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}}


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


\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}


\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.

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