# Geodesic extensions of mechanical systems with nonholonomic constraints

Malika Belrhazi and Tom Mestdag
Department of Mathematics, University of Antwerp,
Middelheimlaan 1, 2020 Antwerpen, Belgium

###### Abstract

For a Lagrangian system with nonholonomic constraints, we construct extensions
of the equations of motion to sets of second-order ordinary differential
equations. In the case of a purely kinetic Lagrangian, we investigate the
conditions under which the nonholonomic trajectories are geodesics of a
Riemannian metric, while preserving the constrained Lagrangian. We interpret
the algebraic and PDE conditions of this problem as infinitesimal versions of
the relation between the nonholonomic exponential map and the Riemannian
metric. We discuss the special case of a Chaplygin system with symmetries and
we end the paper with a worked-out example.

Keywords: Lagrangian systems, nonholonomic constraints, second-order ordinary
differential equations, Riemannian metrics, sode extensions, exponential map,
Chaplygin system.

Mathematics Subject Classification: 37J06, 37J60, 53Z05, 70G45, 70G65.

##  1 Introduction

In many applications involving mechanical systems, the motion is often limited
by velocity-dependent constraints. If these are given by equations, and if
they are non-integrable, they are called nonholonomic constraints. Some
examples of such systems are those that involve wheels that roll without
slipping, or a skate that is prevented to move in a direction perpendicular to
the blade.

Let $L=T-V$ be the Lagrangian of such a mechanical system (given by the
difference in kinetic and potential energy) on a configuration manifold $Q$
and let the nonholonomic constraints (assumed to be linear, throughout the
paper) be represented by a distribution ${\mathcal{D}}$ on $Q$. The equations
of motion for such systems $(L,{\mathcal{D}})$ follow from the so-called
Lagrange-d’Alembert principle (see e.g. [7, 13]). It is well-known that this
is not a purely variational principle (for the Lagrangian functional
associated to $L$), since it only makes use of variations that are required to
follow the constraints. The resulting equations, the so-called
Lagrange-d’Alembert equations of $(L,{\mathcal{D}})$, are therefore not simply
the Euler-Lagrange equations of $L$. Instead, the nonholonomic equations of
motion contain Lagrangian multipliers, and when eliminated, they form a set of
first- and second-order ordinary differential equations (see their expressions
in Section 2).

We will call the solutions of the Lagrange-d’Alembert equations the
nonholonomic trajectories of $(L,{\mathcal{D}})$. In many papers, it has been
recognized that it is often useful if the nonholonomic trajectories can be
interpreted, in one way or another, as part of the solutions of the Euler-
Lagrange equations of some unrelated (and possibly non-mechanical) Lagrangian
$\hat{L}$. This question is often referred to as the ‘Hamiltonization problem
of nonholonomic mechanics’. It was probably first raised in the context of
quantization of classical nonholonomic systems (see [1, 10, 20]), but there
are also other important areas of possible applications, such as e.g. that of
geometric integrators of nonholonomic systems (see e.g. [22, 26]). The
subtlety of the Hamiltonization problem lies in the fact that, in contrast to
the nonholonomic equations of $(L,{\mathcal{D}})$, the Euler-Lagrange
equations of $\hat{L}$ are purely second-order ordinary differential
equations. There have been many different approaches to this question. For
example, in [9, 6] the main idea is to enlarge the mixed first- and second
order system of $(L,{\mathcal{D}})$ to a (chosen) full system of sodes and to
find a Lagrangian $\hat{L}$ for that system using techniques from the so-
called inverse problem of Lagrangian mechanics. In other papers [4, 5, 11, 19]
one rather uses the symmetries of the nonholonomic system to (roughly
speaking) first reduce it to a sode on a quotient manifold. Then, one tries to
find a symplectic or Poisson structure on that quotient for which it is a
Hamiltonian vector field.

The results of the current paper belong to the first approach. In short, we
will investigate the conditions under which it is possible to have a
Hamiltonization by means of the geodesic equations of a Riemannian metric. If
that can be achieved, we will speak of a ‘geodesic extension’ of the
nonholonomic dynamics (even though this question seems to be dubbed the
kinetic Lagrangianization problem in the conclusions of the paper [3]).

In Section 2 we recall the most elementary concepts in this context, in an
attempt to keep the paper self-contained. Here it is important to notice that
we will use a specific technique throughout the paper. Indeed, to suppress the
use of long expressions, and to gain ready geometric insight, it will be
advantageous to express the intrinsic and coordinate-independent tools that we
will use in terms of a so-called anholonomic frame of vector fields. In
particular, we will often choose the frame in such a way that the vanishing of
a part of its corresponding quasi-velocities represents the constraints and
its distribution ${\mathcal{D}}$. In the last sections, it will be of further
advantage to specify the frame in such a way that it also follows the
orthogonal decomposition of the tangent spaces by a given Riemannian metric.

In Section 3 we will first define the notion of a sode extension of a
nonholonomic system as a set of (only) second-order differential equations
whose solutions set contains the nonholonomic trajectories. The name comes
from [3], but the idea can already be found in e.g. [8] as ‘Newton laws with
the restriction property’ and, implicitly, in [9] where the sode extensions
are called ‘associated systems’. In the first part of the paper, we will show
that such sode extensions always exist. We will give two natural constructions
of a sode extension in Propositions 1 and 2. We will motivate our
constructions by showing in Proposition 3 that they represent, in fact, a
generalization of the well-known nonholonomic connection (see e.g.[24, 13]) to
the level of arbitrary Lagrangians. We will say that a nonholonomic system is
purely kinetic, if its potential $V$ is constant and if its kinetic energy can
be derived from a Riemannian metric $g$. Specifically for a purely kinetic
nonholonomic systems, there is also a third construction of a sode extension
at the end of Section 6.

From Section 4 onwards, we restrict the attention to only purely kinetic
nonholonomic systems. We will investigate the conditions under which the
Riemannian metric $g$ of the kinetic energy can be modified to a metric
$\hat{g}$, such that the nonholonomic trajectories become part of its set of
geodesics (and therefore also of the Euler-Lagrange equations that can be
associated to the Lagrangian $\hat{L}$ of the kinetic energy $\hat{g}$). The
specific kind of geodesic extension we envisage here will be such that the
constrained Lagrangian $L_{C}=L|_{\mathcal{D}}$ remains preserved. This means
that we will assume that the restrictions of the Riemannian metrics $g$ and
$\hat{g}$ to the constraint distribution ${\mathcal{D}}$ are the same. We will
say, in that case, that $\hat{g}$ is a ${\mathcal{D}}$-preserving modification
of $g$. We are able to show in Lemma 2 that this has the advantage that the
question narrows to the search for a suitable 1-form (with components
$\theta_{i}$ along the distribution ${\mathcal{D}}$) that satisfies a set of
algebraic and PDE conditions. If these have a solution, we show in Proposition
4 that we can always find a Riemannian metric $\hat{g}$ that satisfies our
purposes. In Section 5 we relate (in Propositions 5 and 6) the conditions of
Lemma 2 to the kind of sode extensions we had encountered in Section 3.

The exponential map of a Riemannian metric plays an important role in
differential geometry and geometric mechanics, for example in results
concerning Jacobi fields, conjugate points, variations of the energy and focal
points. In this paper we will also address the search for a geodesic extension
from the viewpoint of the so-called nonholonomic exponential map. This concept
has been recently introduced in the paper [2], and further developed in [3].
There, among other results, it was used to show that the nonholonomic
trajectories of a purely kinetic system, originating from a fixed point $q$,
can be interpreted as the radial geodesics of a Riemannian metric on a
submanifold. This Riemannian metric should not be confused with the (full)
Riemannian metric $\hat{g}$ we are looking for, since it (and the submanifold
on which it is defined) is different for each different point $q$.
Notwithstanding, like [3], we will be inspired here by the same underlying
result from Riemannian geometry: the so-called Gauss Lemma (see e.g. [18]).
This Lemma states that the exponential map of a Riemannian metric at $q$ is a
radial isometry. In Proposition 7 we show that the algebraic condition of
Lemma 2 follows, as soon as the exponential map and the sought-for modified
metric $\hat{g}$ satisfy a certain Gauss-type condition. Next, in Proposition
8 we show that also the PDE condition can be interpreted as an infinitesimal
condition. Along each nonholonomic trajectory, it can be seen to be equivalent
with a conservation law that involves the parallel transport along a linear
connection whose quadratic spray is, in fact, again a certain sode extension.
For these reasons, we will call the algebraic and PDE conditions infinitesimal
Gauss conditions.

In Section 7 we investigate the presence of symmetry. Chaplygin nonholonomic
systems are such that the Lagrangian and the constraints are both invariant
under the action of a symmetry Lie group $G$. Moreover, it is assumed that
$Q\to Q/G$ is a principal fibre bundle and that the constraint distribution
${\mathcal{D}}$ is the horizontal distribution of a principal connection. We
show in Proposition 10 that the conditions of Lemma 2 can be simplified, and
that the geodesic extension problem can be stated as the search for linear
first integrals that satisfy some further algebraic conditions.

We illustrate the results at the end of the paper with the example of a disk
that rolls vertically without slipping, and we compare the metrics we find
with those that had already appeared in the literature for this example [3,
21, 9].

##  2 Tangent bundle geometry, Lagrangian systems and nonholonomic systems

We will assume throughout that $Q$ is an $n$-dimensional real differentiable
manifold, with local coordinates $(q^{\alpha})$. We will assume that
$\\{X_{\alpha}\\}$ represents an anholonomic frame of vector fields on $Q$.
The adjective anholonomic means here that the Lie brackets of these vector
fields do not necessarily vanish, and satisfy the relation
$[X_{\alpha},X_{\beta}]=R^{\gamma}_{\alpha\beta}X_{\gamma}$. We will call the
components $v^{\alpha}$ in the decomposition $v_{q}=v^{\alpha}X_{\alpha}(q)$
the quasi-velocities of a tangent vector $v_{q}\in T_{q}Q$. We may then use
$(q^{\alpha},v^{\alpha})$ as (non-standard) coordinates on the tangent
manifold $TQ$, where most of the objects of interest live.

Any vector field $X$ on $Q$ can be lifted to vector fields on $TQ$, by means
of the complete and vertical lift, respectively: If
$X=X^{\alpha}\partial/\partial{q^{\alpha}}$, then in natural coordinates
$(q^{\alpha},{\dot{q}}^{\alpha})$,

$X^{C}=X^{\alpha}\frac{\partial}{\partial
q^{\alpha}}+\dot{q}^{\beta}\frac{\partial X^{\alpha}}{\partial
q^{\beta}}\frac{\partial}{\partial\dot{q}^{\alpha}}\quad\text{and}\quad
X^{V}=X^{\alpha}\frac{\partial}{\partial\dot{q}^{\alpha}}.$

One may easily verify that, if $\\{X_{\alpha}\\}$ is a basis for
$\mathcal{X}(Q)$ (the set of vector fields on $Q$), then
$\\{X_{\alpha}^{C},X_{\alpha}^{V}\\}$ forms a basis for $\mathcal{X}(TQ)$. An
important property we will frequently use is that, on the quasi-velocities
$v^{\alpha}$ of the frame,

$X_{\alpha}^{V}(v^{\beta})=\delta_{\alpha}^{\beta},\quad\quad\quad
X_{\alpha}^{C}(v^{\beta})=-R_{\alpha\gamma}^{\beta}v^{\gamma},$

see e.g. [15] (and the references therein) for more details.

We recall some aspects of the calculus of forms along the tangent bundle
projection $\tau:TQ\rightarrow Q$ (see e.g. [25, 29] for a survey). A section
$X$ of the pullback bundle $\tau^{*}TQ=\\{(u,v)\in TQ\times
TQ\,|\,\tau(u)=\tau(v)\\}\to TQ$ is said to be a vector field along $\tau$. We
denote the set of all these vector fields by $\mathcal{X}(\tau)$. We may
interpret any vector field $Y$ on $Q$ as a “basic” vector field $Y\circ\tau$
along $\tau$, but we will no longer make a notational distinction. For that
reason, we may express $X\in{\cal X}(\tau)$ in the given frame
$\\{X_{\alpha}\\}$ as $X=\xi^{\alpha}X_{\alpha}$, with $\xi^{\alpha}\in
C^{\scriptscriptstyle\infty}(TQ)$. It is easy to see that vertical vector
fields on $TQ$ are in 1-1 correspondence with vector fields along $\tau$.
Indeed, each $X=\xi^{\alpha}X_{\alpha}\in\mathcal{X}(\tau)$ can be vertically
lifted to $X^{V}\in\mathcal{X}(TQ)$, given by

$X^{V}=\xi^{\alpha}X_{\alpha}^{V}.$

In particular, there exist a canonical vector field
${\mathbf{T}}=v^{\alpha}X_{\alpha}\in\mathcal{X}(\tau)$. It corresponds to the
section $v_{q}\mapsto(v_{q},v_{q})$ of the pullback bundle, and its vertical
lift gives the Liouville vector field on $TQ$.

A vector field $\Gamma$ on $TQ$ is a second-order ordinary differential
equations field (sode, in short) if all its integral curves
$\gamma:I\rightarrow TQ$ are lifted curves, that is of the type
$\gamma=\dot{c}$. We call $c:I\rightarrow Q$ a base integral curve of
$\Gamma$. When expressed in terms of the frame
$\\{X_{\alpha}^{C},X_{\alpha}^{V}\\}$, a sode $\Gamma$ is of the form

$\Gamma=v^{\alpha}X_{\alpha}^{C}+F^{\alpha}X_{\alpha}^{V},$

for some functions $F^{\alpha}\in\mathcal{C}^{\infty}(TQ)$. The quasi-velocity
components $(v^{\alpha}(t))$ of the integral curves $\gamma(t)$ of $\Gamma$
satisfy $\dot{v}^{\alpha}=F^{\alpha}$.

A sode $\Gamma$ can be used to introduce a horizontal lift
$X^{H}\in\mathcal{X}(TQ)$ of $X\in\mathcal{X}(\tau)$. Its coordinate
expression in the anholonomic frame is

$X^{H}=\xi^{\alpha}\left(X_{\alpha}^{C}-\Gamma^{\beta}_{\alpha}X_{\beta}^{V}\right),\quad\quad\Gamma^{\beta}_{\alpha}=-\frac{1}{2}\Big{(}X_{\alpha}^{V}(F^{\beta})+v^{\gamma}R^{\beta}_{\alpha\gamma}\Big{)}.$

With this, any vector field $Z$ on $TQ$ can be decomposed into the sum of a
horizontal and a vertical lift, $Z=X_{1}^{H}+X_{2}^{V}$ for
$X_{1},X_{2}\in\mathcal{X}(\tau)$. For any vector field $X$ along $\tau$, we
may define

$[\Gamma,X^{V}]=-X^{H}+(\nabla
X)^{V}\quad\text{and}\quad[\Gamma,X^{H}]=(\nabla X)^{H}+(\Phi(X))^{V}.$

The $(1,1)$-tensor field $\Phi$ along $\tau$, the so-called Jacobi
endomorphism, is of no concern for the current paper. Below, we will often use
the dynamical covariant derivative $\nabla$ of $\Gamma$. It acts as a
derivation on $\mathcal{X}(\tau)$. For $f\in C^{\scriptscriptstyle\infty}(TQ)$
and $X\in{\cal X}(\tau)$:

$\nabla(fX)=f\nabla X+\Gamma(f)X\quad\quad\text{and}\quad\quad\nabla
X_{\alpha}=\Gamma^{\beta}_{\alpha}X_{\beta}.$

The equations of motion for an (unconstrained) Lagrangian systems (with
regular Lagrangian $L:TQ\rightarrow\mathbb{R}$) are given by the Euler-
Lagrange equations:

$\frac{d}{dt}\left(\frac{\partial
L}{\partial\dot{q}^{\alpha}}\right)-\frac{\partial L}{\partial q^{\alpha}}=0.$

The solutions of these equations are base integral curves of the Lagrangian
vector field $\Gamma_{L}\in\mathcal{X}(TQ)$, which can be defined by as the
unique sode that satisfies

$\Gamma_{L}(X^{V}(L))-X^{C}(L)=0,\quad\forall X\in\mathcal{X}(Q).$

The main focus of the paper is on nonholonomic mechanical systems.

######  Definition 1.

A nonholonomic system on a manifold $Q$ consists of a pair $(L,\mathcal{D})$,
where $L:TQ\rightarrow\mathbb{R}$ is the Lagrangian of the system and
$\mathcal{D}$ is the $m$-dimensional distribution on $TQ$ that describes the
nonholonomic constraints.

In what follows, we will often use ${\mathcal{C}}$, when we interpret the
distribution ${\mathcal{D}}$ specifically as a submanifold of $TQ$. We will
also use $k$ for $n-m$.

The equations of motion follow from considering an extended principle of
Hamilton (also called Lagrange-d’Alembert principle, see [7, 13]). The
nonholonomic trajectory $c_{v_{q}}(t)=(q^{\alpha}(t))$ that starts at
$v_{q}\in{\mathcal{D}}_{q}$ is a solution of the equations

$\begin{cases}\displaystyle\mu_{\alpha}^{i}\dot{q}^{\alpha}=0,\\\
\displaystyle\frac{d}{dt}\Big{(}\frac{\partial
L}{\partial\dot{q}^{\alpha}}\Big{)}-\frac{\partial L}{\partial
q^{\alpha}}=\sum_{i=1}^{k}{\Lambda_{i}}\mu^{i}_{\alpha}.\end{cases}$

The first set of equations represent the nonholonomic constraints. They ensure
that the velocity ${\dot{c}}_{v_{q}}(t)$ remains in the distribution
${\mathcal{D}}$ during the motion. In the second set of equations, the
functions $\Lambda_{i}(t)$ are called the Lagrange multipliers and are part of
the unknowns.

As before for Lagrangian systems, we will represent the nonholonomic
trajectories as the base integral curves of a vector field
$\Gamma_{(L,\mathcal{D})}$. Suppose that the vector fields $\\{X_{a}\\}$ span
$\mathcal{D}$, and that we complete this set to a frame
$\\{X_{\alpha}\\}=\\{X_{a},X_{i}\\}$ for all vector fields on $Q$. This has
the advantage that the equations for the nonholonomic constraints become very
simple: if $(v^{\alpha})=(v^{a},v^{i})$ are the quasi-velocities with respect
to this frame, then $v_{q}\in D_{q}$ if and only its quasi-velocities satisfy
$v^{i}=0$. In this paper, the indices $\alpha$ run from 1 to $n$, $a$ from 1
to $m$ and $i$ from 1 to $k$.

