Riemann tensor の指標の対称性

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\title{Riemann tensor の対称性}
\author{
　}
\date{1999年07月07日改訂}
\begin{document}
\maketitle
\begin{abstract}

\end{abstract}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Riemann tensor の対称性}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Riemann tensor:
\be
R^{\mu}{}_{\sigma\beta\alpha}=
\partial_{\beta}\Gamma^{\mu}_{\sigma\alpha}-
\partial_{\alpha}\Gamma^{\mu}_{\sigma\beta}+
\Gamma^{\mu}_{\rho\beta}\Gamma^{\rho}_{\sigma\alpha}-
\Gamma^{\mu}_{\rho\alpha}\Gamma^{\rho}_{\sigma\beta}
\ee

ここで
\be
\Gamma^{\sigma}_{\mu\nu}=
\frac{1}{2}g^{\rho\sigma}\left(
\partial_{\mu}g_{\nu\rho}+
\partial_{\nu}g_{\mu\rho}-
\partial_{\rho}g_{\nu\mu}\right)
\ee

\bigskip

自由落下座標系（局所慣性系）では，$\Gamma^{\sigma}_{\mu\nu}$の
全ての成分がゼロになる。

\be
\nabla_{\sigma}g_{\mu\nu}\equiv 0
\ee

より
\be
\Gamma^{\sigma}_{\mu\nu}=0\Leftrightarrow 
\partial_{\sigma}g_{\mu\nu}=0
\ee

つねに自由落下座標系を局所的にとることができる。

また，局所的に
\be
g_{\mu\nu}=\eta_{\mu\nu}
\ee

とすることができる。

\bigskip

この座標系で Riemann tensor （第一指標を下げたもの）は
\bea
R_{\lambda\sigma\beta\alpha}&=&
g_{\lambda\mu}R^{\mu}_{\sigma\beta\alpha} \nn
&=&
g_{\lambda\mu}\partial_{\beta}\Gamma^{\mu}_{\sigma\alpha}-
g_{\lambda\mu}\partial_{\alpha}\Gamma^{\mu}_{\sigma\beta} \nn
&=&
\frac{1}{2}\partial_{\beta}
\left(\partial_{\sigma}g_{\alpha\lambda}+
\partial_{\alpha}g_{\sigma\lambda}-
\partial_{\lambda}g_{\alpha\sigma}
\right)-
\frac{1}{2}\partial_{\alpha}
\left(\partial_{\sigma}g_{\beta\lambda}+
\partial_{\beta}g_{\sigma\lambda}-
\partial_{\lambda}g_{\beta\sigma}
\right)  \nn
&=&
-\frac{1}{2}\left(
\partial_{\lambda}\partial_{\beta}g_{\alpha\sigma}-
\partial_{\lambda}\partial_{\alpha}g_{\beta\sigma}+
\partial_{\sigma}\partial_{\alpha}g_{\lambda\beta}-
\partial_{\sigma}\partial_{\beta}g_{\lambda\alpha}
\right)
\eea

\bigskip

この表式で以下の対称性は明らか。

\be
R_{\alpha\beta\gamma\delta}=R_{\gamma\delta\alpha\beta}
\label{eq:sym1}
\ee

\be
R_{\alpha\beta\gamma\delta}=-R_{\beta\alpha\gamma\delta}=
-R_{\alpha\beta\delta\gamma}
\label{eq:sym2}
\ee

\be
R_{\alpha\beta\gamma\delta}+R_{\alpha\delta\beta\gamma}+
R_{\alpha\gamma\delta\beta}=0
\label{eq:sym3}
\ee

\bigskip

$R_{\alpha\beta\gamma\delta}$はテンソルなので，
この対称性は一般の座標系において成り立つ。

\begin{quote}
\noindent
\underline{exercise}

恒等式
\be
\left[\nabla_{\lambda},
\left[\nabla_{\mu},\nabla_{\nu}\right]\right]\phi+
\left[\nabla_{\nu},
\left[\nabla_{\lambda},\nabla_{\mu}\right]\right]\phi+
\left[\nabla_{\mu},
\left[\nabla_{\nu},\nabla_{\lambda}\right]\right]\phi=0
\ee

