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\def\gua{H.~J.~SUSSMANN}
\def\guo{EULER-LAGRANGE AND COORDINATE INVARIANCE}

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\begin{center}
{\Large{\bf THE EULER-LAGRANGE EQUATION\\
\vspace{1mm}
AND ITS INVARIANCE PROPERTIES}}\\
$\,\,$\\
{\Large{\it H\'ector J. Sussmann}}\\
$\,\,$\\
{\Large November 8, 2000}
\end{center}

\bigbreak



\ssection{THE EULER-LAGRANGE EQUATION}

\ssubsection{Standard calculus of variations problems}
The calculus of variations deals mainly with optimization 
problems involving a function
$$
(x,\dot x)\mapsto L(x,\dot x,t)
$$ 
of ``position, velocity, and time.''  Nowadays, this function is
usually called the {\bem Lagrangian.}  In contemporary mathematics,
the ``position''---or ``configuration,'' or ``state''\Ft{We will point out 
below that one should really distinguish between the {\bem configuration}
$q$ of a physical system and a {\bem coordinate representation}
in terms of an $n$-tuple $x=(x^1,x^2,\ldots,x^n)$ of numbers, but at this 
point we will ignore the distinction}---$x$ is usually
taken to be a vector\Ft{We follow the modern convention of using
superscripts for ``contravariant indices,'' i.e., components of
vectors, and subscripts for ``covariant indices,'' i.e., components of
covectors.}, $x=(x^1,x^2,\ldots,x^n)$ in $n$-dimensional Euclidean
space $\rr^n$, and the velocity $\dot x=(\dot x^1,\dot x^2,\ldots,\dot x^n)$ 
is then also a vector in $\rr^n$.

\begin{remark}\label{ittook} {\rm In the earliest work, $n$ was usually 
one or two or three. Considering higher dimensions is natural in
today's mathematics, but the idea that one might as well write
$(x^1,x^2,\ldots,x^n)$ rather than $x$, or $(x,y)$, or $(x,y,z)$, and
``do everything in $n$-dimensions, with $n$ completely arbitrary,''
was by no means natural to the founders of the subject and took a long
time to evolve. Lagrange, for example, devotes a lot of space in his
{\it M\'ecanique Analytique} to persuading the reader that a system of
$N$ point particles in $3$-dimensional space can be thought of as a single
point evolving in a space of $3N$ coordinates.}\proofend
\end{remark}

\begin{remark} \label{ittook-a} {\rm One can also consider problems 
in which $x$ is a vector in an infinite-dimensional space, and 
a lot of the current activity deals with such questions. 
However, {\bem here we will only
discuss finite-dimensional problems}}.\proofend\end{remark}


\medbreak\noindent The objective of a typical calculus of variations problem
is to find a curve $t\mapsto\xi(t)$ in $x$-space that minimizes the integral 
$$
I(\xi)=\int_a^bL(\xi(t),\dot\xi(t),t)dt
$$
of the Lagrangian, among all curves\Ft{\label{lavre-foot}In twentieth
century mathematics, ``all curves'' means, of course, all curves in some
suitably chosen function space, such as that of all absolutely continuous
curves, or that of all Lipschitz curves.  The choice of function space can
sometimes be crucial, as will be seen when we discuss the ``Lavrentiev
phenomenon'' later on.}


For example, one could specify a time interval
$[a,b]$, an initial point $\bar x$, and a terminal point $\hat x$,
and seek to minimize $I$ in the class of all curves $\xi:[a,b]\to\rr^n$ 
sucht that $\xi(a)=\bar x$ and $\xi(b)=\hat x$. This is a 
``standard calculus of variations problem'':

\medbreak
\begin{equation}\label{standard}
\fbox{\fbox{$\left.\begin{array}{c}{\bf THE}\quad{\bf PROBLEM}\quad\mathcal{P}(L,a,b,\bar x,\hat x)\\
$\,$\quad\\
\left.\begin{array}{rcl}
{\rm Given\,\,}&\,\,& L,\,a,\,b,\,\bar x,\,\hat x\,,\\
\,\,&\,\,&\,\,\\
{\rm minimize\,\,}&\,\,& I=\int_a^bL(\xi(t),\dot\xi(t),t)dt\,,\\
\,\,&\,\,&\,\,\\
{\rm subject\,\,to\,\,}&\,\,&\xi(a)=\bar x\,\,\,{\rm and}\,\,\xi(b)=\hat x\,.
\end{array}\right.\end{array}\right.$}}  
\end{equation} 
\bigbreak\noindent Alternatively, one may think that the use of 
the symbol $\dot x$ as a variable is a little bit confusing, 
and prefer to avoid using it. In order to make it transparently clear 
that $L$ is ``a function of three independent variables''---namely, 
the configuration point $x$, the velocity vector $u$, and the time 
$t$---one would use a letter such as $u$ rather than $\dot x$ 
as the name for the velocity variable.  This leads to regarding the 
Lagrangian as a function
\begin{equation}\label{usingthe}
\Omega\times\rr^n\times[a,b]\ni(x,u,t)\mapsto L(x,u,t)
\end{equation}
and rewriting (\ref{standard}) 
in the equivalent form 
\medbreak
\begin{equation}\label{standardu}
\fbox{\fbox{
$\left.\begin{array}{c}{\bf THE}\quad{\bf PROBLEM}\quad
\mathcal{P}(L,a,b,\bar x,\hat x)\\
$\,$\quad\\
\left.\begin{array}{rcl}
{\rm Given\,\,}&\,\,& L,\,a,\,b,\,\bar x,\,\hat x\,,\\
\,\,&\,\,&\,\,\\
{\rm minimize}&\,\,& I=\int_a^bL(\xi(t),\eta(t),t)dt\,,\\
\,\,&\,\,&\,\,\\
{\rm subject \,\,\,to}&\,\,&\xi(a)=\bar x\,,\,\,\xi(b) =\hat x\,,\\
\,\,&\,\,&\,\,\\
{\rm and}&\,\,&\dot\xi(t)=\eta(t)\,\,{\rm for}\,\,\,
a\le t\le b\,.\end{array}\right.\end{array}\right.$}}
\end{equation}

\bigbreak\noindent The vector variable $x$ is the {\em state}, or 
{\em configuration}, of the problem (\ref{standard}) or (\ref{standardu}).
The set $\Omega$ of all possible values of $x$ is the {\em configuration
space}, or {\em state space}, of the problem. For the time being, we will
require $\Omega$ to be an {\bem open subset of a Euclidean space
$\rr^n$}. Later on, we will allow $\Omega$ to be a more general {\bem 
differentiable manifold}, for example a spherical surface.  

