FINITE-DIMENSIONAL POINTED HOPF ALGEBRAS WITH ALTERNATING GROUPS ARE TRIVIAL

NICOLÁS ANDRUSKIEWITSCH, FERNANDO FANTINO, MATÍAS GRAÑA, LEANDRO VENDRAMIN

Abstract. Any finite-dimensional complex pointed Hopf algebra with group of group-likes isomorphic to the alternating group 𝔸_{m}, m > 4, is a group algebra.

1. Introduction

This paper contributes to the classification of finite-dimensional pointed Hopf algebras over the field ℂ of complex numbers. Our basic reference is [AS]; see loc. cit. for unexplained terminology and notation. If G denotes a finite group, we would like to know all pointed Hopf algebras H with G(H) ≃ G and dim H < ∞. For this, we need to solve the following problem. Let 𝒪 be a conjugacy class of G, σ ∈ 𝒪 fixed, ρ an irreducible representation of the centralizer C_{G}(σ), M(𝒪, ρ) the corresponding irreducible Yetter-Drinfeld module and 𝔅(𝒪, ρ) the associated Nichols algebra. If (V, c) is a braided vector space, that is c ∈ GL(V ⊗V) is a solution of the braid equation, then 𝔅(V) denotes its Nichols algebra; for shortness, we write 𝔅(𝒪, ρ) instead of 𝔅(M(𝒪, ρ)). The problem is:

For which pairs (𝒪, ρ) is the dimension of 𝔅(𝒪, ρ) finite?

We denote by Ĝ the set of isomorphism classes of irreducible representations of a group G. We use the rack notation x ▷ y = xyx^{−1}, x, y ∈ G. See [AG] for information on racks. If σ ∈ G and ρ ∈ { \atop C_{G}(σ)}, then ρ(σ) is a scalar denoted q_{σσ}.

This article is continuation of [AF1], where we began the study of finitedimensional pointed Hopf algebras with group of group-like elements isomorphic to 𝔸_{m}.

Theorem 1.1. Let G = 𝔸_{m}, m ≥ 5. If 𝒪 is any conjugacy class of G, σ ∈ 𝒪 is fixed and ρ ∈ { \atop C_{G}(σ)}, then dim 𝔅(𝒪, ρ) = ∞.

Date: April 21, 2009.

2000 Mathematics Subject Classification. 16W30; 17B37.

This work was partially supported by ANPCyT-Foncyt, CONICET, Ministerio de Ciencia y Tecnología (Córdoba) and Secyt (UNC).

By the Lifting Method [AS], we conclude:

Theorem 1.2. Let G = 𝔸_{m}, m ≥ 5. Any finite-dimensional pointed Hopf algebra ℋ with G(ℋ) ≃ 𝔸_{m}, is isomorphic to ℂ𝔸_{m}. □

This result was known for the particular cases m = 5 and m = 7 [AF1, F]. We prove it for m ≥ 6. Since 𝔸_{3} is abelian, finite dimensional Nichols algebras over it are classified, there are 25 of them. Nichols algebras over 𝔸_{4} are infinite-dimensional except for four pairs corresponding to the classes of (1 2 3) and (1 3 2) and the non-trivial characters of ℤ/3. Actually, these four algebras are connected to each other either by an outer automorphism of 𝔸_{4} or by the Galois group of ℚ(ζ_{3})|ℚ (the cyclotomic extension by third roots of unity). Therefore, there is only one pair to study for 𝔸_{4}.

In a previous version of this paper, we proved Theorem 1.2 for m ≥ 7, but were unable to decide the dimension of the Nichols algebra attached to the pair formed by the class of (1 2 3 4)(5 6) and the character ρ = χ_{(−1)}, corresponding to 𝔸_{6}. It was observed that this class contains a subrack with 18 elements, a union of 2 subracks of order 9, identified as a union of two conjugacy classes in 𝔽_{9} ⋊ ℤ/4. This pair can be discarded now too, and consequently we also finish 𝔸_{6}, by means of [HS1, Th. 8.6].

The paper is organized as follows. In Section 2, we spell out some techniques, based on the analysis of braided vector subspaces of diagonal type, that are needed for our arguments but are also useful elsewhere. In Section 3 we prove Theorem 1.1.

2. Some techniques

2.1. An abelian subrack with 3 elements. We begin by recording a result that is needed in Lemma 2.4 and will be also useful elsewhere.

Lemma 2.1. Let G be a finite group, 𝒪 be the conjugacy class of σ_{1} in G

and (ρ, V) ∈ { \atop C_{G}(σ_{1})}. Let σ_{2} ≠ σ_{3} ∈ 𝒪 − \{σ_{1}\}; let g_{1} = e, g_{2}, g_{3} ∈ G such

that σ_{i} = g_{i}σ_{1}g_{i}^{−1}, for all i. Assume that

• σ_{1}^{h} = σ_{2}σ_{3} for an odd integer h,

• g_{3}g_{2} and g_{2}g_{3} belong to C_{G}(σ_{1}), and

• σ_{i}σ_{j} = σ_{j}σ_{i}, 1 ≤ i, j ≤ 3.

Then dim 𝔅(𝒪, ρ) = ∞, for any ρ ∈ { \atop C_{G}(σ_{1})}.

Proof. Since σ_{i}σ_{j} = σ_{j}σ_{i}, there exists w ∈ V − 0 and λ_{i} ∈ ℂ such that ρ(σ_{i})(w) = λ_{i}w for i = 1, 2, 3. For any 1 ≤ i, j ≤ 3, we call γ_{ij} = g_{j}^{−1}σ_{i}g_{j}. It is easy to see that γ_{ij} ∈ C_{G}(σ_{1}) and that

POINTED HOPF ALGEBRAS OVER 𝔸_{m} 3

γ = (γ_{ij}) = ( \begin{matrix}σ_{1} & σ_{3} & σ_{2} \\ σ_{2} & σ_{1} & σ_{2}^{h}σ_{1}^{−1} \\ σ_{3} & σ_{3}^{h}σ_{1}^{−1} & σ_{1} \end{matrix} ) .

