% File: geometry.tex
% Section: MATH
% Title: Basic geometry
% Last modified: 06.06.2002
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\begin{document}

\begin{center}
\large\textbf{BASIC GEOMETRY}
\end{center}

\vspace{.0cm}

\begin{center}
\large\textbf{Manifolds}
\end{center}

\vspace{.1cm}

An $n$-dimensional differentiable manifold $M$ is a set (actually, topological
space) which locally (in a vicinity of each point) behaves like ordinary
Euclidean space $\mathbb{R}^n$\,. Namely, $M$ is supplied with an atlas
consisting of local coordinate maps
$\varphi_\alpha:U_\alpha\rightarrow \mathbb{R}^n$\,, where subsets
$U_\alpha\subset M$ are open and cover the whole of $M$. On intersections
$U_\alpha\cap U_\beta$\,, the two coordinate systems should be related by
smooth (infinitely differentiable, or $C^\infty$) transformations, i.e.,
both $\varphi_\alpha\circ\varphi_\beta^{-1}$ and
$\varphi_\beta\circ\varphi_\alpha^{-1}\in C^\infty$\,.
In other words, any point $x\in M$ is characterized by (possibly, several)
sets of local coordinates (i.e., sets of $n$ real numbers
$x^i_\alpha=\varphi^i_\alpha(x)\,, \ i=1,\ldots,n$)\,, one set for each map
$\varphi_\alpha$ such that $x\in U_\alpha$\,, different sets being transformed
to each other by $n$ smooth invertible functions of $n$ arguments
$x^{i'}_\beta\,(x^i_\alpha)$\,. Speaking about $x$, we (explicitly or implicitly)
do it in terms of local coordinates provided by one of the maps $\varphi_\alpha$\,.
In what follows, we tacitly use one of existent coordinate maps, omitting index
$\alpha$ everywhere.

A smooth map $\mathbb{R}\rightarrow M$ defines a (smooth) curve $x(t)$ on $M$
parametrized by $t\in\mathbb{R}$ with the corresponding coordinate dependence
$x^i(t)$\,. We may imagine the motion along this curve with $t$ being time.
The corresponding velocity appears under the name of \emph{tangent vector},
which can be defined at each point of the curve. Let, say, $x(0)=x$\,. Then
\begin{equation}
\label{tang}
\xi^i = \frac{dx^i(t)}{dt}|_{t=0} \ \ \ \ \ \ \ \Leftrightarrow \ \ \ \ \ \ \
x^i(t) = x^i + t\xi^i + \mathcal{O}(t^2)
\end{equation}
are components of the tangent vector to the curve $x(t)$ at a point $x$\,.
Of course, $\xi$ is also tangent (at $x$) to any other curve $x_1(t)$
such that $x_1^i(t)-x^i(t)=\mathcal{O}(t^2)$\,. Tangent vectors to $M$ at a point
$x$ form an $n$-dimensional linear space denoted by $T_x M$\,.
Under a change of coordinates (or, rather, of the coordinate map) tangent
vectors transform as follows:
\begin{equation}
\label{tang'}
\xi_\beta^{i'} = \frac{\partial x_\beta^{i'}}
{\partial x_\alpha^i}\,\xi_\alpha^i\,, \ \ \ \ \ \
\xi_\alpha^{i} = \frac{\partial x_\alpha^i}
{\partial x_\beta^{i'}}\,\xi_\beta^{i'}\,, \ \ \ \ \ \
\frac{\partial x_\alpha^i}{\partial x_\beta^{i'}}\,
\frac{\partial x_\beta^{i'}}{\partial x_\alpha^j}
 = \delta^i_j\ .
\end{equation}

The right equality in (\ref{tang}) can be written only for coordinates, not for
the points $x(t)$ themselves, because $M$ is not a linear space and we cannot
add its points. However, we can do it with functions on $M$\,: one can write
an analog of (\ref{tang}) in the coordinate-free form,
\begin{equation}
\label{tang1}
f(x(t)) = f(x) + t\,(\xi f)_x + \mathcal{O}(t^2)\,, \ \ \ \ \ \ \ \ \ \
 (\xi f)_x \doteq \frac{df(x(t))}{dt}|_{t=0} \ .
\end{equation}
Because of $f(x(t))=f(\{x^i(t)\})$\,, we find that
\begin{equation}
\label{action}
(\xi f)_x = \frac{\partial f}{\partial x^i}\,\frac{dx^i(t)}{dt}|_{t=0}
 = \xi^i\,\partial_i f(x)\,, \ \ \ \ \ \ \xi\,x^i = \xi^i\,, \ \ \ \ \
\xi = \xi^i \partial_i\,, \ \ \ \ \ (\partial_i)^j = \delta_i^j\,,
\end{equation}
so that $\partial_i\equiv\partial^x_i\equiv\partial/\partial x^i$ form the
basis of $T_x M$\,. Thus, the action of a tangent
vector $\xi\in T_x M$ on a function $f\in C^\infty(M)$ results in a number
$(\xi f)_x$ which is nothing but the rate of change of $f$ along
any curve having  $\xi$ as its tangent vector (at $x$)\,, or, for short, along
the vector $\xi$ itself.

A smooth map $\varphi:M\rightarrow N\,, \ \varphi(x)=y$\,, \ to some manifold $N$
naturally induces two other maps. The first of them,
$\varphi_*:T_x M\rightarrow T_y N$\,, sends a vector $\xi\in T_x M$ tangent
to a curve on $M$ to a vector $\eta=\varphi_*\xi\in T_y N$ tangent to the
image (on $N$)
of this curve. The second map, a `pullback'
$\varphi^*:C^\infty(N)\rightarrow C^\infty(M)$\,,
simply re-reads a function $f$ on $N$ as a new function
$g$ on $M$ according to $g=\varphi^*f=f\circ\varphi$\,, or $g(x)=f(y(x))$\,.
Since, on any curve, $g(x(t))$ and $f(y(x(t)))$ are identical functions of
$t$, having thus the same small-$t$ expansion, we conclude that
\begin{equation}
\label{pull}
(\eta f)_y = (\xi g)_x \ \ \ \ \ \ \ \text{or} \ \ \ \ \ \ \
(\varphi_*\xi)f = \xi(\varphi^* f)\,.
\end{equation}
In essence, these formulas describe the change of variables:
\begin{equation}
\label{change}
x\rightarrow\varphi(x) = y(x) \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \
\eta^{i'}(y) = \frac{\partial y^{i'}}{\partial x^i}\,\xi^i(x)\,, \ \ \ \ \
\varphi_*\,\partial^x_i = \frac{\partial}{\partial x^i(y)}
 = \frac{\partial y^{i'}}{\partial x^i}\,\partial^y_{i'}\,.
\end{equation}
For a composition of maps we have
\begin{equation}
\label{compos}
M\xrightarrow{\varphi}N\xrightarrow{\psi}K\,: \ \ \ \ \ \ \
(\psi\circ\varphi)^* = \varphi^*\circ\psi^*\,, \ \ \ \ \ \
(\psi\circ\varphi)_* = \psi_*\circ\varphi_*\ .
\end{equation}

\vspace{.3cm}

\begin{center}
\large\textbf{Tensor fields}
\end{center}

\vspace{.1cm}

Tangent vectors are elements of $T_x M$; their components $\xi^i$ carry
upper indices and transform according to (\ref{tang'}) or (\ref{change}).
Elements of the dual space $T^*_x M\doteq(T_x M)^*$\,, i.e., linear functions
on $T_x M$, are called 1-forms; their
components are labelled by lower indices.
More generally, \emph{tensors} $T^r_s\in(T_x M)^{\otimes r}(T^*_x M)^{\otimes s}$
are characterized by sets of components $T^{i_1\ldots i_r}_{j_1\ldots j_s}(x)$
transforming (under $x\rightarrow y$) as
\begin{equation}
\label{tens}
(T')^{i'_1\ldots i'_r}_{j'_1\ldots j'_s}(y)
 = T^{i_1\ldots i_r}_{j_1\ldots j_s}(x)\,
 \frac{\partial y^{i'_1}}{\partial x^{i_1}}
 \cdots\frac{\partial y^{i'_r}}{\partial x^{i_r}}\,
 \frac{\partial x^{j_1}}{\partial y^{j'_1}}\cdots
 \frac{\partial x^{j_s}}{\partial y^{j'_s}}\,.
\end{equation}
We define a \emph{tensor field} $T^r_s(x)$ on $M$ by specifying a tensor of this
type at each point $x\in M$. Explicit $x$-dependence of a tensor field $T(x)$
should not be confused with its transformation properties under change of
coordinates, or mappings to other manifolds; the latter are described
by $T(x)\rightarrow T'(y)$ according to (\ref{tens})\,.

