In Carrolls GR book “Spacetime and Geometry” he comments that

“This is not a situation we can define away; on a curved manifold, a co- ordinate basis will never be orthonormal throughout a neighborhood of any point where the curvature does not vanish.”

Is there a clear proof that the coordinate basis cannot be orthonormal in a curved geometry except for potentially at one point in space?

My intuition says it is because curvature can either make the metric non-diagonalizable negating orthogonality, or it stretches basis vectors negating normality but its not clear this is correct.

  • $\begingroup$ '...except for potentially at one point in space?' I think the correct statement is: '...except for potentially along a curve in space?'. Consider eg the sphere $S^2$ in spherical polars where the coordinate basis is orthogonal everywhere but orthonormal only along the equator. This is probably related to being able to find a local inertial frame at every event along a worldline but can't flesh that out right now. $\endgroup$
    – jacob1729
    Commented Aug 22, 2019 at 9:12
  • $\begingroup$ In principle you could have it on a co-dimension 1 surface. $\endgroup$
    – TimRias
    Commented Aug 22, 2019 at 9:46
  • $\begingroup$ The title, quote from Carroll, and question after the quote all say different things. $\endgroup$
    – user4552
    Commented Aug 22, 2019 at 12:46
  • 1
    $\begingroup$ Prove the contrapositive. Assume the coordinate basis is orthonomal in a neighborhood, and prove the curvature is then zero. $\endgroup$ Commented Aug 22, 2019 at 13:44

1 Answer 1


The excerpt from Carroll refers to the fact that if we write a general metric field as $g = \sum_{jk}g_{jk}(x)dx^j\otimes dx^k$, then we have $g(\partial_j,\partial_k) = g_{jk}(x)$, so the coordinate vector fields $\partial_k := \partial/\partial x_k$ are orthonormal in the given neighborhood only if $g_{jk}(x)=\delta_{jk}$ in the given neighborhood. In other words, they are orthonormal everywhere in the given neighborhood only if the metric is flat in that neighborhood.

I'm using the language of differential forms, where $dx^k$ is a one-form defined such that $dx^k(\partial_j)=\delta^k_j$.


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