Anadijiban Das in his book "Tensors. The Mathematics of Relativity Theroy and Continuum Mechanics", p.39, describes the transformation properties of the Levi-Civita symbol in n dimensions. I have difficulties with a proof, left as an exercise, necessary in the course of this argument. Here is the problem:

We define the Levi-Civita symbol in $ \mathbb{R}^n $ with the help of the generalized Kronecker delta

$ \epsilon_{b_1 \dots b_n} := \delta^{1 \dots n}_{b_1 \dots b_n} \quad $ and $ \quad \epsilon^{a_1 \dots a_n} := \delta_{1 \dots n}^{a_1 \dots a_n}. \quad $ (1)

In addition, we have a number of identities available, both for the Kronecker delta and the Levi-Civita symbol:

$ \det [\lambda^a_b] = \frac{1}{n!} \delta^{a_1 \dots a_n}_{b_1 \dots b_n} \lambda^{b_1}_{a_1} \dots \lambda^{b_n}_{a_n}, \quad $ (2)

$ \epsilon_{a_1 \dots a_n} \epsilon^{a_1 \dots a_n} = n!, \qquad \qquad \qquad $ (3)

$ \epsilon^{a_1 \dots a_n} \epsilon_{b_1 \dots b_n} = \delta^{a_1 \dots a_n}_{b_1 \dots b_n}. \qquad \qquad $ (4)

What we need and what we want to prove is

$ \det [\lambda^c_d] \epsilon_{a_1 \dots a_n} = \lambda^{b_1}_{a_1} \dots \lambda^{b_n}_{a_n} \epsilon_{b_1 \dots b_n}. \quad $ (5)

Then, since the Levi-Civita symbol is supposed to be invariant under any linear transformation, we have

$ \bar{\epsilon}_{a_1 \dots a_n} = \epsilon_{a_1 \dots a_n}, $

and therefore with (5) the transformation rule

$ \bar{\epsilon}_{a_1 \dots a_n} = \frac{1}{\det [\lambda^c_d]} \lambda^{b_1}_{a_1} \dots \lambda^{b_n}_{a_n} \epsilon_{b_1 \dots b_n}. \quad $ (6)

Note that in (2) and (5), $ [\lambda] $ is an arbitrary $ n \times n $ invertible matrix, whereas in (6) it is the specific transformation matrix given by the change of basis

$ \bar{e}_i = \lambda_i^j e_j $

with $ E=\{ e_1,\dots,e_n \}$ and $ \bar{E}=\{ \bar{e}_1,\dots,\bar{e}_n \}$ being two bases of $ \mathbb{R}^n $.

Since in (6) the determinant appears with exponent -1, we find the Levi-Civita symbol to be a covariant (authentic, not pseudo) tensor density of weight -1.

Now, in order to prove (5), with (2) and (4) we try

$ \det [\lambda^c_d] = \frac{1}{n!} \delta^{a_1 \dots a_n}_{b_1 \dots b_n} \lambda^{b_1}_{a_1} \dots \lambda^{b_n}_{a_n} = \frac{1}{n!} \epsilon^{a_1 \dots a_n} \epsilon_{b_1 \dots b_n} \lambda^{b_1}_{a_1} \dots \lambda^{b_n}_{a_n} $.

Seemingly we are close to the solution, by multiplying both sides with $ \epsilon_{a_1 \dots a_n} $ and then using (3) so simplify the right side. But I'm afraid, this is not possible due to Einstein's summation convention. So this seems to be a dead end. Or isn't it? I can't exclude the possibility that others than the above mentioned identities or arguments are necessary.

Can anyone help me with showing (5)?

By the way, you might also want to refer to question Proof the Levi-Civita symbol is a tensor.


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