The Kronecker delta

  • Thread starter newton1
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i need help....:frown:
prove SUM(k) [E(ijk)E(lmk)]= d(il)d(jm) - d(im)d(jl)
where "d" is Kronecker delta symbol and "E" is permutation symbol or
Levi-Civita density
 
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HallsofIvy
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A clarification: the Kronecker delta, d(ij), is 1 if i= j, 0 otherwise.

The Levi-Civita permutation symbol, E(ijk) {real notation is "epsilon"), is 1 if ijk is an even permutation of 123, -1 if ijk is an odd permutation of 123, and 0 otherwise. While d(ij) is defined for all dimensions, E(ijk) implies that i, j, and k can only be 1, 2 ,3. For higher "dimensions" we would need more indices.

SUM(k) [E(ijk)E(lmk)]= E(ij1)E(lm1)+ E(ij2)E(lm2)+E(ij3)E(lm3)
 

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