DECOMPOSITION OF SOME POINTED HOPF ALGEBRAS GIVEN BY THE CANONICAL NAKAYAMA AUTOMORPHISM

M. GRAÑA, J.A. GUCCIONE, AND J.J. GUCCIONE

Abstract. Every finite dimensional Hopf algebra is a Frobenius algebra, with Frobenius homomorphism given by an integral. The Nakayama automorphism determined by it yields a decomposition with degrees in a cyclic group. For a family of pointed Hopf algebras, we determine necessary and sufficient conditions for this decomposition to be strongly graded.

1. Introduction

Let k be a field, A a finite dimensional k-algebra and DA the dual space Hom_{k}(A, k), endowed with the usual A-bimodule structure. Recall that A is said to be a Frobenius algebra if there exists a linear form φ: A → k, such that the map A → DA, defined by x ↦ xφ, is a left A-module isomorphism. This linear form φ: A → k is called a Frobenius homomorphism. It is well known that this is equivalent to say that the map x ↦ φx, from A to DA, is an isomorphism of right A-modules. From this it follows easily that there exists an automorphism ρ of A, called the Nakayama automorphism of A with respect to φ, such that xφ = φρ(x), for all x ∈ A. It is easy to check that a linear form φ̃: A → k is another Frobenius homomorphism if and only if there exists an invertible element x in A, such that φ̃ = xφ. It is also easy to check that the Nakayama automorphism of A with respect to φ̃ is the map given by a ↦ ρ(x)^{−1}ρ(a)ρ(x).

Let A be a Frobenius k-algebra, φ: A → k a Frobenius homomorphism and ρ : A → A the Nakayama automorphism of A with respect to φ.

Definition 1.1. We say that ρ has order m ∈ ℕ and we write ord_{ρ} = m, if ρ^{m} = id_{A} and ρ^{r} ≠ id_{A}, for all r < m.

Let G be a group. Recall that a G-graded algebra is a k-algebra A together with a decomposition A = ⨁_{g∈G} A_{g} of A as a direct sum of subvector spaces, such that A_{g}A_{g′} ⊆ A_{gg′} for all g, g′ ∈ G. When A_{g}A_{g′} = A_{gg′} for all g, g′ ∈ G, the grading is called strong, and the algebra is said to be strongly graded. Assume that ρ has finite order and that k has a primitive ord_{ρ}-th root of unity ω. For n ∈ ℕ, let C_{n} be the group of n-th roots of unity in k. Since the polynomial X^{ord^{ρ}} − 1 has distinct roots ω^{i} (0 ≤ i < ord_{ρ}), the algebra A becomes a C_{ord_{ρ}}-graded algebra

(1.1)  A = A_{ω^{0}} ⊕ ⋯ ⊕ A_{ω^{ordρ −1}}, where A_{z} = {a ∈ A : ρ(a) = za}.

As it is well known, every finite dimensional Hopf algebra H is Frobenius, being a Frobenius homomorphism any nonzero right integral φ ∈ H^{∗}. Let t be a nonzero right integral of H. Let α ∈ H^{∗} be the modular element of H, defined by at = α(a)t (notice that this is the inverse of the modular element α considered in [R1]). For x ∈ H^{∗}, let r_{x} : H → H be defined by r_{x}(b) = bx = x(b_{(1)})b_{(2)}. Then, as follows from [R1, Theorem 3(a)], we have

2000 Mathematics Subject Classification. Primary 16W30; Secondary 16W50.

This work was partially supported by CONICET, PICT-02 12330, UBA X294.

ρ(b) = α^{−1}(b_{(1)})S^{2}(b_{(2)}), i.e. ρ = S^{2} ∘ r_{α^{−1}}.

Since α is a group-like element, S^{2} ∘ r_{α^{−1}} = r_{α^{−1}} ∘ S^{2} and therefore

(1.2) ρ^{l} = S^{2l} ∘ r_{α^{−l}}, i.e. ρ^{l}(h) = α^{−l}(h_{(1)})S^{2l}(h_{(2)}) = α^{l}(S(h_{(1)}))S^{2l}(h_{(2)}),

for all l ∈ ℤ. Now, α has finite order and, by the Radford formula for S^{4} (see [R2] or [S, Theorem 3.8]), the antipode S has finite order with respect to composition. Thus, the automorphism ρ has finite order, which implies that finite dimensional Hopf algebras are examples of the situation considered above.

