168
7 Feynman Transform of a Modular Operad
Then the following three statements are equivalent:
(i) θ is a degree k derivation,
(ii) θ is a degree 0 derivation, and
(iii) Θ is a morphism of odd modular operads.
Proof. Applying the iterated desuspension ↓ k to both sides of the Leibniz rule (7.5),
we obtain
↓ k θ a • b = (−1)
k
↓ k
a
R
• b (↑
k
⊗ 1)(↓
k θ ⊗ 1) + ↓ k
a
L
• b (1 ⊗ ↑ k )(1 ⊗ ↓ k θ)
.
(7.9)
Taking into account that the structure operations a
L
• b of M and the structure
operations a
L
• b of the iterated desuspension s −k M are related by the iteration
of (7.7) as
a
L
• b = (−1)
k
↓
k
a
L
• b (1⊗ ↑
k )
and that, likewise,
a
R
• b = (−1)
k
↓
k
a
R
• b (↑
k
⊗ 1)
we see that (7.9) is satisfied if and only if θ fulfills
θ a • b = a
R
• b (θ ⊗ 1) + a
L
• b (1 ⊗ θ).
(7.10)
Similarly, applying ↓ k to both sides of (7.6) gives
↓ k θ • uv = (−1)
k ↓ k uv ↑ k ↓ k θ.
which is equivalent to
θ • uv = uv θ.
(7.11)
Since (7.10) together with (7.11) means that θ is a degree 0 derivation, we proved
the equivalence of (i) and (ii).
Let us check that (ii) is satisfied if and only if Θ = (1, θ ) is an operad morphism,
i.e. that
t 1 a • b t 2 , θ(t 1 a • b t 2 )
=
t 1 , θ (t 1 )
a • b
t 2 , θ(t 2 )
(7.12)