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

\def\R{{\bf R}}
\def\C{{\bf C}}
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\def\y{{\bf y}}
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\def\f{\vskip 10pt}
\def\u{\vskip -8pt}



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

$\phi(\a\x) = \x- \a$

canonical basis $\{E_{i\a} = \phi^{-1}(e_i) ~|~ i = 1, ..., n \}$

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



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

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

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



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

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

$X_\a^*(f) = \S \xi_i {\partial f \over \partial x_i}_\a$

Let $x_j:  \R^n \rightarrow \R$, $x_j(\x) = x_j$ 

$X_\a^*(x_j) = \S \xi_i {\partial x_j \over \partial x_i}_\a = \xi_i$


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

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

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:

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.

Suppose $f(\x) = 1$. Then $Df = 0$

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 10pt

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

Math 28

Let $f:  X \subset  \R^k \rightarrow \R^m$

Graph of $f = \{ (\x, f(\x) ~|~ \x \in X  \} \subset \R^k \times \R^m$





\end