How to prove a levi-civita symbol and kronecker delta relationship

Question

$\displaystyle \sum_{i=1}^3 \sum_{j=1}^3 \epsilon_{ijk} \epsilon_{ijn} = 2 \delta_{kn}$

When I do the calculations of that I get 3 times the answer, I mean this is easy, but I´m just wrong, Could someone show me the way?

Answer 1

In case $k\neq n$ at least one of the $\varepsilon$ is zero for each term. Consider the case $k=n=1$ The sum reduces to $\sum\sum\varepsilon_{ij1}^2$. The only non-zero terms are $i=2, j=3$ and $i=3, j=2$. The sum is therefore $2$. The same reasoning applies for the other cases $k=n=2$ and $k=n=3$.

Update: The key assumption being that $k,n\in\{1,2,3\}$ as David notes.

Answer 2

Using Einstein notation, $$\varepsilon_{ijk}\varepsilon_{lmn}=\begin{vmatrix} \delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn} \end{vmatrix}=\\\delta_{il}(\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km})-\delta_{im}(\delta_{jl}\delta_{kn}-\delta_{jn}\delta_{kl})+\delta_{in}(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})\\$$ Now, setting $i=l$ and $j=m$ $$\varepsilon_{ijk}\varepsilon_{ijn}=\delta_{ii}(\delta_{jj}\delta_{kn}-\delta_{jn}\delta_{kj})-\delta_{ij}(\delta_{ji}\delta_{kn}-\delta_{jn}\delta_{ki})+\delta_{in}(\delta_{ji}\delta_{kj}-\delta_{jj}\delta_{ki})\\=9\delta_{kn}-3\delta_{kn}-3\delta_{kn}+\delta_{kn}+\delta_{kn}-3\delta_{kn}=2\delta_{kn}$$

Answer 3

Assuming that $k, n \in \{1,2,3\}$. Note that if any two of $a,b,c$ are equal, then $\varepsilon_{abc} = 0$. Consider two cases:

Case 1: $k \neq n$. In this case, we note that if either $i = k$ or $j = n$, the symbol $\varepsilon_{ijk}$ vanishes. Similarly, if either $i = n$ or $j = n$, then $\varepsilon_{ijn}$ vanishes. The remaining terms have $i = j$, for which $\varepsilon_{ijk} = \varepsilon_{ijn} = 0$. So the whole sum is zero when $k \neq n$, which is equal to $\delta_{kn}$.

Case 2: $k = n$. As above, the terms with $i = k$ or $j = k$ all vanish, as well as the ones with $i = j$. So we are left with terms where $\{i,j,k\} = \{1,2,3\}$. There are two elements of $\{1,2,3\} \setminus \{k\}$and therefore two ways to pick $i,j$ as distinct elements of this set. So two terms survive in the sum: $$\sum_{i = 1}^3 \sum_{j = 1}^3 \varepsilon_{ijk}\varepsilon_{ijn} = \varepsilon_{ijk}\varepsilon_{ijk} + \varepsilon_{jik}\varepsilon_{jik} = \varepsilon_{ijk}^2 + \varepsilon_{jik}^2 = 1 + 1 = 2 = 2\delta_{kn}$$