# The Geometry of Tangent Bundles: Canonical Vector Fields

Authors: Tongzhu Li; Demeter Krupka
Journal: Geometry
 (2013)

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## Abstract

A canonical vector field on the tangent bundle is a vector field defined by an invariant coordinate construction. In this paper, a complete classification of canonical vector fields on tangent bundles,
depending on vector fields defined on their bases, is obtained. It is shown that every canonical vector field
is a linear combination with constant coefficients of three vector fields: the variational vector field (canonical lift),
the Liouville vector field, and the vertical lift of a vector field on the base of the tangent bundle.


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## Body

## 1. Introduction

Vector fields on tangent bundles belong to basic concepts of pure and applied differential geometry, global analysis, and mathematical physics. Recent research in geometry extends the well-known correspondence of semisprays, sprays, and geodesic sprays to the classical theory of geodesics and connections (see, e.g., [1, 2]). Vector fields on tangent bundles can be considered as an underlying geometric structure for the theory of second-order differential equations [3–7]. The semispray theory has been used in the calculus of variations on manifolds to characterize extremal curves of a variational functional as integral curves of the Hamilton or Euler-Lagrange vector fields [2, 4, 8, 9]. Sprays and semisprays also provide a natural framework for extension of classical results of analytical mechanics to contemporary mechanical problems and stimulate a broad research in the global theory of nonconservative systems, symmetries, and the constraint theory (see, e.g., [6, 10–12]).This paper is devoted to the structure theory of vector fields on the tangent bundleTX of a manifold X; our aim will be to classify all canonical vector fields on TX, independent of any geometric structure or on the topology of X. Our main theorem says that every canonical vector field is a linear combination with constant coefficients of three independent vector fields: (a) a variational vector field (the natural lift of a vector field, defined on X), (b) the Liouville vector field, and (c) the vertical lift of a vector field on X; this completes the results obtained in [13]. Another result is the method how the main theorem has been formulated and proved; the concepts we use allow generalizations and applications to analogous problems of discovering and describing canonical geometric objects.Our approach to the problem is based on the theory of jets and differential invariants, and on an observation that the coordinate transformations onX naturally define a Lie group Lnr, where n=dimX, the differential group, and its left action on the type fibre of any natural bundle over X [13–17]. The general theory gives us equations that determine a canonical vector field on TX as a differential invariant of the second differential group Lnr.The proof of our result is not straightforward; it relies on the semidirect product structure ofLnr and on the orbit reduction method that has already been applied to the classification problem of differential invariants of a linear connection [18]. The method can also be used in the canonical constructions depending on any geometric objects.Throughout this work,X is a smooth real n-dimensional manifold, TX is the tangent bundle of X, and τX:TX→X is the tangent bundle projection. The second tangent bundle of X is the tangent bundle TTX=T2X over TX with the tangent bundle projection τTX:TTX→TX; its elements are second-order tangentvectors on X. The mapping TτX is the differential TτX:TTX→Tx of τX, satisfying τX∘Tτx=τX∘τTX. The 1-jet prolongation of TX is denoted by J1TX; elements of the set J1TX are the 1-jets Jx1ξ with source x∈X and target ξ(x)∈TxX=τx-1(x); the source and target jet projections are τX1:J1TX→X and τX1,0:J1TX→TX.


