# Constancy of metric tensor components as a test of curvature

1. Sep 24, 2010

### Rasalhague

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:

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

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?

2. Sep 24, 2010

### George Jones

Staff Emeritus
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.

3. Sep 24, 2010

### Rasalhague

Thanks, George. Is a circle really flat then?

4. Sep 24, 2010

### 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.

5. Sep 24, 2010

### 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.

6. Sep 24, 2010

### bcrowell

Staff Emeritus
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.

7. Sep 24, 2010

### 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.

8. Sep 24, 2010

### Rasalhague

Thanks to everyone who chipped in with comments!

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?

9. Sep 24, 2010

### Ben Niehoff

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.

10. Sep 25, 2010

### 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: $\pm \text{diag}(-1,1,1,1)$.