% AMS-LaTeX 1.2 sample file for book proceedings, based on amsproc.cls.
%
\documentclass[twoside]{amsproc}
%\font\mynine=cmr9
\def\quocar{We follow \cite{carath}, pages 210-212.}
\newcounter{gnucount}
\newcommand{\Ft}[1]{\footnote{\mynine\baselineskip 8pt #1}}
\setcounter{gnucount}{0}
\newcommand{\gnu}[1]{\refstepcounter{gnucount}
\medbreak
\begin{center}
\fbox{\begin{minipage}{11cm}
\noindent GNU NO. \thegnucount.\ \ #1\ GNU-END
\end{minipage}}\end{center}
\medbreak}
%Uncomment the following  line to run file without printing the gnus
%\renewcommand{\gnu}[1]{}
\usepackage{epsf}
\input epsf
\usepackage{amssymb}
%\usepackage{makeidx}
\parskip=0.8mm
\def\comment#1{}
\def\snell{Snell}
\def\traj{{\rm Traj\,}}
\def\llist#1{
\vspace{-4.5mm}
\begin{center}
$\quad\quad\,\,$\mbox{\begin{minipage}{12.2cm}
#1
\end{minipage}}\end{center}}

\def\lllist#1{
\vspace{-4.5mm}
\begin{center}
$\quad\quad\quad$\mbox{\begin{minipage}{11.9cm}
#1
\end{minipage}}\end{center}}

\newcommand{\compo}{\circ}
\def\transv{{\cap\kern-5pt\lower -2.2pt\hbox{$|$}\kern-4.9pt\lower
-7pt\hbox{$-$}}} 
\def\baba{|\kern-1.7pt|}
\def\stransv{{\cap\kern-5.3pt\lower -2.2pt\hbox{$\baba$}\kern-5.3pt\lower
-7pt\hbox{$-$}}} 
\newcommand{\eps}{\varepsilon}
\def\verylongrightarrow{\raise2.2pt\hbox to 25pt{\hrulefill}\kern-4pt\longrightarrow}
\def\verylongline{\raise2.2pt\hbox to 37pt{\hrulefill}}
\def\ovarrow#1{{\stackrel{#1}{\verylongrightarrow}}}
\def\ovline#1{{\stackrel{#1}{\verylongline}}}
\def\ovlarrow#1{{\stackrel{#1}{\verylongleftarrow}}}
\def\verylongleftarrow{\longleftarrow\kern-4pt\raise2.2pt\hbox to 25pt{\hrulefill}}
\def\longsearrow{\setlength{\unitlength}{0.5cm}
\begin{picture}(1.2,1.2)
\put(-0.2,1.2){\vector(1,-1){2}}
\end{picture}}
\def\longnearrow{\setlength{\unitlength}{0.5cm}
\begin{picture}(1.2,1.2)
\put(-0.2,-1.2){\vector(1,1){2}}
\end{picture}}
\def\longdownarrow{\setlength{\unitlength}{0.25cm}
\begin{picture}(0.6,1)
\put(0.2,1){\vector(0,-1){2}}
\end{picture}}
\def\longuparrow{\setlength{\unitlength}{0.25cm}
\begin{picture}(1.2,1.2)
\put(0.2,-0.5){\vector(0,1){2}}
\end{picture}}
\def\veceps{\vec{\eps}} 
\def\vic{\pmb}
\def\Is{I^{\#}}
\def\Xis{\Xi^{\#}}
\def\xis{\xi^{\#}}
\def\etas{\eta^{\#}}
\def\zetas{\zeta^{\#}}
\def\qs{q^{\#}}
\def\ts{t^{\#}}
\def\taus{\tau^{\#}}
\def\nus{\nu^{\#}}
\def\restr{\,\lceil\,}
\def\corestr{\,\rceil\,}
\def\equaldef{{\buildrel \rm def \over =}}
\def\mybox{$\lower3.5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{nj}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}$\medbreak}
\def\proofendbox{\enspace\hfill \mybox\bigskip}
\def\proofendsq{\hfill $\square$\medskip}
\def\proofend{\hfill$\diamondsuit\,\,$}
\def\pmb#1{\setbox0=\hbox{$#1$}%
  \kern-.025em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.025em\raise.0433em\box0 }
\def\pmbo#1{\setbox0=\hbox{$#1$}%
  \kern-.055em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.055em\raise.0433em\box0 }
%\def\bem{\bf\em }
\def\bem{\em }
\def\bemu{{\bem U}}
\def\tto{\longrightarrow\kern-9pt\longrightarrow}
\def\zz{\mathbb{Z}}
\def\qq{\mathbb{Q}}
\def\cc{\mathbb{C}}
\def\rr{\mathbb{R}}
\def\nn{\mathbb{N}}
\def\kk{\mathbb{K}}
\def\bmath{\mathbb{B}}
\def\smath{\mathbb{S}}
\def\zmath{\mathbb{Z}}
\newcommand{\nball}{{\bmath}^n}
\font\mynino=cmr9 scaled\magstephalf
\def\tub{\mathcal{T}}
\def\tubx{\mathcal{T}^X}
\def\btub{\bar\mathcal{T}}
\def\btubx{\bar\mathcal{T}^X}
\def\linx{{\rm Lin}(X,X)}
\def\linxxp{{\rm Lin}(X,X')}
\def\id{\kern.2em{\rm I}\kern-.56em{\rm 1}}
\def\idnpball{\id_{\partial\bmath_n}}
\def\so{{\rm So}}
\def\ta{{\rm Ta}}
\def\gr{{\rm Gr}}
\def\dom{{\rm Do}}
\def\im{{\rm Im}}
\def\togr{\,\,{\buildrel \rm igr \over \longrightarrow}\,\,}
\def\caF{\mathcal{F}}
%\def\spaco{\ \ \ \ \ \ }
\newcommand{\ssection}[1]{
%\ifodd \thepage%\clearpage
%\thispagestyle{empty}\hbox{$\,$}%\clearpage
%\thispagestyle{empty}\else%\clearpage
%\thispagestyle{empty}\fi
\hbox{$\,$}\vskip 0.3cm\section{#1}\vskip0.5cm}
\newcommand{\ssubsection}[1]{\subsection{#1}
\addcontentsline{toc}{section}{$\,\,\,\,$\arabic{section}.\arabic{subsection}
\ #1}}
\newcommand{\ben}{\begin{enumerate}}
\newcommand{\een}{\end{enumerate}}
\newtheorem{theorem}{Theorem}[subsection]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{Fact}[theorem]{Fact}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{xca}[theorem]{Exercise}

