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\title{Geometric characterization of linearisable second--order
differential equations}
\author{Eduardo Mart\'{\i}nez$^\dag$\ and Jos\'e F. Cari\~nena$^\ddag$\\
$^\dag$
Dpto. de Matem\'atica Aplicada,
Universidad de Zaragoza\\
E-50015 Zaragoza, Spain\\
$^\ddag$
Departamento de F\'{\i}sica Te\'orica,
Universidad de Zaragoza\\
E-50009 Zaragoza, Spain}
\date{}
\begin{document}
\maketitle
\begin{abstract}
Given an Ehresmann connection on the tangent bundle $\map{\tau}{TM}{M}$
we define a linear connection on the pull--back bundle $\tau^*(TM)$.
With the aid of this tool, necessary and sufficient conditions for the
existence of local coordinates in which a system of second--order
differential equations is linear are derived.
\end{abstract}
\section{Introduction}
In previous papers~\cite{MaCaSa1,MaCaSa2} we developed a theory
of derivations of scalar and vector-valued forms along the tangent
bundle projection $\map{\tau}{TM}{M}$ of a manifold $M$. This theory
was used in order to characterize in a geometric way when
a system of second--order
differential equations is separable, \ie there exists a coordinate
system on the base manifold in which the system decouples as a sum of
one--dimensional independent differential equations (see~\cite{MaCaSa3}).
Two fundamental operators associated with every vector field $X$ along
$\tau$ did appear in these papers. They were denoted $\Dv{X}$ and
$\Dh{X}$ and called the horizontal and vertical covariant derivative
operators. Their usefulness is partially based on the fact that they
are derivations (of degree zero) and that the maps $X\mapsto\Dh{X}$ and
$X\mapsto\Dv{X}$ are $\cinfty{TM}$--linear, so that there is a clear
resemblance with a covariant derivative operator.
In this paper we show that this similarity is not a coincidence but a
consequence of the existence of a covariant derivative on the pull--back
bundle by $\tau$ of the tangent bundle. With the aid of this tool, we
will geometrically characterize those second--order differential
equations which are linearisable under a point transformation of
coordinates.
In order to the paper be selfcontained, we give in Section~1 a summary
of the results of the theory of derivations which are relevant for
subsequent sections. In Section~2 we will introduce the linear
connection associated to a non--linear connection on the tangent bundle
and we will show the relation between the curvature of the linear
connection and the tensors $\Rie$ and $\theta$, that were previously
defined in terms of the horizontal and vertical covariant derivatives.
Some relations between $\Rie$ and $\theta$ which are just the expression
of the Bianchi identities for the linear connection are also given.
Finally, in Section~3 we study the problem of the linearisation
of a second--order differential equations. Our results will be related to
previous ones~\cite{Chern,Thompson}
\section{Preliminaries}
Let $\map{\pi}{E}{M}$ be a fibre bundle and $\map{\phi}{N}{M}$ a
smooth map. A section of~$E$ along~$\phi$ is a map $\map{\sigma}{N}{E}$
such that $\pi\circ\sigma =\phi$. Equivalently, a section along $\phi$
can be considered as a section $\bar{\sigma }$ of the pull--back bundle
$\map{\phi^*\pi}{\phi^*E}{N}$ (see e.\ g.~\cite{Poor}). The relation
between $\sigma$ and $\bar{\sigma}$ is
$\sigma=\phi[\pi]\circ\bar{\sigma}$, and we have the following
commutative diagram
$$
\begin{picture}(100,75)(0,0)
\put(4,0){$N$}
\put(100,0){$M$}
\put(0,60){$\phi^*E$}
\put(100,60){$E$}
\put(55,-5){$\ssize\phi$}
\put(50,70){$\ssize\phi[\pi]$}
\put(-10,35){$\ssize\phi^*\pi$}
\put(15,35){$\ssize\bar{\sigma}$}
\put(53,38){$\ssize\sigma$}
\put(108,35){$\ssize\pi$}
\put(23,4){\vector(1,0){72}}
\put(23,64){\vector(1,0){72}}
\put(11,13){\vector(0,1){41}}
\put(6,54){\vector(0,-1){41}}
\put(105,54){\vector(0,-1){41}}
\put(26,11){\vector(4,3){65}}
\end{picture}
$$
If~$E$ is a vector bundle then the set of sections along~$\phi$ is a
$\cinfty{N}$--module. The most interesting cases are $E=TM$,
$(T^*M)^{\wedge p}$ or any other tensor bundle, and then a section
of~$E$ along~$\phi$ is called a vector field along~$\phi$, a $p$-form
along~$\phi$, or a tensor field along~$\phi$ (respectively). The set of
vector fields along~$\phi$ is denoted by $\vectorfields{\phi}$, and the
set of $p$-forms along~$\phi$ by $\forms{p}{\phi}$. In particular, we
are interested in the case in which~$\phi$ is the projection $\tau$ of
the tangent bundle $\map{\tau}{TM}{M}$. The easiest example of tensor
field along~$\tau$ is the composition $W\circ\tau$ for~$W$ a tensor
field on the base manifold~$M$. We will then say that~$W$ is a basic
tensor field and we will simplify the notation writing just $W$ instead
of $W\circ\tau$. Another example is the canonical vector field
along~$\tau$ denoted by~$\ti$ and defined by the identity map on $TM$
thought of as a section of $TM$ along~$\tau$.
