Translation:On the spacetime lines of a Minkowski world/Paragraph 2

$$n=3$$, $$p=2$$. Theorem of Gauss.
By (1):

$\begin{aligned}\int\int du^{(1)}du^{(2)}\left[\frac{\partial\left(x^{(2)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{23}+\frac{\partial\left(x^{(3)}x^{(1)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{31}+\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{12}\right]=\\ =\int\int\int dv^{(1)}dv^{(2)}dv^{(3)}\frac{\partial\left(x^{(1)}x^{(2)}x^{(3)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}A_{123}, \end{aligned}$,

$A_{123}=\frac{\partial}{\partial x^{(3)}}A_{23}-\frac{\partial}{\partial x^{(2)}}A_{13}+\frac{\partial}{\partial x^{(1)}}A_{23},\quad A_{\alpha\beta}=-A_{\beta\alpha}$|undefined

Introducing the supplements

$\sqrt{\frac{1}{c}}A_{23}=B^{(1)},\quad\sqrt{\frac{1}{c}}A_{31}=B^{(2)},\quad\sqrt{\frac{1}{c}}A_{12}=B^{(3)},$|undefined

$\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(2)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}=N_{1},\quad\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(3)}x^{(1)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}=N_{2},\quad\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}=N_{3},$|undefined

where

$b=\left

is the discriminant of the arc on $$M_{2}$$, thus because of

$dS_{2}=\sqrt{b}du^{(1)}du^{(2)},\quad dS_{3}=\sqrt{c}dx^{(1)}dx^{(2)}dx^{(3)}$:

it follows

$\begin{aligned} & \int\int dS_{2}\left[N_{1}B^{(1)}+N_{2}B^{(2)}+N_{3}B^{(3)}\right]=\\ & \quad=\int\int\int dS_{3}\frac{1}{\sqrt{c}}\left[\frac{\partial}{\partial x^{(1)}}\left(\sqrt{c}B^{(1)}\right)+\frac{\partial}{\partial x^{(2)}}\left(\sqrt{c}B^{(2)}\right)+\frac{\partial}{\partial x^{(3)}}\left(\sqrt{c}B^{(3)}\right)\right] \end{aligned}$|undefined

Here, the normal $$N$$ goes to the exterior. For the generalized divergence:

$\mathrm{div}\left(B^{(1)}B^{(2)}B^{(3)}\right)=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}B^{(\alpha)}\right)$|undefined

or introducing the system reciprocal to $$B^{(\alpha)}$$

$B^{(\alpha)}=\sum_{\beta}c^{(\alpha\beta)}B_{\beta}:$:

it follows

$\mathrm{div}\left(B_{1}B_{2}B_{3}\right)=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\sum_{\beta}c^{(\alpha\beta)}B_{\beta}\right)$|undefined

or eventually, if $$B_{\beta}=\mathrm{B_{\beta}}=\frac{\partial B}{\partial x^{(\beta)}}$$ can be represented as gradient of a scalar (invariant) quantity $$B$$:

$\mathrm{div}\left(\mathrm{B}_{1}\mathrm{B}_{2}\mathrm{B}_{3}\right)=\Delta B=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\sum_{\beta}c^{(\alpha\beta)}\frac{\partial B}{\partial x^{(\beta)}}\right)$ |undefined

$$n=3$$, $$p=1$$. Theorem of Stokes.
By (1):

$\begin{aligned} & \int du\left[\frac{\partial x^{(1)}}{\partial u}A_{1}+\frac{\partial x^{(2)}}{\partial u}A_{2}+\frac{\partial x^{(3)}}{\partial u}A_{3}\right]=\\ & \quad=\int\int dv^{(1)}dv^{(2)}\left[\frac{\partial\left(x^{(1)}x^{(3)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}\mathrm{A}_{23}+\frac{\partial\left(x^{(3)}x^{(1)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}\mathrm{A}_{31}+\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}\mathrm{A}_{12}\right] \end{aligned} $|undefined