Notice that by (1.2), if ρ^{l} = id, then α^{l} = ε and then S^{2l} = id. The converse is obvious. So, the order of ρ is the lcm of those of α and S^{2}. In particular, the number of terms in the decomposition associated with S^{2} divides that in the one associated with ρ. Also, from (1.2) we get that ρ = S^{2} if and only if H is unimodular.

The main aim of the present work is to determine conditions for decomposition (1.1) to be strongly graded. Besides the fact that the theory for algebras which are strongly graded over a group is well developed (see for instance [M]), our interest on this problem originally came from the homological results in [GG].

The decomposition using S^{2} instead of ρ was considered in [RS]. We show below that if S^{2} ≠ id, then this decomposition is not strongly graded. On the other hand, as shown in [RS], under suitable assumptions its homogeneous components are equidimensional. It is an interesting problem to know whether a similar thing happens with the decomposition associated with ρ. For instance, all the liftings of Quantum Linear Spaces have equidimensional decompositions, as shown in Remark 4.4.

2. The unimodular case

Let H be a finite dimensional Hopf algebra with antipode S. In this brief section we first show that the decomposition of H associated with S^{2} is not strongly graded, unless S^{2} = id (this applies in particular to decomposition (1.1) when H is unimodular and ord_{ρ} > 1). We finish by giving a characterization of unimodular Hopf algebras in terms of decomposition (1.1).

Lemma 2.1. Let H be a finite dimensional Hopf algebra. Suppose H = ⨁_{g∈G} H_{g} is a graduation over a group. Assume there exists g ∈ G such that ε(H_{g}) = 0. Then the decomposition is not strongly graded.

Proof. Suppose the decomposition is strongly graded. Then there are elements a_{i} ∈ H_{g} and b_{i} ∈ H_{g^{−1}} such that 1 = ∑_{i} a_{i}b_{i}. Then 1 = ε(1) = ∑_{i} ε(a_{i})ε(b_{i}), a contradiction. □

Corollary 2.2. Assume that S^{2} ≠ id and that

H = ⨁_{z∈k^{∗}} H_{z}, where H_{z} = {h ∈ H : S^{2}(h) = zh}.

Then this decomposition is not strongly graded.

Proof. Since ε ∘ S^{2} = ε, then ε(H_{z}) = 0 for all z ≠ 1. □

Let now φ ∈ ∫_{H^{∗}}^{r} and Γ ∈ ∫_{H}^{l}, such that ⟨φ, Γ⟩ = 1, and let α: H → k be the modular map associated with t = S(Γ). Let ρ be the Nakayama automorphism associated with φ. Assume that k has a root of unity ω of order ord_{ρ}. We consider the decomposition associated with ρ, as in (1.1)

(2.1)  H = H_{ω^{0}} ⊕ ⋯ ⊕ H_{ω^{ordρ −1}}.

Corollary 2.3. If H is unimodular and S^{2} ≠ id, then the decomposition (2.1) is not strongly graded. □

Proposition 2.4. If h ∈ H_{ω^{i}}, then α(h) = ω^{−i}ϵ(h).

Proof. By [R1, Proposition 1(e)], or the proof of [S, Proposition 3.6], ⟨φ, t⟩ = 1. Then

ϵ(h) = ϵ(h)⟨φ, t⟩ = ⟨φ, th⟩ = ⟨φ, ρ(h)t⟩ = ⟨φ, ω^{i}ht⟩ = ω^{i}α(h).

