3
$\begingroup$

I have been looking at some papers on the "restricted Burnside problem". On page 4 of Vaughan-Lee and Zelmanov's survey, "Bounds in the restricted Burnside problem", I think they implicitly are invoking the claim that:

If $G$ is a Lie algebra over $\Bbb{F}_p$ with nilpotency class $C$, generated by elements $g_1,\dotsc,g_m$, then $\lvert G\rvert \le p^{m^C}$.

How does one prove this claim? Is there a nice combinatorial reason for this (I want to guess that this translates to a nice bound on the dimension of $G$)?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ The $k$th term of the lower central series modulo the $(k+1)$th term is generated by the left-normed commutators in $g_1, \dots, g_m$ of weight $k$, of which there are at most $m^{k-1}(m-1)$ for $k > 1$. Now just sum a geometric series and do some estimation, I think. It's obvious it should be $O(m^C)$ anyway. $\endgroup$
    – Sean Eberhard
    Commented May 25, 2023 at 14:46
  • $\begingroup$ ah, okay, so morally $G$ bijects to things generated by words of length $\le C$ in the original generators. okay, that matched my vague hunch, glad to know! $\endgroup$
    – Zach Hunter
    Commented May 25, 2023 at 15:10
  • $\begingroup$ Do you really mean that $G$ is a Lie algebra, not a Lie group? (Also, physical page 4 is logical page 264.) $\endgroup$
    – LSpice
    Commented May 25, 2023 at 16:35
  • $\begingroup$ @LSpice Well he refers to the order $|G|$, so Lie algebra makes sense. It would also make sense to assume $G$ is a finite $p$-group. It hardly makes a difference. $\endgroup$
    – Sean Eberhard
    Commented May 25, 2023 at 16:37
  • $\begingroup$ @SeanEberhard, re, $\lvert G\rvert$ also makes sense for a Lie group $G$ over $\mathbb F_p$, in the sense of the group $\mathbb G(\mathbb F_p)$ of rational points of an affine algebraic $\mathbb F_p$-group $\mathbb G$. But I don't even know enough about the question to see this claim in either form on p. 4 = p. 264, so possibly my reading is absurd in context. (OP's comment, which I hadn't noticed, related to your answer suggests that it probably is.) $\endgroup$
    – LSpice
    Commented May 25, 2023 at 16:43

1 Answer 1

Reset to default
3
$\begingroup$

Converting my comment into an answer:

Let $G = G_1 \ge G_2 \ge \cdots$ be the lower central series. Then $G_k/G_{k+1}$ is spanned by the left-normed commutators $[x_1, \dotsc, x_k]$ with $x_1, \dotsc, x_k \in \{g_1, \dotsc, g_m\}$, of which there are at most $m^{k-1}(m-1)$ nontrivial for $k > 1$ (since $x_1 \ne x_2$). Therefore $$\dim G \le m + \sum_{k=2}^C m^{k-1}(m-1) = m^C.$$

This bound is not sharp. For the free Lie algebra, $G_k/G_{k+1}$ has a basis given by the basic commutators, and the number of these is given by Witt's formula: $$\dim G_k/G_{k+1} = \frac1k \sum_{d \mid k} \mu(d) m^{k/d}.$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .