[Math] Ways in which a manifold can be geodesically incomplete

differential-geometrygeneral-topology

Naively I would have thought that a manifold becomes geodesically incomplete if there are missing points in it or if the geodesics are hitting a boundary.

But I am not sure how to think of geodesic completeness or not for manifolds with a boundary. Like how to think for the closed disk on $\mathbb{R}^2$

I would like to know of other generic ways in which geodesic incompleteness can happen. (Like for pseudo-Riemannian manifolds the Hawking-Penrose theorems pin down many generic scenarios)

Like from $\mathbb{R}^2$ (with the standard metric) if one removes the point $(0,0)$ then there is no geodesic from say the point $(-1,0)$ to $(1,0)$.

  • But I can't visualize how I can actually put a complete metric on this punctured plane and make it geodesically complete? I can try to make the metric hyperbolic near the deleted point so that the geodesics approaching it never actually reach it but go off to infinity but even then I can't see how it will produce a geodesic which connects $(-1,0)$ to $(1,0)$

Also there are two other ways of producing geodesic incompleteness which I know but don't have a good understanding of,

  • Any open subset of a complete connected Riemannian manifold is geodesically incomplete. (I guess this follows from Hopf-Rinow but I can't see it clearly)

  • A noncompact surface which is not diffeomorphic to $\mathbb{R}^n$, and if for some metric every point on this surface has positive curvature, then the metric on it must be incomplete.

I would be happy to see explanations about the above things and also examples for the second case.

Also is it possible to write down as formulas simple examples of incomplete geodesics?

Best Answer

To your first bullet point, you certainly could make it hyperbolic near the point. The (nonunique) geodesic connecting (-1,0) to (1,0) will go around the cusp and look semicircularish (but I don't think it will actually be a semicircle, just approximately one).

Perhaps something easier to visualize is that $\mathbb{R}^2 - \{pt\}$ is diffeomorphic to $S^1\times\mathbb{R}$. If you give this space the product metric of the usual metrics, then you can easily see and work out the details.

To you second bullet point, you should modify the statement slightly (and somewhat pedantically). Any proper nonempty open subset $U$ of a complete connected manifold $M$ is incomplete. To see this, let $p\in U$ and $q\in M-U$. By Hopf-Rinow, since $M$ is complete there is a geodesic $\gamma$ starting at $p$ and ending at $q$. Since $U$ is an open subset, it is totally geodesic: what $U$ thinks are geodesics are precisely what $M$ does. Thus $U$ thinks $\gamma$ is a geodesic which doesn't stay in $U$ for all time, hence $U$ is incomplete.

To the third bullet point, (you say "surface" then use $\mathbb{R}^n$), the theorem is true for $\mathbb{R}^n$ from Cheeger and Gromoll's Soul Theorem together with Perelman's proof of the Soul Conjecture. The soul theorem states that if $M$ is complete and has nonnegative sectional curvature, then $M$ has a compact totally convex totally geodesic submanifold $K$ (called the soul) so that $M$ is diffeomorphic to the normal bundle of $K$. The Soul conjecture asks: If $M$ has nonnegative curvature everywhere and a point with all sectional curvatures positive, must $K$ be a point?

Perelman proved the answer is yes: under these hypothesis, $K$ is a point. But a normal bundle of a point in a manifold is diffeomorphic to $\mathbb{R}^n$.

Finally, I have been told, but I have no idea about references/proofs/etc that every noncompact surface has some metric (necessarily incomplete if the surface isn't $\mathbb{R}^2$ by the above) of positive curvature. I'll try to dig up a reference tomorrow, if I remember to.