10

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?

Please help.

hardmath
  • 37,015
math101
  • 3,065
  • 6
    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). – Sam Nead Aug 21 '15 at 23:38
  • 8
    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. – Brian Tung Aug 21 '15 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 https://leannejdong.github.io/ – math101 Jan 23 '22 at 03:04
  • 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. – hardmath Feb 02 '22 at 01:43
  • The topic of the curvature of a circle was implicitly discussed here under scalar curvature on one-dimensional Riemannian manifold. – hardmath Feb 02 '22 at 01:51

1 Answers1

18

When you look at a circle, you are seeing its extrinsic curvature, which is also what your link is calculating. That is a property of how the circle is imbedded into another manifold, not a property of the circle as a manifold itself.

The curvature being referred to here is the intrinsic curvature, which is defined strictly in terms of the manifold itself, not any imbedding. 1-dimensional manifolds are incapable of supporting any curvature, so the circle is flat.

Paul Sinclair
  • 43,643
  • Many thanks. How to check circle can be regarded as a 1-dimensional manifold? – math101 Aug 22 '15 at 01:34
  • How to show the Riemannian curvature (i.e. the intrinsic curvature) is 0? – math101 Aug 22 '15 at 02:07
  • 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. – Paul Sinclair Aug 22 '15 at 04:37
  • Your first sentence is clear to me. My second question is that, I wish to compute Riemannian curvature. But I have never learnt to do such thing before. Could you show me how the curvature of circle computed as 0? In what books contains preliminary like this? – math101 Aug 22 '15 at 04:42
  • 1
    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. – Paul Sinclair Aug 22 '15 at 04:56
  • Thanks. I shall review Lie bracket, learn how to differentiate tensor field, then get to the riemann tensor. – math101 Aug 22 '15 at 05:41
  • forgot to say...what you said made sense – math101 Aug 22 '15 at 06:34
  • That's good to know. I was afraid I went too high-powered, but I wanted to give the gist of the reason without getting bogged down in details. – Paul Sinclair Aug 22 '15 at 13:41
  • I forgot to ask a thing, the formula you gave for the curvature tensor above...is from which book? It looks different from what is written from my notes... – math101 Aug 22 '15 at 13:49
  • I looked it up (haven't looked at this in many years) in O'Neill's Semi-Riemannian Geometry. I can't recall where I first encountered it. – Paul Sinclair Aug 22 '15 at 13:54
  • http://pentagono.uniandes.edu.co/~jarteaga/trabajo-diario/sub-riemannian/geometric-control-theory/references-add/COL-UNIA-0024062_01.pdf – math101 Aug 28 '15 at 09:35
  • Page 3 above link has a similar formula, but the sign is opposite. Could you reconcile why ? – math101 Aug 28 '15 at 09:36
  • @math101 - I'm afraid not. It appears to be a matter of convention (in both cases, they are defining the curvature tensor by this expression). I did double-check that I had copied O'Neill's version correctly. My own preference would be for the version that gives a positive scalar curvature to the sphere, but that would take some effort to figure out. – Paul Sinclair Aug 29 '15 at 15:35