6
$\begingroup$

Problem. I'm working on an exercise from a book in basic Gaussian geometry, and as a part of my solution, I would like to make the following claim:

Proposition. If $M\subseteq\mathbb{R}^3$ is a complete regular surface, $M$ is a closed subset of $\mathbb{R}^3$.

given the following definition of completeness:

Definition. A regular surface $M\subseteq\mathbb{R}^3$ is said to be complete if for every point $p\in M$ and every tangent vector $Z\in T_pM$, there exists a geodesic $\gamma\colon\mathbb{R}\to M$ defined on all of $\mathbb{R}$ such that $\gamma (0)=p$ and $\gamma'(0)=Z$.

I have tried mainly two ways of proving it:

Idea 1. What seems most natural to me is to do this by contradiction, and to use the limit point characterization of closed sets in metric spaces.

I thus let $M\subseteq \mathbb{R}^3$ be a complete regular surface, and suppose that $M$ is not closed. This means there exists a convergent sequence $(p_n)_{n=1}^\infty$ in $\mathbb{R}^3$ with limit $q$, such that $p_n\in M$ for all $n\in\mathbb{Z}^+$ and $q\in\mathbb{R}^3\setminus M$.

The way I picture this intuitively, is that the surface $M$ is punctured at $q$ or have something like an "edge " at $q$. Since there are $p_n$'s arbitrarily close to $q$, it feels like we should be able to find a $p_n$ and a $Z\in T_{p_n}M$, such that the geodesic $\gamma\colon \mathbb{R}\to M$, with $\gamma (0)=p_n$ and $\gamma'(0)=Z$, would have to pass through $q$, thus giving us the desired contradiction. I don't see any obvious ways to turn this into a formal argument though, and my gut feeling can very well be wrong.

Idea 2. I have also looked a bit at the Hopf-Rinow theorem from Riemannian geometry, which seems to say that a Riemannian manifold $(M, g)$ being geodesically complete implies that $(M,\tilde{d})$ is complete as a metric space, where $\tilde{d}$ is the intrinsic metric.

If something similar holds for regular surfaces in $\mathbb{R}^3$, I think we would be done (no!), because $\tilde{d}$ dominates the standard metric $d$ in $\mathbb{R}^3$ restricted to $M$. This means that if $\rlap{\rule[0.5ex]{2.5em}{0.2ex}}(M,\tilde{d})$ is complete, so is $\rlap{\rule[0.5ex]{3.5em}{0.2ex}}(M,d_{|M})$ (this is wrong, see my comment below!). Since $(\mathbb{R}^3,d)$ is complete, $(M,d_{|M})$ being complete implies that $M$ is closed by elementary point-set topology.

However, my spontanious idea of how do show that geodesically complete implies complete as a metric space would be very similar to Idea 1 (I would seek a contradiction by supposing that we have a Cauchy sequence in $M$ that doesn't converge), so this doesn't seem to take me any further either.

Question. Am I at all on the right track here, or would you recommend some other approach? Is the proposition I'm trying to prove even correct in the first place?

$\endgroup$
5
  • $\begingroup$ One possible result is: if $M$ complete is an open submanifold of $X$, then $M$ must be a union of connected components of $X$. In particular, an open complete submanifold of $\mathbb{R}^n$ is $\mathbb{R}^n$. That may not be enough for what you want to prove. $\endgroup$
    – orangeskid
    Oct 9, 2017 at 10:26
  • $\begingroup$ I made a mistake in my original post. Even if we were able to show that $(M,\tilde{d})$ is complete, that would not necessarily mean that $(M,d_{|M})$ is complete. That $\tilde{d}$ dominates $d_{|M}$ means that every Cauncy squence in $(M,\tilde{d})$ is a Cauchy sequence in $(M,d_{|M})$. This means that it is the completess of $(M,d_{|M})$ that implies the completeness of $(M,\tilde{d})$; not the other way around. $\endgroup$
    – Oskar Henriksson
    Oct 12, 2017 at 21:05
  • $\begingroup$ Yes, that was the problem I faced, did try to prove it and couldn't.... There were these sequences going nowhere in $M$ and still Cauchy in $\mathbb{R}^n$. A good question nevertheless. $\endgroup$
    – orangeskid
    Oct 12, 2017 at 21:19
  • $\begingroup$ @orangeskid Is the reverse implication true? I.e. is every closed regular surface geodesically complete? From point-set topology, we know that $M$ being a closed subset of $\mathbb{R}^3$ means that $(M,d_{|M}$ is a complete metric space, which, as we have seen above, implies that $(M,\tilde{d})$ is also a complete metric space. Can (a variation of) Hopf-Rinow now be applied to conclude that $M$ is geodesically complete in the sense of the definition in my original post? $\endgroup$
    – Oskar Henriksson
    Oct 12, 2017 at 23:05
  • 1
    $\begingroup$ Yes., geodesically complete is equivalent to $\tilde d$ complete. That is part of Hopf-Rinow if I am not mistaken. Of course, it has to be a closed submanifold of a complete manifold. I don't have a good source for Hopf-Rinow. Maybe Bishop and Crittenden? $\endgroup$
    – orangeskid
    Oct 12, 2017 at 23:28

1 Answer 1

Reset to default
7
$\begingroup$

That may not be true. Consider the graph of the function $x\mapsto \sin \frac{1}{x}$ over $(0, \infty)$. This is a complete submanifold of $\mathbb{R}^2$. If you take the product of this curve with $\mathbb{R}$ you get a complete surface in $\mathbb{R}^3$ which is not closed. Note that the surface is isometric to $\mathbb{R}^2$, very creased.

$\endgroup$
9
  • 2
    $\begingroup$ Interesting, so although a the distance from a fixed point to a limit point is finite in $\Bbb R^3$, the intrinsic distance is going to infinity as you approach it. I was going to suggest writing $p_n = \exp_p (t_nv_n)$, where the $v_n$ are unit vectors in $T_pM$. You certainly can assume $v_n\to v$, but the $t_n\to\infty$. Oops. (I think the same thing will happen even with $x\sin(1/x)$, as it is also of infinite arclength.) $\endgroup$
    – Ted Shifrin
    Oct 8, 2017 at 22:01
  • $\begingroup$ @Ted Shifrin; Yes indeed. Would be nice to plot some geodesics. $\endgroup$
    – orangeskid
    Oct 8, 2017 at 22:23
  • $\begingroup$ Well, this is just a cylinder over your plane curve. So we know what the geodesics are on any such creature. $\endgroup$
    – Ted Shifrin
    Oct 8, 2017 at 22:25
  • 1
    $\begingroup$ @Oskar Henriksson: The induced metric is in fact a product metric. As for the factors, one is the standard $\mathbb{R}$, the other one is the long graph ( so again isometric to $\mathbb{R}$). A cylinder in fact, over this long curve. So it's isometric to $\mathbb{R}^2$. If you consider the less wild curve ( still of infinite length) suggested by Ted Shifrin ( graph of $u \sin \frac{1}{u}$, you can take both of the pieces. They are then two $\mathbb{R}^2$, connected at a singular edge. Funny... $\endgroup$
    – orangeskid
    Oct 12, 2017 at 21:36
  • 1
    $\begingroup$ @Oskar Henriksson: then you have a topological manifold, with a singular edge, and the smooth pieces are complete with the induced metric. Perhaps even nicer. And you can never reach the edge if the speed is bounded... $\endgroup$
    – orangeskid
    Oct 12, 2017 at 21:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .