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In scattering theory, one can form a lorentz invariant quantity by $\epsilon_{\mu 1 2\nu}P^{\mu}_{1}P^{\nu}_{2}$ which is really $1\otimes 1$ 's spin 0 state. Is there such a kind of argument to show $\epsilon_{\alpha\beta\mu\nu}$ is also invariant under lorentz transformation without showing the determinant of lorentz transformation being 1. I just want a fancier proof of this.

The invariance of the $\epsilon$ tensor under Lorentz transformations is equivalent to the determinant of the Lorentz transformation being equal to 1, they are the same information. So it is not possible to separate the two statements in any meaningful way, they are essentially the same statement. So no.

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