So we will stipulate that

\bi
\item{} the configuration space $\Omega$ is an open subset of $\rr^n$;
\item{} the variable $u$---the ``velocity''---takes values in $\rr^n$;
\item{} $L$ is a real-valued function on $\Omega\times\rr^n\times[a,b]$. 
%\item{} if $Q$ is a manifold then the values of $u$ are, for 
%each $x\in Q$, the members of the tangent space $T_xQ$ of $Q$ at 
%$x$---so that in fact the pair $(x,u)$ should be thought of as the name 
%of a variable, say $V$, with values in $TQ$, the tangent
%bundle of $Q$---and $L$ is a real-valued function on $TQ\times[a,b]$.
\ei

%\begin{remark}{\rm \label{funcspace} Naturally, neither (\ref{standard})


\begin{remark}\label{pointtopoint} {\rm A minimization problem of the form
(\ref{standard})---or (\ref{standardu})---is a {\em point-to-point}
calculus of variations problem, 
because the endpoint constraint on the trajectory $\xi(\cdot)$ specifies that
$\xi(\cdot)$ has to start at a given point $\bar x$ and end at another given
point $\hat x$.  One can of course consider, both for the calculus of
variations and for optimal control, more general {\em set-to-set}
problems, in which the endpoint constraints are 
\begin{equation}\label{endpoco}
\xi(a)\in S_1\quad{\rm and}\qquad\xi(b)\in S_2\,,
\end{equation}
where $S_1$, $S_2$ are two given subsets of $\Omega$. Even more generally,
one can consider ``mixed'' endpoint constraints of the form 
\begin{equation}\label{endpocomixed}
(\xi(a),\xi(b))\in S\,,
\end{equation}
where $S$ is a given subset of $\Omega\times\Omega$. (For example, the
constraint that $\xi$ be a closed curve---i.~e., that
$\xi(b)=\xi(a)$---would be expressed by taking
$S=\{(x,x):x\in\Omega\}$.)  Such more general problems have been
studied since the very beginning of the calculus of variations, and
give rise to interesting questions about {\bem transversality
conditions}. At this point, {\bem we will only deal with calculus of
variations (and optimal control) point-to-point problems}. This means
that, for the time being, we are leaving out the whole area of
transversality conditions.}\proofend
\end{remark}

\ssubsection{The Euler-Lagrange equation}
We now look for {\bem necessary conditions} for a given
curve $t\mapsto\xi_*(t)$ to be a solution of the problem (\ref{standard}).

So we will be dealing with a problem $\mathcal{P}(L,a,b,\bar x,\hat x)$ as 
before, together with a given curve $\xi_*$. The curve $\xi_*$ 
will be referred to, in this kind of discussion,
as ``the reference curve,'' meaning ``the curve that we are examining 
in order to decide if it is a solution of the minimization problem.''

The general condition known today as the ``Euler-Lagrange equation''
was derived by {\bem  Leonhard Euler}\Ft{Euler entered the University
of Basel in 1720 at the age of
13. There he studied the methods 
developed by the Bernoullis 
for the study of isoperimetric problems, but soon became convinced that
there was a need for a general theory, that would go beyond the solutions
of specific problems. He began to develop his approach in 1731, 
and by 1740 he had achieved his goal and come up with a
general theory. He published his results in 1744, in a book
entitled \textit{Methodus inveniendi maximi minime propriate gaudentes sive
solutio problematis isoperimetrici  latissimo sensu accepti}
(A method for discovering curved lines having a maximum or minimum
property or the solution of the isoperimetric problem taken in its widest
sense). There he gave a 
general procedure for writing down what became known as 
\textit{Euler's equation}. The words ``calculus of variations'' 
do not yet appear in the
\textit{Methodus inveniendi.} They were coined by Euler later,
in 1760, as a name for Lagrange's method that Euler had decided to adopt 
in lieu of his own discretization approach, and appeared in print in 1766.} 
(1707-1793) and {\bem Joseph-Louis Lagrange}\Ft{Lagrange was  
a 19-year-old living in Turin when, on 12 August 1755, he 
wrote to Euler a brief letter to which was attached an appendix containing 
mathematical details of a new idea. He introduced a new method
that could turn out the necessary condition of Euler, and more,  almost 
automatically.  After seeing Lagrange's work, Euler dropped 
his own method, espoused that of Lagrange, and  renamed the subject the
\textit{calculus of variations}.
In the summary to his first paper using variations, Euler says
``Even though the author of this had meditated a long time
and had revealed to friends his desire yet the glory of first discovery was
reserved to the very penetrating geometer of Turin LA GRANGE, who having used
analysis alone, has clearly attained the same solution which the author had
deduced by geometrical considerations.''} (1736-1813).


The Euler-Lagrange equation 
says\Ft{We use the notations that are common today, not Lagrange's.
The symbol $\partial$ for partial derivative was first used by 
Legendre in 1786.} that the identity
{\protect\huge
\begin{equation}\label{eulagfirst}
\fbox{$\frac{d\,\,}{dt}\Big(\frac{\partial L}
{\partial\dot x}\Big)=\frac{\partial L}{\partial x}$}
\end{equation}}
must be satisfied along the curve $\xi_*$.

Equation (\ref{eulagfirst}) makes perfect sense and is a necessary condition
for optimality for a vector-valued variable $x$ as well as for a scalar
one.  It can be written as a system: 
{\protect\Large
\begin{equation}\label{eulagscalar}
\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot x^i}\Biggr)
= \frac{\partial
L}{\partial x^i}\,,\,\,\,\,\,\,\,i=1,\ldots,n\,.
\end{equation}}
Alternatively, we can regard Equation (\ref{eulagfirst}) as a vector
identity, in which 
$$
x=(x^1,\ldots,x^n)
$$ 
denotes an $n$-dimensional vector, and the expressions
$$\frac{\partial L}{\partial x}\,,\qquad\frac{\partial L}{\partial\dot x}
$$ 
stand for the $n$-tuples 
$$
\Big(\frac{\partial L}{\partial x^1},\ldots,
\frac{\partial L}{\partial x^n}\Big)\,,\qquad
\Big(\frac{\partial L}{\partial\dot x^1},\ldots,
\frac{\partial L}{\partial\dot x^n}\Big)\,.
$$
A modern mathematician might be troubled by the use of $\dot x$ both as an
``independent variable'' and as a function of time evaluated along a
trajectory, and might prefer to write (\ref{eulagfirst}) as
{\protect\Large
\begin{equation}\label{eulagvectoru} 
\fbox{$
\frac{d\,\,}{dt}\!\Biggl[\frac{\partial L}{\partial
u}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\Biggr]\!\!= \frac{\partial L}{\partial
x}\Big(\xi_*(t),\dot\xi_*(t),t\Big)\,,\,\,i\!=\!1,\ldots,n\,,$} 
\end{equation}}

