CHAPTER 1 Commutative and Noncommutative Algebraic Geometry

17

3. Quantum Monoids and their Actions on Quantum Spaces

We use the orthogonal product introduced in the previous section as “product” to define the notion of a monoid (some may call it an algebra w.r.t. the orthogonal product). Observe that on the geometric level the orthogonal product consists only of commuting points. So whenever we define a morphism on the geometric side with domain an orthogonal product of quantum spaces f : X ⊥ Y ⟶ Z then we only have to define what happens to commuting pairs of points. That makes it much easier to define such morphisms for noncommutative coordinate algebras.

We are going to define monoids in this sense and study their actions on quantum spaces.

Let E be the functor represented by 𝕂. It maps each algebra H to the one-element set {ι : 𝕂 ⟶ H}.

Definition 1.3.1. Let M be a noncommutative space and let

m : M ⊥ M ⟶ M and e : E ⟶ M

be morphisms in QS such that the diagrams

[figure: M ⊥ M; M; m; M ⊥ M ⊥ M; M ⊥ M; m ⊥ 1; 1 ⊥ m; m]

and

[figure: E ⊥ M ≅ M ≅ M ⊥ E; M ⊥ M; id⊥η; η⊥id; ∇; M ⊥ M; M; ∇; 1_{M}]

commute. Then (M, m, e) is called a quantum monoid.

Proposition 1.3.2. Let M be a noncommutative space with function algebra H. Then H is a bialgebra if and only if M is a quantum monoid.

Proof. Since the functors M ⊥ M, M ⊥ E and E ⊥ M are represented by H ⊗ H resp. H ⊗ 𝕂 ≅ H resp. 𝕂 ⊗ H ≅ H the Yoneda Lemma defines a bijection between the morphisms m : M ⊥ M ⟶ M and the algebra homomorphisms Δ : H ⟶ H ⊗ H and similarly a bijection between the morphisms e : E ⟶ M and the algebra homomorphisms ε : H ⟶ 𝕂. Again by the Yoneda Lemma the bialgebra diagrams in 𝕂-Alg commute if and only if the corresponding diagrams for a quantum monoid commute.

Observe that a similar result cannot be formulated for Hopf algebras H since neither the antipode S nor the multiplication ∇ : H ⊗ H ⟶ H are algebra homomorphisms. In contrast to affine algebraic groups (2.3.2) Hopf algebras in the category 𝕂-Alg^{op} ≅ QR are not groups. Nevertheless, one defines

Definition 1.3.3. A functor defined on the category of 𝕂-algebras and represented by a Hopf algebra H is called a quantum group.

Definition 1.3.4. Let X be a noncommutative space and let M be a quantum monoid. A morphism (a natural transformation) of quantum spaces ρ : M ⊥ X ⟶ X is called an operation of M on X if the diagrams

[figure: M ⊥ X; X; ρ; M ⊥ M ⊥ X; M ⊥ X; m ⊥ 1; 1 ⊥ ρ; ρ]

and

[figure: X ≅ E ⊥ X; M ⊥ X; η⊥id; X; ρ; id_{X}]

commute. We call X a noncommutative M-space.

Proposition 1.3.5. Let X be a noncommutative space with function algebra A = O(X). Let M be a quantum monoid with function algebra B = O(M). Let ρ : M ⊥ X ⟶ X be a morphism in QS and let f : A ⟶ B ⊗ A be the associated homomorphism of algebras. Then the following are equivalent

1. (X , M, ρ) is an operation of the quantum monoid M on the noncommutative space X;

2. (A, H, f) define an H-comodule algebra.

Proof. The homomorphisms of algebras Δ ⊗ 1_{A}, 1_{B} ⊗ f, ϵ ⊗ 1_{A} etc. represent the morphisms of quantum spaces m ⊥ id, id ⊥ ρ, η ⊥ id etc. Hence the required diagrams are transferred by the Yoneda Lemma.

