## The Rips Construction, II

In the last post, we saw how Rips constructed a short exact sequence

$\displaystyle 1\rightarrow N\rightarrow G\stackrel{p}{\rightarrow} Q\rightarrow1$

associated to any finitely presented group ${Q}$. ${N}$ was 2-generated and ${G}$ was small-cancellation (${C'(\lambda)}$ for any desired ${\lambda>0}$). In this post, we will see modifications due to Daniel Wise that impose further conditions on ${G}$. First we will mention an application of the Rips construction relating to subgroup distortion.

1. Subgroup Distortion

If ${H\leq G}$ are finitely generated groups, then the length of an element in the word metric on ${H}$ may be much more than its length in the word metric on ${G}$. One defines the subgroup distortion function

$\displaystyle \delta_H^G(n)=\max\{|h|_H:h\in H,|h|_G\leq n\}.$

Zlil Sela observed that if one applies the Rips construction to a finitely presented group ${Q}$ with unsolvable word problem, then ${\delta_N^G(n)}$ is not bounded from above by any recursive function. Indeed, if such a bound ${f(n)}$ existed, one could determine if a word ${w}$ in the generators of ${Q}$ is equal in ${G}$ to any of the elements in ${N}$ of length at most ${f(|w|)}$. Equality can be tested since ${G}$ has solvable word problem. Hyperbolic groups can therefore have finitely generated subgroups with huge distortion. See Arzhantseva–-Osin for a discussion and the converse. Further, Olshanskii and Sapir show that the set of classes of distortion functions of finitely generated subgroups of a product of two free groups coincides with the set of classes of all Dehn functions of finitely presented groups.

2. Residually Finite Rips Construction

It is unknown whether all ${C'(1/6)}$ groups, or indeed all hyperbolic groups, are residually finite. (For ${C'(1/6)}$ Wise attributes the question to Schupp, and for hyperbolic to Gromov.) It seemed possible that the Rips construction could be used to construct a counterexample, by embedding into ${G}$ some obstruction to it being residually finite. Wise argued that this approach is unlikely to work, since he can modify the Rips construction in a way that makes ${G}$ residually finite (so no obstruction to being residually finite embeds in it).

Wise’s construction produces ${G}$ as an HNN–extension of a free group. It makes use of his earlier result that a graph ${H}$ of free groups is residually finite as long as the edge subgroups ${H_e}$ incident at each vertex group ${H_v}$ satisfy a malnormality condition: any intersection of conjugates of two edge subgroups incident at the same vertex subgroup is trivial, and ${H_e}$ intersects its own conjugate by ${x}$ trivially for every ${x\in H_v\setminus H_e}$. This condition in turn is guaranteed by a small cancellation condition on the generators of the edge subgroups. If ${Q=\langle a_1,\ldots,a_n|r_1,\ldots,r_m\rangle}$, then

$\displaystyle G=\Bigg\langle\begin{array}{cccc} a_1,\ldots,a_n, & | & r_j=*t*t^{-1}, & a_i^{\pm1}xa_i^{\mp1}=*t*t^{-1},\\ x,y,t & | & a_i^{\pm1}ya_i^{\mp1}=*t*t^{-1}, & a_i^{\pm 1}ta_i^{\mp1}=*t* \end{array}\Bigg\rangle$

is an HNN-extension of the free group ${F_{n+2}=\langle a_1,\ldots,a_n,x,y\rangle}$ with stable letter ${t}$. As with the original Rips construction, each occurrence of ${*}$ in each relator is different noise in ${x,y}$, chosen so that the small cancellation requirements are satisfied, making ${G}$ both small cancellation and residually finite. ${N}$ is finitely generated, but this time requiring 3 generators: ${x,y,t}$.

3. Applications

Wise used his construction to produce a residually finite group ${N}$ whose outer automorphism group ${\textrm{Out}(N)}$ is not residually finite. Starting with a group

$\displaystyle H=\langle a,b,c,d|b^a=b^2,c^b=c^2,d^c=d^2,a^d=a^2\rangle$

which Higman showed has no nontrivial finite subgroups, Wise applied the Rips construction to ${Q=H}$. The action of ${G}$ on ${N}$ by conjugation produces a map ${Q=G/N\rightarrow\textrm{Out}(N)}$. This map is injective, since if conjugation by ${g\notin N}$ is the same as conjugation by ${n\in N}$, then ${\gamma:=gn^{-1}}$ is nontrivial and commutes with ${N}$. Since ${G}$ is an HNN-extension of a free group, it is torsion free, so the abelian subgroups ${\langle x,\gamma\rangle}$ and ${\langle y,\gamma\rangle}$ are either ${\mathbb{Z}}$ or ${\mathbb{Z}^2}$. Hyperbolicity of ${G}$ rules out the second case, so ${x^a=\gamma^b}$, ${y^c=\gamma^d}$ for some integers ${a,b,c,d\neq0}$. Then ${x^{ad}=y^{bc}}$, contradicting the fact that the free group ${\langle x,y\rangle}$ embeds in the HNN-extension ${G}$. Since ${Q}$ is a subgroup of ${\textrm{Out}(N)}$, it follows that ${\textrm{Out}(N)}$ is not residually finite.

