I am trying to prove the following as a part of a larger derivation:
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=\nabla^\sigma\gamma^{wn}-N\gamma^{wn}(\nabla^\sigma.N)$$
Where,
$N$ is the unit vector normal to the interface
$\gamma^{wn}$ is a scalar
surface divergence operator, $\nabla^\sigma=\nabla-NN.\nabla$
projected surficial identity tensor, $I^\sigma=I-NN$
Here is my attempt:
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=\gamma^{wn}\nabla^\sigma.I^\sigma+I^\sigma.\nabla^\sigma\gamma^{wn}$$
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=(I-NN).\nabla^\sigma\gamma^{wn}+\gamma^{wn}\nabla^\sigma.(I-NN)$$
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=I.\nabla^\sigma\gamma^{wn}-NN.\nabla^\sigma\gamma^{wn}+\gamma^{wn}(\nabla^\sigma.I-\nabla^\sigma.NN)$$
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=\nabla^\sigma\gamma^{wn}-(N.\nabla\gamma^{wn})N-\gamma^{wn}\nabla^\sigma.NN$$
$$\nabla^\sigma.(\gamma^{wn}I^\sigma)=\nabla^\sigma\gamma^{wn}-(N.\nabla\gamma^{wn})N-\gamma^{wn}(\nabla.N)N-\gamma^{wn}(N.\nabla{N})$$
So terms 1 and 3 are what I needed, but I got two more terms. I am not sure if I did this correctly and what I have missed.
Appreciate any help!