ON NICHOLS ALGEBRAS OVER SL(2, 𝔽_{q}) AND GL(2, 𝔽_{q})

SEBASTIÁN FREYRE, MATÍAS GRAÑA, AND LEANDRO VENDRAMIN

Abstract. We compute necessary conditions on Yetter-Drinfeld modules over the groups SL(2, 𝔽_{q}) and GL(2, 𝔽_{q}) to generate finite dimensional Nichols algebras. This is a first step towards a classification of pointed Hopf algebras with a group of group-likes isomorphic to one of these groups.

1. Introduction

Hopf algebras and their variants appear in many contexts; Quantum Groups [1], Rational and Logarithmic Conformal Field Theories [2, 3], Quantum Field Theories [4] come to mind. This paper is part of the long problem of classifying pointed Hopf algebras. To this end, the best tool known so far is the so called “Lifting procedure” [5]. The main ingredient of the Lifting procedure are Nichols algebras on Yetter-Drinfeld modules over group algebras. The classification of finite dimensional Nichols algebras over group algebras turns out to be a hard problem: when one restricts the attention to abelian groups, it includes the classification of semisimple Lie algebras [6]; when one considers non-abelian groups, only a few (genuine) examples are known [7] and we possess a very limited amount of general results (see [8] for an account).

In last years, two main directions were pursued to classify finite dimensional pointed Hopf algebras over non-abelian groups. One of them is to generate Nichols algebras by means of racks and 2-cocycles [8], getting examples that can be present for many groups. The other one is to concentrate on certain (families of) groups and classify which finite dimensional Nichols algebras can appear. In the last vein, let us cite the recently appeared papers: [9, 10, 11] (where symmetric groups are considered). The present paper can be thought of as an analog of these, but for some finite groups of Lie type: GL(2, 𝔽_{q}) and SL(2, 𝔽_{q}). As in those papers, using the tools in [6] we are able to rule out most of the Yetter-Drinfeld modules over these groups, as they generate infinite dimensional Nichols algebras.

After applying these techniques, the cases we are left with are difficult and one needs stronger tools to compute them. As an example, we point out that when q = 5, the conjugacy classes C_{i} (i = 3, …, 6) in Table 2 are isomorphic as racks to that of the faces of a dodecahedron, while one of the two classes C_{8} is isomorphic to the rack of the faces of an icosahedron (see [7]). The Nichols algebras generated by these racks with a negative 2-cocycle are not yet known.

One of the main results of this paper is the following theorem concerning Nichols algebras over the group SL(2, 𝔽_{q}):

2000 Mathematics Subject Classification. 16W30.

This work was partially supported by CONICET and ANPCyT (Argentina).

Theorem. If the Nichols algebra 𝔅(C) associated to a conjugacy class C of the group SL(2, 𝔽_{q}) is finite dimensional then q is odd and C is one of the classes C_{i} of the table 2 for i = 2, 5, 6, 7, 8.

Necessary conditions over the representations are explicitly given in propositions 3.3, 3.4, 3.5 and 3.9.

In section 4 we analyze Nichols algebras over GL(2, 𝔽_{q}) and as a consequence we obtain necessary conditions over the representations to get finite-dimensional Nichols algebras over this group.

2. Preliminaries and conventions

2.1. Yetter Drinfeld modules over kG. In what follows G is a finite group and k is an algebraically closed field of characteristic 0. Recall that a kG-comodule V is just a G-graded vector space: V = ⊕_{g∈G}V_{g}, where V_{g} = {v ∈ V | δ(v) = g ⊗ v}. A Yetter-Drinfeld module over kG is a left kG-module and a left kG-comodule V satisfying the following compatibility condition

δ(g · v) = ghg^{−1} ⊗ g · v, for v ∈ V_{h}

We denote by _{kG}^{kG}YD the category of Yetter-Drinfeld modules over kG, the morphisms of which are the maps of modules and comodules. This is a braided category, with braiding given by

(1)  c_{M,N} : M ⊗ N → N ⊗ M, c(m ⊗ n) = gn ⊗ m, for m ∈ M_{g}.

Let g ∈ G. We write Z_{g} for the centralizer Z_{g} = {x ∈ G | xgx^{−1} = g}. We write C_{g} for the conjugacy class C_{g} = {xgx^{−1} | x ∈ G}. Let ρ : Z_{g} → GL(W) be a representation. We denote by V (g, ρ) the space Ind_{Z_{g}}^{G}ρ, endowed with the comodule structure δ(h ⊗ w) = ghg^{−1} ⊗ (h ⊗ w). It is folklore that V (g, ρ) is irreducible iff ρ is, that _{kG}^{kG}YD is semisimple iff kG is, and that in this case the simple modules are given by V (g, ρ), letting g run over representatives of conjugacy classes and ρ on irreducible representations of Z_{g}. In this work we need only to consider representations of degree 1 since all the centralizers we have to deal with are abelian.

Let C = C_{g} = {g_{i} | i ∈ I} be the conjugacy class of g, with g_{i} = x_{i}gx_{i}^{−1}, and let v_{i} = x_{i} ⊗ 1 ∈ V (g, χ), where χ = ρ ∈ Ẑ_{g} is a character. Let T ⊆ I be a subset such that g_{i}g_{j} = g_{j}g_{i} for all i, j ∈ T. Let V_{T} ⊆ V (g, χ) be the subspace generated by {v_{i} | i ∈ T }. Then the braiding restricted to V_{T} is of diagonal type, given by

(2)  c(v_{i} ⊗ v_{j}) = q_{ij}v_{j} ⊗ v_{i}, where q_{ij} = χ(x_{j}^{−1}g_{i}x_{j}) ∈ k.