Tensors of the type $T^0_k$ can be considered as $k$-linear functions on
$(T_x M)^{\otimes k}$ through $<\!T,\xi_1\otimes\ldots\otimes\xi_k\!>=
T(\xi_1,\ldots,\xi_k)$\,.
 Let us look more closely at the totally antisymmetric
functions $\omega^k(\xi_1,\ldots,\xi_k)$ ($k$-forms)\,. They form an
\emph{exterior algebra} with respect to the \emph{wedge product} operation
which produces a $(k+m)$-form $\omega^k\wedge\omega^m$ via
\begin{equation}
\label{shuffle}
(\omega^k\wedge\omega^m)(\xi_1,\ldots,\xi_{k+m}) \doteq \sum_{\text{shuffles}}\,
(-)^\sigma\,\omega^k(\xi_{i_1},\ldots,\xi_{i_k})\,
\omega^m(\xi_{j_1},\ldots,\xi_{j_m})\,.
\end{equation}
In particular, the wedge product of $n$ 1-forms is given explicitly by
\begin{equation}
\label{det}
(\omega_1\wedge\ldots\wedge\omega_n)\,(\xi_1,\ldots,\xi_n)
 = \text{det}\,\omega_i(\xi_j)\,.
\end{equation}
Wedge product is associative and skew-commutative:
\begin{equation}
\label{skew}
(\omega^k\wedge\omega^l)\wedge\omega^m
 = \omega^k\wedge(\omega^l\wedge\omega^m)\,, \ \ \ \ \ \ \ \ \
\omega^k\wedge\omega^m = (-)^{km}\,\omega^m\wedge\omega^k\,.
\end{equation}

While a tensor field $T^0_0(x)$ is simply a function on $M$, \,
$T^1_0(x)$ is called a \emph{vector field}, and $T^0_1(x)$ a
\emph{differential form}\,. The last name is justified by the following
definition:
\begin{equation}
\label{diff}
<\!df,\xi\!>_x \ \doteq\ (\xi f)_x \,=\, \xi^i\,\partial_i f(x)
\ \ \ \ \Rightarrow \ \ \ \ \ <\!dx^i,\xi\!> = \xi^i\,, \ \ \ \ \
<\!dx^i,\partial_j\!> = \delta^i_j\ ,
\end{equation}
which agrees with the conventional rule $df(x) = \partial_i f\,dx^i$\,. 1-forms
$dx^i$ constitute a basis in $T^*_xM$ dual to $\{\partial_j\}$\,. Evidently,
they transform like components of vectors, as it should be.
So, any differential form on $M$ (a tensor field $T^0_1$) can
be written down as $\omega=\omega_i(x)\,dx^i$\,.

The operation $d$ here is the $k=0$ case of \emph{exterior differentiation}
which generally sends $k$-forms to $(k+1)$-forms, and is uniquely determined
by the following conditions:
\begin{equation}
\label{d}
d(f\omega) = df\wedge\omega + f\,d\omega\,, \ \ \ \
d(\omega^k\wedge\omega^m) = d\omega^k\wedge\omega^m
 + (-)^k\omega^k\wedge d\omega^m, \ \ \ \ d\circ d = 0\,.
\end{equation}
Since $\{dx^{i_1}\!\wedge\ldots\wedge dx^{i_k}\}$ is the basis in the space
of $k$-forms at $x$, one explicitly obtains
\begin{equation}
\label{}
\omega^k(x)
 = \sum\,\omega_{i_1\ldots i_k}(x)\,dx^{i_1}\!\wedge\ldots\wedge dx^{i_k}
\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \
d\omega^k
 = \sum\,d\omega_{i_1\ldots i_k}\wedge dx^{i_1}\!\wedge\ldots\wedge dx^{i_k}\,.
\end{equation}
There is another general formula for $d\omega^k$ which involves the notion of
the commutator of vector fields (see (\ref{commut}) below)\,:
\begin{multline}
\label{d2}
d\omega\,(\xi_0,\xi_1,\ldots,\xi_k) = \sum_{i=0}^{k}\,(-)^i\,\xi_i
 (\omega\,(\xi_0,\ldots,\xi_{i-1},\xi_{i+1},\ldots,\xi_k)) \\
 + \sum_{i<j}\,(-)^{i+j}\,\omega\,([\xi_i,\xi_j],\xi_0,\ldots,
  \xi_{i-1},\xi_{i+1},\ldots,\xi_{j-1},\xi_{j+1},\ldots,\xi_k)\,.
\end{multline}
In particular,
\begin{equation}
\label{}
d\omega(\xi,\eta) = \xi\omega(\eta) - \eta\,\omega(\xi) - \omega([\xi,\eta])\,.
\end{equation}

Let us now consider the transformation properties of tensor fields induced by
a map $\varphi:M\rightarrow N$. We will use the pullback notation $\varphi^*$
for functions (where it still means $\varphi^*f=f\circ\varphi$) and $k$-forms,
and $\varphi_*$ for vector fields and higher-rank tensors $T^k_0$\,. However,
eq. (\ref{pull}), being essentially true, appears slightly inadequate here:
it deals with the numbers evaluated at different points, $x\in M$
and $y=\varphi(x)\in N$\,, whereas, in the
case of fields, we would like to watch the dependence of both sides
on the same variable (say, $x$)\,. This is achieved by casting (\ref{pull})
into the form
\begin{equation}
\label{pull'}
((\varphi_*\xi)f)\circ\varphi = \xi(f\circ\varphi) \ \ \ \ \ \ \ \ \text{or}
\ \ \ \ \ \ \ \ \varphi^*((\varphi_*\xi)f) = \xi(\varphi^*f)\,.
\end{equation}
Differential forms are treated as dual to vector fields,
\begin{equation}
\label{pull1}
<\!\varphi^*\omega\,,\,\xi\!>_x \ \doteq \ <\!\omega\,,\,\varphi_*\xi\!>_y
\ \ \ \ \ \ \ \ \text{or} \ \ \ \ \ \ \ \ <\!\varphi^*\omega\,,\,\xi\!> \
 = \ \varphi^*\!<\!\omega\,,\,\varphi_*\xi\!>\,.
\end{equation}
As a simple corollary, we deduce the relation \ $\varphi^*(df)=d(\varphi^*f)$\,:
\begin{equation}
\label{}
<\!\varphi^*(df)\,,\,\xi\!> \ = \,\varphi^*\!<\!df\,,\,\varphi_*\xi\!> \
 = \,\varphi^*((\varphi_*\xi)f) \, = \,\xi(\varphi^*f) \,
 = \ <\!d(\varphi^*f)\,,\,\xi\!>.
\end{equation}
Generalization to higher tensors is straightforward:
\begin{equation}
\label{higher}
\varphi_*(\xi\otimes\xi') =
\varphi_*\xi\otimes\varphi_*\xi'\,, \ \ \ \ \ \
\varphi^*(\omega\wedge\omega') = \varphi^*\omega\wedge\varphi^*\omega'\,,
\ \ \ \ \ \ \varphi^*(d\omega)=d(\varphi^*\omega)\,.
\end{equation}