$\mathrm{A}_{23}=\frac{\partial A_{2}}{\partial x^{(3)}}-\frac{\partial A_{3}}{\partial x^{(2)}},\quad\mathrm{A}_{31}=\frac{\partial A_{3}}{\partial x^{(1)}}-\frac{\partial A_{1}}{\partial x^{(3)}},\quad\mathrm{A}_{12}=\frac{\partial A_{1}}{\partial x^{(2)}}-\frac{\partial A_{2}}{\partial x^{(1)}}$|undefined

or

$\begin{aligned} & \int ds\left[\frac{dx^{(1)}}{ds}A_{1}+\frac{dx^{(2)}}{ds}A_{2}+\frac{dx^{(3)}}{ds}A_{3}\right]=\\ & \quad=\int\int dS_{2}\left[N_{1}\frac{1}{\sqrt{c}}\left(\frac{\partial A_{2}}{\partial x^{(3)}}-\frac{\partial A_{3}}{\partial x^{(2)}}\right)+N_{2}\frac{1}{\sqrt{c}}\left(\frac{\partial A_{3}}{\partial x^{(1)}}-\frac{\partial A_{1}}{\partial x^{(3)}}\right)+N_{3}\frac{1}{\sqrt{c}}\left(\frac{\partial A_{1}}{\partial x^{(2)}}-\frac{\partial A_{2}}{\partial x^{(1)}}\right)\right] \end{aligned} $|undefined

Here, the line integral is orbiting around the normal $$N$$ in the negative sense, thus clock-wise if the coordinate system is a right-system; because by § 1 the directions $$\frac{dx}{ds}$$, $$N'$$, $$N$$ are following each other like the coordinate axes, where $$N'$$ is the normal (which is directed outwards of the area framed on $$M_{2}$$) of the framing-$$M_{1}$$. Ordinarily, one prefers a positive sense of circulation and therefore the generalized rotation:

$\mathrm{rot}^{(1)}\left(A_{1}A_{2}A_{3}\right)=\frac{1}{\sqrt{c}}\left(\frac{\partial A_{3}}{\partial x^{(2)}}-\frac{\partial A_{2}}{\partial x^{(3)}}\right)$ etc.|undefined

$\mathrm{rot}^{(1)}\left(A^{(1)}A^{(2)}A^{(3)}\right)=\frac{1}{\sqrt{c}}\left(\frac{\partial}{\partial x^{(2)}}\left(\sum_{\beta}c_{3\beta}A^{(\beta)}\right)-\frac{\partial}{\partial x^{(3)}}\left(\sum_{\beta}c_{2\beta}A^{(\beta)}\right)\right)$ etc.|undefined

$$n=4$$, $$p=3$$.
$\begin{aligned} & \int\int\int du^{(1)}du^{(2)}du^{(3)}\left[\frac{\partial\left(x^{(2)}x^{(3)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)}A_{234}+\frac{\partial\left(x^{(1)}x^{(3)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)}A_{134}+\right.\\ & \left.+\frac{\partial\left(x^{(1)}x^{(2)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)}A_{124}+\frac{\partial\left(x^{(1)}x^{(2)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)}A_{123}\right]=\\ & \quad=\int\int\int dv^{(1)}dv^{(2)}dv^{(3)}dv^{(4)}\frac{\partial\left(x^{(1)}x^{(2)}x^{(3)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}v^{(4)}\right)}\mathrm{A}_{1234} \end{aligned}$

$\mathrm{A}_{1234}=\frac{\partial}{\partial x^{(4)}}A_{123}-\frac{\partial}{\partial x^{(3)}}A_{124}+\frac{\partial}{\partial x^{(2)}}A_{134}-\frac{\partial}{\partial x^{(1)}}A_{234},$|undefined

$A_{123}=-A_{131}=-A_{321}=A_{231}=A_{312}=-A_{213}$ etc.

