# Is a map that preserves torsion and curvature of all smooth curves an isometry?

In my introductory differential geometry class, we learnt that isometries preserve the torsion $$\tau$$ and curvature $$\kappa$$ of curves. I was wondering if the converse of this statement is true: if a map $$F$$ preserves $$\kappa$$ and $$\tau$$ for all curves, is $$F$$ an isometry?

Thanks for any help.

• Hint I haven't followed through. What does such a function do to line segments? Feb 22 at 1:47
• Have you learned the Fundamental Theorem of Curves? If two arclength-parametrized curves have identical curvature and torsion, then they differ by a rigid motion. Feb 22 at 5:48
• @EthanBolker I'm not sure what setting OP is interested in but if you consider this question on manifolds and not just on $\mathbb{R}^n$ then there are no line segments. Feb 22 at 10:04
• @quarague Fair point and an interesting one. There are geodesics but I don't see how they are useful. Feb 22 at 12:11
• @EthanBolker You could rephrase the problem and ask if you have a function on manifolds that maps geodesics to geodesics, is it necessarily an isometry? I think the answer is no because such a map doesn't have to preserve length. Feb 22 at 12:20

Note that this condition implies that all circles of radius $$r$$ must map to circles of radius $$r$$ under $$F$$. We show that if the map is not an isometry, then we can find a circle that maps to a non-circle under $$F$$.
Consider a set of distinct noncollinear points $$P = F(O)$$, $$y_i = F(x_i)$$, $$i = 1,2,3$$ such that $$d(O,x_i) = d(P,y_i) = r>0$$ for $$i = 1,2$$, and $$d(O,x_3) \neq d(P,y_3)$$. Notice that all $$x_i$$'s lie on a circle of radius $$r$$ with center $$O$$. Parameterize this curve by $$\gamma(s)$$, then $$\gamma$$ has constant curvature $$1/r$$ and constant torsion $$\tau = 0$$. The image of $$\gamma$$ will also have zero torsion, so it will be a planar curve.
Now notice that the points $$y_1,y_2$$ are two points equidistant from $$P$$, so they define a circle of radius $$r$$ with center $$P$$, and the image of $$\gamma$$ must parameterize this circle. However, $$y_3$$ does not lie on this circle, so the image of $$\gamma$$ cannot be a circle of radius $$r$$ and hence the curvature is not preserved, a contradiction.
• This assumes we are in $\mathbb{R}^n$ and works only in this setting. On an arbitrary 2-d manifold the collection of points at distance r to some random point is not a circle with constant curvature 1/r. I think OPs statement is still true on manifolds but I don't think this proof works for that setting. Feb 22 at 10:01