What subset of a 3D space can be addressed by a single chart?

[gnnmartin]

TL;DR

We may need more than one chart to fully describe a 3d curved space. Is there some way to specify what can be covered by a particular chart, and what can't?

Both sides of an Einstein Rosen bridge can be covered by a single chart (using isotropic coordinates). If empty space contains two bridges, then I assume (but can't prove) that the space up to the neck of the bridges can be described in a single chart. I'm interested in how much of infinite space can be covered in a single chart, and how to describe the boundaries. While bearing in mind that the description of an event horizon is not simple in a 3d space, where time is not involved, can one say that the whole of space is covered up to the necks of the Einstein Rosen bridges? More generally, can one describe geometrically the limits of a chart?

 

[Antarres]

The way to construct coordinates around a point is to look for $n$(the dimension of the manifold), commuting vector fields which will be your coordinate basis. There's a statement in differential geometry that if you can find $n$ vector fields which all commute, then that is sufficient and necessary to be able to construct a coordinate chart in the said region.

Let's say that you did it. Then in order to see how far you can extend this chart, you will construct geodesics as extensions of coordinate lines stemming from a point around which you constructed the chart. You will find a geodesic congruence and you will observe how this congruence is spreading apart or converging. Once you find points of convergence(caustics), you will see the limits of your chart. This way, for example, you can see that if you construct a chart on a sphere, your coordinate lines will eventually cross at the antipodal point, and so the sphere cannot be covered by this chart. In more general manifold, this analysis can be, of course, more involved.

Also, you can find characteristic geodesics by solving geodesic equations in coordinates already prescribed on spacetime, and look how far they can be extended, thus looking for singularities.
All in all, it is possible to analyze how far a chart is extending, but that is something depending on the chart and a spacetime, you have to perform the analysis on case to case basis, as far as I know.

 

[gnnmartin]

Thanks. Not quite what I hoped, but perhaps what I hoped is not possible. To take an example, we address the surface of the earth using spherical coordinates. The real surface has hills and valleys and overhangs and caves, and we can address the complete surface with some exceptions. By inventing a 3rd dimension (which of course happens to be real) we can describe those parts of the surface that we can't address, and assert we can address all the other parts of the surface.

In the case of my 3d space, a large part can be addressed by a coordinate system suitable for a 3d space with zero curvature, if we do not try to include places inside identifiable closed 2d surfaces. These closed surfaces can be defined without resort to coordinates. I rather hoped that having identified the 2d surfaces, I could say that the coordinate system mapped everywhere except:
1: inside the closed surfaces and
2: other exceptions.
I recognise that there are other curved 3 spaces that still can't be completely mapped with one chart, and you gave the example of a 3 sphere. So I hoped that the 'other exceptions' just mentioned would be identifiable by (say) their being closed 3 surfaces.

 

[jbergman]

The way to construct coordinates around a point is to look for $n$(the dimension of the manifold), commuting vector fields which will be your coordinate basis. There's a statement in differential geometry that if you can find $n$ vector fields which all commute, then that is sufficient and necessary to be able to construct a coordinate chart in the said region.

Let's say that you did it. Then in order to see how far you can extend this chart, you will construct geodesics as extensions of coordinate lines stemming from a point around which you constructed the chart. You will find a geodesic congruence and you will observe how this congruence is spreading apart or converging. Once you find points of convergence(caustics), you will see the limits of your chart. This way, for example, you can see that if you construct a chart on a sphere, your coordinate lines will eventually cross at the antipodal point, and so the sphere cannot be covered by this chart. In more general manifold, this analysis can be, of course, more involved.

Also, you can find characteristic geodesics by solving geodesic equations in coordinates already prescribed on spacetime, and look how far they can be extended, thus looking for singularities.
All in all, it is possible to analyze how far a chart is extending, but that is something depending on the chart and a spacetime, you have to perform the analysis on case to case basis, as far as I know.

I don't think this is correct. Just take the circle with the vector fied of a unit $d/d\theta$. There isn't a single chart for the circle.

Basically, you also need the underlying subset to be homeomorphic to $\mathbb{R}^n$.

 

[Antarres]

@jbergman Well... I never said that you can make a single chart on a circle, I was just describing the way you extend your chart using Riemann normal coordinates around a point. Obviously you wouldn't be able to cover a whole manifold by a single chart.

 

[jbergman]

@jbergman Well... I never said that you can make a single chart on a circle, I was just describing the way you extend your chart using Riemann normal coordinates around a point. Obviously you wouldn't be able to cover a whole manifold by a single chart.

Sorry I misread your post.

 

[Antarres]

@gnnmartin I'm not sure that I understand your construction, but when it comes to spherical coordinates on the earth(considered as a sphere), you don't need a third dimension to describe the terrain. You can use the same spherical coordinates to describe points on hills and valleys, since a sphere with hills and valleys is homeomorphic to a normal sphere.