In components, (\ref{pull1}) gives, in perfect agreement with (\ref{tens})\,,\
$(\varphi^*\omega)_i(x)=\omega_j(y)\frac{\partial y^j}{\partial x^i}$\,. Of
course, it is nothing more than a mere change of variables:
\begin{equation}
\label{}
\omega(y) = \omega_i(y)\,dy^i \ \ \Rightarrow \ \
(\varphi^*\omega)(x) = \omega(y(x)) = \omega_j(y(x))\,dy^j(x)
 = \omega_j(y(x))\,\frac{\partial y^j(x)}{\partial x^i}\,dx^i.
\end{equation}
It is instructive to recognize that underlying transformations of bases
and components are accomplished by the same matrix \
$\partial y^j/\partial x^i$\,:
\begin{equation}
\label{}
\varphi_*\partial^x_i = \frac{\partial y^j}{\partial x^i}\,\partial^y_j\ ,
\ \ \ \ \
(\varphi_*\xi)^j = \frac{\partial y^j}{\partial x^i}\,\xi^i\,, \ \ \ \ \
\varphi^*dy^j = \frac{\partial y^j}{\partial x^i}\,dx^i\,, \ \ \ \ \
(\varphi^*\omega)_i = \frac{\partial y^j}{\partial x^i}\,\omega_j\ .
\end{equation}

\vspace{.3cm}

\begin{center}
\large\textbf{Lie derivative}
\end{center}

\vspace{.1cm}

By means of the action (\ref{action})\,, vector fields send functions to
functions, $\xi:f(x)\rightarrow(\xi f)_x$\,. This fact can be used to endow
the space of vector fields on $M$ with a Lie algebra structure. The appropriate
commutator is defined by
\begin{equation}
\label{commut}
[\xi\,,\,\eta]f \doteq \xi(\eta f) - \eta(\xi f)\,, \ \ \ \ \ \ \ \
[\xi\,,\,\eta]^i = \xi^j\,\partial_j \eta^i - \eta^j\,\partial_j \xi^i\,,
\end{equation}
proves to be a true vector field (a first order differential operator on
functions)\,,
\begin{equation}
\label{dif}
\xi(fg)=(\xi f)\,g+f\,\xi g \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \
[\xi,\eta](fg)=([\xi,\eta]f)\,g+f\,[\xi,\eta]g
\end{equation}
obeys the Jacobi identity, and is respected by the map
$\varphi_*$\,, \ $\varphi_*[\xi\,,\,\eta]=[\varphi_*\xi\,,\,\varphi_*\eta]$\,,
as can be shown by the following computation:
\begin{multline}
\label{}
\varphi^*((\varphi_*[\xi\,,\,\eta])f) = [\xi\,,\,\eta](\varphi^*f)
 = \xi(\eta(\varphi^*f)) - \eta(\xi(\varphi^*f))
 = \xi\varphi^*((\varphi_*\eta)f) - \eta\varphi^*((\varphi_*\xi)f) \\
 = \varphi^*((\varphi_*\xi)(\varphi_*\eta)f)
  - \varphi^*((\varphi_*\eta)(\varphi_*\xi)f)
 = \varphi^*([\varphi_*\xi\,,\,\varphi_*\eta]f)\,.
\end{multline}

It turns out that both (\ref{action}),\,(\ref{commut}) can be interpreted as
particular cases of a certain universal action of vector fields on tensors
of arbitrary rank, called a Lie derivative:
\begin{equation}
\label{Lie}
L_\xi f \doteq \xi f\,, \ \ \ \ \ \ \ \ L_\xi\,\eta \doteq [\xi,\eta]\,.
\end{equation}
Generally, $L_\xi$ is a \emph{derivation} of the corresponding tensor algebra:
it respects the type of a tensor, $L_\xi : T^r_s\rightarrow T^r_s$\,,
satisfies the tensorial Leibniz rule,
\begin{equation}
\label{deriv}
L_\xi\,(T\otimes T') = (L_\xi T)\otimes T' + T\otimes L_\xi T'\,, \ \ \ \ \ \
L_\xi\,(\omega\wedge\omega')
 = (L_\xi\omega)\wedge\omega' + \omega\wedge L_\xi\omega'\,,
\end{equation}
and also commutes with contraction of tensors, e.g.,
\begin{equation}
\label{conv}
L_\xi\!<\!\omega\,,\,\eta\!> \ = \ <\!L_\xi\omega\,,\,\eta\!>
 + <\!\omega\,,\,L_\xi\eta\!>\,,
\end{equation}
whence we obtain, for example, an explicit recipe of the $L_\xi$-action on
$k$-forms\,:
\begin{equation}
\label{Lk}
(L_\xi\omega^k)\,(\eta_1,\ldots,\eta_k) = \xi(\omega^k(\eta_1,\ldots,\eta_k))
 - \sum_{i=1}^{k}\,\omega^k(\eta_1,\ldots,[\xi,\eta_i],\ldots,\eta_k)\,.
\end{equation}
Postulates (\ref{Lie})--(\ref{conv}) unambiguously define the operation $L_\xi$
on arbitrary tensors $T^r_s$\,:
\begin{equation}
\label{Lcomp}
(L_\xi T)^{i_1\ldots i_r}_{j_1\ldots j_s}
 = \xi^n\partial_n T^{i_1\ldots i_r}_{j_1\ldots j_s}
-\sum_{m=1}^{r}\,\partial_n\xi^{i_m}\,T^{i_1\ldots n\ldots i_r}_{j_1\ldots j_s}
+\sum_{m=1}^{s}\,\partial_{j_m}\xi^n\,T^{i_1\ldots i_r}_{j_1\ldots n\ldots j_s}\,.
\end{equation}
The following important properties can be verified:
\begin{equation}
\label{Liehom}
L_\xi L_\eta - L_\eta L_\xi \equiv [L_\xi\,,\,L_\eta] = L_{[\xi,\,\eta]}\ ,
\ \ \ \ \ \ \ L_\xi\circ d = d\circ L_\xi\,.
\end{equation}

At last, the \emph{interior product} \,$\iota_\xi$\,, \,a map which sends
forms $\omega^{k+1}$ to $\omega^k$\,, is defined by
\begin{equation}
\label{inner}
\iota_\xi f = 0\,, \ \ \ \ \ \ \
(\iota_\xi \omega)\,(\eta_1,\ldots,\eta_k) = \omega\,(\xi,\eta_1,\ldots,\eta_k)\,.
\end{equation}
It exhibits the following properties:
\begin{gather}
\label{iota1}
\iota_\xi(\omega^k\wedge\omega^m) = \iota_\xi\omega^k\wedge\omega^m
 + (-)^k\omega^k\wedge \iota_\xi\omega^m, \ \ \ \ \ \ \ \
\iota_\xi\circ\iota_\eta = -\iota_\eta\circ\iota_\xi\,, \\
\label{iota2}
[L_\xi\,,\,\iota_\eta] = \iota_{[\xi,\,\eta]}\,, \ \ \ \ \ \
d\circ\iota_\xi + \iota_\xi\circ d = L_\xi\,, \ \ \ \ \ \
\varphi^*\circ\iota_{\varphi_*\xi} = \iota_\xi\circ\varphi^*\,.
\end{gather}

