I am trying to lorentz-transform the dual electromagnetic tensor $G^{\mu \nu}:= \frac{1}{2} \epsilon ^{\mu \nu \alpha \beta} F_{\alpha \beta}$ and also show (perhaps by using that last result) that $G^{\mu \nu}F_{\mu \nu} = - \frac{4}{c} \vec{E}\vec{B}$ is really (or not really) invariant under Lorentz transformation.
So far i have looked at it in different ways:
1. Idea: Not contracting $\epsilon$ \begin{equation*} {G^{\mu \nu }}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau G^{\sigma \tau} = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \frac{1}{2} \epsilon ^{\sigma \tau \alpha \beta} F_{\alpha \beta} \end{equation*} this seems plausible but what about the levi civita pseudotensor? Shouldn't I Lorentz-transform it fully? As in
2. Idea: Contracting every tensor within \begin{equation*} {G^{\mu \nu}}' = (\frac{1}{2} \epsilon ^{\mu \nu \alpha \beta})' (F_{\alpha \beta})' =\Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) \end{equation*} I apologize for the many indices. But is this the general way to go about the lorentz transformation? It seems to take me to the same last result as in 1.idea if i do the following: \begin{equation*} {G^{\mu \nu}}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \delta ^\xi _\rho \delta ^\omega _\gamma \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} F_{\xi \omega} = \\ \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} F_{\rho \gamma} = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau G^{\sigma \tau} \end{equation*} here i used the identity $\Lambda ^\alpha _\beta \Lambda ^\beta _\gamma = \delta ^\alpha _\gamma$ (is this true for all lorentz transformations?)
So i get the two ways and it seems they're equal, but what if I see it this way:
3. Idea: seeing the Lorentz-transformation of levi-civita as Determinant \begin{equation*} {G^{\mu \nu}}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) =\frac{1}{2} det(\Lambda) \epsilon ^{\mu \nu \alpha \beta} \Lambda _\alpha ^\xi \Lambda _\beta ^\omega F_{\xi \omega} = \\ \pm \frac{1}{2}\epsilon ^{\mu \nu \alpha \beta} \Lambda _\alpha ^\xi \Lambda _\beta ^\omega F_{\xi \omega} = \pm \frac{1}{2}\epsilon ^{\mu \nu \alpha \beta} (F_{\alpha \beta})' \end{equation*} here i used $Det(\Lambda) = \pm 1$ so what is going on here? Where did i do a mistake? Does this have anything to do with non-symmetrical lorentz transformations, as in rotations? If i now solve for the lorentz transformation of $G^{\mu \nu} F_{\mu \nu}$ i get in one instance that they are invariant, on the other instance, they are not; the $\pm$ that comes from the determinant contradicts the equality.
I am inclined to my 2. Idea as being the way to go about it. It seems that one chooses often out of the determinant the +sign, but that is only for proper lorentz transformations, I would really appreciate learning if $G^{\mu \nu} F_{\mu \nu}$ is invariant under Lorentz transformations in general.