Example 1.3.6. 1. The quantum monoid of “quantum matrices”:

We consider the algebra

M_{q}(2) := 𝕂⟨a, b, c, d⟩/I = 𝕂 ⟨\begin{matrix}a & b \\ c & d \end{matrix}⟩ /I

where the two-sided ideal I is generated by the elements

ab − q^{−1}ba, ac − q^{−1}ca, bd − q^{−1}db, cd − q^{−1}dc, ad − da − (q^{−1} − q)bc, bc − cb.

The quantum space M_{q}(2) associated with the algebra M_{q}(2) is given by

\begin{matrix}M_{q}(2)(A) & = 𝕂-Alg(M_{q}(2), A) \\ = \{(\begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix}) |a′, b′, c′, d′ ∈ A; a′b′ = q^{−1}b′a′, …, b′c′ = c′b′\} \end{matrix}

where each homomorphism of algebras f : M_{q}(2) ⟶ A is described by the quadruple (a′, b′, c′, d′) of images of the algebra generators a, b, c, d. The images must satisfy the same relations that generate the two-sided ideal I hence

\begin{matrix}a′b′ = q^{−1}b′a′, a′c′ = q^{−1}c′a′, b′d′ = q^{−1}d′b′, c′d′ = q^{−1}d′c′, \\ b′c′ = c′b′, a′d′ − q^{−1}b′c′ = d′a′ − qc′b′. \end{matrix}

We write these quadruples as 2 × 2-matrices and call them quantum matrices. The unusual commutation relations are chosen so that the following examples work.

The quantum space of quantum matrices turns out to be a quantum monoid. We give both the algebraic (with function algebras) and the geometric (with quantum spaces) approach to define the multiplication.

a) The algebraic approach:

The algebra M_{q}(2) is a bialgebra with the diagonal

Δ (\begin{matrix}a & b \\ c & d \end{matrix}) = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}) ,

i.e. by Δ(a) = a ⊗ a + b ⊗ c, Δ(b) = a ⊗ b + b ⊗ d, Δ(c) = c ⊗ a + d ⊗ c and Δ(d) = c ⊗ b + d ⊗ d, and with the counit

ε (\begin{matrix}a & b \\ c & d \end{matrix}) = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) ,

i.e. ε(a) = 1, ε(b) = 0, ε(c) = 0, and ε(d) = 1. We have to prove that Δ and ε are homomorphisms of algebras and that the coalgebra laws are satisfied. To obtain a homomorphism of algebras Δ : M_{q}(2) ⟶ M_{q}(2)⊗M_{q}(2) we define Δ : 𝕂⟨a, b, c, d⟩ ⟶ M_{q}(2) ⊗ M_{q}(2) on the free algebra (the polynomial ring in noncommuting variables) 𝕂⟨a, b, c, d⟩ generated by the set {a, b, c, d} and show that it vanishes on the ideal I or more simply on the generators of the ideal. Then it factors through a unique homomorphism of algebras Δ : M_{q}(2) ⟶ M_{q}(2) ⊗ M_{q}(2). We check this only for one generator of the ideal I:

Δ(ab − q^{−1}ba) = Δ(a)Δ(b) − q^{−1}Δ(b)Δ(a) = = (a ⊗ a + b ⊗ c)(a ⊗ b + b ⊗ d) − q^{−1}(a ⊗ b + b ⊗ d)(a ⊗ a + b ⊗ c) = aa ⊗ ab + ab ⊗ ad + ba ⊗ cb + bb ⊗ cd − q^{−1}(aa ⊗ ba + ab ⊗ bc + ba ⊗ da + bb ⊗ dc) = aa ⊗ (ab − q^{−1}ba) + ab ⊗ (ad − q^{−1}bc) + ba ⊗ (cb − q^{−1}da) + bb ⊗ (cd − q^{−1}dc) = ba ⊗ (q^{−1}ad − q^{−2}bc + cb − q^{−1}da) ≡ 0 mod (I).

The reader should check the other identities.

