I have seen two version of the curvature definition which are not identical. Curvature $\kappa$ is defined usually with the unit interface normal $\mathbf{n}$.
1) definition. see Eq. 4 here:
\begin{equation}
\kappa = \nabla\cdot\mathbf{n}
\end{equation}
2) definition: see Eq. 16 here:
\begin{equation}
\kappa = \nabla_s\cdot\mathbf{n},
\end{equation}
where $\nabla_s = (\mathbf{I}-\mathbf{n}\mathbf{n})\cdot\nabla$ and $\mathbf{I}$ is a unit matrix.
These two definitions do not seem to the same. I think the first term of the second definition is the same as the first definition, since $\mathbf{I}\cdot\nabla = \nabla$
I will be thankful for any explanation.
Best Answer
Both formula are the same (and missing the conventional $-$ sign and possibly off by a factor of 2 for mean curvature of a surface in $\mathbb{R}^3$), but there are technical details hidden in the formula.
Remember the vector field $\mathbf{n}$ is only defined, initially anyway, on the $(m-1)$-dimensional hypersurface $S\subset\mathbb{R}^m$. In order to apply $\nabla=\nabla^{\mathbb{R}^m}$ on $\mathbf{n}$, you need to extend the definition of $\mathbf{n}$ to a small tubular neighbourhood of $S$ in $\mathbb{R}^3$.
How do you extend $\mathbf{n}$? If you extend $\mathbb{n}$ such that it is constant in direction $\mathbf{n}$, then $\nabla$ sees nothing in the $\mathbf{n}$-direction and so $\nabla\cdot\mathbf{n}$ reduces to taking derivative in the two tangential directions.
However, if you extend it rather arbitrarily, then $\nabla$ will see some change in $\mathbf{n}$ in the direction $\mathbf{n}$. So you need to remove these undesirable arbitrary choice, hence the $(I-\mathbf{n}\mathbf{n}^T)$ applied to $\nabla$ to make sure it only sees the tangential directions.