Then, W = span{g_{1}w, g_{2}w, g_{3}w} is a braided vector subspace of M(𝒪, ρ) of abelian type with Dynkin diagram given by Figure 1. Assume that dim 𝔅(𝒪, ρ) is finite. Then λ_{1} ≠ 1; also λ_{1}^{h} ≠ 1, for otherwise g_{2}w, g_{3}w span a braided vector subspace of Cartan type with Dynkin diagram A_{1}^{(1)}. Thus, we should have λ_{1} = −1 and h even, by [H2, Table 3], but this is a contradiction to the hypothesis on h. □

\begin{aligned} & & & [figure] & [figure] & λ_{1}^{h} & λ_{1}^{h^{2}−2} & λ_{1}^{h} & λ_{1} & λ_{1} & λ_{1} \\ & Figure 1 \end{aligned}

2.2. The technique of a suitable subgroup. Notice that if (V, c) is a braided vector space and W ⊆ V is a subspace such that c(W ⊗W) = W ⊗W, then 𝔅(W) ⊆ 𝔅(V). In particular, if 𝔅(W) is infinite dimensional, also is 𝔅(V).

Let G be a finite group, σ ∈ G, 𝒪_{σ}^{G} = 𝒪 its conjugacy class, C_{G}(σ) its centralizer and ρ ∈ { \atop C_{G}(σ)}. If H is a subgroup of G and σ ∈ H, then 𝒪_{σ}^{H} = 𝒪^{H} denotes the conjugacy class of σ in H.

Lemma 2.2. If dim 𝔅(𝒪^{H}, τ) = ∞ for all τ ∈ { \atop C_{H}(σ)}, then dim 𝔅(𝒪^{G}, ρ) =

∞ for all ρ ∈ { \atop C_{G}(σ)}.

Proof. Since M = Ind_{C_{G}(σ)}^{G}ρ = ⊕_{s∈𝒪^{G}} V_{s}, where V_{s} = {v ∈ V | δ(v) = s⊗v}, we have that M^{H} := ⊕_{s∈𝒪^{H}} V_{s} ⊆ M is a Yetter-Drinfeld module over H. □

Now assume that σ_{1}, σ_{2} ∈ H. Let 𝒪_{i} be the conjugacy class in H of σ_{i}. Assume that 𝒪_{1} ≠ 𝒪_{2}.

Lemma 2.3. If 𝔅(M(𝒪_{1}, τ_{1}) ⊕ M(𝒪_{2}, τ_{2})) has infinite dimension for all pairs τ_{1} ∈ { \atop C_{H}(σ_{1})}, τ_{2} ∈ { \atop C_{H}(σ_{2})}, then 𝔅(𝒪^{G}, ρ) is infinite dimensional.

Proof. As before, M = ⊕_{s∈𝒪^{G}} V_{s} and then ⊕_{s∈𝒪_{1}∪𝒪_{2}} V_{s} ⊆ M is a Yetter-Drinfeld module over H. □

2.3. The group 𝔸_{4} × ℤ/r for r odd. Let r be an odd integer and let G = 𝔸_{4} × ℤ/r. Assume that ℤ/r is generated by τ.

Lemma 2.4. Let 𝒪 be the conjugacy class of σ = ((1 2)(3 4), τ) in G. Then, dim 𝔅(𝒪, ρ) = ∞ for every ρ ∈ { \atop C_{G}(σ)}.

Proof. Apply Lemma 2.1 with σ_{1} = ((1 2)(3 4), τ), σ_{2} = ((1 3)(2 4), τ), σ_{3} = ((1 4)(2 3), τ), g_{1} = e, g_{2} = (1 3 2) × 1, g_{3} = g_{2}^{−1} and h = r + 2. □

Remark 2.5. The case r = 1 of this Lemma is known (see for example [AF1, Prop. 2.4]) and it is used to kill the conjugacy class of involutions in 𝔸_{4}.

2.4. The group SL(2, 3). Let G = SL(2, 3); recall |G| = 24. Here is one presentation of G by generators and relations:

SL(2, 3) ≃ ⟨x, y, z | x^{4} = y^{4} = z^{3} = 1, x^{2} = y^{2}, y^{−1}xy = x^{−1},
z^{−1}xz = y^{−1}, z^{−1}yz = yx^{−1}⟩.

This presentation can be realized by choosing

(1)  x = (\begin{matrix}0 & 2 \\ 1 & 0 \end{matrix}) , y = (\begin{matrix}1 & 1 \\ 1 & 2 \end{matrix}) , z = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) .

Let us consider the conjugacy class of σ = x ∈ G, explicitly

𝒪_{σ} = \{(\begin{matrix}0 & 2 \\ 1 & 0 \end{matrix}) , (\begin{matrix}0 & 1 \\ 2 & 0 \end{matrix}) , (\begin{matrix}1 & 2 \\ 2 & 2 \end{matrix}) , (\begin{matrix}2 & 1 \\ 1 & 1 \end{matrix}) , (\begin{matrix}1 & 1 \\ 1 & 2 \end{matrix}) , (\begin{matrix}2 & 2 \\ 2 & 1 \end{matrix})\} .

We numerate these elements as σ_{1}, …, σ_{6}, in this order. The centralizer C_{G}(σ) is the cyclic group of order 4 generated by σ.

Definition 2.6. The rack underlying the orbit of σ in SL(2, 3) will be denoted 𝒟_{2}^{3}.

