$$ \frac{\partial}{\partial x_{\sigma}} (g^{\mu\alpha} g^{\nu\beta} A_{\mu\nu}) - g^{\mu\alpha} \frac{\partial g^{\nu\beta}}{\partial x_{\sigma}} A_{\mu\nu} - g^{\nu\beta} \frac{\partial g^{\mu\alpha}}{\partial x_{\sigma}} A_{\mu\nu} $$ If we replace _g_^{μα} _g_^{νβ} A_{μνσ} by A_{σ}^{αβ}, _g_^{μα} _g_^{νβ} A_{μν} by A^{αβ} and replace in the transformed first member ∂_g_^{νβ}/∂_x__{σ} and ∂_g_^{μα}/∂_x__{σ} with the help of (34), then from the right-hand side of (27) there arises an expression with seven terms, of which four cancel. There remains (38) $$ A^{\alpha\beta}_{\sigma} = \frac{\partial A^{\alpha\beta}}{\partial x_{\sigma}} + \begin{Bmatrix}\sigma & & \kappa\ \alpha end{Bmatrix} A^{\kappa\beta} + \begin{Bmatrix}\sigma & & \kappa\ \beta&\end{Bmatrix} A^{\alpha\kappa} $$ This is the expression for the extension of a contravariant tensor of the second rank; extensions can also be formed for corresponding contravariant tensors of higher and lower ranks. We remark that in the same way, we can also form the extension of a mixed tensor A_{μ}^{α} (39) $$ A^{\alpha}_{\mu\sigma} = \frac{\partial A^{\alpha}_{\mu}}{\partial x_{\sigma}} - \begin{Bmatrix}\sigma & & \mu\ \tau&\end{Bmatrix} A^{\alpha}_{\tau} + \begin{Bmatrix}\sigma & & \tau\ \alpha&\end{Bmatrix} A^{\tau}_{\mu} $$ By the reduction of (38) with reference to the indices β and σ(inner multiplication with δ_{β}^{σ}), we get a contravariant four-vector $$ A^{\alpha} = \frac{\partial A^{\alpha\beta}}{\partial x_{\beta}} + \begin{Bmatrix}\beta & & \kappa\ \beta&\end{Bmatrix} A^{\alpha\kappa} + \begin{Bmatrix}\beta & & \kappa\ \alpha&\end{Bmatrix} A^{\kappa\beta} $$ On the account of the symmetry of $$ \begin{Bmatrix}\beta & &\kappa\ \alpha&\end{Bmatrix} $$ 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 $$ - \begin{bmatrix}\sigma & & \mu\ \rho&\end{bmatrix} \sqrt{-g} A^{\rho\sigma} $$ If further A^{ρσ} or is symmetrical it is reduced to $$ - \frac{1}{2} \sqrt{-g} \frac{\partial g_{\rho\sigma}}{\partial x_{\mu}} A^{\rho\sigma} $$