On Moment Maps and Jacobi Manifolds

Alexander Leguizam´onRobayo

Universidad Nacional de Colombia Facultad de Ciencias, Departamento de Matem´aticas Bogot´aD.C., Colombia 2021

Alexander Leguizam´onRobayo

Tesis o trabajo de grado presentada(o) como requisito parcial para optar al t´ıtulode: Magister en Ciencias Matem´aticas

Directores: Ph.D. Daniele Sepe Ph.D. Nicol´asMart´ınezAlba

L´ıneade Investigaci´on: Geometr´ıaDiferencial

Universidad Nacional de Colombia Facultad de Ciencias, Departamento de Matem´aticas Bogot´aD.C., Colombia 2021

A mis padres y mi hermana

That is the ﬁrst thing I know for sure: If the questions don’t make sense, neither will the answers.

Kurt Vonnegut, The Sirens of Titan

Agradecimientos

Primeramente, quisiera agradecer a mis supervisores Daniele Sepe y NicolásMartinez, por el apoyo durante este proyecto. Quisiera agradecerle a Nicoláspor introducirme al mundo de la geometr´ıade Poisson y sus gen- eralizaciones. El curso de variedades de Nicolásfue una fuerte base para iniciar con el estudio de este trabajo. Esto sin contar el apoyo de Nicolásal motivarme a participar en distintos eventos y charlas y por siempre ser curioso respecto a los posibles resultados y aplicaciones de la geometr´ıa de Poisson. Estoy agradecido con Daniele por su paciencia para resolver mis dudas sin importar lo básicas que pudieran ser. El curso de geometr´ıasimplectica me ayudómucho a aclarar dudas y tener una visiónmásaterrizada a la hora de aplicar conceptos del curso. Tambiénpor la constante retroal- imentación,esto me permitiócomprender la diferencia entre entender un concepto matemático superficialmente y realmente entenderlo a fondo, sabiendo mostrar todas las sutilezas que hay detrásde un razonamiento riguroso. Agradezco a la Universidad Nacional de Colombia, por apoyar mis estudios a travésde la Beca de exenciónde derechos académicosdurante el primer año,y a travésde la Beca auxiliar docente, durante el segundo año. Tambiénagradezco a todos mis amigos en el Departamento de Matemáticas: a Arturo y Fabio, por su apoyo en el ejercicio de la docencia; a Haly y Jaime, por ser compañerosde viaje; a Lizeth y Yamid, por las discusiones sobre geometr´ıa;y a todos los demáspor las risas y charlas. Agradezco a Gia y a Cuore por darme una excusa para salir a caminar, despejarme, y llegar a trabajar con nuevas ideas. Agradezco a Adriana Isabel, por darme su mano y tener siempre sus brazos abiertos para abrazarme y escucharme. Tu cariñome mantuvo siempre motivado a seguir luchando. Agradezco a mis padres, Yubby y Marvin, y a mi hermana Adriana por todo su apoyo mental y emocional, estando siempre ah´ıpara alentarme, motivarme, y hacerme ver que las cosas siempre pueden mejorar.

Abstract

The main goal of this work is to introduce the idea of a Hamiltonian action in the context of Jacobi structures on line bundles. This work aims to make these construction without relying on the ”Poissonization trick”. Our deﬁnition allows us to recover the notion of (weakly)Hamiltonian action in the context of Poisson, contact, and locally conformally symplectic geometry. Keywords: Jacobi structures; contact manifolds; locally conformally symplectic structures; Hamiltonian actions; moment maps. 2010 Mathematics Subject Classiﬁcation: 53D10;53D17; 53D20, 37J15. Resumen

En el siguiente trabajo introducimos la idea de acciónHamiltoniana en el contexto de la geometr´ıa de Jacobi en fibrados de l´ıneagenerales. Esta construcciónla realizamos de forma intr´ınsecasin necesidad de recurrir al ”truco de Poissonización”. El concepto de acciónHamiltoniana en geometr´ıade Jacobi nos permite recuperar resultados conocidos en geometr´ıade Poisson, contacto, y localmente conformemente simpléctica. Palabras clave: estructuras de Jacobi; variedades de contact; estructuras localmente conformemente simplécticas;acciónHamiltoniana; aplicaciónmomento . Contents

Agradecimientos vii

Abstract ix

1 Introduction 2

2 Vector Bundles 4 2.1 Vector bundles ...... 4 2.1.1 Morphisms ...... 7 2.1.2 Sections of a Vector Bundle ...... 8 2.1.3 Subobjects ...... 12 2.2 Operations on vector bundles ...... 13 2.2.1 Pullback bundles ...... 15 2.3 Algebraic preliminaries ...... 17 2.3.1 Differential Operators on Algebras ...... 17 2.3.2 Derivations and infinitesimal symmetries ...... 19 2.4 Derivations and infinitesimal automorphisms of vector bundles ...... 20 2.4.1 Infinitesimal Automorphisms ...... 23 2.5 Group actions and automorphisms ...... 30 2.5.1 Infinitesimal Actions on Vector Bundles ...... 31

3 Poisson and Symplectic Manifolds 35 3.1 First definitions and properties ...... 35 3.1.1 Multivector fields and the Poisson bivector ...... 37 3.1.2 Hamiltonian vector fields ...... 40 3.1.3 Poisson symmetries ...... 41 3.2 Nondegenerate Poisson manifolds ...... 42 3.2.1 Hamiltonian vector fields ...... 44 3.3 Poisson actions and Hamiltonian Poisson actions ...... 45

4 Jacobi Geometry 52 4.1 Local Lie algebras and Jacobi manifolds ...... 52 4.2 Jacobi morphisms ...... 56 4.3 Nondegenerate Jacobi manifolds ...... 62 4.3.1 Locally conformally symplectic manifolds ...... 63 4.3.2 Contact manifolds ...... 68 Contents 1

5 Group actions on Jacobi manifolds 78 5.1 Jacobi group action ...... 78 5.2 Weakly Hamiltonian action ...... 82 5.3 Hamiltonian actions ...... 85

6 Closing remarks 90 6.1 Further Work ...... 90

Bibliography 91 1 Introduction

Consider a symplectic manifold (M, Ω). The symplectic form Ω allows us to associate a (Hamilto- ∞ nian) vector field Xf to any smooth function f ∈ C (M). The assignment of Hamiltonian vector ∞ fields allows us to endow C (M) with a Lie algebra structure given by {f, g} = Ω(Xf ,Xg). This bracket is of key importance in many areas of mathematics and physics [21, 22]. A similar construction can be done for manifolds with other structures such as: Poisson structures [7, 22], contact structures [9, 3], and locally conformally symplectic (LCS) structures [25, 23]. All of these brackets are particular examples of Jacobi structures, which were defined by Lichnerowicz [15] as a pair (Λ,E) of a bivector Λ and a vector field E such that

[Λ, Λ] = 2E ∧ Λ, [Λ,E] = 0.

The work of Lichnerowicz studied Jacobi structures in a setting, in which the pairs (Λ,E) are locally deﬁned and are related on overlaps by a conformal transformation. This approach leads to the notion of a Jacobi bundle deﬁned by Marle [17]. Earlier, Kirillov [11] studied Lie algebra structures

{·, ·} : Γ(E) × Γ(E) → Γ(E), on the space of sections of a vector bundle E, such that the bracket is local i.e. for any pair of sections u, v ∈ Γ(E) supp({u, v}) ⊂ supp(u) ∩ supp(v). Kirillov proved that when E is a vector bundle with one dimensional fibres (i.e. a line bundle), the Lie algebra structure on Γ(E) is given locally by a Lichnerowicz’ Jacobi structures (Λ,E). On the works of Marle [17], and Vaisman [25], Hamiltonian vector fields are shown to be infinitesimal conformal symmetries of the Jacobi structures induces by the respective brackets. Then the definition of a Hamiltonian vector field must be given in terms of the Jacobi bundle. Here the main question of our work appears: How can the concept of Hamiltonian vector field be defined in the setting of Jacobi bundles? Recent works [24, 26, 28] have introduced a more algebraic language to the study of Jacobi bundles by studying the C∞(M)−module structure of Γ(E). In this setting, the notion of an infinitesimal symmetry is replaced by that of a derivation. A derivation ∆ on Γ(E) is an R−linear operator such that for all u ∈ Γ(E) and f ∈ C∞(M)

∆(fu) = f∆(u) + X∆(f)u, where X∆ ∈ X(M) is the symbol of ∆. In this sense, Hamiltonian vector ﬁelds appear as the symbols of derivations associated to sections of Γ(E). In this context, this work has three aims: 3

1. to provide a more geometric interpretation of the algebraic language of [24] via examples.

2. to propose the main deﬁnitions and prove the main results related to (weakly) Hamilto- nian actions in Jacobi bundles in the spirit of [3] (i.e without resorting to the use of other structures).

3. to recover previous results on Hamiltonian actions, i.e., Stanciu’s results [23] for the LCS case, and Loose’s [16] for the contact case, and the relevant results in Poisson geometry [7, 21].

The structure of this work is given as follows:

• In Chapter 2, we discuss the basic theory of vector bundles. We focus on the constructions that are instrumental to make sense of the theory of Jacobi manifolds.

• In Chapter 3, we discuss the basic theory of Poisson manifolds and their Hamiltonian actions. We recover several results in symplectic geometry as an instance of the results on Poisson manifolds.

• In Chapter 4, we present the theory of Jacobi manifolds. We pay special attention to the transitive cases (contact and locally conformally symplectic manifolds) and present important results in each case.

• In Chapter 5, we introduce the idea of (weakly) Hamiltonian actions. We instantiate it in the trivial bundle case and recover previous results on contact [16] and locally conformally symplectic [23] manifolds.

• In Chapter 6, we close this work by stating some of the main questions that could be explored as a result of this work. 2 Vector Bundles

The following chapter contains a brief introduction to the study of vector bundles. We begin by introducing the basic definitions and constructions on vector bundles. The main references for this section are [12, 7, 2, 18]. In particular, we focus on pullback bundles as they are needed to make sense of morphisms of Jacobi manifolds. Afterwards, we recall the idea of derivations its relationship with vector fields in an algebraic setting following [1, 18]. Then, we introduce the ideas of differential operators, derivations, and infinitesimal automorphisms of vector bundles. We study how these three concepts are related to each other. These concepts are key to understand Jacobi bundles and their symmetries. In this section, we follow [18]. We close this chapter giving the main definitions of group actions on vector bundles and their infinitesimal counterpart.

2.1 Vector bundles

The general idea of a vector bundle is the following: consider a smooth manifold E that is the union of a family of finite dimensional (real) vector spaces {Ep}p∈M parameterized by a smooth manifold M. In order to formalize this idea, we proceed in a similar fashion as in the case for smooth manifolds, that is by defining vector bundle charts, vector bundle atlases, and finally vector bundles.

Deﬁnition 2.1. Let π : E → M be a surjective submersion between smooth manifolds E and M. A trivializing chart or vector bundle chart for π : E → M is a pair (U, φ), where U is an open subset of M and φ : π−1(U) → U × Rr is a diﬀeomorphism, such that the following diagram commutes

φ π−1(U) U × Rr π π1 U,

r where π1 : U × R → U is the projection onto the ﬁrst component. −1 The preimage π (p) of p ∈ M is known as the ﬁbre on p and it is denoted Ep.

As π : E → M is a surjective submersion, each point p ∈ M is a regular value of π. By the submersion level set theorem [14, Corollary 5.3], each of the fibres Ep is an embedded submanifold of E. Given a trivializing chart (U, φ) for π : E → M, for any p ∈ U, the map φ induces a fibrewise p r diffeomorphism φ : Ep → R given by

p φ := π2 ◦ φ|π−1(p). 2.1 Vector bundles 5

Therefore for v ∈ Ep, we have that φ(v) = (p, φp(v)).

p Since φ is a fibrewise diffeomorphism, we can induce a vector space structure on each Ep. We would like that for two distinct trivializing charts whose domain contains p ∈ M induce isomorphic vector space structures. The following definition takes care of this.

Deﬁnition 2.2. Consider two vector bundle charts (U1, φ1) and (U2, φ2) such that U1 ∩ U2 is not empty. We say that these charts are compatible if for all p ∈ U1 ∩ U2 the map

p p −1 r r γ1,2(p) := φ1 ◦ (φ2) : R → R , is a linear isomorphism. r The map g1,2 : U1 ∩ U2 → GL(R ) is called the transition function. Note that a transition function is smooth. Similarly to the case of smooth manifolds, we can deﬁne an atlas using trivializing charts.

Deﬁnition 2.3. A vector bundle atlas {Uα, φα}α∈A for π : E → M is a set of pairwise compatible trivializing charts such that {Uα}α∈A is an open cover of M. We say that two atlases are equivalent if their union is a vector bundle atlas.

Deﬁnition 2.4. A vector bundle is a triple (E, π, M), where E and M are smooth manifolds, and π : E → M is a surjective submersion, with an equivalence class of vector bundle atlases; so we need to know at least one vector bundle atlas.

We say that E is the total space, M the base manifold, and π the projection. If each ﬁbre Ep is an r−dimensional vector space, and the base manifold M is connected, we say that E is of rank r.

From now on, we only work with vector bundles whose base is connected, i.e. those where the notion of rank is well-deﬁned. Vector bundles of rank 1 are know as line bundles. These will be very important in the following sections as a Jacobi structures are structures on line bundles. We denote a vector bundle as follows (π : E → M).

Example 2.5. A vector bundle of rank r of the form (π : M × Rr → M) is a trivial vector r −1 r bundle of rank r, denoted by RM . We know that the preimage of M under π is π (M) = M ×R . There is a global trivializing chart given by (M, id) where id : M × Rr → M × Rr is the identity map.

Example 2.6. Associated to any connected manifold M there is the tangent bundle TM. If dim M = m, the triple π : TM → M is a vector bundle of rank m.

Let {Uα, ϕα}α∈A be an atlas of M. Take (U, ϕ) a chart of M, with coordinate functions x1, . . . , xm. Then all v ∈ TU = π−1(U) is of the form m X ∂ vp = vi . ∂x i=1 i p We deﬁne a trivializing chart (U, φ), where

φ(vp) = (p, (v1, . . . , vn)). 6 2 Vector Bundles

The map φ is linear on ﬁbres and satisﬁes the condition

π1(φ(vp)) = p = π(vp),

m where π1 : U × R → U is the projection onto the ﬁrst component. For two distinct charts (Uα, ϕα) and (Uβ, ϕβ) of M with nonempty intersection, the previous reasoning gives us two distinct vector bundle charts (Uα, φα) and (Uβ, φβ). The transition function m gα,β : Uα ∩ Uβ → GL(R ) is given by

−1 gα,β(p) = dp(ϕα ◦ ϕβ ).

−1 As M is a manifold, the map ϕα ◦ϕβ is a diﬀeomorphism. It follows that its diﬀerential is a linear isomorphism.

By repeating this construction for each chart (Uα, ϕα) ∈ {Uα, ϕα}α∈A, we obtain a vector bundle atlas for TM.

Example 2.7. Let M be a smooth manifold and X ∈ X(M) a nowhere vanishing vector ﬁeld.

There is a vector bundle induced by X, where the ﬁbres are given pointwise by span Xp. We denote the bundle generated by the vector ﬁeld X by π : RX → M. We have that

−1 π (M) = {λXp : λ ∈ R, p ∈ M}.

A trivializing chart (M, φ) is given by

−1 φ : π (M) → M × R, λXp 7→ (p, λ).

It follows that π1(φ(λXp)) = p = π(λXp). Therefore the bundle π : RX → M is a line bundle.

Example 2.8. [2, Example 1.7] Recall that the projective space RP n is the set of equivalence clases of Rn+1 under the equivalence relation x ∼ y if there is some nonzero element λ ∈ R such that x = λy. Associated to RP n is the tautological line bundle or canonical line bundle, where the total space is given as follows

n n+1 E = {([x], v) ∈ RP × R : v = λx, for some λ ∈ R}.

For an element [x] ∈ RP n, the ﬁbre corresponds to the line in Rn+1 that passes through x and the 1 n origin. This vector bundle is denoted by γd or ORP (1). n To prove that ORP (1) is a vector bundle, we construct a vector bundle atlas. n Consider the chart (Ui, ϕi) of RP , where Ui given in homogeneous coordinates by all the elements of the form [x0 : ··· : xi : ··· : xn] such that xi 6= 0 and

n ϕi : Ui → R , [x0 : ··· : xi : ··· : xn] 7→ (x0/xi, . . . , xi−1/xi, xi+1/xi, . . . , xn/xi).

n A trivializing chart (Ui, φi) of π : E → RP is given by

φi([x], (x0, . . . , xi, . . . , xn)) = ([x0/xi : ··· : xi−1/xi : 1 : xi+1/xi : ··· : xn/xi], xi) = ([x], xi). 2.1 Vector bundles 7

For (x, µ) ∈ Ui ∩ Uj × R, we have that

−1 φi ◦ (φj) ([x], µ) = φi([x], µ(x0/xj, . . . , xj−1/xj, 1, xj+1/xj, . . . , xn/xj))

= ([x], µxi/xj).

Then the transition function on Ui ∩ Uj is given by

gij : Ui ∩ Uj → GL(R), [x0 : ··· : xi : . . . xj : ··· : xn] 7→ xi/xj. The transition function is well-deﬁned as it is the quotient of two homogeneous polynomials of degree 1 and xj 6= 0.

2.1.1 Morphisms

Deﬁnition 2.9. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle. A vector bundle morphism is a pair (ϕ, Φ), where ϕ : M1 → M2 is a smooth map and Φ : E1 → E2 is a smooth map such that

• the map Φ covers ϕ, i.e. the following diagram commutes Φ E1 E2

π1 π2 ϕ M1 M2

• the ﬁbrewise map

Φp := Φ(E1)p :(E1)p → (E2)ϕ(p), is a linear transformation for all p ∈ M. Remark 2.10. Composition of vector bundle morphism is given as follows:

For j = 1, 2, 3, let (πj : Ej → Mj) be a vector bundle. Consider the morphisms

(ϕ, Φ) : (π1 : E1 → M1) → (π2 : E2 → M2)

(ψ, Ψ) : (π2 : E2 → M2) → (π3 : E3 → M3).

Then the composition is given by (ψ, Ψ) ◦ (ϕ, Φ) := (ψ ◦ ϕ, Ψ ◦ Φ). If there exists two vector bundle maps

(ϕ, Φ) : (π1 : E1 → M1) → (π1 : E2 → M2)

(ψ, Ψ) : (π2 : E2 → M2) → (π1 : E1 → M1), such that they are inverse to each other, then (ϕ, Φ) is a vector bundle isomorphism. The vector bundle map (ϕ, Φ) is a vector bundle isomorphism, if and only if ϕ is a diﬀeomorphism and

Φp :(E1)p → (E2)ϕ(p) is a ﬁbrewise isomorphism for all p ∈ M1. Vector bundles with vector bundle morphisms form a category that we denote by VB. Additionally, vector bundles over the same manifold M also form a category that we denote VBM . Example 2.11. For any smooth manifold M, there is a canonical vector bundle associated to it the tangent bundle TM. If ϕ : M → N is a smooth map then (ϕ, dϕ): TM → TN is a vector bundle morphism. 8 2 Vector Bundles

Deﬁnition 2.12. We say that a vector bundle (π : E → M) of rank r is trivializable if it is r isomorphic to RM . Any choice of such an isomorphism is a trivialization of E. A manifold M is parallelizable if its tangent bundle TM is a trivializable vector bundle.

r Example 2.13. The trivial vector bundle π : RM → M is trivializable. There is a canonical trivialization given by the identity map (id , id r ). M RM However, there are more trivializations. Consider a map (ϕ, Φ) such that ϕ is a diﬀeomorphism r and the map on ﬁbres Φp ∈ GL(R ). The vector bundle map (ϕ, Φ) is invertible, and so a vector bundle isomorphism.

rj Lemma 2.14. [18, Lemma 12.5] For j = 1, 2, let (πj : RM → M) be a vector bundle and let r1 r2 Φ: RM → RM a map linear on ﬁbres. Then the following are equivalent 1. The map Φ is smooth.

2. The pair (idM , Φ) is a vector bundle morphism.

ˆ r1 j2 3. The map Φ: M → hom(R , R ) given by x 7→ Φx is smooth.

2.1.2 Sections of a Vector Bundle Deﬁnition 2.15. Let (π : E → M) be a vector bundle and U ⊆ M an open set. We say that a

(local) section of E over U is a smooth map s : U → E such that π ◦ s = IdU . If E is of rank r, then a set of sections {s1, . . . , sr} over U is a frame, if for every p ∈ U, the set {s1(p), . . . , sr(p)} is a basis for each Ep. We denote the set of all sections of E over U by ΓU (E) and set of sections of E over U by Γ(E).

Remark 2.16. For any vector bundle (π : E → M), there is a unique vector space structure

defined on each fiber Ep for all p ∈ M. Each fiber has a zero element 0p ∈ Ep. We can define a map 0 : M → E, p 7→ 0p, this map is the zero section. Note that π ◦ 0(p) = π(0p) = p. To prove the smoothness of the zero section, consider an trivializing atlas {(Uα, φα}α∈I of π : E → M. Fixing a Uα, for all p ∈ Uα we have that

(φα ◦ 0)(p) = φα(0p) = (p, 0).

r The map iα : Uα → Uα × R , p 7→ (p, 0) is a smooth map. We see that φ ◦ 0 = i, it follows that the zero section is a smooth map of Uα. Since {Uα}α∈I forms an open cover of M, the zero section is a smooth map.

Moreover, the set O = ∪p∈M 0p is a smooth submanifold of E. To verify this consider an atlas {(Uα, ϕα)}α∈I of M. Then O can be endowed with a smooth structure given by the atlas {(0(Uα), ϕα ◦ π)}α∈I . Note that O is the image of M under the zero section. Note that the inverse of the zero section is given by π|O, which, by smoothness of π, is a smooth map. Then, the zero section is a diﬀeomorphism onto its image. We conclude that O is an embedded submanifold of E. 2.1 Vector bundles 9

Remark 2.17. The set of sections Γ(E) is a vector space, where

• for s1, s2 ∈ Γ(E), vector addition is given by

(s1 + s2)(p) := s1(p) + s2(p),

for all p ∈ M.

• for s ∈ Γ(E) and λ ∈ R, scalar multiplication is given by (λs)(p) := λs(p),

for all p ∈ M. Moreover, any section s ∈ Γ(E) can be multiplied pointwise by any f ∈ C∞(M). The induced map C∞(M) × Γ(E) → Γ(E), (f, s) 7→ fs, endows Γ(E) of a C∞(M)−module structure. Lemma 2.18. Let (π : E → M) be a vector bundle of rank r and U ⊆ M an open set. A frame ∞ {s1, . . . , sr} is a basis for C (U) module of sections ΓU (E).

∞ Proof. To prove that S = {s1, . . . , sr} is a basis for C (U) module of sections ΓU (E), we need to prove that S is a generating set for ΓU (E) and that S is linearly independent. To prove linear independence, we argue by contradiction. Suppose S is not linearly independent, then for the zero section

0 = f1s1 + ··· + frsr, where fi is a smooth function for i = 1, . . . , r and fi need not be zero. It follows that there is a point p ∈ U such that

0 = f1(p)s1(p) + ··· + fr(p)sr(p), where fi(p) 6= 0 for some i = 1, . . . , r. Therefore {si(p)}i=1,...,r are not linearly independent at each p ∈ U, and so cannot be a basis for Ep. This is a contradiction as S is a frame. To prove that S is a generating set, let s ∈ ΓU (E). Then, as S is a frame, at each point p ∈ U, we have p p s(p) = a1s1(p) + · + arsr(p), p where aj ∈ R for j = 1, . . . , r. For i = 1, . . . , r, we deﬁne the functions

p fi : U → R, p 7→ ai . Then any section s can be written as

s = f1s1 + ··· + frsr.

We need to prove that each of the fi for i = 1, . . . , r are smooth. Note that as S is linearly independent, the zero section is only given by fi = 0 for all i = 1, . . . , r. As S is a basis for Ep for all p ∈ U, each si ∈ S is a nonvanishing section. The smoothness of s and each si ∈ S implies that each fi is smooth. It follows that S is a generating set for ΓU (E). 10 2 Vector Bundles

The result of Lemma 2.18, shows that for a vector bundle (π : E → M), its module of sections Γ(E) is a locally free module. This fact is a particular example of the correspondence given by the Serre-Swan theorem. Intuitively, the Serre-Swan theorem [18, Theorem 12.32] can be stated as follows:

Any ﬁnitely generated projective module over the algebra of smooth functions of a compact manifold is the space of smooth sections of a vector bundle.

For a more detailed study of vector bundles and the algebraic objects associated to them, we refer the reader to [18, Chapter 12].

Example 2.19. Consider a smooth manifold M. The set of sections of its tangent bundle Γ(TM) is the set of vector ﬁelds on X(M).

Proposition 2.20. A vector bundle is trivializable if and only if it admits a global frame.

Proof. Let (π : E → M) be a vector bundle of rank r and {s1, . . . , sr} be a global frame. By Pr Lemma 2.18, any section v ∈ Γ(E) can be written as v(p) = i λisi(p), where each λi ∈ R. This induces a map φ : π−1(M) = E → M × Rr given by

φ(v(p)) = (p, λ1, . . . , λr),

for all p ∈ M. By construction the pair (M, φ) defines a global trivializing chart of E, as we wanted to prove. Conversely, let φ : E → M × Rr a global trivializing chart of E. Since φ is a diffeomorphism, for j = 1, . . . , r, we define the following section of E

−1 sj(p) = φ (p, ej),

r where {e1, . . . , er} is the standard basis of R . Note that the set of sections {s1, . . . , sr} forms a frame, as φp is a ﬁbrewise isomorphism for all p ∈ M.

Proposition 2.20 tells us that a vector bundle is trivializable if and only if its module of sections is free.

Example 2.21. Let G be a Lie group. The tangent bundle of a Lie group is a trivializable vector bundle, and it there is a vector bundle isomorphism

TG =∼ G × g.

To construct one of such isomorphisms, recall that the Lie algebra of G is

Lie(G) = g = TeG,

where e is the identity of G. There is a smooth multiplication map m : G × G → G. By ﬁxing a ∈ G, we can deﬁne the left and right translations

La : G → G, g 7→ m(a, g)

Ra : G → G, g 7→ m(g, a). 2.1 Vector bundles 11

Since the multiplication map is smooth, both translations are smooth maps. The diﬀerentials of the translations at g induce vector space isomorphisms

dgLa : TgG → TagG, and similarly for Ra. In particular, there is a vector space isomorphism

deLa : g → TaG. (2-1)

There is an induced map on total spaces

dL : G × g → T G, (g, ζ) 7→ (deLg)(ζ). (2-2)

We claim that the vector bundle map (id, dL) is the desired trivialization of π : TG → G. Note g that dL covers the identity as (deLg)(ζ) ∈ TgG. By the fact that equation 2-1, the map dL = dgLe isomorphism of ﬁbres. Since the identity is smooth, all that is left to prove is the smoothness of the map on the total spaces. To prove so, we can take the diﬀerential of m at (g, h) ∈ G × G

d(g,h)m : TgG × ThG → TghG, (v, w) 7→ d(g,h)(v, w).

This induces a map on the total spaces given by

dm : TG × TG → T G, (v, w) 7→ d(π1(v),π1(w))m(v, w) ∈ Tm(π1(v),π1(w))G.

As shown on Remark 2.16, for any vector bundle there is a zero section. In particular there is zero section 0 : G → TG. Additionally, there is a natural inclusion map i : g → TG. Using these two maps we can construct the map

F : G × g → TG × T G, (g, ζ) 7→ (0(g), i(ζ)).

The map F is smooth as it is smooth on each component. The following computation shows that the map on the total spaces given by equation 2-2 is the map dm ◦ F .

dm ◦ F (g, v) = dm((0(g), i(v)) = d(g,e)m((0(g), i(v)) = deLg(i(v)) + dgRe(0(g)) = deLg(i(v)).

The as both dm and F are smooth, the smoothness of the map on the total spaces given by equation 2-2 follows.

The vector bundle map (idG, dL) is known as the left trivialization. By using Ra instead of La, there is a right trivialization.

Deﬁnition 2.22. Let (π : E → M) be a vector bundle. The support of a section u ∈ Γ(L), denoted supp is the set supp(f) := {x ∈ M|u(x) 6= 0}.

∼ ∞ Example 2.23. Consider the trivial line bundle (π : RM → M), then Γ(RM ) = C (M). Then the support of a section is the same as the support of a function [14, Pag. 43]. 12 2 Vector Bundles

We close the discussion on sections of vector bundles, by mentioning the relationship between vector bundle maps and morphisms in the module of sections.

