## Metric of sphere and geodesics

From the embedding of ${i:\mathbb{S}^n(r)\rightarrow\mathbb{R}^{n+1}}$, we have the induced metric ${i^*g_0}$ on ${\mathbb{S}^n}$, ${g_0}$ is the Euclidean metric on ${\mathbb{R}^{n+1}}$ We can write this metric explicitly under stereographic projection.

$\displaystyle \phi:\mathbb{S}^n(r)\rightarrow \mathbb{R}^n$

$\displaystyle (x^1,x^2,\cdots,x^{n+1})\backslash{N}\rightarrow(u^1,u^2,\cdots,u^n)$

where ${N}$ is the north pole ${(0,0,\cdots,1)}$

$\displaystyle u^i=\frac{rx^i}{r-x^{n+1}}$

and

$\displaystyle \psi=\phi^{-1}:(u^1,u^2,\cdots,u^n)\backslash{N}\rightarrow(x^1,x^2,\cdots,x^{n+1})$

by

$\displaystyle x^i=\frac{2|u|^2u^i}{r^2+|u|^2},\quad x^{n+1}=\frac{(|u|^2-r^2)r}{r^2+|u|^2}$

${g=\psi^*i^*g_0}$ will a metric on ${\mathbb{R}^{n}}$,

$\displaystyle g\left(\frac{\partial}{\partial u^i}, \frac{\partial}{\partial u^j}\right)=g_0\left((i\psi)_*\frac{\partial}{\partial u^i},(i\psi)_*\frac{\partial}{\partial u^j}\right)=\frac{4r^2\delta_{ij}}{(r^2+|u|^2)^2}$

so

$\displaystyle g=\frac{4r^2du^i\otimes du^i}{(r^2+|u|^2)^2}$

From this we define 1-form basis ${\displaystyle w^i=\frac{2rdu^i}{r^2+|u|^2}}$, By the cartan structure equation

$\displaystyle \begin{cases}dw^i=w^j\wedge w_j^i, w^i_j+w^j_i=0\\ dw_j^i=w_j^k\wedge w_k^i+\frac{1}{2}R_{jkl}^i w^k\wedge w^l\end{cases}$

we can find that

$\displaystyle w^i_j=\frac{u^idu^j-u^jdu^i}{r^2+|u|^2}$

$\displaystyle R_{jkl}^i=\frac{1}{r^2}(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$

Also from the metric, we can obtain the christoffel symbols

$\displaystyle \Gamma^k_{ij}=\frac{1}{2}g^{kh}\left\{\frac{\partial g_{ih}}{\partial u^j}+\frac{\partial g_{jh}}{\partial u^i}-\frac{\partial g_{ij}}{\partial u^h}\right\}$

$\displaystyle =\frac{-2}{r^2+|u|^2}\left\{u^j\delta_{ik}+u^i\delta_{jk}-u^k\delta_{ij}\right\}$

We want to verify great circles on sphere are geodesics. Under some rotation, assume the great circle is

$\displaystyle \gamma(\theta)=(r\sin\theta,0,\cdots,-r\cos\theta)$

where ${\theta}$ is the angle of ${\gamma(t)}$ with the ${x^{n+1}}$ axis, thus ${\theta}$ represents the arc length of ${\gamma(\theta)}$. So after stereographic projection, we can get

$\displaystyle \phi\gamma(\theta)=\left(\frac{r\sin\theta}{1+\cos\theta},0,\cdots,0\right)=(r\tan\frac{\theta}{2},0,\cdots)$

Readers can verify it satisfies the equation of geodesics

$\displaystyle \frac{d^2u^k}{d\theta^2}+\Gamma^k_{ij}\frac{du^i}{d\theta}\frac{du^j}{d\theta}=0$

Remark: It is easy to make mistake here. My original intension is using ${\alpha(t)=(t,0,\cdots,0)}$ on ${\mathbb{R}^n}$ and try to verify ${\alpha(t)}$ satisfies the geodesic equation. Since the image of ${\alpha(t)}$ is a line, then ${\phi^{-1}\alpha(t)}$ should be a great circle hence a geodesic.

But one can easily know ${\alpha(t)}$ fails to satisfy the geodesic equation. The subtle error I made is pointed out by Bin Guo, which lies in the parametization of a curve. The geodesics are parametrized by arc length in usual.

To illustrate this, check ${\gamma(t)=(t^2,0)}$ on ${\mathbb{R}^2}$, the image of ${\gamma(t)}$ is a line, which is a geodesic under the Euclidean metric. But ${\gamma(t)}$ does not satisfy the geodesic equation, in which ${\Gamma^k_{ij}=0}$.

For the same reason, in ${\alpha(t)}$, ${t}$ is not such parameter. That is why I use ${\theta}$ to define a curve.

There is an easy way to verify that great arc is geodesic on sphere even if we don’t know what the $\Gamma_{ij}^k$ are.

For an embedded manifold in $\mathbb{R}^n$, suppose $X=(x^1(t),\cdots, x^n(t))$ and $Y=(y^1(t),\cdots, y^n(t))$ are two vector fields. Then

$\displaystyle \nabla_XY=\pi\left(\frac{dY}{dt}\right)=\frac{dY}{dt}-\langle\frac{dY}{dt},\nu\rangle\cdot \nu$

One can use this method to prove that $\nabla_{\gamma'}\gamma'=0$, for any $\gamma$ is a part of great circle.