Curves have no intrinsic geometry Please someone could explain me why the curves have no intrinsic geometry? With surfaces I can see that there are two kind of geometries, i.e. the euclidean one (related to euclidean isometries) and the intrinsic (related to isometries on surfaces), but I am not able to make a comparison/analogy with curves. Thanks.
 A: Curves have the same intrinsic geometry that a straight line has.  If $B$ is between $A$ and $C$ then the distance from $A$ to $B$ (measured along the curve, i.e. the arc length) plus the distance from $B$ to $C$, is the distance from $A$ to $C$.  If a string winds around through space and you pick it up and lay it down unrolled so that it's a straight line, you don't change any of the intrinsic distances; the length of the curve from $A$ to $B$ is still the same.  Hence the curvature doesn't matter.
A: One possible interpretation for the statement «curves have no intrinsic geometry» in the context of Riemannian geometry is the following:


*

*A non-compact $1$-dimensional Riemannian manifold is isometric to an open interval in the line.There is an intrinsic Riemannian invariant of such a manifold: the number $\nu$ of points added by the procedure of completion, which is an element of the set $\{0,1,2\}$. When $\nu=2$, there is moreover a secondary invariant, the length of the curve.

*On the other hand, there is exactly one intrinsic Riemannian invariant of a compact $1$-dimensional Riemannian manifold, its length, which is a positive real number.
Most of this is proved implicitly when one shows that a curve can always be parametrized by arc-length, after classifying $1$-dimensional differentiable manifolds.
A cuter statement, although partial, is that the complete $1$-dimensional manifolds up to isometry are in bijective correspondence with $(0,\infty]$.
Likewise, the local version of this is: any two $1$-dimensional Riemannan manifolds are locally isometric.
