So we came across some new but familiar looking concepts. However in 4 dimensions there should be a quaternary, ternary, binary, unary and 0-ry, object/transform. The ternary will be \[\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ t_1 & x_1 & y_1 & z_1 \\ t_2 & x_2 & y_2 & z_2 \\ t_3 & x_3 & y_3 & z_3 \\ \end{vmatrix} = \textbf{i}\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \\ \end{vmatrix} - \textbf{j}\begin{vmatrix} t_1 & y_1 & z_1 \\ t_2 & y_2 & z_2 \\ t_3 & y_3 & z_3 \\ \end{vmatrix} + \textbf{k}\begin{vmatrix} t_1 & x_1 & z_1 \\ t_2 & x_2 & z_2 \\ t_3 & x_3 & z_3 \\ \end{vmatrix} - \textbf{l}\begin{vmatrix} t_1 & x_1 & y_1 \\ t_2 & x_2 & y_2 \\ t_3 & x_3 & y_3 \\ \end{vmatrix}\]
Therefore for a set of 3-vectors \[_tr_1=x_1\textbf{j}+y_1\textbf{k}+z_1\textbf{l} \\ _xr_1=t_1\textbf{i}+y_1\textbf{k}+z_1\textbf{l} \\ _yr_1=t_1\textbf{i}+x_1\textbf{j}+z_1\textbf{l} \\ _zr_1=t_1\textbf{i}+x_1\textbf{j}+y_1\textbf{k} \\\]
we have
\[T(\textbf{r}_1,\textbf{r}_2,\textbf{r}_3)=\\ \textbf{i}({_t\textbf{r}_1}\cdot({_t\textbf{r}_2} \times {_t\textbf{r}_3}))- \textbf{j}({_x\textbf{r}_1}\cdot({_x\textbf{r}_2} \times {_x\textbf{r}_3}))\\+ \textbf{k}({_y\textbf{r}_1}\cdot({_y\textbf{r}_2} \times {_y\textbf{r}_3}))- \textbf{l}({_z\textbf{r}_1}\cdot({_z\textbf{r}_2} \times {_z\textbf{r}_3}))\]
Which results in a four-vector of the volumes made by various axes with some sign.
The Quaternary will be \[\begin{vmatrix} t_1 & x_1 & y_1 & z_1 \\ t_2 & x_2 & y_2 & z_2 \\ t_3 & x_3 & y_3 & z_3 \\ t_4 & x_4 & y_4 & z_4 \end{vmatrix} =t_1V^{234}_{xyz}-x_1V^{234}_{tyz}+y_1V^{234}_{txz}-z_1V^{234}_{txy}\]
Which can obviously take on other forms by going along the other rows to obtain the triple products according to \(134\),\(124\) or \(123\) sets of vectors, with appropriate signs.
For the binary transform we have \[\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ t_1 & x_1 & y_1 & z_1 \\ t_2 & x_2 & y_2 & z_2 \end{vmatrix} =\textbf{i}({_t\textbf{r}_1} \times {_t\textbf{r}_2} ) -\textbf{j}({_x\textbf{r}_1} \times {_x\textbf{r}_2} ) +\textbf{k}({_y\textbf{r}_1} \times {_y\textbf{r}_2} ) -\textbf{l}({_z\textbf{r}_1} \times {_z\textbf{r}_2} )\]
Therefore making a 4x4 skew symmetric matrix \[\textbf{r}_1 \otimes \textbf{r}_2 = \begin{bmatrix} 0 & y_1z_2-z_1y_2 & z_1x_2-x_1z_2 & x_1y_2-y_1x_2 \\ z_1y_2-y_1z_2 & 0 & t_1z_2-z_1t_2 & y_1t_2-t_1y_2 \\ x_1z_2-z_1x_2 & z_1t_2-t_1z_2 & 0 & t_1x_2-x_1t_2 \\ y_1x_2-x_1y_2 & t_1y_2-y_1t_2 & x_1t_2-t_1x_2 & 0 \end{bmatrix}\]
If one then defines \[y_1z_2-z_1y_2=E_x/c \\ z_1x_2-x_1z_2=E_y/c \\ x_1y_2-y_1x_2=E_z/c \\ \\ t_1x_2-x_1t_2=B_x \\ t_1y_2-y_1t_2=B_y \\ t_1z_2-z_1t_2=B_z \\\]
This matrix will become the covarient form of the electromagnetic tensor \(F_{\mu\nu}\). However this is 6 equations for 8 unknowns, leaving two free parameters that could take any value in the input four-vectors \(\textbf{r}_1,\textbf{r}_2\).
But this shows that \[\textbf{E}={_t\textbf{r}_1} \times {_t\textbf{r}_2} \\ \textbf{B}=\bigg( \begin{vmatrix} t_1 & x_1 \\ t_2 & x_2 \end{vmatrix} , \begin{vmatrix} t_1 & y_1 \\ t_2 & y_2 \end{vmatrix} , \begin{vmatrix} t_1 & z_1 \\ t_2 & z_2 \end{vmatrix} \bigg)\]
That is the electric field corresponds to the cross product of the spatial comonents of the four vectors, and that the magnetic field is the determinants of spatial components with time. The elements of \(B\) are signed areas of the parallellograms made in each basis direction \(x,y,z\).
\[\nabla\cdot\textbf{B}= \frac{\partial}{\partial x}\begin{vmatrix} t_1 & x_1 \\ t_2 & x_2 \end{vmatrix}+ \frac{\partial}{\partial y}\begin{vmatrix} t_1 & y_1 \\ t_2 & y_2 \end{vmatrix}+ \frac{\partial}{\partial z}\begin{vmatrix} t_1 & z_1 \\ t_2 & z_2 \end{vmatrix}\]
... For the unary operation we take a four-vector and make it into a cubic array \[\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ \textbf{i} & \textbf{j} & \textbf{k} & \textbf{l} \\ t_1 & x_1 & y_1 & z_1 \end{vmatrix}\]