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측지선 좌표조각사상의 크리스토펠 심볼

정리¹

$$ \left[ g_{ij} \right] = \begin{bmatrix} 1 & 0 \\ 0 & h^{2} \end{bmatrix} \quad (h \gt 0) $$

그러면 $\mathbf{x}$의 크리스토펠 심볼은 다음과 같으며, 아래의 것들 외에는 모두 $0$이다.

$$ \Gamma_{22}^{1} = -hh_{1},\quad \Gamma_{12}^{2} = \Gamma_{21}^{2} = \dfrac{h_{1}}{h},\quad \Gamma_{22}^{2} = \dfrac{h_{2}}{h} $$

이때 $(u^{1}, u^{2})$는 $U$의 좌표이며, $h_{i} = \dfrac{\partial h}{\partial u^{i}}$이다.

증명

증명에 앞서 필요한 계산을 해두자. $g_{11} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle = 1$이므로,

$$ \dfrac{\partial g_{11}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle}{\partial u_{i}} = 2\left\langle\mathbf{x}_{1i}, \mathbf{x}_{1}\right\rangle = 0 $$

$$ \implies \left\langle\mathbf{x}_{11}, \mathbf{x}_{1}\right\rangle = 0 \quad \text{ and } \quad \left\langle\mathbf{x}_{12}, \mathbf{x}_{1}\right\rangle = 0 \tag{1} $$

마찬가지로 $g_{12} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle = 0$이므로,

$$ \dfrac{\partial g_{12}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle}{\partial u_{i}} = \left\langle\mathbf{x}_{1i}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{2i}\right\rangle = 0 $$

여기서 $i=1$일 때를 보면, $\left\langle \mathbf{x}_{11}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{21}\right\rangle = 0$인데 $(1)$에 의해서 $\left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle = 0$이다. $i=2$일 때는,

$$ \left\langle\mathbf{x}_{12}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{22}\right\rangle = 0 \tag{2} $$

또한 $h = \sqrt{g_{22}} = \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}$이므로,

$$ h_{i} = \dfrac{\partial h}{\partial u^{i}} = \dfrac{\partial \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}}{\partial u^{i}} = \dfrac{1}{2\sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}} 2\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle = \dfrac{1}{h}\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle $$

$$ \implies \left\langle \mathbf{x}_{21}, \mathbf{x}_{2} \right\rangle = hh_{1} \quad \text{ and } \quad \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle = hh_{2}\tag{3} $$

크리스토펠 심볼

$$ \Gamma_{ij}^{k} := \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} $$

이제 위에서 계산한 것을 리스토펠 심볼의 정의에 대입만 해주면 결과를 얻는다.

$$ \begin{align*} \Gamma_{11}^{1} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + 0 \cdot 0 = 0 \end{align*} $$

$$ \begin{align*} \Gamma_{12}^{1} = \Gamma_{21}^{1} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle \cdot 0 = 0 \end{align*} $$

$$ \begin{align*} \Gamma_{22}^{1} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 1 + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle \cdot 0 \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle = - \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle & \text{ by } (2)\\ &= - hh_{1} & \text{ by } (3)\\ \end{align*} $$

$$ \begin{align*} \Gamma_{11}^{2} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + 0 \cdot \dfrac{1}{h^{2}} = 0 \\ \end{align*} $$

$$ \begin{align*} \Gamma_{12}^{2} = \Gamma_{21}^{2} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + hh_{1} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{1}}{h} \\ \end{align*} $$

$$ \begin{align*} \Gamma_{22}^{2} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 0 + hh_{2} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{2}}{h} \\ \end{align*} $$

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