Let $M$ be a connected smooth manifold without boundary. Are the following equivalent?

  1. $M$ is compact
  2. $M$ cannot be realized as a proper open subset $M\subset N$ of another connected manifold $N$. In other words, $M$ cannot be made bigger without adding connected components.
  3. The flow of every vector field on $M$ is complete.

I recall wondering about this as a student. Some of the implications are well know but I never figured out the others. This question reminded of this.

  • 1
    $\begingroup$ Make a infinite chain of tori (gluing with connected sums) so get an infinite chain. This is a non-compact surface: can you embed this as a proper open subset of a connected surface? $\endgroup$ – Mariano Suárez-Álvarez Sep 24 '12 at 0:42
  • 3
    $\begingroup$ @Mariano: for that surface, yes you can. One way to see this is that this surface is diffeomorphic to a sub-surface, the one you get by removing a properly-embedded half-open interval $[0,\infty)$ from the surface, so $\infty$ goes to the end of the surface. The diffeomorphism is obtained by pushing the end into the surface, a type of "finger move". $\endgroup$ – Ryan Budney Sep 24 '12 at 6:16
  • $\begingroup$ @RyanBudney, cool! $\endgroup$ – Mariano Suárez-Álvarez Sep 24 '12 at 7:47
  • $\begingroup$ @Ryan: Can you explain in more detail the "pushing the end into the surface" part of that? I've played around with this answer on other noncompact surfaces and could see how it works, but I'm just not seeing it in this example. $\endgroup$ – Jason DeVito Sep 24 '12 at 20:01
  • $\begingroup$ somewhat similar: math.stackexchange.com/questions/53021/… $\endgroup$ – Ben Crowell Sep 24 '12 at 22:11

That 1 implies 3 is standard and appears in most textbooks on manifolds that include vector field flows, it follows by a standard existence and uniqueness argument.

3 implies 1 is a little more work but it's a fairly standard argument as well. The idea is that if the manifold is non-compact you can take a proper Morse function on the manifold, $f : M \to [0,\infty)$. If you re-scale the gradient like $\frac{f^2}{1+|\nabla f|^2}\nabla f$, its flow lines can't be complete since it hits infinity in a finite time. And technically you don't need any Morse theory to make this conclusion. It's enough to have a proper function $f : M \to [0,\infty)$ then show you can embed an arc $p : [0,\infty) \to M$ in $M$ so that $f(p(x)) = x$ for all $x$. You can define an incomplete vector field whose flow line is the image of $p$, then extend it to $M$.

To relate 1 and 2 notice that in a manifold, a subspace is compact, connected and open means its a path-component. If you take a manifold and collapse each path-component to a point (the path-component space) you get a discrete space. So this gives you 1 implies 2. The converse 2 implies 1 follows from the argument in my comment to Mariano above -- if it's non-compact you can take a properly embedded arc, cut along it and self-embed. In reverse this embeds your manifold in a bigger manifold.

  • $\begingroup$ I don't understand the gradient scaling. If the function is $f(x)=x^2$ on $M=\mathbb R$, the flow lines of $\dot x = 2x/(1+x^4)$ are complete. $\endgroup$ – user31373 Sep 24 '12 at 15:42
  • $\begingroup$ Whoops, I picked the wrong scaling. Will edit. $\endgroup$ – Ryan Budney Sep 24 '12 at 19:34
  • $\begingroup$ Great, in particular the last part is so simple! $\endgroup$ – Michael Bächtold Sep 25 '12 at 10:25

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