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My question is about the Bass-Guivarch formula which relates the growth $d(G)$ of a polynomial growth group $G$ to the torsion-free rank of consecutive quotients in its lower central series: \begin{equation*} d(G) = \sum_{k\geq 0} k\textrm{rank}(G_k/G_{k+1}) \end{equation*} My two questions are

  1. Do the terms $\textrm{rank}(G_k/G_{k+1})$ have to be nonincreasing? It seems intuitive that it does, but I can't seem to prove it. If not, is there an easy example?
  2. From this stack exchange post, Are there any usages for growth rate that are relatively easy to show?, an answer states that we can use the Bass-Guivarch formula to show that groups of quadratic (cubic) growth are virtually $\mathbb{Z}^2$ ($\mathbb{Z}^3$). I understand the case when $\textrm{rank}(G_1/G_2)$ is the only nonzero term in the sum, but not sure how to proceed otherwise. For instance, in relation to my first question, is it possible for the sum to look like $d(G) = 0+2$ in the quadratic case (this is the only other possibility), or $d(G) = 0 + 0 + 3$ in the cubic case. Regardless of my first question, however, the cubic case allows for the sum to look like $d(G) = 1 +2$. I am not sure how to pull out a $\mathbb{Z}^3$ as a finite index subgroup in that case.

Thanks!

Owen Huang
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1 Answers1

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Regarding 1.

This sequence of natural numbers can be quite arbitrary, especially if you consider not necessarily directly indecomposable groups. For example, if $F$ is free of rank $2$, then rank of $F_k/F_{k+1}$ is equal to $n+1$-st term of this sequence; in particular, if you take $G = F/F_s$, it will have same lower central ranks as $G$ up to $s-1$-st one. On the other hand, there are so called filiform groups, for which lower central ranks are $2, 1, 1, 1, 1,..$. Taking direct sums allows you to get quite arbitrary results; but of course, if rank of $G_k/G_{k+1}$ is zero, then all successive lower central quotients will have rank zero as well — see (*).

Regarding 2.

If rank of $G_1/G_2 = G_{ab}$ is either 0 or 1 and your group is torsion free (*), you do not even have a pair of non-commuting elements in your group, so there's no way to have nonzero rank of $G_2/G_3 = [G, G]/[G, [G, G]]$! In general, if rank of $G_{ab}$ is equal to $b$, then rank of $G_2/G_3$ is less then or equal to $b(b-1)/2$ — rank of $G_{ab} \wedge G_{ab}$.

(*): Finite order elements in a (finitely generated) nilpotent group form a (finite) normal subgroup.

xsnl
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  • In your answer to $1$, using the free group of rank $2$ does not address the OP's question about groups of polynomial growth. – Lee Mosher Sep 12 '23 at 01:24
  • In response to your reply to 1) and the comment above, I am specifically looking for groups of polynomial growth (ie., the ones whose central series terminate). Does that change anything? And to 2), if the rank of $G_{ab}$ is 0, that means $[G,G]$ is large, which implies that many elements don't commute, no? Did you mean that we have no pairs of commuting elements? And I'm not sure how the rank =1 case works... – Owen Huang Sep 12 '23 at 03:39
  • @LeeMosher I cannot see a reason why using lower central quotients of free groups does not address the question about nilpotent groups. They are nilpotent. And they are groups. And nilpotent groups have polynomial growth. – xsnl Sep 12 '23 at 07:23
  • @xsni You're right, but perhaps you could add a few words to clarify that the nilpotent group you are working with is $F / F_s$; as written that's hard to discern. – Lee Mosher Sep 12 '23 at 12:41
  • Okay. That makes sense thanks! Could you elaborate on why if the rank of $G_1/G_2$ is 1 and $G$ is torsion free then the rank of $G_2/G_3$ has to be zero? Cheers. – Owen Huang Sep 12 '23 at 14:57
  • Also, following up, what does the $\wedge$ mean in $G_{ab} \wedge G_{ab}$? – Owen Huang Sep 27 '23 at 14:15