Bass--Lubotzky's counterexample to Platonov's conjecture

The representation theory of finite groups over characteristic zero is quite well understood. Representations are semisimple and determined by their characters, and there are only finitely many simples. Many fundamental examples of infinite groups have very different representation theory: consider for instance that $\mathbb{Z}= \langle \begin{matrix}1 & 1 \\0 & 1 \end{matrix}$, and more generally free and surface groups have incredibly rich representation theory in various Lie groups. I don't want to discuss that, but it is reasonable to ask whether there are infinite groups whose representation theory is resembles that of finite groups.

Let \(\Gamma\) be a finitely generated group. By a representation of \(\Gamma\) we mean a finite dimensional complex representation, i.e., essentially a homomorphism \(\rho : \Gamma \to \operatorname{GL}_n(\mathbb{C})\), for some \(n\). We call \(\Gamma\) linear if some such \(\rho\) is faithful (i.e., injective). We call \(\Gamma\) representation rigid if, in each dimension \(n \geq 1\), \(\Gamma\) admits only finitely many isomorphism classes of simple (i.e., irreducible) representations. Recall that, for a fixed field $k$, the pro-algebraic completion (or Hochschild-Mostow completion) is the pro-algebraic group $A_k(G)$ with is universal with respect to finite dimensional representations of $G$, i.e. there is a map $ \phi\colon  G \to A_k(G)$ such that any finite dimensional $k$-representation of $G$ factors through $\phi$. Here we will be working over $\mathbb{C}$ at all times so we will just write $A(G)$. We say further that $\Gamma$ is representation superrigid if the Hochschild-Mostow completion \(A(\Gamma)\) is finite dimensional.

The above terminology is justified as follows: If \(\Gamma = \langle s_1, \ldots, s_d \rangle\) is given with \(d\) generators, then the map \(\rho \mapsto (\rho(s_1), \ldots, \rho(s_d))\) identifies \(R_n(\Gamma) = \operatorname{Hom}(\Gamma, \operatorname{GL}_n(\mathbb{C}))\) with a subset of \(\operatorname{GL}_n(\mathbb{C})^d\). In fact \(R_n(\Gamma)\) is easily seen to be an affine subvariety. It is invariant under the simultaneous conjugation action of \(\operatorname{GL}_n(\mathbb{C})\) on \(\operatorname{GL}_n(\mathbb{C})^d\). The algebraic-geometric quotient \(X_n(\Gamma) = \operatorname{GL}_n(\mathbb{C})\backslash R_n(\Gamma)\) exactly parametrizes the isomorphism classes of semi-simple \(n\)-dimensional representations of \(\Gamma\). It is sometimes called the \(n\)-dimensional ``character variety'' of \(\Gamma\).

With this terminology we see that \(\Gamma\) is representation rigid if and only if all character varieties of \(\Gamma\) are finite (or zero-dimensional). In other words, there are no moduli for simple \(\Gamma\)-representations.

Finally, recall that a group is said to have virtual first Betti number $0$, or $\mathrm{vb}_1=0$ if all its finite index subgroups have finite abelianisation, for example groups with Property (T). We have the following implications between the properties:

Proposition 1: $\text{super-rigid}\Rightarrow\text{rigid}\Rightarrow \mathrm{vb}_1=0$.

Under the assumption $\mathrm{vb}_1=0$, the theory of algebraic groups gives us alternative ways of understanding what it means to be rigid.

Proposition 2:  Assume $\Gamma$ has $\mathrm{vb}_1=0$.  Then given $n$, there exists a finite index subgroup $\Lambda =\Lambda (n)$ of $\Gamma$ such that, for every $n$-dimensional representation $\rho :\Gamma \to GL_n(\mathbb{C})$, $\overline{\rho(\Lambda)}$ is a connected group. If $\rho$ is irreducible then $\overline{\rho (\Lambda)}$ is semisimple.

Proposition 3: A finitely generated $\mathrm{vb}_1=0$ group $\Gamma$ is not rigid iff there exists a finite index subgroup $\Lambda$ of $\Gamma$ and a simple algebraic group $H$ of adjoint type (i.e., $Z(H) =1$) such that $\Lambda$ has infinitely many non-conjugate homomorphisms $\rho: \Lambda\to H$ with $\rho (H)$ Zariski dense in $H$.

