\in {\mathfrak g }'$ one has $[x,y]=0$, which is equivalent to $$\begin{array}{rcll} 
0 & = & \la [x,y], u\ra \qquad & \mbox{ for all } u \in {\mathfrak g }\\
 & = & \la [u, x], y\ra \qquad & \mbox{ for all } y\in {\mathfrak g }'
 \end{array}$$ thus $[u,x]\in [{\mathfrak g }, {\mathfrak g }]^{\perp}=\zz({\mathfrak g })$, and since $x\in {\mathfrak g }'$ can be written as $x=[v,w]$, then $[u,[v,w]]\subseteq \zz({\mathfrak g })$ for all $u,v,w\in {\mathfrak g }$, that is $C^4({\mathfrak g })=0$ and so ${\mathfrak g }$ is at most 3-step nilpotent.

\[corofree\] Let $\tilde{{\mathfrak n }}_{m,k}$ denote a free metabelian nilpotent Lie algebra admitting an ad-invariant metric, then $k\leq 3$.

Remark \[23\] and the previous result says that a free metabelian nilpotent Lie algebra with an ad-invariant metric if it exists, is free nilpotent. Below we determine which free nilpotent Lie algebra admits such a metric.

Whenever ${\mathfrak n }_{m,k}$ is free nilpotent we have that $\dim {\mathfrak n }_{m,k}/C^1({\mathfrak n }_{m,k}) =m$ so that $$\label{e3}
\dim {\mathfrak n }_{m,k}=m+\dim C^1({\mathfrak n }).$$

Hence Equations (\[e1\]) and (\[e3\]) show that if ${\mathfrak n }_{m,k}$ admits an ad-invariant metric then $$\label{e4}
\dim {\mathfrak z }({\mathfrak n }_{m,k})=m.$$