Indeed, by (1), if v_{j} = x_{j} ⊗ w, we have

c(v_{i} ⊗ v_{j}) = g_{i}v_{j} ⊗ v_{i} = (x_{j}x_{j}^{−1}g_{i}x_{j} ⊗ w) ⊗ v_{i}
= (x_{j} ⊗ x_{j}^{−1}g_{i}x_{j}w) ⊗ v_{i} = (x_{j} ⊗ q_{ij}w) ⊗ v_{i},

since x_{j}^{−1}g_{i}x_{j} ∈ Z_{g}.

We write q = q_{T} = (q_{ij}). If T ⊆ I and T′ ⊆ C, T′ = {g_{i} | i ∈ T }, then we abuse notation by calling q_{T′} = q_{T} and V_{T′} = V_{T}.

2.2. Notations. Throughout the paper, p will stand for a prime number and q for a power of p.

If ρ = χ is a character of Z_{g}, we usually write the Nichols algebra generated by the Yetter-Drinfeld module V (g, ρ) by 𝔅(C, χ) or just 𝔅(C) when no confusion can arise.

If n ∈ ℕ, we write R_{n} ⊂ k for the set of primitive n-th roots of unity in k. The order of an element h in a group will be denoted by |h|.

We recall from [6] the Dynkin diagram notation for braidings of diagonal type. If {v_{1}, ⋯, v_{ℓ}} is a basis of the braided vector space V and

c(v_{i} ⊗ v_{j}) = q_{ij}v_{j} ⊗ v_{i},

then the Dynkin diagram of c has one vertex for each i = 1, ⋯, ℓ, and vertices i and j are joined by an edge iff q_{ij}q_{ji} ≠ 1. Moreover, vertices and edges have labels as follows: the vertex i has as label the number q_{ii}, while the edge between i and j (if any) has as label the number q_{ij}q_{ji}.

2.3. Needed lemmas.

Lemma 2.1. Let W be a two-dimensional space with a braiding of diagonal type. Assume that the Dynkin diagram of W is given by [figure: ζ; μ; ζ] and suppose that dim 𝔅(W) < ∞. Then the Dynkin diagram is among the following ones:

Dynkin diagram  fixed parameter
[figure: α; α]  α ∈ k^{×}
[figure: α; α^{−1}; α]  α ≠ 1
[figure: −1; α; −1]  α ≠ ±1
[figure: −α^{−2}; −α^{3}; −α^{−2}]  α ∈ R_{12}
[figure: −α^{−2}; α; −α^{−2}]  α ∈ R_{12}

Proof. This follows at once by inspection on [12, Table 1] □

The following corollary appears in different forms in [9, 10, 11]. We prove it here for the reader’s convenience.

Corollary 2.2.

(1) Assume there exists x ∈ G such that xgx^{−1} = g^{n} ≠ g. Let T = {g, g^{n}}

and write \frac{1}{n} = n^{−1} (mod |g|). Then q = ( \begin{matrix}α & α^{\frac{1}{n}} \\ α^{n} & α \end{matrix} ), α = χ(g). If

dim 𝔅(C, ρ) < ∞ then α = −1 or α ∈ R_{3}. If moreover g^{n^{2}} ≠ g, then α = −1.

(2) As a particular case, if g^{n} = g^{−1} and dim 𝔅(C, ρ) < ∞ then g has even

order and q = ( \begin{matrix}−1 & −1 \\ −1 & −1 \end{matrix} ).

Proof. We consider first the case g^{n^{2}} = g. After using Lemma 2.1 we obtain that α^{n+ \frac{1}{n}} = 1 or α^{n+ \frac{1}{n} +1} = 1. If |α| divides n^{2} + 1 then, since |α| divides n^{2} − 1, α = −1 (α = 1 would imply dim 𝔅(C, ρ) = ∞). If |α| | n^{2} + n + 1 then |α| divides n + 2 and then α ∈ R_{3}. Now we consider the case g^{n^{2}} ≠ g, i.e. g, g^{n} and g^{n^{2}} are different and they belong to Z_{g} ∩ C_{g}. Then, if T = {g, g^{n}, g^{n^{2}}} we have

q_{T} = ( \begin{matrix}α & α^{\frac{1}{n}} & α^{\frac{1}{n^{2}}} \\ α^{n} & α & α^{\frac{1}{n}} \\ α^{n^{2}} & α^{n} & α \end{matrix} ) .

By inspection on [12, Table 2], the only possibilities for q_{T} to produce a finite dimensional Nichols algebra are either

• α = −1, or

• α^{\frac{1}{n} +n} = 1 and α^{\frac{1}{n^{2}} +n^{2}} = 1 (but then α = −1), or
• α^{\frac{1}{n} +n} = 1 and α^{\frac{1}{n^{2}} +n^{2}+1} = 1 (but then α = 1), or
• α^{\frac{1}{n^{2}} +n^{2}} = 1 and α^{\frac{1}{n} +n+1} = 1 (but then α = 1).

This completes the proof. □

Lemma 2.3. (1) Let W be a three-dimensional space with a braiding of diagonal type. Assume that the Dynkin diagram of W is

[figure: α; α; α; β; β; β]

and suppose that dim 𝔅(W) < ∞. Then α = −1 and β ∈ R_{3}.

(2) Suppose that the Dynkin diagram contains a cycle of length ≥ 4. Then dim 𝔅(W) = ∞.

Proof. The first part follows by inspection on [12, Table 2], while the second is [6, Lemma 20]. □

3. Nichols algebras over SL(2, 𝔽_{q})

In this section, E = 𝔽_{q^{2}} will be the quadratic extension of 𝔽_{q} and \overline{x} will be the Galois conjugate of x ∈ E. Recall that the order of SL(2, 𝔽_{q}) is (q − 1)q(q + 1). The conjugacy classes of SL(2, 𝔽_{q}) ar given in Tables 1 and 2 (see [13, §5.2] for q odd and [14] for q even).

