The Rips construction

In some parts we follow the exposition of the blog Here there be dragons.

Using small cancellation techniques to obtain monsters has previously appeared on this blog. Most of the examples there, however, aren't finitely presented. This time we show how to obtain finitely presented, hence hyperbolic, small cancellation groups with rather odd properties. The main tool is a remarkably elementary construction originally due to Rips

Theorem 1 (Rips construction): There exists an algorithm that, given an integer \(m\geqslant 6\) and a finite presentation \(\mathcal{Q}\equiv\langle X\mid R\rangle\) of a group \(Q\), will construct a finite presentation \(\mathcal{P}\equiv\langle X\cup\{a_{1},a_{2}\}\mid\tilde{R}\cup V\rangle\) for a group \(\Gamma\) so that

1.  \(N:=\langle a_{1},a_{2}\rangle\) is normal in \(\Gamma\),
2. \(\Gamma/N\) is isomorphic to \(Q\), 
3. \(\mathcal{P}\) satisfies the small cancellation condition \(C^{\prime}(1/m)\), and
4. \(\Gamma\) is perfect if \(Q\) is perfect.

Proof: One chooses a set of reduced words \(\{u_{r}\mid r\in R\}\cup\{v_{x,i,\varepsilon}\mid x\in X,\,i=1,2,\,\varepsilon =\pm 1\}\) in the free group on \(\{a_{1},a_{2}\}\), all of length at least \(m\,\max\{|r|:r\in R\}\), that satisfies \(C^{\prime}(1/m)\). Then \(\tilde{R}=\{ru_{r}\mid r\in R\}\) and \(V\) consists of the relations \(xa_{i}x^{-1}v_{x,i,\varepsilon}\) with \(x\in X^{\varepsilon},\,i=1,2\), and \(\varepsilon=\pm 1\). Such a choice can be made algorithmic (in many different ways).

To ensure that (4) holds, one chooses the words \(v_{x,i,\varepsilon}\) to have exponent sum \(0\) in \(a_{1}\) and \(a_{2}\). Such a choice ensures that the image of \(N\) in \(H_{1}\Gamma\) is trivial, so if \(\Gamma/N\cong Q\) is perfect then so is \(\Gamma\). One way to arrange that the exponent sums are zero is by a simple substitution: choose \(\tilde{R}\cup V\) as above and then replace each occurrence of \(a_{1}\) by \(a_{1}a_{2}a_{1}^{-2}a_{2}^{-1}a_{1}\) and each occurrence of \(a_{2}\) by \(a_{2}a_{1}a_{2}^{-2}a_{1}^{-1}a_{2}\). If the original construction is made so that the presentation is \(C^{\prime}(1/5m)\) then this modified presentation will be \(C^{\prime}(1/m)\). $\blacksquare$

Wise proved that metric small cancellation groups can be cubulated and, building on work of Wise, Agol proved that cubulated hyperbolic groups are virtually compact special in the sense of Haglund and Wise. This implies that they are virtually subgroups of a RAAG, which are linear, so all groups \(\Gamma\) constructed as above are residually finite. It also follows from Agol's theorem that virtually compact special hyperbolic groups are cohomologically separable, aka good in the sense of Serre, meaning that for every finite \(\mathbb{Z}G\)-module \(M\) and \(p\geqslant 0\), the map \(H^{p}(\widehat{G},M)\to H^{p}(G,M)\) induced by the inclusion of \(G\) into its profinite completion \(G\hookrightarrow\widehat{G}\), is an isomorphism.

Rips originally used his construction to prove

Theorem 2: For all large $m$ there is a $C'(1/m)$ small cancellation group \(G\) such that:

1.  \(G\) has finitely generated subgroups whose intersection is not finitely generated.
2. \(G\) has a finitely generated but not finitely presented subgroup.
3. The subgroup membership problem in \(G\) is not solvable. Indeed, there is a subgroup \((N)\) for which there is no algorithm deciding whether a word in the generating set for \(G\) represents an element of the subgroup.
4. \(G\) contains an infinite ascending chain of \(r\)-generated groups. (\(r\geq 3\) fixed.)

