0

I am having trouble understanding how Lie algebras act. I.e. if I am trying to work with a two dimensional Lie algebra, there isn't much I can do without knowing the Lie Bracket that is defined on the Lie Algebra correct?

glS
  • 6,818
So many hats
  • 450
  • 1
    Well, there are not that many $2$-dimensional Lie algebras (only two over most fields, I didn't really think too hard on whether there might be some more for fields of small characteristic). – Tobias Kildetoft Aug 04 '15 at 13:07
  • @TobiasKildetoft The idea is that I have a basis ${x_1,x_2}$ right, and with no field specified, I am meant to prove that this algebra cannot be simple, but I have no feeling for the Algebra itself. – So many hats Aug 04 '15 at 13:09
  • @TobiasKildetoft Also is the defining factor for having different 2-dim Lie Algebras just having a different bracket? – So many hats Aug 04 '15 at 13:10
  • By different I meant up to isomorphism. – Tobias Kildetoft Aug 04 '15 at 13:11
  • Anyway, with a small bit of fiddling, you can show that it is possible to pick a basis ${x,y}$ such that $[x,y] = x$ (assuming of course that the algebra is not abelian). – Tobias Kildetoft Aug 04 '15 at 13:20
  • @TobiasKildetoft Thank you, this all makes sense now.(11 days later :) ) – So many hats Aug 15 '15 at 02:20
  • related: https://math.stackexchange.com/q/222743/173147 – glS Aug 29 '23 at 00:33

1 Answers1

6

For every field $K$ there are just two different $2$-dimensional Lie algebras. The first one is $K^2$, which is abelian (which means the bracket is zero). The second one is solvable, non-nilpotent, hence non-abelian, and the bracket of the basis $(x,y)$ is an arbitrary nontrivial linear combination of $x$ and $y$, i.e., $[x,y]=\alpha x+\beta y$ for some $\alpha,\beta\in K$ with $(\alpha,\beta)\neq (0,0)$. This Lie algebra is usually denoted by $\mathfrak{r}_2(K)$. All choices $(\alpha,\beta)\neq (0,0)$ give isomorphic Lie algebras, i.e., they are all isomorphic to $\mathfrak{r}_2(K)$. You may choose $[x,y]=x$. Then the adjoint operator $ad(x)$ given by $ad(x)(y)=[x,y]$ is not nilpotent, because one eigenvalue is equal to $1$. Hence $\mathfrak{r}_2(K)$ is not nilpotent by Engel's theroem. Also, it has only inner derivations, i.e. $Der(\mathfrak{r}_2(K))=ad (\mathfrak{r}_2(K))$. Its center is trivial. This gives another proof that it is not nilpotent, because nontrivial nilpotent Lie algebras have a nontrivial center.

The algebra $L=\mathfrak{r}_2(K)$ is solvable, since we have $$ [[L,L],[L,L]]=0. $$ It cannot be simple, since $I=\langle x \rangle$ is a proper Lie ideal.

Dietrich Burde
  • 130,978