Surface Gradient

Question

How to show that $(\vec{n}\times\nabla_s)\times\vec{n}f=\nabla_sf-(\nabla_s\cdot \vec{n})\vec{n}f,$ where $\nabla_s$ is the surface gradient operator, $f$ is a scalar function and $\vec{n}$ is a vector.

Answer 1

Let us start from the vector identity $$ (\vec A \times \vec B) \times \vec C = (\vec A \cdot \vec C)\vec B - (\vec B \cdot \vec C) \vec A\,, $$ which is shown as follows in index notation: $(\vec x \times\vec y)_i=\epsilon_{ijk}x_j y_k$, where the sum over repeated indices is understood, so $$ [(\vec A \times \vec B) \vec C]_l =\epsilon_{lim}(\epsilon_{ijk}A_j B_k)C_m =\epsilon_{iml}\epsilon_{ijk}A_jB_kC_m $$ but $\epsilon_{iml}\epsilon_{ijk}=\delta_{mj}\delta_{lk}-\delta_{mk}\delta_{lj}$ and then indeed $$ [(\vec A \times \vec B) \vec C]_l = (A_j C_j) B_l - (B_k C_k) A_l\,. $$ In our case $\vec A=\vec n$, $\vec B=\nabla_s$ and $\vec C=\vec n$ so $$ (\vec n \times \nabla_s)\times \vec n =(\vec n^2)\nabla_s - (\nabla_s\cdot \vec n) \vec n\,. $$ Assuming $\vec n$ is a constant unit vector, this matches your formula.