\noindent writing the Lagrangian as in (\ref{usingthe}).
This makes it clear that, to compute the left-hand side of (\ref{eulagfirst}), 
we must
\begin{itemize}
\item[1.] first evaluate $\frac{\partial L}{\partial\dot x}$ 
``treating $\dot x$ as an independent variable,'' 
\item[2.] then plug in the curve $\xi_*(t)$ and its velocity 
$\dot\xi_*(t)$ in the slots for $x$ and $\dot x$, 
\item[3.] finally, differentiate with respect to $t$.
\end{itemize}

\ssubsection{The theorem} One possible precise statement of the
theorem about the Euler-Lagrange equation is as follows:
\begin{theorem}\label{eulagthe}
Assume $n$ is a positive integer, $\Omega$ is an open subset of 
$\rr^n$, $a$, $b$ are real numbers such that $a<b$, 
$L$ is a real-valued function on $\Omega\times\rr^n\times[a,b]$, 
and $\bar x$, $\hat x$ are given points in $\Omega$. 
Assume that $L$ is a function of class $C^1$. Let $\xi_*:[a,b]\mapsto\Omega$ 
be a curve of class $C^1$ which is a solution of the minimization problem 
$\mathcal{P}(L,a,b,\bar x,\hat x)$ in the space of all curves of class $C^1$.
Then the Euler-Lagrange equation
(\ref{eulagvectoru}) holds for all $t\in[a,b]$.
\end{theorem}
This is far from being the best possible result. One can prove 
Theorem \ref{eulagthe} under much weaker hypotheses, but in that case
one has to be more careful about the statement. We will discuss such
improvements later, and we will postpone the proof until we have
become more familiar with the result and with its generalizations.

Notice that, under the hypotheses of Theorem \ref{eulagthe},
\begin{itemize}
\item[1.] the curve $t\mapsto(\xi_*(t),\dot\xi_*(t),t)$ is continuous,
\item[2.] the functions 
$$
(x,u,t)\mapsto\frac{\partial L}{\partial x}(x,u,t)
$$
and 
$$
(x,u,t)\mapsto\frac{\partial L}{\partial u}(x,u,t)
$$
are continuous,
\end{itemize}
Therefore the conclusion of Theorem \ref{eulagthe} says in fact that 
the function 
\begin{equation}\label{iscone}
t\mapsto\frac{\partial L}{\partial u}(\xi_*(t),\dot\xi_*(t),t)\,,
\end{equation}
which is only known {\it a priori} to be continuous, is in fact 
differentiable everywhere, and its derivative is the function
\begin{equation}\label{iscone-a}
t\mapsto\frac{\partial L}{\partial x}(\xi_*(t),\dot\xi_*(t),t)\,.
\end{equation}
Hence {\bem the continuous function (\ref{iscone}) is in fact 
continuously differentiable.}

It should be immediately clear from the statement 
of Theorem \ref{eulagthe} the technical conditions can be weakened. 
For example:
\begin{itemize}
\item[(A)] The statement of Theorem \ref{eulagthe} contains absolutely no
reference to any derivatives of $L$ with respect to time. So it ought
to be possible to relax the hypothesis that $L$ is of class $C^1$ and
assume instead that $L$ is just \mbox{differentiable} with respect to $x$ and
$u$, maybe with the $x$- and $u$-derivatives continuous as functions
of $(x,u,t)$.
\item[(B)] Suppose we only assume that the curve $\xi_*$ is
Lipschitz, and that it is a solution of 
$\mathcal{P}(L,a,b,\bar x,\hat x)$ in the space of all Lipschitz curves.  
Then we would of course no
longer know {\it a priori} that the functions (\ref{iscone}) and
(\ref{iscone-a}) are continuous, but we would know that they are
measurable and bounded.  It is not unreasonable to expect
that in this case the conclusion of Theorem \ref{eulagthe} might still follow,
in the sense that {\bf the function (\ref{iscone}) is the indefinite
integral of the function (\ref{iscone-a})} (that is, equivalently, 
the function (\ref{iscone})---which is only known to be bounded and 
measurable---is in fact absolutely continuous, and its derivative 
is equal almost everywhere to the function (\ref{iscone-a}). 
\item[(C)] Suppose we only assume that the curve $\xi_*$ is
absolutely continuous, and that it is a solution of 
$\mathcal{P}(L,a,b,\bar x,\hat x)$ in the space of all absolutely continuous 
curves.  Then we would of course no
longer know {\it a priori} that the functions (\ref{iscone}) and
(\ref{iscone-a}) are continuous or bounded, but we would know that they are
measurable.  Once again, it may not appear  unreasonable to expect
that in this case the conclusion of Theorem \ref{eulagthe} might still follow,
in the sense that {\bf the function (\ref{iscone}) is the indefinite
integral of the function (\ref{iscone-a})} (that is, equivalently, 
the function (\ref{iscone})---which is only known to be and 
measurable---is in fact absolutely continuous, and its derivative 
is equal almost everywhere to the function (\ref{iscone-a}). 
\end{itemize}

\noindent It will turn out that {\bem the conjectures suggested 
by \ (A) and \ (B) are true while, on the other hand, the one
suggested by (C) is false, due to the so-called ``Lavrentiev phenomenon.''}

Moreover, we will see later that Theorem \ref{eulagthe} can be
extended even further, in rather surprising ways. For example, {\bem
no differentiability of $L$ with respect to the velocity $u$ is
needed.}  This should of course look strange to you at this point,
because if we do not assume that $L$ is differentiable with respect to
$u$ then it is not clear at all what the meaning of 
(\ref{eulagvectoru}) might possibly be. It will turn out, however,
that if one interprets things properly, one can get rid of the 
requirement on differentiability with respect to the velocity $u$.