So, α(h) = ω^{−i}ϵ(h), as we want. □

Corollary 2.5. H is unimodular if and only if H_{ω^{i}} ⊆ ker(ϵ), for all i > 0

Proof. ⇒): For h ∈ H_{ω^{i}}, we have ϵ(h) = α(h) = ω^{−i}ϵ(h) and so ϵ(h) = 0, since ω^{i} ≠ 1. ⇐): For h ∈ H_{ω^{i}} with i > 0, we have α(h) = ω^{−i}ϵ(h) = 0 = ϵ(h) and, for h ∈ H_{ω^{0}}, we also have α(h) = ω^{0}ϵ(h) = ϵ(h). □

3. Bosonizations of Nichols algebras of diagonal type

Let G be a finite abelian group, g = g_{1}, …, g_{n} ∈ G a sequence of elements in G and χ = χ_{1}, …, χ_{n} ∈ Ĝ a sequence of characters of G. Let V be the vector space with basis {x_{1}, …, x_{n}} and let c be the braiding given by c(x_{i} ⊗ x_{j}) = q_{ij}x_{j} ⊗ x_{i}, where q_{ij} = χ_{j}(g_{i}). We consider the Nichols algebra R = 𝔅(V) generated by (V, c). We give here one of its possible equivalent definitions. Let T_{c}(V) be the tensor algebra generated by V, endowed with the unique braided Hopf algebra structure such that the elements x_{i} are primitive and whose braiding extends c. Then, R is obtained as a colimit of algebras R = \underrightarrow{lim} R_{i}, where R_{0} = T_{c}(V) and R_{i+1} is the quotient of R_{i} by the ideal generated by its homogeneous primitive elements with degree ≥ 2. See [AG, AS] for alternative definitions and main properties of Nichols algebras. Assume that R is finite-dimensional and let t_{0} ∈ R be a nonzero homogeneous element of greatest degree. Let H = H(g, χ) = R#kG be the bosonization of R (this is an alternative presentation for the algebras considered by Nichols in [N]). We have:

Δ(g_{i}) = g_{i} ⊗ g_{i},  Δ(x_{i}) = g_{i} ⊗ x_{i} + x_{i} ⊗ 1,
S(g) = g^{−1},  S(x_{i}) = −g_{i}^{−1}x_{i},
S^{2}(g) = g,  S^{2}(x_{i}) = g_{i}^{−1}x_{i}g_{i} = q_{ii}^{−1}x_{i}.

The element t_{0} ∑_{g∈G} g is a non zero right integral in H. Let α be the modular element associated with it. Since, for all i, the degree of x_{i}t_{0} ∑_{g∈G} g is bigger than the degree of t_{0}, we have that x_{i}t_{0} ∑_{g∈G} g = 0, and so α(x_{i}) = 0. Moreover, α_{|G} is determined by gt_{0} = α(g)t_{0}g. Thus,

ρ(g) = α(g^{−1})g and ρ(x_{i}) = α(g_{i}^{−1})q_{ii}^{−1}x_{i}.

This implies that the nonzero monomials x_{i_{1}} ⋯ x_{i_{ℓ}}g are a set of eigenvectors for ρ (which generate H as a k-vector space). In particular, ρ is diagonalizable, whence k has a primitive ord_{ρ}-th root of unity. Consider the subgroups

L_{1} = ⟨q_{11}, …, q_{nn}, α(G)⟩ and L_{2} = α(G),

of k^{∗}. Since ρ(x_{i}g_{i}^{−1}) = q_{ii}^{−1}x_{i}g_{i}^{−1} and ρ(g) = α(g), the group L_{1} is the set of eigenvalues of ρ. As in the introduction, we decompose

H = ⨁_{ω∈L_{1}} H_{ω}, where H_{ω} = {h ∈ H : ρ(h) = ωh}.

Proposition 3.1. The following are equivalent:

(1) ⨁_{ω∈L_{1}} H_{ω} is strongly graded,

(2) L_{1} = L_{2},

(3) Each H_{ω} contains an element in G.

(4) H is a crossed product H_{1} ⋉ kL_{1}.

Proof. It is clear that (2)⇒(3) and (4)⇒(1).

