共変微分の幾何代数化 - うべの時空代数

曲線座標系 $(\overset{\iota}{x})$ におけるベクトル場 $A:=\overset{\mu}{A}\underset{\mu}{\gamma}$ に， $(\overset{\iota}{x})$ における1次偏微分商作用素 $\mathrm{D}:=\underset{\mu}{\mathrm{p}}\overset{\mu}{\gamma}$ を作用させる． $\underset{\mu}{\mathrm{p}}$ はすぐ左にある関数値の関数に作用する $\overset{\mu}{x}$ による偏微分であり， $\underset{\mu}{\gamma}$ は $(\overset{\iota}{x})$ における基底ベクトルである．直線座標系 $(\overset{\iota}{X})$ を用意しその1次基底を $\overset{\iota}{\sigma}$ とするならば
$\underset{\mu}{\gamma}=\{\overset{\nu}{X}\underset{\nu}{\sigma}\}\underset{\mu}{\mathrm{p}},$
である．ただし接続がクリストッフェル記号である条件
$\underset{\mu}{\gamma}\underset{\nu}{\mathrm{p}}=\underset{\nu}{\gamma}\underset{\mu}{\mathrm{p}},$
が成り立つものとする．ここで用いる定義や公式を示しておく．
$\underset{\mu}{\gamma}\underset{\nu}{\mathrm{p}}=\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{\gamma},$
$\underset{\mu}{\gamma}\underset{\nu}{\gamma}=\underset{\mu}{\gamma}\vee\underset{\nu}{\gamma}+\underset{\mu}{\gamma}\wedge\underset{\nu}{\gamma},$
$\underset{\mu\nu}{g}=\underset{\mu}{\gamma}\vee\underset{\nu}{\gamma},$
$\underset{\mu\nu}{g}\underset{\rho}{\mathrm{p}}=\overset{\alpha}{\underset{\mu\rho}{\Gamma}}\underset{\alpha\nu}{g}+\overset{\alpha}{\underset{\nu\rho}{\Gamma}}\underset{\alpha\mu}{g},$
$\underset{\mu\alpha}{g}\overset{\nu\alpha}{g}=\underset{\mu}{\gamma}\vee\overset{\nu}{\gamma}=\overset{\nu}{\underset{\mu}{\delta}},$
$\overset{\mu}{\gamma}=\overset{\mu\nu}{g}\underset{\nu}{\gamma},$
$\overset{\mu}{\gamma}\underset{\nu}{\mathrm{p}}=-\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{\gamma},$
$\overset{\mu\nu}{g}\underset{\rho}{\mathrm{p}}=-\overset{\nu}{\underset{\alpha\rho}{\Gamma}}\overset{\mu\alpha}{g}-\overset{\mu}{\underset{\alpha\rho}{\Gamma}}\overset{\alpha\nu}{g}.$
また1次同士に限り
$a\vee b=a\odot b=\frac{ab+ba}{2},\\a\wedge b=a\otimes b=\frac{ab-ba}{2},$
が成り立つ．
$A\mathrm{D}=\{\overset{\mu}{A}\underset{\mu}{\gamma}\}\underset{\nu}{\mathrm{p}}\overset{\nu}{\gamma}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}\underset{\mu}{\gamma}+\underset{\mu}{\gamma}\underset{\nu}{\mathrm{p}}\overset{\mu}{A}\}\overset{\nu}{\gamma}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}\underset{\mu}{\gamma}+\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{\gamma}\overset{\mu}{A}\}\overset{\nu}{\gamma}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}\underset{\mu}{\gamma}+\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\underset{\mu}{\gamma}\overset{\alpha}{A}\}\overset{\nu}{\gamma}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{A}\}\underset{\mu}{\gamma}\overset{\nu}{\gamma}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{A}\}\{\underset{\mu}{\gamma}\vee\overset{\nu}{\gamma}+\underset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}\}\\=\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{A}\}\{\overset{\nu}{\underset{\mu}{\delta}}+\underset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}\}\\=\overset{\mu}{A}\underset{\mu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\mu}{\Gamma}}\overset{\alpha}{A}+\{\overset{\mu}{A}\underset{\nu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{A}\}\underset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}$
0次の部分は曲線座標系における発散であり
$A\vee\mathrm{D},$
と書ける．2次の部分は回転なのだがこのままではクリストッフェル記号が残りわかりにくい．そこで最初に戻り次のように計算を進める．
$A\mathrm{D}=\{\underset{\mu}{A}\overset{\mu}{\gamma}\}\underset{\nu}{\mathrm{p}}\overset{\nu}{\gamma}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}\overset{\mu}{\gamma}+\overset{\mu}{\gamma}\underset{\nu}{\mathrm{p}}\underset{\mu}{A}\}\overset{\nu}{\gamma}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}\overset{\mu}{\gamma}-\overset{\mu}{\underset{\alpha\nu}{\Gamma}}\overset{\alpha}{\gamma}\underset{\mu}{A}\}\overset{\nu}{\gamma}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}\overset{\mu}{\gamma}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\overset{\mu}{\gamma}\underset{\alpha}{A}\}\overset{\nu}{\gamma}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}\}\overset{\mu}{\gamma}\overset{\nu}{\gamma}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}\}\{\overset{\mu}{\gamma}\vee\overset{\nu}{\gamma}+\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}\}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}\}\{\overset{\mu\nu}{g}+\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}\}\\=\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}\}\overset{\mu\nu}{g}+\{\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}\}\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}$
0次の部分は発散なのだが計量が残りわかりにくい．2次の部分は回転であるがまだ簡単にできる． $\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}|\mu\neq\nu$ の基底の成分は
$\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\overset{\alpha}{\underset{\mu\nu}{\Gamma}}\underset{\alpha}{A}-\underset{\nu}{A}\underset{\mu}{\mathrm{p}}+\overset{\alpha}{\underset{\nu\mu}{\Gamma}}\underset{\alpha}{A}\\=\underset{\mu}{A}\underset{\nu}{\mathrm{p}}-\underset{\nu}{A}\underset{\mu}{\mathrm{p}}.$
ここで注意すると $\overset{\nu}{\gamma}\wedge\overset{\mu}{\gamma}=-\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma}$ であるため負で添え字の入れ替わたものも考慮しなければならない．クリストッフェル記号が添字の対称性により消えて直線座標系同様の式になったことに注目してほしい．回転は発散同様
$A\wedge\mathrm{D},$
と書ける．よって曲線座標系においてベクトル場に1次偏微分商作用素を作用させると
$A\mathrm{D}=A\vee\mathrm{D}+A\wedge\mathrm{D}=\overset{\mu}{A}\underset{\mu}{\mathrm{p}}+\overset{\mu}{\underset{\alpha\mu}{\Gamma}}\overset{\alpha}{A}+\underset{\mu}{A}\underset{\nu}{\mathrm{p}}\overset{\mu}{\gamma}\wedge\overset{\nu}{\gamma},$
となる．