In two recent papers~\cite{MaCaSa1,MaCaSa2} we have studied the algebra
of derivations of scalar and vector valued forms along~$\tau$. A
complete classification of the set of derivations of forms along~$\tau$
was given with the aid of an Ehresmann connection in the tangent bundle,
\ie a horizontal subbundle of $T(TM)$ (see~\cite{Poor,Crampin,Grifone}).
In presence of a connection, any vector field $Z\in\vectorfields{TM}$
can be decomposed as a sum $Z=\hlift{X_1}+\vlift{X_2}$ for $X_1$,
$X_2\in \vectorfields{\tau}$, where the indices $H$ and~$V$ mean
horizontal and vertical lift, respectively. In coordinates, if~$Z$ has
the expression $Z=X^i\, \partial/\partial x^i+Y^i\, \partial /\partial v^i$,
then $X_1$ and~$X_2$ are given by $X_1=X^i\, \partial/ \partial x^i$ and
$X_2=(Y^i+\Gamma^i_jX^j)\,\partial /\partial x^i$, respectively. Here,
the functions
$\Gamma^i_j$ are the coefficients of the connection,
which are defined by the
horizontal lift $H_i=\partial/\partial x^i - \Gamma^j_i\, \partial
/\partial v^j$ of the coordinate vector field $\partial /\partial
x^i$. If $W$ is a vertical vector field on $TM$, the vector field $X$
along $\tau$ such that $W=\vlift{X}$ will be denoted $\vdown{W}$. Note
that if $W$ is a horizontal vector field on $TM$ then the vector field
$X$ along $\tau$ such that $\hlift{X}=W$ is just the projection of $W$,
\ie $X=T\tau\circ W$.
With the above decomposition in mind, we define two fundamental
derivations $\Dh{X}$ and~$\Dv{X}$ for every $X\in\vectorfields{\tau}$,
called respectively the horizontal and vertical covariant derivatives,
by mean of the equation
\begin{displaymath}
[\hlift{X},\vlift{Y}]=\vlift{\{\Dh{X}Y\}} - \hlift{\{\Dv{Y}X\}}.
\end{displaymath}
They are called covariant because they are $\cinfty{TM}$-linear in the
subscript argument. We extend the action of these derivations to 1-forms
along $\tau$ by imposing the duality rule
\begin{displaymath}
D\langle X,\alpha\rangle =
\langle DX,\alpha\rangle +\langle X,D\alpha\rangle ,
\end{displaymath}
and to general tensor fields along $\tau$ by imposing the Leibnitz rule
for the tensor product. Derivations satisfying these properties are
called self-dual. If $X$, $Y$ are vector fields along~$\tau$ with local
expressions $X=X^i\, \partial/\partial x^i$ and $Y=Y^i\, \partial/\partial
x^i$, then the coordinate expressions of $\Dh{X}Y$ and $\Dv{X}Y$ are
\begin{eqnarray*}
\Dh{X}Y&=&X^i(H_iY^j+Y^k\Gamma ^j_{ik})\, \xpart{j}\\
\Dv{X}Y&=&X^i(V_iY^j)\,\xpart{j},
\end{eqnarray*}
where $V_i=\partial /\partial v^i$ and $\Gamma ^i_{jk}=V_k(\Gamma
^i_j)$. It can be easily shown that a tensor field~$W$ along $\tau$ is
basic if and only if $\Dv{}W=0$.
