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CHAPTER 4 The Infinitesimal Theory

1. Integrals and Fourier Transforms

Assume for this chapter that 𝕂 is a field.

Lemma 4.1.1. Let C be a finite dimensional coalgebra. Every right C-comodule M is a left C^{∗}-module by c^{∗}m = ∑ m_{(M)}⟨c^{∗}, m_{(1)}⟩ and conversely by δ(m) = ∑_{i} c_{i}^{∗}m⊗ c_{i} where ∑ c_{i}^{∗} ⊗ c_{i} is the dual basis.

Proof. We check that M becomes a left C^{∗}-module

\begin{aligned}(c^{∗}c′^{∗})m & = ∑ m_{(M)}⟨c^{∗}c′^{∗}, m_{(1)}⟩ = ∑ m_{(M)}⟨c^{∗}, m_{(1)}⟩⟨c′^{∗}, m_{(2)}⟩ \\ & = c^{∗} ∑ m_{(M)}⟨c′^{∗}, m_{(1)}⟩ = c^{∗}(c′^{∗}m). \end{aligned}

It is easy to check that the two constructions are inverses of each other. In particular assume that M is a right C-comodule. Choose m_{i} such that δ(m) = ∑ m_{i} ⊗ c_{i}. Then c_{j}^{∗}m = ∑ m_{i}⟨c_{j}^{∗}, c_{i}⟩ = m_{j} and ∑ c_{i}^{∗}m ⊗ c_{i} = ∑ m_{i} ⊗ c_{i} = δ(m).

Definition 4.1.2. 1. Let A be an algebra with augmentation ε : A ⟶ 𝕂, an algebra homomorphism. Let M be a left A-module. Then ^{A}M = {m ∈ M|am = ε(a)m} is called the space of left invariants of M.

This defines a functor ^{A}- : A − Mod ⟶ Vec.

2. Let C be a coalgebra with a group-like element 1 ∈ C. Let M be a right C-comodule. Then M^{coC} := {m ∈ M|δ(m) = m ⊗ 1} is called the space of right coinvariants of M.

This defines a functor -^{coC} : Comod-C ⟶ Vec.

Lemma 4.1.3. Let C be a finite dimensional coalgebra with a group like element 1 ∈ C. Then A := C^{∗} is an augmented algebra with augmentation ε : C^{∗} ∋ a ↦ ⟨a, 1⟩ ∈ 𝕂. Let M be a right C-comodule. Then M is a left C^{∗}-module and we have

C^{∗}M = M^{coC}.

Proof. Since 1 ∈ C is group-like we have ε_{A}(ab) = ⟨ab, 1⟩ = ⟨a, 1⟩⟨b, 1⟩ = ε_{A}(a)ε_{A}(b) and ε_{A}(1_{A}) = ⟨1_{A}, 1_{C}⟩ = ε_{C}(1_{C}) = 1.

We have m ∈ M^{coH} iff δ(m) = ∑ m_{(M)} ⊗ m_{(1)} = m ⊗ 1 iff ∑ m_{(M)}⟨a, m_{(1)}⟩ = m⟨a, 1⟩ for all a ∈ A = C^{∗} and by identifying C^{∗}⊗C = Hom(C^{∗}, C^{∗}) iff am = ε_{A}(a)m iff m ∈ ^{A}M.

Remark 4.1.4. The theory of Fourier transforms contains the following statements. Let H be the (Schwartz) space of infinitely differentiable functions f : ℝ ⟶ ℂ, such that f and all derivatives rapidly decrease at infinity. (f decreases rapidly at infinity if |x|^{m}f(x) is bounded for all m.) This space is an algebra (without unit) under the multiplication of values. There is a second multiplication on H, the convolution

(f ∗ g)(x) = (2π)^{−1/2} ∫_{−∞}^{+∞} f(t)g(x − t)dt.

The Fourier transform is a homomorphism ˆ- : H ⟶ H defined by

f̂(x) = (2π)^{−1/2} ∫_{−∞}^{+∞} f(t)e^{−itx}dt.

It satisfies the identity (f ∗ g)ˆ = f̂ĝ hence it is an algebra homomorphism. We want to find an analogue of this theory for finite quantum groups.

A similar example is the following. Let G be a locally compact topological group. Let μ be the (left) Haar measure on G and ∫ f := ∫_{G} f(x)dμ(x) be the Haar integral.

The Haar measure is left invariant in the sense that μ(E) = μ(gE) for all g ∈ G and all compact subsets E of G. The Haar measure exists and is unique up to a positive factor. The Haar integral is translation invariant i.e. for all y ∈ G we have ∫ f(yx)dμ(x) = ∫ f(x)dμ(x).

If μ is a left-invariant Haar measure then there is a continuous homomorphism mod : G ⟶ (ℝ^{+}, ·) such that ∫ f(xy^{−1})dμ(x) = mod (y) ∫ (f(x)dμ(x). The homomorphism μ does not depend on f and is called the modulus of G. The group G is called unimodular if the homomorphism mod is the identity.

If G is a compact, or discrete, or Abelian group, or a connected semisimple or nilpotent Lie group, then G is unimodular.

Let G be a quantum group (or a quantum monoid) with function algebra H an arbitrary Hopf algebra. We also use the algebra of linear functionals H^{∗} = Hom(H, 𝕂) (called the bialgebra of G in the French literature). The operation H^{∗} ⊗ H ∋ a ⊗ f ↦ ⟨a, f⟩ ∈ 𝕂 is nondegenerate on both sides. We denote the elements of H by f, g, h ∈ H, the elements of H^{∗} by a, b, c ∈ H^{∗}, the (non existing) elements of the quantum group G by x, y, z ∈ G.

Remark 4.1.5. In 2.4.8 we have seen that the dual vector space H^{∗} of a finite dimensional Hopf algebra H is again a Hopf algebra. The Hopf algebra structures are connected by the evaluation bilinear form

ev : H^{∗} ⊗ H ∋ a ⊗ f ↦ ⟨a, f⟩ ∈ 𝕂

as follows:

\begin{matrix}⟨a ⊗ b, ∑ f_{(1)} ⊗ f_{(2)}⟩ = ⟨ab, f⟩, & ⟨∑ a_{(1)} ⊗ a_{(2)}, f ⊗ g⟩ = ⟨a, fg⟩, \\ ⟨a, 1⟩ = ε(a), & ⟨1, f⟩ = ε(f), \end{matrix} ; ⟨a, S(f)⟩ = ⟨S(a), f⟩.

Definition 4.1.6. 1. The linear functionals a ∈ H^{∗} are called generalized integrals on H ([Riesz-Nagy] S.123).

2. An element ∫ ∈ H^{∗} is called a left (invariant) integral on H if

a∫ = ⟨a, 1_{H}⟩∫

or a ∫ = ε_{H^{∗}}(a) ∫ for all a ∈ H^{∗}.

3. An element δ ∈ H is called a left integral in H if

fδ = ε(f)δ

for all f ∈ H.

4. The set of left integrals in H is denoted by Int_{l}(H), the set of right integrals by Int_{r}(H). The set of left (right) integrals on H is Int_{l}(H^{∗}) (Int_{r}(H^{∗})).

5. A Hopf algebra H is called unimodular if Int_{l}(H) = Int_{r}(H).

Lemma 4.1.7. The left integrals Int_{l}(H^{∗}) form a two sided ideal of H^{∗}. If the antipode S is bijective then S induces an isomorphism S : Int_{l}(H^{∗}) ⟶ Int_{r}(H^{∗}).

Proof. For ∫ in Int_{l}(H^{∗}) we have a ∫ = ε(a) ∫ ∈ Int_{l}(H^{∗}) and a ∫ b = ε(a) ∫ b hence ∫ b ∈ Int_{l}(H^{∗}). If S is bijective then the induced map S : H^{∗} ⟶ H^{∗} is also bijective and satisfies S(∫)b = S(∫)S(S^{−1}(b)) = S(S^{−1}(b) ∫) = S(∫)ε(b) hence S(∫) ∈ Int_{r}(H^{∗}).

Remark 4.1.8. Maschke’s Theorem has an extension to finite dimensional Hopf algebras: ε(∫) ≠ 0 iff H^{∗} is semisimple.

Corollary 4.1.9. Let H be a finite dimensional Hopf algebra. Then H^{∗} is a left H^{∗}-module by the usual multiplication, hence a right H-comodule. We have

(H^{∗})^{coH} = Int_{l}(H^{∗}).

Proof. By definition we have Int_{l}(H^{∗}) = ^{H^{∗}}H^{∗}.

Example 4.1.10. Let G be a finite group. Let H := Map(G, 𝕂) be the Hopf algebra defined by the following isomorphism

𝕂^{G} = Map(G, 𝕂) ≅ Hom(𝕂G, 𝕂) = (𝕂G)^{∗}.

This isomorphism between the vector space 𝕂^{G} of all set maps from the group G to the base ring 𝕂 and the dual vector space (𝕂G)^{∗} of the group algebra 𝕂G defines the structure of a Hopf algebra on 𝕂^{G}.

We regard H := 𝕂^{G} as the function algebra on the set G. In the sense of algebraic geometry this is not quite true. The algebra 𝕂^{G} represents a functor from 𝕂-cAlg to Set that has G as value for all connected algebras A in particular for all field extensions of 𝕂.

As before we use the map ev : 𝕂G ⊗ 𝕂^{G} ⟶ 𝕂. The multiplication of 𝕂^{G} is given by pointwise multiplication of maps since ⟨x, ff′⟩ = ⟨∑ x_{(1)} ⊗ x_{(2)}, f ⊗ f′⟩ = ⟨x ⊗ x, f ⊗ f′⟩ = ⟨x, f⟩⟨x, f′⟩ for all f, f′ ∈ 𝕂^{G} and all x ∈ G. The unit element 1_{𝕂^{G}} of 𝕂^{G} is the map ε : 𝕂G ⟶ 𝕂 restricted to G, hence ε(x) = 1 = ⟨x, 1_{𝕂^{G}}⟩ for all x ∈ G. The antipode of f ∈ 𝕂^{G} is given by S(f)(x) = ⟨x, S(f)⟩ = f(x^{−1}).

The elements of the dual basis (x^{∗}|x ∈ G) with ⟨x, y^{∗}⟩ = δ_{x,y} considered as maps from G to 𝕂 form a basis of 𝕂^{G}. They satisfy the conditions

x^{∗}y^{∗} = δ_{x,y}x^{∗} and ∑_{x∈G} x^{∗} = 1_{𝕂^{G}}

since ⟨z, x^{∗}y^{∗}⟩ = ⟨z, x^{∗}⟩⟨z, y^{∗}⟩ = δ_{z,x}δ_{z,y} = δ_{x,y}⟨z, x^{∗}⟩ and ⟨z, ∑_{x∈G} x^{∗}⟩ = 1 = ⟨z, 1_{𝕂^{G}}⟩.

Hence the dual basis (x^{∗}|x ∈ G) is a decomposition of the unit into a set of minimal orthogonal idempotents and the algebra of 𝕂^{G} has the structure

𝕂^{G} = ⊕_{x∈G}𝕂x^{∗} ≅ 𝕂 × … × 𝕂.

In particular 𝕂^{G} is commutative and semisimple.

