϕ ˙(s))2 ds, 2 both with respect to the metric h(λ), reach the extrema for the same functions ϕ : [0, 1] → R3. In particular, the extrema will be geodesics in the metric with (λ) coeﬃcients Rij . It is interesting that, even if the Lagrangian (1.5) has a nonholonomic con- straint, the minimizers still behave as geodesics in a certain metric. This is given in the following result. Theorem 2.5 The Euler-Lagrange equation for the Lagrangian (1.5) is

∇ϕ˙ ϕ˙ = 0, (2.17)

k where ∇∂i ∂j = Γij∂k, with the coeﬃcients given by Ã ! k 1 ks ∂Ris ∂Rjs ∂Rij Γij = R + − . (2.18) 2 ∂xj ∂xi ∂xs

2.2 A more general case We have investigated the case when the vector ﬁelds are given by the formula (1.7). We shall deal in this section with the more general case of the vector ﬁelds X1 = ∂x + A1(y)∂z,X2 = ∂y − A2(x)∂z, (2.19) with Ai(·) smooth functions. The 1-form ω in this case is

ω = dz − A1(y)dx + A2(x)dy. (2.20)

One may check that ω(X1) = ω(X2) = 0. Another important 1-form is

00 00 η = A2dy − A1dx. (2.21) √ A computation shows the vector ﬁelds X1, X2 and ∂t/ λ are orthonormal in the Riemannian metric

 2  1 + λA1(y) −λA1(y)A2(x) −λA1(y) (λ)  2  hij = −λA1(y)A2(x) 1 + λA2(x) λA2(x) . (2.22) −λA1(y) λA2(x) λ

We shall consider the following Lagrangian with a quadratic potential con- straint 1 L = |φ˙|2 − λ[ω(φ˙)2 + ω(φ˙)η(φ˙)]. (2.23) 2 hλ

When A1(x2) = 2x2 and A2(x1) = 2x1, we get the Lagrangian in (1.5). In this case η = 0. The following result is a generalization of Lemma 2.1. We shall denote by 0 0 Ai the derivative with respect to the current variable, i.e. A1 = dA1/dy and 0 A2 = dA2/dx. (λ) Lemma 2.6 If Rij are the components of the Ricci tensor with respect to (λ) the metric hij given in (2.22), then λ R(λ) = R h(−λ) + Q , (2.24) ij ij 2 ij where h(−λ) is obtained by ﬂipping the sign in (2.22),

 00 00 00 00 2A1A1 −(A1A2 + A1A2) −A1  00 00 00 00  Qij = −(A1A2 + A1A2) 2A2A2 A2 , (2.25) 00 00 −A1 A2 0 and R is the Ricci scalar.

Proof: It is just a veriﬁcation taking into account that the Ricci tensor with respect to the metric hλ has the following non-zero components 1 1 1 1 R = λA 0A 0+ λ(A 0)2+λA A 00+ λ(A 0)2− λ2A2(A 0)2−A2λ2A 0A 0− λ2A2(A 0)2, 11 1 2 2 1 1 1 2 2 2 1 1 1 1 2 2 1 2 1 1 1 1 R = − λA 00A + A λ2A A 02+λ2A A 0A A0 + A A λ2A 02− λA A 00, 12 2 1 2 2 2 1 1 1 1 2 2 2 2 1 2 2 1 2 1 1 1 R = − λA 00 + λ2A A 02 + λ2A A 0A 0 + A λ2A 02, 13 2 1 2 1 1 1 1 2 2 1 2 1 1 1 1 R = λA 0A 0+ λA 02+ λA 02+λA A00− λ2A2A 02−A2λ2A 0A 0− λ2A2A 02, 22 1 2 2 1 2 2 2 2 2 2 2 2 2 1 2 2 1 1 1 1 R = λA 00 − λ2A A 02 − λ2A A 0A 0 − A λ2A 02, 23 2 2 2 2 2 2 2 1 2 2 1 1 R = − λ2(A 0 + A 0)2, 33 2 2 1 and the Ricci scalar is λ R = (A 0 + A 0)2. (2.26) 2 1 2 ¤

