📂幾何学

基本形式と座標変換の関係

概要¹

座標変換 $f : V \to U$が与えられたとき、$U$上のメトリック $g$と$V$上のメトリック$\overline{g}$の関係について説明する。

アインシュタインの記法を使う。

公式

座標部分写像 $\mathbf{x} : U \to \mathbb{R}^{3}$のメトリック$g$と$\mathbf{y} = \mathbf{x} \circ f : V \to \mathbb{R}^{3}$のメトリック$\overline{g}$、そして接ベクトル $\mathbf{X} = X^{i}\mathbf{x}_{i} = \overline{X}^{\alpha} \mathbf{y}_{\alpha}$に対して、次の関係が成り立つ。

$$ \begin{align} X^{i} &= \sum\limits_{\alpha} \overline{X}^{\alpha} \dfrac{\partial u^{i}}{\partial v^{\alpha}} \\ g_{ij} &= \sum\limits_{\alpha, \beta} \overline{g}_{\alpha \beta}\dfrac{\partial v^{\alpha}}{\partial u^{i}} \dfrac{\partial v^{\beta}}{\partial u^{j}} \\ g &= \overline{g} \left( \det \begin{bmatrix} \dfrac{\partial v^{\alpha}}{\partial u^{i}} \end{bmatrix} \right)^{2} \\ g^{kl} &= \sum\limits_{\gamma, \delta} \overline{g}^{\gamma \delta}\dfrac{\partial u^{k}}{\partial v^{\gamma}} \dfrac{\partial u^{l}}{\partial u^{\delta}} \\ \end{align} $$

$$ \begin{align} \overline{X}^{\alpha} &= \sum\limits_{i} X^{i} \dfrac{\partial v^{\alpha}}{\partial u^{i}} \\ \overline{g}_{\alpha \beta} &= \sum\limits_{i, j} g_{i j}\dfrac{\partial u^{i}}{\partial v^{\alpha}} \dfrac{\partial u^{j}}{\partial v^{j}} \\ \overline{g} &= g \left( \det \begin{bmatrix} \dfrac{\partial u^{i}}{\partial v^{\alpha}} \end{bmatrix} \right)^{2} \\ \overline{g}^{\gamma \delta} &= \sum\limits_{k, l} g^{kl} \dfrac{\partial v^{\gamma}}{\partial u^{k}} \dfrac{\partial v^{\delta}}{\partial u^{l}} \end{align} $$

説明

$(1) ~ (4)$は一般的に$U$座標系上の情報を、$V$座標系上の情報にどのように表現するかを説明している。その中で、$(1), (4)$は$f : V \to U$のヤコビアンを含み、$(2), (3)$は$g = f^{-1} = U \to V$のヤコビアンを含む。伝統的に、$(1), (4)$のような変換を反変換^{contravariant transformation}と呼ぶ。$J$が$f$のヤコビアンである場合、

$$ \begin{equation} \begin{bmatrix} X^{1} \\ X^{2} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{1}}{\partial v^{2}} \\[1em] \dfrac{\partial u^{2}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \begin{bmatrix} \overline{X}^{1} \\ \overline{X}^{2} \end{bmatrix} = J \begin{bmatrix} \overline{X}^{1} \\ \overline{X}^{2} \end{bmatrix} \tag{1} \end{equation} $$

$$ \begin{equation} \begin{align*} \begin{bmatrix} g^{11} & g^{12} \\[1em] g^{21} & g^{22} \end{bmatrix} &= \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{1}}{\partial v^{2}} \\[1em] \dfrac{\partial u^{2}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \begin{bmatrix} \overline{g}^{11} & \overline{g}^{12} \\[1em] \overline{g}^{21} & \overline{g}^{22} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{1}} \\[1em] \dfrac{\partial u^{1}}{\partial v^{2}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \\ &= J \begin{bmatrix} \overline{g}^{11} & \overline{g}^{12} \\[1em] \overline{g}^{21} & \overline{g}^{22} \end{bmatrix} J^{t} \end{align*} \tag{4} \end{equation} $$