The coassociativity follows from

\begin{matrix}(Δ ⊗ 1)Δ (\begin{matrix}a & b \\ c & d \end{matrix}) = Δ (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}) = ((\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix})) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}) = \\ = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ ((\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix})) = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ Δ (\begin{matrix}a & b \\ c & d \end{matrix}) = (1 ⊗ Δ)Δ (\begin{matrix}a & b \\ c & d \end{matrix}) . \end{matrix}

The reader should check the properties of the counit.

b) The geometric approach:

M_{q}(2) has a rather remarkable (and actually well known) comultiplication that is better understood by using the induced multiplication of commuting points. Given

two commuting quantum matrices (\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) and (\begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix}) in M_{q}(2)(A). Then their

matrix product

(\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) (\begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix}) = (\begin{matrix}a_{1}a_{2} + b_{1}c_{2} & a_{1}b_{2} + b_{1}d_{2} \\ c_{1}a_{2} + d_{1}c_{2} & c_{1}b_{2} + d_{1}d_{2} \end{matrix})

is again a quantum matrix. To prove this we only check one of the relations

(a_{1}a_{2} + b_{1}c_{2})(a_{1}b_{2} + b_{1}d_{2}) = a_{1}a_{2}a_{1}b_{2} + a_{1}a_{2}b_{1}d_{2} + b_{1}c_{2}a_{1}b_{2} + b_{1}c_{2}b_{1}d_{2}
= a_{1}a_{1}a_{2}b_{2} + a_{1}b_{1}a_{2}d_{2} + b_{1}a_{1}c_{2}b_{2} + b_{1}b_{1}c_{2}d_{2}

= q^{−1}a_{1}a_{1}b_{2}a_{2} + q^{−1}b_{1}a_{1}(d_{2}a_{2} + (q^{−1} − q)b_{2}c_{2}) + b_{1}a_{1}b_{2}c_{2} + q^{−1}b_{1}b_{1}d_{2}c_{2}

= q^{−1}(a_{1}a_{1}b_{2}a_{2} + a_{1}b_{1}b_{2}c_{2} + b_{1}a_{1}d_{2}a_{2} + b_{1}b_{1}d_{2}c_{2})
= q^{−1}(a_{1}b_{2}a_{1}a_{2} + a_{1}b_{2}b_{1}c_{2} + b_{1}d_{2}a_{1}a_{2} + b_{1}d_{2}b_{1}c_{2})
= q^{−1}(a_{1}b_{2} + b_{1}d_{2})(a_{1}a_{2} + b_{1}c_{2})

We have used that the two points are commuting points. This multiplication obviously is a natural transformation M_{q}(2) ⊥ M_{q}(2)(A) ⟶ M_{q}(2)(A) (natural in A). It is

associative and has unit (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}). For the associativity observe that by 1.2.14

(((\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) , (\begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix})), (\begin{matrix}a_{3} & b_{3} \\ c_{3} & d_{3} \end{matrix}))

is a pair of commuting points if and only if

((\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) , ((\begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix}) , (\begin{matrix}a_{3} & b_{3} \\ c_{3} & d_{3} \end{matrix})))

is a pair of commuting points.

Since (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}a & b \\ c & d \end{matrix}) = (\begin{matrix}a & b \\ c & d \end{matrix}) = (\begin{matrix}a & b \\ c & d \end{matrix}) (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) for all quantum matrices

(\begin{matrix}a & b \\ c & d \end{matrix}) ∈ M_{q}(2)(B) we see that M_{q}(2) is a quantum monoid.

It remains to show that the multiplication of M_{q}(2) and the comultiplication of M_{q}(2) correspond to each other by the Yoneda Lemma. The identity morphism of

22 1. COMMUTATIVE AND NONCOMMUTATIVE ALGEBRAIC GEOMETRY

M_{q}(2) ⊗ M_{q}(2) is given by the pair of commuting points (ι_{1}, ι_{2}) ∈ M_{q}(2) ⊥ M_{q}(2)(M_{q}(2) ⊗ M_{q}(2)) = 𝕂-Alg(M_{q}(2) ⊗ M_{q}(2), M_{q}(2) ⊗ M_{q}(2)).