Lemma 2.7. [FGV, Subs. 3.2] If (ρ, V) ∈ { \atop C_{G}(σ)}, then dim 𝔅(𝒪_{σ}, ρ) = ∞.

For completeness, we include a proof of this result.

Proof. The class 𝒪_{σ} is real because it contains all elements of order 4 in G; hence, we only need to consider ρ = χ ∈ { \atop C_{G}(σ)} such that χ(σ) = −1, cf. [AZ, 2.2]. Let

(2) g_{1} = id, g_{2} = σ_{3}, g_{3} = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) , g_{4} = σg_{3}, g_{5} = σg_{6}, g_{6} = g_{3}^{−1}.

Then g_{i}▷σ_{1} = σ_{j}. For any i, j, there is an index denoted i▷j and γ_{ij} ∈ C_{G}(σ) such that σ_{i}g_{j} = g_{i▷j}γ_{ij}. A straightforward computation shows that

γ = (γ_{ij}) = ( \begin{matrix}σ & σ^{−1} & 1 & σ^{2} & σ^{2} & 1 \\ σ^{−1} & σ & σ^{2} & 1 & 1 & σ^{2} \\ 1 & σ^{2} & σ & σ^{−1} & σ & σ \\ σ^{2} & 1 & σ^{−1} & σ & σ^{−1} & σ^{−1} \\ σ^{−1} & σ^{−1} & σ & σ & σ & σ^{−1} \\ σ & σ & σ^{−1} & σ^{−1} & σ^{−1} & σ \end{matrix} ) . (3)

Let v ∈ V − 0. We define

(4)  \begin{aligned}u_{1} & := g_{1}v + g_{2}v, & u_{3} & := g_{3}v + g_{4}v, & u_{5} & := g_{5}v + g_{6}v, \\ u_{2} & := g_{1}v − g_{2}v, & u_{4} & := g_{3}v − g_{4}v, & u_{6} & := g_{5}v − g_{6}v. \end{aligned}

By straightforward computations, we can see that, in this basis, M(𝒪, ρ) is a braided vector space of diagonal type with matrix given by

𝒬 = (\begin{matrix}−1 & −1 & 1 & −1 & 1 & −1 \\ −1 & −1 & 1 & −1 & 1 & −1 \\ 1 & −1 & −1 & −1 & −1 & 1 \\ 1 & −1 & −1 & −1 & −1 & 1 \\ −1 & 1 & −1 & 1 & −1 & −1 \\ −1 & 1 & −1 & 1 & −1 & −1 \end{matrix}) .

Hence M(𝒪, ρ) is of Cartan type with Dynkin diagram given by Figure 2; this is not of finite type, and dim 𝔅(𝒪, ρ) = ∞, by [H1]. □

[figure: Figure 2]

2.5. The group SL(2, 3)×ℤ/r for r a prime number. In this subsection, we present a useful variant of the criterium given in 2.4 that will be used in [AFGV]. Let G = SL(2, 3) × ℤ/r. Let us consider the conjugacy class 𝒪 of σ = (x, τ), where x = ( \begin{matrix}0 & 2 \\ 1 & 0 \end{matrix} ) and τ has order r. The centralizer of σ is C_{G}(σ) = ⟨x⟩ × ⟨τ⟩ ≃ ℤ/4 × ℤ/r. We consider Nichols algebras associated to

pairs (𝒪, ρ), where ρ = ρ_{1} ⊗ ρ_{2} ∈ { \atop C_{G}(σ)}, ρ_{1} ∈ ℤ̂/4 and ρ_{2} ∈ ℤ̂/r.

Lemma 2.8. If (ρ, V) ∈ { \atop C_{G}(σ)}, then dim 𝔅(𝒪, ρ) = ∞.

Proof. The braided vector space M(𝒪, ρ) is (V ⊗ W, c_{V ⊗W}), where (V, c_{V}) is the braided vector space associated to (𝒪_{x}, ρ_{1}), (W, c_{W}) is the braided vector space associated to (𝒪_{τ} , ρ_{2}), and

c_{V ⊗W} = (id ⊗ flip ⊗ id)(c_{V} ⊗ c_{W})(id ⊗ flip ⊗ id).

First notice that if r ≥ 3 the conjugacy class 𝒪 is quasi-real of type j, with j an odd number, and g = g^{j^{2}}. Then, by [FGV, Corollary 2.2] we only need to consider ρ(g) = ζ, where ζ = −1 or ζ is a cubic root of 1. Moreover, if r > 3, since r is a prime number, we only need to consider the case ρ(g) = −1. Also, if r = 2, then 𝒪 is a real conjugacy class and by [AZ, Lemma 2.2] we need to consider ρ(g) = −1. Therefore, we only need to consider ρ = χ_{(−1)} ⊗ ϵ or ρ = ϵ ⊗ χ, where χ(g) = ζ is a primitive r-root of 1. If ρ = χ_{(−1)} ⊗ ϵ, the result follows in a analogous way to the proof of Lemma 2.7. Assume then that ρ = ε ⊗ χ for r = 2 or 3. We call ν_{i} = (σ_{i}, τ) and h_{i} = (g_{i}, id), where σ_{i} and g_{i} are as in the proof of Lemma 2.7. Then h_{i} ▷ ν_{1} = ν_{i}, 1 ≤ i ≤ 6, and ν_{i}h_{j} = h_{i▷j}δ_{ij}, where δ_{ij} = (γ_{ij}, τ), with γ_{ij} given by (3), 1 ≤ i, j ≤ 6. Let v ∈ V − 0. We define W as the ℂ-span of {u_{l} | 1 ≤ l ≤ 6}, where

(5)  \begin{aligned}u_{1} & := h_{1}v + h_{2}v, & u_{3} & := h_{3}v + h_{4}v, & u_{5} & := h_{5}v + h_{6}v, \\ u_{2} & := h_{1}v − h_{2}v, & u_{4} & := h_{3}v − h_{4}v, & u_{6} & := h_{5}v − h_{6}v. \end{aligned}

By straightforward computations, we can see that, in this basis, M(𝒪, ρ) is a braided vector space of diagonal type with matrix given by

𝒬 = (\begin{matrix}ζ & ζ & ζ & −ζ & ζ & −ζ \\ ζ & ζ & ζ & −ζ & ζ & −ζ \\ ζ & −ζ & ζ & ζ & ζ & −ζ \\ ζ & −ζ & ζ & ζ & ζ & −ζ \\ ζ & −ζ & ζ & −ζ & ζ & ζ \\ ζ & −ζ & ζ & −ζ & ζ & ζ \end{matrix}) .

