CHAPTER 3 Hopf Algebras, Algebraic, Formal, and Quantum Groups

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3. Dual Objects

At the end of the first section in Corollary 3.1.15 we saw that the dual of an Hmodule can be constructed. We did not show the corresponding result for comodules. In fact such a construction for comodules needs some finiteness conditions. With this restriction the notion of a dual object can be introduced in an arbitrary monoidal category.

Definition 3.3.1. Let (C, ⊗) be a monoidal category M ∈ C be an object. An object M^{∗} ∈ C together with a morphism ev : M^{∗} ⊗ M ⟶ I is called a left dual for M if there exists a morphism db : I ⟶ M ⊗ M^{∗} in C such that

(M ⟶^{db ⊗1} M ⊗ M^{∗} ⊗ M ⟶^{1⊗ev} M) = 1_{M}

(M^{∗} ⟶^{1⊗db} M^{∗} ⊗ M ⊗ M^{∗} ⟶^{ev ⊗1} M^{∗}) = 1_{M^{∗}}.

A monoidal category is called left rigid if each object M ∈ C has a left dual.

Symmetrically we define: an object^{∗}M ∈ C together with a morphism ev : M⊗^{∗}M ⟶ I is called a right dual for M if there exists a morphism db : I ⟶ ^{∗}M ⊗ M in C such that

(M ⟶^{1⊗db} M ⊗^{∗}M ⊗ M ⟶^{ev ⊗1} M) = 1_{M}

(^{∗}M ⟶^{db ⊗1} ^{∗}M ⊗ M ⊗^{∗}M ⟶^{1⊗ev} ^{∗}M) = 1_{^{∗}M}.

A monoidal category is called right rigid if each object M ∈ C has a left dual. The morphisms ev and db are called the evaluation respectively the dual basis.

Remark 3.3.2. If (M^{∗}, ev) is a left dual for M then obviously (M, ev) is a right dual for M^{∗} and conversely. One uses the same morphism db : I ⟶ M ⊗ M^{∗}.

Lemma 3.3.3. Let (M^{∗}, ev) be a left dual for M. Then there is a natural isomorphism

Mor_{C}(- ⊗ M, -) ≅ Mor_{C}(-, - ⊗ M^{∗}),

i. e. the functor - ⊗ M : C ⟶ C is left adjoint to the functor - ⊗ M^{∗} : C ⟶ C.

Proof. We give the unit and the counit of the pair of adjoint functors. We define Φ(A) := 1_{A} ⊗ db : A ⟶ A ⊗ M ⊗ M^{∗} and Ψ(B) := 1_{B} ⊗ ev : B ⊗ M^{∗} ⊗ M ⟶ B. These are obviously natural transformations. We have commutative diagrams

[figure: (A ⊗ M; A ⊗ M ⊗ M^{∗} ⊗ M; FΦ(A)=; 1_{A}⊗db ⊗1_{M}; A ⊗ M) = 1_{A⊗M}; ΨF(A)=; 1_{A}⊗1_{M} ⊗ev]

and

[figure: (B ⊗ M^{∗}; B ⊗ M^{∗} ⊗ M ⊗ M^{∗}; ΦG(B)=; 1_{B}⊗1_{M^{∗}}⊗db; B ⊗ M^{∗}) = 1_{B⊗M^{∗}}; GΨ(B)=; 1_{B}⊗ev ⊗1_{M^{∗}}]

thus the Lemma has been proved by Corollary A.9.11.

The converse holds as well. If - ⊗ M is left adjoint to - ⊗ M^{∗} then the unit Φ gives a morphism db := Φ(I) : I ⟶ M ⊗ M^{∗} and the counit Ψ gives a morphism ev := Ψ(I) : M^{∗} ⊗ M ⟶ I satisfying the required properties. Thus we have

Corollary 3.3.4. If - ⊗ M : C ⟶ C is left adjoint to - ⊗ M^{∗} : C ⟶ C then M^{∗} is a left dual for M.

