# homogeneous algebras

Let $A$ be a finite dimensional algebra over finite field (not necessarily associative). Then $A$ is said to be homogeneous if $Aut(A)$ acts transitively on the one-dimensional subspace of A. If A is homogeneous then either $A^2=0$ or $\text{dim}A=1$. Now I want to check this property for a finite dimensional simple Lie algebra over $GF(2)$. I want to know whether a simple lie algebra over $GF(2)$ is homogeneous or not?

• What is the question? – abx Mar 22 '14 at 15:59
• You probably meean that $A$ is homgeneous if the automorphism group acts transitively on the set of $1$-dimensional subspaces. – Mariano Suárez-Álvarez Mar 22 '14 at 16:02
• yes and I want to know whether a simple lie algebra over GF(2) is homogeneous or not? probabely my question is wrong because simple lie algebras has only two ideal zero and itself. right? – user118746 Mar 22 '14 at 16:04
• A simple Lie algebra $L$ satisfies $L^2=[L,L]=L$. This is impossible with $L^2=0$ or $dim (L)=1$. – Dietrich Burde Mar 22 '14 at 16:37

The simple Lie algebra $W(1,2)^{(2)}$ of dimension 3 over $GF(2)$ obviously contains 7 subspaces of dimension 1, but its automorphism group has order 6 (see e.g. section 5.3 of the paper "B. Eick: Some new simple Lie algebras in characteristic 2: J. Symbol. Comput. 45, 943 -- 951 (2010)"). It is then clear that this Lie algebra is not homogeneous.
Such algebras were studied a lot in the past. According to MR0655406, in: D.N. Ivanov, On homogeneous algebras over $GF(2)$, Vestnik Moskov. Univ. Matematika 37 (1982), N2, 69­-72 (in Russian) it is proved that any homogeneous algebra over $GF(2)$ is isomorphic to one of the algebras in the two-parametric series $A(n,\mu)$ earlier constructed by Kostrikin in: On homogeneous algebras, Izv. AN SSSR Ser. Matem. 29 (1965), 471-484 (in Russian) (http://mi.mathnet.ru/izv2915). Unfortunately, this is difficult to verify as the journal for this year is not available online. Earlier Gross (Finite automorphic algebras over $GF(2)$, Proc. Amer. Math. Soc. 31 (1972), 349-353; DOI:10.2307/2038501; MR:0286856) established the same result under assumption that the automorphism group is solvable.
The algebras $A(n,\mu)$ are defined as the vector space $GF(2^n)$ over $GF(2)$ subject to multiplication $x*y = \mu(xy)^{2^{n-1}}$, where $\mu$ is a nonzero element of $GF(2^n)$. They are obviously not Lie.