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\usepackage{graphicx}
\usepackage{hangul}
\usepackage{amscd,amsmath}
\usepackage{amsfonts}
\usepackage{amssymb,theorem}
\usepackage{longtable}
\newcommand{\wid}{\widetilde}
\newcommand{\ml}{\mathcal{L}}
\newcommand{\hs}{\hspace}
\newcommand{\inv}{^{-1}}
\newcommand{\vphi}{\varphi}
\newcommand{\paxi}{\frac{\partial}{\partial x_i}}
\newcommand{\payj}{\frac{\partial}{\partial y_j}}
\newcommand{\paxj}{\frac{\partial}{\partial x_j}}
\newcommand{\payi}{\frac{\partial}{\partial y_i}}
\newcommand{\li}{[X,Y]}
\newcommand{\dis}{\displaystyle}
\newcommand{\disi}{\displaystyle{\sum_{i=1}^n}}
\newcommand{\disj}{\displaystyle{\sum_{j=1}^n}}
\newcommand{\pa}{\partial}
\newcommand{\Aff}{\mbox{\it Aff}}
\newcommand{\aff}{\mbox{\it aff}}
\newcommand{\cc}{\mathcal{C}}
\newcommand{\dc}{\mathcal{D}}
\newcommand{\ccn}{\mathcal{C}^{\infty}}
\newcommand{\sbb}{\mathbb{S}}
\newcommand{\rb}{\mathbb{R}}
\newcommand{\rc}{\mathcal{R}}
\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}


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\begin{document}
 \parindent=0cm
  \section*{Lie derivative and $\vphi$- related vector fields.}
$\ccn$ vector field $X$�� $X$�� flow $\alp$ �� ���� ������
��������.

$\dis{(\mathcal{L}_Xf)(p):=\lim_{h\rightarrow
0}\frac{1}{h}(f(\alp_h(p))-f(p))}$

$\hs{4.5em}$= Lie derivative of $f$ with respect to $X$.\\

�׷��� $\,\,\frac{d}{dt}|_0f\circ\alp=Xf$ �̰� ���� ���ǰ� ��Ȯ��
$\frac{d}{dt}|_0f\circ\alp$�̹Ƿ�

 $(\mathcal{L}_Xf)(p)=Xf(p)$
�̴�.




flow $\alp_h$(diffeomorphism) �� ���� $\alp_h^*$�� ������ ����
��������.

$\hs{3em}M\hs{1em}\overset{\alp_h}{\rightarrow}\hs{1em}M\hs{1em}\overset{f}{\rightarrow}\rb$

$\hs{3em}p\hs{1.4em}\mapsto\hs{0.6em}\alp_h(p)\hs{0.5em}\mapsto
f\circ\alp_h(p)=\alp_h^*f(p)$\\


�̸� $\alp_h$�� ���� $f$�� pull back�̶�� �ϰ� �Լ� f �ܿ� tensor
field���� ���밡���ϴ�. �̸� �̿��� �Ϲ������� Lie derivative
$\mathcal{L}_X\square=\dis{\lim_{h\rightarrow
0}\frac{1}{h}((\alp_h^*\square)_p-\square_p)}$�� $\square$�� flow
of $X$�� ����
 �̺����� �����Ѵ�. ( $\square=f,$forms,vector fields,tensor fields)\\

 \begin{defn}$(\mathcal{L}_XY)(p):=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}(Y_p-(\alp_{h*}Y)_p)}=\dis{\lim_{k\rightarrow 0}\frac{1}{k}((\alp_k^*Y)_p-Y_p)}$.

 ���⼭  $ \alp_h^*Y:=(\alp_h\inv)_*Y$ �̴�.\\

\end{defn}

{\bf ��.} In $\rb^2$, $X=\frac{\pa}{\pa x_1}$ ,
$Y=(b_1,b_2)=b_1\frac{\pa}{\pa x_1}+b_2\frac{\pa}{\pa x_2}$.

�׷��� $X$�� flow�� $x_1$�� ������ ���� �����̵��̹Ƿ�
$\mathcal{L}_XY$�� $Y$�� $x_1$�� ������ ���� �̺� $(\frac{\pa b_1
}{\pa x_1},\frac{\pa b_2}{\pa x_1})=\frac{\pa b_1}{\pa x_1
}\frac{\pa}{\pa x_1}+\frac{\pa b_2}{\pa x_1}\frac{\pa}{\pa x_2}$�� �ȴ�.\\


\begin{thm}

$X,Y$ are $\ccn$ vector fields on $M\Rightarrow\mathcal{L}_XY=\li$
\end{thm}
\begin{proof}
local �����̹Ƿ�  $p\in M$ �ٹ濡���� ���̸� ����ϴ�. ����
$X$�� $\rb^n$�� �ִٰ� ���� ������ ����1�� ����
$X=\frac{\pa}{\pa x_1}$ �̶�� �����ص� �����ϴ�.