\numberwithin{equation}{section}

%    Absolute value notation
\newcommand{\abs}[1]{\lvert#1\rvert}

%    Blank box placeholder for figures (to avoid requiring any
%    particular graphics capabilities for printing this document).
\newcommand{\blankbox}[2]{%
  \parbox{\columnwidth}{\centering
%    Set fboxsep to 0 so that the actual size of the box will match the
%    given measurements more closely.
    \setlength{\fboxsep}{0pt}%
    \fbox{\raisebox{0pt}[#2]{\hspace{#1}}}%
  }%
}

\setcounter{tocdepth}{1}

\begin{document}


\newcommand{\hoo}{\raise 1.2pt\hbox{$\stackrel{o}{H}$}}
\newcounter{scholco}
\setcounter{scholco}{0}
\newcounter{apco}
\setcounter{apco}{0}
\newcounter{figurco}
\setcounter{figurco}{0}
\renewcommand{\thescholco}{\Roman{scholco}}
\newcommand{\scholium}[1]{\refstepcounter{scholco}
\subsection*{$\!\!\!\!\!\!\!\!$SCHOLIUM \thescholco.\ \ #1}
\addcontentsline{toc}{section}{$\qquad${Scholium \thescholco.\
    \ #1}}}
\newcommand{\figurecap}[1]{\refstepcounter{figurco}\bigskip
\begin{center}{\sc Figure \arabic{figurco}}.\ #1
\end{center}\bigskip}
\newcommand{\ssubsectionap}[1]{\refstepcounter{apco}
\subsection*{$\!\!\!\!\!\!\!\!$A.\arabic{apco}.\ \ #1}
\addcontentsline{toc}{section}{$\qquad$A.\arabic{apco}.\
    \ #1}}

%\section*{This is an unnumbered first-level section head}
%This is an example of an unnumbered first-level heading.

%\specialsection*{This is a Special Section Head}
%This is an example of a special section head%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\footnote{Here is an example of a footnote. Notice that this footnote
%text is running on so that it can stand as an example of how a footnote
%with separate paragraphs should be written.
%\par
%And here is the beginning of the second paragraph.}%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\gua{H.~J.~SUSSMANN}
\def\guo{LIE BRACKETS AND THE MAXIMUM PRINCIPLE}

\pagestyle{myheadings}
\markboth{\gua}{\guo}


\def\restr{\lceil}
\def\pmb#1{\setbox0=\hbox{$#1$}%
  \kern-.025em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.025em\raise.0144em\box0 
  \kern-.025em\raise.0288em\box0 
  \kern-.025em\raise.0433em\box0 }
%\renewcommand{\thesubsubsection}{\S$\,$\arabic{subsubsection}$\,$}
\def\equaldef{\,{\buildrel \rm def \over =}\,}
\def\hham{\mathcal{ H}}
\font\mytenrm=cmr10 scaled\magstephalf
\font\mytensl=cmsl10 scaled\magstephalf
\font\mytenit=cmti10 scaled\magstephalf
\font\forsection=cmbx10 scaled\magstep 0
\font\mynine=cmr9
\font\mynino=cmr9 scaled\magstephalf
\font\mit=cmti9
\font\myninebf=cmbx9
\font\mysmall=cmr7
\font\mysmallit=cmti7
\font\myverysmall=cmr5
\def\eps{\varepsilon}
\def\smalltyperm{\baselineskip=12.2pt \rm}
\def\smalltypesl{\baselineskip=12.2pt \sl}
\newcommand{\bi}{\begin{itemize}}
\newcommand{\ei}{\end{itemize}}
\newenvironment{Quoterm}{
\bigskip\begin{center}\begin{minipage}{11cm}\smalltyperm\noindent}
{\end{minipage}\end{center}\bigskip}
\newenvironment{Quotesl}{
\bigskip\begin{center}
\begin{minipage}{11cm}\smalltypesl\noindent}{\end{minipage}
\end{center}\bigskip}

\newcommand{\bqrm}{\begin{Quoterm}}
\newcommand{\eqrm}{\end{Quoterm}}
\newcommand{\bqsl}{\begin{Quotesl}}
\newcommand{\eqsl}{\end{Quotesl}}
\parindent 6mm
\newtheorem{ttheorem}{Theorem*}
\newtheorem{etheorem}{Theorem*}
\newtheorem{qtheorem}{Theorem*}
\newtheorem{myquestion}[qtheorem]{Open Problem}
\newenvironment{question}{\begin{myquestion}$\!\!\!$.\ \rm}{\end{myquestion}}
\newcommand{\bques}{\begin{question}}
\newcommand{\eques}{\end{question}}
\thispagestyle{empty}
\newcommand{\hdleft}{{\rm H. J. Sussmann}}
\newcommand{\hdright}{{\rm \hfil Hamilton and Weierstrass}}

%\setcounter{page}{-1}

\begin{center}
{\Huge{\bf LIE BRACKETS AND THE\\
$\,$\\
\vspace{-5mm}
MAXIMUM PRINCIPLE}}\\
$\,\,$\\
{\Large{\it H\'ector J. Sussmann} --- December 6, 2000}
\end{center}

\medbreak
\begin{center}
{\Large\bf CONTENTS}
\end{center}

\medbreak

\thispagestyle{empty}
\noindent 1.\ COORDINATE INVARIANCE, LIE BRACKETS, AND THE MAXIMUM\\
\phantom{1.}\  PRINCIPLE \hfill 1


\medbreak
\noindent \ 1.1\ Covariance, invariance, and coordinate democracy \hfill 1

\noindent \ 1.2\ Differentiable manifolds \hfill 2

\noindent \ 1.3\ Lie brackets \hfill 5

\noindent \ 1.4\ Control and Lie brackets \hfill 8

\noindent \ 1.5\ Using Lie brackets to understand optimal trajectories \hfill 9

\noindent \ 1.6\ Implementing Lagrange's covariance idea
in optimal control \hfill 10

\medbreak


%{\large 

\ssection{COORDINATE INVARIANCE, LIE BRACKETS, AND THE MAXIMUM PRINCIPLE}

\noindent In this note we show how the concept of a {\bf Lie
bracket} plays an essential role in the coordinate-free formulation
of the Maximum Principle. We will explain how the important discovery by
Lagrange of the {\bf invariance of the Euler-Lagrange system under 
arbitrary coordinate changes} has an optimal control analgue, 
and why this analogue must involve Lie brackets.

\ssubsection{Covariance, invariance, and coordinate democracy}
While Euler's work was full of 
``geometric'' arguments, and contained many pictures,
Lagrange used no pictures.  (Newton's 1687 {\it Principia} had 250 figures.
Lagrange proudly stated in the preface of his 1788 {\it M\'ecanique
Analytique} that ``no diagrams will be found in this work.'')  

Lagrange was very proud of having freed
mechanics from geometry by doing everything analytically.  Little did he
know that one of his discoveries anticipated Riemann's revolutionary
creation of differential geometry.

