Algebras and regular subgroups. Marco Antonio Pellegrini. Ischia Group Theory Joint work with Chiara Tamburini

Joint work with Chiara Tamburini Università Cattolica del Sacro Cuore Ischia Group Theory 2016

Affine group and Regular subgroups Let F be any eld. We identify the ane group AGL n (F) with the subgroup of GL n+1 (F) consisting of the matrices having (1, 0,..., 0) T as rst column. So, AGL n (F) acts on the right on the set A = {(1, v) : v F n } of ane points. Definition A subgroup (or a subset) R of AGL n (F) is called regular if it acts regularly on A, namely if, for every v F n, there exists a unique element in R having (1, v) as rst row.

Literature The problem of classifying the regular subgroups of AGL n (F) attracted the interest of many authors: Caranti, Dalla Volta, Sala (2006); Catino, Rizzo (2009); Catino, Colazzo, Stefanelli (2015-2016); Childs (2015). These authors use concepts as radical rings, braces and Hopf structures. We prefer to adopt an approach based only on linear algebra. On the other hand this problem is also connected to the classication of nilpotent algebras: Childs (2005); Poonen (2008); De Graaf (2010).

Affine group and Regular subgroups We write every element r of a regular subgroup R of AGL n (F) as where r = ( ) ( ) 1 v 1 v = = µ 0 π(r) 0 I n + δ R (v) R (v), π : AGL n (F) GL n (F), µ R : F n AGL n (F), δ R : F n Mat n (F).

Regular subgroups Example The translation subgroup {( 1 v T = 0 I n ) } : v F n is an example of regular subgroup (δ T = 0).

Regular subgroups Lemma Let R be a regular subset of AGL n (F) containing I n+1 and such that δ R is additive. Then R is a subgroup, if and only if: δ R (vδ R (w)) = δ R (v)δ R (w), for all v, w F n. Also, given v, w F n, Commutativity µ R (v)µ R (w) = µ R (w)µ R (v) vδ R (w) = wδ R (v).

Regular subgroups Theorem Let R be regular subgroup of AGL n (F). Then the following holds: (a) the center Z(R) of R is unipotent; (b) if δ R is linear, then R is unipotent; (c) if R is abelian, then δ R is linear.

Regular subgroups: Heged s (2000) Example Let n 4 and F = F p (p odd). Take the matrix A = diag(a 3, I n 4 ) of GL n 1 (F p ), where A 3 = ( 1 2 2 0 1 2 0 0 1 ). Then A has order p, is orthogonal and its minimum polynomial has degree 3. The subset R of order p 4, dened by R = 1 v h + vjvt 2 0 A h A h Jv T n 1 : v F p, h F p, 0 0 1 is a unipotent regular subgroup of AGL n (F p ), such that R T = {1}. Note that δ R is not linear.

Split local algebras Definition An F-algebra L with 1 is called split local if L/J(L) is isomorphic to F, where J(L) denotes the Jacobson radical of L. In particular L = F 1 + J(L), where F 1 = {α1 L : α F}. We say that L is nite dimensional if J(L), viewed as an F-module, has nite dimension.

Split local algebras Theorem Let L be a nite dimensional split local F-algebra. Then, with respect to the product in L, the subset R = 1 + J(L) = {1 + v : v J(L)} is a group, isomorphic to a regular subgroup of AGL n (F) for which δ R is linear, n = dim F (J(L)).

Split local algebras Conversely we have the following result. Theorem Let R be a regular subgroup of AGL n (F). Set V = R I n+1 = {( ) } 0 v v F n, 0 δ R (v) L R = FI n+1 + V. Then L R is a split local subalgebra of Mat n+1 (F), with J(L R ) = V, if and only if δ R is linear.

Split local algebras and Regular subgroups Our classication of the regular subgroups of AGL n (F) is based on the following proposition (see Caranti, Dalla Volta, Sala (2006)): Proposition Assume that R 1, R 2 are regular subgroups of AGL n (F) such that δ R1 and δ R2 are linear maps. Then R 1 and R 2 are conjugate in AGL n (F) if and only if the algebras L R1 and L R2 are isomorphic.

