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I have been trying to work out the expression for $\varepsilon_{ijkl} \cdot \varepsilon_{ijkl}$. Setting up the determinants, I found that the answer was 6. But, I doubt that I am right. When I had $\varepsilon_{ij} \cdot \varepsilon_{ij}$, my answer was 2. For $\varepsilon_{ijk} \cdot \varepsilon_{ijk}$, my answer was 6, and I am a bit unconvinced that my process was right.

Can anyone give me a hint?

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2 Answers 2

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It should be $4!= 24$ as there are 24 possible $ijkl$'s with all indices distinct.

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  • $\begingroup$ Do you have any hint on how to set up the determinants for this? $\endgroup$ Commented Feb 21, 2017 at 1:04
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    $\begingroup$ Determinants have nothing to do with it. Each of the 24 terms in the sum is unity, so the sum is 24. $\endgroup$
    – mike stone
    Commented Feb 21, 2017 at 1:20
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$$ \varepsilon_{ijkl} \cdot \varepsilon_{mnpq}= \left|\begin{matrix} \delta_{im}& \delta_{in}& \delta_{ip}& \delta_{iq} \\ \delta_{jm}& \delta_{jn}& \delta_{jp}& \delta_{jq} \\ \delta_{km}& \delta_{kn}& \delta_{kp}& \delta_{kq} \\ \delta_{lm}& \delta_{ln}& \delta_{lp}& \delta_{lq} \end{matrix}\right| $$ Or for convincing you may use a brute force method, by starting with $$ \varepsilon_{ijkl} =\delta^0_i\delta^1_j\delta^2_k\delta^3_l- \delta^0_i\delta^1_j\delta^2_l\delta^3_k + \delta^0_i\delta^1_l\delta^2_j\delta^3_k - \delta^0_i\delta^1_l\delta^2_k\delta^3_j + \delta^0_i\delta^1_k\delta^2_l\delta^3_j - \delta^0_i\delta^1_k\delta^2_j\delta^3_l \\ -\delta^0_j\delta^1_i\delta^2_k\delta^3_l+ \delta^0_j\delta^1_i\delta^2_l\delta^3_k - \delta^0_j\delta^1_l\delta^2_i\delta^3_k + \delta^0_j\delta^1_l\delta^2_k\delta^3_i - \delta^0_j\delta^1_k\delta^2_l\delta^3_i + \delta^0_j\delta^1_k\delta^2_i\delta^3_l \\ +\delta^0_k\delta^1_i\delta^2_j\delta^3_l- \delta^0_k\delta^1_i\delta^2_l\delta^3_j + \delta^0_k\delta^1_l\delta^2_i\delta^3_j - \delta^0_k\delta^1_l\delta^2_j\delta^3_i + \delta^0_k\delta^1_j\delta^2_l\delta^3_i - \delta^0_k\delta^1_j\delta^2_i\delta^3_l \\ -\delta^0_l\delta^1_i\delta^2_j\delta^3_k+ \delta^0_l\delta^1_i\delta^2_k\delta^3_j - \delta^0_l\delta^1_k\delta^2_i\delta^3_j + \delta^0_l\delta^1_k\delta^2_j\delta^3_i - \delta^0_l\delta^1_j\delta^2_k\delta^3_i + \delta^0_l\delta^1_j\delta^2_i\delta^3_k $$ You must see that$\varepsilon_{ijkl}\cdot \varepsilon_{ijkl} =24$

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  • $\begingroup$ $\delta^3_k \delta^1_k= \delta^{31}=0$ $\endgroup$ Commented Feb 21, 2017 at 3:33

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