On twisted contact groupoids and on integration of twisted Jacobi manifolds

TL;DR

This paper introduces twisted contact groupoids, explores their role in integrating twisted Jacobi manifolds, and examines their relationships with twisted Poisson and symplectic structures, supported by examples.

Contribution

It extends the theory of contact and symplectic groupoids to include twisted structures and studies their integration of twisted Jacobi manifolds.

Findings

Introduction of twisted contact groupoids

Connection between twisted Jacobi and twisted Poisson manifolds

Examples illustrating each structure

Abstract

We introduce the concept of twisted contact groupoids, as an extension either of contact groupoids or of twisted symplectic ones, and we discuss the integration of twisted Jacobi manifolds by twisted contact groupoids. We also investigate the very close relationships which link homogeneous twisted Poisson manifolds with twisted Jacobi manifolds and homogeneous twisted symplectic groupoids with twisted contact ones. Some examples for each structure are presented.

Equations

$${\Lambda }^{\#}(\zeta )({\alpha }_1,\dots ,{\alpha }_k)=(-1{)}^k\zeta ({\Lambda }^{\#}({\alpha }_1),\dots ,{\Lambda }^{\#}({\alpha }_k)).$$

$$(\Lambda ,E{)}^{\#}(\zeta ,{\zeta }^{\prime })(({\alpha }_1,f_1),\dots ,({\alpha }_k,f_k))=(-1{)}^k(\zeta ,{\zeta }^{\prime })((\Lambda ,E{)}^{\#}({\alpha }_1,f_1),\dots ,(\Lambda ,E{)}^{\#}({\alpha }_k,f_k)).$$

$$(\zeta ,{\zeta }^{\prime })((X_1,f_1),\dots ,(X_k,f_k))=\zeta (X_1,\dots ,X_k)+\sum _{i=1}^k(-1{)}^{i+1}{f}_i{\zeta }^{\prime }(X_1,\dots ,{\hat{X}}_i,\dots ,X_k),$$

$$({\Lambda }^{\#}\otimes 1)(\zeta )({\alpha }_1,\dots ,{\alpha }_{k-1})(X)=(-1{)}^k\zeta ({\Lambda }^{\#}({\alpha }_1),\dots ,{\Lambda }^{\#}({\alpha }_{k-1}),X).$$

$$\frac{1}{2}[(\Lambda ,E),(\Lambda ,E){]}^{(0,1)}=(\Lambda ,E{)}^{\#}(d\omega ,\omega ).$$

$$\left\{\begin{array}[]{l}\frac{1}{2}[\Lambda,\Lambda]+E\wedge\Lambda=\Lambda^{\#}(d\omega)+\Lambda^{\#}(\omega)\wedge E\\ \\ [E,\Lambda]=(\Lambda^{\#}\otimes 1)(d\omega)(E)-((\Lambda^{\#}\otimes 1)(\omega)(E))\wedge E.\end{array}\right.$$

$$\{f,g\}=\Lambda (df,dg)+⟨fdg-gdf,E⟩,f,g\in C^{\infty }(M,\mathbb{R}),$$

$$\{f,\{g,h\}\}+c.p.=(\Lambda ,E{)}^{\#}(d\omega ,\omega )((df,f),(dg,g),(dh,h)),f,g,h\in C^{\infty }(M,\mathbb{R}).$$

$$\{(\zeta ,f),(\eta ,g){\}}^{\omega }=\{(\zeta ,f),(\eta ,g)\}+(d\omega ,\omega )((\Lambda ,E{)}^{\#}(\zeta ,f),(\Lambda ,E{)}^{\#}(\eta ,g),\cdot ),$$

$$\{(df,f),(dg,g){\}}^{\omega }=(d\{f,g\},\{f,g\})+(d\omega ,\omega )((\Lambda ,E{)}^{\#}(df,f),(\Lambda ,E{)}^{\#}(dg,g),\cdot ),$$

$${\Lambda }^a=a\Lambda ,E^a={\Lambda }^{\#}(da)+aE\mathrm{and}{\omega }^a=\frac{1}{a}\omega .$$

$$\{f,g{\}}^a=\frac{1}{a}\{af,ag\},f,g\in C^{\infty }(M,\mathbb{R}).$$

$$i(E)\vartheta =1\mathrm{and}i(E)(d\vartheta +\omega )=0,$$

$${\Lambda }^{\#}(\vartheta )=0\mathrm{and}i({\Lambda }^{\#}(\zeta ))(d\vartheta +\omega )=-(\zeta -⟨\zeta ,E⟩\vartheta ),\mathrm{for}\mathrm{all}\zeta \in \Gamma (T^{\ast }M),$$

$${\Lambda }_2={\psi }_{\ast }{\Lambda }_1,E_2={\psi }_{\ast }{E}_1\mathrm{and}{\omega }_1={\psi }^{\ast }{\omega }_2.$$

$$\frac{1}{2}[\Lambda ,\Lambda ]={\Lambda }^{\#}(d\omega ),$$

$${\mathcal{L}}_Z\Lambda =-\Lambda \mathrm{and}\omega =i(Z)d\omega .$$

$$\{f,g\}$$

$$\{f,g\}=a\varpi^{*}\big{(}\Lambda_{0}(df_{0},dg_{0})+\langle f_{0}dg_{0}-g_{0}df_{0},E_{0}\rangle\big{)}=a\varpi^{*}\{f_{0},g_{0}\}_{0},$$

$$\left\{\begin{array}[]{l}\frac{1}{2}[\Lambda^{a},\Lambda^{a}]+E^{a}\wedge\Lambda^{a}=(\Lambda^{a})^{\#}(d\omega^{a})+(\Lambda^{a})^{\#}(\omega^{a})\wedge E^{a}\\ \\ [E^{a},\Lambda^{a}]=((\Lambda^{a})^{\#}\otimes 1)(d\omega^{a})(E^{a})-(((\Lambda^{a})^{\#}\otimes 1)(\omega^{a})(E^{a}))\wedge E^{a}.\end{array}\right.$$

$$\left\{\begin{array}[]{l}\frac{1}{2}[\Lambda_{0},\Lambda_{0}]+E_{0}\wedge\Lambda_{0}=\Lambda_{0}^{\#}(d\omega_{0})+\Lambda_{0}^{\#}(\omega_{0})\wedge E_{0}\\ \\ [E_{0},\Lambda_{0}]=(\Lambda_{0}^{\#}\otimes 1)(d\omega_{0})(E_{0})-((\Lambda_{0}^{\#}\otimes 1)(\omega_{0})(E_{0}))\wedge E_{0},\end{array}\right.$$

$$\tilde{\Lambda }=e^{-s}(\Lambda +\frac{\partial }{\partial s}\wedge E)and\tilde{\omega }=e^s\omega ,$$

$$\vartheta =\sum _{i=1}^n{p}_{i}d{x}_i\mathrm{and}\omega =\frac{1}{2}\sum _{i,j,k,l=1}^n{p}_i{\lambda }^{ij}{\phi }_{jkl}d{x}_k\wedge d{x}_l,$$

$${\mathcal{L}}_{Z_0}{\Lambda }_0=-{\Lambda }_0-{\Lambda }_0^{\#}(d{\omega }_0(Z_0,\cdot ,\cdot )-{\omega }_0).$$

$$[E,\Lambda ]$$

$$\frac{1}{2}[{\varpi }_{\ast }\Lambda ,{\varpi }_{\ast }\Lambda ]+({\varpi }_{\ast }E)\wedge ({\varpi }_{\ast }\Lambda )$$

$$\frac{1}{2}[{\Lambda }_0,{\Lambda }_0]$$

$$\{f,g\}$$

$$(X_h-1)\{f,g\}=\{h,\{f,g\}\}$$

$$(X_h-1)f={\varpi }^{\ast }({\varpi }_{\ast }(X_h-1)f_0)={\varpi }^{\ast }((Z_0-1)f_0).$$

$$(\Lambda ,E{)}^{\#}(d\omega ,\omega )((dh,h),(df,f),(dg,g))\stackrel{(\text{ref}def-tj)}{=}(\frac{1}{2}[\Lambda ,\Lambda ]+E\wedge \Lambda )(dh,df,dg)$$

Full text

Résumé

On introduit le concept des groupoïdes de contact tordus, comme une extension à la fois des groupoïdes de contact et des groupoïdes symplectiques tordus, et on discute l’ intégration des variétés de Jacobi tordues par groupoïdes de contact tordus. En plus, on établit quelques relations étroites qui unissent les variétés de Poisson homogènes tordues avec celles de Jacobi tordues et les groupoïdes homogènes symplectiques tordus avec ceux de contact tordus. Des exemples pour chaque structure étudiée sont présentés.

**Keywords: **Twisted contact groupoid, homogeneous twisted symplectic groupoid, twisted Jacobi manifold.

A.M.S. classification (2000): 53D10, 53D17, 58Hxx, 53C15.

1 Introduction

The notion of twisted Jacobi manifolds, introduced by J.M. Nunes da Costa and the author of this present paper in [33], is a weakened version of the notion of Jacobi manifolds due to A. Lichnerowicz [29] which, at the same time, generalizes in a natural manner that of twisted Poisson manifolds studied by P. Ševera and A. Weinstein [39]. The latter appeared in Park’s work on string theory [35], as well as in the work [24] on topological field theory by C. Klimčík and T. Ströbl, and are also relevant in the theory of nonholonomic systems [21]. In fact, the associated bracket on the space of smooth functions on the manifold does not satisfy the Jacobi identity whose failure is controlled by a generalized closed $3$-form. Keeping in mind that the Jacobi structures play a central role in the geometric prequantization of Poisson structures [8, 42, 13], one of the motivations behind the study of twisted Jacobi structures is the likely role that they can play in some geometric prequantization process of twisted Poisson structures.

However, the prequantization of Poisson manifolds is also closely related to the integrability of Jacobi structures [13]. For this reason, the present paper is concerned with the search of the global geometric objects that will integrate twisted Jacobi structures.

The corresponding problem for Jacobi manifolds was treated, independently, by P. Libermann [27], C. Albert [1] and Y. Kerbrat and Z. Souici-Benhammadi [23]. For its approach, they introduced various concepts of contact groupoids, as odd-dimensional counterparts of symplectic groupoids [40]. In [17], P. Dazord unified and clarified these concepts and he showed contact groupoids as the central tool for integrating local Lie algebras (it is well known that they are intimately connected with Jacobi structures [19]) and prequantizing Poisson manifolds in Weinstein’s sense. While, in [13], M. Crainic and Ch. Zhu discussed this problem using the method of $A$-paths. The analogous question for twisted Poisson manifolds was studied by A.S. Cattaneo and P. Xu [6] and, in the general context of twisted Dirac structures, by H. Bursztyn et al. [3]. They presented the notion of twisted (pre)symplectic groupoids as a natural extension of the one of symplectic groupoids [40] and they proved that the twisted Poisson manifolds may be regarded as the infinitesimal form of twisted symplectic groupoids.

