204
8 Structures Relevant to Physics
Substituting the last expression into formula (8.39) produces, in the symmetric case,
an alternative formula
δ(v 1 , . . . , v n−1 ) = (−1)
|s
i |+k(|s
i |+|s
i |)
f (v 1 , . . . , v n−1 , s
i )s
i
(8.40)
= (−1)
|s
i |+k
f (v 1 , . . . , v n−1 , s
i )s
i .
Let us finally denote, for n ≥ 1 and g ≥ 0, by δ
g
n−1 : V ⊗n−1 → V the degree 1 map
corresponding to the degree 0 function f
g
n : V n → k. Explicitly
δ
g
n−1 (v 1 , . . . , v n−1 ) :=
f
n
g (s
i , v 1 , . . . , v n−1 )s
i
= (−1)
|s
i |
f
n
g (v 1 , . . . , v n−1 , s
i )s
i
for v 1 , . . . , v n−1 ∈ V .
Let us start to apply our correspondence between functions V n → k and maps
V ⊗n−1 → V to the terms of (8.37). Since they are all of degree 1, we have
k = 1 in (8.39) resp. in (8.40). Using formula (8.38), we conclude that the map
corresponding to the left-hand side of (8.37) is given by
1≤i≤n−1
(−1)
|v 1 |+···+|v i−1 |+|s
i | f
g
n (v 1 , . . . , dv i , . . . , v n−1 , s
i )s
i
−
f
g
n (v 1 , , . . . , v n−1 , ds
i )s
i .
The rightmost term appears with the minus sign because f
g
n (v 1 , , . . . , v n−1 , ds
i ) =
0 only if
|v 1 | + · · · + |v n−1 | + |s
i | + 1 = 0.
Since s is d-closed by assumption, ds
i ⊗ s
i + (−1)
|s
i | s
i ⊗ ds
i = 0, so
−f
g
n (v 1 , . . . , v n−1 , ds
i )s
i = (−1)
|s
i | f
g
n (v 1 , . . . , v n−1 , s
i )ds
i .
By (8.39),
f
g
n (v 1 , . . . , dv i , . . . , v n−1 , s
i )s
i = (−1)
|s
i |
δ
g
n−1 (v 1 , . . . , dv i , . . . , v n−1 )
and
f
g
n (v 1 , . . . , v n−1 , s
i )s
i = (−1)
|s
i |
δ
g
n−1 (v 1 , . . . , v n−1 )