Corollary 3.3.5. (M^{∗}, ev) is a left dual for M if and only if there is a natural isomorphism

Mor_{C}(M^{∗} ⊗ -, -) ≅ Mor_{C}(-, M ⊗ -),

i. e. the functor M^{∗} ⊗ - : C ⟶ C is left adjoint to the functor M ⊗ - : C ⟶ C. The natural isomorphism if given by

(f : M^{∗} ⊗ N ⟶ P) ↦ ((1_{M} ⊗ f)(db ⊗1_{N}) : N ⟶ M ⊗ M^{∗} ⊗ N ⟶ M ⊗ P)

and

(g : N ⟶ M ⊗ P) ↦ ((ev ⊗1_{P})(1_{M^{∗}} ⊗ g) : M^{∗} ⊗ N ⟶ M^{∗} ⊗ M ⊗ P ⟶ P).

Proof. We have a natural isomorphism

Mor_{C}(M^{∗} ⊗ -, -) ≅ Mor_{C}(-, M ⊗ -),

iff (M, ev) is a right dual for M^{∗} (as a symmetric statement to Lemma 3.3.3) iff (M^{∗}, ev) is a left dual for M.

Corollary 3.3.6. If M has a left dual then this is unique up to isomorphism.

Proof. Let (M^{∗}, ev) and (M^{!}, ev^{!}) be left duals for M. Then the functors -⊗M^{∗} and - ⊗ M^{!} are isomorphic by Lemma A.9.7. In particular we have M^{∗} ≅ I ⊗ M^{∗} ≅ I ⊗ M^{!} ≅ M^{!}. If we consider the construction of the isomorphism then we get in particular that (ev^{!} ⊗1)(1 ⊗ db) : M^{!} ⟶ M^{!} ⊗ M ⊗ M^{∗} ⟶ M^{∗} is the given isomorphism.

Problem 3.3.1. Let (M^{∗}, ev) be a left dual for M. Then there is a unique morphism db : I ⟶ M ⊗ M^{∗} satisfying the conditions of Definition 3.3.1.

Definition 3.3.7. Let (M^{∗}, ev_{M}) and (N^{∗}, ev_{N}) be left duals for M resp. N. For each morphism f : M ⟶ N we define the transposed morphism

(f^{∗} : N^{∗} ⟶ M^{∗}) := (N^{∗} ⟶^{1⊗db_{M}} N^{∗} ⊗ M ⊗ M^{∗} ⟶^{1⊗f⊗1} N^{∗} ⊗ N ⊗ M^{∗} ⟶^{ev_{N} ⊗1} M^{∗}).

With this definition we get that the left dual is a contravariant functor, since we have

Lemma 3.3.8. Let (M^{∗}, ev_{M}), (N^{∗}, ev_{N}), and (P^{∗}, ev_{P}) be left duals for M, N and P respectively.

1. We have (1_{M})^{∗} = 1_{M^{∗}}.

2. If f : M ⟶ N and g : N ⟶ P are given then (gf)^{∗} = f^{∗}g^{∗} holds.

Proof. 1. (1_{M})^{∗} = (ev ⊗1)(1 ⊗ 1 ⊗ 1)(1 ⊗ db) = 1_{M^{∗}}.

2. The following diagram commutes

[figure: M; N ⊗ N^{∗} ⊗ M; db_{N} ⊗1; f; 1⊗1⊗f; N; N ⊗ N^{∗} ⊗ N; db_{N} ⊗1; N; 1⊗ev_{N}; g⊗1⊗1; g; P ⊗ N^{∗} ⊗ N; P; 1⊗ev_{N}]

Hence we have gf = (1 ⊗ ev_{N})(g ⊗ 1 ⊗ f)(db_{N} ⊗1). Thus the following diagram commutes

[figure: P^{∗}; P^{∗} ⊗ N ⊗ N^{∗}; 1⊗db; P^{∗} ⊗ P ⊗ N^{∗}; 1⊗g⊗1; 1⊗db; 1⊗db; 1⊗db; P^{∗} ⊗ M ⊗ M^{∗}; P^{∗} ⊗ N ⊗ N^{∗} ⊗ M ⊗ M^{∗}; 1⊗db ⊗1; P^{∗} ⊗ P ⊗ N^{∗} ⊗ M ⊗ M^{∗}; 1⊗g⊗1; 1⊗gf⊗1; P^{∗} ⊗ P ⊗ M^{∗}; P^{∗} ⊗ P ⊗ N^{∗} ⊗ N ⊗ M^{∗}; 1⊗ev ⊗1; 1⊗f⊗1; ev ⊗1; ev ⊗1; ev ⊗1; M^{∗}; N^{∗} ⊗ N ⊗ M^{∗}; 1⊗ev; N^{∗} ⊗ M ⊗ M^{∗}; 1⊗f⊗1; 1⊗g⊗1⊗f⊗1; N^{∗}; ev ⊗1; 1⊗db]

Problem 3.3.2. 1. In the category of ℕ-graded vector spaces determine all objects M that have a left dual.

