ハミルトンのリッチフロー

この記事では, ハミルトンのリッチフローを紹介する.

$\R^2$ に次のリーマン計量を入れる.
\[ g(t)=\dfrac{dx^2+dy^2}{e^{4t}+x^2+y^2}.\] $\d_1=\f{\d}{\d x},\ \ \d_2=\f{\d}{\d y},\ \ R_{ij}:=\Ric(\d_i, \d_j)$ とおくとき, 次が成り立つ. \[ \dfrac{\d g_{ij}}{\d t}=-2R_{ij} \ \cdots (*).\] この偏微分方程式をハミルトンのリッチフローという.
ただし, $g_{ij}$ は $g(t)$ を次のように行列で表したものである.
\[ (g_{ij})= \begin{pmatrix} 1/(e^{4t}+x^2+y^2) & 0 \\ 0 & 1/(e^{4t}+x^2+y^2) \\ \end{pmatrix}. \] ちなみに, $(g_{ij})$ の逆行列 $(g^{ij})$ については,
\[ (g^{ij})= \begin{pmatrix} e^{4t}+x^2+y^2 & 0 \\ 0 & e^{4t}+x^2+y^2 \\ \end{pmatrix}. \] 今から $(*)$ が成り立つことを確かめていく.

まず, $\G^k_{ij}$ を求めるために, 次の公式を思い出そう.
以下, アインシュタインの縮約記法を使う.
\[ \G^k_{ij}=\dfrac{1}{2}g^{kl}(\d_ig_{jl}+\d_jg_{il}-\d_lg_{ij}).\] ここで, $\d_1=\f{\d}{\d x}, \d_2=\f{\d}{\d y}$ として, $\G^k_{ij}$ を計算すると,
\[ \eq{ \G^1_{11} & =-\G^1_{22}=\G^2_{12} =\G^2_{21}=\dfrac{-x}{e^{4t}+x^2+y^2}, \\ \G^1_{12} & =\G^1_{21} =-\G^2_{11} =\G^2_{22} =\dfrac{-y}{e^{4t}+x^2+y^2}. }\]

▼ $\G_{ij}^k$ の計算過程

\[\eq{ \G^1_{11} & =\dfrac{1}{2}g^{11}(\d_1g_{11} +\d_1g_{11}-\d_1g_{11}) +\f{1}{2}g^{12}(\d_1g_{12}+\d_1g_{12}-\d_2g_{11}) \\ & =\dfrac{1}{2}g^{11}\cdot\d_1g_{11}+0 \\ &=\dfrac{1}{2}g^{11}\cdot\dfrac{-2x}{(e^{4t}+x^2+y^2)^2} \\ &=\dfrac{-x}{e^{4t}+x^2+y^2}. }\] $\G^1_{12}=\dfrac{1}{2}g^{11}(\d_1g_{21}+\d_2g_{11}-\d_1g_{12}) +\dfrac{1}{2}g^{12}(\d_1g_{22}+\d_2g_{12}-\d_2g_{12})$
$\ \ \ \ \ \ =\dfrac{1}{2}g^{11}(0+\d_2g_{11}-0)+0 $
$\ \ \ \ \ \ =\dfrac{1}{2}g^{11}\cdot \dfrac{-2y}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ =\dfrac{-y}{e^{4t}+x^2+y^2}$
$\G^1_{22}=\dfrac{1}{2}g^{11}(\d_2g_{21}+\d_2g_{21}-\d_1g_{22}) +\dfrac{1}{2}g^{12}(\d_2g_{22}+\d_2g_{22}-\d_2g_{22})$
$\ \ \ \ \ \ =\dfrac{1}{2}g^{11}(0+0-\d_1g_{22})+0 $
$\ \ \ \ \ \ =\dfrac{1}{2}g^{11}\cdot \dfrac{2x}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ =\dfrac{x}{e^{4t}+x^2+y^2}$
$\G^2_{11}=\dfrac{1}{2}g^{21}(\d_1g_{11}+\d_1g_{11}-\d_1g_{11}) +\dfrac{1}{2}g^{22}(\d_1g_{12}+\d_1g_{12}-\d_2g_{11})$
$\ \ \ \ \ \ =0+\dfrac{1}{2}g^{22}(0+0-\d_2g_{11}) $
$\ \ \ \ \ \ =\dfrac{1}{2}g^{22}\cdot \dfrac{2y}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ =\dfrac{y}{e^{4t}+x^2+y^2}$
$\G^2_{12}=\dfrac{1}{2}g^{21}(\d_1g_{21}+\d_2g_{11}-\d_1g_{12}) +\dfrac{1}{2}g^{22}(\d_1g_{22}+\d_2g_{12}-\d_2g_{12})$
$\ \ \ \ \ \ =0+\dfrac{1}{2}g^{22}(\d_1g_{22}+0-0) $
$\ \ \ \ \ \ =\dfrac{1}{2}g^{22}\cdot \dfrac{-2x}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ =\dfrac{-x}{e^{4t}+x^2+y^2}$
$\G^2_{22}=\dfrac{1}{2}g^{21}(\d_2g_{21}+\d_2g_{21}-\d_1g_{22}) +\dfrac{1}{2}g^{22}(\d_2g_{22}+\d_2g_{22}-\d_2g_{22})$
$\ \ \ \ \ \ =0+\dfrac{1}{2}g^{22}\cdot\d_2g_{22} $
$\ \ \ \ \ \ =\dfrac{1}{2}g^{22}\cdot\dfrac{-2y}{(e^{4t}+x^2+y^2)^2}$
$\ \ \ \ \ \ =\dfrac{-y}{e^{4t}+x^2+y^2}$

