Tensors for Physics

11.2 Δ-Tensors 185

Δ()μ

1 μ 2 ···μ− 1 λ,μ′ 1 μ′ 2 ···μ′− 1 λ

=

2 + 1

2 − 1

Δ(μ−^1 )

1 μ 2 ···μ− 1 ,μ′ 1 μ′ 2 ···μ′− 1

. (11.6)

The total contraction of the fore with the hind subscripts yields

Δ()μ 1 μ 2 ···μ,μ 1 μ 2 ···μ= 2 + 1 , (11.7)

which is the number of the independent components of an irreducible tensor of rank.
In the following product of twoΔ()-tensors, fore and hind subscripts are mixed:

Δ()μ 1 μ 2 ···μ− 1 ν,μν 1 ν 2 ···ν− 1 Δ()νν

1 ν 2 ···ν− 1 ,μ′ 1 μ′ 2 ···μ′

=

1

( 2 − 1 )

Δ()μ

1 μ 2 ···μ,μ′ 1 μ′ 2 ···μ′

.

(11.8)

The special case=2 of this equation corresponds to

Δμλ,νκΔλκ,μ′ν′=

1

6

Δμν,μ′ν′. (11.9)

The total contraction{μ 1 μ 2 ···μ}={μ′ 1 μ′ 2 ···μ′}in (11.8) yields

Δ()μ 1 μ 2 ···μ− 1 ν,μν 1 ν 2 ···ν− 1 Δ()νν 1 ν 2 ···ν− 1 ,μ 1 μ 2 ···μ=

2 + 1

( 2 − 1 )

. (11.10)

The contractionμ′=μ, renamed asμ,in(11.8) and use of (11.6), yields

Δ()μ 1 μ 2 ···μ− 1 ν,μν 1 ν 2 ···ν− 1 Δ()νν 1 ν 2 ···ν− 1 ,μμ′

1 μ′ 2 ···μ−^1

=

2 + 1

( 2 − 1 )^2

Δμ( 1 −μ^12 )···μ− 1 ,μ′

1 μ′ 2 ···μ′− 1

.

(11.11)

The special case=2of(11.11) corresponds to

Δμλ,νκΔλκ,νμ′=

5

18

δμμ′. (11.12)

The next contractionμ′=μgives

Δμλ,νκΔλκ,νμ=

5

6

. (11.13)

The relations (11.12) and (11.13) can also be inferred from (11.9). For comparison,
it is recalled thatΔμλ,νκΔνκ,λμ=Δμλ,λμ=5, cf. (11.7). The order of subscripts
matters!

11.1 Exercise: Contraction Rules for Delta-Tensors
Verify (11.6)for=2.