# Reduction by symmetries of contact mechanical systems on Lie groups

Alexandre Anahory Simoes1, Leonardo Colombo2,
Manuel de León3, Juan Carlos Marrero4,
David Martín de Diego5, Edith Padrón6
1 IE School of Science and Technology, Madrid, Spain
2 Centre for Automation and Robotics (CSIC-UPM), Arganda del Rey, Spain
3 Instituto de Ciencias Matematicas (CSIC) and
Real Academia de Ciencias, Madrid, Spain
4,6 ULL-CSIC Geometría Diferencial y Mecánica Geométrica,
Departamento de Matemáticas, Estadística e Investigación Operativa and
Instituto de Matemáticas y Aplicaciones (IMAULL), Faculty of Sciences,
University of La Laguna, Canary Islands, Spain
5 Instituto de Ciencias Matematicas (CSIC), Madrid, Spain

###### Abstract

We study the dynamics of contact mechanical systems on Lie groups that are
invariant under a Lie group action. Analogously to standard mechanical systems
on Lie groups, existing symmetries allow for reducing the number of equations.
Thus, we obtain Euler-Poincaré-Herglotz equations on the extended reduced
phase space $\mathfrak{g}\times\mathbb{R}$ associated with the extended phase
space $TG\times\mathbb{R}$, where the configuration manifold $G$ is a Lie
group and $\mathfrak{g}$ its Lie algebra. Furthermore, we obtain the
Hamiltonian counterpart of these equations by studying the underlying Jacobi
structure. Finally, we extend the reduction process to the case of symmetry-
breaking systems which are invariant under a Lie subgroup of symmetries.

††AMS Mathematics Subject Classification (2020). Primary 70G45, 70G65, 70G75;
Secondary 53D05, 53D10, 53D17. Keywords: Reduction by symmetries, contact
mechanical systems, Jacobi structures in mechanics, Euler-Poincaré equations,
Lie-Poisson equations, Herglotz principle.

##  1 Introduction

Contact Hamiltonian and Lagrangian systems have deserved a lot of attention in
recent years (see [8, 9], [27, 28], [2], and references therein). One of the
most relevant features of contact dynamics is the absence of conservative
properties contrarily to the conservative character of the energy in
symplectic dynamics; indeed, they have a dissipative behavior. This fact
suggests that contact geometry may be the appropriate framework to model many
physical and mathematical problems with dissipation we find in thermodynamics,
statistical physics, systems with impulsive effect and discontinuities,
quantum mechanics, gravity or control theory, among many others (see [25],
[11], [31], [26], and references therein).

As an illustrative example, consider the motion of a rigid body under a linear
dissipation term. The equations of motion for this kind of systems are of the
general form

$\mathbb{I}\dot{\xi}=\mathbb{I}\xi\times\xi+C\xi,$ (1)

where $\xi\in\mathbb{R}^{3}$ denotes the body angular velocity,
$\mathbb{I}\in\mathbb{R}^{3\times 3}$ is the inertia tensor, and
$C\in\mathbb{R}^{3\times 3}$ denotes a constant positive semi-definite matrix
called the damping matrix (see [35]). In the special case where
$C=\gamma\mathbb{I}$, with $\gamma$ a non-zero real number, we will see that
equations (1) are given by a contact version of Euler-Poincaré equations (see
[17]) for a reduced contact Lagrangian.

As it is noted in [18], in the case of the rigid body on $SO(3)$ we can
rewrite the Lagrangian function $L:TSO(3)\to\mathbb{R}$ for this system as

$L(R,\dot{R})=\frac{1}{2}\operatorname{tr}(\dot{R}\mathbb{I}\dot{R}^{T}),$

with $\mathbb{I}$ symmetric and constant. Moreover, due to the
$SO(3)$-invariance of $L$, we may define the reduced Lagrangian
$l:\mathfrak{so}(3)\rightarrow\mathbb{R}$ given by

$l(\xi)=\frac{1}{2}\langle\mathbb{I}\xi,\xi\rangle.$

The Euler-Poincaré equations applied to the reduced Lagrangian function $l$
give the Euler equations for the rigid body:

$\mathbb{I}\dot{\xi}=\mathbb{I}\xi\times\xi.$

Now, we can consider a slight modification of these constructions and start
with a contact Lagrangian function
$L:TSO(3)\times\mathbb{R}\rightarrow\mathbb{R}$ given by

$L(R,\dot{R},z)=\frac{1}{2}\operatorname{tr}(\dot{R}\mathbb{I}\dot{R}^{T})+\gamma
z,$

where $\gamma\in\mathbb{R}$. This motivates the definition of a contact
reduced Lagrangian function
$l:\mathfrak{so}(3)\times\mathbb{R}\rightarrow\mathbb{R}$ given by

$l(\xi,z)=\frac{1}{2}\langle\mathbb{I}\xi,\xi\rangle+\gamma z.$ (2)

The main purpose of this work is to introduce the Euler-Poincaré-Herglotz
equations as a generalization of Euler-Poincaré equations for contact
Lagrangian functions, which are given by the expression

$\frac{d}{dt}\frac{\delta l}{\delta\xi}=\text{ad}_{\xi}^{*}\frac{\delta
l}{\delta\xi}+\frac{\delta l}{\delta\xi}\frac{\partial l}{\partial z}.$

Applying them to the Lagrangian function (2), we obtain equations (1) with
$C=\gamma\mathbb{I}$.

In this situation, the reduced Legendre transform
$\mathbb{F}l:\mathfrak{g}\times\mathbb{R}\to\mathfrak{g}^{*}\times\mathbb{R}$
is given by the map $\mathbb{F}l(\xi,z)=(\mathbb{I}\xi,z)$. Assuming that
$\mathbb{I}$ is a positive definite symmetric matrix, so that $\mathbb{F}l$ is
a diffeomorphism and we might define the Hamiltonian function

$h(\mu,z)=\frac{1}{2}\mu^{T}\mathbb{J}\mu+\gamma z,$

where $\mathbb{J}=\mathbb{I}^{-1}$. Given a regular reduced Lagrangian $l$, if
$(\xi(t),z(t))$ is a solution of Euler-Poincaré-Herglotz equations and
$\mu(t)=\frac{\delta l}{\delta\xi}(\xi(t),z(t))$ we will see in this paper
that the curve $t\mapsto(\mu(t),z(t))$ satisfies

$\dot{\mu}=\text{ad}^{*}_{\frac{\delta h}{\delta\mu}}\mu-\mu\frac{\partial
h}{\partial z},\,\,\dot{z}=\langle\mu,\frac{\delta
h}{\delta\mu}\rangle-h(\mu,z).$ (3)

In addition, we will define a Lie-Poisson-Jacobi structure and bracket on
$\mathfrak{g}^{*}\times\mathbb{R}$, and derive the corresponding Lie-Poisson-
Jacobi equations (3). The Jacobi structure on the vector space
$\mathfrak{g}^{*}\times\mathbb{R}$ is linear. This kind of structures was
discussed in [22, 23].

Symmetry breaking is common in several physical contexts, from classical
mechanics to particle physics and complex fluids (see [33], [17]). The
simplest example is the heavy top dynamics (the motion of a rigid body with a
fixed point in a gravitational field), where due to the presence of gravity,
we get a Lagrangian which is $\mathrm{SO}(2)$-invariant but not
$\mathrm{SO}(3)$-invariant, contrary to what happens for the free rigid body.
Reduction theory for symmetry breaking with applications to nematic systems
was studied in [16]. Optimal Control for systems with symmetry breaking has
been studied in [15]. In the context of motion planning, the symmetry breaking
appears naturally in the form of a navigation function [7], [12]. More
recently it has been employed to reduce necessary conditions for optimality in
a collision and obstacle avoidance problem (see [36], [37]). Motivated by the
rigid body attitude dynamics studied in [35], [10] for the triaxial attitude
control testbed given in [4], we will study Euler-Poincare-Herglotz reduction
for systems with symmetry breaking and we will also derive the corresponding
Lie-Poisson-Jacobi equations.

This paper is structured as follows: in Section 2, we briefly recall Euler-
Poincaré reduction on Lie groups and in Section 3, we recall the basic
concepts involving contact structures and contact Hamiltonian vector fields.
In section 4, we present the Euler-Poincaré-Herglotz reduction of invariant
contact systems on Lie groups. In section 5, we define the Jacobi bracket on
the extended dual of the Lie algebra $\mathfrak{g}^{*}\times\mathbb{R}$ and
obtain the Hamiltonian counterpart of Euler-Poincaré-Herglotz equations.
Finally, in Section 6, we apply the previous development to systems with a
broken symmetry, through a Euler-Poincaré-Herglotz reduction on a semi-direct
product.

### Notation

In order to be self-contained and use consistent notation, let us establish
the functional derivative notation that we will use without reference all
along the text.

Let $f$ be a smooth function on a finite-dimensional vector space $V$. The
derivative of $f$ at a point $v\in V$ along the direction $u\in V$, may
written in terms of geometric objects as

$Df(v)\cdot u=\left.\frac{d}{dt}\right|_{t=0}f(v+tu)=\langle
df(v),(u)_{v}^{V}\rangle,$

where $df(v)\in T_{v}^{*}V$ is the differential of the function at $V$ and

$(u)_{v}^{V}=\left.\frac{d}{dt}\right|_{t=0}(v+tu)\in T_{v}V$

is the vertical lift of the vector $u$ at $v$.

The vertical lift at $v$ induces a canonical identification between the vector
space $V$ and its tangent space at $v$, i.e., the map
$(\cdot)_{v}^{V}:V\rightarrow T_{v}V$ is an isomorphism. This implies that the
dual map $\varphi:T_{v}^{*}V\rightarrow V^{*}$ is also an isomorphism, so that

$Df(v)\cdot u=\langle\varphi(df(v)),u\rangle.$

The unique vector $\varphi(df(v))\in V^{*}$ is called the functional
derivative of $f$ at $v$. The functional derivative of $f$ at $v$ is usually
denoted by

$\varphi(df(v))=\frac{\delta f}{\delta v}(v).$

In finite dimensions, through a coordinate analysis, we see that the
coordinate expression of $\displaystyle{\frac{\delta f}{\delta v}(v)}$ matches
the component functions of the differential $df(v)$.

##  2 Euler-Poincaré equations

Let $G$ be a Lie group and let $\mathcal{L}_{g}:G\rightarrow G$ be the left
multiplication action. Since $\mathcal{L}_{g}$ is a diffeomorphism, it
naturally induces a “trivialized chart” of $TG$, which more precisely is an
identification between each tangent space $T_{g}G$ with the Lie algebra
$\mathfrak{g}$ through the following map

$TG\rightarrow
G\times\mathfrak{g},\quad(g,\dot{g})\mapsto(g,T_{g}\mathcal{L}_{g^{-1}}(\dot{g})).$
(4)

Consider the Lagrangian function $L$ as a real function
$L:G\times\mathfrak{g}\to\mathbb{R}$ on $G\times\mathfrak{g}$. Then, we can
express the Euler-Lagrange equations in terms of this trivialization as

$\begin{split}\frac{d}{dt}\frac{\delta
L}{\delta\xi}&=\text{ad}_{\xi}^{*}\frac{\delta
L}{\delta\xi}+T_{e}^{*}\mathcal{L}_{g}\left(\frac{\delta L}{\delta
g}\right)\\\ \dot{g}&=T_{e}\mathcal{L}_{g}(\xi).\end{split}$

Here $\operatorname{ad}^{*}_{\xi}:\mathfrak{g}^{*}\rightarrow\mathfrak{g}^{*}$
is the dual of the adjoint operator
$\operatorname{ad}_{\xi}:\mathfrak{g}\rightarrow\mathfrak{g}$ defined by
$\operatorname{ad}_{\xi}(\eta)=[\xi,\eta]$, for $\eta\in\mathfrak{g}$.

Now, if $L:TG\rightarrow\mathbb{R}$ is a left-invariant Lagrangian function,
that is, its composition with the tangent lift of left-translations
$T\mathcal{L}_{g}$ leaves $L$ invariant

$L(T_{h}\mathcal{L}_{g}(v_{h}))=L(v_{h}),\ \forall v_{h}\in T_{h}G,$

then the restriction of $L$ to the tangent space at the identity $e\in G$,
which we identify with the Lie algebra $\mathfrak{g}$, is called the reduced
Lagrangian function and we will denote it by
$l:\mathfrak{g}\rightarrow\mathbb{R}$. Then all the information is contained
in the reduced Lagrangian $l$ since

$l(\eta)=L(T_{e}\mathcal{L}_{g}(\eta)),\quad\forall g\in
G,\eta\in\mathfrak{g}.$

It is well-known that a curve $g:I\rightarrow G$ is a solution of Euler-
Lagrange equations if and only if the curve $\xi:I\rightarrow\mathfrak{g}$
determined by

$\dot{g}=T_{e}\mathcal{L}_{g}(\dot{\xi})$

satisfies the Euler-Poincaré equations given by

$\frac{d}{dt}\frac{\delta l}{\delta\xi}=\text{ad}_{\xi}^{*}\frac{\delta
l}{\delta\xi}.$

If $(y^{i})$ are local coordinates associated to a given basis $\\{e_{i}\\}$
of $\mathfrak{g}$, then its local expression reads

$\frac{d}{dt}\frac{\partial l}{\partial y^{i}}=C_{ji}^{k}y^{j}\frac{\partial
l}{\partial y^{k}},$

where $C_{ij}^{k}$ are the structure constants of the Lie algebra given by

$[e_{i},e_{j}]=C_{ij}^{k}e_{k}$

(see [5, 34] for instance).

##  3 Contact manifolds and contact dynamics

In this section, we will recall the main definition of a contact manifold and
the corresponding Hamiltonian vector fields (see [3, 29, 27] for a more
detailed overview).