3.1. The case q even. There are q + 1 conjugacy classes divided in 4 types:

Table 1: Conjugacy classes in SL(2, 𝔽_{q}), q even.

Type  Representative  Size Number Centralizer
C_{1}  I = c_{1} = ( \begin{matrix}1 \\ & 1 \end{matrix} )  1  1  SL(2, 𝔽_{q})
C_{2}  c_{2} = \binom{1 1}{1}  q^{2} − 1  1  𝔽_{q}
C_{3}  c_{3}(x) = ( \begin{matrix}x \\ & x^{−1} \end{matrix} ) (x ≠ 1)  q(q + 1) \frac{(q−2)}{2}  𝔽_{q}^{×}
C_{4} c_{4}(x) = \binom{1}{1 x + \overline{x}} (x ∈ E \ 𝔽_{q}) (q − 1)q  \frac{q}{2}  cyclic

We first consider the case q = 2. We have SL(2, 𝔽_{q}) ≃ 𝕊_{3}. Nichols algebras generated by irreducible Yetter-Drinfeld modules over 𝕊_{3} were studied in [15]. They are infinite dimensional except for 𝔅(τ, sgn), which is 12-dimensional (here τ is a transposition and sgn is the non-trivial character of its centralizer).

We now consider the case q > 2.

Proposition 3.1. Let q = 2^{n} for n ≥ 2. For each i = 1, …, 4 and for any representation χ of the centralizer of c_{i}, we have dim 𝔅(C_{i}, χ) = ∞.

Proof. We consider each class separately. The class C_{1} gives the trivial braiding.

For C_{2}, we take, for a ∈ 𝔽_{q}^{×}, x_{a} = ( \begin{matrix}a \\ & a^{−1} \end{matrix} ). Then g_{a} = x_{a}c_{2}x_{a}^{−1} = \binom{1 a^{2}}{1}.

The centralizer of c_{2} is the abelian group 𝔽_{q} embedded in SL(2, 𝔽_{q}) as \binom{1 𝔽_{q}}{1}.

If χ : 𝔽_{q} → ℂ is a character of the centralizer, we get q_{ab} = χ(x_{b}^{−1}g_{a}x_{b}) = χ(g_{ab^{−1}}),

whence q_{ab}q_{ba} = χ \binom{1 a^{2}b^{−2} + a^{−2}b^{2}}{1} (we write q_{uv} for q_{g_{u}g_{v}}). Since q = 2^{n},

χ takes values on ±1. If q_{aa} = 1, then dim 𝔅(C_{2}, χ) = ∞, so we may assume q_{aa} = −1.

If there exists a ∈ 𝔽_{q} \ \{0, 1\} such that χ \binom{1 a^{2} + a^{−2}}{1} = −1, then we get

q_{1,a}q_{a,1} = q_{a,a^{2}}q_{a^{2},a} = ⋯ = q_{a^{m},1}q_{1,a^{m}} = −1,

where a^{m+1} = 1. This implies that the space V_{T} contains a cycle of length m ≥ 3 with edges labelled by −1, whence 𝔅(C_{2}, χ) is infinite dimensional by Lemma 2.3.

Assume then that χ \binom{1 a^{2} + a^{−2}}{1} = 1 for all a ≠ 0, 1. But then, for all x in

the subgroup generated by the elements of the form a^{2} + a^{−2} (a ≠ 0, 1), we get

χ \binom{1 x}{1} = 1. Take now any a ∈ 𝔽_{q} \ \{0, 1\}, and let r be the order of a^{2} + a^{−2}.

Since r is odd, 1 = (a^{2} + a^{−2})^{r} is in the subgroup generated by elements a^{2i} + a^{−2i}

for 0 < i ≤ r, but this contradicts the fact that q_{aa} = χ \binom{1 1}{1} = −1.

For the class C_{3}, the centralizer of c_{3}(x) is the cyclic group 𝔽_{q}^{×}. It is easy to see that c_{3}(x) and c_{3}(x^{−1}) = c_{3}(x)^{−1} are conjugate, whence they both have the same odd order. Then by Corollary 2.2 (2), dim 𝔅(C_{3}) = ∞.

Finally, for the class C_{4}, we have that the centralizer of c_{4}(x) is a subgroup of the cyclic group E^{×}, hence it is cyclic. Again, it is easy to see that both c_{4}(x)

and c_{4}(x)^{−1} = \binom{x + \overline{x} 1}{1} are conjugate and they have odd order. Then, by

Corollary 2.2 (2), dim 𝔅(C_{4}) = ∞. □

3.2. The case q odd. There are q + 4 conjugacy classes divided in 8 types, displayed in Table 2.

We first consider the case q = 3. The conjugacy class C_{1} in Table 2 gives infinite dimensional Nichols algebras. Class C_{2} gives either exterior algebras or infinitedimensional algebras. Classes C_{3}, C_{4}, C_{5} and C_{6} give the rack of order 4 associated to the vertices of a tetrahedron (see [16]). Class C_{7} is not present in this case, and class C_{8} gives an infinite dimensional Nichols algebra of diagonal type, after changing the basis in a similar way as in [16, Remark 5.2.1].

Table 2: Conjugacy classes in SL(2, 𝔽_{q}), q odd.