We will say that the Lagrangian is regular with respect to ${\mathcal{D}}$ if
the matrix

$\left(X^{V}_{a}(X^{V}_{b}(L))\right)$

is non-singular everywhere on $TQ$. This is actually a small modification from
the terminology in [15, 16], where the matrix is only assumed to be non-
singular on ${\mathcal{C}}$ (and see also the regularity requirements in [17,
13]). If satisfied, there will exist a unique vector field
${\Gamma_{(L,\mathcal{D})}}$ of the type

$\Gamma_{(L,\mathcal{D})}=v^{a}X_{a}^{C}+f^{a}X_{a}^{V}$

that satisfies, on ${\mathcal{C}}$,

$\Gamma_{(L,\mathcal{D})}(X^{V}(L))-X^{C}(L)=0,\quad\forall X\in{\rm
Sec}(\mathcal{D}).$

We will call this vector field the nonholonomic vector field. Given the
current frame, it can be completely determined from the relation

$\Gamma_{(L,\mathcal{D})}(X^{V}_{a}(L))-X^{C}_{a}(L)=0,\qquad\mbox{on
${\mathcal{C}}$}.$

This equation represents a version of the Lagrange-d’Alembert equations where
the Lagrangian multipliers have already been eliminated. Once
${\Gamma_{(L,\mathcal{D})}}$ is determined, the functions
$\lambda_{i}:=\Gamma_{(L,\mathcal{D})}(X^{V}_{i}(L))-X^{C}_{i}(L)$ on
${\mathcal{C}}$ play the role of the Lagrangian multipliers, in the current
frame.

Let $L$ be a regular Lagrangian. The Hessian $g$ of $L$, with respect to fibre
coordinates, can be thought of as a $(0,2)$-tensor field along the tangent
bundle projection $\tau:TQ\rightarrow Q$ (see [15, 29] for details). To define
it, it is enough to state its action along ‘basic’ vector fields in ${\cal
X}(\tau)$ (i.e. vector fields on $Q$):

$g(X,Y)=X^{V}(Y^{V}(L)),\qquad X,Y\in{\cal X}(Q).$

and extend it to ${\cal X}(\tau)$ by
$C^{\scriptscriptstyle\infty}(TQ)$-linearity. Since there is a 1-1
correspondence between $\tau$-vertical vector fields on $TQ$ and vector fields
along $\tau$, we may also think of it as acting on vertical vector fields.

For what follows, it is important to realize that this Hessian is familiar in
the case of a purely kinetic Lagrangian, i.e. when $L(v)=\frac{1}{2}G(v,v)$
for a Riemannian metric $G$ on $Q$. In this case, we may identify the above
defined Hessian $g$ of $L$ (interpreted as a (0,2) tensor field along $\tau$)
with the ‘basic’ Riemannian metric $G$ (interpreted as a (0,2) tensor field on
$Q$). We will not make a notational distinction between $G$ and $g$ in that
case, and simply write $L(v)=\frac{1}{2}g(v,v)$.

Let’s assume now that $L$ is again an arbitrary regular Lagrangian that is
also regular with respect to ${\mathcal{D}}$. Let $g$ be its Hessian. The
pullback bundle $\tau^{*}{\mathcal{D}}\to TQ$ is a vector subbundle of
$\tau^{*}TQ$ whose set of sections we will denote by ${\cal
X}(\tau,{\mathcal{D}})$. Since $L$ is regular with respect to ${\mathcal{D}}$,
we can consider the set ${\mathcal{X}}^{g}(\tau,{\mathcal{D}})$ of vector
fields along $\tau$ that are orthogonal to ${\cal X}(\tau,{\mathcal{D}})$,
with respect to $g$. That is, $X\in{\mathcal{X}}^{g}(\tau,{\mathcal{D}})$ when

$g(X,Y)=0,\qquad\forall Y\in{\cal X}(\tau,{\mathcal{D}}).$

With this, we can decompose ${\cal X}(\tau)$ as

${\cal X}(\tau)={\cal
X}(\tau,{\mathcal{D}})\oplus{\mathcal{X}}^{g}(\tau,{\mathcal{D}})$

and define two complementary projection operators:

$P:{\cal X}(\tau)\to{\cal X}(\tau,{\mathcal{D}})\quad\mbox{ and }\quad Q:{\cal
X}(\tau)\to{\mathcal{X}}^{g}(\tau,{\mathcal{D}}).$

Starting from a frame $\\{X_{a},X_{i}\\}$ for ${\cal X}(Q)$, we can easily
construct a frame for ${\cal X}(\tau)$ that is adapted to the decomposition.
Indeed, if $g_{ab}=g(X_{a},X_{b})$, $g_{ai}=g(X_{a},X_{i})$ and
$g_{ij}=g(X_{i},X_{j})$, then $g_{ab}$ has an inverse (because of the assumed
regularity with respect to ${\mathcal{D}}$). Then, the set
$\\{X_{a},{\tilde{X}}_{i}\\}$, with

$\tilde{X}_{i}=X_{i}-{g}^{ab}{g}_{bi}X_{a}=X_{i}+K^{a}_{i}X_{a}$

is such a frame. The vector fields $\tilde{X}_{i}\in{\cal X}(\tau)$ are,
however, not basic vector fields.

Next, we introduce the set ${\mathcal{X}}^{T}(TQ)$ of vector fields on $TQ$
that are tangent to the constraint manifold ${\mathcal{C}}$, i.e.

${\mathcal{X}}^{T}(TQ)=\\{Z\in\mathcal{X}(TQ)\,|\,Z(v^{i})=0\\}.$

In view of the relations
$X_{\alpha}^{C}(v^{i})=-R^{i}_{\alpha\gamma}v^{\gamma}$ and
$X_{\alpha}^{V}(v^{i})=\delta^{i}_{\alpha}$, one easily verifies that the
vector fields on $TQ$, given by

$\overline{X_{i}^{C}}=X_{i}^{C}+R^{j}_{i\gamma}v^{\gamma}\tilde{X}_{j}^{V},\quad\overline{X_{a}^{C}}=X_{a}^{C}+R^{j}_{a\gamma}v^{\gamma}\tilde{X}_{j}^{V}\quad\mbox{and}\quad
X_{a}^{V}$

form a basis for ${\mathcal{X}}^{T}(TQ)$. We may augment this basis with the
vector fields $\\{{\tilde{X}_{i}}^{V}\\}$ that span
$({\mathcal{X}}^{g}(\tau,{\mathcal{D}}))^{V}$ to obtain a full basis of ${\cal
X}(TQ)$. From this, we can conclude that we can decompose $\mathcal{X}(TQ)$ as

${\mathcal{X}}(TQ)={\mathcal{X}}^{T}(TQ)\oplus({\mathcal{X}}^{g}(\tau,{\mathcal{D}}))^{V}$

and that we can define two complementary projectors $p$ and $q$,

$p:{\mathcal{X}}(TQ)\rightarrow{\mathcal{X}}^{T}(TQ)\quad\quad\text{and}\quad\quad
q:\mathcal{X}(TQ)\rightarrow({\mathcal{X}}^{g}(\tau,{\mathcal{D}}))^{V}.$

If
$Z=Z^{a}X_{a}^{C}+Z^{i}X_{i}^{C}+\zeta^{a}X_{a}^{V}+\zeta^{i}X_{i}^{V}\in\mathcal{X}(TQ)$,
then

$p(Z)=Z^{a}\overline{X_{a}^{C}}+Z^{i}\overline{X_{i}^{C}}+(\zeta^{a}-\zeta^{i}K_{i}^{a})X_{a}^{V}\quad\mbox{and}\quad
q(Z)=(-Z^{a}R^{j}_{a\gamma}v^{\gamma}-Z^{i}R^{j}_{i\gamma}v^{\gamma}+\zeta^{j})\tilde{X}_{j}^{V}.$

##  3 Extensions of the nonholonomic vector field to a sode

In what follows, we will assume throughout that the Lagrangian $L$ is both
regular and regular with respect to ${\mathcal{D}}$. Let $c_{v_{q}}(t)$ be the
nonholonomic trajectory that starts at $v_{q}\in{\mathcal{D}}_{q}$. We
formalize the notion of a sode extension (implicit in [2, 3]) in the following
definition.

######  Definition 2.

A sode $\Gamma$ is a sode extension of the nonholonomic vector field
$\Gamma_{(L,\mathcal{D})}$, if each nonholonomic trajectory $c_{v_{q}(t)}$ is
a base integral curve of $\Gamma$.

An equivalent way to define a sode extension is that it should satisfy

$\Gamma|_{\mathcal{D}}=\Gamma_{(L,\mathcal{D})}.$

It’s useful to see this property translated into terms of a frame
$\\{X_{\alpha}\\}=\\{X_{a},X_{i}\\}$ on $Q$ whose first elements $\\{X_{a}\\}$
span ${\mathcal{D}}$, and whose corresponding quasi-velocities are
$(v^{\alpha})=(v^{a},v^{i})$. The sode $\Gamma$ and the vector field
$\Gamma_{(L,\mathcal{D})}$ can then be represented, respectively, by

$\Gamma=v^{\alpha}X_{\alpha}^{C}+F^{\alpha}X_{\alpha}^{V}\quad\mbox{and}\quad\Gamma_{(L,\mathcal{D})}=v^{a}X_{a}^{C}+f^{a}X_{a}^{V}.$

Let $v^{a}(t)$ be such that $\dot{v}^{a}=f^{a}$. From the definition, $\Gamma$
is a sode extension if the curve $(v^{a}(t),v^{i}(t)=0)$ is an integral curve
of $\Gamma$, i.e. a solution of $\dot{v}^{a}=F^{a}$ and $\dot{v}^{i}=F^{i}$.
This will be the case, if and only if,

$F^{a}|_{\mathcal{C}}=f^{a},\qquad\mbox{and}\qquad F^{i}|_{\mathcal{C}}=0.$

For the rest of the paper it is important to realize that the Lagrangian
vector field $\Gamma_{L}$ of $L$ is not a sode extension of
$\Gamma_{(L,\mathcal{D})}$. The sode
$\Gamma_{L}=v^{\alpha}X_{\alpha}^{C}+F^{\alpha}X_{\alpha}^{V}$ is uniquely
determined by the relations

$\Gamma_{L}\left(X_{a}^{V}(L)\right)-X_{a}^{V}(L)=0,\qquad\Gamma_{L}\left(X_{i}^{V}(L)\right)-X_{i}^{V}(L)=0.$

From the first set of equations we get, in particular on ${\mathcal{C}}$,

$v^{b}X_{b}^{C}(X_{a}^{V}(L))+F^{b}|_{\mathcal{C}}X_{b}^{V}(X_{a}^{V}(L))+F^{i}|_{\mathcal{C}}X_{i}^{V}(X_{a}^{V}(L))-X_{a}^{C}(L)=0.$

On the other hand, on ${\mathcal{C}}$, the nonholonomic vector field
$\Gamma_{(L,\mathcal{D})}=v^{a}X_{a}^{C}+f^{a}X_{a}^{V}$ satisfies:

$\Gamma_{(L,\mathcal{D})}\left(X_{a}^{V}(L)\right)-X_{a}^{C}(L)=0,$

which becomes when written in full

$v^{b}X_{b}^{C}(X_{a}^{V}(L))+f^{b}X_{b}^{V}(X_{a}^{V}(L))-X_{a}^{C}(L)=0.$

In the assumed regularity with respect to ${\mathcal{D}}$, this last equation
determines the coefficients $f^{b}$ of the nonholonomic vector field. This
means that the relation between the forces $F^{a}$, $F^{i}$ and $f^{a}$ is

$f^{a}=F^{a}|_{\mathcal{C}}-K^{a}_{i}F^{i}|_{\mathcal{C}}\quad\text{on
}\mathcal{C}.$

This is not the required property for $\Gamma_{L}$ to be a sode extension of
${\Gamma_{(L,\mathcal{D})}}$, but it is a useful property for what follows.

We now show that a sode extension of ${\Gamma_{(L,\mathcal{D})}}$ always
exists. The construction we present below is, in essence, also contained in
[16] (Proposition 2) and [17] (formulated as Proposition 2.1 in [2]).

######  Proposition 1.

The vector field $\Gamma^{1}=p(\Gamma_{L})$ is a sode extension of the
nonholonomic vector field $\Gamma_{(L,\mathcal{D})}$. It can, equivalently, be
defined as the unique sode $\Gamma^{1}$ which is tangent to the constraints,
and whose difference from $\Gamma_{L}$ is $g$-orthogonal.

###### Proof.

If we rewrite $\Gamma_{L}$ in terms of the basis
$\\{\tilde{X}_{i}^{V},\overline{X_{i}^{C}},\overline{X_{a}^{C}},X_{a}^{V}\\}$
we get

$\displaystyle\Gamma^{1}$ $\displaystyle=$ $\displaystyle
p(\Gamma_{L})=v^{a}\overline{X_{a}^{C}}+v^{i}\overline{X_{i}^{C}}+(F^{a}-F^{i}K_{i}^{a})X_{a}^{V}$
$\displaystyle=$ $\displaystyle
v^{a}(X_{a}^{C}+R^{j}_{a\gamma}v^{\gamma}\tilde{X}_{j}^{V})+v^{i}(X_{i}^{C}+R^{j}_{i\gamma}v^{\gamma}\tilde{X}_{j}^{V})+(F^{a}-F^{i}K_{i}^{a})X_{a}^{V}$
$\displaystyle=$ $\displaystyle
v^{a}X_{a}^{C}+v^{i}X_{i}^{C}+v^{\mu}R^{j}_{\mu\gamma}v^{\gamma}\tilde{X}_{j}^{V}+(F^{a}-F^{i}K^{a}_{i})X_{a}^{V}$
$\displaystyle=$ $\displaystyle
v^{\alpha}X_{\alpha}^{C}+(F^{a}-F^{i}K_{i}^{a})X_{a}^{V}+0X_{i}^{V}.$

From the first term we may conclude that $\Gamma^{1}$ is indeed a sode. When
restricted to ${\mathcal{C}}$, we know that $F^{a}-F^{i}K_{i}^{a}$ coincides
with $f^{a}$. Therefore, we can conclude that $p(\Gamma_{L})$ is a sode
extension.

To prove the second statement, we set
$\Gamma^{1}=\Gamma_{L}+U^{V}=v^{\alpha}X_{\alpha}^{C}+F^{\alpha}X_{\alpha}^{V}+U^{\alpha}X_{\alpha}^{V}$.
From the first condition, $\Gamma^{1}(v^{i})=0$, we obtain $U^{i}=-F^{i}$. The
second condition can be expressed as
$U\in({\mathcal{X}}^{g}(\tau,{\mathcal{D}}))^{V}$, and thus
$U^{b}=K^{b}_{i}U^{i}=-K^{b}_{i}F^{i}$. With this we see that, again,
$\Gamma^{1}=v^{\alpha}X_{\alpha}^{C}+(F^{a}-F^{i}K_{i}^{a})X_{a}^{V}.$ ∎

In the next paragraph, we give a second construction of a sode extension and
show how it appears in the literature (in a special case).

######  Proposition 2.

Let $\nabla$ be the dynamical covariant derivative of $\Gamma_{L}$ and $Q$ the
projection on ${\mathcal{X}}^{\perp}(\tau,{\mathcal{D}})$, defined by $g$. The
vector field

$\Gamma^{2}=\Gamma_{L}-[\nabla(Q({\mathbf{T}}))]^{V}$

is a sode extension of the nonholonomic vector field
$\Gamma_{(L,\mathcal{D})}$.

###### Proof.

For the canonical vector field ${\mathbf{T}}$ along $\tau$ we can write
${\mathbf{T}}=v^{a}X_{a}+v^{i}X_{i}=\tilde{v}^{a}X_{a}+\tilde{v}^{i}\tilde{X}_{i}$
with $\tilde{v}^{a}=v^{a}+v^{i}K^{a}_{i}$ and $\tilde{v}^{i}=v^{i}$. Remark
that if $v^{i}=0$ (on $\mathcal{C}$) then $\tilde{v}^{i}=0$ and
$\tilde{v}^{a}=v^{a}$. From this, $Q({\mathbf{T}})=v^{i}{\tilde{X}}_{i}$ and

$\nabla(Q({\mathbf{T}}))=\nabla(v^{i}\tilde{X}_{i})=\Gamma_{L}(v^{i})\tilde{X}_{i}+v^{i}\nabla\tilde{X}_{i}=F^{i}\tilde{X}_{i}+v^{i}\nabla\tilde{X}_{i}.$

The last term doesn’t really concern us, since it vanishes on ${\mathcal{C}}$.
Then,

$\Gamma^{2}=v^{\alpha}X_{\alpha}^{C}+F^{a}X_{a}^{V}+F^{i}X_{i}^{V}-F^{i}\tilde{X}_{i}^{V}-v^{i}(\nabla\tilde{X}_{i})^{V}=v^{\alpha}X_{\alpha}^{C}+(F^{a}-F^{i}K^{a}_{i})X_{a}^{V}-v^{i}(\nabla\tilde{X}_{i})^{V},$

from which it is clear that
$\Gamma^{2}|_{{\mathcal{C}}}=\Gamma_{L,{\mathcal{D}}}$. ∎

The above construction is actually familiar, in the case of a purely kinetic
Lagrangian.

######  Definition 3.

A nonholonomic system $(L,{\mathcal{D}})$ is purely kinetic if its Lagrangian
is of the type $L(v)=\frac{1}{2}g(v,v)$, for a Riemannian metric $g$ on $Q$.

We have already remarked before that, in this case, the Hessian of $L$
(interpreted as a (0,2) tensor field along $\tau$) is directly related to the
Riemannian metric $g$. Since $g$ is a Riemannian metric, and therefore
positive-definite, $L$ is both regular and regular with respect to
${\mathcal{D}}$. We first derive an expression for $\Gamma_{L}$ and
${\Gamma_{(L,\mathcal{D})}}$. Next, we introduce the nonholonomic connection
$\nabla^{nh}$, and we show that the sode extension $\Gamma^{2}$ is in fact the
quadratic spray of $\nabla^{nh}$.

We may use the Riemannian metric $g$ to define an orthogonal complement
${\mathcal{D}}^{g}$ of ${\mathcal{D}}$ in $TQ$. With a slight abuse of
notation, we will write $P:{\cal X}(Q)\to{\rm Sec}({\mathcal{D}})$ and
$Q:{\cal X}(Q)\to{\rm Sec}({\mathcal{D}}^{g})$ for the two corresponding
projections, since these operators can clearly be extended to their
counterparts we had encountered in the previous section (with the same
notation) on ${\cal X}(\tau)$.

