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?

  • $\begingroup$ Note that in general relativity, "missing points" basically are what we have at singularities. That is, a singularity such as a black hole singularity is not represented mathematically as a set of points on the manifold. It's represented as a missing set of points. The curvature blows up near these missing points in such a way that it's not possible to extend the manifold to fill in the missing points. $\endgroup$ – Ben Crowell Sep 17 '12 at 14:14
  • $\begingroup$ You can check out constructions of semi-Riemannian manifolds where incompleteness happens for a specific type of geodesic (spacelike, lightlike or timelike) and doesn't happen for the others in O'Neill's book, entitled Semi-Riemannian Geometry with Applications to Relativity. $\endgroup$ – student Sep 17 '12 at 16:10

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.

  • $\begingroup$ thanks for your explanation. Towards the end are you meaning that any non-compact surface (which isn't the whole plane) has an incomplete metric? Also I am not very sure about the intuition that boundaries prevent from geodesic completeness. Isn't a closed disk geodesically complete? But there are geodesics on it which are geodesic also in $\mathbb{R}^2$ and will leave the disk eventually. I am getting confused about how to visualize geodesic incompleteness. $\endgroup$ – Anirbit Jan 20 '11 at 7:46
  • $\begingroup$ Also it will be helpful if you can say show some explicit geodesic equations which are incomplete. $\endgroup$ – Anirbit Jan 20 '11 at 7:47
  • $\begingroup$ At the end, I mean that every non-compact surface which isn't the whole plane not only has an incomplete metric, but has an incomplete metric of positive curvature. About the closed disc question - I'm not really sure what the right definition of "geodesic completeness" is on a manifold with boundary - I've only studied manifolds with no boundary. Finally, I don't actually have a single example of explicit geodesic equations which are incomplete. Part of the problem is that a given set of geodesic equations can be complete on one domain but incomplete on the other $\endgroup$ – Jason DeVito Jan 20 '11 at 17:14
  • $\begingroup$ (continued) In other words, you can't necessarily see geodesic incompleteness in the geodesic equations themselves. $\endgroup$ – Jason DeVito Jan 20 '11 at 17:18

The OP asked in passing what new issues come up when you generalize from a Riemannian space to a semi-Riemannian one. I'll try to sketch the description.

We generally don't want to call something a singularity if it's in some sense infinitely far away so that you can't get to it. To formalize this notion, you need the idea of geodesic incompleteness. In the Riemannian case, you can characterize this by saying that a geodesic can't be extended in a certain direction past a certain length, as measured by the metric. In the semi-Riemannian case, you don't want "length" as measured by the metric, because a null curve always has zero length. Instead you need to talk about extending the geodesic past a certain affine parameter.

As in the Riemannian case, you can have both curvature singularities and conical singularities. In 2+1 dimensions, all singularities in vacuum solutions are conical, i.e., you don't get black holes with event horizons.

You can have singularities that are timelike, spacelike, or null. These notions aren't defined by default, because the singularity isn't a set of points on the manifold. E.g., to define a timelike singularity, you have to talk about a timelike curve (an observer) containing points p and q, such that the singularity is in p's causal future and q's causal past. Black-hole and cosmological singularities are spacelike, not timelike. A timelike singularity is one way of defining a naked singularity (Penrose 1973), which causes Cauchy surfaces not to exist (Geroch 1970).

The way relativists end up mentally classifying all of these things is in terms of their Penrose diagrams: http://en.wikipedia.org/wiki/Penrose_diagram

Penrose, Gravitational radiation and gravitational collapse; Proceedings of the Symposium, Warsaw, 1973. Dordrecht, D. Reidel Publishing Co. pp. 82-91, http://adsabs.harvard.edu/full/1974IAUS...64...82P (not paywalled)

Geroch, J Math Phys 11 (1970) 437


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