Basic notions of Poisson and symplectic geometry in local coordinates, withapplications to Hamiltonian systems.

Alexei A. Deriglazov1, 2, ∗

1Depto. de Matematica, ICE, Universidade Federal de Juiz de Fora, MG, Brazil2Department of Physics, Tomsk State University, Lenin Prospekt 36, 634050, Tomsk, Russia

(Dated: 3 de maio de 2020)

Contents

I. Introduction. 2A. Nonsingular theories. 2B. Singular nondegenerate theories. 2C. Smooth manifolds. 4D. The mapping of manifolds and induced mappings. 6

II. Poisson manifold. 7

III. Hamiltonian dynamical systems on a Poisson manifold. 8A. Hamiltonian vector fields. 8B. Lie bracket and Poisson bracket. 9C. Two basic examples of Poisson structures. 10D. Poisson mapping and Poisson submanifold. 10

IV. Degenerate Poisson manifold. 11A. Casimir functions. 12B. Induced bracket on the Casimir submanifold. 12C. Restriction of Hamiltonian dynamics to the Casimir submanifold. 14

V. First integrals of a Hamiltonian system. 15A. Basic notions. 15B. Hamiltonian reduction to the invariant submanifold. 16

VI. Symplectic manifold and Dirac bracket. 16A. Basic notions. 16B. Restriction of symplectic structure to a submanifold and Dirac bracket. 18C. Dirac’s derivation of the Dirac bracket. 20

VII. Poisson manifold and Dirac bracket. 21A. Jacobi identity for the Dirac bracket. 21B. Some applications of the Dirac bracket. 22C. Poisson manifold with prescribed Casimir functions. 23

VIII. Appendices. 25A. Jacobi identity 25B. Darboux theorem 25C. Frobenius theorem 28

Referencias 32

∗Electronic address: alexei.deriglazov@ufjf.edu.br

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I. INTRODUCTION.

In modern classical mechanics, equations of motion for the most of mechanical and field models can be obtainedas extreme conditions for a suitably chosen variational problem. If we restrict ourselves to mechanical models, theresulting system of Euler-Lagrange equations in the general case contains differential second-order and first-orderequations, as well as algebraic equations. The structure of this system becomes more transparent after the transitionto the Hamiltonian formalism, which studies the equivalent system of equations, the latter no longer contains second-order equations. For the Euler-Lagrange system consisting only of second-order equations, the transition to theHamiltonian formalism was formulated already at the dawn of the birth of classical mechanics. For the systems of ageneral form, the Hamiltonization procedure was developed by Dirac, and is known now as the Dirac formalism forconstrained systems [1–3]. In the Dirac formalism, the Hamiltonian systems naturally fall into three classes, dependingon the structure of algebraic equations presented in the system. According to the terminology adopted in [2], theyare called nonsingular, singular nondegenerate and singular degenerate theories. We briefly describe the structure ofHamiltonian equations for the first two classes. Theories of the third class usually arise if we work within a manifestlycovariant formalism, when basic variables of the theory transform linearly under the action of the Poincare group.Their description can be found in [1–3].

A. Nonsingular theories.

This is the name of mechanical systems that in the Hamiltonian formulation can be described using only first-orderdifferential equations (called Hamiltonian equations)

qa =∂H

∂pa, pa = −∂H

∂qa, (1)

where H(q, p) is a given function and qa = ddτ q

a. The variables qa describe position of the system, while pa are relatedto the velocities, and in simple cases are just proportional to them. The equations show that the function H(q, p),called the Hamiltonian, encodes in fact all the information about dynamics of the mechanical system. The equationscan be written in a more compact form, if we introduce an operation assigning to every pair of functions A(q, p) andB(q, p) a new function, denoted {A,B}, as follows:

{A,B}P =∂A

∂qa∂B

∂pa− ∂B

∂qa∂A

∂pa. (2)

This is called the canonical Poisson bracket of A and B. Then the Hamiltonian equations acquire the form

zi = {zi, H}P , (3)

where zi = (qa, pb), i = 1, 2, . . . , 2n. The equations determine integral lines zi(τ) of the vector field {zi, H} on R2n,created by function H. For smooth vector fields, the Cauchy problem, zi(τ0) = zi0, has unique solution in a vicinityof any point zi0 ∈ R2n. Formal solution to these equations in terms of power series is as follows [3]

zi(τ, zj0) = eτ{zk0 ,H(zi0)}P ∂

∂zk0 zi0. (4)

The functions zi(τ, zj0) depend on 2n arbitrary constants zj0, and hence represent a general solution to the system (3).

B. Singular nondegenerate theories.

Consider the system consisting of differential and algebraic equations

zi = {zi, H}P , (5)

Φα(zi) = 0, α = 1, 2, . . . , 2p < 2n, (6)

where H(zi) and Φα(zi) are given functions. It is supposed that Φα(zi) are functionally independent scalar func-tions1 (constraints), so the equations (6) determine 2n − 2p -dimensional surface N. The system is called singular

1 We recall that the functional independence of functions Φα guarantee that the system (6) can be resolved with respect to 2p variableszα among zi, then zα = fα(zi) are parametric equations of the surface Φα = 0.

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nondegenerate theory, if there are satisfied the following two conditions. The first condition is

det{Φα,Φβ}P∣∣Φα=0

6= 0, (7)

hence the name ”nondegenerate system”. In the Dirac formalism, functions with the property (7) are called second-class constraints. The second condition is that the functions {Φα, H}P (zi) vanish on the surface N

{Φα, H}P |Φα=0 = 0. (8)

To understand the meaning of these conditions, we adopt the followingDefinition 1.1. The system (5), (6) is called self-consistent, if through any point of the surface N passes a solutionof the system.

Fr the self-consistent system, its formal solution can br written as in (4), it is sufficient to take the integrationconstants zi0 on the surface of constraints.

Let us discuss the self-consistency of the system. Given point of the surface N, there is unique solution of equations(5), that passes through this point. It will be a solution of the whole system, if it entirely lies on the surface:

zi = {zi, H}P and Φα(zi(0)) = 0, implies Φα(zi(τ)) = 0 for all τ. (9)

This is a strong requirement, and equations (7) and (8) turn out to be a sufficient conditions for its fulfilment. Theproof with use of special coordinates of R2n can be found in [2]. A more simple proof with use of Dirac bracket willbe presented in Sect. VII B.

An example of a self-consistent system like (5), (6) will be considered in Sect. VII.Let us discuss the necessity of the condition (8).

Affirmation 1.1. Consider the system (5), (6) with functionally independent functions Φα. Then(A) {Φα, H}P |zi(τ) = 0 for any solution zi(τ), that is the algebraic equations {Φα, H}P = 0 are consequences of thesystem.(B) If the system is self-consistent, the conditions (8) hold.

Proof. (A) Let zi(τ) be a solution to (5), (6). We have Φα(zi(τ)) = 0 for all τ , this implies Φα(zi(τ)) = 0. In moredetail, we have the identity

Φα =∂Φα

∂zi[zi − {zi, H}P

]∣∣∣∣zi(τ)

+ {Φα, H}P |zi(τ) = 0, (10)

which implies {Φα, H}P = 0 for any solution zi(τ) (if any), that is the algebraic equation is a consequence of thesystem.

(B) Let zi0 be any point of the surface Φα = 0. Due to the self-consistency, there is a solution zi(τ) thatpasses through this point, zi(0) = zi0. As the equation{Φα, H}P = 0 is a consequence of the system, we have{Φα, H}P (zi(τ)) = 0, in particular {Φα, H}P (zi(0)) = {Φα, H}P (zi0) = 0, that is it vanishes at all points of thesurface N. �

Consider the system (5), (6), and now suppose that some of the functions {Φα, H}P do not vanish identically on thesurface N. As we saw above, this means that the system is not a self-consistent. Then we can look for a sub-surface ofN where the system could be a self-consistent. The procedure is as follows. We separate the functionally independentfunctions among {Φα, H}P , say Ψ1,Ψ2, . . . ,Ψk. As the equations Ψa = 0 are consequences of the system (5), (6), weadd them to the system, obtaining an equivalent system of equations. If the set Φα,Ψa is composed of functionallyindependent functions, we repeat the procedure, analysing the functions {Ψa, H}, and so on. Since the number offunctionally independent functions cannot be more than 2n, the procedure will end at some step. If, in addition tothis, the resulting set of functions satisfies the condition (7), we arrive at the self-consistent system of equations:zi = {zi, H}P , Φα(zi) = 0, Ψa = 0, . . . .

It remains to discuss what happens if, at some stage, the extended system of algebraic equations consists offunctionally dependent functions. Without loss of generality, we assume that the extended system is zi = {zi, H}P ,Φα = 0, Ψ ≡ {Φ1, H} = 0. By construction, it is equivalent to the original system, the function Ψ(zi) does notvanish identically on N, and the functions Φα,Ψ are functionally dependent. As Φα are functionally independent,we present the equations Φα(zi) = 0 in the form zα = fα(zb), and substitute them into the expression for Ψ(zi),obtaining the system zi = {zi, H}P , zα− fα(zb) = 0, Ψ(zb, fα(zb)) = 0, which is equivalent to (5), (6). The functionΨ(zb, fα(zb)) 6= 0 identically. On other hand, it does not depend on zb (otherwise we could write it in the form likez1 = ψ(z2, z3, . . .), then the functions zα − fα(zb), z1 − ψ(z2, z3, . . .) are functionally independent). So, the onlypossibility is Ψ = c = const 6= 0. This means that the system (5), (6) contains the equation c = 0, where c 6= 0.Hence the system is contradictory and has no solutions at all.

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It should be noted that the outlined procedure for obtaining a self-consistent system lies at the corner of the Diracmethod [1].

The study of Hamiltonian systems like (3) and (5), (6) gave rise to a number of remarkable mathematical construc-tions. They are precisely the subject of investigation of Poisson and symplectic geometries [4–8]. In particular, thegeometry behind a singular nondegenerate theory could be summarized by the diagram (133). This will be exploredin Sect. VII B to study the structure of the dynamical system.

In the rest of this section, we introduce our notation and recall some basic notions of the theory of differentiablemanifolds that will be useful in what follows.

C. Smooth manifolds.

Notation. Latin indices from the middle of alphabet are used to represent coordinates zk of a manifold Mn and runfrom 0 to n. If coordinates are divided on two groups, we write yk = (yα, yb), that is Greek indices from the beginningof alphabet are used to represent one group, while Latin indices from the beginning of alphabet represent anothergroup. The notation like Ui(z

j) means that we work with the functions Ui(z1, z2, . . . , zn), where i = 1, 2, . . . , n.

The notation like ∂iA(zk)|zk→fk(yj) means that in the expression ∂iA(zk) the symbols zk should be replaced on the

functions fk(yj). We often denote the inverse matrix ω−1 as ω.Definition 1.2. Vector space V = {V,U, . . .} is called the Lie algebra, if on V is defined the bilinear mapping

[, ] : V× V→ V (called the Lie bracket), with the properties

[V,U ] = −[U, V ] (antisymmetric), (11)

[V, [U,W ]] + [U, [W,V ]] + [W, [V,U ]] ≡ [V, [U,W ]] + cycle = 0 (Jacobi identity). (12)

Due to the bilinearity, we can work with basic vectors T i of the vector space. The Lie bracket for them reads

[T i, T j ] = cijkTk, (13)

where the numbers cijk are called the structure constants of the algebra. The conditions (11) and (12) are satisfied,if the structure constants obey (Exercise)

cijk = −cjik, cijacakb + cycle(i, j, k) = 0. (14)

Example 1.1. For the three-dimensional vector space with elements V = viTi, i = 1, 2, 3, let us define [T i, T j ] =

εijkT k, where εijk is the Levi-Chivita symbol with ε123 = 1. It can be verified, that the set cijk ≡ εijk has theproperties (14), so the vector space turn into a Lie algebra. It is called the Lie algebra of three-dimensional group ofrotations, see Sect. 1.2 in [3] for details.

Let Mn = {z, y, . . .} be n -dimensional manifold, and FM = {A,B, . . .} be space of scalar functions on Mn, that isthe mappings A : Mn → R. Let zi be local coordinates on Mn, that is we have an isomorphism z ∈Mn → zi(z) ∈ Rn.If z′i is another coordinate system, we have the relations

z′i = ϕi(zj), zj = ϕj(z′i), ϕi(ϕj(zk)) = zi. (15)

Let in the coordinates zi and z′i the mapping A is represented by the functions A(zi) : Rn → R and A′(z′i) : Rn → R.They are related by

A′(z′i) = A(zj)∣∣zj→ϕj(z′i) ≡ A(ϕj(z′i)). (16)

We call (16) the transformation law of a scalar function in the passage from zi to z′i. In certain abuse of terminology,we often said ”scalar function A(zi)” instead of ”the function A(zi) is representative of a scalar function A : M→ Rin the coordinates zi”.

Example 1.2. Scalar function of a coordinate. Given coordinate system zi, define the scalar function A1 : z → z1,where z1 is the first coordinate of the point z in the system zi. In the coordinates zi the mapping is representedby the function A1(z1, z2, . . . , zn) = z1. In the coordinates z′i = ϕi(zj) it is represented by A′1(z′1, z′2, . . . , z′n) =z1∣∣zj→ϕj(z′i) = ϕ1(z′1, z′2, . . . , z′n).

We often write z′i(zj) instead of ϕi(zj), zj(z′i) instead of ϕj(z′i), and use the notation z′i ≡ zi′ . In the latter case,i′ and i, when they appear in the same expression, are considered as two different indexes. For instance, in thesenotations the scalar function of z1 -coordinate in the system zi

′= zi

′(zj) is represented by the function z1(zi

′).

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Exercise 1.1. Observe that (15) implies, that derivatives of the transition functions ϕ and ϕ form the inversematrices

∂ϕi

∂z′k

∣∣∣∣z′→ϕ(z)

∂ϕk

∂zj= δij or, in short notation

∂zi

∂zk′∂zk

′

∂zj= δij . (17)

Given curve zi(τ) ∈ Mn, the numbers V i = zi(0) are called components (coordinates) of tangent vector to the

curve at the point zi(0). If V i′

are components of the tangent vector in the coordinates zi′, we have the relation

V i′

= ∂zi′

∂zi

∣∣∣τ=0

V i.