Proposition 2.24. [14, Lemma 10.29]

For j = 1, 2, let (πj : Ej → M) be a vector bundle. A map (id, Φ) : (π1 : E1 → M) → (π2 : E2 → M) induces a map F : Γ(E1) → Γ(E2) on sections given by

F (s)(p) := Φ ◦ s(p), (2-3)

∞ for all p ∈ M, where s ∈ Γ(E1). The map F is C (M)-linear if and only if (id, Φ) is a vector bundle morphism.

2.1.3 Subobjects Deﬁnition 2.25. Let (η : F → M) and (π : E → M) be vector bundles. We say that (η : F → M) is a subbundle of (π : E → M) if

1. the total space F is a submanifold of E;

2. the map η is the restriction of π to F ;

3. for any p ∈ M, the ﬁbre Fp is a subspace of Ex.

Subbundles are the subobjects in VBM . Intuitively, a subbundle (η : F → M) of the vector bundle (π : E → M) can be thought of as a smooth family of vector subspaces Fp ⊂ Ep for all p ∈ M. This idea is formalized in Lemma 2.26.

Lemma 2.26. [18, Lemma 12.19] r Let (π : RM → M) be the trivial vector bundle of rank r over M, and let Fp a k−dimensional ∼ r subspace Fp ⊂ Ep = R given at every point p ∈ M. Consider the map τ : M → Gk,r given by τ(z) = Fz, where the subspace Fz is considered as a point in the Grassmannian manifold Gr,k. Then the family {Fp}p∈M deﬁnes a vector subbundle (η : F → M) if and only if the map τ is smooth.

Example 2.27. For any manifold M, a distribution on M is a vector subbundle of TM.

Deﬁnition 2.28. For j = 1, 2, let (πj : Ej → M) be a vector bundle and (id, Φ) : (π1 : E1 → M1) → (π2 : E2 → M) be a vector bundle morphism. We say that the map Φ has constant rank p k if all linear maps Φ : E1,p → E2,p have the same rank k.

Lemma 2.29. [14, Theorem 10.34] A constant rank morphism deﬁnes the following subbundles over M:

• The kernel of Φ, denoted ker Φ ⊂ E1, is the vector subbundle given by (π1|ker Φ: ker Φ → M), where the total space is given by {v ∈ E1 : Φ(v) = 0}.

• The image of Φ, denoted Im Φ ⊂ E2, is the vector subbundle given by (π2 : Im Φ → M), where the total space is given by {w ∈ E2 : w = Φ(v)}.

If the map Φ is not of constant rank, its image and kernel of Φ need not be vector subbundles. 2.2 Operations on vector bundles 13

Example 2.30. Consider the vector bundle (π : R2 → R), where π is the projection onto the first component. The vector bundle map (id, m):(π : R2 → R) → (π : R2 → R), where id : R → R is the identity map and m : R2 → R2, (u, v) 7→ (u, uv) is not of constant rank. Note that ker m does not define a vector subbundle of (π1 : E1 → M), as the fibre (ker m)0 is one dimensional, while (ker m)p is zero dimensional for all p 6= 0. It follows that ker m is not a smooth manifold and thus it is not a submanifold of R2. The set Im m is not a vector bundle as the fibre (Im m)0 is zero dimensional, while (Im m)p is one dimensional for all p 6= 0.

p Lemma 2.31. Let (ϕ, Φ) : E1 → E2 be a vector bundle map, if Φ is injective for all p ∈ M1 (that p is (ϕ, Φ) is a monomorphism) or if Φ is surjective for all p ∈ M1 (that is (ϕ, Φ) is an epimorphism), then (ϕ, Φ) has constant rank.

Proof. This is a direct consequence of the rank-nullity theorem. p p If ker Φ = {0} for all p ∈ M then rank Φ = dim E1,p is constant. p p On the other hand, if Φ is surjective for all p ∈ M1 then rank Φ = dim E2,ϕ(p) for all p ∈ M1. As a result of Lemma 2.31, the notion of a exact sequence of vector spaces can be extended to vector bundles and morphisms of constant rank.

Example 2.32. For j = 1, 2, let (πj : Ej → M) be a vector bundle such that E1 is a subbundle of E2. Then the inclusion (id, ι): E1 → E2 is a vector bundle monomorphism. Therefore, the ﬁbrewise quotient induces a vector bundle, the quotient bundle, denoted by E2/E1. We can form then a short exact sequence of vector bundles over M

ι pr 0 E1 E2 E2/E1 0, where pr denotes the projection map onto the quotient bundle, and every map is covering the identity.

2.2 Operations on vector bundles

For j = 1, 2, let (πj : Ej → M) be a vector bundle. Then, one can construct new vector bundles from them.

Consider direct sum as the vector bundle (π12 : E1 ⊕ E2 → M) is given on ﬁbres by (E1 ⊕ E2)p = E1,p ⊕ E2,p for all p ∈ M . Let us verify that the direct sum is indeed a vector bundle. To do so, we recall the following result:

Lemma 2.33. For j = 1, 2, let Mj and B be a smooth manifolds. Given a smooth submersion π : M1 → B and a smooth map φ : M2 → B, then the ﬁber product of M1 and M2 with respect to B given by

M1 ×B M2 = {(p, q) ∈ M1 × M2|π(p) = φ(q)}, is a smooth submanifold of M1 × M2. 14 2 Vector Bundles

Proof. Note that the diagonal denoted by ∆ is an embedded submanifold of M1 × M2. As π : M1 → B is a surjective submersion, the map π × φ : M1 × M2 → B × B is transverse to ∆. We recall the fact that if M,N are smooth manifolds and f : N → M is a smooth map transversal to a submanifold S ⊂ M, then f −1(S) is an embedded submanifold of N[14, Theorem 6.30].

It follows that M1 ×B M2 is an embedded submanifold of M1 × M2.

Note that the total space E1 ⊕E2 := {(p, q) ∈ E1 ×E2|π1(p) = π2(q)}. This means that E1 ⊕E2 = E1 ×M E2. By Lemma 2.33, E1 ⊕ E2 is a smooth manifold. The projection π12 : E1 ⊗ E2 → M is given by (p, q) = π1(p) = π2(q). To construct trivializing charts of (π12 : E1 ⊕ E2 → M), we begin by considering trivializations {(Ui, ϕi)} of (π1 : E1 → M) and {(Uj, ϕj} of (π2 : E2 → M). We now 1 deﬁne an open cover {Uk} of M given by the intersections of Ui and Uj. Let gkl : Uk ∩Ul → GL(n1) 2 and gk : Uk ∩ Ul → GL(n2) be a transition function of E1 and E2 respectively. Then the map

1 gkl ⊗ Uk ∩ Ul → GL(n1 + n2) 1 gkl(x) 0 p 7→ 2 , 0 gkl(x)

deﬁnes a transition function on (π12 : E1 ⊕ E2 → M). Similarly, the following vector bundles can be constructed:

• Tensor product (π⊗ : E1 ⊗ E2 → M: is given on ﬁbres by (E1 ⊗ E2)p = E1,p ⊗ E2,p for all 1 2 p ∈ M . The transition functions are gkl ⊗ gkl.

∗ ∗ ∗ • Dual vector bundle (πd : E1 → M): is given on ﬁbres by (E1 )p = (E1,p) for all p ∈ M 1 T and the transition functions are (gkl) .

k k k • Exterior product (π∧k : ∧ E1 → M) is given on ﬁbres by (∧ E1)p : ∧ (E1,p) and the k 1 transition functions are ∧ gkl.

• Hom(E1,E2)-bundles: are given on ﬁbres by (Hom(E1,E2))p = Hom(E1,p,E2,p) As in ∼ ∗ the case of vector spaces, there is a canonical isomorphism Hom(E1,E2) = E1 ⊗ E2. The 1 T 2 transition functions are (gkl) ⊗ gkl. These constructions are very important throughout diﬀerential geometry.

Example 2.34. Let M be a smooth manifold and TM its tangent bundle. The cotangent bundle ∗ • ∗ T M is the dual of TM. Differential forms are sections of (π∧• : ∧ T M → M). Similarly, • multivector fields are sections of (π∧• : ∧ TM → M). Moreover some geometrical structures can be defined in terms of vector bundles. For example, a Riemannian structure g on a manifold is a section of g ∈ Γ(⊗2TM). 2.2 Operations on vector bundles 15

2.2.1 Pullback bundles

Deﬁnition 2.35. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle. Consider a smooth map ∗ ϕ : M1 → M2. Then the pullback bundle of E2 by ϕ is the vector bundle (ˆπ1 : ϕ E2 → M1), where the total space is given by

∗ ϕ E2 = {(p, v) ∈ M1 × E2 : ϕ(p) = π2(v)}, and the projection is given by ∗ πˆ1 : ϕ E2 → M1, (p, v) 7→ p.

Lemma 2.36. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle. Consider a smooth map ϕ : M1 → M2. Then the pullback bundle of E2 by ϕ is a vector bundle. ∗ Proof. We begin by noticing that the total space ϕ E2 is a fibered product of smooth manifolds. ∗ By Lemma 2.33, ϕ E2 is a smooth manifold. It is immediate to see that the projectionπ ˆ1 is a surjective submersion. To verify local triviality, fix a point b ∈ M1 such that a = ϕ(b). Consider a trivializing chart (U, ψ) −1 n of (π2 : E2 → M2), such that a ∈ U. Then ψ :(π1) (U) → U × R is a diffeomorphism. Note −1 that M1 can be covered by open sets of the form ϕ (U). n n Let pr2 : U ×R → R be the projection onto the second component. We define the following map −1 −1 −1 n χU : (ˆπ1) (ϕ (U)) → ϕ (U) × R , (x, v) 7→ (x, (pr2 ◦ ψ)(v)).

It is direct to see thatπ ˆ1 = pr1 ◦ χ. As χU is the identity on the first compontent, and it is the composition of smooth maps on the second component, χU is a smooth map. To prove that χU −1 −1 is a diffeomorphism, it suffices to note that its inverse is given by χU (x, r) = (x, ψ (x, r)). The smoothness of χ−1 is a result of the fact that ψ is smooth.

−1 The previous construction gives a family of trivializing charts {(ϕ (U), χU }. To complete the −1 proof, we need to show that {(ϕ (U), χU } is a vector bundle atlas. For k = i, j, take (Uk, ψk) −1 −1 trivializing charts of (π2 : E2 → M2) such that ϕ (Ui) ∩ ϕ (Uj) 6= ∅, with transition function 2 2 n gij. Note that gij(p) ∈ GL(R ), for all p ∈ Ui ∩ Uj. Then the function −1 −1 2 gij : ϕ (Ui) ∩ ϕ (Uj) → GL(n), x 7→ gij(ϕ(x)), −1 −1 −1 deﬁnes a transition function on ϕ (Ui) ∩ ϕ (Uj) . Therefore {(ϕ (U), χU } is a vector bundle atlas.

Corollary 2.37. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle. Consider a smooth map ∗ ϕ : M1 → M2. Then there is a canonical map Ψ : ϕ E2 → E2 given by Ψ(x, v) = v such that (ϕ, Ψ) is a vector bundle morphism. Proof. Note that Ψ is a smooth map, as it is a projection, and ϕ is a smooth map by hypothesis. As Ψ is the identity on fibres, it is a linear transformation on fibres. Hence, it suffices to prove ∗ that Ψ covers ϕ. Let (x, v) ∈ ϕ E2, then ϕ(x) = π2(v). It follows that

(π2 ◦ Ψ)(x, v) = π2(v) = ϕ(x) = (ϕ ◦ πˆ1)(x). 16 2 Vector Bundles

Proposition 2.38. For j = 1, 2, let πj : Ej → Mj be a vector bundle. Consider a vector bundle morphism Φ : E1 → E1 covering a smooth map ϕ : M1 → M2. Then there exists a unique map ˜ ∗ ˜ Φ: E1 → ϕ E2 covering the identity such that (idM1 , Φ) is morphism of vector bundles and makes the following diagram commute

Φ˜ ∗ Ψ E1 ϕ E2 E2

π1 πˆ1 π2

idM1 ϕ M1 M1 M2,

∗ where the map Ψ : ϕ E2 → E2 is the canonical map given by Corollary 2.37. ˜ Additionally, (idM1 , Φ) is a vector bundle isomorphism if and only if Φx is a ﬁbrewise isomorphism for all x ∈ M1.

∗ −1 Remark 2.39. Since the ﬁbres of ϕ E2 are of the form π2 (ϕ(x)), it follows that for all x ∈ M1, ˜ −1 −1 the map Φx : π1 (x) → π2 (ϕ(x)) is exactly the map Φx, i.e., the restriction of Φ to the ﬁbre Ex.

∗ Proof. We define a map Φ:˜ E1 → ϕ E2, y 7→ (π1(y), Φ(y)). It is immediate to see that Φ˜ is a smooth map and covers the identity. By hypothesis, the map Φ is linear on fibres, i.e., the map −1 −1 Φx :(π1) (x) → (π2) (ϕ(x)) is a linear transformation for all x ∈ M. Remark 2.39, tells us that ˜ ˜ ˜ Φx = Φx, then Φ is linear on fibres. Hence, (idM1 , Φ) is a vector bundle morphism. ˜ ˜ ˜ It is easy to see that Φ = Ψ ◦ Φ. Thus, (ϕ, Ψ) ◦ (idM1 , Φ) = (ϕ, Φ). Uniqueness of Φ follows from the fact that the map Ψ is unique. ˜ To prove the second statement of Proposition 2.38, notice that if (idM1 , Φ) is a vector bundle isomorphism, the map Φ˜ x is a linear isomorphism for all x ∈ M1. Since Φ˜ x = Φx, then Φx is a linear isomorphism for all x ∈ M1. Conversely, suppose that Φx is a linear isomorphism for all x ∈ M1. It follows that Φ˜ x is a linear −1 ∗ isomorphism for all x ∈ M1. Then there is an inverse map (Φ˜ x) :(ϕ E2)x → (E1)x. Fix v ∈ E1 ∗ ˜ −1 such that π(v) = x, then Φ(v) ∈ (E2)ϕ(x) = (ϕ E2)x. Then (Φx) (π(v), Φ(v)) = v. We define the map −1 ∗ −1 Φ˜ : ϕ E2 → E1, (x, y) 7→ (Φ˜ x) (x, y).

Since Φ is a smooth map, the map x 7→ Φx is a smooth map. Hence, x 7→ Φ˜ x deﬁnes a smooth map −1 −1 and so x 7→ (Φ˜ x) is a smooth map. Therefore Φ˜ is a smooth map. By construction, it is inverse ˜ ˜ ˜ −1 to Φ. We conclude that (idM1 , Φ) is a vector bundle isomorphism with inverse (idM1 , Φ ).

For j = 1, 2, let (πi : Ei → N) be a vector bundle and vector bundle map

(idN , Φ) : (π1 : E1 → N) → (π2 : E2 → N).

∗ Consider a smooth map ϕ : M → N, then there is a vector bundle map (idM , ϕ Φ) : (π1,M : ∗ ∗ ϕ E1 → M) → (π2,M : ϕ E2 → M), where

ϕ∗Φ(p, v) = (p, Φ(v)),

∗ ∗ for all (p, v) ∈ ϕ E1. We say that the map (idM , ϕ Φ) is the pullback of the vector bundle morphism (idN , Φ). 2.3 Algebraic preliminaries 17

Proposition 2.40. [6, Prop. 26.3] Let (π : E → N) be a vector bundle and ϕ : M → N be a smooth map. Then 1. The pullback of the trivial vector bundle is canonically isomorphic to the trivial bundle ∗ r ∼ r ϕ RN = RM . 2. If ψ : Q → M is a smooth map then, (ϕ ◦ ψ)∗E = ψ∗(ϕ∗E).

∗ 3. The pullback of the identity morphism is the identity ϕ (idE) = idϕ∗E.

4. If (id, Φ) : E → E1 and (id, Ψ) : E1 → E2 are morphisms of vector bundles over N, then ϕ∗(Ψ ◦ Φ) = ϕ∗(Ψ) ◦ ϕ∗(Φ). Corollary 2.41. Let π : E → M be a vector bundle and (ϕ, Φ) be a vector bundle automorphism. Then (ϕ, Φ) induces a unique map α : E → ϕ∗E covering the identity, such that (Id, α) is a vector bundle isomorphism. We denote χ := α−1 : ϕ∗E → E Proof. Note that by the universal property of the pullback bundle (Proposition 2.38) there exist a unique α : E → ϕ∗E given by α(v)) = (π(v), Φ(v)) with the desired property. As α is induced by a vector bundle automorphism, then it has an inverse map χ : ϕ∗E → E with the desired property. The construction of this map is shown in the proof of Proposition 2.38.

2.3 Algebraic preliminaries

2.3.1 Diﬀerential Operators on Algebras

Deﬁnition 2.42. Consider a commutative R-algebra A with unity. An R-linear map ∂ : A → A such that ∂(ab) = a∂(b) + ∂(a)b, for all a, b ∈ A, is known as a derivation of A. We denote the set of all derivations of A by Der(A).

Lemma 2.43. An R−linear map ∂ : A → A is a derivation of A if and only if ∂(λ) = 0 for all λ ∈ R and ∂a − a∂ ∈ A. This means that there exist a ca ∈ A such that for all b ∈ A,

(∂a − a∂)(b) = cab. Proof. Consider 1 ∈ A, then we have that

∂(1) = ∂(1 · 1) = 1∂(1) + 1∂(1) = 2∂(1).

It follows that ∂(1) = 0. By R−linearity, we have that ∂(λ) = 0 for all λ ∈ R. Conversely, suppose ∂(λ) = 0 for all λ ∈ R and ∂a − a∂ ∈ A for all a ∈ A. Notice that the Leibniz identity for ∂ is equivalent to ∂a − a∂ = ∂(a) for all a ∈ A. Fix a ∈ A, let c := ∂a − a∂ ∈ A. We compute

(∂a − a∂)(1) = (1)c ∂(a) − 0 = c ∂(a) = c. 18 2 Vector Bundles

The for all b ∈ A we have that ∂(a)b = cb = (∂a − a∂)(b), it follows that ∂(ab) = a∂(b) + b∂(a).

Example 2.44. [12, Lemma 3.3] Consider a diﬀerentiable manifold M. Associated to it we have the algebra of smooth functions C∞(M). We have that Der(C∞(M)) =∼ X(M). (2-4)

Recall that Der(C∞(M)) is the set of all R−linear operators D such that

D(fg) = D(f)g + fD(g).

It is evident that every vector ﬁeld is a derivation, since for all p ∈ M,

X(fg)(p) = f(p)X(g)(p) + g(p)X(f)(p).

∞ Conversely, consider a derivation D ∈ Der(C (M)). Then Dp(f) = D(f)(p) is a derivation at p, then it can be proven [12, Section 1.5] that Dp = Xp for some Xp ∈ TpM. Taking this construction for all p ∈ M, we obtain a smooth section of TM, that is a vector ﬁeld.

Definition 2.45. [1, Section 3.1] Let A be a commutative R−algebra with unity. The ring of differential operators on A, denoted by D(A), is defined as a subring of the set End(A) of algebra endomorphisms of A, generated by A and Der(A). We define the set of differential operators of order 0 as D0(A) := A.

Let ca be the linear map given by

ca(Q) = [Q, a] = Qa − aQ,

where a ∈ A, Q ∈ D(A). We deﬁne the set of diﬀerential operators of order n as

n D (A) = {Q ∈ End(A)|ca0 ◦ ca1 ◦ · · · ◦ can (Q) = 0 ∀a0, a1, . . . , an ∈ A}

Proposition 2.46. [1, Lemma 3.1.1] Consider an associative commutative algebra with unity A. Then the set of diﬀerential operators of order at most 1 corresponds to

D1(A) =∼ A ⊕ Der(A). (2-5)

Proof. Let a ∈ A ∩ Der(A). On one hand, as a ∈ A, we have that a(bc) = abc for all b, c ∈ A. On the other hand, as a ∈ Der(A), we have that a(bc) = cab + bac for all b, c ∈ A. Then a = 0. This means that A ∩ Der(A) = {0}. Consider the operator a + δ ∈ A ⊕ Der(A). Then for all b, c ∈ A

[[a + δ, b], c] = [δb − bδ, c]. 2.3 Algebraic preliminaries 19

By Lemma 2.43, we have that there is a unique db ∈ A such that δb − bδ = db for all b ∈ A. As A is a commutative algebra [a, b] = 0 for all a, b ∈ A. We conclude that

[[a + δ, b], c] = [δb − bδ, c] = [db, c] = 0.

Then A ⊕ Der(A) ⊆ D1(A).

Let Q ∈ D1(A). We claim that Q ∈ A ⊕ Der(A). To prove that we write Q = (Q − Q(1)) + Q(1). Let P = Q − Q(1) and q = Q(1). Note that q ∈ A.

We want to prove that P ∈ Der(A). Notice that P (1) = 0. By linearity of P , it follows that P (λ) = 0 for all λ ∈ R. For all a, b ∈ A, we compute

[P, a](b) = P (ab) − aP (b) = Q(ab) − aQ(b) = (Qa − aQ)(b)

Since Q ∈ D1(A), we see that (Qa − aQ) ∈ A. Hence, [P, a] ∈ D0(A) = A for all a ∈ A. By Lemma 2.43, we get that P ∈ Der(A). Therefore D1(A) ⊂ A ⊕ Der(A). We conclude that

D1(A) =∼ A ⊕ Der(A).

2.3.2 Derivations and inﬁnitesimal symmetries

Derivations and infinitesimal symmetries are closely related. In this brief discussion, we will construct infinitesimal symmetries using the algebraic approach of [18, Section 19.20]. Consider a smooth manifold M. We say that a family {ϕt| t ∈ R} is a smooth family of diffeomorphisms if the map ϕ : M × R → M given by ϕ(p, t) = ϕt(p) is smooth. Let {ϕt} be a smooth family of diffeomorphisms of M such that ϕ0 = idM . We would like to compute ϕ − ϕ lim t 0 . t→0 t

The difference ϕt − ϕ0 does not make sense. Nonetheless, note that each ϕt induces a morphism ∗ ∞ ∞ of algebras θt := ϕt : C (M) → C (M). In fact, it can be shown that θt is an automorphism of ∞ the algebra C (M) if and only if ϕt is a diffeomorphism [18, Chapter 7]. As the set of all algebra automorphisms of C∞(M) forms a vector space, it makes sense to define

dθt θh − θ0 Θ := = lim . (2-6) dt t=0 h→0 h

Note that θ0 = idC∞(M), as ϕ0 = idM . The map Θ deﬁned by equation 2-6, is an inﬁnitesimal automorphism of C∞(M). By construction the map Θ is R−linear. We claim that Θ is a derivation of C∞(M). To prove ∞ this, consider two functions f, g ∈ C (M). Since θt is a algebra automorphism, we have that

θt(fg) = θt(f)θt(g). 20 2 Vector Bundles

Taking the derivative with respect to t, we obtain

dθt(fg) dθt(f)θt(g) dθtf dθtg Θ(fg) = = = θ0(g) + θ0(f) = Θ(f)g + fΘ(g). dt t=0 dt t=0 dt t=0 dt t=0 We conclude that Θ is a derivation on C∞(M). Comparing with Example 2.44, we see that infinitesimal automorphisms of C∞(M) are vector fields on the smooth manifold M. To see the inverse correspondence, consider a vector field X. Observe that there is an automorphism ∞ ∗ θt of C (M) given by θt(f) := φt (f) where φt is the flow of X.

2.4 Derivations and inﬁnitesimal automorphisms of vector bundles

Deﬁnition 2.47. Let E,F → M be two vector bundles over M. The set of Diﬀerential operators or order at most k, where k is an integer, is given by

k D (E,F ) = {∆ : Γ(E) → Γ(F )|[[..., [[∆, a0], a1],... ], ak] = 0},

∞ for all a0, a1, . . . , ak ∈ C (M), where ∆ is R-linear and each function ai is thought of as an operator, i.e. multiplication by ai, for i = 1, . . . , k.

∞ Remark 2.48. Note that a ﬁrst order diﬀerential operator ∆ is such that for all a0, a1 ∈ C (M)

∆a0a1 − a0∆a1 − a1∆a0 + a0a1∆ = 0 (2-7) Moreover, ∆ is a local operator. Let u ∈ Γ(E). By equation 2-7, if u(x) = 0 then ∆(u)(x) = 0. It follows that supp ∆u ⊆ supp u.

Definition 2.49. Let E → M be a vector bundle. A first order differential operator ∆ : Γ(E) → Γ(E) such that

∆(fs) = ∂∆(f)s + f∆s, for all f ∈ C∞(M) and s ∈ Γ(E), is a derivation of the vector bundle E. The vector ﬁeld

∂∆ ∈ X(M) is the symbol of the derivation ∆. We denote the set of derivations of E by Der(E).

Remark 2.50. A straightforward computation shows that for a diﬀerential operator ∆, its symbol

∂∆ is unique. 2.4 Derivations and inﬁnitesimal automorphisms of vector bundles 21

Example 2.51. Recall that a connection on a vector bundle [6, Deﬁnition 27.1] (π : E → M) is a map

∇ : X(M) × Γ(E) → Γ(E), (X, s) 7→ ∇X s, (2-8) such that

1. ∇X1+X2 s = ∇X1 s + ∇X2 s

2. ∇X s1 + s2 = ∇X s1 + ∇X s2

3. ∇fX s = f∇X s

4. ∇X (fs) = f∇X s + X(f)s

Note that by condition 4, given a vector ﬁeld X and a connection ∇, there is a derivation ∇X on the module of sections of E. A connection is a way to generalize the exterior derivate.

Deﬁnition 2.52. Let (π : E → M) be a vector bundle. Given a connection ∇ on E, we deﬁne k k+1 the exterior covariant derivative as the operator d∇ :Ω (M; E) → Ω (M; E) given by

k X i ˆ d∇ω(X0,...,Xk) = (−1) ∇Xi (ω(X0,..., Xi,...,Xk)) i=0 X i+j + (−1) ω([Xi,Xj],X0,..., Xˆi,..., Xˆj,...,Xk), i

Proposition 2.53. [6, Proposition 29.1]

The exterior covariant derivative d∇ is the only unique operator such that

1. is R−linear.

k l 2. for ω ∈ Ω (M) and u ∈ Ω (M; E), the operator d∇ satisﬁes the graded Leibniz identity:

k d∇(ω ⊗ u) = d(ω) ⊗ u + (−1) ω ∧ d∇(u),

where d is the deRham diﬀerential.

0 3. For s ∈ Ω (M; E) = Γ(E), (d∇s)(X) = ∇X s.

A direct computation shows that for all sections s ∈ Γ(E) and vector ﬁelds X,Y ∈ X(M)

2 (d∇s)(X,Y ) = R∇(X,Y )s, where R∇ is the curvature of the connection given by

R∇(X,Y )s = ∇X (∇Y s) − ∇Y (∇X s) − ∇[X,Y ]s.

• If the connection is ﬂat (i.e. R∇(X,Y ) = 0 for all X,Y ∈ X(M)), then the pair (Ω (M; E), d∇) • deﬁnes a complex. The cohomology of the complex (Ω (M; E), d∇) is known as the de Rham 22 2 Vector Bundles

cohomology of M with coeﬃcients on E and it is denoted by H•(M; E). Note that the usual de Rham cohomology is the particular case where E = RM . Moreover there is a canonical ﬂat connection ∇ on RM is given by ∇X (f) = X(f), ∞ ∼ for all X ∈ X(M) and for all f ∈ C (M) = Γ(RM ). Proposition 2.54. Let E → M be a vector bundle, then Der(E) forms a Lie algebra under the commutator of operators.

Proof. Let α, β be two derivations of E with symbols δα and δβ, respectively. For s ∈ Γ(E) and f ∈ C∞(M), we compute

[α, β](sf) = α(β(fs)) − β(α(fs))

= δα(δβ(f))s + δβ(f)α(s) + δα(f)β(s) + fα(β(s))

− δβ(δα(f))s − δα(f)β(s) − δβ(f)α(s) − fβ(α(s))

= [δα, δβ](f)s + f[α, β](s).

