## Tag Archives: geodesic

### The thing about Hopf Rinow Theorem

Lemma 1: ${M}$ is a Riemannian manifold. ${\forall p\in M}$, ${\exists\,}$ neighborhood ${N}$ of ${p}$ and ${\epsilon>0}$ such that ${\exp_p:B_\epsilon(0)\rightarrow N}$ is a diffeomorphism and any two points in ${N}$ can ve connected by a unique geodesic with length smaller than ${\epsilon}$.

Lemma 2: ${p\in M}$, ${r}$ is small enough such that ${\exp_p:B_r(0)\rightarrow N_r(p)=\exp(B_r(0))}$ is a differeomorphism, then for any ${q\not \in N_r(p)}$, ${\exists \,q'\in \partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Proof: ${d(\cdot,q)}$ is a continuous function on ${\partial N_r(p)}$. Since ${\partial N_r(p)}$ is compact, then ${\exists\, q'\in \partial N_r(p)}$ such that

$\displaystyle d(q',q)=\inf\{d(\tilde{q},q)|\tilde{q}\in\partial N_r(p)\}$

Suppose ${\gamma_n}$ is a minimizing sequence of ${d(p,q)}$. For any ${\gamma_n}$, there exists ${q_n\in \partial N_r(p)\cap \gamma_n}$, then

$\displaystyle L(\gamma_n)\geq d(p,q_n)+d(q_n,q)\geq r+d(q',q)$

Letting ${n\rightarrow \infty}$, ${d(p,q)\geq r+d(q',q)}$. By the triangle inequality,

$\displaystyle d(p,q)= r+d(q',q)$

$\Box$

Thm(Hopf-Rinow) The following statements are equivalent for Riemannian manifold ${M}$:
(1) ${M}$ is a complete metric space, where the metric is induced by

$\displaystyle d(p,q)=\inf\{L(\gamma)|\gamma\text{ is a curve connects } p,q\}$

(2)The closed and bounded sets of ${M}$ are compact.
(3)${\exists \,p\in M}$ such that ${\exp_p}$ is defined on all of ${T_pM}$.
(4) ${\forall\,p\in M}$, ${\exp_p}$ is defined on ${T_pM}$
Furthermore, each of the statements ${(1-4)}$ implies
(5) Any two points ${p,q\in M}$ can be connected by a geodesic of shortest length. Proof: Let us prove ${(3)\Rightarrow (5)}$ firstly.
By lemma 2, there exists ${r>0}$ and ${q'}$ on ${\partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Suppose ${q'=\exp_ptv}$ for some ${v\in T_pM}$, ${||v||=1}$, then ${\gamma(t)=\exp_ptv}$ is defined on ${[0,\infty)}$ by assumption. Consider ${\{t|t\in [r,d(p,q)]\}}$ satisfies

$\displaystyle d(p,q)=t+d(\gamma(t),q),$

denote such points as ${I}$. ${I}$ is nonempty as ${r\in I}$ and it is closed by the sake of continuity.
Suppose ${t_0=\max_{t\in I}t}$. If ${t_0=d(p,q)}$, we are done. Otherwise consider ${q_0=\gamma(t_0)}$, by lemma 2, ${\exists\,\epsilon>0}$ small enough, ${q''\in\partial N_\epsilon(q_0)}$ such that

$\displaystyle q\not\in N_{\epsilon}(q_0),\quad d(q_0,q)=\epsilon+d(q'',q)$

then

$\displaystyle d(p,q)=t_0+\epsilon+d(q'',q)$

We are going to prove ${q''=\gamma(t_0+\epsilon)}$. From the triangle inequality

$\displaystyle d(p,q'')\geq d(p,q)-d(q'',q)=t_0+\epsilon$

$\displaystyle d(p,q'')\leq d(p,q_0)+d(q_0,q'')=t_0+\epsilon$

then we must have ${d(p,q'')=t_0+\epsilon}$. Then the union of ${\gamma([0,t_0])}$ and the geodesic from ${q_0}$ to ${q''}$ constitutes a shortest curve from ${p}$ to ${q''}$, therefore it must be a geodesic. By lemma 1, such geodesic is unique with given initial values. So it must concider with ${\gamma(t)}$ in the ${N_\epsilon(q_0)}$, which means ${q''=\gamma(t_0+\epsilon)}$. Then