Lagrange allowed the configuration of a collection of $N$ particles to be
represented in terms of completely general systems of $3N$
coordinates. This was a decisive step in the process that would free
geometry from the constraint of having to deal with 3-dimensional Euclidean
space.  ``To appreciate how all was new in this work [the {\em
M\'ecanique Analytique}], let us point out that, when Lagrange described a
motion in terms of a system of Cartesian coordinates in space, this was
such a nontrivial step that he devoted two pages to trying to 
persuade the reader'' (Lochak, \cite{lochak}, p.~49).

He then showed that the  Euler-Lagrange system  has
the same form in ``curvilinear'' or ``generalized'' coordinates, so that 
the system is invariant under 
{\em arbitrary}, nonlinear coordinate changes. This was the
first example of what would later become the cornerstone of differential
geometry, namely, the principle of ``coordinate democracy,'' according to
which all coordinate systems, ``rectangular'' as well as ``curvilinear,''
have to be treated equally. 

Lagrange saw the significance of this invariance property:

\bqrm
It is perhaps one of the principal advantages of our method that 
it \label{beforehand} expresses the equations of every problem in the 
most simple form
relative to each set of variables and that it enables us to see
beforehand which variables one should use in order to facilitate the
integration as much as possible. [Quoted from \cite{yourgrau}, p.~33.]
\eqrm

\begin{remark}{\rm 
For a striking illustration of what Lagrange had in mind here, the
reader should think of how much simpler it is to derive the equations
of motion of a body under an inverse-square attracting force if one
first proves that the motion must take place in a plane, and then does
the whole calculation in polar coordinates, using the invariance of
the Euler-Lagrange equations.} \proofend
\end{remark}



\medbreak\noindent
In 1854, sixty-six years after the publication of Lagrange's treatise,
Riemann gave his famous lecture, sketching the brilliant ideas that would
become the basis of modern differential geometry.    It turns out that one
of these ideas was precisely that of {\em invariance under general
coordinate changes}, whose importance Lagrange
had presciently anticipated.


\ssubsection{Differentiable manifolds}

\begin{definition}
{\rm Let $Q$ be  a set and let $k$ be a nonnegative integer. An
{\em atlas of class $C^k$ on $Q$} is a set $\mathcal{A}$ of 
coordinate charts on $Q$ such that:
\begin{itemize}
\item[1.] Any two charts ${\bf x},\,{\bf X}\in\mathcal{A}$ are 
$C^k$-related.
\item[2.] $\bigcup\Big\{\,D^{{\bf x}}:{\bf x}\in\mathcal{A}\,\Big\}= Q$. 
\end{itemize}
If $n$ is a nonnegative integer such that all the charts belonging to
$\mathcal{A}$ are $n$-dimensional then 
$\mathcal{A}$ is said to be an {\em $n$-dimensional atlas} on $Q$.}\proofend
\end{definition}
\begin{definition}
{\rm Let $k$ be a nonnegative integer. A {\em differentiable manifold of class 
$C^k$} is a pair $Q=(Q_0,\mathcal{A})$ such that $Q_0$ is a set
and $\mathcal{A}$ is an atlas of class $C^k$ on $Q_0$.

If $Q=(Q_0,\mathcal{A})$ is a differentiable manifold of class 
$C^k$, we will refer to $\mathcal{A}$ as the {\em structural atlas},
or the {\em differentiable structure} of $Q$. The set $Q_0$ is the {\em set
of points of $Q$}. Often we will use the same notation $Q$ to 
refer to the set $Q_0$, so we shall talk about 
``a point $q\in Q$'' rather than ``a point $q\in Q_0$.''} \proofend
\end{definition}
If $k$ is a nonnegative integer, 
$Q=(Q_0,\mathcal{A})$ is a differentiable manifold of class 
$C^k$, and $q\in Q$, we use $\mathcal{A}(q)$ to denote the set of 
all the charts ${\bf x}\in\mathcal{A}$ such that 
$q\in D^{{\bf x}}$. It is easy to show that 
all the charts ${\bf x}\in\mathcal{A}(q)$ have the dimension. (This follows 
from a simple rank argument if $k>0$, and from the theorem on invariance 
of domains if $k=0$.) 
This common dimension of all the charts ${\bf x}\in\mathcal{A}(q)$ 
is called the {\em dimension of $Q$ at $q$.}

A manifold $Q$ such that the dimension of $Q$ at $q$ is the same for all
$q\in Q$ is said to be {\em of pure dimension.} In that case the integer
$n$ such that the dimension of $Q$ at $q$ equals $n$ for all $q\in Q$ is 
called {\em the dimension of $Q$}, and $Q$ is said to be 
{\em $n$-dimensional.} 

\begin{remark} 
{\rm If $Q=(Q_0,\mathcal{A})$ is a differentiable manifold, then one can define
a topology on $Q$ by declaring a subset $S$ of $Q$ to be {\em open}
if ${\bf x}(S\cap D^{{\bf x}})$ is open in $R^{{\bf x}}$ for every 
chart ${\bf x}\in\mathcal{A}$. 

One can then show easily that {\em every connected component
of a manifold is a manifold of pure dimension}}. \proofend
\end{remark}

\begin{definition}
{\rm Let $k$ be a strictly positive integer, and let 
$Q=(Q_0,\mathcal{A})$ be a differentiable manifold of class 
$C^k$. Let $q\in Q$ and let $n$ be the dimension of $Q$ at $q$.
A {\em tangent vector for $Q$ at $q$} is a map
$\mathcal{A}(q)\ni{\bf x}\mapsto v^{{\bf x}}\in\rr^n$ that 
assigns to each chart ${\bf x}\in\mathcal{A}(q)$ a column
vector  
$$
v^{{\bf x}}=\left[\begin{array}{c}
v^{{\bf x},1}\\v^{{\bf x},2}\\\vdots\\v^{{\bf x},n}
\end{array}\right]\in\rr^n\,,
$$
in such a way that, if 
${\bf x},{\bf X}$ are any two charts in $\mathcal{A}(q)$, then 
\begin{equation}\label{transfovec}
v^{{\bf X}}=\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)
\cdot v^{{\bf x}}
\end{equation}
or, equivalently, 
\begin{equation}\label{transfovecco}
v^{{\bf X},i}=\sum_{j=1}^n\frac{\partial\Phi^{{\bf x},{\bf X},i}}
{\partial x^j\,\,\,\,}(x)v^{{\bf x},j}\,\,\,
{\rm for}\,\,\,i=1,\ldots,n\,, 
\end{equation}
where $x={\bf x}(q)$ and 
$$
\Phi^{{\bf x},{\bf X}}=
\left[\begin{array}{c}
\Phi^{{\bf x},{\bf X},1}\\\Phi^{{\bf x},{\bf X},2}\\
\vdots\\\Phi^{{\bf x},{\bf X},n}\end{array}\right]\in\rr^n
$$
is the change of coordinates map from
${\bf x}$ to ${\bf X}$}.\proofend 
\end{definition}
Equations (\ref{transfovec}), (\ref{transfovecco})
are the {\em transformation formulas for vectors.}

We use $T_qQ$ to denote the set of all tangent vectors for $Q$ at $q$,
and refer to $T_qQ$ as the {\em tangent space of $Q$ at $q$}.

