# Question in do Carmo's book Riemannian geometry section 7

I have a question. Please help me.

Assume that $M$ is complete and noncompact, and let $p$ belong to $M$. Show that $M$ contains a ray starting from $p$.

$M$ is a riemannian manifold. It is geodesically and Cauchy sequences complete too. A ray is a geodesic curve that its domain is $[0,\infty)$ and it minimizes the distance between start point to each other points of curve.

• It would help if you said what $M$ is (perhaps a manifold?) and also define what it means for such $M$ to be complete. (does it just mean convergent Cauchy sequences have limits, or what?) One more thing: is a ray a geodesic emanating from $p$ ? If so any manifold has such rays; maybe they must extend to infinity... – coffeemath Jun 17 '13 at 12:36
• M is reimannian manifold . it is geodesically and Cauchy sequences complete too.ray is a geodesic curve that its domain is [0.infinite) and it minimizes the distance between start point to each other points of curve – mahdi Jun 17 '13 at 12:50
• Using your definitions (previous comment), the claim is false. Consider the sphere. (It is geodesically complete, but the geodesics [great circles] are not length minimizing on all its domain.) – Willie Wong Jun 17 '13 at 14:07
• Maybe you want $M$ to be non-compact? (Oh, and by the way, it is better to, when responding to comments asking for clarifications, edit the clarifications into the question text itself, instead of just putting it in the comments.) – Willie Wong Jun 17 '13 at 14:08

## 3 Answers

Take a sequence of points $(x_i)$ in the manifold whose distance from $p$ tends to infinity, and connect each of them to $p$ by a minimizing geodesic $\gamma_i(s)$. Choose a convergent subsequence $\gamma'_{i_k}(0)$ at $p$. Then the limit of the sequence is the desired direction of a ray.

• Could you please elaborate on how completeness + noncompactness imply unboundedness? Is it a consequence from the Heine Borel property (which the Riemannian manifold assumes of my memory serves)? – Vim May 24 '17 at 15:19

Otherwise suppose every geodesic emitting from p will fail to be a segment after some distance s. Since the unit sphere in the tangent plane that parameterizing these geodesics is compact, s has a maximum $s_{max}$. This means that the farthest distance from p is $s_{max}$, among all points of the manifold. So the diameter of the manifold is bounded by $2s_{max}$, by the triangle inequality. So the manifold is bounded and complete, by the Hopf–Rinow theorem, it is then compact.

• "So the manifold is bounded and closed, as a metric space, it is then compact." I'm not sure how things are about manifolds, but it's definitely not true for general metric spaces. – Vim May 24 '17 at 15:15
• @Vim: thanks. Edited referencing the Hopf–Rinow theorem. – Xipan Xiao May 24 '17 at 17:26
• I just wanted to point out that this exercise appears before Do Carmo has proved the fact that the distance-to-the-cut-locus function is continuous (as a map to $\mathbb{R}\cup \{\infty\}$ with the natural topology). The existence of the max $s_{max}$ implicitly uses this continuity assumption. That is, you have a map $d:S^{n-1}\rightarrow \mathbb{R}$ with $d(v) = s$ if the geodesic in the direction of $v$ stops minimzing at time $s$. If you know $d$ is continuous, it has a max. But otherwise, more work must be done. So, this answer is correct (+1), but uses a fact not yet available. – Jason DeVito Feb 25 at 15:45

(This is more of a comment to Mikhail's perfectly fine answer, but it got too long.)

I just wanted to append a proof that your suggested $$\gamma$$ really does work.

To be more explicit, we assume $$d(x_i, p) \geq i$$ and that each $$\gamma_i(t) = \exp_p(t\gamma_i'(0))$$ (which connects $$p$$ to $$x_i$$) is unit speed. Since the unit sphere in $$T_p M$$ is compact, the sequence $$\{\gamma_i'(0)\}$$ must have some convergent subsequence, say, with limit $$v$$. I will abuse notation and use $$\gamma_i'(0)$$ to refer to this subsequence. Set $$\gamma(t) = \exp_p(tv)$$.

Lemma: For any fixed $$t$$, we have $$\lim_{i\rightarrow \infty} d(\gamma(t), \gamma_i(t)) = 0$$.

Proof: Because $$d$$ and $$\exp_p$$ are continuous, $$\lim_{i\rightarrow \infty} d(\gamma(t),\gamma_i(t)) = d(\gamma(t), \exp_p(\lim_{i\rightarrow\infty} t\gamma_i'(0))) = d(\gamma(t), \gamma(t)) = 0.$$ $$\square$$

Now, assume for a contradiction that there is some time $$t$$ for which $$\gamma(t)$$ is not minimizing. That is, assume there is a $$t>0$$ for which $$d(p, \gamma(t)) < t$$.

For any $$i > t$$, we know $$d(\gamma_i(t), p) = t$$. From the lemma, we know that there is an $$I$$ with the property that for all $$i > I$$, $$d(\gamma_i(t), \gamma(t)) < t -d(\gamma(t), p)$$.

Then, for any $$i \geq \max\{t,I\}$$, the triangle inequality gives $$t = d(\gamma_i(t0), p) \leq d(\gamma_i(t), \gamma(t)) + d(\gamma(t),p) < t - d(\gamma(t), p) + d(\gamma(t), p) = t$$ which gives a contradiction.