## 2. Second-Order Vectors and Jets of a Vector Field

In this section, we fix basic notation, used throughout this paper. If(U,φ), φ=(xi), is a chart on X, we denote by (τX-1(U),Φ), Φ=(xi,x˙i), the associated chart on the tangent bundle TX; the associated chart on T2X is denoted by (τTX-1(τX-1(U)),Ψ), Ψ=(xi,x˙i,x~i,x˙~i). The tangent bundle projection τX has the chart expression (xi,x˙i)→(xi), and τTX is expressed by (xi,x˙i,x~i,x˙~i)→(xi,x˙i). The tangent mapping TτX is expressed in coordinates as (xi,x˙i,x~i,x˙~i)→(xi,x~i).In the following two lemmas, we recall standard transformation formulas, needed in proofs.Lemma 1.
Let a second-order tangent vectorΞ∈TζTX be expressed in two charts (U,φ), φ=(xi), and (V,ψ), ψ=(yi), as
(1)Ξ=Ξk(∂∂xk)ζ+Ξ˙k(∂∂x˙k)ζ=Θl(∂∂yl)ζ+Θ˙l(∂∂y˙l)ζ.
Then,
(2)Θl=(∂yl∂xk)φ(x)Ξk,Θ˙l=(∂2yl∂xp∂xk)φ(x)x˙p(ζ)Ξk+(∂yl∂xk)φ(x)Ξ˙k.The chart onJ1TX, associated with the chart (U,φ), φ=(xi), on X is denoted by ((τx1)-1(U),Φ1), Φ1=(xi,x˙i,x˙ji). The coordinates of a 1-jet Jx1ξ are xi(Jx1ξ)=xi(x), x˙i(Jx1ξ)=x˙i(ξ(x)), and x˙ji(Jx1ξ)=Dj(x˙iξφ-1)(φ(x)).Lemma 2.
For any two charts(U,φ), φ=(xi), and (V,ψ), ψ=(yi), on X, such that U∩V≠∅, the transformation equations on (τX1)-1(U)∩(τX1)-1(U) are
(3)y˙p=∂yp∂xmx˙m,y˙qp=(∂2yp∂xs∂xmx˙m+∂yp∂xmx˙sm)∂xs∂yq.We need the pullback fibrationτX*J1TX={(ζ,Jx1ξ)∈TX×J1TX|τX(ζ)=x} over TX. We have the commutative diagram
(4)
in which the left vertical arrow is the restriction of the first Cartesian projection pr1:TX×J1TX→TX and the upper horizontal arrow is the restriction of the second Cartesian projection pr2:TX×J1TX→J1TX. We denote
(5)ρX=pr1.
Any chart (U,φ), φ=(xi), on X induces a chart on the pullback manifold τX*J1TX. Denoting by xi(ζ)=xi(x),x˙i(ζ)=ζi the coordinates of a vector ζ∈τX-1(U) at x∈U and by xi(Jx1ξ)=xi(x), x˙i(Jx1ξ)=ξi, x˙ji(Jx1ξ)=ξji the coordinates of a 1-jet Jx1ξ∈(τX1)-1(U), then the induced chart on τX*J1TX, (ρx-1(τX-1(U)),Φ~1), Φ~1=(xi,x˙i,z˙i,z˙ji), is defined by
(6)xi(ζ,Jx1ξ)=xi(x),x˙i(ζ,Jx1ξ)=ζi,z˙i(ζ,Jx1ξ)=ξi,z˙ji(ζ,Jx1ξ)=ξji.
In these coordinates, ρX is the mapping (xi,x˙i,z˙i,z˙ji)→(xi,x˙i) and the second Cartesian projection is (xi,x˙i,z˙i,z˙ji)→(xi,x˙i,z˙ji).


## 3. Differential Groups and Differential Invariants

Recall that for any positive integerr, the rth differential group Lnr is the group of invertible r-jets with source and target at the origin 0, endowed with its natural Lie group structure; the group multiplication in Lnr is the jet composition Lnr×Lnr∋J0rα∘J0rβ=J0r(α∘β)∈Lnr. The first differential group Ln1 is just the group GLn(R). For all r≥s, we denote by πr,s:Lnr→Lns the canonical jet projection and by ι1,r:Ln1→Lnr the canonical injective Lie group morphism. The normal subgroup Knr=Kerπr,1 of Lnr is nilpotent, and Lnr is the interior semidirect product of ι1,r(Ln1) and Knr [13, 17]; we denote Lnr=Ln1×sKnr.The first canonical coordinates onLnr are the functions aj1i,aj1j2i,…,aj1j2⋯jri, where 1≤j1≤j2≤⋯≤jk≤n, k=1,2,…,r, defined as follows. If J0rα∈Lnr is an r-jet and α=(αi) its representative, then aj1j2⋯jki(J0rα)=Dj1Dj2⋯Djkαi(0). Similarly, the formula bj1j2⋯jki(J0rα)=aj1j2⋯jki(J0rα-1) defines the second canonical coordinates bj1i,bj1j2i,…,bj1j2⋯jri; clearly, these coordinates satisfy
(7)ajibkj=δki.
Equations of the subgroup Knr are aji=δji, and equations of the subgroup ι1,r(Ln1) are aj1j2i=0,…, aj1j2⋯jri=0.By a differential invariant, we mean anLnr-equivariant mapping f:P→Q of left Lnr-manifolds [13]. Given the actions of Lnr on P and Q, we get the equation for the differential invariant f:
(8)f(J0rα·p)=J0rα·f(p),
where p∈P and J0rα∈Lnr. Equation (8) splits to an equivalent system:
(9)f(J0rα·p)=J0rα·f(p),f(J0rβ·p)=J0rβ·f(p),
where J0rα∈Ln1 and J0rβ∈Knr (the orbit reduction method [18]).The problem we consider in this paper reduces to solving (9) for some specific left Ln2-manifolds P and Q. In our case, the orbit reduction method simplifies (8) and allows us to obtain its complete solution.