The diagonal of 𝕂^{G} is

Δ(x^{∗}) = ∑_{y∈G} y^{∗} ⊗ (y^{−1}x)^{∗} = ∑_{y,z∈G,yz=x} y^{∗} ⊗ z^{∗}

since

\begin{matrix}⟨z ⊗ u, Δ(x^{∗})⟩ = ⟨zu, x^{∗}⟩ = δ_{x,zu} = δ_{z^{−1}x,u} = ∑_{y∈G} δ_{y,z}δ_{y^{−1}x,u} \\ = ∑_{y∈G}⟨z, y^{∗}⟩⟨u, (y^{−1}x)^{∗}⟩ = ⟨z ⊗ u, ∑_{y∈G} y^{∗} ⊗ (y^{−1}x)^{∗}⟩. \end{matrix}

Let a ∈ 𝕂G. Then a defines a map ã : G ⟶ 𝕂 ∈ 𝕂^{G} by a = ∑_{x∈G} ã(x)x. For arbitrary f ∈ 𝕂^{G} and a ∈ 𝕂G this gives

⟨a, f⟩ = f(∑_{x∈G} ã(x)x) = ∑_{x∈G} ã(x)f(x).

The counit of 𝕂^{G} is given by ε(x^{∗}) = δ_{x,e} where e ∈ G is the unit element.

The antipode is, as above, S(x^{∗}) = (x^{−1})^{∗}.

We consider H = 𝕂^{G} as the function algebra on the finite group G and 𝕂G as the dual space of H = 𝕂^{G} hence as the set of distributions on H.

Then

∫ := ∑_{x∈G} x ∈ H^{∗} = 𝕂G (1)

is a (two sided) integral on H since ∑_{x∈G} yx = ∑_{x∈G} x = ε(y) ∑_{x∈G} x = ∑_{x∈G} yx. We write

∫ f(x)dx := ⟨∫ , f⟩ = ∑_{x∈G} f(x).

We have seen that there is a decomposition of the unit 1 ∈ 𝕂^{G} into a set of primitive orthogonal idempotents {x^{∗}|x ∈ G} such that every element f ∈ 𝕂^{G} has a unique representation f = ∑ f(x)x^{∗}. Since ∫ y^{∗} = ∑_{x∈G}⟨x, y^{∗}⟩ we get ∫ fy^{∗} = ∑_{x∈G}⟨x, fy^{∗}⟩ = ∑ f(x)y^{∗}(x) = f(y) hence

f = ∑(∫ f(x)y^{∗}(x)dx)y^{∗}.

Problem 4.1.1. Describe the group valued functor 𝕂-cAlg(𝕂^{G}, −) in terms of sets and their group structure.

Definition and Remark 4.1.11. Let 𝕂 be an algebraicly closed field and let G be a finite abelian group (replacing ℝ above). Assume that the characteristic of 𝕂 does not divide the order of G. Let H = 𝕂^{G}. We identify 𝕂^{G} = Hom(𝕂G, 𝕂) along the linear expansion of maps as in Example 2.1.10.

Let us consider the set Ĝ := {χ : G ⟶ 𝕂^{∗}|χ group homomorphism}. Since 𝕂^{∗} is an abelian group, the set Ĝ is an abelian group by pointwise multiplication.

The group Ĝ is called the character group of G.

Obviously the character group is a multiplicative subset of 𝕂^{G} = Hom(𝕂G, 𝕂). Actually it is a subgroup of 𝕂-cAlg(𝕂G, 𝕂) ⊆ Hom(𝕂G, 𝕂) since the elements χ ∈ Ĝ expand to algebra homomorphisms: χ(ab) = χ(∑ α_{x}x ∑ β_{y}y) = ∑ α_{x}β_{y}χ(xy) = χ(a)χ(b) and χ(1) = χ(e) = 1. Conversely an algebra homomorphism f ∈ 𝕂-cAlg(𝕂G, 𝕂) restricts to a character f : G ⟶ 𝕂^{∗}. Thus Ĝ = 𝕂-cAlg(𝕂G, 𝕂), the set of rational points of the affine algebraic group represented by 𝕂G.

There is a more general observation behind this remark.

Lemma 4.1.12. Let H be a finite dimensional Hopf algebra. Then the set Gr(H^{∗}) of group like elements of H^{∗} is equal to 𝕂-Alg(H, 𝕂).

Proof. In fact f : H ⟶ 𝕂 is an algebra homomorphism iff ⟨f ⊗ f, a ⊗ b⟩ = ⟨f, a⟩⟨f, b⟩ = ⟨f, ab⟩ = ⟨Δ(f), a ⊗ b⟩ and 1 = ⟨f, 1⟩ = ε(f).

Hence there is a Hopf algebra homomorphism φ : 𝕂Ĝ ⟶ 𝕂^{G} by 2.1.5.

Proposition 4.1.13. The Hopf algebra homomorphism φ : 𝕂Ĝ ⟶ 𝕂^{G} is bijective.

Proof. We give the proof by several lemmas.

Lemma 4.1.14. Any set of group like elements in a Hopf algebra H is linearly independent.

Proof. Assume there is a linearly dependent set {x_{0}, x_{1}, …, x_{n}} of group like elements in H. Choose such a set with n minimal. Obviously n ≥ 1 since all elements are non zero. Thus x_{0} = ∑_{i=1}^{n} α_{i}x_{i} and {x_{1}, …, x_{n}} linearly independent. We get

∑_{i,j} α_{i}α_{j}x_{i} ⊗ x_{j} = x_{0} ⊗ x_{0} = Δ(x_{0}) = ∑_{i} α_{i}x_{i} ⊗ x_{i}.

Since all α_{i} ≠ 0 and the x_{i} ⊗ x_{j} are linearly independent we get n = 1 and α_{1} = 1 so that x_{0} = x_{1}, a contradiction.

Corollary 4.1.15. (Dedekind’s Lemma) Any set of characters in 𝕂^{G} is linearly independent.

Thus φ : 𝕂Ĝ ⟶ 𝕂^{G} is injective. Now we prove that the map φ : 𝕂Ĝ ⟶ 𝕂^{G} is surjective.

Lemma 4.1.16. (Pontryagin duality) The evaluation Ĝ × G ⟶ 𝕂^{∗} is a nondegenerate bilinear map of abelian groups.

Proof. First we observe that Hom(C_{n}, 𝕂^{∗}) ≅ C_{n} for a cyclic group of order n since 𝕂 has a primitive n-th root of unity (char(𝕂)−̸ |G|).

Since the direct product and the direct sum coincide in Ab we can use the fundamental theorem for finite abelian groups G ≅ C_{n_{1}} × … × C_{n_{t}} to get Hom(G, 𝕂^{∗}) ≅ G for any abelian group G with char(𝕂)−̸ |G|. Thus Ĝ ≅ G and ˆ Ĝ = G. In particular χ(x) = 1 for all x ∈ G iff χ = 1. By the symmetry of the situation we get that the bilinear form ⟨., .⟩ : Ĝ × G ⟶ 𝕂^{∗} is non-degenerate.

Thus |Ĝ| = |G| hence dim(𝕂Ĝ) = dim(𝕂^{G}). This proves Proposition 2.1.13.

Definition 4.1.17. Let H be a Hopf algebra. A 𝕂-module M that is a right H-module by ρ : M ⊗ H ⟶ M and a right H-comodule by δ : M ⟶ M ⊗ H is called a Hopf module if the diagram

[figure: M ⊗ H; H; ρ; M ⊗ H; δ; M ⊗ H ⊗ H ⊗ H; M ⊗ H ⊗ H ⊗ H; 1⊗τ⊗1; δ⊗Δ; ρ⊗∇]

commutes, i.e. if δ(mh) = ∑ m_{(M)}h_{(1)} ⊗ m_{(1)}h_{(2)} holds for all h ∈ H and all m ∈ M.

Observe that H is an Hopf module over itself. Furthermore each module of the form V ⊗ H is a Hopf module by the induced structure. More generally there is a functor Vec ∋ V ↦ V ⊗ H ∈ Hopf-Mod-H.

Proposition 4.1.18. The two functors -^{coH} : Hopf-Mod-H ⟶ Vec and - ⊗ H : Vec ∋ V ↦ V ⊗ H ∈ Hopf-Mod-H are inverse equivalences of each other.

Proof. Define natural isomorphisms

α : M^{coH} ⊗ H ∋ m ⊗ h ↦ mh ∈ M

with inverse map

α^{−1} : M ∋ m ↦ ∑ m_{(M)}S(m_{(1)}) ⊗ m_{(2)} ∈ M^{coH} ⊗ H

and

β : V ∋ v ↦ v ⊗ 1 ∈ (V ⊗ H)^{coH}

with inverse map

(V ⊗ H)^{coH} ∋ v ⊗ h ↦ vε(h) ∈ V.

Obviously these homomorphisms are natural transformations in M and V. Furthermore α is a homomorphism of H-modules. α^{−1} is well-defined since

δ(∑ m_{(M)}S(m(1))) = ∑ m_{(M)}S(m_{(3)}) ⊗ m_{(1)}S(m_{(2)})

(since M is a Hopf module) = ∑ m_{(M)}S(m_{(2)}) ⊗ ηε(m_{(1)}) = ∑ m_{(M)}S(m_{(1)}) ⊗ 1

hence ∑ m_{(M)}S(m_{(1)}) ∈ M^{coH}. Furthermore α^{−1} is a homomorphism of comodules since

\begin{aligned}δα^{−1}(m) & = δ(∑ m_{(M)}S(m_{(1)}) ⊗ m_{(2)}) = ∑ m_{(M)}S(m_{(1)}) ⊗ m_{(2)} ⊗ m_{(3)} \\ & = ∑ α^{−1}(m_{(M)}) ⊗ m_{(1)} = (α^{−1} ⊗ 1)δ(m). \end{aligned}

Finally α and α^{−1} are inverse to each other by

αα^{−1}(m) = α(∑ m_{(M)}S(m_{(1)}) ⊗ m_{(2)}) = ∑ m_{(M)}S(m_{(1)})m_{(2)} = m

and

\begin{aligned}α^{−1}α(m ⊗ h) & = α^{−1}(mh) = ∑ m_{(M)}h_{(1)}S(m_{(1)}h_{(2)}) ⊗ m_{(2)}h_{(3)} \\ & = ∑ mh_{(1)}S(h_{(2)}) ⊗ h_{(3)} ( by δ(m) = m ⊗ 1 ) = m ⊗ h. \end{aligned}

Thus α and α^{−1} are mutually inverse homomorphisms of Hopf modules.

The image of β is in (V ⊗ H)^{coH} by δ(v ⊗ 1) = v ⊗ Δ(1) = (v ⊗ 1) ⊗ 1. Both β and β^{−1} are 𝕂-linear maps. Furthermore we have

β^{−1}β(v) = β^{−1}(v ⊗ 1) = vε(1) = v

and

\begin{aligned}ββ^{−1} & (∑ v_{i} ⊗ h_{i}) = β(∑ v_{i}ε(h_{i})) = ∑ v_{i}ε(h_{i}) ⊗ 1 = ∑ v_{i} ⊗ ε(h_{i})1 \\ & = ∑ v_{i} ⊗ ε(h_{i(1)})h_{i(2)} ( since ∑ v_{i} ⊗ h_{i} ∈ (V ⊗ H)^{coH} ) = ∑ v_{i} ⊗ h_{i}. \end{aligned}

Thus β and β^{−1} are mutually inverse homomorphisms.

Since H^{∗} = Hom(H, 𝕂) and S : H ⟶ H is an algebra antihomomorphism, the dual H^{∗} is an H-module in four different ways:

\begin{matrix}⟨(f ⇀ a), g⟩ := ⟨a, gf⟩, & ⟨(a ↼ f), g⟩ := ⟨a, fg⟩, \\ ⟨(f ⇁ a), g⟩ := ⟨a, S(f)g⟩, & ⟨(a ↽ f), g⟩ := ⟨a, gS(f)⟩. \end{matrix} (2)

If H is finite dimensional then H^{∗} is a Hopf algebra. The equality ⟨(f ⇀ a), g⟩ = ⟨a, gf⟩ = ∑⟨a_{(1)}, g⟩⟨a_{(2)}, f⟩ implies

(f ⇀ a) = ∑ a_{(1)}⟨a_{(2)}, f⟩. (3)

Analogously we have

(a ↼ f) = ∑⟨a_{(1)}, f⟩a_{(2)}. (4)

Proposition 4.1.19. Let H be a finite dimensional Hopf algebra. Then H^{∗} is a right Hopf module over H.

