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!

  • $\begingroup$ You don't introduce $W_k$. Also I don't see the meaning of "its $m_1=1$"... Also calling $n_s$ a function of $n$ and $s$ is utterly confusing. $\endgroup$
    – YCor
    Jan 5, 2020 at 23:27
  • $\begingroup$ Sorry. $\{W_k\}_{k=1}^{m}$ are vector fields, so are the $X$ below. The definition of $n_s$ and $m_s$ is the original words in his paper. In the first example, when $s=1$, then at the point $(0,x_2)$ we can see $X=(\partial_1,0)$. Thus $m_1=dim(span\{\partial_1,0\})=1$. The other cases are similar. But I don't understand the definition of $n_s$ in his paper, even don't know how to calculate $n_s$ for this example. $\endgroup$
    – Houa
    Jan 6, 2020 at 1:55
  • $\begingroup$ More explanations. When $s=2$, we have the commutator $[X_1,X_2]=\partial_2$, thus by the definition, we have $m_2=dim(span\{\partial_1,x_1\partial_2,\partial_2\})=2$ at any point in $\mathbb{R}^2$. $\endgroup$
    – Houa
    Jan 6, 2020 at 2:02
  • $\begingroup$ @YCor: The notation is indeed a bit unfortunate. I think in context, R&S are treating $n$ as fixed throughout, and the $n$ in $n_s$ is not a reference to the same $n$. $\endgroup$ Jan 6, 2020 at 8:22

1 Answer 1


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.

  • $\begingroup$ The Witt's formula seems that ns only depends on the number of the generators. By the description in R&S's paper, I try to explain like the following. For given "generator": $Y_1,Y_2,Y_3$, let them generate the "free nilpotent Lie algebra" $\mathfrak{R}_{F,2}$ (let, say $s=2$,) $=\{Y_1,Y_2,Y_3,[Y_1,Y_2],[Y_1,Y_3],[Y_2,Y_3]\}$. Thus $n_2=6$, which is consistent with the result made by Witt's formula. Once we set $Y_i$ to be a specific vector field, like $Y_1=\partial_1$ etc. then we can calculate $m_2$ at some point and obviously $m_2\leq n_2$. $\endgroup$
    – Houa
    Jan 7, 2020 at 8:08
  • $\begingroup$ @xixixi: The value of $n_s$ given by Witt depends on both the number of generators $n$ and the step $s$ (the sum on $k$ is up to $s$). But yes, your explanation is basically right. $n_s$ is the "worst case" dimension, where none of the brackets coincide with each other, and so $m_s$ can be no larger. $\endgroup$ Jan 7, 2020 at 17:01

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