A question on spin algebra

Question

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.

This post imported from StackExchange Physics at 2014-06-29 09:33 (UCT), posted by SE-user user45765

Comments

What's wrong with using that the determinant of lorentz trafos is 1?

This post imported from StackExchange Physics at 2014-06-29 09:33 (UCT), posted by SE-user ACuriousMind

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I wonder whether the similar argument of spin tensor product can be employed here which I have no clue how to implement it at this time. As I have said, I just want the proof look a bit more fancier than using plain definition of the O(1,3). It looks like 0$\in 1\otimes 1\otimes 1\otimes 1$. But I do not have the justification why.

This post imported from StackExchange Physics at 2014-06-29 09:33 (UCT), posted by SE-user user45765

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I doubt it is possible to prove the Lorentz invariance of $\epsilon_{\alpha\beta\mu\nu}$ without using the determinant of the Lorentz transformation.

This post imported from StackExchange Physics at 2014-06-29 09:33 (UCT), posted by SE-user Jonas

Answer 1

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.