The discussion centers on the properties of the Riemann tensor in two-dimensional spaces, specifically addressing the misconception that all 2D spaces are flat. It is established that a 2-dimensional manifold, such as a sphere, can possess non-zero Riemann curvature, contradicting the notion that 2D inherently implies flatness. The Riemann tensor has 16 components in 2D, with only four independent components due to the constraints of the dimensions. The intrinsic curvature of a 2D space is defined independently of any higher-dimensional embedding, emphasizing the distinction between intrinsic and extrinsic curvature.
PREREQUISITESMathematicians, physicists, and students of geometry who are interested in the properties of manifolds, curvature, and the applications of Riemannian geometry in theoretical physics.
mal4mac said:Isn't there still a way to tell the difference? Couldn't an inhabitant of the cylinder walk round the cylinder and get back home? The inhabitant on the infinite plane would never get back home.
stevendaryl said:How would he know he had gotten back at home, if absolutely everything repeated periodically? He would walk a certain distance and find a house that is exactly like his in every way. Is it his house, or an exact copy?
mal4mac said:In reality it could hardly be an exact copy, so your example only holds for a very simple universe, one without life.
A.T. said:You cannot roll out a common doughnut surface onto a flat plane without any stretching/squeezing. But if the radius of revolution goes to infinity, while the radius of the tube stays constant the intrinsic curvature goes to zero.
The key point here is that while a surface with the topology of a torus does not require curvature, it may have curvature (just as bumps on plane leave the topology unchanged, but add curvature). The follow on point is that any torus embedded in Euclidean 3-space has curvature. However, a torus without curvature can be embedded in Euclidean 4-space. This is similar to the limitation that a Klein bottle (a non-orientable analog of a torus) can only be embedded in Euclidean 4-space not Euclidean 3-space.A.T. said:For the cylinder it's simple to visualize: You can roll it out onto a flat plane without any stretching/squeezing (no change of the intrinsic distances).
But for a torus it's less obvious: You cannot roll out a common doughnut surface onto a flat plane without any stretching/squeezing. But if the radius of revolution goes to infinity, while the radius of the tube stays constant the intrinsic curvature goes to zero.