Inspired by the above studies, we introduce the concept of twisted contact groupoids, we establish their fundamental properties (Proposition 3.2), and we prove that these are the global counterparts of twisted Jacobi manifolds (Theorems 3.3 and 4.7). For this, we use the bijection which exists between twisted contact groupoids and homogeneous twisted symplectic ones (Proposition 3.5). The latter are a special case of twisted symplectic groupoids and they integrate homogeneous twisted Poisson manifolds that are linked to twisted Jacobi manifolds with a very close relationships (Propositions 2.3 and 2.5).

The paper is organized as follows. In Section 2 we review basic definitions and results concerning twisted Jacobi manifolds and homogeneous twisted Poisson ones and we give some examples of twisted contact structures. In Section 3, we introduce the notion of contact groupoids, we describe their basic properties and their connection with the homogeneous twisted symplectic groupoids. Finally, we present some examples of such groupoids. In Section 4, after a brief presentation of the integrability problem of a Lie algebroid and of its solution given by M. Crainic and R.L. Fernandes [10], we prove our main theorem: An integrable twisted Jacobi manifold is integrated by a twisted contact groupoid.

Notation: In order to present the concepts of twisted Jacobi and homogeneous twisted Poisson structures on a smooth manifold $M$, we recall that, given a bivector field $\Lambda$ on $M$, the usual homomorphism of $C^{\infty}(M,\mathbb{R})$-modules $\Lambda^{\#}:\Gamma(T^{*}M)\to\Gamma(TM)$, defined, for all $\zeta,\eta\in\Gamma(T^{*}M)$, by $\langle\eta,\Lambda^{\#}(\zeta)\rangle=\Lambda(\zeta,\eta)$, can be extended to a homomorphism, also denoted $\Lambda^{\#}$, from $\Gamma(\bigwedge^{k}T^{*}M)$ to $\Gamma(\bigwedge^{k}TM)$, $k\in\mathbb{N}$, by setting, for any $f\in C^{\infty}(M,\mathbb{R})$, $\Lambda^{\#}(f)=f$, and, for all $\zeta\in\Gamma(\bigwedge^{k}T^{*}M)$ and $\alpha_{1},\ldots,\alpha_{k}\in\Gamma(T^{*}M)$,

[TABLE]

Moreover, for any $(\Lambda,E)\in\Gamma(\bigwedge^{2}(TM\times\mathbb{R}))$, the homomorphism of $C^{\infty}(M,\mathbb{R})$-modules $(\Lambda,E)^{\#}:\Gamma(T^{*}M\times\mathbb{R})\to\Gamma(TM\times\mathbb{R})$ given, for any $(\zeta,f)\in\Gamma(T^{*}M\times\mathbb{R})$, by $(\Lambda,E)^{\#}(\zeta,f)=(\Lambda^{\#}(\zeta)+fE,-\langle\zeta,E\rangle)$, can be extended to a homomorphism $(\Lambda,E)^{\#}:\Gamma(\bigwedge^{k}(T^{*}M\times\mathbb{R}))\to\Gamma(\bigwedge^{k}(TM\times\mathbb{R}))$ by setting, for all $(\zeta,\zeta^{\prime})\in\Gamma(\bigwedge^{k}(T^{*}M\times\mathbb{R}))$ and $(\alpha_{1},f_{1}),\ldots,(\alpha_{k},f_{k})\in\Gamma(T^{*}M\times\mathbb{R})$,

[TABLE]

We indicate [22], for any $(X_{1},f_{1}),\ldots,(X_{k},f_{k})\in\Gamma(\bigwedge^{k}(TM\times\mathbb{R}))$,

[TABLE]

where the hat denotes missing arguments. Also, following [39], we denote by $(\Lambda^{\#}\otimes 1)(\zeta)$ the section of $(\bigwedge^{k-1}TM)\otimes T^{*}M$ that acts on multivector fields by contraction with the factor in $T^{*}M$. Precisely, for all $X\in\Gamma(TM)$ and $\alpha_{1},\ldots,\alpha_{k-1}\in\Gamma(T^{*}M)$,

[TABLE]

2 Twisted Jacobi and homogeneous twisted Poisson manifolds

A twisted Jacobi manifold is a smooth manifold $M$ endowed with a bivector field $\Lambda$, a vector field $E$ and a $2$-form $\omega$ such that

[TABLE]

The bracket on the left hand side is the Schouten bracket of the Lie algebroid $(TM\times\mathbb{R},[\cdot,\cdot],\pi)$ over $M$ modified by the 1-cocycle $(0,1)$ of its Lie algebroid cohomology complex with trivial coefficients [22]. The pair $(d\omega,\omega)\in\Gamma(\bigwedge^{3}(T^{*}M\times\mathbb{R}))$ can be viewed as a closed $3$-form of $TM\times\mathbb{R}$ with respect to the exterior derivative operator on $\Gamma(\bigwedge(T^{*}M\times\mathbb{R}))$ defined by $([\cdot,\cdot],\pi)$ and modified by $(0,1)$ [22]. Writing (2) in terms of the usual Schouten bracket, we obtain [33] its equivalent expression

[TABLE]

As explained in [33], the space $C^{\infty}(M,\mathbb{R})$ of smooth functions on $(M,\Lambda,E,\omega)$ is equipped, just as in the case of ordinary Jacobi manifolds, with the internal composition law

[TABLE]

that is bilinear and skew-symmetric but its Jacobi identity acquires an extra term:

[TABLE]

So, (4) is no more a Lie bracket on $C^{\infty}(M,\mathbb{R})$. However, $(\Lambda,E,\omega)$ produces a Lie algebroid structure $(\{\cdot,\cdot\}^{\omega},\pi\circ(\Lambda,E)^{\#})$ on the vector bundle $T^{*}M\times\mathbb{R}\to M$. The bracket on the space $\Gamma(T^{*}M\times\mathbb{R})$ of smooth sections of $T^{*}M\times\mathbb{R}$ is given, for all $(\zeta,f),(\eta,g)\in\Gamma(T^{*}M\times\mathbb{R})$, by

[TABLE]

$\{\cdot,\cdot\}$ being the Kerbrat-Souici-Benhammadi bracket [23] on $\Gamma(T^{*}M\times\mathbb{R})$, and the anchor map is $\pi\circ(\Lambda,E)^{\#}$, where $\pi:TM\times\mathbb{R}\to TM$ denotes the projection on the first factor. The section $(-E,0)$ of $TM\times\mathbb{R}$ is a $1$-cocycle of the Lie algebroid cohomology complex with trivial coefficients [22] of $(T^{*}M\times\mathbb{R},\{\cdot,\cdot\}^{\omega},\pi\circ(\Lambda,E)^{\#})$. The brackets (4) and (6) are related by

[TABLE]

whence we conclude that the mapping $f\mapsto X_{f}=\Lambda^{\#}(df)+fE=\{f,\cdot\}+\langle df,E\rangle$ from functions to their hamiltonian vectors fields is no longer a homomorphism.

If $E=0$, equations (3) are reduced to $\frac{1}{2}[\Lambda,\Lambda]=\Lambda^{\#}(d\omega)$, which means that $(\Lambda,d\omega)$ defines an exact twisted Poisson structure [39] on $M$.

Two characteristic examples of twisted Jacobi manifolds are the following.

Conformal twisted Jacobi manifolds ([33]): Let $(M,\Lambda,E,\omega)$ be a twisted Jacobi manifold and $a$ an element of $C^{\infty}(M,\mathbb{R})$ that never vanishes on $M$. We set

[TABLE]

Then, $(\Lambda^{a},E^{a},\omega^{a})$ defines a new twisted Jacobi structure on $M$ that is called $a$-conformal to the initially given one. Its associated bracket (4) in $C^{\infty}(M,\mathbb{R})$ is given by

[TABLE]

Twisted contact manifolds ([34]): A twisted contact manifold is a $2n+1$-dimensional smooth manifold $M$ equipped with an $1$-form $\vartheta$ and a $2$-form $\omega$ such that $\vartheta\wedge(d\vartheta+\omega)^{n}\neq 0$, everywhere in $M$. The Reeb vector field $E$ on $M$, defined by

[TABLE]

and the bivector field $\Lambda$ on $M$ whose associated morphism $\Lambda^{\#}$ is given by

[TABLE]

yield a $(d\omega,\omega)$-twisted Jacobi structure on $M$.

The above definition includes that $M$ is an orientable manifold and $\vartheta\wedge(d\vartheta+\omega)^{n}$ is a volume form on $M$. Thus, $\ker\vartheta$ and $\ker(d\vartheta+\omega)$ are complementary subbundles of $TM$, where $\ker(d\vartheta+\omega)$, called the vertical bundle, is of rank 1 and is generated by $E$, however, $\ker\vartheta=\mathrm{Im}\Lambda^{\#}$, called the horizontal bundle and denoted by $\mathcal{H}$, is of rank $2n$. If $M$ is not orientable, a twisted contact structure on $M$ cannot be defined by a single pair $(\vartheta,\omega)$ defined on the whole of $M$. It is described by an open covering $U_{i}$, $i\in I$, of $M$ such that each $U_{i}$ is endowed with a twisted contact structure $(\vartheta_{i},\omega_{i})$ and on the overlaps $U_{i}\cap U_{j}$ there exist nowhere vanishing functions $f_{ij}$ satisfying $\vartheta_{i}=f_{ij}\vartheta_{j}$ and $\omega_{i}=f_{ij}\omega_{j}$. Hence, the family $(U_{i},\vartheta_{i},\omega_{i})$, $i\in I$, defines a locally conformal twisted contact structure on $M$. Of course, in this case, the above decomposition of $TM$ holds locally.

Remark 2.1

We recall that an almost cosymplectic structure [25, 29] on a $2n+1$-dimensional smooth manifold $M$ is defined by a pair $(\vartheta,\Theta)$, where $\vartheta$ is a $1$-form and $\Theta$ is a $2$-form on $M$, such that $\vartheta\wedge\Theta^{n}\neq 0$ everywhere on $M$. So, a twisted contact structure $(\vartheta,\omega)$ on $M$ can be viewed as an almost cosymplectic for which $\Theta=d\vartheta+\omega$. Reciprocally, each almost cosymplectic manifold $(M,\vartheta,\Theta)$ can be considered as twisted contact with $\omega=\Theta-d\vartheta$.**

Some examples of twisted contact manifolds: In [26], P. Libermann gave a list of examples of almost cosymplectic structures: (i) on the sphere $S^{5}$ by embedding it in the almost complex (so, almost symplectic) manifold $S^{6}$ and (ii) on the quadrics homeomorphic to $S^{1}\times\mathbb{R}^{4}$, $S^{2}\times\mathbb{R}^{3}$ and $S^{3}\times\mathbb{R}^{2}$ by embedding they in the quadrics homeomorphic to $S^{2}\times\mathbb{R}^{4}$ and $S^{3}\times\mathbb{R}^{3}$ which are also endowed with almost complex structures. From Remark 2.1, it is clear that the almost cosymplectic structures obtained on $S^{5}$ and on the $5$-dimensional quadratics can be viewed as twisted contact.