Lemma 2.7 Consider the curve φ(s) = (x(s), y(s), z(s)). Then

³ ´³ ´ ˙ ˙ 00 00 ˙ ˙ Q(φ, φ) = 2 A2y˙(s) − A1x˙(s) z˙(s) − A1x˙(s) + A2y˙(s) = 2η(φ) ω(φ). (2.27) Proof: Let φ˙(s) =x ˙(s)X1 +y ˙(s)X2 +z ˙(s)∂t be the tangent vector ﬁeld. Then a straightforward computation provides ˙ ˙ 00 2 00 2 00 00 00 00 Q(φ, φ) = 2A1A1x˙ + 2A2A2y˙ − 2(A1A2 + A1A2)x ˙y˙ − 2A1x˙z˙ + 2A2y˙z˙ 00 00 00 00 = (A2y˙ − A1x˙)(2A2y˙ − 2A1x˙) +z ˙(2A2y˙ − 2A1x˙) 00 00 = (2A2y˙ − 2A1x˙)(A2y˙ − A1x˙ +z ˙). ¤ We note that η and ω measures the departure from the Heisenberg structure and horizontality respectively. When the structure is Heisenberg, η = 0 and when V is horizontal, ω(V ) = 0. In the following we give a global interpretation for the diﬀerence h(−λ) −h(λ) in terms of the horizontal distribution H = ker ω.

Lemma 2.8 (−λ) (λ) hij − hij = 2λKij, (2.28) with  2  −A1(y) A1(y)A2(x) A1(y)  2  Kij = A1(y)A2(x) −A2(x) −A2(x) . (2.29) A1(y) −A2(x) −1 Furthermore, K(V,V ) = −ω(V )2. (2.30)

Proof: Relation (2.28) is obtained by direct computation. For the second part we have

2 2 K(V,V ) = −(V3 − A1V1 + A2V2) = −ω(V ) . ¤ 0 0 (λ) 2 2 Denote ρ = A1 + A2, and let gij = Rij /(λρ ). Denote also ξ = η/ρ .

Theorem 2.9 The actions Z Z 1¯ ¯ g(ϕ ˙(s), ϕ˙(s)) ds and ¯ϕ˙(s)¯ − λ ω(ϕ ˙(s))2 + ξ(φ˙)ω(φ˙) ds, 2 h(λ) reach the extrema for the same functions ϕ : [0, 1] → R3. In particular, the extrema will be geodesics in the metric gij and obeys the ˙ equation ∇φ˙ φ = 0, where ∇ is the Levi-Civita type connection deﬁned by gij. Proof: From Lemma 2.6 and Lemma 2.7 we have

R(λ)(V,V ) = R h(−λ)(V,V ) + λη(V )ω(V ).

Using Lemma 2.8 we get ³ ´ R(λ)(V,V ) = R h(λ)(V,V ) + 2λK(V,V ) + λη(V )ω(V ).

From (2.26), R = λρ2/2. Hence

R(λ)(V,V ) 1 η(V ) = h(λ)(V,V ) + λK(V,V ) + ω(V ). λρ2 2 ρ2

Using (2.31) the last equation can be written also as 1 g(V,V ) = |V |2 − λω(V )2 + ξ(V )ω(V ), 2 h(λ) which completes the proof. ¤

3 Natural Levi-Civita connection on Heisenberg group

We start with the properties of the Levi-Civita connection on H1. For each metric hλ one has a natural Levi-Civita connection ∇λ deﬁned by

∇λ ∂ = Γk (λ)∂ , (3.31) ∂i j ij k where the Christoﬀel symbols are deﬁned by the metric (1.10). They depend linearly on λ and are given by 1 1 1 Γ12 = 4λx2 Γ22 = −8λx1 Γ23 = −2λ 2 2 2 Γ11 = −8λx2 Γ12 = 4λx1 Γ13 = 2λ 3 3 2 2 3 Γ11 = 16λx1x2 Γ12 = 8λ(x2 − x1)Γ13 = −4λx1. 3 3 Γ22 = −16λx1x2 Γ23 = −4λx2 The following result states that the Levi-Civita connection with respect to

∂x1 , ∂x2 , ∂t is always a linear combination of X1 and X2 and hence belongs to the distribution generated by this vectors. Lemma 3.1 For every λ > 0 we have

λ λ λ ∇ ∂x = ∇ ∂x = 4λ (x2X1 + x1X2) , ∇ ∂t = 0, (3.32) ∂x1 2 ∂x2 1 ∂t

λ λ λ ∇ ∂t = ∇ ∂x = 2λX2 , ∇ ∂x = −8λx2 X2, (3.33) ∂x1 ∂t 1 ∂x1 1 λ λ λ ∇ ∂t = ∇ ∂x = −2λX1 , ∇ ∂x = −8λx1 X1. (3.34) ∂x2 ∂t 2 ∂x2 2 Proof: It is just a computation using linear algebra and the expressions for the Christoﬀel symbols given before.