We said that we have a vector field ~V (z) on Mn, if in each coordinate system zj it is defined the set of functionsV i(zj) with the transformation law

V i′(zj

′) =

∂zi′

∂ziV i(zk)

∣∣∣∣∣z→z(z′)

. (18)

The space of vector fields on Mn is denoted TMn . In the tensor analysis, ~V (z) is called the contravariant vector field.We said that we have covariant vector field U(z) on Mn, if in each coordinate system zj it is defined the set of

functions Ui(zj) with the transformation law

Ui′(zj′) =

∂zi

∂zi′Ui(z

k(zj′)). (19)

Gradient of a scalar function A is an example of the contravariant vector field. Its components are Ui = ∂iA.Exercise 1.2. Let A(zi) = 1

2 [(z1)2 + (z2)2 + (z3)2] represent a scalar function in the coordinates zi. Then in the

coordinates zi′, defined by (15), it is represented by A′(zi

′) = 1

2 [(ϕ1(zi′))2 + (ϕ2(zi

′))2 + (ϕ3(zi

′))2]. Gradients of

these functions are Ui = zi and Ui′ = ϕj(zi′)∂ϕ

j(zi′)

∂zi′. Confirm that the two gradients are related by Eq. (19).

Similarly to this, contravariant tensor of second-rank is a quantity with the transformation law

ωi′j′(zk

′) =

∂zi′

∂zi∂zj

′

∂zjωij(zm)

∣∣∣∣∣z→z(z′)

, (20)

and so on.Exercise 1.3. Contraction of ω with covariant vector field grad A gives a quantity with the components V i = ωij∂jA.

Confirm that ~V is a contravariant vector field.Integral line of the vector field V i(zk) on Mn is a solution zi(τ) to dzi(τ)

dτ = V i(zk(τ)). We assume that V i(zk) is a

smooth field, so through each point of the manifold passes unique integral line of ~V .Submanifold of Mn. The k -dimensional submanifold N~ck ∈Mn is often defined as a constant-level surface of the

set of functionally independent scalar functions Φα(z)

N~ck = {z ∈Mn, Φα(zk) = cα, ~c ∈ Rn−k, α = 1, 2, . . . n− k}, (21)

where cα is given set of numbers.We recall that scalar functions Φα(z), α = 1, 2, . . . , n − k are called functionally independent, if for their repre-

sentatives Φα(zi) in the coordinates zi we have: rank (∂iΦα) = n− k. This implies, that contravariant vectors V(α)

with coordinates V(α)i = ∂iΦα are linearly independent. The equations Φα(zi) = cα for the functionally independent

functions can be resolved: zα = fα(za), a = 1, 2, . . . , k. So the coordinates zi are naturally divided on two groups:(zα, za), and za, a = 1, 2, . . . , k, can be taken as local coordinates of the submanifold N~ck. Below we always assume

that the coordinates have been grouped in this way, and det ∂Φα

∂zβ6= 0.

If we have only one function Φ(z), it is functionally independent if it has non vanishing gradient (or, equivalently,if the corresponding covariant vector not vanishes).

The surface of level zero is

Nk = {z ∈Mn, Φα(zk) = 0, α = 1, 2, . . . n− k}. (22)

Consider the curve zi(τ) ∈ Nk ∈ Mn, then the vector V i(z0) = dzi

dτ

∣∣∣0∈ TMn(z0) is called the tangent vector to Nk

at the point zi0 = zi(0) (see also the next subsection). The equality Φα(zi(τ)) = 0 implies V i∂iΦ∣∣z0

= 0 This justifies

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the following terminology: vector field ~V (z) on Mn is called tangent to the submanifold Nk, if V i(z)∂iΦ(z)∣∣z0

= 0 at

all points z0 ∈ Nk.Foliation of Mn. The set {N~ck, ~c ∈ Rn−k} of the submanifolds (21) is called a foliation of Mn. Note that

submanifolds with different ~c do not intercept, and any z ∈Mn lies in one of N~ck.There are coordinates, naturally adapted with the foliation: zk → yk = (yα, ya), with the transition functions

ya = za, yα = Φα(zβ , zb). In these coordinates the sumanifolds N~ck look like hyperplanes:

N~ck = {yi ∈Mn, yα = cα}, (23)

and ya = za can be taken as local coordinates of N~ck. The useful identity is

A(zi(yj))∣∣yα=0

= A(fα(za), za)|za→ya . (24)

Lie bracket (commutator) of vector fields is bilinear operation [ , ] : TMn × TMn → TMn , that with each pair of

vector fields ~V and ~U of TMn associates the vector field [~V , ~U ] of TMn according to the rule

[~V , ~U ]i = V j∂jUi − U j∂jV i. (25)

The quantity [~V , ~U ]i is indeed a vector field, which can be verified by direct computation. We have [~V , ~U ]i =

V j′∂j′(

∂zi

∂zi′U i

′) − (V ↔ U) = ∂zi

∂zi′(V j

′∂j′U

i′ − (V ↔ U)) = ∂zi

∂zi′[~V ′, ~U ′]i

′, in agreement with Eq. (18). The Lie

bracket has the properties (11) and (12), and turns the space of vector fields into infinito-dimensional Lie algebra.

Each vector field determines a linear mapping ~V : FMn → FMn on the space of scalar functions according to therule

~V : A→ ~V (A) = V i∂iA. (26)

Notice that ~V (A) = 0 for all A implies V i = 0. Then the Lie bracket can be considered as a commutator of twodifferential operators

[~V , ~U ](A) = ~V (~U(A))− ~U(~V (A)). (27)

Using this formula, it is easy to confirm by direct computation the Jacobi identity (12) for the Lie bracket (25).

D. The mapping of manifolds and induced mappings.

Given two manifolds Nk = {xa}, Mn = {zi}, the functions zi = φi(xa) determine the mapping

φ : Nk →Mn, xa → zi = φi(xa) ≡ zi(xa). (28)

If φ is an injective function: rank ∂φi

∂xa = k, the image of the mapping is k -dimensional submanifold of Mn: Nk = {zi ∈Mn, z

α − fα(za) = 0}, where the equalities zα = fα(za) are obtained excluding xa from the equations zi = φi(xa).So the manifold Nk can be identified with this submanifold of Mn.

Conversely, the parametric equations zα = fα(za) of a submanifold (22) determine the mapping of embedding

η : Nk = {za} →Mn = {zi}, za → zi = (zα, za), where zα = fα(za). (29)

The mapping φ allows to relate various geometric structures defined on the two manifolds. We start from thespaces of covariant and contravariant tensors at the points x0 and z0 = φ(x0). Taking, for definiteness, the third-ranktensors, φ induces the following mappings

T(0,3)M → T(0,3)

N , Uijk(z0)→ Uabc(x0) =∂zi(x0)

∂xa∂zj(x0)

∂xb∂zk(x0)

∂xcUijk(z0), (30)

T(3,0)N → T(3,0)

M , V abc(x0)→ V ijk(z0) =∂zi(x0)

∂xa∂zj(x0)

∂xb∂zk(x0)

∂xcV abc(x0). (31)

For the case of vector fields, the notion of induced mapping,

V i(z0) =∂zi(x0)

∂xaV a(x0), (32)

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is consistent with the notion of a tangent vector: if V a is tangent vector to the curve xa(τ), then V i is tangent vectorto the image zi(xa(τ))

V i(z(τ)) =d

dτzi(xa(τ)). (33)

Concerning the fields on the manifolds, φ naturally generates the iduced mappings of scalar functions and ofcovariant tensor fields. For the functions the induced mapping

φ∗ : FMn → FNk , A(zi)→ A(xa) = A(zi(xa)) (34)

is just the composition: A = A ◦ φ. For the covariant tensor fields we have

φ∗ : T(0,3)Mn → T(0,3)

Nk , Uijk(zi)→ Uabc(xa) =

∂zi

∂xa∂zj

∂xb∂zk

∂xcUijk(zi(xa)). (35)

II. POISSON MANIFOLD.

Let on the space of functions FM is defined a bilinear mapping {, } : FM × FM → FM (called the Poisson bracket),with the properties

{A,B} = −{B,A} (antisymmetric), (36)

{A, {B,C}}+ cycle = 0 (Jacobi identity), (37)

{A,BC} = {A,B}C + {A,C}B (Leibnitz rule). (38)

When FM is equipped with the Poisson bracket, the manifold Mn is called the Poisson manifold. Comparing (36) and(37) with (11) and (12), we see that the infinite-dimensional vector space FM is equipped with the structure of a Liealgebra.Exercise 2.1. Show that constant functions: A(z) = c for any z, have vanishing brackets (commute) with all otherfunctions.

One of the ways to define the Poisson structure on Mn is as follows.Affirmation 2.1. Let ωij(z) be the contravariant tensor of second rank on Mn. The mapping

{A,B} = ∂iA ωij ∂jB, (39)

determines the Poisson bracket, if the tensor ω obeys the properties

ωij = −ωji (antisymmetric), (40)

ωip∂pωjk + cycle(i, j, k) = 0. (41)

In particular, each numeric antisymmetric matrix determines a Poisson bracket. We call ω the Poisson tensor.Proof. First, we note that (40) implies (36). Second, the mapping (39), being combination of derivatives, is bilinearand automatically obeys the Leibnitz rule. To complete the proof, we need to show that (41) is equivalent to (37).Using (41) we obtain by direct computation

{A, {B,C}}+ cycle(A,B,C) = ∂iA∂jB∂kCωip∂pω

jk + cycle(A,B,C) + ωipωjk∂p [∂iA∂jB∂kC] + cycle(A,B,C)(42)

By direct computation, we can show also that in the first term on r.h.s. the cycle(A,B,C) is equivalent to cycle(i, j, k).So we write the previous equality as

{A, {B,C}}+ cycle(A,B,C) = ∂iA∂jB∂kC[ωip∂pω

jk + cycle(i, j, k)]

+

ωipωjk∂p [∂iA∂jB∂kC + ∂jA∂kB∂iC + ∂kA∂iB∂jC] . (43)

The second line in this equality identically vanishes due to symmetry properties of this term. Indeed, we write thefirst term of the line as follows:

ωipωjk∂p [∂iA∂jB∂kC] = ωipωjk∂p∂i [A∂jB∂kC]− ωipωjk∂p [A∂i∂jB∂kC +A∂jB∂i∂kC] =

ωipωjk∂p [A∂k(∂iB∂jC − [i↔ j])] . (44)

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The two remaining terms of the line we write as

ωipωjk∂p [∂jA∂kB∂iC + ∂kA∂iB∂jC] = ωipωjk∂p [∂kA(∂iB∂jC − [i↔ j])] =

ωipωjk∂p∂k [A(∂iB∂jC − [i↔ j])]− ωipωjk∂p [A∂k(∂iB∂jC − [i↔ j])] . (45)

The last terms in (44) and (45) cancel each other, while the first term in (45) iz zero, being the trace of the productof symmetric Dij ≡ ωipωjk∂p∂k and antisymmetric Eij ≡ A(∂iB∂jC − [i ↔ j]) quantities. Thus we have obtainedthe identity

ωipωjk∂p [∂iA∂jB∂kC + ∂jA∂kB∂iC + ∂kA∂iB∂jC] = 0. (46)

Taking this into account in (43), we see the equivalence of the conditions (37) and (41). �Affirmation 2.2. Let the bracket (39) obeys the Jacobi identity in the coordinates zi. Then the Jacobi identity issatisfied in any other coordinates.

This is an immediate consequence of tensor character of involved quantities. Indeed, the bracket {A,B} =∂iAω

ij∂jB is a contraction of three tensors and so is a scalar function under diffeomorphisms. Then the same istrue for {A, {B,C}}. Let us denote l.h.s. of the Jacobi identity as D(z). Then the Jacobi identity is the coordinate-independent statement that the scalar function D(z) identically vanishes for all z ∈Mn. This can be verified also bydirect computations, see Appendix A. As a consequence, the right hand side of Eq. (41) is a tensor of third rank2.

For the scalar functions of coordinates (see Example 1.2), the Poisson bracket (39) reads

{zi, zj} = ωij . (47)

In classical mechanics these equalities are known as fundamental brackets of the coordinates. Observe that the identity(41) can be written as follows: {zi, {zj , zk}+ cycle(i, j, k) = 0.

The bracket (39) is called nondegenerate if detω 6= 0, and degenerate when detω = 0. Examples will be presentedbelow: (60) is nondegenerate while (64) is degenerate. The structure of the matrix ω depends on its rank, and becomesclear in the so called canonical coordinates specified by the following theorem:Generalized Darboux theorem. Let rank ω = 2k at the point zi ∈ Mn. Then there are local coordinateszi

′= (zβ

′, za

′), a′ = 1, 2. . . . , 2k, β′ = 1, 2, . . . , p = n− 2k such that ω in some vicinity of zi has the form:

ωi′j′ =

0p×p 0 00 0k×k 1k×k0 −1k×k 0k×k

, (48)

or

ωa′b′ =

(0 1−1 0

), ωβ

′j′ = ωj′β′

= 0, where j′ = 1, 2, . . . , n. (49)

Proof is given in Appendix B. We recall that determinant of any odd-dimensional antisymmetric matrix va-nishes, this implies that rank ω is necessary an even number, as it is written above. Let us further denoteza

′= (q1, q2, . . . , qk, p1, p2, . . . , pk). Then, in terms of fundamental brackets, the equalities (49) can be written as

follows:

{qa′, pb′} = δa

′

b′ , {qa′, qb

′} = 0, {pa′ , pb′} = 0, {zj

′, zβ

′} = 0. (50)

III. HAMILTONIAN DYNAMICAL SYSTEMS ON A POISSON MANIFOLD.

A. Hamiltonian vector fields.

Using the Poisson structure (39), with each function H(zi) ∈ FM we can associate the contravariant vector fieldXiH ≡ ωij∂jH = {zi, H} ∈ TM. That is we have the mapping

ω : FM → TM, ω : H → [ω(H)]i = ωij∂jH, we also denote ω(H) ≡ ~XH ∈ TM. (51)

2 This is a non trivial affirmation, since ∂pωjk is not a covariant object.

9

~XH is called the Hamiltonian vector field of H. Then

zi = {zi, H} ≡ ωij∂jH, (52)

are called Hamiltonian equations, the scalar function H is called the Hamiltonian. Solutions zi(τ) of the equations

are called integral lines of the vector field {zi, H} created by H on Mn. We assume that ~XH is a smooth vector field,

so the Cauchy problem for (52) has unique solution in a vicinity of any point of Mn. ~XH at each point is tangentvector to the integral line that passes through this point.