## 4. Canonical Vector Fields on Tangent Bundles

Any diffeomorphismα:U→X, defined on an open set U in X, induces the corresponding lifted diffeomorphisms Tα:TU→TX, J1Tα:J1TU→J1TX, and (Tα,J1Tα):τX*J1TU→τX*J1TX. By a canonical vector field on TX, we mean a morphism Ξ:τX*J1TX→T2X such that
(10)τTX∘Ξ=ρX,
and for all diffeomorphisms α:U→X of X,
(11)Ξ∘(Tα,J1Tα)=T(Tα)∘Ξ.
Condition (10) means that the diagram
(12)
commutes or, which is the same, for all (ζ,Jx1ξ)∈τX*J1TX, Ξ(ζ,Jx1ξ) is a second-order vector on X at the point ζ∈TX,
(13)τTX(Ξ(ζ,Jx1ξ))=ζ.
Since J1Tα(Jx1ξ)=Jα(x)1(Tα∘ξ∘α-1), condition (11) can also be written as
(14)Ξ(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))=TζTα·Ξ(ζ,Jx1ξ).We express conditions (13) and (14) in coordinates. We write in a chart (U,φ), φ=(xi),
(15)Ξ(ζ,Jx1ξ)=Ξ(U,φ)k(ζ,Jx1ξ)(∂∂xk)ζ+Ξ˙(U,φ)k(ζ,Jx1ξ)(∂∂x˙k)ζ.
Then for fixed x, the components are functions of the coordinates ζi,ξi, and ξli; that is,
(16)Ξ(U,φ)k=Ξ(U,φ)k(ζi,ξi,ξli),Ξ˙(U,φ)k=Ξ˙(U,φ)k(ζi,ξi,ξli).Theorem 3.
LetΞ:τX*J1TX→T2X be a morphism over the identity idTX. The following two conditions are equivalent:(1)
Ξ is a canonical vector field.(2)
For any pointsx,y∈X, any charts (U,φ), φ=(xi), at x and (V,ψ), ψ=(yi), at y, and any diffeomorphism α:U→V such that α(x)=y,
(17)Ξ(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=aklΞ(U,φ)k(ζi,ξi,ξji),Ξ˙(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=ajklζjΞ(U,φ)k(ζi,ξi,ξmi)+aklΞ˙(U,φ)k(ζi,ξi,ξmi),
where
(18)akl=(∂(ylαφ-1)∂xk)φ(x),ajkl=(∂2(ylαφ-1)∂xj∂xk)φ(x),bjk=(∂(xkα-1ψ-1)∂yj)ψ(x).Proof.
Consider condition (14). With the abbreviations (18), the coordinates of the vector Txα·ζ and the 1-jet Jα(x)1(Tα∘ξ∘α-1) are
(19)y˙i(Txα·ζ)=akiζk,y˙i(Jα(x)1(Tα∘ξ∘α-1))=akiξk,y˙ji(Jα(x)1(Tα∘ξ∘α-1))=aklibjlξk+akibjlξlk.
The coordinates of the vector TζTα·Ξ(ζ,Jx1ξ) are determined by
(20)TζTα·Ξ(ζ,Jx1ξ)=aklΞ(U,φ)k(ζ,Jx1ξ)(∂∂yl)Txα·ζ+({+aklΞ˙(U,φ)k(ζ,Jx1ξ)}akjlx˙j(ζ)Ξ(U,φ)k(ζ,Jx1ξ)+aklΞ˙(U,φ)k(ζ,Jx1ξ))(∂∂y˙l)Txα·ζ.
Writing
(21)Ξ(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))=Ξ(V,ψ)l(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))(∂∂yl)Txα·ζ+Ξ˙(V,ψ)l(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))(∂∂y˙l)Txα·ζ,
we can express condition (14), with help of (19) and (20), as
(22)Ξ(V,ψ)l(akix˙k(ζ),akix˙k(Jx1ξ),akmibjmx˙k(Jx1ξ)+akibjmx˙mk(Jx1ξ))=aklΞ(U,φ)k(ζ,Jx1ξ),Ξ˙(V,ψ)l(akix˙k(ζ),akix˙k(Jx1ξ),akmibjmx˙k(Jx1ξ)Ξ˙(V,ψ)l++akibjmx˙mk(Jx1ξ){akix˙k(ζ),akix˙k(Jx1ξ),akmibjmx˙k(Jx1ξ)})=ajklx˙j(ζ)Ξ(U,φ)k(ζ,Jx1ξ)+aklΞ˙(U,φ)k(ζ,Jx1ξ).In the well-known sense, the canonical vector fields are completely determined by certain differential invariants, that is, equivariant mappings from the type fibreRn×Rn×Rn2 of τX*J1TX into the type fibre Rn×Rn×Rn of T2X over X with respect to the canonical actions of the differential group Ln2, induced by diffeomorphisms of X. We can characterize these actions explicitly in terms of the first and second canonical coordinates aji, ajki and brp, brsp on the differential group Ln2. Note that
