On the account of the symmetry of {β κ α} 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^{α} = (1/[sqrt](-g)) ([part]([sqrt](-g) A^{αβ})/[part]x_{β})
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) Aμ}?] = [part]([sqrt](-g) A^{σ}_{μ})/[part]x_{σ} - {σ μ τ} [sqrt](-g] A^{α}_{τ}.
If we introduce into the last term the contravariant tensor A^{ρσ} = g^{ρτ} A^{σ}_{τ}, it takes the form
-[σ μ ρ] [sqrt](-g) A^{ρσ}.
If further A^{ρσ} is symmetrical it is reduced to
-1/2 [sqrt](-g) ([part]g_{ρσ}/[part]x_{μ}) A^{ρσ}.
