\documentclass[12pt ]{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*{$\displaystyle{\mathcal{L}_X}$ and $i_X$} $\cdot ~ i_X$ : 임의의 $X \in \mathcal{X}(M)$에 대해 $i_X : \mathcal{E}^p \to \mathcal{E}^{p-1}$는 다음과 같이 정의된다.\\ \hh $i_X\alp |_p := i_{X_p}(\alp_p)$\hh $\forall p \in M$(pointwise operation) \hh or equivalently, $i_X\alp (X_2, \cdots, X_p) = \alp (X,X_2, \cdots, X_p)$\\ \h Recall : $i_X$ is an antiderivation of deg -1 \hhhh\h i.e. $i_X(\alp \wedge \bet) = (i_X \alp)\wedge \bet + (-1)^p \alp \wedge(i_X\bet)$ and $i_X^2 =0$\\ $\cdot ~ \dis{\mathcal{L}_X}$ : 주어진 tensor field $K$에 대해 $X$ 방향으로의 Lie derivative $ \dis{\mathcal{L}_X}$ 은 \hh 다음과 같이 정의된다.\\ \hh $(\dis{\mathcal{L}_X}K)_p := \lim_{t \rightarrow 0} \frac{1}{t} ((\varphi_{t}^{*}K)_p - K_p$) \hh where \{$\varphi_t$\} is a flow of $X$ and $\varphi_t^{*}Y := \varphi_{t*}^{-1}Y$.\\ \h Recall : $\dis{\mathcal{L}_X}f=Xf$ and $\dis{\mathcal{L}_X}Y = [X,Y] $\\ \begin{prop} (a) $\dis{\mathcal{L}_X}: \mathcal{T} \to \mathcal{T}$ is a tensor derivation, \hhh i.e. $\dis{\mathcal{L}_X}(K\otimes L) = (\dis{\mathcal{L}_X}K)\otimes L + K\otimes (\dis{\mathcal{L}_X}L)$.\\ (b) $\forall \alp \in \mathcal{E}^1$ and $\forall Y \in \mathcal{X}, \hspace{1.5em} \dis{\mathcal{L}_X}(\alp(Y)) = (\dis{\mathcal{L}_X}\alp)Y + \alp(\dis{\mathcal{L}_X}Y)$.\\ (c) $\dis{\mathcal{L}_X} : \mathcal{E}(M) \to \mathcal{E}(M)$ is a derivation, i.e. \h $\dis{\mathcal{L}_X}(\alp \wedge \bet) = (\dis{\mathcal{L}_X}\alp)\wedge \bet + \alp\wedge(\displaystyle{\mathcal{L}_X}\bet).$\\ (d) (Cartan formula) $\displaystyle{\mathcal{L}_X}=i_{X} \circ d + d \circ i_{X}$ on $\mathcal{E}(M)$. \end{prop} \newpage \begin{proof} (a) and (c) 보통 미분의 곱에 대한 Leifniz rule 증명과 같다.\\ (b) $\displaystyle{\mathcal{L}_X}(\alp(Y)) = \lim_{t \rightarrow 0}\frac{1}{t}(\varphi_t^*(\alp(Y)) - \alp(Y)) $\\ \hhhh\hh $= \dis\lim_{t \rightarrow 0} \frac{1}{t}((\varphi_t^*\alp)(\varphi_t^*Y) - \alp(\varphi_t^*Y) + \alp(\varphi_t^*Y) - \alp(Y))$\\ \hhhh\hh $= \dis\lim_{t \rightarrow 0} \frac{1}{t}((\varphi_t^*\alp)(\varphi_t^*Y) - \alp(\varphi_t^*Y)) +\lim_{t \rightarrow 0} \frac{1}{t}(\alp(\varphi_t^*Y) - \alp(Y))$\\ \hhhh\hhhh\hh $\downarrow$(by 아래각주\footnote{$\displaystyle \lim_{t \rightarrow 0}\frac{1}{t}(\varphi_t^*\alp(\varphi_t^*Y) - \alp(\varphi_t^*Y))= \lim_{t \rightarrow 0} \frac{1}{t}(\varphi_t^*\alp - \alp)(\varphi_t^*Y)=(\displaystyle{\mathcal{L}_X}\alp)(Y)$}) \hhhh\h $\downarrow$(by 아래각주\footnote{$\displaystyle \lim_{t \rightarrow 0}\frac{1}{t}(\alp(\varphi_t^*Y)-\alp(Y))= \lim_{t \rightarrow 0}\frac{1}{t}\alp(\varphi_t^*Y - Y)= \lim_{t \rightarrow 0}\alp(\frac{1}{t}(\varphi_t^*Y - Y))$ \hhhh\hhhh\hhh = $\alp ( \dis \lim_{t \rightarrow 0}\frac{1}{t}(\varphi_t^*Y - Y))$ = $\alp(\displaystyle{\mathcal{L}_X}Y)$})\\ \hhhh\hh = \hhh$(\displaystyle{\mathcal{L}_X}\alp)(Y)$ \hhh + \hhh$\alp(\displaystyle{\mathcal{L}_X}Y)$\\ (d) It suffices to check on $\displaystyle{\mathcal{F}}$ and $\mathcal{E}^1$, since a derivation is a local operator and \h is completely determined by its action on $\displaystyle{\mathcal{F}}$ and $\mathcal{E}^1$.\\ \h $\displaystyle{\mathcal{L}_X}f = (i_X \circ d + d \circ i_X)f$ \h , $\forall f \in \displaystyle{\mathcal{E}^1} : $ \h $\because(i_X \circ d + d \circ i_X)f = i_X(df)+0 = df(X) = Xf = \displaystyle{\mathcal{L}_X}f $ \h , $\forall f \in \displaystyle{\mathcal{E}^1}$\\ \h $\forall \alp \in \mathcal{E}^1$and $\forall Y \in \mathcal{X},$\\ \h $(\displaystyle{\mathcal{L}_X}\alp)Y=\displaystyle{\mathcal{L}_X}(\alp (Y))-\alp(\displaystyle{\mathcal{L}_X}Y)=X(\alp(Y)) - \alp([X,Y]).$\\ \h $(i_X \circ d + d \circ i_X)(\alp)(Y)=i_X(d\alp)(Y) + d(i_X(\alp))(Y)$ \hhhh\hhhh\hhh$=d\alp(X,Y) + d(\alp(X))Y$ \hhhh\hhhh\hhh$=d\alp(X,Y) + Y(\alp(X))$ \hhhh\hhhh\hhh$=X(\alp(Y)) - \alp([X,Y]).$\\ \h $\therefore(\displaystyle{\mathcal{L}_X}\alp)Y = (i_X \circ d + d \circ i_X)(\alp)(Y).$\\ \end{proof} \newpage \begin{cor} $\displaystyle{\mathcal{L}_X}$ is a total derivation, i.e., $\displaystyle{\mathcal{L}_X}(\alp(X_1, \cdots, X_p)) = (\displaystyle{\mathcal{L}_X}\alp)(X_1, \cdots, X_p) + \alp(\displaystyle{\mathcal{L}_X}X_1,X_2, \cdots, X_p) + \cdots + \alp(X_1, \cdots, X_{p-1}, \displaystyle{\mathcal{L}_X}X_p)\\ ,\forall \alp \in \mathcal{E}^p$ and $X_i \in \mathcal{X}.$\\ \end{cor} \begin{proof} It suffices to show for $\alp=\alp_1 \otimes \cdots \otimes \alp_p , \hspace{1.5em} \alp_i \in \mathcal{E}^1$.\\ $\displaystyle{\mathcal{L}_X}((\alp_1 \otimes \cdots \otimes \alp_p)(X_1, \cdots, X_p))\\ =\displaystyle{\mathcal{L}_X}(\alp_1(X_1) \cdots \alp_p(X_p))\\ =\sum_{i}\alp_1(X_1) \cdots \displaystyle{\mathcal{L}_X}(\alp_i(X_i)) \cdots \alp_p(X_p)\\ \downarrow $\h(by 아래각주\footnote{$\displaystyle{\mathcal{L}_X}(\alp_i(X_i))=(\displaystyle{\mathcal{L}_X}(\alp_i))(X_i)+\alp_i(\displaystyle{\mathcal{L}_X}X_i)$})\\ $=\sum_{i}(\alp_1 \otimes \cdots \otimes \displaystyle{\mathcal{L}_X}\alp_i \otimes \cdots \otimes\alp_p)(X_1, \cdots, X_p) + \sum_{i}(\alp_1 \otimes \cdots \otimes \alp_p)(X_1, \cdots, \displaystyle{\mathcal{L}_X}X_i, \cdots, X_p)$\\ =$\displaystyle{\mathcal{L}_X}(\alp_1 \otimes \cdots \otimes \alp_p)(X_1, \cdots, X_p) + \sum_{i}\alp(X_1, \cdots, \displaystyle{\mathcal{L}_X}X_i, \cdots, X_p)$\\ \end{proof}\\ 숙제 20. Prove invariant formula of $d\ome$ using the above corollary and \hhhh Cartan formula. (Hint : Use induction.) \end{document}