Let $Y=\sum b_i\frac{\pa}{\pa x_i}$. Then clearly
$\mathcal{L}_XY=\sum\frac{\pa b_i}{\pa x_1}\frac{\pa}{\pa x_i}$
and

$[X,Y]=[\frac{\pa}{\pa x_1},\sum b_i\paxi]$

$\hs{3em}=\sum[\frac{\pa}{\pa x_1},b_i\paxi]$

$\hs{3em}=\sum(\frac{\pa b_i}{\pa x_1 }\paxi+b_i\frac{\pa}{\pa x_1
}\paxi-b_i\paxi\frac{\pa}{\pa x_1})$

$\hs{3em}=\sum\frac{\pa b_i}{\pa x_1}\paxi$.\\

���� $\mathcal{L}_XY=[X,Y]$ �̴�.
\end{proof}


\begin{cor}


(1) $\ml_XY=-\ml_YX.$

(2) $\mathcal{L}_XY$ is linear with respect to $X$ and $Y$.

(3) $\ml_X(fY)=(\ml_Xf)Y-f(\ml_XY)$.

(4) $\ml_X[Y,Z]=[\ml_XY,Z]+[Y,\ml_XZ]$.

(5) $\ml_{\li}=[\ml_X,\ml_Y]:=\ml_X\ml_Y-\ml_Y\ml_X$.
\end{cor}

{\bf ����.} Easy exercise.\\

{\bf ���� 11.} $p\in M,$ $X,$ a vector field with flow $\alp$ ��
���� ������ ���̶�.

(1) $F(t):=f(\alp_p(t))$ then $\frac{d^kF}{dt^k}(t)=(X^kf)(t)$ and
show

$\dis{f(\alp_p(t))=f(p)+\sum_{j=1}^{k}}\frac{1}{j!}(X^jf)(p)t^j+G(t)\,\,\,$
 , where $\frac{G(t)}{t^k}\rightarrow 0$ as $t\rightarrow 0$.\\

 (2) Let $Y$ be an another vector field with flow $\beta$ and let

 $\sigma(t)=\overline{\beta}(t,c(t))\,\,$ , $\,\,c(t)=\overline{\alp}(t,b(t))$

 $b(t)=\beta(t,a(t))\,\,\,$ , $\,\,a(t)=\alp(t,p)$  .

���⼭ $\overline{\alp}$�� $-X$�� flow�̴�.\\

Show $\frac{d^2}{dt^2}|_0f(\sigma(t))=2\li_pf\,\,$ or
\,\,$\gamma(t):=\sigma(\sqrt{t})\Rightarrow\frac{d\gamma}{dt}(0)=\li_pf$
.\\

$(Hint)$
$f(\sigma(t))=f(c(t))-(Yf)(c(t))t+\frac{1}{2}(Y^2f)(c(t))t^2+G(t)$,

$g(c(t))=g(b(t))-(Xg)(b(t))t+\frac{1}{2}(X^2g)(b(t))t^2+G_1(t)\cdot\cdot\cdot$
���� �̿��� ���������� �����غ���.\\

\begin{defn}
Let $\vphi:M\rightarrow N,\ccn$, X be a $\ccn$ vector field on $M$
and $Y$ be a $\ccn$ vector field on $N$. $X$ and $Y$ are {\bf
$\vphi-$related }$(X\underset{\vphi}{\sim}Y)$ if
$\vphi_*X_p=Y_{\vphi(p)}\,\,\forall p\in M.$\\

\end{defn}

\begin{prop}
Let $\vphi:M\rightarrow N,\ccn$ and $X$ be a $\ccn$ vector field
on $M$.


(1) $X$ and $\widetilde{X}$ are $\vphi$-related $\Rightarrow
\vphi(\alp(t,p))=\widetilde{\alp}(t,\vphi(p))$ where
$\alp(\wid{\alp}$,resp.) is the flow of $X(\wid{X}$,resp.).

