Elementary Abstract Algebra- Examples and Applications, 2019a

8.6 LEVI-CIVITA SYMBOLS AND APPLICATIONS 347
Let’s focus on the indices i and j. First, if i = j then ɛ ijk = ɛ iik =0,
so S iimn = 0. On the other hand, if i ≠ j, there is only one value of k that
makes ɛ ijk nonzero (because we must have k ≠ i, j). We must also have
m, n ≠ k in order for ɛ kmn ≠ 0. It follows that there are two possibilities for
which S ijmn ≠0:
(A) i ≠ j, m = i and n = j;
(B) i ≠ j, m = j and n = i.
In case (A) we have:
S ijij =
[ ∑
k
ɛ ijk ɛ kij
]
=
[ ∑
k
ɛ 2 ijk
]
=1.
In case (B) we have:
S ijji =
[ ∑
k
ɛ ijk ɛ kji
]
=
− [ ɛ 2 ijk] ] = −1.
[ ∑
k
In summary we have:
S ijmn =1ifm = i, n = j, and i ≠ j;
S ijmn = −1 ifn = i, m = j, and i ≠ j;
S ijmn = 0 otherwise.
Let’s plug this back into our expression for (a × (b × c)) i
.Wecanthen
separate the terms where m = i, n = j from the terms where n = i, m = j.
Notice that there is no longer a sum over 3 indices but only one index, since
m and n are determined by i and j:
∑
a j b i c j
j,j≠i
} {{ }
(terms for m = i, n = j)
−
∑
a j b j c i
j,j≠i
} {{ }
(terms for m = j, n = i)
Now if we add a i b i c i to the first set of terms, and add −a i b i c i to the
second set of terms, then the overall sum doesn’t change but the two expressions
simplify: