\magnification 2000
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein
\voffset -0.5truein
\vsize 10truein

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

\def\R{{\bf R}}
\def\C{{\bf C}}
\def\c{{\bf c}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\d{{\bf d}}
\def\a{{\bf a}}
\def\h{{\bf h}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\0{{\bf 0}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{\vskip -8pt}

Randell 2.1

Let $p \in M$.

Suppose $g_i:  U_i \rightarrow N$, where $p \subset U_i^{open} \subset M$.

$g_1 \sim  g_2$ if $\exists V$ such that $p \in V \subset U_1 \cap U_2$ and $g_1(x) = g_2(x)$ $\forall x \in V$.

The equivalence class [g] is a {\it germ}.

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

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

$G(p)$ is an algebra over $\R$.

\vskip 5pt
\hrule 

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


Note $D_\alpha:  G(p) \rightarrow \R$ is linear and satisfies the Leibniz rule.
\vfill
\eject



Boothby 2.4

$T_\a(\R^n) = \{(\a,\x ) ~|~ \x \in \R^n \}$, $\phi(\a\x) = \x- \a$

canonical basis $\{E_{i\a} = \phi^{-1}(e_i) ~|~ i = 1, ..., n \}$
\vskip 5pt
\hrule
Let $\a \in R^n$

Suppose $f:  X \subset \R^n \rightarrow \R $ where $\a \subset X^{open} \subset M$.

$f \sim g$ if $\exists U^{open}$ s.t. $\a \in U$ and $f(x) = g(x) \forall x \in U$.

Let $C^\infty(a) = \{ [f]:  X \subset \R^n \rightarrow \R \in C^\infty ~|~ a \in dom f 
\}$



$f_i:  U_i \rightarrow \R \in C^\infty(a)$ implies $f_1 + f_2:  U_1 \cap U_2 \rightarrow \R 
\in C^\infty(a)$, $\alpha f_i:  U_i \rightarrow \R \in C^\infty(a)$,  and
 $f_1f_2:  U_1 \cap U_2 \rightarrow \R \in C^\infty(a)$, 

Thus $C^\infty(a)$ is an algebra over $\R$



\vskip 5pt
\hrule 


Let $X_\a = \S \xi_i E_{i\a}$

$X_\a^*:  C^\infty (\a) \rightarrow \R$

$X_\a^*(f) = \S \xi_i {\partial f \over \partial x_i}|_\a$ = 
%%$\del f(\a) \cdot X_\a$
directional derivative of $f$ at $\a$ in the direction of $X_\a$.

Let $x_j:  \R^n \rightarrow \R$, $x_j(\x) = x_j$ 
\hfill
$X_\a^*(x_j) = \S \xi_i {\partial x_j \over \partial x_i}|_\a = \xi_i$


$X_\a^*$ is linear and satisfies the Leibniz rule.

\eject


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

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

$D_\alpha \in T_p(M)$
  

 $T_p(M)$ is closed under addition and scalar multiplication and hence is a vector
space over $\R$


Find a basis for $T_p(M)$:


Let $(U, \phi)$ be a chart at $p$ such that $\phi:U^{open} \rightarrow \phi(U)^{open} \subset R^m$ is a homeomorphism, $\phi(p) = \0 \in R^m$

$\0 \in \phi(U)^{open}$ implies $\exists \ep > 0$ such that $B_\ep(\0) \subset \phi(U)$

Thus if $t \in (-\ep, \ep)$, then $(0, ..., t, ..., 0) \subset \phi(U)$.

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

Let $v_i = D_{\alpha_i}$.

Claim:  $\{v_1, ..., v_m\}$ is a basis for $T_p(M)$.




\eject

Let ${\cal D}(a) = \{ D: C^\infty (\a) \rightarrow \R
~|~ D$ is linear and satisfies the Leibniz rule  $\}$

$D \in {\cal D}(a)$ is called a {\it derivation}

$X_\a^* \in {\cal D}(a)$ 

%%Define $(\alpha D_1 + \beta D_2)(f) = \alpha [D_1(f)] + \beta [D_2(f)]$

${\cal D}(a)$ is closed under addition and scalar multiplication and hence is a vector 
space over $\R$


Let $j: T_\a(\R^n) \rightarrow {\cal D}(a)$, $j(X_\a) =  X_\a^*$


Claim:  $j$ is an isomorphism.

Let $X_\a = \S \xi_i E_{i\a}$ and $Z_\a = \S \zeta_i E_{i\a}$

$j$ is a homomorphism.

$j$ is 1-1:
\vskip -5pt

If $j(X_\a) = j(Z_\a)$, then   $X_\a^*(x_j) = \S \xi_i {\partial x_j \over \partial 
x_i}|_\a = \xi_i = \zeta_i = Z_\a^*(x_j)$

$j$ is onto:  Let $D$ be a derivation.
\vskip -5pt

Suppose $f(\x) = 1$. Then $Df = 0$ by product rule.
\hfil \break
Suppose $g(\x) = c$. Then $Dg = D(cf) = cDf = 0$


Let $h_i(\x) = x_i$.  Let $\xi_i = Dh_i$.  Then $D = X_\a^*$ where $X_\a = \S \xi_i 
E_{i\a}$  (proof:  long calculation, see Boothby).

\vskip 5pt

\hrule


Note since $X_\a^*(f) = \S \xi_i {\partial f \over \partial x_i}|_\a$,
$j(E_{i\a}) = E_{i\a}* = {\partial \over \partial x_i}|_\a$,$




  

  


\end


\hrule
2.5

 Let $U \subset R^n$ 

Let $E_{i\p} = \phi^{-1}_\p(\e_i)$ where $\phi_\p:  T_\p(\R^n) \rightarrow \R^n$, 
$\phi_p(\p\x) = \x - \p$


Defn: A {\it vector field} is a function, $f:  U \rightarrow T(\R^n)$, such that $f(\p) \in 
T_\p(\R^n)$


 

Suppose 

Defn:  A vector field is {\it smooth} if its components relative to the canonical basis are





\end