Let $(\Sigma,g)$ be a surface isometrically embedded in a $3$-dimensional Riemannian manifold $(M,\bar{g})$. I'm particularly interested in the case where the ambient space $(M,\bar{g})$ is not the flat Euclidean space $\mathbb{R}^3$. I would like to seek for a clarification of the definition of Gauss curvature:
Is $K$ defined by
- the determinant of the shape operator $S$? or
- the sectional curvature $R_{1221}=Rm(e_1,e_2,e_2,e_1)$ (where $Rm$ is the Riemann curvature tensor of $g$ and $\{e_1,e_2\}$ is a local orthonormal frame on $\Sigma$)?
When $(M,\bar{g})$ is the flat $\mathbb{R}^3$, these two notions coincide by the Theorema Egregium of Gauss, but in general they certainly do not agree. In fact, by Gauss equation, we have \begin{align} 2R_{1221}=2\bar{R}_{1221}+H^2-|A|^2 \end{align} where
- $\bar{R}_{1221}=\bar{Rm}(e_1,e_2,e_2,e_1)$ ($\bar{Rm}$ is the Riemann curvature tensor of $\bar{g}$, and we regard $e_1,e_2$ as tangent vectors in $TM$),
- $H=\lambda_1+\lambda_2$ is the mean curvature (sum of principal curvatures), and
- $|A|^2=\lambda_1^2+\lambda_2^2$ is the norm-square of 2nd fundamental form.
Thus, it follows that \begin{align} \det(S)=\lambda_1\lambda_2=R_{1221}-\bar{R}_{1221} \end{align} which certainly does not equal to $R_{1221}$.
I'm asking this because $K$ is defined in the 1st way in many text, but at the same time they seems to also claim the Gauss-Bonnet theorem, which holds only if $K$ is defined in the 2nd way. This is pretty confusing. (Of course, it is also possible that I have made some mistakes somewhere around. I welcome any correction.)
Any comment and answer are greatly welcomed and appreciated.