The commutators of two horizontal lifts and two vertical lifts are respectively
given by
\begin{eqnarray*}
[\vlift{X},\vlift{Y}]&=&\vlift{[X,Y]_{\sssize V}} \\{}
[\hlift{X},\hlift{Y}]&=&\hlift{[X,Y]_{\sssize H}}+\vlift{\{R(X,Y)\}}
\end{eqnarray*}
where $[X,Y]_{\sssize V}=\Dv{X}Y-\Dv{Y}X$,
$[X,Y]_{\sssize H}=\Dh{X}Y-\Dh{Y}X - T(X,Y)$ and where $R$ and $T$ are,
respectively, the curvature and the torsion of the non--linear
connection. It is easy to see (e.\ g.\ in coordinates) that if $X$ and
$Y$ are vector fields on the base manifold $M$, then
$[X,Y]=[X,Y]_{\sssize H}$.
The commutators of the covariant derivatives define two important
2-cov\-ar\-iant tensor fields along $\tau$, $\theta $ and $\Rie$,
taking values in the set of (1,1) tensor fields, as follows:
\begin{eqnarray*}
[\Dv{X},\Dv{Y}]Z&=&\Dv{[X,Y]_{\sssize V}}Z\\{}
[\Dv{X},\Dh{Y}]Z&=&\Dh{\Dv{X}Y} - \Dv{\Dh{Y}X}Z+\theta (X,Y)Z\\{}
[\Dh{X},\Dh{Y}]Z&=&\Dh{[X,Y]_{\sssize H}}Z+\Dv{R(X,Y)}Z+\Rie(X,Y)Z.
\end{eqnarray*}
Their coordinate expressions are
\begin{eqnarray*}
\theta &=&\Gamma^k_{jml}\,\xdiff{l}\tensor\xdiff{j}\tensor
\left(\xdiff{m}\tensor\xpart{k}\right)\\
\Rie&=&\frac{1}{2}
\bigl[H_k(\Gamma^i_{lj})-H_l(\Gamma^i_{kj})
+\Gamma^i_{kr}\Gamma^r_{lj}-\Gamma^i_{lr}\Gamma^r_{kj}\bigr]
\xdiff{k}\wedge\xdiff{l}\tensor\left(\xdiff{j}\tensor\xpart{i}\right),
\end{eqnarray*}
where $\Gamma^i_{jkl}=V_l\,\Gamma^i_{jk}$. It follows from
the coordinate expression
of $\Rie$ that
\begin{displaymath}
\Rie= - \Dv{}R ,
\end{displaymath}
\ie if $X$, $Y$, $Z$ are vector fields
along $\tau$, then $\Rie(X,Y)Z=-[\Dv{Z}R](X,Y)$.
The vertical covariant derivative of the canonical vector field along $\tau$
is the identity tensor $\Dv{}\ti=I$, while the horizontal covariant
derivative defines an important (1,1) tensor field associated to the
connection which is called the tension $\ten=-\Dh{}\ti$. Its coordinate
expression is
\begin{displaymath}
\ten=(\Gamma^i_j-\Gamma^i_{jk}v^k)\xdiff{j}\tensor\xpart{i}.
\end{displaymath}
Notice that $\ten=0$ iff the connection is linear, \ie the functions
$\Gamma^i_{jk}$ do not depend on $v^l$. In this case the tensor $\Rie$
coincides with the Riemann curvature tensor of the linear connection.
\section{Induced linear connection}
We remind that a covariant derivative on a vector bundle
$\map{\pi}{E}{N}$ is a $\cinfty{N}$--linear operator mapping every vector
field $V$ of $N$ onto a derivation $\DC{V}$ of the $\cinfty{N}$--module
$\Sec{\pi }$ of sections of the bundle $E$, such that it coincides
precisely with $V$ on $\cinfty{N}$. That is, $\DC{}$ satisfies the
following properties:
\begin{enumerate}
\item $\DC{V+W}\sigma =\DC{V}\sigma +\DC{W}\sigma $
\item $\DC{fV}\sigma =f\DC{V}\sigma $
\item $\DC{V}(\sigma +\rho)=\DC{V}\sigma +\DC{V}\rho$
\item $\DC{V}(f\sigma )=(Vf)\sigma +f\DC{V}\sigma $,
\end{enumerate}
for all $V$, $W\in\vectorfields{N}$, $f\in\cinfty{M}$ and $\sigma $,
$\rho\in\Sec{\pi}$. A covariant derivative on a bundle is equivalent to
a linear connection on such bundle (see~\cite{Poor,KoNo}).
