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In the completeness charpter of book "Riemannian manifold: an introduction to curvature", he constructed two examples (at page 108) of non geodesically complete manifold. 1) any proper open subset of $\mathbb{R}^n$ with standard metric is not geodesically complete, because the geodesics reach the boundary in finite time, so what happens to closed subset? I guess it is complete otherwise it violates the Hopf-Rinow theorem, but what the difference between open and closed set. 2) $\mathbb{R}^n$ with the pullback metric from $\mathbb{S}^n$ by stereographic projection, there are geodesics that escape to infinity in finite time, honestly speaking, I have no idea what it means. Could someone explain a little bit about this. Much appreciated.

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  • $\begingroup$ I figure out first one, I forgot the Hopf Rinow theorem is for manifold without boundary. $\endgroup$ – H-H Sep 21 '18 at 0:04
  • $\begingroup$ Exactly, for the first one, just use Hopf-Rinow, a space is geodesically complete iff it's complete with the distance induced by its metric, No open subset of $\mathbb{R}^n$ is complete with the usual distance, because it's not closed, for example. If a closed subset is also bounded, then it's compact and hence complete again. $\endgroup$ – Laz Sep 21 '18 at 1:58
  • $\begingroup$ I think a bounded closed subset is not complete in geodesic sense, even though complete in distance metric sense. It doesn't violate the Hopf-Rinow theorem. $\endgroup$ – H-H Sep 21 '18 at 2:05
  • $\begingroup$ Take a look in do Carmo's "Riemannian Geometry", page 146 Theorem 2.8. One of the equivalences to the property of being geodesically complete is item b), "every closed bounded set is compact". You have that in $\mathbb{R}^n$, exaclty those are the compact subsets in Euclidean space. $\endgroup$ – Laz Sep 21 '18 at 2:10
  • $\begingroup$ The problem is that a closed subset of $\mathbb R^n$ needs not be a Riemannian manifold. If it is the closure of a smooth open set then it will be a manifold with boundary. (Which is, I think, exactly the content of the first comment) $\endgroup$ – Giuseppe Negro Sep 21 '18 at 21:16
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Remember that if you put a metric on $\mathbb{R}^n$ making a pullback of the metric in $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ by the stereographic projection $\sigma$, then because $\sigma$ is a diffeomorphism (from $\mathbb{S}^n$ minus the North Pole to $\mathbb{R}^n$), it becomes an isometry. Now, isometries take geodesics to geodesics.
Just take a geodesic through the North Pole $N=(0,0, \dots, 0,1)$ (a great circle which is the intersection of $\mathbb{S}^n$ with a plane $P$ through the origin in $\mathbb{R}^{n+1}$ and $N$). Take the piece of it that goes from the South Pole $S=(0,0, \dots, 0,-1)$ to $N$ and parametrize it by acrlength, if you want. This geodesic goes to a straight line through the origin in $\mathbb{R}^n$, and since the one in $\mathbb{S}^n$ is parametrized by arclenght, this straight line $\gamma=\gamma(t)$ is also parametrized by arclenght and it satisfies $\lim_{t\to \pi} \gamma(t)=\infty$. Its lenght between the parameters $0$ and $\pi$ is $\infty$ in the topological sense that in time close $\pi$ it's out of every compact.
I think that's what Lee meant.

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