The proof of (3)⇒(4) is standard. We sketch it for the readers’ convenience. For each ω ∈ L_{1}, pick g_{ω} ∈ H_{ω} ∩ G. Define ρ : L_{1} × H_{1} → H_{1} and f : L_{1} × L_{1} → H_{1} by

ω · a = ρ(ω, a) = g_{ω}ag_{ω}^{−1} and f(ω, ω′) = g_{ω}g_{ω′}g_{ωω′}^{−1}.

Let H_{1} ⋉_{ρ}^{f} kL_{1} be the algebra with underlying vector space H_{1} ⊗ kL_{1} and with multiplication (a ⊗ ω)(b ⊗ ω′) = a(ω · b)f(ω, ω′) ⊗ ωω′. It is easy to see that the map Ψ : H_{1} ⋉_{ρ}^{f} kL_{1} → H given by Ψ(a ⊗ ω) = ag_{ω} is an isomorphism of algebras. In particular, H_{1} ⋉_{ρ}^{f} kL_{1} is associative with unit 1 ⊗ 1, and so ρ is a weak action and f is a normal cocycle that satisfies the twisted module condition.

We now prove that (1)⇒(2). Notice that H is also ℕ_{0}-graded by deg(g) = 0 for all g ∈ G, and deg(x_{i}) = 1. Call H = ⨁_{i∈ℕ} H^{i} this decomposition. Since each H_{ω} is spanned by elements which are homogeneous with respect to the previous decomposition, we have:

H = ⨁_{i≥0, ω∈L_{1}} H_{ω}^{i}, where H_{ω}^{i} = H_{ω} ∩ H^{i}.

So, if ⨁_{ω∈L_{1}} H_{ω} is strongly graded, then each H_{ω} must contain nonzero elements in H^{0}. Since H^{0} ⊆ ⨁_{ω∈L_{2}} H_{ω}, we must have L_{1} = L_{2}. □

Quantum Linear Spaces. If the sequence of characters χ satisfies • χ_{i}(g_{i}) ≠ 1,

• χ_{i}(g_{j})χ_{j}(g_{i}) = 1 for i ≠ j,

then H(g, χ) is the Quantum Linear Space with generators G and x_{1}, …, x_{n}, subject to the following relations:

• gx_{i} = χ_{i}(g)x_{i}g,

• x_{i}x_{j} = q_{ij}x_{j}x_{i},
• x_{i}^{m_{i}} = 0,

where m_{i} = ord(q_{ii}). For these sort of algebras it is possible to give an explicit formula for ρ. In fact, the element t = x_{1}^{m_{1}−1} ⋯ x_{n}^{m_{n}−1} ∑_{g∈G} g is a right integral in H. Using this integral, it is easy to check that α(g) = χ_{1}^{m_{1}−1}(g) ⋯ χ_{n}^{m_{n}−1}(g).

In particular, α(g_{i}) = q_{i1}^{m_{1}−1} ⋯ q_{in}^{m_{n}−1}. A straightforward computation, using that ρ(g) = α(g^{−1})g and ρ(x_{i}) = α(g_{i}^{−1})q_{ii}^{−1}x_{i}, shows that

ρ(x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g) = ∏_{1≤i<j≤n} q_{ij}^{(1−m_{j})r_{i}−(1−m_{i})r_{j}}α(g^{−1})x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g.

Proposition 3.1 applies to this family of algebras.

Example 3.2. Let k be a field of characteristic ≠ 2 and let G = {1, g}. Set g_{i} = g and χ_{i}(g) = −1 for i ∈ {1, …, n}. Then, q_{ij} = −1 for all i, j, and α(g) = (−1)^{n}. In this case, the algebra H is generated by g, x_{1}, …, x_{n} subject to relations

• g^{2} = 1,

• x_{i}^{2} = 0,
• x_{i}x_{j} = −x_{j}x_{i},
• gx_{i} = −x_{i}g.

By Proposition 3.1, we know that H is strongly graded if and only if n is odd.