Let zk(τ) be integral line of ~XA and B be scalar function with grad B 6= 0. Then we can write

d

dτB(zk(τ)) = {B,A}|z(τ) . (53)

Using this equality, and the fact that integral lines pass through each point of Mn, it is easy to prove the three affir-mations presented below. They will be repeatedly used (and sometimes rephrased) in our subsequent considerations.

Affirmation 3.1. Integral line of ~XH entirely lies on one of the surfaces H(zk) = c = const.

Denote ~X(j) the Hamiltonian vector field associated with scalar function of the coordinate zj . Its components are

Xi(j) = ωik∂kz

j = ωij . Hence the Poisson matrix can be considered3 as composed of the columns ~X(j)

ω = ( ~X(1), ~X(2), . . . , ~X(n)). (54)

According to the Affirmation 3.1, integral lines of the vector ~X(j) lie on the hyperplanes zj = const.Affirmation 3.2. Given scalar functions H and Qα, α = 1, 2, . . . , n− k, the following two conditions are equivalent:

(A) Integral lines of ~XH lie in the submanifolds N~ck = {z ∈Mn, Qα = cα, H = c}.(B) All Qα commute with H: {Qα, H} = 0, for all z ∈Mn.

Affirmation 3.3. Let Aα, α = 1, 2, . . . , n − k be functionally independent scalar functions, and denote ~V(α) theHamiltonian field of Aα. The following two conditions are equivalent:

(A) Integral lines of ~V(α) lie in the submanifolds N~ck = {z ∈Mn, Aα = cα}.(B) {Aα, Aβ} = 0 on Mn.

B. Lie bracket and Poisson bracket.

We recall, that the spaces of scalar functions and of vector fields on Mn are the infinito-dimensional Lie algebras:

FM = {A,B, . . . , { , } } and TM = {~V , ~U, . . . , [ , ] }.Affirmation 3.3. The mapping (51) respects the Lie products of FM and TM:

ω({A,B}) = −[ω(A), ω(B)], or, equivalently ~X{A,B} = −[ ~XA, ~XB ]. (55)

According to the last equality, the Hamiltonian vector fields form a subalgebra of the Lie algebra TM.Proof. Using the vector notation (26), we can present the Poisson bracket as follows:

{A,B} = − ~XA(B). (56)

The equality (55) is the Jacobi identity (38) rewritten in the vector notations. Indeed

{{A,B}, C} = {A, {B,C}} − {B, {A,C}}, or ~X{A,B}(C) = ~XA( ~XB(C))− ~XB( ~XA(C)), (57)

for all C, which is just (55). �We also note that in the vector notation, the Jacobi identity (41) states that Hamiltonian fields of coordinates form

the closed algebra

[ ~X(i), ~X(j)] = c(i)(j)(k) ~X(k), (58)

with the structure functions c(i)(j)(k) = −∂kωij .

Exercise 3.1 Show that {Q,H} = const implies [ ~XQ, ~XH ] = 0.

3 Notice that it is an example of coordinate-dependent statement.

10

C. Two basic examples of Poisson structures.

1. Consider the space R2n, denote its coordinates zi = (q1, q2, . . . , qn, p1, p2, . . . , pn) ≡ (qa, pb), a, b = 1, 2, . . . , n, andtake the matrix composed from four n× n blocks as follows:

ωij =

(0 1−1 0

). (59)

In all other coordinate systems zi′, we define components of the matrix ωi

′j′ according to Eq. (20). Then ω is thecontravariant tensor of second rank, which (in the system z) determines the Poisson structure on R2n according toEq. (39):

{A,B}P =∂A

∂qa∂B

∂pa− ∂B

∂qa∂A

∂pa, fundamental (nonvanishing) brackets: {qa, pb}P = δab. (60)

As ω is the numeric matrix, the condition (41) is satisfied in the coordinate system (qa, pb). According to Affirmation2.2, it is then satisfied in all other coordinates. Hamiltonian equations acquire the following form:

qa = {qa, H}P =∂H

∂pa, pa = {pa, H}P = −∂H

∂qa. (61)

It is known (see Sect. 2.9 in [3]) that they follow from the variational problem for the functional

SH : (qa(τ), pa(τ))→ R; SH =

∫ τ2

τ1

dτ [paqa −H(qa, pb)] . (62)

In classical mechanics, R2n equipped with the coordinates (qa, pb) is called the phase space, the bracket (60) is calledthe canonical Poisson bracket, while the functional SH is called the Hamiltonian action.2. Given manifold Mn, let cijk be structure constants of an n -dimensional Lie algebra. We define ωij(z) = cijkz

k.Then the equalities (14) imply (40) and (41), so the tensor ωij determines a Poisson structure on Mn. The corres-ponding bracket

{A,B}LP = ∂iAcijkzk∂jB, fundamental brackets: {zi, zj}LP = cijkz

k, (63)

is called the Lie-Poisson bracket. In particular, the Lie algebra of rotations determines the Lie-Poisson bracket on R3

ωij = εijkzk. (64)

Let Bi are coordinates of a constant vector B ∈ R3. Taking H = ziBi as the Hamiltonian, we obtain the Hamiltonianequations (called the equations of precession)

zi = εijkBjzk, or z = B× z, (65)

where B×z is the usual vector product in R3. For any solution z(τ), the end of this vector lies in a plane perpendicularto B, and describes a circle around B, with an angular velocity equal to the magnitude |B| of this vector. The axisof the gyroscope in the Earth’s gravitational field obeys this equation.

D. Poisson mapping and Poisson submanifold.

Here we discuss the mappings which are compatible with Poisson brackets of the involved manifolds. Intuitively,such a mapping turns the bracket of one manifold into the bracket of another. As an instructive example, we firstconsider the manifolds with the brackets (59) and (63). Introduce the mapping

φ : R2n →Mn, (qa, pb)→ za = φa(q, p) = −cabcpbqc. (66)

Computing the canonical Poisson bracket (60) of the functions φa(q, p), we obtain a remarkable relation between thetwo brackets (Exercise):

{φa(q, p), φb(q, p)}P = cabcφc(q, p), or {φa(q, p), φb(q, p)}P = {za, zb}LP

∣∣z→φ(q,p)

, or (67)

11

∂iφaωij∂jφ

b = ωab(za)∣∣z→φ(q,p)

. (68)

The relation (68) shows that Poisson structures ω and ω are related by the tensor-like law (20). The relations (67)show that Poisson brackets of the special functions φa on R2n are the same as fundamental Lie-Poisson brackets (63)of the manifold Rn. We can made these relations to hold for an arbitrary scalar functions, by using the inducedmapping between the functions A(za) of Mn and A(qa, pb) of R2n

φ∗ : A(za)→ A(qa, pb) ≡ φ∗(A)(q, p) = A(φa(q, p)). (69)

This implies the following relation between the Poisson and Lie-Poisson brackets (Exercise):

{φ∗(A), φ∗(B)}P = φ∗ ({A,B}LP ) . (70)

Formalizing this example, we arrive at the notion of a Poisson mapping.Definition 3.1. Consider the Poisson manifolds Nk = {xa, {A, B}N} and Mn = {zi, {A,B}M}. The mapping (29)is called Poisson mapping if the induced mapping (34) preserves the Poisson brackets

{φ∗(A), φ∗(B)}N = φ∗ ({A,B}M) . (71)

This allows us to compare Poisson brackets of M and N: given two functions A and B of M and their images A andB, we can compare the bracket {A, B}N with the image of scalar function {A,B}M, that is with φ∗ ({A,B}M). Ifthey coincide, we have the mapping (29) that respects Poisson structures of the manifolds. The mapping (66) is anexample of Poisson mapping of the canonical Poisson manifold on the Lie-Poisson manifold.

Poisson submanifold of the Poisson manifold. Let the Poisson manifold Nk be a submanifold of Mn. As wesaw above, there is a natural mapping of projection FN → FM, that can be used to compare the Poisson structures oftwo manifolds. This mapping is used to introduce an important notion of the Poisson submanifold.

Any scalar function A(zi) on Mn is defined, in particular, at the points of Nk, and hence we can consider therestriction of A(zi) on Nk. This is just the projection mapping η∗:

η∗ : FM → FN, A(zi)→ A(za) = A(zi(za)). (72)

The submanifold Nk is called the Poisson submanifold, if the mapping η∗ turn the bracket of M into the bracket of N:

η∗ ({A,B}M ) = {A, B}N . (73)

For the latter use, we write these expressions in local coordinates zi of Mn. Let Nk be the submanifold determinedby functionally independent set of scalar functions Φβ(zi) of Mn

Nk = {zi ∈M; Φβ(zi) = 0}. (74)

Solving Φβ(zi) = 0, we obtain the parametric equations zβ = fβ(za), and take za as local coordinates of Nk. Themapping η : Nk →Mn is the embedding: η : za → (zα, za), where zα = fα(za). The mapping (72) reads as follows:

η∗ : A(zβ , za)→ A(za) = A(fβ(za), za), (75)

while the expression (73) is

{A(zβ , za), B(zβ , za)}M∣∣zβ→fβ(za)

= {A(fβ(za), za), B(fβ(za), za)}N. (76)

Various examples of Poisson mappings and Poisson submanifolds will appear in the analysis of dynamical systems,see section V B .

IV. DEGENERATE POISSON MANIFOLD.

The affirmations discussed above are equally valid for nondegenerate and degenerate manifolds. Now we consi-der some characteristic properties of a Poisson manifold with degenerate Poisson bracket. Nondegenerate Poissonmanifolds will be discussed in Sect. VI.

12

A. Casimir functions.

Poisson manifold with a degenerate bracket has the following remarkable property: in the space FM there arefunctionally independent functions that have null brackets (commute) with all functions of FM. They are called theCasimir functions.Affirmation 4.1. Let Kβ are p functionally independent Casimir functions of a Poisson manifold Mn. Then ω isdegenerated, and rank ω ≤ n− p.Proof. Kβ commutes with any function, in particular, we can write {zi,Kβ} = 0, or ωij∂jKβ = 0. The latter

equation means that ω admits at least p independent null-vectors ~V(β), so rank ω ≤ n− p. �Affirmation 4.2. Consider Poisson manifold with rank ω = n−p. Then there are exactly p functionally independentCasimir functions:

{zi,Kβ} = 0, or ~XKβ = 0, i = 1, 2, . . . , n, β = 1, 2, . . . , p. (77)

Proof. To simplify the proof, let us consider 2n+ 1 -dimensional Poisson manifold with rank ω = 2n. According theDarboux theorem, there are canonical coordinates zi

′such that one of them coomutes with all others, say z1′

commutewith all coordinates, {zi′ , z1′} ≡ ωi

′1′= 0. Let us define a scalar function as follows: at the point z ∈ M2n+1, its

value coincides with the value of first coordinate of this point in the canonical system: K(z) = z1′. In the canonical

coordinates this function is represented by K ′(z1′, z2′

, . . . , z(2n+1)′) = z1′. Then, according to Eqs. (16) and (15), in

the original coordinates it is represented by K(zi) = z1′(z1, z2, . . . , z2n+1). Let us confirm that K(z) is the Casimir

function. Using the transformation laws (16), (19) and (20), we obtain

{zi,K(z)} = ωij(z)∂jK(z) =∂zi

∂zi′

∣∣∣∣z′(z)

ωi′j′ ∂z

j

∂zj′

∣∣∣∣z′(z)

∂zk′

∂zj∂K ′(z′)

∂zk′

∣∣∣∣z′(z)

=[∂zi

∂zi′ωi

′j′ ∂z1′

∂zj′

]∣∣∣∣∣z′(z)

=∂zi

∂zi′

∣∣∣∣z′(z)

ωi′1′

= 0. (78)

So, in the Darboux coordinates the Casimir functions are functions of zβ′, see (50). As the complete set of functionally

independent Casimir functions, we can take the coordinates zβ′

themselves. More than p functionally independentCasimir functions would be in contradiction with Affirmation 4.1. �

The equations (77) have the following remarkable interpretation: all Hamiltonian vector fields XiA = ωij∂jA on Mn

are tangent to the hypersurfaces N~ck = {z ∈ Mn, Kβ(zi) = cβ}, and their integral lines lie in N~ck. Indeed, let zi(τ)

be an integral line of XiA. Consider: d

dτKβ(zi(τ) = XiA∂iKβ(zi)

∣∣z(τ)

= {Kβ , A}|z(τ) = 0. So zi(τ) lies on one of the

surfaces, and then ~XA ∈ TN.Exercise 4.1. Observe: K = zizi is the Casimir function of (64).

B. Induced bracket on the Casimir submanifold.

Below, we always work with a set of functionally independent Casimir functions, assuming that the coordinates

were grouped in two subsets, zi = (zα, za) such that det∂Kβ∂zα 6= 0. Then Eq. (77) reads

ωiα∂αKβ + ωib∂bKβ = 0. (79)

Note that this allows us to restore whole ωij(zk) from the known block ωab(zk) as follows:

ωaα = −ωab∂bKγ(K−1)γα, ωαβ = −ωαb∂bKγ(K−1)γβ . (80)

Geometric interpretation of these relations will be duscussed in Sect. VII C.Consider the submanifold of Mn

N = {z ∈Mn, Kβ(zα, za) = 0}, (81)

determined by Casimir functions (we can take either all functionally independent Casimirs of Mn, or some part ofthem). For shorteness, we call N the Casimir submanifold. Poisson structure ωij(zk) on Mn can be used to generatea natural Poisson structure on the manifold N. To construct it, we first formulate some properties of the set Kβ .