Many papers make use of the so-called nonholonomic connection $\nabla^{nh}$
(see e.g. [24, 13]). It can be defined by the expression

$\nabla_{X}^{nh}Y=P\left(\nabla_{X}^{g}Y\right)+\nabla_{X}^{g}(Q(Y)).$

Herein, $\nabla^{g}$ is the Levi-Civita connection of $g$.

We will assume (until the end of the paper) that the frame
$\left\\{X_{\alpha}\right\\}=\left\\{X_{a},X_{i}\right\\}$ is chosen is such a
way that $\left\\{X_{a}\right\\}$ represents a frame for $\mathcal{D}$ and
that $\left\\{X_{i}\right\\}$ is a frame for $\mathcal{D}^{g}$. This means
that, from now on, $g_{ai}=g\left(X_{a},X_{i}\right)=0$. As before, we denote
the quasi-velocities w.r.t. this frame by $(v^{\alpha})=(v^{a},v^{i})$.

It is well-known that the Levi-Civita connection of $g$ satisfies the Koszul-
formula

$2g(\nabla_{X}^{g}Y,Z)=Xg(Y,Z))+Y(g(X,Z))-Zg(X,Y)+g([X,Y],Z)+g([Z,X],Y)+g([Z,Y],X).$

If we write
$\nabla^{g}_{X_{a}}X_{b}=\Gamma_{ab}^{c}X_{c}+\Gamma_{ab}^{i}X_{i}$,
$[X_{a},X_{b}]=R_{ab}^{c}X_{c}+R_{ab}^{i}X_{i}$, and so forth, we get by
taking $X=X_{a},Y=X_{b},Z=X_{c}$ (among other choices)

$2g_{cd}\Gamma_{ab}^{d}=X_{a}(g_{bc})+X_{b}(g_{ac})-X_{c}(g_{ab})+g_{dc}R_{ab}^{d}+g_{db}R_{ca}^{d}+g_{da}R_{cb}^{d}.$

After multiplying both sides by $v^{a}v^{b}$ and canceling a factor 2, we
obtain

$g_{cd}\Gamma_{ab}^{d}v^{a}v^{b}=X_{a}\left(g_{bc}\right)v^{a}v^{b}-\frac{1}{2}X_{c}\left(g_{ab}\right)v^{a}v^{b}-g_{db}R_{ac}^{d}v^{a}v^{b}.$

######  Lemma 1.

For a purely kinetic Lagrangian
$L=\frac{1}{2}g_{\alpha\beta}v^{\alpha}v^{\beta}=\frac{1}{2}\left(g_{ij}v^{i}v^{j}+g_{ab}v^{a}v^{b}\right)$
the Lagrangian vector field is the geodesic spray $\Gamma_{g}$ of the Levi-
Civita connection. It is given by the expression

$\Gamma_{L}=\Gamma_{g}=v^{\alpha}X^{C}_{\alpha}-\Gamma^{\alpha}_{\beta\gamma}v^{\beta}v^{\gamma}X^{V}_{\alpha}.$

The nonholonomic vector field is given by

$\Gamma_{(L,\mathcal{D})}=v^{d}X_{d}^{c}-\Gamma_{ab}^{d}v^{a}v^{b}X_{d}^{V}.$

The spray of the nonholonomic connection $\nabla^{nh}$ is of the form

$\Gamma_{\nabla^{nh}}=v^{\gamma}X_{\gamma}^{C}-\left(\Gamma_{ab}^{c}v^{a}v^{b}+2\Gamma_{aj}^{c}v^{a}v^{j}+\Gamma_{ib}^{c}v^{i}v^{b}+2\Gamma_{ij}^{c}v^{i}v^{j}\right)X_{c}^{V}-\left(\Gamma_{aj}^{k}v^{a}v^{j}+\Gamma_{ij}^{k}v^{i}v^{j}\right)X_{k}^{V}.$

###### Proof.

We only prove the second expression, since the expression for the first can be
found by a similar reasoning.

Recall that on $\mathcal{C}$ all $v^{i}=0$ and that
$X_{d}^{C}(v^{b})=-R_{de}^{b}v^{e}$ and $X_{d}^{V}(v^{b})=\delta^{b}_{d}$. If
we set $\Gamma_{(L_{g},\mathcal{D})}=v^{d}X_{d}^{c}+f^{d}X_{d}^{V}$, the
coefficients $f^{d}$ satisfy on $\mathcal{C}$

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Gamma_{(L_{g},\mathcal{D})}\left(X_{a}^{V}(L)\right)-X_{a}^{C}(L)$
$\displaystyle=$ $\displaystyle
v^{d}X^{C}_{d}(g_{ab}v^{b})+f^{d}X_{d}^{V}(g_{ab}v^{b})-\frac{1}{2}X_{a}(g_{ij})v^{i}v^{j}-g_{ij}X_{a}^{C}(v^{i})v^{j}-\frac{1}{2}X_{a}(g_{cd})v^{c}v^{d}-g_{cd}X_{a}^{C}(v^{c})v^{d}$
$\displaystyle=$ $\displaystyle
v^{d}X_{d}\left(g_{ab}\right)v^{b}-v^{d}g_{ab}R_{de}^{b}v^{e}+f^{d}g_{ab}\delta_{d}^{b}-\frac{1}{2}X_{a}\left(g_{cd}\right)v^{c}v^{d}+g_{cd}R_{ae}^{c}v^{e}v^{d}$
$\displaystyle=$ $\displaystyle
g_{ab}f^{b}+\left(X_{d}\left(g_{ab}\right)v^{d}v^{b}-g_{ab}R_{de}^{b}v^{d}v^{e}-\frac{1}{2}X_{a}\left(g_{cd}\right)v^{c}v^{d}+g_{cd}R^{c}_{ae}v^{e}v^{d}\right)$
$\displaystyle=$ $\displaystyle g_{ab}f^{b}+g_{ab}\Gamma_{ef}^{b}v^{e}v^{f}.$

We conclude that, indeed, $f^{b}=-\Gamma_{ef}^{b}v^{e}v^{f}$.

If, in the frame $\\{X_{\alpha}\\}$, the connection coefficients of the
nonholonomic connection are given by
$\nabla^{nh}_{X_{\alpha}}X_{\beta}=\tilde{\Gamma}_{\alpha\beta}^{\gamma}X_{\gamma}$,
the corresponding quadratic spray can be expressed as
$\Gamma_{\nabla^{nh}}=v^{\gamma}X_{\gamma}^{C}-\tilde{\Gamma}_{\alpha\beta}^{\gamma}v^{\alpha}v^{\beta}X_{\gamma}^{V}$
(see e.g. [14]). Specifically for the nonholonomic connection, we may compute:

  * •

$\nabla^{nh}_{X_{a}}X_{b}=P(\Gamma_{ab}^{\gamma}X_{\gamma})=\Gamma_{ab}^{c}X_{c}+0X_{i}$,

  * •

$\nabla_{X_{a}}^{nh}X_{j}=P\left(\Gamma_{aj}^{\gamma}X_{\gamma}\right)+\nabla_{X_{a}}^{g}X_{j}=\Gamma_{aj}^{c}X_{c}+\Gamma_{aj}^{k}X_{k}+\Gamma_{aj}^{c}X_{c}=2\Gamma_{aj}^{c}X_{c}+\Gamma_{aj}^{k}X_{k}$,

  * •

$\nabla_{X_{i}}^{nh}X_{j}=P\left(\Gamma_{ij}^{\gamma}X_{\gamma}\right)+\nabla_{X_{i}}^{g}X_{j}=\Gamma_{ij}^{c}X_{c}+\Gamma_{ij}^{k}X_{k}+\Gamma_{ij}^{c}X_{c}=2\Gamma_{ij}^{c}X_{c}+\Gamma_{ij}^{k}X_{k}$,

from which the expression in the statement then follows. ∎

We now show that the second construction of a sode extension is natural in the
following way: it represents a generalization of the quadratic spray
associated to the nonholonomic connection.

######  Proposition 3.

For a purely kinetic Lagrangian $L(v)=\frac{1}{2}g(v,v)$, the spray
$\Gamma_{\nabla^{nh}}$ of the nonholonomic connection is the sode extension
$\Gamma^{2}$ of ${\Gamma_{(L,\mathcal{D})}}$, i.e.

$\Gamma_{\nabla^{nh}}=\Gamma_{g}-[\nabla(Q({\mathbf{T}}))]^{V}.$

###### Proof.

When one restricts the expression of $\Gamma_{\nabla_{nh}}$ in the statement
of Lemma 1 to $v^{i}=0$, one easily verifies that $\Gamma_{\nabla^{nh}}$ is
indeed a sode extension of ${\Gamma_{(L,\mathcal{D})}}$. Besides, we can make
the vector field $\Gamma_{g}$ appear in the expression of
$\Gamma_{\nabla^{nh}}$, as follows:

$\Gamma_{\nabla^{nh}}=\Gamma_{g}-\Gamma^{c}_{\alpha
j}v^{\alpha}v^{j}X_{c}^{V}+\Gamma^{k}_{\alpha a}v^{\alpha}v^{a}X_{k}^{V}.$

In case of the geodesic spray $\Gamma_{g}$, the coefficients of the dynamical
covariant derivative are given by
$\Gamma^{\beta}_{i}=\Gamma^{\beta}_{i\alpha}v^{\alpha}$ and
$\Gamma_{a}^{\beta}=\Gamma_{a\alpha}^{\beta}v^{\alpha}$. A more compact way of
writing the spray of the nonholonomic connection is therefore

$\Gamma_{\nabla^{nh}}=\Gamma_{g}-\Gamma^{c}_{j}v^{j}X_{c}^{V}+\Gamma^{k}_{a}v^{a}X_{k}^{V}.$

For any quadratic spray (like $\Gamma_{g}$) is $\nabla{\mathbf{T}}=0$ (see
e.g. [25]). For the projections $P$ and $Q$ (now interpreted as acting on
$\mathcal{X}(\tau)$), we have

$\displaystyle\nabla(Q({\mathbf{T}}))$ $\displaystyle=$ $\displaystyle(\nabla
Q)({\mathbf{T}})=\nabla(Q({\mathbf{T}}))-Q(\nabla{\mathbf{T}})=\nabla\left(v^{i}X_{i}\right)-Q\left(\Gamma^{g}\left(v^{a}\right)X_{a}+v^{a}\nabla
X_{a}+\nabla\left(v^{i}X_{i}\right)\right)$ $\displaystyle=$ $\displaystyle
P\left(\nabla v^{i}X_{i}\right)-v^{a}Q\left(\nabla
X_{a}\right)=v^{i}P\left(\nabla X_{i}\right)-v^{a}Q\left(\nabla
X_{a}\right)=v^{i}P\left(\Gamma_{i}^{\beta}X_{\beta}\right)-v^{a}Q\left(\Gamma_{a}^{\beta}X_{\beta}\right)$
$\displaystyle=$ $\displaystyle
v^{i}\Gamma_{i}^{b}X_{b}-v^{a}\Gamma_{a}^{k}X_{k}.$

When we take the vertical lift of this expression, it represents the
difference between $\Gamma_{\nabla^{nh}}$ and $\Gamma_{g}$, as required. ∎

##  4 Geodesic extensions

In this section we consider only purely kinetic nonholonomic systems. The sode
extension $\Gamma_{\nabla^{nh}}$ (described in Proposition 3) is a quadratic
spray, but it is not necessarily the geodesic spray of a (pseudo-)Riemannian
metric $\hat{g}$.

######  Definition 4.

A geodesic extension of a purely kinetic nonholonomic system
$(L,{\mathcal{D}})$ is a Riemannian metric $\hat{g}$ whose geodesic spray
$\Gamma_{\hat{g}}$ is a sode extension of ${\Gamma_{(L,\mathcal{D})}}$.

In that case, the nonholonomic trajectories could be interpreted as geodesics
with initial velocities in $\mathcal{D}$. In the rest of the paper, we look at
this problem in more detail.

Consider a distribution ${\mathcal{D}}$ and a Riemannian metric $g$. We may
use the metric $g$ to decompose

$TQ={\mathcal{D}}\oplus{\mathcal{D}}^{g}.$

In what follows we will, again, make use of a frame $\\{X_{a},X_{i}\\}$ that
respects this decomposition and of its corresponding quasi-velocities
$(v^{a},v^{i})$.

If $\hat{g}$ is a pseudo-Riemannian metric (i.e. non-degenerate, but not
necessarily positive-definite), the integral curves of the geodesic spray
$\Gamma_{\hat{g}}$ of $\hat{g}$ satisfy the equations

$\left\\{\begin{array}[]{l}\dot{v}^{a}=-\hat{\Gamma}^{a}_{\alpha\beta}v^{\alpha}v^{\beta}=-\hat{\Gamma}_{bc}^{a}v^{b}v^{c}-\hat{\hat{\Gamma}}^{a}_{j}v^{j},\\\
\dot{v}^{i}=-\hat{\Gamma}^{i}_{\alpha\beta}v^{\alpha}v^{\beta}=-\hat{\Gamma}_{ab}^{i}v^{a}v^{b}-\hat{\hat{\Gamma}}^{i}_{j}v^{j},\end{array}\right.$

where $\hat{\Gamma}^{\alpha}_{\beta\gamma}$ are the connection coefficients of
the Levi-Civita connection $\nabla^{\hat{g}}$ associated to the metric
$\hat{g}$, in the frame $\\{X_{a},X_{i}\\}$. For our purpose, the terms with a
factor $v^{j}$ are of no special interest, since they vanish after restriction
to $\mathcal{C}$. Recall that these equations are equivalent with the Euler-
Lagrange equations of the Lagrangian
$\hat{L}(v)=\frac{1}{2}{\hat{g}}(v,v)=\frac{1}{2}\left(\hat{g}_{ab}v^{a}v^{b}+2\hat{g}_{ai}v^{a}v^{i}+\hat{g}_{ij}v^{i}v^{j}\right)$.

When compared to the nonholonomic equations of motion of $(L,{\mathcal{D}})$,
the integral curves of any sode extension of ${\Gamma_{(L,\mathcal{D})}}$
should satisfy on ${\mathcal{C}}$:

$\left\\{\begin{array}[]{l}\dot{v}^{a}=-\Gamma_{bc}^{a}v^{b}v^{c},\\\
\dot{v}^{i}=0.\end{array}\right.$

We conclude that the pseudo-Riemannian metric $\hat{g}$ we are looking for
must be such that its connection coefficients satisfy the conditions

$(*)\qquad\quad\left\\{\begin{array}[]{l}\hat{\Gamma}_{ab}^{c}v^{a}v^{b}=\Gamma_{ab}^{c}v^{a}v^{b},\\\
\hat{\Gamma}_{bc}^{i}v^{b}v^{c}=0.\end{array}\right.$

The factors $\hat{\Gamma}_{ab}^{c}$ are functions on $Q$ and do not depend on
the quasi-velocities $v^{a}$. We may also write the conditions $(*)$ in a
version without the factors $v^{b}v^{c}$. However, we should keep in mind that
(even though the Levi-Civita connection is torsionless) the Christoffel
symbols are not symmetric in the current anholonomic frame. For example, the
last condition of $(*)$ is equivalent with

$2\hat{\Gamma}_{bc}^{i}+R_{bc}^{i}=0,$

and similar for the first. For compactness of our formulae, we prefer the
first notation.

In what follows, we will make one further assumption:

######  Definition 5.

A symmetric (0,2) tensor field $\hat{g}$ on $Q$ is a
${\mathcal{D}}$-preserving modification of a Riemannian metric $g$ on $Q$ if
$\hat{g}$ and $g$ are related in such a way that

${\hat{g}}|_{{\mathcal{D}}\times{\mathcal{D}}}=g|_{{\mathcal{D}}\times{\mathcal{D}}}.$

In terms of the nonholonomic dynamics, this assumption has the interpretation
that we are looking for an extension $\hat{L}$ of the mechanical Lagrangian
$L$, where we do not wish to alter the constraint Lagrangian
$L_{c}=L|_{\mathcal{C}}={\hat{L}}|_{\mathcal{C}}$. In the current frame
$\\{X_{a},X_{i}\\}$, this assumption means that $g$ and $\hat{g}$ are of the
form

${g}=\left(\begin{array}[]{cc}g_{ab}&0\\\
0&{g}_{ij}\end{array}\right)\quad\mbox{and}\quad\hat{g}=\left(\begin{array}[]{cc}g_{ab}&\hat{g}_{ai}\\\
\hat{g}_{ai}&\hat{g}_{ij}\end{array}\right).$

In this, for example, $({\hat{g}}_{ij})$ is the matrix that corresponds to the
restriction ${\hat{g}}|_{{\mathcal{D}}^{g}\times{\mathcal{D}}^{g}}$, i.e.
${\hat{g}}_{ij}={\hat{g}}(X_{i},X_{j})$.

Recall that the principal restriction $g|_{{\mathcal{D}}\times{\mathcal{D}}}$
of a Riemannian metric remains positive-definite, because $g$ is as a whole.
Among other, this means that the matrix $(g_{ab})$ with
$g_{ab}=g(X_{a},X_{b})$ has an inverse $(g^{ab})$. Under the current
assumptions, this means that there exists also an orthogonal decomposition by
$\hat{g}$,

$TQ={\mathcal{D}}\oplus{\mathcal{D}}^{\hat{g}},$

where ${\rm Sec}({\mathcal{D}}^{\hat{g}})={\rm
span}\\{{\hat{X}}_{i}=X_{i}-{\hat{g}}_{bi}g^{ab}X_{a}\\}$. In the next
proposition, we will not assume from the outset that $\hat{g}$ is a Riemannian
metric. The condition, in the Lemma below, that the restriction
${\hat{g}}|_{{\mathcal{D}}^{\hat{g}}\times{\mathcal{D}}^{\hat{g}}}$ is non-
degenerate is equivalent with asking the matrix
$({\hat{g}}({\hat{X}}_{i},{\hat{X}}_{j}))=({\hat{g}}_{ij}-{\hat{g}}_{ai}{\hat{g}}^{ab}{\hat{g}}_{jb})$
to be invertible.

With these extra assumptions, we will show that we can re-write the conditions
$(*)$ as algebraic and PDE conditions in the unknown functions
${\hat{g}}_{ai}$, which represents the restriction of $\hat{g}$ to
${\mathcal{D}}\times{\mathcal{D}}^{g}$. Besides, the Lemma below shows that
the choice of ${\hat{g}}_{ij}$ is (almost) completely free.

