Is a Riemannian manifold with isometric coordinate charts flat?

Question

Suppose $M$ is a Riemannian Manifold such that each point has a coordinate chart into $\mathbb{R}^n$ that is an isometry, in the sense that the inner products are preserved. Does this imply that $M$ is locally flat in the sense that is vanishing Gaussian curvature?

I really don't know much about differential geometry, but my impression is that this is Gauss' Theorema Egregium (or a corollary thereof), which is sometimes stated as 'Gaussian curvature is invariant under isometry'. Does this mean isometry in the sense that lenghts of paths are preserved (i.e. as metric spaces)? Or in the sense above that inner products are preserved? Or even more?Perhaps, does the coordinate chart need to be smoother than just smooth enough to formulate that it is an isometry? For example, the Wikipedia entry on Gaussian curvature gives a formula that contains second derivatives, so does the version of the Theorema that I cited above only hold for $C^2$ isometries?

Answer 1

An isometry in the sense that the Riemannian metric (scalar product) is preserved is also a local isometry in the sense of geodesic distance, which is not too difficult to see since it preserves locally the length of curves (this does require some reasoning to make it strict, though). Since the Riemannian curvature tensor can be calculated in terms of the metric and the deriviates of the metric up to second order (and is a tensor field, i.e. it depends only on the geometry and is especially independent of choice of coordinates), the curvature tensor will vanish if a Riemannian manifold is locally isometric to Euclidean space in this sense. The converse is also true but harder to show (see, e.g. Spivak, Comprehensive introduction into Differential Geometry, vol 2, theorem 4.13).

Since the Gauss curvature may be expressed using the metric and the curvature tensor, it will also vanish. A global isometry cannot be expected, though, since (e.g.) on a two-dimensional cylinder ($S^1\times \mathbb{R}$) of infinite lenght you can easily write down a metric which makes this locally isometric to $\mathbb{R^2}$

Answer 2

In general, the notion of isometries for Riemannian manifolds is an infinitesimal one. A smooth map $\varphi \colon (M,g) \rightarrow (N,h)$ between smooth Riemannian manifolds is called a local isometry if for each $p \in M$ the differential $d\varphi_p \colon (T_pM, g_p) \rightarrow (T_{\varphi(p)}N, h_{\varphi(p)})$ is an isometry of inner-product spaces. A local isometry is a local diffeomorphism - if it is also a global diffeomorphism then it is called an isometry.

Given this definition, the answer to your question is yes - local isometries preserve the Riemann curvature tensor and since the curvature tensor for $\mathbb{R}^n$ is identically zero the curvature tensor of $M$ vanishes identically.

Note that the relevant term is "the Riemann curvatures" and not "Gaussian curvature" as the Gaussian curvature is usually a scalar that is associated to a surface embedded in $\mathbb{R}^m$ and you are asking about a general $n$-dimensional Riemannian manifold.

Regarding your other questions - local isometries of Riemannian manifolds preserve the length of paths and preserve the distance induced by the Riemannian metrics (the infimum of the length of all paths connecting two points). The definition of a local isometry presented above requires the map $\varphi$ to be at least differentiable. You can also define an isometry as a map that preserves the distances in the sense of metric spaces. Such a map will be continuous but a priori doesn't have to be differentiable. It turns outs that a map that preserves the distances in the sense of metric spaces will automatically be an isometry in the sense defined above and in particular, smooth. This result is called Myers-Steenrod theorem.