I came across the following sentence in a comment on this question: Local existence of parallel vector field

"the existence of a parallel vector field is equivalent to the condition that the metric splits locally into a Riemannian product of a one-dimensional manifold and an (n−1)-dimensional one. This implies, in particular, that the sectional curvatures of planes containing V are all zero."

and I want to know what the poster mean by the metric splitting locally into a Riemannian product and what that has to do with sectional curvatures.


The metric splitting locally means that you can pick local coordinates around any point $p$ such that in those coordinates, the metric tensor looks like a block matrix which is the direct sum of a $1 \times 1$ matrix and an $(n-1) \times (n-1)$ matrix. Observe that this isn't very hard to do for a single point, in fact, at a point, the key information we're getting out of this is that we can write the metric tensor in this form in an entire neighbourhood.

In particular, this implies that if we just restrict our attention to the coordinate neighbourhood, the neighbourhood can be written as a product of a $1$-dimensional Riemannian manifold, and an $(n-1)$-dimensional Riemannian manifold, since the metric decomposes in a nice manner.

Now if you pick any pair of orthogonal vectors at $p$, one of which is the tangent vector corresponding to the $1$-dimensional subspace, and flow along all linear combinations of the two vectors, you'll span out a surface whose scalar curvature is the sectional curvature along the two vectors. The key point to note here is that the Riemannian metric on the spanned surface will essentially be the Euclidean metric, which has $0$ scalar curvature, which means you're done.

  • $\begingroup$ This answer is interesting. Can you please give me a hint how to prove that existence of a parallel field indeed implies the metric is locally a product? (I thought about extending this vector field to a commuting frame, but I don't see how this helps). Thanks. $\endgroup$ – Asaf Shachar May 30 '18 at 5:12

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