Proof. H^{∗} is a left H^{∗}-module by left multiplication hence by 2.1.1 a right Hcomodule by δ(a) = ∑_{i} b_{i}^{∗}a⊗b_{i}. Let f, g ∈ H and a, b ∈ H^{∗}. The (left) multiplication of H^{∗} satisfies

ab = ∑ b_{(H^{∗})}⟨a, b_{(1)}⟩.

We use the right H-module structure

(a ↽ f) = ∑ a_{(1)}⟨S(f), a_{(2)}⟩.

on H^{∗} = Hom(H, 𝕂).

Now we check the Hopf module property. Let a, b ∈ H^{∗} and f, g ∈ H. We apply H^{∗} ⊗ H to its dual H ⊗ H^{∗} and get

δ(a ↽f)(g ⊗ b) = ∑⟨(a ↽ f)_{(H^{∗})}, g⟩⟨b, (a ↽ f)_{(1)}⟩ = ⟨b(a ↽ f), g⟩
= ∑⟨b, g_{(1)}⟩⟨(a ↽ f), g_{(2)}⟩ = ∑⟨b, g_{(1)}⟩⟨a, g_{(2)}S(f)⟩

= ∑⟨b, g_{(1)}ε(f_{(2)})⟩⟨a, g_{(2)}S(f_{(1)})⟩ = ∑⟨(f_{(3)} ⇀ b), g_{(1)}S(f_{(2)})⟩⟨a, g_{(2)}S(f_{(1)})⟩

= ∑⟨(f_{(2)} ⇀ b)a, gS(f_{(1)})⟩ = ∑⟨((f_{(2)} ⇀ b)a) ↽ f_{(1)}, g⟩
= ∑⟨(a_{(H^{∗})}⟨(f_{(2)} ⇀ b), a_{(1)}⟩) ↽ f_{(1)}, g⟩
= ∑⟨(a_{(H^{∗})} ↽ f_{(1)})⟨(f_{(2)} ⇀ b), a_{(1)}⟩, g⟩ = ∑⟨(a_{(H^{∗})} ↽ f_{(1)})⟨b, a_{(1)}f_{(2)}⟩, g⟩

hence δ(a ↽ f) = ∑(a_{(H^{∗})} ↽ f_{(1)}) ⊗ a_{(1)}f_{(2)}.

Theorem 4.1.20. Let H be a finite dimensional Hopf algebra. Then the antipode S is bijective, the space of left integrals Int_{l}(H^{∗}) has dimension 1, and the homomorphism

H ∋ f ↦ (f ⇀ ∫) = ∑_{∫(1)}⟨_{∫(2)}, f⟩ ∋ H^{∗}

is bijective for any 0 ≠ ∫ ∈ Int_{l}(H^{∗}).

Proof. By Proposition 2.1.19 H^{∗} is a right Hopf module over H. By Proposition 2.1.18 there is an isomorphism α : (H^{∗})^{coH} ⊗ H ∋ a ⊗ f ↦ (a ↽ f) = (S(f) ⇀ a) ∈ H^{∗}. Since (H^{∗})^{coH} ≅ Int_{l}(H^{∗}) by 2.1.9 we get

Int_{l}(H^{∗}) ⊗ H ≅ H^{∗}

as right H-Hopf modules by the given map. This shows dim(Int_{l}(H^{∗})) = 1. So we get an isomorphism H ∋ f ↦ (∫ ↽ f) ∈ H^{∗} that is a composition of S and f ↦ (f ⇀ ∫). Since H is finite dimensional both of these maps are bijective.

If G is a finite group then every generalized integral a ∈ 𝕂G can be written with a uniquely determined g ∈ H = 𝕂^{G} as

⟨a, f⟩ = ∫ f(x)S(g)(x)dx = ∑_{x∈G} f(x)g(x^{−1}) (5)

for all f ∈ H.

If G is a finite Abelian group then each group element (rational integral) y ∈ G ⊆ 𝕂G can be written as

y = ∑_{x∈G} ∑_{χ∈Ĝ} β_{χ}⟨x^{−1}, χ⟩x

since

\begin{aligned}⟨y, f⟩ & = ⟨(∫ ↽ ∑_{χ∈Ĝ} β_{χ}χ), f⟩ = ⟨∫ , fS(∑_{χ∈Ĝ} β_{χ}χ)⟩ \\ & = ∑_{x∈G}⟨x, f⟩ ∑_{χ∈Ĝ} β_{χ}⟨x, S(χ)⟩ = ⟨∑_{x∈G} ∑_{χ∈Ĝ} β_{χ}⟨x^{−1}, χ⟩x, f⟩. \end{aligned}

In particular the matrix (⟨x^{−1}, χ⟩) is invertible.

Let H be finite dimensional. Since ⟨∫ , fg⟩ = ⟨(∫ ↼ f), g⟩ as a functional on g is a generalized integral, there is a unique ν(f) ∈ H such that

⟨∫ , fg⟩ = ⟨∫ , gν(f)⟩ (6)

or

∫ f(x)g(x)dx = ∫ g(x)ν(f)(x)dx. (7)

Although the functions f, g ∈ H of the quantum group do not commute under multiplication, there is a simple commutation rule if the product is integrated.

Proposition and Definition 4.1.21. The map ν : H ⟶ H is an algebra automorphism, called the Nakayama automorphism.

Proof. It is clear that ν is a linear map. We have ∫ fν(gh) = ∫ ghf = ∫ hfν(g) = ∫ fν(g)ν(h) hence ν(gh) = ν(g)ν(h) and ∫ fν(1) = ∫ f hence ν(1) = 1. Furthermore if ν(g) = 0 then 0 = ⟨∫ , fν(g)⟩ = ⟨∫ , gf⟩ = ⟨(f ⇀ ∫), g⟩ for all f ∈ H hence ⟨a, g⟩ = 0 for all a ∈ H^{∗} hence g = 0. So ν is injective hence bijective.

Corollary 4.1.22. The map H ∋ f ↦ (∫ ↼ f) ∈ H^{∗} is an isomorphism.

Proof. We have

(∫ ↼ f) = (ν(f) ⇀ ∫)

since ⟨(∫ ↼ f), g⟩ = ⟨∫ , fg⟩ = ⟨∫ , gν(f)⟩ = ⟨(ν(f) ⇀ ∫), g⟩. This implies the corollary.

If G is a finite group and H = 𝕂^{G} then H is commutative hence ν = id.

Definition 4.1.23. An element δ ∈ H is called a Dirac δ-function if δ is a left invariant integral in H with ⟨∫ , δ⟩ = 1, i.e. if δ satisfies

fδ = ε(f)δ  and  ∫ δ(x)dx = 1

for all f ∈ H. If H has a Dirac δ-function then we write

∫ ^{∗} a(x)dx = ∫^{∗}a := ⟨a, δ⟩. (8)

Proposition 4.1.24..

1. If H is finite dimensional then there exists a unique Dirac δ-function δ.

2. If H is infinite dimensional then there exists no Dirac δ-function.

Proof. 1. Since H ∋ f ↦ (f ⇀ ∫) ∈ H^{∗} is an isomorphism there is a δ ∈ H such that (δ ⇀ ∫) = ε. Then (fδ ⇀ ∫) = (f ⇀ (δ ⇀ ∫)) = (f ⇀ ε) = ε(f)ε = ε(f)(δ ⇀ ∫) which implies fδ = ε(f)δ. Furthermore we have ⟨∫ , δ⟩ = ⟨∫ , 1_{H}δ⟩ = ⟨(δ ⇀ ∫), 1_{H}⟩ = ε(1_{H}) = 1_{𝕂}.

2. is [Sweedler] exercise V.4.

Lemma 4.1.25. Let H be a finite dimensional Hopf algebra. Then ∫ ∈ H^{∗} is a left integral iff

a(∑ ∫_{(1)} ⊗ S(∫_{(2)})) = (∑ ∫_{(1)} ⊗ S(∫_{(2)}))a (9)

iff

∑ S(a)∫_{(1)} ⊗ ∫_{(2)} = ∑ ∫_{(1)} ⊗ a∫_{(2)} (10)

iff

∑ f_{(1)}⟨∫ , f_{(2)}⟩ = ⟨∫ , f⟩1_{H}. (11)

Proof. Let ∫ be a left integral. Then

∑ a_{(1)}∫_{(1)} ⊗ S(∫_{(2)})S(a_{(2)}) = ∑(a∫)_{(1)} ⊗ S((a∫)_{(2)}) = ε(a)(∑ ∫_{(1)} ⊗ S(∫_{(2)}))

for all a ∈ H. Hence

(∑ ∫_{(1)} ⊗ S(∫_{(2)}))a = ∑ ε(a_{(1)})(∫_{(1)} ⊗ S(∫_{(2)}))a_{(2)}
= ∑ a_{(1)}∫_{(1)} ⊗ S(∫_{(2)})S(a_{(2)})a_{(3)}
= ∑ a_{(1)}∫_{(1)} ⊗ S(∫_{(2)})ε(a_{(2)}) = a(∑ ∫_{(1)} ⊗ S(∫_{(2)})).

Conversely a(∑ ∫_{(1)} ε(S(∫_{(2)}))) = (∑ ∫_{(1)} ε(S(∫_{(2)})a)) = ε(a)(∑ ∫_{(1)} ε(S(∫_{(2)}))), hence ∫ = ∑ ∫_{(1)} ε(S(∫_{(2)})) is a left integral.

Since S is bijective the following holds

\begin{matrix}∑ S(a) ∫_{(1)} ⊗ ∫_{(2)} = ∑ S(a) ∫_{(1)} ⊗S^{−1}(S(∫_{(2)})) \\ = ∑ ∫_{(1)} ⊗S^{−1}(S(∫_{(2)})S(a)) = ∑ ∫_{(1)} ⊗a ∫_{(2)} . \end{matrix}

The converse follows easily.

If ∫ ∈ Int_{l}(H) is a left integral then ∑⟨a, f_{(1)}⟩⟨∫ , f_{(2)}⟩ = ⟨a ∫ , f⟩ = ⟨a, 1_{H}⟩⟨∫ , f⟩. Conversely if λ ∈ H^{∗} with (11) is given then ⟨aλ, f⟩ = ∑⟨a, f_{(1)}⟩⟨λ, f_{(2)}⟩ =

⟨a, 1_{H}⟩⟨λ, f⟩ hence aλ = ε(a)λ.

If G is a finite group then

δ(x) = \begin{cases}0 if x ≠ e; \\ 1 if x = e. \end{cases} (12)

In fact since δ is left invariant we get f(x)δ(x) = f(e)δ(x) for all x ∈ G and f ∈ 𝕂^{G}. Since G ⊂ H^{∗} = 𝕂G is a basis, we get δ(x) = 0 if x ≠ e. Furthermore ∫ δ(x)dx = ∑_{x∈G} δ(x) = 1 implies δ(e) = 1. So we have δ = e^{∗}.

