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

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!

• $\varepsilon_{\alpha\beta\gamma\delta}\varepsilon^{\alpha\beta\gamma\delta}=(-1)^s n!$ where $s$ is the number of negative eigenvalues in the metric. i.e. $s=0$ in Euclidean space, $s=1$ in Minkowski space. Aug 17 at 2:29

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.
• But how to contract $\varepsilon_{i_1 i_2 i_3 i_4}$ with $\varepsilon^{i_1 i_2 i_3 i_4}$? Does this has something to do with $\eta_{\mu\nu}$? Aug 17 at 8:19
• @Firestar-Reimu : So we always have $\varepsilon_{{i_1}{i_2}{i_3}{i_4}}=-\varepsilon^{{i_1}{i_2}{i_3}{i_4}}$ due to the properties of $\eta_{\mu\nu}$, as explained in my answer for nonvanishing $\varepsilon_{{i_1}{i_2}{i_3}{i_4}}$, and for vanishing components both parts of the above equality are zero. Thus, the tensor contraction in your question is equal to the same contraction in the Euclidian space times $-1$. Aug 17 at 12:39