\[p1\] If $\nn_{m,2}$ is a free 2-step nilpotent Lie algebra endowed with an ad-invariant metric, then $m=3$.

Let ${\mathfrak n }_{m,2}$ be the free 2-step nilpotent Lie algebra on $m$ generators. As we showed in Example \[ex:1\] its center has dimension $m(m-1)/2$. Now if Equation (\[e4\]) holds then $m=3$.

\[p2\] Let $\nn_{m,3}$ be a free 3-step nilpotent Lie algebra provided with an ad-invariant metric, then $m=2$.

As shown in Example \[ex:2\] the center ${\mathfrak z }({\mathfrak n }_{m,3})$ has dimension $d_m(3)=m(m^2-1)/3$. From straightforward calculations, if Equation (\[e4\]) is satisfied then $m= 2$.

\[p3\] No free $k$-step nilpotent Lie algebra ${\mathfrak n }_{m,k}$ on $m$ generators with $k \geq 4$ can be endowed with an ad-invariant metric.

*$\bullet$ 4-step nilpotent case:* In this case $\mathfrak p(m,4)=  {\mathfrak z }({\mathfrak n }_{m,4})$, thus from (\[dim\]): $$\dim{\mathfrak z }({\mathfrak n }_{m,4})\geq d_m(4)=\frac{1}{4}(m^4-d_m(1)-2d_m(2))= \frac{m^2(m^2-1)}{4}.$$ Notice that for $m\geq 2$, one has $m^2(m^2-1)/4>m$.

*$\bullet$ General case, $k\geq 5$:* Let ${\mathfrak n }_{m,k}$ denote the free $k$-step nilpotent Lie algebra in $m$ generators. The goal here is to show that for every $m$ and $k\geq 5$ the dimension of the center of ${\mathfrak n }_{m,k}$ is greater than $m$. In order to give a lower bound 