$$ \begin{Bmatrix}\beta & &\kappa\ \alpha&\end{Bmatrix} $$
with reference to the indices β and κ, the third member of the right
hand side vanishes when A^{αβ} is an antisymmetrical tensor, which we
assume here; the second member can be transformed according to (29a); we
therefore get
(40) $$ A^{\alpha} = \frac{1}{\sqrt{-g}} \frac{\partial(\sqrt{-g}
A^{\alpha\beta})}{\partial x_{\beta}} $$
This is the expression of the divergence of a contravariant six-vector.
_Divergence of the mixed tensor of the second rank._
Let us form the reduction of (39) with reference to the indices α and σ,
we obtain remembering (29a)
(41) $$ \sqrt{-g} A_{\mu} = \frac{\partial(\sqrt{-g}
A^{\sigma}_{\mu})}{\partial x_{\sigma}} - \begin{Bmatrix}\sigma & &
\mu\ \tau&\end{Bmatrix} \sqrt{-g} A^{\sigma}_{\tau} $$
If we introduce into the last term the contravariant tensor A^{ρσ} =
_g_^{ρτ} A^{σ}_{τ}, it takes the form
$$ - \begin{bmatrix}\sigma & & \mu\ \rho&\end{bmatrix} \sqrt{-g}
A^{\rho\sigma} $$