On the other hand, it will turn out, also rather surprisingly, that
the requirement on differentiability with respect to $x$ can be
weakened a bit but not much.  This may seem strange because, if you
look at (\ref{eulagvectoru}), you may get the impression that more
regularity is required for the dependence of $L$ on $u$ than for the
dependence of $L$ on $x$ since, after all, in (\ref{eulagvectoru}) we
are only differentiating $L$ with respect to $x$, whereas $L$ is being
differentiated with respect to $u$ and then the $u$-derivative is
differentiated again. Yet we will see later that some kind of
differentiability with respect to $x$ is crucial no matter what we do,
whereas the differentiability with respect to $u$ can be completely
avoided if one does things right.

\ssubsection{The momentum} The Euler-Lagrange equation 
contains the ``vector''
{\protect\Large
\begin{equation}\label{defmoment}
\fbox{$\pi(t)\equaldef\frac{\partial L}{\partial\dot x}(\xi_*(t),
\dot\xi_*(t),t)\,,$}
\end{equation}}

\noindent known as the {\bem momentum} along the trajectory $\xi_*(\cdot)$. 
So the components $\pi_i$ of the momentum are given by 
{\protect\Large
\begin{equation}\label{defmomentcomp}
\fbox{$\pi_i(t)\equaldef\frac{\partial L}{\partial u^i}(\xi_*(t),
\dot\xi_*(t),t)\,,
\qquad{\rm for}\,\,i=1,\ldots,n\,.$}
\end{equation}}

Naturally, $\pi$ depends on $t$, so it is not a ``vector'' but a 
\mbox{{\bem field of ``vectors'' along the}} 
{\bem trajectory $\xi_*$}, that is, a map
$t\mapsto\pi(t)$ that assigns to each time $t$ in the interval where
$\xi_*$ is defined a ``vector'' $\pi(t)$.  (In modern
differential-geometric terminology, $\pi$ really is a {\bem field of
covectors} along $\xi_*(\cdot)$. As we will see later, $\pi$ shows up
in control theory playing an even more important role, and is often
referred to by control theorists as the ``adjoint vector.'' This is a
somewhat unfortunate choice of terminology, because the $\pi(t)$ are
really {\bem covectors} rather than vectors, but it is standard and we
will follow it.)

\newpage
\ssection{COORDINATE INVARIANCE}

\ssubsection{Coordinate charts}
Now the time has finally come to clarify the distinction between
``configurations'' and ``vectors $x$ in $\rr^n$.''  It is actually
quite simple: if we are trying to describe the behavior of a
physical system (for example, the motion of a point particle), then
this motion will take place in the ``configuration space'' $Q$ of the
system (for example, in three-dimensional physical space for the case
of a point particle). Now, {\bem a point of $Q$ is not a tuple of
numbers}.  We may use tuples of numbers to {\bem represent} points of
$Q$, but these tuples are not the points themselves but, rather, {\bem
representations} of the points.  Choosing a way to represent points
$q$ of $Q$ in a configuration space $Q$ as $n$-tuples
$x=(x^1,\ldots,x^n)$ is called {\bem choosing an $n$-dimensional
coordinate chart} on $Q$. Moreover, this coordinate chart may only be
appropriate to represent {\bem some}, but not necessarily {\bem all}
the points of $Q$.  This leads us to the following
\begin{definition}
{\rm Let $Q$ be a set, and let $n$ be a nonnegative integer. 
An {\bem $n$-dimensional coordinate chart} on $Q$ is a bijective 
map
$$
D^{\bf x}\ni q\mapsto{\bf x}(q)=(x^1(q),\ldots,x^n(q))\in R^{\bf x}
$$
from some subset $D^{\bf x}$ (called the {\bem domain} of the chart
${\bf x}$) onto an open subset $R^{\bf x}$ of $\rr^n$ (called the 
{\bem range} of ${\bf x}$).}\proofend
\end{definition}

It is important to realize that for a physical system there will
always be many different charts. Some of them will be more ``natural''
than others, as we shall soon see, but even among the ``natural'' ones
there will be no such thing as a canonical choice of {\bem the} chart.
For example, even if we want to study the motion of a single particle
in three-dimensional physical space, the possible configurations of
the particle are the {\bem positions}, that is, points in space, but
to represent these positions as triples of numbers we have to choose
an origin and three axes (if we are looking for ``Cartesian''
coordinates), and this can obviously be done in many ways. Moreover,
there are also many other, non-Cartesian, ways to represent points,
e.~g. by means of cylindrical or spherical coordinates.

\ssubsection{Represention of velocity vectors (tangent vectors)} 

Suppose a physical system has configuration space $Q$,
and suppose the system is evolving during a time interval 
$I\subseteq\rr$ in such a way that it describes a curve 
$$
I\ni t\mapsto\xi(t)\in Q\,.
$$
Suppose ${\bf x}$ is an $n$-dimensional chart of our system. 
Assume. moreover, that the curve $\xi$ is entirely contained in the domain 
$D^{\bf x}$ of a chart 
$$
{\bf x}:D^{\bf x}\mapsto R^{\bf x}\subseteq\rr^n\,.
$$
Then our curve $\xi$ admits a {\bem coordinate representation} 
$\xi^{\bf x}$ relative to the chart ${\bf x}$, given by 
$$
\xi^{\bf x}(t)={\bf x}(\xi(t))\quad{\rm for}\quad t\in I\,.
$$
This coordinate representation $\xi^{\bf x}$ is now a curve in 
$\rr^n$, so we may talk, for example, about the curve 
being ``differentiable.''    

What should we mean by the statement that $\xi$ itself (and not just the
coordinate representation $\xi^{\bf x}$!!!) is ``differentiable''
at a particular time $t\in I\,$? One obvious answer is to say that 
its coordinate representation is differentiable at $t$. This has 
the drawback that, as we said before, there are going to be lots of  
different coordinate charts for $Q$, which will give rise to different
coordinate representations of our curve $\xi$.  But let us 
postpone the discussion of this difficulty till later, and, temporarily,
give the following
\begin{definition}\label{asinbef}
{\rm Let $Q$ be a set, and let $I\ni t\mapsto\xi(t)\in Q$ be a 
$Q$-valued map defined on an interval $I\subseteq\rr$. Let
${\bf x}$ be an $n$-dimensional chart on $Q$. 
Assume that $\xi(t)\in D^{\bf x}$ for all $t\in I$.
Let $\bar t\in I$. We say that $\xi$ is {\em differentiable
at time $\bar t$ relative to the chart ${\bf x}$} 
if the coordinate representation 
$\xi^{\bf x}$ is differentiable at $\bar t$.} \proofend
\end{definition}

Now we would like to talk about the ``velocity'' $\dot\xi(\bar t)$
(that is, the derivative of $\xi$ at time $\bar t$). Later we will
give a precise definition of this object, but even without having done
this, it should be reasonable to expect that, whatever $\dot\xi(\bar
t)$ turns out to be, it will be represented, relative to the chart 
${\bf x}$, by the vector $\xi^{\bf x}(\bar t)$, which is a perfectly
well defined $n$-tuple of numbers, i.~e., a member of $\rr^n$. 