A contact manifold $(M,\eta)$ is an $(2n+1)$-dimensional manifold equipped
with a contact form $\eta$, i.e., $\eta$ is a $1$-form on $M$ such that
$\eta\wedge(d\eta)^{n}$ is a volume form. The Reeb vector field
$\mathcal{R}\in\mathfrak{X}(M)$ is the unique vector field that satisfies:

$\iota_{\mathcal{R}}d\eta=0,\quad\eta(\mathcal{R})=1.$ (5)

On a contact manifold $(M,\eta)$, we define the following isomorphism of
vector bundles:

$\displaystyle\flat:$ $\displaystyle TM$ $\displaystyle\longrightarrow$
$\displaystyle T^{*}M,$ (6) $\displaystyle\quad v$ $\displaystyle\longmapsto$
$\displaystyle\iota_{v}d\eta+\eta(v)\eta.$

Notice that $\flat(\mathcal{R})=\eta$.

There is a Darboux theorem for contact manifolds. In a neighborhood of each
point in $M$ one can find local coordinates $(q^{i},p_{i},z)$ such that

$\eta=dz-p_{i}dq^{i}.$ (7)

In these coordinates, we have

$\mathcal{R}=\frac{\partial}{\partial z}.$ (8)

The canonical example of contact manifold is $T^{*}Q\times\mathbb{R}$. Here,
the contact form is given by

$\eta_{Q}=dz-\theta_{Q}=dz-p_{i}dq^{i},$ (9)

where $\theta_{Q}$ is the pullback of the tautological $1$-form of $T^{*}Q$,
$(q^{i},p_{i})$ are bundle coordinates on $T^{*}Q$ and $z$ is the
$\mathbb{R}$-coordinate.

Given a smooth function $f:M\rightarrow\mathbb{R}$, its Hamiltonian vector
field $X_{f}$ is given by

$\flat(X_{f})=df-(f+\mathcal{R}(f))\eta,$ (10)

or, equivalently,

$i_{X_{f}}d\eta=df-R(f)\eta,\quad i_{X_{f}}\eta=-f.$

We call the triple $(M,\eta,H)$ a contact Hamiltonian system, where $(M,\eta)$
is a contact manifold and $H:M\rightarrow\mathbb{R}$ is the Hamiltonian
function. In contrast to their symplectic counterpart, contact Hamiltonian
vector fields do not preserve the Hamiltonian fuction. In fact,

$X_{H}(H)=-\mathcal{R}(H)H.$ (11)

###  3.1 Contact Lagrangian systems

Now we recall the Lagrangian picture of contact systems (see [27, 28] for a
more comprehensive description). Indeed, under a suitable choice of contact
structure on the extended phase space $TQ\times\mathbb{R}$, the contact
Hamiltonian equations with respect to a Lagrangian function on this space are
precisely Euler-Poincaré equations.

Let $Q$ be an $n$-dimensional configuration manifold and consider the extended
phase space $TQ\times\mathbb{R}$ and we call contact Lagrangian function to
any smooth function of the type $L:TQ\times\mathbb{R}\to\mathbb{R}$. In this
paper, we will assume that the Lagrangian is regular, that is, the Hessian
matrix with respect to the velocities $(W_{ij})$ is regular where

$W_{ij}=\frac{\partial^{2}L}{\partial\dot{q}^{i}\partial\dot{q}^{j}},$ (12)

and $(q^{i},\dot{q}^{i},z)$ are bundle coordinates for $TQ\times\mathbb{R}$.
Equivalently, $L$ is regular if and only if the one-form

$\eta_{L}=dz-\theta_{L}$ (13)

is a contact form. Here,

$\theta_{L}=S^{*}(dL)=\frac{\partial L}{\partial\dot{q}^{i}}d{q}^{i},$

where $S$ is the canonical vertical endomorphism $S:TTQ\to TTQ$ extended to
$TQ\times\mathbb{R}$, that is, in local $TQ\times\mathbb{R}$ bundle
coordinates,

$S=dq^{i}\otimes\frac{\partial}{\partial\dot{q}^{i}}.$ (14)

The energy of the system is defined by

$E_{L}=\Delta(L)-L=\dot{q}^{i}\frac{\partial L}{\partial\dot{q}^{i}}-L,$ (15)

where $\Delta$ is the Liouville vector field on $TQ$ extended to
$TQ\times\mathbb{R}$ in the natural way.

The Reeb vector field of $\eta_{L}$, which we will denoted by
$\mathcal{R}_{L}$, is locally given by

$\mathcal{R}_{L}=\frac{\partial}{\partial
z}-(W^{ij})\frac{\partial^{2}L}{\partial\dot{q}^{i}\partial
z}\frac{\partial}{\partial\dot{q}^{j}},$ (16)

where $(W^{ij})$ is the inverse of the Hessian matrix with respect to the
velocities $(W_{ij})$.

The Hamiltonian vector field of the energy $E_{L}$ will be denoted
$\xi_{L}=X_{E_{L}}$, which is a second order differential equation (SODE)
(that is, $S(\xi_{L})=\Delta$) and its solutions are just the ones of the
Herglotz equations for $L$ (see [27, 28]):

$\begin{split}\frac{d}{dt}\frac{\partial
L}{\partial\dot{q}^{i}}&=\frac{\partial L}{\partial q^{i}}+\frac{\partial
L}{\partial\dot{q}^{i}}\frac{\partial L}{\partial z}\\\
\dot{z}&=L.\end{split}$ (17)

Now, we will recall the definition of the _Legendre transformation_ for
contact Lagrangian systems. Given the vector bundle $TQ\times\mathbb{R}\to
Q\times\mathbb{R}$, one can consider the fiber derivative $\mathbb{F}L$ of
$L:TQ\times\mathbb{R}\to\mathbb{R}$, which has the following coordinate
expression in natural coordinates:

$\displaystyle\mathbb{F}L:TQ\times\mathbb{R}$ $\displaystyle\to
T^{*}Q\times\mathbb{R}$ (18) $\displaystyle(q^{i},\dot{q}^{i},z)$
$\displaystyle\mapsto(q^{i},\frac{\partial L}{\partial\dot{q}^{i}},z).$

If the Lagrangian is regular, then $\mathbb{F}L$ is a local diffeomorphism and
one can show that $\eta_{L}$ is simply $(\mathbb{F}L)^{*}\eta_{Q}$.

In the case that $\mathbb{F}L$ is a global contactomorphism, then we say that
$L$ is _hyperregular_. In this situation, we can define a Hamiltonian function
$H:T^{*}Q\times\mathbb{R}\to\mathbb{R}$ such that $E_{L}=H\circ\mathbb{F}L$.
So, the Lagrangian and Hamiltonian dynamics are $\mathbb{F}L$-related, that
is, $(\mathbb{F}L)_{*}(\xi_{L})=X_{H}\circ\mathbb{F}L$ (see [27, 28]).

##  4 Euler-Poincaré-Herglotz equations for contact reduced Lagrangian systems

In this section, we will prove that a similar result to that in Section 2
holds in the contact situation where we consider the extended space
$TG\times\mathbb{R}$ as our initial ambient phase space and we assume that the
Lagrangian function is $G$-invariant. Let
$L:TG\times\mathbb{R}\rightarrow\mathbb{R}$ be a contact Lagrangian function.
Again, the fact that the left translation $\mathcal{L}_{g}:G\to G$ is a
diffeomorphism allows us to consider a “left trivialized chart” of
$TG\times\mathbb{R}$, that is an identification of the product
$T_{g}G\times\mathbb{R}$ with the vector space $\mathfrak{g}\times\mathbb{R}$
given precisely by the map

$TG\times\mathbb{R}\rightarrow
G\times\mathfrak{g}\times\mathbb{R},\quad(g,\dot{g},z)\mapsto(g,T_{g}\mathcal{L}_{g^{-1}}(\dot{g}),z).$
(19)

Thus, if we introduce coordinates $(g,\xi,z)$ on $TG\times\mathbb{R}$ given by
(19), we have the following:

######  Theorem 4.1.

Let $G$ be a Lie group and $L:TG\times\mathbb{R}\rightarrow\mathbb{R}$ a
contact Lagrangian function. If we introduce the coordinates $(g,\xi,z)$ on
$TG\times\mathbb{R}$ given by (19), then the curves $\xi$ and $z$ satisfy the
Herglotz equations on Lie groups

$\begin{split}\frac{d}{dt}\frac{\delta
L}{\delta\xi}&=\emph{ad}_{\xi}^{*}\frac{\delta
L}{\delta\xi}+T_{e}^{*}\mathcal{L}_{g}\left(\frac{\delta L}{\delta
g}\right)+\frac{\delta L}{\delta\xi}\frac{\partial L}{\partial z}\\\
\dot{g}&=T_{e}\mathcal{L}_{g}\left(\xi\right)\\\ \dot{z}&=L.\end{split}$ (20)

###### Proof.

Before starting the proof, let
$\tilde{L}:G\times\mathfrak{g}\times\mathbb{R}\rightarrow\mathbb{R}$ be the
left trivialized Lagrangian function using the coordinates (19). So that

$\tilde{L}(g,\xi,z)=L(g,g\xi,z),$

where we are using the notation $g\xi:=T_{e}\mathcal{L}_{g}(\xi)$. In what
follows, we will find critical values of the action along a trajectory on
$TG\times\mathbb{R}$ of the form $(g(t),\dot{g}(t),z(t))$. But, under the left
trivialized coordinates, this is equivalent to find critical values along
trajectories on $G\times\mathfrak{g}\times\mathbb{R}$ of the form
$(g(t),\xi(t),z(t))$, where

$\xi(t):=T_{g(t)}\mathcal{L}_{g(t)^{-1}}(\dot{g}(t)):=g(t)^{-1}\cdot\dot{g}(t).$

Also, note that for any variation of a solution $g:I\rightarrow G$ of Euler-
Lagrange equations denoted by $c:U\subset\mathbb{R}^{2}\rightarrow G$, the
tangent vectors $\displaystyle{\frac{\partial c}{\partial s}}$ and
$\displaystyle{\frac{\partial c}{\partial t}}$ may be transported to the Lie
algebra $\mathfrak{g}$ in the following way

$\begin{split}\eta:U&\rightarrow\mathfrak{g}\\\ (t,s)&\mapsto
T_{c(t,s)}\mathcal{L}_{c(t,s)^{-1}}\left(\frac{\partial c}{\partial
s}\right)\end{split}\quad\begin{split}\xi:U&\rightarrow\mathfrak{g}\\\
(t,s)&\mapsto T_{c(t,s)}\mathcal{L}_{c(t,s)^{-1}}\left(\frac{\partial
c}{\partial t}\right).\end{split}$ (21)

Moreover, it is a standard fact (see [6, 17]) that these maps satisfy

$\frac{\partial\xi}{\partial s}-\frac{\partial\eta}{\partial t}=[\xi,\eta].$
(22)

Conversely, if there are 2-parameter family of curves $\xi$ and $\eta$ on the
Lie algebra satisfying (22), there exists a smooth function
$c:U\subset\mathbb{R}^{2}\rightarrow G$ such that $\xi$ and $\eta$ are given
by equations (21).

We will often commit a slight abuse of notation with the letters $\xi$ and
$\eta$, in the sense that they represent both a curve and a variation of
curves in $\mathfrak{g}$. The first will be denoted by $\xi(t)$ and the second
by $\xi(t,s)$, respectively $\eta(t)$ and $\eta(t,s)$. The relation between
them is that

$\xi(t)=\xi(t,0)=g^{-1}\cdot\dot{g}\quad\text{and}\quad\eta(t)=\eta(t,0):=g^{-1}\cdot\delta
g$ (23)

Now, find equations (20) are equivalent to Herglotz equations on the contact
extended space $TG\times\mathbb{R}$ let us show that its solutions satisfy
Herglotz principle. Let the action over a curve be given by

$\mathcal{A}(g(\cdot),z_{0})=\int_{0}^{h}L(g(t),\dot{g}(t),z(t))\ dt,$

where the curve $z(t)$ satisfies the equation

$\dot{z}=L(g(t),\dot{g}(t),z(t)),\quad z(0)=z_{0}.$

The action over a variation $c$ of the curve $g$ might be written as

$\mathcal{A}(c(\cdot,s),z_{0})=\int_{0}^{h}L\left(c(t,s),\frac{\partial
c}{\partial t}(t,s),z(t,s)\right)\ dt,$

which in the coordinates (19) is just

$\mathcal{A}(c(\cdot,s),z_{0})=\int_{0}^{h}\tilde{L}\left(c(t,s),\xi(t,s),z(t,s)\right)\
dt,$

where $\xi$ is defined as in (21) and the variation $z(t,s)$ satisfies the
equation

$\dot{z}(t,s)=\tilde{L}(c(t,s),\xi(t,s),z(t,s)),\quad z(0,s)=z_{0}.$

Taking variations $\xi(t,s)$ and $\eta(t,s)$ as in (21) and (22), the first
variation of the action functional gives

$\begin{split}\left.\frac{d}{ds}\right|_{s=0}\int_{0}^{h}&\tilde{L}\left(c(t,s),\xi(t,s),z(t,s)\right)\
dt\\\ &=\int_{0}^{h}\left[\frac{\partial\tilde{L}}{\partial g}\delta
g+\frac{\partial\tilde{L}}{\partial\xi}\delta\xi+\frac{\partial\tilde{L}}{\partial
z}\delta z\right]\ dt=z(h,0).\end{split}$

Observe now that the infinitesimal variation of the curve $z$, that is,
$\delta z$ must satisfy the inhomogeneous linear differential equation

$\frac{\partial\delta z}{\partial t}(t)=\frac{\partial\tilde{L}}{\partial
g}\delta
g+\frac{\partial\tilde{L}}{\partial\xi}\delta\xi+\frac{\partial\tilde{L}}{\partial
z}\delta z.$