λ ∇ ∂x = 4λx2(∂x + 2x2∂t) + 4λx1(∂x − 2x1∂t) = 4λ(x2X1 + x1X2), ∂x1 2 1 2

λ ∇ ∂t = 2λ(∂x − 2x1∂t) = 2λX2, ∂x1 2 λ ∇ ∂x = −8λx2(∂x − 2x1∂t) = −8λx2X2, ∂x1 1 2 λ ∇ ∂t = −2λ∂x − 4λx2∂t = −2λX1, ∂x2 1 λ ∇ ∂x = −8λx1∂x − 16λx1x2∂t = −8λx1X1. ∂x2 2 1 As all Γk = 0, for k = 1, 3, it follows that ∇λ ∂ = 0. 33 ∂t t ¤ The following proposition shows the vector ﬁelds X1 and X2 are geodesic vector ﬁelds. Lemma 3.2 For every λ > 0,

λ λ ∇X1 X1 = 0, ∇X2 X2 = 0, (3.35)

λ λ ∇X2 X1 = 2∂t, ∇X1 X2 = −2∂t. (3.36) Proof: Using Lemma 3.1

λ ∇ X1 = −8λx2X2 + 2x2 2λX2 = −4λx2 X2, ∂x1 and next we compute that ∇λ X = ∇λ ∂ + 2x ∇λ ∂ = 2λ X . ∂t 1 ∂t x1 2 ∂t t 2 Then

λ λ λ ∇ X1 = ∇ X1 + 2x2∇ X1 = −4λx2X2 + 2x2 2λX2 = 0. X1 ∂x1 ∂t For the second identity we have

λ λ λ λ ∇ X1 = ∇ ∂x + 2x2∇ ∂t + 2∂t − 2x1∇ ∂x X2 ∂x2 1 ∂x2 ∂t 1 = 4λx2X1 + 4λx1X2 − 4x2λX1 − 4x1λX2 + 2∂t = 2∂t. Similar for the others. ¤ Another property of the Levi-Civita connection is given in the following result. Proposition 3.3 The restriction of the connection ∇λ on H × H does not depend on λ. Proof: Let U, V ∈ H be two horizontal ﬁelds which have the decomposition

1 2 1 2 U = U X1 + U X2 ,V = V X1 + V X2,

λ λ 1 λ 2 ∇U V = ∇U (V X1) + ∇U (V X2) 1 2 1 λ 2 λ = U(V ) X1 + U(V ) X2 + V ∇U X1 + V ∇U X2. (3.37) The third and the fourth term in the last relation will be computed using Lemma 3.2. For the third term,

λ 1 λ 2 λ 2 λ 2 ∇U X1 = U ∇X1 X1 + U ∇X2 X1 = U ∇X2 X1 = 2 U ∂t, and for the fourth term,

λ 1 λ 2 λ 1 λ 1 ∇U X2 = U ∇X1 X2 + U ∇X2 X2 = U ∇X1 X2 = −2 U ∂t. Hence the relation (3.37) becomes

λ 1 2 1 2 1 2 ∇U V = U(V ) X1 + U(V ) X2 + 2V U ∂t − 2U V ∂t µ ¶ V 1 V 2 = U(V 1) X + U(V 2) X + 2 · det ∂ . (3.38) 1 2 U 1 U 2 t

λ Hence ∇|H×H is independent on λ and the proposition is proved. λ Denote by DU V the horizontal component of ∇U V , namely 1 2 DU V = U(V ) X1 + U(V ) X2. (3.39)

D has the properties of a linear metric connection. The determinant has the geometrical signiﬁcance of a parallelogram area generated by the vectors V and U. In the following we state the main result of this section. It is the second fundamental formula (Gauss’ formula) for the case of the Heisenberg group. Theorem 3.4 For all λ > 0 there is a mapping D : H × H → H with the properties DfU V = fDU V, (3.40)

3 DU (fV ) = fDU V + U(f) V, f ∈ F(R ), (3.41)

DU h(V,W ) = h(DU V,W ) + h(V,DU W ), ∀U, V, W ∈ H, (3.42) such that the following decomposition takes place

λ ∇U V = DU V + 2 · area(U, V ) ∂t, ∀U, V ∈ H. (3.43) Proof: Choose D as in relation (3.39). Remains to show that D has the properties (3.41)-(3.43). The ﬁrst one is obvious. For the second one

1 2 DU (fV ) = U(fV )X1 + U(fV )X2

1 2 = U(f)(V X1 + V X2) + f DU V = U(f)V + fDU V. For the last property

i i i i i i DU h(V,W ) = U(V W ) = U(V )W + V U(W )

i i = U(V ) h(Xi,W ) + h(Xi,V )U(W ) i i = h(U(V )Xi,W ) + h(U(W )Xi,V )

= h(DU V,W ) + h(DU W, V ), where i = 1, 2. The proof of the theorem is complete. ¤

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Ovidiu Calin Eastern Michigan University Mathematics Department Ypsilanti, MI, 48197 USA [email protected] Vittorio Mangione Universit´adegli Studi di Parma 43100, Parma, Via M. D’Azeglio, 85/A, Italia [email protected] Eingegangen am 1. March 2002