# Gauss curvature derived from unit normal vector

I want to know more about the differential geometry of surfaces, especially Gaussian curvature. Obviously, we can get the mean curvature of a surface from the divergence of the unit normal vector of the surface. However, can the Gaussian curvature be derived from the divergence or curl of the unit normal vector of the surface? Perhaps there is also some historical / background information about their importance? Thank you in advance.

Supplement: The mean curvature of a surface specified by an equation $$\displaystyle\,\!F(x,y,z)=0$$ can be calculated by using the gradient $$\displaystyle\nabla F=\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)$$ and the divergence of the unit normal. A unit normal is given by $$\displaystyle\frac{\nabla F}{|\nabla F|}$$ and the mean curvature is $$\displaystyle H = -{\frac{1}{2}}\operatorname{div}\left(\frac{\nabla F}{|\nabla F|}\right)$$

• If you view the Gauss map as a map from the surface to the unit sphere, then the Gaussian curvature is the determinant of the Jacobian of this map. In other words, it measures how much the Gauss map distorts area infinitesimally. Commented Jul 3, 2020 at 17:30

$$\displaystyle K_G = \dfrac{1}{2}\operatorname{div}\left[\frac{\nabla F}{|\nabla F|}\cdot\operatorname{div}\left(\frac{\nabla F}{|\nabla F|}\right)+\frac{\nabla F}{|\nabla F|}\times\operatorname{curl}\left(\frac{\nabla F}{|\nabla F|}\right)\right]$$
• from this post math.stackexchange.com/q/2888749/694687 it seems that $\nabla\times\hat{n}=\nabla\times\frac{\nabla F}{\lVert\nabla F\rVert}$ is zero all the time. Am I just missing something? Commented Oct 13, 2022 at 13:45