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analogy.
✷
The following simple lemma will be useful in many places:
Lemma 2.5.7.
Let H be a Hopf algebra with invertible antipode. Then
S −1 (a (2) ) ∙ a (1) = a (2) ∙ S −1 (a (1) ) = 1 H ɛ(a) for all a ∈ H .
Proof.
The following calculation shows the claim:
S −1 (a (2) ) ∙ a (1) = S −1 ◦ S ( S −1 (a (2) ) ∙ a (1)
)
= S −1 ( S(a (1) ) ∙ a (2)
)
= S −1 (1 H )ɛ(a) = 1 H ɛ(a)
[ S is antialgebra morphism]
The other identity is proven analogously.
✷
Remark 2.5.8.
Let H be a bialgebra. An endomorphism ˜S : H → H such that
∑
S(x (2) )x (1) = ∑ x (2) S(x (1) ) = ɛ(x)1 H for all x ∈ H
x
x
is also called a skew antipode. We will usually avoid requiring the existence of an antipode
and of a skew antipode and impose instead the stronger condition on the antipode of being
invertible.
Corollary 2.5.9.
1. If H is either commutative or cocommutative, then the identity S 2 = id H holds.
2. If H and K are Hopf algebras with antipodes S H and S K , respectively, then any bialgebra
map ϕ : H → K is a Hopf algebra map, i.e. ϕ ◦ S H = S K ◦ ϕ.
Proof.
1. If H is commutative, then
x (2) ∙ S(x (1) ) = S(x (1) ) ∙ x (2)
defn. of S
= ɛ(x)1 H .
From the previous proposition, we conclude that S 2 = id H . If H is cocommutative, then
Again we conclude that S 2 = id H .
x (2) ∙ S(x (1) ) = x (1) ∙ S(x (2) )
defn. of S
= ɛ(x)1 H .
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