2. In the category of chain complexes 𝕂-Comp determine all objects M that have a left dual.

3. In the category of cochain complexes 𝕂-Cocomp determine all objects M that have a left dual.

4. Let (M^{∗}, ev) be a left dual for M. Show that db : I ⟶ M ⊗ M^{∗} is uniquely determined by M, M^{∗}, and ev. (Uniqueness of the dual basis.)

5. Let (M^{∗}, ev) be a left dual for M. Show that ev : M^{∗} ⊗ M ⟶ I is uniquely determined by M, M^{∗}, and db.

Corollary 3.3.9. Let M, N have the left duals (M^{∗}, ev_{M}) and (N^{∗}, ev_{N}) and let f : M ⟶ N be a morphism in C. Then the following diagram commutes

[figure: N ⊗ N^{∗}; N ⊗ M^{∗}.; 1⊗f^{∗}; I; M ⊗ M^{∗}; db_{M}; db_{N}; f⊗1]

Proof. The following diagram commutes

[figure: M; N ⊗ N^{∗} ⊗ M; db ⊗1; N; N ⊗ N^{∗} ⊗ N; db ⊗1; f; 1⊗1⊗f; N; 1⊗ev; 1]

This implies (f ⊗ 1_{M^{∗}}) db_{M} = ((1_{N} ⊗ ev_{N})(1_{N} ⊗ 1_{N^{∗}} ⊗ f)(db_{N} ⊗1_{M}) ⊗ 1_{M^{∗}}) db_{M} = (1_{N} ⊗ev_{N} ⊗1_{M^{∗}})(1_{N} ⊗1_{N^{∗}} ⊗f ⊗1_{M^{∗}})(db_{N} ⊗1_{M} ⊗1_{M^{∗}}) db_{M} = (1_{N} ⊗ev_{N} ⊗1_{M^{∗}})(1_{N} ⊗ 1_{N^{∗}} ⊗ f ⊗ 1_{M^{∗}})(1_{N} ⊗ 1_{N^{∗}} ⊗ db_{M}) db_{N} = (1_{N} ⊗ (ev_{N} ⊗1_{M^{∗}})(1_{N^{∗}} ⊗ f ⊗ 1_{M^{∗}})(1_{N^{∗}} ⊗ db_{M})) db_{N} = (1_{N} ⊗ f^{∗}) db_{N}.

Corollary 3.3.10. Let M, N have the left duals (M^{∗}, ev_{M}) and (N^{∗}, ev_{N}) and let f : M ⟶ N be a morphism in C. Then the following diagram commutes

[figure: N^{∗} ⊗ N; I.; ev_{N}; N^{∗} ⊗ M; M^{∗} ⊗ M; f^{∗}⊗1; 1⊗f; ev_{M}]

Proof. This statement follows immediately from the symmetry of the definition of a left dual.

Example 3.3.11. Let M ∈ _{R}M_{R} be an R-R-bimodule. Then Hom_{R}(M., R.) is an R-R-bimodule by (rfs)(x) = rf(sx). Furthermore we have the morphism ev : Hom_{R}(M., R.) ⊗_{R} M ⟶ R defined by ev(f ⊗_{R} m) = f(m).

(Dual Basis Lemma:) A module M ∈ M_{R} is called finitely generated and projective if there are elements m_{1}, …, m_{n} ∈ M und m^{1}, …, m^{n} ∈ Hom_{R}(M., R.) such that

∀m ∈ M : ∑_{i=1}^{n} m_{i}m^{i}(m) = m.

Actually this is a consequence of the dual basis lemma. But this definition is equivalent to the usual definition.

Let M ∈ _{R}M_{R}. M as a right R-module is finitely generated and projective iff M has a left dual. The left dual is isomorphic to Hom_{R}(M., R.).