Introducing the supplements

$-\sqrt{\frac{1}{c}}A_{234}=B^{(1)},\quad\sqrt{\frac{1}{c}}A_{134}=B^{(2)},\quad-\sqrt{\frac{1}{c}}A_{124}=B^{(3)},\quad\sqrt{\frac{1}{c}}A_{123}=B^{(4)},$|undefined

$\begin{aligned}-\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(2)}x^{(3)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)} & =N_{1}, & \sqrt{\frac{c}{b}}\frac{\partial\left(x^{(1)}x^{(3)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)} & =N_{2},\\ -\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(1)}x^{(2)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)} & =N_{3}, & \sqrt{\frac{c}{b}}\frac{\partial\left(x^{(1)}x^{(2)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}u^{(3)}\right)} & =N_{4}, \end{aligned} $|undefined

where $$b$$ is the discriminant of the arc-element on $$M_{3}$$, so because of

$dS_{3}=\sqrt{b}du^{(1)}du^{(2)}du^{(3)},\quad dS_{4}=\sqrt{c}dx^{(1)}dx^{(2)}dx^{(3)}dx^{(4)}$:

it is given

$\begin{aligned} & \int\int dS_{3}\left[N_{1}B^{(1)}+N_{2}B^{(2)}+N_{3}B^{(3)}+N_{4}B^{(4)}\right]=\\ & \quad=\int\int\int dS_{4}\frac{1}{\sqrt{c}}\left[\frac{\partial}{\partial x^{(1)}}\left(\sqrt{c}B^{(1)}\right)+\frac{\partial}{\partial x^{(2)}}\left(\sqrt{c}B^{(2)}\right)+\frac{\partial}{\partial x^{(3)}}\left(\sqrt{c}B^{(3)}\right)+\frac{\partial}{\partial x^{(4)}}\left(\sqrt{c}B^{(4)}\right)\right] \end{aligned} $|undefined

where $$N$$ goes to the exterior. Therefore it is given for the generalized four-dimensional divergence:

$\mathrm{Div}\left(B^{(1)}B^{(2)}B^{(3)}B^{(4)}\right)=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}B^{(\alpha)}\right),$|undefined

$$\mathrm{div}\left(B_{1}B_{2}B_{3}B_{4}\right)=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\sum_{\beta}c^{(\alpha\beta)}B_{\beta}\right),$$,

$\mathrm{div}\left(\mathrm{B}_{1}\mathrm{B}_{2}\mathrm{B}_{3}\mathrm{B}_{4}\right)=\square B=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\sum_{\beta}c^{(\alpha\beta)}\frac{\partial B}{\partial x^{(\beta)}}\right).$|undefined

$$n=4$$, $$p=2$$.
$\begin{aligned} & \int\int du^{(1)}du^{(2)}\left[\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{12}+\frac{\partial\left(x^{(1)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{13}+\frac{\partial\left(x^{(1)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{14}+\right.\\ & \quad\left.+\frac{\partial\left(x^{(2)}x^{(3)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{23}+\frac{\partial\left(x^{(2)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{24}+\frac{\partial\left(x^{(3)}x^{(4)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}A_{34}\right]=\\ & =\int\int\int dv^{(1)}dv^{(2)}dv^{(3)}\left[\frac{\partial\left(x^{(2)}x^{(3)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}\mathrm{A}_{234}+\frac{\partial\left(x^{(1)}x^{(3)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}\mathrm{A}_{134}+\right.\\ & \quad\left.+\frac{\partial\left(x^{(1)}x^{(2)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}\mathrm{A}_{124}+\frac{\partial\left(x^{(1)}x^{(2)}x^{(3)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}\mathrm{A}_{123}\right] \end{aligned}$

Introducing the supplements

$\begin{aligned} & \sqrt{\frac{1}{c}}A_{12}=B^{(34)},\quad\sqrt{\frac{1}{c}}A_{13}=B^{(42)},\quad\sqrt{\frac{1}{c}}A_{14}=B^{(23)},\\ & \sqrt{\frac{1}{c}}A_{23}=B^{(14)},\quad\sqrt{\frac{1}{c}}A_{24}=B^{(31)},\quad\sqrt{\frac{1}{c}}A_{34}=B^{(12)}, \end{aligned} $|undefined