Since ι_{1} = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ 1 = (\begin{matrix}a ⊗ 1 & b ⊗ 1 \\ c ⊗ 1 & d ⊗ 1 \end{matrix}) and ι_{2} = 1 ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}) = (\begin{matrix}1 ⊗ a & 1 ⊗ b \\ 1 ⊗ c & 1 ⊗ d \end{matrix})

we have id = (ι_{1}, ι_{2}) = ((\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ 1, 1 ⊗ (\begin{matrix}a & b \\ c & d \end{matrix})). The Yoneda Lemma defines the diagonal as the image of the identity under 𝕂-Alg(M_{q}(2)⊗M_{q}(2), M_{q}(2)⊗M_{q}(2)) ⟶ 𝕂-Alg(M_{q}(2), M_{q}(2) ⊗ M_{q}(2)) by the multiplication. So Δ((\begin{matrix}a & b \\ c & d \end{matrix})) = Δ = ι_{1} ∗ ι_{2} =

((\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ 1) ∗ (1 ⊗ (\begin{matrix}a & b \\ c & d \end{matrix})) = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}).

Thus M_{q}(2) defines a quantum monoid M_{q}(2) with

M_{q}(2)(B) = \{(\begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix})|a′, b′, c′, d′ ∈ B; a′b′ = q^{−1}b′a′, …, b′c′ = c′b′ \} .

This is the deformed version of M_{2}^{×} the multiplicative monoid of the 2 × 2-matrices of commutative algebras.

2. Let A_{q}^{2|0} = 𝕂⟨x, y⟩/(xy − q^{−1}yx) be the function algebra of the quantum plane 𝔸_{q}^{2|0}. By the definition 1.2.5 we have

𝔸_{q}^{2|0}(A′) = \{(\begin{matrix}x \\ y \end{matrix})|x, y ∈ A′; xy = q^{−1}yx\} .

The set

M_{q}(2)(A′) = \{(\begin{matrix}u & x \\ y & z \end{matrix})|u, x, y, z ∈ A′; ux = q^{−1}xu, …, xy = yx\}

operates on this quantum plane by matrix multiplication

M_{q}(2)(A′) ⊥ 𝔸_{q}^{2|0}(A′) ∋ ((\begin{matrix}a & b \\ c & d \end{matrix}) , (\begin{matrix}x \\ y \end{matrix})) ↦ (\begin{matrix}a & b \\ c & d \end{matrix}) · (\begin{matrix}x \\ y \end{matrix}) ∈ 𝔸_{q}^{2|0}(A′).

Again one should check that the required equations are preserved. Since we have a matrix multiplication we get an operation as in the preceding proposition. In particular A_{q}^{2|0} is a M_{q}(2)-comodule algebra.

As in example 1. we get the comultiplication as δ((\begin{matrix}x \\ y \end{matrix})) = δ = ((\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ 1) ∗

(1 ⊗ (\begin{matrix}x \\ y \end{matrix})) = (\begin{matrix}a & b \\ c & d \end{matrix}) ⊗ (\begin{matrix}x \\ y \end{matrix}) .

3. Let A_{q}^{0|2} = 𝕂⟨ξ, η⟩/(ξ^{2}, η^{2}, ξη+qηξ) be the function algebra of the dual quantum plane 𝔸_{q}^{0|2}. By the definition 1.2.5 we have

𝔸_{q}^{0|2}(A′) = {(a′ b′)|a′, b′ ∈ A′; a′^{2} = 0, b′^{2} = 0, a′b′ = −qb′a′ } .

The quantum monoid M_{q}(2) also operates on the dual quantum plane by matrix multiplication

𝔸_{q}^{0|2}(A′) ⊥ M_{q}(2)(A′) ∋ ((ξ η) , (\begin{matrix}a & b \\ c & d \end{matrix})) ↦ (ξ η) · (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ 𝔸_{q}^{0|2}(A′).

This gives another example of a M_{q}(2)-comodule algebra A_{q}^{0|2} ⟶ A_{q}^{0|2} ⊗ M_{q}(2) with

δ((ξ η)) = δ = ((ξ η) ⊗ 1) ∗ (1 ⊗ (\begin{matrix}a & b \\ c & d \end{matrix})) = (ξ η) ⊗ (\begin{matrix}a & b \\ c & d \end{matrix}) .

