Levi-Civita Alternating Symbol

Hello purakanui,

the Levi-Civita-Symbol is generally a tensor:

$\epsilon_{ijk} : \mathcal{V} \times \cdots \times \mathcal{V} \rightarrow \mathbb{K}$

mapping the n-fold cartesian product of a vector space to its underlying field (usually the real or complex numbers).

$\epsilon_{ijk} = \left\{ \begin{matrix} 1 & , \quad \text{for even permutations} \\ -1& , \quad \text{for odd permutations}\\ 0&, \quad \text{else} \end{matrix}<br />$

However if you represented it as matrix, it'd have to have only 2 indices. Of course one can define the Levi-Civita-Symbol with 2 indices:

$\epsilon_{ij} := 1 \ \text{for ij=12,} -1 \ \text{for ij=21,} \ 0$ else.

Then a matrix representing this would look like this: $E = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

However, I have to warn you: This E is actually NOT really a matrix.
At least it doesn't behave like a matrix under the usual multiplication of a matrix with vectors.

So better leave the tensor notation of the Levi-Civita-symbol as it is instead of writing it as an array.

If you want to express the Levi-Civita-symbol in another basis, then you have to apply basis transformations on it,

$\epsilon_{ijk} = a^p_i b^q_j c^r_k \epsilon_{pqr}$ , using the summation convention, where

$a^p_i, \ b^q_j, \ c^r_k$ are now really matrices,

e.g. $a^p_i = \begin{pmatrix} a^1_1 & \cdots & a^3_1 \\ \vdots & \ddots & \vdots \\ a^3_1 & \cdots & a^3_3 \end{pmatrix}$

But use the summation convention to calculate the values!

I have to remark that it might well be that the Levi-Civita-Symbol is defined as it is, in every basis, but I don't know.
Here is how the Levi-Civita-Symbol with 3 indices may be represented:

Visualization of the Levi-Civita symbol as a 3×... - Visual Hints - Quora