To define a tangent vector $v\in T_qQ$, it suffices to specify its component
representation $v^{{\bf x}}$ relative to some coordinate chart 
${\bf x}$ belonging to the structural atlas of $Q$, because once this is
done the representation 
$v^{{\bf X}}$ relative to any other chart 
${\bf X}$ is determined by (\ref{transfovec}).

\begin{definition}
{\rm Let $k$ be a strictly positive integer, and let 
$Q=(Q_0,\mathcal{A})$ be a differentiable manifold of class 
$C^k$. Let $q\in Q$ and let $n$ be the dimension of $Q$ at $q$.
A {\em cotangent vector}---or {\em covector}---{\em for $Q$ at $q$} is a map
$\mathcal{A}(q)\ni{\bf x}\mapsto w^{{\bf x}}\in\rr^n$ that 
assigns to each chart ${\bf x}\in\mathcal{A}(q)$ a row
vector  
$$
w^{{\bf x}}=\left[\begin{array}{c}
w^{{\bf x}}_1\\w^{{\bf x}}_2\\\vdots\\w^{{\bf x}}_n
\end{array}\right]\in\rr_n\,,
$$
in such a way that, if 
${\bf x},{\bf X}$ are any two charts in $\mathcal{A}(q)$, then 
\begin{equation}\label{transfocovec}
w^{{\bf x}}=w^{{\bf X}}\cdot
\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)
\end{equation}
or, equivalently, 
\begin{equation}\label{transfocovecco}
w^{{\bf x}}_j=\sum_{i=1}^nw^{{\bf X}}_i\frac{\partial\Phi^{{\bf x},{\bf X},i}}{\partial x^j\,\,\,\,}(x)\,\,\,{\rm for}\,\,\,j=1,\ldots,n\,,
\end{equation}
where $x={\bf x}(q)$ and 
$$
\Phi^{{\bf x},{\bf X}}=
\left[\begin{array}{c}
\Phi^{{\bf x},{\bf X},1}\\\Phi^{{\bf x},{\bf X},2}\\
\vdots\\\Phi^{{\bf x},{\bf X},n}\end{array}\right]\in\rr^n
$$
is the change of coordinates map from
${\bf x}$ to ${\bf X}$}.\proofend 
\end{definition}
Equations (\ref{transfocovec}), (\ref{transfocovecco}), 
are the {\em transformation formulas for covectors.}

We use $T_q^*Q$ to denote the set of all covectors of $Q$ at $q$,
and refer to $T_q^*Q$ as the {\em cotangent space of $Q$ at $q$}.

To define a tangent vector $w\in T_q^*Q$, it suffices to specify its component
representation $w^{{\bf x}}$ relative to some coordinate chart 
${\bf x}$ belonging to the structural atlas of $Q$, because once this is
done the representation 
$w^{{\bf X}}$ relative to any other chart 
${\bf X}$ is determined by (\ref{transfocovec}).

\begin{proposition}\label{propoinva} 
Let $k$ be a strictly positive integer, and let 
$Q=(Q_0,\mathcal{A})$ be a differentiable manifold of class 
$C^k$. Let $q\in Q$, and let $v\in T_qQ$,  $w\in T_q^*Q$. Define 
the inner product 
$$
w\cdot v\equaldef w^{{\bf x}}\cdot v^{{\bf x}}\,,
$$
where ${\bf x}$ is any chart belonging to $\mathcal{A}(q)$. Then 
$w\cdot v$ does not depend on the choice of ${\bf x}$.
\end{proposition}
\begin{proof}
Let ${\bf x}$, ${\bf X}$ be two charts belonging to $\mathcal{A}(q)$. Then
\begin{eqnarray*}
w^{{\bf X}}\cdot v^{{\bf X}}&=&
\Biggl(
w^{{\bf x}}\cdot
\Big(\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)\Big)^{-1}
\Biggr)\cdot\Biggl(\frac{\partial\Phi^{{\bf x},{\bf X}}}
{\partial x\,\,\,\,}(x)v^{{\bf x}}\Biggr)\\
&=&w^{{\bf x}}\cdot v^{{\bf x}}\,,
\end{eqnarray*}
using (\ref{transfovec}), (\ref{transfocovec}), and the fact that the 
square matrix
$\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)$
is invertible. (The invertibility of 
$\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)$
follows from (\ref{transfovec}), because (\ref{transfovec}) implies
that 
\begin{eqnarray*}
v^{{\bf X}}&=&
\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)v^{{\bf x}}
v^{{\bf x}}\\
&=&\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)
\frac{\partial\Phi^{{\bf X},{\bf x}}}{\partial x\,\,\,\,}(x)v^{{\bf X}}
\end{eqnarray*}
for every $v\in T_qQ$, so 
$$
v=\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)
\frac{\partial\Phi^{{\bf X},{\bf x}}}{\partial x\,\,\,\,}(x)v
$$
for every $v\in\rr^n$. This implies that 
$$
\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,}(x)
\frac{\partial\Phi^{{\bf X},{\bf x}}}{\partial x\,\,\,\,}(x)=
{\rm identity\,\,matrix}\,,
$$
and the conclusion follows.) 
\end{proof}

It follows from Proposition \ref{propoinva} that {\em the cotangent space
$T_q^*Q$ can be regarded in a natural way as the dual of the tangent 
space $T_qQ$.}

\ssubsection{Lie brackets}

\medbreak What do we need to make the Maximum Principle completely 
coordinate-free?
To answer this question we recall that a control system
is, basically, a family of vector fields, so it is natural to ask 
oneself what are the ``natural'' coordinate-free operations 
among vector fields. 

