# flat manifold, curvature and the circle

A Riemannian manifold is said to be flat if the curvature is 0 everywhere. An example in dimension 1 is the circle. However, I cannot see how the curvature of the circle could be 0. See for instance:

Curvature of a circle

How could the curvature of a circle be 0? How to show the Riemannian curvature is 0?

• You are confusing two notions of curvature. There is the geodesic curvature of a circle (which depends on its embedding in the plane) and the Riemannian curvature of the circle (which is intrinsic). Commented Aug 21, 2015 at 23:38
• To add a bit of intuition onto the other responses: A sphere is intrinsically curved in the sense that you can't cut it in half, say, and "flatten it out" without distorting lengths, angles, or both. That's more or less why any map of the Earth is a compromise. A cylinder, on the other hand, can be cut and rolled out into a flat sheet, so it has no intrinsic curvature, in exactly the same way that you can cut a circle and roll it out to a line segment. Commented Aug 21, 2015 at 23:46
• I no longer work on PDE (for many years). Answers received appeared to be way too late and so I could not credit the helpfulness leannejdong.github.io Commented Jan 23, 2022 at 3:04
• A standard result is that the geodesic curvature of a circle is the reciprocal of its radius (plus or minus, depending on the direction of parametric embedding in the plane, so extrinsic). The Riemann curvature tensor is defined intrinsically from the "distances" (metric) between points of a smooth manifold. If you have no objection, I'll update the links in the body of your Question. Commented Feb 2, 2022 at 1:43
• The topic of the curvature of a circle was implicitly discussed here under scalar curvature on one-dimensional Riemannian manifold. Commented Feb 2, 2022 at 1:51

• It is obviously one dimensional. $(\cos \theta, \sin \theta)$ provides smooth 1-d maps. What more do you want? As for the curvature, look at the definition. It requires two idependent coordinates to have anything that isn't 0. Commented Aug 22, 2015 at 4:37
• If $D$ is the Levi-Civita connection, then the curvature tensor is given by: $$R_{XY}Z = D_{[X, Y]}Z - [D_X, D_Y]Z$$ Since the Lie bracket is an anticommutator ($[U, V] = - [V, U]$), it is 0 if both vectors, or both operators are the same, or even parallel. In a 1D manifold, all vectors are parallel (since there is only one direction for them to point), so the Lie brackets are always 0, and so is the curvature. Commented Aug 22, 2015 at 4:56