The proof works by embedding recursively presented groups exhibiting the relevant behavior in a finitely presented group and applying the Rips construction. Using covering space theory, one shows that \(\{b^{i}ab^{-i}\}\) is a free basis for the subgroup of the free group \(F_{2}=\langle a,b\rangle\) it generates. Therefore the amalgamated free product
\[
\langle a,b,c,d \mid b^{i}ab^{-i}=d^{i}cd^{-i},i\in\mathbb{Z}\rangle
\]
of two copies of \(F_{2}\) along this subgroup has two finitely generated free subgroups \(\langle a,b\rangle\) and \(\langle c,d\rangle\) with infinitely generated intersection. Applying the Rips construction to a finitely presented group \(Q\) into which this recursively presented group embeds, one sees that \(p^{-1}(\langle a,b\rangle)=\langle a,b,x,y\rangle\) and \(p^{-1}(\langle c,d\rangle)\) are \(4\)-generated, but \(p^{-1}(\langle a,b\rangle\cap\langle c,d\rangle)\) is not finitely generated.

Parts (2)-(4) are proved similarly. (2) uses a finitely generated, recursively presented but not finitely presented group, such as the Lamplighter group. (3) uses a recursively presented group with unsolvable word problem (for instance
\[
\langle a,b,c,d \mid b^{i}ab^{-i}=d^{i}cd^{-i},i\in S\rangle,
\]
where \(S\) is recursively enumerable but not recursive). (4) uses a finitely presented group with an ascending chain of \(t\)-generated subgroups, such as the Baumslag-Solitar group \(BS(1,2)=\langle t,a \mid ta^{2}t^{-1}=a\rangle\).

One obtains a group \(G\) exhibiting all these pathologies simultaneously by simply taking \(Q\) to be the direct product presentation of various finite presentations that would produce the various pathologies individually. 

The theme of the Rips construction is to pass pathologies from the quotient into the hyperbolic group, and this works remarkably well. Some things, however, can't be passed up. For instance, it remains open whether there exists a residually finite hyperbolic group. At the time of the Rips construction, Agol hadn't proved his spectacular theorem, so it was unknown whether we could impose residual finiteness as one of the conditions in Rips' theorem. This inspired Wise to modify the Rips construction to get a more explicit description.

HNN extensions

Wise's modification to the Rips construction produces \(G\) as an HNN-extension of a free group. It is still somewhat surprising to me that there are so many exotic HNN-extensions of a free group, even though we have seen in a previous post that HNN-extensions of Fuchsian groups can be arbitrarily weird in some sense. 

The construction makes use of Wise's 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 \(Q=\langle a_{1},\ldots,a_{n} \mid r_{1},\ldots,r_{m}\rangle\), then
\[
G=\left\langle
\begin{array}{c}
a_{1},\ldots,a_{n}, \\
x,y,t
\end{array}
\;\middle|\;
\begin{array}{c}
r_{j}=*t*t^{-1},\quad a_{i}^{\pm 1}xa_{i}^{\mp 1}=*t*t^{-1}, \\
a_{i}^{\pm 1}ya_{i}^{\mp 1}=*t*t^{-1},\quad a_{i}^{\pm 1}ta_{i}^{\mp 1}=*t*
\end{array}
\right\rangle
\]
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: \(x,y,t\). Wise used this to produce residually finite, hyperbolic small cancellation groups with non residually finite outer automorphism groups. Let us now introduce the quotient group $Q$ that we will stick in.

Higman's group

 Higman's group is the ingeniously constructed group with presentation.
$$G=\langle a,b,c,d | a^{-1}ba=b^2, b^{-1}cb=c^2, c^{-1}dc=d^2, d^{-1}ad=a^2\rangle.$$

It is an infinite, finitely presented group with no finite quotients whatsoever.

To see it has no finite quotients, suppose that there is a finite group $G$ with elements $g_1, g_2, g_3, g_4$ which are the images of $a,b,c,d$ respectively, i.e. satisfying the relations given by the presentation above. Let the order of \(g_{i}\) be \(n_{i}\). Taking indices modulo \(4\) we find 

\[g_{i+1}=g_{i}^{-n_{i}}g_{i+1}g_{i}^{n_{i}}=g_{i+1}^{2^{n_{i}}}\quad\text{for all $i$}\]

so that \(n_{i+1}\mid 2^{n_{i}}-1\).  It is a fun fact of elementary number theory that the smallest prime factor of $n$ is strictly smaller than the smallest prime factor of $2^n-1$. Hence every prime factor of \(n_{i+1}\) is a prime factor of \(2^{n_{i}}-1\) and is bigger than the smallest prime factor of \(n_{i}\) unless \(n_{i}=1\). This is true for all \(i\), giving a contradiction unless all \(n_{i}\) equal \(1\).