There is also a formulation of the conditions that solely makes use of unknown
functions $\theta_{i}={\hat{g}}_{ai}v^{a}$. In that case, also the Lagrangian
multipliers $\lambda_{i}=\Gamma_{(L,\mathcal{D})}(X_{i}^{V}(L))-X_{i}^{C}(L)$
(restricted to $\mathcal{C}$) make their appearance in the equations. These
are both components of a semi-basic 1-form on $TQ$. This is a 1-form that
vanishes when applied to vertical lifts and which, for this reason, can be
fixed by its action on complete lifts. Relying on the decomposition
$TQ={\mathcal{D}}\oplus{\mathcal{D}}^{g}$, we may consider the semi-basic
1-form defined by

$\theta(X^{C})=0,\quad\forall X\in{\rm
Sec}({\mathcal{D}}),\qquad\theta(X^{C})={\hat{g}}({\mathbf{T}},X),\quad\forall
X\in{\rm Sec}({\mathcal{D}}^{g}).$

In the above, we have interpreted the Riemannian metric $\hat{g}$ again as the
Hessian of its corresponding Lagrangian, i.e. as a (basic) (0,2) tensor field
along $\tau$. We will be mainly interested in $\theta|_{\mathcal{C}}$, with
components $\theta_{i}={\hat{g}}_{ai}v^{a}$, in the dual frame of
$\\{X_{a}^{C},X_{i}^{C},X_{a}^{V},X_{i}^{V}\\}$.

Next, we consider the semi-basic 1-form, defined by

$\lambda(X^{C})=0,\quad\forall X\in{\rm
Sec}({\mathcal{D}}),\qquad\lambda(X^{C})=\Gamma_{(L,\mathcal{D})}(X^{V}(L))-X^{C}(L),\quad\forall
X\in{\rm Sec}({\mathcal{D}}^{g}),$

and we consider its restriction to ${\mathcal{C}}$, $\lambda|_{\mathcal{C}}$.
Again, its components are the restrictions of the multipliers
$\lambda_{i}=\Gamma_{(L,\mathcal{D})}(X_{i}^{V}(L))-X_{i}^{C}(L)$ to
$\mathcal{C}$ we had encountered before.

######  Lemma 2.

Suppose that $g$ is a Riemannian metric and that $\hat{g}$ is a
${\mathcal{D}}$-preserving pseudo-Riemannian modification of $g$ whose
restriction
${\hat{g}}|_{{\mathcal{D}}^{\hat{g}}\times{\mathcal{D}}^{\hat{g}}}$ is non-
degenerate. Then, the equations (*) are equivalent with the equations

$\displaystyle(A)$ $\displaystyle\hat{g}_{bk}R^{k}_{ac}v^{a}v^{b}=0,$
$\displaystyle(B)$
$\displaystyle\hat{g}_{di}{\Gamma}^{d}_{ab}v^{a}v^{b}=(g_{ki}\Gamma^{k}_{ab}+X_{a}(\hat{g}_{bi})-\hat{g}_{bk}R^{k}_{ai})v^{a}v^{b},$

in the unknown restriction
${\hat{g}}|_{{\mathcal{D}}\times{\mathcal{D}}^{g}}=({\hat{g}}_{ai})$. When
written in terms of the 1-form $\theta=(\theta_{i})$ the algebraic condition
$(A)$ and the PDE condition $(B)$ become

$\displaystyle(A)$ $\displaystyle\theta_{k}R^{k}_{ac}v^{a}=0,$
$\displaystyle(B)$
$\displaystyle\Gamma_{(L,\mathcal{D})}(\theta_{i})+\theta_{k}R^{k}_{ia}v^{a}+\lambda_{i}=0.$

###### Proof.

In the frame $\\{X_{a},X_{i}\\}$, we have $g_{ai}=0$. Koszul’s formula (see
Section 3) for the metric $g$ leads to

$\left\\{\begin{array}[]{l}2g_{cd}\Gamma_{ab}^{d}v^{a}v^{b}=(2X_{a}\left(g_{bc}\right)-X_{c}\left(g_{ab}\right)-2g_{db}R_{ac}^{d})v^{a}v^{b},\\\
2g_{ki}\Gamma_{ab}^{k}v^{a}v^{b}=(-X_{i}\left(g_{ab}\right)-2g_{db}R_{ai}^{d})v^{a}v^{b}.\end{array}\right.$

Likewise, when applied on the metric $\hat{g}$ Koszul’s formula gives

$\left\\{\begin{array}[]{l}2(\hat{g}_{cd}\hat{\Gamma}_{ab}^{d}+\hat{g}_{ck}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2X_{a}\left(\hat{g}_{bc}\right)-X_{c}\left(\hat{g}_{ab}\right)-2\hat{g}_{db}R_{ac}^{d}-2\hat{g}_{bk}R^{k}_{ac})v^{a}v^{b},\\\
2(\hat{g}_{di}\hat{\Gamma}_{ab}^{d}+\hat{g}_{ki}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2X_{a}\left(\hat{g}_{bi}\right)-X_{i}\left(\hat{g}_{ab}\right)-2\hat{g}_{db}R_{ai}^{d}-2\hat{g}_{bk}R^{k}_{ai})v^{a}v^{b}.\end{array}\right.$

Since $\hat{g}_{ab}=g_{ab}$, we can recognize some terms of the first set in
the second to obtain the following identities

$\begin{cases}2({g}_{cd}\hat{\Gamma}_{ab}^{d}+\hat{g}_{ck}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2{g}_{cd}{\Gamma}_{ab}^{d}-2\hat{g}_{bk}R^{k}_{ac})v^{a}v^{b},\\\
2(\hat{g}_{di}\hat{\Gamma}_{ab}^{d}+\hat{g}_{ki}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2X_{a}\left(\hat{g}_{bi}\right)+2g_{ki}\Gamma_{ab}^{k}-2\hat{g}_{bk}R^{k}_{ai})v^{a}v^{b}.\end{cases}$

When conditions $(*)$ hold, the first identity above leads to condition $(A)$
and the second to condition $(B)$. We only need the extra assumptions in the
statement to prove the converse.

If we suppose that the conditions $(A)$ and $(B)$ hold, we get from the
identities that

$\begin{cases}({g}_{cd}\hat{\Gamma}_{ab}^{d}+\hat{g}_{kc}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}={g}_{cd}{\Gamma}_{ab}^{d}v^{a}v^{b},\\\
(\hat{g}_{di}\hat{\Gamma}_{ab}^{d}+\hat{g}_{ki}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=\hat{g}_{di}{\Gamma}_{ab}^{d}v^{a}v^{b}.\end{cases}$

We may rewrite the first as
$\hat{\Gamma}_{ab}^{d}v^{a}v^{b}=({\Gamma}_{ab}^{d}-{g}^{cd}\hat{g}_{ck}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}$.
When plugged in the second, this gives
$(-\hat{g}_{di}{g}^{cd}\hat{g}_{ck}+\hat{g}_{ki})\hat{\Gamma}^{k}_{ab}v^{a}v^{b}=0$.
Under the current assumptions, we get indeed
$\hat{\Gamma}^{k}_{ab}v^{a}v^{b}=0$ and therefore also
$\hat{\Gamma}_{ab}^{d}v^{a}v^{b}={\Gamma}_{ab}^{d}v^{a}v^{b}$.

For the statements involving $\theta_{i}$, it is clear that we may immediately
rewrite $(A)$ in the desired form. For $(B)$, recall first that, on
${\mathcal{C}}$,

$\displaystyle\lambda_{i}$ $\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}(X_{i}^{V}(L))-X_{i}^{C}(L)=\Gamma_{(L,\mathcal{D})}(g_{ij}v^{j})-\frac{1}{2}X_{i}(g_{ab})v^{a}v^{b}-g_{ab}X_{i}^{C}(v^{a})v^{b}$
$\displaystyle=$ $\displaystyle
g_{ij}\Gamma_{(L,\mathcal{D})}(v^{j})-\frac{1}{2}X_{i}(g_{ab})v^{a}v^{b}-g_{ab}R^{a}_{ic}v^{b}=g_{ki}\Gamma^{k}_{ab}v^{a}v^{b}+g_{db}R^{d}_{ai}v^{a}v^{b}+g_{ab}R^{a}_{ic}v^{c}v^{b}$
$\displaystyle=$ $\displaystyle-g_{ki}\Gamma_{L}(v^{k}).$

In the before-last equality, we have again made use of Koszul’s formula for
the metric $g$. Condition $(B)$ can then be expressed as

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\hat{g}_{di}\Gamma^{d}_{ab}v^{a}v^{b}+g_{ki}\Gamma^{k}_{ab}v^{a}v^{b}+X_{a}(\hat{g}_{bi})v^{a}v^{b}-\hat{g}_{bk}R^{k}_{ai}v^{a}v^{b}$
$\displaystyle=$
$\displaystyle\hat{g}_{di}\Gamma_{(L,\mathcal{D})}(v^{d})-g_{ki}\Gamma_{L}(v^{k})+\Gamma_{(L,\mathcal{D})}(\hat{g}_{bi})v^{b}-\hat{g}_{bk}R^{k}_{ai}v^{a}v^{b}$
$\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}(\hat{g}_{bi}v^{b})-g_{ki}\Gamma_{L}(v^{k})-\hat{g}_{bk}R^{k}_{ai}v^{a}v^{b}$
$\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}(\hat{g}_{bi}v^{b})+\lambda_{i}-\hat{g}_{bk}v^{b}R^{k}_{ai}v^{a}$
$\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}(\theta_{i})+\lambda_{i}+\theta_{k}R^{k}_{ia}v^{a}.$

∎

The next proposition shows that the problem of finding a geodesic extension by
means of a Riemannian metric is solved, once we have found a solution for the
algebraic condition $(A)$ and the PDE equation $(B)$.

######  Proposition 4.

Consider the equations $(A)$ and $(B)$ from the previous Lemma.

  1. 1.

If $\hat{g}$ is a Riemannian metric that is a ${\mathcal{D}}$-preserving
modification of $g$ and a geodesic extension of $(L,{\mathcal{D}})$, then its
restriction
${\hat{g}}|_{{\mathcal{D}}\times{\mathcal{D}}^{g}}=({\hat{g}}_{ai})$ satisfies
the algebraic equations $(A)$ and the PDE equations $(B)$.

  2. 2.

If $\hat{g}$ is (0,2)-tensor field that is a ${\mathcal{D}}$-preserving
modification of $g$ and whose restriction
${\hat{g}}|_{{\mathcal{D}}\times{\mathcal{D}}^{\perp}}=({\hat{g}}_{ai})$
satisfies the algebraic equations $(A)$ and the PDE equations $(B)$, then
there exists a geodesic extension of $(L,{\mathcal{D}})$ by a Riemannian
metric.

###### Proof.

Given what we have said in the proof of Lemma 2, the first statement is
obvious. To prove the second, we only need to check that a restriction
$\hat{g}|_{{\mathcal{D}}\times{\mathcal{D}}^{g}}=({\hat{g}}_{ai})$ (solving
$(A)$ and $(B)$) can always be completed to a full Riemannian metric
$\hat{g}$, by an appropriate choice of the freedom in
$\hat{g}|_{{\mathcal{D}}^{\perp}\times{\mathcal{D}}^{g}}=({\hat{g}}_{ij})$. In
this proof, we will use the shorthand notation
$\hat{g}=\begin{pmatrix}{g}_{ab}&\hat{g}_{ai}\\\
\hat{g}_{ai}&\hat{g}_{ij}\end{pmatrix}=\begin{pmatrix}A&B\\\
B^{T}&D\end{pmatrix}$. Here, $A$ is a positive-definite $m\times m$-matrix,
$B$ a $m\times k$-matrix and $D$ a $k\times k$-matrix.

First, we show that we can choose $D$ such that $\hat{g}$ is a pseudo-
Riemannian metric (i.e. non-degenerate). Let $C:=B^{T}A^{-1}B$. Since

$\det(\hat{g})=\det(A)\det(D-B^{T}A^{-1}B)=\det(A)\det(D-C),$

with $\det(A)\neq 0$, we need to find a symmetric and invertible $D$ in such a
way that $\det(D-C)\neq 0$. These requirements can be achieved by considering
an $\alpha\neq 0$ which is not an eigenvalue of $C$ and by choosing $D=\alpha
I_{k}$ (where $I_{k}$ is the identity matrix of order $k\times k$). Since
$\det(D-B^{T}A^{-1}B)\neq 0$ is, in the current notations, equivalent with the
condition that the restriction
${\hat{g}}|_{{\mathcal{D}}^{\hat{g}}\times{\mathcal{D}}^{\hat{g}}}$ is non-
degenerate, all the conditions of Lemma 2 are now satisfied.

We now check whether we can specify $\alpha$ further, in such a way that
$\hat{g}$ is also positive-definite. This means that for every non-zero $v\in
T_{q}Q$ we must have

$\hat{g}_{q}(v,v)=g_{ab}v^{a}v^{b}+2\hat{g}_{ai}v^{a}v^{i}+\alpha\delta_{ij}v^{i}v^{j}>0.$

Since $g$ is positive-definite we already know that $g_{ab}v^{a}v^{b}>0$. If
we assume, for our convenience, that the frame $\\{X_{a}\\}$ of
${\mathcal{D}}$ is an orthogonal eigenbasis for the symmetric $g_{ab}$, then
we may write this as

$g_{ab}v^{a}v^{b}=\tau_{1}(v^{1})^{2}+\tau_{2}(v^{2})^{2}+\ldots+\tau_{m}(v^{m})^{2}>0,$

where $\tau_{i}$ are the (positive) eigenvalues of $g_{ab}$. Let
$\tau_{\text{min}}=\min\\{\tau_{1},\tau_{2},\ldots,\tau_{m}\\}>0$. Then

$g_{ab}v^{a}v^{b}\geq\tau_{\text{min}}\left((v^{1})^{2}+\ldots+(v^{m})^{2}\right)=\tau_{\text{min}}\delta_{ab}v^{a}v^{b}.$

If we can show that there exists an $\alpha$ such that

$\tau_{\text{min}}\delta_{ab}v^{a}v^{b}+2\hat{g}_{ai}v^{a}v^{i}+\alpha\delta_{ij}v^{i}v^{j}>0$

for every $(v^{a},v^{i})\neq(0,0)$, we also get that

$\hat{g}_{q}(v,v)=g_{ab}v^{a}v^{b}+2\hat{g}_{ai}v^{a}v^{i}+\alpha\delta_{ij}v^{i}v^{j}>0.$

In conclusion, we can finish the proof if we show that there exist a value for
$\alpha$ for which the matrix $\begin{pmatrix}\tau_{\text{min}}I_{m}&B\\\
B^{T}&\alpha I_{k}\end{pmatrix}$ is positive-definite, i.e. all its
eigenvalues $\lambda$ are strictly positive.

If $\lambda$ is an eigenvalue of the above matrix, it will satisfy

$\det\begin{pmatrix}(\tau_{\text{min}}-\lambda)I_{m}&B\\\
B^{T}&(\alpha-\lambda)I_{k}\end{pmatrix}=0.$

We distinguish between two cases. Suppose, first, that
$\lambda=\tau_{\text{min}}$, then it is clearly strictly positive. Next, if
$\lambda\neq\tau_{\text{min}}$, then it will satisfy the equation:

$\det((\tau_{\text{min}}-\lambda)I_{m})\det((\alpha-\lambda)I_{m}-B^{T}\frac{1}{\tau_{\text{min}}-\lambda}I_{k}B)=0$

and therefore also

$(\tau_{\text{min}}-\lambda)^{m-k}\det((\alpha-\lambda)(\tau_{\text{min}}-\lambda)I_{m}-B^{T}B)=0.$

Since the first factor is not zero, we see that $\lambda\neq\tau_{\text{min}}$
will be an eigenvalue of $\begin{pmatrix}\tau_{\text{min}}I_{m}&B\\\
B^{T}&\alpha I_{k}\end{pmatrix}$ if and only if
$\mu=(\alpha-\lambda)(\tau_{\text{min}}-\lambda)$ is an eigenvalue of
$B^{T}B$. This last relation can also be written as

$\lambda^{2}-(\tau_{\text{min}}+\alpha)\lambda-\mu+\tau_{\text{min}}\alpha=0.$

Of all eigenvalues $\mu$ of $B^{T}B$, take $\mu_{\text{max}}=\max\\{\mu\\}$.
To end the proof, we will assume that $\alpha\neq 0$ is not an eigenvalue of
$C$, and is such that $\alpha>-\tau_{\text{min}}$ and
$\alpha>\frac{\mu_{\text{max}}}{\tau_{\text{min}}}$. In that case, the two
values $\lambda$ that satisfy the above quadratic equation are indeed strictly
positive, because their sum and product satisfy, respectively,

$\tau_{\text{min}}+\alpha>0,\qquad\mbox{and}\qquad-\mu+\tau_{\text{min}}\alpha\geq-\mu_{\text{max}}+\tau_{\text{min}}\alpha>0.$

∎

In Section 8.1 we will derive and integrate the conditions $(A)$ and $(B)$
explicitly for the example of a vertically rolling disk. We will also provide
Riemannian metrics $\hat{g}$ which are geodesic extensions of the nonholonomic
system.

The conditions $(A)$ and $(B)$, when written in terms of $\theta_{i}$ are
familiar from a completely different context. If we set $\phi_{i}=-\theta_{i}$
(and forget that here $\theta_{i}$ is assumed to be linear in the $v^{a}$),
the conditions

$\phi_{k}R^{k}_{ac}v^{a}=0\quad\mbox{and}\quad\Gamma_{(L,\mathcal{D})}(\phi_{i})+\phi_{k}R^{k}_{ia}v^{a}-\lambda_{i}=0$

also appear in [15] (Corollary 1). The correspondence with our conditions is
only ostensibly, however, since there the conditions have a completely
different interpretation and role in the theory: there, they are related to
conditions for the nonholonomic problem to be consistent with a vakonomic
problem. It would lead us too far to explain here deeper the relation between
this consistency problem and the geodesic extension problem we are now dealing
with.