Hence, M(𝒪, ρ) is of Cartan type with Dynkin diagram of infinite type (if r = 2 the Dynkin diagram is A_{5}^{(1)}). Thus dim 𝔅(𝒪, ρ) = ∞. □

3. Nichols algebras over 𝔸_{m}

3.1. Notations on symmetric groups. Let σ ∈ 𝕊_{m}. We say that σ is of type (1^{n_{1}}, 2^{n_{2}}, …, m^{n_{m}}) if the decomposition of σ as product of disjoint cycles contains n_{j} cycles of length j, for every j, 1 ≤ j ≤ m. Let A_{j} = A_{1,j} ⋯ A_{n_{j},j} be the product of the n_{j} ≥ 0 disjoint j-cycles A_{1,j}, …, A_{n_{j},j} of σ. Then

σ = A_{1} ⋯ A_{m}; (6)

we shall omit A_{j} when n_{j} = 0. The even and the odd parts of σ are

σ_{e} := ∏_{j even} A_{j}, σ_{o} := ∏_{1<j odd} A_{j}. (7)

Thus, σ = A_{1}σ_{e}σ_{o} = σ_{e}σ_{o}; we need to define σ_{o} in this way for simplicity of some statements and proofs. We say also that σ has type (1^{n_{1}}, 2^{n_{2}}, …, σ_{o}), for brevity.

The centralizer 𝕊_{m}^{σ} = T_{1} × ⋯ × T_{m}, where

T_{j} = ⟨A_{1,j}, …, A_{n_{j},j}⟩ ⋊ ⟨B_{1,j}, …, B_{n_{j}−1,j}⟩ ≃ (ℤ/j)^{n_{j}} ⋊ 𝕊_{n_{j}}, (8)

1 ≤ j ≤ m. We describe the irreducible representations of the centralizers. If ρ = (ρ, V) ∈ { \atop C_{𝕊_{m}}(σ)}, then ρ = ρ_{1} ⊗ ⋯ ⊗ ρ_{m}, where ρ_{j} ∈ T̂_{j} has the form

(9)  ρ_{j} = Ind_{(ℤ/j)^{n_{j}} ⋊𝕊_{n_{j}}^{χ_{j}}}^{(ℤ/j)^{n_{j}} ⋊𝕊_{n_{j}}} (χ_{j} ⊗ μ_{j}),

with χ_{j} ∈ { \atop (ℤ/j)^{n_{j}}} and μ_{j} ∈ 𝕊̂_{n_{j}}^{χ_{j}} – see [S, Section 8.2].

Remark 3.1. Let σ ∈ 𝔸_{m}. Then σ = A_{1}σ_{e}σ_{o}, see (7); clearly, σ_{e}, σ_{o} ∈ 𝔸_{m}. Since σ_{e}, σ_{o} ∈ Z(C_{𝔸_{m}}(σ)), the center of C_{𝔸_{m}}(σ), ρ acts by a scalar on σ_{e} and σ_{o}, i. e. ρ(σ_{e}) = λ Id and ρ(σ_{o}) = λ̃ Id. Hence, q_{σσ} = λλ̃. Notice that if the orders of σ_{e} and σ_{o} are relatively prime and q_{σσ} = −1, then λ = −1 and λ̃ = 1.

We introduce some elements of 𝕊_{m} attached to a cycle α that will be used later. Let α = (i_{1} i_{2} i_{3} ⋯ i_{4n}) be a 4n-cycle in 𝔸_{m}. We define

g_{α} := ∏_{l=1}^{2n} (l 4n − l + 1) (10)

Thus, g_{α} ∈ 𝔸_{m} is an involution and g_{α} ▷ α = α^{−1}.

3.2. Scheme of the proof of Theorem 1.1. We proceed to the strategy of the proof of Theorem 1.1, postponing to later subsections the consideration of some particular cases. Let G = 𝔸_{m}, with m ≥ 6, σ ∈ G of type (1^{n_{1}}, 2^{n_{2}}, …, m^{n_{m}}), 𝒪 its conjugacy class and ρ ∈ { \atop C_{G}(σ)}. Assume that dim 𝔅(𝒪, ρ) < ∞. Then σ is real with even order and q_{σσ} = −1 by [AF1, 2.3]; 𝒪_{σ}^{𝔸_{m}} = 𝒪_{σ}^{𝕊_{m}} and [C_{𝕊_{m}}(σ) : C_{𝔸_{m}}(σ)] = 2 (see for instance [JL, Proposition 12.17]). Hence, any subrack of 𝒪_{σ}^{𝕊_{m}} is obviously a subrack of 𝒪_{σ}^{𝔸_{m}} and we may apply the techniques from [AF2].

(a) If j ≥ 6 is even and has an odd divisor, then n_{j} = 0. Otherwise, 𝒪_{σ}^{𝔸_{m}} contains a subrack of type 𝒟_{p}^{(2)}, with p odd prime, by [AF2, 2.11] and dim 𝔅(𝒪, ρ) = ∞ by [AF2, 2.9].

