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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.

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  • $\begingroup$ 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. $\endgroup$
    – DanielC
    Feb 14 at 20:34

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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.

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If it is not clear from the context that all index are summation ones, we can raise some of them to make it explicit. It is possible to choose them freely because the metric is the identity:

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

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