I have a worked example of finding the lower central series but can't get my head around it.
So L= Hn, the nth Heisenberg Lie Algebra. Then L has basis: {u1,...,un, v1,...vn, z} and:
[ui, uj]=[vi, vj]=0,
[ui, vj]= -[vj, ui] = (delta ij)z and
z is in the centre of L, that is [z, a]=[a, z] for all a element of L.
The following working is then given to find the lower central series:
L^2=span{[x,y] / x,y elements of L}
=span{ [ui, vj], [ui, uj], [vi, vj], [z,x] }
=span{z} = kz
Im not sure how it gets from the first line to the second line. Why does taking all possible pairs of elements from the basis work?
Then:
L^3
=[L,L^2]
=[L, z] ={0}
So we have L^3={0} and we are done.
For both L^2, L^3 i dont understand the logic of each step. Thanks!
U mentioned that L has a $\underline{basis}$: {u1,...,un, v1,...vn, z}. For $x\in L$, you can write : $x=x^zz+\sum_{i=1}^{n} x^u_iu_i +x^v_iv_i$ where $x^z, x^u_i$ and $x^v_i$ are scalars. Do the same for $y$, and span $[x,y]$ using the bilinearity (and skew symmetry) of the bracket : $$[x,y]= \sum_{i,j} \alpha^z_{i}[z,u_i]+\alpha^z_{j}[z,u_j] + \beta_{ij}[u_i,v_j] $$.
where $\alpha$'s and $\beta$'s are scalars. This shows that $L_2=span\{ [u_i, v_j],[z,u_i],[z,v_i] \}=span\{z\}$ ...
$L_3=\{[x,y]; x\in L\: ,\: y\in L2 \}$. Since $L_2=span\{z\}$, $y \in L_2$ can be written $y=\alpha z$, with $\alpha$ a scalar. Since z is central you get :
$$L_3=\{[x,\alpha z];x\in L ,\:\alpha\: scalar\}=\{0\}$$.