# Two Dimensional Lie Algebra

I read in mark wildon book "introduction to lie algebras" "Let F be any field. Up to isomorphism there is a unique two-dimensional nonabelian Lie algebra over F. This Lie algebra has a basis {x, y} such that its Lie bracket is described by [x, y] = x"

How can i proof this bracket [x,y] = x satisfies axioms of Lie algebra such that [a,a] = 0 for $a \in L$ and satisfies jacoby identity

and can some one give me an example of two dimensional nonabelian Lie algebra

-
Well, an arbitrary element looks like $ax+by$ (for $a,b\in F$), so if nothing else you can just check the axioms by hand. –  Aaron Mazel-Gee Oct 28 '12 at 14:23
@AaronMazel-Gee if $l_1 \in L$ then $l_1 = ax + by$ $[l_1,l_1] = [ax + by,ax + by] = aa[x,x] + ab[x,y] + ba [y,x] + bb[y,y] = aa[x,x] + abx + ba(-x) + bb[y,y]$ how can that be $0 \in L$ –  user46309 Oct 28 '12 at 14:31
We always have $[x,x]=[y,y]=0$, and then the other two terms cancel. –  Aaron Mazel-Gee Oct 29 '12 at 17:47
k i already understand it now, –  user46309 Nov 2 '12 at 2:54

can u explain about it more? we are only know about bracket of basis of L such that [x,y] = x, how can we check with this bracket like this satisfies axioms of Lie algebra if $l_1 \in L$ then $l_1 = ax + by$ $[l_1,l_1] = [ax + by,ax + by] = aa[x,x] + ab[x,y] + ba [y,x] + bb[y,y] = aa[x,x] + abx + ba(-x) + bb[y,y]$ how can we got 0 from that? –  user46309 Oct 28 '12 at 14:40
In general $[a,a]=0$ for any $a \in L$. This follows from skew-symmtry. Since $[x,y]=-[y,x]$, then you set $x=y=a$ and (if the characteristic of the field is not 2) you get $[a,a]=0$. –  PAD Oct 29 '12 at 6:25
so, it can hold if the char F $\neq$ 2 ok, can u give me a example of two dimensional non abelian lie algebra, i already search in the internet, but i didnt find any example of two dimensional non abelian lie algebra –  user46309 Oct 29 '12 at 13:33
You just gave one example. Take a two dimensional vector space $V$ and a basis $\{x, y \}$. Define $[x,y]=x$ then extend by linearity to any two vectors in $V$. If you prefer matrices then use the adjoint representation. Since ad$x\, x=0$ and ad$x\, y =x$ you get the matrix of adx to be $$\begin{pmatrix} 0 & 1 \cr 0&0 \end{pmatrix}$$ and a similar matrix for ady. These two matrices generate a two dimenional subalgebra of $gl(2, F)$. –  PAD Oct 31 '12 at 17:15