(23)aplbrp=δrl,apqlbrp+aplaqsbrsp=0.
If A=(aji,ajki) is an element of the differential group Ln2, denote by αA (resp., βA) the transformation of Rn×Rn×Rn2 (resp., Rn×Rn×Rn), defined by A. For any points (ζi,ξi,ξji)∈Rn×Rn×Rn2 and (Zi,Ξi,Ξ˙i)∈Rn×Rn×Rn, we denote
(24)(ζ-i,ξ-i,ξ-ji)=αA(ζi,ξi,ξji),(Z-i,Ξ-i,Ξ˙-i)=βA(Zi,Ξi,Ξ˙i).
The following lemma defines the points (24) explicitly.Lemma 4.
LetA=(aji,ajki) be an element of the differential group Ln2.(a)
The canonical group action ofLn2 on Rn×Rn×Rn2 is given by the equations:
(25)ζ-i=akiζk,ξ-i=akiξk,ξ-ji=akmibjmξk+akibjmξmk.(b)
The canonical group action ofLn2 on Rn×Rn×Rn is given by the equations:
(26)Z-i=akiZk,Ξ-i=akiΞk,Ξ˙-i=ajkiζjΞk+akiΞ˙k.Proof.
The leftLn2-actions on the type fibres of τX*J1TX and T2X can be derived from the general theory of differential groups and differential invariants [13]. Note that these Ln2-actions can also be defined by transformation properties of components of jets and tangent vectors; compare with Lemmas 1 and 2 with the substitution
(27)akl=∂yl∂xk,akpl=∂2yl∂xp∂xk,brp=∂xp∂yr,brsp=∂2xp∂yr∂ys.Since a canonical vector field is always a morphism over the identity mappingidTX, the corresponding differential invariant F:Rn×Rn×Rn2→Rn×Rn×Rn has an expression (ζi,ξi,ξji)→F(ζi,ξi,ξji)=(ζi,Fj(ζi,ξi,ξji),F˙j(ζi,ξi,ξji)); we denote F=(Fi,F˙i).Theorem 5.
A mappingF=(Fi,F˙i) is a differential invariant if and only if
(28)Fi(aspζs,aspξs,asmpbqsξm+ampbqsξsm)=akiFk(ζs,ξs,ξsm),F˙i(aspζs,aspξs,asmpbqsξm+ampbqsξsm)=akpiζpFk(ζs,ξs,ξsm)+akiF˙k(ζs,ξs,ξsm),
for all (ζs,ξs,ξsm)∈Rn×Rn×Rn2 and (aji,ajki)∈Ln2.Proof.
Equations (28) are direct consequences of (25) and (26).Remark 6.
Note that our definition of the canonical vector field differs from the lifting of a vector field from a base manifold to its tangent bundle, which is defined by means of the lifting of diffeomorphisms and 1-parameter groups of diffeomorphisms to the tangent bundle [2, 9].Remark 7.
We can specify Theorem3 to diffeomorphisms α, preserving a given point x∈X, such that α(x)=x, and to charts (U,φ) and (V,ψ) such that (U,φ)=(V,ψ) and φ(x)=0. Then, the components Ξ(U,φ)l and Ξ˙(U,φ)k of a canonical vector field Ξ satisfy
(29)Ξ(U,φ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=aklΞ(U,φ)k(ζk,ξk,ξmk),Ξ˙(U,φ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=ajklζjΞ(U,φ)k(ζk,ξk,ξmk)+aklΞ˙(U,φ)k(ζk,ξk,ξmk),
where
(30)akl=(∂(xlαφ-1)∂xk)0,ajkl=(∂2(xlαφ-1)∂xj∂xk)0,bjk=(∂(xkα-1ψ-1)∂xj)0.


