%%\input eplain

\magnification 1400
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\voffset -0.3truein
\vsize 9.8truein

\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{\psi}
\def\f{\phi}
\def\s{\sigma}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{ \vskip -5pt }
\def\h{\vskip -5pt \hskip 20pt}
\def\l{\vskip 5pt
\hrule
\vskip -5pt}



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

$G(p) = G(p, \R)$


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

$C^{\infty}(p) =  \{g ~|~ g^{smooth}:   U \rightarrow N$, for some $U^{open}$ such that $p 
\in U
\subset M \}$



\l

$D$ 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
$D(fg) = f(p)Dg + g(p)Df$



\vskip 5pt
\hrule
\vskip -5pt

Let $\alpha:  I \rightarrow M$ where $I =$ an interval $\subset \R$, $\alpha(0) = 
p$.
Note $[\alpha] \in G[0, M]$

Directional derivative of $[g]$ in direction $[\alpha]$ = $$D_\alpha g = {d(g \circ 
\alpha) \over dt}|_{t = 0} 
\in \R$$

$D_\alpha: G(p) \rightarrow \R$ is a derivation.

\vskip 5pt
\hrule
\vskip -5pt

$T_p(M) = \{v:  G(p) \rightarrow \R ~|~ v $ is linear and satisfies the Leibniz 
rule $ \}$

 $v \in T_p(M)$ is called a {\it derivation}



\vskip 5pt
\hrule
\vskip -5pt

Given a chart  $(U, \phi)$  at $p$ where $\phi(p) = \0$,

 the {\it standard  basis} for $T_p(M)$ =  
$\{( {\partial \over \partial x_1} )_p, ..., ( {\partial \over \partial x_m} )_p  \}$, where  $({\partial \over \partial x_i} )_p= D_{\alpha_i}$ 

and  for some $\ep > 0$, $\alpha_i: (-\ep, \ep) 
\rightarrow M$, 
$\alpha_i(t) = \phi^{-1}(0, ..., t, ..., 0) $

If $v \in T_p(M)$, then $v = \Sm a_i ({\partial \over \partial x_1} )_p$ where $a_i = v([\pi_i \circ 
\phi])$


\l

\vfill
\eject

$TM= \cup_{p \in M} T_p(M) = \left\{  \left(  p,v\right)  \mid p\in M,v\in T_{p}M\right\}  $, 

let $\pi$:$~TM\rightarrow M$ be defined by $\pi(p,v)=p$. 

Let $(\phi, U)$ be a chart for $M$.

If $q \in U$, let $\{ ( {\partial \over \partial x_1} )_q, ..., ( {\partial \over \partial x_m} )_q  \}$ be the standard basis (w.r.t $(\phi, U)$)
for $T_q(M) = T_q$


$t_{\phi}$:$~\pi^{-1}(U) \rightarrow \phi(U)\times{\R}^{m} \subset \R^{2m}$,

$t_{\phi}(q, v) = (\phi(q), a_1, ..., a_m)$ where $v = \Sm a_i ( {\partial \over \partial x_i} )_q$


Let ${\cal A}$ be a maximal atlas for $M$.

Basis for topology on $TM:$  \bh
$\{W ~|~ \exists (\phi, U) \in {\cal A}$  s.t. $W \subset \pi^{-1}(U)$ and 
$ t_\phi(W)$ open in $\R^{2m} \}$

Claim:  $TM$ is a $2m-$manifold and \bh
${\cal C} = \{(t_\phi, \pi^{-1}(U)) ~|~ (\phi, U) \in {\cal A}\}$ is a 
pre-atlas for $TM$. 

 $\pi$:$~TM\rightarrow M$, $\pi(p,v)=p$ is smooth 

$df:  TM \rightarrow TN$ defined by $df(p,v) = (f(p), d_pf(v))$ is smooth if $f:  M \rightarrow N$ is smooth.

