Show that $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$

Question

Question: Show that $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$ where $\epsilon_{ijk}$ is the Levi-civita symbol and $\delta_{ij}$ is the Kronecker delta symbol.

My attempt: $\epsilon_{ijk}$ assumes non-zero values only when $i\not = j\not =k$ i.e.where $i,j,k$ is a permutation of $1,2,3$.

1. If $i \not = l$ then $l=j$ or $l=k$, and then $\epsilon_{ijk}\epsilon_{ljk}=0$
2. When $i=l$, then $\epsilon_{ijk}\epsilon_{ljk}=\epsilon_{ijk}\epsilon_{ijk}=(\epsilon_{ijk})^2=6$ since $i,j,k$ can have $3!=6$ permutations.

Then we have finally $\epsilon_{ijk}\epsilon_{ljk}=6\delta_{il}$ and not $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$.

Is the question itself wrong? Or is there a flaw in my attempt to solve?

Please help.

Answer 1

The summation convention says that you sum any letter that appears twice.
So $\epsilon_{ijk}\epsilon_{ljk}$, it really means $\sum_j\sum_k\epsilon_{ijk}\epsilon_{ljk}$ You are right that $\epsilon_{ijk}\epsilon_{ijk}=6$, but then $\epsilon_{ijk}\epsilon_{ijk}=\sum_i\sum_j\sum_k\epsilon_{ijk}\epsilon_{ijk}$
For the same reason, $\delta_{ii}=3,\delta_{11}=1$

Answer 2

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

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

With facts that $\delta_{ii}=3$ and $\delta_{ik}\delta_{kj}=\delta_{ij}$, then equation $(*)$ becomes $$ \begin{aligned} \epsilon_{ijk}\epsilon_{ljk}&=\delta_{il}\delta_{jj}-\delta_{ij}\delta_{jl}\\ &=3\delta_{il}-\delta_{il}\\ &=2\delta_{il} \end{aligned} $$

Therefore $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$.