$\sqrt{\frac{c}{a}}\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(u^{(1)}u^{(2)}\right)}=N_{34},$ etc.|undefined

$-\sqrt{\frac{c}{b}}\frac{\partial\left(x^{(2)}x^{(3)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}v^{(3)}\right)}=N_{1},$ etc.|undefined

where $$a$$ or $$b$$ are the discriminants of the arc-element of $$M_{2}$$ or $$M_{3}$$:

$\begin{aligned}\int\int dS_{2}\left[N_{12}B^{(11)}+N_{13}B^{(13)}+N_{14}B^{(14)}+N_{23}B^{(23)}+N_{24}B^{(24)}+N_{34}B^{(34)}\right]=\\ =-\int\int\int dS_{3}\left\{ N_{1}\frac{1}{\sqrt{c}}\left[^{\ast}+\frac{\partial}{\partial x^{(2)}}\left(\sqrt{c}B^{(12)}\right)+\frac{\partial}{\partial x^{(3)}}\left(\sqrt{c}B^{(13)}\right)+\frac{\partial}{\partial x^{(4)}}\left(\sqrt{c}B^{(14)}\right)\right]\right.\\ N_{2}\frac{1}{\sqrt{c}}\left[\frac{\partial}{\partial x^{(1)}}\left(\sqrt{c}B^{(21)}\right)+{}^{\ast}+\frac{\partial}{\partial x^{(3)}}\left(\sqrt{c}B^{(23)}\right)+\frac{\partial}{\partial x^{(4)}}\left(\sqrt{c}B^{(24)}\right)\right]\\ N_{3}\frac{1}{\sqrt{c}}\left[\frac{\partial}{\partial x^{(1)}}\left(\sqrt{c}B^{(31)}\right)+\frac{\partial}{\partial x^{(2)}}\left(\sqrt{c}B^{(32)}\right)+{}^{\ast}+\frac{\partial}{\partial x^{(4)}}\left(\sqrt{c}B^{(34)}\right)\right]\\ \left.N_{4}\frac{1}{\sqrt{c}}\left[\frac{\partial}{\partial x^{(1)}}\left(\sqrt{c}B^{(41)}\right)+\frac{\partial}{\partial x^{(2)}}\left(\sqrt{c}B^{(42)}\right)+\frac{\partial}{\partial x^{(3)}}\left(\sqrt{c}B^{(43)}\right)+{}^{\ast}\right]\right\} \end{aligned} $|undefined

where the normal plane $$N_{\alpha\beta}$$ is given by $$[N'N]$$ and $$N'$$, which is the normal of $$M_{2}$$ directed outwards of the area limited on $$M_{3}$$ as mentioned in § 1. By that, the generalized vector divergence becomes:

$\mathfrak{Div}^{(1)}\left(B^{(12)}B^{(13)}B^{(14)}B^{(23)}B^{(24)}B^{(34)}\right)=\frac{1}{\sqrt{c}}\sum\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}B^{(1\alpha)}\right)$ etc.|undefined

$\mathfrak{Div}^{(1)}\left(B_{12}B_{13}B_{14}B_{23}B_{24}B_{34}\right)=\frac{1}{\sqrt{c}}\sum_{\alpha}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\sum_{\beta,\gamma}c^{(1\beta)}c^{(\alpha\gamma)}B_{\beta\gamma}\right)$ etc.|undefined

The system

$B_{\alpha_{3}\alpha_{4}}=\sum_{\beta_{3},\beta_{4}}c_{\alpha_{3}\beta_{3}}c_{\alpha_{4}\beta_{4}}B^{\left(\beta_{3}\beta_{4}\right)}=\frac{1}{\sqrt{c}}\sum c_{\alpha_{3}\beta_{3}}c_{\alpha_{4}\beta_{4}}A_{\beta_{1}\beta_{2}}$|undefined

shall be called the system dual to $$A_{\alpha\beta}$$ and be denoted as $$A_{\alpha_{3}\alpha_{4}}^{\ast}$$. So it follows