13

Affirmation 4.3. Let Kβ(zα, zb) are Casimir functions, and zα = fα(zb) is a solution to the equations Kβ(zα, zb) = 0.Thena) zα − fα(zb) are Casimir functions;b) The Poisson tensor of Mn satisfies the identity

ωiα = ωia∂afα. (82)

Proof. a) Contracting the expression {zi, zα−fα} = ωiα−ωia∂afα with ∂αKβ and using (79), we obtain ωiα∂αKβ−ωia∂af

α∂αKβ = −ωia(∂aKβ + ∂afα∂αKβ) = −ωia∂aKβ(fα, za) = 0 since Kβ(fα, za) ≡ 0. As ∂αKβ is an invertible

matrix, the equality {zi, zα − fα}∂αKβ = 0 implies {zi, zα − fα} = 0.b) Let Kα are Casimir functions. According to Item a), zα − fα(zb) also are the Casimir functions, then {zi, zα −

fα} = 0, or ωiα = ωia∂afα. �

Affirmation 4.4. For any function B(zi) and Casimir functions zβ − fβ(zb), there is the identity

ωip∂pB∣∣zβ=fβ(zc)

= ωia(zc, fβ(zc))∂aB(zc, fβ(zc)), (83)

where p = (1, 2, . . . , n), while a = (1, 2, . . . , n− p). Note the geometric interpretation of this equality: if two functions

B and B′ of FM coincide on N, their Hamiltonian vector fields also coincide on N: B|N = B′|N implies ~XB |N = ~XB′ |N.Proof. Let us write

ωip∂pB∣∣zβ=fβ(zc)

= ωia∂aB∣∣zβ=fβ(zc)

+ ωiβ∂βB∣∣zβ=fβ(zc)

. (84)

Using the identity (82) we have ωiβ∂βB = ωid∂dfβ(zc)∂βB(zc, zβ) ≡ ωid

[∂dB(zc, fβ(zc))− ∂dB(zc, zβ)

∣∣zβ=fβ(zc)

].

Using this expression for the term ωiβ∂βB in (84), we arrive at the desired identity (83). �We are ready to construct the induced Poisson structure. We take za as local coordinates of N, and using ωab -block

of ωij , introduce the antisymmetric matrix

ωab(zc) = ωab(fβ(zc), zc). (85)

Let us confirm that ω obeys the condition (41). We write the condition (41), satisfied for ωij , taking the indices i, j, kequals to a, b, c, and substitute zβ = fβ(zc). This gives us the identity

ωap∂pωbc∣∣zβ=fβ(zc)

+ cycle(a, b, c) = 0. (86)

Using the identity (83), we immediately obtain

ωad(fβ(zc), zc)∂dωbc(fβ(zc), zc) + cycle(a, b, c) = 0, (87)

which is just the Jacobi identity for the tensor ωab. Thus the bracket

{A(za), B(za)} = ∂aAωab∂bB, (88)

defined on N, obeys the Jacobi identity. In an another coordinate system on N, say za′, we define the components

ωa′b′ according the rule (20):

ωa′b′ = ∂az

a′∂bzb′ ωab

∣∣∣za(za′ )

. (89)

Then ω is a tensor of N, while the expression (88) is a scalar function, as it should be for the Poisson bracket.It remains to confirm, that the obtained bracket does not depend on the coordinate system chosen for its cons-

truction. Let zi = φi(zj′) be transition functions between two coordinate systems. For a point of N, this implies the

following relation between its local coordinates za and za′:

za = φa(fα′(za

′), za

′). (90)

Using these finctions in the expression (89), we obtain the components ωa′b′(za

′) of the tensor ωab(za) in the co-

ordinates za′. On the other hand, using the Poisson tensor ωi

′j′ in coordinates zi′, we could construct the matrix

ωa′b′(fα

′(za

′), za

′) according the rule (85). The task is to show that ωa

′b′ coincides with ωa′b′ .

14

As the functions ωa′b′(fβ

′(za

′), za

′) are components of the tensor ωi

′j′ of M, we use the transformation law (20),and write

ωa′b′(fβ

′(za

′), za

′) = ωa

′b′(zβ′, za

′)∣∣∣N

= ∂kza′ωkp∂pz

b′∣∣∣zi(zi′ )

∣∣∣∣N. (91)

In the last expression we have a quantity D(zi), and need to replace the coordinates zi by the transition functions

zi(zi′) taken at the point of N. Equivalently, we can first to restrict D on N, replacing zβ on fβ(za), and then to

replace za on its expression (90) through coordinates za′. Making this, and then using the identity (83), we obtain

ωa′b′(fβ

′(za

′), za

′) = ∂kz

a′(zi)ωkp(zi)∂pzb′(zi)

∣∣∣zβ→fβ(za)

∣∣∣∣za(za′ )

=

∂aza′(zβ(za), za)ωab(zβ(za), za)∂bz

b′(zβ(za), za)∣∣∣za(za′ )

. (92)

Comparing this expression with (89), we arrive at the desired result: ω = ω.Exercise 4.2. Confirm that the Poisson manifold N is the Poisson submanifold of M in the sense of definition

(73).Consider the Poisson manifold Mn with rank ωij = n − p, and let the submanifold (81) is determined by a

complete set of p functionally independent Casimir functions. Then the induced Poisson structure is non degenerate:det ωab 6= 0. To see this, suppose an opposite, det ωab = 0, and let zi be canonical coordinates of M. Then ω is anumeric degenerate matrix, so it has a numeric null-vector, ωabcb = 0. As a consequence, the function A(zi) = zacacommutes with all coordinates (50), and hence is a Casimir function of M. It depends only on the variables za, soit is functionally independent of the Casimir functions zβ − fβ(za) = 0. This is in contradiction with the conditionrank ωij = n− p.

Let us resume the obtained results. Let M be Poisson manifold with degenerate Poisson bracket ω. Then on thesubmanifold N ∈ M, determined by any set of Casimir functions, there exists the Poisson bracket ω, such that thePoisson manifold N turns into the Poisson submanifold of M. In the coordinates zi = (zβ , za), divided on two groupsaccording to the structure of Casimir functions (81), components of the matrix ω coincide with fundamental bracketsof coordinates za restricted on N:

ωab = {za, zb}M∣∣zβ→fβ(za)

. (93)

C. Restriction of Hamiltonian dynamics to the Casimir submanifold.

The degeneracy of a Poisson structure implies that integral lines of any Hamiltonian system on this manifold havespecial properties: any solution started in a Casimir submanifold entirely lies in it. So the dynamics can be consistentlyrestricted on the submanifold, and the resulting equations are still Hamiltonian. To discuss these properties, we willneed the notion of an invariant submanifold.Definition 4.1. The submanifold N ∈Mn is called an invariant submanifold of the Hamiltonian H, if any trajectoryof (52) that starts in N, entirely lies in N

zi(0) ∈ N, → zi(τ) ∈ N for any τ. (94)

Affirmation 4.5. A Casimir submanifold of (Mn, {, }) is invariant submanifold of any Hamiltonian H ∈ FM.Proof. Let the solution zi(τ) of (52) cross N at τ = 0: Kα(zi(0)) = 0. We need to show that Kα(zi(τ)) = 0 for anyτ . Compute:

d

dτKα(zi(τ)) = ∂iKαz

i = {Kα, H}|z(τ) = 0, (95)

so Kα(zi(τ)) = c = const, then Kα(zi(0)) = 0 implies c = 0. �Affirmation 4.6. Solutions of (52): zi = ωij∂jH, that belong to the Casimir submanifold (81), obey the Hamiltonianequations

za = ωab∂bH(zc, fα(zc)), (96)

where za are local coordinates on N and ωab is the restriction of ωij on N

ωab = ωab(zc, fα(zc)). (97)

15

Proof. According to Affirmation 4.5, we can add the algebraic equations zβ = fβ(za) to the system (52), thusobtaining a consistent equations with solutions living on N. In the equations for za we substitute zβ = fβ(za), andusing the identity (83), obtain closed system (96) and (97) for determining za. Then the equations for zβ can beomitted from the system. Jacobi identity for ω has been confirmed above. �

V. FIRST INTEGRALS OF A HAMILTONIAN SYSTEM.

A. Basic notions.

If z(τ) is a solution of (52), for any function Q(z) we have: Q(z(τ)) = {Q(z), H(z)}|z(τ). In other words, functions

Q(z) follow the Hamiltonian dynamics together with z(τ). The function Q(z) (with non-vanishing gradient) is called

the integral of motion, if it preserves its value along the trajectories of (52): Q(z(τ)) = const, or Q(z(τ)) = 0 (notethat the value of Q(z(τ)) can vary from one trajectory to another).Affirmation 5.1. Q(z) is an integral of motion of the system (52), if and only if its bracket with H vanishes

{Q,H} = 0. (98)

Since {H,H} = 0, the Hamiltonian himself is an example of the integral of motion. So, any Hamiltonian systemadmits at least one integral of motion. The Casimir functions obey to Eq. (98) for any H, so they represent firstintegrals of any Hamiltonian system on a given manifold. As a consequence, a Hamiltonian system on the manifoldMn with rank ω = n− p has at least p+ 1 integrals of motion.Exercise 5.1. a) Confirm the Affirmation 5.1. (hint: take into account, that integral lines of (52) cover all themanifold).b) Observe: if Q1 and Q2 are integrals of motion, then c1Q1 + c2Q2, f(Q1) and {Q1, Q2} are the integrals of motionas well. The integral of motion {Q1, Q2} may be functionally independent on Q1 and Q2.

Using the first integrals, we can restrict the dynamics to the surface of their level. This method, called the reductionprocedure, allows to reduce the number of differential equations that we need to solve. It is based on the followingaffirmations.Affirmation 5.2. Let Qα(z), α = 1, 2, . . . , p are functionally independent integrals of motion of H. Then Nc = {z ∈Mn, Qα(z) = cα = const} is n− p -dimensional invariant submanifold of H.

Indeed, given solution with z(0) ∈ Nc, that is Qα(z(0)) = cα, we have Qα(z(τ)) = Qα(z(0)) = cα for any τ , so thetrajectory z(τ) entirely lies in Nc. The manifolds Nc and Nd with c 6= d do not intercept. So the Poisson manifoldMn is covered by p -parametric foliation of the invariant submanifolds Nc.

Since the Casimir function is an integral of motion of any Hamiltonian, the Affirmation 5.2 implies, once again, ageometric interpretation of Eq. (77): integral lines of all Hamiltonian vector fields of Mn lie on the surfaces of Casimirfunctions.Affirmation 5.3. Let the Hamiltonian system

zi = {zi, H}, (99)

admits p functionally independent integrals of motion Qα(z) = cα. We present them in the form zα = fα(zb, cα).Then the system of n differential equations (99) is equivalent to the system

zb = {zb, H}, zα = fα(zb, cα), (100)

composed of n− p differential and p algebraic equations.Proof. Adding the consequences zα = fα(zb, cα) to the equations (99), we write the resulting equivalent system asfollows

zα = {zα, H}, zb = {zb, H}, zα = fα(zb, cα). (101)

To prove the equivalence of (100) and (101), we need to show that the equation zα = {zα, H} is a consequence of thesystem (100). Let zα(τ), zb(τ) be a solution to (100). Computing derivative of the identity zα(τ) ≡ fα(zb(τ), cα) wehave zα(τ) = ∂bf

α(zb, cα)∣∣z→z(τ)

zb = ∂bfα(zb, cα){zb, H}

∣∣z→z(τ)

= {fα, H}|z→z(τ) = {zα, H}|z→z(τ). On the last

step we used (98). Hence, the desired equation is satisfied by any solution to the system (100). �Example 5.1. Using the reduction procedure, any two-dimensional Hamiltonian system can be completely

integrated, that is solving the equations is reduced to evaluation of an integral. Indeed, consider the system

16

x = {x,H(x, y)} ≡ h(x, y), y = {y,H(x, y)}. We assume that grad H 6= 0 (otherwise H = const and the sys-tem is immediately integrated). Let y = f(x, c) is a solution to the equation H(x, y) = c. As H is an integral ofmotion, we use the Affirmation 5.3 to present the original system in the equivalent form: x = h(x, y), y = f(x, c).Replacing y on f(x, c) in the differential equation, the latter can be immediately integrated. The general solutionx(τ, c, d), y(τ, c, d) in an implicit form is as follows:∫

dx

h(x, f(x, c))= τ + d, y = f(x). (102)

There is a kind of multi-dimensional generalization of this example, see Affirmation B1 in Appendix B.

B. Hamiltonian reduction to the invariant submanifold.

When a dynamical system admits an invariant submanifold, its dynamics can be consistently restricted on thesubmanifold. Then it is natural to ask, whether the resulting equations form a Hamiltonian system? For instance,according to Affirmation 5.2, we can add the algebraic equations4 Qα(z) = 0 to the Hamiltonian system (99), thusobtaining a consistent equations with solutions living on the invariant submanifold N = {z ∈Mn, Qα(z) = 0}. UsingAffirmation 5.3, we exclude zα and obtain differential equations on the manifold N with the local coordinates zb

zb = hb(zc) ≡ {zb, H}∣∣zα→fα(zc)

. (103)

They already don’t know anything about the ambient space Mn. Hence we ask, if the resulting equations representa Hamiltonian system on N? That is we look for the Hamiltonian equations

zb = ωba(zc)∂aH(zc), (104)

that could be equivalent to (103).Let us list some known cases of the Hamiltonian reduction.1. Reduction of non singular theory (3) on the surface of constant Hamiltonian gives a Hamiltonian system with

time-dependent Hamiltonian. The method is known as Maupertuis principle, see [3] for details.2. Hamiltonian reduction on the surface of Casimir functions, see Affirmation 4.6. The particular example is a

Hamiltonian system with Dirac bracket, see Eq. (151) below.3. Hamiltonian reduction of non singular theory on the surface of first integrals Φα with the property det{Φα,Φβ} 6=

0, see Eq. (159) below.4. Singular non degenerate theory (5)-(8) is equivalent to the theory of Item 2, , see Affirmations 7.5 and 7.3 below.

Hence it admits the Hamiltonian reduction to the surface of constraints.5. According to Gitman-Tyutin theorem, singular degenerate theory admits Hamiltonian reduction on the surface

af all constraints, see [2] for details.

VI. SYMPLECTIC MANIFOLD AND DIRAC BRACKET.

A. Basic notions.

As we saw in Sect. II, Poisson manifold can be defined choosing a contravariant tensor with the properties (40)and (41). Here we discuss another way, which works for the construction of a nondegenerate Poisson structures oneven-dimensional manifolds. Let on the even-dimensional manifold M2n is defined the covariant tensor ωij(z

k) (calledthe symplectic form) with the properties

ωij = −ωji (antisymmetric), (105)

det ω 6= 0 (nondegenerate), (106)

∂iωjk + cycle = 0 (closed). (107)

M2n equipped with a symplectic form is called the symplectic manifold.

4 Without loss of generality, we have taken cα = 0.

17

We recall that determinant of any odd-dimensional matrix vanishes, so (106) implies that we work on the even-dimensional manifold. Some properties of a symplectic form are in order.Affirmation 6.1 The inverse matrix ωij of the matrix ωij obeys the properties (40) and (41). So it determines thePoisson structure (39) on M2n. In other words, any symplectic manifold locally is a Poisson manifold.Exercise 6.1. Prove that (107) implies (41).