As we mentioned in the last section, the operators $\Dh{}$ and $\Dv{}$
are similar to actual covariant derivatives. Taking into account that
sections of the bundle $\map{\tau^*\tau}{\tau^*(TM)}{TM}$ are
(canonically identified with) vector fields along $\tau$, it is clear
that the horizontal and vertical covariant derivatives must be related
with a true covariant derivative on such bundle. Indeed, for every
vector field $W$ on $TM$ we define the operator $\TDC{W}$ by
\begin{displaymath}
\qquad\qquad
\TDC{W}Z=\vdown{\{P_V[P_H(W),\vlift{Z}]\}}+\T{\tau}{}\circ[P_V(W),\hlift{Z}],
\qquad\qquad(*)
\end{displaymath}
where $Z$ is a vector field along $\tau$ and $P_H,P_V$ are the
horizontal and vertical projectors defined by the connection. The
$\real$--linearity of $[\cdot\,,\cdot]$ implies that $\TDC{}$ satisfies
properties (1) and~(3). We now prove that $\TDC{}$ satisfies~(2): if
$F$ is a function on $TM$, then using
$[FP_H(W),\vlift{Z}]=-\vlift{Z}F\,P_H(W)+F[P_H(W),\vlift{Z}]$ and
$[FP_V(W),\hlift{Z}]=-\hlift{Z}F\,P_V(W)+F[P_V(W),\hlift{Z}]$ we have
\begin{eqnarray*}
\TDC{FW}Z
&=&-\vlift{Z}F\,P_VP_H(W)+F\vdown{\{P_V[P_HW,\vlift{Z}]\}}
+F\T{\tau}{}\circ[P_V(W),\hlift{Z}]\\
&=&F\,\TDC{W}Z,
\end{eqnarray*}
because $P_VP_H=0$. Furthermore,
$[P_H(W),F\vlift{Z}]=P_H(W)F\,\vlift{Z}+F[P_H(W),\vlift{Z}]$ and
$[P_V(W),F\hlift{Z}]=P_V(W)F\,\hlift{Z}+F[P_V(W),\hlift{Z}]$, so that
\begin{eqnarray*}
\TDC{W}(FZ)
&=&\vdown{\{P_H(W)F\,\vlift{Z}+FP_V[P_HW,\vlift{Z}]\}}\\
&&\quad{}+\T{\tau}{}\circ\bigl((P_V(W)F)\hlift{Z}+F[P_V(W),\hlift{Z}]\bigr)\\
&=&(P_H(W)F+P_V(W)F)Z
+F\vdown{(P_V[P_H(W),\vlift{Z}])}\\
&&\quad{}+F\T{\tau}{}+([P_V(W),\hlift{Z}]),\\
\end{eqnarray*}
and taking into account that $P_H+P_V=I$, we find that $\TDC{}$
satisfies (4). Thus, $\TDC{}$ is a covariant derivative in the pull--back
bundle by~$\tau$.
As we said above, every vector field $W$ on $TM$ can be decomposed as a
sum of a vertical plus a horizontal lift of a vector field along $\tau$,
$W=\hlift{X}+\vlift{Y}$. The following relation holds
\begin{displaymath}
\TDC{\hlift{X}+\vlift{Y}}=\Dh{X}+\Dv{Y}.
\end{displaymath}
Indeed, taking into account the equation defining $\Dh{}$ and $\Dv{}$,
we obtain
\begin{eqnarray*}
\TDC{\hlift{X}+\vlift{Y}}Z
&=&\vdown{(P_V[\hlift{X},\vlift{Z}])}+\T{\tau}{}+([\vlift{Y},\hlift{Z}])\\
&=&\vdown{(\vlift{(\Dh{X}Z)})}+\T{\tau}{}+(\hlift{(\Dv{Y}Z)})\\
&=&\Dh{X}Z+\Dv{Y}Z.
\end{eqnarray*}
We have proved the following result:
\begin{theorem}
Given a connection on the tangent bundle $TM$ to a manifold $M$, there
exists one linear connection $\TDC{}$ on the pull--back bundle
$\map{\tau^*\tau}{\tau^*(TM)}{TM}$ such that
\begin{displaymath}
\TDC{\hlift{X}+\vlift{Y}}=\Dh{X}+\Dv{Y}.