The rigidity properties seems like very strong properties for an infinite group to have. However, it is a guiding principle that higher rank lattices are rigid in almost every possible sense, so it makes sense to start looking for examples there, and indeed we have the following:

1.  Let \(K\) be a number field (finite field extension of \(\mathbb{Q}\)), \(S\) a finite set of places containing all archimedean places, and \(K(S)\) the ring of \(S\)-integers in \(K\). Let \(G\) be a linear algebraic group over \(K\), and \(G(K(S))\) the group of \(S\)-integral points in \(G(K)\). Under certain general conditions, for semi-simple \(G\), the Margulis superrigidity theorem applies here, and it implies in particular that \(G(K(S))\) is representation rigid.
2. If \(\Gamma' \leq \Gamma\) is a subgroup of finite index then \(\Gamma\) is representation rigid if and only if \(\Gamma'\) is representation rigid. Call groups \(\Gamma\) and \(\Gamma_1\) (abstractly) commensurable if they have finite index subgroups \(\Gamma' \leq \Gamma\) and \(\Gamma'_1 \leq \Gamma_1\) which are isomorphic. In this case \(\Gamma\) is representation rigid if and only if \(\Gamma_1\) is.
   A group \(\Gamma\) of ``arithmetic type'' if \(\Gamma\) is commensurable with a product\[\prod_{i=1}^n G_i(K_i(S_i)),\]where each factor is as above. These are thus representation rigid.
3. Call \(\Gamma\) of ``Golod-Shafarevich representation type'' if \(\Gamma\) is residually finite (the finite index subgroups have trivial intersection) and \(\rho(\Gamma)\) is finite for all representations \(\rho\). Such groups are representation rigid. (More generally, \(\Gamma\) is representation rigid if and only if \(\rho(\Gamma)\) is so for all representations \(\rho\). On the other hand, such groups are linear if and only if they are finite. 
   Any torsion residually finite \(\Gamma\) is of Golod-Shafarevich representation type (see Burnside's proof of the Burnside conjecture for linear groups). 

A preliminary search through the 'usual' groups which come up in nature reveals no further examples, so Platonov conjectured that a rigid linear group is of arithmetic type.

Bass--Lubotzky constructed counterexamples to this, inspired by the Platonov--Tavgen construction of a Grothendieck pair, which has previously been discussed on the blog. Lubotzky later simplified and generalised the construction, and it is this that I would like to present.

The proof goes via taking a suitable fibre product, guided by the heuristic that one can often pass up properties from a quotient. For this we need to know when fibre products of (super-)rigid groups are again (super-)rigid.

For $i=1,2$, let $L_i$ be a finitely generated group with epimorphisms $\rho_i$  from $L_i$ onto the same finitely presented group $D$, and $R_i = \mathrm{Ker} ~ \rho_i$.  Let $$ \Gamma = L_1 \times_{D} L_2 = \{(x,y)\in L_1\times L_2\mid \rho_1(x) = \rho_2(y)\}
$$ be the fibre product of $L_1$ and $L_2$ over $D$.

The projections $\pi_i$ of $\Gamma$ to $L_i$ are onto with kernels $(1,R_2)$ and $(R_1,1)$.  Also, one can easily see that $\Gamma$ is finitely generalized  since $D$ is finitely presented.

Proposition 4:  If $L_1$ and $L_2$ are rigid and $\Gamma$ has $\mathrm{vb}_1=0$, then $\Gamma$ is also rigid.

Proof:  If not, by Proposition 3 there exists a finite index subgroup $\Lambda$ of $\Gamma$ and a simple algebraic group $H$ that $\Lambda$ has infinitely many  non-equivalent homomorphisms $\rho:\Lambda\to H$ with dense image.  Let $R'_1 =(R_1,1)\cap\Lambda$ and $R'_2 = (1,R_2)\cap\Lambda$.  Then $[R'_1,R'_2] = 1$ and both are normal in $\Lambda$.  Thus $\overline{\rho (R'_1)}$ and $\overline{\rho (R'_2)}$ are commuting normal subgroups of $H$, so one of them must be trivial. Hence for infinitely many $\rho$'s, either $\rho (R'_1)$ is trivial or $\rho (R'_2)$ is.   In the first case we have that $\rho$ factors through $\pi_2$, i.e., we have a representation of
$\Lambda /R'_1 = \Lambda /(R_1,1)\cap\Lambda\simeq \Lambda (R_1,1)/(R_1,1)$ which is commensurable to $L_2$.  Hence, this group is not rigid and since rigidity is a commensurability invariant, $L_2$ is not rigid, a contradiction. The case with $\rho (R'_2) = 1$ is treated similarly. $\blacksquare$


It is somewhat more difficult to prove, but under an additional hypothesis, it is also true that when we start with $L_i$ super-rigid and reductive, then $\Gamma$ is super-rigid.

Recall that a group is called reductive if every representation of it is reductive.

Proposition 5: Assume $L_1$ and $L_2$ are super-rigid and reductive, $D$ has no finite index subgroup and $\Gamma$ is $\mathrm{vb}_1=0$ then $\Gamma$ is super-rigid and reductive, in fact $A(\Gamma )^0 = A(L_1)^0 \times A(L_2)^0$

Now that we know we can upgrade $\mathrm{vb}_1=0$ to rigidity, we just need to figure out when we can get $\mathrm{vb}_1=0$.

