Transformation law for the Levi-Civita symbol under a change of basis

Question

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?

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.