This is a first order ordinary differential equation whose solution is

$\delta z(\tau)=e^{\int_{0}^{\tau}\frac{\partial\tilde{L}}{\partial
z}ds}\left(\int_{0}^{\tau}e^{-\int_{0}^{t}\frac{\partial\tilde{L}}{\partial
z}ds}\left[\frac{\partial\tilde{L}}{\partial g}\delta
g+\frac{\partial\tilde{L}}{\partial\xi}\delta\xi\right]\ dt\right),$

since $\delta z(0)$ vanishes. Now, observe that we have seen above that the
first variation of the action is equal to $\delta z(h)$. Hence,

$\left.\frac{d}{ds}\right|_{s=0}\mathcal{A}(c(\cdot,s),z_{0})=e^{\int_{0}^{h}\frac{\partial\tilde{L}}{\partial
z}ds}\left(\int_{0}^{h}e^{-\int_{0}^{t}\frac{\partial\tilde{L}}{\partial
z}ds}\left[\frac{\partial\tilde{L}}{\partial g}\delta
g+\frac{\partial\tilde{L}}{\partial\xi}\delta\xi\right]\ dt\right),$

where from (22) we have that $\delta\xi=\dot{\eta}+[\xi,\eta]$ and from (23)
we have that $\delta g=g\cdot\eta$. Defining the auxiliary function

$\sigma(t)=e^{-\int_{0}^{t}\frac{\partial\tilde{L}}{\partial z}ds},$

we obtain

$\left.\frac{d}{ds}\right|_{s=0}\mathcal{A}(c(\cdot,s),z_{0})=\frac{1}{\sigma(h)}\left(\int_{0}^{h}\sigma(t)\left[\frac{\partial\tilde{L}}{\partial\xi}(\dot{\eta}+\text{ad}_{\xi}\eta)+\frac{\partial\tilde{L}}{\partial
g}g\cdot\eta\right]\ dt\right).$

Integrating by parts, in order to get rid of time derivatives of $\eta$, and
noting that the boundary terms vanish since $\eta$ is zero at the endpoints we
conclude that the action reduces to the expression below

$\frac{1}{\sigma(h)}\left(\int_{0}^{h}\left[\sigma(t)\text{ad}_{\xi}^{*}\frac{\partial\tilde{L}}{\partial\xi}-\frac{d}{dt}\left(\sigma(t)\frac{\partial\tilde{L}}{\partial\xi}\right)+\sigma(t)T_{e}^{*}\mathcal{L}_{g}\left(\frac{\partial\tilde{L}}{\partial
g}\right)\right]\eta\ dt\right).$

Since $\eta$ is arbitrary and due to the fundamental theorem of calculus of
variations, the curve $(g(t),\xi(t),z(t))$ is a critical point of the action
if and only if it satisfies the equation

$\sigma(t)\text{ad}_{\xi}^{*}\frac{\partial\tilde{L}}{\partial\xi}-\sigma(t)\frac{d}{dt}\left(\frac{\partial\tilde{L}}{\partial\xi}\right)-\dot{\sigma}\frac{\partial\tilde{L}}{\partial\xi}+\sigma(t)T_{e}^{*}\mathcal{L}_{g}\left(\frac{\partial\tilde{L}}{\partial
g}\right)=0,$

which finishes the proof since this is equivalent to

$\sigma(t)\left[\text{ad}_{\xi}^{*}\frac{\partial\tilde{L}}{\partial\xi}-\frac{d}{dt}\left(\frac{\partial\tilde{L}}{\partial\xi}\right)+\frac{\partial\tilde{L}}{\partial
z}\frac{\partial\tilde{L}}{\partial\xi}+T_{e}^{*}\mathcal{L}_{g}\left(\frac{\partial\tilde{L}}{\partial
g}\right)\right]=0$

and the auxiliary function $\sigma$ is nowhere zero. The last two equations in
(20) follow by construction. ∎

Next, let us define first the lifted action of $G$ on $TG\times\mathbb{R}$
whose translation by $g\in G$ is the map
$\hat{\mathcal{L}}_{g}:TG\times\mathbb{R}\rightarrow TG\times\mathbb{R}$ given
by

$\hat{\mathcal{L}}_{g}(v_{h},z)=(T_{h}\mathcal{L}_{g}(v_{h}),z).$ (24)

Suppose that $L:TG\times\mathbb{R}\rightarrow\mathbb{R}$ is a left invariant
contact Lagrangian function, thus satisfying $L\circ\hat{\mathcal{L}}_{g}=L$.
We define the reduced contact Lagrangian to be the function
$l:\mathfrak{g}\times\mathbb{R}\rightarrow\mathbb{R}$ which is the restriction
of the contact Lagrangian function $L$ to the space $T_{e}G\times\mathbb{R}$.

Observe that in this case

$l(\xi,z)=L(e,\xi,z)=L(g,g\cdot\xi,z)$

by the left-invariant property. The results of the previous theorem are
further simplified using this additional assumption and we obtain in
consequence the Euler-Poincaré-Herglotz equations on the product space
$\mathfrak{g}\times\mathbb{R}$.

######  Theorem 4.2.

Let $G$ be a Lie group and $L:TG\times\mathbb{R}\rightarrow\mathbb{R}$ a left-
invariant contact Lagrangian function. If
$l:\mathfrak{g}\times\mathbb{R}\rightarrow\mathbb{R}$ is the corresponding
reduced Lagrangian, $g:I\rightarrow G$ is a curve on the Lie group and
$\xi:I\rightarrow\mathfrak{g}$ a curve on the Lie algebra, then the following
are equivalent:

  1. 1.

The curves $g$ and $z:I\rightarrow\mathbb{R}$ satisfy Herglotz equations for
$L$;

  2. 2.

The Herglotz principle

$\delta\int_{0}^{h}L(g(t),\dot{g}(t),z(t))\
dt=0,\quad\dot{z}=L(g(t),\dot{g}(t),z(t))$

holds for variations of $g$ with fixed endpoints;

  3. 3.

The curves $\xi(t)=T_{g(t)}\mathcal{L}_{g^{-1}(t)}(\dot{g}(t))$ and $z$
satisfy the Euler-Poincaré-Herglotz equations

$\frac{d}{dt}\frac{\delta l}{\delta\xi}=\emph{ad}_{\xi}^{*}\frac{\delta
l}{\delta\xi}+\frac{\delta l}{\delta\xi}\frac{\partial l}{\partial
z}\quad\text{and}\quad\dot{z}=l.$ (25)

  4. 4.

The reduced Herglotz variational principle

$\delta\int_{0}^{h}l(\xi(t),z(t))\ dt=0,\quad\dot{z}=l(\xi(t),z(t))$ (26)

holds using variations of the form $\delta\xi=\dot{\eta}+[\xi,\eta]$, where
$\eta$ is a curve on $\mathfrak{g}$ vanishing at the endpoints.

###### Proof.

The equivalence of items $1.$ and $2.$ is general for all contact manifolds,
so in particular for $TG\times\mathbb{R}$ (see [2]).

Next, we prove the equivalence of both variational principles, that is, items
$2.$ and $4.$ Observe that $l:\mathfrak{g}\times\mathbb{R}\to\mathbb{R}$
determines uniquely the function $L:TG\times\mathbb{R}\to\mathbb{R}$ by left-
translations and viceversa. Thus one only needs to show that variations
$(\delta g,\delta z)\in TG\times\mathbb{R}$ with $\delta g$ having fixed
endpoints induces and are induced by variations $(\delta\xi,\delta z)$ with
$\delta\xi$ of the form $\delta\xi=\dot{\eta}+[\xi,\eta]$, where $\eta(t)$
vanishes at end points. But, this is the content in the proof of Theorem 4.1.

To conclude, we show the equivalence between $3$. and $4$. Indeed by using the
definition of the auxiliary function $\sigma(t)$, the calculations in the
proof of Theorem 4.1 and integration by parts, we deduce that the reduced
Herglotz variational principle holds if and only if

$\sigma(t)\left[\text{ad}_{\xi}^{*}\frac{\partial
l}{\partial\xi}-\frac{d}{dt}\left(\frac{\partial
l}{\partial\xi}\right)+\frac{\partial l}{\partial z}\frac{\partial
l}{\partial\xi}\right]=0.$ (27)

So, since the auxiliary function $\sigma$ is nowhere zero, the result follows.
∎

######  Remark 4.3.

If we have determined a trajectory $(\xi(t),z(t))$ in
$\mathfrak{g}\times\mathbb{R}$ solving Euler-Poincaré-Herglotz equations, we
may obtain the corresponding trajectory on the original phase space
$TG\times\mathbb{R}$ through the reconstruction procedure, which amounts to
solving the reconstruction equation

$\dot{g}=g\cdot\xi.$

######  Example 4.4.

Consider now the matrix Lie group $SO(3)$, and a Lagrangian function
$L:TSO(3)\times\mathbb{R}\rightarrow\mathbb{R}$ of the form

$L(R,\dot{R},z)=\frac{1}{2}\langle\langle\dot{R},\dot{R}\rangle\rangle-\gamma
z,$

with

$\langle\langle\dot{R},\dot{R}\rangle\rangle=\int_{B}\rho(X)\|\dot{R}X\|^{2}\
d^{3}X,$

where $B$ is the subset of $\mathbb{R}^{3}$ occupied by some rigid body. In
this case, the left multiplication map is defined by
$\mathcal{L}_{R_{1}}:SO(3)\rightarrow SO(3)$

$\mathcal{L}_{R_{1}}(R)=R_{1}R$

and so its tangent map is just

$T_{R}\mathcal{L}_{R_{1}}(Y)=R_{1}Y,\quad Y\in T_{R}SO(3).$

Moreover, the Lie algebra of $SO(3)$, denoted by $\mathfrak{so}(3)$ is
composed by $3\times 3$ skew-symmetric matrices and the adjoint map is given
by matrix commutator, i.e.,

$\text{ad}_{\xi}\eta=\xi\eta-\eta\xi,\quad\xi,\eta\in\mathfrak{so}(3).$

Now, the expression of $L$ on left trivialized coordinates is well-known to be

$L(R,\xi,z)=\frac{1}{2}\langle\xi,\xi\rangle-\gamma z,$

where the inner product on $\mathfrak{so}(3)$ is given by

$\langle\xi_{1},\xi_{2}\rangle=\frac{1}{2}\text{tr}\
(\xi_{1}^{T}\mathbb{I}\xi_{2}),$

with $\mathbb{I}$ the inertia tensor. Many of the computation are easily
carried using the hat map
$\hat{(\cdot)}:\mathbb{R}^{3}\rightarrow\mathfrak{so}(3)$ defined by

$(\omega_{1},\omega_{2},\omega_{3})\mapsto\left(\begin{matrix}0&\omega_{1}&\omega_{2}\\\
-\omega_{1}&0&\omega_{3}\\\ -\omega_{2}&-\omega_{3}&0\end{matrix}\right)$

which is a Lie algebra isomorphism. Under these definitions the Lagrangian
function is written as

$L(R,\xi,z)=\frac{1}{2}\xi^{T}\mathbb{I}\xi-\gamma z$

and the equations (20) give

$\begin{split}\mathbb{I}\dot{\xi}&=-\xi\times\mathbb{I}\xi-\gamma\mathbb{I}\xi\\\
\dot{\xi}&=R\xi\\\ \dot{z}&=L.\end{split}$

$\diamond$

##  5 Lie-Poisson-Jacobi reduction

In this section, we will introduce a Jacobi structure on the space
$\mathfrak{g}^{*}\times\mathbb{R}$ and we will see that for a regular
$G$-invariant Lagrangian function the reduced dynamics is Hamiltonian with
respect to this Jacobi structure.

###  5.1 Contact dynamics on a Lie group

Now, assume that the configuration space $Q$ is a Lie group $G$ and that the
regular Lagrangian function $L:TG\times\mathbb{R}\to\mathbb{R}$ is
$G$-invariant. Then, similarly to what happens between the contact manifold
$TG\times\mathbb{R}$ and the canonical contact manifold
$T^{*}G\times\mathbb{R}$, we may also identify $\mathfrak{g}\times\mathbb{R}$
and $\mathfrak{g}^{*}\times\mathbb{R}$ through the reduced Legendre
transformation
$\mathbb{F}l:\mathfrak{g}\times\mathbb{R}\rightarrow\mathfrak{g}^{*}\times\mathbb{R}$
given by

$\mathbb{F}l(\xi,z)=\left(\frac{\delta l}{\delta\xi},z\right).$

Note that,

$\mathbb{F}L(e,\xi,z)=(e,\mathbb{F}l(\xi,z)),$

that is, the reduced Legendre transform is essentially the restriction of the
Legendre transform to $\mathfrak{g}\times\mathbb{R}$. Thus, the requiring $L$
to be regular is sufficient in order to make $\mathbb{F}l$ a local
diffeomorphism.