What is now the reason for the remarkable relations on M_{q}(2)? It is based on a fact that we will show later namely that M_{q}(2) is the universal quantum monoid acting on the quantum plane 𝔸_{q}^{2|0} from the left and on the dual quantum plane 𝔸_{q}^{0|2} from the right. This however happens in the category of quantum planes represented by quadratic algebras. Here we will show a simpler theorem for finite dimensional algebras.

Problem 1.3.1. Determine the ℍ-points of the quantum plane 𝔸_{q}^{2|0} where ℍ is the ℝ-algebra of the quaternions.

Definition 1.3.7. 1. Let X be a quantum space. A quantum space M(X) together with a morphism of quantum spaces μ : M(X) ⊥ X ⟶ X is called a quantum space acting universally on X (or simply a universal quantum space for X) if for every quantum space Y and every morphism of quantum spaces f : Y ⊥ X ⟶ X there is a unique morphism of quantum spaces g : Y ⟶ M(X) such that the following diagram commutes

[figure: M(X) ⊥ X; X .; μ; f; Y ⊥ X; g⊥1_{X}]

2. Let A be a 𝕂-algebra. A 𝕂-algebra M(A) together with a homomorphism of algebras δ : A ⟶ M(A) ⊗ A is called an algebra coacting universally on A (or simply a universal algebra for A) if for every 𝕂-algebra B and every homomorphism of 𝕂-algebras f : A ⟶ B ⊗ A there exists a unique homomorphism of algebras g : M(A) ⟶ B such that the following diagram commutes

[figure: A; M(A) ⊗ A; δ; f; B ⊗ A; g⊗1_{A}]

By the universal properties the universal algebra M(A) for A and the universal quantum space M(X) for X are unique up to isomorphism.

Proposition 1.3.8. 1. Let A be a 𝕂-algebra with universal algebra M(A) and δ : A ⟶ M(A) ⊗ A. Then M(A) is a bialgebra and A is an M(A)-comodule algebra by δ.

2. If B is a bialgebra and if f : A ⟶ B ⊗ A defines the structure of a B-comodule algebra on A then there is a unique homomorphism g : M(A) ⟶ B of bialgebras such that the following diagram commutes

[figure: A; M(A) ⊗ A; δ; f; B ⊗ A; g ⊗ 1_{A}]

The corresponding statement for quantum spaces and quantum monoids is the following.

Proposition 1.3.9. 1. Let X be a quantum space with universal quantum space M(X) and μ : M(X) ⊥ A ⟶ A. Then M(X) is a quantum monoid and X is an M(X)-space by μ.

2. If Y is another quantum monoid and if f : Y ⊥ X ⟶ X defines the structure of a Y-space on X then there is a unique morphism of quantum monoids g : Y ⟶ M(X) such that the following diagram commutes

[figure: M(X) ⊥ X; X .; μ; f; Y ⊥ X; g⊥1_{X}]

Proof. We give the proof for the algebra version of the proposition. Consider the following commutative diagram

[figure: M(A) ⊗ A; M(A) ⊗ M(A) ⊗ A; 1_{M(A)}⊗δ; A; M(A) ⊗ A; δ; δ; Δ⊗1_{A}]

where the morphism of algebras Δ is defined by the universal property of M(A) with respect to the algebra morphism (1_{M(A)} ⊗ δ)δ. Furthermore there is a unique morphism of algebras ϵ : M(A) ⟶ 𝕂 such that

[figure: A; M(A) ⊗ A; δ; 1_{A}; A ≅ 𝕂 ⊗ A; ϵ⊗1_{A}]

commutes.

The coalgebra axioms arise from the following commutative diagrams

[figure: A; M(A) ⊗ A; δ; δ; Δ⊗1_{A}; M(A) ⊗ A; M(A) ⊗ M(A) ⊗ A; 1_{M(A)}⊗δ; Δ⊗1_{A}; 1_{M(A)}⊗δ; Δ⊗1_{M(A)}⊗1_{A}; 1_{M(A)}⊗Δ⊗1_{A}; M(A) ⊗ M(A) ⊗ A; M(A) ⊗ M(A) ⊗ M(A) ⊗ A; 1_{M(A)}⊗1_{M(A)}⊗δ]