$(2)$のような変換は共変換^{covariant transformation}と呼ばれる。

$$ \begin{equation} \begin{align*} \begin{bmatrix} g_{11} & g_{12} \\[1em] g_{21} & g_{22} \end{bmatrix} &= \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{1}} \\[1em] \dfrac{\partial v^{1}}{\partial u^{2}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \begin{bmatrix} \overline{g}_{11} & \overline{g}_{12} \\[1em] \overline{g}_{21} & \overline{g}_{22} \end{bmatrix} \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{1}}{\partial u^{2}} \\[1em] \dfrac{\partial v^{2}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \\ &= (J^{-1})^{t} \begin{bmatrix} \overline{g}_{11} & \overline{g}_{12} \\[1em] \overline{g}_{21} & \overline{g}_{22} \end{bmatrix} J^{-1} \end{align*} \tag{2} \end{equation} $$

$(3)$は行列の積を含まないので変換ではない。 $(5), (6), (8)$も行列の積で表すと、

$$ \begin{equation} \begin{bmatrix} \overline{X}^{1} \\ \overline{X}^{2} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{1}}{\partial u^{2}} \\[1em] \dfrac{\partial v^{2}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \begin{bmatrix} \overline{X}^{1} \\ \overline{X}^{2} \end{bmatrix} = J^{-1} \begin{bmatrix} \overline{X}^{1} \\ \overline{X}^{2} \end{bmatrix} \tag{5} \end{equation} $$

$$ \begin{equation} \begin{align*} \begin{bmatrix} \overline{g}_{11} & \overline{g}_{12} \\[1em] \overline{g}_{21} & \overline{g}_{22} \end{bmatrix} &= \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{1}} \\[1em] \dfrac{\partial u^{1}}{\partial v^{2}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \begin{bmatrix} g_{11} & g_{12} \\[1em] g_{21} & g_{22} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{1}}{\partial v^{2}} \\[1em] \dfrac{\partial u^{2}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \\ &= J^{t} \begin{bmatrix} g_{11} & g_{12} \\[1em] g_{21} & g_{22} \end{bmatrix} J \end{align*} \tag{6} \end{equation} $$

$$ \begin{equation} \begin{align*} \begin{bmatrix} \overline{g}^{11} & \overline{g}^{12} \\[1em] \overline{g}^{21} & \overline{g}^{22} \end{bmatrix} &= \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{1}}{\partial u^{2}} \\[1em] \dfrac{\partial v^{2}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \begin{bmatrix} g^{11} & g^{12} \\[1em] g^{21} & g^{22} \end{bmatrix} \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{1}} \\[1em] \dfrac{\partial v^{1}}{\partial u^{2}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \\ &= J^{-1} \begin{bmatrix} g^{11} & g^{12} \\[1em] g^{21} & g^{22} \end{bmatrix} (J^{-1})^{t} \end{align*} \tag{8} \end{equation} $$

導出

としよう。

$$ g_{ij} = \left\langle \mathbf{x}_{i}, \mathbf{x}_{j} \right\rangle $$

座標部分写像$\mathbf{y} = \mathbf{x} \circ f : V \to \mathbb{R}^{3}$について、としよう。

$$ \mathbf{y}_{\alpha} = \dfrac{\partial \mathbf{y}}{\partial v^{\alpha}},\quad \overline{g}_{\alpha \beta} = \left\langle \mathbf{y}_{\alpha}, \mathbf{y}_{\beta} \right\rangle $$

$$ \overline{g} = \det \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix},\quad \begin{bmatrix} \overline{g}^{\gamma \beta} \end{bmatrix} = \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix}^{-1} $$

連鎖法則により、次を得る。

$$ \mathbf{x}_{i} = \dfrac{\partial \mathbf{y}}{\partial v^{1}} \dfrac{\partial v^{1}}{\partial u^{i}} + \dfrac{\partial \mathbf{y}}{\partial v^{2}} \dfrac{\partial v^{2}}{\partial u^{i}} = \sum \limits_{\alpha} \dfrac{\partial \mathbf{y}}{\partial v^{\alpha}} \dfrac{\partial v^{\alpha}}{\partial u^{i}} = \mathbf{y}_{\alpha} \dfrac{\partial v^{\alpha}}{\partial u^{i}} $$

$$ \mathbf{y}_{\alpha} = \dfrac{\partial \mathbf{x}}{\partial u^{1}} \dfrac{\partial u^{1}}{\partial v^{\alpha}} + \dfrac{\partial \mathbf{x}}{\partial u^{2}} \dfrac{\partial u^{2}}{\partial v^{\alpha}} = \sum \limits_{i} \dfrac{\partial \mathbf{x}}{\partial u^{i}} \dfrac{\partial u^{i}}{\partial v^{\alpha}} = \mathbf{x}_{i} \dfrac{\partial u^{i}}{\partial v^{\alpha}} $$