Conversely, take a Poisson manifold with the nondegenerated bracket, detω 6= 0, and let ω is its inverse. Contractingthe condition (41) with ωniωmjωpk we immediately obtain (107). We obtainedAffirmation 6.2. Poisson manifold with non degenerate bracket is a symplectic manifold.Darboux theorem. In a vicinity of any point, there are coordinates yk where ωij(zk) acquires the form

ω′ij(yk) =

(0 1−1 0

), then ω′ij(y

k) =

(0 −11 0

). (108)

Proof is given in Appendix B.Poincare lemma. In a vicinity of any point, the symplectic form ω can be presented through some covariant vectorfield aj as follows:

ωij = ∂iaj − ∂jai. (109)

(Terminology: closed form is locally an exact form). Conversely, the tensor ω, constructed from given aj accordingto Eq. (109), obeys the condition (107).Proof. According to Darboux theorem, there are coordinates yi = (x1, . . . , xn, p1, . . . , pn) where ωij ac-quires the canonical form (108), and then can be written as ω′ij = ∂ia

′j − ∂ja

′i, where a′i(y

k) =12 (−p1, . . . ,−pn, x1, . . . , xn). Returning to the original coordinates, we write ai(z

k) = ∂yj

∂zi a′j(y(z)), where a′j(y(z)) =

12 (−p1(zk), . . . ,−pn(zk), x1(zk), . . . , xn(zk)). This contravariant vector field satisfy the desired property: ∂iaj−∂jai =∂yn

∂zi∂ym

∂zi ω′nm = ωij .

The field ai can equally be obtained by direct integrations

ai = − 1

n− 1

n∑j=1

∫ωij(z

k)dzj . � (110)

Due to the Poincare lemma, it is easy to construct examples of closed and non-constant form ω, and then the tensorω, that will automatically obey a rather complicated equation (41). Note also that in the Darboux coordinates yk,the Poisson bracket acquires the canonical form (60).

Since any symplectic manifold is simultaneously a Poisson manifold, it has all the properties discussed in Sect. III.In particular, we have the mapping

ω : FM → TM, ω : A→ XiA = [ω(A)]i = ωij∂jA, (111)

and the basic relation between the Lie and Poisson brackets

ω({A,B}) = −[ω(A), ω(B)], or, equivalently X{A,B} = −[ ~XA, ~XB ]. (112)

Symplectic form can be used to determine the mapping ω : TM × TM → FM as follows

ω : ~X, ~Y → ω( ~X, ~Y ) ≡ ωijXiY j , then ω( ~X, ~X) = 0. (113)

Then the Poisson bracket can be considered as a composition5 of the mappings (113) and (111)

{A,B} = −ω(ω(A), ω(B)) ≡ −ω( ~XA, ~XB). (114)

Exercise 6.2. a) Prove that {Q,H} = c = const if and only if [ ~XQ, ~XH ] = 0. b) Confirm (114).By analogy with Riemannian geometry, on the symplectic manifold there is a natural possibility of raising and

lowering the indices of tensor quantities. It is achieved with use of symplectic tensor and its inverse. For instance,

5 In the coordinate-free formulation of Poisson geometry, the equality ω(ω(A), ω(B)) = −{A,B} is taken as the definition of the symplecticform ω.

18

the mapping Ui = ωijVj and its inversion V i = ωijUj establish an isomrphism between the spaces of covariant and

contravariant vector fields.Affirmation 6.3. V i is a Hamiltonian vector field if and only if Ui = ωijV

j obeys the condition

∂iUj − ∂jUi = 0. (115)

Proof. The equation ∂iA = ωijVj ≡ Ui for determining of A implies (115) as a necessary condition. Conversely,

when (115) is satisfied, the function

A =1

n

n∑j=1

∫Uj(z

k)dzj , (116)

generates the field V i: V i = ωij∂jA. �As an application of the developed formalism, we mention the following

Affirmation 6.4. Consider the Poisson manifold M2n with non-degenerated Poisson structure: detω 6= 0. Let ω isthe corresponding symplectic form, and ai is the contravariant vector field defined in (109). Then the Hamiltonianequations (52) follow from the variational problem

SH =

∫dτ[ai(z)z

i −H(z)]. (117)

Exercise 6.3. Prove the affirmation6.

B. Restriction of symplectic structure to a submanifold and Dirac bracket.

We recall that the mapping of manifolds Nk = {xa} → Mn = {zi}, given by xa → zi(xa), induces the mapping

T(0,m)M → T(0,m)

N of covariant tensor fields, see (35). Let Mn = {zk, ωij(zk)} is a symplectic manifold and Nk is asubmanifold determined by the functions Φα(zk) = 0 (see (22)), n and k are even numbers. Consider the embeddingNk →Mn, given by xa → zi = (fα(xa), xa). Then the induced mapping

ωfab(xc) =

∂zi

∂xa∂zj

∂xbωij(f

α(xc), xc), (118)

is called restriction of symplectic form ωij(zk) on Nk. If ωf obeys the properties (106) and (107), Nk turns into a

symplectic manifold. The inverse matrix then determines a Poisson bracket on Nk. Here we discuss the necessary andsufficient conditions under which this occurs. We will need the following matrix identity.Affirmation 6.5. Consider an invertible antisymmetric matrix

A =

(a b−bT c

), and its inverse A−1 =

(α β−βT γ

). (119)

Then the matrix γ is invertible if and only if a is invertible. In addition, we have

γ−1 = c+ bTa−1b, (120)

a−1 = α+ βγ−1βT . (121)

Proof. Eqs. (120) and (121) immediately follow from the identity AA−1 = 1, written in terms of the blocks. �Affirmation 6.6. The matrix (118) obeys the properties (106) and (107) if and only if

det{Φα,Φβ} ≡ det4αβ 6= 0, (122)

6 While variation of (117) formally leads to (52), the following difficulty should be noted. Formulating a variational problem, we fix twopoints in phase space and then look for an extremal trajectory between them. The first-order system (52) has unique solution for thegiven initial ”position”: zi(τ1) = zi1. This implies, that position at a future instant τ2 is uniquely determined by the initial position ofthe system. So, if we look for the extremal trajectory between two arbitrary chosen points zi(τ1) = zi1 and zi(τ2) = zi2, the variationalproblem (117) generally will not have a solution.

19

on Nk.Proof. Consider the problem in the coordinates of Mn

yk = (yα, ya), yα = Φα(zβ , zb), ya = za, a = 1, 2, . . . , k, (123)

adapted with the functions Φα. Denote ωij(zk) the Poisson tensor of Mn. Using the transformation law (20), weobtain

ω′ij(yk) =

({Φα,Φβ} {Φα, zb}{za,Φβ} {za, zb}

)∣∣∣∣zi(yj)

, denote its inverse as ω′ij(yk) =

(ω′αβ(yk) ω′αb(y

k)

ω′aβ(yk) ω′ab(yk)

). (124)

For the latter use we make the following observation. The symplectic matrix ω′ij(yα, ya) obeys the identity (107).

In particular, we have ∂aω′bc(y

α, ya) + cycle = 0 for any fixed yα. Considering ω′bc(ya, yα) as a function of ya, and

applying the Affirmation 6.1, we conclude that its inverse, say ωabD , obeys the identity ωadD ∂dωbcD + cycle(a, b, c) = 0.

Using Affirmation 6.5 for the matrices (124), the explicit form of the inverse matrix is

ωabD (yα, yc) =({za, zb} − {za,Φα}4αβ{Φβ , zb}

)∣∣∣zi(yj)

. (125)

Let us return back to the proof. In adapted coordinates, the embedding Nk → Mn is given by xa → yi = (yα, ya),where yα = 0 and ya = xa. The equation (118) reads

ωfab(xa) = ω′ab(y

α, ya)|yα=0,ya→xa , (126)

that is the restriction of ω′ij(yk) on Nk reduces to the setting yα = 0 in a, b -block of the matrix ω′ij(y

k). We needto confirm that ωf is a nondegenerate and closed form. The symplectic matrix ω′ij(y

α, ya) obeys the identity (107).In particular, we have ∂aω

′bc(y

α, ya) + cycle = 0 for any fixed yα. Taking yα = 0, we conclude that ωf is closed.Further, using Affirmation 6.5 for the matrices (124), we conclude that the matrix ωf is invertible if and only ifdet{Φα,Φβ} 6= 0. �

As the restriction (118) determines a symplectic structure on Nk, its inverse gives a Poisson bracket. Its explicitexpression in terms of the original bracket can be obtained using the representation (126) for ωfab. Using Affirmation6.5 for the matrices (124) and Eq. (24), we can write for the inverse of ωfab the expression

ωabf (xc) =({za, zb} − {za,Φα}4αβ{Φβ , zb}

)∣∣∣zi(yj)

∣∣∣∣yα=0,ya→xa

(127)

=({za, zb} − {za,Φα}4αβ{Φβ , zb}

)∣∣∣zα→fα(za)

∣∣∣∣za→xa

. (128)

Thus we obtained the following result.Affirmation 6.7. Let ωij = {zi, zj} be non degenerate Poisson tensor and Φα be the functionally independentfunctions with det{Φα,Φβ} 6= 0. Then the matrix

ωabf (zc) =({za, zb} − {za,Φα}4αβ{Φβ , zb}

)∣∣∣zα=fα(zc)

, (129)

where zα = fα(zc) are parametric equations of the surface Φα = 0, obeys the Jacobi identity, and determines a nondegenerate Poisson bracket on Nk.

Let us find a bracket on Mn whose restriction to Nk gives the bracket (129). The equality (129) prompts to consider

{A,B}D = {A,B} − {A,Φα}4αβ{Φβ , B} = ∂iA[{zi, zj} − {zi,Φα}4αβ{Φβ , zj}

]∂jB ≡ ∂iA ωijD ∂jB. (130)

This is the famous Dirac bracket [1]. The tensor ωijD(zk) obeys the Jacobi identity (see below), and hence turn Mn

into the Poisson manifold (Mn, {, }D). The Dirac bracket has the desired property. Indeed, for any function A(zi),Eq. (130) implies

{A,Φα}D = 0, (131)

so Φα are Casimir functions of the Dirac bracket. As we saw in Sect. IV, this implies that all Hamiltonian fieldsV iA = ωijD∂jA are tangent to the surfaces Φα = cα. In addition, we can restrict the Dirac tensor ωijD on the submanifold

20

Nk, see Eq. (85). This gives the Poisson bracket (129) on Nk and turns it into a Poisson submanifold of the Poissonmanifold (Mn, {, }D).

It remains to prove the Jacobi identity for the Dirac bracket.Affirmation 6.8. Consider Poisson manifold Mn = {zi, ωij(zk)} with a non degenerate tensor ω. Let Φα(zk)

be functionally independent functions which obey the condition (122). Then the Dirac tensor ωijD(zk), specified in(130), satisfies the identity (41), and hence the Dirac bracket (130) satisfies the Jacobi identity: {A, {B,C}D}D +cycle (A,B,C) = 0.Proof. It is convenient to consider the problem in the coordinates (123) adapted with the functions Φα. Using Eqs.(20) and (131), we obtain the Dirac tensor in these coordinates

ω′ijD (yk) =

(0 00 ωabD (zk)

∣∣z(y)

), (132)

where ωabD (zk) is a, b -block of the Dirac tensor ωijD(zk) in original coordinates, then ωabD (zk)∣∣z(y)

is just the expression

written in (125). The desired Jacobi identity ω′inD ∂nω′jkD + cycle = 0 will be fulfilled, if the matrix (125) obeys the

identity ωadD ∂d(ωbcD ) + cycle = 0. But this was confirmed above, see the discussion below of Eq. (124). �

The results of this subsection can be presented in the form of diagram (133), that relates geometrical structures onthe manifold Mn (top line), and on its submanifold Nk (bottom line).

ωij ←→ ωij −→ ωijD↓ ↓

ωfab ←− −− −→ ωabf

(133)

C. Dirac’s derivation of the Dirac bracket.

Dirac arrived at his bracket in the analysis of a variational problem for singular nondegenerate theories like (5),(6). Consider the variational problem

S =

∫dτ [paq

a −H0(qa, pb) + λαΦα(qa, pb)] , (134)

for the set of independent dynamical variables zi ≡ (qa, pb), i = (1, 2, . . . , 2n), and λα, α = (1, 2, . . . , 2p < 2n). H0

and Φα are given functions, where Φα obey the conditon (122). Variation of the action with respect to zi and λα

gives the equations of motion7

zi = {zi, H0}+ λα{zi,Φα}, Φα = 0, (135)

where {, } is the canonical Poisson bracket on R2n. Let zi(τ), λ(τ) be a solution of the system. Computing derivativeof the identity Φα(zi(τ)) = 0 , we obtain the algebraic equations

{Φα, H0}+ {Φα,Φβ}λβ = 0. (136)

that must be satisfied for all solutions, that is it is a consequence of the system. According to this equation, allvariables λβ are determined algebraically, λβ = −4βα{Φα, H0}, where 4 is the inverse matrix of 4. Adding theconsequences to the system, we obtain its equivalent form

zi = {zi, H0} − {zi,Φα}4αβ{Φβ , H0} ≡[ωij − {zi,Φα}4αβ{Φβ , zj}

]∂jH0, (137)

Φα = 0, λβ = −4βα{Φα, H0}, (138)

7 It is instructive to compare the systems (135) and (100). The constraints Φα = 0 should not be confused with the first integrals.Indeed, first integrals represent the first-order differential equations which are consequences of a special form of the original equations,cαi[z

i−{zi, H}] = ddτQα(z) = 0, whereas constraints are the algebraic equations. As a consequence, solutions of the systems (135) and

(100) have very different properties. Solutions of the system (100) pass through any point of R, while all solutions of (135) live on thesubmanifold Φα = 0.

21

where the sectors zi and λβ turn out to be separated. The expression appeared on r.h.s. of (137) suggests to introducethe new bracket on M2n

{A,B}D = {A,B} − {A,Φα}4αβ{Φβ , B}, (139)

which is just the Dirac bracket. Then equations (137) represent a Hamiltonian system with the Dirac bracket

zi = {zi, H0}D, (140)

and with the Hamiltonian being H0.