レビチビタ接続のとき, $\G^k_{ij}=\G^k_{ji}$ であることを思い出して, 以上の計算結果をまとめると,
\[\eq{ \G^1_{11} & =-\G^1_{22}=\G^2_{12} =\G^2_{21}=\dfrac{-x}{e^{4t}+x^2+y^2}, \\ \G^1_{12} & =\G^1_{21}=-\G^2_{11} =\G^2_{22}=\dfrac{-y}{e^{4t}+x^2+y^2}. }\]

$\G^k_{ij}$ が求まったので, 次は $R_{ijk}^{\ \ \ \ \ l}$ を求める. 計算すると,
$R_{111}^{\h 1}=R_{222}^{\ \ \ \ \ 2}= R_{112}^{\ \ \ \ \ 1}=R_{212}^{\ \ \ \ \ 2}=0,$
$R_{211}^{\h 2}=R_{122}^{\ \ \ \ \ 1}= \dfrac{2e^{4t}}{(e^{4t}+x^2+y^2)^2}$
となる.

▼ $R_{ijk}^{\ \ \ \ \ l}$ の計算過程

$R_{ijk}^{\ \ \ \ \ l}=-R_{jik}^{\ \ \ \ \ l}$ より,
$i=j$ のとき, $R_{ijk}^{\ \ \ \ \ l}=0$ である.
従って,
$\ \ \ R_{111}^{\ \ \ \ \ 1}=R_{222}^{\ \ \ \ \ 2} =R_{112}^{\ \ \ \ \ 1}=0$
となる.

$R_{211}^{\ \ \ \ \ 2},\ R_{122}^{\ \ \ \ \ 1},\ R_{212}^{\ \ \ \ \ 2}$ に関しては, 公式
$\ \ \ R_{ijk}^{\ \ \ \ \ l}= \d_i\G^l_{jk}-\d_j\G^l_{ik}+\G^m_{jk}\G^l_{im} -\G^m_{ik}\G^l_{jm}$
を用いて計算すると,

$R_{211}^{\ \ \ \ \ 2}=\d_2\G^2_{11}-\d_1\G^2_{21} +(\G^1_{11}\G^2_{21} + \G^2_{11}\G^2_{22}) -(\G^1_{21}\G^2_{11} + \G^2_{21}\G^2_{12})$
$\ \ \ \ \ \ \ =\dfrac{e^{4t}+x^2-y^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{e^{4t}-x^2+y^2}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ \ \ \ \ \ + (\dfrac{x^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{-y^2}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ \ \ \ \ - (\dfrac{-y^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{x^2}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ =\dfrac{2e^{4t}}{(e^{4t}+x^2+y^2)^2}. $

$R_{122}^{\ \ \ \ \ 1}=\d_1\G^1_{22}-\d_2\G^1_{12} +(\G^1_{22}\G^1_{11} + \G^2_{22}\G^1_{12}) -(\G^1_{12}\G^1_{21} + \G^2_{12}\G^1_{22})$
$\ \ \ \ \ \ \ =\dfrac{e^{4t}-x^2+y^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{e^{4t}+x^2-y^2}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ \ \ \ \ \ + (\dfrac{-x^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{y^2}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ \ \ \ \ - (\dfrac{y^2}{(e^{4t}+x^2+y^2)^2} + \dfrac{-x^2}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ =\dfrac{2e^{4t}}{(e^{4t}+x^2+y^2)^2}. $

$R_{212}^{\ \ \ \ \ 2}=\d_2\G^2_{12}-\d_1\G^2_{22} + (\G^1_{12}\G^2_{21}+\G^2_{12}\G^2_{22}) - (\G^1_{22}\G^2_{11}+\G^2_{22}\G^2_{12})$
$\ \ \ \ \ \ \ =\dfrac{-2xy}{(e^{4t}+x^2+y^2)^2} - \dfrac{-2xy}{(e^{4t}+x^2+y^2)^2} $
$\ \ \ \ \ \ \ \ \ \ \ + (\dfrac{xy}{(e^{4t}+x^2+y^2)^2} + \dfrac{xy}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ \ \ \ \ - (\dfrac{xy}{(e^{4t}+x^2+y^2)^2} + \dfrac{xy}{(e^{4t}+x^2+y^2)^2} )$
$\ \ \ \ \ \ \ =0. $

これで, リッチテンソルを求めるための準備ができた. 次の公式を思い出そう.
$R_{ij}=\Ric(\d_i, \d_j)=R_{ijk}^{\ \ \ \ \ i}.$
これにより,
$R_{11}=R_{i11}^{\ \ \ \ \ i} =R_{111}^{\ \ \ \ \ 1}+R_{211}^{\ \ \ \ \ 2}= \dfrac{2e^{4t}}{(e^{4t}+x^2+y^2)^2} ,$
$R_{22}=R_{i22}^{\ \ \ \ \ i} =R_{122}^{\ \ \ \ \ 1}+R_{222}^{\ \ \ \ \ 2}= \dfrac{2e^{4t}}{(e^{4t}+x^2+y^2)^2} ,$
$R_{12}=R_{i12}^{\ \ \ \ \ i}= R_{112}^{\ \ \ \ \ 1}+R_{212}^{\ \ \ \ \ 2}=0.$
リッチテンソル $\Ric(\cdot ,\ \cdot)$ は対称テンソルであるので,
$R_{21}=R_{12}=0.$

以上の結果と,
$\ \ \ \dfrac{\d g_{11}}{\d t}=\dfrac{\d g_{22}}{\d t}= \dfrac{-4e^{4t}}{(e^{4t}+x^2+y^2)^2}$
より,
$\dfrac{\d g_{ij}}{\d t}=-2R_{ij}.$