から，(\ref{eq:sym3}) を導け。
\end{quote}

\begin{quote}
\noindent
\underline{exercise}

ある点$X^{\mu}_0$で局所慣性系をとる。
このときその近傍$X^{\mu}_0+\xi^{\mu}$での計量は
\be
g_{\mu\nu}(X^{\mu}_0+\xi^{\mu})=\eta_{\mu\nu}-\frac{1}{3}
R_{\mu\lambda\nu\sigma}(X^{\mu}_0)\xi^{\lambda}\xi^{\sigma}+O(\xi^3)
\ee

となる。これを確かめよ。
\end{quote}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Bianchi identity}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Bianchi identity:
\be
\nabla_{\lambda}R_{\alpha\beta\mu\nu}+
\nabla_{\nu}R_{\alpha\beta\lambda\mu}+
\nabla_{\mu}R_{\alpha\beta\nu\lambda}=0
\label{eq:Bianchi}
\ee

この恒等式も，自由落下座標系において具体的に示すことができる。

\begin{quote}
\noindent
\underline{exercise}

恒等式
\be
\left[\nabla_{\lambda},
\left[\nabla_{\mu},\nabla_{\nu}\right]\right]A^{\alpha}+
\left[\nabla_{\nu},
\left[\nabla_{\lambda},\nabla_{\mu}\right]\right]A^{\alpha}+
\left[\nabla_{\mu},
\left[\nabla_{\nu},\nabla_{\lambda}\right]\right]A^{\alpha}=0
\ee

および(\ref{eq:sym3})から，(\ref{eq:Bianchi}) を導け。
\end{quote}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{divergence-free tensor}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Bianchi identity において，$g^{\alpha\mu}$をかけて縮約をとると
\be
\nabla_{\lambda}R_{\beta\nu}-\nabla_{\nu}R_{\beta\lambda}+
\nabla_{\mu}R^{\mu}{}_{\beta\nu\lambda}=0
\ee

ここで
\be
R_{\beta\delta}\equiv g^{\alpha\gamma}R_{\alpha\beta\gamma\delta}
\ee

は Ricci tensor。

\bigskip

さらに，$g^{\beta\nu}$をかけて縮約をとると
\be
\nabla_{\lambda}R-\nabla_{\nu}R^{\nu}{}_{\lambda}-
\nabla_{\mu}R^{\mu}{}_{\lambda}=0
\ee

すなわち
\be
\nabla_{\lambda}R-2\nabla_{\mu}R^{\mu}{}_{\lambda}=0
\ee

ここで
\be
R\equiv g^{\mu\nu}R_{\mu\nu}
\ee

はスカラー曲率。

\bigskip

書き換えると
\be
\nabla_{\mu}\left(R^{\mu}{}_{\lambda}-
\frac{1}{2}\delta^{\mu}_{\lambda}R\right)=0
\ee

あるいは
\be
\nabla_{\mu}\left(R^{\mu\nu}-
\frac{1}{2}g^{\mu\nu}R\right)=0
\ee

\bigskip

Einstein tensor $G_{\mu\nu}$ を次のように定義する。
\be
G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R
\ee

このとき
\be
\nabla_{\mu}G^{\mu\nu}=0
\ee

を満たすことがわかる。

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Riemann tensor の独立な成分}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Riemann tensor の独立な成分の数：四次元の場合}

四次元の場合，地道に数えてみる。

\bigskip

$R_{\alpha\beta\gamma\delta}$において，
$\alpha$と$\beta$は反対称なので，
$\alpha$と$\beta$の組のとりうる場合の数は，
\be
\frac{4\times 3}{2}=6
\ee

