# Indicial notation proof

1. Sep 6, 2010

### ftarak

1. The problem statement, all variables and given/known data

Use the identity
$$\epsilon_{ijk} \epsilon_{klm} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}$$
as a shortcut to obtain the following results:
a) $$\epsilon_{ilm} \epsilon_{jlm}= 2\delta_{ij}$$
b) $$\epsilon_{ijk} \epsilon_{ijk} = 6$$
2. Relevant equations

3. The attempt at a solution
I tried to solve that by solving the determinant $$\epsilon_{ilm} \epsilon_{jlm}= [\delta]$$ but the result just became zero. or even, I tried to change i=k and j=l in the first equation but the result was zero as well. I don't know what should I do, I'm really stuck.

2. Sep 9, 2010

### gabbagabbahey

This is incorrect, on the LHS only the indices $i$, $j$, $l$, and$m$ are free indices ($k$ is being summed over, so it is called a dummy index), so only those indices should appear on the RHS side of the identity. Instead, you should have:

$$\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$$

Start by using the fact that $\epsilon_{ijk}$ is invariant under cyclic transposition of indices so that $\epsilon_{jlm}=\epsilon{mjl}$ and hence, $\epsilon_{ilm} \epsilon_{jlm}=\epsilon_{ilm} \epsilon_{mjl}$, which is in the same form as your identity with $j \to l$, $k\to m$, $l\to j$ and $m\to l$... What does that give you (show your calculatiuons)?