A finite-diameter Riemannian manifold with properly embedded infinite geodesics

Question

I'm working on the following problem:

Let $S$ be the unit $2$-sphere minus the north and south poles, and let $M = (0,\pi) \times \mathbb R$. Define $q : M \to S$ by $q(\phi,\theta) = (\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi)$, and let $g$ be the metric on $M$ given by pulling back the round metric $g = q^*\mathring{g}$. Prove the Riemannian manifold $(M,g)$ has diameter $\pi$, but contains infinitely long properly embedded geodesics.

Constructing the geodesics in question is simple: they are the preimages of great circles that do not pass through the north and south poles. The equator, for example, is one: $\gamma(t) = (\frac\pi 2, t)$ is a properly embedded geodesic in $M$ with infinite length. But this specific example seems to disprove the assertion that the diameter of $M$ is finite. What am I overlooking to show the diameter is finite?

Answer 1

That this problem illustrates that for some Riemannian manifolds $(M, g)$ the length-minimizing paths are not geodesics.

Example (Up to reparameterization) the only geodesic that passes through $(\frac{\pi}{2}, 0)$ and $(\frac{\pi}{2}, \frac{5\pi}{4})$ is the preimage of the the equator, that is, $t \mapsto (\frac{\pi}{2}, t)$, but the length of this geodesic between those two points is $\frac{5 \pi}{4}$, which (as the problem shows) is larger than the diameter $\pi$ of $M$, hence it is not length-minimizing.

(In fact, for this particular example there are pairs of points that are not connected by any geodesic, despite that the space is connected.) Indeed, this reinforces the point that geodesics are locally length-minimizing, but globally they need not be. (There are simpler examples of this fact, too: For any nonantipodal points $x, y \in S^n$, the minor and major arcs of the great circle passing through them are both geodesics, but the major arc is longer than the minor arc and hence not minimizing.)

Hint The key observation is that the metric in the preimage of the polar regions is small, so a strategy for building a short path between two points is to travel from your starting point to that region, do your lateral movement there, and then drop down to the point of interest.

Additional hint Suppose we want to travel from $(a, b)$ to $(a', b')$. Then we have at least one of $a + a' \leq \pi$ or $(\pi - a) + (\pi - a') \leq \pi$, and by symmetry (reflection $M$ across the preimage of the equator) we may as well assume the former. Consider the path $\gamma_\epsilon$ defined to be the concatenation of the paths $\gamma_\epsilon^1$ connecting $(a, b)$ vertically (along a "meridian") from to $(\epsilon, b)$ where $0 < \epsilon \leq \min\{a, a'\}$, $\gamma_\epsilon^2$ connecting $(\epsilon, b)$ horizontally (along a "parallel") to $(\epsilon, b')$, and $\gamma_\epsilon^3$ connecting $(\epsilon, b')$ to $(a', b')$ vertically. Then, length of $\gamma$ satisfies \begin{align}L(\gamma_\epsilon) &= L(\gamma_1) + L(\gamma_2) + L(\gamma_3)\\ &= (a - \epsilon) + |b' - b| \sin \epsilon + (a' - \epsilon)\\ &\leq (a + a') + (|b' - b| - 2) \epsilon \leq \pi + (|b' - b| - 2) .\epsilon\end{align} So, the distance from $(a, b)$ to $(a', b')$ satsfies$$d((a, b), (a', b')) = \inf_\gamma L(\gamma) \leq \inf_{\epsilon > 0} L(\gamma_\epsilon) \leq \pi ,$$ where in the second expression $\gamma$ varies over all curves from $(a, b)$ to $(a', b')$. But those points were arbitrary, so $$\operatorname{diam}(M, d) = \sup d((a, b), (a', b')) \leq \pi ,$$ where the supremum is taken over $M \times M$. To show equality, by definition we need to produce sequences $(a_i, b_i)$ and $(a_i', b_i')$ such that $$\sup_i d((a_i, b_i), (a_i', b_i')) = \pi .$$ Of course, it would suffice to find points $(a, b), (a', b')$ such that $d((a, b), (a', b')) = \pi$. One such pair is $(\frac{\pi}{2}, 0)$ and $(\frac{\pi}{2}, \pi)$.