isometry preserves the coefficient of 1st fundamental form. How about the coefficient of 2nd fundamental form? Is there any counterexample? Thanks for your answer in advance.
Indeed, if F is isometry map,
the coefficient of the 1st fundamental form, such as E ⇒ $E = <x_u,x_u> = <F^{*}(x_u),F^{*}(x_u)> = \bar{E}$ But the coefficient of the 2nd fundamental form, ⇒ $L = - <x_u,U_u> = <F^{*}(x_u),F^{*}(U_u)> = \bar{L}$
Since asking, I notice that $F^{*}(U_u)$ may not the derivative of normal vector of the isometry image.
When is it equal?
In general the "coefficients" are not preserved. For example, a cylinder is isometric to a plane, but the second fundamental form contains one non-zero radius of curvature, whereas the plane does not. It is the determinant of the second fundamental form - also known as the Gaussian curvature -- which is preserved. Think about the "coefficients" of the second fundamental form as entries of a diagonal matrix whose determinant needs to be constant.