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.
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.