Geodesics Through a Singularity

Question

A singularity on a manifold with metric is defined to be a point at which some geodesic cannot be continued through. For example in Schwarzchild spacetime, $r=0$ defines such a point.

Is it the case that any geodesic which hits a singularity cannot be continued past it? This is obviously true for the Schwarzchild solution. However I worry that perhaps that's just an artifact of the symmetry of the situation. There's nothing in the definition of a singularity that refers to an arbitrary geodesic through that point.

I've tried to think about this myself, but I'm not sure I have the requisite differential geometry to solve it. I can't see how (say) the Hopf-Rinow theorem, or any related results would help!

It would be great to get a mathematician's perspective on this question. Apologies if the solution if obvious and I'm just missing something!

Answer 1

You can consider product spaces:

For example let $C$ be a two dimensional cone with a singularity at the tip, denoted p. If you consider the product space $\mathbb R \times C$ equipped with the product metric (where $\mathbb R$ carries the standard metric) the point $(0,p)$ will be singular but for example the geodesic $t \mapsto (t,p)$ crosses it.

Answer 2

A metric in the plane with a conical singularity at say the origin with total angle $>2\pi$ around it illustrates the point well. The issue is the non-uniqueness of the continuation of the geodesic hitting the origin. Any continuation that forms an angle of $\geq\pi$ on "both sides" can legitimately be considered a geodesic (as far as locally minimizing properties are concerned, for example).