An obvious example  of such an operation is 
{\bf addition}. 
It turns out that, if $f$ and $g$ are two vector fields, then there is 
another natural covariant object derived from them,  namely,
the {\bf Lie bracket} $[f,g]$.  This is defined in coordinates 
by  the formula 
\begin{equation}
[f,g](x)=\frac{\partial g}{\partial x}(x)\cdot f(x)-
\frac{\partial f}{\partial x}(x)\cdot g(x)\,.
\end{equation}
Here we are writing $f$ and $g$ as columns of functions, that is,
\begin{equation}
f(x)=\left[\begin{array}{c}
f^1(x)\\
f^2(x)\\
\vdots\\
f^n(x)\end{array}\right]\,,
\qquad
g(x)=\left[\begin{array}{c}
g^1(x)\\
g^2(x)\\
\vdots\\
g^n(x)\end{array}\right]\,.
\end{equation}
In addition, if $h$ is a vector field, written as a column of functions as above, 
then $\frac{\partial h}{\partial x}$ is
the Jacobian matrix of $h$, which is a square matrix of
functions:   
\begin{equation}
\frac{\partial f}{\partial x}=\left[\begin{array}{ccc}
\frac{\partial f^1}{\partial x^1}&\cdots&\frac{\partial f^1}{\partial x^n}\\
\frac{\partial f^2}{\partial x^1}&\cdots&\frac{\partial f^2}{\partial x^n}\\
\vdots&\ddots&\vdots\\
\frac{\partial f^n}{\partial x^1}&\cdots&\frac{\partial f^n}{\partial x^n}
\end{array}\right]\,.
\end{equation}
It follows that, if $f$, $g$ are vector fields, and we let $h=[f,g]$, then
the components $h^i$ of $h$ are given by
$$
h^i=\sum_{j=1}^n\frac{\partial g^i}{\partial x^j}f^j-
\sum_{j=1}^n\frac{\partial f^i}{\partial x^j}g^j\,.
$$
\begin{remark}
If $f$, $g$ are vector fields, then $[f,g]$ is a vector field.
{\rm The precise meaning of this is that ``$[f,g]$ transforms like a vector 
field under a change of coordinates.'' To prove this, let us 
recall the way vector fields transform. Suppose $h$ is a vector field,
given with respect to a coordinate chart ${\bf x}:D^{{\bf x}}\mapsto\rr^n$
by functions $h^{{\bf x},1},\ldots,h^{{\bf x},n}$. Suppose 
${\bf X}$ is another chart,
and $h$ is given in this new chart by functions 
$h^{{\bf X},1},\ldots,h^{{\bf X},n}$. Suppose 
$$
\Phi^{{\bf x},{\bf X}}:{\bf x}(D^{{\bf x}}\cap D^{{\bf X}})
\mapsto{\bf X}(D^{{\bf x}}\cap D^{{\bf X}})
$$
and
$$
\Phi^{{\bf X},{\bf x}}:{\bf X}(D^{{\bf x}}\cap D^{{\bf X}})
\mapsto{\bf x}(D^{{\bf x}}\cap D^{{\bf X}})
$$
be the coordinate change maps.
}

{\rm The fact that $h$ is a vector field says that, at each point $q\in
D^{{\bf x}}\cap D^{{\bf X}}$, if we take a curve $\xi$ such that
$\xi(0)=q$, whose tangent
vector at time $0$ is $h(q)$, and let $\xi^{\bf x}$, $\xi^{\bf X}$, 
be the coordinate representations, so that 
$$
\xi^{\bf x}(t)={\bf x}(\xi(t))\quad{\rm and}\quad
\xi^{\bf X}(t)={\bf X}(\xi(t))\,,
$$
then, if we let $x={\bf x}(q)$, $X={\bf X}(q)$, we have
\begin{eqnarray*}
h^{{\bf X}}(X)&=&\frac{d\,\,}{dt}\Big|_{t=0}\Big(\xi^{{\bf X}}(t)\Big)\\
&=&\frac{d\,\,}{dt}\Big|_{t=0}\Big({\bf X}(\xi(t))\Big)\\
&=&\frac{d\,\,}{dt}\Big|_{t=0}\Big(\Phi^{{\bf x},{\bf X}}\Big({\bf x}(\xi(t))
\Big)\Big)\\
&=&\frac{d\,\,}{dt}\Big|_{t=0}\Big(\Phi^{{\bf x},{\bf X}}\Big(\xi^{{\bf x}}(t)
\Big)\Big)\\
&=&\frac{\partial\Phi^{{\bf x},{\bf X}}}{\partial x\,\,\,\,\,\,\,}(x)\cdot
h^{{\bf x}}(x)\,,
\end{eqnarray*}
so that the components $h^{{\bf X},i}$, $h^{{\bf x},j}$ are related by
\begin{eqnarray*}
h^{{\bf X},i}(X)&=&\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(x)
h^{{\bf x},j}(x)\\
&=&\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(\Psi(X))
h^{{\bf x},j}(\Psi(X))
\end{eqnarray*}
for $i=1,\ldots,n$, if we write
$$
\Phi=\Phi^{{\bf x},{\bf X}}\,,\quad\Psi=\Phi^{{\bf X},{\bf x}}\,.
$$
Now suppose $f$, $g$ are two vector fields, and $q$ is a point belonging
to $D^{{\bf x}}\cap D^{{\bf X}}$. Then 
$$
f^{{\bf X},i}(X)=\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(\Psi(X))
f^{{\bf x},j}(\Psi(X))=
\sum_{j=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^j}f^{{\bf x},j}
\Biggr)(\Psi(X))\,,
$$
so 
$$
\frac{\partial f^{{\bf X},i}}{\partial X^k}(X)=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}f^{{\bf x},j}
+\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial f^{{\bf x},j}}{\partial x^\ell}
\Biggr)(\Psi(X))\cdot
\frac{\partial\Psi^\ell}{\partial x^k}(X)\,.
$$
Similarly,
$$
\frac{\partial g^{{\bf X},i}}{\partial X^k}(X)=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}g^{{\bf x},j}
+\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}
\Biggr)(\Psi(X))\cdot
\frac{\partial\Psi^\ell}{\partial x^k}(X)\,.
$$
Therefore
\begin{eqnarray*}
&\,&\sum_{k=1}^n\frac{\partial g^{{\bf X},i}}{\partial X^k}f^{{\bf X},k}(X)\\
&\,&\,\,=\sum_{k=1}^n\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}g^{{\bf x},j}
+\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}
\Biggr)(\Psi(X))\cdot\frac{\partial\Psi^\ell}{\partial x^k}(X)
\cdot f^{{\bf X},k}(X)\\
&\,&\,\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}g^{{\bf x},j}
+\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}
\Biggr)(\Psi(X))\cdot\Biggl(\sum_{k=1}^n
\frac{\partial\Psi^\ell}{\partial x^k}(X)
\cdot f^{{\bf X},k}(X)\Biggr)\\
&\,&\,\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}g^{{\bf x},j}
+\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}
\Biggr)(x)\cdot f^{{\bf x},\ell}(x)\\
&\,&\,\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}g^{{\bf x},j}
f^{{\bf x},\ell}\Biggr)(x)
+\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}f^{{\bf x},\ell}\Biggr)(x)\,.
\end{eqnarray*}
Similarly, 
\begin{eqnarray*}
&\,&\sum_{k=1}^n\frac{\partial f^{{\bf X},i}}{\partial X^k}g^{{\bf X},k}(X)\\
&\,&\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}f^{{\bf x},j}
g^{{\bf x},\ell}\Biggr)(x)
+\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial f^{{\bf x},j}}{\partial x^\ell}g^{{\bf x},\ell}\Biggr)(x)\\
&\,&\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^\ell\partial x^j}f^{{\bf x},\ell}
g^{{\bf x},j}\Biggr)(x)
+\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^\ell}\frac
{\partial f^{{\bf x},\ell}}{\partial x^\ell}g^{{\bf x},j}\Biggr)(x)\\
&\,&\,=\sum_{j=1}^n\sum_{\ell=1}^n
\Biggl(\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}f^{{\bf x},\ell}
g^{{\bf x},j}\Biggr)(x)
+\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^\ell}\frac
{\partial f^{{\bf x},\ell}}{\partial x^\ell}g^{{\bf x},j}\Biggr)(x)\,,
\end{eqnarray*}
if we assume that ${\bf x}$ and ${\bf X}$ are $C^2$-related, so that 
$$
\frac{\partial^2\Phi^i}{\partial x^\ell\partial x^j}=
\frac{\partial^2\Phi^i}{\partial x^j\partial x^\ell}\,.
$$
Then 
\begin{eqnarray*}
h^{{\bf X},i}(X)&=&
\sum_{k=1}^n\frac{\partial g^{{\bf X},i}}{\partial X^k}f^{{\bf X},k}(X)
\sum_{k=1}^n-\frac{\partial f^{{\bf X},i}}{\partial X^k}g^{{\bf X},k}(X)\\
&=&\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^j}\frac
{\partial g^{{\bf x},j}}{\partial x^\ell}f^{{\bf x},\ell}\Biggr)(x)
-\sum_{j=1}^n\sum_{\ell=1}^n\Biggl(\frac{\partial\Phi^i}{\partial x^\ell}\frac
{\partial f^{{\bf x},\ell}}{\partial x^\ell}g^{{\bf x},j}\Biggr)(x)\\
&=&\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(x)
\Biggl(\sum_{\ell=1}^n\frac{\partial g^{{\bf x},j}}{\partial x^\ell}(x)
f^{{\bf x},\ell}(x)-\frac{\partial f^{{\bf x},\ell}}{\partial x^\ell}(x)
g^{{\bf x},j}(x)\Biggr)(x)\\
&=&\sum_{j=1}^n\frac{\partial\Phi^i}{\partial x^j}(x)
h^{{\bf x},j}(x)\,,
\end{eqnarray*}
which is the desired transformation formula.}\proofend
\end{remark}

