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 $1,2,3$.
- If $i \not = l$ then $l=j$ or $l=k$, and then $\epsilon_{ijk}\epsilon_{ljk}=0$
- 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.
Best Answer
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, $\delta_{ii}=3,\delta_{11}=1$