######  Proposition 5.1.

If $L$ is $G-$invariant, the Legendre transform is $G$-equivariant, that is,

$\mathbb{F}L(g\cdot v,z)=g\cdot\mathbb{F}L(v,z),\quad(v,z)\in
TG\times\mathbb{R}.$

Suppose that the Legendre transformation is a diffeomorphism and consider the
function $h:\mathfrak{g}^{*}\times\mathbb{R}\rightarrow\mathbb{R}$ given by

$h(\mu,z)=\langle\mu,\xi\rangle-l(\xi,z),$

where $\displaystyle{\mu=\frac{\delta l}{\delta\xi}}$. Then it is easy to see
that

$\frac{\delta h}{\delta\mu}=\xi.$

By inserting these equalities in Euler-Poincaré-Herglotz equations we get the
following result:

######  Proposition 5.2.

Given a regular reduced Lagrangian function $l$, if $(\xi(t),z(t))$ is a
solution of Euler-Poincaré-Herglotz equations and $\mu(t)=\frac{\delta
l}{\delta\xi}(\xi(t),z(t))$ then the curve $t\mapsto(\mu(t),z(t))$ satisfies

$\begin{split}\dot{\mu}&=\text{ad}^{*}_{\frac{\delta
h}{\delta\mu}}\mu-\mu\frac{\partial h}{\partial z}\\\
\dot{z}&=\langle\mu,\frac{\delta h}{\delta\mu}\rangle-h(\mu,z).\end{split}$

###### Proof.

By differentiating the curve $t\mapsto\mu(t)$ we get

$\dot{\mu}=\frac{d}{dt}\frac{\delta l}{\delta\xi}=ad^{*}_{\xi}\frac{\delta
l}{\delta\xi}+\frac{\delta l}{\delta\xi}\frac{\partial l}{\partial z},$

where we used Euler-Poincaré-Herglotz equations in the last equality. Using
the equalities preceding the proposition statement, we deduce

$\dot{\mu}=\text{ad}^{*}_{\frac{\delta h}{\delta\mu}}\mu-\mu\frac{\partial
h}{\partial z}.$

The equation for $\dot{z}$ follows from the definition of $h$. ∎

These equations will be called Lie-Poisson-Jacobi equations. In the next
subsections, we will show that this is a Hamiltonian system with respect to a
Jacobi structure on $\mathfrak{g}^{*}\times\mathbb{R}$.

######  Example 5.3.

Let us consider again the matrix Lie group $SO(3)$ and consider the
hyperregular $G$-invariant Lagrangian function

$L(R,\dot{R},z)=\frac{1}{2}\langle\langle\dot{R},\dot{R}\rangle\rangle-\gamma
z,$

so that we might consider the reduced Lagrangian which is the restriction of
$L$ to $\mathfrak{g}\times\mathbb{R}$

$l(\xi,z)=\frac{1}{2}\xi^{T}\mathbb{I}\xi-\gamma z.$

Under all the assumptions taken in Example 4.4, the Legendre transform is
given by the map

$\mathbb{F}l(\xi,z)=(\mathbb{I}\xi,z).$

Suppose that $\mathbb{I}$ is a positive definite symmetric matrix, so that
$\mathbb{F}l$ is a diffeomorphism and we might define the Hamiltonian function

$h(\mu,z)=\mu^{T}\mathbb{I}^{-1}\mu-l(\mathbb{I}^{-1}\mu,z)=\mu^{T}\mathbb{I}^{-1}\mu-\frac{1}{2}(\mathbb{I}^{-1}\mu)^{T}\mu+\gamma
z,$

and letting
$\mathbb{J}=(2\mathbb{I}^{-1}-(\mathbb{I}^{-1})^{T})=\mathbb{I}^{-1}$ we get

$h(\mu,z)=\frac{1}{2}\mu^{T}\mathbb{J}\mu+\gamma z.$

The Lie-Poisson-Jacobi equations are then given by

$\begin{split}\dot{\mu}&=\text{ad}^{*}_{\frac{\delta
h}{\delta\mu}}\mu-\gamma\mu\\\ \dot{z}&=\frac{1}{2}\mu^{T}\mathbb{J}\mu-\gamma
z.\end{split}$

$\diamond$

###  5.2 Lie-Poisson-Jacobi bracket

Let us recall first the definition of a Jacobi structure (see [24] and [30]).

######  Definition 5.4.

(Jacobi structure) A Jacobi structure on a manifold $M$ is a tuple
$(\Lambda,E)$, where $\Lambda$ is a bi-vector field and $E$ is a vector field,
satisfying the following equations

$[\Lambda,\Lambda]=2E\wedge\Lambda,\quad[E,\Lambda]=0,$

with $[\cdot,\cdot]$ the Schouten-Nijenhuis bracket.

######  Definition 5.5.

(Jacobi bracket) A Jacobi bracket on a manifold $M$ is a bilinear, skew-
symmetric map $\\{\cdot,\cdot\\}:C^{\infty}(M)\times C^{\infty}(M)\rightarrow
C^{\infty}(M)$ satisfying the Jacobi identity and the following weak Leibniz
rule

$\text{supp}(\\{f,g\\})\subseteq\text{supp}(f)\cap\text{supp}(g).$

A Jacobi manifold is a manifold possessing either a Jacobi structure or a
Jacobi bracket since these two definitions are equivalent (see [24], [30],
[29], [19]). However, it is much more convenient to introduce a Jacobi
structure for practical purposes.

Now, from the Jacobi structure we can define an associated Jacobi bracket as
follows:

$\\{f,g\\}=\Lambda(df,dg)+fE(g)-gE(f),\quad f,g\in C^{\infty}(M,\mathbb{R})$

In this case, the weak Leibniz rule is equivalent to the generalized Leibniz
rule

$\\{f,gh\\}=g\\{f,h\\}+h\\{f,g\\}+ghE(f),$ (28)

In this sense, this bracket generalizes the well-known Poisson brackets.
Indeed, a Poisson manifold is a particular case of Jacobi manifold in which
$E=0$.

Given a Jacobi manifold $(M,\Lambda,E)$, we consider the map
$\sharp_{\Lambda}:\Omega^{1}(M)\rightarrow\mathfrak{X}(M)$ defined by

$\sharp_{\Lambda}(\alpha)=\Lambda(\alpha,\cdot).$

We have that $\sharp_{\Lambda}$ is a morphism of $C^{\infty}$-modules, though
it may fail to be an isomorphism. Given a function $f:M\rightarrow\mathbb{R}$,
we define the Hamiltonian vector field $X_{f}$ by

$X_{f}=\sharp_{\Lambda}(df)+fE.$

Contact structures are examples of Jacobi structures. Given a contact manifold
$(M,\eta)$, we may associate it a natural Jacobi structure. Indeed, we define
the bivector $\Lambda$ as

$\Lambda(\alpha,\beta)=-d\eta(\flat^{-1}(\alpha),\flat^{-1}(\beta)),\qquad\alpha,\beta\in\Omega^{1}(M)\;.$
(29)

So that the pair $(\Lambda,E=-\mathcal{R})$ is a Jacobi structure (see [30],
[14]).

In Darboux coordinates, the bivector $\Lambda$ reads as

$\Lambda=\frac{\partial}{\partial p_{i}}\wedge\left(\frac{\partial}{\partial
q^{i}}+p_{i}\frac{\partial}{\partial z}\right).$ (30)

In addition, given a function $h:M\rightarrow\mathbb{R}$, the contact
Hamiltonian vector field is given by

$X_{h}=\sharp_{\Lambda}(dh)-h\mathcal{R}$

and along its integral curves the following equation is satisfied with respect
to the associated Jacobi bracket

$\dot{f}=\\{h,f\\}-f\frac{\partial h}{\partial z},\quad\forall f\in
C^{\infty}(M).$ (31)

The preceding equation implies in particular that the Hamiltonian function is
not conserved along its integral curves since

$\dot{h}=-h\frac{\partial h}{\partial z}.$

In Darboux coordinates, the bracket is given by

$\displaystyle\\{f,g\\}=\frac{\partial f}{\partial q^{i}}\frac{\partial
g}{\partial p_{i}}-\frac{\partial f}{\partial p_{i}}\frac{\partial g}{\partial
q^{i}}-\frac{\partial f}{\partial z}\left(p_{i}\frac{\partial g}{\partial
p_{i}}-g\right)+\frac{\partial g}{\partial z}\left(p_{i}\frac{\partial
f}{\partial p_{i}}-f\right)$

and the Hamiltonian vector field is given in canonical coordinates by:

$X_{h}=\frac{\partial h}{\partial p_{i}}\frac{\partial}{\partial
q^{i}}-\left(\frac{\partial h}{\partial q^{i}}+p_{i}\frac{\partial h}{\partial
z}\right)\frac{\partial}{\partial p_{i}}+\left(p_{i}\frac{\partial h}{\partial
p_{i}}-h\right)\frac{\partial}{\partial z}$

So, given a manifold $Q$, the contact manifold
$(T^{*}Q\times\mathbb{R},\eta_{Q})$, where $\eta_{Q}$ is the contact form
defined in (9), has a canonical Jacobi structure and an associated Jacobi
bracket defined as above, which we will denote by $\\{\cdot,\cdot\\}_{can}$
from now on.

Let us now introduce a Jacobi structure on $\mathfrak{g}^{*}\times\mathbb{R}$.

######  Proposition 5.6.

The structure given by

$\Lambda(\mu,z)(df,dg)=\left\langle\mu,\left[\frac{\delta
f}{\delta\mu},\frac{\delta
g}{\delta\mu}\right]\right\rangle+\left\langle\mu,\frac{\delta
f}{\delta\mu}\right\rangle\frac{\partial g}{\partial
z}-\left\langle\mu,\frac{\delta g}{\delta\mu}\right\rangle\frac{\partial
f}{\partial z}$ (32)

and $\displaystyle{E=-\mathcal{R}=-\frac{\partial}{\partial z}}$ is a Jacobi
structure on $\mathfrak{g}^{*}\times\mathbb{R}$ and induces a Jacobi bracket
given by

$\\{f,g\\}(\mu,z)=\Lambda(\mu)(df,dg)-f\frac{\partial g}{\partial
z}+g\frac{\partial f}{\partial z}.$ (33)

###### Proof.

We just have to prove that $\Lambda$ is a Jacobi structure, since then the map
$\\{\cdot,\cdot\\}$ would be the associated Jacobi bracket. Using the fact
that

$\Lambda=\text{pr}_{1}^{*}\Lambda_{0}+\Delta\wedge\mathcal{R},$

where $\Lambda_{0}$ is the Lie-Poisson structure on $\mathfrak{g}^{*}$ defined
by

$\Lambda_{0}(\mu)(df,dg)=\left\langle\mu,\left[\frac{\delta
f}{\delta\mu},\frac{\delta g}{\delta\mu}\right]\right\rangle,\quad f,g\in
C^{\infty}(\mathfrak{g}^{*}),$

$\text{pr}_{1}:\mathfrak{g}^{*}\times\mathbb{R}\rightarrow\mathfrak{g}^{*}$ is
the projection onto the first factor, $\Delta$ is the vector field on
$\mathfrak{g}^{*}\times\mathbb{R}$ defined by

$\Delta(\mu,z)=\left.\frac{d}{dt}\right|_{t=0}(\mu+t\mu,z)$

and $\mathcal{R}=\frac{\partial}{\partial z}$, we may deduce after some
computations involving the Schouten-Nijenhuis bracket and interior products
that (see [32])

$[\Lambda,\Lambda]=2[\Lambda_{0},\Delta\wedge\mathcal{R}]=2\Lambda_{0}\wedge\mathcal{R},$

where we used that
$[\Delta\wedge\mathcal{R},\Lambda_{0}]=[\Lambda_{0},\Delta\wedge\mathcal{R}]$
(since $\Lambda_{0}$ and $\Delta\wedge\mathcal{R}$ are both $(2,0)$-tensors)
and the fact that $[\Lambda_{0},\Delta]=\Lambda_{0}$. The previous equality is
equivalent to

$[\Lambda,\Lambda]=2\Lambda\wedge\mathcal{R},$

using linearity and skew-symmetry of the wedge product.

In addition, we also have that $[\mathcal{R},\Lambda]$ vanishes:

$[\mathcal{R},\Lambda]=[\mathcal{R},\Lambda_{0}]+[\mathcal{R},\Delta\wedge\mathcal{R}].$

the first term vanishes since $\Lambda_{0}$ is pulled-back from
$\mathfrak{g}^{*}$ and so
$[\mathcal{R},\Lambda_{0}]=\mathcal{L}_{\mathcal{R}}\Lambda_{0}=0$. The second
term also vanishes since
$[\mathcal{R},\Delta\wedge\mathcal{R}]=[\mathcal{R},\Delta]\wedge\mathcal{R}+\Delta\wedge[\mathcal{R},\mathcal{R}]=[\mathcal{R},\Delta]\wedge\mathcal{R}$
and the Lie bracket of $[\mathcal{R},\Delta]$ is zero. Hence,
$(\Lambda,E=-\mathcal{R})$ is indeed a Jacobi structure. ∎

######  Remark 5.7.

The Jacobi structure $(\Lambda=\text{pr}_{1}^{*}\Lambda_{0}+\Delta\wedge R$,
$R=-\partial/\partial z)$ on $\mathfrak{g}^{*}\times\mathbb{R}$ is linear on
the vector space $\mathfrak{g}^{*}\times\mathbb{R}\to\mathbb{R}$. In fact, it
is a particular case of a more general class of linear Jacobi structures on
vector bundles which were previously considered in [22] (see also [23, 21]).

######  Definition 5.8.

The Jacobi structure and bracket defined in the previous Proposition are
called the Lie-Poisson-Jacobi structure and bracket on
$\mathfrak{g}^{*}\times\mathbb{R}$, respectively.

In the next proposition, we will define the Hamiltonian vector fields
associated with this Jacobi structure and we will see what is the expression
of the corresponding Hamilton equations.