To consider the (group) flows on manifolds, we have to examine the case of
\emph{invertible} maps $\varphi : M\rightarrow N$ in more details. In this
case, we may study both $\varphi_*$ and $\varphi^*$ uniformly, as if directed
from $M$ to $N$, by introducing a map
$\tilde\varphi : T^r_s(M)\longrightarrow T^r_s(N)$ as follows:
\begin{equation}
\label{phitild}
\tilde\varphi f = (\varphi^*)^{-1}f\,, \ \ \ \ \
\tilde\varphi\,\xi = \varphi_*\,\xi\,, \ \ \ \ \
\tilde\varphi\,\omega = (\varphi^*)^{-1}\omega\,, \ \ \ \ \
\tilde\varphi\,(a\otimes b) = (\tilde\varphi\,a\otimes\tilde\varphi\,b)\,.
\end{equation}
Actually, the component form of $\tilde\varphi$ is given by (\ref{tens})\,,
being merely a change of coordinates:
\begin{equation}
\label{}
(\tilde\varphi f)(y) = f(x(y))\,, \ \ \ \ \
\tilde\varphi\,dx^j = \frac{\partial x^j}{\partial y^i}\,dy^i\,, \ \ \ \ \
(\tilde\varphi\,\omega)_i = \frac{\partial x^j}{\partial y^i}\,\omega_j\ .
\end{equation}
It follows from (\ref{compos}) that
$\widetilde{\varphi\circ\rho} = \tilde\varphi\circ\tilde\rho$, and from
(\ref{pull1}) that $\tilde\varphi$ commutes with the contraction:
\begin{equation}
\label{}
\tilde\varphi<\!\omega\,,\,\xi\!> \ = \ (\varphi^*)^{-1}\!<\!\omega\,,\,\xi\!>
\ = \ <\!(\varphi^*)^{-1}\omega\,,\,\varphi_*\xi\!>
\ = \ <\!\tilde\varphi\omega\,,\,\tilde\varphi\xi\!>\,.
\end{equation}

Let now $\xi$ be a vector field on a manifold $M$. By solving differential
equations, we can find its \emph{integral curves} which at each point
$x$ have $\xi(x)$ as tangent vector. A \emph{flow} $\xi^t$ generated by
$\xi$ is a one-parameter Abelian group of transformations on $M$ which
describes the travelling along these integral curves:
\begin{equation}
\label{flow}
\xi^t : \ x(s)\,\rightarrow \ \xi^t x(s) = x(s+t)
\ \ \ \ \ \Longrightarrow \ \ \ \ \ \
\xi^s\xi^t = \xi^{s+t}, \ \ \ \ \xi^{-t} = (\xi^t)^{-1}\,.
\end{equation}
This definition implies $\xi^t_*\xi=\xi$ and
\begin{equation}
\label{fxit}
\frac{d}{dt}f(\xi^t x) = (\xi f)(\xi^t x) \ \ \ \ \ \ \ \Longrightarrow
\ \ \ \ \ \ \ \frac{d^n}{dt^n}f(\xi^t x) = (\xi^n\!f)(\xi^t x)\,,
\end{equation}
which results in the Taylor expansion characteristic to an exponential:
\begin{equation}
\label{expon}
f(\xi^t x) = \sum_{n=0}^{\infty}\frac{t^n}{n!}\ (\xi^n\!f)_x \ \ \ \ \ \ \
\text{or} \ \ \ \ \ \ \
(\xi^t)^* f = \sum_{n=0}^{\infty}\frac{t^n}{n!}\ \xi^n\!f\,.
\end{equation}
These expressions for $\xi^t$ generalize the well-known exponential relations
between elements of matrix groups and their Lie algebras. As a corollary, we
find that
\begin{equation}
\label{comflows}
[\xi\,,\,\eta] = 0 \ \ \ \ \ \ \Longleftrightarrow \ \ \ \ \ \
\xi^t\circ\eta^s = \eta^s\circ\xi^t\,.
\end{equation}

Now we can give a universal definition of the Lie derivative $L_\xi$ in terms
of the flow $\xi^t$:
\begin{equation}
\label{Lie2}
L_\xi = -\frac{d}{dt}\,\tilde{\xi^t}|_{t=0}\ .
\end{equation}
In particular, \,$\tilde{\xi^t} T=T \ \Leftrightarrow \ L_\xi T=0$ \,for
any tensor field $T$. Owing to the properties of $L_\xi$ and $\tilde\varphi$,
it suffices to verify (\ref{Lie2}) on functions and vectors. From
(\ref{expon}) one obtains
\begin{equation}
\label{Lie2f}
-\frac{d}{dt}\,\tilde{\xi^t}f|_{t=0} = -\frac{d}{dt}\,(\xi^{-t})^*f|_{t=0}
 = \xi f = L_\xi f\,,
\end{equation}
and, invoking also (\ref{pull'})\,,
\begin{multline}
\label{Lie2v}
(-\frac{d}{dt}\,\tilde{\xi^t}\eta)f|_{t=0}
 = -\frac{d}{dt}\,(\xi^t_*\eta)f|_{t=0}
 = -\frac{d}{dt}\,(\xi^{-t})^*\,(\eta\,(\xi^t)^*f)|_{t=0} \\
 = -\frac{d}{dt}\,(1-t\xi+\mathcal{O}(t^2))\,\eta\,
 (1+t\xi+\mathcal{O}(t^2))f|_{t=0} = \xi\eta f - \eta\xi f = (L_\xi \eta)f\,.
\end{multline}

At last, consider a map $\varphi : M\rightarrow N$\,. By definition, it sends
integral curves (on $M$) of a vector field $\xi$ to those of $\varphi_*\xi$
(on $N$)\,. This entails
\begin{equation}
\label{phiflow}
\varphi\circ\xi^t = (\varphi_*\xi)^t\circ\varphi\,, \ \ \ \ \
\tilde\varphi\circ L_\xi = L_{\varphi_*\xi}\circ\tilde\varphi\,, \ \ \ \ \
\tilde{\xi^t}\circ L_\xi = L_\xi\circ\tilde{\xi^t}\,.
\end{equation}

\vspace{.3cm}

\begin{center}
\large\textbf{Lie groups}
\end{center}

\vspace{.1cm}

A Lie group $G$ \,is a manifold supplied with a differentiable group
multiplication law. We denote a unity in a group by $e$, left and right shifts
by $L_h$ and $R_h$\,,
\begin{equation}
\label{shifts}
L_h : g\rightarrow hg\,, \ \ \ \ R_h : g\rightarrow gh\,, \ \ \ \
L_{gh} = L_g\circ L_h\,, \ \ \ \ R_{gh} = R_h\circ R_g\,, \ \ \ \
L_g\circ R_h = R_h\circ L_g\,.
\end{equation}
A vector field $A$ on $G$ is called left-invariant if $(L_g)_*A=A$\,. Such
fields constitute a subalgebra of the Lie algebra of all vector fields on
$G$\,. This subalgebra $\mathfrak{g}$ is called a Lie algebra of
the Lie group $G$\,. Equally well we may think of $\mathfrak{g}$ as a tangent
space $T_e G$ at unity: translating the vectors of this space by left shifts
just recovers left-invariant fields.

Let $\varphi : G\rightarrow H$ be a homomorphism of Lie groups. Then
$\varphi_* : \mathfrak{g}\rightarrow\mathfrak{h}$ is a Lie algebra
homomorphism. Indeed, $\varphi_*$ commutes with Lie bracket and sends
$A\in\mathfrak{g}$ to a left-invariant field also:
\begin{multline}
\label{hom}
\varphi(gh) = \varphi(g)\,\varphi(h) \ \ \ \Longrightarrow \ \ \
\varphi\circ L_g = L_{\varphi(g)}\circ\varphi \ \ \ \Longrightarrow \ \ \
\varphi_*\circ (L_g)_* = (L_{\varphi(g)})_*\circ\varphi_* \\
\ \ \ \Longrightarrow \ \ \ (L_{\varphi(g)})_*\,\varphi_*A
 = \varphi_*(L_g)_* A = \varphi_*A
\ \ \ \Longrightarrow \ \ \ \varphi_*A \in\mathfrak{h}\,.
\end{multline}
In particular, an automorphism of $G$ (isomorphism $\varphi:G\rightarrow G$)
induces an automorphism (invertible endomorphism)
$\varphi_* : \mathfrak{g}\rightarrow\mathfrak{g}$\,.

