Theorema Egregium and coeficcients of the second fundamental form The Theorema Egregium says that Gaussian curvature $K$ of a regular surface $S$ is invariant under local isometries. We have a local description of the Gaussian curvature as follows
$$K = \dfrac{eg-f^2}{EG-F^2}$$
where $E$, $F$, and $G$ are the coefficients of the first fundamental form of $S$ and $e$, $f$ and $g$ the coefficients of the second fundamental form of $S$.
This implies that $eg-f^2$ is invariant under local isometries as well. I am following the proof of Manfredo's Differential Geometry of Curves and Surfaces for the theorema Egregium, but the end of the proof seems less natural to me that the proof of the fact that Christoffel's symbols are invariant under local isometries.
My question:
Is there another way of proving the quantity $eg-f^2$ depends only on Christoffel's symbols and the coefficients of the fundamental form, thus it is also invariant under local isometries?
Thank you in advance.
EDIT :
Maybe I should add the following, it is not hard to see that, if $\chi: U \to S$ is a local parametrization compatible with some orientation $N$ on $S$ then
$$eg-f^2 = \langle \chi_{u,u},\chi_{v,v} \rangle - \langle \chi_{u,v},\chi_{u,v} \rangle  + \mathrm{Christoffel's ~symbols}$$
Therefore my question reduces to showing that $\langle \chi_{u,u},\chi_{v,v} \rangle - \langle \chi_{u,v},\chi_{u,v} \rangle$ can be expressed in terms of Christoffel's symbols and the coefficients of the first fundamental form.
 A: In a course of a colleague, I found the following result that might be useful to you. The idea is to look at the tangent parts of $x_{uu}, x_{uv}, x_{vv}$, etc.
Recall that the tangent part of the derivative of a tangent vector is the covariant derivative on the surface.

Let $M$ be a surface in $\mathbb{E}^3$ with Gauss curvature $K$ and $x\colon U\subset \mathbb{R}^2\to M$ a regular parametrisation. Then
  $$ \bigl((x_{vv}^T)_u - (x_{uv}^T)_v\bigr) \cdot x_u = (K\circ x)\bigl((x_u\cdot x_u)(x_v\cdot x_v)-(x_u\cdot x_v)^2\bigr),$$
  where the superscript ${}^{T}$ denotes the component tangent to the surface.

Proof: If $N$ is a normal unit vector field along $x$, then
$$ 
\begin{align*}
x_{vv}^T &= x_{vv}+(x_v\cdot N_v)N \\
x_{uv}^T &= x_{uv}+(x_u\cdot N_v)N.
\end{align*}
$$
Deriving gives
$$ 
\begin{align*}
(x_{vv}^T)_u &= x_{uvv}+(x_v\cdot N_v)_u N + (x_v\cdot N_v) N_u \\
(x_{uv}^T)_v &= x_{uvv}+(x_u\cdot N_v)_v N + (x_u\cdot N_v) N_v,
\end{align*}
$$
and thus 
$$
(x_{vv}^T)_u - (x_{uv}^T)_v=\bigl((x_v\cdot N_v)_u-(x_u\cdot N_v)_v\bigr)N
+(x_v\cdot N_v)N_u - (x_u\cdot N_v)N_v.
$$
Take the inproduct with $x_u$. We use the definition of the shape operator, the functions $E$, $F$, \ldots, $f$ and $g$ and the Gauss curvature $K$ to obtain
$$ \begin{align*}
\bigl((x_{vv}^T)_u - (x_{uv}^T)_v\bigr)\cdot x_u &= (x_v\cdot N_v)(N_u\cdot x_u)-(x_u\cdot N_v)(N_v\cdot x_u)\\
&= eg - f^2 = (K\circ x) (EG-F)^2.
\end{align*}$$
To prove the Theorema Egregium, one should also use the following result:

If $\phi$ is a local isometry of $M$, then $\phi$ preserves the tangent part of $x_{uu}$, $x_{uv}$ and $x_{vv}$:
  $$ \phi_* \bigl((x_{uu}^T)\bigr) =\bigl((\phi\circ x)_{uu} )\bigr)^T, \text{ etc. for $x_{uv}$ and $x_{vv}$}. $$

