Clearly $\widetilde{\mathcal U}$ and $\mathcal U$ have the same cardinal. Also, $|\mathcal U|=\sum_{j=1}^m(m-j+1)(m-j)$ since for every fixed $j=1,\ldots,m$, the amount of possibilities to choose $k\geq j$ and $i>j$ is $(m-j+1)$ and $(m-j)$ respectively.

Straightforward computations give $ |\widetilde{\mathcal U}|=1/3\, m^3+m^2+2/3\,m$ which combined with (\[u\]) proves that for any $m$ and $k\geq 5$ $$\dim {\mathfrak z }({\mathfrak n }_{m,k})\geq 1/3 \,m^3+m^2+2/3\,m .$$ The right hand side is greater than $m$ for all $m\geq 2$. According to (\[e4\]), the free $k$-step nilpotent Lie algebra ${\mathfrak n }_{m,k}$ does not admit an ad-invariant metric if $k\geq 5$.

\[t1\] Let ${\mathfrak n }_{m,k}$ be the free $k$-step nilpotent Lie algebra in $m$ generators. Then ${\mathfrak n }_{m,k}$ admits an ad-invariant metric if and only if $(m,k)=(3,2)$ or $(m,k)=(2,3)$.

Propositions (\[p1\]) (\[p2\]) and (\[p3\]) prove that if ${\mathfrak n }_{m,k}$ admits an ad-invariant metric then $(m,k)=(3,2)$ or $(m,k)=(2,3)$. Let us show the resting part of the proof.

The Lie algebra ${\mathfrak n }_{3,2}$ has a basis $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ with non zero brackets $$\label{b32}
[e_1,e_2]=e_4,\qquad [e_1,e_3]=e_5,\qquad [e_2,e_3]=e_6.$$ Since $C^1({\mathfrak n }_{3,2})=\zz({\mathfrak n }_{3,2})$, if the metric $\la \,,\,\ra$ is ad-invariant the center of ${\mathfrak n }_{3,2}$ is totally isotropic, so it must hold $$\la e_i, e_j\ra=0 \qquad 