Free Lie algebra and nilpotent groups in Rothschild and Stein's paper In 

Rothschild, Linda Preiss; Stein, Elias M., Hypoelliptic differential operators and nilpotent groups, Acta Math. 137(1976), 247-320 (1977). ZBL0346.35030. PDF at archive.ymsc.tsinghua.edu.cn

they mentioned the free nilpotent Lie algebra $\mathfrak{R}_{F,s}$ in Part I, section 3, example 4. And they introduced $n_s$ and $m_s$ in Part II, section 7, where


*

*$n_s$ is the dimension of the free nilpotent Lie algebra $\mathfrak{R}_{F,s}$ of step $s$ on $n$ generators,

*$m_s$ is the dimension of the linear space spanned by all commutators of the vector fields $\{W_k\}_{k=1}^n$ ($n$ is the number of vector fields) of lengths $\leq s$ restricted to $\xi$.


I can understand $m_s$ easily but It's difficult to understand $n_s$ to me. I think I need some example for $n_s$. Consider the vector fields: 
$$X=(X_1,X_2)=(\partial_1,x_1\partial_2)$$
Its $m_1=1$ restricted to point $(0,x_2)$ and $m_1=2$ at others, while $m_2=2$ at any point in $\mathbb{R}^2$. What is $n_1$ and $n_2$ for this example? why?
And by the theorem 4 and introduction in this paper, if we lift the vector fields as
$$\widetilde{X}=(\widetilde{X_1},\widetilde{X_2})=(\partial_1,x_1\partial_2+\partial_3)$$
then, $m_2=3$ and $m_2=n_2$. Why does $m_2=n_2$?
Thanks for your help! 
 A: First, the notation is a little confusing.  I think you are supposed to understand that $n$ is always the number of vector fields in the fixed set $\{X_1, \dots, X_n\}$.  But the symbol $n$ in the notation $n_s$ is just an arbitrary letter and isn't a reference to the number of vector fields.  So in your example, $n$ is $2$, and $n_1$ just means "the dimension of the free nilpotent algebra of step 1 on 2 generators".  $n_2$ is the dimension of the free nilpotent algebra of step 2 on 2 generators.  If we were working with a set of 47 vector fields, then $n_2$ would refer to the dimension of the free nilpotent algebra of step 2 on 47 generators.
In small examples, this is easy to compute.  $n_1$ will always equal $n$, because the free nilpotent Lie algebra of step $1$ on $n$ generators, call them $Y_1, \dots, Y_n$, is simply the abelian Lie algebra spanned by $Y_1, \dots, Y_n$ with all brackets vanishing, and its dimension is $n$.  In your example with $n=2$, we have $n_2 = 3$; the free nilpotent Lie algebra of step $2$ on $2$ generators is the Heisenberg Lie algebra spanned by $Y_1, Y_2, Z$, where $[Y_1, Y_2]=Z$ and $[Y_1, Z]=[Y_2, Z]=0$.
When $s \ge 3$ it gets harder.  You can't just count all possible brackets of $Y_1, \dots, Y_n$ of order up to $s$, because Jacobi's identity implies some linear dependence among them.  But in general, the value of $n_s$ can be found from Witt's formula; see Corollary 4.14 of

Reutenauer, Christophe, Free Lie algebras, London Mathematical Society Monographs. New Series. 7. Oxford: Clarendon Press. xvii, 269 p. (1993). ZBL0798.17001.

Indeed, we have $$n_s = \sum_{k=1}^s \frac{1}{k} \sum_{d \mid k} \mu(d) n^{k/d}$$
where $\mu$ is the Möbius function.
