In 



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> *Rothschild, Linda Preiss; Stein, Elias M.*, [**Hypoelliptic differential operators and nilpotent groups**](http://dx.doi.org/10.1007/BF02392419), Acta Math. 137(1976), 247-320 (1977). [ZBL0346.35030](https://zbmath.org/?q=an:0346.35030). [PDF at archive.ymsc.tsinghua.edu.cn](http://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6213-11511_2006_Article_BF02392419.pdf)
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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! 



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



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> *Reutenauer, Christophe*, Free Lie algebras, London Mathematical Society Monographs. New Series. 7. Oxford: Clarendon Press. xvii, 269 p. (1993). [ZBL0798.17001](https://zbmath.org/?q=an:0798.17001).
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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](https://en.wikipedia.org/wiki/M%C3%B6bius_function).