$\mathfrak{Div}^{(1)}\left(A_{12}^{\ast}A_{13}^{\ast}A_{14}^{\ast}A_{23}^{\ast}A_{24}^{\ast}A_{34}^{\ast}\right)=\frac{1}{\sqrt{c}}\mathrm{A}_{234},$|undefined

$\mathfrak{Div}^{(2)}\left(A_{12}^{\ast}A_{13}^{\ast}A_{14}^{\ast}A_{23}^{\ast}A_{24}^{\ast}A_{34}^{\ast}\right)=-\frac{1}{\sqrt{c}}\mathrm{A}_{134}$ etc.|undefined

and

$\mathfrak{Div}^{(1)}\left(A_{12}^{\text{ }}A_{13}A_{14}A_{23}A_{24}A_{34}\right)=\frac{1}{\sqrt{c}}\mathrm{A}_{234}^{\ast},$|undefined

$\mathfrak{Div}^{(2)}\left(A_{12}^{\text{ }}A_{13}A_{14}A_{23}A_{24}A_{34}\right)=-\frac{1}{\sqrt{c}}\mathrm{A}_{134}^{\ast}$ etc.|undefined

where $$\mathrm{A}_{234}^{\ast}$$ etc. are to be formed from $$A_{\alpha\beta}^{\ast}$$, as $$\mathrm{A}_{234}$$ etc. from $$A_{\alpha\beta}$$.

In the case

$\mathrm{Div}\left(A_{\alpha\beta}^{\ast}\right)\equiv0$

it therefore follows

$A_{\alpha\beta}=\mathrm{A}_{\alpha\beta}=\frac{\partial A_{\alpha}}{\partial x^{(\beta)}}-\frac{\partial A_{\beta}}{\partial x^{(\alpha)}}$.|undefined

$$n=4$$, $$p=1$$.
$\begin{aligned} & \int du\left(\frac{\partial x^{(1)}}{\partial u}A_{1}+\frac{\partial x^{(2)}}{\partial u}A_{2}+\frac{\partial x^{(3)}}{\partial u}A_{3}+\frac{\partial x^{(4)}}{\partial u}A_{4}\right)=\\ & \quad=\int\int dv^{(1)}dv^{(2)}\left(\frac{\partial\left(x^{(1)}x^{(2)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{12}+\frac{\partial\left(x^{(1)}x^{(3)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{13}+\right.\\ & \qquad\left.\frac{\partial\left(x^{(1)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{14}+\frac{\partial\left(x^{(2)}x^{(3)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{23}+\frac{\partial\left(x^{(2)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{24}+\frac{\partial\left(x^{(3)}x^{(4)}\right)}{\partial\left(v^{(1)}v^{(2)}\right)}A_{34}\right) \end{aligned}$|undefined

or

$\int ds\sum\frac{dx^{(\alpha)}}{ds}A_{\alpha}=\int\int dS_{2}\sum N_{\alpha_{3}\alpha_{4}}\frac{1}{\sqrt{c}}A_{\alpha_{1}\alpha_{2}}$|undefined

with the corresponding orientation (by § 1). Therefore, it is given for the generalized rotation with the common signs:

$\mathrm{Rot}^{(12)}\left(A_{1}A_{2}A_{3}A_{4}\right)=\frac{1}{\sqrt{c}}\left(\frac{\partial A_{4}}{\partial x^{(3)}}-\frac{\partial A_{3}}{\partial x^{(4)}}\right)$ etc.|undefined

$\mathrm{Rot}_{12}\left(A_{1}A_{2}A_{3}A_{4}\right)=\frac{\partial A_{2}}{\partial x^{(1)}}-\frac{\partial A_{1}}{\partial x^{(2)}}$ etc.|undefined