The fact that $[f,g]$ is a vector field suggests that it should have
an intrinsic, coordinate-free characterization.  In other words:
{\bf rather than define what $[f,g]$ means using coordinates, and then
prove that when we change coordinates $[f,g]$ transforms as it should,
it ought to be possible to say directly what $[f,g]$ without invoking
coordinates at all.} 

This is indeed true.  The precise {\bf coordinate-free
characterization of the Lie bracket} is:
\begin{equation}
[f,g](q)=\lim_{t\to 0}\frac{\delta(t)-q}{t^2}\,,
\end{equation}
where
\begin{equation}
\delta(t)=qe^{tf}e^{tg}e^{-tf}e^{-tg}\,.
\end{equation}
(Here $t\mapsto qe^{th}$ denotes, for a vector field $h$, the integral curve
of $h$ that goes through $q$ when $t=0$.) 

\ssubsection{Control and Lie brackets} Lie brackets play a
crucial role in optimal control theory in many ways and, more
generally, they are the key objects in the theory of nonlinear
finite-dimensional control.  They play a central role in nonlinear
controllability, feedback linearization, and system equivalence.  

Just to give a simple illustration having nothing to with optimality,
let us consider a control theory problem totally analogous to that of
the characterization of flatness for a metric.  In control theory,
``linear control systems''---i.e., systems of the form $\dot q=Aq+Bu$,
evolving in $\rr^n$, with $u$ taking values in $\rr^m$, and $A$, $B$
matrices of the appropriate sizes---play a crucial role, comparable in
importance to that of the flat metrics within the class of general
Riemannian metrics.  So it is important to know when a system that
looks nonlinear because it is presented in ``curvilinear
coordinates,'' is in fact linear, in the sense that one can make it
linear by changing coordinates.  The answer turns out to be a
criterion involving Lie brackets.

Precisely, suppose we are studying control systems
$$
\dot q=f(q)+u_1g_1(q)+\ldots+u_mg_m(q)\,,\quad q\in\rr^n\,,
$$
with $C^\infty$ vector fields $f$, $g_1,\ldots,g_m$, and a point $\bar q$ in 
$\rr^n$.

Is there a way to know if such a system is ``a linear system in disguise,''
that is, a system obtained from a linear system by changing 
coordinates? 

For example, look at the two systems
\begin{eqnarray}
\dot x_1&=&x_2+2x_2x_3\,,\nonumber\\
\dot x_2&=&x_3-x_1x_2+x_2^3\,,\label{syslin}\\
\dot x_3&=&u+x_2^2+x_1x_3-x_1^2x_2+2x_1x_2^3\,,\nonumber
\end{eqnarray}

\begin{eqnarray}
\dot x_1&=&u_1\,,\nonumber\\
\dot x_2&=&u_2\,,\label{sysnon}\\
\dot x_3&=&u_1x_2-u_2x_1\,.\nonumber
\end{eqnarray}
Which one is ``more nonlinear''?  We will see later that 
(\ref{syslin}), which superficially looks very nonlinear, is in fact 
a linear system in disguise, whereas (\ref{sysnon}), 
whose nonlinearity appears to be much milder, is in fact highly nonlinear.
And the answer will be obtained by studying the Lie bracket structures
associated to both systems.   It will turn out that
a necessary condition for the system to be equivalent near $\bar q$ to
a linear system is that all the iterated brackets of members of the
set $\{f,g_1,\ldots,g_m\}$ involving two $g_i$'s should vanish
near $\bar q$. Under an extra technical condition (``strong 
accessibility,'' that is, the requirement that the Lie brackets 
${\rm ad}_f^k(g_i)(\bar q)$, $k\ge 0$, $i=1,\ldots,m$ span 
the whole space) this condition is sufficient as well.

