\documentclass[10pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \usepackage{graphicx} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} \newcommand{\dis}{\displaystyle} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^m}} \newcommand{\pa}{\partial} \newcommand{\Aff}{\mbox{\it Aff}} \newcommand{\aff}{\mbox{\it aff}} \newcommand{\cb}{\mathbb{C}} \newcommand{\cc}{\mathcal{C}} \newcommand{\ccn}{\mathcal{C}^{\infty}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\rb}{\mathbb{R}} \newcommand{\rc}{\mathcal{R}} \newcommand{\hhhh}{\hspace{4em}} \newcommand{\hhh}{\hspace{3em}} \newcommand{\hh}{\hspace{2em}} \newcommand{\h}{\hspace{1em}} \newcommand{\alp}{\alpha} \newcommand{\bet}{\beta} \newcommand{\del}{\delta} \newcommand{\gam}{\gamma} \newcommand{\vep}{\varepsilon} \newcommand{\eps}{\epsilon} \newcommand{\lam}{\lambda} \newcommand{\kap}{\kappa} \newcommand{\sig}{\sigma} \newcommand{\ome}{\omega} \newcommand{\Gam}{\Gamma} \newcommand{\Ome}{\Omega} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{proof}{{\bf 증명}}{\hfill\framebox[2mm]{}} \newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \section*{Exterior derivative} 1. $d : \mathcal{F} = \mathcal{E}^0 (M) \rightarrow \mathcal{E}^1 (M)$ is defined by $d f_p (X_p ) = X_p f$, $\forall f \in \mathcal{F}, X_p \in T_p M $ \hhhh\h$f$\hh$\mapsto$\h$df$\\ \\ {\bf Check} : $df$ 는 $\ccn$ 1-form이다.\\ \hhhh Clearly $df_p \in T_p^*M$ \h ($\because df_p(X_p + cX'_p) = df_p(X_p) + c\cdot df_p(X'_p))$ \hhhh $\Longrightarrow df = \dis{\sum} a_j dx^j $ locally on (U,x) \hhhh $\Longrightarrow a_i =df(\frac{\pa}{\pa x^i}) = \frac{\pa f}{\pa x^i} ~~: \ccn~~~(\because f ~~: \ccn)$ \hhhh $\Longrightarrow df =\dis{\sum_j} \frac{\pa f}{\pa x^j}dx^j~~~ : \ccn$ \hhhh $\therefore df \in \mathcal{E}^1(M)$\\ \\ \\ {\bf Remark}\\ (1) $d (x^i) = d x^i$ :\\ \h $d(x^i)(\frac{\pa}{\pa x^j })= \frac{\pa}{\pa x^j} (x^i) = \del_{ij} = d x^i (\frac{\pa}{\pa x^j})$ 또는 $d(x^i)(X) = X(x^i) = dx^i(X)$\\ \\ (2) $f: M \rightarrow \mathbb{R}$, $\ccn$일 때 본질적으로 $df$는 $f_*$와 같다. 보다 정확하게\\ \h $df$ :TM $\overset{^{f_*}}{\rightarrow} T\mathbb{R}\rightarrow\mathbb{R}$이 된다. \hhh $X \mapsto f_* X $ \hhhh\hh $\shortparallel$ \hhhh\hh $\square\frac{d}{dt}\mapsto \square$(This map is a canonical translation.) \h $\square = f_*(X)(t) = X(t\cdot f)=Xf=df(X)$ where $\mathbb{R} \overset{t} {\rightarrow} \mathbb{R}$ is the identity map.\\ (3) $d(f + g) = df + dg$ \h $d (fg) = g(df) + f(dg)$\\ \begin{math} \left( \begin{array}{l} d(f + g)(X_p) = X_p (f + g ) = X_p f + X_p g\\ d(fg)(X_p) = X_p (fg) = g(X_p f) + f (X_p g) = (g(df) + f(dg))(X_p) \end{array} \right) \end{math} \newpage 2. 지금까지 $d: \mathcal{E}^0 \rightarrow \mathcal{E}^1$를 정의 하였다. 이제 이를 일반화시킨 $d: \mathcal{E}^p \rightarrow \mathcal{E}^{p+1}$가 존재함을 \h 보이자.\\ \begin{thm} $d: \mathcal{F} \rightarrow \mathcal{E}^1$ can be extended uniquely to an anti-derivation of degree 1 on $\mathcal{E}$ s.t. \h $d^2 = 0$. i.e. $\exists ! \;\; d: \mathcal{E}^p \rightarrow\mathcal{E}^{p+1}$, $\forall p = 0,1,2, \cdots $s.t. \h (1) $d(\alp +\bet) = d \alp + d \bet$ \h (2) $d(\alp \wedge \bet) = d \alp \wedge \bet + (-1)^p \alp \wedge d \bet $\hh $p = deg(\alp)$ \h (3) $df$ are as in 1 for $f \in \mathcal{F}$ \h (4) $d^2 = 0$ \end{thm} \begin{proof} 우선 local하게 d를 정의하고 위의 네 가지 property와 uniqueness를 check한 다음 global하게 확장하자.\\ Local existence:\\ \h On $(U, x)$, let $\alp = \dis{\sum_I} f_I d x^I$ , where $d x^I = d x^{i_1}\wedge\cdots\wedge d x^{i_p}$. \h Define $ d \alp := \dis{\sum_I} d f_I \wedge d x^I $\\ \h (1)$\sim$(4) property를 만족하는 operator가 있다면 이렇게 정의될 수 밖에 없다. 왜냐하면\\ \h $d\alpha = \dis{\sum_I} d(f_I d x^I) = \dis{\sum_I}(df_I\wedge d x^I + f_I\wedge d(dx^I)) = \dis{\sum_I} d f_I \wedge d x^I$ \hh $\smile$(by (1)) \hh $\smile$(by (2))\hhhh \hhhh \h $\smile$(by 아래각주\footnote {$d(dx^I)=d(d x^{i_1}\wedge\cdots\wedge d x^{i_p})=\Sigma(-1)^{j+1}d x^{i_1}\wedge\cdots dd(x^{j+1})\cdots\wedge d x^{i_p}=0$ ($\because dd(x^{j+1})=0$ by (4)) }).\\ \\ \h 위에서 정의된 d는 $d^2 f = 0$, $\forall f \in \mathcal{F}$을 만족한다. \hh $\because d(d f) = d(\dis{\sum_i} \frac{\pa f}{\pa x^i} d x^i)= \dis{\sum_i}d(\frac{\pa f}{\pa x^i}) \wedge d x^i$\hh (by definition) \hhhh\hhhh\hh $= \dis{\sum_{i,j}} \frac{\pa}{\pa x^j} (\frac{\pa f}{\pa x^i}) d x^j\wedge d x^i$ \hhhh\hhhh\hh $= \dis{\sum_{i