rotations and translations conserve principal curvatures of a regular surface

Question

Even though it is intuitively clear, I get stuck on showing that a rotation or a translation of a regular surface conserve the principal curvatures in all the points of the surface ( while a general local isometry of a surface conserves only the Gaussian curvature). I think the reason is the conservation of the euclidean scalar product but I can't link this fact with the definition of principal curvature

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$If $T$ is an isometry of $\Reals^{3}$, then there exists a unique orthogonal $3 \times 3$ matrix $A$ and a unique vector $\Vec{b} = T(0)$ such that $$ T(\Vec{x}) = A\Vec{x} + \Vec{b}\quad\text{for all $\Vec{x}$ in $\Reals^{3}$.} $$ The crucial points are, $DT = A$ is constant, and orthogonality of $A$ implies $$ \left. \begin{aligned} \Brak{A\Vec{u}, A\Vec{v}} &= \Brak{\Vec{u}, \Vec{v}}, \\ (A\Vec{u}) \times (A\Vec{v}) &= \pm A(\Vec{u} \times \Vec{v}), \end{aligned}\right\} \quad\text{for all $\Vec{u}$ and $\Vec{v}$ in $\Reals^{3}$,} $$ with the sign in the cross product formula equal to $\det A$. (For a rotation or translation, $\det A = 1$; for a reflection or glide reflection, $\det A = -1$.)

If $\Phi$ is a parametrization of a regular surface $S$, and if $\Psi = T \circ \Phi$ is the induced parametrization of $T(S)$, then with $(u, v)$ denoting coordinates in the domain (surface parameters) and subscripts denoting partial derivatives, \begin{align*} &\begin{aligned} \Psi_{u} &= A\Phi_{u}, \\ \Psi_{v} &= A\Phi_{v}; \end{aligned} &&\begin{aligned} \Psi_{uu} &= A\Phi_{uu}, \\ \Psi_{uv} &= A\Phi_{uv}, \\ \Psi_{vv} &= A\Phi_{vv}. \end{aligned} \end{align*}

It's now straightforward to check that every differential-geometric quantity for $S$ defined in terms of partials of $\Phi$, and the Euclidean dot and cross products, is preserved by $T$.