4. Liftings of Quantum Linear Spaces

In this section we consider a generalization of Quantum Linear Spaces: that of their liftings. As above, G is a finite abelian group, g = g_{1}, …, g_{n} ∈ G is a sequence of elements in G and χ = χ_{1}, …, χ_{n} ∈ Ĝ is a sequence of characters of G, such that

χ_{i}(g_{i}) ≠ 1, (4.1) χ_{i}(g_{j})χ_{j}(g_{i}) = 1, for i ≠ j. (4.2)

Again, let q_{ij} = χ_{j}(g_{i}) and let m_{i} = ord(q_{ii}). Let now λ_{i} ∈ k and λ_{ij} ∈ k for i ≠ j be such that

λ_{i}(χ_{i}^{m_{i}} − ε) = λ_{ij}(χ_{i}χ_{j} − ε) = 0.

Suppose that λ_{ij} + q_{ij}λ_{ji} = 0 whenever i ≠ j. The lifting of the quantum affine space associated with this data is the algebra H = H(g, χ, λ), with generators G and x_{1}, …, x_{n}, subject to the following relations:

gx_{i} = χ_{i}(g)x_{i}g, (4.3) x_{i}x_{j} = q_{ij}x_{j}x_{i} + λ_{ij}(1 − g_{i}g_{j}), (4.4) x_{i}^{m_{i}} = λ_{i}(1 − g_{i}^{m_{i}}). (4.5)

It is well known that the set of monomials {x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g : 0 ≤ r_{i} < m_{i}, g ∈ G} is a basis of H. It is a Hopf algebra with comultiplication defined by

Δ(g) = g ⊗ g, for all g ∈ G, (4.6) Δ(x_{i}) = g_{i} ⊗ x_{i} + x_{i} ⊗ 1. (4.7)

The counit ε satisfies ε(g) = 1, for all g ∈ G, and ε(x_{i}) = 0. Moreover, the antipode S is given by S(g) = g^{−1}, for all g ∈ G, and S(x_{i}) = −g_{i}^{−1}x_{i}. We note that S^{2}(g) = g and S^{2}(x_{i}) = q_{ii}^{−1}x_{i}.

Let 𝕊_{n} be the symmetric group on n elements. For σ ∈ 𝕊_{n} let

t_{σ} = x_{σ_{1}}^{m_{σ_{1}} −1} ⋯ x_{σ_{n}}^{m_{σn} −1} ∑_{g∈G} g.

Note that t_{σ} ≠ 0.

Lemma 4.1. The following hold:

(1) λ_{ji}g_{i}g_{j} lies in the center of H(g, χ, λ) for i ≠ j.

(2) λ_{i}g_{i}^{m_{i}} lies in the center of H(g, χ, λ).
(3) t_{σ}g = t_{σ}, for all g ∈ G.
(4) t_{σ}x_{σ_{n}} = 0.

Proof. For (1) it is sufficient to see that λ_{ji}g_{i}g_{j} commutes with x_{l}. If λ_{ji} = 0 the result is clear. Assume that λ_{ji} ≠ 0. Then, χ_{i} = χ_{j}^{−1}, and thus

λ_{ji}g_{i}g_{j}x_{l} = λ_{ji}χ_{l}(g_{i})χ_{l}(g_{j})x_{l}g_{i}g_{j}
= λ_{ji}χ_{i}(g_{l})^{−1}χ_{j}(g_{l})^{−1}x_{l}g_{i}g_{j}
= λ_{ji}x_{l}g_{i}g_{j}.

The proof of (2) is similar to that of (1) and (3) it is immediate. For (4), we have:

t_{σ}x_{σ_{n}} = x_{σ_{1}}^{m_{σ_{1}} −1} ⋯ x_{σ_{n}}^{m_{σn} −1} ∑_{g∈G} gx_{σ_{n}}

= x_{σ_{1}}^{m_{σ_{1}} −1} ⋯ x_{σ_{n}}^{m_{σn} −1}x_{σ_{n}} ∑_{g∈G} χ_{σ_{n}}(g)g

