Proof of Indicial Notation Identity: Shortcut and Results

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ftarak
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Homework Statement



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

Homework Equations




The Attempt at a Solution


I tried to solve that by solving the determinant [tex]\epsilon_{ilm} \epsilon_{jlm}= [\delta][/tex] 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.
 
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ftarak said:

Homework Statement



Use the identity
[tex] \epsilon_{ijk} \epsilon_{klm} = \delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}[/tex]


This is incorrect, on the LHS only the indices [itex]i[/itex], [itex]j[/itex], [itex]l[/itex], and[itex]m[/itex] are free indices ([itex]k[/itex] 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:

[tex]\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}[/tex]

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

Homework Equations




The Attempt at a Solution


I tried to solve that by solving the determinant [tex]\epsilon_{ilm} \epsilon_{jlm}= [\delta][/tex] 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 [itex]\epsilon_{ijk}[/itex] is invariant under cyclic transposition of indices so that [itex]\epsilon_{jlm}=\epsilon{mjl}[/itex] and hence, [itex]\epsilon_{ilm} \epsilon_{jlm}=\epsilon_{ilm} \epsilon_{mjl}[/itex], which is in the same form as your identity with [itex]j \to l[/itex], [itex]k\to m[/itex], [itex]l\to j[/itex] and [itex]m\to l[/itex]... What does that give you (show your calculatiuons)?