Type  Representative  Size Number Centralizer
C_{1}  c_{1} = ( \begin{matrix}1 \\ & 1 \end{matrix} )  1  1  SL(2, 𝔽_{q})
C_{2}  c_{2} = ( \begin{matrix}−1 \\ & −1 \end{matrix} )  1  1  SL(2, 𝔽_{q})
C_{3}  c_{3} = \binom{1 1}{1}  \frac{q^{2}−1}{2}  1  ℤ_{2} × 𝔽_{q}
C_{4}  c_{4} = \binom{1 x}{1} (\sqrt{x} ∉ 𝔽_{q})  \frac{q^{2}−1}{2}  1  ℤ_{2} × 𝔽_{q}
C_{5}  c_{5} = \binom{−1 1}{−1}  \frac{q^{2}−1}{2}  1  ℤ_{2} × 𝔽_{q}
C_{6}  c_{6} = \binom{−1 x}{−1} (\sqrt{x} ∉ 𝔽_{q})  \frac{q^{2}−1}{2}  1  ℤ_{2} × 𝔽_{q}
C_{7}  c_{7}(x) = ( \begin{matrix}x \\ & x^{−1} \end{matrix} ) (x ≠ ±1)  q(q + 1) \frac{(q−3)}{2}  𝔽_{q}^{×}
C_{8} c_{8}(x) = \binom{−1}{1 x + \overline{x}} (x ∈ E \ 𝔽_{q}) (q − 1)q  \frac{q−1}{2}  cyclic

We now consider the case q > 3.

Remark 3.2. Conjugation by ( \begin{matrix}x \\ & 1 \end{matrix} ) ∈ GL(2, 𝔽_{q}) gives an automorphism of

SL(2, 𝔽_{q}) which switches C_{3} with C_{4}, and C_{5} with C_{6}. Hence, when classifying the irreducible Yetter-Drinfeld modules which produce finite dimensional Nichols algebras over C_{3}, one automatically also classifies those over C_{4}. Ditto for C_{5} and C_{6}.

Furthermore, if one is only interested in the conjugacy classes as racks, it turns out that C_{i} (i = 3, …, 6) are all isomorphic. Indeed, consider the projection SL(2, 𝔽_{q}) → PSL(2, 𝔽_{q}); it is injective when restricted to these classes. Further, it maps C_{3} and C_{5} to the same conjugacy class in PSL(2, 𝔽_{q}) when −1 is a square in 𝔽_{q}, and it maps C_{3} and C_{6} to the same conjugacy class when −1 is not a square in 𝔽_{q}.

Proposition 3.3. dim 𝔅(C_{1}) = ∞. If dim 𝔅(C_{2}) < ∞, then 𝔅(C_{2}) is the exterior algebra.

Proof. The first statement is clear since in this case the braiding is the usual flip x ⊗ y ↦ y ⊗ x, and therefore the Nichols algebra is the symmetric algebra. In the second case, if χ(c_{2}) = 1 we get again an infinite dimensional Nichols algebra. Thus, we must have χ(c_{2}) = −1, and the braiding is given by x ⊗ y ↦ −y ⊗ x, wich yields an exterior algebra. □

Proposition 3.4. If dim 𝔅(C_{7}) < ∞, then q_{T} = ( \begin{matrix}−1 & −1 \\ −1 & −1 \end{matrix} ) and x has even

order, where T = {c_{7}(x), c_{7}(x^{−1})} (notice that c_{7}(x^{−1}) and c_{7}(x) are conjugate).

Proof. It follows from Corollary 2.2 (2). □

Proposition 3.5. Let x ∈ E \ 𝔽_{q}. Then c_{8}(x) and c_{8}(x)^{−1} are conjugate, and

we take T = \{c_{8}(x), c_{8}(x)^{−1}\}. If dim 𝔅(C_{8}) < ∞, then q_{T} = ( \begin{matrix}−1 & −1 \\ −1 & −1 \end{matrix} ) (in

particular, c_{8}(x) has even order).

Proof. The proof follows from Corollary 2.2 (2), once we see that c_{8}(x) and c_{8}(x)^{−1} are conjugate. To this end, we consider the quadratic form

ϕ(u, v) = u^{2} + uv(x + \overline{x}) + v^{2}

over 𝔽_{q}. It is easy to see that this form has rank 2. Then, by [17, Proposition 4 (1.7)], it represents all elements of 𝔽_{q}^{×} and in particular it represents −1. Thus, there exist a, c ∈ 𝔽_{q} such that a^{2} + ac(x + \overline{x}) + c^{2} = −1. This implies that

( \begin{matrix}a & c + a(x + \overline{x}) \\ c & −a \end{matrix} ) ∈ SL(2, 𝔽_{q}). Furthermore,

( \begin{matrix}a & c + a(x + \overline{x}) \\ c & −a \end{matrix} ) \binom{−1}{1 x + \overline{x}} ( \begin{matrix}a & c + a(x + \overline{x}) \\ c & −a \end{matrix} )^{−1} = \binom{x + \overline{x} 1}{−1}

finishing the proof. □

Proposition 3.6. dim 𝔅(C_{3}) = ∞. The same holds for 𝔅(C_{4}).

Proof (case q ≠ 3^{2n+1}). Let α = χ(c_{3}) (χ ∈ ℤ̂_{2} ×𝔽̂_{q}). First of all note that |c_{3}| = p. Thus, if dim 𝔅(C_{3}) < ∞, we may suppose that |α| = p (if α = 1 then 𝔅(C_{3}) is infinite dimensional). Let us consider the case q ≡ 1 (mod 4). In this case there exists an element a ∈ 𝔽_{q} such that a^{2} = −1. Then

( \begin{matrix}a \\ & a^{−1} \end{matrix} ) \binom{1 1}{1} ( \begin{matrix}a^{−1} \\ & a \end{matrix} ) = \binom{1 −1}{1} = \binom{1 1}{1}^{−1}.

Now the Proposition follows from Corollary 2.2 (2), since c_{3} must have an even order.