= λ_{σ_{n}}x_{σ_{1}}^{m_{σ_{1}} −1} ⋯ x_{σ_{n−1}}^{m_{σ_{n−1}} −1}(1 − g_{σ_{n}}^{m_{σn}}) ∑_{g∈G} χ_{σ_{n}}(g)g,

and the result follows by noticing that

(1 − g_{σ_{n}}^{m_{σn}}) ∑_{g∈G} χ_{σ_{n}}(g)g = ∑_{g∈G} χ_{σ_{n}}(g)g − ∑_{g∈G} g_{σ_{n}}^{m_{σn}} χ_{σ_{n}}(g)g

= ∑_{g∈G} χ_{σ_{n}}(g_{σ_{n}}^{m_{σn}} g)g_{σ_{n}}^{m_{σn}} g − ∑_{g∈G} χ_{σ_{n}}(g)g_{σ_{n}}^{m_{σn}} g = 0,

since χ_{σ_{n}}(g_{σ_{n}}^{m_{σn}}) = q_{σ_{n}σ_{n}}^{m_{σn}} = 1. □

Proposition 4.2. t_{σ} is a right integral.

Proof. Let M = (m_{1} − 1) + ⋯ + (m_{n} − 1). Let

A = {f : {1, …, M} → {1, …, n} : #f^{−1}(i) = m_{i} − 1 for all i}.

For f ∈ A, let x_{f} = x_{f(1)}x_{f(2)} ⋯ x_{f(M)}. We claim that if f, h ∈ A, then x_{f} ∑_{g∈G} g = βx_{h} ∑_{g∈G} g for some β ∈ k^{∗}. To prove this claim, it is sufficient to check it when f and h differ only in i, i + 1 for some 1 ≤ i < M, that is, when h = f ∘ τ_{i}, where τ_{i} ∈ S_{M} is the elementary transposition (i, i + 1). But, in this case, we have:

x_{f} ∑_{g∈G} g = x_{h∘τ_{i}} ∑_{g∈G} g

= q_{h(i+1)h(i)}x_{h} ∑_{g∈G} g + λ_{h(i+1)h(i)}x_{h,î,{ \atop i+1}}(1 − g_{h(i)}g_{h(i+1)}) ∑_{g∈G} g

= q_{h(i+1)h(i)}x_{h} ∑_{g∈G} g

where x_{h,î,{ \atop i+1}} = x_{h(1)} ⋯ x_{h(i−1)}x_{h(i+2)} ⋯ x_{h(M)}. The second equality follows from relation (4.4) and item (1) in the previous Lemma. The Proposition follows now using items (3) and (4) in the Lemma. □

Now we see that • α(x_{i}) = 0,

• α(g) = χ_{1}^{m_{1}−1}(g) ⋯ χ_{n}^{m_{n}−1}(g). For the first assertion, by Proposition 4.2, we can take σ such that σ_{1} = i. Then,

x_{i}t_{σ} = x_{i}^{m_{i}}x_{σ_{2}}^{m_{σ_{2}} −1} ⋯ x_{σ_{n}}^{m_{σn} −1} ∑_{g∈G} g

= x_{σ_{2}}^{m_{σ_{2}} −1} ⋯ x_{σ_{n}}^{m_{σn} −1}λ_{i}(1 − g_{i}^{m_{i}}) ∑_{g∈G} g = 0,

where the second equality follows from item (2) of Lemma 4.1. In particular, α(g_{i}) = q_{i1}^{m_{1}−1} ⋯ q_{in}^{m_{n}−1}. Since ρ(h) = α(S(h_{(1)}))S^{2}(h_{(2)}), we have:

• ρ(g) = α(g^{−1})g,

• ρ(x_{i}) = α(g_{i}^{−1})q_{ii}^{−1}x_{i} = ∏_{\begin{matrix}1≤j≤n \\ j≠i \end{matrix}} q_{ij}^{1−m_{j}}x_{i}.

Thus, as ρ is an algebra map,

ρ(x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g) = q_{11}^{−r_{1}} ⋯ q_{nn}^{−r_{n}}α(g_{1}^{−r_{1}} ⋯ g_{n}^{−r_{n}}g^{−1}) x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g
= ∏_{1≤i<j≤n} q_{ij}^{(1−m_{j})r_{i}−(1−m_{i})r_{j}}α(g^{−1}) x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g.

