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The Kronecker's Delta is defined as

$$\delta_{ij}= \begin{cases} 1 & i=j \\ 0 & i \ne j \end{cases}$$

Also, the Permuation Symbol known as Levi Cevita's Symbol is introduced as

$$\varepsilon_{ijk}= \begin{cases} 1 & \text{$ijk$ is an even permutation of $123$} \\ -1 & \text{$ijk$ is an odd permutation of $123$} \\ 0 & \text{$ijk$ has two same indices} \end{cases}$$

where $i$, $j$, and $k$ are natural numbers $1,2,3$. These symbols are widely used in vector and tensor analysis and in differential geometry. There is a relation between them as the following theorem states.

Theorem. The following relation holds between the Kronecker's Delta and permutation symbol $$\varepsilon_{ijk}\varepsilon_{pqr}= \begin{vmatrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \\ \end {vmatrix}$$ where $|\cdot|$ denotes the determinant.

I am looking for different proofs of this theorem. Also, I don't want to prove it by just investigating that the equality holds for different choices of the indices one by one!

I don't have any idea to take a first step. Any hint or help is appreciated. :)

Hosein Rahnama
  • 15,554

1 Answers1

6

For each $i \in \left\{1,2,3\right\}$, let $e_i$ be the column vector in $\mathbb{Z}^3$ whose $i$-th entry is $1$ and whose other two entries are $0$. (In other words, $e_i$ is the $i$-th column of the $3\times 3$ identity matrix.)

For any $i, j, k \in \left\{1,2,3\right\}$, let $K_{ijk}$ be the $3\times 3$-matrix whose columns are $e_i, e_j, e_k$ (in this order).

Observe the following:

  1. For any $i, j, k \in \left\{1,2,3\right\}$, we have $\det K_{ijk} = \varepsilon_{ijk}$.

  2. For any $u, v \in \left\{1,2,3\right\}$, we have $\delta_{uv} = e_u^T e_v$. Thus, for any $i, j, k, p, q, r \in \left\{1,2,3\right\}$, we have

$$ \left(\begin{matrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \\ \end{matrix}\right) = K_{ijk}^T K_{pqr} ,$$ so that $$\det\left(\begin{matrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \\ \end{matrix}\right) = \det\left(K_{ijk}^T K_{pqr}\right) = \det K_{ijk} \det K_{pqr} = \varepsilon_{ijk} \varepsilon_{pqr} $$

darij grinberg
  • 19,173