We can thus give the following
\begin{definition}
{\rm Let $Q$, $I$, $\xi$, ${\bf x}$, $\bar t$ be as in Definition
\ref{asinbef}, and assume that $\xi$ is differentiable at $\bar t$ 
relative to the chart ${\bf x}$. Then the {\em ${\bf x}$-representation
of $\dot\xi(\bar t)$} is the vector
$(\dot\xi(\bar t))^{\bf x}\in\rr^n$ given by}
$$
\qquad\qquad\qquad\qquad\qquad\qquad
(\dot\xi(\bar t))^{\bf x}\in\rr^n=\dot{\overbrace{\xi^{\bf x}}}(\bar t)\,.
\qquad\qquad\qquad\qquad\qquad\qquad\diamondsuit
$$
\end{definition}

Notice that {\bem we have not yet defined what the ``velocity vector''
$\dot\xi(\bar t)$ is}. We have only said what its representation
$(\dot\xi(\bar t))^{\bf x}$ relative to a particular chart ${\bf x}$ is.
Later, we will assign a precise meaning to the expression 
$\dot\xi(\bar t)$, as a ``tangent vector.''

\ssubsection{Invariance of the Euler-Lagrange equation}
We have now come to the main point of this discussion: the important 
discovery, made by Lagrange, that {\bem the Euler-Lagrange equation
is invariant under arbitrary changes of coordinates}.  

Let us first show what this means by doing a straightforward 
calculation.  Suppose we have a curve 
$$
I\ni t\mapsto\xi_*(t)\in Q
$$
and two different coordinate charts 
$$
{\bf x}:D^{\bf x}\mapsto R^{\bf x}\subseteq\rr^n\,,\qquad
{\bf X}:D^{\bf X}\mapsto R^{\bf X}\subseteq\rr^n\,.
$$
Suppose our curve $\xi_*$ is contained in both domains, that is,
$$
\xi_*(t)\in D^{\bf x}\cap D^{\bf X}\quad{\rm for}\quad{\rm all}\quad t\in I\,.
$$
Suppose our two coordinate charts are {\bem $C^2$ related}, in the following 
sense:
\begin{definition}\label{defrelated}
{\rm Let $Q$ be a set, and let ${\bf x}$, ${\bf X}$, be two 
$n$-dimensional charts
on $Q$. Let $k$ be a nonnegative integer. 
We say that ${\bf x}$ and ${\bf X}$ are {\em $C^k$ related} if
\begin{itemize}
\item[1.] the images ${\bf x}(D^{\bf x}\cap D^{\bf X})$,
${\bf X}(D^{\bf x}\cap D^{\bf X})$ of the ``overlap set''
$D^{\bf x}\cap D^{\bf X}$ under the coordinate maps 
${\bf x}$, ${\bf X}$, are open in $R^{\bf x}$, $R^{\bf X}$, respectively;
\item[2.] the ``change of coordinates'' maps
\begin{eqnarray*}
\Phi^{{\bf x},{\bf X}}:{\bf x}(D^{\bf x}\cap D^{\bf X})&\mapsto&
{\bf X}(D^{\bf x}\cap D^{\bf X})\,,\\
\Phi^{{\bf X},{\bf x}}:{\bf X}(D^{\bf x}\cap D^{\bf X})&\mapsto&
{\bf x}(D^{\bf x}\cap D^{\bf X})\,,
\end{eqnarray*}
defined by the conditions
$$
\Phi^{{\bf x},{\bf X}}(x)=X\quad{\rm whenever}
\quad x={\bf x}(q)\,,\,X={\bf X}(q)\quad{\rm for\,\,\,some}\quad
q\in D^{\bf x}\cap D^{\bf X}\,,
$$
$$
\Phi^{{\bf X},{\bf x}}(X)=x\quad{\rm whenever}
\quad x={\bf x}(q)\,,\,X={\bf X}(q)\quad{\rm for\,\,\,some}\quad
q\in D^{\bf x}\cap D^{\bf X}\,,
$$
are of class $C^k$.\proofend
\end{itemize}}
\end{definition}

\noindent(The map $\Phi^{{\bf x},{\bf X}}$ is the ``coordinate change from 
${\bf x}$ to ${\bf X}$,'' that is, the map that assigns, to each 
point $x\in R^{\bf x}$ which is the ${\bf x}$-coordinate representation
of some $q\in D^{\bf x}\cap D^{\bf X}$, the point $X$ that represents
the same $q$ in the ${\bf X}$-chart. It is clear that 
$\Phi^{{\bf x},{\bf X}}$ is well defined, because, if 
$x\in R^{\bf x}$ is the ${\bf x}$-coordinate representation
of some $q\in D^{\bf x}\cap D^{\bf X}$, then this $q$ is unique,
because ${\bf x}$ is injective, and ${\bf X}(q)$ exists, because
$q\in D^{\bf X}$. Naturally, similar remarks apply to the map 
$\Phi^{{\bf X},{\bf x}}$.)

Now suppose $L$ is a ``Lagrangian on $Q$.''  The precise meaning of
this is that $L$ is a function of points $q\in Q$, ``velocity
vectors'' (i.~e., tangent vectors) $\dot q$, and time, that is,
$L$ is a function
$$
Q|\dot Q\times[a,b]\ni(q,\dot q,t)\mapsto L(q,\dot q,t)\in\rr\,,
$$
defined at points $(q,\dot q)$ in the ``configuration-velocity space''
$Q|\dot Q$ and times $t$ in some interval $[a,b]$. More precisely, 
what we are now temporarily calling ``configuration-velocity space''
will later be called ``the tangent bundle of configuration space,'' and the 
mysterious ``velocity vectors'' will be interpreted as 
``tangent vectors.'' Since we have
not yet defined the notion of a tangent vector, we shall for the time
being assume that $L$---whatever it may be---is specified 
by giving its representation
$(x,\dot x,t)\mapsto L^{\bf x}(x,\dot x,t)$, for each coordinate 
chart ${\bf x}$, by means of a function $L^{\bf x}$. 

