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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.

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    $\begingroup$ Hint I haven't followed through. What does such a function do to line segments? $\endgroup$ Feb 22 at 1:47
  • $\begingroup$ 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. $\endgroup$ Feb 22 at 5:48
  • $\begingroup$ @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. $\endgroup$
    – quarague
    Feb 22 at 10:04
  • $\begingroup$ @quarague Fair point and an interesting one. There are geodesics but I don't see how they are useful. $\endgroup$ Feb 22 at 12:11
  • $\begingroup$ @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. $\endgroup$
    – quarague
    Feb 22 at 12:20

1 Answer 1

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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.

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    $\begingroup$ 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. $\endgroup$
    – quarague
    Feb 22 at 10:01
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    $\begingroup$ Definitely not. I imagine that investigating the image of curves with constant geodesic curvature and torsion would still be key, but characterizing such curves in general is difficult I'm sure. $\endgroup$
    – whpowell96
    Feb 22 at 15:42

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