(2) $X\underset{\vphi}{\sim}\wid{X}$,
$Y\underset{\vphi}{\sim}\wid{Y}\Rightarrow[X,Y]\underset{\vphi}{\sim}[\wid{X},\wid{Y}]$.
\end{prop}

\begin{proof}

(1)
$X\underset{\vphi}{\sim}\wid{X}\Rightarrow\vphi_*(X_p)=\wid{X}_{\vphi(p)}$
�̰� $p$�� ������ integral curve $\alp_p(t)$ �� ����

$\vphi(\alp_p(t))$�� $\vphi(p)$�� ������ $\widetilde{X}$��
integral curve�� �ȴ�. �ֳ��ϸ�


$\frac{d}{dt}(\vphi\circ\alp_p(t))=\wid{X}(\vphi(\alp_p(t)))$����
���̸� �Ǵµ�\\

$\frac{d}{dt}(\vphi\circ\alp_p)=\vphi_*\frac{d\alp}{dt}\hs{3em}$

$\hs{4.5em}=\wid{X}_{\vphi(p)}\hs{3em}$ since
$\frac{d\alp}{dt}=X_p$\\

���� uniqueness�� ����
$\vphi(\alp_p(t))=\wid{\alp}(t,\vphi(p))$�̴�.\\

$\hs{0.5em}i.e.,\hs{4em}\rb\times
M\supset\dc\,\,\,\,\,\,\overset{id\times\vphi}
{\longrightarrow}\,\,\,\,\,\,\wid{\dc}\subset\rb\times\wid{M}$

$\hs{10em}\alp\downarrow\hs{5em}\downarrow\wid{\alp}$

$\hs{10.5em}M\,\,\,\,\,\,\underset{\vphi}{\longrightarrow}\,\,\,\,\,\,\wid{M}$\\


(2) $\vphi_*(\li_p)=[\wid{X},\wid{Y}]_{\vphi(p)}$ �� ������. ����
$\forall f\in\ccn$�� ���� \\

$\vphi_*(\li_p)(f)=\li_p(f\circ\vphi)$

$\hs{6.6em}=X_pY(f\circ\vphi)-Y_p(X(f\circ\vphi))$

$\hs{6.6em}=X_p(\vphi_*Y)(f)-Y_p(\vphi_*X)(f)$

$\hs{6.6em}=X_p((\wid{Y}f)\circ\vphi)-Y_p((\wid{X}f)\circ\vphi)$

$\hs{6.6em}=\vphi_*(X_p)(\wid{Y}f)-\vphi_*(Y_p)(\wid{X}f))$

$\hs{6.6em}=\wid{X}_{\vphi(p)}(\wid{Y}f)-\wid{Y}_{\vphi(p)}(\wid{X}f)$

$\hs{6.6em}=[\wid{X},\wid{Y}]_{\vphi(p)}f$
\end{proof}

\begin{cor}
$\hs{3em}$\\(1) Let $\vphi:M\rightarrow N$ be a diffeomorphism and
$X$ be a vector field on $M$ with flow $\{\alp_t\}$. Then
$\{\vphi\circ\alp_t\circ\vphi\inv\}$ is a flow of $\vphi_*X$.

(2) Let $\vphi:M\rightarrow M$ be a diffeomorphism, then
$\vphi_*X=X\Leftrightarrow\vphi\circ\alp_t=\alp_t\circ\vphi\,\,,\,\,\forall\,t.$
\end{cor}

\begin{proof}
(1) �� �������� $\vphi(p)$������ flow $\wid{\alp}$ ��
$\vphi(\alp(t,p))$ ��
�����Ƿ�

 $\hs{1.5em}\wid{\alp}_t\circ\vphi=\vphi\circ\alp_t$ �̴�.\\

(2) $(\Rightarrow)$ $X\underset{\vphi}{\sim}X$ �̰� flow�� �� ��
$\alp_t$�� �����Ƿ� (1)�� ���� �ڸ��ϴ�

$\hs{1.5em}(\Leftarrow)$ $\vphi\circ\alp_t\circ\vphi\inv$��
$\vphi_*X$�� flow�̰�  $\alp_t$ �� $X$�� flow�ε� �� vector
field��

$\hs{1.5em}$flow�� ���� �����Ƿ�  $\vphi_*X=X$ �̴�.
\end{proof}

\begin{lem}
 $X$ generates $\{\alp_t\}$ and $Y$  generates $\{\beta_t\}$. Then

$\li=0\Leftrightarrow\alp_t\circ\beta_s=\beta_s\circ\alp_t\,\,\,\forall\,s,t.$
\end{lem}
\begin{proof}
$({\bf \Leftarrow})$
$\alp_t\circ\beta_s\circ\alp_t\inv=\beta_s$�̰� ���� ��������
(2)�� ���� $(\alp_t)_*Y=Y$ �̴�. ($\alp_t$�� diffeomorphism����
$Y$�� vector field�� ���� �ȴ�.) �׷���

$\li=\ml_XY=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}(Y-(\alp_{t*}Y))}=0$.\\

 $({\Rightarrow})$ For $\forall q\in M$, $\dis{\lim_{h\rightarrow
 0}\frac{1}{h}(Y_q-(\alp_{h*}Y)_q)}=0$. Let $c(t)=(\alp_{t*}Y)_p\,\,\,,$
 then

