Free (nilpotent) Lie algebras

Question

If $F$ is a free Lie algebras of finite rank $n\gt 1$. When $\dim (F) $ and $\dim (F/\gamma_{n+1}(F))$ is infinity, in where $\gamma_n (X)$ is the $n-$term of lower central series $X$.

Thanks for your comments

Answer 1

The free Lie algebra on a "basis" $x_1,\ldots,x_n$ contains all words in $x_i$, i.e., say for $n=3$, $$ x_1 ,x_2, x_3 \\ [x_1,x_2], [x_1,x_3], [x_2,x_3] \\ [x_1,[x_1,x_2]], [x_1,[x_1,x_3]], [x_2,[x_1,x_2]], [x_2,[x_1,x_3]], [x_3,[x_1,x_2]], [x_3,[x_1,x_3]], $$ and so on. So it is infinite-dimensional. The free-nilpotent Lie algebras are finite-dimensional, because then words which are longer than $c$ become zero. The Jacobi identity gives further restrictions. It is well-known that the dimension of a $c$-step nilpotent Lie algebra on $n$ generators is $$ \dim L(c,n)=\sum_{m=1}^c \frac{1}{m}\sum_{d\mid m}\mu(d)n^{m/d}. $$ For example, $L(2,3)$ is given by $x_1,x_2,x_3$ and $x_4=[x_1,x_2]$, $x_5=[x_1,x_3]$ and $[x_2,x_3]=x_6$, and no restriction by the Jacobi identity, because it is $2$-step nilpotent. So $\dim L(2,3)=6$. A bit more interesting is to see that $\dim L(3,3)=14$.