##  5 Relation to sode extensions

Let $(L=\frac{1}{2}g(v,v),{\mathcal{D}})$ be a purely kinetic nonholonomic
system. If we take a basis $\\{X_{a},X_{i}\\}$ that respects the orthogonal
decomposition by $g$ (i.e. with $\text{span}\\{X_{a}\\}=\mathcal{D}$ and
$\text{span}\\{X_{i}\\}=\mathcal{D}^{g}$), we have already seen that the
nonholonomic vector field is

${\Gamma_{(L,\mathcal{D})}}=v^{a}X_{a}^{C}+f^{a}X_{a}^{V}=v^{a}X_{a}^{C}-\Gamma_{bc}^{a}v^{b}v^{c}X_{a}^{V}.$

Let $\hat{g}$ be a Riemannian metric with $\hat{g}_{ab}=g_{ab}$. We have
already mentioned that we may also use the metric $\hat{g}$ to decompose $TQ$
into $TQ=\mathcal{D}\oplus{{\mathcal{D}}}^{\hat{g}}$. A basis
$\\{\hat{X}_{i}\\}$ that spans ${\mathcal{D}}^{\hat{g}}$ is given by the
vector fields $\hat{X}_{i}=X_{i}-{g}^{ab}\hat{g}_{bi}X_{a}$. As in Section 3,
this decomposition will lead to a decomposition of the type

${\mathcal{X}}(TQ)={\mathcal{X}}^{T}(TQ)\oplus({\mathcal{X}}^{\hat{g}}(\tau,{\mathcal{D}}))^{V}$

where $({\mathcal{X}}^{\hat{g}}(\tau,{\mathcal{D}}))^{V}={\rm
span}\\{\hat{X}_{i}^{V}\\}$. Let’s denote the projections that correspond to
this decomposition by

$\hat{p}:\mathcal{X}(TQ)\rightarrow{\mathcal{X}}^{T}(TQ)\quad\text{ and
}\quad\hat{q}:\mathcal{X}(TQ)\rightarrow({\mathcal{X}}^{\hat{g}}(\tau,{\mathcal{D}}))^{V}.$

The geodesic spray $\Gamma_{\hat{g}}$ (i.e. the Lagrangian vector field
$\Gamma_{\hat{L}}$ of $\hat{L}(v)=\frac{1}{2}\hat{g}(v,v)$) is given in terms
of the basis $\\{X_{a},X_{i}\\}$ by

$\Gamma_{\hat{g}}=v^{\alpha}X_{\alpha}^{C}+\hat{F}^{a}X_{a}^{V}+\hat{F}^{i}X_{i}^{V}=v^{\alpha}X_{\alpha}^{C}-{\hat{\Gamma}}^{a}_{\alpha\beta}v^{\alpha}v^{\beta}X_{a}^{V}-{\hat{\Gamma}}^{i}_{\alpha\beta}v^{\alpha}v^{\beta}X_{i}^{V},$

where $\hat{\Gamma}^{\gamma}_{\alpha\beta}$ are the Christoffel symbols of
$\hat{g}$, as before.

Similarly as in Propositions 1 and 2, we may construct from $\Gamma_{\hat{g}}$
two new sodes. The first is

$\Gamma_{\hat{g}}^{1}=\hat{p}(\Gamma_{\hat{g}})=v^{\alpha}X_{\alpha}^{C}+({\hat{F}}^{a}+g^{ab}\hat{g}_{bi}\hat{F}^{i})X_{a}^{V}+0X_{i}^{V}.$

The second is (according to Proposition 3) the quadratic spray of the
nonholonomic connection (now the one associated to $\hat{g}$ and
${\mathcal{D}}$). To derive its expression, we see that the orthogonal
projection
$\hat{Q}:\mathcal{X}(\tau)\rightarrow\mathcal{X}^{\hat{g}}(\tau,\mathcal{D})$
has the property

$\hat{Q}({\mathbf{T}})=\hat{Q}(v^{a}X_{a}+v^{i}X_{i})=\hat{Q}((v^{a}+v^{i}g^{ab}\hat{g}_{bi})X_{a}+v^{i}\hat{X}_{i})=v^{i}{\hat{X}}_{i}.$

With that, we may define

$\displaystyle\Gamma_{\hat{g}}^{2}$ $\displaystyle=$
$\displaystyle\Gamma_{\hat{L}}-(\hat{\nabla}(\hat{Q}({\mathbf{T}})))^{V}$
$\displaystyle=$
$\displaystyle\Gamma_{\hat{g}}-(\Gamma_{\hat{g}}(v^{i})\hat{X}_{i}+v^{i}\hat{\nabla}(\hat{X}_{i}))^{V}$
$\displaystyle=$ $\displaystyle
v^{\alpha}X_{\alpha}^{C}+({\hat{F}}^{a}+g^{ab}\hat{g}_{bi}\hat{F}^{i})X_{a}^{V}-v^{i}(\hat{\nabla}(\hat{X}_{i}))^{V}.$

In the next proposition, we relate the condition $(A)$ to sode extensions of
${\Gamma_{(L,\mathcal{D})}}$ by one of these vector fields.

######  Proposition 5.

Let $g$ and $\hat{g}$ be two Riemannian metrics which are a
${\mathcal{D}}$-preserving modification of each other. The following
statements are equivalent:

  1. (i)

The condition $(A)$ is satisfied for $\hat{g}$,

  2. (ii)

The vector field $\Gamma_{\hat{g}}^{1}$ is a sode extension of
$\Gamma_{(L_{g},\mathcal{D})}$,

  3. (iii)

The vector field $\Gamma_{\hat{g}}^{2}$ is a sode extension of
$\Gamma_{(L_{g},\mathcal{D})}$,

  4. (iv)

The nonholonomic vector field $\Gamma_{(L_{g},\mathcal{D})}$ satisfies on
${\mathcal{C}}$,

$\Gamma_{(L,\mathcal{D})}(X^{V}(\hat{L}))-X^{C}(\hat{L}))=0,\qquad\forall
X\in{\rm Sec}(\mathcal{D}).$

###### Proof.

We have already remarked in the proof of Lemma 2 that the Christoffel symbols
${\hat{\Gamma}}^{\gamma}_{\alpha\beta}$ of $\hat{g}$ and
$\Gamma^{\gamma}_{\alpha\beta}$ of $g$ are always related as follows:

$2({g}_{cd}\hat{\Gamma}_{ab}^{d}+\hat{g}_{kc}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2{g}_{cd}{\Gamma}_{ab}^{d}-\hat{g}_{bk}R^{k}_{ac}-\hat{g}_{ak}R^{k}_{bc})v^{a}v^{b}.$

The condition $(A)$ is satisfied if, and only if, we have

$2({g}_{cd}\hat{\Gamma}_{ab}^{d}+\hat{g}_{kc}\hat{\Gamma}^{k}_{ab})v^{a}v^{b}=(2{g}_{cd}{\Gamma}_{ab}^{d})v^{a}v^{b}.$

We may, equivalently, rewrite this as

$g_{dc}\hat{F}^{d}|_{{\mathcal{C}}}+\hat{g}_{ck}\hat{F}^{k}|_{{\mathcal{C}}}=g_{dc}f^{d}.$

Since the matrix $g_{cd}$ is invertible we get

$\hat{F}^{d}|_{{\mathcal{C}}}+g^{dc}\hat{g}_{ck}\hat{F}^{k}|_{{\mathcal{C}}}=f^{d}.$

Since the steps are reversible, this condition is in fact equivalent with the
condition in (i). We continue the proof by showing that also all of the other
statements are equivalent to this.

Indeed, given that $v^{i}=0$ on ${\mathcal{C}}$, it is easy to see from the
expressions we had obtained before that

$\Gamma_{\hat{g}}^{1}|_{\mathcal{C}}=v^{a}X_{a}^{C}+f^{a}X_{a}^{V}={\Gamma_{(L,\mathcal{D})}}$

if and only if
$\hat{F}^{d}|_{{\mathcal{C}}}+g^{dc}\hat{g}_{ck}\hat{F}^{k}|_{{\mathcal{C}}}=f^{d}$,
and similarly for $\Gamma_{\hat{g}}^{2}$.

The statement $(iv)$ says that the coefficients $f^{a}$ of the nonholonomic
vector field ${\Gamma_{(L,\mathcal{D})}}$ satisfy, on $\mathcal{C}$,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Gamma_{(L_{g},\mathcal{D})}(X_{a}^{V}(\hat{L}))-X_{a}^{C}(\hat{L})$
$\displaystyle=$ $\displaystyle
v^{b}X_{b}^{C}X_{a}^{C}(\hat{L})+f^{b}X_{b}^{V}X_{a}^{V}(\hat{L})-X_{a}^{C}(\hat{L})$
$\displaystyle=$ $\displaystyle
v^{b}X_{b}^{C}X_{a}^{C}(\hat{L})+f^{b}{g}_{ab}-X_{a}^{C}(\hat{L}).$

Now, recall that the Lagrangian vector field
$\Gamma_{\hat{L}}=\Gamma_{\hat{g}}$ always satisfies

$\Gamma_{\hat{L}}(X^{V}(\hat{L}))-X^{C}(\hat{L})=0,\quad\forall
X\in\mathcal{X}(Q).$

In particular, we have (after restriction to ${\mathcal{C}}$)

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Gamma_{\hat{L}}(X_{a}^{V}(\hat{L}))-X_{a}^{C}(\hat{L})=v^{b}X_{b}^{C}X_{a}^{V}(\hat{L})+\hat{F}^{b}|_{\mathcal{C}}X_{b}^{V}X_{a}^{V}(\hat{L})+F^{i}|_{\mathcal{C}}X_{i}^{V}X_{a}^{V}(\hat{L})-X_{a}^{C}(\hat{L})$
$\displaystyle=$ $\displaystyle
v^{b}X_{b}^{C}X_{a}^{V}(\hat{L})+\hat{F}^{b}|_{\mathcal{C}}{g}_{ab}+F^{i}|_{\mathcal{C}}\hat{g}_{ai}-X_{a}^{C}(\hat{L}).$

If we subtract this from the previous, the statement in $(iv)$ is indeed
equivalent with
$\hat{F}^{b}|_{\mathcal{C}}+F^{i}|_{\mathcal{C}}\hat{g}_{ai}g^{ab}=f^{b}$. ∎

We have already shown in Proposition 4 that $\Gamma_{\hat{g}}$ is a sode
extension of $\Gamma_{(L_{g},\mathcal{D})}$ if and only if $(A)$ and $(B)$ are
satisfied. From the previous we know that, if only $(A)$ is satisfied, it is
not enough to guarantee that $\Gamma_{\hat{g}}$ is a sode extension:
Proposition 5 says that in that case only $\Gamma_{\hat{g}}^{1}$ or
$\Gamma_{\hat{g}}^{2}$ is one. Therefore, we need to add the condition that
also the restriction $\hat{\nabla}(\hat{Q}({\mathbf{T}}))|_{\mathcal{C}}$
vanishes. The proof of the next proposition is now obvious.

######  Proposition 6.

Let $g$ and $\hat{g}$ be two Riemannian metrics which are a
${\mathcal{D}}$-preserving modification of each other. The following
statements are equivalent:

  1. (i)

$\Gamma_{\hat{g}}$ is a SODE extension of $\Gamma_{(L,\mathcal{D})}$,

  2. (ii)

The conditions $(A)$ and $(B)$ are satisfied,

  3. (ii)

(A) is satisfied and $\hat{\nabla}(\hat{Q}({\mathbf{T}}))|_{\mathcal{C}}=0$,

  4. (iii)

$\Gamma_{\hat{g}}|_{\mathcal{C}}=\Gamma_{\hat{g}}^{1}|_{\mathcal{C}}=\Gamma_{\hat{g}}^{2}|_{\mathcal{C}}$.

##  6 Infinitesimal Gauss-type conditions

Throughout this section, given $v_{q}\in T_{q}Q$, we will often make use of
the canonical identification of $T_{v_{q}}(T_{q}Q)$ as the tangent space
$T_{q}Q$:

$(w_{q})_{{v_{q}}}^{V}=w^{\alpha}X_{\alpha}^{V}(v_{q})\,\,\longleftrightarrow\,\,w_{q}=w^{\alpha}X_{\alpha}(q).$

Let $\phi_{t}^{nh}:{\mathcal{D}}\to{\mathcal{D}}$ be the flow of the
nonholonomic vector field ${\Gamma_{(L,\mathcal{D})}}$. The nonholonomic
exponential map is defined in [2] as

$\exp^{nh}_{q}(v_{q})=\tau(\phi_{1}^{nh}(v_{q})),$

where $\tau:TQ\to Q$ is the tangent bundle projection. Among other properties,
it has been shown in [3] specifically for a purely kinetic nonholonomic system
$(L(v)=\frac{1}{2}g(v,v),{\mathcal{D}})$ that for each fixed $q\in Q$, there
exists a submanifold $M_{q}$ of $Q$ (with $q\in M_{q}$) such that the
nonholonomic exponential map restricts to a diffeomorphism
$\exp^{nh}_{q}:U_{0}\subset{\mathcal{D}}_{q}\to M_{q}$, where $U_{0}$ is a
starshaped open subset of $D_{q}$ around $0_{q}\in U_{0}$. Moreover,
$\exp^{nh}_{q}(0_{q})=q$ and the morphism
$T_{0_{q}}(\exp^{nh}_{q}):{\mathcal{C}}_{q}\to T_{q}Q$ (under the
identification between ${\mathcal{C}}_{q}$ and $T_{0_{q}}U_{0}$) is the
canonical inclusion. Important for us, is that it has also been shown in [3]
that, for every $v_{q}\in U_{0}$,

$\exp^{nh}_{q}(tv_{q})=c_{v_{q}}(t),$

with $c_{v_{q}}:[0,1]\to M_{q}$ the nonholonomic trajectory satisfying
$c_{v_{q}}(0)=q$ and ${\dot{c}}_{v_{q}}(0)=v_{q}$.

In what follows we will use a description of the nonholonomic exponential in
quasi-velocities. Given a frame $\\{X_{\alpha}\\}=\\{X_{a},X_{i}\\}$, adapted
to $TQ={\mathcal{D}}\oplus{\mathcal{D}}^{g}$, as before, we set in local
coordinates

$\exp^{nh}_{q}:v_{q}=v^{a}X_{a}(q)\mapsto\tilde{q}^{\alpha}=\epsilon^{\alpha}(v).$

We will also use $((E^{\alpha}_{\beta})(v))$ for the matrix representation of
the tangent map of $\exp^{nh}_{q}$ in $v_{q}$ in quasi-velocities: For
$u_{q}\in{\mathcal{D}}_{q}$, we set

$T_{v_{q}}(\exp^{nh}_{q})(u_{q})=u^{a}\left(E^{b}_{a}(v)X_{b}(\tilde{q})+E^{i}_{a}(v)X_{i}(\tilde{q})\right)\in
T_{\tilde{q}}Q.$

######  Lemma 3.

At $0_{q}\in T_{q}Q$, we have $E_{a}^{b}(0)=\delta_{a}^{b}$, $E_{a}^{i}(0)=0$,

$\frac{\partial E^{d}_{a}}{\partial
v^{b}}(0)=\frac{1}{2}R^{d}_{ab}(q)-\frac{1}{2}(\Gamma_{ab}^{d}(q)+\Gamma^{d}_{ba}(q))\quad\text{and}\quad\frac{\partial
E^{i}_{a}}{\partial v^{b}}(0)=\frac{1}{2}R^{i}_{ab}(q).$

###### Proof.

The first two properties follow from the fact that, when $v_{q}=0_{q}$,
$T_{0_{q}}(\exp^{nh}_{q})$ can be regarded as the inclusion. We will compute
the other expressions in two steps.

Let $v_{q}=v^{a}X_{a}(q)\in T_{q}Q$. Consider a nonholonomic trajectory
$c_{v_{q}}(t)$ and its derivative
$\dot{c}_{v_{q}}(t)=V^{a}(t)X_{a}\left(c_{v_{q}}(t)\right)\in\mathcal{D}_{c_{v_{q}}(t)}$.
The functions $V^{a}(t)$ then satisfy
$\dot{V}^{a}(t)=-\Gamma_{bc}^{a}\left(c_{v_{q}}(t)\right)V^{b}(t)V^{c}(t)$.
Since in the purely kinetic case $c_{v_{q}}(t)=\exp\left(tv_{q}\right)$, we
may also write this as

$\dot{c}_{v_{q}}(t)=E_{a}^{\alpha}(tv)v^{a}X_{\alpha}\left(c_{v_{q}}(t)\right).$

We get from this

$E_{a}^{d}(tv_{q})v^{a}=V^{d}(t)\quad\text{and}\quad
E_{a}^{i}(tv_{q})v^{a}=0.$

By applying $\frac{d}{dt}|_{t=0}$ on both sides of both equations we get

$\frac{\partial E_{a}^{d}}{\partial
v^{\mu}}(0)v^{\mu}v^{a}=\dot{V}^{d}(0)=-\Gamma^{d}_{ef}(q)v^{e}v^{f}$

and after symmetrization

$\frac{\partial E_{a}^{d}}{\partial v^{c}}(0)+\frac{\partial
E_{c}^{d}}{\partial v^{a}}(0)=-(\Gamma_{ac}^{d}(q)+\Gamma_{ca}^{d}(q)).$

Likewise, it follows from ${\partial E_{a}^{i}}/{\partial
v^{\mu}}(0)v^{\mu}v^{a}={\partial E_{a}^{i}}/{\partial v^{b}}(0)v^{b}v^{a}=0$
that ${\partial E_{a}^{i}}/{\partial v^{b}}(0)+{\partial E_{b}^{i}}/{\partial
v^{a}}(0)=0$.