and

[figure: A; M(A) ⊗ A; δ; δ; Δ⊗1_{A}; 1_{M(A)}⊗1_{A}; M(A) ⊗ A; M(A) ⊗ M(A) ⊗ A; 1_{M(A)}⊗δ; 1_{M(A)}⊗1_{A}; M(A) ⊗ A ≅ M(A) ⊗ 𝕂 ⊗ A; 1_{M(A)}⊗ϵ⊗1_{A}]

and

[figure: A; M(A) ⊗ A; δ; 1_{A}; δ; Δ⊗1_{A}; M(A) ⊗ A; M(A) ⊗ M(A) ⊗ A; 1_{M(A)}⊗δ; ϵ⊗1_{A}; ϵ⊗1_{M(A)}⊗1_{A}; A; M(A) ⊗ A ≅ 𝕂 ⊗ M(A) ⊗ A.; δ]

In fact these diagrams imply by the uniqueness of the induced homomorphisms of algebras (Δ ⊗ 1_{M(A)})Δ = (1_{M(A)} ⊗ Δ)Δ, (1_{M(A)} ⊗ ϵ)Δ = 1_{M(A)} and ϵ ⊗ (1_{M(A)})Δ = 1_{M(A)}. Finally A is an M(A)-comodule algebra by the definition of Δ and ϵ.

Now assume that a structure of a B-comodule algebra on A is given by a bialgebra B and f : A ⟶ B⊗A. Then there is a unique homomorphism of algebras g : M(A) ⟶ B such that the diagram

[figure: A; M(A) ⊗ A; δ; f; B ⊗ A; g⊗1_{A}]

commutes. Then the following diagram

[figure: A; M(A) ⊗ A; δ; M(A) ⊗ A; M(A) ⊗ M(A) ⊗ A; Δ⊗1_{A}; 1_{M(A)}⊗δ; f; g⊗1_{A}; g⊗g⊗1_{A}; B ⊗ A; B ⊗ B ⊗ A; Δ_{B}⊗1_{A}; 1_{B}⊗f]

implies ((g ⊗ g)Δ ⊗ 1_{A})δ = (g ⊗ g ⊗ 1_{A})(Δ ⊗ 1_{A})δ = (g ⊗ g ⊗ 1_{A})(1_{M(A)} ⊗ δ)δ = (g ⊗ (g ⊗ 1_{A})δ)δ = (1_{B} ⊗ (g ⊗ 1_{A})δ)(g ⊗ 1_{A})δ = (1_{B} ⊗ f)f = (Δ_{B} ⊗ 1_{A})f = (Δ_{B} ⊗ 1_{A})(g ⊗ 1_{A})δ = (Δ_{B}g ⊗ 1_{A})δ hence (g ⊗ g)Δ = Δ_{B}g. Furthermore the diagram

[figure: A; M(A) ⊗ A; δ; B ⊗ A; f; g⊗1_{A}; ϵ⊗1_{A}; A ≅ 𝕂 ⊗ A; 1_{A}; ϵ_{B}⊗1_{A}]

implies ϵ_{B}g = ϵ. Thus g is a homomorphism of bialgebras.

Since universal algebras for algebras A tend to become very big they do not exist in general. But a theorem of Tambara’s says that they exist for finite dimensional algebras (over a field 𝕂).

Definition 1.3.10. If X is a quantum space with finite dimensional function algebra then we call X a finite quantum space.

The following theorem is the quantum space version and equivalent to a theorem of Tambara.

Theorem 1.3.11. Let X be a finite quantum space. Then there exists a (universal) quantum space M(X) with morphism of quantum spaces μ : M(X) ⊥ X ⟶ X.

The algebra version of this theorem is

Theorem 1.3.12. (Tambara) Let A be a finite dimensional 𝕂-algebra. Then there exists a (universal) 𝕂-algebra M(A) with homomorphism of algebras δ : A ⟶ M(A) ⊗ A.