VII. POISSON MANIFOLD AND DIRAC BRACKET.

A. Jacobi identity for the Dirac bracket.

While our discussion of the Dirac bracket in previous section was based on a symplectic manifold, the prescription(130) can equally be used to generate a bracket {A,B}D starting from a degenerate Poisson bracket {A,B}. We showthat {A,B}D still satisfies the Jacobi identity. To prove this, we will need the following auxiliary statement.Affirmation 7.1. Consider the Poisson manifold Mm+n = {xK = (xα, xi), ωIJ(xK), rank ω = n}. Let Kα(xI),α = 1, 2, . . . ,m be functionally independent Casimir functions, and Φα(xI), α = 1, 2, . . . , p < n be functionallyindependent functions which obey the condition (122). Then(A) The m+ p functions Kα,Φβ are functionally independent.(B) In the coordinates

zI = (zα, zi), where zα = Kα(xI), zi = xi, (141)

the functions Φα(zα, zi) obey the condition rank ∂Φα

∂zi = p. In other words Φα, considered as functions of zi, arefunctionally independent.Proof. (A) In the coordinates (141), our functions are zα and Φα(zα, zi). We will show that functional dependenceof the set implies that the matrix {Φα,Φβ} is degenerate. Then non degeneracy implies functional independence ofthe set - the desired result.

Consider (m+ p)× (m+ n) matrix

J =∂(zα,Φα(zα, zi))

∂(zα, zi)=

(1m×m 0∂Φα

∂zβ∂Φα

∂zi

). (142)

If zα,Φα(zα, zi) are functionally dependent, we have rank J < m + p, then some linear combination of rows of the

matrix J vanishes: cαδαI + cα

∂Φα

∂zI= 0 for all I. This equation, together with explicit expression (142) for J , implies

cα∂Φα

∂zi= 0, ~c 6= 0. (143)

Consider now the matrix {Φα,Φβ} in the coordinates (141). Using the Poisson tensor

ω′IJ(zK) ≡(ω′αβ ω′αj

ω′iβ ω′ij

)=

(0m×m 0

0 ωij(xK)|x(z)

), (144)

we obtain {Φα,Φβ} = ∂iΦαω′ij∂jΦ

β . Then (143) implies the degeneracy of the matrix: {Φα,Φβ}cβ = 0.

(B) Item (A) implies: rank J = m+ p. Then from explicit form (142) for J it follows, that rank ∂Φα

∂zi = p. �Affirmation 7.2. The Dirac bracket (130), constructed on the base of a degenerate Poisson bracket {A,B}, satisfiesthe Jacobi identity.Proof. We use the notation specified in Affirmation 7.1. The original Poisson tensor in the coordinates (141) iswritten in Eq. (144). According to Affiration 4.2, its block ω′ij is a non degenerate matrix. ω′IJ(zK) satisfies theJacobi identity, that due to special form (144) of this tensor reduces to the expression

ω′in∂

∂znω′jk + cycle = 0. (145)

22

Using the prescription (130), we use ω′IJ(zK) to write the Dirac tensor

ω′IJD (zK) =

(0m×m 0

0 ω′ijD

), where ω′ijD = ω′ij − ω′in∂nΦα4αβ∂kΦβ . (146)

Jacobi identity for ω′IJD (zK) will be satisfied, if

ω′inD∂

∂znω′jkD + cycle = 0. (147)

Note that zα enter into the expressions (145)-(147) as the parameters. In particular, the derivative ∂∂zα falls out of

all these expressions. According to Item (B) of Affirmation 7.1, the functions Φα(zβ , zi), considered as functions ofzi, are functionally independent. Taking this into account, we can apply the Affirmation 6.8 to the matrices specifiedby (145) and (146), and conclude that (147) holds. �

B. Some applications of the Dirac bracket.

Given scalar function A, we associate with it the function

Ad = A− {A,Φα}4αβΦβ . (148)

The two functions coincide on the surface Φα = 0. There is a remarcable relation between the Dirac bracket of originalfunctions and the Poisson bracket of deformed functions:

{A,B}D = {Ad, Bd}+O(Φα), (149)

which means that the two brackets also coincide on the surface. This property can be reformulated in terms of vectorfields as follows. Given scalar function A, integral lines of Hamiltonian field V i = ωij∂jAd that cross the surfaceΦα = 0, entirely lie on it.

Below we use the Dirac bracket to analyse some Hamiltonian systems consisting of both dynamical and algebraicequations.

1. Consider the Hamiltonian system zi = {zi, H}D on the Poisson manifold (M, {, }D). As {Φα, H}D = 0, thefunctions Φα are integrals of motion of the system. According to Affirmation 5.2, all the submanifolds N~ck = {z ∈Mn, Φα(z) = cα} are invariant submanifolds, that is any trajectory that starts on N~ck entirely lies on it. In particular,we haveAffirmation 7.3. The equations

zi = {zi, H}D, Φα = 0, (150)

form a self-consistent system in the sense of Definition 1.1.Further, according to Affirmation 5.3, these equations are equivalent to the system: zα − fα(za) = 0, zb ={zb, H(zi)}D

∣∣zα→fα(zb)

. We replace zα on fα(zb) using Eqs. (83) and (85), this gives

zα = fα(zb), zb = {zb, H(zb)}D(ω), where H(zb) = H(zb, zα(zb)). (151)

This shows that the variables zb obey the Hamiltonian equations on the submanifold N. Observe, that the Affirmation7.3 equally follows from Eqs. (151).

2. Let us rewrite the system (150) in terms of original bracket as follows: zi = {zi, H − Φα4αβ{Φβ , H}} +

Φα{zi, 4αβ{Φβ , H}}, Φα = 0, or, equivalently

zi = {zi, H − Φα4αβ{Φβ , H}}, Φα = 0. (152)

Note that the functions Φα are not the Casimir functions of the original bracket. As the systems (152) and (150) areequivalent, we obtained an example of a self-consistent theory of the type (5), (6):Affirmation 7.4. Given Poisson manifold (Mn, {, }), let H be given function and Φα is a set of functionally inde-pendent functions that obey the condition det{Φα,Φβ}

∣∣Φα=0

≡ det4αβ 6= 0. Then the equations

zi = {zi, H}, Φα = 0. (153)

23

with the Hamiltonian H = H − Φα4αβ{Φβ , H} form a self-consistent system.3. Affirmation 7.5. Singular nondegenerate theory defined by the equations (5), (6) with the properties (7) and

(8) is a self-consistent, and is equivalent to (150).Proof. Using (7), we rewrite the system (5), (6) in the equivalent form as follows

zi = {zi, H}D + {zi,Φα}4αβ{Φβ , H}, Φα = 0. (154)

Take any point of the submanifold Φα = 0. According to Affirmation 7.3, there is a solution zi(τ) of (150) that passesthrough this point. Due to the condition (8) we have {Φβ , H}

∣∣zi(τ)

= 0, then the direct substitution of zi(τ) into

(154) shows that it is a solution of this system. �4. Example of Hamiltonian reduction. Let the Hamiltonian system zi = ωij∂jH(zk) with detω 6= 0 admits

the first integrals Φα(zk), α = 1, 2, . . . , n− k with the properties

{Φα, H} = 0, det{Φα,Φβ} ≡ 4αβ 6= 0. (155)

Then dynamics can be consistently restricted on any one of invariant surfaces N~ck = {zk ∈ Mn, Φα = cα}. Without

loss of generality, we consider reduction on N~0k, then Φα = 0 implies zα = fα(za), while the independent variaveisobey the equations

za = ωaj∂jH∣∣zα=fα(za)

. (156)

We do the substitution indicated in this equation and show that the result is a Hamiltonian system. Consider theproblem in the adapted coordinates (123). Then Φα = 0 turn into yα = 0 while instead of (156) we have

ya = ω′aj∂jH∣∣yα=0

= ω′ab∂bH′∣∣yα=0

+ ω′aβ∂βH′∣∣yα=0

, (157)

where explicit form of ω′ is given by (124), and H ′(yi) = H(zk(yi)).The equation {Φα, H} = 0 in the coordinates yk

gives: 0 = {yα, H ′} = ω′αa∂aH′ +4αβ∂βH ′, or ∂βH

′ = −4βγω′γa∂aH ′. Using this expression in (157) we obtain

ya =[{za, zb} − {za,Φβ}4βγ{Φβ , zb}

]∂bH(zi)

∣∣∣zi(yj)

∣∣∣∣yα=0

. (158)

Now note that A(zi(yj))∣∣yα=0

= A(fα(za), za)|za→ya , so the equations of motion reads

za = ωabD (fα(za), za)∂bH(fα(za), za), (159)

where ωabD (fα(za)za) is (a, b) -block of the Dirac tensor, see (129). According to Affirmation 6.7, it obeys the Jacobiidentity, so the equations (159) represent a Hamiltonian system, which is equivalent to (156).

C. Poisson manifold with prescribed Casimir functions.

Let Kα(zβ , zb) with det ∂Kα

∂zβ6= 0 be representatives of the scalar functions in local coordinates zi = (zβ , zb) of the

manifold Mn, where β = 1, 2, . . . , p, b = 1, 2, . . . , n − p. Without loss of generality, we assume that n − p is an evennumber: n− p = 2k. The task is to construct a Poisson bracket on Mn, that has Kα as the Casimir functions. Onepossible solution of this task can be found by using of coordinate system where the functions Kα turn into a part ofcoordinates.

Introduce the following coordinates on Mn:

zj′

= ϕj′(zi) = (Kα(zi), za). (160)

Construct the matrix a with elements aij′ = ∂zj

′

∂zi = ∂iϕj′ , its inverse is denoted as a ≡ a−1. In the local coordinates

zj′, define the bracket

{A,B} = ∂i′A W i′j′

0 ∂j′B, W i′j′

0 (zi′) =

(0p×p 0

0 ω0(zj′)

), (161)

24

where ω0 is a 2k × 2k matrix with the elements ωa′b′

0 (zα′, zc

′) satisfying the identity (41) with respect to zc

′. As this

matrix we can also take any Poisson structure ωa′b′

0 (zc′) on the submanifold Kα(zβ , zb) = 0. For instance, we could

take it in the canonical form

ωa′b′

0 =

(0k×k 1k×k−1k×k 0k×k

). (162)

According to Eq. (20), in the original coordinates zi the bracket reads

{A,B} = ∂iAωij∂jB, ωij =

[aTW0(Kα(zi), za)a

]ij. (163)

The Affirmation 2.2 guarantees that it satisfies the Jacobi identity, hence it turn Mn into a Poisson manifold.Affirmation 7.6. Kα are Casimir functions of the bracket (163).Proof. Consider, for instance, {A,K1} = ∂iA(aTW0a)ij∂jK

1. Compute the term: (W0a)ij∂jK1 = (W0a)ijaj

1 =

W ij0 δj

1 = W i10 = 0. �

In resume, the set of functionally independent functions Kα(zi) turns out to be the set of Casimir functions of thePoisson manifold with the bracket (163).

Denoting ∂αKβ = bα

β , ∂aKβ = ca

β , the Poisson structure (163) can be written in the following form:

ω =

((cb−1)Tω0cb

−1 (ω0cb−1)T

−ω0cb−1 ω0

). (164)

Blocks of this matrix can be compared with Eqs. (80). We can restrict the bracket (163) on the Casimir submanifold,obtaining the bracket (see Eq. (85))

{A(za), B(za)} = ∂aAωab∂bB, ωab = ωab0 (fα(za), za). (165)

In particular, if ω0 in Eq. (161) was originally chosen to be independent of the coordinates zα, we have ωab = ωab0 .Casimir submanifold with the bracket (165) is the Poisson submanifold of Mn (163) in the sense of definition (73).Example 7.1. Consider M3 and the function K(z1, z2, z3) with ∂1K 6= 0. Then

a =

∂1K 0 0∂2K 1 0∂3K 0 1

, a =1

det a

1 0 0−∂2K 1 0−∂3K 0 1

. (166)

Taking

W =

0 0 00 0 10 −1 0

, (167)

we obtain the Poisson structure on M3 that has K(z) as the Casimir function

ω = aTWa =1

det a

0 ∂3K −∂2K−∂3K 0 1∂2K −1 0

, or ωij =1

det aεijk∂kK. (168)

If Vi and Uj are contravariant vectors, the quantity 1det aε

ijk∂kK Vi Uj is a scalar function under the diffeomorphisms

(15). So ωij of Eq. (168) is a second-rank covariant tensor, as it should be. Restriction of the bracket (168) on theCasimir submanifold K = 0 gives the canonical Poisson bracket: {z2, z3} = 1.Example 7.2. SO(3) Lie-Poisson bracket. Chosing K = 1

2 [(z1)2 + (z2)2 + (z3)2] − 1 (see Example 1.2) in theexpressions of previous example, we obtain diffeomorphism-covariant form of the Lie-Poisson bracket:

{zi, zj} =1

det aεijkzk. (169)

25

VIII. APPENDICES.

A. Jacobi identity

Affirmation A 1. Let the bracket (39) obeys the Jacobi identity in the coordinates zi. Then the Jacobi identity issatisfied in any other coordinates.Proof. We need to show that the validity of the identity (B) for the bracket (39) with ωij(z) implies its validity for

this bracket with ωi′j′ defined in (20).