If q ≡ 3 (mod 4) we have (since p ≠ 3),

c_{3} ≠ c_{3}^{4} = \binom{1 4}{1} = ( \begin{matrix}2 \\ & 2^{−1} \end{matrix} ) \binom{1 1}{1} ( \begin{matrix}2^{−1} \\ & 2 \end{matrix} )

and then, by Corollary 2.2 (1), we get dim 𝔅(C_{3}) = ∞.

The statement about C_{4} follows from Remark 3.2. □

We consider now the case q = 3^{2n+1}. For this, we need Lemma 3.7 and Corollary 3.8 below, which we could not find in the literature. Let φ denote the Euler function, φ(n) = #{m ∈ ℕ | m < n and m is coprime to n}.

Lemma 3.7. If n ∈ ℕ is such that 3 ∤ n and 4 ∤ n, then φ(n) > (\frac{n}{2})^{\frac{3}{4}} .

Proof. If n is even, n = 2p_{1}^{r_{1}}p_{2}^{r_{2}} ⋯ p_{N}^{r_{N}}, where 5 ≤ p_{1} < p_{2} < ⋯ < p_{N} are prime numbers. Then

\frac{φ(n)^{2}}{n} = \frac{1}{2} ∏_{i} \frac{(p_{i} − 1)^{2}p_{i}^{2r_{i}−2}}{p_{i}^{r_{i}}} = \frac{1}{2} ∏_{i}(p_{i} − 1)^{2}p_{i}^{r_{i}−2}.

Since (p_{i} − 1)^{2} > p_{i}^{3/2} ≥ p_{i}^{2− \frac{r_{i}}{2}} for r_{i} ≥ 1, we have

\frac{φ(n)^{2}}{n} > \frac{1}{2} ∏ pi^{\frac{r_{i}}{2}} = \frac{\sqrt{n}}{2\sqrt{2}}.

If n is odd, φ(n) = φ(2n) > n^{\frac{3}{4}} > (\frac{n}{2})^{\frac{3}{4}} . □

Corollary 3.8. \frac{φ(3^{p}−1)}{p} > 3^{\frac{p−1}{2}} for all odd prime number p.

Proof. If p = 3, 5 or 7, it follows immediately. Otherwise, by the Lemma,

\frac{φ(3^{p} − 1)}{p} > \frac{(3^{p} − 1)^{3/4}}{2^{3/4}p} > \frac{3^{3p/4}}{2p} > 3^{\frac{p−1}{2}} .

□

We finish now with the proof of Proposition 3.6.

End of proof of Proposition 3.6 (case q = 3^{2n+1}). Assume first that r = 2n+1 is a prime number. For z ∈ 𝔽_{q}, we denote f_{z} = f_{z}(X) ∈ 𝔽_{3}[X] the minimal polynomial of z. We consider the sets

I = \{f_{x^{2}} ∈ 𝔽_{3}[X] | x^{2} ∈ 𝔽_{3^{r}} \ 𝔽_{3}, |x^{2}| = \frac{3^{r} − 1}{2}\} , and
S = {X^{r} + a_{r−1}X^{r−1} + ⋯ + a_{1}X − 1 ∈ 𝔽_{3}[X] | ∀i = 1, …, r, a_{r−i} + a_{i} = 0} .

Note that #I = ^{\frac{φ( \frac{3^{r} −1}{2} )}{r}} = \frac{φ(3^{r}−1)}{r} (because all minimal polynomials of elements in 𝔽_{3^{r}} \ 𝔽_{3} have degree r, since r is a prime number). We have that #S = 3^{\frac{r−1}{2}} < #I by Corollary 3.8. Then there exists f ∈ I \ S, and we chose such an f. Since f(0) = (−1)^{r} · norm(x^{2}) = −1, in order for f not to be in S, there exists an i such that a_{r−i} + a_{i} ≠ 0.

Consider now x ∈ 𝔽_{3^{r}}^{×} of order \frac{3^{r}−1}{2}, and let X = \{x, x^{3}, …, x^{3^{r−1}}\} be the orbit of x by the action of the Galois group of the extension 𝔽_{3^{r}}/𝔽_{3}. Then ∏_{y∈X} y = 1 (because x is a square). Also, if ∅ ≠ S ⊆ X and ∏_{y∈S} y ∈ 𝔽_{3}^{×}, then S = X. Indeed,

∏_{y∈X} y = x^{∑_{i=0}^{r−1} 3^{i}} = x^{\frac{3^{r} −1}{2}} = 1. Now, if z = ∏_{y∈S} y, and S ≠ X, then z = x^{s} for some s < \frac{3^{r}−1}{2}. Note that x is a square and s < |x|, which implies that x^{s} ≠ ±1. Let

A = {a^{2} + a^{−2} | a ∈ 𝔽_{3^{r}}^{×} \ (R_{1} ∪ R_{2} ∪ R_{3} ∪ R_{4} ∪ R_{6})} = {a^{2} + a^{−2} | a ∈ 𝔽_{3^{r}} \ 𝔽_{3}}.

Now,

a_{i} = ∑_{\begin{matrix}Y⊂X \\ #Y=r−i \end{matrix}} ∏_{y∈Y} y, a_{r−i} = ∑_{\begin{matrix}Y⊂X \\ #Y=i \end{matrix}} ∏_{y∈Y} y = ∑_{#Y=r−i} ∏_{y∈Y} \frac{1}{y} .

Thus, a_{r−i} + a_{i} ∈ (A), the subgroup generated by A. Then, 1 ∈ (A).

Let now χ be a character of the centralizer of c_{3}. If χ \binom{1 t}{1} ≠ 1 for some

t ∈ A, then we get a Dynkin diagram with a cycle of length ≥ 4, which is of infinite type. Otherwise, since 1 ∈ (A), we get q_{11} = 1, which also yields an infinite dimensional Nichols algebra by Lemma 2.3.