Therefore [α, β] ∈ Der(E) and its symbol is given by [δα, δβ]. Proposition 2.55. [26, Remark 2.1] Let L → M be a line bundle then every first order differential operator is a derivation. Proof. Let ∆ be a first order differential operator on L. Consider a trivialization on U ⊂ M given ∞ by a frame s ∈ ΓU (L). Then any section u ∈ ΓU (L) can be written as u = fus, where ∈ C (M). By equation 2-7, we have that

∆(a0a1u) − a0∆(a1u) − a1∆(a0u) + a0a1∆(u) = 0,

∞ for all a0, a1 ∈ C (U) and for all u ∈ ΓU (E). Without loss of generality, let u = a0s, then

∆(a1u) = a0∆(a1s) − a0a1∆(s) + a1∆(u). (2-9)

∞ ∞ Using the fact that s is a frame for ΓU (L), we deﬁne the map X : C (M) → C (M), where X(f) is given by ∆(fs) − f∆(s) = X(f)s, for all f ∈ C∞(M). Then equation 2-9 becomes

∆(a1u) = a0(X(a1)s + a1∆(u) = X(a1)u + a1∆(u).

We claim that X is a derivation of C∞(M). To prove this consider f, g ∈ C∞(M), using equation 2-7 we compute

X(fg) = ∆(fgs) − fg∆(s) = f∆(gs) + g∆(fs) − 2fg∆(s) = f(∆(gs) − g∆(s)) + g(∆(fs) − f∆(s)) = fX(g) + gX(f).

It follows that X is indeed a derivation of C∞(U), and so X ∈ X(U). 2.4 Derivations and inﬁnitesimal automorphisms of vector bundles 23

To verify that X ∈ X(M) , we need to show that X does not depend on the trivialization used. To prove this consider a new frame v = as where a ∈ C∞(U) is a nowhere vanishing function. In terms of the new frame v, we compute

Xv(f)v = ∆(fv) − f∆(v) = ∆(fas) − f∆(as) = f∆(as) + a∆(fs) − fa∆(s) − f∆(as) = a∆(fs) − af∆(s)

= aXs(f)s

= Xs(f)v.

It follows that X is a well-deﬁned derivation on C∞(M). Replacing on equation 2-9, we get that

∆(fu) = X(f)u + f∆(u), that is ∆ is a derivation of (π : L → M).

Example 2.56. Consider the trivial line bundle (π : RM → M). Then any derivation ∆ ∈ ∞ Der(RM ) is of the form X + a where X ∈ X(M) and a ∈ C (M). To see this recall that there is a global frames ˜ given bys ˜(x) = (x, 1) for all x ∈ M. Canonically, we identifys ˜ with the smooth function s(x) = 1. For all f ∈ C∞(M) we have that

∆(fs˜) = X(f)˜s + ∆(˜s).

Ass ˜ is a global frame, we deﬁne a by ∆(˜s) = as˜. Therefore

∆(fs˜) = (X(f) + a)˜s.

Then we see that ∆ : C∞(M) → C∞(M) by ∆(f) = X(f) + a.

2.4.1 Infinitesimal Automorphisms Definition 2.57. Let (π : E → M) be a vector bundle (π : E → M) and let X ∈ X(E) be a vector field on the total space E. We say that X is an infinitesimal automorphism of (π : E → M) if

1. there exists some δX ∈ X(M) such that for all v ∈ E

(dvπ)(X(v)) = δX (π(v)),

we say that δX is the symbol of X;

2. for all v ∈ E and for all λ ∈ R

(dvmλ)(X(v)) = X(λv),

where mλ is the induced scalar product map mλ : E → E, v 7→ λv; 3. for all v, w ∈ E such that π(v) = π(w)

(d(v,w)σ)(X,X) = X(σ(v, w)) = X(v + w),

where σ is the addition map given by σ : E × E → E, (u, v) 7→ u + v. 24 2 Vector Bundles

We denote the set of all inﬁnitesimal automorphisms of (π : E → M) by InfAut(π : E → M).

Let M be a smooth manifold. We say that a family of diﬀeomorphisms {ϕt}t∈I is a smooth family of diﬀeomorphisms if the map ϕ : M × I → M, (x, t) 7→ ϕt(x) is smooth.

Definition 2.58. Let (π : E → M) be a vector bundle. We say that {(ψt, Ψt)}t∈I is a smooth family of vector bundle automorphisms, if {ψt}t∈I is a smooth family of diffeomorphisms of M and {Ψt}t∈I is a smooth family of diffeomorphisms of E, such that (ψt, Ψt) is a vector bundle isomorphism for all t ∈ I.

If {(ψt, Ψt)}t∈I is a smooth family of vector bundle automorphisms of (π : E → M). Then for each t ∈ I,Ψt is linear on ﬁbres and the following diagram commutes

E Ψt E π π ψ M t M.

Corollary 2.59. The vector ﬁeld X ∈ X(E) is an inﬁnitesimal automorphism of (π : E → M) with

symbol δX if and only if the family {(ψt, Ψt)}t∈I is a smooth family of vector bundle automorphisms of (π : E → M), were Ψ and ψ denote the ﬂows of X and δX , respectively.

Before proving Corollary 2.59, we recall the following result:

Lemma 2.60. [14, Proposition 9.13] Suppose M and N are smooth manifolds, ϕ : M → N, X ∈ X(M), and Y ∈ X(N). Let θ and η be the ﬂows of X and Y , respectively. If X and Y are ϕ-related, then for each t ∈ I:

ηt ◦ ϕ = ϕ ◦ θt.

Proof of Corollary 2.59. Let X ∈ InfAut(π : E → M). By Lemma 2.60, for each t ∈ I, we have that

π ◦ Ψt = ψt ◦ π (2-10)

mλ ◦ Ψt = Ψt ◦ mλ

σ ◦ (Ψt, Ψt) = Ψt ◦ σ.

For each t ∈ I:

• the ﬁrst equality means that Ψt covers ψt,

• the second and third equality mean that Ψt is linear on ﬁbres.

We conclude that {(ψt, Ψt}t∈I is a smooth family of vector bundle automorphisms of (π : E → M). The converse follows by taking derivatives of equation 2-10 with respect to t at t = 0.

∗ ∗ ∞ ∞ Corresponding to the map ψt, the pullback ψt is an automorphism ψt : C (M) → C (M). The following deﬁnition gives us a notion of pullback when working with sections of E. 2.4 Derivations and inﬁnitesimal automorphisms of vector bundles 25

Definition 2.61. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle and (ψ, Ψ) : (π1 : E1 → M1) → (π2 : E2 → M2) a vector bundle map, such that it is an isomorphism on fibres. For a section u ∈ Γ(E2), we define its pullback via (ψ, Ψ) as

∗ −1 (Ψ u)(x) := (Ψx ◦ u ◦ ψ)(x), (2-11) for all x ∈ M1.

Corollary 2.62. For j = 1, 2, let (πj : Ej → Mj) be a vector bundle and (ψ, Ψ) : (π1 : E1 → M1) → (π2 : E2 → M2) a vector bundle map, such that it is an isomorphism on ﬁbres. Then the ∗ ∗ ∗ pullback Ψ u of a section u ∈ Γ(E) is a smooth section Ψ u ∈ Γ(ϕ E2). ˜ ˜ Proof. Note that Φx = Φx. By Proposition 2.38, there is a vector bundle morphism (idM1 , Φ). As ˜ ˜ (idM1 , Φ) covers the identity, we can apply Lemma 2.14 to show that the map x 7→ Φx = Φx is smooth. For each x ∈ M, the map Φx because Φ is smooth by hypothesis. The section u is smooth and the map ϕ is also smooth by hypothesis. It follows that Ψ∗u is smooth.

∗ Lemma 2.63. The pullback on sections Ψ := Γ(E2) → Γ(E1) is a linear map.

Proof. Let u, v ∈ Γ(E) and a, b ∈ R, then

∗ −1 (Ψ (au + bv))(x) = (Ψx ◦ (au + bv) ◦ ψ)(x) = (Ψ−1 ◦ (au) ◦ ψ)(x) + (Ψ|−1◦(bv) ◦ ψ)(x) x Ex = a(Ψ−1 ◦ u ◦ ψ)(x) + b(Ψ|−1◦v ◦ ψ)(x) x Ex = a(Ψ∗u)(x) + b(Ψ∗v)(x), where the second to last equality follows form the fact that Ψ|−1 is a linear map. Ex Lemma 2.64. Let (π : E → M) be a vector bundle and (ψ, Ψ) : (π : E → M) → (π : E → M) a vector bundle isomorphism. Then the map Ψ∗ : Γ(E) → Γ(E) is a vector space isomorphism.

Proof. By Lemma 2.63, we have that Ψ∗ is a linear map. To prove injectivity consider u, v ∈ Γ(E) such that Ψ∗u = Ψ∗v, then for all x ∈ M:

(Ψ∗u)(x) = (Ψ∗v)(x) −1 −1 (Ψx ◦ u ◦ ψ)(x) = (Ψx ◦ v ◦ ψ)(x).

Given that Ψx is an isomorphism on ﬁbres, and using the fact that ψ is a diﬀeomorphism, we get

(u ◦ ψ)(x) = (v ◦ ψ)(x) u(x) = v(x).

So Ψ∗ is injective. To prove surjectivity not that for any section v ∈ Γ(E) can be written as Ψ∗u where u is given by

u(x) = (Ψ−1| ◦v ◦ ψ−1)(x). Eψ−1(x)

And thus Ψ∗ is a vector space isomorphism. 26 2 Vector Bundles

Proposition 2.65. Let (π : E → M) be a vector bundle and X and inﬁnitesimal automorphism

of (π : E → M). Then there is a unique derivation ∆X ∈ Der(E) given by

d ∗ ∆X (u) = Ψt u , dt t=0

where Ψt is the ﬂow of X. Proof. To prove this claim, we reason in a similar fashion as in Section 2.3.2. Consider a family of ∞ ∞ maps {(ht,Ht)} of (π : E → M), such that ht : C (M) → C (M) is an algebra automorphism, the map Ht = Γ(E) → Γ(E) is a vector space automorphism and (h0,H0) = (idM , idE). For all ∞ sections u ∈ Γ(E) and functions f ∈ C (M), the automormophisms ht and Ht are related as follows:

Ht(fu) = ht(f)Ht(u). (2-12) Taking derivatives we obtain ∆(fu) = δ(f)u + f∆(u), (2-13)

dHt dht(f) where∆ := dt |t=0∈ Der(E), and δ ∈ X(M) is the symbol of ∆ given by δ(f) := dt |t=0∈ X(M), for all f ∈ C∞(M).

Consider an infinitesimal automormophism X ∈ X(E). We denote the flow of X by Ψt. Let ψt : M → M be the unique diffeomorphism covered by Ψt. Note that (ψt, Ψt) is a vector bundle ∗ automorphism. By Lemma 2.64, the pullback map Ψt : Γ(E) → Γ(E) is a vector space isomor- ∗ ∞ ∞ ∗ phism. Moreover, the map ψt : C (M) → C (M) is an algebra isomorphism. By letting Ht = Ψt ∗ and ht = ψt and replacing in equation 2-13, there is a derivation ∆X ∈ Der(E) given by

d ∗ ∆X (u) = Ψt u , dt t=0 ∞ where u ∈ Γ(E). For all f ∈ C (M), the symbol of ∆X is given by ∗ dψt f δX (f) = . dt t=0

To show uniqueness note that the flow Ψt of X is unique. Also there is a unique map ψt covered by Ψt. Therefore the symbol δX is unique and by Remark 2.50, the derivation ∆X is unique. The correspondence shown on Proposition 2.65 appears when studying infinitesimal symmetries of distributions. Example 2.66. Consider a distribution D ⊂ TM. To construct an infinitesimal symmetry of a

distribution, we start by considering a smooth family of diffeomorphisms {ϕt} such that ϕ0 = idM and it preserves the distribution D. Recall that dϕt : TM → TM is uniquely defined by ϕt, then for any vector field Y ∈ X(M) and x ∈ M, we can pull Y back as follows

∗ −1 ((dϕt) Y )(x) = ((dxϕt) ◦ Y ◦ ϕt)(x).

∗ We say that ϕt preserves the distribution if (dϕt) Y ∈ Γ(D) for all Y ∈ Γ(D). As Γ(D) is a vector subspace of X (M) then

∗ d ∗ ((dϕt) Y )(x) − Y (x) ((dϕt) Y )(x) = lim ∈ Γ(D). (2-14) t→0 dt t=0 t 2.4 Derivations and inﬁnitesimal automorphisms of vector bundles 27

The limit on the right-hand side of equation 2-14 is exactly the definition of the Lie derivative of ∗ dϕt f ∞ Y with respect to X ∈ X(M), where X(f) := dt |t=0 for all f ∈ C (M). Then a vector field X is an infinitesimal symmetry of D if for all Y ∈ D

LX Y = [X,Y ] ∈ Γ(D).

By construction {(ϕt, dϕt)}t∈I is a smooth family of vector bundle isomorphisms . Associated to it, there is the derivation of (πD : D ⊂ TM → M) given by the Lie derivative with respect to X, as for all f ∈ C∞(M) and for all Y ∈ Γ(D) we have

LX fY = X(f)Y + fLX Y.

Let (π : E → M) be a vector bundle. Consider X ∈ InfAut(π : E → M), then for all v ∈ E:

dvπX(v) = δX (π(v)).

The previous calculation, induces following short exact sequence of vector bundles over E.

0 ker Dπ TE Dπ π∗TM 0

This means that TE π∗TM =∼ . (2-15) ker Dπ In other words, we have that TE =∼ ker Dπ ⊕ π∗TM. (2-16) A choice of representatives of the equivalence clases of equation 2-15 induces an isomorphism in equation 2-16. Without any additional structure, we do not have a canonical choice of representatives in equation 2-15. Therefore, there is no canonical isomorphism in equation 2-16. Any such choice is known as an Ehresmann connection. We refer the reader to [12, Chapter 8, 9, 17] for a more detailed study of connections.

Lemma 2.67. Let π : E → M be a vector bundle. Then there is a canonical isomorphism

π∗E =∼ ker Dπ.

Proof. Recall that by the deﬁnition of pullback bundle, we have that

∗ π E = {(w1, w2) ∈ E × E : π(w1) = π(w2)}.

∗ Let (w1, w2) ∈ π E, then π(w1) = π(w2). We denote by p := π(w1). Consider trivializing chart (U, φ) with coordinates (p, v1, . . . , vn), such that p ∈ U. An element w ∈ Ep can be written as

w = (p, w1, . . . , wn).

There is a vector space structure on Ep were addition is given by

1 1 n n w1 + w2 = (p, w1 + w2, . . . , w1 + w2 ), for all w1, w2 ∈ Ep, and the scalar multiplication is given by

tw = (p, tw1, . . . , twn), 28 2 Vector Bundles

for all t ∈ R and for all w ∈ Ep.

We can construct a vector Xw2 ∈ Tw1 E as follows. Consider the curve γ : I → Ep given by

γ(t) = w1 + tw2.

Let (∂x1, . . . , ∂xm, ∂v1, . . . , ∂vn) be a base of Tw1 E. Taking derivatives at t = 0

m n X X i Xw2 :=γ ˙ (0) = 0∂xi + w2∂vi. i=1 i=1

It is immediate to see that Xw2 ∈ ker Dw1 π.

Conversely, consider a vector X ∈ ker Dw1 π, then X is of the form

m n X X i X = 0∂xi + (X )∂vi. i=1 i=1

We can construct an element wX ∈ E given by

1 n wX = (p, X ,...,X ).

These constructions only used the vector space structure on ﬁbres, and so they do not depend on

the choice of trivializing chart. Finally, note that the map given by w 7→ Xw is linear with inverse given by the map X 7→ wX . The map given by w 7→ Xw is the desired isomorphism.

Proposition 2.68. A section s : M → E induces a canonical isomorphism

s∗TE =∼ E ⊕ TM, (2-17)

of vector bundles over M.

Proof. Choose a section s ∈ Γ(E), i.e. a smooth map s : M → E such that π ◦ s = IdM . Note that the diﬀerential Ds : TM → s∗TE gives us a way to push vector ﬁelds on M forward to smooth sections of s∗TE. We have the following split short exact sequence of vector bundles over M.

0 s∗ ker Dπ s∗TE Dπ TM 0

Ds By Lemma 2.67, it follows that s∗TE =∼ E ⊕ TM, canonically.

Corollary 2.69. Let ∆ be a derivation of (π : E → M), then there is a unique X∆ ∈ X(E).

Proof. Let u ∈ Γ(E) and x ∈ M. Then ∆(u)(x) ∈ E, also the symbol δ∆(x) ∈ TM. By Proposition 2.68, there is a unique X∆ ∈ TE such that X∆(u(x)) = −∆(u)(x) + (Dxu)(δ∆(x)). Smoothness of X∆ follows from the fact that it is the addition and composition of smooth maps.

Theorem 2.70. Let (π : E → M) be a vector bundle. Then there is a vector space isomorphism

InfAut(E) =∼ Der(E). 2.4 Derivations and inﬁnitesimal automorphisms of vector bundles 29

Proof. Consider the linear maps given by InfAut(E) =∼ Der(E)

d ∗ X 7→ Ψt dt t=0

X∆ ← ∆, [ where X∆(u(x)) = −∆(u)(x) + (Du(x)π)(δ∆(x) for all u ∈ Γ(E). Proposition 2.65 takes care of one side of the correspondence. Given a derivation ∆, by Corollary

2.69 there is a unique X∆. We need to prove that X∆ is indeed an infnitesimal automorphism. For that we need to verify that the ﬂow Ψt of X covers the ﬂow ψt of δ∆. To do so, we compute

X∆(u(x)) = −∆(u)(x) + Dxuδ∆(x)

(Du(x)π)(X∆(u(x))) = −(Du(x)π)(∆(u)(x) + (Dxu)(δ∆(x))

(Du(x)π)(X∆(u(x))) = (Du(x)π)(Dxuδ∆(x))

(Du(x)π)(X∆(u(x))) = δ∆(π(u(x)))

Dπ ◦ X∆ = δ∆ ◦ π. Then at the level of ﬂows we have that

d d (π ◦ Ψt) = (ψt ◦ π). dt t=0 dt t=0

It follows that Ψt covers ψt for all t ∈ I. Then X∆ ∈ InfAut(E). To verify the fact that these constructions are inverse to one another, let (π : E → M) is a vector bundle and X ∈ X(E) is an infinitesimal automorphism of (π : E → M). Then the flow Ψt of X covers the flow ψt of δX . For all u ∈ Γ(E) and x ∈ M, we have that (Ψ∗u)(x) = (Ψ |−1◦u ◦ ψ )(x). t t Ex t Taking derivatives on both sides we get d dϕ (∆ (u))(x) = Ψ |−1 (u(x)) + (D u) t (x) = −X(u(x)) + (D u)δ (x). X t Ex x x X dt t=0 dt t=0

Note that ∆X (u)(x) ∈ ker Du(x)π. To verify this we compute

(Du(x)π)(∆(u)(x)) = (Du(x))(−X(u(x))) + (Du(x))(Dxu)δX (x).

Since Dπ ◦ X = δX ◦ π, we get

(Du(x)π)(∆(u)(x)) = −δX (x) + δX (x) = 0.

Consider the inﬁnitesimal automorphism X∆X associated to ∆X . By Proposition 2.68, we have that d X (u(x)) = −∆ (u) + (D u)δ (x) = Ψ |−1 (u(x)) = X(u(x)). ∆X X x X t Ex dt t=0

We conclude that X = X∆X , where ∆X is the derivation associated to X. Corollary 2.71. Let (π : L → M) be a line bundle, then InfAut(L) =∼ D1(L) (2-18) Proof. This is a direct consequence of Theorem 2.70 and Proposition 2.55. 30 2 Vector Bundles

2.5 Group actions and automorphisms

Deﬁnition 2.72. Let π : E → M be a vector bundle. We deﬁne a vector bundle action of a Lie group G on E as a pair of smooth maps (a, A) given by

A : G × E → E, (g, v) 7→ Ag(v),

a : G × M → M, (g, p) 7→ ag(p)

such that, for all g ∈ G:

1. the map a deﬁnes a smooth group action on M.

2. the map A deﬁnes a smooth group action on E.

3. the map Ag covers ag, i.e. ag ◦ π = π ◦ Ag. (2-19)

4. the map Ag is compatible with scalar product. That is for λ ∈ R,the following identity holds

mλ ◦ Ag = Ag ◦ mλ, (2-20)

where mλ is the induced scalar product map mλ : E → E, v 7→ λv.

5. the map Ag is compatible with vector addition. That is

Ag ◦ σ = σ ◦ (Ag,Ag), (2-21)

where σ is the addition map given by σ : E × E → E, (u, v) 7→ u + v.

Remark 2.73. Since (ag,Ag) is a group action, we have that

• (ae,Ae) = (idM , idE).

• For all g, h ∈ G and using remark 2.10:

(ag,Ag) ◦ (ah,Ah) = (ag ◦ ah,Ag ◦ Ah) = (agh,Agh).

Corollary 2.74. There is an induced action α : G × Γ(E) → Γ(E), given by

α(g, u(x)) := αgu(x) = Ag−1 u(agx), (2-22)

for all x ∈ M, u ∈ Γ(E), and g ∈ G.

Proof. Since (a, A) is a group action, we have that ae = idM and Ae = idE where e ∈ G is the identity element. It follows that αeu(x) = u(x). Let g, h ∈ G, we compute

αh(αgu(x)) = Ah−1 (Ag−1 u(agahx)) = Ah−1g−1 u(aghx) = A(gh)−1 u(aghx) = αgh(u(x)).

Therefore α is a group action. 2.5 Group actions and automorphisms 31

Let (π : E → M) be a vector bundle and let (a, A) be a vector bundle action of a Lie group g. We say that the action is trivial on ﬁbres if for all g ∈ G and x ∈ M the map on ﬁbres

(Ag)x = idEx . Similarly, we say that the action is trivial on the base if for all g ∈ G and for all x ∈ M the map ag(x) = x.

Example 2.75. Let G be a Lie group. By Example ??, the tangent bundle TG =∼ G × g, where g = Lie(G). We can induce a vector bundle action of G on TG =∼ G × g, as follows

Ag(h, ζ) = (agh, adgζ), where agh = gh, adg is the adjoint action of g, for all h ∈ G and all ζ ∈ g. Note that the adjoint action is linear on g. It is immediate to see that Ag covers ag. Hence the pair (a, A) is a vector bundle action.

Example 2.76. Suppose M is a smooth manifold and G a Lie group acting on M via diﬀeomor- phisms with action a : G × M → M. For all g ∈ G, the map ag is a diﬀeomorphism. There is a map dag : TM → TM. There is an induced vector bundle action on TM given by (a, da), where da : G × TM → TM is a smooth action on TM. We call this action (a, da) the lift of a to TM. Similarly there is a lifted action on T ∗M given by (a, (da∗)−1).

2.5.1 Infinitesimal Actions on Vector Bundles Definition 2.77. Let a be a Lie group G action on a manifold M. For ζ ∈ g := Lie(G), the fundamental vector field associated to ζ is the vector field given by

M d Xζ (p) := aexp(−tζ)(p), dt t=0 for all p ∈ M.

Lemma 2.78. Let M,N be smooth manifolds and let aM : G × M → M, aN : G × N → N be smooth actions of a Lie group G on M and N respectively. If a map F : M → N is G−equivariant M N , i.e., F ◦ ag = ag ◦ F for all g ∈ G, then

M N Xζ ∼F Xζ ,

M N where Xζ ,Xζ are the fundamental vector ﬁelds corresponding to the inﬁnitesimal action of ζ ∈ g = Lie(G) on M and N, respectively.

M N Proof. As F is G−equivariant, we have that F ◦ ag = ag ◦ F for all g ∈ G. Let g = exp(−tζ). Then for all x ∈ M M N F (aexp(−tζ)(x)) = aexp(−tζ)(F (x)). Taking derivatives with respect to t at t = 0, we obtain

M N (DxF )(Xζ (x)) = Xζ (F (x)).

The result follows. 32 2 Vector Bundles

Corollary 2.79. Let (a, A) be a vector bundle action of a Lie group G on a vector bundle E → M and ζ ∈ g. Then the following hold

E M Xζ ∼π Xζ E E Xζ ∼mλ Xζ E E E (Xζ ,Xζ ) ∼σ Xζ ,

where:

E M • Xζ , Xζ are the fundamental vector ﬁelds associated to ζ , on E and M respectively

• mλ is the scalar product on ﬁbres by λ ∈ R.

• σ is the vector addition on ﬁbres.

Proof. As (a, A) is a vector bundle action the maps π, mλ and σ are G−equivariant (Deﬁnition 2.72). By Lemma 2.78 the result follows.

Corollary 2.80. The fundamental vector ﬁelds form a Lie subalgebra of (X(E), [·, ·]).

Proof. This result follows from the following fact [14, Prop 8.30]:

Consider a smooth map ϕ : M → N, where M and N are smooth manifolds. Let X1,X2 ∈ X(M) and Y1,Y2 ∈ X(N) such that for j = 1, 2, Xi ∼ϕ Yi. Then [X1,X2] ∼ϕ [Y1,Y2].

Corollary 2.81. Let (π : E → M) be a vector bundle and let (a, A) be a vector bundle action of a Lie group G on (π : E → M). Then there is an induced action on sections S : g × Γ(E) → Γ(E) given by

S(ζ, u) = Sζ (u) := ∆ E u, Xζ

E for all u ∈ Γ(E), where ∆ E is the derivation associated to the inﬁnitesimal generator X . Xζ ζ Moreover for all x ∈ M and u ∈ Γ(E)

d ∆ E (u)(x)k = Aexp(tζ)u(aexp(−tζ)x) Xζ dt t=0

E M E Proof. By Corollary 2.79, we have that Xζ ∼π Xζ . It follows that Xζ is an inﬁnitesimal automorphism of (π : E → M). Then by Theorem 2.70, there is a unique derivation ∆ E . Xζ E ζ Note that the ﬂow of Xζ is given by Ψt (u(x)) = Aexp(tζ)u(x). Additionally, Ψt covers the map ζ ψt (x) = aexp(−tζ)x. Then

d ζ ∗ d ∆ E (u)(x) = ((Ψ ) u)(x) = Aexp(tζ)u(aexp(tζ x) . Xζ t dt t=0 dt t=0 2.5 Group actions and automorphisms 33

2.5.1.1 Trivial Line Bundles

Proposition 2.82. Let (a, A) be a Lie group action on a trivial line bundle RM . Then the following holds

Ag(x, t) = (ag(x),Bg(x)t), (2-23) where Bg(x) := pr2Ag(x, 1), Bg(x) is a nowhere vanishing function on M and pr2 is the projection onto the second component.

Before proving the proposition, we prove an auxiliary lemma

Lemma 2.83. Let (a, A) be a Lie group action on a trivial line bundle RM . We can see Ag as a −1 map Ag(x, t) := (ah(x), bg(x, t)). Then, the following identities hold for all λ ∈ R, t, s ∈ π (x), and g, h ∈ G.

be(x, t) = t

bgh(x, t) = bg(ah(x), bh(x, t))

bg(λt) = λbg(t)

bg(t + s) = bg(t) + bg(s).

Proof. Recall that A is a group action. Therefore Ae(x, t) = (ae(x), be(x, t)) = (x, t). The ﬁrst identity follows. For the second, we have that

Agh(x, t) = Ag◦Ah(x, t) = Ag(ah(x), bh(x, t)) = (ag◦ah(x), bg(ah(x), bh(x, t))) = (agh(x), bgh(x, t)).

Now consider λ ∈ R, we have

Ag(x, λt) = λAg(x, t)

(ag(x), bg(λt)) = (ag(x), λbg(x))

bg(λt) = λbg(t).

Similarly, we compute

Ag(x, t + s) = Ag(x, t) + Ag(x, s)

(ag(x), bg(x, t + s)) = (ag(x), bg(x, t)) + (ag(x), bg(x, s))

(ag(x), bg(x, t + s)) = (ag(x), bg(x, t) + bg(x, s))

bg(x, t + s) = bg(x, t) + bg(x, s). 34 2 Vector Bundles

Proof of Proposition 2.82. Using the previous lemma let us compute the derivative of bg.

∂b b (x, t + h) − b (x, t) g (x, t) = lim g g ∂t h→0 h b (x, t) + b (x, h) − b (x, t) = lim g g g h→0 h b (x, h) = lim g h→0 h hb (x, 1) = lim g h→0 h

= bg(x, 1).

Note that bg(x, t) = Bg(x)t + Cg(x). Since bg(x, 0) = 0, it follows that Cg = 0. We have then that bg(x, t) = Bg(x)t. Using the second expression from the previous lemma we have that

bgh(x, t) = bg(ah(x), bh(x, t))

Bgh(x)t = Bg(ah(x))bh(x, t)

Bgh(x)t = Bg(ah(x))Bh(x)t

Bgh(x) = Bg(ah(x))Bh(x).