To see that Higman's group is infinite, Higman exhibits it as the amalgamated product of two groups $K_{1,2}=\langle a_1, a_2, b_2\mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2\rangle$ and $K_{3,4}=\langle a_3, a_4, b_4\mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2\rangle$ with free amalgamated subgroups $\langle a_1, b_2\rangle$ and $\langle a_3, b_4\rangle$ with $a_1=b_4, b_2=a_3$. Each of $K_{1,2}$ and $K_{3,4}$ is in turn an amalgamated product of two copies of $BS(1,2)$ over cyclic subgroups. This decomposition shows that the group is an amalgamated product of two aspherical groups with free associated subgroups. So the group is (combinatorially) aspherical, hence its presentation complex is a classifying space. One can compute from this that Higman's group is acyclic. 

Entirely analogous arguments apply to show the same properties for the groups $$H_n=\langle a_1, \dots a_n | a_i^{-1}a_{i-1}a_i=a_{i-1}^2 \rangle$$ when $n \geq 4$.

Although it has no finite quotients, Higman's group is far from simple. Take $K_{1,2}$ and add a relation $w(a_1,b_2)=w(b_2,a_1)$. For example, $(a_1b_2)^3=(b_2a_1)^3$.  You get a non-trivial group $K$ (to see this impose further relations $a_2=b_2=1$ on $K$ to get an infinite cyclic group). Similarly imposing $(a_3b_4)^3=(b_4a_3)^3$ on $K_{3,4}$ you get a non-trivial group $K'$. Then there is a homomorphism from the Higman group to the amalgamated product of $K$ and $K'$ with a non-trivial kernel. The fact that $K$ (and, equally, $K'$) is not trivial is not obvious but it is not very difficult to prove. (I found this argument on MO.)

More is true: Schupp proved that Higman's group is SQ-universal, so every countable group embeds into one of its quotients. Baumslag observed that since there are no finite quotients, any maximal normal subgroup must be of infinite index, so quotienting out by such a normal subgroup must give rise to a finitely generated infinite simple group. From the point of view of analytic group theory, it is also rather strange: it doesn't have Property (T) or (FA) since it admits an amalgamated product decomposition. Rivas and Triestino showed that Higman's group is left orderable, but it is perfect since it admits no finite quotients, so it can't be amenable by Witte-Morris' result that left orderable amenable groups are locally indicable (this also follows from the amalgamated product decomposition).

Higman's group doesn't admit a free splitting. Here is an argument due to Agol:

Suppose that $G \cong A\ast B$, a free product of non-trivial groups $A, B$. Consider the subgroup $H=\langle a,b\rangle$. Clearly  the relation $a^{-1}ba=b^2$ holds in the subgroup $H$, and in fact one can show that this is the only relation needed, so $H\cong BS(1,2)$, a soluble Baumslag-Solitar group. $H$ is non-trivial and not a non-trivial free product or free group since it is solvable and non-cyclic. Then by the Kurosh subgroup theorem, $H$ is conjugate into $A$ or $B$, let's say into $A$. Now we may take a non-trivial quotient of $G\cong A\ast B \twoheadrightarrow B$ by killing $A$. In turn, this kills the subgroup $H$ which is conjugate into $A$, so kills $a, b$. But one sees directly from the presentation that killing $a$  and $b$, one kills $c$ and $d$ as well, giving the trivial group, a contradiction. 

The existence of finitely presented acyclic groups such as Higman's group has topological consequences. Here is a construction of a non-aspherical, acyclic space $X$ whose fundamental group is also acyclic, due to Ebert:

Take any finitely presented infinite acyclic group $G$, for example, Higman's group. It is a theorem of Kervaire (''Smooth homology spheres and their fundamental groups'') that for each $n \geq 5$, there is an integral homology sphere $M^n$ with fundamental group $G$. Consider $X=M-\ast$. The integral homology of $X$ is trivial, the fundamental group of $X$ is $G$. Suppose that $X$ were  aspherical. Observe that $M\simeq X \cup_f D^n$ is obtained by attaching an $n$-cell. Since $\pi_{n-1} (X)=0$, $f$ is nullhomotopic, and hence $M \simeq S^n \vee X$. It follows that the universal cover of $M$ has the homotopy type of $\tilde{X}$ with infinitely many $S^n$'s attached (since $G$ is infinite). So $H_n (\tilde{M};\mathbb{Z}) \neq 0$, a contradiction, because $\tilde{M}$ is a noncompact $n$-manifold.

