# Surfaces with the same constant Gaussian curvatures are locally isometric?

Question:

I want to know whether surfaces with the same constant Gaussian curvatures are locally isometric or not. If the answer is yes, why?

Definitions:

Gaussian curvature of a surface at some point is the product of the eigenvalues of the shape operator of the surface at that point$.^1$

Shape operator of a surface is the minus derivative of the unit normal vectors on the surface. Formally speaking, let $f:U \to \mathbb {R}^3$ be a surface element with unit normal vector map $\nu$, $\nu: U \to S^2$ is defined by $$\nu (u_1,u_2):=\frac{\frac{\partial f}{\partial u_1} \times \frac{\partial f}{\partial u_2}}{\left \Vert \frac{\partial f}{\partial u_1} \times \frac{\partial f}{\partial u_2} \right \Vert},$$

then for every $u\in U$ we have the linear map $$D\nu|_u:T_uU \to T_uf,$$ where $T_uU=\{u\} \times \mathbb R^2$ and $T_uf=Df|_u\left(T_uU\right)$, and $$Df|_u:T_uU \to T_uf$$ is a linear isomorphism. Then the shape operator $L:=-D\nu \circ (Df)^{-1}$ is defined pointwise by $$L_u:=-\left(D\nu|_u \right) \circ \left(Df|_u\right)^{-1}:T_uf \to T_uf\,.^2$$

Two surface elements $f:U \to \mathbb R^3,g:V \to \mathbb R^3$ are said to be isometric, if there is a parameter transformation $\Phi:U \to V$ such that $$\left \langle \frac{\partial f}{\partial u^i},\frac{\partial f}{\partial u^j}\right \rangle =\left \langle \frac{\partial (g \circ \Phi)}{\partial u ^i}, \frac{\partial (g \circ \Phi)}{\partial u ^j} \right \rangle$$ for all $i,j\,.^3$

[1], [2], [3] Wolfgang Kühnel, "Differential Geometry Curves-Surfaces-Manifolds", Second Edition, American Mathematical Society, 2006.