Now we compute the expression for the differences ${\partial
E_{a}^{d}}/{\partial v^{c}}(0)-{\partial E_{c}^{d}}/{\partial v^{a}}(0)$ and
${\partial E_{a}^{i}}/{\partial v^{b}}(0)-{\partial E_{b}^{i}}/{\partial
v^{a}}(0)$. We know that in natural coordinates and in quasi-velocities, we
get (respectively)

$T_{v_{q}}(\exp^{nh}_{q})(u_{q})=u^{a}\frac{\partial\epsilon^{\alpha}}{\partial
v^{a}}(v)\frac{\partial}{\partial
q^{\alpha}}|_{\tilde{q}}=E_{a}^{\beta}(v)u^{a}X_{\beta}(\tilde{q}).$

Since $\\{{\partial}/{\partial q^{\alpha}}|_{\tilde{q}}\\}$ is a basis for
$T_{\tilde{q}}Q$ there exists an invertible matrix
$(A_{\beta}^{\alpha}(\tilde{q}))$ of functions such that
$X_{\beta}(\tilde{q})=A_{\beta}^{\alpha}(\tilde{q}){\partial}/{\partial
q^{\alpha}}|_{\tilde{q}}$. Therefore ${\partial\epsilon^{\alpha}}/{\partial
v^{a}}(v)=E_{a}^{\beta}(v)A_{\beta}^{\alpha}(\epsilon(v))$, and in particular
${\partial\epsilon^{\alpha}}/{\partial
v^{a}}(0)=E_{a}^{\beta}(0)A_{\beta}^{\alpha}(q)=A_{a}^{\alpha}(q)$. When we
take a $\partial/\partial v^{b}$-derivative on both sides of the first
relation, we get

$\frac{\partial^{2}\epsilon^{\alpha}}{\partial v^{a}\partial
v^{b}}(v)=\frac{\partial E_{a}^{\beta}}{\partial
v^{b}}(v)A_{\beta}^{\alpha}(\epsilon(v))+E_{a}^{\beta}(v)\frac{\partial
A_{\beta}^{\alpha}}{\partial
q^{\mu}}(\epsilon(v))\frac{\partial\epsilon^{\mu}}{\partial v^{b}}(v),$

and at $v=0$:

$\frac{\partial^{2}\epsilon^{\alpha}}{\partial v^{a}\partial
v^{b}}(0)=\frac{\partial E_{a}^{\beta}}{\partial
v^{b}}(0)A_{\beta}^{\alpha}(q)+\delta_{a}^{\beta}\frac{\partial
A_{\beta}^{\alpha}}{\partial q^{\mu}}(q)A^{\mu}_{b}(q).$

If we change in this last equality $a$ and $b$, and subtract we get the
identity

$0=\left(\frac{\partial E_{a}^{\beta}}{\partial v^{b}}(0)-\frac{\partial
E_{b}^{\beta}}{\partial
v^{a}}(0)\right)A_{\beta}^{\alpha}(q)+\left(\frac{\partial
A_{a}^{\alpha}}{\partial q^{\mu}}(q)A^{\mu}_{b}(q)-\frac{\partial
A_{b}^{\alpha}}{\partial q^{\mu}}(q)A^{\mu}_{a}(q)\right).$

The last term is, in fact, related to the Lie bracket
$R^{\beta}_{ba}X_{\beta}=[X_{b},X_{a}]=(A^{\mu}_{b}{\partial
A_{a}^{\alpha}}/{\partial q^{\mu}}-A^{\mu}_{a}{\partial
A_{b}^{\alpha}}/{\partial q^{\mu}}){\partial}/{\partial q^{\alpha}}$. The
difference now becomes

$0=\left[\frac{\partial E_{a}^{\beta}}{\partial v^{b}}(0)-\frac{\partial
E_{b}^{\beta}}{\partial
v^{a}}(0)-R_{ab}^{\beta}(q)\right]A^{\alpha}_{\beta}(q).$

For $\beta=d$ and $\beta=i$, we get, respectively,

$\frac{\partial E_{a}^{d}}{\partial v^{b}}(0)-\frac{\partial
E_{b}^{d}}{\partial v^{a}}(0)=R_{ab}^{d}(q),\qquad\frac{\partial
E_{a}^{i}}{\partial v^{b}}(0)-\frac{\partial E_{b}^{i}}{\partial
v^{a}}(0)=R_{ab}^{i}(q).$

We conclude that

$\begin{cases}\displaystyle\frac{\partial E_{a}^{d}}{\partial
v^{b}}(0)-\frac{\partial E_{b}^{d}}{\partial
v^{a}}(0)=R_{ab}^{d}(q),\\\\[5.69054pt] \displaystyle\frac{\partial
E_{a}^{d}}{\partial v^{c}}(0)+\frac{\partial E_{c}^{d}}{\partial
v^{a}}(0)=-(\Gamma_{ac}^{d}(q)+\Gamma_{ca}^{d}(q)),\end{cases}\begin{cases}\displaystyle\frac{\partial
E_{a}^{i}}{\partial v^{b}}(0)-\frac{\partial E_{b}^{i}}{\partial
v^{a}}(0)=R_{ab}^{i}(q),\\\\[5.69054pt] \displaystyle\frac{\partial
E_{a}^{i}}{\partial v^{b}}(0)+\frac{\partial E_{b}^{i}}{\partial
v^{a}}(0)=0.\end{cases}$

This leads to the expressions in the statement of the Lemma. ∎

We can now relate condition $(A)$ to the Gauss Lemma of Riemannian geometry.
Let $h$ be a Riemannian metric on a manifold $Q$ with exponential map
$\exp^{h}_{q}$. Suppose that $v_{q}\in T_{q}Q$ is such that
$\exp_{q}^{h}(v_{q})$ is defined. The Gauss Lemma says, in essence, that
$\exp_{q}^{h}:T_{q}Q\to Q$ is a radial isometry (under appropriate
identification). This means that ${\mathcal{H}}^{0}=(\exp^{h}_{q})^{*}h$
satisfies the ‘Gauss condition’:

$\mathcal{H}^{0}_{v_{q}}(v_{q},u_{q})=\mathcal{H}^{0}_{0_{q}}(v_{q},u_{q}),$

for all $u_{q}\in T_{v}T_{q}Q\cong T_{q}Q$. The expression for
$\mathcal{H}^{0}$ can equivalently be written as

$h_{\exp^{h}_{q}(v_{q})}(T_{v_{q}}\exp^{h}_{q}(v_{q}),T_{v_{q}}\exp^{h}_{q}(u_{q}))=h_{q}(v_{q},u_{q}).$

We refer to e.g. [18] for the details.

We do not need the Gauss Lemma as such. In [3] it is shown that a certain
Gauss-type condition is necessary and sufficient for the existence of a
Riemannian metric $g^{nh}_{q}$ on the submanifold $M_{q}$ with the property
that the radial kinetic nonholonomic trajectories departing from $q$ are
minimizing geodesics of $g^{nh}_{q}$. This is also not the property that we
will be interested in here, but our result has a similar flavour. In the next
proposition we give a sufficient condition under which the condition $(A)$
(that appears in Lemma 2 and Proposition 4) always holds.

######  Proposition 7.

Let $\exp^{nh}_{q}$ be the nonholonomic exponential map of a purely kinetic
$(L,{\mathcal{D}})$ and let $\hat{g}$ be a Riemannian metric on $Q$ which is a
${\mathcal{D}}$-preserving modification of $g$. If
$\mathcal{G}^{0}=(\exp^{nh}_{q})^{*}\hat{g}$ satisfies the following Gauss-
type condition

$\mathcal{G}^{0}_{v_{q}}(v_{q},u_{q})=\mathcal{G}^{0}_{0_{q}}(v_{q},u_{q}),\quad\quad\forall
v_{q},u_{q}\in\mathcal{D}_{q},$

or, equivalently,

$\hat{g}_{\exp^{nh}_{q}(v_{q})}\left(T_{v_{q}}\left(\exp_{q}^{nh}\right)\left(v_{q}\right),T_{v_{q}}\left(\exp_{q}^{nh}\right)\left(u_{q}\right)\right)=\hat{g}_{q}(v_{q},u_{q}),$

then condition $(A)$ is satisfied for $\hat{g}$.

###### Proof.

The equivalent condition is the first written in full, after having taken into
account that the tangent map of the exponential map in zero is the inclusion
(with the identification between $T_{q}Q$ and $T_{v_{q}}(T_{q}Q)$).

Given that $v_{q},u_{q}\in\mathcal{D}_{q}$, we can write them as
$v_{q}=v^{a}X_{a}(q)$ and $w_{q}=u^{a}X_{a}(q)$. With this the Gauss condition
becomes

$\hat{g}_{\alpha\beta}(\exp^{nh}_{q}(v_{q})){E_{a}^{\alpha}(v)}v^{a}E_{b}^{\beta}(v)=g_{ab}(q)v^{a}.$

Now, we substitute $v_{q}$ by $tv_{q}$ and cancel a factor $t$ on both sides,

$\hat{g}_{{\alpha\beta}}(c_{v_{q}}(t))E_{a}^{\alpha}(tv)v^{a}E_{b}^{\beta}(tv)=g_{ab}v^{a},$

where we have also used $\exp^{nh}_{q}(tv_{q})=c_{v_{q}}(t)$. If we apply
$\left.\frac{d}{dt}\right|_{t=0}$ on both sides, we obtain

$\displaystyle\frac{\partial\hat{g}_{\alpha\beta}}{\partial
q^{\mu}}(q)(\dot{c}_{v_{q}})^{\mu}(0){E_{a}^{\alpha}(0)}v^{a}{E_{b}^{\beta}(0)}+\hat{g}_{{\alpha\beta}}(q)\frac{\partial
E_{a}^{\alpha}}{\partial
v^{\mu}}(0)v^{\mu}v^{a}{E_{b}^{\beta}(0)}+\hat{g}_{{\alpha\beta}}(q){E_{a}^{\alpha}(0)}v^{a}\frac{\partial
E_{b}^{\beta}}{\partial v^{\mu}}(0)v^{\mu}=0.$

In the first term, we may rewrite
$(\dot{c}_{v_{q}})^{\mu}(0){\partial}/{\partial q^{\mu}}|_{q}$ as
$v^{f}X_{f}(q)+v^{i}X_{i}(q)=v^{f}X_{f}(q)$, since the vector
$v_{q}={\dot{c}}_{v_{q}}(0)\in\mathcal{D}_{q}$ has components $v^{i}=0$. If we
also insert the expressions that we had found in Lemma 3, we get

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
v^{f}X_{f}(\hat{g}_{ab})v^{a}+\hat{g}_{{db}}\left(-(\Gamma^{d}_{af}+\Gamma_{fa}^{d})\right)v^{f}v^{a}+\frac{1}{2}\hat{g}_{{ad}}\left(-(\Gamma^{d}_{bf}+\Gamma^{d}_{jb})+R^{d}_{bf}\right)v^{f}v^{a}+\frac{1}{2}\hat{g}_{{ai}}R^{i}_{bf}v^{f}v^{a}$
$\displaystyle=$
$\displaystyle\left(X_{f}(\hat{g}_{ab})-\frac{1}{2}\hat{g}_{{db}}\big{(}2\Gamma^{d}_{af}+0\big{)}-\frac{1}{2}\hat{g}_{{ad}}\big{(}2\Gamma^{d}_{fb}\big{)}+\frac{1}{2}(\frac{1}{2}(\hat{g}_{ai}R^{i}_{bf}+\hat{g}_{fi}R^{i}_{ba}))\right)v^{a}v^{f}$
$\displaystyle=$
$\displaystyle\left(X_{f}(\hat{g}_{ab})-\hat{g}_{db}\Gamma^{d}_{af}-\hat{g}_{ad}\Gamma^{d}_{fb}+\frac{1}{4}(\hat{g}_{ai}R^{i}_{bf}+\hat{g}_{fi}R^{i}_{ba})\right)v^{a}v^{f}.$

Recall first that for the metric $g$, $\nabla^{g}_{X_{f}}g=0$. In the current
frame this gives

$\left(X_{f}({g}_{ab})-\hat{g}_{db}\Gamma^{d}_{af}-{g}_{ad}\Gamma^{d}_{fb}\right)v^{a}v^{f}=0.$

Since we assume that $\hat{g}_{ab}=g_{ab}$, we may use this to see the first
terms in the previous expression vanish. What remains is then indeed condition
$(A)$ from Lemma 2,

$\left(\hat{g}_{ai}R^{i}_{bf}+\hat{g}_{fi}R^{i}_{ba}\right)v^{a}v^{f}=0,$

or, equivalently, $\hat{g}_{ai}R^{i}_{bf}v^{a}v^{f}=0$. ∎

We end this paragraph with an equivalent characterization of the condition
$(B)$. It will be most convenient to come back to the version of $(B)$, in the
form where we regard it as a PDE in the unknowns
$\theta_{i}={\hat{g}}_{ai}v^{a}$ (i.e. the restriction of some semi-basic
1-form $\theta$ to ${\mathcal{C}}$). We now show that $(B)$ can be interpreted
as the infinitesimal version of a geometric property along $c_{v_{q}}(t)$. To
do so, we introduce yet an other sode extension $\Gamma_{\overline{\nabla}}$
of ${\Gamma_{(L,\mathcal{D})}}$, which is the quadratic spray of a linear
connection $\overline{\nabla}$ on $Q$, different from the nonholonomic
connection $\nabla^{nh}$. The construction of the connection
$\overline{\nabla}$ is inspired by a result that can be found in [28].

We have already introduced the two projection operators
$P:TQ\rightarrow\mathcal{D}$ and $Q:TQ\rightarrow\mathcal{D}^{g}$, related to
$g$. Consider the Levi-Civita connection $\nabla^{g}$ of $g$. We may use it to
define the operator $\nabla^{P}$ as

$\nabla^{P}:{\rm Sec}(\mathcal{D})\times{\rm
Sec}(\mathcal{D})\rightarrow\mathcal{X}(Q),\quad(X,Y)\mapsto\nabla^{P}_{X}Y=\mathcal{P}(\nabla^{g}_{X}Y).$

Proposition 2.1 in [28] states that we may extend this to a covariant
derivative

$\overline{\nabla}^{P}:\mathcal{X}(Q)\times{\rm
Sec}(\mathcal{D})\rightarrow\mathcal{X}(Q),\quad(X,Y)\mapsto\nabla^{P}_{P(X)}Y+P([Q(X),Y]).$

We can do the same for the projection operator $Q$. Next, from Theorem 2.2 in
[28], we obtain that, for $X,Y\in\mathcal{X}(Q)$, the operator

$\overline{\nabla}:\mathcal{X}(Q)\times\mathcal{X}(Q)\rightarrow\mathcal{X}(Q),\quad(X,Y)\mapsto\overline{\nabla}^{P}_{X}({P}(Y))+\overline{\nabla}^{Q}_{X}({Q}(Y))$

is a linear connection on $Q$. In the current frame, the coefficients of this
linear connection are

${\overline{\nabla}}_{X_{a}}X_{b}=\Gamma_{ab}^{c}X_{c},\quad{\overline{\nabla}}_{X_{a}}X_{i}=R^{k}_{ai}X_{k},\quad{\overline{\nabla}}_{X_{i}}X_{a}=R^{c}_{ia}X_{c},\quad{\overline{\nabla}}_{X_{i}}X_{j}=\Gamma_{ij}^{k}X_{k},$

and its corresponding spray is

$\Gamma_{\overline{\nabla}}=v^{\alpha}X_{\alpha}^{C}+(\Gamma_{ab}^{c}v^{a}v^{b}+R^{c}_{ia}v^{i}v^{a})X_{c}^{V}+(R^{k}_{ai}v^{a}v^{i}+\Gamma_{ij}^{k}v^{i}v^{j})X_{k}^{V}.$

By restricting it to $v^{i}=0$ we can easily see that it is, indeed, a sode
extension of ${\Gamma_{(L,\mathcal{D})}}$.

Let $c_{v_{q}}(t)$ be a nonholonomic trajectory, with
${\dot{c}}_{v_{q}}(t)=V^{a}(t)X_{a}(c_{v_{q}}(t))\in{\mathcal{D}}_{c_{v_{q}}(t)}$.
Since $\Gamma_{\overline{\nabla}}$ is a sode extension of
${\Gamma_{(L,\mathcal{D})}}$, we may interpret $c_{v_{q}}(t)$ as one of its
base integral curves. We may then use the linear connection
${\overline{\nabla}}$ to parallel transport a vector $w_{q}\in T_{q}Q$ along
$c_{v_{q}}(t)$. In this way, we obtain a vector field $W(t)$ along
$c_{v_{q}}(t)$ whose components along $\\{X_{a},X_{i}\\}$ are solutions of the
initial value problem

$\dot{W}^{\alpha}+{\overline{\Gamma}}^{\alpha}_{\gamma\beta}V^{\gamma}W^{\beta}=0,\qquad
W^{\alpha}(0)=w^{\alpha}.$

For our specific case, we get

$\begin{cases}\dot{W}^{c}=-{\Gamma}_{ab}^{c}V^{a}W^{b},\\\
\dot{W}^{i}=-R_{aj}^{i}V^{a}W^{j}.\end{cases}$

The above differential equations are separated. If we take the initial value
$w_{q}\in{\mathcal{D}}_{q}^{g}$, we have $w^{a}=0$ and therefore is the
solution of the first simply $W^{a}(t)=0$. This means that
$W(t)\in{\mathcal{D}}_{c_{v_{q}}(t)}^{g}$.

For a 1-form $h(t)$ along the curve $c_{v_{q}}(t)$, its covariant derivative
${\overline{\nabla}}_{{\dot{c}}_{v_{q}}(t)}h(t)$ is again a 1-form along
$c_{v_{q}}(t)$. It can be defined below by means of its action on vector
fields $X(t)$ along $c_{v_{q}}(t)$:

$\Big{(}{\overline{\nabla}}_{{\dot{c}}_{v_{q}}(t)}h(t)\Big{)}\Big{(}X(t)\Big{)}=\frac{d}{dt}\Big{(}\Big{(}h(t)\Big{)}\Big{(}X(t)\Big{)}\Big{)}-\Big{(}h(t)\Big{)}\Big{(}{\overline{\nabla}}_{{\dot{c}}_{v_{q}}(t)}X(t)\Big{)}.$

Consider $\\{X^{i},X^{a}\\}$ the basis of 1-forms, dual to the basis
$\\{X_{i},X_{a}\\}$ of vector fields. Then ${\rm span}\\{X^{i}\\}$ is the
annihilator of $\mathcal{D}$, say $\mathcal{D}^{o}$. Since
$\dot{c}_{v_{q}}(t)$ has only components in $\mathcal{D}_{c_{v_{q}}(t)}$, we
can compute that, along $c_{v_{q}}(t)$, the covariant derivative of
$h(t)=h_{i}(t)X^{i}(c_{v_{q}}(t))+h_{b}(t)X^{b}(c_{v_{q}}(t))$ is given by

${\overline{\nabla}}_{\dot{c}_{v_{q}}(t)}h=\left(\dot{h}_{c}-{\Gamma}^{b}_{ac}V^{a}h_{b}\right)X^{c}+\left(\dot{h}_{i}-{R}^{k}_{ai}V^{a}h_{k}\right)X^{i}.$

Now, consider the 1-form $\lambda(t)=\lambda({\dot{c}}_{v_{q}}(t))$ along
$c_{v_{q}}(t)$, i.e. the restriction of the semi-basic 1-form $\lambda$ that
represented the Lagrangian multipliers (in Section 4). For this, we can define
a unique unique 1-form $h(t)$ along $c_{v_{q}}(t)$ that satisfies

${\overline{\nabla}}_{\dot{c}_{v_{q}}}h(t)=\lambda(t),\qquad h(0)=0_{q}\in
T^{*}_{q}Q.$

This $h(t)$ is the unique solution of the initial value problem

$\begin{cases}\dot{h}_{c}(t)=\Gamma^{b}_{ac}(c_{v_{q}}(t))V^{a}(t)h_{b}(t),\\\
\dot{h}_{i}(t)=R^{j}_{ai}(c_{v_{q}}(t))V^{a}(t)h_{j}(t)+\lambda_{i}(t),\end{cases}\qquad\begin{cases}h_{c}(0)=0,\\\
h_{i}(0)=0.\end{cases}$

From the first relation, we may conclude that $h_{b}(t)=0$.

