<s>[INST] Covariant Maxwell equations invariant under parity transformation I tried to proof that the Maxwell equations are invariant under parity transformations. Therefore I used the covariant formulation of the Maxwell equations 
\begin{align}
\partial_{\nu}F^{\nu\mu} &= \frac{4\pi}{c}j^{\mu}\\
\partial_{\nu}\tilde{F}^{\nu\mu} &= 0
\end{align}
and the parity transformation given by 
\begin{align}
P = \begin{pmatrix} 1 & 0 & 0 & 0  \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
\end{align}
Regarding only the first equation $\partial_{\nu}F^{\nu\mu} = \frac{4\pi}{c}j^{\mu}$ we have
\begin{align}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ \vec{\nabla} \end{pmatrix} = \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ -\vec{\nabla} \end{pmatrix} 
\end{align}
as well as 
\begin{align}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} c\rho \\ \vec{j} \end{pmatrix} = \begin{pmatrix} c\rho \\ -\vec{j} \end{pmatrix} 
\end{align}
and 
\begin{align}
P \cdot F^{\nu\mu} &= \begin{pmatrix} 1 & 0 & 0 & 0  \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ E^1 & 0 & -B^3 & B^2 \\ E^2 & B^3 & 0 & -B^1 \\ E^3 & -B^2 & B^1 & 0  \end{pmatrix} = \begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ -E^1 & 0 & B^3 & -B^2 \\ -E^2 & -B^3 & 0 & B^1 \\ -E^3 & B^2 & -B^1 & 0  \end{pmatrix}
\end{align}
Based on these calculations, is there a way to see that Maxwell equations are invariant under parity transformations and if so how do I see it? 
 [/INST] As current is a vector, it is not invariant under parity. Therefore neither is Ampère's law. 
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