$\gamma$と$\delta$の組についても同様で，$6$通り。

\bigskip

$R_{\alpha\beta\gamma\delta}$を$R_{(\alpha\beta)(\gamma\delta)}$
とみると，$(\alpha\beta)$と$(\gamma\delta)$について対称なので，
可能な場合の数は
\be
\frac{6\times 7}{2}=21
\ee

\bigskip

これでおわりか？まだまだ。

\bigskip

ここまでで可能な指標は
\bea
(\alpha\beta\gamma\delta)&=&
(0101),(0102),(0103),(0112),(0113),(0123),\nn
& &(0202),(0203),(0212),(0213),(0223),
(0303),\nn
& &(0312),(0313),(0323),
(1212),(1213),(1223),\nn
& &(1313),(1323),(2323)
\eea

の$21$個であるが，このうち
\be
R_{0123}-R_{0213}+R_{0312}=0
\ee

なので，一つは独立ではない
（また，これ以外には従属関係はない）。

\bigskip

したがって，四次元の場合，Riemann tensor の独立な成分の数は$20$個である。


\subsection{Riemann tensor の独立な成分の数：$D$次元の場合}

効率よく数えよう。

\bigskip

条件(\ref{eq:sym2})だけからは，可能な数は
\be
\frac{D(D-1)}{2}\times\frac{D(D-1)}{2}=\frac{D^2(D-1)^2}{4}
\ee

だけある。

条件式(\ref{eq:sym3})の数は，
\be
D\times\frac{D(D-1)(D-2)}{3!}=\frac{D^2(D-1)(D-2)}{6}
\ee

個ある。したがって，独立な成分の個数は
\be
\frac{D^2(D-1)^2}{4}-\frac{D^2(D-1)(D-2)}{6}=\frac{D^2(D^2-1)}{12}
\ee

である。

\bigskip

実は，性質(\ref{eq:sym1})は(\ref{eq:sym2})と(\ref{eq:sym3})から
導くことができる。

\begin{quote}
\noindent
\underline{exercise}

このことを示せ。
\end{quote}


結局，$D$次元においては，Riemann tensor の独立成分の個数は
\be
\frac{D^2(D^2-1)}{12}
\ee

である。

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{2次元におけるRiemann tensor}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\mu, \nu=1, 2$とする。

\bigskip

２次元において独立な Riemann tensor の成分は
\be
R_{1212}
\ee

だけである。

\bigskip

Ricci tensor の成分は
\bea
R_{11}&=&g^{22}R_{2121}=g^{22}R_{1212} \nn
R_{22}&=&g^{11}R_{1212} \nn
R_{12}&=&R_{21} \nn
&=&g^{21}R_{2112}=-g^{12}R_{1212}
\eea

スカラー曲率は
\bea
R&=&g^{11}R_{11}+g^{22}R_{22}+2g^{12}R_{12} \nn
&=&2g^{11}g^{22}R_{1212}-2g^{12}g^{12}R_{1212} \nn
&=&2\left(g^{11}g^{22}-g^{12}g^{12}\right)R_{1212}
\eea

ところで
\bea
g^{11}&=&\frac{1}{g}g_{22} \nn
g^{22}&=&\frac{1}{g}g_{11} \nn
g^{12}&=&-\frac{1}{g}g_{12}
\eea

ここで
\be
g=g_{11}g_{22}-g_{12}g_{12}
\ee

なので
\bea
R_{11}&=&\frac{1}{g}g_{11}R_{1212} \nn
R_{22}&=&\frac{1}{g}g_{22}R_{1212} \nn
R_{12}&=&\frac{1}{g}g_{12}R_{1212}
\eea

また
\be
g^{11}g^{22}-g^{12}g^{12}=\frac{1}{g}
\ee

なので
\be
R=\frac{2}{g}R_{1212}
\ee

\bigskip

以上のことから

\be
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=0
\ee

\be
R_{\mu\nu\rho\sigma}=\frac{1}{2}
R\left(g_{\mu\rho}g_{\nu\sigma}-
g_{\mu\sigma}g_{\nu\rho}\right)
\ee


