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\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}
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\newcommand{\pa}{\partial}
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\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}
}
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\newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}}

\begin{document}
 \parindent=0cm
  \section*{Lie Bracket.}
  \begin{thm}
  Let $X$ be a $\ccn$ vector field on $M$ with $X(p)\neq 0. $ Then

  $\exists$coordinate  chart $(U,x)$ around $p$ such that $X=\frac{\pa}{\pa
  x_1}$ on $U$.
  \end{thm}

\begin{proof}
Since this is a local problem, we may assume $X$ is a $\ccn$
vector field on a neighborhood of $0\in\rb^n$ and
$X(0)=\frac{\pa}{\pa u_1}(0)$.

Define $\vphi:\rb\rightarrow\rb^n$ by
$\vphi(a_1,\cdot\cdot\cdot,a_n)=\alp(a_1,p)$ where
$p=(0,a_2,\cdot\cdot\cdot,a_n)$ and $\alp(t,p)$ is the flow of $X$
so that $\vphi_*(\frac{\pa}{\pa u_1}(a))=X(\vphi(a))$.

In particular $\vphi(\frac{\pa}{\pa
u_1}(0))=X(\vphi(0))=X(0)=\frac{\pa}{\pa u_1}(0).$

Note that $\vphi|_{\{0\}\times \rb^{n-1}}=id$ and hence
$\vphi_*(\frac{\pa}{\pa u_i}(0))=\frac{\pa}{\pa
u_i}(0),\,i=1,2,\cdot\cdot\cdot n.$

$\therefore\,\vphi_*|_0=id$ and locally $\vphi^{-1}$ gives a
desired coordinate chart.
\end{proof}\\

\begin{defn}{\it Let $X,Y$ be $\ccn$ vector fields on $M$. The Lie bracket of $X$ and $Y$, denoted by $[X,Y]$
 is a $\ccn$ vector field on $M$ defined by \\$\hs{3em}[X,Y]_pf=X_p(Yf)-Y_p(Xf)\,\,\forall p \in\ccn(p)$.}

\end{defn}


사실 $[X,Y]$가 잘 정의됨을 보이기 위해서는 다음내용을 체크해야
한다.\\

{\bf Check.} (1) $[X,Y]_p$ is a linear derivation:\\

$[X,Y]_p(f+cg)=X_p(Y(f+cg))-Y_p(X(f+cg))$

$\hs{7em}=X_p(Yf+cYg)-Y_p(Xf+cXg)$

$\hs{7em}=X_p(Yf)+cX_p(Yg)-Y_p(Xf)-cY_p(Xg) \,\,\,\,$

$\hs{7em}=[X,Y]_p(f)+\li_p(g)$.

$\li_p(fg)=X_p(Y(fg))-Y_p(X(fg))$

$\hs{5.2em} =X_p((Yf)g+f(Yg))-Y_p((Xf)g+f(Xg))$

$\hs{5.2em}=g(p)X_p(Yf)-g(p)Y_p(Xf)+f(p)X_p(Yg)-f(p)Y_p(Xg)\,\,\,$

$\hs{5.2em}=g(p)\li_p(f)+f(p)\li_p(g)$\\

(2) $\li$ is a $\ccn$ vector field :

각 $f$에 대해 $\li(f)=X(Yf)-Y(Xf)$ 에서 $f,X,Y$ 가 모두
$\ccn$이므로 $\li(f)$  역시 $\ccn$이다. 모든 $\ccn f$에 대해
$\ccn$
이므로 $\li$는 $\ccn$ vector field이다.\\

\begin{prop}
(1) $[X,Y+cZ]=\li+c[X,Z],c\in\rb.$

(2) $\li=-[Y,X].$

(3) $[fX,gY]=f(Xg)Y-g(Yf)X+fg\li.$

(4) $(Jacobi\,\,identity)\,\,\,[\li,Z]+[[Y,Z],X]+[[Z,X],Y]=0.$
\end{prop}

(증명)  $"\li=XY-YX"$로부터 직접 계산에 의해 네가지 모두 쉽게 증명할 수 있다.\\

{\bf Note.} In $\rb^n$, $[\frac{\pa}{\pa u_i},\frac{\pa}{\pa
u_j}]=0.$


  \end{document}