Notice that Bg must be invertible. Therefore Bg cannot vanish. 3 Poisson and Symplectic Manifolds

This chapter gives an summary of Poisson geometry. We start by showing the basic deﬁnitions of Poisson manifolds, Poisson maps, both in terms of brackets and of the Poisson bivector. We then introduce Hamiltonian vector ﬁelds and symmetries of Poisson manifolds. Later, we present how these concepts appear in the non-degenerate case, i.e., that of symplectic manifolds. We close the chapter by studying group actions on Poisson manifolds.

3.1 First deﬁnitions and properties

Deﬁnition 3.1. A Poisson algebra is a triple (A, ·, {·, ·}), where (A, ·) is an associative algebra over a ﬁeld F such that char F = 0, and {·, ·} : A × A → A is a bracket such that

1.( A, {·, ·}) is a Lie algebra, i.e. {·, ·} is skew symmetric and for all f, g, h ∈ A, the Jacobi identity: {f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0,

holds.

2. {·, ·} is a derivation of (A, ·) in each entry, i.e. for all f, g, h ∈ A, the Leibniz rule

{f, gh} = {f, g}h + g{f, h},

holds.

A bracket {·, ·} that satisﬁes the previous conditions is a Poisson bracket.

Deﬁnition 3.2. A Poisson Manifold is a pair (P, {·, ·}) where P is a smooth manifold and (C∞(P ), {·, ·}) is a Poisson algebra. A Poisson structure is any bracket {·, ·} : C∞(M)×C∞(M) → C∞(M) such that (C∞(M), {·, ·}) is a Poisson algebra.

Deﬁnition 3.3. For j = 1, 2, let (Pj, {·, ·}j) be a Poisson manifold. A smooth map ϕ : P1 → P2 ∞ such that for all f, g ∈ C (P2)

{f ◦ ϕ, g ◦ ϕ}1 = {f, g}2 ◦ ϕ, is a Poisson map.

Example 3.4. Any manifold M can be endowed with the zero Poisson structure. That is (M, {·, ·}), where {f, g} = 0 for all f, g ∈ C∞(M). 36 3 Poisson and Symplectic Manifolds

2n Example 3.5. On R with linear coordinates (x1, . . . , xn, y1, . . . , yn) there is the canonical Poisson structure, whose bracket is given by

n X ∂f ∂g ∂f ∂g {f, g} = − . ∂y ∂x ∂x ∂y i=1 i i i i

n Example 3.6. Consider R with linear coordinates (x1, . . . , xn) and an n × n matrix C = (cij) ∞ n such that cij = −cji and each cij ∈ C (R ). The bracket given by

n X ∂f ∂g {f, g} = c , ij ∂x ∂x i,j=1 i j

deﬁnes a Poisson bracket if and only if

n X ∂cij ∂cjk ∂cki clk + cli + clj = 0, (i, j, k = 1, . . . , n), (3-1) ∂xl ∂xl ∂xl l=1

as shown on [13, Proposition 1.36].

When cij is a linear function for i, j = 1, . . . , n, we say that the bracket {·, ·} is a linear Poisson structure.

Example 3.7. [7, Example 1.4] Let g be a finite dimensional Lie algebra. An important example of a linear Poisson structure on g∗ is given as follows: ∞ ∗ ∗ ∗ Consider a function f ∈ C (g ) and an element ζ ∈ g . Then the differential dζ f : Tζ g → R corresponds to an element of g∗∗ via the following identification

d hdζ f, vi = f(ζ + tv) , dt t=0

for all v ∈ g∗. ∗∗ ∼ Since g is ﬁnite dimensional we have that g = g. We can then think of dζ f as an element of g. We can deﬁne a bracket on C∞(g∗) as follows:

{f, g}(ζ) := h[dζ f, dζ g]g, ζi, (3-2)

where [·, ·] is the Lie bracket on g. Note that skew symmetry and the Jacobi identity follow from the fact that (g, [·, ·]) is a Lie algebra. The Leibniz rule follows from the fact that we are taking diﬀerentials of functions. The Poisson structure given by (C∞(g∗, {·, ·}), with the bracket given by equation 3-2 is known as the Lie-Poisson structure on g∗.

Lemma 3.8. For j = 1, 2, let gj be a ﬁnite dimensional Lie algebra. If ψ : g1 → g2 is a Lie algebra ∗ ∗ ∗ morphism, the transpose map ψ :(g2, {·, ·}2) → (g1, {·, ·}1), where, for j = 1, 2, the bracket {·, ·}j ∗ is the Lie-Poisson bracket on gj , is a Poisson map. 3.1 First deﬁnitions and properties 37

∞ ∗ ∗ Proof. Let f, g ∈ C (g1). For every ζ ∈ g , we compute

∗ ∗ ∗ ∗ {f ◦ ψ , g ◦ ψ }2(ζ) = h[dζ (f ◦ ψ ), dζ (g ◦ ψ )], ζi

= h[ψdψ∗(ζ)f, ψdψ∗(ζ)g], ζi

= hψ[dψ∗(ζ)f, dψ∗(ζ)g], ζi ∗ = h[dψ∗(ζ)f, dψ∗(ζ)g], ψ (ζ)i ∗ = {f, g}1(ψ (ζ)).

A ﬁnal, but very important result for this work, is the locality of the Poisson bracket. Proposition 3.9. Let (P, {·, ·}) be a Poisson manifold. For any smooth functions f, g ∈ C∞(P ):

supp({f, g}) ⊂ supp(f) ∩ supp(g).

Proof. Take f, g ∈ C∞(P ). As the bracket {·, ·} is antisymmetric, it suﬃces to prove that supp({f, g}) ⊂ supp(f). We argue by contrapositive, i.e., we need to show that x∈ / supp(f) implies that x∈ / supp({f, g}). Fix x ∈ M \supp(f). Consider two open neighbourhoods U, V of x such that V ⊂ U ⊂ M \supp(f).

Let ϕ be a bump function such that ϕ|V := 1 and ϕ|M\U := 0. It follows that (ϕf)(x) = 0 for all x ∈ M. Using the Leibniz rule, we have that

0 = {0, g} = {ϕf, g} = {ϕ, g}f + ϕ{f, g}.

Restricting the previous expression to V

0 = {ϕ, g}f|V +ϕ{f, g}|V = {ϕ, g}|V 0 + {f, g}|V 1 = {f, g}|V .

We conclude that x∈ / supp({f, g}). Therefore supp({f, g}) ⊂ supp(f). The result follows by the antisymmetry of the Poisson bracket {·, ·}.

3.1.1 Multivector fields and the Poisson bivector There is another more practical way to encode the information of a Poisson manifold in terms of tensors. Recall that for a smooth manifold M, vector fields correspond to smooth sections of its tangent bundle TM. Also differential k−forms correspond to smooth sections of ∧kT ∗M. We can make a similar construction to that of differential forms but using the tangent bundle instead of the cotangent one. That is the bundle ∧kTM → M, whose smooth sections we call k-multivector fields and denote by Xk(M). These are the covariant counterpart to differential forms. Proposition 3.10. [7, Proposition 1.16] Let M be a manifold and ν : C∞(M) × · · · × C∞(M) → C∞(M) an alternating, R−multilinear map of degree k, that is a derivation on each entry. Then there is a one to one correspondence with k−multivector fields ν given by

ν(f1, . . . , fk) = ν(df1, . . . , dfk), (3-3)

∞ where f1, . . . , fk ∈ C (M). 38 3 Poisson and Symplectic Manifolds

Corollary 3.11. Let M be a smooth manifold then there is a one to one correspondence between bivector ﬁelds and skew-symmetric biderivations of the R−algebra C∞(M). The bijection is given by {f, g} := π(df, dg), for all f, g ∈ C∞(M), where {·, ·} : C∞(M) × C∞(M) → C∞(M) is the skew-symmetric biderivation associated to the bivector π ∈ X2(M).

For multivector fields there is no analogue of the de Rham operator of differential forms. However, there is an operator that generalises the Lie derivative of vector fields.

Deﬁnition 3.12. Let η ∈ Xk(M) and let ν ∈ Xl(M) be multivector ﬁelds. The Schouten- Nijenhuis bracket is the multivector [η, ν] ∈ Xk+l−1(M) given by

[η, ν] = η ◦ ν − (−1)(k−1)(l−1)η ◦ ν, (3-4)

where

X σ η ◦ ν(α1, . . . , αk+l−1) := (−1) η(d(ν(ασ(1), . . . , ασ(k))), ασ(k+1), . . . , ασ(k+l−1)), σ∈S(k,l−1)

1 S(k, l − 1) are the (k, l − 1) shuffles, and αi ∈ Ω (M) for i = 1, . . . , k + l − 1. A(k, l)−shuffle is a permutation σ ∈ Sk+l, where Sk+l is the symmetric group of k + l integers, such that σ is increasing on the first k integers and on the last l integers:

σ(1) < σ(2) < ··· < σ(k), σ(k + 1) < σ(k + 2) < ··· < σ(k + l).

The set of all (k, l)−shuﬄes is denoted by S(k, l).

Remark 3.13. The Schouten-Nijenhuis bracket generalises the following operations:

• The Lie bracket of vector ﬁelds. To see this note that for all X,Y ∈ X(M), the Schouten- Nijenhuis bracket is given by

[X,Y ] = X ◦ Y − (−1)(1−1)(1−1)Y ◦ X = X ◦ Y − Y ◦ X.

• The Lie derivative of multivector ﬁelds. It can by shown by a direct computation that

[X,Y ] = LX Y,

for all X ∈ X(M) and for all Y ∈ Xl(M).

Proposition 3.14. [13, Proposition 3.7]. • k Let M be a manifold and let X (M) = ⊕kX (M) its multivector ﬁelds. The Schouten-Nijenhuis bracket deﬁnes a shifted graded Lie algebra structure on X•(M). That is, for P ∈ Xp(M), Q ∈ Xq(M), and R ∈ Xr(M):

• the bracket is shifted graded, i.e. [P,Q] ∈ Xp+q−1(M)

• the bracket is graded skew symmetric, i.e. [P,Q] = −(−1)(p−1)(q−1)[Q, P ] 3.1 First deﬁnitions and properties 39

• the graded Jacobi identity holds:

(−1)(p−1)(r−1)[P, [Q, R]] + (−1)(q−1)(p−1)[Q, [R,P ]] + (−1)(r−1)(q−1)[R, [P,Q]] = 0

• the graded Leibniz rule holds:

[P,Q ∧ R] = [P,Q] ∧ R + (−1)(p−1)qQ ∧ [P,R]

There are several differences between multivector fields and differential forms. We refer the inter- ested reader to [7, Chapter 1], for a Poisson geometric perspective, and [13, Chapter 3], for a more algebraic approach. Let {·, ·} : C∞(P ) × C∞(P ) → C∞(P ) be a bilinear, antisymmetric bracket on the algebra of smooth functions. We define the Jacobiator J(f1, f2, f3) as

J(f1, f2, f3) = {f1, {f2, f3}} + {f2, {f3, f1}} + {f3, {f1, f2}}. (3-5)

Lemma 3.15. For f, g ∈ C∞(M), let {f, g} = π(df, dg), where π ∈ X2(M). Then the Jacobiator

Jπ of {·, ·} satisﬁes 1 J = − [π, π]. π 2

∞ Proof. For j = 1, 2, 3, let fj ∈ C (M). By equation 3-4, we have that

[π, π](df1, df2, df3) = 2(π ◦ π)(df1, df2, df3)

= 2 (π(d(π(df1, df2)), df3) + π(d(π(df2, df3)), df1) + π(d(π(df3, df1)), df2)) .

Using the deﬁnition of the bracket

[π, π](df1, df2, df3) = −2 ({f1, {f2, f3}} + {f2, {f3, f1}} + {f3, {f1, f2}})

The result follows.

Corollary 3.16. Let (P, {·, ·}) be a Poisson manifold. Then there is a unique bivector π ∈ X2(P ) such that [π, π] = 0 and {f, g} = π(df, dg), for all f, g ∈ C∞(P ).

Proof. This is a direct consequence of Proposition 3.10, and Lemma 3.15.

In other words, Corollary 3.16 tells us that a Poisson manifold can be given as a pair (P, π) where π is a bivector self vanishing under the bracket of multivector ﬁelds. In terms of bivectors Poisson maps are described by the following result

Proposition 3.17. For j = 1, 2, let (Pj, πj) be a Poisson manifold and ϕ :(P1, {·, ·}1) → (P2, {·, ·}2) be a Poisson map. Then ϕ is a Poisson map if and only if π1 ∼ϕ π2. 40 3 Poisson and Symplectic Manifolds

∞ Proof. Using the deﬁnition of Poisson map, we compute for f, g ∈ C (P2)

{f, g}2 ◦ ϕ(x) = {f ◦ ϕ, g ◦ ϕ}1(x) ∗ ∗ π2,ϕ(x)(dϕ(x)f, dϕ(x)g) = π1,x(ϕ dxf, ϕ dxg)

π2,ϕ(x)(dϕ(x)f, dϕ(x)g) = (dϕπ1)x(dϕ(x)f, dϕ(x)g).

Remark 3.18. Since composition of Poisson maps is well defined, we can define the Poisson category Poisson, whose • objects are Poisson manifolds. • morphisms are Poisson maps. Similarly we can define a subcategory of Poisson structures over the same manifold M denoted by

PoissonM , whose • objects are Poisson manifolds (M, π) with same base manifold. • morphisms are Poisson maps. ] ∗ We say that a Poisson manifold (P, π) is non-degenerate at x ∈ P if π : Tx P → TxP is an isomorphism. A non-degenerate Poisson manifold is one that is non-degenerate for all x ∈ P .

3.1.2 Hamiltonian vector fields Consider a Poisson manifold (P, {·, ·}). An immediate consequence of the Leibniz rule for the Poisson bracket and the fact that every derivation is a vector field (Example 2.44) is the fact that we can assign a vector field to a function H as follows

H 7→ XH := {H, ·}.

We say that XH is the Hamiltonian vector ﬁeld associated to H, the Hamiltonian function. In terms of the Poisson bivector, this assignment corresponds to the map H 7→ π](dH).

We denote by XHam(P, {·, ·}) the set of all Hamiltonian vector ﬁelds.

Proposition 3.19. Let (P, {·, ·}) be a Poisson manifold. Then XHam(P, {·, ·}) is a Lie subalgebra of X(P ). ∞ ∞ Proof. Let f, g ∈ C (P ) and Xf ,Xg their respective Hamiltonian vector ﬁelds. For h ∈ C (P ), we compute

X{f,g}(h) = {{f, g}, h} = {{f, h}, g} + {f, {g, h}} = {f, {g, h}} − {g, {f, h}}

= Xf (Xg(h)) − Xg(Xf (h))

= [Xf ,Xg](h).

This proves that XHam(P, {·, ·}) is closed under the Lie bracket. 3.1 First deﬁnitions and properties 41

3.1.3 Poisson symmetries Deﬁnition 3.20. Let (P, π) be a Poisson manifold. Then a diﬀeomorphism ϕ : P → P such that

ϕ∗π = π is Poisson automorphism.

Note that since ϕ is a diffeomorphism, pushforwards are well defined. Next, we define its infinitesimal counterpart.

Definition 3.21. Let (P, π) be a Poisson manifold. A vector field X such that LX π = 0 is an infinitesimal Poisson automorphism. We denote by XP oisson(P, π) the set of all infinitesimal Poisson automorphisms.

Proposition 3.22. Let (P, π) be a Poisson manifold. A vector ﬁeld X ∈ X(P ) is an inﬁnitesimal Poisson automorphism if and only if X is a derivation on the Poisson algebra (C∞(P ), {·, ·}).

Proof. Let f, g ∈ C∞(P ). We compute

(LX π)(df, dg) = X(π(df, dg)) − π(LX df, dg) − π(df, LX dg),

0 = X{f, g} − π(d(ιX df), dg) − π(df, d(ιX dg)), X{f, g} = {X(f), g} + {f, X(g)}.

Using Proposition 3.22, it is easy to show that XP oisson(P, π) is a Lie subalgebra of X(P ). Let ∞ X,Y ∈ XP oisson(P, π) and f, g ∈ C (P ). We have

[X,Y ]{f, g} = X(Y ({f, g})) − Y (X({f, g})) = X({Y (f), g} + {f, Y (g)}) − Y ({X(f), g} + {f, X(g)}) = {X(Y (f)), g} + {Y (f),X(g)} + {X(f),Y (g)} + {f, X(Y (g))} − {Y (X(f)), g} − {X(f),Y (g)} − {Y (f),X(g)} − {f, Y (X(g))} = {X(Y (f)), g} − {Y (X(f)), g} + {f, X(Y (g))} − {f, Y (X(g))} = {[X,Y ]f, g} + {f, [X,Y ]g}.

Note that by the Jacobi identity XHam(P, π) ⊂ XP oisson(P, π).

Lemma 3.23. For a Poisson manifold (P, π) the Hamiltonian vector ﬁelds XHam(P, π) form an ideal of the Lie algebra of Poisson vector ﬁelds XP oisson(P, π).

∞ Proof. Let XH ∈ XHam(P, π) and X ∈ XP oisson(P, π). For f ∈ C (P ), we have

[XH ,X](f) = XH (X(f)) − X(XH (f)) = {H,X(f)} − X({H, f}) = {H,X(f)} − ({X(H), f} + {H,X(f)}) = {−X(H), f}.

So [XH ,X] is a Hamiltonian vector ﬁeld with Hamiltonian function −X(H). 42 3 Poisson and Symplectic Manifolds

3.2 Nondegenerate Poisson manifolds

] ∗ Recall that a Poisson manifold (P, π) is non-degenerate if for all x ∈ P the map πx : Tx P → TxP is an isomorphism. This induces an antisymmetric bilinear form

∗ ∗ π|x: Tx P × Tx P → R.

It can be proven that this form is non-degenerate only if the dimension of TxP is even. We get that only an even dimensional Poisson manifold can be non-degenerate. Note that there is an induced map π] :Ω1(P ) → X(P ) given by

] ] (π (η))(x) := πx(η(x)),

for all x ∈ P and η ∈ Ω1(P ).

Similarly, given a ω ∈ Ω2(M), we say that ω is non-degenerate if the map

[ ∗ ωx : TxM → Tx M,X(x) 7→ ιX(x)ωx,

is an isomorphism for all x ∈ M. There is an induced map ω[ : X(M) → Ω1(M) given by

[ [ (ω (X))(x) := ωx(X(x)),

for all x ∈ M and X ∈ X(M). We see that for a non-degenerate Poisson manifold (P, π) there is a unique 2-form ω such that π] = (ω[)−1. Smoothness of ω follows from the fact that π is smooth and the maps π] and ω[ are inverse to each other.

Lemma 3.24. Let (P, π) be a non-degenerate Poisson manifold then [π, π] = 0 if and only if its associated 2-form ω is closed.

Proof. First note that the Poisson bracket can be computed in terms of ω, for all f, g ∈ C∞(P ) as follows ] ] {f, g} = π(df, dg) = ω(π (df), π (dg)) = ω(Xf ,Xg), (3-6)

where Xf ,Xg are the Hamiltonian vector ﬁelds associated to f and g respectively. Let X,Y,Z ∈ X(P ). We compute

dω(X,Y,Z) = X(ω(Y,Z)) − Y (ω(Z,X)) + Z(ω(X,Y )) − ω([X,Y ],Z) − ω([Y,Z],X) − ω([Z,X],Y ).

Letting X = Xf , Y = Xg, Z = Xh, the Hamiltonian vector ﬁelds associated to f, g, h respectively, and using equation 3-6, we get

dω(Xf ,Xg,Xh) = {f, {g, h}} + {g, {h, f}} + {h, {f, g}} − {{f, g}, h} − {{g, h}, f} − {{h, f}, g} = 2 ({f, {g, h}} + {g, {h, f}} + {h, {f, g}}) . 3.2 Nondegenerate Poisson manifolds 43

Using Lemma 3.15, we get that

dω(Xf ,Xg,Xh) = [π, π](df, dg, dh), for all f, g, h ∈ C∞(P ). As T ∗M is C∞(P )−generated by closed forms, the result follows.

A symplectic manifold (M, ω) is a manifold M together with a closed (dω = 0) non-degenerate 2-form ω. The previous discussion allows us to conclude that

Proposition 3.25. There is a one to one correspondence between symplectic manifolds and non- degerate Poisson manifolds.

Proof. By the previous discussion, associated to a non degenerate bivector π there is a non- degenerate 2−form ω. Such that (π])−1 = ω[. By Lemma 3.24, π is a Poisson bivector if and only if ω is closed. The result follows.

2n Example 3.26. If we take R with coordinates (x1, . . . , xn, y1, . . . , yn), then the form

n X ω0 = dxi ∧ dyi i=1

2n is a symplectic form and (R , ω0) is a symplectic manifold.

n Example 3.27. Now consider M = C with linear coordinates (z1, . . . , zn), then the form

n i X ω = dz ∧ dz¯ 0 2 i i i=1 is a symplectic form.

Example 3.28. Another important example is the cotangent bundle T ∗M of any manifold M, with coordinates (x1, . . . , xm, ξ1, . . . , ξm). Observe that this manifold is always even dimensional of dimension dim T ∗M = 2 dim M. We can deﬁne the form

m X ω = dxi ∧ dξi i=1

It can be proven that this deﬁnition is independent to the coordinates [21, Section 2] so it deﬁnes a global symplectic 2-form.

Deﬁnition 3.29. A diﬀeomorphism ϕ :(M1, ω1) → (M2, ω2) between symplectic manifolds such ∗ that ϕ ω2 = ω1 is a symplectomorphism.

Example 3.30. It is easy to see that the ﬁrst two examples are symplectomorphic. The symplectomorphism is given by

2n n ϕ : R 7→ C ;(xi, yi) 7→ xi + iyi = zi.

Note that this deﬁnition of symplectomorphism is not the same as that of a Poisson map. 44 3 Poisson and Symplectic Manifolds

4 Example 3.31. Let (x1, y1, x2, y2) be Cartesian coordinates for R and let (x1, y1) for be Cartesian coordinates for R2. The canonical symplectic form for each manifold is

4 2 ω0 = dx1 ∧ dy1 + dx2 ∧ dy2, and ω0 = dx1 ∧ dy1.

The corresponding Poisson bivectors are

4 2 π0 = ∂y1 ∧ ∂x1 + ∂y2 ∧ ∂x2, and π0 = ∂y1 ∧ ∂x1.

4 2 Note that the projection pr : R → R , (x1, y1, x2, y2) 7→ (x1, y1) is not a symplectomorphism, since it is not a diﬀeomorphism. However, the map pr is a Poisson map since

4 2 pr∗(π0) = π0.

3.2.1 Hamiltonian vector ﬁelds

Let (M, ω) be a symplectic manifold and consider a smooth function H : M → R. Then, its diﬀerential dH is a 1-form. Because ω is non-degenerate, there exists a unique vector ﬁeld XH

such that ιXH ω = dH. Associated to the vector field XH , there is a flow ϕt : M → M for t ∈ R, assuming that XH is complete. This flow is in fact a family of diffeomorphisms generated by XH , such that for all t such that d ϕ∗ω = ϕ∗L ω = ϕ∗(dι ω + ι dω) = ϕ∗(ddH + ι dω) = 0 dt t t XH t XH XH t XH

∗ for all t, where dω = 0 due to the closedness of the two form. As ϕ0ω = ω0, it follows that ϕt ∗ preserves the symplectic form ω, i.e., ϕt ω = ω for all t.

We call the function H the Hamiltonian function and the vector field XH its Hamiltonian vector field. We denote by XHam(M, ω) the set of all Hamiltonian vector fields of (M, ω). Vector fields that preserve the symplectic form are known as symplectic vector fields, denoted by

XSymp(M, ω).

Lemma 3.32. Let (M, ω) be a sympletic manifold. Then X ∈ XSymp(M, ω) if and only if the form ιX ω is closed.

∗ Proof. Denote the ﬂow of X by ϕt . As X ∈ XSymp(M, ω), we have that ϕt ω = ω. It follows that d ϕ∗ω = ϕ∗L ω = ϕ∗(dι ω) = 0, dt t t X t X

for all t. We see that the form ιX ω is closed. As this argument works in the opposite direction, the result follows.

The diﬀerence between a symplectic and Hamiltonian vector ﬁeld X is that ιX ω is closed for X symplectic and exact for X Hamiltonian.

1 Remark 3.33. Let (M, ω) a symplectic manifold. If HdR(M) = {0}, then every symplectic vector ﬁeld is Hamiltonian. 3.3 Poisson actions and Hamiltonian Poisson actions 45

Proposition 3.34. Let (M, ω) be a symplectic manifold with associated Poisson manifold (M, π = ω−1).

• A vector ﬁeld X is symplectic if and only if it is Poisson.

• A vector ﬁeld is Hamiltonian for (M, ω) if and only if it is Hamiltonian for (M, π).

Proof. To prove the ﬁrst claim, consider X ∈ XSymp(M) a symplectic vector ﬁeld, then

LX ω = 0 −1 LX π = 0

LX π = 0. Since this argument works in both directions, the result follows. ] ∞ For the second statement, let X ∈ XP oisson(M, π) then X = π (dfX ) for some fX ∈ C (M). We compute −1 ] ιX ω = ιπ (fX )π = dfX .

So X ∈ XSymp(M, ω). We can follow this reasoning on the opposite direction to obtain the converse statement.

3.3 Poisson actions and Hamiltonian Poisson actions

Deﬁnition 3.35. Let (P, π) be a Poisson manifold and let G be a Lie group. A Poisson group action is a smooth action a : G × P → P such that for all g ∈ G, the map ag : P → P is a Poisson automorphism. Associated to the action of G on (P, π) there is the inﬁnitesimal action φ : g → X(P ) given by the map ζ 7→ Xζ , where

d Xζ (x) = exp(−tζ) · x . dt t=0 This map is a Lie algebra homomorphism.

Proposition 3.36. Let (P, π) be a Poisson manifold and G y (P, π) a Poisson action. Then the infinitesimal generators of the action are Poisson vector fields. Conversely, given an action of G on (P, π) (not necessarily Poisson), if G is connected and all infinitesimal generators of the action are Poisson vector fields, then the action is Poisson.

Proof. Let ζ ∈ g. The fundamental vector field Xζ has flow φt(p) := φ(t, p) = exp(−tζ) · p. As the action of G is Poisson, the flow φt(p) is a Poisson map. By definition of the Lie derivative we have that

d d LXζ π = ((φ−t)∗)(πφt(p)) = πp = 0. dt t=0 dt t=0

To prove the second statement, consider G a connected group and Xζ ∈ XP oisson(P, π). As G is connected, it is determined by a neighbourhood U of the identity e [6, Proposition 14.1]. This means that given a neighbourhood U of e then

n G = ∪n∈NU , 46 3 Poisson and Symplectic Manifolds

n where U = {g1 ··· gn|gi ∈ U, i = 1, . . . , n}. As a result, it suﬃces to prove that ag is a Poisson map for all g ∈ U, where U is a neighbourhood of e, in order to verify that G y (P, π) is Poisson. As the exponential map is a diﬀeomorphism in a neighbourhood U of the identity, there is a gt ∈ U

such that exp(−tζ) = gt. Given that LXζ π = 0, it follows that

d ∗ aexp(−tζ)π = 0. dt t=0 ∗ ∗ ∗ ∗ Then aexp(−tζ)π is constant. As aexp(−0ζ)π = idP π = π, we get that aexp(−tζ)π = π for all t ∈ I and ζ ∈ g. So the ﬂow φt(p) = exp(−tζ) · p = gt · p is a Poisson map for all g ∈ G. We conclude that the action G y (P, {·, ·}) is Poisson. Deﬁnition 3.37. Let (P, π) be a Poisson manifold and let g be a Lie algebra acting on P . We say that the action of g is weakly Hamiltonian if for all ζ ∈ g:

Xζ ∈ XHam(P, π).

