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Let $c: I \to \Bbb{R}^3$ be an arc length parametrized curve with curvature $k \gt 0$ and torsion τ. Let $F(x) = Ax +b$, where $A \in SO(3)$ and $b \in \Bbb{R}^3$. Show that the curve $\tilde{c} := F \circ c$ has curvature $\kappa$ and torsion τ.

(Note that $SO(3)$ refers to special orthonormal and 3x3).

I'm confused as to why $c$ and $\tilde{c}$ have the same curvature and torsion ($\kappa$, τ).

I'm also unsure of the notation for $SO(3)$ and what properties $A$ has since $A \in SO(3)$. Does this just mean that $A$ is orthonormal (all vectors are orthogonal and have length 1) (i.e. what is "special orthonormal")?

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    $\begingroup$ So what are $\tilde T, \tilde N, \tilde B$ in terms of $T, N, B$? $\endgroup$ – Ted Shifrin Feb 16 '17 at 2:48
  • $\begingroup$ Notice that $F$ is a rigid transformation, i.e. it preserves distances between any pair of points in curve $c$ as mapped to curve $\overline{c}$. So you should expect quantities like $\kappa$ and $\tau$ to be conserved. Perhaps if the technical details of showing this are a challenge, treating the mapping as two steps, a rotation and a translation, would make the details easier for you. $\endgroup$ – hardmath Feb 16 '17 at 16:29
  • $\begingroup$ Special orthonormal means orthonormal with determinant $1$. The idea here is that the transformation you are given is a general isometry of $\mathbb{R}^3$ with the standard Riemannian metric, thus all intrinsic properties of geometric objects will be preserved. $\endgroup$ – Daniel Robert-Nicoud Feb 16 '17 at 16:51

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