I'm having trouble understanding how a space can be locally Euclidean, with zero curvature, and globally non-Euclidean, with curvature. If the space had locally approximately zero curvature, I see how global non-zero curvature could arise. If it is exactly zero, however, how can it accumulate to be non-zero?

My professor said my issue was due to me imagining the local region to be finite in size, whereas it was actually infinitesimal. I understand infinitesimals to be arbitrarily small, but finite, so I don't see how this makes a difference. Is the local geometry only Euclidean in the sense it approaches Euclidean geometry as the size/area/volume of the region approaches 0, but for any finitely sized region, the curvature would be non-zero?

In this case, would it then be true that there is no space with global curvature^ where the curvature is zero for any finitely sized region (and thus every region that is physical)?

^except maybe at any stationary point that existed, if you had an open ball containing only that point?

If my question isn't clear, it's probably due to my inexperience with this field and these terms; please let me know what doesn't make sense and I'll try to rephrase!

  • $\begingroup$ If you gave an example, you question would be much clearer. $\endgroup$
    – G. Smith
    Feb 25, 2020 at 7:47
  • 3
    $\begingroup$ I think that language is misleading and that is what confuses you. Take for example a sphere, it isn't flat, but at any point the tangent plane is a normal Euclidean plane. So the sphere is locally Euclidean. $\endgroup$
    – MBN
    Feb 25, 2020 at 9:52

2 Answers 2


It is widely stated that in general relativity spacetime is locally flat, but this is simply not true. Spacetime is flat if and only if the Riemann tensor is zero, and in general this is not the case and there is no coordinate transformation that will make the Riemann tensor zero.

But it is always possible to choose coordinates at a point in spacetime in which the Christoffel symbols are zero, and this is the sense in which spacetime appears locally flat. These coordinates are called normal coordinates, and in GR we are usually interested in the Fermi normal coordinates. These are just the rest frame coordinates of a freely falling observer. In the Fermi normal coordinates the geodesic equation reduces to Newton's second law, so in these coordinates Newton's laws of motion apply i.e. to the observer spacetime appears to be flat.

But while we can always take a point in spacetime and find the normal coordinates at this point, if we move away from this point the Christoffel symbols will cease to be zero, and if we move far enough away the deviation from Newton's laws will become measurable. This is the sense in which the flatness is only local.

  • $\begingroup$ Given a Riemann manifold $(M.g)$, pick a point $p$ on $M$, then the tensors are only defined in the tangent plane at point $p$, i.e., $T(M)_{p}$. It's possible you're conflating extrinsic with intrinsic curvature. $\endgroup$ Mar 8, 2020 at 9:22
  • $\begingroup$ As I unterstood: If the Christoffel smybols are zero - an infinit. parallel transported vector on a loop will be the same as in the beginning. Thus the curvature in this region is zero, so spacetime is flat? Whats wrong with this statement $\endgroup$
    – nuemlouno
    Jul 16, 2020 at 15:39
  • $\begingroup$ @nuemlouno at any point you can find coordinates (the normal coordinates) that make the Christoffel symbols zero at that point. As soon as you move away from that point, in space or time, the Christoffel symbols are no longer zero. $\endgroup$ Jul 16, 2020 at 15:43

It's hard to be sure what you're asking, since you seem to conflate "flat" with "globally euclidean", and these are not the same thing. A circle is flat but not globally euclidean; likewise for a flat torus. A manifold is globally flat by definition if and only if it is everywhere locally flat.

It also seems that you might be conflating the flatness of a manifold with the flatness of an embedding. These are different concepts. The unit circle is flat, but it is not flatly embedded in the plane.


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