A Lie algebra with two of its Lie bracket are zero while the third is not.

Question

Is it possible to construct a system of three vector field u,v and w an $R^{3}$ such that [u,v]=0=[u,w], but [v,w]$\neq$0? I tried to solve it by applying the Jacobi identity property which makes the 2nd and 3rd term vanish so I am left with [u[v,w]]=0. I am stuck here. Any help please. Thanks in advance

Answer 1

The Heisenberg Lie algebra with basis $(u,v,w)$ satisfies $[u,v]=[u,w]=0$ and $[v,w]=u$. Since this Lie algebra is $2$-step nilpotent, all terms $[a,[b,c]]$ in the Jacobi identity are automatically zero.

Answer 2

Yes, consider the 2 dimensional non commutative Lie algebra, $[a,b]=a$, add $c$, in the center $[c,a]=[c,b]=0$