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I came across this proof while looking for hints on my homework, and I think it's only gotten me more confused. This is from Global Lorentzian Geometry.

Lemma 5.4 If $(H,h)$ is a Riemannian manifold, then $(H,h)$ is complete.

Proof. By the Hopf-Rinow Theorem, it suffices to show that $(H,h)$ is geodesically complete. Thus suppose that $c:[a,1) \rightarrow H$ is a unit speed geodesic which is not extendible to $t=1$. Choosing any $p \in H$, we may find a constant $\alpha > 0$ such that any unit speed geodesic starting at $p$ has length $\ell \geq \alpha$. Set $\delta = \min\{\alpha/2, (1-a)/2\} > 0$. Since isometries preserve geodesics, it follows from the homogeneity of $(H,h)$ that any unit speed geodesic starting at $c(1-\delta)$ may be extended to a geodesic of length $\ell \geq 2 \delta$. In particular, $c$ may be extended to a geodesic $c:[a, 1+\delta) \rightarrow H$, in contradiction to the inextendibility of $c$ to $t=1$.

How exactly does it "follow from homogeneity" that we are able to extend unit speed geodesics? Isn't that what we're trying to show in the lemma? And how does considering geodesics based at $c(1-\delta)$ lead us to extending $c$ to $c:[a,1+\delta) \rightarrow H$?

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    $\begingroup$ " we may find a constant α>0 such that any unit speed geodesic starting at p has length ℓ≥α", how to show there exists a uniform lower bound $\alpha$ for all initial velocities $\ell$ may take? $\endgroup$ – Vim May 23 '17 at 10:45
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Homogeneity implies that all metric balls of the same radius are isometric. Therefore if one can extend a geodesic at a point $p$ in each direction by a distance of $\delta$, then one can extend by the same $\delta$ at every point of the manifold.

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  • $\begingroup$ Ah, okay, that makes sense. Thank you. $\endgroup$ – JoeDub Nov 26 '13 at 19:02
  • $\begingroup$ Does the proof work also for locally homogeneous spaces? Or is the statement false in that case? $\endgroup$ – W. Rether Jan 10 '18 at 10:28
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    $\begingroup$ If $M$ is locally homogeneous then its universal cover will be homogeneous, and geodesics are "the same" for both. @W.Rether $\endgroup$ – Mikhail Katz Jan 10 '18 at 10:39
  • $\begingroup$ Thank you. And why is the universal cover homogeneous? $\endgroup$ – W. Rether Jan 10 '18 at 10:45
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    $\begingroup$ @W.Rether, that's part of the standard theory. Did you look this up in Helgason for example? $\endgroup$ – Mikhail Katz Jan 10 '18 at 10:48
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Another way to prove this is by Hopf Rinow's Theorem. I'll prove that a homogeneous Riemannian manifold is a complete metric space.

Let $\{x_n\}\subset M$ be a Cauchy sequence and fix $p\in M$. For each $n\in\mathbb{N}$ there is a $g_m\in \operatorname{Iso}(M)$ s.t. $g_m(x_m)=p$. Let $B_\epsilon (p)$ a normal ball around $p$. There is a $N\in\mathbb{N}$ s.t. for every $n\geq N$, $$\epsilon>d(x_N,x_n)=d(g_Np,g_np)=d(p,g_N^{-1}g_np)$$ So $d(p,x_n)=d(p,g_np)\in g_N(B_\epsilon(p))$. Then $\{x_n\}$ admits a convergente subsequence. As $x_n$ is a cauchy sequence, $x_n$ converges to the limit of the subsequence.

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