Consider a flow $A^t$ on $G$ generated by $A\in\mathfrak{g}$\,. Due to
(\ref{phiflow}) it commutes with $L_g$\,. Denoting $A^t e\doteq e^{tA}$, we
find that
\begin{equation}
\label{etA}
A^t g = A^t L_g e = L_g A^t e = g e^{tA} = R_{e^{tA}}g\,, \ \ \ \ \ \ \
e^{tA} e^{sA} = A^s e^{tA} = A^s A^t e = e^{(t+s)A}.
\end{equation}
So, an integral curve $e^{tA}$ (an orbit of unity $e$ under the flow $A^t$)
is a 1-parameter Abelian subgroup of $G$\,. In matrix groups, it turns out
to be a true exponential. For any group homomorphism $\varphi:G\rightarrow H$
one obtains
\begin{equation}
\label{phietA}
\varphi(e^{tA}) = \varphi\circ A^t e = (\varphi_* A)^t\varphi(e)
 = (\varphi_* A)^t e = e^{t\varphi_* A}\,.
\end{equation}

A homomorphism $\rho$ of a Lie group $G$ to the group of automorphisms of a
vector space $V$ is called a \emph{representation} of $G$ in $V$, or,
in other words, $G$ acts on $V$ ($G\triangleright V$)\,:
\begin{equation}
\label{reprG}
\rho(gh) = \rho(g)\circ\rho(h) \ \ \ \ \ \ \text{or} \ \ \ \ \ \
(gh)\triangleright x = g\triangleright(h\triangleright x)\,, \ \ \ \ \
x\in V\,.
\end{equation}
Due to
(\ref{hom})\,, $\rho_*$ is then a \emph{representation} of $\mathfrak{g}$
in $V$\,. Let us consider \emph{adjoint} representations of both
$G$ and $\mathfrak{g}$ \,in $\mathfrak{g}$\,. We begin with a homomorphism
A\!D\,:\,$G\rightarrow\text{Aut}G$ which associates with each $h\in G$ an
\emph{inner automorphism}
\begin{equation}
\label{int}
\text{A\!D}_{h} : g\rightarrow hgh^{-1} \ \ \ \ \ \ \Longrightarrow
\ \ \ \ \ \ \text{A\!D}_{\,h} = R_{h^{-1}}\circ L_h\,, \ \ \ \ \
\text{A\!D}_{gh} = \text{A\!D}_g\circ\text{A\!D}_{\,h}\,.
\end{equation}
The corresponding maps
\begin{equation}
\label{Adj}
(\text{A\!D}_{h})_*\doteq\text{Ad}_{\,h}
 : \mathfrak{g}\rightarrow\mathfrak{g}\,,
\ \ \ \ \ \text{Ad}_{\,h} = (R_{h^{-1}})_*\circ(L_h)_*
 = (R_{h^{-1}})_*\,, \ \ \ \ \
\text{Ad}_{\,gh} = \text{Ad}_{\,g}\circ\text{Ad}_{\,h}
\end{equation}
establish the adjoint representation of $G$ in $\mathfrak{g}$ as a
homomorphism Ad\,:
$G\rightarrow\text{Aut}\mathfrak{g}$ given explicitly by
$g\rightarrow\text{Ad}_g$ (in the matrix case, $\text{Ad}_{\,g}\,A$ is
simply $gAg^{-1}$)\,. From (\ref{phietA}) one finds
\begin{equation}
\label{getA}
g\,e^{tA} g^{-1} \equiv \text{A\!D}_g\,e^{tA} = e^{t\text{Ad}_g A}\,.
\end{equation}
Further, $\text{Ad}_*\doteq\text{ad}
:\mathfrak{g}\rightarrow\text{End}\mathfrak{g}$ \,is nothing but the
adjoint representation of $\mathfrak{g}$ given by $\text{ad}_A B=[A,B]$\,.
Indeed,
\begin{equation}
\label{adj}
\text{ad}_A = \frac{d}{dt}\,\text{Ad}_{e^{tA}}|_{t=0}
 = \frac{d}{dt}\,(R_{e^{-tA}})_*|_{t=0} = \frac{d}{dt}\,A^{-t}_*|_{t=0}
 = -\frac{d}{dt}\,\tilde{A^t}|_{t=0} = L_A = [A,\cdot\,]\,.
\end{equation}
We differentiate directly in $\text{End}\mathfrak{g}$ because it is a
vector space where addition is allowed (and tangent spaces are naturally
identified with the underlying space)\,. Owing to the property
$\text{Ad}_{e^{tA}}\circ\text{Ad}_{e^{sA}}=\text{Ad}_{e^{(t+s)A}}$\,,
linear transformations $\text{Ad}_{e^{tA}}$ with fixed $A$ form
a 1-parameter matrix group (an orbit of $\mathbf{1}$ under the
corresponding flow on $\text{Aut}\mathfrak{g}$) which can be written
down as (ordinary) exponential of its tangent vector $\text{ad}_A$\,:
\begin{equation}
\label{adjexp}
\text{Ad}_{e^{tA}} = e^{t\,\text{ad}_A}
 = \mathbf{1} + t[A,\cdot\,] + \frac{t^2}{2}[A,[A,\cdot\,]] + \ldots\ .
\end{equation}

In a dual space $\mathfrak{g}^*$ of left-invariant 1-forms $\omega$ such
that $(L_g)^*\omega=\omega$\,, the group $G$ acts (from the right) via
\emph{coadjoint} (anti)representation
$\text{Ad}^*:G\rightarrow\text{Aut}\mathfrak{g}^*$, according to
\begin{equation}
\label{Ad*}
<\!\text{Ad}^*_g\,\omega\,,\,A\!> \ \doteq \ <\!\omega\,,\,\text{Ad}_g\,A\!>\,,
\ \ \ \ \ \ \ \ \ \text{Ad}^*_{gh} = \text{Ad}^*_h\circ\text{Ad}^*_g\ .
\end{equation}
Of course, $(\text{Ad}^*)_*\doteq\text{ad}^*
:\mathfrak{g}\rightarrow\text{End}\mathfrak{g}^*$ is a coadjoint
(anti)representation of $\mathfrak{g}$\,:
\begin{equation}
\label{ad*}
\text{ad}^*_A = \frac{d}{dt}\,\text{Ad}^*_{e^{tA}}|_{t=0} \ \ \ \ \ \
\Longrightarrow \ \ \ \ \ \
<\!\text{ad}^*_A\,\omega\,,\,B\!> \ = \ <\!\omega\,,\,\text{ad}_A B\!>\,.
\end{equation}

