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

Advertisements

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'<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.