 
Proof:  See Hitchin 4.1 (in Chapter 1 of\bh
http://www2.maths.ox.ac.uk/~hitchin/hitchinnotes/hitchinnotes.html

\eject

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 $s(p) = (p, ({d\over dx})_p)$

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

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



Define $s_f:  \R \rightarrow \R$, $s_f(p) = {df \over dx}(p)$. ~~~~~ I.e., $s_f = {df \over dx}$

Note $s_f$ is smooth.


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


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

\l


Ex:  If $M = \R$, then $s(p) = a(p)({d\over dx})_p$
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)({df 
\over 
dx})_p
= a(p){df
\over
dx}(p)
$



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

Note $s_f$ is smooth.


 


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


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

\eject
In the above we used the charts $\phi_p: \R \rightarrow \R, \phi_p(x) = x - p$.  

Thus ${d(g (\phi_p^{-1}(x)) ) \over dx}|_{x = 0} $
 $ = {d(g (x + p) ) \over dx}|_{x = 0} = {dg \over dx}(p) $


Note $\phi_0(x) = \phi_p (x + p)$.  

Thus $ {d(\phi_p (\phi_0^{-1}(x) ) \over dx}|_{x = 0} 
= {d(\phi_p( \phi_p^{-1}(x+p) )) \over dx}|_{x = 0} 
= {d(x+p) \over dx}|_{x = 0} 
  = 1$


\vfil
\l

If we use the chart $\psi_q: \R \rightarrow \R, \psi_q(x) = q - x$.  

Then ${d(g (\psi_p^{-1}(x)) ) \over dx}|_{x = 0} $
 $ = {d(g (p-x) ) \over dx}|_{x = 0} = {-dg \over dx}(p) $





Note ${d(\psi_q (x + p) ) \over dx}|_{x = 0} = {d\psi_q \over dx}|_p= 
{d(q-x) \over dx}|_p = -1   $



Example of a non-smooth vector field on $\R$:


If $p \geq 0$, let $s(p) = (p, ({d\over dx})_p)$ \bh
[i.e., the basis element of $T_p(\R)$ from $\phi_p$] 

If $p < 0$, let $s(p) = (p, (-{d\over dx})_p)$ \bh
[i.e., the basis element of $T_p(\R)$ from $\psi_p$] 


\l
\vfil


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.


Ex:  Let $\{( {\partial \over \partial x_1})_p, ..., ( {\partial \over \partial x_m})_p\}$ be a basis for $T_p(M)$.


Let $s:  M \rightarrow TM$, 
$s(p) = (p, \Sm a_i(p) ({\partial \over \partial x_i})_p)$ 

\vfil
\l

Defn:  $s$ is {\it never zero} if $s(p) \not = (p, \0)$ for all $p \in M$.

Prop:  Let $G$ be a Lie group. Then $G$ admits a never-zero
vector field.

Note:  $S^n$ admits a never-zero
vector field iff $n$ odd.

\vfil
\l


\eject

Let $p_2(s(p)) = p_2(p, v_p) = v_p$

Defn:  The vector fields $s_1, ..., s_k$ are {\it linearly independent} iff for all $p 
\in M$, \bh $p_2(s_1(p)), ..., p_2(s_k(p))$ are linearly independent.



Defn:  $M$ is parallelizable (or equivalently the \emph{``tangent 
bundle }$\pi$%
\emph{:~}$TM\rightarrow M$\emph{ is trivial''}) iff $TM$ admits $m$ 
linearly
independent vector fields.

Suppose $M$ is parallelizable.  Thus for each $p \in M$, let 
$\{ v_{1, p}, ..., v_{m, p}\}$ be ANY basis for $T_p(M)$ such that
$s_i: M \rightarrow TM$, $s_i(p) = (p,  v_{i, p} )$ is a SMOOTH vector field. 


NOTE:  We can form $m$ vector fields using basis elements iff $M$ is parallelizable.  