(b) n_{2^{k}} ≤ 2, for all k ≥ 2. Otherwise, 𝒪_{σ}^{𝔸_{m}} contains a subrack of type 𝒟_{3} by the proof of [AF2, 3.10] and dim 𝔅(𝒪, ρ) = ∞ by [AF2, 3.8].

(c) The type of σ_{e} is (2^{n_{2}}, 4^{n_{4}}), by Proposition 3.5.

So far, we have that

σ = A_{1}σ_{e}σ_{o}

where A_{1} is of type (1^{n_{1}}), σ_{e} is of type (2^{n_{2}}, 4^{n_{4}}), with n_{2} + n_{4} even and n_{4} ≤ 2, and σ_{o} is of type (3^{n_{3}}, 5^{n_{5}}, …).

(d) n_{4} > 0. Otherwise, σ is of type (1^{n_{1}}, 2^{n_{2}}, σ_{o}); here n_{2} is even, because (1^{n_{1}}, 2^{n_{2}}, σ_{o}) ∉ 𝔸_{m} if n_{2} is odd. Then we conclude by Prop. 3.6.

(e) n_{2} ≤ 2. Otherwise, 𝒪_{σ}^{𝔸_{m}} contains a subrack of type 𝒟_{3} by the proof of [AF2, 3.12] and dim 𝔅(𝒪, ρ) = ∞ by [AF2, 3.8] – note that σ ≠ σ^{−1} because n_{4} > 0.

(f) If n_{2} > 0, then n_{1} = 0. Otherwise, 𝒪_{σ}^{𝔸_{m}} contains a subrack of type 𝒟_{3} by the proof of [AF2, 3.9] and dim 𝔅(𝒪, ρ) = ∞ by [AF2, 3.8] – note that σ ≠ σ^{−1} because n_{4} > 0.

(g) σ_{o} is trivial by Prop. 3.7.

(h) The remaining types are: (2, 4), excluded by Prop. 3.9; (1^{n_{1}}, 4^{2}), excluded by Prop. 3.2; and (2^{2}, 4^{2}), excluded by Prop. 3.3.

3.3. The classes (1^{n_{1}}, 4^{2}) and (2^{2}, 4^{2}). We now apply the technique of the subgroup with H = SL(2, 3).

Proposition 3.2. Let G = 𝔸_{m} or 𝕊_{m}, σ ∈ G, 𝒪 the conjugacy class of σ and ρ ∈ { \atop C_{G}(σ)}. If the type of σ is (1^{n_{1}}, 4^{2}), then dim 𝔅(𝒪, ρ) = ∞.

Proof. The group SL(2, 3) acts faithfully on 𝔽_{3} × 𝔽_{3}, and also on 𝔽_{3} × 𝔽_{3} \ {(0, 0)}, which consists of 8 elements. Therefore, we get an injective morphism ψ : SL(2, 3) → 𝕊_{8} ⊆ 𝕊_{m}. Using a particular labelling of the elements, this map is given by x ↦ (1 3 2 6)(4 5 8 7), y ↦ (1 4 2 8)(3 7 6 5), z ↦ (1 4 7)(2 8 5), whence the image lies in 𝔸_{8} ⊆ 𝔸_{m}. By Lemma 2.7, the claims follows. □

Proposition 3.3. Let σ ∈ 𝔸_{12}, 𝒪 the conjugacy class of σ and ρ ∈ { \atop C_{𝔸_{12}}(σ)}. If the type of σ is (2^{2}, 4^{2}), then dim 𝔅(𝒪, ρ) = ∞.

Proof. As before, we have a faithful permutation action of SL(2, 3), which is the product ψ × φ, where ψ is the morphism in the proof of Proposition 3.2, and φ : SL(2, 3) → 𝔸_{4} is given by

x ↦ (9 10)(11 12), y ↦ (9 11)(10 12), z ↦ (9 11 12)

(notice that the group generated by (9 10)(11 12), (9 11)(10 12), (9 11 12) is isomorphic to 𝔸_{4}). The image of ψ × φ lies in 𝔸_{8} × 𝔸_{4} ⊆ 𝔸_{12} and the type of (ψ × φ)(x) is (2^{2}, 4^{2}). By Lemma 2.7, the claims follows. □

Remark 3.4. This argument applies also to the class of σ ∈ 𝕊_{m} with type (2^{2}, 4^{2}). This was dealt with transversal subracks in [AF2].

3.4. Remaining cases.

Proposition 3.5. Let σ ∈ 𝔸_{m} be of type (1^{n_{1}}, 2^{n_{2}}, 4^{n_{4}}, …, (2^{k})^{n_{2^{k}}} , σ_{o}), with k ≥ 3 and n_{2^{k}} > 0, 𝒪 the conjugacy class of σ in 𝔸_{m} and ρ = (ρ, V) ∈

{ \atop C_{𝔸_{m}}(σ)}. Then dim 𝔅(𝒪, ρ) = ∞.

Proof. As explained in Subsection 3.2 (b), we may assume n_{2^{k}} = 1 or 2.

(I) Assume that n_{2^{k}} = 1. Let α = (i_{1} i_{2} ⋯ i_{2^{k}}) be the 2^{k}-cycle appearing in the decomposition of σ as product of disjoint cycles, and we call

I := (i_{1} i_{3} i_{5} ⋯ i_{2^{k}−1}) and P := (i_{2} i_{4} i_{6} ⋯ i_{2^{k}}).

In the proof of [AF2, Lemma 2.11], it was shown that (a) I and P are disjoint 2^{k−1}-cycles,

(b) α^{2} = IP,

(c) αIα^{−1} = P, (hence σIσ^{−1} = P),

(d) P^{t}αP^{t} = α^{2t+1}, for all integer t.

