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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

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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
The book is by Karin Erdmann and Mark Wildon. – m_t_ Oct 16 '14 at 8:01
up vote 2 down vote accepted

By linearity, it is enough to check the Jacobi identity on the basis elements. When there are repetitions on the Jacobi identity it is satisfied automatically. Therefore you have to check nothing!

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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
thanks sir for your answer – user46309 Nov 2 '12 at 2:53

In $2$ dimensional case, we have $[x,y]=0$ or $[x,y]=z=ax+by$ if $a$ is zero then by changing the variables you get what you looking for but if $a$ was not zero then divide both sides by $a$ so that it becomes $$[x,y/a]=x+by/a$$ now change the $x+by/a$ variable to $z$. $$[ z-by/a,y/a]=[z,y/a]=z$$ then change $y/a$ to $u$ so that you get $[z,u]=z$

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