If M_{R} is finitely generated projective then we use db : R ⟶ M ⊗_{R} Hom_{R}(M., R.) with db(1) = ∑_{i=1}^{n} m_{i} ⊗_{R} m^{i}. In fact we have (1 ⊗_{R} ev)(db ⊗_{R}1)(m) = (1 ⊗_{R} ev)(∑ m_{i} ⊗_{R} m^{i} ⊗_{R} m) = ∑ m_{i}m^{i}(m) = m. We have furthermore (ev ⊗_{R}1)(1 ⊗_{R} db)(f)(m) = (ev ⊗_{R}1)(∑_{i=1}^{n} f ⊗_{R} m_{i} ⊗_{R} m^{i})(m) = ∑ f(m_{i})m^{i}(m) = f(∑ m_{i}m^{i}(m)) = f(m) for all m ∈ M hence (ev ⊗_{R}1)(1 ⊗_{R} db)(f) = f.

Conversely if M has a left dual M^{∗} then ev : M^{∗} ⊗_{R} M ⟶ R defines a homomorphism ι : M^{∗} ⟶ Hom_{R}(M., R.) in _{R}M_{R} by ι(m^{∗})(m) = ev(m^{∗} ⊗_{R} m). We define ∑_{i=1}^{n} m_{i} ⊗ m^{i} := db(1) ∈ M ⊗ M^{∗}, then m = (1 ⊗ ev)(db ⊗1)(m) = (1 ⊗ ev)(∑ m_{i} ⊗ m^{i} ⊗ m) = ∑ m_{i}ι(m^{i})(m) so that m_{1}, …, m_{n} ∈ M and ι(m^{1}), …, ι(m^{n}) ∈ Hom_{R}(M., R.) form a dual basis for M, i. e. M is finitely generated and projective as an R-module. Thus M^{∗} and Hom_{R}(M., R.) are isomorphic by the map ι.

Analogously Hom_{R}(.M, .R) is a right dual for M iff M is finitely generated and projective as a left R-module.

Problem 3.3.3. Find an example of an object M in a monoidal category C that has a left dual but no right dual.

Definition 3.3.12. Given objects M, N in C. An object [M, N] is called a left inner Hom of M and N if there is a natural isomorphism Mor_{C}(- ⊗ M, N) ≅ Mor_{C}(-, [M, N]), i. e. if it represents the functor Mor_{C}(- ⊗ M, N).

If there is an isomorphism Mor_{C}(P ⊗ M, N) ≅ Mor_{C}(P, [M, N]) natural in the three variable M, N, P then the category C is called monoidal and left closed.

If there is an isomorphism Mor_{C}(M ⊗ P, N) ≅ Mor_{C}(P, [M, N]) natural in the three variable M, N, P then the category C is called monoidal and right closed.

If M has a left dual M^{∗} in C then there are inner Homs [M, - ] defined by [M, N] := N ⊗ M^{∗}. In particular left rigid monoidal categories are left closed.

Example 3.3.13. 1. The category of finite dimensional vector spaces is a monoidal category where each object has a (left and right) dual. Hence it is (left and right) closed and rigid.

2. Let Ban be the category of (complex) Banach spaces where the morphisms satisfy ∥ f(m) ∥≤∥ m ∥ i. e. the maps are bounded by 1 or contracting. Ban is a monoidal category by the complete tensor product M ˆ⊗N. In Ban exists an inner Hom functor [M, N] that consists of the set of bounded linear maps from M to N made into a Banach space by an appropriate topology. Thus Ban is a monoidal closed category.

3. Let H be a Hopf algebra. The category H-Mod of left H-modules is a monoidal category (see Example 3.2.4 2.). Then Hom_{𝕂}(M, N) is an object in H-Mod by the multiplication

(hf)(m) := ∑ h_{(1)}f(mS(h_{(2)})

as in Proposition 3.1.14.

