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I'm giving a lecture about growth rates in a seminar. the chapter in the book I'm working with mainly focuses on computing growth rates and states Gromov's polynomial growth theorem (and the fact that it's way beyond the scope of the chapter). I feel like the chapter, and therefore my lecture, lacks concrete results that can be implied from certain growth rates. is there anything trivial or at least intuitively understandable of this sort? most of what I see on the Wikipedia page is calculations of growth rates.

Aviv
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  • Which book are you referring to? – Shaun May 06 '22 at 14:54
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    @Shaun office hours with a geometric group theorist – Aviv May 06 '22 at 15:08
  • Well, Gromov's result is rather concrete - see wikipedia. You can talk about it without giving a proof. Certainly you should ask your adviser for the seminar for more details on your topic. It seems that you need to present some of the basics on geometric group theory, too. – Dietrich Burde May 06 '22 at 15:16
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    @DietrichBurde what is there to talk about Gromov's result apart from a proof? is there an intuition of it apart from the fact that both polynomial growth and being virtually nilpotent are both characteristics of "small" groups? – Aviv May 06 '22 at 15:21
  • For example one could discuss easy consequences of Gromov's theorem, see for example the book by Drutu and Karpovic, Chapter $16$, or here. Or saying more on the gap conjecture. – Dietrich Burde May 06 '22 at 16:06
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    Generally speaking, I think that applications of a great theorem are just as important for understanding it as its proof. – Lee Mosher May 06 '22 at 19:40
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    I've nominated this question for re-opening. It may be brief but the discussion of Gromov's theorem is good enough, and there is a possibility for some good and unique answers. – Lee Mosher May 06 '22 at 19:49
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    Please do not answer closed questions in the comments in order to bypass closure. – Xander Henderson May 06 '22 at 21:27
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    I'll repeat, I think this is a good candidate for re-opening. It would be an even better candidate if the OP made some improvements, of course: some definitions, an actual statement of Gromov's polynomial growth theorem, any possible implications of it that the OP may know, no matter how trivial. There is a lot of good mathematics to be said in answer to the question. – Lee Mosher May 06 '22 at 21:40

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If you already know something about growth rates from that Wikipedia page, in particular the Bass-Guivarch formula for the polynomial growth degree of a finitely generated nilpotent group, then there are a lot of reasonably quick applications of Gromov's polynomial growth theorem. Here are some of them.

Let $G$ be a finitely generated group whose growth is polynomial of degree $d \ge 1$ (if $G$ has polynomial growth degree $d=0$ then $G$ is finite; you don't need Gromov's Theorem for that).

By applying Gromov's Theorem, it follows that $G$ has a nilpotent subgroup $N$ of finite index. And then, by applying the Bass-Guivarch formula to $N$, you can make the following additional implications for low values of $d$:

  1. If $d=1$ then $G$ has a finite index subgroup isomorphic to $\mathbb Z$.
  2. If $d=2$ then $G$ has a finite index subgroup isomorphic to $\mathbb Z^2$.
  3. If $d=3$ then $G$ has a finite index subgroup isomorphic to $\mathbb Z^3$.
  4. If $d=4$ then $G$ has a finite index subgroup isomorphic to either $\mathbb Z^4$ or to a finite index subgroup of the integer Heisenberg group $$H_3(Z) = \left\{ \begin{pmatrix} 1 & K & M \\ 0 & 1 & L \\ 0 & 0 & 1 \end{pmatrix} \, : \, K, L, M \in \mathbb Z \right\}. $$

You can see that this begins to get complicated when $n=4$, and although one can continue onward it does get more and more complicated as $n$ increases.

In each case, one makes these calculations by using finite index of $N < G$ to conclude that $N$ and $G$ have the same polynomial growth degree $d$, then writing out the lower central series $$N = N_1 \, \triangleright \, N_2 \,\, \triangleright \, ... \, \triangleright \, N_K = \{\text{Id}\}, \quad \quad N_{k+1} = [N,N_k] $$ and then applying the Bass-Guivarch formula \begin{align*} d &= \sum_{k \ge 1} k \cdot \text{rank}(N_k / N_{k+1}) \\ &= 1 \cdot \text{rank}(N_1 / N_2) + 2 \cdot \text{rank}(N_2 / N_3) + 3 \cdot \text{rank}(N_3 / N_4) + \cdots \end{align*} where $\text{rank}(N_k / N_{k+1})$ denotes the rank of the free abelian summand of the finitely generated abelian group $N_k / N_{k+1}$. In the course of the calculations one will need to work with elements of the theory of lower central series, finitely generated abelian groups, and finitely generated nilpotent groups, to make the final conclusions.

For a few details in the simplest case $d=1$, by applying the Bass-Guivarch theorem one concludes that all higher commutator subgroups $N_k$ for $k \ge 2$ are finite (because $\text{rank}(N_k/N_{k+1})=0$ for $k \ge 2$), and that $N/[N,N]$ is a finitely generated abelian group of rank $1$ (because $\text{rank}(N_1/N_2) = \text{rank}(N/[N,N])=1$). It quickly follows (from the classification of finitely generated abelian groups) that $N/[N,N]$ has an infinite cyclic subgroup of finite index, hence so does $N$ and finally so does $G$.

Lee Mosher
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  • Thank you, very informative answer! I have a question: How do we go from an infinite cyclic subgroup of finite index in $N/[N,N]$ to such a subgroup in $N$? – Yuxiao Xie Sep 27 '22 at 10:55
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    Once you know that $[N,N]$ is finite (for reasons explained in my last paragraph), simply pull the generator of the infinite cyclic subgroup of $N/[N,N]$ back to an element of $N$ itself, and check that it generates a finite index subgroup of $N$. – Lee Mosher Sep 27 '22 at 21:41