A Lie group $G$ \,acts on a manifold $M$ from the right,
$M\triangleleft G$\,, if
\begin{equation}
\label{act}
x(gh) = (xg)h \ \ \ \ \ \text{or} \ \ \ \ \ R_{gh} = R_h\circ R_g
\ \ \ \ \ \ \ \ \ (x\triangleleft g \equiv xg \equiv R_g\,x\,, \ \ x\in M)\,.
\end{equation}
Each $A\in\mathfrak{g}$ induces a \emph{fundamental vector field} $A^*$ on
$M$ via
\begin{equation}
\label{fund}
(A^*)^t x = x e^{tA} \equiv R_{e^{tA}}\,x\ .
\end{equation}
In other words, the flow $(A^*)^t$ produces the same $x$-orbit
as the right action of $e^{tA}$
(equivalently, the vectors $A^*$ are tangent to orbits $x e^{tA}$ generated
by $A$)\,. Hence, $A\rightarrow A^*$ is a homomorphism of Lie algebras.
If $M=G$ (when $G$ acts on itself by right shifts), $A^*\equiv A$\,.
E.g., for $\text{Ad}^*$ (it acts from the right!) we can find the value of
the field $A^*$ at some $\omega\in\mathfrak{g}^*$ simply by differentiating
the orbit:
\begin{equation}
\label{ad*om}
A^*(\omega) = \frac{d}{dt}\,\text{Ad}^*_{e^{tA}}\,\omega\,|_{t=0}
 = \text{ad}^*_A\,\omega\ , \ \ \ \ \ \ \ \
A^*_i(\omega) = -C^k_{ij}\,\omega_k A^j\,.
\end{equation}
$G$-invariant functions on $M$ (they are constant on orbits, $f(xg)=f(x)$)
are annihilated by the fundamental vector fields,
$A^*f\equiv 0 \ \ \forall A\in\mathfrak{g}$\,, because, due
to (\ref{fxit})\,,
\begin{equation}
\label{A*1}
(A^*f)_x = \frac{d}{dt}\,f((A^*)^t x)|_{t=0}
 = \frac{d}{dt}\,f(x e^{tA})|_{t=0} = 0\,.
\end{equation}
Under the (right) group action, the fields $A^*$ transform as follows:
\begin{equation}
\label{RgA*}
(R_g)_*\,A^* = ((R_g)_*A)^* = (\text{Ad}_{g^{-1}}A)^*\,.
\end{equation}
Specifically, if $A^*$ is tangent to the orbit $x e^{tA}$ at a point $x$,
then $(R_g)_*\,A^*$ is tangent at $xg$ to the orbit
$(xe^{tA})g=(xg)(g^{-1}e^{tA}g)=(xg)e^{t\text{Ad}_{g^{-1}}A}$
generated by $\text{Ad}_{g^{-1}}A$\,.

%\newpage
\vspace{.3cm}

\begin{center}
\large\textbf{Hamiltonian dynamics}
\end{center}

\vspace{.1cm}

Hamiltonian dynamics on manifolds is formulated in terms of flows generated by
\emph{Hamiltonian vector fields}. The corresponding \emph{Hamilton equations}
are written as follows:
\begin{equation}
\label{dyn}
 \dot{F}(x) \doteq \frac{d}{dt}\,F(\bar{H}^t x)|_{t=0} = (\bar{H}F)_x
 \equiv \{H,F\}_x\,, \ \ \ \ \ \
\dot{x} \doteq \frac{d}{dt}\,\bar{H}^t x|_{t=0} = \bar{H}(x)\,.
\end{equation}
Here $x$ is a point of some manifold $M$ (phase space), $H$ and $F$ are
functions on $M$, and $\{\,,\,\}$ is a bilinear operation
(\emph{Poisson brackets}) producing a Hamiltonian vector field $\bar{H}$
from a given function $H(x)$. In the last relation of (\ref{dyn}) we slightly
abuse notation by differentiating points instead of their coordinates.

A natural setting for this construction is provided by \emph{symplectic
manifolds} carrying a $T_2^0$ tensor field (2-form) $\Omega$ which is closed
($d\Omega=0$) and non-degenerate:
\begin{equation}
\label{Omega}
\Omega = \frac{1}{2}\,\Omega_{ij}(x)\,dx^i\wedge dx^j\,, \ \ \ \ \
\Omega_{ij} = -\Omega_{ji}\,, \ \ \ \ \
\exists\,\Lambda^{ij}(x) : \ \Lambda^{ik}\Omega_{kj} = \delta^i_j\,, \ \ \ \
\Lambda^{ij} = -\Lambda^{ji}\,.
\end{equation}
We define $\bar{H}$ and $\{\,,\,\}$ by
\begin{equation}
\label{barH}
(dH)_i = \Omega_{ij}\bar{H}^j \ \ \ \ \ \text{or} \ \ \ \ \
\bar{H}^i = \Lambda^{ij}\,\partial_j H\,, \ \ \ \ \ \ \
\{H,F\} = \Lambda^{ij}\,\partial_i F\,\partial_j H\,,
\end{equation}
which can be cast in coordinate-free form through the use of interior
product $\iota$ (\ref{inner})\,:
\begin{equation}
\label{barH2}
dH = -\iota_{\bar H}\Omega\,, \ \ \ \ \ \ \
\{H,F\} = \bar{H}F = \ <\!dF,\bar{H}\!> \
 = -<\!\iota_{\bar F}\Omega,\bar{H}\!> \ = \ \Omega(\bar{H},\bar{F})\,.
\end{equation}
It is non-degeneracy of the symplectic form $\Omega$ (invertibility of
$\Omega_{ij}$) that enables us to associate with $dH$ a vector field
$\bar{H}$\,. Moreover, it establishes isomorphism between tangent and
cotangent spaces on $M$ relating (bijectively) 1-forms $\iota_\xi\,\Omega$
with vectors $\xi$\,:
\begin{equation}
\label{iso}
(\iota_\xi\,\Omega)_i = (\iota_\xi\,\Omega)(e_i) = \Omega(\xi,e_i)
 = -\Omega_{ij}\xi^j\,,
\ \ \ \ \ \ \xi^j = -\Lambda^{ji}(\iota_\xi\,\Omega)_i\,.
\end{equation}
Also, it follows from $d\Omega=0$ and (\ref{iota2}) that Hamiltonian flows
respect symplectic form,
\begin{equation}
\label{resp}
L_{\bar H}\Omega = (d\circ\iota_{\bar H} + \iota_{\bar H}\circ d)\,\Omega
 = -d^2 H = 0 \ \ \ \ \ \ \Longrightarrow \ \ \ \ \ \
 (\bar{H}^t)^*\Omega = \Omega\,.
\end{equation}
Any function $H(x)$ on a symplectic manifold may serve as a Hamiltonian and
generate the corresponding flow. Thus, the equality $\{H,F\} = 0$ means that
$F$ is invariant under the flow  $\bar{H}^t$\,, and vice versa.

Poisson brackets evidently obey
\begin{equation}
\label{Leib}
\{H,F\} = -\{F,H\}\,, \ \ \ \{H,FG\} = \{H,F\}G + F\{H,G\}\,,
\ \ \ \{H,\Phi(F)\} = \{H,F\}\frac{d\Phi}{dF}\,.
\end{equation}
The Jacobi property of Poisson brackets stems from $d\,\Omega=0$
via (\ref{barH2}) and (\ref{d2})\,,
\begin{multline}
\label{Jacobi}
\{F,\{H,G\}\} + \{G,\{F,H\}\} +\{H,\{G,F\}\}
 \equiv \{F,\{H,G\}\} + \text{cycle} = \bar{F}\bar{H}G + \text{cycle} \\
\!\! = \bar{H}\bar{F}G - [\bar{H},\bar{F}]G + \text{cycle}
 = \bar{H}\Omega(\bar{F},\bar{G}) - \Omega([\bar{H},\bar{F}],\bar{G})
 + \text{cycle} = d\Omega(\bar{H},\bar{F},\bar{G}) = 0
\end{multline}
and additionally shows that the Poisson structure is respected by
Hamiltonian flows,
\begin{equation}
\label{resp2}
L_{\bar H}\{F,G\} = \{L_{\bar H}F,G\} + \{F,L_{\bar H}G\}\,,
\end{equation}
and that the map $H\rightarrow\bar H$ is a homomorphism from a Poisson
algebra to a Lie algebra of Hamiltonian vector fields,
\begin{equation}
\label{homo}
\overline{\{H,F\}} = [\bar{H},\bar{F}]
\end{equation}
(acting by both parts of (\ref{homo}) on any function $G$ reproduces
the Jacobi identity)\,.