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{3次元におけるRiemann tensor}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\mu, \nu=1, 2, 3$とする。

\bigskip

３次元において独立な Riemann tensor の成分は
\be
R_{1212},~~R_{1223},~~R_{1231},~~
R_{2323},~~R_{2331},~~R_{3131}
\ee

の$6$個である。

\bigskip

独立な Ricci tensor の成分は
\bea
R_{11}&=&g^{22}R_{2121}+g^{33}R_{3131}+g^{23}R_{2131}+g^{32}R_{3121} \nn
&=&g^{22}R_{1212}+g^{33}R_{3131}-2g^{23}R_{1231} \nn
R_{22}&=&g^{11}R_{1212}+g^{33}R_{3232}+g^{31}R_{3212}+g^{13}R_{1232} \nn
&=&g^{11}R_{1212}+g^{33}R_{2323}-2g^{31}R_{1223} \nn
R_{33}&=&g^{11}R_{1313}+g^{22}R_{2323}+g^{12}R_{1323}+g^{21}R_{2313} \nn
&=&g^{11}R_{3131}+g^{22}R_{2323}-2g^{12}R_{2331} \nn
R_{12}&=&g^{21}R_{2112}+g^{23}R_{2132}+g^{31}R_{3112}+g^{33}R_{3132} \nn
&=&-g^{12}R_{1212}+g^{23}R_{1223}+g^{13}R_{1231}-g^{33}R_{2331} \nn
R_{23}&=&g^{12}R_{1223}+g^{11}R_{1213}+g^{32}R_{3223}+g^{31}R_{3213} \nn
&=&g^{12}R_{1223}-g^{11}R_{1231}-g^{23}R_{2323}+g^{13}R_{2331} \nn
R_{31}&=&g^{22}R_{2321}+g^{12}R_{1321}+g^{13}R_{1331}+g^{23}R_{2331} \nn
&=&-g^{22}R_{1223}+g^{12}R_{1231}-g^{13}R_{3131}+g^{23}R_{2331}
\eea

\bigskip

自由落下座標系（局所慣性系）を用いる。
さらに，$g_{\mu\nu}(P)=\delta_{\mu\nu}$とする。

\be
\left(\begin{array}{c}
R_{11} \\
R_{22} \\
R_{33} \\
R_{12} \\
R_{23} \\
R_{31}
\end{array}\right)
=
\left(\begin{array}{cccccc}
0 & 1 & 1 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1
\end{array}\right)
\left(\begin{array}{c}
R_{2323} \\
R_{3131} \\
R_{1212} \\
R_{2331} \\
R_{3112} \\
R_{1223}
\end{array}\right)
\ee

\be
\left(\begin{array}{c}
R_{2323} \\
R_{3131} \\
R_{1212} \\
R_{2331} \\
R_{3112} \\
R_{1223}
\end{array}\right)
=
\left(\begin{array}{cccccc}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\
\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1
\end{array}\right)
\left(\begin{array}{c}
R_{11} \\
R_{22} \\
R_{33} \\
R_{12} \\
R_{23} \\
R_{31}
\end{array}\right)
\ee

\bigskip

このことから，次のことがわかるか？
\bea
R_{\alpha\beta\gamma\delta}&=&
R_{\alpha\gamma}g_{\beta\delta}+
R_{\beta\delta}g_{\alpha\gamma}-
R_{\alpha\delta}g_{\beta\gamma}-
R_{\beta\gamma}g_{\alpha\delta} \nn
&-&\frac{1}{2}R\left(
g_{\alpha\gamma}g_{\beta\delta}-
g_{\alpha\delta}g_{\beta\gamma}
\right)
\eea


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\begin{thebibliography}{99}

\bibitem{Sasaki}
佐々木節　一般相対論　産業図書

\end{thebibliography}

\end{document}
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