rotations and translations conserve principal curvatures of a regular surface 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 
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$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 $T$.
