# Levi-Civita tensor

The Kronecker delta can be represented by a two dimensional matrix:

\begin{gather} \delta_{ij}=\mathbb{I}= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}. \end{gather}

Similarly, can the Levi-Civita tensor \begin{gather} \epsilon_{ijk} \end{gather} be represented by a three dimensional matrix, and if so, what does the matrix look like?

• $\uparrow$ Yes. – AccidentalFourierTransform Apr 30 '18 at 14:39
• Wow! Short comment for a short question! Make a sandwich pile of three antisymmetric matrices. Three different antisymmetric matrices : one of bread, one of cheese, and one of peanut butter! ;-) – Cham Apr 30 '18 at 14:44
• I'm reasonably certain that the Wikipedia entry in the LC tensor had exactly what you're looking for... – Kyle Kanos Apr 30 '18 at 15:58

It might be a little messy, but imagine the layers as having the $i$ index, the rows having the $j$ index, and the columns having the $k$ index.
Chiefly, the Levi-Civita Tensor gives $1$ for cyclic permutations and $-1$ for anti-cyclic permutations. In the image above, the front most layer (blue) has index $i=1$. Hence, the indices in the layer is ${1,j,k}$.
For the cyclic combination ${1,2,3}$, that is, at layer $1$, row $2$, column $3$, the value is $1$. Similarly, for the anti-cyclic combination ${1,3,2}$, the value is $-1$ appears at layer $1$, row $3$, column $2$. There are repeats indices in other rows and columns combinations in this layer, and hence the other values are $0$ in this layer.
Similar arguments can be done for the middle layer (red), i.e. for indices ${2,j,k}$ and for the back most layer (green), i.e. for indices ${3,j,k}$, which is left as an exercise for the reader.