Let $(M_{1},\Lambda_{1},E_{1},\omega_{1})$ and $(M_{2},\Lambda_{2},E_{2},\omega_{2})$ be two twisted Jacobi manifolds. A smooth map $\psi:M_{1}\to M_{2}$ is called twisted Jacobi map if it satisfies the following conditions

[TABLE]

Moreover, we say that $\psi:M_{1}\to M_{2}$ is a conformal twisted Jacobi map if there exists a non vanishing smooth function $a$ on $M_{1}$ such that $\psi$ is a twisted Jacobi map between $(M_{1},\Lambda_{1}^{a},E_{1}^{a},\omega_{1}^{a})$ and $(M_{2},\Lambda_{2},E_{2},\omega_{2})$.

A homogeneous twisted Poisson manifold [33] is an exact twisted Poisson manifold $(M,\Lambda,d\omega)$ [39],

[TABLE]

endowed with a vector field $Z$, called the homothety vector field of $M$, such that

[TABLE]

Now, we establish a one to one correspondence between twisted Jacobi and homogeneous twisted Poisson manifolds, as in the non-twisted framework [16].

Proposition 2.2

Let $(M,\Lambda,d\omega,Z)$ be a homogeneous twisted Poisson manifold and $M_{0}$ a $1$-codimensional submanifold of $M$ transverse to $Z$. Then, $M_{0}$ receives an induced twisted Jacobi structure $(\Lambda_{0},E_{0},\omega_{0})$ characterized by one of the following properties:

For any pair $(f,g)$ of homogeneous functions of degree $1$ with respect to $Z$, defined on an open $\mathcal{O}$ of $M$, the bracket $\{f,g\}$ is also a homogeneous function of degree 1 with respect to $Z$. 2. 2.

Let $\varpi:U\to M_{0}$ be the projection, along the integral curves of $Z$, of a tubular neighborhood $U$ of $M_{0}$ in $M$ onto $M_{0}$ and $a$ a function on $U$ that is equal to $1$ on $M_{0}$ and homogeneous of degree $1$ with respect to $Z$. Then, the projection $\varpi$ is an $a$-conformal twisted Jacobi map.

Proof. We have that any homogeneous function $f\in C^{\infty}(U,\mathbb{R})$ of degree $1$ with respect to $Z$ is of type $f=a\varpi^{*}f_{0}$, where $f_{0}\in C^{\infty}(M_{0},\mathbb{R})$. Hence, $df=a\varpi^{*}df_{0}+\varpi^{*}f_{0}da$. Let $(f,g)$ be a pair of homogeneous functions of degree $1$ with respect to $Z$, defined on an open $\mathcal{O}\subset U$ of $M$, and $(f_{0},g_{0})$ the corresponding pair of functions on $M_{0}$, i.e., $f=a\varpi^{*}f_{0}$ and $g=a\varpi^{*}g_{0}$. Then,

[TABLE]

Since $\mathcal{L}_{Z}(a\Lambda)=0$ and $\mathcal{L}_{Z}(\Lambda^{\#}(da))=0$, the tensors fields $a\Lambda$ and $\Lambda^{\#}(da)$ are projectable along the integral curves of $Z$. Let $\Lambda_{0}=\varpi_{*}(a\Lambda)$ and $E_{0}=\varpi_{*}(\Lambda^{\#}(da))$ be their projections and $\{\cdot,\cdot\}_{0}$ the bracket (4) defined by $(\Lambda_{0},E_{0})$. Thus, (10) can be written as

[TABLE]

which means that $\{f,g\}$ is also homogeneous with respect to $Z$.

We regard $(\Lambda,d\omega)$ as the twisted Jacobi structure $(\Lambda,0,\omega)$ on $M$ and we consider its $a$-conformal structure $(\Lambda^{a},E^{a},\omega^{a})$, $\Lambda^{a}=a\Lambda$, $E^{a}=\Lambda^{\#}(da)$ and $\omega^{a}=\frac{1}{a}\omega$, which is also twisted Jacobi. Thus, the following equations hold.

[TABLE]

Since $i(Z)d\omega=\omega$ and $\mathcal{L}_{Z}a=a$, we have $\mathcal{L}_{Z}\omega^{a}=0$ and $i(Z)\omega^{a}=0$, which mean that $\omega^{a}$ is projectable along the integral curves of $Z$. Let $\omega_{0}$ be the $2$-form on $M_{0}$ for which $\omega^{a}=\varpi^{*}\omega_{0}$. So, $d\omega^{a}=\varpi^{*}d\omega_{0}$. Also, the equation $[Z,[\Lambda^{a},\Lambda^{a}]]=0$ means that $[\Lambda^{a},\Lambda^{a}]$ is projectable onto $M_{0}$ with respect to $Z$. Its projection is the Schouten bracket of the projection of $\Lambda^{a}$ with itself. Hence, by projecting the system (11) along the integral curves of $Z$ and taking into account that $\Lambda_{0}=\varpi_{*}(a\Lambda)$ and $E_{0}=\varpi_{*}(\Lambda^{\#}(da))$, we obtain

[TABLE]

that signify that $(\Lambda_{0},E_{0},\omega_{0})$ is a twisted Jacobi structure on $M_{0}$ and $\varpi:(M,\Lambda,0,\omega)\to(M_{0},\Lambda_{0},E_{0},\omega_{0})$ is an $a$-conformal twisted Jacobi map. $\blacklozenge$

Proposition 2.3** ([33])**

Let $(\Lambda,E,\omega)$ be a twisted Jacobi structure on a smooth manifold $M$. Its twisted poissonization defines a homogeneous twisted Poisson structure on $\tilde{M}=M\times\mathbb{R}$ whose the bivector field $\tilde{\Lambda}$ and the $2$-form $\tilde{\omega}$ are given, respectively, by

[TABLE]

and the homothety vector field is $\frac{\partial}{\partial s}$, where $s$ is the canonical coordinate on $\mathbb{R}$. The canonical projection $\varpi:M\times\mathbb{R}\to M$ is a $e^{s}$-conformal twisted Jacobi map and the induced twisted Jacobi structure on $M$, viewed as the submanifold $M\times\{0\}$ of $\tilde{M}$, in the sense of Proposition 2.2, is the initial given one.

Corollary 2.4

When the structure $(\Lambda,E)$ on $M$ comes from a $\omega$-twisted contact form $\vartheta$ on $M$, then $\tilde{\Lambda}$ is the inverse of the twisted symplectic form $\tilde{\Omega}=d(e^{s}\varpi^{*}\vartheta)+e^{s}\varpi^{*}\omega$ on $\tilde{M}$, which is homogeneous with respect to $\frac{\partial}{\partial s}$, and reciprocally.

By applying the above results, we may construct another example of twisted contact manifold.

The cosphere bundle of a twisted Poisson manifold. Let $(M,\Lambda,\varphi)$ be a twisted Poisson manifold. As D. Roytenberg has observed [38], the cotangent bundle $T^{\ast}M$ of $M$ is endowed with a nondegenerate $d\omega$-twisted Poisson structure which corresponds to a $d\omega$-twisted symplectic structure $\Omega=d\vartheta+\omega$, $\vartheta$ being the Liouville form on $T^{\ast}M$. In a local coordinates system $(x_{1},\ldots,x_{n},p_{1},\ldots,p_{n})$ of $T^{\ast}M$,

[TABLE]

where $\lambda^{ij}$ and $\varphi_{jkl}$ are, respectively, the local components of $\Lambda$ and $\varphi$. We remark that $(\Omega,\omega)$ is homogeneous with respect to the Liouville vector field $Z=\sum_{i=1}^{n}p_{i}\frac{\partial}{\partial p_{i}}$ of $T^{\ast}M$. Consider the action $\Phi$ of the multiplicative group $\mathbb{R}_{+}=(0,+\infty)$ by dilations on the fibers of $T^{\ast}M\setminus\{0\}$, i.e., for any $x\in M$ and $z\in T_{x}^{\ast}M$, $\Phi(s,z)=sz$. The cosphere bundle $S^{\ast}M$ of $M$ is the quotient manifold $(T^{\ast}M\setminus\{0\})/\mathbb{R}_{+}$. By working as in [37], we prove that the cone $S^{\ast}M\times\mathbb{R}_{+}$ over $S^{\ast}M$ equipped with the homogeneous twisted symplectic structure $(d(s\vartheta)+s\omega,s\frac{\partial}{\partial s})$, $s\in\mathbb{R}_{+}$, is twisted symplectomorphic to $(T^{\ast}M\setminus\{0\},\Omega,Z)$. Thus, $S^{\ast}M$ receives a twisted contact structure.

We continue with the study of the projection of a twisted Jacobi structure along the integral curves of its vector field.

Proposition 2.5

Let $(M,\Lambda,E,\omega)$ be a twisted Jacobi manifold and $M_{0}$ a submanifold of $M$, of codimension $1$, transverse to $E$. We denote by $\varpi:U\to M_{0}$ the projection onto $M_{0}$ of a tubular neighbourhood $U$ of $M_{0}$ in $M$ along the integral curves of $E$, i.e., for any $x_{0}\in M_{0}$, $\varpi^{-1}(x_{0})$ is a connected arc of the integral curve of $E$ through $x_{0}$, and by $\eta$ the $1$-form along $M_{0}$ that verifies $i(E)\eta=1$ and $i(X)\eta=0$, for any vector field $X$ on $M$ tangent to $M_{0}$, and we set $Z_{0}=\Lambda^{\#}(\eta)$. If $\omega=\varpi^{*}\omega_{0}$, where $\omega_{0}$ is a $2$-form on $M_{0}$, then, there exists on $M_{0}$ a unique exact twisted Poisson structure $(\Lambda_{0},d\omega_{0})$ such that $\varpi:U\to M_{0}$ is a twisted Jacobi map and

[TABLE]

Moreover, $(\Lambda_{0},d\omega_{0},Z_{0})$ is homogeneous if and only if $\omega_{0}=d\omega_{0}(Z_{0},\cdot,\cdot)$.