If G is a finite Abelian group we get δ = α ∑_{χ∈Ĝ} χ for some α ∈ 𝕂. The evaluation gives 1 = ⟨∫ , δ⟩ = α ∑_{x∈G,χ∈Ĝ}⟨χ, x⟩. Now let λ ∈ Ĝ. Then ∑_{χ∈Ĝ}⟨χ, x⟩ = ∑_{χ∈Ĝ}⟨λχ, x⟩ = ⟨λ, x⟩ ∑_{χ∈Ĝ}⟨χ, x⟩. Since for each x ∈ G \ {e} there is a λ such that ⟨λ, x⟩ ≠ 1 and we get

∑_{χ∈Ĝ}⟨χ, x⟩ = |G|δ_{e,x}.

Hence ∑_{x∈G,χ∈Ĝ}⟨χ, x⟩ = |G| = α^{−1} and

δ = |G|^{−1} ∑_{χ∈Ĝ} χ. (13)

Let H be finite dimensional for the rest of this section. In Corollary 1.22 we have seen that the map H ∋ f ↦ (∫ ↼ f) ∈ H^{∗} is an isomorphism. This map will be called the Fourier transform.

Theorem 4.1.26. The Fourier transform H ∋ f ↦ f̃ ∈ H^{∗} is bijective with

f̃ = (∫ ↼ f) = ∑⟨∫_{(1)}, f⟩∫_{(2)} (14)

The inverse Fourier transform is defined by

ã = ∑ S^{−1}(δ_{(1)})⟨a, δ_{(2)}⟩. (15)

Since these maps are inverses of each other the following formulas hold

\begin{matrix}⟨f̃, g⟩ = ∫ f(x)g(x)dx & ⟨a, b̃⟩ = ^{∫ ∗}S^{−1}(a)(x)b(x)dx \\ f = ∑ S^{−1}(δ_{(1)})⟨f̃, δ_{(2)}⟩ & a = ∑⟨∫_{(1)}, ã⟩ ∫_{(2)} . \end{matrix} (16)

Proof. We use the isomorphisms H ⟶ H^{∗} defined by f̂ := f̃ = (∫ ↼ f) = ∑⟨∫_{(1)}, f⟩ ∫_{(2)} and H^{∗} ⟶ H defined by â := (a ⇀ δ) = ∑ δ_{(1)}⟨a, δ_{(2)}⟩. Because of

⟨a, b̂⟩ = ⟨a, (b ⇀ δ)⟩ = ⟨ab, δ⟩ (17)

and

⟨f̃, g⟩ = ⟨(∫ ↼ f), g⟩ = ⟨∫ , fg⟩ (18)

we get for all a ∈ H^{∗} and f ∈ H

\begin{aligned}⟨a, ˆ f̂⟩ & = ⟨af̂, δ⟩ = ∑⟨a, δ_{(1)}⟩⟨f̂, δ_{(2)}⟩ = ∑⟨a, δ_{(1)}⟩⟨∫ , fδ_{(2)}⟩ ( by Lemma 1.25 ) \\ & = ∑⟨a, S(f)δ_{(1)}⟩⟨∫ , δ_{(2)}⟩ = ⟨a, S(f)⟩⟨∫ , δ⟩ = ⟨a, S(f)⟩. \end{aligned}

This gives ˆf̂ = S(f). So the inverse map of H ⟶ H^{∗} with f̂ = (∫ ↼ f) = f̃ is H^{∗} ⟶ H with S^{−1}(â) = ∑ S^{−1}(δ_{(1)})⟨a, δ_{(2)}⟩ = ã. Then the given inversion formulas are clear.

We note for later use ⟨a, b̃⟩ = ⟨a, S^{−1}(b̂)⟩ = ⟨S^{−1}(a), b̂⟩ = ⟨S^{−1}(a)b, δ⟩.

If G is a finite group and H = 𝕂^{G} then

f̃ = ∑_{x∈G} f(x)x.

Since Δ(δ) = ∑_{x∈G} x^{−1}^{∗} ⊗ x^{∗} where the x^{∗} ∈ 𝕂^{G} are the dual basis to the x ∈ G, we get

ã = ∑_{x∈G}⟨a, x^{∗}⟩x^{∗}.

If G is a finite Abelian group then the groups G and Ĝ are isomorphic so the Fourier transform induces a linear automorphism ˜- : 𝕂^{G} ⟶ 𝕂^{G} and we have

ã = |G|^{−1} ∑_{χ∈Ĝ}⟨a, χ⟩χ^{−1}

By substituting the formulas for the integral and the Dirac δ-function (1) and (13) we get

\begin{matrix}f̃ = ∑_{x∈G} f(x)x, & ã = |G|^{−1} ∑_{χ∈Ĝ} a(χ)χ^{−1}, \\ f = |G|^{−1} ∑_{χ∈Ĝ} f̃(χ)χ^{−1}, & a = ∑_{x∈G} ã(x)x. \end{matrix} (19)

This implies

f̃(χ) = ∑_{x∈G} f(x)χ(x) = ∫ f(x)χ(x)dx (20)

with inverse transform

ã(x) = |G|^{−1} ∑_{χ∈Ĝ} χ(a)χ^{−1}(x). (21)

Corollary 4.1.27. The Fourier transforms of the left invariant integrals in H and H^{∗} are

δ̃ = εν^{−1} ∈ H^{∗}  and  ˜∫ = 1 ∈ H. (22)

Proof. We have ⟨δ̃, f⟩ = ⟨∫ , δf⟩ = ⟨∫ , ν^{−1}(f)δ⟩ = εν^{−1}(f)⟨∫ , δ⟩ = εν^{−1}(f) hence δ̃ = εν^{−1}. From ˜1 = (∫ ↼ 1) = ∫ we get ˜∫ = 1.

Proposition 4.1.28. Define a convolution multiplication on H^{∗} by

⟨a ∗ b, f⟩ := ∑⟨a, S^{−1}(δ_{(1)})f⟩⟨b, δ_{(2)}⟩.

1. INTEGRALS AND FOURIER TRANSFORMS 129

Then the following transformation rule holds for f, g ∈ H: f̃g = f̃∗ g̃. (23) In particular H^{∗} with the convolution multiplication is an associative algebra with unit ˜1_{H} = ∫, i.e. ∫ ∗ a = a ∗ ∫ = a. (24)

Proof. Given f, g, h ∈ H^{∗}. Then

⟨f̃g, h⟩ = ⟨∫ , fgh⟩ = ⟨∫ , fS^{−1}(1_{H})gh⟩⟨∫ , δ⟩
= ∑⟨∫ , fS^{−1}(δ_{(1)})gh⟩⟨∫ , δ_{(2)}⟩ = ∑⟨∫ , fS^{−1}(δ_{(1)})h⟩⟨∫ , gδ_{(2)}⟩
= ∑⟨f̃, S^{−1}(δ_{(1)})h⟩⟨g̃, δ_{(2)}⟩ = ⟨f̃∗ g̃, h⟩.

From (22) we get ˜1_{H} = ∫. So we have f̃ = ˜1f = ˜1 ∗ f̃ = ∫ ∗f̃.

If G is a finite Abelian group and a, b ∈ H^{∗} = 𝕂Ĝ. Then

(a ∗ b)(μ) = |G|^{−1} ∑_{χ,λ∈Ĝ,χλ=μ} a(λ)b(χ).

In fact we have

\begin{aligned}(a ∗ b)(μ) & = ⟨a ∗ b, μ⟩ = ∑⟨a, S^{−1}(δ_{(1)})μ⟩⟨b, δ_{(2)}⟩ \\ & = |G|^{−1} ∑_{χ∈Ĝ}⟨a, χ^{−1}μ⟩⟨b, χ⟩ = |G|^{−1} ∑_{χ,λ∈Ĝ,χλ=μ} a(λ)b(χ). \end{aligned}

One of the most important formulas for Fourier transforms is the Plancherel formula on the invariance of the inner product under Fourier transforms. We have

Theorem 4.1.29. (The Plancherel formula) ⟨a, f⟩ = ⟨f̃, ν(ã)⟩. (25)

Proof. First we have from (16)

⟨a, f⟩ = ∑⟨∫_{(1)}, ã⟩⟨∫_{(2)}, S^{−1}(δ_{(1)})⟩⟨f̃, δ_{(2)}⟩ = ∑⟨∫ , ãS^{−1}(δ_{(1)})⟩⟨f̃, δ_{(2)}⟩
= ∑⟨∫ , S^{−1}(δ_{(1)})ν(ã)⟩⟨f̃, δ_{(2)}⟩ = ∑⟨∫ , S^{−1}(S(ν(ã))δ_{(1)})⟩⟨f̃, δ_{(2)}⟩
= ∑⟨∫ , S^{−1}(δ_{(1)})⟩⟨f̃, ν(ã)δ_{(2)}⟩ = ∑⟨∫ , S^{−1}(δ)_{(2)}⟩⟨f̃, ν(ã)S(S^{−1}(δ)_{(1)})⟩
= ⟨∫ , S^{−1}(δ)⟩⟨f̃, ν(ã)⟩.

Apply this to ⟨∫ , δ⟩. Then we get

1 = ⟨∫ , δ⟩ = ⟨∫ , S^{−1}(δ)⟩⟨δ̃, ν(˜∫)⟩ = ⟨∫ , S^{−1}(δ)⟩εν^{−1}ν(1) = ⟨∫ , S^{−1}(δ)⟩. Hence we get ⟨a, f⟩ = ⟨f̃, ν(ã)⟩.

Corollary 4.1.30. If H is unimodular then ν = S^{2}.

Proof. H unimodular means that δ is left and right invariant. Thus we get

⟨a, f⟩ = ∑⟨∫_{(1)}, ã⟩⟨∫_{(2)}, S^{−1}(δ_{(1)})⟩⟨f̃, δ_{(2)}⟩
= ∑⟨∫ , ãS^{−1}(δ_{(1)})⟩⟨f̃, δ_{(2)}⟩ = ∑⟨∫ , S^{−1}(δ_{(1)}S(ã))⟩⟨f̃, δ_{(2)}⟩
= ∑⟨∫ , S^{−1}(δ_{(1)})⟩⟨f̃, δ_{(2)}S^{2}(ã)⟩ ( since δ is right invariant)
= ⟨∫ , S^{−1}(δ)⟩⟨f̃, S^{2}(ã)⟩ = ⟨f̃, S^{2}(ã)⟩.

Hence S^{2} = ν.

We also get a special representation of the inner product H^{∗} ⊗ H ⟶ 𝕂 by both integrals:

Corollary 4.1.31.

⟨a, f⟩ = ∫ ã(x)f(x)dx = ∫ ^{∗} S^{−1}(a)(x)f̃(x)dx. (26)

Proof. We have the rules for the Fourier transform. From (18) we get ⟨a, f⟩ = ⟨∫ , ãf⟩ = ∫ ã(x)f(x)dx and from (17) ⟨a, f⟩ = ⟨S^{−1}(a)f̃, δ⟩ = ∫^{∗} S^{−1}(a)(x)f̃(x)dx.

The Fourier transform leads to an interesting integral transform on H by double application.

Proposition 4.1.32. The double transform f̆ := (δ ↼ (∫ ↼ f)) defines an automorphism H ⟶ H with

f̆(y) = ∫ f(x)δ(xy)dx.

Proof. We have

⟨y, f̆⟩ = ⟨y, (δ ↼ (∫ ↼ f))⟩ = ⟨(∫ ↼ f)y, δ⟩
= ∑⟨(∫ ↼ f), δ_{(1)}⟩⟨y, δ_{(2)}⟩ = ∑⟨∫ , fδ_{(1)}⟩⟨y, δ_{(2)}⟩
= ∑⟨∫_{(1)}, f⟩⟨∫_{(2)}, δ_{(1)}⟩⟨y, δ_{(2)}⟩ = ∑⟨∫_{(1)}, f⟩⟨∫_{(2)} y, δ⟩
= ∑⟨∫_{(1)}, f⟩⟨∫_{(2)}, (y ⇀ δ)⟩ = ⟨∫ , f(y ⇀ δ)⟩
= ∫ f(x)δ(xy)dx

since ⟨x, (y ⇀ δ)⟩ = ⟨xy, δ⟩.