$\displaystyle d(p,q)=t_0+\epsilon+d(\gamma(t_0+\epsilon),q)$

This contradicts the fact ${t_0}$ is maximal. So ${t_0}$ must be ${d(p,q)}$, and the curve ${\gamma([0,t_0])}$ is just the minimal geodesic connects ${p}$ and ${q}$.

${(4)\Rightarrow (3)}$ Evidently

${(3)\Rightarrow (2)}$ Suppose ${K}$ is a bounded set in ${M}$. From ${(5)}$ we know, ${K}$ can be contained in ${\exp_p B_r(0)}$ where ${B_r(0)\subset T_pM}$ for some ${r}$ large enough. Since ${B_r(0)}$ is compact and ${\exp_p}$ is a continuous mapping, then ${\exp_p B_r(0)}$ is compact, therefore its closed subset ${K}$ is compact.

${(2)\Rightarrow(1)}$ Any cauchy sequence in ${M}$ is bounded, its closure is compact by ${(2)}$. Then it must have a converge subsequence and being cauchy, it has to converge itself.

${(1)\Rightarrow (4)}$ For any ${p\in M}$, we need to prove ${\exp_ptv}$ is defined on ${[0,\infty)}$ for any ${v\in T_pM}$. Consider ${t_n\nearrow T<\infty}$ and ${\exp_p tv}$ is defined on each ${t_n}$, denote ${\exp_pt_nv=q_n}$.

Since ${d(q_n,q_m)\leq |t_n-t_m|}$, ${\{q_n\}}$ is a cauchy sequence, ${\exists\, q\in M}$ such that ${q_n\rightarrow q}$. Since there exists a neighborhood of ${q}$, say ${N_\epsilon(q)}$, such that ${\forall z\in N_\epsilon(q)}$, any geodesic starting from ${z}$ can be extended at least up to length ${\rho_0>0}$.

For sufficiently large ${n}$, ${q_n\in N_\epsilon(p)}$ and ${d(q_n,q)<\rho_0}$. ${\exp_ptv|_{[t_n,t_{n+1}]}}$ is a curve starting from ${\gamma(t_n)}$, then it can be extended at least up to length ${\rho_0}$. By the uniqueness of geodesic, ${\exp_ptv}$ can be extened beyond ${q}$, namely ${q=\exp_pTv}$ is well defined. So ${\exp_p}$ is defined on whole ${[0,\infty)}$.

$\Box$

Remark:  Should thank Bin Guo’s picture

### Meyer’s Thm

Thm:(Meyers) Suppose ${M}$ is a ${n-}$dimensional complete manifold and

$\displaystyle \text{Ric}_M\geq (n-1)K>0$

Then ${\text{diam}M<\frac{\pi}{\sqrt{K}}}$ hence ${M}$ is compact and ${\pi_1(M)}$ is finite.

This theorem is proved by the following lemma.

Lemma: Suppose ${\text{Ric}_M\geq (n-1)K>0}$, then every geodesic which is longer than ${\displaystyle \frac{\pi}{\sqrt{K}}}$ on ${M}$ must have conjugate points hence not a minimal geodesic.

Proof: Suppose ${\gamma(t):[0,l]\rightarrow M}$ is a mininal normal geodesic with length ${\displaystyle l>\frac{\pi}{\sqrt{K}}}$, ${t}$ is the arc length parameter. Then there exists ${\displaystyle K'=\left(\frac{\pi}{l}\right)^2}$ such that ${K'.

