Is their an elegant way to prove the product of two Levi Cevita tensors is equivalent to a determinant of a matrix of Kronecker deltas? I know that the anti-symmetry and cyclic nature should be easily proved via determinant row-interchange laws, but is the proof inelegant when it comes to showing it zeroes out when two indices are repeated?

The property in mind is: $$\epsilon_{ijk}\epsilon_{lmn}=\det\left| \begin{array}{cccc} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \end{array} \right| $$

  • $\begingroup$ Does antisymmetry not prove that it zeroes out when two indices are repeated? $\endgroup$ – user149874 Sep 4 '14 at 9:04
  • $\begingroup$ I didn't mean to say antisymmetry, I only meant the sign flipping. $\endgroup$ – user82004 Sep 4 '14 at 17:28

The determinante of $\left(A_{ij}\right)$ is defined by Leibniz as $$\det \left(A_{ij}\right) := \sum_{\sigma}\text{sign}(\sigma)\Pi_i A_{i,\sigma(i)} $$ With permutations $\sigma$ and their sign $\text{sign}(\sigma)$ ($=+1$ if Permutation is an even of $1,2,3,4,...$, $-1$ if pertutation is odd, $0$ else)

This can be simplified with $$\det \left(A_{ij}\right)=\epsilon_{i_i,i_2,..}A_{1,i_1}A_{2,i_2}A_{3,i_3}\cdots $$

Introducing base vectors $e_i$ of a $n$ dimensional vector space with $i=1,2,...,n$ and $e_ie_j=\delta_{ij}$ (Here $n=3$) Than you can write your Matrix $$\begin{pmatrix} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn}\end{pmatrix}= \begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix}\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}$$

The determinant of a matricies build of vectors is now $$\det\begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix}=\epsilon_{ijk}\qquad \det\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}=\epsilon_{lmn}$$ Using the product rule of determinats $$\det\begin{pmatrix} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn}\end{pmatrix}=\det\begin{pmatrix}e_i^\top\\e_j^\top\\e_k^\top\end{pmatrix} \cdot\det\begin{pmatrix}e_l&e_m&e_n\end{pmatrix}= \epsilon_{ijk}\epsilon_{lmn} $$

PS. I am using the notation $\left(A_{ij}\right)$ is the matrix with bound indicies and $A_{ij}$ are their elements. PPS. This calculation can also be found on http://de.wikipedia.org/wiki/Levi-Civita-Symbol


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