I'm trying to prove that the Levi-Civita symbol $\epsilon_{i_1 ... i_n}$ is a tensor density of weight $w=-1$. For this purpose, it has to be shown that the transformation law for the components of this pseudo-tensor under a change of basis, $\hat \epsilon_{j_1 ... j_n}$, is given by

$$\hat \epsilon_{j_1 ... j_n} =\big(\det(C)\big)^{-1} \epsilon_{i_1 ... i_n} C^{i_1}_{j_1}...C^{i_n}_{j_n} \tag{1}$$

With $C=(C^a_b)_{n \times n}$ being the change-of-basis matrix. I have seen in this related post that this is done using the expression of the determinant of a matrix through the Levi-Civita symbol:

$$ \tilde{\epsilon}_{\mu_1' \mu_2' \dots \mu_n'} |M| = \tilde{\epsilon}_{\mu_1 \mu_2 \dots \mu_n} M^{\mu_1}{}_{\mu_1'}M^{\mu_2}{}_{\mu_2'} \dots M^{\mu_n}{}_{\mu_n'} \tag{2.66}$$

However, I have not clear what is the relation of this expression and the expression of the determinant,

$$|M|=\epsilon_{i_1 ... i_n}M^{i_1}_{1}...M^{i_n}_{n} \tag{2}$$

How could equation (2) be modified in order to get (1) and complete the demonstration?

  • $\begingroup$ It's built-in because when you permute the rows $\mu_i$ you get the additional minus signs tracked by 2.66. $\endgroup$
    – mike stone
    Commented Aug 22, 2021 at 18:21
  • 1
    $\begingroup$ The key part is removing the explicit indices in Eq. (2) and you will get $\left(\det{M}\right)\epsilon_{i_1 \ldots i_n} = \epsilon_{j_1 \ldots j_n} M^{j_i}_{i_1} \ldots M^{j_n}_{i_n}$. This is because the explicit indices are themselves proportional to the Levi-Civita symbol. Try switching two of them and you will see why. $\endgroup$ Commented Aug 23, 2021 at 0:42
  • $\begingroup$ @Vincent Thacker I'm sure it's something simpler than I think, but I still don't see it clearly... Could you please explain this reasoning in a little more detail? $\endgroup$
    – Invenietis
    Commented Aug 23, 2021 at 7:55

1 Answer 1


It is necessary to eliminate the explicit indices $1\ldots n$ in Eq. (2). To do that, we first replace them with regular indices $i_1 \ldots i_n$ and observe the behavior of the expression when we try different values for $i_1 \ldots i_n$. Then a suitable general expression for arbitrary indices $i_1 \ldots i_n$ can be determined.

So substituting arbitrary test values for $i_1 \ldots i_n$, we find that

  1. The expression vanishes when two indices are equal (because it equals its own negative).
  2. The expression changes sign when two indices are switched.

The above deductions completely characterize the behavior of $i_1 \ldots i_n$. They tell us that the expression must be proportional to the Levi-Civita symbol itself. With the explicit indices eliminated, it is a straightforward task to arrive at the desired formula.


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