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!

  • 1
    $\begingroup$ $\varepsilon_{\alpha\beta\gamma\delta}\varepsilon^{\alpha\beta\gamma\delta}=(-1)^s n!$ where $s$ is the number of negative eigenvalues in the metric. i.e. $s=0$ in Euclidean space, $s=1$ in Minkowski space. $\endgroup$
    – Aiden
    Commented Aug 17, 2023 at 2:29

1 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.

  • $\begingroup$ 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 $\eta_{\mu\nu}$? $\endgroup$ Commented Aug 17, 2023 at 8:19
  • $\begingroup$ @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 $-1$. $\endgroup$
    – akhmeteli
    Commented Aug 17, 2023 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.