the free metabelian $k$-step nilpotent Lie algebra on $m$ generators is defined as $$\tilde{{\mathfrak n }}_{m,k}:= \mathfrak f_m/(C^{k+1}(\mathfrak f_m)+ \mathfrak f_m''),$$

-   the *free $k$-step nilpotent* Lie algebra on $m$ generators ${\mathfrak n }_{m,k}$ is defined as the quotient algebra $${\mathfrak n }_{m,k}:=\mathfrak f_m/C^{k+1}(\mathfrak f_m).$$

In particular free metabelian nilpotent of any degree are 2-step solvable.

\[23\] For $k=2,3$ any $k$-step nilpotent Lie algebra is 2-step solvable, which follows from the Jacobi identity. Thus for the free nilpotent ones we get $\tilde{{\mathfrak n }}_{m,k}={\mathfrak n }_{m,k}$ for $k=2,3$.

Let ${\mathfrak n }_{m,k}$ be a free $k$-step nilpotent Lie algebra and let $\{e_1,\ldots,e_m\}$ be an ordered set of generators. The construction of a *Hall* basis associated to this sets of generators is explained below (see [@Ha; @GG]).

Start by defining the *length* $\ell$ of each generator as $1$. Take the Lie brackets $[e_i, e_j]$ for $i>j$, which by definition satisfy $\ell([e_i,e_j])=2$. Now the elements $e_1, \dots , e_m$, $[e_i, e_j]$, $i>j$ belong to the Hall basis. Define a total order in that set by extending the order of the set of generators and so that $E>F$ if $\ell(E)>\ell(F)$. They allow the construction of the elements of length $3$ and so on.