I now attempt to rein myself in and go back to the Rips construction.

Non-residually finite outer automorphisms

Wise applied the Rips construction to \(Q=H\). The action of \(G\) on \(N\) by conjugation produces a map
\[
Q=G/N\to\operatorname{Out}(N).
\]
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 \(\langle a_1,\gamma\rangle\) and \(\langle a_2,\gamma\rangle\) are either \(\mathbb{Z}\) or \(\mathbb{Z}^{2}\). Hyperbolicity of \(G\) rules out the second case, so \(a_1^{a}=\gamma^{b}\), \(a_2^{c}=\gamma^{d}\) for some integers \(a,b,c,d\neq 0\). Then \(a_1^{ad}=a_2^{bc}\), contradicting the fact that the free group \(\langle a_1,a_2\rangle\) embeds in the HNN-extension \(G\). Since \(Q\) is a subgroup of \(\operatorname{Out}(N)\), it follows that \(\operatorname{Out}(N)\) is not residually finite. (We note that this argument is now known to work for any small cancellation group from the original Rips construction.)

Mel'nikov's question

Recall that the profinite topology on \(G\) has a base 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 \(\delta(G)\leq\operatorname{rank}(G)\), 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 \(\delta(G)\leq 3\), 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 \(H\).

Explicit 2-complexes

Since we know that all small cancellation groups have aspherical presentations, it would be nice if we could describe their classifying spaces explicitly. Wise also modified the Rips construction to make \(G\) the fundamental group of a locally \(\mathrm{CAT}(-1)\) \(2\)-complex whose \(2\)-cells are metrised as congruent right-angled hyperbolic pentagons. 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 \(Q=\langle x_{1},\ldots,x_{n} \mid r_{1},\ldots,r_{m}\rangle\), 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 \(\pi_{1}K\) to have \(r_{i}\) and \(x_{j}^{\pm 1}a_{i}x_{j}^{\mp 1}\) 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 \(\mathrm{CAT}(-1)\) (respectively \(\mathrm{CAT}(0)\) or \(\mathrm{CAT}(1)\)) iff the link of each vertex is \(\mathrm{CAT}(1)\). For a 2-complex, where the link is a graph, this link condition has a particularly nice formulation: \(K\) is locally \(\mathrm{CAT}(-1)\) iff no link of a vertex contains an embedded loop of length less than \(2\pi\). Our complex \(K\) satisfies this condition iff:

1.  For any two adjacent sides of any pentagon, the corresponding letters \(a_{i}\) and \(a_{i+1}\) are distinct, and
2. 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 \emph{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. Observe that if each letter \(s\) in a set \(A\) is replaced by 5 letters \(s^{0},\ldots,s^{4}\) and each word \emph{abcde} replaced by the 25 words \(\{a^{i}b^{i+j}c^{i+2j}d^{i+3j}e^{i+4j}\}_{0\leq i,j\leq 6}\) 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 \(m\).

Further applications

This post has gotten quite long, and the further applications get somewhat technical, so I'll lazily wrap up by just listing a few. A lot of work used the construction to exhibit countable groups as outer automorphism groups, but this has been discussed before so we omit such applications.

1. Building on ideas of Gromov, Ollivier--Wise use a variant of the Rips construction to produce non-Hopfian groups with Property (T)
2. Belegradek--Osin use small cancellation theory over relatively hyperbolic groups to study a variation on the Rips construction and produce:
   
   • for any finitely presented group $\Gamma$ and for all $n$, a non-elementary hyperbolic group $G$ and an epimorphism $G\to \Gamma$ such that every representation $G \to \rm{GL}_n(\mathbb C)$ that factors through $\Gamma$.
   • a large, torsion free, hyperbolic group $G$ and an element $g \in G$ such that $G/\langle \langle g^n\rangle \rangle $ is not large for infinitely many values of $n$.  
     a torsion free hyperbolic group $G$ with positive first Betti number, but with $g \in G$ such that $G/\langle \langle g^n\rangle \rangle $ has property (T) for all $n$. 
3. Arzhantseva--Steenbock produce hyperbolic groups with subgroups that don't have the unique product property but do have Property (T), and also prove the existence of torsion-free Tarski monsters without the unique product property (in particular, any proper subgroup has unique products).