So, the basis {x_{1}^{j_{1}} ⋯ x_{n}^{j_{n}}g} is made up of eigenvectors of ρ. Consider the groups

k^{∗} ⊇ L_{1} = ⟨q_{11}, …, q_{nn}, α(G)⟩ ⊇ L_{2} = α(G).

Using that ρ(x_{i}g_{i}^{−1}) = q_{ii}^{−1}x_{i}g_{i}^{−1} and ρ(g) = α(g^{−1})g, it is easy to see that L_{1} is the set of eigenvalues of ρ and that the order of ρ is the l.c.m. of the numbers m_{1}, …, m_{n} and the order of the character α_{|G} ∈ Ĝ (in particular, k has a primitive ord_{ρ}-th root of unity). As before, we decompose H = H(g, χ, λ) as

H = ⨁_{ω∈L_{1}} H_{ω}, where H_{ω} = {h ∈ H : ρ(h) = ωh}.

The following result is the version of Proposition 3.1 for the present context.

Theorem 4.3. The following are equivalent:

(1) ⨁_{ω∈L_{1}} H_{ω} is strongly graded,

(2) L_{1} = L_{2},

(3) Each component H_{ω} contains an element in G,

(4) H is a crossed product H_{1} ⋉ kL_{1}.

Proof. Clearly (2) and (3) are equivalent and (4)⇒(1). The proof of (3)⇒(4) is the same as for Proposition 3.1. Next we prove that (1)⇒(3). Let ω ∈ L_{1}. By Lemma 2.1, we know that ε(H_{ω}) ≠ 0. Since H_{ω} has a basis consisting of monomials x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}g and ε(x_{i}) = 0, there must be an element g ∈ G inside H_{ω}. □

Remark 4.4. We next show that for liftings of Quantum Linear Spaces, the components in the decomposition H = ⨁_{ω∈L_{1}} H_{ω} are equidimensional. In fact, in this case we can take the basis of H given by

{(x_{1}g_{1}^{−1})^{r_{1}} ⋯ (x_{n}g_{n}^{−1})^{r_{n}}g : 0 ≤ r_{i} < m_{i}, g ∈ G}.

Since ρ(x_{i}g_{i}^{−1}) = q_{ii}^{−1}x_{i}g_{i}^{−1}, the map

θ: ℤ_{m_{1}} × ⋯ × ℤ_{m_{n}} × G → k^{∗},

taking (r_{1}, …, r_{n}, g) to the eigenvalue of (x_{1}g_{i}^{−1})^{r_{1}} ⋯ (x_{n}g_{n}^{−1})^{r_{n}}g with respect to ρ, is a well defined group homomorphism. From this it follows immediately that all the eigenspaces of ρ are equidimensional.

5. Computing H_{1}

Assume we are in the setting of the liftings of QLS. Suppose H is a crossed product or, equivalently, that L_{1} = L_{2}. Then, there exist elements γ_{1}, …, γ_{n} ∈ G, such that α(γ_{i}) = q_{ii}. Set γ̃_{i} = g_{i}^{−1}γ_{i}^{−1} and let y_{i} = x_{i}γ̃_{i}. It is immediate that y_{i} ∈ H_{1}. Let N = ker(α_{|G}) ⊆ G. It is easy to see that H_{1} has a basis given by {y_{1}^{r_{1}} ⋯ y_{n}^{r_{n}}g : g ∈ N}. Furthermore, H_{1} can be presented by generators N, y_{1}, …, y_{n} and relations

• gy_{i} = χ_{i}(g)y_{i}g,

• y_{i}y_{j} = q_{ij}χ_{j}(γ̃_{i})χ_{i}^{−1}(γ̃_{j})y_{j}y_{i} + χ_{j}(γ̃_{i})λ_{ij}(γ̃_{i}γ̃_{j} − γ_{i}^{−1}γ_{j}^{−1}),
• y_{i}^{m_{i}} = λ_{i}χ_{i}^{m_{i}(m_{i}−1)/2}(γ̃_{i})(γ̃_{i}^{m_{i}} − γ_{i}^{−m_{i}}).