In addition, we shall assume that 
``the values of $L^{\bf x}(x,\dot x,t)$ and $L^{\bf X}(X,\dot X,t)$ are 
the same whenever $x$, $X$ are coordinate representations 
of the same point $q$, and $\dot x$, $\dot X$ are coordinate representations 
of the same `velocity' $\dot q$ relative to two charts 
${\bf x}$, ${\bf X}$.''   First of all, we have to decide what this means. 
Clearly, what it should mean is that 
\begin{equation}\label{reloder}
L^{\bf x}({\bf{\hat x}}(q,\dot q),t)=L^{\bf X}({\bf{\hat X}}(q,\dot q),t)
\quad{\rm whenever}\quad (q,\dot q)\in D^{\bf{\hat x}}\cap D^{\bf{\hat X}}\,,
\end{equation}
where ${\bf{\hat x}}(q,\dot q)$, ${\bf{\hat X}}(q,\dot q)$,
are the coordinate representations of a 
``configuration-velocity point'' $(q,\dot q)$ with respect to the charts 
${\bf x}$, ${\bf X}$. Equivalently, (\ref{reloder}) should say that
\begin{equation}\label{reloder-a}
\left.\begin{array}{l}
L^{\bf x}(\hat x,t)=L^{\bf X}(\hat X,t)
\quad{\rm whenever}\\
\hat x={\bf{\hat x}}(q,\dot q)\,\,{\rm and}\,\,
\hat X={\bf{\hat X}}(q,\dot q)\,\,\,{\rm for\,\,some}\,\,
\in D^{\bf{\hat x}}\cap D^{\bf{\hat X}}\,.\end{array}\right.
\end{equation}
The first thing we have to do is figure out what (\ref{reloder-a})
means directly in terms of the functions 
$L^{\bf x}$ and $L^{\bf X}$, without going through the intermediate 
point $(q,\dot q)$. Clearly, the vectors $\hat x$, $\hat X$ are 
coordinate representations of a position and velocity, so they should 
belong to $\rr^{2n}$ and be given as pairs $(x,\dot x)$,
$(X,\dot X)$, where $x,X$ represent $q$ and $\dot x,\dot X$ represent 
$\dot q$.  Moreover, as long as $q\in
D^{\bf{\hat x}}\cap D^{\bf{\hat X}}$, we already know how to express
$X$ in terms of $x$, using the map $\Phi^{{\bf x},{\bf X}}$, and 
there ought to be a similar way to ``express $\dot x$ in terms 
of $\dot X$.'' (As we will soon see, this is not 100\%{} accurate. 
We will express $\dot x$ in terms of $X$ and $\dot X$, not just in terms of 
$\dot X$ alone.) Let us find the formula for this. Using 
$\Phi$ for $\Phi^{{\bf x},{\bf X}}$,  
and  $\Psi$ for $\Phi^{{\bf X},{\bf x}}$, 
we have 

\begin{eqnarray*}
X^i&=&\Phi^i(x^1,\ldots,x^n)\,\\ x^j&=&\Psi^j(X^1,\ldots,X^n)\,,
\end{eqnarray*}
for $i=1,\ldots,n$, $j=1,\ldots,n$. Then we can express 
$\dot X$ as a function $\check\Phi(x,\dot x)$, and 
$\dot x$ as a function $\check\Psi(X,\dot X)$, by differentiating. 
The result is

\begin{eqnarray}
\dot X^i&=&\check\Phi^i(x^1,\ldots,x^n,\dot x^1,\ldots,\dot x^n)
=\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(x^1,\ldots,x^n)\dot x^j\,,
\label{defxX}\\
\dot x^j&=&\check\Psi^j(X^1,\ldots,X^n,\dot X^1,\ldots,\dot X^n)
=\sum_{i=1}^n\frac{\partial\Psi^j}{\partial X^i}(X^1,\ldots,X^n)\dot X^i\,,
\label{defXx}
\end{eqnarray}
for $i=1,\ldots,n$, $j=1,\ldots,n$.  So now we are able to write 
the complete formulae relating $\hat x=(x,\dot x)$  and
$\hat X=(X,\dot X)$:
\begin{eqnarray}
X^i&=&\Phi^i(x^1,\ldots,x^n)\,,\label{funa} \\ 
\dot X^i&=&\check\Phi^i(x^1,\ldots,x^n,\dot x^1,\ldots,\dot x^n)\,,
\label{funb} \\
x^j&=&\Psi^j(X^1,\ldots,X^n)\,,\label{func} \\
\dot x^j&=&\check\Psi^j(X^1,\ldots,X^n,\dot X^1,\ldots,\dot X^n)\,,
\label{fund} 
\end{eqnarray}
for $i=1,\ldots,n$, $j=1,\ldots,n$, where the functions
$\check\Phi^i$, $\check\Psi^j$ are defined by (\ref{defxX}) and
(\ref{defXx}). (Notice that, as we announced before,
we cannot express $\dot X$ just in terms of $\dot x$; the formulas actually
give $\dot X$ as a function of $x$ and $\dot x$.)   

The identities (\ref{defxX}) and (\ref{defXx}) are the {\bem transformation
formulas for velocity vectors}. They tell us how to find the
components of such a vector in one coordinate system if we know the
components in another coordinate system.  

