What is the relationship between the 2nd Fundamental Form and Gaussian Curvature? I am looking at the proof for this Lemma:

If $S \subset \mathbb{R}^3$ is a regular, compact orientable surface,
  then $S$ has an elliptic point.

The proof concludes with stating that since $II_p$ has a fixed sign, then that implies that the Gaussian Curvature of $p$ on $S$ is positive.
I am having trouble seeing why this is so. I know that both the 2nd fundamental form and Gaussian curvature involves $dN_p$, but one is an Inner product and the other is determinant. I also know that the prinicpal curvatures are the eigenvalues of $dN_p$, but I can't see how they are related more explicitly.
Would really appreciate some help is developing a better understanding of this. Thanks!
 A: Gaussian curvature is $$K(p):={\rm det}\ dN_p $$
If $x(u,v)$ is a parametrization, and if $$ N_u= dN x_u ,\ N_v=dN
x_v
$$ then we have first and second fundament form :
 $$ (x_u,x_u)=E,\ (x_v,x_v)=G,\ (x_u,x_v)=F $$
$$ (N_u ,x_u)=e,\ (N_v,x_v)=g,\ (N_u,x_v)= f $$ and $$ {\rm det} dN
=\frac{eg-f^2}{EG-F^2} $$
Here recall a normal curvature : $$k_n(p)= -(dN c'(0),c'(0)) $$ for
a curve on $S$ passing through $p$. In further $dN$ is self adjoint
so that there exist eigenvalues $dN v_i=\lambda_i v_i $. That is,
$-\lambda_i$ are normal curvatures. That is sign of $K$ is equal to
$\lambda_1\lambda_2$.
If $S$ is a compact then we choose some inner point $o$ which is not in $S$.
 So the farthest point $x\in S$ from $o$ has
 normal curvatures of same sign. Hence we are done.
A: Only for elliptic points the sign is positive and for saddle/hyperbolic points it is negative. 
By Euler relation in the fiber tangent bundle normal curvature sign passes from positive via zero to negative if anti-clastic type $ k_1k_2< 0 $ of surface and for syn-clastic $ k_1k_2> 0 $ both are on the positive side.
$$ k_n = k_1 \cos^{2}\psi +  k_2 \cos^{2}\psi $$
The second fundamental form determinant can be expressed in terms of the first fundamental form coefficients & derivatives thereby making scalar Gauss curvature entirely isometric mapping independent, demonstrated in the proof of Gauss Theorema Egregium.