We have now gathered all the tools to define an equivalent characterization of
the PDE condition $(B)$ from Proposition 4.

######  Proposition 8.

Consider $q\in Q,v_{q}\in\mathcal{D}_{q}$ and the 1-form $h(t)$, as defined
above. A ${\mathcal{D}}$-preserving modification $\hat{g}$ of $g$ satisfies
$(B)$ if and only if

$\hat{g}_{c_{v_{q}}(t)}\left({\dot{c}}_{v_{q}}(t),W(t)\right)+\Big{(}h(t)\Big{)}\Big{(}W(t)\Big{)}=\hat{g}_{q}(v_{q},w_{q}),\qquad\forall
w_{q}\in\mathcal{D}^{g}_{q},$

where $W(t)$ is the parallel transport of $w_{q}\in\mathcal{D}^{g}_{q}$ with
respect to ${\overline{\nabla}}$.

Apart from the term in $h(t)$ this condition is again reminiscent to a Gauss-
type condition. If we multiply the relation in the proposition on both sides
with an extra factor $t$, we can express it equivalently in terms of the
exponential map $\exp^{nh}_{q}(tv_{q})=c_{v_{q}}(t)$, as follows:

$\hat{g}_{\exp_{q}^{nh}(tv_{q})}\left(T_{tv_{q}}(\exp^{nh}_{q})(tv_{q}),W(t)\right)+t\Big{(}h(t)\Big{)}\Big{(}W(t)\Big{)}=\hat{g}_{q}(tv_{q},w_{q}).$

###### Proof.

From Lemma 2 we know that the expression $(B)$ can be written as

$\Gamma_{(L,\mathcal{D})}(\theta_{i})+\lambda_{i}+\theta_{k}R^{k}_{ib}v^{b}=0,$

or, after expanding and multiplying both sides with $w^{i}$:

$X_{f}(\hat{g}_{ai})v^{f}v^{a}w^{i}+{\hat{g}}_{ai}f^{a}w^{i}+\lambda_{i}w^{i}+\hat{g}_{ka}v^{a}R^{k}_{ib}v^{b}w^{i}=0.$

This expression is valid for arbitrary choices of $q,v^{a}$ and $w^{k}$. If we
replace them, specifically, by $c_{v_{q}}(t)$, $V^{a}(t)$ and $W^{k}(t)$,
respectively, we get, for all $t$:

$X_{f}(\hat{g}_{ai})V^{f}(t)V^{a}(t)W^{i}(t)+{\hat{g}}_{ai}{\dot{V}}^{a}(t)W^{i}(t)+\lambda_{i}(t)W^{i}(t)+\hat{g}_{ak}V^{a}(t)R^{k}_{ib}V^{b}(t)W^{i}(t)=0$

or,

$\frac{d}{dt}\left(\hat{g}_{ai}V^{a}(t)\right)W^{i}(t)+\left(\dot{h}_{i}(t)-R^{j}_{ai}(c_{v_{q}}(t))V^{a}(t)h_{j}(t)\right)W^{i}(t)+\hat{g}_{ak}V^{a}(t){\dot{W}}^{k}(t)=0.$

We may simplify this to

$\frac{d}{dt}\left(\hat{g}_{ai}V^{a}(t)W^{i}(t)\right)+\left(\dot{h}_{i}(t)W^{i}(t)+{\dot{W}}^{j}(t)h_{j}(t)\right)=0,$

from which

$\frac{d}{dt}\left(\hat{g}_{ai}V^{a}(t)W^{i}(t)+{h}_{i}(t)W^{i}(t)\right)=0.$

This means that

$\hat{g}_{ai}V^{a}(t)W^{i}(t)+{h}_{i}(t)W^{i}(t)=\hat{g}_{ai}v^{a}w^{i}+{h}_{i}(0)w^{i}.$

Since $h_{i}(0)=0$, we get the result.

The other direction follows if we start from the expression above, take a
derivative by $t$ and set $t=0$, afterward. ∎

From Propositions 4, 7 and 8 we may conclude:

######  Proposition 9.

Let $\exp_{q}^{nh}$ be the nonholonomic exponential map of a purely kinetic
$(L,{\mathcal{D}})$ and let $\hat{g}$ be a Riemannian metric which is a
${\mathcal{D}}$-preserving modification of $g$. If, for all
$v_{q}\in{\mathcal{D}}_{q}$, they satisfy both

$\hat{g}_{\exp^{nh}_{q}(v_{q})}\left(T_{v_{q}}\left(\exp_{q}^{nh}\right)\left(v_{q}\right),T_{v_{q}}\left(\exp_{q}^{nh}\right)\left(w_{q}\right)\right)=\hat{g}_{q}(v_{q},w_{q}),\qquad\forall
u_{q}\in{\mathcal{D}}_{q}$

and

$\hat{g}_{\exp_{q}^{nh}(tv_{q})}\left(T_{tv_{q}}(\exp^{nh}_{q})(tv_{q}),W(t)\right)+t\Big{(}h(t)\Big{)}\Big{(}W(t)\Big{)}=\hat{g}_{q}(tv_{q},w),\qquad\forall
w_{q}\in{\mathcal{D}}^{g}_{q},$

then ${\hat{g}}$ is a geodesic extension of $(L,{\mathcal{D}})$.

Because of the similarity with the statement of the Gauss Lemma, we call the
conditions $(A)$ and $(B)$ the infinitesimal version of the above Gauss-type
conditions.

##  7 Chaplygin systems

Chaplygin systems (also called generalized Chaplygin systems or nonholonomic
systems of principle type, see e.g. [7, 12, 23]) are essentially nonholonomic
systems with extra symmetry properties. The goal of this section is to use
that symmetry to simplify the conditions $(A)$ and $(B)$.

Suppose that we have a purely kinetic Chaplygin system. This means that both
the Riemannian metric $g$ (and its Lagrangian $L$) and the constraint manifold
${\mathcal{C}}$ are invariant under the action $G\times Q\to Q$ of a Lie group
$G$ on the configuration manifold $Q$ (when lifted to $TQ$). In addition, we
assume that $\pi:Q\to Q/G$ is a principal fibre bundle and that the
nonholonomic distribution $\mathcal{D}$ is the horizontal distribution of a
principal connection $\omega$ on $\pi$. Let $\\{E_{i}\\}$ be a basis for the
Lie algebra of $G$ and $\\{(E_{i})_{Q}\\}$ its corresponding infinitesimal
generators. If $(x^{a})$ are coordinates on $Q/G$, the horizontal lifts
$X_{a}$ of the vector fields $\partial/\partial x^{a}$ on $Q/G$ (by means of
$\omega$) are invariant vector fields on $Q$ that span ${\mathcal{D}}$. The
set $\\{X_{a},(E_{i})_{Q}\\}$ is then a basis for ${\cal X}(Q)$. If we
construct the basis in this way, we get

$[X_{a},(E_{i})_{Q}]=0,\quad[X_{a},X_{b}]=B_{ab}^{j}(E_{j})_{Q},\quad[(E_{i})_{Q},(E_{j})_{Q}]=-C_{ij}^{k}(E_{k})_{Q}.$

Here, $B_{ab}^{j}$ can be interpreted as the curvature of the connection
$\omega$ and $C_{ij}^{k}$ as the structure constants of the Lie algebra.

Let, for now, $G_{ab}=g(X_{a},X_{b})$, $G_{ai}=g(X_{a},({E}_{i})_{Q})$ and
$G_{ij}=g(({E}_{i})_{Q},({E}_{j})_{Q})$. Similarly as in Section 2, we can
create a basis for $\mathcal{D}^{g}$, by defining the vector fields
${\tilde{X}}_{i}$ as

${\tilde{X}}_{i}=({E}_{i})_{Q}-{G}^{ab}{G}_{ia}X_{b}=({E}_{i})_{Q}+{K}^{b}_{i}X_{b}.$

In the orthogonal basis $\\{X_{a},{\tilde{X}}_{i}\\}$, we have
$g_{ab}=g(X_{a},X_{b})=G_{ab}$, $g_{aj}=g(X_{a},\tilde{X}_{j})=0$ and
$g_{ij}=g(\tilde{X}_{i},\tilde{X}_{j})$. We first compute the bracket
coefficients $R^{k}_{ia}$ and $R^{k}_{bc}$ in terms of the current bracket
coefficients $B^{k}_{ab}$ and $C^{k}_{ij}$.

On the one hand, we have
$[X_{a},{\tilde{X}}_{i}]=[X_{a},({E}_{i})_{Q}+{K}^{b}_{i}X_{b}]=X_{a}({K}^{b}_{i})X_{b}+{K}^{b}_{i}B_{ab}^{j}({E}_{j})_{Q}$,
while on the other hand is
$[X_{a},{\tilde{X}}_{i}]=R_{ai}^{b}X_{b}+R_{ai}^{j}{\tilde{X}}_{j}=(R_{ai}^{b}+{K}^{b}_{j}R_{ai}^{j})X_{b}+R_{ai}^{j}({E}_{j})_{Q}$.
Therefore, $R_{ai}^{j}={K}^{b}_{i}B_{ab}^{j}$. Likewise, from
$[X_{a},X_{b}]=B_{ab}^{j}({E}_{j})_{Q}$ and
$[X_{a},X_{b}]=R_{ab}^{c}X_{c}+R_{ab}^{j}{\tilde{X}}_{j}=(R_{ab}^{c}-K^{c}_{j}R_{ab}^{j})X_{c}+R_{ab}^{j}({E}_{j})_{Q}$,
we can conclude that $R_{ab}^{j}=B_{ab}^{j}$.

The Lagrangian is still
$L=\frac{1}{2}g_{ab}v^{a}v^{b}+\frac{1}{2}g_{ij}v^{i}v^{j}$. We now give an
expression for the multipliers $\lambda_{i}$ in the Chaplygin case. From the
invariance of the Lagrangian it follows that $({E}_{i})_{Q}^{C}(L)=0$ (see
e.g. [27]). Moreover, on ${\mathcal{C}}$,
$X_{b}^{C}(L)=\Gamma_{(L,D)}\left(X_{b}^{V}(L)\right)$. With that, we have, on
${\mathcal{C}}$,

$\displaystyle\lambda_{i}$
$\displaystyle=\Gamma_{(L,D)}\left({\tilde{X}}_{i}^{V}L\right)-{\tilde{X}}_{i}^{C}(L)=\Gamma_{(L,D)}\left(g_{ij}v^{j}\right)-{K}^{b}_{i}X_{b}^{C}(L)-{\dot{K}^{b}_{i}}X_{b}^{V}(L)$
$\displaystyle=-{K}^{b}_{i}{\Gamma_{(L,\mathcal{D})}}\left(X_{b}^{V}(L)\right)-{\dot{K}^{b}_{i}}X_{b}^{V}(L)=-\Gamma_{(L,D)}\left({K}^{b}_{i}X_{b}^{V}(L)\right)=\Gamma_{(L,\mathcal{D})}\left(G_{ai}v^{a}\right).$

We may now re-express the conditions $(A)$ and $(B)$. The $(A)$ condition,
$\theta_{k}R^{k}_{ac}v^{a}=0$, becomes here $\theta_{k}B^{k}_{ac}v^{a}=0$. For
the $(B)$ condition, we obtain after plugging in $(A)$ and the expression for
$\lambda_{i}$,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}(\theta_{i})+\lambda_{i}+\theta_{k}R^{k}_{ia}v^{a}=\Gamma_{(L,\mathcal{D})}(\hat{g}_{ai}v^{a})+\Gamma_{(L,\mathcal{D})}\left(G_{ai}v^{a}\right)-\theta_{k}K^{b}_{i}B_{ab}^{k}v^{a}$
$\displaystyle=$
$\displaystyle\Gamma_{(L,\mathcal{D})}\left((\hat{g}_{ai}+G_{ai})v^{a}\right).$

As a consequence, $\mu_{i}=(\hat{g}_{ai}+G_{ai})v^{a}$ are first integrals of
the nonholonomic vector field ${\Gamma_{(L,\mathcal{D})}}$.

Recall that $(G_{ai})$ in the Proposition below stands for the restriction
$g|_{{\mathcal{D}}\times V\pi}$, where $V\pi$ is the vertical distribution of
the principal fibre bundle $\pi:Q\to Q/G$.

######  Proposition 10.

Let $(L,{\mathcal{D}})$ be a purely kinetic Chaplygin nonholonomic system. If
${\Gamma_{(L,\mathcal{D})}}$ has linear first integrals
$\mu_{i}=\mu_{ai}v^{a}$ that satisfy

$\mu_{k}B^{k}_{ac}v^{a}=G_{bk}B^{k}_{ac}v^{b}v^{a},$

then there exists a geodesic extension of $(L,{\mathcal{D}})$ by a Riemannian
metric $\hat{g}$.

###### Proof.

The proof follows from Proposition 4, since the conditions ensure that a (0,2)
tensor field ${\hat{g}}$ with ${\hat{g}}(X_{a},X_{b})=g_{ab}$ and
${\hat{g}}(X_{a},{\tilde{X}}_{i})=\mu_{ai}-G_{ai}$ can be extended to a
Riemannian metric. ∎

##  8 An example: the vertically rolling disk

The purpose of this section is to determine a geodesic extension $\hat{g}$, on
a concrete example. We have chosen the example of the disk (rolling vertically
without slipping), not only because it is well-known and simple, but also
because it is involved enough to enlighten some of the constructions and
results we have obtained in the previous sections. Moreover, for this example,
already, some quadratic Lagrangians $\hat{L}$ are known in the literature [9,
3] and at the end of the section we will compare them with the ones we find
through our methods.

We will use the following coordinates, see the figures in [21] or in [26]. The
couple $(x,y)$ stands for the position of the center of the mass $C$,
$\varphi$ is the orientation angle of the disk (measured between the tangent
of the contact point and the positive $x$-axis) and $\theta$ is the angle that
a point fixed on the disk makes with respect to the vertical. In this example
the configuration manifold is $Q=\mathbb{R}^{2}\times S^{1}\times S^{1}$.
Because we assume the disk to remain vertical during its motion, the potential
is constant and the Lagrangian is of purely kinetic type. If we choose all
relevant constants to be unit, the Lagrangian function is given by

$L=\frac{1}{2}g(v,v)=\frac{1}{2}(\dot{x}^{2}+\dot{y}^{2}+\dot{\varphi}^{2}+\dot{\theta}^{2})$

and the linear nonholonomic constraints are

$\begin{cases}\dot{x}=\cos(\varphi)\dot{\theta},\\\
\dot{y}=\sin(\varphi)\dot{\theta}.\end{cases}$

This means that the distribution $\mathcal{D}$ can be spanned by

$\mathcal{D}=\text{span}\\{X_{a}\\}=\textrm{span}\left\\{X_{\varphi}=\frac{\partial}{\partial\varphi},X_{\theta}=\frac{\partial}{\partial\theta}+\cos(\varphi)\frac{\partial}{\partial
x}+\sin(\varphi)\frac{\partial}{\partial y}\right\\}.$

The constrained Lagrangian is

$L_{c}=L|_{\mathcal{C}}=\frac{1}{2}(\dot{\varphi}^{2}+2\dot{\theta}^{2}).$

With this input one may write down the Lagrange-d’Alembert equations. It is
well-known (see e.g. [7, 13, 9]) that the center of mass will make either a
circular movement, or a motion along a straight line.

###  8.1 The vertically rolling disk as a purely kinetic nonholonomic system

Remark first that the metric $g$ is such that $g({\partial}/{\partial
q^{i}},{\partial}{/\partial q^{j}})=\delta_{ij}$, with
$(q^{i})=(x,y,\theta,\varphi)$. In this subsection we choose the following
basis for the orthogonal complement $\mathcal{D}^{g}$:

$\mathcal{D}^{g}=\text{span}\\{X_{i}\\}=\text{span}\left\\{X_{x}=\frac{\partial}{\partial
x}-\cos(\varphi)\frac{\partial}{\partial\theta},X_{y}=\frac{\partial}{\partial
y}-\sin(\varphi)\frac{\partial}{\partial\theta}\right\\}.$

It is important to realize that none of our results depend on the specific
choices we have made for the frames. We will make this clear in the next
subsection.

