\magnification 1900
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\voffset -0.4truein
\vsize 10truein

\def\emph{}

\def\S{\Sigma_{i=1}^n}
\def\Sm{\Sigma_{i=1}^m}

\def\s{{\sigma}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\h{\hskip 10pt}
\def\u{\vskip -8pt}
\def\bh{\hfil \break}

\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}

\def\0{{\bf 0}}
\def\Z{{\bf Z}}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\p{{\bf p}}
\def\f{\phi}
\def\s{\sigma}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{ \vskip -5pt }
\def\h{\vskip -5pt \hskip 20pt}

$C^{\infty}(M) =  \{g ~|~ g^{smooth}:  M \rightarrow \R \}$

$D$ is a derivation iff $ D: C^\infty (p) \rightarrow \R$ and
 $D$ is linear and satisfies the Leibniz rule.

That is $D$ is a derivation if $D(f) \in \R$,\bh
 $D(cf) = cD(f)$, $D(f+g) = D(f) + D(g)$, \bh
$D(fg) = f(p)Dg + g(p)Df$



Defn: A {\it vector field} or {\it section of the tangent bundle} $TM$ is a smooth function
\bh
$s{:~}M\rightarrow TM$
so that $\pi\circ s={id}$ [i.e., $s(p) = (p, v_p)]$.

Ex:  If $M = \R$, let $s(p) = (p, ({\partial \over \partial x})_p)$

Sometimes we will drop the p and write $s(p) = ({\partial \over \partial 
x})_p$

Let $f \in  C^{\infty}(\R)$.  For all $p \in \R$,  $s(p)(f) = ({\partial f 
\over 
\partial x})_p
= {\partial f
\over
\partial x}(p)
$



Define $s_f:  \R \rightarrow \R$, $s_f(p) = {\partial f \over \partial
x}(p)$. ~~~~~ I.e., $s_f = {\partial f \over \partial x}$

Note $s_f$ is smooth.

Lemma 3.4.1:  For any vector field $s$ and smooth functions $f$ and $g$ on 
$M$, we have

\centerline{$s_{fg}(p)=f(p)\cdot s_{g}(p)+s_{f}(p)\cdot g(p)$}

Proof:  ${\partial (fg) \over \partial x}(p) = f(p){\partial g \over \partial 
x}(p) + {\partial f \over \partial x}(p)g(p)$ 
 


Thus we can think of a vector field as a function \bh $S: C^{\infty}(M) 
\rightarrow C^{\infty}(M)$, $S(f) = s_f$


Ex:   $S: C^{\infty}(\R) 
\rightarrow C^{\infty}(\R)$, $S(f) = {\partial f \over \partial
x}$.~~~~~~ I.e., $S = {\partial  \over \partial
x}$



Ex:  If $M = \R$, then $s(p) = a(p)({\partial \over \partial x})_p$
where $a: \R \rightarrow \R$ is a smooth function.


Let $f \in  C^{\infty}(\R)$. \bh For all $p \in \R$,  $s(p)(f) = 
a(p)({\partial f 
\over 
\partial x})_p
= a(p){\partial f
\over
\partial x}(p)
$



Define $s_f:  \R \rightarrow \R$, $s_f(p) = a(p){\partial f \over \partial
x}(p)$. ~~~~~ I.e., $s_f = a {\partial f \over \partial x}$

Note $s_f$ is smooth.

Lemma 3.4.1:  For any vector field $s$ and smooth functions $f$ and $g$ on 
$M$, we have

\centerline{$s_{fg}(p)=f(p)\cdot s_{g}(p)+s_{f}(p)\cdot g(p)$}

Proof:  ${a(p)\partial (fg) \over \partial x}(p) = a(p)f(p){\partial g \over 
\partial 
x}(p) + a(p){\partial f \over \partial x}(p)g(p)$ 
 


Thus we can think of a vector field as a function \bh $S: C^{\infty}(M) 
\rightarrow C^{\infty}(M)$, $S(f) = s_f$


Ex:   $S: C^{\infty}(\R) 
\rightarrow C^{\infty}(\R)$, $S(f) = a{\partial f \over \partial
x}$  I.e., $S = a{\partial  \over \partial
x}$

We can also write this as:

Ex:  If $M = \R^2$, then 
 $s(\p) = a(\p) ({\partial \over \partial x})_\p +  b(\p) ({\partial 
\over \partial y})_\p$
where $a, b: \R^2 \rightarrow \R$ are  smooth functions.



\end
\eject

Lemma 3.4.2:  Let $S:C^{\infty}(M)\rightarrow C^{\infty}(M)$ be linear, and suppose
$S(fg)(p)=f(p)\cdot S(g)(p)+S(f)(p)\cdot g(p)$. Then $S$ is a vector field.

Proof:  Define $S(p): C^{\infty}(M) \rightarrow \R$ to be
  the function which sends $f \in C^{\infty}(M)$
 to $S(f)(p)$, i.e, the function $S(f)$ evaluated at $p$.

Note that the hypothesis implies that $S(p)$ is linear and satisfies the Liebniz rule and hence is a derivation.



Defn:  If $A, B$ are vector fields, let $AB = A \circ B$ 

Defn:  The {\it Lie Bracket} of vector fields $A$ and $B$ is $[A, B] = AB - BA: C^{\infty}(M)\rightarrow C^{\infty}(M)$.

Thm: The Lie bracket of vector fields is a vector field.

Randell 3.3

Defn:  A {\it flow} on $M$ is a smooth
action of the Lie group ${\R}^{1}$ on $M$, $\sigma$:~${\R}%
^{1}\times M\rightarrow M$.

A flow is also called a {\it dynamical system}.


$ \sigma(t, m) = \sigma_t(m) $

$\sigma_0(m) = m $, \hfil $\sigma_t \circ \sigma_s (m) = \sigma_{t+s} (m) = \sigma_s \circ \sigma_t (m)$
 $

$\sigma_{-t} = \sigma_t^{-1}$  

$\sigma_t:  M \rightarrow M$ is a diffeomorphism.


Ex: $\sigma:  \R \times \R^2 \rightarrow \R^2$, $\sigma_t(x, y) = (x, y) + t(1, 2)$

\vfill

Defn:  The {\it orbit} of $x \in M$ = \hfil \break
$\R(x) = \{ y \in M ~|~ \exists t \in \R$ such that $y = tx\}$

\vskip 10pt
\line{A {\it flow line} is the smooth path $\alpha_p: \R \rightarrow M$, 
$\alpha_p(t) = \sigma(t, p)$.}

Prop: each $q \in M$ lies on a unique flow line.

\eject

``differentiating along the flow'':  Given a flow $\sigma$ on $M$, define $s_{\sigma}$:~$M\rightarrow TM$ by
$s_{\sigma}(p)=\left(  p,d\alpha_{p}/dt|~_{t=0}\right)  $

\vfill



Proposition 3.3.3:  $s_{\sigma}$ is a section of $TM$.






\end



$G(p) = \{[g] ~|~ g^{smooth}:  U \rightarrow \R$, for some $U^{open}$ such that $p 
\in U
\subset M \}$

$C^{\infty}(M) =  \{g ~|~ g^{smooth}:  M \rightarrow \R \}$

$Z_a = \{g ~|~ g^{smooth}:  M \rightarrow \R, Dg_a = 0 \}$

The cotangent space $T^*_a = C^{\infty}(M)/ Z_a$. 

$T^*_a$ is an $m-$dimensional vector space:




$(df)_a