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

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?

• It's built-in because when you permute the rows $\mu_i$ you get the additional minus signs tracked by 2.66. Aug 22, 2021 at 18:21
• 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. Aug 23, 2021 at 0:42
• @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? Aug 23, 2021 at 7:55

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
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.