Let $\pi$ be a finitely presented group and let $\phi:\pi\to \Bbb{Z}$ be an epimorphism. Given $n\in \Bbb{N}$ we denote by $\pi_n$ the kernel of $\pi\xrightarrow{\phi} \Bbb{Z}\to \Bbb{Z}/n$ and given a group $\Gamma$ we denote by $\operatorname{rank}(\Gamma)$ its rank, i.e. the size of a minimal generating set.

The rank gradient of $(\pi,\phi)$ is then defined as $ \liminf \frac{\operatorname{rank}(\pi_n)}{n}.$ This definition, I think, basically goes back to Marc Lackenby. If $\ker(\phi)$ is finitely generated, then $\mbox{rank}(\pi_n)\leq \mbox{rank}(\ker(\phi))+1,$ i.e. the rank gradient is zero. Now suppose that $(\pi,\phi)$ is represented by an ascending HNN-extension, i.e. there exists an isomorphism $\pi\cong \langle A,t|tAt^{-1}=\varphi(A)\rangle$ where $\varphi$ is a monomorphism and such that $\phi$ is given by the obvious surjection on the HNN-extension onto $\Bbb{Z}$. In this case we also have $\operatorname{rank}(\pi_n)\leq \mbox{rank}(A)+1,$ i.e. the rank gradient is zero. The same evidently also holds for descending HNN-extensions.

My question is now, whether there exist such pairs $(\pi,\phi)$ where the ranks of the groups $\pi_n$ are not bounded but for which the rank gradient is zero.


Here are 2 remarks.

1) Since you assume $\pi$ finitely presented, by Bieri-Strebel 78, $\phi$ is the structural epimorphism of an HNN extension over a f.g. subgroup. More precisely, we can write $\pi=\langle H,t\mid txt^{-1}=\kappa(x),x\in K\rangle$ for some injective homomorphism $\kappa$ of a finitely generated subgroup $K$ of $H$ into $H$, and $\phi:\pi\to\mathbf{Z}$ maps $t$ to 1 and $H$ to 0. A convenient way to view $\pi$ is as the semidirect product $\Lambda\rtimes\mathbf{Z}$, where $\Lambda$ is the iterated amalgamated sum $$\cdots H\ast_K H\ast_K H\cdots,$$ where each $K$ embeds into its left neighbor $H$ by the inclusion, and embeds into its right neighbor $H$ by $\kappa$. So a closely related question is to determine, in such a setting, whether the rank of $\Lambda_n$ grows linearly, where $\Lambda_n$ is the amalgamated product (with the same inclusions) $H\ast_K\dots \ast_K H$ of $n$ copies of $H$. Indeed, $\pi_n$ is generated by $\Lambda_n$ and $t$. In the ascending case you mention, $\Lambda_n$ is isomorphic to $H$ and hence has bounded rank, so $\pi_n$ has bounded rank as well.

2) I know that Lackenby and you are motivated by topology, but from a group theoretic point of view it is natural to ask the question in the case of finitely generated groups. (I first expected the (standard restricted) wreath product $\pi=F\wr\mathbf{Z}$, where $F$ is a nontrivial perfect group, to provide an example (here $\pi_n=F^n\wr\mathbf{Z}$), but it seems that for such groups $\text{rank}(\pi_n)=2$ although $\text{rank}(F^n)$ grows logarithmically.)

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  • $\begingroup$ Thanks Yves for your comments. I like the example the example where $\operatorname{rank}(F^n)$ grows logarithmically. I was aware of the relationship to the groups $\Lambda_n$, in fact in a paper with Jason DeBlois and Stefano Vidussi we played such games to show that the rank gradient is non-zero if the HNN-extension is the fundamental group of a 3-manifold and where the HNN-extension does not correspond to a fiber bundle. We used a lot of heavy 3-manifold machinery (Agol, Wise et al) to show that the rank gradient is non-zero, and I was wondering whether all this is really necessary. $\endgroup$ – Stefan Friedl Apr 30 '13 at 6:43

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