2. Derivations

Definition 4.2.1. Let A be a 𝕂-algebra and _{A}M_{A} be an A-A-bimodule (with identical 𝕂-action on both sides). A linear map D : A ⟶ M is called a derivation if

D(ab) = aD(b) + D(a)b.

The set of derivations Der_{𝕂}(A,_{A}M_{A}) is a 𝕂-module and a functor in _{A}M_{A}. By induction one sees that D satisfies

D(a_{1} … a_{n}) = ∑_{i=1}^{n} a_{1} … a_{i−1}D(a_{i})a_{i+1} … a_{n}.

Let A be a commutative 𝕂-algebra and _{A}M be an A-module. Consider M as an A-A-bimodule by ma := am. We denote the set of derivations from A to M by Der_{𝕂}(A, M)_{c}.

Proposition 4.2.2. 1. Let A be a 𝕂-algebra. Then the functor Der_{𝕂}(A, -) : A-Mod-A ⟶ Vec is representable by the module of differentials Ω_{A}.

2. Let A be a commutative 𝕂-algebra. Then the functor Der_{𝕂}(A, -)_{c} : A-Mod ⟶ Vec is representable by the module of commutative differentials Ω_{A}^{c}.

Proof. 1. Represent A as a quotient of a free 𝕂-algebra A := 𝕂⟨X_{i}|i ∈ J⟩/I where B = 𝕂⟨X_{i}|i ∈ J⟩ is the free algebra with generators X_{i}. We first prove the theorem for free algebras.

a) A representing module for Der_{𝕂}(B, -) is (Ω_{B}, d : B ⟶ Ω_{B}) with

Ω_{B} := B ⊗ F (dX_{i}|i ∈ J) ⊗ B

where F (dX_{i}|i ∈ J) is the free 𝕂-module on the set of formal symbols {dX_{i}|i ∈ J} as a basis.

We have to show that for every derivation D : B ⟶ M there exists a unique homomorphisms φ : Ω_{B} ⟶ M of B-B-bimodules such that the diagram

[figure: B; Ω_{B}; d; D; M; φ]

commutes. The module Ω_{B} is a B-B-bimodule in the canonical way. The products X_{1} … X_{n} of the generators X_{i} of B form a basis for B. For any product X_{1} … X_{n} we define d(X_{1} … X_{n}) := ∑_{i=1}^{n} X_{1} … X_{i−1} ⊗ dX_{i} ⊗ X_{i+1} … X_{n} in particular d(X_{i}) = 1⊗dX_{i} ⊗1. To see that d is a derivation it suffices to show this on the basis elements:

d(X_{1} … X_{k}X_{k+1} … X_{n})
= ∑_{j=1}^{k} X_{1} … X_{j−1} ⊗ dX_{j} ⊗ X_{j+1} … X_{k}X_{k+1} … X_{n}
+ ∑_{j=k+1}^{n} X_{1} … X_{k}X_{k+1} … X_{j−1} ⊗ dX_{j} ⊗ X_{j+1} … X_{n}
= d(X_{1} … X_{k})X_{k+1} … X_{n} + X_{1} … X_{k}d(X_{k+1} … X_{n})

Now let D : B ⟶ M be a derivation. Define φ by φ(1 ⊗ dX_{i} ⊗ 1) := D(X_{i}). This map obviously extends to a homomorphism of B-B-bimodules. Furthermore we have

\begin{aligned}φd(X_{1} … X_{n}) & = φ(∑_{j} X_{1} … X_{j−1} ⊗ dX_{j} ⊗ X_{j+1} … X_{n}) \\ & = ∑_{j} X_{1} … X_{j−1}φ(1 ⊗ dX_{j} ⊗ 1)X_{j+1} … X_{n} = D(X_{1} … X_{n}) \end{aligned}

hence φd = D.

To show the uniqueness of φ let ψ : Ω_{B} ⟶ M be a bimodule homomorphism such that ψd = D. Then ψ(1 ⊗ dX_{i} ⊗ 1) = ψd(X_{i}) = D(X_{i}) = φ(1 ⊗ dX_{i} ⊗ 1). Since ψ and φ are B-B-bimodules homomorphisms this extends to ψ = φ.

b) Now let A := 𝕂⟨X_{i}|i ∈ J⟩/I be an arbitrary algebra with B = 𝕂⟨X_{i}|i ∈ J⟩ free. Define

Ω_{A} := Ω_{B}/(IΩ_{B} + Ω_{B}I + Bd_{B}(I) + d_{B}(I)B).

We first show that IΩ_{B}+Ω_{B}I+Bd_{B}(I)+d_{B}(I)B is a B-B-subbimodule. Since Ω_{B} and I are B-B-bimodules the terms IΩ_{B} and Ω_{B}I are bimodules. Furthermore we have bd_{B}(i)b′ = bd_{B}(ib′) − bid_{B}(b′) ∈ Bd_{B}(I) + IΩ_{B} hence IΩ_{B} + Ω_{B}I + Bd_{B}(I) + d_{B}(I)B is a bimodule.

Now IΩ_{B} and Ω_{B}I are subbimodules of IΩ_{B} + Ω_{B}I + Bd_{B}(I) + d_{B}(I)B. Hence A = B/I acts on both sides on Ω_{A} so that Ω_{A} becomes an A-A-bimodule.

Let ν : Ω_{B} ⟶ Ω_{A} and also ν : B ⟶ A be the residue homomorphisms. Since νd_{B}(i) ∈ νd_{B}(I) = 0 ⊆ Ω_{A} we get a unique factorization map d_{A} : A ⟶ Ω_{A} such that

[figure: A; Ω_{A}; d_{A}; B; Ω_{B}; d_{B}; ν; ν]

commutes. Since d_{A}(\overline{b}) = \overline{d_{B}(b)} it is clear that d_{A} is a derivation.

Let D : A ⟶ M be a derivation. The A-A-bimodule M is also a B-B-bimodule by bm = \overline{b}m. Furthermore Dν : B ⟶ A ⟶ M is again a derivation. Let φ_{B} : Ω_{B} ⟶ M be the unique factorization map for the B-derivation Dν. Consider the following diagram

[figure: B; Ω_{B}; d_{B}; A; Ω_{A}; d_{A}; ψ; φ; M; D]

We want to construct ψ such that the diagram commutes. Let iω ∈ IΩ_{B}. Then φ(iω) = \overline{i}φ(ω) = 0 and similarly φ(ωi) = 0. Let bd_{B}(i) ∈ Bd_{B}(I) then φ(bd_{B}(i)) = \overline{b}φd_{B}(i) = \overline{b}D(\overline{i}) = 0 and similarly φ(d_{B}(i)b) = 0. Hence φ vanishes on IΩ_{B} + Ω_{B}I + Bd_{B}(I)+d_{B}(I)B and thus factorizes through a unique map ψ : Ω_{A} ⟶ M. Obviously ψ is a homomorphism of A-A-bimodules. Furthermore we have Dν = φd_{B} = ψνd_{B} = ψd_{A}ν and, since ν is surjective, D = ψd_{A}. It is clear that ψ is uniquely determined by this condition.

2. If A is commutative then we can write A = 𝕂[X_{i}|i ∈ J]/I and Ω_{B}^{c} = B ⊗ F (dX_{i}). With Ω_{A}^{c} = Ω_{B}^{c}/(IΩ_{B}^{c} + Bd_{B}(I)) the proof is analogous to the proof in the noncommutative situation.

Remark 4.2.3. 1. Ω_{A} is generated by d(A) as a bimodule, hence all elements are of the form ∑_{i} a_{i}d(a′_{i})a′′_{i}. These elements are called differentials.

2. If A = 𝕂⟨X_{i}⟩/I, then Ω_{A} is generated as a bimodule by the elements \{\overline{d(X_{i})}\}.

3. Let f ∈ B = 𝕂⟨X_{i}⟩. Let B^{op} be the algebra opposite to B (with opposite multiplication). Then Ω_{B} = B ⊗ F (dX_{i}) ⊗ B is the free B ⊗ B^{op} left module over the free generating set {d(X_{i})}. Hence d(f) has a unique representation

d(f) = ∑_{i} \frac{∂f}{∂X_{i}} d(X_{i})

with uniquely defined coefficients

\frac{∂f}{∂X_{i}} ∈ B ⊗ B^{op}.

In the commutative situation we have unique coefficients

\frac{∂f}{∂X_{i}} ∈ 𝕂[X_{i}].

4. We give the following examples for part 3:

\frac{∂X_{i}}{∂X_{j}} = δ_{ij},

\frac{∂X_{1}X_{2}}{∂X_{1}} = 1 ⊗ X_{2},

\frac{∂X_{1}X_{2}}{∂X_{2}} = X_{1} ⊗ 1,

\frac{∂X_{1}X_{2}X_{3}}{∂X_{2}} = X_{1} ⊗ X_{3},

\frac{∂X_{1}X_{3}X_{2}}{∂X_{2}} = X_{1}X_{3} ⊗ 1.

This is obtained by direct calculation or by the product rule

\frac{∂fg}{∂X_{i}} = (1 ⊗ g)\frac{∂f}{∂X_{i}} + (f ⊗ 1)\frac{∂g}{∂X_{i}} .

The product rule follows from

d(fg) = d(f)g + fd(g) = ∑((1 ⊗ g)\frac{∂f}{∂X_{i}} + (f ⊗ 1)\frac{∂g}{∂X_{i}})d(X_{i}).

Let A = 𝕂⟨X_{i}⟩/I. If f ∈ I then \overline{d(f)} = d_{A}(\overline{f}) = 0 hence

∑ \frac{∂f}{∂X_{i}} d_{A}(\overline{X_{i}}) = 0.

These are the defining relations for the A-A-bimodule Ω_{A} with the generators d_{A}(\overline{X_{i}}).

For motivation of the quantum group case we consider an affine algebraic group G with representing commutative Hopf algebra A. Recall that Hom(A, R) is an algebra with the convolution multiplication for every R ∈ 𝕂-cAlg and that G(R) = 𝕂-cAlg(A, R) ⊆ Hom(A, R) is a subgroup of the group of units of the algebra Hom(A, R).

Definition and Remark 4.2.4. A linear map T : A ⟶ A is called left translation invariant, if the following diagram functorial in R ∈ 𝕂-cAlg commutes:

[figure: G(R) × Hom(A, R); Hom(A, R); ∗; G(R) × Hom(A, R); Hom(A, R); ∗; 1⊗Hom(T,R); Hom(T,R)]

i. e. if we have

∀g ∈ G(R), ∀x ∈ Hom(A, R) : g ∗ (x ∘ T) = (g ∗ x) ∘ T.

This condition is equivalent to

Δ_{A} ∘ T = (1_{A} ⊗ T) ∘ Δ_{A}. (27)

In fact if (27) holds then g ∗ (x ∘ T) = ∇_{R}(g ⊗ x)(1_{A} ⊗ T)Δ_{A} = ∇_{R}(g ⊗ x)Δ_{A}T = (g ∗ x) ∘ T.

