## 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))

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