Given functions A(z), B(z), C(z), let us consider the auxiliary functions A(z) ≡ A(z′(z)) and so on. Since theJacobi identity is satisfied in the coordinates z, we can write

∂iA(z′(z))ωip(z)∂p[∂jB(z′(z))ωjk(z)∂kC(z′(z))

]+ cycle(A,B,C) = 0. (170)

Computing the derivatives, we present this identity as follows

∂i′A|z′(z)∂zi

′

∂ziωip(z)∂p

∂j′B ∂zj

′

∂zj

∣∣∣∣∣z(z′)

ωjk(z(z′)∂zk

′

∂zk

∣∣∣∣∣z(z′)

∂k′C

∣∣∣∣∣∣z′(z)

+ cycle(A,B,C) = 0. (171)

Using the identity ∂p

[D(z′)|z′(z)

]= ∂zp

′

∂zp ∂p′D(z′)|z′(z), we obtain[∂i′A(z′)ωi

′p′(z′)∂p′[∂j′B(z′)ωj

′k′(z′)∂k′C(z′)]

+ cycle(A,B,C)]∣∣∣z′(z)

= 0, (172)

which is just the Jacobi identity for the bracket {A(z′), B(z′)} = ∂i′Aωi′j′(z′)∂j′B. �

B. Darboux theorem

Lemma B1. (On rectification of a vector field). Let V i(zj) be vector field, nonvanishing at the point zj0 ∈Mn. Then

there are coordinates yi such that V i(yj) = (1, 0, . . . , 0) at all points yj in some vicinity of zj0. In other words, integral

lines of the field coincide with coordinate lines of the coordinate y1: V i(yj) = dyi

dτ , where yi(τ) = (y1 = τ, ya = Ca),

and C2, . . . , Cn are fixed numbers8.Proof. Without loss of generality we take z0 = (0, 0, . . . , 0) and V 1(z0) 6= 0, then V 1(z) 6= 0 in some vicinity of z0.Write the equations for integral lines

dzi

dτ= V i(zj(τ)), (173)

and solve them with the following initial conditions on the hyperplane z1 = 0:

z1(0) = 0, z2(0) = y2, . . . , zn(0) = yn, where y2, . . . , yn are fixed numbers. (174)

Denote by

zi(τ) = f i(τ, y2, . . . , yn), (175)

the integral line that at τ = 0 passes through the point (0, y2, . . . , yn). This determines the nondegenerate mapping

f : (τ, y2, . . . , yn) → zi = f i(τ, y2, . . . , yn). (176)

The nondegeneracy follows from (174) and (176) as follows:

det∂(z1, . . . , zn)

∂(τ, y2, . . . , yn)

∣∣∣∣z0

= det

(V 1(z0) 0 . . . , 0V a(z0) δab

)= V 1(z0) 6= 0, (177)

8 In this section we use the notation V i(zj) and V i(yj) instead of V i and V ′i to denote components of the vector ~V in different coordinatesystems

26

so we can take the set

y1 = τ, y2, . . . , yn, (178)

as new coordinates of Mn, then the transition functions are given by Eq. (176). For the latter use, we note that

∂z2

∂y1

∣∣∣∣z0

= V 2(z0),∂z2

∂y2

∣∣∣∣z0

= 1,∂z2

∂yα

∣∣∣∣z0

= 0, α = 3, 4, . . . , n. (179)

According to (175), integral line of the field ~V in the new system is yi(τ) = (y1 = τ, ya = Ca), that is it coincides

with the coordinate line of y1, then V i(yj) = dyi

dτ = (1, 0, . . . , 0). �Lemma B2. Let Mn = {zk, ωij(zk)} be Poisson manifold with rank ω(z0) 6= 0. Then there is a pair of scalar

functions, say q ∈ FM and p ∈ FM, with the property {q, p} = 1. Their Hamiltonian fields ~Vq and ~Up are linearly

independent and have vanishing Lie bracket, [~Vq, ~Up] = 0.Proof. Without loss of generality we take ω12 6= 0. As the function q(zi) we take the scalar function of the coordinatez2, its representative in the system zi is

q(zi) = z2, then its Hamiltonian field is V iq (zk) = ωik∂

∂zkz2 = ωi2 = (ω12, 0, ω32, . . . , ωn2). (180)

In particular, V 1q = ω12 6= 0. We rectificate this field according to Lemma B1, then its components in the system yj

are

V iq (yj) = (1, 0, . . . , 0). (181)

The representative of the function q in the system yj is q(yj) = z2(yj), so its bracket with any other function reads

{q(yj), B(yj)} = V iq (yj)∂

∂yiB =

∂

∂y1B. (182)

Taking as the function p the scalar function of the coordinate y1: p(yj) = y1, we obtain the desired pair of functions

{z2(yi), y1} = 1, or, in initial coordinates, {z2, y1(zj)} = 1. (183)

In the coordinate system yj the Hamiltonian fields of these functions are

V iq (yj) = (1, 0, . . . , 0), U ip(yj) = ω′ik

∂

∂yky1 = ω′i1 = (0, ω′21, ω′31, . . . , ω′n1). (184)

From their manifest form they are linearly independent. Besides, as the Hamiltonian field of a constant vanishes, we

have [~Vq, ~Up] = − ~W{q,p} = − ~W1 = 0. �Lemma B3. (On existence of a pair of canonical coordinates). Let Mn = {zk, ωij(zk)} be Poisson manifold withrank ω(z0) 6= 0. Then there are coordinates q, p, ξ3, . . . , ξn with the properties

{q, p} = ω′12 = 1, {q, ξα} = ω′1α = 0, {p, ξα} = ω′2α = 0, (185)

{ξα, ξβ} = ω′αβ(ξγ), that is ω′αβ do not depend on q, p. (186)

In addition, Jacobi identity for ωij and Eqs. (185) and (186) imply the Jacobi identity for ω′αβ : ω′αρ∂ρω′βγ+cycle = 0.

Proof. (A) We take q(zi) = z2, and rectify the vector field V iq using the Lemma B1. In the process, we obtain

the coordinates yi, the components of the field V iq (yj) = (1, 0, . . . , 0) in these coordinates, and the scalar function

p(yj) = y1 which obeys

{q, p} = 1. (187)

(B) Let U ip(yj) be components of Hamiltonian vector field of the function p in the coordinates yj . According the

Lemma B2, ~Vq and ~Up are commuting fields, then

0 = [~Vq, ~Up]i = V kq

∂

∂ykU ip − Ukp

∂

∂ykV iq =

∂U ip∂y1

, implies U ip = U ip(y2, . . . , yn), (188)

27

that is ~Up does not depend on q and p. Consider integral lines of the field ~Up. Taking into account that U1p (yj) = 0,

we have

dy1

dλ= 0, then y1 = C = const, (189)

dya

dλ= Uap (y2(λ), . . . , yn(λ)). (190)

For definiteness, we assume U2p (z0) 6= 0. We apply the Lemma B2 to the field Uap (yb), with a, b = 2, 3, . . . , n, that is

we solve Eqs. (190) with initial conditions on the surface y2 = 0:

y2(0) = 0, y3(0) = ξ3, . . . , yn(0) = ξn, (191)

Denote solution of the problem as

ya(λ) = ga(λ, ξ3, . . . , ξn), a = 2, 3, . . . , n. (192)

These equations are invertible, since (190)-(192) imply (here α, β = 3, 4, . . . , n)

det∂(y2, . . . , yn)

∂(λ, ξ3, . . . , ξn)

∣∣∣∣z0

= U2p (z0) det

∂yα(λ = 0)

∂ξβ= U2

p (z0) det 1 = U2p (z0) 6= 0. (193)

We denote the inverse formulas as follows:

λ = g(y2, . . . , yn), ξ3 = g3(y2, . . . , yn), . . . , ξn = gn(y2, . . . , yn), (194)

and introduce the new coordinates

(y1, y2, y3, . . . , yn) → (y1, λ(ya), ξα(ya)), a = 2, 3, . . . n, α = 3, 4, . . . , n, (195)

with the transition functions (194). Integral lines of the fields U and V in the new coordinates are (C, λ, ξ3, . . . ξn)and (y1 = τ, g(y2, . . . , yn), gα(y2, . . . , yn). Along the integral lines of U only the second coordinate λ changes. Alongthe integral lines of V changes the first coordinate, y1 = τ , while λ and ξα, being functions of y2, . . . , yn, remainconstants. Therefore, in these coordinates both fields are straightened: V iq = (1, 0, 0, . . . , 0), U ip = (0, 1, 0, . . . , 0).

(C) The Poisson brackets of q and p with scalar functions of the coordinates ξα, α = 3, 4, . . . , n vanish

{q, ξα} = Vq(ξα) =

∂ξα

∂τ= 0, {p, ξα} = Vp(ξ

α) =∂ξα

∂λ= 0. (196)

So, the functions q p, and ξα obey the equation (185).(D) The last step is to introduce the mapping

(y1, y2, y3, . . . , yn) → (q = z2(yj), p = y1, ξα = g(y2, . . . , yn)). (197)

Its invertibility follows from the direct computation

det∂(q, p, ξ3, . . . , ξn)

∂(y1, y2, y3, . . . , yn)

∣∣∣∣z0

= det

∂z2

∂y1∂z2

∂y2 . . . ∂z2

∂yn

∂y1

∂y1∂y1

∂y2 . . . ∂y1

∂yn

. . . . . .∂ξα

∂y1∂ξα

∂y2∂ξα

∂yβ

. . . . . .

∣∣∣∣∣∣∣∣∣∣∣z0

= det

0 1 0 . . . 01 0 0 . . . 0. . . . . .

0 ∂ξα

∂y2 1

. . . . . .

= −1. (198)

In the computation we used the equations (179), (180), (194) and (193). In particular: ∂z2

∂y1

∣∣∣z0

= ∂f2(τ,y2,...,yn)∂τ

∣∣∣z0

=

V 2q |z0 = ω22 = 0. Therefore we can take q, p, ξα as a coordinate system on Mn. As we saw above, the coordinates

obey the desired property (185). To confirm (186), we use {q, ξα} = 0 in the Jacobi identity, obtaining

{q, {ξα, ξβ}} = −{ξα, {ξβ , q}} − {ξβ , {q, ξα}} = 0, or∂

∂p{ξα, ξβ} = 0. (199)

28

So {ξα, ξβ} ≡ ω′αβ does not depend on p. Similar computation of {p, {ξα, ξβ}} implies, that ω′αβ does not dependon q. �

If rank ω′αβ(ξγ) 6= 0, the manifold Mn−2 = {ξγ , ω′αβ(ξγ)}, in turn, satisfies the conditions of the Lemma B3.Generalized Darboux theorem. Let Mn = {zk, ωij(zk)} be Poisson manifold with rank ω = 2k at the pointzi0 ∈Mn. Then there are local coordinates, where ω has the form:

ω′ =

0p×p 0 00 0k×k 1k×k0 −1k×k 0k×k

, p = n− 2k, (200)

at all points in some vicinity of zi0.Proof. The proof is carried out by induction on the pairs of canonical coordinates constructed in Lemma B3. Afterk steps, we get the coordinates ξα, qb, pc, α = 1, 2, . . . , n − 2k, b, c = 1, 2, . . . , k, in which the tensor ω has theblock-diagonal form

ω′ =

ω′αβ 0 00 0k×k 1k×k0 −1k×k 0k×k

, (201)

and ω′αβ = {ξα, ξβ}. From the rank condition and from the manifest form (201) of the matrix ω′, we have 2k =rank ω′ = rank ω′αβ + 2k, or rank ω′αβ = 0. This implies ω′αβ = 0 for all α and β. �

Affirmation B1. Let Q(zi) be first integral of the Hamiltonian system zi = ωij∂jH with a non degenerate tensorωij . Then solution of this system of n equations reduces to the solution of a Hamiltonian system composed by n− 2equations.Proof. Introduce the coordinates z′i: z′1 = z1, z′2 = Q(zi), z′3 = z3, . . . , z′n = zn, thus turning Q into the secondcoordinate of the new system. Applying the Lemmas B2 and B3, we construct the coordinates q, p, ξα with q = Q.Poisson tensor in these coordinates has the form

ω′ =

0 1 0−1 0 00 0 ω′αβ(ξγ)

. (202)

Consider our Hamiltonian equations in these coordinates. The equation q = ∂pH′ together with q = c2 = const

implies that H ′ does not depend on p: H ′ = H ′(q, ξγ). Then on the surface q = c1 = const, the original system isequivalent to

p = − ∂qH′(q, ξγ)|q=c2 , (203)

ξα = ω′αβ∂βH′(c2, ξ

γ), α = 3, 4, . . . , n. (204)

The n − 2 Hamiltonian equations (204) can be solved separately from (203), let ξα(τ, c2, . . . , cn) be theirgeneral solution. Using these functions in Eq. (203), the latter is solved by direct integration: p =−∫dτ ∂qH

′(q, ξγ)|q=c2,ξ=ξ(τ,c2,...,cn). �It should be noted that the range of applicability of this affirmation in applications is rather restricted. Indeed, to

find manifest form of the equations (204), we need to rectify two vector fields. And for this, it is necessary to solvetwice a system of equations like the original system!

C. Frobenius theorem

The equation ∂xX(x, y, z) = 0 has two functionally independent solutions: X1 = y and X2 = z. Frobenius theoremcan be thought as a generalization of this result to the case of the system of first-order partial differential equationsAia(zk)∂iX(zk) = 0. The theorem can also be reformulated in a purely geometric language, see the end of this section.

We will need some properties of vector fields and their integral lines on a smooth manifold Mn = {zi, i = 1, 2, . . . , n}.We recall that integral line of the vector field V i(zk) on Mn is a solution zi(τ) to dzi(τ)

dτ = V i(zk(τ)). As before, we

29

assume that through each point of the manifold passes unique integral line of ~V . The following result is an analogueof Affirmation 5.1 for the case of non Hamiltonian vector fields9.Affirmation C1. Let ~V (zk) be vector field on Mn and F (zk) be scalar function. The following two conditions areequivalent:

(A) ~V is tangent10 to the surface F (zk) = const: V i∂iF = 0 at each point zk ∈ Mn. (B) Integral lines of ~V lie onthe surfaces F (zk) = c = const.

Proof. Let zk(τ) be an integral line of ~V . Then the Affirmation follows immediately from the equality

d

dτF (zk(τ)) = V i(zk)∂iF (zk)

∣∣z(τ)

. � (205)

Evidently, the same is true for a set of vector fields:

Affirmation C2. Let ~A1(zk), . . . , ~Ak(zk) be linearly independent vector fields on Mn, and {N~ck, ~c ∈ Rn−k} be afoliation of Mn. The following two conditions are equivalent:

(A) ~Aa are tangent to N~ck at each point zk ∈Mn: ~Aa(zk) ∈ TNck(zk), and hence form a basis of TNck(zk).

(B) Each integral line of each ~Aa lies in one of the submanifolds N~ck.Lemma C1. Let {N~ck, ~c ∈ Rn−k}, where

N~ck = {zi = (zβ , zb) ∈Mn, Fα(zβ , zb) = cα, ~c ∈ Rn−k, a = 1, 2, . . . , k, α = 1, 2, . . . n− k}, (206)

be foliation of Mn (see Sect. I). Then there is a set of k linearly independent vector fields ~Ua(zk) on Mn with thefollowing properties.

(A) For any zk ∈Mn, the vectors ~Ua(zk) are tangent to the submanifold N~ck that passes through this point:

U ia∂iFα = 0 (or, equivalently, in the coordinate-free notation: ~Ua(Fα) = 0). (207)

In particular, at each point they form a basis of tangent space to the submanifold.

