Let$\displaystyle S_1$ and $\displaystyle S_2$ be regular surfaces in $\displaystyle \mathbb{R}^3$. Let $\displaystyle \phi: S_1 \rightarrow S_2$ be an isometry, that is to say

$\displaystyle I_p^{S_1} (x,y) = I_{\phi(p)}^{S_2} (d_p \phi x, d_p \phi y)$ for all $\displaystyle x,y \in T_p S_1$, the tangent plane at$\displaystyle p\in S_1$, where I is the first fundamental form on the appropriate tangent plane, and $\displaystyle d_p$ is the differential map.

Let $\displaystyle (U_1,F_1,V_1)$ and $\displaystyle (U_2,F_2,V_2)$ be local parametrisations of $\displaystyle S_1$ at $\displaystyle p$ and $\displaystyle S_2$ at $\displaystyle \phi(p)$ respectively.

Let $\displaystyle u = F^{1}(p).$

I am trying to show that the gaussian curvature is the same at $\displaystyle p$ and $\displaystyle \phi(p). $I guess that this means I am trying to show that the first fundamental forms at $\displaystyle p$ and $\displaystyle \phi(p)$ are the same when i put the appropriate basis vectors in, i.e.

$\displaystyle I_p^{S_1} (X_i, X_j)= I_{\phi(p)}^{S_2} (X_i ', X_j ')$, where $\displaystyle X_i = DF_1(u) e_i$ and $\displaystyle X_i = DF_2 (F_2^{-1} \circ \phi \circ F_1 (u)) e_i$. I figured showing this would be sufficient because then the gaussian curvature is completely determined by the first fundamental form (the notation I've used means you seperate the two columns of the jacobian and use them as basis vectors).

So if I try to show this, then I end up with

$\displaystyle

I_p^{S_1} (X_i, X_j) = I_{\phi(p)}^{S_2} (D(\phi \circ F_1)(u)e_i, D(\phi \circ F_1)(u)e_j )

$and then I am totally stuck. Can anyone offer any advice? I understand this question is a bit long winded.