# Is a bounded (Riemannian) manifold always totally bounded?

Let $(M,g)$ be a finite dimensional Riemannian manifold. Suppose it is bounded as a metric space, i.e. $$\sup \{ d(x,y) | x,y \in M \} < \infty$$ where $d$ is the distance induced by $g$. Is $M$ also totally bounded?

I know bounded spaces are not totally bounded in general, but the only examples I know are in some infinite dimensional Banach space. I think the answer should be yes for manifolds, since by Nash Embedding Theorem they are (isometric to) subspaces of some Euclidean space, and bounded subspaces of $\mathbb{R}^n$ are also totally bounded. If this is true (which assumes I correctly understand the statement of Nash Theorem), is there a more concrete proof?

• The Nash Embedding theorem is a very, very, very big gun, which you almost certainly will not need here. I may be wrong, but the difference between finite and infinite dimensional B-spaces which, in my opinion, is relevant here, is the fact that for finite dimensional manifolds you have local compactness, which you inherit from $\mathbb{R}^n$ through the coordinate patches. I'll try to think of a proof, but don't know whether I'll find the time. (But yes, the Nash embedding theorem will allow you to give a proof as well ;-) Apr 11, 2014 at 17:27
• Actually, after thinking of it for some more time, I don't think the Nash Embedding theorem will help you. If I recall correctly, by increasing the dimension, you can isometrically embed any Riemannian manifold into a subset of Euclidean space of prescribed arbitrarily small size. I would have to search for a reference for this, though. But consider a half line embedded isometrically (i.e. parametrized by arclength) as a spiral spiralling to $0$ in $\mathbb{R}^2$. This is certainly not totally bounded. Don't confuse intrinsic distance and extrinsic distance. Apr 11, 2014 at 17:37
• If $(M, g)$ is connected and complete, the Hopf-Rinow theorem guarantees that closed, bounded subsets of $(M, g)$ are compact, hence totally bounded. Offhand, it looks possible to adapt this into a proof that "bounded implies totally bounded" for a connected (second-countable) Riemannian manifold, though I haven't actually tried to fill in the details. Apr 11, 2014 at 21:14
• Thanks for pointing out Hopf-Rinow! this positively answers my question in the case of a connected complete manifold, since $M$ is trivially a closed and bounded subset of $M$. The punctured disk example below is of course not complete.
– Lor
Apr 12, 2014 at 10:44

This is false. Take universal cover $M$ of the open 2 dimensional Euclidean unit disk $D^2$ with center removed, equipped with pull back flat metric. I let $d$ denote the distance function on $M$ induced by the pull-back Riemannian metric. I claim that the diameter of $M$ is $\le 2$ (actually, it is exactly 2 but we do not need this). To see this it is convenient to identify $M$ with the complex half-plane $\{z: Re(z)<0\}$ and the universal cover $M\to D^2\setminus 0$ with the complex exponential function. Note also that for every $y, t\in {\mathbb R}$, $$\lim_{x\to-\infty} d(x+ iy, x+it)=0.$$ Furthermore, the length of every path $(x+iy), -\infty<x<\infty$ equals $1$. Therefore, to connect any two points $p, q\in M$ by a path of length $<2$, first connect $p$ and $q$ by horizontal lines to some points $p', q'$ with $Re(p')=Re(q')\ll 0$ and then connect $p'$ to $q'$ by a vertical interval.
• +1. But I think you should indicate how to verify that the condition on the $\sup$ is satisfied in this example. Apr 12, 2014 at 6:24
• Are you suggesting this universal cover with the pullback metric has finite diameter? I have some difficulties understanding this. Please correct me if I'm wrong: the universal cover can be identified with the upper half plane with projection $F(x,y)=e^{-y}(\cos(x),\sin(x))$. Therefore the induced metric on it is $g(x,y)=e^{-2y}(dx^2+dy^2)$ and for this metric horizontal lines (i.e. $y=c$) have infinite length.
• Great example! ${}$ Apr 12, 2014 at 15:41
• I think the identification with the half plane my actually be more confusing. It may be easier to say to think about the polar coordinate representation of $D$ as $(r,\theta) \in (0,1) \times \mathbb{R}$, except now we don't identify $\theta$ with $\theta + 2\pi$. Then you can say that to get from a point $(r,\theta)$ to a point $(R,\Theta)$ with $\Theta - \theta$ between $2k\pi$ and $(2k+2)\pi$, you first go from $(r,\theta)$ to very close to the origin, then spin around the origin $k$ times, and then shoot out radially. Mechanically this is the same thing you wrote, but I think is... Jun 19, 2020 at 18:12