Page 17 - Intro to Tensor Calculus

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Figure 1.1-3. Mnemonic device for position of subscripts.

Another form of this identity employs both subscripts and superscripts and has the form

k
j
j k
e ijk e imn = δ δ − δ δ . (1.1.5)
m n n m
One way of proving this identity is to observe the equation (1.1.5) has the free indices j, k, m, n. Each
of these indices can have any of the values of 1, 2or 3. There are 3 choices we can assign to each of j, k, m
4
or n and this gives a total of 3 = 81 possible equations represented by the identity from equation (1.1.5).
By writing out all 81 of these equations we can verify that the identity is true for all possible combinations
that can be assigned to the free indices.
An alternate proof of the e − δ identity is to consider the determinant

1 δ 1 δ 1 10 0
δ
1 2 3
2 δ 2 δ 2 = 01 0 =1.
δ

1 2 3
3 δ 3 δ 3 00 1
δ
1 2 3
By performing a permutation of the rows of this matrix we can use the permutation symbol and write
i i
δ δ δ i
1 j 2 j 3 ijk
j
δ δ δ = e .
1 2 3
k 1 δ k 2 δ 3 k
δ
By performing a permutation of the columns, we can write
i i
δ δ δ i
r s t
j δ j δ j = e ijk e rst .
δ
r s t
k δ k s δ t k
δ
r
Now perform a contraction on the indices i and r to obtain
i i i
δ δ
i s δ
j j t ijk
δ δ δ j = e e ist .
i s t
k δ k δ k
δ
s
t
i
2
i
1
3
Summing on i we have δ = δ + δ + δ = 3 and expand the determinant to obtain the desired result
1
i
2
3
j k
j k
δ δ − δ δ = e ijk e ist .
t s
s t