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I read in book written by Karin Erdmann and Mark J. Wildon's "Introduction to Lie algebras" "Let F be in 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 defined by [x, y] = x"

How to prove that Lie bracket [x,y] = x satisfies axioms of Lie algebra such that [a,a] = 0 for $a \in L$ and satisfies jacobi identity and can some one give me an example of two dimensional nonabelian Lie algebra

glS
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user46309
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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 02 '12 at 02:54
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    The book is by Karin Erdmann and Mark Wildon. – Matthew Towers Oct 16 '14 at 08:01
  • This is called a one-dimensional affine algebra. You may consider $x=d/dz$ and $y=z d/dz$. Then $\exp(ay) \exp(bx) z= \exp(a) ~z+b$. – Cosmas Zachos Apr 04 '17 at 19:01
  • Why does one need to check the axioms at all? In the book, they take a 2-dimensional non abelian Lie algebra and show that this Lie Algebra has a basis ${x,y}$ such that $[x,y]=x$. We are not defining a Lie bracket on a 2-dimensional vector space, but we are simply showing that the Lie bracket we already have satisfies $[x,y]=x$. So why do you need to check the axioms? – mathStudent Apr 26 '20 at 15:30

2 Answers2

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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 $[z,u]=z$

karvens
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Reza Naderi
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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!

PAD
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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 06:25
  • 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 02 '12 at 02:53
  • For future readers, see http://math.stackexchange.com/questions/24601/classsifying-1-and-2-dimensional-algebras-up-to-isomorphism . $[x, y] = x$ is a bit misleading since you really need $[x, y] = z$ for some $z$ independent of $x, y$. – muaddib Jul 27 '15 at 17:54