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
    Aug 17 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 $\eta_{\mu\nu}$? $\endgroup$ Aug 17 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
    Aug 17 at 12:39

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