 $c'(t)=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}(c(t+h)-c(t))}=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}((\alp_{(t+h)*}Y)_p-(\alp_{t*}Y)_p)}$

$\hs{2em}=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}((\alp_{t}\circ\alp_h)_*Y)_p-(\alp_{t*}Y)_p)}=\dis{\lim_{h\rightarrow
 0}\frac{1}{h}((\alp_{t_*}\circ\alp_{h_*})Y)_p-(\alp_{t*}Y)_p)}$

 $\hs{2em}=\alp_{t*}\{\dis{\lim_{h\rightarrow
 0}\frac{1}{h}((\alp_{h*}Y)_{\alp_{-t}(p)}-Y_{\alp_{-t}(p)})}\}=0$\\

 $\therefore c(t)=c(0)\,\,and\,\alp_{t*}Y=Y$�̹Ƿ�  ���������� (2)����
 ���� $\alp_t\circ\beta_s=\beta_s\circ\alp_t$ �̴�.
\end{proof}\\

{\bf Recall.} $\rb^n$ ���� $[\paxi,\paxj]=0$ �̴�.\\

���� �����κ���
���� ���� ������ ���� �� �ִ�.\\

\begin{thm} Let $X_1,X_2,\cdot\cdot\cdot,X_k$ be linearly
independent $\ccn$ vector fields in a neighborhood of $p\in M$. If
$[X_i,X_j]=0\,\,,\,\,\forall i,j=1,2,\cdot\cdot, k$. Then

$\exists$ a coordinate chart $(U,x)$ around $p$ such that
$X_i=\paxi\,\,,\,\,i=1,2,\cdot\cdot\cdot,k.$
\end{thm}

\begin{proof}
local problem �̹Ƿ� $\rb^n$�󿡼� ���� �ȴ�. $p=0$���� �ΰ�
linear change of coordinate�� ���� $0$������
$X_i(0)=\frac{\pa}{\pa u_i}(0),i=1,\cdot\cdot,k$ �� �ǰ� �� ��
�ִ�.

�� $X_i$��  flow��  $\{\alp_t^i\}$ �� ����. �׸��� �Լ�
$\vphi:\rb^n\rightarrow\rb^n$�� ������ ���� �ش�.

$\vphi(x_1,x_2,\cdot\cdot\cdot,x_n)=\alp_{x_1}^1(\alp_{x_2}^2(\cdot\cdot(\alp_{x_k}^k(0,\cdot\cdot,0,x_{k+1},\cdot
\cdot x_n))\cdot\cdot))=\alp_{x_1}^1(\overline{x})$, ���⼭
$\overline{x}=\alp_{x_2}^2(\cdot\cdot(\alp_{x_k}^k(0,\cdot\cdot,0,x_{k+1},\cdot
\cdot x_n))\cdot\cdot)$.  �׷���

 $\vphi_*\frac{\pa}{\pa u_1
}(x_1,\cdot\cdot\cdot,x_n)=\frac{d}{dt}|_{t=x_1}\alp_t^1(\overline{x})=X_1(\alp^1(x_1,\overline{x}))=X_1(\vphi(x))$.

$[X_1,X_2]=0$ �̹Ƿ� ���� ���������� ����
$\vphi(x)=\alp_{x_1}^1(\alp_{x_2}^2\cdot\cdot)=\alp_{x_2}^2(\alp_{x_1}^1\cdot\cdot)$


�̴�. ���� ���� ������� $\vphi_*\frac{\pa}{\pa
u_2}(x)=X_2(\vphi(x))$ �̴�.

�� �Ϲ������� $\vphi_*\frac{\pa}{\pa
u_i}(x)=X_i(\vphi(x))\,\,,\,\,\forall i=1,2,\cdot\cdot, k$��
�����Ѵ�. ���� $\vphi_*\frac{\pa}{\pa
u_i}(0)=X_i(0)=\frac{\pa}{\pa u_i}(0),i=1,\cdot\cdot,k
 $ .

 �׸���
$i=k+1,\cdot\cdot,n$�� ���ؼ��� $\vphi|_{0\times\rb^{n-k}}=id$��
 �ǹǷ�

    $\vphi_*\frac{\pa}{\pa u_j}(0)=\frac{\pa}{\pa
    u_j}(0)\,,\,j=k+1,\cdot\cdot,n$.

 ���� $ \vphi_*(0)=id$�� �Ǿ�  $\vphi\inv$ �� $p$�ٹ濡����  coordinate
 chart�� �ְ� �� �ٹ濡�� $\paxi=X_i\,,\,i=1,2,\cdot\cdot,k$ �̴�.

\end{proof}
\end{document}
