Non-compact and complete manifold

Thm: A normal geodesic {\gamma:[0,\infty)\rightarrow M} is called shortest if and only if {\gamma|_{[0,t]}} is shortest. A complete Riemannian manifold is non-compact if and only if for any point {p\in M}, there is a shortest geodesic {\gamma:[0,\infty)\rightarrow M} such that {\gamma(0)=p}

Proof: If there is a shortest geodesic {\gamma:[0,\infty)\rightarrow M}, then {\{\gamma(n), n=1,2,\cdots\}} is a sequence of points with no converging subsequence, because {|\gamma(i)-\gamma(j)|\geq 1}, {i\neq j}. So {M} is non-compact.

If {M} is complete and non-compact, from the Hopf-Rinow theorem, there exists {p_n}, {n\geq 0} such that {d(p,p_n)\rightarrow \infty} as {\infty}. Suppose {\gamma_n=\exp_p{tv_n}:[0,a_n]\rightarrow M} is the shortest normal geodesic connecting {p} and {p_n}, {v_n\in B_1(0)\subset T_pM}. Then {\exists\, v\in B_1(0)} such that {v_n\rightarrow v}, going to subsequence if necessary. Consider {\exp_ptv:[0,\infty)\rightarrow M}, if it is not shortest for some {t_0}, then there exists {t_0>d(p,\gamma(t_0))+\epsilon_0}. Since {\gamma_n(t)\rightarrow \gamma(t)} for any fixed {t>0},

\displaystyle t_0=d(p,\gamma_n(t_0))\leq d(p,\gamma(t_0))+d(\gamma(t_0),\gamma_n(t_0))<t_0-\epsilon_0-d(\gamma(t_0),\gamma_n(t_0))

when {n} is large enough, we get contradiction. So {\gamma} is shortest. \Box


Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: