I was wondering whether the Gauß, mean curvature and shape operator of a surface actually depend on the chosen parametrization?
Under a reparametrization of $f: \Omega \subset \mathbb{R}^2 \rightarrow S \subset \mathbb{R}^3$ I understand a diffeomorphism $\Phi : M \subset \mathbb{R}^2 \rightarrow \Omega$ such that our surface is given by $f \circ \Phi$.
Intuitively, I would say that neither of them is conserved under reparametrization, as every surface is locally an embedding and therefore all of them vanish, am I correct about this?
The shape operator is given by $L = -DN Df^{-1},$ where $N$ is the Gauß map.
$\newcommand{\Reals}{\mathbf{R}}$In the usual sense of "depend",[*] no, the Gauss curvature, mean curvature, and shape operator of a (locally oriented) regular surface in $\Reals^{3}$ do not depend on parametrization; that's what's meant by saying these are "geometric" data. :)
Depending on your definition of the shape operator (e.g., O'Neill's: If $U$ is a continuous unit normal field on $M$ in a neighborhood of $p$ and $\nabla$ denotes the covariant derivative, then $S_{p}(v) = -\nabla_{v}(U)$ for each tangent vector $v$ to $M$ at $p$), it's more or less clear the shape operator doesn't depend on the parametrization (i.e., depends only on the set of points in $M$ and the ambient Euclidean geometry of space).
The Gauss and mean curvatures are the determinant and trace of the shape operator, so they also do not depend on parametrization.
[*] The coordinate representations do generally change under reparametrization, as would the representation of any non-constant function.