CHAPTER 4 The Infinitesimal Theory

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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}. (1)

In fact if (1) 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 (1) 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 (1).

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).