######  Proposition 5.9.

Let $(\Lambda,R)$ be the Jacobi structure on
$\mathfrak{g}^{*}\times\mathbb{R}$ defined above. If
$h:\mathfrak{g}^{*}\times\mathbb{R}\rightarrow\mathbb{R}$ is the reduced
Hamiltonian function, the integral curves of the Hamiltonian vector field
$X_{h}$ satisfy the Lie-Poisson-Jacobi equations

$\begin{split}\dot{\mu}&=\emph{ad}^{*}_{\frac{\delta
h}{\delta\mu}}\mu-\mu\frac{\partial h}{\partial z}\\\
\dot{z}&=\langle\mu,\frac{\delta h}{\delta\mu}\rangle-h(\mu,z).\end{split}$

###### Proof.

Note that if $(\mu(t),z(t))$ is an integral curve of the contact Hamiltonian
vector field $X_{h}$, then we have that for all function $f$

$\dot{f}=\langle
df(\mu,z),(\dot{\mu},\dot{z})\rangle=\left\langle\dot{\mu},\frac{\delta
f}{\delta\mu}\right\rangle+\frac{\partial f}{\partial z}\dot{z}.$

Also, using the definition of the Jacobi bracket we have that

$\\{h,f\\}(\mu,z)=\left\langle\mu,\left[\frac{\delta
h}{\delta\mu},\frac{\delta
f}{\delta\mu}\right]\right\rangle+\left(\left\langle\mu,\frac{\delta
h}{\delta\mu}\right\rangle-h\right)\frac{\partial f}{\partial
z}-\left(\left\langle\mu,\frac{\delta
f}{\delta\mu}\right\rangle-f\right)\frac{\partial h}{\partial z}.$

Thus using (31) we deduce that

$\dot{f}=\\{h,f\\}(\mu,z)-f\frac{\partial h}{\partial
z}=\left\langle\text{ad}_{\frac{\delta h}{\delta\mu}}^{*}\mu-\mu\frac{\partial
h}{\partial z},\frac{\delta
f}{\delta\mu}\right\rangle+\left(\left\langle\mu,\frac{\delta
h}{\delta\mu}\right\rangle-h\right)\frac{\partial f}{\partial z}.$

Since the function $f$ is arbitrary the result follows. ∎

The following result relates the new bracket on the reduced product space
$\mathfrak{g}^{*}\times\mathbb{R}$ with the canonical Jacobi bracket on
$T^{*}G\times\mathbb{R}$. Recall how the the left action of $G$ on itself
lifts to the cotangent space $T^{*}G$. The cotangent left action is given by
the map

$\phi_{g}:T^{*}G\rightarrow T^{*}G,\quad\alpha_{h}\in T_{h}^{*}G\mapsto
T_{gh}^{*}\mathcal{L}_{g^{-1}}(\alpha_{h})\in T_{gh}^{*}G.$

######  Proposition 5.10.

Let $F$ and $H$ be $G$-invariant functions on $T^{*}G\times\mathbb{R}$ and let
$f$ and $h$ be their restrictions to $\mathfrak{g}^{*}\times\mathbb{R}$,
respectively. Then

$\\{F,H\\}_{can}(\alpha,z)=\\{f,h\\}(T_{e}^{*}\mathcal{L}_{g}(\alpha),z),\quad\alpha\in
T_{g}^{*}G,$

where $\\{\cdot,\cdot\\}_{can}$ is the canonical Jacobi bracket on
$T^{*}G\times\mathbb{R}$ and $\\{\cdot,\cdot\\}$ is the Lie-Poisson-Jacobi
structure on $\mathfrak{g}^{*}\times\mathbb{R}$.

###### Proof.

The canonical Jacobi bracket on $T^{*}G\times\mathbb{R}$ has the following
form

$\\{F,H\\}_{can}=\\{F\circ i_{z},H\circ i_{z}\\}_{T^{*}G}-F\frac{\partial
H}{\partial z}+H\frac{\partial F}{\partial z},$ (34)

where $\\{\cdot,\cdot\\}_{T^{*}G}$ is the canonical Poisson bracket on
$T^{*}G$ and $i_{z}:T^{*}G\hookrightarrow T^{*}G\times\mathbb{R}$ is the
inclusion mapping $i_{z}(\alpha)=(\alpha,z)$. Observe that it is a standard
fact that

$\\{F,H\\}_{T^{*}G}(\alpha)=\\{f,h\\}_{0}(T_{e}^{*}\mathcal{L}_{g}(\alpha)),$

where $\\{\cdot,\cdot\\}_{0}$ is the Lie-Poisson bracket on $\mathfrak{g}^{*}$
(see [34], for instance). Moreover, since $F$ and $H$ are invariant functions
we have that also their derivative with respect to the contact variable $z$.
Therefore, (34) equals

$\displaystyle\\{F,H\\}_{can}(\alpha,z)=$ $\displaystyle\\{f\circ i_{z},h\circ
i_{z}\\}_{0}(T_{e}^{*}\mathcal{L}_{g}(\alpha))$
$\displaystyle-f(\alpha,z)\frac{\partial h}{\partial
z}(\alpha,z)+h(\alpha,z)\frac{\partial f}{\partial z}(\alpha,z),$

which is exactly the bracket $\\{f,h\\}(T_{e}^{*}\mathcal{L}_{g}(\alpha),z)$
(see Proposition 5.6). ∎

######  Example 5.11.

The reduced space of $T^{*}SO(3)\times\mathbb{R}$ is
$\mathfrak{so}(3)^{*}\times\mathbb{R}$, which we can identify with
$\mathbb{R}^{4}$ under the suitable Lie algebra isomorphism. The corresponding
Lie-Poisson-Jacobi bracket is

$\begin{split}\\{h,f\\}(\mu,z)&=\mu\cdot\left(\frac{\delta
h}{\delta\mu}\times\frac{\delta
f}{\delta\mu}\right)+\left(\mu\cdot\frac{\delta
h}{\delta\mu}-h\right)\frac{\partial f}{\partial z}-\left(\mu\cdot\frac{\delta
f}{\delta\mu}-f\right)\frac{\partial h}{\partial z}\\\ &=\frac{\delta
f}{\delta\mu}\cdot\left(\mu\times\frac{\delta h}{\delta\mu}-\mu\frac{\partial
h}{\partial z}\right)+\left(\mu\cdot\frac{\delta
h}{\delta\mu}-h\right)\frac{\partial f}{\partial z}+f\frac{\partial
h}{\partial z}\end{split}$

Given the Hamiltonian function
$h(\mu,z)=\frac{1}{2}\mu^{T}\mathbb{J}\mu+\gamma z$, the corresponding Lie-
Poisson-Jacobi equations are then

$\begin{split}\dot{\mu}&=\mu\times\mathbb{J}\mu-\gamma\mu\\\
\dot{z}&=\frac{1}{2}\mu^{T}\mathbb{J}\mu-\gamma z.\end{split}$

###  5.3 Lie-Poisson-Jacobi reduction theorem

The objective of this section is two-fold: we are going to prove that the
standard momentum map of the Lie group $T^{*}G$ induces a Jacobi map from
$T^{*}G\times\mathbb{R}$ to $\mathfrak{g}^{*}\times\mathbb{R}$. Then we prove
that Jacobi maps allow to perform a reduction and to find a Jacobi structure
on $\mathfrak{g}^{*}\times\mathbb{R}$. If we start with the canonical Jacobi
structure then the reduced structure will be precisely the Lie-Poisson-Jacobi
structure we defined in the last section. Moreover, we will prove that Jacobi
maps project Hamiltonian vector fields onto Hamiltonian vector fields. This is
the Jacobi version of the Poisson reduction Theorem given in [34].

It is well-known (see [1], [34]) that the map
$J:T^{*}G\rightarrow\mathfrak{g}^{*}$ defined by

$\langle J(\alpha_{g}),\xi\rangle=\langle
T_{e}^{*}\mathcal{L}_{g}(\alpha_{g}),\xi\rangle,\quad\alpha_{g}\in
T_{g}^{*}G,\,\,\xi\in\mathfrak{g},$ (35)

is a momentum map on the cotangent bundle of the Lie group. Moreover it is a
Poisson map with respect to the standard Poisson structure and Lie-Poisson
structure on $T^{*}G$ and $\mathfrak{g}^{*}$, respectively.

######  Definition 5.12.

Given two Jacobi manifolds $(M_{1},\Lambda_{1},E_{1})$ and
$(M_{2},\Lambda_{2},E_{2})$, with Jacobi brackets $\\{\cdot,\cdot\\}_{1}$ and
$\\{\cdot,\cdot\\}_{2}$, respectively, and a smooth map
$\varphi:M_{1}\rightarrow M_{2}$, then the map $\varphi$ is said to be a
Jacobi map if for every $f$ and $h$ in $C^{\infty}(M_{2})$ we have that

$\\{f\circ\varphi,h\circ\varphi\\}_{1}=\\{f,h\\}_{2}\circ\varphi.$

The following theorem establishes that in the presence of a Jacobi action, the
Jacobi bracket reduces to a Jacobi bracket on the space of orbits of the
action.

######  Theorem 5.13 (Jacobi reduction theorem by a Lie Group action).

Let $G$ be a Lie group acting on a Jacobi manifold $(P,\\{\cdot,\cdot\\})$ by
Jacobi maps. Suppose that $P/G$ is a smooth manifold and the projection
$\pi:P\rightarrow P/G$ is a submersion. Then, there is a unique Jacobi bracket
on $P/G$ denoted by $\\{\cdot,\cdot\\}_{red}$ called the reduced Jacobi
bracket such that $\pi$ is a Jacobi map.

Moreover, suppose that $H:P\rightarrow\mathbb{R}$ is $G$-invariant and define
$h:P/G\rightarrow\mathbb{R}$ by $H:=h\circ\pi$. If $\phi_{t}^{X_{H}}$ and
$\phi_{t}^{X_{h}}$ are the Hamiltonian flows of $H$ and $h$, respectively,
then they satisfy the equation

$\pi\circ\phi_{t}^{X_{H}}=\phi_{t}^{X_{h}}\circ\pi.$

###### Proof.

Let $f,h\in C^{\infty}(P/G)$. Notice that the Jacobi bracket of the functions
$f\circ\pi$ and $h\circ\pi$ is also $G$-invariant. Indeed if
$\phi_{g}:P\rightarrow P$ denotes the action then

$\\{f\circ\pi,h\circ\pi\\}\circ\phi_{g}=\\{f\circ\pi\circ\phi_{g},h\circ\pi\circ\phi_{g}\\}=\\{f\circ\pi,h\circ\pi\\},$

where the first equality holds since $\phi_{g}$ acts by Jacobi maps and the
second holds since $\pi$ is invariant under composition with the action by
definition. Now, observe that a function $\Phi$ on $P$ is $G$-invariant if and
only if there exists a function $\varphi:P/G\rightarrow\mathbb{R}$ such that
$\Phi=\varphi\circ\pi$. Thus, denote by $\\{f,h\\}_{red}$ the function on
$P/G$ defined by

$\\{f\circ\pi,h\circ\pi\\}=\\{f,h\\}_{red}\circ\pi.$

We must check that $\\{f,h\\}_{red}$ is indeed a Jacobi bracket. It is clearly
bilinear and skew-symmetric, since $\\{f\circ\pi,h\circ\pi\\}$ also is. It
satisfies Jacobi identity since $\\{\cdot,\cdot\\}$ also does. And finally, we
must check if it satisfies the weak Leibniz condition. But note that if $k\in
C^{\infty}(P/G)$ and $(\Lambda,E)$ is the Jacobi structure associated to
$\\{\cdot,\cdot\\}$

$\displaystyle\\{f,hk\\}_{red}\circ\pi=$
$\displaystyle\\{f\circ\pi,(h\circ\pi)(k\circ\pi)\\}$ $\displaystyle=$
$\displaystyle(h\circ\pi)\\{f\circ\pi,k\circ\pi\\}+(k\circ\pi)\\{f\circ\pi,h\circ\pi\\}$
(36) $\displaystyle+(h\circ\pi)(k\circ\pi)E(f\circ\pi)$

Now, since $E(\Phi)=\\{1,\Phi\\}$, for $\Phi\in\mathcal{C}^{\infty}(P)$, we
deduce that the vector field $E$ is $\pi$-projectable and, thus, there exists
a vector field $E_{red}$ over $P/G$ such that
$E_{red}(f)\circ\pi=E(f\circ\pi)$, $\forall$ $f\in\mathcal{C}^{\infty}(P/G)$.
Therefore, from (36), it follows that

$\\{f,hk\\}_{red}\circ\pi=(h\\{f,k\\}_{red}+k\\{f,h\\}_{red}+hkE_{red}(f))\circ\pi.$

This implies that $\\{\cdot,\cdot\\}_{red}$ satisfies the weak Leibniz
condition. In fact, following the same argument we could prove that the Jacobi
structure associated to $\\{\cdot,\cdot\\}_{red}$, denoted by
$(\Lambda_{red},E_{red})$, by uniqueness is given by

$\Lambda_{red}\circ\pi=T\pi(\Lambda)\quad\text{and}\quad
E_{red}\circ\pi=T\pi(E).$

Now, note that given $p\in P$

$\frac{d}{dt}(f\circ\pi\circ\phi_{t}^{X_{H}}(p))=\\{H,f\circ\pi\\}\circ\phi_{t}^{X_{H}}(p)+(f\circ\pi)E(H)\circ\phi_{t}^{X_{H}}(p).$

Since $H=h\circ\pi$ the later is equivalent to

$\frac{d}{dt}(f\circ\pi\circ\phi_{t}^{X_{H}}(p))=\\{h,f\\}_{red}\circ\pi\circ\phi_{t}^{X_{H}}(p)+fE_{red}(h)\circ\pi\circ\phi_{t}^{X_{H}}(p).$
(37)

Similarly, on the quotient space $P/G$, letting $[p]:=\pi(p)$, we have that

$\frac{d}{dt}(f\circ\phi_{t}^{X_{h}}([p]))=\\{h,f\\}_{red}\circ\phi_{t}^{X_{h}}([p])+fE_{red}(h)\circ\phi_{t}^{X_{h}}([p]).$

This last equation, uniquely defines the Hamiltonian flow $\phi_{t}^{X_{h}}$.
Therefore, by comparison with equation (37), we must have

$\pi\circ\phi_{t}^{X_{H}}=\phi_{t}^{X_{h}}\circ\pi.$

∎

######  Remark 5.14.

The first part of this theorem may be deduced of a more general construction
for Jacobi manifolds in the literature (see [13] and also [20]). Anyway, we
have included a proof of the result to make the paper more self-contained

######  Proposition 5.15.

The smooth submersion
$\hat{J}:T^{*}G\times\mathbb{R}\rightarrow\mathfrak{g}^{*}\times\mathbb{R}$
given by

$\hat{J}(\mu_{g},z)=(J(\mu_{g}),z)$

is a Jacobi map between the canonical Jacobi manifolds
$T^{*}G\times\mathbb{R}$ and $\mathfrak{g}^{*}\times\mathbb{R}$, where $J$ is
given by (35).