\end{displaymath}
for all $X$, $Y\in\vectorfields{\tau}$. Such connection is explicitly
given by equation~{\rm(*)} and will be said to be the linear
connection induced by the non--linear connection on $TM$.
\end{theorem}
We remind that if $\map{\phi}{N}{M}$ is a smooth map and $H(E)$ is an
Ehresmann connection on the bundle $\map{\pi}{E}{M}$, then
$(\T{\phi[\pi]}{})^{-1}(H(E))$ is an Ehresmann connection on the bundle
$\map{\phi^*\pi}{\phi^*E}{N}$. This connection is known as the induced
conection by $\phi$. We remark that in the case $\phi=\tau$ the induced
connection is different from the induced linear connection here
defined. In particular, if the original connection is non linear, then
the induced connection is non linear too.
If we choose local coordinates $(x^i,v^i,w^i)$ on $\tau ^*(TM)$ such that
\begin{displaymath}
\tau^*\tau (x^i,v^i,w^i)=(x^i,v^i)\quad\text{and}\quad
\tau [\tau ](x^i,v^i,w^i)=(x^i,w^i),
\end{displaymath}
then the expressions of the horizontal lifts of the coordinate vector
fields with respect to the induced linear connection are
\begin{displaymath}
\left(\xpart{i}\right)^{\sssize\widetilde{H}}
=\xpart{i}-w^k\,\Gamma^j_{ik}\pd{}{w^j}
\qquad\quad
\left(\vpart{i}\right)^{\sssize\widetilde{H}}=\vpart{i}.
\end{displaymath}
Consider now the Riemann curvature tensor
\begin{displaymath}
\widetilde{\Rie}(V,W)Z=[\TDC{V},\TDC{W}]Z-\TDC{[V,W]}Z.
\end{displaymath}
of the connection $\TDC{}$. We will show that the tensors $\Rie$ and $\theta$
are but components of $\widetilde{\Rie}$.
\begin{theorem}
The curvature of the connection $\TDC{}$ is given by
\begin{eqnarray*}
\widetilde{\Rie}(\hlift{X},\hlift{Y})Z&=&\Rie(X,Y)Z\\
\widetilde{\Rie}(\vlift{X},\hlift{Y})Z&=&\theta (X,Y)Z\\
\widetilde{\Rie}(\vlift{X},\vlift{Y})Z&=&0
\end{eqnarray*}
for $X,Y,Z\in\vectorfields{\tau}$.
\end{theorem}
\begin{proof}
For two horizontal vector fields $\hlift{X}$ and $\hlift{Y}$, we have
\begin{eqnarray*}
\widetilde{\Rie}(\hlift{X},\hlift{Y})Z
&=&[\Dh{X},\Dh{Y}]Z - \TDC{\hlift{\{\h[X,Y]\}}+\vlift{R(X,Y)}}Z\\
&=&[\Dh{X},\Dh{Y}]Z - \Dh{\h[X,Y]}Z - \Dv{R(X,Y)}Z\\
&=&\Rie(X,Y)Z.
\end{eqnarray*}
Similarly, for $\vlift{X}$ and $\hlift{Y}$, we obtain
\begin{eqnarray*}
\widetilde{\Rie}(\vlift{X},\hlift{Y})Z
&=&[\Dv{X},\Dh{Y}]Z - \TDC{\hlift{\{\Dv{X}Y\}} - \vlift{\{\Dh{Y}X\}}}Z\\
&=&[\Dv{X},\Dh{Y}]Z - \Dh{\Dv{X}Y}Z+\Dv{\Dh{Y}X}Z\\
&=&\theta (X,Y)Z.
\end{eqnarray*}
Finally, for two vertical vector fields we find
\begin{eqnarray*}
\widetilde{\Rie}(\vlift{X},\vlift{Y})Z
&=&[\Dv{X},\Dv{Y}]Z - \TDC{\vlift{\{\v[X,Y]\}}}Z\\
&=&[\Dv{X},\Dv{Y}]Z - \Dv{\v[X,Y]}Z\\
&=&0.
\end{eqnarray*}
\end{proof}
Having a linear connection, we have at our disposal a menagerie of results.