If the action of Lie(G) = g is weakly Hamiltonian, then it is a Poisson action, since XHam(P, π) ⊂ XP oisson(P, π). By Proposition 3.36, if G is connected, it follows that the action of G is Poisson. A direct consequence of the Jacobi identity is the following: Corollary 3.38. Let (P, π) be a Poisson manifold, then the operator

n n n+1 δπ := [π, ·]: X (P ) → X (P ), n+1 n is homological (i.e., δπ ◦ δπ = 0 for all n) and induces a complex

δn−1 δn δn+1 ... Xn−1(P ) π Xn(P ) π Xn+1(P ) π ... called the Poisson cohomology complex of P . The n−th Poisson cohomology space of (P, π) corresponds to n n n−1 Hπ (P ) := Ker δπ /Im δπ . Proof. This is a direct consequence of Proposition 3.14. Consider Q ∈ Xq(P ), then

δπ ◦ δπ(Q) = [π, [π, Q]]. By the graded Jacobi identity, and the fact that [π, π] = 0 (−1)q−1[π, [π, Q]] − [π, [Q, π]] + (−1)q−1[Q, [π, π]] = 0 (−1)q−1[π, [π, Q]] − [π, [Q, π]] = 0 The graded skew symmetry of the Schouten-Nijenhuis bracket tells us that [π, Q] = −(−1)q−1[Q, π]. Then −[π, [Q, π]] = (−1)q−1[π, [π, Q]]. Replacing on the Jacobi identity, we get 2(−1)q−1[π, [π, Q]] = 0. The result follows. 3.3 Poisson actions and Hamiltonian Poisson actions 47

Proposition 3.39. [13, Proposition 4.5]. Let (P, π) be a Poisson manifold. The map π] : T ∗M → TM, induces a chain map ∧•π] : • • (Ω (M), d) → (X (M), δπ). In other words, the following diagram commutes

Ωq(M) d Ωq+1(M)

∧q π] ∧q+1π] Xq(M) δπ Xq+1(M) for all q ∈ N. q q There is an induced map HdR(M) → Hπ(M). Moreover if π is non-degenerate the induced map is an isomorphism.

] ∗ • ] Proof. Note that (idM , π ): T M → TM is a vector bundle morphism. Hence, the pair (idM , ∧ π ): ∧•T ∗M → ∧•TM is a vector bundle map. It follows that there is a well deﬁned map ∧•π] : Ω•(M) → X•(M) at the level of sections. We need to check that ∧•π] induces a morphism of complexes. To prove so, note that Ωn(M) is generated by elements of the form α = A(x1, . . . , xn)df1 ∧ · · · ∧ dfn. Then, it is suﬃcient to make the computation using α. On one hand, we get

q+1 ] q+1 ] ∧ π (dα) = ∧ π (dA ∧ df1 ∧ · · · ∧ dfq) = XA ∧ Xf1 ∧ · · · ∧ Xfq .

On the other hand,

• ] dπ(∧ π (α)) = δπ(AXf1 ∧ · · · ∧ Xfq )

= [π, AXf1 ∧ · · · ∧ Xfq ]

= [π, A] ∧ Xf1 ∧ · · · ∧ Xfq

= XA ∧ Xf1 ∧ · · · ∧ Xfq ,

∞ where the third equality follows from the fact that [π, Xf ] = 0 for all f ∈ C (M). We conclude that ∧•π] is indeed a morphism of complexes. Finally since (π])−1 = ω[, using similar reasoning we get that ∧•ω[ induces an inverse morphism of chains. We conclude that for non-degenerate Poisson manifolds, the map ∧•π] :Ω•(M) → X•(M) induces an isomorphism of chains.

In particular, we will focus on the 0−th and 1−st cohomology spaces. Recall that X0(P ) = C∞(P ). 1 0 Similarly, X (P ) = X(P ). Note that Hπ(P ) corresponds to all the functions f such that {f, h} = 0 for all h ∈ C∞(P ). These functions are called Casimir functions [22] . We can also see that 1 XP oisson(P, π) Hπ(P ) = . (3-7) XHam(P, π)

1 Remark 3.40. One can immediately see that if Hπ(P ) = 0 then XHam(P, π) = XP oisson(P, π). In this case, any Poisson action g y (P, π) is weakly Hamiltonian. 1 ∼ 1 In the case where π is non-degenerate we have that Hπ(P ) = HdR(P ). Hence, Remark 3.33 is recovered through th language of Poisson geometry. 48 3 Poisson and Symplectic Manifolds

Lemma 3.41. Let (P, π) be a Poisson manifold and let g y (P, π) be a weakly Hamiltonian action. Then there exists a smooth map µ : M → g∗ such that for all ζ ∈ g

Xζ = {hµ, ζi, ·}.

The map µ is known as a moment map. Associated to µ, there is a map µ : g → C∞(P ) given by µ(ζ) = µζ = hµ, ζi. The map µ is known as the comoment map associated to µ.

Proof. Since g y (P, π) is weakly Hamiltonian, then for all ζ ∈ g, the infinitesimal generator Xζ ∈ XHam(P, π) is Hamiltonian. By definition of Hamiltonian vector field, there is a function ∞ fζ ∈ C (P ) such that ] Xζ = π (dfζ ).

ζ ∗ ζ By ﬁxing x ∈ P , there is an element µx ∈ g such that hµx, ζi = fζ (x). Since fζ is smooth, the ∗ ζ map µ : M → g given by hµ(x), ζi = µx is also smooth. The result follows.

Corollary 3.42. An action g y (P, π) is weakly Hamiltonian if and only if there exists a moment map.

Proof. The result is direct as hµ, ζi is the Hamiltonian function associated to Xζ .

Deﬁnition 3.43. Let (P, π) be a Poisson manifold, an action g y (P, π) is Hamiltonian if there exists a smooth map µ : P → g∗ such that

] ζ 1. for all ζ ∈ g, Xζ = π (dµ ), i.e. g y (P, π) is weakly Hamiltonian. 2. for all ζ, η ∈ g, the map µ :(g, [·, ·]) → (P, {·, ·}) is a Lie algebra morphism, i.e. µ([ζ, η]) = {µ(ζ), µ(η)}, for all ζ, η ∈ g.

Proposition 3.44. Let (P, π) be a Poisson manifold. A map µ : g → C∞(P ) is a Lie algebra homomorphism if and only if µ : M → g∗ is a Poisson map, where we are considering the Lie- Poisson structure on g∗.

Proof. Let µ : M → g∗ be a smooth map. We can identify ζ ∈ g with the linear map ζ : g∗ → R. Then for ζ, η ∈ g and x ∈ P , using the deﬁnition of the Lie-Poisson structure, we get

{µ(ζ), µ(η)}(x) − µ([ζ, η])(x) = {ζ ◦ µ, η ◦ µ}P (x) − hµ(x), [ζ, η]i

= {ζ ◦ µ, η ◦ µ}P (x) − {ζ, η}g∗ (µ(x)).

Therefore, µ is a Lie algebra morphism if and only if

{ζ ◦ µ, η ◦ µ}P = {ζ, η}g∗ ◦ µ.

But the last equation is the deﬁnition of Poisson map, as desired.

Corollary 3.45. Let (P, π) be a Poisson manifold. A weakly Hamiltonian action g y (P, π) is Hamiltonian if and only if µ : M → g∗ is a Poisson map, with the Lie-Poisson structure on g∗.

The following example illustrates how a weakly Hamiltonian action need not be Hamiltonian. 3.3 Poisson actions and Hamiltonian Poisson actions 49

Example 3.46. [20]. 2 2 Let g = R with the trivial bracket. Denote (R , ω0) by (M, ω), where x ∈ M has coordinates x = (q, p). Recall that

ω0 = dq ∧ dp. Consider the action R2 y (M, ω) given by

2 2 2 a : R × R → TM = T R

((u, v), x) 7→ u∂q(x) + v∂p(x).

This action is weakly Hamiltonian. To see this notice that for (u, v) ∈ R2 the inﬁnitesimal generator is given by

X(u,v) = u∂q + v∂p. We contract with the symplectic form to get

ιX(u,v) ω = udp − vdq We see that udp − vdq = d(up − vq + c)

2 where c is a constant. Therefore X(u,v) is a Hamiltonian vector ﬁeld for all (u, v) ∈ R . Hence the action is weakly Hamiltonian. Note that the only constant that makes the function f(u, v) = up(x) − vq(x) + c a linear map is c = 0. The moment map is given by

2 ∗ µ : M → (R ) x 7→ (p(x), −q(x)).

We see that the comoment map µ is given by

µ(u, v)(x) = (up − vq)(x) = up(x) − vq(x), for all x ∈ M. To prove that this action is not Hamiltonian, we need to verify that µ is not a Lie algebra morphism. It suﬃces to check on the generators (1, 0) and (0, 1) of R2. We compute µ(1, 0)(x) = p(x), µ(0, 1)(x) = −q(x), for all x ∈ M. On one hand, we get that [(1, 0), (0, 1)] = 0. On the other hand, we get that

{p, q} = ω0(∂q, ∂p) = 1. It follows that µ is not a Lie algebra homomorphism. We conclude that this action is not Hamil- tonian.

Deﬁnition 3.47. Let (P, π) be a Poisson manifold and G a Lie group with Lie algebra g acting on P . We say that the action of G is Hamiltonian if 50 3 Poisson and Symplectic Manifolds

• the action G y (P, π) is weakly Hamiltonian, that is 1. the action G y (P, π) is Poisson. 2. the action of g = Lie(G) y (P, π) is weakly Hamiltonian.

• there exists a G− equivariant moment map µ : M → g∗ for the action g y (P, π). That is for all g ∈ G ∗ µ(g · x) = Adgµ(x). (3-8)

Proposition 3.48. [7, Proposition 4.8] . Let (P, π) be a Poisson manifold and let G y (P, π) a weakly Hamiltonian action. Then 1. If µ is G-equivariant, then µ is a Poisson map.

2. If µ is a Poisson map and G is connected then µ is G-equivariant.

Proof. Suppose µ : M → g∗ is G−equivariant. Then

∗ µ(g · x) = Adgµ(x).

Let gt = exp(−tζ) and η. We compute

∗ hµ(exp(−tζ) · x), ηi = hAdexp(−tζ)µ(x), ηi = hµ(exp(−tζ) · x), ηi = hµ(x), Adexp(tζ)ηi

Diﬀerentiating both sides with respect to t and evaluating at t = 0, we get that

hdxµXζ , ηi = hµ(x), [ζ, η]i.

Using the moment condition and the deﬁnition of the Lie-Poisson structure we get

] ζ hdxµ · π (dµ ), ηi = {ζ, η}g∗ (µ(x)) ] ζ ∗ hπ (dµ ), dx · µηi = {ζ, η}g∗ (µ(x))

{ζ ◦ µ, η ◦ µ}P (x) = {ζ, η}g∗ (µ(x)),

where ζ ◦ µ(x) = hµ(x), ζi. This means that µ is a Poisson map. To prove the converse assume that G is connected. Using the previous reasoning in the opposite direction, we get that µ is a Poisson map if

∗ µ(exp(−tζ) · x) = Adexp(−tζ)µ(x),

for all ζ ∈ g and x ∈ P . Since the exponential map is a local diﬀeomorphism, then there is a neighbourhood of the identity Usuch that

∗ µ(g · x) = Adgµ(x),

n for all g ∈ U. By connectedness of G, G = ∪n∈NU . So it holds for all g ∈ G.

Corollary 3.49. Let (P, π) be a Poisson manifold and let G y (P, π) be a weakly Hamiltonian action. Then 3.3 Poisson actions and Hamiltonian Poisson actions 51

1. If G y (P, π) is Hamiltonian, then g y (P, π) is Hamiltonian.

2. If g y (P, π) is Hamiltonian and G is connected, then G y (P, π) is Hamiltonian. Proof. This is a direct consequence of Proposition 3.48 and Corollary 3.45.

In classical mechanics, Noether’s principle states that associated to every symmetry in a mechan- ical system, there is a conserved quantity. The concept of a moment map associated to a group action (a symmetry) provides a way to formalize Noether’s principle, as well as a generalisation of the Hamiltonian function [21].

The name moment map comes from the fact that the function µ is a generalisation of the angular and linear momenta of classical mechanics, as we see in the following example.

Example 3.50. [21, Section 22.4] Consider the action of SO(3) on M = R3 by rotations. This action lifts to a symplectic action on the cotangent bundle T ∗M =∼ R6 (Example 2.76) . Let x ∈ M and (x, y) ∈ T ∗MR6.

Recall that so(3) is isomorphic to (R3, ×), where × is the cross product via the following isomorphism   0 −v3 v2 3 Φ: R → so(3), Φ((v1, v2, v3)) =  v3 0 −v1 . −v2 v1 0 Note that Φ(v)w = v × w. For ζ ∈ so(3) such that ζ = Φ(v), its inﬁnitesimal action on R3 is

ζ · w = v × w, for all w ∈ R3. Then the inﬁnitesimal action of ζ ∈ so(3) on T ∗M =∼ R6 is given by

av(x, y) = (v × x, v × y), where v = Φ(ζ). The function 6 3 ∗ µ : R → (R ) , (x, y) 7→ x × y is a moment map, where hµ(x, y), ai = (x × y) · a. In this case, the map µ is known as angular momentum. 4 Jacobi Geometry

This chapter contains an introduction of Jacobi geometry. We begin by explaining Kirillov’s local Lie algebras [11] and its correspondence with Lichnerowicz’ Jacobi structures as presented in [17]. We study Jacobi geometry from the point of view of Jacobi bundles [17, Deﬁnition 3.1], when working over trivial bundles we refer to Jacobi manifolds. We then show how Poisson manifolds appear as particular cases of Jacobi manifolds. We conclude by exploring in detail the non- degenerate cases of Jacobi bundles: contact bundles and locally conformally symplectic bundles. The main reference for this section is [24]. For the study of local Lie algebras, we used the works of [11, 17, 28]. For the study of locally conformally symplectic manifolds, we follow [24] for the general line bundle approach, as for the trivial bundle case we use [25, 23]. In this part the approach is straightforward as the main examples (Poisson and symplectic manifolds) have been studied at length throughout Chapter 3. For the study of contact manifolds,the discussions are more detailed as we present in detail how the approaches of [3] and [24] recover the classical results found in references such as [9, 8].

4.1 Local Lie algebras and Jacobi manifolds

Deﬁnition 4.1. Let (π : E → M) be a vector bundle. We say that a bracket [·, ·] : Γ(E)×Γ(E) → Γ(E) deﬁnes a local Lie algebra structure if

1. the pair (Γ(E), [·, ·]) is a Lie algebra, and

2. the bracket [·, ·] is local, i.e., for all u, v ∈ Γ(E):

supp([u, v]) ⊂ supp(u) ∩ supp(v).

The pair (Γ(E), [·, ·]) is known as a local Lie algebra.

Example 4.2. Let M be a connected smooth manifold and let g be a ﬁnite dimensional Lie algebra. Consider the vector bundle given by (π : M × g → M). Then the Lie bracket on g induces a local Lie algebra structure on Γ(M × g), given for all u, v ∈ Γ(M × g):

[u, v](p) = [u(p), v(p)], for all p ∈ M. This bracket is evidently a Lie bracket. Locality follows form the fact that if [u, v](p) 6= 0, then u(p) 6= 0 and v(p) 6= 0.

Example 4.3. Consider a connected smooth manifold M. Then there is a local Lie algebra given by (X(M), [·, ·]), where [·, ·] is the Lie bracket on vector ﬁelds. This bracket is local as it is a diﬀerential operator. 4.1 Local Lie algebras and Jacobi manifolds 53

Local Lie algebras on line bundles are of particular interest to us.

Example 4.4. Consider M a subset of R2n and the trivial line bundle over M. We know that ∞ Γ(RM ) = C (M). For coordinates x1, . . . , xn, y1, . . . , yn, the bracket given by

n X ∂f ∂g ∂f ∂g {f, g} = − , ∂y ∂x ∂x ∂y i=1 i i i i for all f, g ∈ C∞(M), is a local Lie algebra. Note that this bracket is the canonical Poisson structure of Example 3.5. In fact, due to Proposition 3.9, any Poisson manifold (M, {·, ·}) induces a local Lie algebra on the trivial line bundle over M.

Example 4.5. Consider a smooth manifold M and the trivial line bundle RM over M. Fix a vector ﬁeld X ∈ X(M), then the bracket

{f, g} = hfdg − gdf, Xi = fX(g) − gX(f), where f, g ∈ C∞(M), is a local Lie algebra.

Deﬁnition 4.6. A Jacobi bundle (π : L → M, {·, ·}) is a line bundle (π : L → M) with a bracket on sections {·, ·} such that (Γ(L), {·, ·}) is a local Lie algebra. We say that the pair (Γ(L), {·, ·}) is a Jacobi structure, where {·, ·} is the Jacobi bracket. A Jacobi manifold (M, {·, ·}), is a manifold M with a local Lie bracket {·, ·} : C∞(M) × C∞(M) → C∞(M)

Remark 4.7. Recall that for any trivial line bundle (π : RM → M) there is a canonical isomor- ∞ ∼ phism C (M) = Γ(RM ). Then a Jacobi manifold is the same as a Jacobi bundle whose underlying vector bundle is the trivial one.

Note that examples 4.4 and 4.5 are Jacobi manifolds. In both cases, locality of the bracket follows from the fact that the bracket is a diﬀerential operator. The following result tells us that the converse is true in the case of Jacobi bundles.

Proposition 4.8. Let (π : L → M, ·, ·) be a Jacobi bundle. Then the following statements are equivalent:

1. The Lie bracket {·, ·} is a ﬁrst order diﬀerential operator in both entries.

2. The Lie bracket {·, ·} is local.

Proof. As every diﬀerential operator is local, then (1) implies (2). To prove the converse requires more work. We recall the following result

Theorem 4.9 (Peetre’s Theorem). [12, Section 19.1].

For j = 1, 2, let Ej → M be a vector bundle. Then a linear operator D : Γ(E1) → Γ(E2) such that supp(Du) ⊂ supp(u), i.e., D is local, is a diﬀerential operator. 54 4 Jacobi Geometry

Fix u ∈ Γ(E), then the operator {u, ·} : Γ(E) → Γ(E) is such that for all v ∈ Γ(E)

supp({u, v}) ⊂ supp(v).

We apply Peetre’s theorem to find that {u, ·} is a differential operator for all u ∈ Γ(E). A similar argument, shows that {·, u} is a differential operator for all u ∈ Γ(E)(see [11, Lemma 1]). The fact that {·, ·} is a first order differential operator follows from the local representation of the bracket and the application of the Jacobi identity. The details of this proof are in [11, Lemma 2].

Remark 4.10. There is no similar result for local Lie algebras whose rank is greater or equal to 2.

In examples 4.4 and 4.5, the bracket is given by a bivector and by a vector ﬁeld respectively. In fact, these are extreme examples of Jacobi manifolds.

Definition 4.11. A Jacobi pair (Λ,E) on a manifold M, is given by a bivector field Λ ∈ X2(M) and a vector field E ∈ X(M) such that

[Λ,E] := LEΛ = 0, [Λ, Λ] = 2E ∧ Λ.

Proposition 4.12. [24, Proposition 2.9] Let (π : L → M, {·, ·}) be a Jacobi bundle and let U ⊆ M −1 ∼ be an open subset such that π (U) = U × R. Then there exists a Jacobi pair (ΛU ,EU ), such that the local Lie algebra structure on C∞(U) is given by

{f, g} = hfdg − gdf, EU i + ΛU (df, dg), (4-1)

for all f, g ∈ C∞(U). In particular, a Jacobi manifold (π : M × R → M, {·, ·}) is equivalent to the existence of globally deﬁned Jacobi pair.

We denote Jacobi manifolds by (M, {·, ·}) where (C∞(M), {·, ·}) is a local Lie algebra on the trivial line bundle.

Corollary 4.13. Let (M, {·, ·}) be Jacobi manifold. Then there is a unique Jacobi pair (Λ,E) such that {f, g} = hfdg − gdf, Ei + Λ(df, dg),

Proof. This is a direct consequence of Proposition 4.12.

Following Corollary 4.13, we denote a Jacobi manifold by (M, Λ,E), where (Λ,E) is a Jacobi pair. Jacobi structures given in terms of Jacobi pairs were studied by Lichnerowicz. For a more detailed summary of Lichnerowicz’s approach, we refer the reader to [17] and the references within. Jacobi structures were independently studied by Kirillov [11]. A proof of Proposition 4.12 can be computed using the local representation of a diﬀerential operator. This approach was originally used by Kirillov [11, Lemma 3]. Recently, Tortorella [24, Proposition 2.9] proved this result using the language of Lie algebroids and Maurer-Cartan forms. However the higher language of Lie algebroids is beyond the scope of this work. 4.1 Local Lie algebras and Jacobi manifolds 55

In light of Proposition 4.12, we see that examples 4.4 and 4.5 are extreme examples of Jacobi manifolds given by Jacobi pairs (Λ,E). Example 4.4 corresponds to the case where E ≡ 0. Similarly, example 4.5 corresponds to the case where Λ ≡ 0. Moreover, Jacobi bundles generalize other structures such as:

Example 4.14. Associated to a Locally conformally symplectic bundle (π : L → M, Ω, ω) is there is a Jacobi bundle. The detailed construction of the bracket related to this structure and the appropriate deﬁnitions are presented in Section 4.3.1.

Example 4.15. Associated to a contact manifold (M, H) there is a Jacobi bundle. The detailed construction of the bracket related to this structure and the appropriate deﬁnitions are presented in Section 4.3.2.

Lemma 4.16. Let (π : L → M, {·, ·}) be a Jacobi bundle and let τ : L|U → U × R be a local trivialization, with associated Jacobi pair (ΛU ,EU ). Then the subspace

∞ ∞ Cadm(U) := {f ∈ C (U)|df(E) = 0} ⊂ C (U) of admissible functions, endowed with the bracket induced by (ΛU ,EU ) is a Poisson algebra.

Proof. We know from the local description of Jacobi bundles (Proposition 4.12) that there is a Jacobi pair (ΛU ,EU ) such that [ΛU , ΛU ] = 2ΛU ∧ EU and [ΛU ,EU ] = 0. We claim that (Cadm(U), {·, ·}), where {·, ·} is the Jacobi bracket restricted to admissible functions, is a Pois- son algebra. Note that for all f, g ∈ Cadm(U) we have that

{f, g} = hfdg − gdf, EU i + ΛU (df, dg) = ΛU (df, dg).

Therefore, this bracket is a derivation of the ring of admissible functions in each entry. By Lemma 3.15, to verify that the bracket restricted to admissible functions meets the Jacobi identity we evaluate the Schouten-Nijenhuis bracket on admissible functions, as follows

[ΛU , ΛU ]|adm = (ΛU ∧ EU )|adm = 0

Therefore, the induced bracket restricts to a Poisson bracket and so we have a locally deﬁned Poisson algebra.

There is a one-to-one correspondence between Jacobi bundles and homogeneous Poisson structures. This correspondence explored by [17, Section 4] and the references within. In particular, it was studied in more detail in [5].

The works[10, 27], show that Jacobi structures on (πL : L → M) are in correspondence with ∗ ∗ Poisson structures on the total space of (πL∗ : L \ 0 → M), where 0 is the zero section of L . The correspondence between Jacobi and Homogeneous Poisson structures has been widely used as a tool to prove results in the Jacobi world by proving them ﬁrst in the Poisson world and bringing them back. This strategy is known as the ”Poissonisation trick”. For example in the work of [4], it is used to prove results regarding the integrability of Jacobi manifolds. The approach of the present work follows in spirit the approaches of [19, 24], in their aim to prove results about Jacobi geometry intrinsically, that is without resorting to the Poissonisation trick. 56 4 Jacobi Geometry

4.2 Jacobi morphisms

Deﬁnition 4.17. For j = 1, 2 , let (π : Lj → Mj, {·, ·}j) be a Jacobi bundle. A pair (ϕ, F ) of a ∗ smooth map ϕ : M1 → M2 and a vector bundle isomorphism F : ϕ L2 → L1 is called a Jacobi morphism if for all smooth sections u, v ∈ Γ(L2) the following relation holds

∗ ∗ ∗ {F ◦ ϕ u, F ◦ ϕ v}1 = F ◦ ϕ {u, v}2. (4-2)

We denote Jacobi morphisms by the pair (ϕ, F ).

Lemma 4.18. For j = 1, 2, 3, let (πj : Lj → Mj, {·, ·}j) be a Jacobi manifold and let (ϕ, F ): L1 → L2 and (ψ, G): L2 → L3 be Jacobi morphisms. The composition of Jacobi morphisms is given by (η, H) := (ψ, G) ◦ (ϕ, F ) = (ψ ◦ ϕ, F ◦ ϕ∗G),

∗ ∗ ∗ ∗ where ϕ G : ϕ ψ L3 → ϕ L2 is given by (x, r) 7→ (x, G(r)).

Proof. Note that by deﬁnition of Jacobi morphism there is a vector bundle morphism

∗ (idM2 ,G):(πψ,3 : ψ L3 → M2) → (π2 : L2 → M2).

As ϕ : M1 → M2 is a smooth map, then the vector bundle morphism (idM2 ,G) induces a vector bundle morphism:

∗ ∗ ∗ ∗ (idM1 , ϕ G):(πψ◦ϕ,3 : ϕ ψ L3 → M1) → (πϕ,2 : ϕ L2 → M1). (4-3)

−1 ∗ −1 As G is an isomorphism, it has an inverse G . Then there is an induced map ϕ G : L2 → ∗ ∗ ∗ ∗ −1 ϕ ψ L3. By Proposition 2.40(4), it follows that ϕ G is an isomorphism with inverse ϕ G .

∗ ∗ ∗ ∗ By Proposition 2.40(2), we have that ϕ ψ L3 = (ψ ◦ ϕ) L3 = η L3. Using the fact that (ϕ, F ) is a Jacobi morphism, there is a vector bundle map

∗ (idM1 ,F ):(πϕ,2 : ϕ L2 → M1) → (π1 : L1 → M1). (4-4)

Combining equations 4-3 and 4-4, we obtain a vector bundle morphism over M1, given by:

∗ ∗ (idM1 ,F ◦ ϕ G):(πη,3 : η L3 → M1) → (π1 : L1 → M1).

∗ Let H := F ◦ ϕ G. Then by the previous reasoning, (idM1 ,H) is a vector bundle isomorphism as it is the composition of vector bundle isomorphisms. In other to prove that (η, H) is a Jacobi morphism, we need an additional result.

Consider section r ∈ Γ(L2) and a point p ∈ M1, then

(ϕ∗G ◦ ϕ∗ ◦ G−1 ◦ r)(p) = ϕ∗G(p, G−1(r(ϕ(p)))) = (p, r(ϕ(p))).

It follows that ϕ∗G ◦ ϕ∗ ◦ G−1 = ϕ∗ (4-5) 4.2 Jacobi morphisms 57

Let u, v ∈ Γ(L3), using equation 4-5, we compute

∗ ∗ ∗ ∗ H ◦ η {u, v}3 = F ◦ ϕ G ◦ ϕ ◦ ψ {u, v}3 ∗ ∗ −1 ∗ = F ◦ ϕ G ◦ ϕ ◦ G ◦ G ◦ ψ {u, v}3 ∗ ∗ −1 ∗ ∗ = F ◦ ϕ G ◦ ϕ ◦ G ◦ {G ◦ ψ u, G ◦ ψ v}2 ∗ ∗ ∗ = F ◦ ϕ {G ◦ ψ u, G ◦ ψ v}2 ∗ ∗ ∗ ∗ = {F ◦ ϕ ◦ G ◦ ψ u, F ◦ ϕ ◦ G ◦ ψ v}1 ∗ ∗ −1 ∗ ∗ ∗ −1 ∗ = {F ◦ ϕ G ◦ ϕ ◦ G ◦ G ◦ ψ u, F ◦ ϕ G ◦ ϕ ◦ G ◦ G ◦ ψ v}1 ∗ ∗ ∗ ∗ ∗ ∗ = {F ◦ ϕ G ◦ ϕ ◦ ψ u, F ◦ ϕ G ◦ ϕ ◦ ψ v}1 ∗ ∗ = {H ◦ η u, H ◦ η v}1.

Therefore (η, H) is a Jacobi morphism.

Proposition 4.19. Jacobi bundles (not necessarily over the same manifold) form a category where the objects are Jacobi bundles (L → M, {·, ·}), morphisms are Jacobi morphisms (as in Deﬁnition 4.17), and morphism composition is given by Lemma 4.18. We denote this category by Jac.

Proof. Note that Lemma 4.18 tells us that composition of Jacobi morphisms is well deﬁned, i.e., it yields a Jacobi morphism.

Additionally, we have an identity element given by (IdM , IdL). All that is left is to check associativity on morphism. For j = 1,..., 4, let (Lj → Mj, {·, ·}j) be a Jacobi manifold. For j = 1, 2, 3, let (ϕj,Fj) be a Jacobi morphism from Lj → Lj+1. We compute

(ϕ3,F3) ◦ ((ϕ2,F2) ◦ (ϕ1,F1)) = (ϕ3,F3) ◦ (ϕ2 ◦ ϕ1,F1 ◦ F2,ϕ1 )

= (ϕ3 ◦ ϕ2 ◦ ϕ1,F1 ◦ F2,ϕ1 ◦ F3,(ϕ2◦ϕ1)).