For notational convenience, we set

r := 2^{k−3},  g̃_{l} := P^{2^{r}l}, 1 ≤ l ≤ 4.

Notice that (i) if k ≥ 4, then g̃_{l} = (P^{2^{r−1}l})^{2} ∈ 𝔸_{m}. (ii) if k = 3, then g̃_{4} = id and g̃_{2} = P^{2} are in 𝔸_{m}, whereas g̃_{1} = P and

g̃_{3} = P^{3} are not in 𝔸_{m}. For every 1 ≤ l ≤ 4, we define g_{l} = g̃_{l} in the case (i) or in the case (ii) with l = 2 or 4, and g_{l} = g̃_{l}α in the case (ii) with l = 1 or 3. Then, g_{l} ∈ 𝔸_{m}, 1 ≤ l ≤ 4. We define α_{l} := g_{l} ▷ α and

σ_{l} := g_{l} ▷ σ. (11)

Notice that σ_{l} = (g_{l} ▷ σ_{e}) σ_{o}, for all l. Then (σ_{l})_{1≤l≤4} is a subrack of 𝒪 of type 𝒟_{4} in the sense of [AF2, Def. 2.2]. Notice that α_{4} = α, α_{2} = α_{4}^{2^{k−1}+1} and α_{3} = α_{1}^{2^{k−1}+1}. Thus, σ_{2} = σ_{e}^{2^{k−1}+1}σ_{o} because σ_{e}^{2^{k−1}} = α^{2^{k−1}}. If we define τ_{l} := (g_{l} ▷ σ_{e})^{−1} σ_{o}, for all l, then (σ_{l})_{1≤l≤4} ∪ (τ_{l})_{1≤l≤4} is a subrack of 𝒪 of type 𝒟_{4}^{(2)}. Let

g := ∏_{t=2}^{k} ∏_{s=1}^{n_{2^{t}}} g_{A_{s,2^{t}}} ∈ 𝔸_{m},

see (10). Then g is an involution in 𝔸_{m} such that g ▷ σ = σ_{e}^{−1}σ_{o}. Let

h_{l} := g_{l}g, 1 ≤ l ≤ 4;

clearly, h_{l} ▷ σ = τ_{l}, 1 ≤ l ≤ 4. By straightforward computations, we have the following relations:

\begin{matrix}· & g_{4} & g_{1} & g_{2} & g_{3} \\ σ_{4} & g_{4} σ & g_{3} σα^{2r} & g_{2} σ_{2} & g_{1} σα^{−2r} \\ σ_{1} & g_{2} σα^{−2r} & g_{1} σ & g_{4} σα^{2r} & g_{3} σ_{2} \\ σ_{2} & g_{4} σ_{2} & g_{3} σα^{−2r} & g_{2} σ & g_{1} σα^{2r} \\ σ_{3} & g_{2} σα^{2r} & g_{1} σ_{2} & g_{4} σα^{−2r} & g_{3} σ \\ τ_{4} & g_{4} σ_{e}^{−1}σ_{o} & g_{3} σ_{e}^{−1}σ_{o}α^{2r} & g_{2} σ_{e}^{−2^{k−1}−1}σ_{o} & g_{1} σ_{e}^{−1}σ_{o}α^{−2r} \\ τ_{1} & g_{2} σ_{e}^{−1}σ_{o}α^{−2r} & g_{1} σ_{e}^{−1}σ_{o} & g_{4} σ_{e}^{−1}σ_{o}α^{2r} & g_{3} σ_{e}^{−2^{k−1}−1}σ_{o} \\ τ_{2} & g_{4} σ_{e}^{−2^{k−1}−1}σ_{o} & g_{3} σ_{e}^{−1}σ_{o}α^{−2r} & g_{2} σ_{e}^{−1}σ_{o} & g_{1} σ_{e}^{−1}σ_{o}α^{2r} \\ τ_{3} & g_{2} σ_{e}^{−1}σ_{o}α^{2r} & g_{1} σ_{e}^{−2^{k−1}−1}σ_{o} & g_{4} σ_{e}^{−1}σ_{o}α^{−2r} & g_{3} σ^{−1} \end{matrix}

\begin{matrix}· & h_{4} & h_{1} & h_{2} & h_{3} \\ σ_{4} & h_{4} σ_{e}^{−1}σ_{o} & h_{3} σ_{e}^{−1}σ_{o}α^{−2r} & h_{2} σ_{e}^{−2^{k−1}−1}σ_{o} & h_{1} σ_{e}^{−1}σ_{o}α^{2r} \\ σ_{1} & h_{2} σ_{e}^{−1}σ_{o}α^{2r} & h_{1} σ_{e}^{−1}σ_{o} & h_{4} σ_{e}^{−1}σ_{o}α^{−2r} & h_{3} σ_{e}^{−2^{k−1}−1}σ_{o} \\ σ_{2} & h_{4} σ_{e}^{−2^{k−1}−1}σ_{o} & h_{3} σ_{e}^{−1}σ_{o}α^{2r} & h_{2} σ_{e}^{−1}σ_{o} & h_{1} σ_{e}^{−1}σ_{o}α^{−2r} \\ σ_{3} & h_{2} σ_{e}^{−1}σ_{o}α^{−2r} & h_{1} σ_{e}^{−2^{k−1}−1}σ_{o} & h_{4} σ_{e}^{−1}σ_{o}α^{2r} & h_{3} σ_{e}^{−1}σ_{o} \\ τ_{4} & h_{4} σ & h_{3} σ α^{−2r} & h_{2} σ_{2} & h_{1} σ α^{2r} \\ τ_{1} & h_{2} σα^{2r} & h_{1} σ & h_{4} σ α^{−2r} & h_{3} σ_{2} \\ τ_{2} & h_{4} σ_{2} & h_{3} σ α^{2r} & h_{2} σ & h_{1} σ α^{−2r} \\ τ_{3} & h_{2} σ α^{−2r} & h_{1} σ_{2} & h_{4} σ α^{2r} & h_{3} σ \end{matrix}

Notice that α ∈ Z(C_{𝕊_{m}}(σ)) and α^{2} ∈ 𝔸_{m}; thus, α^{2r} ∈ Z(C_{𝔸_{m}}(σ)), and ρ(α^{2r}) acts by a scalar κ, with κ^{4} = 1 because

Id = ρ(id) = ρ(α^{2^{k}}) = ρ((α^{2r})^{4}) = κ^{4} Id .