In particular, the Bianchi identities
\begin{displaymath}
\sum_{{\fam0 cyclic}} \TDC{U}\widetilde{\Rie}(V,W)=0,
\end{displaymath}
implies some relations that $\Rie$ and $\theta$ must satisfy. They are:
\begin{displaymath}
\begin{array}{c}
\theta(X,Y)Z-\theta(Z,Y)X =0,\\[3pt]
\Dh{Z}\theta(X,Y)-\Dh{Y}\theta(X,Z)-\theta(X,T(Y,Z))+\Dv{X}\Rie(Y,Z)=0,\\[3pt]
\sum_{{\fam0 cyclic}}\left\{\Dh{X}\Rie(Y,Z)+\Rie(X,T(Y,Z))-
\theta(R(X,Y),Z)\right\}=0.
\end{array}
\end{displaymath}
The proof is a matter of straightforward calculation and will be
omited. This Bianchi identities were first found in~\cite{tesis} by
calculating the Jacobi identities for $\Dh{}$ and $\Dv{}$. We also
mention that in~\cite{MaCaSa1} we found another set of relations
between the torsion and the curvature of the non--linear connection,
called the Bianchi identities for the non--linear connection.
\section{Characterization of linearisable \sode}
In this section we will consider the following problem: Given a
system of second--order differential equations
\begin{displaymath}
\ddot{x}{}^i=f^i(x^1,\ldots,x^n,\dot{x}{}^1,\ldots,\dot{x}{}^n),
\qquad i=1,\ldots,n,
\end{displaymath}
we ask under what circumstances there exists a coordinate transformation
\begin{displaymath}
\bar{x}{}^i=\xi^i(x^1,\ldots,x^n),
\qquad i=1,\ldots,n,
\end{displaymath}
such that in the coordinates $\bar{x}{}^i$ the system of second--order
equations turns out to be linear:
\begin{displaymath}
\ddot{\bar{x}}^i=A^i_j\dot{\bar{x}}^j+B^i_j\bar{x}^j+C^i,\qquad i=1,\ldots,n,
\end{displaymath}
for some contants $A^i_j$, $B^i_j$ and $C^i$.
In geometric terms, a system of second--order equations (\sode) on a
manifold~$M$ is interpreted as a vector field $\Gamma$ in the tangent
bundle to $M$, such that $\T{\tau}{v}(\Gamma (v))=v$ for every $v\in
TM$. Every \sode\ defines a connection on the tangent bundle (see~\cite
{Grifone,Crampin}). In coordinates, if $f^i$ are the forces defined by a
\sode~$\Gamma$, $f^i=\Gamma v^i$, then the coefficients of the
connection are $\Gamma^i_j=-\frac{1}{2}\partial f^i/\partial v^j$. It
follows that the functions $\Gamma^i_{jk}=\partial\Gamma^i_j/\partial
v^k$ are symmetric in the subscripts $j,k$, that is, the torsion of the
connection vanishes. In fact, the vanishing of the torsion is a
necessary and sufficient condition for the connection to be defined by a
\sode.
More directly related to the \sode, there are a derivation $\nabla$
called the dynamical covariant derivative and a (1,1) tensor field
$\Phi$ called the Jacobi endomorphism. They can simultaneously be
defined by the equation
\begin{displaymath}
[\Gamma ,\hlift{X}]=\hlift{\{\nabla X\}}+\vlift{\{\Phi (X)\}},
\end{displaymath}
and contain the whole information about the \sode. In fact, on
functions $\nabla $ coincides with $\lie{\Gamma }$. The action of
$\nabla$ over coordinate vector fields is determined by the coefficients
of the connection in such coordinates
\begin{displaymath}
\nabla \left(\xpart{j}\right)=\Gamma ^i_j\,\xpart{i},
\end{displaymath}
so that the dynamical covariant derivative of $X=X^i\,\partial/\partial
x^i$ has the expression
\begin{displaymath}
\nabla X=(\Gamma X^i+\Gamma ^i_jX^j)\xpart{i}.
\end{displaymath}
In terms of the induced linear connection we have $\nabla
X=\TDC{\Gamma}X+\ten(X)$. The coordinate expression of the Jacobi
endomorphism is
\begin{displaymath}
\Phi =\left(-\pd{f^i}{x^j}-\Gamma ^i_k\Gamma ^k_j-\Gamma(\Gamma^i_j)\right)
\,\xdiff{j}\tensor\xpart{i},
\end{displaymath}
and it satisfies the following relation
\begin{displaymath}
\Phi (X)=R(\ti,X)-\Dh{X}\nabla \ti.
