Recall that two Riemannian manifolds are isometric if there exists a diffeomorphism that preserves the metric on the tangent spaces. The curvature tensor (associated to the Levi-Civita connection) measures the deviation from flatness, where a manifold is **flat** if it is locally isometric to a neighborhood of .

**Theorem 1 (The Test Case)** *The Riemannian manifold is locally isometric to if and only if the curvature tensor vanishes. *

One way is easy to check by computation. We show that zero implies flatness.

First of all, note that on with the usual metric (or an open submanifold thereof), there is a family of vector fields () which at each point form an orthonormal basis. Moreover, they commute. By contrast, if vector fields with this condition exist on a Riemannian manifold, then it is necessarily locally isometric to . For the commutation condition means that we can locally choose coordinates such that . Then by orthonormality the metric must be given by . So if , we have to construct such vector fields. Pick and forming an orthonormal basis for . I claim that these can be extended to vector fields as above. First of all, we may assume that as a manifold is an open subset of , just not necessarily with the standard metric, and that . Define first along the line by parallel translation of along the -axis. Then define by starting with and parallel translating along the line from Keep using parallel translation in this manner to make these into vector fields.

I claim now that for all . By linearity over smooth functions, it is sufficient to show the case of . We treat the case . Then we already have

since was obtained by parallel translation along “vertical” lines and

Thus

where we have used the Clairaut-like theorem for connections whose curvature tensor vanishes! Now this implies is parallel in , hence identically zero because it vanishes on the horizontal line . This proof can be extended to higher .

We do the same for for . Then the are orthonormal because parallel translation preserves inner products. I claim that . To see this, note that

by symmetricity. Since the are parallel in every direction, we get . Consequently the test case is proved.

(The term “test case” is borrowed from Michael Spivak.)

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