We show that κ = ±1. If we call σ̃ = σ_{e}α^{−1}, then σ_{e}^{2r} = σ^{2r}α^{2r} and σ^{2r} ∈ Z(C_{𝔸_{m}}(σ)); thus, ρ(σ^{2r}) acts by a scalar κ̃. Now

Id = ρ(σ_{e}^{2r}) = ρ(σ^{2r})ρ(α^{2r}) = κ̃κ Id .

That is, 1 = κ̃κ. Now, σ̃ is product of 2^{t}-cycles with t ≤ k − 1. Then, σ^{2r} = σ^{2^{k−2}} and (σ^{2r})^{2} = σ^{2^{k−1}} = id. Hence, κ^{2} = 1, and κ = ±1.

Let v ∈ V − 0. We define W := ℂ-span of {u_{l}, w_{l} | 1 ≤ l ≤ 4}, where

(12)  \begin{aligned}u_{1} & := g_{4}v + g_{2}v, & w_{1} & := h_{4}w + h_{2}w, \\ u_{2} & := g_{4}v − g_{2}v, & w_{2} & := h_{4}w − h_{2}w, \\ u_{3} & := g_{1}v + g_{3}v, & w_{3} & := h_{1}w + h_{3}w, \\ u_{4} & := g_{1}v − g_{3}v, & w_{4} & := h_{1}w − h_{3}w. \end{aligned}

By straightforward computations, we can see that W is a braided vector subspace of M(𝒪, ρ) of Cartan type with matrix of coefficients given by

(\begin{matrix}Q & Q \\ Q & Q \end{matrix}) , where Q = (\begin{matrix}−1 & −1 & −κ & κ \\ −1 & −1 & −κ & κ \\ −κ & κ & −1 & −1 \\ −κ & κ & −1 & −1 \end{matrix}) ,

and Dynkin diagram given by Figure 3 which is not of finite type. Therefore, dim 𝔅(𝒪, ρ) = ∞, by [H1].

[figure: Figure 3]

(II) Assume that n_{2^{k}} = 2. Let A_{1,2^{k}} = (i_{1} i_{2} ⋯ i_{2^{k}}) and A_{2,2^{k}} = (i_{2^{k}+1} i_{2^{k}+2} ⋯ i_{2^{k+1}}) the two 2^{k}-cycles appearing in σ, and let I = I_{1}I_{2} and P = P_{1}P_{2}, with

I_{1} := (i_{1} i_{3} i_{5} ⋯ i_{2^{k}−1}),  I_{2} := (i_{2^{k}+1} i_{2^{k}+3} i_{2^{k}+5} ⋯ i_{2^{k+1}−1}),
P_{1} := (i_{2} i_{4} i_{6} ⋯ i_{2^{k}}),  P_{2} := (i_{2^{k}+2} i_{2^{k}+4} i_{2^{k}+6} ⋯ i_{2^{k+1}}).

For every 1 ≤ l ≤ 4, we define g_{l} = g̃_{l} in the case k ≥ 4 or in the case k = 3 with l = 0 or 2, and we define g_{l} = g̃_{l}A_{1,2^{k}} in the case (ii) with l = 1 or 3. Then, g_{l} ∈ 𝔸_{m}, 1 ≤ l ≤ 4. Now, we take σ_{l} as in (11), τ_{l}, h_{l}, 1 ≤ l ≤ 4, as in the case (I) above and we proceed in an analogous way. □

Proposition 3.6. Let σ ∈ 𝔸_{m} be of type (1^{n_{1}}, 2^{n_{2}}, σ_{o}), 𝒪 the conjugacy class of σ in 𝔸_{m} and ρ = (ρ, V) ∈ { \atop C_{𝔸_{m}}(σ)}. Then dim 𝔅(𝒪, ρ) = ∞.

Proof. Notice that n_{2} = 2k is even. Assume first that σ_{o} = e. For every l, 1 ≤ l ≤ k, we define

C_{l} = (4l − 3 4l − 2)(4l − 1 4l),
D_{l} = (4l − 3 4l − 1)(4l − 2 4l),
α_{l} = (4l − 2 4l − 1)(4l − 3 4l − 2) = (4l − 1 4l − 2 4l − 3).

It is easy to see that the group generated by C_{l}, D_{l} and α_{l} is isomorphic to 𝔸_{4}. Moreover, the group generated by

C = C_{1} ⋯ C_{k}, D = D_{1} ⋯ D_{k} and α = α_{1} ⋯ α_{k}

is also isomorphic to 𝔸_{4} and C is an involution, conjugate to σ in 𝔸_{m}. Then, the Nichols algebra 𝔅(𝒪, ρ) is infinite dimensional. Now, if σ_{o} ≠ e, as before, we have that σ belongs to a subgroup isomorphic to 𝔸_{4} × ⟨σ_{o}⟩. Then, the result follows from Lemma 2.4. □

In our next Proposition, we apply the technique of the octahedral subrack 𝔒 introduced in [AF2, Sec. 4], and based in results of [AHS].