\end{displaymath}
Moreover, $\Phi$ is related with the curvature by mean of
\begin{displaymath}
(\Dv{X}\Phi)(Y)-(\Dv{Y}\Phi)(X) =3R(X,Y).
\end{displaymath}
Furthermore, we have the following commutators:
\begin{eqnarray*}
[\nabla ,\Dv{X}]Y&=&\Dv{\nabla X}Y - \Dh{X}Y,\\{}
[\nabla ,\Dh{X}]Y&=&\Dh{\nabla X}Y+\Dv{\Phi (X)}Y-[\Dv{Y}\Phi ](X)-R(X,Y),
\end{eqnarray*}
for $X$ and $Y$ vector fields along $\tau$.
\medskip
We now turn to the question of the linearisability of second--order
ordinary differential equations. In geometric terms, a \sode~$\Gamma$
is said to be linearisable in velocities ($v$--linearisable) if for
each point $m\in M$ there exists a set of local coordinates $(x^i)$ in a
neighborhood ${\cal U}$ of~$m$ such that the coordinate expression of
the forces defined by~$\Gamma$ is of the form
\begin{displaymath}
f^i(x,v)=A^i_j(x)v^j+b^i(x)
\end{displaymath}
for some functions $A^i_j$, $b^i\in\cinfty{{\cal U}}$.
\begin{theorem}
A \sode~$\Gamma $ is $v$--linearisable if and only if the
connection~$\TDC{}$ is flat.
\end{theorem}
\begin{proof}
Assume that $\Gamma $ is $v$--linearisable. In the coordinates in which
$\Gamma $ is linear in the velocities the coefficients of the connection
defined by $\Gamma $ are $\Gamma^i_j= - \frac{1}{2}A^i_j(x)$. Thus
$\Gamma ^i_{jk}=0$, from where it follows $\widetilde{\Rie}=0$.
Conversely, if $\TDC{}$ is flat, then in a neighbourhood of each point
there exists a basis $\{X_i\}$ of $\vectorfields{\tau}= \Sec{\tau
^*\tau}$ such that $\TDC{W}X_i=0$ for every vector field $W$ on $TM$.
Since $\TDC{\vlift{X}}=\Dv{X}$ we have that $X_i$ are basic vector
fields. Moreover, taking into account that $\TDC{\hlift{X}}=\Dh{X}$
and that $[X,Y]=\Dh{X}Y - \Dh{Y}X$ we find that $[X_i,X_j]=0$. Thus,
there exist local coordinates $(x_i)$ for $M$ such that
$X_i=\partial/\partial x^i$. In these coordinates the functions $\Gamma
^i_{jk}$ vanish, from where it follows that $\Gamma ^i_j$ are basic
functions and that the expression of the forces is an affine function of
the velocities.
\end{proof}
It is worthy of note that the forces depend only on the positions iff,
being linearisable, the tension of the connection vanishes. Also note
that the existence of a $v$--linearisable \sode\ imposes severe
restrictions to the topology of the manifold $M$. Indeed, since
$\theta=0$, we have that if $X$ and $Y$ are vector fields on $M$,
then $\Dh{X}Y$ is also a basic vector field, so that it defines a
linear connection $\DC{}$ on $M$. Since the curvature tensor of this
linear connection is just the $\Rie$ tensor, we have that $\DC{}$ is
flat. In particular, $M$ is locally isomorphic to an ordinary affine
space (endowed with the natural connection on it).
A \sode~$\Gamma $ is said to be linearisable if for each point $m\in M$
there exist local coordinates $(x^i)$ in a neighbourhood ${\cal U}$
of~$m$ such that the coordinate expressions of the forces defined
by~$\Gamma $ are of the form
\begin{displaymath}
f^i(x,v)=A^i_jv^j+B^i_jx^j+C^i,
\end{displaymath}
for some constants $A^i_j$, $B^i_j$ and $C^i$.
\begin{theorem}
A \sode~$\Gamma $ is linearisable if and only if the connection $\TDC{}$
is flat and the Jacobi endomorphism is parallel.
\end{theorem}
\begin{proof}
If $\Gamma $ is linearisable, then it is $v$--linearisable, and
therefore $\TDC{}$ is flat. In coordinates in which~$\Gamma $ is linear
we have that the components of the Jacobi endomorphism are constant
$\Phi^i_j= - B^i_j - \frac{1}{4} A^i_k A^k_j$, from where
$\TDC{}\Phi=0$ readily follows.