## 5. Canonical Vector Fields: Classification

We find all solutionsF=(Fi,F˙i) of the equations for differential invariants, associated with canonical vector fields Ξ:τX*J1TX→T2X (Theorem 5). These equations can be written in coordinates as
(31)Fi(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=akiFk(x˙s,z˙s,z˙sm),(32)F˙i(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=akpix˙pFk(x˙s,z˙s,z˙sm)+akiF˙k(x˙s,z˙s,z˙sm).
The following lemma solves (31).Lemma 8.
The functionsFi satisfy condition (31) if and only if
(33)Fi(x˙j,z˙j,z˙kj)=Ax˙i+Bz˙i,
for some constants A,B∈R.Proof.
(1) First we consider (31) for the group elements, belonging to the subgroup Kn2 of Ln2; equations of Kn2 are
(34)asp=δsp,
and we get the system
(35)Fi(x˙p,z˙p,aqmpz˙m+z˙qp)=Fi(x˙s,z˙s,z˙sm).
This equation is obviously satisfied at all points where z˙m=0. On the other hand, suppose that there exists at least one index j such that z˙j≠0; then to every point y˙qp one can find the group parameters aqmp such that aqmpz˙m+z˙qp=y˙qp. Indeed, if, for example, z˙1≠0, we set
(36)aq,1p=y˙qp-z˙qpz˙1,aq,2p,aq,3p,…,aq,np=0.
This choice of the group parameters yields
(37)Fi(x˙p,z˙p,y˙qp)=Fi(x˙p,z˙p,z˙sm),
which shows that Fi(x˙j,x˙j,x˙kj) is independent of x˙qp.
(2) In view of (37), we can write Fi(x˙j,x˙j) instead of Fi(x˙j,x˙j,x˙kj). Turning back to conditions (31), we have the following equations for the functions Fi:
(38)Fi(asjx˙s,asjz˙s)=akiFk(x˙j,z˙j).
Properties of Fi are completely determined by the subgroup Ln1 of Ln2. If asp=λδsp, λ≠0, then by (38) Fi satisfies the positive homogeneity condition
(39)Fi(λx˙j,λz˙k)=λFi(x˙j,z˙k).
We suppose, however, that the functions Fi are defined at the origin x˙p=0, z˙p=0; then it is easily seen that (39) also holds for λ=0. Indeed, in this case, we have Fi(λ·0,λ·0)=Fi(0,0)=λFi(0,0) for all λ≠0, hence Fi(0,0)=0. On the other hand, the points λx˙j and λz˙k are always defined for λ=0; then Fi(0·x˙j,0·z˙k)=Fi(0,0)=0=0·Fi(x˙j,z˙k), which proves (39) for λ=0.
Then, we have from (28) D1,qFi(λx˙l,λz˙l)x˙q+D2,qFi(λx˙l,λz˙l)z˙q=Fi(x˙l,z˙l) by differentiation with respect to λ; we see that the expression on the left does not depend on λ. For λ=0, we get D1,qFi(0,0)x˙q+D2,qFi(0,0)z˙q=Fi(x˙l,z˙l), showing that Fi is linear in x˙q and z˙q; that is,
(40)Fi(x˙j,z˙j)=Sqix˙q+Tqiz˙i.
Substituting now into (38), we get
(41)Sqiasqx˙s+Tqiasqz˙s=aki(Sqkx˙q+Tqkz˙q).
That is, Sqiasq=aqiSsq and Tqiasq=aqiTsq. Thus, Sqialqδsl=δliaqlSsq, hence Sqiδsl=δqiSsl and by the trace operation, nSqi=δqiSll. Analogously, nTqi=δqiTll. These expressions together with (40) prove formula (33).
(3) If condition (33) is satisfied, then we get (31) by immediate substitution.Now, we wish to solve (32). In view of Lemma 8, these equations are of the form
(42)F˙i(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=akpix˙p(Ax˙k+Bz˙k)+akiF˙k(x˙s,z˙s,z˙sm).We prove separately the following lemma.Lemma 9.
The following two conditions are equivalent:(a)
The functionsQpiq=Qpiq(x˙j,z˙j) satisfy
(43)Qpil(asjx˙s,asjz˙s)amp=aklasiQmsk(x˙j,z˙j).(b)
The functionsQpiq are of the form
(44)Qmik(x˙j,z˙j)=(P(δjkδmi+δmkδji)+Q(δjkδmi-δmkδji))x˙j+(R(δjkδmi+δmkδji)+S(δjkδmi-δmkδji))z˙j,
where P,Q,R,S∈R.Proof.