So {\bem Lie brackets help us organize our knowledge of nonlinear
systems in terms of their deep structural properties.} This 
way of classifying systems is as superior
to the na\"\i{}ve approach based on a superficial counting of the
number of nonlinearities as the animal taxonomy of modern zoology
is to the classifications found in medieval bestiaries.  Exactly as a
dolphin is structurally closer to a zebra than to a shark---in spite of
the very visible ways, such as  shape and habitat, in which a dolphin
resembles a shark more than a zebra---System (\ref{syslin}) is 
much closer to a linear system than System (\ref{sysnon}).  Lie brackets
enable us to discover this, and then one can find abundant confirmation by
looking at the properties of both systems. (For example, (\ref{syslin}) is
globally stabilizable by a smooth state feedback, but (\ref{sysnon}) is
not even locally stabilizable near any point.) 

\ssubsection{Using Lie brackets to understand optimal
trajectories}For optimal control problems, the effort to understand the
properties  
of optimal trajectories in terms of reasonable invariants has begun 
only recently, and most of the work remains to be done. 

Just to give the
flavor of some of these results, we concentrate on one problem that has 
attracted a lot of attention in recent years, namely, that of minimum time
control.  Suppose, for example, that we are looking at minimum time control
of systems in $\rr^n$ of the form $\dot x=f(x)+ug(x)$, with $|u|\le 1$,
where $f$ and $g$ are smooth vector fields.
Here the control space $U$ is the interval $[-1,1]$.  
The Hamiltonian is $H=\varphi+u\psi-p_0$, where $\varphi=\langle p,f(x)\rangle$
and $\psi=\langle p,g(x)\rangle$.  Then $u$ will equal $1$ when 
$\psi>0$, and $-1$ when $\psi<0$.  A na\"\i{}ve genericity argument may
suggest the wrong conclusion that $\psi=0$ is a ``rare'' event, which will 
happen at most at isolated points, so that in fact the optimal trajectories
will be ``bang-bang,'' i.e., such that $u$ is piecewise constant with 
values $1$ and $-1$ and finitely many jumps.  This, however, is quite wrong.
\begin{example}{\rm \label{nobang}
Let $Q=\rr^2$, and assume that $q\!=\!(x,y)\!\in\! Q$ evolves according to 
$\dot x\!=\!1\!-\!y^2$, $\dot y\!=\!u$, $|u|\!\le\! 1$, so 
$f$, $g$ are the vector fields $(x,y)\mapsto(1-y^2,0)$, $(x,y)\mapsto(0,1)$, 
respectively.  (Equivalently, $f=(1\!\!-\!\!y^2)\frac{\partial\,\,}{\partial x}$ 
and $g=\frac{\partial\,\,}{\partial y}$.)  Suppose  we want a trajectory  
that goes from $(0,0)$ to  $(1,0)$ in minimum time.  It is obvious 
that the solution is given by $u(t)\equiv 0$, $y(t)\equiv 0$, 
$x(t)\equiv t$, $0\le t\le 1$, and the optimal time is $1$.  
So the optimal control is ``singular''---i.e., takes values in the 
open interval $]\!\!-\!\!1,1[\,\,$---and therefore drastically fails 
to be bang-bang. And one can easily verify  
that this example is ``stable under perturbations'' in any reasonable sense
of the word.  So it is not true that optimal controls are always, or even 
generically, bang-bang.  \proofend
}\end{example}

In view of our previous remarks, it is to be expected that the {\em
Lie brackets} of the vector fields $f$ and $g$ will play a crucial
role in determining the properties of a solution.  Not surprisingly, a
close analysis of the problem, applying (MP), shows that this is
indeed so.

For example, the ``singular'' control of Example \ref{nobang} is closely
related to a property of the Lie brackets of the vector fields 
$f$ and $g$. Precisely, {\em the singular trajectory exists and is contained 
in the $x$ axis because $[f,g]$ vanishes on the $x$ axis}. 

Since the computation of Example \ref{nobang}, as presented,  does not
explicitly display any
Lie brackets, we will now make the Lie brackets stand out by looking, more
generally, at systems of the form $\dot x=f(x)+ug(x)$, with $|u|\le 1$,
without specializing to a particular pair of vector fields $f$, $g$. 

If one computes the derivative of $\psi$ along an extremal $\gamma$,
the result is $\rho$ evaluated along $\gamma$, where $\rho=\langle
p,[f,g]\rangle$.  So, if $n$ happens to be $2$, and $g$ and $[f,g]$
are linearly independent at each point, then $\psi$ and $\dot\psi$
cannot vanish simultaneously, for otherwise $p$ would be orthogonal to
$g(x)$ and $[f,g](x)$, so $p$ would vanish, which is a
contradiction. (Recall that the Maximum Principle for variable
time-interval problems has the extra condition that $H$ must
vanish. Since $H=\langle p,f(x)\!+\!u\,g(x)\rangle\!-\!p_0$, it is
clear that $p\!=\!0$ implies $p_0\!=\!0$, contradicting
nontriviality.)  Now, if a function and its derivative do not vanish
simultaneously, it follows that the zeros of the function are
isolated. In our case, this implies that the optimal controls must be
bang-bang.  So we have proved a simple theorem on the structure of
optimal trajectories: {\em for a minimum time problem in $\rr^2$
arising from a control system $\dot x=f(x)+ug(x)$, with $|u|\le 1$, if
$g$ and $[f,g]$ are linearly independent at each point, then all
optimal trajectories are bang-bang.}  The reader can verify that in
Example \ref{nobang} the Lie bracket $[f,g]$ vanishes along the $x$
axis, so the trajectory of that example is contained in the ``singular
set'' of points where $g$ and $[f,g]$ fail to be linearly independent,
in perfect agreement with our theorem.

\ssubsection{Implementing  Lagrange's covariance idea in optimal 
control} Our elementary arguments about Lie brackets and minimum time
control provide a good illustration of how Lagrange's principle 
of ``coordinate democracy'' applies in the optimal control setting,
and  why a truly intrinsic formulation of the maximum principle, on
manifolds, is needed. 

To understand optimal trajectories, one needs to study certain
functions, such as the ``switching function'' $\psi$.  We can think of
$\psi$ as a ``generalized component'' of the (co)vector $p$, in the sense
that the ordinary components $p_i$ of $p$ are the inner products of $p$
with the coordinate vector fields $\frac{\partial\,\,}{\partial q^i}$, and
$\psi$ is the product of $p$ with the vector field $g$. The adjoint
equation gives us an explicit formula for the functions $\dot p_i$.  A
truly ``democratic'' version of the maximum principle should treat all
generalized components equally, that is, should give us in one swoop the
derivative of {\em every} function $\langle p,X(q)\rangle$, for {\em every}
vector field $X$.