Similarly, we get

∗ ((ϕ3,F3) ◦ (ϕ2,F2)) ◦ (ϕ1,F1) = (ϕ3 ◦ ϕ2,F2 ◦ ϕ2F3) ◦ (ϕ1,F1) ∗ = (ϕ3 ◦ ϕ2 ◦ ϕ1,F1 ◦ F2,ϕ1 ◦ F3,ϕ1,ϕ2 )

= (ϕ3 ◦ ϕ2 ◦ ϕ1,F1 ◦ F2,ϕ1 ◦ F3,(ϕ2◦ϕ1))

It follows that ϕ3 ◦ ϕ2 ◦ ϕ1 is well-deﬁned. Therefore Jac is a category.

We now apply Deﬁnition 4.17 to the case of Jacobi manifolds.

Proposition 4.20. For i = 1, 2, let ((πj : Ej → Mj, {·, ·}i) be a Jacobi bundle. Let (ϕ, F ) be a Jacobi morphism, and let (U, φ1) be a trivializing chart of (π1 : E1 → M1) such that ϕ(U) ⊂ V ⊆ M2, where (V, φ2) is a trivializing chart of (π2 : E2 → M2). Then there exists a nowhere vanishing function a ∈ C∞(U) such that for all f, g ∈ C∞(ϕ(U)),

∗ ∗ s ∗ r {a(ϕ f), a(ϕ g)}1s = a(ϕ {f, g}2)s,

s r where s is a frame of Γ(L1)U , r is a frame of Γ(L2)ϕ(U), and {·, ·}1, {·, ·}2 are the respective brackets on functions induced by each frame. 58 4 Jacobi Geometry

∞ Proof. Fix a frame s ∈ Γ(L1)U . Then for all u ∈ Γ(L1)U , there exists a function fu ∈ C (U), such that u = fus. ∗ ∗ ∗ Let r ∈ Γ(L2)ϕ(U) be a local frame of Γ(L2)2. Then ϕ r ∈ Γ(ϕ L2)U . Since F : ϕ L2 → L1 is an isomorphism, by Proposition 2.38, the map Fx is a linear isomorphism. It follows that ∗ F ◦ ϕ r ∈ Γ(L1)U is a nowhere vanishing section, and thus it is a frame of Γ(L1)U . Then there exists a nowhere vanishing function a ∈ C∞(U) such that F ◦ ϕ∗r = as.

As r is a frame of Γ(L2)ϕ(U), then

r {u, v}2 = {fur, fvr}2 = {fu, fv}2r,

∞ where fu, fu ∈ C (ϕ(U)) are functions such that u = fur and v = fvr, for all u, v ∈ Γ(L2)ϕ(U)). As (ϕ, F ) is a Jacobi morphism we have that for all u, v ∈ Γ(L2)ϕ(U):

∗ ∗ ∗ {F ◦ ϕ u, F ◦ ϕ v}1 = F ◦ ϕ {u, v}2 ∗ ∗ ∗ r {F ◦ ϕ (fur),F ◦ ϕ (fvr)}1 = F ◦ ϕ {fu, fv}2r ∗ ∗ ∗ ∗ ∗ r ∗ {ϕ fuF ◦ ϕ (r), ϕ fvF ◦ ϕ (r)}1 = ϕ {fu, fv}2F ◦ ϕ r ∗ ∗ ∗ r {ϕ fu(as), ϕ fv(as)}1 = ϕ {fu, fv}2(as) ∗ ∗ s ∗ r {aϕ fu, aϕ fv}1 = aϕ {fu, fv}2.

Corollary 4.21. For j = 1, 2, let (Mj, {·, ·}j) be a Jacobi manifold. Then a Jacobi morphism is ∞ given by a pair (ϕ, a) of a smooth function ϕ : M1 → M2 and a ∈ C (M1) a nowhere vanishing ∞ function, such that for all f, g ∈ C (M2)

∗ ∗ ∗ {a ◦ ϕ f, a ◦ ϕ g}1 = a ◦ ϕ {f, g}2. (4-6)

Proof. This the particular case of Proposition 4.20 where U = M1.

For Jacobi manifolds, we say that a Jacobi morphism is a pair (ϕ, a) with the conditions stated on Corollary 4.21. This deﬁnition corresponds to the conformal Jacobi morphism given by [17] . If a = 1, then this deﬁnition is a strict Jacobi morphism. There is another characterization of a Jacobi manifolds, that of Jacobi pairs. The following result shows how a Jacobi morphism relates the Jacobi pairs of two Jacobi manifolds.

Proposition 4.22. For j = 1, 2, let (M, Λj,Ej) be a Jacobi manifold. A pair (ϕ, a) of a smooth ∞ map ϕ : M1 → M2 and of a nonvanishing function a ∈ C (M1) deﬁnes a Jacobi morphism if and only if a aΛ1 ∼ϕ Λ2,E1 ∼ϕ E2,

a ] where E1 = aE1 + Λ1(da).

∞ Proof. Using equation 4-6 for a Jacobi morphism (a, ϕ), then for f, g ∈ C (M2), we have

∗ ∗ ∗ {a(ϕ f), a(ϕ g)}1 = a(ϕ {f, g}2). 4.2 Jacobi morphisms 59

On the right hand side, we have

∗ ∗ a(ϕ {f, g}2) = a (ϕ (hfdg − gdf, E2i + Λ2(df, dg))) (4-7) On the left hand side, we have

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ {a(ϕ f), a(ϕ g)}1 = haϕ gd(aϕ f) − aϕ fd(aϕ g),E1i + Λ1(d(aϕ f), d(aϕ g)) (4-8) Note that d(aϕ∗f) = ϕ∗fda + ad(ϕ∗f). The term consisting of a contraction with a vector ﬁeld on the right hand side of equation 4-8 becomes

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ haϕ gd(aϕ f) − aϕ fd(aϕ g),E1i = haϕ f (ϕ gda + ad(ϕ g)) − a(ϕ g)(ϕ fda + ad(ϕ f)) ,E1i 2 ∗ ∗ 2 ∗ ∗ = ha (ϕ f)d(ϕ g) − a (ϕ g)d(ϕ f),E1i ∗ ∗ ∗ ∗ = ah(ϕ f)d(ϕ g) − (ϕ g)d(ϕ f), aE1i Let us consider now the term consisting of a contraction with a bivector ﬁeld on equation 4-8.

∗ ∗ ∗ ∗ ∗ ∗ Λ1(d(aϕ f), d(aϕ g)) = Λ1(ϕ fda + ad(ϕ f), ϕ gda + ad(ϕ g)) ∗ ∗ ∗ ∗ ∗ ∗ = Λ1((ϕ f)da, ad(ϕ g)) + Λ1(ad(ϕ f), (ϕ g)da) + Λ1(ad(ϕ f), ad(ϕ g)) ∗ ∗ ∗ ∗ 2 ∗ ∗ = a(ϕ f)Λ1(da, d(ϕ g)) − a(ϕ g)Λ1(da, d(ϕ g)) + a Λ1(d(ϕ f), d(ϕ g)) ∗ ∗ ∗ ∗ ] 2 ∗ ∗ = h(ϕ f)d(ϕ g) − (ϕ g)d(ϕ f), aΛ1(da)i + a Λ1(d(ϕ f), d(ϕ g))

Replacing and rearranging, we have that equation 4-8 becomes

∗ ∗ D ∗ ∗ ∗ ∗ ] E ∗ ∗ {a(ϕ f), a(ϕ g)}1 = a (ϕ f)d(ϕ g) − (ϕ g)d(ϕ f), aE1 + Λ1(da) + aΛ1(d(ϕ f), d(ϕ g)) . (4-9)

a Notice that if aΛ1 ∼ϕ Λ2 and E1 ∼ϕ E2, then (a, ϕ) is a Jacobi morphism. On the other hand suppose that (a, ϕ) is a Jacobi morphism, then there is still some work to do. Suppose f is a nonzero constant function, then by equations 4-7 and 4-9, we have

∗ ∗ a ∗ h(ϕ f)d(ϕ g),E1 i = ϕ hfdg, E2i ∗ ∗ a ∗ (ϕ f)hd(ϕ g),E1 i = (ϕ f)hdg, E2i ∗ a hd(ϕ g),E1 i = hdg, E2i. a Therefore E1 ∼ϕ E2. Now, if we take f and g as any pair of smooth functions on M2. Using the previous result and equations 4-7 and 4-9, we get

∗ ∗ ∗ aΛ1(d(ϕ f), d(ϕ g)) = ϕ Λ2(df, dg). The desired result follows.

Corollary 4.23. For i = 1, 2, let (Mi, {·, ·}i) be a Jacobi manifold with associated Jacobi pair (Λi,Ei). If a Jacobi morphism (ϕ, a) is such that ϕ is a diﬀeomorphism. Then (ϕ, a) deﬁnes a Jacobi morphism if and only if

a (ϕ∗)aΛ1 = Λ2, (ϕ∗)E1 = E2, (4-10) a ] where E1 = aE1 + Λ1(da). 60 4 Jacobi Geometry

Proof. The pushforward of vectors and bivector fields is well defined by the fact that ϕ is a diffeomorphism. By Proposition 4.22 and the surjectivity of ϕ,

(ϕ∗)aΛ1 = Λ2|ϕ(M1)= Λ2.

a The result follows from a similar reasoning for E1 .

∞ Lemma 4.24. For j = 1, 2, let (Mj, Λj,Ej) be a Jacobi manifold. For all f ∈ C (M2), let ] ∞ Xf = Λ2(df) + fE2 and similarly for all g ∈ C (M1). ∞ A pair (ϕ, a) of a smooth map ϕ : M1 → M2 and of a nonvanishing function a ∈ C (M1) deﬁnes ∞ a Jacobi morphism if and only if for all ∈ C (M2), the vector ﬁelds Xaϕ∗f and Xf are ϕ−related Proof. Suppose (ϕ, a) is a Jacobi morphism, then by Proposition 4.22, we have that

a aΛ1 ∼ϕ Λ2,E1 ∼ϕ E2,

a ] where E1 = aE1 + Λ1(da). 1 Fix α ∈ Ω (M2), we evaluate

∗ ∗ ] ∗ ϕ (α(Xf )) = ϕ (α(Λ2(df))) + ϕ (α(fE2)) ∗ ] ∗ ∗ ∗ ] = (ϕ α)(aΛ1(dϕ f)) + (ϕ α)(ϕ f(Λ1(da) + aE1)) ∗ ] ∗ ∗ ] = (ϕ α)(aΛ1(dϕ f) + ϕ f(Λ1(da) + aE1)). ∗ = (ϕ α)(Xaϕ∗f ).

∞ To prove the converse suppose that Xaϕ∗f ∼ϕ Xf for all f ∈ C (M2). First assume that f is ∗ ] ∗ 1 constant. It follows that Xf = fE2 and Xaϕ∗f = ϕ fΛ1(da) + ϕ f(aE1). Fix α ∈ Ω (M2), we compute

∗ ∗ ϕ (α(Xf )) = (ϕ α)(Xaϕ∗f ) ∗ ∗ ∗ ∗ ] ϕ fϕ (α(E2)) = ϕ f(ϕ α)(Λ1(da) + (aE1)) ∗ ∗ ] ϕ (α(E2)) = (ϕ α)(Λ1(da) + (aE1)).

a ] It follows that E1 = Λ1(da) + (aE1) is ϕ− related to E2. Using the previous result, a direct calculation shows that aΛ1 is ϕ−related to Λ2 as we wanted to prove.

Proposition 4.25. Let (π : L → M, {·, ·}) be a Jacobi bundle. Consider the trivializations given 0 ∞ by the frames s0 and s0 over U ⊂ M , where U is an open set. Then for all f1, f2 ∈ C (U) 1 {f1, f2}s = {af1, af2}s0 , 0 a 0 ∞ where {·, ·} and {·, ·} 0 are the brackets written in the respective trivialization, and a ∈ C (M) s0 s0 0 is a nowhere vanishing function such that s0 = as0.

Proof. Let s1, s2 ∈ Γ(L). Recall that choosing a trivialization is equivalent to choosing a frame (Proposition 2.20). Let s0 be a local frame on U. Take s1, s2 ∈ Γ(L)U , then for j = 1, 2, there ∞ exists fj ∈ C (M) such that sj = fjs0. The Jacobi bracket on U is given by

{s1, s2} = {f1, f2}s0 s0. 4.2 Jacobi morphisms 61

0 Similarly, by choosing a frame s0, we have that

0 {s , s } = {g , g } 0 s , 1 2 1 2 s0 0

0 where si = gis0 for i = 1, 2. 0 0 ∞ As both s0 and s0 are local frames on U, we have that s0 = as0, where a ∈ C (M) is a nonvanishing −1 function. Then for i = 1, 2, si = fis0 = a gis0. We compute

0 {s , s } = {g , g } 0 s 1 2 1 2 s0 0 1 {f1, f2}s s0 = {af1, af2}s0 s0. 0 a 0

We say that two brackets on functions are conformally equivalent if they diﬀer by a conformal factor, as in Proposition 4.25.

Example 4.26. An interesting example to study is that of Poisson manifolds. For a Poisson manifold, a Jacobi morphism (Corollary 4.21) is known as a conformal Poisson morphism. In general, the image of a Poisson bracket under a conformal Poisson morphism need not be a Poisson bracket.

Proposition 4.27. Consider a Poisson manifold (M, π) and a ∈ C∞(M) a nowhere vanishing function. Then

1. the bivector given by aπ is a Poisson bivector if and only if a is a Casimir function.

] 2. the pair (aπ, Xa), where Xa = π (da), is a Jacobi pair. Moreover, there is a Jacobi morphism (idM , a):(M, π, 0) → (M, aπ, Xa).

3. the brackets induced by (π, 0) and (aπ, Xa) are conformally equivalent. Proof. Using the properties of the Schouten-Nijenhuis bracket we compute

[aπ, aπ] = [aπ, a] ∧ π + (−1)0a[aπ, π] = a ∧ π](da) ∧ π + aπ](da) ∧ π = 2π](da) ∧ aπ.

] Note that π (da) is the Hamiltonian vector ﬁeld Xa associated to a. Then

[aπ, aπ] = 2Xa ∧ aπ.

Note that a is Casimir if and only if Xa = 0. It follows that [aπ, aπ] = 0 if and only if a is Casimir, proving (1).

As Xa is a Hamiltonian vector ﬁeld, we have that

[Xa, aπ] = LXa (aπ)

= (LXa a)π + a(LXa π) = 0 + 0 = 0. 62 4 Jacobi Geometry

It follows that (aπ, Xa) is a Jacobi pair, with bracket given by:

{f, g}a = aπ(df, dg) + hfdg − gdf, Xai.

The fact that (idM , a) is a Jacobi morphism follows from Corollary 4.23. By Proposition 4.22,the ∞ Jacobi morphism given by (idM , a), means that for all f, g ∈ C (M)

{af, ag}π = a{f, g}a.

Then the brackets are conformally equivalent.

This result shows how extra care should be taken when considering a Poisson manifold (M, π) as a Jacobi one, since conformal morphisms do not lead to other Poisson structures, as noted in [4].

4.3 Nondegenerate Jacobi manifolds

Recall that in the case of Poisson manifolds there is a correspondence between non-degenerate Pois- son manifolds and symplectic manifolds. To see what would be the analogue of this construction in Jacobi geometry, we must ﬁrst deﬁne non-degeneracy in this new context.

Deﬁnition 4.28. Let (π : L → M, {·, ·}) be a Jacobi manifold and let (U, ΛU ,EU ) a local trivialization. The anchor map is given by

∗ ] ρU : T U ⊗ R → TU;(η, f) 7→ ΛU (η) + fEU .

Let U = {Uα} be an open covering of M. We say that the Jacobi bundle is non-degenerate, if for all x ∈ M we can ﬁnd a trivialization (U, ΛU ,EU ) such that x ∈ U and ρU is surjective.

The Definition 4.28 coincides with the standard definition of non-degeneracy of Jacobi bundles using the anchor of the associated Lie algebroid found in the literature [3]. However, the study of Lie algebroids associated to Jacobi manifolds is beyond the scope of this work. In contrast to the Poisson case, non-degeneracy of Jacobi manifolds appears in two different structures: contact structures (odd dimensional case) and locally conformally symplectic structures (even dimensional case). Recall that a Poisson manifold (M, π) can be seen as a Jacobi (M, π, 0). For an open set U ⊂ M, ] it follows that ρU (η, f) = π (df). Suppose that ρU is surjective for all U, i.e. that (M, π, 0) is a non-degenerate Jacobi manifold. By counting dimensions, we see that π] is an isomorphism. And so the dimension of M must be even. Then there is an associated symplectic manifold (M, π−1). We see how Definition 4.28, recovers the usual notion of non-degeneracy in Poisson manifolds.

Remark 4.29. The importance of studying non-degenerate Jacobi bundles lies in the fact that every Jacobi bundle has a characteristic foliation (whose leaves might not be all of the same dimension), where each leaf has a unique non-degenerate Jacobi structure such that its canonical injection is a Jacobi map (see [11, Theorem 1]). 4.3 Nondegenerate Jacobi manifolds 63

4.3.1 Locally conformally symplectic manifolds

Definition 4.30. [24, Section 2.5.2]. Let (π : L → M) be a line bundle. A pair (Ω, ∇) given by an L−valued differential 2−form Ω, and a flat connection ∇ on (π : L → M), such that Ω is nongenerate and d∇Ω = 0 is a locally conformally symplectic (LCS) structure structure on (π : L → M). We say that (π : L → M, Ω, ∇) is a locally conformally symplectic (LCS) bundle, if (Ω, ∇) is an LCS structure on (π : L → M).

Lemma 4.31. An LCS structure (Ω, ∇) on the trivial bundle over a manifold M corresponds to a pair (Ω, ω), where Ω ∈ Ω2(M) is non-degenerate and a closed form ω ∈ Ω1(M) such that

dΩ − ω ∧ Ω = 0.

The form ω is known as the Lee form [25].

Proof. Let (π : L → M, Ω, ∇) be an LCS bundle where (π : L → M) is a trivializable bundle. Fix a frame s ∈ Γ(L), for all u ∈ Γ(L) there exists an fu such that u = fus. Recall that Ω: X(M) × X(M) → Γ(L) is an antisymmetric, non-degenerate, bilinear map. Then for all X,Y ∈ X(M), there is a unique f(X,Y ) such that

Ω(X,Y ) = f(X,Y )s.

∞ There is a unique map Ω : X(M) × X(M) → C (M) given by Ω(X,Y ) = f(X,Y ). It is evident that Ω is smooth, bilinear, antisymmetric, and non-degenerate. Hence, Ω ∈ Ω2(M).

Using a similar reasoning, for each X ∈ X(M) there exists a unique fX such that

∇X s = fX s.

We can deﬁne a ω ∈ Ω1(M) by

fX (x) = ω(X)(x), for all x ∈ M and for all X ∈ X(M).

We claim that d∇ corresponds to dω := d + ∧ω. This follows by a direct computation using k n Deﬁnition 2.52. Fix α ∈ Ω (M), and let X0,...,Xk ∈ X(M), then there is a unique α ∈ Ω (M), such that

α(X1,...,Xk) = α(X1,...,Xk)s. Then

k+1 X i d∇α(X0,...,Xk) = (−1) (α(X0,..., Xî,...,Xk))ω(Xi)s i=0 k+1 X i + (−1) Xi(α(X0,..., Xî,...,Xk))s i=0 X i+j + (−1) α([Xi,Xj],X0,..., Xî,..., Xˆj,...,Xk)s. i

Using the invariant formula for the exterior derivative [14, Proposition 14.32], it follows that

d∇α(X0,...,Xk) = dα(X0,...,Xk) + (α ∧ ω)(X0,...,Xk).

Hence d∇Ω = 0 becomes dΩ + Ω ∧ ω = 0.

The result follows by noticing that (π : RM → M) is a trivializable bundle with a canonical frame given by x 7→ (x, 1) for all x ∈ M.

Deﬁnition 4.32. Following Lemma 4.31, we say that an LCS manifold is a triple (M, Ω, ω), where Ω is a nongenerate 2−form and ω is a closed form such that

dΩ − ω ∧ Ω = 0.

Note that Deﬁnition 4.32 is the usual deﬁnition of LCS manifolds (see [25, 23]). The name locally conformally symplectic comes from the fact that an LCS manifold (M, Ω, ω) can

be covered by open sets {Uα} such that

d(fαΩ) = 0,

for all α, where fα is a nowhere vanishing function (see [25]).

Example 4.33. Symplectic manifolds correspond to LCS manifolds such that ω = 0. A manifold is globally conformally symplectic(GCS) if it is LCS and the Lee form is exact. This means that there exists some f such that df = ω, therefore

d(fΩ) = 0.

The non-degeneracy of Ω induces a map

Ω[ : X(M) → Γ(T ∗M ⊗ L),X 7→ Ω(X, ·),

and an inverse map Ω] : T ∗M ⊗ L → TM. We can associate a vector ﬁeld to any section u ∈ Γ(L) by ] Xu = Ω (d∇u).

We say that Xu is the Hamiltonian vector ﬁeld associated to the section u ∈ Γ(L). Note that for X ∈ X(M) and all f ∈ C∞(M), we have that

d∇(fu)(X) = f∇X u + X(f)u = fd∇u(X) + df(X)u.

It follows that ] ] Xfu = Ω (fd∇u + df ⊗ u) = fXu + Ω (df ⊗ u).

Proposition 4.34. [24, Proposition 2.34] There is a Jacobi structure associated to any LCS bundle (π : L → M, Ω, ∇). 4.3 Nondegenerate Jacobi manifolds 65

Proof. We deﬁne the Jacobi bracket on sections by

{u, v} = ∇Xu v = Ω(Xv,Xu), (4-11) where Xu,Xv are the Hamiltonian vector ﬁelds corresponding to the sections u, v ∈ Γ(L). As ∇uv = {u, v}, it follows that {·, ·} is a derivation of (π : L → M) on each entry.

To prove the Jacobi identity, let u, v, w ∈ Γ(L). We compute

d∇Ω(Xu,Xv,Xw) = ∇Xu (Ω(Xv,Xw)) − ∇Xv (Ω(Xu,Xw)) + ∇Xw (Ω(Xu,Xv))

− Ω([Xu,Xv],Xw) + Ω([Xu,Xw],Xv) − Ω([Xv,Xw],Xu)

= ∇Xu (∇Xv (w)) + ∇Xv (∇Xw (u)) + ∇Xv (∇Xu (v))

− ∇[Xu,Xv ]w + ∇[Xu,Xw]v − ∇[Xv ,Xw]u. Using the fact that the connection is ﬂat, we get that

d∇Ω(Xu,Xv,Xw) = ∇Xu (∇Xv (w)) + ∇Xw (∇Xu (v)) + ∇Xv (∇Xw (u))

+ ∇Xv (∇Xu (w)) + ∇Xu (∇Xw (v)) + ∇Xw (∇Xv (u))

− ∇Xu (∇Xv (w)) − ∇Xw (∇Xu (v)) − ∇Xv (∇Xw (u))

= ∇Xv (∇Xu (w)) + ∇Xu (∇Xw (v)) + ∇Xw (∇Xv (u)) = {v, {u, w}} + {u, {w, v}} + {w, {v, u}}.

2 As d∇ = 0, the Jacobi identity follows.

Remark 4.35. Let (π : L → M, Ω, ∇) be an LCS bundle. Note that for a section u ∈ Γ(L), there is a derivation given by {u, v} = ∇Xu v. The Hamiltonian vector field is precisely the symbol of the derivation {u, ·} = ∇Xu . Example 4.36. Let (M, Ω, ω) be an LCS manifold. Using Lemma 4.31 and the correspondences ∞ used in its proof, for all f ∈ C (M) its Hamiltonian vector field Xf is uniquely defined by

ιXf Ω = dωf = df − fω. (4-12)

It follows that the bracket is given by

{f, g} = Ω(Xf ,Xg) = Xf (g) − gω(Xf ) = −Xg(f) + fω(Xg) (4-13) for all f, g ∈ C∞(M). Equations 4-12 and 4-13 recover equations (2.3) and (2.4) of [25]. We have the following construction of a Jacobi bracket on LCS manifolds: Proposition 4.37. Let (M, Ω, ω) be an LCS manifold. Then the Lie bracket deﬁned by equation 4-11 is given by the Jacobi pair (Λ,E) deﬁned by

ιΛ](β)Ω = β, iEΩ = ω, for all β ∈ Ω1(M). 66 4 Jacobi Geometry

Proof. By the non-degeneracy of Ω there is an isomorphism X(M) =∼ Ω1(M) given by the map [ ] 1 Ω (X) = ιX Ω for all X ∈ X(M) with inverse denoted by Ω :Ω (M) → X(M). Then we can deﬁne a unique bivector Λ by Λ(η, β) := Ω(Ω](η), Ω](β)), for all η, β ∈ Ω1(M). By construction Λ is such that

ιΛ](β)Ω = β, (4-14) for all β ∈ Ω1(M).

Recall that a Jacobi bracket is given in terms of Jacobi pairs by equation 4-1. We can see the bracket of f, g ∈ C(M) as follows

{f, g} = (Λ](df) + fE)(g) − gE(f).

Note that the bracket of equation 4-11 can be written as

{f, g} = Xf (g) − gω(Xf ),

where Xf is the Hamiltonian vector ﬁeld associated to f given by

ιXf Ω = df − fω. (4-15) As the symbol of the derivation {f, ·} is unique, we have that

] Λ (df) + fE = Xf . (4-16)

Replacing equation 4-16 in 4-15 we get

] ιXf Ω = ιΛ (df)Ω + ιfEΩ = df + fω. Using equation 4-14, it follows that

ιfEΩ = fω,

ιEΩ = ω,

as we wanted to prove.

Remark 4.38. As we have deﬁned non-degeneracy locally (Deﬁnition 4.28), we need to check it on local trivializations or equivalently in the trivial bundle. For a smooth function f and a 1−form η, the anchor of an LCS manifold (M, Ω, ω) is given by

ρ(η, f) = (Ω−1)](η) + fE.

As Ω is non-degenerate, (Ω−1)] : T ∗M → TM is an isomorphism. Therefore, ρ is surjective. Note that in the LCS case the surjectivity of the anchor does not depend on the vector ﬁeld E.

Proposition 4.39. For j = 1, 2, let (Mj, Ωj, ωj) be an LCS manifold. A pair (ϕ, a) of a smooth ∞ map ϕ : M1 → M2 and a nowhere vanishing function a ∈ C (M1) is a Jacobi morphism if and only if ∗ ∗ aϕ Ω2 = Ω1 , aω1 = aϕ ω2 + da. 4.3 Nondegenerate Jacobi manifolds 67

∞ ˜ ∗ ∞ Proof. Suppose (ϕ, a) is a Jacobi morphism. For f ∈ C (M2) , let f = aϕ f ∈ C (M1). Then, ∞ for all f, g ∈ C (M2) ˜ ∗ {f, g˜}1 = aϕ {f, g}2.

Suppose g 6= 0 is a constant function. Using equation 4-13, and Lemma 4.24 we have that

∗ ∗ ∗ aϕ {f, g}2 = aϕ (h−gω2,Xf i = ahϕ (gω2),Xf˜i.

On the other hand, we have that

∗ ˜ ∗ ∗ aϕ {f, g}2 = {f, g˜}1 = Xf˜(˜g) − gω˜ 1(Xf˜) = h(ϕ g)da − a(ϕ g)ω1,Xf˜i.

It follows that

∗ ∗ ∗ h(ϕ g)da − a(ϕ g)ω1,Xf˜i = ahϕ (gω2),Xf˜i, ∗ ∗ ∗ ∗ (ϕ g)da − a(ϕ g)ω1 = a(ϕ g)(ϕ ω2).

As g 6= 0, it follows that ∗ da − aω1 = aϕ ω2.

The fact that (ϕ, a) is a Jacobi morphism and equation 4-13 imply that

∗ Ω1(Xg˜,Xf˜) = aϕ (Ω2(Xg,Xf )) [ ∗ [ hΩ1(Xg˜),Xf˜i = aϕ (hΩ2(Xg),Xf i [ ∗ [ hΩ1(Xg˜),Xf˜i = ahϕ (Ω2(Xg)),Xf˜i, [ ∗ [ Ω1(Xg˜) = aϕ (Ω2(Xg)) where the second to last equality follows by Lemma 4.24 applied to Xf . A second use of Lemma 4.24, this time on Xg tells us that

∗ ∗ ιXg˜ Ω1 = aϕ (ιdϕXg˜ Ω2) = ιXg˜ (aϕ Ω2).

We conclude that ∗ Ω1 = aϕ Ω2.

