Proof of Indicial Notation Identity: Shortcut and Results

[ftarak]

Homework Statement

Use the identity
<br /> \epsilon_{ijk} \epsilon_{klm} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}<br />
as a shortcut to obtain the following results:
a) \epsilon_{ilm} \epsilon_{jlm}= 2\delta_{ij}
b) \epsilon_{ijk} \epsilon_{ijk} = 6

Homework Equations

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.

 

[gabbagabbahey]

Homework Statement

Use the identity
<br /> \epsilon_{ijk} \epsilon_{klm} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}<br />

This is incorrect, on the LHS only the indices i, j, l, andm 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}<br />

as a shortcut to obtain the following results:
a) \epsilon_{ilm} \epsilon_{jlm}= 2\delta_{ij}
b) \epsilon_{ijk} \epsilon_{ijk} = 6

Homework Equations

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.

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)?