Hom_{𝕂}(M, N) is an inner Hom functor in the monoidal category H-Mod. The isomorphism ϕ : Hom_{𝕂}(P, Hom_{𝕂}(M, N)) ≅ Hom_{𝕂}(P ⊗ M, N) can be restricted to an isomorphism

Hom_{H}(P, Hom_{𝕂}(M, N)) ≅ Hom_{H}(P ⊗ M, N),

because ϕ(f)(h(p ⊗ m)) = ϕ(f)(∑ h_{(1)}p ⊗ h_{(2)}m) = ∑ f(h_{(1)}p)(h_{(2)}m) = ∑(h_{(1)}(f(p)))(h_{(2)}m) = ∑ h_{(1)}(f(p)(S(h_{(2)})h_{(3)}m)) = h(f(p)(m)) = h(ϕ(f) (p ⊗ m)) and conversely (h(f(p)))(m) = ∑ h_{(1)}(f(p)(S(h_{(2)})m)) = ∑ h_{(1)}(ϕ(f) (p ⊗ S(h_{(2)})m)) = ∑ ϕ(f)(h_{(1)}p ⊗ h_{(2)}S(h_{(3)})m) = ϕ(f)(hp ⊗ m) = f(hp)(m). Thus H-Mod is left closed.

If M ∈ H-Mod is a finite dimensional vector space then the dual vector space M^{∗} := Hom_{𝕂}(M, 𝕂) again is an H-module by (hf)(m) := f(S(h)m). Furthermore M^{∗} is a left dual for M with the morphisms

db : 𝕂 ∋ 1 ↦ ∑_{i} m_{i} ⊗ m^{i} ∈ M ⊗ M^{∗}

and

ev : M^{∗} ⊗ M ∋ f ⊗ m ↦ f(m) ∈ 𝕂

where m_{i} and m^{i} are a dual basis of the vector space M. Clearly we have (1⊗ev)(db ⊗1) = 1_{M} and (ev ⊗1)(1⊗db) = 1_{M^{∗}} since M is a finite dimensional vector space. We have to show that db and ev are H-module homomorphisms. We have

(h db(1))(m) = (h(∑ m_{i} ⊗ m^{i}))(m) = (∑ h_{(1)}m_{i} ⊗ h_{(2)}m^{i})(m) = ∑(h_{(1)}m_{i})((h_{(2)}m^{i})(m)) = ∑(h_{(1)}m_{i})(m^{i}(S(h_{(2)})m)) = ∑ h_{(1)}S(h_{(2)})m = ε(h)m = ε(h)(∑ m_{i} ⊗ m^{i})(m) = ε(h) db(1)(m) = db(ε(h)1)(m) = db(h1)(m),

hence h db(1) = db(h1). Furthermore we have

\begin{matrix}h ev(f ⊗ m) = hf(m) = ∑ h_{(1)}f(S(h_{(2)})h_{(3)}m) = ∑(h_{(1)}f)(h_{(2)}m) = \\ ∑ ev(h_{(1)}f ⊗ h_{(2)}m) = ev(h(f ⊗ m)). \end{matrix}

4. Let H be a Hopf algebra. Then the category of left H-comodules (see Example 3.2.4 3.) is a monoidal category. Let M ∈ H-Comod be a finite dimensional vector space. Let m_{i} be a basis for M and let the comultiplication of the comodule be δ(m_{i}) = ∑ h_{ij} ⊗ m_{j}. Then we have Δ(h_{ik}) = ∑ h_{ij} ⊗ h_{jk}. M^{∗} := Hom_{𝕂}(M, 𝕂) becomes a left H-comodule δ(m^{j}) := ∑ S(h_{ij}) ⊗ m^{i}. One verifies that M^{∗} is a left dual for M.

Lemma 3.3.14. Let M ∈ C be an object with left dual (M^{∗}, ev). Then

1. M ⊗ M^{∗} is an algebra with multiplication

∇ := 1_{M} ⊗ ev ⊗1_{M^{∗}} : M ⊗ M^{∗} ⊗ M ⊗ M^{∗} ⟶ M ⊗ M^{∗}

and unit

u := db : I ⟶ M ⊗ M^{∗};

2. M^{∗} ⊗ M is a coalgebra with comultiplication

Δ := 1_{M^{∗}} ⊗ db ⊗1_{M} : M^{∗} ⊗ M ⟶ M^{∗} ⊗ M ⊗ M^{∗} ⊗ M

and counit

ε := ev : M^{∗} ⊗ M ⟶ I.