If r = 2n + 1 is not a prime number, one can repeat the same argument with r′, a prime factor of r, taking 𝔽_{3^{r′}} as a subfield of 𝔽_{3^{r}}. □

Proposition 3.9. If p ≠ 3 and dim 𝔅(C_{5}) < ∞, then χ = sgn × ε, where sgn is the non-trivial representation of ℤ_{2} and ε is the trivial representation of 𝔽_{q}. The same statement goes for 𝔅(C_{6}).

Proof. As before, for a ∈ 𝔽_{q}^{×}, let

x_{a} = ( \begin{matrix}a \\ & a^{−1} \end{matrix} ) , g_{a} = x_{a}c_{5}x_{a}^{−1} = \binom{−1 a^{2}}{−1} .

Then q_{ab} = χ \binom{−1 a^{2}b^{−2}}{−1}, and q_{a1}q_{1a} = χ \binom{1 −(a^{2} + a^{−2})}{1}. For t ∈ 𝔽_{q},

let n_{t} = \binom{1 t}{1}. Let

A = {−(a^{2} + a^{−2}) | a ∈ 𝔽_{q}^{×} \ R_{1} ∪ R_{2} ∪ R_{3} ∪ R_{4} ∪ R_{6}}.

If there exists t ∈ A with χ(n_{t}) ≠ 1, then dim 𝔅(C_{5}) = ∞ by Lemma 2.3, as in the proof of Proposition 3.1. Thus, we may suppose that χ(n_{t}) = 1 for all t ∈ A. Now, notice that #A ≥ \frac{q−9}{4}. When q ∉ \{5, 25, 7, 11, 13\}, we get #A > \frac{q}{p}. In this case, A generates (as an abelian group) the whole 𝔽_{q}, which implies the statement. If q = 11 or q = 13 an easy computation shows that A generates 𝔽_{q}, whence we are done.

Let q = 25, 𝔽_{q} = 𝔽_{5}[s]/(s^{2} − 2). Let first a = 1 + s, then |a| = 12 (hence a ∉ R_{1} ∪ R_{2} ∪ R_{3} ∪ R_{4} ∪ R_{6}), and −(a^{2} + a^{−2}) = −((3 + 2s) + (3 + 3s)) = 4. Now, take a = 1 + 2s, then |a| = 24 and −(a^{2} + a^{−2}) = 2s. Therefore, A generates 𝔽_{q}, and we are done.

We deal now with the cases q ∈ {5, 7}. Let q = 5 and a = 2 ∈ 𝔽_{q}^{×}. With

g_{1}, g_{a}, we get a Dynkin diagram as in Lemma 2.1, with ζ = χ \binom{−1 1}{−1}, and

μ = χ \binom{1 2}{1}. Since this implies that |ζ| divides 10 and μ = ζ^{8}, by the Lemma

we get that 𝔅(C_{5}) is infinite dimensional unless either • ζ ≠ 1 and ζμ = 1, or

• ζ ≠ 1 and μ = 1, or
• ζ = −1.

The first case is impossible, while in the second and third case we arrive to the statement.

Let q = 7. Let a = 3, α = χ \binom{−1 1}{−1} and β = χ \binom{1 1}{1}. Then if β ≠ 1,

we have a Dynkin diagram as in Lemma 2.3, which yields an infinite dimensional Nichols algebra (notice that β ∉ R_{3}). If β = 1, as α^{6} = β and |α| divides 14, this implies α = −1, which is the statement again.

Remark 3.2 gives the statement about C_{6}. □

Remark 3.10. If p = 3 one can prove, using the same techniques as in Proposition 3.6 (case q = 3^{2n+1}), that χ = sgn × ε when restricted to each subfield 𝔽_{3^{r}} where r is an odd prime.

4. Nichols algebras over GL(2, 𝔽_{q})

We proceed in this section with the groups GL(2, 𝔽_{q}). Recall that the order of GL(2, 𝔽_{q}) is (q − 1)^{2}q(q + 1). Again, E will be the quadratic extension of 𝔽_{q} and \overline{x} will be the Galois conjugate of x ∈ E. Since GL(2, 𝔽_{2}) ≃ 𝕊_{3}, we consider only the case q > 2.

There are q^{2} − 1 conjugacy classes divided in 4 types:

Table 3: Conjugacy classes in GL(2, 𝔽_{q}).

Type  Representative  Size  Number Centralizer
C_{1}  c_{1}(x) = ( \begin{matrix}x \\ & x \end{matrix} )  1  q − 1  GL(2, 𝔽_{q})
C_{2}  c_{2}(x) = \binom{x 1}{x}  q^{2} − 1  q − 1  𝔽_{q}^{×} × 𝔽_{q}
C_{3}  c_{3}(x, y) = ( \begin{matrix}x \\ & y \end{matrix} ) (x ≠ y)  q(q + 1) \frac{(q−1)(q−2)}{2} 𝔽_{q}^{×} × 𝔽_{q}^{×}
C_{4} c_{4}(x) = \binom{−x\overline{x}}{1 x + \overline{x}} (x ∈ E \ 𝔽_{q}) (q − 1)q \frac{q(q−1)}{2}  cyclic

For the next proposition we need to recall the character table of the non-abelian group GL(2, 𝔽_{q}) from [13, §5.2]. We know that there are q^{2} − 1 irreducible representations. The relevant information for our purposes is contained in the following table:

