- #1

Rasalhague

- 1,387

- 2

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 S

^{2}, the unit 2-sphere:

[tex]g = \begin{pmatrix}1 & 0 \\ 0 & \sin^2(\theta) \end{pmatrix}[/tex]

where [itex]\theta[/itex] 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 S

^{2}, the circle defined in these polar coordinates by [itex]\theta = \pi / 2[/itex]. Is this a submanifold of S

^{2}? On the equator, [itex]g[/itex] has constant components:

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

1 & 0\\ 0 & 1 \end{pmatrix}[/tex]

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 [itex]g[/itex] are constant. But there's nothing special about the equator; with a suitable choice of charts, [itex]g[/itex] could be given constant components over any circle on S

^{2}. 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?