구면(sphere)

리만 곡률 텐서

$\begin{array}{ll} \begin{array}{ll} R_{111}^1 & 0 \ R_{112}^1 & 0 \end{array} & \begin{array}{ll} R_{121}^1 & 0 \ R_{122}^1 & 0 \end{array} \ \begin{array}{ll} R_{211}^1 & 0 \ R_{212}^1 & 1 \end{array} & \begin{array}{ll} R_{221}^1 & -1 \ R_{222}^1 & 0 \end{array} \ \begin{array}{ll} R_{111}^2 & 0 \ R_{112}^2 & -\sin ^2(v) \end{array} & \begin{array}{ll} R_{121}^2 & \sin ^2(v) \ R_{122}^2 & 0 \end{array} \ \begin{array}{ll} R_{211}^2 & 0 \ R_{212}^2 & 0 \end{array} & \begin{array}{ll} R_{221}^2 & 0 \ R_{222}^2 & 0 \end{array} \end{array}$
covariant tensor

$\begin{array}{ll} \begin{array}{ll} R_{1111} & 0 \ R_{1112} & 0 \end{array} & \begin{array}{ll} R_{1121} & 0 \ R_{1122} & 0 \end{array} \ \begin{array}{ll} R_{1211} & 0 \ R_{1212} & R^2 \sin ^2(v) \end{array} & \begin{array}{ll} R_{1221} & -R^2 \sin ^2(v) \ R_{1222} & 0 \end{array} \ \begin{array}{ll} R_{2111} & 0 \ R_{2112} & -R^2 \sin ^2(v) \end{array} & \begin{array}{ll} R_{2121} & R^2 \sin ^2(v) \ R_{2122} & 0 \end{array} \ \begin{array}{ll} R_{2211} & 0 \ R_{2212} & 0 \end{array} & \begin{array}{ll} R_{2221} & 0 \ R_{2222} & 0 \end{array} \end{array}$

측지선 이 만족시키는 미분방정식

$\frac{d^2\alpha_k }{dt^2} + \Gamma^{k}_{~i j }\frac{d\alpha_i }{dt}\frac{d\alpha_j }{dt} = 0$
풀어쓰면,

$\frac{d^2 u}{dt^2} + 2\Gamma^{1}_{~1 2 }\frac{du }{dt}\frac{dv }{dt} = 0$

$\frac{d^2 v}{dt^2} + \Gamma^{2}_{~1 1 }\frac{du }{dt}\frac{du }{dt} = 0$

가우스곡률 항목 참조

$K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right)$
반지름 R인 구면의 가우스곡률

$K=\frac{1}{R^2}$

위의 좌표계에서 $u=\phi,v=\theta$ 로 생각하자.
라플라시안

$\Delta f = {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial f \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 f \over \partial \phi^2}={1 \over r^2 }({\partial^2 f \over \partial \theta^2} +\cot\theta {\partial f \over \partial \theta} + \frac{1}{ \sin^2 \theta} {\partial^2 f \over \partial \phi^2})$