# How to view black holes / horizon from a manifold perspective

1. Apr 3, 2014

### clerk

Hi friends,
I was wondering about the following - in GR texts we always see these penrose diagrams and some line representing the horizon and all these timelike , spacelike curves and all that ... but the picture that I have of GR is just that of a smooth 4 manifold endowed with a metric . Can I view black holes in such a picture? For example, this event horizon is a submanifold and the singularity as some point which cannot be covered by any coordinate patch ? Thanks for any help ..

2. Apr 3, 2014

### WannabeNewton

Well yes the black hole is just a submanifold with boundary embedded in the space-time manifold. The singularity being complemented out of the manifold is a bit more subtle an issue.

However this is rather standard material that is covered in advanced GR texts, see for example chapters 9 and 12 of Wald and chapters 8 and 9 of Hawking and Ellis.

3. Apr 3, 2014

### bcrowell

Staff Emeritus
Yes, but the standard way to define this is that singularity is not part of the manifold. If you did want to define it as part of the manifold, it wouldn't be a point. It would be a spacelike surface.

4. Apr 3, 2014

### PAllen

But not an ordinary 3-surface. On approach to the SC singularity, the 3-surfaces approach a collapsed 3-cylinder: 2-sphere X line, with 2-sphere area going to zero. The approach surfaces can be embedded in Euclidean 4-space but not in Euclidean 3-space.

5. Apr 3, 2014

### pervect

Staff Emeritus
I thought the singularity would be a line, r=0, t=anything. WHich is spacelike because of the nature of t inside the event horizon...

6. Apr 3, 2014

### PAllen

If you construct a sequence of spacelike surfaces approaching the singularity (which is normally not part of the metric) every such 3-surface embeds 2-spheres; the closer to the singularity, the smaller (in area) the 2-spheres. Each of these 3-surfaces is not embeddable in Euclidean 3-space, so just calling the singularity a line is abstracting from these topological issues.

7. Apr 3, 2014

### bcrowell

Staff Emeritus
The dimensionality of the set missing from a particular coordinate space can depend on the choice of coordinates; the dimensionality of the same set can be different if you pick a different set of coordinates . For example, the origin of a Euclidean plane in Cartesian coordinates is a set in the coordinate space with dimension zero, but it you switch to polar coordinates it's a set with dimension 1.

In the example of the Euclidean plane, we can resolve this ambiguity because we have a metric that is well defined at the origin. But in the Schwarzschild spacetime we don't have a metric at the singularity, so it's not obvious to me that this ambiguity can be resolved.

Say we have a black hole that forms by collapse from a spherically symmetric shell of mass. The shell is two-dimensional before the event horizon forms. If it's two-dimensional the instant before the event horizon forms, then I think it's clearly two-dimensional the instant after. I think it then remains two-dimensional up until the instant before the singularity forms. But does it then retain its original dimensionality once it becomes the singularity? I don't see any obvious way to answer this question, because we don't have a metric at the singularity.

This is reminiscent of the difficulties involved in defining whether a singularity is timelike or spacelike. It's not a point-set on which the metric is defined, so there is no obvious, simple way to define it. To define it, you have to go through various complicated indirect ideas like discussing light-cones near the singularity. See http://adsabs.harvard.edu/full/1974IAUS...64...82P . Maybe this approach can be adapted in order to define the dimensionality of a singularity.