1

Let suppose we have contact lie algebra K(3,(2,2,2)) over GF(3), according to the book "Modular lie algebras and their representations" from Helmut Strade, K(3,(2,2,2)) is simple Lie algebra ( it means that it has no ideal except trivial and itself) and its dimension is equal to $729$. I want to investigate K(3,(2,2,2)) over GF(2), How can we check that it simple or not? what is its dimension ( 64 or 63) ? If it is not simple how can we find nilpotent radical of K(3,(2,2,2))?

Nil
  • 1,306

1 Answers1

1

We have $K(m,\underline{n})=K(m,\underline{n})^{(1)}$ unless $m\equiv -3\mod p$. But for $m=3$ and $p=2$ this congruence holds. So the algebra $K(3,\underline{n})$ over $GF(2)$ is not simple. In general, the derived algebra $K(m,\underline{n})^{(1)}$ is simple. Its dimension is $p^{\sum_in_i}-1$ if $m\equiv -3 \mod p$, and $p^{\sum_in_i}$ otherwise. For $m=3$ and $p=2$ we obtain that $\dim K(3,(2,2,2))^{(1)}=2^{2+2+2}-1=63$.

Dietrich Burde
  • 130,978
  • thank you for your answer, therefore for $m >= 3$ then $K(m,\underline{n})$ is not simple over GF(2). But you explained that $K(m,\underline{n})^{(1)}$ is simple in general, ( for $K^{(1)}$ the nilpotent radical of that $N(K^{(1)})$ is meaningless, right? – Nil Sep 26 '14 at 18:59
  • I need a proof that shows that $K(m,\underline{n})^{(1)}$ is simple? please. – Nil Sep 26 '14 at 19:19
  • A proof is given in Strade-Farnsteiner's book, Theorem $5.5$. But be careful, they denote $K(m,\underline{n})$ by $K'(m,\underline{n})$, and rename $K(m,\underline{n})^{(1)}$ by $K(2r+1,\underline{n})$. – Dietrich Burde Sep 29 '14 at 08:35