Conversely if the diagram commutes, then take R = A, g = 1_{A} and we get ∇_{A}(1_{A} ⊗ x)(1_{A} ⊗ T)Δ_{A} = 1_{A} ∗ (x ∘ T) = (1_{A} ∗ x) ∘ T = ∇_{A}(1_{A} ⊗ x)Δ_{A}T for all x ∈ Hom(A, A). To get (27) it suffices to show that the terms ∇_{A}(1_{A}⊗x) can be cancelled in this equation. Let ∑_{i=1}^{n} a_{i}⊗b_{i} ∈ A⊗A be given such that ∇_{A}(1_{A}⊗x)(∑ a_{i}⊗b_{i}) = 0 for all x ∈ Hom(A, A) and choose such an element with a shortest representation (n minimal). Then ∑ a_{i}x(b_{i}) = 0 for all x. Since the b_{i} are linearly independent in such a shortest representation, there are x_{i} with x_{j}(b_{i}) = δ_{ij}. Hence a_{j} = ∑ a_{i}x_{j}(b_{i}) = 0 and thus ∑ a_{i} ⊗ b_{i} = 0. From this follows (27).

Definition 4.2.5. Let H be an arbitrary Hopf algebra. An element T ∈ Hom(H, H) is called left translation invariant if it satisfies

Δ_{H}T = (1_{H} ⊗ T)Δ_{H}.

Proposition 4.2.6. Let H be an arbitrary Hopf algebra. Then Φ : H^{∗} ⟶ End(H) with Φ(f) := id ∗u_{H}f is an algebra monomorphism satisfying

Φ(f ∗ g) = Φ(f) ∘ Φ(g).

The image of Φ is precisely the set of left translation invariant elements T ∈ End(H).

Proof. For f ∈ Hom(H, 𝕂) we have u_{H}f ∈ End(H) hence id ∗u_{H}f ∈ End(H). Thus Φ is a well defined homomorphism. Observe that

Φ(f)(a) = (id_{H} ∗ u_{H}f)(a) = ∑ a_{(1)}f(a_{(2)}).

Φ is injective since it has a retraction End(H) ∋ g ↦ ε_{H} ∘g ∈ Hom(H, 𝕂). In fact we have (εΦ(f))(a) = ε(∑ a_{(1)}f(a_{(2)})) = ∑ ε(a_{(1)})f(a_{(2)}) = f(∑ ε(a_{(1)})a_{(2)}) = f(a) hence εΦ(f) = f.

The map Φ preserves the algebra unit since Φ(1_{H^{∗}}) = Φ(ε_{H}) = id_{H} ∗u_{H}ε_{H} = id_{H}. The map Φ is compatible with the multiplication: Φ(f ∗ g)(a) = ∑ a_{(1)}(f ∗

g)(a_{(2)}) = ∑ a_{(1)}f(a_{(2)})g(a_{(3)}) = ∑(id ∗u_{H}f)(a_{(1)})g(a_{(2)}) = Φ(f)(∑ a_{(1)}g(a_{(2)})) = Φ(f)Φ(g)(a) so that Φ(f ∗ g) = Φ(f) ∘ Φ(g).

For each f ∈ H^{∗} the element Φ(f) is left translation invariant since ΔΦ(f)(a) = Δ(∑ a_{(1)}f(a_{(2)})) = ∑ a_{(1)} ⊗ a_{(2)}f(a_{(3)}) = (1 ⊗ Φ(f))Δ(a).

Let T ∈ End(H) be left translation invariant then S ∗T = ∇_{H}(S ⊗1)(1⊗T)Δ_{H} = ∇_{H}(S ⊗ 1)Δ_{H}T = u_{H}ε_{H}T. Thus Φ(εT) = id ∗u_{H}ε_{H}T = id ∗S ∗ T = T, so that T is in the image of Φ.

Proposition 4.2.7. Let d ∈ Hom(H, 𝕂) and Φ(d) = D ∈ Hom(H, H) be given. The following are equivalent:

1. d : H ⟶ _{ε}𝕂_{ε} is a derivation.

2. D : H ⟶ _{H}H_{H} is a (left translation invariant) derivation. In particular Φ induces an isomorphism between the set of derivations d : H ⟶ _{ε}𝕂_{ε} and the set of left translation invariant derivations D : H ⟶ _{H}H_{H}.

Proof. Assume that 1. holds so that d satisfies d(ab) = ε(a)d(b) + d(a)ε(b). Then we get D(ab) = Φ(d)(ab) = ∑ a_{(1)}b_{(1)}d(a_{(2)}b_{(2)}) = ∑ a_{(1)}b_{(1)}ε(a_{(2)})d(b_{(2)}) + ∑ a_{(1)}b_{(1)}d(a_{(2)})ε(b_{(2)}) = aD(b) + D(a)b. Conversely assume that D(ab) = aD(b) + D(a)b. Then d(ab) = εD(ab) = ε(a)εD(b) + εD(a)ε(b) = ε(a)d(b) + d(a)ε(b).

3. The Lie Algebra of Primitive Elements

Lemma 4.3.1. Let H be a Hopf algebra and H^{o} be its Sweedler dual. If d ∈ Der_{𝕂}(H,_{ε}𝕂_{ε}) ⊆ Hom(H, 𝕂) is a derivation then d is a primitive element of H^{o}. Furthermore every primitive element d ∈ H^{o} is a derivation in Der_{𝕂}(H,_{ε}𝕂_{ε}).

Proof. Let d : H ⟶ 𝕂 be a derivation and let a, b ∈ H. Then (b ⇀ d)(a) = d(ab) = ε(a)d(b) + d(a)ε(b) = (d(b)ε + ε(b)d)(a) hence (b ⇀ d) = d(b)ε + ε(b)d. Consequently we have Hd = (H ⇀ d) ⊆ 𝕂ε + 𝕂d so that dim Hd ≤ 2 < ∞. This shows d ∈ H^{o}. Furthermore we have ⟨Δd, a ⊗ b⟩ = ⟨d, ab⟩ = d(ab) = d(a)ε(b) + ε(a)d(b) = ⟨d ⊗ ε, a ⊗ b⟩ + ⟨ε ⊗ d, a ⊗ b⟩ = ⟨1_{H^{o}} ⊗ d + d ⊗ 1_{H^{o}}, a ⊗ b⟩ hence Δ(d) = d ⊗ 1_{H^{o}} + 1_{H^{o}} ⊗ d so that d is a primitive element in H^{o}.

Conversely let d ∈ H^{o} be primitive. then d(ab) = ⟨Δ(d), a ⊗ b⟩ = d(a)ε(b) + ε(a)d(b).

Proposition and Definition 4.3.2. Let H be a Hopf algebra. The set of primitive elements of H will be denoted by Lie(H) and is a Lie algebra. If char(𝕂) = p > 0 then Lie(H) is a restricted Lie algebra or a p-Lie algebra.

Proof. Let a, b ∈ H be primitive elements. Then Δ([a, b]) = Δ(ab − ba) = (a⊗1+1⊗a)(b⊗1+1⊗b)−(b⊗1+1⊗b)(a⊗1+1⊗a) = (ab−ba)⊗1+1⊗(ab−ba) hence Lie(H) ⊆ H^{L} is a Lie algebra. If the characteristic of 𝕂 is p > 0 then we have (a ⊗ 1 + 1 ⊗ a)^{p} = a^{p} ⊗ 1 + 1 ⊗ a^{p}. Thus Lie(H) is a restricted Lie subalgebra of H^{L} with the structure maps [a, b] = ab − ba and a^{[p]} = a^{p}.

Corollary 4.3.3. Let H be a Hopf algebra. Then the set of left translation invariant derivations D : H ⟶ H is a Lie algebra under [D, D′] = DD′ − D′D. If char = p then these derivations are a restricted Lie algebra with D^{[p]} = D^{p}.

Proof. The map Ψ : H^{o} ⟶ H^{∗} ⟶^{Φ} End(H) is a homomorphism of algebras by 4.2.6. Hence Ψ(d ∗ d′ − d′ ∗ d) = Φ(d ∗ d′ − d′ ∗ d) = Φ(d)Φ(d′) − Φ(d′)Φ(d). If d is a primitive element in H^{o} then by 4.2.7 and 4.3.1 the image D := Ψ(d) in End(H) is a left translation invariant derivation and all left translation invariant derivations are of this form. Since [d, d′] = d ∗ d′ − d′ ∗ d is again primitive we get that [D, D′] = DD′ − D′D is a left translation invariant derivation so that the set of left translation invariant derivations Der_{𝕂}^{H}(H, H) is a Lie algebra resp. a restricted Lie algebra.

Definition 4.3.4. Let H be a Hopf algebra. An element c ∈ H is called cocommutative if τΔ(c) = Δ(c), i. e. if ∑ c_{(1)} ⊗ c_{(2)} = ∑ c_{(2)} ⊗ c_{(1)}. Let C(H) := {c ∈ H|c is cocommutative }.

Let G(H) denote the set of group like elements of H.

Lemma 4.3.5. Let H be a Hopf algebra. Then the set of cocommutative elements C(H) is a subalgebra of H and the group like elements G(H) form a linearly independent subset of C(H). Furthermore G(H) is a multiplicative subgroup of the group of units U(C(H)).

Proof. It is clear that C(H) is a linear subspace of H. If a, b ∈ C(H) then Δ(ab) = Δ(a)Δ(b) = (τΔ)(a)(τΔ)(b) = τ(Δ(a)Δ(b)) = τΔ(ab) and Δ(1) = 1 ⊗ 1 = τΔ(1). Thus C(H) is a subalgebra of H.

The group like elements obviously are cocommutative and form a multiplicative group, hence a subgroup of U(C(H)). They are linearly independent by Lemma 2.1.14.

Proposition 4.3.6. Let H be a Hopf algebra with S^{2} = id_{H}. Then there is a left module structure

C(H) ⊗ Lie(H) ∋ c ⊗ a ↦ c · a ∈ Lie(H)

with c · a := ∇_{H}(∇_{H} ⊗ 1)(1 ⊗ τ)(1 ⊗ S ⊗ 1)(Δ ⊗ 1)(c ⊗ a) = ∑ c_{(1)}aS(c_{(2)}) such that

c · [a, b] = ∑[c_{(1)} · a, c_{(2)} · b].

In particular G(H) acts by Lie automorphisms on Lie(H).

Proof. The given action is actually the action H ⊗ H ⟶ H with h · a = ∑ h_{(1)}aS(h_{(2)}), the so-called adjoint action.

We first show that the given map has image in Lie(H). For c ∈ C(H) and a ∈ Lie(H) we have Δ(c · a) = Δ(∑ c_{(1)}aS(c_{(2)})) = ∑ Δ(c_{(1)})(a ⊗ 1 + 1 ⊗ a)Δ(S(c_{(2)})) = ∑ Δ(c_{(1)})(a⊗1)Δ(S(c_{(2)}))+∑ Δ(c_{(2)})(1⊗a)Δ(S(c_{(1)})) = ∑ c_{(1)}aS(c_{(4)})⊗c_{(2)}S(c_{(3)})+ ∑ c_{(3)}S(c_{(2)}) ⊗ c_{(4)}aS(c_{(1)}) = c · a ⊗ 1 + 1 ⊗ c · a since c is cocommutative, S^{2} = id_{H} and a is primitive.

We show now that Lie(H) is a C(H)-module. (cd) · a = ∑ c_{(1)}d_{(1)}aS(c_{(2)}d_{(2)}) = ∑ c_{(1)}d_{(1)}aS(d_{(2)})S(c_{(2)}) = c · (d · a). Furthermore we have 1 · a = 1aS(1) = a.

To show the given formula let a, b ∈ Lie(H) and c ∈ C(H). Then c · [a, b] = ∑ c_{(1)}(ab−ba)S(c_{(2)}) = ∑ c_{(1)}aS(c_{(2)})c_{(3)}bS(c_{(4)})−∑ c_{(1)}bS(c_{(2)})c_{(3)}aS(c_{(4)}) = ∑(c_{(1)}· a)(c_{(2)}·b)−∑(c_{(1)}·b)(c_{(2)}·a) = ∑[c_{(1)}·a, c_{(2)}·b] again since c ∈ C(H) is cocommutative.