Regular subgroups and Jordan Forms Theorem Let R be a regular subgroup of AGL n (F) and 1 z be an element of the center Z(R) of R. Then, up to conjugation of R under AGL n (F), we may suppose that z = J z, where J z = diag (J m1,..., J mk ) is the Jordan form of z having Jordan blocks J mi eigenvalue 1, of respective sizes associated to the m i m i+1 for all i 1.

Our method We restrict our attention to regular subgroups U of AGL n (F) such that the map δ U is linear.

Our method We restrict our attention to regular subgroups U of AGL n (F) such that the map δ U is linear. Since U is unipotent, there exists 1 z Z(U) and, up to conjugation, we may assume that z is a Jordan form.

Our method We restrict our attention to regular subgroups U of AGL n (F) such that the map δ U is linear. Since U is unipotent, there exists 1 z Z(U) and, up to conjugation, we may assume that z is a Jordan form. Our classication is obtained working inside C AGLn(F)(z) using the following parameters d, r and k, considered in this order. Let H be any unipotent subgroup of AGL n (F). d(h) = max{deg min F (h I n+1 ) h H}; r(h) = max{rk(h I n+1 ) h H}; k(h) = dim F {w F n : wπ(h) = w, h H}.

Two general cases Lemma Let U be a unipotent regular subgroup of AGL n (F). If d(z(u)) = n + 1 then, up to conjugation, U = C AGLn(F)(J n+1 ). Moreover U is abelian.

Two general cases Lemma Let U be a unipotent regular subgroup of AGL n (F). If d(z(u)) = n + 1 then, up to conjugation, U = C AGLn(F)(J n+1 ). Moreover U is abelian. Lemma Let U be a regular subgroup of AGL n (F), n 2. If d(z(u)) = n then, up to conjugation, for some xed α F: U = R α =. 1 x 1 x 2... x n 2 x n 1 x n 0 1 x 1... x n 3 0 x n 2 0 0 1... x n 4 0 x n 3.... 0 0 0... 1 0 x 1 0 0 0... 0 1 αx n 1 0 0 0... 0 0 1... In particular U is abelian and r(u) = n 1. Furthermore, R 0 and R α are not conjugate for any α 0.

Two general cases If n 4 is even, then R α is conjugate to R 1 for any α 0 and an epimorphism Ψ : F[t 1, t 2 ] FI n+1 + R α is obtained setting Ψ(t 1 ) = αx 1, Ψ(t 2 ) = α n 2 2 Xn 1. In this case Ker (Ψ) = t 1 n 1 t 22, t 1 t 2. If n is odd, R α and R β (α, β F ) are conjugate if and only if β/α is a square in F.

Our method: an example Lemma Let U be a regular subgroup of AGL 3 (F). If d(z(u)) = r(z(u)) = 2, then U is abelian, conjugate to and char F = 2. 1 x 1 x 2 x 3 0 1 0 x 2 0 0 1 x 1 0 0 0 1

Our method: an example Proof. We may assume z = diag(j 2, J 2 ) Z(U), whence δ(v 1 ) = E 2,3. From the structure of the centralizer of z and the unipotency of U we obtain δ(v 2 ) = E 1,3 + αe 2,1 + βe 2,3. It follows v 1 δ(v 2 ) = v 3. Now, we apply δ R (vδ R (w)) = δ R (v)δ R (w) to v 1, v 2, which gives δ(v 3 ) = δ(v 1 )δ(v 2 ) = 0. Direct calculation shows that U is abelian. Hence d(z(u)) = d(u) = 2. In particular, (µ(v 2 ) I 4 ) 2 = 0 gives α = β = 0. Finally (µ(v 1 + v 2 ) I 4 ) 2 = 0 gives char F = 2.

Standard regular subgroups We can identify the direct product AGL m1 (F) AGL m2 (F) with the subgroup: 1 v w 0 A 0 : v F m 1, w F m 2, A GL m1 (F), B GL m2 (F). 0 0 B Clearly, in this identication, if U i are respective regular subgroups of AGL mi (F) for i = 1, 2 then U 1 U 2 is a regular subgroup of AGL m1 +m 2 (F).

