Is curvature possible for a 2D metric?

[stevendaryl]

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.

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?

What I said was that there is no observable difference between Situation A: The universe is an infinite 2D plane in which absolutely everything repeats at regular intervals in one specific direction. Situation B: The universe is a cylinder.

 

[mal4mac]

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?

In reality it could hardly be an exact copy, so your example only holds for a very simple universe, one without life.

 

[ChrisVer]

In reality it could hardly be an exact copy, so your example only holds for a very simple universe, one without life.

What does "could hardly be an exact copy" mean in mathematics? Periodicity doesn't distinguish whether it's a copy or the same object. Perfect crystal structure is an example to that...
The main point is in Steven's post#30

 

[A.T.]

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.

Related video presenting another method:

 

[PAllen]

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.

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.

[Edit: seeing AT's last post, I am referring to smooth embeddings; curvature is preserved at every point, among other things. Nash also proved a theorem related to the one described in the video on smooth embeddings, that any conceivable manifold can be smoothly embedded in some higher dimensional Euclidean manifold.}