When compared to the expressions in the previous sections, the index $a$ will
run over $\varphi$ and $\theta$, while the index $i$ is here either $x$ or
$y$. If we compute the metric coefficients with respect to this basis, we get
$g_{ai}=0$, meaning here that $g_{\theta x}=g_{\theta y}=g_{\varphi
x}=g_{\varphi y}=0$. Moreover

$(g_{ab})=\begin{pmatrix}g_{\varphi\varphi}&g_{\varphi\theta}\\\
g_{\varphi\theta}&g_{\theta\theta}\end{pmatrix}=\begin{pmatrix}1&0\\\
0&2\end{pmatrix}\quad\mbox{and}\quad(g_{ij})=\begin{pmatrix}g_{xx}&g_{xy}\\\
g_{xy}&g_{yy}\end{pmatrix}=\begin{pmatrix}1+\cos^{2}(\varphi)&\cos(\varphi)\sin(\varphi)\\\
\cos(\varphi)\sin(\varphi)&1+\sin^{2}(\varphi)\end{pmatrix}.$

With this, the Lagrangian in quasi-velocities
$(v_{x},v_{y},v_{\theta},v_{\varphi})$ becomes

$L=\frac{1}{2}\left(2v_{\theta}^{2}+v_{\varphi}^{2}\right)+\frac{1}{2}\left(\left(1+\cos^{2}(\varphi)\right)v_{x}^{2}+2\cos(\varphi)\sin(\varphi)v_{x}v_{y}\right.\left.+\left(1+\sin^{2}(\varphi)\right)v_{y}^{2}\right).$

We may read off the coefficients $R_{\alpha\beta}^{\gamma}$ from the Lie
brackets
$\left[X_{\alpha},X_{\beta}\right]=R_{\alpha\beta}^{c}X_{c}+R_{\alpha\beta}^{k}X_{k}$.
Here they are

$\displaystyle\left[X_{\varphi},X_{\theta}\right]=-\sin(\varphi)X_{x}+\cos(\varphi)X_{y},\qquad\left[X_{x},X_{y}\right]=0,\qquad\left[X_{x},X_{\theta}\right]=0,\qquad\left[X_{y},X_{\theta}\right]=0,$
$\displaystyle\left[X_{\varphi},X_{x}\right]=\frac{1}{2}\sin(\varphi)X_{\theta}-\frac{1}{2}\sin(\varphi)\cos(\varphi)X_{x}-\frac{1}{2}\sin^{2}(\varphi)X_{y},$
$\displaystyle\left[X_{\varphi},X_{y}\right]=-\frac{1}{2}\cos(\varphi)X_{\theta}+\frac{1}{2}\cos^{2}(\varphi)X_{x}+\frac{1}{2}\cos(\varphi)\sin(\varphi)X_{y}.$

Since all $g_{ab}$ are constant and $R^{a}_{db}=0$, we get from the Koszul
formula,

$2g_{cd}\Gamma_{ab}^{d}=X_{a}\left(g_{bc}\right)+X_{b}\left(g_{ac}\right)-X_{c}\left(g_{ac}\right)+g_{dc}R_{bc}^{d}-g_{db}R_{ac}^{d}-g_{ad}R_{bc}^{d},$

that all $\Gamma_{ab}^{d}=0$.

We are now ready to find a solution for the equations $(A)$ and $(B)$ of Lemma
2. For $(A)$, we get the following two algebraic conditions that should be
satisfied by $\hat{g}_{\varphi x},\hat{g}_{\varphi y},\hat{g}_{\theta x}$ and
$\hat{g}_{\theta y}$:

$\begin{cases}\hat{g}_{\varphi x}\sin(\varphi)-\hat{g}_{\varphi
y}\cos(\varphi)=0,\\\ -\hat{g}_{\theta x}\sin(\varphi)+\hat{g}_{\theta
y}\cos(\varphi)=0.\end{cases}$

Condition $(B)$, written in a version without $v^{a}v^{b}$ (and with
$\Gamma^{d}_{bc}=0$) is here:

$0=X_{a}(\hat{g}_{bi})+X_{b}(\hat{g}_{ai})-\hat{g}_{bk}R^{k}_{ai}-\hat{g}_{ak}R^{k}_{bi}.$

If we first take $a=b=\varphi$, this gives
$X_{\varphi}\left(\hat{g}_{\varphi_{i}}\right)-\hat{g}_{\varphi
k}R_{\varphi_{i}}^{k}=0$, or:

$X_{\varphi}\left(\hat{g}_{\varphi x}\right)-\hat{g}_{\varphi x}R_{\varphi
x}^{x}-\hat{g}_{\varphi y}R_{\varphi x}^{y}=0\quad\mbox{and}\quad
X_{\varphi}\left(\hat{g}_{\varphi y}\right)-\hat{g}_{\varphi x}R_{\varphi
y}^{x}-\hat{g}_{\varphi y}R_{\varphi y}^{y}=0.$

After substituting the values of $R_{\varphi x}^{y}$, etc, and after
substituting the relations we had derived from the algebraic condition $(A)$,
we may write this as

$\frac{\partial\hat{g}_{\varphi
x}}{\partial\varphi}+\frac{1}{2}\hat{g}_{\varphi
x}\tan(\varphi)=0\quad\mbox{and}\quad\frac{\partial\hat{g}_{\varphi
y}}{\partial\varphi}-\frac{1}{2}\hat{g}_{\varphi y}\cot(\varphi)=0.$

The solutions of these equations are $\hat{g}_{\varphi
x}=C\sqrt{\cos(\varphi)}$ and $\hat{g}_{\phi y}=D\sqrt{\sin(\varphi)}$ (with
$C,D$ possibly functions of $(x,y,\theta)$), but the only ones satisfying
again the algebraic conditions are $\hat{g}_{\varphi x}=\hat{g}_{\varphi
y}=0$.

When we put $a=\theta$ and $b=\varphi$, we get:

$0=X_{\varphi}\left(\hat{g}_{\theta i}\right)-g_{\theta\theta}R_{\varphi
i}^{\theta}-\hat{g}_{\theta k}R_{\varphi i}^{k},$

which becomes here:

$\frac{\partial\hat{g}_{\theta
x}}{\partial\varphi}-\sin(\varphi)+\frac{1}{2}\hat{g}_{\theta
x}\sin(\varphi)\cos(\varphi)+\frac{1}{2}{\hat{g}}_{\theta
y}\sin^{2}(\varphi)=0$

and

$\frac{\partial\hat{g}_{\theta
y}}{\partial\varphi}+\cos(\varphi)-\frac{1}{2}\hat{g}_{\theta
x}\cos^{2}(\varphi)-\frac{1}{2}{\hat{g}}_{\theta
y}\cos(\varphi)\sin(\varphi)=0,$

from which (with $E,F$ functions of $(x,y,\theta)$)

$\hat{g}_{\theta
x}=E\sqrt{\cos(\varphi)}-2\cos(\varphi)\quad\mbox{and}\quad\hat{g}_{\theta
y}=F\sqrt{\sin(\varphi)}-2\sin(\varphi).$

Again, only for $E=0$ and $F=0$ the algebraic conditions are satisfied.
Finally, we get for $a=b=\theta$ that $X_{\theta}\left(\hat{g}_{\theta
i}\right)=0$ (since $R_{\theta i}^{\alpha}=0$). This is automatically
satisfied since the $\hat{g}_{\theta i}$ we had found only depend on
$\varphi$.

We conclude that the trajectories of the vertically rolling disk with initial
velocities in $\mathcal{D}$ are geodesics of all metrics $\hat{g}$ with metric
coefficients in the frame $\\{X_{a},X_{i}\\}$ given by

$\hat{g}_{ab}=g_{ab},\quad\quad\hat{g}_{\theta
x}=-2\cos(\varphi),\quad\quad\hat{g}_{\varphi x}=0,\quad\quad\hat{g}_{\theta
y}=-2\sin(\varphi),\quad\quad\hat{g}_{\varphi y}=0,$

and where the coefficients $\hat{g}_{ik}$ can be anything as long as $\hat{g}$
is positive-definite. We may look at the constructive proof of Proposition 4
for one possible (but not exclusive) way to realize this: we only need to find
a constant $\alpha$ such that $\hat{g}=\begin{pmatrix}A&B\\\ B^{T}&\alpha
I\end{pmatrix}$ is positive-definite. The matrices $A$ and $B$ are here given
by

$A=\begin{pmatrix}2&0\\\ 0&1\end{pmatrix}\quad\text{ and }\quad
B=\begin{pmatrix}-2\cos(\varphi)&-2\sin(\varphi)\\\ 0&0\end{pmatrix}.$

The constant $\alpha\neq 0$ is not allowed to be an eigenvalue of
$C=B^{T}A^{-1}B$, and should be larger than $-\tau_{\text{min}}$ and
$\frac{\mu_{\text{max}}}{\tau_{\text{min}}}$, where
$\tau_{\text{min}}=\text{min}\\{\text{eigenvalues of $A$}\\}$ and
$\mu_{\text{max}}=\text{max}\\{\text{eigenvalues of $B^{T}B$}\\}$. Here, the
eigenvalues of $A$ are $1$ and $2$, the eigenvalues of $B^{T}B$ are $0$ and
$4$ and the eigenvalues of $C$ are $0$ and $2$. We see that any $\alpha>4$
fulfills all requirements. We can take for example $\alpha=42$ and then

$\hat{L}=\frac{1}{2}\left[v_{\varphi}^{2}+2v_{\theta}^{2}-4\cos(\varphi)v_{\theta}v_{x}-4\sin(\varphi)v_{\theta}v_{y}+42v_{x}^{2}+42v_{y}^{2}\right].$

The relation between the quasi-velocities
$\\{v_{x},v_{y},v_{\theta},v_{\varphi}\\}$ and the natural fibre bundle
coordinates $\\{\dot{x},\dot{y},\dot{\varphi},\dot{\theta}\\}$ is

$\dot{x}=v_{x}+v_{\theta}\cos(\varphi),\quad\dot{y}=v_{y}+v_{\theta}\sin(\varphi),\quad\dot{\varphi}=v_{\varphi},\quad\dot{\theta}=v_{\theta}-v_{x}\cos(\varphi)-v_{y}\sin(\varphi),$

or conversely

$\begin{cases}v_{\varphi}=\dot{\varphi},\\\
v_{\theta}=\frac{1}{2}(\dot{\theta}+\cos(\varphi)\dot{x}+\sin(\varphi)\dot{y}),\\\
v_{x}=\frac{1}{2}(-\cos(\varphi)\dot{\theta}+(\sin^{2}(\varphi)+1)\dot{x}-\cos(\varphi)\sin(\varphi)\dot{y}),\\\
v_{y}=\frac{1}{2}(-\sin(\varphi)\dot{\theta}-\cos(\varphi)\sin(\varphi)\dot{x}+(\cos^{2}(\varphi)+1)\dot{y}).\end{cases}$

After substituting this, one obtains an expression for $\hat{L}$ in natural
coordinates.

|  On the left, we have plotted some geodesics for $\hat{g}$ with $\alpha=42$
(their projections for the center of mass $(x,y)$). The black circle through
the origin is the nonholonomic trajectory through the origin with initial
velocity
$({\dot{x}}_{0}=1,{\dot{y}}_{0}=0,{\dot{\theta}}_{0}=1,{\dot{\varphi}}_{0}=1)\in{\mathcal{D}}_{(0,0,0,0)}.$
The nearby grey geodesics are perturbations in either ${\dot{x}}_{0}$,
${\dot{y}}_{0}$ and ${\dot{\theta}}_{0}$, with $\epsilon=\pm 0.01$.
---|---

We can now compare our results with the Lagrangians that have been derived in
the literature, as a solution of the Hamiltonization problem. A second attempt
to obtain a Riemannian metric $\hat{g}$ from the same solution
$(g_{ab},{\hat{g}}_{ai})$ is to choose ${\hat{g}}_{ij}$ in such a way that it
is a multiple of the original $g_{ij}$, i.e. ${\hat{g}}_{ij}=\beta g_{ij}$. In
[9] (see also [21]) we can find the following regular quadratic Lagrangian
(among other non-quadratic Lagrangians):

${\hat{L}}_{\cite[cite]{[\@@bibref{}{BFM}{}{}]}}=\frac{1}{2}({\dot{\theta}}^{2}+{\dot{\varphi}}^{2})-\frac{1}{2}({\dot{x}}^{2}+{\dot{y}}^{2})+\dot{\theta}(\dot{x}\cos(\varphi)+\dot{y}\sin(\varphi)).$

When written in our current quasivelocities, we find for the corresponding
metric

$({\hat{g}}_{ab})_{\cite[cite]{[\@@bibref{}{BFM}{}{}]}}=({\hat{g}}_{ab}),\quad({\hat{g}}_{ai})_{\cite[cite]{[\@@bibref{}{BFM}{}{}]}}=({\hat{g}}_{ai}),\quad({\hat{g}}_{ij})_{\cite[cite]{[\@@bibref{}{BFM}{}{}]}}=-\begin{pmatrix}1+\cos^{2}(\varphi)&\cos(\varphi)\sin(\varphi)\\\
\cos(\varphi)\sin(\varphi)&1+\sin^{2}(\varphi)\end{pmatrix},$

meaning that it corresponds with $\beta=-1$. This metric has therefore the
same constrained Lagrangian $L_{c}$ as $L$ and as our Lagrangian $\hat{L}$.
However, the choice $\beta=-1$ leads only to a pseudo-Riemannian metric.

The following Lagrangian is mentioned in [3]:

${\hat{L}}_{\cite[cite]{[\@@bibref{}{Simoes}{}{}]}}=\frac{1}{2}(2{\dot{\theta}}^{2}+{\dot{\varphi}}^{2}+{\dot{x}}^{2}+{\dot{y}}^{2})-\dot{\theta}(\dot{x}\cos(\varphi)+\dot{y}\sin(\varphi)).$

It comes from a Riemannian metric, but it does not preserve the constrained
Lagrangian (i.e. it is not a ${\mathcal{D}}$-preserving modification of $g$),
since

$(g_{ab})_{\cite[cite]{[\@@bibref{}{Simoes}{}{}]}}=\begin{pmatrix}1&0\\\
0&1\end{pmatrix}.$

Finally, there are also some results that can be related to the geodesic
extension problem in [8]. However, the authors of [8] do not make use of a
${\mathcal{D}}$-preserving modification, and they mention that they are unable
to find a geodesic extension by their methods, specifically, for the
vertically rolling disk.

###  8.2 The vertically rolling disk as a Chaplygin system

It is well-known ([7, 13]) that the configuration space $Q=S^{1}\times
S^{1}\times\mathbb{R}^{2}$ of the vertically rolling disk is the total space
of a principal fibre bundle with structure group $G=\mathbb{R}^{2}$. Both the
Lagrangian and the constraints are invariant under this symmetry group, and
the constraints can be thought of as the horizontal distribution of a
principal connection. This means that the vertically rolling disk can be
regarded as a Chaplygin system.

For the vertical distribution of $\pi:Q\rightarrow Q/G$ we can take
$V\pi=\text{span}\\{({E}_{x})_{Q}=\frac{\partial}{\partial
x},({E}_{y})_{Q}=\frac{\partial}{\partial y}\\}$, and for the horizontal
distribution we may set
$\mathcal{D}=\text{span}\\{X_{\varphi}=\frac{\partial}{\partial\varphi},X_{\theta}=\cos(\varphi)\frac{\partial}{\partial
x}+\sin(\varphi)\frac{\partial}{\partial
y}+\frac{\partial}{\partial\theta}\\}$, as before. We may compute that

$G_{x\varphi}=0,\quad G_{x\theta}=\cos(\varphi),\quad G_{y\varphi}=0,\quad
G_{y\theta}=\sin(\varphi),$

and

$\left[X_{\varphi},X_{\theta}\right]=-\sin(\varphi)\frac{\partial}{\partial
x}+\cos(\varphi)\frac{\partial}{\partial y}.$

It is important to remark, however, that the result in Proposition 10 makes
use of a frame of ${\mathcal{D}}^{g}$ that is different from the frame
$\\{X_{x},X_{y}\\}$ we had encountered earlier. The vector fields
${\tilde{X}}_{i}=({E}_{i})_{Q}-G^{bc}G_{ci}X_{b}=({E}_{i})_{Q}+K^{b}_{i}X_{b}$
are here

${\tilde{X}}_{x}=\frac{\partial}{\partial
x}-\frac{1}{2}\cos(\varphi)\left(\frac{\partial}{\partial\theta}+\cos(\varphi)\frac{\partial}{\partial
x}+\sin(\varphi)\frac{\partial}{\partial y}\right)$

and

${\tilde{X}}_{y}=\frac{\partial}{\partial
y}-\frac{1}{2}\sin(\varphi)\left(\frac{\partial}{\partial\theta}+\cos(\varphi)\frac{\partial}{\partial
x}+\sin(\varphi)\frac{\partial}{\partial y}\right),$

and the relation between $\\{{\tilde{X}}_{i}\\}$ and the frame $\\{{X}_{i}\\}$
of the previous subsection is

$\begin{cases}{\tilde{X}}_{x}=(1-\frac{1}{2}\cos^{2}(\varphi))X_{x}-\frac{1}{2}\cos(\varphi)\sin(\varphi)X_{y},\\\
{\tilde{X}}_{y}=-\frac{1}{2}\sin(\varphi)\cos(\varphi)X_{x}+(1-\frac{1}{2}\sin^{2}(\varphi))X_{y}.\end{cases}$

According to Proposition 10 we should look for linear first integrals
$\mu_{x}$ and $\mu_{y}$ of $\Gamma_{(L,\mathcal{D})}$ that satisfy

$\left(\mu_{x}-G_{x\theta}v_{\theta}-G_{x\varphi}v_{\varphi}\right)B_{\varphi\theta}^{x}v_{\varphi}+\left(\mu_{y}-G_{y\theta}v_{\theta}-G_{y\varphi}v_{\phi}\right)B_{\varphi\theta}^{y}v_{\varphi}=0,$

which is here

$-\sin(\varphi)\mu_{x}+\cos(\varphi)\mu_{y}=0.$

If we take a ${\Gamma_{(L,\mathcal{D})}}$-derivative we also get
$\cos(\varphi)\mu_{x}+\sin(\varphi)\mu_{y}=0$ (because
${\Gamma_{(L,\mathcal{D})}}(\sin(\varphi))=\cos(\varphi)\dot{\varphi}$). We
can, therefore conclude that the first integrals should be
$\mu_{x}=\mu_{y}=0$. With that, we may get the coefficients of the metric
$\hat{g}$ with respect to the frame $\\{X_{a},{\tilde{X}}_{i}\\}$ (see the
proof of Proposition 10),

${\hat{g}}(X_{a},X_{b})=g_{ab},\qquad\widetilde{{\hat{g}}_{ai}}={\hat{g}}(X_{a},{\tilde{X}}_{i})=\mu_{ai}-G_{ai},$

which are here

$\begin{cases}\widetilde{\hat{g}_{x\varphi}}=0\qquad\mbox{and}\qquad\widetilde{\hat{g}_{x\theta}}=-\cos(\varphi)\\\
\widetilde{\hat{g}_{y\varphi}}=0\qquad\mbox{and}\qquad\widetilde{\hat{g}_{y\theta}}=-\sin(\varphi)\end{cases}.$

We see that this method is way shorter and more efficient since we had already
solved the involved differential equations.

We end this section by showing that the new coefficients we had found are
compatible with the old ones. Indeed, given the relation between the two bases
$\\{X_{i}\\}$ and $\\{{\tilde{X}}_{i}\\}$ one easily verifies that, for
example,

$\widetilde{\hat{g}_{\theta
y}}=-\frac{1}{2}\sin(\varphi)\cos(\varphi)\hat{g}_{\theta
x}+\left(1-\frac{1}{2}\sin^{2}(\varphi)\right)\hat{g}_{\theta y},$

as it should.

#### Acknowledgements.

The authors thank the Research Fund of the University of Antwerp (BOF) for its
support through the DOCPRO project 49747.

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