For ${T_{\gamma(0)}M}$, there exists an orthonormal basis ${\{\frac{\partial }{\partial t}, V_i, i=1,\cdots,n-1\}}$. Let ${V_i(t)}$ be the parallel displacement of ${V_i}$ along ${\gamma}$, ${T}$ is the one of ${\frac{\partial }{\partial t}}$.

Define ${W_i=V_i(t)\sin \sqrt{K'}t }$. Then the second variation of ${\gamma}$ along ${W_i}$ is

$\displaystyle I(W_i,W_i)=\int_{0}^l\langle\nabla_TW_i, \nabla_TW_i \rangle+R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l-\langle W_i, \nabla_T\nabla_TW_i\rangle+ R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l K'\sin^2\sqrt{K'}t+\sin^2\sqrt{K'}tR(T,V_i,T,V_i)dt$

So

$\displaystyle \sum_1^{n-1} I(W_i,W_i)=\int_0^l[(n-1)K'-\text{Ric}_M(V_i)]\sin^2\sqrt{K'}tdt<0$

So there exists a ${I(W_i,W_i)<0}$, for such ${i}$, this means ${\gamma}$ is locally maximal. This is contradiction, because we choose ${\gamma}$ is a minimal geodesic.

### Gauss lemma and geodesics

From the Gauss lemma to the minimality of geodesics

Lemma:(Gauss) Suppose ${p\in M}$, ${v\in T_u(T_pM)}$

$\displaystyle \langle u,v\rangle=\langle(d\exp_p)_uu,(d\exp_p)_uv\rangle$

Here we make natural equivalence of ${T_u(T_pM)}$ and ${T_pM}$. This lemma means ${d\exp}$ presvers the inner product.

For any ${p\in M}$ there exists a neighborhood ${U}$ of ${p}$ such that the exponential map is diffeomorphism from ${B_\epsilon(0)\subset T_p(M)}$ to ${U_p}$.

Thm: ${p\in M}$, for such ${U_p}$, suppose ${B=\{\exp_pv|||v||<\epsilon\}}$ is any geodesic ball contained in ${U_p}$. Then the shortest length connecting a point on ${\partial B}$ and ${p}$ is the geodesic ray from ${p}$.

Proof: Suppose ${\gamma(t)=\exp_p\sigma(t):[0,1]\rightarrow M}$ connects ${p}$ and ${q\in \partial B}$, where ${\sigma(t)}$ is a curve in ${T_pM}$. Suppoe ${\displaystyle \alpha(t)=\frac{\sigma(t)}{|\sigma(t)|}}$ is the unit normal along ${\sigma(t)}$, then ${\dot{\sigma}(t)=\langle \dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)}$ with ${\beta(t)}$ is perpendicular to ${\alpha(t)}$. Then by Gauss lemma

$\displaystyle \dot{\gamma}(t)=(d\exp_p)_{\sigma(t)}\dot{\sigma}(t)=(d\exp_p)_{\sigma(t)}\left[\langle\dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)\right]$

$\displaystyle =\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)+(d\exp_p)_{\sigma(t)}\beta(t)$

$\displaystyle l(\gamma)=\int_0^1 |\dot{\gamma}(t)|dt\geq \int_0^1|\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)|dt$

$\displaystyle =\int_0^1 |\langle\dot{\sigma}(t),\alpha(t)\rangle|dt(\text{ since }||\alpha(t)||=1)$

$\displaystyle \geq \int_0^1 \langle\dot{\sigma}(t),\alpha(t)\rangle dt=\int_0^1 \langle {\sigma}(t),\alpha(t)\rangle'dt=\langle {\sigma}(1),\alpha(1)\rangle$

$\displaystyle =||\sigma(1)||=\text{length of }\exp t\sigma(1)= \text{geodesic ray from }p \text{ to } q$

We use the fact ${\langle{\sigma}(t),\dot{\alpha}(t)\rangle=0}$, because ${||\alpha(t)||=1}$ which implies ${\langle{\alpha}(t),\dot{\alpha}(t)\rangle=0}$.

Remark: I own this proof to Prof. Xiaochun Rong.

### 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.