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Since S 2 χ H = χ H and since S 2 is a an algebra morphism of H ∗ , we have by lemma 3.3.17.2
for any β ∈ H ∗
S 2 (χ H β) = S 2 (χ H ) ∙ S 2 (β) = χ H ∙ S 2 (β)
so that S 2 restricts to an endomorphism of the linear subspace χ H H ∗ ⊂ H ∗ .
Lemma 3.3.18.
Let H be a Hopf algebra. Let γ ∈ H ∗ be a nonzero right integral and Γ ∈ H be a left integral,
normalized such that 〈γ, Γ〉 = 1. Then
Tr H ∗(S 2 ) = 〈ɛ, Γ〉〈γ, 1〉 = (dim H)TrS 2 | χH H ∗ .
✷
Proof.
By applying proposition 3.3.3 to H opp,copp , we find
Tr H S 2 = 〈γ, 1〉 ∙ 〈ɛ, Γ〉 .
We denote by ˜Γ ∈ H ∗∗ the image of Γ ∈ H in the bidual of H. Now γ is a Frobenius form
with dual bases (Γ (1) , S(Γ (2) )) which implies that ˜Γ is a Frobenius form for H ∗ with dual bases
(S(γ (1) ), γ (2) ).
Now lemma 3.3.14 applies to e := χ H with α = dim H, thus yielding
dim H ∙ Tr(S 2 )| χH H ∗ = 〈˜Γ, S 2 (χ H γ (2) )S(γ (1) )〉
= 〈˜Γ, S 2 (χ H )S 2 (γ (2) )S(γ (1) )〉 [S 2 algebra morphism]
= 〈˜Γ, χ H S(γ (1) ∙ S(γ (2) )〉 [S 2 χ H = χ H ]
= 〈γ, 1〉 ∙ 〈χ H , Γ〉
By the same lemma 3.3.14, taking f = L Γ and e = 1 with α = 1, we have for the second
factor
χ H (Γ) = 〈γ, S(Γ (2) )ΓΓ (1) 〉
Combining the two results yields
= 〈γ, ɛ(Γ (2) )ΓΓ (1) 〉 [Γ is a left integral of H]
= 〈γ, ΓΓ〉 = ɛ(Γ)〈γ, Γ〉 = ɛ(Γ)
dim H ∙ Tr(S 2 )| χH H ∗ = 〈γ, 1〉 ∙ 〈χ H, Γ〉 = 〈γ, 1〉 ∙ 〈ɛ, Γ〉
which finishes the proof of the lemma.
✷
Theorem 3.3.19 (Larson-Radford, 1988).
Let K be a field of characteristic zero. Let H be a finite-dimensional K-Hopf algebra. Then the
following statements are equivalent:
1. H is semisimple.
2. H ∗ is semisimple.
3. S 2 = id H .
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