Rank gradients and HNN-extensions 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.
 A: Here are 2 remarks.

*

*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.


*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: see Jacques Thévenaz, Maximal subgroups of direct products.)