Now let g ∈ G(H). Then g · a = gaS(g) = gag^{−1} since S(g) = g^{−1} for any group like element. Furthermore g · [a, b] = [g · a, g · b] hence g defines a Lie algebra automorphism of Lie(H).

Problem 4.3.2. Show that the adjoint action H ⊗H ∋ h⊗a ↦ ∑ h_{(1)}aS(h_{(2)}) ∈ H makes H an H-module algebra.

Definition and Remark 4.3.7. The algebra 𝕂(δ) = 𝕂[δ]/(δ^{2}) is called the algebra of dual numbers. Observe that 𝕂(δ) = 𝕂 ⊕ 𝕂δ as a 𝕂-module.

We consider δ as a ”small quantity“ whose square vanishes.

The maps p : 𝕂(δ) ⟶ K with p(δ) = 0 and j : 𝕂 ⟶ 𝕂(δ) are algebra homomorphism satisfying pj = id.

Let 𝕂(δ, δ′) := 𝕂[δ, δ′]/(δ^{2}, δ′^{2}). Then 𝕂(δ, δ′) = 𝕂 ⊕ 𝕂δ ⊕ 𝕂δ′ ⊕ 𝕂δδ′. The map 𝕂(δ) ∋ δ ↦ δδ′ ∈ 𝕂(δ, δ′) is an injective algebra homomorphism. Furthermore for every α ∈ 𝕂 we have an algebra homomorphism φ_{α} : 𝕂(δ) ∋ δ ↦ αδ ∈ 𝕂(δ).

These algebra homomorphisms induce algebra homomorphisms H ⊗ 𝕂(δ) ⟶ H ⊗ 𝕂(δ) resp. H ⊗ 𝕂(δ) ⟶ H ⊗ 𝕂(δ, δ′) for every Hopf algebra H.

Proposition 4.3.8. The map

e^{δ-} : Lie(H) ⟶ H ⊗ 𝕂(δ) ⊆ H ⊗ 𝕂(δ, δ′)

with e^{δa} := 1 + a ⊗ δ = 1 + δa is called the exponential map and satisfies

e^{δ(a+b)} = e^{δa}e^{δb},
e^{δαa} = φ_{α}(e^{δa}),
e^{δδ′[a,b]} = e^{δa}e^{δ′b}(e^{δa})^{−1}(e^{δ′b})^{−1}.

Furthermore all elements e^{δa} ∈ H ⊗ 𝕂(δ) are group like in the 𝕂(δ)-Hopf algebra H ⊗ 𝕂(δ).

Proof. 1. e^{δ(a+b)} = (1 + δ(a + b)) = (1 + δa)(1 + δb) = e^{δa}e^{δb}.

2. e^{δαa} = 1 + δαa = φ_{α}(1 + δa) = φ_{α}(e^{δa}).

3. Since (1+δa)(1−δa) = 1 we have (e^{δa}) = 1−δa. So we get e^{δδ′[a,b]} = 1+δ[a, b] = 1 + δ(a − a) + δ′(b − b) + δδ′(ab − ab − ba + ab) = (1 + δa)(1 + δ′b)(1 − δa)(1 − δ′b) = e^{δa}e^{δ′b}(e^{δa})^{−1}(e^{δ′b})^{−1}.

4. Δ_{𝕂(δ)}(e^{δa}) = Δ(1 + a ⊗ δ) = 1 ⊗_{𝕂(δ)} 1 + (a ⊗ 1 + 1 ⊗ a) ⊗ δ = 1 ⊗_{𝕂(δ)} 1 + δa ⊗_{𝕂(δ)} 1 + 1 ⊗_{𝕂(δ)} δa + δa ⊗_{𝕂(δ)} δa = (1 + δa) ⊗_{𝕂(δ)} (1 + δa) = e^{δa} ⊗_{𝕂(δ)} e^{δa} and ε_{𝕂(δ)}(e^{δa}) = ε_{𝕂(δ)}(1 + δa) = 1 + δε(a) = 1.

Corollary 4.3.9. (Lie(H), e^{δ-}) is the kernel of the group homomorphism p : G_{𝕂(δ)}(H ⊗ 𝕂(δ)) ⟶ G(H).

Proof. p = 1 ⊗ p : H ⊗ 𝕂(δ) ⟶ H ⊗ 𝕂 = H is a homomorphism of 𝕂-algebras. We show that it preserves group like elements. Observe that group like elements in H ⊗ 𝕂(δ) are defined by the Hopf algebra structure over 𝕂(δ). Let g ∈ G_{𝕂(δ)}(H ⊗ 𝕂(δ)). Then (Δ_{H} ⊗ 1)(g) = g ⊗_{𝕂(δ)} g and (ε_{H} ⊗ 1)(g) = 1 ∈ 𝕂(δ).

Since p : 𝕂(δ) ⟶ 𝕂 is an algebra homomorphism the following diagram commutes

[figure: (H ⊗ 𝕂) ⊗ (H ⊗ 𝕂); H ⊗ H ⊗ 𝕂.; ≅; (H ⊗ 𝕂(δ)) ⊗_{𝕂(δ)} (H ⊗ 𝕂(δ)); H ⊗ H ⊗ 𝕂(δ); ≅; (1⊗p)⊗(1⊗p); 1⊗p]

We identify elements along the isomorphisms. Thus we get (Δ_{H} ⊗ 1_{𝕂})(1_{H} ⊗ p)(g) = (1_{H⊗H} ⊗ p)(Δ_{H} ⊗ 1_{𝕂(δ)})(g) = ((1_{H} ⊗ p) ⊗_{𝕂(δ)} (1_{H} ⊗ p))(g ⊗_{𝕂(δ)} g) = (1_{H} ⊗ p)(g) ⊗ (1_{H} ⊗ p)(g), so that 1_{H} ⊗ p : G_{𝕂(δ)}(H ⊗ 𝕂(δ)) ⟶ G(H). Now we have (1_{H} ⊗ p)(gg′) = (1_{H} ⊗ p)(g)(1_{H} ⊗ p)(g′) so that 1_{H} ⊗ p is a group homomorphism.

Now let g = g_{0} ⊗ 1 + g_{1} ⊗ δ ∈ G_{𝕂(δ)}(H ⊗ 𝕂(δ)) ⊆ H ⊗ 𝕂 ⊕ H ⊗ 𝕂δ. Then we have (1_{H} ⊗ p)(g) = 1 iff g_{0} = 1 iff g = 1_{H} ⊗ 1_{𝕂(δ)} + g_{1} ⊗ δ. Furthermore we have

Δ_{H⊗𝕂(δ)}(g) = g ⊗_{𝕂(δ)} g ⟺

1_{H} ⊗ 1_{H} ⊗ 1_{𝕂(δ)} + Δ_{H}(g_{1}) ⊗ δ = (1_{H} ⊗ 1_{𝕂(δ)} + g_{1} ⊗ δ) ⊗_{𝕂(δ)} (1_{H} ⊗ 1_{𝕂(δ)} + g_{1} ⊗ δ)

= 1_{H} ⊗ 1_{H} ⊗ 1_{𝕂(δ)} + (g_{1} ⊗ 1_{H} + 1_{H} ⊗ g_{1}) ⊗ δ ⟺

Δ_{H}(g_{1}) = g_{1} ⊗ 1_{H} + 1_{H} ⊗ g_{1}.

Similarly we have ε_{𝕂(δ)})(g) = 1 iff 1 ⊗ 1 + ε(g_{1}) ⊗ δ = 1 iff ε(g_{1}) = 0.

4. Derivations and Lie Algebras of Affine Algebraic Groups

Lemma and Definition 7.4.1. Let G : 𝕂-cAlg ⟶ Set be a group valued functor. The kernel Lie (G)(R) of the sequence

[figure: 0; Lie (G)(R); G(R(δ)); G(p); G(j); G(R); 0]

is called the Lie algebra of G and is a group valued functor in R.

Proof. For every algebra homomorphism f : R ⟶ S the following diagram of groups commutes

[figure: 0; Lie (G)(R); G(R(δ)); G(p); G(j); G(R); 0; G(f(δ)); G(f); 0; Lie (G)(S); G(S(δ)); G(p); G(j); G(S); 0]

Proposition 4.4.2. Let G : 𝕂-cAlg ⟶ Set be a group valued functor with multiplication ∗. Then there are functorial operations

G(R) × Lie (G)(R) ∋ (g, x) ↦ g · x ∈ Lie (G)(R)
R × Lie (G)(R) ∋ (a, x) ↦ ax ∈ Lie (G)(R)

such that

g · (x + y) = g · x + g · y,
h · (g · x) = (h ∗ g) · x,
a(x + y) = ax + ay,
(ab)x = a(bx),
g · (ax) = a(g · x).

Proof. First observe that the composition + on Lie (G)(R) is induced by the multiplication ∗ of G(R(δ)) so it is not necessarily commutative.

We define g·x := G(j)(g)∗x∗G(j)(g)^{−1}. Then G(p)(g·x) = G(p)G(j)(g)∗G(p)(x)∗ G(p)G(j)(g)^{−1} = g ∗ 1 ∗ g^{−1} = 1 hence g · x ∈ Lie (G)(R).

Now let a ∈ R. To define a : Lie (G)(R) ⟶ Lie (G)(R) we use u_{a} : R(δ) ⟶ R(δ) defined by u_{a}(δ) := aδ and thus u_{a}(b+cδ) := b+acδ. Obviously u_{a} is a homomorphism of R-algebras. Furthermore we have pu_{a} = p and u_{a}j = j. Thus we get a commutative diagram

[figure: 0; Lie (G)(R); G(R(δ)); G(p); G(j); G(R); 0; a; G(u_{a}); id; 0; Lie (G)(R); G(R(δ)); G(p); G(j); G(R); 0]

that defines a group homomorphism a : Lie (G)(R) ⟶ Lie (G)(R) on the kernel of the exact sequences. In particular we have then a(x + y) = ax + ay.

Furthermore we have u_{ab} = u_{a}u_{b} hence (ab)x = a(bx).

The next formula follows from g · (x + y) = G(j)(g) ∗ x ∗ y ∗ G(j)(g)^{−1} = G(j)(g) ∗ x ∗ G(j)(g)^{−1} ∗ G(j)(g) ∗ y ∗ G(j)(g)^{−1} = g · x + g · y.

We also see (h∗g)·x = G(j)(h∗g)∗x∗G(j)(h∗g)^{−1} = G(j)(h)∗G(j)(g)∗x∗G(j)(g)^{−1}∗ G(j)(h)^{−1} = h · (g · x). Finally we have g · (ax) = G(j)(g) ∗ G(u_{a})(x) ∗ G(j)(g^{−1}) = G(u_{a})(G(j)(g) ∗ x ∗ G(j)(g^{−1})) = a(g · x).

Proposition 4.4.3. Let G = 𝕂-cAlg(H, -) be an affine algebraic group. Then Lie (G)(𝕂) ≅ Lie(H^{o}) as additive groups. The isomorphism is compatible with the operations given in 4.4.2 and 4.3.6.

Proof. We consider the following diagram

[figure: 0; Lie (G)(𝕂); 𝕂-cAlg(H, 𝕂(δ)); p; 𝕂-cAlg(H, 𝕂); 0; e; ω; ≅; 0; Lie(H^{o}); G_{𝕂(δ)}(H^{o} ⊗ 𝕂(δ)); e^{δ-}; p; G(H^{o}); 0]

We know by definition that the top sequence is exact. The bottom sequence is exact by Corollary 4.3.9.

