152
6 Operads
Example 6.30 Recall [11, Definition 13] that an anti-associative algebra is a couple
A = (A, ,) consisting of a graded vector space A and a degree +1 operation :
A ⊗ A → A which is anti-associative, i.e.,
a a (b b c) + (−1)
|a| (a a b) ) c = 0,
for all a, b, c ∈ A. An odd modular operad with step s = 0 such that T (S; g) = 0
only if card(S) = 2 and g = 0 is precisely an anti-associative algebra A with
an involution τ : A → A such that τ (ab) = τ (b)τ (a) for all a, b ∈ A, cf.
Example 6.23.
Example 6.31 Let V be a graded vector space and s ∈ V ⊗ V a symmetric
degree +1 tensor. The construction of the modular endomorphism operad given in
Examples 6.6 and 6.28 translate verbatim, though the operations a • b and • uv now
have degree +1.
One must however be careful. While in the case of ordinary modular operads
both constructions of the a ◦ b operation, i.e., the one via composition (6.15) and the
one via composition (6.16) lead to the same results, now the results are different.
The reason is that, while for |s| = 0 the above compositions are dual to each
other, if |s| = 1 they are not, because the duality (2) acquires a nontrivial sign.
One immediately sees that the resulting f a • b g’s differ by (−1) |f |+|g| .
Likewise, compositions (6.62) and (6.64) are in the |s| = 1 case not dual to each
other and the resulting • uv (f )’s differ by (−1) |f | . What happens is so surprising that
we formulate it as a proposition; recall that Lin S and Lin M denote the two versions
of the category Lin discussed in Remark 6.11.
Proposition 6.11 For |s| = 1 compositions (6.16) and (6.64) lead to an odd
modular operad in Lin S , while compositions (6.15) and (6.62) to an odd modular
operad in Lin M .
Proof. Let us show that the a • b -operations defined by the odd version of (6.16)
satisfy (6.86). Since |¯ s| = | ¯ ¯
s|, one must be cautious. We have
f a • b (g c • d h) =
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
#
f ⊗ (g c • d h)
=
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
#
f ⊗ (1 S 2 ⊗ ¯ ¯
s ⊗ 1 S 3 )
# (g ⊗ h)
= (−1)
|f |
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
# 1 S 1 {a,b}}S 2 ⊗ ¯ ¯
s ⊗ 1 S 3
# (f ⊗ g ⊗ h),
while
(f a • b g) c • d h =
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
#
(f a • b g) ⊗ h
=
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
#
(1 S 1 ⊗ ¯
s ⊗ 1 S 2 )
# (f ⊗ g) ⊗ h
=
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
# 1 S 1 ⊗ ¯
s ⊗ 1 S 2 {c,d}}S 3
# (f ⊗ g ⊗ h).
6 Operads
Example 6.30 Recall [11, Definition 13] that an anti-associative algebra is a couple
A = (A, ,) consisting of a graded vector space A and a degree +1 operation :
A ⊗ A → A which is anti-associative, i.e.,
a a (b b c) + (−1)
|a| (a a b) ) c = 0,
for all a, b, c ∈ A. An odd modular operad with step s = 0 such that T (S; g) = 0
only if card(S) = 2 and g = 0 is precisely an anti-associative algebra A with
an involution τ : A → A such that τ (ab) = τ (b)τ (a) for all a, b ∈ A, cf.
Example 6.23.
Example 6.31 Let V be a graded vector space and s ∈ V ⊗ V a symmetric
degree +1 tensor. The construction of the modular endomorphism operad given in
Examples 6.6 and 6.28 translate verbatim, though the operations a • b and • uv now
have degree +1.
One must however be careful. While in the case of ordinary modular operads
both constructions of the a ◦ b operation, i.e., the one via composition (6.15) and the
one via composition (6.16) lead to the same results, now the results are different.
The reason is that, while for |s| = 0 the above compositions are dual to each
other, if |s| = 1 they are not, because the duality (2) acquires a nontrivial sign.
One immediately sees that the resulting f a • b g’s differ by (−1) |f |+|g| .
Likewise, compositions (6.62) and (6.64) are in the |s| = 1 case not dual to each
other and the resulting • uv (f )’s differ by (−1) |f | . What happens is so surprising that
we formulate it as a proposition; recall that Lin S and Lin M denote the two versions
of the category Lin discussed in Remark 6.11.
Proposition 6.11 For |s| = 1 compositions (6.16) and (6.64) lead to an odd
modular operad in Lin S , while compositions (6.15) and (6.62) to an odd modular
operad in Lin M .
Proof. Let us show that the a • b -operations defined by the odd version of (6.16)
satisfy (6.86). Since |¯ s| = | ¯ ¯
s|, one must be cautious. We have
f a • b (g c • d h) =
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
#
f ⊗ (g c • d h)
=
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
#
f ⊗ (1 S 2 ⊗ ¯ ¯
s ⊗ 1 S 3 )
# (g ⊗ h)
= (−1)
|f |
1 S 1 ⊗ ¯
s ⊗ 1 S 2 3
# 1 S 1 {a,b}}S 2 ⊗ ¯ ¯
s ⊗ 1 S 3
# (f ⊗ g ⊗ h),
while
(f a • b g) c • d h =
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
#
(f a • b g) ⊗ h
=
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
#
(1 S 1 ⊗ ¯
s ⊗ 1 S 2 )
# (f ⊗ g) ⊗ h
=
1 S 1 2 ⊗ ¯ ¯
s ⊗ 1 S 3
# 1 S 1 ⊗ ¯
s ⊗ 1 S 2 {c,d}}S 3
# (f ⊗ g ⊗ h).