Table 4: Representations of GL(2, 𝔽_{q}) in scalar matrices

Representation  Dimension  Number  c_{1}(x) = ( \begin{matrix}x \\ & x \end{matrix} )
U_{α} (α ∈ 𝔽̂_{q}^{×})  1  q − 1  α(x)^{2}
V_{α} (α ∈ 𝔽̂_{q}^{×})  q  q − 1  α(x)^{2}
W_{α,β} (α, β ∈ 𝔽̂_{q}^{×} y α ≠ β)  q + 1  \frac{1}{2}(q − 1)(q − 2)  α(x)β(x)
X_{γ} (α ∈ Ê_{q}^{×})  q − 1  \frac{1}{2}q(q − 1)  γ(x)

Proposition 4.1. If dim 𝔅(C_{1}, χ) < ∞ then:

(1) −1 = α(x)^{2}, where α ∈ 𝔽̂_{q}^{×}; or
(2) −1 = α(x)β(x), where α, β ∈ 𝔽̂_{q}^{×}; or
(3) −1 = α(x), where α ∈ Ê^{×}; or

Proof. If follows from [16, Lemma 3.1]. □

Proposition 4.2. If dim 𝔅(C_{3}) < ∞ then q_{T} = ( \begin{matrix}α & β \\ β & α \end{matrix} ) where α = χ(c_{3}(x, y)),

β = χ(c_{3}(y, x)), and T = {c_{3}(x, y), c_{3}(y, x)}. Furthermore, one of the following conditions is satisfied:

• β^{2} = 1 and α ≠ 1
• β^{2} ≠ 1 and αβ^{2} = 1
• β^{2} ≠ 1 and αβ^{2} ≠ 1 and α = −1
• β^{2} ∈ R_{12} and α = −β^{4} ∈ R_{3}

Proof. Notice that c_{3}(x, y) and c_{3}(y, x) are conjugated in GL(2, 𝔽_{q}) by the involu-

tion ( \begin{matrix}& 1 \\ 1 \end{matrix} ). The result now follows from Lemma 2.1. □

Proposition 4.3. Let T = \{c_{4}(x), \binom{x + \overline{x} x\overline{x}}{−1}\}. If dim 𝔅(C_{4}) < ∞ then q_{T}

is either ( \begin{matrix}−1 & −1 \\ −1 & −1 \end{matrix} ) or ( \begin{matrix}ω & ω \\ ω & ω \end{matrix} ), where ω ∈ R_{3}.

Proof. Note that the centralizer of c_{4}(x) is the cyclic group

\{( \begin{matrix}a & −cx\overline{x} \\ c & a + c(x + \overline{x}) \end{matrix} ) : a, c ∈ 𝔽_{q}^{×}\} ,

and this group is isomorphic to E^{×} by a + cx ↦ ( \begin{matrix}a & −cx\overline{x} \\ c & a + c(x + \overline{x}) \end{matrix} ). Take the

involution \binom{1 x + \overline{x}}{−1} to get

\binom{x + \overline{x} x\overline{x}}{−1} = \binom{1 x + \overline{x}}{−1} \binom{−x\overline{x}}{1 x + \overline{x}} \binom{1 x + \overline{x}}{−1} ∈ C_{c_{4}(x)}.

Since E^{×} is cyclic, there exists n ∈ ℕ such that c_{4}(x)^{n} = \binom{x + \overline{x} x\overline{x}}{−1}. We get

the result by using Corollary 2.2(1). □

Proposition 4.4. If p = 2 then 𝔅(C_{2}) is infinite dimensional. If p ≠ 2, q ≠ 9 and

dim 𝔅(C_{2}) < ∞ then χ \binom{x^{i} a}{x^{i}} = (−1)^{i}.

Proof. Similarly to the proof of Proposition 3.9, we take x_{a} = ( \begin{matrix}a \\ & 1 \end{matrix} ) and we

get g_{a} = \binom{x a}{x}. Then q_{ab} = χ(g_{ab^{−1}}), q_{a1}q_{1a} = χ \binom{x^{2} x(a + a^{−1})}{x^{2}}. As

before, if q_{1a}q_{a1} ≠ 1 for some a ∈ 𝔽_{q}^{×} \ (R_{1} ∪ R_{2} ∪ R_{3}), then we get a Dynkin diagram of infinite type, since

q_{1,a}q_{a,1} = q_{a,a^{2}}q_{a^{2},a} = ⋯ = q_{a^{m},1}q_{1,a^{m}}, |a| = m + 1.

Assume then that ∀a ∈ 𝔽_{q}^{×} \ (R_{1} ∪ R_{2} ∪ R_{3}), q_{1a}q_{a1} = 1. We define χ_{1} ∈ 𝔽̂_{q}^{×} and

χ_{2} ∈ 𝔽̂_{q} by χ \binom{τ σ}{τ} = χ_{1}(τ)χ_{2}(τ^{−1}σ). Let

A = {x^{−1} (a + a^{−1}) | a ∈ 𝔽_{q}^{×} \ (R_{1} ∪ R_{2} ∪ R_{3})} .

Thus we assume χ_{1}(x^{2})χ_{2}(b) = 1 for all b ∈ A. Since the orders of 𝔽_{q}^{×} and 𝔽_{q} are coprime, we get χ_{2}(b) = 1 for all b ∈ A. Note that, since #(R_{1} ∪ R_{2} ∪ R_{3}) ≤ 4, #A ≥ \frac{q−5}{2}. If #A > \frac{q}{p} = p^{n−1} then A is not contained in any 𝔽_{p}-hyperplane of 𝔽_{q}, and thus it generates 𝔽_{q} as an abelian group. Therefore, χ_{2} = 1 if p^{n−1}(p − 2) > 5. Furthermore, we get χ_{1}(x^{2}) = 1, from where we conclude with the statement.