(1) Ifaji=λδji, Lemma 9, equation (43) λQmil(λx˙j,λz˙j)=λ2Qmil(x˙j,z˙j) yields(45)Qmil(x˙j,z˙j)=Smjilx˙j+Tmjilz˙j,
for some Smjil,Smjil∈R (Lemma 8). Substituting back to (43), we get conditions for the coefficients Smjil,Smjil∈R:
(46)(Ssjilakjx˙k+Tsjilakjz˙k)ams=atlaki(Smjktx˙j+Tmjtkz˙j),
or, which is the same, (Ssjilx˙t+Tsjilz˙t)δmkatjaks=δjjδsi(Smrktx˙r+Tmrtkz˙r)atjaks; that is,
(47)Spjilatjamp=aklasiSmtsk,Tpjilatjamp=aklasiTmtsk.
To determine the constantsSpjil, we write Spjilδtsδmkasjakp=akjaspδjlδpiSmtsk from which it follows that
(48)Spjilδtsδmk+Sjpilδtkδms=δjlδpiSmtsk+δplδjiSmtks.
We apply to this formula various trace operations. Contractions in s,t and then in k,m yield nSpjilδmk+Sjpilδmk=δjlδpiSmssk+δplδjiSmsks and
(49)n2Spjil+nSjpil=δjlδpiSmssm+δplδjiSmsms.
Contractions in s,m and in k,t yield Spjilδtk+nSjpilδtk=δjlδpiSmtmk+δplδjiSmtkm and nSpjil+n2Sjpil=δjlδpiSmtmt+δplδjiSmttm. From these formulas, we find
(50)n(n+1)(Spjil+Sjpil)=(δjlδpi+δplδji)(Smsms+Smttm),n(n-1)(Spjil-Sjpil)=(δjlδpi-δplδji)(Smttm+Smtmt),Spjil+Sjpil=1n(n+1)(δjlδpi+δplδji)(Smsms+Smttm),Spjil-Sjpil=1n(n-1)(δjlδpi-δplδji)(Smttm+Smtmt),
hence
(51)Spjil=12n(n+1)(δjlδpi+δplδji)(Smsms+Smttm),+12n(n-1)(δjlδpi-δplδji)(Smttm+Smtmt).
The same computation applies to Tpjil. Thus,
(52)Spjil=P(δjlδpi+δplδji)+Q(δjlδpi-δplδji),Tpjil=R(δjlδpi+δplδji)+S(δjlδpi-δplδji),
for some constants P,Q,R,S∈R. Formula (44) now follows from (52) and (45).
(2) Conversely, suppose thatQmik satisfies (44). Writing
(53)Qpiq=(P(δjqδpi+δpqδji)+Q(δjqδpi-δpqδji))x˙j+(R(δjqδpi+δpqδji)+S(δjqδpi-δpqδji))z˙j,Qmsk=(P(δjkδms+δmkδjs)+Q(δjkδms-δmkδjs))x˙j+(R(δjkδms+δmkδjs)+S(δjkδms-δmkδjs))z˙j
and substituting these expressions into (44), we have
(54)Qpiq(asjx˙s,asjz˙s)ampbqk=Pamix˙k+Pδmkatix˙t+Qamix˙k-Qδmkatix˙t+Ramiz˙k+Rδmkatiz˙t+Samiz˙k-Sδmkatiz˙t,asiQmsk(x˙j,z˙j)=Pamix˙k+Pδmkasix˙s+Qamix˙k-Qδmkasix˙s+Ramiz˙k+Rδmkasiz˙s+Samiz˙k-Sδmkasiz˙s,
proving (13).Now, we are in position to give a solution to (42).Lemma 10.
The functionsF˙i satisfy condition (42) if and only if
(55)A=0,(56)F˙i(x˙j,z˙j,z˙kj)=Kx˙i+Lz˙i+Bx˙rz˙ri
for some constants K,L∈R.Proof.
(1) Suppose thatF˙i satisfy (42). Then if asp=δsp, we have
(57)F˙i(x˙p,z˙p,aqmpz˙m+z˙qp)=akpix˙p(Ax˙k+Bz˙k)+F˙i(x˙s,z˙s,z˙sm).
Differentiating with respect to z˙sr, we have
(58)(∂F˙i∂z˙sr)(x˙p,z˙p,aqmpz˙m+z˙qp)=(∂F˙i∂z˙sr)(x˙p,z˙p,z˙qp),
which shows that the derivative ∂F˙i/∂z˙sr does not depend on z˙qp. Then, however,
(59)F˙i(x˙j,z˙j,z˙kj)=Pi(x˙j,z˙j)+Qsir(x˙j,z˙j)z˙rs,
where the functions Pi and Qsir do not depend on z˙qp. Substituting from (59) back to (57), we have
(60)Pi(x˙p,z˙p)+Qsir(x˙p,z˙p)(armsz˙m+z˙rs)=akpix˙p(Ax˙k+Bz˙k)+Pi(x˙p,z˙p)+Qsir(x˙p,z˙p)z˙rs,
hence Qsir(x˙p,z˙p)armsz˙m=δsiarmsx˙r(Ax˙m+Bz˙m). Thus, the functions Pi and Qsir in (59) must satisfy
(61)Qsir(x˙p,z˙p)z˙m+Qsim(x˙p,z˙p)z˙r=δsix˙r(Ax˙m+Bz˙m)+δsix˙m(Ax˙r+Bz˙r).
Note that the trace operation ini and s yields
(62)1n(Qssr(x˙p,z˙p)z˙m+Qssm(x˙p,z˙p)z˙r)=x˙r(Ax˙m+Bz˙m)+x˙m(Ax˙r+Bz˙r).
(2) We now use (56) for the group parameters ajki=0. We have the condition
(63)F˙i(aspx˙s,aspz˙s,ampbqsz˙sm)=akiF˙k(x˙s,z˙s,z˙sm),
and, from (59),
(64)Pi(asjx˙s,asjz˙s)+Qsir(asjx˙s,asjz˙s)amsbrtz˙tm=akiPk(x˙j,z˙j)+akiQskr(x˙j,z˙j)z˙rs.
These equations split to the system
(65)Pi(asjx˙s,asjz˙s)=akiPk(x˙j,z˙j),Qsir(asjx˙s,asjz˙s)amsbrt=akiQskr(x˙j,z˙j).