When $M$ is parallelizable, we can define:

$t: TM \rightarrow M \times R^m $, $t(p, v) = (p, a_1, ..., a_m)$ where $v = \Sm a_i  
v_{i, p} $


Let $\rho_1:  M \times R^m  \rightarrow M$, $\rho_1(p, \x) = p$.  


$\rho_2:  M \times R^m  \rightarrow R^m$, $\rho_1 (p, \x) = \x$.  

 $\rho_1 \circ t:  TM \rightarrow M$,  $(\rho_1 \circ t)(p, v) = \pi(p, v) = p$

Recall $\pi^{-1}(p) = T_p(M) $

Prop:  $t|_{T_p(M) }: T_p(M) \rightarrow \{p\} \times R^m $ is a linear isomorphism for all $p$.

or equivalently,

$\rho_2 \circ t|_{T_p(M) }:  T_p(M)  \rightarrow R^m$ is a linear isomorphism for all $p$.  

since  $\rho_2 \circ t|_{T_p(M) } (\Sm a_i v_{i, p} )= (a_1, ..., a_m)$ 

HENCE:  $t: TM \rightarrow M \times R^m $ is a diffeomorhism.

Note that for all $p \in M$, given a basis $\{v_{1, p}, ..., v_{m, p}\}$ be a basis for $T_p(M)$, we can always define a linear isomorphim:
\vskip 5pt
\centerline{$t_p:   T_p(M)  \rightarrow R^m, ~T(\Sm a_i v_{i, p} )= (a_1, ..., a_m)$}

However, $t: TM \rightarrow M \times R^m$, $t(p, v) = (p, t_p(v))$ may not be smooth (recall example of non-smooth vector field on p. 4).
   
In general $TM$ may not be diffeomorphic to $M \times R^m $. 

\eject

Randell 3.4 The bracket of two vector fields.

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., $s(p) = (p, v_p)]$.


I.e,  $s$ takes $p \in M$ to the derivation $v_p:  C^{\infty}(M) \rightarrow \R$

Let $f \in  C^{\infty}(M)$ 


Define $s_f:  M \rightarrow \R$, $s_f(p) = v_p([f])$ where $s(p) = (p, v_p)$

Note $s_f$ is smooth.


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





\vfill

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 $s:  M \rightarrow TM$, $s(p) = (p, S_p)$ where

Define $S_p: C^{\infty}(M) \rightarrow \R$, 
$S_p(f) = S(f)(p)$, i.e, the function $S(f)$ evaluated at $p$.

Claim $S_p$ is a derivation.

Show $S_p$ is linear and satisfies the Leibniz rule.

Claim $s$ is smooth.




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.

\eject

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}.

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


\l
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.

\vfill

``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 $TM$.


\eject

Let $\alpha:  I \rightarrow M$ where $I =$ an interval $\subset \R$, $\alpha(0) = 
p$.
Note $[\alpha] \in G[0, M]$

Directional derivative of $[g]$ in direction $[\alpha]$ = $D_\alpha g = {d(g \circ 
\alpha) \over dt}|_{t = 0} 
\in \R$

$D_\alpha: G(p) \rightarrow \R$ is a derivation.


\vskip 5pt
\hrule
\vskip -5pt

Given a chart  $(U, \phi)$  at $p$ where $\phi(p) = \0$,

 the {\it standard  basis} for $T_p(M)$ =  
$\{( {\partial \over \partial x_1} )_p, ..., ( {\partial \over \partial x_m} )_p  \}$, where  $({\partial \over \partial x_i} )_p= D_{\alpha_i}$ 

and  for some $\ep > 0$, $\alpha_i: (-\ep, \ep) 
\rightarrow M$, 
$\alpha_i(t) = \phi^{-1}(0, ..., t, ..., 0) $

If $v \in T_p(M)$, then $v = \Sm a_i ({\partial \over \partial x_1} )_p$ where $a_i = v([\pi_i \circ 
\phi])$


\l



Let $(U, \phi)$ be a chart for $M$ such that $\0 \in \phi(U)$.

