8.2 Loop Homotopy Algebras
203
We are going to rewrite (8.37) into an equivalent form closer to Equation (33)
of [9]. Our translation will be based on the correspondence that assigns, for n ≥ 1,
to a degree k function f : V ⊗n → k the degree k + 1 map δ : V ⊗(n−1) → V by the
formula
δ(v 1 , . . . , v n−1 ) =:
(−1)
k|s
i | f (s
i , v 1 , . . . , v n−1 )s
i ,
(8.39)
for v 1 , . . . , v n−1 ∈ V . The correspondence f → δ is one-to-one when s is
non-degenerate. Let us explain the sign factor (−1)
k|s
i | . We started from the
composition 7
(1 ⊗ f )(s
i ⊗ s
i ⊗ v 1 , . . . , v n−1 )
which, according to the Koszul sign convention, equals
(−1)
k|s
i | s
i ⊗ f (s
i , v 1 , . . . , v n−1 ).
Next we commuted the degree |s
i | vector s
i with the scalar f (s
i , v 1 , . . . , v n−1 ) of
degree 0 and obtained
(−1)
k|s
i | f (s
i , v 1 , . . . , v n−1 ) ⊗ s
i .
Finally, we multiplied the vector s
i with the scalar f (s
i , v 1 , . . . , v n−1 ). The result
was the right-hand side of (8.39).
Notice that when f is fully symmetric in homogeneous v 1 , . . . , v n , δ is fully
symmetric in v 1 , . . . , v n−1 . In this case also
f (s
i , v 1 , . . . , v n−1 ) = (−1)
|s
i |(|v 1 |+···+|v n−1 |) f (v 1 , . . . , v n−1 , s
i ).
Since f is of degree k, f (v 1 , . . . , v n−1 , s
i ) = 0 only if
|v 1 | + · · · + |v n−1 | + |s
i | + k = 0,
so |v 1 | + · · · + |v n−1 | ≡ |s
i | + k mod 2. We therefore have
f (s
i , v 1 , . . . , v n−1 ) = (−1)
|s
i |(|s
i |+k) f (v 1 , . . . , v n−1 , s
i )
= (−1)
|s
i |+k|s
i | f (v 1 , . . . , v n−1 , s
i ).
7 Notice that
s
i ⊗ s
i = s
i ⊗ s
i because s is symmetric and |s
i | + |s
i | = 1.
203
We are going to rewrite (8.37) into an equivalent form closer to Equation (33)
of [9]. Our translation will be based on the correspondence that assigns, for n ≥ 1,
to a degree k function f : V ⊗n → k the degree k + 1 map δ : V ⊗(n−1) → V by the
formula
δ(v 1 , . . . , v n−1 ) =:
(−1)
k|s
i | f (s
i , v 1 , . . . , v n−1 )s
i ,
(8.39)
for v 1 , . . . , v n−1 ∈ V . The correspondence f → δ is one-to-one when s is
non-degenerate. Let us explain the sign factor (−1)
k|s
i | . We started from the
composition 7
(1 ⊗ f )(s
i ⊗ s
i ⊗ v 1 , . . . , v n−1 )
which, according to the Koszul sign convention, equals
(−1)
k|s
i | s
i ⊗ f (s
i , v 1 , . . . , v n−1 ).
Next we commuted the degree |s
i | vector s
i with the scalar f (s
i , v 1 , . . . , v n−1 ) of
degree 0 and obtained
(−1)
k|s
i | f (s
i , v 1 , . . . , v n−1 ) ⊗ s
i .
Finally, we multiplied the vector s
i with the scalar f (s
i , v 1 , . . . , v n−1 ). The result
was the right-hand side of (8.39).
Notice that when f is fully symmetric in homogeneous v 1 , . . . , v n , δ is fully
symmetric in v 1 , . . . , v n−1 . In this case also
f (s
i , v 1 , . . . , v n−1 ) = (−1)
|s
i |(|v 1 |+···+|v n−1 |) f (v 1 , . . . , v n−1 , s
i ).
Since f is of degree k, f (v 1 , . . . , v n−1 , s
i ) = 0 only if
|v 1 | + · · · + |v n−1 | + |s
i | + k = 0,
so |v 1 | + · · · + |v n−1 | ≡ |s
i | + k mod 2. We therefore have
f (s
i , v 1 , . . . , v n−1 ) = (−1)
|s
i |(|s
i |+k) f (v 1 , . . . , v n−1 , s
i )
= (−1)
|s
i |+k|s
i | f (v 1 , . . . , v n−1 , s
i ).
7 Notice that
s
i ⊗ s
i = s
i ⊗ s
i because s is symmetric and |s
i | + |s
i | = 1.