###### Proof.

We must prove that for any $f$ and $h$ in
$C^{\infty}(\mathfrak{g}^{*}\times\mathbb{R})$ the following identity holds

$\\{f\circ\hat{J},h\circ\hat{J}\\}_{can}=\\{f,h\\}\circ\hat{J},$

where $\\{\cdot,\cdot\\}_{can}$ is the Jacobi bracket on
$T^{*}G\times\mathbb{R}$ and $\\{\cdot,\cdot\\}$ is the Lie-Poisson-Jacobi
bracket on $\mathfrak{g}^{*}\times\mathbb{R}$.

Note that the function $f\circ\hat{J}$ is $G$-invariant for any function $f$.
Indeed, given $\alpha_{h}\in T_{h}^{*}G$, we have that

$f\circ\hat{J}\circ\phi_{g}(\alpha_{h},z)=f\circ\hat{J}(T_{gh}^{*}\mathcal{L}_{g^{-1}}(\alpha_{h}),z)$

where $\phi_{g}$ denotes the cotangent lifted action of $G$ on
$T^{*}G\times\mathbb{R}$. Then applying the definition of $\hat{J}$, we deduce

$f\circ\hat{J}\circ\phi_{g}(\alpha_{h},z)=f(T_{e}^{*}\mathcal{L}_{gh}(T_{gh}^{*}\mathcal{L}_{g^{-1}}(\alpha_{h})),z).$

Finally, applying the composition rule of cotangent maps we conclude

$f\circ\hat{J}\circ\phi_{g}(\alpha_{h},z)=f(T_{e}^{*}\mathcal{L}_{h}(\alpha_{h})),z)=f\circ\hat{J}(\alpha_{h},z).$

Moreover, note that for any $f$ the restriction of $f\circ\hat{J}$ to the
identity is just $f$ itself since

$f\circ\hat{J}(\alpha_{e},z)=f(T_{e}^{*}\mathcal{L}_{e}(\alpha_{e}),z)=f(\alpha_{e},z).$

Hence, we may apply Proposition 5.10 to deduce that

$\\{f\circ\hat{J},h\circ\hat{J}\\}_{can}(\alpha_{h},z)=\\{f,h\\}(T_{e}^{*}\mathcal{L}_{h}(\alpha_{h}),z)$

which is equivalent to

$\\{f\circ\hat{J},h\circ\hat{J}\\}_{can}(\alpha_{h},z)=\\{f,h\\}\circ\hat{J}(\alpha_{h},z).$

∎

######  Theorem 5.16.

Let $H:T^{*}G\times\mathbb{R}\rightarrow\mathbb{R}$ be a $G$-invariant
function. Then the function
$h:\mathfrak{g}^{*}\times\mathbb{R}\rightarrow\mathbb{R}$ given as
$h:=H|_{\mathfrak{g}^{*}\times\mathbb{R}}$ satisfies

$H=h\circ\hat{J}.$

Moreover, the contact Hamiltonian vector field $X_{H}$ on
$T^{*}G\times\mathbb{R}$ with flow $\phi_{t}^{X_{H}}$ and the Jacobi
Hamiltonian vector field $X_{h}$ on $\mathfrak{g}^{*}\times\mathbb{R}$ with
flow $\phi_{t}^{X_{h}}$ are related by $\hat{J}$ or, equivalently, their flows
satisfy the equation

$\hat{J}\circ\phi_{t}^{X_{H}}=\phi_{t}^{X_{h}}\circ\hat{J}.$

###### Proof.

First note that for any $(\alpha_{g},z)\in T^{*}G\times\mathbb{R}$ we have
that

$H(\alpha_{g},z)=H\circ\hat{J}(\alpha_{g},z),$

since $H$ is $G$-invariant. Then, since
$\hat{J}(\alpha_{g},z)\in\mathfrak{g}^{*}\times\mathbb{R}$, we have that

$H(\alpha_{g},z)=h\circ\hat{J}(\alpha_{g},z),$

which proves the first statement in the theorem.

The second statement is a consequence of Theorem 5.13. Indeed, note that the
map

$\begin{split}\psi:(T^{*}G\times\mathbb{R})/G&\longrightarrow\mathfrak{g}^{*}\times\mathbb{R}\\\
[\alpha_{g},z]&\mapsto\hat{J}(\alpha_{g},z)\end{split}$

is a diffeomorphism. Moreover, by construction we have that
$\psi\circ\pi=\hat{J}$, where
$\pi:T^{*}G\times\mathbb{R}\rightarrow(T^{*}G\times\mathbb{R})/G$ is the
quotient map. Now, the map $\psi$ and the reduced Jacobi brackets on
$(T^{*}G\times\mathbb{R})/G$ given by Theorem 5.13 induce a Jacobi bracket
$\\{\cdot,\cdot\\}_{*}$ on $\mathfrak{g}^{*}\times\mathbb{R}$ that makes
$\psi$ a Jacobi map so that

$\\{f,h\\}_{*}\circ\psi=\\{f\circ\psi,h\circ\psi\\}_{red},\quad f,g\in
C^{\infty}(\mathfrak{g}^{*}\times\mathbb{R}).$

By definition of the reduced bracket, $\pi$ is also a Jacobi map so that

$\\{f,h\\}_{*}\circ\psi\circ\pi=\\{f\circ\psi\circ\pi,h\circ\psi\circ\pi\\}_{can},\quad
f,g\in C^{\infty}(\mathfrak{g}^{*}\times\mathbb{R}),$

which by construction gives

$\\{f,h\\}_{*}\circ\hat{J}=\\{f\circ\hat{J},h\circ\hat{J}\\}_{can},\quad
f,g\in C^{\infty}(\mathfrak{g}^{*}\times\mathbb{R}).$

But by Proposition 5.10 and since $f\circ\hat{J},h\circ\hat{J}$ are
$G$-invariant functions on $T^{*}G\times\mathbb{R}$ we must have that

$\\{f,h\\}_{*}=\\{f,h\\},$

where $\\{\cdot,\cdot\\}$ is the canonical bracket on
$\mathfrak{g}^{*}\times\mathbb{R}$ defined before. So, we may conclude that

$\hat{J}\circ\phi_{t}^{X_{H}}=\psi\circ\pi\circ\phi_{t}^{X_{H}}=\psi\circ\phi_{t}^{X_{h_{red}}}\circ\pi,$

where we used the last statement of Theorem 5.13, relating the Hamiltonian
flow of $H$ with that of $h_{red}$, the function defined by
$H:=h_{red}\circ\pi$. Now observe that, since $\psi$ is a Jacobi map and
$h_{red}=h\circ\psi$, we have that

$h_{red}\circ\pi=h\circ\hat{J}=h\circ\psi\circ\pi,$

the following relation holds between Hamiltonian flows

$\psi\circ\phi_{t}^{X_{h_{red}}}=\phi_{t}^{X_{h}}\circ\psi.$

Therefore, we deduce

$\hat{J}\circ\phi_{t}^{X_{H}}=\phi_{t}^{X_{h}}\circ\psi\circ\pi=\phi_{t}^{X_{h}}\circ\hat{J}.$

∎

######  Example 5.17.

Consider the Lie group $SO(3)$ where the left action is given by
$\mathcal{L}_{R}(R_{0})=RR_{0}$ for any $R,R_{0}\in SO(3)$. Therefore,
$T_{e}\mathcal{L}_{R}(\xi)=R\xi$ for $\xi\in\mathfrak{so}(3)$ and
$T_{e}^{*}\mathcal{L}_{R}(\mu_{R})=R^{T}\mu_{R}$, where $\mu_{R}\in
T_{R}^{*}SO(3)$. Thus, we introduce the Jacobi map
$\hat{J}:T^{*}SO(3)\times\mathbb{R}\rightarrow\mathfrak{so}(3)^{*}\times\mathbb{R}$
given by

$\hat{J}(\mu_{R},z)=(R^{T}\mu_{R},z).$

Let the Hamiltonian $H:T^{*}SO(3)\times\mathbb{R}\rightarrow\mathbb{R}$ be

$H(\mu_{R},z)=\frac{1}{2}\operatorname{tr}(\mu_{R}^{T}\mathbb{J}\mu_{R})+\gamma
z.$

Let $h:\mathfrak{so}^{*}(3)\times\mathbb{R}\rightarrow\mathbb{R}$ be given by
$h(\mu,z)=\frac{1}{2}\mu^{T}\mathbb{J}\mu+\gamma z$, where we are identifying
$\mathfrak{so}^{*}(3)$ with $\mathbb{R}^{3}$ through the map
$\breve{(\cdot)}:\mathbb{R}^{3}\rightarrow\mathfrak{so}^{*}(3)$ satisfying
$\langle\breve{\Pi},\xi\rangle=\Pi_{i}\xi_{i}$, where
$\Pi=(\Pi_{1},\Pi_{2},\Pi_{3})\in\mathbb{R}^{3}$ and $\xi\in\mathfrak{so}(3)$
is identified with $(\xi_{1},\xi_{2},\xi_{3})$ through the hat map given in
Example 4.4. Then, it is straightforward to check that
$h\circ\hat{J}(\mu_{R},z)=H(\mu_{R},z)$.

##  6 Euler-Poincaré-Herglotz and Lie-Poisson-Jacobi equations with symmetry
breaking

The purpose of this section is to obtain the reduced Euler-Poincaré-Herglotz
equations as a generalization of the procedure followed in [17]. Let $X$ be a
finite-dimensional vector space. Consider a Lagrangian system evolving on a
Lie group $G$ and assume the Lagrangian is not symmetry invariant. At the same
time, we consider a representation of $G$ on a dual vector space $X^{*}$.
Coupling to the Lagrangian a parameter depending on vectors in $X^{*}$, we
obtain a symmetry invariant Lagrangian function under the action of $G$.
Loosely speaking, the Lagrangian system is coupled with vectors in $X^{*}$
that are acted by a Lie group symmetry. The associated action considered at
this stage restores the full Lie group symmetry and allows us to apply the
semi-direct product reduction theory [33], [17] (see also [16], [15]), to
obtain the corresponding Euler-Poincaré-Herglotz system on the semi-direct
product Lie algebra $(\mathfrak{g}\ltimes X^{*})\times\mathbb{R}$. This gives
rise to a new reduced system that finds no analogs in classical reduced-order
models in contact mechanical systems.

Assume (i) the Lagrangian function
$L_{\alpha_{0}}:TG\times\mathbb{R}\to\mathbb{R}$ is not $G$-invariant but
depends on a parameter $\alpha_{0}$ that may be considered to be an element of
the dual space $X^{*}$, where $X$ is a vector space. Hence, we define the
extended Lagrangian function as $L_{\textnormal{ext}}:TG\times\mathbb{R}\times
X^{*}\to\mathbb{R}$, with
$L_{\textnormal{ext}}(\cdot,\alpha_{0})=L_{\alpha_{0}}$. Assume (ii) there is
a left representation $\rho$ of $G$ on $X$, i.e., $\rho:G\to\mathrm{GL}(X)$ is
a Lie group morphism. Hence, there is a left representation $\rho^{*}$ of $G$
on $X^{*}$, i.e., $\rho^{*}:G\to\mathrm{GL}(X^{*})$, defined using adjoint
linear maps. Indeed, for any $g\in G$, we set
$\rho^{*}(g)=\rho({g^{-1}})^{*}\in\mathrm{GL}(X^{*})$, where the right-hand
side is the adjoint map of $\rho({g^{-1}})\in\mathrm{GL}(X)$, i.e., for any
$x\in X$ and $\alpha\in X^{*}$

$\displaystyle\langle\rho({g^{-1}})^{*}(\alpha),x\rangle=\langle\alpha,\rho({g^{-1}})(x)\rangle.$

From now, we will denote the image of $g$ by $\rho$ and $\rho^{*}$ by
$\rho_{g}$ and $\rho^{*}_{g}$, respectively, and $\rho^{*}$ will be called the
adjoint representation.

The adjoint representation induces a left action of $G$ on the extended phase
space $TG\times\mathbb{R}\times X^{*}$ defined as

$\displaystyle\Phi:G\times(TG\times\mathbb{R}\times X^{*})$
$\displaystyle\longrightarrow TG\times\mathbb{R}\times X^{*},$
$\displaystyle(g,(v_{h},z,\alpha))$
$\displaystyle\longmapsto(\hat{\mathcal{L}}_{g}(v_{h},z),\rho_{g}^{*}(\alpha)),$
(38)

where $\hat{\mathcal{L}}_{g}$ has been defined in (24). Assume also that (iii)
the extended Lagrangian function is $G$-invariant under (38), i.e.,
$L_{\textnormal{ext}}\circ\Phi_{g}=L_{\textnormal{ext}}$, for any $g\in G$, or
more explicitly

$L_{\textnormal{ext}}(\hat{\mathcal{L}}_{g}(v_{h},z),\rho^{*}_{g}(\alpha))=L_{\textnormal{ext}}(v_{h},z,\alpha),$

for any $v_{h}\in TG$ and $\alpha\in X^{*}$.

