가우스-보네 정리

가우스-보네 정리

Gauss-Bonnet Theorem

가우스 보네 정리 p185-186

$\mathbf{x} : U \to M$을 simply connected geodesic coordinate patch라고 하자.(사실 geodesic이라는 조건은 global thm이 있어서 필요없는데, 여기에서는 쓴다.)

$\gamma a piecewise regular curves s.t. \gamma(I) \subset \mathbf{x}(U)$ rmflrh $\gamma$가 $R$을 바운드한다고 하자.

$\alpha_{1}, \dots, \alpha_{n}$을 jump angle이라고하자. 그러면

$$ \iint_{R} K dA + \int_{\gamma} K_{g} ds + \sum \alpha_{i} = 2\pi $$

증명

$\mathbf{x}$가 측지 좌표이므로, $\left[ g_{ij} \right] = \begin{bmatrix} 1 & 0 \\ 0 h^{2} \end{bmatrix}$

Compute the angular variation $\alpha(t) := \angle ( \mathbf{x}_{1}, T=\gamma^{\prime}(t))$.

Assume $\gamma$ is a unit speed curve.

$$ T(t) = \gamma^{\prime}(t) \\ P(t) = $\text{parallel vector field starting from a juction point s.t. } \dfrac{P \tiems T}{\left\| P \times T \right\|} = \mathbf{n} $$

$\phi(t) = \angle(\mathbf{x}_{1}, P)$, $\theta(t) = \angle(P, T)$,

$\delta_{\alpha} = \int_{\gamma} \alpha^{\prime}(t) dt$, $\alpha(t) = \phi(t) + \theta(t) + 2n\pi$ $= \int_{\gamma} \phi^{\prime}(t) dt + \int_{\gamma} \theta^{\prime}(t)dt$ $= \iint_{R} K dA + \int_{\gamma}k_{g}ds$

$\mathbf{x}_{1}(\gamma(t)) \cdot P(t) = \cos\phi(t)$

미분 $\implies -\sin \phi(t) \dfrac{d \phi}{d t}(t) = \dfrac{d \mathbf{x}_{1}(\gamma(t))}{d t} \cdot P(t) + \mathbf{x}_{1} \cdot \dfrac{d P}{d t}(t)$ 뒤에 항은 $M$과 수직한다.

$= \dfrac{d}{dt}(\mathbf{x}_{1}(\gamma(t))) \cdot P(t) = \left( \mathbf{x}_{11}(\gamma(t)) \gamma_{1}^{\prime}(t) + \mathbf{x}_{12}(\gamma(t)) \gamma_{2}^{\prime}(t) \right) \cdot P(t)$ Chain rule

$= \left( \left[ L_{11}\mathbf{n} + \Gamma_{11}^{1}\mathbf{x}_{1} + \Gamma_{11}^{2}\mathbf{x}_{2} \right]\gamma_{1}^{\prime}(t) + \left[ L_{12}\mathbf{n} + \Gamma_{12}^{1}\mathbf{x}_{1} + \Gamma_{12}^{2}\mathbf{x}_{2} \right]\gamma_{2}^{\prime}(t) \right) \cdot P(t)$ 0되는 거 정리하면 $= \dfrac{h_{1}}{h}\gamma_{2}^{\prime}(t)\mathbf{x}_{2} \cdot P(t)$

$\implies -\sin\phi(t) \phi^{\prime}(t) = \dfrac{h_{1}}{h}\mathbf{x}_{2} \cdot P(t)$

$\mathbf{x}_{1}, \dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|}$는 정규직교기저.

$P = (\mathbf{x}_{1} \cdot P)\mathbf{x}_{1} + (\dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|}\cdot P) \dfrac{\mathbf{x}_{2}}{\left\| \mathbf{x}_{2} \right\|} = \cos\phi \mathbf{x}_{1} + \sin\phi \dfrac{x_{2}}{h}$

$\mathbf{x}_{2} \cdot P = \cos\phi \mathbf{x}_{1} \cdot \mathbf{x}_{2} + \dfrac{\sin\phi}{h}\dfrac{\mathbf{x}_{2}\cdot \mathbf{x}_{2}}{h^{2}} = h\sin \phi$

$\delta \phi = \int_{0}^{L} \phi^{\prime}(t)dt = \int_{0}^{L} - h_{1}\gamma_{2}^{\prime}dt = \int_{\gamma}-h_{1}d\gamma_{2} = \int_{\gamma}-h_{1}du_{2} = \text{(Green’s thm)} \iint_{R} h_{11}du_{1}du_{2} = -\iint_{R} \dfrac{h_{11}}{h}\sqrt{g}du_{1}du_{2} = \iint_{R} K dA$

we want $\delta \theta = \int_{\gamma} k_{g}ds$.

$\theta (t) = \angle(P, T)$, $\cos\theta(t) = P \cdot T$를 미분하면, $\implies -\sin\theta(t)\theta^{\prime}(t) = \dfrac{d P}{d t} \cdot T + P \cdot \dfrac{d T}{d t} = P \cdot T^{\prime}$(첫항의 $P$미분이 $\mathbf{n}$이랑 평행하니까.)

$k_{g} = \mathbf{S} \cdot T^{\prime} = (\mathbf{n}\times \mathbf{T}) \cdot T^{\prime} = \mathbf{n}\cdot(T\ times T^{\prime}) = \dfrac{P \times T}{\sin \theta} \cdot (T\times T^{\prime}) = \dfrac{1}}{\sin\theta} P \cdot (T\times (T\times T^{\prime})) = \dfrac{-1}{\sin\theta} P \cdot T^{\prime}$(BAC-CAB 쓴 거)

$\therefore \theta^{\prime}(t) = K_{g}$

$\implies \delta \theta = \int_{\gamma} \theta^{\prime}(t) dt = \int_{\gamma}K_{g}dt(유닛이라고 가정했으니까?)$

$$ 2\pi = \delta\alpha + \alpha_{i} $$

매끄러운 부분이랑, 점프하는 부분 다 더하면 $2\pi$

응용

구면위의 삼각형의 넓이를 계산하기 개 쌉 편해진다.

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