Immutability is not uniformly decidable in hyperbolic groups

TL;DR

This paper proves that there is no general algorithm to determine whether a finitely generated subgroup of a torsion-free hyperbolic group is immutable, highlighting fundamental limits in algorithmic group theory.

Contribution

It establishes the non-existence of a uniform algorithm for recognizing immutability in hyperbolic groups, answering a previously open question.

Findings

No uniform algorithm exists for recognizing immutability.

Immutability is not a decidable property in this context.

The result impacts understanding of subgroup properties in hyperbolic groups.

Abstract

A finitely generated subgroup H of a torsion-free hyperbolic group G is called immutable if there are only finitely many conjugacy classes of injections of H into G. We show that there is no uniform algorithm to recognize immutability, answering a uniform version of a question asked by the authors.

Equations

$$1\to N\to G\to Q\ast \mathbb{Z}\to 1,$$

$$c_u=tu{t}^{-2}ut,$$

$$d_u=utu{t}^{-1}u.$$

$${\Gamma }_u=G{\ast }_{\{g_u=a,h_u=b\}}{\Gamma }_0.$$

$$K_u\cong N{\ast }_{\{g_u=a,h_u=b\}}{\Gamma }_0,$$

$$K_u\cong N\ast {\Gamma }_0.$$

Full text

The work of the first author was supported by the National Science Foundation and by a grant from the Simons Foundation (#342049 to Daniel Groves)

The third author is partially funded by EPSRC Standard Grant number EP/L026481/1. This paper was completed while the third author was participating in the Non-positive curvature, group actions and cohomology programme at the Isaac Newton Institute, funded by EPSRC Grant number EP/K032208/1.

In [4] we introduced the following notion which is important for the study of conjugacy classes of solutions to equations and inequations over torsion-free hyperbolic groups, and also for the study of limit groups over (torsion-free) hyperbolic groups.

We gave the following characterization of immutable subgroups.

The following corollary is immediate.

If $\Gamma$ is a torsion-free hyperbolic group then the immutable subgroups of $\Gamma$ form some of the essential building blocks of the structure of $\Gamma$–limit groups. See [4] and [5] for more information.

In [4, Theorem 1.4] we proved that given a torsion-free hyperbolic group $\Gamma$ it is possible to recursively enumerate the finite tuples of $\Gamma$ which generate immutable subgroups. This naturally lead us to ask the following

We are not able to answer this question, but we can answer the uniform version of this question in the negative, as witnessed by the following result. It is worth remarking that the algorithm from [4, Theorem 1.4] is uniform, in the sense that one can enumerate pairs $(\Gamma,S)$ where $\Gamma$ is a torsion-free hyperbolic group (given by a finite presentation) and $S$ is a finite subset of words in the generators of $\Gamma$ so that $\langle S\rangle$ is immutable in $\Gamma$.

Consider a finitely presented group $Q$ with unsolvable word problem (see [7]), and let $G$ be a hyperbolic group that fits into a short exact sequence

[TABLE]

where $N$ is finitely generated and has Kazhdan’s Property (T). Such a $G$ can be constructed using [2, Corollary 1.2], by taking $H$ from that result to be a non-elementary hyperbolic group with Property (T), and recalling that having Property (T) is closed under taking quotients.

Let $t$ be the generator for the second free factor in $Q\ast\mathbb{Z}$. Given a word $u$ in the generators of $Q$, define words

[TABLE]

and

[TABLE]

We lift the elements $c_{u},d_{u}\in Q\ast\mathbb{Z}$ to elements $\bar{c}_{u},\bar{d}_{u}\in G$.

By considering geodesic words representing $\bar{c}_{u}$ and $\bar{d}_{u}$, it is possible to find long words in the generators of $N$ as in the statement of the claim so that any concatenation of $(w_{u}\bar{c}_{u}x_{u})^{\pm}$ and $(y_{u}\bar{d}_{u}z_{u})^{\pm}$ is such a quasigeodesic. From this, it follows immediately that the free group $\langle w_{u}\bar{c}_{u}x_{u},y_{u}\bar{d}_{u}z_{u}\rangle$ is quasi-isometrically embedded and has free image in $G$. This can be done algorithmically because the word problem in $G$ is (uniformly) solvable, so we can compute geodesic representatives for words and calculate Gromov products. ∎

Let $g_{u}=w_{u}\bar{c}_{u}x_{u}$ and $h_{u}=y_{u}\bar{d}_{u}z_{u}$, and let $J_{u}=\langle g_{u},h_{u}\rangle$. Note that the image of $J_{u}$ in $Q$ is either trivial (if $u=_{Q}1$) or free of rank $2$ (otherwise). Therefore, if $u=_{Q}1$ then $J_{u}\cap N=J_{u}$ and otherwise $J_{u}\cap N=\{1\}$.

Now consider the group

[TABLE]

Since $\langle a,b\rangle$ is malnormal and quasiconvex in $\Gamma_{0}$ and $\langle g_{u},h_{u}\rangle$ is quasiconvex in $G$, the group $\Gamma_{u}$ is hyperbolic by the Bestvina–Feighn Combination Theorem [3].

Let $K_{u}=\langle N,\Gamma_{0}\rangle\leq\Gamma_{u}$. We remark that a presentation for $\Gamma_{u}$ and generators for $K_{u}$ as a subgroup of $\Gamma_{u}$ can be algorithmically computed from the presentations of $G$ and $\Gamma_{0}$ and the word $u$.

[TABLE]

whereas in case $u\neq_{Q}1$ we have

[TABLE]

Thus, if $u\neq_{Q}1$ then $K_{u}$ splits nontrivially as a free product, as required.

On the other hand, suppose that $u=_{Q}1$, and suppose that $K_{u}$ acts on a tree $T$ with trivial or cyclic edge stabilizers. Since Property (T) groups have Property (FA) [8], $N$ and $\Gamma_{0}$ must act elliptically on $T$. However, if they do not have a common fixed vertex, then their intersection (which is free of rank $2$) must fix the edge-path between the fixed point sets for $N$ and for $\Gamma_{0}$, contradicting the assumption that edge stabilizers are trivial or cyclic. Thus, there is a common fixed point for $N$ and $\Gamma_{0}$, and so $K_{u}$ acts on $T$ with global fixed point. It follows from Corollary 3 that $K_{u}$ is immutable, as required. ∎

An algorithm as described in the statement of the theorem would (when given the explicit presentation of $\Gamma_{u}$ and the explicit generators for $K_{u}$) be able to determine whether or not $K_{u}$ is immutable. In turn, this would decide the word problem for $Q$, by Claim 3. Since this is impossible, there is no such algorithm, and the proof of Theorem 5 is complete. ∎