We can now obtain the reduced augmented Lagrangian
$\ell_{ext}:\mathfrak{g}\times\mathbb{R}\times X^{*}\to\mathbb{R}$, which is
given by

$\displaystyle\ell_{ext}(\xi,z,\alpha):=L_{ext}(e,\xi,z,\alpha),$ (39)

Critical points of the reduced variational principle for $\ell_{ext}$ satisfy
the following Euler-Poincaré-Herglotz equations (see Theorem 6.1 below for a
proof)

$\displaystyle\frac{d}{dt}\left(\frac{\delta\ell_{ext}}{\delta\xi}\right)$
$\displaystyle=\hbox{ad}^{*}_{\xi}\left(\frac{\delta\ell_{ext}}{\partial\xi}\right)+\mathbf{J}_{X}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha},\alpha\right)+\frac{\delta\ell_{\textnormal{ext}}}{\delta\xi}\frac{\partial\ell_{\textnormal{ext}}}{\partial
z},$ (40) $\displaystyle\dot{\alpha}$
$\displaystyle={-\rho_{\xi}^{\prime*}(\alpha),\quad\alpha(0)=\alpha_{0}},$
(41) $\displaystyle\dot{z}$ $\displaystyle=\ell_{\textnormal{ext}}$ (42)

where $\mathbf{J}_{X}:T^{*}X\cong X\times X^{*}\to\mathfrak{g}^{*}$ is the
momentum map corresponding to the left action of $G$ on $X$ defined using the
left representation $\rho$ of $G$ on $X$, i.e., for any $x\in X$,
$\xi\in\mathfrak{g}$ and $\alpha\in X^{*}$

$\displaystyle\langle\mathbf{J}_{X}(x,\alpha),\xi\rangle=\langle\alpha,\xi_{X}(x)\rangle,$
(43)

with $\xi_{X}$ being the infinitesimal generator of the left action of $G$ on
$X$ and $\rho^{\prime*}:\mathfrak{g}\to\mathfrak{gl}(X^{*})$ is defined as the
adjoint of $\rho^{\prime}$, which is the infinitesimal representation induced
by the left representation $\rho$ of $G$ on $X$. Note that the solution to
(41) is given by $\alpha(t)=\rho^{*}_{g^{-1}(t)}(\alpha_{0})$. Also, note that
using the notation of [17], [18] we have
$\mathbf{J}_{X}(x,\alpha)=x\diamond\alpha$. To summarize, we have the
following theorem.

######  Theorem 6.1.

Let $g:I\rightarrow G$ and $z:I\rightarrow\mathbb{R}$ be smooth curves on an
interval $I$. Define the curves
$\xi(t)=T_{g(t)}\mathcal{L}_{g^{-1}(t)}(\dot{g}(t))$ and
$\alpha(t)=\rho^{*}_{g^{-1}(t)}(\alpha_{0})$, for $\alpha_{0}\in X^{*}$ Let
also $g_{0}=g(0)\in G$. Under assumptions (i)-(iii), the following are
equivalent:

  1. 1.

The variational principle

$\delta\int_{0}^{h}L_{\alpha_{0}}(g(t),\dot{g}(t),z(t))\
dt=0,\quad\dot{z}=L_{\alpha_{0}}(g(t),\dot{g}(t),z(t))$

holds for variations of $g$ with fixed endpoints.

  2. 2.

The curves $g:I\rightarrow G$ and $z:I\rightarrow\mathbb{R}$ satisfy Euler-
Poincaré-Herglotz equations for $L_{\alpha_{0}}$;

  3. 3.

The reduced constrained variational principle

$\delta\int_{0}^{h}\ell_{\textnormal{ext}}(\xi(t),z(t),\alpha(t))\
dt=0,\quad\dot{z}=\ell_{\textnormal{ext}}(\xi(t),z(t),\alpha(t))$ (44)

holds using variations of $\xi$ and $\alpha$ of the form

$\displaystyle\delta{\xi}$ $\displaystyle=\dot{\eta}+\hbox{ad}_{\xi}\eta,$
$\displaystyle\delta{\alpha}$
$\displaystyle={-\rho_{\eta}^{\prime*}(\alpha)}.$

  4. 4.

The curves $\xi(t)=T_{g(t)}\mathcal{L}_{g^{-1}(t)}(\dot{g}(t))$ and $z$
satisfy the Euler-Poincaré-Herglotz equations

$\displaystyle\frac{d}{dt}\left(\frac{\delta\ell_{ext}}{\delta\xi}\right)$
$\displaystyle=\hbox{ad}^{*}_{\xi}\left(\frac{\delta\ell_{ext}}{\partial\xi}\right)+\mathbf{J}_{X}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha},\alpha\right)+\frac{\delta\ell_{\textnormal{ext}}}{\delta\xi}\frac{\partial\ell_{\textnormal{ext}}}{\partial
z},$ (45) $\displaystyle\dot{\alpha}$
$\displaystyle={-\rho_{\xi}^{\prime*}(\alpha)},\ \
\alpha(0)=\rho^{*}_{g_{0}^{-1}}(\alpha_{0}),\hskip
5.0pt\dot{z}=\ell_{\textnormal{ext}}$ (46)

###### Proof.

The proof follows arguments similar to the ones given in Theorem 4.1 and
Theorem 4.2. The equivalence of items $1.$ and $2.$ is general for all contact
manifolds, so in particular to $TG\times\mathbb{R}$ (see [2]).

We will show that the first variational principle implies the second
constrained variational principle, that is, we prove the equivalence between
$1.$ and $3$. Notice that, since $L_{\textnormal{ext}}$ is $G$-invariant under
$\Phi_{g}$, and $\alpha=\rho^{*}_{g^{-1}}(\alpha_{0})$ the integrand in the
first variational principle is equal to the integrand in the second
constrained variational principle. Indeed, using the defintion of $\ell_{ext}$
and the $G$-invariance, we get

$\ell_{\textnormal{ext}}(\xi,z,\alpha)=L_{ext}(e,\xi,z,\alpha)=L_{ext}(e,T_{g}\mathcal{L}_{g^{-1}}(\dot{g}),z,\rho^{*}_{g^{-1}}(\alpha_{0}))=L_{\alpha_{0}}(g,\dot{g},z),$

where we dropped the time dependence to ease the notation.

Moreover, all variations of $g$ vanishing at the endpoints induce and are
induced by variations of $\xi$ of the form
$\delta{\xi}=\dot{\eta}+\operatorname{ad}_{\xi}\eta$, with
$\eta(0)=\eta(T)=0$.

Recall that the infinitesimal variations $\eta$ are given by
$\eta=T_{g}\mathcal{L}_{g^{-1}}(\delta{g})$. This fact is used to write the
variations of the curve $\alpha$ as
$\delta{\alpha}=-\rho_{\eta}^{\prime*}(\alpha)$ (see [18]). Hence, the first
variational principle implies the second constrained variational principle.

To conclude, we show the equivalence between $3$. and $4$. Indeed by using the
definition of the auxiliar variable $\sigma(t)$, the calculations for Theorem
4.1, integration by parts, and the fact that

$\displaystyle\left\langle\delta{\alpha},\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha}\right\rangle=\left\langle-\rho_{\eta}^{\prime*}(\alpha),\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha}\right\rangle=\left\langle\alpha,\rho_{\eta}^{\prime}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha}\right)\right\rangle=\Big{\langle}\mathbf{J}_{X}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha},\alpha\right),\eta\Big{\rangle},$

where we used the infinitesimal version of the definition of the adjoint
representation,i.e.,

$\left\langle\rho_{\eta}^{\prime*}(\alpha),x\right\rangle=-\left\langle\alpha,\rho_{\eta}^{\prime}\left(x\right)\right\rangle,\quad\forall
x\in X,\alpha\in X^{*},\eta\in\mathfrak{g},$

we deduce that the reduced Herglotz variational principle holds if and only if

$\sigma(t)\left[\text{ad}_{\xi}^{*}\frac{\delta
l}{\partial\xi}-\frac{d}{dt}\left(\frac{\delta
l}{\delta\xi}\right)+\mathbf{J}_{X}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha},\alpha\right)+\frac{\partial
l}{\partial z}\frac{\delta l}{\delta\xi}\right]=0.$ (47)

By taking the time derivative of $\alpha$, we get
$\dot{\alpha}=-\rho_{\xi}^{\prime*}(\alpha),\hskip
5.0pt\alpha(0)=\rho^{*}_{g_{0}^{-1}}(\alpha_{0})$ and, since the auxiliary
function $\sigma$ is nowhere zero, the result follows. ∎

There is a special cases of Theorem 6.1 that is of particular interest and we
state it as corollary.

######  Corollary 6.2.

Let $X=\mathfrak{g}$ and $\rho$ be the adjoint representation of $G$ on $X$,
i.e., $\rho_{g}=\hbox{Ad}_{g}$, for any $g\in G$. Then, the Euler–Poincaré-
Herglotz equations (45)–(46) give the following equations

$\displaystyle\frac{d}{dt}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\xi}\right)$
$\displaystyle=\operatorname{ad}^{*}_{\xi}\left(\frac{\delta\ell_{\textnormal{ext}}}{\delta\xi}\right)-\operatorname{ad}_{\frac{\delta\ell_{\textnormal{ext}}}{\delta\alpha}}^{*}\alpha+\frac{\delta\ell_{\textnormal{ext}}}{\delta\xi}\frac{\partial\ell_{\textnormal{ext}}}{\partial
z},$ $\displaystyle\dot{\alpha}$
$\displaystyle={-\operatorname{ad}_{\xi}^{*}\alpha,\hskip
5.0pt\alpha(0)=\operatorname{Ad}_{g_{0}^{-1}}^{*}\alpha_{0}},\,\dot{z}=\ell_{\textnormal{ext}}.$

###### Proof.

We first begin by noting that $\alpha\in X^{*}=\mathfrak{g}^{*}$ and
$\rho^{*}$ is the coadjoint representation of $G$ on $X^{*}$, i.e.,
$\rho^{*}_{g}=\operatorname{Ad}_{g}^{*}$, for any $g\in G$. We also have
$\rho_{\xi}^{\prime}=\operatorname{ad}_{\xi}$, for any $\xi\in\mathfrak{g}$
and it follows that $\xi_{X}(x)=\operatorname{ad}_{\xi}x$, for any $x\in X$.
From (43), we have

$\displaystyle\langle\mathbf{J}_{X}(x,\alpha),\xi\rangle$
$\displaystyle=\langle\alpha,\operatorname{ad}_{\xi}x\rangle$
$\displaystyle=\langle\alpha,-\operatorname{ad}_{x}\xi\rangle$
$\displaystyle=\langle-\operatorname{ad}_{x}^{*}\alpha,\xi\rangle,$

which gives $\mathbf{J}_{X}(x,\alpha)=-\operatorname{ad}_{x}^{*}\alpha$. ∎

###  6.1 Reduced Legendre transformation

If we assume that the reduced Lagrangian $\ell_{\textnormal{ext}}$ is hyper-
regular, then we can obtain the reduced Hamiltonian
$h_{\textnormal{ext}}:\mathfrak{g}^{*}\times\mathbb{R}\times
X^{*}\to\mathbb{R}$ (by using the reduced Legendre transformation) given by

$\displaystyle
h_{\textnormal{ext}}(\mu,z,\alpha)=\langle\mu,\xi\rangle-\ell_{\textnormal{ext}}(\xi,z,\alpha),$

where $\mu=\frac{\partial\ell_{\textnormal{ext}}}{\partial\xi}$, with
$\mu(\cdot)\in C^{1}([0,T],\mathfrak{g}^{*})$. The Euler–Poincaré-Herglotz
equations (45)–(46) can now be written as the Lie–Poisson–Jacobi equations,
which are given by

$\displaystyle\dot{\mu}$ $\displaystyle=\emph{ad}^{*}_{\frac{\delta
h_{\textnormal{ext}}}{\delta\mu}}\mu-\mu\frac{\partial
h_{\textnormal{ext}}}{\partial z}-\mathbf{J}_{X}\left(\frac{\delta
h_{\textnormal{ext}}}{\delta\alpha},\alpha\right),$ (48)
$\displaystyle\dot{z}$ $\displaystyle=\langle\mu,\frac{\delta
h_{\textnormal{ext}}}{\delta\mu}\rangle-h_{\textnormal{ext}}(\mu,z,\alpha).$
(49) $\displaystyle\dot{\alpha}$ $\displaystyle={-\rho^{\prime*}_{\frac{\delta
h_{\textnormal{ext}}}{\delta\alpha}}(\alpha),\hskip
5.0pt\alpha(0)=\rho^{*}_{g_{0}^{-1}}(\alpha_{0})}.$ (50)

###  6.2 Lie-Poisson-Jacobi equations for systems with symmetry breaking

We can define a Jacobi bracket on the space of smooth functions on
$\mathfrak{g}^{*}\times\mathbb{R}\times X^{*}$ with respect to which the
previous equations are just the Hamiltonian equations. Consider the bracket

$\begin{split}\\{f,g\\}_{X^{*}}(\mu,z,\alpha)=\left\langle\mu,\left[\frac{\delta
f}{\delta\mu},\frac{\delta
g}{\delta\mu}\right]\right\rangle&+\left\langle\mu,\frac{\delta
f}{\delta\mu}\right\rangle\frac{\partial g}{\partial
z}-\left\langle\mu,\frac{\delta g}{\delta\mu}\right\rangle\frac{\partial
f}{\partial z}\\\ &+\left\langle\alpha,\frac{\delta f}{\delta\mu}\frac{\delta
g}{\delta\alpha}-\frac{\delta f}{\delta\alpha}\frac{\delta
g}{\delta\mu}\right\rangle\end{split}$

This bracket falls into the same definition of Lie-Poisson-Jacobi bracket as
the one given in Definition 5.8. In this case, the associated Lie algebra is
$\mathfrak{g}\ltimes X$ (see [18] for more details on the subject of semi-
direct products).