∗ ∗ To prove the converse, suppose that Ω1 = aϕ Ω2 and aϕ ω2 = da − aω1. On one hand we have ∞ that for all f ∈ C (M2)

ι Ω = df˜− fω˜ = adϕ∗f + aϕ∗(fω ) = aϕ∗(df − fω ). Xf˜ 1 1 2 2

∞ On the other hand for all f ∈ C (M2)

ι Ω = aϕ∗(ι Ω ) = aϕ∗(df − fω ) Xf˜ 1 dϕXf˜ 2 2

It follows that dϕXf˜ is the Hamiltonian vector field associated to f. As Hamiltonian vector fields are uniquely defined, we get that Xf˜ ∼ϕ Xf . The result follows by Lemma 4.24. 68 4 Jacobi Geometry

Example 4.40. [17] Consider an LCS manifold (M, Ω, 0). Then (M, Ω) is symplectic since Ω is non-degenerate and dΩ + 0 ∧ Ω = dΩ = 0. The Hamiltonian vector ﬁelds are exactly the same since

ιXf Ω = df.

And so is the Poisson bracket

{f, g} = Ω(Xf ,Xg). (4-17) ∞ Recall that functions C (M) can be seen as sections of the trivial bundle RM . By a change of frame of RM and Proposition 2.20, we obtain that in other coordinates the Poisson bracket of equation 4-17 is conformally related to the bracket in the new frame. That is 1 {f, g} = {af, ag} , 0 a 1

where {·, ·}0 is the Poisson bracket, {·, ·}1 is the bracket on the new frame, and a is a nonvanishing function on M. ] By Proposition 4.22, we have that Λ1 = aΛ0 and E1 = Λ0(da) + aE0. Since E0 = 0, we have that ] E1 = aΛ0(da). The new pair (Λ1,E1) is associated to the LCS structure (M, Ω1, ω1) given by 1 1 Ω = Ω, ω = da. 1 a 1 a Since a is nowhere vanishing, we can write a = e−f for some f ∈ C∞(M). Then

f Ω1 = e ω, ω1 = df.

Therefore the Poisson bracket of a symplectic manifold and the Jacobi bracket of an GCS manifold are conformally equivalent.

4.3.2 Contact manifolds Deﬁnition 4.41. A contact manifold is a pair (M, H), where M is a smooth manifold of odd dimension 2n + 1 and H ⊂ TM is a hyperplane distribution of codimension 1 such that the

curvature cH given by

cH : Γ(H) × Γ(H) → Γ(L),

TM where L := H , is non degenerate. For two X,Y ∈ Γ(H), the curvature is given by

cH(X,Y ) = [X,Y ] mod H.

As the codimension of H is 1, we have that the vector bundle (π : L → M) is a line bundle. We can think of the projection θ : TM → L as a 1-form with values on L i.e θ ∈ Ω1(M,L). It is evident that H = ker θ. We say that θ is a diﬀerential form of contact type or a contact diﬀerential form. 4.3 Nondegenerate Jacobi manifolds 69

When L is the trivial bundle and the distribution H is given by ker θ, θ ∈ Ω1(M), we say that (M, θ) is a cooriented contact manifold. The form θ corresponds to the standard notion of contact manifolds defined by a 1-form found in the literature [9]. When working with coorientable contact manifolds we will denote the contact form by α instead of θ, to emphasize that it is not an L−valued differential form. Note that the hyperplane distribution H is not only represented by α but also by any form λα, where λ is a nonvanishing smooth function on M. We say that two forms α, η such that both represent the same hyperplane distribution are (conformally) equivalent. This defines a conformal class of equivalent contact forms that represent the same distribution.

Remark 4.42. The usual deﬁnition of a cooriented contact manifold (M, H) says that H is a codimension 1 distribution that is maximally nonintegrable [8, Section 1.1]. For a hyperplane distribution H = ker α, nonintegrability means that

α ∧ (dα)n 6= 0.

This is equivalent to Deﬁnition 4.41.

To see this claim consider two vector ﬁelds X,Y ∈ Γ(H). We compute

dα(X,Y ) = X(α(Y )) − Y (α(X)) − α([X,Y ]) = −α([X,Y ]).

We have that

dα|H= −α ◦ cH. (4-18)

First note that for all x ∈ M, the pair (Hx, dα|H) is a symplectic vector space. Then dα|H is n n non-degenerate if and only if (dα|H) 6= 0. Since H = ker α, it follows that α ∧ (dα) 6= 0.

n n Conversely, if α ∧ (dα) 6= 0, then (dα|H) 6= 0. It follows that dα|H is non-degenerate, which implies that the curvature is non-degenerate. We prefer the approach using the curvature as it does not depend on the choice of a contact form.

2n+1 Example 4.43. On R with Cartesian coordinates (x1, y1, . . . , xn, yn, z) the 1−form

n X α = dz + xjdyj (4-19) j=1 is a contact form. This form is known as the standard contact form. By Darboux theorem [8, Theorem 2.5.1], for a contact manifold (M, H), there are local coordinates on an open set U ⊂ M, such that H|U = ker α|U ,where α is given by equation 4-19.

2n+1 Example 4.44. Similarly, on R \{0} with polar coordinates (r1, ϕ1, . . . , rn, ϕn, z) the 1−form

n X 2 αp = dz + rj dϕj (4-20) j=1 is a contact form. 70 4 Jacobi Geometry

Example 4.45. [8, Example 2.1.11] n+1 n n+1 Let M = R × RP . Denote Cartesian coordinates on R by (x0, . . . , xn) and homogenous n coordinates on RP by (y0, . . . , yn). Then

 n  X H := ker  yjdxj j=0 is a well-defined hyperplane distribution on M, because the right-hand side is 1−form that is well- defined up to multiplication by a nowhere vanishing function. It can be proven that if n is even then (M, H) is not a coorientable contact manifold [8, Proposition 2.1.13]. Definition 4.46. Let (M, H) be a contact manifold. A vector field R ∈ X(M) such that

LRY = [R,Y ] ∈ Γ(H), is a Reeb vector ﬁeld.

We denote the set of all Reeb vector ﬁelds of (M, H) by XReeb(M, H). Lemma 4.47. The set of Reeb vector ﬁelds forms a Lie subalgebra of (X(M), [·, ·]).

Proof. Let R1,R2 ∈ XReeb(M, H) and Y ∈ Γ(H). By the Jacobi identity of [·, ·] we have

[[R1,R2],Y ] + [[R2,Y ],R1] + [[Y,R1],R2] = 0.

Using the fact that R1,R2 ∈ XReeb(M, H), we obtain

[[R2,Y ],R1] ∈ Γ(H) and [[Y,R1],R2] ∈ Γ(H). Evaluating θ

θ([[R1,R2],Y ]) + θ([[R2,Y ],R1]) + θ([[Y,R1],R2]) = θ(0)

θ([[R1,R2],Y ]) + 0 + 0 = 0.

Therefore [[R1,R2],Y ] ∈ Γ(H), as we wanted to prove. Proposition 4.48. [3, Lemma 2.2]. Every vector ﬁeld X ∈ X(M) can be uniquely written as the X = R + Y , where R is a Reeb vector ﬁeld, and Y ∈ Γ(H). This means that we have the following isomorphism ∼ X(M) = XReeb(M, H) ⊕ Γ(H). (4-21)

Proof. First recall that cH is non-degenerate. Let X ∈ XReeb(M, H) be a vector ﬁeld such that X ∈ Γ(H). Then for any Y ∈ Γ(H), the curvature cH(X,Y ) = [X,Y ] mod H = 0. Then X is the zero vector ﬁeld. So the intersection XReeb(M, H) ∩ Γ(H) = {0} . Therefore the sum XReeb(M, H) + Γ(H) is direct. To show the isomorphism consider the following map

[X, ·] : Γ(H) → Γ(L); W 7→ [X,W ] mod H,

where X ∈ X(M). Because cH is non-degenerate, there exists V ∈ Γ(H) such that [X, ·] = cH(V, ·). Therefore the vector ﬁeld R = X −V ∈ XReeb(M, H). Thus, X = R+V as we wanted to prove. 4.3 Nondegenerate Jacobi manifolds 71

Corollary 4.49. The contact form θ restricted to XReeb induces a vector space isomorphism between XReeb and smooth sections of L.

As a consequence of Corollary 4.49, for every u ∈ Γ(L) there is a unique vector ﬁeld Ru ∈ XReeb(M, H) known as the Reeb vector ﬁeld associated to u given by

θ(Ru) = u.

In the case of cooriented contact manifolds there is a Reeb vector ﬁeld of particular importance.

Example 4.50. Recall that for a contact form α there is a unique vector ﬁeld Rα such that

α(Rα) = 1 and iRα dα = 0.

This vector field is an infinitesimal automorphism of α and it is called the Reeb vector field associated to α.

Lemma 4.51. The vector spaces Γ(H) and Γ(H∗ ⊗ L) are isomorphic.

Proof. There is a map

∗ β : Γ(H) → Γ(H ⊗ L),X 7→ ϕX := cH(X, ·), where ϕX (Y ) := cH(X,Y ) for all Y ∈ Γ(H). The map β is injective as cH is non-degenerate. Counting dimensions we have that dim H = dim H∗ = dim H∗ ⊗ L, since L is a line bundle. Then β is a vector space isomorphism.

Corollary 4.52. There is a vector space isomorphism

X(M) =∼ Γ(L) ⊕ Γ(H∗ ⊗ L). ∼ Proof. By Proposition 4.48, there is a vector space isomorphism X(M) = XReeb(M, H) ⊕ Γ(H). ∼ ∼ By Corollary 4.49, we have that Γ(L) = XReeb(M, H). By Lemma 4.51, we also have that Γ(H) = Γ(H ⊗ L). The result follows by combining these three results.

Lemma 4.53. Let (M, H = ker α) be a contact manifold, then every β ∈ Ω1(M) can be uniquely written as β = fα + η, (4-22) where f ∈ C∞(M), and η ∈ Γ(H∗).

∞ ∼ Proof. By Corollary 4.49, there is an isomorphism C (M) = XReeb(M, H). Then by Proposition 4.48, we have that X(M) =∼ C∞(M) ⊕ Γ(H). ∞ ∼ Canonically there is an isomorphism C (M) = RRα given by f 7→ fRα, where RRα ⊂ X(M) is the set of vector ﬁeld generated by Rα. Hence ∼ X(M) = RRα ⊕ Γ(H). Taking duals, we have that 1 ∼ ∗ ∗ Ω (M) = (RRα) ⊕ Γ(H ). 72 4 Jacobi Geometry

∗ ∼ Recall that α(Rα) = 1, therefore (RRα) = Rα. It follows that

1 ∗ Ω (M) =∼ Rα ⊕ Γ(H ).

Lemma 4.54. Let u ∈ Γ(L) and f ∈ C∞(M). Then the Reeb vector ﬁeld associated to fu ∈ Γ(L) is given by −1 Rfu = fRu − β ((df)|H ⊗ u), (4-23)

where Ru is the Reeb vector ﬁeld associated to u, β is the isomorphism deﬁned on the proof of Lemma 4.51, and (df)|H is the restriction of df to H.

Proof. By Proposition 4.48, we can write

H Rfu = R + X , (4-24)

H where R ∈ XReeb(M, H) and X ∈ Γ(H). By deﬁnition of Rfu, we have that fu = θ(Rfu). On the other hand, u = θ(Ru). Then fθ(Ru) = θ(fRu) = θ(Rfu). As θ restricted to Reeb vector ﬁelds is an isomorphism (Corollary 4.49), we have that

R = fRu. (4-25)

As Rfu ∈ XReeb(M, H), we have that θ([Rfu,Y ]) = 0 for all Y ∈ Γ(H). Using equation 4-24 for all Y ∈ Γ(H):

H H H 0 = θ([fRu + X ,Y ]) = −θ(Y (f)Ru + f[Y,Ru]) + θ([X ,Y ]) = −df(Y ) ⊗ u + θ([X ,Y ]).

Using the deﬁnition of curvature (see Deﬁnition 4.41), we get that

H −df(Y ) ⊗ u = cH(X ,Y )

−(df)|H ⊗ u = ϕXH

Via the map β we have that −β−1(df ⊗ u) = XH . (4-26) The result follows by replacing equations 4-25 and 4-26 in equation 4-24.

Remark 4.55. Let (M, α) be a cooriented contact manifold. In this case isomorphism β is given

by dα|H. Suppose X is the Reeb vector ﬁeld associated to the function f, then the condition of equation 4-26 corresponds to

ιXH dα|H= −df|H. (4-27) The condition given by equation 4-27, is similar to that of a Hamiltonian vector field in symplectic geometry. This was a key insight used in the work of Loose [16] to define Hamiltonian vector fields in the contact setting. The approach of this work is more general (as shown in Chapter 5), and recovers this insight as a particular case of Theorem 5.31.

It is easy to see that Definition 4.46 indeed recovers the usual definition of Reeb vector fields for cooriented contact manifolds in terms of contact forms [8, Lemma 1.5.8]. 4.3 Nondegenerate Jacobi manifolds 73

Lemma 4.56. Let X ∈ X(M) and (M, H = ker α) a cooriented contact manifold. Then X is a Reeb vector ﬁeld of H if and only if there exists a function λ ∈ C∞(M) such that

LX α = λα.

Proof. For all X ∈ X(M) and all Y ∈ Γ(H) we have that

α([X,Y ]) = X(iY α) − Y (iX α) − dα(X,Y )

α([X,Y ]) = −iY d(iX α) − dα(X,Y )

It follows that X is a Reeb vector ﬁeld if and only if

iY diX α = −dα(X,Y ), (4-28) for all Y ∈ Γ(H). Let Y ∈ X(M) and ﬁx a Reeb vector ﬁeld X, we compute

ιY (LX α) = ιY (diX α + iX dα) = ιY dιX α + ιY ιX dα.

By equation 4-28, we have that if Y ∈ Γ(H), then ιY (LX α) = 0. Therefore, ker α ⊆ ker(LX α). Suppose there exists some Y ∈ Γ(ker(LX α)) such that Y/∈ Γ(H). By Proposition 4.48, it suﬃces H to assume that Y ∈ XReeb(M, H). Then by Lemma 4.54, we have that Y = fRα + Y . Fix X a Reeb vector ﬁeld, we have that

H α([X,Y ]) = α([X, fRα + Y ]) = α([X, fRα]).

As Y ∈ Γ(ker(LX α)) we have that

LX α(Y ) = X(α(Y )) − α([X,Y ]) = X(α(Y )) − α([Y, fRα]) = −fα([X,Rα]) = 0.

Let g = α(X), then

−fα([X,Rα]) = −f(X(ιRα α) − Rα(ιX α) − ιRα ιX dα) = −f(X(1) − Rα(g)) = fRα(g) = 0,

∞ for all g ∈ C (M). It follows that Y ∈ Γ(H). Then Γ(ker(LX α)) = Γ(ker α). Therefore, LX α must be a multiple of α, i.e. there is a λ ∈ C∞(M) such that

LX α = λα.

To prove the converse, suppose X is such that there is some function λ ∈ C∞(M) such that

LX α = λα. Then for all Y ∈ H

ιY (LX α) = ιY (λα) = 0.

This implies that iY diX α = −dα(X,Y ) for all Y ∈ Γ(H). Therefore X is a Reeb vector ﬁeld. 74 4 Jacobi Geometry

Proposition 4.57. There is a Jacobi structure associated to any contact manifold (M, H).

Proof. By Corollary 4.49, we endow Γ(L) with a Lie algebra structure given by

{u, v} := θ([Ru,Rv]), (4-29)

for all u, v ∈ Γ(L), and Ru,Rv are their respective Reeb vector fields. ∞ Consider a smooth function f ∈ C (M). Then, by Proposition 4.48 Rfu = R + Y where R ∈ XReeb(M, H) and Y ∈ Γ(H). On one hand, we see that θ(Rfu) = fu = fθ(Ru) = θ(fRu) so R = fRu. On the other hand θ([Rfu,X]) = 0 for all X ∈ Γ(H). To verify that the bracket defined by equation 4-29 defines a Jacobi manifold, we need to check that it is a differential operator on each of its entries. Consider u, v ∈ Γ(L) and f ∈ C∞(M), we compute

{u, fv} = θ([Ru,Rfv]).

H H ] By Lemma 4.54, we know that Rfv = fRv + Rfv, where Rfv = Ω ((df)|H ⊗ v). It follows that

H {u, fv} = θ([Ru, fRv + Rfv]) H = θ([Ru, fRv]) + θ([Ru,Rfv])

= θ(Ru(f)Rv) + fθ([Ru,Rv])

= Ru(f)v + f{u, v}.

By antisymmetry, we get that the Reeb bracket is indeed a diﬀerential operator on both entires. The bracket given by equation 4-29 is known as the Contact or Reeb bracket associated to θ. Therefore (π : L → M, {·, ·}) where {·, ·} is the Reeb bracket is a Jacobi manifold.

As our main example, we shall now construct the Reeb bracket for a cooriented Jacobi manifold.

Example 4.58. Consider a cooriented contact manifold (M, H = ker α). Let f, g ∈ C∞(M), then

associated to them are the following Reeb vector ﬁelds Xf ,Xg. We denote by R the Reeb vector ﬁeld associated the form α as in Example 4.50. By Lemma 4.54 we can write

H H Xf = fR + Xf ; Xg = gR + Xg .

Replacing in equation 4-29,

H H α([Xf ,Xg]) = α([fR + Xf , gR + Xg ]) H H H H = α([fR, gR]) + α([fR, Xg ]) − α([gR, Xf ]) + α([Xf ,Xg ]).

Let us consider the ﬁrst term

α([fR, gR]) = fRα(gR) − gRα(fR) − dα(fR, gR) = fRg − gRf. 4.3 Nondegenerate Jacobi manifolds 75

Using the fact that R is the Reeb vector ﬁeld associated to α, we have that the second term becomes

H H H H α([fR, Xg ]) = Xg α(fR) − fRα(Xg ) − dα(fR, Xg )) H = Xg (f). We have an analogous result for the third term

H H α([gR, Xf ]) = gXf . For the fourth term we have

H H H H H H H H α([Xf ,Xg ] = Xf α(Xg ) − Xg αXf − dα(Xf ,Xg ) H H = −dα(Xf ,Xg ). Using Lemma 4.54 in the cooriented case we get that for all Y ∈ Γ(H)

H −iY df = dα(Xf ,Y ). Replacing all terms and we get that

H H H H α([Xf ,Xg]) = fRg − gRf + Xf (g) − Xg (f) − dα(Xf ,Xg ) H H = fRg − gRf + i H dg − i H df − dα(X ,X ) Xf Xg f g

= fRg − gRf + i H dg − i H df + i H df Xf Xg Xg

= fRg − gRf + i H dg Xf H H = fRg − gRf + dα(Xf ,Xg ). As 4-29 is a Jacobi bracket on functions, there is a Jacobi pair that describes it. The last expression is indeed very close to a Jacobi pair, however the second term is not a bivector.

In the cooriented case, the isomorphism β becomes dα] : H∗ → H, we can reinterpret dα as an element Λ˜ ∈ ∧2H, given by Λ(˜ γ, η) = dα(dα]γ, dα]η), (4-30) for γ, η ∈ Γ(∧2H∗). By Lemma 4.53, Ω1(M) =∼ Rα ⊗ Γ(H∗), where Rα is the set of diﬀerential 1−forms generated by 2 α. Then there is a unique Λ ∈ X (M) such that Λ|H= Λ˜ and vanishes outside. Finally, we can write the Reeb bracket as follows

{f, g} = fRg − gRf + Λ(df, dg). (4-31)

Note that this bracket depends on the contact form chosen to represent the contact structure and it corresponds to the form of a Jacobi bracket. Remark 4.59. Now that we have a local characterization of contact manifolds, we shall prove that they are non-degenerate Jacobi manifolds. The anchor of a contact manifold over a trivialization is given by ] ρ(f, η) = Λ (η) + fRα, 76 4 Jacobi Geometry

where α is a locally deﬁned coorientation and Rα its associated Reeb vector ﬁeld. We get that ρ is surjective because Λ] : T ∗M → TM restricts to an isomorphism of H∗ → H (Lemma 4.51) and f 7→ fRα is surjective onto RRα (Corollary 4.49), the other direct summand. 2n+1 Example 4.60. Consider R with Cartesian coordinates (x1, y1, . . . , xn, yn, z) and let α be the standard contact form (Example 4.43).

Note that ker α is generated by the vector ﬁelds Xj = ∂xj , and Yj = ∂yj − xj∂z. Pn As dα = j=1 dxj ∧ dyj, we have that dα(Xj) = dyj and dα(Yj) = −dxj. Let Z = ∂z, then the bracket generated by the standard contact form is given by the Jacobi pair n X Λ = Xj ∧ Yj,E = Z. j=1

2n+1 Example 4.61. Let us now consider R \{0} with polar coordinates (r1, ϕ1, . . . , rn, ϕn, z) the standard contact form in polar coordinates αp (Example 4.44). 2 We see that ker αp is generated by Rj = ∂rj and Fj = ∂ϕj − rj ∂z. Pn As dαp = 2 j=1 rjdrj ∧ dϕj, we have that dαp(Rj) = 2rjdϕj and dαp(Fj) = −2rjdrj. By letting Z = ∂z, then the bracket generated by the standard contact form is given by the Jacobi pair n 1 X Λ = R ∧ F ,E = Z. p 2r j j j j=1 We devote the last part of this section to show that the deﬁnition of Jacobi morphisms indeed recovers the usual notion of contact morphisms in the case of cooriented contact manifolds [8, Deﬁnition 2.1.2].

Proposition 4.62. Let (Mi, Hi = ker αi) be a contact manifold for i = 1, 2. Then the pair (a, ϕ) ∞ of a nowhere vanishing function a ∈ C (M1) and a smooth map ϕ : M1 → M2, is a Jacobi ∞ morphism if and only if given f ∈ C (M2) we have that

Xaϕ∗f ∼ϕ Xf , (4-32)

where Xf is the Reeb vector ﬁeld associated to f.

Proof. We denote by (Ri, Λi) the Jacobi pair induced by the contact form αi, for i = 1, 2. Note that

] Xf = fR2 + Λ2(df) ∗ ] ∗ Xa(ϕ∗f) = a(ϕ f)R1 + Λ1(d(a(ϕ f))) ∗ a ] ∗ = (ϕ f)R1 + aΛ1(d(ϕ f)).

a By Proposition 4.22, we have that (a, ϕ) is a Jacobi morphism if and only if R1 ∼ϕ R2 and aΛ1 ∼ϕ Λ2. This means that the first (second) term of Xa(ϕ∗f) is ϕ-related to the first (second) term of Xf . From which the result follows. Corollary 4.63. Under the same hypotheses of Proposition 4.62, if ϕ is a diffeomorphism, if (a, ϕ) is a Jacobi morphism then ∗ aα1 = ϕ α2. (4-33) 4.3 Nondegenerate Jacobi manifolds 77

Proof. Recall that the contact bracket for contact manifold (M, H = ker α) is given by

{f, g} = α([Xf ,Xg]),

∞ where f, g ∈ C (M) and Xf ,Xg are the Reeb vector ﬁelds associated to f and g. Then equation 4-6 becomes ∗ aα1([Xa(ϕ∗f),Xa(ϕ∗f)]) = ϕ (α2([Xf ,Xg]))

By Proposition 4.62, we have that Xa(ϕ∗f) ∼ϕ Xf and similarly for g. Then their Lie brackets are ϕ−related. As the Reeb vector ﬁelds form a Lie subalgebra of X(M) (Lemma4.47), the result follows.

Example 4.64. Consider the contact manifolds of Example 4.44 and the restriction of 4.43 to R2n+1 \{0}. The map f : R2n+1 \{0} → R2n+1 given by n X f(x1, y1, . . . , xn, yn, z) = ((x1 + y1)/2, (y1 − x1)/2,..., (xn + yn)/2, (yn − xn)/2, z + xjyj/2), j=1 is such that ∗ f αp = α. And so by Corollary4.63, the map (f, 1) is a Jacobi morphism.

The last example of this chapter shows us how the language of Jacobi structures provides a way to relate contact and symplectic structures.

Example 4.65. Consider the contact manifold (R2n+1, H = ker α), where α is the standard 2n contact form in Cartesian coordinates and a symplectic manifold (R , ω0), where ω0 is the standard symplectic form. Associated to (R2n+1, α) we have the Jacobi pair n X Λ = Xj ∧ Yj,E = Z, j=1

where Xj = ∂xj ,Yj = ∂yj − xj∂z and Z = ∂z. 2n Meanwhile, associated to (R , ω0) there is the bivector n X Π = ∂xj ∧ ∂yj . j=1

The projection map π : R2n+1 → R2n given by

π(x1, y1, . . . , xn, yn, z) = (x1, y1, . . . , xn, yn), is a Jacobi morphism. First, note that π is surjective, so pushforwards are well deﬁned. We also see that π∗Yj = ∂yj , π∗Xj = ∂xj , and π∗Z = 0. It follows that

Λ ∼π Π,E ∼π 0.

By Proposition 4.22, we have that (π, 1) is a Jacobi morphism. 5 Group actions on Jacobi manifolds

In this chapter we give the main definitions related to Hamiltonian actions on Jacobi manifolds. We start this chapter by discussing the infinitesimal symmetries of a Jacobi manifold. In a similar fashion as the Poisson case, we define (weakly) Hamitonian actions of a Lie group and of a Lie algebra. Throughout this chapter, we show how the results for Poisson and symplectic Hamiltonian actions can be recovered using the language of Jacobi manifolds. Moreover, we show how the results of [23] (for LCS manifolds) and those of [16] (for contact manifolds), are particular cases of Hamiltonian actions on Jacobi manifolds. We close the chapter showing how any contact action is Hamiltonian.

5.1 Jacobi group action

Let (π : L → M, {·, ·}) be a Jacobi bundle and let G be a Lie group acting on (π : L → M) with action (a, A). Then for all g ∈ G the following diagram commutes

A L g L π π a M g M.

As (a, A) is a vector bundle action, for all g ∈ G the morphism (ag,Ag) is a vector bundle ∗ automorphism. By Corollary 2.41, there is a unique map χg : L → agL, such that (idM , χg) is a −1 ∗ vector bundle isomorphism. Let Fg−1 := (χg) : agL → L.

Deﬁnition 5.1. We say that (a, A) is a Jacobi action if (ag,Fg−1 ) is a Jacobi morphism for all g ∈ G.

Recall that Jacobi bundles whose underlying line bundle over M is trivial are given by (M, Λ,E) where (Λ,E) is a Jacobi pair. By Deﬁnition 5.1, a Jacobi action on (M, Λ,E) preserves the conformal class of the bracket associated to (Λ,E). We say that this action is a conformal Jacobi action. A conformal Jacobi action on (M, Λ,E) need not preserve the Jacobi pair. Let (M, Λ,E) be a Jacobi manifold and let G act on M by Jacobi morphisms. Then the action of g is ∞ given by Jacobi morphism (ag, fg), where ag : M → M is a diﬀeomorphism and fg ∈ C (M) is a smooth nowhere vanishing function. By Corollary 4.23, we have that

∗ ∗ ∗ {(fg ◦ ag)(h1), (fg ◦ ag)(h2)}g = (fg ◦ ag)({h1, h2}(Λ,E)),

∞ for all h1, h2 ∈ C (M) and all g ∈ G, where {·, ·}(Λ,E) is the bracket associated to (Λ,E) and ] {·, ·}g is the bracket associated to (fgΛ, Λ (dfg)+fgE). Note that the brackets {·, ·}(Λ,E) and {·, ·}g are conformally equivalent. 5.1 Jacobi group action 79

When working with the trivial line bundle (π : RM → M), if ϕ : M → M is a diﬀeomorphism, ∗ ∼ then ϕ RM = RM canonically. We say that a Jacobi action of G on (M, Λ,E) is strict or trivial on ﬁbres if (χg)x := (idRM )x for all g ∈ G and for all x ∈ M. This means that

∗ ∗ ∗ {ag(h1), ag(h2)}(Λ,E) = ag({h1, h2}(Λ,E)),

∞ for all h1, h2 ∈ C (M) and all g ∈ G, where {·, ·}(Λ,E) is the bracket associated to (Λ,E). Example 5.2. Let (M, Λ,E) be a Jacobi manifold and let G act on M by Jacobi morphisms. In the case where the action (ag, fg) of g ∈ G is trivial on ﬁbres for all g ∈ G, we have that fg = 1 for all g ∈ G. Then the Jacobi pair is preserved under the action, that is (ag)∗Λ = Λ and (ag)∗E = E. In the case where E := 0, we have a Poisson manifold (M, Λ). So a Jacobi action trivial on ﬁbres on a Jacobi manifold (M, Λ, 0) is the same as a Poisson action on the Poisson manifold (M, π).

Example 5.3. Let (M, α) be a cooriented contact manifold and let G act on (M, α) via Jacobi morphisms. As (ag, fg) is a Jacobi morphism then by Corollary 4.63

∗ agα = fgα.

