Divergence of cross product, using contra/covariant index notation

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.

• Well, on flat space is useless to use „up” and „down” indices. But you can do it, if you want to consider curved space. However, the „vector/cross product” makes sense only in 3 or 7 dimensions. You can, however, use the exterior differential and codifferential to make sense of curved space. Feb 14 at 20:34

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.
$$\nabla \cdot (A \times B) = [\epsilon^i{_{jk}} A^j B^k]_{,i} \\ = \epsilon^i{_{jk}} A^j{_{,i}} B^k + \epsilon^i{_{jk}} A^j B^k,_i \\ = B^k\epsilon_k{^i}{_j} A^j{_{,i}} - A^j\epsilon_j{^i}{_k} B^k,_i\\ = B \cdot (\nabla \times A) - A \cdot ( \nabla \times B )$$