Let us now compute the partial derivatives of the functions given by 
(\ref{funa}), (\ref{funb}), (\ref{func}), (\ref{fund}). We have

\begin{eqnarray*}
\frac{\partial X^i}{\partial x^j}
&=&\frac{\partial\Phi^i}{\partial x^j}(x^1,\ldots,x^n)\,,\\
\frac{\partial\dot X^i}{\partial x^j}
&=&\sum_{k=1}^n\frac{\partial^2\Phi^i}{\partial x^j\partial x^k}
(x^1,\ldots,x^n)\dot x^k\,,\\
\frac{\partial\dot X^i}{\partial\dot x^j}
&=&\frac{\partial\Phi^i}{\partial x^j}(x^1,\ldots,x^n)\,,\\
\frac{\partial x^j}{\partial X^i}
&=&\frac{\partial\Psi^j}{\partial X^i}(X^1,\ldots,X^n)\,,\\
\frac{\partial\dot x^j}{\partial X^i}
&=&\sum_{k=1}^n\frac{\partial^2\Psi^j}{\partial X^i\partial X^k}
(X^1,\ldots,X^n)\dot X^k\,,\\
\frac{\partial\dot x^j}{\partial\dot X^i}
&=&\frac{\partial\Psi^j}{\partial X^i}(X^1,\ldots,X^n)\,.
\end{eqnarray*}
Now let us study how the components of the momentum 
are represented in our two charts ${\bf x}$, ${\bf X}$. 
Let us use $p$, $P$ to denote the two representations, so 
$$
p=(p_1,\ldots,p_n)=
\Big(\frac{\partial L}{\partial\dot x^1},\ldots,
\frac{\partial L}{\partial\dot x^n}\Big)
$$
and 
$$
P=(P_1,\ldots,P_n)=
\Big(\frac{\partial L}{\partial\dot X^1},\ldots,
\frac{\partial L}{\partial\dot X^n}\Big)\,.
$$
Repeated use of the Chain Rule gives
\begin{eqnarray*}
\frac{\partial L}{\partial X^i}
&=&\sum_{k=1}^n\frac{\partial L}{\partial x^k}
\cdot\frac{\partial x^k}{\partial X^i}
+\sum_{k=1}^n\frac{\partial L}{\partial\dot x^k}
\cdot\frac{\partial\dot x^k}{\partial X^i}\\
&=&\sum_{k=1}^n\frac{\partial L}{\partial x^k}
\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^n\frac{\partial L}{\partial\dot x^k}
\cdot\frac{\partial^2\psi^k}{\partial X^i\partial X^\ell}
\dot X^\ell\,,\\
P_i&=&\frac{\partial L}{\partial\dot X^i}\\
&=&\sum_{j=1}^n\frac{\partial L}{\partial x^j}
\cdot\frac{\partial x^j}{\partial\dot X^i}
+\sum_{j=1}^n\frac{\partial L}{\partial\dot x^j}
\cdot\frac{\partial\dot x^j}{\partial\dot X^i}\\
&=&\sum_{j=1}^n\frac{\partial L}{\partial\dot x^j}
\cdot\frac{\partial\dot x^j}{\partial\dot X^i}\\
&=&\sum_{j=1}^n\frac{\partial L}{\partial\dot x^j}
\cdot\frac{\partial\Psi^j}{\partial X^i}\\
&=&\sum_{j=1}^np_j\cdot\frac{\partial\Psi^j}{\partial X^i}\,.
\end{eqnarray*}
So we got the formula 
\begin{equation}\label{tramompP}
P_i=\sum_{j=1}^np_j\cdot\frac{\partial\Psi^j}{\partial X^i}\,,
\end{equation}
and an identical argument also yields the identity
\begin{equation}\label{tramomPp}
p_j=\sum_{i=1}^nP_i\cdot\frac{\partial\Phi^i}{\partial x^j}\,.
\end{equation}
Let us look at these transformation formulae together with the identities
we derived earlier for velocity vectors:
\begin{eqnarray}
P_i&=&\sum_{j=1}^n\frac{\partial\Psi^j}{\partial X^i}p_j\,,\label{deftrapP}\\
p_j&=&\sum_{i=1}^n\frac{\partial\Phi^i}{\partial x^j}P_i\,,
\label{deftraPp}\\
\dot X^i&=&\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}\dot x^j\,,
\label{deftraxX}\\
\dot x^j&=&\sum_{i=1}^n\frac{\partial\Psi^j}{\partial X^i}\dot X^i\,.
\label{deftraXx}
\end{eqnarray}
Notice that {\bem the transformation formulae for the momentum 
are different from the formulae for the velocities.} 
The formula giving you $\dot X$ in terms of $\dot x$ is the same 
as that giving you $p$ in terms of $P$. In other ways, 
{\bem the momentum and the velocity transform in opposite ways.}
The same rule that enables you to go from $\dot x$ to $\dot X$ 
will take you from $P$ to $p\,$, not from $p$ to $P\,$!!

For historical reasons, objects such as the momentum are called 
{\bem covariant vectors}, or, simply, {\bem covectors}. 
Objects such as the velocity, which transform in the opposite way, are called
{\bem contravariant vectors}, or, simply, {\bem vectors}, or
{\bem ordinary vectors}, or {\bem tangent vectors}.   

\begin{remark}
{\rm You may have noticed that we used {\bem superscripts} for the components 
of velocity vectors, and {\bem subscripts} for the components of momentum
``vectors.'' This is common practice in differential geometry and
physics, and there are very good reasons for it, which will be discussed 
later.}\proofend
\end{remark}