Proof. Since $(\Lambda,E,\omega)$ defines a twisted Jacobi structure on $M$, (3) holds on $M$. Thus, because of $\omega=\varpi^{*}\omega_{0}$ and $\varpi_{*}E=0$,

[TABLE]

which means that $\Lambda$ is projectable along the integral curves of $E$ onto $M_{0}$. Let $\Lambda_{0}=\varpi_{*}\Lambda$ be its projection. Furthermore, (13) implies that $[E,[\Lambda,\Lambda]]=0$, whence we get that $[\Lambda,\Lambda]$ is also projectable onto $M_{0}$. Its projection $\varpi_{*}[\Lambda,\Lambda]$ is the Schouten bracket of the projection $\varpi_{*}\Lambda$ of $\Lambda$ with itself. Thus, by projecting the first equation of the system (3) parallel to the integral curves of $E$, we obtain

[TABLE]

whence we deduce that $(\Lambda_{0},d\omega_{0})$ endows $M_{0}$ with an exact twisted Poisson structure and that $\varpi:U\to M_{0}$ is a twisted Jacobi map.

Now, by integrating along the integral curves of $E$ and by restricting, if necessary, the tubular neighbourhood $U$ of $M_{0}$ in $M$, we can construct a function $h$ on $U$ such that $h|_{M_{0}}=0$ and $i(E)dh=1$, hence $dh|_{M_{0}}=\eta$. Let $X_{h}=\Lambda^{\#}(dh)+hE$ be the hamiltonian vector field of $h$ with respect to $(\Lambda,E)$. Since $[E,\Lambda](dh,\cdot)=0$, $[E,X_{h}]=E$, which says that $X_{h}$ is projectable onto $M_{0}$ along the integral curves of $E$. Let $Z_{0}=\varpi_{*}(X_{h})=\varpi_{*}(\Lambda^{\#}(dh))$ be its projection which coincides with the restriction $X_{h}|_{M_{0}}=\Lambda^{\#}(\eta)$ of $X_{h}$ on $M_{0}$.

In order to establish (12), we denote by $\{\cdot,\cdot\}$ the composition law (4) on $C^{\infty}(M,\mathbb{R})$ determined by $(\Lambda,E)$ and by $\{\cdot,\cdot\}_{0}$ the composition law in $C^{\infty}(M_{0},\mathbb{R})$ defined by $\Lambda_{0}$. Let $(f_{0},g_{0})$ be a pair of smooth functions on $M_{0}$ and $(f,g)=(\varpi^{*}f_{0},\varpi^{*}g_{0})$ the associated pair of smooth functions on $M$ that are constant along the integral curves of $E$. We have

[TABLE]

On the other hand,

[TABLE]

But, for any $f\in C^{\infty}(M,\mathbb{R})$ of type $f=\varpi^{*}f_{0}$ with $f_{0}\in C^{\infty}(M_{0},\mathbb{R})$,

[TABLE]

Also,

[TABLE]

Thus, taking into account (15), (17) and (18), (16) takes the form

[TABLE]

whence we deduce (12). Also, according to (9) and (12), we get that $(\Lambda_{0},d\omega_{0})$ is homogeneous with respect to $Z_{0}$ if and only if $\omega_{0}=d\omega_{0}(Z_{0},\cdot,\cdot)$. $\blacklozenge$

3 Twisted contact and homogeneous twisted symplectic groupoids

We recall some basic results about the Lie groupoids which are needed in below and we fix our notations. For more detailed information, we suggest the following standard references [31, 4, 18].

Let $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ be a Lie groupoid with source map $\alpha$ and target map $\beta$. We denote by

$m:\Gamma_{2}\to\Gamma$ the product map on the set of composable pairs $\Gamma_{2}=\{(g,h)\in\Gamma\times\Gamma\,/\,\alpha(g)=\beta(h)\}$ of $\Gamma\times\Gamma$, $(g,h)\mapsto m(g,h)=g\cdot h$, which is compatible with $\alpha$ and $\beta$, i.e., for any $(g,h)\in\Gamma_{2}$, $\alpha(g\cdot h)=\alpha(h)$ and $\beta(g\cdot h)=\beta(g)$, and associative, i.e., for any triple $(g,h,k)$ of composable elements, $(g\cdot h)\cdot k=g\cdot(h\cdot k)$;

• -

$\varepsilon:\Gamma_{0}\hookrightarrow\Gamma$ the embedding of $\Gamma_{0}$ into $\Gamma$, called the unit section of $\Gamma$, which associates a unit with each element of $\Gamma_{0}$222For this reason, $\Gamma_{0}$ called, also, manifold of units., i.e., for any $g\in\Gamma$, $g\cdot\varepsilon(\alpha(g))=g=\varepsilon(\beta(g))\cdot g$, and satisfies $\alpha\circ\varepsilon=\beta\circ\varepsilon=\mathrm{Id}_{\Gamma_{0}}$;

• -

$\iota:\Gamma\to\Gamma$ the inversion map, $g\mapsto\iota(g)=g^{-1}$, which has the following properties: for any $g\in\Gamma$, $\alpha(g^{-1})=\beta(g)$, $\beta(g^{-1})=\alpha(g)$, $g^{-1}\cdot g=\varepsilon(\alpha(g))$ and $g\cdot g^{-1}=\varepsilon(\beta(g))$.

For each $g\in\Gamma$, the maps $L_{g}:\beta^{-1}(\alpha(g))\to\beta^{-1}(\beta(g))$, $h\mapsto L_{g}(h)=g\cdot h$, and $R_{g}:\alpha^{-1}(\beta(g))\to\alpha^{-1}(\alpha(g))$, $h\mapsto R_{g}(h)=h\cdot g$, are smooth diffeomorphisms and they are called the left translation and the right translation by $g$, respectively. We denote by $C_{L}^{\infty}(\Gamma,\mathbb{R})$ (resp. $C_{R}^{\infty}(\Gamma,\mathbb{R})$) the set of left (resp. right) invariant functions on $\Gamma$, i.e., the set of smooth functions on $\Gamma$ that are invariant by the left (resp. right) translations. It is easy to verify that $C_{L}^{\infty}(\Gamma,\mathbb{R})=\alpha^{\ast}C^{\infty}(\Gamma_{0},\mathbb{R})$ (resp. $C_{R}^{\infty}(\Gamma,\mathbb{R})=\beta^{\ast}C^{\infty}(\Gamma_{0},\mathbb{R})$). Also, we say that a vector field $X$ on $\Gamma$ is a left invariant vector field (resp. right invariant vector field) if it is tangent to $\beta$-fibers, so $\beta_{\ast}(X)=0$, (resp. tangent to $\alpha$-fibers, so $\alpha_{\ast}(X)=0$) and invariant under the left translations, i.e., $L_{g\ast}(X(h))=X(g\cdot h)$, (resp. the right translations, i.e., $R_{g\ast}(X(h))=X(h\cdot g)$).

Let $A(\Gamma)=\ker\beta_{\ast}\cap T_{\Gamma_{0}}\Gamma$ be the vector bundle over $\Gamma_{0}$ consisting of tangent spaces to $\beta$-fibers at the points of $\Gamma_{0}$. By left translations, each section of $A(\Gamma)$ gives rise to a unique left invariant vector field on $\Gamma$. Hence, the space of smooth sections $\Gamma(A(\Gamma))$ of $A(\Gamma)$ may be identified with the space of smooth left invariant vector fields on $\Gamma$ which is closed with respect the usual Lie bracket $[\cdot,\cdot]$ on $T\Gamma$. So, $\Gamma(A(\Gamma))$ inherits a Lie bracket $[\cdot,\cdot]$. $A(\Gamma)$ equipped with the above bracket and the restriction to $A(\Gamma)$ of the linear bundle map $T\alpha:T\Gamma\to T\Gamma_{0}$ as anchor map, becomes a Lie algebroid over $\Gamma_{0}$. $(A(\Gamma),[\cdot,\cdot],T\alpha)$ called the Lie algebroid of the Lie groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$.

Recall that a Lie algebroid [31, 18] over a smooth manifold $M$ is a real vector bundle $A\to M$ together with a Lie bracket $[\cdot,\cdot]$ on the space of smooth sections $\Gamma(A)$ and a bundle map $\rho:A\to TM$, called the anchor map, whose extension to sections satisfies the Leibniz identity

[TABLE]

Moreover, the presence of a multiplicative function $r$ on a (Lie) groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$, i.e., for any $(g,h)\in\Gamma_{2}$, $r(g\cdot h)=r(g)+r(h)$, permits us to define a left groupoid action of $\Gamma$ on the canonical projection $\varpi:\tilde{\Gamma}_{0}=\Gamma_{0}\times\mathbb{R}\to\Gamma_{0}$ defined by the map

[TABLE]

where $\Gamma\star\tilde{\Gamma}_{0}=\{(g,(x,s))\in\Gamma\times\tilde{\Gamma}_{0}\,/\,\alpha(g)=\varpi(x,s)=x\}$. Then, $\Gamma\star\tilde{\Gamma}_{0}$ itself has a (Lie) groupoid structure [31, 18] with manifold of units $\tilde{\Gamma}_{0}$ and it is called the action groupoid associated to $\mathbf{ac}^{r}$. It is easy to see that $\Gamma\star\tilde{\Gamma}_{0}$ may be identified with the product manifold $\tilde{\Gamma}=\Gamma\times\mathbb{R}$. Under this identification, the structural maps $\tilde{\alpha}$ (source) and $\tilde{\beta}$ (target) of $\tilde{\Gamma}$ are given, respectively, by

[TABLE]

The set $\tilde{\Gamma}_{2}$ of composable pairs of $\tilde{\Gamma}\times\tilde{\Gamma}$ is identified with $\Gamma_{2}\times\mathbb{R}$ via the map

[TABLE]

and the multiplication $\tilde{m}$ on $\tilde{\Gamma}_{2}$ is defined by

[TABLE]

While, the inversion map of $\tilde{\Gamma}$ is $\tilde{\iota}(g,s)=(\iota(g),s-r(g))$ and the unit section of $\tilde{\Gamma}$ is $\tilde{\varepsilon}(x,s)=(\varepsilon(x),s)$.

3.1 Twisted contact groupoids

Definition 3.1

A twisted contact groupoid is a Lie groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ of dimension $2n+1$ equipped with a $\omega$-twisted contact form $\vartheta$, $\omega$ being a $2$-form on $\Gamma$ of type $\omega=\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}$, where $\omega_{0}\in\Gamma(\bigwedge^{2}T^{\ast}\Gamma_{0})$ and $r$ is a smooth function on $\Gamma$, such that

[TABLE]

holds on the set of composable pairs $\Gamma_{2}$ of $\Gamma\times\Gamma$. In (20), $pr_{i}:\Gamma_{2}\subset\Gamma\times\Gamma\to\Gamma$ denotes the projection on the $i$-factor, $i=1,2$, of $\Gamma_{2}$.

In the next proposition we establish the basic properties of a twisted contact groupoid.

Proposition 3.2

Let $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\vartheta,\omega,r)$ be a twisted contact groupoid and $(A(\Gamma),[\cdot,\cdot],\alpha_{\ast})$ its Lie algebroid. We denote by $(\Lambda_{\Gamma},E_{\Gamma},\omega)$ the $(d\omega,\omega)$-twisted Jacobi structure on $\Gamma$ defined by $(\vartheta,\omega)$. Then

1. i)

$r$* is multiplicative, i.e., for any $(g,h)\in\Gamma_{2}$, $r(g\cdot h)=r(g)+r(h)$, it is annulated on the units of $\Gamma$, i.e., $r\circ\varepsilon=0$, and $r\circ\iota=-r$.* 2. ii)

$\iota^{*}\vartheta=-e^{r}\vartheta$. 3. iii)

$\Gamma_{0}$, identified with $\varepsilon(\Gamma_{0})$, is a Legendrian submanifold of $\Gamma$, i.e., $\dim\Gamma_{0}=n$ and $\varepsilon^{*}\vartheta=0$. 4. iv)

$E_{\Gamma}$* is a left invariant vector field and it has $r$ as first integral.* 5. v)

For any section $X_{0}^{l}$ of $A(\Gamma)\to\Gamma_{0}$, we write with $X^{l}$ (resp. $X^{r}$) the left (resp. right) invariant vector field on $\Gamma$ generated by $X_{0}^{l}$ (resp. $-\iota_{\ast}X_{0}^{l}$). Let $E_{0}^{l}$ be the section of $A(\Gamma)\to\Gamma_{0}$ that corresponds to $E_{\Gamma}$, i.e., $E_{\Gamma}=E^{l}$. Then

[TABLE] 6. vi)

$\iota:\Gamma\to\Gamma$* is a $-e^{-r}$-twisted conformal Jacobi map.* 7. vii)

A vector field $Y$ on $\Gamma$ is left (resp. right) invariant if and only if there exists $(\zeta_{0},f_{0})\in\Gamma(T^{*}\Gamma_{0}\times\mathbb{R})$ such that

[TABLE] 8. viii)

For any pair $(f_{0},g_{0})$ of smooth functions on $\Gamma_{0}$, $\{\alpha^{\ast}f_{0},e^{-r}\beta^{\ast}g_{0}\}=0$, where $\{\cdot,\cdot\}$ denotes the bracket (4) on $C^{\infty}(\Gamma,\mathbb{R})$ defined by $(\Lambda_{\Gamma},E_{\Gamma})$.

Proof. i) We consider the tangent groupoid $T\Gamma\overset{T\alpha}{\underset{T\beta}{\rightrightarrows}}T\Gamma_{0}$ of $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ [32]. The set of composable pairs in $T\Gamma\times T\Gamma$ is $T\Gamma_{2}$ with composition law $\oplus$ the tangent map $Tm$ of $m$, i.e., for any pair $(X,Y)\in T_{(g,h)}\Gamma_{2}$, where $X=\frac{dg(t)}{dt}|_{t=0}\in T_{g}\Gamma$, $Y=\frac{dh(t)}{dt}|_{t=0}\in T_{h}\Gamma$, and $(g(t),h(t))$, $t\in\mathbb{R}$, is a path in $\Gamma_{2}$ with $(g,h)=(g(0),h(0))$,

[TABLE]

From the associativity of $Tm$ and (20) we deduce the multiplicativity of $r$. Also, for any $g\in\Gamma$, we have $r(g)=r(g\cdot\varepsilon(\alpha(g)))=r(g)+r(\varepsilon(\alpha(g)))$ and $r(g)=r(\varepsilon(\beta(g))\cdot g)=r(\varepsilon(\beta(g)))+r(g)$, whence, taking into account the surjectivity of the submersions $\alpha$ and $\beta$, we get $r\circ\varepsilon=0$. Furthermore, from the relation $g^{-1}\cdot g=\varepsilon(\alpha(g))$ and the above results, we take $r(g^{-1}\cdot g)=r(\varepsilon(\alpha(g)))\Leftrightarrow r(g^{-1})+r(g)=0\Leftrightarrow(r\circ\iota)(g)=-r(g)$, for any $g\in\Gamma$. Thus, $r\circ\iota=-r$.

ii) For any $g\in\Gamma$ and $X\in\Gamma(T\Gamma)$,

[TABLE]

since $g=g\cdot\varepsilon(\alpha(g))=\varepsilon(\beta(g))\cdot g$. So, from the multiplicativity (20) of $\vartheta$, we get

[TABLE]

and

[TABLE]

whence, taking into consideration of $r\circ\varepsilon=0$ and of surjectivity of $T\alpha$ and $T\beta$, we obtain $\varepsilon^{\ast}\vartheta=0$. Moreover, from the properties of $\iota$ and (20), we get

[TABLE]

which means that $\iota^{\ast}\vartheta=-e^{r}\vartheta$.

iii) In (ii) we proved $\varepsilon^{\ast}\vartheta=0$. So, in order to establish that $\Gamma_{0}\equiv\varepsilon(\Gamma_{0})$ is a Legendrian submanifold of $\Gamma$, it is enough to show that $\dim\Gamma_{0}=n$. We consider the horizontal bundle $\mathcal{H}$ on $\Gamma$ which is of type $\mathcal{H}=\ker\vartheta$ and, for any $g\in\Gamma$, we denote by $\mathcal{H}_{g}$ the hyperplane through $g$. Because $\iota^{\ast}\vartheta=-e^{r}\vartheta$, $\mathcal{H}$ is invariant by inversion. Furthermore, for any $(X,Y)\in\mathcal{H}_{g}\times\mathcal{H}_{h}$ with $(g,h)\in\Gamma_{2}$, $m^{\ast}\vartheta_{(g,h)}(X,Y)\stackrel{{\scriptstyle(\ref{mult-cond})}}{{=}}e^{-r(h)}\vartheta_{g}(X)+\vartheta_{h}(Y)\Leftrightarrow\vartheta_{g\cdot h}(X\oplus Y)=0\Leftrightarrow X\oplus Y\in\ker\vartheta_{g\cdot h}=\mathcal{H}_{g\cdot h}$, which means that $\mathcal{H}$ is closed with respect to $Tm$. Hence, by developing the argumentation of P. Dazord [17] for the pair $(\mathcal{H},d\vartheta+\omega)$, we get $\dim\Gamma_{0}=n$.

iv) At every point of $\varepsilon(\Gamma_{0})\equiv\Gamma_{0}$, the vector field $E_{\Gamma}$ can be decomposed as $E_{\Gamma}=E_{0}^{l}+\varepsilon_{\ast}(\beta_{\ast}(E_{\Gamma}))$, where $E_{0}^{l}$ is the tangent component to the $\beta$-fibre through the considered point. Let $E^{l}$ be the left invariant vector field on $\Gamma$ generated by $E_{0}^{l}$. We have, at each $g\in\Gamma$, $E_{g}^{l}=0_{g}\oplus E^{l}_{\varepsilon(\alpha(g))}$. Therefore, $\vartheta_{g}(E^{l}_{g})=\vartheta_{g\cdot\varepsilon(\alpha(g))}(0_{g}\oplus E^{l}_{\varepsilon(\alpha(g))})\stackrel{{\scriptstyle(\ref{mult-cond})}}{{=}}e^{-r(\varepsilon(\alpha(g)))}\vartheta_{g}(0_{g})+\vartheta_{\varepsilon(\alpha(g))}(E^{l}_{\varepsilon(\alpha(g))})=\vartheta_{\varepsilon(\alpha(g))}(E^{l}_{\varepsilon(\alpha(g))})$. But, $1=\vartheta_{\varepsilon(\alpha(g))}(E_{\Gamma\varepsilon(\alpha(g))})=\vartheta_{\varepsilon(\alpha(g))}(E_{\varepsilon(\alpha(g))}^{l}+\varepsilon_{\ast}(\beta_{\ast}(E_{\Gamma}))_{\varepsilon(\alpha(g))})=\vartheta_{\varepsilon(\alpha(g))}(E_{\varepsilon(\alpha(g))}^{l})$. Consequently,

[TABLE]

By exterior differentiation of (20), we obtain

[TABLE]

and we apply (25) to the pair $(E^{l}_{g},E_{\Gamma g})$ of vectors at the point $g=g\cdot\varepsilon(\alpha(g))$. Because $r\circ\varepsilon=0$, $E_{g}^{l}=0_{g}\oplus E^{l}_{\varepsilon(\alpha(g))}$ and $E_{\Gamma g}=E_{\Gamma g}\oplus\varepsilon_{\ast}(\alpha_{\ast}(E_{\Gamma}))_{\varepsilon(\alpha(g))}$, we take

[TABLE]

We compare $d\vartheta_{\varepsilon(\alpha(g))}(E^{l}_{\varepsilon(\alpha(g))},\varepsilon_{\ast}(\alpha_{\ast}(E_{\Gamma}))_{\varepsilon(\alpha(g))})$ with $d\vartheta_{g}(E^{l}_{g},E_{\Gamma g})$. From (24) and the fact $\varepsilon^{\ast}\vartheta=0$, we get

[TABLE]

and

[TABLE]

But,

[TABLE]

So, $\vartheta_{g}([E^{l}_{g},E_{\Gamma g}])\stackrel{{\scriptstyle(\ref{mult-cond})(\ref{rel []})}}{{=}}\vartheta_{\varepsilon(\alpha(g))}([E^{l}_{\varepsilon(\alpha(g))},\varepsilon_{\ast}(\alpha_{\ast}(E_{\Gamma}))_{\varepsilon(\alpha(g))}])$, which yields

[TABLE]

Consequently, (3.1) can be written as

[TABLE]

whence, taking into account (7), we deduce that $de^{-r(\varepsilon(\alpha(g)))}(E^{l}_{\varepsilon(\alpha(g))})=0$. Hence, from the last equation and (29), we take

[TABLE]

which means that $E_{g}^{l}$ and $E_{\Gamma g}$ are parallel vectors of $T_{g}\Gamma$. Also, we have $\vartheta_{g}(E_{\Gamma g})\stackrel{{\scriptstyle(\ref{reeb})}}{{=}}1\stackrel{{\scriptstyle(\ref{theta-left})}}{{=}}\vartheta_{g}(E^{l}_{g})$. Hence, we conclude that, for any $g\in\Gamma$, $E_{\Gamma g}=E^{l}_{g}$, which means that $E_{\Gamma}$ is left invariant.

