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1 Preliminaries
If exchange the columns of the determinant, then there is
A 1r A 1s A 1t
A 2r A 2s A 2t
A 3r A 3s A 3t
= ε rst
A i j
If exchange rows and columns at the same time, then there is
A ir A is A it
A jr A js A jt
A kr A ks A kt
= ε i jk ε rst
A i j
Let A i j = δ i j , then the determinant
A i j
=
δ i j
= 1, or
δ ir δ is δ it
δ jr δ js δ jt
δ kr δ ks δ kt
= ε i jk ε rst
A i j
= ε i jk ε rst
Quod erat demonstrandum.
When two or three pairs of dummies appear, Eq. (1.6.22) is reduced to
ε i jk ε i jt = 2δ kt
(1.6.23)
ε i jk ε i jk = 6
(1.6.24)
Example 1.6.2 Prove ε − δ identity ε i jk ε ist = δ js δ kt − δ ks δ jt .
Proof From Eqs. (1.6.3) and (1.6.18), we obtain
e j × e k = ε jki e i = ε i jk e i , e s × e t = ε str e r = ε rst e r
(1)
(e j × e k ) · (e s × e t ) = ε i jk e i · ε rst e r = ε i jk ε rst e i · e r = ε i jk ε rst δ ir = ε i jk ε ist (2)
According to the mixed product formula of trivector a· b× c = c· a× b = b· c× a,
the left side of Eq. (2) can be expressed as
(e j × e k ) · (e s × e t ) = e s · [e t × (e j × e k )] = −e s · [(e j × e k ) × e t ]
(3)
According to the triple vector product formula of trivector (a × b) × c = (c ·
a)b − (b · c)a and Eq. (1.6.5), the right side of Eq. (3) can be expressed as
− e s · [(e j × e k ) × e t ] = −e s · [(e j · e t )e k − (e k · e t )e j ]
= δ kt e s · e j − δ jt e s · e k = δ kt δ s j − δ jt δ sk = δ js δ kt − δ ks δ jt
(4)