Notice that if λ_{i} ≠ 0, then χ_{i}^{m_{i}(m_{i}−1)/2}(γ̃_{i}) = ±1. We claim that

λ_{ij}γ̃_{i}γ̃_{j}, λ_{ij}γ_{i}γ_{j}, λ_{i}γ̃^{m_{i}} and λ_{i}γ^{m_{i}}

belong to kN. It is clear that γ^{m_{i}} ∈ N, since α(γ^{m_{i}}) = q_{ii}^{m_{i}} = 1. We now prove the remaining part of the claim. Assume that λ_{ij} ≠ 0. Then χ_{i}χ_{j} = ε. Hence,

• If l ≠ i, j, then χ_{l}(g_{i}g_{j}) = q_{il}q_{jl} = q_{li}^{−1}q_{lj}^{−1} = χ_{i}χ_{j}(g_{l}^{−1}) = 1.

• q_{ii} = χ_{i}(g_{i}) = χ_{j}(g_{i}^{−1}) = q_{ij}^{−1} = q_{ji} = q_{jj}^{−1}.

Thus, m_{i} = ord(q_{ii}) = ord(q_{jj}) = m_{j}, and then

χ_{i}^{m_{i}−1}(g_{i}g_{j})χ_{j}^{m_{j}−1}(g_{i}g_{j}) = (q_{ii}q_{ij}q_{ji}q_{jj})^{m_{i}−1} = 1 and α(γ_{i}γ_{j}) = q_{ii}q_{jj} = 1.

It is now immediate that α(g_{i}g_{j}) = χ_{1}^{m_{1}−1}(g_{i}g_{j}) ⋯ χ_{n}^{m_{n}−1}(g_{i}g_{j}) = 1, and so

α(γ̃_{i}γ̃_{j}) = α(g_{i}^{−1}γ_{i}^{−1}g_{j}^{−1}γ_{j}^{−1}) = α(g_{j}g_{i})^{−1}α(γ_{j}γ_{i})^{−1} = 1.

It remains to check that λ_{i}γ̃^{m_{i}} ∈ kN. Assume now that λ_{i} ≠ 0. Then χ_{i}^{m_{i}} = ε. Thus,

• If l ≠ i, then χ_{l}^{m_{l}−1}(g_{i}^{m_{i}}) = q_{il}^{(m_{l}−1)m_{i}} = q_{li}^{(1−m_{l})m_{i}} = χ_{i}^{m_{i}}(g_{l}^{1−m_{l}}) = 1.

Since χ_{i}^{m_{i}−1}(g_{i}^{m_{i}}) = q_{ii}^{m_{i}(m_{i}−1)} = 1, this implies that

α(g_{i}^{m_{i}}) = χ_{1}^{m_{1}−1}(g_{i}^{m_{i}}) ⋯ χ_{n}^{m_{n}−1}(g_{i}^{m_{i}}) = 1,

and so

α(γ̃_{i}^{m_{i}}) = α(γ_{i}g_{i})^{−1} = α(γ_{i})^{−1}α(g_{i})^{−1} = 1.

References

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[R2] D.E. Radford, The order of the antipode of a finite dimensional Hopf algebra is finite, Amer. J. Math. 98 (1976), no. 2, 333–355.

[RS] D. E. Radford and H.-J. Schneider, On the even powers of the antipode of a finitedimensional Hopf algebra, J. Algebra 251 (2002), no. 1, 185–212.

[S] H-J. Schneider, Lectures on Hopf algebras, Trabajos de Matemática, 31/95, Universidad Nacional de Córdoba, FaMAF, Córdoba, 1995.

M.G., J.A.G., J.J.G.: Depto de Matemática - FCEyN Universidad de Buenos Aires Pab. I - Ciudad Universitaria 1428 - Buenos Aires - Argentina E-mail address, M.G.: matiasg@dm.uba.ar E-mail address, J.A.G.: vander@dm.uba.ar E-mail address, J.J.G.: jjgucci@dm.uba.ar