Metrics and Gauss curvature

The first and second fundamental form in a surface are symmetric bilinear forms, in particular they are metrics. Of course, the first fundamental form is positive definite because is the Euclidean metric in the tangent plane. We consider the second fundamental form $$\sigma_p:T_pS\times T_pS\rightarrow{\mathbb R}$$ $$\sigma_p(v,v)=\langle -dN_p(v),v\rangle.$$

If $X=X(u,v)$ is a parametrization of the surface, the matrix of $\sigma_p$ with respect to the basis $\{X_u,X_v\}$ is $$\left(\begin{array}{cc}e&f\\ f&g\end{array}\right).$$

We study the type of $\sigma_p$. However, it is better to choose a more suitable basis of $T_pS$, indeed, a basis of principal directions because in such a case, $$\sigma_p\rightarrow \left(\begin{array}{cc}\kappa_1(p)&0\\ 0&\kappa_2(p)\end{array}\right).$$ Because this basis diagonalizes the metric $\sigma_p$, we conclude:

1.  The point is elliptic ($K(p)>0$) is equivalent to $\kappa_i(p)>0$ for $i=1,2$, or $\kappa_i(p)<0$ for $i=1,2$ and this means that   $\sigma_p$ is  definite. 
2. The point is hyperbolic ($K(p)<0$) is equivalent to $\kappa_1(p)<0<\kappa_2(p)$. Then $\sigma_p$ is a non-degenerate indefinite metric with signature $(1,1)$.
3. $K(p)=0$, that is, some principal curvature is $0$. We have two subcases. 
• If the point is parabolic, then the non-zero principal curvature is positive or negative. This is equivalent to say that $\sigma_p$ is positive semidefinite or negative semidefinite and also, the metric is degenerate. Here the radical of $\sigma_p$ is the vector subspace spanned by the principal direction of the zero principal curvature.
• If the point is flat, then the metric is null.