Why the geodesic curvature is invariant under isometric transformations? As I know the geodesic curvature
$$
\kappa_g = \sqrt{\det~g} \begin{vmatrix}  \frac{du^1}{ds} & \frac{d^2u^1}{ds^2} + \Gamma^1_{\alpha\beta}  \frac{du^\alpha}{ds} \frac{du^\beta}{ds} \\  \frac{du^2}{ds} & \frac{d^2u^2}{ds^2} + \Gamma^2_{\alpha\beta}  \frac{du^\alpha}{ds} \frac{du^\beta}{ds} \end{vmatrix}, 
$$
where $g$ is the metric tensor, $\Gamma^v_{\alpha\beta}$ is the Christoffel symbols of the second kind.
And the first fundamental form of the surface $I = (du^1, du^2) g (du^1, du^2)^T$. I think $I$ is invariant under isometric transformations but not the metric tensor $g$. So why $\kappa_g$ is invariant under isometric transformations?

$ \kappa_g$ depends purely on the coefficients of the first fundamental form ( of surface theory FFF) and their derivatives, second fundamental form SFF coefficients are not involved.
It is invariant in isometric mappings ( bending transformations) like lengths,angles,  $K$ Gauss curvature , integral curvature etc. Liouville's theorem gives the expressions. Reference of text books of Differential geometry.
$K$ is an exception where the determinants of SFF and FFF can be used to  derive it in  the Gauss Egregium theorem.
