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This is how I began,

Proof.

Using

$$ \epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \tag{1}\label{eq1} $$

I wrote it as

$ \epsilon_{ijk}\epsilon_{ijk} = \delta_{jj}\delta_{kk} - \delta_{jk}\delta_{kj} \tag{2}\label{eq2}$

and

$ \delta_{jj} = 3 \tag{3}\label{eq3} $

$\therefore \delta_{jj} \delta_{kk} = 9 $ and $$ \delta_{jk}\delta_{kj} = ? \tag{4}\label{eq4}$$ I'm confused on how to simplify $\ref{eq4}$. My first thoughts would be that it's $\delta_{jk}\delta_{kj} = \delta_{kk}\delta_{kk} = 3*3 = 9$ since it only has a value when $j=k$ but after explicitly writing out the terms I can see that

If $j=1$ then

$$\begin{equation}\begin{aligned} \delta_{1k}\delta_{1k} = 1 \text{ when k=1, but 0 otherwise} \\ \delta_{2k}\delta_{2k} = 1 \text{ when k=2, but 0 otherwise} \\ etc ... \end{aligned}\end{equation}\tag{5}\label{eq5}$$

$\therefore \delta_{jk}\delta_{kj} = 3 $

But I don't quite understand why $ \delta_{jk}\delta_{kj} = 3 $ without explicitly writing out the terms. Can you help me rationalize this or look at it a different way so that I can understand it more intuitively?

Micheal S. Bingham
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    You have nine $(j,,k)$ pairs, of which three (those with $j=k$) contribute. Alternatively, the original equation is a sum over $i,,j,,k$, scoring $(\pm1)^2=1$ for each of $3!=6$ permutations. You can even write it as $\epsilon^2=6$, which is like squaring a vector viz. $v^2:=v\cdot v=v_iv_i$. – J.G. Sep 15 '21 at 14:41
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    Lets view these symbols as components of matrices. The matrix with components $\delta_{ij}$ is the identity matrix. Next a combination $A_{ij}B_{jk}$ is the $ik$ component of the matrix that you get from multiplying the matrices $A$ and $B$. And further $A_{ij}B_{ji}$ is the trace of this matrix. So $\delta_{jk}\delta_{kj}$ is the trace of the square of the identity matrix = the trace of the identity matrix which is just $1+1+1=3$. – Winther Sep 15 '21 at 14:56

3 Answers3

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Hint:

Due to Einstien's convention the repeated idicies in $S=\epsilon_{ijk} ~\epsilon_{ijk}$ mean summation over them so. $$S=\sum_{i=1}^{3} \sum_{j=1}^{3}\sum_{k=1}^{3}\epsilon_{ijk}~ \epsilon_{ijk}=\epsilon_{123} \epsilon_{123}+ \epsilon_{231} \epsilon_{231}+\epsilon_{312}\epsilon_{312}+\epsilon_{321} \epsilon_{321}+ \epsilon_{213} \epsilon_{213}+\epsilon_{132}\epsilon_{132}+.......=1+1+1+1+1+1+(1-1)+(1-1)..+0+0_...=6$$

Z Ahmed
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Use the Kronecker delta contraction

$\delta_{ik}\delta_{kj}=\delta_{ij}$

where we get terms of $1=\delta_{jj}$ iff $i=j$. Then in three dimensions summing these unities over three values of $j$ gives $3$.

Oscar Lanzi
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First to know that $\epsilon_{ijm}\epsilon_{klm}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk} \, (*)$, the proof of this equation is in Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory.

Then for the $\epsilon_{ijk}\epsilon_{ljk}$, we can make some substitutions in the equation $(*)$ as follows

$$ \begin{aligned} m&:= i \\ i&:=j \\ j&:=k \\ k&:=j \\ l&:=k \end{aligned} $$

With facts that $\delta_{ii}=3$ and $\delta_{ik}\delta_{kj}=\delta_{ij}$, then equation $(*)$ becomes $$ \begin{aligned} \epsilon_{ijk}\epsilon_{ijk}&=\delta_{jj}\delta_{kk}-\delta_{jk}\delta_{kj}\\ &=\delta_{jj}\delta_{kk}-\delta_{jj}\\ &=3\cdot3-3 \\ &=6 \end{aligned} $$

Therefore $\epsilon_{ijk}\epsilon_{ijk} = 6$.

D.L.
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