Tensor Calculus – Divergence of Cross Product Using Contra/Covariant Index Notation

Question

On this answer about divergence of a cross product, the following proof using Einstein notation appears:

\begin{eqnarray*}
\nabla \cdot (A \times B) &=& [\epsilon_{ijk} A_j B_k],_{i} \\
&=& \epsilon_{ijk} A_{j_{,i}} B_k + \epsilon_{ijk} A_j B_{k_{,i}} \\
&=& B_k ( \epsilon_{kij} A_{j_{,i}}) – A_j ( \epsilon_{jik} B_{k_{,i}}) \\
&=& B \cdot (\nabla \times A) – A \cdot ( \nabla \times B )
\end{eqnarray*}

is it possible to rewrite it using supreindex (contravariant vector/tensor components) and subindex (covariant ones) ?

Could be it starts with something like this:

\begin{eqnarray*}
\nabla \cdot (A \times B) &=& [\epsilon^i_{jk} A^j B^k]_{,i} \\
&=& \epsilon^i_{jk} A^j_{,i} B^k + \epsilon^i_{jk} A^j B^k_{,i} \\
\end{eqnarray*}

but I do not known how to continue.

Answer 1

Sure! However when you mix up contravariant and covariant indices on the orientation tensor $\epsilon$, you can get very confused very quickly on its antisymmetry properties.

A more elegant convention: $$\begin{align} \mathbf u&=\nabla\times\mathbf v\\&\leftrightarrow\\ u^\alpha &=\epsilon^{\alpha\mu\nu} ~ \nabla_\mu v_\nu.\end{align}$$ Once you have that the rest of the expression is quite natural: $$ \nabla\cdot(A\times B) = \nabla_\lambda (\epsilon^{\lambda\mu\nu} A_\mu B_\nu)\\ =\epsilon^{\lambda\mu\nu}(\nabla_\lambda A_\mu) B_\nu + \epsilon^{\lambda\mu\nu} A_\mu (\nabla_\lambda B_\nu)\\ =B_\nu \epsilon^{\nu\lambda\mu}\nabla_\lambda A_\mu - A_\mu\epsilon^{\mu\lambda\nu} \nabla_\lambda B_\nu $$ and at this point it's just relabeling, if you even feel the need to do that.