Proof. We are going to construct the 𝕂-algebra M(A) quite explicitly. First we observe that A^{∗} = Hom_{𝕂}(A, 𝕂) is a coalgebra (cf. problem A.6.8) with the structural morphism Δ : A^{∗} ⟶ (A ⊗ A)^{∗} ≅ A^{∗} ⊗ A^{∗}. Denote the dual basis by ∑_{i=1}^{n} a_{i} ⊗ ā^{i} ∈ A ⊗ A^{∗}. Now let T (A ⊗ A^{∗}) be the tensor algebra of the vector space A ⊗ A^{∗}. Consider elements of the tensor algebra

xy ⊗ ζ ∈ A ⊗ A^{∗},

x ⊗ y ⊗ Δ(ζ) ∈ A ⊗ A ⊗ A^{∗} ⊗ A^{∗} ≅ A ⊗ A^{∗} ⊗ A ⊗ A^{∗},
1 ⊗ ζ ∈ A ⊗ A^{∗},
ζ(1) ∈ 𝕂.

The following elements

xy ⊗ ζ − x ⊗ y ⊗ Δ(ζ) (1)

and

1 ⊗ ζ − ζ(1) (2)

generate a two-sided ideal I ⊆ T (A ⊗ A^{∗}). Now we define

M(A) := T (A ⊗ A^{∗})/I

and the cooperation δ : A ∋ a ⟶ ∑_{i=1}^{n}(a ⊗ ā^{i}) ⊗ a_{i} ∈ T (A ⊗ A^{∗})/I ⊗ A. This is a well-defined linear map.

To show that this map is a homomorphism of algebras we first describe the multiplication of A by a_{i}a_{j} = ∑_{k} α_{ij}^{k}a_{k}. Then the comultiplication of A^{∗} is given by Δ(ā^{k}) = ∑_{ij} α_{ij}^{k}ā^{i} ⊗ ā^{j} since (Δ(ā^{k}), a_{l} ⊗ a_{m}) = (ā^{k}, a_{l}a_{m}) = ∑_{r} α_{lm}^{r}(ā^{k}, a_{r}) = α_{lm}^{k} = ∑_{ij} α_{ij}^{k}(ā^{i}, a_{l})(ā^{j}, a_{m}) = (∑_{ij} α_{ij}^{k}ā^{i} ⊗ ā^{j}, a_{l} ⊗ a_{m}). Now write 1 = ∑ β^{k}a_{k}. Then we get ϵ(ā^{i}) = β^{i} since ϵ(ā^{i}) = (ā^{i}, 1) = ∑_{j} β^{j}(ā^{i}, a_{j}) = β^{i}. So we have δ(a)δ(b) = (∑_{i=1}^{n}(a ⊗ ā^{i}) ⊗ a_{i}) · (∑_{j=1}^{n}(b ⊗ ā^{j}) ⊗ a_{j}) = ∑_{ij}(a ⊗ b ⊗ ā^{i} ⊗ ā^{j}) ⊗ a_{i}a_{j} = ∑_{ijk} α_{ij}^{k}(a ⊗ b ⊗ ā^{i} ⊗ ā^{j}) ⊗ a_{k} = ∑_{k}(a ⊗ b ⊗ Δ(ā^{k})) ⊗ a_{k} = ∑_{k}(ab ⊗ ā^{k}) ⊗ a_{k} = δ(ab). Furthermore we have δ(1) = ∑_{i}(1 ⊗ ā^{i}) ⊗ a_{i} = ∑_{i} ā^{i}(1) ⊗ a_{i} = 1 ⊗ ∑_{i} ā^{i}(1)a_{i} = 1 ⊗ 1. Hence δ is a homomorphism of algebras.

Now we have to show that there is a unique g for each f. First of all f : A ⟶ B⊗A induces uniquely determined linear maps f_{i} : A ⟶ B with f(a) = ∑_{i} f_{i}(a) ⊗ a_{i} since the a_{i} form a basis. Since f is a homomorphism of algebras we get from ∑_{k} f_{k}(a) ⊗ a_{k} = f(ab) = f(a)f(b) = ∑_{ij}(f_{i}(a) ⊗ a_{i})(f_{j}(b) ⊗ a_{j}) = ∑_{ij} f_{i}(a)f_{j}(b) ⊗ a_{i}a_{j} = ∑_{ijk} α_{ij}^{k}f_{i}(a)f_{j}(b) ⊗ a_{k} by comparison of coefficients

f_{k}(ab) = ∑_{ij} α_{ij}^{k}f_{i}(a)f_{j}(b).