With such a version one could, in each case, work with the generalized
components of $p$ that are most natural for the problem under study,
such as, for example, the ``switching function'' $\langle
p,g(q)\rangle$ for a system $\dot q=f(q)+ug(q)$. 
This would be, we believe,
true in spirit to Lagrange's idea that it is good to have a method
that ``{\bf\em expresses the equations of every problem in the most
simple form relative to each set of variables and ... enables us to
see beforehand which variables one should use in order to facilitate
the integration as much as possible}.''

An intrinsic version, treating all ``generalized components''
(sometimes known as ``momentum functions'') equally, can be formulated in 
a number of ways, e.g. using Poisson brackets. 
For a general version of the maximum principle on manifolds, 
using generalized components as above (and also for an alternative 
version involving connections along curves), we refer the reader
to \cite{susscofree}. The crucial point of the version given in
\cite{susscofree} is---considering, for simplicity, the case
of the minimum time problem for a control system of the form
$\dot q=f(q,u)$---that the adjoint equation is rephrased as the
following statement (the ``intrinsic adjoint equation''):
\bi
\item[(IAE)] {\sl 
for every smooth vector field $X$, the derivative 
with respect to $t$ of the function 
$$
t\mapsto \langle\pi(t),X(\xi_*(t))\rangle
$$
is the function
$$
t\mapsto\langle\pi(t),[f_{\eta_*(t)},X](\xi_*(t))\rangle
$$
where we define
$$
f_u(q)=f(q,u)\,.
$$}
\ei
The point of using (IAE) is that it treats the  inner product 
of with $X$---i.e., ``the $X$ component of $\pi(t)$''---equally 
for all vector fields $X$, thus enabling us to follow Lagrange's
prescription (cf. Page \pageref{beforehand}) 
that one should express 
``the equations of every problem in the most simple
form relative to each set of variables and ... enables us to see beforehand
which variables one should use in order to facilitate the integration as
much as possible,'' by working in each case with the 
functions of the form $\langle p,X\rangle$ that are most suited to the problem.


\ssubsection{Why the language of Lie brackets is natural
for optimal control} The preceding discussion, based on rather
trivial examples, is meant 
to illustrate what we have in mind when we talk about
``understanding the qualitative properties of trajectories'' and why
we say that the appropriate language for
discussing this question and formulating conditions is that of Lie
brackets of vector fields.  

The reader should think of the following analogy.  When one tries to solve
a simple $2\times 2$ system of linear equations
such as $2x+3y=5$, $6x-2y=4$, one may
fail to notice that the number $2.(-2)-3.6$ is important.  If, however, one
``does it with letters'' (as a Calculus student would say), 
i.e., sets out to solve a {\bem general} system
$a_{11}x+a_{12}y=b_1$, $a_{21}x+a_{22}y=b_2$, then one discovers
immediately that the number $a_{11}a_{22}-a_{12}a_{21}$ determines the
properties of the solutions, so it becomes natural to gives this number a
name such as ``determinant,'' and to look for a generalization of the
concept to higher dimensions.  Similarly, when one applies the maximum
principle to study a particular problem such as the one of Example
\ref{nobang}, one is in fact computing a Lie bracket, although this fact
may go unnoticed.  But when one ``does it with letters,'' e.g. by working
with a 
general problem of the form $\dot x=f(x)+ug(x)$, leaving $f$ and $g$
unspecified, then the Lie bracket $[f,g]$ stands out immediately as the
object that ``determines'' what happens, exactly as the expression
$a_{11}a_{22}-a_{12}a_{21}$ did in the linear algebra example.

It is a general fact that, when we apply the maximum principle to minimum
time problems, we get conclusions of the form ``if such and such thing is
true of the Lie brackets of the vector fields involved, then the optimal
trajectories have such and such properties.''  This can be ascertained in
all kinds of examples, provided that one makes sure that the computations
are always done ``with letters.''  But the occurrence of the brackets is due
to a profound {\em a priori} reason, namely, Statement (INV) of the
previous section.

In recent years, ever since the advent of ``differential geometric control
theory,''  much has been done to determine, in a systematic way, how
structural properties of a system, embodied in their Lie bracket relations,
relate to properties of the optimal trajectories.  For example, it has been
known for a long time that for linear minimum time optimal control 
the ``bang-bang property'' holds. From a general nonlinear perspective,
one can show that the bang-bang property holds when certain Lie brackets 
$B_i$ can be expressed as linear combinations of some other brackets. Linear
systems just happen to be those for which the $B_i$ vanish.  This provides 
an explanation of the linear bang-bang theorem from the nonlinear
perspective, as well as a generalization.

\bibliographystyle{amsalpha}
\begin{thebibliography}{A}

\bibitem{berkovitz}{Berkovitz,~L.~D., \textit{Optimal Control Theory.}
Springer-Verlag, New York, 1974.}

\bibitem{carath}{Carath\'eodory,~C., \textit{Calculus of Variations
and Partial Differential Equations of the First Order. Part II:
Calculus of Variations.} Eng. transl. by R.~B.~Dean, 2nd edition,
Holden Day, Inc., San Francisco, 1967. (Originally published as
\textit{Variationsrechnung und Partielle Differentialgleichungen
erster Ordnung.} B.~G.~Teubner, Berlin, 1935.)}

\bibitem{cesari}{Cesari, L., \textit{Optimization---Theory and Applications}.
Springer-Verlag, New York, 1983.}

\bibitem{hestenesbook}{Hestenes,~M.~R., \textit{Calculus of Variations
and Optimal Control Theory.} Wiley, 1966.}

%\bibitem{hestenes}{Hestenes,~M.~R., \textit{A general problem in the calculus
%of variations with applications to paths of least time.}  Rand Corporation
%RM-100, ASTIA Document No. AD-112382, 1950.}

\bibitem{leemarkus}{Lee,~E.~B., and L.~Markus, \textit{Foundations of
Optimal Control Theory.} Wiley, New York, 1967.}

\bibitem{lochak}{Lochak,~G., \textit{La G\'eom\'etrisation de la 
Physique.} Flammarion, Paris, 1994.}

\bibitem{pontrybook}{Pontryagin,~L.~S., V.~G.~Boltyanskii, R.~V.~Gamkrelidze,
and E.~F.~Mischenko, \textit{The Mathematical Theory of Optimal Processes.} 
Wiley, New York, 1962.}

\bibitem{susscofree}{Sussmann,~H.~J., \textit{An introduction to the
coordinate-free maximum principle.} In \textit{Geometry of Feedback and 
Optimal Control,} B.~Jakubczyk and W.~Respondek Eds., Marcel Dekker, 
New York, 1997, pp.~463--557.}

\bibitem{yourgrau}{Yourgrau,~W., and S.~Mandelstam,
\textit{Variational Principles in Dynamics and Quantum Theory.}
W.~B.~Saunders \&{} Co., Philadelphia, 1968.}


\end{thebibliography}

%}

\end{document}