Then, however,
(66)Pi(x˙j,z˙j)=Kx˙i+Lz˙i,
for some constants K,L∈R (Lemma 8), and
(67)Qsir(x˙j,z˙j)=(P(δjrδsi+δsrδji)+Q(δjrδsi-δsrδji))x˙j+(R(δjrδsi+δsrδji)+S(δjrδsi-δsrδji))z˙j,
where P,Q,R,S∈R (Lemma 9). These functions satisfy (61); that is,
(68)(P(δjrδsi+δsrδji)+Q(δjrδsi-δsrδji))x˙jz˙m=δsi(2Ax˙mx˙r+B(x˙mz˙r+x˙rz˙m)).
The trace in i and s yields
(69)(P(n+1)+Q(n-1))x˙rz˙m+(R(n+1)+S(n-1))z˙rz˙m+(P(n+1)+Q(n-1))x˙mz˙r+(R(n+1)+S(n-1))z˙mz˙r=n(2Ax˙mx˙r+B(x˙mz˙r+x˙rz˙m)).
Consequently, A=0 and
(70)(P(n+1)+Q(n-1))(x˙rz˙m+x˙mz˙r)=nB(x˙mz˙r+x˙rz˙m)(R(n+1)+S(n-1))z˙mz˙r=0,
hence
(71)A=0,P(n+1)+Q(n-1)=nB,R(n+1)+S(n-1)=0.
Then from (68),
(72)(P(δjrδsi+δsrδji)+Q(δjrδsi-δsrδji))x˙jz˙m+S(1-nn+1(δjrδsi+δsrδji)+(δjrδsi-δsrδji))z˙jz˙m+(P(δjmδsi+δsmδji)+Q(δjmδsi-δsmδji))x˙jz˙r+S(1-nn+1(δjmδsi+δsmδji)+(δjmδsi-δsmδji))z˙jz˙r=δsiB(x˙mz˙r+x˙rz˙m).
The terms containing z˙jz˙m should vanish separately. Since these terms are
(73)1n+1S(-(n-1)δsiz˙rz˙m-(n-1)δsrz˙iz˙m+(n+1)δsiz˙rz˙m-(n+1)δsrz˙iz˙m)+1n+1S(-(n-1)δsiz˙mz˙r-(n-1)δsmz˙iz˙r+(n+1)δsiz˙mz˙r-(n+1)δsmz˙iz˙r)=-2nn+1S(δsrz˙iz˙m+δsmz˙iz˙r),
we have, from (71) and (73),
(74)S=0,R=0.
Analogously, the terms with x˙jz˙m should vanish separately; that is,
(75)(P(δjrδsi+δsrδji)+Q(δjrδsi-δsrδji))x˙jz˙m+(P(δjmδsi+δsmδji)+Q(δjmδsi-δsmδji))x˙jz˙r=δsiB(x˙mz˙r+x˙rz˙m).
Since this equation can be written as
(76)(P+Q)δsi(x˙rz˙m+x˙mz˙r)+(P-Q)(δsrx˙iz˙m+δsmx˙iz˙r)=δsiB(x˙mz˙r+x˙rz˙m),
we get P+Q=B and P-Q=0; thus,
(77)P=Q=12B.
Summarizing, we see that condition (42) implies, from (59), (66), (67), (71), (74), and (77),
(78)A=0,
(48) and
(79)F˙i(x˙j,z˙j,z˙kj)=Kx˙i+Lz˙i+Bx˙rz˙ri.
(3) It remains to prove that conditions (55) and (56) imply (42). The left-hand side of (42) is
(80)F˙i(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=Kasix˙s+Lasiz˙s+Bx˙t(atmiz˙m+amiz˙tm),
and the right-hand side is
(81)akpix˙p(Ax˙k+Bz˙k)+akiF˙k(x˙s,z˙s,z˙sm)=akiKx˙k+akiLz˙k+Bx˙p(akpiz˙k+akiz˙pk).
These formulas already verify condition (42). The proof is complete.We can now summarize our results in the following theorem.Theorem 11.
LetX be a manifold and let Ξ:τX*J1TX→T2X be a morphism over idTX. The following two conditions are equivalent:(a)
Ξ is a canonical vector field.(b)
For any chart(U,φ), φ=(xi), on X(82)Ξ=B(z˙i∂∂xi+x˙rz˙ri∂∂x˙i)+(Kx˙i+Lz˙i)∂∂x˙i,
where B,K,L∈R are arbitrary constants.Proof.
(1) We show that (a) implies (b). Suppose that we have a canonical vector fieldΞ. Then, in any chart on X the 2nd-order vector field Ξ has an expression
(83)Ξ(ζ,Jx1ξ)=Ξ(U,φ)k(ζ,Jx1ξ)(∂∂xk)ζ+Ξ˙(U,φ)k(ζ,Jx1ξ)(∂∂x˙k)ζ
such that for any points x,y∈X, any charts (U,φ), φ=(xi), at x and (V,ψ), ψ=(yi), at y, and any diffeomorphism α:U→V such that α(x)=y,
(84)Ξ(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=aklΞ(U,φ)k(ζk,ξk,ξmk),Ξ˙(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=ajklx˙j(ζ)Ξ(U,φ)k(ζk,ξk,ξmk)+aklΞ˙(U,φ)k(ζk,ξk,ξmk).
(Theorem 3). If x=y and (U,φ)=(V,ψ), the components Ξ(U,φ)l and Ξ˙(U,φ)l define a differential invariant F=(Fi,F˙i) (Theorem 5); then, however, Ξ must be of the form (82) (Lemmas 9 and 10).
(2) To prove that (b) implies (a), we first show that any two members of the family of vector fields (82) agree on intersection of their domains. Let (U,φ), φ=(xi), and (V,ψ), ψ=(yi), be two charts on X such that U∩V≠∅, let (xi,x˙i,z˙i,z˙ji) and (yi,y˙i,u˙i,u˙ji) be the corresponding coordinates on τX*J1TX. We want to show that
