GEOMETRIC FORMULATION OF MECHANICS ON LIE ...

GEOMETRIC FORMULATION OF MECHANICS ON LIE ALGEBROIDS 5It is easy to see that complete and vertical lifts satisfy the relations[σ V , η V ] = 0 [σ V , η C ] = [σ, η] V [σ C , η C ] = [σ, η] C ,for σ, η ∈ Sec(E). In fact, this relations are equivalent to the definition of thebracket given above.4. Geometry of T E EIn addition to the basic constructions explained in the last section, there areother important geometrical objects on T E E. This are the Liouville section, thevertical endomorphism and second-order differential equations.The Liouville section. The Liouville section ∆ is the section of τ 1 whose valueat the point a is the vertical lift to the point a of a itself, that is,∆(a) = ξ V (a, a) = (a, 0, a V a ).The coordinate expressions of ∆ and ρ 1 (∆) are∆ = y α V α ρ 1 (∆) = y α ∂∂y α .The vertical endomorphism. The vertical endomorphism S is the endomorphismof the bundle τ 1 : T E E → E defined by projection followed by vertical liftingS(a, b, v) = (a, 0, b V a ). The local expression of S isS = V α ⊗ X α ,where {X α , V α } is the dual base of {X α , V α }.The vertical endomorphism satisfies the following properties• S(σ V ) = 0 and S(σ C ) = σ V• S 2 = 0.• Im S = Ker S = Ver(T E E)• The Nijenhuis torsion of S vanishes.Second-order differential equations. In the case E = T M a second-orderdifferential equation on a manifold M is a vectorfield Γ on T M such that theirintegral curves are the natural prolongation of curves on the base manifold M.In the case of a general Lie algebroid the notion that replaces that of naturalprolongation is the notion of admissible curve.A tangent vector v to E at a point a is called admissible if T a τ(v) = ρ(a). Inother words, v is admissible if and only if (a, a, v) is in T E E. The set of admissibleelements of T E E will be denotes by Adm(E). A curve in E is admissible if itstangent vector is admissible at every point. A second-order differential equation(sode) on a Lie algebroid E is a section of T E E which takes values on Adm(E).It is easy to prove that Γ is a sode on E if and only if τ 2 ◦ Γ = id E if and only ifS(Γ) = ∆.In local coordinates, the expression of Γ and ρ 1 (Γ) areΓ(x, y) = y α X α + f α (x, y)V α ρ 1 (Γ) = ρ i αy α ∂∂x i + f α ∂∂y α .The integral curves of the sode Γ, i. e. the integral curves of ρ 1 (Γ), are the solutionsof the differential equationsdx idt = ρi α(x)y αdy αdt= f α (x, y).

6 EDUARDO MARTÍNEZNotice that the name sode must be used with care since the differential equationof its integral curves are not really second-order.5. Lagrangian systemsVariational principle. Let m 0 and m 1 be two points in the manifold M. Givena Lagrangian L ∈ C ∞ (E) we look for the critical points of the functionJ(η) =∫ baL(η(t)) dton the set of admisible curves η : [t 0 , t 1 ] → E such that γ(t 0 ) = m 0 and γ(t 1 ) = m 1for some fixed t 0 , t 1 ∈ R, and where γ(t) = τ(η(t)).We consider admissible variations η s (t) of the actual solution η 0 (t), and we denoteby γ s (t) = τ(η s (t)) the projection to M. The infinitesimal variation Z(t) =dη s (t)/ds ∣ s=0and its projection X(t) = dγ s (t)/ds ∣ s=0must satisfy the differentialequationdX i= y α ∂ρi α∂x j Xj + ρ i αZ α ,dttogether with the boundary conditions X(t 0 ) = X(t 1 ) = 0.Since the above boundary value problem is difficult to solve, we will consideronly a special kind of variations which will be enough for our purposes. For everysection σ of E such that σ(m 0 ) = σ(m 1 ) = 0 we set Z(t) = ρ 1 (σ C (η 0 (t))), andtherefore X(t) = ρ(σ(γ 0 (t))). Then Z satisfies the specified conditions and thevariation of J along Z isd∫ t1∫ t1∫ds J(η s) ∣ = Z(t)L dt = (ρ(σ C )L)(η 0 (t)) dt = d σ C L.s=0 t 0 t 0 η 0Taking into account the local expression of a complete lift, we have{d σ C L = ρ i ∂Lα∂x i − d ( )} ∂Ldt ∂y α − C γ ∂Lαβyβ ∂y γ σ α + d ( ) ∂Ldt ∂y α σα ,from where it follows that the curve η 0 is a solution of the Euler-Lagrange equationsdx i( )dt = ρi αy α d ∂Ldt ∂y α = ρ i ∂Lα∂x i − ∂LCγ αβyβ ∂y γ .The Lagrangian L is said to be regular if the Euler-Lagrange equations can bewritten in normal form. It is easy to see that L is regular if and only if thesymmetric matrix g αβ = ∂ 2 L/∂y α ∂y β is regular.Symplectic formalism. The Euler-lagrange equations given above can be obtainedas the solution of a symplectic equation. By a symplectic form on a Liealgebroid E → M we mean a section ω of E ∗ ∧ E ∗ → M such that it is nondegenerateas a bilinear form and closed, dω = 0. Given a regular LagrangianL ∈ C ∞ (E) we can construct a symplectic form ω L on the Lie algebroid T E E → Eas follows. We define the Cartan 1-section θ L by θ L = S(dL) and the Cartan2-section as (minus) the differential of θ L , that is, ω L = −dθ L .In local coordinates θ L and ω L are given byθ L = ∂L∂y α X α .ω L =∂2 L∂y α ∂y β X α ∧ V β + 1 ( ∂ 2 L2 ∂x i ∂y α ρi β −∂2 L∂x i ∂y β ρi α + ∂L )∂y γ Cγ αβX α ∧ X β

GEOMETRIC FORMULATION OF MECHANICS ON LIE ALGEBROIDS 7From this expression it is clear ω L is symplectic if and only if L is regular.The energy function E L defined by L is E L = d ∆ L−L, which in local coordinatesis E L = y α ∂L/∂y α − L.In terms of this objects we set the symplectic equationi Γ ω L = dE L ,for a section Γ of T E E. As we will readily show, the Euler-Lagrange equations forL are the equations for the integral curves of Γ.Indeed, if we put Γ = g α X α + f α V α theni Γ ω L = g β ∂ 2 [L∂y α ∂y β Vα − f β ∂ 2 (L ∂ 2∂y α ∂y β + Lgβ∂x i ∂y α ρi β −∂2 L∂x i ∂y β ρi α + ∂L )]∂y γ Cγ αβX αand(dE L =∂2 L∂y α ∂y β yβ V α − ρ i ∂Lα∂x i − ∂ 2 )Lρi α∂x i ∂y β yβ X α .The equality of the V α components implies g αβ (y α − g α ) = 0. If the Lagrangian isregular, then this equation has a unique solution g α = y α , which implies that Γ issode. Taking this into account, the X α components are equal ifρ i βy β ∂2 L∂x i y α + f β ∂2 L∂y α y β = ∂Lρi α∂x i − ∂LCγ αβyβ ∂y γ .In the left hand side of this equation we recognize the derivative of∂L∂y αthat it can be written in the form( ) ∂Ld Γ∂y α= ρ i ∂Lα∂x i − Cγ αβyβ∂L∂y γ ,along Γ, sowhich is the second of the Euler-Lagrange equations.Noether’s theorem. Noether’s theorem can be easily extended to Lagrangiansystems on Lie algebroids. We consider a regular Lagrangian L and we will denoteby Γ the sode defined by L.Theorem: Let σ be a section of E and f a function on M, such thatd σ C L = f. ˙Then the function G = 〈 θ L , σ 〉 − f is a first integral of Γ and σ C is a symmetry ofΓ, that is [σ C , Γ] = 0. Moreover i σ C ω L = dG.6. Hamiltonian formalismThere exists an anologous Hamiltonina formulation on the dual E ∗ of a Liealgebroid E. We just indicate a few details on how this is constructed.The dual of a Lie algebroid admits also a prolongation π 1 : T E E ∗ → E ∗ with astructure of Lie algebroid. The manifold T E E ∗ isT E E ∗ = { (µ, b, w) ∈ E ∗ × E × T E ∗ | π(µ) = τ(b), v ∈ T µ E ∗ and ρ(b) = T µ π(w) }and the bundle projection is π 1 (µ, b, w) = w. A base of sections of T E E ∗ is()( )X α (µ) = µ, e α , ρ i ∂α∂x i P α ∂(µ) = µ, 0, ,∂µ αand the asociated coordinates will be denoted (x i , µ α , z α , w α ).The Lie algebroid structure of T E E ∗ is defined in terms of projectable sectionsas in the case of T E E, and the anchor is ρ 1 (µ, b, w) = w. In this case no other