Recall that the profinite topology on ${G}$ has a basis of closed sets consisting of cosets of finite index normal subgroups. The rank of a group ${G}$ is the minimum number of generators, and its topological rank ${\delta(G)}$ is the minimal number of elements needed to generate a group whose closure in the profinite topology is ${G}$. So ${\delta(G)\leq\textrm{rank}(G)}$, and O.V. Mel’nikov asked if fixing ${\delta(G)}$ bounds the rank of a residually finite group ${G}$. Wise answered the question in the negative by applying the Rips construction to a free product ${Q}$ of ${k}$ copies of Higman’s group ${H}$ mentioned in the previous paragraph. The rank of ${G}$ is at least the rank of ${Q}$, which is ${k}$ times the rank of ${H}$ by Grushko’s Theorem. But ${\delta(G)\leq3,}$ since any nontrivial finite index subgroup of ${G}$ containing ${N}$ would give rise to a proper finite quotient of ${Q}$ and hence (by restriction on some factor) to a proper finite quotient of ${H}$.

Wise also constructed a residually finite group ${G}$ with closed supgroups ${H,N}$ whose double coset ${HN}$ is not residually finite.

4. Compact Negatively Curved 2-Complexes

Wise also modified the Rips construction to make ${G}$ the fundamental group of a locally ${CAT(-1)}$ 2-complex whose 2-cells are metrized as congruent right-angled hyperbolic pentagons. We will mostly follow the presentation in Bridson-Haefliger. Unlike the previous constructions, the number ${M}$ of generators for ${N}$ depends on the presentation ${Q},$ but it is still finite. The construction does not use the number five significantly, so an analogous procedure with heptadecagons or such can be used to satisfy an arbitrary small cancellation condition.

Given a finite presentation ${Q=\langle x_1,\ldots,x_n|r_1,\ldots,r_m\rangle}$, one forms the desired 2-complex ${K}$ as follows, and sets ${G=\pi_1(K)}$. Construct ${K}$ using a single vertex, and an edge (loop) for each old generator ${x_1,\ldots,x_n}$ and each new generator from a set ${A}$, where ${M=|A|}$ is to be chosen later. The 2-cells of ${K}$ will be right-angled pentagons and we label the sides of the pentagons with words of length ${\rho=\max\{|r_i|+2,5\}}$ that determine the attaching maps. We want the relators for ${\pi_1K}$ to have ${r_i}$ and ${x_j^{\pm 1}a_ix_j^{\mp 1}}$ as substrings of words whose other letters are chosen from ${A}$, so we take a pentagon labelled as follows (shown for the ${r_i}$, the other relators get similar pentagons). The choice of ${a_0,\ldots,a_4\in A}$ is deferred.

Gromov showed that a complex formed from finitely many isometry classes of hyperbolic (or euclidean, or spherical) cells is locally ${CAT(-1)}$ (respectively ${CAT(0)}$ or ${CAT(1)}$) iff the link of each vertex is ${CAT(1)}$. For a 2-complex, where the link is a graph, this link condition has a particularly nice formulation: ${K}$ is locally ${CAT(1)}$ iff no link of a vertex contains an embedded loop of length less than ${2\pi}$. Our complex ${K}$ satisfies this condition iff:

• For any two adjacent sides of any pentagon, the corresponding letters ${a_i}$ and ${a_{i+1}}$ are distinct, and
• No ordered pair ${(a_i,a_{i+1})}$ of cyclically adjacent edge letters occurs twice among all pentagons (or in the same pentagon twice).

Call a subset of 5-letter words in an alphabet ${A}$ of ${m}$ letters admissible if it can be used to build pentagons satisfying the above criteria. If we show that largest size of an admissible set grows at least quadratically (hence superlinearly) with ${m}$, then ${M}$ can be chosen to make the above construction work. (This can all be done effectively.) Here we depart from Bridson-Haefliger’s presentation, and observe that if each letter ${s}$ in a set ${A}$ is replaced by 5 letters ${s^0,\ldots,s^4}$ and each word ${abcde}$ replaced by the 25 words ${\{a^ib^{i+j}c^{i+2j}d^{i+3j}e^{i+4j}\}_{0\leq i,j<5}}$ with indices taken mod 5, then we obtain a new admissible set for the new alphabet. This shows the largest size of admissible sets grows at least quadratically with ${m}$.