Lie algebra with $\mathfrak{g}/\mathcal{C}_2\mathfrak{g}$ not the free 2-step nilpotent Lie algebra I am looking for examples of finite-dimensional complex nilpotent Lie algebras $\mathfrak{g}$ where the quotient $\mathfrak{g}/\mathcal{C}_{2}\mathfrak{g}$ of the Lie algebra with its 2nd lower central series is not free-nilpotent. I am aware that $\mathfrak{g}$ cannot be a free nilpotent Lie algebra.
 A: The free-nilpotent Lie algebra $F_{g,c}$ of class $c$ in $g$ generators
is by definition the finite-dimensional quotient
$$
\mathfrak{f}(M)/\mathfrak{f(M)}^{c+1},
$$
where $\mathfrak{f}(M)$ is the free Lie algebra with generator set $M=\{x_1,\ldots x_g\}$.
For example, for $c=2$ and $g=2$ we obtain the Heisenberg Lie algebra $F_{2,2}$, with generators $M=\{x_1,x_2\}$ of nilpotency class $2$, with vector space basis $\{x,y,[x,y]\}$.
To answer your question: Let $\mathfrak{g}$ be the standard graded filiform nilpotent Lie algebra with basis $\{x_1,\ldots ,x_n\}$ and brackets $x_1,x_i]=x_{i+1}$ for $2\le i\le n-1$. Then $\mathfrak{g}/\mathfrak{g}^2$ need not be free-nilpotent. 
A: You should define $C^1\mathfrak{g}=\mathfrak{g}$ (so that formulas such as $[C^i\mathfrak{g},C^j\mathfrak{g}]\subset C^{i+j}\mathfrak{g}$ are valid), and hence you mean $\mathfrak{g}/C^3\mathfrak{g}$. For $\mathfrak{g}$ with $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ $n$-dimensional with $n<\infty$, we have $\dim(C^2\mathfrak{g}/C^3\mathfrak{g})\le n(n-1)/2$, with equality iff
$\mathfrak{g}/C^3\mathfrak{g}$ is free 2-step-nilpotent.
Hence it's enough to find cases where there is not equality. For $n\le 2$ strict inequality is impossible (unless $n=2=\dim(g)$). For $n\ge 3$ it is very easy to find examples. The simplest would be modding out the free 2-step-nilpotent Lie algebra on $n$ generators by a nonzero subspace of its derived subalgebra. Practically, the smallest non-abelian example obtained this way is the 4-dimensional (Heisenberg)$\times$(1-dim abelian).

Addendum:
Say that $\mathfrak{g}$ is $s$-step nilpotent if $C^{s+1}\mathfrak{g}=0$, and say that the class of $\mathfrak{g}$ is the smallest such $s$.
The collection of $s$-step nilpotent Lie algebras is a variety, in the sense (from universal algebra) that it is closed under taking subalgebras, quotients, and arbitrary direct products.
[For instance, $\mathfrak{g}$ is abelian iff it is 1-step nilpotent, and then its class is 1 unless $\mathfrak{g}=0$ in which case its class is $0$; $\mathfrak{g}$ is 2-step nilpotent means that $[\mathfrak{g},\mathfrak{g}]$ is contained in the center, and class $2$ means non-abelian and 2-step nilpotent.]
The free $s$-step nilpotent Lie algebra on a set $X$ is then the quotient $\mathfrak{f}_{X,s}$ of the free one $\mathfrak{f}_X$ by $C^{s+1}\mathfrak{f}_X$.
It is then inconsistent to call the "free nilpotent $n$-generated Lie algebra of class $s$". Indeed for a single generator or for $n=1$, its class is smaller than $s$.
(A few people rather mean by $s$-step nilpotent that the class is $=s$ rather than equal to $s$, but I don't see the point in having 3 words (class, length, step) for the same definition and none simpler than "class at most $s$" for another even simpler one.)
A: If I'm not crazy a free Lie algebra is necessarily infinite dimensional so any finite dimensional nilpotent Lie algebra will do.
Perhaps you don't mean 'non-free' but 'non-Abelian'?