8 EDUARDO MARTÍNEZanchor is compatible with the bracket. The brackets of the elements of the localbase {X α , P α } are[X α , X β ] = C γ αβ X γ [X α , P β ] = 0 [P α , P β ] = 0.On T E E ∗ we can also define vertical and complete lifts. If σ = σ α e α is a sectionof E and θ = θ α e α is a section of E ∗ then the local expression of σ C and θ V areσ C = σ α X α − µ γ(ρ i ∂σ γ )α∂x i + Cγ αβ σβ P α and θ V = θ α P α .The brackets of complete and vertical lifts are[σ C , η C ] = [σ, η] C [σ C , θ V ] = (d σ θ) V [β V , θ V ] = 0.In the Lie algebroid T E E ∗ → E ∗ there exists a canonical symplectic form ω 0defined as follows. We define the section θ 0 of (T E E ∗ ) ∗ → E ∗ byand then ω 0 = −dθ 0 . In coordinates〈 θ 0µ , (µ, b, w) 〉 = 〈 µ , b 〉 ,θ 0 = µ α X α and ω 0 = X α ∧ P α + 1 2 µ γC γ αβ X α ∧ X β .Every function H on E ∗ defines a section σ H of T E E ∗ by the equationi σH ω 0 = dH,and hence a dynamical system on E ∗ by means of the associated vectorfield X H =ρ 1 (σ H ) ∈ X(E ∗ ). In coordinatesσ H = ∂H ()X α − ρ i ∂Hα∂µ α ∂x i + µ γC γ ∂HαβP α∂µ βand thus the equations for the integral curves of X H aredx idt = ∂H dµ αρi α∂µ α dt= −ρ i ∂Hα∂x i − µ γC γ ∂Hαβ.∂µ βThe canonical Poisson bracket which is known to exists on E ∗ can be obtainedby means of ω 0 as{F, G} = −ω 0 (σ F , σ G ).7. ExamplesWe consider a Lie algebra g acting on a manifold M, that is, we have a Liealgebra homomorphism g → X(M) mapping every element ξ of g to a vectorfieldξ M on M. The bundle E is E = M × g with the projection onto the first factor.The anchor is the map ρ(m, ξ) = ξ M (m). The bracket is defined by declaring ρ tobe the anchor and defining the bracket of constant sections as the constant sectioncorresponding to the bracket on g, that is, if σ(m) = (m, ξ) and η(m) = (m, ζ) aretwo constant sections, then [σ, η](m) = (m, [ξ, ζ] g ).By identifying T E ≡ T M × T g ≡ T M × g × g, an element of T E E is of the form(a, b, v) = ( (m, ξ), (m, η), (v m , ξ, ζ) )and the condition T τ(v) = ρ(b) implies that v m = η M (m). Therefore, we canidentify T E E with M × g × g × g,and the various maps areτ 1 (m, ξ, η, ζ) = (m, ξ), τ 2 (m, ξ, η, ζ) = (m, η), ρ 1 (m, ξ, η, ζ) = (η M (m).ξ, ζ)

GEOMETRIC FORMULATION OF MECHANICS ON LIE ALGEBROIDS 9If σ is a section of E, it is of the form σ(m) = (m, λ(m)) for a function λ: M → g.Then, the vertical and complete lift of σ areσ V (m, ξ) = (m, ξ, 0, λ(m)) σ C (m, ξ) = (m, ξ, λ(m), ξ M λ(m) + [ξ, λ(m)] g ),where ξ M λ = ξ i M ∂λ∂x i is the differential of λ along ξ M .We consider a Lagrangian of mechanical type L(m, ξ) = (1/2)g(ξ, ξ) − V (m),where g is an inner product on the Lie algebra g and V is a function on M. TheCartan sections and the differential of the energy are given by〈 θ L , (m, ξ, η, ζ) 〉 = g(ξ, η 1 )ω L ((m, ξ, η 1 , ζ 1 ), (m, ξ, η 2 , ζ 2 )) = g(η 1 , ζ 2 ) − g(ζ 1 , η 2 ) + g(ξ, [η 1 , η 2 ] g )Therefore if Γ(m, ξ) = (m, ξ, ξ, ζ) thendE L (m, ξ, η 2 , ζ 2 ) = g(ξ, ζ 2 ) + 〈 dV (m) , η 2 〉(i Γ ω L − dE L )(z 2 ) = −g(ζ, η 2 ) + g(ξ, [ξ, η 2 ] g ) − 〈 dV (m) , η 2 〉 .If we define ad † ξand the gradient of V byg(ad † ξ η 1, η 2 ) = g(η 1 , [ξ, η 2 ] g ), and g(grad V (m), η) = 〈 dV (m) , η 〉 ,then (i Γ ω L − dE L )(z 2 ) = −g( ˙ξ − ad † ξ ξ + grad V (m), η 2), from where we get thatthe sode Γ isΓ(m, ξ) = (m, ξ, ξ, ad † ξξ − grad V (m)).The integral curves (m(t), ξ(t)) of Γ are the solution of the diferential equationsṁ = ξ(t) M(m) ˙ξ − ad†ξξ = − grad V (m).As a particular example we can consider the heavy top, where g = so(3) andM = S 2 . An element of M will be considered as an unit vector γ in R 3 (representingthe direction of the gravity), and an element of so(3) will also be considered as avector ω in R 3 (representing the angular velocity in body coordinates). The metricg is given by the inertia tensor of the top, g(ω 1 , ω 2 ) = ω 1 · Iω 2 , and the potentialis V (γ) = Mglγ · e, where e is the unit vector from the fixed point to the center ofmass. The anchor map and the bracket are given byThenρ(γ, ω) = γ × ω ≡ (γ, γ × ω) ∈ T γ S 2 , and [ω 1 , ω 2 ] so(3) = ω 1 × ω 2 .ad † ω 1ω 2 = I −1 (Iω 2 × ω 1 ) and grad V = I −1 ( ∂V∂γ × γ) = −MglI−1 (γ × e).and the equations of motion are˙γ + ω × γ = 0 I ˙ω + ω × Iω = Mglγ × e,which are the Euler-Arnold equations.We now apply Noether’s theorem to find some symmetries. We consider rotationsarround the gravity axis. The generator is the section σ(γ) = (γ, γ),whose completelift is σ C (γ, ω) = (γ, ω, γ, 0). Therefored σ C L = ρ(γ) · ∂L∂γ = 0,

10 EDUARDO MARTÍNEZsince ρ(γ) = 0. Thus σ is a symmetry of the system and the constant of motion isJ γ = 〈 θ L , σ 〉 = ∂L · γ = (Iω) · γ,∂ωwhich is the component of the angular momenta in the direction of the gravity. Itcan be shown that σ is a symmetry for any Lagrangian system defined on S 2 ×so(3),because σ is in the center of the Lie algebroid, i. e. commutes with every section.If the body is symmetric, that is Ie = I 3 e and I 1 = I 2 , then the section η(γ) =(γ, e) is also a symmetry, and the associated first integral isJ e = 〈 θ L , η 〉 = ∂L · e = (Iω) · e.∂ωSimilarly, the prolongation of E ∗ can be identified with M × g ∗ × g × g ∗ wherethe maps π 1 , π 2 and ρ 1 areπ 1 (m, µ, η, ν) = (m, µ), π 2 (m, µ, η, ν) = (m, η), ρ 1 (m, µ, η, ν) = (η M (m).µ, ν)and the canonical symplectic form ω 0 is given byω 0((m, µ, η1 , ν 1 ), (m, µ, η 1 , ν 1 ) ) = 〈 ν 2 , η 1 〉 − 〈 ν 1 , η 2 〉 + 〈 µ , [η 1 , η 2 ] g 〉 .If we consider a Hamiltonian H of mechanical typeH(m, µ) = 1 G(µ, µ) + V (m),2where G is an inner product on the dual g ∗ of the Lie algebra g and V is a functionon M, the equations of motion areṁ = ρ(G(µ))In the case of the heavy top this equations are˙γ + I −1 µ × γ = 0˙µ − ad ∗ G(µ) µ = −dV.˙µ + (I −1 µ × µ) = Mgl(γ × e),which are the Euler-Arnold equations in Hamiltonian form. The function ˆσ = γ · µis a constant of the motion for every Hamiltonian function defined on S 2 × so(3) ∗ ,i. e. it is a Casimir for the Poisson bracket. It is easy to see that sections on thecenter of a Lie algebroid corresponds to linear Casimir functions on the dual of theLie algebroid.References[1] A. Weinstein, Lagrangian Mechanics and groupoids, Fields Inst. Comm. 7 (1996) 207–231.[2] K. Mackenzie, Lie groupoids and Lie algebroids in Differential Geometry, LMS Lecture NotesSeries, 124, Cambridge University Press, 1987.[3] J. Klein, Espaces variationnels et mécanique, Ann. Inst. Fourier 12 (1962) 1–124.[4] M. Crampin, Tangent bundle geometry for Lagrangian dynamics, J. Phys. A: Math. Gen. 16(1983) 3755–3772.[5] E. Martínez, Lagrangian Mechanics on Lie Algebroids, (preprint, 1999).[6] E. Martínez, Hamiltonian Mechanics on Lie Algebroids, (preprint, 2000).Departamento de Matemática Aplicada, Universidad de Zaragoza, María de Luna 3,50015 Zaragoza, SpainE-mail address: emf@unizar.es