Next, we will show that $r$ is a first integral of $E_{\Gamma}$. At every point of $\varepsilon(\Gamma_{0})\equiv\Gamma_{0}$, the vector field $E_{\Gamma}$ can be decomposed as $E_{\Gamma}=E_{0}^{r}+\varepsilon_{\ast}(\alpha_{\ast}(E_{\Gamma}))$, where $E_{0}^{r}$ is the tangent component to the $\alpha$-fibre through the considered point. Let $E^{r}$ be the right invariant vector field on $\Gamma$ generated by $E_{0}^{r}$. Because, for any $g\in\Gamma$, $g=\varepsilon(\beta(g))\cdot g$ and $E^{r}_{g}=E^{r}_{\varepsilon(\beta(g))}\oplus 0_{g}$, $\vartheta_{g}(E^{r}_{g})=\vartheta_{\varepsilon(\beta(g))\cdot g}(E^{r}_{\varepsilon(\beta(g))}\oplus 0_{g})\stackrel{{\scriptstyle(\ref{mult-cond})}}{{=}}e^{-r(g)}\vartheta_{\varepsilon(\beta(g))}(E^{r}_{\varepsilon(\beta(g))})$. But, $1\stackrel{{\scriptstyle(\ref{reeb})}}{{=}}\vartheta_{\varepsilon(\beta(g))}(E_{\Gamma\varepsilon(\beta(g))})=\vartheta_{\varepsilon(\beta(g))}(E^{r}_{\varepsilon(\beta(g))})$. So, $\vartheta_{g}(E^{r}_{g})=e^{-r(g)}$. Also,

[TABLE]

whence we conclude $\mathcal{L}_{E_{\Gamma}}r=0$.

v) We consider the hamiltonian vector field $X_{e^{-r}}=\Lambda_{\Gamma}^{\#}(de^{-r})+e^{-r}E_{\Gamma}$ of $e^{-r}$ and the right invariant vector field $E^{r}$ on $\Gamma$ defined in the proof of (v). We apply (25) to the pair $(E^{r},X_{e^{-r}})$ at a point $g=\varepsilon(\beta(g))\cdot g$ of $\Gamma$ and we prove, as in (iv), that $X_{e^{-r}}=E^{r}$. Since $E_{\Gamma}=E^{l}$ at the points of $\varepsilon(\Gamma_{0})\equiv\Gamma_{0}$, we have $E_{0}^{l}=E_{0}^{r}+\varepsilon_{\ast}(\alpha_{\ast}(E_{0}^{l}))$. Thus, $0=\beta_{\ast}(E_{0}^{r}+\varepsilon_{\ast}(\alpha_{\ast}(E_{0}^{l})))=\beta_{\ast}(E_{0}^{r})+\alpha_{\ast}(E_{0}^{l})=\beta_{\ast}(E_{0}^{r})+\beta_{\ast}(\iota_{\ast}E_{0}^{l})$. Because $E_{0}^{r}+\iota_{\ast}E_{0}^{l}$ is tangent to $\alpha$-fibers, the above equation yields that $E_{0}^{r}+\iota_{\ast}E_{0}^{l}=0$, whence we deduce that $E^{r}$ coincide with the right invariant vector field on $\Gamma$ generated by $-\iota_{\ast}E_{0}^{l}$. Hence, (21) holds.

vi) It is enough to prove that $\iota_{\ast}(-e^{-r}\Lambda_{\Gamma})=\Lambda_{\Gamma}$, $\iota_{\ast}X_{-e^{-r}}=E_{\Gamma}$ and $\iota^{\ast}\omega=-e^{r}\omega$. Since $\alpha\circ\iota=\beta$, $\beta\circ\iota=\alpha$ and $r\circ\iota=-r$, we have $\iota^{\ast}\omega=\iota^{\ast}(\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0})=\beta^{\ast}\omega_{0}-e^{-r\circ\iota}\alpha^{\ast}\omega_{0}=-e^{r}\omega$. Also, $\iota_{\ast}X_{-e^{-r}}=\iota_{\ast}(-E^{r})=E^{l}$ and, for any $\zeta\in T^{\ast}_{g^{-1}}\Gamma$,

[TABLE]

whence we obtain $T\iota\circ(-e^{-r}\Lambda_{\Gamma})^{\#}\circ\,^{t}T\iota=\Lambda_{\Gamma}^{\#}$, since $\mathcal{H}=\mathrm{Im}\Lambda_{\Gamma}^{\#}$ is invariant by $\iota$.

vii) We establish (22) as in Proposition 4.3 of [17]. For the proof we use the pair $(\mathcal{H},d\vartheta+\omega)$ and the fact that $\omega=\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}$.

viii) We consider a pair $(f_{0},g_{0})$ of smooth functions on $\Gamma_{0}$ and we compute the bracket $\{\alpha^{\ast}f_{0},e^{-r}\beta^{\ast}g_{0}\}$. Since $T\beta\circ\Lambda_{\Gamma}^{\#}\circ\,^{t}T\alpha=0$ (see [17]), $\mathcal{L}_{E_{\Gamma}}r=0$ and $\beta_{\ast}(E_{\Gamma})=0$, we have

[TABLE]

But, in (v) we proved that $\Lambda_{\Gamma}^{\#}(dr)-E_{\Gamma}=-e^{r}X_{e^{-r}}$ and $X_{e^{-r}}$ is right invariant. So, $\alpha_{\ast}(\Lambda_{\Gamma}^{\#}(dr)-E_{\Gamma})=0$ and $\{\alpha^{\ast}f_{0},e^{-r}\beta^{\ast}g_{0}\}=0$. The last equation means that the twisted contact hamiltonian vector fields corresponding to the left invariant functions of $\Gamma$ have as first integrals the right invariant functions of $\Gamma$ multiplied by $e^{-r}$, and reciprocally. $\blacklozenge$

Now, we can formulate the following theorem.

Theorem 3.3

Let $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\vartheta,\omega,r)$, $\omega=\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}$ with $\omega_{0}\in\Gamma(\bigwedge^{2}T^{\ast}\Gamma_{0})$, be a twisted contact groupoid, $(A(\Gamma),[\cdot,\cdot],T\alpha)$ its associated Lie algebroid, and $(\Lambda_{\Gamma},E_{\Gamma},\omega)$ the twisted Jacobi structure on $\Gamma$ defined by the $\omega$-twisted contact form $\vartheta$. Then

$(\Lambda_{\Gamma},E_{\Gamma},\omega)$* induces on $\Gamma_{0}$ a twisted Jacobi structure $(\Lambda_{0},E_{0},\omega_{0})$ such that $\alpha:\Gamma\to\Gamma_{0}$ is a twisted Jacobi map and $\beta:\Gamma\to\Gamma_{0}$ is a $-e^{-r}$-conformal twisted Jacobi map.* 2. 2)

The map

[TABLE]

induces a Lie algebroid isomorphism between $(T^{*}\Gamma_{0}\times\mathbb{R},\{\cdot,\cdot\}^{\omega_{0}},\pi\circ(\Lambda_{0},E_{0})^{\#})$, which is the Lie algebroid over $\Gamma_{0}$ determined by $(\Lambda_{0},E_{0},\omega_{0})$, and $(A(\Gamma),[\cdot,\cdot],T\alpha)$.

Proof.

1. Let $(f_{0},g_{0})$ be a pair of smooth functions on $\Gamma_{0}$ and $(\alpha^{\ast}f_{0},\alpha^{\ast}g_{0})$ the corresponding pair of left invariant functions on $\Gamma$. We have

[TABLE]

Obviously, the function $\Lambda_{\Gamma}(\alpha^{\ast}df_{0},\alpha^{\ast}dg_{0})$ is left invariant. On the other hand, since $E_{\Gamma}$ is left invariant, it acts by differentiation on $C^{\infty}_{L}(\Gamma,\mathbb{R})=\alpha^{\ast}C^{\infty}(\Gamma_{0},\mathbb{R})$ [4]. Thus, $\langle\alpha^{\ast}f_{0}\alpha^{\ast}dg_{0}-\alpha^{\ast}g_{0}\alpha^{\ast}df_{0},E_{\Gamma}\rangle$ is also a left invariant function on $\Gamma$. So, $\{\alpha^{\ast}f_{0},\alpha^{\ast}g_{0}\}$ is left invariant. Consequently, there exists $h_{0}\in C^{\infty}(\Gamma_{0},\mathbb{R})$, $h_{0}=\varepsilon^{\ast}\{\alpha^{\ast}f_{0},\alpha^{\ast}g_{0}\}$, such that $\{\alpha^{\ast}f_{0},\alpha^{\ast}g_{0}\}=\alpha^{\ast}h_{0}$. The map $(f_{0},g_{0})\mapsto h_{0}$ endows $C^{\infty}(\Gamma_{0},\mathbb{R})$ with an internal composition law $\{\cdot,\cdot\}_{0}$ which is the bracket (4) defined by a twisted Jacobi structure $(\Lambda_{0},E_{0},\omega_{0})$ on $\Gamma_{0}$. Effectively, for any $u_{0},v_{0}\in C^{\infty}(\Gamma_{0})$,

[TABLE]

where $\Lambda_{0}$ is the bivector field on $\Gamma_{0}$ associated to $\Lambda_{0}^{\#}=T\alpha\circ\Lambda^{\#}_{\Gamma}\circ\,^{t}T\alpha$ and $E_{0}=\alpha_{\ast}(E_{\Gamma})$. Since $\alpha$ is a submersion surjective, the last equation means that $\{\cdot,\cdot\}_{0}$ is the bracket (4) defined by $(\Lambda_{0},E_{0})$. Moreover, for any $u_{0},v_{0},w_{0}\in C^{\infty}(\Gamma_{0})$,

[TABLE]

The left side of (30) is equal to \alpha^{\ast}\big{(}\{u_{0},\{v_{0},w_{0}\}_{0}\}_{0}+c.p\big{)}. However, taking into consideration that $\omega=\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}$, $\beta_{\ast}E_{\Gamma}=0$ and $\mathcal{L}_{E_{\Gamma}}r=0$, the right side of (30) can be written as

[TABLE]

From the above results and the surjectivity of the submersion $\alpha$, we deduce that, for any $u_{0},v_{0},w_{0}\in C^{\infty}(\Gamma_{0},\mathbb{R})$,

[TABLE]

which means that $(\Lambda_{0},E_{0},\omega_{0})$ defines a twisted Jacobi structure on $\Gamma_{0}$. By construction, $\alpha$ is a twisted Jacobi map.

Since $\alpha$ is a twisted Jacobi map and $\iota$ is a $-e^{-r}$-conformal twisted Jacobi map (see Proposition 3.2), their composition $\beta=\alpha\circ\iota$ is a $-e^{-r}$-conformal twisted Jacobi map.