The problem with spherical coordinates is that they're underfined at the poles(because asimuthal angle is undefined there as you know). That's why this coordinate system, a degenerate one, is not considered a valid way to cover a manifold. In case of a more complicated manifold, even a spherical coordinate system would not suffice to cover it. But generally the coordinate system, that is, a chart that you use to cover a part of a manifold must be defined in all the points that you're aiming to cover. As far as I know, no curved space can be covered by one such coordinate system. Your first question was about the domain of validity of a picked coordinate system, and I mentioned that by seeing how far you extend your coordinate lines, you can see how far it can reach. But depending on the particular space that you cover, and particular coordinates that you use, you'll have to perform this analysis to see what region your charts are covering.

Sometimes in some manifolds, if you can manage to deal with degenerate points, like poles on the sphere, you'll still use a coordinate system that's degenerate and pay attention to those points when you have to deal with them. But in general case, every manifold is covered by multiple charts that you prescribe.

 

[gnnmartin]

I'm not sure that I understand your construction, but when it comes to spherical coordinates on the earth(considered as a sphere), you don't need a third dimension to describe the terrain. You can use the same spherical coordinates to describe points on hills and valleys, since a sphere with hills and valleys is homeomorphic to a normal sphere.

You don't need a third dimension, but it can enable you to define a large chart. Forget about the earth, and just consider a part of the earth, or indeed any large non-flat surface with mountains, valleys and caves. The imaginary 3rd dimension enable us to define the parts of the surface (the caves) that, when excluded from the chart, enable us to include everywhere else on the surface in the chart.

You say that
As far as I know, no curved space can be covered by one such coordinate system. But if you call a non-flat surface a curved surface, then the 'bumpy but flat on a large scale' surface can (I say diffidently) be entirely mapped with a single chart as long as there are no 'caves' as defined with the help of an imaginary 3rd dimension. I had hoped the same logic applied to a 3 dimensional non-flat open space.

I think I ought to withdraw now. I had hoped there was something uncontroversial that I could state that would make an argument I am constructing easier to accept, but that now seems unlikely. Thank you again for your help. I'll continue to monitor the thread in case I am asked a question, but won't otherwise try to pursue the thread further.

 

[gnnmartin]

I think I ought to withdraw now.

But find my problem melting away, thanks to this threads promptings. If you can project the 3 space onto a flat 3 space, then you can map it with a single chart. If you have an algorithm that excises volumes to resolve ambiguity, then you can map the remaining space in a single chart. The problem I have is touched on by https://www.physicsforums.com/threads/global-coordinate-chart-on-a-2-sphere.1007558/, does the topology of the 3 space prevent the mapping in some way that sidesteps the algorithm?

I don't know enough about topology to make use of that, but I am happy enough to assume a 'simple' topology.

 

[Antarres]

Well, to answer your last question, when you are making a projective transformation, what you're doing is that you're basically taking points in one space and using a congruence of lines(usually straight lines, that you can imagine as rays) you're connecting points from one space to the points in another space. If this type of transformation is a bijection, then you can say that every point of one space has a one to one correspondence with a point in another space. If you can make such a transformation onto a flat space, then you can effectively construct a single chart on the base space that you're projecting. However, it is not always possible to create a bijection in this way, often projections are maps that are not injective(they're still useful despite this fact).

Indeed, the topological structure of the space is helpful in determining whether this is possible or not. The dimensionality of the space you're in doesn't play a huge role here. It's obvious that using a projection you can map a higher dimensional space onto a lower dimensional one, but this is irrelevant.

For example, a sphere is a 2-dimensional surface. Maybe you imagine it as a 2-dimensional surface that is immersed into a 3d space, but intrinsically you don't have to imagine it like this, a sphere doesn't have to be immersed into a higher dimensional space in order to be imagined or analyzed.

When you want to see whether two spaces have the same topology, you're looking for transformations that are bijective and continuous both ways(meaning both the direct and inverse transformation are continuous). Spaces connected by such transformations are called homeomorphic, and they have the same topological structure. For example, a 2-sphere doesn't have the same topology as flat 2-space, and so it cannot be continuously mapped onto flat 2-space. However parts of the sphere can be mapped continuously onto flat space. This is something that is the definition of a manifold. On a manifold, parts of the space can be continuously mapped onto flat space and those mappings are called charts. In order to be able to construct a single chart on the whole manifold, that manifold needs to be homeomorphic to flat space. This is not possible for every manifold, and it doesn't depend on the dimensionality of the manifold, those problems appear in all dimensions.

Dimension of the manifold is precisely defined by the dimension of the flat space which you map onto using your charts.

 

[WWGD]

Ultimately, if a single chart suffices, the space is itself ( globally, not just locally)homeomorphic to 3D, or, equivalent, to an open 3-ball. So it must share all topological characteristics of $\mathbb R^3$.