######  Proposition 6.3.

The Hamiltonian equations with respect to the Lie-Poisson-Jacobi bracket
$\\{\cdot,\cdot\\}_{X^{*}}$ and the Hamiltonian function $h$ on
$\mathfrak{g}^{*}\times\mathbb{R}\times X^{*}$ are

$\displaystyle\dot{\mu}$ $\displaystyle=\emph{ad}^{*}_{\frac{\delta
h}{\delta\mu}}\mu-\mu\frac{\partial h}{\partial
z}-\mathbf{J}_{X}\left(\frac{\delta h}{\delta\alpha},\alpha\right),$
$\displaystyle\dot{z}$ $\displaystyle=\langle\mu,\frac{\delta
h}{\delta\mu}\rangle-h(\mu,z,\alpha).$ $\displaystyle\dot{\alpha}$
$\displaystyle={-\rho^{\prime*}_{\frac{\delta h}{\delta\mu}}(\alpha)}.$

###### Proof.

Let $f\in C^{\infty}(\mathfrak{g}^{*}\times\mathbb{R}\times X^{*})$. Then,

$\dot{f}=\langle
df(\mu,z,\alpha),(\dot{\mu},\dot{z},\dot{\alpha})\rangle=\left\langle\dot{\mu},\frac{\delta
f}{\delta\mu}\right\rangle+\frac{\partial f}{\partial
z}\dot{z}+\left\langle\dot{\alpha},\frac{\delta
f}{\delta\alpha}\right\rangle.$

Also, using the definition of the Jacobi bracket we have that

$\begin{split}\\{h,f\\}_{X^{*}}(\mu,z,\alpha)=\left\langle\mu,\left[\frac{\delta
h}{\delta\mu},\frac{\delta
f}{\delta\mu}\right]\right\rangle&+\left(\left\langle\mu,\frac{\delta
h}{\delta\mu}\right\rangle-h\right)\frac{\partial f}{\partial
z}-\left(\left\langle\mu,\frac{\delta
f}{\delta\mu}\right\rangle-f\right)\frac{\partial h}{\partial z}\\\
&+\left\langle\alpha,\frac{\delta h}{\delta\mu}\frac{\delta
f}{\delta\alpha}-\frac{\delta h}{\delta\alpha}\frac{\delta
f}{\delta\mu}\right\rangle.\end{split}$

Thus using (31) we have that

$\begin{split}\dot{f}=\\{h,f\\}_{X^{*}}-f\frac{\partial h}{\partial
z}&=\left\langle\text{ad}_{\frac{\delta
h}{\delta\mu}}^{*}\mu-\mu\frac{\partial h}{\partial z},\frac{\delta
f}{\delta\mu}\right\rangle+\left(\left\langle\mu,\frac{\delta
h}{\delta\mu}\right\rangle-h\right)\frac{\partial f}{\partial z}\\\
&+\left\langle\alpha,\rho_{\frac{\delta
h}{\delta\mu}}^{{}^{\prime}}\frac{\delta
f}{\delta\alpha}\right\rangle-\left\langle\alpha,\rho_{\frac{\delta
f}{\delta\mu}}^{{}^{\prime}}\frac{\delta h}{\delta\alpha}\right\rangle\\\
&=\left\langle\text{ad}_{\frac{\delta h}{\delta\mu}}^{*}\mu-\mu\frac{\partial
h}{\partial z},\frac{\delta
f}{\delta\mu}\right\rangle+\left(\left\langle\mu,\frac{\delta
h}{\delta\mu}\right\rangle-h\right)\frac{\partial f}{\partial z}\\\
&+\left\langle\rho_{\frac{\delta
h}{\delta\mu}}^{{}^{\prime}*}\alpha,\frac{\delta
f}{\delta\alpha}\right\rangle-\left\langle\mathbf{J}_{X}\left(\frac{\delta
h}{\delta\alpha},\alpha\right),\frac{\delta f}{\delta\mu}\right\rangle\\\
\end{split}$

Since the function $f$ is arbitrary, the result follows. ∎

###  6.3 The Heavy Top with dissipation

Consider $G=\mathrm{SO}(3)$, with Lagrangian function given by

${L_{\mathbf{e}_{3}}(g,\xi,z)=\frac{1}{2}\langle\xi,\mathbb{I}\xi\rangle-m\mathbf{g}l\langle\mathbf{e}_{3},R\boldsymbol{\chi}\rangle-\gamma
z},$

where $R\in SO(3)$, $\langle\xi,\xi\rangle=\operatorname{tr}(\xi^{T}\xi)$, for
any $\xi\in\mathfrak{g}=\mathfrak{so}(3)$,
$\xi(\cdot)=\sum_{i=1}^{3}\xi^{i}(\cdot)e_{i}$, with the elements of the basis
of $\mathfrak{g}$ given by

$\displaystyle e_{1}=\begin{bmatrix}0&\phantom{-}0&\phantom{-}0\\\
0&\phantom{-}0&-1\\\ 0&\phantom{-}1&\phantom{-}0\end{bmatrix},\hskip
5.0pte_{2}=\begin{bmatrix}\phantom{-}0&\phantom{-}0&\phantom{-}1\\\
\phantom{-}0&\phantom{-}0&\phantom{-}0\\\
-1&\phantom{-}0&\phantom{-}0\end{bmatrix},\hskip
5.0pte_{3}=\begin{bmatrix}0&-1&\phantom{-}0\\\ 1&\phantom{-}0&\phantom{-}0\\\
0&\phantom{-}0&\phantom{-}0\end{bmatrix},$

which satisfy

$\displaystyle[e_{1},e_{2}]=e_{3},\hskip 5.0pt[e_{2},e_{3}]=e_{1},\hskip
5.0pt[e_{3},e_{1}]=e_{2},$

$\mathbb{I}:\mathfrak{g}\to\mathfrak{g}^{*}=\mathfrak{so}(3)^{*}$ is the
inertia tensor of the top (it is calculated with respect to the pivot, which
is not, in general, the center of mass), $m$ is the mass of the body,
$\mathbf{g}$ is the acceleration due to gravity, $\mathbf{e}_{3}$ is the
vertical unit vector, $\boldsymbol{\chi}\in\mathbb{R}^{3}$ is the unit vector
from the point of support to the direction of the body’s center of mass
(constant) in body coordinates, $l$ is the length of the line segment between
these two points and $\gamma\in\mathbb{R}$.

Under the dual pairing, where
$\langle\alpha,\xi\rangle=\operatorname{tr}(\alpha\xi)$, for any
$\xi\in\mathfrak{g}$ and $\alpha\in\mathfrak{g}^{*}$, the elements of the
basis of $\mathfrak{g}^{*}$ are given by

$\displaystyle e^{1}=\begin{bmatrix}0&\phantom{-}0&\phantom{-}0\\\\[4.0pt]
0&\phantom{-}0&\phantom{-}\dfrac{1}{2}\\\\[4.0pt]
0&-\dfrac{1}{2}&\phantom{-}0\end{bmatrix},\hskip
5.0pte^{2}=\begin{bmatrix}\phantom{-}0&\phantom{-}0&-\dfrac{1}{2}\\\\[8.0pt]
\phantom{-}0&\phantom{-}0&\phantom{-}0\\\\[4.0pt]
\phantom{-}\dfrac{1}{2}&\phantom{-}0&\phantom{-}0\end{bmatrix},\hskip
5.0pte^{3}=\begin{bmatrix}\phantom{-}0&\phantom{-}\dfrac{1}{2}&\phantom{-}0\\\\[4.0pt]
-\dfrac{1}{2}&\phantom{-}0&\phantom{-}0\\\\[8.0pt]
\phantom{-}0&\phantom{-}0&\phantom{-}0\end{bmatrix}.$

It is easy to verify that the Lagrangian function is invariant under the left
action of the isotropy group

$\displaystyle G_{\alpha_{0}}=\\{g\in
G\mid\operatorname{Ad}_{g}^{*}\alpha_{0}=\alpha_{0}\\}\cong\mathrm{SO}(2),$

i.e., rotations about the vertical axis $\mathbf{e}_{3}$, but not
$G$-invariant. The potential function breaks the symmetry partially. However,
the framework we introduced before will help us to restore the full symmetry.

Let $X=\mathfrak{g}$ and $\rho$ be the adjoint representation of $G$ on $X$,
i.e., $\rho_{g}=\operatorname{Ad}_{g}$, for any $g\in G$. So, $\rho^{*}$ is
the coadjoint representation of $G$ on $X^{*}=\mathfrak{g}^{*}$, i.e.,
$\rho^{*}_{g}=\operatorname{Ad}_{g}^{*}$, for any $g\in G$. Let the extended
Lagrangian function $L_{\textnormal{ext}}:G\times\mathfrak{g}\times
X^{*}\to\mathbb{R}$ be given as follows

$\displaystyle{L_{\textnormal{ext}}(g,\xi,\breve{\alpha})=\frac{1}{2}\langle\xi,\mathbb{I}\xi\rangle-m\mathbf{g}l\langle
R^{-1}\alpha,\boldsymbol{\chi}\rangle-\gamma z},$

where $\alpha\in\mathbb{R}^{3}$ is identified with
$\breve{\alpha}\in\mathfrak{g}^{*}$. If we set $\breve{\alpha}_{0}=-2e^{3}$ in
the extended Lagrangian function, we recover our original Lagrangian function.
It is easy to verify that the extended Lagrangian function is $G$-invariant
under (38), i.e., $L_{\textnormal{ext}}\circ\Phi_{g}=L_{\textnormal{ext}}$,
for any $g\in G$.

Thus, we can define the reduced extended Lagrangian function to be

$\ell_{ext}(\xi,\alpha,z)=\frac{1}{2}\langle\xi,\mathbb{I}\xi\rangle-m\mathbf{g}l\langle\alpha,\boldsymbol{\chi}\rangle-\gamma
z,$

and by Corollary 6.2, the Euler–Poincaré-Herglotz equations (under the
identifications $\mathfrak{g}\cong\mathbb{R}^{3}$ and
$\mathfrak{g}^{*}\cong\mathbb{R}^{3}$) give the following equations

$\displaystyle\mathbb{I}\dot{\xi}$
$\displaystyle={\operatorname{ad}_{\xi}^{*}\mathbb{I}\xi-\operatorname{ad}_{\frac{\partial
L_{\textnormal{ext}}}{\partial\alpha}}^{*}\alpha-\gamma\mathbb{I}\xi},$ (51)
$\displaystyle\dot{\alpha}$
$\displaystyle={-\operatorname{ad}_{\xi}^{*}\alpha,\hskip
5.0pt\alpha(0)=\operatorname{Ad}_{g_{0}^{-1}}^{*}\alpha_{0}},\quad\dot{z}=\ell_{\textnormal{ext}}.$
(52)

Using the expressions (for more details, see [18], [34])

$\displaystyle\operatorname{ad}_{\xi}^{*}\mathbb{I}\xi=\mathbb{I}\xi\times\xi,\hskip
5.0pt\operatorname{ad}_{\frac{\partial
L_{\textnormal{ext}}}{\partial\alpha}}^{*}\alpha=-m\mathbf{g}l\hskip
2.0pt\alpha\times\boldsymbol{\chi},\hskip
5.0pt\operatorname{ad}_{\xi}^{*}\alpha=\alpha\times\xi,\hskip
5.0pt{\operatorname{Ad}_{g_{0}^{-1}}^{*}\alpha_{0}=g_{0}\alpha_{0}}$

and $\mathbb{I}\in\mathbb{R}^{3\times 3}$ is the inertia matrix, (51)–(52)
give the following equations

$\displaystyle\mathbb{I}\dot{\xi}$
$\displaystyle=\mathbb{I}\xi\times\xi-m\mathbf{g}l\hskip
2.0pt\boldsymbol{\chi}\times\alpha-\gamma\mathbb{I}\xi,$
$\displaystyle\dot{\alpha}$ $\displaystyle={-\alpha\times\xi,\hskip
5.0pt\alpha(0)=g_{0}\alpha_{0}},\,\,\dot{z}=\ell_{\textnormal{ext}}.$

In the case $C=\gamma\mathbb{I}$ we obtain the equations for rigid body
attitude dynamics studied in [35], [10] for the triaxial attitude control
testbed given in [4].

We do not give all the details and leave it up to reader to verify that the
Lie–Poisson–Jacobi equations (6.3)–(50) are given by

$\displaystyle\dot{\mu}$
$\displaystyle=\mu\times\mathbb{I}^{-1}\mu-m\mathbf{g}l\hskip
2.0pt\boldsymbol{\chi}\times\alpha-\gamma\mu,$ $\displaystyle\dot{\alpha}$
$\displaystyle={-\alpha\times\mathbb{I}^{-1}\mu,\hskip
5.0pt\alpha(0)=g_{0}\alpha_{0}},\,\,\quad\dot{z}=\frac{1}{2}\mu^{T}\mathbb{I}\mu-m\mathbf{g}l\langle
g^{-1}\alpha,\boldsymbol{\chi}\rangle-\gamma z.$

##  7 Conclusions and Future Work

In this paper, we have established a geometric framework to study reduced
contact dynamics on Lie groups. In future work we wish to employ this
framework to study

  * •

the stability of reduced contact systems,

  * •

Noether’s theorems and dissipated quantities - the analogous notion of
conserved quantities in symplectic dynamics.

  * •

contact structure-preserving discretization schemes of contact dynamics on Lie
groups and semi-direct products. Discretization schemes complying with the
reduction process could also be used to find and numerically implement optimal
control policies in systems with dissipation.

## Acknowledgments

The authors acknowledge financial support from Grant PID2019-106715GB-C21
funded by MCIN/AEI/ 10.13039/501100011033. J.C. Marrero and E. Padrón
acknowledge financial support from the Spanish Ministry of Science and
Innovation under grant PGC2018-098265-B-C32.

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