Let f ∈ 𝕂-cAlg(H, 𝕂). Since Ker(f) is an ideal of codimension 1 we get f ∈ H^{o}. The map f is an algebra homomorphism iff ⟨f, ab⟩ = ⟨f ⊗ f, a ⊗ b⟩ and ⟨f, 1⟩ = 1 iff Δ_{H^{o}}(f) = f ⊗ f and ε_{H^{o}}(f) = 1 iff f ∈ G(H^{o}). Hence we get the right hand vertical isomorphism 𝕂-cAlg(H, 𝕂) ≅ G(H^{o}).

Consider an element f ∈ 𝕂-cAlg(H, 𝕂(δ)) ⊆ Hom(H, 𝕂(δ)). It can be written as f = f_{0}+f_{1}δ with f_{0}, f_{1} ∈ Hom(H, 𝕂). The linear map f is an algebra homomorphism iff f_{0} : H ⟶ 𝕂 is an algebra homomorphism and f_{1} satisfies f_{1}(1) = 0 and f_{1}(ab) = f_{0}(a)f_{1}(b) + f_{1}(a)f_{0}(b). In fact we have f(1) = f_{0}(1) + f_{1}(1)δ = 1 iff f_{0}(1) = 1 and f_{1}(1) = 0 (by comparing coefficients). Furthermore we have f(ab) = f(a)f(b) iff f_{0}(ab) + f_{1}(ab)δ = (f_{0}(a) + f_{1}(a)δ)(f_{0}(b) + f_{1}(b)δ) = f_{0}(a)f_{0}(b) + f_{0}(a)f_{1}(b)δ + f_{1}(a)f_{0}(b)δ iff f_{0}(ab) = f_{0}(a)f_{0}(b) and f_{1}(ab) = f_{0}(a)f_{1}(b) + f_{1}(a)f_{0}(b).

Since f_{0} is an algebra homomorphism we have as above f_{0} ∈ H^{o}. For f_{1} we have (b ⇀ f_{1})(a) = f_{1}(ab) = f_{0}(a)f_{1}(b) + f_{1}(a)f_{0}(b) = (f_{1}(b)f_{0} + f_{0}(b)f_{1})(a) hence (b ⇀ f_{1}) = f_{1}(b)f_{0} +f_{0}(b)f_{1} ∈ 𝕂f_{0} +𝕂f_{1}, a two dimensional subspace. Thus f_{1} ∈ H^{o}.

In the following computations we will identify (H^{o} ⊗ 𝕂(δ)) ⊗_{𝕂(δ)}(H^{o} ⊗ 𝕂(δ)) with H^{o} ⊗ H^{o} ⊗ |K(δ).

Let f = f_{0} + f_{1}δ = f_{0} ⊗ 1 + f_{1} ⊗ δ ∈ H^{o} ⊕ H^{o}δ = H^{o} ⊗ 𝕂(δ). Then f is a homomorphism of algebras iff f(ab) = f(a)f(b) and f(1) = 1 iff f_{0}(ab) = f_{0}(a)f_{0}(b) and f_{1}(ab) = f_{0}(a)f_{1}(b)+f_{1}(a)f_{0}(b) and f_{0}(1) = 1 and f_{1}(1) = 0 iff Δ_{H^{o}}(f_{0}) = f_{0} ⊗f_{0} and Δ_{H^{o}}(f_{1}) = f_{0}⊗f_{1}+f_{1}⊗f_{0} and ε_{H^{o}}(f_{0}) = 1 and ε_{H^{o}}(f_{1}) = 0 iff (Δ_{H^{o}} ⊗id_{𝕂(δ)})(f_{0}⊗ 1+f_{1} ⊗δ) = f_{0} ⊗f_{0} ⊗1+f_{0} ⊗f_{1} ⊗δ+f_{1} ⊗f_{0} ⊗δ = (f_{0} ⊗1+f_{1} ⊗δ)⊗_{𝕂(δ)}(f_{0} ⊗1+f_{1} ⊗δ) and (ε_{H^{o}} ⊗ id_{𝕂(δ)})(f_{0} ⊗ 1 + f_{1} ⊗ δ) = 1 ⊗ 1 iff (Δ_{H^{o}} ⊗ id_{𝕂(δ)})(f) = f ⊗_{𝕂(δ)} f and (ε_{H^{o}} ⊗ id_{𝕂(δ)})(f) = 1 iff f ∈ G_{𝕂(δ)}(H^{o} ⊗ 𝕂(δ)).

Hence we have a bijective map ω : 𝕂-cAlg(H, 𝕂(δ)) ∋ f = f_{0} + f_{1}δ ↦ f_{0} ⊗ 1 + f_{1} ⊗ δ ∈ G_{𝕂(δ)}(H^{o} ⊗ 𝕂(δ)). Since the group multiplication in 𝕂-cAlg(H, 𝕂(δ)) ⊆ Hom(H, 𝕂(δ)) is the convolution ∗ and the group multiplication in G_{𝕂(δ)}(H^{o}⊗𝕂(δ)) ⊆ H^{o} ⊗ 𝕂(δ) is the ordinary algebra multiplication, where the multiplication of H^{o} again is the convolution, it is clear that ω is a group homomorphism. Furthermore the right hand square of the above diagram commutes. Thus we get an isomorphism e : Lie(H^{o}) ⟶ Lie (G)(𝕂) on the kernels. This map is defined by e(d) = 1 + dδ ∈ 𝕂-cAlg(H, 𝕂(δ)).

To show that this isomorphism is compatible with the actions of 𝕂 resp. G(H^{o}) let α ∈ 𝕂, a ∈ H, and d ∈ Lie(H^{o}). We have e(αd)(a) = ε(a) + αd(a)δ = u_{α}(ε(a) + d(a)δ) = (u_{α} ∘ (1 + dδ))(a) = (u_{α} ∘ e(d))(a) = (αe(d))(a) hence e(αd) = αe(d).

Furthermore let g ∈ G(H^{o}) = 𝕂-cAlg(H, 𝕂), a ∈ H, and d ∈ Lie(H^{o}). Then we have e(g · d)(a) = e(gdg^{−1})(a) = (1 + gdg^{−1}δ)(a) = ε(a) + gdg^{−1}(a)δ = ∑ g(a_{(1)})ε(a_{(2)})gS(a_{(3)}) + ∑ g(a_{(1)})d(a_{(2)})gS(a_{(3)})δ = ∑ g(a_{(1)})e(d)(a_{(2)})gS(a_{(3)}) = (j ∘ g ∗ e(d) ∗ j ∘ g^{−1})(a) = (g · e(d))(a) hence e(g · d) = g · e(d).

Proposition 4.4.4. Let H be a Hopf algebra and let I := Ker(ε). Then Der_{ε}(H, -) : Vec ⟶ Vec is representable by I/I^{2} and d : H ⟶^{1−ε} I ⟶^{ν} I/I^{2}, in particular

Der_{ε}(H, -) ≅ Hom(I/I^{2}, -)  and  Lie(H^{o}) ≅ Hom(I/I^{2}, 𝕂).

Proof. Because of ε(id −uε)(a) = ε(a) − εuε(a) = 0 we have Im(id −ε) ⊆ I. Let i ∈ I. Then we have i = i−ε(i) = (id −ε)(i) hence Im(id −ε) = Ker(ε). We have I^{2} ∋ (id −ε)(a)(id −ε)(b) = ab − ε(a)b − aε(b) + ε(a)ε(b) = (id −ε)(ab) − ε(a)(id −ε)(b) − (id −ε)(b). Hence we have in I/I^{2} the equation (id −ε)(ab) = ε(a)(id −ε)(b) + (id −ε)(a)ε(b) so that ν(id −ε) : H ⟶ I ⟶ I/I^{2} is an ε-derivation.

Now let D : H ⟶ M be an ε-derivation. Then D(1) = D(1 1) = 1D(1) + D(1)1 hence D(1) = 0. It follows D(a) = D(id −ε)(a). From ε(I) = 0 we get D(I^{2}) ⊆ ε(I)D(I) + D(I)ε(I) = 0 hence there is a unique factorization

[figure: H; I; id −ε; I/I^{2}; ν; D; D; M.; f]

Corollary 4.4.5. Let H be a Hopf algebra that is finitely generated a s an algebra. Then Lie(H^{o}) is finite dimensional.

Proof. Let H = 𝕂⟨a_{1}, …, a_{n}⟩. Since H = 𝕂 ⊕ I we can choose a_{1} = 1 and a_{2}, …, a_{n} ∈ I. Thus any element in i ∈ I can be written as ∑ α_{J}a_{j_{1}} … a_{j_{k}} so that I/I^{2} = 𝕂\overline{a_{2}} + … + \overline{a_{n}}. This gives the result.

Proposition 4.4.6. Let H be a commutative Hopf algebra and _{H}M be an Hmodule. Then we have Ω_{H} ≅ H ⊗ I/I^{2} and d : H ⟶ H ⊗ I/I^{2} is given by d(a) = ∑ a_{(1)} ⊗ \overline{(id −ε)(a_{(2)})}.

Proof. Consider the algebra B := H ⊕ M with (a, m)(a′, m′) = (aa′, am′ + a′m). Let G = 𝕂-cAlg(H, -). Then we have G(B) ⊆ Hom(H, B) ≅ Hom(H, H) ⊕ Hom(H, M). An element (φ, D) ∈ Hom(H, B) is in G(B) iff (φ, D)(1) = (φ(1), D(1)) = (1, 0), hence φ(1) = 1 and D(1) = 0, and (φ(ab), D(ab)) = (φ, D)(ab) = (φ, D)(a)(φ, D)(b) = (φ(a), D(a))(φ(b), D(b)) = (φ(a)φ(b), φ(a)D(b) + D(a)φ(b), hence φ(ab) = φ(a)φ(b) and D(ab) = φ(a)D(b) + D(a)φ(b). So (φ, D) is in G(B) iff φ ∈ G(H) and D is a φ-derivation. The ∗-multiplication in Hom(H, B) is given by (φ, D) ∗ (φ′, D′) = (φ ∗ φ′, φ ∗ D′ + D ∗ φ′) by applying this to an element a ∈ H. Since (φ, 0) ∈ G(B) and (uε, D) ∈ G(B) for every ε-derivation D, there is a bijection Der_{ε}(H, M) ≅ {(uε, D_{ε}) ∈ G_{ε}(B)} ≅ {(1_{H}, D_{1}) ∈ G_{1}(B)} ≅ Der_{𝕂}(H, M) by (uε, D_{ε}) ↦ (1, 0) ∗ (uε, D_{ε}) = (1, 1 ∗ D_{ε}) ∈ G_{1}(B) with inverse map (1, D_{1}) ↦ (S, 0) ∗ (1, D_{1}) = (uε, S ∗ D_{1}) ∈ G_{ε}(B). Hence we have isomorphisms Der_{𝕂}(H, M) ≅ Der_{ε}(H, M) ≅ Hom(I/I^{2}, M) ≅ Hom_{H}(H ⊗ I/I^{2}, M).

The universal ε-derivation for vector spaces is \overline{id −ε} : A ⟶ I/I^{2}. The universal ε-derivation for H-modules is D_{ε}(a) = 1 ⊗ \overline{(id −ε)(a)} ∈ A ⊗ I/I^{2}. The universal 1-derivation for H-modules is 1 ∗ D_{ε} with (1 ∗ D_{ε})(a) = ∑ a_{(1)} ⊗ \overline{(id −ε)(a_{(2)})} ∈ A ⊗ I/I^{2}.