Suppose $q \in U$.  Choose $\ep > 0$ such that 
$B(\phi(q), \ep) \subset \phi(U)$ and $B(\0, \ep) \subset \phi(U)$.

 
Let $\tau_q:  B(\phi(q), \ep) \rightarrow B(\0, \ep) $, $ \tau_q(\x) = \x - \phi(q)$.


 the {\it standard  basis} for $T_q(M)$ with respect to  $(U, \phi)$ = 
 
\rightline{the standard  basis for $T_q(M)$ with respect to  $(\phi^{-1}(B(\0, \ep)), \tau_q \circ \phi)$  
}

Hence the standard basis (w.r.t. $(U, \phi)$) = 
$\{( {\partial \over \partial x_1} )_q, ..., ( {\partial \over \partial x_m} )_q  \}$, where  $({\partial \over \partial x_i} )_q= D_{\alpha_i}$ 


$\alpha_i: (-\ep, \ep) 
\rightarrow M$, 
$\alpha_i(t) = \phi^{-1}(\tau^{-1}(0, ..., t, ..., 0)) $
\f
\rightline{$=
\phi^{-1}(\phi_1(q), ..., \phi_{i-1}(q), \phi_{i}(q) + t, \phi_{i+1}(q), ..., \phi_{m}(q) ) $ ~~~~~~
where $\phi_i = \pi_i \circ \phi$.
}



\l

Suppose $f:  M \rightarrow \R$ is smooth.  

Recall $f$ is smooth iff for all $p \in M$, there exists a chart $(U, \phi)$ such that $p \in U$ and $f \circ \phi^{-1}:  \phi(U) \subset \R^m \rightarrow \R$ is smooth.

Claim:  ${\partial f \over \partial x_i}:  U \rightarrow \R$,  
${\partial f \over \partial x_i}(q) = ( {\partial  \over \partial x_i} )_q(f)$   
is smooth.

NOTE:  ${\partial f \over \partial x_i}:  U \rightarrow \R$ is only defined on $U$, and is NOT a globally defined function on $M$.


We will show that for $p \in U$ and w.r.t the chart $(\phi^{-1}(B(\0, \ep)), \tau_p \circ \phi)$  

 ${\partial f \over \partial x_i}  \circ (\tau_p \circ \phi)^{-1}:  B(\0, \ep)  
\subset \R^m \rightarrow \R$ is smooth. 


Let $q = (\tau_p \circ \phi)^{-1}(\x) $  %%$( \phi)^{-1}(\x + \phi(p))$


${\partial f \over \partial x_i}  \circ (\tau_p \circ \phi)^{-1}(\x) = $
%%${\partial f \over \partial x_i}(( \phi)^{-1}(\x + \phi(p)))$ 
${\partial f \over \partial x_i}(q) = $
$( {\partial  \over \partial x_i} )_q(f) = $   
$D_{\alpha_i}(f)  = {d(f \circ \alpha_i) \over dt}|_{t = 0} =  $

${d((f \circ  \phi^{-1})(\phi_1(q), ..., \phi_{i-1}(q), \phi_{i}(q) + t, \phi_{i+1}(q), ..., \phi_{m}(q) ) ) \over dt}|_{t = 0} =  $
${\partial ( f \circ  \phi^{-1}) \over \partial x_i}|_{\phi(q)} $



\vfill





















\end


  $$\commdiag{Y&\mapright^f&E\cr \mapdown&\arrow(3,2)\lft{f_t}&\mapdown\cr
     Y\times I&\mapright^{\bar f_t}&X}$$


\vskip 5pt
\hrule
\vskip -5pt

$T_p(M) = \{v:  G(p) \rightarrow \R ~|~ v $ is linear and satisfies the Leibniz 
rule $ \}$

 $v \in T_p(M)$ is called a {\it derivation}





$G(p) = \{[g] ~|~ g^{smooth}:  U \rightarrow \R$, 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