Conversely, assume that $\TDC{}$ is flat and $\TDC{}\Phi=0$. Since
$\theta =0$, we have $\Dh{}\circ\Dv{}=\Dv{}\circ\Dh{}$ and thus
$\Dv{}\ten= - \Dv{}\Dh{}\ti=- \Dh{}\Dv{}\ti=- \Dh{}I=0$. Moreover, since
$\Phi$ is basic the curvature vanishes, from where we have
\begin{displaymath}
0=\Dv{}\Phi= - \Dv{}\Dh{}\nabla\ti= - \Dh{}\Dv{}\nabla\ti= - \Dh{}\Dh{}\ti=
\Dh{}\ten.
\end{displaymath}
Thus the tension is a parallel tensor field. In coordinates in which the
coefficients $\Gamma ^i_{jk}$ vanish we have that $\ten^i_j=\Gamma^i_j$
and $\Phi^i_j$ are constant. It follows that $\partial f^i/\partial
v^j$ and $\partial f^i/\partial x^j$ are constant too, and thus $\Gamma
$ is linear in these coordinates.
\end{proof}
As a consequence of the relations between the Jacobi endomorphism, the
curvature and the $\Rie$ tensor we have the following result:
\begin{theorem}
A \sode~$\Gamma $ is linearisable if and only if $\theta =0$, $\Dh{}\Phi=0$
and $\Dv{}\Phi=0$.
\end{theorem}
\begin{proof}
>From $\Dv{}\Phi=0$ we have $R=0$, and thus $\Rie= - \Dv{}R=0$.
\end{proof}
\null\bigskip
To finish the paper we consider some previous results in the literature.
In~\cite{Thompson} the following characterization was given: A \sode\
$\Gamma$ is linearisable iff:
\begin{enumerate}
\item $\Gamma=\Lambda+\vlift{A_0}+\vlift{A_1(\ti)}$, where $\Lambda$ is
a spray, $A_0$ is a vector field on $M$ and $A_1$ is a (1,1) tensor
field on $M$.
\item The curvature of the connection $\DC{}$ defined by $\Lambda$ is
flat.
\item $\DC{}A_1$ and $\DC{}^2 A_0$ vanishes.
\end{enumerate}
In order to relate it with our results, we note that the vanishing of
$\theta$ is equivalent to (1) and that the connection defined by the
spray $\Lambda$ is given by $\DC{X}Y=\Dh{X}Y$ for
$X,Y\in\vectorfields{M}$. Thus $\DC{}$ is flat iff $\Rie$ vanishes.
Finally, since $A_1=-2\ten$ we have $\DC{}A_1=0$ iff the tension is
parallel. In such a case, $\DC{}A_0=-\Phi-\ten^2$, from where we deduce
that $\DC{}^2A_0=0$ if and only if the Jacobi endomorphism is parallel.
\smallskip
Kamran~\cite{Kamran} has also studied the problem of
linearisation by using Cartan's method of equivalence. Kamran dealt with
time--dependent coordinate transformations and therefore their results
are not applicable to our case. On the other hand, Chern~\cite{Chern}
has solved the equivalence problem under spatial diffeomorphism. The
invariants founded there correspond to $\ti$, $\nabla\ti$, $\Phi$, $R$ and
$\theta$, and the covariant derivatives of $\nabla\ti$, one of them
being the tension, $\ten=-\Dh{}\ti=\Dv{}\nabla\ti$. If follows that if
$R$ and $\theta$ vanish and moreover $\Phi$ and $\ten$ are constant (\ie
parallel), then the \sode\ is equivalent to a linear one. From our
results it follows that some of this conditions are redundant.
\medskip
Finally it is worthy to mention that a diferent kind of linearisability
problem has been treated in~\cite{GrThWi,SaMaLe}. There, the problem is
to linearise a non--autonomous \sode\ by coordinate transformations of the
form $\bar{t}=\tau(t,x)$, $\bar{x}=\xi(t,x)$. These coordinate changes
induce projective transformations on $\real\times TM$
so that the formalism used here is not directly applicable. We hope
that further research will allow us to solve that problem.
\bigskip
\noindent{\bf Acknowledgements: }We wish to thank Willy Sarlet and Frans
Cantrijn for illuminating suggestions. Partial financial support from
DGICYT and the NATO Collaborative Research Grants Progamme is
acknowledged.
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\end{document}