Let $L$ be a finitely generated group, $\rho:L\to D$ a homomorphism onto a finitely presented group $D$ with kernel $R$, and $\Gamma = L\times_{D} L$ the fibre product of $L$ over $D$,
$$ \Gamma = \{(x,y)\in L\times L\mid \rho (x) = \rho (y)\}. $$ Note that $\Gamma = (R,1)\Delta (L) = (1,R)\Delta (L)$ when $\Delta (L)$ is the diagonal embedding of $L$ into $\Gamma$, $\Delta (L) = \{(x,x)\mid x\in L\}$.

Lemma 6:  If $L$ is $\mathrm{vb}_1=0$ and if for every finite index normal subgroup $L_0$ of $L$, and every finite index subgroup $R_0$ of $R$ which is normal in $L$, $[L_0,R_0]$ is of finite index in $R_0$, then $\Gamma$ is $\mathrm{vb}_1=0$.

Proof: Let $\Gamma_0$ be a finite index normal subgroup of $\Gamma$, $L_0 = \Gamma_0\cap\Delta (L)$ and $R_0 = (R,1)\cap\Gamma_0$.  By abuse of notation we will consider $R_0$ as a subgroup of $L$ and at the same time a subgroup of $(L, 1)$.
It follows from the assumptions that $[R_0,L_0]$ is of finite index in $R_0$ and hence in $(R,1)$. As $L$ is $\mathrm{vb}_1=0$, $[L_0,L_0]$ is of finite index in $\Delta (L)$. This implies that $[R_0L_0, R_0L_0]$ is of finite index in $\Gamma$. As $\Gamma_0\supseteq
R_0L_0$, we get that $[\Gamma_0,\Gamma_0]$ is of finite index in $\Gamma$. $\blacksquare$

Let now $L$ be a finitely generated group with $\mathrm{vb}_1=0$.  Assume $L$ has an infinite finitely  presented quotient $D$ with kernel $R$, such that $\hat D = \{ 1\}$ when $\hat D$ is the profinite completion of $D$, i.e., $D$ has no proper finite index subgroup.
Clearly $D^{ab} = \{ 1\}$, i.e, $D$ is a perfect group.  It therefore has a universal central extension \begin{equation}1\rightarrow H_2(D)\rightarrow \bar D\rightarrow D\rightarrow 1
\end{equation} (i.e. for any other central extension by $\mathbb Z$, say $D'$, there is a map $\bar D \to D'$ compatible with the maps to $D$).  

For an abelian group $A$, we denote $rk(A) = \dim_{\mathbb{Q}} (A\otimes_{\mathbb{Z}}\mathbb{Q})$. A bit of algebra then gives

Lemma 7: Let
\begin{equation}1\rightarrow C\rightarrow E\rightarrow D\rightarrow 1
\end{equation} be a central extension of $D$ such that $E^{ab}$ is finite.  Then $rk(C)\leq rk(H_2(D))$. 

Lemma 8: For a finite index normal subgroup $L_0$ of $L$ denote $R_0 =R\cap L_0$.  Then $rk(R_0/[L_0,R_0])\leq rk(H_2(D))$ for every $L_0\vartriangleleft L$ of finite index.

Proof: The map $L_0\to D$ is onto since $\hat D=1$ and so:
$$
1\to R_0/[L_0,R_0] \rightarrow L_0/[L_0,R_0] \rightarrow D\rightarrow 1
$$
is a central extension of $D$.  As $L$ is $\mathrm{vb}_1=0$, $L^{ab}_0$ is finite and our Lemma follows from Lemma 7. $\blacksquare$

Given $L$ and $\rho :L\to D$ with $R = \mathrm{Ker} ~\rho$ as before, choose now $L_0$ of finite index with $R_0 = R\cap L_0$ and $rk(R_0/[L_0,R_0])$ maximal among all possible such $L_0$.  As $\hat D = \{ 1\}$, it follows that $L_0/R_0\simeq D$ and we have a central extension
\begin{equation}1\rightarrow R_0/[L_0,R_0] \rightarrow D_0:=L_0/[L_0,R_0]\rightarrow D=L_0/R_0\rightarrow 1
\end{equation}
 Now, $R_0/[L_0,R_0]$ is dense in the profinite topology of $D_0$ since $\hat D = \{ 1\}$, hence $\hat D_0$ is abelian.  But, $L_0$ is $\mathrm{vb}_1=0$, so $\hat D_0$ is finite.  Replace $L_0$ by a finite index subgroup $L'_0$ so that $D_1=L'_0/([L_0,R_0]\cap L'_0)$ satisfies $\hat D_1 = \{ 1\}$ and in particular $D_1^{ab} = \{ 1\}$.  Let us now rename and call $L:= L'_0$, $\tilde D:=D_1$ and $\tilde R = \mathrm{Ker} (L\to \tilde D)$. So $\tilde D$ is a central (possibly infinite!) extension of the original $D$ and $\Hat{\Tilde D} = \{ 1\}$.  We get the exact sequence:
\begin{equation}
1\rightarrow\tilde R\rightarrow L\rightarrow\tilde D\rightarrow 1.
\end{equation}