2. In order to show that $\mathfrak{I}:\Gamma(T^{*}\Gamma_{0}\times\mathbb{R})\longrightarrow\Gamma(A(\Gamma))$ induces a Lie algebroid isomorphism, it suffices to prove that $\mathfrak{I}$ is compatible with the brackets and the anchors maps and that $\ker\mathfrak{I}=(0,0)$. We denote also by $\alpha^{\ast}$ the map from $\Gamma(T^{\ast}\Gamma_{0}\times\mathbb{R})$ to $\Gamma(T^{\ast}\Gamma\times\mathbb{R})$ given, for any $(\zeta_{0},f_{0})\in\Gamma(T^{\ast}\Gamma_{0}\times\mathbb{R})$, by $\mathbf{\alpha^{\ast}}(\zeta_{0},f_{0})=(\alpha^{\ast}\zeta_{0},\alpha^{\ast}f_{0})$. Being $E_{\Gamma}=E^{l}$, we have $\mathfrak{I}=\pi\circ(\Lambda_{\Gamma},E_{\Gamma})^{\#}\circ\,^{t}T\alpha$. Since $\alpha$ is a twisted Jacobi map, $T\beta\circ\Lambda_{\Gamma}^{\#}\circ\,^{t}T\alpha=0$ and $E_{\Gamma}$ is left invariant, it is easy to check that $\alpha^{\ast}$ is a Lie algebras homomorphism, i.e., for all $(\zeta_{0},f_{0}),(\eta_{0},g_{0})\in\Gamma(T^{\ast}\Gamma_{0}\times\mathbb{R})$,

[TABLE]

Thus, by applying the anchor map $\pi\circ(\Lambda_{\Gamma},E_{\Gamma})^{\#}$ to (31), we obtain

[TABLE]

which signifies that $\mathfrak{I}$ is compatible with the brackets on $\Gamma(T^{*}\Gamma_{0}\times\mathbb{R})$ and $\Gamma(A(\Gamma))$. As well, we have $T\alpha\circ\mathfrak{I}=T\alpha\circ\pi\circ(\Lambda_{\Gamma},E_{\Gamma})^{\#}\circ\,^{t}T\alpha=\pi\circ T\alpha\circ(\Lambda_{\Gamma},E_{\Gamma})^{\#}\circ\,^{t}T\alpha=\pi\circ(\Lambda_{0},E_{0})^{\#}$, whence we get the compatibility of $\mathfrak{I}$ with the anchors of $T^{*}\Gamma_{0}\times\mathbb{R}$ and $A(\Gamma)$. Moreover, since $\ker\vartheta=\mathrm{Im}\Lambda_{\Gamma}^{\#}$ and $\ker(d\vartheta+\omega)=\langle E^{l}\rangle$ are complementary subbundles of $T\Gamma$ and $\alpha$ is a submersion surjective, $\mathfrak{I}(\zeta_{0},f_{0})=\Lambda_{\Gamma}^{\#}(\alpha^{*}\zeta_{0})+\alpha^{*}f_{0}E^{l}=0$ if and only if $(\zeta_{0},f_{0})=(0,0)$. So, $\ker\mathfrak{I}=(0,0)$. $\blacklozenge$

3.2 Homogeneous twisted symplectic and twisted contact groupoids

We complete this section by investigating, in the groupoid framework, the relationship which links homogeneous twisted symplectic structures with twisted contact structures. For this reason, we recall the following concepts.

A multiplicative vector field on a Lie groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ [31] is a pair of vectors fields $(Z,Z_{0})$, where $Z\in\Gamma(T\Gamma)$ and $Z_{0}\in\Gamma(T\Gamma_{0})$, such that $Z:\Gamma\to T\Gamma$ is a morphism of Lie groupoids over $Z_{0}:\Gamma_{0}\to T\Gamma_{0}$, when $T\Gamma\overset{T\alpha}{\underset{T\beta}{\rightrightarrows}}T\Gamma_{0}$ is endowed with the tangent Lie groupoid structure. The above condition is equivalent to the property that the flows $\Psi_{t}$ of $Z$ are (local) Lie groupoid automorphisms over the flows $\psi_{t}$ of $Z_{0}$.

A twisted symplectic groupoid [6] is a Lie groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ equipped with a nondegenerate multiplicative $2$-form $\Omega$ on $\Gamma$, i.e., $m^{\ast}\Omega=pr_{1}^{\ast}\Omega+pr_{2}^{\ast}\Omega$, where $pr_{i}:\Gamma_{2}\to\Gamma$ is the projection on the $i$-factor, $i=1,2$, of $\Gamma_{2}$, and a closed $3$-form $\phi_{0}$ on $\Gamma_{0}$ such that $d\Omega=\alpha^{*}\phi_{0}-\beta^{*}\phi_{0}$. If the closed $3$-form $\phi_{0}$ is exact, i.e., $\phi_{0}=d\omega_{0}$ with $\omega_{0}\in\Gamma(\bigwedge^{2}T^{\ast}\Gamma_{0})$, we say that $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\Omega,d\omega_{0})$ is an exact twisted symplectic groupoid. Then, $\Omega$ has the particular type $\Omega=\alpha^{*}\omega_{0}-\beta^{*}\omega_{0}+d\eta$, where $\eta$ is an $1$-form on $\Gamma$, and it is multiplicative if and only if $d\eta$ is.

Taking into consideration the first footnote, we introduce the notion of a homogeneous twisted symplectic groupoid as follows.

Definition 3.4

A homogeneous twisted symplectic groupoid is an exact twisted symplectic groupoid $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\Omega,d\omega_{0})$ endowed with a multiplicative vector field $(Z,Z_{0})$ such that $\mathcal{L}_{Z}\Omega=\Omega$ and $i(Z_{0})d\omega_{0}=\omega_{0}$.

Hence, we have

Proposition 3.5

If $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\vartheta,\omega,r)$, $\omega=\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}$ with $\omega_{0}\in\Gamma(\bigwedge^{2}T^{\ast}\Gamma_{0})$, is a $2n+1$-dimensional twisted contact groupoid, then its action groupoid $\tilde{\Gamma}\overset{\tilde{\alpha}}{\underset{\tilde{\beta}}{\rightrightarrows}}\tilde{\Gamma}_{0}$, which is defined by (19), endowed with the structure $(\tilde{\Omega},d\tilde{\omega}_{0},(\frac{\partial}{\partial s},\frac{\partial}{\partial s}))$, where $\tilde{\Omega}=d(e^{s}\vartheta)+e^{s}\omega$, $\tilde{\omega}_{0}=e^{s}\omega_{0}$ and $s$ is the canonical coordinate on $\mathbb{R}$, is a homogeneous twisted symplectic groupoid, and reciprocally.

Proof. By construction, $\dim\tilde{\Gamma}=2n+2$. So, $\tilde{\Omega}$ is of maximal rank on $\tilde{\Gamma}$, i.e., $\tilde{\Omega}^{2n+2}\neq 0$, if and only if $\vartheta\wedge(d\vartheta+\omega)^{n}\neq 0$, i.e., $(\vartheta,\omega)$ is a twisted contact structure on $\Gamma$. Furthermore, $d\tilde{\Omega}=d(e^{s}\omega)=d(e^{s}(\alpha^{\ast}\omega_{0}-e^{-r}\beta^{\ast}\omega_{0}))=d(e^{s}\alpha^{\ast}\omega_{0})-d(e^{s-r}\beta^{\ast}\omega_{0})=\tilde{\alpha}^{\ast}(d\tilde{\omega}_{0})-\tilde{\beta}^{\ast}(d\tilde{\omega}_{0})$ and the multiplicativity of $\tilde{\Omega}$ is equivalent with that of $(\vartheta,r)$. Effectively, let $\tilde{pr}_{i}$ be the projection on the $i$-factor, $i=1,2$, of $\tilde{\Gamma}_{2}$ and $((g_{1},s_{2}-r(g_{2})),(g_{2},s_{2}))$ an element of $\tilde{\Gamma}_{2}$. Since $\tilde{\Omega}=e^{s}ds\wedge\vartheta+e^{s}d\vartheta+e^{s}\omega$, we have

[TABLE]

and

[TABLE]

Hence, $\tilde{m}^{\ast}\tilde{\Omega}_{((g_{1},s_{2}-r(g_{2})),(g_{2},s_{2}))}=(\tilde{pr}_{1}^{\ast}\tilde{\Omega}+\tilde{pr}_{2}^{\ast}\tilde{\Omega})_{((g_{1},s_{2}-r(g_{2})),(g_{2},s_{2}))}$ if and only if the components of (32) and (33) containing $ds$ are equals and as also ones containing $\omega$. The first equality gives us $\vartheta_{m(g_{1},g_{2})}=e^{-r(g_{2})}\vartheta_{g_{1}}+\vartheta_{g_{2}}$, whence we obtain (20). From the second equality, taking into account that $\alpha(g_{1}\cdot g_{2})=\alpha(g_{2})$, $\beta(g_{1}\cdot g_{2})=\beta(g_{1})$ and $\alpha(g_{1})=\beta(g_{2})$, we get $\omega_{m(g_{1},g_{2})}=e^{-r(g_{2})}\omega_{g_{1}}+\omega_{g_{2}}\Leftrightarrow r(g_{1}\cdot g_{2})=r(g_{1})+r(g_{2})$.

Finally, it is enough to show that $(\frac{\partial}{\partial s},\frac{\partial}{\partial s})$ is a multiplicative vector field on $\tilde{\Gamma}\overset{\tilde{\alpha}}{\underset{\tilde{\beta}}{\rightrightarrows}}\tilde{\Gamma}_{0}$ and $(\tilde{\Omega},\tilde{\omega}_{0})$ is homogeneous with respect to $(\frac{\partial}{\partial s},\frac{\partial}{\partial s})$. By considering the identifications $T\tilde{\Gamma}\equiv T\Gamma\times T\mathbb{R}$ and $T\tilde{\Gamma}_{0}\equiv T\Gamma_{0}\times T\mathbb{R}$, we can easily check that $T\tilde{\alpha}\circ\frac{\partial}{\partial s}=\frac{\partial}{\partial s}\circ\tilde{\alpha}$, $T\tilde{\beta}\circ\frac{\partial}{\partial s}=\frac{\partial}{\partial s}\circ\tilde{\beta}$ and $T\tilde{m}(\frac{\partial}{\partial s},\frac{\partial}{\partial s})=\frac{\partial}{\partial s}\circ\tilde{m}$. Hence, $(\frac{\partial}{\partial s},\frac{\partial}{\partial s})$ is a multiplicative vector field on $\tilde{\Gamma}$. The homogeneity of $(\tilde{\Omega},\tilde{\omega}_{0})$ with respect to $(\frac{\partial}{\partial s},\frac{\partial}{\partial s})$ is obvious. $\blacklozenge$

Moreover, by adapting Theorem 2.6 of A. Cattaneo and P. Xu [6] in the homogeneous case, we obtain

