$\nabla\cdot(f\vec{g})=f\nabla\vec{g}+\vec{g}\cdot\nabla f$ using Levi-Civita

Question

I need to prove the following equality: $\nabla\cdot(f\vec{g})=f\nabla\vec{g}+\vec{g}\cdot\nabla f$.

I know that proving this equality using the properties of $\nabla$ and $(\cdot)$ is easy, what the exercise asks me to do is to use the symbol of Levi-Civita, I'm new to working with tensors and that's why I'm a bit confused. My question is, is there a formula/relationship between $\varepsilon$ and $\nabla$, $\cdot$, that can help me prove equality? (At the moment I have found a relationship between $\varepsilon$ and $\nabla$, $\times$, but it does not help me to solve this exercise)

Answer 1

Let $f:\mathbb{R}^n\to\mathbb{R}$, $\vec{g}:\mathbb{R}^n\to\mathbb{R}^n$ be arbitrary.

The identity $\nabla \cdot (f\vec{g})=(\nabla f)\cdot\vec{g}+f(\nabla \cdot \vec{g})$ can then be verified using index notation. In index notation, the vector $\vec{g}$ is expressed as $\vec{g}=g_i\hat{e}_i$, where $\hat{e}_i$ is the standard Cartesian basis vector in the direction of increasing $x_i$ and $g_i$ is the component of $\vec{g}$ along $\hat{e}_i$ (since the Cartesian basis vectors are orthonormal, $g_j=\vec{g}\cdot \hat{e}_j$ as can be verified by taking the dot product of $\vec{g}=g_i\hat{e}_i$ with $\hat{e}_j$). It is important to note that implicit on the right hand side is the sum over $i\in \{1,2,...,n\}$. It is then natural to express the differential vector operator $\nabla$ as $\nabla=\hat{e}_i \partial_i$, where $\partial_i=\frac{\partial}{\partial x_i}$ is the partial derivative operator in the $\hat{e}_i$ direction. Then an object like the divergence of $\vec{g}$ can be understood as $$\nabla \cdot \vec{g}=(\hat{e}_i \partial_i) \cdot (g_j \hat{e}_j)=(\hat{e}_i \cdot \hat{e}_j) \partial_i g_j=\delta_{ij}\partial_i g_j=\partial_i g_i$$ where $\delta_{ij}$ is the Kronecker delta that is defined to be $1$ if $i=j$ and $0$ if $i\neq j$ and we have used the fact that $\partial_i \hat{e}_j$ vanishes for all $i,j\in \{1,2,...,n\}$. Proceed analogously by writing the left hand side of the identity in index notation to get started. Make sure to use the product rule in the form $\partial_i a_j b_k = a_j \partial_i b_k + b_k \partial_i a_j$.

The Levi-Civita symbol $\epsilon_{ijk}$ is defined to be equal to $1$ if $(ijk)$ is an even (cyclic) permutation of $(123)$, $-1$ if $(ijk)$ is an odd (acyclic) permutation of $(123)$, and $0$ otherwise. It is not needed for this identity since it does not involve cross products, which would be expressed as $\hat{e_i} \times \hat{e_j}=\epsilon_{ijk} \hat{e_k}$ in index notation. Dot products, on the other hand, can be handled exclusively using the Kronecker delta: $\hat{e_i} \cdot \hat{e_j} = \delta_{ij}$.