Contraction of Levi-Civita tensor in Minkowski space-time (or any space-time)

Question

I know that $\varepsilon_{i_1 i_2\cdots i_n}\varepsilon^{i_1 i_2\cdots i_n} = n!$ in Euclid space.

But in Minkowski space-time, the metric tensor is: $$ \eta_{\mu\nu} = \mathrm{diag}\{-1, 1, 1, 1\} $$ Levi-Civita tensor will be an 0-4 tensor $\varepsilon_{\alpha\beta\gamma\delta}$

How to calculate $\varepsilon_{\alpha\beta\gamma\delta}\varepsilon^{\alpha\beta\gamma\delta}$ in Minkowski space-time

or broadly when there is a certain metric $g_{\mu\nu}$ of any space-time, thanks!

Answer 1

When you raise four indices of a component $\varepsilon_{i_{1} i_{2} i_{3} i_{4}}$ for 4-dimensional space, then, if the component does not vanish, all indices $i_{1},i_{2},i_{3},i_{4}$ are different, so, with Minkowski metric, raising one of them gives factor $-1$, raising each of the other indices gives factors $1$, so $\varepsilon_{i_{1} i_{2} i_{3} i_{4}}=-\varepsilon^{i_{1} i_{2} i_{3} i_{4}}$, so we get a factor $-1$ compared to what we get in Euclidian 4D space.