6.1 Cyclic Operads
115
While
τ (ω 1 ⊗ ω 2 ⊗ ω 3 ⊗ ω 4 ) = (τ ) · (ω 1 ⊗ ω 4 ⊗ ω 3 ⊗ ω 2 ),
isomorphism (6.27) brings ω 1 ⊗ ω 4 ⊗ ω 3 ⊗ ω 2 into ω 1 ⊗ ω 4 ⊗ 1 ⊗ ω 3 ⊗ ω 2 , and
(1
⊗m
⊗ s ⊗ 1
⊗n )(ω 1 ⊗ ω 4 ⊗ 1 ⊗ ω 3 ⊗ ω 2 )
=
(ω 1 ⊗ ω 4 ⊗ s
i ⊗ s
i ⊗ ω 3 ⊗ ω 2 ).
(6.28)
The Koszul sign rule gives
ρ(ω 1 ⊗ ω 4 ⊗ s
i ⊗ s
i ⊗ ω 3 ⊗ ω 2 )
=
(−1)
|s
i ||ω 4 |+|s
i ||ω 3 | (ω 1 ⊗ s
i ⊗ ω 3 ⊗ ω 4 ⊗ s
i ⊗ ω 2 )
and, finally,
(f ⊗ g)
(ω 1 ⊗ s
i ⊗ ω 4 ⊗ ω 3 ⊗ s
i ⊗ ω 2 )
=
(−1)
(|ω 1 |+|s
i |+|ω 4 |)|g| f (ω 1 ⊗ s
i ⊗ ω 4 )g(ω 3 ⊗ s
i ⊗ ω 2 )
=
(−1)
|f ||g| f (ω 1 ⊗ s
i ⊗ ω 4 )g(ω 3 ⊗ s
i ⊗ ω 2 ),
where we used that |ω 1 | + |s
i | + |ω 4 | + |f | = 0. The accumulated contribution of
the sign factors above is precisely )(−1) κ as claimed.
For i = m + 1 and j = 1 (6.25) acquires a particularly nice form, namely
(f m+1 ◦ 1 g)(v 1 , . . . , v m+n )
(6.29)
=
(−1)
|f ||g| f (v 1 , . . . , v m , s
i )g(s
i , v m+1 , . . . , v m+n ).
Example 6.9 Assume that s ∈ V ⊗ V is non-degenerate, i.e., that there exists a
graded symmetric bilinear degree 0 form B : V ⊗ V → k satisfying (6.23). Then B,
considered as an element of End V (2), plays a rôle of a two-sided unit. It is indeed
easy to verify that
f n ◦ 1 B = B 2 ◦ 1 f = f
for f ∈ End V (n).
Example 6.10 The symmetric element s =
s
i ⊗ s
i ∈ V ⊗ V defines for each
n ≥ 1 a linear map
Lin(V
⊗n+1 , k) −→ Lin(V
⊗n , V ), f −→ f ,
(6.30)