The cylinder is a smooth surface with zero Gaussian curvature and so can be flattened down into a plane with no distortion of distances.
Does this mean that Euclidean space $\mathbb{E}^2$ and the cylinder $S^{1} \times \mathbb{R} $ are equivalent as metric spaces equipped with the same distance function? Also, what is the radius of this cylinder obtained from the general space $\mathbb{E}^2$?
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They share the same first fundamental theorem. The plane and cylinder are isometrically equivalent with corresponding mappable regions. Isometry is bending and preserves all parameters that are functions of common Christoffel symbols.
The second fundamental form is different... and normal curvature of lines changes in bending..
Gaussian curvature is a local property... having zero Gaussian curvature means that sufficiently small neighborhoods of each point can be flattened faithfully into the plane. Certainly the Euclidean plane and the cylinder are not globally equivalent, even topologically, let alone as metric spaces. For instance, note that there are closed geodesic curves on the cylinder but not in the plane.