전역 가우스-보네 정리

전역 가우스-보네 정리

Global Gauss-Bonnet Theorem

Global Gauss Bornet Thm

정리1

$R \subset M$을 oriented surface $M$의 정칙 영역이라고 하자. $C_{1}, \dots, C_{n}$을 닫힌 단순 피쓰와이즈 레귤러 커브스라고 하자. s.t. $\bigcup\limits_{j=1}^{n}C_{j} = \partial R$

$\forall C_{i}$는 positive orienteation

$\theta_{p}$는 바깥각도 $C_{i}$들의

$$ \implies \sum_{i=1}^{n} \int_{C_{i}}K_{g}ds + \iint_{R} K dA + \sum\theta_{i} = 2\pi \chi(R) $$

증명

지오대식 코디네이트로 리전을 다 덮는다.

$\left\{ \mathbf{x}_{\alpha} \right\}$ family of geodesic coordinate s.t. $\bipcup _{\alpha} \supset R$

$\mathsrc{T}$를 $R$의 삼각화라고 하자. such that $\forall T \in \mathscr{T}$, $\exists \alpha, T \subset \mathbf{x}(U_{\alpha})$

Given an orientation on each triangles. By the local thm,

$$ \int_{\partial T_{i}}K_{g}ds + \iint_{T_{i}}K_{ds} + \sum_{k=1}^{3}\theta_{ik} = 2\pi $$

이걸 다 더하면,

$$ \sum\limits_{i=1}^{F}\left(\int_{\partial T_{i}}K_{g}ds + \iint_{T_{i}}K_{ds} + \sum_{k=1}^{3}\theta_{ik}\right) = 2\pi F $$

$$ \sum_{i}^{n} \int_{\partial C_{i}} K_{g}ds + \iint_{R} K dA + \sum_{i=1}^{F}\sum_{k}^{3}\theta_{ik} $$

$\theta_{ik} = $ external angles of $T_{i}$ $\phi_{jk} = \pi - \theta_{jk} = $ interior angles

$\sum \sum \theta_{ik} = \sum \sum (\pi - \phi_{ik}) = 3\pi F - \sum\sum\phi_{ik}$

$E_{e} = $ # of external edges of $\mathscr{T}$ $E_{i} = $ # of inernal edges of $\mathscr{T}$ $V_{e} = $ # of external vertices of $\mathscr{T}$ $V_{i} = $ # of inernal vertices of $\mathscr{T}$

$C_{i} $ is closed $\implies $E_{e} = V_{e}$

$3F = 2E_{i} + E_{e}$

$\sum\sum \theta_{ik} = 3\pi F - \sum\sum\phi_{ik} = 2\pi(2E_{i} + E_{e}) - \sum\sum\phi_{ik} = 2\piE_{i} + \pi E_{e} - \sum\sum\phi_{ik} = 2\pi E_{i} + \pi E_{e} - (2\pi V_{i} + \pi V_{et} + \sum(\pi - \theta_{i}))$ $\theta_{i}$는 $C_{i}$들의 외각들.

$2\pi E_{i} + \pi E_{e} - 2\pi V_{i} - \pi V_{et} - \sum _{j=1}^{V_{ec}}(\pi - \theta_{i})$

뒤에항은 $\pi V_{ec} - \sum_{i=1}^{p}\theta_{i}$

$E_{e} = V_{e}$니까 더해서 치환

Corollary

$M$ is compact oriented surface. 컴패트이면 바운더리가 없고, 계산이 쉬워져서 좋은 경우이다.

$$ \iint_{M} dA = 2\pi \chi(M) $$

example $\mathbb{S}^{2}$

$$ 2\pi \chi(\mathbb{S}^{2}) = \iint_{\mathbb{S}^{2}} 1 dA = 4\pi \cdot 1^{2} = 2\pi $$

$$ \chi(\mathbb{S}^{2}} = 2 $$


  1. Richard S. Millman and George D. Parker, Elements of Differential Geometry (1977), p ↩︎

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