したがって$g_{ij}$は、次のように表される。

$$ g_{ij} = \left\langle \mathbf{x}_{i}, \mathbf{x}_{j} \right\rangle = \left\langle \mathbf{y}_{\alpha}, \mathbf{y}_{\beta} \right\rangle \dfrac{\partial v^{\alpha}}{\partial u^{i}} \dfrac{\partial v^{\beta}}{\partial u^{j}} = \overline{g}_{\alpha \beta}\dfrac{\partial v^{\alpha}}{\partial u^{i}} \dfrac{\partial v^{\beta}}{\partial u^{j}} $$

行列の積で表すと、以下のようになる。

$$ \begin{bmatrix} g_{11} & g_{12} \\[1em] g_{21} & g_{22} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{1}} \\[1em] \dfrac{\partial v^{1}}{\partial u^{2}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} \begin{bmatrix} \overline{g}_{11} & \overline{g}_{12} \\[1em] \overline{g}_{21} & \overline{g}_{22} \end{bmatrix} \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{1}}{\partial u^{2}} \\[1em] \dfrac{\partial v^{2}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} $$

$J$を$f : V \to U$のヤコビアンとしよう。

$$ J = \begin{bmatrix} \dfrac{\partial u^{1}}{\partial v^{1}} & \dfrac{\partial u^{1}}{\partial v^{2}} \\[1em] \dfrac{\partial u^{2}}{\partial v^{1}} & \dfrac{\partial u^{2}}{\partial v^{2}} \end{bmatrix} \quad \text{and} \quad J^{-1} = \begin{bmatrix} \dfrac{\partial v^{1}}{\partial u^{1}} & \dfrac{\partial v^{1}}{\partial u^{2}} \\[1em] \dfrac{\partial v^{2}}{\partial u^{1}} & \dfrac{\partial v^{2}}{\partial u^{2}} \end{bmatrix} $$

すると、

$$ \begin{bmatrix} g_{ij} \end{bmatrix} = (J^{-1})^{t} \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix} J^{-1} $$

$$ \begin{align*} g = \det \begin{bmatrix} g_{ij} \end{bmatrix} = \det \Big( (J^{-1})^{t} \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix} J^{-1} \Big) &= \det (J^{-1})^{t} \det \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix} \det J^{-1} \\ &= \det \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix} (\det J^{-1})^{2} \\ &= \overline{g} \left( \det \begin{bmatrix} \dfrac{\partial v^{\alpha}}{\partial u^{i}} \end{bmatrix} \right)^{2} \end{align*} $$

また、逆行列は、

$$ \begin{bmatrix} g^{kl} \end{bmatrix} = \begin{bmatrix} g_{ij} \end{bmatrix}^{-1} = \Big( (J^{-1})^{t} \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix} J^{-1} \Big)^{-1} = J \begin{bmatrix} \overline{g}_{\alpha \beta} \end{bmatrix}^{-1} J^{t} = J \begin{bmatrix} \overline{g}^{\gamma \delta} \end{bmatrix} J^{t} $$

$$ \implies g^{kl} = \overline{g}^{\gamma \delta}\dfrac{\partial u^{k}}{\partial v^{\gamma}} \dfrac{\partial u^{l}}{\partial u^{\delta}} $$

接ベクトルを$\mathbf{X} = X^{i}\mathbf{x}_{i} = \overline{X}^{\alpha} \mathbf{y}_{\alpha}$とすると、

$$ X^{i}\mathbf{x}_{i} = \overline{X}^{\alpha} \mathbf{y}_{\alpha} = \overline{X}^{\alpha} \dfrac{\partial u^{i}}{\partial v^{\alpha}} \mathbf{x}_{i} \implies X^{i} = \overline{X}^{\alpha} \dfrac{\partial u^{i}}{\partial v^{\alpha}} $$

$U$、$V$を逆に考えると、残りの結果を得る。

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