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\newcommand{\bet}{\beta}
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\newtheorem{thm}{정리}
\newtheorem{cor}[thm]{따름정리}
\newtheorem{lem}[thm]{보조정리}
\newtheorem{prop}[thm]{명제}
\newtheorem{cl}{Claim}

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\newtheorem{que}{질문}
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\newtheorem{defn}{정의}
\newtheorem{rem}{주}
\newtheorem{note}{Note}
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\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}
