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제1 기본형식과 좌표변환의 관계

개요¹

좌표변환 $f : V \to U$가 주어져있을 때, $U$ 위의 메트릭 $g$와 $V$ 위의 메트릭 $\overline{g}$의 관계에 대해 설명한다.

아인슈타인 표기법을 사용한다.

공식

좌표조각사상 $\mathbf{x} : U \to \mathbb{R}^{3}$의 메트릭 $g$와 $\mathbf{y} = \mathbf{x} \circ f : V \to \mathbb{R}^{3}$의 메트릭 $\overline{g}$, 그리고 탄젠트 벡터 $\mathbf{X} = X^{i}\mathbf{x}_{i} = \overline{X}^{\alpha} \mathbf{y}_{\alpha}$에 대해서 다음의 관계가 성립한다.

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

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

설명

$(1) ~ (4)$는 공통적으로 $U$ 좌표계 위에서의 정보를, $V$ 좌표계 위에서의 정보로 어떻게 표현하는지를 말해준다. 그 중에서 $(1), (4)$는 $f : V \to U$의 자코비안을 포함하고 있고, $(2), (3)$은 $g = f^{-1} = U \to V$의 자코비안을 포함한다. 전통적으로 $(1), (4)$와 같은 변환을 반변 변환^{contravariant transformation}이라한다. $J$를 $f$의 자코비안이라고 할 때,

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

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

$(2)$와 같은 변환은 공변 변환^{covariant transformation}이라 한다.

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

$(3)$은 행렬곱을 포함하지 않으므로 변환이 아니다. $(5), (6), (8)$도 행렬곱으로 나타내면,

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

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

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

유도

에 대해서, 다음과 같다고 하자.

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

좌표조각사상 $\mathbf{y} = \mathbf{x} \circ f : V \to \mathbb{R}^{3}$에 대해서, 다음과 같다고 하자.

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

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

연쇄법칙에 의해 다음을 얻는다.

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

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

따라서 $g_{ij}$는 다음과 같이 표현된다.

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

이를 행렬곱으로 나타내면 아래와 같다.

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

$J$를 $f : V \to U$의 자코비안이라고 하자.

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

그러면,

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

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

또한 역행렬은,

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

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

탄젠트 벡터를 $\mathbf{X} = X^{i}\mathbf{x}_{i} = \overline{X}^{\alpha} \mathbf{y}_{\alpha}$라고하면,

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

$U$, $V$를 바꿔서 생각하면 나머지 결과를 얻는다.

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