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In a previous question, I asked about hyperbolic groups in which every finitely generated subgroup is hyperbolic. I am now curious about the reverse question: what are some examples of hyperbolic groups containing finitely generated non-hyperbolic groups?

Any non-coherent hyperbolic group will give an example (e.g. via the Rips construction). Do there exist examples of coherent hyperbolic groups containing f.g. non-hyperbolic subgroups?

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    $\begingroup$ Off the top of my head I'm not sure about examples where the hyperbolic group is coherent, but just for some more examples of hyperbolic groups with finitely generated non-hyperbolic subgroups, Brady found the first examples where the subgroup is even finitely presented (londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/…) and Italiano-Martelli-Migliorini found an example where the subgroup is even of type F (arxiv.org/abs/2105.14795). $\endgroup$ Jul 11 at 11:10

2 Answers 2

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It is an open problem to find a coherent hyperbolic group with a finitely generated, non-hyperbolic subgroup. See Wise's survey article for the state of the art on coherent groups.

Wise, Daniel T. (3-MGL) An invitation to coherent groups. What's next?—the mathematical legacy of William P. Thurston, 326–414, Ann. of Math. Stud., 205, Princeton Univ. Press, Princeton, NJ, 2020.

I suspect this is listed as a question there, but if not, it's very much in that spirit.

The theorem of Gersten that I mentioned in answer to your last question showed that there is no 2-dimensional example. All known coherent 3-dimensional groups are essentially 3-manifold groups. Sam Nead's answer to that question explained why finitely generated subgroups of hyperbolic 3-manifold groups are also hyperbolic.

As Matt Zaremsky mentions in comments, the key examples of finitely presented non-hyperbolic subgroups of hyperbolic groups are given by the original example of Noel Brady and the recent work of Italiano--Martelli--Migliorini. There are also related constructions by Fujiwara, Groves--Manning, Kropholler, Llosa Isenrich--Martelli--Py and Lodha.

It would be very surprising if any of these examples were to turn out to be coherent. In fact, it is reasonable to conjecture that these examples are all incoherent, but I believe it is unknown in each case. Proving that Brady's example (say) is incoherent wouldn't answer your question, but would be a nice thing to do.

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    $\begingroup$ In the first sentence you probably want "finitely generated" before "non-hyperbolic" $\endgroup$ Jul 11 at 12:54
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    $\begingroup$ @MattZaremsky: I certainly do! Fixed. $\endgroup$
    – HJRW
    Jul 11 at 12:55
  • $\begingroup$ Brady's example is noncoherent. $\endgroup$ Jul 11 at 17:23
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    $\begingroup$ @MoisheKohan: I certainly believe it, but something needs to be said. Are you envisaging using combinatorial Morse theory on some other map to Z? $\endgroup$
    – HJRW
    Jul 12 at 5:55
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    $\begingroup$ @HJRW: I wrote a sketch of the proof as an answer. $\endgroup$ Jul 12 at 17:48
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It's a bit too long for a comment: Brady's example is not coherent. Below is a sketch of the proof. The example constructed by Brady (see the reference in HJRW's answer) is obtained as a certain ramified covering over the space $X=\Theta^3$, where $\Theta$ is the theta-graph (two vertices and three edges). This graph is the union of two circles $C_1, C_2$. In particular, the space $X$ contains two 3-dimensional tori $C_1^3, C_2^3$, whose intersection contains a 2-dimensional torus $T$. Taking a ramified covering $Y\to X$, we obtain lifts $M_1, M_2$ of the tori $C_1^3, C_2^3$ to $Y$, such that $M_1, M_2$ are 3-dimensional compact hyperbolic manifolds intersecting in a subset containing a compact hyperbolic surface $S$ which is incompressible in both $M_1, M_2$ (the surface $S$ is a lift of the torus $T$). Furthermore, the manifolds $M_1, M_2$ are $\pi_1$-injective in $Y$. In particular, $\pi_1(Y)$ contains the amalgam $\pi_1(M_1)\star_H \pi_1(M_2)$, where $H$ is a certain quasiconvex subgroup of infinite index in both $\pi_1(M_1), \pi_1(M_2)$. Now, one uses the fact that both manifolds $M_1, M_2$ fiber over the circle, let $G_1, G_2$ denote surface fiber subgroups in $\pi_1(M_1), \pi_1(M_2)$. The intersection $G_1\cap H=G_2\cap H=F$ is a free group of infinite rank. Thus, the subgroup $$ G=\langle G_1, G_2\rangle= G_1\star_F G_2 $$ is finitely generated but not finitely presentable (this can be checked, for instance, by looking at the 2-nd cohomology group of $G$ using the Mayer-Vietoris sequence).

I omitted many details in this proof but they are not hard to check once one understands details of Brady's construction.

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    $\begingroup$ I think you can also do as @HJRW suggested and build a different map from Brady's group to $\mathbb{Z}$ whose kernel is f.g. but not f.p. Brady builds his Morse function starting with a map from $\Theta^3$ to $S^1$ that wraps the edges around so as to ensure every ascending and descending link is a 2-sphere. If you instead "kill" one of the pairs of ascending directions and one of the pairs of descending directions, making them "flat", I believe now every ascending and descending link (of either a vertex or a flat edge) is a circle. (And then the kernel thing follows.) $\endgroup$ Jul 12 at 20:05
  • $\begingroup$ Very nice! ${}{}$ $\endgroup$
    – HJRW
    Jul 13 at 9:53

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