The integral forms $$n=4$$ in the notation of the absolute differential calculus.
As appendix, the methods of the already mentioned absolute differential calculus shall be demonstrated, because it will be applied later; while it is less suited for the transformation of the actual integral form, it can hardly be avoided in connection with other vectorial formations which are more combined. In the mentioned work, shows, based on the differential equations for the second derivative $$\frac{\partial^{2}x^{(\alpha)}}{\partial\bar{x}^{(\lambda)}\partial\bar{x}^{(\mu)}}$$, that from a covariant system $$A_{\alpha_{1}\alpha_{2}\dots\alpha_{p}}$$ of $$p$$-the order, a system of $$p+1$$-th order emerges as follows:

$A_{\alpha_{1}\alpha_{2}\dots\alpha_{p}/\alpha_{p+1}}=\frac{\partial}{\partial x^{\left(\alpha_{p+1}\right)}}A_{\alpha_{1}\alpha_{2}\dots\alpha_{p}}-\sum_{h=1}^{p}\sum_{\beta=1}^{n}\left\{ \begin{matrix}\alpha_{p+1}\alpha_{h}\\ \beta \end{matrix}\right\} A_{\alpha_{1}\dots\alpha_{h-1}\beta\alpha_{h+1}\dots\alpha_{p}},$|undefined

where the symbols of second order with triple-indices arise, which are defined as follows:

$\left\{ \begin{matrix}\alpha\beta\\ \gamma \end{matrix}\right\} =\sum_{\delta=1}^{n}c^{(\gamma\delta)}\begin{bmatrix}\alpha\beta\\ \delta \end{bmatrix}=\sum_{\delta=1}^{n}c^{(\gamma\delta)}\cdot\frac{1}{2}\left(\frac{\partial c_{\alpha\delta}}{\partial x^{(\beta)}}+\frac{\partial c_{\beta\delta}}{\partial x^{(\alpha)}}-\frac{\partial c_{\alpha\beta}}{\partial x^{(\delta)}}\right).$|undefined

and denote this as the covariant differential quotient of $$A_{\alpha_{1}\alpha_{2}\dots\alpha_{p}}$$ with respect to $$x^{\left(\alpha_{p+1}\right)}$$. The prime separates the indices added by differentiation from the others. For the contravariant differential quotient it is given:

$A^{\left(\alpha_{1}\alpha_{2}\dots\alpha_{p}/\alpha_{p+1}\right)}=\sum_{\beta=1}^{n}c^{\left(\alpha_{p+1}\beta\right)}\left[\frac{\partial}{\partial x^{(\beta)}}A^{\left(\alpha_{1}\alpha_{2}\dots\alpha_{p}\right)}+\sum_{h=1}^{p}\sum_{\gamma=1}^{n}\left\{ \begin{matrix}\beta\gamma\\ \alpha_{h} \end{matrix}\right\} A^{\left(\alpha_{1}\alpha_{2}\dots\alpha_{h-1}\gamma\alpha_{h+1}\dots\alpha_{p}\right)}\right].$|undefined

Then we have, as it can be easily shown:

$\begin{aligned}\mathrm{A}_{1234} & =A_{123/4}-A_{124/3}+A_{134/2}-A_{234/1}\\ & =\sqrt{c}\mathrm{Div}\left(B^{(1)}B^{(2)}B^{(3)}B^{(4)}\right)=\sqrt{c}\underset{\alpha,\beta}{\sum}c_{\alpha\beta}B^{(\alpha/\beta)}=\underset{\alpha}{\sum}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}B^{(\alpha)}\right),\\ & =\mathrm{div}\left(B_{1}B_{2}B_{3}B_{4}\right)=\sqrt{c}\underset{\alpha,\beta}{\sum}c^{(\alpha\beta)}B_{\alpha/\beta}=\underset{\alpha}{\sum}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\underset{\beta}{\sum}c^{(\alpha\beta)}B_{\beta}\right) \end{aligned}$|undefined

with the connection

$-\frac{1}{\sqrt{c}}A_{234}=B^{(1)}=\underset{\beta}{\sum}c^{(1\beta)}B_{\beta}$,|undefined

