Constancy of metric tensor components as a test of curvature

[Rasalhague]

Later, we shall see that constancy of the metric components is sufficient for a space to be flat, and in fact there always exists a coordinate system on any flat space in which the metric is constant (Carroll: Lecture Notes on General Relativity.)

In the previous section he derived the components, with respect to the coordinate bases associated with a polar coordinate system, of the Riemannian metric tensor field on S², the unit 2-sphere:

$$g = \begin{pmatrix}1 & 0 \\ 0 & \sin^2(\theta) \end{pmatrix}$$

where $\theta$ is the zenith angle as shown in the diagram above the paragraph quoted.

Earlier in the chapter, he talks about a circle as a manifold. Let's consider the equator of S², the circle defined in these polar coordinates by $\theta = \pi / 2$. Is this a submanifold of S²? On the equator, $g$ has constant components:

$$g=\begin{pmatrix}1 & 0\\ 0 & \sin^2(\pi / 2) \end{pmatrix}=\begin{pmatrix}<br /> 1 & 0\\ 0 & 1 \end{pmatrix}$$

Apparently the equator is flat. At least, every subset of points on the equator is flat, since it can be covered with a set of charts with respect to each of whose coordinate bases the components of $g$ are constant. But there's nothing special about the equator; with a suitable choice of charts, $g$ could be given constant components over any circle on S². If every point in a Riemannian manifold belongs to a flat submanifold, doesn't this make the whole manifold flat?

Obviously I'm missing something pretty basic here. Is there a fuller way to word "constancy of the metric components is sufficient for a space to be flat" that would make clear how I'm misunderstanding it? Are extra conditions implicit? Do the components have to be constant in one single chart that covers the whole manifold?

 

[George Jones]

If every point in a Riemannian manifold belongs to a flat submanifold, doesn't this make the whole manifold flat?

No. As another counterexample, think of a flat Friedmann-Robertson-Walker universe. This has flat 3-dimensional submanifolds (the "space" part of sapcetime) that cover (foliate) the whole spacetime, but spacetime, in this case, itself is not flat.

 

[Rasalhague]

Thanks, George. Is a circle really flat then?

 

[Rasalhague]

Perhaps I should say: is S¹ flat in the sense that it has no intrinsic curvature? (Although it may be mapped to circles embedded in some other manifold, such as E², with extrinsic curvature 1/r.) I suppose it could be thought of as a piece of string and straightened out without gaps or wrinkles, so I guess maybe so.

 

[t_barlow]

That's right, any 1-dimensional manifold has zero intrinsic curvature. In order to calculate the Riemann tensor you have to parallel transport in two different directions, and you can't do that in one dimension - or rather your two directions are actually the same direction, so they just cancel. The other way to see it is, the curvature form is a 2-form, and there are no 2-forms in 1 dimension.

 

[bcrowell]

That's right, any 1-dimensional manifold has zero intrinsic curvature. In order to calculate the Riemann tensor you have to parallel transport in two different directions, and you can't do that in one dimension - or rather your two directions are actually the same direction, so they just cancel. The other way to see it is, the curvature form is a 2-form, and there are no 2-forms in 1 dimension.

Yet another way to say it: curvature measures the failure of Euclid's parallel postulate, but we can't express a parallel postulate in one dimension.

 

[Ben Niehoff]

Yet a third way to say it is that in 1 dimension, we can always find a coordinate transformation to coordinates where the metric is constant*. Since curvature depends on derivatives of the metric, it automatically vanishes.

* In general, the metric takes the form

$$ds^2 = f(x)^2\,dx^2.$$

So we can bring this to a constant metric if we solve

$$dy = f(x) \, dx,$$

which always has a solution, because it is a first-order, linear differential equation.

 

[Rasalhague]

Thanks to everyone who chipped in with comments!

Yet a third way to say it is that in 1 dimension, we can always find a coordinate transformation to coordinates where the metric is constant*. Since curvature depends on derivatives of the metric, it automatically vanishes.

I guess implicit in your first sentence is the condition that the metric (tensor field) must have constant coefficients over the whole chart. And I guess a curve on S² couldn't count as the domain of a chart because it has less dimensions than the whole manifold, S².

On a Riemannian manifold, given any point, we can find a chart (not necessarily the same chart for each point) with respect to whose coordinate bases the metric tensors have constant coefficients, part of the local (extrinsic?) flatness theorem. But for a manifold to qualify as (intrinsically) flat, it just needs to be possible to chose a set of charts which cover the manifold such that the coefficients of the metric tensors in the coordinate bases of each chart are the same at all points in that chart. Is that right?

 

[Ben Niehoff]

On a Riemannian manifold, given any point, we can find a chart (not necessarily the same chart for each point) with respect to whose coordinate bases the metric tensors have constant coefficients

No, this is false, and maybe this is part of the confusion. Given any point, you can find a chart such that the metric is equal to the flat-space metric, and the first derivatives of the metric all vanish. But these are true only at that particular point, and the second derivatives do not vanish in general! The metric cannot be made "constant" in general, as that would imply ALL derivatives vanish.

The second derivatives of the metric carry information about the curvature.

When we say a metric is "constant", we mean not just at a point, but in a neighborhood.

 

[Rasalhague]

Okay, I'll try again:

An intrinsically flat Riemannian manifold is one that can be covered by a set of charts such that, in each of these charts, the coefficient matrix of the metric tensor field is everywhere constant in the coordinate bases, $\mathrm{d}x^a \otimes \mathrm{d}x^b \, (p)$, associated with that particular chart. (Where $p$ ranges over all points in the domain of the chart.)

For each point in a general Riemannian manifold, there exists some chart such that the coefficient matrix of the metric tensor at this particular point, $p_0$, in the coordinate basis, $\mathrm{d}x^a \otimes \mathrm{d}x^b (p_0)$, associated with this chart, is the identity matrix, $I = \text{diag}(1,...,1)$, and the values of the first derivatives, with respect of each coordinate, of each coefficient function of the metric tensor field, in the coordinate bases associated with this chart, are zero.

Likewise for a pseudo-Riemannian manifold, except replace the identity matrix with the canonical form of the flatspace pseudometric tensor, e.g. in the famous case, that Minkowski space: $\pm \text{diag}(-1,1,1,1)$.