Proof. 1. The associativity is given by (∇ ⊗ 1)∇ = (1_{M} ⊗ ev ⊗1_{M^{∗}} ⊗ 1_{M} ⊗ 1_{M^{∗}})(1_{M} ⊗ ev ⊗1_{M^{∗}}) = 1_{M} ⊗ ev ⊗ ev ⊗1_{M^{∗}} = (1_{M} ⊗ 1_{M^{∗}} ⊗ 1_{M} ⊗ ev ⊗1_{M^{∗}})(1_{M} ⊗ ev ⊗1_{M^{∗}}) = (1⊗∇)∇. The axiom for the left unit is ∇(u⊗1) = (1_{M} ⊗ev ⊗1_{M^{∗}})(db ⊗ 1_{M} ⊗ 1_{M^{∗}}) = 1_{M} ⊗ 1_{M^{∗}}.

2. is dual to the statement for algebras.

Lemma 3.3.15. 1. Let A be an algebra in C and left M ∈ C be a left rigid object with left dual (M^{∗}, ev). There is a bijection between the set of morphisms f : A ⊗ M ⟶ M making M a left A-module and the set of algebra morphisms f̃: A ⟶ M ⊗M^{∗}.

2. Let C be a coalgebra in C and left M ∈ C be a left rigid object with left dual (M^{∗}, ev). There is a bijection between the set of morphisms f : M ⟶ M ⊗ C making M a right C-comodule and the set of coalgebra morphisms f̃: M^{∗} ⊗ M ⟶ C.

Proof. 1. By Lemma 3.3.14 the object M ⊗ M^{∗} is an algebra. Given f : A ⊗ M ⟶ M such that M becomes an A-module. By Lemma 3.3.3 we associate f̃ := (f ⊗ 1)(1 ⊗ db) : A ⟶ A ⊗ M ⊗ M^{∗} ⟶ M ⊗ M^{∗}. The compatibility of f̃ with the multiplication is given by the commutative diagram

The unit axiom is given by Conversely let g : A ⟶ M ⊗ M^{∗} be an algebra homomorphism and consider g̃ := (1 ⊗ ev)(g ⊗ 1) : A ⊗ M ⟶ M ⊗ M^{∗} ⊗ M ⟶ M. Then M becomes a left A-module since

[figure: A ⊗ A; A; ∇; A ⊗ A ⊗ M ⊗ M^{∗}; A ⊗ M ⊗ M^{∗}; ∇⊗1⊗1; A ⊗ M ⊗ M^{∗}; M ⊗ M^{∗}; f⊗1; A ⊗ M ⊗ M^{∗} ⊗ M ⊗ M^{∗}; A ⊗ M ⊗ M^{∗}; 1⊗1⊗ev ⊗1; M ⊗ M^{∗} ⊗ M ⊗ M^{∗}; M ⊗ M^{∗}; 1⊗ev ⊗1; f̃⊗f̃; 1⊗f⊗1; 1⊗db ⊗1⊗1; f⊗1; f̃; 1⊗1⊗db; 1⊗f̃; f̃⊗1⊗1; f⊗1⊗1; 1; 1⊗db; 1; f⊗1]

[figure: I; M ⊗ M^{∗}; db; A; A ⊗ M ⊗ M^{∗}; 1⊗db; M ⊗ M^{∗}; f⊗1; u; u⊗1; 1⊗1]

[figure: A ⊗ A ⊗ M; A ⊗ M; ∇⊗1; A ⊗ M ⊗ M^{∗} ⊗ M; M ⊗ M^{∗} ⊗ M ⊗ M^{∗} ⊗ M; g⊗1⊗1⊗1; M ⊗ M^{∗} ⊗ M; 1⊗ev ⊗1⊗1; A ⊗ M; M; g̃; 1⊗g̃; M ⊗ M^{∗} ⊗ M; 1⊗1⊗1⊗ev; g̃; 1⊗g⊗1; 1⊗1⊗ev; g⊗g⊗1; g⊗1; 1⊗ev; 1⊗ev; g⊗1]

and

[figure: A ⊗ M; M ⊗ M^{∗} ⊗ M; g⊗1; M; 1⊗ev; M; u⊗1; db ⊗1; 1]

commute.

2. (M^{∗}, ev) is a left dual for M in the category C if and only if (M^{∗}, db) is the right dual for M in the dual category C^{op}. So if we dualize the result of part 1. we have to change sides, hence 2.