# “Arclength” parameterization of a surface

Given an arbitrary parameterization of a differentiable curve $$r: R\rightarrow R^n$$ in terms of some variable $$t$$, we can re-parameterize (although perhaps not in closed-form) $$r$$ in terms of its arclength $$s$$. This gives a handy representation where as $$s$$ increases by 1, we move 1 unit along $$r$$ in $$R^n$$.

Is is 1) possible and 2) comparably straightforward to do this for surfaces? I.e. if $$r: R^m \rightarrow R^n$$ is differentiable can we reparameterize $$r$$ so that moving 1 unit in $$R^m$$ moves us 1 unit along the surface $$r$$ in $$R^n$$?

• I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface. – MisterRiemann Oct 29 '18 at 17:39

## 1 Answer

No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $$1$$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.

• The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration. – Scott Oct 29 '18 at 18:06
• I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas. – Balloon Oct 29 '18 at 18:14
• This is infinitely more clear after having read the pizza article. :) – Scott Oct 29 '18 at 20:49
• Glad to hear that! :) – Balloon Oct 29 '18 at 20:50