Standard regular subgroups We can identify the direct product AGL m1 (F) AGL m2 (F) with the subgroup: 1 v w 0 A 0 : v F m 1, w F m 2, A GL m1 (F), B GL m2 (F). 0 0 B Clearly, in this identication, if U i are respective regular subgroups of AGL mi (F) for i = 1, 2 then U 1 U 2 is a regular subgroup of AGL m1 +m 2 (F). For every partition λ of n + 1 dierent from (1 n+1 ), we dene one or two abelian regular subgroups S λ, S λ of AGL n(f).

Standard regular subgroups We start with the abelian subgroup: S (1+n) = C AGLn(F)(J 1+n). More generally, for a partition λ of n + 1 such that: λ = (1 + n 1, n 2,..., n s ), s 1, n i n 1+i 1, 1 i s 1, we dene an abelian regular subgroup S λ of AGL n (F), setting s S λ = S (1+n 1,...,n s) = S (1+n j ). j=1 In particular, if λ = (2, 1 n 1 ), then S λ = T.

Standard regular subgroups Lemma Given a partition (1 + n 1, n 2 ) of n + 1, with 1 + n 1 n 2 > 1, set: 1 u v S (1+n 1,n 2 ) = 0 τ 1+n 1 (u) w Du T : u F n 1, v F n 2, 0 0 τ n2 (v) where w = (0,..., 0, 1) F n 2, D = antidiag(1,..., 1) GL n1 (F). Then S (1+n 1,n 2 ) is an indecomposable regular subgroup of AGL n (F). Definition A subgroup H of AGL n (F) is indecomposable if there exists no decomposition of F n as a direct sum of non-trivial π(h)-invariant subspaces.

Standard regular subgroups Next, consider a partition µ of n + 1 such that: µ = (1 + n 1, n 2,..., n s ), s 2, 1 + n 1 n 2 > 1, n i n 1+i 1, 2 i s 1. We dene the abelian regular subgroup S µ in the following way: s Sµ = S (1+n 1,n 2,n 3...,n = S s) (1+n 1,n 2 ) S (1+n j ). The regular subgroups S λ, S µ associated to partitions as above will be called standard regular subgroups. As already mentioned S λ and S µ are always abelian. j=3

Standard regular subgroups Theorem Let λ 1 = (1 + n 1,..., n s ), λ 2 = (1 + m 1,..., m t ) be two partitions of n + 1 as in the rst case and let µ 1 = (1 + a 1,..., a h ), µ 2 = (1 + b 2,..., b k ) be two partitions of n + 1 as in second case. Then (a) S λ1 is not conjugate to S µ 1 ; (b) S λ1 is conjugate in AGL n+1 (F) to S λ2 if and only if λ 1 = λ 2 ; (c) S µ 1 is conjugate in AGL n+1 (F) to S µ 2 if and only if µ 1 = µ 2.

The classification for AGL 1 (F) As shown by Tamburini (2012), the only regular subgroup of AGL 1 (F) is the translation subgroup ( ) x0 x T = S (2) = 1. 0 x 0 However, already for n = 2, a description becomes much more complicated. Also, restricting to the unipotent case, one may only say that every unipotent regular subgroup is conjugate to: 1 x 1 x 2 0 1 σ(x 1 ) 0 0 1 where σ Hom Z (F, F).

The classification for AGL 2 (F) A complete set of representatives for the conjugacy classes of regular subgroups U of AGL 2 (F) with linear δ, for any eld F: U FI n+1 + U Ker (Ψ) x 0 x 1 x 2 S (3) 0 x 0 x 1 t 13 indec. S (2,1) 0 0 x 0 x 0 x 1 x 2 0 x 0 0 t 1, t 2 2 0 0 x 0

The classification for AGL 3 (F) A complete set of representatives for the conjugacy classes of abelian regular subgroups U of AGL 3 (F), for any eld F. U FI 4 + U char F Ker (Ψ) x 0 x 1 x 2 x 3 S (4) 0 x 0 x 1 x 2 0 0 x 0 x 1 any t 14 S (3,1) R λ, λ F 0 0 0 x 0 x 0 x 1 x 2 x 3 0 x 0 x 1 0 0 0 x 0 0 any t 13, t 22, t 1 t 2 0 0 0 x 0 x 0 x 1 x 2 x 3 0 x 0 0 x 1 0 0 x 0 λx 2 any t 1 2 λt 22, t 1 t 2 0 0 0 x 0