$\frac{1}{\sqrt{c}}A_{134}=B^{(2)}=\underset{\beta}{\sum}c^{(2\beta)}B_{\beta}$ etc.|undefined

$\begin{aligned}\mathrm{A}_{234} & =A_{23/4}-A_{24/3}+A_{34/2}\\ & =\sqrt{c}\mathfrak{Div}^{(1)}\left(B^{(12)}B^{(13)}B^{(14)}B^{(23)}B^{(24)}B^{(34)}\right)=\sqrt{c}\underset{\alpha,\beta}{\sum}c_{\alpha\beta}B^{(1\alpha/\beta)}=\underset{\alpha}{\sum}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}B^{(1\alpha)}\right),\\ & =\sqrt{c}\mathfrak{Div}^{(1)}\left(B_{12}B_{13}B_{14}B_{23}B_{24}B_{34}\right)=\sqrt{c}\underset{\alpha,\beta,\gamma}{\sum}c^{(1\alpha)}c^{(\beta\gamma)}B_{\alpha\beta/\gamma}=\underset{\alpha}{\sum}\frac{\partial}{\partial x^{(\alpha)}}\left(\sqrt{c}\underset{\beta,\gamma}{\sum}c^{(1\beta)}B_{\beta\gamma}\right) \end{aligned}$|undefined

with the connection

$\frac{1}{\sqrt{c}}A_{12}=B^{(34)}=\underset{\alpha,\beta}{\sum}c^{(3\alpha)}c^{(4\beta)}B_{\alpha\beta}$ etc.,|undefined

where we could write, following the things stated above, also $$A_{\alpha\beta}^{\ast}$$ instead of $$B_{\alpha\beta}$$.

1685 Thus

$\frac{1}{\sqrt{c}}\mathrm{A}_{234}=\mathfrak{Div}^{(1)}\left(A_{\alpha\beta}^{\ast}\right)=\underset{\alpha}{\sum}c^{(1\alpha)}\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\alpha\beta/\gamma}^{\ast}$ etc.|undefined

$\frac{1}{\sqrt{c}}\mathrm{A}_{234}^{\ast}=\mathfrak{Div}^{(1)}\left(A_{\alpha\beta}\right)=\underset{\alpha}{\sum}c^{(1\alpha)}\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\alpha\beta/\gamma}$ |undefined

For $$\mathfrak{Div}\left(A_{\alpha\beta}^{\ast}\right)\equiv0$$ we have, as already mentioned above,

$A_{\alpha\beta}=\mathrm{A}_{\alpha\beta}=A_{\alpha/\beta}-A_{\beta/\alpha}$

and

$\begin{aligned}\mathfrak{Div}\left(A_{\alpha\beta}\right) & =\underset{\alpha}{\sum}c^{(1\alpha)}\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\alpha/\beta\gamma}-\underset{\alpha}{\sum}c^{(1\alpha)}\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\beta/\alpha\gamma}=\\ & =\underset{\alpha}{\sum}c^{(1\alpha)}\left[\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\alpha/\beta\gamma}-\frac{\partial}{\partial x^{(\alpha)}}\left(\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\beta/\gamma}\right)\right]. \end{aligned}$|undefined

since in Euclidean space (vanishing of the Riemann symbols) the permutation of the differentiation order $$A_{\beta/\alpha\gamma}=A_{\beta/\gamma\alpha}$$ is allowed, and $$\sum c^{(1\alpha)}\sum c^{(\beta\gamma)}A_{\beta/\gamma\alpha}$$ represents the contravariant differential quotient of $$\underset{\beta,\gamma}{\sum}c^{(\beta\gamma)}A_{\beta/\gamma}$$ with respect to $$x^{(1)}$$.