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As stated in the title, with $i,j,k=1,...,N$.

I expanded $\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{ijk}$ as follows:

$$\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{ijk}=\underbrace{\mathcal{\epsilon}_{1jk}\mathcal{\epsilon}^{1jk}+\mathcal{\epsilon}_{2jk}\mathcal{\epsilon}^{2jk}+...+\mathcal{\epsilon}_{(N-1)jk}\mathcal{\epsilon}^{(N-1)jk}+\mathcal{\epsilon}_{Njk}\mathcal{\epsilon}^{Njk}}_{\text{$N$ terms (without expanding $j$ and $k$)}} \tag{1}$$

Considering only the leading term of $(1)$:

$$\mathcal{\epsilon}_{1jk}\mathcal{\epsilon}^{1jk}=\underbrace{\mathcal{\epsilon}_{12k}\mathcal{\epsilon}^{12k}+\mathcal{\epsilon}_{13k}\mathcal{\epsilon}^{13k}+...+\mathcal{\epsilon}_{1(N-1)k}\mathcal{\epsilon}^{1(N-1)k}+\mathcal{\epsilon}_{1Nk}\mathcal{\epsilon}^{1Nk}}_{\text{($N-1$) terms (without expanding $j$)}} \tag{2}$$

where $j \neq 1$.

Now for $\mathcal{\epsilon}_{12k}\mathcal{\epsilon}^{12k}$ only:

$$\mathcal{\epsilon}_{12k}\mathcal{\epsilon}^{12k}=\underbrace{\mathcal{\epsilon}_{123}\mathcal{\epsilon}^{123}+\mathcal{\epsilon}_{124}\mathcal{\epsilon}^{124}+...+\mathcal{\epsilon}_{12(N-1)}\mathcal{\epsilon}^{12(N-1)}+\mathcal{\epsilon}_{12N}\mathcal{\epsilon}^{12N}}_{\text{$(N-2)$ terms}} \tag{3}$$

where $k \neq 1,2$

Combining the results from $(1),(2)$ and $(3)$, I got

$$\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{ijk}=N(N-1)(N-2) \tag{4}$$


But from the identity that$$\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{klm}=\delta^{l}_{i}\delta^{m}_{j}-\delta^{m}_{i} \delta^{l}_{j} \tag{5}$$

$\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{ijk}$ gives

$$\mathcal{\epsilon}_{ijk}\mathcal{\epsilon}^{ijk}=\mathcal{\epsilon}_{jki}\mathcal{\epsilon}^{ijk}=\delta^{j}_{j}\delta^{k}_{k}-\delta^{k}_{j}\delta^{j}_{k}=N^2-N=N(N-1) \tag{6}$$

which is different to $(4)$.

Can someone please explain where my conceptual errors lie from $(1)$ to $(4)$ (if not from $(5)$ to $(6)$)?

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    $\begingroup$ Why are you considering $\epsilon^{ijk}$ with $N\neq 3$? If you want $\epsilon$ to represent the Levi-Civita symbol, then you need as many indices as they can take on values. $\endgroup$ – jacob1729 May 30 at 16:35
  • $\begingroup$ @fqq All non-zero terms are positive. $(-1)(-1)=1$ $\endgroup$ – floccinaucinihilipilificator May 30 at 17:39
  • $\begingroup$ Your formula works fine for $N=3$ and the entire calculation makes no sense for $N\ne 3$, as jacob1729 pointed out. $\endgroup$ – G. Smith May 30 at 18:04

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