Then this action only preserves the contact form α if the action is trivial on ﬁbres. In this way, a Jacobi action recovers the notions of conformal and strict contact actions found in the literature [8, Section 7.7]. A Jacobi action on a contact manifold is the same as the notion of a conformal contact action and if the Jacobi action is trivial on ﬁbres, then it is the same as a strict contact action.

Example 5.4. Let (M, Ω, ω) be an LCS manifold and let G act on (M, Ω, ω) via Jacobi morphisms.

As (ag, fg) is a Jacobi morphism, by Proposition 4.39 we have that

∗ ∗ agΩ = fgΩ, fgω = (fg)(agω1) + dfg.

rg ∞ Let fg = e where rg ∈ C (M). Then a direct computation shows us that

∗ rg ∗ agΩ = e Ω, ω = (agω1) + drg. (5-1)

Equation 5-1, shows us that a Jacobi action is in fact the same as a twisted symplectic action as deﬁned in [23, Deﬁnition 3.3].

Deﬁnition 5.5. Let (π : L → M, {·, ·}) be a Jacobi bundle and let ∆ ∈ Der(L). We say that ∆ is a Jacobi derivation if for all u, v ∈ Γ(L)

∆{u, v} = {∆u, v} + {u, ∆v}. (5-2)

We denote the set of all Jacobi derivations of (π : L → M, {·, ·}) by DerJac(L, {·, ·}). We say that X ∈ InfAut(π : L → M) is an infinitesimal Jacobi automorphism if its associated derivation ∆X is a Jacobi derivation. We denote the set of all infinitesimal Jacobi automorphisms of (π : L → M, {·, ·}) by InfAutJac(L, {·, ·}). Proposition 5.6. Let (π : L → M, {·, ·}) be a Jacobi bundle. A derivation ∆ is a Jacobi derivation, if and only if the flows (ψt, Ψt) of its associated infinitesimal automorphism X∆ and of its symbol generate a smooth family of Jacobi automorphisms. 80 5 Group actions on Jacobi manifolds

Proof. Let X ∈ InfAutJac(L, {·, ·}) and let Ψt be the ﬂow of X and ψt the unique ﬂow covered by Ψt . Note that if (ψt, Ψ˜ −t) is a Jacobi morphism then for all u, v ∈ Γ(L)

Ψ˜ −t ◦ ψt{u, v} = {Ψ˜ −t ◦ ψtu, Ψ˜ −t ◦ ψtv}.

˜ ˜ ∗ By Remark 2.39, the map Ψ−t|E(x) is precisely Ψ−t|E(x). Therefore Ψ−t ◦ ψt : Γ(L) → Γ(L) is the ∗ pullback Ψt . Then ∗ ∗ ∗ Ψt {u, v} = {Ψt u, Ψt v}. dΨ∗u t By taking derivatives with respect to t at t = 0, since ∆X (u) = dt for all u ∈ Γ(L), we get t=0 that

∆X {u, v} = {∆X u, v} + {u, ∆X v},

for all u, v ∈ Γ(L). Then ∆X ∈ DerJac(L, {·, ·}). Conversely, let ∆ ∈ DerJac(L, {·, ·}). Then, by Theorem 2.70, there is a unique X∆ ∈ InfAut(L) d(Ψ∗u) t such that ∆(u) = dt , where Ψt is the ﬂow of X∆. Replacing in equation 5-2, we get that t=0 d ∗ d ∗ d ∗ (Ψt {u, v}) = Ψt u , v + u, Ψt v . dt t=0 dt t=0 dt t=0 On the other hand, we have that d ∗ d ∗ d ∗ ∗ Ψt u , v + u, Ψt v = ({Ψt u, Ψt v}) dt t=0 dt t=0 dt t=0 Hence, ∗ ∗ ∗ Ψt {u, v} = {Ψt u, Ψt v}.

We conclude that X∆ ∈ InfAutJac(L, {·, ·}). Example 5.7. Let us consider a Jacobi manifold (M, Λ,E). As shown in Example 2.56, a derivation on the trivial line bundle is given by a pair (X, r) ∈ X(M) × C∞(M). The following result gives us necessary and suﬃcient conditions for a derivation (X, r) to be a Jacobi derivation of (M, Λ,E) Proposition 5.8. A derivation (X, r) of the trivial line bundle is a Jacobi derivation of (M, Λ,E) if and only if

] LX Λ = rΛ, LX E = Λ (dr) + rE.

Proof. By Deﬁnition 5.5, the pair (X, r) is a Jacobi derivation of the bracket {·, ·} associated to (Λ,E) if for all f, g ∈ C∞(M)

(X + r)({f, g}) = {(X + r)f, g} + {f, (X + r)g}. (5-3)

] We set δf = Λ (df) + fE. As the Jacobi bracket is a derivation in each of its entries, we have that

{f, rg} = r{f, g} + δf (r)g. (5-4)

Replacing Equation 5-4 in Equation 5-3 gives us

X({f, g}) − r{f, g} = {X(f), g} + {f, X(g)} + δf (r)g − δg(r)f. 5.1 Jacobi group action 81

Let fg = fdg − gdf. Expanding the brackets we have that

X(Λ(df, dg)) + X(hfg, Ei) − r(Λ(df, dg)) − r(hfg, Ei) = Λ(dX(f), dg) + hX(f)g, Ei + Λ(df, dX(g)) + hfX(g),Ei (5-5)

+ δf (r)g − δg(r)f.

First consider the case where f ∈ C∞(M) is constant. Then the bivector terms in equation 5-5 vanish. We have that

X(hfg, Ei) − r(hfg, Ei) = hX(f)g, Ei + hfX(g),Ei + δf (r)g − δg(r)f. (5-6)

On one hand, a direct computation shows that

X(hfg, Ei) − hX(f)g, Ei + hfX(g),Ei = [X,E](fg) = (LX E)(fg). (5-7)

On the other hand, a direct computation shows that

] δf (r)g − δg(r)f = hfg, Λ (dr)i.

Using the fact that ] hfg, δri = hfg, Λ (dr)i + hfg, aEi, and replacing equations 5-6 and 5-7 in equation 5-5, we get that

(LX E)(fg) = δr(fg).

Suppose that neither f nor g are constant. Using the previous result, we can take only the bivector terms from equation 5-5, we have that

X(Λ(df, dg)) − r(Λ(df, dg)) = Λ(dX(f), dg) + Λ(df, dX(g)) r(Λ(df, dg)) = X(Λ(df, dg)) − Λ(dX(f), dg) − Λ(df, dX(g))

(rΛ)(df, dg) = (LX Λ)(df, dg).

We conclude that equation 5-3 holds if and only if

LX Λ = rΛ, and LX E = δr.

Lemma 5.9. Let (L → M, {·, ·}) be a Jacobi manifold. The set DerJac(L, {·, ·}) of all Jacobi derivations forms a Lie subalgebra of Der(L).

Proof. For i = 1, 2, let ∆i ∈ DerJac(E, {·, ·}). Consider u, v ∈ Γ(L), we compute

[∆1, ∆2]{u, v} = ∆1∆2{u, v} − ∆2∆1{u, v}

= ∆1 ({∆2u, v} + {u, ∆2v}) − ∆2 ({∆1u, v} + {u, ∆1v})

= {∆1∆2u, v} − {∆2∆1u, v} + {u, ∆1∆2v} − {u, ∆2∆1v}

= {[∆1, ∆2]u, v} + {u, [∆1, ∆2]v}.

Then [∆1, ∆2] ∈ DerJac(L, {·, ·}), as we wanted to prove. 82 5 Group actions on Jacobi manifolds

Corollary 5.10. Let (π : L → M, {·, ·}) be a Jacobi bundle , let (a, A) be a Jacobi action of a Lie group G and let g = Lie(G). Then there is an inﬁnitesimal action on sections S : g × Γ(L) → Γ(L) given by

d ζ ∗ ∆ζ (u) := ((Ψt ) u) , (5-8) dt t=0 ζ for all u ∈ Γ(L) and for all ζ ∈ g, where Ψt is the ﬂow of the fundamental vector ﬁeld Xζ ∈ X(E). Moreover ∆ζ ∈ DerJac(L, {·, ·}) for all ζ ∈ g. Proof. By Corollary 2.81, there is an induced action on sections given by equation 5-8. As the action

is Jacobi, the ﬂows (ψt, Ψt) of Xζ and δζ are Jacobi morphisms for all t ∈ I. Then by Proposition 5.6, the derivation ∆ζ associated to Xζ is a Jacobi derivation of (π : L → M, {·, ·}).

Deﬁnition 5.11. We say that the action g y (Γ(L), {·, ·}) given by Corollary 5.10 is the in- ﬁnitesimal Jacobi action of g on sections of the Jacobi bundle (π : L → M, {·, ·}). Note that on a Jacobi manifold (M, Λ,E) the derivations associated to a strict Jacobi action are of the form (X, 1). This means that

LX Λ = Λ, LX E = E.

5.2 Weakly Hamiltonian action

Deﬁnition 5.12. Let (π : L → M, {·, ·}) be a Jacobi manifold. Consider a section u ∈ Γ(L). Then this section induces a Jacobi derivation by the Jacobi identity given by

u 7→ ∆u := {u, ·}.

A derivation that is induced by the bracket of a section is a Hamiltonian derivation. We denote

the set of all such derivations as HamJac(L). We denote the symbol of ∆uby δu, . We say that δu is a Hamiltonian vector field whose Hamiltonian section is u. We denote the set of all Hamiltonian vector fields by XHam(M). Example 5.13. Consider an LCS bundle (π : L → M, Ω, ∇). Then for a Hamiltonian derivation ] ∆u(v) = {u, v}, its symbol is given by δu = Ω (d∇u). Similarly for a contact manifold (M, H) the symbol of a Hamiltonian derivation ∆u is δu = Ru where Ru is the associated Reeb vector field. Proposition 5.14. Let (L → M, {·, ·}) be a Jacobi manifold. Then for all u, v ∈ Γ(L)

∆{u,v} = [∆u, ∆v], δ{u,v} = [δu, δv]. Proof. The ﬁrst Lie algebra morphism is a direct result of the Jacobi identity of the bracket on sections of L. To verify the second Lie algebra morphism, let f ∈ C∞(M) and u, v, w ∈ Γ(L). On one hand, we get that

{u, {v, f · w}} = {u, f{v, w} + δv(f)w}

= {u, f{v, w}} + {u, δv(f)w}

= f{u, {v, w}} + δu(f){v, w} + δv(f){u, w} + δu(δv(f))w 5.2 Weakly Hamiltonian action 83

On the other hand, using the Jacobi identity

{u, {v, f · w}} = {{u, v}, f · w} + {v, {u, f · w}}

= f{{u, v}, w} + δ{u,v}(f) · w + {v, f ·{u, w} + δu(f) · w}

= f{{u, v}, w} + δ{u,v}(f) · w + f{v, {u, w}}

+ δv(f){u, w} + δu(f){v, w} + δv(δu(f)) · w

Equaling both sides we get

δu(δv(f))w = δ{u,v}(f) · w + δv(δu(f)) · w

δu(δv(f)) − δv(δu(f)) = δ{u,v}(f)

[δu, δv] = δ{u,v}.

Corollary 5.15. The set HamJac(L) is a Lie sublagebra of DerJac(L).

Proof. This is a direct consequence of Proposition 5.14.

Similarly, the symbols of Hamiltonian Jacobi derivations form a Lie subalgebra of X(M).

Remark 5.16. In the case of Jacobi manifolds (M, Λ,E), the Hamiltonian derivation associated ∞ to f ∈ C (M) corresponds to the pair (X, r) where X = δf and r = E(f). This is corresponds to the second statement of [17, Proposition 2.8].

Note that δu is the map φ in [17]. Proposition 5.14 recovers the ﬁrst statement of [17, Proposition 2.8].

Proposition 5.17. The set of Hamiltonian derivations HamJac(L) forms an ideal of DerJac(L)

Proof. Let ∆u ∈ HamJac(L) with associated section u and ∆ ∈ DerJac(L). Then for a section v ∈ Γ(L) we have

[∆u, ∆](v) = ∆u(∆(v)) − ∆(∆u(v)) = {u, ∆(v)} − ∆({u, v}) = {u, ∆(v)} − {∆(u), v} − {u, ∆(v)} = {−∆(u), v}.

Then [∆u, ∆] ∈ HamJac(L).

Deﬁnition 5.18. Let (π : L → M, {·, ·}) be a Jacobi bundle and let g be a Lie algebra. We say that the action g y (Γ(L), {·, ·}) is weakly Hamiltonian if for all ζ ∈ g

∆ζ ∈ DerJac(L, {·, ·}).

Example 5.19. Consider a strict Jacobi action of g on (M, Λ, 0) that is weakly Hamiltonian, then ∞ ] the derivations are of the form (δf , 1) for f ∈ C (M). Note that δf = Λ (df). Then the resulting action is the same as a weakly Hamiltonian Poisson action of g on (M, Λ). 84 5 Group actions on Jacobi manifolds

Proposition 5.20. Let (π : L → M, {·, ·}) be a Jacobi manifold. The action g y (Γ(L), {·, ·}) is ∗ weakly Hamiltonian if and only if then there exists a section µ of (πg∗⊗L : g ⊗ L → M) such that ∆ζ (v) = {hµ, ζi, v} for all v ∈ Γ(L). Deﬁnition 5.21. The map µ of Proposition 5.20 is called a moment map.

Proof. Since g y (Γ(L), {·, ·}) is weakly Hamiltonian, for all ζ ∈ g there is a section uζ ∈ Γ(L) such that for all sections v ∈ Γ(L)

∆ζ (v) = {uζ , v}. ∗ Let µ : M → g ⊗ L be given by uζ (x) = hµ(x), ζi. As the pairing h·, ·i is a smooth map, it follows that µ is a smooth. ∗ Conversely, suppose there exists a moment map µ : M → g ⊗ L such that ∆ζ (v) = {hµ, ζi, v} for all v ∈ Γ(L). Then ∆ζ ∈ HamJac(L). Given a moment map µ : M → g∗ ⊗ L, there is a map µ : g → Γ(L) given by

µ(ζ)(x) := hµ(x), ζi,

for all ζ ∈ g and for all x ∈ M. We say that µ is a comoment map associated to µ. Example 5.22. In the case of a weakly Hamiltonian action of a Lie algebra g on a contact manifold (M, H), with projection θ : TM → L, we have that for ζ ∈ g there is a section µ(ζ) such that

∆ζ v = {µ(ζ), v} = θ([Rµ(ζ),Rv]),

for all v ∈ Γ(L), where Rµ(ζ) is the Reeb vector ﬁeld associated to µ.

Denote the symbol of ∆ζ by δζ , then by Example 5.13, we get that

δζ = Rµ(ζ). It follows that, for all x ∈ M:

θ(δζ (x)) = µ(ζ)(x) = hµ(x), ζi. Hence we recover the notion of contact moment map deﬁned in [16]. Example 5.23. Consider an LCS bundle (π : L → M, Ω, ∇), a Lie algebra g, and a weakly Hamiltonian action g y (Γ(L), {·, ·}) with moment map µ. For ζ ∈ g, we have a derivation ∆ζ with symbol δζ . On the other hand ∆ζ v = {µ(ζ), v} and by construction of the bracket, the symbol ] of ∆ζ is given by Ω (d∇µ(ζ)). It follows that

] δζ = Ω (d∇µ(ζ)), for all ζ ∈ g. Proposition 5.24. Let (π : L → M, {·, ·}) be a Jacobi manifold. Consider a weakly Hamiltonian action g y (Γ(L), {·, ·}) with moment map µ. Given a local frame s0 in an open set U then the ∗ moment map in terms of s0 is a map µs0 : U → g . Moreover, if there is a new frame s1 = hs0 where h ∈ C∞(U) is a nowhere vanishing function, then the moment maps on each frame are related by

hµs1 = µs0 . 5.3 Hamiltonian actions 85

Proof. Fixing a frame s0 on U, we have that for all sections u ∈ Γ(E)U there is a unique function ∞ fu ∈ C (U) such that u = fus0. Fix an element ζ ∈ g, then hµ, ζi ∈ Γ(L). It follows that there is ∞ ∗ a unique function fhµ,ζi ∈ C (U) such that hµ, ζi = fhµ,ζis0. Then there is a map µs0 : M → g given by fhµ,ζi(x) = {µs0 (x), ζ} for all x ∈ U. Consider a new frame s1 = hs0. Then for all u ∈ Γ(L)U , we have that u = gus1. It follows that fu = hgu, in particular fhµ,ζi = hghµ,ζi . Deﬁning µs1 in the same way as µs0 , we get that

fhµ,ζi(x) = hghµ,ζi

hµs0 (x), ζi = hhµs1 (x), ζi

µs0 = hµs1 .

Note that a weakly Hamiltonian action is deﬁned in a way that is independent of the trivialization. Proposition 5.24 shows how the moment maps in diﬀerent trivializations are related. This result applied to LCS manifolds coincides with [23, Remark 3.10].

5.3 Hamiltonian actions

Deﬁnition 5.25. Let (π : L → M, {·, ·}) be a Jacobi manifold and let g be a ﬁnite dimensional Lie algebra. We say that the action g y (Γ(L), {·, ·}) is Hamiltonian if

1. the action g y (Γ(L), {·, ·}) is weakly Hamiltonian, with moment map µ, and

2. the comoment map µ is a Lie algebra morphism, i.e.,

µ([ζ, η]) = {µ(ζ), µ(η)},

for all ζ, η ∈ g.

Directly from Deﬁnition 5.25, we see that an action g y (Γ(L), {·, ·}) is Hamiltonian if the following diagram commutes

ϕ g DerJac(L) µ ∆•

(Γ(L), {·, ·}) where ϕ is the inﬁnitesimal action, and ∆• is the map u 7→ ∆u = {u, ·}.

Remark 5.26. Note that (π : g∗ ⊗L → M) is a vector bundle. Then a moment map µ is a section of (π : g∗ ⊗ L → M). There is an induced action of G on (π : g∗ ⊗ L → M), given by

g∗⊗L ∗ Ag (β ⊗ l) = Adgβ ⊗ Agl, for all g ∈ G, β ∈ g∗, and l ∈ L. 86 5 Group actions on Jacobi manifolds

Lemma 5.27. Let π(L → M, {·, ·}) be a Jacobi bundle and let g y (L → M, {·, ·}) be a weakly Hamiltonian action . Then for all g ∈ G and ζ ∈ g

L g∗⊗L Ag hAg µ(x), ζi = hµ(x), Adg−1 ζi,

L g∗⊗L ∗ where Ag and Ag are the action of g ∈ G on sections of L and Γ(g ⊗ L), respectively. Pn ∗ Proof. Fix x ∈ M, let µ(x) = j βj(x) ⊗ u(x), where βj ∈ Γ(g ) for all j = 1, . . . , n, n = dim g, ∗ such that {β1(x), . . . , βn(x)} form a base of g for all x ∈ M, and u ∈ Γ(L). It suﬃces to prove the result on µ(x) = β(x) ⊗ u(x), as the same reasoning can be applied to µ(x)j = βj(x) ⊗ u(x) for j = 1, . . . , n. Fix g ∈ G and ζ ∈ g, we get

g∗⊗L ∗ hAg µ(x), ζi = hAdgβ(x), ζi ⊗ Ag1 (u(x)). We consider the action of g−1 on both sides

L g∗⊗L L ∗ Ag−1 hAg µ(x), ζi = Ag−1 hAdgβ(x), ζi ⊗ Ag−1 (u(x)) . Using the linearity of the action of g−1:

L g∗⊗L ∗ ∗ Ag−1 hAg µ(x), ζi = hAdg−1 Adgβ(x), ζi ⊗ AgAg−1 (u(x)) = hβ(x), Adg−1 ζi ⊗ u(x). We conclude that L g∗⊗L Ag hAg µ(x), ζi = hµ(x), Adg−1 ζi.

Deﬁnition 5.28. Let (L → M, {·, ·}) be a Jacobi manifold. We say that this action G y L is Hamiltonian if

1. it is weakly Hamiltonian that is a) the action G y L is Jacobi. b) the induced action g y L is weakly Hamiltonian.

2. there exists a G-equivariant moment map µ : M → g∗ ⊗ L associated to the action g y L. That is for all g ∈ G and for all x ∈ M:

g∗⊗L µ(agx) = Ag µ(x).

Corollary 5.29. A moment map µ : M → g∗ ⊗ L is G−equivariant, in the sense of Deﬁnition 5.28, if and only if µ ∈ Γ(g∗ ⊗ L) is invariant under the action of G y Γ(g∗ ⊗ L). Proof. To verify this, note that for all x ∈ M,

g∗⊗L µ(agx) = Ag µ(x) g∗⊗L Ag−1 µ(agx) = µ(x) g∗⊗L (Ag µ)(x) = µ(x). The result follows as this reasoning works in both directions. 5.3 Hamiltonian actions 87

Proposition 5.30. Let (L → M, {·, ·}) be a Jacobi manifold let G y (L → M, {·, ·}) be a Jacobi action,

1. if G y (L → M, {·, ·}) is Hamiltonian, then the induced action g y (Γ(L), {·, ·}) is Hamilto- nian.

2. if g y (Γ(L), {·, ·}) is Hamiltonian and G is connected then G y (L → M, {·, ·}) is Hamilto- nian.

Proof. Since G y (L → M, {·, ·}) is Hamiltonian, the map µ : M → g∗ ⊗ L is G− equivariant. Let ζ, η ∈ g. By Lemma 5.27, we have that

L g∗⊗L Aexp(tζ)hAexp(−tζ)µ(x), ηi = hµ(x), Adexp(tζ)ηi.

Using the G− equivariance of µ, we have

L Aexp(tζ)hµ(aexp(−tζ)x), ηi = hµ(x), Adexp(tζ)ηi.

Taking derivatives of t at t = 0, we get that

∆ζ hµ(x), ηi = hµ(x), [ζ, η]i.

It follows that

∆ζ µ(η) = {µ(ζ), µ(η)} = µ([ζ, η]). We conclude that µ : g → Γ(L) is a Lie algebra homomorphism. For the second statement, suppose that µ is a Lie algebra homomorphism. Since exp : g → G is a local diﬀeomorphism then we can consider an open set U of the identity of G such that g ∈ U. Consider the induced action on sections of exp(−tζ), note that when restricted to U we can follow the previous reasoning in the opposite direction. So g = exp(−tζ) acts in a Hamiltonian fashion. n By the fact that G is connected, we have that G = ∪n∈NU . It follows that the action of G is Hamiltonian.

We close this chapter by showing how every contact action is Hamiltonian.

Theorem 5.31. Let (M, H) be a contact manifold. Then

∼ DerJac(L → M, {·, ·}) = (Γ(L), {·, ·}), as Lie algebras.

Proof. Associated to (M, H) there is a Jacobi manifold (L → M, {·, ·}). Let ∆ ∈ DerJac(L → M, {·, ·}).

The proof proceeds in two steps. First we show that every ∆ ∈ DerJac(L → M, {·, ·}) is Hamilto- nian. This induces a bijection given by ∆ 7→ s∆ ∈ Γ(L). Then, we show that this assignment is a Lie algebra morphism. Since {f, ·} is a derivation on the module structure on Γ(L), it follows that for all u, v ∈ Γ(L) and f ∈ C∞(M)

{u, fv} = {u, v} + Ru(f)v, 88 5 Group actions on Jacobi manifolds

where Ru is the Reeb vector ﬁeld associated to u, as given by Corollary 4.49. Let u, v ∈ Γ(L) and f ∈ C∞(M), on one hand we have that:

∆{u, fv} = ∆(f){u, v} + {Ru(f), v}

= f∆{u, v} + δ∆(f){u, v} + Ru(f)∆v + δ∆(Ru(f))v

= f ({∆u, v} + {u, ∆v}) + δ∆{u, v} + Ru(f)∆v + δ∆(Ru(f))v. On the other hand, we get that ∆{u, fv} = {∆u, fv} + {u, ∆fv}

= f{∆u, v} + R∆u(f)v + {u, f∆v} + {u, δ∆(f)v}

= f{∆u, v} + R∆u(f)v + f{u, ∆v} + Ru(f)∆v + δ∆(f){u, v} + Ru(δ∆(f))v. By equaling both expressions we obtain

δ∆(Ru(f))v = R∆u(f)v + Ru(δ∆(f))v

δ∆(Ru(f)) − Ru(δ∆(f)) = R∆u(f)

[δ∆,Ru] = R∆u.

By Proposition 4.48 δ∆ = Rs∆ + Y , where Rs∆ is the Reeb vector ﬁeld associated to a section s∆ and Y ∈ Γ(H). We compute the Jacobi bracket using the fact that Ru is a Reeb vector ﬁeld

{s∆, u} = [Rs∆ + Y,Ru] mod H

= ([Rs∆ ,Ru] + [Y,Ru]) mod H

= [Rs∆ ,Ru] mod H

= R∆u mod H. It follows that

{s∆, u} = ∆u. Therefore every Jacobi derivation of a contact manifold is Hamiltonian.

The assignment ∆ 7→ s∆ is surjective, as there is a Hamiltonian derivation associated to any section. To check injectivity, let s∆, r∆ ∈ Γ(L) be two sections associated with the same derivation ∆. We compute

{s∆ − r∆, u} = {s∆, u} − {r∆, u} = ∆u − ∆u = 0. Then the given assignment is bijective.

All that is left is to prove that it is a Lie algebra isomorphism. For j = 1, 2, let ∆j ∈ DerJac(L → M, {·, ·}), and let sj be its associated section. Using the Jacobi identity of {·, ·}, we get for all u ∈ Γ(L)

{{s∆1 , s∆2 }, u} = {s∆1 , {s∆2 , u}} − {s∆2 , {s∆1 , u}}

= {s∆1 , ∆2u} − {s∆2 , ∆1u}

= ∆1∆2u − ∆2∆1u

= [∆1, ∆2]u. 5.3 Hamiltonian actions 89

The map ∆ 7→ s∆ is then a Lie algebra isomorphism.

The proof of Theorem 5.31 lies on the fact that there is in isomorphism between Reeb vector ﬁelds and sections of L.

Corollary 5.32. Every Jacobi action of a connected Lie group G on a contact manifold (M, H) is Hamiltonian.

Proof. By virtue of Theorem 5.31, we have an isomorphism H : DerJac(L → M, {·, ·}) 7→ (Γ(L), {·, ·}). Let ϕ : g 7→ DerJac(L → M, {·, ·}), the inﬁnitesimal action of the Lie algebra. Then there is a moment map given by µ := H ◦ ϕ. Since µ : g → (Γ(L), {·, ·}) is the composition of Lie algebra homomorphisms, for all ζ, η ∈ g

µ([ζ, η]) = {µ(ζ), µ(η)}.

Using Proposition 5.30 and the fact that G is connected, we conclude that the action of G is Hamiltonian. 6 Closing remarks

The present work shows how Jacobi geometry generalizes Poisson, LCS, and Contact manifolds. This generalization uses the language of line bundles to account for any conformal symmetries, without a choice of frame. There are many advantages to this abstract line bundle approach:

• many of the results on Poisson geometry can be easily generalized, as we did in Chapter 5.

• there is no need to choose a particular frame to work on. For example, in contact geometry, the line bundle approach allows us work without worrying about the choice of contact form.

However, when studying Jacobi manifolds, one should be careful as the choice of frame shows up as a conformal factor in the symmetries and group actions. Finally, this work presents many examples that aim to bridge the more abstract language of [24] and [28], with the classical language of [17], as shown in chapters 4 and 5.

6.1 Further Work

There are some questions beyond the scope of this work:

1. Obstructions: in Poisson (and symplectic) geometry there is a ﬁrst obstruction to Hamil- 1 tonian actions given by Hπ(M). A natural question is how to recover this obstruction using the language of Jacobi bundles with an arbitrary line bundle.

2. Lie-Poisson: in Poisson geometry, there is a canonical Poisson structure on C∞(g∗). Al- though we managed to use the Lie-Poisson construction to relate Hamiltonian actions on Jacobi manifolds on trivial bundles, we did propose a similar construction in the general case. There is the problem of deﬁning a canonical Jacobi manifold structure related to g∗ ⊗L .

3. Use of Jets: in Poisson geometry the bracket can be completely encoded by a section π ∈ ∧2TM. The works of [24] and [28] show that there is an analogue way to encode the information of the Jacobi bracket using Jets and the Atiyah algebroid. In a similar fashion as the Poisson bivector was instrumental to study of obstructions, a ”Jacobi biderivation” might be key to study obstructions related to Hamiltonian actions in the general Jacobi setting. Bibliography

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