This formulation would require \[\eta^{\mu \nu}=\frac{1}{1+i^2+j^2+k^2} \begin{bmatrix} 1 & i & j & k \\ i & -1 & -k & j\\ j & k & -1 & -i\\ k & -j & i & -1 \end{bmatrix}\]
This would require perhaps a normalisation of \(\frac{1}{\sqrt{2}}\) on each matrix.
For the potential 4 vector \[A^{\mu}=(\varphi/c,A_x,A_y,A_z)\]
Then \[A_{\mu}=\eta_{\mu \nu}A^{\nu}\]
Which gives \[A_\mu = \begin{bmatrix} \frac{\varphi}{c} +A_xi+A_yj+A_zk \\ \frac{\varphi i}{c} -A_x +A_yk -A_zj \\ \frac{\varphi j}{c} -A_xk -A_y +A_zi \\ \frac{\varphi k}{c} +A_xj +A_yi-A_z \end{bmatrix} \\ = \begin{bmatrix} \frac{\varphi}{c} \\ -A_x \\ -A_y \\ -A_z \end{bmatrix} + \begin{bmatrix} A_x \\ \frac{\varphi}{c} \\ A_z \\ -A_y \end{bmatrix} i + \begin{bmatrix} A_y \\ -A_z \\ \frac{\varphi}{c} \\ A_x \end{bmatrix} j + \begin{bmatrix} A_z \\ A_y \\ -A_x \\ \frac{\varphi}{c} \end{bmatrix} k\]
There are now hypercomplex four potentials from the real potential such that \(A_\mu \to A_\mu + B_\mu i + \Gamma_\mu j + \Delta_\mu k\). Now, as usually defined we can create the tensor \[H_{\mu \nu} = (dA)_{\mu \nu}= \partial_\mu A_\nu - \partial_\nu A_\mu\]
But with the hypercomplex components it is clear that \(H_{\mu \nu} = F_{\mu \nu} +I_{\mu \nu}i + J_{\mu \nu}j + K_{\mu \nu}k\) where \[F_{\mu \nu}=Re[H_{\mu \nu}]=\partial_\mu A_\nu - \partial_\nu A_\mu \\ I_{\mu \nu}=Im_i[H_{\mu \nu}]=\partial_\mu B_\nu - \partial_\nu B_\mu \\ J_{\mu \nu}=Im_j[H_{\mu \nu}]=\partial_\mu \Gamma_\nu - \partial_\nu \Gamma_\mu \\ K_{\mu \nu}=Im_k[H_{\mu \nu}]=\partial_\mu \Delta_\nu - \partial_\nu \Delta_\mu\]
This means that the real component \(F_{\mu \nu}\) is equal to th regular EM tensor \[F_{\mu \nu} = \frac{1}{c} \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -cB_z & cB_y \\ -E_y & cB_z & 0 & -cB_x \\ -Ez & -cB_y & cB_x & 0 \end{bmatrix}\]
However, \[F^{\alpha \beta}=\eta^{\alpha \gamma}\eta^{\beta \delta}F_{\gamma \delta}\]
Through explicit calculation this results in \[Re[F^{\mu \nu}]=\frac{1}{2c} \begin{bmatrix} 0 & cB_x & cB_y & cB_z \\ -cB_x & 0 & E_z & -E_y \\ -cB_y & -E_z & 0 & E_x \\ -cB_z & E_y & -E_x & 0 \end{bmatrix} \\ =\frac{1}{2}G_{\gamma \delta} = \frac{1}{4} \varepsilon_{\alpha \beta \gamma \delta}F^{\alpha \beta}\]