If instead of A^{ρσ}, we introduce in a similar way the symmetrical co-variant tensor A_{ρσ} = _g__{ρα} _g__{σβ} A^{αβ}, then owing to (31) the last member can take the form $$ \frac{1}{2} \sqrt{-g} \frac{\partial g_{\rho\sigma}}{\partial x_{\mu}} A_{\rho\sigma} $$ In the symmetrical case treated, (41) can be replaced by either of the forms (41a) $$ \sqrt{-g} A{\mu} = \frac{\partial (\sqrt{-g} A^{\sigma}_{\mu})}{\partial x_{\sigma}} - \frac{1}{2} \frac{\partial g_{\rho\sigma}}{\partial x_{\mu}} \sqrt{-g} A^{\rho\sigma} $$ or (41b) $$ \sqrt{-g} A{\mu} = \frac{\partial (\sqrt{-g} A^{\sigma}_{\mu})}{\partial x_{\sigma}} + \frac{1}{2} \frac{\partial g_{\rho\sigma}}{\partial x_{\mu}} \sqrt{-g} A_{\rho\sigma} $$ which we shall have to make use of afterwards. §12. The Riemann-Christoffel Tensor. We now seek only those tensors, which can be obtained from the fundamental tensor _g_^{μν} by differentiation alone. It is found easily. We put in (27) instead of any tensor A^{μν} the fundamental tensor _g_^{μν} and get from it a new tensor, namely the extension of the fundamental tensor. We can easily convince ourselves that this vanishes identically. We prove it in the following way; we substitute in (27) $$ A_{\mu\nu} = \frac{\partial A_{\mu}}{\partial x_{\nu}} - \begin{Bmatrix}\mu & & \nu\ \rho&\end{Bmatrix} A_{\rho} $$ _i.e._, the extension of a four-vector. Thus we get (by slightly changing the indices) the tensor of the third rank $$ A_{\mu\sigma\tau} = \frac{\partial^2 A_{\mu}}{\partial x_{\sigma} \partial x_{\tau}} - \begin{Bmatrix}\mu & & \sigma\ \rho&\end{Bmatrix} \frac{\partial A_{\rho}}{\partial x_{\tau}} - \begin{Bmatrix}\mu & & \tau\ \rho&\end{Bmatrix} \frac{\partial A_{\rho}}{\partial x_{\sigma}} - \begin{Bmatrix}\sigma & & \tau\ \rho&\end{Bmatrix} \frac{\partial A_{\mu}}{\partial x_{\rho}} + \begin{bmatrix} - \frac{\partial}{\partial x_{\tau}} \begin{Bmatrix}\mu&&\sigma\ \rho&\end{Bmatrix} + \begin{Bmatrix}\mu&&\tau\ \alpha\end{Bmatrix} \begin{Bmatrix}\alpha&&\sigma\ \rho&\end{Bmatrix} + \begin{Bmatrix}\sigma&&\tau\ \alpha\end{Bmatrix} \begin{Bmatrix}\alpha&&\mu\ \rho&\end{Bmatrix} \end{bmatrix} A_{\rho} $$ We use these expressions for the formation of the tensor A_{μστ} - A_{μτσ}. Thereby the following terms in A_{μστ} cancel the corresponding terms in A_{μτσ}; the first member, the fourth member, as well as the member corresponding to the last term within the square bracket. These are all symmetrical in σ, and τ. The same is true for the sum of the second and third members. We thus get (43) $$ A_{\mu\sigma\tau} - A_{\mu\tau\sigma} = B^{\rho}_{\mu\sigma\tau} A_{\rho} $$