Proposition 3.7. Let σ ∈ 𝔸_{m} be of type (1^{n_{1}}, 2^{n_{2}}, 4^{n_{4}}, σ_{o}), with n_{4} > 0 and σ_{o} ≠ id, 𝒪 the conjugacy class of σ and ρ ∈ { \atop C_{𝔸_{m}}(σ)}. Then dim 𝔅(𝒪, ρ) = ∞.

Proof. We can assume 0 < n_{4} ≤ 2 by Subsection 3.2 (b). We have two possibilities.

(i) Case n_{4} = 1. We assume σ = A_{2}(1 2 3 4)σ_{o}; so σ_{e} = A_{2}(1 2 3 4). The condition q_{σσ} = −1, implies that ρ acts by λ = −1 on σ_{e} and by λ̃ = 1 on σ_{o} –see Remark 3.1. We define

α_{1} = (1 2 3 4),  α_{2} = (1 2 4 3),  α_{3} = (1 3 2 4),
α_{4} = (1 3 4 2),  α_{5} = (1 4 2 3),  α_{6} = (1 4 3 2),

σ_{l} = A_{2}α_{l}σ_{o} and τ_{l} = A_{2}α_{l}σ_{o}^{−1}, 1 ≤ l ≤ 6. It is easy to see that the family (σ_{l}, τ_{l})_{1≤l≤6} is a subrack of 𝒪 of type 𝔒^{(2)}. Let g ∈ 𝔸_{m} such that g ▷ σ_{o} = σ_{o}^{−1} and g ▷ σ_{e} = σ_{e}; thus g ▷ σ = τ_{1}. Also g^{−1} ▷ σ_{o} = σ_{o}^{−1}. We check the conditions (H4)-(H7) of [AF2, Th. 4.11]:

ρ(σ_{6}) = ρ(A_{2}α_{6}σ_{o}) = ρ(σ_{e}^{−1}σ_{o}) = λ^{−1}λ̃ = −1,
ρ(τ_{1}) = ρ(A_{2}α_{1}σ_{o}^{−1}) = ρ(σ_{e}σ_{o}^{−1}) = λλ̃^{−1} = −1,
ρ(g^{−1}σ_{1}g) = ρ(A_{2}α_{1}σ_{o}^{−1}) = −1,
ρ(g^{−1}σ_{6}g) = ρ(A_{2}α_{6}σ_{o}) = ρ(σ_{e}^{−1}σ_{o}^{−1}) = q_{σσ} = −1.

Now the result follows from [AF2, Th. 4.11].

(ii) Case n_{4} = 2. We take σ = A_{2}(1 2 3 4)(5 6 7 8)σ_{o} and we define

α_{1} = (1 2 3 4)(5 6 7 8), α_{2} = (1 2 4 3)(5 6 8 7), α_{3} = (1 3 2 4)(5 7 6 8),
α_{4} = (1 3 4 2)(5 7 8 6), α_{5} = (1 4 2 3)(5 8 6 7), α_{6} = (1 4 3 2)(5 8 7 6).

Now we proceed in an analogous way to the previous case. □

To deal with de conjugacy class of type (2, 4) of 𝔸_{6} we need to recall a very useful theorem.

Theorem 3.8. [HS1, Theorem 8.6] Let g, h ∈ G and V = ⨁_{s∈𝒪_{g}} V_{s}, W = ⨁_{t∈𝒪_{h}} W_{t} be irreducible objects in _{G}^{G}𝒴𝒟. If dim 𝔅(V ⊕ W) is finitedimensional, then for all s ∈ 𝒪_{g} and t ∈ 𝒪_{h}, (st)^{2} = (ts)^{2}.

Proposition 3.9. Let σ ∈ 𝔸_{6} be of type (2, 4), 𝒪 the conjugacy class of σ in 𝔸_{6} and ρ = (ρ, V) ∈ { \atop C_{𝔸_{6}}(σ)}. Then dim 𝔅(𝒪, ρ) = ∞.

Proof. Let H be the subgroup of 𝔸_{6} generated by σ_{1} = (1 2)(3 5 4 6), σ_{2} = (1 2 5 3)(4 6) and σ_{3} = (2 3 4). Notice that H ≃ (ℤ/3 × ℤ/3) ⋊ ℤ/4 has order 36. Also, it is easy to see that σ_{1} and σ_{2} are not conjugate in H. If 𝒪_{i} is the conjugacy class of σ_{i} in H, then 𝒪_{i} has 9 elements. Let V_{i} be an irreducible object in _{H}^{H}𝒴𝒟 of the form M(𝒪_{i}, ρ_{i}), for ρ_{i} ∈ { \atop C_{H}(σ_{i})}. Since (σ_{1}σ_{2})^{2} ≠ (σ_{2}σ_{1})^{2} then, by Theorem 3.8, dim 𝔅(V_{1} ⊕ V_{2}) = ∞. Now, the result follows from Lemma 2.3. □

Acknowledgements. We have used [GAP] to perform some computations. We thank István Heckenberger and Hans-Jürgen Schneider for interesting discussions.

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N. A., F. F. : Facultad de Matemática, Astronomía y Física, Universidad Nacional de Córdoba. CIEM – CONICET. Medina Allende s/n (5000) Ciudad Universitaria, Córdoba, Argentina

M. G., L. V. : Departamento de Matemática – FCEyN, Universidad de Buenos Aires, Pab. I – Ciudad Universitaria (1428) Buenos Aires – Argentina

L. V. : Instituto de Ciencias, Universidad de Gral. Sarmiento, J.M. Gutierrez 1150, Los Polvorines (1653), Buenos Aires – Argentina

   E-mail address: (andrus, fantino)@famaf.unc.edu.ar
   E-mail address: (matiasg, lvendramin)@dm.uba.ar