Theorem 3.6

Let $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\Omega,d\omega_{0},(Z,Z_{0}))$ be a homogeneous twisted symplectic groupoid, $(A(\Gamma),[\cdot,\cdot],T\alpha)$ its associated Lie algebroid, and $(\Lambda,d\omega,Z)$, $\omega=\alpha^{\ast}\omega_{0}-\beta^{\ast}\omega_{0}$, the homogeneous twisted Poisson structure on $\Gamma$ defined by $(\Omega,d\omega)$. Then

$(\Lambda,d\omega,Z)$* induces on $\Gamma_{0}$ a homogeneous twisted Poisson structure $(\Lambda_{0},d\omega_{0},Z_{0})$ such that $\alpha:\Gamma\to\Gamma_{0}$ is a twisted Poisson map and $\beta:\Gamma\to\Gamma_{0}$ is a twisted anti-Poisson map.* 2. 2)

If $(T^{*}\Gamma_{0},[\![\cdot,\cdot]\!]^{d\omega_{0}},\Lambda_{0}^{\#})$ is the Lie algebroid over $\Gamma_{0}$ determined by $(\Lambda_{0},d\omega_{0})$, i.e., for any $\zeta_{0},\eta_{0}\in\Gamma(T^{\ast}\Gamma_{0})$,

[TABLE]

the map $\mathfrak{J}:\Gamma(T^{*}\Gamma_{0})\longrightarrow\Gamma(A(\Gamma))$, $\zeta_{0}\mapsto\Lambda^{\#}(\alpha^{*}\zeta_{0})$333Using the same argumentation as in the case of symplectic groupoids [9], we can show that the left invariant vectors fields on a twisted symplectic groupoid $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\Omega,\phi_{0})$ is of type $\Lambda^{\#}(\alpha^{*}\zeta_{0})$, where $\zeta_{0}\in\Gamma(T^{\ast}\Gamma_{0})$., induces a Lie algebroid isomorphism between $(T^{*}\Gamma_{0},[\![\cdot,\cdot]\!]^{d\omega_{0}},\Lambda_{0}^{\#})$ and $(A(\Gamma),[\cdot,\cdot],T\alpha)$.

Proof. Taking into account the results of [6] and our data $i(Z_{0})d\omega_{0}=\omega_{0}$, we have only to check the homogeneity of $\Lambda_{0}$ with respect to $Z_{0}$, where $\Lambda_{0}=\alpha_{\ast}\Lambda=-\beta_{\ast}\Lambda$. Since $\Lambda$ is the inverse of $\Omega$ and $\mathcal{L}_{Z}\Omega=\Omega$, we get $\mathcal{L}_{Z}\Lambda=-\Lambda$. Also, from the multiplicativity of $(Z,Z_{0})$ on $\Gamma$, we have $T\alpha\circ Z=Z_{0}\circ\alpha$ and $T\beta\circ Z=Z_{0}\circ\beta$. So, because $\alpha$ is a submersion surjective, $\mathcal{L}_{Z_{0}}\Lambda_{0}=\mathcal{L}_{\alpha_{\ast}Z}(\alpha_{\ast}\Lambda)=\alpha_{\ast}(\mathcal{L}_{Z}\Lambda)=\alpha_{\ast}(-\Lambda)=-\Lambda_{0}$. $\blacklozenge$

Using Proposition 3.5 and Theorem 3.6, we can formulate the following

Proposition 3.7

Let $(\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0},\vartheta,\omega,r)$ be a twisted contact groupoid, $(\Lambda_{0},E_{0},\omega_{0})$ its induced twisted Jacobi structure on $\Gamma_{0}$ (Theorem 3.3) and $(\tilde{\Gamma}\overset{\tilde{\alpha}}{\underset{\tilde{\beta}}{\rightrightarrows}}\tilde{\Gamma}_{0},\tilde{\Omega},d\tilde{\omega}_{0},(\frac{\partial}{\partial s},\frac{\partial}{\partial s}))$ its associated homogeneous twisted symplectic groupoid whose the induced homogeneous twisted Poisson structure on $\tilde{\Gamma}_{0}$ (Theorem 3.6) is $(\tilde{\Lambda}_{0},d\tilde{\omega}_{0},\frac{\partial}{\partial s})$. Then, the twisted Poissonization of $(\Lambda_{0},E_{0},\omega_{0})$ coincides with $(\tilde{\Lambda}_{0},d\tilde{\omega}_{0},\frac{\partial}{\partial s})$.

3.3 Examples of twisted contact and homogeneous twisted symplectic groupoids

1. A twisted contact groupoid constructed by a twisted contact manifold. Let $(\Gamma_{0},\vartheta_{0},\omega_{0})$ be a $2n+1$-dimensional, simply connected, twisted contact manifold and $(\Lambda_{0},E_{0})$ the corresponding twisted Jacobi structure on $\Gamma_{0}$ characterized by (7) and (8). We consider the pair groupoid $\Gamma_{0}\times\Gamma_{0}\overset{p_{2}}{\underset{p_{1}}{\rightrightarrows}}\Gamma_{0}$ of $\Gamma_{0}$, $p_{i}$, $i=1,2$, being the projection on the $i$-factor, and the vector bundle groupoid $\mathbb{R}\overset{\pi}{\underset{\pi}{\rightrightarrows}}\{0\}$. By identifying $\Gamma_{0}\times\{0\}$ with $\Gamma_{0}$, we construct the product groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ of $\Gamma_{0}\times\Gamma_{0}$ with $\mathbb{R}$, i.e., $\Gamma=\Gamma_{0}\times\Gamma_{0}\times\mathbb{R}$. The structure maps $\alpha$ and $\beta$ of $\Gamma$ are, respectively, the projections on the second and first factor of $\Gamma$, that are $\alpha(x,y,t)=y$ and $\beta(x,y,t)=x$. The multiplication map $m$ on $\Gamma_{2}=\{((x,y,t),(y,z,s))\,/\,x,y,z\in\Gamma_{0}\;\;\mathrm{and}\;\;t,s\in\mathbb{R}\}\subset\Gamma\times\Gamma$ is given by $m((x,y,t),(y,z,s))=(x,z,t+s)$, the inversion map $\iota:\Gamma\to\Gamma$ by $\iota(x,y,t)=(y,x,-t)$ and the embedding $\varepsilon:\Gamma_{0}\hookrightarrow\Gamma$ by $\varepsilon(x)=(x,x,0)$. The triple $(\vartheta,\omega,r)$, where

[TABLE]

$p_{3}$ being the projection $\Gamma\to\mathbb{R}$, defines a $r$-multiplicative twisted contact structure on $\Gamma$ whose the Reeb vector field is $E_{\Gamma}=0+E_{0}+0$ and its bivector field (8) is $\Lambda_{\Gamma}=-e^{r}\Lambda_{0}+\Lambda_{0}+0$. It is easy to verify that

[TABLE]

on $\Gamma$, where $c$ is a nonzero constant dependent of $n$, and that $(\vartheta,r)$ satisfies (20). Clearly, the induced twisted Jacobi structure on $\Gamma_{0}$ by $(\Lambda_{\Gamma},E_{\Gamma},\omega)$ is the initial given one.

2. The gauge groupoid of a principal line bundle over a twisted contact manifold. Let $(\Gamma_{0},\vartheta_{0},\omega_{0})$ be a $2n+1$-dimensional twisted contact manifold, $(\Lambda_{0},E_{0})$ the corresponding twisted Jacobi structure on $\Gamma_{0}$ defined by (7) and (8), and $f_{0}$ a first integral of $E_{0}$. Let, also, $\pi:P\to\Gamma_{0}$ be a principal line bundle over $\Gamma_{0}$, i.e., its structure group $G$ is a $1$-dimensional Lie group that acts freely on $P$ on the right and $P/G=\Gamma_{0}$, endowed with a connection $\mathfrak{H}$. Precisely, $\mathfrak{H}$ is a $G$-invariant distribution on $P$, complementary to its vertical bundle $\mathfrak{V}=\ker\pi_{\ast}$. We consider the gauge groupoid $\Gamma\overset{\alpha}{\underset{\beta}{\rightrightarrows}}\Gamma_{0}$ of $P$ which is the quotient $P\times P/G\rightrightarrows P/G$ of the pair groupoid $P\times P\overset{\pi_{2}}{\underset{\pi_{1}}{\rightrightarrows}}P$ of $P$, $\pi_{i}$, $i=1,2$, being the projection on the $i$-copy of $P$, by the right diagonal action of $G$ on $P\times P$. The natural projection $\varpi:P\times P\to P\times P/G$, $(p_{1},p_{2})\overset{\varpi}{\mapsto}[(p_{1},p_{2})]=\{(p_{1}g,p_{2}g)\,/\,g\in G\}$, is a Lie groupoid morphism over $\pi$. Thus, the source map $\alpha$ and the target map $\beta$ of $\Gamma$ are, respectively, the maps defined by $\alpha\circ\varpi=\pi\circ\pi_{2}$ and $\beta\circ\varpi=\pi\circ\pi_{1}$. Hence, for each $[(p_{1},p_{2})]\in\Gamma$, $\alpha([(p_{1},p_{2})])=\pi(p_{2})$ and $\beta([(p_{1},p_{2})])=\pi(p_{1})$, while the inversion $\iota:\Gamma\to\Gamma$ and the embedding $\varepsilon:\Gamma_{0}\hookrightarrow\Gamma$ are given, respectively, by $\iota([(p_{1},p_{2})])=[(p_{2},p_{1})]$ and $\varepsilon(x)=[(p,p)]$, where $p\in\pi^{-1}(x)$. Moreover, the product of two composable elements $[(p_{1},p_{2})]$, $[(q_{1},q_{2})]$ of $\Gamma$, that means that $\alpha([(p_{1},p_{2})])=\pi(p_{2})=\pi(q_{1})=\beta([(q_{1},q_{2})])$, is $m([(p_{1},p_{2})],[(q_{1},q_{2})])=[(p_{1},q_{2}g)]$, where $g$ is the unique element of $G$ for which $p_{2}=q_{1}g$. The triple $(\vartheta,\omega,r)$, where

[TABLE]

defines a $r$-multiplicative twisted contact structure on $\Gamma$, i.e., (34) and (20) hold on $\Gamma$ and $\Gamma_{2}$, respectively. In order to present the tensors fields $\Lambda_{\Gamma}$ and $E_{\Gamma}$ that define the corresponding $(d\omega,\omega)$-twisted Jacobi structure on $\Gamma$, we recall that given a $q$-vector field $Q_{0}$ on $\Gamma_{0}$ there exists a unique $q$-vector field $Q_{0}^{h}$ on $P$, called the horizontal lift of $Q_{0}$ on $P$ with respect to $\mathfrak{H}$, such that, for any $p�$