\vspace{.1cm}

A canonical example of symplectic manifold is given by a cotangent bundle
$T^*M$\,, the set of all cotangent spaces (spaces of 1-forms) over a
manifold $M$. Its points are denoted by $(p,q)$\,, with $q$ a point
on $M$ and $p=p_i\,dq^i$ a 1-form $\in T^*_qM$, and coordinates by
$(p_i,q^j)$\,. A symplectic form here is defined by
\begin{equation}
\label{can}
\Omega = dp_i\wedge dq^i\,.
\end{equation}
Obviously, it is closed and non-degenerate.
Its component presentation generalizes (\ref{Omega})\,,
\begin{gather}
\Omega = \frac{1}{2}\,(\Omega^{ij}\,dp_i\wedge dp_j
 + \Omega^i{}_j\,dp_i\wedge dq^j + \Omega_i{}^j\,dq^i\wedge dp_j
 + \Omega_{ij}\,dq^i\wedge dq^j) = dp_i\wedge dq^i\,, \\
\Omega^{ij} = \Omega_{ij} = \Lambda^{ij} = \Lambda_{ij} = 0\,, \ \ \
\Omega^i{}_j = -\Omega_j{}^i = \Lambda^i{}_j = -\Lambda_j{}^i
 = \delta^i_j\ ,
\end{gather}
with matrix indices being summed over like
$(MN)^i_j=M^{ik}N_{kj}+M^i{}_k N^k{}_j$\ .
We can now obtain explicit formulas for the Hamiltonian
vector field $\bar{H}=(\bar{H}_i,\bar{H}^j)$ (like any vector tangent
to $T^*M$, it is parametrized by doubled number of components)\,,
\begin{equation}
\label{vect}
\bar{H}_i = \Lambda_i{}^j\,\partial_j H = -\partial_i H
 \equiv -\frac{\partial H}{\partial q^i}\,, \ \ \ \ \ \ \ \
\bar{H}^j = \Lambda^j{}_i\,\partial^i H = \partial^j H
 \equiv \frac{\partial H}{\partial p_j}\ ,
\end{equation}
and for the Poisson brackets:
\begin{equation}
\label{Pois}
\{H,F\} = \bar{H}F = \bar{H}_i\,\partial^i\!F + \bar{H}^i\,\partial_i F
 = \frac{\partial H}{\partial p_i}\,\frac{\partial F}{\partial q^i}
  - \frac{\partial H}{\partial q^i}\,\frac{\partial F}{\partial p_i}\ .
\end{equation}
Now eq.\,(\ref{dyn}) produces the Hamilton equations in canonical form:
\begin{equation}
\label{Ham}
\dot{p}_i = -\frac{\partial H}{\partial q^i}\,, \ \ \ \
\dot{q}^i = \frac{\partial H}{\partial p_i}\,, \ \ \ \ \ \
\dot{F} = \{H,F\}\,.
\end{equation}

Let us now consider a more general situation when the tensor $\Lambda^{ij}(x)$
is not invertible (Poisson brackets are degenerate)\,. A symplectic form
$\Omega$ does not exist, but the Poisson structure does, and Hamiltonian
vector fields can be constructed from differentials of functions, as before
($\Lambda^{ij}$ is assumed to guarantee the
properties (\ref{Leib})--(\ref{homo}) of Poisson brackets)\,. It turns out
in this case, unlike the symplectic one, that Hamiltonian vectors span not
the entire tangent bundle $TM$\,, but only certain subspaces of each $T_xM$,
being therefore tangent to certain submanifolds
(\emph{symplectic leaves})
where the induced Poisson structure is no longer degenerate, and its inverse
is symplectic. These submanifolds constitute a symplectic
foliation on $M$. Hamiltonian vector fields span entire tangent bundles of
each symplectic leaf, and only those functions $F$ on $M$ that are constant on
each leaf result in $\{F,\cdot\}\equiv0$ (it is these functions that made the
original Poisson structure degenerate: they annihilated Poisson brackets
without being constant throughout)\,. The whole picture is consistent owing to
(\ref{homo})\,: a commutator of two Hamiltonian vector fields
tangent to a leaf is also tangent to this leaf (Hamiltonian flows do not take
out things from the leaf)\,. Therefore, symplectic leaves prove to be
common integral submanifolds for all Hamiltonian flows. They carry an induced
symplectic structure given by $\Omega(\bar{H},\bar{F})=\{H,F\}$\,.
Invertibility of the corresponding tensors when restricted to the leaves is
the result of quotienting by kernels of the relevant Poisson brackets.

A famous example of a degenerate Poisson structure is provided by the
(Lie-Berezin-Kostant-)\,Kirillov bracket on a dual space $\mathfrak{g}^*$ to
a Lie algebra $\mathfrak{g}$\,. Any element $A\in\mathfrak{g}$ by definition
specifies on $\mathfrak{g}^*$ a linear function
\begin{equation}
\label{A()}
A(\omega)= \ <\!\omega\,,A\!> \ = A^i\omega_i\,, \ \ \ \ \ \ \
e_i(\omega) = \omega_i\,.
\end{equation}
Let us define Poisson brackets of such linear functions as their Lie algebra
commutator,
\begin{equation}
\label{Kir0}
\{A(\omega),B(\omega)\} \equiv \{A,B\}(\omega) \doteq [A,B]\,(\omega)
 = \ <\!\omega\,,[A,B]\!>\,.
\end{equation}
Then the coordinate functions $\omega_i$ have the following brackets:
\begin{equation}
\label{Kir1}
\{\omega_i,\omega_j\}\doteq \ <\!\omega,[e_i,e_j]\!> \ =C^k_{ij}\,\omega_k\,.
\end{equation}
This (Kirillov) bracket extends to arbitrary functions via the last equality
in (\ref{Leib})\,,
\begin{equation}
\label{Kirf}
\{H,F\}(\omega) = C^k_{ij}\,\omega_k\,\frac{\partial H}{\partial\omega_i}
 \frac{\partial F}{\partial\omega_j}\,,
\end{equation}
whence a Hamiltonian vector field $\bar H$ is
\begin{equation}
\label{KirH}
\bar{H}_i(\omega)=-C^k_{ij}\,\omega_k\,\frac{\partial H}{\partial\omega_j}\,.
\end{equation}

The last two formulas can be also rewritten in coordinate-free form.
Identifying $T_\omega\mathfrak{g}^*$ with $\mathfrak{g}^*$, we note the following
simple equality for the differential of a linear function $A(\omega)$\,:
\begin{equation}
\label{}
<\!d(A(\omega))\,,\lambda\!> \ = A^i<\!d\omega_i\,,\lambda\!> \
 = A^i\lambda_i = \ <\!\lambda\,,A\!> \ \ \ \ \ \ \
(\lambda\in\mathfrak{g}^* \simeq T_\omega\mathfrak{g}^*)
\end{equation}
Similarly, differential of any function $H$ on $\mathfrak{g}^*$ can be
treated as a $\mathfrak{g}$-valued field $H'(\omega)$\,,
\begin{equation}
\label{F'}
<\!dH\,,\lambda\!>_\omega \ \doteq \ <\!\lambda\,,H'(\omega)\!>
\ \ \ \ \ \ \Longrightarrow \ \ \ \ \ \
H'(\omega) = \frac{\partial H}{\partial\omega_i}\,e_i\,, \ \ \ \
\omega'_i = e_i\,, \ \ \ \ A'(\omega) = A\,.
\end{equation}
In these terms, the Kirillov bracket and Hamiltonian vector field are
written as
\begin{equation}
\label{Kirf2}
\{H,F\}(\omega) = \ <\!\omega\,,[H'(\omega),F'(\omega)]\!>\,, \ \ \ \ \ \ \ \
\bar{H}(\omega) = \text{ad}^*_{H'(\omega)}\,\omega\,.
\end{equation}
The last relation is verified as follows:
\begin{equation}
\notag
<\!\bar{H},F'\!>_\omega \ = \ <\!dF,\bar{H}\!>_\omega \
 = (\bar{H}F)_\omega = \{H,F\}(\omega) = \ <\!\omega\,,[H',F']\!> \
 = \ <\!\text{ad}^*_{H'}\omega\,,F'\!>.
\end{equation}

