$[[x,y],z]=[x,[y,z]] \Rightarrow [x,y]=0$?

Question

I got the next problem:

Let $A$ be a Lie algebra, prove that if the bracket associates $([[x,y],z]=[x,[y,z]]$) then the bracket is zero $([x,y]=0)$.

Can't get the result using the properties (alternating, Jacobi identity, anticommutativity), i think that the result is false. Any suggestions? Thanks.

score 3 · Accepted Answer · answered Nov 22 '13 at 07:03

3

The Lie algebra $N$ formed by strictly upper triangular $3\times 3$ matrices satisfies

$$[z, [x, y]]= 0\ \ \ \forall x, y, z\in N.$$

But $[x, y] \neq 0$.

answered Nov 22 '13 at 07:03

how are you defining the bracket? – Lix Nov 22 '13 at 07:30
For any two matrix $A$ and $B$, $[A, B] = AB-BA$. – Nov 22 '13 at 07:31

$[[x,y],z]=[x,[y,z]] \Rightarrow [x,y]=0$?

1 Answers1