(85)B(z˙i∂∂xi+x˙rz˙ri∂∂x˙i)+(Kx˙i+Lz˙i)∂∂x˙i=B(u˙i∂∂yi+y˙ru˙ri∂∂y˙i)+(Ky˙i+Lu˙i)∂∂y˙i;
that is, in the notation of Lemma 1,
(86)Bu˙l=∂yl∂xkBz˙k,Ky˙l+Lu˙l+By˙ru˙rl=∂2yl∂xp∂xkBx˙pz˙k+∂yl∂xk(Kx˙k+Lz˙k+x˙rz˙rk).
Each element of the family of vector fields (82) defines a differential invariant F=(Fi,F˙i), where
(87)Fi=Bz˙i,F˙i=Kx˙i+Lz˙i+Bx˙rz˙ri.
Recall that
(88)Fi(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=akiFk(x˙s,z˙s,z˙sm),F˙i(aspx˙s,aspz˙s,asmpbqsz˙m+ampbqsz˙sm)=akpiζpFk(x˙s,z˙s,z˙sm)+akiF˙k(x˙s,z˙s,z˙sm),
(Theorem 5). Now the right-hand sides of (86) can be written as
(89)aklFk(x˙i,z˙i,z˙ri),akplFk(x˙i,z˙i,z˙ri)z˙p+aklF˙k(x˙i,z˙i,z˙ri),
and the left-hand sides are
(90)Fl(y˙i,u˙i,u˙ri)=Fl(asix˙s,asiz˙s,(asmix˙m+amix˙sm)brs),F˙l(y˙i,u˙i,u˙ri)=F˙l(y˙i,u˙i,(asmix˙m+amix˙sm)brs),
where
(91)akl=(∂yl∂xk)Ψ(ζ),akpl=(∂2yl∂xp∂xk)Ψ(ζ),brp=(∂xp∂yr)Φ(ζ).
Expressions (89) and (90) prove (86) as well as existence of Ξ.
(3) To complete the proof, it remains to show that the vector fieldΞ is a canonical vector field; to this purpose, we verify condition (2) of Theorem 3. Express the vectors Ξ(ζ,Jx1ξ) and Ξ(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1)) as in formula (15), Section 4,
(92)Ξ(ζ,Jx1ξ)=Ξ(U,φ)k(ζ,Jx1ξ)(∂∂xk)ζ+Ξ˙(U,φ)k(ζ,Jx1ξ)(∂∂x˙k)ζ,Ξ(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))=Ξ(V,ψ)k(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))(∂∂yk)ζ+Ξ˙(V,ψ)k(Txα·ζ,Jα(x)1(Tα∘ξ∘α-1))(∂∂y˙k)ζ.
We have already proved that
(93)Ξ=Bz˙k∂∂xk+(Kx˙k+Lz˙k+Bx˙rz˙rk)∂∂x˙k,Ξ=Bu˙k∂∂yk+(Ky˙k+Lu˙k+By˙ru˙rk)∂∂y˙k,
on the corresponding coordinate neighbourhoods. Thus, the components of these vector fields are
(94)Ξ(U,φ)i=Bz˙i,Ξ˙(U,φ)k=Kx˙k+Lz˙k+Bx˙rz˙rk,Ξ(V,ψ)i=B~u˙i,Ξ˙(V,ψ)k=K~y˙k+L~u˙k+B~y˙ru˙rk.
We substitute these expressions into formulas (17) of Theorem 3,(95)Ξ(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=aklΞ(U,φ)k(ζi,ξi,ξji),Ξ˙(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=ajklζjΞ(U,φ)k(ζi,ξi,ξmi)+aklΞ˙(U,φ)k(ζi,ξi,ξmi).
Then since
(96)Ξ(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=B~aklξk,aklΞ(U,φ)k(ζi,ξi,ξji)=aklBξk,Ξ˙(V,ψ)l(akiζk,akiξk,akmibjmξk+akibjmξmk)=K~alkζl+L~alkξl+B~aprζp(almkbrmξl+alkbrmξml),ajklζjΞ(U,φ)k(ζi,ξi,ξmi)+aklΞ˙(U,φ)k(ζi,ξi,ξmi)=Bajlkζjξl+alk(Kζl+Lξl+Bζrξrl),
we have
(97)B~aklξk=Baklξk,K~alkζl+L~alkξl+B~aprζp(almkbrmξl+alkbrmξml)=Bajlkζjξl+alk(Kζl+Lξl+Bζrξrl).
But these conditions are equivalent to B~=B, K~=K, and L~=L as required.Having in mind that the canonical constructions are geometric constructions independent of charts, we can also state our theorem in an equivalent way as follows. LetX be an n-dimensional manifold. Then for any vector field ζ on X, there are exactly three independent canonical vector fields Z on TX that can canonically be constructed from ζ. If in a chart
(98)ζ=ζi∂∂xi,
then
(99)Z=B(ζi∂∂xi+x˙r∂ζi∂xr∂∂x˙i)+(Kx˙i+Lζi)∂∂x˙i,
where B,K,L∈R are arbitrary constants. Taking K,L=0 and B=1, we get the variational vector field; if B,L=0 and K=1, we get the Liouville vector field, and, if B,K=0 and L=1, we have the vertical lift vector field.



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