Comparing (\ref{Kirf2}) to (\ref{ad*om})\,, we notice that $\bar{H}(\omega)$
at a point $\omega\in\mathfrak{g}^*$ coincides with the value of
fundamental vector field of coadjoint action induced by the element
$H'(\omega)\in\mathfrak{g}$\,. For linear Hamiltonians $A(\omega)$\,, when
$A'(\omega)$ is independent of $\omega$\,, the picture becomes even more
impressive: Hamiltonian vector fields $\bar{A}(\omega)$ produced from
$A(\omega)$ by the Kirillov bracket, and fundamental vector fields
$A^*(\omega)$ generated by the coadjoint action, coincide identically on
$\mathfrak{g}^*$,
\begin{equation}
\label{A'}
A\in\mathfrak{g} \ \ \ \ \Longrightarrow \ \ \ \
A(\omega)= \ <\!\omega\,,A\!>\,, \ \ \ \ \ A'(\omega) = A\,, \ \ \ \ \
\bar{A}(\omega) = A^*(\omega) = \text{ad}^*_A \omega\,,
\end{equation}
spanning the tangent spaces to coadjoint orbits.
The $\text{Ad}^*$-\,invariant functions (they are constant on the orbits)
are thus annihilated by any $\bar H$, have zero Poisson brackets with any
function, and produce zero Hamiltonian vector fields:
\begin{equation}
\label{}
F(\text{Ad}^*_g\,\omega) = F(\omega) \ \ \ \ \Longrightarrow \ \ \ \
\bar{H}F = \{H,F\} = - \bar{F}H = 0\,, \ \ \
\bar{F}(\omega) = \text{ad}^*_{F'(\omega)}\,\omega = 0\,.
\end{equation}
Due to these functions, the Kirillov bracket is degenerate on the whole
$\mathfrak{g}^*$. However, restricting to the orbits removes the
degeneracy, and a symplectic (Kirillov) form can be defined on symplectic
leaves (which are exactly coadjoint orbits) in a standard fashion:
\begin{equation}
\label{Kirom}
\Omega^{\text K}(\bar{H},\bar{F})_\omega = \{H,F\}(\omega)
 = \ <\!\omega\,,[H',F']\!>\,.
\end{equation}

Coadjoint orbits provide an important (actually, the basic) example of
\emph{Hamiltonian action} of a group on a symplectic manifold: the flows
induced by the group action prove to be the Hamiltonian ones. Namely, there
exists a Lie algebra homomorphism $A\rightarrow\tilde{A}(x)$ from
$\mathfrak{g}$ to the Poisson algebra of functions on a manifold $M$, and
the Hamiltonian $\tilde{A}(x)$ generates exactly the fundamental vector
field $A^*(x)$\,:
\begin{equation}
\label{Hamact}
\widetilde{[A,B]}\,(x) = \{\tilde{A},\tilde{B}\}(x)\,, \ \ \ \ \ \
A^*(x) = \bar{\tilde{A}}(x)\,.
\end{equation}
Unlike the $\text{Ad}^*$ example above, where explicit formulas (\ref{A'})
for $\tilde{A}$ and $A^*$ were simple enough, in the general case it may not
be so. However, for a given $x\!\in\! M$, \ $\tilde{A}(x)$ can be viewed as
a linear function on $\mathfrak{g}$\,, or an element of $\mathfrak{g}^*$.
This motivates the following definition of the \emph{moment map} \
$\mu:M\rightarrow\mathfrak{g}^*$\,:
\begin{equation}
\label{mu}
\tilde{A}(x) \doteq \ <\!\mu(x)\,,A\!> \ = A(\mu(x))
\ \ \ \ \ \ \Longrightarrow \ \ \ \ \
\tilde{A} = A\circ\mu = \mu^*\!A
\end{equation}
(in the last equality we denote the function $A(\omega)$ simply by $A$)\,.
By (\ref{A'}), on coadjoint orbits $\mu(\omega)=\omega$\,, \,
or \,$\mu=\text{id}$\,.

Consider some properties of the moment map. By definition, it relates
the Hamiltonians $A(\omega)$ and $\tilde{A}(x)$ for each $A\in\mathfrak{g}$\,,
as well as the corresponding Poisson structures\,:
\begin{equation}
\label{muP}
\mu^*\{A,B\} = \mu^*[A,B] = \widetilde{[A,B]}
 = \{\tilde{A},\tilde{B}\} = \{\mu^*\!A\,,\,\mu^*\!B\}\,.
\end{equation}
Also, it is \emph{equivariant} (commutes with the group action)\,,
\begin{equation}
\label{equivar}
\mu(xg)=\text{Ad}^*_g\ \mu(x) \ \ \ \ \ \ \text{or} \ \ \ \ \ \
\mu\circ g = \text{Ad}^*_g\circ\mu\,,
\end{equation}
because it sends fundamental vector fields (and thus the corresponding
flows) acting on $M$ to those acting on $\mathfrak{g}^*$ via \,
$\mu_*:A^*(x)\rightarrow A^*(\mu(x))=\bar{A}(\mu(x))$\,. \,Indeed, for any
$B\in\mathfrak{g}$\,,
\begin{equation}
\notag
((\mu_*A^*)B)(\mu(x)) = (A^*(\mu^*\!B))(x) = (A^* \tilde{B})(x)
 = \{\tilde{A},\tilde{B}\}(x) = [A,B](\mu(x)) = (\bar{A}B)(\mu(x))
\end{equation}
Another simple corollary of (\ref{muP}) shows that $\mu^*$ relates symplectic
forms $\Omega$ and $\Omega^{\text K}$ restricted to the corresponding orbits
on $M$ and $\mathfrak{g}^*$:
\begin{multline}
\label{muom}
\Omega\,(A^*,B^*)_x = \Omega\,(\bar{\tilde A},\bar{\tilde B})_x
 = \{\tilde{A},\tilde{B}\}\,(x) = [A,B]\,(\mu(x))
 = \Omega^{\text K}(\bar{A},\bar{B})_{\mu(x)} \\
 = \Omega^{\text K}(\mu_* A^*,\mu_* B^*)_{\mu(x)}
 = (\mu^*\Omega^{\text K})(A^*,B^*)_x \ \ \ \ \ \Longrightarrow
\ \ \ \ \mu^*\Omega^{\text K}=\Omega \ \ \text{on orbits.}
\end{multline}
Thus, the moment map $\mu$ sends $G$-\,orbits on $M$ to those on
$\mathfrak{g}^*$, together with all the Hamiltonian dynamics defined on them.

The following statement is a Hamiltonian analog of Noether's theorem. If the
dynamics on $M$ is governed by some $G$-invariant Hamiltonian $H(x)$\,, then
the moment $\mu$\,, taken as a $\mathfrak{g}^*$-\,valued function on $M$,
is conserved: $\dot\mu=0$\,. Indeed, (\ref{A*1}) means that $A^*\!H=0$
for any $A\in\mathfrak{g}$\,. Therefore,
\begin{equation}
\label{Noether}
<\!\bar{H}\mu,A\!> \ = \bar{H}<\!\mu,A\!> \ = \bar{H}\tilde{A}
=-\bar{\tilde A}H = -A^* H =0
\ \ \ \ \Longrightarrow \ \ \ \ \dot\mu = \bar{H}\mu = 0\,.
\end{equation}

\end{document}