(B) Integral lines of ~Ua that pass through zk ∈Mn, lye in N~ck that passes through this point.

(C) ~Ua are commuting fields

[~Ua, ~Ub] = 0. (208)

Proof. Introduce the coordinates, adapted with the foliation: zk → yk = (yα, ya), with the transition functionsya = za, yα = Fα(zβ , zb). In these coordinates the sumanifolds N~ck look like hyperplanes:

N~ck = {yi ∈Mn, yα = cα}, (209)

and ya can be taken as local coordinates of N~ck. Consider the vector fields ~Ua on Mn, which in the system yk havethe following components: U ia(yk) = δa

i. Their integral lines are just lines of the coordinates ya of the submanifoldsN~ck. Evidently, the fields obey the conditions (A)-(C) of the Lemma. Their explicit form in the original coordinatesis as follows:

U ia(zk) =

[∂zi

∂yjU ja(yk)

]∣∣∣∣y(z)

= (U ba, Uβa ) =

(δab,

∂fβ(zc, yγ)

∂za

∣∣∣∣yγ→Fγ(zb,zβ)

), (210)

where fβ(zc, yγ) is solution to the system F β(fβ , zc) = yγ . Since (207) and (208) are covariant equations, the fields(210) satisfy them in the original coordinates zk. �Lemma C2. Linear combinations of vector fields with closed algebra also form a closed algebra under the followingconditions:

if ~Va = bab~Ub, det b 6= 0, [~Ua, ~Ub] = cab

c~Uc, then [~Va, ~Vb] = γabc~Vc. (211)

9 This can also be thought as a paraphrase of the followig property: vector grad F (x, y, z) in R3 is orthogonal to the surfaces of levelF (x, y, z) = c of the scalar function F (x, y, z).

10 See the definition of a vector field tangent to a submanifold on page 6.

30

Proof. The equality (211) follows from direct computation, and gives

γabc = ba

dbbecde

f bfc + V ja (∂jbb

f )bfc − (a↔ b), (212)

where b is inverse for b. �Lemma C3. Let ~A1, . . . , ~Ak is a set of linealy independent vector fields on Mn, with closed algebra of commutators

[ ~Aa, ~Ab]i = cab

d(zk)Aid. (213)

Then there is a set of linearly independent fields ~Va, which are linear combinations of ~Aa and have vanishing commu-tators

[~Va, ~Vb] = 0. (214)

Proof. The components Aia = (aab, ba

β) of linearly independent fields form k×n matrix with rank equal k. Without

loss of generality we assume det aab 6= 0, and let aa

b be the inverse matrix. We show that ~Va ≡ aab ~Ab are the desiredfields.

The expressions (213) with components i = c can be solved with respect to cabd as follows: [ ~Aa, ~Ab]

c = cabdad

c

implies cabc = [ ~Aa, ~Ab]

dadc. Using this equality, we exclude cab

c from the expressions (213) with i = β, obtaining

[ ~Aa, ~Ab]β = [ ~Aa, ~Ab]

dadcbc

β . In more detail, this reads

Aai∂ib

βb − (a↔ b) = Aa

i(∂iabd)ad

cbcβ − (a↔ b) =

Aai∂i(ab

dadcbc

β)−Aaiabd∂i(adcbcβ)− (a↔ b) = Aai∂ib

βb −Aa

iabd∂i(ad

cbcβ)− (a↔ b), (215)

which implies Aaiab

d∂i(adcbc

β) − (a ↔ b) = 0. Contraction of this equality with aeaaf

b gives the following relationbetween components of the fields with closed commutator algebra:

aacAc

i∂i(abdbd

β)− (a↔ b) = 0. (216)

Now, the fields ~Va ≡ aac ~Ac with the components Va

i = (Vab, Va

β) = (δab, aa

cbcβ) satisfy the conditions of the

Lemma. Indeed, [~Va, ~Vb]c = Va

i∂iδcb − (a↔ b) = 0, and [~Va, ~Vb]

β = Vai∂i(ab

dbdβ)− (a↔ b) = 0 due to (216). �

Given vector field V i(zk), let us denote ϕi(τ, z0) the unique solution to the problem

dzi

dτ= V i(zk(τ)), zi(0) = zi0. (217)

For any fixed value of τ , the integral lines ϕi(τ, z), z ∈Mn determine the transformation

ϕτ : Mn →Mn, zi → ϕi(τ, zk). (218)

Sometimes we will also use the coordinate-free notation ϕτ (z) for the integral line ϕi(τ, zk). Composition of twotransformations has the property

ϕτ ◦ ϕs = ϕτ+s. (219)

Indeed, ϕi(τ, ϕj(s, zk)) and ϕi(τ + s, zk) as functions of τ obey the problem (217) with zi0 = ϕi(s, zk). Since theproblem has unique solution, they coincide. So, the set of transformations {ϕτ , τ ∈ R} is a one-parametric Lie groupwith the group product being the composition law (219).

Let ϕτ and ψλ be the one-parametric groups criated by linearly independent fields V i(zk) and U i(zk). There is aremarkable relation between commutativity of the transformations and of the vector fields.Lemma C4. The following two conditions are equivalent: (A) ϕτ ◦ ψλ(zk) = ψλ ◦ ϕτ (zk) for all τ , λ and zk. (B)

[~V (z), ~U(z)] = 0 for all z.Proof. (A) → (B). Expanding in series of Taylor, we obtain [ϕτ ◦ ψλ(zk) − ψλ ◦ ϕτ (zk)]i = ϕi(τ, ψj(λ, zk)) −ψi(λ, ϕj(τ, zk)) = [~V (z), ~U(z)]iτλ + O2(τ) + O2(λ) + O3(τ, λ). Since l.h.s. vanishes for any τ and λ, we conclude

[~V (z), ~U(z)] = 0.

(B) → (A). Consider the fields ~V and ~U in the coordinates yk of the Lemma B1. Then ~V (yk) = (1, 0, . . . , 0) andits integral line through the point yk is

ϕi(τ, yk) = (y1 + τ, y2, . . . , yn). (220)

31

Besides, the condition (B) reads 0 = [~V , ~U ]i = ∂Ui

∂y1 , that is the field ~U does not depend on y1. Consider ψλ ◦ ϕτ (zk)

and ϕτ ◦ ψλ(zk) in the system yk as functions of λ. Using (220), we can write

ψi(λ, ϕj(τ, yk)) = (ψ1, ψ2, . . . , ψn), then ψi(0, ϕj(τ, yk)) = ϕi(τ, yk) = (y1 + τ, y2, . . . , yn), (221)

ϕi(τ, ψj(λ, yk)) = (ψ1 + τ, ψ2, . . . , ψn), then ϕi(τ, ψj(0, yk)) = (y1 + τ, y2, . . . , yn). (222)

By construction, ψi(λ) satisfy the equation

dxi

dλ= U i(x2, x3, . . . , xn). (223)

As the r.h.s. of this equation does not depend on x1, the function ϕi(λ) also satisfy this equation. Besides, ψi(λ) andϕi(λ) satisfy the same initial conditions, see (221) and (222). Hence they coincide. �

Any set of coordinate lines, say the lines of the coordinates z1, z2, . . . , zk, can be used to construct a set of commutingvector fields. They are the tangent fields to the coordinate lines. The following Lemma is an inversion of this statement.It also generalizes the Lemma B1 to the case of several fields.

Lemma C5. (On rectification of the commuting vector fields). Let ~V1, ~V2, . . . , ~Vk be linearly independent and

commuting vector fields in vicinity of z0 ∈Mn: [~Va, ~Vb] = 0. Then:

There are coordinates yi = (ya, yα), α = k + 1, . . . , n, where the fields ~Va are tangent to the coordinate lines ya:V ia (yj) = δia, a = 1, 2, . . . , k.

Notice the immediate consequences of the Lemma: through each point z1 ∈ Mn passes a surface Nk such that~V1(z), ~V2(z), . . . , ~Vk(z) form a basis of the tangent spaces TN(z) at any point z ∈ Nk. Integral lines of the fields~Va, that cross Nk, entirely lie in Nk. Evidently, in the coordinates yk these surfaces are given by the equationsyα = cα = const.Proof. Without loss of generality we assume that the point z0 has null coordinates. Selecting the appropriate n− kvectors among the basic vectors of coordinate lines, say ~eα, with coordinates eiα = δiα, α = k + 1, . . . , n, we complete

the vectors ~Va(z0) up to a basis of TM(z0). Then determinant of the matrix composed from components of the basicvectors is not equal to zero at z0

det(~V1, . . . , ~Vk, ~ek+1, . . . , ~en)|z0=0 6= 0. (224)

Denote ϕτa the one-parametric group (218) criated by the field ~Va. Consider the mapping h : O(~0) ∈ Rn → Mn

defined according the rule

z = h(τ1, . . . τk, y1, . . . , yn−k) = ϕτ1 ◦ . . . ◦ ϕτk(0, . . . , 0, yk+1, . . . yn). (225)

Derivatives of this function at the point τa = yα = 0 are dhdτa

∣∣∣0

= ddτa

ϕτa(0, . . . , 0, 0, . . . , 0)∣∣∣τa=0

= ~Va(0) and

dhdyα

∣∣∣0

= ddzα (0, . . . , 0, 0, . . . , yα, . . . , 0)

∣∣zα=0

= (0, . . . , 0, 0, . . . , 1, . . . , 0) = ~eα. Then det ∂(z1,z2,...,zn)∂(τ1,...τk,yk+1,...,yn)

∣∣∣0

=

det(~V1, . . . , ~Vk, ~ek+1, . . . , ~en)|0 6= 0, see (224). So the mapping (225) is invertible, and we can take yi ≡ (τa, yα)

as a coordinate system on Mn. The transition functions are given by Eq. (225).

Consider the integral line ϕsa(z) of the field ~Va through some point z. According to Lemma C4, commutativity ofthe fields implies the commutativity of their one-parametric groups, so we have

ϕsa(z) = ϕsa ◦ ϕτ1 ◦ . . . ϕτa . . . ◦ ϕτk(0, yα) = ϕτ1 ◦ . . . ϕτa+sa . . . ◦ ϕτk(0, yα) = h(τ1, . . . , τa + sa, . . . , τk, yα). (226)

This shows that integral lines of ~Va are the coordinate lines of ya -coordinate of the new system. Hence the integrallines lie in the submanifolds Nk = {yk ∈Mn, y

α = cα = const}.To find equations of these surfaces in the original coordinates, denote h the inverse mapping of (225). Let the point

z1 has coordinates τ1, . . . , τk, ck+1, . . . cn in the system yi. Then the submanifold is Nk = {z ∈Mn, h

α(zi) = cα}. �Frobenius theorem. Let Aia(zk), a = 1, 2, . . . k be a set of functions with rank A = k. The system of first-orderpartial differential equations

Aia(zk)∂iX(zk) = 0, (227)

has n− k functionally independent solutions, if and only if the vectors ~Aa form a set with closed algebra

[ ~Aa(zk), ~Ab(zk)] = cab

c(zk) ~Aa(zk). (228)

32

Proof. Let the functions Fα(zk), α = 1, 2, . . . , n− k represent the solutions:

Aia(zk)∂iFα(zk) = 0. (229)

Consider the foliation {N~ck, ~c ∈ Rn−k} determined by Fα according to Eq. (206), and let ~Ua(zk) be vector fieldsdescribed in Lemma C1.

Denoting zia(τ) integral lines of ~Aa(zk), we have ddτ F

α(zia(τ)) = Aia(zk)∂iFα(zk)

∣∣zia(τ)

= 0 according to (229). Then

Fα(zia(τ)) = cα = const, that is integral lines of ~Aa(zk) lie in N~ck, and ~Aa(zk) are tangent vectors to this submanifold

at each point. Then we can present them through the basic vectors ~Ub: ~Aa = bab~Ub of Lemma C1. According to

Lemma C1, [~Ua, ~Ub] = 0. According to Lemma C2, this implies (228).Let (228) is satisfied. Assuming Aia = (aa

b, baβ) with det a 6= 0 (see Lemma C3), we write the system (227) in the

equivalent form aabAib(z

k)∂iX(zk) ≡ V ia (zk)∂iX(zk) = 0. According to Lemma C3, we have [~Va, ~Vb] = 0. Accordingto Lemma C5, there are coordinates yk where V ia (yk) = δa

i. In these coordinates our system acquires the form∂∂yaX

′(yβ , yb) = 0. The functions F β(yβ , yb) = yβ give n− k functionally independent solutions. �

Frobenius theorem, geometric formulation. Let ~A1(zk), . . . , ~Ak(zk) be linearly independent vector fields on Mn.The following two conditions are equivalent:

(A) The fields ~Aa form closed algebra:

[ ~Aa(zi), ~Ab(zi)] = cab

c(zi) ~Ac(zi). (230)

(B) There is a foliation {N~ck, ~c ∈ Rn−k} of Mn such that ~Aa(zk) are tangent to N~ck at each point zk ∈Mn, and henceform a basis of TNck(zk).

Proof. (B) → (A). Consider z0 ∈ Mn and let z0 ∈ N~ck, where N~ck is one of submanifolds specified in (B). Let zi(τ)

be integral line of the field [ ~Aa, ~Ab]i, which at τ = 0 passes through z0. We write

d

dτFα(zi(τ)) = [ ~Aa, ~Ab]

i∂iFα∣∣∣zi(τ)

=[~Aa( ~Ab(F

α))− (a↔ b))∣∣∣zi(τ)

= 0, (231)

since ~Ab(Fα) = Aib∂iF

α = 0. The equality (231) implies that integral line of the field [ ~Aa, ~Ab]i through z0 entirely

lies in N~ck, so the vector [ ~Aa, ~Ab]i(z0) lies in TNck(z0). Hence it can be presented through the basic vectors ~Aa, which

gives the desired result (230).(A) → (B). Let (230) is satisfied. Using Lemma C3, we construct k linearly independent and commuting fields

~Va. According to Lemma C5, there are coordinates yk where V ia (yk) = δai. Consider the foliation {N~ck, ~c ∈ Rn−k}

where N~ck = {zk ∈Mn, yα = cα = const}. By construction, ~Va ∈ TNck and form a basis of TNck at each point zk ∈Mn.

According to Lemma C3, the linearly independent vectors ~A are linear combinations of ~Va, so they also form a basisof TNck at each point zk ∈Mn. �

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