The crucial point is: 

Lemma 9:  For every normal subgroup $L_1$ of finite index in $L$ and every $R_1$ of finite index in $\tilde R$ which is normal in $L$, $[L_1,R_1]$ is of finite index in $R_1$.

Proof:  The group $\tilde R/[L_1,R_1]$ is a finitely generated virtually abelian group.  Hence $L/[L_1,R_1]$ is an extension of the form $1\to C\to L/[L_1,R_1]\to D\to 1$ where $C$ is virtually abelian and $\hat D = \{ 1\}$.  Moreover, $L$ is $\mathrm{vb}_1=0$. So, all this implies that $\widehat{(L/[L_1,R_1])}$ is finite. Hence, whenever $L/[L_1,R_1]$ is mapped into a residually-finite group, its image is finite.  This applies, in particular, to the image of $L/[L_1,R_1]$ in its action (by conjugation) on the finitely generated virtually abelian group $\tilde R/[L_1,R_1]$. Thus for some finite index subgroup $L_2$ of $L_1$, $[L_2,\tilde R]\subseteq [L_1,R_1]$. By the maximality choice of $L_0$, it follows that $[L_2,\tilde R]$ is of finite index in $\tilde R$ and hence $[L_1,R_1]$ is also of finite index there. $\blacksquare$

To summarise, by replacing $D$ by $\tilde D$ (and the original $L$ by a subgroup of finite index), we get an exact sequence:
\begin{equation}1\rightarrow\tilde R\rightarrow L\rightarrow\tilde D\rightarrow 1
\end{equation}
which satisfies the claim.  Hence by Lemma 6, we have:

Corollary 10:  The fibre product $\Gamma = L\times_{\tilde D} L$ is $\mathrm{vb}_1=0$.

Propositions 4 and 5 show how to get new (super) rigid groups from old ones.  The standard examples of rigid groups are irreducible lattices in semisimple groups of higher rank, but these do not have infinite quotients, so fibre products of them are commensurable to other arithmetic groups.  This leads us to lattices in $Sp(n,1)$ and $F_4^{(-20)}$. They are super-rigid by work of Corlette and Gromov--Schoen, and at the same time have many infinite quotients.

So, from now on, let $L$ be a torsion-free uniform (=cocompact) lattice in one of the groups $Sp(n,1)$ or $F_4^{(-20)}$.  Ol'shanskii and Rips independently proved that a non-elementary hyperbolic group has a finitely presented infinite quotient $\rho :L\to D$ where $D$ has no proper finite index subgroups. This uses the same sort of small cancellation methods that Ol'shanskii is famous for, for example he used these to show that non-elementary hyperbolic groups are SQ-universal.

Replace $D$ now by $\tilde D$ (and $L$ by a finite index subgroup, also  called $L$) as above.  Let $\tilde\rho :L\to\tilde D$ be the new map and $\Gamma = L\times_{\tilde D} L$ the fibre product. 

Theorem 11: $\Gamma$ is a super-rigid group. In fact, $A(\Gamma)^0 = A(L)^0\times A(L)^0$.

Proof:  By Corollary 10, $\Gamma$ is $\mathrm{vb}_1=0$.  Hence Proposition 5 can be applied to deduce that $\Gamma$ is super-rigid (and hence rigid) and $A(\Gamma)^0 = A(L)^0\times A(L)$. $\blacksquare$

The proof actually shows that every representation of $\Gamma$ can be extended, on a finite index subgroup, to a representation of $L\times L$.

To finish the disproof of Platonov's conjecture, it remains to show that \(Q\) is not isomorphic to a lattice in any product of linear algebraic groups over archimedean and non-archimedean fields. The point here is again superrigidity of lattices in quaternionic and octonionic hyperbolic spaces, which means that if the fibre product is a lattice then it forces there to be a map on ambient Lie groups, so any product of linear algebraic groups containing the fibre product as a lattice all of them must be linear over an Archimedean local field (over non-Archimedean fields everything is totally disconnected). Since everything is simple, we end up getting that the product of linear algebraic groups has to be a product of the Lie group in which $L$ is a lattice, but $\Gamma$ isn't a lattice there.