Recursively each element of the Hall basis of ${\mathfrak n }_{m,k}$ is defined as follows. The generators $e_1,\ldots,e_m$ are elements of the basis of length 1. Assume we have defined basic elements of lengths $1,\ldots, r-1\leq k-1$, with a total order satisfying $E > F$ if $\ell(E) > \ell(F)$.

If $\ell(E) =s$ and $\ell(F) = t$ and $r = s + t\leq k$, then $[E, F]$ is a basic element of length $r$ if both of the following conditions hold:

-   $E$ and $F$ are basis elements and $E>F$, and

-   if $\ell(E)>1$ and $E=[G,H]$ is the unique decomposition with $G,H$ basic elements, then $F\geq H$.

This gives rise to a natural graduation of ${\mathfrak n }_{m,k}$: $${\mathfrak n }_{m,k}=\bigoplus_{s=1}^k \mathfrak p(m,s),$$ where $\mathfrak p(m,s)$ denotes the subspace spanned by the elements of the Hall basis of length $s$. Notice that

-   $C^r({\mathfrak n }_{m,k})=\oplus_{s=r+1}^k\mathfrak p(m,s)$,

-   $\mathfrak p(m,k) = {\mathfrak z }({\mathfrak n }_{m,k})$

The first assertion follows from the fact that every bracket of $r + 1$ elements of ${\mathfrak n }_{m,k}$, is a linear combination of brackets of $r + 1$ elements in the Hall basis (see proof of Theorem 3.1 in [@Ha]). This implies $C^r({\mathfrak n 