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\begin{document}
\parindent=0cm
\section*{ 2. Stokes' theorem.}

\begin{thm}
For $(M^n,\mu)$ , $(\pa M,\pa \mu)$, $i^*:\pa M\hookrightarrow M$
and $\ome\in\mathcal{E}_c^{n-1}(M)$,


$ \hs{8em}\dis{\int_Md\ome=\int_{\pa M}\ome}$


\end{thm}

{\bf Remark.} 여기서 $\int_{\pa M}\ome$는 보다 엄밀하게 말하면
$\int_{\pa M}i^*\ome$이고 $\pa M=\emptyset$이면 $0$으로
정의한다.\\

\begin{proof}
1. (Easy case)

Suppose $supp\,\ome\subset(U,x)$, a positive coordinate chart
(single chart). Then $\ome$ can be written as
$\dis{\ome=\sum^{n}_{i=1}(-1)^{i-1}f_idx_1\we\cdot\cdot\cdot
\we\overset{\we}{dx_i}\we\cdot\cdot\cdot\we dx_n}$ with
$supp\,f_i\subset U.$

Then $\dis{d\ome=\sum_i(-1)^{i-1}\sum_j\frac{\pa f_i}{\pa
x_j}dx_j\we dx_1\we\cdot\cdot\cdot\we\overset{\we}{dx_i}
\we\cdot\cdot\cdot\we dx_n}$

여기서 $j=i$일 때만 항이 없어지지 않으므로 따라서\\

$\dis{d\ome=\sum_i\frac{\pa f_i}{\pa x_i}dx_1\we\cdot\cdot\cdot\we
dx_n}$


$\dis{\int_Md\ome=\int_{\hb^n}\sum_{i=1}^n\frac{\pa(f_i\circ x\inv
)}{\pa u_i}du_1\cdot\cdot\cdot du_n }$

$\dis{\hs{3em}=\sum_{i=1}^n \underbrace{\int_{-\infty}^{\infty}
\cdot\cdot\cdot
\int_{-\infty}^{\infty}}\underbrace{\int_{-\infty}^{0}}\frac{\pa(f_i\circ
x\inv)}{\pa u_i} du_1\cdot\cdot\cdot du_n}$

$\hs{6em}u_n,\cdot\cdot\cdot,u_2\hs{2em}u_1$

$\dis{\hs{3em}=\int_{-\infty}^{\infty}\cdot\cdot\cdot\int_{-\infty}^{\infty}\int_{-\infty}^{0}\frac{\pa(f_1\circ
x\inv)}{\pa u_1} du_1du_2\cdot\cdot\cdot du_n}$

$\dis{\hs{3em}=\int_{-\infty}^{\infty}\cdot\cdot\cdot\int_{-\infty}^{\infty}[f_1\circ
x\inv ]_{-\infty}^0du_2\cdot\cdot\cdot du_n}$

$\dis{\hs{3em}=\int_{-\infty}^{\infty}\cdot\cdot\cdot\int_{-\infty}^{\infty}(f_1\circ
x\inv )(0,u_2,\cdot\cdot\cdot,u_n)du_2\cdot\cdot\cdot du_n}$

$\dis{\hs{3em}=\int_{\pa M}i^*\ome}$\\

위 식에서 $supp\,f_i$는 $U$안에 있다는 사실로부터
$-\infty,\infty$에서의 적분값 $(f_2\circ
x\inv)(\pm\infty)=0,(f_1\circ x\inv)(-\infty)$은 모두 0임을
이용하였다.\\

2. (General case)

Choose a finite cover $\{U_i\}$ of positive coordinate chart such
that $supp\,\ome\subset\bigcup U_i$ and let $\{\rho_i\}$ be a
partition of unity subordinate to $\{U_i\}$. Then

$\dis{\int_Md\ome=\int_M\sum_i\rho_id\ome=\int_M\sum_id(\rho_i\ome)}$

$\dis{\hs{3em}=\sum_i\int_Md(\rho_i\ome)=\sum_i\int_{\pa
M}\rho_i\ome=\int_{\pa M}\sum_i\rho_i\ome=\int_{\pa M}\ome}.$
\end{proof}\\

{\bf Remark.} $\phi:N\rightarrow M$ 에 대해 integral along
$\phi$를 $\int_{\phi}\ome=\int _N \phi^*\ome $ 로 정의한다. 또한

$\phi:N^n\rightarrow M^n$에 대해 $\phi\inv(compact)=compact$일 때
$\phi$를 proper라고 한다.

$\phi$가 proper일 때 $\pa \phi=\phi\circ i:\pa N\rightarrow M $
이라 두면 $\ome\in\mathcal{E}_c^{n-1}(M)$에 대해

$\dis{\int_{\phi}d\ome=\int_{\pa \phi}\ome}$ 이 성립한다.\\\\


(증명) $\dis{\int_{\phi}d\ome=\int_N\phi^*
d\ome=\int_Nd(\phi^*\ome) =\int_{\pa N}i^*\phi^*\ome=\int_{\pa
N}(\phi\circ i )^*\ome=\int_{\pa \phi}\ome}$


\end{document}