I Geometries from Line Element $$dl^2=d\theta^2 + \sin^2\theta\, d\varphi^2$$

[Demystifier]

Consider the line element
$$dl^2=d\theta^2 + \sin^2\theta\, d\varphi^2$$
where $\theta\in [0,\pi]$. The standard interpretation of this line element is to take $\varphi\in [0,2\pi)$, in which case the line element represents the standard metric of the sphere $S^2$. However, from the line element itself I don't see why I could not take $\varphi\in (-\infty,\infty)$. Is it consistent to interpret $\varphi$ as $\varphi\in (-\infty,\infty)$, and if it is, what geometrical object the line element represent in this case? Is it a smooth manifold, or does it contain some singularities?

 

[PeterDonis]

Is it consistent to interpret $\varphi$ as $\varphi\in (-\infty,\infty)$, and if it is, what geometrical object the line element represent in this case?

I don't know if there is a unique answer to this question; however, one possible answer is a cylinder, $S^1 \times R$. The $\theta$ coordinate would wrap around the cylinder, and the $\phi$ coordinate would be a weirdly scaled axial coordinate along the cylinder, with the scaling dependent on $\theta$ instead of being constant. The chart would not cover the entire cylinder since the locus $\theta = 0$ (or $\theta = \pi$) would be a coordinate singularity.

 

[DrGreg]

I suppose you could consider it embedded in 3D space with a radial coordinate $r=e^\varphi$, but the metric wouldn't be Euclidean. The cross-section in the equatorial xy-plane would be a logarithmic spiral. The cross-section in any vertical plane that included the z-axis would be two infinite sets of concentric semicircles.

If you restricted yourself to $\varphi\in (0,\infty)$ you could use $r = \varphi$ instead.

 

[Orodruin]

I don't know if there is a unique answer to this question; however, one possible answer is a cylinder, $S^1 \times R$. The $\theta$ coordinate would wrap around the cylinder, and the $\phi$ coordinate would be a weirdly scaled axial coordinate along the cylinder, with the scaling dependent on $\theta$ instead of being constant. The chart would not cover the entire cylinder since the locus $\theta = 0$ (or $\theta = \pi$) would be a coordinate singularity.

The thing is though, the coordinate chart where $\theta$ and $\phi$ are valid coordinates already describe a cylindrical topology where the geometry is that of a sphere. You can remove the coordinate singularity but you will then just recover the sphere.

With the line element of the standard sphere and the Levi-Civita connection, you recover the sphere. You may of course consider other connections, which would describe different geometry, but that opens up an entirely new can of worms.

 

[martinbn]

The same manifold can have many different metrics, so a metric cannot determine the manifold. Strictly speaking you need the manifold first. The question is a bit like "I am think of a manifold with this metric, guess what I have in kind."

 

[Orodruin]

The same manifold can have many different metrics, so a metric cannot determine the manifold. Strictly speaking you need the manifold first. The question is a bit like "I am think of a manifold with this metric, guess what I have in kind."

The relevant question in this question is: given a coordinate chart with a particular form for the metric, what are the possible global geometries?

The chart here obviously carries the same geometry as part of the sphere - in essence the sphere minus the poles and the question becomes if it would be possible to construct any other manifold than the sphere with the standard metric that has such a chart. I suspect this is not possible.

 

[PAllen]

The relevant question in this question is: given a coordinate chart with a particular form for the metric, what are the possible global geometries?

The chart here obviously carries the same geometry as part of the sphere - in essence the sphere minus the poles and the question becomes if it would be possible to construct any other manifold than the sphere with the standard metric that has such a chart. I suspect this is not possible.

Well, smoothness comes into play. The obvious choice for the OP suggestion is an infinite chain of spheres connected at the missing pole points. This would be topologically a cylinder, as has already been noted. I guess a flaw in this is that you would have to declare that a countable set of theta coordinate values are excluded from the manifold.

 

[PeterDonis]

The thing is though, the coordinate chart where $\theta$ and $\phi$ are valid coordinates already describe a cylindrical topology where the geometry is that of a sphere.

This is true for the standard range of the $\varphi$ coordinate, but the OP has specified that the range of $\varphi$ is changed from the standard range.

You can remove the coordinate singularity but you will then just recover the sphere.

I don't understand. The coordinate singularity in $\theta$ is not removable; it is there, heuristically, because $\theta$ is an angular coordinate covering a dimension whose topology is $S^1$, which is impossible to cover with a single coordinate chart.

 

[martinbn]

The relevant question in this question is: given a coordinate chart with a particular form for the metric, what are the possible global geometries?

This is how i understood it. In general nothing can be said. For example if we take an open disc in the plane with the Euclidean metric. It can be extanded in any two dimensional manifold with many different metrics. Of course there are restrictions but definitely no uniqueness.

The chart here obviously carries the same geometry as part of the sphere - in essence the sphere minus the poles and the question becomes if it would be possible to construct any other manifold than the sphere with the standard metric that has such a chart. I suspect this is not possible.

Yes, i guess in this example one could answer the question the way demystifier hopes.

Ps Is the question motivated by GR?

 

[Demystifier]

Ps Is the question motivated by GR?

Yes, but I wanted first to understand a simpler (yet nontrivial) case.

 

[Demystifier]

The same manifold can have many different metrics, so a metric cannot determine the manifold. Strictly speaking you need the manifold first. The question is a bit like "I am think of a manifold with this metric, guess what I have in kind."

It makes sense to me, but in practice physicists always work backwards. They first find a solution of Einstein (or some similar) local differential equations, and then try to find out what kind of global geometry this solution represents. What I want to understand is to what extent such a practice is justified.

 

[martinbn]

It makes sense to me, but in practice physicists always work backwards. They first find a solution of Einstein (or some similar) local differential equations, and then try to find out what kind of global geometry this solution represents. What I want to understand is to what extent such a practice is justified.

Is that the case? My impression was that they emphasize that the equations are local and do not determine the global properties.

 

[Demystifier]

Is that the case? My impression was that they emphasize that the equations are local and do not determine the global properties.

See e.g. the Wald's book, Sec. 6.4 The Kruskal Extension.

 

[Orodruin]

The FLRW solution on the other hand starts from the global properties in terms of spatial symmetry and works the other way around.

It is also a fact that any spacetime satisfies the Einstein field equations. The only question is what the stress energy tensor is and if it can be interpreted as pertaining to something that actually exists.

 

[martinbn]

See e.g. the Wald's book, Sec. 6.4 The Kruskal Extension.

There is no claim for uniqueness. This of course is very special if you take it to be vacuum and spherecally symmetric. I wouldn't say that this is what happens ussually based on this example.

 

[PeterDonis]

See e.g. the Wald's book, Sec. 6.4 The Kruskal Extension.

As @martinbn notes, this is a special case, because Birkhoff's Theorem says that the only spacetime geometry that is vacuum and spherically symmetric is the Schwarzschild geometry. There is no such uniqueness theorem for any other case that I'm aware of except Reissner-Nordstrom (spherical symmetry plus electrovacuum--the only stress-energy is a source free electromagnetic field).

Also note that any maximal extension has to assume that the same local conditions continue to hold globally, i.e., everywhere. For a vacuum region surrounding a spherically symmetric planet or star, the Kruskal extension is not valid, because the vacuum condition ceases to hold at the boundary of the planet or star. So in any real solution we cannot rely solely on maximal extension to tell us what the global geometry is. We have to actually go and look at what the conditions are.