Furthermore we define g(a ⊗ ā) := (1 ⊗ ā)f(a) ∈ B. Then we get in particular g(a ⊗ ā^{i}) = (1 ⊗ ā^{i})(∑_{j} f_{j}(a) ⊗ a_{j}) = f_{i}(a). We can extend the map g to a homomorphism of algebras g : T (A ⊗ A^{∗}) ⟶ B. Applied to the generators of the ideal we get g(ab ⊗ ā^{k} − a ⊗ b ⊗ Δ(ā^{k})) = (1 ⊗ ā^{k}) ∑_{l} f_{l}(ab) ⊗ a_{l} − ∑_{rsij} α_{ij}^{k}(1 ⊗ ā^{i})(f_{r}(a) ⊗ a_{r}) · (1 ⊗ ā^{j})(f_{s}(b) ⊗ a_{s}) = f_{k}(ab) − ∑_{ij} α_{ij}^{k}f_{i}(a)f_{j}(b) = 0. We have furthermore g(1 ⊗ ζ − ζ(1)) = (1 ⊗ ζ)f(1) − ζ(1) = (1 ⊗ ζ)(1 ⊗ 1) − ζ(1) = 1ζ(1) − ζ(1) = 0. Thus the homomorphism of algebras g vanishes on the ideal I so it may be factored through M(A) = T (A)/I. Denote this factorization also by g. Then the diagram commutes since (g ⊗ 1_{A})δ(a) = (g ⊗ 1_{A})(∑_{i}(a ⊗ ā^{i}) ⊗ a_{i}) = ∑_{i}(1 ⊗ ā^{i})f(a) ⊗ a_{i} = ∑_{ij} f_{j}(a)(ā^{i}, a_{j}) ⊗ a_{i} = ∑_{i} f_{i}(a) ⊗ a_{i} = f(a).

We still have to show that g is uniquely determined. Assume that we also have (h ⊗ 1_{A})δ = f then ∑_{i} h(a ⊗ ā^{i}) ⊗ a_{i} = (h ⊗ 1_{A})δ(a) = f(a) = ∑_{i} f_{i}(a) ⊗ a_{i} hence h(a ⊗ ā^{i}) = f_{i}(a) = g(a ⊗ ā^{i}), i.e. g = h.

Definition 1.3.13. Let A be a 𝕂-algebra. The universal algebra M(A) for A that is a bialgebra is also called the coendomorphism bialgebra of A.

Problem 1.3.2. 1. Determine explicitly the dual coalgebra A^{∗} of the algebra

A := 𝕂⟨x⟩/(x^{2}). (Hint: Find a basis for A.)

2. Determine and describe the coendomorphism bialgebra of A from problem 1.1. (Hint: Determine first a set of algebra generators of M(A). Then describe the relations.)

3. Determine explicitly the dual coalgebra A^{∗} of A := 𝕂⟨x⟩/(x^{3}).

4. Determine and describe the coendomorphism bialgebra of A from problem 1.3.

5. (*) Determine explicitly the dual coalgebra A^{∗} of A := 𝕂⟨x, y⟩/I where the ideal I is generated as a two-sided ideal by the polynomials

xy − q^{−1}yx, x^{2}, y^{2}.

6. (*) Determine the coendomorphism bialgebra of A from problem 1.5.

7. Let A be a finite dimensional 𝕂-algebra with universal bialgebra A ⟶ B ⊗ A.

Show

i) that A^{op} ⟶ B^{op} ⊗ A^{op} is universal (where A^{op} has the multiplication ∇τ : A ⊗ A ⟶ A ⊗ A ⟶ A);

ii) that A ≅ A^{op} implies B ≅ B^{op} (as bialgebras); iii) that for commutative algebras A the algebra B satisfies B ≅ B^{op} but that

B need not be commutative. iv) Find an isomorphism B ≅ B^{op} for the bialgebra B = 𝕂⟨a, b⟩/(a^{2}, ab+ba).

(compare problem 1.2 2).