We now return to the issue of the invariance of the Euler-Lagrange equation.
The equation says that a certain ``vector'' is equal to zero.  This 
``Euler-Lagrange vector'' is the one having components
$$
e_i=a_i-b_i
$$
relative to the ${\bf x}$ chart, where 
\begin{eqnarray*}
a_i&=&\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot x^i}\Biggr)\,,\\
b_i&=&\frac{\partial L}{\partial x^i}\,.
\end{eqnarray*}
In the ${\bf X}$ system, the ````Euler-Lagrange vector'' will 
have components
$$
E_i=A_i-B_i
$$
where 
\begin{eqnarray*}
A_i&=&\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot X^i}\Biggr)\,,\\
B_i&=&\frac{\partial L}{\partial X^i}\,.
\end{eqnarray*}
We now have to find the transformation formula relating the $E_i$ 
to the $e_i$. For this purpose, we first study the transformation
of the $a_i$ to the $A_i$ and of the $b_i$ to the $B_i$.
We have
\begin{eqnarray*}
A_i&=&\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot X^i}\Biggr)\\
&=&\sum_{k=1}^n\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot x^k}
\Biggr)\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\frac{\partial L}{\partial\dot x^k}
\cdot\frac{d\,\,}{dt}\Biggl(\frac{\partial\Psi^k}{\partial X^i}\Biggr)\\
&=&\sum_{k=1}^n\frac{d\,\,}{dt}\Biggl(\frac{\partial L}{\partial\dot x^k}
\Biggr)\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^n\frac{\partial L}{\partial\dot x^k}
\cdot\frac{\partial^2\Psi^k}{\partial X^i\partial X^\ell}\dot X^\ell\\
&=&\sum_{k=1}^na_k\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\Psi^k}{\partial X^i\partial X^\ell}\dot X^\ell\,,
\end{eqnarray*}
\begin{eqnarray*}
B_i&=&\frac{\partial L}{\partial X^i}\\
&=&\sum_{k=1}^n\frac{\partial L}{\partial x^k}
\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^n\frac{\partial L}{\partial\dot x^k}
\cdot\frac{\partial^2\psi^k}{\partial X^i\partial X^\ell}
\dot X^\ell\\
&=&\sum_{k=1}^nb_k\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\psi^k}{\partial X^i\partial X^\ell}
\dot X^\ell\,,
\end{eqnarray*}
and then
\begin{eqnarray*}
E_i&=&A_i-B_i\\
&=&\sum_{k=1}^n a_k\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\Psi^k}{\partial X^i\partial X^\ell}\dot X^\ell\\
&\,&\,\,-\sum_{k=1}^nb_k\cdot\frac{\partial\Psi^k}{\partial X^i}
-\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\Psi^k}{\partial X^i\partial X^\ell}\dot X^\ell\\
&=&\sum_{k=1}^na_k\cdot\frac{\partial\Psi^k}{\partial X^i}
-\sum_{k=1}^nb_k\cdot\frac{\partial\Psi^k}{\partial X^i}\\
&=&\sum_{k=1}^n(a_k-b_k)\cdot\frac{\partial\Psi^k}{\partial X^i}\\
&=&\sum_{k=1}^n e_k\cdot\frac{\partial\Psi^k}{\partial X^i}\,.
\end{eqnarray*}
So we have derive the formula
\begin{equation}
E_i=\sum_{k=1}^n e_k\cdot\frac{\partial\Psi^k}{\partial X^i}\,,
\end{equation}
from which it follows that if all the the $e_i$ vanish then all the 
$E_i$ vanish as well. An identical argument works in the other direction
as well, so we have in fact shown that
\begin{equation}
e_1=e_2=\ldots=e_n=0\quad\Longleftrightarrow
\quad E_1=E_2=\ldots=E_n=0\,.
\end{equation}
In other words

\bigbreak
\begin{center}
\fbox{\fbox{\fbox{\begin{minipage}{7.6cm}
\protect\huge{\sl The Euler-Lagrange equation holds in the ${\bf x}$ chart
if and only if it holds in the ${\bf X}$ chart.}
\end{minipage}}}}\end{center}

This is the desired invariance statement.

%,\qquad i=1,\ldots,n\,.

\ssubsection{A comment on vectors, covectors, and tensors} Let us compare 
the transformation formulae for the velocity, the momentum, the 
``Euler-Lagrange vector,'' and other objects we have introduced, 
such as the two parts whose difference is the Euler-Lagrange vector.
The formulas say that
\begin{eqnarray}
\dot X^i&=&\sum_{k=1}^n\frac{\partial\Phi^i}{\partial x^k}\dot x^k\,,
\label{fla-a}\\
\dot x^j&=&\sum_{k=1}^n\frac{\partial\Psi^j}{\partial X^k}\dot X^k\,,
\label{fla-b}\\
P_i&=&\sum_{k=1}^n\frac{\partial\Psi^k}{\partial X^i}p_k\,,
\label{fla-c}\\
p_j&=&\sum_{k=1}^n\frac{\partial\Phi^k}{\partial x^j}P_k\,,
\label{fla-d}\\
E_i&=&\sum_{k=1}^n\frac{\partial\Psi^k}{\partial X^i}e_k\,,
\label{fla-e}\\
e_j&=&\sum_{k=1}^n\frac{\partial\Phi^k}{\partial x^j}E_k\,.
\label{fla-f}\\
A_i&=&\sum_{k=1}^na_k\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\Psi^k}{\partial X^i\partial X^\ell}\dot X^\ell\,,
\label{fla-h}\\
a_j&=&\sum_{k=1}^nA_k\cdot\frac{\partial\Phi^k}{\partial x^i}
+\sum_{k=1}^n\sum_{\ell=1}^nP_k
\cdot\frac{\partial^2\Phi^k}{\partial x^i\partial x^\ell}\dot x^\ell\,,
\label{fla-i}\\
B_i&=&\sum_{k=1}^nb_k\cdot\frac{\partial\Psi^k}{\partial X^i}
+\sum_{k=1}^n\sum_{\ell=1}^np_k
\cdot\frac{\partial^2\psi^k}{\partial X^j\partial X^\ell}
\dot X^\ell\,.\label{fla-j}\,,\\
b_j&=&\sum_{k=1}^nB_k\cdot\frac{\partial\Phi^k}{\partial x^j}
+\sum_{k=1}^n\sum_{\ell=1}^nP_k
\cdot\frac{\partial^2\Phi^k}{\partial X^i\partial x^\ell}
\dot x^\ell\,.\label{fla-k}
\end{eqnarray}

\vspace{-2mm}
\noindent These formulas show that:

\vspace{-1mm}
\begin{itemize}
\item[1.] The transformation law for the $e$s and $E$s is the same
as that for the $p$s and $P$s. In other words: {\bem the Euler-Lagrange
``vector'' is really a covector, like the momentum.}
\item[2.] The transformation laws for the $x$s and $X$s are ``opposite''
from those for the $p$s and $P$s. This why velocities are called
{\bem contravariant vectors}. 
\item[3.] In all three cases (that is, in the transformation laws for 
the $e$s and $E$s, the $p$s and $P$s, and the $x$s and $X$s), the 
laws are such that {\bem if the object under consideration vanishes
in one coordinate chart then it vanishes in all of them.} That is, 
\begin{itemize}
\item[(T)] {\bem the vanishing of the object is an invariant property, not 
depending on the choice of a coordinate chart.} 
\end{itemize}
Later, we will define other quantities, called {\bem tensors}, which are 
more general than vectors and covectors, but still have Property (T).
Property (T) will turn out to be a manifestation of the {\bem tensor
nature} of the velocity and the momentum.
\item[4.] The transformation laws for the $a$s and $A$s and the $b$s and $B$s
are quite different. In particular, {\bem Property (T) does not hold.}
So {\bem the two parts whose difference is the Euler-Lagrange covector
are not themselves covectors, and are not even tensors of any kind.}
\end{itemize}

\end{document}