The classification for AGL 3 (F) U 3 1 x 0 x 1 x 2 x 3 0 x 0 0 x 2 0 0 x 0 x 1 2 t 1 2, t 22 S (2,1 2 ) 0 0 0 x 0 x 0 x 1 x 2 x 3 0 x 0 0 0 0 0 x 0 0 any t 1, t 2, t 3 2 0 0 0 x 0

The classification for AGL 3 (F) Lemma U is conjugate to exactly one of the following subgroups: 1 x 1 x 2 x 3 0 1 0 0 (a) N 1 =, if k(u) = 2; 0 0 1 x 1 0 0 0 1 1 x 1 x 2 x 3 0 (b) N 2 = 1 0 x 2, if k(u) = 1, d(u) = 2 and 0 0 1 x 1 0 0 0 1 char F 2; 1 x 1 x 2 x 3 0 1 0 x (c) N 3,λ = 1 + x 2 0 0 1 λx, λ F, if k(u) = 1 and 2 0 0 0 1 d(u) = 3.

The classification for AGL 4 (F) A complete set of representatives for the conjugacy classes of indecomposable abelian regular subgroups U of AGL 4 (F), when F has no quadratic extensions U FI 5 + U char F Ker (Ψ). x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 x 2 x 3 0 0 x 0 x 1 x 2 0 0 0 x 0 x 1 any t 5 1 S (5) S (3,2) 0 0 0 0 x 0 x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 0 x 2 0 0 x 0 0 x 1 0 0 0 x 0 x 3 any t 3 1 t 22, t 1 t 2 0 0 0 0 x 0

The classification for AGL 4 (F) U FI 5 + U char F Ker (Ψ). x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 0 x 3 0 0 x 0 0 0 0 0 0 x 0 x 1 any t 1 3, t 2 1 t 2, t 2 2 U 4 1 U 4 2 0 0 0 0 x 0 x 0 x 1 x 2 x 3 x 4 0 x 0 0 0 x 1 0 0 x 0 0 x 3 0 0 0 x 0 x 2 0 0 0 0 x 0 any t1 2 t 2t 3, t 2 2, t 2 3, t 1t 2, t 1t 3

The classification for AGL 4 (F) Representatives for the conjugacy classes of decomposable abelian regular subgroups U of AGL 4 (F), when F has no quadratic extension U FI 5 + U char F Ker (Ψ) x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 x 2 0 S (4,1) 0 0 x 0 x 1 0 0 0 0 x 0 0 any t 4 1, t 22, t 1 t 2 S (3,2) 0 0 0 0 x 0 x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 0 0 0 0 x 0 0 0 0 0 0 x 0 x 3 any t 3 1, t 23, t 1 t 2 0 0 0 0 x 0

The classification for AGL 4 (F) U FI 5 + U char F Ker (Ψ) x 0 x 1 x 2 x 3 x 4 0 x 0 x 1 0 0 S (3,1,1) 0 0 x 0 0 0 0 0 0 x 0 0 any t 3 1, t 22, t 32, t 1 t 2, t 1 t 3, t 2 t 3 S (2,2,1) 0 0 0 0 x 0 x 0 x 1 x 2 x 3 x 4 0 x 0 0 x 1 0 0 0 x 0 x 2 0 0 0 0 x 0 0 any t 2 1 t 22, t 32, t 1 t 2, t 1 t 3, t 2 t 3 0 0 0 0 x 0

The classification for AGL 4 (F) U FI 5 + U char F Ker (Ψ) x 0 x 1 x 2 x 3 x 4 U 3 S 0 x 0 0 x 2 0 1 (1) 0 0 x 0 x 1 0 t 12, t 22, t 32, 2 0 0 0 x 0 0 t 1 t 3, t 2 t 3 S (2,1 3 ) 0 0 0 0 x 0 x 0 x 1 x 2 x 3 x 4 0 x 0 0 0 0 0 0 x 0 0 0 0 0 0 x 0 0 any t 1, t 2, t 3, t 4 2 0 0 0 0 x 0

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