8
$\begingroup$

Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.

Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$ and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:

Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.

$\endgroup$
  • 3
    $\begingroup$ You might like to look at some basic notions of geometric group theory. The set of ends is a quasi-isometry invariant, and the Svarc--Milnor Lemma says that the universal cover of a compact manifold $M$ is quasi-isometric to a Cayley graph of $\pi_1M$. So it suffices to find a group with uncountably many ends, and the free group of rank 2 suffices. Curiously, it's actually impossible for a universal cover to have countably infinitely many ends. $\endgroup$ – HJRW Dec 18 '15 at 20:35
  • $\begingroup$ @FanZheng: Second countable spaces can have infinitely many ends. Take the universal cover of a wedge of two circles, for example. $\endgroup$ – Lee Mosher Dec 18 '15 at 20:44
  • $\begingroup$ @HJRW: The ends of the fundamental group aren't all that relevant to this question, although simply connected examples can be interesting. It's about whether $M$ has uncountably many ends (as in the answer of @SebastianGoette), not the universal cover of $M$. $\endgroup$ – Lee Mosher Dec 18 '15 at 20:49
  • 1
    $\begingroup$ @LeeMosher, I used bad notation, but I think my point stands (unless I misread the question). If $\pi_1N$ is free of rank 2 (say) then taking $M$ to be the universal cover of $N$ provides the example. $\endgroup$ – HJRW Dec 18 '15 at 21:08
  • 2
    $\begingroup$ @HJRW: Sure, for example $N=$ the connected sum of two copies of $S^2 \times S^1$. $\endgroup$ – Lee Mosher Dec 18 '15 at 21:19
16
$\begingroup$

The answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.

Fix $p$ in one of the pairs of pants $Y_0$. For all other pairs of pants $Y$, pick a point $q\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y_0$.

For each sequence $(a_n)_n\in\{0,1\}^{\mathbb N}$, we construct a ray from $p$ leaving $Y_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you very much for your nice edit of my question and also for your clever solution. In this afternoon, I also found a very similar example called as the Cantor tree surface, inspired by your construction. $\endgroup$ – Laiyuan Gao Dec 19 '15 at 10:34
5
$\begingroup$

Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set.

You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set.

In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.