We study now the cases q = 2^{n}(n ∈ ℕ) and q ∈ {3, 5, 7}. Let q = 2^{2n}. We write 𝔽_{4} ⊆ 𝔽_{q}, 𝔽_{4} = 𝔽_{2}[s]/(s^{2} + s + 1). We get s + s^{−1} = 1, |s| = 3. Let then

β := q_{1s}q_{s1} = χ \binom{x^{2} x}{x^{2}}, α := χ(c_{2}) = χ \binom{x 1}{x}. Notice that α^{2|x|} = 1

and β = α^{|x|+2}. Then either β = 1 (which implies α = 1 since |x| is odd), or, by using Lemma 2.3, α = −1, but then β = −1. In any case, we get an infinitedimensional Nichols algebra.

The case q = 2^{2n+1}. Since 2^{2n+1} ≡ 2 (mod 3), then R_{3} = {1}. Therefore, #A = \frac{2^{n}−2}{2} = 2^{n−1} − 1. It is easy to see that 1 ∉ A and 0 ∉ A. Let (A) be the subgroup generated by A. We will prove that 1 ∈ (A), which will imply (A) = 𝔽_{q}, and we shall be done. Let r be a prime number such that r | 2n + 1. If ξ ∈ 𝔽_{2^{r}} \ 𝔽_{2}, we denote by f_{ξ} its minimal polynomial. On the one hand there exist exactly \frac{2^{r}−2}{r} irreducible polynomials in 𝔽_{2}[X] of degree r. On the other hand,

f_{ξ} = X^{r} + a_{r−1}X^{r−1} + ⋯ + a_{1}X + 1 = Π_{i=1}^{r}(X − ξ_{i}).

As done in the proof of Proposition 3.6 (case q = 3^{2n+1}), one can prove that a_{r−k} + a_{k} ∈ (A) ∩ 𝔽_{2}. It suffices then to prove that there exists k such that a_{r−k} + a_{k} = 1. If a_{r−k} + a_{k} = 0 for all k, then f_{ξ} ∈ S, where

S = {X^{r} + b_{r−1}X^{r−1} + ⋯ + b_{1}X + 1 | b_{r−k} = b_{k} ∀k}.

Thence, we have I = {f ∈ 𝔽_{2}[X] | f irreducible} ⊆ S, which is a contradiction because #S = 2^{\frac{r−1}{2}} < \frac{2^{r}−2}{r} for any r > 3. If r = 3, f = (X + 1)^{3} ∈ S \ I, and since #S = #I, we also get I \ S ≠ ∅. Then #(A) ≥ 2^{r−1} + 1 and then (A) = 𝔽_{2^{r}}.

The case q = 3. If x = 1, we use Corollary 2.2 (2) to see that dim 𝔅(C_{2}) = ∞ for any representation (notice that |c_{2}(x)| = 3). If x = −1, c_{2}(x) generates its centralizer and again by Corollary 2.2 (2) we get χ(c_{2}(x)) = −1.

The case q = 5. Note that |c_{2}(x)| = 5|x|. If x = 1 then 𝔅(C_{2}) is infinite dimensional by Corollary 2.2 (2). Assume now that x = 2 or x = 3. We take a = 2 and then a + a^{−1} = 0. Let

α = χ(g) = q_{11} = q_{22} = q_{33} = q_{44},
β = q_{12}q_{21} = q_{24}q_{42} = q_{34}q_{43} = q_{31}q_{13} = χ ( \begin{matrix}4 \\ & 4 \end{matrix} ) = ±1 and
γ = q_{23}q_{32} = q_{14}q_{41} = χ \binom{4 1}{4} .

If β = −1 we get a length-four cycle, thence β = 1. Note that α^{10} = 1 and then |α| = 2, 5, 10. But α^{−2} = γ. If γ = 1 then α = −1 and the representation is completely determined as in the statement. If γ ≠ 1 we have a Dynkin diagram as in Lemma 2.1 with ζ = α and μ = α^{−2}, which is of infinite type. If x = 4,

take again a = 2. We have that q_{12}q_{21} = ⋯ = β = χ ( \begin{matrix}1 \\ & 1 \end{matrix} ) = 1 and that

γ = q_{23}q_{32} = q_{14}q_{41} = χ \binom{1 2}{1}. Again, α^{−2} = γ, and we get the statement as

for x = 2, 3.

The case q = 7. Note that |c_{2}(x)| = 7|x|. The case x = 1 follows as before from Corollary 2.2 (2). The cases x = 3 and x = 5 follow by taking a = 3. In

fact, let α = q_{11} = χ(c_{2}(x)) and β = q_{13}q_{31} = χ \binom{2 3}{2}. If β ≠ 1, we get a

Dynkin diagram with a cycle of length 6. Then β = 1, but β = α^{8} and since |α| divides 42, we must have α = ±1. If α = 1 we again have an infinite dimensional Nichols algebra, whence α = −1. But c_{2}(x) generates the centralizer, and we get the statement. The cases x = 2 and x = 4 are easier: we take as before a = 3, α = q_{11}, β = q_{13}q_{31}. Then we get that α^{8} = β = ±1 but |α| divides 21. If x = 6 again we use a = 3, α = q_{11} = χ(c_{2}(x)), β = q_{13}q_{31} = χ \binom{1 −1}{1}. As before,

we must have α^{36} = β = 1 but |α| divides 14, which implies the result. □

Acknowledgement. We benefited from discussions with N. Andruskiewitsch, F. Fantino, C. Sánchez and A. Pacetti, to whom we thank. We used GAP [18] to do some of the computations.

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E-mail address: (sfreyre|matiasg|lvendramin)@dm.uba.ar

Departamento de Matemática - FCEyN, Universidad de Buenos Aires, Pab I - Ciudad Universitaria, (1428) Buenos Aires - Argentina